Electrical Machines, Drives, and Power Systems 5E (Theodore Wildi)_text.pdf

Electrical

Drives,

Machines,

and Power Systems /

Theodore Wildi

Electrical Machines,

and Power Systems

Drives,

Fifth Edition

Theodore Wildi Professor Emeritus, Laval University

Prentice Hall

Upper Saddle

River,

New Jersey

Columbus, Ohio

Library of Congress Cataloging-in-Publication Data

page 136 by Weston Instilments; pages 204, 239, 251, 312, by ABB; page 207 by Hammond; pages 209, 232, 777, 778, 796, 797 by 339, 344, 370, 583, 584, 626, 700, 701, 782

Theodore.

Wildi,

Electrical

machines,

Theodore Wildi.— 5th p.

and

drives,

power systems

/

ed.

cm.

Includes bibliographical references and index.

ISBN 1.

3.

0-1 3-093083-0 (alk.

Electric

paper)

machinery.

Electric driving.

I.

2.

Electric

power systems.

Title.

TK2182.W53 2002 621.31'

042— dc21

2001051338

Editor in Chief: Stephen Helba Assistant Vice President and Publisher: Charles E. Stewart, Jr.

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This

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Pages

21, 86, 87, 88, 107, 115, 371, 643, 640,

651, 652, 691, 701, 702, 708, 713, 731, 741, 743, 762, 773

General Electric; pages 1 1 7,

Westinghouse; pages 232, 250 by Ferranti-Paekard; pages 704 by Montel, Sprecher and Schnh; page 235 by American Superior Electric; pages 252, 386, 644, 668, 669, 670, 689, 691, 703, 706, 708, 764, 765 by Hydro-Quebec; pages 265, 354, 392, 628, 737, 839, 840 by Lab Volt; pages 2(57, 301 by Brook Crompton-Parkinson Ltd.; page 290 by ElectroMecanik; pages 294, 295, 315, 354, 441, 442, 452, 547, 744 by Siemens; pages 300, 301, 391, 404 by Gould; page 304 by Reliance Electric; pages 337, 338, 340, 645 by Marine Industrie; pages 342, 344, by Allis-Chalniers Power Systems, Inc.; page 345 by Air France; pages 424, 425, 433 by Pacific Scientific, Motor and Control Division, Rockford, IL; pages 425, 426 by AIRPAX Corporation; pages 440, 442, 447, 450, 452, 458, 785 by Square D; pages 440, 441, 448, 449 by Klockner-Moeller; page 441 by Potter and Brumfield; pages 443, 451 by Telemeeanique, Group Schneider; page 454 by Hubbel; pages 477, 493, 500 by International Rectifier; page 595 by Robicon Corporation; page 627 by Carnival Cruise Lines; page 645 by Les Ateliers d'Ingeniere Dominion; page 647 by Tennessee Valley Authority; page 649 by Novenco, Inc.; page 694 by Pirelli Cables Limited; page 650 by FosterWheeler Energy Corporation; page 651 by Portland General Electric; pages 653, 654 by Electricity Commission of New South Wales; page 658 by Connecticut Yankee Atomic Power Company Georges Betancotirt; page 659 by Atomic Energy of Canada; page 667 by Canadian Ohio Brass Co., Ltd.; page 674 by IREQ; page 700 by Canadian General Electric; pages 703, 704, 714 by Dominion Cutout; pages 703, 704, 705, 707 by Kearney; page 731 by Sangamo; page 735 by Gentec Inc.; page 738 by Service de la C.I.D.E.M., Ville de Montreal; page 758 by GEC Power Engineering Limited, England; page 759 by Manitoba Hydro; page 762 by New Brunswick Electric Power Commission; pages 763, 764 by United Power Association; page 772 by EPRI; page 289 by Services Electromecaniques Roberge; page 830 by Fluke Electronics Canada, Inc.; pages 842, 843, 844, 849 by Oinron Canada Inc.; pages 851, 852, 853, 855 by St. Lawrence Stevedoring; pages 854, 855, 856, 857 by Schneider Electric; page 133 by Leroy Somer and Emerson Electric; page 436 by 233,

09, 100,

264, 300, 402, 59 1, 605, 619

by

289 by H. Roberge; pages

by Baldor Electric Company;

—

Emerson

Electric.

Copyright © 2002, 2000, 1997, 1991, 1981 by Sperika Enterprises Ltd. and published by Pearson Education, Inc., Upper Saddle River, New Jersey 07458. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

Prentice Hall

10

98765432 ISBN 0-13-093083-0

Preface

This

fifth

edition

was prompted

in part

by the great

in-

is

crease of computers in industrial controls and au-

in

tomation,

which has produced computer programs

can simulate relays and relay contacts. These

that

no longer pertinent isolation

to discuss

dc and ac machines

because wherever they are being

age. Consequently, the term drive

motor alone but the

now

on/off discrete controls have eliminated the wiring

the

and installation of hardware components

torque and speed of the machine. This

virtual relays

a keyboard.

and contacts

The devices

that

that

in

favor of

can be programmed on

perform these operations

are called

Programmable Logic Controllers (PLCs),

or simply

programmable

were

initially

controllers.

These devices

stand-alone computers that controlled a

specific robot or

manufacturing operation. However,

with the advent of the Internet, they have

now been

integrated with the overall manufacturing process,

leading seamlessly to integration with sales,

procurement, and consumer satisfaction.

The 20

is

management,

relay control of

machines covered

now supplemented by coverage Chapter 3

in

of PLC controls

Chapter 3

of

PLCs and shows, by way of example, how they

used

.

1

covers the basic principles are

running the activities of a large service enter-

in

prise.

1

This

new

chapter illustrates

computer-based

setting

how

these trend-

activities involving controls

and automation are being integrated with other business activities, including e-commerce.

As

I

mentioned

in the last edition, similar up-

heavals have occurred in ply

amazing

influence on the

involves not

entire unit that directs the

way

power technology. It is simof power elec-

to witness the entrance

tronics into every facet of industrial drives. Thus,

it

having a

is

machinery

electrical

courses are being taught.

How

has this dramatic change

come about?

mainly due to the high-power solid

state

It is

switching

devices, such as insulated gate bipolar transistors

(IGBTs), which can operate

at

frequencies of up to

20 kHz. The change has also been driven by tors

and gate turn-off thyristors (GTOs)

is

thyris-

that

handle currents of several thousand amperes ages of up to 5 kV. Another key element

Chapter

in

direct

in-

an electronic control forms part of the pack-

stalled,

can

at volt-

the

com-

puting power of microprocessors that can process signal data in real time with incredible speed.

The high switching frequencies of IGBTs permit the use of pulse-width-modulation techniques in

power

converters. This, in turn, enables torque and

speed control of induction motors

down

to zero

was not feasible in rectangular-wave converters that were employed only a few years ago. Most industrial drives are in the fractional horsepower to the 500 hp range. That is precisely the range now available for control by IGBTs. The result has speed. This

been an explosion

in the retrofitting

Lower maintenance

costs,

of existing drives.

higher efficiency,

and

PREFACE

economically

attractive.

pedagogical quality. As a

made such changeovers

greater productivity have

Thus, dc drives are being

placed by induction motor drives, which require less

•

Every sector of is

•

industrial

this

revolutionary con-

verter technology. Electric elevators, electric tives, electric transit vehicles,

and

ing, ventilating

air

locomo-

•

is

— an

power

distribution of electric

Most

I

distorted wave.

are

wave in

Power Research

(EPR1)

Institute

in

Palo

Alto, California, in collaboration with several electrical

manufacturers, has also resulted

in

the cre-

ation of high-power static switches, thyristor-eon-

and converters

trolled series capacitors,

that

electric power.

method

that enables

into

its

Once they know how

harmonic components,

harmonics quickly

in a

unravel a

to

their interest

rises.

at all.

All the important changes first introduced in

Important development work, carried out by the Electric

they affect the behav-

students to calculate the harmonic content

seeing large rotating machines, such as synchronous

have no moving pails

how

also devised a simple

condensers and frequency changers, being replaced by solid-state converters that

have added a new chapter

I

generated and

and the quality of

and

we

importantly,

ics are

industry that has

been relatively stable for over 50 years. Here,

im-

this

ior of capacitors, inductors, cables, transformers,

new technology.

utilize this

also affecting the transmission

make

on harmonics. Chapter 30 reveals how harmon-

servomechanisms, heat-

conditioning systems, fans,

being modified to

The change

to

portant topic easier to understand.

compressors, and innumerable industrial production lines are

their solutions

Chapter 7 on Active, Reactive, and Apparent

Power was completely revised

and commercial activity

by

therefore being affected

The end-of-chapter problems and were revised and double-checked.

maintenance while offering equal and often superior

dynamic performance.

more than 20

result,

percent of the pages were altered.

re-

can

fill

previous editions have been kept tion.

in this fifth edi-

Thus, the writing of circuit equations, the

discussion of higher frequency transformers, and the equivalent circuit

diagram of the single-phase

induction motor have

all

been retained.

the role of phase-shift transformers.

These new methods of power flow

FACTS

by the acronym

(Flexible

control,

AC

known

Systems) will permit existing transmission and bution lines to carry more

demand

creasing

extremely bilize a

power

for electricity.

to

On

that

It is

sta-

is

may suddenly be menaced by an

in electric

Most students

in

many

formulating them.

I

particularly easy to follow. Readers will be re-

minder of the circuit-solving procedure. on

all rest

•

motor drives

is

Chapter

panded

similar to

ers.

employed to control the flow of power in elecAs a result, everything falls neatly and coherently into place. The teaching and learning of electric machines, drives, and power systems are

1

1

on Special Transformers was ex-

to include higher

The reader

is

frequency transform-

guided through the reasoning

that

behind the design of such transformers, and

tric utilities.

they

thereby

made much

ex-

dis-

glad to refer to this section as a convenient

base. In other words, the converter tech-

nology used

2.

such equations, but

close an ac/de circuit-solving methodology that

account of their

remarkable that these innovations

common

to solve

perience difficulty

unexpected disturbance.

a

section covering the writing of circuit

know how

distri-

can also

A new

equations was added to Chapter

meet the ever-in-

fast response, the converters

network

•

Transmission

become smaller

why

as the frequency increases.

High-frequency transformers are directly related to the

higher frequencies encountered

in

switch-

ing converters.

easier.

The following changes have been made

in

the

•

Chapter 16 on Synchronous Generators has been

expanded

fourth and fifth editions:

to

show why an

increase

in size in-

evitably leads to higher efficiencies and greater •

Every page of the original work was examined for clarity of expression

and reviewed as

to its

outputs per kilogram. This fundamental aspect of

machine design

will interest

many

readers.

PREFACE

•

A new

section

was added

Chapter

to

1

phase induction motor.

many worked-out problems,

to solve the circuit,

presents a rigorous, yet

It

which permits

a better under-

Chapter 2

1

,

Electronics,

Fundamental Elements of Power was revised and expanded to in-

modulation

(PWM)

techniques.

how

made

they can be

to

converters and

generate almost any

motors operating

major addition

Electric Utility

to Part

It

power

sags, swells,

It

power becomes

become

visit

students

made

may in

to establish

how the

lines.

book requires

user-friendly treatment of even

topics, this

book

broad range of readers.

will

First,

meet it

is

the needs of a

appropriate for

students following a two-year electrical

community versities.

in

colleges, technical institutes, and uni-

Owing

to its

very broad coverage, the text

can also be incorporated

program.

program

Many

for their electric

in

a 4-year technology

universities have adopted the

book

power service courses.

wealth of practical information that can be di-

rectly applied to that greatest laboratory

electrical industry itself.

actual use.

at close

hand the

The photographs help convey

the

1

chapters, a conscious effort

was

coherence, so that the reader can see

various concepts

fit

together. For

example, the

to those

found

lines, in turn,

And

reactive

transmission

in

bring up the question

power

is

an important

aspect in electronic converters. Therefore, knowledge

it

in is

one sector applied

is

strengthened and broadened

in another.

As

a result, the learning

of electrical machines, drives, and HPwer syêms be-

comes a challenging, thought-prov^^fêxperience. In order to convey the real-w|ftd aspects df machinery and power systems, particular. attentio|i has

been paid

to the inertia

of revolving masses, the

physical limitations of materials, ai^l .the problems

created by heat. This

appro^kfjal^^^

multidisciplinary programs of

many

fftfe

and

eo^lleês

technical institutes.

Instructors responsible for industrial training will find a

in

the transmission and distribution of

Transmission

when

complex

the impor-

not have had the opportunity to

machines are similar

increasingly important.

gained

its

book shows

terminology and power equations for synchronous

and some trigonometry. to

the

an industrial plant or to see

only a background in basic circuit theory, algebra,

Owing

and Websites

articles,

by diagrams and pictures, showing

illustrated

Throughout the 3

also

as regards

a reality, these

subject matter covered in this

will also

invited to con-

is

various stages of construction or

in

of reactive power.

The

also avail-

magnificent size of these devices and machines.

methods of controlling the quality of

electricity will

is

end of most chapters

at the

electrical energy.

harmonics, and brownouts. As dereg-

ulation of electric

electronic

power

Manual

Industrial Application prob-

of books, technical

equipment used

explains the

electronically.

discusses the quality of electric

appear

The

Reference section toward the end of the book.

Some

vector control.

IV dealing with

Power Systems.

practical, inter-

tance given to photographs. All equipment and sys-

them

technologies that are being developed to control the flow of electric

that

tems are

at

Chapter 29, Transmission and Distribution represents a

lems

A Solutions

A quick glance through

special section explains the

PWM drives and flux

basics of

—

appeal to hands-on users. The reader

in the

Chapter 23, Electronic Control of Alternating

A

the end of each chapter are di-

at

end of the book.

sult the list

/^Current Motors, was greatly expanded to cover

variable speeds.

exercises

mediate, and advanced. Furthermore, to encourage

the

waveshape and frequency.

the properties of induction

The

vided into three levels of learning

able for instructors.

illustrates the

It

oflGBT

its

particularly suitable

the reader to solve the problems, answers are given at

clude switching converters and pulse width

extraordinary versatility

is

for self-study.

programmed

standing of this ubiquitous single-phase machine.

•

being de-

is

voted to continuing education, this book, with

motor. Hand-held computers can be

•

effort

velop the equivalent circuit diagram of a single-

simple approach, based on the 3-phase induction

•

when much

Finally, at a time

8 to de-

v

of

all

—

the

In

summary,

I

employ

a tlicM>Bk&$ ipfefeAal,

multidisciplinary approach to give a broad under-

standing of modern electric power. Clearly, longer the staid subject

it

was considered

it

is

to

no be

PREFACE

vi

some years ago. There this

is

dynamic, expanding

good reason to believe that field will open career op-

and Bernard Oegema of Schneider Canada; Carl Tobie of Edison Electric

Institute;

Damiano Esposito

portunities for everyone.

and Vance

make a final remark concerning the As mentioned previously, power technology has made a quantum jump in the past

Scott Lindsay of Daiya Control

I

would

like to

use of this book.

Belisle,

machines, drives, and power systems, there will

now

be a long period of consolidation during which existing

machines and devices

will

be replaced by newer

Systems; Louis

and Jean Lamontagne of Lumen; Benoit

Arsenault and Les Halmos of Allen Bradley.

eight years, mainly on account of the availability of fast-acting semiconductors. In the field of electrical

E. Gulliksen of Carnival Cruise Lines;

extend a special note of thanks to Professor

I

Thomas

Young

of

Rochester

the

Technology, to Dr. Robert University,

College, and to Jean Anderson of Lab- Volt Ltd. for

models. But the basic technology covered herein will

having extensively reviewed and commented

not change significantly in the foreseeable future.

on various aspects of this book and

Consequently, the reader will find that

this

book can

of

Peros of Seneca

Professor Martin

to

Institute

H. Alden of McMaster

T.

their valued viewpoints.

also

I

in

depth

for having offered

want

to

acknowledge

also be used as a valuable long-term reference.

the contribution of Professor Stephane Montreuil of

Acknowledgments

end-of-chapter problems and the solutions manual.

CEGEP Levis-Lauzon for having gone over all the

the

want

I

In preparing this edition

and previous editions of

my

acknowledge

book,

1

would

like to

the impor-

tant contribution of the following persons.

Consultant; David Krispinsky, Rochester Institute of

Technology; Athula Kulatunga, Southeast Missourri State University; Rick Miller, Ferris State University;

Nehir,

Montana

State University; Martin

University;

Chandra

James

E,

Sekhai;

University;

Gerald Sevigny, Southern Maine Technical College; Philippe Viarouge, Laval University; Stacy Wilson,

Western

Kentucky

Rochester

Institute

A& M and

University;

Sri R. Kolla,

University;

Thomas

of Technology; Dr.

P

Texas

Ted James, Pasadena City College;

Bowling Green

State University.

Commercial, industrial and institutional contributors:

Andre Dupont, Raj Kapila, G. Linhofer,

Katherine Sahapoglu of ABB; Roger Bullock, Gerry

McCormick, James Nanney, Darryl J. Van Son, and Roddy Yates of Baldor Electric Company; Jacques Bedard, Guy Goupil, and Michel Lessard of Lab-Volt Ltd.; Richard B. Dube of General Electric Company; Abdel-Aty Edric and Ashock Sundaram of Electric Power Research Institute; Neil H. Woodley of Westinghouse Electric

Rene Poulin of Inc. in

of

Lawrence

in the application

and

to

of pro-

photographer Hughes

also

also hereby acknowledged.

is

want

E. Stewart,

Editor;

the Centre de Robotique Industrielle

reviewing and describing the essential features

PLCs

and

Jr.,

to

to express

my

appreciation to Charles

Publisher; to Delia Uherec, Associate

Alexandrina B. Wolf, Senior Production

Editor, of Prentice Hall, for planning, coordinating,

and administrating

As to

in

this text.

previous editions,

my

provide his valuable help

art,

in

son Karl continued preparing the line

photographs, and word processing of this

latest

edition.

My

thanks also go to

ing supported

me

thor, consultant,

Goyette, Jim

Corporation; Maurice Larabie, Jean-Louis Marin,

controllers,

St.

providing industrial ex-

in

Chicoine for his work. The important contribution of

Young,

Enjeti,

know-how

perience and

grammable

M.

Roach, Bob Jones

Purdue

appreciation to Jean- Serge

and Giles Campagna of

Electric,

I

Peros, Seneca College;

my

Stevedoring for their help

Professors and reviewers: Robert T. H. Alden, McMaster University; Ramon E. Ariza, Delgado Community College; Fred E. Eberlin, Educational

M. H.

to express

Lamirande of Omron, Pierre Juteau of Schneider

I

tors

in

my

my

wife, Rachel, for hav-

continuing vocation as au-

and teacher.

also wish to voice

my

gratitude to the instruc-

and students, practicing engineers, and techni-

cians

who

asked questions and made suggestions by

e-mailing their messages to [email protected].

You

are cordially invited to

do the same. Theodore Wildi

PART

I.

FUNDAMENTALS

Distinction between sources and

2.2

loads 1.

UNITS 3 l.O

Introduction 3

LI

Systems of units 3

1.2

Getting used to SI 4

1.3

Base and derived units of the SI 4

\A

Definitions of base units 5

1.

Definitions of derived units 5

1.6

Double-subscript notation for

units 7

Conversion charts and

their use 8

Per-unit system with one base

l.ll

Per-unit system with

Sign notation for voltages

2.6

Graph of an

2.7

Positive and negative currents

18

19

Sinusoidal voltage

2.9

Converting cosine functions into sine

2.

10

2.

1

two bases

10

1

19

Effective value of an ac voltage 20

Phasor representation 2

2.12

Harmonics 23

2.13

Energy

in

14

Energy

in a

2.15

Some

an inductor 25 capacitor 25

useful equations 26

1

12

ELECTROMAGNETISM 2.

MAGNETISM, AND CIRCUITS 15 2.0

Introduction

2.

Conventional and electron current flow 15

17

alternating voltage

2.8

1

6

FUNDAMENTALS OF ELECTRICITY,

1

17

2.5

2.

l.IO

Questions and Problems

17

functions 20

The per-unit system of measurement 9

1.

2.4

Multiples and submultiples

Commonly used

1.

Sign notation

voltages

of SI units 7 1.

16

2.3

15

Magnetic density

2.17 2.

1

8

2.19

field intensity

H and

B 27

B-H curve B-H curve

of

f ACULTAD

vacuum 27

DE

q r ~n of a magnetic material" 27

Determining the relative permeability 28

*

WIN A _ -

-

BlBi-iO I fcw*

CONTENTS

i

2.20

Faraday's law of electromagnetic

3.9

induction 29 a conductor 30

2.21

Voltage induced

2.22

Lorentz force on a conductor 3

2.23

2.24

2.25

Kinetic energy of rotation, inertia

in

3.10

Torque,

inertia,

and change

Direction of the force acting on a

3.11

Speed of a motor/load system 57

straight conductor 3

3.12

Power flow

Residual flux density and coercive

in a

mechanically coupled

system 58

force 32

3.13

Motor driving a load having

Hysteresis loop 33

3.14

Electric motors driving linear motion

Hysteresis loss 33 Hysteresis losses caused by

3.15

Heat and temperature 60

rotation 33

3.16

Temperature scales 61

Eddy Eddy

3.17

currents 34

3.18

Eddy-current losses

in a revolving

core 35 2.31

Current

in

an inductor 36

Transmission of heat 62

Heat transfer by conduction 62

3.20

Calculating the losses by

convection 63

Kirchhoffs voltage law 40

2.33

Kirchhoffs voltage law and double-

3.22

Heat transfer by radiation 64

3.23

Calculating radiation losses 64

Questions and Problems 65

40

2.34

Kirchhoffs current law 41

2.35

Currents, impedances, and associated

PART

II.

voltages 41

2.36

Kirchhoffs laws and ac

2.37

KVL and

2.38

Solving ac and dc circuits with sign

2.39

circuits

4.

44

Circuits and hybrid notation

ELECTRICAL MACHINES AND TRANSFORMERS

43

sign notation 43

45


DIRECT-CURRENT GENERATORS 4.0

Introduction 71

4.1

Generating an ac voltage 71

4.2

Direct-current generator 72

4.3

Difference between ac and dc generators 73

FUNDAMENTALS OF MECHANICS AND HEAT 50

4.4

Improving the waveshape 73

4.5

Induced voltage 75 Neutral zones 76

3.0

Introduction 50

4.6

3.1

Force 50

4.7

Value of the induced voltage 76

3.2

Torque 51

4.8

Generator under load: the energy

3.3

4.9

3.5

Mechanical work 5 Power 52 Power of a motor 52

3.6

Transformation of energy 53

3.7

Efficiency of a machine 53

4.11

commutation 78 Commutating poles 79

3.8

Kinetic energy of linear motion 54

4.12

Separately excited generator 79

3.4

58

temperature

Heat transfer by convection 63

3.21

2.32

notation

to raise the

3.19

CIRCUITS AND EQUATIONS

subscript notation

Heat required of a body 6

currents in a stationary iron

core 35

2.30

inertia

loads 59

2.26

2.28

in

speed 57

2.27

2.29

moment of

54

conversion process 77

4.

10

Armature reaction 77 Shifting the brushes to improve

71

CONTENTS

4. 13

No-load operation and saturation

5.18

curve 79 4.14

Shunt generator 80

5.

4.15

Controlling the voltage of a shunt

5.20

1

9

generator 81 4.16

Dynamic braking and mechanical constant

1

Armature reaction

5.21

1 1

Flux distortion due to armature reaction

Equivalent circuit 82

1

1

Commutating poles 113 Compensating winding 114

Separately excited generator under

5.22

load 82

5.23

Basics of variable speed control

4. 18

Shunt generator under load 83

5.24

Permanent magnet motors

4.19

Compound

4.20

Differential

4.

1

7


generator 83

compound

1

1

Load

4.22

Generator specifications 84

84

6.

EFFICIENCY AND HEATING OF ELECTRICAL MACHINES 120 120

Introduction

6.1

Mechanical losses

Field 84

6.2

Electrical losses

4.24

Armature 85

6.3

Losses as a function of load

4.25

Commutator and brushes 86

6.4

Efficiency curve

123

4.26

Details of a multipole generator 88

6.5

Temperature

125

4.27

The The

6.6

Life expectancy of electric

6.7

Thermal

4.28

ideal

commutation process 91

practical

commutation process 92

1

20

20

1

rise

DIRECT-CURRENT MOTORS 96 Introduction 96

classification of

126

6.8

Maximum

ambient temperature and

6.9

Temperature

6.10

Relationship between the speed and

Counter-electromotive force

5.2

Acceleration of the motor 97

Mechanical power and torque 98

5.4

Speed of rotation 100

5.5

Armature speed control 101

rise

size of a

machine

1

Shunt motor under load 103

ACTIVE, REACTIVE, POWER 134 7.0

Introduction

7.1

Instantaneous power

7.2

Active power 136

7.3

Reactive power 137

7.

102

Series motor Series

5.10

Applications of the series motor

104

motor speed control 105 1

5.11

Compound motor

5.12

Reversing the direction of rotation

5.13

Starting a shunt

5.14

Face-plate starter 108

5.15

Stopping a motor 109

06

106 1

07

7.4

motor 108

5.16

Dynamic braking 109

5.17

Plugging 110

7.6

3

134 134

Definition of reactive load and reactive source

7.5

!

AND APPARENT

Field speed control

5.7

5.8

27

30


5.6

5.9

1

by the resistance

method 129

(cemf) 96

5.3

23

insulators

hot-spot temperature rise 5.1

1

equipment 126


5.0

1

1

6.0

CONSTRUCTION OF DIRECT-CURRENT GENERATORS 4.23

1

17

generator 84

4.21

characteristics

tin

1

138

The capacitor and power 139

reactive

Distinction between active and reactive

power 140

CONTENTS

Combined active and apparent power 141

7.7

reactive loads:

Relationship between

7.9

7.10

Power Power

7.11

Further aspects of sources and loads

7.12 7.

3

1

1

4

Q, and S

141

8.20

Varmeter 177

8.2

A remarkable

1

144

7.

1

6

17

Systems comprising several loads 146

9.

148

Solving

AC

circuits using the

method 148 Power and vector notation

power

Rules on sources and loads (sign

154

Rules on sources and loads (double subscript notation)

Introduction

9.

Voltage induced

1

Introduction

83

Elementary transformer

9.4

Polarity of a transformer

9.5

Properties of polarity marks

9.6

Ideal transformer at no-load; voltage

1

84 1

85 1

86 186

187

ratio

9.7

Ideal transformer under load; current

9.8

Circuit

188

symbol

for an ideal

transformer 191

158

Polyphase systems 158

9.9

Single-phase generator 159

9.

Impedance

10

Power output of a single-phase

ratio

191

Shifting impedances from secondary to

primary and vice versa 192


60

Two-phase generator 160 Power output of a 2-phase

8.5

1

9.3

8.1

1

a coil

Applied voltage and induced

8.2

generator

in

9.2

ratio

8.4

80

1

183

9.0

154

THREE-PHASE CIRCUITS 158

8.3

178

THE IDEAL TRANSFORMER 183

voltage 151


8.0

single-phase to 3-phase

144

notation) 7.

3-phase,


triangle

7.15

in

177

transformation

Reactive power without magnetic fields

7.

P,

143

triangle

Power measurement 4-wire circuits

7.8

factor

19

8.

10.

195

PRACTICAL TRANSFORMERS 197

generator 16

10.0

Introduction

8.6

Three-phase generator 162

10.

Ideal transformer with an imperfect

8.7

Power output of generator

1

a 3 -phase 10.2

8.8

Wye

Voltage relationships

connection

164

8.10

Delta connection

Power transmitted by line

8.12

8.13

97 199

Primary and secondary leakage

10.3

167

reactance 200 a 3-phase

1

0.4

168

Equivalent circuit of a practical transformer 202

Active, reactive and apparent

3-phase circuits

1

197

Ideal transformer with loose

coupling 165

8.

1

core

62

8.9

1

1

power

in

1

Construction of a power

0.5

169

transformer 203

Solving 3-phase circuits

170

10.6

Standard terminal markings 204

8.14

Industrial loads

171

10.7

Polarity tests

8.15

Phase sequence 174

10.8

Transformer taps 205

10.9

Losses and transformer rating 206

8.

1

6

Determining the phase sequence

8.

1

7

8.

1

8

Power measurement Power measurement 3-wire circuits

176

in

ac circuits

in

3-phase,

1

75 1

76

204

10.10

No-load saturation curve 206

10.11

Cooling methods 207

10.

1

2

Simplifying the equivalent circuit 209

CONTENTS

Voltage regulation 211

10.14

Measuring transformer

transformers 260

impedances 212


10.15

Introducing the per unit method 215

1

11.

0.

6

1

Impedance of a transformer 2

Typical per-unit impedances 216

10.18

Transformers

1

THREE-PHASE INDUCTION

MOTORS 13.0

Introduction 263

13.1

Principal

13.2

Principle of operation

13.3

The

TRANSFORMERS

225

Introduction 225

1

1

1

.2

Autotransformer 226

1

.3

Conventional transformer connected

1

1

.5

1

1

.6

Number

3.5

1

1

cage motor 273

228

Acceleration of the rotor-slip 274

13.7

Motor under load 274 and slip speed 274

Voltage transformers 230

13.8

Current transformers 23

13.9

Slip

13.10

Voltage and frequency induced

Opening

the secondary of a

CT can

be

rotor

Toroidal current transformers 234

.7

11.8

Variable autotransformer 235

11.9

High-impedance transformers 236

1

1

.

1

0

1

1

.

1

1

1

2.

1

Active power flow 278

Torque versus speed curve 28

13.15

Effect of rotor re s stance 282

13.16

Wound-rotor motor 284

Introduction 243

13.17

Three-phase windings 285

Basic properties of 3-phase

13.18 1

2.4

Wye-delta connection 247

12.6

Open-delta connection 248

Three-phase transformers 249

12.8

Step-up and step-down

1

2.

1

1

0

1

12. 12

19

3.2

1

i

Sector motor 288

Linear induction motor 289 Traveling waves 291 Properties of a linear induction

motor 291 13.22

Wye-wye connection 248

12.7

3.

13.20

244

1

1

an induction

13.14

Delta-wye connection 246

2.

in

13.13

2.3

1

Estimating the currents

High-frequency transformers 238

1

12.9

13.12


Delta-delta connection

2.5

Characteristics of squirrel-cage

motor 277

transformer banks 243

1

13.11

Induction heating transformers 237

2.2

1

in the

275

induction motors 276

THREE-PHASE TRANSFORMERS 243 12.0

of poles-synchronous

Starting characteristics of a squirrel-

3.6

dangerous 233 1

264

265

rotating field

speed 271

1

as an autotransformer

components 263

Direction of rotation 270

13.4

transformer 225

11. 4

263


in parallel

Dual-voltage distribution

1

.

Polarity

219

SPECIAL 1

13.

1

10.17

11.0

12.

marking of 3-phase

10.13

12.13

Magnetic

levitation

293


autotransformer 25

SELECTION AND APPLICATION OF THREE-PHASE INDUCTION

Phase-shift principle 253

MOTORS

14.

Three-phase to 2-phase

14.0

transformation 254

1

4.

1

4.2

1

Calculations involving 3-phase trans-

299

Introduction 299

Standardization and classification of induction motors 299

Phase-shift transformer 256

formers 258

xi

Classification according to environ-

ment and cooling methods 299

CONTENTS

Classification according to electrical

14.3

16.

16.0

Introduction 335

16.

Commercial synchronous

335

14.4

Choice of motor speed 303

14.5

Two-speed motors 303

14.6

Induction motor characteristics under

16.2

Number

various load conditions 305

16.3

16.4

Main Main

Field excitation and exciters 342

14.7

Starting an induction

1

generators 335

motor 308

of poles 335

336

features of the stator features of the rotor

340

14.8

Plugging an induction motor 308

16.5

14.9

Braking with direct current 309

16.6

Brushless excitation 343

14.10

Abnormal conditions 310 Mechanical overload 310

16.7

Factors affecting the size of

16.8

No-load saturation curve 345

14.11 1

4.

1

2

Line voltage changes 3 Single-phasing 310

14.14

Frequency variation 311

4.

1

5

synchronous generators 344

1

14.13

1

circuit

of an ac generator 346

Induction motor operating as a

16.10

Determining the value of

generator 3

16.11

Base impedance, per-unit

1

Complete torque-speed characteristic of an induction machine 314

14.17

Features of a wound-rotor induction

Start-up of high-inertia loads 315

14.19

Variable-speed drives 315

Frequency converter 3

s

348

s

349

350

1

6.

1

2

Short-circuit ratio

1

6.

1

3

Synchronous generator under

16.14

14.18

X X

load 350

motor 315

4.20

Synchronous reactance-equivalent

16.9

14.16

1

SYNCHRONOUS GENERATORS

and mechanical properties 301

Regulation curves 352

1

6.

1

5

Synchronization of a generator 353

1

6.

1

6

Synchronous generator on an

infinite

bus 355

1


16.17

1

Infinite bus-effect of varying the

exciting current 355

15.

EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR 322 15.0 15.1

16.18

Infinite bus-effect

of varying the

mechanical torque 355

Introduction 322 The wound-rotor induction motor 322 Power relationships 325

16.19

Physical interpretation of alternator

behavior 357 1

6.20

Active power delivered by the

motor 326

1

6.2

Control of active power 359

15.4

Breakdown torque and speed 327

16.22

5.5

Equivalent circuit of two practical

1

15.2 1

1

5.3

Phasor diagram of the

nfl, i

generator 358

iction 1

6.23

motors 327 15.6

15.7

5.8

15.9

Power

transfer

between two

sources 361

Calculation of the breakdown

16.24

Efficiency, power, and size of

torque 328

electrical

Torque-speed curve and other


characteristics 1

Transient reactance 359

machines 362

329

Properties of an asynchronous

17.

SYNCHRONOUS MOTORS

generator 330

17.0

Introduction 369

Tests to determine the equivalent

17.1

Construction 370

circuit

33

l


1

7.2

17.3

Starting a synchronous Pull-in torque

372

369

motor 372

CONTENTS

]

1

Motor under load-general

7.4

7.5

18.19

Deducing

xiii

the circuit diagram of a

description 372

single-phase motor 411

Motor under load-simple


1

calculations 373

Power and torque 376

17.7

Mechanical and

17.8

Reluctance torque 378

19.1

Elementary stepper motor 417

Losses and efficiency of a

19.2

Effect of inertia

1

7.9

19.

377

electrical angles

1

synchronous motor 379 17.10 1

7.

1

1

power 380

Excitation and reactive

Power

1

9.0

418

Effect of a mechanical load 4

9.3

19.5

Start-stop stepping rate

19.6

Slew speed 421

420

V-curves 382

17.13

Stopping synchronous motors 383

19.7

Ramping 422

The synchronous motor versus induction motor 385 Synchronous capacitor 385

19.8

Types of stepper motors 422

14

7.

17.15

the

Motor windings and associated 424 19.10 High-speed operation 427 Modifying the time constant 428 19.11 19.12 Bilevel drive 428 19.13 Instability and resonance 434 19.14 Stepper motors and linear drives 434 Questions and Problems 434 19.9

drives


SINGLE-PHASE 1

8.

MOTORS

391

Introduction 39

18.0

Construction of a single-phase

1

induction motor 39 18.2

Synchronous speed 393

18.3

Torque-speed characteristic 394

18.4

Principle of operation

PART

III.

394

ELECTRICAL AND ELECTRONIC DRIVES

Locked-rotor torque 396

1

8.5

1

8.6

Resistance split-phase motor 396

1

8.7

Capacitor-start motor 398

1

8.8

Efficiency and

1

1

Torque versus current 420

19.4

factor rating 38

Introduction 41

17.12

1

18.

STEPPER MOTORS 417

17.6

8.9

18.10 1

8.

1

1

8.

12

1

power

20.

factor of single-

BASICS OF INDUSTRIAL

CONTROL

MOTOR

439

phase induction motors 399

20.0

Introduction 439

Vibration of single-phase motors 40!

20.1

Control devices 439

Capacitor-run motor 402

20.2

Normally-open and normally-closed contacts 443

Reversing the direction of

403

20.3

Relay

Shaded-pole motor 403

20.4

Control diagrams 445

rotation

coil exciting current

443

18.13

Universal motor 404

20.5

Starting

methods 446

18.14

Hysteresis motor 405

20.6

Manual

across-the-line starters

18.15

Synchronous reluctance motor 407

20.7

Magnetic across-the-line

18.16

Synchro drive 408

20.8

Inching and jogging 450

EQUIVALENT CIRCUIT OF A SINGLE-PHASE

MOTOR

Reversing the direction of

20.9

rotation 451 1

8.

1

7

18.18

Magnetomotive force Revolving

motor 410

mmfs

in

distribution

a single-phase

409

20.10 20.

1

1

20.12

447

starters

Plugging 453

Reduced-voltage starting 454

Primary resistance starting 454

448

CONTENTS

xiv

20.

1

3

20.14

Autotransformer starting 458

21.17

Power gain of

Other starting methods 460

21.18

Current interruption and forced

20.15

Cam

20.16

Computers and controls 462

a thy ristor

494

commutation 495

switches 461

21.19

Basic thyristor power circuits 496

2 .20

Controlled rectifier supplying a

1

ELECTRIC DRIVES

passive load (Circuit 7

Fundamentals of electric drives 462

1

8

Typical torque-speed curves 463

1

9

Shape of the torque-speed

20.

1

20.

20.

curve 464 20.20

Current-speed curves 466

20.21

Regenerative braking 467

21 D)

Controlled rectifier supplying an ac-

21 .22

Line-commutated inverter (Circuit

tive load (Circuit 2,

21.0

2 .23 1

472

Potential level

Voltage across some circuit

.2

21 .25

2

1

.4

2

1

.5

Battery charger with series

.7

6,

Table 2 D) 502 1

Delayed triggering-rectifier

21.29

Delayed triggering-inverter mode 507

21.30

Triggering range 508

21.31

Equivalent circuit of a

2 1 .32

Currents in a 3-phase, 6-pulse converter 5

Single-phase bridge rectifier 480

481

Filters

2 .9

Three-phase, 3-pulse diode

21.10

Three-phase, 6-pulse rectifier 485

21.11

Effective line current, fundamental

1

Power factor 511 21.34 Commutation overlap 514 21.35 Extinction angle 514 21.33

483

489 Distortion power

line current

21.12

Three-phase, 6-pulse controllable

mode 505

476

Battery charger with series

rectifier

Table

converter 509

21.8 1

5,

Three-phase, 6-pulse rectifier feeding

1

inductor 478 21

Cycloconverter (Circuit

an active load 504

The diode 475 Main characteristics of a diode 476 resistor

Table

2 .27

THE DIODE AND DIODE CIRCUITS

.6

4,

Basic principle of operation 503

1

1

switch (Circuit

500

21 .26

2 .28

2

3,

21D) 501

elements 474

21.3

static

converter (Circuit

21.1 1

AC

2 ID)

Introduction 472

2

Table 2 ID) 497

Table 2 ID) 498

2 1 .24

FUNDAMENTAL ELEMENTS OF POWER ELECTRONICS 472

Table

21.21


21.

1,

496

factor

490

Displacement power factor, total power factor 490 21.14 Harmonic content, THD 491 21.13

DC-TO-DC SWITCHING CONVERTERS

21.36

DC-to-DC switching converter

21.38

Rapid switching 519

21.39

Impedance transformation 522

2 .40

Basic 2-quadrant dc-to-dc

21.41

Two-quadrant electronic

2 1 .42

Four-quadrant dc-to-dc

1

converter 522

THE THYRISTOR

converter 525

AND THYRISTOR CIRCUITS 21.15

Thethyristor 492

21.16

Principles of gate firing

Semiconductor switches 515

21.37

converter 526

492

21.43

Switching losses 528

5

1

CONTENTS

Dc-to-ac rectangular wave

ELECTRONIC CONTROL OF ALTERNATING CURRENT MOTORS 575

converter 529

23.0

Introduction 575

Dc-to-ac converter with pulse-width

23.1

Types of ac drives 575

modulation 530

23.2

DC-TO-AC SWITCHING CONVERTERS 2 .44 1

2 .45 1

23.

21.46

Dc-to-ac sine wave converter 532

21.47

Generating a sine wave 533

PWM pulse train

Synchronous motor drive using current-source dc link 577

Synchronous motor and

23.3

cycloconverter 580

534

21.48

Creating the

21.49

Dc-to-ac 3-phase converter 535

21.50

xv

Cycloconverter voltage and frequency

23.4

control

Conclusion 537

580

Squirrel-cage induction motor with


23.5

ELECTRONIC CONTROL OF DIRECT-

23.6

cycloconverter 582 22.

CURRENT MOTORS

Squirrel-cage motor and static voltage controller

541

quadrant speed control 541

22.1

First

22.2

Two-quadrant control-field reversal

544

SELF-COMMUTATED INVERTERS 23.8

Self-commutated inverters for cage

23.9

Current-source self-commutated

motors 592

Two-quadrant control-armature

22.3

reversal

545

frequency converter (rectangular

Two-quadrant control-two

22.4

wave) 593

converters 545

Four-quadrant control-two converters

22.5

23.10

Two-quadrant control with positive

22.6

23.

torque 549 22.7

Four-quadrant drive 549

22.8

Six-pulse converter with freewheeling

23.

1

1

1

2

diode 551 Half-bridge converter 556

22.10

Detraction 558

Motor drive using

a dc-to-dc

22. 12

Introduction to brushless dc

22.

Commutator replaced by reversing

Recovering power

in a

wound-rotor

23.14

Pulse-width modulation and ind

modulation 602 motors 604

Synchronous motor

22.

1

1

5

6

U^'COi"

TORQUE AND SPEED CONTROL as a brushless dc

OF INDUCTION MOTORS

machine 568 22.

wound-

Review of pulse-width

switches 566 22. 14

motor 597

23.13

motors 565 3

control of a

rotor induction

PULSE-WIDTH MODULATION DRIVES

switching converter 560

1

wave) 594 Chopper speed

induction motor 599

22.9

1

Voltage-source self-commutated

frequency converter (rectangular

with circulating current 546

1

590

Introduction 54

22.0

22.

589

Soft-starting cage motors

23.7

BGi'l. VAO Q[ ^!3 L fTRM

Standard synchronous motor and

23.15

Dc motor and

brushless dc machine 569

23.16

Slip speed, flux orientation,


604

23.17

Features of variable-speed controlconstant torque

mode 607

A

aBdBwiOiECA

torque 605

Practical application of a brushless dc

motor 569

flux orientation

CONTENTS

xvi

23. 8 1

Features of variable-speed controlconstant horsepower

23.19

24.

Features of variable-speed control-

1

Induction motor and

its

24.

1

2

Equivalent circuit of a practical

Volts per hertz of a practical

Speed and torque control of induction motors 614 Carrier frequencies 615

23.25

Dynamic

Condenser 650

24.15

Cooling towers 650

24. 6

Boiler- feed

24. 7

Energy flow diagram for a steam

24.

Thermal

1

8

stations

and the

environment 652

616

Principle of flux vector control

23.27

Variable-speed drive and electric

NUCLEAR GENERATING STATIONS

618

Principal

23.29

Operating

mode mode

Composition of an atomic nucleus;

24.20

The source of uranium 655

isotopes 655

of the 3-phase

converter 622

Operating

24. 9 1

components 621

23.28

of the single-phase

converter 624

Conclusion 629 Questions and Problems 629

IV.

pump 65

plant 651

control of induction

23.26

PART

thermal generating

648

Turbines 650

motors 615

23.31

646

24.14

1

23.24

23.30

installations

24.13

1

traction

of a hydropower plant 644

Makeup of a station

motor 613 23.23

Makeup

Pumped-storage

THERMAL GENERATING STATIONS

equivalent

motor 612 23.22

0

24.11

circuit 61

23.2

1

mode 610

generator

23.20

Types of hydropower stations 643

24.9

mode 610

ELECTRIC UTILITY POWER

24.21

Energy released by atomic fission 656

24.22

Chain reaction 656

24.23

Types of nuclear reactors 657

24.24 24.25

Example of a light-water reactor 658 Example of a heavy-water reactor 659

24.26

Principle of the fast breeder reactor

SYSTEMS 24.27

660

Nuclear fusion 661 Questions and Problems 661

24.

GENERATION OF ELECTRICAL

ENERGY

635

25.

24.0

Introduction 635

24.

Demand

1

TRANSMISSION OF ELECTRICAL

ENERGY

of an electrical system 635

24.2

Location of the generating station 637

24.3

Types of generating

24.4

Controlling the power balance

24.5

between generator and load 638 Advantage of interconnected

stations

25.0 25.

1

637

Conditions during an outage 641

Frequency and

electric clocks

642

components of a power

664

25.2

Types of power

25.3

Standard voltages 667

25.4

Components of a line

24.7

lines

HV

665 transmission

667

25.5

Construction of a line 668

25.6

Galloping lines 669 Corona effect-radio interference 669 Pollution 669 Lightning strokes 670

HYDROPOWER GENERATING STATIONS

25.7

Available hydro power 642

25.9

25.8 24.8

Principal

distribution system

systems 639 24.6

664

Introduction 664

CONTENTS

25.10

Lightning arresters on buildings 671

25.11

Lightning and transmission lines 671

25.12

Basic impulse insulation level

26 A]

Low-voltage distribution 709 PROTECTION OF MEDIUM-VOLTAGE DISTRIBUTION SYSTEMS

(BIL) 672

25.14

Ground wires 673 Tower grounding 673

25.15

Fundamental objectives of a

25.13

26. 2 1

Coordination of the protective devices 714

26.

transmission line 675

1

3

26.14

25.16

Equivalent circuit of a line 676

25.17

Typical impedance values 676

25.18

Simplifying the equivalent circuit 678

25.19

xvii

Fused cutouts 7

1

Reclosers 716

26.15

Sectionalizers 716

26.16

Review of

Voltage regulation and power-

MV protection

717

LOW-VOLTAGE DISTRIBUTION

transmission capability of

26.17

transmission lines 680 25.20

Resistive line

680

25.21

Inductive line 681

25.22

Compensated inductive

25.23

25.24 25.25

25.26

25.27

25.28

25.29

26.

line

683

1

8

LV

distribution system

Grounding

717

electrical installations 7

Electric shock

26.20

Grounding of 120

V

240V/120V

and

systems 720

Inductive line connecting two

systems 685

26.21

Equipment grounding 721

Review of power transmission 686 line voltage 687 Methods of increasing the power capacity 689 Extra-high-voltage lines 689 Power exchange between power centers 692 Practical example of power exchange 693

26.22

Ground-fault circuit breaker 723

26.23

Choosing the

Rapid conductor heating: 2

I

t

factor

724

26.24

The

26.25

Electrical installation in

role of fuses

725

buildings 725

26.26

Principal

components of an

electrical

725

installation



THE COST OF ELECTRICITY 729 26.

DISTRIBUTION OF ELECTRICAL

Introduction 729

ENERGY 698

Tariff based

upon energy 730

Tariff based

upon demand 730

26.0

Introduction 698

Demand

meter 730

SUBSTATIONS Tariff based 26.1

Substation equipment 698

upon power

factor

732

Typical rate structures 733

Demand

698

26.2

Circuit breakers

26.3

Air-break switches 702

26.4

Disconnecting switches 702

26.5

Grounding switches 702

26.6

Surge arresters 702

1

719

26.19

Power

^giiTAD Of ILF

controllers 733

factor correction

Measuring

\

737

electrical energy, the

watthourmeter 740 27.9

Operation of the watthourmeter 741

26.7

Current-limiting reactors 705

27.10

Meter readout 742

26.8

Grounding transformer 706

27.11

Measuring three-phase energy and

26.9

Example of a substation 707

power 743

26.10

Medium-voltage distribution 709


CONTENTS

xviii

28.

DIRECT-CURRENT TRANSMISSION 746

Why

Distribution system 785

Compensators and

28.0

Introduction 746

28.

Features of dc transmission 746

29. 10

1

28.2

Basic dc transmission system 747

28.3

Voltage, current, and

Power

fluctuations

29.

power

1

1

characteristic

28.6

Power

on a dc

line

29.13

752

28.9

Power reversal 755 Components of a dc transmission line

30.

755

Inductors and harmonic

28.12

Converter transformers 756

28. 13

Reactive power source 757

compensator: principle of

Conclusion 796

Harmonic filters on the ac side 757 28.15 Communications link 757 28.16 Ground electrode 757 28. 7 Example of a monopolar converter station 757 28. 8 Thyristor converter station 758 28.19 Typical installations 760

30.0

Introduction 799

Harmonics and phasor diagrams 799 Effective value of a distorted

wave 800 Crest factor and total harmonic

30.3

distortion

30.5

Displacement power factor and

power

Non-linear loads 804

Generating harmonics 805

30.8

Correcting the power factor 807

30.9

Generation of reactive power 808 EFFECT OF HARMONICS

30. 10

30.

29.

Thyristor-controlled series capacitor

1

1

30.12

Harmonic current in a capacitor 809 Harmonic currents in a conductor 810 Distorted voltage and flux in a

810 Harmonic currents

coil

30.13

in

a 3-phase,

4-wire distribution system 812

(TCSC) 769 29.2

Vernier control 771

29.3

Static

29.4

Eliminating the harmonics 776

29.5

Unified power flow controller

Harmonics and resonance 813 Harmonic filters 818 30.16 Harmonics in the supply network 819

(UPEC) 776

30.17

29.6

Static

synchronous compensator 773

total

804

30.6

SOLID-STATE CONTROLLERS 768

Introduction 768

factor

circuits

30.7

TRANSMISSION AND DISTRIBUTION

29.0

802

Harmonics and


TRANSMISSION POWER FLOW CONTROLLERS

(THD) 801

30.4

1

1

799

30.1

on the

28. 14

1

series

HARMONICS

30.2 filters

dc side (6-pulse converter) 756

29.

The

754

Bipolar transmission line 754

1

principle of


control 753

Effect of voltage fluctuations

1

The shunt compensator:

operation 793

28.7

28.

29. 12

75

28.8

28. 10

circuit

operation 787

Typical rectifier and inverter

28.5

784

analysis 787

relationships 748

28.4

PWM converters?

29.8

29.9

30.14

30.15

Transformers and the

K

factor 821

frequency changer 780

HARMONIC ANALYSIS DISTRIBUTION

CUSTOM POWER PRODUCTS 30.

29.7

1

8

Procedure of analyzing a periodic

Disturbances on distribution

wave 823

systems 782


CONTENTS

31.

PROGRAMMABLE LOGIC

31.18

Getting to

CONTROLLERS

831

31.19

Linking the PLCs 853

Introduction 83

31.20

Capacity of industrial

PLCs 83

31.21

Programming the PLCs 853 The transparent enterprise 855

Elements of

system 832

31.0 3

1

.

3

1

.2

3

1

.3

3

l

.4

1

a control

31.10 31.11

Conventional control circuits and

31.6 3

1

.7

3

1

.8

3

1

.9

circuits

Appendixes 865

AXO

Conversion Charts 865

AX1

Properties of Insulating

Materials 869

AX2

PLC

Mechanical and Thermal Properties of Some

Electrical,

Common

Conductors (and

Insulators) 870

Security rule 847

31.13

Programming the PLC 847 Programming languages 847

31.14

References 859

844

31.12

31.15

851


Examples of the use of a PLC 835 The central processing unit (CPU) 838 Programming unit 838 The I/O modules 839 Structure of the input modules 839 Structure of the output modules 840 Modular construction of PLCs 84 Remote inputs and outputs 84

31. 5

know PLCs

AX3

Properties of

Round Copper

Conductors 871

Advantages of PLCs over relay

Answers

cabinets 848

MODERNIZATION OF AN INDUSTRY

PLCs 850

31.16

Industrial application of

31.17

Planning the change 850

to

Problems 873

Answers to Industrial Application Problems 877 Index 879

xix

1

To Rachel

Part One Fundamentals

Chapter

1

Units

example,

Introduction

1.0

in

measuring length some people use the

inch and yard, while others use the millimeter and

an important role

Units play

everything

effect,

we buy and

thing

we

familiar that

sell is

Some

means of units.

we

our daily

how

lives. In

meter. Astronomers

them

become

for granted,

they started, or

deal with the rod and chain. But these units of length

so

can be compared with great accuracy because the

seldom

why

is based upon the speed of light. Such standards of reference make it possible to compare the units of measure in one country, or in

standard of length

they

were given the sizes they have.

was defined

as the length

of 36 barleycorns strung

end

and the yard

was the distance from the

tip

Centuries ago the foot

to the

end of

we have come

measure more

now based upon

This

way

precisely.

in

improvement in

in

in

defining

Most

units are

which

terms of the speed of

our standards of measure has

hand with the advances

in

1.1

Systems

Over

the years systems of units

of units have been devised

to

A

may be

described as one

in

which

the

units bear a direct numerical relationship to each

technology,

other, usually expressed as a

whole number. Thus

the English system of units, the inch, foot,

the other.

are related to each other by the

Although the basic standards of reference are recall

any

meet the needs of commerce, industry, and science. system of units

and the one could not have been achieved without

ognized by

in

the anchors that tie together the units used in the

and time by the duration of atomic vibrations.

gone hand

measure

Standard units of length, mass, and time are

world today.

and reproducible. Thus the me-

and yard are measured

light,

a long

specialty, with the units of

other.

of King Edgar's nose

the physical laws of nature,

are both invariable ter

one

his outstretched hand.

Since then our units of

to end,

employ the parsec, physicists some surveyors still have to

use the angstrom, and

measured and compared by

of these units have

often take

stopping to think

in

see and feel and every-

The same

countries of the world, the units of

numbers

1

in

and yard

2, 3,

and

36.

correlation exists in metric systems,

except that the units are related to each other by

everyday measure are far from being universal. For

multiples of ten.

3

Thus

the centimeter, meter,

and

FUNDAMENTALS

4

kilometer are related by the numbers 1

00 000.

1

00,

1

000, and

4.

It

centimeters than to convert yards into feet, and this

decimal approach

scientist, the

layman, thereby blending the theoretical and

one of the advantages of the

is

can be used by the research

technician, the practicing engineer, and by the

therefore easier to convert meters into

It is

the practical worlds.

metric system of units.*

Despite these advantages the SI

Today

the officially recognized metric system

the International

System of Units, for which the is SI. The SI was formally

universal abbreviation

introduced

1960,

in

the

at

"Systeme international d

to everything. In specialized areas of

not the answer

and even

in

atomic physics,

The

used to

official introduction

of Units, and

its

be

measure

will continue to

plane angles in degrees, even though the SI unit radian. Furthermore,

1

day and hour

will

is

the

be used,

still

unites.

despite the fact that the SI unit of time

1.2 Getting

may

day-to-day work, other units

more convenient. Thus we

General

Eleventh

Conference of Weights and Measures, under the official title

is

is

SI

Base and derived units

1.3

of the International System

adoption by most countries of the

is

the second.

of the SI

The foundation of the International System of Units rests upon the seven base units listed in Table A. l

world, did not, however, eliminate the systems that

were previously employed.

become

habits, units

cannot readily

from yards

to

let

go.

lates to the

It is

because long familiarity with a

an idea of

magnitude and how

its

(particularly in the electrical it

it

re-

growing importance of SI

the

and mechanical

know

necessary to

TABLE 1A

BASE UNITS

Quantity

Symbol

Unit

And

physical world.

Nevertheless,

makes

we

not easy to switch overnight

meters and from ounces to grams.

this is quite natural,

unit gives us

Just like well-established

a part of ourselves, which

Length

meter

m

Mass Time

kilogram

kg

second

s

Electric current

ampere

Temperature

kclvin

A K

Luminous

candela

cd

mole

mol

fields)

the essentials of this

Amount

intensity

of substance

measurement system. Consequently, one must be able to convert ple,

from one system

unambiguous way.

to

another

in a

sim-

From

In this regard the reader will

these base units

we

derive other units to

discover that the conversion charts listed in the

express quantities such as area, power, force, mag-

Appendix are particularly

netic flux,

and so on. There

number of

units

The

SI possesses a

tures shared

helpful.

number of remarkable

by no other system of

fea-

we can

is

really

derive, but

no

limit to the

some occur so

frequently that they have been given special names.

units:

Thus, instead of saying that the unit of pressure 1

.

2.

It is

It

a decimal system.

employs many

dustry and

pere, kilogram, 3.

It is

units

commerce;

commonly used

some of the most

ships in electricity, mechanics,

The metric

Canada

am-

that

have special names are

listed in

Table

l

B.

and watt.

a coherent system that expresses with star-

tling simplicity

*

in in-

for example, volt,

unit of length

is

the official spelling

heat.

spelled either meter or metre. In is

metre.

TABLE 1B

DERIVED UNITS

basic relation-

and

is

newton per square meter, we use a less cumbersome name, the pascal. Some of the derived units the

Quantity

Unit

Symbol

Electric capacitance

farad

F

Electric charge

coulomb

C

Electric conductance

Siemens

S

UNITS

A

TABLE 1 B

5

tuned to the resonant fre-

epiartz oscillator,

(continued)

quency of cesium atoms, produces a highly accuQuantity

Symbol

Unit

and stable frequency. The ampere (A) is that constant current which, if maintained in two straight parallel conductors of inrate

Electric potential

volt

V

Electric resistance

ohm

tt

Energy

joule

J

finite length,

Force

newton

N

placed

Frequency

hertz

Hz

tween these conductors a force equal

newton per meter of

Illumination

lux

lx

iicni y

n

Luminous flux

lumen

lm

Magnetic flux

weber

Wb

Magnetic flux density

tesla

T

1

1 1

LI LI<_

LuIlLC

Plane angle

radian

The kelvin ature,

watt

W

pascal

Pa

steradian

in

units illustrate the

ated with this

water,

ice,

italics is

sr

and water vapor

definition.

The

The candela

definitions of the SI base

extraordinary precision associunits.

The

text in

explanatory and does not form part of the

by light

in

the length of the path travelled

vacuum during

a time interval of 1/299

until

A

coexist

equal

point

is

to

called the

is

273.16

equal

kelvins,

to 0.01 de-

temperature ofO °C

is

therefore

(cd)

1983 the speed of light was defined 792 458 m/s exactly.

to

he 299

the luminous intensity, in a

is

given direction, of a source that emits monochromatic radiation of frequency 540

X

10

12

hertz and

that has a radiant intensity in that direction of 1/683

watt per steradian. is

the

amount of substance of

system that contains as many elementary

a

entities as

there are atoms in 0.012 kilogram of carbon 12.

Note:

792 458 of a second. ///

is

triple

The mole (mol) is

cooled

cell is

equal to 273.15 kelvins, exactly

definition:

The meter (m)

of water.

triple point

an evacuated

point of water and

triple

base units

modern system of

thermodynamic temperthermodynamic

begins to form. The resulting temperature where

ice

gree Celsius (°C).

official

length.

the fraction 1/273.16 of the

is

temperature of the

by

The following

10~ 7

X

to 2

rad

Power

1.4 Definitions of

vacuum, would produce be-

in

(K), unit of

Pure water

Pressure Solid angle

of negligible circular cross-section, and

meter apart

1

tities

When

the

mole

is

used, the elementary en-

must be specified and may be atoms, mole-

cules, ions, electrons, other particles, or specified

The kilogram (kg) is the unit of mass; it is equal to the mass of the international prototype of

groups of such particles.

the kilogram.

The international prototype of the kilogram

is

a

1.5 Definitions of derived units

particular cylinder of platinum-iridium alloy that is

preserved

in

International

a vault at Sevres, France, by the

Bureau of Weights and Measures.

Duplicates of the prototype exist in all important standards laboratories

in the

world. The platinum-

iridium cylinder (90 percent platinum, 10 percent iridium)

is

about 4 cm high and 4 cm

The second

(s) is the

in

diameter.

duration of 9 192 63

1

770

periods of the radiation corresponding to the transition state

between the two hyperfine levels of the ground of the cesium- 33 atom. 1

Some

of the more important derived units are de-

fined as follows:

The coulomb (C) transported in

(Hence

1

l

is

the quantity of electricity

second by a current of

coulomb =

/

The degree Celsius (°C) and

is

used

in

l

ampere.

ampere second.) is

equal to the kelvin

place of the kelvin for expressing

Celsius temperature (symbol

t)

defined by the equa-

= T — Ta where T is the thermodynamic perature and Tn = 273.15 K, by definition. tion

t

tem-

FUNDAMENTALS

6

The farad

(F)

is

the capacitance of a capacitor

between the plates of which there appears a ence of potential of

volt

I

when

quantity of electricity equal to

force

coulomb.

farad coulomb per volt) The henry (H) is the inductance of a closed circuit in which an electromotive force of volt is produced when the electric current in the circuit varies uniformly at a rate of ampere per second. (Hence

=

1

(I

/

=

I

volt

tromotive force. (Hence

second per ampere.)

angle with

that force

is

which gives

kilogram an acceleration of

second per second. (Hence

I

newton

=

to

I

a

angle with

kilogram

mass and an acceleration,

defined

in

The

terms of a

Exponent form 24

000 000 000 000 000 000 000 000

1()

000 000 000 000 000 000 000

10

000 000 000 000 000 000 000 000 000 000 000

10

l

I

I

I

in length to the radius.

is

the unit of electric conduc-

vertex at the center of a sphere and en-

tesla (T)

000 000

10

I

000 00 1

l()

0.1

0.01

2

IN

0.000 000 001 0.000 000 000 001

6 3

1()

10

2

10'

10-'

fi

l()" 9

10

10"

0.000 000 000 000 000 001

10

i2 15

Symbol

yotta

Y

zetta

Z E

peta

P

tera

T

tr't-

mana

G

mega

M

kilo

k

hecto

h

dec a

da

deci

d

centi

c

milli

m

micro

nano

n

pico

P

fern to

f

alto

a

~ lx

10" 21 10

SI

2

10

10

0 000 000 000 000 000 000 000 001

Prefix

12

uf

0.000 000 000 000 001

0.000 000 000 000 000 000 001

UNITS

exa

0.001

0.000 001

SI

'

l()

10

l

in length to the radius.

the unit of magnetic flux density

,s

000 000 000 000 000 000 000 I

is

equal to one weber per square meter.

also applies to sta-

it

Multiplier

I

its

PREFIXES TO CREATE MULTIPLES AND SUBMULTIPLES OF

TABLE 1C

(S)

of a square with sides equal

is

ampere. )

closing an area of the spherical surface equal to that

meter per second squared.

Although the newton

J volt per

ohm. (The Siemens was formerly named the mho. ) The steradian (sr) is the unit of measure of a solid

meter per

I

ohm =

tance equal to one reciprocal

newton meter)

1

I

vertex at the center of a circle and sub-

The Siemens

1

The newton (N)

its

tended by an arc equal

1

mass of

am-

The pascal (Pa) is the unit of pressure or stress equal to one newton per square meter. The radian (rad) is the unit of measure of a plane

I

I

I

pere, this conductor not being the source of any elec-

The hertz (Hz) is the frequency of a periodic phenomenon of which the period is second. The joule (J) is the work done when the point of application of newton is displaced a distance of meter in the direction of the force. (Hence J joule

=

every application where a

to

1

1

henry

and

involved.

points, produces in this conductor a current of

1

I

is

The ohm (il) is the electric resistance between two points of a conductor when a constant difference of potential of volt, applied between these two

charged by a

is

it

tionary objects

differ-

24

zepto

z

yocto

y

UNITS

the difference of electric poten-

electricity.

They contain notes

between two points of a conducting wire carry-

the reader

who

The volt (V) tial

is

ing a constant current of

(Hence

I

=

volt

is

power

the

is

(Hence

I

=

watt

The weber (Wb) ing a circuit of

one

motive force of

uniform rate in

J joule

watt.

1

in

Quantity

second. {Hence

I

an electro-

it

reduced

is

it

to zero at a

weber

=

I

volt

second.

Multiples

1.6

and submultiples

of SI units

by adding appropriate prefixes

Thus prefixes such as

to the units.

mega, nano, and centi

kilo,

multiply the value of the unit

by factors

radian

rad

square meter

nr

Energy (or work)

joule

J

Force

newton

N

Length

meter

m

Mass Power

kilogram

kg

watt

W

Pressure

pascal

Pa

Speed

meter per second

m/s

1

rad/s

Nm

Volume Volume

cubic meter

m

liter

L

rotation

.

Although the radian

ID, IE, and IF

2.

Most

countries, including

COMMON

UNITS

3.

units some common

IN

A

The newton

is

4.

The pascal

5.

In this

is

units

6.

in

spelled liter or

litre.

(

1

rad/s

= 9.55

The

W

joule per (kilogram kelvin)

J/kg-Kor J/kg °C

Temperature

kelvin

K

Temperature difference

kelvin or degree Celsius

Thermal conductivity

watt per (meter-kelvin)

Kor °C W/m-Kor W/m°C

is

organi-

1

N/nr.

Canada

It

is litre.

Note

Specific heat

K

some

r/min).

official spelling in

J

I

we

57.3°).

mainly used for liquids and gases.

is

Symbol

SI unit

exactly equal to a temperature difference of

I

°C.

I

2

The °C

is

l

I

a recognized SI unit and, in practical

often used instead of the kelvin.

Thermodynamic, or absolute, temperature °C.

is

This unit of volume

watt

is

~

the revolution per minute (r/min) to

joule

it

(as well as

a very small pressure equal to

book we use

Heat

temperature difference of

rad

(I

a very small force, roughly equal to the

Thermal power

calculations, 2.

Canada

book

THERMODYNAMICS

Quantity

1.

in this

force needed to press a doorbell.

encountered in mechanics, thermodynamics, and

TABLE 1E

6

meter.

10 watts.

list

5

the SI unit of angular measure,

is

designate rotational speed

Tables

4

radian per second

6

Commonly used

1.7

3

zations in the United States), use the spelling metre instead of

seconds,

10

megawatt —

1

newton meter

amperes,

-9

=

Note

2

Torque

use the degree almost exclusively

/âmpere = 1000 nanosecond

Symbol

SI unit

Area

J

1

MECHANICS

listed in

Table 1C. For example, 1

IN

Angle

Speed of Multiples and submultiples of SI units are generated

UNITS

per second. )

produces

turn,

COMMON

TABLE 1D

joule per sec-

the magnetic flux that, link-

is

volt as

1

1

1

that gives rise to the

production of energy at the rate of ond.

equal to

watt per ampere.)

I

The watt (W)

particularly useful to

not yet familiar with the SI.

ampere, when the power

1

between these points

dissipated

is

1

The absolute temperature

T is

is

expressed

in kelvins.

On

related to the Celsius temperature

t

the other hand, the temperature of objects

by the equation

T=

t

+

273.

1

5.

is

usually expressed

FUNDAMENTALS

8

COMMON

TABLE 1F

UNITS

ELECTRICITY AND

IN

MAGNETISM Quantity

Symbol

SI unit

Capacitance

farad

F

Conductance

Siemens

c

Electric charge

c

Electric current

coulomb ampere

F nprtrv

joule

j

Erecjuencv

hertz

Hz

Inductance

henry

Potential difference

volt

i

i

A Figure

2

watt

Resistivity

ohm ohm

Magnetic

ampere

Drawn from and Conversion Charts" by Theodore

Enterprises Ltd. All rights reserved.

n

"Metric Units

Om

meter

for units of length.

Conversion chart adapted and reproduced with permission. Copyright © 1991, 1995 by Sperika

W

Power

1.1

Conversion chart

H V

Resistance

field strength

Note

Wildi.

A/m

IEEE

Press, Piscataway NJ, 08855-1331.

3

per meter

Magnetic flux

weber

Wb

Magnetic flux density

tesla

T

4

Magnetomotive force

ampere

A

5

1

.

2. 3.

4. 5.

listed in

descending order of size, and the

Formerly called mho.

of the connected units: the yard

Hz = cycle per second. A/m = ampere turn per meter. T - Wb/nr. What was formerly called an ampere

than the inch, the inch

I

I

ampere:

I

A

I

ampere

turn

is

now simply

simple method.

Conversion charts and their use

Unfamiliar units can be converted to units

is

an arithmetic process that often leaves us

wondering

if

from yard

move downward

in the

Appendix eliminate this problem because they show the relative size of a unit by the position it occupies on the page. The charts in the

in

units are

we reach we want to

to

two arrows

millimeter.

convert from millime-

we

move

up-

ward against

the direction of the arrows until

we

start at

millimeter and

we

apply

the following rules: 1

If, in

.

traveling

move

in the

from one

unit to another,

direction of the arrow,

we

we

multi-

ply by the associated number.

between. Conversely,

2.

The

if

to mil-

we have

ters to yards,

largest unit is at the top, the smallest at the bottom,

and intermediate units are ranked

direction of the

(36 and 25.4) until

Conversely,

from yards

in Fig. 1.1,

reach yard. In making such conversions

our calculations are correct.

The conversion

to convert

limeters. Starting

we know

well by using standard conversion tables. But this strictly

we can

turn.

Suppose we wish

1.8

36 times larger

25.4 times larger than the

convert from one unit to any other by the following

I

called

is

is

millimeter, and so on. With this arrangement

1

1

1

lines join-

them bear an arrow that always points toward the smaller unit. The numbers show the relative size ing

connected by arrows, each of which

if

we move

against the arrow,

we

divide.

bears a number. to the smaller

hence,

its

The number

is

the ratio of the larger

of the units that are connected and,

value

is

always greater than

row always points toward the smaller In Fig.

I.I, for

unity.

The

ar-

unit.

example, five units of length

the mile, meter, yard, inch,

and millimeter

—

are

Because the arrows point downward,

this means when moving down the chart we multiply, and when moving up, we divide. Note that in moving from one unit to another, we can follow any path we

that

please; the conversion result

is

always the same.

UNITS

The rectangles bearing SI units extend toward the

Each rectangle bears

from other units. for the unit as

ENERGY

slightly

of the chart to distinguish them

left

name

well as the

the

9

TNT

kilotonne of J

symbol

1.167 x 10 6

of the unit written .

out in full.

|

I kilowatt hour

.

kW-h |

| 3.6

mega joule

MJ|

|

Example

I- 1

1000

Convert 2.5 yards to millimeters.

British thermal unit

Btu |

|

| 1.055

Solution

kJ

fkilojoule

|

Starting (Fig.

from yard and moving toward millimeter

we move downward

1.1),

the arrows.

1000 in the

calorie

direction of

|

1

| 4.184

We must therefore

multiply the numbers

joule |

associated with each arrow: N-m

|n ewton-meter

2.5

-

yd

2.5

(

= 2286

X

36)

X

(

25.4) millimeters van second

mm

I

I

6.24 x 10 18

Example 1-2

eV

[electronvolt

|

Convert 2000 meters into miles.

Figure 1.2 Solution Starting

move

See Example from meter and moving toward mile, we

and then against, the direction of

with,

first

the arrows.

Consequently,

2000 meters

= 2000 X

1

(

= 2000 X

we

1 995 by Sperika Enterprises Drawn from "Metric Units and Conversion Charts" by Theodore Wildi. IEEE Press,

sion.

Copyright© 1991,

Ltd. All rights reserved.

obtain

.0936) (- 1760) miles

Piscataway, NJ, 08855-1331.

1.0936

1760

=

1-3.

Conversion chart adapted and reproduced with permis-

The per-unit system of measurement

1-9

1.24 mi

The

SI units just described enable us to specify the

Example 1-3

magnitude of any quantity. Thus mass

Convert 777 calories to kilowatt-hours.

kilograms, volts.

Solution

Referring to the chart on

moving from calorie travel

downward

to

ENERGY

(Fig. 1.2)

kilowatt-hour,

we

and first

(with the arrow 4.184) and then

upward (against the arrows 1000, 1000, and Applying the conversion rule,

we

power

in watts,

and

However, we can often

size of

expressed

in

get a better idea of the

something by comparing

thing similar. In effect,

is

electric potential in

it

to the size of

some-

we can create our own unit and

specify the size of similar quantities

compared

to this

arbitrary unit. This concept gives rise to the per-unit 3.6).

find

method of expressing

the

magnitude of a quantity.

For example, suppose the average weight of

777 calories

= 777 (X 4.184) (- 1000)

adults in

1000) (- 3.6)

New

any individual

=

9.03

X

10"

4

kW

h

York

is

130

lb.

Using

this arbitrary

weight as a base, we can compare the weight of

a

in

terms of

person weighing 160

lb

this

base weight. Thus

would have

a per-unit

FUNDAMENTALS

10

weight of 160

weighing

mb/ 130

I

The

lb/

lb

=

130 lb

-

Another person

1.23.

would have

15 lb

1

n

3500

weight of

a per-unit

a

4800

0.88.

measurement has

per-unit system of

vantage of giving the size of a quantity

±

the ad*2

terms of

in

450

xv 3000

n

n

a particularly convenient unit, called the per-unit in

reference to our previ-

a football

player has a per-unit

base of the system. Thus,

ous example,

weight of

1.7

if

we immediately know

above average. Furthermore,

far 1.7

X

130

Note

=

that

221

1

.7

per-unit,

To

where

it

is

Figure 1.3

weight

is

Conventional

are given, they

would be absurd

weighs

state that the football player is

weight

1

.7 lb.

to

His weight

the selected base unit

generalize, a per-unit system of

consists of selecting one or

is

130

lb.

Fig.

1

.3,

composed of several

we

them. In this book

decide to use an impedance of base, the per-unit

measuring

*,(pu)

and impedance.

The base may be power

is

40

ratings of 25 hp,

hp,

The

1

the base

=

3. is

of 15 hp. In

40 hp/50 hp

-

0.8 and

world

in this per-unit

and 150 hp/ 15 hp

-

^50

n 4800 n

if

expressed

_ "

=

1.67, 40 hp/ 15 hp

3.2

3000

a

1500

ft

in

vector notation

shown

per-unit values.

in Fig.

1

is

same

impedances are

We can

solve this

other circuit. For example,

used, the per-unit circuit

is

that

.5.

power

-

2.33(pu)

3.2(pu)

-nrrv

2.67,

10.

therefore important to

value, the actual values of

per-unit

_ ~

real circuit, but the

we would any

R 2 (pu)

know the magnitude If we do not know the quantities we are

0.30

i

dealing with cannot be calculated.

The

_ 0.30 ~

150011

of the base of the per-unit system. its

12

1500

and 3 pu, respectively.

well have selected a base

case the respective per-unit rating

this

2.33

50 hp, the three motors have

would be 25 hp/ 15 hp It is

0.5;

ratings of 0.5, 0.8,

We could equally

=

per-unit circuit (Fig. 1.4) contains the

now

Thus,

=

elements as the

circuit as

of 50 hp. The corresponding per-unit ratings

hp

XL (pu)

said to have

PH

where power

a

3500

a power, a voltage,

and 150 hp. Let us select an arbitrary base power

=

=

Xc (pu)

a current, or a velocity. For example, suppose that

50 hp/50 hp

=

R 2 (pu)

one quantity as our

system

stick, the per-unit

are then 25 hp/50

we

150011

system with one base

three motors have

If

as the

impedances are as follows:

more convenient mea-

select the size of only

a single base.

500 ohms

are particularly interested in

current, power, torque,

we

1

measurement

selecting convenient measuring sticks for voltage,

1.10 Per-unit

resistors, capacitors,

and inductors having the impedances shown.

suring sticks and comparing similar things against

If

circuit.

lb.

whenever per-unit values

always pure numbers. Thus

are

his

his actual

method can

also be applied to im-

pedances. Consider, for example, the circuit

in

Figure 1.4 Per-unit circuit.

2(pu)

UNITS

2.33

3.2

1

1

In order to understand the significance of this re-

j

sult, the

reader should study the two following ex-

amples. The bases are the same as before, namely

4kV

0.30

500

Figure 1.5

=

/B

kW

ZB =

A

125

32 ft

Example 1-4

Per-unit circuit with

notation.

i

A 400 II resistor carries a current of 60 A. above base values, a.

system with two bases Per-unit

1.11

b. c.

when two bases

particularly useful

are

bases are usually a base voltage £" B

P B Thus

power 4

becomes used. The

electrotechnology the per-unit system

In

.

kV and

d. e.

The per-unit resistance The per-unit current The per-unit voltage across the resistor The per-unit power dissipated in the resistor The actual E and P of the resistor

and a base

the selected base voltage

may

Solution

be a.

the selected base

power 500 kW.

The

per-unit resistance

The two base values can be selected quite indeb.

unit

The

per-unit current

system

is

that

it

automatically establishes a cor-

/>w

For example, is

base power

PB

base voltage

EH

impedance Z B

and the base

if

—

is

d.

base voltage

The

the base voltage

/

4

is

500 kW, the base current

ZB = £ b /'b = In effect,

system

impedance.

per-unit

is

kV

and the base

is

1

The

0.48 is

V/125

A =

32

ft

also get a base current and a base

Consequently,

the

so-called

=

0.48

=

6

=

X X

tf(pu)

12.5

is

E(pu)

=

6

-

2.88

X

X

/(pu)

0.48

is

E — E B X E(pu) = 4 kV X 6 = 24 kV

is

4( )00

/(pu)

actual voltage across the resistor

A

25

=

power

P(pu)

by selecting the voltage/power per-

we

The

n

= ^b/^b = 500 000/4000 =

The base impedance

system.

12.5

per-unit voltage across the resistor

EH

—

base current

e.

'b

per-unit

=

/ B is

£(pu)

unit

ft/32 ft

= 60 A/ 125 A =

/(pu)

Thus c.

base current

power

is

interesting feature of the voltage/power per-

responding base current and base impedance. the

= 400

tf(pu)

pendently of each other.

One

Using the

calculate:

2-base

system really gives us a 4-base per-unit

The

actual

power

dissipated in the resistor i$^;^'as''

P = PB x

^(pu)

= 500 kW X = 1440kW

BCOLTAD

2.88

Df.

RU,]?\

vlii TA A

©IBUCUfcCA

FUNDAMENTALS

12

Example 1-5

A

7.2

/|.(Pu)

.

= 2.844

kV

source delivers power to a 24 II resistor

and a 400

kW

electric boiler (Fig. 1.6).

Draw

equivalent per-unit circuit diagram. Use the

base values as

Example

in

the

same

/i(pu)

/,(pu)

= 0.444

= 2.4

1-4. R{pu)

boiler

[

0.75

j

Calculate

The The The The The The

a.

b. c.

d. e. f.

per-unit E(pu), /?(pu), PCpu) per-unit current

P(pu)

0.8

/ 2 (pu)

per-unit line current

(pu)

l {

power absorbed by the resistor power absorbed by the resistor

per-unit actual

Figure 1.7 Per-unit version of Figure

1

.6.

actual line current

The

per-unit line current

/t

(pu)

ft

24

n

r 400 kW

The

d.

per-unit

is

=

/,(pu)

+

/ 2 (pu)

=

0.444

+

2.4

=

2.844

power

P(pu) Figure 1 .6 See Example

/L

in the resistor is

= =

E(pu)

=

4.32

X

X

1.8

/ 2 (pu)

2.4

1-5.

The

e.

actual

power

in the resistor is

Solution a.

The

per-unit line voltage

£,(pu)

P 2 = P a X «pu) = 500 kW X 4.32

is

- 7.2kV/4kV =

1.8

= 2160 kW The

per-unit resistance

is

The

f.

R(pu)

= 24

n=

fi/32

actual line current

=

12

The

per-unit

power of the

P(pu)

We b.

can

The

boiler

=

is

= 400 kW/500 kW =

now draw

c.

The

=

£(pu)//?(pu)

=

1.8

-

2.4

-

-

1

/,

125

Name

/,

X

(pu)

2.844

=

355.5

A

the seven base units of the

International

0.75 1

-2

1-3

per-unit current

X


is 1

/ 2 (pu)

IB

0.8

the per-unit circuit (Fig. 1.7)

per-unit current f2

is

0.75

Name

System of

Units.

five derived units of the SI.

Give the symbols of seven base

units,

paying

is

particular attention to capitalization. /,(pu)

=

P(pu)/E(pu)

-

0.444

=

0.8/1.8

1-4

Why

are

some derived

units given special

UNITS

1-5

What ergy,

1-6

1-58

revolution

1

-60

oersted

1-59

degree

1

-6

ampere

are the SI units of force, pressure, en-

power, and frequency?

Give the appropriate prefix for the following multipliers: 100, 1000, 10

1/1000,

10" 6

,

Make

the following conversions using the conver-

1/10, 1/100,

,

io'-\

1-62

10 square meters to square yards

1-63

250

1-64

1645 square millimeters

Express the following SI units in symbol form: 1-7

megawatt

1-21

millitesla

1-8

terajoule

1-22

millimeter

1-9

millipascal

turn

sion charts:

10" 9 ,

6

1

1

1-23

1

revolution

1-10

kilohertz

1-24

megohm

1-11

gigajoule

1-25

megapascal

1-12

milliampere

1-26

millisecond

-65

1

3

MCM to square millimeters

000

circular mils to square millimeters

1-66

640 acres

1-67

81

1-68

to square inches

square kilometers

to

000 watts

to

Btu per second

33 000 foot pound-force per minute to kilowatts

1-13

microweber

1-27

picofarad

1-14

centimeter

1-28

kilovolt

1-15

liter

1-29

megampere

1-16

milligram

1-30

kiloampere

1-17

microsecond

1-31

kilometer

1-69 1

-70

1-71

1-18

millikelvin

1-19

milliradian

1-20

terawatthour

1-32

1-33

1

-34

flow

I

-38

density

1-39

power

1-36

plane angle

1-40

temperature

1-41

microjoules

10 pound-force to kilogram-force

60 000

-72

1

-73

1

-74

50 ounces

1

-75

76 oersteds

1

-76

5

1

-77

1

lines per square inch to teslas

kilogauss

.2 teslas to

to

kilograms

to

000 meters

amperes per meter

to miles

and

frequency

magnetic flux

to

meters

milliliter

1-35

1-37

0 foot pound-force

1

symbol:

rate of

1

feet to cubic

nanometer

State the SI unit for the following quantities

write the

250 cubic

mass

80 ampere hours

1-78

25 pound-force

1-79

25 pounds

1-80

3 tonnes to

1-81

1-82

Give the names of the SI units that eorresponcl to

1

00 000

0.3

to

to

coulombs

newtons

to

kilograms

pounds

lines of force to

webers

pounds per cubic inch

kilograms per

to

cubic meter

the following units:

1-83 1-42

Btu

1-51

bar

1-43

horsepower

1-52

pound-mass

1-44

line

of flux

1-53

1

-84

1

-85

pound-force

2 inches of mercury to millibars

200 pounds per square inch

to pascals

70 pounds-force per square inch

to

newtons

per square meter 1-45

inch

1-54

kilowatt-hour

1-46

angstrom

1-55

gallon per

1-47

cycle per second

1-48

gauss

1-49

line per

1-50

°F

1

1

square inch 1-57

mho

-87

1-88

pound-force per 1

square inch

1

5 revolutions per minute to radians per

second

minute 1-56

-86

-89

A temperature of 20 °C to kelvins A temperature of 200 °F to kelvins 1

A temperature kelvins

difference of

1

20° Celsius to

14

I

-90

FUNDAMENTALS

A resistance of 60 ft

is

Industrial application

selected as the base

resistance in a circuit. If the circuit contains 1

-94

A

motor has an efficiency of 92.6%. What

is

the efficiency in per-unit?

A

variable-speed motor having a nameplate

three resistors having actual values of 100 fl,

3000

II,

and 20 O, calculate the per-unit 1-95

value of each resistor. 1-91

A power of 25 kW

rating of 15 hp,

and a voltage of 2400

are selected as the base

voltage of a

V

power and base

power system. Calculate

890 r/min develops a

torque of 25 newton meters

at

1260 r/min.

Calculate the per-unit values of the torque,

the

speed, and power.

value of the base impedance and the base

1-96

Three

resistors

have the following

ratings:

current. 1

-92

A resistor

has a per-unit value of 5.3.

base power 12

is

470

is

250

If the

resistor

resistance

A

10012

kW and the base voltage ohmic value of

V, calculate the

the resistor.

1-93

A

length of 4

m

is

selected as a base unit.

b.

the per-unit length of

1

mile

C

300

II

the per-unit length of

1

foot

of resistors 1

d. e.

the per-unit value of a

the magnitude of the base area (in

m f.

ft

W W 40 W 24 75

and voltage

values of the resistance, power,

m2 3 the magnitude of the base volume (in m

c.

50

Using resistor A as a base, determine the per-unit

Calculate a.

B

power

3

the per-unit value of an area of 2 square

A

rating

C, respectively.

30 hp cage motor has the following cur-

rent ratings:

)

volume of 6000

-97

B and

)

FLA:

full-load current

LRA: locked

NLA:

36

A

rotor current 2 8 1

A

no-load current 14 A.

miles

Calculate the per-unit values of LRA and NLA.

Chapter 2 Fundamentals of Electricity, Magnetism, and Circuits

Introduction

2.0

positive

^

negative (

terminal (+) -fli-ftr

This

chapter briefly reviews

some of

t

ermina

the funda-

mentals of electricity, magnetism, and circuits.

We assume

the reader

already familiar with the

is

dry

cell

including the solution of electric circuits.

basics,

However, a review

useful because

is

it

focuses on

those items that are particularly important in

technology. Furthermore,

used throughout this

Some

currents.

it

book

power

establishes the notation

and

to designate voltages

Figure 2.1

of the topics treated here will also

Dry

cell,

provide the reader with a reference for subjects covered in later chapters. electron current flow

Conventional and electron

2.1

current flow Consider the dry cell positive

(

shown

+ and one

difference of potential volts) is tive

terminal

If

(

—

)

1

,

having one

terminal.

The

between them (measured

in

to an excess of electrons at the nega-

compared

we connect

tential in

due

in Fig. 2.

negative

)

to the positive terminal.

a wire across the terminals, the po-

difference causes an electric current to flow is composed of a steady come out of the negative

the circuit. This current

stream of electrons that terminal,

move along

the wire, and reenter the cell

Figure 2.2

by the positive terminal (Fig. 2.2),

Electron flow.

15

'

<-)

FUNDAMENTALS

16

Before the electron theory of current flow was fully understood, scientists of the 17th century arbi-

decided

trarily

that current in a

from the positive terminal

to the negative terminal

This so-called conventional current flow

(Fig. 2.3).

used today and

is still

conductor flows

is

the accepted direction of

power technology. book we use the conventional current

current flow in electric In this

flow, but

flow

is

it is

worth recalling

that the actual electron

opposite to the conventional current flow.

In

order to establish a general rule, consider two

black boxes

A and B

ally

changing

between sources

/ that is

in direction (Fig. 2.4).

drop along the wires

assumed

is

that are

connected

to be zero.

some way

in

A2

to the external ter-

.

Under such highly

are also continually changing.

is

how can we

tell

To answer

sometimes important

and loads

in

an electric

to identify the sources

circuit.

By

definition, a

source delivers electrical power whereas a load absorbs

it.

Every

electrical device (motor, resistor,

thermocouple, battery, capacity, generator,

How

can

we

tell

the

A or B

the question,

suppose we have appro-

instantaneous polarity the terminals

(

+ )( —

)

determine the

of the voltage across

and the instantaneous direction of

conventional current flow. The following rule then applies:

etc.) that

carries a current can be classified as either a source

or a load.

whether

a source or a load?

priate instruments that enable us to is

Each

and B h B 2 A variable voltage exists across the terminals, and its magnitude and polarity minals A],

and loads It

continu-

The voltage

box contains unknown devices and components

variable conditions,

2.2 Distinction

connected by a pair of

that are

wires carrying a variable current

•

A device

is

a source

whenever current flows out

of the positive terminal.

one from the other? •

A device

is

a load

whenever current flows

into a

positive terminal.

conventional current flow

If the

instantaneous polarities and instantaneous

current flow are as

from the load.

rule that

However,

if

shown

box

A

is

in

and box

The above is

A the

follows

B

is

a

box B would become

load.

rule for establishing

a source or load

it

the current should reverse while

the polarity remains the same, the source

Fig. 2.4,

a source and box

is

whether a device

very simple, but

it

has impor-

tant applications, particularly in alternating current circuits.

Figure 2.3

Some

devices, such as resistors, can behave only

as loads.

Other devices, such as photocells, can act

only as sources. However,

many

Conventional current flow. either as sources or as loads.

/

Figure 2.4 between a source and a

Distinction

load.

devices can behave

Thus when a

battery

FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS

power,

delivers electric

flows out of the recharged, (4

)

(

+

)

it

when

being

is

it

acts as a load (current flows into the

it

The

acts as a source (current

terminal);

ties

•

on a system, but they can briefly behave

generators

When

is

thing

when

is

like

£AB = + tween

100 V, which reads: The voltage be-

A and B

is

100 V, and

A

is

(

the capacitor

it

+ is

acts as a source

On

terminal.

)

charging up,

and current flows into the

+

(

)

it

•

positive with

the

£ BA = 100 V, which reads: The voltage between A and B is 100 V, and B is negative with respect to A.

acts

As another example,

terminal.

if

we know

ator voltage in Fig. 2.6 has a value

that the gener-

E2 = — \

100 V,

we know that the voltage between the terminals is 00 V and that terminal 2 is negative with respect to

Sign notation

2.3

relative polari-

true of capacitors.

discharging

and current flows out of the

as a load

and the

can be designated by the

respect to B.

The same

a capacitor

other hand,

B

and

the electromechanical conditions are

if

appropriate.

A

double-subscript notation, as follows:

terminal). Similarly, electric motors usually act

as loads

potential difference

of terminals

1

1

In

arithmetic

we

scribe addition

mechanics, direction

)

to de-

etc.,

to indicate the

compared

+

that the direction

This interpretation of

(

+

to an ar-

the speed

if

—400 r/min,

100 r/min to

of rotation has reversed. )

and

—

(

terminal

2.5 Sign notation for voltages Although we can represent the value and the polarity of voltages by the double-subscript notation (£, 2

£ AB

in the

signs

is

fre-

we

etc.),

£2 V

etc.) (

+

sign.

)

in

marked with

to

G

having a positive terminal

terminal B. Terminal

A is positive

minal B. Similarly, terminal spect to terminal A. tive

a positive

(

+

)

by

sign.

is

itself:

it

is

Note

B

is

A and

The other terminal

2

shows

a

a negative

with respect to ter-

negative with re-

that terminal

A is

not posi-

only positive with respect to B.

Figure 2.6 = 100 If E 21 terminal

V,

terminal 2

is

a

arbitrarily

describe a system of notation that enables us

indicate the polarity of voltages. Fig. 2.5

source

(£,,

For example. Fig. 2.7 shows

which one of the terminals

Double-subscript notation for voltages

We now

symbol

,

It

and identifying one of the terminals by a

chapters that follow.

source £,

2.4

often prefer to use the sign notation.

consists of designating the voltage by a ,

)

,

positive

met

quently

1.

and

of an electric current, of a mechanical

motor changes from

means

)

In electricity

meaning

the

(

chosen direction. For example,

bitrary

it

we broaden

+ and —

(

and subtraction.

of a rotational speed,

force,

of a

use the symbols

negative with respect

1.

Figure 2.5

Figure 2.7

Double-subscript notation to designate a voltage.

Sign notation to designate a voltage.

to

FUNDAMENTALS

18

unmarked, but

is

is

automatically assumed to be neg-

ative with respect to the

With •

If

(

we

this notation the

+

= +

state that £,

)

indicated in the diagram.

+

(

)

sign

that the

The terminal bearing

then actually positive and the

is

other terminal

is

negative. Furthermore, the

magnitude of the voltage across the terminals

Figure 2.9

islOV.

Solution of

Conversely,

•

means

10 V, this

of the terminals corresponds to that

real polarity

the

terminal

following rules apply:

if

the terminals

£,

= -

shown on

diagram. The terminal bearing the

(

+

)

sign

actually negative, and the other terminal

is

The magnitude of the voltage across

tive.

terminals

is

the

is

the

10 V.

circuit of Fig. 2.8 consists of three sources

7

V

V2 and V? ,

,,

with a positive in series to

— each (

+

)

—

having a terminal marked

The sources are connected using jumper wires A, B, C,

sign.

a resistor R,

Determine the actual value and polarity of the voltage across each source, V,

spectively. In effect, point

V2 = +

10 V, and

knowing

V? = -40

that

V

=

ues and polarities are as in

—

That

is

a negative sign.

2.6

Graph

of

an alternating voltage

In the chapters that follow,

whose voltages change alternating voltages

voltage

at

2.

1

may be

0).

we encounter

sources

polarity periodically.

The

Such

represented by means

vertical axis indicates the

each instant, while the horizontal axis

in-

dicates the corresponding time. Voltages are positive

when they when

are

E2

above the horizontal axis and nega-

they are below. Figure 2.10 shows the \

produced by the generator of Fig,

2.6.

find that the true val-

shown in Fig. 2.9. However, jumper A, it seems im-

2

directing our attention to

possible that (

re-

negative with respect to point C.

voltage

we

only

{

V.

Solution stated,

is

B and positive with respect why A carries both a positive and

tive

Using the rules just

A

It

jumpers B and C,

to point

of a graph (Fig.

and D.

-4

inherently positive nor inherently negative.

has a polarity with respect to

posi-

Example 2-1 The

2-1

10 V, the real polarity of

the reverse of that

is

Example

).

it

can be both positive

However, we must remember

Figure 2.8 Example

Circuit of

2-1.

(

+

)

and negative

that

A

is

neither

Figure 2.10 Graph of an alternating voltage having a peak

of

1

00

V.


from zero,

Starting taining

+ 100 V

falls to

zero

E2]

gradually increases,

after 0.5 second.

spect to terminal

because

1

E2

During the interval from

1

spect to terminal polarities

onds are

1

.

2.7 Positive

is

insets

at 0.5,

I, IT,

E2X

is

negative with re-

The instantaneous

of the generator

shown by

positive.

to 2 seconds,

negative; therefore, terminal 2

0

this

positive with re-

is is

\

at-

then gradually

It

end of one second. During

at the

one-second interval, terminal 2

1

and

voltages and

.5,

and

TIT

of Fig. 2.10.

2.

1

7 sec-

and negative currents 0

We

also

indicate

1

make use of positive and negative signs to the direction of current flow. The signs are

*-

time

>v

X

|

allocated with respect to a reference direction given

on the circuit diagram. For example, the current a resistor (Fig. 2.11)

Y to

X.

One

be positive

may flow from X

of these two directions

+

(

is

in

Y or from

considered to

and the other negative

)

to

(

—

Figure 2.13

and the corresponding graph

Electric circuit

The arrow

).

of current.

indicates the positive direction of current flow.

Solution

According zero to

Because

may

is

the interval

flow from

X

to

Y

or from

Y

to X.

interval

from

from

+2 A to

in the resistor.

to

l

it

to

l

second.

from B

to

A

of the arrow). During the

2 seconds, the current decreases

zero, but

circulates

it still

Between

increases from zero to ative,

from 0

positive, the current flows

in the resistor (direction

Figure 2.11 Current

it

from

to the graph, the current increases

+2 A during

from B

to

A

2 and 3 seconds, the current

—2 A and,

because

it

is

neg-

really flows in a direction opposite to that

of the arrow; that

is,

from

A

to

B

in the resistor.

Figure 2.12 Circuit

element showing positive direction

of current flow.

2.8 Sinusoidal voltage The

The positive direction means of an arrow (Fig. 2

A flows from X to Y,

and if

is

it

is

2.

1

shown

2).

flows

Thus,

of the arrow),

from is

it

if

by

a current of

+2

Y to X (direction opposite to that

designated by the symbol

graph

a resistor

in

shown

of this graph.

in

Figure

2.

R 1

—2

A.

varies according to the

3.

It

may

therefore

be expressed by the equation e

= Em

cos

{lirft

+

6)

(2.1)

A. Conversely,

Example 2-2 The current

ac voltage generated by commercial alternators

very nearly a perfect sine wave.

in the positive direction


current flows

arbitrarily

is

Interpret the

meaning

where e

—

Em — / =

instantaneous voltage [V]

peak value of the sinusoidal voltage [V] frequency [Hz

t

=

time

6

=

a fixed angle [radj

[s]

FUNDAMENTALS

20

The expression lirft and 6 are angles, expressed However, it is often more convenient to

The voltage

in radians.

express the angle

in

e ab

degrees, as follows:

e

= Em

cos (360 ft

*

= £ ni

cos

+

In these

(2.3)

0)

equations the symbols have the same sig-

s is

100 cos 488 622°

X 50 X 27.144 +

30°)

V

moment

at this

terminal a

Note

21 AAA-

100 cos (360

-20.8 Thus,

+

=

= or <<|>

t

=

(2.2)

6)

at

the voltage

is

V

—20.8

and

negative with respect to terminal b.

is

that an angle of

488 622° corresponds

to

488

622/360 = 357 complete cycles plus 0.2833 cycle. The latter corresponds to 0.2833 X 360° = 102°, and 100 cos 102° = -20.8v 1

nificance as before, and the time-dependent angle

(= 360

ft) is

Example 2-3 The sine wave

_ in Fig.

2.14 represents the voltage

£. lb across the terminals a

operates at 50 Hz.

and b of an ac motor

Knowing

that 6

100 V, calculate the voltage

27 .1 44

also expressed in degrees.

at

t

2.9 Converting cosine functions into sine functions

that

= 30°, and E m = = 0 and at t =

We can convert a cosine function of voltage or current into a sine function

Em Solution

The voltage

at

t

e, xh

=

At

this

is

moment

0

Em

is

= E m cos (360 ft + 9) - 100 cos (360 X 50 X = 100 cos 30°

-

a

by adding 90°

to the angle 6.

Thus,

s.

86.6

Similarly,

0

+

30°)

+86.6

V

and terminal

B

+

= 90°)

we can convert a sine function

/m

is

0)

(360 ft

cos (360 ft

+

6)

(2.4)

into a cosine

function by subtracting 90° from the angle

V

the voltage

+

(360 ft

sin

/ m sin

therefore positive with respect to terminal b.

+

cos (360 ft

6.

Thus,

=

+ 6-90)

(2.5)

2.10 Effective value of an ac voltage Although the properties of an ac voltage are known when its frequency and peak value Em are specified, it is much more common to use the effective value £e(j .

For a voltage

E

between

cfi

-

that varies sinusoidally, the relationship

and

£ni

is

given by the expression

Ecn = EJ,2

(2.6)

-

The

effective value of an ac voltage

is

some-

times called the

RMS

the voltage.

a measure of the heating effect of

It is

(root

compared

the ac voltage as

mean square) value of to that

of an equivalent

dc voltage. For example, an ac voltage having an fective value of

1

ing effect in a resistor as does a dc voltage of

Figure 2.14 Sinusoidal voltage having a peak value of 100

expressed by e ab =

Em

cos (360

ft

+

30°).

V and

ef-

35 volts produces the same heat1

35 V.

The same remarks apply to the effective value of an ac current. Thus a current that varies sinusoidally and whose peak value is / m possesses an effective value

/C

given by |

t


U

n

2

(2.7)

Most alternating current instruments are brated to rent

show

c.

e

cali-

the effective value of voltage or cur-

and not the peak value (Fig. 2.15).

value of an alternating voltage or current

When

given

is

339

it

understood

that

it

the

is

effective

E^-and

Furthermore, the subscript in

Owing

E

given by

sin 21

600

t

t

to the phase lag of 30°, the current

is

value.

given by /eir is

dropped /

and the effective values of voltage and current are simply represented by the symbols

is

= E m sin 360 ft = 339 sin 360 X 60

the d.

is

Let us assume the voltage

21

and

/.

= /m sin (360 ft - 30) = 14.1 sin (21 600 f

=

14.1 sin

(

-

30)

30)

Example 2-4

A 60 Hz source having an effective voltage of 240 V delivers an effective current of 10 A to a circuit. The current lags the voltage by 30°. Draw the waveshape for

E and

Em = E b.

/

In

The peak voltage x

2

m =

/

1

Phasor representation

most power studies the frequency

we simply

is

= 240

x

2

val-

1

take

it

is

fixed,

for granted. Furthermore,

and so

we

n

2

is

magnitudes and phase angles.

= 10x2=

14.1

are

not particularly concerned with the instantaneous

= 339 V

voltages and currents but more with their

The peak current /

The waveshapes giving the instantaneous ues of e and are shown in Figure 2. 6.

2.1

/.

Solution a.

e.

A

ages are measured

in

And because the

RMS volt-

terms of the effective values

E

Figure 2.15

Commercial voltmeters and ammeters are graduated ing

up

to

2500 A and 9000

V.

(Courtesy of General Electric.)

in

effective values. This

range

of instruments

has scales rang-

FUNDAMENTALS

22

\

16.67

\

30°

14

•

A

\

\

(a)

ms

A

1

'*

9'

1

t

E

h-

1.39

E

/

V

339

I

ms

Figure 2.17 current phasor

The

/

and voltage phasor

E are

in

phase.

Figure 2.16 Graph showing the instantaneous values of voltage and current. The current lags 30° behind the voltage. The effective voltage is 240 V and the effective current is 10 A.

E and

interested in

E m we

peak values

rather than the

are really only

,

Figure 2.18 Phasor / lags behind phasor

0.

E by an

angle of

0

degrees.

This line of reasoning has given rise to the phasor method of representing voltages and currents.

The

basic purpose of phasor diagrams

sense that

it

A

phasor

is

phasor

show

represents.

the

same angle

If a

phasor

bears an arrow, and

its

length

is

The angle between two phasors

to the electrical phase angle

propor-

3.

is

it

equal

between the quantities.

1

.

Two

phasor

E is / is

phasors are said to be

are parallel to each other

them

in

phase when they

and point

in the

same

Two

one

The phase angle between

direction as phasor

said to lead phasor

/.

same

Fig. 2.

is

1

8,

it

/ lags

behind

make

then

Conversely, a

E if phasor make

has to be rotated counterclockwise to

point in the

/,

it

direction. Thus, referring to

clear that phasor

£ by

we could

E

leads phasor /

equally well say that

0 degrees.

then zero.

is

phasors are said to be out of phase

they point in different directions. gle

to be rotated clockwise to

same

by 6 degrees. But

4. 2.

rotate

line up.

said to lag behind phasor

rules apply to phasors:

direction (Fig. 2.17).

we

sweep through

to

make them

to

E has

point in the

phasor /

The following

we have

similar to a vector in the

tional to the effective value of the voltage or current it

Consequently, whether

/.

phasor or the other,

between voltages

the magnitudes and phase angles

and currents.

to

is

between them

is

point in the

same

The phase

the angle through

one of the phasors has

when

to be rotated to

Referring

phasor

/

now

to Fig. 2.

1

9

we could rotate (3 to make it

clockwise by an angle

an-

which

make

it

direction as the other. Thus,

referring to Fig. 2. 1 8, phasor / has to be rotated

counterclockwise by an angle 6 to in the

same

phasor

E has

gle 6 to

make

it

point

direction as phasor E. Conversely,

make

to be rotated it

clockwise by an an-

point in the

same

direction as

Figure 2.19 Phasor / leads

£ by

B

E by

degrees.

(3

degrees. But phasor

/

also lags


same

point in the

direction as phasor E.

could then say that phasor

23

We E by

leads phasor

/

P degrees. But this is the same as saying that phasor / lags phasor E by 0 degrees. In practice,

we always

select the smaller phase angle

between the two phasors

to designate the lag

or lead situation. 5.

common

Phasors do not have to have a

may be shown that

in Fig. 2.20.

£|

same

from each

entirely separate

By

phase with

in

is

applying rule

3,

we can

because they point

l

see

in the

x

direction. Furthermore,

sor £,

origin but

other, as

by 90°, and

E2

phasor

/ 2 leads

lags behind /2 by

1

pha-

35°.

Figure 2.21 Different ways of showing the phase relationships between three voltages that are mutually displaced at 120°.

Solution

To draw trary

equivalent to 240 V. Phasor

Figure 2.20 Phasors do not have to start from a

show

their

the phasor diagram,

common

origin to

magnitudes and phase relationships.

it

we

direction for phasor E,

E with a length

lags 30° behind

(Fig. 2.22).

/ is

select any arbimaking its length then drawn so that

Knowing

equivalent to

frequency

that the

is

1

0A

60 Hz,

between the positive peaks

the time interval

is

given by In the in Fig.

same way,

£bc

the three phasors

2.2 1 a can be rearranged as

without affecting the

shown

,

and

£ca

in Fig. 2.2

1

= 360 ft 30 = 360 X 60 = 1.39 ms 9

b

phase relationship between

t

£ab in Fig. 2.21b still points in the as Eab in Fig. 2.21a, and the same is

them. Note that

same direction true for the

Fig. 2.2

three

240 V

other phasors. 1

c

shows

still

t

another arrangement of the

phasors that does not

in

any way

alter their

magnitude or phase relationship.

The angle

0

between two phasors is a measure peak positive val-

of the time that separates their ues.

Knowing

the frequency,

we can

calculate the

of the voltage

and current given

in

Figure 2.16.

time.

Example 2-5 Draw the phasor diagram of the voltage and current in Fig. 2. 16. Calculate the time interval between the positive

Figure 2.22 Phasor diagram

peaks of

E

and

/.

2.12

Harmonics

The voltages and

currents in a

power The

quently not pure sine waves.

circuit are freline

voltages

FUNDAMENTALS

24

usually have a satisfactory

waveshape but

Fig. 2.23. This distortion

distorted, as

netic saturation in the cores of transformers or

switching action of thyristors or

IGBTs

In order to understand the distorting effect of a

the cur-

shown in can be produced by mag-

sometimes badly

rents are

by the

in electronic

drives.

harmonic,

us consider two sinusoidal sources in series (Fig. 2.24a).

quencies are respectively 60

Hz and

1

Their

c,

fre-

80 Hz. The

V and 20 V. The fundamental (60 Hz) and the third harmonic 80 Hz) voltages are assumed to pass through zero at the same time, and both are perfect sine waves. Because the sources are in series, the terminal corresponding peak amplitudes are 100

(

1

voltage

—4

let

and e 2 connected

is

equal to the

sum of

the instantaneous

The resulting wave (Fig. 2.24b).

voltages produced by each source.

— A

0

4- V

terminal voltage

is

a flat-topped

-

Thus, the sum of a fundamental voltage and a har-

monic voltage yields a nonsinusoidal waveform whose degree of distortion depends upon the mag-

—

nitude of the harmonic (or harmonics)

60

1

20

1

300

240

80

it

contains.

420

360

Figure 2.23 This severely distorted 60

Hz

current obtained on an

electronic drive contains the following harmonics: funda-

mental (60 Hz) = 59 A; fifth harmonic (300 Hz) - 15.6 A; seventh harmonic (420 Hz) = 10.3 A. Higher harmonics are also present, but their amplitudes are small.

(Courtesy of Electro-Mecanik.)

The to the

distortion of a voltage or current can be traced

harmonics

it

contains.

A harmonic

age or current whose frequency

is

is

any

volt-

an integral multi-

ple of (2. 3, 4. etc., times) the line frequency.

Consider a

set

of sine waves in which the lowest

frequency is/ and multiples of / the lowest

the other

By

all

frequency

waves

other frequencies are integral

definition, the sine is

wave having

called the fundamental and

are called harmonics. For example,

waves whose frequencies are 20, 40, and 380 Hz is said to possess the following

a set of sine 1

00.

components: (b)

fundamental frequency: 20

Hz

(the lowest

frequency)

Figure 2.24

second harmonic: 40 Hz (2 fifth

harmonic: l()0Hz(5

X 20 Hz)

a.

X 20 Hz) Hz

(

19

X 20 Hz)

sinusoidal sources having different frequen-

cies connected b.

nineteenth harmonic: 380

Two

in

series.

A fundamental and

third

harmonic voltage can

gether produce a flat-topped wave.

to-


We

can produce a periodic voltage or current of

any conceivable shape. All gether a fundamental

we have

to

do

component and an

of harmonic components. For example,

to

is

add

to-

power

we can

distorted

gen-

wave having an amplitude of 00 V and a frequency of 50 Hz by connecting the following sine wave sources in series, as shown in Table 2A.

In

1

100 V

SQUARE WAVE

waveshapes

Freq.

[VI

[Hz|

that are rich in

mental

voltage

harmonics.

produce

together

power. This fundamental power

motor

a

is

fundamental

power

the useful

and an arc furnace

to rotate

The product of

heat up.

Amplitude

in

ac circuits the fundamental current and funda-

the corresponding Harmonic

produced whenever voltages

electronic circuits. All these circuits produce

that causes a

TABLE 2A

are also

and currents are periodically switched, such as

arbitrary set

square

erate a

They

circuits.

25

to

harmonic voltage times

harmonic current also produces a

Relative

amplitude

harmonic power. The

latter is

usually dissipated as

heat in the ac circuit and, consequently, does no 127.3

50

third

42.44

150

1/3

fifth

25.46

250

1/5

fundamental

useful

1

work.

Harmonic

and

currents

voltages

should therefore be kept as small as possible. It

should be noted that the product of a funda-

mental voltage and a harmonic current yields zero

seventh

18.46

350

1/7

ninth

14.15

450

1/9

net power.

Harmonics are covered

in

greater detail

in

Chapter 30.

2.13 Energy in an inductor lh

I27

1.00

6350

1/127

A

energy

coil stores

carries a current

/.

W 127.3/n

A tal

square

wave

wave and an

is

thus

infinite

50 n

1/n

W=

composed of a fundamennumber of harmonics. The

and they are consequently

less

wave.

and pointy corners of the square

In practice,

square waves are not produced by

adding sine waves, but the example does

l

-

field

when

it

given by

,

LI

(2.8)

energy stored

L = inductance of /

=

in the coil [J]

the coil [H|

current [A]

important.

However, these high-frequency harmonics produce the steep sides

=

is

where

higher harmonics have smaller and smaller amplitudes,

magnetic

in its

The energy

show

that

any waveshape can be built up from a fundamental

wave and an appropriate number of harmonics.

we can decompose a distorted periwave into its fundamental and harmonic components. The procedure for decomposing a distorted wave is given in Chapter 30. Harmonic voltages and currents are usually undesirable, but in some ac circuits they are also unavoidable. Harmonics are created by nonlinear

If

the current varies, the stored energy rises and falls

in step

with the current. Thus, whenever the current

increases, the coil absorbs energy and

current

The cussed

falls,

energy

is

properties of an inductor are in

Section 2.3

whenever

the

released.

more

fully dis-

1

Conversely,

odic

loads,

such as electric arcs and saturated magnetic

2.14 Energy

in

A capacitor stores ever a voltage

energy

is

E

a capacitor

energy

in its electric field

appears across

its

when-

terminals.

The

given by

W='c£

2

(2.9)

FUNDAMENTALS

26

W=

where

W=

energy stored

C— E=

capacitance of the capacitor [F]

in the

capacitor

fJ]

8

The energy

Example 2-6 having an inductance of 10

in series

with a 100

capacitor.

jjlF

current in the circuit

is

40

A

voltage across the capacitor

energy stored this

X

1/2

10

10" 3

X

X 40 2

J

stored in the capacitor

is

voltage [V)

W= A coil

1 Ml LI =

in

the electric

mH

= is

connected

2

1/2

32

CE =

1/2

X

100

X 10" 6 X 800 2

J

The instantaneous

Some

2.15

and the instantaneous

800 V. Calculate the and magnetic fields at

is

useful equations

We terminate this tions (Table

moment.

2B)

section with a

list

of useful equa-

that are frequently required

solving ac circuits.

The equations

when

are given without

proof on the assumption that the reader already pos-

Solution

The energy

stored in the coil

sesses a

is

knowledge of ac

circuits in general.

IMPEDANCE OF SOME COMMON AC CIRCUITS

TABLE 2B

Impedance

Circuit diagram

XL =

Equation

(2-I0)

2TTJL

(2-ll)

2irfC

Z = \R 2 + X

—

o

o

1|

o-m—it—— Ac

H

Z=

\

R~

2

(2-12)

+ Xc r

(2-

1

3)

XL

Z = VR- +

(X}

- XC Y

(2-14)

RX,

a*

(2-15)

\R 2 + X

: (

RX C

(2-16)

Vr 2 + Xc 2

Xc \ R +_XL 2 1

2

VR +

(X L

- Xcf

(2-17)


ELECTROMAGNETISM

X

by def10~ 7 or

approximately 1/800 000. This enables us

to write

In the SI, the inition.

Magnetic

2.16

and Whenever

a

component,

H

field intensity

B

magnetic flux

exists in a

it

field intensity

is

due

H, given by

H=U/I

(2.18)

in the

approximate form:

//= 800 000 3

magnetic

field intensity

a flux density of

H— U= /

=

magnetic

field intensity

f

rubber, and air have

length of the

that

component [m]

The resulting magnetic flux density

is

1)

A/m

of 800

produces

millitesla.

I

Nonmagnetic materials such

A/m]

magnetomotive force acting on the component [A] (or ampere turn)

(2.2

The B-H curve of vacuum is a straight line. A vacuum never saturates, no matter how great the flux density may be (Fig. 2.25). The curve shows that a

where

fixed,

is

has a numerical value of 4tt

body or

presence of a magnetic

to the

It

Eq. 2-20

flux density cf>

magnetic constant

27

B-H

as copper, paper,

curves almost identical to

of vacuum.

mT 2.0,

given by

B = $IA

r

.

—

1

,

1

(2.19)

where

B =

flux density [T]

§ = A =

flux in the

There

component WbJ [

cross section of the is

density (B)

component [m

2.17 In

]

a definite relationship between the flux

and the magnetic

field intensity

any material. This relationship graphically

2

by the

B-H curve

B-H curve

of

is

(H) of

usually expressed

Figure 2.25 of the material.

B- /-/curve of

of

nonmagnetic materials.

vacuum

vacuum, the magnetic flux density

B

is

directly

2.18

expressed by the equation

B =

The il 0

H

(2.20)

B-H curve

of a magnetic

material

proportional to the magnetic field intensity H, and is

vacuum and

flux density in a magnetic material also de-

pends upon the magnetic is

subjected.

value

Its

is

field intensity to

which

it

given by

where

B =

B =

flux density [T]

H=

magnetic

|x 0

—

field intensity

where

[A/m

magnetic constant [= 4tt

X

7

I0~ l*

B,

|jl

before, and

0

,

and

(jl,.

is

H have

ijL

oK /y

the

(2.22)

same significance

as

the relative permeability of the

material.

The value of

ijl,.

is

not constant but varies with

the flux density in the material. Consequently, the Also called the permeability of vacuum. The complete expression for

(jl

0 is

4

tt

X

K)

7

henry/meter.

relationship between this

makes Eq. 2.22

B and

H

is

not linear, and

rather impractical to use.

We

FUNDAMENTALS

28

prefer to

show

the relationship by

saturation curve. Thus, Fig. 2.26

means of a B-H shows typical

saturation curves of three materials

used

in electrical

and cast

steel.

field intensity

of 1.4

sity

T

commonly

machines: silicon iron, cast

iron,

ial, it is

The curves show that a magnetic of 2000 A/m produces a flux den-

in cast steel

but only 0.5

T

jjl,-

~

800 000

(2.23)

fl///

in cast

where

2.19 Determining the relative

permeability

tio

easy to calculate the relative permeability

using the approximate equation

iron.

The

would be produced in vacuum, under the same magnetic field intensity H. Given the saturation curve of a magnetic matersity that

relative permeability

of the flux density

in

B =

flux density in the magnetic material [TJ

H

corresponding magnetic field intensity [A/ml

—

Example 2-7 jjl,.

of a material

is

Determine the permeability of

the ra-

flux density of

the material to the flux den-

1

silicon iron

.4 T.

teslas

0

1000

2000

3000

4000

*H Figure 2.26

B-H saturation curves

of three

magnetic materials.

5000

6000

A/m

( 1

%)

at

a


the magnetic materials saturate

Solution

Referring to the saturation carve (Fig. 2.26), that

a flux density of

intensity

jlil..

1

.4

see

a magnetic field

eventually

all

the

B-H

more and more and

B-H curve

curves follow the

of vacuum.

of 1000 A/m. Consequently,

=

2.20 Faraday's law of

800 000 B/H

= 800 000 X At

T requires

we

29

1.4/1000

this flux density, silicon iron is

more permeable than vacuum (or

electromagnetic induction

=1120 1120 times

air).

shows the saturation curves of a broad from vacuum to Permalloy®, one most permeable magnetic materials known.

Fig. 2.27

In

1831, while pursuing his experiments, Michael

Faraday made one of the most important discoveries in

electromagnetism.

Now known

as

Faraday's

it

revealed a

range of materials

law of electromagnetic induction,

of the

fundamental relationship between the voltage and

Note that as the magnetic field intensity increases,

flux in a circuit. Faraday's law states:

Figure 2.27 Saturation curves of magnetic

curve of

vacuum where

H is

and nonmagnetic materials. Note

high.

that

all

curves

become asymptotic

to the

B-H

FUNDAMENTALS

30

function of time, a voltage

tween

its

of the induced voltage

tional to the rate of

By 1

definition,

when

units,

its

E =

propor-

is

and according

terminals. Consequently,

V

if

induced be-

is

is

The induced voltage

(2.24)

in

A/

where

In

= —

At

coil

[Wb]

fixed in space.

is

common)

(although

case,

it

is

induction

with

relative

is

induced ac-

in this special

easier to calculate the

induced voltage with reference electromagnetic

move

The

in the flux linking the

cording to Faraday's law. However,

[s]

of

generators, the coils

coils and, consequently, a voltage

time interval during which the flux

law

many motors and

motion produces a change

in the coil

change of flux inside the

Faraday's

induced a conductor

respect to a flux that

induced voltage [V]

changes

zero as soon as the flux

falls to

ceases to change.

2.21 Voltage

A

A

X1/10

1000

given by

N~—

= number of turns

3

the flux varies in-

side a coil of /V turns, the voltage induced

/V

in

is

= 60 V

system of

the flux inside a loop varies at the rate of

E=

AO = 2000 X N— At

to the SI

1

change takes place uniformly

this

1/10 of a second (A/), the induced voltage

change of flux.

weber per second, a voltage of

tween

induced be-

is

terminals.

The value

2.

Because

linking a loop (or turn) varies as a

If the flux

1.

to the conductors,

rather than with reference to the coil

itself. In effect,

established the basis of operation of transformers,

whenever a conductor cuts a magnetic field, a voltage is induced across its terminals. The value of the

generators, and alternating current motors.

induced voltage

opened the door to a host of practical applications and

is

given by

E=

Example 2-8

A coil of 2000 turns surrounds a flux of 5 mWb produced by a permanent magnet net

is

(Fig. 2.28).

The mag-

suddenly withdrawn causing the flux inside

the coil to drop uniformly to 2

second.

What

is

mWb

in

1/10 of a

the voltage induced?

flux

E— B= = /

induced voltage [VJ flux density [Tl

active length of the conductor in the

magnetic

change

AO

(5

v

is

mWb -

N=

2

mWb)

3

(2.25)

where

Solution

The

Blv

—

field

[m]

relative speed of the conductor fm/s|

mWb Example 2-9 The stationary conductors of a

2000

an active length of 2 N

S

\

0.6 teslas.

,>

moving

at

m

large generator

and are cut by a

have

field

Calculate the voltage induced in each conductor. 5 4> 2

mWb

=2mWb

Solution A/ = 1/10

s

According to Eq. 2-25. we find

Figure 2.28 Voltage induced by a moving magnet.

See Example

2-8.

E=

Blv

=

0.6

X

of

a speed of 100 m/s (Fig. 2.29).

2

X

100

-

120

V


120

3

V

Figure 2.30 Force on a conductor.

Figure 2.29 Voltage induced

Example

in

a stationary conductor.

See

2-9.

-

on a conductor

2.22 Lorentz force

When

conductor

a current-carrying

magnetic field,

it

is

is

placed

force

is

of fundamental importance because

stitutes the

and of

tors,

it

con-

basis of operation of motors, of genera-

many

nitude of the force

electrical instruments.

The force

is

parallel to

it

=

0.

The mag-

when

greatest

the conductor

perpendicular to the field (Fig. 2.30) and zero is

Figure 2.31 Force

depends upon the orientation of

conductor with respect to the direction of the

field.

it

we

electromagnetic force, or Lorentz force. This

call

the

1

N

in a

subjected to a force which

—

(Fig. 2.3

l

).

is

when

Calculate the force on the conductor

BII

tremes, the force has intermediate values.

tor is

straight

perpen-

it is

Solution

Between these two ex-

The maximum force acting on a

if

dicular to the lines of force (Fig. 2.30).

X

0.5

X 200 = 300 N

3

conduc-

given by

2.23 Direction of the force acting F=

BII

on a straight conductor

(2.26)

where

Whenever

F=

force acting on the conductor [N]

B =

flux density of the field [T]

a conductor carries a current,

rounded by a magnetic into the

field.

For

page of this book, the circular

have the direction shown

in

it

a current

lines

Figure 2.32a.

of force

=

active length of the conductor [m]

figure

/

-

current in the conductor A]

N, S poles of a powerful permanent magnet.

[

the magnetic field created

The magnetic shape shown

Example 2-10

A conductor

3

m long carrying a current of 200 A is

placed in a magnetic field

whose density

is

0.5 T.

in

field

sur-

The same

/

shows

is

flowing

between

does not, of course, have

the

the

the figure because lines of force

never cross each other. What, then, the resulting field?

is

the shape of

FUNDAMENTALS

32

N

//

H cross-section

.

Xi

-

- ct>

I

.

I

(a)

Y turns

length

-

/

Figure 2.33a Method of determining the B-H properties

of

a mag-

netic material.

curve oa

(b)

a value

Figure 2.32 a. Magnetic field due to magnet and conductor. b. Resulting magnetic field pushes the conductor downward.

If

in

Bm

Figure 2.33b. The flux density reaches for a magnetic field intensity

the current

the flux density

curve, but

To answer

the question,

we observe

that the lines

of force created respectively by the conductor and

permanent magnet

the

act in the

above the conductor and low

in

same

direction

opposite directions be-

Consequently, the number of lines above the

it.

B

we

the magnetic

H nv

gradually reduced to zero,

does not follow the original

moves along

oa. In effect, as sity,

now

is

a curve

ab

situated

above

reduce the magnetic field inten-

domains

that

were lined up under

Hm tend to retain their origiphenomenon is called hystereConsequently, when H is reduced to zero, a sub-

the influence of field nal orientation. This sis.

stantial flux density remains.

It

is

called residual

flux density or residual induction (B r ).

conductor must be greater than the number below.

The

resulting magnetic field therefore has the shape

given

in

Figure 2.32b.

Recalling that lines of flux act like stretched elastic bands,

upon

it

is

easy to visualize that a force acts

the conductor, tending to

push

it

downward.

2.24 Residual flux density

and coercive force Consider the

coil

of Figure 2.33a, which surrounds

a magnetic material

formed

in the

shape of a

ring.

A

magnetic

field intensity

current source, connected to the coil, produces a current

whose value and direction can be changed from zero, we gradually increase /,

Figure 2.33b

H

Residual induction and coercive force.

at will. Starting

so that

and B increase. This increase traces out

//

FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS we wish

If

to eliminate this residual flux,

33

we

have to reverse the current in the coil and gradually

H

increase

in the

opposite direction.

we move along curve

As we do

so,

be. The magnetic domains

gradually change their previous orientation until the

becomes zero

flux density

point

at

The magnetic

c.

required to reduce the flux to zero

field intensity

is

called coercive force In

also

B

reducing the flux density from

have

to furnish

overcome the

The energy supplied

A

to zero,

magnetic

in orientation.

dissipated as heat

is

we

used to

is

frictional resistance of the

domains as they oppose the change

material.

r

energy. This energy

the

in

very sensitive thermometer would in-

dicate a slight

temperature

rise as the ring is

being

Figure 2.34 Hysteresis loop.

If

B

expressed

is

amperes per meter, the area

demagnetized.

dissipated per cycle,

in

in

and

teslas

of the loop

is

H in

the energy

joules per kilogram.

2.25 Hysteresis loop Transformers and most electric motors operate on

such devices the flux

alternating current. In iron tion.

changes continuously both

The magnetic domains

in

depends upon the frequency. Thus,

pressed

if

at

a rate that

the flux has a

in

J/nr

)

is

equal to the area

(in

T-A/m) of

the hysteresis loop.

value and direc-

are therefore oriented

one direction, then the other,

first in

in the

To reduce

hysteresis losses,

we

select

magnetic

materials that have a narrow hysteresis loop, such as the grain-oriented silicon steel used in the cores

of alternating-current transformers.

frequency of 60 Hz, the domains describe a com-

every 1/60 of a second, passing succes-

plete cycle

through peak flux densities +Z? m and —B m the peak magnetic field intensity alternates be-

sively as

tween

+ // m and — // m

as a function

of H,

.

we

we

If

v

B

obtain a closed curve called

hysteresis loop (Fig. 2.34).

B and coercive force

plot the flux density

H

c

The

residual induction

have the same

signifi-

cance as before.

2.27 Hysteresis losses caused

by rotation Hysteresis losses are also produced

for example, an armature

volves

in

AB, made of

S (Fig. 2.35). The magnetic domains

armature rotates, the

describing a hysteresis loop, the flux

and

+# m

.

+ 5 m +B n ,

0,

and

a, of

moves

— B m —B n ,

corresponding respectively

b, c, d, e, f,

energy

that the

is

0,

to points a,

Figure 2.34. The magnetic

material absorbs energy during each cycle this

iron, that re-

in the

armature

field, irrespective

of the position of the armature. Consequently, as the

2.26 Hysteresis loss

successively from

a piece of

a field produced by permanent magnets N,

tend to line up with the magnetic

In

when

iron rotates in a constant magnetic field. Consider,

dissipated as heat.

We

and

can prove

amount of heat released per cycle

(ex-

first

sal

toward

A and

N

poles of the domains point

then toward B.

A complete

rever-

occurs therefore every half-revolution, as can be

seen

in

Fig.

2.35a and 2.35b. Consequently, the

magnetic domains cally,

in

the armature reverse periodi-

even though the magnetic

field is

constant.

Hysteresis losses are produced just as they are in an ac magnetic field.

FUNDAMENTALS

34

Figure 2.37 Concentric conductors carry ac currents due to ac

flux

<1>.

currents are progressively smaller as the area of the (b)

loops surrounding the flux decreases. In Fig. 2.38 the ac flux passes

Figure 2.35 Hysteresis losses due to rotation.

through a solid

metal plate.

It is

packed

of rectangular conductors touching

set

basically equivalent to a densely

each other. Currents swirl back and forth inside

Eddy currents

2.28

Consider an ac flux

that links a rectangular-shaped

conductor (Fig. 2.36). According to Faraday's law, an ac voltage If the

E

is {

tor to heat up. If a

the

first,

induced across

conductor

alternating current

is

/,

current

U

short-circuited, a substantial

second conductor is

is

conduc-

placed inside

induced because

centric

it

is the power disshows four such conloops carrying currents I 7 2 7 3 and / 4 The

is

less than

penetrated by an ac flux can this regard, special care

/,

and

so, too,

x

,

,

,

The

flux

c|)

in Fig.

be increasing. As a the

flux

<1>

in trans-

2.37 and 2.38

result,

eddy currents flow

in

is

assumed

to

due to the Lenz's law,

such a

way

as to

oppose

the increasing flux.

.

conductor

Figure 2.36

has to be taken

is

hot. In

sections of the enclosing tanks to overheat.

eddy currents

An ac

become very

formers so that stray leakage fluxes do not cause

Consequently, the short-circuit

sipated in this loop. Fig. 2.37

in the figure.

These so-called eddy currents (or Foucault currents) can be very large, due to the low resistance of the plate. Consequently, a metal plate that

terminals.

its

will flow, causing the

a smaller voltage

links a smaller flux.

shown

the plate, following the paths

V

—

metal plate

Figure 2.38 induces voltage

Ev

Large ac eddy currents are induced

in

a

solid

metal plate.


Eddy currents

2.29

iron

35

a stationary

in

core

insulation

The eddy -current problem becomes particularly important

when

iron has to carry an ac flux. This

the case in all ac

2.39a

shows a

coil carrying

A large

of the core.

ally

become red hot (even

due

to these

at a

its

Figure 2.39b Eddy currents are reduced by

splitting the

core

in half.

in

two The voltage of what it was

length, taking care to insulate the

from each other

induced

in

each section

(Fig. 2.39b). is

one half

before, with the result that the

eddy currents, and

corresponding losses, are considerably reduced.

we

If

continue to subdivide the core,

we

composed of stacked laminations, usually

of a millimeter thick. Furthermore, a small silicon is

find that

decrease progressively. In practice, the core

the losses

ity,

/'•-•''

core could eventu-

frequency of 60 Hz)

can reduce the losses by splitting the core

two along

is

currents

eddy-current losses.

sections

the

Eddy

up as shown and they flow throughout the

entire length

We

i.-A

an ac current that pro-

duces an ac flux in a solid iron core. are set

is

motors and transformers. Figure

alloyed with the steel to increase

thereby reducing the losses

still

more

a fraction

amount of its

(Fig. 2.39c).

The cores of ac motors and generators are fore

always laminated.

A

eddy current in one lamination

resistiv-

there-

Figure 2.39c Core built up of

thin,

insulated laminations.

thin coating of insulation

covers each lamination to prevent electrical contact

between them. The stacked laminations are tightly held in place

by

bolts

and appropriate end-pieces.

For a given iron core, the eddy-current losses decrease in proportion to the square of the

laminations.

number of

2.30 Eddy-current losses in a revolving core The

stationary field in direct-current motors and gen-

erators produces a constant dc flux. This constant flux induces

eddy currents

To understand how they

in the

revolving armature.

are induced, consider a solid

cylindrical iron core that revolves between the poles

of a magnet (Fig. 2.40a). As lines and,

duced along

Owing

it

its

its

is in-

length having the polarities shown.

to this voltage, large

core because

resistance

eddy currents flow is

These eddy currents produce are immediately converted is

turns, the core cuts flux

according to Faraday's law, a voltage

in the

very low (Fig. 2.40b). large

FR

into heat.

losses

which

The power

loss

proportional to the square of the speed and the

square of the flux density.

To reduce

Figure 2.39a Solid iron core carrying

an ac

flux.

the eddy-current losses,

we

laminate

the armature using thin circular laminations that are

FUNDAMENTALS

36

rotation

(b)

(b)

Figure 2.41 Armature

a.

Figure 2.40

Much

b.

up

of thin laminations.

a revolving armature.

a.

Voltage induced

b.

Large eddy currents are induced.

in

built

smaller eddy currents are induced.

of current. However,

known and we want insulated from each

other.

The laminations

tightly stacked with the flat side

are

running parallel to

the flux lines (Fig. 2.41).

rent

/.

We

To

often happens that e

knowledge of advanced mathemat-

get around this problem,

we can

use a

graphical solution, called the volt-second method.

2,31 Current in It is

well

known

an inductor

yields the

that in an inductive circuit the volt-

known

Ai

= L~~

how

=

instantaneous voltage induced in the circuit

|

V]

L = inductance of the A/7 A/

=

E

/,

time

.

e,

later, after

carries a current

an interval At.

circuit [H]

/, at

a time

From Eq. 2-27 we

the rate of

change

=

1

eAt

L

of change of current [A/s]

when we know

which a

We want to determine the current a very short

Ai rate

in

appears across an inductance L.

can write

This equation enables us to calculate the instanta-

neous voltage

=

response to

in

applied voltage.

Suppose the inductance /

e

the current in an induc-

and decreases with time,

variable voltage

where

It

and has the advantage of en-

Consider, for example, Fig. 2.42,

(2.27)

At

results

abling us to visualize

a e

same

tor increases

age and current are related by the equation

is

can use the same equation, but the solu-

tion requires a ics.

it

to calculate the resulting cur-

which means

that the

short interval At

is

change

given by

in

current Ai during a


Volt-seconds are gained (and

it

-«

voltage

1

is

when a

lost)

37

variable

applied across an inductor.

Figure 2.42 Variable voltage applied across sulting

change

in

current.

Initial

an inductor and current

re-

is A,.

mental changes

—

(f 2

f|).

As

U =

initial

A =

/,

in

current A/ during the long period

we

a result

current

find current

/,

+

+

e 2 At 2

(A/,

+

A/ 2

I 2 at

+

time

+

A/ 3

.

t

.

2

.)

average voltage e during the interval A/

-

X

duration Af of the interval

I

L

2

=

+ | +

/,

(Âf,

-

=

3

I

curve during the interval

.)

.

.

.)

.

/ algebraic sum of

all

+

between

and

t

A under

t

2

the voltage

-seconds across the induccurve between

tance during the interval At

The values of is

tive

(

+

+

Af)

=

initial

current

+

A/

e2

or negative

)

eas AAj, (f,

and

t

f

{

Therefore, the current in the inductance after the in-

/at time

little

areas under the voltage curve

net area volt

the

I

4 =

Am

x

Af

.

2

\

terval

+

Af 3

+ AA 2 + AA 3 +

(A/4,

/ area A/\ under the voltage A/

+

AA 2

The sum of

,

A/\ 3

these

(

and so on may be posi-

,

)

and, therefore, the

and so on may be

,

(

—

+

)

2

and

(

—

)

(

+

)

little ar-

or

(

—

).

values of the A/\s

gives the net area under the voltage curve between /,

+

AA

fj

are usually

current at a time ter

f,

f2

(Fig. 2.43).

more ,

2

.

in Fig.

2.44 the net area/\ after time inter-

val Tis equal to (A

interested in calculating the

when

We

f

Thus,

L

We

and

f

2 is

many Af

ize, the

,

— A2

)

volt-seconds.

To general-

current after an interval Tis always given by

intervals af-

then have to add the incre-

/

=

/,

+ AIL

(2.28)

FUNDAMENTALS

38

volts

Figure 2.44

The

T is A and A 2

net volt-seconds during interval

algebraic

sum

areas

of the

equal

the

to

.

A

where /(

=

current at start of interval

T

/

=

current after lime interval

T\A]

A —

under the volt-time curve dur-

net area

T

ing lime /.

—

[

V-s]

inductance [HI

Consider, for example an inductor

L,

having negli-

gible resistance, connected to a source

whose

volt-

age varies according to the curve of Fig. 2.45a. the initial current

is

zero, the value at instant

=A

F

t\

If

Figure 2.45

is

/L l

As time goes

However, the current reaches time

/->

because

at this

its

moment

maximum

value

at

the area under the

any more. Beyond becomes negative and, consequently,

vol age curve ceases to increase l

t,

the voltage

the net area begins to diminish.

ample, the net area

equal to

is

the corresponding current

f=(A + i

At instant

/4 ,

At instant ^, for ex+ A 2 — A$) and

(/\,

stant

is /

the negative area 04 3

words,

it

the current

is

to

is

A

is

ex-

+ A 2 The

net

4-

4)

).

l

the inductor.

onds during the

interval

from 0

to

t

2

.

As

it

becomes

direct proportion to the volt-seconds received.

during the discharge period from

t

2

in

Then

to /4 the inductor

loses volt- seconds and the current decreases accordingly.

An

inductor, therefore, behaves very

a capacitor.

in

known

much

like

However, instead of storing ampere-secstores volt-seconds.

a capacitor having a capacitance that the voltage

E across

its

C

For it

is

terminals

is

given by

also zero. After inin

other

where

changes direction. at

in

charged up with volt-seconds, the current increases

well

becomes negative;

Another way of looking

inductor stores volt-seconds.

example,

A 2 - A y )IL

zero and so the current 4,

Current

onds (coulombs), an inductor

is

actly equal to the positive area (A

area

An

b.

in-

and so does the current.

progressively

creases

under the curve

by, the area

a.

the situation (Fig. 2.45),

consider that the inductor accumulates volt-sec-

in

E

is }

the initial voltage and

Q

.

L

is

the charge

coulombs (ampere seconds, positive or negative)

the capacitor received during a given interval.


In the

same way,

for an inductor having an in-

ductance L, the current

/ it

carries

Example 2-11 The voltage across

given by

is

39

the terminal s of an inductor of 2

H

varies according to the curve given in Fig. 2.46. a.

Calculate the instantaneous current cuit,

where in

/] is

the initial current and

knowing

the cir-

/ in

that the initial current

is

zero.

1

QL

is

the "charge'

b.

Repeat the calculations for an

current

initial

of 7 A.

volt-seconds (positive or negative) that the in-

ductor received during a given interval. It

1

interesting to note that

is

volt-second.

turns

coil of

1

weber-turn

600 turns

of 20 milliwebers has stored

a flux netic

Thus a

12V-s/3H

tained

equal to

mag-

a total 12

000mWb

12 volt-seconds. If the inductor has an induc-

tance of 3 henries,

Fig.

it

X 20mWb =

charge of 600 turns

=

in

is

that surrounds

it

is

carrying a current of

Q

{

/L

=

=4 A. the voltage of Fig. 2.45a

an inductance of 100 H.

The

and the current rises to a

maximum

initial

is

applied to

current

of 6.9

zero,

is

A

before

again dropping to zero after a time interval of

Important Note: an interval

T

is

value to all the

If the

27

s.

current at the beginning of

not zero,

we

simply add the

initial

ampere values calculated by the

volt-second method.

Figure 2.46 See Example 2-11.

a.

Interval

from zero

to 3

s:

During

this interval

the area in volt-seconds increases uniformly

and progressively. Thus, area

A

and so

is

4

V

forth.

s;

after

after

one second, the

two seconds

8

it is

Using the expression

=

/

V

s;

AIL, the

current builds up to the following respective

2.45b shows the instantaneous current ob-

when

Solution

values: 2 A, 4 A, and so on, attaining a final

value of 6 Interval

A after three

from 3

s to

5

seconds. s:

The area continues

to

increase but at a slower rate, because voltage is

smaller than before.

When = t

5

surface starting from the beginning therefore the current

Interval

from 5

s to

by 4 squares, which

is

7 is

16 V-s/2 s:

s,

is

H =

The surface

equivalent to 8

E

the total

16

V

s;

8 A.

increases

V

s.

FUNDAMENTALS

40

means that the sum of the voltage rises is equal to sum of the voltage drops. In our methodology it

Consequently, the current increases by 4 A, thus reaching 12 A. Note that the current no

the is

age rise" or a "voltage drop/'

is

not constant during this interval.

from 7

Interval

s to

8

We

The voltage sud-

s:

denly changes polarity with the result that the

V

8

during this interval subtract from the

s

ously.

The

from the beginning

net area

therefore 24

V

-

s

8

V

=

s

Consequently, the current interval

7=16

is

Interval voltage

from 8

is

s

at the

=

V-s/2 77 to 10

s:

V

16

ence.

from

end of

2.33 Kirchhoff's voltage law and double-subscript notation

we assumed

is

may

elements

zero

in

which

six circuit

The

s:

—

at t

14

is

Beyond

zero.

current of

+7

s,

this point

be sources or loads, and the conI

to 4.

we have

A,

to

add

we can start with any node cw or ccw direction until we starting point. In so doing, we

in either a

come back

to the

This ordered voltage

The new current wave is simply 7 A above the curve shown in Fig. 2.46. Thus, the

=

1

1

s is

CIRCUITS When

+

6

7

2.7.

We

1

set

of labels

subscripts.

We

is

used to establish the

write the

voltage sub-

scripts in sequential fashion, following the

it is

order as the nodes

cw around

in

presume the reader

same

meet.

loop

ABCD, we

successively encounter

nodes 2-4-3-1-2. The resulting

KVL

equation

is

therefore written

essential to fol-

E24 + £43 + E31 + E l2 =

based upon the voltage

rules that are

we

For example, starting with node 2 and moving

3 A.

AND EQUATIONS

and current notations covered

and

=

writing circuit equations,

low certain

going

elements A, E, and D,

and move

ously.

at t

In

encounter the labeled nodes one after the other.

each of the currents calculated previ-

current

elements

around a circuit loop, such as the loop involving

negative volt-

equal to the positive area,

and so the net current

A to

Consider Fig. 2.47

A, B, C, D, E, and F are connected together. The

the current changes direction,

7

by the sign notation.

later

nections (nodes) are labeled

the negative area

initial

followed

this

8 A.

seconds continue to accumulate and

With an

a matter of individual prefer-

will begin with the double-subscript nota-

Because the terminal

that

10 s to 14

in

The choice

tion,

coil resistance).

Interval

can be expressed

that voltages

is

a "volt-

is

second area does not change and neither does

(remember

We

is

s.

zero during this interval, the net volt-

the current

have seen

either double-subscript or sign notation.

of one or the other

volt-seconds that were accumulated previ-

b.

not necessary to specify whether there

longer follows a straight line because the volt-

age

0

Sections 2.4, 2.5, familiar with the

is

solution of such equations, using linear and vector

method

will

review only

the writing of these equations, using

Kirchhoff s

algebra. Consequently, our

voltage law

(KVL) and

By following

a

current law (KCL).

few simple

rules,

solve any circuit, ac ordc, no matter

it

is

possible to

how complex.

We

begin our explanation of the rules regarding voltages.

2.32 Kirchhoff's voltage law Kirchhofrs voltage law

sum

states that the algebraic

of the voltages around a closed loop

Thus,

in a

3

is

zero.

closed circuit, Kirchhoff's voltage law

Figure 2.47 Rule

for writing

notation.

KVL

equations using double-subscript

FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS If we select loop CEF and start with node 4 and move ccw, we successively encounter nodes 4-2-3-

4.

KVL equation

The resulting

+

£42

The

set

usually

0

ac or dc. If they are ac, the voltages will

to

it is

essential to equate

all

we have done so far and We do not recommend attempts

equations to zero as

continue to do.

equate voltage rises to voltage drops. In

£,

set

even represent instantaneous values.

order to prevent errors,

will

and so

be expressed as phasors, having certain

of voltages can

KVL

= -10 V

KVL equa-

magnitudes and phase angles. In some cases the

In

= ~E ]2 ~~ E 2 $ = -E X2 + E yl = -40 + 30

£31

of voltages designated by the

may be

tions

Transposing terms,

is

+ em

£23

= +10 V

3

indicating that terminal

is

1

positive with respect to

terminal 3 and that the voltage between the two

is

10 V.

2.34 Kirchhoff s current law

finding the solution to such double-subscript

equations,

pressed as

it

useful to

is

EXY

remember that

a voltage ex-

can always be expressed as

— E YX

Kirchhoff s current law

sum

means

zero. This

rents that leave

Example 2-12

shows two sources connected in series, having terminals (nodes) 2, and 3. The magnitude 2.48

1

+40 V and £ 32

Fig. 2.49

and

is

equal to the

sum of

the cur-

shows

five currents arriving at a

The sum of

common

the currents flowing

,

E l2 and E^ 2 are specified as E i2 = = +30 V. We wish to determine the I

equal to

is

the currents that

it.

terminal (or node). into the is (I 2

+

node I4

+

is (/,

7 S ).

magnitude and polarity of the voltage between the open terminals

point

at a

sum of

that the

flow into a terminal

and polarity of

states that the algebraic

of the currents that arrive

,

and vice versa.

Fig.

41

/,

+

/ 3 ),

while the

sum

that leaves

it

Applying KCL, we can write

+

/3

=

I2

+

I4

+

/,

3.

Solution In

writing the loop equation, let us start at terminal

and

move ccw

The resulting

we

till

again

come back

KVL equation £,

2

to terminal

1

1

is

+ £ 23 + E M =

0

Figure 2.49

?3

1t

Rule

for writing

KCL

equations.

2.35 Currents, impedances,

and associated voltages Consider an impedance 2

Z

carrying a current A

connected between two terminals marked Figure 2.48

(Fig. 2.50).

See Example 2.12.

will

A

1

and 2

voltage E, 2 having a magnitude ,

/Z,

appear across the impedance. The question of

FUNDAMENTALS

42

Let us write the circuit equations for Fig. 2.5

Loop 2312,

starting with

node 2 and moving cw:

=0

+ /4 Z4 + E 3I -/,Z, Figure 2.50 E12 = + /Z.

Voltage

/4

Z4

is

preceded by a

+

(

)

sign,

going around the loop we are moving

now

polarity

The

arises: Is

question

is

across an impedance

direction as the current flow

voltage IZ

or

— IZ?

/,

Z

in

write £, 2

= +/Z

the

the associated

Conversely,

moving across an impedance against ative sign.

Thus,

or dc, and the

ductive In

(

jX,

most

),

E2]

the direction

resistive (R), in-

(—jXc ). impossible to predict the

or capacitive

circuits

it

is

On

the other hand, voltage I{Z

Loop 3423,

starting with

+ /3Z3 Voltages /3Z3 and

because

in

Fig. 2.5

and Zj,

1

in

,

I2

node 3 and moving ccw:

Z2 +

/4

Z4 =

0

/4

Z4

are preceded by a

(

+

sign,

)

the direction of the respective currents. Voltage is

preceded by a negative sign because

ing against current

Loop 242,

I2

are

Z2

mov-

.

starting with

£24

we

in

/2

node 2 and moving cw:

~ hî =

0

Consider for example the circuit of

which two known voltage sources E, 3

E24 are connected to four known impedances Z 2 Z3 and Z4 Because the actual directions of ,

pre-

going around the loop we are moving

actual direction of current flow in the various circuit elements.

is X

against the direction of/,.

in

~IZ. The current can be ac

impedance can be

.

in

when

of current flow, the voltage /Zis preceded by a neg-

=

tion of /4

because

in the direc-

ceded by a negative sign because we are moving

preceded by a positive sign. Thus,

is

we

Fig. 2.50,

+ /Z

equal to

resolved by the following rule:

When moving same

£ l2

1

KCL

node

at

2:

/s

=

U +h

/.

.

unknown, we simply assume arbitrary directions as shown in the figure. It is a remarkable fact that no matter what directions are assumed, the final outcome after solving

current flows are presently

the equations (voltages, currents, polarities, phase

angles, power, etc.)

is

always correct.

KCL

node

at

3:

h+

/.

= h

Example 2-13 Write the circuit equations and calculate the currents flowing in the circuit of Fig. 2.52, that

£AD - + 108 V

and

ECD = + 48

knowing

V.

Solution

We first and

/3

select arbitrary directions for currents

and write the 6

108

Figure 2.51 Writing

KVL and KCL

V-—-

See Example

a

Ail

12

Figure 2.52 equations.

2.13.

/,,

circuit equations as follows:

Q

^

48 V

/2

,


For the loop

DABCD composed of the two sources we

and the 6 fi and 4 fl resistors,

For the loop

DCBD

Applying

1

composed of

the

(cw)

48

V

source

2 II resistors:

E DC +

47 2

+

-48 +

4/ 2

+

KCL at

12/ 3

=

12/,

=

0

(ccw)

Figure 2.53

See Example Solution a.

We conclude and

73

72

To solve

= -3 A

we

73

that the directions

flows from

To

= +5 A

establish an arbi-

first

we assumed

assumed because

Thus, suppose

between points a and

write the circuit equation,

let

/

b.

move cw

us

c.

This

yields for

I\

2 is

bears a nega-

/2

left to right

around the loop, starting from terminal

However, the actual direction of

opposite to that

we

find

ECd +

were correct because both currents carry a

positive sign.

the circuit,

trary direction of current flow.

*3

Solving these simultaneous equations,

= +8 A

2.14.

0

node B, we get

h + h =

/,

763

obtain

£da + 6'i - 4/ 2 + ECD = 0 -108 + 67, - 47 2 + 48 - 0 and the 4 il and

16

43

7(16

+

j

Substituting the values of tion

+ £bc =

63)

Eac and Ehi

and combining the terms

in

.

we

/,

0

in this

equa-

get

tive sign.

-200 2.36 Kirchhoff s

laws and ac circuits

The same basic rules of writing double-subscript equations can be applied to ac circuits, including

3-phase circuits. sistive sistive,

elements

The only difference in

all

Solving b.

765

A.

+

75.8°

we

100

find that

I

=

L 1

150°

.9 Z.

=

0

20.5°.

To determine £. lb we write the following equation, moving cw around the loop: ,

that the re-

£ua + £

;ib

+ £bc =

0

Transposing terms,

three. Furthermore, the volt-

magnitudes and phase angles. is

+

this equation,

ages and currents are expressed as phasors, having

sor equations

120°

dc circuits are replaced by reor capacitive elements, or a

inductive,

combination of

is

A.

The

more time-consuming, but

equations themselves can be written by inspection. Let us consider

= £ac — £ bc = 200 L 120° -

solution of phathe

down almost

100

l_

150°

Using vector algebra, we find

two examples.

E

ilb

=

123.9

L

96.2°

Example 2-14 In the circuit

ate the

of Fig. 2.53, two sources A,

B

gener-

2.37

KVL and

Voltages E, c

sign notation

following voltages:

= 200 L

E bc =

120°

100

L

150°

in

ac and dc circuits are frequently indi-

cated with sign notation and designated by symbols

such as

E. v e m

for such circuits,

Calculate a.

The value of the current

b.

the value of

£ ab

and

its

/ in

the circuit

phase angle

,

and so on. To write the equations

we employ

As we cruise around a larity (+ or — ) of the first

the following rule:

loop,

we observe

the po-

terminal of every voltage

FUNDAMENTALS

44

(£,, E. v e m

etc.)

,

we

meet.

of the voltage source

minal

If

only the

(

+

terminal

)

marked, the unmarked

is

ter-

2.38 Solving ac and dc circuits with sign notation

taken to be negative. This observed polar-

is

(+ or — precedes the respective voltages as we write them down in the KVL equation. The following example illustrates the application of this rule. ity

In circuits using sign notation,

wc

treat the

IZ

volt-

)

ages

in

same way

the

subscript notation. In other words, the IZ voltage

across an impedance sign

Example 2-15 In Fig. 2.54, it is

known

wish age

Z

preceded by a positive

is

whenever we move across

the

impedance

in

the direction of current flow. Conversely, the IZ

,

1

determine the value and polarity of the volt-

to

Ec

EA and £ B E A = + 37 V and E R - - 5 V. We

given the polarity marks of

that

as in circuits using double-

voltage

move

is

preceded by a negative sign whenever

The following example

across the open terminals. to

we

against the direction of current flow. illustrates the

procedure

be followed.

Solution First,

we

assign an arbitrary polarity

minal voltage

Ec We .

then proceed

+ to the tercw around the )

(

loop in Fig. 2.54, starting with voltage

EA

.

This

yields the following equation:

Example 2-16 The circuit of Fig.

E =

1600

/L

60°.

2.55 is powered by an ac source The values of the respective im-

pedances are indicated

-EA + Ec -E B =

(cw)

0

Ec

Calculate a.

b.

T2

T1

in the figure.

The current flowing in each element The voltage Ex across the 72 ohm capacitive reactance.

Solution a.

To solve to

flow

this

problem, the currents are assumed

in the arbitrary directions

shown.

We

then write the following equations.

Moving cw around -

E-

40)

Figure 2.54 Rule

Note

for writing

KVL

that the sign

sponds

equations using sign notations.

preceding each voltage correof the terminal that was

to the polarity

encountered

in

going

cw around

first

the loop.

Transposing terms.

Ec ~ E A + E H

= + 37 = +22 V Thus, the magnitude of of terminal Tl

is

to

is

22 V, and the polarity

indeed positive with respect to

minal T2. The polarity

happens

Ec

5

we assumed

have been the correct one.

at

ter-

the outset

Figure 2.55

See Example

2.15.

the loop

-

7,(30)

BDAB, we

+

I2

(-

j

obtain

37)

=

0 (cw)


moving ccw around

Then,

we

the second loop

45

ABCA,

obtain

/2 (-j

Finally,

-

37)

(-

/3

KCL at

applying

+

/,

/2

j

-

72)

=

21 / 3

0

(ccw)

N

node A, we have

+ h =

o

Upon solving these equations, we obtain the

fol-

lowing results:

=

/,

44.9

zl

215°

=

/3

b.

I2

=

30.3

Figure 2.56

L

40°

See Example

2.16.

14.9/1 24°

We can think of E x as being a voltmeter connected across the capacitor. As a result, Solution the 'Voltmeter"

and capacitor together

form a closed loop for which

we can

To meet

write ing

a circuit equation, as for

Thus, in traveling

we

any other loop.

cw around

requirement,

this

KVL

we

which

equations,

write the followthe

should

reader

verify:

the loop

write

-/,(-j72) + E x =

£ l2 + Eb - E, = £23 + Ec - Eh = E M + £a - E c =

0

Thus

0 0

0

Transposing terms, we obtain

Ex = =

/3

(-j72)

14.9 zl 24°

(-

j

E 12 = Ea - Eb = 26

72)

45 and so

£23 = Eb Ex =

1073

zl

Ec =

26

£ 31 - Ec ~ E = iX

45

and hybrid notation

2.39 Circuits In

tation

We

employ both sign noand double-subscript notation as shown in the

some

circuits

it

useful to

is

Ll

I

notation).

We

wish

to

determine the values of

£ 23 and £ 31 (double-subscript ,

notation).

E ]2

*

20°

-26/1

240°

- 26

0°

ing the loop created by

E

Therefore E, N

=

x

=

240°

Z_

=

210°

can even express the sign notation

in

and terminals

N

terms of

in

follow-

and

1

,

we

KVL equation £ N! +

Example 2-17 Fig. 2.56 shows a 3-phase system in which E — 26 L 0°, Eb = 26 zl 20°, and Ec = 26/1 240° (sign

1

L

L

-

90°

double-subscript notation. For example,

can write the

following example.

zl

26

120°

-30°

zl

45

-66°

-26/1

0°

zl

zl

E,

E N = — £,,, which ,

0 can be expressed as

E. v

This completes our review on the writing of dc

and ac

circuit equations.

46

FUNDAMENTALS


2-6

What

2-1

Three dc sources G h G 2 and generate voltages as follows: ,

£34

= -100 V = -40 V

£56

= +60 V

£,2

Show

the actual polarity

minals 2-2

In

in

G3

(Fig. 2.57)

c.

d.

Magnetomotive force

b.

2-7

Problem

(

+ )( — )

of the

ter-

2-8

2-3

if

2-9

the

show

the voltage

and the actual polarity of the generator minals 2-4

at instants

A conductor 2 m

1

,

2, 3,

long

and

moves

60 km/h through a magnetic

2-10

is

moved, and

the coil falls to 1.2

Figure 2.57

2-3.

to

A.

Draw

the force on the

moving N

pole.

N pole act in the

direction as the direction of rotation?

the

waveshape of a sinusoidal

V

and a frequency of 5 Hz.

at a

b.

speed of

If the

voltage

is

=

5

voltage at field

t

having a

the flux linking

mWb in 0.2

the average voltage induced.

Figure 2.58

Calculate the force on the

Does same

a.

carries a cur-

2-11

A sinusoidal

zero

at

t

=

0,

what

is

the

ms? 75 ms? 150 ms?

current has an effective value

of 50 A. Calculate the peak value of current.

The magnet

See Problem

T

mm.

Calculate the force on the conductor.

b.

4.

A coil having 200 turns links a flux of 3 mWb, produced by a permanent magnet.

2-1

produce a flux density of 0.6

voltage having a peak value of 200

voltage.

See Problems

to

gap having a length of 8

c.

ter-

flux density of 0.6 T. Calculate the induced

2-5

air

mmf required. Conductor AB in Figure 2.29 rent of 800 A flowing from B a.

Terminals 1-4 and 3-6 Terminals 1-3 and 4-6

Referring to Figure 2.58,

want

an

at 0.2 T, 0.6 T,

T.

Calculate the

determine the voltage

following terminals are connected together. a. Terminals 2-3 and 4-5

c.

We in

and polarity across the open terminals

b.

of cast iron

tive permeability

2-1, if the three sources are

in series,

Referring to Figure 2.26, calculate the rela-

and 0.7

each case.

connected

the SI unit of

is

Magnetic flux Magnetic flux density Magnetic field intensity

a.

and

2-2.

s.

Calculate

2-12

A sinusoidal

voltage of 120

V

is

applied to

a resistor of 10 O.

Calculate a.

the effective current in the resistor

b.

the peak voltage across the resistor


2-13

c.

the

d.

the

power dissipated by the resistor peak power dissipated by the resistor

A distorted

voltage contains an

th 1

2- 8 1

fre-

2-

1

9

quency of the fundamental. 2-14

The current lags

in

a 60

Hz

(

—

)

,

and curthe

is

each case, indicate which phasor

is

(

)

)

{

36 degrees behind the voltage. 2-20

The

alternating voltage e 2 in Fig. 2.24a

is


angle between the following phasors and,

c.

is

B 2 which box

The resistance of the conductors joining the two boxes in Figure 2.4 is zero. If A, is + with respect to A 2 can B be — with respect to B 2 ?

Referring to Fig. 2.59, determine the phase

b.

terminal A,

A 2 to

,

positive peaks of voltage and current.

a.

if

from

(

single-phase motor

Calculate the time interval between the

2-15

Figure 2.4,

source?

har-

monic of 20 V, 253 Hz. Calculate the

In

rent flows

47

e2

= 20

cos (360 ./f- 9)

in

If 6

lagging.

=

1

In d it stria I

2-2

1

and/ =

50°

value of e 2

/, and / 3 li and / 3 E and /,

at

t

=

0,

1

80 Hz, calculate the

and

at

/

=

262.37

s.

Appl icatio n

In Fig. 2.60 write the

KVL circuit equa-

tions for parts (a), (b), (c),

and

(d).

(Go cw

around the loops.)

6

A

Figure 2.59 See Problem 2-15.

2-

1

6

The voltage applied

to an ac

the current

is /

= 20

magnet

is

E — 60 sin - 60°), all sin


cj>

1

and an-

(c|>

gles being expressed in degrees. a.

Draw

the phasor

diagram for

E and

/,

Draw

the

function of c.

1

a.

/ as

In Fig. 2.6

1

write the

for parts (a), (b),

<\>.

and

KCL circuit equations (c),

and determine the

true direction of current flow.

peak negative power

in the circuit.

2-23

leads of the third

if

cw around

the

harmonic source are


equations for parts

Referring to Fig. 2.24, draw the wave-

shape of the distorted sine wave,

b.

a

Calculate the peak positive power and the

2- 7

2-22

waveshape of E and

2.21.

us-

ing effective values. b.

Figure 2.60

See Problem

2-24

An

KVL and KCL circuit

(a), (b), (c),

and

(d).

(Go

the loops.)

electronic generator produces the out-

shown

reversed.

put voltage pulses

Calculate the peak voltage of the result-

this voltage is applied across a 10 fl resis-

ing waveshape.

tor,

calculate

in Fig. 2.63. If

1

FUNDAMENTALS

48

a.

the fundamental frequency of the

2-25

current

Repeat the calculations of Problem 2-24 for the

b.

the peak power, in watts

c.

the energy dissipated per cycle, in

2-26


power per cycle

the average

e.

the value of the dc voltage that

in the

—

the effective value of the voltage in the

figure

o

1

0

the average voltage

8

4

V

+ 100

resistor

7A

y*4A

2

in parts

the loops.)

would

produce the same average power

g.

KVL and KCL circuit

(Go cw around

(a) to (g).

d.

in Fig. 2.64.

equations for the ac circuits shown

joules

f.

waveshape shown

A

A

3

A

/

^

2

4

6

8

s

Figure 2.63

See Problem

2.24.

^4_A

A

1'

+ 100 (b)

(a)

v

(c)

0

4

-

Figure 2.61 See Problem 2.22.

2

-100 V

Figure 2.64

See Problem 'l

2

5 LI

R

11

/1

|

(b)

(a)

6"

I

—

T

4 LI

1

7Q I— ',1

T

(c)

Figure 2.62

See Problem

2.23.

CSJ

6

1

0

2.25.

seconds


E l2 = 100

20

0°

>50U

12

tf

(a)

(b)

(c)

£-13

£ ab =

49

£ B = 50 150_°

101.30°

E A = 20

= 30 1-3 0°

Eba =100L0 c

[45_°

7

£ 3 = 100(0°

12

J

40

U

'l

A

60

12

) i

30

*

(Ô

3' 1

24

12

^

40

12

^

"1 (d)

Figure 2.65

See Problem 2.26.

(e)

(g)

30

hr

12

Chapter 3 Fundamentals of Mechanics and Heat

3.0 Introduction In

order to get a thorough grasp of electrical

technology,

is

it

mechanics and large motors

is

essential to

power

know something about

For example, the starting of

heat.

determined not only by the magnitude

of the torque, but also by the inertia of the revolving

And

parts.

the overload capacity of an alternator

determined not only by the size of also by the temperature that

its

its

the

—

—

is

we know

exert a muscular effort to

is

overcome

tional force that continually pulls

the force of

lift

it

a stone,

we

the gravita-

downward.

There are other kinds of forces, such as the force

by exploding dynamite. All these forces are ex-

windings can safely

And we could mention many more cases where

comprehensive approach

familiar force

For example, whenever we

exerted by a stretched spring or the forces created

of the conductors as by the currents they

chanical/thermal approach

gravity.

is

And the stringing of a transmission line is determined as much by the ice-loading and mechancarry.

The most

conductors, but

withstand.

ical strength

Force

3.1

pressed

in

terms of the newton (N), which

is

the SI

unit of force.

The magnitude of

the force of gravity depends

upon the mass of a body, and

is

given by the ap-

proximate equation

the electrical/me-

F = 9.8w

essential to a thor-

(3. 1)

ough understanding of power technology. For

this reason, this introductory

chapter covers

certain

fundamentals of mechanics and heat. The top-

ics are

not immediately essential to an understanding

of the chapters which follow, but they constitute a

may wish Consequently, we rec-

valuable reference source, which the reader to consult

ommend

from time

a quick

to time.

first

reading, followed by a closer

study of each section, as the need arises.

where

F= m = 9.8

=

force of gravity acting on the body [N]

mass of the body

[kg]

an approximate constant that applies

when

objects are relatively

close to the surface of the earth

(within 30 km).

FUNDAMENTALS OF MECHANICS AND HEAT

Example 3-1

51

F

Calculate the approximate value of the force of

on a mass of

gravity that acts

1

2 kg.

Solution

The force of gravity

m = = 17.6 N

F=

9.8

is

X

9.8

=

12

Figure 3.1 Torque T=

17.6 newtons

1

Fr.

1

When using the make

to

the

equal to 0.453

hand, a

between the pound

a distinction

pound-force

A pound

(lbf).

592 37 kg,

pound-force

is

On

exactly.

and

(lb)

T= F= r =

the other

X

equal to (9.806 65

0.453

592 37) newtons exactly, or about 4.448 N. If the

pulley

around

Example 3-2

(3.2)

where

mass

a unit of

is

T = Fr

we have

English system of units,

is

torque |N m| force [N]

radius [mj

move,

free to

it

will begin to rotate

axis.

its

Calculate the approximate value of the force of gravity that acts

on a mass of 140

newtons and

sult in

Express the

lb.

re-

Example 3-3

A

in pounds-force.

motor develops

a starting torque of 150

pulley on the shaft has a diameter of Solution

the braking force needed to prevent the

Using the conversion charts in Appendix

massif 140 Eq.

3.

1

-

lb

9.8

=

140 (- 2.205)

the force of gravity

F=

m =

9.8

X

AXO,

= 622.3(^4.448) =

1

=

63.5

622.3

39.9 pound-force

=

39.9 lbf

1

mass of 140

entirely different

The

N

F =

lb.

139.9 is

N

lbf.

radius T/r

action

is

from a mass of 140

al-

lb.

3.3 Mechanical

it

rotate.

Mechanical work

pose a string /•

is

(Fig. 3.

1

Torque

If

we

pull

on the

cl

is

done whenever a force

in the direction

the pulley will tend to rotate.

The torque exerted is

F moves

of the force. The work

Fd

(3.3)

where

W= F= cl =

work

(

J]

force [N]

distance the force

moves [m]

a ra-

string with a force

on the pulley by the tangential force

be

is

of the force. For example, sup-

wrapped around a pulley having ).

N would

given by

between the axis of rotation and the

point of application

required. If the ra-

work

W=

on a body, tending to make

is

sufficient to prevent rotation.

equal to the product of the force times the perpen-

F

= 300 N

almost

However,

produced when a force exerts a twisting

dicular distance

motor from

m; consequently, a braking force

50/0.5

1

dius were 2 m, a braking force of 75

is

dius

0.5

is

=

Torque

Torque

If a

turning.

a distance

3.2

m.

63.5 kg. Using

though the numbers are nearly the same, a force of is

N

m, calculate

Solution

Note that the force of gravity of exactly equal to the

a

is

Again using the conversion charts, a force of 622.3

140 lbf

l

given by

Example 3-4

A

mass of 50 kg

3.2).

is

lifted to a height

Calculate the work done.

of 10

m

(Fig.

1

FUNDAMENTALS

52

500 kg

Figure 3.2

W=

Work

Fd Figure 3.3 Power P - Wit

Solution

The force of

gravity acting on the 50 kg

F=

m =

9.8

The work done

X

9.8

50

mass

is

F=

= 490 N

The work done

X

490

10

W=

= 4900 J

The power

Power

Power

is

9.8

X 500 = 4900 N

is

is

W= Fd = 3.4

m=

9.8

P =

the rate of doing work.

It is

given by the

=

Fd = 4900 X 30 = 147 000

J

is

Wit 147 000/12

=

12

250

W=

12.25

kW

equation:

Expressed

P=

Wit

(3.4)

in

horsepower,

P=

12 250/746

=

16.4

hp

where

P = power [W] W = work done

=

t

The

unit of

power

is

the watt (W).

corresponds

is

use the

sometimes expressed

to the

is

motor

equal

average power output

height of 30

m

a mass of

lifts

in

power developed by

12

s

500 kg through a

the motor, in kilowatts and in

Solution

ity

in the

acting on the

cable

mass

is

The mechanical power output of a motor depends upon its rotational speed and the torque it develops. The power is given by

where

P = T— n = 9.55 =

(Fig. 3.3). Calculate the

horsepower.

The tension

equal to the force of grav-

that

is

being

motor

in

Example 3-5 electric

of a

to

of a dray horse.

An

Power

1000 W. The

One horsepower

units.

[s]

We often

equal to

is

power output of motors horsepower (hp) It

[J]

time taken to do the work

kilowatt (kW), which

746 W.

3.5

lifted:

mechanical power [WJ torque [N-m]

speed of rotation [r/min] a constant to take care of units

(exact value

=

30/tt)

We can measure the power output of a motor by means of a prony brake. It consists of a stationary flat belt that presses against a pulley mounted on the motor shaft. The ends of the belt are connected to two spring

scales,

and the

belt pressure

is

adjusted


V (Fig. 3.4). As the motor turns,

by tightening screw

53

3.6 Transformation of energy

we can increase or decrease the power output by adpower developed by the motor into heat

When register

The mechanical

tension of the belt.

justing the

by the the

entirely converted

is

it

not running, the spring scales

equal pulls and so the resulting torque

(as

P 2 The

re-

does

in Fig. 3.4), pull

Thermal energy (heat released by

ley

is

therefore

a radius ters.

late

f>,

exceeds pull

— P 2 newtons. If the pulley has T = (P — P 2 )r newton me-

(/>,

Knowing

a stove, by

by the sun)

Chemical energy (energy contained

3.

in

dyna-

mite, in coal, or in an electric storage battery)

)

the net torque

r,

.

friction,

on the circumference of the pul-

in

mov-

ing car)

is

However, when the motor turns clockwise

sultant force acting

one of the following forms:

a coiled spring or the kinetic energy of a

2.

zero.

exist in

Mechanical energy (potential energy stored

1.

rubbing against the pulley.

belt

motor

is

Energy can

produced by

Electrical energy (energy

4.

a gen-

x

the speed of rotation,

we can

erator, or

calcu-

the power, using Eq. 3.5.

by lightning)

Atomic energy (energy released when

5.

cleus of an atom

is

the nu-

modified)

Although energy can be neither created nor destroyed,

it

can be converted from one form to an-

other by means of appropriate devices or machines.

For example, the chemical energy contained

in

coal

can be transformed into thermal energy by burning

The thermal energy contained

the coal in a furnace. in

steam can then be transformed into mechanical

energy by using a turbine. Finally, the mechanical energy can be transformed into electrical energy by

means of a In the

generator.

above example, the furnace, the turbine,

Figure 3.4

and the generator are the machines

Prony brake.

ergy transformation.

that

Unfortunately, whenever energy the output

Example 3-6 During a prony brake test on an electric motor, the

25

spring scales indicate (Fig. 3.4).

turns at 0.1

N

and 5 N, respectively

Calculate the power output

if

the

motor

1700 r/min and the radius of the pulley

is

always

transformed,

because

all

ma-

chines have losses. These losses appear in the form of heat, causing the temperature

of the machine to

rise.

Thus, the electrical energy supplied to a motor

is

partly dissipated as heat in the windings. Furthermore, is

some of its mechanical energy

m.

is

less than the input

do the en-

is

also

lost,

due

to bear-

ing friction and air turbulence created by the cooling Solution

fan.

The torque

T= = The power

The mechanical

losses are also transformed into

put of a motor

Fr (25

-

5)

X

0.1

=

2

Nm

is

less than the electrical input.

3.7 Efficiency of a

machine

is

The efficiency of a machine

P =

ai

mechanical power out-

heat. Consequently, the useful

is

779.55

=

1700

X

2/9.55

= 356

W

The motor develops 356 W, or about 0.5 hp.

is

given by the equation

P

t\

= p X

100

(3.6)

FUNDAMENTALS

54

where

Kinetic energy

—

Tl

P0 =

The energy

power

input

\/2mv

power of the machine fW]

output

efficiency is

is

machine

to the

particularly

[

W]

low when thermal

from 25

the efficiency of steam turbines ranges that of internal

gines (automobile engines, diesel motors)

we must remember it

lies

how low that a

having an efficiency of 20 percent loses, of heat, 80 percent of the energy

Ek = kinetic energy [JJ m = mass of the body [kg] v = speed of the body fm/s]

to

be-

A bus having a mass of 6000 kg moves at a speed of

machine

100 km/h. If it carries 40 passengers having a total mass of 2400 kg, calculate the total kinetic energy of the loaded vehicle. What happens to this energy

in the

form

receives.

efficiently.

Example 3-8

these

Electric motors transform electrical energy into

mechanical energy much more

when

the bus

Solution

pending on the size of the motor.

Total

150

kW

percent

when

mass of

it

motor has an efficiency of 92

operates

at full-load.

The speed

+ 2400 = 8400

=

100 km/h

v

100

refers to the

The mechanical output power

=

163

kW

To stop the

is

is

\/2mv 3

2

1/2

245 928

J

X 8400 X 3.25 MJ

27.

2

bus, the brakes are applied and the re-

sulting frictional heat

is

entirely

produced

at the

ex-

pense of the kinetic energy. The bus will finally

come

losses are

150

27.8 m/s

kinetic energy

Ek = =

P 0 = 150kW

163

m

s

mechanical

The

150/0.92

kg

X 1000 3600

The 150 kW rating always power output of the motor. The input power is

The

6000

is

is

Solution

{

to a stop?

Calculate the

losses in the machine.

P = PJi] =

braked

the loaded bus

m =

_

electric

is

Their ef-

ficiency ranges between 75 and 98 percent, de-

Example 3-7

(3.7)

combustion en-

tween 15 and 30 percent. To realize efficiencies are,

2

where

converted into mechanical energy. Thus,

40 percent, while

A

a form of mechanical energy

efficiency [percent]

—

Pi

is


13

to rest

when

all

the kinetic energy (3.25

MJ)

has been dissipated as heat.

kW

Considering the high efficiency of the motor, the losses are quite moderate, but they

be enough to heat a large

home

in the

would

still

winter.

3.8 Kinetic

3.9 Kinetic

A

energy of linear motion

A falling stone or a swiftly sess kinetic energy,

which

moving automobile posis

energy due to motion.

energy of

moment

middle of

Its

rotation,

of inertia

revolving body also possesses kinetic energy.

magnitude depends upon the speed of rotation

and upon (he mass and shape of the body. The netic energy of rotation

on page 56.

is

ki-



TABLE 3A

MOMENT OF

INERTIA J AROUND AN AXIS

OF ROTATION

0

Figure 3.5

Mass

m revolving at a distance

J

r

around axis

0.

= mr

(3.9)

Figure 3.6 Solid disc of

mass

m and

radius

r.

(3.10)

Figure 3.7 Annular

ring of

mass

J

m having a

rectangular cross-section.

= -(Rf + R 2 I

(3.11)

)

Figure 3.8 Straight bar of

mass

m pivoted on

its

center.

J

(3.12)

12

Figure 3.9 Rectangular bar

of

mass

J

m revolving around

= -(Rf + /V

axis 0.

(3.13)

55

1

FUNDAMENTALS

56

5.48

X

10"'V/7-

(3.8)

The

b.

Ek = =

where

E k = kinetic energy [J] J = moment of inertia [kg-rrrl = rotational speed [r/minj X 1()~ 3 = constant to take care of units 2 Lexact value = tt /1800] //

5.48

The moment of inertia./ (sometimes simply called depends upon the mass and shape of the body.

inertia )

may be

value

Its

calculated for a

shapes by using Eqs. 3.9 to

3.

1

3 given in Table 3 A. If

body has a complex shape,

the

number of simple

can always be bro-

it

ken up into two or more of the simpler shapes given in the table.

are then

The

individual Js of these simple shapes

added to give the

kinetic energy

= Note as

5.48

X

5.48

X

3.

1

is

l0"-V/r

X

(3.8)

175

X

1800

2

MJ

that this relatively small flywheel possesses

much

kinetic energy as the loaded bus

Example

tioned in

men-

3-8.

Example 3-10

A

flywheel having the shape given

in Fig. 3.1

1

is

composed of a ring supported by a rectangular hub. The ring and hub respectively have a mass of 80 kg and 20 kg. Calculate the moment of

inertia of the

flywheel.

J of the body.

total

Inertia plays a very important part in rotating

machines; consequently,

it

is

worth our while to

solve a few problems.

80 kg

Example 3-9

A 1

1400 kg

solid

m

flywheel has a diameter of

steel

and a thickness of 225

mm (Fie.

225

3. 10).

mm 1400 kg

V

1800 r/m

Figure 3.11 Flywheel in Example 3.10.

Figure 3.10 Flywheel in Example

3.9.

Calculate Its

b.

The at

Solution

moment

a.

of inertia

kinetic energy

For the

when

ring,

the flywheel revolves

J

I800r/min

]

Solution a.

Referring to Table 3 A, inertia

we

find the

moment

of

= m (Rr + R 2 2 )/2 = 80 (0.4 2 + 0.3 2 )/2 =

(3.1 1)

10 kg

m

2

For the hub,

is

J2

J

(3.10)

1400

X

0.5

The

2

total

= mL 2 /\2 = 20 X (0.6) 2 /12 = moment

= 175ks-m 2 J

=

J

(3. 12)

0.6 kg

m

2

of inertia of the flywheel

X

+

J2

=

10.6

kg

m

is

2

k,


3.10 Torque, inertia,

volving body, and that a

change the speed of a

to is

to subject

The

given period of time.

depends upon the

Speed

1

In electric

way

only one

is

3-1

speed

in There

and change

it

re-

to a torque for

of change of speed

rate

inertia, as well as

on the torque.

A

=

A/7

9.55

power technology,

a system there are three

erted by the load, and the speed.

(3.14)

T=

The load

a shaft (Fig. 3. 2).

interval of time during

torque

J 9.55

is

— moment = constant

applied

creasing

or

Suppose

the

T

[sj

.

Y

to take care

=

in a

We now

explain

exerts a constant torque

counterclockwise direction.

TM developed by

the

can be varied by

in-

the torque

acts clockwise,

and

decreasing

system

it

the

current

electric rest

is initially at

and

that

/.

7M =

Because the torques are equal and opposite, the

net torque acting on the shaft

2

zero, and so

is

it

has

]

no tendency

to rotate.

of units Load

mm

30/ttJ

the torque acts in the direction of rotation, the

speed rises. Conversely, tion

which the

of inertia fkg-m

[exact value If

motor

[Nm]

torque

=

Ar

speed [r/min]

in

to consider:

they interreact.

1

change

main factors

the torque developed by the motor, the torque ex-

Tu that always acts On the other hand,

=

often happens that

Consider a load coupled to a motor by means of

TMJ

where A/?

it

system

an electric motor drives a mechanical load. In such

how

simple equation relates these factors:

of a motor/load

57

if

of rotation, the speed

acts against the direc-

it

The term An may

falls.

therefore represent either an increase or a decrease in

speed.

'

M

Example 3-11 The to

fly wheel

increase

torque of

20

of Fig. 3.11 turns

Nm.

at

60 r/min.

We wish

Motor

600 r/min by applying a For how long must the torque be

speed

its

to

Figure 3.12

applied?

Shaft

is

stationary

TM

71-

Solution

The change

in

A/?

speed

-

The moment of

(600

-

inertia

J

As

is

=

60)

= 540

a result of the

twists and

r/min

becomes

shaft

deformed, but other-

wise nothing happens.

Suppose we want

is

10.6 kg-

opposing torques, the

slightly

/z, To do so, we increase the motor current so TM exceeds T The net torque on the shaft acts

speed

2

that

.

.

x

clockwise, and so

Substituting these values in Eq. 3.14

the load to turn clockwise at a

it

begins to rotate clockwise. The

speed increases progressively with time but as soon 9.55 7Af/./

A/7

(3.14) as the desired

540

=

9.55

x 20

A/7 10.6

TL The .

yielding

speed

motor current so

30

s

is

TM

reached, is

let

us reduce the

again exactly equal to

net torque acting on the system

and the speed Ar

/z,

that

any more (Fig.

/z,

is

now

zero

will neither increase or decrease

3. 13).

FUNDAMENTALS

58

Load

Load

Figure 3.13 Shaft turns

Figure 3.14

cw 7M

Shaft turns

71-

Whenever

This brings us to a very important conclusion.

The speed of a mechanical load remains constant when the torque TM developed by the motor is equal and opposite to the torque T L exerted by the load.

At

first, this

we

accept, because

when TM =

7^

rather difficult to

is

the

opposite

to

a motor remains constant motor torque is exactly equal and

load

motor/load system

is

torque.

In

the

effect,

then in a state of dynamic

equilibrium.

With the load now running clockwise suppose we reduce

TM

so that

it

is

at

a speed

TL

less than

.

on the shaft now acts counterclock-

net torque

wise. Consequently, the speed decreases and will

continue to decrease as long as

T

imbalance between

the

enough, the speed

we

then reverse. If

TM — TL when

motor torque 7M and load torque and opposite, the speed will

change. The rate of change depends upon the inerof the rotating parts, and this aspect

tia

more

is

covered

in

detail in Section 3. 13.

Power flow

3.12

our reasoning (and reality) shows.

the

the

are not exactly equal

{

T,

will eventually

exceeds

TM

and

TM

.

If

long

lasts

become zero and

control the motor torque so that

the reverse speed reaches a value n 2

,

in

a mechanically

coupled system

The speed of

repeat:

whenever

The

TL

.

are inclined to believe that

the system should simply stop. But

,

this is not so, as

We

conclusion

ccw TM = TL

Returning again to Fig.

3. 13,

we

torque

TM

acts in the

torque

T

acts opposite to speed n

see that motor

same direction (clockwise) as speed n lt This means that the motor delivers mechanical power to the shaft. On the other hand, load {

x

the load receives mechanical

.

Consequently,

power from

the shaft.

We can therefore state the following general rule: When the torque developed by a motor acts in same direction as the speed, the motor delivpower to the load. For all other conditions, the motor receives power from the load. the

ers

In Fig. 3.14, for example, the

power from

the load because

7M

motor receives

acts opposite to n 2

.

the system will continue to run indefinitely at this

Although

new speed

brief periods in electric trains and electric hoists.

(Fig. 3. 14).

In conclusion, torques

Figs. 3.12, 3.13,

TM and T

{

are identical in

and 3.14, and yet the shaft may be

turning clockwise, counterclockwise, or not

The

actual

steady-state

whether TM was greater or

speed less than

depends

7L

this is

an unusual condition,

it

occurs for

The behavior of the motor under these conditions will be examined in later chapters.

at all.

upon

3.13 Motor driving a load

for a certain

having inertia

period of time before the actual steady-state condi-

was reached. The reader should ponder moments over this statement.

tion

a

few

When

a motor drives a mechanical load, the speed

is

usually constant. In this state of dynamic equilibrium.


the torque

TM developed by the motor is exactly equal T imposed by the load.

and opposite to the torque

The

inertia

of the revolving parts does not

However,

play under these conditions.

torque

speed

is

raised so that

it

increase,

will

come

if

we have

already

when

less than that

of

speed drops. The increase or decrease

in

speed (An)

motor torque

is

TM

torque Tis now replaced by the net torque (TM — 7L

=

A/7

- T )AtIJ

9.55 (rM

x

stays con-

paper remains

must be greater than the load torque

It

.

We

in

have

given by the Eq. 3.14, except that

is still

Nm)

in the

order for the speed to increase.

seen.

Conversely,

because the tension

unchanged. Let the required motor torque be

exceeds the load torque, the

as

the speed increases from 120 r/min to 160

stant

motor

the

As

r/min, the load torque (5400

into

the load, the

the

b.

x

59

-

A/7

):

J

(3.15)

At

where

160

-

= 4500 = 5 s

120

m

kg

r/min

- TL )At

9.55 (7\,

An =

= 40

2

J

= change in speed [r/minj TM = motor torque [N-m] 7L = load torque [N m] At = time interval during which TM and TL are acting [s] J = moment of inertia of all A/7

_

40

-

9.55

4500 Thus,

TM - 5400 = 3770 rM = 9170

revolving parts [kg-irrj

The motor must Example 3-12

A

large reel

torque of 9170

of paper installed

machine has

a

diameter of

moment of

and a

1

inertia of

at the

end of

a

paper

kgm 2

.

It is

therefore develop a constant

Nm

during the acceleration

period.

m, a length of 5.6 m,

.8

4500

5400) 5

The mechanical power of the

driven

reel

motor

160 r/min

at

accelerating

is

by a directly coupled variable-speed dc motor turn-

The paper

ing at 120 r/min.

tension of a.

160 r/min

c.

in 5

c.

from 120 r/min

to

seconds, calculate the torque that

power of

the

this interval.

motor

after

it

As soon

7 = Fr = 6000 X

1

reel

.8/2

The power developed by the

P =

nT

120

P =

is

is

N

m). The power

nT _ ~

160

X 5400

9.55

9.55

90.5

kW (equivalent to

121 hp)

is

= 5400 N m reel

motor

3.14 Electric motors driving linear

motion loads

is

Rotating loads such as fans, pumps, and machine

X 5400 9.55

kW

160 r/min)

therefore reduced to

tools are well suited for direct mechanical coupling to electric motors.

67.85

(

equal to the load torque (5400

has

(3.5)

9.55

kW (equivalent to 206 hp)

as the desired speed

of the motor

Solution

The torque exerted on the

9.55

reached, the motor only has to develop a torque

reached the desired speed of 160 r/min.

a.

160X9170-

-

153.6

the reel

speed of 120 r/min.

motor must develop during

Calculate the

-

power of the motor when

the speed has to be raised

the

nT

P =

6000 N.

Calculate the

If

kept under a constant

9.55

turns at a constant b.

is

(about 91 hp)

move

in

On

the other hand, loads that

a straight line, such as hoists, trains, wire-

FUNDAMENTALS

60

drawing machines,

must be equipped with a

etc.,

where

motion converter before they can be connected to a

rotational speed [r/min]

T=

torque [N-m]

F= v = 9.55 =

we seldom

converters are so utterly simple that

=

n

The motion converter may be a rope-pulley arrangement, a rack and pinion mechanism, or simply a wheel moving over a track. These rotating machine.

force [N] linear speed [m/s|

a constant [exact value

=

30/tt]

think of the important part they play. Straight-line motion involves a linear speed v

and a force

while rotary motion involves a rota-

speed n and a torque

tional tities

F,

related

when

T.

How

are these quan-

a motion converter

is

used?

Consider a jack driven by a motor that rotates

T (Fig.

a linear speed

plied in raising the load

is

v.

motive turns

the other hand, the

needed

is

at

1

to pull an electric train

The motor on board

the loco-

200 r/min. Calculate the torque de-

at

F

Solution

The power sup-

given by

power

kN

This

nT=

9.55fY

1200 T =

X

P a = Fv On

of 25

a speed of 90 km/h.

veloped by the motor. 3. 15).

causes a vertical ram to exert a powerful force at

A force at

a speed n while exerting a torque

while moving

Example 3.13

9.55

(3.16)

25 000

j = 4974 N-m = input to the jack

5

X

(90 000/3600)

kN

m

is

given by

= nT

P:

3.15 Heat and temperature (3.5)

1

9.55

Assuming verter,

Whenever

there are no losses in the motion con-

we have

the SI unit p. 1

1

heat

applied to a body,

is

mal energy. Heat is

'

energy?

Consequently,

nT=

9.55Fv

a

body receives

this type

of

atoms of the body vibrate more

First, the

intensely. Second,

we can

receives ther-

the joule.

What happens when

= po

it

therefore a form of energy and

is

its

temperature increases, a fact

verify by touching

it

or by observing the

(3.16)

reading of a thermometer.

For a given amount of heat, the increase

tem-

in

perature depends upon the mass of the body and the material of which

100 kJ of heat to

is

it

made. For example,

if

by 24°C. The same amount of heat supplied of copper raises

we add

kg of water, the temperature

1

rises

to

temperature by 263°C.

its

therefore obvious that heat and temperature are

kg

1

It

is

two

quite different things. If

we remove

heat from a body,

its

temperature

drops. However, the temperature cannot

low a lower zero.

or

It

limit.

corresponds

— 273.1 5 °C. At

This limit to a

fall

be-

called absolute

is

temperature of 0 kelvin

absolute zero

all

and the only motion

Figure 3.15

tions cease

Converting rotary motion into linear motion.

that of the orbiting electrons.

atomic vibra-

that subsists

is


1806

T

iron melts

T

450 K

i

1083

aluminum melts

933

660

1220

lead melts

600

327

621

water boils water freezes

373 273

100

212 32

0

-273.15

®

®

1981

-459.67

Fahrenheit scale

Celsius scale

Kelvin scale

2791

810°F

450 °C

1, 356

copper melts —

T

1533

61

Figure 3.16 Temperature scales.

3.16

The

Temperature scales

given

specific heat capacity of several materials

Table

in

AX2

in

the

is

Appendix.

The kelvin and the degree Celsius are the SI units of temperature. Fig. 3. 6 1

tionships

between

shows some

interesting rela-

Kelvin,

the

Celsius,

and

Fahrenheit temperature scales. For example, iron melts at

1

806

K

or

1

533°C or 279 °F 1

Example 3-14 Calculate the heat required to raise the temperature

of 200

L of water from 10°C

tank

perfectly insulated (Fig. 3.17).

is

heat capacity of water

3.17

Heat required to raise the temperature of a body

The temperature heat

it

terial.

receives,

rise

its

weighs

1

kg.

of a body depends upon the

mass, and the nature of the ma-

The relationship between these

quantities

is


Q = mcAt

(3. 17)

where

Q =

quantity of heat added to (or

m = c

=

removed from)

mass of the body

=

body

[JJ

specific heat capacity of the

material

At

a

|kg]

making up

the

body

[J/(kg-°C)l

Figure 3.17

change

Electric

in

temperature [°C]

water heater.

is

to

70°C, assuming the

The

specific

41 80 J/kg °C, and one

liter

n FUNDAMENTALS

62

Solution

quired

convection

convection

The mass of water

is

200 kg, and so the heat

re-

is

Q ~ mcAt = 200 X 4180 X = 502 MJ

(70

-

10)

Referring to the conversion table for Energy (see

Appendix),

we

find that 50.2

MJ

is

equal to 13.9

kW-h.

3.18 Transmission of heat Many problems

in electric

power technology

lated to the adequate cooling of devices

are re-

and ma-

knowledge of the mechanism by which heat is transferred from one body to another. In the sections that follow, we

Figure 3.18 Heat transmission by convection, conduction, and

ra-

diation.

chines. This, in turn, requires a

briefly

review the elementary physics of heat trans-

We

mission.

also include

some simple

but useful

Referring to Fig. 3.19,

we can

equation

equations, enabling us to determine, with reason-

P =

able accuracy, the heat loss, temperature rise, and so

on of

electrical

equipment.

its

we

-

—

atoms

(Fig. 3. 18). This

A =

—

atomic vibration

]

t

2)

—

end of the

bar.

atom

is

to the next, to the other

at

d = thickness of the body [m]

Consequently, the end opposite the

flame also warms up, an observation

one time or another.

ferred along the bar

we have

In effect, heat

is

all

trans-

by a process called conduction.

The rate of heat transfer depends upon the thermal conductivity of the material. Thus, copper is a better thermal

conductor than

steel

is,

and plastics

and other nonmetallic materials are especially poor conductors of heat.

The SI

unit of thermal conductivity

per meter degree Celsius conductivity of several in

Tables

AX

1

and

AX2

[W/(m

common in the

°C)J.

is

the watt

The thermal

materials

Appendix.

is

given

2

body [m difference of temperature between

surface area of the

opposite faces [°C] transmitted from one

made

(3.18)

[W/(m-°C)J bring a hot flame near one end of an iron bar,

temperature rises due to the increased vibration its

]

P = power (heat) transmitted [W] X = thermal conductivity of the body

(t

of

-

where

3.19 Heat transfer by conduction If

calculate the ther-

mal power transmitted through a body by using the

Figure 3.19 Heat transmission by conduction.

]


Example 3-15

63

tank

The temperature difference between two sides of a

mica

sheet of

50°C

is

cm" and thickness ing

3

is

through the sheet,

(Fig. 3.20). If

mm,

its

area

is

200

calculate the heat flow-

in watts.

Sol ui ion

AX1,

According to Table of mica

ducted

is,

W/m

0.36

is

the thermal conductivity

The thermal power con-

°C.

therefore.

-

\A(t

P

t

l

2)

(3.18)

0.36

d

Figure 3.21

X

Convection currents

0.02 (120

70)

120

in oil.

W

0.003

The warm 0.02

m

2

it

comes

in

contact with the cooler tank,

X = 0.36

chills,

nal tank.

120'

oil

becomes heavier, sinks to the bottom, and moves upward again to replace the warmer oil now moving away. The heat dissipated by the body is, therefore, carried away by convection to the exter-

mica

The

tank, in turn, loses

convection to the surrounding

x

its

heat by natural

air.

70°C

3

3.21

mm

Calculating the losses

by convection

Figure 3.20 Mica sheet,

The

Example 3-15.

heat loss by natural convection in air

is

given

by the approximate equation

P =

3.20

Heat transfer by convection

In Fig.

3.18 the air in contact with the hot steel bar

3/Uf,

-

t

2

y

(3.19)

where

warms up and, becoming a

chimney.

As

the bar,

smoke

moves upward, it is which, in turn, also warms

the hot air

placed by cooler air

A continual

lighter, rises like

current of air

removing

its

is

P — A — f, = t2 =

in

re-

up.

therefore set up around

heat by a process called nat-

heat loss by natural convection [W|

surface of the body [m] surface temperature of the body [°C]

ambient temperature of the surrounding air l°CJ

ural convection.

The convection process can be accelerated by employing a fan

to create a rapid circulation

In the

case of forced convection, such as that

produced by a blower, the heat carried away

is

of given approximately by

fresh in

air.

Heat transfer by forced convection

is

used

P = 1280 Va {t 2

most electric motors to obtain efficient cooling.

when oil. The

Natural convection also takes place

body

is

immersed

contact with the currents

in a liquid,

such as

(3.20)

where oil in

body heats up, creating convection

which follow the path shown

/,)

a hot

in Fig. 3.21.

P — heat loss by forced convection [W] — volume of cooling air |m Vs]

V.

x

FUNDAMENTALS

64

=

i\

temperature of the incoming (cool) air

3.22 Heat transfer by radiation

[°C| f

—

2

We

temperature of the outgoing (warm)

have

air|°C]

same

when hydrogen,

Surprisingly, Eq. 3.20 also applies a

much

lighter gas,

basked

all

is

used as the cooling medium.

properties

Example 3-16 area of 1.2 nr.

When

operates

it

at full-load,

60°C

surface temperature rises to

20°C

in

the

an ambient of

(Fig. 3.22). Calculate the heat loss

by natural

as

energy

light,

P =

=

diate

when

only converted to heat

is

have discovered

even those

heat,

as the physi-

that

that all bodies ra-

very cold. The

are

amount of energy given off depends upon

the tem-

perature of the body.

On L25

-

3A{t

t2

}

the other hand,

all

bodies absorb radiant en-

3

X

(60

1.2

ergy absorbed depends upon the temperature of

)

-

20)

L25

W

= 362

the surrounding objects. There tinual ial

362 i

f

)

\

W H

bodies,

as

convection

body

the

is

same

when

in

as that of

body then radiates a

body

as

the temperature of a

surroundings. The

its

much energy

is

hotter than

is

tinually lose heat in

therefore, a con-

each were a miniature sun.

if

Equilibrium sets

\

is,

exchange of radiant energy between mater-

and the net radiation

zero.

its

On

as

it

receives

the other hand,

environment,

by radiation, even

if

it it

will is

if

con-

located

vacuum.

3.23 Calculating radiation losses

Figure 3.22 Convection and radiation losses

a

in

totally

enclosed

motor.

The heat

that a

body

loses by radiation

is

given by

the equation

P = kA

Example 3-17

kW

fan rated at 3.75

through a 750

kW

motor

the inlet temperature is

3

1

is

blows 240 rrrVmin of to carry

22°C and

away

air

away by

P = A = T, = T2 =

the outlet temper-

the circulating

air.

lx

=

1280

(t 2

-

k

=

X 240/60

heat radiated

(31

-

22)

(3.21)

= 46 kW

IW]

surface area of the body \m'\

absolute temperature of the body [K]

absolute temperature of the surround-

a constant, ture of the

f,)

(approximate)

)

ing objects IK]

Consequently, the losses are

P = 1280 V

- T2 4

where

°C. estimate the losses in the motor.

losses are carried

4

(7,

the heat. If

Solution

The

passes

readily

it

ergy from the objects that surround them. The en-

Solution

ature

and

meet a solid body, such

earth. Scientists

convection.

A

the

and living things on the surface of the

cal objects

enclosed motor has an external surface

totally

warmth produced by

through the empty space between the sun and the earth. Solar

the sun's rays

A

in the

sun's rays. This radiant heat energy possesses the

Table

3B

which depends upon the na-

body surface

gives the values of k for surfaces

monly encountered

in electrical

equipment.

com-

FUN DA MENTA LS OF MFC HA NfCS A NI) HFA I

3-3

RADIATION CONSTANTS

TABLE 3B

Give

the SI unit

symbol Type of surface polished silver bright

oxidized copper

3

aluminum paint

3

Nichrome

2

tungsten

2

oxidi/ed iron

4

insulating materials paint or nonmetallic

perfect emitter

5

enamel

5

5.669

(blackbody)

)

X l()" X 10 s X 10~ s X 10 s X 10" s X 10 8 X 10'"* X 10" s X 10 s X 10'"*

1

force

work

pressure

area

mass

temperature

thermal energy

thermal power

3-4

200

force of

An

tion,

coated with a non-

knowing

surrounding objects are

that all

mechanic exerts a

end of a wrench hav-

automobile engine develops a torque of

at

an

at

a speed of

power output

4000

r/min. Calculate

watts and in horse-

in

power.

by radia-

lost

at the

ing a length of 0.3 m. Calculate the torque

The motor

is

N

power

he exerts.

the

Example 3-16

electrical

In tightening a bolt, a

Example 3-18 in

energy

electrical

600 N-m

enamel. Calculate the heat

mechanical power

mechanical energy

3-5

metallic

and the corresponding SI

for the following quantities:

4

s

0.2

copper

oxidized

W/(m 2 -K

Constant k

65

3-6

ambient temperature of 20°C.

A crane

lifts

200

ft in

and

in

15

a mass of 600 lb to a height of s.

Calculate the power

watts

in

horsepower.

Solution

An

3-7

=

= 60°C or (273. 15 + 60) = 333 K T2 = surrounding temperature = 20°C or (273. 15 4- 20) - 293 K

7,

surface temperature

From Table 3B. k The power

=

5

X

10

by radiation

lost

s

W/(m 2 -K 4

is,

line

4

=

5

-

7\

10" 8

X

(7,

X

3-8

(3.21)

)

4

-

293

20 kW.

c.

A

large flywheel has a

it

(hp]

J

lb-fr. Calculate

rotates at

its

moment

of inertia of

kinetic energy

when

60 r/min.

4

The rotor of an induction motor has a moment of inertia of 5 kg-nr. Calculate the

3-9

)

W (approximate)

= 296

kW from the

20

The power output of the motor [kW] and The efficiency of the motor The amount of heat released Btu/h]

b.

therefore,

(333

1

to

).

4

1.2

motor draws

and has losses equal

Calculate a.

500

P = kA

electric

energy needed to bring the speed It

is

interesting to note that the

almost as

much

motor dissipates

heat by radiation (296

W)

as

it

does

by convection (362 W). 3-

1

0

Questions and Problems 3-11 3-

1

A cement is

from zero

b.

from 200 r/min

c.

from 3000 r/min

Name ried

Practical level

What What

block has a mass of 40 kg.

the force of gravity acting on it?

3-2

is

needed

How much

to

energy

to

the three

to

400 r/min

to

400 r/min

ways whereby heat

from one body

A motor develops a cw 50 N-m.

If this situation persists for

some

time, will the

direction of rotation eventually be

needed

to

lift

a sack

of flour weighing 75 kg to a height of 4

m?

car-

torque of 60 N-m,

lift it?

is

is

to another.

and the load develops a ccw torque of

a.

force

200 r/min

a.

b.

What

value of motor torque

the speed constant?

is

cw

needed

or to

ccw? keep

FUNDAMENTALS

66

3-12

A

motor drives

r/min.

Nm, and 15 Nm.

12

of

cw speed of 1000 a cw torque of exerts a ccw torque

a load at

3- 9 1

The motor develops the load

a.

Will the speed increase or decrease?

b.

If this situation persists for

some

1

negligible, calculate the following:

what

time, in

a.

direction will the shaft eventually rotate'?

3-13


what 3-14

is

the

if


=50

TM = 40 N

power delivered by if

the

b.

m,

power

3-20

motor?

TM = 40 N m

r/min, calculate the

= 50

Calculate the heat [MJJ required to raise the

temperature of 100 kg of copper from 20°C

deliv-

3-2

1

Repeat Problem 3-20 for

1

00 kg of

alu-

minum.

Referring to Fig. 3.14, n2

The torque developed by the motor [N-m] The force opposing the motion of the bus [N]

to 100°C.

and

ered by the motor.

3-15

The electric motor in a trolley bus develops a power output of 80 hp at 200 r/min as the bus moves up a hill at a speed of 30 miles per hour. Assuming that the gear losses are

if

TM = 40 N m and


3-22

power received

by the motor.

The motor in Fig. 3.23 drives a hoist, raising a mass m of 800 kg at a uniform rate of 5 m/s. The winch has a radius of 20 cm. Calculate the torque [N-m] and speed

Intermediate level

[r/min] of the motor. 3- 6 1

During a prony brake

test

on a motor (see and speed

Fig. 3.4), the following scale

readings were noted:

P2 = 5 n =

1

If

P,

lbf

=28

lbf

160 r/min

the diameter of the pulley

calculate the

12 inches,

is

power output of

the

motor

in

kilowatts and in horsepower. 3- 7 1

A

motor drives a flywheel having a moment

of inertia of 5 kg-nr.

The speed

increases ij

from 1600 r/min

to

1800 r/min

in 8

Motor:

s.

Calculate a.

The torque developed by

b.

The energy

c.

The motor power [W]

d.

The power

in the

input

the

flywheel

[W]

at

at

motor [N-m] 1800 r/min

Figure 3.23 |kJ|

to the flywheel at

A

3-23

dc motor coupled

velops 120 hp

at

a constant speed of 700

The moment of 2 parts is 2500 lb-ft

ing a.

|

inertia of the revolv-

1

Problem 3-22

m/s, calculate the

r/min] and torque

[ft-

lbf]

is

re-

new speed

of the motor.

i

Calculate the torque [N-m] developed by the

3-24

How many

Btus are required to raise the

temperature of a 50 gallon (U.S.) reservoir of water from 55°F to

Calculate the motor torque IN-m] needed so that the

{Note;

to

Industrial appl cat ion

.

motor. b.

Tf the hoisting rate in

duced

to a large grinder de-

r/min.

Problem 3-22.

1750

r/min

3-18

Electric hoist,

1600 r/min

speed will increase

to

The torque exerted by

mains the same.)

750 r/min

in 5

the grinder re-

s.

the tank will

it

is

take

1

80°F, assuming that

perfectly insulated. if

the tank

electric heater?

is

How

heated by

long

a 2

kW

FUNDAMENTALS OF MECHANICS AND HEA T

3-25

A

large indoor transformer

using an

is

painted a non-

It is

proposed to refurbish

aluminum

paint. Will this affect

metallic black.

it

the temperature of the transformer? If so, will

3-26

An

it

electrically heated

perature

is

m X

cement floor covers

30 m. The surface tem-

25 °C and the ambient tempera-

ture

is

23 °C. Approximately

heat

is

given

the point of is

off, in

The cable and other

electrical

how much

kilowatts? Note: from

view of heat radiation, cement

considered to be an insulator.

components

inside a sheet metal panel dissipate a total

of 2 kW.

A blower

inside the panel keeps

the inside temperature at a uniform level

throughout. The panel

run hotter or cooler?

an area of 100

3-27

67

high, and 2

Assuming tion

ft

is

4

ft

wide, 8

ft

deep, and totally closed.

that heat

is

radiated by convec-

and radiation from

all

sides except the

bottom, estimate the temperature inside the panel

if

the ambient temperature

The panel enamel.

is

is

30 °C.

painted with a nonmetallic

Part Two Electrical

Machines and

Transformers

Chapter 4 Direct-Current Generators

This

Introduction

4.0

We

begin our study of rotating machinery with

the direct-current generator.

generators are not as

common

because direct current,

when

are discussed.

as they used to be,

required,

is

mainly

without

current

any

using

moving

important because tion to the

and

it

is

4.1

motor and vice versa. Owing

similar construction, the

reason

fundamental properties of

erator after

we

We show

when

operates

meant by the neutral

fine

what

how

the induced voltage

termines

it

at

is

its

is

point.

it

is

(ac) generator.

The

that the voltage generated in

any dc gen-

inherently alternating and only

becomes dc

is

has been rectified by the commutator. 1

begin with the basic principles

the importance of brush position

the study of a direct-

Fig. 4. shows an elementary ac generator composed of a coil that revolves at 60 r/min between the N, S poles of a permanent magnet. The

applied to a dc motor.

chapter

may seem,

edge of the alternating-current

to their

iearn about a dc generator can be di-

of a 2-pole generator

it

current (dc) generator has to begin with a knowl-

generators and motors are identical. Consequently,

In this

Generating an ac voltage

Irrelevant as built

same way; consequently, any dc generator can

we

actual

including multipole designs.

many dc

Commercial dc generators and motors are

rectly

their voltage-regulation characteristics.

physical construction of direct-current machines,

brief periods.

anything

next.

of dc generators

represents a logical introduc-

behavior of dc motors. Indeed,

operate as a

commutating poles and

The chapter ends with a description of the

industry actually operate as generators for

in

for

We then discuss the major types

parts.

Nevertheless, an understanding of dc generators

The need

problem of pole-tip saturation are covered

the

produced by electronic rectifiers. These rectifiers

the

followed by a study of the behavior of the

current flow, and the importance of armature reaction

Direct-current

can convert the current of an ac system into direct

motors

is

generator under load. Mechanical torque, direction of

is due to an external driving force, such motor (not shown). The coil is connected to two slip rings mounted on the shaft. The slip rings are connected to an external load by means of two

no-load.

rotation

and de-

as a

We show

generated and what de-

stationary brushes x and

magnitude.

71

y.

1 ELECTRICA L MA CHINES AND TRANSFORMERS

72

rotation (Fig. 4.2).

60 r/min

The waveshape depends upon

We

shape of the N, S poles. designed

The

assume

the

were

the poles

generate the sinusoidal wave shown.

to

coil

in

our example revolves

uniform

at

speed, therefore each angle of rotation corresponds to

makes

a specific interval of time. Because the coil

one

turn per second, the angle of

360°

4.2 cor-

in Fig.

responds to an interval of one second. Consequently,

we can

also represent the induced voltage as a func-

tion of time (Fig. 4.3).

v

1

+20

cycle -

N \

F Al> c

an elementary ac generator

of

1

revolution per second.

\

V

y_

\

\

\

/

\

/

\

f

\

/

\

turning at

\

/ \

1

\

/ /

\

Figure 4.1 Schematic diagram

\

/

\

\

/ \

\

/

\ s

1

As

the coil rotates, a voltage

A

is

induced (Eq. 2-25)

between

its

between

the brushes and, therefore, across the load.

terminals

The voltage

is

time

1.25

/ /

and D. This voltage appears

/

-

1

cycle

generated because the conductors of

the coil cut across the flux produced by the N, S poles.

The induced voltage

(20 V, say)

when

the coil

is

zontal position, as shown. coil

is

momentarily

quently the voltage

is

No

maximum

therefore

momentarily flux

is

in the hori-

cut

when

in the vertical position;

at

these instants

feature of the voltage

is

that

its

is

zero.

Figure 4.3 Voltage induced as a function of time.

the

conse-

Another

polarity changes

4.2 Direct-current generator If

the brushes in Fig. 4.1 could be switched from

every time the coil makes half a turn. The voltage can

one

therefore be represented as a function of the angle of

about to change,

slip ring to the other

every time the polarity was

we would

obtain a voltage of con-

stant polarity across the load. V

Brush x would always

be positive and brush y negative. We can obtain this result by using a commutator (Fig. 4.4), A commu-

+ 20

tator in that

is

its

simplest form

composed of

is

cut in half, with each

a slip ring

segment insulated from

t-'.A I)

the other as well as \

180 -

360 degrees 450

angle

connected

from

to coil-end

the shaft.

The commutator revolves with age between the segments tionary brushes x and

One segment

A and the other to coil-end is

the coil and the volt-

picked up by two

The voltage between brushes nating voltage in the coil

Figure 4.2 in

the ac generator as a function of

the angle of rotation.

sta-

y.

x

and y pulsates

but never changes polarity (Fig. 4.5).

Voltage induced

is

D.

is

rectified

The

alter-

by the com-

mutator, which acts as a mechanical reversing switch.

DIRECT-CURRENT GENERATORS

Due

60 r/min

to the constant polarity

73

between the brushes,

Hows in The machine represented in Fig.

the current in the external load always

the

same

4.4

is

direction.

called a direct-current generator, or

4.3 Difference

dynamo.

between ac

and dc generators The elementary

ac and dc generators in Figs. 4.1

and 4.4 are essentially and an ac voltage chines only differ

same way.

built the

between the poles of

case, a coil rotates

is

induced

in the

way

We

commutator

Elementary dc generator

machines which carry both

is

simply an ac generator rectifier called

a

commu-

(Fig. 4.6a).

tator (Fig. 4.6c).

generators carry

while dc generators require a

Figure 4.4 equipped with a mechanical

each

magnet

The ma-

the coil.

in

In

the coils are connected

to the external circuit (Fig. 4.6): ac

slip rings (Fig 4.6b)

a

sometimes build small

slip rings

and a

commu-

Such machines can function

si-

multaneously as ac and dc generators.

tator.

4.4 Improving the Returning +20

to the

waveshape

dc generator,

we can improve

the

pulsating dc voltage by using four coils and four

segments, as shown

shape but

it

is

given

never

The resulting waveThe voltage still pulsates

in Fig. 4,7.

in Fig. 4.8.

falls to zero;

it

is

much

closer to a steady

dc voltage.

By increasing the number of coils and segments, we can obtain a dc voltage that is very smooth.

0 0

90

180

270

360 degrees

Modern dc generators produce The

angle 9

ripple of less than 5 percent.

Figure 4.5

voltages having a coils are lodged in

the slots of a laminated iron cylinder.

The elementary dc generator produces a pulsating dc

the cylinder constitute the

voltage.

The

coils

The percent ripple is the ratio of the RMS value of the ac component of voltage to the dc component, expressed

(b)

(a)

in percent.

(

C

)

Figure 4.6

The three armatures rings or

and

armature of the machine.

(a), (b),

and

(c)

have

identical windings.

a commutator), an ac or dc voltage

is

obtained.

Depending upon how they are connected

(to slip


74

rotation

rotation

A

Figure 4.7 Schematic diagram of a dc generator having 4 and 4 commutator bars. See Fig. 4.9.

Figure 4.9 coils

The actual physical construction of the generator shown in Fig. 4.7. The armature has 4 slots, 4 coils, and 4 commutator bars.

v

A

schematic diagram such as Fig. 4.7

where the

tips,

and so on. But we must remember

coil sides (a h a 2

;

b,,

b2

ally located at 180° to

side as Figure 4.7

ol 0

I

I

90

180

I

I

|_

I

360

270

degrees

*0

Figure 4.8

The voltage between the brushes than

in Fig.

that

4.5.

one

of Fig. 4.7, because

we

in

will

be using similar drawings

of dc machines. The four coils

the figure are identical to the coil

At the instant shown,

coil

A

is

shown

in Fig. 4.

1

not cutting any flux

The reason is that the coil sides of these two coils are midway between the poles. On the other hand, coils B and D are cutting flux coming and neither

is

coil C.

from the center of the

N

and S poles. Consequently,

the voltage induced in these coils

possible value (20 V, say). That

is at its

is

maximum

also the voltage

across the brushes at this particular instant.

each other and not side by

seems

to indicate.

coil side is at the

is at

a, is in

important to understand the physical meaning

to explain the behavior

that the

of each coil are actu-

etc.)

The actual construction of this armature is shown in Fig. 4.9. The four coils are placed in four slots. Each coil has two coil sides, and so there are two coil sides per slot. Thus, each slot contains the conductors of two coils.

other

It is

;

For reasons of symmetry, the coils are wound so

more uniform

is

us

between the poles, under the poles, near the

cated:

pole

tells

coil sides of the individual coils are lo-

the top of slot

bottom of

bottom of a

slot 3.

The

1,

while coil side a 2

coil

the

is in

connections to the com-

mutator segments are easy to follow armature.

and the

slot

the top. For example, in Fig. 4.7 coil side

The reader should compare

in this

simple

these connec-

tions with those in Fig. 4.9 to verify that they are the

same. Note also the actual position and schematic position of the brushes with respect to the poles. the position of the coils when moved the armature has through 45°. The sides a,, A are now sweeping past pole tip and a 2 of coil Fig. 4.10

shows

1

pole

same

tip 4.

The

sides of coil

C

are experiencing the

flux because they are in the

A. Consequently, the voltage

e.A

same

slots as coil

induced

in coil

A is


rotation

15

rotation

Figure 4.10

when

Position of the coils rotated

the armature of Fig. 4.9 has

through 45°.

Figure 4.11a Physical construction of a dc generator having 12

exactly the

same

as the voltage e c induced in coil C.

A

C

while coil

coils,

12

slots,

and 12 commutator

moving downward, moving upward. The polarities of e a

however, that coil

Note,

is

bars.

is

rotation

and ec are, therefore, opposite as shown.

The same reasoning leads us and? are equal and opposite Ll

+

+

+

=

conclude that e h

to

in polarity.

This means

at all times.

Consequently,

no current will flow in the closed loop

formed by the

that e.d

eh

ec

This

four coils.

is

circulating current

ed

0

most fortunate, because any such

would produce

2

I

R

The voltage between the brushes c c (or e a

+

c\\)

at the instant

minimum

to the

shown.

shown

voltage

losses.

is

equal to e b

It

in Fig. 4.8.

The armature winding we have just discussed called a

lap winding.

the

It is

+

corresponds

most

common

is

type of

winding used in direct-current generators and motors.

4.5

Figure 4.11b Schematic diagram

Induced voltage

induced Figures 4.

1

and

la

4.

1

show

lb

a

more

in

When

the armature rotates, the voltage

E induced

each conductor depends upon the flux density

which

it

cuts.

This fact

is

E= Because the density point to point, coil

depends

based upon the equation Blv

in the air

(2.25)

gap varies from

the value of the induced voltage per

upon

its

of the

armature and the voltages

the 12 coils.

realistic ar-

mature having 12 coils and 12 slots instead of only 4.

in

instantaneous

position.

Consider, for example, the voltages induced

armature when Fig. 4.11.

it

in

the

occupies the position shown

The conductors

in slots

in

and 7 are ex-

1

actly

between the poles, where the

flux density

is

zero.

The voltage induced

coils

lodged

in

slots

1

and 7

is,

in the

therefore, zero.

two

On

the other hand,

the conductors in slots 4 and 10 are directly under

the center of the poles, greatest.

where the

The voltage induced

in

flux density

the

two

is

coils


76

lodged

in

Finally,

due

duced

in

as that

slots

induced

in the coils

lodged

and

in slots 5

0V

same

the

is

rotation

in-

1

coil

A

/

1

shows the instantaneous voltage ineach of the 2 coils of the armature. They 18, and 20 V, respectively. Note that the

in

4.

1

lb

1

are 0, 7,

brushes short-circuit the coils is

maximum.

therefore,

is,

magnetic symmetry, the voltage

to

the coils lodged in slots 3 and 9

Figure

duced

these

which the voltage

in

momentarily zero. Taking polarities into account, we can see that

between the brushes

the voltage 18

+

to

brush

7)

=

70

and brush x

V,

is

+

(7

is

18

+ 20 +

positive with respect

y.

This voltage remains essentially con-

stant as the

armature rotates, because the number of

coils

between the brushes

always the same,

is

irre-

spective of armature position.

Note

4.11b straddles two

that brush x in Fig.

commutator segments Consequently,

the

that are

brush

connected

However, since the induced voltage momentarily zero, no current brush.

sitioned on the

to

brush

If

That

we were

the case in Figs. 4.

is

(0

+

7

+

1

1

is

a

momen-

and

4.

1

1

b.

18

+ 20 +

18)

=

63

continually short-circuit coils that generate 7 V.

Large currents will flow

in the short-circuited coils

and brushes, and sparking

with coils that are momentarily

age between the brushes and

the

at

same time

sparking occurs, there

is

poor commutation.

a neutral zone.

4.7 Value of the induced voltage The voltage induced winding

is

in a

dc generator having a

lap


E0 =

E0 = Z= n = (J> =

Z/?*/60

(4J)

voltage between the brushes [V] total

number of conductors on

the armature

speed of rotation [r/min] tlux per pole

|Wb]

This important equation shows that for a given generator the voltage

only holds true

is

zones

Neutral zones are those places on the surface of the

armature where the tlux density

is

zero.

When

the

generator operates at no-load, the neutral zones are located exactly between the poles.

No

voltage

is

in-

is

directly proportional to the flux

per pole and to the speed of rotation.

tion. If the

4.6 Neutral

in

We

in contact

will result. Thus, shifting

the brushes off the neutral position reduces the volt-

When

through the neutral zone.

brushes so they are

where

V.

decreases. Furthermore, in this position, the brushes

said to be

try to set the

between the brushes would be-

Thus, by shifting the brushes the output voltage

causes sparking.

in a coil that cuts

always

brush yoke by 30° (Fig.

to shift the

4.12), the voltage

come

which

y,

The brushes are neutral position when they are pocommutator so as to short-circuit

duced

coil B.

those coils in which the induced voltage tarily zero.

A.

coil

in this coil is

flow through the

will

The same remarks apply

momentarily short-circuits said to be in the

to coil A.

short-circuits

Figure 4.12 Moving the brushes off the neutral point reduces the output voltage and produces sparking.

if

the brushes are

Example 4-1 The armature of a slots.

Each

Wb.

equation

brushes are shifted off neutral, the

equivalent to reducing the

0.04

The

on the neutral

posieffect

number of conductors Z

_

6-pole,

coil has

600 r/min generator, has 90

4 turns and the tlux per pole

is

Calculate the value of the induced voltage.


The current delivered by

Solution

Each turn corresponds to two conductors on the armature, and

The

coils arc required to

fill

the

number of armature conductors

total

Z = 90

90

coils

X

4 turns/coil

X

90

slots.

flows through

2 conductors/turn

The speed

we

in the

same

discover

direction in

The same

is

N

mo-

true for conductors that are

mentarily under a S pole. However, the currents unis

n

= 600

der the

r/min

N

pole flow

the opposite direction to

in

those under a S pole. Referring to Fig.

Consequently,

En = Z//4V60 = 720 X 600 X = 288 V The voltage between the brushes fore

always flows

we would

those conductors that are momentarily under a pole.

= 720

generator also

the

the armature conductors. If

could look inside the machine, that current

is

all

77

at

4. 13, the ar-

mature conductors under the S pole carry currents 0.04/60 that

flow into the page, away from the reader.

Conversely, the armature currents under the

no-load

is

there-

N

pole

flow out of the page, toward the reader.

Because the conductors

288 V, provided the brushes are on neutral.

lie in

a magnetic field,

they are subjected to a force, according to Lorentz's

4.8

Generator under load: the energy conversion process

When

a direct-current generator is

under load, some

fundamental flux and current relationships take place that are directly related to

the mechanical-electrical

energy conversion process. Consider for example, a 2-pole generator

that

is

while delivering current

/

driven counterclockwise to a load (Fig. 4.

1

3).

law (sections 2.22 and 2.23).

find that the individual forces all

we examine

F on

the di-

we

act clockwise. In effect, they

the conductors

produce a torque

which the gen-

that acts opposite to the direction in

erator

is

we must

being driven. To keep the generator going, exert a torque on the shaft to

overcome

this

opposing electromagnetic torque. The resulting mechanical

which rotation

If

rection of current flow and the direction of flux,

how

is

power

is

converted into electrical power,

delivered to the generator load. That

is

the energy conversion process takes place.

torque due to F

4.9

Armature reaction

Until now,

we have assumed

motive force (mmf) acting to the field.

in a

that the only

dc generator

However, the current flowing

ture coils also creates a powerful that distorts

we we

and

field

weakening takes place

reaction.

the impact of the armature

mmf.

return to the generator under load (Fig. 4.13). If

consider the armature alone,

magnetic

chanical torque.

magnetomotive force

mmf is called armature

To understand

Figure 4.13 The energy conversion process. The electromagnetic torque due to Fmust be balanced by the applied me-

due

arma-

both motors and generators. The effect produced by

the armature

load

that

in the

and weakens the flux coming from the

poles. This distortion in

magnetois

at right

field as

shown

it

in Fig. 4.14.

will

produce a

This

field acts

angles to the field produced by the N, S poles.

The intensity of the armature flux depends upon its mmf, which in turn depends upon the current carried by the armature. Thus, contrary armature flux

is

to the field flux, the

not constant but varies with the load.


78

armature flux

rotation

neutra zone ,

neutral zone

Figure 4.14 Magnetic field produced by the current flowing armature conductors.

We can

in

the

immediately foresee a problem which the

armature flux will produce. Fig.

zone

flux in the neutral

is

shows

4. 14

that the

Figure 4.15 Armature reaction S poles.

may

sparking

by the brushes. As

will

The

occur.

depend upon

a result, severe

intensity of the sparking

the armature flux

the load current delivered

that

is

it

distorts the flux

mmf

and

field

whose shape

is

mmf

in the direction

The

all

The

neutral

still

another effect:

the higher flux density in pole tips 2. 3 causes saturation to set

in.

Consequently, the increase

in flux

decrease

in flux

under pole

tips 2, 3 is less than the

under pole

tips

1,

4.

As

a result, the total flux pro-

duced by the N, S poles

less than

may be

1

as

.

was when

induced voltage given

as 10 percent.

in flux

under load,

we

could

zone when the gener-

move

the brushes to re-

For generators, the brushes are shifted

to the

new

zone by moving them

tation.

For motors, the brushes are shifted against

in

the direction of ro-

the direction of rotation.

As soon However,

We

as the brushes are

improves,

tion

forth

it

For large machines, the decrease

much

ro-

neutral

no-load. This causes a cor-

at

in the

does not

duce the sparking.

rises

responding reduction 4.

to the shift in the neutral is

the

is

generator was running

by Eq.

Due ator

of rotation of the

dc generators.

flux distortion produces

it

improve commutation

to

produced by the

illustrated in Fig. 4.15.

zones have shifted

in space;

4.10 Shifting the brushes

produces a magnetic field

armature. This occurs in

important to note that the orientation of the

with the armature.

tate

the armature

combination of the armature

poles. In effect, the

It is

armature flux remains fixed

and hence upon

by the generator.

The second problem created by

mmf

N,

no longer zero and, con-

sequently, a voltage will be induced in the coils that are short-circuited

produced by the

distorts the field

and

if

meaning

moved, the commuta-

there

is

less

sparking.

the load fluctuates, the armature

falls

and so the neutral zone

shifts

mmf

back and

between the no-load and full-load positions. would therefore have to move the brushes back

and forth

to

cedure

not practical and other

is

obtain sparkless commutation. This pro-

means

are used to

DIRECT- CURRENT GENERA TORS

resolve the problem. ever, the to

For small dc machines, how-

brushes are set

ensure reasonably

in

an intermediate position

good commutation

at all loads.

mmf of the

the

commutating poles

is

greater than the armature

mmf. This

flux in the neutral zone,

which aids

made

19

slightly

creates a small the

commuta-

tion process (see Section 4.28).

Commutating poles

4.11

Fig. 4. 16

shows how

the

commutating poles of

a 2-pole machine are connected. Clearly, the direc-

To counter the

of armature

effect

reaction

in

medium- and large-power dc machines, we always place a set of

commutating poles* between the main These narrow poles carry wind-

poles (Fig. 4. 16). ings that are

connected

with the armature.

in series

The number of turns on the windings that the

equal

and opposite

mmfa

of the armature.

the

designed so

As

exactly

bucking each other

ing the

armature

mmf

flowing through the windings

in-

opposite to the

mmf of the commutating poles acts mmf of the armature and, therefore,

neutralizes

effect.

is

its

restricted to the

However,

the neutralization

narrow brush zone where com-

mutation takes place. The distorted flux distribution

under the main poles, unfortunately, remains

the same.

the load current varies, rise

at all

in this

between the main poles

we no longer have

mmfc

magnetomotive force

the

to

two magnetomotive forces

space

is

poles develop a magnetomotive force

tion of the current

dicates that the

and

fall

times.

together,

By

way, the flux is

4.12 Separately excited generator

nullifyin the

always zero and so

to shift the brushes. In practice,

Now

that

we have learned some basic facts about we can study the various types and

dc generators,

their properties.

magnets

Thus, instead of using permanent

to create the

magnetic

field,

we can use a shown

pair of electromagnets, called field poles, as

O (+)

in Fig. 4. 17.

When

generator

supplied by an independent source

is

the dc field current in such a

(such as a storage battery or another generator, called an exciter), the generator

is

said to be sepa-

rately excited. Thus, in Fig. 4. 7 the dc source con1

nected to terminals a and b causes an exciting current / x to flow. If the armature

is

or a diesel engine, a voltage

£0

brush terminals x and

driven by a motor

appears between

y.

O (-) Figure 4.16

Commutating poles produce an the

mmf,

of the

mmf c

that

opposes

armature.

Commutating poles are sometimes called mierpoles.

Figure 4.17 Separately excited 2-pole generator. The N, S

field

poles are created by the current flowing

field

windings.

in

the

1 ELECTRICAL MACHINES AND TRANSFORMERS

80

How does the saturation curve relate to the induced £n ? If we drive the generator at constant speed,

4.13 No-load operation

voltage

and saturation curve

E0 is directly proportional to the flux When

a separately excited dc generator runs at no-

load (armature circuit open), a change

the excit-

in

ing current causes a corresponding change in the in-

We now examine

duced voltage.

whose shape 4.

8a.

1

which increases the

If

we

f|>

plot

as a function of I X1

per pole.

is

1

is

mmf of the

obtain the satu-

ration curve of Fig. 4. 8a. This curve

whether or not the venerator

obtained

turning.

identical to the saturation curve of Fig.

result

is

shown

in Fig. 4.

1

8b;

rated voltage of a dc generator

above the knee of the curve.

usually a

is

In Fig. 4.

example, the rated (or nominal) voltage varying the exciting current,

called the

it is

we can

is

1

1

8b. for

20

V.

By

vary the

in-

duced voltage as we please. Furthermore, by

re-

versing the current, the flux will reverse and so, too, will the polarity of the induced voltage.

Induced voltage

vs speed.

For a given exciting

current, the induced voltage increases in direct pro-

rated flux

portion to the speed, a result that follows from Eq. 4.

A

.

obtain a curve

little

flux

we

Consequently,

we

Let us gradually

so that the

,

is

,

no-load saturation curve of the generator.

The

raise the exciting current / x field increases,

The

as a function of 7 X

the relationship

between the two. Field flux vs exciting current.

Ea

by plotting

If

we

of the induced voltage also reverses. However,

ity

we

1

reverse the direction of rotation, the polar-

reverse both the exciting current

and

if

the direc-

tion of rotation, the polarity of the induced voltage

remains the same.

Figure 4.18a Flux per pole versus exciting current.

When the flux

the exciting current

relatively small,

is

small and the iron in the machine

is

saturated. Very

little

mmf

is

needed

un-

is

to establish

the flux through the iron, with the result that the

mmf developed

by the

field coils

is

available to drive the flux through the air gap.

Because the permeability of

air

3

A

almost entirely

is

constant, the

Figure 4.18b Saturation curve of a dc generator.

flux increases in direct proportion to the exciting

current, as

shown by

the linear portion

0a of the

4.14 Shunt generator

saturation curve.

However, as we continue current, the iron in the field to saturate.

A

to raise the exciting

and the armature begins

large increase in the

quired to produce a small increase

mmf is now in flux, as

by portion be of the curve. The machine to

is

re-

shown

now

said

be saturated. Saturation of the iron begins to be

important

when we

reach the so-called "knee" ab of

the saturation curve.

A

shunt-excited generator

shunt-field winding

is

is

connected

a

machine whose

in parallel

with the

armature terminals, so that the generator can be self-excited (Fig. 4. 19). this

connection

is

that

it

The

principal advantage of

eliminates the need for an

external source of excitation.

How

is

generator

self-excitation achieved?

is

When

started up, a small voltage

is

a shunt

induced

in

i


81

field rheostat

y

Figure 4.20 Controlling the generator voltage with a field rheostat.

rheostat

is

a resistor with an adjustable

To understand how suppose that

p

is in

tact

Ea

is

1

the output voltage varies,

V when

20

movable contact

the

we move the conresistance R between

the center of the rheostat. If

toward extremity m, the

points

A

sliding contact.

{

p and b diminishes, which causes

the excit-

ing current to increase. This increases the flux and,

£tv On

consequently, the induced voltage

hand,

if

(b)

we move

the other

R

the contact toward extremity n,

l

increases, the exciting current diminishes, the flux

Figure 4.19

b.

diminishes, and so

Self-excited shunt generator.

a.

Schematic diagram field is

of

one designed

a shunt generator.

to

be connected

in

A

We

shunt

shunt

(alter-

know points

the

armature, due to the remanent flux in the poles.

produces a small exciting current

This voltage the

shunt field.

The

resulting small

mmf acts

/ x in

in the

same direction as the remanent flux, causing the flux per pole to increase.

which increases

/x ,

The increased

flux increases

which increases

more, which increases

E0

the flux

even more, and so

This progressive buildup continues until a

maximum

E0

forth.

reaches

value determined by the field resistance

and the degree of saturation. See next section.

p and

R

E0

if

we

of the shunt field circuit between

{

We draw

b.

ing to the slope of

a straight line correspond-

R and superimpose

on the

it

{

uration curve (Fig. 4.2

sat-

This dotted line passes

1).

through the origin, and the point where

it

intersects

the curve yields the induced voltage.

Ea

still

will fall.

the saturation curve of the generator and the

total resistance

nate term for parallel) with the armature winding.

£0

can determine the no-load value of

For example,

50

12

if

the shunt field has a resistance of

and the rheostat

R = 50

12.

{

The

line

is

m,

then

R must E = 50 V, / =

pass

set at

extremity

corresponding

to

{

through the coordinate point

I

A.

This line intersects the saturation curve where the voltage

is

150

V

(Fig. 4.21).

That

is

the

maximum

voltage the shunt generator can produce.

By changing

4.15 Controlling the voltage

of a

shunt generator

to It

is

easy to control the induced voltage of a shunt-

excited generator. rent

We

simply vary the exciting cur-

by means of a rheostat connected in series with

the shunt field (Fig. 4.20).

the setting of the rheostat, the total

resistance of the field circuit increases, causing

decrease progressively. For example,

creased to 120

12,

A\

is

E0 in-

the resistance line cuts the satu-

ration curve at a voltage

we continue

if

Ea

of 120 V. a critical value will

be

reached where the slope of the resistance line

is

If

to raise

A*,,


82

Figure 4.22 Equivalent circuit of a dc generator.

revolving conductors. Terminals

l ,

2 are the external

armature terminals of the machine, and F,, F 2 are the field

now

winding terminals. Using study the more

common

we

this circuit,

will

types of direct-current

generators and their behavior under load.

4.17 Separately excited generator under load

Figure 4.21

The no-load voltage depends upon the resistance the shunt-field

of

Let us consider a separately excited generator that

circuit.

driven

equal to that of the saturation curve region.

When

unsaturated

in its

this resistance is attained, the

induced

voltage suddenly drops to zero and will remain so for any

R

v

greater than this critical value. In Fig. 4.2

the critical resistance corresponds to

200

at

a battery (Fig. 4.23).

and so

is

The

When

load, terminal voltage

E0

resistance

whose field

the

E {2

zero.

is

constant

machine operates is

However,

is

excited by

The induced voltage £0

is

at no-

equal to the induced

because the voltage drop is

is

exciting current

the resultant flux.

therefore fixed.

voltage

il.

constant speed and

if

in the

armature

we connect

a load

4.16 Equivalent circuit

across the armature (Fig. 4.23), the resulting load

We

R Kr Terminal voltage E l2 is now less than the induced voltage £ As we increase the load, the terminal

current / produces a voltage drop across resistance

set

have seen

that the

of identical coils,

all

armature winding contains a of which possess a certain re-

()

sistance.

which

The

exists

machine

the

armature resistance R^

total

between the armature terminals when is

stationary.

It is

measured on the com-

mutator surface between those segments that der the ally

(

+

)

that

is

and

(

—

)

The

brushes.

resistance

lie is

.

voltage diminishes progressively, as 4.24.

shown

as a function of

cun e

of the generator.

load current

is

called the load

y

un-

usu-

very small, often less than one-hundredth of an

ohm.

Its

value depends mainly upon the

voltage of the generator. circuit,

we can

represent

one of the brushes.

If

To simplify machine has

resistance of these windings

The equivalent

circuit

posed of a resistance R„ (Fig. 4.22).

The

is

in series

latter is the

with

interpoles, the

included

of a generator in series

power and

the generator

R 0 as if it were

the

in

R ir

is

thus

com-

with a voltage

voltage induced

Ea

in the

in Fig.

The graph of terminal voltage

Figure 4.23 Separately excited generator under load.


V

tions

100

produce corresponding changes

in

83

the genera-

tor terminal voltage, causing the lights to flicker.

Compound generators eliminate this problem. A compound generator (Fig. 4.25a) is similar to a shunt generator,

field coils

connected

except that in series

has additional

it

with the armature.

These series field coils are composed of a few turns of heavy wire, big enough to carry the armature cur0

10

5

rent.

A

The

4.25b

showing the shunt and Figure 4.24

of the series coils

total resistance

fore, small. Figure

When

In practice,

ration

the induced voltage

Ev also decreases

with increasing load, because pole-tip satu-

slightly

tends to decrease the field flux. Consequently,

voltage

the terminal

E l2

falls off

more

rapidly than

can be attributed to armature resistance alone.

the series coils

series field connections.

As

the generator

off more sharply with increasing load than that of

a separately excited generator. field

The reason

current in a separately excited

constant,

whereas

citing current falls

in a self-excited

load to full-load is

drop

coils acts in the

is

said to

15% and 10%,

Compound

its

raises the value of

original

Ea

.

If

practically constant

from no-load

it

respectively.

generator prevent

voltage of a dc generator from de-

we

can usually tolerate a reasonable drop in terminal voltage as the load increases, this has a serious effect

on lighting circuits. For example, the distribu-

tion

system of a ship supplies power to both dc ma-

chinery

and

incandescent

lamps.

The

current

mmf

no-load value,

properly designed, the terminal voltage remains

regulation

creasing with increasing load. Thus, although

flows

developed

the series coils are

voltage from no-

to

now

direction as the

machine remains

The voltage

however,

Consequently, the field flux un-

generator the ex-

The compound generator was developed the terminal

same

/c

mmf

The

that the

whereas for a separately excited generator

usually less than 10 percent.

be

field.

coils.

about 15 percent of the full-load

is

4.19

in

coils,

loaded, the terminal volt-

as the terminal voltage drops. For a

self-excited generator, the

voltage,

is

is

through the series field

der load rises above

The terminal voltage of a self-excited shunt generator

The shunt

zero.

age tends to drop, but load current

which falls

is

flux, just as in a standard self-excited shunt generator.

by these

Shunt generator under load

there-

diagram

carry exciting current / x which produces the field

of the shunt

4.18

is,

the generator runs at no-load, the current

Load characteristic of a separately excited generator. in

a schematic

is

Figure 4.25

delivered by the generator fluctuates continually, in

a.

Compound

response to the varying loads. These current varia-

b.

Schematic diagram.

generator under load.

to full-load.

The


84

rise in the

induced voltage compensates for the

ar-

Load characteristics

4.21

mature IR drop. In

the

some cases we have

to

compensate not only for

armature voltage drop, but also for the IR drop

in

between the generator and the load.

the feeder line

The generator manufacturer then adds one or two extra turns

on

winding so

the series

that the terminal

ma-

voltage increases as the load current rises. Such

chines are called over-compound generators.

compounding placed

is

too strong, a low resistance can be

with the series

in parallel

the current in the series field as reducing the

If the

field.

This reduces

The load characteristics of some shunt and compound generators are given in Fig. 4.26. The voltage of an over-compound generator increases by 1

value of the diverter resistance

is

if

the

equal to that of the

series field, the current in the latter

is

reduced by

half.

compound

flat-compound generator remains constant.

tor

15 percent

is

compound generator the

mmf of the As

a re-

terminal voltage falls drastically with in-

We

can make such a generator by

simply reversing the series field of a standard com-

pound

generator. Differential

were formerly used

in

below

its

the

is

30 percent

lower.

4.22 Generator specifications The nameplate of

a generator indicates the power,

voltage, speed, and other details about the machine. ratings, or

nominal characteristics, are the

following information

compound

nameplate of a 100

Power

100

Voltage

250 20

generators

kW

V A 50° C

Exciting current

Temperature

rise

These specifications

tell

dc arc welders, because they

tended to limit the short-circuit current and to stabi-

punched on

is

the

kW generator: Speed

1200r/min

Type

Compound B

Class

us that the machine can

power of 100

deliver, continuously, a lize the arc

On

no-load value, while that

of a differential-compound generator

ple, the

series field acts opposite to the shunt field.

creasing load.

applied, whereas that of a

values guaranteed by the manufacturer. For exam-

generator

sult, the

is

other hand, the full-load voltage of a shunt genera-

These

In a differential

full-load

and has the same effect

number of turns. For example,

4.20 Differential

when

percent

kW at a volt-

age of 250 V, without exceeding a temperature

rise

during the welding process.

The voltage regulation of the differential compound generator in Fig. 4.26 is (no-load — full00-70)770 = 42.9%. load )/full- load = (

of 50°C.

can therefore supply a load current of

It

100 000/250

= 400 A.

and the current

in the

It

possesses a series winding,

shunt field

is

20 A.

In practice,

1

the terminal voltage its

% 100

rating of

overcom pound

power from

compound

ceed

1

separate excitation

class

B

used

in the

80

shunt

60

differential

00

250

V.

is

adjusted to a value close to

We may draw

any amount of

the generator, as long as

kW and the current

is

it

does not ex-

400 A. The

less than

designation refers to the class of insulation

machine.

compound

40

CONSTRUCTION OF

20

DIRECT-CURRENT GENERATORS We have

0-

50

100

described the basic features and properties

% of direct-current generators.

We now

look

at

the

Load current

mechanical construction of these machines,

direct-

Figure 4.26

ing our attention to the field, the armature, the

Typical load characteristics of dc generators.

mutator, and the brushes.

com-


85

4.23 Field The

field

chine.

It

produces the magnetic flux is

composed of side

in the

ma-

basically a stationary electromagnet a set

of salient poles bolted to the

in-

of a circular frame (Figs. 4.27 and 4.28). Field

mounted on the poles, carry the dc exciting

coils,

current.

The frame

is

usually

stacked iron laminations. In

solid cast

our discussions so far

2-pole generators. ator or poles.

some generators

the

created by permanent magnets.

flux is

In

made of

whereas the pole pieces are composed of

steel,

we have considered only

However,

motor may have

in practice a

2, 4, 6,

or as

many

The number of poles depends upon

flux

dc generas

24

Figure 4.29 Adjacent poles

of multipole

generators have opposite

magnetic

polarities.

ical size

of the machine; the bigger

the phys-

poles

frame

will have.

it

By using

is,

it

the

more

a multipole design,

we

can reduce the dimensions and cost of large mafield

chines, and also

The

improve

field coils

their

performance.

of a multipole machine are con-

nected together so that adjacent poles have oppo-

commutator

site

magnetic polarities (Fig. 4.29). The shunt

composed of

coils are

field

several hundred turns of

wire carrying a relatively small current. The coils

armature

are insulated

from the pole pieces

to

prevent short-

circuits.

Figure 4.27

mmf

The

Cross section of a 2-pole generator.

developed by the coils produces a

magnetic flux that passes through the pole pieces, the frame, the armature, is

the short space

pieces.

It

and the

air gap.

The

air

gap

between the armature and the pole

ranges from about

1

erator rating increases from

1

Because the armature and

.5 to

kW

mm as the gen-

5 to

100 kW.

field are

composed of

magnetic materials having excellent permeability,

most of the

mmf

produced by the

field

is

used to

drive the flux across the air gap. Consequently, by

reducing

its

shunt field

made fect

length, coils.

we can

becomes too

must be

enough so

Figure 4.28

Cutaway view

does not overheat when

It

has 3

field, the coils are

top of the shunt-field coils.

tor size

a 4-pole shunt generator.

gap cannot be

great.

generator has a series

wound on of

air

too short otherwise the armature reaction ef-

If the

brushes per brush set.

diminish the size of the

However, the

large

rent of the generator.

it

The conduc-

that the

winding

carries the full-load cur-


86

4.24 Armature The armature

the rotating part of a dc generator.

is

It

consists of a commutator, an iron core, and a set of

The armature

coils (Fig. 4.30).

keyed

is

revolves between the field poles.

composed of slotted, to

form a

to a shaft

The

and

iron core

is

iron laminations that are stacked

solid cylindrical core.

The laminations

are

individually coated with an insulating film so that

they do not

As

come

in electrical

contact with each other.

The

a result, eddy-current losses are reduced.

are lined

up

to provide the space

needed

slots

to insert the

armature conductors.

The armature conductors carry the load current They are insulated from the iron core by several layers of paper or mica and delivered by the generator.

are firmly held in place by fiber slot sticks. If the ar-

mature current

is

below 10 A, round wire

is

Figure 4.31 Armature lamination with tapered

but for currents exceeding 20 A, rectangular con-

ductors are preferred because they

make

slots.

used; iron teeth

fiber slot stick

better use

of the available slot space. The lamination of a small armature tion

view of the

is

shown

slot

in Fig.

4.31.

A

of a large armature

cross sec-

is

shown

in

Fig. 4.32.

Figure 4.32 Cross-section of a slot containing 4 conductors.

4.25

Commutator and brushes

The commutator

is

composed of an assembly of

ta-

pered copper segments insulated from each other by

mica

sheets,

and mounted on the shaft of the ma-

chine (Fig. 4.33). The armature conductors are con-

nected to the commutator

in

a

manner we

will ex-

plain in Section 4.26.

Great care

Figure 4.30 Armature of a dc generator showing the commutator, stacked laminations, slots, and shaft. (Courtesy of General Electric Company, USA)

is

taken

in

building the commutator

because any eccentricity will cause the brushes bounce,

producing

unacceptable

sparking.

to

The

sparks burn the brushes and overheat and carbonize the commutator.


yoke

that permits the entire brush

tated through an angle position. In going

87

assembly to be

and then locked

ro-

in the neutral

around the commutator, the suc-

cessive brush sets have positive and negative polarities.

Brushes having the same polarity are connected

together and the leads are brought out to one positive

and one negative terminal

(Fig. 4.34b).

The brushes are made of carbon because it has good electrical conductivity and its softness does not score the commutator. To improve the conductivity, a small amount of copper is sometimes mixed with the carbon. The brush pressure is set by means of adjustable springs. friction

tator

A

of

is

too great, the

and brushes; on the other hand,

if

commuit

is

too

a dc machine.

2-pole generator has two brushes fixed dia-

metrically opposite to slide

pressure

weak, the imperfect contact may produce sparking.

Figure 4.33

Commutator

If the

produces excessive heating of the

each other (Fig. 4.34a). They

on the commutator and ensure good electrical

contact

between the revolving armature and the

sta-

tionary external load.

Multipole machines possess as they

have poles. The brush

of one or that

many brush

sets, in turn, are

more brushes, depending upon

sets as

composed

the current

has to be earned. In Fig. 4.35c, for example,

brushes

mounted side-by-side make up the brush

The brush sets are spaced

at

two set.

equal intervals around the

commutator. They are supported by a movable brush

+

(a)

9

6 (b)

Figure 4.34 a.

Brushes of a 2-pole generator.

b.

Brushes and connections of a 6-pole generator.

Figure 4.35 Carbon brush and ultraflexible copper lead. b. Brush holder and spring to exert pressure. c. Brush set composed of two brushes, mounted on a.

rocker arm.

(Courtesy of General Electric Company, USA)

88


The pressure

is

usually about 15

kPa

2 lb/in"),

and the permissible current density mately 10

A/cm

2

65 A/in

is

In order to get a better

2 ).

Thus, atypical brush

cm X cm of 4.5 N

having a cross section of 3

1.2 in

1

X

lb)

and can

the construction of a

modern

0.4 in) exerts a pressure

1

carry a current of about 30 A. Fig. 4.36

shows

4-pole dc generator.

In

4.26 Details of a multipole generator

approxigenerators,

was

1

2-

the schematic diagram

is

of such a machine having 72 slots on the armature,

72 segments on the commutator, and 72

coils.

The

armature has a lap winding, and the reader should note

erator that

understanding of multipole

us examine the construction of a

pole machine. Fig. 4.38a

order to appreciate the

progress that has been made. Fig. 4.37 shows a gen-

let

how

similar

it is

2-pole machine (Fig.

built in 1889.

Figure 4.36 Sectional view of a 100 kW, 250 V, 1750 r/min 4-pole dc generator. (Courtesy of General Electric Company, USA)

to the 4.

1

1

schematic diagram of a

b).

Coils

A and C

are

mo-


89

The voltage generated between brushes x and y

sum of the

equal to the

connected

coils

commutator segments

to

brush sets are similarly generated by five (

form the

+ +

-2, 2-3,

)

(

coils.

brush sets are connected together

The

terminal.

)

(

—

connected to form the

larly

I

and 5-6. The voltages between the other

3-4, 4-5,

The

is

voltages generated by the five

to

brush sets are simi-

)

—

(

terminal. These

)

connections are not shown on the diagram. For similar

reasons of clarity,

that are placed Fig.

diagram. Coil

is

B

stalled in

1889

250 A

at

a voltage of

1 1

0

It

deliv-

Other prop-

V.

pioneering machine include the following:

Speed Total

first in-

to light the streets of Montreal.

ered a current of erties of this

Thompson generator was

1

300 r/min

2390 kg

weight

mm 330 mm

292

Armature diameter Stator internal diameter

Number of commutator

Only

A has its coil sides in slots

B

are in slots

is

connected

the three

4 and

10.

I

and

7,

while

Furthermore, coil

is

poles.

segments 3 and

B

The voltage

I,

4.

A

are

between the poles. Consequently,

induced

the coil sides of

to

shown, the coil-sides of coil

the neutral zone

no voltage

Figure 4.37

y.

C are shown so as not to complicate the

In the position

This direct-current

the interpoles

connected to commutator segments 72 and

while coil

in

show

between brushes x and

A, B, and

those of coil

A

not

4.38b gives a detailed view of the armature

coils lying coils

we do

between the N, S poles.

A.

in coil

On

the other hand,

are directly under the in coil

B

is

maximum

N

and S

at this

mo-

ment. Consequently, the voltage between adjacent

commutator segments 3 and 4 is maximum. The voltage in coil C is also zero because its coil sides are sweeping across the neutral zone. Note that the positive circuit coils

and negative brushes each short-

having zero induced voltage.

76

bars

#4

Armature conductor size

#

Shunt field conductor size

14

Example

4-2

The generator

in Fig.

4.38 generates 240

V

between

adjacent brushes and delivers a current of 2400

A modern generator having the same power and speed weighs 7 times less and occupies only 1/3 the floor

Calculate

space.

a.

b.

mentarily in the neutral zone, while coil

coming from

the flux

The

coil

width (known as coil pitch)

the coil sides

tween poles

l.

B

and the center of pole

of coil

is

cutting

A

is

such that

coming from adjacent N,

Thus, the coil sides of coil

center of pole 2

B

c.

The current delivered per brush set The current flowing in each coil The average voltage induced per coil

the center of the poles.

the coil sides cut the flux

S poles.

A

to the load.

lie

under the

3.

Similarly,

Solution a.

A current of 2400 A flows out of the + (

and back

into the

(

There are 12 brush

The

—

)

sets,

2. 3.

/

terminal

6 positive and 6 negative.

current per brush set

is

are in the neutral zones be-

2 and poles

)

terminal of the generator.

2400/6

= 400 A

90


Figure 4.38b Closeup view of the armature

coils

between adjacent brushes.

DIRECT-CURRENT GENERA TORS

b.

Each positive brush coils to the right

/

c.

-

400/2

each

in

200

coil

is

A

71

There are six coils between adjacent brush

The average voltage per

Eavge

The

4.27

coil

240/6

a generator

|

{

The

commutation 72

I

currents flowing

the

If

80 A, the

shown

carry

all

40

versal takes place

other.

left. If

In Fig. 1

,

the current is

how commutation

4.39a the brush

and the 40

short distance,

now is

A from

brush unite

is

The

to

U

re-

|

(c) 80

di-

we

right

A

and

output.

1

40

40

contact with segment

2, 1

.

Owing

the total current,

72

|

1

|

only one-fourth of

namely 0.25 X 80

V

|

3

|

4040

7 (e)

80

between the

only one-fourth of the total contact area, and so is

I

to the

commutator is proportional to the conThe area in contact with segment 2 is from segment 2

40

©| |© 71

while 75 per-

contact resistance, the conductivity

40

40

V.

and 25 percent of the brush surface

contact with segment

the current

3

|

middle of seg-

on the

give the 80

in

tact area.

© ,2

|

to the

changes

takes place,

in the

the coils

in

brush and

N

40 40

4.39b the commutator has moved a

Fig.

1

called commutation.

brush produces a voltage drop of about

is

|

j

The contact resistance between the segment and

cent

72

4.39a to 4.39e.

the left of the

In

40

40

during the millisecond interval that

rection in this brief interval

ment

I

means

move from one end of the brush

To understand

|

(b)

© 71

on the right-hand side of the

The process whereby

refer to Figs.

3

I

80

0

40 A.

current in these coils must reverse.

a coil takes to

© 2

1/

T

moving from

are

40

40

in Fig.

brush will soon be on the left-hand side. This that the

|

flow toward

and the

the right

coils

commutator segments

right to left, the coils

3

|

6020

the armature

in

that the currents in the coils

the load current is

2

I

©

V

|

under load, the individual coils

coming both from

the brush,

/

20

40

40

Note

1

= 40 V

windings next to a positive brush are 4.39a.

® (a)

on the armature carry one-half the load current carried by one brush.

©

40

sets.

71

is

40

is

process When

72

|

40

ideal

40

40

40

to the left of the brush.

Consequently, the current

=

from the

set gathers current

and

9

=

20 A. By

Figure 4.39 Commutation

of the current

in coil 1.

are neglected and current reversal

brush contact resistance.

is

Inductive effects

caused by the

ELECTRICA L MA CHINES A ND TRA NS FORMERS

92

same token,

the

brush

is

we now

If

the current

X 80 - 60

0.75

from segment

cover that the current flowing

and 2

now

are the

ities

means

dis-

must be 20 A.

1

A to

dropped from 40

and the brush area

is

in coil

X

1/10

a

little

in

L =

the currents are equal. This is

zero

Segment 2

left.

is

now

still

far-

contact with 75

in

A from

from segment 2 and 20

If coil

I

again 20 A, but

is

the opposite direction to

now understand how

what

it

segments In Fig.

plete

slide

e

and the current

in coil 2 is

1

is

com-

about to be reversed. it

is

important

(amperes per square

centimeter) remains the same

at

the brush face. Thus, the heat

produced by the con-

resistance

is

every point across

spread uniformally across the

brush surface. Unfortunately, such ideal commutanot possible in practical machines, and

tion

is

now

investigate the reason why.

induced voltage [V]

change of current [A/s]

rate of

LA//A/

= It

we

The

practical

X

6

10

X

5.75

+ 40 -

[

X

V

the presence of this induced voltage (attribut-

is

that

flow

in coil

1

coil is considered.

when

We

ues for these currents

1,

and the currents

creasing or decreasing. tance does not

come

that

that

it

takes place

very short time; consequently, the current can-

The reason

is

the

armature coils have inductance and

it

it

should.

strongly opposes a rapid change in current. in

Fig.

4.39 has 72 bars and that the armature turns

at

600

r/min.

in 1/10

One

revolution

that the

is,

therefore,

The

currents

in Fig. 4.39.

the middle of seg-

is in

in the coils are neither in-

As

a result, the coil induc-

into play.

4.40b the current

in coil

1

is

changing due

However, the

in-

value of 20 A. Suppose the coil current

35 A. From Kirchhoff s current law, the currents

flowing from segments then respectively 75

20 A. Note

completed

of a second and during this short period 72

form over the brush touches segment

into the brush are

5 A, instead of

face.

2,

60

A and

no longer uniis

low where

and high where

it

1

In Fig. 4.40c the brush

1

is

The density

segment

cally placed as regards rent in coil

and 2

1

A and

that the current density

the brush touches

commutator

Suppose, for example,

val-

order to determine the re-

4.40a the brush

In Fig.

ment

currents

have assumed plausible in

should be compared with those

is

not reverse as quickly as

new

the self-inductance of the

sulting current flows in the brush.

to its ideal

is

40)]

duced voltage e prevents the current from dropping

process in a

-

(

3

10

able to L), that opposes the change in current.

In Fig.

commutation

The problem with commutation

in-

is

to the contact resistance effect.

4.28

given by (4.2)

Figs. 4.40a to 4.40e illustrate the in coil

commutation process,

is

LM&t

1.39

over the brush.

to note that the current density

tact

=

100

We

the brush contact resis-

4.39e the current reversal

In this ideal

self-induction

=

seg-

tance forces a progressive reversal of the current as the

only

inductance of the coil [H|

Hows

did before!

it

is

has an inductance of, say, 100 fxH. the

1

Applying Kirchhoff" s current law, we find

1.

that the current in coil

can

=

duced voltage

percent of the brush, and so the currents divide ac-

A

1

at this instant.

moved

4.39d the commutator has

cordingly: 60

in

=

e

1

in coil

which

1

the same. Consequently, the conductiv-

same and so

past the brush. Thus, the

or 1.39 ms!

s

The voltage induced by

A//A/

In Fig.

1/720

e

contact with segments

in

that the current in coil

ther to the

ment

-

1/72

20 A.

moved

4.39c the commutator has

In Fig.

further,

we

contact with the brush, the cur-

in

rent in this coil has

commutator bars sweep

time available to reverse the current

apply Kirchhoff s current law,

Thus, by coming

to the

1

A.

is

momentarily symmetri-

segments

1

and

2.

has not fallen to zero, and

But the curis still,

say.


30 A. As a 40

40

40

40

40

while that sity

© 71

72

|

segment 2

2

3

|

f

still

35

40

2

/

|

3

I

I

has

left-

moved beyond

it

in coil

1

the

has

has a value of 20 A,

from segment

to the brush

is is

The

1

resulting high current den-

sity

causes the brush to overheat

720

coils are being

)

at the tip.

Because

commutated every second,

overheating raises the brush

75 5

71

therefore, 7

40

©

V

|

A

despite the fact that the contact area

getting very small.

72

1

Assuming

not reversed.

now 60 A,

|

is,

midpoint of the brush and the current

the current flowing

71

70

is

1

0 A. The current den-

will tend to overheat.

4.40d segment

In Fig.

(a) 80

40

1

on the left-hand side of the brush

hand side of the brush

4040

40

only

is

times greater than on the right-hand side. The

vl /l

|

segment

result, the current in

in

93

tip to the

this

incandescent

point and serious sparking will result. In

80

designing dc motors and generators, every

is made to reduce the self-inductance of the One of the most effective ways is to reduce the number of turns per coil. But for a given output voltage, this means that the number of coils must be increased. And more coils implies more

effort

coils.

40

40

40

30

40

© 71

|

72

1^1/2

|

3

I

1

commutator

7010

(c) 80

bars. Thus, in practice, direct-current

generators have a large

number of

mutator bars — not so much the output voltage but to

coils

and com-

to reduce the ripple in

overcome

problem of

the

commutation. 20

40

40

40

40

Another important factor

© 71

72

|

J

1

|

2

7

is

made

3

|

that the

(d) 80

zone.

As

© |

72

1

|

x

2

,

|

the brush 3

I

4

40 4C

(e) 80

tance

opposes the reversal

always

is

commutation induced

is

1

.5 V.

the brush

.

The

of current.

in the

to the self-

measures, the composition of

carefully chosen.

It

affects the brush

V

to as

much

This drop occurs between the surface of

and the commutator surface.

A large brush

drop helps commutation, but unfortunately

1

mmf.

created in the neutral

coil.

creases the losses. of the current in coil

is

voltage drop, which can vary from 0.2 as

Figure 4.40 Commutation

is

armature

the

which opposes the voltage due

In addition to these

71

than

this tlux, a voltage

inductance of the

40

40

aiding commutation

the coil side undergoing

sweeps through coil

40

greater

Therefore, a small tlux

60 20

40

slightly

in

commutating poles

4

|

tht\

40

mmf of the

As

a result, the

it

in-

commutator and

brushes become hotter and the efficiency of the coil

induc-

generator

is

slightly reduced.


94


resistance the

Practical level

Sketch the main components of a dc gener-

4-1

ator.

4- 2 1

4-2

Why

are the brushes of a de

ways placed

at

machine

al-

100

is

<},

calculate the

machine operates

a.

At no-load

b.

At full-load

Fig. 4.

1

mmf when

rated voltage

at

8b shows the no-load saturation

curve of a separately excited dc generator

the neutral points?

when

it

revolves

1500 r/min. Calculate

at

4-3

Describe the construction of a commutator.

the exciting current needed to generate

4-4

How

120

is

the induced voltage of a separately

excited dc generator affected a.

the speed increases?

b.

the exciting current

How

4-5

do we adjust

4- 3

if

1

is

reduced?

tion

the voltage of a shunt

is

4. 10, the

induced voltage

momentarily 18

V, in the posi-

shown. Calculate the voltages induced

C

A, B, and

same

at the

Referring to Fig. 4.

1

age induced

A when

decreases with increasing load. Explain.

has rotated by 90°; by 120°.

over-

4- 5 1

load

and

differential

compound generators

as to construction

b.

as to electrical properties

1

instant.

lb, calculate the volt-

the armature

positive with respect to brush y 1

b.

Show

Does

the polarity of each of

the polarity reverse

when

a coil turns through 180°?

4-16

The generator of

Fig. 4.38 revolves at

r/min and the flux per pole

is

20

960

mWb.

Calculate the no-load armature voltage

if

each armature coil has 6 turns.

Intermediate level separate excited dc generator turning at

4-17

How many

a.

127 V. The armature resistance

machine delivers

is

brush sets are needed for the gen-

erator in Fig. 4.38?

1400 r/min produces an induced voltage of

the

is

in coil

the 12 coils.

a.

A

Brush x

in Fig. 4,

Explain the difference between shunt, compound,

4-9

330 r/min.

The terminal voltage of a shunt generator

increases.

4-8

1

D

in coils

4-14

Explain why the output voltage of an compound generator increases as the

4-7

at

Referring to Fig. in coil

generator?

4-6

V

If the

b.

2 12 and

machine delivers

a total load current of

1800 A. calculate the current flowing

a current of 12 A.

armature

in

each

coil.

Calculate

Advanced

a.

The terminal voltage |V]

b. c.

The heat dissipated in the armature [W] The braking torque exerted by the armature

A

separately excited dc generator pro-

4-18

[Nm| 4-10

duces a no-load voltage of 115

V.

segments 3 and 4 must be greater than

40 V?

What

happens if a. The speed is increased by 20 percent? b. The direction of rotation is reversed'.' c. The exciting current is increased by 10 percent? d. The polarity of the field is reversed?

level

The voltage between brushes x and y is 240 V in the generator shown in Fig. 4.38. Why can we say that the voltage between

4- 9 1

Referring to Fig. ity

of

£ xv when

4.

1

0,

determine the polar-

the armature turns counter-

clockwise.

4-20

a.

In Fig. 4.38

determine the polarity of

tween commutator segments 3 and 4-

1

1

Each pole of a

1

00 kW, 250

V flat-compound

generator has a shunt field of 2000 turns and a series field of 7 turns.

If

the total shunt-field

ing that the armature b.

At the same

instant,

is

4.

£ 34

be-

know-

turning clockwise,

what

segment 35 with respect

is

to

the polarity of

segment 34?


4-2]

The armature shown

5.4 (Chapter 5)

in Fig.

has 81 slots, and the commutator has 243

segments. lap

It

will be

winding having

flux per field pole

is

wound

/ n du stria I

4-24

to give a 6-pole

30

mWb,

The induced voltage

at a

calculate the

is 60 ohms and 2 The I R loss in

are 15

mm

wide and

commutator 4-22

A 200 W,

1

20

is

V,

that the

diameter of the

1

4-23

dc generator has a

winding on the armature.

The

rated armature current

The The

I

total losses in the 2

R

effi-

is

is

5 A.

0.023 pu.

machine

losses in the armature

The generator in Problem 4-24 weighs 2600 lb. Calculate the output in watts per

4-26

In

Problem 4-24 calculate

The full-load current of the generator The current carried by the armature coils

the torque re-

quired to drive the generator

(The shunt

field

is

at

1750 r/min.

powered by a separate

source.)

4-27

A 4-pole 2 8 A.

dc generator delivers a current of

The average brush voltage drop on

each of the four brush sets

Calculate

b.

the rated current the armature

a.

1

a.

r/min separately

kilogram.

equal to 1.33 ms.

A 4-pole 250 kW, 750V lap

4-25

800 r/min dc generator

The brush width is such as to cover 3 commutator segments. Show that the duration of the commutation is

1

b. c.

450 mm.

has 75 commutator bars.

process

V 750

Calculate

The average flux density per pole The time needed to reverse the current in each armature coil, knowing that the brushes

c.

500

speed of

1200 r/min b.

A 240 kW,

ciency of 94%. The shunt field resistance

following: a.

Appl tea t ion

excited dc generator has an overall

turn per coil. If the

1

95

is

found

to

be

0.6 V. Calculate the total brush loss in the

machine, neglecting friction

loss.

1

Chapter 5 Direct-Current Motors

Today,

5.0 Introduction

this

general statement can be challenged

because the availability of sophisticated electronic

Now

thai

erators,

we have we can

a

good understanding of dc gen-

motors transform

Direct-current

drives has

begin our study of de motors.

energy

electrical

hoists, fans, cars.

vari-

ment has given

it

service and

force (cemf)

has to drive, and this require-

rise to three basic types

Direct-current motors are built the

of motors:

erate either as a trate,

Series motors

3.

Compound motors

motor or

means of a switch

all

electric utility sys-

electric trains,

it

is

in steel mills,

As soon

form the alternating current der to use dc motors.

flows

mines, and

sometimes advantageous

low.

to trans-

is

To

connected to a dc source (Fig. 5.

1

).

illus-

which the armature,

E

The armature has is

s

by

a re-

created by a set

in

The

as the switch

is

closed, a large current

the armature because

its

resistance

individual armature conductors are

is

very

imme-

diately subjected to a force because they are im-

into direct current in or-

The reason

as

of permanent magnets.

tems furnish alternating current. However, for speapplications such as

in

sistance R, and the magnetic field

Direct -current motors are seldom used in ordinary

because

as a generator.

consider a dc generator

initially at rest, is

industrial applications

same way

generators are; consequently, a dc machine can op-

Shunt motors

2.

mersed

that the torque-

in

the magnetic field created by the perma-

nent magnets. These forces add up to produce a

speed characteristics of dc motors can be varied over a

in

Counter-electromotive

5.1

The torque-

speed characteristic of the motor must be adapted to the type of the load

cial

still

thousands more are being produced every year.

definite torque-speed

pump or fan) or a highly

able one (such as a hoist or automobile).

.

possible to use alternating current

pumps, calendars, punch-presses, and

These devices may have a

characteristic (such as a

1

it

there are millions of dc motors

mechanical energy. They drive devices such as

into

made

motors for variable speed applications. Nevertheless,

wide range while retaining high efficiency.

powerful torque, causing the armature

96

to rotate.

DIRECT-CURRENT MOTORS

motor

5.2 Acceleration of the The

net voltage acting in the armature circuit in Fig.

(£ s

—£

The

resulting armature current

5.2

is

/ is

limited only by the armature resistance R, and so

()

volts.

)

=

/

(£ s

- EJ/R

(5.1)

When the motor is at rest, the induced £0 = 0, and so the starting current is Figure 5.1 Starting a

line.

The

On

the other hand, as

gins to turn, a

cut a

in

soon as the armature be-

second phenomenon takes place: the

We know

generator effect.

duced

that a voltage

Ea

is in-

the armature conductors as soon as they

magnetic field (Fig.

5.2).

This

is

always

true,

The value and the same as those

no matter what causes the rotation. polarity of the

obtained

induced voltage are

when

the

machine operates

The induced voltage the

E0

is

as a generator.

therefore proportional to

speed of rotation n of the motor and to the flux

per pole, as previously given by Eq. 4.

tf>

£0 in

or the circuit-breakers to

the type

the

(4.

polarity .

is

equal to

It

£

(1

is

force (cemf) because

always acts against the source voltage

acts against the voltage in the sense that the

net voltage acting in the series circuit

equal to (£ s

— £

()

)

volts

and not (£ s

of Fig. 5.2

+ £n

)

they are

consequent rapid acceleration of the armature.

As

the speed increases, the counter-emf

creases, with the result that the value of (£ s

diminishes.

It

follows from Eq.

5.

)

armature

that the

1

£0 in— £u

current / drops progressively as the speed increases.

maximum

volts.

is

to accelerate until

()

voltage

£

s

.

voltage (£s

to act ical

In effect,

if

£0 were

reaches a def-

less than the source

equal to

— E0 would become )

the current

The driving

/.

£

s

,

the net

zero and so, too,

forces

would cease

on the armature conductors, and the mechan-

drag imposed by the fan and the bearings would

immediately cause the motor

to

slow down. As the

speed decreases the net voltage (Es

and so does the current fall

it

speed. At no-load this speed pro-

duces a counter-emf £ slightly

would

case of a motor, the induced voltage

called counter-electromotive

£s

the armature and

Z

if

ductors produce a powerful starting torque and a

1

a constant that de-

of winding. For lap windings

However,

trip.

mo-

blow

Although the armature current decreases, the

number of armature conductors. In the

its

Z is

the fuses to

absent, the large forces acting on the armature con-

motor continues

number of turns on

be 20 to 30 times

would cause

tor. In practice, this

1

Z//4V60

the case of a generator.

pends upon the

may

current

starting

greater than the nominal full-load current of the

inite,

As

voltage

= EJR

/

dc motor across the

97

as soon as the torque

current

is

/.

— £0

The speed

)

increases

will cease to

developed by the armature

equal to the load torque. Thus,

motor runs

when

a

no-load, the counter-emf must be

at

£

slightly less than

s?

to flow, sufficient to

so as to enable a small current

produce the required torque.

Example 5-7 The armature of a permanent-magnet dc generator has a resistance of

when

1

H is

and generates a voltage of 50

500

r/min. If the armature

is

V

con-

nected to a source of 150 V, calculate the following:

Figure 5.2 Counter-electromotive force (cemf)

the speed

in

a dc motor.

a.

The

starting current

"1


98

b.

The counter-emf when

the

motor runs

at

1000

Mechanical power and torque

5.3

r/min. At 1460 r/min. c.

The armature current

at

1000 r/min. At 1460

The power and torque of

r/min.

motor are two of

a dc

most important properties.

its

We now derive two sim-

ple equations that enable us to calculate them.

Solution a.

moment of start-up,

At the

tionary, so

rent

E0 = 0 V

1

the armature

(Fig. 5.3a).

The

is

.

sta-

According

to Eq. 4.

wound armature

starting cur-

£ =

limited only by the armature resistance:

is

(t

- EJR =

/

150 V/l

n=

A

150

the

1

cemf induced

in a lap-

given by

is

Z/i$/60

(4.1)

Referring to Fig. 5.2, the electrical power P. supx

b.

Because

the generator voltage

r/min, the

cemf of

the

1000 r/min and 146 c.

The

V

motor at

is

50

V

will be

at

100

V

at

E

s

P*

However, drop s

150

-

100

E

in the

= EJ

s

is

equal to the

sum of E0

E = Ea + s

=

(E S

=

50/1

When

the

A

(Fig. 5.3b)

= (E0 + = EJ +

motor speed reaches 1460 r/min, the

cemf will be 146 V, almost equal to the source voltage. Under these conditions, the armature current is

The

2 I

R term

IR)I 2 I

R

only

electrical

=

=

(Es 4

- EJ/R =

(150

-

146)/1

(5.4)

represents heat dissipated in the ar-

mature, but the very important term

/

(5.3)

EJ

P*

50

IR

follows that

- EJ/R -

plus the IR

is It

I

(5.2)

armature:

= 50 V

The corresponding armature current

I:

1000

is

E - Eu =

equal to the supply voltage

is

multiplied by the armature current

1460 r/min.

net voltage in the armature circuit at

r/min

500

plied to the armature

power

that is

EJ

is

the

converted into mechanical

power. The mechanical power of the motor therefore exactly equal to the product of the

A

is

cemf

multiplied by the armature current

and

the

corresponding

motor torque

smaller than before (Fig. 5.3c).

is

much

P = EJ

(5.5)

DIRECT- CURRENT MOTORS

where

where

P —

E0 -

mechanical power developed by the

T

motor [W]

Z

induced voltage

in

=

total

torque (N-mJ

number of conductors on the armature [Wb|*

-

effective flux per pole

<&

=

/

=

armature current A|

6.28

=

constant, to take care of units

the armature (cemf)

IV] /

[

current supplied to the armature [A] [exact value

2.

99

Turning our attention to torque that the

mechanical power

P

is

T,

=

2tt

1

we know

given by the

Eq. 5.6 shows that

motor

expression

either

we can

raise the torque of a

by raising the armature current or by

raising the flux created by the poles.

P = n 7/9.55 where n

is

(3.5)

Example 5-2 The following

the speed of rotation.

Combining Eqs. /?779.55

3.5, 4.1,

and

5.5,

we

obtain

- EJ = Z/i3>//6()

hp),

250

V,

details are given

on a 225 k\V (~ 300

1200 r/min dc motor (see Figs. 5.4 and

5.5):

243

armature coils turns per coil

and so

1

type of winding

T = Z*//6.28 The torque developed by a lap-wound motor therefore given

armature slots

81

commutator segments

243

is

field poles

by the expression

T = Z$//6.28

lap

(5.6)

6

diameter of armature

559

axial length of armature

235

The

effective flux

is

niven by

mm mm

- 60 EJZn.

Figure 5.4

225 kW, 250 V, 1200 r/min. The armature core has a diameter and an axial length of 235 mm. It is composed of 400 stacked laminations 0.56 mm thick. The armature slots and the commutator has 243 bars. (H. Roberge)

Bare armature and commutator of a dc motor rated of

559

has 81

mm


00

1

Figure 5.5 Armature

a.

of Fig. 5.4 in the

process

b.

One

c.

Connecting the

d.

Commutator connections ready

of the 81 coils coil

of

ready to be placed

ends

to the

being wound; coil-forming machine gives the coils the desired shape. in

the slots.

commutator

bars.

for brazing. (H.

Roberge)

T=

Calculate a.

The

b.

The number of conductors per

c.

The

rated armature current slot

9.55 Pin

=

9.55

X 225 000/1200

=

1791

N-m

flux per pole

The

flux per pole

is

Solution a.

=

6.28 TiZI

nearly equal to the applied voltage (250 V).

=

(6.28

The

=

25.7

We

can assume

that the

induced voltage

rated armature current /

- F/E0 =

E0

d>

is

is

225 000/250

5.4

= 900 A

Speed

When b.

Each

coil

is

made up of 2 conductors,

gether there are 243

X

2

= 486

so alto-

conductors on

Conductors per

slot

Coil sides per slot

The motor torque

1790)/(486

is

= =

486/81 6

=

IR drop due

always small compared

On

900)

a dc motor drives a load between no-load and

full-load, the

equal to 6

X

of rotation

to

armature resistance

to the supply voltage

This means that the counter-emf

the armature.

c.

X

mWb

E

s

£0

is

is

E

s

.

very nearly

.

the other hand,

may be expressed by

we have

already seen that

E0

the equation

En =

Z/7*/60

(4.1)


101

(variable)

!

motor armature

motor

field

(fixed)

O

(

j

O

3-phase motor

Figure 5.6

Ward-Leonard speed control system.

Replacing

E0

by

£ we s

3-phase

obtain

,

as

Zn*/60,

the

line. This method of speed control, known Ward-Leonard system, is found in steel

and paper

'A mills, high-rise elevators, mines,

That

RKTR

is

60£

v

(approx)

A

'SiaJU3$6Â

modern

In

installations the generator

placed by a high-power electronic converter that

changes the ac power of the

electrical utility to dc,

The Ward-Leonard system

= speed of rotation [r/min] E = armature voltage fVJ Z = total number of armature

armature of a dc motor.

s

shows

This important equation is

conductors speed of the

that the

directly proportional to the armature supply

and inversely proportional

voltage

We will now

study

more than

is

simple way of applying a variable dc voltage

//

pole.

mills.

often re-

by electronic means.

where

motor

is

to the flux per

how this equation is

applied.

It

just a to the

can actually force the mo-

tor to

develop the torque and speed required by the

load.

For example, suppose

slightly

higher than the

Current will then flow 5.6,

E

is

s

E

cemf

i}

adjusted to be

of the motor.

in the direction

and the motor develops

shown

in Fig.

a positive torque.

The

armature of the motor absorbs power because

Armature speed control

5.5

According to Eq. 5.7, constant

with

Now, suppose we reduce Es by reducing the gen<1> G As soon as E s becomes less than £u current / reverses. As a result, the motor torque

speed depends only upon the

reverses and (2) the armature of the motor delivers

if

the flux per pole

(permanent magnet

fixed excitation), the

E By

armature voltage

s

.

In

practice,

we can

tion

or field

fall in

vary

E

s

of the motor

s,

the

proportion.

by connecting the

is

G

(Fig. 5.6).

The

field excita-

kept constant, but the generator

can be varied from zero

and even reversed.

to

maximum

The generator output voltage

f can therefore be varied from zero s

mum, with

£

kept

M to a separately excited variable-

voltage dc generator

excitation / x

field

is

raising or lowering

motor speed will rise and

motor armature

to

either positive or negative

maxi-

polarity.

Consequently, the motor speed can be varied from zero to

maximum

generator

is

/

flows into the positive terminal.

in either direction.

Note

erator excitation

.

(

,

1

)

power to generator G. In effect, the dc motor suddenly becomes a generator and generator G suddenly becomes a motor. The electric power that the dc motor

now netic its

the

delivers to

G

is

derived

at the

expense of the

connected mechanical load. Thus, by reducing

motor

is

£

s,

suddenly forced to slow down.

What happens to the dc power received by genG? When G receives electric power, it operates as a motor, driving its own ac motor as an asynchronous generator!* As a result, ac power is fed erator

that the

driven by an ac motor connected to a

ki-

energy of the rapidly decelerating armature and

The asynchronous generator

is

explained

in

Chapter

14.

1


02

back into the

P = EJ = 380 X 3000 =

normally feeds the ac motor.

line that

The tact that power can be recovered this way makes the Ward-Leonard system very efficient, and constitutes another of

Example 5-3 A 2000 kW, 500 by a 2500

kW

control system

generator, using a

shown

T = 9.55/Vw

motor

V, variable-speed

in Fig. 5.6.

is

driven

Ward-Leonard The total resis-

tance of the motor and generator armature circuit

The motor turns when E0 is 500 V.

10 mi}. r/min,

at a

is

nominal speed of 300

The motor torque and speed when

The speed of of

this

(400

=

n

with the armature (Fig. 5.7). The current

It is

E

s

,

in the

which subtracts

yielding a smaller

lot

below

its

nominal

recommended for small motors beof power and heat is wasted in the rheoonly

and the overall efficiency is

is

low. Furthermore,

poor, even for a fixed setting

increases as the armature current increases. This

produces a substantial drop

motor armature

speed with increasing

in

mechanical load.

is

P = EJ = 380 X 2000 - 760 The motor speed

to control

to place a rheostat in se-

is

380)/0.01

A

to the

connected me-

of the rheostat. In effect, the /R drop across the rheo-

-

stat

The power

motor

the speed regulation

is

s

its

electromechanical braking torque.

cause a stat,

= (E - EJ/R =

motor and

Rheostat Speed Control Another way

speed.

Solution

2000

the

ables us to reduce the speed

s


kN-m

supply voltage across the armature. This method en-

E = 350 V and £0 = 380 V

-=

47.8

140 000)/228

1

from the fixed source voltage

The motor torque and speed when

/

=

X

chanical load will rapidly drop under the influence

ries

s

a.

(9.55

rheostat produces a voltage drop

E = 400 V and E0 = 380 V b.

=

the speed of a dc

Calculate a.

kW

140

Braking torque developed by the motor:

advantages.

its

1

kW

is

X 300 = 228

(380 V/500 V)

The motor torque

r/min

is

T = 9.55P/R

b.

Because

-

(9.55

=

31.8

X 760

000)/228

kN-m

E a — 380

V, the

motor speed

is still

228 r/min.

The armature current /

is

= (E EJ/R = = - 3000 A s

The current

is

(350

-

Power returned by

38())/0.01

it

flows

in re-

motor torque also reverses.

the

and the 10 mi} resistance:

5,6 Field According

negative and so

verse; consequently, the

Figure 5.7 Armature speed control using a rheostat.

motor

to the

generator

a dc

speed control

to Eq. 5.7

we

can also vary the speed of

motor by varying the

keep the armature voltage

numerator

in

Eq. 5.7

is

field flux

E

s

.

Let us

now

constant so that the

constant. Consequently, the

motor speed now changes

in

inverse proportion to


the flux

drop,

cr>:

if

we

increase the flux the speed will

This

method of speed control

frequently used

is

when

the

called

base speed. To control the flux (and hence,

motor has

the speed),

to run

we connect

above

a rheostat

its

R

rated speed,

in series

To understand

at

constant speed.

less

method of speed

this

motor

in Fig.

5.8a

is

The counter-emf

running slightly

ZTn

is

£

due

s,

we suddenly

to the

increase the

resistance of the rheostat, both the exciting current /x

and the flux

reduces the

cemf

jump

a

to

to

will diminish. This

immediately

causing the armature current

much

is

again almost equal to

E

/

higher value. The current

.

£

will accelerate until

It

same E0 with

Clearly, to develop the

motor must turn

raise the

motor speed above

shunt-wound motors,

£0

and

s

()

s.

flux, the

troducing a resistance

E

motor develops a

the

field,

greater torque than before.

with

control, sup-

initially

than the armature supply voltage

IR drop in the armature. If

Despite the weaker

value depends

its

between

the very small difference

{

the field (Fig. 5.8a).

pose that the

changes dramatically because

upon

and vice versa.

103

its

weaker

a

can therefore

nominal value by with the

in series

this

We

faster.

in-

For

field.

method of speed control

enables high-speed/base-speed ratios as high as 3 to 1.

Broader speed ranges tend

produce

to

instability

and poor commutation.

Under

abnormal conditions, the

certain

exciting current of a shunt motor dentally, the only flux

manent magnetism that the

motor has

remaining

in the poles.*

is

is

may

flux

drop to dangerously low values. For example,

if

the

interrupted acci-

due

that

This flux

to the re-

is

so small

dangerously high

to rotate at a

speed to induce the required cemf. Safety devices are introduced to prevent such runaway conditions. (a)

5-7

Shunt motor under load

Consider a dc motor running chanical load

at

no-load. If a

suddenly applied

is

me-

to the shaft, the

small no-load current does not produce enough

torque to carry the load and the motor begins to

slow down. This causes the cemf to diminish, p.u.

—/

Tvsn

2

— "-7

/ —

/

/

/

/

higher torque.

*

Tvs

motor

I

is

When

the torque developed by the

exactly equal to the torque imposed by the

mechanical load, then, and only then, will the speed

/

remain constant (see Section

t

3.1 1).

To sum

up, as

the mechanical load increases, the armature current

•rated load

(b)

re-

sulting in a higher current and a corresponding

rises

and the speed drops.

"—7

The speed of

/

stant

2

1

^ speed

a shunt

from no-load

motor stays

only drops by 10 to 15 percent

when

it

is

p. u.

/

The term residual magnetism

Figure 5.8

also used. However, the

is

IEEE Standard Dictionary of Electrical and a shunt motor including the

a.

Schematic diagram

b.

Torque-speed and torque-current characteristic of

induction;

a shunt motor.

will be less than the residual induction."

of

full-load

n

armature current

field

relatively con-

to full-load. In small motors,

rheostat.

Terms

states.

.

netie eireuit, the if

.

If

Electronics

there are no air gaps ... in the inag-

remanent induction

there are air gaps

.

.

.

will equal the residual

the remanent induction

"1

1


04

applied. In big machines, the drop in part, to

is

even

less,

due

By

the very low armature resistance.

A

51

ad-

justing the field rheostat, the speed can, of course,

be kept absolutely constant as the load changes.

Typical torque-speed and torque-current charac-

of a shunt motor are shown in Fig. 5.8b. The speed, torque and current are given in per-unit values. The torque is directly proportional to the arteristics

mature current. Furthermore, the speed changes only from

from 0 pu

pu

1.1

pu as the torque increases

to 0.9

to 2 pu.

Example 5-4

A

shunt motor rotating

120

V

1500 r/min

at

and the shunt-field resistance mature resistance a.

b. c.

is

fed by a

source (Fig. 5.9a). The line current

is 0.

is

II. If

The

=

Figure 5.9

See Example

120 V/120

P = 6000 - 250 - 5750 = 7.7

motor

(equivalent to 5750/746

=

/

b.

5

The voltage across

W

O= A 1

= 50 A

I

the armature

E=

and

in

armature iron losses.

120

5.8 Series is

A

series

motor

motor

identical in construction to a

is

shunt motor except for the

V

Voltage drop due to armature resistance

nected

= 50X0.1 =

The cemf generated by

5

V

field.

The

field

con-

is

This series field

is

armature current (Fig. 5.10a).

composed of a few

turns of wire

having a cross section sufficiently large the armature

is

with the armature and must, there-

in series

fore, carry the full

IR

hp)

is

-

I

is

The actual mechanical output is slightly less than 5750 because some of the mechanical power is dissipated in bearing friction losses, in windage losses,


5.4.

W

the

field current (Fig. 5.9b) is /x

A

Mechanical power developed by the armature

The current in the armature The counter-emf The mechanical power developed by

Solution: a.

51

the ar-

calculate the following:

ii,

1

120

is

to carry

is

the current.

E0 =

-

120

5

=

V

115

Although the construction ties

c.

The

total

power supplied

to the

motor

is

similar, the proper-

of a series motor are completely different from

is

those of a shunt motor. In a shunt motor, the flux F P,

=

El

=

120

Power absorbed by

X

= 6120

51

W

per pole field

the armature

is

is

constant

P n = El = 120 X 50 = 6000

W

armature

is

IR

2

=

50

2

X

0.

1

in

a series motor

When

the current

is

large and vice versa. Despite these

is

differences, the

P =

But

depends upon the armature current

and. hence, upon the load. large, the flux

in the

loads because the shunt

line.

is

the flux per pole

Power dissipated

at all

connected to the

- 250

W

same

basic principles and equations

apply to both machines.


105

series field

(b)

Figure 5.10 a.

Series motor connection diagram.

b.

Schematic diagram of a series motor.

When

a series motor operates at full-load, the

per pole

flux

identical

same

as that of a shunt

starts up, the

armature current

3 p.u.

motor of

power and speed. However, when

motor

series

the

is

speed

/;

armature current

the

/

higher

is

Figure 5.11 than normal, with the result that the flux per pole also greater than

normal,

torque of a series

motor

ft

is

is

follows that the starting

of

ing the

On

T versus

/

curves of Figs. 5.8 and

the other hand, if the

than full-load, the pole are

5.

1

Conversely, the speed

motor operates

necting an external resistor

at less

ture

The weaker field same way as it would

smaller than normal.

shunt motor with a

weak shunt

the load current of a series

For ex-

field.

motor drops

ample,

if

half

normal value, the flux diminishes by half and

its

so the

speed doubles. Obviously,

if

the load

is

to

ate at

no-load,

ft

tends to run away, and the resulting

could tear the windings out of the

centrifugal forces

and the

sistor

and

field.

The

field

reduces

be lowered by con-

in series

total

with the arma-

IR drop across the armature

the

voltage, and so the speed must

re-

supply

fall.

Typical torque-speed and torque-current characteristics are

ferent

shown

in Fig. 5.

1

1

.

They

are quite dif-

from the shunt motor characteristics given

in

Fig. 5.8b.

small,

may rise to dangerously high values. For reason we never permit a series motor to oper-

speed

this

may

1

armature current and the flux per

causes the speed to rise in the for a

and current-torque characteristic

a series motor.

considerably greater than

of a shunt motor. This can be seen by compar-

that

the

Typical speed-torque

Example 5-5

A

15 hp,

240

V,

1780 r/min dc series motor has

full-load rated current of teristics are

54 A.

Its

a

operating charac-

given by the per-unit curves of Fig. 5.11.

armature and destroy the machine.

Calculate

5.9

When

Series motor speed control a series

motor carries

a load,

its

speed

a.

The current and speed when 24 N m

b.

The efficiency under

may

the load torque

is

these conditions

have to be adjusted slightly. Thus, the speed can be

low resistance

increased by placing a with the

series

field.

smaller than before,

and an increase

in

The

field

in parallel

current

which produces a drop

speed.

is

then

in flux

Solution a.

We

first

establish the base power, base speed,

and base current of the motor. They correspond to the full-load ratings as follows:


106

PH = //

=

hp

15

X 746 =

15

11

190

W

are also used in electric cranes and hoists: light loads are lifted quickly

= 1780r/min

H

= 54A

/B

Compound motor

5.11

The base torque

therefore,

is,

A compound dc

P vh

9.55

Tv =

=

X

9.55

11

780

190/1

mmf of the Fig. 5.

N-m

torque of 24

corresponds to a per-

two

runs

=

T(pu)

=

24/60

2

1

at

Referring to Fig.

5.

tained at a speed of

From

the

X

/j(pu)

T vs

/

1

.4

1

winding

0.4

1

,

a torque of 0.4

pu

pu. Thus, the speed

nB

=

1.4

shows

low and the

is

However,

is at-

by current machine:

is

As

X 1780

/x

it

curve, a torque of 0.4 pu re-

is

The

total

current

b.

X

/(pu)

-

/B

0.6

To calculate the and

X 54 =

away

like a shunt

no-load.

at

mmf of the

series field

remains flux per

falls

with increasing load and the to full-load is

generally

between 10 percent and 30 percent.

A

we have

efficiency,

to run

to

know P

32.4

x

P0 =

/2779.55

= 7776

= 2492 X

()

W

24/9.55

= 6263 W = PJP, = 6263/7776 =

0.805 or 80.5%

5.10 Applications of the series

motor Series motors are used on equipment requiring a high starting torque.

They

which must run series

motor

is

at

are also used to drive devices

high speed

at light

loads.

The

particularly well adapted for traction

purposes, such as in electric trains. Acceleration rapid

because the torque

is

high

at

is

low speeds.

Furthermore, the series motor automatically slows

down

as the train goes up a grade yet turns at top

speed on

flat

ground. The power of a series motor

tends to be constant, because high torque

is

accom-

panied by low speed and vice versa. Series motors

Figure 5.12 Connection diagram of a dc compound motor. b. Schematic diagram of the motor.

a.

is

fully excited

Pr

P = EI = 240 X

T|

32.4

is

therefore greater under load than at no-load.

speed drop from no-load

=

the series

of the series field

mmf of the shunt field mmf (and the resulting

The motor speed

is

the motor

/ in

and so the motor behaves

the load increases, the

constant.

pole)

When

the shunt field

does not tend

increases but the

r/min

mmf

quires a current of 0.6 pu. Consequently, the load

/

and

the connection and schematic

no-load, the armature current

negligible.

- 2492

fields add.

diagrams of a compound motor.

unit torque of

=

a series field

compound motor, the The shunt field is always

stronger than the series field.

= 60 N-m

//

motor carries both

a shunt field. In a cumulative

"is

A load

and heavy loads more slowly.


If 1.6

the series field

the shunt field,

we

is

connected so

increasing load. 1.2

creases, and this

it

The speed

may

opposes

compound

obtain a differential

motor. In such a motor, the total

1.4

that

107

mmf decreases with

rises as the

lead to instability.

load in-

The

differ-

-r-+ential

1.0

j a

Fig. 5.

a x>

compound motor

differential

0.8

compound

compound

a> 0)

a

of shunt,

1

3

shows

the typical torque-speed curves

compound and

basis. Fig. 5. 14

w

has very few applications.

shows

series

motors on a per-unit

a typical application of dc

0.6

motors

in a steel mill.

0.4

5.12 Reversing the direction 0.2

of rotation 0 0

0.2

0.4

0.6

0.8

Torque

1.2

1.0

1.4

1.6

(per-unit)

the direction of rotation of a dc motor,

must reverse either (2) both the shunt

Figure 5.13 Typical

To reverse

speed versus torque characteristics

of various

and

l

)

in

we

the armature connections or

series field connections.

terpoles are considered to

The change

dc motors.

(

connections

form is

The in-

part of the armature.

shown

in Fig. 5. 15.

Figure 5.14 Hot

strip finishing mill

to the

runout table

composed

(left

of

6 stands, each driven by a 2500

kW

dc motor. The wide steel

foreground) driven by 161 dc motors, each rated 3 kW.

(Courtesy of General Electric)

strip is

delivered

"1 1


OS

(-)

(+)

(->

(+)

<+>

(0

(b)

(a)

|

I

Figure 5.15

compound

a.

Original connections of a

b.

Reversing the armature connections to reverse the direction of rotation.

c.

Reversing the

field

motor.

connections to reverse the direction of

5.13 Starting a shunt motor [f

we apply

full

rotation.

5.14 Face-plate starter

voltage to a stationary shunt motor,

Fig. 5. 16

shows

the schematic diagram of a manual

the starting current in the armature will be very high

face-plate starter for a shunt motor. Bare copper

and we run the

contacts are connected to current-limiting resistors

risk of

/?,,

Burning out the armature;

a.

Damaging

b.

the

commutator and brushes, due

to

*

and

R4 Conducting arm .

when

it

of insulated handle

is

1

sweeps

across

pulled to the right by means

2. In the position

shown,

the

M

heavy sparking; c.

R2

the contacts

Overloading the feeder;

d.

Snapping off the shaft due

e.

Damaging

the driven

to

mechanical shock;

equipment because of the

sudden mechanical hammerblow.

arm touches dead copper contact and the motor circuit is open. As we draw the handle to the right the conducting arm first touches fixed contact N. The sii; y voltage Es immediately causes full field current / x to flow, but the

armature current

/ is

limited by the four resistors in the starter box. The All dc motors must, therefore, be provided with

a

means

to limit the starting current to

values, usually between rent.

One

solution

is

to

1

.5

reasonable

and twice full-load cur-

connect a rheostat

in series

motor begins

to turn and, as the

cemf E0

the armature current gradually falls.

speed ceases

builds up,

When

any more, the arm

the motor

is

pulled to

the

next contact, thereby removing resistor

R from

the

to rise

}

with the armature. The resistance

is

duced as the motor accelerates and eliminated entirely, full

when

the

gradually reis

eventually

machine has attained

Today, electronic methods are often used to

control.

and

to

provide speed

circuit.

The

current immediately jumps to

higher value and the motor quickly accelerates to next higher speed.

we move

speed.

limit the starting current

armature

arm

When

the speed again levels off

to the next contact,

and so

finally touches the last contact.

netically held in this position

net 4,

which

is in

a

the

forth, until the

The arm

is

mag-

by a small electromag-

series with the shunt field.


109

Figure 5.16 Manual face-plate starter for a shunt motor. If the

supply voltage

the field excitation

is

suddenly interrupted, or

electromagnet releases the arm, allowing to

it

to return

dead position, under the pull of spring

its

3.

This

prevents the motor from restarting un-

safety feature

expectedly

if

should accidentally be cut, the

when

the supply voltage

is

reestablished.

When

the

tion of the

cemf Ea

motor

are as

shown

armature IR drop, If

is

running normally, the direc-

armature current

£

()

is

we suddenly open

motor continues

I x

and the polarity of the Neglecting the

in Fig. 5.17a.

equal to

E

s.

the switch (Fig. 5.17b), the

to turn, but

its

speed will gradually

drop due to friction and windage losses.

5.15

Stopping a motor

One

inclined to believe that stopping a dc

other hand, because the shunt field

induced voltage is

almost

a simple,

always

trivial,

not

a heavy inertia load,

the

true.

system to

come

is

same

it

a generator a large dc motor

may

to a halt. For

many

coupled

is

take an hour or

more

Ea

whose armature

is

motor

under these circumstances,

is

motor

is

by simple mechanical

armature

is

suddenly connected to

reasons such the external resistor (Fig. 5. 7c). Voltage

way we stop a

car.

this current

motor

original current

same

A more elegant method consists of

electrically.

flows

in the

brake the

circulating a reverse current in the armature, so as to

brake the

mediately produce an armature current

a braking

friction, in the

Two methods

are

/,. It

(

nected to a source nected to the

throw switch.

£

s,

whose

field is directly

and whose armature

is

either the line or to

an external resistor

R (Fig.

very smooth stop.

concon-

5. 17).

im-

However,

/2

.

is

The

reverse torque brings the machine to a rapid, but

l

same source by means of a doubleThe switch connects the armature to

.

developed whose magnitude depends upon

Dynamic braking

Consider a shunt motor

E0 will

opposite direction to the

dynamic braking and (2) plugging.

5.16

/2

follows that a reverse torque

em-

ployed to create such an electromechanical brake:

now

Let us close the switch on the second set of contacts so that the

we must apply One way to

at the

is

for

often unacceptable and,

torque to ensure a rapid stop.

the

on open-circuit.

1

a lengthy deceleration time

On

excited,

continues to exist, falling

rate as the speed. In essence, the

operation. Unfortunately, this

When

is

to

motor

is still

Figure 5.17a Armature connected

to a

dc source

Es

.

1

1


0

Figure 5.17b Armature on open

circuit

generating a voltage

E0

\x

Time

.

Figure 5.18 Speed versus time curves

for

various braking methods.

reversing the armature current by reversing the

ter-

minals of the source (Fig. 5.19a).

Under normal motor conditions, armature rent

is

/,

/,

Dynamic braking.

where

In practice, resistor

R

R0

cur-

given by

is

=

(E s

- £0 )//? 0

the armature resistance. If

we

suddenly

reverse the terminals of the source, the net voltage is

chosen so that the

initial

acting on the armature circuit becomes (E 0 + E s The so-called counter-emf E0 of the armature is no ).

braking current rent.

The

initial

is

about twice the rated motor cur-

braking torque

is

then twice the nor-

longer counter to anything but actually adds to the

mal torque of the motor.

As

E0

in

E

supply voltage the

motor slows down, the gradual decrease

produces a corresponding decrease

in

/2

becoming zero when

finally

the arma-

The speed drops quickly

This net voltage would produce

greater than the full-load armature current. This

current tor,

ture ceases to turn.

.

.

Consequently, the braking torque becomes smaller

and smaller,

s

an enormous reverse current, perhaps 50 times

would

initiate

an arc around the commuta-

destroying segments, brushes, and supports,

at first

and then more slowly, as the armature comes

even before the

line circuit breakers

to a

halt.

The speed decreases exponentially, somewhat

like

the

voltage across a discharging capacitor.

Consequently, the speed decreases by half intervals of time

Ta To

dynamic braking,

.

in

equal

illustrate the usefulness

Fig. 5. 18

of

compares the speed-

time curves for a motor equipped with dynamic braking and one that simply coasts to a stop.

5.17 Plugging

We a

can stop the motor even more rapidly by using

method called plugging.

It

consists of suddenly

Figure 5.19a Armature connected

to

dc source

Es

.

could open.


value.

However,

much

is

it

draw the

easier to

speed-time curves by defining a new time con-

T0 which

stant

the time for the speed to de-

is

crease to 50 percent of

its

original value. There

is

a direct mathematical relationship between the

conventional time constant

T0

constant

T and

the half-time

given by

It is

.

Ta = 0.6937

We

Figure 5.19b is

Plugging.

can prove that

this

given by

Jnr

Tn =

(5.9)

°

To prevent such a catastrophe, we must by introducing a

reverse current

resistor

R

in series

As

in

dynamic

with the reversing circuit (Fig. 5. 9b). 1

braking, the resistor

U

braking current

With

oped even effect, at

when

Ta =

the armature has

E0 =

= moment

J

devel-

=

is

braking

motor

P =

cuit,

otherwise

interruption

it

the armature cir-

will begin to run in reverse. Circuit

mounted on

the

motor

The curves of Fig. 5.18 enable us

dynamic braking

plugging and

for the

to

completely after an interval if

dynamic braking

the

its

is

2T0 On .

it

same

initial

is still

25 per-

original value at this time. Nevertheless,

more popular

in

motor

to the braking resistor

a constant (exact value

=

a constant

This equation

is

braking effect pated is

in the

exact value

=

log c 2]

due

that the

energy

dissi-

resistor. In general, the

motor

entirely

braking

to the

subjected to an extra braking torque due to

windage and

friction,

and so the braking time

be less than that given by Eq.

hp),

250

V,

1

280 r/min dc motor

mechanical time constant

drives a large flywheel and the total

We

can therefore speak of a mechanical

much

same way we speak of

time constant

T

the electrical

time constant of a capacitor that dis-

in

the

charges into a resistor.

essence,

motor

T

is

it

is

speed

is

its

initial

connected

to a

210

V

is

kW.

moment 177 kg

of

m2

.

It

in-

The

dc source, and

1280 r/min just before the armature

switched across a braking resistor of 0.2

its is

il.

Calculate

The mechanical time constant T0 of

the braking

system

takes for the speed

36.8 percent of

of the flywheel and armature

motor

a.

the time

to fall to

will

5.9.

has windage, friction, and iron losses of 8

ertia

of the

|

Dynamic braking and

braking.

[W]

=

most applications.

We mentioned that the speed decreases exponentially with time when a dc motor is stopped by dynamic

In

the

based upon the assumption

is

Example 5-6 A 225 kW (- 300 5.18

starts [r/min]

power delivered by

2

0.693

comparative simplicity of dynamic braking ren-

ders

shaft

compare

the other hand,

used, the speed

motor

(30/77) /log c 2]

shaft.

braking current. Note that plugging stops the motor

cent of

=

131.5

usually controlled by an automatic

is

null-speed device

of inertia of the rotating

initial

l

we must immediately open

stops,

fall to |sj

speed of the motor when

initial

i

as the

previous value

lkg-nr] /?

which

its

parts, referred to the

to a stop. In

= EJR,

As soon

value.

its initial

come

0, but I2

is

time for the motor speed to one-half

to limit the initial

reverse torque

circuit, a

zero speed,

about one-half

designed

131.5 7,

where

about twice full-load current.

to

plugging

this

is

limit the

(5.8)

mechanical time constant

b.

The time

for the

motor speed

to

drop

to

20 r/min


c.

The time

for the speed to drop to

only braking force friction,

due

that

is

20 r/min

to the

if

The stopping time increases

the

windage,

20 r/min

and iron losses

Solution a.

We

;

note that the armature voltage

the speed

When

the armature

210V. The

the resistor

P,

210

V

is

(276/10)

=

28 min

This braking time

switched to the brak-

initial

dynamic braking

very

is still

power delivered

to

X 60 =

1656

2

/R

2

= 210

/0.2

= 220 500

The time constant Ta

s

Ta =

7//,

177 131.5

-

10

28 times longer than when

used.

to a

is

dynamically

complete stop.

In practice,

however, we can assume that the machine stops

W

is 2

is is

a motor which

Theoretically,

braked never comes

ter

an interval equal to 5 If the

/(131.5 Py)

(5.9)

motor

af-

seconds.

plugged, the stopping time has a

is

2

X 1280 X 220 500

Tn

definite value given by ts

= 2T0

(5.10)

where

s

—

ts

b.

=

and

is

= E

approximately

is

1280 r/min.

is

ing resistor, the induced voltage

close to

is

proportion to the

in

time constant. Consequently, the time to reach

The motor speed drops by 50 percent every The speed versus time curve follows the se-

10

s.

stopping time using plugging

Tn — time

constant as given

in

[

sj

Eq. 5.9

[sj

quence given below:

speed (r/min)

time

Example 5-7 The motor in Example 5-6

(s)

1280

0

640 320

10

ing resistor

30

Calculate

80

40

a.

40 20

50

b.

an interval of 60

60 to

20 r/min

The initial braking The stopping time

after

The

We

ro -

=

.///,

/>,

/(131.5 P,

X = 276 s = (177

The

initial

/,

The

,

initial

power

is

s

braking current

- EiR =

initial

is

420/0.4

braking power

=

1050

= 8000

According

to Eq. 5.9,

Tn

A

is

X

{

220.5

kW

has the same value as

before: is

T0 =

10

s

)

2

1280 )/(131.5 4.6

current and braking

P = EJ = 210 X 1050 =

have

1280

The new time constant 2

so that the

fl,

as before.

E = E + E = 210+ 210 - 420 V

s.

of the braking time.

=

same

net voltage acting across the resistor

(

x

the

Solution

The initial windage, friction, and iron losses are 8 kW. These losses do not vary with speed in exactly the same way as do the losses in a braking resistor. However, the behavior is comparable, which enables us to make a rough estimate

n

is

20

160

The speed of the motor drops c.

plugged, and the brak-

increased to 0.4

is

braking current

.

.

is

min

X

8000)

The time

to

come ts

to a

complete stop

= 2T0 = 20

s

is


Armature reaction

5.19

now we have assumed

Until

ing in a dc

motor

current flowing

is

that

due

the flux

that the only

However,

to the field.

N

mmf actthe

armature conductors also cre-

in the

magnetomotive force

ates a

113

that distorts

and weakens

coining from the poles. This distortion and zone

field

weakening takes place

in

motors as well as

We recall that the magnetic action mmf is called armature reaction.

generators.

armature

in

of the

Figure 5.20 Flux distribution

a motor running

in

at no-load.

due

5.20 Flux distortion

to armature reaction When

motor runs

a

at no-load, the

small current

flowing in the armature does not appreciably affect the flux

when

coming from

<2>,

the

the poles (Fig. 5.20). But

armature carries

its

normal current,

it

duces a strong magnetomotive force which,

would create a

acted alone,

superimposing

and

<1>|

increases

under the

we

2,

(Fig. 5.2

if

it

By

1).

obtain the resulting

Figure 5.21 Flux created by the full-load armature current.

our example the flux density

(Fig. 5.22). In

flux

02

flux

pro-

half of the pole and

left

de-

it

ceases under the right half. This unequal distribution

produces two important effects. First the neutral zone shifts toward the rotation).

The

left

result

is

(against the direction of

poor commutation with

sparking at the brushes. Second, flux

density

in

pole

tip

A,

due

to the

saturation

higher

sets

Consequently, the increase of flux under the hand side of the pole the right-hand side.

is

Flux 4> 3

than flux

slightly less

less than the

chines the decrease in flux percent and

therefore

no-load. For large

may

left-

decrease under

at full- load is

be as

much

zone

in.

Figure 5.22 Resulting flux distribution

a motor running at

full-

main the

as 10

case of a dc generator, these narrow poles de-

velop a magnetomotive force equal and opposite

causes the speed to increase with load.

it

in

load.

the

mmf of the armature

to

so that the respective mag-

Such a condition tends to be unstable; to eliminate the

problem,

we sometimes add

netomotive forces a series field of

rise

and

fall

one current varies. In practice, the

or

two turns to increase the flux under load. Such tating poles

motors are said to have a stabilized-shunt winding.

Commutating poles

5.21

set

of

improve commutation, we always place a

commutating poles between the main poles of

medium- and large-power dc motors

made

(Fig. 5.23).

As

slightly greater than that of the

armature. Consequently, a small flux subsists

in the

region of the commutating poles. The flux

is

signed to induce

To counter the effect of armature reaction and thereby

is

together as the load

mmf of the commu-

in the coil

tion a voltage that

is

equal and opposite to the self-

induction voltage mentioned in Section 4.28. result,

commutation

de-

undergoing commuta-

is

As

a

greatly improved and takes

place roughly as described in Section 4.27.


Figure 5.24 Six-pole

dc motor having a compensating winding distributed

in

slots in the

main poles. The machine also has 6

commutating poles. (Courtesy of General Electric

Company)

base speed. In so doing, the rated values of armature current,

armature voltage, and field flux must not be

may be we assume an

exceeded, although lesser values In rately

making our

E. v

is

negligible (Fig. 5.25).

the

armature current

exciting current /f in

per-unit values.

£a happens /a is

that

it

1

.

The advantage of the

Thus, the per-unit torque

used. ideal sepa-

unit flux

f

to

,

The armature

/a ,

the flux

and the speed n are

Thus,

be 240

if

all

is

T

is

given by the per-

times the per-unit armature current

T=$> r I

ll

/.,

(5.11)

voltf,

the

expressed

the rated armature voltage

By

the

voltage

Ea

same reasoning, is

1

shunt field flux O, has a per-unit

the per-unit armature

equal to the per-unit speed n times the

per-unit flux O,

=

V and the rated armature current

600 A, they are both given a per-unit value of

Similarly, the rated

per-unit approach

renders the torque-speed curve universal.

excited shunt motor in which the armature re-

sistance

age

analysis,

value of

The

(5.12)

logical starting point of the torque-speed

curve (Fig. 5.26),

is

the condition

where the motor

1

1

6


(T=

develops rated torque

'a

The

rated speed

)

—

speed (n

at rated

I

).

order to reduce the speed below base speed,

In

we

l

often called base speed.

is

gradually reduce the armature voltage to zero,

and

while keeping the rated values of at their per-unit

value of

l

.

3> r constant

Applying Eq.

(5.

7=1

corresponding per-unit torque

X

1

1

the

),

=

1

1.

Furthermore, according to Eq. (5.12), the per-unit

Figure 5.25

£a = n X £a /a

voltage

Per-unit circuit diagram

the state of

operation,

,

= n. Figures 5.27 and 5.28 show

]

and

known

<£>,

during this phase of motor

mode.

as the constant torque

Next, to raise the speed above base speed, alize that the armature voltage

anymore because The only solution

T

already at

it is

is

to

keep

means

=

base speed, the per-unit flux

OH

1

0

'

2.0

1.0

^

speed

n

is

Eq. (5.12),

1.

1

this

Thus, above

\/n.

equal to the recipro-

of the per-unit speed. During this operating

cal

mode, the armature current can be kept of

level

T= Figure 5.26

=

and so

1,

rated level of

its

flux. Referring to

{

re-

E,d at its rated level of

and reduce the that n<$

we

cannot be increased

ct»

1.

=

/

r a

Recalling Eq. (5.11),

(\/n)

X

it

at its rated

follows that

-\ln. Consequently, above base

1

speed, the per-unit torque decreases as the reciprocal of the per-unit speed.

motor

1

is

clear that since the per-

and armature voltage are both

unit armature current

equal to

It is

during this phase, the power input to the

equal to

1

.

Having assumed an

chine, the per-unit mechanical

equal to is

why

1,

which corresponds

the region

ideal

power output

is

maalso

to rated power. That

above base speed

is

named

the

constant horsepower mode.

We

conclude that the ideal dc shunt motor can

operate anywhere within the limits of the torque-

speed curve depicted

in Fig. 5.26.

In practice, the actual torque-speed

fer considerably

from

that

shown

curve

may

in Fig. 5.26.

dif-

The

curve indicates an upper speed limit of 2 but some

machines can be pushed

to limits

of 3 and even 4, by

reducing the flux accordingly. However,

speed

is

raised

lems develop and centrifugal forces

'

0

1.0 1.25 >-

speed

n

L2.0

dangerous.

When

the ventilation

when

the

above base speed, commutation probthe motor runs

becomes poorer and

tends to rise above

its

may become

below base speed, the temperature

rated value. Consequently, the

armature current must be reduced, which reduces the

Figure 5.28

torque. Eventually,

when

the speed

is

zero,

all

forced


ventilation ceases

and even the

field current

must be

117

magnet motors

5-24 Permanent

reduced to prevent overheating of the shunt field coils.

As a

result, the

permissible stalled torque

only have a per-unit value of 0.25. tical

torque-speed curve

The ishes

The resulting pracspeed dimin-

can be largely overcome by using an external

of air, no matter

what

It

delivers a constant stream

the speed of the

Under these conditions,

approaches that

shown

We

have seen

and a

that shunt-field

field current to

motors require coils

produce the

flux.

consumed, the heat produced, and

in Fig. 5.29.

drastic fall-off in torque as the

blower to cool the motor.

to be.

shown

is

may

motor happens

the torque-speed curve

large space taken

up by the

vantages of a dc motor.

By

The energy

the relatively

field poles are disad-

using permanent mag-

nets instead of field coils, these disadvantages are

overcome. The

result

is

a smaller

motor having a

higher efficiency with the added benefit of never risking run-away due to field failure.

in Fig. 5.26.

A further advantage is

that the effective air

The reason that

that the

is

many

increased

times.

magnets have a permeability

nearly equal to that of

is

mature sible

is

of using permanent magnets

gap

air.

As

a result, the ar-

mmf cannot create the intense field that is pos-

when

soft-iron

pole

pieces

are

employed.

Consequently, the field created by the magnets does not

become distorted,

armature reaction

is

as

shown

in Fig. 5.22.

Thus, the

reduced and commutation

is

im-

proved, as well as the overload capacity of the motor. 0

0.2 0.4 0.6 0.8 1.0

—

A further advantage is that the long air gap reduces the

2.0

speed

inductance of the armature and hence

n

much more

quickly to changes

in

it

responds

armature current.

Permanent magnet motors are particularly advan-

Figure 5.29

Torque-speed curve of a typical dc motor.

tageous

in capacities

below about

5 hp.

The magnets

Figure 5.30

Permanent magnet motor rated slots 20;

1

.5 hp,

commutator bars: 40; turns per

0.34 a.

(Courtesy of Baldor Electric

Company)

90

V,

2900

coil: 5;

r/min, 14.5 A.

conductor

Armature diameter: 73 mm; armature length: 115 mm; 17 AWG, lap winding. Armature resistance at 20°C:

size: No.


are ceramic or rare-earth/cobalt alloys.

shows

PM

motor.

and

tia

fast

Its

the series winding, per pole.

5.30

Fig.

2900 r/min elongated armature ensures low iner-

the construction of a

response

hp,

.5

when used

The only drawback of tively high cost

1

has a

V,

motors

motor

of the magnets and the inability to

a.

b.

obtain higher speeds by field weakening.

5-12

Questions and Problems Name

1

make

Explain what

is

a 5- 3 1

meant by the generator

What determines

ef-

ity

in

If the

line, calcu-

following: full-load

when

V

the armature

is

connected

200

1

to a

source. Calculate the armature voltat

1500

The following details are known about a 250 hp, 230 V, 435 r/min dc shunt motor:

A

nominal full-load current: 862

the magnitude and polar-

of the counter-emf

V

r/min. At 100 r/min.

fect in a motor.

5-3

23 A.

separately excited dc motor turns at

115

three types of dc motors and

is

field

and the

age required so that the motor runs

sketch of the connections. 5-2

A

The shunt

15 (1,

1

connected to a 230

mmf per pole at mmf at no-load

The The

r/min

Practical Level 5-

is

late the

the rela-

is

of

total resistance

nominal armature current

servo applications.

in

PM

90

H

insulation class:

a dc motor?

weight: 3400 kg

The counter-emf of a motor

5-4

is

always

mm

external diameter of the frame: 915

slightly less than the applied armature volt-

length of frame: 1260

mm

age. Explain.

and efficiency

a. Calculate the total losses

Name two methods

5-5

that are

used to vary

the speed of a dc motor.

5-6

why

Explain

ing current

a starting resistor

motor up

needed

to bring a

percent of the total losses

Show one way to reverse the direction rotation of a compound motor.

5-8

A 230 V

5-9

20 per-

Calculate the value of the armature resistance

knowing

as well as the counter-emf.

speed?

to

field excit-

the shunt field causes

if

cent of the total losses. c.

is

approximate shunt

b. Calculate the

the armature current of a shunt

motor decreases as the motor accelerates.

Why

5-7

at

full-load.

of

that

at full-load are

50 due

to armature resistance. d.

If

we wish

to attain a

what should be

shunt motor has a nominal arma-

the

speed of

1

100 r/min,

approximate exciting

current? ture current of

tance

0.15

11,

If the

armature

c.

The mechanical power developed by [kW] and

the armature

if

the

motor

across the 230 b.

V

the

mo-

inal

is

initial starting

shown

The value of

20 hp, 240

V,

400

nom400 A, calculate

in Fig. 5. 17. If the is

the braking resistor

to limit the

125 percent of

maximum

its

R

if

we

braking current

to

nominal value

line,

to limit the initial current to

1

b.

The braking power [kW] when the motor has decelerated to 200 r/min, 50 r/min, 0 r/min.

a.

The motor

15 A.

5-15

in

Problem

5-

1

4

is

now

stopped

by using the plugging circuit of Fig. 5.19.

The compound motor of Fig. 5.12 has 1200 turns on the shunt

1

armature current

want

directly connected

Intermediate level 1

wish to stop a

the following: a.

Calculate the value of the starting resistor

needed

5-1

We

ing circuit

[WJ

[hp]

a. In Problem 5-9 calculate the

current

5-14

r/min motor by using the dynamic brak-

The counter-emf [V| The power supplied to tor,

resis-


b.

a.

5-10

is

60 A.

winding and 25 turns on

Calculate the the

maximum

new braking

resistor

braking current

is

R

so that

500 A.


Calculate the braking power [kW]

b.

motor has decelerated

to

when

the

5-20

200 r/min, 50 r/min,

ing:

0 r/min.

Compare

c.

the braking

power developed

at

200

r/min to the instantaneous power dissipated in resistor R.

Advanced 5-16

5-2

1

tor has a

225 kW,

a

diameter of 559

length of 235

mm.

1

The The

c.

at a

The value of

c.

The

the armature

the counter-cmf at

full

flux per pole, in milliwebers

A standard

20 hp, 240

V.

1

load

[mWb]

500 r/min

self-

it

200 r/min

Calculate the following:

ing.

total kinetic

ings and

commutator

is

the

decided to cool the machine by

the air by

tor

equal to the J calcu-

in-

blower and channeling

means of an

The highest 30°C and exits the mo-

air duct.

the temperature of the air that

J of the wind-

from

1500 r/min without overheat-

expected ambient temperature

energy of the revolving parts if

Ft is

to

stalling an external

turns at 1200 r/min

speed of 600 r/min,

requirement has arisen whereby the mo-

tor should run at speeds ranging

kinetic energy of the armature alone

when

The number of conductors on

b.

A

200 r/min mo-

mm and an axial

The approximate moment of inertia, knowing 3 that iron has a density of 7900 kg/m

b.

a.

cooled dc motor has an efficiency of 88%.

level

The armature of

a.

Referring to Fig. 5.30, calculate the follow-

is

should not exceed 35°C. Calculate the

capacity of the blower required,

in

cubic

lated in (a)

feet per minute. (Hint: see Section 3.21.)

5-17

If

we reduce

the normal exciting current of

a practical shunt

motor by 50 percent,

speed increases, but

it

5-22

the

A

250

hp,

nominal

never doubles.

500 V dc shunt motor draws

field current of 5

A

a

under rated

The field resistance is 90 12. Calculate ohmic value and power of the series re-

load.

Explain why, bearing

in

mind

the saturation

the

of the iron under normal excitation. 5- 8 1

The speed of

a series

motor drops with

ing temperature, while that of a shunt

ris-

mo-

tor increases. Explain.

Industrial

5-19

5-23

magnetism per 100°C increase

The motor runs

at

3% in

of

its

tempera-

a no-load speed of

2500 r/min when connected

to a 150

A5

500

hp dc motor draws a

In each case, calculate the

V

field as a

What

source in an ambient temperature of 22°C.

room where

that the field current drops

the shunt field and resistor

V

source.

field current

of

A when the field is connected to a 150 V source. On the other hand, a 500 hp motor draws a field current of 4.3 A when the field is connected to a 300 V dc source.

magnet motor equipped with

Estimate the speed

A when

0.68

cobalt-samarium magnets loses

ture.

needed so

are connected to the

A pplication

A permanent

sistor

to 4.5

if

the

motor

is

placed


is

in

a

40°C.

power required

for the

percentage of the rated power of the motor.

conclusions can you draw from these results?

Chapter 6 Efficiency and Heating of Electrical Machines

6.0 Introduction

Whenever

a

one form loss.

The

6.1

machine transforms energy from

to another, there

is

loss takes place in the

causing

(

duction

in efficiency.

1

)

an increase

in

Mechanical losses are due

machine

bearing friction,

to

brush friction, and windage. The friction losses

always a certain

depend upon

itself,

temperature and (2) a

Mechanical losses

the speed of the

machine and upon

the design of the bearings, brushes, commutator,

re-

and

Windage

depend on

the

speed and design of the cooling fan and on the

tur-

slip

rings.

losses

From the standpoint of losses, electrical machines may be divided into two groups: those that

bulence produced by the revolving

have revolving parts (motors, generators,

sence of prior information,

those that do Electrical

(transformers,

not

etc.)

reactors,

and mechanical losses are produced

and

tests

of these mechanical losses.

In this

chapter

chines, but the

we analyze

same

internal fan

the losses in dc

losses are also found in

mamost

in

us a clue as to

We

how

is

they

important because

may be

it

determine the value

mounted on

the

motor

shaft.

It

cool air from the surroundings, blows

the windings, and expels

machines operating on alternating current. The study of power losses

parts. In the ab-

usually conduct

Rotating machines are usually cooled by an

machines.

in stationary

itself to

etc.).

in ro-

tating machines, while only electrical losses are

produced

on the machine

we

it

draws it

over

again through suitable

vents. In hostile environments, special cooling

methods are sometimes used,

gives

reduced.

as

illustrated in

Fig. 6.1.

also cover the important topics of tempera-

ture rise

and the service

life

of electrical equipment.

We show

that both are related to the class

of insula-

tion used

and

have been

that these insulation classes

6.2 Electrical losses

standardized.

Electrical losses are

120

composed of the following:

EFFICIENCY AND HEATING OF ELECTRICAL MACHINES

1.

Conductor

/

R

losses (sometimes called

copper

R =

losses) 2.

Brush losses

3.

Iron losses

1.

Conductor Losses The

in

resistance, in turn,

depends upon

the length, cross section, resistivity,

and tempera-

A =

rent

ture

it

carries.

of the conductor.

The following equations

determine the resistance

able us to ture

resistance and the square of the cur-

The

at

at)

(6.2)

which

losses in a conductor de-

R = L =

its

(6.1)

P A

= pod +

p

pend upon

121

en-

p

resistance of conductor [il]

cross section of conductor

length of conductor [m]

=

resistivity

[

m2

]

of conductor

at

temperature

conductor

at

0°C

/

any tempera-

and for any material:

= a =

resistivity of

Po

m]

temperature coefficient of resistance

0°C t

[il

=

at

l/°C]

[

temperature of conductor [°C]

The values of p and a for different materials are listed in Appendix AX2. In dc motors and generacopper losses occur

tors,

shunt

field, the

armature, the series

in the

field, the

commutating

poles,

and

compensating winding. These I~R losses show

the

up as rise

conductor temperatures

heat, causing the

to

above ambient temperature.

Instead of using the

2

I

R

we sometimes

equation,

number

prefer to express the losses in terms of the

of watts per kilogram of conductor material. The losses are then given by the equation

P c = l00Oy 2 p/£

(6.3)

where

Pc — J =

Figure 6.1 Totally tor for

enclosed, water-cooled, 450 kW, 3600 r/min mo-

use

machine

is

in

Warm

a hostile environment.

air inside

Westinghouse nameplate. After releasing

its

water-cooled pipes, the cool

air

reenters the

chine by

pipes located diagonally

as cooling-water inlet

on the heat exchanger serve

and

outlet respectively.

(Courtesy of Westinghouse)

(W/kg|

]

p £

=

density of the conductor [kg/m

1000

=

constant, to take care of units

conductor [nll-m) 1 ]

heat to a

maway of two rectangular pipes leading into the end bells. The cooling air therefore moves in a closed circuit, and the surrounding contaminated atmosphere never reaches the motor windings. The circular capped set of

[A/mm

loss

2

resistivity of the

the

above the

current density

power

=

blown upward and through a water-cooled

heat exchanger, situated immediately

specific conductor

According

mass

is

density. For

tween

1

to this equation, the

loss per unit

proportional to the square of the current

.5

copper conductors, we use densities be-

A/mm 2

losses vary

and 6

A/mm 2

from 5 W/kg

to

.

The corresponding

90 W/kg

(Fig. 6.2).

The

higher densities require an efficient cooling system to prevent

an excessive temperature

rise.


122

9.6

W/kg

cm

1

carbon brush

2

200 80° C copper conductor

commutator

Figure 6.2 Copper losses may be expressed

in

watts per

Figure 6.3 Brush contact voltage drop occurs between the brush face and commutator.

kilo-

gram.

2.

Brush Losses The

2

I

R

losses in the brushes are

negligible because the current density

A/mm", which

0.1 per.

is

only about

used

and eddy currents, as previously explained

in

Sections 2.27 and 2.30. Iron losses depend upon the

cop-

magnetic flux density, the speed of rotation, the

However, the contact voltage drop between the

quality of the steel, and the size of the armature.

is

far less than that

brushes and commutator

may produce

in

significant

pending on the type of brush, the applied pressure,

They typically range from 0.5 W/kg to 20 W/kg. The higher values occur in the armature teeth, where the flux density may be as high as .7 T. The

and the brush current (Fig.

losses in the armature core are usually

losses.

3.

The drop

varies from 0.8

V

to

1.3 V,

de-

6.3).

Iron Losses Iron losses are produced

in the ar-

mature of a dc machine. They are due to hysteresis

1

The

losses can be

much

lower.

minimized by annealing the

steel

(Fig. 6.4).

Figure 6.4

kW electric oven is used to anneal punched steel laminations. This industrial process, carried out in a conatmosphere of 800°C, significantly reduces the iron losses. The laminations are seen as they leave the oven. (Courtesy of General Electric)

This 150 trolled


Some

iron losses are also

They

faces.

are

due

produced

in

the pole

to flux pulsations created as suc-

and

cessive armature teeth

sweep across

slots

the

Strange as chanical drag

it

may seem,

on the armature, producing the same

mechanical

effect as

impose a me-

iron losses

power

the line to continue to rotate. This no-load

overcomes

the friction, windage, and iron losses,

and provides for the copper losses 2

pole face.

123

The I R losses in commutating field load current

in the

shunt

field.

the armature, series field,

seldom more than 5 percent of

is

and

are negligible because the nothe

nominal full-load current.

friction.

As we load

machine the current increases

the

PR

in

Example 6-1

the armature circuit. Consequently, the

Adc machine turning at 875 r/min carries an armawinding whose total weight is 40 kg. The cur-

the armature circuit (consisting of the armature and

ture

A/mm 2

rent density is 5 ture

is

80°C. The

amount

to

1

1

and the operating tempera-

total iron losses in the

armature

00 W.

all

the other windings in series with

the other hand, the no-load losses

losses in

will rise.

it)

On

mentioned above

remain essentially constant as the load increases, unless the speed of the machine changes appreciably.

It

follows that the total losses increase with

Calculate load. a.

b.

The copper losses

Because they are converted

into heat, the tem-

perature of the machine rises progressively as the

The mechanical drag [N

inJ

due

to the iron losses

load increases.

However, the temperature must not exceed

Solution a.

Referring to Table sistivity

p

of copper

=

p0

=

15.88

=

21.3

(

at

in the

80°C

Appendix, the

re-

maximum used

is

the

+

1

AX2 at)

machine. Consequently, there

in the

power

that the

+ 0.00427 X

80)

The specific power

loaded beyond is

loss

8890 kg/m

deliver.

is

a limit to

This temper-

inal or rated

nllm

The density of copper

machine can

power enables us to establish the nompower of the machine. A machine

ature-limited (1

the

allowable temperature of the insulation

heat.

3

The

its

nominal rating

inevitably shortens the service

will usually over-

more

insulation deteriorates

life

rapidly,

which

of the machine.

is

If

a

machine runs

intermittently,

it

can carry heavy

2

P L = 100O/ p/£ = 1000 X 5 2 X

(6.1)

21.3/8890

overloads without overheating, provided that the operating time rating of

= 60 W/kg

1

is

copper loss

is is

P = 60x 40 = 2400

W

The braking torque due

to iron losses

is

it

can be for

P = «779.55

= 875

T=

12

A

(

3.5)

779.55

N-m

or approximately 8.85 ft-lbf

Losses as a function of load

physically impossible for a generator

kW to deliver an output of

100 kW, even

6.4 Efficiency curve The efficiency of ful

output

power

it

must absorb some power from

machine

is

the ratio of the use-

to the input

=

P -

-

X

power

P- (Section x

plus the losses p.

ti

However,

a

power PG

Furthermore, input power

dc motor running at no-load develops no useful

power.

kW for

one millisecond.

3.7).

6.3

2

from

calculated

1100

1

for higher loads the capacity

limited by other factors, usually electrical. For in-

stance,

rated at 10 b.

Thus, a motor having a nominal

However,

short periods. Total

short.

kW can readily carry a load of

0

100

We =

is

equal to useful

can therefore write

P °

X

100

(6.4)


124

The

where

=

T|

ful

output power [W]

P =

input

power [W]

zero

is

losses

the

25 percent of

rating, the

nominal

its

copper losses

we have

V,

in the 2

=

50 r/min, 230

loaded

is

to

armature curof

1/4)

its full-

the following:

to calculate the

efficiency of a dc machine.

Example 6-2 A dc compound motor having

motor

approximately 25 percent (or

is

square of the current,

1

no-load because no use-

load value. Because the copper losses vary as the

|W1

The following example shows how

1

at

developed by the motor.

25 percent load When rent

x

=

is

efficiency [%]

P0 —

p

efficiency

power

armature circuit

W

X 595 = 37

(1/4)

no-load losses a rating of 10

kW,

50 A, has the following losses

W

= 830

at

total losses full

load:

=

bearing friction loss

brush friction loss

windage

loss

mechanical losses

(1)

total

(2)

iron losses

(3)

copper

loss in the shunt field

copper losses at in the

b.

in the series field

c.

in

the

total

(4)

commutating winding

copper

W 50 W 200 W 290 W 420 W 120 W

= = =

500

load

at full

—

load

at 25, 50, 75, 100,

595

Pa =

at

no-load

of the machine.

Draw

a graph

showing

10

kW X

power supplied P,

830

In the

(1/4)

to the

=

867

W

= 2500

at

motor

W

25 percent

same way, we

At 50 percent load 2

(6.2)

100

= 74%

find the losses at 50, 75,

100, and 150 percent of the

(1/2)

W

is

= (PJPJ X 100 = (2500/3367) X

efficiency as a function of mechanical load (neglect

3.35 hp)

is

= 2500 + 867 - 3367

and the efficiency

W

and 150 percent of the nom-

+

is

T!

Calculate the losses and efficiency

inal rating

W 25 W 70 W

37

Useful power developed by the motor

loss in the

armature circuit

and

40

full load:

armature

a.

= = = = = =

nominal load:

the losses are

X 595 + 830 = 979

W

the losses due to brush contact drop).

At 75 percent load Solution

2

No-load The copper losses

(3/4) in the

At 100 percent load

sum of the mechanical

losses (1), the iron losses (2),

595

W

the losses are

+ 830 =

1425

W

and the shunt-field

At 150 percent load

losses (3):

2

(1.5)

no-load losses

1165

armature circuit

are negligible at no-load. Consequently, the no-

load losses are equal to the

the losses are

X 595 + 830 =

= 290 + 420 +

120

- 830

the losses are

X 595 + 830 = 2169

W

W The

efficiency calculations for the various loads

6A and

These losses remain essentially constant as the load

are listed in Table

varies.

graphically in Fig. 6.5.

the results are

shown


125

Figure 6.5 Losses and efficiency as a function of mechanical power.

See Example It is

LOSSES AND EFFICIENCY OF A DC

TABLE 6A

when Total

Output

Load

losses

power P

m

[Wl

[W]

75

1

100

1

l

r /r]

830

0

74

979

5

000

5

980

83.6

165

7

500

8 665

86.5

10 15

000

1

000

1

selecting a motor to

425

87.5

17 170

roughly equal to the load

We

3 367

425

The efficiency curve

mature

rises sharply as the load in-

fall.

This

is

typical of the

efficiency curves of all electric motors, both ac dc. Electric

motor designers usually

and

try to attain the

above calculation of efficiency we could

have included the losses due to brush voltage drop.

Assuming a constant drop, brush loss at full-load

brushes

=

say,

amounts

losses,

V per brush, the to 0.8 V X 50 A X 2

of 0.8

80 W. At 50 percent load, the brush loss

would be 40 W. These losses,

do

loads the

Consequently,

a particular job,

we

W

(

1

6.5

it

has to drive.

maximum

our example

(830

+

at that

load where the ar-

copper losses are equal to the no-load this

corresponds to a

= 660 W,

830)

1

an output of

1

total

8

1

may wish

to

check these

Temperature

The temperature

rise

1

The

5.8 hp) and an efficiency of 87.68 percent. results.

rise

of a machine or device

difference between the temperature of

its

is

the

warmest

accessible part and the ambient temperature.

It

may

be measured by simply using two thermometers.

peak efficiency at full-load. In the

circuit

losses. In

loss of

87.4

over a broad range of power,

and then slowly begins to

at light

poor.

can prove that the efficiency of any dc ma-

chine reaches a

reader

creases, flattens off

is

Efficiency

P-

[Wj

0

2 169

150

power

t)

motor

should always choose one having a power rating

500

830 867

50

Input

2

0

important to remember that

efficiency of any

MOTOR

25

6.2.

when added

modify the efficiency curve only

to the other

slightly.

However, due

to the practical difficulty of placing a

to the really warmest spot inside method is seldom used. We usuupon more sophisticated methods, de-

thermometer close the machine, this ally

rely

scribed in the following sections.

Temperature

rise

has a direct bearing on the

power rating of a machine or device.

It

also has a di-


126

rect bearing

on

its

temperature rise

useful service

is

life.

Consequently,

In crystallizing, organic insulators

and

a very important quantity.

brittle.

ical vibration will

expectancy of equipment

6.6 Life

electric

a

Apart from accidental electrical and mechanical

is

its

insulation:

higher the temperature, the shorter

made on many

its

insulating materials have

cause them

to break.

life.

The

Low

1

200°C

for the

same length of time.

temperatures are just as harmful as high

temperatures are, because the insulation tends

that

tors

their flexibility at temperatures as

life

means

ture of 105°C,

will

it

have a service

years at a temperature of 115°C, of

125°C, and of only one year

The

at

factors that contribute

at

life

a tempera-

6.7

of only four

two years

6.6).

Because of these

lation tallize

it

upon

their ability to withstand heat.

correspond to the

(6) time (Fig.

maximum

slowly begins to crys*

rapidly as the temperature rises.

Such

as

IEEE. Underwriters Laboratories. Canadian

Standards Association.

humidity

high temperature

chemicals

fungus

(UU so

dust

noxious gases

rodents

Figure 6.6 Factors that

may

These classes

temperature levels

shorten the service

life

of

an

of:

105°C, 130°C, 155°C, 180°C. and 220°C (formerly

and the transformation takes place more

time

that set standards*

have grouped insulators into five classes, depending

factors, the state of the insu-

changes gradually;

Thermal classification

Committees and organizations

to the deteriora-

and

retain

— 60°C.

of insulators

tion of insulators are (1) heat, (2) humidity, (3) vi-

bration, (4) acidity, (5) oxidation,

low as

at

135°C!

most

have been developed, however, which

motor has a

that if a

expectancy of eight years

to

freeze and crack. Special synthetic organic insula-

approximately by half every time the temperature increases by 10°C. This

that

On the other

synthetic polymers can withstand temper-

atures as high as

Tests

shown

some

does not exceed 00 °C.

the service life of electrical apparatus diminishes

normal

Under normal

expectancy of eight to ten years provided

life

hand,

expectancy of electrical apparatus

limited by the temperature of

hard

conditions of operation, most organic insulators have

their temperature

failures, the life

become

Eventually, the slightest shock or mechan-

insulator.

vibration


represented

by the

letters

A, B,

thermal classification (Table

6B)

F, is

H, and R). This

40°C, This standardized temperature was estab-

a cornerstone in

lished for the following reasons:

design and manufacture of electrical apparatus.

the

1

6.8

Maximum ambient temperature and hot-spot temperature rise

maximum ambient

TABLE 6B Class

l()5°C

A

temperature, which

Illustrative

B

F

when immersed

H

be included life at

220°C

R 240°C S

above

240°C

C

in this class if

such as

and paper when suitably impregnated or

silk, oil.

by experience or accepted

Other materials or combinations of materials tests

shown

they can be

shown

to

may

be included

have comparable thermal

life at

to

have comparable

substances. Other materials or combinations of materials

shown

to

have comparable

may life at

with suitable bonding

etc.,

in this class if

by experience

130°C.

Materials or combinations of materials such as mica, glass fiber, asbestos,

with suitable bonding

etc.,

be included in this class

if

by experience

155°C.

Materials or combinations of materials such as silicone elastomer, mica, glass fiber, asbestos,

etc.,

with suitable bonding substances such as appropriate silicone resins. Other materials or combinations of

may

they can be

shown

to

have

Materials or combinations of materials which by experience or accepted tests can be

shown

to

have

shown

to

have

shown

to

have

be included

life at

in this class if


tests

180°C.

the required thermal life at 200°C.

Materials or combinations of materials which by experience or accepted tests can be the required thermal life at 22()°C.

Materials or combinations of materials which by experience or accepted tests can be the required thermal life at 240°C.

Materials consisting entirely of mica, porcelain, glass, quartz, and similar inorganic materials. Other materials or combinations of materials

they can be

implies that electrical

sive,

performance guarantees.

I05°C.

shown

to

continuously

at

may

be included

have the required thermal

The above insulation classes indicate a normal

ated

to give

Materials or combinations of materials such as mica, glass fiber, asbestos,

materials

N

likely to encounter.

enables them to standardize the size of their

substances. Other materials or combinations of materials

comparable 200°C

It

machines and

in a dielectric liquid

or accepted tests they can be

80°C

machines are

examples and definitions

or accepted tests they can be

I55°C

2.

Materials or combinations of materials such as cotton,

coated or

may 130°C

enables electrical manufacturers to foresee

their

usually

is

It

CLASSES OF INSULATION SYSTEMS

thermal

1

.

the worst ambient temperature conditions that

Standards organizations have also established a

equipment insulated with

105°C. Note that

life

life at

in this class if

expectancy of 20 000 h

a class

A

this classification

insulation system

assumes


tests

temperatures above 240°C. to

40 000

h at the stated temperature. This

would probably

that the insulation

system

last for is

2 to 5 years

if

oper-

not in contact with corro-

humid, or dusty atmospheres.

For a complete explanation of insulation classes, insulation systems, and temperature indices, see the

127

companion IEEE Standards Publications Nos. 96. 97, 98, 99, and 101. See also IEEE Std

Underwriters Laboratories publication on insulation systems

UL

1446, 1978.

1

IEEE

Std 1-1969 and

17-1974 and


128

The temperature of a machine point, but there are places

warmer

motor, relay, and so forth, he intends to put on the

varies from point to

where the temperature

market. Thus, for class

is

than anywhere else. This hottest-spot tem-

perature must not exceed the

maximum

To show how

allowable

and

Figure 6.7 shows the hot-spot temperature limits F.

and

H

insulation (curve

1).

built a 10

test the

They

are the temperature limits previously mentioned

maximum

insulation, the is

(130

-

40)

kW motor using class B

motor he places

it

in

temperature of 40°C and loads

in

10

The maximum ambient temperature of 40°C is also shown (curve 3). The temperature difference between curve and curve 3 gives the maxSection 6.7.

kW

insulation.

it

up

until

This limiting temperature

delivers

detectors, located at strategic points inside the ma-

chine, record the temperature of the windings. After the temperatures have stabilized (which

rise

to establish the physical size

may

take

is

called the hot-spot temperature. If the hot-

Class

say

is,

H

180°C

Class F

155°C

165°C

r / Class

/

© ©

105°C

//

//

B

130°C

A

/

J / / j

145°C

/

hot-spot

120°C

temperature rise by

//

J/

J 100°C

embedded average temperature by the

thermocouple rise

resistance method

©

40°C

40°C limiting

ambient temperature

Figure 6.7 Typical limits of

Shows Shows Shows

the the

noted, and

of the spot temperature so recorded

Class

is

enables the this

manufacturer

(3)

it

permissible temperature rise for each insula-

tion class.

(2)

To

of mechanical power. Special temperature

several hours), the hottest temperature

(1)

90°C.

a constant ambient

1

imum

=

the temperature rise affects the size

of a machine, suppose a manufacturer has designed

temperature of the particular class of insulation used.

for class A, B,

B

allowable temperature rise

some dc and ac industrial machines, according to the insulation class: maximum permissible temperature of the insulation to obtain a reasonable maximum permissible temperature using the resistance method

the limiting ambient temperature

service

life

147°C, the


manufacturer would not be permitted to

The reason

product.

-

(147°

=

40°)

On

B

the hottest-spot tempera-

if

is

(

100°

The manufacturer immediately

6()°C.

ceives that he can still

per-

make

rise limits.

For instance, he can reduce the conducthe hot-spot temperature rise

90°C. Obviously,

close to

this

very

is

reduces the weight

and cost of the windings. But the manufacturer also realizes that the

him

ables turn,

reduces the amount of iron.

ing the

with a

now

reduced conductor size

reduce the size of the

to

By

en-

thus redesign-

motor, the manufacturer ultimately ends up

machine

The hot-spot temperature measure because

that operates within the permissible

it

rise is rather difficult to

has to be taken

at

the very inside

of a winding. This can be done by embedding a small temperature detector, such as a thermocouple or thermistor. However, this direct method of measuring hot-spot temperature

is

costly,

and

only

is

justified for larger machines.

To simplify

This, in

slots.

Temperature rise by the resistance method

6,9

per-

remain within the permissible temperature

tor size until

the size

product.

—

more economical design

a

The manufacturer could reduce

standards.

of the motor and thereby market a more competitive

insulation.

only 100°C, the temperature rise

=

and

for class

the other hand,

ture is

40°)

maximum

1()7°C exceeds the

90°C

missible rise of

his

sell

that the temperature rise

is

129

matters, accepted standards permit a

second method of determining temperature

rise.

It

is

based upon the average winding temperature, measured by resistance, rather than the hot-spot temper-

The maximum allowable average winding

ature.

temperature rise limits and has the smallest possible

temperatures for the various insulation classes are physical size, as well as lowest cost. In practice,

it

is

shown

in

not convenient to carry out per-

formance of 4()°C.

tests in a

is

usually loaded to

rated ca-

its

much lower (and more comfortable) am-

pacity in

bient temperatures.

by

tablished testing

controlled ambient temperature

The motor

Toward

this end,

it

bodies

for

that,

purposes, the ambient temperature

ture rise

is

recorded as before.

under these conditions

90°C

than

is

B

(for class

allowed to

sell his

temperature to

standards-setting

may

is

If the

tempera-

insulation), the manufacturer

product.

is

assumed

in the

to

correspond to a hot-spot

-

rise

of ( 120°

-

40°)

- 80°C

40°)

=

is

rise

assumed of ( 30° 1

90°C.

The average temperature of by the resistance method.

It

a

winding

is

A

75

kW

known winding temagain when the machine is hot. For example, if the winding is made of copper, we can use the following equation (dethe

winding resistance

at a

perature, and measuring

it

F,

operates If

U =

(125°

(6.5)

The motor

F

234 =

= 93°C

permissible hot-spot tem-

insulation

average temperature of the winding

when

rise is

32°)

to Fig. 6.7, the

perature rise for class

15°C.

- 234

where

Solution

1

/,)

125°C, does the motor meet the

The hot-spot temperature

=

U = ^(234 +

the hot-

mpe rat Lire stand ard s ?

According

av-

at full-

R,

ambient temperature of 32°C.

spot temperature is te

its

erage temperature:

motor, insulated class

load in an

found

consists of measuring

rived from Eqs. 6.1 and 6.2) to determine

Example 6-3

case

average winding tempera-

correspond to a hot-spot temperature

lie

equal to or less

120°C

For example,

2, Fig. 6.7.

insulation, an

temperature of 130°C. Consequently, an average

has been es-

anywhere between 10°C and 40°C. The hottest-spot temperature

ture of

curve

B

of class

is

(155°

-

40°)

easily meets the temperature

R2 = R = ]

t\

—

hot f°C]

a constant equal to 1/a

=

1/0.004 27

hot resistance of the winding

cold resistance of the winding

temperature of the winding [°ci

when cold

1


30

Knowing

winding temperature by the

the hot

we can immediately

sistance method,

Alternatively,

re-

insulation.

calculate the

temperature rise

If this

within the permissible limit (80°C for class lation), the

product

is

B

falls

insu-

acceptable from a standards

when performance

point of view. Note that

tests are

carried out using the resistance method, the ambient

temperature must again If the

lie

winding happens

wire, Eq. 6.3 can

must be changed

between 10°C and 40°C,

to be

be used,

still

made of aluminum but the number 234

to 228.

A dc

motor

be rewound using class F

a very last resort,

A

word of

final

may

caution: temperature rise stan-

also on the type of apparatus (motor, transformer, relay, etc.), the type of construction (drip-proof, totally

enclosed,

etc.),

and the

of application of

field

the apparatus (commercial, industrial, naval,

be consulted before conducting a heat-run

machine or device

test

been

idle for several is

days

found

to

in

an

il. The motor then opwhen temperatures have staresistance is found to be 30 il. The

maximum

Although

bilized, the field

basic physical size depends

corresponding ambient temperature built with class

B

is

24°C.

If

the

allowable temperature rise

power

shown

in Fig. 6.8.

Suppose we have

to build

another generator having the same power and the winding, at full-

To generate

dards

we

either

have

the to

same voltage

at

half the speed,

double the number of conductors

on the armature or double the flux from the Consequently,

we must

practice,

The average temperature of the shunt full-load t2

increase both.

We

conclude that for a

(R 2 /R

i

)

true for both ac

- 234 + 19) - 234

+

(234

(30/22) (234

/,)

The average temperature

its

rise at full-load is

c.

- 24° = 87°C. The maximum allowable

temperature

resistance for class

B

insulation

is

rise

(120°

-

kW, 2000 r/min motor

by 40°)

80°C. Consequently, the motor does not

meet the standards. Either

its

rating will

have

be reduced, or the cooling system improved, can be put on the market.

is

machine depends uniquely

torque. Thus, a 100

111°

it

always

and dc machines.

Basically, the size of a

upon

b.

before

is

bigger than a high-speed machine (Fig. 6.9). This

is

= =

we

given power output, a low-speed machine

field at

= 111°C

=

poles.

either increase the size of

the armature, or increase the size of the poles. In

Solution a.

volt-

age, but running at half the speed.

load

The full-load temperature rise by the resistance method Whether the motor meets the temperature stan-

its

upon the power and

Consider the 100 kW, 250 V, 2000 r/min generator

The average temperature of

es-

rating of a machine,

speed of rotation.

insulation, calculate the

following:

c.

a

and size of a machine

have a

tablishes the nominal

b.

on

(Fig. 6.10).

shunt-field resistance of 22

a.

etc.).

Consequently, the pertinent standards must always

erates at full-load and,

is

size

6.10 Relationship between the speed

that has

ambient temperature of 19°C,

motor

its

dards depend not only on the class of insulation, but

specific

Example 6-4

may

have to be increased.

corresponding temperature rise by subtracting the

ambient temperature.

As

it

to

Figure 6.8 100 kW, 2000 r/min motor; mass: 300

kg.


// 1 te

nn eel la te

A

6-9

1

3

level

V

dc motor connected to a 240

line pro-

duces a mechanical output of 160 hp.

Knowing 6-10

A

1

V

5

1

kW,

that the losses are 12

late the input

power and

calcu-

the line current.

dc generator delivers

20

1

A

to a

load. If the generator has an efficiency of

power

81 percent, calculate the mechanical

needed 6-

Figure 6.9

1

1

it

[hp].

Calculate the full-load current of a 250 hp,

230

100 kW, 1000 r/min motor; mass: 500 kg.

to drive

V

dc motor having an efficiency of 92

percent.

6-12

same physical size as a 10 kW motor 200 r/min because they develop the same

A machine

having class B insulation attains

has about the

a temperature of

running at

torrid

torque.

a.

Low-speed motors are therefore much more costly than high-speed for

b.

it

is

often cheaper to use a small

6-

1

3

high-speed motor with a gear box than to use a large

low-speed motor directly coupled to

its

load.

Practical level

6-15

What causes

the losses in a dc motor. iron losses

Explain

and how can they

why

the temperature of a

What determines

If

the

power

electric

tor in

ma-

the vents in a motor,

its

by

always low of

6-16

it

delivers an output of

motor driving

m

a skip hoist with-

of minerals from a

1.5 metric tons

deep every 30 seconds.

Thermocouples

power must be reduced. Explain.

per-

the

mo-

in kilowatts.

are used to

measure the

winding temperature of insulated class

runs at full-load, what

If the

94

power output of

horsepower and

kW ac motor,

out-

nomi-

its

hp.

An

cent, calculate the

machine

rating of a

so,

if

Calculate the efficiency of the motor in

nal hot-spot

we cover up

put

is

hoist has an overall efficiency of

chine? 6-5

the temperature rise?

efficiency of a motor

trench 20

increases as the load increases. 6-4

machine running too hot and,

draws

be reduced? 6-3

Is the

The

1

6-2

8()°C.

is

Example 6-2 when

Questions and Problems Name

1

What

when it operates at 10 percent nal power rating. Explain. 6-14

6-1

resistance) in a

how much?

motors of equal power. Consequently,

low-speed drives,

208°C (by

ambient temperature of

is

the

F. If

a

inter-

1200

the motor

maximum

tem-

perature these detectors should indicate in an 6-6

If a

motor operates

may we

load

it

in

above

a cold environment, its

rated

ambient temperature of 40°C? 3()°C? 14°C?

power? 6-17

Why? 6-7

Name some

A

60 hp ac motor with class F insulation

has a cold winding resistance of 12 of the factors that contribute to

23°C.

the deterioration of organic insulators.

When

it

runs

at rated

load

in

11 at

an am-

bient temperature of 3 °C, the hot winding 1

6-8

A motor is built with class H insulation. What maximum hot-spot temperature can withstand?

resistance it

is

found

to be 17.4 il.

a.

Calculate the hot winding temperature.

b.

Calculate the temperature

rise

of the motor.

1


132

Could the manufacturer increase

c.

plate rating of the

6-18

An

motor has a normal

electric

when

eight years is

30°C.

If

it

is

6-24

life

of

is

service

A No.

1

6-25

of the

life

Knowing

The current density [A/mm

b.

The

specific copper losses

is

the conductor

No. 4

A/mm

at

Express the current density

is

its

in the

rise

its

is

losses.

6-27

On

the

reasonably

is

is

tion

1

is

What

is its

rewound using

F

ft

cable

dc circuit to carry

48 A. Assuming a

The power

maximum

loss, in watts, in the

The approximate voltage

6-28

In

Problem 6-27,

if

2-conductor

at the

243

V when

rying a current of 60 A, what

insu-

load end is

if

V.

the voltage drop in the

cable must not exceed 10

motor having class B insula-

Assume

would normally have a service life of h, provided the winding tempera-

of 70°C.

by resistance does not exceed 120°C.

By how many hours duced

RW 75. A 420 V

a max-

No. 6 gauge cop-

the voltage at the service panel

expected

class

Code allows

in a

cable

situ-

20 000 ture

in

it is

car-

minimum

conductor size would you recommend?

kW ac

1

that

the following: a.

lation?

An

found

operating temperature of 70°C, calculate

deliver at a temperature rise

of two years.

A

being used on a 240

b.

it

By

25°C.

at

it is

AWG wire size, and

Electrical

current of 65

a current of

6.5).

ated in a particularly hot location has a ser-

if

ohms

of 56

per conductor, type

nominal rating

its

electromagnet (insulated class A)

span

27 meters long.

is

of a 4-pole dc motor has a

Determine the

The National

imum

Based on these facts, if a 20 kW motor has a fullload temperature rise of 80°C, what

life

field

calculate the weight of the wire per pole,

range between 50 percent

example, Fig.

it

AWG that

The shunt

inches.

in circular mils

of a motor

efficiency

and 150 percent of

power can of 105°C?

6.2, calculate the resistance un-

the bare copper wire has a diameter of 0.04

120°C, calcu-

fW/kg].

roughly proportional to

(see, for

be as high as 70°C.

kilograms.

The temperature other hand,

AWG conductor in

scraping off the insulation,

b.

constant

may

total resistance

a cur-

2

conductor temperature

it is

der these conditions of a 2-conductor cable

]

per ampere.

6-23

the prop-

lists

of commercially available copper

an area where the operating temperature of

105°C, cal-

conductor operates

If the

life

pounds.

in

AX3

Appendix

proposed to use a No. 4

[W/kgl

a.

vice

a temper-

at

that the

6-26

An aluminum

An

ohms

35

long

level

late the specific losses

6-22

in

Using Eq.

a.

rent density of 2

1

1

conductors. In an electrical installation,

temperature of the conductor

6-2

The Table erties

culate the following:

6-20

0.

weight of the conductor

60°C,

210m

0 round copper wire

carries a current of 12 A.

Advanced

of No. 2/0 single copper conductor

reel

ature of 25°C. Calculate the approximate

motor? 6-19

A

has a resistance of

installed in a location

new probable

the

is



where the ambient temperature what

name-

the

motor? Explain.

if

the

is

motor runs

ature (by resistance) of

the service life re-

6-29

A dc

a

maximum

operating temperature

busbar 4 inches wide, 1/4 inch thick,

and 30

feet long carries a current of

2500 A. Calculate

the voltage drop

for 3 h at a temper-

temperature of the busbar

200°C?

the

power

loss per

meter?

is

1()5°C.

if

the

What

is


6-30

Equation 6.3 gives the resistance/temperature relationship of

brush pressure:

copper conductors,

1

33

1.5 lbf

brush contact drop:

1

.2

V

namely, coefficient of friction: 0.2 t2

= R 2 IR

(234

X

+

t x

)

- 234 Calculate the following:

Using the information given

AX2, deduce

minum 6-3

1

in

Appendix

a.

a similar equation for alu-

b.

conductors.

c.

The commutator of a .5 hp, 2-pole, 3000 r/min dc motor has a diameter of 63 mm.

the contact voltage drop

1

d.

Calculate the peripheral speed in feet per

minute and 6-32

in

miles per hour.

e.

The following information is given on the brushes used in the motor of Problem 6-3

newtons)

number of brushes: 2 1

5

3/4 in long.

resistivity

The

frictional

brushes

A

(The 5/16

in

when

energy expended by the two the

commutator makes one revo-

lution (in joules)

brush dimensions: 5/8 in wide, 5/16 in thick,

contact with the

The total electrical power loss (in watts) due to the two brushes The frictional force of one brush rubbing against the commutator surface (in lbf and in

1

f.

current per brush:

The resistance of the brush body in ohms The voltage drop in the brush body The total voltage drop in one brush, including

X

5/8 in area

commutator)

of brush: 0.0016 Il.in

g.

is in

h.

The power loss due of 3000 r/min The total brush loss motor rating

to friction,

given the speed

as a percent of the

1

.5

hp

Chapter 7 Active, Reactive,

and Apparent Power

7.0 Introduction

The

7.1

The instantaneous power supplied

concept of active, reactive, and apparent

power plays

a major role in electric

power

Instantaneous power

nology. In effect, the transmission of electrical energy

across

and the behavior of ac machines are often easier

that

to

encouraged

to

The terms

The reader

pay particular attention active, reactive,

is

which

the

positive value

always expressed

is

means

may be that

in

The

A

positive or negative.

power flows

into the de-

vice. Conversely, a negative value indicates that

in

power

is

flowing out of the device.

to describe transient-state be-

we apply them to dc circuits. Our study begins with an analysis of the instantaneous power in an ac circuit. We then go on to define the meaning of active and reactive power and how to identify sources and loads. This is followed by a definition of apparent power, power factor, and the power triangle. We then show how ac circuits can be solved using these power concepts. In conhavior, nor can

clusion, vector notation

active and reactive

it.

instantaneous power

to this chapter.

and apparent power

voltages and currents are sinusoidal.

They cannot be used

is

terminals times the instantaneous current

watts, irrespective of the type of circuit used.

therefore

apply to steady-state alternating current circuits

its

flows through

Instantaneous power

understand by working with power, rather than dealing with voltages and currents.

to a device

simply the product of the instantaneous voltage

tech-

power

is

in

Example

A

sinusoidal voltage having a peak value of 162

and a frequency of 60 Hz

is

V

applied to the terminals

of an ac motor. The resulting current has a peak value of 7.5 a.

A and

b.

134

in

terms of the

.

Calculate the value of the instantaneous voltage

and current

circuit.

lags 50° behind the voltage.

Express the voltage and current electrical angle

used to determine the an ac

7-1

at

an angle of 120°.

ACTIVE, REACTIVE,

c.

Calculate the value of the instantaneous

p =

power

ei

=

AND APPARENT POWER X

140.3

= + 989

7.05

1

35

W

at 120°. d.

Plot the

Because the power

curve of the instantaneous power deliv-

positive,

is

flows

it

at this

instant into the motor,

ered to the motor. d.

a.

In order to plot the

power,

Solution

Let us

assume

that the voltage starts at zero

We

increases positively with time.

and

can therefore

we

for angles ranging

Table

7A

curve of instantaneous

repeat procedures (b) and (c) above

lists

from

(J)

=

0

to

=

360°.

part of the data used.

write

e

= Em

sin

§ =

1

62

sin

TABLE 7A (f>

= /

b.

At

50°, consequently,

= /m

(J)

e

/

c.

=

sin

(<(>

-

G)

=

162 sin 120°

=

140.3

=

7.5 sin (120° 7.5

=

7.05

Voltage

7.5 sin

(c|>

162 sin

-

degrees

50°)

=

X

162

-

50°)

=

0.866

7.5 sin 70°

0.94

A

The instantaneous power

at

120°

p

USED TO PLOT

in

an ac

circuit.

(]j

Power

Current 7.5 sin

-

volts

amperes

50°)

P watts

0

-5.75

0

-3.17

-218

50

124.1

0

75

156.5

3.17

497

0

115

146.8

6.8

1000

155

68.5

7.25

497

180

0

5.75

0

3.17

-218

205

-68.5

230

-124.1

is

power

AND

68.5

Figure 7.1 Instantaneous voltage, current and

i,

25

0

V

X

Angle

can write

we have

120°

=

=

we

e,

FIG. 7.1

The current lags behind the voltage by an angle 0

VALUES OF

(See Example

7-1

.)

0

0


136

The of

power

voltage, current, and instantaneous

The power

plotted in Fig. 7.1.

attains a positive

+ 000 W and a negative peak of - 2 1

ative

power means

that

power is

1

are

peak

W. The neg-

8

actually flowing

0-50°,

180° -230°,

360° -410°.

and

Although a power flow from a device considered

to

be impossible,

it

happens often

in

We

is

given

The simple

to be

20

power cycle

may seem The

the sections that follow.

quency of

s.

120 Hz, which

the voltage

and current

is

to an ac generator.

and as we would expect

E and

/

The

line

E and

a reading

To

/

are effective values.

the instantaneous values of voltage

did

power *

con-

the sinusoidal curves of

The peak values are and V2/ amperes because, as

(Fig. 7.2d).

^2E volts ously, E and we

we

will give

watts (Fig. 7.2c).

we have drawn

circuit, /

it

get a better picture of what goes on in such a

produce the and

respec-

/,

are in phase (Fig. 7.2b). If

P = El

re-

effective

in a resistive circuit,

nect a wattmeter (Fig. 7.3) into the line,

twice the fre-

that

quite normal: the

voltage and current are designated

phasors

This means that the frequency of

is

is

always twice the

ac circuit of Fig. 7.2a consists of a

connected

ac circuits.

also note that the positive peaks occur at in-

tervals of 1/1

the

in

is

power*

7.2 Active

tively,

reason

phenomenon

frequency.

sistor

a load to a device considered to be a source

this

from

the load (motor) to the source. This occurs during the intervals

power. Again,

frequency of ac power flow

in

Section 7.1,

we

E

respectively stated previ-

By multiplying and current

as

obtain the instantaneous

in watts.

Many persons refer to active power as real power or true power considering it to be more descriptive. In this book we use the term active power, because conforms to the IEEE designation. it

Figure 7.2 a.

An ac voltage E produces an ac current

/

in this re-

Figure 7.3 Example of a high-precision wattmeter rated 50 V, 200 V; 1 A, 5 A. The scale ranges from 0-50

sistive circuit.

c.

Phasors E and / are in phase. A wattmeter indicates El watts.

d.

The

b.

tive

active

power

power pulses.

is

composed

V,

100

W to

of

a series of posi-

1000W. (Courtesy of Weston Instruments)

0-

ACTIVE, REACTIVE,

The power wave consists of a

from zero

pulses that vary

to a

= 2EI = 2P

value of

The

fact that

X

(V2/)

power

is

always positive reveals

watts.

that

always

it

flows from the generator to the resistor. This

pow

properties of what

although

er:

imum,

it

137

series of positive

maximum

(i2E)

of the basic

AND APPARENT POWER

is

is

one

called active

pulsates between zero and

max-

never changes direction. The direction of

it

power flow

is

shown by an arrow P (Fig. 7.2c). is clearly midway between

The average power IP and zero, and so That

watts.

is

=

EI

power indicated by

the

value

its

precisely the

P =

is

2E//2

wattmeter.

The two conductors leading

to the resistor in Fig.

2EJ

power. However, unlike cur-

7.2a carry the active rent flow,

power does

not flow

down one conductor

Power flows over both conconsequently, as far as power is con-

and return by the other. ductors and, cerned, line,

In

as

we can replace the conductors by shown in Fig. 7.2c.

general, the line represents any transmission

number of conductors

The generator an active load.

and the unit

megawatt

is

is

it

may

/

Reactive power consists of a series of positive and

lags 90° behind

E.

are frequently used multiples of

back and forth in this manner is power (symbol Q), to distinguish it from the unidirectional active power mentioned before. The reactive power in Fig. 7.4 is also given by the product EL However, to distinguish this power from active power, another unit is used the van Its

identical to the resistive

is

7.2a) except that the resistor

by a reactor X

E

.

x

As

is

a result, current

/

now

re-

therefore again

is

twice the line frequency.

Power

that surges

drawn the waveforms for

in

such a

E and

/

power

circuit,

we

and, by again

we

obtain

(Fig. 7.4c).

This

power p consists of a series of identical positive and negative pulses.

instantaneous the reactor

The

positive

waves correspond

power delivered by

the generator to

and the negative waves represent instan-

power delivered from

generator.

to

the reactor to the

The duration of each wave corresponds The

one quarter of a cycle of the line frequency.

megavar (Mvar).

Special instruments, called vanneters, are available to measure the reactive

multiplying their instantaneous values,

curve of instantaneous

—

multiples are the kilovar (kvar) and

lags 90°

(Fig. 7.4b).

To see what really goes on

to

Phasor

frequency of the power wave

behind the voltage

taneous

b.

c.

called reactive

circuit (Fig.

the

this in-

The symbol for active power is P The kilowatt (kW) and

The circuit of Fig. 7.4a

have

/ in

negative power pulses.

Reactive power

placed

current

circuit.

have.

the watt.

7.3

An ac voltage E produces an ac ductive

an active source and the resistor

the watt (W).

(MW)

Figure 7.4 a.

connecting two devices, irrespective of the

line

is

a single

(C)

7.5).

A

line voltage sin 8

if

E

(where B

reading

power

in

a circuit (Fig.

varmeter registers the product of the effective

is

times the effective line current is

the phase angle

only obtained

between

when E and

/

/

times

E and

/).

A

are out of phase;

they are exactly in phase (or exactly 180° out of

phase), the varmeter reads zero.

Returning pulse

is

to Fig. 7.4, the dotted area

under each

the energy, in joules, transported in

direction or the other. Clearly, the energy

is

one

deliv-

ered in a continuous series of pulses of very short duration, every positive pulse being followed by a


138

By

definition*, a reactor

is

considered to be

a re-

active load that absorbs reactive power.

Example

A

7-2

reactor having an inductive reactance of 4

connected

of a 120

to the terminals

V

0

is

ac generator

(Fig. 7.6a). a.

Calculate the value of the current in the reactor

b.

Calculate the power associated with the reactor

c.

Calculate the power associated with the ac

generator d.

Figure 7.5 Varmeter with a zero-center scale. or negative reactive power flow up

It

to

Draw

the phasor

diagram for the

circuit

indicates positive

100 Mvars. ,

4

Q

30 A|

negative pulse. The energy flows back and forth

between the generator and the inductor without ever being used up.

What tive

is

A

the reason for these positive

and nega-

,

energy surges? The energy flows back and forth

because magnetic energy

+

rr^>

120V

alternately being stored

is

3.6 kvar

positive, the magnetic field

is

inside the coil.

A moment

is

when

later

L

l

4J

30 A

up and released by the reactor. Thus, when the

power

/

building up

the

power

negative, the energy in the magnetic field

is

is

E

de-

120

V

creasing and flowing back to the source.

We now tive

have an explanation for the brief nega-

power pulses

in Fig. 7.1. In effect,

they repre-

30 A

(c)

sent magnetic energy, previously stored up in the

motor windings,

that

is

being returned to the source.

Figure 7.6

See Example

7.4 Definition of a reactive load

7.2.

and

reactive source

Solution a.

Current

in

the circuit:

Reactive power involves real power that oscillates

back and forth between two devices over a transmission

line.

Consequently,

is

it

whether the power originates

at

/,=

impossible to say

one end of the

line

assume

that

b.

or the other. Nevertheless,

some devices generate absorb

it.

In other

it is

useful to

reactive

power while others

E = XL

120

4

Power associated with

Q=

EI

=

120

V

= 30 A

f1

the reactor:

X 30 - 3600

var

=

3.6 kvar

words, some devices behave like

reactive sources and others like reactive loads.

This definition

is in

agreement with IEEE and 1EC conventions.

ACTIVE, REACTIVE,

power

This reactive

is

AND APPARENT POWER

equal to the current

is

Because the reactor absorbs 3.6 kvar of reactance

times the voltage across

ries

power, the ac generator must be supplying

Consequently, the generator power:

delivers 3.6 kvar.

it

The

flows therefore in the direction

reactive

shown

power

Q

The

=

EIC

120

reactive

pressed

car-

it

namely

V X

A=

30

power delivered by

in vars

=

3600 var

3.6 kvar

the capacitor

power

or kilovars. Reactive

is

ex-

Q now

flows from the capacitor to the reactor.

We

This phasor

diagram applies

(the reactor)

and the reactive source (the ac genera-

tor) as

Q =

(Fig. 7.6b).

The phasor diagram is shown in Fig. 7.6c. Current // lags 90° behind voltage E.

d.

terminals,

its

it.

a source of reactive

is

39

source of reactive power. The reactive power deliv-

absorbed by the reactor.

ered by the capacitor c.

1

to the reactive load

have arrived

very important conclusion:

at a

a source of reactive power. a reactive power source whenever it is a capacitor

is

It

acts as

part of a

sine-wave-based, steady-state circuit.

well as the line connecting them.

Let us take another step and remove the reactor

7.5

The capacitor and reactive power

from the

circuit in Fig. 7.7a, yielding the circuit of

Fig. 7.8a.

now

Suppose

that

we add

a capacitor having a reac-

tance of 4 11 to the circuit of Fig. 7.6. This yields the circuit

of Fig. 7.7a.

pacitor

is /c

expect,

it

=

The

current

120 V/4 il

=

30

/L

.

drawn by

A and,

as

the ca-

we would

leads the voltage by 90° (Fig. 7.7b).

rent of

capacitor

is

now

alone, connected to It still

30 A, leading the voltage

7.8b). Consequently, the capacitor

E

carries a cur-

by 90°

still

this

power go? The answer power

delivers reactive

is

to the

(Fig.

acts as a

source of reactive power, delivering 3.6 kvar.

does

The vector sum of I L and / c is zero and so the ac is no longer supplying any power at all to

The

the terminals of the ac generator.

Where

that the capacitor

very generator to

generator

the circuit.

not

30

However, the current

changed; consequently,

AX

1

Where can only

V =

20 is

it

in the reactor

has

continues to absorb

3.6 kvar of reactive power.

power coming from? capacitor, which acts as

this reactive

come from

the

c

It

3.6 kvar

a

-4/

30 A

(a)

'c

30 A

(b)

120 V varmeter

30 A

(c)

LTi capacitor

120 V

(b)

Q

= FJ C

Figure 7.8 30 A

Figure 7.7

See Example 7.3.

a.

Capacitor connected to an ac source.

b.

Phasor

c.

Reactive power flows from the capacitor to the

l

c

generator.

leads

Eby

90°.

1


40

which

it

is

connected! For most people, this takes a

time to accept.

little

How, we may

ask, can a passive

device like a capacitor possibly produce any power?

The answer energy

is

that reactive

power

really represents

a pendulum, swings back and forth

that, like

without ever doing any useful work. The capacitor acts as a

temporary energy-storing device repeatedly

accepting energy for brief periods and releasing

as a reactor does, a capacitor stores electrostatic en-

we connect

7.8c), var,

it

that reactive

from the capacitor

power

is

to the generator.

El

= -3600

The generator

X

= 400

3.5

1

prefer to call

a receiver of reactive power, which, of

same

thing. In

cerned the generator acts as a load.

The

active

power absorbed by

active

power =

1

circuit

W)

power (1304 power (812 var). tive

7.9).

The

power associated with

L,

C

respective elements

is

a basic difference

remember

is

between active and

most important

re-

thing

one cannot be converted

and reactive powers function

independently of each other and, consequently, in electric

a

burden on the transmission

line that

whereas active power eventually

3./

produces a tangible result (heat, mechanical power, light, etc.), reactive

14

A

}=

A

that oscillates

4

-3.5./

power only represents power

back and

forth.

All ac inductive devices such as magnets, trans-

Q

formers, ballasts, and induction motors, absorb

20 A

re-

power because one component of the current they draw lags 90° behind the voltage. The reactive power plays a very important role because it proactive

Figure 7.9

duces the ac magnetic field

See Example

a source of ac-

they can be treated as separate quantities

carries them, but,

-+*

16.12

that the

into the other. Active

Both place

-

is

and a receiver of reactive

circuits.

the generator.

2Q 1=3

a source of

between active and reactive power

There

carry the currents shown. Calculate the active and reactive

is

7.6 Distinction

to

connected to a group of R,

elements (Fig.

it

304 W.

active power, and perhaps the

is

the resistors must

be supplied by the generator; hence

summary, a ca-

pacitive reactance always generates reactive power.

Example 7-3 An ac generator G

var

is

we sometimes

like reactive load, but

course, amounts to the

2

indeed flowing

now behaving it

20

In conclusion, the ac generator

a varmeter into the circuit (Fig.

will give a negative reading of

showing

XC =

1

ergy (see Section 2.14). If

3.5 il capacitor generates reactive power: 2

1

The R, L, C circuit generates a net reactive power of 400 - 588 = 8 12" var This reactive power must be absorbed by the generator; hence, as far as reactive power is con-

it

However, instead of storing magnetic energy

again.

The

Qc =

7.3.

A building,

in

these devices.

shopping center, or

city

may

be con-

sidered to be a huge active/reactive load connected

an electric

Solution

to

The two

tain

resistors absorb active

P =

2

1

2

power given by

+ (16.12 2 X 784 + 520 = 304 W

R =

(14

X

4)

1

The

3 II reactor absorbs reactive

Q L = 1% =

l4

2

X

3

=

588 var

power:

2)

=

utility

system. Such load centers con-

thousands of induction motors and other

elec-

tromagnetic devices that draw both reactive power (to sustain their (to

magnetic fields) and active power

do the useful work). This leads us to the study of loads that absorb

both active and reactive power.

AND APPARENT POWER

ACTIVE, REACTIVE,

7.7

Combined

and reactive loads: apparent power

The

active

flow

in the

the arrows in Fig.

P

Loads that absorb both active power

power tance ple,

Q may

be considered to be

and reactive

made up of a

the circuit

actor are

of Fig.

7.

1

Oa

in

which a

resistor

and

p

,

while the reactor draws a current

/ q

According to our definitions, the resistor active load

while the reactor

Consequently, behind.

I

p

is in

is

phase with

The phasor diagram

an

a reactive load.

E while

(Fig. 7.

the resultant line current / lags

.

is

1

/t] lags

90°

Ob) shows that

behind

Furthermore, the magnitude of

E

/ is

and a varmeter

by an an-

7.

into the circuit, the readings will

both be positive, indicating

£/q

re-

connected to a source G. The resistor draws

a current /

gle 6.

resis-

and an inductive reactance. Consider, for exam-

power components P same direction, as shown by 10c. If we connect a wattmeter

active and reactive

Q both

and

141

P = £V p

Q=

watts and

vans, respectively.

Furthermore,

if

we connect

an ammeter into the

will indicate a current of /

line,

it

sult,

we

amperes. As a

plied to the load

is

equal to EI watts. But

viously incorrect because the

an active component (watts)

is

nent (vars). For this reason the product EI

Apparent power

is

ob-

this is

composed of and a reactive compopower

called

is

apparent power. The symbol for apparent power

given by

re-

power sup-

are inclined to believe that the

expressed neither

in

is

5*.

watts nor

but in voltamperes. Multiples are the kilo-

in vars,

voltampere (kVA) and megavoltampere (MVA).

7.8 Relationship

between P Q, 3

and S Consider the single-phase circuit of Fig.

7.

1

1

a

com-

source]

posed of a source, a load, and appropriate meters. Let us assume that

(b)

E

•

the voltmeter indicates

•

the

•

the wattmeter indicates

•

the varmeter indicates

ammeter

Knowing

indicates /

that

volts

amperes

+P watts

+Q vars

P and Q are

positive,

we know

that

the load absorbs both active and reactive power.

Consequently, the line current an angle

Current nents

I

p

/ lags

behind

E. lb

by

8.

/

and

rature, with

values of

/

can be decomposed into two compo/C]

,

respectively in phase, and

phasor

and

/ L|

E

(Fig. 7.1 lb).

in

quad-

The numerical

can be found directly from the

in-

strument readings / l

Figure 7.10 a.

Circuit consisting of

a source feeding an active and

reactive load.

Phasor diagram of the voltage and currents.

c.

Active

and reactive power flow from source

to load.

q

= PIE = QIE

(7.1) (7.2)

Furthermore, the apparent power 5 transmitted over the line

b.

P

is

given by S

El,

I

=

from which

S/E

(7.3)

1

ELECTRICAL MACHINES AND TRANS EORMERS

42

Figure 7.11 Instruments used to measure £, P, and Q in a circuit. The phasor diagram can be deduced from the instrument readings.

a.

/,

b.

Referring to the phasor diagram (Fig. 7.1 lb),

it

is

Example

7-5

A wattmeter and varmeter are connected into a

obvious that r-

2

=

/

p

single-phase line that feeds an ac motor.

2

+

/

q

spectively indicate

1

W and 960

800

20 V

1

They

re-

var.

Consequently, Calculate ->

P E

s

E That

2

Q +

The in-phase and quadrature components

a.

E

and

c.

in

0

2

(7.4)

d.

line current

which

= P = Q= S

We

Solution

apparent power [VA]

Referring to Fig.

power [Wl reactive power [var] active

tor,

can also calculate the value of the angle 0 be-

cause the tangent of 6

Thus,

is

obviously equal to

/ //

p

/

a.

/

.

b.

p q

=

arctan / // q p

=

arctan

QiP

1

1

,

where the load

is

now

= PIE =

1

= QIE =

960/120

800/120

= -

15

8

A

A

From

the phasor

= V7 2 +

Example 7-4 alternating-current motor absorbs

40

diagram we have

(7.5) /

tive

7.

/

2

= Vl5 2 +

8

2

= 17A

kW of ac-

power and 30 kvar of reactive power. Calculate power supplied to the motor.

c.

The apparent power

is

the apparent

S =

EI=

120

X

17

= 2040 VA

Solution

S

= VP^Vq 2 ~

V40 r + 30 2

= 50 kVA

d.

a

mo-

we have

we have 0

An

/

q

The line current / The apparent power supplied by the source The phase angle between the line voltage and

b. is,

2 2 S = P +

/

The phase angle 6 between E and

/ is

(7.4)

6

=

arctan

=

28.1°

QIP =

arctan 960/1800

(7.1) (7.2)

ACTIVE, REACTIVE,

The power

Example 7-6

A voltmeter and ammeter connected tive circuit

into the induc-

of Fig. 7.4a give readings of 140

V

and

AND APPARENT POWER

factor of a resistor

cause the apparent power tive

power.

On

it

equal to the ac-

is

the other hand, the

an ideal coil having no resistance

20 A, respectively.

100 percent be-

is

draws

143

power

factor of

zero, because

is

it

does not consume any active power. Calculate

To sum

The apparent power of the load The reactive power of the load

a.

b.

is

parent

The active power of the load

c.

up, the

power factor of a circuit or device

simply a way of stating what fraction of

power

is real,

In a single-phase circuit the

The apparent power

S

and current. Thus, referring

is

Q =

factor

also

is

to Fig. 7.11,

PIS

=

El p IEI

is

=

v

=

=

cos B

El

= 2800

would give a reading of 2800

factor

X 20

140 var

power

=

2.8 kvar

Consequently,

vanneter were connected into the circuit,

If a

power

=

= EI= 140 X 20 = 2800 VA = 2.8 kVA

The reactive power

b.

ap-

a measure of the phase angle 6 between the voltage

Solution a.

its

or active, power.

power

it

factor

=

cos 6

=

P/S

(7.7)

var.

where The active power

c.

is

zero.

power If

a

wattmeter were connected into the

circuit,

factor

0

but

because the current

voltage,

7.9

it

is

is

also equal to

is

2800 VA,

2800

voltage and current

is

the ratio of the active

power

S. It is

power P

to the ap-

said to be lagging

factor

=

leading

(7.6)

power

factor

is

said to be

the current leads the voltage.

in

Example

7-5 and the phase angle between the line voltage

power delivered or absorbed by circuit or device [W] active

apparent

Power factor a

if

Calculate the power factor of the motor

and

=

the current lags behind the

Example 7-7 PIS

where

S

if

voltage. Conversely, the


power

P =

phase angle between the

If we know the power factor, we automatically know the cosine of the angle between E and / and, hence, we can calculate the angle. The power factor

var.

The power factor of an alternating-current device or

parent

=

90° out of phase with the

Power factor

circuit is

factor of a single-phase

circuit or device

would read zero.

To recapitulate, the apparent power

= power

it

is

power of

the circuit or device

line current.

the

Solution

[VA]

power

factor

expressed as a simple number, or as

=

PIS

=

1800/2040

-

0.882 or 88.2%

percentage.

Because the active power apparent

power

S,

it

P can

(lagging)

never exceed the

follows that the power factor

can never be greater than unity (or

100 percent).

cos 0

=

0.882

therefore, 0

=

28.1°


144

Example 7-8

A single-phase 20 V, 60 Hz

1

motor draws a current of 5 A from a The power factor of the motor is

line.

65 percent. Calculate a.

b.

The The

power absorbed by the motor power supplied by the line

active

reactive

Solution a.

The apparent power drawn by

=

5m

EI

=

120

X

5

the

motor

is

- 600 VA ^(390 W)

The

power absorbed by

active

the motor

is

P m = S m cosQ

= 600 X b.

The

0.65

- 390

power absorbed by

reactive

(7.7)

Figure 7.12 Power triangle

is

4.

the

motor

VSI^

the

that

power from

amount of nonpro-

Power

Active power is

considered

P absorbed by to

Active power device

is

P

a circuit or device

be positive and

that

is

drawn

hori-

Reactive power vice

is

delivered by a circuit or

considered to be negative and

drawn horizontally 3.

is

Q

upwards

is

to the left

when

is

drawn

circuit is

Example 7-8

and

Q

shown in The power

is

rules.

look like phasors, but they

The concept of

the

power

as convetriangle

is

solving ac circuits that comprise sev-

and reactive power components.

Further aspects of sources and loads

1

Let us consider Fig. 7.13a in which a resistor and capacitor are connected to a source.

The

circuit

similar to Fig. 7. 10 except that the capacitor

As

power flows from

is

a

is

re-

a result, reactive

power flows

G

while active

to the source

the source

active and reactive

G

to the resistor.

power components

The

therefore

flow

in

opposite directions over the transmission

line.

A

wattmeter connected into the circuit

give a positive reading ter will

absorbed by a circuit or de-

considered to be positive and

vertically

S, P,

from the capacitor

zontally to the right 2.

triangle for

However, we can think of them

nient vectors. useful

by a

downwards

accordance with these

active source.

convention, the following rules apply: .

are not.

7.1

triangle

2 2 2 The S = P + Q relationship expressed by Eq. 7.4, brings to mind a right-angle triangle. Thus, we can show the relationship between S, P, and Q graphically by means of a power triangle. According to

1

in

eral active

ductive power.

7.10

2

1

that is delivered

considered to be negative and

vertically

components

the line than active power. This burdens

the line with a relatively large

is

The power Fig. 7.

motor draws even more reactive

Q

Reactive power

drawn

(7.4)

= V600 2 - 390 2 = 456 var Note

a motor. See Example 7-8.

W or device

Qm =

of

G

P = EI p

power Q. Thus,

Q—

ElLy The source power P but receives reactive

give a negative reading

delivers active

will

watts, but a varme-

G

is

simultaneously an active

source and a reactive load.

ACTIVE, REACTIVE,

AND APPARENT POWER

145

device (or devices) connected to the receptacle.

If

the device absorbs active power, the receptacle will provide

G

(a)

E

(source)

if

it;

the device delivers active power,

the receptacle will receive 7

p

simple receptacle outlet

7 g

Y

liver

tive

—

or accept

power

nected to

Q

—

In other words, a

it.

is at all

either active

times ready to de-

power P or

reac-

accordance with the devices con-

in

it.


to

any 3-phase 480

V

service entrance to a factory or the terminals of a (b)

kV

high-power 345

Example

A 50

jjiF

transmission

line.

7-9

paper capacitor

terminals in

Example

is

placed across the motor

7-8.

Calculate a.

(c)

b. c.

d.

The reactive power generated by the capacitor The active power absorbed by the motor The reactive power absorbed from the line The new line current

Solution a.

Figure 7.13 a.

The impedance of

Source feeding an active and reactive (capacitive)

Xc =

load. b.

Phasor diagram

c.

The

active

of the circuit.

and reactive powers flow

in

opposite

di-

the capacitor

is

1/(2 77/C)

=

1/(217

=

53

(2.11)

x 60 x 50 x

1()"

6 )

il

rections.

The current It

in

may seem unusual

to

we must

line,

but

P

not the

is

/

have two powers flowing

opposite directions over the

remember that active power a reactive power Q and that

again

same

as

The

reactive

each flows independently of the other.

tacle in

a

home, also deserves our

such outlets are ultimately alternators that

power

Qc

active also

20

V

attention. All

connected

may seem,

an re-

source (as

we would

as

it

expect), but

behave as an active or reactive load.

factors

huge

to the

can act not only as an active or

Odd

determine whether

way or the other?

It all

it

will

behave

A

£/ -

q

-

120

X

the capacitor

2.26

271 var

recep-

the electrical transmission

and distribution systems. electrical outlet

1

2.26

is

120/53

power generated by

Speaking of sources and loads, a deceptively simple electrical outlet, such as the

capacitor

= EIXC =

=

same transmission

in the

it

may What

in

one

depends upon the type of

b.

The motor continues to draw the same power because it is still fully loaded.

active

Consequently,

P m = 390 The motor

also draws the

W same

reactive

power

as before, because nothing has taken place to

change

its

magnetic

field.

Qm =

Consequently,

456 var

is


146

The motor draws 456 var from

c.

the line, but the

same power drawn from the

capacitor furnishes 271 var to the

line.

The

line

net reactive

is,

therefore,

Ql = Q m - Q c

= 456 = The

active

185 var

power drawn from P\.

d.

271

the line

= 390 W

=

The apparent power drawn from

- V390 + - 432 VA 2

The new

1

the line

/W390 W)

is

Figure 7.14 Power triangle of a motor and capacitor connected an ac line. See Example 7-9.

2

85

432/120

nected

in

7. 15a).

We wish

A to 3.6 A by

This represents a big improvement because the line smaller and the operation of the motor has

factor of the line 4),

= /ys L = =

<|>

L

=

390/432

We

concerned) of active and reactive power flow

arcos 0.903

=

delivers reactive

arrow

active and reactive

active

powers

power

P.

We rather

complex

inductive,

cir-

Consider, for example, a group of loads con-

On

to the system.

it

ar-

the other

represents a capacitor,

power

it

The 16 kvar

in

independent) nature of the

powers enables us

to

add

all the

a circuit to obtain the total active

same way, we can add the reactive power Q. The reapparent power S is then found by

In the

to obtain the total reactive

S

The concept of active and reactive power enables us

some

C

distinct (and

Systems comprising

simplify the solution of

is

directed therefore toward the source.

is

The

powers

the line.

several loads

to

A

the source to the load.

25.5°

sulting total

cuits.

Thus, because load

hand, because load

observe the effect of the capacitor on the ap-

7.12

to

simply draw a block diagram of the individual

row flows from

0.903 or 90.3%

power supplied by

ab-

absorbs reactive power; consequently, the 5 kvar

The power triangle is shown in Fig. 7,14. The repower Qc generated by the capacitor is drawn vertically downward. By comparing this power triangle with that in Fig. 7.12, we can visuparent

power

Using the power approach, we do not have

is is

active

ally

V source (Fig.

worry about the way the loads are interconnected.

(Fig. 7.15b).

cos

380

to a

loads, indicating the direction (as far as the source

not been changed in the least.

The new power

way

to calculate the apparent

by the source.

placing the capacitor in parallel with the motor.

is

a very unusual

sorbed by the system as well as the current supplied

Thus, the line current drops from 5

current

to

line current is

= SJE = = 3.6 A

/j

is

recall that

=

\

P2

-\

when adding

Q

1

(7.4)

reactive powers,

we

assign a positive value to those that are absorbed by the system

and a negative value

to those that are

generated (such as by a capacitor). In the same way,

AND APPARENT POWER

ACTIVE, REACTIVE,

Reactive power absorbed by the system:

2.

380 V>

14kW

D

A

8 kvar

9 kvar

(a)

0

+

(5

+

7

Q 2 = (-9 Q

Net reactive power

4.

- -25

16)

Q = + 20

kvar

kvar

absorbed by the system:

=

25)

(

16 kvar

= +20

8)

Reactive power supplied by the capacitors:

3.

5 kvarf

=

G,

2kW

147

5 kvar

kW

8

Apparent power of the system:

5.

S

=

\

P2

7Q

-

24.5

kVA

V

Because the 380

6.

=

1

V24^+7~ 5?

source furnishes the appar-

ent power, the line current

=

I 7.

The power cos

=

4> L

power, but

it

tem of the netic

PIS

=

=

500/380

=

24/24.5

64.5

A

is

0.979 (leading)

kW

source delivers 24

of active

receives 5 kvar of reactive power. This

power flows

reactive

comes

= 24

factor of the system

V

The 380

(b)

S/E

is

into the local distribution sys-

company, where

electrical utility

available to create magnetic fields.

fields

may be

it

be-

The mag-

associated with distribution

transformers, transmission lines, or even the elec-

8 kvar

tromagnetic relays of customers connected to the Figure 7.15 a.

b.

Example of active and reactive loads connected to a 380 V source. All loads are assumed to be directly connected to the

same distribution system. The power triangle for the system 7.15c.

It

is

we

Thus, starting with the 5 kvar load,

380 V receptacle.

is

shown

move from one

progressively

device to the next around the system.

While so doing, we draw the magnitude and we

assign a positive value to active

powers

that are

absorbed and a negative value to those that are generated (such as

by an

Note that usually powers

we cannot add

power

total

apparent

their

power factors are

Let us

I.

now

Active

the apparent

various parts of a circuit to obtain the

in

S.

We

can only add them

power absorbed by

if

meet.

7.

1

5:

(2

+

8

+

14)

right) of

point,

direc-

each power vector,

tail

accordance with the power of each device

When

we

can

starting point to the

end

the selection

draw a power vector from the

is

complete,

which yields the inclined vector having a value

of 24.5 kVA. The horizontal component of

the right,

24

we know

kW and, because

that

by the system. The

it

represents

vertical

it is

this vec-

directed to

power absorbed

component of

5 kvar

is

the system:

directed

P =

left,

tor has a value of

identical.

solve the circuit of Fig.

to head, in

we

alternator).

down,

tion (up,

in Fig.

the graphical solution to our problem.

= +24

kW

tive

downward: consequently,

power generated by

it

the system.

represents reac-

1

48


5 kvar

starting

point

Figure 7.15c Power triangle

of the system.

7.13 Reactive

magnetic We

with the rapidly operating switch rather than with

power without

the resistor

fields

sometimes encounter situations where loads

absorb reactive power without creating any magnetic field at all. This can happen in electronic power circuits when the current flow is delayed by means of a rapid switching device, such as a

itself.

Nevertheless, reactive power

consumed just

as surely as

in the circuit.

This switching circuit will be

cussed

in detail in

if

is

a reactor were present dis-

Chapter 30.

7.14 Solving AC circuits using the power triangle method

thyristor.

Consider, for example, the circuit of Fig.

7.

16 in

We have seen that active and reactive powers can be

60 Hz source is connected to a resistive load of 10 (2 by means of a sy nchronous mechanical switch. The switch opens and closes its

rather

contacts so that current only flows during the latter

tation.

which a 100

V,

We can

some

to vector (j) no-

calculate the active and reactive powers

associated with each circuit element and deduce the

corresponding voltages and currents. The following

see,

if

we connected

a

wattmeter and varmeter between the source and the

(sometimes called displacement power factor)

of 84.4 percent. The reactive power

example demonstrates the usefulness of

power

this

triangle approach.

W

would respectively read +500 and corresponds to a lagging power fac-

+ 318 var. This tor

draw a phasor diagram or resorting

We

to solve

ac circuits without ever having to

almost by intu-

behind the voltage. Indeed,

switch, they

complex

forced delay causes the current to lag

part of each half cycle. ition, that this

added algebraically. This enables us

is

associated

Example 7-10 In Fig. 7.17a, the voltage is

60

V.

between terminals

1

and

3

ACTIVE, REACTIVE,

AND APPARENT POWER

149

0)^ 7.07

A

(eff

5ft (a)

12H R

60 V

T ©6 (a)

141

V S

14.1

©

—

A

9-

700

=

r

VA

S

--^

*l

=

780

VA

1

r^rs_+_

I

(b)

60 V

Figure 7.16 a.

Active

and

reactive

power flow

in

a switched

resis-

tive load. b.

Figure 7.17

The delayed current flow is the cause tive power absorbed by the system.

of the reac-

a.

Solving ac circuits by the power triangle method.

b.

Voltages and currents

in

the

circuit.

See Example

7-10.

from which the active power absorbed

Calculate a.

The current

b.

The voltage between terminals

c.

The impedance between terminals

in

each circuit element I

P =

and 2 l

S

Solution

We know the impedances of the elements and (Fig. 7. 60 V exists between terminals 3 and We now proceed in logical steps, as follows: a.

The current

in

the capacitor

Ic

=

60/5

=

The current

12

=

X 60 - -720

60/12

=

7b).

The

current

5

A

/,

=

must, therefore, be

=

S/E 3]

The voltage across

A

in the resistor is

/R

= VP + Q = V300 + (-720) 2 = 780 VA

/1

from which the reactive power generated

Qc =

1

that

E 23 = IXL

780/60

=

A

13

the inductive reactance

=13X8=

104

is

V

is

The var

terminals 1-3:

2

1

2

is

12

W

The apparent power associated with

and 2

I

X 60 = 300

5

is

reactive

reactance

power absorbed by

the inductive

is

G L = E23 X = +1352

/,

=

var

104

X

13

1


50

The

power absorbed by

total reactive

Qi. + Qc = = +632 var

Q=

the circuit

1.25 Mvar

is

1352

0.2

MW

T

- 720

]—

substation

The

total active

power absorbed by

the circuit

15

is

12.47 kV 3

The apparent power absorbed by

2.8

MW

0.75 Mvar

2 Mvar

2

The voltage of

£2 = ,

c.

is

C

10.03 kV

289 A

MW

= \P + Q 2 = V300 2 + 632 2 = 700 VA

S

b.

the circuit

load

Q

2,4

W

P = 300

L2

the line

=

SII {

10.03 kV

therefore

is

700/13

=

V

53.9

The impedance between terminals 2-1

Z=

E 2\U\. =

12.47 kV

53.9/13

=

289 A

is

4.15 II

289 A

Figure 7.18 Voltages, currents and power.

See Example

7-1

1

Example 7-11

A

kV

single-phase 12.47

transmission line several

C from

kilometers long feeds a load (Fig. 7.18).

The

reactance of 15

Instruments

at the

inputs to

6

=

arccos 0.833

power dissipated

b) Active

MW and 2 Mvar, respectively.

P = RI2 =

=

Calculate the line current and

phase angle with respect

its

X

0.2

=

33.6°

in the line:

2.4

Y

a)

at the

substation:

substation in-

power

dicate that the active and reactive the line are 3

Phase angle between the voltage and current

has a resistance of 2.4 il and a

line il.

a substation

10

6

X 289 2 = 0.2 MW

Active power absorbed by the load:

to the line voltage at the substation

b)

the active

—

power absorbed by the load power absorbed by the load

c) the reactive

rL

^sub

=

2.8

MW - 0.2 MW

3

MW

d) the line voltage at the load e) the

and

phase angle between the voltage

at the load

c)

Q L = XJ =

Solution a)

Reactive power absorbed by the 2

that at the substation

15

X

289

2

=

1.25

X

line:

10

6

=

1.25

Reactive power absorbed by the load:

Apparent power delivered

VP

S

to the line:

V3 +

+ Q2 =

2

3.60

2

2

Qc

2

=

Qsub

- Gl =

=

0.75

Mvar

2

Mvar -

1.25

Mvar

MVA d)

Apparent power

at the load:

Line current: /

=

S

=

3

600 000

E Power

Mvar

12

470

VA

Sc

=

V

289,4 Voltage

factor at the substation

FP =

P S

=

3

3.6

MW MVA

=

0.833

= -

+ Ql = \fl& 2.90

at the

Ec=

0.75

2

MVA

load end of the line:

Sc /

2.90

=

MVA

289

A

10.03

kV

AND APPARENT POWER

ACTIVE, REACTIVE,

Power factor

1

5

load end of the line:

at the

P,

2.8

S,

2.90

FP

MW MVA

=

0.965 ou 96.5%

p

e

—

s=

E.dh

r

*|b

Phase angle between the voltage and current at the rest of circuit

j

load:

(a)

=

9C It

=

arccos 0.965

15.2°

follows that the phase angle between the voltage the substation

at

15.2°)

=

We

that at the load

is

—

(33.6°

18.4°.

Fig. 7. sis.

and

summarizes the

8

1

S = + E\I*

results of this analy-

could have found the same values using

However, on account of its simpower method of solving this problem

vector algebra.

the

plicity, is

(b)

very appealing.

7.15 If

rest of circuit

Power and vector notation

vector notation

is

used to solve an ac

can readily determine the active

with

associated

We

sources.

component,

any

current that

P

is

the active

value for

rest of circuit

or

Q

(c)

The vector product

it.'

power S in terms of P + jQ, power and Q the reactive

P

E4 I*

E

power absorbed (or delivered) by the component.

A positive

S=-

the

the conjugate (/*) of the

flows through

£/* gives the apparent

where

including

simply multiply the phasor voltage

component by

across the

we

circuit,

and reactive power

means

same sequence ab

com-

that the

Figure 7.19 Method of writing power equations.

Z

associated with

The apparent power S

(not ba).

is

therefore written

ponent absorbs active or reactive power. Negative values

mean

reactive

component

that the

power.

When

It

calculating the £7* vector product,

procedure

very important to follow a standard der to obtain the correct result. plies to circuits that

tion or the

is

culate the active

nal

a

Z We

to

7.

The procedure

ap-

1

9a

in

which

a circuit element

of circuit."

We

want

to cal-

and reactive power associated with

note that current

terminal

Consequently, the

in or-

would be incorrect In Fig. 7.

that

b,

when

i.e.

in

/

flows from termithe

sequence ab.

calculating the product

subscripts of voltage

E

must be written

in the

a

current has a value

/^B,

its

conjugate /*

= IL~ B.

S = £ ba /.*

to write

9b, sign notation

current

enters

/

Z

is

by

used, and

the

Consequently, the apparent power

S

element

In the case of Fig. 7. 19c,

we want

to,

)

is

seen

given by

is

a

(

+

)

(

+

the

in Fig. 7.

1

we

Zby

write

the

(

—

sign because )

terminal of

S = — £4 /* be)

terminal.

we can determine

power associated with from b

(

it

terminal.

Z.

cause the current enters If

+

= +£,/*

The £]/* product is preceded by current / is shown as entering by

Thus, 'If

1

use the double subscript nota-

part of a larger "rest

element

is

sign notation (see Sections 2.4 and 2.5).

Consider Figure

Z

it

= £ah /*

S

delivers active or

the apparent

the "rest of circuit" (roc).

9a, because the current circulates

to a in the roc,

we would

write:

1


52

we would

Similarly, in Fig. 7.19c,

write

Let us illustrate the procedure by a few examples.

Example 7-12 In the circuit

2

of Fig.

the following values are

7. 19c,

given

Figure 7.20

E4 = 70 Z25°

See Example

=

I

7-13.

4 Z40°

Calculate the active and reactive power associated

with element

Z.

The voltage across

Solution

We

have

/

= 4 Z40°;

therefore I*

Since the current flows into the

power equation must bear 5

a

—

(

—

(

)

=

4

the capacitor

E y2 + /(— IO7) =

Z-40°

-£4/*

~

= -7()Z25° X 4 Z.-40 0 = -280 Z-15°

Current

=

terminal

+7

28()(cos(-15°)

= -270.5 +7

sin

(-15°))

in the

We active

10./

x

=

24.6

Z(-47° +

=

24.6

Z43°

2.46

Z -47° 90°)

capacitor flows from terminal 2 to

with the capacitor

is

72.5

S =

W and Q

~ -270.5

1

Consequently, the power associated

3.

= P+.iQ Thus P

0 10 J

sign:

)

given by

= =

£32

terminal, the

is

= +72.5

var

W

that element Z delivers 270 .4 of power and absorbs 72,5 var of reactive

conclude

£23 /*

= -24.6 Z43° X = -60.5 Z90°

2.46

- -60.5 (cos 0 - 60.5 7

+

90°

Z47°

7 sin 90°)

=

power.

= P +JQ Example 7-13 Given the

circuit

.

of Fig. 7.20

which

in

E l2 =

30 Z78°,

determine the power associated with the capacitor

whose reactance

is

Hence P — 0 and Q = —60.5; Consequently, the acpower associated with the capacitor is zero, and

tive it

delivers 60.5 var of reactive power.

10 O.

Example 7-14

Solution

Going cw around

the circuit,

we can

write (see

The

circuit in Fig. 7.21

sistor

Sections 2.32 to 2.39)

connected

actance.

E2] ~

7

1(1

E2{ _ - 10/ ~

= -2.46 Z

-

10 7)

12.5

133°

-

30

Z

0

a 45 i!

rere-

The source generates

by the phasor

a voltage described

E ab = 159Z65°.

78°

Z -55°

= +

composed of

with a 28 (1 inductive

is

in series

2.46

Z-47°

Calculate a.

The magnitude and phase of the current

/

AND APPARENT POWER

ACTIVE, REACTIVE,

4512

a

Voltage across the reactance

Ecb = = 159

[65_°

£\ib

'

28

j28

c.

is

/

X 3^33.11°

j28

= 84^(33.11° + = 84^123.11°

n

153

90°)

The conjugate /* of the current

/ is

= 3^-33.11°

/*


the resistor

is

Figure 7.21

S r = E.J*

Solving an ac circuit using vector notation.

= (135^33. 11°) (3^-33.11°) = 405^0° b.

and across the reactance

the resistor c.

The active and reactive power associated with

Applying Kirchhoff 's voltage law (see Section 2.32),

we

because there

sin 0°)

jO)

£ba + E ac + Ecb =

-£ab + -159^65° +

we

phase angle

/

=

+ -

j28

159

53

L

45/

+

j28/

=

0

/(45

+

j28)

-

0

/

-

S

r

=

159

L

45

45

+

j28

45

=

252/190°

-

252 (cos 90°

= 252

-

+

(0

+

j

sin 90°)

jl)

j252

Thus, the reactance absorbs only reactive power

(252

var).

=

65° 3

/ (65°

31.89°

£ ac = 45

(84 Z_123. 11°) (3^-33.11°)


- 53^31.89°

Z_

the reac-

65°

+ 28 = 53 = arctan 28/45 = 31.89° \

W)

.

=

=

Voltage across the resistor

=

in

2

2

~

~EJ

A

31.89°)

= -477^(65° is

/

X 3^33.11°

135zl33.ll

E\vJ''~

the source

*

= -(159^65°) (3^-33.11°)

= 3^33.11°

=

power (405

real

0

obtain

amplitude

hence 45

no j component

is

Transforming the denominator into polar coordinates,

is


obtain

tance

b.

+

(1

j

Thus, the resistor absorbs only

Solution

and so

+

(cos 0°

= 405

and the source

the resistor, the reactance,

a.

= 405 - 405

The magnitude and phase of the voltage across

0

33.11°)

= -477^31.89° = -477

(cos 31.89°

- -477

(0.849

= -405 -

j

+

252

j

+

j

sin 31.89°)

0.528)

is


154

The active and reactive powers are both negative, which proves that the source delivers an active power of 405 and a reactive power of 252 var.

W

on sources and loads

7.16 Rules

(sign notation)

We

are often interested in determining whether a de-

vice

an active/reactive source or an active/reactive

is

load without

such as

sis,

making

a complete mathematical analy-

we performed

Section

in

To enable

7. 15.

Figure 7.22 us to positively identify the nature of the source or load, consider Fig. 7.22 in line current

/.

The device

The voltage between

which a device is

part of a circuit.

the terminals

+

A carries a

(

is

£,

and

that are respectively parallel to, to E. Let /

E.

p

be the component of

out of phase with E.

behind or 90° ahead of

The

1

.

A

voltage

b.

the line current

+

)

A device

when double

/,

re-

is

is

an active source

subscript notation

used. Consider Figure 7.23 in which a device ries a current /

flowing

in the direction

b

is

enable us to state 3.

/ is

shown

in

phase and

A device

is

shown. The

£ab The .

voltage

b.

line current /

shown

is

in

phase and

as entering terminal

Otherwise, the device

is

an active source.

A device a.

is

current

/

is

an active source.

rule also applies:

a reactive load

when

lags 90° behind voltage E. xh

rule also applies:

a reactive load /c]

when

lags 90° behind voltage

line current /

is

shown

E and

as entering the

(

+)

terminal.


Based on these

rules,

tionships in Fig. 7.22, active load because

A is

/

p

is

a reactive source.

and observing the phasor

we deduce is in

that device

A

relais

an

phase with E. Also, device

a reactive source because /

q

is

90° ahead of E.

Figure 7.23

Same These

rules are in

fol-

an active load when:

£ab and component / p are

a.

The following

as entering the

agreement with IEEE and IEC conventions.

is

A car-

lowing rule applies:

4.

component

b.

whether a device

or active load

terminal.

The following a.

We can also tell

rule applies*:


2.

on sources and loads (double subscript notation)

an active load

a.

(

and

rela-

/.

voltage between terminals a and

when E and component / are

is

between Eand

7.17 Rules

an active load or an active

is

The following

device

in

E.

E

between

whether a device source.

angles

diagram, together with the phasor

circuit

lationships

at right

/ that is parallel to

phase with, or 180° Similarly, / can be either 90°

be either

will therefore

It

active/reactive source or depending upon the phasor

and one

sign.

)

A may be an

active/reactive load

tionship

The phase angle between E and / can have any value. As a result, / can be decomposed into two components, I p and / of the terminals bears a

Device

circuit

as

in Fig.

subscript notation

is

7.22 except that double-

used.

and

a.

ACTIVE, REACTIVE,

b.

shown

line current / is

nal

as entering by termi-

c.


a reactive source.

is

we deduce

tionships in Fig. 7.23, active

source because

A

Also, device

90° behind

Eab

is

I

p

is

1

that device

A

an

is

80° out of phase with E, lb

a reactive load because /

q

e.

7-1

1

Using the rules given 7. 17,

.

lags

Sections 7.16 and

in

determine which of the devices

in Fig.

7.24a through 7.24f acts as an active (or

power

active)

.

7-12


55

The peak power output of the reactor The duration of each positive power pulse

d.

Based on these rules, and observing the phasor rela-

1

The reactive power absorbed by the reactor The apparent power absorbed by the reactor The peak power input to the reactor

a. b.

a.

AND APPARENT POWER

re-

source.

A single-phase motor draws a current of 2 A at a power factor of 60 percent. 1

Calculate the in-phase and quadrature comPractical level 7-1

What

ponents of current

power? reactive

the unit of active

is

power? apparent power? 7-

7-2

7-3

A

capacitor of 500 kvar

7-5

placed

1

3

in parallel

A

240

V,

60 Hz

line.

A

1

wattmeter

line gives a reading of

the

Name

motor and

a static device that

can generate reac-

power.

Name

7- 14

a static device that absorbs reactive

If a

the reactive

in parallel

Problem 7-13,

What

a.

is

the approximate

power

factor, in

The current

in

What

is

the

power

large

The apparent power of The line current

e.

The power

7- 5 1

motor absorbs 600

kW at a power

is

An

a.

b. is

connected across a

1

20

A

10 12 reactance

60 Hz

line.

is

V,

connected to a 120 V,

Calculate:

induction motor absorbs an apparent at

a

power

factor of 80

percent. Calculate:

generates. 10 (2 resistor

and without drawing any

power of 400 kVA

60 Hz source. Calculate the reactive power

A

14)

circuits in Fig. 7.25.

connected to a 240 V,

60 Hz source. Calculate: a. The active power absorbed by the resistor b. The apparent power absorbed by the resistor c. The peak power absorbed by the resistor d. The duration of each positive power pulse

7.

phasor diagrams, find the impedance of the

7-16

machine.

it

factor of the motor/capacitor

Using only power triangle concepts (Section

90 percent. Calculate the apparent power and reactive power absorbed by the

p,F capacitor

the ac line

combination

factor of

A 200

calculate:

active

d.

motor?

Intermediate level

A

is

with the motor of

power reading of the wattmeter The total reactive power absorbed by the capacitor and motor The

b.

a single-phase motor lags

power factor of the power it absorbs.

capacitor having a reactance of 30 (2

connected

power.

factor of the

7-10

a

2765 W. Calculate

50° behind the voltage.

7-9

to

connected into the

c.

7-8

with respect

apparent power of the group.

candescent lamp?

1-1

/

single-phase motor draws a current of

A from

percent, of a capacitor? of a coil? of an in-

7-6

and

with an inductor of 400 kvar. Calculate the

tive

7-4

is

/

the line voltage.

c.

7-17

A

power absorbed by the motor power absorbed by the motor What purpose does the reactive power serve

The The

active

reactive

circuit

composed of a

1

2 ft resistor in

series with an inductive reactance of 5 (2

carries an ac current of 10 A. Calculate: a.

The

active

b.

The

reactive

c.

d.

power absorbed by the resistor power absorbed by the inductor The apparent power of the circuit The power factor of the circuit

(0

(b)

(a)

G

E

F

/ /

(d)

(e)

Figure 7.24

See Problem

7-11.

Hf-

1 40 V[

^

ion

4I2(

]

en

5A

J

X (c)

(b)

(a)

Figure 7.25

See Problem

7-18

7-15.

A coil

having a resistance of 5 ft and an

ductance of 2

H

7-2

in-

carries a direct current of

20 A. Calculate: a. The active power absorbed b. The reactive power absorbed

Advanced 7-19

7-20

A

ply voltage

is

connected

200

is

ft. If

the sup-

V, calculate:

a.

The

reactive

power absorbed by

The

reactive

power generated by

the coil

the capacitor

Problem

a

7- 3, if 1

in parallel

we

d.

7_22

power absorbed by

The

The apparent power of the

c.

The power

The power

V

the system

is

0.6 lagging (Fig. 7.26).

calculate:

sy stem

factor of the system

156

coil

factor at the terminals of a

source

Without using phasor diagrams,

with the motor, calculate:

a.

The active power dissipated by the The apparent power of the circuit

120

place a capacitor of

b.

total active

c.

a.

The value of E

b.

The impedance of

the load

re-

in parallel with

b.

level

motor having

500 var

having a reactance of 1011 and a

a capacitive reactance of 10

power factor of 0.8 absorbs an active power of 1200 W. Calculate the reactive power drawn from the line. In

A coil

sistance of 2 ft

Z

ACTIVE, REACTIVE,

AND APPARENT POWER

157

29.

E=

120

V

»E= 120 V Figure 7.26 /= 5 A

See Problem 7-22.

7-23

In Figs.

7.27a and 7.27b, indicate the mag-

150°

(b)

(a)

nitude and direction of the active and reactive

power flow.

and

/

and

,

q

treat

(Hint:

Decompose

/ into /

5

them independently.)

Figure 7.27

See Problem


7-24

A single-phase

A

p

7-23.

capacitor has a rating of

30 kvar, 480 V, 60 Hz. Calculate

its

capacib.

tance in microfarads.

The

active and reactive

power consumed by

the line

7-25

In

Problem 7-24

calculate:

c.

a.

The peak voltage across the capacitor when it is connected to a 460 V source

b.

The

active, reactive

and apparent power ab-

sorbed by the load d.

resulting energy stored in the capacitor

7-28 at that instant, in

The

joules

The voltage across

A2

hp,

230

V,

1

the load

725 r/min 60 Hz single-

phase washdown duty motor, manufactured 7-26

Safety rules state that one minute after a capacitor

is

disconnected from an ac

voltage across

discharge that

is

it

must be 50

V

or less.

done by means of a

by Baldor Electric Company, has the

line, the

The Full load current:

resistor

permanently connected across the

is

fol-

lowing characteristics:

efficiency:

capacitor terminals. Based on the discharge

power

1

1.6

A

75.5%

factor:

74%

curve of a capacitor, calculate the discharge resistance required, in

ohms,

tor in

Problem 7-24. Knowing the

tance

is

when

weight: 80 lb

for the capaci-

a.

resis-

sorbed by this machine

subjected to the service voltage

the capacitor

is in

operation, calculate b.

7-27

A

3.2 kV,

1

60 Hz single-phase

line con-

(1.

The metering equipment

at

the substation indicates that the line voltage is

12.5

kV and

that the line

of active power and 2

is

drawing 3

Mvar of reactive

power. Calculate: a.

The current flowing

operates

at

If a

40 microfarad capacitor

is

connected

current feeding the motor.

The

c.

MW

Will the presence of the capacitor affect the

temperature of the motor?

has a resistance of 2.4 II and a reac-

tance of 12

it

across the motor terminals, calculate the line

nects a substation to an industrial load. line

when

full load.

wattage rating.

its

Calculate the active and reactive power ab-

7-29

A single-phase heater absorbs 4 kW on a 240 V line. A capacitor connected in parallel with the resistor delivers 3 kvar to the line. a.

Calculate the value of the line current.

b. If the capacitor in the line

new

is

line current.

removed, calculate the

Chapter 8 Three-Phase

8.0 Introduction

8.1

Electric power

We

is

tributed in the

generated, transmitted, and dis-

form of 3-phase power. Homes

and small establishments are wired power, but

this

power

ferred over single-phase

is

can gain an immediate preliminary understand-

piston

pre-

is

comparable

to a single-phase

the other hand, a 2-cylinder engine

for several impor-

is

Three-phase motors, generators, and transformers are simpler, cheaper,

and more

down move

efficient

b.

Three-phase transmission lines can deliver

c.

more power for a given weight and cost The voltage regulation of 3-phase transmission

inside

identical

They

in unison.

power

to deliver

rather than at the

running engine and a

A cuits

knowledge of 3-phase power and 3-phase is,

cir-

the

basic

different times.

circuit

will see that

most 3-phase

circuits

we

identical,

can be reduced to

time.

the reader

is

As

this

the reader

may know

produces a smoother

much smoother

output torque.

a 3-phase electrical system, the

As

a result, the total

one phase may be used

behavior of

power at power flow is

all

to represent the

three.

Although we must beware of carrying analogies

elementary single-phase diagrams. In this regard,

we assume

not

way as

very smooth. Furthermore, because the phases are

techniques used to solve single-phase circuits can be directly applied to 3-phase circuits. Furthermore,

do

cylinders, but they

three phases are identical, but they deliver

therefore, essential to an understanding of

power technology. Fortunately,

a

and

to the shaft in successive pulses

same

lines is inherently better in

In

move up

are staggered in such a

from personal experience,

Similarly,

to

6-cylinder

engine could be called a 6-phase machine. 6-cylinder engine identical pistons

a.

machine. On

comparable

The more common

a 2-phase machine.

tant reasons:

common

A single-cylinder engine having one

gasoline engine.

merely represents a tap-off from the

power

Polyphase systems

ing of polyphase systems by referring to the

for single-phase

basic 3-phase system. Three-phase

Circuits

familiar with the previous

too

chapters dealing with ac circuits and power.

far,

system 158

the is

above description reveals

basically

that a 3-phase

composed of three

single-phase

THREE-PHA SE CIRCUITS

systems that operate in sequence.

much

fact is realized,

Once

this basic

A

winding

stator

59

1

of the mystery surrounding

3-phase systems disappears.

Single-phase generator

8.2

Consider a permanent magnet stant

NS

revolving

con-

at

speed inside a stationary iron ring (Fig. 8.1).

The magnet

is

driven by an external mechanical

source, such as a turbine.

The

Figure 8.2 ring (or stator) re-

duces the reluctance of the magnetic circuit; consequently, the flux density in the air if

having terminals a, insulated ductors,

A

were absent.

the ring

from

one

in

it.

1

is

Each

each

Ea1 =

At this instant

gap

is

flux

does not cut

greater than

multiturn rectangular coil

mounted

0 because the

the conductors of winding A.

inside the ring but

voltage E. xi

maximum when

is

position of Fig.

turn corresponds to

two con-

8.

the poles are in the

because the flux density

1

On

greatest at the center of the pole.

slot.

the voltage

when

zero

is

is

the other hand,

the poles are in the position

of Fig. 8.2 because flux does not cut the conductors winding

stator

A

at this If

tion,

moment.

we

plot E. xl as a function of the angle of rota-

and provided the N, S poles are properly

shaped,

we

obtain the sinusoidal voltage

Suppose

Fig. 8.3. *

shown

in

the alternating voltage has a peak

value of 20 V. Machines that produce such voltages are called alternating-current generators or syn-

chronous generators. The particular machine shown Figure 8.1

in Fig. 8.

embedmaximum ( + ).

A single-phase generator with a multiturn ded

in

two

slots.

At this instant

Ea1

is

called a single-phase generator

is

1

coil

v + 20

As

the

magnet

turns,

it

sweeps across the con-

inducing a voltage

in

them according

/ / /

—y

\

X \ \

10

ductors,

/

V

to the

/ /

\

equation:

/

\

£al =

B/v

(2.25)

wherein

-

/

KJO

\— —

90

0

10

2

(

B = instantaneous flux density cutting across conductors I

= =

the

\ \ \

\ \ \

3(i0 degrees 4S >0

/

\ \

—

\

f

/

/ \

\ -

\

/

\

/

20

/

in the slots [T]

Figure 8.3

length of conductors lying in the magnetic field

v

V]

\

/

\

instantaneous voltage induced in the coil

'0

\

/

igle

\

=

/

N

Voltage induced

in

winding A.

[m]

peripheral speed of the revolving poles [m/s]

The poles shown

The sum of the voltages induced

in all the

ductors appears across the terminals.

con-

The terminal

voltage

in Fig. 8.

composed of

negative pulses.

1

would generate an

alternating

rather brief t'lat-toppped positive and


160

Power output of a single-phase generator

8.3

If

a resistor

rent will

is

connected across terminals

The current

/,,

is in

1

a cur-

phase with the voltage and, con-

sequently, the instantaneous series of positive pulses, as

average power

power

trical

a,

flow and the resistor will heat up (Fig. 8.4).

is

is

power is composed of a shown in Fig. 8.5. The

one-half the peak power. This elec-

derived from the mechanical power

provided by the turbine driving the generator. As a result, the turbine

ergy

in pulses, to

must deliver

mechanical en-

its

match the pulsed

electrical output.

This sets up mechanical vibrations whose frequency is

twice the electrical frequency. Consequently, the

generator will vibrate and tend to be noisy.

8.4

Two-phase generator

Using the same single-phase generator, a

second winding (B) on the

let

us

mount

Figure 8.5 Graph of the generator

stator,

is

voltage, current,

and power when the

under load.

identical to

voltage E. tl becomes zero and voltage

maximum

therefore out

by curves load on phase

Note

A

in Fig.

£ aj

that

positive value

8.6b and by phasors

Eh2 because before E h2 does.

stator

Figure 8.4 resistor.

its

is

it

in Fig, 8.6c.

reaches

its

peak

called a two-phase generator,

windings are respectively called

phase A and phase

Single-phase generator delivering power to a

attains

value.

leads

This machine

and the

E b2

The two voltages are of phase by 90°. They are represented

positive

B.

Example 8-1 The generator shown

in Fig.

8.6a rotates

at

6000

r/min and generates an effective sinusoidal voltage

winding A, but displaced from

it

by a mechanical

of 170

V

per winding.

angle of 90° (Fig. 8.6a).

As

the

magnet

rotates, sinusoidal voltages are in-

Calculate

each winding. They obviously have the

a.

same magnitude and frequency but do not reach maximum value at the same time. In effect, at

b.

duced

in

their

moment when the magnet occupies the position shown in Fig. 8.6a. voltage passes through its

c.

positive value, whereas voltage

zero. This

is

conductors

Eh2

is

an-

gle of 90°

the

maximum

The peak voltage across each phase The output frequency The time interval corresponding to a phase

Solution a.

The peak voltage per phase

is

because the flux only cuts across the

in slots 1

after the rotor has

and a

at this instant.

made one

However,

quarter-turn (or 90°),

£ ni = \2E = = 240 V

1.414

X

170

(2.6}

THREE-PHASE CIRCUITS

b.

One cycle makes one

is

completed every time the magnet

The period of one cycle

turn.

T= = =

s

=

Power output now

Let us 0.01

will flow in

phase with

/= c.

A phase terval

is

\/T=

means

Hz

100

1/0.01

angle of 90° corresponds to a time in-

Consequently, phasor

behind phasor

A and B

each

E. {i

(Fig. 8.7a). Currents

and

/.,

/h

They are respectively in The currents are, therefore,

resistor.

and

Eh2

-

90° out of phase with each other (Fig. 8.7b). This

of one quarter-revolution, or 10 ms/4

2.5 ins.

connect two identical resistive loads

across phases

s

10 ins

The frequency

of a 2-phase

generator

1/6000 min

60/6000

8.5

is

161

£a]

E h2

lags 2.5

=

that

/.,

reaches

period before

now produces

/b

its

maximum value one quarter-

does. Furthermore, the generator

a 2-phase

power

output.

The instantaneous power supplied

ms

to

each

resis-

tor is equal to the instantaneous voltage times the

.

instantaneous current. This yields the two power

waves shown of phase

in Fig. 8.8.

A is maximum,

Note

that

that

when the power B is zero, and

of phase

we add the instantaneous powers of we discover that the resultant power is and equal to the peak power P m of one

vice versa. If

both phases, constant

(a)

1

load on

(a)

phase

\

-h (b)

0-

90.

if

i

360

21

—t

an

jle

of

450

_

r ot ation

6

i load on phase B

(c)

n

(b)

Figure 8.6 a.

Schematic diagram of a 2-phase generator.

b.

Voltages induced

c.

Phasor diagram of the induced voltages.

in

a 2-phase generator.

Figure 8.7 Two-phase generator under load. b. Phasor diagram of the voltages and currents.

a.

A

ELECTRICAL MACHINES AND TRA NSEORMERS

62

1

phase.* In other words, the total

2-phase generator

also constant.

is

benefit,

it

peak power =

it

is

to drive the

less noisy.

'I

1

instantaneous power of phase

in size,

i

except for the addition

of an extra winding.

Three-phase generator

A 3-phase generator is similar to a 2-phase generator, except that the stator has three identical windings stead of two.

placed

at

The

three windings a-1, b-2,

120° to each other, as

When

magnet

the

position

in-

maximum Voltage

in Fig. 8.9a,

E b2

Consequently,

only voltage

will reach

its

Ea

is in ,

is

The The

i

i

i

i

;

180

270

360

the

*-

angle of rotation

Total instantaneous

power output

it

is

used

to

current

is

one-third of a turn). Similarly, voltage

designate different things.

has to be read

in

context to be understood.

ways

the

in

Figure 8.8 Power produced by a 2-phase generator.

which

out of phase with the voltage (relets to pha-

tain its positive

peak

Ec3

will at-

after the rotor has turned

through 240° (or two-thirds of a turn) from

Consequently, the three stator voltages

three phases of a transmission line (the three conduc-

tors of the line)

£b2

and

,

The phase-to-phase voltage

4.

The phase sequence

(the line voltage)

(the order in

Ec3

They

120°.

3.

its ini-

position.

tial

sor diagram) 2.

90

0

at its

positive peak after

The following examples show some of the word phase is used. .

r i

positive value.

The icrm phase

1

i

windings have the

the rotor has turned through an angle of 120° (or

:;:

i

in Fig. 8.9a.

At the moment when the magnet

shown

i

Instantaneous power of phase B

and c-3 are

effective values, but the peaks occur at differ-

ent times.

Ib =

rotated at constant speed,

is

the voltages induced in the three

same

Eb

. pov rer = .peak

/MA, i

shown

A

|

j

8.6

Pm

=

fa

A 2-phase generator does

i

without any increase

Ea

every instant. As a

As an important produces twice the power output

not vibrate and so

added

power output of the

at

mechanical power needed

result, the

generator

same

the

is

—

are

and as phasors

—

al

,

are respectively out of phase by

shown

as sine

waves

in Fig. 8.9b,

in Fig. 8.9c.

which the phasors follow

each other) 5.

The burned-out phase

(the burned-out

winding of

a

3-phase

8.7

machine) 6.

The 3-phase voltage

(the line voltage of a 3-phase system)

7.

The 3-phase currents

are unbalanced (the currents in a 3-phase

8.

line

or machine are unequal and not displaced

The

pliase-shifi transformer (a device that

at

120°)

(a short-circuit

between two

line

and

E52

conductors)

Phase-to-ground fault

six wires to deliver

phase loads (Fig.

The phase-to-phase fault

(a short-circuit

between

a line or

/c ,

8.

1

power to the individual singleThe resulting currents /.„ / b

0a).

,

are respectively in phase with voltages

and

EcV

Because the

currents have the

same

1.

The phases are unbalanced

(the line voltages, or the line

£al

,

resistors are identical, the

effective values, but they are

winding and ground) 1

to

can change the

voltage)

10.

Let us connect the three windings of the generator

three identical resistors. This arrangement requires

phase angle of the output voltage with respect to the input

9.

Power output of a 3-phase generator

mutually out of phase by

1

20° (Fig.

8.

1

that they are out of

other)

reach their positive peaks

The

0b).

phase simply means

currents, are unequal or not displaced at 120° to each

fact

that they

at different times.


The instantaneous power supplied to each resisis again composed of a power wave that surges

tor

between zero and a (a)

163

maximum

power peaks in the the same time, due to

P nv However,

value

do not occur

the

three resistors

at

the phase angle

between the

we add the instantaneous powers of all resistors, we discover that the resulting power

voltages. If

three is T

1

Eai

constant, as in the case of a 2-phase generator.

However, the

1

^c3

SN

\

put

\

(b)

0

740

3iiO

50

\/

J

output of a 3-phase generator has

P m Because

is

1

the electrical out-

.

constant, the mechanical

power required

generator does not vibrate. Furthermore, the power

erator to the load,

is

line,

connecting the gen-

constant.

Example 8-2 The 3-phase generator shown

in Fig.

nected to three 20 12 load resistors. voltage induced in each phase 120°

(c)

I

120°

a.

c.

a.

d.

Three-phase generator. in a 3-phase generator.

b.

Voltages induced

c.

Phasor diagram

of the

a.

Each

resistor to

behaves as a single-phase load

in

each resistor

(b)

Three-phase, 6-wire system.

Corresponding phasor diagram.

120 V. calculate

an effective voltage of 120 V. The

power dissipated

b.

con-

The power dissipated in each resistor The power dissipated in the 3-phase load The peak power P lu dissipated in each resistor The total 3-phase power compared to P lu

connected

Figure 8.10

is

the effective

Solution

induced voltages.

a.

is

8.10a

If

the following:

b.

Figure 8.9

to

also constant, and so a 3-phase

flow over the transmission

-N

X

is

drive the rotor

t

J

total

a magnitude of 1.5

is,

therefore,

1

64


I i

u

*

U

'b

AI

360

240

0

480 "\

t

x

I

(a)

Figure 8.11 a.

Three-phase, 4-wire system.

b.

Line currents

in

a 3-phase, 4-wire system.

P = E 2 IR =

= 720 b.

The (all

total

l20

The peak power

/2()

W

power dissipated

three resistors)

2

in the

3-phase load d.

3

The

ratio of

X 720

is

Pm

to

8.485

is

2 60/1440 1

=

1.5

Thus, whereas the power

The peak voltage across one

resistor

sates is

tal

E m = <2E = V2 X - 169.7 V The peak current /m

X

69.7

*

is

absolutely constant from instant

to instant. c.

PT

PT /P m =

= 2160 This power

each resistor

= £ m /m = = 1440W

is

P v = 3P =

in

in

120

each resistor

= EJR = = 8.485 A

is

169.7/20

between 0 and a

power

for

all

in

maximum

each resistor of 1440

three resistors

is

W,

pul-

the

to-

unvarying and

equal to 2 160 W.

Wye

8-8 The

connection

three single-phase circuits of Fig. 8.10 are

electrically

we

independent. Consequently,

can

connect the three return conductors together

form

a single return

conductor (Fig.

to

8.1 la). This

reduces the number of transmission line conduc-

The

tors

from 6

tral

conductor (or simply neutral), carries the sum

to 4.

of the three currents

(/ a

+

/b

+

/c ).

At

first

it

seems

that the cross section of this

conductor should be

three times that of lines

and

diagram of Fig. 'c

return conductor, called neu-

a, b,

8,1 lb clearly

c. However, the shows that the sum

of the three return currents is zero at every instant. For example, at the instant corresponding to

= / max and I h = / a = -0.5 / max making + 4 = 0- We arrive at the same result (and

Figure 8.12

240°, I c

Three-phase, 3-wire system showing source and load.

4+

A>

,


I

much more simply) by taking the sum of the phaOb. The sum is clearly (/., + / b + I c ) in Fig. 8.

sors

1

zero.

We

can, therefore,

together without in

remove

the neutral wire al-

any way affecting the voltages one stroke

or currents in the circuit (Fig. 8.12). In

we accomplish a great saving because the number of

conductors

line

However, the loads in

order to

remove

not identical, the

from

drops

six

three!

to

8.11a must be

in Fig.

identical

the neutral wire. If the loads are

absence of the neutral conductor

produces unequal voltages across the three loads.

The

circuit of Fig.

—composed and load —

8.12

generator, transmission line, n

3-phase, 3 -wire system.

The generator, connected

load, are said to be

as the

of the

is

called

as well

in wye, be-

cause the three branches resemble the letter Y. For equally obvious reasons, to

some people

prefer

use the term connected In star.

The

circuit of Fig. 8.1 la

4-wire system.

system

is

The

usually the

than the line

is

called a 3-phase,

neutral conductor in such a

same

size or slightly smaller

conductors. Three-phase, 4-wire sys-

tems are widely used to supply electric

power

to

commercial and industrial users. The line conductors are

often called phases,

which

is

the

same

term applied to the generator windings.

8.9

Voltage relationships

Consider the wye-connected armature windings of a 3-phase generator (Fig. 8. 13a). in

The induced voltage

each winding has an effective value

sented by the length of each phasor Fig. 8.

are

1

3b.

Knowing

represented

tion

is,

in the

repre-

diagram

in

that the line-to-neutral voltages

by phasors

E

ail ,

Eblv and Ecn the quesE. lb Ehc and

what are the line-to-line voltages

£ca ? Referring

E lN

to Fig. 8.13a,

lowing equations, based on

we

,

,

can write the

fol-

Kirchhoff s voltage law:

Eab = Ean + E nh = ân ~ £bn £ bc = £ bn + E nc = E hn - Ecn Eca = Ecn + E na = Ecn - Ean

(8.1)

(8-0

Figure 8.13 Wye-connected

a.

(8.2)

Line-to-neutral voltages of the generator.

c.

Method

d.

Line voltages

to

determine

line

voltage

£ab £ bC) and £ca

placed at 120°. (8.3)

3-phase gen-

b.

(8.2)

(8.3)

stator windings of a

erator.

,

£ab

.

are equal and dis-


166

Referring

we draw phasor

Eq. 8.1,

first to

ex-

E.db

a

actly as the equation indicates:

The

diagram shows that line by 30° (Fig. 8.13c). Using

resulting phasor

voltage

E

Ean

leads

ilb

simple trigonometry, and based upon the fact that the length of the line-to-neutral phasors

is

£j N

we

,

have the following:

E

length

x

£ah = =

of phasor

2

X

2

X £ LN V3/2

£, N cos 30°

Figure 8.14 Voltages induced

in

a wye-connected generator.

- V3 £ LN Calculate

The

line-to-line

therefore

V3

voltage (called line voltage)

is

times the line-to-neutral voltage:

- V3£ LN

a.

b.

(8.4)

c.

The line-to-neutral voltage The voltage induced in the individual windings The time interval between the positive peak

A and

voltage of phase

where

phase

E = x

£ LN — V3 = Due

d.

effective value of the line-to-neutral

Solution

voltage [V|

a.

=

a constant [approximate value

to the

equal to

V3 E

t

N

.

The

of

truth

this

=

can

be seen by referring to Fig. 8.13d, which shows three phasors: £. l]v

drawn according

The

line

£"

hc

,

and

phasors are

and

8.3, respec-

.

is

The windings

1

3

V

800

are connected in wye; conse-

quently, the voltage induced in each winding

all

£ca The

to Eqs. 8.1, 8.2,

line voltage

line-to-neutral voltage

symmetry of a 3-phase system, we con-

is

peak of

£ln = î/^ 3 = 23 900/V3

b.

tively.

The

1.73]

clude that the line voltage across any two generator terminals

The peak value of the

effective value of the line voltage [V]

the positive

B

13 c.

800

is

V.

One complete

cycle (360°) corresponds to 1/60

s.

Consequently, a phase angle of 120° corresponds

voltages are equal in magnitude and

to an interval of

mutually displaced by 120°.

To

further clarify these results. Fig.

the voltages

8.

between the terminals of

generator whose line-to-neutral voltage

The

line

voltages are

all

equal to

14

is

3-phase system, but the voltage between b,

c\

b and n, etc.)

is

Example 8-3 3-phase 60

Hz

generator, connected in wye, gen-

erates a line (line-to-line) voltage of 23

900

V.

1

X

1/180

s

60

5.55

ms

positive voltage peaks are, therefore, sepa-

rated by intervals of 5.55 ms.

lines a, b, c, consti-

nevertheless an ordinary single-phase voltage.

A

The

100 V3, or

tute a

b and

=

100 V.

The voltages between

120

360

a 3-phase

173 V.

any two lines (a and

T=

shows

d.

The peak

line voltage is

E m = V2E L = 1.414 X = 33 800 The same voltage

23 900

(2.6)

relationships exist in a wye-

connected load, such as

that

shown

in Figs. 8.1

THREE- PHASE CIRCUITS

and

8.

1

2. In

other words, the line voltage

is

1

67

V3

times the line-to-neutral voltage.

Example 8-4 The generator

in Fig.

8.12 generates a line voltage

of 865 V, and each load resistor has an impedance of 50

fl.

Calculate

The voltage across each resistor The current in each resistor The total power output of the generator

a.

b. c.

Solution

The voltage across each

a.

resistor

is

£ L n = £| A/3 = 865/V3 = 500 V The current

b.

in

/

each resistor

= ELN /R =

(8.4)

is

500/50

10A All the line currents are, therefore, equal to 10

Power absorbed by each

c.

resistor

P = ElN I = 500 X 5000

A. (b)

is

10

W

The power delivered by the generator

to all

three resistors is

P =

3

X 5000 =

15

kW

Figure 8.15 Impedances connected in delta. b. Phasor relationships with a resistive

a.

8.10 Delta

connection

A

is

3-phase load

said to be

voltages are equal

and the

This corresponds

to

three

balanced when the

line

line currents are equal.

identical

impedances

connected across the 3-phase line, a condition that is

usually

encountered

in

The three impedances (as

we

The

3-phase

may

the line; consequently, resistor currents

£bc

,

and

ECiV

h

/2 ,

and

/3

£ab

law, the line currents are given

by

be connected

8.

-h

wye

in 1

(8.5)

5a).

(8.6)

h -

(8.7)

12

shown).

Let us determine the voltage tionships in

and current

rela-

such a delta connection,* assuming a

resistive load.

The

resistors are

,

Furthermore, according to Kirchhoff's

voltages are produced by an external gen-

erator (not

/

are in phase with the respective line voltages

circuits.

already have seen) or in delta (Fig.

line

load.

connected across

The connection letter A.

is

so

named because

it

resembles the Greek


68

1

Let the current

each branch of the delta-connected

in

which corresponds

load have an effective value the length of phasors

/2 , /3

.

Furthermore,

let

to

the

have an effective value / L which corresponds to the length of phasors /.,, / b /c Referring line currents

,

,

equation

shows

.

we draw phasor exactly as the indicates. The resulting phasor diagram

Eq. 8.5,

first to

that

/

a

/.,

leads

ple trigonometry,

/,

by 30° (Fig.

we can now

8.

1

5b).

Using sim-

write

A

— 10

lx

=

2

X

I,

cos 30°

=

2

X

I,

V 3/2

= V3/

c

o— Figure 8.15c

See Example

y

.

The

line current

is

therefore

V3

8.5.

times greater than

the current in each branch of a delta-connected load: /,.

= V3/Z

(8.8)

= l(W3 =

Iy

b.

5.77

A

The voltage across each impedance

is

550

V.

Consequently,

where

= —

/,

/,

Z= -

effective value of the line current [AJ effective value of the current in one

EIIy

=

550/5.77

95 ft

branch of a delta-connected load [A]

—

"V3

a constant [approximate value

The reader can

=

1.73]

position of phasors /b and /c and thereby observe that 20°. ,

the three line currents are equal and displaced by

Table

8A summarizes

1

the basic relationships bein

by a 3-phase

magnitude and

readily determine the

tween the voltages and currents

wye-connected

and delta-connected loads. The relationships are

motor winding, generator wind-

ing, etc.) as long as the

elements

in the three

line

The apparent power supplied by is

a single-phase

equal to the product of the line voltage

line current

/.

The question now

arises:

line

E times the What

is

the

apparent power supplied by a 3-phase line having line voltage

valid for any type of circuit element (resistor, capacitor, inductor,

Power transmitted

8.11

If

we

8. 16a,

E and

a

a line current I?

refer to the

the apparent

wye-connected load of

power supplied

to

Fig.

each branch

is

phases

are identical. In other words, the relationships in

Table

8A

apply to any balanced 3-phase load.

The apparent power supplied is

Example 8-5 Three identical impedances are connected across a 3-phase

a.

b.

is

550

V

apparent power

in delta

line (Fig. 8.15c). If the line

S

=

E ,

X

/

X

3

=

Solution

The current

:!:

In

3-phase balanced circuits,

factors. If the

in

each impedance

is

\ 3

El

\3

10 A. calculate the following:

The current in each impedance The value of each impedance [ft]

is

we can add

ers of the three phases because they

a.

the

.

total

current

to all three branches

obviously three times as great. * Consequently,

power

the apparent pow-

have identical power

factors are not identical, the apparent

powers cannot be added.


VOLTAGE AND CURRENT RELATIONSHIPS

TABLE 8A

Wye

IN

169

3-PHASE CIRCUITS Delta connection

connection

7/1.73

Figure 8.16a Impedances connected •

The current current

•

•

in

in

Figure 8.16b Impedances connected

wye.

each element

is

equal to the line

•

The voltage across each element line voltage E divided by V 3.

The voltages across

is

equal to the

•

the elements are 120° out of

in the

elements are 120° out of

•

•

case of a delta-connected load (Fig.

the apparent

power supplied S

= E X

each branch

to

is

equal to the line

The voltage across each element

The The

8.

1

6b),

is

equal to the

voltages across the elements are 120° out of

currents in the elements are 120° out of

power

/

is

the

same

Consequently, the

as for a

total

in

The relationship between active power P, reactive power Q, and apparent power S is the same for bal-

wye-connected

apparent power

and apparent 3-phase circuits

8.12 Active, reactive,

is

V3

is

load.

also the

anced 3-phase

We

circuits as for single-phase circuits.

therefore have 2 S = \'P +

same.

We

each element

phase.

phase.

which

in

divided by V3.

phase.

The currents

In the

/

delta

line voltage E.

phase. •

The current current

/.

in

therefore have

Q

2

(8. JO)

and

S =

V 3 El

cos 6

(8.9)

=

PIS

(8.

where

where

S

=

total

apparent power delivered by a

3-phase line

E— / = V3 =

[VA]

effective line voltage [V]

effective line current [A) a constant [approximate value

=

1.73]

= total 3-phase apparent power VA| = P total 3-phase active power [W| Q = total 3-phase reactive power |var| cos 0 = power factor of the 3-phase load 0 = phase angle between the line current S

|

and the line-to-neutral voltage

|°|

1

1


70

Example 8-6

A

3-phase motor, connected

power

a line current of 5 A. If the

motor a.

is

The

440V

to a

line,

factor of the

The

total active

The

total reactive

60 Hz

power power absorbed by

Solution

The

= V3

= V3 X 440 X

EI

=

3811

VA

=

3.81

kVA

total active

See Example

is

O

1

R

power

=

3.05

total reactive

is

X

3.81

The The

=

\ S

current in each line

is

8

1

V =

also

3.

resistance of each element

is

is

A

3. 15

1

5 A.

0.80

R =

- P2 =

2

each resistor 1

kW power

in

= P/E= 000 W/3

/

EII

=

-

318/3.15

101 12

Example 8-8

is

In the circuit

Q =

8-7.

The current

P = ScosO =

The

JT

5

b.

c.

O

Figure 8.17 apparent power

total

S

b.

w

line

machine

the

3000

=

9"

V

550

3-phase

apparent power

total

c.

The

A o

80 percent, calculate the following:

b.

a.

P

draws

\ 3.81

2

-

3.05

2

a.

b.

2.28 kvar

of Fig.

8.

1

8,


The current in each line The voltage across the inductor terminals

Solution

8.13 Solving 3-phase circuits

A

a.

balanced 3-phase load may be considered

composed of

three

identical

Consequently, the easiest cuit

is

to consider

examples

way

to

be

single-phase loads.

method

4

{

to

is

composed of an

inductive reactance

fl in series with a resistance

R=

3

II.

Consequently, the impedance of each branch

Z

to solve such a cir-

only one phase. The following

illustrate the

Each branch

X =

~~~

V4-

3^

=

5

The voltage across each branch

a

(2.i:

is

be employed.

ELN = ELN3 = 440 V/V3 - 254 V Example 8-7

The current

Three identical

3000 550

V

W

resistors dissipating a total

are connected in

wye

in

/

across a 3-phase (50.8

A

is

- E LN /Z -

The current

The value of each

in

each

line

resistor

440

Solution a.

The power dissipated by each

resistor

V

3-phase line

is

P = 3000 W/3 = 1000W The voltage across

the terminals of each resistor

is

Figure 8.18

E — 550 VA/3 -318 V

See Example

254/5

=

50.8

also the line current.)

Calculate

b.

is

power of

line (Fig. 8. 17).

a.

each circuit element

8-8.

is

A


The voltage across each inductor

b.

E = IX = 50.8 X = 203.2 V V

In a

is

wye connection

1

impedance per phase

the

7

is

understood to be the line-to-neutral impedance. The

4

voltage per phase

is

simply the

line voltage divided

by V3. Finally, the current per phase

equal to the

is

line current.

Example 8-9

A

identical capacitors If

Hz

3-phase 550 V, 60

the line current

is

line

is

connected

connected to three

in delta (Fig. 8.19).

22 A, calculate the capacitance

of each capacitor.

The assumption of a wye connection can be made not only for individual loads, but for entire load centers such as a factory containing motors,

lamps, heaters, furnaces, and so forth.

We

assume

in

that the load center

is

connected

simply

wye and

proceed with the usual calculations.

Solution

The current /

each capacitor

in

=

/ L /V3

=

22 A/V3

Voltage across each capacitor

Capacitive reactance

X

Xc = EJI =

is

c

-

Example 8-10

A

A

= 550 V

of each capacitor

550/12.7

-

The capacitance of each capacitor

C=

12.7

1/(2tt

=

61.3

V

(line-to-line)

8.20a). If the plant

is

power

factor

total

of 415

kVA

3-phase line (Fig. is

87.5 percent lag-

ging, calculate the following:

43.3 il

b.

The impedance of the plant, per phase The phase angle between the line-to-neutral

c.

The complete phasor diagram

a. is

voltage and the line current

1/2tt/Xc

=

manufacturing plant draws a

from a 2400

X 60 X

43.3)

(2.11)

for the plant

Solution

|jlF

a.

We

assume a wye connection composed of impedances Z(Fig. 8.20b).

three identical

The voltage per branch

E= = The

Figure 8.19

See Example

8-9.

8.14 Industrial loads In

is

connected

=

in delta or in

capacitors, and so on, often have only three external

and there

is

no way

to tell

how

the inter-

connections are made. Under these circum-

we simply assume (A wye connection is

connection

stances,

that the

wye.

slightly easier to handle

than a delta connection.)

(8.9)

000/(2400 V 3

=

A

100

= b.

E/I

=

)

is

1386/100

13.9 12

The phase angle 0 between the line-to-neutral voltage 386 V) and the corresponding line current ( 100 A) is given by (

nal

is

S/(EV3)

Z=

wye. For ex-

V

= 415

ample, 3-phase motors, generators, transformers,

terminals,

1386

The impedance per branch

most cases, we do not know whether a particular

3-phase load

2400/V3

current per branch /

is

1

is in

cos 6 0

= power factor = = 29°

0.875

(8.11)


172

4000

V

3 phase

2400V

y

1

?\

3- phase line

—1((a)

1800 kvar 'Tan /

a = 100

1386

A

V

Figure 8.21 Industrial motor and

capacitor.

A delta-connected capacitor

bo-

is

See Example

bank rated

also connected to the line. If the

at

8-1

1

1800 kvar

motor produces an

output of 3594 hp at an efficiency of 93 percent and a (b)

power

factor of

90 percent

(lagging), calculate the

following:

c.

The The The

d.

The apparent power supplied by

a.

b.

power absorbed by the motor power absorbed by the motor reactive power supplied by the transmisactive

reactive

sion line the transmis-

sion line e. f.

g.

The transmission line current The motor line current

Draw

the complete phasor diagram for one

phase Solution a.

Figure 8.20 a. Power input

to

a

See Example

wye connection

b.

Equivalent

c.

Phasor diagram

The

factory.

current

in

P 2 = 3594 X

8-10.

and

is

0.746

equivalent to

= 268 kW 1

Active power input to motor:

of the factory load.

of the voltages

Power output of 3594 hp

currents.

= p 2^ = 2681/0.93 = 2883 kW

each phase lags 29° behind the b.

(3.6)

Apparent power absorbed by the motor:

line-to-neutral voltage.

Sm c.

The complete phasor diagram is shown in Fig. 8.20c. In practice, we would show only one phase; for example,

Ean

,

/.„

Q m = \Si=

A

1

P- n

= V3203 T^r 2883 2

395 kvar

.

5000 hp wye-connected motor

4000

2883/0.90

Reactive power absorbed by the motor:

and the phase angle

between them.

Example 8-11

= PJcos 0 = = 3203 kVA

V (line-to-line),

3-phase, 60

is

Hz

connected to a line (Fig. 8.2

1

).

c.

Reactive power supplied by the capacitor bank (see section 7.5):


Qc = -1800kvar Total reactive

cause the capacitors are connected

power absorbed by

assumed a wye connection

the load:

in delta,

1

and

1800

+

1395

we

try to

This

is

is

pacitor bank.

an unusual situation because reactive

being returned to the

the capacitor

most cases

line. In

bank furnishes no more than

Qm

The

Active power supplied by the line

P L = P m = 2883

5L

e.

= VPl + Qi = V2883 2 + = 2911 kVA

Transmission line current h.

f.

Motor

/m

g.

The

to recognize that

is

in

wye

same reactive power), the line current would lead £ LN by 90°. Consequently,

A

90° ahead of £, N That is the correct position for phasor /c no matter how the capacitor /c

.

is

bank

is

connected

internally.

Phase angle 0 L between the transmission current and

line is

(

-

405)

£ LN

cos6,,

2


= PJS L = 0.99

eL

=

2883/291

8°

(8.9)

4000)

4000) 462

A

is

£LN = 4000/V3 = 2309 V

420

A

462

A

420

A

462

A

420

A

462

A

Phase angle 9 between the motor current and the line-to-neutral voltage

4000

is:

V

3-phase

6

line

is:

is

= 5 m /(£L V3) = 3 203 000/(V3 X = 462 A

cos 0

/'/

(while gen-

is

= SJ(E^3) = 2911 000/(V3 X = 420 A

line current

of 260

kW

Apparent power supplied by the

solution

were connected

erating the

we draw

kilovars of reactive power, d.

if

follow the actual currents inside the ca-

the capacitors

power

we

for the motor. This can

create unnecessary phase-angle complications

QL = Qc + Q m = = -405 kvar

73

= power factor = = 25.8°

0.9

260

260

260

A

A

A

(The motor current lags 25.8° behind the voltage, as

shown

in Fig. 8.22a.)

Line current drawn by the capacitor bank /c

= Gc/(£lV3) = 800 000/(V3 X - 260 A 1

(b) is

—IH 1800 kvar

4000)

Figure 8.22 Phasor relationships

a.

for

one phase. See Example

8-11.

the

Where should phasor current /c be located on phasor diagram? The question is important be-

b.

Line currents. Note that the motor currents exceed the currents of the source.

"1

1


74

The

line current

(420 A) leads

Eln by

8° be-

cause the kvars supplied by the capacitor bank ex-

ceed the kvars absorbed by the motor.

The phasor diagram

one phase

tor

is

shown

in

Fig. 8.22a.

The circuit diagram and shown in Fig. 8.22b.

current flows are

We want to emphasize the ing a tual

wye

importance of assum-

connection, irrespective of what the ac-

connection

nection for

all

may

By assuming a wye conwe simplify the

be.

circuit elements,

Figure 8.23

The

sequence

are observed

in

the

Figure 8.24 The letters are observed

in

the sequence a-c-b.

in

the

letters

a-b-c.

calculations and eliminate confusion.

As

a final remark, the reader has no doubt no-

ticed that the solution of a 3-phase

problem

in-

volves active, reactive, and apparent power. The

impedance value of devices such

as resistors,

mo-

tors,

and capacitors seldom appears on a nameplate.

This

is

to

be expected because most industrial loads

involve electric motors, furnaces, lights, and so on,

which are seldom described

in

terms of resistance

and reactance. They are usually represented as de-

draw a given amount of power

vices that

power

The

situation

is

somewhat

3-phase transmission sistances fixed.

at a

given

factor.

lines.

different in the case of

Here we can define

re-

and reactances because the parameters are


to equivalent circuits

describing the behavior of individual machines such as induction motors

and synchronous machines.

In conclusion, the solution of

may L,

3-phase circuits

involve either active and reactive power or R,

C elements — and

8.15

sometimes both.

Phase sequence

Figure 8.25

The In addition to line voltage

letters

are observed

sequence

a-c-b.

and frequency, a 3-phase

system has an important property called phase sequence. Phase sequence

is

important because

it

de-

termines the direction of rotation of 3-phase motors

and whether one 3-phase system can be connected in parallel

with another. Consequently, in 3-phase

systems, phase sequence

quency and voltage

is

as important as the fre-

understanding of phase

sequence by considering the following analogy. Suppose the letters a, b, c are printed at 20° 1

tervals

on a slowly revolving disc

in-

(Fig. 8.23). If the

disc turns counterclockwise, the letters appear

in

the sequence a-b-c-a-b-c. Let us call this the posi-

are.

Phase sequence means the order three line voltages

We can get a quick intuitive

in

which

become successively

the

positive.

tive

sequence.

It

can be described

three ways: abc, bca, or cab.

in

any one of


If the same disc turns clockwise, the sequence (Fig. 8.24). We call this becomes a-c-b-a-c-b .

.

.

and

the negative sequence,

it

can be described by

rule:

When

175

using the double-subscript notation, the

sequence of the

first

subscripts corresponds to the

phase sequence of the source.

any one of three forms: acb ? cba, or bac. Clearly, there

a difference between a positive sequence

is

and a negative sequence.

In Fig. 8.17, the

Suppose we interchange any two in Fig. 8.23,

the result

is

letters

on the disc

while retaining the same counterclock-

wise rotation.

the letters

If

as

shown

becomes c-b-a-c-b-a

a and c

.

.

,

The sequence now which is the same as the

negative sequence generated by the disc in Fig. 8.24.

We

to

phase sequence of the source

be A-C-B.

Draw

sequence can be converted into a

negative sequence by simply interchanging two

the line voltages.

Solution

The voltages follow the sequence A-C-B, which is the same as the sequence AC-CB-BA-AC ....

ECB -E BA and shown

in Fig. 8.27.

is

We

can reverse the phase

line

by interchanging any two

se-

let-

quence of a 3-phase

conductors. Although this verted into a positive sequence by interchanging

change,

it

may appear to

be a

trivial

can become a major problem when large

letters.

Let us

now

busbars or high-voltage transmission lines have to consider a 3-phase source having

ter-

minals a, b ? c (Fig. 8.26a). Suppose the line voltages

£ AC -

is

the corresponding phasor diagram

Similarly, a negative sequence can be con-

any two

is

the phasor diagram of

Consequently, the line voltage sequence

conclude that for a given direction of rota-

tion a positive

ters.

known

are interchanged,

in Fig. 8.25. .

Example 8-12

that

£ab E hc E ,

,

Cil

revolving phasors

sweep past

be interchanged. In practice, measures are taken so

are correctly represented

shown

in

by the

As they

Fig. 8.26b.

be

such drastic mechanical changes do not have to

made

at

the last minute.

major distribution systems

The phase sequence of all is

known

in

advance, and

the horizontal axis in the conventional

any future connections are planned accordingly. counterclockwise direction, they follow the se-

quence If

Eab -E bc -Eca -Eab -E bc ....

we

direct our attention to the first letter in each

subscript,

we find that the sequence

The source shown sequence a-b-c.

in Fig.

8.26a

is

We can, therefore,

is

a-b-c-a-b-c

.

sequence

.

said to possess the state the

8.16 Determining the phase

following

Special instruments are available to indicate the

phase sequence, but ing

we can

also determine

two incandescent lamps and a

it

by us-

capacitor.

three devices are connected in wye. If

The

we connect

the circuit to a 3-phase line (without connecting the neutral), 3-phase

one lamp

source

der:

(a)

—

—

^/^^^^

Determining the phase sequence of a 3-phase

£"cb

source. b.

always burn brighter than

(b)

Figure 8.26 a.

will

The phase sequence is in the following bright lamp dim lamp capacitor

the other.

Phase sequence depends upon the order

in

the line voltages reach their positive peaks.

which

Figure 8.27

See Example

8-12.

£ ac

or-


176

3-phase (a) line

lamp

Figure 8.29 Method of connecting a single-phase wattmeter.

moves upwhen the connections between source and load made as shown in Fig. 8.29. Note that the ± cur-

In single-phase circuits the pointer (b)

scale are

rent terminal nal.

Figure 8.28 a.

When

is

connected to the

the wattmeter

upscale reading

Determining phase sequence using two lamps and

means

supply terminals

1,

is

that

±

potential termi-

connected

power

is

way. an

this

flowing from

2 to load terminals

3, 4.

a capacitor. b.

Resulting phasor diagram.

8,18 Suppose, for example, cuit

is

connected to a 3-phase

8.28a. If the brightly, the

lamp connected phase sequence

ages follow each other

which

is

to say in the

line, as

to is

phase

shown

C

sequence is

CB

line volt-

CB-BA-AC,

E BA EAC The

£"

,

.

,

shown

in

in Fig.

burns more

C-B-A. The

sequence

in the

corresponding phasor diagram

8.28b.

in Fig.

In a 3-phase,

3-wire system, the active power sup-

may be measured by two shown in Fig. equal to the sum of the two

single-phase wattmeters connected as 8.30.

The

total

power

is

wattmeter readings. For balanced loads, is

less than

1

if

the

power

00 percent, the instruments

give different readings. Indeed,

Power measurement in

3-phase, 3-wire circuits

plied to a 3-phase load

factor

8.17

Power measurement

that a capacitor/lamp cir-

if

the

will

power factor is

ac circuits

Wattmeters are used

measure active power

to

in

single-phase and 3-phase circuits.

Owing is built,

to its external

a wattmeter

meter and ammeter combined Consequently, rent terminals.

it

way

connections and the

may be considered in

it

to be a volt-

same box.

the

has 2 potential terminals and 2 cur-

One of

the potential terminals

one of the current terminals bears a signs are polarity

marks

that

±

sign.

and

The

±

determine the positive

when same time

or negative reading of the wattmeter. Thus, the

±

voltage terminal

as current

is

is

entering the

positive at the

±

current terminal, then

the wattmeter will give a positive (upscale) reading.

The maximum voltage and current the instrument can tolerate are shown on the nameplate.

Figure 8.30 Measuring power in a 3-phase, 3-wire two-wattmeter method.

circuit

using the


less

than

50 percent, one of the wattmeters will give We must then reverse the con-

177

Power measurement

8.19

a negative reading.

in

3-phase, 4-wire circuits

nections of the potential coif so as to obtain a reading of this negative quantity. In this case, the

of the 3-phase circuit

is

power

equal to the difference be-

The two-wattmeter method gives

the

active

balanced or un-

is

single-phase

three

that the

±

±

as

current terminal

potential terminal.

shown

When

is

total

in Fig. 8.3

power. 1

.

Note

again connected to the

the wattmeters are con-

nected this way, an upscale reading means that active

balanced.

Example 8-13 A full-load test on

a 10 hp, 3-phase

following results: P, the current in line

circuits,

wattmeters are needed to measure the

The connections are made

tween the two wattmeter readings.

power absorbed whether the load

4-wire

3-phase,

In

voltage

motor yields the

= +5950 W; P 2 = + 2380 W;

each of the three lines

600

is

V. Calculate the

10 A; and the

is

power

factor of

power is flowing from source A, B, C, N to the load. The total power supplied to the load is equal to the sum of the three wattmeter readings. The threewattmeter method gives the active power for both balanced and unbalanced loads.

Some

wattmeters, such as those used on switch-

boards, are specially designed to give a direct read-

the motor.

out of the 3-phase power. Figure 8.32 shows a Solution

megawatt-range wattmeter

Apparent power supplied to the motor

^3EI = V3 X 600 X

S =

= Active

10 390

power

is

circuit that

a generating station.

The

measures the

current trans-

formers (CT) and potential transformers (PT) step

10

down

VA

power supplied

in

the line currents and voltages to values

com-

patible with the instrument rating.

to the

motor

is

SSI

P = 5950 + 2380 = 8330 W cosO = P/S= 8330/10 390

=

0.80, or 80 percent p.

Example 8-14 >

When

the

the line

motor

Example 8-13 runs

in

current drops to 3.6

readings are

P,

A

at

p

and the wattmeter

- +1295 W; P 2 =

Calculate the no-load losses

no-load,

and power

LOAD

.

..

• :

p>

845 W. factor.

Solution

Apparent power supplied to motor

)

S = V3 EI = V3 X 600 X = 3741 VA

3.6

Figure 8.31 Measuring power

in

a 3-phase, 4-wire

circuit.

No-load losses are

P =

P,

+ P2 =

= 450

1295

-

8,20 Varmeter 845

A varmeter indicates

W

It is

Power factor

=

P/S

=

450/3741

=

0.12

= \2%

built the

the reactive

same way

power

as a wattmeter

ternal circuit shifts the line voltage

in a circuit.

is,

but an in-

by 90° before

it


178

Figure 8.32 Measuring active power

is

in

applied to the potential

employed tions

in the control

a high-power

coil.

circuit.

Varmeters are mainly

rooms of generating

and the substations of

electrical utilities

load resistance (Fig. 8.33). Furthermore, given

and

the

In 3-phase, 3-wire

balanced

We

circuits,

the

we

can cal-

two wattmeter

simply multiply the

differ-

ence of the two readings by V 3. For example,

two wattmeters

indicate

=

6 76 vars. Note that 1

surement

is

if

is

this

(5950

—

2380)

,

phase sequence of the

1

it

is

line voltages

£ l2

connected as indicated.

capacitive and

If the

in-

ductive reactances are interchanged, the 3-phase

system becomes completely unbalanced.

X

method of var mea-

only valid for balanced 3-phase circuits.

Example 8-15

A

800

phases

kW 1

= 440 Z

single-phase load

connected between

is

and 2 of a 440 V, 3-phase 0,

E23 = 440 Z -

line,

E ]2

wherein

E3l = 440 Z

120,

12(X

Calculate the load currents and line currents

8.21

It

A

remarkable single-phase to 3-phase transformation

sometimes happens

unity

power

3-phase

line.

that a large single-phase

a.

This can create a badly unbalanced it

three phases perfectly

is

possible to balance the

by connecting

the single-phase load

b.

When

is

connected alone

line

balancing reactances are added across

the remaining lines, as

shown

in Fig.

8.34

Solution a.

The

resistance of the single-phase load

a capacitive

reactance and an inductive reactance across the other two lines.

When

on the 3-phase

factor load has to be connected to a

system. However,

,

essential that the three impedances be

the

+5950 W and +2380 W repower

spectively, the reactive

V3

-2-3-

,

power from

readings (Fig. 8.30).

l

£23 E3l

large industrial consumers.

culate the reactive

impedances V 3 times greater than the value of the

sta-

The reactances must each have

R

E2 P

= 800 000

0.242

n

is


The current

in the

1

79

load and in two of the three

lines is

The

current in the third line

3-phase system b.

By

is

and so the

zero,

is

badly unbalanced,

introducing capacitive and inductive reac-

tances having an impedance of 0.242

we

0.4 19 fl,

V3 =

obtain a balanced 3-phase

line, as

demonstrated below. Taking successive loops

around the respective Section 2.32),

E ]2 -

elements

in Fig.

2.38

£ 3] -

j

obtain the following results:

= 0 /. /, = 4A3E l2 = X 440Z0 = 1817Z0

4.13

£23 +

we

0.242/,

£12

2.38

circuit

and using Kirchhoff's voltage law (see

8.34,

= 0 j 0.419 U X 440Z(-120 + 0.419

/3

-

0

= j 2.38 £ 23 = 1047^-30

U =

.'.

90) /3

X 440Z(120 + 90 -

=

£ 31 = = 1047Z30

-j 2.38

180)

£23

Applying Kirchhoff's current law and

Figure 8.33

A single-phase resistive load can be transformed a balanced 3-phase load.

3,

we

to

nodes

1

,

2,

obtain

into

U

=

/,

~

/3

= 1817Z0 - 1047Z.30 = 1817 - 907 - j 523 = 1047/1-30 /b

Ic

= = = = =

h ~ h 1047Z-30 - 1817Z0 907 - j 523 - 1817 -907 — j 523 1047Z210

= h ~ h = 1047Z30 - 1047Z.-30 = 907 + j 523 - 907 + j 523 =1047j = 1047Z90

Figure 8.34

Thus, / A / B /c make up a balanced 3-phase system because they are equal and displaced at 120°

See Example 8-14.

to

,

,

each other (Fig. 8.35).

180


Tm

e.

8-6

The ohmic value of each

a.

What

b.

Could we reverse

is

resistor

the phase sequence in Fig. 8.10? it

by changing the direction

of rotation of the magnet?

8-7

A 3-phase draws a

motor connected

line current

to a

600

V

line

of 25 A. Calculate the

apparent power supplied to the motor. 8-8

Three incandescent lamps rated 60 W, 120 are connected in delta.

needed so

£ 23

8-9

8-14. b.

fl resistors are

new power


8-10

If

one

cut,

is

Practical level 8-

1

wye-connected generator

V

in

each of

its

windings.

8-11

voltage of 100

terminals is

in Fig. 8.9

V

What

is

1,

a

to a

8-3

8-12

at

240°

each of these

b.

If

c.

R =

is

£al Could we ahead of£ al ? .

Eb2

is

120°

lines a-b-c of Fig.

15 (2,

what

resistors are

line voltage is 13.2 is

a.

b. c.

d.

the line current if the resistors are in delta?

known

to

be connected

in

wish to apply full-load

100 kVA,

to a

each resistance

the elements are connected

wye

a.

In

b.

In delta

The windings of a 3-phase motor nected

in delta. If the resistance

two terminals

is

0.6

II,

what

is

are con-

between

the resis-

the voltage across each resistor? is

the current in

tance of each winding?

each line?

Calculate the power supplied to the 3-phase

Three

the resistors are

also say that

8-14

Three 24 fl resistors are connected across a

load.

8-5

is

if

wye?

the resistors are

If

We

if

1

What

the line current in

kW when

V, 3-phase line.

load. Calculate the value of

The voltage between 8.12 is 620 V. a.

What

208

4 kV, 3-phase generator using a resistive

8- 3

8-4

is

wye, calculate the resistance of each.

instants?

Referring to Fig. 8.9c, phasor

is

c.

the instantaneous value of the volt-

behind phasor

Eh2

240°, and 330°.

each of these instants?

is

connected

the polarity of terminal a with re1 at

What

connected b.

at 0°, 90°, 120°,

age across terminals 2, b

same

conductor of a 3-phase line

the load then supplied by a single-

heater dissipates 15

generates a peak

per phase.

spect to terminal c.

3-phase load?

supplied to the load.

A 3-phase

Calculate the instantaneous voltage between

What

to the

in delta

burns out, calculate the

line

connected a.

The generator

b.

connected

in-

Calculate the line voltage.

a.

line

V

is

phase voltage or a 2-phase voltage?

A 3-phase

duces 2400

8-2

one

the fuse in

If

line voltage

lamps burn normally?

on a 208 V, 3-phase line. a. What is the power supplied

Figure 8.35

See Example

Three 10

that the

What

connected

kV and

in delta

V, 3-phase line. Calculate the

resistance of three elements connected in

in delta. If the

wye

the line current

1202 A, calculate the following: The current in each resistor The voltage across each resistor The power supplied to each resistor The power supplied to the 3-phase load

600

8-15

that

A 60 hp

would

dissipate the

3-phase motor absorbs 50

from a 600 V, 3-phase

b.

kW

line. If the line cur-

is 60 A, calculate the following: The efficiency of the motor The apparent power absorbed by the motor

rent a.

same power.


The reactive power absorbed by The power factor of the motor

c.

d.

8-16

Three 15

voltage

If the line

530

is

motor

Q reac-

in Fig. 8.18.

The

8-24

V, calculate the

An

industrial plant

2.4

kV

8-

1

7

The voltage across each

W lamps and a

Two 60

connected

in

wye. The

to the terminals outlet.

nal Y,

What

a.

1

resistor

b.

0

|jlF

circuit

capacitor are

8-

1

8

connected

is

V

X-Y-Z of a 3-phase 20 1

8-25

wye

1

Two kW,

the phasor diagram for the line voltages.

b.

The

0 jxF capacitors are connected

An

In

Problem

to terminal

8-20

Hz

line.

Three

1

5

If the

power generated

1

is

X, which lamp will be brighter?

connected

resistors

to a in

in 3

8-27

absorb 60

3-phase

line. If

a. b.

H resistors (R) and three 8 fl reline.

Without draw-

8-28

R and X in series, connected in wye R and X in parallel, connected in delta R connected in delta and X connected in wye

8-23

In

is

in

wye and

that the

motor

R

with an inductive reactance X. Calculate the values of

R and

X.

power

output.

energy does the motor consume

h?

The The

load

power

factor;

line current if the line voltage

630

is

V.

A 20 H resistor is connected A and B of a 3-phase, 480 V

between line.

lines

Calculate

Two 30

il resistors are

AB

and

BC

connected between

of a 3-phase 480

V

line.

Calculate the currents flowing in lines A, B, and C, respectively.

is

8-30

A

1

50 kW, 460

V, 3-phase heater

stalled in a hot water boiler.

that each branch can

be represented by a resistance

a.

motor has an efficiency of 85 percent,

balanced, calculate the following:

phases

50 Hz instead of 60 Hz.

Problem 8-15, assume

connected

V

from a 600

respectively.

8-29

In Fig. 8.19, calculate the line current if the

frequency

A

line.

the currents that flow in lines A, B, and C,

rent for each of the following connections:

8-22

the load


ways

ing a phasor diagram, calculate the line cur-

c.

.5

1

16 A,

is

The wattmeters in Fig. 8.30 register +35 kW and —20 kW, respectively. If the load is

wye, calculate the

absorbed.

across a 530 V, 3-phase

b.

kW and

motor having a cos 0 of 82 per-

How much

connected

actors (X) are connected in different

a.

line indicate 3.5

calculate the mechanical

8- 7, if the capacitor

new power 1

V

Calculate the active power supplied to the

a.

Calculate

b.

Three delta-connected

kW when

deter-

8,

motor.

they are reconnected

8-2

electric

3-phase

in

c.

8-19

1

respectively. If the line current

The apparent power The power factor of

b.

line current

reactive

8.

the resistance and reactance.

cent draws a current of 25

across a 2300 V, 60

The

can be represented by

calculate the following: a.

the following: a.

that the plant

wattmeters connected into a 3-phase,

3-wire 220

level

Three

a

imped-

the equivalent line-to-neutral

mine the values of

8-26

Advanced

kVA from

factor of 80 percent

an equivalent circuit similar to Fig.

the phase sequence?

is

is

Assuming

b.

The capacitor is connected to termiand the lamp that burns brighter is

What Draw

draws 600

power

ance of the plant?

connected to terminal X. a.

line at a

lagging.

and apparent power sup-

active, reactive,

plied to the 3-phase load b.

the phase angle between the line cur-

is

and the corresponding line-to-neutral

voltage?

following: a.

What

b.

rent

and three 8

ft resistors

connected as shown

tors are

the

181

does

in series

8-3

1

it

produce

Three 5

if

the line voltage

il resistors are

across a 3-phase 480

V

connected line.

is

in-

What power is

in

470 V?

wye

Calculate the

1

82

ELECTRICA L MACHINES AND TRANSFORMERS

has a full-load efficiency of 93.6% and a

current flowing in each. If one of the resistors

is

that flows in the

8-32

One

power

disconnected, calculate the current

remaining two.

of the three fuses protecting a 3-phase

200 kW, 600 V is removed so as to reduce the heat produced by the boiler. What power does the heater de-

factor of

83%. Calculate

the following:

b.

The active power drawn by the motor; The apparent power drawn by the motor;

c.

The

a.

full-load line current;

electric heater rated at

A 450 kW, duces

1

300

a.

575 lb

V,

3-phase steam boiler pro-

8-34

is

612

produced

if

A 40 hp, 460 V, TEFC, premium

1 1

Company

V, 3-phase,

Assuming

the

is

32

in,

450 kg Square

D

used to drive a 1600 hp,

60 Hz squirrel-cage motor.

motor has

a

minimum

96%

effi-

and 90%,

What

is

the reactive

power drawn from

the

line at full-load?

80 r/min, 3-phase, 60 Hz

manufactured by Baldor Electric

X

in

delivered by the controller. b.

efficiency induction motor

X 24

respectively, calculate the full-load current

the line

V.

in

ciency and power factor of

of steam per hour. Estimate

the quantity of steam

voltage

A 92

motor controller

2400

velop under these conditions?

8-33

8-35

c.

What

is

the phase angle

neutral voltage

and the

between the

line current?

line-to-

Chapter 9 The Ideal Transformer

9.0 Introduction frequency/

The

transformer

ful electrical

is

probably one of the most use-

devices ever invented.

It

can raise

or lower the voltage or current in an ac circuit, isolate circuits

from each

other,

and

it

it

can

can increase

or decrease the apparent value of a capacitor, an inductor, or a resistor Furthermore, the transformer

enables us to transmit electrical energy over great distances and to distribute

it

safely in factories

and

homes.

We

will study

transformers

some of

the basic properties of

in this chapter. It will

help us under-

stand not only the commercial transformers covered in later

chapters but also the basic operating princi-

ple of induction motors, alternators,

and synchro-

nous motors. All these devices are based upon the

Figure

laws of electromagnetic induction. Consequently,

a.

we encourage

9.1

b.

covered here.

Voltage induced

in

a coil

A

itive

9.1

voltage

able

the reader to pay particular attention

to the subject matter

A

is

induced

in

a

coil

when

coil of Fig. 9. la,

links) a variable flux

soidally at a

The

which surrounds

(or

links a vari-

sinusoidal flux induces a sinusoidal voltage.

and negative peaks

$ max

.

The

induces a sinusoidal ac voltage

Consider the

it

flux.

effective value

is

alternating flux

in the coil,

whose

given by

flux alternates sinu-

E=

frequency/ periodically reaching pos-

183

4.44/7V
(9. 1)

1


84

where

9.2 Applied voltage

E =

effective voltage induced [V]

/=

frequency of the flux [Hz]

= number of turns

/V

^max = peak value of

=

4.44 It

on the

and induced voltage

and draws a current / m

[Wb]

the flux


=

is

It

even by an ac current that flows

— N

is

As

in

any inductive is in

The

Thus,

derived from Faraday's law equation

which AcjVAr

A(|)/Ar in

of flux and c

b,

1

when

and so the voltage

AcJ)/Ar

is

is

change

the instantaneous induced voltage.

is

in Fig. 9.

the flux

the rate of

is

the flux

time, the rate of change Ac|)/Ar is

is

is

increasing with

greater than zero

positive. Conversely,

when

decreasing with time, the rate of change

less than zero; consequently, the voltage

negative. Finally,

when

the flux

is

neither in-

creasing nor decreasing (even for one microsecond), the rate of change Act>/A/

voltage

given by

is

circuit, / m lags

90° behind

phase with the current (Fig. 9.2b).

detailed behavior of the circuit can be ex-

c|>

density

sinusoidal current ,

which

Consequently,

I

is

peak flux

Bm

.

]X

is

which,

use the peak

rms value? The reason

m

induces an effective voltage

of the

coil,

whose value

E

E must be

is

cIJ

milx

given by Eq.

is

The

.

E a and

flux

9.

1

.

On

the

the induced

identical because they appear be-

tween the same pair of conductors. Because

we can

.

across the terminals

other hand, the applied voltage

voltage

a sinusoidal

sinusoidal flux

called the magnetizing cur-

£u =

= 4.44/MP max

from which we obtain

is

(9.2)

4.44 JN

determines the

level of saturation.

Example 9-1 The coil in Fig.

.

9.1

possesses 4000 turns and links

an ac flux having a peak value of 2

quency

is

mWb.

If

the fre-

60 Hz, calculate the effective value and (a)

frequency of the induced voltage E. Sol til ion

4.44/W* inax

(9.1)

4.44

X 60 X 4000 X

2131

V

0.002 (b)

The induced voltage has an of 2131

age

is

V

effective or

RMS

Figure 9.2 value

a.

and a frequency of 60 Hz. The peak volt-

2131 V2

= 3014

V.

£,

write

proportional to the peak flux in iron cores,

m produces

rent.The peak value of this ac flux

zero, and so the

why do we

also arises:

/

in turn creates a

zero.

max instead of the

that the

is

The

mmf /V7 m

E, is

The question flux

Xm

of the coil

plained as follows:

Eq. 9.1

is

If the resistance

.

negligible, the current

and

in the coil

itself.

e

to a si-

2 tt/V2)

be created by a moving magnet, a nearby ac

coil, or

connected

coil of /V turns

coil

does not matter where the ac flux comes from:

may

shows a

Fig. 9.2a

nusoidal ac source £„. The coil has a reactance

The voltage E induced plied voltage

b.

Eg

in

a

coil is

equal

to the

ap-

.

Phasor relationships between

£fg! E,

/

m and ,

.

THE IDEAL TRANSFORMER

This equation shows that for a given frequency and a given to the

number of

turns,

(I>

lllax

varies in proportion

applied voltage E„. This means that

if

£g

is

kept constant, the peak flux must remain constant.

For example, suppose we gradually

a.

b. c.

d.

The The The The

185

peak value of flux

mmf

peak value of the

inductive reactance of the coil

inductance of the coil

insert an

Solution iron core into the coil while

keeping E„ fixed (Fig.

The peak value of

the ac flux will remain ab-

solutely constant during this operation, retaining original value

$ max

even when the core

is

its

com-

pletely inside the coil. In effect, if the flux increased (as

we would

also increase. at

But

this is

we

said,

E„

is

is

therefore the same.

netizing current

core

is

flux, a

The peak current

is

lm

is

much

Wik) = ^2/ = =

kept fixed.

For a given supply voltage E„, the ac flux

and 9.3

b.

E would impossible because E — E„

(9.2)

90)

expect), the induced voltage

every instant and, as

9.2

= EJ(4MJN) = 120/(4.44 X 60 X = 0.005 - 5 mWb

a.

9.3).

in Figs.

The peak

However, the mag-

smaller

when

the iron

U=

same

-

inside the coil. In effect, to produce the

smaller magnetomotive force

is

in Fig. 9.3 is

much

X 4

A

5.66

is

NI m = 90 X 5.66 509.

1

needed with

The

an iron core than with an air core. Consequently, the

magnetizing current

mmf U

>/2

smaller than c.

in Fig. 9.2.

flux

is

equal to 5

mmf is

the coil

509.

1

mWb at the

The inductive reactance

d.

The inductance

when

is

X m = EJI m = = 30

instant

ampere-turns.

120/4

SI

is

L = XJ2irf

=

30/(2tt

(2.10)

X

60)

- 0.0796

= 9,3

79.6

mH

Elementary transformer

In Fig. 9.4, a coil

having an

air

core

an ac source E„. The resulting current (b)

a total flux

around the

<1>,

which

we

coil. If

Figure 9.3

The

a.

is

flux in the coil

remains constant so long as

Eg

first,

flux.

constant.

An

second

Phasor relationships.

b.

the

it

A

coil

Hz

having 90 turns

is

in

produces the space

bring a second coil close to

ac voltage

and

its

E2

is

therefore induced

in

the

value can be measured with a

The combination of the two coils is The coil connected to the

called a transformer. is

connected

to a

120 V, 60

source. If the effective value of the magnetizing

current

dispersed

excited by

/m

will surround a portion 4> ml of the total

coil

voltmeter.

Example 9-2

is

is

4 A, calculate the following:

source

is

called the primary winding (or primary)

and the other one (or secondary).

is

called the secondary winding

1

ELECTRICAL MACHINES AND TRANSF ORMERS

86

Figure 9.5 Terminals having the

Figure 9.4

marked

Voltage induced

m1

leakage

;

in

a secondary winding. Mutual

flux is

f1

flux is

.

value

voltage exists only between primary terminals is

and secondary terminals 3-4, respectively.

voltage exists between primary terminal

ondary terminal trically isolated

The

flux

up into two

3.

The secondary

is

No

and sec-

1

therefore elec-

from the primary.

parts: a

mutual flux

4> ml

,

which

links the

same

does. Suppose, during

instant as

apart, the

mutual flux

total flux

,

two

we

coils

is

is

very small compared to the

then say that the coupling between

weak.

We

two

coils closer together.

two

coils touch, the

)

by bring-

However, even

mutual flux will

small compared to the total flux pling

is

weak, voltage

E2

is

.

When

still

be

the cou-

relatively small and,

when

still,

load

connected across the secondary terminals. In

most

it

collapses almost completely

industrial transformers, the

and

is

a

primary and sec-

ondary windings are wound on top of each other

positive with respect to sec-

(Fig. 9.5).

Terminals

1

and

3 are

terminal

1

and another large dot beside secondary

The dots are called polarity marks. The polarity marks in Fig. 9.5 could equally

terminal

3.

well

be placed beside terminals 2 and 4 because, as the voltage alternates, they too,

become simultaneously

positive, every half-cycle. Consequently, the polar-

marks may be shown beside terminals

beside terminals 2 and

1

and 3 or

4.

9,5 Properties of polarity

A

transformer

is

marks

usually installed in a metal enclo-

sure and so only the primary and secondary terminals are accessible, together with their polarity marks. But

although the transformer

may

lowing rules always apply

to

not be visible, the fol-

to polarity

marks;

improve the coupling between them.

A current entering a polarity- marked terminal produces a mmf that acts in a "positive'' direc-

9.4 Polarity of a transformer

tive" direction* (Fig. 9.6). Conversely, a current

1

.

tion.

In Fig. 9.4 fluxes

On

and

3> ml are both

stant.

It

They

also pass through zero at the

follows that voltage

E2

will reach

same its

in-

peak

As

a result,

it

produces a flux

in the "posi-

flowing out of a polarity-marked terminal pro-

produced by

magnetizing current /nr Consequently, the fluxes are in phase, both reaching their peak values at the same instant.

1

that

if

worse is

primary terminal

then said to possess the same polarity. This sameness

ity

E2

bring the secondary right up to the primary so

that the

secondary terminal 3

ondary terminal 4

can obtain a better cou-

pling (and a higher secondary voltage ing the

that

positive with respect to primary terminal 2

which

links only the turns of the primary. If the coils are far

we

at the

can be shown by placing a large dot beside primary

created by the primary can be broken

turns of both coils; and a leakage flux

the

instantaneous polarity are

one of these peak moments,

A 1-2

same

with a dot.

*

"Positive" and "negative" are

cause

we can

shown

in

quotation marks be-

rarely look inside a transformer to see in

which direction

the flux

is

actually circulating.


187

9.6 Ideal transformer at no-load;

voltage ratio Before undertaking the study of practical, commer-

we

cial transformers,

shall

examine

the properties

of the so-called ideal transformer. By definition, an ideal transformer has

no losses and

nitely permeable. Furthermore,

by the primary ondary,

is

its

core

is infi-

any flux produced

completely linked by the sec-

and vice versa. Consequently, an

ideal

transformer enclosure

transformer has no leakage flux of any kind.

current entering a polarity-marked terminal pro-

duces a

have properties which ap-

Practical transformers

Figure 9.6

A

flux in

a "positive" direction.

proach those of an ideal transformer. Consequently, our study of the ideal transformer will help us understand the properties of transformers Figure 9.8a shows an ideal transformer

duces a mint and flux

in the

"negative" direction.

Thus, currents that respectively flow into and out

2.

turns.

The primary

of polarity-marked terminals of two coils pro-

E„ and

The

If

one polarity-marked terminal

is

momentarily

positive, then the other polarity-marked terminal to

momentarily positive (each with respect

is

its

in

primary and secondary respectively possess

duce magnetomotive forces that buck each

other.

in general.

is

4

the

and

N2

connected to a sinusoidal source

the magnetizing current / U1 creates a tlux

flux

is

m

it

is

a mutual flux.

flux varies sinusoidally, and reaches a peak value

)

mrtx

.

According

to Eq. 9.

1,

we

can therefore write:

other terminal). This rule enables us to re-

late the

*m-.^.

phasor voltage on the secondary side

ti

with the phasor voltage on the primary side.

For example, with phasor

in Fig. 9.7,

£ ab

phasor

£dc

is in

phase

.

(a)

(a) t:

g

,

A',

(b)

(b) 4>rr

Figure 9.7 a.

Instantaneous polarities

b.

Phasor

rent

is

increasing. relationship.

when

the magnetizing cur-

Figure 9.8 The ideal transformer at no-load. Primary and secondary are linked by a mutual flux. b. Phasor relationships at no-load.

a.

.

completely linked by the primary and sec-

ondary windings and, consequently,

The

which /V,

1


88

=

£,

4.44/JV, lllilx

(9.3)

Calculate: a.

The

effective voltage across the secondary

and terminals

E2 = From

4.44./yV2

we deduce

these equations,

the voltage ratio

O max

(9.4)

b.

The peak voltage across

the expression for

c.

The instantaneous voltage across the secondary when the instantaneous voltage across the primary is 37 V

and turns ratio a of an ideal trans-

former:

the secondary terminals

Solution: a.

The

turns ratio

N2 /N

where £|

The secondary voltage

N2 =

numbers of turns on

voltage induced

secondary [V]

in the

on the primary

turns

25 times more

£ 2 = 25 X

E

Furthermore, because the primary

If

the transformer has fewer turns

b.

The voltage

on the sec-

As

is

£,

shorter

any inductor, current / m lags 90° behind applied voltage Zs The phasor reprethan phasor

£",.

obviously

phase with magne-

senting flux

is

which produces

However, because magnetic circuit

this is

is

.

in

varies sinusoidally; consequently,

P cak)=

V 2£ = V 2 X 3000

=4242 V c.

The secondary voltage at

e 2 = 25

permeable and so

diagram of such a transformer 9.8b except that phasor / m

is

is

identical to Fig.

infinitesimally small.

primary and 2250 turns on the secondary source.

tween the primary and secondary is

4 A.

x

37

E

x

V

Pursuing our analysis,

I 2 will I2

con-

Does

The coupling be-

swer

is

is

perfect, but the

let

us connect a load

Z across A

the secondary of the ideal transformer (Fig. 9.9).

transformer having 90 turns on the

magnetizing current

=

e

current ratio

secondary current

Hz

when

X 37 = 925 V

Example 9-3

nected to a 120 V, 60

25 times greater than

9.7 Ideal transformer under load;

required to produce the

is

is

every instant. Consequently,

it.

Thus, under no-load conditions, the phasor

A not quite ideal

is:

an ideal transformer,

infinitely

no magnetizing current

$m

V

E 2 = 3000 V

.

tizing current I m

flux

90/2250

in

tT

the

apply Eq. 9.5:

l

=

the peak secondary voltage

2(

E2

we can

/E 2 =N /N 2

\20/E2

180° out of phase) as indicated by the polarity

marks.

120

in Fig.

phase with phasor £, (and not

ondary than on the primary, phasor

]

which again yields

,

E2

X

25

Instead of reasoning as above,

equal to the ratio of the

is

and secondary voltages are induced by the same

9.8b. Phasor

=

Ei

-3000 V

and secondary voltages turns.

Consequently:

turns.

turns ratio

number of

therefore 25 times greater

is

than the primary voltage because the secondary has

the secondary

This equation shows that the ratio of the primary

mutual

(9.5)

voltage induced in the primary [V]

numbers of

=

-2250/90

x

= 25

=

E2 = N\ = a

is:

E2

immediately flow, given by:

= E2 IZ

we connect the load? To anwe must recall two facts. First, in

change when

this question,

an ideal transformer the primary and secondary windings are linked by a mutual flux

m and by no ,

other

THE IDEA L TRANSFORMER

same

the

<*W:

when

time. Thus,

maximum +

is

be

E

To

N2

N,

{

mark on

when

effect,

on the primary

/,, )

/,

lags behind

behind

Ee

,

we can now draw

facts,

polarity

a resistive-inductive load, current

E2

by an angle

Flux

6.

but no magnetizing current

produce

the phasor

under load (Fig.

ideal transformer

Assuming

9.9b).

must flow out of the

the secondary side (see Fig. 9.9a).

Using these

to

(

flows into a polarity mark

/[

side, I 2

diagram of an

I2

maximum +

phase. Furthermore, in order to produce the

in

bucking

(a)

I 2 is

89

In other words, the currents must

).

(

goes through zero,

/2

goes through zero, and when

1

this flux

because

this is

/

lags 90°

m

m

is

needed

an idea trans-

former. Finally, the primary and secondary currents

(b)

According

are in phase.

to Eq. 9.6, they are related

by the equation:

Figure 9.9 a.

b.

AT,

under load. The mutual mains unchanged. Phasor relationships under load. Ideal transformer

(9.7)

flux re-

where

f = primary flux. In

other words, an ideal transformer, by defini-

tion,

has no leakage flux. Consequently, the voltage

ratio

under load

is

the

same

/Vj

as at no-load, namely:

the

if

E„

the supply voltage

is

E

induced voltage

primary

E2

E2

also remains fixed.

remains fixed whether a load Let us

remains fixed. {

now examine

the

is

It

mmf would produce

in current

not.

the mutual flux

change under

m But we just saw .

load.

only remain fixed

We if

conclude

If

it

acted

that

in

m does not

m can

the primary develops a

mmf

which exactly counterbalances Thus, a primary current

/j

that flux

N2 I 2

at

every

must flow so

AVi = N 2 I 2 To obtain

.

profound change

instant.

f and

I2

what we gain

and vice versa. This

is

see that the

in voltage,

the primary

E2 l 2

we

lose

consistent with the

Ef

to x

must equal the apparent power output

power inputs and outwould mean that the transformer itself absorbs power. By definition, this of the secondary.

If the

puts were not identical,

is

impossible

in

it

an ideal transformer.

Example 9-4 ideal transformer

having 90 turns on the primary

and 2250 turns on the secondary

is

connected

to a

(9.6)

200

the required instant-to-instant bucking ef-

fect, currents

ratio. In effect,

we

the inverse of the volt-

is

requirement that the apparent power input

An

that:

9.5 and Eq. 9.7,

transformer current ratio

age

mmf N2 I2 a

turns ratio

Comparing Eq.

We conclude that

magnetomotive forces

current I 2 produces a secondary

—

connected or

created by the primary and secondary windings. First,

alone, this

turns on the secondary

kept fixed, then

Consequently, mutual flux

number of

on the primary

x

a Second,

secondary current [A]

= number of turns

N2 =

- N /N2

£,/£ 2

=

I2

current |A)

must increase and decrease

at

V,

50 Hz source. The load across the secondary

draws a current of 2 cent lagging (Fig.

A

at

9. 10a).

a

power

factor of 80 per-

90

1


Figure 9.10 See Example 9-4. b. Phasor relationships.

a.

current in the secondary. Therefore

Calculate: a.

The

b.

The instantaneous

effective value of the primary current

current

in the

100

mA,

primary when

/,

1/

100

is

mA

c.

The peak

d.

Draw

flux linked

by the secondary winding

the phasor diagram

c.

a.

The

is

The peak

turns ratio

= ?s / — '2 instantaneous = 25 X 0. = 2.5 A

the

same

as that linking the primary.

flux in the secondary

=

1/25

current ratio

is

= E^/(4MfN ) = 200/(4.44 X 50 X = 0.01 ]

therefore 25 and because the

10

primary has fewer turns, the primary current

25 times greater than the secondary current.

25

X

2

d.

by means of Eq. N,/, 90/, /,

b.

To draw

= 50 A

Instead of reasoning as above, the current

mWb

the phasor diagram,

lows: Secondary voltage

=

we can

The instantaneous current

E2

is

in

Phase angle between

in the

power primary

is

al-

ways 25 times greater than the instantaneous

as fol-

25

X 200

phase with £, indicated by the polarity

marks. For the same reason, 2

we reason

is

E2 = 25 X Ei = = 5000 V

calculate

9.6.

= N 2 l2 - 2250 X = 50 A

90)

is

Consequently: /,

is

is:

^=yv,/yv 2 -90/2250

The

=

In an ideal transformer, the flux linking the sec-

ondary Solution:

l2

is:

instantaneous

1

the instantaneous current in the secondary

when

/j

is

in

E 2 and

l 2 is

factor

=

cos 6

0.8

=

cos e

9

=

36.9°

phase with

I2

.

THE IDEA L TRA NS FORMER

Figure 9.11 Symbol for an b. Symbol for an

a.

ideal transformer ideal transformer

and phasor diagram using sign notation. and phasor diagram using double-subscript

The phase angle between £, and /, is also 36.9°. The mutual tlux lags 90° behind E„ (Fig. 9.10b).

In

and

for

an

1

To highlight the bare former,

it is

9.1 lb),

ideal transformer

best to

/,

la,

draw

in

it

symbolic form. Thus,

in-

and the mutual tlux ing primary

4> m

,

we

(Fig.

9.

1

1).

to indicate the

direction of current flow as well as the polarities of

E

voltages into

,

and

£2 F° r example, -

a current

one polarity-marked terminal

panied by a current

is

ity-marked terminal. Consequently,

ways

in

{

flowing

always accom-

/,

and

U

are al-

phase.

/N 2 =

a,

we

and

E2

are always in phase,

and so are

double-subscript notation

£ab

/2

and

Ecd

are

always

(which

in

used (Fig.

is

phase and so are

.

may sometimes

the nature of the load

be a source) connected to

the secondary side.

9.9

Impedance

ratio

Although a transformer

is

generally used to trans-

form a voltage or current,

it

also has the impor-

tant ability to transform an

impedance. Consider,

for example, Fig. 9. 12a in

which an

former

T

load Z.

The

is

ideal trans-

connected between a source E„ and a ratio of transformation

is

a,

and so we

can write

Furthermore,

N

/[

flowing out of the other polar-

I2

x

simply show a box hav-

and secondary terminals

marks are added, enabling us

Polarity

and

E

The angle a depends upon

essentials of an ideal trans-

stead of drawing the primary and secondary windings

notation.

/ 2 .*

If the

symbol

9.8 Circuit

9

an ideal transformer, and specifically referring to

Fig. 9. /,

1

if

we

let

the ratio of transformation

obtain

£,

*

= a£ 2

Some

texts

show

the respective voltages

and currents as being

180° out of phase. This situation can arise depending upon

how

the behavior of the transformer

is

described, or

how

the

By

us-

voltage polarities and current directions are assigned.

and

ing the methodology /,

=

/ 2 /a

never any doubt as to

we have adopted how the phasors

in this

book, there

should be drawn.

is

92

1


impedance across

the actual

tical to

the secondary

terminals multiplied by the square of the turns

ratio.

The impedance transformation is real, and not illusory like the image produced by a magnifying

An

glass.

(a)

ideal transformer can

of any component, be ductor. For example,

modify the value

a resistor, capacitor, or in-

it

a 1000 12 resistor

if

placed

is

across the secondary of a transformer having a pri-

mary

secondary turns ratio of

to

across the primary as

X

(1/5)"

= 40

12.

if

it

Similarly,

a reactance of 1000 fl

ondary,

it

1

:5,

will appear

it

had a resistance of 1000

is

if

a capacitor having

connected to the sec-

appears as a 40 il capacitor across the

primary. However, because the reactance of a capacitor

—

(X c Figure 9.12 a. Impedance transformation using a transformer. b. The impedance seen by the source differs from

is

inversely proportional to

l/2ir/'C), the

the primary terminals

actual value.

Z

We

its

capacitance

apparent capacitance between is


its

can therefore artificially increase

(or decrease) the microfarad value of a capacitor

by means of a transformer.

EJE,

impedances from secondary to primary and vice versa

9.10 Shifting and

LIU = As

far as the

ance

source

Z x between

is

l/a

concerned,

it

sees an imped-

the primary terminals given by:

Zx = EJ1

As

a further illustration of the impedance-changing

properties of an ideal transformer, consider the circuit

of Fig.

9.

1

3a.

It is

composed of

a source E„, a

!

transformer T, and four impedances Z, to

On the other hand, Z given by

the secondary sees an

impedance

transformer has a turns ratio

We can progressively to 9.l3e.

Zx

can be expressed

^

=

a£\

in

another way:

/ 2 /a

2

Z

As

the

impedances are shifted

x

1

3b

way,

the

impedances are transferred

a~.

to the pri-

A,

mary

side, the ideal

this position the

a~Z

(9.8)

This means that the impedance seen by the source times the real impedance (Fig.

ideal transformer has the

9.

in this

transformer ends up

treme right-hand side of the circuit (Fig.

Z = 2

secondary imped-

impedance values are multiplied by

If all

Consequently,

a

shift the

the circuit configuration remains the same, but the shifted

a~£\ a

/,

.

ances to the primary side, as shown in Figs.

Z = E 2 /I 2 However,

Z4 The

a.

amazing

9.

1

is

2b). Thus, an

ability to increase

is

iden-

1

3d). In

secondary of the transformer

is

on

open-circuit. Consequently, both the primary and

secondary currents are zero.

move In

der

We

can therefore

re-

the ideal transformer altogether, yielding the

equivalent circuit

or decrease the value of an impedance. In effect, the

impedance seen across the primary terminals

at the ex-

9.

comparing

how

a circuit

shown

Figs. 9.

1

in Fig. 9.l3e.

3a and

9.

which contains a

1

3e,

real

we may wonT

transformer


Z,

a

2

193

Z3

(e)

Figure 9.13 a.

The

and

actual circuit showing the actual voltages

currents.

Impedance Z 2

b.

is

c.

Impedance Z3

Impedance Z4

is

rents All

in

/2

.

change

E3 and

in

/3

Note

.

Note .The cur-

shifted to the primary side.

the corresponding

e.

E 2 and

in


is

the corresponding d.

Note


the corresponding changes

T are now

change

E4 and

in

/4

zero.

the impedances are

now

transferred to the

mary side and the transformer

pri-

no longer needed.

is

can be reduced to a circuit which has no transformer

any meaningful relationship

at all. In effect, is there

between the two circuits? The answer

is

yes

—

there

is

a useful relationship between the real circuit of Fig.

9.

13a and the equivalent circuit of Fig. 9.l3e. The

reason

that the voltage

is

£ across each element in the

secondary side becomes

a£ when

the element

shifted to the primary side. Similarly, the current

each element the element

On

in the

is

secondary side becomes

account of this relationship,

simply reduce

in Fig. 9.

rents.

by

I

1

it

Z4

easy to solve

is

it

shown

in Fig. 9.

to the equivalent

3e and solve for

3a.

the voltages and cur-

all

/a

and by

a,

which yields the actual voltages

illustrate,

in Fig. 9.

14

in the

suppose that the is

£4

volts

and

amperes. Then,

secondary

that the real current

it

cuit, the

voltage across the a"Z4 impedance

£4 X

is /4

a volts.

On

through the impedance

side.

real voltage across

through

to

1

form shown

These values are then respectively multiplied

and currents of each element

To

when

//a


a real circuit such as the one

We

is

/ in

in

the equivalent ciris

equal

the other hand, the current is

equal to

/4

^

a amperes


194

Figure 9.14 Actual voltage and current

in

Z4

impedance

a

2

.

Z4

a£ 4

Figure 9.15 Equivalent voltage and current

(c)

.

z, 1?

words, whenever an impedance

(Fig. 9. 15). In other is

Z4

in

transferred to the primary side, the real voltage

across the impedance increases by a factor a, while the real current decreases by the factor a.

general,

In

ferred

whenever an impedance

from one side of a transformer

the real voltage across the turns ratio. If the side

it

changes

impedance

is

is

trans-

to the other,

transferred to the

where the transformer voltage

is

higher, the

voltage across the transferred impedance will also

be higher. Conversely, ferred to the side

if

the

J

proportion to

in

impedance

is

Figure 9.16 a.

where the transformer voltage

is

b.

lower than the

the ratio of the In

real voltage

—

again, of course, in

c.

some cases

it

is

useful to shift is

Impedance Z

1

is

The source

is

impedances

secondary side (Fig. 9.16a). The procedure

in

to the is

the

d.

and

in £-,

A,.

transferred to the secondary side. in

£g Note .

also

T are zero. All the impedances and even the source are now on the secondary side. The transformer is no longer needed because its currents are zero.

that the currents in

from the primary side

and

transferred to the secondary side.

Note the corresponding change

number of turns.

the opposite way, that

actual circuit, showing the real voltages

Note the corresponding change

lower, the voltage across the transferred impedance is

The

currents on the primary side.

trans-

THE IDEA L TRA NS FORMER

same, but

all

vided by a

2

impedances so transferred are now (Fig. 9.16b).

We

can even

source E„ to the secondary side, where source having a voltage is

now

Eg /a. The

it

Z-

becomes a

ideal transformer

\

R 2 + (X L - Xc

= V4 2 +

primary of

circuit (Fig. 9.16c). In this position the

circuit in Fig. 9. 18

95

is

shift the

located at the extreme left-hand side of the

the transformer

+

\ 16

-

(5

-

2

(2. 17)

)

2

2)

9

on open-circuit. Consequently,

is

both the primary and secondary currents are zero. before,

The impedance of the

di-

1

we can remove

=

As

the transformer completely,

The

s

n


is

leaving us with the equivalent circuit of Fig. 9.16d.

= E/Z =

/

Example 9-5

The voltage across

Calculate voltage Fig. 9.17,

E

knowing

and current

/ in the circuit

that ideal transformer

primary to secondary turns ratio of

1

:

T

£7 100

has a

100.

The

the

impedances

to solve this to the

problem

is

,

= 8V

therefore,

100

- 800 V


impedance values are

or 10 000. Voltage

E becomes

£7100, but current / remains unchanged because

9-

1

The

already on the primary side (Fig. 9.

1

500 turns and

coil in Fig. 9.2a has

a re-

actance of 60 il but negligible resistance.

it it

is

X

is,

is

primary side of the trans-

turns than the secondary, the 2

E

A

to shift all

former. Because the primary has 100 times fewer

divided by 100

8

2

2X4

=

//?

actual voltage

E= way

easiest

=

=

the resistor

of

Solution

The

10/5

8).

is

connected to a 120 V, 60 Hz source

If

£u

,


The

effective value of the magnetizing cur-

rent / m

9-2

b.

The peak value of / ni

c.

d.

The peak and mm!' produced by The peak flux 4> m:ix

In

Problem

to

40

9-3

,

if

the voltage E„

What

coil is

new

reduced

is

mmf developed

and the peak flux

(I)

I1UIX

.

meant by mutual flux? by leakage

flux?

Figure 9.17

See Example

1

V, calculate the

by the 1:100

9-

the coil

9-5.

9-4

The

ideal transformer in Fig. 9.9 has

turns on the primary and

secondary.

of 12

Figure 9.18 Equivalent circuit of Fig. 9.17.

a voltage

a resistance


The voltage E 2

b.

The

c.

e.

The current /, The power delivered to The power output from

the

secondary

In

Problem

the

impedance seen

d.

9-5

11.

Z is

500

turns on the

The source produces

E„ of 600 V, and the load

a.

300

current I 2

9-4,

by the source

what

is

the primary

[

WJ W] |

196

9-6


In Fig. 9. 17, calculate the voltage across

the capacitor

through

9-9

and the current flowing

Two

coils are set

up as shown

in Fig. 9.4.

Their respective resistances are small and

may be

it.

1,

neglected.

The

having terminals

coil

2 has 320 turns while the coil having

ter-


minals 9-7

The nameplate on a 50 kVA transformer shows a primary voltage of 480 V and a sec-

when

1

is

mine the approximate number of turns on the primary and secondary windings. Toward

this

wound around

the

end, three turns of wire are

and a voltmeter

external winding,

across this 3-turn coil.

then applied to the

1

connected

is

V winding,

voltage across the 3-turn winding

found

to

How many turns are there on the 480 V and 120 V windings (approximately)? A coil with an air core has a resistance of is connected to a 42 V, 60 14.7 fl. When it

Hz

ac source,

it

draws a current of

1

.24 A.

Calculate the following: a.

b. c.

The impedance of the coil The reactance of the coil, and its inductance The phase angle between the applied voltage (42 V) and the current

(

1

.24 A).

Hz

It is

voltage

is

found

that

applied to

ter-

voltage across terminals 3-4

V. Calculate the

peak values of <)>,

<(>, ,,

.

|jlF,

600

V

paper capacitor

we need one having jjlF. It is

is

available,

a rating of about

proposed to use a transformer

modify the 40

fxF so that

it

appears as

fxF. The following transformer ratios are available: 120 V/330 V; 60 V/450 V; 480 V/1 50 V. Which transformer is the most

300

be 0.93 V.

9-8

1-2, the

4> ml

A 40 but

to

and the

is

^

300

A voltage of 76 V is

20

22

and

4 has 160 turns.

a 56 V, 60

minals

We wish to deter-

ondary voltage of 20 V.

3,

appropriate and what

is

the reflected value

of the 40 fxF capacitance? To which side of the transformer should the

be connected?

40 p,F capacitor

Chapter

1

Practical Transformers

10.0 Introduction In

we

Chapter 9

discovered real

its

with an imperfect core

studied the ideal transformer and

basic properties.

However,

in the

The

world transformers are not ideal and so our

happens

account. Thus, the windings of practical transform-

have resistance and the cores are not

iron

infinitely

is

not completely captured by the sec-

And

losses,

a practical

we

primary

which con-

The

transformer can be described by an

resistance

current

is in

is

phase with

E

current

its

£p

that

1

0.

1

a).

The

produces a

x

(Fig. 10.1b).

Xm

a

is

measure of

the permeability of the transformer core. Thus,

age regulation and the behavior of transformers that

also used to illustrate

with the primary

To furnish these drawn from the line. This

The magnetizing reactance

cir-

The per-unit method mode of application.

We

represents the iron losses and

/,

the permeability

in parallel.

Rm

losses a small current

developed from fundamental concepts. This

connected

in parallel

excited by a source

is

enables us to calculate such characteristics as volt-

are

low?

the resulting heat they produce.

equivalent circuit comprising an ideal transformer

cuit is

rather

is

voltage £],

discover that the properties of

and resistances and reactances. The equivalent

and eddy-current

hysteresis

whose permeability

R m and X m

elements

tribute to the temperature rise of the transformer.

In this chapter

What

replaced by an

terminals of the ideal transformer (Fig.

finally, the iron cores pro-

duce eddy-current and hysteresis

having

core

is

can represent these imperfections by two circuit

ondary. Consequently, the leakage flux must be taken into account.

the previous

in

such a perfect core

if

losses and

permeable. Furthermore, the flux produced by the

primary

transformer studied

ideal

chapter had an infinitely permeable core.

simple analysis must be modified to take this into

ers

Ideal transformer

10.1

/m

is

flowing

X m is relatively through X m represents low,

if

The mag-

low. the

netizing current needed to create the flux 3> m in the

is

core. This current lags 90° behind

197

E

.

}

1


98

= 0

/1

Jo

/o=0

A

5

120 V 60 Hz A

An

ideal

X

T

Figure 10.1a

An

imperfect core represented by a reactance

a resistance

Rm

Xm and

Figure 10.2a

See Example

10-1

.

Xm

The values of the impedances R m and

can be

found experimentally by connecting the trans-

transformer

is

shown

former to an ac source under no-load conditions and

measuring the active power and reactive power absorbs.

m =

is

The peak value

again given by Eq. 9.2:

E f(AMfN x

x

(9.2)

)

it

The following equations then apply:

R m = E* 2 /P m Xm = Er/Q m

in Fig. 10.1b.

of the mutual flux <£ m

Example 10-1 (10-1)

A

(10.2)

exciting current 70 of 5

large transformer operating at no-load

A when

draws an

the primary

is

where

a wattmeter test

R [u =

Xm =

resistance representing the iron losses [Cl]

equal to 180

magnetizing reactance of the primary

Calculate

winding Ei

=

Pm — Q in = The

[flj

primary voltage [V]

c.

iron losses

reactive

power needed IvarJ

total

/ in

.

[W]

flux

m

to set

up the mutual

It is

ally a small

is

equal to the phasor

called the exciting current 7 0

.

m

at

no-load for

The reactive power absorbed by The value of R m and X m The value of 7f 7 m and 7

sum

of 7f

It is

usu-

,

,

the core

t)

The apparent power supplied Sm

in

percentage of the full-load current. The

phasor diagram

W.*

Solution a.

current needed to produce the flux

an imperfect core

and

a.

b.

it is

=

E,7C

=

120

to the core

X

is

5

= 600 VA The

iron losses are

P m = 80 W

this less-than-ideal

1

The

reactive

Qm =

power absorbed by

^^~Pi

= 572

Figure 10.1b Phasor diagram

of

a practical transformer

at no-load.

con-

Hz source (Fig. 10.2a). From known that the iron losses are

nected to a 120 V, 60

var

the core

= V600 r^780I

is

PRACTICAL TRANSFORMERS

The impedance corresponding

b.

99

1

to the iron losses

is

2

Rm = E

=

{

120 /180

The magnetizing reactance

c.

2 }

= 0

is

2 /Q m = 120 /572

25.2 ft

The current needed If

1

a

so

Xm = E

/

2

=

IP m

supply the iron losses

to

=

E\/R m

=

1.5

=

is

120/80

Figure 10.3

A

Transformer with

The magnetizing current /m

= E IX m = = 4.8 A

across the primary

120/25.2

The exciting current

-

V/? 5

is

x

$m its

h =

infinitely

+

/t)

i

peak value

Vl.5

2

+

4.8

2

cause

it

A

E2

to the current is

given

in Fig. 10.2b.

Ep

and

it

sets

up

a

mutual flux

given by ^> nikl

is

Ep and = £ p /(4.44 /TV,).

infinitely

is

permeable and be-

/, = 0. E2 = (N2 /N E p Owing zero, no mmf is available to

has no losses, the no-load current

The voltage

The phasor diagram

at no-load.

This flux lags 90° behind

a in the core.

Because the core

is

=

II

permeable core

is

is

given by

being

drive flux through the

air;

{

)

.

consequently, there

is

no

leakage flux linking with the primary. If

= 1.5

A 120

V

Let us

now connect

a load

Z

across the sec-

ondary, keeping the source voltage

A'L^Si

= 4.8

/m

10.4).

/«

~ 5

A

Ep

fixed (Fig.

This simple operation sets off a train of

events which

we

list

as follows:

Figure 10.2b Phasor diagram.

10.2 Ideal transformer

with loose coupling

We have just seen how an ideal transformer behaves when it has an imperfect core. We now assume a transformer having a perfect core but rather loose

coupling between ings.

We

also

its

primary and secondary wind-

assume

that the

primary and sec-

ondary windings have negligible resistance and the turns

areN ,N2

Consider the transformer to a

source E„ and operating

Figure 10.4 Mutual fluxes and leakage fluxes produced by a trans-

.

l

in Fig. 10.3 at

no-load.

connected

The voltage

former under load. The leakage fluxes are due to the imperfect coupling between the

coils.


200

1

Currents

.

and

/,

2.

I2

flow

to

= N2 /N

hence

: {

produces an

mmf /V,/,. equal and

/V,/|

mmf N2 I 2

= N2 I2 while

-

/|

produces an

These magnetomotive forces are in direct

when

opposition because

flows into the polarity-marked terminal flows out of polarity-marked terminal

The mmfiV 2 / 2 produces

3.

portion of

C I>

(

2

(

,

.

mmf /V]/,

A portion Flux

,

of

( t 2 -

produces a

( ml

is

,

)

called the

The magnetomotive the magnetic field

)

does

we

can

Figure 10.5

not.

A

$

ni

links with the sec-

P n ) does

analyze

this

and

/,

In general,

1

of two parts; a

same

the

new mutual

flux

(The mutual flux

,-,.

inl

total flux

posed of a mutual flux

A voltage E n

.

composed

and

we combine

is

£n

A voltage E

2.

produced by

mutual flux

m

ni

,

and

(Fig. 10.5). This

/2

com-

is

is

we

flux

n

is

mutual flux

is

NJ

phase with 2

1

.

n and

induced by mutual flux

x

(10.5) 4> in

and

=

4.44./N

l

Ep =

six basic facts,

(10.6)

m

applied voltage E^.

we now proceed

to

note that the primary leakage flux

/ lr

and leakage flux

O

l2

10-3 Primary

E

We can better identify the four induced voltages E2 £ n and £ t2 by rearranging the transformer cir-

induced

composed of two

A voltage E [2

,

,

s

in the

secondary

is

cuit as parts:

induced by leakage flux

winding I2

and secondary

leakage reactance

is in

.

Fifth, the voltage actually

Using these

.

I

primary

in the

=4.44.^,*,-,

Sixth, induced voltage

cre-

h while the secondary leakage created by /V 2 / 2 Consequently, leakage flux

is in

induced

parts:

develop the equivalent circuit of the transformer.

created by

phase with

are not in phase.

Ep

3> m2 into a single

ated by the joint action of the primary and sec-

Fourth,

(10.4)

given by

.

ondary mmfs.

(J>,-,

E2

#m

given by

not

m2 and a leakage flux <£ (2

4.44/7V2

induced by leakage flux

£,

Third,

m and

and a leakage

(p mJ in Fig. 10.4

as ( F mla in Fig. 10.3.)

Second, the

E [2

composed of two

reason as follows: is

tp

induced by mutual flux

Similarly, the voltage is

/]

E2

E2 =

how

is,

situation?

we

voltage

upset

/2

that existed in the core be-

produced by

flux.

given by

(

new

First, the total flux

A

2.

c (

primary leakage flux.

forces due to

|.


flux

transformer possesses two leakage fluxes and a

ac flux

total

was connected. The question

fore the load

A

.

^ m 2) links with the primary

ondary winding, while another portion not.

3.

mutual

Similarly, the

4.

t

called the secondary leakage flux.

2 is

,

/

1, 1 2

a total ac flux 4> 2

winding while another portion Flux

in

by the ideal-transformer equation

related /,// 2

immediately begin

I2

and secondary windings. They are

the primary

and

10.6.

to

.

4.44/N 2
(10.3)

by two fluxes,

clearly

r2

and

This rearrangement does not change the value

of the induced voltages, but

Er

Thus, the secondary

show even more

that the jV 2 turns are linked

$m

given by

shown in Fig. is drawn twice

age stand out by

itself.

it

Thus,

does make each it

becomes

volt-

clear that


201

Figure 10.6 Separating the various induced voltages due to the mutual flux and the leakage fluxes.

Figure 10.7 Resistance and leakage reactance

E l2

is

of the

primary and secondary windings.

drop across a reactance. This

really a voltage

X

secondary leakage reactance

-

{

2 is

given by

Example 10-2 The secondary winding of 180 turns.

X = Ef2 /I2

(1

l2

0.7)

When

a transformer possesses

the transformer

is

under load, the

secondary current has an effective value of 8 A, 60 1

The primary winding

also

is

shown

twice, to

separate £, from E n Again, it is clear that E n is simply a voltage drop across a reactance. This pri.

mary leakage reactance

Xn

Xn

is

=£

fl

shown

in

Figure 10.7.

We

(10.8)

// l

tive

which, of course, act windings.

mWb. The

m has a peak value of

secondary leakage flux

,

2

has a

mWb.

Calculate a.

in series

and

b.

with the respec-

c.

/?,


have also added the

primary and secondary winding resistances Ri,

20

peak value of 3

given by

The primary and secondary leakage reactances are

Hz. Furthermore, the mutual

its

in the

secondary winding

leakage flux

The value of the secondary leakage reactance The value of E 2 induced by the mutual


202

—Oi Eo 1

'lm

Figure 10.8 Complete equivalent

/f

X

a practical transformer. The shaded box T

circuit of

If

Solution a.

The

effective voltage induced by the secondary

leakage flux

an

is

we add

ideal transformer.

elements

circuit

sent a practical core,

we

Xm

and

Rm

to repre-

obtain the complete equiv-

alent circuit of a practical transformer (Fig. 10.8).

is

In this circuit

E l2 = =

4.44/yV 2

f2

X 60 X

4.44

=

O

143.9

X

180

T

is

an ideal transformer, but only the

(10.3)

primary and secondary terminals

0.003

cessible;

I

-2

and 3-4 are ac-

other components are "buried" inside

transformer

the

V

all

However, by appropriate

itself.

we can find the values of all the circuit elements that make up a practical transformer. Table 10A shows typical values of /?,, R 2 X n X r2 X m and R m for transformers ranging from kVA tests

b.

The secondary leakage reactance

is

X = EnJI 2

(10.7)

{2

=

143.9/18

,

,

I

,

400 MVA. The nominal primary and secondary £np and £ ns range from 460 V to 424 000 V. The corresponding primary and secondary currents / np and / ns range from 0.4 7 A to 29 000 A. The exciting current /0 for the various transto

= 8ft c.


voltages

the mutual flux

is

1

E2 = 4.44^2*. =

(10.4)

X 60 X

4.44

1

X

80

0.02

formers

also shown.

is

is

It

always much smaller

than the rated primary current

= 959 V

Note

that in

where S n

is

the rated

/,

EnpInp = En J ns —

each case

5n

power of the transformer.

10.4 Equivalent circuit TABLE 10A

of a practical transformer The

circuit of Fig.

1

0.7

and inductive elements

is

composed of

(/?,,

R 2 X n X i2 ,

,

resistive ,

Z) cou-

m which links the primary and secondary windings. The leakagepled together by a mutual flux

free

in

same properties and obeys

For example, we can

primary

(Ay/V 2 )

side

2 ,

as

we

by

shift

multiplying

did before.

I

10

2400

2400

1

400000

100

1000

2470

69000

13800

V A A

460

347

600

6900

424000

0.4 17

4.17

8.02

14.5

29000

2.17

28.8

167

145

943

a

58.0

5.16

11.6

27.2

0.0003

pos-

R,

fi

1.9

0.095

0.024

0.25

0.354

X~n

a

32

4.3

39

151

0.028

Chapter

Xa xm

Q.

1.16

0.09

0.09

1.5

27

150000

460

Km A»

52.9

impedances their

'ns

kVA V

same

It

the

rules as the ideal transformer discussed in 9.

/„p

the dotted

actually an ideal transformer.

is

sesses the

Ens

,

magnetic coupling enclosed

square

£n P

ACTUAL TRANSFORMER VALUES

to the

values

by

a

200000 29000 400000 51000

220000

505000 432000

A

0.0134 0.0952

0.101

0.210

317

,


10.5 Construction of a

power

Winding

resistances 2

to reduce the J

transformer

R

loss

R and R 2 :

1

0.9a

Power transformers

are usually designed so that

a

characteristics

approach those of an ideal

ondary are wound on one

their

transformer. Thus, to attain high permeability, the

made of

The resulting magnetizing current / m is at least 5000 times smaller than it would be if an air core were used. Furthermore, to keep the iron losses down, the core core

is

is

iron (Fig.

1

0.9a).

laminated, and high resistivity, high-grade silicon

steel is used.


supply the iron losses than

/m

is

/r

needed

to

usually 2 to 4 times smaller

power transformer

and secondary

leg. In practice, the

amount of copper. For

built

up so as

Figure

1

to

is

0.9b shows

how

2. 10a).

small transformer are stacked to build up the core.

Figure

1

0.9c shows the primary winding of a

voltages.

them as closely together as insulation considera-

The

coils are carefully insulated

from each other and from the core. Such coils is

means

that the

tight

when

It

also guarantees

a load

is

connected

secondary terminals.

cou-

secondary

almost exactly equal to

times the primary voltage. voltage regulation

1

the laminations of a

ondary coils on top of each other, and by spacing

voltage at no-load

the

not square (as shown) but

be nearly round. (See Fig.

ondary windings depends upon

between the

sec-

primary

reason, in larger transformers the cross section

Leakage reactances X n and X r2 are made as small as possible by winding the primary and sec-

pling

ensure

much

bigger transformer.

.

tions will permit.

to

a simplified version of

coils are distributed over both core

of the laminated iron core is

is

which the primary and

in

legs in order to reduce the

same

are kept low, both

and resulting heat and

high efficiency. Figure

203

N2 /N

i

good to the

The number of

turns on the primary and sectheir

than a low-voltage winding. current in a

HV

winding

is

On

the other hand, the

much

smaller, enabling

us to use a smaller size conductor.

amount of copper windings

respective

A high-voltage winding has far more turns

is

in

As

a result, the

the primary and secondary

about the same. In practice, the outer

coil (coil 2, in Fig. I0.9a)

length per turn

is

greater.

weighs more because the

Aluminum

ductors are used.

laminated iron core

Figure 10.9a

Figure 10.9b

Construction of a simple transformer.

Stacking laminations inside a

coil.

or copper con-


204

subtractive polarity

additive polarity

x2 «

Figure 10.10 Additive and subtractive polarity

H

cation of the

1

-X

additive polarity

1

depend upon the

lo-

terminals.

when

terminal H,

is

diagonally

opposite terminal X,. Similarly, a transformer has subtractive polarity

when

terminal Hj

terminal X, (Fig. 10.10). If

we know

is

adjacent to

that a

power

transformer has additive (or subtractive) polarity,

we do

not have to identify the terminals by symbols.

Subtractive polarity

is

standard for

all

single-

phase transformers above 200 kVA, provided the high-voltage winding

is

rated

above 8660

V. All

other transformers have additive polarity.

Figure 10.9c Primary winding 290 A.

of

a large transformer;

rating

128

kV,

10.7 Polarity tests To determine whether

(Courtesy ABB)

a transformer possesses ad-

ditive or subtractive polarity,

A ther

transformer

is

reversible in the sense that ei-

winding can be used as the primary winding, 1

where primary means the winding that

is

we proceed

as follows

(Fig. I0.ll):

.

connected

Connect

the high-voltage

winding

to a

low-

voltage (say 120V) ac source E„.

to the source. 2.

10.6 Standard terminal saw

in

markings

Section 9.4 that the polarity of a trans-

a

jumper

J

between any two adjacent

HV and LV terminals. 3.

We

Connect

Connect a voltmeter adjacent

HV

and

LV

Ex

between the other two

terminals.

former can be shown by means of dots on the primary

and secondary terminals. This type of marking on instrument transformers.

On power

is

used

j

transformers,

however, the terminals are designated by the symbols H, and H 2 for the high-voltage (HV) winding and by X and X 2 for the low-voltage (LV) winding. By con(

vention, H, and Xi have the

Although the

H2

bols H,,

,

polarity

X,, and

power transformers the four

dard

X2

it is

is

same

polarity.

known when

the

sym-

are given, in the case of

common

practice to

mount

terminals on the transformer tank in a stan-

way

so that the transformer has either additive

or subtractive polarity.

A transformer is said to have

Figure 10.11 Determining the polarity of a transformer using an ac source.

PRA CTICA L TRA NS FORMERS

Connect another voltmeter

4.

winding. polarity

Ex

If

Ep across

additive. This tells us that H,

is

HV

the

gives a higher reading than

Ep

ing. ,

the

and X,

to the

On the other hand, if £x than E the polarity is subp

are diagonally opposite.

gives a lower reading

jumper J

In this polarity text,

E

secondary voltage

the

Ep

voltage tracts

s

E

E

s

with the primary

terminal of the voltmeter is

Ex = Ep + £ or Ex = the polarity. We can now see

is

marked H,

.

10,8 Transformer taps Due

to

age

in a particular

may

consistently be lower than normal. Thus, a dis-

voltage drops in transmission lines, the volt-

s

region of a distribution system

tribution transformer having a ratio of

may be connected voltage

the polarity test, an ordinary 120 V,

60 Hz source can be connected

2400 V/120

to a transmission line

V

where the

never higher than 2000 V. Under these con-

is

its

to the

HV

winding,

nominal voltage may be several

During a polarity

test

on a 500 kVA, 69 kV/600

motors

may

1

1

V

stall

10. 13).

this

problem taps are provided on the

Taps enable us

8 V,

Ex =

We

percent.

119 V. Determine

markings of the terminals.

is

additive because

HV

may

be

4'/>, 9,

Ex LV

is

greater than

voltage

is

only 2076

would use terminal

(or

H 2 and

and

Some

X,).

Figure 10.12 shows another circuit that

may

used to determine the polarity of a transformer. source, in series with an open switch,

LV winding

a voltage

A dc

connected

of the transformer. The trans-

former terminal connected is

is

to the positive side

marked X,. A dc voltmeter

is

of the

Determining the polarity

/,, 9,

ratio so

or

13'/>

is

1

V

1

0.

(instead of

and tap

1

3, if the line

2400 V), we

5 to obtain 120

V

whenever

above or below a preset

the secondary voltage

level.

Such tap-chang-

ing transformers help maintain the secondary volt-

age within

±2

percent of

its

rated value throughout

the day.

connected 2 O-

Figure 10.13 of

a transformer using a dc

on

transformers are designed to change the

taps automatically

HV terminals. When the switch is closed, is momentarily induced in the HV wind-

Figure 10.12 source.

be

1

or 13'/2 percent below normal. Thus,

the secondary side.

source

by 4

secondary voltage, even though the primary voltage

terminals con-

Consequently, the

across the

change the turns

can therefore maintain a satisfactory

nected by the jumper must respectively be labelled

to the

to

referring to the transformer of Fig.

Solution

X2

electric

under moderate overloads.

as to raise the secondary voltage

the following readings

10.11),

Ep =

were obtained:

H, and

cook food, and

electric stoves take longer to

To correct

transformer (Fig.

polarity

consider-

primary windings of distribution transformers (Fig.

Example 10-3

the polarity

is

ably less than 120 V. Incandescent lamps are dim,

hundred kilovolts.

.

H2

marked

ditions the voltage across the secondary

making

even though

Ep

)

the terms additive and subtractive originated.

In

The

+

(

and the other

either adds to or sub-

In other words,

,

the pointer of the voltmeter

upscale, the transformer terminal connected

effectively connects

in series

Consequently,

.

from

Ep — Es depending on how

,

and terminals H, and X, are adjacent.

tractive,

moment,

this

If, at

moves

205

Distribution transformer with taps at

2184

V,

and 2076

V.

2400

V,

2292

V,


206

Losses and transformer

10.9

rating

Like any electrical machine, a transformer has

They

losses.

are

composed of

The nameplate of a distribution transformer indicates 250 kVA, 60 Hz, primary 4160 V, secondary 480 V.

the following:

Calculate the nominal primary and secondary

a. 2 1.

2. 3.

I

R

__

Example 10-4

losses in the windings

currents.

we apply 2000 V to the 4 60 V primary, can we still draw 250 kVA from the transformer?

If

b.

Hysteresis and eddy-current losses in the core

1

Stray losses due to currents induced in the tank

and metal supports by the primary and sec-

Solution

ondary leakage fluxes

a.

The duce

l)

form of heat and pro-

losses appear in the

an increase

efficiency.

temperature and 2) a drop

in

Under normal operating conditions,

efficiency of transformers

is

very high;

it

may

Nominal current of

L

np

nominal S

=

99.5 percent for large power transformers.

—

ns

The heat produced by the iron losses depends upon the peak value of the mutual flux m which in turn depends upon the applied voltage. On the ,

E np

4160

b.

we

...

apply 2000

V

to the primary, the flux

rent should not

we must

set limits to

at

an accept-

two

limits

determine the nominal voltage / np

E np

The power rating of a transformer

is

product of the nominal voltage times the nominal current

of the

However, the

primary or secondary

result

is

winding.

not expressed in watts, be-

cause the phase angle between the voltage and current

may have any

voltage

and

equal to the

value

at all,

depending on the na-

ture of the load. Consequently, the

capacity of a transformer

is

power-handling

expressed

in

voltam-

S = 2000

V X

is

directly related to the

through

it.

apparent power

This means that a 500

will get just as hot feeding a

as a

500

kW

kVA

that

flows

transformer

500 kvar inductive load

resistive load.

The rated kVA, frequency, and voltage are always shown on the nameplate. In large transformers the

corresponding rated currents are also shown.

A=

120

kVA

Ep

on the

pri-

mary of a transformer, with the secondary open-circuited.

creases

As

the voltage rises, the mutual flux

in direct

Om

in-

proportion, in accordance with Eq.

9.2. Exciting current / u will therefore increase but,

when

the iron begins to saturate, the magnetizing

current

/m

has to increase very steeply to produce

/Q ,

of a transformer

60

Let us gradually increase the voltage

voltamperes (MVA), depending on the size of the rise

using this far lower

10.10 No-load saturation curve

the required flux. If

The temperature

cur-

nominal value, other-

is

peres (VA), in kilovoltamperes (kVA) or in mega-

transformer.

However, the load its

maximum power output

the

of the transformer winding

(primary or secondary).

exceed

and

and

wise the windings will overheat. Consequently,

both the applied

voltage and the current drawn by the load. These

nominal current

A

521

.

the iron losses will be lower than normal

Consequently,

keep the transformer temperature

is

n

the core will be cooler.

to

= 60 A

s

other hand, the heat dissipated in the windings de-

able level,

winding

is

S — 250 X 1000 — — — £ 480 £ns

pends upon the current they

carry.

V

480

n..

nominal If

the

nominal S

—

/

winding

250 X 1000

Nominal current of

the

reach

V

4160

Sn

~

£p

nominal

in

the

we

pass

we draw

a graph of

Ep

versus

see the dramatic increase in current as

the

normal

operating

point

Transformers are usually designed

peak flux density of about

1

.5 T,

(Fig.

we

10.14).

to operate at a

which corresponds

roughly to the knee of the saturation curve. Thus,

when nominal

voltage

is

applied to a transformer,

the corresponding flux density

is

about

1.5 T.

We

can exceed the nominal voltage by perhaps 10 percent, but if

we were

to

apply twice the nominal


207

kV 20 18

16 -

no 'mat oper atincjpoi it

14 En 12

10

—

nc>min al cu rrem

8 6 4 2

0 0

0,5

6

5

4

1

A

exciting current /Q

Figure 10.14 No-load saturation curve of a 167 kVA, 14.4 kV/480

V,

60 Hz transformer.

Figure 10.15 Single-phase dry-type transformer, type AA, rated voltage, the exciting current could

become even

greater than the nominal full-load current.

The nonlinear shows

Xm

relationship between

that the exciting branch

in Fig. 10. la) is

effect,

although

Rm

not as constant as is

Ep

and

/0

(composed of R m and it

Xm

de-

creases rapidly with increasing saturation. However,

and so

R m and X m

terials inside

close to rated voltage,

remain essentially constant.

10.11 Cooling To prevent rapid

at

methods

deterioration of the insulating

ma-

kVA

air.

The

metallic housing

is

fitted

the

can be

di-

with ventilating

may

flow over

windings and around the core (Fig.

it

away

in

a

to the tank,

dissipated by radiation and convection to Oil

1

consequently,

is;

0.

1

6).

is

a

much it

is

better in-

invariably

used on high-voltage transformers.

As

the

power

1

0. 15).

filled tank (Fig.

rating increases, external radiators

10.17). Oil circulates

around the

transformer windings and moves through the radiators, air.

where

For

the heat

still

is

again released to surrounding

higher ratings, cooling fans blow

over the radiators (Fig. For transformers

may be

Such dry-type transformers are used inside build-

oil

hostile atmospheres.

is

are

and enclosed

oil

sulator than air

Larger transformers can be built the same way, but

away from

mineral

in

the outside air (Fig.

forced circulation of clean air must be provided.

ings,

for

are added to increase the cooling surface of the oil-

by the natural flow of the surrounding

louvres so that convection currents

immersed

steel tank. Oil carries the heat

a transformer, adequate cooling of the

Indoor transformers below 200

200 kVA

Distribution transformers below

usually

where

windings and core must be provided.

rectly cooled

V,

appears. In

reasonably constant,

most transformers operate

at

60 Hz, insulation class 150°C indoor use. Height: 600 mm; width: 434 mm; depth: 230 mm; weight: 79.5 kg. (Courtesy of Hammond) 15 kVA, 600 V/240

10.

in the

1

air

8).

megawatt range, cooling

effected by an oil-water heat exchanger. Hot

drawn from

the transformer tank

heat exchanger where

it

is

pumped

to a

flows through pipes that are

ELECTRICA L MA CHINES AND TRANSFORMERS

208

Figure 10.16

Two single-phase

transformers, type OA, rated 75

60 Hz, 55°C temperature rise, pedance 4.2%. The small radiators at the side inkVA, 14.4 kV/240

V,

im-

Figure 10.17

crease the effective cooling area.

Three-phase, type

OA

grounding transformer, rated

1900 kVA, 26.4 kV, 60 Hz. The power of this transformer is 25 times greater than that of the transformers in

contact with cool water. Such a heat exchanger

is

very effective, but also very costly, because the water itself

used. Thus, a transformer 18 000/24 000/32

.

is still

self-cooled. Note,

the transformer

much room as

itself.

method of cooling

may have

a triple rating of

000 kVA depending on whether

by the natural circulation of

it

The type of transformer cooling

is

designated by

the following symbols:

air

(AO)

(

1

8

AA-dry-type, self-cooled

000

AFA-dry-type, forced-air cooled

or

OA-oil-immersed, self-cooled

by forced-air cooling with fans (FA) (24 000

kVA) 3.

it

cooled

kVA) 2.

10.16, but

big transformers are designed to have a

multiple rating, depending on the

1

in Fig.

has to be continuously cooled and recirculated.

Some

is

shown

however, that the radiators occupy as

OA/FA-oi I- immersed,

or

by the forced circulation of forced-air cooling

(FOA)

oil

(32

air

accompanied by

000 kVA).

self-cooled/forced-

cooled

A O/FA/FOA-oil -immersed,

self-cooled/forced-

air cooled/forced-air, forced-oil

These elaborate cooling systems are nevertheless

cooled

economical because they enable a much bigger output from a transformer of a given size and weight

The temperature

(Fig. 10.19).

transformers

is

rise

by resistance of oil-immersed

either

55°C or 65° C. The tempera-


209

Figure 10.19 Three-phase, type OA/FA/FOA transformer rated 36/48/60 MVA, 225 kV/26.4 kV, 60 Hz, impedance

Figure 10.18

7.4%. The circular tank enables the

Three-phase, type FOA, transformer rated 1300 MVA,

the temperature rises

24.5 kV/345 kV, 60 Hz 65°C temperature

oil in

ance:

1 1

imped-

contact with

air.

Other

nuclear power generating station,

The

is

one

forced-oil circulating

below the cooling (Courtesy of Westinghouse) just

of the largest

weight of core and

turc

must be kept low

By

fans.

weight of

to preserve the quality

coils:

37.7

t

weight of tank and accessories: 28.6

pumps can

coil

(44.8

Total weight: 104.5

oil.

details:

.5%. This step-up transformer, installed at a

units ever built.

be seen

rise,

oil to expand as and reduces the surface of the

m3

):

38.2

t

t

t

of the

contrast, the temperature rise of a dry-type

transformer

may be

as high as

1

80°C, depending on

the type of insulation used.

10.12 Simplifying the equivalent circuit

N,

The complete equivalent circuit of the transformer as shown in Fig. 0.8 gives far more detail than is needed in most practical problems. Consequently, 1

let

us try to simplify the circuit

former operates I

.

l

)

at

At no-load (Fig. 10.20) because

T

is

when

no-load and 2) 1 2 is

/n

flows

a transformer

at no-load.

the trans-

drop across them

zero and so

These impedances are so small

circuit of

at full-load.

current in

is /,

an ideal transformer. Consequently,

only the exciting current

Figure 10.20 Complete equivalent

N2

in

R and }

Xn

.

that the voltage

is

negligible. Furthermore, the

R 2 and X [2

is

zero.

We

can, therefore,

neglect these four impedances, giving us the

much simpler tio,

a

= N /N2 ]

circuit of Fig. 10.2 ,

is

1

obviously equal

.

The

turns ra-

to the ratio

of


210

primary

the

to

secondary voltages

Ep/E meas

sured across the terminals. 2.

hi full-load

I

p

Consequently,

least

is at

we can

20 times larger than

sponding magnetizing branch. The resulting

shown

cuit

is

cuit

may

/()

a£ s

in Fig. 10.22.

This simplified

when

be used even

the load

is

a

cir-

2

Z

J

cir-

only 10

percent of the rated capacity of the transformer.

We

1

.

neglect /0 and the corre-

Figure 10.23 Equivalent circuit with impedances shifted to the

can further simplify the circuit by

shift-

mary

pri-

side.

ing everything to the primary side, thus elimi-

nating transformer

T (Fig.

nique was explained

in

10.23). This tech-

Section 9.10. Then, by

summing the respective resistances and reacwe obtain the circuit of Fig. 10.24. In

tances,

this circuit

Rp =

fl t

+ crR 2

Xn = X n +

a

2

(10.9)

Xr

(10.10)

.

where

Rp =

total

transformer resistance referred to the

total

of

a large transformer

is

mainly reactive.

primary side

Xp =

Figure 10.24 internal impedance

The

transformer leakage reactance referred

to the

primary side

The combination of R p and X constitutes the totransformer impedance Z referred to the prip mary side. From Eq. 2. 2 we have

tal

1

\Rcn

VR- + X;

L,

Impedance Z p

a:1

N,

N2

when

one of the important parameters of

is

the transformer.

It

produces an internal voltage drop

the transformer

is

loaded. Consequently,

Zp

affects the voltage regulation of the transformer.

Figure 10.21

Transformers above 500

Simplified circuit at no-load.

Xp

reactance

Rp *f2

*2

as

.

In such transformers

voltages

tance

Xp

10.25).

a transformer

at full-load.

we can

neglect

Rp

,

is

as far

and currents are concerned.* The is

thus reduced to a simple reac(Fig.

quite remarkable that the relatively

circuit of Fig.

simple reactance

Figure 10.22

possess a leakage

between the source and the load

It

complex

kVA

that is at least five times greater than

equivalent circuit

Simplified equivalent circuit of

(I0.ll)

10.8 can be reduced to a

in series

with the load.


Solution a.

Rated primary current /np

= 5 n /Enp =

3

000 000/69 000 = 43.5

A

Rated secondary current Ais

Figure 10.25 The internal impedance

b. of

a large transformer

= SJEns =

3

000 000/4160 = 721

A

Because the transformer exceeds 500 kVA, the windings have negligible resistance compared to

is

their leakage reactance;

mainly reactive.

we can

therefore write

Zp = X p = 127(1 Referring to Fig. 10.26a, the approximate imped-

ance of the 2000

10.13 Voltage regulation An

important attribute of a transformer

is its

Z = E~IP = 4160 = 8.65 il

volt-

age regulation. With the primary impressed voltage held constant

at its rated value, the

tion, in percent, is

voltage regula-

Load impedance

defined by the equation:

a voltage regulation

=

— E

l

— E

- X

100

2

Z=

/

p

2

secondary voltage

X

(69/4. 16)

/2

secondary side

000 000

8.65

= 2380

12

we have

28.95

A

no-load [V]

at

a£ at full-load

s

[V]

2

(a Z) /

p

= 2380 X

28.95

= 68 902 V

The voltage

regulation depends upon the

power

power

factor

factor of the load. Consequently, the

must be specified. load voltage

If the

load

may exceed

is

capacitive, the no-

the full-load voltage, in

which case the voltage regulation

E = s

negative.

at

3000 kVA, 69

X

Because the primary voltage it

=

(4.16/69) is

4154

held constant

follows that the secondary voltage

4160

is

68 902

at

at

single-phase transformer rated

kV/4. 6 kV, 60 1

of 127

(1,

Hz

has a total internal impedance

V.

x9

= 127

n

Ij_

Zp

referred to the primary side. \

kV

Calculate a.

b.

I

The rated primary and secondary currents The voltage regulation from no-load to fullload for a 2000 kW resistive load, knowing that the

primary supply voltage

is

fixed

.

a = t-t^ =

at

4.16

69 kV c.

The primary and secondary ondary

is

currents

if

accidentally short-circuited.

the sec-

Figure 10.26a

See Example

10-7.

16.58

V 69 kV,

no-load

Example 10-5

A

is

= 69 000/V127 2 + 2380 2 =

secondary voltage

2

referred to primary side:

Referring to Fig. 10.26b

(10.12)

where

£ NL = £ hl =

kW load on the

1 z

T

is


2 2 1

Voltage regulation

is;

X

voltage regulation

4160

^

100

41 54

X

(10.12)

100

4154

0.14% The voltage regulation c.

is

excellent,

Referring again to Fig. 10.26b, is

accidentally short-circuited, /

p

= E p/Xp = = 543 A

if

the secondary

aEs =

0 and so

Figure 10.27 Open-circuit test and determination of

69 000/127

The corresponding current

/s

turns

Rm Xm ,

,

and

ratio.

on the secondary and 10.24 by means of an open-circuit and a

side:

short-

circuit test.

=

/s

a/

p

=

(69/4.16)

X 543

During the open-circuit

= 9006 A

test,

rated voltage

and current

plied to the primary winding

/0

,

is

ap-

voltage

E p and active power P m are measured (Fig. 10.27). The secondary open-circuit voltage E s is also mea,

These

sured.

test results

give us the following

in-

formation: active

power absorbed by core =

apparent power absorbed by core

power absorbed by core

reactive

where

Figure 10.26b

See Example

short-circuit currents in both the primary

secondary windings are rated values. 1

2

The

I

R

1

Rm

R m = Ep

losses are, therefore, 12.5

2

or

imme-

diately to prevent overheating. Very powerful elec-

Turns

a

ratio

(10.1)

X m = Ep 2 /Q m

(10.2)

Xm

is

a

= NJNo

56 times greater than normal and, unless the wind-

may be

During the short-circuit is

or torn apart.

is

For a given transformer, values of

Xm

,

/?

,

less than 5 percent

much lower

/ sc

should be less than

its

than

of rated voltage)

The primary

nominal value

to

prevent overheating and, particularly, to prevent a

we can determine

m R p and

the secondary winding

applied to the primary (Fig. 10.28).

current

impedances

test,

Ep/E,

short-circuited and a voltage E„

normal (usually

10.14 Measuring transformer

is

is

tromagnetic forces are also set up. They, too, are

ings are firmly braced and supported, they

2

/P m

Magnetizing reactance

56 times greater than normal. The circuit-breaker

damaged

P~m

corresponding to the core loss

and

2.5 times greater than the

or fuse protecting the transformer must open

1

-

= S m = Ep Ia = Qm

10-7.

Resistance

The

Qm =

f* in

X p shown

the actual

in Figs.

10.21

rapid change in winding resistance while the test

being made.

is


The voltage E sc current

/ sc

,

measured on the primary side

,

and power

(Fig. 10.28)

P sc

are

Transformer impedance referred to the primary

and the

Zp = £ sc //sc = - 650 (2

following calculations made: Total transformer

mary side

impedance referred

to the pri-

is

RJtJ =

Rp

1

is

2600/4

Resistance referred to the primary (10.13)

is

2400/16

= ison

Total transformer resistance referred to the primary side

2

is

Leakage reactance referred

to the

primary

is

(10.14)

primary side

V650 2 - 150 2

Xp

Total transformer leakage reactance referred to the

- 632

a

is

VZ; Example 10-6

(10.11)

__

During a short-circuit

test

on a transformer rated

500 kVA, 69 kV/4. 6 kV, 60 Hz, the following 1

age, current, and

Terminals X,,

2600 V

£\

(

volt-

power measurements were made. were in short-circuit (see Fig,

X2


10.28):

2600

V

4A 2400

Example 10-7

W

An

Calculate the value of the reactance and resistance of the transformer, referred to the

HV

side.

Solution

shop, a 69

Referring to the equivalent circuit of the trans-

former under short-circuit conditions (Fig. 10.29),

we

was conducted on the transExample 10-6. The following results were obtained when the low-voltage winding was excited. (In some cases, such as in a repair open-circuit test

former given

in

kV

voltage

may

not be available and the

open-circuit test has to be done by exciting the

winding

LV

at its rated voltage.)

find the following values:

E = 4160 V

/n

K

Using

this

acteristics a.

= 2A

5000

W

information and the transformer char-

found

in

Example

the values of X m and

Rm

10-6, calculate:

on the primary side

(Fig. 10.21) b. the

efficiency of the transformer

plies a load of

80

%

when

250 kVA, whose power

it

(lagging).

Solution a.

Applying Eq.

10.1 to the

secondary side:

Figure 10.28 Short-circuit test to

winding resistance.

determine leakage reactance and

= 4160 2 /5000 =

3461 (1

sup-

factor

is


214

The apparent power S m

= E

5m

s

= 4160 X

I {)

2

and R p Let us assume

= 8320 VA

load

is

4 60

6650

The load current

11

/?

Xm

m and

000/4 160)

2

11

referred the primary side

=

275 times

greater.

The

Rm =

X 2602

275

275 X

=

II

=

3461 fl

715

X

10

952 X

10

3

turns ratio

=

715

kH

(1

- 952

k(l

the

time. Thus,

when we

=

with cos 6 is

about 250

about

that

loads and voltages fluctuate

b. Industrial

0.8,

it

all

state that a load is

is

kVA and

power

The

total

swer, even

in

if

iron loss

is

=

3.62

X

is

150

1966

same

the

rated voltage on the

is

LV

as that

measured

at

side of the transformer.

Total losses are

certain

it.

= 6966 The

circuit of the transformer

and

5000

bosses

Knowing this, assumptions that make it much able to give

active

+

1966

W = 7 kW

power delivered by

the transformer

its

add the

The values of R p known, and so we only have to magnetizing branch. To simplify the calcu-

lations,

we

is

Rp

2

is

calculating efficiency, there

we were

The equivalent

Xp

Il

= The

is

easier to arrive at a solution.

load

A

3.62

arriving at a precise mathematical an-

we can make

and

-

P iron = 5000W

Consequently, in

60/16.59

2

=

about 69 kV.

no point

=

is:

copper loss (primary and secondary)

250 kVA,

factor

the primary side

/2 /a

Copper

the

Furthermore, the primary voltage

0.8.

=

16.59

if

understood that the load that the

is

The current on /,

would have been found primary had been excited at 69 kV.

These are the values

000/4160

= 69W/4I60 V -

a

(1 3

is

The

values on the primary side are therefore:

Xm =

calculate the efficiency of

= SIE, = 250 = 60 A

12

= 4160 2 /6650 = 2602

will be (69

Xp

the transformer.

= V8320 2 - 5000 2

The values of

We now

V.

1

greater than

that the voltage across the

.

^s^pI

Qm =

much

cause these impedances are

is:

to the load

is

represented by Fig. 10.30.

PG = S cosG = 250 X

are already

shifted

input terminals

1

1

1,

50f2

X m and R m from points 3, 4 to the 2.

632

This change

a

is

/,

justified be-

=

200

The

active

power received by

60 A

Note

efficiency

that in

is

7

therefore

= PJP - 200/207 = 0.966 or 96.6% X

making

the calculations,

sider the active power.

transformer and 10-7.

is

3

ti

Figure 10.30

the transformer

= + bosses = 200 + = 207 kW The

See Example

0.8

kW

its

ciency calculations.

The

reactive

we

only con-

power of the

load does not enter into

effi-


The

10.15 Introducing the per-unit

best approach

calculate is

often encountered

when

dealing

with transformers and other electrical machines.

reason

is

relative

rents

employ

to

The

that per-unit values give us a feel for the

its

ohmic value and use

For example,

former

listed in

_ E ns _ " L ~

and kilowatts, we simply work

volts

we don't have to carry when per-unit values are used. The per-unit method as applied to transformers

V 8 A

28

'

12.012

Using

value of the secondary resistance

who are

easy to understand. However, readers

useful to read Sections 1.9 to 1.11 in Chapter

1

tual values

R

of

{

,

at

Table

be-

1

OA which

,

,

,

transformers ranging from

scanning through the table,

1

kVA

we

is

400

to

ances vary from 505 000 17 to 0.0003

MVA.

In

12, a

range

over

all

that the various voltages,

currents and impedances are expressed in actual

values using volts, amperes and ohms.

ACTUAL TRANSFORMER VALUES

/„s

*i

Ri

*n

Q Q n

n

4.17

/ np

Using

this load

V A

2400

,

impedance

576

as a reference, the rela-

value of the primary resistance R\

/?,(pu)

is

5.16(2

=

0.0090

576

The

l

1000

100

10

12

and R 2 (pu) are pure

relative values R\(p\i)

numbers because they are the ratio of two quantities that bear the same unit. Circuit elements on the primary side are always

compared with

the nominal load

impedance Z np on

400000

2400

2400

12470

69000

13800

460

347

600

6900

424000

0.417

4.17

8.02

14.5

29000

167

145

943

.6

27.2

0.0003

17

28.8

58.0

5.16

1.9

0.095

0.024

0.25

0.354

32

4.3

39

151

0.028

0.09

0.09

1.5

27

2.

1.

16

1

1

a Q

200000

29000

150000

505000

460

Rm

400000

51000

220000

432000

317

A,

A

0.0134

0.0952

0.101

0.210

52.9

R\,R 2 X n X l2 X tn and R m in ohms, we could express them relative to another ohmic value. The question is: what value should we ,

,

on

the

secondary side are compared with the nominal load

impedance Z ns on the secondary Proceeding of the 10

kVA

values X, |(pu),

The

relative

in this

side.

for the other impedances

we

transformer,

obtain the relative

m (pu), etc. displayed in Table 10B. impedances of the other transformers /?

are calculated the tive

way

same way.

In

nominal load impedances

as the reference impedances.

each case, the respec-

Z np and Z ns are chosen

Using the rated voltage

and power of the transformer, they are given

E _np Instead of expressing

pri-

12

the primary side. Similarly, circuit elements

kVA

/np

E

no

recognizable pattern to the values; they are

impedance on the

is:

in

is

TABLE 10A

12.017

~ np

tive

is

0.0079

(pu)

of five

excess of a billion to one. Furthermore, there

is

side

R2

0.095 12 2

re-

see that the imped-

as a reference, the relative

Similarly, the nominal load

displays the ac-

It

R 2 Xn X (2 X m and R m

map. The reason

R

mary

Let us begin by looking

ohmic value

it

fore proceeding further.

produced here for convenience.

this

not

yet familiar with per-unit calculations will find

V V A A

trans-

347

with numbers. Consequently,

Ens

can

is

along units

£„p

kVA

case of the 10

in the

We

as a reference.

Table 10A, the nominal load im-

pedance on the secondary side

and powers. Thus, instead of dealing with

ohms, amperes,

the

it

magnitudes of impedances, voltages, curZ,ls

is

the nominal load

(voltage and current) of the transformer.

method Per-unit notation

is

215

*«P.

- E

by:

1

(HU5a)

sa

,

En, (10.15/;)

choose as a basis of comparison?

s

tt

ELECTRICAL MA CHINES AND TRANSFORMERS

2 6 1

symbols #|(pu),

7\?

2 (pu),

R2 X

values of/?,,

In practice, the relative

<

tl

,

etc.,

and are designated by the

are called per-unit values

X n (pu),

The

etc.

quantities

Example 10.8^

A

transformer rated 250 kVA, 4 60 V/480 V, 60 Hz 1

has an impedance of 5.1 %. Calculate:

used as references are called base quantities. Thus,

Z np Z ns S n E ,

a.

£ ns

.

,

,

/

/ ns listed in

,

Table 10B are

impedance on

the base

base quantities.

all

examining Table OB. the reader I

will note that

former referred for a given transformer, the values of

X, ,(pu)

and

X r2 (pu)

kVA V V A

2400

100

1000

400000

2400

12470

69000

13800

460

347

600

6900

424000

0.417

4.17

8.02

14.5

29000

17

28.8

167

145

943

5760

576

1555.

4761

0.4761

211.6

12.0

3.60

47.61

449.4

R\ (pu)

0.0101

0.0090

0.0075

0.0057

0.00071

/? 2

(PU)

0.0090

0.0079

0.0067

0.0053

0.00079

X X rj X ni

(pu)

0.0056

0.0075

0.0251

0.0317

0.0588

(pu)

0.0055

0.0075

0.0250

0.0315

0.0601

(pu)

34.7

50.3

96.5

106

966

/? ni

Znp = E p - 69

2

/S n

= 4160

2.

69.4

(pu)

0.032

A,
88.5

141.5

0.023

0.013

90.7

0.015

666 0.0018

2

is

/250 000

tt

Base impedance on the secondary side 2

n Q

zns

Base impedance on the primary side

Table 10A.

in

10

I

A

A,s

z„ P

ri

of the trans-

Solution a.

PER UNIT TRANSFORMER VALUES

TABLE 10B

Aip

Zp

primary side

to the

AVpu) and

are nearly the same. This pat-

does not show up

tern of similarity

£„»

and sec-

are nearly the same. Similarly, the values of

/? 2 (pu)

t„p

impedance

the total internal

b.

In

the primary

ondary side

Zns = E /S n = 480 = 0.92 n

2

s

b.

The

actual value of

Zp =

Zp

% X Z np =

5.1

is

/250 000

on the primary side

0.051

X 69

(1

=

is:

3.52 11

10,17 Typical per-unit impedances

We

have seen

ative

that

we can

get a better idea of the

rel-

magnitude of the winding resistance, leakage

actance,

etc.,

re-

of a transformer by comparing these im-

pedances with the base impedance of the transformer.

There

is

even a similarity between the per-unit

whose

values of transformers ferent.

ratings are quite dif-

For example, the R\(pu) of the

former (0.0101

)

is

as the /?,(pu) of the

I

kVA trans-

of the same order of magnitude

1000

kVA transformer

despite the fact that the latter

is

(0.0057)

1000 times more

powerful and the voltages are vastly different. Clearly, the per-unit

method

offers insights that

In

making

the comparison, circuit elements located

on the primary side are compared with the primary base impedance. Similarly, circuit elements on the

secondary side are compared with the secondary base

impedance. The comparison can be made either on a percentage or on a per-unit basis; ter.

kVA

transformers ranging from 3

example, the table shows

would otherwise not be evident.

we

shall use the lat-

Typical per-unit values are listed in Table IOC for to 100

MVA.

For

that the per-unit resistance

of the primary winding of a transformer ranges from

10,16 Impedance of a transformer The total internal impedance Zp of a transformer was defined in Section 10. 12 and highlighted in Fig. 10.24. In power and distribution transformers its

value

However,

is it

always indicated on the nameplate. is

expressed as a percent of the nominal

load impedance. Thus, 3.6

%. the per

if

the nameplate

unit value of

Zp

is

0.036.

is

marked

0.009

to

and 100

0.002 for

all

MVA. Over

the per-unit resistance

power this 7?,

ratings

between 3

kVA

tremendous power range,

of the primary or secondary

windings varies only from 0.009 to 0.002 of the base

impedance of

the transformer.

Knowing

the base im-

pedance of either the primary or the secondary winding,

we can

readily estimate the order of magnitude of

the real values of the transformer impedances. Table

IOC

is,

therefore, a useful source of information.


TABLE 10C

217

TYPICAL PER-UNIT VALUES OF TRANSFORMERS

Figure 10.31 Equivalent circuit of a transformer. Typical per-unit values

Circuit element (see Fig.

/?,

0.3

1

3

1

orR 2

.

X n orX n

kVA

to

250

kVA

l

MVAio I00MVA

0.009-0.005

0.005-0.002

0.008-0.025

0.03-0.06

20-30

50-200

20-50

100-500

0.05-0.03

0.02-0.005

Example 10-9 Using the information given the

approximate

0.35

SI

1.7 SI

Table 10C, calculate

in

23

ma

4.6 mSl

impedances of a

real values of the

250 kVA, 4160 V/480 V, 60 Hz distribution

trans-

former.

Solution

We

first

determine the base impedances on the

mary and secondary

Example

10-8,

we

From

side.

the

results

pri-

of

Figure 10.32

have

See Example

Z np = Z

lls

=

69

0.92

n

This example shows the usefulness of the per-

We now calculate the real impedances by multiplying Znp and Z ns by the per-unit values given in Table IOC. This yields the following results:

=

0.005

X 69

R2 = Xn = X r_ = Xm = Rm =

0.005

X

0.025

X 69

0.025

X

/?,

30 50

(2

0.92 tt

10-9.

tt

=

0.35 12

H= =

0.92 (2

1

=

4.6

mil

alent circuit of the

range of transformers.

.7 12

23 m()

X 69 fl = 2070 12 = X 69 n - 3450 (2 -

method of estimating impedances. The equiv250 kVA transformer is shown in Fig. 10.32. The true values may be 20 to 50 percent higher or lower than those shown in the figure. The reason is that the per-unit values given in Table 10C are broad estimates covering a wide unit

2 kil 3.5 kI2

Example 10.10 The 500 kVA, 69 kV/4160 V, 60 Hz transformer shown in Fig. 10.30 has a resistance R of 150 (2


218

and a leakage reactance unit a.

X p of 632 f2.

Using the per-

G(pu)

method, calculate:

= V0.5 2 when

the voltage regulation

kVA

tween zero and 250

= VS 2 (pu) - P 2 (pu)

the load varies be-

= The

80%

b.

the actual voltage across the

c.

the actual line current I

250

kVA

R The

Solution clear that the pres-

is

it

R p and Xp

voltage drop across

magnetizing branch does not affect the voltage

We now draw

regulation. in

To determine fer all voltages,

HV

(69

kV)

terminals

1

is

We

assume the voltage between 69 kV, and that it remains fixed.

The base power P H The base voltage E

is

is

ti

kVA 69 kV

500

Consequently, the base current 'b

= Pb/Ek = 300 = 7.25 A

and the base impedance

Z B = E B /I H = 69 The

will re-

impedances, and currents to the

side.

2

,

we

the voltage regulation,

per-unit value of

Rp

Figure

Note

terminals

per-unit value of

Xp (pu) The

Xp

it

4 of the

circuit

250

=

4

The

Z ]2 (pu) =

primary

Figure 10.30.

The

per-unit im-

is

2/136.87° 1.2

0.0158

+

1.6

+

j(1.2

+

1,

0.0664

0.0664)

is

= 69 000/69 kV =

A-p(pu)

flp(pu)

1.0

j

0.0664

power absorbed

= 250 kVA/500 kVA =

A'.(pu)

0.5 j

per-unit value of the active

the load

3.33

power absorbed by

is

P(pu)

=

5(pu) cos 9

=

0.5

X

0.8

-

0.4

Figure 10.33

The

per-unit value of the reactive

by the load

is

is

5(pu)

The

2

= 1.616 + j 1.266 = 2.053Z 38.07°

is

E i2

in

impedance between terminals

per-unit

is

=

not

X j 3.33 + j 3.33

= 1.6+j

per-unit value of the apparent

by the load

3,

2.50

Z34 (pu)

0.0158

The

shown

equivalent circuit diagram.)

is

632/9517

is

does not enter into the calcula-

is

ft

shown

(These terminals are not accessible; they exist only in the

= 9517

3.333

The magnetizing branch

pedance between terminals

000/7.25

=

Q is

0.3

that the load appears across the

3,

000/69 000

per-unit value of voltage

E, 2 (pu)

Uf_

the equivalent per-unit circuit

R p (pu) = 150/9517 = 0.0158 The

corresponding to

v

" 0(pu)

10.33.

shown because tions.

0.4

£2 (pu) _

X L (pu)

Consequently, the

,

X

2

2.50

~

P(pu)

P is

corresponding to 1.0

)

(pu)

v

per-unit load reactance

ence of the magnetizing branch does not affect the

/?,

£ 2 (pu _

.

Fig. 10.30,

0.3

per-unit load resistance

load

{

examining

2

lagging power factor

at a

of

In

0.4

is

power absorbed

Per-unit equivalent circuit of a

feeding a 250

kVA

load.

500 kVA transformer


The per-unit current

I

219

ensure proper load-sharing between the two trans-

is {

formers, they must possess the following: /i(pu)

.0

-

Z 12 (pu)

L

2.053

a.

38.07°

b.

The same primary and secondary voltages The same per-unit impedance

= 0.4872^-38.07° Particular attention must be paid to the polarity

The

per-unit voltage

£34 (pu) =

/j(pu)

£34

across the load

of each transformer, so that only terminals having

is

10.34).

- (0.4872/1-38.07°) (2^36.87°)

=

same

the

X Z34 (pu)

circuit as

0.9744/1 -1.20°

polarity

An

soon as the transformers are excited.

In order to calculate the currents flowing in each

transformer

The

per-unit voltage regulation

E 34 (pu)

at

£34 (pu)

is

must

— £34 (pu)

no-load

at

Consider

0.0263

is

former referred

therefore 2.63

is

first

mary voltage

0.9744

The voltage regulation

when

we

they are connected in parallel,

determine the equivalent circuit of the

the equivalent circuit

when

a single

transformer feeds a load Z, (Fig. 10.35a). The

1^ -j0.9744_

a.

first

system.

full-load

full-load

at

connected together (Fig.

are

error in polarity produces a severe short-

£p

pri-

and the impedance of the trans-

to the

primary side

is

Zpl

.

If the ratio

%. 2

We

can

now

calculate the actual values of the volt-

O-

age and current as follows: Voltage across terminals

4

3,

A

is

O-

£ 34 = EM (pu) X £ B = 0.9744 X 69 000

= b.

=

£34

X

67.23

is

B

(4160/69 000)

X

10

3

X

>H 2

Actual line current /,

Figure 10.34 Connecting transformers

a

..

in parallel to

share a load.

is

=

/,(pu)

=

0.4872

=

3.53

X X

/B

7.246

A

10.18 Transformers When

X 1#

0.0603

= 4054 V c.

* 2# .

2

kV

67.23

Actual voltage across the load

£ S6 =

*H

1

in parallel

growing load eventually exceeds the power

rating of an installed transformer,

connect a second transformer

we sometimes

in parallel

with

it.

To

Figure 10.35a Equivalent circuit of a transformer feeding a load

ZL

.


220

of transformation that

shown

in Fig.

is 1

a, the circuit

can be simplified to

0.35b, a procedure

we are

already

portional to the respective

we want

kVA ratings. Consequently,

to fulfill the following condition:

familiar with. If is

a second transformer having an impedance

connected

circuit

the

in parallel

becomes

with the

shown

that

first,

spectively

/,

and

/2

.

(I0.18)

2

in Fig. 10.35c. In effect,

impedances of the transformers are

The primary currents

Z

the equivalent

in the

From

Eqs.

and 10.18

10. 17

can readily be

it

in parallel.

proved

transformers are re-

Because the voltage drop £, 3 is the same, we can write

across the impedances

that the desired condition

is

met

if

the trans-

formers have the same per-unit impedances. The following example shows what happens

when

the

per-unit impedances are different.

(10.16)

Example 10-11 that

is.

A

1

00 kVA transformer

an existing 250 (10.17)

The

of the primary currents

is

connected

in parallel

with

supply a load of

330 kVA. The transformers are rated 7200 V/240 but the

ratio

is

kVA transformer to

1

V,

kVA unit has an impedance of 4 percent 250 kVA transformer has an impedance of

00

therefore de-

while the termined by the magnitude of the respective primary

impedances

—and

not by the ratings of the

two

6 percent (Fig. 10.36a). trans-

formers. But in order that the temperature rise be the

Calculate

same

a.

for both transformers, the currents

must be pro-

The nominal primary

current of each trans-

former

c.

The impedance of the load referred to the primary side The impedance of each transformer referred to

d.

The

b.

the primary side actual primary current in each transformer

Solution a.

Nominal primary current of former

the

250 kVA

trans-

is

Figure 10.35b Equivalent circuit with

all

impedances

referred to the

=

250 000/7200

34.7

A

primary side.

X

t H, -

1:250

kVA

-d H 7

1

> 240

j:

V

z

a Z

y

,

<\

H,

X,

100 kVA:

Z p2

d :H 2

=

4% X2

Figure 10.35c Equivalent circuit of two transformers

a load

Zv All

impedances

in parallel

feeding

referred to the primary side.

load

;=

X 2 b-

:

7200

6%

::|Zpr

Figure 10.36a Actual transformer connections.

b

V

330

kVA


Nominal primary current of the 100 kVA former

=

and

A = 46 -

side, is

im-

0.36b,

we

justified be-

46

find that the

Load impedance referred

primary side

to the

is

=

A

28.8

17.2

which

is

A seriously overloaded

is

25 percent above unit

is

its

A

28.8

20.7)

carries a primary current of

it

The 250 kVA

cause the transformers are fairly big.

+

20.7/(12.4

The 100 kVA transformer because

are considered to be enis

= 46 X

/,

that transformer

This assumption

tirely reactive.

A

primary

to the

Note

Zp2

13.9

of the two transformers

circuit

in Fig. 10.35c.

Zpl

=

100 000/7200

and the load, referred

pedances

1

1

load current divides in the following way:

The equivalent given

Referring to Fig.

d.

is

/ n2

b.

trans-

22

17.2 A,

rated value of 13.9 A.

not overloaded because

carries a current of 28.8

A

versus

its

it

only

rated value of

34.7 A. Clearly, the two transformers are not carry2

2 = E p /S| olld = 7200 = 157 n The approximate

2

/330 000

ing their proportionate share of the load.

The 100 kVA transformer of

load current

its

c.

=

JE p

S hr

= 330 000/7200 = 46 A

The base impedance of

the

kVA

207

unit

its

Transformer impedance referred

tends to carry

proportionate share of the load.

If the

is

percent impedances were equal, the load would be

shared between the transformers

il

power

their respective to the

in

proportion to

ratings.

primary

is

Zpl =

X 207 =

0.06

Base impedance of the 100


12.4(1

kVA

Practical level unit

is 1

Znp2 = 7200 2 /10() 000 = 518 Transformer impedance referred side

to the

transformer (6 percent).

A low-impedance transformer always more than

250 kVA

Z npt = 7200 2 /250 000 -

side

overloaded because

is

impedance of the 250 A.

is

low impedance (4 percent), compared

0-

1

12

to the

Name

the principal parts of a

transformer.

primary

10-2

Explain

how

a voltage

is

induced

in the

secondary winding of a transformer.

is

Z p2 =

0.04

X 518 =

10-3

20.7 il

The secondary winding of a transformer has twice as

many

turns as the primary.

Is

the secondary voltage higher or lower

than the primary voltage? /„,

/,

= 34.7 A = 28.8

10-4

A

Which winding

is

connected

to the load:

the primary or secondary?

ma a

z pi

=

Z pl

= 20.7

46

A

10-5

State the voltage and current relationships

between the primary and secondary wind-

7200

V

n 157

I 2 = 17.2 /n ,

A

7

= 13.9 A

Figure 10.36b Equivalent

circuit.

transformer

is

Calculations

show

seriously overloaded.

that the

ings of a transformer under load. The primary and secondary windings have /V|

a

100 kVA

and

N2

turns, respectively.

10-6

Name

10-7

What purpose does

the no-load current of

a transformer serve

?

the losses

produced

in a

transformer.

222

10-8


Name in

three conditions that must be

met

order to connect two transformers

10-15

What

the purpose of taps

is

on a

trans-

Name

10-16

three

methods used

10-17

The primary of a transformer is connected to a 600 V, 60 Hz source. If the primary has 1200 turns and the secondary has 240,

the

peak flux

a 60

in

Hz

ac supply voltage

10-18

The transformer by a

20

1

V,

fixed.

is

10.37

in Fig.

is

excited

60 Hz source and draws

a no-

load current / 0 of 3 A. The primary and secondary windings respectively possess

low-voltage winding

200 and 600

turns. If

mary

linked by the secondary, cal-

V

is

excited by a

source, calculate the voltage

across the

A 6.9 kV

HV

winding.

transmission line

a transformer having

is

a.

connected to

b.

1500 turns on the

c.

If

d.

the load across the secondary has an im-

pedance of 5 H, calculate the following: a. The secondary voltage b. The primary and secondary currents

The primary of a transformer has twice as many turns as the secondary. The primary voltage is 220 V and a 5 17 load is con-

as the primary

Figure 10.37

See Problem

10-18.

is

40 percent of

the pri-

the transformer, as well

and secondary currents.

The voltage indicated by the voltmeter The peak value of flux 4> The peak value of m

Draw

the phasor

/ tr 4> m

10-19

In Fig.

1

,

and

0.38,

diagram showing

when 600 V

is

applied to

V

is

measured

H2

across terminals X,, a.

What and

is

E h £2

the voltage

,

80

X2

.

between terminals H,

X2 ?

b. If terminals H,,

X, are connected together,

calculate the voltage across terminals c.

,

<£,

terminals H] and

nected across the secondary. Calculate the

power delivered by

tlux

culate the following:

primary and 24 turns on the secondary.

10-14

why

Explain

in the core.

The windings of a transformer respectively have 300 and 7500 turns. If the 2400

10-13

Problem 10-11, calculate the peak

transformer remains fixed as long as the

calculate the secondary voltage.

10-12

In

value of the tlux

to cool trans-

formers. 10-11

nominal

Intermediate level

former? 10-10

a ratio of

to 2.4 kV. Calculate the

current of each winding.

parallel.

10-9

A 3000 kVA transformer has 60 kV

in

Does

H2 X2 ,

the transformer have additive or sub-

tractive polarity?

.

PR A CTICA L TRA NS FORMERS

^ = 300 /,

TV,

223

= 1200

(96 A)

Figure 10.38

See Problem 10-20

10-19.

what would happen

a. Referring to Fig. 10.34, if

we

reversed terminals H, and

H2

of trans-

Would

X2

The primary

if

H h H2

terminals

of transformer

B were

and

Advanced 10-27

level

Referring to Fig. 10.39, calculate the peak value of flux

Explain

leg

reversed?

Explain.

10-21

wound on one

is

other.

the operation of the transformer

bank be affected X,,

10-33.

and the secondary on the

former B? b.

Figure 10.39

See Problem

why

the secondary voltage of a

in the

supplied by a 50

is

core

Hz

if

the transformer

source.

practical transformer decreases with in1

creasing resistive load.

The impedance of a transformer

0-28

as the coupling

10-22

10-23

What

is

meant by

the following terms:

a.

Transformer impedance

b.

Percent impedance of a transformer

The transformer

in

0-24

A 2300 V

the im-

is

R\

=

18 12

R 2 = 0.005

xn =

connected

to terminals

40

10-25

A

66.7

kVA,


The percent impedance of the transformer The impedance [12] referred to the sec-

d.

The percent impedance

in the

f.

transformer un-

Calculate the losses and efficiency the transformer delivers 66.7

when

MVA to a

load having a power factor of 80 percent.

10-26

If the

transformer shown

were placed ture rise

Explain.

in a

tank of

would have

to

The total copper losses at full load The percent resistance and percent

reac-

tance of the transformer

der these conditions. b.

referred to the sec-

ondary side e.

Calculate the losses

referred to

ondary side

percent. a.

[12]

c.

an efficiency

when it delivers full power load having a power factor of 100

of 99.3 percent to a

The transformer impedance the primary side

connected across the secondary

MVA transformer has

a

s

1

b. is

12

E p = 14.4 kV (nominal) £ = 240 V (nominal) x n_ = o.oi n

the transformer has a nominal rating of 75

and 4 in Fig. 10.13. Calculate the following: a. The voltage between terminals X, and X 2 b. The current in each winding, if a 12 kVA load

given for the

is

transformer circuit of Fig. 10.22.

If

line

The following information

10-29

pedance 1(2] referred to: a. The 60 kV primary b. The 2.4 kV secondary 1

increases

reduced between the

primary and secondary windings. Explain.

Problem 10-15 has an

impedance of 6 percent. Calculate

is

in Fig. oil,

10.15

the tempera-

be reduced to 65°.

10-30

During a short-circuit

test

on

a 10

66 kV/7.2 kV transformer (see the following results

were obtained:

Eg = 2640 V / sc - 72 A R = 9.85 kW .

(

MVA,

Fig. 10.28),



The


and the

total resistance

actance referred to the 66 b.

total

leakage

kV primary

The nominal impedance of

re-

10-34

side

A

transformer has a rating 200 kVA,

400 V/277

14

the transformer

The high-voltage windWhat is the

V.

ing has a resistance of 62 H.

referred to the primary side c.

The percent impedance of

V

approximate resistance of the 277

the transformer

winding? 10-31

In

Problem 10-30,

if

rated voltage are 35

the iron losses at

kW,

calculate the full-

load efficiency of the transformer

power

factor of the load

is

if

10-35

The primary winding of the transformer Problem 10-34

the

is

wound

with No.

1

in

gauge

1

AWG wire. Calculate the approximate

85 percent.

cross section (in square millimeters) of the

10-32

a.

The windings of a transformer operate current density of 3.5

made of copper and ture of

A/mm

2 .

If

at a

conductors 1

operate

0-36

tempera-

at a

If

75°C, calculate the copper loss per

aluminum windings were

0-33

If

cording to Fig. 10.39,

it

would have very

poor voltage regulation. Explain propose a method of improving

why and it.

kVA

weighs

1

transformer rated

whereas a 100

18 kg,

of the same kind weighs in

watts per kilogram in each case.

10-37

a transformer were actually built ac-

10

445 kg. Calculate the power output

same condi-

tions.

1

secondary winding.

oil-filled distribution

kVA transformer

used, calculate

the loss per kilogram under the

An at

kilogram. b.

in the

they are

The transformer shown in Fig. 10. 13 has rating of 40 kVA. If 80 V is applied between terminals X, and will

X

2,

a

what voltage

appear between terminals 3 and 4?

If

a single load

is

applied between terminals

what

is

the

3 and 4

current that can be

maximum drawn?

allowable

Chapter

1

Special Transformers

windings, each rated

11.0 Introduction

connected

Many transformers are designed to meet specific we

industrial applications. In this chapter

at

1

20

V.

The windings

and so the

in series,

total

tween the lines is 240 V while that between the lines and the center tap is 20 V (Fig. ). The center tap, called neutral, is always connected to

study

1

some of the

special transformers that are used in dis-

tribution systems,

neon

signs, laboratories, induction

furnaces, and high-frequency applications.

they are special, they

1

.

2.

The ampere-turns of

I

.

I

on the high-voltage winding

to the neutral terminal

is

usu-

of the secondary

windings are connected

that both

to

ground.

the transformers are under load:

The voltage induced in a winding is proportional to the number of turns, quency, and the flux in the core.

bonded

winding so

a result, the following approximations can be

made when

H2

Terminal ally

of the standard transformers discussed in Chapter 10.

As

I

ground.

Although

possess the basic properties

still

are

voltage be-

The nominal

rating of these distribution trans3

mounted on poles of

the fre-

pany

(Fig.

1

1

.2) to

kVA

500 kVA. They are comsupply power to as many as 20

formers ranges from

directly

to

the electrical utility

customers.

the primary are equal and

The load on

opposite to the ampere-turns of the secondary.

distribution

transformers varies

greatly throughout the day, depending on customer 3.

4.

The apparent power input to the transformer equal to the apparent power output. The exciting current

may be

in the

is

demand.

In residential districts a

peak occurs

morning and another peak occurs

primary winding

noon. The power peaks never

neglected.

in the

in the late after-

last for

more than one

or two hours, with the result that during most of the

24-hour day the transformers operate far below

11.1

their

Dual-voltage distribution transformer

normal

rating.

tem, every effort

Transformers that supply electric power to dential

areas

generally

have

Because thousands of such

transformers are connected to the public

small. This

resi-

two secondary

is

is

made

utility sys-

keep the no-load losses

achieved by using special low-loss

silicon-steel in the core.

225

to


226

—

I

A

Figure 11.3 Autotransformer having

N2 turns

A/-,

turns on the primary

and

on the secondary.

H2

11.2 Autotransformer

Figure 11.1 a.

Distribution transformer with

ondary. b.

Same

The

1

central conductor

20 V/240 V secis

distribution transformers

the neutral.

turns,

mounted on an

ing

connected

reconnected to give

only 120 V.

N

Consider a single transformer winding having

is

iron core (Fig.

1 1

to a fixed-voltage ac

.3).

{

The windand

source

the resulting exciting current /(> creates an ac flux <& m in the core.

the flux

is

As

in

Suppose a tap there are

any transformer, the peak value of

fixed so long as E\

N2

C

is

is

fixed (Section 9.2).

taken off the winding, so that

turns between terminals

A

and C.

Because the induced voltage between these terminals

is

proportional to the

number of

turns,

E2

is

given by

E 2 = (N2 /N0 X

£,

(ll.l)

Clearly, this simple coil resembles a transformer

having a primary voltage £, and a secondary voltage

E 2 However, .

A and the A are no longer isolated from

the primary terminals B,

secondary terminals C,

common

each other, because of the If

we connect

terminal A.

a load to secondary terminals

CA,

the resulting current I2 immediately causes a pri-

mary current /] to flow (Fig. 1 .4). The BC portion of the winding obviously carries current /,. Therefore, according to Kirchhoff s 1

CA portion carries a current (I 2 — /]). Furthermore, the mmf due to /, must be equal and opposite to the mmf produced by (7 2 — /]). As current law, the

Figure 11.2

a result,

we have

Single-phase pole-mounted distribution transformer rated:

100 kVA, 14.4 kV/240 V/120

V,

60 Hz.

(N

I ]

l

- N2 = )

(I 2

- I,)N2

SPECIAL TRANSFORMERS

221

B

O—

:

(N,-N 2

c

)

(h

-fx)

E2

A -o-

load

Figure 11.4 Autotransformer under load. The currents flow

which reduces

opposite directions

in

3.6

to I ]

assuming

N = I2 N 2

that both the transformer losses b.

and exciting current are negligible, the apparent c.

power drawn by the load must equal the apparent power supplied by the source. Consequently,

1

1

.

1

,

1

1

1

1

.3

CA

and

Solution

= E 2 I2

and

.2,

connected across the secondary,

is

The secondary voltage and current The currents that How in the w inding The relative size of the conductors on windings

BC

(U.3) a.

Equations

load

(ll. 2)

]

£,/,

kW

the upper and lower windings.

calculate: a.

Finally,

in

The secondary voltage

are identical to those

of a standard transformer having a turns ratio

N /N2

is

E2 = 80% X 300 =

240

V

.

i

However,

in

winding

actually part of the primary winding. In

effect,

is

this

autotransformer the secondary

separate secondary winding.

As

difference in size

cal

a result, autotrans-

becomes

power

and

2.

On

isolation

windings

is

a

E

= PIE 2 = 3600/240 =

The current supplied by

]

/E 2

The

/,

important

lies

=

=

PIE,

the current in

between

between the primary and secondary serious drawback

in

Autotransformers are used to

some start

secondary

applications.

induction

mo-

voltage of transmission lines,

and, in general, to transform voltages to

is

15

A

the source

(Fig.

1

1.5).

is

the current in

=

A winding BC = 12 A winding CA = 15 — 3600/300

12

12

=

3

A

the other hand, the absence of electri-

tors, to regulate the

mary

output.

particularly


b.

current

and cheaper than

lighter,

standard transformers of equal

0.5

I2

an autotransformer eliminates the need for a

formers are always smaller,

when

The secondary

ratio

Example 11-1 The autotransformer

is

close to

in Fig.

when

the pri-

The conductors

in the

secondary winding

can be one-quarter the size of those ing

BC

because the current

(see Fig.

winding

1

1.5).

BC

is

However,

is

in

CA

wind-

4 times smaller

the voltage across


tween the primary and secondary voltages,

I

11.4 has an 80 per-

cent tap and the supply voltage £,

c.

is

300

V. If a

namely (300 - 240) = 60 V. Consequently, winding CA has four times as many turns as winding BC. Thus, the two windings require essentially the same amount of copper.


228

-12A

/,

B

-O 2 =

/

G

:

15A

300 V

1

3A A

240

<\

V

Figure 11.5 Autotransformer of Example 11-1.

11.3 Conventional transformer

6.

The voltages add when terminals of opposite

X2

polarity (H, and

connected as an autotransformer conventional

changed

into

may add to,

is

in series.

Depending

made, the secondary

volt-

or subtract from, the primary voltage.

basic operation and behavior of a transformer

unaffected by a mere change

in

is

external connections.

Consequently, the following rules apply whenever a conventional transformer

is

connected as an auto-

transformer:

1

.

2.

3.

in

If

nominal voltage

1

1

.6

We

Hz.

wish to reconnect

in three different

age a.

shown in kVA, 600 V/l 20 V, 60

ways

it

as an autotransformer

to obtain three different volt-

ratios:

600 600 20 1

to

720

V V

secondary

primary

V primary

to

480

V

secondary

V V

primary to 480

maximum

secondary

load the transformer can

carry in each case.

rating.

rated current flows in one winding, rated cur-

ing (reason:

4.

If

The ampere-turns of the windings

always equal).

rated voltage exists across

one winding, rated

voltage automatically exists across the other

winding (reason: The same mutual flux links 2

both windings). 5.

If the

i

current in a winding flows from Hj to

the current in the other

X 2 to X,

and

5

1

rent will automatically flow in the other wind-

are

H2

single-phase transformer

has a rating of a

Calculate the

rating.

The voltage across any winding should not exits

Fig.

c.

any winding should not exceed

nominal current

ceed

(or

are connected together.

)

Example 11-2 The standard

b.

The current its

and Xj) are con-

an autotransformer by connecting the

upon how the connection age

X2

two-winding transformer can be

primary and secondary windings

The

H2

or

when H, and X,

voltages subtract

A

,

nected together by means of a jumper. The

H2

winding must flow from

and vice versa.

— 120 v-ô

o ,

Figure 11.6 Standard 15 kVA, 600 V/120 V transformer.


229

Solution

Nominal current of the 600 V winding

= SIE =

/,

=

000/600

15

{

is

Nominal current of the 120 V winding I2

a.

=

=

S/E 2

To obtain 480

15

000/120

=

A

25

is

125

A

secondary voltage (120 V)

V, the

X2

between terminals X,,

must subtract from

we

the primary voltage (600 V). Consequently,

connect terminals having the same polarity

shown

gether, as

in Fig.

ing schematic diagram

Note

.7.

1

The correspond-

given

that the current in the 120

same

the

1

is

as that in the load.

to-

in Fig.

V

1

1

winding

Because

this

Figure 11.7 Transformer reconnected as an autotransformer to

.8.

give a ratio of

600 V/480

V.

is

wind-

ing has a nominal current rating of 125 A, the

maximum

load can draw a S.,

The

.

A X

25

480

V=

60

kVA

currents flowing in the circuit at full-load

shown

are 1

1

power.

If

in Fig.

we assume

from X, 25

A

to

1

1

Note the following:

.8.

that the current

X2

in the

of

1

25

A flows

winding, a current of

H2

must flow from

to

H,

in the

other

winding. All other currents are then found

by applying Kirchhoff 's current law. 2.

The apparent power supplied by is

S b.

=

100

To obtain a

= 720

V.

site polarity

gether, as

same

600 V/720

=

V, the

of Fig.

1 1

.7

showing voltages and

secondary

primary voltage: 600

to the

Figure 11.8 Schematic diagram current flows.

V = 60kVA

600

+

Consequently, terminals of oppo-

(H, and

X2

in the

must be connected

)

in Fig.

1

1

secondary winding

load

125

is

is

again

and therefore the

is

again 125 A.

720

V = 90kVA

load current

to-

.9.

as that in the load,

maximum maximum Sb

add

shown

The current the

A X

ratio of

voltage must

120

the source

equal to that absorbed by the load:

The

now

A X

The previous examples show ventional transformer

is

that

when

a con-

connected as an auto-

Figure 11.9 Transformer reconnected to give a

ratio of

600 V/720

V.


230

69 kV

line

primary

\C

c/__

25

A

distributed

\^

capacitance

voltmeter

Oto 150 V secondary grounded

Figure 11.10 Transformer reconnected

to give

a

ratio of

120 V/480

V.

Figure 11.11 on a 69 kV line. Note between the windings.

Potential transformer installed

transformer,

it

can supply a load far greater

As

than the rated capacity of the transformer.

mentioned

earlier, this

is

the distributed capacitance

one of the advantages

of using an autotransformer instead of a con-

most exactly

ventional transformer. However, this

is

The nominal secondary voltage

ways

example

the case, as the next part of our

not al-

be.

To obtain

we

the desired ratio of 120

to

480

X,X 2

is

now connected

same

transmission

600

V

winding

maximum

ilar to that

load current cannot exceed 25 A.

AX

and

Sc

=

25

This load

is

less than the

480

the

isolate

V = 12kVA

full line

nominal rating (15

winding

voltage on the

In this regard, is

HV

to withstand the

one terminal of the secondary

always connected

to

ground

ing these three autotransformer connections.

to be isolated

The temperature rise of the transformer is the same in each case, even though the loads are respectively 60 kVA, 90 kVA, and 12 kVA. The

pacitance between the two windings

that the currents in the

windings and

to eliminate

when touching one of the

secondary leads. Although the secondary appears

from the primary, the distributed ca-

visible connection

between

voltage

ground.

the

secondary

By grounding one of

nals, the highest voltage

and so the losses are the same.

lines

and ground

The nominal usually less than insulation

is

is

makes an

in-

which can produce a very high

the flux in the core are identical in each case

11.4 Voltage transformers

sim-

side.

the danger of a fatal shock

is

is

between the primary and secondary

windings must be particularly great

kVA) of the standard transformer. We want to make one final remark concern-

reason

metering

of conventional transformers. However,

the insulation

therefore,

is,

to

lines (Fig. 11.11).

The construction of voltage transformers

is

as that in the load; consequently, the

The corresponding maximum load

to

measure or monitor the voltage on

lines

equipment from these in the

ir-

may

This permits standard instruments and relays

ers are used to

a),

to terminals

(Fig. 11. 10).

This time, the current

usually 115 V,

be used on the secondary side. Voltage transform-

V,

again connect H, and X, (as in solution

but the source

the

V

is

respective of what the rated primary voltage

shows. c.

phase with the primary voltage.

in

winding

and

the secondary termi-

between

the secondary

limited to 115 V.

rating of voltage transformers

500 VA. As

a result, the

often far greater than the

volume of

Voltage transformers (also called potential trans-

formers) are high-precision transformers the ratio of primary voltage to

a

known

load."'

constant,

in

which

secondary voltage

which changes very

little

Furthermore, the secondary voltage

steel.

is

with is

copper or

al-

In the

case of voltage transformers and current transformers,

the load

is

called burden.

is

volume of


Voltage transformers installed on

ways measure

the

line-to-neutral

eliminates the need for two

one side of the primary

is

late the

kV

lines al-

to

ground. For

transformer

shown

11.12 has one large porcelain bushing to iso-

HV line from

the grounded case.

The

latter

houses the actual transformer.

The

basic impulse insulation (BIL) of

Current transformers

1 1 .5

This

bushings because

connected

example, the 7000 VA, 80.5 in Fig.

HV

HV

voltage.

23

650 kV

expresses the transformer's ability to withstand

Current transformers are high-precision transform-

which the

ers in

rent

a

is

of primary to secondary cur-

ratio

known

constant that changes very

little

with the burden. The phase angle between the

pri-

mary and secondary current is very small, usually much less than one degree. The highly accurate current ratio

and small phase angle are achieved by

keeping the exciting current small. Current transformers are used to measure or

lightning and switching surges.

monitor the current

ondary

in a line

and

to isolate the

equipment connected

tering and relay

me-

to the sec-

The primary is connected in series with shown in Fig. 1. 3. The nominal sec-

side.

the line, as

1

ondary current

mary current

1

usually 5 A, irrespective of the pri-

is

rating.

Because current transformers (CTs) are only used for measurement and system protection, their

power

rating

small

is

and 200 VA. As

—generally

in the

formers, the current ratio to the

number of

ondary windings. tio

of 150 A/5

A

between 15

VA

case of conventional transis

inversely proportional

turns on the primary and sec-

A current transformer having a rahas therefore 30 times more turns

on the secondary than on the primary. For safety reasons current transformers must

ways be used when measuring currents mission lines. The insulation between

in

HV

the primary

and secondary windings must be great enough withstand the line surges.

stand

is

full line-to-neutral

The maximum voltage

the

CT can

line

load'

primary

C

:-j-:c / distributed capacitance

Figure 11.12 7000 VA, 80.5 kV, 50/60 Hz potential transformer having an accuracy of 0.3% and a BIL of 650 kV. The primary terminal at the top of the bushing is connected to the HV line while the other is connected to ground. The secondary is composed of two 115 V windings each tapped at 66.4 V Other details: total height: 2565 mm; height of porcelain bushing: 1880 mm; oil: 250 L; weight: 740 kg.

Figure 11.13

(Courtesy of Ferranti- Packard)

Current transformer installed on a 69 kV

secondary

secondary grounded

to

voltage, including

always shown on the nameplatc. 69 kV

al-

trans-

-

line.

with-


232

the

As in the case of voltage transformers (and for same reasons) one of the secondary terminals is

always connected

to

ground.

ing serves to isolate the

HV

1

kV

line

line.

from

The

A

current

large bush-

the ground.

The

terminals that are connected in series with the

The

line current

Hows

HV

current

c.

CT

a typical installation

is

shown

By way of comparison, former shown because

it

is

in Fig.

shown

is

1

1

.

17

in Fig. 11.15

is

is

50

in-

a.

The current

in Fig.

trans-

The

turns ratio

=

400/5

N^/Nn

11,17 has a rating

I2

=

80

is

smaller, mainly

It is

imped-

ratio is

and

The secondary current

of 50 VA, 400 A/5 A, 36 kV. 60 Hz.

total

the transmission-line

Solution

insulated for only 36 kV.

Example 1 1-3 The current transformer

If

il.

The secondary current The voltage across the secondary terminals The voltage drop across the primary

/,// 2

VA current

much

2

280 A, calculate

in Fig. 11.16.

the

1.

the

up the bushing and out by the other terminal. The of a

a.

line.

bushing, through the primary of the transformer, then

ternal construction

3.

wires on the secondary side possess a

b.

one terminal, down

into

1

ance (burden) of

CT is housed in the grounded steel case at the lower end of the bushing. The upper end of the bushing has two

manner similar to that shown in The ammeters, relays, and connecting

4.4 kV, in a

Fig. II.

Figure 11.14 shows a 500 VA, 100 A/5

transformer designed for a 230

into an ac line, having a line-to-neutral voltage of

=

=

1/80

is

280/80

=

3.5

A

connected

!

Figure 11.14 500 VA, 100 A/5 A, 60 Hz current transformer, insulated for a 230 kV line and having an accuracy of 0.6%. (Courtesy of Westinghouse)

N

Figure 11.15 Current transformer

in

the

final

(Courtesy of Ferranti-Packaro)

process

of construction.


M

"

233

Mm

'

Figure 11.17 Epoxy-encapsulated current transformer rated 50 VA, A, 60 Hz and insulated for 36 kV. (Courtesy of Montel, Sprecher & Schuh)

400 A/5

/,

Current transformer

220

kV,

3-phase

in

one phase

series with

line inside

The voltage across

c.

IR

=

X

3.5

The secondary voltage The primary voltage is E,

=

4.2/80

=

is

1

.2

=

4.2

V

than normal.

is

=

52.5

a miniscule voltage drop,

the 14.4

kV

cle.

mV

The

the

is

compared

to

line-to-neutral voltage.

Opening the secondary of a CT can be dangerous

and

falls

much

s

first

falls,

for

but

it

most of

remains

at a

the time.

The

thing happens during the second half-cycle.

During these saturated

intervals, the induced volt-

age across the secondary winding cause the flux changes very

little.

is

negligible be-

However, during

tremely high

primary

flowing

is

accidentally opened, the primary current

in the

circuit. If the sec-

to-

primary current

the unsaturated intervals, the flux changes

is

is

half cycle, flux
secondary circuit of a current transformer while

ondary

higher

of every half cy-

II. 18, as the

core also rises and

in the

1

so large that the core

during the

fixed, saturation level

secondary

may be 00

Every precaution must be taken to never open the current

to that

normal exciting cur-

core reaches peaks

flux

Referring to Fig.

rises

same

11-6

to the

tally saturated for the greater part

/,

This


rent, the flux in the

therefore 4.2 V.

0.0525

no further bucking effect due

ampere-turns. Because the line current

is

to

E2 =

compared

the exciting current of the transformer because there

a substation.

the burden

negligible

is

of the electrical load. The line current thus becomes

of a

is

b.

continues to flow unchanged because the imped-

ance of the primary

Figure 11.16

rate,

at

an ex-

inducing voltage peaks of several

hundred volts across the open-circuited secondary. This

is

a

dangerous situation because an unsuspect-


234

ing operator could easily receive a bad shock.

voltage

is

The

particularly high in current transformers

having ratings above 50 VA.

view of

In

ondary

must

the above,

first

if

CT

circuit of a

a meter or relay in the sec-

we

has to be disconnected,

secondary winding and

short-circuit the

then remove the component. Short-circuiting a current transformer

does no harm because the primary

current remains unchanged and the secondary current

can be no greater than that determined by the turns

ra-

winding may be

re-

tio.

The

moved

short-circuit across the

after the secondary circuit

is

again closed.

11-7 Toroidal current transformers Figure 11.18

When

the line current exceeds 100 A,

we can some-

times use a toroidal current transformer.

It

consists

Primary current,

CT

is

flux,

and secondary voltage when a

open-circuited.

of a laminated ring-shaped core that carries the sec-

ondary winding. The primary

is

composed of a

sin-

gle conductor that simply passes through the center

of the ring (Fig.

conductor

is

1

1

.

1

9).

The

unimportant as long as

it

is

secondary possesses

less centered. If the


is

having a ratio of 1000 A/5

200 turns

position of the primary

more

N

N. Thus, a toroidal

A

or

turns,

CT

has 200 turns on the

secondary winding. Toroidal

widely used

(MV)

CTs are

indoor installations. They are also incorporated

in circuit-breaker

(Fig.

simple and inexpensive and are

low-voltage (LV) and medium-voltage

in

1

bushings to monitor the line current

1.20). If the current

limit, the

CT causes the

Figure 11.19 Toroidal transformer having a ratio of

nected to measure the current

in

a

1000 A/5

A, con-

line.

exceeds a predetermined

circuit-breaker to

trip.

Example 11-4

A potential

transformer rated 14 400 V/l 15

current transformer rated 75/5

V

and a

A are used to measure

the voltage and current in a transmission line. If the

voltmeter indicates

1

1

V

1

and the ammeter reads

3 A, calculate the voltage and current in the line.

Solution

cm

The voltage on

E= The current

111

the line

x

is

(14 400/115)

=

13

900V Figure 11.20

in the line is

Toroidal transformer surrounding a conductor inside a

/=

3

X

(75/5)

=

45

A

bushing.


11.8 Variable autotransformer

1

1.22).

a fixed

A

variable autotransformer

is

often used

when we

wish to obtain a variable ac voltage from a fixedvoltage ac source. a

single-layer

The transformer

toroidal iron core.

A

composed of

movable carbon brush

in slid-

winding serves as a variable

ing contact with the tap.

is

winding wound uniformly on a

The brush can be

set in

235

The input voltage £, is usually connected to 90 percent tap on the winding. This enables

E 2 to vary from 0 to

1

10 percent of the input voltage.

Variable autotransformers are efficient and provide

good voltage regulation under variable ondary

line

loads.

The

sec-

should always be protected by a fuse or

circuit-breaker so that the output current

/2

never ex-

ceeds the current rating of the autotransformer.

any position between 0

and 330°. Manual or motorized positioning

may

be

used (Figs. 11.21 and 11.23).

As

the brush slides over the bared portion of the

winding, the secondary voltage portion to the

Figure 11.21 Cutaway view

number of

of a

E2

turns

increases in pro-

swept out

(Fig.

manually operated 0-140 V, 15 A showing (1) the laminated

variable autotransformer

toroidal core; (2) the single-layer winding; (3) the

mov-

able brush.

{Courtesy of American Superior Electric)

Figure 11.23

200 A, 0-240 V, 50 A, 120 V units, connected in series-parallel. This motorized unit can vary the output voltage from zero to 240 V in 5 s. Dimensions: 400 mm x 1500 mm. {Courtesy of American Superior Electric) Variable autotransformer rated at

50/60 Hz.

Figure 11.22 Schematic diagram ing a fixed

90%

tap.

of

a variable autotransformer hav-

It

is

composed

of eight

236


11.9

High-impedance transformers

The primary winding P is connected to a V ac source, and the two secondary windings

240

The transformers we have studied so far are alt designed to have a relatively low leakage reactance, ranging perhaps from 0.03 to

0.

S are connected tube.

per unit (Section

1

10.

1

However, some

3).

applications require

and commercial

industrial

much higher reactances, some-

across the long neon

the secondary voltage

of the transformer (Fig. circuit voltage (20

following

in the

in series

to the large leakage fluxes

E2

tp.,

falls rapidly

as

typical applications:

kV)

1

soon as the neon tube

current

is

1

.24c).

The high open-

initiates the discharge, but

lights up, the

automatically limited to 15

secondary

mA. The

electric toys

arc welders

corresponding voltage across the neon tube

fluorescent lamps

electric arc furnaces

to

neon signs

reactive

oil

and

with

increasing current, as seen in the regulation curve

times reaching values as high as 0.9 pu. Such high-

impedance transformers are used

b

,

Owing

falls

The power of these transformers ranges from 50 VA to 1500 VA. The secondary voltages

power regulators

1

5 kV.

burners Let us briefly examine these special applications.

1

.

A

toy transformer

is

often accidentally short-

by children

circuited, but being used

ther practical nor safe to protect

Consequently, the transformer that

its

leakage reactance

is

is

it

it

nei-

is

with a fuse.

designed so

so high that even a

permanent short-circuit across the low- voltage secondary will not cause overheating.


to

some

bell trans-

formers that provide low-voltage signalling

power throughout a home.

If

a short-circuit oc-

curs on the secondary side, the current

is

auto-

matically limited by the high reactance so as not to burn out the transformer or

damage

the

fragile annunciator wiring. 2.

Electric arc furnaces

and discharges

in

gases

possess a negative E/I characteristic, meaning that

once the arc

struck, the current increases

is

as the voltage falls.

To maintain

a uniform discharge,

ance

in series

ance

may

a steady arc, or

we must add

an imped-

with the load. The series imped-

be either a resistor or reactor, but

prefer the latter because

it

consumes very

we

little

active power.

However, the load,

it

is

if

a transformer

usually

is

used

to

more economical

porate the reactance in the transformer

designing

it

itself,

have a high leakage reactance.

to

typical

example

shown

in Fig.

1

is 1

the neon-sign transformer

.24.

0

supply to incor-

by

A

Figure 11.24 a. Schematic diagram

— 15

of a

30

mA

neon-sign transformer.

b.

Construction of the transformer.

c.

Typical E-I characteristic of the transformer.


range from 2

kV

to

20 kV, depending mainly

ings are very loosely coupled.

mary windings

upon the length of the tube. Returning to Fig.

1

1

.24a,

we

note that the

center of the secondary winding

(typically

is

only one-half the voltage across the

neon tube. As a

result, less insulation is

have properties similar

to

neon-sign

HV

1

.25).

The

controller permits

more

or less sec-

causing the leakage flux

to flow,

A change

in the

flux produces a corresponding

active

power absorbed by

leakage

change

in

transformer, incorporated in a static var

improve the power factor of the

pensator,

is

the re-

The com-

the transformer.

transformers. Capacitors are usually added to total circuit.

line

765 kV) while

windings (typically 6 kV)

to vary accordingly.

Fluorescent lamp transformers (called ballasts)

1

ondary current

needed

for the high-voltage winding.

kV and

between 230

are connected to an electronic controller (Fig.

This ensures that the secondary line-to-ground voltage

three pri-

are connected to the

the three secondary

grounded.

is

The

237

further discussed in Section 25.27.

Oil-burner transformers possess essentially the

same

about 10

A

kV

the oil jet.

3.

Some

oil

Q

secondary open-circuit voltage of

The

M

'4

arc continually ignites the va-

while the burner

is

in

operation.

between two carbon

taining an intense arc

A relatively

low secondary voltage

leakage flux

elecis

used tertiary

and the large secondary current

is

^

primary

electric furnaces generate heat by main-

trodes.

kV

Q

M

two closely immediately above

creates an arc between

spaced electrodes situated

porized

3-phase primary input 230

characteristics as neon-sign trans-

formers do.

winding

limited by the

leakage reactance of the transformer. Such trans-

formers have ratings between 100

MVA.

In

kVA and 500

secondary leakage flux

very big furnaces, the leakage reactance

of the secondary, together with the reactance of the conductors,

usually sufficient to provide

is

the necessary limiting impedance. 4.

Arc- welding transformers are also designed to

have a high leakage reactance so as to stabilize the arc during the welding process.

The open-

about 70 V, which

facilitates

circuit voltage

striking the arc

is

when

the electrode touches the

work. However, as soon as the arc lished, the

secondary voltage

15 V, a value that

the arc 5.

As

and the

is

falls to

about

we mention

the

units that absorb reactive

transmission

line.

enormous 3-phase

power from a 3-phase

These transformers are

tionally designed to

11.10 Induction heating

transformers

welding current.

a final example of high-impedance trans-

formers,

leak-

estab-

depends upon the length of

intensity of the

Figure 11.25 Three-phase static var compensator having high age reactance.

inten-

produce leakage flux and,

consequently, the primary and secondary wind-

High-power induction furnaces also use the former principle to produce high-quality other alloys.

The induction

stood by referring to Fig.

frequency 500 that

Hz

1

trans-

steel

and

principle can be under1.26.

ac source

is

A

relatively high-

connected to a

coil

surrounds a larse crucible containing molten


238

primary

molten iron

Figure 11.26 Coreless induction furnace. The currents

in

the molten metal.

the reactive

The

iron.

coil is the primary,

itself.

O

produces eddy

capacitor furnishes

power absorbed by the

coil.

and the molten iron

secondary turn, short-circuited

acts like a single

upon

flux

The

Consequently,

carries a very large

it

secondary current. This current provides the energy that

keeps the iron

scrap metal as

15

it is

melting other

in a liquid state,

added

to the pool.

Such induction furnaces have ratings between kVA and 40 000 kVA. The operating frequency

becomes progressively lower

power

as the

Figure 11.27 Channel induction furnace and

is

required to drive the flux through

the molten iron and through the

must remember far

air.

In this regard,

that the temperature of

above the Curie

point,

far as permeability

is

and so

it

we

molten iron

The magnetizing

is

like air as

why

these

Capacitors are installed close to the coil to sup-

In

power

it

known

as a

channel fur-

nace, a transformer having a laminated iron core

made

to link with a channel of

in Fig.

1

1

ted to the

.27.

The channel

Hows

molten

iron, as

Hz

is fit-

coil is

channel and through the molten

iron in the crucible. In effect, the channel lent to a single turn short-circuited

coil.

is

On

is

the

large because the sec-

obviously not tightly coupled to the

Nevertheless, the power factor II. 26,

and 80 percent. As a

is

higher

being typically between 60

result, a

smaller capacitor bank

required to furnish the reactive power.

Owing

to the

very high ambient temperature, the

primary windings of induction furnace transformers are

always made of hollow, water-cooled copper

on

is

itself.

aluminum, copper, and other metals, 1

1.28

shows

as well as iron.

a very special application of

is

source, and the secondary current

in the liquid

primary

is

than that in Fig.

Figure

shown

a ceramic pipe that

bottom of the crucible. The primary

excited by a 60 12

is

other hand, the leakage flux

ondary turn

low because the flux

is

permeable iron core.

conductors. Induction furnaces are used for melting

absorbs.

another type of furnace,

current

to a highly

is

furnaces are often called coreless induction furnaces.

ply the reactive

confined

is

behaves

concerned. That

water-cooled

rating

60 Hz is used when the power exceeds about 3000 kVA. The power factor of coreless induction furnaces is very low (typically 20 percent) because a large magincreases. Thus, a frequency of

netizing current

its

transformer.

equiva-

the induction heating principle.

11.11 High-frequency transformers In electronic

power supplies there is often a need to from the input and to reduce the

isolate the output

weight and cost of the such as

in

aircraft,

unit. In

there

is

other applications,

a strong incentive to


239

order to illustrate the reason for this phenomenon, limit

we how

avoid a cumbersome theoretical analysis, take a practical transformer and observe

when

haves

the frequency

Consider Fig. tional

20 V/24

1

ing of 36

in the

II. 29,

be-

1

.5

T.

a rat-

0.5 kg

The

flux

core attains a peak of 750 fjiWb. The lami-

nated core

is

about

made of 1

ordinary silicon steel having

mm

(12 mils) and the

W. The

and

the primary

1

.5

A

current rating

silicon

A >~

1 t

core

mA for

12 mil 1.5

600

total

300

cm

mA

120 V 60 Hz

is

for the secondary.

core: 6 x 5 x 2.5

300

it

which shows a conven-

peak flux density of

at a

to

will

raised.

60 Hz transformer having

V,

a thickness of 0.3 is

is

VA. This small transformer weighs

and operates

loss

we

our discussion to transformers. Furthermore,

120

24 V

t

T

.

J

P = 36 VA B=1.5T 0max = 75OAWb core loss =

1

W

Figure 11.29

Figure 11.28 Special application of the transformer effect. This picture of

shows one stage

in

a steam-turbine generator.

the diameter of a 5

t

It

consists of expanding

coil-retaining ring.

A

coil of

as-

its

temperature up to 280°C

expansion enables the ends, where

it

ring to

in

h.

The

main

resulting

be slipped over the

rise of the large

this

it

in-

mass.

(Courtesy of ABB)

means

that the

can be increased

E= = = which

4.44,/TV]

400 Hz, while

in

electronic

to

to,

typically

power supplies

quency may range from 5 kHz

An

is

the

at a fre-

same peak

^ max

will re-

to Eq. 9.3,

corresponding primary voltage

to

the fre-

50 kHz.

increase in frequency reduces the size of such

devices as transformers, inductors, and capacitors. In

(9.3)

max

X 6000 X 600 X 750 X 000 V

12

is

6 l()

100 times greater than before! The sec-

becoming 2400

frequency

Assuming

4.44

by using a relatively high frequency compared in aircraft the

it

00 times higher than

750 |xWb. However, according

at

ondary voltage

60 Hz. Thus,

1

follows that the flux

minimize weight. These objectives are best achieved

say,

for.

is

coil-

clean and produces a very uniform

is

was designed

it

flux density,

the ring, bringing

about 3

cools and contracts. This method of

duction heating

temperature

in

to the transformer,

us consider the effect of operating

what

is

which induces large eddy currents

let

quency of 6000 Hz, which

wound around the ring and connected to a 35 kW, 2000 Hz source (left foreground). The coil creates a 2000 Hz magnetic field, bestos-insulated wire

Without making any changes

the construction of the rotor

will likewise be 100 times greater,

The operating conditions are The primary and secondary currents remain unchanged and so the power of the transformer is now 3600 VA, 00 times greater than

shown

in Fig.

V.

11.30.

1

in Fig.

1

1

.29. Clearly, raising the

a very beneficial effect.

frequency has had


240

However, the advantage because

at

6000 Hz

700 W), due

to the increase in

is

eddy current and hys-

Thus, the transformer

teresis losses.

not feasible because

To

it seems enormous (about

it

in Fig.

we

of 12 mil silicon

cording

to

1

.5

T to 0.04 T. As

be

a result, ac-

Eq. 9.3, the primary and secondary volt-

ages will have to be reduced to 320 spectively.

as

reduction in

steel, this requires a

from

the flux density

The new power of

P = 320 X

0.3

=

96

V

and 64 V,

re-

the transformer will

VA (Fig.

11.31

).

This

is al-

most 3 times the original power of 36 VA, while

same temperature rise. By using thinner laminations made of

re-

it

is

special

possible to raise the flux density

above 0.04 T while maintaining the same core losses.

Thus,

if

we

replace the original core with

this special material, the flux density to 0.2 T.

=

cl>

max of

750 |xWb X

(0.2 T/1.5 T)

means

primary voltage can be raised to

that the

100

£ = 4.4477V, * nlllx = 4.44 X 6000 X 600 X = 1600 V core: 6 x 5 x 2.5

12 mi

silicon

300

120/

silicon

12

600

Figure 11.31

interested, of course, in maintaining the 1

V to 24 V. This

20

read-

is

number of turns on the primary will be reduced to 600 t X (120 V/1600 V) = 45 turns,

while the secondary will have only 9 turns. Such a drastic reduction in the

wire

the

Bearing

former

can

size

mind

in

is still

number of turns means

be

increased

that the capacity of the trans-

480 VA,

it

follows that the rated pri-

mary current can be raised to 4 A while that secondary becomes 20 A. This rewound

same

size

in

the

trans-

11.33) has the

special core (Fig.

its

that

significantly.

and weight as the one

in

Fig.

11.29.

Furthermore, because the iron and copper losses

same

are the

both cases, the efficiency of the

in

high frequency transformer It is

now obvious

is better.

that the increase in

has permitted a very large increase pacity of the transformer.

It

in the

frequency

power

ca-

follows that for a given

A

cm

special core J '

= 3600

VA

300

mA

1

ff=1.5T

2400 V

^max^SOjUWb

1600^

core loss = 700

6 kHz ^

W

600

r

120

5

p = 480 VA B = 0.2T

A

t+

t

«W= lOO^Wb

320 V

r

core: 6 x 5 x 2.5

cm

core loss =

1

W

t

120/

cm

special core 1.5

.

is

.32).

from 600

mil

mA

6 kHz

1

Figure 11.32

core: 6 x 5 x 2.5

32*0V^

1

power output a high frequency transformer is much smaller, cheaper, more efficient, and lighter than a 60 Hz transformer.

Figure 11.30

300

VA (Fig.

480

core: 6 x 5 x 2.5

—

/

I(T

A=

the

cm

15A 600

X

100

.5

achieved by rewinding the transformer. Thus,

I

mA

12 kV 6 kHz

^Wb, which

V,

ily

can be raised

This corresponds to a peak flux

1

former with

taining the

nickel-steel,

V X We are

320

original voltage ratio of

same

320

is

and so the enhanced capacity of the transformer

is

Based upon the properties

1.29.

1

.30

can reduce the

flux density so that the core losses are the in Fig.

1

1

will quickly overheat.


they were

The corresponding secondary voltage

not as great as

is

the core loss

64

A

vh Y

P = 96 VA # = 0.04T 0max = 2OAiWb core loss =

1

W

4

20 A

A

B= 120~\T 6 kHz

1

45/

Figure 11.33

9/

24

V T

0.2

T

<^max = lOO^Wb core loss = 1

W

SPECIA L TRA NS FORMERS


Calculate the voltage across the secondary

a.

winding Practical level 11-1

What

is

What

the difference between an auto-

is

former? 1

1

-3

Why

1

-4

11-5

Calculate the voltage drop the transformer

c.

If the

produces on the

Of a

is

looped four

times through the toroidal opening, calcu-

current transformer?

new

late the

current ratio.


why the secondary winding PT must be grounded.

Explain

CT or A toroidal

of a

11-12

The nameplate of a small transformer dicates 50 VA, 120 V, 12.8 V When 1

current transformer has a ratio of

How many

turns does

have?

it

current transformer has a rating of 10

VA, 50 A/5 A, 60 Hz,

1

V

8.8

would

2.4 kV. Calculate

V

13.74

If

V were

120

at

no-load

available,

what

Why

the secondary voltage be?

this voltage

the nominal voltage across the primary

in-

applied to the primary, the

is

voltage across the secondary is

A

line conductor.

primary conductor

must we never open the secondary of

1500 A/5 A. 11-6

ammeter has an impedance

the

b.

the purpose of a voltage trans-

a current transformer? 1

if

of 0.1 5 n.

transformer and a conventional transformer? 11-2

24

is

higher than the indicated

nameplate voltage?

winding. 11-13

Intermediate level 11-7

A

single-phase transformer has a rating of

100 kVA, 7200 V/600 V, 60 Hz.

If

re-

it is

connected as an autotransformer having a ratio it

1

1-8

In

can carry.

Problem

11-14 1

1-7,

how

should the trans-

H2

,

X,,

X2

)

be con-

in

Advanced 11-10

A

of 6.6 k V/600

how

is

Problem

1

1-7

is

V

What

load can

it

120

V

winding wound

Many

airports use series lighting systems in

rent

winding, or vice versa?

sion line

it

where

is

installed

numin

60 Hz

kept constant

at

20 A. The secondary

A

to a

incandescent lamp.

a.

Calculate the voltage across each lamp.

b.

The 0.07

resistance of the secondary winding

0

while that of the primary

Knowing

It

that the

is

is

0.008 U.

magnetizing current and

the leakage reactance are both negligible,

has a primary to secondary capacitance If

is

100 W, 6.6

should the connections be made?

of 250 pF.

large

windings are individually connected

carry

100 VA, 2000 A/5 A, 60 Hz, 138 kV.

calculate the voltage across the primary

on a transmis-

winding of each transformer. the line-to-neutral voltc.

If

140 lamps, spaced

138 kV, calculate the capacitive vals, are

leakage current that flows to ground (see

11-11

V

the 12.8

current transformer has a rating of

is

il. Is the

upon

source. In one installation, the primary cur-

recon-

level

age

0.306

the pri-

5.2 11 and that of the secondary

1

ber of current transformers are connected

nected again as an autotransformer having a

and

mary

series across a constant current,

The transformer ratio

However, the resistance of

which the primary windings of a

nected? 1-9

epoxy and cannot be

seen.

is

of 7800 V/7200 V, calculate the load

former terminals (H,,

1

Referring to Problem 11-12, the windings are encapsulated in

at

in series

power

The

a temperature of 105°C.

toroidal current transformer of Fig. 1

000 A/5 A. The

conductor carries a current of 600 A.

line

in inter-

using

wire, calculate the

Fig. M.13).

11.19 has a ratio of

every 50

No 14 minimum voltage of the Assume the wire operates at

connected

11-15

A

source.

no-load

test

on a

1

5

kVA. 480 V/l 20

60 Hz transformer yields

the following

V,

242


saturation curve data

winding

is

The primary a.

Draw

is

If the

the

1

20

V

c.

known

to

have 260

Draw

the saturation curve at

flux in

60 Hz (peak

mWb versus current in

mA). At

what point on the saturation curve does

turns.

uration

the saturation curve (voltage versus

become important?

sat-

Is the flux dis-

torted under these conditions?

mA).

current in b.

when

excited by a sinusoidal source.

experiment were repeated using a

50 Hz source, redraw the resulting saturation curve.

E

14.8

31

A)

59

99

49.3

144

66.7

210

90.5

430

110

120

130

136

142

V

700

1060

1740

2300

3200

mA

Chapter 12 Three-Phase Transformers

connected

12.0 Introduction is distributed throughout North America by means of 3-phase transmission lines. In order to transmit this power efficiently and economically, the

Power

vice versa.

kV

3.8

power

to

that has to be transmitted

and the distance is

also

A

it

uniform, ranging from

systems to 600

V

1

20/240

shift

from one

this re-

ratio

of the

the primaries and secon-

enables us to change the number of

if

there

were a

12-phase system.

practical application for

it,

we

could even convert a 3-phase system into a 5-phase

can be achieved by using three

system by an appropriate choice of single-phase transformers and interconnections.

form a 3-phase transformer bank.

In

making

tant to

When

how

into a 2-phase, a 6-phase, or a

Indeed,

single-phase transformers connected together to

12.1

The amount of phase

depends again upon the turns

phases. Thus, a 3-phase system can be converted

windings mounted on a 3-legged core. However, result

between the 3-phase input voltage and

shift feature

level to another.

The transformers may be inherently 3-phase,

same

shift

daries are interconnected. Furthermore, the phase-

having three primary windings and three secondary

the

are connected.

transformers, and on

quires the use of 3-phase transformers to transform the voltages

may

wye, or

3-phase transformer bank can also produce a

the 3-phase output voltage.

V single-phase

3-phase systems. Clearly,

in

a result, the ratio of the 3-phase input

upon how they

phase

the appropriate

voltage levels used in factories and homes. These are fairly

As

and the secondaries

only upon the turns ratio of the transformers, but

765 kV) depend upon the amount of

has to be carried. Another aspect

in delta

voltage to the 3-phase output voltage depends not

voltages must be at appropriate levels. These levels (1

ways. Thus, the primaries

in several

be connected

the various connections,

observe transformer

polarity

may produce

polarities.

it

is

An

imporerror in

a short-circuit or unbalance

Basic properties of 3-phase transformer banks

the line voltages

three single-phase transformers are used to

former banks can be understood by making the

The

transform a 3-phase voltage, the windings can be

and currents.

basic behavior of balanced 3-phase trans-

lowing simplifying assumptions:

243

fol-


244

1

.

2.

3.

The exciting currents

nected

are negligible.

The transformer impedances, due

to the resis-

is

Terminal H, of each trans-

H2

connected to terminal

of the next

X2

tance and leakage reactance of the windings,

transformer. Similarly, terminals X, and

are negligible.

cessive transformers are connected together.

The

total

tual physical layout of the transformers

apparent input power to the trans-

former bank

Fig. I2.l. is

put power.

Furthermore,

when

single-phase transformers

are connected into a 3-phase system, they retain their basic single-phase properties,

voltage ratio, and flux

ratio,

polarity

marks X,,

X2

in

and Hj,

between primary and secondary that

Ex

x

is in

Fig.

phase with

EH

the core.

H2 is

,

Given the

C

AO

the phase shift

Ht

X

Ho

X;

-O-i

1

2

balanced

BOH

.

three-phase

CO R

of

to a level appropriate for

the outgoing transmission line is

The schematic diagram is drawn in such a way to show not only the connections, but also the phasor

H2

X;

Hi

X,

Ho

X2

load

I

transform the voltage of the incoming

transmission line A, B,

line

l,

2, 3.

The incoming

connected to the source, and the outgoing

connected

to the load.

The transformers

\ 3/„

line

are con-

Figure 12.1 Delta-delta connection of three single-phase trans-

formers.

The incoming

lines (source) are A, B,

the outgoing lines (load) are

V3/

1

,

2, 3.

s

load


is

in Fig. 12.2.

zero, in the sense

three single-phase transformers P, Q, and 12.1

The corresponding schematic diagram

all

such as current

12.2 Delta-delta connection The

is

of suc-

The acshown in

equal to the total apparent out-

given

is

in delta-delta.

former

of

a delta-delta connection and associated phasor diagram.

C and

THREE-PHASE TRANSFORMERS

relationship between the primary and secondary

c.

drawn

d.

corresponding primary winding to

e.

voltages. Thus, each secondary parallel to the

which

is

it

winding

coupled. Furthermore,

duces voltages

£AB £ BC ECA ,

,

if

is

G

source

in-

current in the

HV

current in the

LV

lines

lines

currents in the primary and secondary

windings of each transformer

pro-

according to the

The The The

245

f.

The load

carried by each transformer

dicated phasor diagram, the primary windings are

same way, phase by phase. For example, the primary of transformer P between lines A and B is oriented horizontally, in the same direction

Solution

oriented the

£AB

as phasor

a.

The apparent power drawn by S

.

Because the primary and secondary voltages

EH

and

Hn

phase,

Ex

En

(secondary voltage of

b.

E AB (primary £2 3 s m phase

transformer P) must be in phase with of the same transformer). Similarly,

with

£ BC

,

and

E3i

ECA

with

i-

tween the respective incoming and outgoing mission lines are If

in

small.

is

c.

connected to lines

1

The

-2-3, the

incoming

=

(24.4

=

102

in the

/

V3 times greater than the respective

and

/s

ondary windings the transformer

flowing (Fig.

bank

in the

The

rating of

I2

single transformer.

Note tutes

that

arrangement, each transformer,

considered alone, acts as

phase

H2

circuit.

in the

rent / s

if

it

were placed

Thus, a current

primary winding

flowing from

X2

to

is

X|

/

p

MVA.

is

(8.9)

X

10 )/(V3

X

138 000)

A LV

lines is

=

5/(V3 E)

=

(24 A

6

X

10 )/(V3

X 4160)

= 3386 A

although the transformer bank consti-

a 3-phase

also 24.4 line

6

current in the

three times the rating of a

is

is

HV

primary and sec-

The power

12.2).

HV line

current in each

/,=5/(V3£)

d.

currents

MVA

follows that the apparent power fur-

in any delta connection, the line

produces balanced line currents

A-B-C. As

It

nished by the

trans-

phase.

a balanced load

lines

24.4

(7.7)

The transformer bank itself absorbs a negligible amount of active and reactive power because the 1~R losses and the reactive power associated

resulting line currents are equal in magnitude. This

currents are

21/0.86

=

is

with the mutual flux and the leakage fluxes are

,

such a delta-delta connection, the voltages be-

In

P/cos 0

of a given transformer must be in

Xi

follows that

it

= =

the plant

e.

Referring to Fig.

mary winding

in a single-

flowing from H] to

/

p

1

2.2, the current in

each

pri-

is

=

1

02/V 3

=

58.9

A

associated with a curin the

The

secondary.

current in each secondary winding /s

= 3386/V3 =

1955

is

A

Example 12-1 Three single-phase transformers are connected delta-delta to step

down

a line voltage of

4160 V to supply power The plant draws 21

MW

1

38

kV

in

to

to a

manufacturing plant.

at a

lagging power factor

f.

Because the plant load

is

balanced, each trans-

former carries one-third of the 24.4/3

The

-

8.13

total load, or

MVA.

individual transformer load can also be

obtained by multiplying the primary voltage

of 86 percent.

times the primary current:

Calculate a.

b.

The apparent power drawn by the plant The apparent power furnished by the HV

S = line

=

E p Ip = 8.13

138 000

MVA

X

58.9


246

Note

that

we can

do not know how the fect, the plant load

3-phase load

(shown

composed of hundreds of which

are

connected

as a

is

connected. In ef-

box

in Fig.

individual loads,

in

delta,

we

others

12.2)

is

some of wye.

in

Furthermore, some are single-phase loads operating at

much lower

voltages than

4160

V,

powered by

smaller transformers located inside the plant.

sum total of these

The

calculate the line currents and the

currents in the transformer windings even though

The

relative values of the currents in the trans-

former windings and transmission Fig.

C

1

are

The

2.4.

Thus,

the line currents in

V 3 times the currents

line currents in

in the

phases

lines are given in

phases A, B, and

primary windings.

2, 3 are the

l,

same

A shift

delta-wye connection produces a 30° phase

between the

of the incoming and

line voltages

outgoing transmission

lines.

Thus, outgoing

line

loads usually results in a reasonably

well-balanced 3-phase load, represented by the box.

AO-

12.3 Delta-wye connection

When

the transformers are connected in delta-wye,

the three primary windings are connected the

way as

in Fig.

1

2.

1

.

together, creating a

BO

same

H,

X

H2 M

X;2 X f

1

—

C—-—

balanced

2

:

I

H,

Xt

H2

X;

H,

X

H2

x 2 |—

-O(

•

three-phase load

However, the secondary windings

are connected so that

all

the

common

X2

terminals are joined

neutral

N

CO

(Fig. 12.3). In

3

-O--I

such a delta- wye connection, the voltage across each primary winding age.

is

equal to the incoming line volt-

However, the outgoing

line voltage is a/3

the secondary voltage across each transformer.

times

as

the currents in the secondary windings.

Figure 12.3 Delta-wye connection of three single-phase transformers.

load

Figure 12.4 Schematic diagram of a delta-wye connection and associated phasor diagram. (The phasor diagrams on the mary and secondary sides are not drawn to the same scale.)

pri-

THREE-PHA SE TRANSFORMERS

375 A

3932 A

247

1

80 kV 90

MVA

13.2kV«

375 A

3932 A

Figure 12.5

See Example

E i2

voltage

£ AB

,

12-2.

is

30° ahead of incoming

as can be seen

The voltage across

line voltage

from the phasor diagram.

The voltage between

outgoing line feeds an isolated group of loads, the phase

shift creates

no problem. But,

if

and 3

the outgoing

be connected in parallel with a line comfrom another source, the 30° shift may make

such a parallel connection impossible, even line voltages are

One

that

is

it

s

b.

1,

2,

The load

C

is

kV

139

carried by each transformer

S

=

90/3

is

= 30 MVA

wye conThe

reduces the amount of insulation

needed inside the transformer. The to

the

therefore,

the outgoing lines

E = 80 V3 =

otherwise identical.

of the important advantages of the

nection

if

is,

is

line has to

ing

the secondary

80 kV.

If the

current in the primary winding

HV winding has

/

be insulated for only 1/V3, or 58 percent of the

The

p

is

= 30 MVA/ 13.2 kV = 2273 A

current in the secondary winding

is

line voltage. /s

Example 12-2

c.

Three single-phase step-up transformers rated

MVA, a 13.2 a 90

1

3.2

kV

kV/80 kV

at

The current

If

in

each incoming

/=

they feed

MVA load, calculate the following:

The

2273 V 3

b. c.

The secondary line voltage The currents in the transformer windings The incoming and outgoing transmission

line

A, B,

= 3937 A

current in each outgoing line /

a.

A

375

40

are connected in delta-wye on

transmission line (Fig. 12.5).

= 30 MVA/80 kV =

=

375

1

,

2, 3 is

A

12.4 Wye-delta connection line

The currents and voltages

currents

in a wye-delta connection

are identical to those in the delta-wye connection of

Solution

The

easiest

way

Section to solve this

the windings of only

former

problem

is

one transformer,

say, trans-

P.

The voltage across ously 13.2 kV.

2.3.

The primary and secondary connec-

tions are simply interchanged. In other words, the

H 2 terminals are connected together to create a neuand the X X 2 terminals are connected in delta.

tral, a.

1

to consider

the primary winding

is

obvi-

[

,

Again, there results

a

30° phase

shift

between the

voltages of the incoming and outgoing lines.


248

ac source |

Figure 12.6

Wye-wye connection

ac source

with neutral of the primary connected to the neutral of the source.

'

Figure 12.7

Wye-wye connection using a

12.5

tertiary winding.

Wye-wye connection

to a delta-delta connection, except that

former

When cial

transformers are connected

in

wye-wye, spe-

precautions have to be taken to prevent severe

One way

distortion of the line-to-neutral voltages. to prevent the distortion is to

connect the neutral of

the primary to the neutral of the source, usually

way

of the ground (Fig. 12.6). Another

way

is

by to

provide each transformer with a third winding, called tertiary winding. three 1

2.7).

The

tertiary

transformers are connected

They

windings of the in

delta

(Fig.

often provide the substation service

is

delta connection

is

is

ample,

if

two 50 kVA transformers are connected

bank as

it

is

obviously 2

may seem,

it

X 50 =

100 kVA. But, strange

can only deliver 86.6

The open-delta connection emergency

situations.

Thus,

if

mainly used

is

b

n^-Mj Q

I

possible to transform the voltage of a 3-phase

The open-delta arrangement

is

in

identical

Figure 12.8a Open-delta connection.

o

[h

x~|

in

three transformers

A O-

12.6 Open-delta connection

open-delta.

kVA before

the transformers begin to overheat.

0—4

It is

in

open-delta, the installed capacity of the transformer

transformer.

system by using only 2 transformers, connected

only 86.6 per-

cent of the installed transformer capacity. For ex-

Note that there is no phase shift between the incoming and outgoing transmission line voltages of

wye-wye connected

trans-

seldom used because the load

capacity of the transformer bank

voltage where the transformers are installed.

a

one

absent (Fig. 12.8). However, the open-


are connected in delta-delta and

comes defective and has ble to feed the load

to

one of them be-

be removed,

it

is

249

Thus, the ratio

possi-

maximum

on a temporary basis with the

kVA 300 kVA 260

load

installed transformer rating

two remaining transformers.

= Example 12-3

Two

single-phase 150

formers are connected

maximum

V

kVA, 7200 V/600 in

0.867, or

86.7%

12.7 Three-phase transformers trans-

open-delta. Calculate the

3-phase load they can carry.

A transformer bank transformers

former Solution

may

(Fig.

1

composed of three single-phase

be replaced by one 3-phase trans-

2.9).

The magnetic core of such

a

transformer has three legs that carry the primary and

Although each transformer has a rating of 50 k VA, 1

two together cannot carry a load of 300 kVA. The following calculations show why:

secondary windings of each phase. The windings are

the

The nominal secondary current of each former

connected

internally, either in

wye

the result that only six terminals

or in delta, with

have to be brought

trans-

outside the tank. For a given total capacity, a 3-phase

is

transformer

=

/s

1

50 kVA/600

V=

250

A

is

single-phase

always smaller and cheaper than three transformers.

Nevertheless,

single-

phase transformers are sometimes preferred, partic-

The current ceed 250 A

mum

/s in lines 1, 2, 3

(Fig.

cannot, therefore, ex-

12.8b). Consequently, the maxi-

load that the transformers can carry

when

a replacement unit

one 3-phase 5000 (8.9)

259 800

Figure 12.8b Associated schematic and phasor diagram.

VA

is

essential.

For ex-

kVA. To guarantee continued service we can

is

S = ^3EI

= V3 X 600 X 250 = = 260 kVA

ularly

ample, suppose a manufacturing plant absorbs 5000

ond one as

kVA

a spare. Alternatively,

we can

single-phase transformers each rated plus one spare.

install

transformer and keep a sec-

at

install three

1667 kVA,

The 3-phase transformer option

is

250


more expensive (total capacity: 2 X 5000 = 10 000 kVA) than the single-phase option (total capacity: 4 X 1667 = 6667 kVA). Fig. 12.10 shows successive stages of construction of a

3-phase

1

10

MVA,

222.5 kV/34.5

changing transformer/ Note that 1

three

A

main

legs, the

'

in

kV

tap-

addition to the

tional lateral legs.

They enable

the designer to re-

duce the overall height of the transformer, which simplifies the

problem of shipping.

In effect,

when-

ever large equipment has to be shipped, the designer

is

faced with the problem of overhead clear-

ances on highways and

rail lines.

magnetic core has two addi-

1 £

lap-changing transformer regulates the secondary voltage

by automatically switching from one tap to another on the pri-

mary winding. The tap-changer

is

a motorized device under the

control of a sensor that continually monitors the voltage that

hns to be held constant.

3 I 1—t w V

Figure 12.10a Core of a 1 1 0 MVA, 222.5 kV/34.5 kV, 60 Hz, 3-phase transformer. By staggering laminations of different widths, the core legs can be made almost circular. This reduces the coil diameter to a minimum, resulting in 2 less copper and lower l R losses. The legs are tightly bound to reduce vibration. Mass of core: 53 560 kg.

Figure 12.9 Three-phase transformer for an electric arc furnace, rated 36 MVA, 1 3.8 kV/1 60 V to 320 V, 60 Hz. The secondary voltage is adjustable from 1 60 V to 320 V by means of 32 taps on the primary winding (not shown). The three large busbars in the foreground deliver a current of 65 000 A. Other characteristics: impedance: 3.14%; diameter of each leg of the core: 71 mm; overall height of core: 3500 mm; center line distance between adjacent core legs: 1220 mm. (Courtesy of Ferranti- Packard)

Figure 12.10b

Same

transformer with coils


in

in

place.

wye and

The primary

the secondaries

in

Each primary has 8 taps to change the voltage in steps of ±2.5%. The motorized tap-changer can be seen in the right upper corner of the transformer. Mass of copper: 1 5 230 kg. delta.


The 34.5 kV windings (connected in delta) are to the core. The 222.5 kV windings (connected in wye) are mounted on top of the 34.5 kV mounted next

windings.

A space of several centimeters separates the

two windings cool

251

to

ensure good isolation and to allow

flow freely between them. The

oil to

ings that protrude

nected to a 220

bushings are

kV

much

HV

bush-

from

the oil-filled tank are con-

line.

The medium voltage (MV)

smaller and cannot be seen

in the

photograph (Fig. 12.10c).

12.8 Step-up

and step-down

autotransformer When the

voltage of a 3-phase line has to be stepped

up or stepped down by a moderate amount, nomically advantageous

to

it

is

eco-

use three single-phase

transformers to create a wye-connected autotrans-

The

former. in Fig.

1

agram

is

2.

1

actual physical connections are 1

a,

given

shown

and the corresponding schematic in Fig.

12. lib.

The respective

di-

line-

to-neutral voltages of the primary and secondary

are obviously in phase. Consequently, the incoming

Figure 12.10c

Same

has been subjected to a 1050 kV impulse test on the HV side and a similar 250 kV test on the LV side. Other details: power rating: 110 MVA/146.7 MVA (OA/FA); total

mass

transformer ready

including

oil:

158.7

for shipping.

t;

It

overall height: 9 m; width:

and outgoing transmission phase. tral,

The

neutral

is

line

connected

voltages are

to the

otherwise a tertiary winding must be added

to

prevent the line-to-neutral voltage distortion mentioned previously (Section 12.5).

8.2 m, length: 9.2 m.

{Courtesy of ABB)

A

B

O2

-o

H

Figure 12.11a Wye-connected autotransformer.

in

system neu-

TN

Figure 12.11b Associated schematic diagram.

load

252


Figure 12.11c Single-phase autotransformer (one of a group of three) connecting a 700 kV, 3-phase, 60 Hz transmission line to an existing 300 kV system. The transformer ratio is 404 kV/173 kV, to give an output of 200/267/333 MVA per transformer, at a temperature rise of 55°C. Cooling is OA/FA/FOA. A tertiary winding rated 35 MVA, 1 1.9 kV maintains balanced and distortion-free line-to-neutral voltages, while providing power for the substation. Other

and windings: 132 t; tank and accessories: 46 t; oil: 87 t; total 265 t. BIL rating is 1950 kV and 1050 kV on the HV and LV side, respectively. Note the individual 700 kV (right) and 300 kV (left) bushings protruding from the tank. The basic impulse insulation (BIL) of 1950 kV and 1050 kV expresses the transformer's ability to withstand lightning and switching surges. (Courtesy of Hydro-Quebec)

properties of this transformer: weight of core weight:


For a given power output, an autotransformer

former (see Section

1

1

.2).

This

is

the ratio of the

incoming

line voltage lies

between 0.5 and

particularly true

winding rated

1

1

.9

kV.

kV

an existing 300

kV

The cerned)

=

3

66.3

with a tertiary

MVA.

basic transformer rating (as far as size is

considerably less than

MVA.

This

lies

between 0.5 and

kV

to

kV

transmission

230 kV

line has to be

supply a load of 200

MVA.

totransformers are to be used. Calculate the basic

power and voltage rating of each transformer, assuming they are connected as shown in Fig.

Solution

199

stations

ing

is

and

in

-

199

H2

between H, and

230/V3

to

I,

SN3E

=

(200

X

=

335

lines

1

is

and

lx

power flow over

trans-

the phase shifting principle, con-

between phases B and

12. 12).

As we

slide contact

A o £

kV

AP

P

A is

= 66 kV

—

—

o to

B

: rheostat

line

C

pri-

66 kV. line

is

(8.9) 6 1()

)/(V3

X 345

000) output

A

voltage

The power associated with winding X,X 2 S

converter

special electric controls. Phase shift-

QMne a

voltage

each phase of the outgoing

=

Such multi-

is

kV

133

secondary voltage rating of 133 in

line.

in large electronic

also used to control

output

=

133

H2

This means that each transformer has an effective

The current

phase an-

shifting en-

mission lines that form part of a power grid.

line

mary

.5)

kV

The voltage of winding X,X 2 between

=

1

us consider only

between X, and

=

345/V3


£, A

=

say).


£AN =

let

shift the

Such phase

systems from an ordinary 3-phase phase systems are used

of a 3-phase line (Fig.

To simplify the calculations,

The

(345/230

ables us to create 2-phase, 6-phase, and 12-phase

sider a rheostat connected

Em =

keeping with the

2.0.

3-phase system enables us to

To understand

12.11b.

The

con-

12.9 Phase-shift principle

A

Three single-phase transformers connected as au-

one phase (phase A,

is

load-carrying

system.

a 3-phase,

stepped up to 345

in

fact that the ratio of transformation

gle of a voltage very simply.

Example 12-4 The voltage of

is

its

part of a 3-phase trans-

It is

former bank used to connect a 700 line to

basic rating of the 3-phase transformer bank

X

22.1

capacity of 200

2.

Figure 12.11c shows a large single-phase auto-

transformer rated 404 kV/173

is

if

outgoing

line voltage to

The

is

smaller and cheaper than a conventional trans-

253

= 66 000 X

335

-

22.

1

Winding H,H 2 has the same power

is

MVA rating.

sic rating

of each single-phase transformer

fore 22.1

MVA.

The is

ba-

there-

Figure 12.12 EAP can be phase -shifted by means of a potentiometer. Voltage

with respect to

EAC

C P


254

from phase B toward phase C, voltage both

amplitude and phase.

in

moving from one end of

shift in

to the other.

changes

a 60° phase

the potentiometer

we move from B

Thus, as

E AP

We obtain

C, voltage

to

EAP gradually advances in phase with respect to £AB At the same time, the magnitude of E AP varies slightly, from E (voltage between the lines) to 0.866 .

E when

the contact

Such circuits

draws

is

in the

middle of the rheostat.

a simple phase-shifter can only be used in

where the load between terminals few milliamperes.

a

If a

IR drop

plied, the resulting

pletely changes the voltage

A

and P

heavier load

ap-

is

in the rheostat

com-

and phase angle from

what they were on open-circuit.

we connect a mulautotransformer between phases B and C 12.13). By moving contact P, we obtain the

To get around titap

(Fig.

this

nals

shifts as be-

but this time they remain essentially un-

changed when a load

A and

P.

Why

is

is

connected between termi-

this

so? The reason

flux in the autotransformer fixed.

mains

As

is

is

that the

E BC

fixed because

(both

in

magnitude

and

phase)

whether the autotransformer delivers a current

3 tapped autotransformers con-

nected between lines A, B, and C. Contacts P,, P 2

P 3 move

in

obtain a

maximum

tandem as we switch from one

set

We now

to source

of 60° as

ABC. We we move

the autotransformers to the

some

practical applications

of the phase-shift principle.

12.10 Three-phase to 2-phase transformation 2-phase system are equal but dis-

in a

placed from each other by 90°. There are several

ways

to create a

source.

One

2-phase system from a 3-phase

of the simplest and cheapest

single-phase autotransformer having taps cent and 86.6 percent.

We

two phases of a 3-phase

line,

If the

A

discuss

shift

,

of

taps to the next. This arrangement enables us to cre-

line

phase

from one extremity of other.

The voltages

shows

,

to

the load or not.

Fig. 12. 14

P h P 2 P 3 whose phase angle

ate a 3-phase source

changes stepwise with respect

is

a result, the voltage across each turn re-

fixed

shifter.

problem,

same open-circuit voltages and phase fore,

Figure 12.14 Three-phase phase

connect as

voltage between lines A,

is

to use a

at

50 per-

between any

it

shown

B,C

is

in Fig.

100

1

2.

1

5.

V, voltages

EAT and £ NC are both equal to 86.6 V. Furthermore, they are displaced from each other by 90°. This relationship can be seen by referring to the phasor dia-

gram 1

.

(Fig.

1

Phasors

2.

1

5c) and reasoning as follows:

EAB £ BC ,

,

and

ECA

are fixed by the

source. 2.

UÂ-^^Â/->^^Â/'•^l

O line

B

1

1

3

4

b

b

Figure 12.13 Autotransformer used as a phase-shifter.

7

n line

Phasor

E AN

cause the

is in

phase with phasor

same ac

£ AB

be-

flux links the turns of the au-

<

totransformer. 3.

Phasor

same

E AT

is

reason.

in

phase with phasor

£ AB

for the


and the other an 86.6 percent tap on

tap

the

255

primary

The transformers are connected as shown in 12.16. The 3-phase source is connected to termi-

winding. Fig.

nals A, B,

C

secondary

and the 2-phase load

windings.

The

ratio

connected to the

is

of transformation

(3-phase line voltage to 2-phase line voltage)

by

EAB /E l2

.

The

is

given

Scott connection has the advantage of

isolating the 3-phase

and 2-phase systems and provid-

ing any desired voltage ratio between them. load

Except for servomotor applications, 2-phase

1

systems are seldom encountered today.

Example 12-5

A 2-phase,

kW (10 hp), 240 V, 60 Hz motor has

7.5

an efficiency of 0.83 and a power factor of 0.80. is

to

It

be fed from a 600 V, 3-phase line using a Scott-

connected transformer bank

(Fig. 12.16c).

Calculate a.

b. c.

The apparent power drawn by the motor The current in each 2-phase line The current in each 3-phase line

Solution a.

-NC

The

power drawn by

active

the

The apparent power drawn by Figure 12.15 a. Simple method to obtain a 2-phase system from a 3-phase line, using a single transformer winding. b. Schematic diagram of the connections. c. Phasor diagram of the voltages.

From KirchhofPs

ECA = Loads

I

E AN + £ NC + phasor £ NC must have

and direction shown

=

S

=

The

ratio

b.

The current

in

1

00/86.6

-

the

fixed and given by

EAH /EAV =

to

is

9036/0.8

VA

295/2

is

= 5648 VA

each 2-phase

line is

c.

The transformer bank tle

itself

consumes very

lit-

active and reactive power; consequently, the

3-phase line supplies only the active and reac-

1.15.

Another way

295

=

motor

f=S/E = 5648/240 = 23.5 A

other,

of transformation (3-phase voltage to is

1

<|>

the

in the figure.

and 2 must be isolated from each

2-phase voltage)

1

5=11

such as the two windings of a 2-phase induction motor.

P/cos

The apparent power per phase

voltage law,

Consequently,

0.

the value

is

P = PJt) = 7500/0.83 = 9036 W

(c)

4.

motor

produce a 2-phase system

Scott connection.

It

consists of

two

is lo

use

identical

single-phase transformers, the one having a 50 percent

power absorbed by the motor. The power furnished by die 3-phase therefore, 295 VA. tive

total ap-

parent

line

1

1

is,


256

86.6%

phase

AO

1

OB

Figure 12.16c

See Example

12-5.

The 3-phase

=

/

line current

5/(V3 E)

=

l

1

is

295/(V3

X

600)

= 10.9A Figure

1

2.

line voltages

12.1

1

1

6c shows the power circuit and the

and currents.

Phase-shift transformer

A phase-shift transformer is a special type of 3-phase autotransformer that shifts the phase angle between

incoming and outgoing

the

lines without

changing

the voltage ratio.

Consider a 3-phase transmission to the terminals (b)

b.

Scott connection.

Phasor diagram

(Fig.

incoming

Figure 12.16 a.

former

of the Scott connection.

1

A, B,

2.

1

7).

C

The transformer

line voltages

however, changing their

1

,

2, 3 are shifted

connected

twists

a magnitude. The

through an angle

that all the voltages of the line

line

of such a phase-shift transall

the

without, result

is

outgoing transmission

with respect to the voltages of

the incoming line A, B, C.

The angle may be

lead-


257

Example 12-6 phase-shift

A phase-shift transformer is designed

transformer

MVA on a 230 kV,

3-phase

±

variable between zero and a.

to control

50

1

The phase angle

line.

is

15°.

Calculate the approximate basic power rating

of the transformer. Calculate the line currents

b.

outgoing transmission Tap changer

incoming and

in the

lines.

Solution a.

Figure 12.17a

The basic power

Phase-shift transformer.

ST

Note

0.025 S L a max

=

56

The

X

0.025

(12.1)

X

150

15

MVA

power

rating

is

much

that the transformer carries.

feature of b.

is

= =

that the

power

rating

less than the

This

a

is

autotransformers.

all

line currents are the

same

both lines, be-

in

cause the voltages are the same. The line current

is

I

Figure 12.17b Phasor diagram showing the range over which the phase angle of the outgoing line can be varied.

SL

=

(150

=

377

Fig. 12,18a

ing or lagging,

and

is

usually variable between zero

and ±20°

The phase angle is sometimes varied in discrete means of a motorized tap-changer. The basic power rating of the transformer

(which determines

power

its

size)

depends upon the ap-

carried by the transmission line, and

upon the phase

shift.

For angles

less than 20°,

it

is

given by the approximate formula

ST

=

0.025 S L

a max

leg.

PN

=

basic

power

rating of the 3-phase

000)

A

an example of a 3-phase transformer

with a tap brought out

shift of, say,

a, 3.

S,

0.025

=

The incoming

line

and the outgoing result

Similarly,

is

£2 n ECN

l

is

connected

to terminals

line to terminals

1,

A, B,

a£ s

C

£, N lags 20° behind £ AN 20° behind E HN and £ 1N lags

that

.

,

(Fig. 12.18c).

principle of obtaining a phase shift

connect two voltages

in

E ]b

apparent power carried by the trans-

mission line [VA]

generated by phase A. The values of E,> N and

B

is

is

to

series that are generated

ated by phase

an approximate coefficient

A and

2, 3.

by two different phases. Thus, voltage

shift |°]

has one

The wind-

transformer bank VA]

a max = maximum transformer phase

20

ings of the three phases are interconnected as shown.

[

=

A

terminal

at

a second winding having terminals

The basic Sv

X 230

The transformer has two windings on each

20° behind

where

(8.9) h

10 )/(V3

Thus, the leg associated with phase

winding

The (I2.l)

is

X

could be used to obtain a phase

degrees.

steps by

parent

that

N3E

=

connected

in series

selected so that the output voltage

is

gener-

E PN E ]b are

with

equal to the in-

put voltage while obtaining the desired phase angle

_\-SS


coming terminals are

A, B, C; the outgoing terminals

Figure 12.18c Phasor diagram

are 1,2,3.

a transformer that gives a phase-

of

shift of 20°.

- E

EPN

=

\.\4E

= 0.40E

In practice, the internal circuit

phase-shift transformer

However,

it

rests

upon the

The purpose of such transformers

just discussed. will

of a tap-changing,

much more complex. basic principles we have

is

be covered

in

Chapter 25.

12.12 Calculations involving

3-phase transformers The behavior of a 3-phase transformer bank is calculated the same way as for a single-phase transformer. In making the calculations, we proceed as Figure 12.18b Schematic diagram

follows: of the transformer in Fig. 12.18a. 1.

We

assume

that the

primary and secondary

windings are both connected between them.

In

our particular example,

line-to-neutral voltage of the

incoming

if

E is

the

line, the re-

spective voltages across the windings of phase

A are

in

wye, even

if

they are not (see Section 8.14). This eliminates the

problem of having

to deal with delta-wye

and delta-delta voltages and currents.


2.

We

consider only one transformer (single phase)

of this assumed 3.

The primary voltage of former

is

coming 4.

wye-wye transformer

259

Z T (pu)

MP")

= tO.115

bank.

this hypothetical trans-

the line-to-neutral voltage of the in-

line.

The secondary voltage of

this

transformer

is

the

line-to-neutral voltage of the outgoing line. 5.

The nominal power

rating of this transformer

is

one-third the rating of the 3-phase transformer

bank. 6.

Figure 12.19

The load on

transformer

this

is

See Example

one-third the

12-7.

load on the transformer bank.

This

Example 12-7 The 3-phase step-up transformer shown 10.18 (Chapter 10)

is

rated 1300

impedance

kV, 60 Hz,

MVA,

a.

kV

11.5 percent.

Fig.

in

It

steps

The

power

a

S,

power

810

therefore, j

0.115

The voltage

=

81 0/3

= 270

1

2.

9.

1

MVA

£, across the load

is

ter-

HV side of the transformer MVA at 370 kV with a lagging

minals when the delivers

is,

The equivalent circuit is shown in Fig. b. The power of the load per phase is

circuit of this trans-

Calculate the voltage across the generator

almost entirely reactive.

is

impedance

ZT (pu) =

former, per phase. b.

per-unit

up

line.

Determine the equivalent

a very large transformer; consequently, the

24.5 kV/345

the voltage of a generating station to

345

is

transformer impedance

£,

The

= 370 kV/V3 = 213.6W

per-unit

power of

the load

is

factor of 0.90.

MVA =

5 L (pu) = 270 MVA/433.3

0.6231

Solution a.

First,

we

By

note that the primary and second-

ary winding connections are not specified.

We

don't need this information. However,

we assume in

selecting £, as the reference phasor, the

per-unit voltage across the load

that both

£ L (pu) =


We this

shall use the per-unit

We

problem.

the secondary

select the

method

EB =

The

to solve

£B

/.(pu) L F

=

199.2

kV

The power /u

Ratio of transformation

a

=

per-unit current in the load

=

is

345/24.5

=

5,(pu) 11

£ L (pu)

is

345/V3

=

is

0.6231

£L

=

0.581

1300/3

=

I

1.0723 is

0.9.

Consequently.

by an angle of arcos 0.90

=

25.84°.

Consequently, the amplitude and phase of the

14.08

is

given by

rating of the transformer

will be used as the base

SB

=

factor of the load

lags behind

per-unit load current

The nominal power

kV

0

1.0723Z.0

nominal voltage of

winding as our base voltage,

The base voltage

213.6 kV/1 99.2

=

wye.

is

power 5 B Thus,

/,

.

433.3

MVA

The

(pu)

-

per-unit voltage

0.581 1Z. -25.84°

£

s

(Fig. 12.19)

is


260

E

s

(pu)

= EJpu) +

/ L (pu)

X ZT (pu)

Hi

= 1.0723^0° + (0.5811/1 -25.84°) X (0.1 15/190°)

=

1.0723

+

0.0668Z.64.16

=

1.0723

+

0.0668(cos 64.16°

j

X1

0

+

sin 64.16°)

=

1.1014

=

1.103/13. 12°

+

j

0.0601

Therefore,

E = s

The

1.103

X 345 kV =

per-unit voltage on the primary side

Ep = The

381 kVZ.3.12°

1.

is

also

103/13. 12°

effective voltage across the terminals of the

generator

is,

therefore,

£g = = =

£p(pu)

X £B

1.103

X

27.02

kV

(

primary)

24.5

kV

Figure 12.20 Polarity

12.13 Polarity marking

marking

of

3-phase transformers.

of 3-phase transformers The HV terminals of a 3-phase transformer are marked H h H 2 H 3 and the LV terminals are marked X h X 2 X 3 The following rules have been stan-

ondary

dardized:

are

,

.

,

1

.

If the

wye-wye

terminals on the

or delta-delta, the

voltages between similarly-marked terminals are in phase. Thus, is

in

phase with

E xx

£ HiH|

is

in

phase with

Exx

EH

is in

phase with

Ex

£H

H

H

x

If the

,

HV

side

LV

side.

Thus,

E HH

leads

£ X|Xi

by 30°

EH

H|

leads

£ XiX|

by 30°

£H

Hi

leads

Ex

30° x by

Fig. 12.20

shows two ways of representing

the

delta-wye terminal markings.

primary and secondary windings are con-

nected

internal connections

and so on.

and so on. 2.

The

so that the voltages on the

always lead the voltages of similarly-marked

primary windings and secondary wind-

ings are connected

line voltages.

made

in

wye-delta or delta-wye, there results

a 30° phase shift

between the primary and sec-

3.

These rules are not affected by the phase sequence of the line voltage applied to the primary

side.



1

2-7

In order to

261

meet an emergency, three

single-phase transformers rated Practical level 1

2-

1

that the transformer terminals

H2

have polarity marks H,,

X,,

,

X2

,

make

wye-delta on a 3-phase 18

a.

What

Delta-wye

b.

Open-delta

1

2-2

250 kVA, 7200 V/600 in

V,

at

load

450 kVA,

is

In the

a.

12-9

incoming and outgoing transmis-

In the

of 36

MVA,

in Fig. 12.9

13.8

nominal currents ondary 12-4

kV/320 in the

has a rating

V Calculate the

lines.

in Fig.

mode

ates in the forced-air

225

current

during the

600

to

is

1,

V.

kV

6.9

2,

and 3

Then,

in

a

P

are

by mistake con-

Determine the voltages measured between

lines 1-2, 2-3,

Draw

the

and

3-1.

new phasor diagram.

Three

1

50 kVA, 480 V/4000

V,

60 Hz

sin-

is

if

the primary line volt-

kV and

12-11

when

The

exciting

the transformers are

The core

loss in a

300

kVA

distribution transformer

the primary line

is

3-phase

estimated to

be 0.003 pu. The copper losses are

150 A.

in

line.

operating at no-load.

0.0015 pu.

the transformer overloaded?

If

the time, and the cost of electricity

4.5 cents per

kWh,

is

calculate the cost of

the no-load operation in the course of

connected?

Calculate the line currents for a 600 load.

the transformer operates

effectively at no-load 50 percent of

Problem 12-2 are

V line to 7.2 kV. How must they be

kVA

load the bank

current has a value of 0.02 pu. Calculate

600

c.

balanced and equal

on a 4000 V, 3-phase

10.19 oper-

used to raise the voltage of a 3-phase

b.

maximum

lines

the line current

The transformers

a.

the


Calculate the currents in the sec-

ondary lines

12-6

is

A-B-C

b.

morning peaks.

Is

What

gle-phase transformers are to be installed

The transformer shown

b.

400 kVA.

and the voltage between

a.

Intermediate level

is

250 kVA,

the transformers overloaded?

ings of transformer

12-10

age

at

are connected in open-

Referring to Figs. 12.3 and 12.4, the line

Calculate the nominal currents in the pri-

a.

V

nected in reverse.

mary and secondary windings of the transformer shown in Fig. 10.18, knowing that

12-5

transformer bank?

to the

similar installation the secondary wind-

primary and sec-

the windings are connected in delta-wye.

line.

the outgoing line voltage?

transformers rated

Are

kV

load that can be

voltage between phases

primary and secondary windings

The transformer

maximum

can carry on a continuous basis?

is

12-3

is

kV/600

b.

sion lines b.

Two

calculate the

following currents:

the

delta to supply a load of

60 Hz, are con-

wye-delta on a 12 470 V, 3-phase

line. If the

a.

2-8

2.4

Three single-phase transformers rated nected

What

b.

a.

is

connected

nections:

are connected

in

schematic drawings of the following con-

1

kV

100 kVA, 13.2 kV/2.4

Assuming

12-12

one

year.

The

bulletin of a transformer manufacturer

kVA, 230 V/208

Calculate the corresponding primary

indicates that a 150

and secondary currents.

60 Hz, 3-phase autotransformer weighs

V,

262


310

lb,

whereas a standard 3-phase trans-

12-14

former having the same rating weighs

1220 12-13

1b.

Why

1

kVA, 480 V/l 20

5

in delta to

formers on a 600

H2 X h X2 ,

V,

source.

60 Hz are con-

12-15

Then

can be drawn from the 600

calculate the

maximum

that the autotransformer can carry.

You wish

to operate a

40

hp,

460

V,

3-phase motor from a 600 V, 3-phase sup-

V

ply. The full-load current of the motor 42 A. Three 5 kVA, 20 V/480 V,

3-phase

line.

The H,,

marks appear on the

is

1

single-phase transformers are available.

the transformers should be

How would

you connect them? Are they

connected.

able to furnish the load current

b.

Determine the 3-phase voltage output

the

c.

Determine the phase

of the transformer. shift

between the

3-phase voltage output and the 600 V, 3-phase input.

V

load

function as autotrans-

polarity

Show how

(kVA)

at

metal housing. a.

Problem 12-13 calculate the maximum

line current that

this difference 9

Three single-phase transformers rated nected

In

motor without overheating?

drawn by

Chapter 13 Three-Phase Induction Motors

that ranges

13.0 Introduction Three-phase

induction

motors are the motors

most frequently encountered

and easy

are simple, rugged, low-priced,

They run

tain.

to

constant speed from

The speed

frequency-dependent

consequently,

these

is

motors

are

not

supports

main-

at essentially

zero to full-load. and,

They

in industry.

to control the

the 3-phase induction

its

We

the basic principles of

its

behavior.

general construction and the

We way

use two types of rotor windings: (1) conven-

3-phase windings made of insulated wire and

then

cage induction motors (also called cage motors)

the

and wound-rotor induction motors.

A squirrel-cage

is

composed of bare cop-

into the slots. The opposite ends are welded two copper end-rings, so that all the bars are short-circuited together. The entire construction

pushed

thousand horsepower permit the reader to see that

same

rotor

per bars, slightly longer than the rotor, which are

motors ranging from a few horsepower to several operate on the

The type of winding

gives rise to two main classes of motors: squirrel-

Squirrel-cage, wound-rotor, and linear induction

all

circumference of the lam-

(2) squirrel-cage windings.

windings are made.

they

A number of evenly spaced slots,

internal

of rotor slots to provide space for the rotor winding.

motor and develop the funda-

mental equations describing discuss

that

tions.

speed of commercial induction

we cover

frame

made up of

The rotor is also composed of punched laminaThese are carefully stacked to create a series

fre-

tional

chapter

steel

core

inations, provide the space for the stator winding.

motors. In this

hollow, cylindrical

stacked laminations.

quency electronic drives are being used more and

more

a

punched out of the

easily

adapted to speed control. However, variable

mm to 4 mm, depending on the

from 0.4

power of the motor. The stator (Fig. 13.2) consists of a

to

basic principles.

(bars and end-rings) resembles a squirrel cage,

from which the name

Principal

13.1

A 3-phase

components

induction motor (Fig.

parts: a stationary stator

rotor

is

and

1

3.

1

)

small and

in-

aluminum, molded

The

show progressive

13.3a). Figs.

to

form an

13.3b and 13.3c

stages in the manufacture of a

squirrel-cage motor.

263

In

are

die-cast

tegral block (Fig.

a revolving rotor.

derived.

made of

has two main

separated from the stator by a small air gap

is

medium-size motors, the bars and end-rings


264

A wound

rotor has a 3-phase winding, similar

the one on the stator.

The winding

tributed in the slots and

is

is

uniformly

usually connected in 3-

wire wye. The terminals are connected to three rings,

which

revolving

turn with the rotor (Fig.

slip-rings

and

13.4).

associated

slip-

The

stationary

brushes enable us to connect external resistors ries

to

dis-

in se-

with the rotor winding. The external resistors are

mainly used during the start-up period; under normal running conditions, the three brushes are short-circuited.

13.2 Principle of operation

Figure 13.1 Super-E, premium efficiency induction motor rated 10 hp, 1760 r/min, 460

V,

3-phase, 60 Hz. This

to-

tally-enclosed fan-cooled motor has a full-load current of tor of

1

2.7 A, efficiency of 91 .7%, and

81%. Other

power

fac-

characteristics: no-load current:

5 A; lockedrotor current: 85 A; locked rotor torque:

breakdown torque: 3.3 pu; service factor 90 kg; over-all length including shaft: 491 mm; overall height: 279 mm. (Courtesy of Baldor Electric Company) 2.2 pu; 1

.15; total weight:

The operation of a 3-phase induction motor is based upon the application of Faraday's Law and the Lorentz force on a conductor (Sections 2.20, 2.2

1,

and 2.22). The behavior can readily be understood

by means of

the following example.

Consider a series of conductors of length

/,

whose extremities are short-circuited by two bars A and B (Fig. 3.5a). A permanent magnet placed above this conducting ladder, moves rapidly to the 1

right at a

speed

r,

so that

across the conductors.

its

magnetic

field

B sweeps

The following sequence

of

events then takes place:

Figure 13.2

Exploded view of the cage motor of Fig. 13.1, showing the stator, rotor, end-bells, cooling and terminal box. The fan blows air over the stator frame, which is ribbed to improve heat (Courtesy of Baldor Electric Company)

fan, ball bearings,

transfer.

THREE-PHASE INDUCTION MOTORS

moving magnet field

is

replaced by a rotating

windings, as

in the stator

13.3

is

produced by the 3-phase currents

The

we

now

will

265

field.

The flow

that

explain.

rotating field

Consider a simple stator having 6 salient poles, each of which carries a coil having 5 turns (Fig.

1

3.6).

Coils that are diametrically opposite are connected in series

by means of three jumpers

connect terminals

a-a, b-b,

and

that respectively

c-c.

This creates

AN, BN, CN,

that

are mechanically spaced at 120° to each other.

The

three identical sets of windings

Figure 13.3a Die-cast aluminum squirrel-cage rotor with integral cooling fan.

(Courtesy of Lab-Volt)

1

A

.

E—

voltage

while

it

Blv

is

induced

in

each conductor

being cut by the flux (Faraday's law).

is

The induced voltage immediately produces a /, which flows down the conductor un-

2.

current

derneath the pole-face, through the end-bars,

and back through the other conductors.

Because the current-carrying conductor

3.

the magnetic field of the

lies in

permanent magnet,

it

experiences a mechanical force (Lorentz force). 4.

The force always

acts in a direction to drag the

conductor along with the magnetic

field

(Section 2.23). If the

celerate

move,

conducting ladder

is

toward the

However, as

right.

free to

it

it

will ac-

picks up

speed, the conductors will be cut less rapidly by the

moving magnet, with age

E and

the current

the force acting If

the ladder

magnetic

the result that the induced volt/

will diminish. Consequently,

on the conductors

were

field, the

to

move

at the

will also decrease.

same speed

and the force dragging the ladder along would

come

as the

induced voltage E, the current all

/,

be-

zero.

In an itself to

Figure 13.3b

(1),

induction motor the ladder

is

form

I

a squirrel-cage (Fig.

closed upon

3.5b) and the

in the manufacture of stator and Sheet steel is sheared to size punched (3), blanked (4), and

Progressive steps rotor laminations.

blanked

punched

(2),

(5).



266

upper

Figure 13.3c Progressive steps a.

in

Molten aluminum

the injection molding of a squirrel-cage is

poured

into

a cylindrical

cavity.

rotor.

The laminated

rotor stacking

is

firmly held

between

two molds. b.

Compressed

rams the mold assembly into the cavity. Molten aluminum is forced upward through the and into the upper mold. Compressed air withdraws the mold assembly, now completely filled with hot (but hardened) aluminum. The upper and lower molds are pulled away, revealing the die-cast rotor. The cross section view shows that the upper and lower end-rings are joined by the rotor bars. (Lab-Volt) air

rotor bar holes

c.

d.

two

coils in

each winding produce magnetomotive

forces that act in the

The

same

three sets of windings are connected in

thus forming a perfectly

common

symmetrical

always flow

neutral N.

Owing

arrangement,

the

wye,

to the

line-to-

bers,

suppose that the peak current per phase

Thus,

when

/a

= +7

pere-turns

balanced 3-phase system.

Because the current

ings.

The

displaced

/.„

/h ,

and

/c will

currents will have the in

flow

In

sume

B, C,

flux.

It is

this flux

we

same value but

in turn,

are interested

will

be

create a

we

corresponding

a

as-

by the arrows)

A will

is

1

0 turns

= 70 amof

value

positive, the flux

is

flux.

directed

upward, according to the right-hand

As time goes

by,

we

rule.

can determine the instanta-

neous value and direction of the current

each

in

winding and thereby establish the successive flux patterns. Thus, referring to Fig. rent /a has a value of

in.

order to follow the sequence of events, that positive currents (indicated

vertically

the wind-

time by an angle of 120°. These currents

produce magnetomotive forces which, magnetic

in

and

10 A.

is

A, the two coils of phase

mmf of 7 A X

together produce an

we connect a 3-phase source to terminals A,

to

Furthermore, to enable us to work with num-

neutral impedances are identical. In other words, as

If

line to neutral.

line.

regards terminals A, B, C, the windings constitute a

alternating currents

windings from

in the

Conversely, negative currents flow from neutral

direction.

A X

1

0 turns

1

3.7 at instant

10 A, whereas

fb

and

1

/c

,

cur-

both

—5 A. The mmf of phase A is = 00 ampere-turns, while the mmf

have a value of 10

+

1

Figure 13.4b of the slip-ring end of the rotor. (Courtesy of Brook Crompton Parkinson Ltd)

Close-up

267


268

length

/

Figure 13.5a Moving magnet cutting across a conducting ladder.

ength

Figure 13.5b Ladder bent upon

of phases

B and C

rection of the

B

/

form a squirrel-cage.

itself to

are each 50 ampere-turns.

mmf depends

upon

current flows and, using the right-hand rule, that the direction of the resulting

shown

in Fig. I3.8a.

Note

The

di-

the instantaneous

magnetic

we

find is

as

that as far as the rotor

is

field

concerned, the six salient poles together produce a

magnetic

having essentially one broad north

field

pole and one broad south pole. This

means

that the

6-pole stator actually produces a 2-pole field.

combined magnetic At instant tains a

2,

peak of

field points

one-sixth cycle

—

10 A, while

/a

The

upward. later,

and

current

/b

/ L at,

both have a

value of

+5 A

new

has the same shape as before, except that

it

field

has

(Fig.

1

3.8b).

moved clockwise by an

We

discover that the

angle of 60°. In other

words, the flux makes 1/6 of a turn between instants l

and

cycle,

we find that the magnetic

plete turn during

field

one cycle (see Figs.

makes one com1

3.8a to I3.8f).

The rotational speed of the field depends, therefore, upon the duration of one cycle, which in turn depends on the frequency of the source.

2.

Proceeding

Figure 13.6 Elementary stator having terminals A, B, C connected to a 3-phase source (not shown). Currents flowing from line to neutral are considered to be positive.

in this

instants 3, 4, 5, 6,

and

way 7,

for each of the successive

quency

is

60 Hz, the resulting

separated by intervals of 1/6

in 1/60

s,

that

is,

field

If the fre-

makes one

turn

3600 revolutions per minute. On

Figure 13.7 Instantaneous values of currents and position of the flux

269

in Fig.

13.6.


270

A

A

g

O I'-

N

N

Figure 13.8d

Figure 13.8c

Flux pattern at instant

Flux pattern at instant 3. A

4.

A

g

N

Figure 13.8e

Figure 13.8f

Flux pattern at instant 5.

Flux pattern at instant

the other hand, if the frequency were 5 Hz, the field would make one turn in 1/5 s, giving a speed of only 300 r/min. Because the speed of the rotating field is

change any two of the the

new phase sequence

necessarily synchronized with the frequency of the

the

same

we

find that the field

source,

it

is

called synchronous speed.

produces a

speed

field that rotates clockwise. If

line

positive crests of the currents in Fig. in the

we

connected to the

will be

interstator,

A-C-B. By following

of reasoning developed in Section

in the opposite,

therefore, reverse

The

lines

now

revolves

at

1

3.7 follow

order A-B-C. This phase sequence

its

3.3,

or counterclockwise direction. will,

direction of rotation.

Although early machines were poles, the stators of

1

synchronous

Interchanging any two lines of a 3-phase motor

13.4 Direction of rotation

each other

6.

built with salient

modern motors have

internal di-


ameters that are smooth. Thus, the salient-pole stator of Fig. 13.6 as

shown

is

1

3.6, the

are replaced

Note

that

two

each

coils of phase

A (Aa and An)

slots

coil

in Fig.

coil pitch is

covers

more

duces more flux per turn, terminal

shown

A to

the neutral

coils of phases

1

80° of the circum-

A

current

only 60°.

because

efficient

/a

pro-

it

flowing from

N yields the flux distribution are identical to those Fig.

120° to each other.

The

netic field

at

due

1

3.9b, they are

resulting

mag-

to all three phases again consists of

In practice, instead

shown

in Fig.

two, three or

more

of using a single coil per pole

13.9a, the coils

coil

lodged

in

is

subdivided into

adjacent slots.

The

staggered coils are connected in series and constitute

what

known as a phase group. Spreading the coil way over two or more slots tends to create a

is

in this

sinusoidal flux distribution per pole, the

in 5 suc-

in Fig. 13.20.

of poles

Soon after the invention of the induction motor, it was found that the speed of the revolving flux could be reduced by increasing the number of poles. To construct a 4-pole stator, the coils are distributed as shown in Fig. I3.l0a. The four identical

A now span only 90° of the stator cir-

cumference. The groups are connected in

such a

way

that adjacent

tomotive forces acting

when

other words,

winding of phase

two poles. as

be placed

in series to

synchronous speed

groups of phase

B and C

of phase A and, as can be seen in

displaced

Number

13.5

13.9a.

in the figure.

The

shown

is

on the inner surface of the

ference whereas the coils in Fig. 13.6 cover

The 180°

gered coils connected cessive slots

and 13.24a.

two

by the two coils shown

are lodged in

stator.

replaced by a smooth stator such

in Figs. 13.2

In Fig.

They

now

271

which improves

performance of the motor and makes

it

less noisy.

A phase group (or simply group) composed of 5

stag-

A

in

in series

and

groups produce magneopposite directions. In

a current

flows

/,,

(Fig. 13. 10a),

it

in the stator

creates four al-

N-S poles. The windings of the other two phases

ternate

cal but are displaced

are identi-

from each other (and from

When

phase A) by a mechanical angle of 60°.

wye-connected windings are connected

source, a revolving field having four poles (Fig.

13. 10b).

This field rotates

speed of the 2-pole will shortly explain

field

why

shown

at

the

to a 3-phase is

created

only half the

in Fig.

13.9b.

We

this is so.

group

phase

Figure 13.9a

Phase group 1 is composed of a single coil lodged two slots. Phase group 2 is identical to Phase group 1 The two coils are connected in series. In practice, a phase group usually consists of two or more staggered coils. in

.

(/ c

1

C

= - 5 A)

Figure 13.9b Two-pole,

magnetic

A and

/

b

full-pitch,

=

/c

lap-wound stator and resulting

when the = -5 A.

field

current

in

phase A = +10


272

phase group

group

1

group

1

phase B

1

Figure 13.10a

The

phase groups

four

pole magnetic

group

of

phase A produce a

4-

field.

1

group

1

-5

A.

rent flow in the three phases, let us restrict our at-

tention to phase A. In Fig. 13.11 each phase group

covers

a

mechanical angle of 360/8

=

45°.

Suppose the current in phase A is at its maximum positive value. The magnetic flux is then centered on phase A, and the N-S poles are located as

shown

in Fig.

13.12a. One-half cycle later, the

current in phase tive value.

Figure 13.10b

fore,

Four-pole, full-pitch, lap-wound stator

magnetic

field

when

/a

= +10 A and

and

/b

=

The

A

will reach

its

maximum

except that

all

N

the

poles will

resulting lc

= -5

A.

S poles and vice versa (Fig. 13.12b). In comparing the

two

figures,

it is

netic field has shifted

clear that the entire

by an angle of 45°

this gives us the clue to finding the

We we

can increase the number of poles as

much

as

please provided there are enough slots. Thus,

Fig.

13. II

shows a 3-phase, 8-pole

stator.

phase consists of 8 groups, and the groups of phases together produce an 8-pole rotating

When

connected to a 60

like the

Hz

spokes of a wheel,

Each all

the

field.

source, the poles turn,

at

a synchronous speed

of 900r/min.

How will

can

we

tell

what the synchronous speed

be? Without going into

all

nega-

same as behave become

flux pattern will be the

the details of cur-

tion.

cles

mag-

— and

speed of rota-

The flux moves 45° and so it takes 8 half-cy(= 4 cycles) to make a complete turn. On a

60 Hz system the time fore 4

X

1/60

=

1/15

to s.

make one

turn

is

there-

Consequently, the flux

r/s or 900 r/min. The speed of a rotating field depends therefore upon the frequency of the source and the number of poles on the stator. Using the same reasoning as

turns at the rate of 15

above,

we

can prove that the synchronous speed

always given by the expression

is


phase group

phase group

1

maximum

1

Figure 13.12b

Figure 13.12a Flux pattern

273

when

the current

in

phase A

is

at

when

Flux pattern

its

maximum

positive value.

in Fig.

the current

negative value.

13.12a but

it

The

phase A

in

pattern

is

is

the

at

its

same as

has advanced by one pole

pitch.

ing field created by the stator cuts across the rotor bars and induces a voltage in

P

This

where

is

cut, in rapid succession,

= synchronous

ns

/=

speed fr/min]

the

creases with frequency and decreases with the

N

is

pole followed by a

number of

N

and S poles

that

sweep across

always equal

is at rest,

it

a is

frequency of the source.

to the

in-

Because the rotor bars are short-circuited by the

num-

end-rings, the induced voltage causes a large cur-

ber of poles.

rent to flow

— usually

bar in machines of

Example 13-1 duction motor having 20 poles

when

it is

connected

several hundred amperes per

medium power.

The current-carrying conductors

Calculate the synchronous speed of a 3-phase in-

Hz

by a

conductor per second; when the rotor

This equation shows that the synchronous speed

a 50

of them.

S pole. The frequency of the voltage depends upon

frequency of the source [Hz]

p = number of poles

to

all

an ac voltage because each conductor

the flux created

by the

stator,

are in the path of

consequently, they

all

experience a strong mechanical force. These forces source.

tend to drag the rotor along with the revolving Solution

In

ns

= 120///7 = 120 X = 300 r/min

50/20 1

.

field.

summary:

A

revolving magnetic field

3-phase voltage

is

is

set

up when

a

applied to the stator of an

induction motor.

13.6 Starting characteristics

of a squirrel-cage

2.

motor

Let us connect the stator of an induction

motor

The revolving

field

induces a voltage

in the ro-

tor bars. to a

3-phase source, with the rotor locked. The revolv-

3.

The induced voltage rents

which flow

creates large circulating cur-

in the rotor bars

and

end-rinizs.


274

The current-carrying

4.

magnetic

the

5.

rotor bars are

by the

field created

immersed

in

they are

stator;

equal to the load torque.

therefore subjected to a strong mechanical force.

constant

The sum of the mechanical forces on

motor only turns

which tends

tor bars produces a torque

the rotor along in the

volving

same

the ro-

all

to drag

direction as the re-

at

—slip

load.

The moment this state of equilibrium

as the rotor

released,

is

it

rapidly acceler-

ates in the direction of the rotating field.

up speed,

As

picks

it

the relative velocity of the field with re-

spect to the rotor diminishes progressively. This

causes both the value and the frequency of the

duced voltage

to

decrease because the rotor bars are

more slowly. The

cut

first,

in-

will continue to increase, but

never catch up with the revolving

it

rotor bars

field. In effect, if

same speed as the field (synflux would no longer cut the

and the induced voltage and current

fall to

to

produce a current

overcome

tor bars sufficiently large to

usually less than

machines That

(

motors

field (called slip),

0.1% of synchronous

in

speed

is

small:

motor

will begin to slow

is initially

down and

at

greater motor torque.

running

at no-load. If

the revolving field will rate.

The

in-

at synchronous speed, they sometimes called asynchronous machines.

never actually turn

and

13.9 Slip The

state

are

speed

slip

slip s

of an induction motor

is

the difference be-

tween the synchronous speed and the rotor speed, expressed as a percent (or per-unit) of synchronous speed.

The

per-unit slip

= = —

s

ns

n

The


is

motor comes

slip

synchronous speed [r/minl rotor speed fr/minj

zero

slip is practically

equal to

l

100%) when

(or

no-load and

at

the rotor

is

is

locked.

.

motor

hp, 6-pole induction

phase, 60

Hz

source.

excited by a 3-

is

the full-load speed

If

is

1

140

r/min, calculate the slip.

Solution

The synchronous speed of ns

producing a greater and

The question

is,

for

how

long

can this go on? Will the speed continue to drop un-

the

seldom exceeds 5%.

the

motor

is

resulting current in the bars

will increase progressively,

No;

it

constant speed machines. However, because they

A 0.5

a higher and higher

duced voltage and the

the

and more) rarely

speed.

apply a mechanical load to the shaft, the motor

cut the rotor bars

til

).

induction motors are considered to be

Example 13-2

the

kW

000

(1

kW and less),

10

why

is

the ro-

in

13.8 Motor under load we

1

motors run very

loads, induction

exceeds 0.5% of synchronous speed, and for small

the braking

At no-load the percent difference

between the rotor and

Suppose

1

zero.

chronous speed so as

torque.

upset, the

is

3.

close to synchronous speed. Thus, at full-load, the

Under these conditions the force acting on the rotor bars would also become zero and the friction and windage would immediately cause the rotor to slow down. The rotor speed is always slightly less than synwould

is

will

the rotor did turn at the

chronous speed), the

torque

its

rotor current, very large at

decreases rapidly as the motor picks up speed.

The speed

change (Section

will start to

slip for large

As soon

when

constant speed

exactly equal to the torque exerted by the mechanical

Under normal

13.7 Acceleration of the rotor

a

at

very important to understand that a

rate. It is

motor speed

field.

When this state is reached, the

speed will cease to drop and the motor will turn

The

= =

\20flp

=

120

X

60/6

(13.1)

1200 r/min

difference between the synchronous speed of

to a halt?

motor and the mechanical load

the revolving flux and rotor speed

is

the slip speed:

will reach a

of equilibrium when the motor torque

is

exactly

ns

-

n

=

1200

-

1140

=

60 r/min


The

slip is

c.

s

=

(« s

=

0.05 or

-

500 r/min

at

60/1200

(13.2) d.

5%

Motor turning

2000 r/min

at

From Example

The voltage and frequency induced in the rotor both slip. They are given by the follow-

depend upon the

a.

At

K-

=

s

(13.3)

sf

(approx.)

=

motor speed n

0.

slip is

=

n)/n s

-

(1200

0)/1200

the induced current)

h in b.

When

=

=

4'

X 60 = 60 Hz

1

motor turns

the

same

in the

motor speed n

field, the

frequency of the source connected to s

the stator [Hz]

=

E2 = Eoc =

slip

at rest

would be induced bars were disconnected from

the

£oc

is

c.

When

= tf=

the

=

direction as the

positive.

-

(1200

the end-rings. In

n

in the

motor speed

= —500. The

-

(-500)1/1200

should be noted that Eq. 13.3 always holds

(1200

+

500)/ 1200

but Eq. 13.4

=

1.417

valid only

if

the revolving flux

(expressed in webers) remains absolutely constant.

A slip

However, between zero and full-load the actual value of

E2

is

Hz

negative; thus,

is

[1200

—

35

slip is

=

is

1

opposite direction

=

/V 3 times the voltage between the open-

circuit slip-rings. It

500)/ 200

X 60 =

0.583

(n s

1

slip is

is

—

is

The

0.583

motor turns

to the field, the

s

voltage

true,

=

n)/n s

in the rotor bars

wound-rotor motor the open-circuit

the case of a

700/1200

fi

the voltage that if

=

[V]

cage motor, the open-circuit voltage

In a

(« s

the rotor current)

open-circuit voltage induced in the ro-

when

-

=

is

The frequency of the induced voltage (and of

voltage induced in the rotor at slip s

tor

1

is

the rotor [Hz]

s

=

The frequency of the induced voltage (and of

(13.4)

frequency of the voltage and current

synchronous speed of the

13-2, the

standstill the

where

/=

direc-

1200 r/min.

is

Consequently, the

ing equations:

f2 =

same

Solution

motor

E 2 = sEoc

in the

tion as the revolving field

13.10 Voltage and frequency induced in the rotor

f2 =

opposite

in the

direction to the revolving field

=

n)/n,

Motor turning

215

/?)//? s

greater than

=

1700/1200

implies that the motor

1

is

operating as a brake.

only slightly less than the value given

The frequency of the induced voltage and

by the equation.

current

f2 =

Example 13-3

rotor

is

f=

s

X 60 =

1.417

85 Hz

The 6-pole wound-rotor induction motor of Example 13-2 the

is

excited by a 3-phase 60

Hz

d.

source. Calculate

frequency of the rotor current under the follow-

ing conditions:

The motor speed turns in the same n = +2000. The s

a.

At

b.

Motor turning

500 r/min

tion as the revolving field

in the

same

direc-

positive because the rotor

direction as the field: slip is

=

(n s

=

(1200

standstill at

is

-

n)ln s

-

2000)/ 1200

= -800/1200 = -0.667


276

A negative

slip implies that the

motor

actually

is

1.

operating as a generator.

The frequency of current

the induced voltage

means

negative frequency

reversed. Thus,

is

in the rotor

wind-

stator

quency

is

mutual flux

is

can say that the frequency

is

a transformer (Fig.

1

lists

I

kW

simply 40 Hz.

no-load

therefore low;

is

machines

efficiency

that the current

rents are is

the full-load current

is

compared

to

it.

is

to

it.

and

all

Finally, the base

it

to 0.05 for large

the

is

is

zero.

under load,

mmf which

tends

m This sets up an opposthe stator. The opposing mmfs of

change the mutual flux

4>

.

in a transformer.

are created, in

The total m power needed to produce these three fluxes is slightly greater than when the motor is operating at no-load. However, the active power (kW) absorbed addition to the mutual flux

(Fig. 13.14).

reactive

the syn-

is

to create

within ac-

machines. The

motor

mmfs of the secondary and primary As a result, leakage fluxes n and

other torques are

speed

it

the rotor and stator are very similar to the opposing

other cur-

all

3).

needed

zero because the output power

ing current flow in

and

Similarly, the base torque

the full-load torque and

compared

to

The base

torque are expressed in per-unit values. current

1

links both the

similar to the

is

Motor under load. When

2.

power range between

in the

and 20 000 kW. Note

m

chanical tolerances will permit.

the typical properties of squirrel-

cage induction motors

3.

is

the current in the rotor produces a

Table I3A

it is

made as short as meThe power factor at ranges from 0.2 (or 20%)

ceptable limits, the air gap

we

Characteristics of squirrelcage induction motors

1

in

flux

the revolving field and, in order to keep

for small

13.1

The

consequently

rotor;

Considerable reactive power

concerned, a negative frequency gives the same reading as a positive frequency. Consequently,

and the

that sup-

friction losses in the rotor plus

the iron losses in the stator.

the phase sequence

itive,

windage and

plies the

A-B-C when the frequency is posis A-C-B when the frenegative. As far as a frequency meter is

rotor voltages

component

ing flux 3> m and a small active

the phase sequence of the

if

(of

of a magnetizing component that creates the revolv-

phase se-

that the

the motor runs at no

between 0.5 and 0.3 pu

full-load current).

is

quence of the voltages induced ings

When

at no-load.

The no-load current is similar to the exciting current in a transformer. Thus, it is composed

and rotor

f2 = sf= -0.667 X 60 = -40 Hz

A

Motor

load, the stator current lies

chronous speed of the motor. The following explanations will clarify the meaning of the values given

by the motor increases

in

in the table.

the mechanical load.

follows that the power factor

TABLE 13A

size

—

No-load

*

Small

Current

Torque

Slip

(per-unit)

(per-unit)

(per-unit)

means under

factor

Big*

Small

Big

Small

Big

Small

Big

Small

Big

l

1

1

1

0.03

0.004

0.7

0.96

0.8

0.87

to

to

to

to

0.9

0.98

0.85

0.9

-0

0

0

0.2

0.05

1

0

0

0.4

0.1

0.5 rotor

Power

Efficiency

Small*

Full-load

Locked

almost direct proportion

TYPICAL CHARACTERISTICS OF SQUIRREL-CAGE INDUCTION MOTORS

Loading

Motor

It

0

0.3

0

5

4

to

to

to

to

6

6

3

1

11

kW

(15 hp); big

1

.5

0.5

means over

1

1

120

kW

(1500 hp) and up

to

25 000

hp.

to


Figure 13.13

Figure 13.14

At no-load the flux tual flux tive

(t>

power

the motor

in

m To create .

is

this flux,

is

mainly the mu-

At full-load the mutual flux decreases, but stator

considerable reac-

needed.

of the motor improves dramatically as the mechanical load increases. At full-load

it

98% 3.

and rotor leakage fluxes are created. The reactive power needed is slightly greater than in Fig. 13.13.

where

ranges from 0.80 for

small machines to 0.90 for large machines.

ciency

211

at full-load is particularly

high;

it

The

/ = Ph = E= 600 =

effi-

can attain

for very large machines.

full-load current [A]

output power [horsepower) rated line voltage (V)

empirical constant

Locked-rotor characteristics. The locked-rotor

current the I

The

2

is

R

5 to 6 times the full-load current,

losses 25 to

rotor

36 times higher than normal.

must therefore never remain locked for

more than a few seconds. Although the mechanical power is

needed

Recalling that the starting current that

the

no-load current

we can

0.3 pu,

lies

is

5 to 6 pu

and

between 0.5 and

readily estimate the value of these

currents for any induction motor. at standstill is

motor develops a strong torque. The power

zero, the

factor

making

low because considerable reactive power

to

produce the leakage flux

stator windings.

and

Example 13-4 a.

3-phase induction motor having a rating of

because the stator and

500 hp, 2300

windings are not as tightly coupled (see

Section 10.2).

Calculate the approximate full-load current, locked-rotor current, and no-load current of a

These leakage fluxes are much

larger than in a transformer the rotor

in the rotor

is

V.

b.

Estimate the apparent power drawn under

c.

State the nominal rating of this motor,

locked-rotor conditions.

expressed

13.12 Estimating the currents in an induction motor

Solution a.

The

full-load current of a 3-phase induction

may be

motor

means of

the following ap-

= 600 P h /E

(13.5)

calculated by

proximate equation: I

in kilowatts.

The

full-load current /

is

= 600 P h /E - 600 X 500/2300 = 130 A (approx.)

(13.5)


278

The no-load

current

=

/0

=

0.3/

X

0.3

1

energy

30

39

A

starting current

=

/ LR

(approx.)

6/

-

X

6

The remaining

the

watts,

it

and not of

this

fol-

power

stator.

f

js

is

owing

active

power P v

gap and transferred

Due

dissi-

windings. Another portion

dissipated as heat in the stator core,

P

Due

00

kVA

power of a motor

always

(8.9)

is

relates to the

motor expressed

in

expressed

car-

is

to the rotor

jr

to the

PR

losses in the rotor, a third portion

dissipated as heat, and the remainder

in kilo-

The nominal

SI units

is

is,

rating

therefore,

windage and bearing-friction tain

PL

,

the mechanical

is

finally

power P m By .

f v representing losses, we finally ob-

subtracting a small fourth portion

mechanical output

to the electrical input.

P

available in the form of mechanical

(approx.)

,

power available

at the shaft

to drive the load.

The power flow diagram of Fig. 13.15 enables

P = 500/1.34 = 373 kW (see Power Appendix AX0)

to identify

conversion chart in

ties

Voltages, currents, and phasor diagrams enable us

understand the detailed behavior of an induction

Figure 13.15 Active

power flow

in

a 3-phase induction motor.

and

Efficiency.

motor

is

us

to calculate three important proper-

of the induction motor: (1)

power, and (3)

L 13.13 Active power flow

to

is {

to the

by electromagnetic induction. EI

= V3 X 2300 X 780

When

electrical

flows through the ma-

in the

iron losses.

1

it

pated as heat

ried across the air

3

how

copper losses, a portion

130

is

=

easier to see

to the stator

tions

= V3

is

flows from the line into the 3-phase

The apparent power under locked-rotor condi-

S

c.

it

converted into mechanical energy by

chine. Thus, referring to Fig. 13.15, active

Pc

is

= 780 A (approx.) b.

is

lowing the active power as

= The

motor. However,

is

its

its

efficiency, (2)

its

torque.

By

definition, the efficiency of a

the ratio of the output

power

to the input

power: efficiency Cn)

- PJP e

(13.6)


R losses in the rotor. It can rotor rR losses P-. are related

2.

/

power P T by

Motor

be shown * that the

4.

to the rotor input

motor

„

the equation

The torque Tm developed by

torque.

any speed

at

9.55

279

is

the

given by

Pm (3.5)

Pjv = sP

n

(13.7)

T

9.55

where rotor

]V

—

s

PR

losses

[W]

P

r

9.55 transmitted to the rotor

T

Equation 13.7 shows that as the

PR

A

nous speed

=

(s

slip increases, the

0.5) dissipates in the

locked

is

ted to the rotor

is

(s

=

1), all

/n s

it

the

(13.9)

s

P

power transmitted

r

"s

9.55

When

= =

power transmit-

The

Mechanical power. The mechanical power P m is equal to the power transmitted to the rotor minus its PR losses. Thus,

developed by the motor

any

[W]

to the rotor

synchronous speed [r/minj multiplier to take care of units [exact value: 60/2tt|

dissipated as heat.

3.

at

speed [N-ml

form of heat

receives.

PJn

torque developed by the motor

rotor turning at half synchro-

50 percent of the active power

r

where

consume a larger and larger propower P r transmitted across the air

to the rotor.

the rotor

[

W]

losses

portion of the

P

therefore.

slip

P = power

gap

-v)

s)

9.55

P =

rotor

(1

« s (l

actual torque 7,

slightly less than

overcome

the

available

Tm due ,

windage and

at

the shaft

friction losses.

most calculations we can neglect

in

is

to the torque required to

However, this

small

difference.

Pm = P 1

1

= P

r

Equation 13.9 shows that the torque

J'*

sP r

r

(13.7)

proportional to the active tor.

whence

The

actual mechanical

the load

is

needed

to

losses.

In

(13.8)

power available

P m due

to drive

power overcome the windage and friction most calculations we can neglect this slightly less than

,

must absorb a large amount of active power. The form of

latter is dissipated in the

to the

small loss. mechanical

electromagnetic

power

heat, consequently,

the temperature of the rotor rises very rapidly.

Example 13-5

A

3-phase induction motor having a synchronous

speed of 1200 r/min draws 80

power output

directly

to the ro-

Thus, to develop a high locked-rotor torque, the

rotor

s)P T

is

power transmitted

kW

from a 3-phase

electrical

X

speed of flux

losses

transferred

electromagnetic torque

p,

of rotor

to rotor

Pm - P 1

9.55

in rotor

i

9.55

but from Eq. 3.5

_

(iii)

p,

(i)

1

rotor speed

Pm ~

Tm must

but the mechanical torque

X mechanical

torque

the electromagnetic torque

9.55

Timc

equal

.

Thus

Hence,

rm = Tmia

_ nT^ Also from Eq. 3.5 we can write

tn)

Substituting

Py

=

(ii), (iii),

SP,

(iv)

and

(iv) in

(i).

we

find


280

feeder. tor

The copper losses and iron losses in the stato 5 kW. If the motor runs at 52 r/min,

amount

1

e.

The efficiency

is

= PJPC =

1

T)


70/80

0.875 or 87.5% a.

b. c.

d.

The active power transmitted to the rotor The rotor ER losses The mechanical power developed The mechanical power delivered to the load, knowing that the windage and friction losses are equal to 2

e.

kW

The efficiency of

Example 13-6

A

3-phase, 8-pole squirrel-cage induction motor,

connected

line,

possesses a synchronous

and iron losses

the copper

motor

the

60 Hz

to a

speed of 900 r/min. The motor absorbs 40 kW, and

5

kW

and

1

kW,

in the stator

amount

to

respectively. Calculate the torque

developed by the motor. Solution a.

Active power to the rotor

Solution

is

The power transmitted across the

= b.

The

-

80

=

5

75

kW

Pr

slip is

=

s

=

(n s

-

Tm =

n)/n s

-

(1200

Rotor

I

R

Note torque)

]v

c.

v

X

75

=

3

The mechanical power developed P^n

d.

0.04

=

P\

=

75

~ -

2 I

3

kW

windage

rotor

P

9.55

X

=

361

N-m

T

P>„

-

P,

P mt due

=

is

72

13-7.

this

problem

(the

at

a standstill or running at

power P transmitted v

full

to the

equal to 34 kW, the motor develops a torque

Example 13-7 to the load

to the friction

2

to

independent of the speed of rotation. The

and

A 3-phase induction motor having a nominal rating of 100 hp (-75

= 70 kW

is

kW)

and a synchronous speed of 1800

connected to a 600

V

source (Fig. 13.16a).

The two-wattmeter method shows a

HP(^75kW)

1783 r/min

Figure 13.16a

(13.9)

34 000/900

solution

the

that is

100

See Example

kW

fn s

losses.

=

is

of 361 N-m.

r/min Pi

1

9.55

motor could be rotor

P losses in - 72 kW x

slightly less than

5

speed, but as long as the

is

The mechanical power P delivered is

to the rotor

0.04

losses are

P = sP =

~ Pi - = 34

gap

Pjs

= 52)/ 1200

11

= 48/1200 2

= Pc ~ = 40 -

air

total

power con-


sumption of 70 kW, and an ammeter indicates a

measurements give

current of 78 A, Precise

line

b.

The

slip

is

a rotor s

known about

stator iron losses

windage and

P =

between two

resistance

kW

2

r

= (1800 = 0.0205

the motor:

friction losses

Px =

.2

1

kW

stator terminals

Rotor

=

0.34 (2

2

J

R

b.

Rotor I~R losses

c.

Mechanical power supplied

Mechanical power developed

= to the load, in

Py

~

P)v

=

e.

Torque developed

Solution

Power supplied

to the stator

P c = 70

is

d.

kW

R

0.34/2

wye con-

e.

Torque

63.5

kW

=

62.3

kW =

=

83.5 hp

-

/\

63.5

62.3

-

X

1.2

1.34 (hp)

is

62.3/70

-

0.89 or

89%

1763 r/min:

at

=

T0.17 ft

9.55

R =

=

3.1

kW

P =

2

]s

Iron losses

{

Power supplied

=

Pc

=

(70

-

X

3

2

(78)

X

0.17

y

in s

=

9.55

X 64 900/1800

The torque developed by

to the rotor:

^js 3.1

-

summarized

a

Pi'

-

in

motor depends upon

speed, but the relationship between the

2)

Figure 13.16b

Example

calculations are

Fig. 13.16b.

13.14 Torque versus speed curve

kW

13-7.

its

two cannot

be expressed by a simple equation. Consequently,

=

64.9

3.1

in

P

= 344 N m The above

2

3 I

Power flow

-

= PJP e -

=

1.33

losses are

P =

Pr

is

is

R = 2

kW

to the load:

Efficiency of the motor T]

Stator resistance per phase (assume a

PL

63.5

=

1763 r/min

at

-

"4.9

Mechanical power P,

Efficiency

Stator f

1.33

to the rotor

d.

nection)

=

64.9

r

horsepower

a.

763)/ 1800

P = sP = 0.0205 X c.

Power supplied

1

losses:

]r

Calculate a.

n)fn s

(n s

speed of 1763 r/min. In addition, the following characteristics are

281

kW

kW

we

2kW

prefer to

show

the relationship in the

form of a


282

curve. Fig. 13.17

shows

inal

full-load torque

T and

1.5

torque)

the

is

is

following example illustrates the changes that occur.

starting torque

is

the it

breakdown torque. motor runs at a speed

is

rings.

is

Figure

accelerat-

ing from rest to the

10 n. If

motor torque

torque.

As soon

motor

will

as the

two torques

breakdown

if

torque), the

motor

Small motors (15 hp and

breakdown torque

at

20

breakdown torque

(

at

develop

of about

/z d

1

98%

that

it

from no-load

to full-load,

N-m (-73.7

fHbf).

The

full-load current is

is

100 A. The ro-

This can be achieved by using a material of

in

Figure

1

3.

1

8b.

It

can be seen that the

is

start-

ing torque doubles and the locked-rotor current de-

00 A to 90 A. The motor develops its breakdown torque at a speed /Vd of 500 r/min, compared to the original breakdown speed of 800 r/min. creases from

If

we

becomes

imum

es-

is

motor having a syn-

and the locked-rotor current

shown

of syn-

rotor resistance of a squirrel-cage rotor

V

and end-rings. The new torque-speed curve

of

13.15 Effect of rotor resistance

sentially constant

380

higher resistivity, such as bronze, for the rotor bars

chronous speed.

The

A

2.5.

500 hp and more)

about

8a shows the torque-speed curve of a

Let us increase the rotor resistance by a factor of

T

their

80%

1

tor has an arbitrary resistance R,

lower

will quickly stop.

less)

a speed

synchronous speed. Big motors attain their

of 100

are in balance, the

the load torque exceeds 2.5

3.

1

again equal to the load

turn at a constant but slightly

speed. However, (the

is

1

kW (13.4 hp), 50 Hz,

chronous speed of 000 r/min and a full-load torque

me-

the

chanical load increases slightly, the speed will drop until the

and end-

minimum

The

torque developed by the motor while

At full-load the

in the rotor bars

torque (called breakdown

Pull-up torque

T.

nom-

The torque-speed curve is greatly affected by such a change in resistance. The only characteristic that remains unchanged is the breakdown torque. The

T.

maximum

2.5

aluminum, or other metals

the torque-speed curve of a

conventional 3-phase induction motor whose

of 70

except

A

increases with temperature. Thus, the resis-

1

again double the rotor resistance so that 5 R, the locked-rotor torque attains a

value of 250

A (Fig.

N-m

it

max-

for a corresponding current

13.18c).

further increase in rotor resistance decreases

tance increases with increasing load because the

both the locked-rotor torque and locked-rotor cur-

temperature

rent.

rises.

For example,

if

the rotor resistance

is

in-

In designing a squirrel-cage motor, the rotor resis-

creased 25 times (25 R), the locked-rotor current

tance can be set over a wide range by using copper,

drops to 20 A, but the motor develops the same

breakdown torque 2.5

T

2

T ,

|

locked-rotor torque

1.5 Ti

o

^pull-up

0.5

full

torq ue

nominal

T

T

20

60

40

—

p~

rotational

speed

Figure 13.17 Typical torque-speed curve of a

3-phase squirrel-cage induction motor.

80

100

%

load

Nm

N-m

Figure 13.18 Rotor resistance affects the motor characteristics.

283


284

Nm),

torque (100

starting

as

locked-rotor current was 100 In

summary,

because

it

relatively

A

it

when

did

the

1

.

a high rotor resistance

is

The locked-rotor current can be

rotor torque will

low

squirrel-cage motor.

Unfortunately,

it

current

(Fig.

resistors in

series with the rotor. Nevertheless, the locked-

desirable

produces a high starting torque and a starting

drastically re-

duced by inserting three external

(Fig. 13.1 8d).

13.18c).

also produces a rapid fall-off in 2.

still

be as high as that of a

The speed can be varied by varying

the exter-

speed with increasing load. Furthermore, because the slip at rated torque

The

losses are high.

and the motor tends

is

high, the

efficiency

is

R

therefore low

3.

it is

preferable to have

up

and the

slip

We

is

is

small. Consequently, the effi-

high and the motor tends to run cool.

which require

is

a diagram of the circuit used to start

nected to three wye-connected external resistors by set

of slip-rings and brushes. Under

locked-rotor (LR) conditions, the variable resistors

and a low running resistance by designing the rotor bars in a special

However,

are set to their highest value.

way

(see Fig. 14.5,

Chapter

14).

up, the resistance

the rotor resistance has to be varied

if

over a wide range,

it

may

a long time to bring

wound-rotor motor. The rotor windings are con-

means of a

can obtain both a high starting resistance

ideally suited to accelerate high-

to speed.

Fig. 13.19

at rated torque

is

The speed de-

less with increasing load,

a

ciency

The motor

inertia loads,

a low rotor resistance (Fig. 13.18a).

much

I

nal rotor resistors.

to overheat.

Under running conditions creases

motor

2

load speed

be necessary to use a

is

is

As

the

motor speeds

gradually reduced until

full-

reached, whereupon the brushes are

By properly selecting the resistance we can produce a high accelerating torque

short-circuited.

wound-rotor induction motor. Such a motor envalues,

ables us to vary the rotor resistance at will by

means of an external

with a stator current that never exceeds twice

full-

rheostat.

load current.

To

13.16 Wound-rotor motor

We

explained the basic difference between a

squirrel-cage motor and a wound-rotor motor in

Section

1

3.

more than

1

.

Although a wound-rotor motor costs

a squirrel-cage motor,

it

start

large

motors,

we

offers the fol-

lowing advantages:

a large thermal capacity.

A

liquid

electrolyte.

To vary

its

resistance,

trodes.

The

is

com-

a suitable

we simply

vary

large thermal capacity of the electrolyte

Figure 13.19 to the three slip-rings of

in

the level of the electrolyte surrounding the elec-

speed controller

connected

use

liquid rheostat

posed of three electrodes immersed

starting rheostat

External resistors

often

rheostats because they are easy to control and have

a wound-rotor induction motor.

and


limits the

temperature

rise.

For example, in

conjunction bring a large

synchronous machine up

We

used

one ap-

kW wound-rotor motor to

plication a liquid rheostat

with a 1260

in

is

to speed.

can also regulate the speed of a wound-rotor

motor by varying the resistance of the rheostat. As

we

increase the resistance, the speed will drop. This

method of speed control has the disadvantage of heat

lot

ciency

is

that a

dissipated in the resistors; the effi-

is

therefore low. Furthermore, for a given

rheostat setting, the speed varies considerably

if

the

rating of a self-cooled wound-rotor

motor depends upon the speed

at

Thus, for the same temperature

rise,

40

which

it

operates.

a motor that can

lOOkWat 800 r/min can deliver only

develop

kW at 900 r/min. it

60

lodged

coils,

However,

the

if

can deliver 50

motor

is

in

60

slots.

total

The

X

of (4

and are staggered

intervals (Fig.

The

13.20).

X

coils in each

are connected in series

may

3

at

5)

group

one-slot

coils are identical

possess one or more turns.

=

The width

and

of each

coil is called the coil pitch.

Such

a distributed

winding

is

obviously more

costly to build than a concentrated winding having

only one coil per group. However, starting torque

it

improves the

and reduces the noise under running

conditions.

from a

the stator windings are excited

3-phase source, a multipolar revolving

field

duced. The distance between adjacent poles the pole pitch.

It

is

is

is

pro-

called

equal to the internal circumfer-

about

ence of the stator divided by the number of poles. For

cooled

example, a 12-pole stator having a circumference of

1

by a separate fan,

group must have a

When

mechanical load varies.

The power

coils per

285

kW at 900 r/min.

600

mm has a pole-pitch of 600/12 or 50 mm.

In practice, the coil pitch

is

between

80% and

100% of the pole pitch. The coil pitch is usually made less than the pole pitch in order to save copper

13,17 Three-phase windings named

and

to

Nikola Tesla invented the 3-phase induction motor.

The

shorter coil width reduces the cost and weight

In

883 a 27-year-old Yugoslav

1

His

first

model had a

similar to the one

scientist

salient-pole stator winding

shown

in Fig.

1

3.6.

Since then the

design of induction motors has evolved consider-

modern machines

ably;

are built with lap windings

distributed in slots around the stator.

A

lap

winding consists of a

set

tribution

improves the torque during

much

To

Fig. 13.21a.

=

poles

X

coils

is

it

follows that the

equal to the

number of groups.

A

stator

must have

at least

2, 3.

or

more

coils per

The number of tion.

coils

12 slots.

group

and

in

coils are held upright, with

one

at least

of

12 coils.

many 5 coils per group

has coils. Consequently, a 4-pole, 3-phase

designers have discovered that

us

4-pole,

Furthermore, in a lap winding the stator has as it

let

shown

at least

minimum number

3-phase stator must therefore have

slots as

laid out flat as

4X3=12

phase groups. Because a group must have coil,

The 24

is

phases

Thus, a 4-pole, 3-phase stator must have

one

and

easier to insert in the slots.

get an overall picture of a lap winding,

The number of groups groups

start-up,

2-pole machines, the shorter pitch also makes the

suppose a 24-slot stator


the air gap.

often results in a quieter machine. In the case of

evenly distributed around the stator circumference. is

in

of the windings, while the more sinusoidal flux dis-

coils

of phase groups

improve the flux distribution

it

is

However, motor preferable to use

rather than only one.

slots increases in propor-

For example, a 4-pole. 3-phase stator having 5

Figure 13.20

The five coils are connected phase group.

in

series to create

one


286

coil side set in

each

the windings are

slot. If

now

laid

c.

down so that all the other coil sides fall into the slots, we obtain the classical appearance of a 3-phase lap winding having two

The

coils are

coil sides per slot (Fig.

1

3.2 b). 1

Number of groups poles

d.

The pole

one for each phase. Each wind-

ing consists of a

One

ber of poles.

(say) to slot 13.

around the circumference of the

rically distributed

The following examples show how

stator.

e.

this is

done.

of

= 40

4-

=

10

4.

pitch corresponds to

pole pitch

number of groups equal to the numThe groups of each phase are symmet-

= number

per phase

10

Coils per group

connected together to create three

identical windings,

—

= -

=

slots/poles

120/10

12 slots

pole pitch extends therefore from slot

1

The coil pitch covers 10 slots (slot to slot The percent coil pitch = 10/12 = 83.3%. 1

The next example shows

in

greater detail

1

1

).

how

the coils are interconnected in a typical 3-phase sta-

Example 13-8 The stator of a possesses 120

tor winding.

3-phase, 10-pole induction motor

slots. If a lap

winding

is

used, calcu-

a.

b. c.

d. e.

The total number of coils The number of coils per phase The number of coils per group The pole pitch The coil pitch (expressed as a percentage of the pole pitch), 1

to slot

if

Example 13-9

A

late the following:

the coil width extends from slot

stator

having 24 slots has to be

1

.

2.

The connections between The connections between

are standing upright, with

We

bution for phase

Solution a.

A

b.

Coils per phase

120

-

3

-

the coils the phases

coils.

one

Assume

coil side in

that they

each

slot

will first determine the coil distri-

A and

then proceed with the con-

nections for that phase. Similar connections will

120-slot stator requires 120 coils.

=

with a

Solution

The 3-phase winding has 24 (Fig. 13.22).

1

wound

3-phase, 4-pole winding. Determine the following:

then be

40.

made

for phases

B and C. Here

is

the line of

reasoning: a.

The revolving

p

field creates

4 poles; the motor

4 groups per phase, or 4

therefore has

phase groups in

all.

Each rectangle

X

3

in Fig.

=

1

13.22a

I

represents one group. Because the stator contains

lln iVr

1/2

20 21 22 23 24

3

4

5

6

24

7

/

slot

number

coils,

each group consists of 24/12

2

b.

The groups

(poles) of each phase

24

must be uni-

The group distribution for phase A is shown in Fig. 13.22b. Each shaded rectangle represents two

Figure 13.21a in

2 con-

secutive coils.

formly spaced around the

Coils held upright

=

stator slots.

stator.

upright coils connected in series, producing the

two terminals shown. Note

that the

me-

chanical distance between two successive 20 21 22 23 24

1

2

3

4

5

6

groups always corresponds to an electrical

7

phase angle of 180°. Figure 13.21b Coils laid

down

c.

to

make a

typical lap winding.

Successive groups of phase site

magnetic

polarities.

A must have oppo-

Consequently, the four

êach each group

one group of one phase

coils are

ot two composed of

coils in series

AAA

Figure 13.22a

The 24

is

grouped two-by-two

to

make 12

groups.

BtJtfMtzftlMt (—180°

Figure 13.22b The four groups

of

(electrical)—

phase A are selected so as

start of

phase

to

be evenly spaced from each

other.

A

Figure 13.22c

The groups

of

phase A are connected

start of

in

series to create alternate

phase

,

start of

N-S

poles.

phase C

1

1

cil

:b;

120°

240°— Figure 13.22d

The

start of

phases B and

A

C

C

begins 120° and 240°, respectively, after the start of phase A.

'4

Figure 13.22e

When

all

phase groups are connected, only

six leads remain.

287


288

Figure 13.22f

The phase may be connected

A are

groups of phase

wye

in

connected

duce successive N-S-N-S poles Phase

A now

1

A2 B and C

same way around

the stator.

and three leads are brought out

delta,

pro-

f.

are spaced

(Fig.

The groups series in the

in

phases B and

same way

C

A are A,A 2

,

.

sulting 3 wires corresponding to the 3 phases

are brought out to the terminal

box of

the

ma-

chine (Fig. 13.22f). In practice, the connections

made, not while the coils are upright

shown) but only in

the slots.

according to Fig.

450

after they

first coil

of

1

and connections

3.22e.

and 13.24b show the

kW (600 hp)

and

coil

induction motor.

procedure used

kW (50 hp)

in

stator.

are connected in

as those of phase

B|B 2 and C)C 2 They may be connected either in wye or in delta inside the machine. The re,

Thus, the

shortened

the first and sixth slots

winding a smaller 37.5

(Fig. 13.22e). This yields six terminals:

are

in

Fig. 13.25 illustrates the

(Fig. 13.22d). e.

to slot 6).

1

lodged

Figs. 13.24a

A

of phase (

suit

stator of a

located at 120° and 240° (electrical) with re-

A

is

span of

to a

may be

3.23). All the other coils

1

follow

However, the

the terminal box.

6 slots, the coil pitch

A

phase

.

starting terminals B, and C, are respectively

spect to the starting terminal

=

to 5 slots (slot

ter-

to

Because the pole pitch corresponds 24/4

3.22c).

has two terminals, a starting

The phase groups of phases the

in

in series to

(Fig.

minal A, and a finishing terminal d.

or

have been

(as

laid

down

13.18 Sector motor Consider a standard 3-phase, 4-pole, wye-connected

motor having a synchronous speed of 1800 r/min. Let us cut the stator is

left

in

in half,

so that half the winding

removed and only two complete (per phase). Next,

let

N

and S poles are

us connect the three phases

wye, without making any other changes to the ex-

isting coil connections. Finally,

nal rotor

gap

above

we mount

this sector staton leaving a

(Fig. 13.26).

the origi-

small

air


should be reduced to half the stator winding nal

now

289

original value because

its

has only one-half the origi-

number of turns. Under

these conditions, this re-

markable truncated sector motor still develops about

20 percent of

The moves

original rated power.

its

sector motor produces a revolving field that the

at

same peripheral speed as However,

the original 3-phase motor.

making

the flux in

instead of

a complete turn, the field simply travels

continuously from one end of the stator to the other.

13.19 Linear induction motor It is

flat,

obvious that the sector stator could be

laid out

without affecting the shape or speed of the

field. Such a flat stator produces a field moves at constant speed, in a straight line. Using the same reasoning as in Section 3.5, we can

magnetic Figure 13.24a

that

Stator of a 3-phase,

Hz

induction motor.

108 preformed

One

coil

side

450 kW, 1 1 80 r/min, 575 V, 60 The lap winding is composed of

coils

having a pitch from slots

falls into

the bottom of a slot

other at the top. Rotor diameter: 500

460 mm. (Courtesy mecaniques Roberge) length:

mm;

1

to 15.

and the

1

prove that the flux travels

at a linear

synchronous

speed given by

axial

vs

= 2wf

(13.10)

of Services Eiectro-

where \\

=

linear

vr

—

width of one pole-pitch

/= Note

synchronous speed |m/s]

frequency [Hz]

that the linear

speed does not depend upon the

number of poles but only on is

[in]

the pole-pitch. Thus,

moving

at

the

stator (say), If

a

flat

flat stator,

along with

same speed

as that of a 6-pole linear

provided they have the same pole-pitch.

squirrel-cage winding

is

brought near the

the travelling field drags the squirrel cage it

use a simple

(Section 13.2). In practice,

aluminum

we

generally

or copper plate as a rotor (Fig.

13.27). Furthermore, to increase the

power and

duce the reluctance of the magnetic path, two tors are usually

sides of the

Figure 13.24b

Close-up view of the preformed

coil in Fig.

13.24a.

it

possible for a 2-pole linear stator to create a field

to re-

flat sta-

mounted, face-lo-face, on opposite

aluminum

plate.

a linear induction motor.

The combination

The

is

direction of the

called

motor

can be reversed by interchanging any two stator leads. If

we connect

the stator terminals to a 3-phase,

60 Hz source, the rotor will again turn 1800

r/min.

To prevent

saturation,

at

close to

the

voltage

In

many

practical applications, the rotor

is

sta-

tionary while the stator moves. For example, in

some high-speed

trains, the rotor is

composed of

a

Figure 13.25

50

Stator winding of a 3-phase,

carrying 48 coils connected a.

Each

composed

coil is

in

hp,

575

V,

60 Hz, 1764

r/min induction motor.

of 5 turns of five No. 15

copper wires connected

with a high-temperature polyimide insulation. Five No. 15 wires b.

c.

One

coil

fore,

from

Each

coil

side

threaded

stator

possesses 48

slots

into slot

1

(say)

in parallel.

in parallel is

and the other side goes

The wires are covered

equivalent to one No. 8 wire.

into slot 12.

The

coil pitch

is,

there-

to 12.

1

side

side placed

is

The

wye.

in

and 4 empty

fills

the

does not touch the second coil shows 3 empty and uninsulated slots a composition paper liner. The remaining 10 slots each carry one coil

a

slot

and

same

slot.

Starting from the top, the photograph

half

slots insulated with

is

covered with a paper spacer so that

it

side. d.

A

varnished cambric cloth, cut

in

the shape of a triangle, provides extra insulation between adjacent phase

groups.

(Courtesy of Services Efectromecaniques Roberge)

290


13.20 Traveling

We

sometimes

are

291

waves with the impression that

left

when

the flux reaches the

there

must be a delay before

more

at

end of

the beginning. This

a linear stator,

returns to restart once

it

is

not the case.

ear motor produces a traveling

wave of

The

lin-

which

flux

moves continuously and smoothly from one end of shows how the flux moves from left to right in a 2-pole linear motor. The flux cuts off sharply at extremities A, B of the stator to the other. Figure 13.28

Figure 13.26 Two-pole sector induction motor.

However,

the stator.

pears

at

the right,

it

N

as fast as a

or S pole disap-

builds up again at the

left.

linear rotor

Properties of a linear induction motor

13.21

(aluminum, copper or iron plate)

The

properties of a linear induction motor are al-

most

identical to those of a standard rotating

ma-

chine. Consequently, the equations for slip, thrust,

power,

L

etc.,

are also similar.

Slip

Slip,

is

defined by

=

s

(v s

-

(I3.ll)

v)/v s

where s \\

Figure 13.27

Components

of

a 3-phase linear induction motor.

aluminum

thick

tending over the stator

is

plate fixed to the full

v

ground and ex-

length of the track.

The

linear

bolted to the undercarriage of the train and

straddles the plate. Train speed

is

varied by chang-

ing the frequency applied to the stator (Fig.

1

3.3

2.

1

slip

synchronous linear speed [m/s] speed of rotor (or stator) [m/s]

Active power flow. With reference to Fig.

active

way

it

3.7,

power flows through

a linear

motor

and 13.8 apply

tween consecutive phase groups of phase

mm,

A

is

300

calculate the linear speed of the magnetic field.

(13.6)

(\

-s)P

is

given by:

F= P pitch vs

is

300 mm. Consequently,

=

2vv/

=

2

=

45 m/s or

X

X 1

75

62 km/h

T

(13.12)

/v s

where (13.10)

0.3

(13.8)

r

thrust or force developed by a lin-

ear induction motor

Solution

The pole

The

13.6,

(13 .7)

T

Pm = Thrust.

that the

to both types of machines:

= PJP c P}r = sP

3.

same

Consequently, Eqs.

flat.

t\

Example 13-10 The stator of a linear induction motor is excited from a 75 Hz electronic source. If the distance be-

3. 15,

1

in the

does through a rotating motor, except

and rotor are

stator 1

).

— = =

F = thrust [Nl P = power transmitted to the rotor W] v = linear synchronous speed |m/s| v

s

(

292


Example 13-11

An

overhead crane

in a factory is driven horizon-

by means of two linear induction motors

tally

whose

two

rotors are the

I-beams upon which

steel

The 3-phase, 4-pole

the crane rolls.

linear stators

(mounted on opposite sides of the crane and facing the respective pitch of 8

webs of

the I-beams) have a pole

cm and are driven by a variable frequency

electronic source. During a test tors, the

on one of the mo-

following results were obtained:

Hz

stator frequency: 15

power

kW

to stator: 5

copper loss

+

iron loss in stator:

1

kW

crane speed: 1.8 m/s Calculate a.

Synchronous speed and

b.

Power

c.

I

d.

Mechanical power and thrust

2

R

slip

to the rotor

loss in rotor

Solution a.

Linear synchronous speed

= 2wf - 2 X 0.08 X = 2.4 m/s

vs

The

Power

=

(v s

=

0.25

- v)/v = (2.4 - 1.8)/2.4

to the rotor

/>

js

r

= c.

!

R

s

(13.11)

is

P = Pc - 5 -

2

15

slip is

j

b.

(13.10)

- P

f

(see Fig. 13.15)

1

4kW

loss in the rotor

is

Figure 13.28

Shape of the magnetic field created by a 2-pole, 3-phase linear stator, over one complete cycle. The successive frames are separated by an interval of time equal to 1/6 cycle or

60%.

Pir

= *P = 0.25 X = kW

(13.7)

r

1

4


d.

Mechanical power

cause the flux density

is

The

(Fig- 13.15)

}r

sulting

value

induced current reaches

nnn

and

1

time. This current, re-

and

creates magnetic

3,

shown

as

sss

the re-

maximum

its

Fig.

in

13.29.

is

According

F = PJv s = 4000/2.4 =

same

at virtually the

turning by conductors

poles thrust

greatest at the center of

magnet moves very slowly,

the pole. If the

Pr-P 4-1 = 3kW

Pm = =

is

293

1667

(13.12)

to the

laws of attraction and repulsion,

while the rear half

N=

1.67

magnet

the front half of the

attracted

is

repelled upward

is

downward. Because

the distribution of the nnn and sss poles

kN (-375

is

sym-

lb)

metrical with respect to the center of the magnet, the vertical forces of attraction and repulsion are

13.22 Magnetic levitation In

we saw

Section 13.2

moving permanent

that a

magnet, sweeping across a conducting ladder, tends to

drag the ladder along with the magnet.

now show

We

that this horizontal tractive force

accompanied by a

is

also to

Referring to Fig. 13.29, suppose that conduc-

1,2,3 are three conductors of ladder. The center of the N pole of

the stationary

the

sweeping across the top of conductor age induced

in

this

conductor

is

2.

magnet is The volt-

maximum

be-

But suppose now rapidly.

magnet

(stationary)

\

s

s

s

s

/

/

that the

nil.

magnet moves very

its

maximum

con-

value a fraction of a

after the voltage has attained

Consequently, by the time the current

its

in

maximum.

conductor 2

maximum, the center of the magnet is already some distance ahead of the conductor (Fig. 13.30). The current returning by conductors and 3 again is

1

creates nnn and sss poles; however, the is

now

directly

N

pole of the

above an nnn pole, with the

low

result that a strong vertical force tends to

magnet upward.* This

effect

push the

called the principle

is

of magnetic levitation.

777TTT\ n n

ss

is

to its inductance, the current in

speed

N

conducting ladder

Owing

ductor 2 reaches

magnet ::

force

vertical

only a horizontal tractive

is

force.

second

push the magnet away from the ladder.

tors

Consequently, there

will

which tends

vertical force,

and the resulting

equal,

\

1

Magnetic n n n

0'

levitation

is

used

in

some

ultra-high-

speed trains that glide on a magnetic cushion rather than on wheels.

A powerful electromagnet fixed un-

derneath the train moves above a conducting

ducing currents Figure 13.29

ladder.

Currents and magnetic poles at low speed.

in the rail in the

The force of

levitation

same way

is

rail in-

as in our

always accompa-

nied by a small horizontal braking force which

must, of course, be overcome by the linear motor magnet

high

that propels the train.

See Figs. 13.31 and 13.32.

speed

N

The

current

is

always delayed (even

terval of time A/,

m

of the

rotor.

This delay

the

is

low speeds) hy an the

in-

IJR time constant

so brief that, at low speeds, the

maximum at virtually the same lime and voltage does. On the other hand, at high speeds,

current reaches

place as the

at

which depends upon

its

same delay At produces

a significant shift in

space be-

Figure 13.30

tween the points where the voltage and current reach

Currents and magnetic poles at high speed.

respective

maximum

values.

their

Figure 13.31 This 17

t

track.

is driven by a linear motor. The motor consists of a stationary rotor and a flat stator undercarriage of the train. The rotor is the vertical aluminum plate mounted in the center of the

electric train

fixed to the

The 3-tonne

stator

is

varied from zero to 105 Hz.

energized by a 4.7 MVA electronic dc to ac inverter whose frequency can be The motor develops a maximum thrust of 35 kN (7800 lb) and the top speed

200 km/h. Direct-current power at 4 kV is fed into the inverter by means with 6 stationary dc busbars mounted on the left-hand side of the track.

means

of a

brush assembly

in

is

contact

a superconducting electromagnet. The magnet

is 1300 magnet are maintained at a 2 temperature of 4 K by the forced circulation of liquid helium. The current density is 80 A/mm and the resulting flux density is 3 T. The vertical force of repulsion attains a maximum of 60 kN and the vertical gap between the magnet and the reacting metallic track varies from 100 mm to 300 mm depending on the current.

Electromagnetic levitation

nun long, 600

mm wide,

is

obtained by

and 400

mm deep,

of

and weighs 500

kg.

The

coils of the

,

(Courtesy of Siemens)

294


295

superconducting electromagnet

conducting track

guide and support

wheels

linear

motor

(stator)

conducting plate (rotor)

brush assembly for power input,

from 4 kVdc source

Figure 13.32 Cross-section view of the main components of the high-speed train (Courtesy of Siemens)


shown

in Fig.

1

3.31

Give two advantages of a wound-rotor

13 -8

motor over a squirrel-cage motor. Practical level 13-1

Name

the principal

components of an

Both the voltage and frequency induced

13-9

in

in-

the rotor of an induction

duction motor.

motor decrease

as the rotor speeds up. Explain.

13-2

Explain

how

a revolving field

is

set

up

in

13-10

a 3-phase induction motor. 1

3-3

If

we double

the

number of poles on

stator of an induction motor, will

its

A 3-phase, connected

the

syn-

chronous speed also double?

20-pole induction motor to a

a.

What

is

b.

If the

voltage

600

V.

is

60 Hz source.

the synchronous speed? is

reduced to 300 V, will the

synchronous speed change? 13-4

The

rotor of an induction should never be c.

locked while to the stator. 1

3-5

Why

full

voltage

is

13-

Explain.

13-7

1

1

groups are there, per phase?

Describe the principle of operation of a linear induction motor.

does the rotor of an induction motor

turn slower than the revolving field?

13-6

How many

being applied

13- 12

Calculate the approximate values of the

What happens to the rotor speed and rotor current when the mechanical load on an

starting current, full-load current, and

induction motor increases?

575

Would you recommend using

a 50 hp

induction motor to drive a 10 hp load? Explain.

no-load current of a 150 horsepower,

13- 3 1

V,

Make

3-phase induction motor.

a drawing of the magnetic field cre-

ated by a 3-phase, 12-poIe induction motor.

296


13-14

How

wc change

can

the direction of rota-

the

motor?

tion of a 3-phase induction

mmf developed by the windings, the resulting mmf point in a direction

Does

c.

intermediate between the

Intermediate level

13-15

sponding

Calculate Ihe synchronous speed of a

a.

1

3-phase, 12-pole induction motor that

b.

cited

by

What

is

load

a

3-2

1

ex-

is

if

produces

the slip at full

synchronous speed

a

when connected to number of

source. Calculate the

6 percent?

is

slots

of 900 r/min

nominal speed

mmf s corre-

and 4?

3-phase lap- wound stator possessing

72

60 Hz source,

the

A

to instants 3

60 Hz

a

coils

per phase group as well as the probable 1

3-

1

6

A

3-phase 6-pole induction motor

Hz

nected to a 60

induced

supply.

in the rotor bars

is

con-

The voltage is 4 V when the

is

locked. If the motor turns in the

13-17

its

13-22

The 3-phase, 4-pole

stacking (axial length) of 200

At 1000 r/min

maximum

At 1500 r/min


a.

Calculate the approximate values of

b.

a.

full-

The

kW. 4000

V, 3-phase.

1

9

A

900 b. c.

3-phasc, 75 hp, 440

V

2 percent.

1

3-23

when

the stator

induction motor

A large

4000

3-phase,

A

and a

385

and a power factor of 83 percent.

when

Calculate the nominal current per phase.

sponding speed

open-circuit voltage of

is

240 V appears

total active

found

to

is

wye and

tween two

stator terminals

total iron losses are

windage and

motor turns

same

direction as the

r/min. in the

same

direction as the

rotating field b.

At 900

rotating field c.

a.

of b.

field

Referring to Fig. 13.7, calculate the instan-

taneous values of I5()

/.,,

/h ,

and

/c

for an angle

u .

Determine the actual direction of current flow in the

three phases at this instant

10

friction losses are 12

il.

The

kW.

c.

d.

The load mechanical power [kW|, torque [kN-m], and efficiency

1

At 3600 r/min. opposite to the rotating

is 0.

kW and the

23.4

The power factor at full-load The active power supplied to the 2 The total I R losses in the rotor

a. b.

in the

At 600 r/min,

is

the resistance be-


open-circuit voltage and frequency across

a.

kW

corre-

be 709.2 r/min. The stator

duction motor when the rotor is locked. The stator has 6 poles and is excited by 60 Hz source. If the rotor is driven by a

a

The

accurately measured and

in

the dc

squirrel-

power of 2344

operating at full-load.

connected

if

bars

in the rotor

60 Hz

V,

across the slip-rings of a wound-rotor in-

the slip-rings

0.7 T,

connected to a

is

The peak voltage induced The pole-pitch

has a full-load efficiency of 91 percent

An

is

a

the

cage induction motor draws a current of

variable-speed dc motor, calculate the

13-20

flux density per pole

If

60 Hz source

Calculate the nominal full-load speed and is

mm and

mm.

peripheral speed |m/s] of the revolving

field

induction motor,

60Hz

torque knowing that the slip

3-

stator of Fig. 13.25

has an internal diameter of 250

frequency:

c.

r/min,

1

Fig. 13.22.

b.

current of a 75

con-

coil

ro-

load current, starting current, and no-load

13-18

complete

same

direction as the flux, calculate the approxi-

mate voltage induced and a. At 300 r/min

the

nection diagram, following steps (a) to (f) in

tor

Draw

coil pitch.

and calculate

3-24

If

we

rotor

( \

7c\

slightly increase the rotor resistance

of an induction motor, what effect does this a.

have (increase or decrease) upon

Starting torque

b.

Starting current

c.

Full- load speed

d. Efficiency


e.

Power

f.

Temperature

13-30

factor

of the motor

rise

rated

at its

power output 1

3-25

13-26

voltage

Explain the principle of magnetic levitation.

Advanced

In Fig. 13.5a the

The

permanent magnet has and moves at 30 m/s.

mm

flux density in the air

gap

is

0.5

T

1

3-3

1

and the effective resistance per rotor bar is

mil. Calculate the current

1

3-27

If the

conducting ladder

in Fig.

3.5a

1

is

pulled along with a force of 20 N, what

1

3-28

A

5000

3-phase,

hp,

6000

V,

60 Hz

1

What

r/min.

I~R losses 1

3-29

at

The motor

at

0.0073

3-28 has the

1

=

13-33

17°C

=

V

1600

= 6000 V = 100 A

line-to-line stator voltage

6.

active

7.

windage and

8.

iron losses in the stator

9.

locked-rotor current

2207

power supplied

to

motor

at

450

3.

1

9).


Voltage between the slip rings

Rotor resistance (per phase) and the

tolal

Approximate

rotor current, per phase

The train shown in Fig. 13.31 moves at 200 km/h when the stator frequency is

By supposing

a negligible slip,

motor [mm].

A 3-phase,

300 kW, 2300

V,

at

=

The motor has

load efficiency and

power

1

V =

2760

terminal voltage rises to

kW

effect (increase or decrease)

1800

A

to stator with rotor locked

=

kW

a full-

factor of 92 per-

cent and 86 percent, respectively.

kW

39

6000

5

used to

drive a compressor.

motor operates

=

friction losses

60 Hz, is

no-

kW

power

a speed of

1

develop a torque of 20

1780 r/min induction motor

no-load stator current, per phase

10. active

at

the linear

5.

91

kN-m

to

calculate the length of the pole-pitch of

=

17°C

4.

=

motor has

105 Hz.

rings with rotor locked

load

inserting resis-

fol-

open-circuit voltage induced between slip-

3.

Problem 13-29 by

in

If the

c.

Problem

11 at

wish to control the speed of the motor

at

13-32

dc resistance between rotor slip-rings

2.

We

are the approximate rotor

dc resistance between stator terminals

0.112H

Torque developed by the rotor

power dissipated

lowing characteristics: 1.

Mechanical power output

e.

a.

2-

rated load?

in

the stator

in

d.

b.

pole wound-rotor induction motor turns

594

Active power supplied to the rotor

following:

is

on the magnet?

the braking force exerted

I~R losses

c.

tors in series with the rotor (see Fig.

tractive force. 1

(locked-rotor) conditions:

b.

given

and the

/

LR

in

full-

Reactive power absorbed by the motor

a.

level

a width of 100

Referring to the motor described

Problem 13-29, calculate under

297

at full-load,

V

If

the

while the

determine the

upon

a.

Mechanical power delivered by the motor

b.

Motor torque

c.

Rotational speed

d. Full-load current e.

Power

f.

Starting torque

factor and efficiency

Calculate a.

Rotor and stator resistance per phase

75°C (assume b.

it

turns at

at

g.

wye connection)

Voltage and frequency induced

when c.

a

200 r/min and

Breakdown torque

i.

Motor temperature

in the rotor at

Starting current

h.

rise

594 r/min j.

Reactive power absorbed by the motor to

k.

Flux per pole Exciting current

create the revolving field, at no-load 2

d.

I

at e.

1.

R

losses in the stator

when

the

Iron losses

motor runs

no-load (winding temperature 75°C)


at

no-load

13-34

A

3-phase, 60

Hz

linear induction

motor

has to reach a top no-load speed of 12 m/s

298


and

must develop a

it

standstill thrust

of

Calculate

10 kN. Calculate the required pole-pitch

and the

minimum

2

I

R

a. b.

loss in the rotor, at

c.

standstill.

d.

Industrial application 1

3-35

A

1

e.

0 hp, 575 V,

1

1

60 r/min, 3-phase, 60 Hz

induction motor has a rotor

made of alushown in Fig.

minum, similar to the rotor 3.3a. The end-rings are trimmed 1

f.

is

1

3-37

What

making them

effect will this

a.

The

The starting torque The temperature rise

c.

full

hp,

standstill

less thick.

1

1

83 r/min, 575 V, 3-phase, 60

320

V

is

3-36

The

stator of a

600

that the

RMS

about 0.6 V. Estimate

the no-load speed of the motor. 1

3-38

The

rotor of a

60 hp,

1

760 r/min, 60 Hz

at full-load

induction motor has 1

known

It is

brush voltage drop

load speed of the motor

Hz

at

between the open-circuit

lines of the rotor.

have on the following:

b.

A 25

density

T

0.54

wound-rotor induction motor produces

in a lathe,

cutting off the cooling fins and also a portion of the rings,

The number of coils on the stator The number of coils per phase The number of coils per group The coil pitch (in millimeters) The area of one pole The flux per pole if the average flux

hp,

1

1

60 r/min, 575

V,

3-phase, 60 Hz induction motor has 90 slots, an internal diameter of 20 inches, and an axial length of 16 inches.

ter

of

1

1

1

17 bars

and a diame-

inches. Calculate the average

force on each bar (in newtons)

motor

is

running

at full-load.

when

the

Chapter 14 Selection

and Application

of

Three-Phase Induction Motors

be replaced by that of any other manufacturer,

14.0 Introduction

When purchasing a 3-phase induction motor we

a particular application, several types can

have to

make

fill

The

for

shaft height, or the type of coupling.

often discover that

selection

is

establishes limiting values for electrical,

generally

simplified because the manufacturer of the lathe, fan,

pump, and so

forth indicates the type of

motor

best suited to drive the load. Nevertheless, ful to

know something about

it

that is

is

use-

must

satisfy

also cover

some

to

motors

Motors are grouped

Standardization and classification of

1.

*

Manufacturers

Motors and Generators. Standards

publication in

Canada

Drip-proof motors. The frame

motors

of one manufacturer can

(NEMA)

which they have

We limit our discussion to five important

and solid particles which

in a

MG-I

fall at

drip-proof

liquid drops

any angle between

vertical. The means of a fan directly couCool air, drawn into the motor

are cooled by

pled to the rotor.

Standards in the United Slates are governed by National Electrical

in

0 and 15 degrees downward from the

motors under 500 hp

have standardized dimensions. Thus, a 25 hp,

Hz motor

categories, de-

into several

motor protects the windings against

induction motors*

725 r/min, 60

environment and

pending upon the environment

in general.

to operate.

1

rise.

cooling methods

classes.

industrial

to starting

14.2 Classification according

special appli-

interesting devices will enable the reader to gain a better understanding of induction

all

requirements as

in-

nous generators and frequency converters. These

The frames of

minimum

locked-rotor current, overload capacity,

and temperature

cations of induction machines, such as asynchro-

14.1

mechan-

and thermal characteristics. Thus, motors

the basic construction

and characteristics of the various types of 3-phase

we

stan-

ical,

torque,

duction motors that are available on the market. In this chapter

The

dardization covers not only frame sizes, but also

we

the need. Consequently,

a choice.

without having to change the mounting holes, the

governed by Canadian Standards Association (CSA) publication

titled

are similarly

299

C

154.

The two standards

arc essentially identical.


300

through vents

in the

frame,

is

and then expelled. The

ings

temperature

rise

resistance)

ing

blown over

the wind-

4.

maximum

allowable

and high-power motors

(measured by the change

may

be 6()°C

in

wind-

80°C 105°C

125°C, depending on the type of insulation used the windings. Drip-proof motors (Fig. 14.1

used 2.

in

)

ribbed motor frame (Fig.

in

a splash-

proof motor protects the windings against liquid fall at

any angle be-

tween 0 and 100° downward from the temperature

rise

These motors are mainly used 3.

vertical,

similar to that in drip-proof motors and

maximum

the

to the shaft,

in

is

An

external

air

over the

1

4.3).

A

concentric outer

shield prevents physical contact with the fan and

serves to channel the airstream.

drops and solid particles that

is

coupled

in

can be

air.

blows

usually cooled by an external blast of fan, directly

Medium-

enclosed are

that are totally

or

most locations.

Splash-proof motor. The frame

Cooling

Totally enclosed, fan-cooled motors.

perature rise 5.

is

the

same

The permissible tem-

as for drip-proof motors.

Explosion-proof motors. Explosion-proof mo-

tors are

used

highly inflammable or explosive

in

surroundings, such as coal mines, grain elevators.

They

oil refineries,

and

are totally enclosed (but not

also the same.

wet locations.

Totally enclosed, nonventilated motors.

These

motors have closed frames that prevent the free

exchange of

air

between the inside and the outside

They are designed for very wet and dusty locations. Most are rated below 10 kW beof the case.

cause

it

is

difficult to get rid

of the heat of larger

machines. The motor losses are dissipated by natural

convection and radiation from the frame. The

permissible 1

temperature

rise

is

65°C,

85°C,

10°C, or I30°C, depending on the class of insu-

lation (see Fig. 14.2).

Figure 14.2

Two

enclosed nonventilated (TENV) 2 hp, 1725 cage motors are shown in foreground and two 30 hp, 1780 r/min totally enclosed blower-cooled motors (TEBC) in background. These 3-phase, 460 V motors are intended to operate at variable speeds totally

r/min

ranging from a few revolutions per minute to about 3

times rated speed.

The 2 hp motors have a full-load current of 2.9 A, and power factor of 76 per-

efficiency of 84 percent

cent. Other characteristics: no-load current:

locked-rotor current: pu; tal

26

1

.7 A;

A; locked-rotor torque: 4.2

breakdown torque: 5.0

pu; service factor: 1.0; to-

weight: 39 kg; overall length including shaft:

mm;

overall height:

The 30 hp motors have a efficiency of

377

235 mm. 34 A, 84 per-

full-load current of

93 percent, and power factor

of

cent. Other characteristics: no-load current: 12 A;

tion

Energy efficient drip-proof, 3-phase squirrel-cage inducmotor rated 230 V/460 V, 3 hp, 1750 r/min, 60 Hz.

locked-rotor current: 214 A; locked rotor torque: 1 .6 pu; breakdown torque: 2.84 pu; service factor: 1 .0; total weight: 200 kg; overall length including shaft: 834 mm; overall height: 365 mm.

{Courtesy of Gould)

(Courtesy of Baldor Electric

Figure 14.1

Company

)

SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS

301

14.3 Classification according to electrical

and mechanical

properties enclosures just mentioned,

In addition to the various

3-phase squirrel-cage motors can have special electrical

and mechanical characteristics, as

listed

below.

Motors with standard locked-rotor torque (NEMA Design B). Most induction motors belong to this group. The per-unit locked-rotor torque decreases as the size of the motor increases. Thus, it ranges from .3 to 0.7, as the power increases from 20 hp to 200 hp (15 kW to 150 kW). The corre/.

l

Figure 14.3 Totally

enclosed fan-cooled induction motor rated 350

440 (Courtesy of Gould hp,

1760

r/min,

V,

sponding locked-rotor current should not exceed 6.4 times the rated full-load current. These general-

3-phase, 60 Hz.

purpose motors are used to drive fans, centrifugal

pumps, machine 2.

and so

tools,

forth.

High starting-torque motors

(NEMA Design C).

These motors are employed when

Pumps and

are difficult. that

have

to start

starting conditions

piston-type compressors

under load are two typical applica-

tions. In the

range from 20 hp to 200 hp, the locked-

rotor torque

is

200%

of full-load torque, which cor-

responds to a per-unit torque of

The locked-rotor

2.

current should not exceed 6.4 times the rated full-

load current.

equipped with

In general, these motors are

double-cage

rotor.

double-cage rotor

lowing Figure 14.4

a.

Totally enclosed, fan-cooled, explosion-proof motor.

Note the particularly rugged construction

of this type

b.

facts:

The frequency of the rotor as the motor speeds up

A conductor that

of motor.

(Courtesy of Brook Crompton-Parkinson Ltd

(cage )

airtight)

the

and the frames are designed

enormous pressure

that

may

to an internal explosion.

the flanges

on the end-bells are made extra long

Furthermore, in

order to cool any escaping gases generated by such

may

permissible temperature rise tally

the

is

enclosed motors (see Fig.

1

be initiated by

same

4.4).

lies

close to the rotor surface

has a lower inductive reactance than

The

as for to-

The conductors of cage

1

are

When the motor is connected

build up inside the

motor due

an explosion. Such explosions

)

much

2)

smaller

than those of cage 2

to withstand

the spark or a short-circuit within the windings.

1

current diminishes

one buried deep inside the iron core (cage c.

a

The excellent performance of a (Fig. 14.5) is based upon the fol-

to the line with the

rotor at standstill, the frequency of the rotor current is

equal to line frequency.

Owing

ductive reactance of squirrel-cage

to the high in2, the rotor cur-

rent flows mainly in the small bars of cage

effective

motor resistance

is

essentially equal to that of cage

high starting torque

is

1.

The

therefore high, being 1.

developed.

Consequently,

a


302

Figure 14.5 Typical torque-speed curves of

NEMA

design B, C, and

D

minimum NEMA Hz squirrel-

motors. Each curve corresponds to the

values of locked-rotor torque, pull-up torque, and breakdown torque of a 3-phase 1800 r/min, 10 hp, 60

cage induction motor. The cross-section

As falls,

the

of the respective rotors indicates the type of rotor bars

motor speeds up, the rotor frequency

with the result that the inductive reactance of

tors are usually

windings

that the reactance of both

The

rotor current

tance of cage

1

so low (typically

is

is

Hz)

I

negligible.

then limited only by the resis-

is

and cage 2 operating

in parallel.

The

rated speed

is

1

,

the effective rotor resistance at

much lower than

at standstill.

For

this

reason the double-cage rotor develops both a high starting torque

and a low

slip at full-load.

Despite their high torque. Design

recommended

not

reason

is

that

punch holes

for starting high-inertia loads.

3.

in

cage

1

.

Owing to its small may melt.

The

start-

size,

tends to overheat and the bars

High-slip

(NEMA

motors

rated speed of high-slip, Design lies

between

85%

and

95%

These motors are used

in sheet metal.

ates the operation, a clutch

When

the

Punching a hole requires a tremendous amount son

is

that the

punching energy

fraction of a second.

punch does

The energy

delivered in a

is

is

furnished by the

in a

itself.

As

the

work, the speed of the flywheel lot

of kinetic

very short time. The speed of the motor

also drops considerably, along with that of the fly-

wheel. However, the Class

D

will not

The high-remade of brass, and the mo-

its

drops immediately, thus releasing a

energy

the

drawn from

exceed

As soon

of synchronous speed.

to accelerate high -inertia

initi-

engages the flywheel,

of power, sometimes exceeding 1000 hp. The rea-

the current

motors usually

worker

is

that

causing the punch to descend and pierce the sheet.

Design D). The

loads (such as centrifugal dryers), which take a rel-

motor

its

design ensures that

rated value.

as the hole

is

D

the line at the lower speed

is

pierced, the only load on

the flywheel,

which

is

now

gradually

brought back up to speed. During the acceleration

atively long time to reach full speed.

period, the

sistance squirrel cage

thus restoring the energy

is

machine tools

flywheel rather than by the motor

motors are

most of the rotor tR losses during

up are concentrated it

C

to

large drop in speed with increasing load

also ideal to drive impact-type

Because the conductors of cage 2 are much larger than those of cage

designed for intermittent duty

prevent overheating.

both squirrel-cage windings diminishes. At rated

speed the rotor frequency

used.

motor delivers energy it

lost

to the flywheel,

during the impact.

A


powerful motor will quickly accelerate the

motor, and

fly-

wheel, permitting rapid, repetitive operation of the

punch is

press.

low, a

On

much

the other hand,

if

2. it

and and

D it

the characteristics of

motors.

The

(as a percentage of full-load

Fig. 14.5 enable us to

NEMA

rotor construction

of equal power.

Design B, C, is

if

increased (by using brass instead

of copper or aluminum), the locked-rotor torque creases, but the speed at rated torque

is

By way

also shown,

obtained by changing the rotor design. For example, is

always greater

torque) than that of a similar low-speed motor

can be seen that the distinguishing properties are

the rotor resistance

in-

two 10

motor speed

The choice of motor speed

synchronous speeds. The difference

would

is

rather limited because

by quantum jumps, depending upon the frequency and the number of poles. For example, sible to build a

and running

ing an acceptable efficiency say,

it

impos-

is

conventional induction motor hav-

of 2000 r/min on a 60

Hz

supply.

at a

speed,

totally en-

in price

justify the use of a high-speed

alone

motor and a

900

to drive a load operating at, say,

speeds (100 r/min or

r/min.

has to operate at very low a gearbox

less),

mandatory.

is

The gears are often an integral part of the motor, making for a very compact unit (Fig. 14.6).

synchronous speed of induction motors changes

the

60 Hz,

hp, 3-phase,

closed, fan-cooled induction motors having different

gearbox

lower.

of example. Table 14A compares the

properties of

When equipment 14.4 Choice of

factor are

The locked-rotor torque of a high-speed motor is

The torque-speed curves of

power

will

only take longer to bring the flywheel up to speed.

compare

efficiency and

higher.

the repetition rate

smaller motor will suffice;

its

303

A

gearbox

is

when equipment

also mandatory

has to run above 3600 r/min. For example, in one industrial application a large gear unit

is

used to

5000 r/min centrifugal compressor a 3560 r/min induction motor.

drive a 1200 hp,

coupled to

Such a motor

would require 2 poles and a corresponding synchronous speed of 3600 r/min. The

2000)/3600

=

slip

of (3600

-

0.444 means that 44.4% of the power

14,5 The

Two-speed motors

stator of a squirrel-cage induction

supplied to the rotor would be dissipated as heat.

designed so that the motor can operate

(See Section

ent speeds.

1

3. 13.)

The speed of a motor the speed of the for

machine

low-speed machines,

a high-speed rectly

is

it

obviously determined by it

has to drive. However,

is

motor and a gearbox instead of

.

di-

coupling a low-speed motor to the load.

There are several advantages 1

often preferable to use

to using a

is

to

wind the

pumps. One way stator with

that only

one winding

only half the copper

gearbox:

is

in

Power

being utilized.

However, special windings have been invented

whereby the speed

TABLE 14A

is

changed by simply changing

The synchronous

COMPARISON BETWEEN TWO MOTORS OF DIFFERENT SPEEDS Synchronous

Power

speed

factor

drill

obtain two

operation at a time and so

the external stator connections.

of a low-speed

to

two separate windThe problem is

in the slots is

high-speed motor

less than that

differ-

ings having, say, 4 poles and 6 poles.

For a given output power, the size and cost of a is

two

Such motors are often used on

presses, blowers, and

speeds

motor can be at

Efficiency

Locked-rotor

Mass

torque

Price

(2002)

hp

kW

r/min

%

9c

%

kg

U.S. $

l()

7.5

3600

89

90

150

50

650

10

7.5

900

75

85

1

70

2000

25

1


304

ac source

Figure 14.6 Gear motor rated at 2.25 kW, 1740 r/min, 60 Hz. The output torque and speed are respectively 172 N-m and 125 r/min. (Courtesy of Reliance

speeds

obtained

Electric)

usually

are

speed

the

in

(3600/1800 r/min, 1200/600 r/min,

etc.).

ratio

2:1

The lower

produced by the creation of consequent

is

poles.

Consider, for example, one phase of a two-pole,

3-phase motor (Fig, 14.7a).

connected

in series to a

flows into terminal of terminal

As

2.

When

and current

1

two poles

the

are

60 Hz ac source, current

a result, one

N

I2

(=

l\)

I {

tlows out

pole and one S pole

Figure 14.7

Two

a.

are created

and the flux has the pattern shown. The

synchronous speed

=

ns

is

120f/p

that

stator

This

120

X

60/2

percent of the pole-pitch.

shown rent

/,

minal

now

in Fig.

As a

windings.

When

the coils are connected

motor

is

produced. Two

Because every

in parallel, as

14.7b. In this case, at the instant cur-

result,

two

S pole, the

connect the two poles

flows into terminal 2.

series produce a

of the

in parallel,

a 4-pole

poles are conse-

each pole covers only one-quarter of the

achieved by using a coil pitch equal to 50

Let us

in

r/min

circumference instead of the usual one-half.

is

connected

quent poles.

= 3600 Note

b.

-

short-pitch coils

two-pole motor.

1

N

,

current

I2

flows into

ter-

poles are created by the

it

two

nious

N

pole must be accompanied by a

follows that two S poles will appear between

N poles. The

way

south poles created in this inge-

are called consequent poles.

nection produces 4 poles in

speed

is

all,

The new con-

and the synchronous

1800 r/min. Thus, we can double the number

of poles by simply changing the stator connections. is

upon

this principle that

2-speed motors are

built.

It


now

again produces 4 poles per phase, but

possess the same polarity (Fig.

Two-speed motors have a

1

they

all

4.8b).

relatively lower effi-

ciency and power factor than single-speed motors

They can be designed

do.

develop

to

both

(at

speeds) either constant power, constant torque, or variable torque. that has to

The choice depends upon

The 2-speed motors described so tios

of

2:

l

.

the load

be driven.

If

the

motor drives a

The reason

big a change in speed.

far

have pole

fan. this is

may

that the

ra-

be too

power of

a fan varies as the cube of the speed. Consequently, if

the speed

is

reduced by

one-eighth, which

To overcome

power drops

half, the

to

often too low to be of interest.

is

this

problem, some 3-phase wind-

ings are designed to obtain lower pole ratios such as 8/10,

1

4/ 6, 26/28,

10/ 4,

1

1

larly useful in driving

and 38/46. These pole

PAM, motors

amplitude modulation, or

2-speed fans

horsepower range and more. moderate reduction

in fan

PAM

are particuthe

in

hundred

motors enable a

power by simply recon-

necting the windings to give the lower speed.

motor characteristics under

14.6 Induction

various load conditions The complete torque-speed curves displayed 1

Figure 14.8 a. High-speed connection of a 3-phase stator, yielding 4 poles. b. Low-speed connection of same motor yielding

4.5 are important, but

most of the time a motor runs

Figure

1

4.8

shows

the stator connections for a

2-speed, 4-pole/8-pole,

ing.

to

I

to 6, are

3-phase motor. Six leads,

brought out from the stator wind-

For the high-speed connection, power

is

applied

terminals 1-2-3, and terminals 4-5-6 are open.

resulting delta connection produces

having two

Tn

two S poles connected

(Fig.

.

It

so happens that between

circuiting terminals 1-2-3

The

1

4.8a).

Note

is

made by

and applying power

resulting

4.9).

is

essentially a

The slope of

pends mainly upon the rotor resistance

the line de-

—

the lower

the resistance, the steeper the slope. In effect,

the slip

tance

R

s,

it

can be shown that

torque

T,

at rated

line voltage E,

frequency,

and rotor

resis-

are related by the expression

4 poles per phase

= kTR/E 2

(14.

1)

that

where k

in series.

The low-speed connection minals 4-5-6.

1

s

N and

the four poles are

The

that

close to synchro-

nous speed, supplying a torque that varies from zero to full-load torque

straight line (Fig.

numbered

at

these limits the torque-speed curve

8 poles.

in Fig.

must be recognized

it

shortto ter-

double-wye connection

is

a constant that

depends upon the con-

struction of the motor.

This expression enables us

formula showing

how

to establish a

simple

the line voltage and rotor


306

Figure 14.9

The torque-speed curve

is

resistance affect the behavior of the load. In effect,

motor

once

between the no-load and rated torque operating

essentially a straight line

we know

motor under

the characteristics of a

for a given load condition,

we can

predict

its

speed, torque, power, and so on, for any other load condition. These quantities are related by the formula

an accuracy of better than 5 percent which cient for

most

suffi-

Example 14-1

A

V

3-phase, 208

induction motor having a syn-

chronous speed of 1200 r/min runs to a

V

215

line

stant torque load. Calculate the

A-

is

practical problems.

when connected

Ax

points.

increases to

240

at

1

140 r/min

and driving a con-

speed

if

the voltage

V.

where n

=

Solution subscript referring to the

initial,

conditions (the given conditions

spond

to the

=

subscript referring to the

s

=

slip

E=

may

The

slip at

215 Vis

corres

nominal rating of the motor)

x

T= R =

or given load

new

load conditions

=

(n s

=

(1200

=

0.05

n)ln s

-

11

40)/ 1200

torque [N-m]

When the voltage

rotor resistance

rises to

stator voltage [V]

applying the formula, the only restriction

.

V, the load torque

we can

write

is

new torque Tx must not be greater than Tn (EJEn ) 2 Under these conditions Eq. 14.2 yields

that the

240

and

rotor resistance remain the same. Consequently, in

applying Eq. 14.2, In

—

.v

x

=

sn

=

0.04

(E n /Ex )

2

=

0.05 (215/240)

2 -


The

slip

speed

Example 14-3

therefore,

is,

X 1200 = 48

0.04

A 3-phase, 4-pole

r/min

rating of

The new speed nx

=

240

at

1200

V

- 48 =

1152 r/min

Under

is

Hz

line.

The

it

wye

across the

Nm

speed of 1000 r/min.

at a

N

Calculate the speed for a torque of 400

b.

Calculate the value of the external resistors so

kW

at

200

1000)/ 1800

=

0.444

motor develops 10

that the

m.

r/min.

Solution

to a.

The given conditions

are

machine has run for several

after the

Tn = 300 N

hours.

.v

Calculate

The hot

in

is

initial

23°C. The speed drops

connected

these conditions the motor de-

a.

com-

8-pole induction motor driving a

connected to a fixed 460 V, 60

rotor resistance in terms of the cold re-

n

torque of 400

The approximate hot temperature of the bars, knowing they are made of copper

m

- (1800 -

All other conditions being fixed,

sistance b.

wound-rotor induction motor has a

kWJ 760 r/min, 2.3 kV, 60 Hz. Three ex-

velops a torque of 300

cold rotor temperature

a.

0

1

rotor slip-rings.

pressor runs at 873 r/min immediately after

864 r/min

1

ternal resistors of 2 f i are

is

Example 14-2

A 3-phase,

307

we have

for a

N m the following:

rotor -v

x

= -

sn

(TJTn ) = 0-444 (400/300)

0.592

Solution a.

The synchronous speed ns

The

=

120

initial

=

///;

and

120

X

= 900

slip

speed

(900

-

873)/900

=

0.03

sx

=

(900

-

864)/900

=

0.04

speed change

is

in rotor resistance.

x

=

0.04

=

.s

due

entirely

.v

n

to the

change

b.

is

The hot

R,

734 r/min

1066

kW at 200

is

9.55 Pin 9.55

X

(3.5)

10 000/200

= 478 N-m

)

The greater than the

rated torque 7Vaicd

=

is

9.55 Pin

-

R

(234

9.55 X 110 000/1760 = 597 N m

is

+ T

)

- 234

(6.5)

Because Tx

is

less than r, atcd

,

we can

apply

x

Eq. 14.2.

\

=

1066 r/min.

is

=

rotor temperature

t,=

=

to 10

r/min

cold rotor resistance. b.

1800

load.

=

33%

-

=

The torque corresponding

Tx =

1.33 /?„

hot rotor resistance

-

1800

734 r/min with increasing

therefore write

(RJR n

X

speed drops from 1000 r/min to

that the

(R x /R n )

0.03

Rx =

Note

are fixed; consequently,

We can

0.592

r/min n

=

=

Consequently, the motor speed

final slips are

The voltage and torque

The

60/8

n

.v

the

The

is:

1.33 (234

= 108°C

+

23)

- 234

The

slip is sx

= (1800 -

200)/ 1800

=

0.89


308

we

All other conditions being fixed,

have, from

Rule

1

The

-

heat dissipated in the rotor during

the starting period (from zero speed to final rated

Eq. 14.2: sx

=

*n

speed)

(TJTn ) (RJR n )

all

0.89

=

0.44 (478/300) (RJ2)

is

equal to the final kinetic energy stored

This rule holds

true, irrespective

of the stator volt-

age or the torque-speed curve of the motor. Thus,

and so

in

the revolving parts.

if

a motor brings a massive flywheel up to speed, and

rx = Three 2.5

ft

n

wye-connected

tor circuit will

10

2.5

if

the energy stored in the flywheel

joules, the rotor will have dissipated

resistors in the ro-

enable the motor to develop

kW at 200r/min.

the

form of heat. Depending upon the

tor

and

its

is

then 5000

5000

joules in

size of the ro-

cooling system, this energy could easily

produce overheating.

14.7 Starting an induction

motor 14.8 Plugging an induction motor

High-inertia loads put a strain on induction motors

some

because they prolong the starting period. The starting

In

current in both the stator and rotor

and

is

high during

this

that the revolving field

lem. For motors of several thousand horsepower, a

prolonged starting period

may even

is

installed.

many

loads.

To

The

line voltage

may

fall

overload the

motor

riod, the It

below normal

chanical

relieve the problem, induction motors are

rotor.

often started on reduced voltage. This limits the

During

its

speed to

power P m

is

fall.

The associated me-

entirely dissipated as heat in the

Unfortunately, the rotor also continues to re-

P

r

the line voltage drop as well as the heating rate of the

which

windings. Reduced voltage lengthens the start-up

Consequently, plugging produces

time, but this

is

whether the start-up time

remembering

is

long or short,

the following rule for a

it

is

motor

also

from the

heat /"/?

(Fig.

is

plugged, the rotor

2 l

stator,

14. 10).

losses in the

may

is

locked.

melt the rotor bars or

overheat the stator winding. In this regard

Figure 14.10 a 3-phase induction motor

as

even exceed those when the rotor

high rotor temperatures

is

not loaded mechanically:

When

dissipated

Motors should not be plugged too frequently because

worth that

is

rotor that

usually not important. However,

oppo-

in the

plugging pe-

acts as a brake.

ceive electromagnetic power

the motor, and consequently reduces

this

absorbs kinetic energy from the still-revolving

load, causing

seconds, thus affecting other connected

power drawn by

suddenly turns

site direction to the rotor.

transmission line feeding the plant where the motor

for

motor

industrial applications, the induction

load have to be brought to a quick stop. This

can be done by interchanging two stator leads, so

becomes a major prob-

interval so that overheating

its

R losses

are very high.

it is

worth


remembering the following tions for a

Rule 2

motor

that

The

-

is

heat dissipated in the rotor during

is

three times the original kinetic energy of

the revolving parts.

all

ergy dissipated original

it

produces

In effect, the en-

only equal to the

in the rotor is

energy stored

kinetic

that

is

does plugging.

far less heat than

the plugging period (initial rated speed to zero

speed)

The advantage of dc braking

rule for plugging opera-

not loaded mechanically:

309

in

the revolving

masses, and not three times that energy. The energy dissipated

in the rotor is

independent of the

magnitude of the dc current. However, a smaller

Example 14-4 A 100 kW, 60 Hz,

dc current increases the braking time. The dc cur175 r/tnin motor

1

is

coupled

to a

flywheel by means of a gearbox. The kinetic energy of

is 300 kJ when the motor The motor is plugged to a stop

the revolving parts

all

runs at rated speed.

and allowed

to run

up

to

1

175 r/min in the reverse

two or three times the rated current of Even larger values can be used, prothe stator does not become too hot. The

rent can be

the motor.

vided that

braking torque

proportional to the square of the

is

dc braking current.

direction. Calculate the energy dissipated in the rotor if the

flywheel

is

the only load.

Example 14-5

A

Solution

During the plugging period, the motor speed drops

from

175 r/min to zero. The heat dissipated

1

rotor

is

3

erates to

X 300

kJ

= 900

nominal speed

kJ.

The motor then

kJ.

this

+ 300 =

1200

is

The

period

By reversing the speed this way, the total

dissipated in the rotor from start to finish

900

accel-

in the reverse direction.

energy dissipated in the rotor during

300

in the

25 kg

62 A. 24

heat

therefore

a.

14.9 Braking with direct current high-inertia load can

also be brought to a quick stop by circulating dc

current in the stator winding.

Any two

having a

is

The dc 0.32

We

moment

of inertia of

between two

stator ter-

and the rated motor current

12,

want

total

resistance

to stop the

is

motor by connecting a

battery across the terminals.

Calculate

b.

its

V

.

is

kJ.

induction motor and

m2

minals

c.

An

50 hp, 1760 r/min, 440 V, 3-phase induction mo-

tor drives a load

The dc current in the stator The energy dissipated in the rotor The average braking torque if the stopping time is 4 min

Solution a.

The dc current

stator termi/

is

— EIR —

24/0.32

= 75 A

nals can be connected to the dc source.

The in

direct current produces stationary N, S poles

This current

The number of poles created is equal number of poles which the motor develops

to the

how

the

motor

ter-

When

the rotor

an ac voltage

is

sweeps past the stationary

induced

in the rotor bars.

The

field,

volt-

age produces an ac current and the resulting rotor 2 I

R

losses are dissipated at the expense of the

kinetic energy stored in the revolving parts.

motor

finally

comes

to rest

when

all

b.

The

kinetic energy in the rotor

and load

at

1760

is

5.48

X 10" 3

5.48

X

10" 3

- 424

kJ

Ek = =

Jn

2

X 25 X 1760

(3.8) 2

The

the kinetic en-

ergy has been dissipated as heat in the rotor.

is

short.

r/min

minals are connected to the dc source.

slightly higher than the rated

not overheat, because the braking time

normally. Thus, a 3-phase, 4-pole induction motor

produces 4 dc poles, no matter

is

current of the motor. However, the stator will

the stator.

Consequently, the rotor absorbs 424 kJ during the braking period.


310

c.

The average braking torque

7"

can be calculated

An = 1760

-

factor 1.15. The allowable temperature

10°C higher than

from the equation

tors operating at

9.55 TAt/J 9.55

T=

19.2

T X

(3.14)

X

(4

vided. This

is

because even

Abnormal conditions

14.10

in the stator,

to internal

overheating of

the bearings, etc.) or to external conditions. External

problems may be caused by any of the following: 1

.

not

much

as 125 percent, as

if

recommended

the external frame

perature of the windings

Abnormal motor operation may be due problems (short-circuit

overloads as

to carry

long as supplementary external ventilation

N-m

may

pro-

is

for long periods is

cool, the tem-

be excessive.

14.12 Line voltage changes The most important consequence of a line voltage change is its effect upon the torque-speed curve of the motor. In effect, the torque at any speed

Mechanical overload

mo-

normal load.

During emergencies a drip-proof motor can be

made

60)/25

then

rise is

that permitted for drip-proof

pro-

is

portional to the square of the applied voltage. Thus, 2.

Supply voltage changes

3.

Single phasing

4.

Frequency changes

the stator voltage decreases by

if

voltage drop to the

We

will

examine

the nature of these

problems

in

the sections that follow.

According

motor

shall

±10%

of the nominal voltage, and for any frequency within

of the nominal frequency.

If

the voltage and

frequency both vary, the sum of the two percentage

changes must not exceed 10 percent. Finally, designed

tors are

up

to

1000

to the

mo-

much

On when

sea level. At higher altitudes the the permissible limits

poor cooling afforded by the thinner

due

air.

Mechanical overload

as twice their rated

power

for short periods,

As may

the line.

less than its rated value.

the motor

is

is

too high

running, the flux per pole will be at full-load, this

current, with the result that the temperature increases slightly

and the power factor

is

slightly reduced.

the 3-phase voltages are unbalanced, they can

produce a serious unbalance of the three rents.

line cur-

This condition increases the stator and rotor

losses, yielding a higher temperature.

Although standard induction motors can develop as

much

drawn from

the other hand, if the line voltage

A voltage

un-

3.5% can cause the temperature to increase by 15°C. The utility company should be notified whenever the phase-to-phase line voltages differ by more than 2 percent. balance of as

14.11

often produced during start-up, due

increases both the iron losses and the magnetizing

If

may exceed

is

starting current

above normal. For a motor running

to operate satisfactorily at altitudes

m above

temperature

all

heavy

at

A line

a result of the lower voltage, the starting torque

be

to national standards, a

operate satisfactorily on any voltage within

±5%

10%, the torque

every speed will drop by approximately 20%.

little

as

they should not be allowed to run continuously be-

yond

their rated capacity.

heating,

Overloads cause over-

which deteriorates the insulation and

duces the service

life

of the motor. In practice, the

overload causes the thermal overload relays starter

fore

its

box

re-

to trip, bringing the

motor

in the

to a stop be-

temperature gets too high.

Some

drip-proof motors are designed to carry a

continuous overload of 15 percent. This overload capacity

is

shown on

the nameplate by the service

14.13 Single-phasing If

one

or

if

line

of a 3-phase line

a fuse

ning, the

is

accidentally opened,

blows while the 3-phase motor

machine

is

run-

will continue to run as a single-

phase motor. The current drawn from the remaining

two

lines will

almost double, and the motor will be-

gin to overheat.

motor

will

The thermal

eventually

trip

relays protecting the the

circuit-breaker,

thereby disconnecting the motor from the

line.


%

311

breakdown torque

250

'X

200 3-phase

P"

^

150

i

^pull-up torq ue nominal torque T

100

full

—

load

i\ i

50

sirigle-

phase

—

i i

i

\

\ \

\

\

!

\

i

40

20

80

60

100

%

speed \

Figure 14.11 Typical torque-speed curves

when

a 3-phase squirrel-cage motor operates normally and

when

it

operates on

single-phase.

The torque-speed curve a 3-phase

down

seriously affected

torque decreases to about

value, all.

is

when

motor operates on single phase. The break-

40%

and the motor develops no

of

its

original

starting torque at

Consequently, a fully loaded 3-phase motor

simply stop

if

one of its

lines

is

resulting locked-rotor current

normal 3-phase

enough

LR

current.

may

suddenly opened. The is

about

It

is

to trip the circuit breaker or to

90%

of the

therefore large

blow the

1

4. II

each other closely single-phase

until the

breakdown

is

50 Hz may cause problems when they

nected to a 60

torque approaches the

in

some

to gear

1

large distribution system, except during a

is

major dis-

However, the frequency may vary electrical

signif-

energy

generated by diesel engines or gas turbines. The in a hospital, the electrical

system on a ship, and the generators

camp, are examples of

this

is

in

a

lumber

type of supply.

The most important consequence of change

may

20%

not be acceptable

we

either have

motor speed or supply an expen-

50 Hz source. well on a 60

Hz

line,

but

terminal voltage should be raised to 6/5 (or

its

of the nameplate rating. The

20%)

torque

is

new breakdown

then equal to the original breakdown

torque and the starting torque

is

only slightly re-

remain satisfactory.

A 60 Hz

Important frequency changes never take place on a

emergency power supply

this

duced. Power factor, efficiency, and temperature

torque.

on isolated systems where

and

A 50 Hz motor operates

rise

icantly

the

are con-

system. Everything runs

applications. In such cases

down

sive auxiliary

14.14 Frequency variation

turbance.

Hz

faster than normal,

fuses.

shows the typical torque-speed curves of a 3-phase motor when it runs normally and when Note that the curves follow it is single-phasing. Fig.

Machine tools and other motor-driven equipment imported from countries where the frequency

a frequency

the resulting change in motor speed:

if

the

frequency drops by 5%, the motor speed drops by 5%.

but

its

motor can also operate on

83%) of its nameplate and fore,

a

50 Hz

line,

terminal voltage should be reduced to 5/6 (or value.

The breakdown torque same as be-

starting torque are then about the

and the power

ture rise

factor, efficiency,

and tempera-

remain satisfactory.

14.15 Induction motor operating as a generator Consider an

electric train

cage induction motor that wheels.

As

powered by a is

directly

the train climbs up a

hill,

squirrel-

coupled the

to the

motor

will


312

Figure 14.12

The

makes

electric train

the round

trip

be-

tween Zermatt (1604 m) and Gornergrat (3089 m) in Switzerland. The drive is provided by four 3-phase wound-rotor induction motors, rated 78 kW, 1470 r/m, 700 V, 50 Hz. Two aerial conductors constitute phases A and B, and the rails provide phase C. A toothed gear-wheel 573 mm in diameter engages a stationary rack on the roadbed to drive the train up and down the steep slopes. The speed can be varied from zero to 14.4 km/h by means of variable resistors

in

the rotor

circuit.

The

rated thrust

is

78 kN. (Courtesy of ABB) run

than synchronous speed, devel-

at slightly less

overcome both

oping a torque sufficient

to

and the force of

At the top of the

gravity.

level ground, the force of gravity

friction hill,

to overcome the The motor runs at

and the motor has only

into play

friction of the rails light load

and the

and very close

What happens when

to

air.

move

begins

it

to

rotate

is

to a

it

gasoline engine (Fig.

as the engine speed exceeds the

synchronous speed, the motor becomes a generator, delivering active it

is

power P

to the electrical

system

connected. However, to create

its

to

mag-

in-

speed. Thus, a higher engine speed produces a

mo-

develops a counter torque that opposes the

same

As soon

effect as a

coupled to the

speed. However, as soon as this takes place, the tor

to

above synchronous

crease in speed. This torque has the

and coupling

line

14.13).

motor has to absorb reactive power power can only come from the ac line, with the result that the reactive power Q flows in the opposite direction to the active power P (Fig. 14. 3). The active power delivered to the line is directly proportional to the slip above synchronous

the train begins to

accelerate and because the motor

phase

which

synchronous speed.

downhill? The force of gravity causes the train

wheels,

on

no longer comes

We can make an asynchronous generator by connecting an ordinary squirrel-cage motor to a 3-

netic field, the

Q. This

1

brake. However, instead of being dissipated as heat,

mechanical braking power

the

3-phase line

in the

is

3-phase system

returned to the

form of electrical energy.

An

in-

P

duction motor that turns faster than synchronous

speed

acts, therefore, as a generator,

it

^

gasoline

converts the

engine

mechanical energy

and

this

chine

is

energy

receives into electrical energy,

returned to the

induction

line are rarely

motors

Such

a

ma-

a

motor

In cranes, for

to run

running

off

it

to the line.

the

G

squirrel-cage

induction motor

applications that

above synchronous speed.

motor receives power from

f

a

example, during the lowering cycle,

"load" and returns

ooo QHill

used to drive trains (Fig.

14. 12), there are several industrial

may cause the

line.

called an asynchronous generator.

Although 3-phase

is

it

mechanical

Figure 14.13 Gasoline engine driving an asynchronous generator

connected

to a

3-phase

opposite directions.

line.

Note that

Pand Qflow

in


313

Solution a.

The apparent power drawn by when it operates as a motor is

= = =

S

1.73

X 440 X

31.2

kVA

=

for

put

reached

is

very small

at

3%. The reactive power may be supplied by a group of capacitors connected to the terminals of the moWith

this

is

slightly

When least

r/min

/=

produces

pni\2{)

=

4

a

at

frequency

a speed of

slightly

X 2400/120 = 80

The terminal voltage of with the capacitance, but

its

less

the

less

connected

reactive

when

power

as

2400 than

Consequently, the capaci-

/c

phase

is

= QIE = 5700/440 =

13

A

magnitude

is

limited by is

The capacitive reactance per phase

insuffi-

Xc ~ Ell = = 34 il

at least as

the machine normally ab-

is

440/13

The capacitance per phase must be

at least

operating as a motor. l/2ir/X c

=

1/(2tt

We

=

78 [xF

40 hp, 1760 r/min, 440 V, 3-phase squirrel-cage induction motor as an asynchronous to use a

The

motor 84%.

rated current of the

and the full-load power factor

is

is

41 A,

Figure 14.15 shows

connected. Note that tive

a.

The voltage

in delta.

Example 14-6

generator.

2

because the capacitors are

C= wish

26.2

the generator increases

capacitor bank must be able to supply

much

2

5.7 kvar per phase.

440

tive current per

generator voltage will not build up. The

sorbs

is

V

is

Hz.

saturation in the iron. If the capacitance cient, the

power absorbed

machine operates as an asynchronous

17^-3 =

per phase

than that corresponding to the speed of rotation.

Thus, a 4-pole motor driven

reactive

generator, the capacitor bank must supply at

arrangement we can supply a 3-phase

The frequency generated

kW

1

load without using an external 3-phase source (Fig. 14. 14).

is

0.84

= V31.2 2 = 7 kvar

slips, typically less

than

tor.

26.2

Q = \ S2 - P

However, the rated out-

41

power absorbed

active

P = S cos B = 31.2 X

The corresponding greater electrical output.

machine

V3 EI

The corresponding Figure 14.14 Capacitors can provide the reactive power for any asynchronous generator. This eliminates the need a 3-phase external source.

the

Calculate the capacitance required per phase

X 60 X

how if

34)

the generating system

is

the load also absorbs reac-

power, the capacitor bank must be increased to

if

provide

it.

the capacitors are connected in delta. b.

At what speed should the driving engine run generate a frequency of 60

Hz?

to

b.

The driving engine must

turn at slightly

more

than synchronous speed. Typically, the slip


314

14.16 Complete torque-speed characteristic of an induction machine 30

= =

Q

7kVar

kW t

^

OOP 4D=

1840 r/min

I

^>25 kW

On

j^r

B

We

have seen

that a

3-phase squirrel-cage induction

motor can also function as a generator or

as a brake.

— motor,

generator,

These three modes of operation and brake

— merge

into each other, as

can be seen

from the torque-speed curve of Fig. curve, together with the adjoining


grams,

14. 16.

power flow

illustrates the overall properties

This dia-

of a 3-phase

squirrel-cage induction machine.

should be equal to the full-load

machine operates as slip

=

when

the

a motor. Consequently,

1800

= 40 The engine should

slip

-

1760

r/min

We see,

for example, that when the shaft turns in same direction as the revolving field, the induction machine operates in either the motor or the generator mode. But to operate in the generator the

mode, the

therefore run at an approxi-

must turn

mate speed of n

shaft

must turn

faster than

at less

than synchronous speed.

Finally, in order to operate as a brake, the shaft

- 800 + 40 = 1840 r/min 1

must

turn in the opposite direction to the revolving

flux.

V-

u BRAKE

synchronous

speed. Similarly, to operate as a motor, the shaft

n m

r

|

speed

stator

rotor

+ 2 h

i rotor

stator

T = torque developed

stator

n = speed of rotation

by the machine

Figure 14.16 Complete torque-speed curve

of

a 3-phase induction machine.

n s = synchronous speed of the revolving field


14.17 Features of a wound-rotor induction motor

sult,

it

imum

possible for the motor to develop

is

far,

we have directed our attention

cage induction motor and generator and brake.

motor

is

the

its

to the squirrel-

that this type

is

of

in industry.

However, the wound-rotor induction motor has cer-

trial

1

applications.

make

attractive in special indus-

it

These may be

listed as follows:

final

We

have already seen

that for a given load, an in-

crease in rotor resistance will cause the speed of an

induction motor to

we

Thus, by varying the exter-

fall.

wound-rotor motor we can obtain

want, so long as

nous speed. The problem

Frequency converter

We now examine

in the shortest

14.19 Variable-speed drives

any speed Variable-speed drives

3.

speed can be reached

possible time.

nal resistors of a

Start-up of very high-inertia loads

.

2.

max-

related properties as a

The reason

one most frequently used

tain features that

its

torque during the entire acceleration period.

Thus, the

So

315

as heat in the resistors

it

is

that the

is

makes

for a very inefficient

system, which becomes too costly

these applications.

below synchro-

power dissipated

when

the motors

have ratings of several thousand horsepower.

14.18 Start-up of high-inertia loads

around

this

problem by connecting the

We get

slip-rings to

an electronic converter. The converter changes the

We

recall that

whenever a load

is

brought up to

speed by means of an induction motor, the energy dissipated in the rotor

is

equal to the kinetic energy

power

at

system (Fig.

this

14. 17).

imparted to the load. This means that a high-inertia

speed control system

load will produce very high losses in the rotor, caus-

that

become excessively

ing

it

the

wound-rotor motor

in the

to

is

hot.

itself

is

dissipated

is

is

a result, such a variable-

very efficient,

is lost in

the

form of

in the

sense

heat.

that the external resistors

can be varied as the motor picks up speed.

14.20 Frequency converter

A conventional

remains cool.

Another advantage

power

As

The advantage of

that the heat

external resistors connected to the slip-rings.

Thus, the rotor

little

into power at line frepower back into the 3-phase

low rotor frequency

quency and feeds

As

a re-

wound-rotor motor may be used as

a frequency converter to generate a frequency different

from

that

of the

utility

company. The

stator of

Figure 14.17

The water supply

in

the City of Stuttgart, Germany,

provided by a pipeline that 11 0

km

long.

Constance is

in

is

1

.6

m

in

is

diameter and

The water is pumped from Lake the Alps. The pump in the background

driven by a wound-rotor induction motor rated at

3300 kW, 425 to 595 r/min, 5 kV, 50 Hz. The variable speed enables the water supply to be varied according to the needs of the city. The enclosed motor housing seen in the foreground contains an air/water heat exchanger that uses the 5°C water for cooling purposes. During

connected to the slipup to speed the sliprings are connected to an electronic converter which feeds the rotor power back into the line. (Courtesy of Siemens) start-up, liquid rheostats are rings, but

when

the motor

is


316

R

n

R

R

wound-rotor induction

"motor"

Figure 14.18 Wound-rotor motor used as a frequency converter.

Figure 14.19 Power flow in a frequency converter when the output frequency

machine

the wound-rotor line,

and the rotor

by a motor

is

M (Fig.

is

driven

14.

1

connected at

rotor supplies

power

E 2 and

to the 3 -phase load at a voltage

both of which depend upon the

slip.

frequency / 2 Thus, accord-

Calculate a.

,

b.

we have

ing to Eqs. 13.3 and 13.4,

greater than the line frequency.

an appropriate speed

The

8).

is

to the utility

The The is

turns ratio of the stator to rotor windings

when the rotor same direction as

rotor voltage and frequency

driven

720 r/min

at

in the

the revolving field

h

=

(13.3)

sf

c.

(13.4) In general, the desired

times that of the 13.3, in

utility

frequency

is

company. According

be positive and greater than

1

.

As

is

to Eq.

rotor voltage and frequency

driven

a.

The

turns ratio

a result, the shaft

a

ing flux.

the frequency converter

is

when

to the

the rotor

revolving

Solution

must be rotated against the direction of the revolv-

The operation of

720 r/min opposite

at

field

two or three

order to attain this frequency the slip must

The

b.

The

slip at

is

= E]/Eoc = 2300/500 = 4.6

720 r/min

is

-

(1

then identical to that of an induction motor operating as a brake.

However,

the

dissipated as heat in the rotor,

power P ]n usually is

now

.v

=

available to

supply power to the load. The converter acts as a generator, and the active in

Fig.

14.19.

power flow

Note how similar

is

as

shown

this

is

to the

power flow when an induction motor runs

=

The

(;? s

n)ln,

=

0.6

rotor voltage at

720 r/min

E 2 = sEoc = = 300 V

as a

800 - 720)/l80()

0.6

is

X 500

brake (Fig. 14.16).

The

Example 14-7 A3-phase wound-rotor induction motor has of 150 hp

(—110 kW), 1760

r/min, 2.3 kV,

60 Hz. ro-

The

ro-

tor voltage

tor

is

between the slip-rings

is

500

V.

driven by a variable-speed dc motor.

is

f2 = sf= 0.6 X 60 = 36 Hz

a rating

Under locked-rotor conditions, the open-circuit

rotor frequency

c.

The motor speed is considered to be negative ( — ) when it turns opposite to the revolving field. The slip at —720 r/min is


=

(n - n)ln = (1800 - (-720))/1800 = (1800 + 720)/ 1800

s

s

The converter must

s

at a

speed of 1800 r/min. The negative sign indicates that the rotor must run opposite to the re-

volving

1.4

The

therefore be driven

317

field.

The induction motor driving

the

converter must, therefore, have a synchronous

rotor voltage and frequency at


720 r/min b.

are

E2 =

=

sE„ c

1

X 500

.4

The

rotor delivers an output of

corresponds to Ppated in the rotor,

= 700 V f2 = sf= 1.4 X 60 = 84 Hz

P

useful power delivered The power P r transferred

is

]V

to a load (Fig. 14.20).

from the

stator to the rotor

P,

60 kW. This

but instead of being dissi-

= P vh = 20

is

60/3

(13.7)

kW

Example 14-8

We

wish

rotor

60

to

use a 30

kW, 880

motor as a frequency converter

kW at an

The power

60 Hz wound-

r/min,

converter

(F) to generate c.

frequency

is

60 Hz,


The speed of the induction motor (M)

b.

The

Fig. 14. 19

active

power delivered

to the stator of the

c.

d.

Will the frequency converter overheat under

Solution

To generate

1

80 Hz the

slip

the small cop-

Fig. 14.20,

we can

see

how

must be

rotor receives 20 kW of power from the stator and 40 kW of mechanical power from the driving motor M. The rotor converts this power into 60 kW of electrical power at a frequency of 180 Hz. Induction motor M must therefore have a rating of 40 kW, 60 Hz, 800 r/min.

summary, the

1

f2 =

(13.3)

sf

180

X 60

s

load

from which

f

s

The

stator

is still

=

= 60 Hz

3

fed from the 60

Hz

line,

con-

sequently, the synchronous speed of the converter

ven

is

at a

900

r/min.

The converter must be

dri-

speed n given by s

=

(/? s

3

=

(900

-

h)//? s

-

(13.2)

frequency converter

driving

«)/900

motor

from which Figure 14.20 n

= -1800

r/min

the

into (and out of) the con-

electrical

these conditions

a.

kW plus

the stator.

verter.

In

The power of the induction motor (M)

and

power flows

active

frequency converter

in

The remaining power input to the rotor amounting to (60 - 20) = 40 kW, is derived from the mechanical input to the shaft. By referring to

that dri-

ves the frequency converter

equal to 20

per and iron losses

approximate frequency of 180 Hz (see

Fig. 14.18). If the supply-line

input to the stator of the frequency

is

See Example

14-8.

3

1

d.

ELECTRICA L MA CHINES A ND TRA NSEORMERS

8

The

because the 20 its

14-10

stator of the converter will not overheat

kW

it

absorbs

much

is

nominal rating of 30 kW. The rotor

overheat either, even though

The increased power

arises

from the

the voltage induced in the rotor

higher than

at standstill.

The

is

and the rotor stator

how much

60 kW.

fact that

14-11

more

at

A

is

By

breakdown torque and

are the

nected to a 520

180

V

line.

Explain

is

con-

how

the

following parameters are affected:

twice

effective,

a.

Locked-rotor current

b.

Locked-rotor torque

c.

No-load current

d.

No-load speed

e.

Full-load current

f.

Full-load

iron losses in the stator are normal.


g.

14-12

Practical level

50 r/min

1

line.

3-phase squirrel-cage induction motor

60 Hz, consequently, the

is

1

having a rated voltage of 575 V,

iron losses in the is

V,

locked-rotor torque reduced?

probably not overheat. The

will

frequency

is

60 Hz,

will not

rotor will be high because the frequency

normal speed, the cooling

50 hp, 440

connected to a 208 V, 3-phase

three times

Hz; however, because the rotor turns

standard squirrel-cage induction motor

less than

delivers

it

A

rated at

a.

power

factor

Full-load efficiency


if

we

eliminated the

gearbox and used another motor directly 14-1

What

is

between a drip-proof

the difference

coupled

motor and an explosion-proof motor? 14-2

What

is

the approximate

life

expectancy

of a motor? 14-3

Explain

a

NEMA Design D

unsatisfactory for driving a

14-4

Identify the

b.

14-13

why

to the load,

motor

How many

Draw

poles would the motor have?

the

pump.

motor components shown

14.5).

in

Give the values of the LR, pull-up,

and breakdown torques and the correflow of active power

3 -phase induction a. b.

14-6

As As

a

motor when

it

operates

14-14

motor

A 300

b.

14-8

14-9

to start

If the line

on such a line?

b. c.

14-15

relates to induction motors.

can bring an induction motor to a

quick stop either by plugging

from method produces the

citing the stator

in the

60 Hz

it

motor? Explain.

Which

amount of

2

I

R

the

losses.

voltage then drops to 1944 V,

The new speed, knowing

We

The new power output 2 The new I R losses in the

heat

that the load

rotor

wish to make an asynchronous generator

using a standard squirrel-cage induction tor rated at

40

(Fig. 14. 14).

or by ex-

a dc source. least

590 r/min. Calculate

torque remains the same

for the following applications:

Give some of the advantages of standard-

We

V, 3-phase,


A saw in a lumber mill A variable speed pump it

2300

if

type of ac motor would you recom-

ization as

and r/minl.

approximate value of the rotor

a.

a.

hp,

full-load speed of

Will a 3 -phase motor continue to rotate

What mend

lbf

squirrel-cage induction motor turns at a

a brake

motor be able

[ft

in a

one of the lines becomes open? Will the

14-7

squirrel-cage induction

motor, rated at 30 hp, 900 r/min (see Fig.

sponding speeds

Show

power

its

the typical torque-speed curve of a

NEMA Design C

is

Fig. 14.3.

14-5

what would

output have to be [hp]?

hp,

208

V,

mo-

870 r/min, 60 Hz

The generator

is

driven

at

2100

r/min by a gasoline engine, and the load consists of three 5 12 resistors connected in

wye. The generator voltage builds up when three 100 fxF capacitors are connected in

SELECTION AND APPLICATION OF THREE PHASE INDUCTION MOTORS

wye is

a.

across the terminals.

520

c.

d.

The

tor

e.

voltage

14-19

is

A 30 000

hp—

Advanced 14-20

best suited to drive the

60 Hz

hp, 13.2 kV, 3-phase,

A

plant.

The motor runs

at

a.

an

it

has an efficiency of 98.

LR

are respectively 0.7

1

comThe motor power

drives the


% and a

c.

d. e.

14-21

in

Which

A

The

is

of the two rotors will be the hottest,

60 Hz, drives

rotor

is

in

speed?

a belt conveyor.

connected

the slip rings

in

wye and

a. b.

is

530

V. Calculate the

The The

rotor

winding resistance per phase

resistance that must be placed in series

with the rotor (per phase) so that the motor

knowing 14-22

A

150 hp,

40 hp

at a

speed of 600 r/min,

that the line voltage

water tempera-

is

load, close to

its

is

running

The motor and compressor

in

Problem

at 1.3

14-16 are started on reduced voltage,

the acceleration period

compressor has

a

The squirrel-cage 2 000 lb -ft

a J of 18

How What

long will

is

clocked

(

1

.2 pu),

is

equal to

calculate the

b.

The magnitude of the plugging torque The moment of inertia of the rotor

In

Problem 14-22 calculate the energy

a.

motor

rotor alone has

is

that the torque exerted

14-23

.

it

dissipated in the rotor during the plugging

take to bring the motor

and compressor up b.

no-

following:

of inertia of

lb ft~ referred to the

Assuming

the starting torque

0.25 pu. The

is

moment

s.

during the plugging interval

and the average starting torque during

a.

at

synchronous speed of

reversed, and the stopping time

shaft.

kV

1200 r/min. The stator leads are suddenly

exchanger.

130 000

2.4

165 r/min, 440 V, 60 Hz,

1

3-phase induction motor

ture as the water flows through the heat

14-18

the

following;

cooled by

350 gallons (U.S.) of water through the heat exchanger per minute.

has the shortest acceleration

nominal open-circuit voltage between

circulating

Calculate the increase

motor and,

3-phase, wound-rotor induction motor

2.3 kV,

pu and 4.7 pu.

Problem 14-16

D

class

.4

this

having a rating of 150 hp, 1760 r/min,

torque and current

The full-load current The total losses at full load The exact rotor TR losses if the windage and friction losses amount to 62 kW The LR current and torque The torque developed at the compressor shaft

The motor

1

1800 r/min. Could

to

Which motor

will deliver

14-17

inertia load of

after reaching the no-load


B induction

hp. squirrel-cage. Design

1

time from zero to 1200 r/min? b.

by means of a gearbox,

The

motor deliver

if so,

exact full-load speed of 1792.8 r/min and

factor of 0.90.

[hp] can the

motor be replaced by a

a turbo compressor in a large oxygen-

at a

should be used, and what

level

kgnr, from 0

air-to-water cooled induction motor drives

pressor

line.

motor accelerates an

manufacturing

kW,

without overheating?

following gasoline engines are

which one

10

at

to be con-

approximate speed of the motor?

What power

b.

Hz

line voltage

will be the

generator?

b.

What

a.

stator current

If the

induction motor rated

nected to a 60

bank

available— 30 hp, 100 hp, and 150

14-16

A 3-phase

1450 r/min, 380 V, 50 Hz has

The approximate frequency generated active power supplied to the load reactive power supplied by the capaci-

The The

b.

If the line

V, calculate the following:

to speed, at

interval.

no-load?

the energy dissipated in the rotor

during the starting period [Btu]?

14-24

A 3-phase, rating of

8-poIe induction motor has a

40

hp, 575 V.

60 Hz.

It

drives a

320


steel

flywheel having a diameter of 3

inches and a thickness of 7.875

1

.5

The

in.


14-27

torque-speed curve corresponds to that of

D

a design a.

in Fig. 14.5.

of inerlia

d.

Calculate the rated speed of the motor and

Motor B: 75

Draw

[ft-lbf].

hp,

900

day.

How

erates about 6 hours per day. In

b.

Using Eq.

time.

r/min; lubricate

Motor A runs continually, 24 hours per Motor B drives a compressor and op-

at 0, 180,

Problem 14-24 calculate the average

a.

lubricate

every 10 000 hours of running time.

360.540. 720, and810r/min.

14-25

to be

40 hp

the torque-speed curve of the

motor and give the torques [N-m]

motor have

Motor A: 75 hp, 3550 r/min; every 2200 hours of running

|lb-ft~|.

Calculate the loeked-rotor torque

a

its

the corresponding torque [ft-lbf]. c.

in

following schedule applies to two motors:

Calculate the mass of the flywheel and

moment b.

motor given

The bearings

greased regularly, but not too often. The

often

should the bearings of each motor be

torque between 0 and 180 r/min. 3.

r/min, c.

greased per year?

14 calculate the time required

to accelerate the

flywheel from 0 to 180

14-28

mium

the flywheel at 180 r/min.

540

that this

N-m

in

14-29

A standard 40

hp motor, similar

mass of

line current

The energy required to

to

is

14-30

the av-

if

A constant

has windings similar to those shown

60 kg

minals to

make

tion

the

min]

when

is

The

resistance

V

in Fig.

is

and 2

1

measured between

12

il.

in the

ter-

What

high-speed connec-

resistance

would you

expect to measure between terminals 4 and

Assuming energy

horsepower, 2-speed induction

motor rated 2 hp, 1760/870 r/min, 460

Gornergrat |MJ]

|

mo-

climb from Zermatt

The minimum time required trip

g.

one

the energy sav-

ings that accrue to the high-efficiency

14.8.

f.

to the

Problem 14-28, costs $1723

82%. Calculate

factor of

at full-load

the loaded train

erage weight of a passenger e.

in

tor during the 3-year period.

The approximate transmission when the motors are operating total

is

and has an efficiency of 90.2% and power

The speed of rotation of the gear wheel when the train moves at 9 miles per hour The speed ratio between the motor and the

The

at full-

$0.06/kWh.

cost of energy

described

gear wheel

d.

$2243, runs

during a 3-year period, knowing that the


c.

at

Calculate the cost of driving the motor

addition to

The train in Fig. 14. 12 has a mass of 78 500 lb and can carry 240 passengers.

b.

and an efficiency of 93.6%.

load 12 hours per day, 5 days a week.

the flywheel load.

a.

pre-

time the load exerts a fixed

counter-torque of 300

14-26

86%

The motor, priced

knowing

r/min.

V, 3-phase,

energy induction motor has a power

factor of

d. Calculate the time required to accelerate

the flywheel from 0 to

1780 r/min, 460

60 Hz, drip-proof Baldor Super E

assuming no other load on the motor.

Using Eq. 3.8 calculate the kinetic energy in

A 40 hp,

that

80 percent of the

the train

is

going uphill and that 80

percent of the mechanical energy verted to electrical energy

6

electrical

converted into mechanical energy

is

14-31

A

in the

low-speed connection?

150 hp

7

1

175 r/min, 460 V, 3-phase,

60 Hz induction motor has the following

recon-

properties:

when going

downhill, calculate the total electrical en-

ergy consumed during a round

trip

[kW-h].

no-load current: 71

A

full-load current: 183

A


locked-rotor current:

1

550

A full-load torque:

886

breakdown

2552

A

a.

ft-lbf b.

torque:

ft-lbf

A

power

factor:

from the main panelboard

motor, 850

elboard

is

ft

away. The voltage

480

V

ture of the cable

to the

at the

mo-

the

motor

is

impedance

is

purely re-

approximate current

started up across the line.

Estimate the resulting starting torque.

d.

Compare

it

with the rated starting torque,

percentagewise.

pan-

and the average temperais

the cable

c.

3-conductor 250 kcmil copper cable

stretches

Assuming

when

32%

circuit of the

under locked-rotor conditions.

sistive, calculate the

locked-rotor torque: 1205 ft-lbf

locked-rotor

Determine the equivalent tor

321

estimated to be 25°C.

i

-32

In

Problem 14-3

1

express the currents and

torques in per-unit values.

Chapter 15 Equivalent Circuit of the Induction Motor

15.0 Introduction

secondary windings

cal

—one

The preceding three chapters have shown that we

single primary winding

can describe the important properties of squir-

we want

if

pensible. In this chapter'" circuit

and observe

tor

Finally,

We

indis-

We

low-power and high-power mo-

we develop

its

proper-

is

the

same in

equivalent

cir-

as that of a transformer,

Chapter

10, Fig. 10.20.

assume a wye connection

E„

=

r\

—

stator

a-,

=

stator leakage reactance

x2

=

rotor leakage reactance

r2

—

rotor winding resistance

is

very simi-

for the stator

The

and

circuit para-

identical

who

winding resistance

externa] resistance, effectively connected slip-ring

and the neutral of the

rotor

primary windings and 3 identi-

This chapter can be skipped by the reader

source voltage, line to neutral

between one

construction to a 3-phase transformer. Thus, the

to

)

meters, per phase, are identified as follows:

the equivalent circuit of an

3-phase wound-rotor induction motor

have lime

1

acts exactly like

its

their basic differences.

The wound-rotor induction motor

motor has 3

5.

the rotor, and a turns ratio of 1:1.

Rx =

lar in

1

previously developed

under load.

15.1

A

cuit (Fig.

then analyze the

asynchronous generator and determine ties

it

a conventional transformer, and so

we develop the equivalent

from basic principles.

characteristics of a

is

On

consider a

and a single secondary wind-

When the motor is at standstill,

to

gain even a better understanding of the properties of the motor, an equivalent circuit diagram

we can

ing in analyzing the behavior of the motor.

rel-cage and wound-rotor induction motors without

using a circuit diagram. However,

each phase.

set for

account of the perfect symmetry,

Xm =

magnetizing reactance

Rm =

resistance

does not

iron losses

whose

losses correspond to the

and windage and

friction losses

study the more theoretical aspects of induction

T=

motor behavior.

322

ideal transformer

having a turns ratio of

1

:

EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR

323

Figure 15.1 Equivalent circuit of a wound-rotor induction motor at standstill.

Figure 15.2 Approximation of the equivalent

for

3-phase trans-

the case of a conventional

In

acceptable

circuit is

load current true: / 0

may

/

.

p

compared

negligible

is

However,

in a

motor this

be as high as 40 percent of

the air gap. Consequently,

we cannot

to the

no longer

is

shown

we

in Fig.

can 1

E2 U ,

on the primary and secondary side of

in the

secondary winding will become

these

new

to Ei (the turns ratio

This greatly simplifies the equa-

induced

is

Fig.

1

5.2

motor when the rotor

How

when

affected

the

Suppose the motor runs the rotor speed

is

ns

(

I

—

motor

starts

a slip

at s),

s,

where

meaning /7

S

is

The frequency

locked.

is

turning? that

the syn-

chronous speed. This will modify the values of

be used.

hp the exact

circuit of Fig.

1

5.

1

if

the

slip

is

v,

l

;

is

sfand

I

be equal

motor were

sta-

the actual voltage

sEi this

changes

the

imped-

ance of the secondary leakage reactance from

j.\ 2

to

jsx 2 Because resistors are not frequency-sensitive, .

the values of r2 and the

two together

tance For motors under

is

shows

5.3

E2 would

)

E2 =

a true rep-

is

resentation of the is it

is

But because the

compromising accuracy.*

/

Directing our attention to the secondary side, the

amplitude of the induced voltage

tions that describe the behavior of the motor, with-

out

1

where

operating conditions.

tionary.

it

sfi

the frequency of the source E„. Fig.

to the input terminals, as

shift

5.2.

and

the ideal transformer T. Furthermore, the frequency

because of

p

eliminate the

magnetizing branch. However, for motors exceeding 2 hp,

I x

we would be justified in removing the magnetizing branch composed of jX m and R m because

former,

the exciting current /0

motors above 2 hp.

R2

,

R K remain

to

the same. Let us

form a single secondary

lump resis-

given by

should

R,

=

(I5.l)


324

external resistance

frequency

frequency

f

sf

Figure 15.3 Equivalent circuit of a wound-rotor motor the stator

in

The

is

details of the

Fig. I5.4a,

it

is

running at a

secondary circuit are shown

and the resulting current

^~ =

when

But the frequency of the voltages and currents

f.

sE, Z_

a-£,

„

A-

=

•

R2

VRj +

+js.x 2

slip s. in

The frequency

the rotor

of the voltages

in

I 2 is

^

- B

/U'2

(15.2)

2

Cvx,)

where

total

(3

=

resistance

arctan sx 2 IR 2

(

The corresponding phasor diagram Fig.

5.4b.

1

and currents

is sf.

diagram

It is

1

5.3)

circuit

shown

is

of rotor

in

important to realize that this phasor

relates to the

frequency sf Consequently,

frequency = sf

it

cannot be integrated into the phasor diagram on the

primary there

is

side,

fect, the

.

/,

and

/,

(frequency /)

absolute value of

that of U.

/2

is

a direct relationship between

in the rotor

£, and

where the frequency

Nevertheless,

I2

(frequency

Furthermore, the phase angle is

exactly the

same

(a)

sf)

in the stator. In ef-

exactly the

is

/,

/

as that

(3

same

as

between

between

E2

and

This enables us to draw the phasor diagram for £,

and

/,

as

shown

in Fig. 15.5.

To summarize: 1

.

The

effective value of

tive value of

2,

is

/,

equal to the effec-

even though

their frequencies

(b)

are different. 2.

The

effective value of

tive value of 3.

E2

E

equal to the effec-

is {

divided by the slip

The phase angle between as that between E 2 and I 2

E .

and ]

/]

Figure 15.4 a.

,s.

is

the

same

b.

Equivalent circuit of the rotor; E2 and / 2 have a frequency sf Phasor diagram showing the current lagging behind the voltage by angle

(3.


we can

Thus, on the primary side

= U =

/,

R2 +

325

write

(15.4)

jsx 2

Therefore,

(15.5)

R2

Zn jx2

S

The impedance Z2 seen between nals

1,

2 of the ideal transformer

the primary termiis,

therefore,

Figure 15.6 Equivalent

circuit of

a wound-rotor motor referred to

the primary (stator) side.

Z2 =

1

=

+

jx 2

(15.6)

/,

As

a result,

we can

This equivalent circuit of a wound-rotor induc-

simplify the circuit of Fig.

motor

The leakage reactances jx Jx 2 can now be lumped together to create

tion

a single total leakage reactance jx.

motor

15.3 to that

shown

in Fig. 15.6.

it

is

is

so similar to that of a transformer that

not surprising that the wound-rotor induction

{

total

It is

equal to the

leakage reactance of the motor referred to the

stator side.

is sometimes called a rotary transformer. The equivalent circuit of a squirrel-cage induction motor is the same, except that R 2 is then equal to the

equivalent resistance r2 of the rotor alone referred to the stator, there being

15.2

no external

resistor.

Power relationships

some power relationships for the 3-phase induction motor. The following equations The equivalent

circuit enables us to arrive at

basic electromechanical

can be deduced by visual inspection of the equiva-

V P =

lent circuit of the

wound-rotor motor

(Fig. 15.7):

(sx 2 h

arctan sx 2 IR 2

Figure 15.5 voltage and current in the stator are separated by the same phase angle p, even though the frequency

The is

different.

The

final

equivalent circuit of the wound-rotor

induction motor

is

shown

in Fig. 15.7. In this di-

agram, the circuit elements are fixed, except for the resistance

R 2 /s.

Its

value depends upon the slip

and hence upon the speed of the motor. Thus, the value of

R 2 ls

will vary

from

motor goes from start-up speed

(s

=

0).

(.v

R 2 to infinity as the = 1) to synchronous

Figure 15.7

The primary and secondary leakage reactances x-, and x2 are combined to form an equivalent total leakage reactance x.


326

1

.

Active power absorbed by the motor

Efficiency of the motor

10.

is

PJP

n = 2.

Reactive power absorbed by the motor

Note: The preceding quantities are "per phase";

is

some must be

Q = EÎX m + Irx 3.

motor

Apparent power absorbed by the motor

=

S

2

\

P + Q

is:

multiplied by 3 to obtain the total

quantities.

is

15,3 Phasor diagram

2

of the induction 4.

Power

factor of the

motor

If

cos 6

=

PIS

we use current /, in Fig. 5.7 as the reference phawe obtain the complete phasor diagram of the 1

sor,

wound-rotor motor shown in Fig. 5.8. In this diagram (and also in future calculations) it is useful to define an impedance Z, and angle a as follows:

5.

Line current

6.


1

is

is

r

Power cuit

dissipated as

I

R

losses in the rotor cir-

V>~ +jc 2

a —

arctan xlr

In these equations r,

is

{

]V

2

I

R 2 -sP

Because

stator. v

and

Mechanical power developed by the motor

is

a are fixed,

r,

and jx are

than r and so the angle x

= 9.

P,

(

1

-

lation of

Torque developed by the motor j

9.55

-

Pm 111

9.55

P

-

I

ns

n

_

9.55 Pl\ ^

is

(\

s)'

s)

(1

5.7b)

motor referred

fixed,

is

to the

follows that

it

1

000

hp, jx

is

a approaches

In the equivalent circuit .V)

5.7a)

Z

of Fig.

1

much

larger

90°.

5.7, the calcu-

mechanical power, torque, and speed de-

and R 2 is. The magnetizing branch

pends upon

r jx,

R m and jX m

does not come into play. Consequently,

{

for these calculations the equi valent circuit and cor-

responding phasor diagram can be simplified to that

shown

in Figs. 15.9

and

15. 10.

v

Figure 15.9 As far as mechanical power, Figure 15.8 Phasor diagram of the voltages and currents in Fig. 1 5.7. The power factor of the motor is cos tt.

{

irrespective of the speed of the motor.

motors above

In large

(1

the stator resistance and jx

the total leakage reactance of the

is

P = 8.

2

=

Z,

P = ICR 2 /s 7.

motor

is

torque,

and speed are

concerned, we can neglect the magnetizing branch

Xm

Rm This yields a simpler motor behavior.

and

sis of

.

circuit for

the analy-


1

Phasors

.

AB

angle

2.

BC

and

have the same length and

between them

the angle

CAB =

angle

327

is

(

80

1

ACB =

—

a)°.

a/2

Consequently, /t

hR^s

/,r,

Figure 15.10 Phasor diagram of the circuit of Fig. 1 5.9. Note that phasor / Z is always a degrees ahead of phasor 1

15.4

Breakdown torque and speed

We have seen that the torque developed by the motor is given by T = 9.55 P /n where P is the power der

livered to the resistance to a basic

imum

power

s

v

R 2 is

transfer theorem, the

(and therefore the torque

the value of R 2 /s

is

is

breakdown torque

slip at

= R 2 IZ

sh

The current

is

(15.9)

X

breakdown torque

at the

is

According

power

is

max-

maximum) when

=

/, b

£,/(2Z, cos a/2)

The breakdown torque

(I5.10)

is

equal to the absolute value of im-

pedance Z|. Thus, for

maximum

R 2 /s = Z Under these circumstances, voltage drop across Z,

We

(Fig. 15.9).

The

= EJ2

cos ^

/,Z,

/-,.

1

is

torque

-

(15.8)

l

the

magnitude of the

R 2 /s.

We

note that the magnitudes of both the breakdown

Tb and

torque

circuit resistance

However, the

R2

pends upon

The phasor diagram corresponding to condition is shown in Fig. 5. It is a 1

1

.

breakdown current

the

/ ib

are fixed,

sense that they are independent of the rotor

in the

can therefore write

(I5.ll)

%

(4Z, cos- a/2) S

/?

equal to that across

1

9.55

Th =

.

R2

.

breakdown torque de-

slip at the

Indeed,

if

R2 =

Z,, the

breakdown

torque coincides with the starting torque because s b this special

special case

of the phasor diagram of Fig. 15.10. Simple

geom-

etry yields the following results:

then equal to

l

.

These conclusions are

all

is

borne out by

the torque-speed curves in Fig. 13. 18 (Chapter 13). In the case of squirrel-cage motors, the resistance

R 2 becomes equal

to r 2

,

which

is

the resistance of the

rotor alone reflected into the stator. In practice, the

angle

a

lies

correspond tors. In

between 80° and 89°. The larger angles to

medium- and high-power cage mo-

R 2 /Z

such machines the ratio

as 0.02. Consequently, the at slips as

small as 2 percent.

15.5 Equivalent circuit of practical Figure 15.11 Phasor diagram when the motor develops its maximum torque. Under these conditions R2 ls = Z,.

can be as low }

breakdown torque occurs

two

motors

The impedances and resulting equivalent circuits of two squirrel-cage motors, rated 5 hp and 5000 hp are given in Figs.

1

5.

1

2 and

1

5.

1

3,

together with the


328

motor

The motors

ratings.

wye and

are both connected in

3.

a/2

=

4.

The

slip at

impedances are given per phase.

the

15.6 Calculation of the

We

now

will

calculate the

2 l

+

.5

6

2

=

6.

=

76°

1

8

nb

for

arclan

.v/r,

arctan 6/1.5

l. 2/6.

The

-

n s (\

=

1450r/min

.v

440V/V 3

0.194)

is

= a/2

6900 V/V'3

Motor

rating:

60 Hz,

1

800 r/min, 440

full-loiid cuitciu: 7

V,

5000

3-phase

A

lneked-rotor current: 39

A

hp,

60 Hz. 600

r/min,

358

r2

rotor resistance

j.x

-

total

1

1

.5 12

.2 12

reactance

1

6 6 1

A

r2

rotor resistance 0.080 (2

j.x

=

total

=

magnetizing reactance 46

leakage reactance 2.6 12 12

no-load losses resistance 600 (2

(The no-load losses include the iron losses plus

The no-load

windage and

15

friction losses.)

1

stator resistance 0.083 fi

Rm =

no-load losses resistance 900 12

3-phase

=

/X m

10 (1

V,

=

/*,

leakage reactance 6 12

6900

A

locked-rotor current:

stator resistance

jXm = magnetizing

rating:

full-load current:

=

Rm =

b)

-

1800(1

2 Z, cos

5 hp,

0.194

is

= =

breakdown

current at

/,K

Motor

=

8

1

O 6.

a

is

breakdown

hp motor.

V I.

at

breakdown torque Th /, b

0.788

]

The speed n b

5.

and the corresponding speed n h and current the 5

breakdown

= R 2 /Z =

sb

breakdown torque

-

38°, cos a/2

losses of 26.4

kW for windage

and

kW

(per phase) consist of

friction

and

1

1

.4

kW

for the

iron losses.

Figure 15.13

Figure 15.12 Equivalent circuit of a 5 hp squirrel-cage induction

Because there R2 ^ r2

motor. rotor,

.

is

no external resistor

in

the

Equivalent circuit of a 5000 hp squirrel-cage induction motor.

Although

this

powerful than the motor

gram remains the same.

motor in Fig.

is

1000 times more

15.12, the circuit dia-


-

7.

A3 — = 2X6.18X0.788

The power

to the rotor 2

P = =

I

r

{

R 2 ls = 2

26.

X

26.1

A

2 I\

Z

6.18

9.55

P

9.55

X 4210

W

is

r

200

0

800

600

400

The

total

is

torque

r/min

SPEED

the torque developed is,

therefore, 3

X

22.3

per phase.

= 67 N-m.

Figure 15.14 Torque-speed curve

The same

15.7 Torque-speed curve

We can determine the complete torque-speed curve of hp motor by selecting various values of slip and

solving the circuit of Fig. 15. 12.

Table 15 A, and the curve

in

is

The

given

5 hp,

440

V,

r/min,

5000 hp 1

5. 15

to the

breakdown torque. These

characteristics are

typical for large squirrel-cage induction motors. in Fig. 15.14.

TORQUE-SPEED AND LOAD CHARACTERISTIC

5000

60 Hz squirrel-cage induction

hp,

6900

V,

600

r/min,

60 Hz squirrel-cage induction

motor

/,

IWj

[A]

s

T

n

[N-ml

[r/min J

Torque

IkN-ml

0.0125

2.60

649

3.44

1777

5.09

1243

6.60

1755

0.026

5.29

1291

6.85

1753

0.6

0.05

9.70

2256

12.0

1710

0.4

2 1

1.49

/

p

4.9

1617

0

1616

8.2

23.4

1614 1610

1120

360

10.6

40.8

2921

480

17.7

64.7

1593

26.8

6095

540

30.8

80.4

1535

42.1

10114

570

51.7

89.5

1

0.03077

47.0

11520

581.5

68.2

93.1

1133

0.02

43.1

10679

588

79.8

95.1

878

0.0067

19.9

5000

596

90.

96.6

358

0.0033

10.2

2577

598

85.1

95.4

198

7.39

0.2

26.4

4196

22.3

1440

0.1

0.4

33.9

3441

18.3

1080

0.05

0.6

36.6

2674

14.2

720

0.8

37.9

2150

11.4

360

38.6

1788

0.026.

-377 -600

6.3

14.4

=

[A|

[9*1

0

0.2

0

|

240

1620

9.49

[r/min

[hpl

0

18.8

at s

ElTcy

500

3547

developed

cos a

2.98

17.2

s

Speed

4.95

0.1

The rated power of 5 hp

Total

power

0.025

1

for the

results are listed

motor s

made

the results and Fig.

the near-synchronous speed from no-load right up

TORQUE-SPEED CHARACTERISTIC 1800

lists

shows the torque-speed curve. Note the relatively low starting torque for this large motor, as well as

TABLE 15B TABLE 15A

a 5 hp motor.

of

calculations are

motor. Table 15B

and other characteristics the 5

1000 1200 1400 1600 1800

N-m

1800 that this

N-m

X

= 4210

22.3

Note

22.3

(per

is

The breakdown torque Tb

8.

breakdown torque phase )

440

329

363


330

breakdown torque = 47 kN

m

(per phase)

254 V

j110

Figure 15.16 Equivalent circuit of a 5 hp motor operating as an asynchronous generator. Note that a negative resis-

tance

SPEED Figure 15.15 Torque-speed curve

1

.

is

reflected into the primary circuit.

Net resistance of branch

R n = -48 + of

a 5000 hp motor. 2.

Z = \Rl +

asynchronous generator have already learned

tion

motor can

act as a generator if

lent circuit for the 5

hp motor,

it

-

it

is

driven above 3.

Now

power

= V

that a squirrel-cage induc-

synchronous speed.

that

Current

its

= -46.5

(

-

46.88

/,

other prop-

is

H

is

.r

46.5)

2

+

6

2

0

branch 1-2-3-4

in

we have the equivawe can calculate the

can generate, together with

-2-3-4

Impedance of branch 1-2-3-4

15-8 Properties of an

We

I.5

l

is

- EIZ = 254/46.88 - 5.42 A

erties as a generator.

Let us connect the motor to a 440 V, 3-phase line

and drive

it

at a

speed of

1

845 r/min, which

4.

Active power delivered to the rotor

P = IcR 2 /s =

is

is

5.42- (-48)

{

45 r/min above synchronous speed. The s

=

(/? s

=

(1800

-

slip is

= -1410

n)ln s

-

This negative power means that 1410

1845)/ 1800

flowing from the rotor

- -0.025 5.

The value of /? 2 Av

in the

equivalent circuit

W

is,

The

2

l

R

there-

P - Ifr 2 = ]r

= 6.

The negative resistance indicates that power is flowing from the rotor to the stator rather than from the stator to the rotor. Referring to Fig. 15.16

make

the following calculations:

we

35.2

5.42"

P

v

X

1.2

W

The mechanical power equal to

is

losses in the rotor are

fore,

R 2 ls = 1.2/(- 0.025) = -48 n

W

to the stator.

input to the shaft

is

plus the losses P-]r in the rotor:

p

= p + p. = 1410 + = 1445 W

35.2


7.

The

2

I

R

2

F =

l,

js

=

=

r,

44.

5.42

2

X

1.5

W

1

= PJS = 1294/1502 0,861 = 86.1%

cos 0

losses in the stator are

16.

The

efficiency of the asynchronous generator

useful electric 8.

The

windage and

iron plus

mechanical input

9.

The

to the line feeding

rotor to stator

1

.

3

=

phases

3

9.55

=

X 1

X

3

1445

2

=

/,

=

176 var

a-

=

5.42

2

X

No-load

2

1

5,6, this

/X m

= 254

at

/110

power absorbed by the motor

X nv R m

The power

p

that the value of

/, is

negligible

Xm R m ,

at

at

no-

R 2 /s

is

compared

very togh

to / 0 Thus, .

.

Their values can be de-

no-load, as follows:

Measure the

stator resistance

r,

R u between any

is

generator terminals A,

762

2

VA

'"I

Run

the

motor

is

£ NL

load current

/ NI

=

RlA.ll

no-load using rated line-to(Fig. 15.

and the

1

7).

total

Measure

the no-

3-phase active

.

The following calculations of total apparent power total reactive power Q N] are then made:

S NL and

1502/254

factor at the generator terminals

at

line voltage,

power P Nl p

an induction motor runs

exceedingly small. Referring to Fig.

means

value of

b.

= S/E= = 5.91 A

in

the

no-load the circuit consists essentially of the

power

762 vars

2

line current /

and x

two terminals. Assuming a wye connection, the

VP 2 + Q 2 = V1294 + 1502

,

by means of

is

a.

at the

,

termined by measuring the voltage, current, and

Qi

/

When is

magnetizing branch

var

is

The

test

load, the slip

2

r2

tests.

and so current

Apparent power

=

N-m


following

is

Total reactive

S

22.3

6

Reactive power absorbed by the magnetizing

B

-

is:

854

the equivalent circuit can be found

+ Qi = 176 + 586 =

15.

hp

The approximate values of

Q =

14.

X P

9.55

is

1445/746

Torque exerted by the driving motor

X 1294 - 3882 W)

is

= 586

3.

5.81

X

3

Reactive power absorbed by the leakage reac-

Q2 = E

1

=

=

tance

reactance

12.

3

to drive the generator

15,9 Tests to determine

2,

1

=

n

W

(P c for the 10.

The horsepower needed

T=

v

1294

8.

1

= P - P]S - P - P v = ]410 - 44.1 - 71.7 =

7.

1

minus

losses

v

= 89.5%

0.895

is

Pc = power delivered from

ir

1445

power delivered

active

motor

the

1294

W

71.7

is

Pc P

power

friction losses are

P f + P v = E2 /R m = 254 2 /900

=

331

is

ML


332

Figure 15.17

A

Figure 15.18

no-load test permits the calculation of

Xm

and

tfm of

the magnetizing branch.

A

locked-rotor test permits the calculation of the total

leakage reactance

From these

P + Pv

windage,

t

The

Rm

resistance

representing

circuit of

and iron losses

friction,

P + r

/\

results

xand the total resistance + r2 we can determine the equivalent

the induction motor.

Hence,

is

/WO 'ud ~ The magnetizing reactance

).

More

is:

r,

elaborate tests are conducted on large ma-

chines, but the above-mentioned procedure gives

adequate

results that are

Locked-rotor

Under

test

motor

the rotor of an induction

current

I

almost

is

p

Furthermore, the slip that r2 /s

is

s is

locked, the stator

is

times

six

its

rated

value.

equal to one. This means

equal to r2 where r 2 ,

the rotor reflected into the stator.

is

the resistance of

Because

greater than the exciting current 70

,

/ is p

we can

1

5.9,

composed of the leakage reactance and the reflected rotor

x, the stator resistance r h

sistance

R 2 /s =

r 2 /\

mined by measuring

—

re-

r2 Their values can be deter.

the voltage, current, and

most cases.

Example 15-1

A no-load test conducted on a 30 hp, 835 r/min, 440 V, 3-phase, 60

Hz

squirrel-cage induction motor yielded

the following results:

much

neglect

the magnetizing branch. This leaves us with the circuit of Fig.

in

when

rated line voltage,

No-load voltage

(line-to-line):

No-load current: 14

440

V

A

No-load power: 1470

W

Resistance measured between two terminals: 0.5 (2

power

The locked-rotor

under locked-rotor conditions, as follows:

test,

conducted

at

reduced

volt-

age, gave the following results: a.

Apply reduced 3-phase voltage that the stator current

is

to the stator so

about equal to

its

rated

Take readings of total

E

l

R (line-to-line),

3-phase power P K (Fig. ]

/,

R and the ,

Locked-rotor current:

JS' LR

are then

= F LR I

60 A

15. 18).


The following calculations

V

W

Locked-rotor power: 7200

value. b.

Locked-rotor voltage (line-to-line): 163

circuit of the motor.

made: Solution

R

[

\v J 3

Assuming

the stator windings are connected

wye, the resistance per phase

= Cir/3/ lr 3/T R (r, + r 2 = P LR

r,

*

)

From

-

0.5

the no-load test

a/2

we

=

find

is

0.25 II

in

EQUIVALENT CIRCUIT OE THE INDUCTION MOTOR

Rx =

=

'

10 669

P m = 1470

2

5-

1

e Il /p nl = 440 = i46a

2

/(1470

^lr'lr V3

/10 568

-3 X

14

2

X

0.25)

A

we

/,

R

X 60V

163

=

60

A

2

- 7200

_

G,.r

15

c.

-

0.67 (2

=

0.25

H

=

0.67

-

The value of Z, and the angle a The speed when the breakdown torque

is

The current

I

\

at the

breakdown torque

(see

The value of

a.

Problem 15-2, draw the equivalent circuit if the motor runs at 950 r/min in

is

/

2

breakdown torque [N-m]

ator? Calculate the torque of the

machine.

is

„= 7200/(3 X 60 2

,

the

In

the same direction as the revolving flux. Does the machine operate as a gener-

b.

r/3

V, calculate the

d.

1.42(2

2

Total resistance referred to stator

+ ^ =

346

reached

2

333

X 60

3

3/iLR

is

5 12 and the line-

is

Fig. 15.9)

Total leakage reactance referred to stator

—

voltage

to- neutral

15-3

x

leakage reactance

following:

Pl R = V16 939

333 var

the

12. If

find

b.

15

and

wye-connected squirrel-cage motor

equivalent rotor resistance of 0.5

a.

=

to the text, explain the

the impedances, currents,

having a synchronous speed of 900 r/min

VA

VS V °LR

in Fig. 15.19.

voltages in Fig. 15.2.

W

2

shown

is

Without referring

total

939

because

has a stator resistance of 0.7 il and an

the locked-rotor test

16

circuit

meaning of

15-2

7200

r2

15. 1.)


10 568 var

x m = ^\[/Q\l = 440 = 18.3 n

From

seeEq.

The equivalent

W

1

R,

0;

VA V10 669 2 - 1470 2

-

R2 =

(In a squirrel-cage motor,

440 X 14V 3

333

Draw runs

)

flux.


if

the

motor

950 r/min opposite to the revolving Does the machine operate as a genera-

at

tor? Calculate the torque.

r {

1

0.25

0.25

=

0.42

5-4

n

Q

A 550

780 r/min, 3-phase, 60 Hz squirrel-cage induction motor running no-load draws a current of 12 A and a total power of 1500 W. Calculate the V,

value of

1

Xm

and

Rm

at

per phase (see

Fig. 15.2).

15-5 440

18.3

#n

146

f3

Q

The motor rent of

X

30

in

Problem 15-4 draws

A and

when connected

a

power of 2.43

to a

90

a cur-

kW

V, 3-phase line

under locked-rotor conditions. The

resis-

tance between two stator terminals

is

Figure 15.19

0.8 fl. Calculate the values of r h r 2 and

Determining the equivalent

x and

induction motor (see

,

circuit of

Example

15-1).

a squirrel-cage

the locked-rotor torque [N

rated voltage.

m|

at

334

1

5-6


If

motor in 6200 V, calculate

the line voltage for the

Fig. 15.15

the

dropped

new breakdown

to

/ nd us trio I

15-9

torque and starting

5-7

Consider the 5 hp motor whose equivalent circuit

torque. 1

a.

A 440 V,

3-phase,

motor has the following

characteristics:

If the

=

1

x

=

6

.2 il c.

n

15-8

In

compare

magnetizing branch can be neglected,

connected

in series

if

a 4.5 fl resis-

with each

Problem 15-7 calculate the

is

connected

in series

line.

starting

torque and the breakdown torque reactor

Calculate the 50 to obtain the

calculate the value of the starting torque

is

Determine the values of the leakage reacat

a

frequency of 50 Hz.

and the breakdown torque tor

in Fig. 15.12.

tance and the magnetizing reactance

1.5 fi

x

r2

shown

ing reactances. b.

=

r

is

Calculate the values of the inductances (in millihenries) of the leakage and magnetiz-

800 r/min squirrel-cage

1

application

if

a 4.5 ft

with each

15-10

The

5

V


with the voltage

at

60 Hz.

hp motor represented by the equiva-

lent circuit

503

it

Hz

same magnetizing current and

of Fig.

1

5.

1

2

is

connected

(line-to-line), 3-phase,

to a

80 Hz

source. The stator and rotor resistances are assumed to remain the same. a. Determine the equivalent circuit when the

line.

motor runs b.

at

2340 r/min.

Calculate the value of the torque [N-m] and

power

[hp]


Chapter

1

Synchronous Generators

some

16.0 Introduction

other source of motive power.

3-phase voltage

Three-phase synchronous primary source of

all

generators

consume. These machines are the verters in the world.

are

the

the electrical energy largest

energy con-

into electrical energy, in

powers ranging up

will study the construction

In this chapter

we

and characteristics of these are based

upon

large,

modern

in

view

this materia]

upon the dc exciting The frequency of the speed and the number of

used when the power output

However,

generators.

may wish

rotates, a

the speed of rotation and

voltage depends upon the

for greater outputs,

and more practical

the elementary principles cov-

Section 8.6, and the reader

ered

it

poles on the field. Stationary-field generators are

1500

to

As

induced, whose value depends

current in the stationary poles.

They convert mechanical energy

MW. They

upon

we

is

A

to re-

to

is

less than 5

employ a revolving dc

tor

field.

revolving-field synchronous generator has a

stationary armature called a stator.

before proceeding further.

kVA.

cheaper, safer,

is

it

winding

is

The 3-phase

sta-

directly connected to the load, with-

out going through large, unreliable slip-rings and

16.1

A

Commercial synchronous

brushes.

generators

insulate the windings because they are not sub-

stationary stator also

makes

jected to centrifugal forces. Fig. 16.

Commercial synchronous generators

are built with

The

The field is excited by a dc generator, mounted on the same shaft. Note that the brushes on the commutator have to be connected to

cut by a re-

another set of brushes riding on slip-rings to feed

as a dc generator.

which

is

usually

The armature possesses a 3-phase winding whose terminals are connected to three slip-rings mounted on the shaft. A set of volving

armature.

the dc current I x into the revolving field.

16.2

brushes, sliding on the slip-rings, enables the armature to

is

Number

of poles

an external 3-phase load.

The number of poles on a synchronous generator de-

driven by a gasoline engine, or

pends upon the speed of rotation and the frequency

be connected

The armature

a schematic

alternator.

A stationary -field synchronous generator has the salient poles create the dc field,

is

easier to

diagram of such a generator, sometimes called an

either a stationary or a rotating dc magnetic field.

same outward appearance

1

it

to

335


3 36

ils

pilot exciter

25

kW

3-phase alternator

500 MW, 12 kV, 60 Hz

Figure 16.1 Schematic diagram and cross-section view of a typical 500 synchronous generator and its 2400 kW dc exciter. The dc exciting current lx (6000 A) flows through the commutator and two slip-rings. The dc control current lc from the pilot exciter permits variable field control of the main exciter, which, in turn, controls /x

MW

.

we wish

to produce. Consider, for

conductor

that

is

p =

example, a stator

successively swept by the

N

and S

120.////

= 20 X

60/200

1

poles of the rotor.

when an N ilar

If

a positive voltage

is

induced

pole sweeps across the conductor, a sim-

negative voltage

is

crosses the conductor, the induced voltage goes

through a complete cycle. The same other conductor on the stator; that the alternator

is

true for every

we can

therefore de-

frequency

is

given by

From an

(,6I) 120

is

frequency of the induced voltage [Hz]

that carry a

is

is

always connected

connected

to

.

The voltage per phase

is

between the

the voltage

200 r/min

nected to a synchronous generator. voltage has a frequency of 60 Hz,

If

is

con-

the induced

how many

poles

in

16.3).

wye and

the

A wye connection is

only 1/V3 or lines.

between a

line voltage.

amount of

We can

58%

of

This means that stator

is

only

conductor

58%

of the

therefore reduce the

insulation in the slots which, in turn,

enables us to increase the cross section of the conductors.

A

larger conductor permits us to in-

crease the current and, hence, the power output

Solution Eq. 16.

ground.

and the grounded stator core at

composed of

3-phase lap winding (Figs. 16.2,

the highest voltage

hydraulic turbine turning

It is

preferred to a delta connection because 1

Example 16-1

From

S poles

identical to that of a 3-phase

induction motor (Section 13. 17).

The winding

p = number of poles on the rotor n = speed of the rotor [r/min]

does the rotor have?

N and

electrical standpoint, the stator of a syn-

chronous generator

neutral

A

8 pairs of

16.3 Main features of the stator

where

/=

1

a cylindrical laminated core containing a set of slots

pn

f=

36 poles, or

induced when the S pole

speeds by. Thus, every lime a complete pair of poles

duce

=

1

,

we have

of the machine.


337

Figure 16.2a Stator of a 3-phase,

500 MVA, 0.95 power factor, 15 2350 mm; 378

effective axial length of iron stacking:

kV,

60 Hz, 200 r/min generator.

Internal diameter:

9250 mm;

slots.

(Courtesy of Marine Industrie)

2.

When

a synchronous generator

voltage induced

in

and the waveform distortion

is

is

under load, the

each phase becomes distorted, is

no longer sinusoidal. The

mainly due

to

an undesired third

harmonic voltage whose frequency that

is

three times

of the fundamental frequency. With a

wye

connection, the distorting line-to-neutral harmonics

do not appear between the

lines

because they

effectively cancel each other. Consequently, the line voltages

remain sinusoidal under

conditions. Unfortunate! v.

when

all

load

a delta connec-

tion

is

used, the harmonic voltages

but add up. Because the delta

is

do not cancel,

closed on

itself,

they produce a third-harmonic circulating current,

which increases the

The nominal erator depends

greater the

However, exceeds 25

its

losses.

kVA

synchronous gen-

rating. In general, the

rating, the higher the voltage.

nominal line-to-line voltage seldom

kV because

takes up valuable space

conductors.

R

line voltage of a

upon

power

the

I

the increased slot insulation at

the

expense of the copper

Figure 16.2b

The copper bars connecting successive 19 250 A per phase.

stator poles are

designed

to carry a current of

3200

A.

The

total

output

is


Figure 16.2c The stator is built up from toothed segments silicon-iron steel laminations (0.5

an insulating varnish. The

mm

slots are 22.3

mm deep. The salient poles of the much

thicker (2

mm)

thick),

of high-quality

covered with

mm wide

and 169

composed of These laminations

rotor are

iron laminations.

are not insulated because the dc flux they carry does not vary.

the

The width of the poles from tip to tip is 600 mm and gap length is 33 mm. The 8 round holes in the face

air

of the salient pole carry the bars of

338

a squirrel-cage winding.

Figure 16,3

MVA, 3600 r/min, 19 kV, 60 Hz steam-turbine generator during the construction phase. The windings are water-cooled. The stator will eventually be completely enclosed in a metal housing (see background). The housing contains hydrogen under pressure to further improve the cooling. {Courtesy of ABB) Stator of a 3-phase, 722

339

ELECTRICA L MA CHINES A ND ERA NS FORMERS

340

16.4 Main features of the rotor

made

ing, the field coils are

of bare copper bars,

with the turns insulated from each other by strips of

Synchronous generators are rotors: salient-pole rotors rotors.

built with

two types of

mica

and smooth, cylindrical

Salient-pole rotors are usually driven by

have

to turn at

Most hydraulic

rotors.

(Fig.

turbines

dc field winding,

in series,

embedded

1

6.6).

Under normal

we

in the

often add

pole-faces

conditions, this winding

does not carry any current because the rotor turns

at

generator changes suddenly, the rotor speed begins

is

directly coupled to

required, a large

number of poles

to fluctuate,

50 Hz

producing momentary speed variations

above and below synchronous speed. This induces a

are re-

voltage

in

the squirrel-cage winding, causing a large

quired on the rotor. Low-speed rotors always pos-

current to flow therein.

sess a large diameter to provide the necessary space

magnetic

for the poles.

connected

maximum power from

the waterwheel, and because a frequency of is

coils are

synchronous speed. However, when the load on the

a waterfall. Because the rotor

Hz

The

low speeds (between 50 and 300

r/min) in order to extract the

or 60

6.5).

a squirrel-cage winding,

are driven by high-speed steam turbines.

Salient-pole

1

In addition to the

low-speed hydraulic turbines, and cylindrical rotors

/.

(Fig.

with adjacent poles having opposite polarities.

The

large circular steel

salient poles are

frame which

ing vertical shaft (Fig.

1

6.4).

is

mounted on a

dampen

fixed to a revolv-

The

field of the stator,

the oscillation of the rotor. For this reason,

the squirrel -cage winding

To ensure good cool-

current reacts with the

producing forces which

is

sometimes called a

damper winding.

Figure 16.4 This 36-pole rotor

is

being lowered into the stator shown

a 330 V, electronic rectifier. Other details are: mass: 600 (Courtesy of Marine Industrie)

in Fig. t;

16.2.

moment

The 2400 A dc

of inertia: 41

40

exciting current

tm 2

;

air

gap: 33

is

supplied by

mm.

Figure 16.5 tor is

made

width of 89

Figure 16.6

a 250 MVA salient-pole generaof 18 turns of bare copper bars having a

This rotor winding

mm

for

and a thickness

of 9

Salient-pole of a

250

MVA

generator showing 12 slots

t0 carry the squirre |. cage winding,

mm.

Figure 16.7a Rotor of a 3-phase steam-turbine generator rated 1530 MVA, 1500 r/min, 27 kV, 50 Hz. The 40 longitudinal slots are being milled out of the solid steel mass. They will carry the dc winding. Effective axial magnetic length: 7490 mm; diameter: 1800 mm. (Courtesy of Allis-Chalmers Power Systems Inc., West Mis, Wisconsin) 34


342

Figure 16.7b Rotor with rent of

1 1

its

.2

4-pole dc winding. Total mass: 204

kA

is

supplied by a 600

V dc

Inc.,

lines,

even

anced load conditions. 2.

Cylindrical rotors.

;

It

is

well

known

than low-speed turbines.

The same

is

that high-

true of high-

speed synchronous generators. However, to generfrequency

we cannot

air

gap: 120

of the

N

main

mm. The dc

exciting cur-

shaft.

use less than

to

and S poles.

The high speed of rotation produces strong cenwhich impose an upper

trifugal forces,

speed steam turbines are smaller and more efficient

ate the required

2

and retained by high-strength end-rings, * serve create the

due to unbal-

the line currents are unequal

85 t-m the end

of inertia:

West Allis, Wisconsin)

also tends to maintain bal-

anced 3-phase voltages between the

when

moment

brushless exciter bolted to

(Courtesy of Allis-Chalmers Power Systems

The damper winding

t;

diameter of the at

rotor. In the

3600 r/min, the

limit

on the

case of a rotor turning

elastic limit

of the

steel requires

the manufacturer to limit the diameter to a maxi-

mum of erful

m.

.2

l

1000

On

the other hand, to build the

MVA to

1500

MVA

pow-

generators the vol-

2 poles, and this fixes the highest possible speed.

ume

On

high-power, high-speed rotors have to be very long.

a

60 Hz system

speed

is

it

is

3600

r/min.

The next lower

It

follows that

1800 r/min, corresponding to a 4-pole ma-

chine. Consequently, these steam-turbine generators possess either 2 or

The

of the rotors has to be large.

4 poles.

rotor of a turbine-generator

The dc is

cylinder which contains a series of longitudinal

slots

milled out of the cylindrical mass (Fig. I6.7).

wedged

field excitation

and exciters

of a large synchronous

a long, solid

steel

Concentric field coils, firmly

16.5 Field excitation

into the slots

generator

is

*

1.

See Fig.

1

an important part of

28 (Chapter

II).

its

overall design.


stationary field

—o A

rrrr

air

3-phase

alternator

-OB -oC

gap

343

terminals

pole

bridge rectifier /

pilot exciter

^.exciting coil

3\ 3-phase rotor

3-phase stator winding

r—T main exciter

y alternator

Figure 16.8 Typical brushless exciter system.

The reason

is

must ensure not only

that the field

stable ac terminal voltage, but to

sudden load changes

tem

order to maintain sys-

in

Quickness of response

stability.

a

must also respond one of the

is

voltage

may have

as

as

little

300

to

to rise to twice

normal value

its

400 milliseconds. This

in

represents a

very quick response, considering that the power of the exciter

may be

several thousand kilowatts.

important features of the field excitation. In order to attain

it,

two dc generators

and a pilot

citer

no rotating parts

The main

slip-rings.

voltage

lies

main ex-

at all are also

involve

employed.

exciter feeds the exciting current to the

of the synchronous generator by

field

and

are used: a

exciter. Static exciters that

way

Under normal conditions

between 125

V

and 600

V.

of brushes

/c ,

produced by the

1000

kVA

kW

alternator

exciter

needed

(2.5% of

its

1

6.

1

).

whereas

a

2500 kW exciter suffices for an alternator of (only 0.5% of its rating). 500 Under normal conditions the excitation is varied

MW

automatically.

It

responds to the load changes so as to

maintain a constant ac line voltage or to control the active

power delivered

re-

to the electric utility system.

serious disturbance on the system

may produce

a

A

sud-

den voltage drop across the terminals of the alternator.

The

exciter must then react very quickly to keep

the ac voltage

from

falling.

wear and carbon

dust,

we

constantly

and replace brushes, slip-rings,

and commutators on conventional dc excitation

sys-

systems have been developed. Such a system con-

to excite a

rating)

to clean, repair,

regulated

It is

pilot exciter (Fig.

is

to brush

have

tems. To eliminate the problem, brushless excitation

The power rating of the main exciter depends upon the capacity of the synchronous generator. Typically, a 25

Due

the exciter

manually or automatically by control signals that vary the current

16.6 Brushless excitation

For example, the exciter

sists

of a 3-phase stationary-field generator whose

ac output

is

rectified

by a group of

output from the rectifiers

rectifiers.

The dc

fed directly into the field

is

of the synchronous generator (Fig.

1

6.8).

The armature of the ac exciter and the are mounted on the main shaft and turn

rectifiers

together

with the synchronous generator. In comparing the excitation system of Fig.

we

can

see

they are

1

6.8 with that of Fig. 16.

3-phase rectifier replaces the commutator, rings,

and brushes.

(which

is

really a

by an electronic

In other

slip-

words, the commutator

mechanical rectifier.

1,

except that the

identical,

rectifier) is replaced

The

result

is

that

the

brushes and slip-rings are no longer needed.

The dc

control current

/c

from the

regulates the main exciter output

/v .

pilot exciter

as in the case of

ELECTRICA /. MA CHINES A ND ERA NS FORMERS

344

Figure 16.9 This brushless exciter provides the dc current for the rotor of

4 1

is*

shown in Fig. 16.7. The exciter consists kVA generator and two sets of diodes.

a 7000

Each set, corresponding respectively to the posiand negative terminals, is housed in the circular rings mounted on the shaft, as seen in the center of the photograph. The ac exciter is seen to the right. The two round conductors protruding tive

from the center

of the shaft (foreground) lead the

exciting current to the

1530

MVA

generator.

(Courtesy of Allis-Chalmers Power Systems

Inc.,

West All is, Wisconsin)

The frequency of

a conventional dc exciter.

main exciter

is

the

generally two to three limes the syn-

chronous generator frequency (60 Hz). The crease

in

frequency

is

in-

obtained by using more poles

on the exciter than on the synchronous generator. Fig.

1

shows

6.9

the rotating portion of a typical

brushless exciter. Sialic exciters that involve no rotating parts at

all

employed.

are also

16.7 Factors affecting the size of

synchronous generators

The prodigious amount of energy generated by elecutility companies has made them very con-

trical

scious about the efficiency of their generators. For

example, station

if

the efficiency of a

improves by only

1

%,

1

it

000

MW generating

represents extra rev-

enues of several thousand dollars per day. gard, the size of the generator tant

because

as the

power

its

is

In this re-

particularly impor-

efficiency automatically improves

increases. For example,

if a

small

l

kilowatt synchronous generator has an efficiency of

50%, l()

a larger, but similar

model having a capacity of

MW inevitably has an

This improvement

in

efficiency of about

efficiency with size

why synchronous

is

MW and

000 up possess efficiencies of the order of 99%. Another advantage of large machines is son

generators of

1

power output per kilogram increases increases. For example,

if

a

l

20 kg (yielding 1000W/20 kg

90%.

the rea-

Figure 16.10 Partial

view

of

87 MVA, 428

a 3-phase, salient-pole generator rated

r/min,

are water-cooled. that the

as the

power

kW generator weighs = 50 W/kg). a lOMW

and the use

50 Hz. Both the

The high

rotor

resistivity of

of insulating plastic tubing

water to be brought

into direct

parts of the machine.

{Courtesy of ABB)

and

stator

pure water

enables the

contact with the

live

S YNCHRONO US

GENERA TORS

345

generator of similar construction will weigh only

ticated cooling techniques (Figs. 16.10

20 000 kg, thus yielding 500 W/kg. From a power

Other technological breakthroughs, such as better

standpoint, large machines

materials,

weigh

relatively less than

small machines; consequently, they are cheaper.

Section 16.24

why

at

the end of this chapter explains

Everything, therefore, favors the large machines. in size,

we

run into seri-

ous cooling problems. In effect, large machines herently produce high

power

in-

losses per unit surface

area (W/m~); consequently, they tend to overheat.

To prevent an unacceptable temperature must design

efficient cooling

we become

rise,

systems that

ever more elaborate as the power increases. For ex-

ample, a circulating cold-air system cool synchronous generators but between 50

the 1000

hollow, point

MW

is

MW

range have to be equipped with

water-cooled

conductors.

Ultimately,

exceeds the savings made elsewhere, and

upper limit to

the

a

reached where the increased cost of cooling

is

To sum

1).

in

modifying the design of early ma-

chines (Fig. 16. 12).

As

regards speed, low-speed generators are

al-

power. Slow-speed bigness simplifies the cooling

problem; a good air-cooling system, completed with a heat exchanger, usually suffices. For example, the large,

slow-speed 500

generators installed are air-cooled

500

MVA, 200 r/min

in a typical

1

to

plant

whereas the much smaller high-speed

MVA, 800 r/min units installed

have

synchronous

hydropower in a

steam plant

be hydrogen-cooled.

adequate to

whose rating is below and 300 MW, we have

hydrogen cooling. Very big generators

to resort to in

1

ways bigger than high-speed machines of equal

However, as they increase

MW,

16.

and novel windings have also played a

major part

the efficiency and output per kilogram increase

with size.

50

and

this fixes

size.

16.8 No-load saturation curve Fig. 16.

3a shows a 2-pole synchronous generator It

is

driven

at

constant speed

by a turbine (not shown). The leads from the 3-phase, wye-connected stator are brought out to terminals A, B. C, N, and a variable exciting current /x

up, the evolution of big alternators has

mainly been determined by the evolution of sophis-

1

operating at no-load.

produces the flux

in the air

gap.

Let us gradually increase the exciting current

while observing the ac voltage

£ 0 between terminal

Figure 16.11

The

electrical-

200/1 15 the is

V,

energy needed on board the Concord

12 000 r/min, 400 Hz. Each generator

enormous power developed by

used

to cool the

generator and

(Courtesy of Air France)

is

is

aircraft is

supplied by four 3-phase generators rated 60 kVA,

driven by a hydraulic motor, which absorbs a small portion of

the turboreactor engines.

The

then recycled. The generator

hydraulic fluid streaming from the hydraulic motor itself

weighs only 54.5

kg.

346


Figure 16.12

was

This rotating-field generator

system. The alternator

was

first

in 1888. It was used in a 1000-lamp street lighting steam engine and had a rated output of 2000 V, 30 A at a frewhich represents 26 W/kg. A modern generator of equal speed and power

installed

driven by an 11

00

in

North America

r/min

quency of 1 10 Hz. It weighed 2320 kg, produces about 140 W/kg and occupies only one-third the

A, say, and the neutral N. For small values of / x the ,

floor

space.

Fig. 16. 13c

ator

ing current. However, as the iron begins to saturate,

phases on the

the voltage rises /x

.

If

we

much

less for the

plot the curve of

£u

same increase

versus

the no-load saturation curve of the

generator.

It

/x ,

we

a schematic diagram of the gener-

the

revolving rotor and the three

stator.

in

obtain

16.9

synchronous

Synchronous reactance equivalent circuit of an ac generator

similar to that of a dc generator

is

is

showing

voltage increases in direct proportion to the excit-

(Section 4. 13). Fig. 16.13b

curve of a 36

shows

the actual no-load saturation

Consider a 3-phase synchronous generator having

MW,

3-phase generator having a

terminals A, B,

nominal voltage of 12

kV

(line to neutral).

about 9 kV, the voltage increases

in

Up

put of 12 kV, but

if

1

age rises only to 15 kV.

is

driven by a turbine (not

proportion to

shown), and

is

chine and

load are both connected in wye, yield-

00 A produces an out-

the current

feeding a balanced 3-phase load

(Fig. 16. 14).

the current, but then the iron begins to saturate.

Thus, an exciting current of

C

The generator

to

doubled, the volt-

its

is

excited by a dc current

Ix

.

The ma-

ing the circuit of Fig. 16.15. Although neutrals N,

and

N2

tential

are not connected, they are at the

because the load

is

same po-

balanced. Consequently,


\

I;

B

:

•

I

i

347

|

C

i

i

!

i

Figure 16.13c Electric circuit representing the generator of Fig. 16.13a. is

an alternating-current machine, the inductance

X

manifests itself as a reactance

X =

s

,

given by

277/L

s

where

X = s

= E — /'

synchronous reactance, per phase [H] generator frequency [Hz]

apparent inductance of the stator winding, per

phase HJ [

The synchronous reactance of ternal

Figure 16.13 Generator operating

a. b.

impedance, just

like

its

a generator

is

an

in-

internal resistance R.

The impedance is there, but it can neither be seen nor touched. The value of X s is typically 10 to 100 times greater than R; consequently,

at no-load.

No-load saturation curve 3-phase generator.

of

a 36 MVA, 21

kV,

neglect the resistance, unless

we

we can always

are interested in

efficiency or heating effects.

We

can simplify the schematic diagram of

16.16 by showing only one phase of the

we could connect them

together (as indicated by the

fect, the

two other phases

Fig.

stator. In ef-

are identical, except that

short dash line) without affecting the behavior of the

their respective voltages (and currents) are out of

voltages or currents in the circuit.

phase by

The

field carries

an exciting current which pro-

duces a flux O. As the

duces

in the stator three

field revolves, the flux in-

equal voltages

Eu

that are

120° out of phase (Fig. 16.16).

Each phase of the stator winding possesses a resistance R and a certain inductance E. Because this

1

20°. Furthermore,

tance of the windings, cuit

of Fig.

1

6.

1

7.

we

if

we

neglect the resis-

obtain the very simple

fore be represented

by an equivalent

of an induced voltage

E0 in

circuit

O

which induces

composed

series with a reactance

In this circuit the exciting current / x

flux

cir-

A synchronous generator can thereX

s

.

produces the

the internal voltage

E
For a

348


/

A

\

B load

C load

alternator

Figure 16.14 Generator connected

to

a load.

Figure 16.17 Equivalent only

circuit of

a 3-phase generator, showing

one phase.

E

given synchronous reactance, the voltage

E0

terminals of the generator depends upon

E0

load Z. Note that

ages and

/ is

E

and

are line-to-neutral volt-

16,10 Determining the value of

X

We can determine the unsaturated value of X

by the

During the open-circuit at rated

test the

sponding exciting current

En

are recorded.

The

excitation

age

is

/ xn

is

is

driven

raised until

The

attained.

is

s

s

test.

generator

speed and the exciting current

the rated line-to-line voltage

Electric circuit representing the installation of Fig. 16.14.

the

the line current.

following open-circuit and short-circuit

Figure 16.15

at

and the

corre-

and line-to-neutral

volt-

then reduced to zero and the

three stator terminals are short-circuited together.

With the generator again running exciting current

value

/ xn

The tor

is

at rated

gradually raised to

speed, the

its

original

.

resulting short-circuit current / sc in the sta-

windings

is

measured and

X

s

is

calculated by us-

ing the expression

X = £ n // s

(16.2)

Sc

where

X =

synchronous reactance, per phase

En —

rated open-circuit line-to-neutral voltage

s

[11]*

LV|

Figure 16.16 Voltages and impedances its connected load.

:;:

in

a 3-phase generator and

This value of reactance.

behavior.

It

X

is

h

corresponds

widely used

to the direct-axis

to describe

synchronous

synchronous machine


/ sc

=

When

short-circuit current, per phase, using

the

same

exciting current

En

required to produce

The synchronous reactance

/ Xll that

the terminals are short-circuited, the only

impedance limiting

was

[A]

to the

the current flow

X - EJ1 = - 5n

is

its

heavily saturated, the value of X s

may

the iron

be only half

unsaturated value. Despite this broad range

usually take the unsaturated value for

due

that

4000/800

s

When

is

synchronous reactance. Consequently,

not constant, but

is

varies with the degree of saturation.

349

X

s

The synchronous reactance per phase

we

because

fore 5

it

yields sufficient accuracy in most cases of interest.

The equivalent

b.

is

there-

il.

phase

circuit per

shown

is

in

Fig. 16.18a.


Example 16-2

A

3-phase

synchronous generator produces an

open-circuit line voltage of citing current

is

6928

50 A. The ac terminals

=

are then

short-circuited, and the three line currents are

be 800 A.

a.

Calculate the synchronous reactance per phase.

b.

Calculate the terminal voltage

The

resistors are

connected

if

wye

in

current /

three 12 il

across the

= EJZ=

minals.

E=

1

X;

\

2

+

5

(2.12) 2

n

3

is

The voltage across

ter-

2

\ 12

=

found

to

\R

Z

V when the dc ex-

circuit is

1R

=

4000/13

=

A

308

the load resistor

X

308

1

2

is

- 3696 V

Solution a.

The

line-to-neutral induced voltage

The

line voltage

under load E,

E0 = E L /V3 = 6928/V3

(8.4)

= 4000 V

- V3 E = V3 X 3696 - 6402 V

The schematic diagram of x

s

= 5

is

is

sualize

a

what

16.11

We

is

happening

Fig. 16.18b helps us viin the actual circuit.

Base impedance,

recall that

first select

when using

a base voltage

per-unit

we

EH

power per phase as the base power.* lows that the base impedance Z B is given by

the rated

1

line voltage =

=

we

In the

use the rated

line-to-neutral voltage as the base voltage

t

s

the per-unit system

and a base power.

case of a synchronous generator,

Z,

X

^

It

and fol-

(16.3)

6394 V

alternator :;:

Figure 16.18 a. See Example 16-2. b.

Actual line voltages and currents.

In

many power studies the base power is selected to be equal power of the generator and the base voltage is

to the rated

the linc-to-linc voltage. This yields the

base impedance.

same value Z n

for the


350

where

ZB =

base impedance (line-to-neutral) of the generator [ft]

EB =

Note

base voltage (line-to-neutral) fV] [

The

d.

for other

may

X

s

(pu)

lies

between

Note

P =

nous reactance of

60 Hz ac generator has a synchro1

.2

from

are

2

/

line to

(pu) R(pu)

pu and a resistance of 0.02

2 l

X

=

0.02

0.02

that at full-load the per-unit value

equal to

Example 16-3 5 kV,

=

be expressed as

=

of

I is

1

The copper

1

7.5

per-unit copper losses at full-load are

=

depending upon the design of the machine.

A 30 MVA,

X

0.02

that the generator possesses.

.

2,

ZB =

impedance values

that all

P(pu)

impedances

a per-unit value of ZB In general,

and

0.15 ft

used as a basis of comparison

is

Thus, the synchronous reactance

0.8

0.02

=

neutral.

S B — base power per phase VA]

The base impedance

=

losses for

0.02 S B

600

=

all

0.02

3 phases are

X

30

=

0.6

MW

kW

pu.

16.12 Short-circuit ratio Calculate a.

b. c.

d.

The base

voltage, base

power and base imped-

Instead of expressing the synchronous reactance as

ance of the generator

a per-unit value of

ZB

The The The

sometimes used.

the ratio of the field current / x!

actual value of the synchronous reactance actual

needed

winding resistance, per phase

age

copper losses

total full-load

EH

a.

is

SH

Eq.

000/V3

15

= 30MVA/3 = = 10 7 VA

10

is

1

6.2.

Thus,

if

X

as defined in

s

X

s

is

l

.2,

the

1/1.2 or 0.833.

is

many

=

(16.3)

types of loads, but they can

all

be reduced to

two basic categories:

7.5 ft


is

X = X (pu) X ZB = 1.2ZB = 1.2 X = 9(2 s

The resistance per phase

7.5

is

Figure 16.19

R =

The

short-circuit.

exactly equal to the

the per-unit value of is

volt-

produce

The behavior of a synchronous generator depends upon the type of load it has to supply. There are

ti

s

|// X 2)

to

is

16.13 Synchronous generator under load

MVA

Z = E H 2 /S H - 8660 2 /10 7

c.

on a sustained

short-circuit ratio

is

The base impedance

b.

,

armature

needed

reciprocal of the per-unit value of

E B = £,/V3 = = 8660 V The base power

the short-circuit ratio

to the field current I x2

short-circuit ratio (7 X

The base voltage

,

to generate rated open-circuit

rated current / B

Solution

It is

R(pu)

X ZB

Equivalent

circuit of

a generator under load.


1

.

2.

The

Isolated loads, supplied by a single generator

The

infinite

bus

diagram

resulting phasor

Note

16.20.

E

that

t)

given

is

E

leads

by

5

begin our study with isolated loads, leaving greater than the terminal voltage, as

the discussion of the infinite bus to Section 16. 16. In

Consider a 3-phase generator that supplies power to a load

having a lagging power

some cases

the load

is

somewhat

that current / leads the terminal voltage factor. Fig. 16. 19

represents the equivalent circuit for one phase. In or-

What

effect

The answer

does

.

E0

in Fig.

capacitive, so

by an angle

The

16.21.

across the synchronous reactance the following facts:

list

Current

/

lags behind terminal voltage

E by

0.

voltage

EK

to the pha-

sum of £ and Ex However, the terminal voltage is now greater than the induced voltage £0 which is a sor

angle

0.

90° ahead of

is still

£0 is again equal

the current. Furthermore,

an

is

have on the phasor diagram?

this

found

is

Fig.

we would expect.

der to construct the phasor diagram for this circuit,

1

in

degrees.

Furthermore, the internally-generated voltage

We

we

35

.

,

2. 3.

Cosine 6 Voltage

= power

Ex

4.

EK

Voltage

the phasor 5.

Both the

£0

by 90°.

/

- jIX s

E0

very surprising result. In effect, the inductive reac-

across the synchronous reactance

leads current sion

factor of the load.

It is

given by the expres-

sum of £

Ex

plus

£x

O

is

equal to

rent / x

is

that

s

enters into partial resonance with the capacit

may appear we

something for nothing, the higher terminal

more power.

voltage does not yield any

load

is

entirely capacitive, a very high ter-

.

minal voltage can be produced with a small excitare voltages that exist inside

synchronous generator windings and cannot

Flux

X

reactance of the load. Although

If the

ing current.

However,

in later

that such under-excitation

be measured directly. 6.

itive

are getting

.

generated by the flux

and

tance

is

chapters,

we

will see

undesirable.

produced by the dc exciting cur-

Example 16-4 .

A

36

MVA,

20.8 kV, 3-phase alternator has a syn-

chronous reactance of 9 1

Cl

and a nominal current of

kA. The no-load saturation curve giving the

En

tionship between If the excitation is

age remains fixed

and

/ x is

given

rela-

in Fig. 16.13b.

adjusted so that the terminal voltat

current required and

21 kV. calculate the exciting

draw

the phasor

diagram for

the following conditions:

Figure 16.20 Phasor diagram

for

a lagging power factor load.

a.

No-load

b.

Resistive load of 36

c.

Capacitive load of 12

MW Mvar

Solution

We

/

shall

immediately simplify the circuit

to

show

only one phase. The line-to-neutral terminal voltage for

all

cases

is

fixed at

E= E

a.

20.8/V3

At no-load there

is

=

12

kV

no voltage drop

chronous reactance; consequently, Figure 16.21 Phasor diagram

for

a leading power factor load.

E = £ = tl

12

kV

in the

syn-


352

Q=

E,E 0

1

2/3

-

4 Mvar

X

10 712

^12kV The Figure 16.22a Phasor diagram

line current

= Q/E = - 333 A

/

at no-load.

is

The voltage across The exciting current /x

b.

=

The /

36/3

100

-

(see Fig. I6.l3b)

£ x =JIX = 1000 X ,/

This voltage

The voltage £ generated by ()

phasor sum of sor diagram,

£o =

\

E and £ x

its

value

.

1000

A

is

equal to the

/ x is

kV

Figure 16.22c Phasor diagram with a capacitive

E0 generated £

The voltage

£<,

1

6.22c).

/x

kV

by

E

»

Note

that

£0

is

=

kV

12

load.

/ x is

equal to the

.

12

+ (-3)

70

A

is

(see Fig. 16.13b)

again less than the terminal volt-

age £.

The required exciting current

is

The phasor diagram

= 200 A

/x

c.

kV/!90 o

The corresponding exciting current

given by 15

x

= £ + £x = = 9kV

Referring to the pha-

T~£; = Vl2 2 + 9 2 =

3

»

kV^90°

9

-

E0 9 kV

phasor sum of £ and

=

9

i

90° ahead of/.

is

is

Ex

the terminal voltage.

9

00Q

(333 A

is

The current is in phase with The voltage across Xs is S

s

(

leads / by 90° (Fig.

6.22a.

3

X K/712 000 =

12

1

As before £x

MW

12

full-load line current

= PIE =

X

£x = JIX,=j333 X

is

The phasor diagram is given in Fig. With a resistive load of 36 MW: The power per phase is

P =

4

(see Fig. 16.13b)

The phasor diagram is given in Fig. 16.22b. With a capacitive load of 12 Mvar: The reactive power per phase is

given

for this capacitive load

is

in Fig. 16.22c.

16.14 Regulation curves When a single load,

we

voltage

The

synchronous generator feeds a variable

are interested in

£ changes

relationship

knowing how

the terminal

as a function of the load current

between

£ and / is called the

/.

regula-

tion curve. Regulation curves are plotted with the field

excitation fixed and for a given load Fig.

MVA, E

1 1

Figure 16.22b Phasor diagram with a

kA

unity

12

power

kV

factor load.

1

power

factor.

6.23 shows the regulation curves for the 36

21

kV,

3-phase

generator

discussed

in

Example 16-4. They are given for loads having unity power factor, 0.9 power factor lagging* and 0.9 power factor leading, respectively. These curves were derived using the method of Example 16-4,


povver factor

"-^

The percent regulation

x

0.9 lacigincJ

%

.

regulation

353

is

=

X

100

rated load 10 Lr\/ mnn a

/ y lea dine

(\5-

=

12)

X

100

- 25%

12

I

We

note that the percent regulation of a synchro-

nous generator erator.

much

is

The reason

is

greater than that of a dc gen-

the high impedance of the syn-

chronous reactance.

750

500

250

Load current

1250

1000

16.15 Synchronization of a generator

/

Figure 16.23

We often

Regulation curves of a synchronous generator at three different load

power

E 0 was

kept fixed instead of E. In each

power requirements of

connected

starting point for all the curves

Later,

terminal voltage

000 A). The change

rent

load

(

(

was set so that the was the rated line2 kV) at rated line cur-

1

is

voltage between no-load and

full-

expressed as a percent of the rated terminal

voltage.

load.

The percent regulation

is

given by the

a large utility system

to the

when

system to provide the extra power.

power demand

falls,

selected gen-

from the sys-

until

the

power again builds up

the following day.

Synchronous generators are therefore regularly being connected and disconnected from a large grid in response to

En

"

regulation

many

X

100

power

customer demand. Such a grid

said to be an infinite bus because

equation

%

in

For example, as

erators are temporarily disconnected

tem

1

in

common

build up during the day, generators are successively

of the three cases, the value of E0

to- neutral

connect two or more generators

to

factors.

the

except that

have

parallel to supply a

it

is

contains so

generators essentially connected

in parallel

that neither the voltage nor the frequency of the grid

can be altered.

where

Before connecting a generator to an

=

EH =

(or in parallel with another generator),

no-load voltage VJ [

synchronized. rated voltage [VI

nized 1.

Example 16-5 power

factor curve in Fig. 16.23.

it

A

generator

meets

all

is

it

bus

must be

said to be synchro-

the following conditions:

The generator frequency

is

equal to the system

frequency.

Calculate the percent regulation corresponding to the unity

when

infinite

2.

The generator voltage

is

equal to the system

voltage.

Solution

The

rated line-to-neutral voltage at full-load

is

3.

The generator voltage

is in

phase with the sys-

tem voltage.

EB =

12

kV 4.

The no-load terminal voltage 15

is

kV

The phase sequence of the generator same as that of the system.

To synchronize an

alternator,

is

we proceed as

the

follows:


1

.

Adjust the speed regulator of the turbine so that the generator frequency

is

3.

ment has

frequency. 2.

En

is

E0 and E

(Fig. 16.24).

by

This

instru-

a pointer that continually indicates the

phase angle between the two voltages, covering

Adjust the excitation so that the generator volt-

age

Observe the phase angle between

means of a synchroscope

close to the system

the entire range

equal to the system voltage E.

from zero

to

360 degrees.

Although the degrees are not shown, the

dial has a

when the voltages are in when we synchronize an alter-

zero marker to indicate phase. In practice,

nator, the pointer rotates slowly as

it

tracks the

phase angle between the alternator and system voltages. If the generator frequency

is

slightly

higher than the system frequency, the pointer rotates clockwise, indicating that the generator has a

tendency to lead the system frequency. Conversely,

if

the generator frequency

is

slightly

low, the pointer rotates counterclockwise.

bine speed regulator

is

The

A fi-

that the pointer barely creeps across the dial.

nal

check

is still

4.

The

made

to see that the alternator voltage

equal to the system voltage. Then,

moment Figure 16.24 Synchroscope.

is

the pointer crosses the zero

line circuit

tur-

fine-tuned accordingly, so

breaker

is

at the

marker

.

.

closed, connecting

the generator to the system.

{Courtesy of Lab- Volt)

Figure 16.25 This floating r/min,

oil

derrick provides

60 Hz supply

all

its

own energy needs. Four

board are thyristor-controlled dc motors. (Courtesy of Siemens)

diesel-driven generators rated

the electrical energy. Although ac power

is

1200 kVA, 440 V, 900 all the motors on

generated and distributed,

~


In is

modern generating

stations, synchronization

usually done automatically.

If

will

We

seldom have except

parallel

in isolated locations (Fig,

As mentioned previously, to

it

is

in

16.25).

An

bus

own

its

many

alternators

is

system so powerful

a

that

its

terminals.

Once con-

nous generator becomes part of a network comprising hundreds of other generators that deliver

thousands of loads.

It is

power

impossible, therefore, to

specify the nature of the load (large or small, resistive or capacitive)

connected

to the terminals of this

What,

then, determines the

particular generator.

tion,

=

(E0

-

E)/X s

Because the synchronous reactance

£x

the current lags 90° behind

(Fig.

is 1

therefore 90° behind E, which

is

inductive,

6.26b). The

machine delivers? To answer

we must remember

means

that

that both the value

we

were an induc-

it

we

Consequently, when

tive reactance.

a synchronous generator, to the infinite bus.

The

it

over-excite

power power increases as

supplies reactive

reactive

we raise the dc exciting current. Contrary we might expect, is impossible to make a it

tor deliver active

now

Let us

power by

raising

its

ques-

and the

can vary only two

genera-

excitation.

decrease the exciting current so that

Ea becomes smaller than E. As a result, phasor E = £0 — £" becomes negative and therefore points to the left (Fig. 6.26c). As always, current / = EJX lags 90° behind E x However, this puts / 90° ahead of E, x

S

.

which means it

that the alternator sees the

system as

if

were a capacitor. Consequently, when we under-

excite an alternator,

machine parameters:

what

to

1

this

frequency of the terminal voltage across the generator are fixed. Consequently,

if

it

nected to a large system (infinite bus), a synchro-

the

therefore circulate in the circuit

the generator sees the system as

voltage and frequency upon any

apparatus connected to

power

E

£o

it.

infinite

imposes

to

/

current to

/ will

s

much more common

bus) that already has

(infinite

current

s

X

given by

given by

connect a generator to a large power system

connected

E

experience a difference of potential

A

connect only two generators

to

increase the exciting current, the volt-

increase and the synchronous reactance

=

Synchronous generator on an infinite bus

16.16

we now

Ea will

age

355

it

draws reactive power from the

system. This reactive power produces part of the 1

.

2.

The

exciting current

magnetic

/x

The mechanical torque exerted by Let us see

fects the

how

der

the turbine

field required

by the machine; the remain-

supplied by exciting current

is

—

16.18 Infinite bus effect of varying the mechanical torque

—

16.17 Infinite bus effect of varying the exciting current

Let us return to the situation with the synchronous

generator floating on the

connect is

equal

it

after

we synchronize

to an infinite bus, the

to,

and

E of the system

in

a generator

and

induced voltage

£

()

phase with, the terminal voltage

(Fig.

1

6.26a). There

is

no difference

of potential across the synchronous reactance and,

consequently, the load current the generator

no power;

it

is

is

.

a change in these parameters af-

performance of the machine.

Immediately

/x

/ is

zero.

connected to the system,

said to float

on the

line.

Although it

delivers

and

in

phase. If

we open

line,

the

E0

and

E being

equal

steam valve of the

tur-

bine driving the generator, the immediate result

an increase

in

mechanical torque (Fig.

1

is

The

6.27a).

E0 will atmaximum value a little sooner than before. £0 will slip ahead of phasor E, leading by

rotor will accelerate and, consequently, tain

its

Phasor

a phase angle 8.

same

it

Although both voltages have

the

value, the phase angle produces a difference

Figure 16.26a Generator

floating

on an

infinite

bus.

Figure 16.26b Over-excited generator on an

infinite

bus.

Figure 16.26c Under-excited generator on an

infinite

bus.

turbine

Figure 16.27 a.

Turbine driving the generator.

b.

Phasor diagram showing the torque angle

5.

356


E = £ — E across the

of potential

x

()

synchronous

re-

actance (Fig. 16.27b).

A

current / will flow (again lagging 90° behind

£J, but

this

time

it is

almost

in

phase with E.

It

fol-

power delivered by

the generator also increases.

To

understand the physical meaning of the diagram,

let

us

tor will

continue to accelerate, the angle 8 will con-

tinue to diverge,

system

the

and the

power delivered to up. However, as soon

electrical

will gradually build

power delivered to the system is power supplied by the turwill cease to accelerate. The generator

examine the

and position of the

currents, fluxes,

poles inside the machine.

Whenever 3-phase

lows that the generator feeds active power into the system. Under the driving force of the turbine, the ro-

currents flow in the stator of a

generator, they produce a rotating magnetic field identical to that in an induction motor. In a synchrothis field rotates at the same speed same direction as the rotor. Furthermore, same number of poles. The respective

nous generator

and

in the

has the

as the electrical

it

equal to the mechanical

fields

bine, the rotor

stationary with respect to each other.

again run

will

angle 8 between It

is

synchronous speed, and the Torque

at

Ea

and

£ will

remain constant.

important to understand that a difference of

potential

is

created

when two equal

voltages are out

of phase. Thus, in Fig. 16.27, a potential difference of 4

kV

exists

between

Ea

and

E,

although both

produced by the rotor and

When

tween them.

the stator current /

veloped.

The only

16,19 Physical interpretation of alternator behavior

ator (by admitting

it

the

phase angle between

value of

E

x

E0

and

E

increases, the

increases and, hence, the value of / in-

creases. But a larger current

means

that the active

may

be set up be-

the generator floats on the line,

zero and so no forces are de-

is

flux

that created

is

induces the voltage

If a

6.27b shows that when

Depending on

hand and the rotor poles on the other hand, powerful forces of attraction and repulsion

and

1

stator are, therefore,

the relative position of the stator poles on the one

voltages have a value of 12 kV.

The phasor diagram of Fig.

357

£

mechanical torque

()

is

(Fig.

1

by the

rotor,

6.28a).

applied to the gener-

more steam

to the turbine), the

rotor accelerates and gradually advances by a

mechanical angle a, compared sition (Fig.

begin

1

to its original po-

6.28b). Stator currents immediately

to flow,

owing

to the electrical

5 between induced voltage

E 0 and

phase angle

terminal volt-

age E. The stator currents create a revolving field

I:

I

/

Figure 16.28a The N poles of the of the stator.

rotor are lined

up with the S poles

Figure 16.28b The N poles of the the stator.

rotor are

ahead

of the

S poles

of


358

and a corresponding

set

of

N

and S poles. Forces

P =

of attraction and repulsion are developed be-

-

sin

8

(16.5)

tween the stator poles and rotor poles, and these magnetic forces produce a torque that opposes

where

P — En = E= X = 8 =

mechanical torque exerted by the turbine.

the

When

the electromagnetic torque

is

equal to the

mechanical torque, the mechanical angle will no longer increase but will remain

at

a constant

active power, per phase

There

is

a direct relationship

a and

chanical angle

between the me-

the torque angle 8, given

b

=

pa/2

by

(16.4)

[

V

]

terminal voltage, per phase [V]

synchronous reactance per phase

s

value a.

[W]

induced voltage, per phase

torque angle between

En

This equation can be used under tions, including the case

when

and all

E

[il]

[°]

load condi-

the generator

is

con-

nected to an infinite bus.

where

To understand 8

=

torque angle between the terminal voltage

E0

E and

the excitation voltage

meaning, suppose a generator

infinite

generator

is

bus having a voltage E.

kept constant so that

The term E0 E/X S is then power which the alternator

the generator

mechanical angle between the centers

vary directly with sin

of the stator and rotor poles [mechanical

its

connected to an

Furthermore, suppose that the dc excitation of the

[electrical degrees)

p — number of poles on

a ~

is

degrees]

we admit more

Thus, as

and

so, too, will the active

is

constant.

delivers to the bus will

8, the sine

gle.

£0

fixed, and the active

of the torque an-

steam, 8 will increase

power output. The relatwo is shown graphically in Note that between zero and 30° the

tionship between the

Example 16-6 The rotor poles shift

by

1

.

Fig.

of an 8-pole synchronous generator

0 mechanical degrees from no-load

to full-

load.

gle of 30°.

Calculate the torque angle between

a.

16.29.

power increases almost linearly with the torque angle. Rated power is typically attained at an an-

E0

and the

E at full-load. E or £u is leading?

terminal voltage

Which

b.

voltage,

,

Solution a.

The torque angle 8

b.

When

is:

= pa/2 = = 40°

8

X

10/2

a generator delivers active power,

ways leads

E0

al-

E.

16.20 Active power delivered by the generator

We

can prove (see Section 16.23) that the active

power delivered by a synchronous generator given by the equation

is

Figure 16.29 Graph showing the relationship between the active power delivered by a synchronous generator and the torque angle.


=

However, there is an upper limit to the active power the generator can deliver. This limit is reached when 8 is 90°. The peak power output is then

P m AX — E„E/X .

.

S

If

we

the infinite bus.

The

tion

is

trip as

flow

in

try to

exceed

and

X

(3

13.3)

= 4()MW

the stator. In practice, this condi-

soon as synchronism

16.21 Control of active

large, pulsating cur-

never reached because the circuit breakers

power

is,

this limit

rotor will turn faster than the

is lost.

We

resynchronize the generator before liver

the alternator

therefore,

and lose synchronism with

rotating field of the stator, rents will

MW

The peak power output of

(such as by admitting more steam to the turbine), the rotor will accelerate

13.3

359

it

then have to

can again de-

When

synchronous generator

a

system,

speed

its

is

connected

to a

kept constant by an extremely

is

sensitive governor. This device can detect speed

changes as small as 0.01%. system sensitive

to the grid.

power

An

automatic control

such small speed changes im-

to

mediately modifies the valve (or gate) opening of

Example 16-7

the turbine so as to maintain a constant speed and

A

constant

36

MVA,

21 kV, 1800 r/min, 3-phase generator

On

connected to a power grid has a synchronous reactance of 9 (1 per phase. If the exciting voltage

is

kV (line-to-neutral), and the system voltage 17.3 kV (line-to-line), calculate the following:

12

is

in

b.

before

We

network

have

is

is

done

as

more elaborate systems

under the control of a computer.

individual overspeed detectors are

always ready

to

particularly

a generator, for

if

station so that

and transmission of energy

efficiently as possible. In

the entire

stations.

communicate with each other

modify the power delivered by each

In addition,

Solution a.

station operators

the generation

of step (loses synchronism)

falls out

it

power delivered by

advance between the various generating

The

The active power which the machine delivers when the torque angle 8 is 30° (electrical) The peak power that the generator can deliver

output.

each generator depends upon a program established

to a.

power

a big utility network, the

respond to a large speed change,

one reason or another,

should suddenly become disconnected from the

Ea = E= 8 =

kV

12

17.3

system. Because the steam valves are

kV/V3 =

10

kV

attain a

30°

onds.

The

active

power delivered

P = (E„EIXS)

= The

total

6.67

X

10/9)

X

grid

is

X

centrifugal forces at synchronous speed

any excess speed can quickly create a very situation.

Consequently, steam valves

gencies. At the

- 20

all

three phases

is

burners must be shut

MW

sin

90

10/9)

X

same

is

attained

time, the pressure build-up

off.

16.22 Transient reactance synchronous generator connected

to a

system

subject to unpredictable load changes that 1

in

the steam boilers must be relieved and the fuel

A X

may

to 5 sec-

must immediately be closed off during such emer-

(E0 EIX S ) (12

4

0.5

The maximum power, per phase, when 8 = 90°.

P = =

The

in

wide

are already close to the limit the materials can with-

dangerous

MW

6.67)

speed 50 percent above normal

stand, so

8

power delivered by (3

b.

(12

sin

power

to the

still

open, the generator will rapidly accelerate and

times occur very quickly.

In

is

some-

such cases the simple


360

X'

short-

direct bearing

circuit

at the

on the capacity of the

circuit breakers

generator output. In effect, because they must short-circuit

a

interrupt cycles,

it

three

in

to

six

follows that they have to interrupt a very

normal

high current.

load

On

the other hand, the low transient reactance

simplifies the voltage regulation

problem when the

load on the generator increases rapidly. First the

drop due

ternal voltage -short-circuit-

load

would be ing.

if

to X' ci

the synchronous reactance

X

Second,

f

in-

smaller than

is

X

s

were

below

stays at a value far

X

it

act-

for a

s

sufficiently long time to quickly raise the exciting

current

/x

.

which helps

•V'd

-

/

Raising the excitation

increases

E0

.

to stabilize the terminal voltage.

•

Example 16-8

A 250 MVA,

time

25 kV, 3-phase steam-turbine gener-

ator has a synchronous reactance of 1.6 pu and a

Figure 16.30

transient reactance X' d of 0.23 pu.

Variation of generator reactance following a short-

rated output at a circuit.

circuit

power

factor of

suddenly occurs on the

It

delivers

100%.

line,

A

its

short-

close to the

generating station. equivalent circuit flect the

shown

in Fig.

does not

16. 17

re-

behavior of the machine. This circuit

only valid under steady-state conditions or

is

when

the load changes gradually.

tance X'

X

s

X

f

varies

when

a generator

is

suddenly short-circuited. Prior to the short-circuit, the synchronous reactance

is

X

simply

However,

s

much lower value X' d

creases gradually until

upon 100

T.

it

is

kVA

it

only

in

lasts a fraction

the 1000

It

.

again equal to

The duration of the

the size of the generator. For

machines

the circuit breakers should fail to

a.


then in-

X

interval

s

after a

depends

it

The

sy

2

=

25 000 /(250

=

2.5 fl

nchronous reactance

X

10

6 )

is

may

last as

X =X s

=

long

s

(pu)

1.6

X

ZB 2.5

= 4il

The reactance X' d of the alternator.

It

is

may

called the transient reactance

be as low as

synchronous reactance. short-circuit current to the

is

machines below

of a second, but for

MVA range

the generator

at

as 10 seconds.

sponding

open

Solution

immedi-

the instant of short-circuit, the reactance

time interval

The induced voltage E0 prior to the short-circuit The initial value of the short-circuit current The final value of the short-circuit current if

varies as a function of time.

shows how

ately falls to a

c.

must be replaced by another reac-

whose value

Fig. 16.30

a.

b.

For sudden load current changes, the synchronous reactance

Calculate

is

1

5 percent of the

Consequently,

much higher

the

The

initial

rated line-to- neutral voltage per phase

E = 25/\3 =

14.4

kV

than that corre-

synchronous reactance

JVS

.

This has a

The

rated load current per phase

is

is


36

rated

—-load —I—

short-circuit

kA

,47.3

-i

Change Figure 16.31 See Example 16-8.

current

in

5/V3

E 6

= 250 X

X

10 /(1.73

which

5

6

s

short-circuit occurs across

See Example

drop

Ex

shows

and during the

is

23.1

T of

terval

= 5774 X

IX S

= EJX, = = 6.8 kA

16-8.

27.2/4

only 1.2 times rated current.

is

Fig. 16.32

internal voltage

£x = =

1

1

1

4

25 000)

= 5774 A The

when a

the terminals of a generator.

/

=

1

2 3 * time

1

Figure 16.32

5780 A

/

1

1

0

4

We

assume

to,

a time in-

5 seconds. Note that in practice the cir-

would

cuit breakers

kV

the generator current prior

short-circuit.

certainly trip within 0.

1

s

after

the short-circuit occurs. Consequently, they have to

The current is in phase with E because power factor of the load is unity. Thus,

interrupt a current of about

the

ring to the phasor diagram (Fig. 16.31),

47 kA.

refer-

E0

is

16.23 Power transfer between E = i}

\

E2 +

E\

2

V14.4 +

The

2

The

kV

27.2 b.

two sources 23.1

transient reactance

circuit of Fig.

because

it is

0.23

6.33a

is

particularly important

in the

study of generators,

synchronous motors, and transmission

is

we

such circuits

=

1

encountered

X

In

lines.

are often interested in the active

power transmitted from a source A to a source B vice versa. The magnitude of voltages £, and E 2

2.5

<

= The

0.575

initial short-circuit

n

or as

well as the phase angle between them, are quite ar-

current

bitrary.

Applying Kirchhoff s voltage law

to this

is

= EJX = 27.2/0.575

circuit,

we

obtain the equation

l1

=

47.3

If

kA

we assume

angle 6 and

which c.

If

is

8.2 times rated current.

the short-circuit

tion

is

is

sustained and the excita-

unchanged, the current

E

that / lags

leads ]

E2

behind £\ by an arbitrary

by an angle

phasor diagram shown (Fig. leads

/

1

we

obtain the

6.33b). Phasor IX

by 90°. The active power absorbed by B

is

will eventually

level off at a steady-state value:

P = E2 I cos

(16. 6)


362

The active power alway s flows from the lagging voltage. In Fig.

leads

E2

;

1

6.33,

it

hence power flows from

is

the leading to

obvious

that

E

{

left to right.

Example 16-9 Referring to Fig. 16.33a, source

age age

A generates

a volt-

E = 20 kV Z 5° and source B generates a E 2 — 15 kV Z 42°. The transmission line

volt-

{

con-

necting them has an inductive reactance of 14

11.

Calculate the active power that flows over the line

and specify which source

is

actually a load.

Solution

The phase angle between the two sources is 42° — 5° = 37°. The voltage of source B leads that of source A because its phase angle is more positive. Consequently, power Hows from B to A and so A is actually a load. The active power is given by:

E,E2

X

Figure 16.33 Power flow between two voltage sources.

.

- sin

o

20 kV X

(16.8)

kV

15

sin

37°

14

From

the sine law for triangles,

we have 20 000

/X/sin 8

E,/sin

=

£,/sin(90

= EJcos Consequently, Substituting

(

/

cos 0

=

P =

i\f

+

9)

12.9

Note

we

E = E2 = ]

8

=

X= active

active

(16.7)

E2

-

sin 8

(16.8)

power transmitted [W] 1

The physical

[VJ

phase angle between

E

and

x

E2

f°l

reactance connecting the sources

may seem, power flows (

15

kV)

to

cost,

[11

The magnitude off

E

x

and

does not have

is

is

E2

machines machine has a pro-

power output, relaand temperature rise. The following its

why

efficiency,

these characteristics are

]

Let us consider a small ac generator having the

equal to that

:

determined by the angle 6 be-

to be specified.

inti-

mately related.

following characteristics:

power P received by B

between

it

size of an electrical

found effect upon analysis reveals

/

6

16.24 Efficiency, power, and size of

tive

and

10

the one having the higher voltage (20 kV).

find

delivered by A, because the reactance consumes no

tween

that, strange as

from the source having the lower voltage

voltage of source 2 [V]

the phase angle

X

12.9

1

X

voltage of source

active power.

=

MW

electrical

P =

000

B

where

The

15

14

£, sin h/X

16.7) in Eq. 16.6,

X

0.602

=

kW

power output

I

rated voltage

120 V, 3 phase

rated current

4.8

rated speed

1

A

800 r/min


efficiency

73%

input torque

7.27

morhent of

0.0075 kg*m

inertia

Under these conditions, we can

N-m

of the generator as

erties 2

its

size

For example, suppose that

m m

80

predict the prop-

is

increased.

dimen-

the linear

all

The volume

363

external diameter

0.

external length

0.15

mass

20 kg

mass

power output/mass

50 W/kg

Using

we can

The mass of the bigger machine will therefore be 27 X 20 kg = 540 kg. The losses will rise to 27 X 0.37 kW = lOkW. The slots are 3 times wider and 3 times deeper. As a result, the cross section of the conductors is 9 times greater which means they can carry 9 times more current. The larger machine can therefore de-

this information,

1

sions are tripled.

crease by a factor of 3

P"

=

-

X

calculate the losses

100

kW

x

73

eq. 3.6

100

liver a current

As power

input

=

P,

-

losses

1.37

kW

is

kW -

1.37

1

.0

losses comprise the

kW

that

kW

0.37

PR losses

in the

and eddy-current losses

the hysteresis

and the windage and

windings,

in the iron

way

that

actly the

same proportion, while keeping

materials throughout. Thus, iron lamination

type

is

used

was used

in the larger

of insulation

is

if

such

in

dimensions are raised

linear

its

the

the

same

keep the same

current densities (A/itT) as in the original machine.

las)

in the

flux densities (tes-

various parts of the magnetic circuit

(core, air gap, stator teeth, etc.).

As

a result, the

losses per

cm

3

R

will be

original machine.

and

2

I

It

losses per

cm 3

B

is

flux density in the larger

windage and

assume

unchanged.

However,

before.

it

recall

machine

length

the

/

is

the

has

it.

same

tripled.

As

has the same

number of conductors

and be-

as before

cause they are connected the same way, the genera-

X

produce a voltage of 9

tor will

120

V =

1080

V.

Thus, by tripling the linear dimensions, the voltage and current both increase by a factor of

9.

This

power output increases 9X9 = 81 times. The power output of the new generator is therefore 81 X kW = 81 kW. The power input needed to drive the ac generathat the

1

tor is P,

:

=

81

kW +

losses

kW. The new efficiency

= is

81

kW +

10

kW =

therefore:

in the

71

=

°

vol-

81

friction

91

its

X

100

eq. 3.6

P;

=

kW x kW

0.89

100

- 89%

that the

(

left

the speed at

We

Blv.

the length of the conduc-

/ is

The

is

E =

which the flux cuts across

v

follows that the copper losses

same way. number of slots, conductors and interconnections remain the same as before and that the speed of rotation 800 r/min) is further

the flux density,

and

losses also increase the

We

43.2 A.

and the iron

everywhere the same as

that the

A=

times because the diameter of the rotor has tripled.

91

iron losses will increase in proportion to

ume. Let's assume

4.8

regards the generated voltage per conductor,

means

same

X

creases by a factor of 9. Because the larger generator

nuts and bolts. will

of 9

same

magnifying everything, including the bearings,

We

as

also used, thereby duplicating and

will also maintain the

so, too, will

a result, the voltage generated per conductor also in-

machine. The same type

we

27 and

ex-

in

a particular type of

in the stator,

In this larger generator

27. Consequently, the

Furthermore, the peripheral speed v has increased 3

friction losses.

Let us increase the size of the machine a

=

a factor of

determined by equation (2.25)

tor

The

by

the losses.

of the machine: r\

will increase

3

will therefore in-

1

The which the

efficiency has increased from is

a dramatic improvement.

power output has increased

73%

to

The reason

89%

is that

81 times, while the


364

losses increased only

27 times. Consequently, the

bound to

ciency of the machine was

effi-

increase with size.

about 20()°C. Consequently, the cooling of large

machines

a very important matter.

is

The original machine produced an output of 50 W/kg. The larger machine has a mass of 540 kg and produces 81 kW. Consequently, it produces 81

regarding physical size, power output, efficiency,

kW/540 kg = 150 W/kg which

including ac and dc motors and transformers.

3 times greater

is

In

conclusion, the general principles covered here

temperature

rise

and so

forth,

apply to

machines,

all

than before.

The proof,

generator

larger

if

eighty-one

produce 81 kW,

X 20

therefore

is

relatively


and cheaper than the smaller machine. As

lighter

=

kg

kW

1

their

1620 kg.

Practical level

generators were used to

combined mass would be 81 This generating center would

1

6-

1

Why

obviously be more costly and take up more floor space than the single 81

As another matter of

moment of mass and

inertia

,/

kW

of a rotor

the square of

we

is

proportional to

Hence, when linear dimensions are

= nir —

243.

The moment of

therefore 243

The

X

inertia

tripled,

X

27

2

3

J

=

3

5

=

1.8

kgm 2

16-3

is

the original

1

kW

1080 V,

rated voltage

43.2

rated current

3

A

1800r/min

89%

moment

of inertia

Nm

1.8

kg-nr

external length

0.45

1

m m

linear

dimensions are

temperature

rise.

the losses increase 27 times. Hence, the

power

dis-

sipated per square meter increases by a factor of better

prevent

damage

is

directly-coupled genera-

If the

must generate a frequency of 60 Hz,

The number of poles on The exaci turbine speed

An

the rotor

isolated 3-phase generator produces a

a load

If

is

cooling

hound

maintain the same line voltage?

What

conditions must be met before a gener-

Calculate the

number of poles on

system?

the genera-

means

to the insulating materials, the

3.

are

Calculate the

number of poles on shown in Fig. 16.

6-8

A 3-phase

generator turning

Hz.

How

fected to

its

if

200 r/min

will the terminal voltage be af-

terminals?

Resistive load

tem-

b.

Inductive load

c.

Capacitive load

of

1

1

the following loads are connected

a.

maximum

at

the air1

generates a no-load voltage of 9 kV, 60

To

to be hotter.

perature rise has to be limited to a

6-7

craft generator 1

unless

6-6

When

tripled, the heat-dissipating

used, the larger machine

found

close to

tor in Fig. 16.12 using the information given.

150W/ks is

is

it

350 r/min.

at

ator can be connected to a 3-phase

surface area of the machine increases 9 times but

Consequently,

6-5

540 kg

big problem

site,

should turn

to

1

The one

analyzing a hydropower

connected to the machine, must the exci-

1

mass power output/mass

the larger?

tation be increased or decreased in order

483 0.54

is

having a lagging power factor of 0.8

rated speed

external diameter

For a given power output, which of

no-load line voltage of 13.2 kV.

phase

efficiency input torque

In

a.

16-4

kW

main differences between steam-

that the turbines

b.

81

wye?


are in striking contrast to

machine.

power output

State the

tors

.

characteristics of the larger generator are

summarized below. They

synchronous generators?

in large

the stator always connected in

these machines

=

of the larger machine

0.0075 kg-nr

is

ators.

its

will

are the advantages of using a stationary

turbine generators and salient-pole gener-

recall that the

radius (see Table 3A).

its

increase by a factor of J

16-2

generator.

interest,

What

armature


16-9

In

Problem

16-8,

if

the field current

Calculate

is

kept constant, calculate the no-load volt-

a.

age and frequency

b.

the speed

if

is

The synchronous impedance Z per phase The total resistance of the circuit, per s

lOOOr/min

a.

phase

5 r/min.

b.

365

The

c.

total reactance

of the circuit, per

phase

Intermediate level

16-10

What

meant by the synchronous

is

tance of a 3-phase generator?

16-11

reac-

Draw

nator

meaning of

all

the parameters.

State the advantages of brushless excita-

Using a schematic

how

circuit diagram,

the rotor in Fig. 16.7

16-17

show

needed

a

a.

24.2 12.1

A 3-phase

£0 16-18

Ev

kV

16

generator possesses a synchro-

is

100

(1,

calculate the value of

per phase.

,

The generator

in Fig.

is

3

kV

connected to an

Q and the excitation

per phase

to

load having a lagging

1

6.2 has a synchro-

nous reactance of 0.4 H, per phase.

nous reactance of 6 voltage

KVA,

factor of 0.8. If the synchronous re-

actance

kV kV

generator rated 3000

2400 KVA,

power

to generate a no-

load line voltage of

b.

A 3-phase

20 kV, 900 r/min, 60 Hz delivers power

excited.

is

the volt-

age across the load

Referring to Fig. 16.13, calculate the exciting current

16-13

The phase angle between E0 and

f.

the

tion systems over conventional systems.

16-12

h.

e.

equivalent circuit of a generator and explain the

g.

The line current The line-to-neutral voltage across the load The line voltage across the load The power of the turbine driving the alter-

d.

It is

bus having a

infinite

line

voltage of 14 kV, and the excitation volt-

(ref. Fig.

age

is

adjusted to

1

.

14 pu.

16.19). Calculate the line-to-neutral volt-

age

E for

a resistive load of 8 (2 and draw

Calculate

the phasor diagram. a.

16-14

a.

In

Problem 16-13, draw

the curve of

E

for the following resistive loads: infin-

sus

/

ity,

24. 12, 6,

3,0 ohms. power P per phase

b. Calculate the active

b. c.

in

inside stator circumference) corresponding

each case. c.

Draw

to this

the curve of

E

versus

value of load resistance a

is

P.

the

For what

power output

16-19

Referring to Fig.

1

6.2, calculate the length

of one pole-pitch measured along the

16-16

on the 500

[in].

M VA alternator of

Fig. 16.2 yields the following results:

maximum?

ternal

displacement angle

A test taken 1.

16-15

The torque angle 8 when the generator delivers 420 The mechanical displacement angle a The linear pole shift (measured along the

MW

ver-

circumference of the

in-

2.

stator.

The 3-phase generator shown

Open-circuit line voltage

is

15

kV

for a dc

exciting current of 1400 A.

Using the same dc current, with the armature short-circuited the resulting ac line current is 21

in Fig.

000 A.

16.16 has the following characteristics:

Calculate

E0 = 2440 V

X = s

R = load impedance

Z=

a.

144(1 17 12

175 (2 (resistive)


the generator, per

phase

c.

The value of the synchronous reactance The per-unit value of X s

d.

The

b.

short-circuit ratio

366


Advanced 16-20

16-25

level

b.

d.

tween going

incoming

the cool air, if

the air flow

air is

of 500 A. 1

.3

b.

be-

280 nrVs 16-26

gap

(See Section

16-22

2.

1

the line circuit breakers suddenly trip, cal-

second

1

later,

assum-

By how many mechanical degrees do 1

how many

degrees?

A 400 Hz

electrical

By

alternator has a 2-hour rating of

nal diameter of

mmf re-

length of 9.5

80

1

V,

7).

and has an

slots

22 inches and an

in.

The

rotor

for a field current of 3

Referring to Fig. 16.17, the following

the

-second interval?

position) during the

stator possesses

quired for the iron portion of the magnetic circuit.

.

80

percent power factor (Fig. 16.34a). The

gap length

no-load. Neglect the

at

lb

kg nr.

75 kVA, 1200 r/mim 3-phase, 450

inches, calculate the flux density in

the air

6

poles advance (with respect to their normal

and warm out-

that the air

10

10

culate the speed of the generating unit (tur-

and carries a dc current

Knowing

X

6

driven

is

whose moment of 2 ft The rotor has a J

ing that the wicket gates remain wide open.

Referring to Fig. 16.4, each coil on the rotor has 21 .5 turns,

If

X

54

bine and alternator)

The torque developed by the turbine The average difference in temperature

c.

is

of 4.14 a.

Problem 16-20

in

a hydraulic turbine

inertia is

The total losses in the machine The copper losses in the rotor

a.

16-21

The generator by

The synchronous generator in Fig. 16.2 has an efficiency of 98.4% when it delivers an output of 500 MW. Knowing that the dc exciting current is 2400 A at a dc voltage of 300 V, calculate the following:

1

is

A at

inter-

axial

designated 1

1

5 V.

in-

Calculate

formation

is

given about a generator:

c.

The number of poles on the rotor The number of coils on the stator The number of coils per phase group on

d.

The

a.

Ea = E=

12

kV

14

kV

x =

2

b.

the stator s

Ea

leads

a

E

e.

a.

length of one pole pitch, measured

along the circumference of the stator

by 30°

Calculate the total active power output of

The tor

resistance of the dc winding on the ro-

and the power needed to excite

it

the generator.

1

6-23

b.

Draw

c.

Calculate the power factor of the load.

the phasor

diagram for one phase.

The steam-turbine generator shown 16.3 has a

The

excitation voltage

pu and the machine

is

E0

is

16-27

in Fig.

synchronous reactance of

1

.3

pu.

adjusted to 1.2

connected

to

an

hid List rial a pplica tion

A 33.8

kVA, 480

operate

at

a

bus of 19 kV.

If

the torque angle 5

16-24

active

b.

The

line current

c.

Draw

is

given:

83.4%

Weight: 730

Wk

2

lb

(moment of

Insulation: class the phasor diagram, for

80 percent. The

is

power output

The

factor of

diesel-

designed to

infi-

20°, calculate the following: a.

power

60 Hz is

following additional information Efficiency:

nite

V, 3-phase,

driven emergency alternator

one phase

In Problem 16-23. calculate the active power output of the generator if the steam valves are closed. Does the alternator receive or deliver reactive power and how much?

inertia)

:

15.7

2

lb. ft

B

Calculate a.

The minimum horsepower

rating of the

diesel engine to drive the generator b.

The maximum allowable temperature of windings, using the resistance method

the


16-28

A 220 MVA, 500 0.9

power

r/min, 13.8 kV,

factor, water-turbine

50 Hz,

Efficiency

nous generator, manufactured by Siemens,

Insulation class:

mass of

Total (t

=

stator:

Unsaturated synchronous reactance: 1.27 pu

Runaway speed

158

2

mass of

rotor:

Static excitation

t

current

270

in

generator mode:

890 r/min

metric ton)

Total

factor:

Transient reactance: 0.37 pu

F

of inertia: 525 t*m

power

98.95%

synchro-

has the following properties:

Moment

at full-load, unity

367

is

is

used and the excitation

2980 A under an

excitation volt-

age of 258 V.

t

Figure 16.34a Rotor and stator of a 75 kVA, 1200 r/min, 3-phase, 450 driven by a 100 hp, 1200 r/min synchronous motor.

V,

400 Hz

alternator for shipboard use.

The

alternator

is

Figure 16.34b Stator and rotor of the 100 hp, 1200 r/min, 60

serves as a base

for

the alternator.

The

Hz synchronous motor. The

rotor is

duction motor. (Courtesy of Electro-Mecanik)

equipped with a

squirrel

stator is mounted on a bedplate that also cage winding to permit starting as an in-

368


The generator

also designed to oper-

is

16-29

In industry application

Problem 16-28,

ate as a motor, driving the turbine as a


pump. Under these conditions,

a.

The horsepower

b.

when it runs as a pump motor The kinetic energy of the rotor when

develops an output of 145

the

motor

MW.

Both the stator and rotor are water-

at rated

cooled by passing the water through the c.

hollow current-carrying conductors. The

The

is

treated so that

less than 5 (JiS/cm.

through the stator

its

conductivity

a rate of 8.9

second and through the rotor

at

liters

runs

when

kinetic energy of the rotor its

maximum

it

allowable runaway

is

speed

The pure water Hows

at

it

speed

reaches

water

rating of the generator

d.

per

The time

to reach the

runaway speed

event that a short-circuit occurs

5.9 liters

erator

is

delivering

its

when

rated load,

in the

the gen-

and assum-

per second. Given the above information,

ing that the water continues to flow unchecked

make

through the turbine (gates wide open)

a.

the following calculations:

The

rated active

unity

power

power

output, in

MW

factor and at 0.9 lagging

at

power

factor b. c.

d.

The rated reactive power output, in Mvar The short circuit ratio The value of the line-to-neutral synchronous reactance, per phase

e.

16-30

In

Problem 16-28 calculate the power

dis-

sipated in the rotor windings and the

power

loss per pole.

Knowing

water flow and that the is

inlet

the rate of

temperature

26°C, calculate the temperature of the

water flowing out of the rotor windings.

minimum

the total losses of the generator at full load

What

and unity power factor

the circulating water?

is

the

resistivity

(H-m) of

Chapter 17 Synchronous Motors

17.0 Introduction

motor speed stays constant, irrespective

fixed, the

of the load or voltage of the 3-phase

The synchronous generators described in the previous chapter can operate either as generators or as motors.

When

they run

operating as motors (by con-

field.

The speed of

Most synchronous motors

synchronism with the revolving rotation

is

150

therefore tied to the

frequency of the source. Because the frequency

constant speed but because they possess

We

will study

these features in this chapter.

synchronous motors. As the name implies, synchroin

at

other unique electrical properties.

necting them to a 3-phase source), they are called

nous motors run

However,

line.

synchronous motors are used not so much because

kW (200 hp) and

speeds ranging from

is

1

15

are

rated

between

MW (20 000 hp) and turn

at

50 to 800 r/min. Consequently, 1

these machines are mainly used in heavy industry

Figure 17.1 Three-phase, unity power factor synchro-

nous motor rated 3000 hp (2200 kW), 327 r/min, 4000 V, 60 Hz driving a compressor used in a pumping station on the Trans-

Canada

pipeline.

Brushless excitation

provided by a 21 kW, 250 fier,

which

is

V

mounted on the

shaft

the bearing pedestal and the main

{Courtesy of General

369

is

alternator/recti-

Electric)

between rotor.


370

(Fig.

we in

1

7.

1

).

At the other end of the power spectrum,

find tiny single-phase synchronous motors used

They

control devices and electric clocks.

cussed

17.1

in

Chapter

are dis-

rent

is

fed into the winding from an external exciter.

Slots are also

punched out along

ing similar to that in a 3-phase induction motor. This

18.

damper winding serves to start the motor. Modern synchronous motors often employ

Construction

less excitation, similar to that

Synchronous motors are

identical in construction to

salient-pole ac generators.

The

stator

is

composed

of a slotted magnetic core, which carries a 3-phase lap winding.

Consequently, the winding

is

also

identical to that of a 3-phase induction motor.

The rotor has by a dc current connected

a set of salient poles that are excited (Fig.

in series to

the circumference

of the salient poles. They carry a squirrel-cage wind-

17.2).

two

The exciting

slip-rings,

coils are

and the dc cur-

used

in

brush-

synchronous

generators. Referring to Fig. 17.3, a relatively small

3-phase generator, called fier are

mounted

at

dc current / x from the salient-pole

exciter,

and a 3-phase

one end of the motor rectifier

windings,

is

recti-

shaft.

The

fed directly into the

without

going

through

brushes and slip-rings. The current can be varied by controlling the small exciting current Ic that flows in the stationary field winding of the exciter. Fig. 17.4

Figure 17.2 Rotor of a 50 Hz to 1 6 2/3 Hz frequency converter used to power an electric railway. The 4-pole rotor at the left is associated with a single-phase alternator rated 7000 kVA, 1 6 2/3 Hz, PF 85%. The rotor on the right is for a 6900 kVA, 50 Hz, 90% PF synchronous motor which drives the single-phase alternator. Both rotors are

equipped with squirrel-cage windings. (Courtesy of ABB)

Figure 17.3 Diagram showing the main components chronous generator.

of

a brushless exciter

for

1

-

dc control source

2

-

stationary exciter poles

3

-

alternator (3-phase exciter)

4

-

3-phase connection

5

-

bridge rectifier

6

-

dc

7

-

rotor of synchronous

8

-

stator of

9

-

3-phase input to stator

line

a synchronous motor.

motor

synchronous motor

It

is

similar to that of a syn-

SYNCHRONOUS MOTORS

shows how

mounted

The

the exciter, rectifier,

in a

and

kW synchronous

salient poles are

motor.

rotor and stator always have the

ber of poles. the

3000

As

number of

in the

same num-

poles determines the synchronous

Example 17-1 Calculate the number of salient poles on the rotor of the synchronous motor shown in Fig. 17.4a. Solution

120

/ P

=

120,///?

frequency of the source [HzJ poles

s

200 = p =

speed fr/min]

p — number of

at

200 r/min;

(17.1) /z

/=

60 Hz and runs

consequently,

where

= motor

at

The motor operates

-

s

J

case of an induction motor,

speed of the motor:

//

37

The

rotor possesses

1

(120

X

60)//?

36 poles

8 north poles and

1

8 south poles.

Figure 17.4a Synchronous motor rated 4000 hp (3000 kW), 200 r/min, 6.9 kV, 60 Hz, 80% power factor designed to drive an ore crusher.

The brushless exciter (alternamounted on the overhung

tor/rectifier) is

shaft

and

is

rated

50 kW, 250

(Courtesy of General

V.

Electric)

Figure 17.4b Close-up

of the

50

kW

exciter,

showing

the armature winding and 5 of the 6

diodes used

to rectify the


ac current.

Electric)


372

synchronous motor

17.2 Starting a

A

synchronous motor cannot

quently, the rotor

is

When

slightly is

can

it

the stator

start

of stator

conse-

up as an induction

connected

is

motor accelerates

line, the

itself;

axis of S pole

usually equipped with a squirrel-

cage winding so that motor.

by

start

axis of N pole of rotor

until

to the

3-phase

reaches a speed

it

below synchronous speed. The dc excitation

suppressed during

While

this starting period.

the rotor accelerates, the rotating flux cre-

ated by the stator sweeps across the slower salient poles.

moving

Because the coils on the rotor possess

number of turns, a high voltage is rotor winding when it turns at low

a relatively large

induced

in the

speeds. This voltage appears between the slip-rings

Figure 17.5

The poles

of the rotor are attracted to the opposite

poles on the stator. At no-load the axes of the poles coincide.

and

decreases as the rotor accelerates, eventually

it

becoming negligible when synchronous speed. To

we

prove the starting torque, slip-rings or connect

the rotor approaches

and

limit the voltage,

them

either short-circuit the

to

an auxiliary resistor

power capacity of

we sometimes have

ited,

to the stator.

As

in the

the supply line

is

lim-

reduced voltage

to apply

case of induction motors,

we

use either autotransformers or series reactors to

Chapter 20). Very

limit the starting current (see

synchronous motors (20

large

sometimes brought up called a

tor,

lations the

more) are

speed by an auxiliary mo-

to

pony motor.

MW and some

Finally, in

motor may be brought up

to

big instal-

pull-in torque of a

right

moment. For example,

motor

will

immediately slow

breakers will

chronous speed, the rotor

trip. In

This produces alternate

N

moment happen

at this

tion

is set

on the

to

at close to

syn-

excited with dc current.

and S poles around the

circumference of the rotor (Fig.

site polarity

1

strong magnetic attrac-

up between them. The mutual

attraction

locks the rotor and stator poles together, and the rotor

is literally

field.

yanked

the circuit

ment when excitation should be applied. The motor with the revolving

Once is

carries

the

induced

no

field.

motor turns

at

synchronous speed, no

in the squirrel-cage

winding and so

current. Consequently, the behavior of a is

entirely different

from

that of

into step with the revolving

The torque developed

because of the magnetic attraction between the

poles of the rotor and the opposite poles of the

To reverse

the direction of rotation,

we

interchange any two lines connected to the

stator.

simply stator.

7.5). If the poles

be facing poles of oppo-

stator, a

down and

practice, starters for synchro-

an induction motor. Basically, a synchronous motor

running is

re-

then pulls automatically and smoothly into step

rotates is

magnetic

nous motors are designed to detect the precise mo-

voltage

17.3 Pull-in torque motor

should happen that

stator, the resulting

synchronous motor

as the

it

pulsion produces a violent mechanical shock. The

it

As soon

if

is

at the

emerging N, S poles of the rotor are opposite the

N, S poles of the

speed by a

variable-frequency electronic source.

synchronous motor

powerful, but the dc current must be applied

the

during the starting period. If the

The

im-

to

at this

propriately called the pull-in torque.

moment

is

ap-

17.4 Motor under load general description

When

a synchronous motor runs at no-load, the ro-

tor poles are directly opposite the stator poles and

their axes coincide (Fig.

1

7.5).

However,

ply a mechanical load, the rotor poles

if

we

ap-

fall slightly

SYNCHRONOUS MOTORS

373

a bigger mechanical load, and the increased power

can only

come from

the 3-phase ac source.

17.5 Motor under load

simple calculations We

can get a better understanding of the operation

of a synchronous motor by referring to the equiva-

shown

lent circuit

in Fig.

1

7.7a.

It

phase of a wye-connected motor.

represents one

It

identical to

is

the equivalent circuit of an ac generator, because

both machines are built the same way. Thus, the

Figure 17.6 The rotor poles are displaced with respect to the axes of the stator poles when the motor delivers mechani-

flux

<£>

rent

cal power.

/x

As

.

E0

Consequently,

a between

the poles increases progressively as

increase the load (Fig.

more powerful torque is

to the

and the motor develops an ever

field,

But there

we

Nevertheless, the

17.6).

magnetic attraction keeps the rotor locked revolving

to

The mechanical angle

E0 is in

motor comes

step creates a

away from

to a halt.

the stator poles

circuit breakers

and

A motor that pulls out of

major disturbance on the immediately

trip.

line,

and the

This protects the

motor because both the squirrel-cage and

stator

windings overheat rapidly when the machine ceases to run at

The

Under

7.7b).

E0 =

If,

we

in addition,

adjust the

motor "floats" on the

E, the

line current /

practically zero. In ef-

is

is

to

supply the small

friction losses in the motor,

and so

it is

negligible.

the shaft?

if

we

The motor

apply a mechanical load to will begin to

causing the rotor poles to poles by an angle a.

E0

reaches

its

Due

fall

to this

maximum

electrical degrees

mechanical

value a

behind

E.

slow down,

behind the stator

E

is

i}

E0

tial

Ex

than

now

The mechanical

placement a produces an electrical phase

between

shift,

little later

before. Thus, referring to Fig. I7.7c,

8

dis-

shift 5

and E.

The phase upon the magneto-

these conditions, in-

phase with the line-to-neutral

only current needed

windage and

synchronous speed.

pull-out torque depends

1

What happens

ceeds the pull-out torque of the motor, the rotor

the

and the

fect, the

as the angle increases.

a limit. If the mechanical load ex-

poles suddenly pull

voltage £(Fig.

excitation so that line

varies with the excitation.

already mentioned, the rotor and stator poles

duced voltage turn at synchronous speed.

in

depends on the dc exciting cur-

are lined up at no-load.

behind the stator poles, but the rotor continues

£0

created by the rotor induces a voltage

the stator. This flux

shift

produces a difference of poten-

across the synchronous reactance

X

s

given by

motive force developed by the rotor and the stator poles.

The

mmf of the

dc excitation

/x ,

rotor poles

while that of the stator depends

upon the ac current flowing pull-out torque

is

in

E x — E — E0

depends upon the

The nom-

the windings.

usually 1.5 to 2.5 times the

Consequently, a current

j'X,

inal full-load torque.

The mechanical angle a between stator poles has a direct bearing

As to

the rotor and

/

must flow

given by

=

E*

from which

on the stator current.

the angle increases, the current increases. This

is

be expected because a larger angle corresponds to

/

= -jEJX = -j(E -

s

E„)/X x

in the circuit,

374


Example 17-2a

A 500 hp, 720 r/min synchronous motor connected to a

3980

age

V, 3-phase line generates an excitation volt-

£n of

1790

ing current II

is

V

(line-to-neutral)

when

the dc excit-

25 A. The synchronous reactance

and the torque angle between

Ea

and

E

is

22

30°.

is

Calculate a.

b.

Figure 17.7a

c.

Equivalent circuit of a synchronous motor, showing

d.

one phase.

The value of £ x The ac line current The power factor of the motor The approximate horsepower developed by

the

motor e.

E Eo

The approximate torque developed

at the shaft

Solution

This problem can best be solved by using vector no-

Figure 17.7b Motor

tation.

at no-load, with

EQ

adjusted to equal E. a.

The voltage E

(line-to-neutral) applied to the

motor has a value

Ex

—

E - E0

E = EjV3 = 3980/V3

4

= 2300 V Let us select

E

E

as the reference phasor,

whose

angle with respect to the horizontal axis

sumed

is

as-

to be zero. Thus,

E = 2300Z0 0 It

follows that

E0

is

given by the phasor

Ea = 1790/1-30° Figure 17.7c Motor under load 17.7b, but

it

The equivalent

EQ

has the same value as lags behind E.

circuit per

phase

Moving clockwise around

The current lags 90° behind E x because Xs is inductive. The phasor diagram under load is shown in Because

/ is

nearly in phase with E, the

motor absorbs active power. This power

is

entirely

transformed into mechanical power, except for the relatively small

copper and iron losses

In practice, the excitation voltage to

in the stator.

E0

is

adjusted

be greater or less than the supply voltage E.

value depends upon the

and the desired power

power output of

factor.

the

given

the circuit

plying KirchhofTs voltage law

Fig. I7.7c.

is

in Fig.

17.8a.

in Fig.

-E + E + E0 = x

EX =

E-

write

0

E0

= 2300^0° - 1790Z-30 0 = 2300 (cos0° + ;sin0°) 1790 (cos -30°

= 2300 -

1550

= 750 + j

895

Its

motor

and ap-

we can

- 1168Z50 0

+j

+7

sin

895

-30°)

SYNCHRONOUS MOTORS

375

Ex 1168 V

Figure 17.8a Equivalent circuit of a synchronous motor connected to a

source

E.

Thus, phasor leads phasor b.

The

Ex has a value E by 50°.

of

1

1

68

V

and

it

1790

Figure 17.8b

fl2I = Ex

_ ~

1

See Example

A 50 ° A 90°

22

Approximate torque:

e.

9.55

T=

A and

Thus, phasor / has a value of 53 40° behind phasor E.

The power

motor

factor of the

it

=

is

current

E across /.

Hence,

power

X

9.55

3715

X

1

10

3

720

N-m

Example 17-2b The motor in Example 17-2a

has a stator resistance

of 0.64 Q. per phase and possesses the following

factor

=

cos 6

=

=

0.766, or

losses:

cos 40°

2

I

76.6%

R

losses in the rotor:

3,2

Stator core loss:

The power

280.

given by the

motor terminals and the

the

X P

n

lags

cosine of the angle between the line-to-neutral voltage

17-2.

168

= 53^-40°

c.

V

given by

line current / is

factor

is

lagging because the current

Windage and

3.3

friction loss:

1

.5

kW kW kW

lags behind the voltage.

The complete phasor diagram

is

shown

in Fig.

17.8b. d.

Calculate a.

Total active

power input

to the stator:

b. c.

Pi

=

3X

-

3

L

Solution

cos 40° a.

W=

142

280.

Power

input to the stator

lkW Stator/

Neglecting the

shaft

E, N / cos 6

X 2300 X 53 X

= 280

The actual horsepower developed The actual torque developed at the The efficiency of the motor

2 I

R

2 /?

losses

-

3

X

is

53

2

280.

X

Total stator losses

=

power transmitted across is 280. kW.

Power transmitted

to the rotor

stator, the electrical

the airgap to the rotor

1

=

271.4

+

kW

Q = 5.4 kW = 8.7 kW = 280. - 8.7

0.64

losses and iron losses in the

5.4

1

3.3

1

kW

Approximate horsepower developed:

P - 280.1 X

3

10 /746

=

375 hp

The power at the shaft minus the windage and

is

the

power

to the rotor

friction losses.

The

rotor


376

IR

losses are supplied by an external dc source

and so they do not affect the mechanical power.

chanical power.

by

Power

available at the shaft:

P0 =

271.4

-

269.9

X

l(r

-

Example The corresponding torque is:

value calculated in b.

T -

X P

9.55

P=

361.8 hp H

-

therefore expressed by

"

sin 8

(17.2)

of the motor, per

phase [Wl

X

269.9

10*

720

//

E E

P = mechanical power

17-2a.

X

9.55

is

where

very close to the approximate

is

synchronous motor

kW

269.9

746 This power

a

the equation

=

1.5

is available in the form of meThe mechanical power developed

across the air gap

En = E—

line-to-neutral voltage of the source [V]

X =

synchronous reactance per phase

s

= 3580 N-m

=

8

line-to-neutral voltage induced by

torque angle between

£

(>

and

[V]

/x

E

[electrical degrees] c.

=

Total losses

Total Total

4-

3.3

4

3.2

4

1.5

=

13.4

power input = 280.1 4- 3.2 = 283.3 power output = 269.9 kW

Efficiency

Note

5.4

=

=

269.9/283.3

=

0.9527

that the stator resistance of 0.64

compared

small

to

95.3

Q

Consequently, the true phasor diagram close to the phasor diagram of Fig.

kW

is

% very

of 22

reactance

the

kW

1

is

Q.

This equation shows that the mechanical power increases with the torque angle, and its maximum value is reached when 8 is 90°. The poles of the rotor are then midway between the N and S poles of the stator. The peak power P max (per-phase)

very

far as torque

tional to the

Power and torque

speed

When

a synchronous motor operates under load,

draws active power from given by the same equation

The power

the line.

we

t>

As

in the

s )

is

sin 8

voltage E, the excitation voltage

PR

power

is

This

is

If

we

and iron losses

to

the

is

directly propor-

fixed.

The torque

is

derived from Eq. 3.5:

where

T= P =

(16.5)

power

£

()

,

and the phase

neglect the relatively in the

stator,

all

the

transmitted across the air gap to the rotor.

analogous

it

is

torque, per phase [N-m]

mechanical power, per phase [W]

ns

=

synchronous speed [r/min]

9.55

=


absorbed by the motor depends upon the supply

small

concerned,

previously used for

case of a generator, the active

angle 8 between them.

is

mechanical power because the rotor

it

the synchronous generator in Chapter 16:

P = (£ £/X

given by

7.8b.

As

17.6

is

The maximum torque

=

60/2 tt|

motor can develop

is

called the pull-out torque, mentioned previously.

It

occurs

when 8 = 90°

the

(Fig. 17.9).*

power P r transmitted

across the air gap of an induction motor (Section 13.13). tor

PR

However, losses

in

are

a synchronous motor, the roentirely

source. Consequently,

all

supplied by the dc

the

power transmitted

The remarks rotors.

in ihis

section apply to motors having

Most synchronous motors have

case the pull-out torque occurs

at

smooth

salient poles: in this

an angle of about 70°.

S YNCHRONO US

MOTORS

P

T

8

90

800

100

120

86.6

693

150

50

400

180

0

0

These values are plotted

377

in Fig. 17.9.

The torque curve can be found by applying Eq.

b.

17.4:

T= Figure 17.9 Power and torque per phase as a function of the torque angle 8. Synchronous motor rated 150 kW (200 hp), 1200 r/min, 3-phase, 60 Hz. See Example 17-3.

1

1

If

9.55 /VI 200

=

/VI 25 the

output:

7mux = 800 N-m r/min.

460

synchronous

V, 3-phase

motor has a synchronous reactance of 0.8 phase.

=

The pull-out torque TnvAX coincides with

c.

maximum power Example 17-3 A 50 kW, 200

9.55 Pin s

the excitation voltage

En

is

fixed

at

12,

per

300

V,

The the

per phase, determine the following:

actual pull-out torque

N-m) because

this is a

3 times as great

is

(2400

3-phase machine. Similarly,

power and torque values given

in Fig.

1

7.9 must

also be multiplied by 3. Consequently, this 150 a.

b. c.

The power versus 8 curve The torque versus 8 curve The pull out torque of the motor

maximum

motor can develop a or about

400

kW

output of 300 kW,

hp.

Solution a.

The


17.7 Mechanical

is

As

corresponding values of

is

a

and the number of poles

p.

It

given by

(17.2)

selecting different values for 8,

late the

case of synchronous generators, there

a, the torque angle 8

is

sin 5

= (266 X 300/0.8) = 99 750 sin 8 [W] = 100sin8[kWl By

in the

precise relationship between the mechanical angle

is

P = (E0 E/XJ

electrical

angles

E = EjV3 = 460/ V3 = 266 V The mechanical power per phase

and

8

=

/x*/2

(1

7.5)

sin 8

Example 17-4 A 3-phase, 6000

we

P and

can calcuT,

per phase.

tor

P

T

PI

[kW]

[N*m

0

0

0

30

50

400

60

86.6

693

?

4 kV, 180 r/min, 60 Hz mo1.2

° 1

from

At

their no-load position. If

the line-to-neutral excitation late the

£0 =

2.4 kV, calcu-

mechanical power developed.

Solution

The number of poles (continued)

(1.

full-load the rotor poles are displaced by a me-

chanical angle of

5

kW

has a synchronous reactance of

\20fln s

is

=

120

X 60/180 = 40


378

The

axis of

electrical torque angle is

N

pote

of rotor

8

Assuming

a

motor

to the

=

pa/2

wye

=

X

(40

)/2

1

=

axis of S pole

!

of stator

;

20°

connection, the voltage

E

applied

is

E = E,/V3 =4kV/V3 = 2.3 kV - 2309 V and the excitation voltage

is

Ea = 2400 V

Figure 17.10a

The mechanical power developed per phase

P = (E E/X

=

power =

Total

(y

s)

(2400

X

=

1

=

1573

X

3

5

sin

The flux produced by the stator gap through the salient poles.

is

flows across the air

(17.2)

2309/1.2) sin 20°

axis of N pole of rotor

573 300

J

axis of S pole

!

of stator

\

kW

1573

= 4719 kW (-6300

hp)

17.8 Reluctance torque If

we

gradually reduce the excitation of a synchro-

nous motor when that the

it

is

running

motor continues

to

no-load,

at

speed even when the exciting current reason

is

that the flux

to cross the short

produced by the

gap between the

much

the stator rather than the

tween the poles.

In other

Fig.

1

7. 10a.

On

is

find

is

zero.

The

stator prefers

is

less in the axis

this

of the

phenomenon,

in

the

tract the salient poles in 1

7. 10c).

tains

a mechanical load

is

applied to the shaft, the

have the shape shown

maximum

Fig. 17,11

rotor poles will fall behind the stator poles, and the stator flux will

its

in Fig.

1

7.

1

0b.

positive value at 8

maximum

is

namely

at

8

=

90°.

the reluctance torque as a func-

The torque reaches a maximum

=

45°. For larger angles

negative value

at

8

=

it

attains

135°. Obviously,

Thus, a considerable reluctance torque can be de-

to run as a reluctance-torque motor, the angle

veloped without any dc excitation

lie

at all.

The reluctance torque becomes zero when the midway between the stator poles. The reason is that the N and S poles on the stator atrotor poles are

is

zero

where the regular torque T at-

value,

shows

tion of the angle 8.

a

opposite directions (Fig.

Consequently, the reluctance torque

precisely at that angle

motor develops a reluctance torque. If

salient poles are attracted to the stator poles,

thus producing a reluctance torque.

and

salient poles

concentrated as shown

account of

Figure 17.10b

The

longer air gap be-

words, because the reluc-

tance of the magnetic circuit salient poles, the flux

we

run at synchronous

must

between zero and 45°. Although a positive torque still developed between 45° and 90°, this is an un-

stable region of operation.

The reason

angle increases the power decreases.

is

that as the

S YNCHRONO US

MOTORS

379

Figure 17.12

Figure 17.10c

The reluctance torque is zero when the are midway between the stator poles.

salient poles

In

a synchronous motor the reluctance torque

plus

(1)

the smooth-rotor torque (2) produce the resultant torque (3).

Torque

can be seen is

due

(2) is

to the

dc excitation

in Fig. 17.12.

of the rotor.

However, the difference

not very great, and for this reason

we

shall con-

tinue to use Eqs. 17.2 and 17.5 to describe synchro-

nous motor behavior.

Losses and efficiency synchronous motor

17.9 Figure 17.11

of a

Reluctance torque versus the torque angle.

In order to give the reader a sense of the order of

As

magnitude of the pull-out torque, resistance, reaccase of a conventional synchronous

in the

tance,

motor, the mechanical power curve has exactly the

and losses of a synchronous motor, we have

drawn up Table 7A. It shows the characteristics of a 2000 hp and a 200 hp synchronous motor, respectively labeled Motor A and Motor B. 1

same shape sence

as the torque curve. Thus, in the ab-

of dc

Does

8

at

=

mechanical

the

excitation,

reaches a peak

power

45°.

power The answer is

the saliency of the poles modify the

and torque curves shown

in Fig. 17.9?

The following points should be noted: 1.

yes. In effect, the curves

shown

in Fig.

1

7.9 are those

27° and 37°.

of a smooth-rotor synchronous motor. The torque of a salient-pole rotor

motor

is

1

7.

1

1

.

It

at full-load

corresponds

ranges between

to the electrical an-

gle 8 mentioned previously.

equal to the

component and

nent of Fig.

sum of the smoothreluctance-torque compo-

The torque angle

the

2.

(4.2

Thus, the true torque curve of a syn-

chronous motor has the shape

(3)

The peak reluctance torque

given is

of the peak smooth-rotor torque.

the

in Fig. 17. 12.

As is

about 8 per-

3.

The

is

to excite the

2000 hp motor

only about twice that needed for

200 hp motor

per-unit

a result, the

cent greater than that of a smooth-rotor motor, as

kW)

larger the

about 25 percent

peak torque of a salient-pole motor

The power needed

(2.

1

kW).

In general, the

synchronous motor the smaller

power needed

total losses

to excite

of Motor

four times those of Motor

A (38 kW) B

(9.5

is

the

it.

are only

kW)

despite the


380

CHARACTERISTICS OF TWO

TABLE 17A

SYNCHRONOUS MOTORS MOTOR A

NAMEPLATE RATING power

[hp]

power

[

k

W

2000 hp 492 k

W

1

]

4000 V 220 A

line voltage line current

speed

MOTOR

B

440 V

A

208

90()r/min

60 Hz

3

3

.0

l

pull-out torque (pu)

dc exciter power

17.10 Excitation and reactive

power Consider a wye-connected synchronous motor connected to a 3-phase source whose line voltage fixed (Fig. 17.13).

voltage

wye

wye

kW

gap

V

25

10

mm

V

25

1

mm

6

stray losses

stator/ rotor I

2

2

kW kW kW 3.5 kW 2 kW 9.5 kW

97.5%

94.0%

1

/?

R

total losses

efficiency

IMPEDANCES AND VOLTAGES stator

X

s

XJR

0.0638

l

2

phase voltage

fact that

E E0

is

therefore created by the

U

v

E.

induces a line-to-

in the stator. If in the stator,

However, because

E

com-

.

is

it

fixed,

we

neglect the

follows that E.d

is

=

also fixed, as

case of a transformer (see Section 9.2).

mmf

needed to create the constant flux

may be produced by both.

either

by the

<4>

stator or the rotor or

If the rotor exciting current / x is zero, all the

Ex.

0.0262

254 285

V V

ten times as powerful. This

another property of large motors: the more

horsepower they develop, the smaller the tive losses are.

As

in

power. Compare the

effi+

two motors: 97.5% versus 94.0%.


much

rela-

a result, the efficiencies im-

prove with increase ciencies of the 4.

produce

On the other

23

V 2873 V

is

line currents /

O H

0.62

il

2309

Motor A

The

l

122

S

phase voltage

is

Rs

is x

(line-to-neutral values)

7.77(2

stator resistance ratio

1

total flux

very small IR drop

in the

kW kW 4 kW 10.3 kW 4.2 kW 38 kW 8.5

friction

also fixed.

neutral voltage

The

stator core loss

is

E

follows that the line-to-neutral

Pursuing our reasoning, flux

kW

2.1

The

r.

LOSSES windage and

E

It

bined action of U.d and

2.2

27°

1

U

.0

36.7°

4.2

dc exciter voltage

l

.4

1

at full-load

connection

air

motor

hand, the rotor produces a dc magnetomotive force

factor

torque angle

as far as

s

a magnetomotive force U.d in the stator.

LOAD CHARACTERISTICS power

R

concerned.

49 k

60 Hz

phases

the effect of is

200 hp 1

I800r/min

frequency

ways neglect performance

X

s

per phase

is

122 times larger than

R As s

.

a result,

X

6-

Figure 17.13

larger than the resistance of the stator

winding. Note that for the 2000 hp motor

6

s

we can

total flux 4> is due to the mmf produced by the ro(U ) plus the mmf produced by the stator (Ua ). For

The is

tor

al-

a given

r

EL

,

the flux

is

essentially fixed.

S YNCHRONO US

produced by the

flux has to be

The stator must power from the

stator.

then absorb considerable reactive

we

3-pfiase line (see Section 7.9). But

if

rotor with a dc current / x

mmf helps

the rotor

,

duce part of the flux O. Consequently,

power

drawn from

is

the ac line. If

The ical

active power absorbed power of the motor.

is

MOTORS

38

equal to the mechan-

excite the

pro'p

less reactive

we

gradually

raise the excitation, the rotor will eventually pro-

duce

the required flux

all

by

The

itself.

stator then

draws no more reactive power, with the

power

the

What happens tive

motor becomes unity

factor of the

critical level?

we

if

The

of reactive power, just as

if

it

motor

this

we can cause

power

to correct the

same time

the

at

agram

the

current

q

,

and

motor designed

p

/

line

36.87°. This cur/

p

and

power P

q

is

=

0.8/ s

(17.7)

=

0.6/ s

(17.8)

given by

P = £ab / p = The

=

broken up into two components

/

active

The

in Fig. 17.14.

clear that

power facme-

power

for a

power equal

factor of 0.8 (leading).

power

reactive

0.8

£ab /

power delivered by

Q = £ab /q =

0.6

the

£ab /

(17.9)

s

motor

is

(17.10)

s

to

can supply

A

factor of 0.8 can deliver

75 percent of

kW

rated

its

me-

motor shown

75% X 3000 = 2250

in

kvar to

same time as it develops its rated mekW. Motors designed to opcrate at leading power factors are bigger and more costly than unity power factor motors are. The reathe line at the

chanical output of 3000

is

is

rating

chanical load. Thus, the 3000

son

it

motor

factor motor

develops the same me-

leads E. lb by arcos 0.8

/s

rent can be /

as the

It

power, they are usually designed to op-

erate at a full-load

Fig. 17.4

power

chanical

as they furnish

Most synchronous motors are designed to operate at unity power factor. However, if they also have to de-

reactive

shows an 80% power

15

to the load they are driving.

Power factor

liver reactive

7.

1

also operating at full-load.

The

17.11

di-

at full-load.

Fig.

to the

source

important property, synchronous

motors are sometimes used

chanical

like a

absorb or deliver reactive power.

to either

of a plant

this

were a capacitor.

Thus, by varying the dc excitation

Because of

.0).

Figure 17.14 Unity power factor synchronous motor and phasor

of absorbing reac-

The motor then behaves

line.

1

motor above

excite the

stator, instead

(

power, actually delivers reactive power

3-phase

tor

result that

that for a

given horsepower rating, both the dc

Figure 17.15 80 percent power factor synchronous motor and phasor diagram at full-load.

It

follows from Eqs. 17.9 and 17. 10 that

exciting current and the stator current are higher.

Q=

This can be explained as follows. Fig. 17. 14

is

the schematic

power factor motor operating to-neutral voltage

The

active

is E,db

and the

is,

of rated mechanical output

line-

was

as line current

power absorbed per phase

P = Eay Jp

The

P

— 15%

diagram of a unity

at full-load.

0.75

is /

p

stated previously.

.

If

we compare

therefore,

I

p

with

/s

,

we

find that / s

=

i

.25 /

p

.

Thus, for the same mechanical power output, a mo(17.6)

tor

designed for a leading power factor of

80%

has


382

25%

to carry a line current that is

power

that operates at unity

greater than one

,'s2

factor.

's2

\36.9°

17.12 V-curves

200

P = 800

Suppose a synchronous motor rated mechanical load.

We

havior as the excitation

is

in

wish

is

operating

to

examine

power remains

power

1

6.

will

We

assume

we reduce

If

S

=

/x

power.

1000 kVA. As a

is

(b)

shown

Figure 17.18

in

200 A but

Field excitation raised to

a.

with

same me-

chanical load. Motor delivers reactive power to the

unity,

line.

Fig. b.

= 00 A and P = 800 kW.

Phasor diagram shows current leading the voltage.

1

the excitation to

draw reactive power from

to the active

mechan-

factor

thus yielding the phasor diagram 7.

S = 1000 kVA (a)

fixed. Let us begin by adjusting

the excitation I x so that the

1

its

be-

its

Because a change

varied.

excitation does not affect the speed, the

ical

at

A

kW

We

assume

70 A,

the

motor

the line in addition that

S increases

to

result, the line current will in-

crease from

ponent of

/

sl

p

to / sl (Fig.

1

phase with

in

because the motor

is still

7.

1

7).

£. lb

is

Note the

that the

same

com-

as before

developing the same me-

chanical power.

Current

E b

o

—v I

M

y

P = 800 kW S = 800 kVA

A

4' 100

A

/s)

lags behind £"ab

motor

the rotor

smaller than before, but the apparent

is

is

power S absorbed by If (b)

(a)

Figure 17.16 Synchronous motor operating at unity power factor with a mechanical ioad of 800 kW. Field excitation is 100 A. b. Phasor diagram shows current in phase with the voltage.

we

lagging.

the stator

is

field current I x in

greater.

increase the excitation to

motor delivers reactive power is

a.

and so the power

,

The

factor of the

/x

= 200 A,

to the line to

the

which

connected (Fig. 17.18). The apparent power

again greater than

We

assume S comes /s2 and

=

in the unity

power

1000 kVA. The

it

is

factor case.

line current be-

However, the in-phase it leads ELXh component of /s2 is still equal to I p because the mechanical power is the same. .

By varying the excitation this way, we can power of the synchronous motor

the apparent

plot

as a

function of the dc exciting current. This yields a

V-shaped curve Ix

=

70

A

36.9°

P = 800 kW S = 1000 kVA (a)

tive (b)

the b.

reduced to 70 A but with same meMotor absorbs reactive power from

Field excitation

chanical load. line.

Phasor diagram shows current lagging behind the voltage.

The V-curve

the V-curve corresponds to full-load.

V-curve

Figure 17.17 a.

(Fig. 17.19).

is

always

displayed for a fixed mechanical load. In our case,

is

power

The no-load

also shown, to illustrate the large reacthat

can be absorbed or delivered by

simply changing the excitation.

Example

A

1 7-5

4000 hp (3000 kW), 6600

synchronous motor operates

power 1

1

(2,

V,

60 Hz, 200 r/min

at full-load at a

leading

factor of 0.8. If the synchronous reactance


is

SYNCHRONOUS MOTORS

383

kVA ^

1000

full-load

x O+

'

source E

yrio-load

400

O-

200

Figure 17.20 synchronous motor connected to a source E. Note the arbitrary (+) polarity marks and arbitrary direction of current flow. See Example 17-5. Circuit of a

200

120

80

40

-'x

Figure 17.19 No-load and full-load V-curves nous motor.

E = 3815^0° of

a 1000 hp synchroIt

c.

The apparent power of the motor, per phase The ac line current The value and phase of E0

d.

Draw

e.

Determine the torque angle 8

a.

b.

the phasor

=

given by

328/136.9°

Writing the equation for the circuit

-E+jlX + E0 =

diagram

s

shall

find

0

thus

immediately change the given values

correspond

The

to

E0 = E-jIX = 3811/10° - j (328/136.9°) 11 = 3811^0° - 3608/1(36.9° + 90°) = 3811 (cos 0° + y sin0°) -

active

to

one phase of a wye-connected motor.

power per phase

P=

3000/3

is

= lOOOkW

The apparent power per phase

3608 (cos 126.9°


(8.11)

d.

Consequently,

is

EQ

lags 26° behind E,

complete phasor diagram

E= The

line current I

/

EjV3 =

leads

the value ,

\

1

1

V

e.

The torque angle

8

is

is

shown

and the

in Fig. 17.21.

26°.

17,13 Stopping synchronous

motors =

Owing 36.9°.

to the inertia

of the rotor and

its

load, a large

synchronous motor may take several hours

and phase of the excita-

£ we draw (>

38

1250 X 1000/3811

an angle of arcos 0.8

To determine tion voltage

=

C

is

= S/E= = 328 A

E by

6600/ V3

sin 126.9

3811 + 2166 -; 2885 = 5977 -y2885 = 6637/1-26°

is

= 1250kVA The

+;

=

S = P/cosd = 1000/0.8

c.

we

s

We

b.

/ is

/

Solution

a.

follows that

after

the equivalent circuit time,

being disconnected from the

we

line.

to stop

To reduce

the

use the following braking methods:

of one phase (Fig. 17.20). This will enable us to write the circuit equations. Furthermore, lect

E as

the reference phasor

and so

we

se-

I.

Maintain

full

short-circuit.

dc excitation with the armature

in


384

e.

The time required for the speed 600 r/min to 150 r/min

to fall

from

Solution a.

In Fig. 17.22a the

motor has just been discon-

nected from the line and generator

is

now

operating as a

The speed frequency is 60 Hz.

in short-circuit.

r/min, and the

is still

Consequently, the impedance per phase

Z=VR

2

+ Xi

\ 0.2

6637 V

=

2

600

is

(2.X 12)

+

16

2

16 ft

Figure 17.21

See Example

The current per phase

17-5.

/

is

= EJZ=

2400/16

= 150A 2.

Maintain

full

dc excitation with the armature

connected to three external 3.

Apply mechanical braking.

In

methods

it

I

and

2, the

The power

resistors.

r/min

P=

motor slows down because

functions as a generator, dissipating

the resistive elements of the circuit.

its

energy

is

Mechanical

usually applied only after the motor has

reached half-speed or

less.

A

lower speed prevents

undue wear of the brake shoes.

Example 17-6 A 1500 kW, 4600

2

3I

R =

13.5

in b.

braking

dissipated in the 3 phases at

3

X

150

E0

2

X

is

fixed, the in-

0.2

kW

Because the exciting current

duced voltage

is

proportional to the speed.

Consequently, when the speed has dropped to

150 r/min,

Ea = 2400 X

600 r/min, 60 Hz synchronous motor possesses a synchronous reactance of

The frequency

16 ft and a stator resistance of 0.2 ft, per phase.

and so

= 600 V

(150/600)

V,

is

also proportional to the speed,

The excitation voltage E 0 is 2400 V, and the moment of inertia of the motor and its load is 275 kg-irr. We wish to stop the motor by short-cir-


cuiting the armature while keeping the dc rotor

the frequency; consequently,

/=

60

X

(15/60)

= is

15

Hz

proportional to

current fixed.

16U Calculate a.

The power dissipated

in the

armature

at

600

in the

armature

at

150

r/min b.

The power dissipated r/min

c.

d.

The The

600

is

kinetic energy at

600 r/min

kinetic energy at 150 r/min

Figure 17.22a Motor turning at 600 r/min (Example

17-6).

S YNCHRONO US

4

P =

LI

13.5

150 A

600 V

whence

0.2 Q.

150 r/min

Note

that the

t

MOTORS

3 85

Wit

(3.4)

=

508.6//

=

37.7

s

motor would stop much sooner

if

external resistors were connected across the stator terminals.

Figure 17.22b Motor turning at 150 r/min (Example

17-6).

The synchronous motor

17.14

versus the induction motor X,

=

X

16

=

(15/60)

4

0 We

Referring to Fig. 17.22b the

phase

at

150 r/min

new impedance

per

But

is

have already seen

that induction

motors have

600

excellent properties for speeds above at

r/min.

lower speeds they become heavy, costly, and

have relatively low power factors and efficiencies.

Z = V0.2 2 +

4

2

=

a

4

Synchronous motors

are particularly attractive

for low-speed drives because the

The current phase

power

factor can

is

always be adjusted

/= £0 /Z =

600/4

=

A

150

to

1

.0

Although more complex

and the efficiency

is

high.

weight and

to build, their

cost are often less than those of induction motors of

Thus, the short-circuit current remains un-

changed

r/min to 150 r/min. The power dissipated 3 phases

therefore the

is

equal power and speed. This

motor decelerates from 600

as the

P =

same

13.5

is

particularly true for

speeds below 300 r/min.

in the

as before:

A synchronous motor can tor

kW

of

a

plant

Furthermore,

its

while

improve the power

carrying

starting torque

its

rated

fac-

load.

can be made consid-

The

erably greater than that of an induction motor. c.

The

kinetic energy at

5.48

d.

The

A

10

-3

=

5.48

=

542.5 kJ

X

10

"

is

reason

Jrr

(3.8)

X 275 X

of the squirrel -cage wind-

at

5.48

X

10

3

150

2

motors in

decelerating from

-

=

542.5

=

508.6 kJ

The time

in

and

kilns,

A synchronous capacitor is essentially a synchronous motor running armature

for the speed to drop is

ultra-low speeds. Thus, huge motors

33.9

lost as heat in the

r/min to 150 r/min

at

MW range drive crushers, rotary

17.15 Synchronous capacitor

kl

is

the 10

variable-speed ball mills.

W = E -Ek2

This energy

rating.

very low frequencies enable us to run synchronous

33.9 kJ

The loss in kinetic energy 600 r/min to 150 r/min is

motor and

same nominal

The biggest difference is in the starting torque. High-power electronic converters generating

is

X 275 X

effi-

synchronous speed. Figure 17.23 compares

a synchronous motor having the

kinetic energy at 150 r/min

tance.

that the resistance

ing can be high without affecting the speed or

ciency

600-

is

the properties of a squirrel-cage induction

E k2 = = e.

X

600 r/min

given by

resis-

from 600

at

no-load.

Its

only purpose

is

to ab-

sorb or deliver reactive power on a 3-phase system, in

order to stabilize the voltage (see Chapter 25).

machine

acts as an

The

enormous 3-phase capacitor

(or


386

I

i

i

s ynchrcDnous

motor

97 induction

>

motor

/J

96

rf // // //

o c

a)

it

95

!y

if if if if

(a)

if

94 93

92 91,

25

50 *-

75

100

125

%

mechanical power

Figure 17.24a

%

Three-phase, 16

250

rated

-200 Mvar

kV,

900

r/min

synchronous capacitor

(supplying reactive power) to

+300

Mvar (absorbing reactive power). It is used to regulate the voltage of a 735 kV transmission line. Other characteristics: mass of rotor: 143 t; rotor diameter: 2670 mm; axial length of stator iron: 3200 mm; air gap

synchronous motor

200

= 150

length: 39.7

mm.

(b)

100 induction motor

^

50

60

40

20

80

100

%

speed

Figure 17.23 Comparison between the torque

(b) of

efficiency (a)

synchronous motor, both rated r/min, 6.9 kV, 60 Hz. inductor)

and

starting

a squirrel-cage induction motor and a

whose

reactive

at

4000

hp,

1800

power can be varied by

changing the dc excitation.

Most synchronous capacitors have ratings that range from 20 Mvar to 200 Mvar and many are hydrogen-cooled (Fig. 17.24). They are started up like synchronous motors. However, if the system cannot furnish the required starting power, a pony motor is used to bring them up to synchronous speed. For example, in one installation, a 160

Mvar

mm Figure 17.24b Synchronous capacitor enclosed in its steel housing containing hydrogen under pressure (300 kPa, or" 2 about 44 lbf/in ). {Courtesy of Hydro-Quebec)

SYNCHRONOUS MOTORS


is

and brought up

started

speed by means of a 1270

387

to

kW wound-rotor motor.

Example 17-7

A


16 kV,

actance of 0.8 pu and

is

Calculate the value of

EQ

a.

Absorbs 160 Mvar

b.

Delivers 120

It

160 Mvar,

rated at

is

1200 r/min, 60 Hz.

has a synchronous

connected

to a 16

kV

re-

line.

so that the machine

Q=

120 Mvar
Mvar /

Solution a.

i 4335

The nominal impedance of the machine

Zn = En 2 /S n = 16 000 2 /(160 X =

(16.3)

10

6

5550 V

9250

V

14 800

V

)

a

1.6

A

is

Figure 17.25b Over-excited synchronous capacitor delivers reactive

The synchronous reactance per phase

X = X s (pu)Zn = s

= The

i.28

0.8

X

is

power (Example

1.6

From

Fig.

17-7).

7.25a

1

n

we can

-£+ 77X

line current for a reactive load

of 160

Mvar

s

write

+ £o =

0

hence

is

E„ /n

= 5 n /(V3£n -

160

X

c

X

10 7( .73 1

16 000)

= 5780 A The drop across the synchronous reactance

is

b.

Ex =

IX S

= 5780 X

=

E- jIX

1.28

Note

that the excitation voltage less than the line voltage

The load current when ing 120 IU


E = EjV3 = = 9250 V Selecting

E as

E= The current machine

is

/

This time

we have

0°

lags 90° behind

the

absorbing reactive power; conse-

quently, /

= 5780^-90°

850 V)

10 7(1.73

machine

X

/

E by

leads

is

deliver-

16 000)

90° and so

= 4335Z90 0

Fig. 17.25b

we can

write

En = E-jIX = 9250^0° - 4335 X 1.28/180° = (9250 + 5550)Z()° s

=

14 800/10°

is

(9250 V).

)

(

X

/

From

E because

1

90°)

is

= Q/(V3EU 120

the

(

-

= 4335 A

16 000/1.73

9250,/

Mvar

-

is

the reference phasor,

1.28^(90°

much

= 7400 V The

s

= 9250^0° - 5780 X = 1850^0°

)


388

5780

motor [hp] knowing

A

it

has an efficiency

of 95%. 17-8

A synchronous operates

at

a

motor driving a pump power factor of 100%.

What happens

if

the dc excitation

is

in-

creased? 17-9

mm^ Q= 160 Mvar 7400

A 3-phase,

to a

4 kV, 60 Hz

current of

320

A and

line

draws a

absorbs 2000 kW.

E

V

y

> I

9250

Calculate

V

a.

1850

V

b. c.

5780

225 r/min synchronous motor

connected

A

d.

Figure 17.25a Under-excited synchronous capacitor absorbs reactive power (Example 17-7).

17-10

The apparent power supplied to the motor The power factor The reactive power absorbed The number of poles on the rotor

A synchronous 3-phase

motor draws

line. If the

1

50

A from

exciting current

raised, the current drops to 140 A.

a

is

Was

the

motor over- or under-excited before the

was changed?

excitation

The

excitation voltage (14

800 V)

is

now

con-

siderably greater than the line voltage (9250 V).


a. Calculate the

approximate full-load current

of the 3000 hp motor


in Fig. 17.1, if

it

has

an efficiency of 97%.

Practical level 1

1

7-

1

7-2

Compare

7-3

the construction of a synchro-

17-12

7-4


how

Explain

a synchronous motor starts up.

17-13

should the dc excitation be applied?

Why does the Name some

What

A 3-phase

a.

What

b.

If

is

factor.

speed of a synchronous motor

1

.8

is

we

used for?

meant by an under-excited syn-

17-14

does 1

7-7

over-excite a synchronous motor, its

power

line voltage

at

motor draws 2000

factor of

the approximate

90%

leading. Calculate

the

power

kV, but the exciting current remains

how

the following

a.

Motor speed and mechanical power output

b.

Torque angle 8

c.

Position of the rotor poles

d.

Power

e.

Stator current

factor

A synchronous

motor has the following

parameters, per phase (Fig. 17.7a):

kVA at

power developed by

unity

suddenly drops to

mechanical power output increase?

A synchronous a

60 Hz operates

quantities are affected:

to a squirrel-

chronous motor?

The

unchanged. Explain

of the advantages of a syn-

it

synchronous motor rated 800

hp, 2.4 kV,

meant by a synchronous capacitor

is

and what 7-6

what speed must

quencies?

cage induction motor.

1

at

a squirrel -cage induction motor.

chronous motor compared

17-5

the value of the field resistance?

is

the rotor turn to generate the indicated fre-

remain constant even under variable load? 1

What

nous generator, a synchronous motor, and

When 1

b.

E= x = s

/

2.4 kV;

2

a

= 900 A.

Ea =

3

kV

SYNCHRONOUS MOTORS Draw

the phasor diagram and determine:

17-19

nous motor rated 400 hp, 2300

b.

Active power, per phase

r/min, 80 A,

c.

Power

The

d.

Reactive power absorbed (or delivered),

factor of the

a. In

motor

is

new

the mechani-

citation

its

power

excitation is

is

adjusted

17-20

0.6

increased without making any

is

is

the effect

The synchronous capacitor (I,

phase

upon the

per phase.

0.007

is

to a stop,

it

by the motor

dc excitation initial line

motor

the

tor

line current

reactive

power absorbed

level

10 H, per phase.

The

stator is

a.

at full-load b.

(4000 hp) with a leading power factor of 0.89. If the efficiency

is

97%,

calculate c.

the following:

c.

The apparent power The line current The value of E0 per phase The mechanical displacement of the poles The

their no-load position

total reactive

electrical f.

power supplied

power

b.

A so

that the

its

is

rated value, or 1600 V, at

d.

The total braking power and braking torque at 900 r/min The braking power and braking torque at 450 r/min The average braking torque between 900 r/min and 450 r/min The time for the speed to fall from 900 r/min to 450 r/min, knowing that the moment of inertia

of the rotor

is

1

.7

6

X

1()

lbfr.

17-21

A 500 hp,

3-phase, 2200 V, unity power

factor synchronous motor has a rated current of 103 A.

It

can deliver

its

rated out-

put so long as the air inlet temperature 1

7- 7 1

we wish

40°C or

less.

The manufacturer

is

states that

to adjust the

the output of the

motor must be decreased

factor to unity.

by

1

percent for each degree Celsius

above 40°C.

Calculate a.

250

voltage across the resistors

system

The approximate maximum power the motor can develop, without pulling out of

Problem

fixed at

H

wye. The

Industrial application to the

step [hp] In

is

in

r/min.

,

from e.

connected to three large 0.6

Calculate

connected

wye, and the motor operates

machine coasts

in Fig.

17.4 possesses a synchronous reactance of

d.

is

900

kV motor shown

hp, 6.9

resistance per

the

braking resistors connected

(or delivered)

one-tenth of

The 4000

The

11. If

der to shorten the stopping time, the sta-

The torque angle

power absorbed by

17.24

will run for about 3 h. In or-

d.

active

in Fig.

possesses a synchronous reactance of

unity. If the ex-

c.

a.

to

synchronism

The The The

b.

adjusted

,

c.

following:

in

is

iy

The value of Xs and of £ 0 per phase The pull-out torque [ft-lbfj The line current when the motor is about

a. b.

power absorbed

reactive

factor

other change, what

Advanced

E

hp synchronous motor drives a

so that the

a.

synchronous reactance of

Calculate

pull out of

compressor and

b.

450

the line current if

(or delivered) by the motor, per phase.

17-18

stator has a

0.88 pu, and the excitation

suddenly removed,

b. Calculate the

A 500

V,

60 Hz, drives a compressor.

to 1.2 pu.


cal load

17-17

factor synchro-

Torque angle 8

and the new torque angle 5

17-16

power

unity

a.

per phase

17-15

A 3-phase,

389

The exciting voltage £0 The new torque angle

required, per phase

If the air inlet

46°C, calculate the

motor

current.

temperature

maximum

allowable

is

390

1

7-22


An 8800 kW.

6.0 kV,

1

500 r/min, 3-phase,

50 Hz, 0.9 power factor synchronous motor

manufactured by Siemens has the

fol-

sing the above information, calculate the fol-

wing: a.

1.

962

Rated current:

A

Rated torque: Pull-out torque:

4.

Locked-rotor current:

5.

Excitation voltage:

56.0

b.

kNm

2.

3.

c.

1.45 pu

160

V

387

A

Full-load efficiency, excluding excitation losses:

9.

Temperature

11.

f.

520 kg-m~

Maximum

g.

12.

13.

14.

its

in

gallons (U.S.)

total

moment

which the motor can

of inertia

(in

pull into syn-

The The

total losses

of the motor

total efficiency

at full-load

of the motor

at full-

moment

2

The

reactive

power delivered by

the

motor

full-load

If the iron losses are

equal to the stator cop-

per losses, calculate the approximate resis-

465 L/min

permissible external

1370 kg-m

The maximum

at

25°C

of cooling water:

32°C Flow of cooling water:

inertia:

motor including

load

of inertia of rotor: rise

e.

97.8%

to

10.

the

chronism d.

Excitation current:

Moment

mass of

The flow of cooling water

lb-ft'),

4.9 pu

7.

8.

total

per minute

6.

system

The

enclosure, in metric tons

lowing properties:

of

tance between two terminals of the

stator.

h. Calculate the resistance of the field circuit.

Mass of rotor: 6. 10 t (t = metric Mass of stator: 7.50 Mass of enclosure: 3.97 t

t

ton)

Chapter

1

Single-Phase Motors

18.0 Introduction

Fig. 18.2

shows

the progressive steps in wind-

ing a 4-pole, 36-slot stator. Starting with the lami-

Single-phase motors electric

are the

most familiar of all

motors because they are used

in

nated iron stator, paper insulators

home

ers

—

are

first

inserted

in

—called

the

slots.

r/min,

60 Hz

slot lin-

The main

appliances and portable machine tools. In general,

employed when 3-phase power

they are

is

not

available.

There are many kinds of single-phase motors on the market, each designed to cation.

However, we

meet a specific appli-

will limit our study to a

few

basic types, with particular emphasis on the widely

used split-phase induction motor.

18.1

Construction of a singlephase induction motor

Single-phase induction motors are very similar to

3-phase induction motors. They are composed of a squirrel-cage rotor (identical to that in a 3-phase

motor) and a stator (Fig.

1

8. 1).

main winding, which creates a

The set

stator carries a

of N, S poles.

It

also carries a smaller auxiliary winding that only

when

the motor

auxiliary winding has the

same num-

operates during the brief period starts up.

The

Figure 18.1 Cutaway view of a 5 hp, 1725 phase capacitor-start motor.

ber of poles as the main winding has.

(Courtesy of Gould)

391

single-


392

Figure 18.2a Bare, laminated stator of a 1/4 hp squirrel-cage rotor

is

(1

87 W), single-phase motor. The 36 3-phase motor.

slots are insulated with a

paper

liner.

The

identical to that of a

Figure 18.2c

Figure 18.2b Four poles of the main winding are inserted

in

the slots.

Four poles

of the auxiliary

winding straddle the main

winding.


winding 1

is

then

laid

the

in

8.2b). Next, the auxiliary

that (Fig.

its 1

slots

winding

is

(Figs.

I8.2a,

embedded

so

poles straddle those of the main winding 8.2c).

The reason

be explained shortly.

for this

arrangement will

Each pole of the main winding consists of group of four concentric (Fig.

1

8.3a).

coils,

connected

Adjacent poles are connected so as

produce alternate N, S

polarities.

the center of each pole

The empty

(shown as

a

in series

to

slot in

a vertical dash

SINGLE-PHASE MOTORS

393

one pole pitch

—(180°)

—H

10 20

25 30 turns No. 16 wire

Figure 18.4 Main and auxiliary windings motor. iary

The

winding opens

mounted on the nous speed. line)

in

stationary contact

and the

when

shaft,

a 2-pole single-phase series with the auxil-

in

the centrifugal switch,

reaches 75 percent of synchro-

partially filled slots

on either side of

are used to lodge the auxiliary winding.

The

has only two concentric coils per pole (Fig. Fig.

1

shows a 2-pole

8.4

8.3c).

1

main

stator; the large

winding and the smaller auxiliary winding are at right

18.2

Synchronous speed

in the

speed of

dis-

angles to each other.

placed

As

it

latter

case of 3 -phase motors, the synchronous all

single-phase induction motors

is

given

by the equation ns

=

120/ 1

(17.1)

P where ns

90°-*! center of

main winding

=

/=

center of auxiliary

p

winding

The

synchronous speed [r/min] frequency of the source [Hz]

= number of poles

rotor turns at slightly less than synchronous

speed, and the full-load slip 5 percent for fractional

Figure 18.3 Main winding of a 4-pole, 36-slot motor showing the number of turns per coil. b. Mmfs produced by the main winding. a.

c.

laid

out

typically 3 percent to

flat,

Position of the auxiliary winding with respect to the

main winding.

is

horsepower motors.

Example 18-1 Calculate the speed of the 4-pole single-phase

shown in percent. The tor

Fig. 18.1 if the slip at full-load line

frequency

is

60 Hz.

is

mo3.4


394

Solution

produces an ac

to the stator. the resulting current / s

The motor has 4

=

ns

poles, consequently, 120///?

=

(120

X

flux

//

is

s

.

The

flux pulsates back

and forth

but, unlike

the flux in a 3-phase stator, no revolving field

60)/4

duced. The flux induces an ac voltage

= 1800r/min The speed

<£>

is

pro-

in the station-

ary rotor which, in turn, creates large ac rotor currents. In effect, the rotor

given by:

behaves

like the short-circuited

secondary of a transformer; consequently, the motor s

=

(n s

0.034

=

(1800

n

=

1739 r/min

-

n)!n s

-

(17.2)

has no tendency to

h)/1800

However, the other,

As

spin.

18.3 Torque-speed characteristic

if

we

start

by

itself (see Fig.

1

8.5

a schematic diagram of the rotor and main

is

will continue to rotate in the direction of

it

ates until

tor

reaches a speed slightly below synchro-

it

winding of a 2-pole single-phase induction motor.

turn. Fig.

when

is

locked. If an ac voltage

is

applied

that the

develops a positive torque as soon as

Suppose the rotor

8.6).

a matter of fact, the rotor quickly acceler-

nous speed. The acceleration indicates Fig.

1

spin the rotor in one direction or

8.6

1

the

shows the

is

mo-

begins to

typical torque-speed curve

main winding

starting torque

it

is

zero, the

excited.

Although the

motor develops a power-

rotor current

ful

torque as

it

approaches synchronous speed.

18.4 Principle of operation The

principle of operation of a single-phase induc-

tion

motor

is

quite complex, and

by the cross-field theory

As soon

E

120 V, 60 Hz

is

may be

explained

*

as the rotor begins to turn, a speed

induced

in the rotor

emf

conductors as they cut the

ac source

stator flux
Figure 18.5 Currents

The is

in

the rotor bars

when

resulting forces cancel

the rotor

is

locked.

each other and no torque

produced.

*6

ac source

Figure 18.7 Currents induced

They produce a

in

the rotor bars

due

to rotation.

flux
angles to the

stator flux

speed *

Figure 18.6 Typical torque-speed curve of a single-phase motor.

The double revolving Held theory (diseussed 1

8.

18)

is

in

Section

also used to explain the behavior of the single-

phase motor.

SINGLE-PHASE MOTORS

= r/4

t

/

395

=

3774

Figure 18.8 Instantaneous currents and

flux in a single-phase momain winding excited. The duration of one T seconds, and conditions are shown at suc-

tor with the

cycle

is

cessive quarter-cycle intervals. a.

Stator current

b.

Stator current

however, c.

d. e.

/s is

is

maximum,

rotor current

zero, rotor current

is

smaller than

S

is

zero.

is

/r

maximum;

.

maximum, but negative. Rotor current is maximum, but negative. After one complete cycle (t = T) the conditions Stator current

is

re-

peat. f.

Resulting flux S to a minimum r

the rotor speed increases.

flow

currents produce an ac flux

same time behind

does not reach

/,

to

r

which

acts at right

its

is

the

maximum value at the

as 3> s does. In effect,
due

s ,

to the

The combined volving magnetic motor.

Equally important

angles to the stator flux ,.

causes currents

It

bars facing the stator poles. These

in the rotor

fact that

.

inductance of the

action of

4\ and

4> r

rotor.

produces a

field, similar to that in a

The value of

r

re-

3-phase

increases with increasing

speed, becoming almost equal to 3> s

why

speed. This explains in part

at

synchronous

the torque

in-

creases as the motor speeds up.

We can

understand

duced by referring the currents tor /s

= + 10

A

r

=

T

and

sume

how

the revolving field

to Fig. 18.8.

It

and fluxes created respectively by

stator, at

that the

successive intervals of time.

motor

is

is

pro-

gives a snapshot of the ro-

We

as-

running far below synchronous


396

speed, and so

much

is

,.

smaller than

.

is

obvious that the combination of S and

a revolving field. Furthermore, the flux izontally

and

shown wise it

in Fig.

18.8f.

The

strong hor-

Thus, the

elliptic pattern

flux rotates counterclock-

iary

rotor.

As

motor approaches syn-

the

tioned previously)

is

and

standstill

torque

T— =

=

/s

Fig.

1

8.9.

sina

*/ a / s

(1 8. 1)

locked-rotor torque [N mJ locked-rotor current

in the

auxiliary

in the

main

]

locked-rotor current

winding [A|

a starting torque in a single-phase motor,

When the main

at

low speeds. The locked-rotor

at

winding [A

create a revolving field. This

(men-

,.

result,
where

18.5 Locked-rotor torque we must somehow

weak. As a

T=

produced.

done by adding an auxiliary winding, as shown

during the

a

given by

is

/.,

To produce

is

the rotor flux

thereby producing a powerful torque both

ens

chronous speed, CI\ becomes almost equal to

when

acceleration period

direction as the rotor. Furthermore,

speed of the

immediately see that the auxil-

will

winding produces a strong flux

synchronous speed, irrespective of the ac-

rotates at

tual

vertically.

low speed follows the

same

in the

weak

relatively

field strength at

it

produces

is

The reader

By observ-

ing the flux in the successive pictures of Fig. 18.8,

is

a = phase angle between

in

k

=

and auxiliary windings are

and

/s

/.,

[°|

a constant, depending on the design

of the motor

connected to an ac source, the main winding pro-

duces a flux duces a flux

,

the

If

two fluxes

so that <1\ either lags or leads set up.

manner

To

while the auxiliary winding pro-

S ,

The 2-phase revolving

are out of phase,

a rotating field

field

is

obtain the desired phase shift between

(and hence between

pedance is

created in a

similar to the revolving field of a 3-phase

motor (see Section

/.,

Z in

and

4> a ),

The

rise to various types

many cases

pedance

in the

self, as

is

incorporated

the desired im-

auxiliary winding

it-

explained below.

special switch

is

also connected in series with

the auxiliary winding.

when

and

depending upon the desired starting torque.

of split-phase motors. In

A

/s

an im-

resistive, inductive, or capaci-

The choice of impedance gives

13.3).

we add

series with the auxiliary winding.

impedance may be tive,

the

chronous

It

disconnects the winding

motor reaches about 75 percent of synspeed.

A

speed-sensitive

switch mounted on the shaft

purpose (Fig.

1

is

centrifugal

often used for this

8. 10).

18.6 Resistance split-phase motor

i

i

The main winding of a single-phase motor is always 2 made of relatively large wire, to reduce the I R losses (Fig. 18. la). The winding also has a relatively large number of turns. Consequently, under 1

o-

Figure 18.9 Currents and fluxes

ac source

locked-rotor conditions, the inductive reactance at standstill

when the main and An elliptical revolv-

auxiliary windings are energized. ing field

is

«-o

produced.

high and the resistance locked-rotor current

applied voltage

/s

E (Fig.

is

low.

As

a result,

is

the

lags considerably behind the 18.11 b).

SINGLE-PHASE MOTORS

397

Figure 18.10 a.

Centrifugal switch position.

b.

The

in

Centrifugal switch position.

Due

the closed, or stopped,

stationary contact in

is

closed.

the open, or running,

to centrifugal force, the rectan-

gular weights have

swung

out against the

straining tension of the springs. This

caused the

plastic collar to

move

re-

has

to the

left

along the shaft, thus opening the stationary contact

series with the auxiliary winding.

In a resistance split-phase

centrifugal

motor (often simply

called split-phase motor), the auxiliary winding has

switch

)

E

in

) )

auxiliary winding

70 turns per pole No. 22 wire

a relatively small

resistance

number of

higher and

is

turns of fine wire.

Its

reactance lower than that

its

of the main winding, with the result that the lockedrotor current

/ a is

more nearly in phase with E. The a between /a and / s produces

resulting phase angle the starting torque.

The main winding

/s

120 turns per pole

and

line current /.,.

At

/ L is

start-up,

it

equal to the phasor is

sum

of

usually 6 to 7 times the

nominal current of the motor.

No. 16 wire

Owing

to the

small wire used on the auxiliary

winding, the current density

is

high and the winding

heats up very quickly. If the starting period lasts for

a=

more than

25°

5 seconds, the

and may burn

winding begins

out, unless the

motor

is

to

smoke

protected by

a built-in thermal relay. This type of split-phase tor is well

mo-

suited for infrequent starting of low-

inertia loads.

Example 18-2

A

resistance split-phase

motor

is

rated at

1/4

hp

(187 W), 1725 r/min, 115 V, 60 Hz. When the rotor is locked, a test at reduced voltage on the main and (b)

auxiliary windings yields the following results:

Figure 18.11 a.

b.

Resistance split-phase motor (1/4 hp, 115 V, 1725 r/min, 60 Hz) at standstill. Corresponding phasor diagram. The current in the auxiliary winding leads the current

winding by 25°.

in

the main

main

auxiliary

winding applied voltage current active

power

E= / =

V 4 A 23

60

W

winding

£= /, = P., =

23 1

.5

30

V A

W


398

Calculate

v s 2 - Pi

G = s

a.

b.

The phase angle between / a and / The locked-rotor current drawn from s

= V92 2 - 60 2 =

the line at

V

115

= VSJ -

Oa

69.7 var

Pi

Solution

17.0 var

We first calculate the phase E of the main winding. a.

The apparent power

=

5S

-

£V S

angle

<$>

s

between

Is

and

The

Q=

is

X

23

4

=

The

The power cos

factor

<|> s

+ Ga 69.7 + 17.0 =

- PJS =

-

60/92

S

S 0.65

= VP 2 + Q 2 V

=

4> s

We now calculate the phase angle and E of the auxiliary winding. =

The power cos

EI,

=

factor

=

4> a

=

/,

a

between

r

SIE

-V$6J* = 125 V

V

current at 23

=

125/23

-

=

5.44

X

-

(115/23)

is

5.44

The locked-rotor current drawn /,

The apparent power

W

The locked-rotor

49.6°

lags 49.6° behind the voltage E.

5a

is

is

thus,

/s

is

86.7 var

apparent power absorbed

total

motor

the

Qs

=

VA

92

power absorbed by

total reactive

at

A 1

V

15

is

A

27.2

is

23

X

1.5

=

34.5

VA

Due

to their

low

cost, resistance split-phase in-

duction motors are the most popular single-phase motors. They are used where a moderate starting

is

PJS.A

=

30/34.5

=

torque 0.87

is

required and where the starting periods are

infrequent.

They drive

fans,

chines, oil burners, small

thus,

pumps, washing ma-

machine

tools,

and other

devices too numerous to mention. The power rating

=

*a /.,

29.6°

usually lies between

a =

=

cj)

s

-

<\>.

A

=

/s

and

49.6°

-

W and 250 W (1/12 hp

to

1/3 hp).

lags 29.6° behind the voltage.


60

Ia is

29.6°

18.7 Capacitor-start motor The

20.0°

capacitor-start

motor

is

identical to a split-phase

motor, except that the auxiliary winding has about as b.

To determine

we first caloiP and Q drawn by both

the total line current,

culate the total value

windings and then deduce the

power S. The total

total

apparent

many

turns as the

series with the auxiliary

The capacitor active

power absorbed

is

main winding

has. Furthermore, a

capacitor and a centrifugal switch are connected in

80°,

which

is

is

winding

chosen so that

(Fig. 18. 2a). 1

/.,

leads

/s

by about

considerably more than the 25° found

in

a split-phase motor. Consequently, for equal starting

= The

reactive

powers

60

Q

auxiliary windings are

s

+ 30 = 90 W and Q.A of the main and

torques, the current in the auxiliary winding

about half that

in a split-phase motor.

It

is

only

follows that

during the starting period the auxiliary winding of a capacitor motor heats up less quickly. Furthermore,

SINGLE-PHASE MOTORS

voltage, electrolytic capacitors are

centrifugal

much

399

smaller

and cheaper than paper capacitors. However,

elec-

can only be used for short

peri-

trolytic capacitors

ods

ac circuits whereas paper capacitors can op-

in

on ac

erate

indefinitely. Prior to the

development of

electrolytic capacitors, repulsion-induction

had

motors

be used whenever a high starting torque was

to

Repulsion-induction

required.

motors possess a

commutator and brushes that require considerable maintenance. Most motor manufacturers special

have stopped making them.

when a high They are built in sizes kW (~I/6 hp to 10 hp).

Capacitor-start motors are used starting torque

ranging from

1

is

required.

20

W to 7.5

Typical loads are compressors, large fans, pumps,

and high-inertia loads. Table

motor

1

8A

gives the properties of a capacitor-start

having

a

rating

W(l/3

250

of

1760 r/min, 115 V, 60 Hz. Fig.

18.

1

hp),

shows

3

the

torque-speed curve for the same machine. Note that during the acceleration phase (0 to 1370 r/min) the

main and auxiliary windings together produce a very high starting torque.

When

the rotor reaches 1370

r/min, the centrifugal switch snaps open, causing the

motor

to operate

along the torque-speed curve of the

main winding. The torque suddenly drops from 9.5

(b)

Nm to 2.8 Nm, but the motor continues to accelerate Figure 18.12

until

a.

Capacitor-start motor.

b.


reaches 1760 r/min, the rated full-load speed.

it

and power factor of single-phase induction motors

18,8 Efficiency the locked-rotor line current /L cally

is

smaller, being typi-

4 to 5 times the rated full-load current.

Owing to the

high starting torque and the relatively

low value of /a the capacitor- start motor to applications

is

well suited

involving either frequent or prolonged

starting periods.

Although the starting characteristics

of this motor are better than those of a split-phase tor,

mo-

both machines possess the same characteristics

under load. The reason identical circuit

is

that the

main windings

and the auxiliary winding

when

the

is

motor has come up

no longer

are

in the

is

full-load a 186

a direct

W motor (1/4 hp) has an efficiency

and power factor of about 60 percent. The low power factor

is

mainly due to the large magnetizing current,

which ranges between 70 percent and 90 percent full-load

Consequently, even

current.

at

these motors have substantial temperature

The

relatively

these motors

to speed.

The wide use of capacitor-start motors

The efficiency and power factor of fractional horsepower single-phase motors are usually low. Thus, at

horsepower

is

rises.

low efficiency and power a

consequence of

ratings.

Integral

of

no-load

factor of

their fractional

horsepower

single-

result of the availability

of small, reliable, low-cost

phase motors can have efficiencies and power fac-

electrolytic capacitors.

For given capacitance and

tors

above 80 percent.


400

1370 r/min a

auxiliary

S

winding' in circuit

c

300 r/min

;entrifuga

svvitch oper s

\

centrifuga :

\ I

i

clos es

S\/vitch

2.8

no minal

ill Figure 18.13 Torque-speed curves

^*"*

"

400

600

ki.

full

load

/bu r/mi n

\\

'///// V////. 800

1000 > speed n

of a capacitor-start motor, rated 1/3

1200

1400

hp (250 W), 1760

r/min,

1800 r/min

1600

1

15

V,

60 Hz, class A

insulation.

CHARACTERISTICS OF A CAPACITOR -START MOTOR

TABLE 18A Rating:

N-m

V////

200

0

^

tore

m 1.35

ji

250 W, 1760

r/min,

115

V,

60 Hz, Insulation Class 105°C

No-load

Full -load

115

V

voltage

115

power

250

W

current

4.0

V A

current

5.3

A

losses

105

W

RF.

64%

voltage

Locked

63.9%

efficiency

speed

1760 r/min

torque

1.35

Nm

115

23

7a

current 7 L

torque

3.4

current

Nm

1600 r/min

speed %

13

A

V A 19 A 29 A

voltage current 7 S

current

Breakdown

rotor

torque capacitor

6Nm 320

[jlF

*

SINGLE-PHASE MOTORS

18.9 Vibration of single-

phase motors If

we

we

touch the stator of a single-phase motor,

note that

it

vibrates rapidly, whether

full-load or no-load.

operates

it

These vibrations do not

at

exist in

2-phase or 3-phase motors; consequently, single-

phase motors are more noisy.

What causes electric

to the fact

the deceleration intervals coincide with the negative

peaks. Consequently, the acceleration/deceleration

power whereas

it

properties given in Table 18A. is

5.3

draw

A and the

it

+ 1000 the

W

The

We

line.

it

is

current,

power P supplied

find that

218 W.

When

P

we

to the

we can

mo-

oscillates

between

power

positive

the

motor receives energy from the

when

full-load current

lags 50° behind the line voltage. If

18. 14).

W and

mechan-

motor having the

waveshapes of voltage and

plot the instantaneous tor (Fig.

is

It

delivers constant

power. Consider the 250

ical

due

power is positive, negative, or zero, the mechanical power delivered is a steady 250 W. The motor will slow down during the brief periods when the electric power it receives is less than 250 W. On the other hand, it will accelerate whenever the electric power exceeds the mechanical output plus the losses. The acceleration intervals coincide with the positive peaks of the power curve. Similarly,

motor always receives pulsating

this vibration?

that a single-phase

40

line.

is

Conversely,

negative the motor returns energy to the

However, whether the instantaneous

electric

occur twice per cycle, or 120 times per sec-

intervals

ond on

60 Hz system. As a

a

and rotor vibrate

The

stator

at

vibration and noise. is

18.15).

both the stator

are

transmitted to the

in turn,

generates additional

vibrations

mounting base which, motor

result,

twice the^line frequency.

To eliminate

the problem, the

often cradled in a resilient mounting (Fig.

It

consists of

two

soft rubber rings placed

between the end-bells and a supporting metal bracket. Because the rotor also vibrates, a tubular

rubber isolator

is

sometimes placed between

Figure 18.14

The instantaneous power absorbed by a single-phase motor varies between +1000 output is constant at 250 W; consequently, vibrations are produced.

W and

-218 W. The power

the


402

shaft

and the mechanical load, particularly when

the load

a fan.

is

Two-phase and 3-phase motors do not because the

total

from

phases

all the

is

vibrate

power they receive

instantaneous

constant (see Section 8.7).

18,10 Capacitor-run motor The capacitor-run motor is essentially a 2-phase motor that receives its power from a single-phase source. It has two windings, one of which is directly connected to the source. The other winding is also connected to the source, but capacitor (Fig. 18.

Figure 18.15 Single-phase capacitor-start motor supported

in

a

re-

and noise

silient-mount cradle to reduce the vibration

transmitted to the mounting surface. Motor rated at 1/3 hp,

1725

r/min,

230

V,

60 Hz has a

3.0 A, efficiency of 60 percent,

full-load current of

and power

factor of

60

percent. Other characteristics: no-load current: 2.6 A;

has a large

compared

1

with a paper

in series

The capacitor-fed winding

6).

number of turns of relatively small

wire,

connected winding.

to the directly

This particularly quiet motor

is

used to drive fixed

loads in hospitals, studios, and other places where silence

is

important.

It

has a high power factor on

account of the capacitor and no centrifugal switch

is

locked-rotor current: 13 A; locked rotor torque: 3.6 pu;

breakdown torque: 3.0

pu; service factor:

required. 1

278 mm;

232 mm.

(Courtesy of Baldor Electric Company)

The motor acts when it operates at

low.

Capacitor-run motor having a NEMA rating Corresponding phasor diagram at full load.

of

30 millihorsepower.

motor only

full-load (Fig. 18.16b).

*

Figure 18.16 b.

is

as a true 2-phase

these conditions, fluxes

(a)

a.

starting torque

.35; total

weight: 10 kg; overall length including shaft: overall height:

However, the

ct>

s

a

and

Under by the

SINGLE-PHASE MOTORS

two windings are equal and out of phase by 90°. The motor

is

The shaded-pole motor is very popular for ratings below 0.05 hp (—40 W) because of its extremely

below 500 W.

simple construction (Fig.

Reversing the direction

winding

order to reverse the direction of rotation of the

In

we have

discussed so

far,

we have

to inter-

change the leads of either the auxiliary winding or

main winding.

the

However,

if

a single-phase motor

with a centrifugal switch,

versed while the motor

is

its

is

equipped

same

In the case

running because both windings are

times. In the case of very small motors, the ro-

in Fig.

1

8.

1

7. In

in position

1,

while winding

winding

B

is in

A

is

When

the switch

is

thrown

is

directly across the line,

series with the capacitor.

connection the motor turns clockwise.

switch is

the circuit

such a motor, the main and

auxiliary windings are identical.

this

in

can be reversed by using a double-throw switch

shown

to position 2, the role

reversed and the motor will

a that lags

behind

tor is

as

and

!, 2 ,

3

all in

,

come

3)

<£>

and

starts the

and

2

a

3

This current produces a flux .

x

Consequently,

The combined

.

motor.

The

+

(i

/ b in

3)

<1>

2

.

and

As

is

from the

(I>

b

the

produces

rotation

and then

run up to speed in the opposite direction.

copper ring (auxiliary winding) /b

main winding r\

%J

•

q n n o

\J

xJ

ac source

Figure 18.17 Reversible single-phase motor using a 2-pole switch

Figure 18.18a

and capacitor.

Fluxes

in

a shaded-pole motor.

—

sJ

-

right.

and the

a

revolving field that drives the rotor clockwise.

o

+

which

before, the

of the windings

to a halt

( 2

field,

the ring,

With

When

also lags

up by the pole on the

is set

b lags behind

bined action of

a

shaded (ring side) of the pole.

Flux 4> 2 induces a current sulting flux

action of

direction of rotation

to the

similar torque

links the

pole, inducing a

produces a weak revolving

unshaded side

A

phase. Flux

on the left-hand <&

at all

tation

nents

/.,.

18. 16)

a copper ring surrounding

The main winding is a simple coil connected to The coil produces a total flux <£> that may be considered to be made up of three compo-

behind

can be changed while the mo-

basically a

is

the ac source.

rather large current

of a capacitor-run motor (Fig.

It

which the auxiliary

in

a portion of each pole.

the main wind-

If

direction.

the direction of rotation

composed of

rotation cannot be re-

ing leads are interchanged, the motor will continue to turn in the

is

short-circuited ring

running.

18. 18).

small squirrel-cage motor

of rotation

motors

18.12 Shaded-pole motor

then essentially vibration-free. Capacitor-

run motors are usually rated

18.11

403

re-

comweak


404

TABLE 18B Properties of a Shaded-Pole Motor, having 2 poles,

Rated 6 W, 115

V,

60 Hz.

No-load

input

A

0.26

current

power

W

15

3550 r/min

speed

Locked rotor

input

A

0.35

current

power

torque

24

W

10

mN-m

Full-load

input

power

2900

torque

2600 r/min

breakdown speed

power

18.13 Universal motor

starting

torque,

efficiency,

and

factor are very low, the simple construction

and absence of

a centrifugal switch give this

marked advantage

in

8.

1

9.

entire

very similar to basic construc-

motor

is

is

shown

in Fig.

laminated to

re-

low-power applications. The it

fixed by the position of the copper rings. Table

18B gives the

The

is

The

magnetic circuit

tion of a small universal

motor

direction of rotation cannot be changed, because is

universal motor

a dc series motor (Section 5.8).

1

a

mN-m

21

V,

The single-phase Although the

mN-m

W

6

breakdown torque 5 millihorsepower, 115

r/min

19

mechanical power

at

W

21

speed

Figure 18.18b Shaded-pole motor rated 60 Hz, 2900 r/min. (Courtesy of Gould)

A

0.33

current

ac flux

typical properties of a 2-pole shaded-

pole motor having a rated output of 6

W.

Example 18-3 Calculate the full-load efficiency and slip of the

shaded-pole motor whose properties are listed

in

Table USB. Solution

The

efficiency

is

(PJP,)

X X

(6/21)

100

(3.6)

100

28.6% (/7 S

—

(3600 0.194

n)/n s

- 2900)/3600 = 19.4%

Figure 18.19 Alternating-current series motor, also called universal

motor.

SINGLE-PHASE MOTORS

duce eddy-current losses. Such a motor can operate on either ac or dc, and the resulting torque-speed about the same

in

each case. That

is

why

it

is

is

called

Series motors are built in

motors formerly used

When

the motor

is

connected

to

an ac source, the

ac current flows through the armature and the series field.

The

field

in

produces an ac flux

electric locomotives.

performance curves of a

8000 r/min, universal motor

115 V,

The

hp.

different sizes,

very large traction

to

some

Fig. 18.20 gives the ac

a universal motor.

many

from small toy motors

starting

full-load current

405

175

is

rated at 1/100

mA.

that reacts

with the current flowing in the armature to produce

18.14 Hysteresis motor

a torque. Because the armature current and the flux

reverse simultaneously, the torque always acts in the

same

in this

direction.

No

revolving field

is

produced

type of machine; the principle of operation

is

same as that of a dc series motor and it possesses the same basic characteristics. The main advantage of fractional horsepower the

universal motors ing torque.

is

speed and high

their high

They can

start-

therefore be used to drive

high-speed centrifugal blowers

in

vacuum

cleaners.

The high speed and corresponding small size for a given power output is also an advantage in driving portable tools, such as electric saws and drills. Noload speeds as high as 5000 to 5 000 r/min are pos-

To understand sis

motor,

the operating principle of a hystere-

us

let

first

consider Fig.

1

8.2

1

.

It

shows

a

stationary rotor surrounded by a pair of N, S poles that

can be rotated mechanically

rection. ial

The

rotor

is

in

composed of

of high coercive force. Thus,

magnet material whose

As

in

it

resistivity

of an insulator. Consequently,

up eddy currents

a

such a

it

is

a clockwise di-

ceramic materis

a permanent

approaches that impossible to

set

rotor.

the N, S field rotates,

it

magnetizes the rotor;

consequently, poles of opposite polarity are contin-

uously produced under the moving N, S poles. In effect, the

revolving field

is

continuously reorient-

1

sible but, as in

any series motor, the speed drops

ing the magnetic individual

rapidly with increasing load.

domains

in the rotor. Clearly, the

domains go through

a

complete cycle (or

hysteresis loop) every time the field

makes one

complete revolution. Hysteresis losses are therefore

produced

in the rotor,

proportional to the area of the

r/mm hysteresis loop (Section 2.26).

14 000

These losses are

dis-

sipated as heat in the rotor.

Let us assume that the hysteresis loss per revo-

12 000

lution

%

is

Eh

joules and that the field rotates

50-

10 000 efficie

ncy

mA 40-

8000

\

Spe

400

o

stationary rotor

30-|- 300

6000 /

CI rrent

^

5= LU

4000

2000

—

-

rate

j

20-

200

10-

100

torque

010

30

20

40

mN m

Torque

Figure 18.20 Characteristics of a small tor

1

15

having a full-load rating of

60 Hz universal mo1/100 hp at 8000 r/min. V,

Figure 18.21 Permanent magnet revolving

field.

rotor

and a mechanically-driven

at


406

n revolutions per minute. the rotor per minute

The energy

The corresponding power

nE h (dissipated as heat)

Ph =

Wit

=

the mechanical

the N, S poles. This

power

P = Because P

[W] in the rotor

power used

can

to drive

given by

is

/z779.55

= P h we

is

(3.4)

/z£ h /60

However, the power dissipated

come from

hysteresis

motor

W=

only

dissipated in

is

(3.5)

have

,

=

/z779.55

speed

Figure 18.22 Typical torque-speed curves of two capacitor-run

/?£ h /60

motors:

whence

T= £

h /6.28

a.

Hysteresis motor

b.

Induction motor

(18.2)

where

T = Eh —

torque exerted on the rotor [N-m] hysteresis energy dissipated in the ro-

per turn

tor,

=

6.28

[J/r]

=

constant [exact value

2tt]

Equation 18.2 brings out the remarkable feature that the torque

18.21)

is

tion. In

needed

to drive the

magnets

other words, whether the poles just barely

creep around the rotor or whether they

move

speed, the torque exerted on the rotor

is

same.

It

(Fig.

constant, irrespective of the speed of rota-

is

this basic

at

high

always the

property that distinguishes hys-

Figure 18.23 teresis

motors from

all

other motors.

In practice, the revolving field

3-phase

stator, or

auxiliary winding.

When

inside such a stator, it

is

by a single-phase

it

Single-phase hysteresis clock motor having 32 poles

produced by a

stator

a hysteresis rotor

is

is

essentially constant as

(a) in Fig.

18.22. This

is

until

Thanks

it

the curve

entirely different

squirrel-cage induction motor,

toward zero as

shown by

from a

whose torque

falls

approaches synchronous speed.

to the fixed

frequency of large distribution

systems, the hysteresis motor

is

employed

ferrite rotor.

placed

immediately accelerates

reaches synchronous speed. The accelerating

torque

and a

having an

in electric

clocks, and other precise timing devices (Fig.

1

8.23).

It

is

also used to drive tapedecks, turntables, and

other precision audio equipment. In such devices the constant speed

looking

for.

is,

However,

of course, the feature the hysteresis

ticularly well suited to drive

of their high

inertia. Inertia

motor

we is

are

par-

such devices because

prevents

many

synchro-

nous motors (such as reluctance motors) from coming up to speed because to reach synchronism, they

have

to

suddenly lock

in

with the revolving

field.

SINGLE-PHASE MOTORS

No

such abrupt transition occurs

motor because to

in the hysteresis

When

c.

develops a constant torque right up

it

at

synchronous speed. In

some

W=


function as a vibration-free capacitor-run motor. is

accelerating,

its full

torque

P =

available to carry the mechanical load and to

overcome

Once

inertia.

the

motor runs

reaches synchronous

it

speed, the rotor poles are

still

There

d.

magnetized and so

is

=

Wit

no energy

motor runs

at

to the rotor.

therefore,

0.8

=

180

J

in the rotor is

180/60

=

3

W

loss in the rotor

when

the

synchronous speed because the

magnetic domains no longer reverse.

an ordinary permanent-magnet

like

is,

X

225

moves The en-

the rotating field

stalls,

ergy loss per minute

enhanced by designing the motor to

While the motor

motor

225 r/min with respect

turntable audio equipment these fea-

tures are further

is

the

407

sychronous motor. The rotor poles will lag behind the stator poles

by a certain angle, whose magni-

18.15

tude depends upon the mechanical torque exerted

Synchronous reluctance motor

by the load.

We Example 18-4 A small 60 Hz poles. In

hysteresis clock

motor possesses 32

making one complete

turn with respect to

to 0.8

b.

c.

the

motor

when when

rotor losses

the motor

is

stalled

motor runs

the

at

syn-

it

pull-in

up as a standard

salient poles lock with the re-

and so the motor runs

at

synchronous

speed. Both the pull-in and pull-out torques are to those of a hysteresis

motor of

accelerate high-inertia loads to synchronous speed.

The

and pull-out torques are about

T= £ =

poles

s

-

=

120///;

=

225 r/min

The maximum power

P =

/?

779.55

= 3W

=

0.8/6.28

at

stator poles are

X

(18.2)

60/32

is

(225

(or 3/746

=

X

0.127)/9.55

1/250 hp)

1

).

a rate that corresponds to the slip. If the

is

120

18.22.

corre-

slipping past the rotor

Nm

0.127

The synchronous speed /?

=

h /6.28

18.24

approaches syn-

sponding to full-load torque (operating point

equal in a hysteresis motor:

b.

starts

when

The reason can be seen by referring to Fig. Suppose the motor has reached a speed //,

Solution

The

field,

weak, compared

rotor losses

the stator. Fig.

equal size. Furthermore, reluctance motors cannot

chronous speed

a.

Such a reluctance motor

volving

number

poles must be equal

rotor milled out to create four salient poles.

chronous speed, the

The pull-in and pull-out torques The maximum power output before The The

number of poles on

shows a

squirrel-cage motor but,

stalls

d.

number of

J.

Calculate a.

of salient poles. The to the

the revolving field, the hysteresis loss in the rotor

amounts

can build a synchronous motor by milling out a

standard squirrel-cage rotor so as to create a

Figure 18.24 Rotor of a synchronous reluctance motor.


408

rotor

so

is

to

in the

lock with the revolving field,

time

it

this interval {At),

it

chronous speed

//

s

will

,

sweep

It is

particularly well adapted to variable-frequency

electronic speed control. Inertia

is

then no problem

because the speed of the revolving field always

never be achieved. The

tracks with the speed of the rotor. Three-phase re-

is

going from speed

that in

is

must do

not achieved during

past a rotor pole. If pull-in

problem

it

takes for one stator pole to

/?,

to syn-

the kinetic energy of the re-

motors of several hundred horsepower

luctance

have been

using this approach.

built,

volving parts must increase by an amount given

by Eq. 3.8:

AE = k

5.48

X \0~'J(n

18.16 Synchro drive

2

-

s

2 /;,

(18.3)

)

In

where

./ is

moment of

the

inertia.

Furthermore, the time interval

At

=

60/(/z s

-

is

some remote-control systems we may have

move the position of a small given by

meters away. This problem (18.4)

/?,)/?

flexible shaft.

But

if

must develop an accelerating power P d of at

tor

P = AEJAt

X

1.8

10

4

n s (n s

- n\Y Jp

power P

will

never pull into

step. In essence, a reluctance

motor can only synchronize when the small and the

moment

How does

of inertia J

is

slip

speed

is

impractical.

Two

rotors are also connected in parallel a single-phase source.

about

this

arrangement

is

and energized

The remarkable

that the rotor

other.

Thus,

if

we slowly turn rotor A clockwise B will move clockwise through

17°.

Obviously, such a system enables us to control

a rheostat

from

a

remote

'location.

receiver

a synchro system.

feature

on one ma-

chine will automatically track with the rotor on the

»/b

of

in

phases of the respective

~U

Figure 18.25

then

rheostat

motors whose 3-phase stators are connected

from

is

cheaper than any other type of synchronous motor.

Components and connections

We

knob and

through 17°, rotor

low.

Despite this drawback, the reluctance motor

away, the

becomes

100

such a shaft work?

parallel (Fig. 18.25).

demanded by the load. If the sum of P u + P L exceeds the power capacity of the motor, it

by using a

m

is

Consider two conventional wound-rotor induction

x

easily solved

electrical shaft to tie the

(approx)

Furthermore, the motor must continue to supply the

together.

(18.5)

Ll

=

least

employ an

one or two

the rheostat

flexible-shaft solution

Consequently, to reach synchronous speed, the mo-

is

to

is

rheostat that

SINGLE-PHASE MOTORS

Two One

409

miniature wound-rotor motors are required.

the stator voltages are again in balance (phase by

coupled to a control knob,

phase), and the torque-producing currents disappear.

(the transmitter)

is

and the other (the receiver)

is

coupled

Synchros are often employed

to the rheo-

to indicate the po-

The 5-conductor cable (conductors a-b-c-1-2) linking the transmitter and receiver constitutes the

with the result that the torque requirements are

flexible electrical shaft.

small.

stat.

The behavior of system

this selsyn or synchro control

Assume

explained as follows.

is

that the

sition of an antenna, a valve, a

gun

turret,

Such transmitters and receivers

and so on,

are built with

watch-like precision to ensure that they will track

with as

error as possible.

little

transmitter and receiver are identical and the rotors are in identical positions. cited, they

behave

When

the rotors are ex-

like the primaries of

two

EQUIVALENT CIRCUIT OF A SINGLE-PHASE MOTOR

trans-

formers, inducing voltages in the respective stator

windings. The voltages induced

in the three stator

windings of the transmitter are always unequal

Chapter

In

(Fig.

we developed

15

5.6) for

1


one phase of a 3-phase induction mo-

because the windings are displaced from each other

tor.

by 120°. The same

exception that the magnetizing branch has been

in the stator

is

true for the voltages induced

moved

of the receiver.

Nevertheless, no matter what the respective stator voltages of the transmitter and receiver are identical in both

the rotors

may

be, they

synchros (phase by phase) when

occupy the same

position.

The

stator volt-

ages then balance each other and, consequently, no

The

current flows in the lines connecting the stators. rotors,

however, carry a small exciting current

Now

if

we

This circuit

/0

They

will

reproduced

with the

in Fig. 18.26,

between

to the technically correct position

points

1

and

2.

The reason

for the

change

is

that

most single-phase motors are fractional horsepower

machines for which the exact needed this

circuit

diagram

to get reasonably accurate results.

we now develop

model,

is

Using

a similar equivalent

circuit for a single-phase motor.

.

turn the rotor of the transmitter,

three stator voltages will change.

is

its

18.17 Magnetomotive force

no longer

distribution

balance the stator voltages of the receiver; consequently, currents

necting the

/.„ / b , / L

,

will

flow

in the lines

con-

two devices. These currents produce a

torque on both rotors, tending to line them up. Since the rotor of the receiver

with the transmitter.

is

free to

As soon

move,

it

will line

up

as the rotors are aliened.

In order to optimize the starling torque, efficiency,

power tor,

tor

and noise

factor,

level of a single-phase

mo-

magnetomotive force produced by each stapole must be distributed sinusoidally across the the

pole face. That

number of

is

the reason for using the special

turns (10, 20, 25, and 30) on the four

concentric coils

shown

Let us examine the poles

when

I

8.3(a).

the concentric coils carry a peak current

2 amperes. Table

of, say,

mmf, using

of the

in Fig.

mmf created by one of the four 1

8C shows

the distribution

numbers

as a measure of

the slot

distance along the pole. For example, the 25-turn coil

lodged

in slots

these slots an

2 and 8 (Fig.

mmf of 25 X

8.27). produces

1

2

=

between

50 amperes

(or

am-

pere-turns). Similarly, the 10-turn coil in slots 4 and 6

2

The

Figure 18.26 Equivalent circuit of one phase of a 3-phase cage motor referred to the

produces between these

primary (stator) side.

Fig.

1

8.27.

the pole

slots

distribution of these

is

an

mmf of 20 A.

mmfs

is

illustrated in

The total mmf produced in the middle of 60 + 50 + 40 + 20 = 70 A and it drops 1


4 0 1

soidal, but the

TABLE 18C

only 0.4

1-9

slot slot

3-7

slot

4-6

9

30

2-8

slot

Mml'

Turns

Coil pitch

25

2

20

2

X

10

40

-

10

mmf in 85 turns

the center of the pole will be

=

the current reverses and

mmf

X 30 = 60 A X 25 = 50 A X 20

A X

20

equal

to, say,

—

1

,2

A, the

mmf will

However, the

still

be distributed sinusoidally but with a peak value the center of

A

We

A

l

.2

A X

in

102 A.

conclude that the ac current produces a pul-

mmf, which is distributed each pole and whose amplitude 3 - phase stator. the

mmf

sinusoidally across varies sinusoidally

mmf

Thus, unlike the

time.

in

= -

85 turns

sating

170 ampere turns

85 turns

will also reverse.

34 A. Subsequently, when

is

does not rotate but remains fixed

18.18 Revolving

produced by

a

of a single-phase stator

mmfs

in place.

a single-

in

phase motor It

can be proved mathematically that a stationary pul-

mmf

sating

having a peak amplitude

placed by two

volving

in

M can

be

re-

mmfs having a fixed amplitude Mil

re-

opposite directions

synchronous speed.

at

Referring to our previous example, a 4-pole pulsat-

mmf that reaches positive and negative peaks of A at a frequency of 60 Hz can be replaced by two 4- pole mmfs having a constant amplitude of 85 A roing

123456789 -

-

1

1

^

pole pitch

70

tating in opposite directions at

1800 r/min. The

re-

mmfs are also distributed sinusoidally in space. As the oppositely moving mmfs take up successive positions, the sum of their magnitudes at any volving

Figure 18.27 Distribution of the

pole

when

current

magnetomotive force across one is

2 A.

point in space

is

equal to the pulsating

point. This can be seen

mmf at

that

by referring to Fig. 18.28,

which shows a portion of the forward and backward off in steps on either side of center. Adjacent poles

have the same magnetic

mmf

distribution but with opposite

polarities.

We have superposed upon this figure a smooth mmf having a perfectly sinusoidal distribution. It reveals that the stepped mmf produced by the four concentric coils tracks the sine Indeed, soidal

we could

mmf without

The

wave very

replace the stepped

closely.

mmf by

a sinu-

introducing a significant error.

current flowing in the four coils alternates

sinusoidally (in time) at the line frequency of 60 Hz.

Consequently, as the current varies, the in

proportion. For example,

mentarily 0.4 A, the

when

mmf varies

the current

mmf distribution

is

mo-

remains sinu-

revolving fields

(mmfF and mmfB

the stationary but pulsating

sweeping

),

past

mmf.

The revolving mmfs respectively produce the same effect as the revolving mmf created by a 3-phase stator. Consequently, we would expect the circuit diagram of a single-phasor motor to resemble that of a 3-phase motor. However, since the

mmfs slip

.v

the rotor has a

if

with respect to the forward-moving mmf,

will automatically

spect to the

The ing

on

rotate in opposite directions, their effect

the rotor will be different. Thus,

slip

of (2

s)

with

it

re-

backward-moving mmf.

circuit

mmf

have a

diagram as regards the forward-mov-

having a

slip s

is

shown

in Fig.

18.29a.

SINGLE-PHASE MOTORS

Figure 18.29 50

mmfB ^

r'

a.

/\

v

\^

~

mmf f

Equivalent circuit as regards the forward-moving

mmf. b.

Equivalent circuit as regards the backward-moving

mmf.

-90°

0 6

.

+90° Similarly, the circuit diagram for the backward-

~

revolving Fig.

1

mmf having

8.29b. For the

the physical to

^0A -90°

/0\ /t=

+90°

—

a slip (2

moment we

meaning of

r,, r 2 ,

Vj,

s) is

shown

in

will not define .\

2 , etc.,

except

say that they are related to the stator and rotor

resistances and reactances.

How

should

wc merge

these two diagrams into a single diagram to represent the single-phase motor?

4"

Figure 18.28 pulsating mmf having a peak amplitude of 170 A can be represented by a forward and backward revolving mmf having a fixed amplitude of 85 A. Shown are successive positions of mmf F and mmf B and the corresponding amplitude of the stationary, pulsating mmf.

18.19 Deducing the circuit diagram of a single-phase motor

The

First,

the / jF

we know

that the oppositely rotating

same magnitude. Consequently, and

/

cuits can

,

B are identical, which

be connected

means

in series.

mmfs have

the stator currents that the

two

cir-

Second, the forward


412

2

2jx 2

2./.V,

r,

2

Figure 18.31 Equivalent circuit of a single-phase motor at standstill.

In practice

we assume x = x 2

.

x

The above analysis ances r,,.Vj, etc., shown Thus,

Equivalent circuit of a single-phase motor.

of r

E¥

is

associated with

EB

backward voltage Because the

mmfK

associated

is

circuits are in series, the

voltages must be equal to the voltage stator.

It

,

while the

with

mmfB

sum of

r h r2

,

-V], .v 2 ,

etc.,

which case the

meaning of the

E applied to the

circuit

suppose the motor

slip s

=

1

.

is

Example 18-5

A test

motor behaves

like a

(the

represent the following physical elements:

Draw stator resistance

2r 2

=

rotor resistance referred to the stator

2/.X-,

=


2jx 2

=

rotor leakage reactance referred to the

2jX m

—

and iron

resistance corresponding to the

and iron losses


4

il

3 11

600

losses:

O

60

il

the equivalent circuit diagram and determine

the

factor of

Solution circuit

the values of the listed friction,

single-

results:

power output, efficiency, and power motor when it turns at 1725 r/min.

The equivalent

stator

windage,

725 r/min

2 11

magnetizing reactance:

the

2R m =

1

n

tion,

reveals that the parameters

etc.,

=

60 Hz,

V,

resistance corresponding to the windage, fric-

sim-

r,

2r,

3

in

short-circuit.

,

20

rotor leakage reactance referred to the stator:

cir-

is in

.V|

1

stator leakage reactance:

the

rotor) ,

on a 1/4 hp,

phase motor reveals the following

stationary, in

which the secondary winding It

imped-

the equivalent circuit.

stator resistance:

of Fig. 18.30 therefore reduces to that shown

ple transformer in

10 ohms, the value

is

forth, for the other

rotor resistance referred to the stator:

Under these conditions,

Fig. 18.31. In essence, the

ohms, and so

parameters

forward and backward circuits are identical. The cuit

in

the stator resistance 5

.

these

follows that the equivalent circuit of the sin-

interpret the

if is

{

ances

gle-phase motor can be represented by Fig. 18.30

To

18.29 to 18.31 are

in Figs.

equal to one-half of the actual physical quantities.

Figure 18.30

stator voltage

indicates that the imped-

The

slip is s

=

(1800

We first determine circuit

between points

diagram

(Fig. 18.32)

shows

impedances divided by .two.

the

1

725)/ 1800

=

0.0417.

impedance of the forward

1, 3:

SINGLE-PHASE MOTORS

4

1

1.5

J

1

1

+ j

.

=

1

=

14.89

+

+

1.5

j

+

j

300

30

+

13.89

j

1

.5

19.53

j

21.03

The impedance of the backward points 3, 2

+

48

between

circuit

120 V

is

1

+

1

1.5

j

1

1

1

+ j

=

+

1

=

j

1.93

The current /

=

1

+

j

+

0.93

+

j

+

1.02

j

1.5

.45

1

2.95

in the stator is

£/(Z F

/

.5

300

30

+ ZB = =

+j

16.82

12()/(

)

23.98)

Figure 18.32

120/(29.29^54.95)

See Example

= 4.097Z -54.95

The forward voltage between points

1,

3

18-5.

is

E v = IZ ¥ = (4.097Z -54.95) X (14.89 + = 4.097/1-54.95 X 25.77Z54.7

Backward

rotor current:

21.03)

j

l

In

=

I I

= 105.6Z-0.25 The backward voltage between

points 3, 2

E B = IZ H = 4.097^-54.95 X (1.93 + = 4.097^-54.95 X 3.52Z56.8 =

is

j

l

2.95)

14.42/1 1.85

=

3.89

' 1

30

4.097

300

Z

+

48

1

.5

.5

1

(0.93

+

j

1

.45)

Z - 54.9 5 X 1,72 Z 57 .32 1.81 Z 55.78

+

j

Pv =

1.5

54.95

(

1

Z- 54.95

X

3.89

+

23.96

48.02/1 1.79 2.16

to rotor:

1.5

j

1

9.53)

2

lv

9.55 P.

Z 54.58

X

Forward torque 1

2.044/1

j

j

Z- 53.4

Forward power j

48.02/1 1.79 4.097

+

+

1

1

48

.02

.02

L81Z55.78

Forward rotor current:

=

1

Z - 54.95

4 .097

4.09 7

'f

I

300

30

j

/i

48 7,

=

1.02

2

X 48 =

W

200.5

:

X

9.55

s

X

2.044

1

Backward power I pj'~

H

=

to rotor

=

200.5

=

1.064

N m

800

3.89

PB 2

X

:

1.02

=

15.4

W

1


414

Backward torque Tn

Pn

9.55 //

motors?

X

9.55

-

15

such a mounting necessary on 3-phase

:

15.4

-

0.082

N-m

1

800

1

s

What

1

- Tu =

1.064

-

0.082

-

0.982

8-8

N-m

X

1725

0.982 177

~

9.55

b.

W

c.

9.55

d.

Horsepower:

e.

177

f.

0.24 hp

746 Active power input to

=

X

120

is

in this

best suited to drive the follow-

A small portable drill A 3/4 hp air compressor A vacuum cleaner A 1/100 hp blower A 1/3 hp centrifugal pump A 1/4 hp fan for use in a hospital

g.

An

h.

A

ward

electric timer

hi-fi turntable

stator:

=

4.097 cos 54.95

282.3

W

Intermediate level 1

Power

main advantage of a capacitor-

ing loads: a.

nT _

e

the

Which of the motors discussed chapter

Mechanical power output P:

£/cos

is

run motor?

Net torque: 7V

8-7

8-9

factor:

Referring to Fig.

1

8.

1

1 ,

the effective im-

pedance of the main and auxiliary wind-

cos 54.95

-

0.57

ings under locked-rotor conditions are

= 57%

given as follows: Efficiency:

177

0.627

= 62.7%

Effective

Effective

resistance

reactance

282

Main winding

4

Auxiliary winding

7.5 il

fl

4

7.5 11

il

Questions and Problems If the line

Practical level 18-1

A 6-pole to a

single-phase motor

60 Hz source. What

connected

a.

synchronous

b.

is

is its

speed? 1

8-2

What

is

the purpose of the auxiliary wind-

c.

d.

The power

the rotation of such a

State the

18-10

What

8-5

main difference between a

18-6

The palm of the human hand can just

how

1/4

a shaded-pole motor

the properties

hp motor

is

64°C

and advan-

a.

Can

b.

Is

a person keep his

the

Referring to Fig.

18. 13, if the

stant at

havior of the motor

some single-phase motors resilient

mounting?

Is

hand on the frame?

motor

connected to a load whose torque

Why

equipped with a

an ambient tem-

motor running too hot?

tages of a universal motor. are

in

perature of 76°F,

18-11

some of

If

split-

are their relative advantages?

Explain briefly

List

factor under locked-rotor con-

barely tolerate a temperature of 130°F.

operates. 1

/s

the full-load temperature of the frame of a

phase motor and a capacitor-start motor.

18-4

/,,

ditions

motor? 8-3

119 V, calculate the

is

The magnitude of and / s The phase angle between /u and The line current I L

ing in a single-phase induction motor?

How can we change 1

voltage

following:

on the

is

is

con-

4 N-m, explain the resulting be-

line.

when

it

is

switched

SINGLE-PHASE MOTORS

18-12

a.

A single-phase

moior vibrates

quency of 100 Hz. What *

power

the

'

b.

c.

a.

motor does not have

set in a resilient

mounting.

to

be

b.

Why?

at

c.

1600 r/min. Calculate the hys-

Referring to the 6

Table a. b. c.

d.

W shaded-pole motor

18-18 in

and the voltage

at the

The

[N m]

starting torque

A

3 hp,

1

725 r/min 230 V, totally-enclosed,

fan-cooled, capacitor-start, capacitor- run,

single-phase motor manufactured by

The rated power output in millihorsepower The full-load power factor The slip at the breakdown torque The per unit no-load current and locked-ro-

Baldor Electric

Company

has the follow-

ing properties:

A

no-load current: 5

tor current

18-14

resistance of the transmission line starting current

Industrial appl ication

8B, calculate the following:

1

the Appendix, calcu-

in

following:

when

in-lb

teresis loss per revolution [J],

18-13

The The

AX3

1

motor terminals

60 Hz single-phase hysteresis

motor develops a torque of 6 running

late the

line?

A capacitor-run A 4-pole,

Using Table

at a fre-

the frequency of

is

4

locked-rotor current: 90

Referring again to Fig. 18.13, calculate

full -load current:

A

A

15

the following: a.

The locked-rotor torque

b.

The

per-unit value of the

c.

The

starting torque

winding d.

The

e.

How

is

locked-rotor torque: 30 lbfft

(t't-Ibf |

LR

torque

when only

the

main

breakdown torque: 20 full-load

breakdown torque

are the lorque-speed curves affected

the line voltage falls

from

1

V

15

to

100

if

power

18-15

Table

In

service factor:

1.15

V?

a.

mass: 97 1

9 lbf

The voltage across

1b

Using the above information, calculate the

the capacitor under

following:

The corresponding phase angle between and

18-16

ft


locked-rotor conditions b.

87%

factor:

full-load torque: level

lbfft

excited

per-unit

Advanced

79%

full-load efficiency:

a.

/s

The

per-unit values of locked-rotor torque,

locked-rotor current, and breakdown torque /.,


if

b.

the capacitor-

The

full-load torque expressed in

run motor operates at full-load, calculate c.

The capacitor

could be added across

that

the following: the stator so that the full- load a.

The

line current

/,

rises b. c.

d.

18-17

The power factor of the motor The active power absorbed by each winding The efficiency of the motor

The motor described in Table 8A has an LR power factor of 0.9 lagging. It is installed in a workshop situated 600 ft from a home, where the main service entrance is located. The line is composed of a 2conductor cable made of No. 12 gauge copper. The ambient temperature is 25°C

newton-

meters

18-19

A

from

87%

to

power factor

90%

3/4 hp, 1725 r/min, 230 V, totally-

enclosed, fan-cooled, capacitor-start, single-phase motor manufactured by

1

and the service entrance voltage

is

122 V.

Baldor Electric

Company

has the fol-

lowing properties: no-load current: 4.4

A

locked-rotor current: f u 1 1 - load c u rre nt

:

5 3 .

locked-rotor torque:

30

A

A 9.5 lbf

ft

416


locked-rotor full- load

power

factor:

efficiency:

breakdown torque:

58%

long and

fed from the service entrance

is

where the voltage

66%

is

230 V ±5%.

Using the above information, determine 6.

1

lbfft

the following: full-load

power

service factor:

factor: 1

68%

a.

b.

full-load torque:

mass:

29

The motor

ft

We

wish

motor to

lb

starting torque (newton-meters),

a cable temperature of

to raise the

90%

pacitor across

power

at full-load its

25

n

C

factor of the

by installing a ca-

terminals. Calculate the

approximate value of the capacitance, is

copper cable

Code

2.25 lbf

The lowest assuming

.25

fed by a 2-conductor No. 12 that has a National Electrical

rating of

20 A. The cable

is

240

feet

microfarads.

in

Chapter 19 Stepper Motors

writers, tapedecks, valves,

Stepper

motors are special motors

used

that are

In this chapter

when motion and position have to be precisely controlled. As their name implies, stepper motors rotate in discrete steps,

a pulse that

is

each step corresponding

supplied to one of

Depending on

its

its

18°, or

by

as

little

cover the operating princi-

will

and

By varying

the pulse rate, the

Elementary stepper motor

A very

simple stepper motor

is

made of

a 2-pole rotor

Stepper motors can turn clockwise or counter-

in Fig. 19.

1

soft iron.

The windings can

means of three switches A, B, C.

When the switches are open,

clockwise, depending upon the sequence of the pulses that are applied to the windings.

any position. However,

The behavior of a stepper motor depends greatly upon the power supply that drives it. The power sup-

sulting magnetic field created

ply generates the pulses,

a

which

microprocessor.

The pulses

while counterclockwise (ccw) pulses are (-).

number of steps is known exactly follows that the number of revolutions

(

rotate

As

the rotor can take up

switch

will line

A

is

closed, the re-

by pole

up with pole

60°, Next,

if

2. In

we open

we now

so doing,

it

will

B and siwill turn ccw

switch

3.

in

60° steps by closing and opening the switches

in

this

we can make

the rotor advance

the sequence A, B, C, A, B,

417

If

ccw

Clearly,

step.

will attract

time lining up with pole

by an additional 60°,

This permits the motor to be used as a precise posi-

1

up as shown.

multaneously close switch C, the rotor

a

is al-

ccw by

if

and simultaneously close switch B,

the rotor will line

)

times.

accuracy of one

A

+

at all

to an

open switch

it

are

result, the net

ways precisely known

the rotor and so

in turn are usually

counted and stored, clockwise (cw) pulses being

It

shown

consists of a stator having three salient poles and

be successively connected to a dc power supply by

r/min.

by

will also discuss

19.1

It

a time, or to rotate stepwise at speeds as high as

initiated

We

limitations.

the types of drives used to actuate these machines.

as a fraction

advance very slowly, one step

4000

plotters, type-

printers.

stator windings.

motor can be made

to

we

and

more common stepper motors, together with

their properties

to

of a degree per pulse.

at

ple of the

design, a stepper motor can ad-

vance by 90°, 45°,

X-Y

tioning device in machine tools,

19.0 Introduction

C

Furthermore,


418

opposite (cw) direction. Picking up speed, again overshoot the center line of pole

upon

will

it

where-

2,

the magnetic field will exert a pull in the

ccw

direction.

The rotor will therefore oscillate like a pendulum around the center line of pole 2. The oscillations will gradually die out because of bearing friction. Fig.

1

9.2

shows the angular position of The rotor starts at

the rotor as a function of time.

0° (center of pole

Figure 19.1 rotor

which each step moves the

in

switches

to a halt (at 3 ms).

line at

reverse

by operating the

rotation

the

reverse sequence A, C, B, A, C,

in the

B

the last switch that

was closed

in a

switching se-

quence must remain closed. This holds the rotor its last

position and prevents

it

in

from moving under

the influence of external torques. In this stationary state the

motor

will

remain locked provided the

external torque does not exceed the holding torque

moving from one

tion of the rotor will

position to the next, the

mo-

be influenced by the inertia and

the frictional forces that

amine

/

>

come

into play.

We now ex-

amplitude

in

rotor has a

low

ing friction.

inertia

It is

at

facing pole

l.

Let this cor-

respond to the zero degree (0°) angular position. At the

moment

switch

A opens and switch B closes,

rotor will start accelerating

ccw toward pole

the

2.

It

rapidly picks up speed and soon reaches the center line

of pole

2,

where

However, the rotor able speed and it

does

so, the

it

is

it

should

now moving

come

to

rest.

with consider-

will overshoot the center line.

magnetic

field

of pole 2 will pull

As

it

in

the opposite direction, thereby braking the rotor.

The

rotor will

come

to a halt

comes

to rest

The reader will note that in Fig. 19.2 we have drawn the instantaneous speed of the rotor as a function of time. The speed can be given in revolutions per second, but for stepper motors it is more meaningful to speak of degrees per second. The also

speed

momentarily zero

is

at

t

=

and becomes permanently zero is

ter line

greatest

of pole

whenever the

2.

3 ms, 5 ms, 7 ms, at

/

>

10 ms. The

rotor crosses the cen-

Clearly, the oscillations last a rel-

down.

atively long time before the rotor settles

Without making any other changes, suppose we

when no-load and that the

and a small amount of bear-

initially

until the rotor

10 ms.

shaft.

We

discover that both the pe-

riod and the amplitude of the oscillations increase

19.2 Effect of inertia motor operates

rotor

4 ms.

wheel on the

the

The

reverse and again crosses the center

in

increase the inertia of the rotor by mounting a fly-

the nature of these forces.

Suppose

line

overshoots the center line

oscillations continue this way, gradually di-

minishing at

speed

of the motor.

=

t

The

.... In order to fix the final position of the rotor,

In

by 30° before coming

by 60°.

we can

and reaches 60° (center It

now moves

Simple stepper motor

l)

of pole 2) after 2 ms.

and

start

moving

in the

the inertia increases. In Fig. 19.3, for

exam-

time to reach the 60° position has

ple, the

creased from 2

ms

to

in-

4 ms. Furthermore, the am-

plitude of the oscillations has increased. also takes a longer time to settle

down

The

(20

rotor

ms

in-

stead of 10 ms).

The

oscillations can be

For example,

the friction.

damped by

if

raised sufficiently, the oscillations 19. 3

increasing

the bearing friction

shown

is

in Fig.

can be suppressed so as to give only a single

overshoot,

shown

damping

accomplished by using an eddy-current

is

in

Fig.

brake or a viscous damper. fluid

such as

oil

19.4.

In

practice,

the

A viscous damper uses a

or air to brake the rotor

whenever

STEPPER MOTORS

deg

,

ing effect

60

zero

30

proportional to speed;

is

when

the rotor

10

—

*-

12

14

16

19.3 Effect of a mechanical load

18

overshoot.

Let us return to the condition

shown

where the rotor has low

and a small amount

inertia

of viscous damping due to bearing rotor

is

coupled

moving, the effect

would expect,

it

deg

w)

to

I

angu lar p os

i

tic

ms

4

)n

in Fig.

shown

is

t

8

10

12

14

16

18

ms

ms

in Fig.

greater.

19.2, except that the iner-

in Fig.

The overshoot

is

greater and the rotor

9.2 with is

the mechanical load and the

The

oscillations

down.

order to obtain fast stepping response,

in

its

load) should be as

pressed by using a viscous damper.

The time

to

move from one

position to the next

can also be reduced by increasing the current

deg

in the

winding. However, thermal limitations due to l~R

90 -antjular pos

lion

losses dictate the

60

CO

1

small as possible and the oscillations should be sup-

takes longer to settle down.

g

is

damped more

are

oscillations

the inertia of the rotor (and

Figure 19.3 tia is

it

As we

also prolong the time before the rotor settles

Therefore,

conditions as

9.5.

Furthermore, the overshoot

inertia increase the stepping time.

—» time

8.

19.5).

summary, both

In

!

Same

1

quickly.

pee(

0

W

in Fig.

takes longer for the motor to attain

smaller and the 60

19.2,

friction. If the

90

»

in Fig.

mechanical load while

to a

the 60° position (compare 2 »

therefore

is

at rest.

is

time

Figure 19.2 In moving from pole 1 to pole 2, the rotor oscillates around its 60° position before coming to rest. The speed is zero whenever the rotor reaches the limit of

?

that the brakit

speed

0

its

moving. Viscous damping means

it is

angular position

90

419

~*

maximum

current that can be

used.

speed

30

Returning to Fig. I9.l,

1 .

-

0

8 i

10

12

—*

time

14

16

18

ms

Same

shows

and the

in-

stantaneous position of the rotor (as well as

its

1

Figure 19.4

9.6

conditions as

in Fig.

19.3, except that viscous

We

assume

and that 90

at the

ngul ar po sitio n

c

V)

to

it

have

/.,,

/h

that the

driving a mechanical load. Note that

is

is

zero

—"

time

14

16

18

ms

a duration of 8 ms. Consequently, the step-

Same

conditions as

coupled

to

is,

0.048

is

1

in Fig.

19.2, except that the rotor

a mechanical load.

s to

000/8

=

125 steps per second.

complete one

therefore,

minute.

Figure 19.5 is

=

beginning and

at the

In this figure the pulses

revolution requires 6 steps, and so 12

/c

makes one-half revolution. stepper motor has some inertia

end of each pulse.

ping rate 10

,

the motor

the speed of the rotor

deg

a.

the current pulses

when

speed)

damping has been added.

us excite the wind-

let

ings in succession so that the motor rotates. Fig.

60/0.048

However,

start-stop

the

=

turn. 1

takes 6

One

+ 25 1

The average speed

250 revolutions per

stepper motor rotates

jumps and not smoothly

motor would.

it

in

as an ordinary

420


current

71 a,

10A

S777777777Z7A

-fr77777777777i^.

J777777,

i

10A

deg 240

—

/

© 180 « 120

60 0

settling time

H

6

-

ms

8

0

24

16 -

ms

time

24

16 time

Figure 19.6 Graph of current pulses, angular position, and instantaneous speed (24 ms) produce one half-revolution.

of rotor during the

When the motor continue to flow

is

first

four steps.

at rest, a

in the last

Three steps

holding current must

winding

that

was excited

so that the rotor remains locked in place.

19,5 Start-stop stepping rate

When

motor inches along

the stepping

stop fashion

shown

in Fig.

1

9.6, there

is

in the start-

an upper limit

to the permissible stepping rate. If the pulse rate of the

current in the windings

is

too

fast, the rotor is

unable

to accurately follow the pulses, and steps will be -

lost.

current

This defeats the whole purpose of the motor, which

Figure 19.7 Graph of pull-over torque versus current motor; diameter: 3.4 inches; length: 3.7

to correlate of

in;

19.4 Torque versus current

this

As mentioned

rate

motor depends upon the current. shows the relationship between the two

stepper

Fig. 19.7

for a typical stepper motor. 8 A, the

motor develops

the torque that the

When

a torque

of 3 N-m. This

it

is

pulses. In order to maintain

is

settle

down

before ad-

that the interval

at least

between successive

6 ms, which means

limited to a

maximum

of

1

9.6,

=

167 steps

per second (sps).

Bearing

in

mind what was

clear that the

said in Section 19.2,

maximum number

of steps

it

is

second depends upon the load torque and the

called the

1

steps

that the stepping

000/6

is

motor can exert while moving

from one position to the next, so pull-over torque.

the current

)

to the next position. Referring to Fig.

means

must be

previously, the torque developed by

—

synchronism, the rotor must

vancing

is

instantaneous position (steps) with the

number of net ( + and

a stepper weight: 5.2

Ibm.

a

its

tia

is

pet-

iner-

of the system. The higher the load and the greater

the inertia, the lower will be the allowable

of steps per second.

number

STEPPER MOTORS

deg

1

mode

start-stop

300

42

J

240 CD

/

C

slew

CO

mg nnode

180

slew curve I

120

/ 60

/

0

4

2

8

6

10

time

speed

start-stop

k 0

6

4

2

10

8

ms

16

14

12 *-

s,ewspeed

ms

16

14

12 time

800 sps

400 500 600

200

0

Figure 19.9

speed

»-

Angular position versus time curve when the stepper motor operates in the start-stop mode and the

a.

Figure 19.8 and slewing characteristic of a typical stepper motor. Each step corresponds to an advance of 1 .8

slewing mode. Stepping rate

is

same

the

in

both

Start-stop

cases.

Instantaneous speed versus time curve

b.

degrees.

curve

1

curve

2:

:

start-stop curve with only stepper

same

conditions as curve

tional load inertia of

curve

3:

2 kg-cm

1

motor

stepper motor operates inertia

ping

slewing curve

start-stop stepping

mode

is

ferred to as the start-without-erwr

without-error characteristic Fig.

19.8.

It

shows

is

imum 500 But

sometimes mode.

A

that if the stepper 1

.4

re-

start-

shown by curve

alone, under a load torque of, say,

is

1

in

motor runs

N-m, the max-

possible stepping rate, without losing count,

rate, the

if

the

motor drives

some

a device having

in-

A stepper motor can

the

this

way

be made to run

it is

at

at

uniform speed

every step.

When the

said to be slewing.

motor runs essentially

tia effect is

shows

at

500

steps per second.

should exceed 2.2 sps, the

net

slewing. For example, the

motor

However,

N-m when

will fall out

Fig. 19.9

if

the pulse rate

to

its

500

the startis

an average speed of 250 steps per second will therefore

to the

windings.

slewing. Suppose the motor

The motor

is

no longer correspond

shows the difference between

mode and

slews

it

the load torque

of step and the position

number of pulses provided

stop

the re-

turning in

both

cover the same numl

step every 4 ms.

However, the angle (position) increases smoothly

without starting and stopping

motor runs

is

ber of steps per second, namely

Slew speed

9.8

motor can develop a torque of 2.2 N-m when

at

2).

1

between the load torque and the steps per

second when the motor

cases.

19.6

motor can carry a greater load torque

slewing. Curve 3 in Fig.

is

lationship

permissible start-stop rate drops to about

400 steps per second for the same load torque (curve

it

(steps) of the rotor will

steps per second.

ertia, the

the

2

when The

when

the start-stop and the

slewing mode.

but with an addi-

,

in

at

Because

uniform speed, the

with time

responding slew speed

On

iner-

absent. Consequently, for a given step-

when the motor is slewing, and this is shown

by the uniform slope of line

gle

is

OA (Fig.

I9.9a).

constant (Fig.

1

The

cor-

9.9b).

the other hand, in the start-stop mode, the an-

increases stepwise. Consequently, the speed


422

maximum and

continually oscillates between a

and

average value

its

is

zero

equal to the slew speed

the pole-faces are also slotted so as to create a

The

the stator.

(Fig. 19.9b).

8-pole stator

typical construction of a toothed

shown

is

teeth (salient poles)

When

in the circular insert

19.13. For a given drive system,

Ramping

19.7

a stepper motor

is

carrying a load,

suddenly go from zero to a stepping

same way, a motor

it

cannot

5000

rate of, say,

num-

ber of teeth. These teeth are the real salient poles on

it

is

on the rotor and

of Fig.

number of

the

stator that de-

termines the angular motion per step. Steps of 18°, 1

5°, 7.5°, 5°,

and

1

.8° are

common.

5000

Permanent magnet stepper motors are similar

sps cannot be brought to a dead stop in one step. Thus,

to variable reluctance motors, except that the rotor

sps. In the

to bring a

motor up

to speed,

gradually. Similarly, to stop a

high speed,

slewing

that is

must be accelerated

it

motor

that

is

running

must be decelerated gradually

it

at

at

—always

subject to the condition that the instantaneous position of the rotor

pulses.

must correspond

The process whereby

and decelerated

is

number of

to the

a motor

is

accelerated

called ramping. During the accel-

ramping consists of a progressive

eration phase,

in-

in the number of driving pulses per second. The ramping phase is usually completed in a fraction of a second. The ramp is generated by the power

has permanent N and S poles. Fig. 19.10 shows a permanent magnet motor having 4 stator poles and

6 rotor poles, the

Due

is

programmed

the

motor and

19.8

to retain precise position control

its

it

over

being permanent magnets.

lined up with the last pair of stator poles that

were

excited by the driver. In effect, the motor develops a detent torque

when no

crease

supply that drives the stepper motor. Furthermore,

latter

permanent magnets, the rotor remains

to the

Coils

B

1,

which keeps the rotor

place even

in

current flows in the stator windings.

A

1

,

A2

are connected in series, as are coils

B2. Starting from the position shown,

are excited, the rotor will

move through

if

coils

30°.

However, the direction of

upon

the direction of current flow. Thus,

rotation if

depends the cur-

load.

Types of stepper motors

There are 3 main types of stepper motors; •

variable reluctance stepper motors

•

permanent magnet stepper motors

•

hybrid stepper motors Variable reluctance stepper motors are based

upon

the principle illustrated in Fig.

to obtain small

1

9.

1

.

However,

angular steps, of the order of 1.8°

(instead of the 60°

jumps shown

in the figure), the

structure of the stator and rotor has to be modified to create

many more

circular rotor riphery.

The

salient poles

many as As to

1

poles. This

and milling out

is

done by using a

slots

around

its

pe-

teeth created thereby constitute the

of the rotor, of which there

may

be as

00.

the stator,

it

often has four, five, or eight

main poles, instead of the three shown. However,

Figure 19.10 Permanent magnet stepper motor per step.

that

B

an angle of

advances Ô 0

STEPPER MOTORS

B produces N and S

rent in coils

423

shown

poles as

in

Fig. 19.10, the rotor will turn ecw. Stepper motors that

have to develop considerable power are usually

equipped with permanent magnets.

Hybrid stepper motors have two

mounted on

iron armatures

identical soft-

same

the

The

shaft.

matures are indexed so that the salient poles lap. Fig.

1

9.

1

1

shows two 5-pole armatures that

a

driven by a 4-pole

motor look

the

ar-

inter-

are

This arrangement makes

stator.

like a variable reluctance motor.

However, a permanent magnet between the armatures

(Fig. 19.

PM 1

sandwiched

is

lb).

It

produces a

unidirectional axial magnetic field, with the result

on armature

that all the poles

1

are

N

poles, while

those on armature 2 are S poles.

A2

Stator coils Al,

are connected in series, and

(a)

so are stator coils Bl, B2.

Figure 19.11a Hybrid motor having a 4-pole stator and two 5-pole armatures mounted on the same shaft. The salient poles on the first armature are all N poles, while those on the second armature are all S poles. Each step produces an advance of 18°.

and the rotor

net,

in Fig.

will rotate

poles B.

upon

A1

will

19.1 la. If

by

The

remain

we now

18°,

shows an exploded view of

embedded

which permanent magnets

Fig. 19.

^

armature 2

rotor.

14a shows another type of hybrid motor

and Fig. 19.14b struction.

1

a hybrid

in the stator slots, in addition to the

permanent magnet on the

armature

depend

the direction of current flow in coils B.

stepping motor. Fig. 19.13 shows the special con-

are

PM

shown

in the position

thereby lining up with stator

struction of a stator in

shaft

a

excite coils B, the rotor

direction of rotation will again

Fig. 19.12 stator

J_

The motor develops

small detent torque because of the permanent mag-

is

a cross-section

19.14c

Figs.

and

view of

I9.14d

its

con-

respectively

show

the specifications

istics

of this motor. Note that the pull-out charac-

teristic

and torque-speed character-

corresponds to the slewing curve while the

pull-in characteristic corresponds to the start-with-

A2 out-error curve. It

should be noted that the number of poles on

the stator of a stepper (b)

different

Figure 19.11b showing the permanent magnet sandwiched between the two armatures. The

Side view of the

PM

4-pole stator

is

number of poles on

rotor,

common

to both armatures.

from

have studied so

that in far.

number of poles they do.

motor

is

never equal to the

the rotor. This feature

is

totally

any other type of motor we

Indeed,

it

is

the difference in the

that enables the

motors

to step as

424


Figure 19.12 Exploded view

of

of

is composed of two soft-iron armatures having 50 sandwiched between the armatures. The stator has 8 poles, each

a standard hybrid stepping motor. The rotor

salient poles each.

A

short permanent

which has 5 salient poles

(Courtesy of Pacific

is

the pole face. Outside diameter of motor: 2.2

in

Scientific,

magnet

Motor and Control

in;

axial length: 1.5

in;

weight: 0.8

lb.

Division, Rockford, IL)

19.9 Motor windings

are represented

by the contacts Ql

transistors are used as switches

and associated drives

to

Q8.

In practice,

because they can turn

the current on and off at precise instants of time.

Stepper motors use either a bipolar or a unipolar

winding on the

stator.

Bipolar Winding.

In a 4-pole stator, the bipolar

ing consists of the

such as shown

two

in Fig.

schematically in Fig. reverses periodically, I

b in coil set

B.

The

1

coil sets

I9.ll.

Al,

They

A2 and

are represented

1

5.

/.,

coils are excited

by a

source, and because the current pulses ternate,

wind-

coils can

a switching means

is

required.

in three dif-

normal drive, and

wave drive only one set of coils is excited The switching sequence for cw rotation

given

Table

/h

in

are

shown

dc

duced by

al-

In the

The switches

drive, (2)

In the

must

Ib

wave

a time.

common

/.,,

be excited sequentially

(3) half-step drive.

Bl, B2

The current in coil set A and the same is true for current 9.

The

ferent ways: (l)

a time.

/a

1

9A and the resulting current pulses

in Fig.

and

1

9. 16.

/b rotates

Note

sets

is i.

r

that the flux pro-

by 90° per

normal drive, both

at

step.

of coils are excited

The switching sequence for cw

rotation

is

at

given

STEPPER MOTORS

425

'Ilk

Figure 19.14a External view of a hybrid stepper motor.

It

equipped

is

with bipolar windings rated to operate at 5 V. External

diameter of motor:

1

.65

in;

axial length:

0.86

in;

weight: 5.1 oz. DETAIL

OF

STACK POLE FACE WITH MAGNf IS IN PL. ACL

[Courtesy

ofAIRPAX©

Corporate)

Figure 19.13 and construction of an enhanced motor stator lamination stack assembly. Rare earth permanent magnets are fitted into the stator slots in addition to the permanent magnet of the hybrid rotor. (Courtesy of Pacific Scientific, Motor and Control Division) Stator lamination details

in

Table

shown

1

9B, and the resulting current pulses

in Fig.

way between rotates

1

9.

1

7.

Note

step.

I b are

/.,,

that the flux is oriented

the poles at each step.

by 90° per

The normal

However,

COIL A

COIL B

midstill

it

drive develops a

wave drive. The half-step drive is obtained by combining wave drive and the normal drive. The switching slightly greater torque than the

quence for cw rotation

is

given

in

the se-

Table 19C, and

Figure 19.14b Cross-section view of the hybrid stepper motor shown in Fig.

the resulting current pulses / a 7 b are ,

19. 18.

The

flux

now

shown

in Fig.

rotates only 45° per step.

main advantage of the half-step drive proves the resolution of position and

is

it

that

it

19.14a.

(Courtesy

ofAIRPAX©

Corporate)

The im-

tends to re-

set

I

A, 2A. Consequently,

when

sequence, an alternating flux

they are operated

is

duce the problem of resonance.

vantage of the unipolar winding

Unipolar Winding. The unipolar winding consists

switching transistors drops from 8 to 4, and

of two coils per pole instead of only one (Fig.

sient response

19.19a).

Unipolar means

ing always flows in the

A

1

,

A2 produces flux

that the current in a

same

in the

direction.

The

wind-

coil set

opposite direction to coil

the

is

in

produced. The ad-

is

number of

that the

slightly faster. Fig.

19.

1

the tran-

9b shows

schematic diagram of the windings and the

switching sequence for a wave drive. The flux rotates in exactly the

same way

as

shown

in Fig.

19. 16.

Specifications L82402

L82401

Ordering Part No. (Add Suffix)

Unipolar Suffix Designation

-P2

-P1

5

12

5

12

9.1

52.4

9.1

52.4

46.8

14.3

DC Operating Voltage Res. per Winding O Ind.perWinding

mH

7.5

Holding Torque mNm/oz-in* Rotor Moment of Inertia g *

m

Bipolar

-P1

-P2

77.9 87.5/12.4

73.4/10.4 12.5 x 10-"

2

Detent Torque mNm/oz-in

9.2/1.3

Step Angle

7.5°

Step Angle Tolerance*

.5°

Steps per Rev.

48

Max Operating Temp

100°C

Ambient Temp Range - 20°C to 70°C

Operating Storage

- 40°C

Bearing Type

to

85°C

Bronze sleeve

Insulation Res. at

SOOVdc

100

megohms max

650 - 50 VRMS 60 Hz for

Dielectric Withstanding Voltage

Weight g/oz

144/5.1

Lead Wires

26AWG

1

seconds

to 2

'Measured with 2 phases energized.

Figure 19.14c Specifications of the hybrid stepper motor

shown

operation at a rated driving voltage of either 5

V

in Fig.

The motor can be

19. 14a.

built for either

unipolar or bipolar

or 12 V.

(Courtesy of AIR PAX© Corporate)

UNIPOLAR TORQUE L8240I

70 0

L/ft

vs

BIPOLAR

SPEED

TORQUE

PHASE DRIVE

2

SPEED

vs

PHASE ORivE

2

70 0

60 0

8.50

60.0

B

50

SO.O

708

50 0

7

08

5

66

991

T

E

z

-

L82402 L/R

991

z

40 0

566

UJ

PULL OUT

3

—

Z

"

o gsoo

40 0

Z>

O g30

. 25

Pul

1

.

Out

«25

0 PULL IN^

PULL IN^

20 0

2

lOO

83

20 0

?

100

".42

'

50

lOO

150

200

SPEED

NOTE: The above curves

250

300

330

400

0

50

100

150

200

SPEED

(PPS)

42

0

0 0

83

250

300

350

« 30

(PPS)

are typical.

Figure 19.14d Typical torque-speed characteristics of the hybrid stepper motor

sponds

to the slewing characteristics; the pull-in

(Courtesy

ofAlRPAX©

shown in Fig. 19.14a. The pull-out curve correcurve corresponds to the start-without-error characteristic. ,

Corporate)

426

Z

S

STEPPER MOTORS

19.10 High-speed operation

ductance of the windings.

All

a winding has an in-

If

ductance of L henrys and a resistance of R ohms,

So

tar,

we have assumed

winding

that the current pulse in a

immediately

rises

to

its

rated value

/ at

beginning of the pulse and drops immediately zero

at

the

end of the pulse

interval

Tp (Fig.

9.20a).

1

does not happen because of the

In practice, this

the to

in-

1

I

E volts

Let the coil be connected to a dc source of (Fig.

1

connected across the windings

9.20b).

A diode (D) is

to prevent the high in-

duced voltage from destroying the switching

moment

its

equal to LIR seconds.

it

transis-

interrupts the current flow.

The

resulting current has the shape given in Fig. I9.20d.

i Q4

is

by means of a transistor

tor at the

+

T0

time constant

A2

Q2

r r

How can we transistor

is

only reaches

When

explain this pulse shape?

the

switched on, the transient current its

rated value

time constants, namely 3

/

T0

= EIR

/,

after about 3

seconds. Then,

when

the transistor turns the line current off, the transient

current

—

/

2

continues to flow

seconds (Fig.

£

1

in the coil for

about 3

9.20c). If this current pulse

pared with the ideal current pulse shown

x_r

Q8

B2

B1

Q6

T t

1

1

9.20a, .

we observe two

value

final

when

in Fig.

important facts:

Because the current does not immediately its

T0

com-

is

the transistor

is

rise to

turned on,

moAs a result, the rotor quickly as we would expect.

the initial torque developed by the stepping

Figure 19.15 Schematic diagram showing how the stator coils A1 A2 and B1 B2 are connected to the common dc source by means of switches Q1 to Q8. The dc source

tor

is

smaller than normal.

does not

move

as

,

is

shown twice

TABLE 19A

to simplify the connection diagram.

Qi

Q3 Q5 Q7

Q2 Q4 Q6 Q8

When

ROTATION

I

on

2

3

—

on

on

the transistor

is

continues to circulate

turned off, current

As

on

W//,

— step

1

2

4

3

u

1

u

3

n 1

2

m,

4] on

step

i

in the coil/diode loop.

'/////

WAVE SWITCHING SEQUENCE FOR

CW Step

2.

—m n

step 2

step 4

step 3

Figure 19.16 Current pulses

in

a wave drive and the resulting

flux positions at

each

step.

See Table 19A

for switching

sequence.


428

UJ

lnJ

H

ill

H/K H\K 3/E m step

1

\

3

R

R

Fl

step 2

step 3

step 4

E

Figure 19 .17 Current pu Ises

in

a normal drive and resulting

CW

//////

on

on on

on

on

W/W//,

on

on

on

on

on

step

3

T

()

instead of

Tp The .

we cannot switch from one coil to the quickly as we would have thought. shortest possible pulse that

current to rise to

T0

6

its

rated value

seconds (Fig. I9.20e).

rent rises to rent drops

its

It

still

/

/

to zero).

that

l

ms

has a length of

T0 T0

6 ms. This corresponds to a

rate

of about

1

000/6

=

(cur-

T0

maximum

X

I

166 steps per second. Such

to

quicken the stepping

the time constant

Ta

.

is

to reduce

motor windings and

ing the dc voltage so that the

same

Such an arrangement

is

The

external resistor has a value 4 times that

E

E

to 5

As

volts.

maximum

same

factor.

is

raised

a result, the time constant

stepping rate can be increased by

Thus, stepping rates of the order of

1000 per second become

The only drawbacks 1.

feasible.

to this solution are the following:

The power supply

is

more expensive because it much power (the volt-

has to deliver 5 times as

age

A

is

lot

5

El

instead of E).

of power

which means

is

wasted

in the external resistor,

that the efficiency of the

Low

system

is

very low.

in

small stepping motors that develop only a

efficiency

is

not too important

in

the

1

00

W range must be

driven by other means.

19.12 Bilevel drive

rais-

rated current

shown

2.

stepper motors

This can be done by adding an

external resistance to the

will flow.

rate

.

few watts of mechanical power. But fast-acting

Modifying the time constant

One way

the

ms

ous means are used to speed them up.

9.11

1

stepping

stepping rates are considered to be slow, and vari-

1

1

that the

Thus, the duration

of one step can be no shorter than about 6

=

|

(cur-

so happens that the

to 8 ms.

4

|

drops by a factor of 5 (LIR to L/5 R). This means

windings of stepper motors have time constants ranging from about

9.2

from

permits the

consists of 3

It

3

|

of the coil resistance R, and the dc voltage

next as

rated value) plus another 3

from

2

|

|

1

T0 means

1 |

is

pulse being thus

prolonged by the component 3

The

switching sequence.

I

a result, the effective duration of the pulse

Tp +

for

W/a 4

3

2

I

Q2 Q4 Q6 Q«

See Table 19B

ROTATION

Step

Qi

step.

NORMAL SWITCHING SEQUENCE FOR

TABLE 19B

Q3 Q5 Q7

each

flux positions at

/

in Fig.

Bilevel drives enable us to obtain fast rise and

fall

times of current without usins externa] resistors. The

STEPPER MOTORS

HALF-STEP SWITCHING SEQUENCE FOR

TABLE 19C 'Step

Ql

Q2 Q4 Q6 Q8

Q3 Q5 Q7

2

I

ROTATION

4

3

6

5

— —

—

—

—

on

on

on

on

on

on

on

— —

CW

429

8

7

l

on

on

on

—

on

on

on

y ////////////

V/////MW//, step

i

3

2

1 |

I

I

4

|

i

step

5

1

step 2

1

6

|

7

|

|

8

1

|

i

I

1

1

1

step 4

step 3

u

!

n

step 5

jH

step 6

—

IE

j3 1

\

X

n

(*1

step 7

step 8

Figure 19.18 Current pulses

in

a half-step drive and resulting

flux positions at

principle of a bilevel drive can be understood ferring to Fig. 19.22a. Switches transistors that

Ql and Q2

open and close the

by

re-

represent

circuit in the

man-

ner explained below. Numerical values will be used to

each

explain

step.

how

assumed

to

See Table 19C

for switching

the circuit behaves. Thus, the winding

have a resistance of 0.3

of 2.4

mH and a rated current of

ply

60

is

V

sequence.

with a tap

at

1

(2,

is

an inductance

0 A. The power sup-

3 V. Thus,

if

the voltage

were

430

ELECTRICAL MACHINES AND TRANSF ORMERS

B

WAVE SWITCHING SEQUENCE FOR

TABLE 19D

CW ROTATION Step

Ql

Q2 Q3 Q4 step

2

1 I

4

3

—

3

4

— — on —

—

— — —

on

— —

1

on

— — —

on

1 I

I

I

I

on

— —

2

I

(c)

Figure 19.19 Coil arrangement in a 4-pole unipolar winding. b. Schematic diagram of coils, switches, and power supply

a.

c.

Current pulses ing.

in

See Table 19D

applied permanently,

for switching

the

=

200 A. This

in is

the

much

initiated

is initially

closed.

The

current pulse

by closing Q2. Current then

starts

flow-

shown in Fig. 19.22b. The time constant of this electronic circuit is T0 = 2.4 mH/0.3 il = 8 ms. The initial rate of rise

ing as

flux rotates in the

same way as

in

a bipolar wind-

of current corresponds to a straight line reaches 200

A

in 8

rises at a rate of

greater than the rated current of 10 A.

is

a unipolar drive.

sequence.

resulting current

winding would be 60 Y70.3 fi

Switch Ql

in

a wave drive using a unipolar winding. The

to reach

10

A

ms. Thus, the current

200 A/8 ms is,

=

OP

that

in the coil

25 000 A/s. The time

therefore, 10/25

000 = 0.4 ms

(Fig. 19.22c),

As soon as the current reaches this rated value, Ql opens, which forces the current to follow the new path shown in Fig. 19.22d. The current is switch

time

3rn Figure 19.20a

Figure 19.20d

a winding.

Ideal current pulse in

Real current pulse.

I

= h

Figure 19.20e Shortest possible current pulse that rated current

-

6

0 ^

t

still

attains the

/.

^ Tp

Figure 19.20b Typical circuit of a switching transistor

and

coil

con-

nected to a dc source. The diode protects the transistor

against overvoltage.

i

+

= 0

Q (a)

O"0

(b)

Figure 19.21 Figure 19.20c Transient current

switched

a. in coil

and diode when

transistor

Circuit to increase the rate of growth

current

is

b.

off.

431

in

the

and decay

of

coil.

Resulting current pulse.

Compare

with Fig.

1

9.20d.


432

D2 60 V

~

Q1 200

0.3

n

2.4

mH

10

Q2

(b)

(a)

57 V

A--

A

-

—

~ x

(c)

r.

D1

D1

D2 D2

10

0.3

n

2.4

mH

n

2.4

mH

A (e)

(d) )

3

0.3

V

L Figure 19.22 a.

Circuit of a bilevel drive

b.

Equivalent circuit

Rate

Equivalent circuit

e.

Equivalent

now

circuit

fed by the 3

3 V/0.3

The

H-

I

current

in coil is

in coil is

zero.

increasing.

when when

V

current

in coil is

constant.

current

in coil is

decreasing.

source and remains fixed

after 5

open Q2, which forces I9.22e.

drive a current through the coil that

Consequently,

current will stay at this value until

in Fig.

at

OA.

end the pulse, say

shown

when

current

increase of current and time to reach 10 A.

c.

d.

of

when

we want to we

ms. To terminate the pulse

the current to follow the path

The 57

V

source

now

tries to

circuit

is

/

will decrease.

is

opposite to

/.

The time constant of the

again 8 ms, and so the current will decrease

X 25 000 - 23 750 A/s. It will therebecome zero after a time interval of 10/23 750 = 0.42 ms. The moment the current reaches zero. Ql at

a rate of 57/60

fore

STEPPER MOTORS

closes. This forces the current to

next pulse

shown

is

in Fig.

The

initiated. 1

remain zero

9.22f, together with the

ing sequence that produces

In addition to bilevel drives,

method, except

constant during the

flat

Q Q2 switch1

,

chopper drives are is

similar to

portion of the pulse by re-

by using a low fixed dc voltage

Choppers are described

in

sophisticated.

board drives are shown

ms-^|

'2 '3

^ -5

|-

ms-

-*|

|—

0.42

Some

in Figs.

of these circuit-

19.23 and 19.24, to-

Figure 19.23

and the stepper motors they

(Courtesy of Pacific Scientific, Motor

Q1

X

X

Q2

0

0

Figure 19.22f

gether with the motors they control.

Typical electronic drives

0.4

(3 V).

Chapter 21.

Electronic drives for stepper motors have be-

come very

o

that the current is kept

peated on-off switching of the high voltage (60 V) rather than

A

r_

is

it.

also used. Their principle of operation the bilevel

10

until the

resulting pulse shape

433

and Control

control.

Division)

Pulse waveshape using a switching sequence of (x

=

closed, o

bilevel drive.

Q1 and Q2

= open).

Note the

that creates

it

\


434

19.13 Instability and resonance

19-5

Explain what

wave

When

a stepper

ing speeds,

it

motor

is

operating

may become

at certain

unstable.

The

is

due

instability, often called

to the natural vibration

which manifests

may

is

between 2000 sps and 8000

possible to

ramp through

1

9-6

19-7

may

sps. Nevertheless,

this

it

19-8

range without los-

1

9-9

of 5 threads per inch. The motor has to to

in.

19-10

the pulses precisely,

position a machine tool,

X-Y

why

ramping or

is

to a precise angle of rotation.

1

How

A stepper

long will

motor

motor

1

0

is

it

driven

take for the rotor

revolution?

rotates

1

.8°

per step.

The

It

lead screw, in turn,

much does 19-11

the cutting tool

A stepper motor advances If its

by

movement.

move? 7.5° per pulse.

torque-speed characteristic

Fig. 19.8, calculate the it

is

tool.

how

pulsed 7 times, by

is

develops when is

9.

1

having a duration of

make one complete

If the

inch.

we can

in Fig.

series of pulses

threads per inch.

arm, and so on, to a

This great precision without feedback son

motor

produces a linear motion of a cutting

precision of one-thousandth of an inch over the full length of the desired

new angu-

drives a lead screw having a pitch of 20

make 200 X

produce a linear motion of

By counting

a stepper

The stepper motor 20 ms.

Consequently, each step produces a displacement of 0.001

replaced

is

1

viscous damping employed in

is

When

to

coupled to a lead screw having a pitch

1000 steps

Why

by a

ample, that a stepper motor having 200 steps per

=

9.

motion per pulse.

sponds

converted to a linear displacement. Suppose, for ex-

5

1

Calculate the

True or false?

Most stepper motors are coupled to a lead screw of some kind which permits the rotary motion to be

is

rotor.

slewing properly, every pulse corre-

sps.

19.14 Stepper motors and linear drives

revolution

rotor in Fig.

by a 4-pole

stepper motors?

ing step and thereby attain stable slewing speeds

between 8000 and 15 000

The 2-pole lar

of the stepper motor,

speeds. For example, the range of instability lie

Intermediate Level

resonance,

one or more range of

itself at

drive,

drive.

slew-

rotor

turn erratically or simply chatter without rotating

any more. This

meant by normal

is

and half-step

drive,

is

power

given

[watts]

it

slewing

a.

At 500 steps per second

b.

At 200

the rea-

stepper motors are so useful in control

19-12

systems.

steps per second

A stepper motor similar to that

shown

Fig. 19.14 has a unipolar winding.


ates in the start-stop

mode

at a

Jt

in

oper-

pulse rate

of 150 per second, (see Fig. 19.14d). Practical Level 1

9-

1

9-2

1

What What

is

is

the

main use of stepper motors?

a.

What

b.

How much

and a permanent magnet stepper motor? Describe the construction of a hybrid stepper motor. 19-4

the

maximum

How much produce

19-13

torque

it

can develop?

mechanical power (millihorse-

power) does

the difference between a reluctance c.

19-3

is

it

develop?

mechanical energy

in 3

[J]

does

it

seconds?

For a given load torque, the stepping

rate

can be increased by increasing the rate of

A stepper motor advances 2.5° per step. How many pulses are needed to complete

windings.

8 revolutions?

complished.

rise

and rate of

fall

of the current in the

Name two ways

this

can be ac-

STEPPER MOTORS

19-14

Referring to Fig.

maximum

what

19. 14d,

The motor

the

is

slew speed of the unipolar

it

chopper-driven

is

develops a torque of 2.2

at

N-m

motor, expressed in revolutions per

sps. Calculate the following:

minute?

a.

435

65

V

at

10 000

and

The speed [r/min] and power [hp] of the motor when it is running at 0 000 sps The time constant of the windings [msj The time to reach 3 A when 65 V is ap1

Advanced Level 19-15

a.

b.

Referring to the stepper motor properties listed in Fig. 19.14c, calculate the

winding rated

stant of a bipolar b.

If

V

the 12

how

current to reach

What

is

1

long will final

its

it


take for the

19-19

value?

the final value of the current in the

motor each have 50

the

a hybrid stepper

The angle between two successive

teeth

on

in

and the next tooth on the other armature

when

it is

when mode?

slewing than

operating in the stop-start

19-18

9-20

can a stepper motor develop a larger

torque

it

permanent magnet stepper

motor used

for positioning a valve has the

following specifications:

is

33

mH. What

1

9-2 1

60 mCl 0.77

6

N-m

9.5

N-m

0.

holding torque:

1

8

sps:

mH

N-m

steps per revolution: rotor inertia:

19-22

X

0.7

Motor diameter:

value of resistance should

25°C

200 10~ 3 kg-m 2

4.2 in

axial length:

9 kg

7.0 in

at

proposed

It is

with each winding

0°C and 100°C. The time is

1

.32 ms. Calculate the

100°C.

to use the 12

V

stepper

in Fig. 19.14c, to drive a

disc having a 6

moment

metal

of inertia of 80

X

2

g-m The desired speed is 250 r/min. The disc rubs against a stationary member, which exerts a constant friction 10

.

TF How many

torque a.

b.

.

pulses are required per second

produce a speed of 250 r/min?

We

want the motor

stop

Motor weight:

A

have a hold-

motor L82402, whose characteristics are

to

Motor

to

A unipolar stepper motor is designed to

shown detent torque:

50

is

stepper motor possess a

in series

time constant

winding inductance:

at

it

croseconds?

constant at

winding resistance:

torque

when

so that the time constant becomes 400 mi-

A

3

is

SI units (N-m).

be connected

bipolar 1

known

The windings of a

operate between

current:

detent torque

resistance of 26 II and an inductance of

is

A powerful

winding:

is

torque of 11 oz-in. Express these values

tooth on one arma-

The angle of advance per pulse

Why

motor

can exert

torque that a nonexcited

stepper motor

1

19-17

it

ing torque of 74 oz-in and a detent

The angle between one ture

The

excited.

maximum

a stepper

torque

equipped with a permanent magnet.

an armature

c.

it is

static

stepper motor can exert

salient poles (teeth).


b.

maximum

when

The two armatures on

a.

The holding torque of the

winding?

19-16

1

plied to the winding [|xS]

2 V.

are applied to the winding, ap-

proximately

c.

at

c.

time con-

(a).

to operate in the start-

mode using the pulse rate calculated How much pull-in torque does it de-

velop under these conditions?

in

436


c.

d.

What torque

is

needed

to accelerate the

We

wish to drive a ma-

chine tool at a speed of 500 r/min.

What

this objective

is

the largest admissible friction

torque TV?

19-23

pulses per second.

metal disc from zero to 250 r/min?

device directly to the motor?

A stepper motor advances pulse,

and

its

slew rate

is

1

.8°

per im-

Can

be achieved by coupling the If not,

can

you suggest a solution?

limited to 1200

Figure 19.24 Rotor, stator, and electronic speed controller of a switched reluctance (SR) motor. The 8-pole rotor and 12-pole stator are used to drive a horizontal washing machine. The SR motor has a nominal rating of 3/4 hp, a peak rating of 1 .5 hp, and can reach speeds of 13 000 r/min. The stator has an outside diameter of 140 mm and a stacking of 50 mm. The electronic controller supplies pulsed 3-phase power to the stator. The entire unit is designed to operate from a 120 V, 60 Hz, single-phase source. Switched reluctance motors operate on the same principle as stepper motors, (Courtesy of Emerson Electric)

Part Three Electrical

and Electronic Drives

437

Chapter 20 Basics of Industrial Motor Control

systems that are very complex. The basic compo-

20.0 Introduction

nents are the following: Industrial control, in

passes

all

mance of an chinery,

it

to control the perfor-

When

applied to ma-

starting,

acceleration,

electrical system.

involves

reversal, deceleration, its

broadest sense, encom-

its

methods used

the

the

1.

we

will study the electrical

(but not electronic) control of 3-phase alternating-

circuits

2.

Manual

3.

Cam

4.

Pushbuttons

5.

Relays

Our study

is

limited to elementary

6.

Magnetic contactors

7.

Thermal relays and fuses

8.

Pilot lights

9.

Limit switches and other special switches

because industrial circuits are usually too

intricate to explain briefly.

However, the basic

ciples covered here apply to

no matter

how complex

it

prin-

any system of control,

may appear

to be.

10.

20.1 Control devices Every control basic

circuit

is

composed of

varies with the

The

Resistors, reactors, transformers, and capacitors

The ensuing a

number of

size of the

components

power of the motor, but

the principle

list

of Basic Components for Control

Circuits illustrates these devices, and states their

components connected together to achieve the

desired performance.

circuit breakers

switches

and stopping of a motor and

load. In this chapter

current motors.

Disconnecting switches

main purpose and application. Fuses are not

in-

cluded here because they are protective devices rather than control devices.

They

are discussed in

of operation remains the same. Using only a dozen

Chapter 26. The symbols for these and other de-

basic components,

vices are given in Table 20A.

it

is

possible to design control

439


440

COMPONENTS FOR CONTROL CIRCUITS

BASIC

Disconnecting switches

A disconnecting switch power

source.

isolates the

motor from

the

consists of 3 knife-switches and 3 line

It

The knife-switches can means of an An interlocking mechanism prevents

fuses enclosed in a metallic box.

be opened and closed simultaneously by external handle. the

hinged cover from opening when the switch

is

closed. Disconnecting switches (and their fuses) are selected to carry the nominal full-load current of the tor,

and

mo-

to withstand short-circuit currents for brief in-

tervals.

Figure 20.1 Three-phase, fused disconnecting switch rated

600

30

V,

A.

(Courtesy of Square D)

Manual

circuit

breakers

A manual circuit breaker opens and a toggle switch.

It

trips

closes a circuit, like

(opens) automatically

when

the

current exceeds a predetermined limit. After tripping,

it

can be reset manually. Manual circuit breakers are often used instead of disconnecting switches because no fuses

have

to

be replaced.

Figure 20.2 Three-phase circuit breaker, 600 (Courtesy of Square D)

V,

100 A.

Cam switches A cam

switch has a group of fixed contacts and an equal

number of moveable to

open and close

dle or knob. tion

Cam

contacts.

in a preset

The contacts can be made

sequence by rotating a han-

switches are used to control the

mo-

and position of hoists, callenders, machine tools,

and so on.

Figure 20.3 Three-phase surface-mounted

2kW. (Courtesy of Klockner-Moefler)

cam

switch,

230

V,

BASICS

OF INDUSTRIAL MOTOR CONTROL

441

Pushbuttons

A pushbutton Two

or

a switch activated by finger pressure.

is

more contacts open or close when

the button

is

depressed. Pushbuttons are usually spring loaded so as to return to their

normal position when pressure

is

re-

moved.

Figure 20.4 Mechanical-interlocked pushbuttons with

NO

(nor-

and NC (normally closed) contacts; rated to interrupt an ac current of 6 A one million times. {Courtesy of Siemens) mally open)

Control relays

A control

relay

and closes a

set

gized.

The

which

attracts a

is

an electromagnetic switch that opens

of contacts

when

the relay coil

is

ener-

relay coil produces a strong magnetic field

movable armature bearing

the contacts.

Control relays are mainly used in low-power circuits.

They include time-delay

relays

whose contacts open

or

close after a definite time interval. Thus, a time-delay

closing relay actuates

been energized. relay actuates

On

its

its

contacts after the relay coil has

the other hand, a time-delay opening

contacts

some time

after the relay coil

has been de-energized.

Figure 20.5 Single-phase relays: 25 A,

1

15/230

V and 5

A,

1

15

V.

{Courtesy of Potter and Brumfield)

Thermal relays A thermal relay (or

overload relay)

is

a temperature-

whose contacts open or close when the exceeds a preset limit. The current flows

sensitive device

motor current

through a small, calibrated heating element which raises the temperature of the relay.

Thermal relays are inherent

time-delay devices because the temperature cannot fol-

low the instantaneous changes

in current.

Figure 20.6 Three-phase thermal relay with variable current ting, 6 A to 10 A. (Courtesy of Klockner-Moeller)

set-

(continued)

BASIC

COMPONENTS FOR CONTROL CIRCUITS

Magnetic contactors

A magnetic contactor is

basically a large control relay

designed to open and close a power relay coil and a magnetic plunger,

movable

When

contacts.

tracts the

the relay coil

magnetic plunger, causing

against the force of gravity. in contact

circuit.

which

it

is

It

possesses a

carries a set of

energized,

it

at-

to rise quickly

The movable contacts come

with a set of fixed contacts, thereby closing

power circuit. In addition to the power contacts, one more normally open or normally closed auxiliary con-

the

or

When

tacts are usually available, for control purposes.

the relay coil

de-energized, the plunger

is

falls,

thereby

opening and closing the respective contacts. Magnetic contactors are used to control motors ranging from 0.5

hp

to several

hundred horsepower. The

size,

dimensions,

and performance of contactors are standardized.

Figure 20.7 Three-phase magnetic contactor rated 50 hp, 575 V, 60 Hz. Width: 158 mm; height: 155 mm; depth: 107

mm;

weight: 3.5 kg.


Pilot lights

A pilot light indicates the on/off state of a remote component

in a control

Figure 20.8 Pilot light, 120

3

V,

system.

W mounted

in

a start-stop push-

button station.


Limit switches and special switches

A limit switch is

a

low-power snap-action device

that

opens or closes a contact, depending upon the position of a mechanical part. Other limit switches are sensitive to pressure, temperature, liquid level, direction tion,

of rota-

and so on.

Figure 20.9a Limit switch with

one

NC

contact; rated for ten million

operations; position accuracy: 0.5


Figure 20.9b Liquid level switch.


mm.

BASICS


443

Proximity detectors Proximity detectors are sealed devices that can detect objects without

Their service tions.

They

coming

in direct

contact with them.

independent of the number of opera-

life is

are wired to an external dc source

erate an alternating nal oscillator.

magnetic

When

field

and gen-

by means of an

inter-

comes within

few

a metal object

a

millimeters of the detector, the magnetic field decreases,

which

in turn

causes a dc control current to flow. This

current can be used to activate another control device,

such as a relay or a programmable logic controller. Capacitive proximity detectors, based on a similar principle but generating an ac electric field, are able to detect

nonmetallic objects, including liquids.

Figure 20.10 Proximity detector to monitor the loading of a conveyor

belt.

(Courtesy of Telemecanique, Groupe Schneider)

In order to understand the sections that follow,

20A

the legends in Table

proceeding

should be read before

it

excited. This

is

places a heavier than expected duty on auxiliary

further.

contacts that energize the coil.

open and normally

20.2 Normally

Example 20-1

A 3-phase NEMA size 5 magnetic contactor rated at

closed contacts show components

Control circuit diagrams always in

moment

the relay coil at the

a state of rest, that

is,

when

they are not energized

270 A, 460 V possesses a 120 V, 60 Hz relay coil. The coil absorbs an apparent power of 2970 VA and 212 VA, respectively, in the open and closed con-

(electrically) or activated (mechanically). In this

tactor position. Calculate the following: state,

some

electrical contacts are

They

open while others

are respectively called normally

a.

open contacts (NO) and normally closed contacts

b.

(NC) and

c.

are closed.

are designated by the following symbols:

normally open contact (NO)

~~

|~~

The inrush exciting current The normal, sealed exciting current The control power needed to actuate the relay coil compared to the power handled by the contactor

|

normally closed contact (NC)

Solution

~^f~

a.

The inrush

20.3 Relay coil exciting current

When

a magnetic contactor

open position, the magnetic air

gap,

compared

Consequently,

when

to

in the

is

in its

the contactor

the contactor

Because the

is

is

coil is excited

the magnetizing current

is

it

is

closed.

by a fixed ac voltage,

much

higher

in the

is

=

2970/120

=

relay coil current

24.75

A

when

the contac-

tor is sealed (closed) is

/= SIE= c.

The

drawn by

212/120

ent

=

1.77

steady-state apparent control

to actuate the relay coil

power S

= =

is

2 2 1

A

power needed

VA. The appar-

that the contactor can handle

open

than in the closed contactor position. In other

words, a considerable inrush current

5/£

The normal

in-

much lower

open than when

=

is

closed.

case of an ac contactor the

is

b.

circuit has a very long

ductive reactance of the relay coil

when

/

de-energized or

current in the relay coil

EI

VT

215 120

460

VA

X

270

VT

is


444

GRAPHIC SYMBOLS FOR ELECTRICAL DIAGRAMS

TABLE 20A

1

o

2

—

33

32 15

or

or

[]

[j]

16

or

r^H

or J]

4 5

—

17

0

18

—

19

nnnn

or

\

or

h~

or

^ws^

APPLICATION 6

*

i

i 8

i

—/ — 1

r -i

T

22 10

r

*

o

• )

rwj^

23

t

identified

o

by an appropriate

39

- _Q-

letter

<

or

12

40

13

14 41

< CD

connection 2. conductors crossing 3. conductors connected 4. three conductors 5. plug; recepseparable connector 7. ground connection; arrester 8. disconnecting switch 9. normally open con11. pushbutton NO; NC 12. circuit-breaker 13. single-pole switch; tact (NO) 10. normally closed contact (NC) three-way switch 14. double pole double throw switch 15. fuse 16. thermal overload element 17. relay coil 18. resistor 19. winding, inductor or reactor 20. capacitor; electrolytic capacitor 21. transformer 22. current 1. terminal;

tacle

6.

26. shunt winding 23. potential transformer 24. dc source (general) 25. cell commutating pole or compensating winding 28. motor; generator (general symbols) 29. dc motor; dc generator (general symbols) 30. ac motor; ac generator (general symbols) 32. 3-phase squirrel-cage induction motor; 3-phase wound-rotor motor 33. synchronous motor; 3-phase alternator 34. diode 35. thyristor or SCR 36. 3-pole circuit breaker with magnetic overload device, drawout type 37. dc shunt motor with commutating winding; permanent magnet dc generator 39. NPN 38. magnetic relay with one NO and one NC contact.

transformer; bushing type 27. series winding;

transistor

40.

For a complete

Symbols

PNP transistor list

of

for Electrical

and

and American National Standard Graphic

Electronics Diagrams" (ANSI Y32.2/IEEE No. 315) published by the Institute of Electrical and

Electronics Engineers, Inc., countries.

41. pilot light

graphic symbols and references see "IEEE Standard

New York, NY

10017. Essentially the

same symbols

are used

in

Canada and

several other

BASICS

Thus, the small control power (212 control a load

whose power

is

A block diagram

VA) can

215 120/212


=

gles,

1015 times greater.

with a brief description of

20.4 Control diagrams

rection of

power

control system can be represented by four types

•

wiring diagram

of rectan-

function.

The

rectan-

that indicate the di-

or signal flow (Fig. 20. 11). is

similar to a block diagram,

bols rather than by rectangles.

The symbols give us

an idea of the nature of the components; consequently, one-line diagrams yield

block diagram one-line diagram*

set

except that the components are shown by their sym-

of circuit diagrams. They are listed as follows, in or-

•

its

by arrows

A one-line diagram

der of increasing detail and completeness: •

composed of a

each representing a control device, together

gles are connected

A

is

445

A list of typical The

lines

symbols

is

more information.

displayed in Table 20A.

connecting the various components repre-

•

schematic diagram

two or more conductors (Fig. 20. 2). A wiring diagram shows the connections between the components, taking into account the

•

Also called single-line diagram.

physical location of the terminals and even the color

sent

1

pilot light

(motor running)

600 V 3-phases

fused disconnecting switch

thermal overload

magnetic contactor

motor

relay

pushbutton station

start-stop

Figure 20.11 Block diagram

of

a combination

starter.

600 V 3-phase

c

cN> disconnecting switch

f

use

NO

contact start

pushbutton

/PB1

(

A

)

relay coil

PB2

|

o

o

pilot light

stop

A auxiliary

contact

Figure 20.12 One-line diagram of a combination starter.

pushbutton


446

600 V 3-phase

Figure 20.13 Wiring diagram of a combination starter.

s

1

L1

os

L2

2

600 V 3-phase

s

3

T

T2

T

J3

4T11-0

oN>-

*

L3

Ostart

stop 7.1

-O

1

_L

6.2

O-

%o-L-T Figure 20.14 Schematic diagram

of

a combination

starter.

of wire. These diagrams are employed stalling

when

equipment or when troubleshooting a

incir-

circuit.

The symbols used

components

to designate the various

are given in Table

20B.

cuit (Fig. 20.13).

A

schematic diagram shows

the electrical

all

connections between components, without regard

arrangement.

to their physical location or terminal

This type of diagram

is

when troumode of oper-

indispensable

bleshooting a circuit or analyzing

its

ation (Fig. 20. 14). In the sections that follow, this the kind of

diagram we

will

The reader should note

is

be using.

methods

Three-phase squirrel-cage motors are started

by connecting them directly across the applying reduced voltage to the

method depends upon

stator.

either

line or

The

starting

power capacity of

the

by

the

supply line and the type of load.

that the four

Figs. 20.11 to 20.14 all relate to the

20.5 Starting

diagrams

in

same control

Across-the-line starting sive.

is

The main disadvantage

is

simple and inexpenthe high starting cur-

BASICS

which

rent, It

is

5 to 6 times the rated full-load current.

can produce a significant

line voltage drop,

which

may

affect other

line.

Voltage-sensitive devices such as incandescent

customers connected to the same

lamps, television tools

sets,

Mechanical shock not be overlooked.

aged

and high-precision machine

respond badly to such voltage dips. is

another problem that should

Equipment can be seriously dam-

if full- voltage starting

torque.

Conveyor

sudden starting

produces a hammerblow

belts are another

may

example where

times tolerate across-the-line starting even for

circuit

up

to 10

A motor control circuit contains

two basic components: a disconnecting switch and a starter. The disconnecting switch is always placed between the supply line and the starter. The switch and starter sometimes mounted

make

a combination starter.

Figure 20.15 Manual starters

for

same enclosure to The fuses in the dis-

in the

connecting switch are rated

at

about 3.5 times

single-phase motors rated

waterproof enclosure.


to protect the

1

motor and supply

line

motor or starter or a failure to start Under normal start-up conditions, the fuses do

circuit in the

have time

to blow,

even though the

initial

6 to 7 times full-load current. The fuse

up.

not

current

is

rating, in

amperes, must comply with the requirements of the National Electric Code.

some cases

the disconnecting switch and

its

fuses are replaced by a manual circuit breaker.

20.6 Manual across-the-line starters Manual 3-phase

ing current during the acceleration period.

are

is

against catastrophic currents resulting from a short-

mo-

000 hp. Obviously, the fuses and breakers must be designed to carry the start-

tors rated

motor against sustained overloads. Their primary function

In

we can some-

447

load current; consequently, they do not protect the

not be acceptable.

In large industrial installations


full-

starters are

composed of

a circuit

breaker and either two or three thermal relays,

mounted

in

all

an appropriate enclosure. Such starters

are used for small motors

trip the circuit

V

(

10 hp or less) at voltages

600 V. The thermal breaker whenever the current

ranging from 120

to

relays in

one

of the phases exceeds the rated value for a significant length of time.

hp (0.75 kW);

left:

surface mounted; center: flush mounted;

right:

ELECTRICA LAND ELECTRONIC DRIVES

448

Single-phase manual starters (Fig. 20. 5) are built

2.

1

along the same principles but they contain only one

The thermal

thermal relay. particular

motor

that

is

*

starters across-the-line

They

The

the motor against

relay comprises

The thermal

in series

relay

is

with the three phases.

often designated by the letters

A

OL

(overload).

whenever a motor has location.

T protects

three individual heating elements, respectively

connected

starter.

20.7 Magnetic across-the-line

Magnetic

relay

sustained overloads.*

relays are selected for the

connected to the

The thermal

are also used

kW. 20.16 shows

are

starters

employed

from a remote

to be controlled

whenever the power rat-

ing exceeds 10 Fig. its

a typical magnetic starter and

associated schematic diagram.

ing switch three

is

The disconnectThe starter has

external to the starter.

main components: a magnetic contactor, a

thermal relay, and a control station. scribe these 1

.

The magnetic contactor A possesses

A and one

contacts

We now

de-

components. three heavy

small auxiliary contact

A

x.

As

can be seen, these contacts are normally open. Contacts

A must be

big

enough

to carry the start-

ing current and the nominal full-load current

without overheating. Contact

because

The

it

relay coil

is

x

x

is

much

represented by the

bol (A) as the contacts

A

A

smaller

only carries the current of relay coil A.

it

controls. Contacts

remain closed as long as the coil

is

Figure 20.16a Three-phase across-the-line magnetic 600 V, 60 Hz. {Courtesy of Klockner-Moeller)

same sym-

A and

energized.

contactor

disconnecting switch 1

4

thermal overload relay

A

r

Iti

A

A

iT3

s t

pilot light

a?rl

control station

-reset

stop

Figure 20.16b Schematic diagram of a 3-phase across-the-line magnetic

starter.

pushbutton

starter,

30

hp,

BASICS


449

multiple of the current setting

Figure 20.17 Typical curve of a thermal overload relay,

ping time versus line current.

The

sured from cold-start conditions.

showing

tripping time

If

is

trip-

mea-

the motor has been

operating at full-load for one hour or more, the tripping time

is

small, normally closed contact

relay gets too hot and stays is

manually

The

T forms

part of

opens when the thermal

It

open

until the relay

the start button. Coil

current rating of the thermal relay

opens

is

after a period of

ter

At rated current (multiple

1),

s.

The thermal

it

relay

close contact

T

following an overload.

3.

The

control station,

buttons,

may be

away from

It

re-

when

The

is

its

on.

nor-

remains excited be-

now closed.

Contact A x

T

pro-

that a thermal relay will

the ambient temperature around the starter

We

is

can remedy the situation by changing

the location of the starter or

by replacing the relay by

another one having a higher current rating. Care

must be taken before making such a change, because if

the starter.

returns to

no apparent reason. This condition can occur

too high.

It is

down.

to,

coil

is

it

effect.

sometimes happens

trip for

is

composed of start-stop push-

located either close

released

sustained overload, the opening of contact

preferable to wait a few minutes before pushing the button to allow the relay to cool

is

said to be a self-sealing contact.

duces the same

trips af-

equipped with a reset button enabling us to

full line volt-

To stop the motor, we simply push the stop butwhich opens the circuit to the coil. In case of a

the relay

but at twice rated current,

an interval of 40

the

in

ton,

overload current. Thus, Fig. 20.17 shows the

trips,

pushbutton

cause auxiliary contact A x

tripping time as a multiple of the rated relay

never

set

age appears across the motor and the pilot light

time that depends upon the magnitude of the

current.

is

immediately energized

is

mal position, but the relay

is

chosen to protect the motor against sustained

T

A

causing contacts A and A x to close. The

When the

reset.

overloads. Contact

connecting switch; the pushbutton station transparent polycarbonate cover.

(Courtesy of Klockner-Moeller)

reduced about 30 percent.

the relay assembly.

Figure 20.18 Three-phase across-the-line combination starter, 150 hp, 575 V, 60 Hz. The protruding knob controls the dis-

the ambient temperature around the

or far

too high, the occasional tripping

may

motor

is

also

actually serve

pilot light is optional.

as a warning.

Referring to Fig. 20.16b, to first

start the

motor

we

close the disconnecting switch and then depress

Fig. 20.18 Fig.

20.19

shows a shows

typical combination starter.

another

combination

starter


450

Figure 20.19 Three-phase across-the-line combination starter rated 100 hp, 575 V, 60 Hz. The isolating circuit breaker is controlled by an external handle. The magnetic contactor is mounted in the bottom left-hand corner of the waterproof enclosure. The small 600 V/120 V transformer

in

the lower right-hand corner supplies low-

voltage power for the control

circuit.


equipped with a small step-down transformer to excite the control circuit.

Such transformers

are always

used on high-voltage starters (above 600 V) because they permit the use of standard control components,

Figure 20.20 Three-phase 5 kV starter for a 2500 hp cage motor. The medium- and low-voltage circuits are completely isolated from each other to ensure safety. The compartment is 2286 mm high, 610 mm wide, and 813 mm deep. The entire starter weighs 499 kg. (Courtesy of Square D, Groupe Schneider)

such as pushbuttons and pilot lights while reducing the shock hazard to operating personnel.

fuses.

shows a medium-voltage across-theline starter for a 2500 hp, 41 60 V, 3-phase, 60 Hz squirrel-cage motor. The metal compartment Fig. 20.20

houses three fuses and a 3-phase vacuum contactor. The contactor can perform 250 000 operations at full -load before maintenance is required. The 20 V 1

holding coil draws 21.7

A

current drops to 0.4

A

during pull-in, and the

during normal operation.

Closing and opening times of the main contactor are respectively 65 Fig. 20.21

shows a

special combination starter that

can be reset remotely following a short-circuit. distinguishing feature requires

no

fuses.

The

is

that

it

is

programmable and

sophisticated contactor

signed to interrupt short-circuit currents 3 ms,

which

is

comparable to

Its

is

in less

that offered

by

de-

than

HRC

acts also as a disconnecting

switch and consequently the overall smaller than

size

more conventional combination

is

much

starters.

20.8 Inching and jogging In

some mechanical systems, we have

to adjust the

position of a motorized part very precisely. To ac-

complish so that

A

ms and 130 ms.

The contactor

it

this,

we

energize the motor in short spurts

barely starts

before

it

again comes to a

double-contact pushbutton

J is

shown

in

start/stop

as

circuit,

added Fig.

halt.

to the usual

20.22. This

arrangement permits conventional start-stop control as

well as jogging, or Inching.

description If the

shows how

jog button

pressed) relay coil

The following

the control circuit operates.

J is in its

normal position (not de-

A is excited as soon as the start but-

BASICS


stop

st a j

80

n

I

o-

-o

o-

A

451

— O

/

start

Qstop

Off

t

4h

L3 0-

Figure 20.22 Control circuit and pushbutton station for start-stop job operation. Terminals 8, L3 correspond to terminals

L3

in Fig.

8,

20.13.

estimated that each impulse corresponds to 30

It is

normal

start-stop operations. Thus, a contactor that

can normally

start

and stop a motor 3 million times,

can only jog the motor

00 000 times, because the

1

contacts have to be replaced. * Furthermore, jog-

ging should not be repeated too quickly, because the intense heat of the breaking arc

may

cause the main

contacts to weld together. Repeated jogging will also overheat the motor.

Figure 20.21 Special self-protected starter rated at 40 hp, 460

42 kA

addition to a short-circuit capability of

Hz.

In

460

V,

V,

features adjustable thermal and magnetic

it

settings. Overall dimensions:

mm

wide, 179

243

mm

high,

90

60 at

the contactor

and so the motor button

ter the start

mm

20.9 Reversing the direction of rotation

is

Ax

in the

main con-

will continue to turn af-

released. Thus, the control cir-

same way as in Fig. 20.16b. Suppose now that the motor is stopped and we

cuit operates in the

depress the jog button. This closes contacts

A

relay coil tacts

1 ,

effect.

excited. Contact

now open and

The motor

jog button leased, coil tor

is

2 are

A will

contacts

is

4 and

closes, but con-

the closure of

A x has no

up speed so long

depressed. However,

when

it

as the is

re-

A will become de-energized and contacA x to open. Thus, when

2 are again bridged, the motor will

,

to a halt. Thus,

button

will pick

Ax

3,

drop out, causing

1

we

come

by momentarily depressing the jog

can briefly apply power to the motor.

Jogging imposes severe duty on the main power contacts

A

NEMA

normal duty.

trip

We tactor closes

required,

is

deep.

depressed. Sealing contact

is

When jogging

usually selected to be one

size larger than that for

(Courtesy of Telemecanique, Groupe Schneider)

ton

is

because they continually make and

break currents that are 6 times greater than normal.

can reverse the direction of rotation of a 3 -phase

motor by interchanging any two

manual 3-position cam switch

When

contactor

connected contactor

A is closed,

to terminals

B

is

A, B,

closed, the

as

A and B and a

shown

lines

This can be

lines.

done by using two magnetic contactors

in Fig. 20.23.

LI L2, and L3 are ,

C of the motor.

same

lines are

But when

connected

to

motor terminals C, B, A. In the forward direction, the

contact

1,

contactor

A

cam

which energizes relay

switch engages

coil

A, causing

A to close.**

magnetic contact has an estimated mechanical

life

of

about 20 million open/close cycles, but the electrical contacts

should be replaced after 3 million normal cycles.

The contacts and relay

coils

may

propriate letters. Thus, the letters

be designated by any ap-

F and R

are often used to

designate forward and reverse operating components. In this

book we have adopted

the letters

A and B mainly

reasons of continuity from one circuit to the next.

for


452

A

LI

1

o2

A

L2

disconnecting

O-

switch

3

A

L3

O

B

B

cam U

switch

i

^

(AJ

forward 1

oO

q_Lq emergency pushbutton

Figure 20.23a Simplified schematic diagram of a reversible magnetic starter.

Figure 20.23c

Figure 20.23b Three-position

cam

switch

in Fig.

Emergency stop pushbutton

20.23a.

To reverse to position 2.

the rotation,

However,

in

we move the cam switch so, we have to move

doing

past the off position (0). Consequently,

ble

to

energize

coils

A

and

B

it

is

impossi-

de-energized. This

is

because when the other contactor closes, a shortcircuit

results

across the line.

current could easily be

50

to

The

short-circuit


normal, and both contactors could be severely dam-

may

To eliminate this danger, the contactors are mounted side by side and mechanically interlocked,

prevent a contactor from dropping out, even after is

20.23a.

simultaneously.

Occasionally, however, a mechanical defect

relay coil

in Fig.



its

a serious situation,

aged.

so as to

make

it

physically impossible for both to be

BASICS


453

L1

O-

L2

disconnecting

switch

L3

0-

B

-HIB

-

-j

A

B

start

stop

®—

"Hf0-

o

-oi

8

zero speed

switch

-± - o1

o

-

•

iB xl

I

R

F

T

Figure 20.24a Simplified schematic diagram of a starter with plugging control.

closed at the same time. The interlock steel bar,

is

a simple

2.

pivoted at the center, whose extremities are

tied to the

movable armature of each

NO contact

The

start

one

NC contact 3, 4 which operate together. Thus,

pushbutton has one

contact 3, 4 opens before contact

contactor.

During an emergency, pushbutton U, equipped

3.

Contactor

B

is

l

,

l ,

2 and

2 closes.

used to stop the motor.

It is

with a large red bull's-eye, can be used to stop the

identical to contactor A, having 2 auxiliary

motor

contacts

(Fig. 20.23c). In practice, operators find

easier to hit a large button than to turn a

cam

it

switch

to the exposition.

B xl and B x2

4.

The

stop pushbutton

pushbutton. Thus, 7,

20.10 Plugging 5.

We have already

seen that an induction motor can be

soon as the machine has come

cuit of Fig.

to rest.

The

The

cir-

.

Contactor

A is

tion to

3

iliary

its

start the

main contacts A,

contacts

Ax

,

Contacts

and

A x2

.

it

it

6 closes. is

normally

closes as soon as the motor turns in

A xl

Bx

and

Contacts

A x2

,

A and B

and

B x2

motor. In addi-

prevent the relay coils

has 2 small aux-

cited at the

same

running, contact

have only

start or

to

be

stop the motor.

are electrical interlocks to

A and B

time. Thus,

A x2

of coil B.

are sealing contacts so

pressed momentarily to

circuit operates as follows:

used to

5,

Contact F-C of the zero-speed switch

that pushbuttons

7. l

depressed contact

it is

circuit for the eventual operation 6.

20.24a shows the basic elements of such

a plugging circuit.

when

the forward direction. This prepares the plugging

(Section 14.8). However, to prevent the motor from

line as

identical to the start

is

8 opens before contact

open, but

brought to a rapid stop by reversing two of the lines

running in reverse, a zero-speed switch must open the

addition to the 3 main

in

contacts B.

is

from being ex-

when

the

motor

open. Consequently,

is

454


B cannot become excited by deB until such time as con-

relay coil

drag cup

pressing pushbutton

A has

tactor

dropped

out, causing contact

A x2

to reclose.

Several types of zero-speed switches are on the

market and Fig. 20.24b shows one

on

that operates

the principle of an induction motor.

It

consists of a

small permanent magnet rotor N, S and a bronze ring or cup supported on bearings,

pivot between stationary contacts

manent magnet motor.

is

As soon

coupled

as the

which

F and

to the shaft

in the

position.

_M

is

same

When

motor stops turning, the brass ring returns bearing

per-

of the main

direction, thereby closing contacts F-C.

bearing

free to

The

motor turns clockwise, the

permanent magnet drags the ring along drag cup

is

R.

the

to the off

Because of its function and shape, the ring

often called a drag-cup. Fig. 20.24c

that operates

shows another zero-speed switch

on the principle of centrifugal

force.

20.11 Reduced-voltage starting contact c

Figure 20.24b Typical zero-speed switch for use

Some

industrial

gradually. in Fig.

20.24a.

presses,

loads have to be started very

Examples are coil winders, printing conveyor belts, and machines that

process fragile products. In other industrial applications, a

motor cannot be directly connected

the line because the starting current all

these cases

we have

to

is

to

too high. In

reduce the voltage ap-

plied to the motor either by connecting resistors (or reactors) in series with the line or

by employ-

ing an autotransformer. In reducing the voltage,

we 1

.

recall the following:

The locked-rotor

current

is

proportional to the

voltage: reducing the voltage

the current 2.

by

by half reduces

half.


is

proportional to the

square of the voltage: reducing the voltage by half reduces the torque by a factor of four.

20.12 Primary resistance starting Primary resistance starting consists of placing three 4

Figure 20.24c Zero-speed switch, centrifugal type. (Courtesy of Hubbel)

resistors in series with the

motor during the

period (Fig. 20.25a). Contactor

A

closes

start-up

first

and


BASICS

455

L2

disconnecting switch

L3

r.

Figure 20.25a diagram

Simplified schematic

L1

stop

of the

power section

of

a reduced-voltage primary resistor

stator.

RA

L3

st art

L1

|

RT

RT

RA

0RT

L3

stop

RA

^-v

®_

^ ®-

RT

iÔ

Figure 20.25b

Figure 20.25c

Control circuit of Fig. 20.25a.

Control circuit of Fig. 20.25a using an auxiliary relay RA.

when

the

motor has nearly reached synchronous

speed, a second contact

B

short-circuits the resistors.

This method gives a very smooth

start

with complete

absence of mechanical shock. The voltage drop across the resistors

is

high

at first,

but gradually di-

minishes as the motor picks up speed and the current

Consequently, the voltage across the motor

falls.

ter-

minals increases with speed, and so the electrical and

mechanical shock

is

negligible

when

full

finally applied (closure of contactor B).

are short-circuited after a delay that

voltage

The

is

resistors

depends upon

the setting of a time-delay relay.

The schematic control diagram

(Fig.

20.25b)

re-

veals the following circuit elements:

As soon

as the start pushbutton

is

depressed, relay

A and RT are excited. This causes the contacts A and A x to close immediately. However, the contact RT coils

only closes after a certain time delay and so the relay coil of contactor If the

B

is

only excited a few seconds

large, the inrush exciting currents start

pushbutton contacts

shown

later.

magnetic contactors A, B are particularly

in Fig.

if

could damage the

they are connected as

20.25b. In such cases,

it

is

better to

add an auxiliary relay having more robust contacts. Thus,

in Fig.

RA is

to carry the exciting currents

and B. Note

20.25c, the purpose of auxiliary relay

that the start

of relay coils A

pushbutton contacts carry

only the exciting current of relay coils

RA and

RT.

Other circuit components are straightforward, and A,

B

magnetic contactor relay coils

the reader should have

A

A,

auxiliary contact associated with

RT:

time-delay relay that closes the circuit of coil

B

after a preset interval of

time

no

difficulty in analyzing the

operation of the circuit.

How when

are

the

starting

characteristics

affected

resistors are inserted in series with the stator?


456

r V

T

-v-

-2?

1200 800 ^ speed

400

¥ \ 16001800 r/min

800 1200 ^ speed

400

16001800 r/min

Figure 20.26b

Figure 20.26a Typical torque-speed curves of a

3-phase squirrel-cage

Typical current-speed curves of

a 3-phase squirrel-cage

induction motor: (1) full-voltage starting; (2) primary re-

induction motor: (1) full-voltage starting; (2) primary re-

sistance starting with voltage reduced to 0.65 pu.

sistance starting with voltage reduced to 0.65 pu.

Fig.

20.26a shows the torque-speed curve

full

voltage

r/min

is

1

when

applied to a typical 3-phase, 1800

induction

motor.

shows what happens when series with the line.

The

c.

Corresponding curve 2 resistors are inserted in

resistors are

chosen so

The apparent power drawn from

the line, with

the resistors in the circuit d.

The locked-rotor torque developed by

the motor

that

Solution the locked-rotor voltage across the stator

is

0.65 pu. 2

The locked-rotor torque is, therefore, (0.65 ) = 0.42 pu or only 42 percent of full-load torque. This means that the motor must be started at light load. Fig.

full

voltage

is

applied to the

when

stator.

the current

When

the speed reaches about 1700 r/min, the resis-

tors are short-circuited.

pu

.8

full

voltage the locked-rotor apparent

S

to 2.5 pu,

which

the resistors are in the circuit.

b.

The voltage across

E=

The current jumps from about is

The

a very moderate jump.

power

= V3EI

current

0.65

the

1400

motor

at

0.65 pu

is

X 460 - 299 V

drawn by

the

motor decreases

proportion to the voltage:

Example 20-2

A

1

50

/

kW (200 hp), 460 V, 3-phase 3520 r/min, 60

Sm =


c.

the

motor un-

der full-voltage, locked-rotor conditions b.


when

the

the resistors are in the circuit

motor

X

1400

= 910 A motor

is

the line

is

the

V3£/

= V 3 X 299 X 910 = 471 kVA

0.65 pu.

Calculate

0.65


Hz induction motor has a locked-rotor torque of 600 N*m and a locked-rotor current of 1400 A. Three resistors are connected in series with the line so as to reduce the voltage across the motor to

a.

=

The apparent power drawn from SL =

is

(8.9)

- V 3 X 460 X = 1114kVA

Curve 2

shows

1

At

20.26b shows the current versus speed curve

when

1

a.

V3E1

= V3 X 460 X 910 = 724 kVA

in

BASICS OF INDUSTRIAL

SL

kVA

= 724

Sm

kVA

= 471

Jl1

rotor

series

460

resistors

V

CO-

///

910 A

Ql

P m = 165 kW

441 kvar

= 441 kvar

Qm

series

299 V

m

A

910

locked rotor

resistors

460 V

910 A

910

A

Figure 20.28

Thus, percentagewise, the apparent power only 724 kVA/1 1 14

kVA = 65%

See Example

20-3.

The

can only absorb active power

is

of the appar-

power under full-voltage conditions, The torque varies as the square of the voltage:

ent

cuit.

T=

0.65

=

0.42

2

resistors

in the cir-

Consequently, the reactive power supplied by

the line must be equal to that absorbed by the motor:

X 600 X 600

QL =

= 252 N m (—186 The

=

Hi


d.

kW

574

=

299 V

457

S m = 471 kVA

724 kVA

S locked

MOTOR CONTROL

ft-lbf)

results of these calculations are

The

summarized

kvar

1

power supplied by the

active

in

44

P L = ^Sl-Ql = V724

2

line

-44l

is 2

Fig. 20.27.

- 574 kW Example 20-3 In Example 20-2,

The if

of the motor alone the series resistors

the locked-rotor is

power

power absorbed by

active

=

dissipate.

-

574

165

- 409 kW

We will

solve this problem by considering active and

reactive

powers and using the power

triangle

The apparent power drawn by the motor

at

The

power per

active

method.

P = p R /3 = The current

=

471

kVA

resistor

is

= l36kW

409/3

reduced

is

Sm

in

each resistor /

(from Example 20-2)

The value of each The corresponding apparent power drawn by line

is

0.35, calculate the value of

and the power they

Solution

voltage

the three resistors

factor

is

= 910 A

resistor

(from Example 20-2)

is

the

P =

is

2

I

R

= 9\0 2 R R = 0.164 n

136 000

5 L = 724 kVA (from Example 20-2) The

active

power drawn by

P m = 5 m cos6 = = 165 kW

the

motor

471

X

is

The

reactive

power absorbed by

G m = Vs>-/* = 441 kvar

0.

1

64

The physical than

The

three resistors

tance of

0.35

the

motor

2 2 V471 -165

is

if

fl

must therefore each have a

resis-

and a short-term rating of 36 kW. 1

size of these resistors

is

much

smaller

they were designed for continuous duty.

This

is

an interesting example of the usefulness

of the power triangle method difficult

in

solving a relatively

problem. The results are summarized

Fig. 20.28.

in


458

20.13 Autotransformer starting Compared

to a resistance starter, the

an autotransformer starter it

draws a much lower

tage

is

is

The disadvan-

line current.

more, and the

that autotransformers cost

transition

advantage of

that for a given torque

from reduced-voltage

to full-voltage

is

not quite as smooth.

Autotransformers usually have taps to give output voltages of 0.8, 0.65, and 0.5 pu.

The

corre-

mi

sponding starting torques are respectively 0.64, 0.42, and 0.25 of the full-voltage starting torque.

-»

H

Furthermore, the starting currents on the line side are also reduced to 0.64, 0.42, and 0.25 of the full-

Fig.

cuit It

in

contacts

contactor riod

open

diagram of such a

has two contactors

NO

PJ

20.29 shows a starter using two autotrans-

formers connected

when

A

is in

the

delta.

starter is

A and

A

NO

in Fig. 20.30.

A has five A x This

contact

.

Figure 20.29 Reduced-voltage autotransformer

operation only during the brief pe-

motor

is

starting up.

575

V,

60 Hz.


65%

65% start

stop J

'

clLo—f-O

O-

RT2 RT1

Mfo RT3

Figure 20.30 Simplified schematic diagram

of

jP

simplified cir-

given

B. Contactor

and one small

— ^ *

voltage locked-rotor current.

an autotransformer

starter.

(*>r interlock

—

starter,

100

hp,

BASICS

Contactor B has 3 while the motor

NO contacts

B.

The autotransformers are set on the 65 percent The time-delay relay RT possesses three contacts RT1, RT2, RT3. The contact RT1 in parallel with the start button closes as soon as coil

The

other two contacts RT2,

A and B

RT is en-

RT3

operate

that the locked-rotor torques are identical, but the

locked-rotor line current

as

soon as the

start

button

is

higher torque because the terminal voltage

is

higher than the 65 percent value that existed

at the

ment of start-up.

duced voltage appears across the motor terminals.

all

A few seconds later, contact RT2 in series with coil A opens, causing contactor A to open. At the same RT3

A

causes contactor B to close.

drops out, followed almost im-

mediately by the closure of contactor B. This action applies full voltage to the

In transferring

motor

is

smaller

when using an

autotransformer.

Because the autotransformers operate for very short periods, they can be

wound with much

smaller

wire than continuously rated devices. This enables us to drastically reduce the size, weight, and cost of

these components.

from contactor

A to

Example 20-4

line.

contactor B,

A

200 hp (150 kW), 460

V, 3-phase,

3520

r/min,

60 Hz induction motor has a locked-rotor torque of

problem because

600 N*m and a locked-rotor current of 1400 A (same motor as in Example 20-2). Two autotransformers,

when contactor B drawn from the For

is

mo-

the other hand, the line current at

disconnected from the line for a frac-

tion of a second. This creates a

the contacts

speeds

On

slightly

motor and simultaneously

disconnects the autotransformer from the

the

au-

However, when the motor reaches about 90 percent

depressed. This excites the autotransformer and re-

Thus, contactor

much lower using an

of synchronous speed, resistance starting produces a

A closes

time, contact

is

totransformer (2.7 versus 4.2 pu).

RT relay setting.

prevent them from closing simultaneously.

Contactor

and

the torque

are mechanically interlocked to

depends upon the

after a delay that

Contactors

and 20.31b compare

when autotransformer starting (3) and resistance starting (2) is used. The locked-rotor voltage in each case is 0.65 pu. The reader will note line current

tap.

ergized.

459

Figs. 20.31a

service

It is in

running.

is


closes, a large transient current

line.

This transient surge

is

is

hard on

and also produces a mechanical shock.

this reason,

rate circuits in

we sometimes employ more

which

in open delta, and having a 65 percent tap, employed to provide reduced-voltage starting.

connected are

elabo-

Calculate the

motor

is

never completely a.

disconnected from the

line.

b.

The apparent power absorbed by the motor The apparent power supplied by the 460 V line

p.u.

6

_ 2

y

0

400

800

1200 ^ speed

0

1600 1800 r/min

400

800

1200

1600 1800 r/min

speed

Figure 20.31a

Figure 20.31b

Typical reduced voltage (0.65 pu) torque-speed curves

Typical reduced voltage (0.65 pu) current-speed curves

of

a 3-phase squirrel-cage induction motor:

(2)

primary

resistance starting; (3) autotransformer starting.

of

a 3-phase squirrel-cage induction motor:

(2)

primary

resistance starting; (3) autotransformer starting.


460

c.

d.

The current supplied by The locked-rotor torque

the

460

V

The

line

results of these calculations are

rized in Fig. 20.32.

them with

It is

summa-

worthwhile comparing

the results in Fig. 20.27.

Solution a.

The voltage across

E= The

current /

the motor

X

0.65

the

motor

Sm =

In addition to resistors

is

the

and torque when

motor

V3EI

= V3 X =

471

299

X 910

is

current

negligible (Section 12.1).

pedance

method can be used two identical 3-phase

starting

in parallel

when

As

used.

is

the motor

a result, the im-

two windings were

in parallel. After the

motor has picked up

if

speed, the second 3-phase winding

is

brought into

two windings operate in parallel. 20.33 shows how two 3-pole contactors A and

471

drawn from

kVA

the line

B

= SL /(V3£) = 471 000/(1.73 X = 592 A

that this current

l,

(8.9)

be

2, 3.

A

arranged closes

Shortly

windings

7, 8,

9

for

first,

after,

part-winding

L1

O

L2

O

L3

O

starting.


varies as the square of

contactor

in parallel

B

closes, bringing

with windings

the motor voltage:

T=

0.65

=

0.42

2

X 600

X 600

= 252 N-m 5 m =471 kVA

S L = 471 kVA

locked rotor

299 V

460 V 592

A

£

910

A

autotransformer

Figure 20.32

See Example

Figure 20.33 20-4.

starting.

thus energizing windings

460)

considerably smaller

is

can

Contactor

is

than the line current (910 A) with resistance

d.

is

the

higher than

is

connected

Fig.

= Sm =

/

Note

the induction motor has

service so that the

iSj

The

The part-winding

these 3-phase windings the

Consequently,

c.

tions.

Some

winding connec-

running. During the starting phase, only one of

The apparent power supplied by the line is equal to that absorbed by the motor because active and reactive power consumed by the autotransformers

in the stator

when

windings that operate

kVA

to limit the current

starting induction motors.

only require a change

is

and autotransformers, sev-

methods are employed

eral other

= 910 A

1400


b.

20.14 Other starting methods

X 460 = 299 V

0.65

drawn by

=

is

Part-winding starting of an induction motor.

1, 2, 3.

BASICS

There are many different types of part-winding

OF IND USTRIAL MOTOR CONTROL

forward

connections and some larger motors have specially

stop

reverse

Hh

*

3

i

46

:

designed windings so that the starting performance is

optimized. In

wye -delta

starting,

all

stator leads

six

are

brought out to the terminal box. The windings are

connected

in

wye during

start-up,

and

in delta dur-

ing normal running conditions. This starting

gives the

same

having a 58 percent

tap.

The reason

that the volt-

is

age across each wye-connected winding 1

/

V3

(

=

method

results as an autotransformer starter

0.58) of

its

only

is

rated value.

Figure 20.35 Schematic diagram of a cam switch permitting forward-reverse and stop operation of a 3-phase motor.

wound-rotor motors, we pro-

Finally, to start

gressively short-circuit the external rotor resistors in

one, two, or

more

steps.

The number of

steps de-

pends upon the size of the machine and the nature knob, some contacts are closed while others are open.

of the load (see Fig. 13.19).

This information

20.15 Some

Cam

switches

given

in

have

to

be under the

lifting

closed and contacts

and low-

ering rate, and the load has to be carefully set at the proper place.

Such

is

down

l

and 3 are open.

turned to the stop position,

Fig.

quence can be done with cam switches.

cam

all

to

connect the

cam

signed for the forward, reverse, and stop operation of

state of the contacts

rectly

forward

Figure 20.34 a.

Cam

b.

Detail of the

cam

c.

Table

the on-off state of the five contacts.

switch external appearance.

listing

controlling contact

1

in

the stop position.

shows how

is

shown

stop

di-

reverse

X

X

X

3

off position

that controls

(Fig. 20.35)

1

in

knob

i

(open or closed)

2

cam shown

the

contacts are open.

on the diagram for each position of the knob.

contact

(b)

and 5 are

switch to a 3-phase motor. The

a 3-phase induction motor. For each position of the

1

2, 4,

When

20.34b shows the shape of the cam

The schematic diagram switch de-

contact

an open contact. In the

the opening and closing of contact

a supervised control se-

20.34 shows a 3-position

is

forward position, for example, contacts

continuous control of an operator. In hoists, for ex-

ample, an operator has to vary the

a table, usually glued to

A cross (X) designates a closed

contact, while a blank space

industrial operations

Fig.

is

the side of the switch.

4

X

5

X

X


462

The 3-phase

line

appropriate

cam-switch

jumpers

J

1

,

and motor are connected

J2, J3, J4 are also required to

the connections.

+

to the

Note

terminals.

that

motor

generator or brake

complete

The reader should analyze

quadrant 2

the cir-

quadrant

torque

1

-

connections and resulting current flow for

cuit

each position of the switch. For example, when the switch

is

in the

L

are closed and

L3

forward position, contacts is

1

connected to

T

L2

1 ,

to

i

0

2, 4, 5

^ speed

T2, and

to T3.

Some cam

switches are designed to carry several

hundred amperes, but

we

often prefer to use

quadrant 3

quadrant 4

motor

generator or brake

mag-

netic contactors to handle large currents. In such

cases a small

cam

switch

is

employed

to control the

relay coils of the contactors. Very elaborate control

schemes can be designed with multicontact cam

Figure 20.36 Electric drives

can operate

in

four distinct quadrants.

switches.

20.16 Computers and controls

However, some

industrial drives require a

motor

to

function at various torques and speeds, both in for-

The

control devices

we have covered

are used throughout the industry.

advent of computers, the behavior of

many

it is

now

in this

chapter

However, with the

possible to simulate

relay coils

and relay contacts.

Furthermore, the connections between these devices

can also be simulated. As a

make very complex

result,

it is

possible to

control circuits by simply using

ward and

of using real relays, contacts, and time-delay dash-

we simply program

these devices (and their

The computers used for this Programmable Logic Controllers

wiring) on a computer.

purpose are called

(PLCs). Their construction and basic principle of operation are covered in Chapter 3

ample, the motor

may

run clockwise or counter-

clockwise, and the torque

may

speed and torque

may be

positive or negative.

In describing industrial drives, the various operat-

ing

modes can

best be

shown

in graphical

a vertical axis (Fig. 20.36). This gives rise to four operating quadrants, labelled respectively quadrants

and

in the

a machine operates in quadrant

*

An

is

used to

seen the basic control equip-

start

same

direction. Consequently, a

such,

it

delivers mechanical

electric

is

a

and stop induction motors.

to

machine op-

power in

to the load.

As

quadrant

3,

The ex-

A

machine

that operates in quadrant 2 develops its

speed

is

negative. In other

system consisting of one or several

motors and of the entire

designed

both the

cept that both the torque and speed are reversed.

a positive torque but electric drive

1,

erating in this quadrant functions as a motor.

machine also operates as a motor that

1,

4.

torque and speed are positive, meaning that they act

20.17 Fundamentals of electric drives* ment

hori-

zontal axis, and the positive and negative torques on

If

we have

form. The

and negative speeds are plotted on a

positive

2, 3,

In this chapter

act either with or

against the direction of rotation. In other words, the

1

ELECTRIC DRIVES

to function for brief periods as

a generator or brake. In electric locomotives, for ex-

a keyboard, a monitor, and a computer. Thus, instead

pots,

reverse. In addition to operating as a motor,

machine often has

the

electric control

equipment

govern the performance of these motors. (IEEE

Standard Dictionary of Electrical and Electronics Terms)

words, the torque acts clockwise while the machine turns counterclockwise. In this quadrant, the

ma-

chine absorbs mechanical power from the load;

BASICS

consequently,

it

Depending on the way

it

when

also function as a brake 2.

ately

Jy

latter is

and unavoidably converted into

^brake

\

T

\

®t

.

©

is

again

motor

\

©

(3)

heat. In effect,

—

cuit

brake

very inefficient. Consequently, the

when

tions as a generator

Quadrant 4 the torque

is

operating in quadrant

^

Figure 20.37 tion

motor operating

brake—^\

\© \

2.

identical to quadrant 2, except that

generator

—

T

^\

apply.

0

\\

motor

0

\

20.18 Typical torque-speed curves

now

machine

acts as a

quadrant

2,

14.

1

speed curve

in

is

it

1,

as a brake in

quadrant 4 (Section

leads are reversed, another torque-

shows

obtained. This dash-line curve

that the

generator or brake hand,

in

machine now operates as a motor

Note

in

and as a brake

machine can function

in

v

©\\

\ brake

V

\

\

quadrants 2 and

always runs as a motor

To give another example,

Figure 20.38 Typical torque-speed curve of a dc motor.

in Fig. 20.37, the

quadrant

as a generator in quadrant 2,

rant 4.

X

®

generator

14. 16.

curve

and as a generator

6). If the stator

that the 3,

to the solid

motor

+

\

1

with particular attention to Fig. Referring

—

\

\ motor

1

and frequency.

at fixed voltage

+

cir-

The torque-speed curve of a 3-phase induction motor is an excellent example of the motor-generator-brake behavior of an electrical machine. We first examined it in Chapter 14, Section 4. 6. The reader is encouraged to take a few moments to review this section,

\

Typical torque-speed curve of a squirrel-cage induc-

and speed are reversed; consequently, the

same remarks

\

generator^

usually arranged so that the machine func-

is

+

\

immedi-

absorbs mechanical power from the shaft. Both power inputs are dissipated as heat often inside the machine itself. For example, whenever a machine is plugged, it operates as a brake. In larger power drives we seldom favor the brake mode of operation it is

\

0

when a machine functions as a brake, it absorbs electric power from the supply line at the same time as it

because

motor

\

~T

operating in

The mechanical power absorbed

converted to electric power, but the

+

^.generator

dynamic braking. is connected, a machine

external resistor, such as in

quadrant

463

functions basically as a generator.

The mechanical power is converted into electric power and the latter is usually fed back into the line. However, the electric power may be dissipated in an

may


in

4.

quadrant in

quad-

either as a

On the other

quadrants

Fig. 20.38

1

and

shows

3.

the

generator-brake

modes

are again apparent. If the

armature leads are reversed,

we

obtain the dotted

torque-speed curve. In designing variable-speed electric drives,

vary the speed and torque

try to

in

we

a smooth, contin-

uous way to satisfy the load requirements. This

is

usually done by shifting the entire torque-speed characteristic axis.

back and forth along the horizontal

For example, the torque-speed curve of the dc

motor

(Fig. 20.38)

may

be shifted back and forth by

we can

complete torque-speed curve of a dc shunt motor

varying the armature voltage. Similarly,

when

the curve of an induction motor by simultaneously

the armature voltage

is

fixed.

The motor-

shift


464

Varying the voltage and frequency

varying the voltage and frequency applied to the

To

we

will first

show how

motor operation. By keeping the

rule" of

better understand the basic principles of vari-

able speed control,

variable

we

ied,

induction motor.

close to

Shape

its

rated value.

20%

Fig. 20.39

The torque-speed curve of a 3-phase squirrel-cage induction motor depends upon the voltage and frequency applied to the stator. We already know that

(

now

affected

when both

how

arises,

is

kW)

1

is

1

We also know that the

60 If

The

V,

full -load

and 60

speed and torque are respec-

Nm;

the torque-speed curve

new torque-speed curve

the left.

The curve

tor voltage

is

doubled, the

doubled. Under these conditions, the

curve

shape of the torque-speed curve remains the same, but

its

is

shifted to the right

nous speed

Even

position along the speed axis shifts with the

N m.

is

if

is

we

V

and

shifted toward the

if

we

raise the voltage

2700

V,

90 Hz),

the

and the new synchro-

r/min.

bring the frequency

down to zero (dc),

the torque-speed curve retains essentially the

frequency.

to

15 Hz),

same shape, but crosses synchronous speed of 1800/4 = 450

and frequency by 50 percent (690

sta-

80

retains the

the axis at a

is

breakdown torque

reduce both the voltage and frequency

r/min (Fig. 20.40). Similarly,

the frequency

the

one-fourth their original value (115

the voltage and frequency are

hp,

60 Hz squirrel-cage induc-

varied? In practice, they are varied in the same pro-

when

com-

stator. 1

460

portion so as to maintain a constant flux in the air gap. Thus,

frequencies be-

N m and the locked-rotor torque is

we

var-

is

always

shows the torque-speed curve of a 5

3-phase,

tion motor.

synchronous speed depends on the frequency. The question

1

tively 1725 r/min

fixed, the torque varies as the

square of the applied voltage.

at

is

of rated frequency, the volts per

pensate for the IR drop in the

is

motor

in the

However,

volts per

frequency

hertz ratio has to be progressively increased to

of the torque-speed

curve

the frequency

level while the

ensure that the flux

low about

20.19

same

hertz at the

frequency affects the behavior of a squirrel-cage

if

same

in the

proportion has given rise to the "volts per hertz

stator.

same

N-m r-160

-+-120

I rated load

80 torque

60

H

5

r

/m in

40

2100 18 00 15 00 12 00

9()0

6C)0

300

()

3()0

6()0

—

9()0

\

12 00 15 00 18 00 21 00 24 00 2700 3000 3300 36 00 3900

i

i

*0

r/min

speed

80

\

+ -160

-200 J


of

a 15 hp, 460

V,

60 Hz, 3-phase squirrel-cage induction motor.


BASICS

N m 115 V, 15 Hz -160 tsi

460 V, 60 Hz

465

690 V, 90 Hz

y

-200


and voltages.

at three different frequencies

Because the shape of the torque-speed curve the

same

frequencies,

at all

it

developed by an induction motor ever the slip speed (in r/min)

is

follows that the torque is

the

same when-

the same.

is

Example 20-5

A standard 3-phase,

NEMA

class

D

velops a torque of If

the

motor

is

1

0 hp, 575 V, 750 r/min, 60 Hz 1

squirrel-cage induction motor de1

10

N m at a speed of

1

440

r/min.

excited at a frequency of 25 Hz, cal-

culate the following: a.

b.

The required voltage to maintain the same in the machine The new speed at a torque of 0 N m 1

tlux

1

Solution a.

Figure 20.41

To keep duced

Stator excited by dc current.

tor

creases with increasing speed, reaching a in

The mo-

develops a symmetrical braking torque that

both directions, as shown

ure, the

dc current

in the

in Fig.

windings

duce the rated breakdown torque.

20.4 is

1

.

in-

maximum

the

b.

same

flux, the voltage

(25/60)

is

re-

X 575 = 240 V

The synchronous speed of the motor

must be

proportion to the frequency:

E=

shape. Current can be circulated in any two lines of the stator while leaving the third line open.

in

4-pole, 60

Hz

obviously 1800 r/min. Consequently,

the slip speed at a torque of

1

10

Nm

In this fig-

adjusted to pro1

800

-

1440

- 360

r/min

is

.


466

The

slip

speed

is

same

the

speed

at

25

=

ns

Hz

current-speed curve retains the same shape, no mat-

at

what the synchronous speed happens

ter

Thus, as the synchronous speed

is

(25/60)

The new speed

same torque, The synchronous

for the

irrespective of the frequency.

1

X

1800

10

N-m

= 750

rent-speed curve shifts along the horizontal axis

r/min

with the

minimum

nous speed. In

is

= 750 - 360 = 390

r/min

frequency

current following the synchro-

torque-speed and current-

effect, the

move back and

speed curves n

to be.

varied, the cur-

is

forth in unison as the

varied.

is

Suppose, for example, that the voltage and

quency are reduced by 75 percent

20.20 Current-speed curves The current-speed

The locked-rotor mo-

characteristic of an induction

synchronous speed. The

minimum value minimum current is equal


needed

tor

is

a V-shaped curve having a

the machine.

Because the

460

15 hp,

V,

at

N

to

reducing the frequency,

is

be

120

to the full

breakdown

we

torque. Thus, by

obtain a larger torque

with a smaller current (Fig. 20.43). This

to create the flux in

is

one of

is the same at all speeds. shows the current-speed curve of the 60 Hz squirrel-cage induction motor

progressively increasing the voltage and frequency.

we

all

quency can be varied automatically so

speeds; consequently, the

the corresponding torque

As in the case of the torque-speed shown that if the stator flux is held

is

tor

its

load by

its

that the

breakdown torque

all

fre-

mothe

zero to rated speed. This ensures a rapid

acceleration at practically constant current.

can

it

develops close to

way from

80 N-m.

curve,

can gradually accelerate a motor and

During the start-up period, the voltage and

We have plotted the effective

always positive. The locked-rotor current

A and

m, equal

the big advantages of frequency control. In effect,

values of current for is

current decreases to 80 A, but the

stator flux is kept constant,

mentioned previously. current

fre-

115V, 15 Hz.

corresponding torque (Fig. 20.40) increases to 160

the magnetizing current Fig. 20.42

to

In conclusion, an induction

motor has excellent

characteristics under variable frequency conditions.

constant, the

N-m

A

s

y-160

brake (60)-.

1

I

1

1

-h -120

—

y

Tim \

f

Milld

generc to

5)

Mill f

gen era to

(6 0)

-u

o (O

w

—

D ~E •o

—80

no cor

I

40

2100 1800 1500 1200

9()0

6()0

V-

(11

'

3()0

t ()

40

A

115 V, 15Hz

300 600 450

460 V,

f

50 Hz

900 1200 1500 1800 2100 24 00 2700 3000 3300 3600 39 00 r/min

\44-

speed

30

4 -1 20

/

Lan

+ i

A

\ ./

-200 1

Figure 20.42 Current-speed curve at 60 Hz and 15 Hz. Also T-n curve

—? f

\

at

460

V,

60 Hz.

BASICS


The corresponding N-m

E=

stator voltage

467

is

X 460 = 856 V

(111.7/60)

The 60 Hz current-speed and torque-speed curves (Fig. 20.42) show that the stator current is 40 A when the torque

curve

is

current

is

100 N-m. Because the current-speed

along with the torque-speed curve, the

shifts

again 40

A at

3200 r/min and 100 N-m.

20.21 Regenerative braking * speed

A

further advantage of frequency control

Figure 20.43 The starting torque increases and the current de-

permits

creases with decreasing frequency.

Hz

regenerative

20.44, suppose the motor line. It is

running

constant torque

For a given frequency the speed changes very with increasing load. In

many ways,

braking.

little

the torque-

we

Referring

100

N-m

(operating point

percent, the motor will immediately operate along

230

V

torque-speed curve. Because the

motor with variable armature- voltage control.

speed cannot change instantaneously (due to

inertia),

we

suddenly find ourselves

Example 20-6

the

new torque-speed curve. The motor torque

Using the information revealed by the 60 Hz

ative; consequently, the

and current-speed curves of

Fig.

1). If

suddenly reduce the frequency and voltage by 50

the 30 Hz,

20.42, calculate the voltage and frequency required

operating point 2 on

at

following the 50-percent curve until

rL

(operating point 4)

is

neg-

speed will drop very quickly,

The

we

reach torque

interesting feature

is

that

so that the machine will run at 3200 r/min while de-

What

veloping a torque of 100 N-m.

is

the corre-

460 V, 60 Hz

230 V, 30 Hz

sponding stator current? Solution

+

We first have to find the slip speed corresponding to

-1 )0-

r-1601

-T

i

a torque of 100 N-m.

.j.

?™

-

i

J0+*torq

According

to Fig. 20.42,

when

the

motor operates

60 Hz and a torque of 100 N-m, the speed r/min. Consequently, the slip speed

is

at

*0

1650

is

6C)0

=

ns

=

1800

«]

—

n

3()0

()

w-

-

1650

=

3(M)

6()0

*-

V

-
9()0-

12 00 15 00|1800

r/ mir

V

speed

150 r/min 50

The

slip

100

N-m

speed at

is

the

same when

the

ft

motor develops

3200 r/min. Consequently, the synchro-

= 3200 +

v 150

- 3350

The corresponding frequency

/=

k

-1 20

nous speed must be ns

(3350/1800)

is,

X 60 =

r/min

-200-

\

therefore,

111.7

Hz

it

Fig.

connected to a 460 V, 60

is

speed characteristic resembles that of a dc shunt

torque-speed

that

to

1650 r/min, driving a load of

at

Tv =

is

Figure 20.44 Effect of suddenly changing the

stator frequency.


468

in

moving along the curve from point 2 to point 3, en-

ergy

is

as an

returned to the ac

asynchronous generator during

The

ability to

20-

1

1

because the motor acts

line,

runs at

20- 2 1

tive braking,

is

economy of regenera-

We

becoming so

controlled induction motor drives are

in

3-phase, 4-pole squirrel-cage

20- 3 1

same

flux in the air gap. Calculate

Referring to Fig. 20.42, what

Without referring

knowing that 460 V, 60 Hz?

conditions,

four types of circuit diagrams and

ergized

describe the purpose of each.

20-2

stator.

a.

at

Machine running

as a

the stator

motor

at

to the text, describe the

and developing a torque of 100

shown

operation of the starter

in Fig. b.

20. 6b, and state the use of each component. 1

20-3

Give the symbols for a tact,

and

NO and a NC

c.

con-

for a thermal relay.

Identify

all

the

components shown

20.23a using the equipment Table 20A.

A situated 20-5

If the start

Where

list

are contact

in Fig.

given

a.

c.

T and

coil

20- 5

physically?

1

and stop pushbuttons

in Fig.

will

happen?

effect

if

contact

Ax

in

pushbutton were

parallel with the start

If a short-circuit

20. 14,

20-8

would

it

20-16

will

short-circuit

a.

between the turns of

M in Fig. 20. 14

b.

20- 7

50%

a.

increase in the line current

Under what circumstances

down

clockwise

in

quad-

develop a clockwise or

is

reduced-

b.


in

+

b.

+3150r/min, - 100 N-m

1650 r/min,

+

100

N-m

the

motor current

is

of Fig. 20.22

If the control circuit

motor

shown

will start

in Fig. 20.14,

and continue

is

used

show

in

that

to run if

we momentarily press the start button, Show that if we press the jog button, the is

depressed.

which quad20- 1 8

A magnetic normal

a.

if

60 A? 240 A?

motor only runs for as long as the button

the following torque-speed oper-

ating points occur?

1

V, 3-phase,

the

voltage starting required?

do

20 N-m

relay having the tripping curve

place of that

the motor?

rants

it

in Fig. 20.

take to trip

circuit?

of one phase. Which device will shut

20-10

1

7 has to protect a 40 hp, 720 r/min induction motor having a nominal current rating of 40 A. If the relay is set to 40 A, how long will it

1

20-9

Does

A thermal 575

M of Fig.

open the

the stator winding of motor

produces a

of

which quadrants a machine operates

A machine is turning

given

re-

have on the

occurs in motor

which device

A partial

N-m

counterclockwise torque?

operation of the starter?

20-7

1650 r/min

Intermediate level

Referring to Fig. 20. 14,

moved, what

en-

As a brake As a motor As a generator

rant 3.

20.24a are pushed simultaneously, what

20-6

State in

b.

in

is

Machine running as a brake at 300 r/min Machine driven as an asynchronous generator at a torque

20- 4 1

20-4

the cur-

is

rent in the stator under the following

Practical level

Name

to turn at a no-load

the required voltage and frequency to be

applied to the

1

60 Hz.

rated at 208 V,

is

want the motor

ing the

Chapter 23.

Questions and Problems 20-

it

speed of about 225 r/min while maintain-

popular. These electronically-controlled drives are

covered

A standard

induction motor

main reason why frequency-

the

power [hp] of the motor when 450 r/min.

chanical

this interval.

develop a high torque from zero to

speed, together with the

full

Referring to Fig. 20.39, calculate the me-

tacts

contactor can

make

3 million

circuit interruptions before its con-

need

to

be replaced.

jogs the motor so

that

it

If

an operator

starts

and stops

BASICS


once per minute, after approximately

how

the

many working days

have

high-efficiency motor;

will the contacts

assuming the operator

to be replaced,

a.

b. a.

Referring to Fig. 20.24a and assuming that the

motor

initially at rest,

is

eration of the circuit

when

motor

If the

happens

if

is

a.

is

20-26

b.

20-27

the start button in Fig. 20.25c

is

adjusted for a delay of 10

1

is

quence of events start

button

b.

that relay

20-28

depressed.

knowing in

s.

that takes place

that relay

Draw

RT is

set for a

when

A

is

c.

d.

120 A, and the thermal

e.

if

the

that the

20-29

[hp] at

Draw

the torque-speed curve that passes 1

,

2,

and 4 (see

a.

In

Problem 20-28 draw if

60

V

is

the torque-speed

applied to the armature, while

maintaining the same field excitation,

motor had been running b.

What

is

the frequency of the current in the at

a speed of 300 r/min?

Referring to Fig. 20.39 and neglecting

windage and

friction losses, calculate the

power P r supplied machine runs a.

b. c.

As As As

a

motor

at

a brake at

to the rotor

when

the

Advanced 20-30

a.

In

level

The curves 460

V,

1650 r/min

in Fig.

20.26 relate to a 100 hp,

1765 r/min, 3-phase, 60

tion motor,

whose

750 r/min 2550 r/min

curves

1

and 2

|

Problem 20-23, calculate the value of

Calculate the torque developed resistors are in the circuit

the rotor I^R losses in each case.

Referring to Fig. 20.39, calculate the volt-

inducis

ft-lbf].

a generator at

age and frequency that must be applied to

Hz

full-load current

120 A. Calculate the breakdown torque for

b.

20-25

negligible.

The torque [N m] at 300 r/min The starting torque [ft-lbf]

armature coils

20-24

is

The mechanical power output

curve

for several hours at full-load).

20-23

fl. It is

source, and the

Fig. 20.38)

load current suddenly rises to 240 A.

(Assume

V dc

1800 r/min; the corre-

is

through quadrants

given by Fig. 20.17, cal-

culate the approximate tripping time

240

1200 r/min

relays are set to this value. If the relay

tripping curve

to a

be neglected, calculate the following: a. The armature current at 900 r/min

its fi-

20.26a and 20.26b. The is

1

Assuming constant field excitation and assuming that armature reaction effects can

possesses the characteristics given by

full-load current

quadrant

instantaneously from quadrant

sponding armature current

the actual circuit connections,

in Figs.

in

Why?

shunt-wound dc motor has an

no-load speed

the

b.

1

2.

quadrant 4?

A 4-pole,

100 hp, 460 V, 3-phase induction motor

curve

quadrant

move

it

to a

connected

delay of

sequence, until the motor reaches

40 A.

armature circuit resistance of 4

nal speed.

20-22

Can 1

s.

sta-

100

impossible for a machine to instanta-

to point in

momentarily depressed,

is

a current of

at

It is

neously change from a point

Referring to Fig. 20.30, describe the se-

5

a.

is

With the motor running, explain what happens when the stop button

20-2

Referring to Fig. 20.42, calculate the volt-

Nm

press the stop

momentarily depressed, knowing

RT

a

Nm

tor so that the locked-rotor torque is

running normally, what

we momentarily

when

Nm

At a speed of 2400 r/min, developing

age and frequency to be applied to the

Explain the sequence of events that takes place

runs as a relatively

explain the op-

the start button

button?

20-20

it

speed of 1200 r/min, developing a

a

torque of 60

momentarily depressed, b.

At

that

torque of 100

works an 8-hour day? 20-19

machine so

469

rent

20-3

1

is

480

A

in Fig.

the

line cur-

[ft-lbf].

The motor having given

when

and the

the T-n characteristic

20.39

is

running

at

a no-load

470


speed of inertia

1

800 r/min. The

of the rotor and

total

its

load

moment is

of

64 min

2

90

lb -ft

.

! I

I

The speed has

16 min

value of 1200 r/min by suddenly changing

and frequency applied

the voltage

8 min

to the

stator.

4 min

c.

The voltage and frequency required The initial kinetic energy stored in the moving parts The final kinetic energy in the moving

d.

Is all

a. b.

E

1

IME

VERSUS TRIP PIN(3C UR RE:n T

i-

460

3-ph ase uurv e

-

32 s

16s

/

parts

1-pheise

8s

the lost kinetic energy returned to the

5 hp,

given

curv e

4s

2s

60 Hz induction

V, 3-phase,

1

s 0.6

0.8

3

2

1.5

1

4

5

8910

6 7

in Fig. 20.39.

What

new shape of the curve if we Hz to the stator? the new breakdown torque fft-lbf]-

->-

Multiple of current setting

the

is

apply 230 V, 60

20-33

i

THEF3 MAI- REL AY

min

1

motor has the torque-speed characteristic

a.

i

I

I—

3-phase line? Explain.

A

I

!

2 min

Calculate

20-32

\-\ —

32 min

be reduced to a no-load

to

b.

Calculate

In


Figure 20.45

See Problem 20-34.

stator volt-

age needed to reduce the breakdown tactor? (The thermal characteristic corre-

torque to 60 N-m.

sponds Industrial appl i cat ion

20-34

A 30 hp,

1

20-35

780

200

r/min,

V, 3-phase cage

motor driving a compressor

is

relates to

shown

rent of

Curve 3

when

the

motor runs

electrician set the relay at

load current of the motor.

line voltage

Under normal ammeter

Due

the

to a fault

200

V

motor

mal relay What is

circuit, the

phase motor. As a phases

A and B

What

is

a.

C of the compressor motor sud-

denly blew, causing

the

it

1

20-36

35 A. possible time

took for the thermal relay to

trip the

fall to

1

55

reduced torque causes the

full

and

it

speed before the ther-

trips out. the per-unit starting current and

mal conditions? b.

result, the current in

rose to

per-

per-unit starting torque under these abnor-

to run as a single-

maximum

The

2.20.

to accelerate very slowly

doesn't reach

fuse in the distribution panel associated

with phase

V, 3-phase.

is

drop caused by the large

V. In turn, the

line.

on another

200

across the motor terminals to

indicated that the motor draws a current

A from

A at

starting current, causes the voltage

full-

operating conditions, a hook-on

of 71

465

larly

phase.

The plant

is

The motor is started during a particulow voltage-sag in the electric utility system. This sag, combined with the

single-

82 A, which corresponds to the rated

it

unit starting torque

normal 3-phase operation, and

curve 2 applies

to the manufacturer's specifi-

known that the motor in Problem 20-34 draws a locked-rotor cur-

protected by

in Fig. 20.45.

According cations,

a thermal relay having the time/current characteristic

to cold start conditions.)

it

con-

Estimate the time

it

took for the relay to

trip.

The

stator winding of the motor in Problem 20-34 has a line-to-neutral

tance of 23 mfl.

resis-

BASICS

a.

Calculate the stator copper losses

when

motor runs normally on the 3-phase

the

line,

driving the compressor. b. Calculate the stator

3-phase 60

coil of a

Hz

1

3

kW, 230

V,

when

the contactor

is

in the

VA at

a

power

We the

want

230

V

W and

1

1.5

VA.

to excite the coil directly off

line.

To achieve this result, calpower rating of

culate the resistance and

the resistor that should be connected in

contactor has a rating of

120 V. According to the manufacturer's catalog,

draws 100

47

factor of 0.75. In the holding position, the coil absorbs 3

single-phasing tend to overheat the motor?

The holding

position, the coil

copper losses when the

motor runs as a single-phase motor. Does

20-37


open

series with the coil a) is

open and

closed.

b)

when

when

the contactor

the contactor

is

Chapter

21

Fundamental Elements of Power Electronics

taining these

21.0 Introduction

they are not

Electronic systems and controls have gained wide acceptance in power technology; consequently, it has become almost indispensable to know something about power electronics. Naturally, we cannot cover all

components and devices are simple

—but

their behavior

can be understood

without having an extensive background in semi-

conductor theory.

Potential level

21.1

aspects of this broad subject in a single chapter.

Nevertheless,

we can

In Chapter 2, Sections 2.4 and 2.5, we described two ways of representing voltages in a circuit. We now introduce a third method that is particularly useful in circuits dealing with power electronics. The method is based upon the concept of potential levels. To understand the operation of electronic cir-

explain in simple terms the be-

havior of a large number of electronic power circuits, including those most

As

commonly used

far as electronic devices are

will first cover diodes in all electronic

and

thyristors.

today.

we

concerned,

They

are found

systems that involve the conversion

We then go more recent devices

of ac power to dc power and vice versa.

cuits,

on

have a potential level with respect to a reference

to discuss the application of

such as gate turn-off thyristors (GTOs), bipolar

is

of a thyristor and

its

In

power

electronics

basically

no

MOSFETs),

different

from

points

all

shown

so, that

power electronics can be explained by and closing of circuits

at

much

the opening

that circuits

then measured with respect to this zero ref-

with an ac source

Of the

con-

nal

472

1

is

as a horizontal line having a potential of 0 V.

composed of an 80

of

precise instants of time.

However, we should not conclude

is

Consider, for example, the circuit of Fig. 21.

these devices act basically

much

ter-

erence terminal. In graphs, the reference level

that

associated switching circuitry.

as high-speed switches; so

useful to imagine that individual terminals

The reference terminal is any convenient point chosen in a circuit; it is assumed to have zero electric potential. The potential level of all other

and insulated gate bipolar transistors (IGBTs). Their action on a circuit

is

minal.

junction transistors (BJTs), metal oxide semicon-

ductor field effect transistors (power

it

V

1,

battery connected in series

E having

a peak voltage of

three possible terminals,

as the reference point.

let

The

1

00

V.

us choose termi-

potential level of

FUNDAMENTAL ELEMENTS OF POWER ELECTRONICS

473

^"reference potential

Figure 21.1

method

Potential level

of representing voltages.

Figure 21.2 this terminal is therefore

designated 1

Consider

shown by

a horizontal line,

now

is

always 80

and terminal 2

V,

spect to terminal

The

1.

is

and

1

positive with re-

level of this terminal

is

therefore indicated by a second horizontal line 2

V above line 1. Now consider terminal

placed 80

minals

1

and 3

value

initial

is

is

E

Voltage

between

and we assume

ter-

that

1

.

E

Because is first

with respect to terminal

1.

alternating, the

is

negative, then positive,

The changing

shown by curve 3. Thus, during

the interval

level

is

from 0

to

h the level of point 3 is below the level of point 1, which indicates that terminal 3 is negative with re-

t

1

.

During the interval

and so the

larity reverses,

line 1.

,

and

2,

3.

Terminal

1

is

could have chosen another terminal as a

3 and, as before,

f

,

to f4 the po,

level of curve

3

is

now

therefore negative with re-

spect to terminal 3, because line 1

This potential-level method

is

below curve

now

3.

enables us to

we

21.4).

Knowing

to terminal

1

that

that

it is

E

.3

is initially

line

polarities.

100

to r 3 2,

,

terminal 3

is

because curve 3

we can draw curve 1. we know

the level of terminal 2,

always 80

V

positive with respect to termi-

we draw curve 2 so that it is always 80 V above curve 1. By so doing, we autonal

1

.

Consequently,

matically establish the level of terminal 2 with respect to terminal Figs. 21.2

pearance; however, larities

at

every instant, the relative po-

and potential differences between terminals

are identical.

two

3.

and 21.4 do not have the same ap-

From an

electrical point of view, the

figures are identical.

We

invite the reader to

check by comparing the voltages and polarities at various instants in the

is

2

80 V

—

their relative

two

figures.

reference potential

positive with respect to terminal is

above

line 2.

tween these terminals reaches a during

t

this interval.

Then, from

The voltage

maximum f

3

to

f

6,

be-

of 20

V

terminal 3

negative with respect to terminal 2 and the volt-

age between them reaches a

180

V

at instant

f

v

maximum

value of

(Fig.

V negative with respect

in a circuit, as well as their relative

For example, during the interval from

3

an alternating voltage and

determine the instantaneous voltages between any

two terminals

ref-

we chose terminal

by a horizontal is

(as in Fig. 21 .2),

To determine

1

represent the zero potential of

this reference terminal

its

100 V, with terminal 3 negative with

potential of terminal 3

above

1

erence terminal. Thus, in Fig. 2

that terminal 3 3.

alternating

respect to terminal

spect to terminal

We

the potential level of terminal 2.

The difference of potential between terminals 2

Potential levels of terminals

in Fig. 21.2.

Figure 21.3 Changing the reference

terminal.


474

Figure 21.4

The

relative potential levels are the

same as

in Fig.

21

.2.

In analyzing electronic circuits the reference terminal

may

be selected anywhere; however,

it

easy to observe the waveshapes of the voltages interested

we are

circuit first

look

elements

at the

known. This simple

also

voltage levels that appear across

commonly we examine

active and passive circuit elements in electronic circuits. Specifically,

sources, switches, resistors, coils, and capacitors.

By

definition, ideal ac

and dc voltage

a circuit can modify these levels.

On the other

hand, ac and dc current sources have infinite internal

impedance. Consequently, they deliver a constant levels in the circuit

must

Potential Across a Switch.

open

When

(Fig. 21 .5), the voltage across

its

a switch

Potential Across a Resistor. If no current flows a resistor,

potential,

its

is

3,

4 must be

if

we happen

to

is

known. On

a current

On

the

when

make up

the circuit.

switch

closed the potential level of both terminals

is

the other hand,

same

know

the potential

/,

the other hand, if the resistor car-

the IR drop produces a correspond-

ing potential difference between the terminals. For

example,

if

current actually flows in the direction

I

Q

terminals dethat

at the

zero (Fig. 21.6).

of one of the terminals, the level of the other

also

ries

terminals

because the IR drop

is

pends exclusively upon the external elements

is

rule also applies to ideal-

adapt themselves accordingly. 2.

the 1

in

level

and the voltage

know

3.

Consequently,

current,

to

ized thyristors and diodes, because they behave like

sources have zero internal impedance. Consequently,

in

we happen

perfect (albeit one-way) switches.

they impose rigid potential levels; nothing that hap-

pens

if

level of terminal 2, then the level of terminal

some

Sources.

2

Figure 21.5

must be the same. Thus,

some

found

1.

y

Potential across a switch.

in.

21.2 Voltage across

Let us

1^

should be

|

Figure 21.6 Potential across a resistor.

R

O |


shown above 4.

of terminal 3

in Fig. 21.6, the potential

that of terminal 4,

by an amount equal

those

at the

moments when

same

THE DIODE AND DIODE CIRCUITS

to IR.

The

Potential Across a Coil or Inductance.

minals of a coil are

is

ter-

21.3

The diode

potential only during

the current

is

not changing. If

the current varies, the potential difference

is

given by

A diode

is

an electronic device possessing two

Thus,

if

flowing

(A//Af)

the current in Fig. 21.7

amount equal

is

to

above

L

increasing while

shown, the potential

in the direction

of terminal 5

is

(2.27)

that of terminal

A//Af. Conversely,

creasing while flowing potential of terminal 5

in the direction

is

below

level

6 by an

if / is

de-

shown, the

that of terminal 6.

ter-

minals, respectively called anode (A) and cathode

(K) (Fig. 21. 9). Although

E=L

475

it

has no moving parts, a

diode acts like a high-speed switch whose contacts

open and close according

Rule it

1.

When no

voltage

open between terminals 2.

If

is

applied across a diode,

open switch. The

acts like an

Rule

to the following rules:

we apply an

circuit is therefore

A and K (Fig.

inverse voltage

diode so that the anode

is

2 .9a). 1

E2 across the

negative with respect to

the cathode, the diode continues to act as an open

switch (Fig. 2

/

1

We

.9b).

say that the diode

is

reverse

biased.

Rule

3.

If a

momentary forward voltage E of 0.7 V x

or

Figure 21.7 Potential across

more

is

A is slightly positive with respect to the cathode, the

an inductor.

terminals 5.

Potential Across a Capacitor The terminals of

a capacitor are at the

capacitor

is

same

potential only

when

the

completely discharged. Furthermore,

the potential difference

applied across the terminals so that anode

between the terminals

become

and a current

/

say that the diode

is

forward biased.

re-

is

zero (Fig. 2

1

.8).

i-ff-S /

Figure 21.8 Potential across a capacitor.

6.

Initial Potential

potential levels

know

is

otherwise,

Level

A

final rule regarding

worth remembering. Unless

we assume

we

the following initial

conditions: a.

All currents in the circuit are zero and none are in the

b.

process of changing.

All capacitors are discharged. / =

These assumed

alyze the behavior of any circuit from the

power

is

o

starting conditions enable us to an-

applied.

moment

Figure 21.9 Basic rules governing diode behavior.

acts

immediately be-

gins to flow from anode to cathode (Fig. 21.9c).

mains unchanged during those intervals when the current /

The diode

short-circuited.

like a closed switch

We


476

while the diode conducts, a small

In practice,

drop

voltage

appears

across

terminals.

its

However, the voltage drop has an upper value of about

1.5 V, so

tronic circuits.

drop that

can be neglected

it

most elec-

in

precisely because the voltage

It is

small with respect to other circuit voltages

is

we can assume

the diode

when it conducts. Rule 4. As long as current

essentially a closed

is

even as

little

returns

to

Conduction

becomes

as 10

its

flows, the diode acts

However,

(jls,

if

it

stops flowing for

the ideal diode immediately

open

original

state

(Fig.

21.9d).

resume when the anode again

will only

slightly positive with respect to the cath-

ode (Rule

diode behaves like a nor-

mally open switch whose contacts close as soon as the

anode voltage becomes

spect to the cathode.

Its

entirely converted into heat.

ceed the permissible

limits,

slightly positive with re-

contacts only reopen

when

The

re-

must never ex-

otherwise the diode will

be destroyed. Most silicon diodes can operate

satis-

factorily provided their internal temperature lies be-

tween -50°C and +200°C. The temperature of a diode can change very quickly, due to

its

small size

transfer, diodes

mounted on thick metallic supports,

are usually

called heat sinks. Furthermore, in large installations, the diodes

may be cooled by

fans,

by

gives the specifications of

some

or

oil,

by a continuous flow of deionized water. Table 2

1

typical diodes. Fig.

21.10 shows a range of low power to very high

power

3).

In conclusion, a perfect

is

sulting temperature rise of the diode

and small mass. To improve heat

switch

like a closed switch.

which

loss

diodes.

Diodes have many applications, some of which are found again and again, in one in electronic

we

follow,

power

form or another,

circuits. In the sections that

will analyze a

few

circuits that involve

the current (not the voltage) has fallen to zero. This

only diodes. They will illustrate the methodology

simple rule

of power circuit analysis while revealing some ba-

is

crucially important to an understand-

ing of circuits involving diodes and thyristors.

Symbol For a Diode. The symbol

sic principles

for a diode (Fig.

21.9) bears an arrow that indicates the direction of

conventional current flow

when

the diode conducts.

2

1

much

Voltage.

A diode

can withstand only

breaks down. The peak inverse voltage (PIV) ranges from 50 V to 4000 V, depending on the construction. If the rated so

PIV

inverse voltage before

it

exceeded, the diode begins to conduct

is

many

verse and, in

Maximum

cases,

is

2

1

.7

Battery charger with series inductor

Single-phase bridge rectifier Filters

2

Three-phase, 3-pulse diode rectifier

1

.9

21.10 Three-phase, 6-pulse diode

rectifier

21.11 Effective line current; fundamental line in re-

current

21.12 Distortion power factor is

also a limit

to the

average current a diode can carry. The maxi-

mum

current

may range from

industrial applica-

21.8

immediately destroyed.

Average Current. There

many

21.14 cover the following

Battery charger with series resistor

.5

1

Peak Inverse

to

topics:

2 .6

21.4 Main characteristics of a diode

common

tions. Sections 21.5 to

a few hundred mil-

21.13 Displacement power factor

21.14 Harmonic content

liamperes to over 4000 A, depending on the construction rating

which,

and

and size of the diode. The nominal current

depends upon the temperature of the diode, in turn,

how

it

Maximum

is

depends upon the way

it

is

with series resistor

mounted

cooled.

The

Temperature. The

diode times the current

21 .5 Battery charger

it

carries

voltage is

across

a

equal to a power

circuit of Fig.

2l.lla represents a simplified

battery charger. Transformer T, connected to a

1

20

V

ac supply, furnishes a sinusoidal secondary volt-


TABLE

21

Relative

PROPERTIES OF SOME TYPICAL DIODES

A

power

low high

12

0.6

100

0.6

1000

1.1

E2

/,rlA|

30 240

0.8

l

very high

[V]

d [mm]

[mAJ

/

mm

1000

0.05

175

1000

0.6

200

11

4.5

200 200

25

54

47

26

600

1000

10 000

2000

1

I2

50

3.8

4.6

32

- average dc current

E ()

£nivi

A)[A|

medium

/0

All

voltage drop corresponding to

/()

- peak value of surge current for one cycle E 2 - peak inverse voltage I 2 - reverse leakage current corresponding to E 2 /cr

T - maximum d - diameter

junction temperature (inside the diode)

}

/

- length

Figure 21.10 Average current: b. Average current: c. Average current: d. Average current: a.

4 A; PIV: 400 V; body length: 10 mm; diameter: 5.6 mm. 15 A; PIV: 500 V; stud type; length less thread: 25 mm; diameter: 17 mm. 500 A; PIV: 2000 V; length less thread: 244 mm; diameter: 40 mm. 2600 A; PIV: 2500 V; Hockey Puk; distance between pole-faces: 35 mm; diameter: 98 mm.

(Photos courtesy of International Rectifier)

age having a peak of resistor,

ries across the

To explain choose point tential

1

00

A 60 V battery, D are connected

V.

and an ideal diode

a

I

(1

in se-

the operation of the circuit, 1 as the reference terminal. is,

soidally

let

us

The po-

therefore, a straight hori-

The

level

terminal

The

The

potential of terminal 2 swings sinu-

above and below point

whether 2

secondary.

of this terminal

zontal line.

1,

is

1,

according to

positive or negative with respect to

of terminal 3

is

always 60

because the battery voltage

potential levels are

shown

is

V

1.

above

constant.

in Fig. 21.11b.


478

©

1

n

i

100 V (peak)

60

5

(a)

Figure 21.11 a. b.

Simple battery charger circuit. Corresponding voltage and current waveforms.

drop

Circuit analysis a.

Prior to

—

t

0,

we assume

The potential of point 4 is, therefore, at same level as point 3, and 2 is at the level of

zero.

the

point b.

g.

f,,

anode 2

is

neg-

The

the diode cannot conduct (Rule 2).

At

instant

respect to

(Rule

3).

this

moment

sistor gradually builds up,

d.

of 40 From

V before f

,

to

U

now be

at the

The voltage £43 across

level as point 2.

it

falls to

reaching a

zero

same

The

+40

As long

is

t2

At

instant

t2

21.6 Battery charger with series inductor The current flow

40

is

zero,

in the battery

A when £43 =

2

duces

f

and the diode 4).

1

R

by an inductor, as shown

in

mind the behavior of an

As

in the

in Fig. 2

circuit,

1

.

12a.

bearing

inductor, previously ex-

1

example of

Fig. 2

1

1

.

1

,

the ideal diode

when anode 2 bebegins to conduct t positive with respect comes to cathode 4. From this moment on, point 4 follows point 2, and at instant

{

moment on, point 4 must follow point 3. From t2 to f4 point 2 is negative with respect ,

1

losses and a corresponding poor effi-

Let us analyze the operation of this

a.

From

this f.

.

We can get around the problem by replacing

plained in Section 2.3

immediately opens the circuit (Rule

charger of Fig. 2

limited by resistor R. Unfortunately, this pro-

the resistor

as current flows, the diode be-

the current

7.75 A.

-

given by

haves like a closed switch; e.

is

charges up.

is

n

current reaches a peak of V.

one cycle

pulsating current always flows into the

the latter receives energy and progressively

ciency.

- £43 /i

v

the re-

maximum

at instant

the current in the circuit /

f

on, the ideal diode

acts like a closed switch.

Consequently, point 4 must

instant

positive terminal of the battery. Consequently,

u terminal 2 becomes positive with 4, and the diode begins to conduct t

From

across the diode

V at

itself. The resulting current is a truncated wave having a peak value of 40 A. Its calcu-

lated average value during

ative with respect to cathode 4; consequently,

c.

PTV

of 160

Finally, at instant t4 the cycle (a to f above) re-

sine to

maximum

peats

1.

During the interval from 0

in the resistor), the

reaches a

that all currents are

to

point 3. Because point 4 follows point 3 (no IR

voltage

£43

appears across the inductor (Fig.


479

Figure 21.12 a. b.

Battery charger using a series inductor. Corresponding voltage and current waveforms.

2

1

1

.

2b).

The

latter

begins to accumulate volt-

point 2 to the level of point 3.

seconds, and the current increases gradually un-

maximum

reaches a

til it

/ max

=A

level until instant t4

given by

It

whereupon

+ /L

(2.28)

)

is

During the interval from

ergy.

i

=

dotted area between

Note t2 ,

and

whereas

it

was

was zero

used. This

fact that current

at this

the voltage

£43

at instant

a

consistent with the

is

it

is

maximum)

is

be-

rent decreases f3

between

when dotted A +

dotted area

As soon

to

f 3 , it

returns

the inductor dis-

and

area

A

t

3

,

becoming is

current

if

is

60

the line frequency

V. Calculate the

is

Solution

equal to

To

calculate the peak current,

we must

find the

value of area/l + This can be done by integral calculus, but we will employ a much simpler .

(

)

graphical method. Thus, referring to Fig.

2

1

.

1

we have redrawn the voltage levels usThe 60 Hz voltage cycle is di-

2c,

ing graph paper.

{

vided into 24 equal parts, each representing is

mH

peak

60 Hz.

)

as the current

the circuit,

2

has an inductance of 3.3

and the battery voltage

.

{

c.

t

Example 21-1 The coil in Fig. 21.12

a.

zero.

charges volt-seconds. Consequently, the cur-

at

,

2

no

across the inductor

becomes negative and so

zero

t

again to the circuit (see Section 2.13).

it

moment when

through an inductor

cause the voltage £43 across t2

peak

its

longer changing (has reached a

After

[V-s]

2

to t 2 the in-

r,

inductance [H]

that the current reaches

resistor

b.

t

)

L =

whole cy-

an interesting example of the use of

ductor stores energy and, from

+

the

an inductor to store and release electrical en-

where

A

stays at this

cle repeats.

This (

,

zero, the diode

opens

whereupon point 4 must jump from

time interval A/ equal to

At

=

(1/24)

X

(1/60)

=

1/1440

s


480

2 during

conduction

this

interval.

Conduction

when a falls to zero at instant t (Fig. 21.13b). The polarity then reverses and E 2] becomes positive, meaning that terminal 2 is positive with respect to terminal 1. Current / b now flows ceases

z

{

R in the same direction as before, but this way of diodes Bl and B2. Consequently, 3 now follows point 2 while point 4 follows 1. Voltage E 34 across the load is, therefore,

through time by point point

composed of a

between zero and a

sates

value of this rectified voltage

V

0.006944

By counting

s

=

(

1/1440

6.944

mV

we

squares,

X

s)

{

+

19

)

X

s.

E=

find that

A

(

+ contains its

area

0.90

=

132

mV = s

0.132

V

s

curve

The peak current 'ma*

is,

therefore,

>W

Thus,

mH

as

did with a resistor of

it

However, the big advantage of the inductor has essentially no losses. to

dc power

is

E l2 E2

\.

therefore

is

1

the positive to supply

gether

is

The conversion of ac power much more efficient.

first

E ]2

is

positive, terminal

R by way

of diodes

is

1

/

a

1

or 2 as zero reference potentials.

select as reference level terminal 2 dur-

half-cycle, terminal

nique, terminal

during the second

1

By

4 always remains

using this tech-

at

zero potential

while terminal 3 follows the positive portion of the sine waves. is

It

E34

then becomes evident that

a pulsating dc voltage. Figure 2

In addition to

its

and current of the

1

.

1

across

3c shows

load.

dc value, load voltage

E34

con-

an ac component whose fundamental fre-

tains

quency

called a single-phase bridge

respect to terminal 2 and current

drawn

is

twice the line frequency. In effect, the

to-

available in a single package.

When

also

This enables us to use the potential levels

the rectified voltage

3a enables us to rectify both

circuit operates as follows:

drawing the

we have

so forth, on alternate half-cycles.

it

and negative half-cycles of an ac source,

make up what

voltage

1

3b, in addition to

11.

that

dc power to a load R. The four diodes

rectifier. It is

The

.

1

half-cycle, terminal 2 during the third half-cycle, and

21.7 Single-phase bridge rectifier circuit of Fig. 21

.

(2V2)/ir]

in-

the load

The

1

=

for the source voltage,

we can

ing the

Thus, the current reaches the same peak with an ductor of 3.3


of either terminals

L = 0.132/0.0033 = 40 A

recti-

[V]

Referring to Fig. 2

6.944

(21.1)

effective value of the ac line voltage [V]

=

}

curve

=

given by

is

dc voltage of a single-phase bridge fier

corresponds to

=

Ed =

V =

10

approximately 19 squares; consequently,

A

E m equal

where

V

Consequently, each small square rep-

resents an area of

b.

value

£d -0.90£

Similarly, the ordinates are scaled off in 10 intervals.

maximum

peak voltage of the source. The average

to the

Figure 21.12c See Example 21-1.

waves that The voltage pul-

series of half-cycle sine

are always positive (Fig. 21.1 3c).

source

positive with

flows through

Al and A2. Consequently,

point 3 follows point 1, and point 4 follows point

voltage across the load pulsates between zero and

+£ m

,

twice per cycle. Consequently, the peak-to-

peak ripple

is

equal to

Em

.

In the case of a resistive load, the current has the

same waveshape value,

is

as the voltage;

given by /„

= EJR

its

average, or dc


peak voltage -

£*

481

m

Figure 21.13 a.

Single-phase bridge

b.

Voltage levels.

rectifier.

Figure 21.13c Voltage and current waveforms

load R.

in

Example 21-2 The ac source of

1

b.

in Fig. 2

]

-

13a has an effective voltage

20 V, 60 Hz. The load draws a dc current of 20 A.

The dc current

in the

load

is

known

to

be 20 A,

but the diodes only carry the current on alternate half-cycles. Consequently, the average dc current in each diode

is:

Calculate a.

b.

/

The dc voltage across the load The average dc current in each diode

= IJ2 =

20/2

=

10

A

21.8 Filters Solution a.

The dc voltage across

the load

is

given by Eq.

The

rectifier circuits

we have

studied so far produce

pulsating voltages and currents. In

21.1:

£d = = =

0.90

E

0.90

X

108

V

loads,

we

must be used 120

some

types of

cannot tolerate such pulsations, and filters to

smooth out the valleys and peaks. The

basic purpose of a dc filter

power flow

is

to

produce a smooth

into a load. Consequently, a filter

must


482

peak voltage =

Em

inductor absorbs energy

peak voltage =

ia

£m

ib

(3)

B1

Figure 21.15 Current and voltage waveforms with inductive

©

A2

The load current

14a is much more conThe voltage between ter-

in Fig. 21.

stant than in Fig. 21.13a. (b)

(a)

Em as be-

minals 3 and 4 pulsates between zero and

a.

Rectifier with inductive

b.

Rectifier with capacitive

smooth

filter. filter.

and

it

must release energy whenever

the voltage or current tends to

fall.

In this

way

(Fig. 21.15).

across the load

The dc voltage across

given by Eq. 2

is still

absorb energy whenever the dc voltage or current rise,

E54

fore, but the voltage

Figure 21.14

tends to

1

.

1

.

This

Bridge

rectifiers are

coil is sufficient to

are inductors and ca-

They tend

is

negligi-

and many other mag-

provide good

though the voltage across a

pacitors. Inductors store energy in their

be expected be-

most cases the self-inductance of the

rent in the load.

field.

very

used to provide dc current for

the netic devices. In

filters

to

is

the load

bly small.

tends to maintain a constant voltage and cur-

The most common

is

cause the dc IR drop across the inductor

relays, electromagnets, motors, filter

filter.

coil

magnetic strongly, the

filtering.

may

Thus,

al-

pulsate very

dc current can be smooth. Consequently,

to maintain a constant current; con-

the magnetic field pulsates very

little.

sequently, they are placed in series with the load (Fig. 2

1

.

tric field.

14a). Capacitors store

They tend

energy

in their elec-

to maintain a constant voltage;

Example 21-3 35 V, 20 A dc power supply us-

consequently, they are placed in parallel with the

We wish to build a

load (Fig. 21. 1 4b).

ing a single-phase bridge rectifier and an inductive

The

greater the

the better

filter,

is

amount of energy

stored in the

the filtering action. In the case of

a bridge rectifier using an inductor, the peak-to-

peak ripple

in

percent

is

given by

ripple

=

filter.

late the following: a.

(21.2) c.

where

d.

ripple

WL

=

The effective value of the ac voltage The energy stored in the inductor The inductance of the inductor The peak-to-peak current ripple

peak-to-peak current as a percent of

Solution

the dc current [%]

a.

dc energy stored inductor

P = /= 5.5 =

The peak-to-peak current ripple should be If a 60 Hz ac source is available, calcu-

about 10%.

b.

5.5 J*,

in the

smoothing

[JJ

dc power drawn by the load [W] frequency of the source [Hz] coefficient to take care of units

1

The

effective ac voltage

E can

Eq. 21.1:

Ed =

0.9

=

0.9

E £ E = 150 V

135

be derived from


The dc power output of the

b.

= The energy is

rectifier is

X 20 = 2700

135

483

to be stored in the inductor or

W

"choke"

given by 5.5

P

/ npple _ 5.5 X 2700 60 X 10

=

24.75 J

Consequently, to obtain a peak-to-peak current ple of 10 percent, the inductor in its

magnetic

must

rip-

store 24.75 J

field.

The inductance of the choke can be calculated

c.

from 1

=

WL

LI]

(2.8)

2

= ~L(20) 2

24.75

L = 0.124 H The peak-to-peak

d.

Figure 21.16 ripple

is

about 10 percent of

the dc current:

A and

19

=

0.1

X 20 =

2

current therefore pulsates between

introduction to 3-phase rectifiers in general.

now 1.

analyze

point, the

levels rectifier is

composed of three

diodes connected in series with the secondary wind-

2

1

1

.

The

line-to-neutral voltage has a

peak value

Em A large filter inductance L is connected in .

series

with the load, so that current fd remains essentially ripple-free.

Although the load

is

represented by a

3

shown

has

may be

a dc motor, a large

some

serious drawbacks, but

it

mag-

rectifier

provides a good

.

moment we apply power,

voltages

E3U

appear. Consequently, at

=

point 1 suddenly

becomes

Dl (Section

an electroplating bath. This simple

Em

energized, points K, 4,

N are at the same level because /d is zero.

a heat-dissipating resis-

Thus, the load

is

the

consuming device and not net, or

by the ac source and they

Before the transformer

K. This immediately

tor.

These potential

in Fig. 21.17.

successively reach a peak value

always a useful energy-

it

the

secondary terminals follow the voltage

1, 2,

is

resistance R, in reality

By choosing

the Load.

levels are rigidly fixed

of a 3-phase, delta-wye transformer (Fig. 6).

We

behavior.

Voltage Across

rectifier

The simplest 3-phase

its

transformer neutral as the zero potential reference

21.9 Three-phase, 3-pulse

ings

fed

A

21 A.

diode

rectifier with inductive filter

by a 3-phase transformer.

/peak-to-peak

The dc output

Three-phase, 3-pulse

21.3,

3).

conduction Current

rapidly, attaining a final value /

upon load

R.

During

level as point 1

,

,

0 the potential of


initiates

Rule

t

However,

E lN E2N

this interval

because the diode

/,

in

diode

increases

which depends

K is

is at

the

same

conducting.


484

Figure 21.17 Voltage and current waveforms

As

points

K

and

move

1

in

a 3-phase, 3-pulse

rectifier.

to an angle 6 () of 60° (Fig. 21.17). critical

because immediately

comes


to

Rule

that that

it

starts in

is

terminal 2 be-

K and

begins to carry current

conduction

The moment

later,

conduction

3, this initiates

in

1.

According

diode D2, so

At the same time

/.

diode D2,

ceases in diode

it

Dl. Consequently, beyond 60°, point

K follows the

level of point 2.

is

called commutation.

When

takes place automatically (as ple),

it

it

to an-

the switchover

does

in

our exam-

called natural commutation, or line

is

mutation. In this book mutation, because

it

we

is

prefer the term line

comcom-

the line voltage that forces

from one diode to the next. Commutation from one diode to another does not

the transfer of current

really take place instantaneously, as

cated.

Owing

we have

D2

while

occurs

at

1

80°, because

becomes positive with respect to point 2 (and point K). Commutation again takes place as the load current switches from diode

diode D3. Point

K

peaks of waves

2,

1 ,

D2

to

therefore follows the positive

and

and each diode carries the

3,

full-load current for equal intervals of time (120°).

The diode currents shapes composed of

i

2,

/

3

have rectangular wave-

positive current intervals of

20° followed by zero current intervals of 240°. Voltage

EKN

across the load and inductor in Fig.

21.17 pulsates between 0.5 voltage

is

Em

and

E m The .

ripple

therefore smaller than that produced by a

single-phase bridge rectifier (Fig. 21.15). Moreover, the fundamental ripple frequency

is

supply frequency, which makes

easier to achieve

good

filtering.

The dc voltage

it

three times the

across the load

is

given by

indi-

Ed =

to transformer leakage reactance, the

current gradually increases in diode

moment

critical

terminal 3 then

1

The sudden switchover from one diode other

The next

together in time, they

eventually reach a critical moment, corresponding

it

de-

0.675

E

(21.3)

where

creases in diode Dl. This gradual transition contin-

ues until

all

the load current

is

carried by diode D2.

However, the commutation period ically less than

purposes,

we

2

ms on

will

a 60

assume

it

Hz

is

Ed =

very short (typ-

system) and, for our

occurs instantaneously.

average or dc voltage of a 3-pulse- rectifier

E= 0.675 =

[V]

effective ac line voltage [V]


=

3/(tt

V2)]


Note

that if

we

reverse the diodes in Fig. 2 1

1

.

6, the

same way, except that the load current reverses. Voltage E KN becomes negative and point K follows the negative peaks of waves 1,

485

21.10 Three-phase, 6-pulse rectifier*

rectifier operates the

2,

2.

and

T

Line Currents Currents

transformer.

in the

/,

,

/

2

,

/

3

that

flow

(identical to the

supplies

power

loads

and

3.

diodes also flow

a

Consider the circuit of Fig. 2

former

/?,

in the

secondary windings of the

As we have

.

8 in

1

which a

upper

.

L and

load

x

3-phase, 3-pulse rectifier

set of

diodes together

are identical to the

/?,

we have

studied.

just

seen, these currents have

chopped rectangular waveshape which is quite from the sinusoidal currents we are famil-

different

Thus, load current

/d

shown. The lower

set

,

flows

in the neutral line, as

of diodes, together with

and L 2 also constitute a 3-phase, 3-pulse ,

iar with.

Furthermore, the currents flow for only

one-third of the time in a given winding. this intermittent flow, the

output power

is

maximum

less than the

transformer. For example,

if

Due

possible dc

nominal rating of the

load current /d2 flows in the neutral, as shown. The two 3-phase rectifiers operate quite independently

of each other,

the transformer in Fig.

K

following the positive peaks of

A

points 1, 2, 3 while

All diodes conduct during 120° intervals. If

we make

R]

The chopped secondary

= R2

current in the neutral currents are reflected

we can remove

circuit of Fig. 21.19.

feeding the transformer also change

very abruptly. The sudden jumps in currents

field

/c

produce rapid fluctuations

in the

/a , / b ,

surrounding the feeder. These fluctuations can

and

L, respectively.

a

3-phase,

start at

more than one-third of

try to

rectifier.

design

The 6 diodes 6-pulse

source

® Figure 21.18 Dual 3-phase, 3-pulse

rectifier.

the two inshown as R

constitute

rectifier.

However, each diode

for only 120°.

is

Also called a 3-phase bridge

© © ©

3-phase

and the dc

/ d2

It

is

what

is

called

6 different moments during each cycle of the

line frequency.

the time. This

achieved by using 3-phase, 6-pulse

=

The two loads and

so that the transformer windings carry

current for

,

zero. Consequently,

6-pulse because the currents flowing in the 6 diodes

tele-

lines.

rectifiers

/d

ductors are simply combined into one,

called

Because of these drawbacks, we

then

magnetic

induce substantial voltages and noise in nearby

phone

,

becomes

the neutral conductor, yielding the

into the primary windings, with the result that the

and

follows the negative peaks.

1

heating.

line currents

R2

rectifier

The corresponding

but with the polarity reversed. to

00 k VA, we can show that it can only deliver 74 kW of dc power without over21.16 has a rating of

trans-

in Fig. 21.16),

6 diodes and their associated dc

to

R 2 The

with inductor

1

one shown

LA

rectifier.

still

conducts


486

D1

D2

/2

D3

/,[

|

© 3-phase

source

© D4

®

D5

/,

D6

Figure 21.19 Three-phase, 6-pulse

The

line currents

rectifier with inductive filter.

/.,,

/ b , /c

supplied by the trans-

former are given by Kirchhoff s law:

£ 12

ages,

,

£23, £31 (and

£2j £32 E l3 ,

,

)

rather than

the line-to-neutral voltages used in Fig. 21.20.

K

level of

h = U A, = h le = ?

A

waves while voltage

put

The

follows the tops of the successive sine

remains

1.225 E, where

E

is

zero potential.

at

between

fluctuates

The

out-

E

1.414

and

the effective value of the line

i

The average value of

voltage.

They

consist of three identical rectangular

that are out of

rents

phase by 120° (Fig. 21.20). The cur-

now flow

for two-thirds of the time in the sec-

ondary windings. As a 100

kVA

waves

result,

can be shown that a

it

transformer can deliver 95

kW

of dc

power without overheating. Figs. 21.18

output voltage rectifier. Its

is

value

that the

=

0.

89

1

£ and

E

is

only

six times the line frequency.

ple

much

is

.35

—

is

ripple

easier to

filter.

ripple

=

(

1

.4

1

4

1

Consequently, the

rip-

The approximate peak-

average dc

given by

1.35 E, as

.225)

twice that of a 3-phase, 3-pulse is

is

the fundamental ripple frequency

The peak-to-peak

to-peak current ripple in percent

and 21.19 reveal

E KA

given by Eq. 21.4.

is

given by

P

0.17— fWL

(21.5)

where

E

(21.4)

peak-to-peak current as a percent of

ripple

the dc current [%\

where

WL

£d ~ E=

= P =

dc voltage of a 6-pulse rectifier [V]

/=

effective line voltage [VJ

dc energy stored

in the

inductor

[J]

dc power drawn by the load [W] frequency of the 3-phase, 6-pulse source [Hz]

1

.35

=


=

3

V2 /tt] Fig.

The instantaneous output voltage is equal to the intercept between levels K and A in Fig. 21.20. However, it is much easier to visualize the waveshape of

E KA

by using terminal

point. Thus, in Fig. 21.21,

A

as a reference

we show

the line volt-

21.21

shows

whenever the value

£d

.

that the

inductor stores energy

rectifier voltage

This energy

is

exceeds the average

then released during the brief

when the rectifier voltage is less than £d The peak inverse voltage across each diode

interval

equal to the peak value of the line voltage, or

.

is

V2£


240

120

•

360

487

£ d = 1.654 £

/d.

0-

0-

+

'a

'd-

+ /d

.

0

+ /d J 0

-/h

Figure 21.20 Voltage and current waveforms

The 3-phase, 6-pulse ment over

in Fig.

21 .19.

rectifier is a big

the 3-phase, 3-pulse rectifier.

tutes the basic building block of

most

improveIt

consti-

large rectifier

installations.

Another way of looking tifier is to

21 .22).

at the

3-phase bridge rec-

imagine the diodes to be

The box

is

output terminals

in

a box (Fig.

fed by three ac lines and

K

it

has two

and A. The diodes act

like

automatic switches that successively connect these

terminals to the ac lines.

made

in six distinct

The connections can be

ways, as shown

in Fig. 21.22.

It

follows that the output voltage

EKA

is

composed of

segments of the ac

That

is

why we draw

line voltages.

line voltages in Fig, 21.21 instead

of line-to-neutral

voltages.

Each dotted connection a diode that

is

conducting.

in Fig.

21.22 represents

The successive 60-de-

gree intervals correspond to those in Fig. 2

1

.20.

For


488

L

stores energy

L

releases energy

1.35

£

1414 E

Figure 21.21 Another way of showing

EKA

using

line

voltage potentials. Note also the position of

E2N

with respect to the line volt-

ages.

0-60

(

180° -240*

(d)

60°

-

120

(

240°

-

300'

(e)

120°- 180'

300°

-

360°

(0

Figure 21.22 Successive diode connections between the 3-phase input and dc output terminals of a 3-phase, 6-pulse

rectifier.


example, from 300° to 360°, because

Dl and D5

flowing, diodes

lows from Fig. 21.19 that to line

while

1

A

K

is

are

5

Fundamental

-

we can

small,

is

required in

Does

b.

if

the peak-to-peak

not to exceed 5 percent.

choke modify

the presence of the

the

E KA ?

Solution

3-phase bridge rectifier has to supply power to a

V

dc load.

If

600

a

Using Eq. 21.5, we have

a.

60 Hz

V, 3-phase,

=

npple

available, calculate the following:

a.

Voltage rating of the 3-phase transformer

b.

DC current per diode

c.

PIV

d.

Peak-to-peak ripple its

21-4, is

peak-to-peak ripple in the output voltage

Example 21-4

is

Example

ripple in the current

from the 3-phase source.

feeder

X 60 Hz = 360 Hz

Calculate the inductance of the smoothing choke

a.

line represents a loss-free

The dc power absorbed by the load must therefore be equal to the active power drawn

360 kW, 240

6

Example 21-5

connection.

A

ripple frequency

fol-

effectively connected to line 2.

is

each dotted

that

i

It

effectively connected

Because the diode voltage drop

assume

and

z,

are conducting.

489

_

—

0.17P

360 000

0.]_7_x

WL

60 X

WL =

across each diode

204

J

and

in the output voltage

Consequently, the inductor must store 204

frequency

magnetic

field.

The inductance

J in its

found from

is

Solution a.

Secondary

line voltage

wL

is

E = Ed /1.35 = = 177 V ratio

of 600 V/l 77

V

L =

dc load current

b.

/d

dc current per diode peak current

in

each diode

= =

360 kW/240

=

1500

1500/3

The presence of

b.

the

= 500 A

33

1

PIV

1.414

K and A.

affect the

remains

It

at

V peak-to-peak.

= 500 A 21

A

Effective line current,

.1 1

.225

E

and

1

.4

1

4

E

X

We

saw

sist

of

177

fluctuates

(Fig.

21

.2

1

).

between In other

words, the voltage fluctuates between

line current

1.225

X

177

1.414

X

177

The peak-to-peak

= 217 V = 250 V

and

in Fig.

21.20 that the ac

line currents con-

20-degree rectangular waves having an am/d ,

where

/ d is

the dc current flowing in the

load. Let us direct our attention to the current I b

flowing

in line

neutral voltage

and

E min = E max =

1

plitude

The output voltage E KA 1

mH

across each diode

- \J2E = = 250 V d.

4

10

choke does not

voltage ripple between

fundamental c.

X

1.8

0.18

satisfactory.

The primary and secondary windings may be connected either in wye or in delta.

ul

= ^L(1500) 2

204

would be

]

2

240/1.35

Thus, a 3-phase transformer having a line volt-

age

=

is

it

2 and to the corresponding line-to-

E 2 n- They

are

shown

in Fig.

21.23

can be seen that the rectangular current wave

symmetrically located with respect to the sinu-

soidal voltage

maximum.

In other words, the center

of the positive current pulse coincides with the peak ripple

is,

therefore,

of the positive voltage wave. Thus,

Wto-pcak = 250 - 217 =

33

V

sidered to be

''in

phase" with

£ 2N

.

Ib

can be con-


490

Thus, owing to the presence of harmonics,* the

^2N

fundamental component

/

/f

K

'7 60

3

/ F is slightly less

effective value of the line current

*\\ 240

y

360

21.12 Distortion power factor

180 Vf,

We

have just seen

/ F is in

that the


phase with the corresponding line-to-neutral

voltage (Fig. 2 .23). Consequently, 1

Figure 21.23 and

current

line

in

phase 2

of

Fig. 21.20.

The

6-pulse rectifier ition,

effective value / of the rectangular line cur-

rent can be

power

clined to say that the

Line-to-neutral voltage

power

power

factor

is

factor

is


—

power

apparent power

power

active 1

X

- // X

180°

120° effective voltage

therefore

= V 20 /1 80

/

I

= This effective current

rms component

As we have

nents.

/F

0.816 is

EI (21.6)

d

composed of

plus

the

all

seen, /F

is in

is

phase with the

the value of F ?To calculate

\

EI

3

But according

line-

it

we

effective current

/

to Eq. 2

1

=

.8, Iv

factor

=

Thus, the actual power factor

95.5%. The reason

is

0.955

reason

95.5%

is

due

to the load is

As

is

100%

not

that the line current

a result,

but only is

power

rectan-

factor of

to distortion in the current.

Although the power factor of our

The dc power

/,

0.955

gular and not sinusoidal. Thus, the

as follows:

X V3

/,

\ 3

power

a fundamen-

harmonic compo-

X

*\ _ «,\3 =

^ /d

to-neutral voltage.

What

in-

100 percent. However, by defin-

active

=

we would be

factor of the 3-phase,

deduced from the relationship I

tal

than the

/.

rectifier

is

less

than 100%, the fundamental component of current is

nevertheless in phase with the line-to-neutral

voltage. Consequently, this ideal rectifier absorbs

The

active ac

load)

power supplied

P Because no power fier,

to the rectifier (and

its

no reactive power from the

line.

is

it

follows that

ilc

= V3

is lost

£/F

(8.9)

21.13 Displacement power factor,

or stored in our ideal recti-

P ac = Pd We can .

total

therefore write

power factor

In Fig. 21.23, the

V3£/F = £d /d = .35 1

and so

Combining Eqs. 2

/F 1

.6

=

and 2 /F

=

0.78 1

.7

rent

fundamental component of cur-

phase with the line-to-neutral voltage.

in

However, E!d

in later circuits

rectangular current

we

wave can

will discover that the shift so that

it

lags be-

hind the line-to-neutral voltage. This causes the /d

we

0.955

is

(21.7)


/ F to shift

along

find /

(21.8)

Harmonics

are discussed in detail in Chapter 30.

wi,th

it.


This angular shift of the fundamental component of current with respect to the line-to-neutral voltage

is

called displacement, and the cosine of the angle

is

called displacement

ment power factor

power factor of

power

in Fig.

The

factor.

21.23

is

unity.

rectangular

5th, 7th,

The

other words,

total

a load or electrical installation

power

The

is

of

3.

=

all

is

(2

1. 9a)

EI L

21.23 contains the

in Fig.

harmonics, and so

odd harmonics

feature of these harmonic

fundamental

I¥

order of the harmonic. For example,

com-

divided by the

is

given by:

17th harmonic has an

Displacement power factor

=

—P

rent

the funda-

if

mental component has an rms value of

1

500 A,

rms value of 1500/17

The degree of distortion of an (21 .9b)

forth; in

that are not multiples

that their respective amplitudes are equal

factor

The displacement power

are the rms values of the

in the line current.

wave

to the amplitude of the

—P

etc.,

1th, 13th, 17th,

1

The remarkable

ponents

factor

,

,

,

displace-

given by the expression:

Total

in which / HA /HB /HC harmonic components

491

=

the

88 A.

ac voltage or cur-

defined as the ratio of the rms value of all the

is

harmonics divided by the rms value of the fundamental component. This total harmonic distortion

In these equations,

P = active power per phase [W] E = effective value of voltage per phase [V] /L = effective value of line current including =

THD

THD

Example 21-6

wave of Fig. 2 .23 occurs very frequently in power electronics. It is therefore worthwhile to examine it more closely, parcurrent

ticularly as regards

its

1

harmonic content.

First,

The 3-phase, 6-pulse rectifier in Fig. 21.19 furnishes a dc current of 400 A to the load. Estimate, for line 1

a.

periodic current in a line can be expressed by the

b.

2

c.

=I 2F +/ 2H

(21.10) d.

which / fF

=

rms value of the

= rms

/H

= rms

line current

value of the fundamental com-

ponent of

mea-

The

effective value of the fundamental

compo-

a.

line current

value of

all

the

equal to the

ual harmonics.

sum

Thus,

total

2

write

The

=

effective or

0.816

c.

Id

=

0.955

/H 7

(2i.ua)

/

=

the line current

X 400 = 326 A

0.816

the fundamental

-

0.955

The rms value of the

.

'iWiY\+'riB+'ric+/ HD+--

=

rms value of

The rms value of 7h

harmonic content

of the squares of the individ-

we can

1

/

harmonic comb.

can also be shown that the

The peak value of the 7th harmonic The rms value of the 7th and 1th harmonics combined

Solution

ponents combined

Ip\ is


nent of line current I

It

The

sured by an rms hook-on ammeter

any

equation

in

(21.11b)

should refer to Chapter 30.

line current [AJ

21.14 Harmonic content and The rectangular

7,1

/ F and / H are defined as before. For more information on harmonics, the reader

effective value of fundamental

component of

given by the formula

is

where

the fundamental and harmonics [AJ /h

(THD)

X 326 =

7th harmonic

V? =

311/7

(21.6)

is

31

is

1

is

= 44A

The peak value of the / H7 = 44 V2 = 62A

A


492

The rms value of the

d.

=

311/11

28

1th

1

is

/ F /l

=

1

pulse

£e

to the gate (Fig. 2

plications,

The rms value of the 7th and bined

—

harmonic

A 1

harmonics com-

1th

J 7

(H7 + H1

1)

=

+ ^ = 44 + f 1

H7

7

2

consequently, {H7 + H) n

HI

1

2

28

ode current

= 2720

/ falls to zero, after

when

all

an-

which the gate again

exerts control.

= V2720 = 52 A

same way a

Basically, a thyristor behaves the

diode does except that the gate enables us to

THETHYRISTOR AND THYRISTOR CIRCUITS

ap-

as conduction starts, the gate loses

further control. Conduction will only stop

2

2

some

.25b). In

1

useful to prolong the pulse for

several milliseconds.

As soon

given by 2

is

it

conduction precisely when ingly slight advantage

is

we want

initiate

This seem-

to.

of profound importance.

It

enables us not only to convert ac power into dc

power, but also to do the reverse: convert dc power

21.15 The thyristor*

into ac

A

thyristor

an electronic switch, similar to a

is

diode, but wherein the instant of conduction can be controlled. Like a diode, a thyristor possesses

anode and a cathode, plus a

third terminal called a

gate (Fig. 21.24). If the gate

is

connected

The 21.25a). To

positive.'

is

blocked (Fig.

thyristor initiate

is

said

power. Thanks to the development of

SCRs we have

change

if

the

to

be

conduction, two

lists

Consider Fig. 21 .27a connected

number of The anode must be

b.

A current / a

positive.

must flow

some of

the properties of typical

See also Figure 21

.26.

21-16 Principles of gate firing

sistor are

conditions have to be met: a.

reli-

fundamental

a

the control of large blocks of power.

in

Table 21 B thyristors.

witnessed

to the

cathode, the thyristor will not conduct, even

anode

an

able

few microseconds. In practice, the current is by applying a short, positive voltage

which a

amplitude

provided the anode

is

thyristor

and a

re-

across an ac source.

short positive pulses

gate, of sufficient

into the gate for at least

in

in series

£

is

to initiate

positive.

A

applied to the

conduction,

These pulses may be

generated by a manual switch or an electronic con-

injected

trol circuit.

A

Referring to Fig. 2 .27b, the gate pulses occur 1

anode

I

angles 0 h 6 2 B 3 6 4 and 0 5 Table 2 1C explains ,

,

.

,

the circuit reacts to these pulses.

or

The reader should

cathode

I

follow the explanations carefully.

K

Figure 21 .24 Symbol of a thyristor,

at

how

SCR. /

*

Thyristor lable

is

a generic term that applies to

to refer specifically to the

commonly This

is in

the terms

called

When

SCR

SCR

the

When

(semiconductor controlled

the wording,

anode

the gate

it

in this

book

rectifier).

in the literature to

(b)

(a)

use

and thyristor interchangeably.

is

is

we

adopt the following terms:

positive with respect to the cathode,

simply say the anode 2.

4-layer control-

reverse-blocking triode thyristor.

response to a general trend

To simplify 1.

all

semiconductor devices. However, we use

is

is

Figure 21.25 A thyristor does not conduct when the gate

a.

positive.

is

con-

nected to the cathode.

positive.

positive with respect to the cathode,

simply say the gate

we

we

b.

A

thyristor

conducts when the anode

and a current pulse

is

is

positive

injected into the gate.


TABLE

21

Relative

B

power

PROPERTIES OF SOME TYPICAL THYRISTORS /,

[A]

medium high very high /j

/cr

[A]

60

500

I

500

1200

000

1200

8

no 1200

- maximum effective - peak value of surge

E2

/C r

10

[V]

Ep

[V]

-10 -5 -20

/G

[mA]

£r,[V]

7j [°C]

d [mm]

peak inverse gate voltage gate current corresponding to

2.5

105

11

33

1.25

125

27

62

50

1.5

125

58

27

peak inverse anode voltage positive gate voltage to initiate conduction

EG

T - maximum junction temperature }

d - diameter - length /

Figure 21.26 of SCRs from medium to very high power capacity. Average current: 50 A; voltage: 400 V; length less thread: 31mm; diameter: 17 mm. Average current: 285 A; voltage: 1200 V; length less thread: 244 mm; diameter: 37 mm. Average current: 1000 A; voltage: 1200 V; distance between pole-faces: 27 mm; overall diameter: 73 mm.

Range b. c.

(Photos courtesy of International

Rectifier)

493

[mm]

50

current for one cycle

Eq — Iq -

/

50

current during conduction

E2 Ep -

a.

493


494

(b)

(a)

Figure 21.27 Thyristor and

a. b.

TABLE

connected to an ac source. depends on the timing of the gate

resistor

Thyristor behavior

DESCRIPTION OF THYRISTOR BEHAVIOR (SEE

21 C

Angle or time

]

Although the anode thyristor

angle 0! 180°

FIG. 21.27)

Explanation of circuit operation

interval

zero to 0

e, to

pulses.

behaves

Conduction

starts

positive, conduction

is

like

is

impossible because the gate voltage

Conduction continues even though the gate voltage has

we

zero.

The

because both the anode and gate are positive.

ther effect once the thyristor conducts.

consequently,

is

an open switch.

The anode

to

fallen to zero.

Gate pulses have no

cathode voltage drop

is less

than

1

.5

fur-

V;

can consider that the anode and cathode are shorted. The thyristor behaves like

a closed switch.

angle 180° 180° to 360°

The

62

360° to 540°

thyristor current

Conduction ,

it

is

produces no

Conduction

zero, conduction ceases,

is

impossible because the anode effect.

and the gate regains control.

negative. Although the gate

is

triggered at angle

thyristor experiences an inverse voltage during this half-cycle.

and ceases again as soon as the current

starts at 6 3

more than during

The

is

is

the first positive half-cycle. Consequently, the

zero.

The

gate pulse

is

delayed

anode current flows for a

shorter time.

720°

to

900°

Conduction

starts at

angle 6 5 but the resulting anode current ,

is

very small because of the long

delay in firing the gate.

To summarize Table 21C, we can

control the

21 .17

Power gain

of a thyristor

current in an ac circuit by delaying the gate pulses

with respect to the If the

start

of each positive half-cycle.

pulses occur at the very beginning of each

half-cycle, conduction lasts for tor

behaves

1

80°,

like an ordinary diode.

hand, with a resistive load layed, say, by 150°

—

if

and the

On

thyris-

the other

the pulses are de-

—current only flows during

remaining 30° of each half-cycle.

the

When

a voltage pulse

is

applied to the gate, a certain

gate current flows. Because the pulses last only a few

microseconds, the average power supplied to the gate is

very small, in comparison to the average power sup-

plied to the load.

The

ratio

of the two powers, called

power gain, may exceed one gate input of only

1

million. Thus, an average

W may control a load of 1000 kW.

FUNDAMENTAL ELEMENTS OF POWER ELECTRONICS An SCR

The

does not, of course, have the magical

two solutions

first

property of turning one watt into a million watts.

amine the

The large power actually comes from an appropriate power source, and the SCR gate only serves to control the power flow. Thus, in the same way that a small power input to the accelerator of an auto-

current source

mobile produces a tremendous increase

in

motive

SCR

power, so does a small input to the gate of an

produce a tremendous increase

in electrical

power.

third

72 , the net

control only after the anode current falls to zero. quite naturally (as

did at the end of each cycle in Fig. 2 1 .27) or force

it

we

can

Such forced commutasome circuits where the anode

current has to be interrupted at a specific instant.

Consider Fig. 21. 28a load resistor

source E.

If

in

which a

definitely thereafter. tion in the

1

.

thyristor

and a

R are connected in series across a dc we apply a single positive pulse to the

gate, the resulting dc load current

SCR

in

7,

will

However, so long

flow

in-

one of 3 ways:

Momentarily reduce the dc supply voltage

unchanged. But

E to

zero.

Open

3.

Force the anode current to zero for a brief period.

by means of a switch.

Ql

we

if

For example,

thyristor.

is

conducting and

circuit has

been

is

not.

state values.

Fig. 2

1

We assume that the

It

full

voltage appears across load R. Thus,

and

72

-

(Fig. 2

1

Ql,

in

way and

b.

Forced commutation.

1

= EIR

7,

we

trigger thyristor

charge the capacitor

7,

=

high resistance restart

in the

oppo-

so point 1 will eventually reach the

When

the transients have sub-

Current

0.

smaller than load current

a dc source.

in

.29b). This causes the capacitor to dis-

will quickly

R

{)

shown

72 7,

in Fig.

2 .29d, 1

can be made much

by using a relatively

.

conduction

in the load, in

we Fig.

fire

Ql,

21.29e.

Discharge current 7C now causes the extinction of Q2, and the capacitor charges up with the opposite polarity as

to

Q

charge in the circuit formed by C, Ql, and Q2. The discharge current 7C forces Ql to stop conducting and produces the condition shown in Fig. 21.29c. The level of point 1 drops to E volts below the level of point K, with the result that 7, reaches a momentary peak of 7, = 2 EIR. Current

To

Figure 21.28 Thyristor connected

shown

0.

To stop conduction

producing the condition shown

a.

C is charged to

follows that capacitor

Neglecting the voltage drop across

.29a.

but with

R

firing

long enough so that the

in operation

level of point 3.

®

is

the circuit operates, suppose

Q2

sided, the circuit appears as

(a)

it

in Fig. 21 .29a, a load

the supply voltage E, with polarities as

site

Ql

increase 7 2 until

voltages and currents have reached their steady-

7]

®

as the net cur-

the thyristor stops conducting, and the

Ql and Q2. To understand how

Q2

2.

Ql

in-

in the

thyristors

However, we can stop conduc-

the load circuit

flowing

72 )

can be switched on and off by alternately

it

to zero artificially.

tion is required in

connected

with the result that the current flowing in the resistor is

ond

may cease flowing

—

(/,

is

gradually

current pulse, usually supplied by triggering a sec-

ceases to conduct and the gate regains

current

current

As we

gate regains control. In practice, 7 2 can be a brief

21.18 Current interruption and forced commutation

The

us ex-

not zero, the thyristor continues to conduct,

is

equal to

A thyristor

C delivering a current 72

thyristor decreases.

rent

let

method. In Fig. 2 1. 28b, a variable

with thyristor Ql.

in parallel

crease

are trivial, so

495

become

shown

in Fig. 21.29f.

When

conditions

stable, the circuit reverts to the

started with,

namely

Fig. 21.29a.

one we


496

©

©

T

I

I

©

r

Q2

Q2

1

Q1

©

©TlL—

©

©

©

® (c)

(b)

(a)

©

©

©

I

r E -±-

©

it

1

/,

©

©

©

©

Q2

©

Q1

©

•I

J-

(0

(e)

Figure 21 .29

A discharging rent

in

capacitor

Cand an

auxiliary thyristor

load flcan be switched on and

off

Q2 can

This type of forced commutation, using a com-

mutating capacitor,

is

employed

ers* that generate their the availability of GTOs, largely eliminated the

in

some

convert-

own frequency. However, MOSFETs, and IGBTs has

need to use thyristors

in

such

force-corn mutated applications. For this reason, in the following discussion of thyristor

we

power

circuits,

consider only those involving line commutation.

21.19 Basic thyristor

power Thyristors

are

circuits

used

in

many

different

ways.

power electronics, six basic circuits cover about 90 percent of all industrial applications. These circuits, and some of their applications, are However,

force-commutate the main

by triggering Q1 and

in

Q2

in

listed in

To explain circuits,

is

into

power of another frequency, including zero

quency

(dc).

a rotating

any device

that converts

A converter may

machine.

be a

power of one

rectifier, inverter,

.

.

I,

cir-

circuit 6.

the principle of operation of these ba-

we

will

use

single-phase

sources.

single-phase examples are less complex, and they

enable us to focus attention on the essential principles involved.

21.20 Controlled rectifier supplying a passive load (Circuit 1, Table 21 D) By

definition, a passive load

inherent source of energy.

load

is

is

one

that contains

no

The simplest passive

a resistor.

connected

A converter

.

In practice, 3-phase sources are mainly used, but

Fig. 21 .30a

quency

Ql.Thus, the cur-

Table 2 ID. They are labeled circuit

cuit 2, circuit 3,

sic

thyristor

succession.

shows a

in series

resistive load

and a

thyristor

across a single-phase source. The

fre-

source produces a sinusoidal voltage having a peak fre-

or even

value line

Em

.

The

gate pulses are synchronized with the

frequency and,

in

our example, they are delayed


lAbLb

21

oUMb

V

bJAblO

H YHIb

1

Circuit No.

UH KUWbH

1

L/IHUUI

I

497

o

Thyristor circuit

Typical applications

1

Controlled rectifier supplying a passive load

Electroplating, dc arc welding, electrolysis

L

Controlled rectifier supplying an active load

Battery charger, dc motor control, dc transmission line

•\

LlIlC-LUIllIllUldlCU lIlvcILCl Slippiyillg all

dCUVC dC lUdU

r\\^ IIIOIOI COI1LIU1,

WUUIIU-IOLOI

II1UIOI

SpccU

control, dc transmission line

AC

4

static

Spot welding, lighting control, ac motor speed

switch

control, ac starter

Low-speed synchronous motor

Cycloconverter

5

control, elec-

troslag refining of metals

High-voltage dc transmission, synchronous mo-

Three-phase converter

6

tor drive

by an angle of 90°. Conduction

is

every time the ac voltage reaches

therefore initiated

its

maximum

posi-

tive value.

Based upon explanations given

21.16,

obvious that current will flow for 90°.

it

is

In Fig. 21 .30b,

it

is

90 degrees. This

Section

seen that the current lags be-

hind the voltage because nal

in

it

only flows during the

lag produces the

same

fi-

effect as

an inductive load. Consequently, the ac source has to

supply reactive power

power P (see Section power factor decreases pulse.

On

Q

in

addition to the active

The displacement

7.13).

as

the other hand,

we

if

delay the triggering

the

SCR

is

triggered at

21.21 Controlled rectifier supplying

an active load (Circuit

2,

Table 21 D) Fig. 21 .31

Em and a dc load Ed SCR in series with an inductor. The

shows an ac source

connected by an

load (represented by a battery) receives energy be-

cause when the thyristor conducts, current the positive terminal.

/ enters

Smoothing inductor L

limits

the peak current to a value within the

SCR

Gate pulses E„"

an angle

initiate

conduction

at

rating.

(Fig. 21.31b).

zero degrees (the start of the cycle), no reactive

power

is

absorbed by the

rectifier.

/

®

!

T

/T"\

(A

tt

S""""^

'90

©

360f""iî terminal© chosen /

180

V

\ \

as reference

/

delay

1 v

(a)

Figure 21.30 a.

SCR

b.

Voltage and current waveforms.

supplying a passive load.

(b)

terminal® chosen as reference

8,


498

Using terminal

as a zero reference potential,

1

follows that the potential of terminal 2

above

it.

If the

above and below the level of

SCR

were replaced by a diode, conduction

at

the anode

angle 0 O because this

becomes

positive.

is

the instant

However,

in

our

example, conduction only begins when the gate

As soon

K jumps from

and voltage latter

A

The current then gradually decreases and becomes As at angle 6 3 where A _ is equal to A + soon as conduction 'stops, point K jumps from level

zero

A

EA2

is

as conduction starts,

level 2 to the level of point

A,

appears across the inductor. The

begins storing volt-seconds, and current / in-

at

02

i

The volt-seconds reach a maxwhere area A + is maximum. The cor-

,

(

responding peak current / max

is

)

given by

=A

(

+ /L

As

ted line. lags

K

I

is

)

until the

shown by

the dot-

(Fig. 21.30), the load current

behind

Em

voltage

;

conse-

quently, the source again has to supply reactive

power

Q as

rent

/.

power P

well as active

A

firing angle a, area

We can

i

+

)

=

(a

we reduce

therefore vary the active

where a

0),

From

If

is

the

increases, and so does cur-

plied to the load from zero (a

measured

=

power

a,) to a

starting

sup-

maximum

from 9 0

.

a practical point of view, the circuit could

be used as a variable battery charger. Another application

(2.28)

in circuit

displaced)

(is

{

)

of point 2 and stays there

to the level

creases accordingly.

imum

.

,

next gate pulse. The level of

fired at 6, degrees.

point

it

volts

1.

would begin

when

Ed

Furthermore, the potential of terminal

oscillates sinusoidally

terminal

lies

is

to control the

motor. In this case, the armature, and

speed and torque of a dc

Ed represents the counter-emf of

L

the armature inductance.

21.22 Line-commutated inverter (Circuit 3, Table 21 D) An

inverter,

power. fier,

are 1

.

It

by

changes dc power into ac

which converts ac power

into dc power.

recti-

There

two main types of inverters: Self-commutated inverters (also called /o/x^-

commutated 2.

definition,

performs the reverse operation of a

inverters) in

which the commutation

means are included within the power inverter Line-commutated inverters, wherein commutation is effected by virtue of the line voltages on the ac side of the inverter In this section

we examine

the operating principle

of a line-commutated inverter. The circuit of such an inverter

is

identical to that of a controlled rectifier,

except that the battery terminals are reversed (Fig. 21.32a). Thus, the potential of terminal 2 lies below that of terminal

1

(Fig. 21 .32b).

Because current can

only flow from anode to cathode, the dc source delivers

On

power whenever

Ed

the thyristor conducts.

power P must be abwe assume no

the other hand, this

sorbed by the ac terminals because

Figure 21.31 a.

SCR

b.

Voltage and current waveforms.

supplying an active load.

losses in the inductor or the thyristor." Consequently,

the circuit of Fig. 21 .32a vert dc

power

is

potentially able to con-

into ac power.


499

/

/

©

—

© Q\

^>

,

e

.

3

EA

k

®

©

(a)

(b)

Figure 21.32 Line-commutated inverter. b. Voltage and current waveforms.

a.

The power converter

consists therefore of a sim-

ple thyristor and inductor that connect the dc source

However,

to the ac load.

the ac side erates is

its

must be an existing ac system

own ac

in operation.

that gen-

voltage, whether or not the inverter

For example, the ac system could be

an electric

that of

important to note that

it is

utility

company composed of

hundreds of generators and thousands of loads that are easily able to absorb the additional

livered

by the

power P de-

To increase the current and hence the active power flow, we simply advance the firing angle 9 This causes A + to increase. However, this process ,

(

cannot be carried too cease,

A

(

To achieve power conversion,

the peak ac volt-

far.

mum area that A

In order for conduction to

A

However, the maxithat bounded by the wave between G 0 and 6 3 and the

must equal

_>

+

(

.

)

_ can have

i

s

)

(

trough of the sine

,

horizontal line of point 2 (Fig. 21. 32b). ing angle

is

larger; but if

inverter.

)

able value of

advanced, it

A

(

+

becomes

_ (

thyristor has to be triggered within a precisely de-

reached.

tially at

the level of terminal

must be applied either prior

K. Since

to

2,

the triggering pulse

must be triggered prior

Suppose the

SCR

is

A

is ini-

to 6 0 or after 6 3 (Fig.

become

2 1. 32b). For reasons that will soon the gate

K

to 0 O

clear,

The dc

,

angle 0 3

still

£d /d

at

0 O equal to area ,

A

(

+ /L. The current then gradubegin to build

up.

Conduction stops

at 0 2

,

when

A

(

_ }

=A

(

+

.

)

For the

initiated af-

inverter conditions, the current

has to supply reactive power

the inverter. Consequently, site directions in

ally falls as the negative volt-seconds

is

peaks lag behind the positive voltage peaks, and so

inductor accumulates volt-seconds until angle 0 o

reached. Thus, the resulting current reaches a peak

63

.

Under normal

K immediately jumps from level 2 to level A and the is

trip.

same reason, conduction must never be

the ac source

.

triggered at 6 degrees. Point

when angle

current will then build up with

each cycle, until the circuit breakers

ter

and

avail-

conduction will never stop. In

,

essence, current / will not be zero

must be positive with respect

maximum

}

age has to be greater than the dc voltage, and the

fined range. First, to initiate conduction, anode

the fir-

larger

)

should exceed the

A

As

where

P and Q

/d is the

average value of current

is

Q to

oppo-

P =

/.

into the ac terminals

from sinusoidal and a

ance) ac system

in

an inverter. In our example,

The current pulses flowing are far

flow

stiff

(low-imped-

needed so as not

to distort the

sinusoidal voltage. However, the pulses do contain


500

component that is in phase with the The effective value of this given by 7 p = EJJE, where E is the

a fundamental

sinusoidal voltage £, A

component

is

.

effective value of the ac voltage. In practice, ap-

added

propriate filters are

to ensure that the cur-

rent flowing into the ac line soidal.

We

should

also

is

reasonably sinu-

bear

mind

in

that

current can flow in both directions (Fig. 21.33).

The ac current flowing

of gates gl and g2. Thus,

the line, even

Fig. 21.32.

voltage.

4,

is

ac static switch

connected

is

composed of two

thyristors

in antiparallel (back-to-back), so that

However,

The

the load

is

purely resistive.

that the current

is

displaced behind the

if

The well-known triac used in domestic lightdimming controls is an example of such an electronic switch. If the gates are fired at

An

the gate pulses are syn-

such delayed firing will draw reactive power from reason

static switch (Circuit Table 21 D)

if

lesser ac current will flow in the load.

line-commutated inverters always involve 3-phase

AC

R can be a

chronized with the line frequency, a greater or

systems and not the simple single-phase circuit of

21.23

load resistor

in the

precisely controlled by varying the phase angle

the static switch the other hand, in the

if

0° and

in the fully

is

neither gate

open position. Thus, a

1

80° respectively,

closed position.

is fired,

static

On

the switch

is

switch can be

used to replace a magnetic contactor. In contrast to

©

magnetic contactors, an electronic contactor solutely silent and

A

ab-

is

contacts never wear out.

\

/ i

its

'

\

360

\\y (b)

CDFigure 21.33c potential levels

Single-phase, water-cooled contactor

two Hockey Puk

CD-

1200 A (RMS)

at

2000

V;

4.5 L/min at 35° max. For intermittent

welding applications,

20 cycles. Width:

Figure 21.33 a.

Electronic contactor.

b.

Waveforms

with a resistive load.

composed

of

Continuous current rating: cooling water requirements:

thyristors.

1

this unit

(10%

duty) spot

can handle 2140

A for

75 mm; length: 278 mm; depth:

mm. (Photo Courtesy of International Rectifier)

1

14


21,24 Cycloconverter (Circuit Table 21 D)

ing).

5,

,

g

A cycloconverter produces low-frequency ac power directly from a higher- frequency ac source. A simple cycloconverter

is

shown

in Fig. 21.34. It consists

three groups of thyristors,

of

power

To understand pose

all

1

,

g2, g3, g

circuit

behaves

Figure 21.35 Typical voltage output of a cycloconverter.

g6 g4

g1

N

(Fig. 21.35).

The

identical to that of Fig. 21.17.

T thyristors Q4, Q5, by 4 similar pulses g4, g5, g6, g4. This makes terminal 4 negative with respect to N. The f

firing process is then repeated for the

1 cycle

g5

is

result, the

and terminal

fired

Figure 21.34 Elementary cycloconverter.

g4

As a

like a 3-pulse rectifier


Q6, are

the operation of the circuit, sup-

gT

such a way that the thyristors func-

During the next interval

to a resistive load R.

g2 g3

in

if

is

the gates of thyristors

by 4 successive pulses

waveshape of E4N

thyristors are initially blocked (nonconduct-

g1

1 ,

T,

are triggered

they were ordinary diodes.

tion as

4

mounted back-to-back

and connected to a 3-phase source. They jointly supply single-phase

Then, for an interval

Ql Q2, and Q3

501

g2

g3 g1

g4

g5

Ql, Q2, Q3


502

and so on, with the

thyristors,

result that a low-

frequency ac voltage appears across the load. The duration of sine

cycle

1

is

2

T

seconds.

Compared

wave, the low-frequency waveshape

poor.

when

However,

means

to a

rather

flat-topped and contains a large 180

It is

ripple

is

3-phase

the

this is

frequency

is

of secondary importance because

are available to

improve

it.

Returning to Fig. 21.35 and assuming a 60 source,

Hz

60 Hz.

we can show

sponds to 540°, on a 60 therefore, (540/360)

X

that

Hz

each half-cycle corre-

Hz base. The duration of T is, (1/60)

=

responds to a frequency of 1/(2

0.025

X

s,

0.025)

which

cor-

= 20 Hz.

Obviously, by repeating the firing sequence gl, g2, g3, gl,

.

.

.

,

we

could keep terminal 4 positive

21.25 3-phase, 6-pulse controllable converter (Circuit 6, Table 21 D) The 3-phase, 6-pulse the

thyristor converter

most widely used

one of

is

units

rectifier/inverter

in

Due to its practical importance, we will explain how it operates in some detail. As in all 3-phase converters, the waveforms become power

electronics.

rather complex, although not particularly difficult to

understand.

Even

the

simplest circuits yield

chopped voltages and currents each other, and everything that

it

is

taxes the

that pile

mind

to

on top of

keep track of

going on. Consequently,

we

will

keep the waveforms as simple as possible, so as

to

we wish, followed by an equally long negative period, when g4, g5, g6, g4 are fired. In this way we can generate frequencies as low as we

connected to the secondary winding of a 3-phase

The high end of the frequency spectrum is 40 percent of the supply frequency.

cal to the rectifier circuit of Fig. 21.19, except that the

for as long as

.

please.

.

.

limited to about

The reader should

also note that this cycloconverter

can supply a single-phase load from a 3-phase system, without unbalancing the 3-phase lines. Later,

we

will

encounter cycloconverters that

highlight the basic principle of operation.

Three-phase, 6-pulse converters have 6 thyristors

transformer (Fig. 21.36). The arrangement

diodes are replaced by thyristors. Because initiate

conduction whenever

tors enable us to vary the

we

identi-

is

we can

please, the thyris-

dc output voltage when the

converter operates in the rectifier mode.

The con-

can produce a sinusoidal, low-frequency 3-phase

verter can also function as an inverter, provided that

output from a 60 Hz, 3-phase input.

a dc source

Figure 21.36 Three-phase, 6-pulse thyristor converter.

is

used

in

place of the load resistor R.


21.22. There

21.26 Basic principle of operation

We

works

in the rectifier

2 .37. In this figure, the 1

enclosed

box, where they successively switch

in a

the output terminals K, 3.

how the conmode by referring to Fig. six SCRs are assumed to be

can gain a basic understanding of

verter

The load

is

A to the

ac supply lines

l

,

2,

represented by a resistor in series with

an inductor L. The inductor

is

assumed

to

have a very

large inductance, so that the load current Id remains

two thyristors Q 1 Q5 located between terminals K- and A-2 are conducting. constant. In Fig. 2 1 .37a, the

,

1

A moment

later,

and A-l conduct are similarly

steps

Q4

the thyristors Q2, (Fig. 21.37b).

between K-2

The other

switched, in sequence.

have been completed, the

thyristors

When

these

entire switching cy-

The reader will note that the dc current 7d the ac lines. However, Fig. 2 .37 shows that

cle repeats.

flows in

it

is

a true ac current of amplitude /d

that the current in

brief intervals.

one of the three

For example,

.

It is

and so

also evident

lines

is

zero for

in Fig. 21.37, there is

however, an important difference.

The thyristors can be made to conduct at precise moments on the ac voltage cycle. Thus, conduction can be initiated when the instantaneous voltage between the ac age

is

high or low.

lines is either

If the volt-

low, the dc output voltage will also be low.

Conversely,

if

line voltage is

when

the thyristors conduct

momentarily near

its

the ac

peak, the dc

output voltage will be high. In effect, the output voltage E KA is composed of short 60-degree segments of the ac line voltage. The average value of

E KA becomes In

£d

the dc output voltage

examining Fig. 2

1

.37,

it

positive. This tive

e l2

power is

must be so because the

to the load.

positive

when

Knowing how

/d

For example,

flows

.

can be seen that the

current always flows out of a line that

1

the current in each line reverses periodically,

is,

503

is

line

momentarily

line delivers acin Fig. 21.37a,

in the direction

shown.

the thyristor converter behaves as

a rectifier, the question arises; to operate as an inverter?

how

can

it

be made

Three basic conditions

have to be met. First, we must have a source of dc current /d Such a current source can be provided if a voltage .

momentarily no current

in line 3.

The switching sequence we have just described is

similar to that of the diode bridge rectifier of Fig.

source

E0

is

connected

in series

tance (Figs. 2 .38a and 2 1 .38b). 1

fb)

(b)

Figure 21.37

mode (see Fig. 21 .36) Q1 and Q5 conducting. Q2 and Q4 conducting.

Figure 21.38

Rectifier

Inverter

a.

a.

b.

b.

mode

(see Fig. 21.36)

Q1 and Q5 conducting. Q2 and Q4 conducting.

with a large induc-

504


Second, the converter must be connected to a

voltage polarities in the inverter

3-phase line that can maintain an undistorted sinu-

even when the line current is nonsiThe voltage may be taken from a power

The reader can ^see

soidal voltage,

nusoidal. utility,

as before,

or generated by a local alternator.

Third, to force

power

must be switched so

Figs. 21.37

into the line, the thyristors

that current / d flows into

and

an ac

when

low. Conversely,

inverter operation can best be understood

The SCRs enclosed in the box are arranged the same way as in Fig. 2 .37. In other words, the converters in the two figures are absolutely identical. Looking first at the dc side, the dc current / d must flow in the same direction as to Fig. 21.38.

cannot conduct

shown.

In other

E must be connected {)

the ac side, the 3-phase line

terminals

1

,

is

is

2 .38, because current

/d

1

is

if

E

is

{)

the dc voltage

low, the thyristors

is

is

high, the thyris-

its

line voltage

is

peak.

wish

make one

to

final

Consequently,

whose average value

voltage must be equal to

drop across the inductor

keeping current

On

important observa-

/d

£0

is

£ KA is is £ d .

a fluctuating

This average

because the dc voltage

negligible. In addition to

constant and almost ripple free,

the inductor serves as a buffer between the fluctuat-

simply connected to

must be properly timed so

receives power. This

minal that

in

ing voltage

£ KA

and the constant voltage

£0

.

2, 3.

We are now ready to fire the thyristors. the firing

Indeed,

same

the instantaneous ac voltage

must be triggered when the ac

line voltages.

words, the posiA.

fired

near

voltage

in reverse.

to terminal

.

The voltage that appears between terminals K and A is composed of 60-degree segments of the ac

{)

tive side of

/d

the

tion.

On the other hand, because we want the dc source E to deliver power, I d must flow out of the positive terminal, as

is

only the instantaneous

tors

We

1

SCRs

is

it

dc supply voltage

If the

line frequency.

before because

line currents

line voltages that differ.

must be

The

that the line current alternates

and 21.38;

must therefore be precisely synchronized with the

by referring

are consis-

mode.

has a peak value equal to

it

waveshape of the ac

the

momentarily positive. The gate firing

line that is

mode

tently opposite-to those in the rectifier

However,

consistently

done

in Fig.

always flows into an ac

momentarily positive. Note

—

21.27 Three-phase, 6-pulse rectifier feeding an active load

that the ac line

Consider the

ter-

circuit

Q1

k

Q3

k

Q5

/a

i

1

3-phase

3-phase

/b

Transline

1

former Ic

-d>1

'4

Figure 21.39 Three-phase, 6-pulse

rectifier.

of Fig. 21.39

in

which a

3-phase, 6-pulse converter supplies power to a load.

that the line

Q4

.06

<.|:

.02


The load sistor

R

composed of

is

converter

is

a dc voltage

and a

Ql

Q6

to

The

some

because anode

time, so that

hand,

Ql

is

Then,

triggered by gate pulse gl.

curs and

Ql

At 60°

that

thyristor

Q2

fired

is

successive waves.

Q6

SCR's conduct therefore,

rectifier

which

to Fig. 21.39,

we can

thyristors are conducting at

The converter

Let us

fired.

Two

til

K

and

A

a glance

Ed =

all

Q

1

6 0 will continue to flow in

at

1

.

,

K

Commutation jumps from

A similar switching action

Note

that the triggering delay

an inductor, the dc voltage between points 4 and

A

for a full 120° and each voltage

also 1.35 E. Consequently, the dc current ld

is

ration of

is

Ql

Q2

Q3

Q4

Q5

Q6

Q1

fires

fires

fires

fires

fires

fires

fires

-Q5,

Q6—

conducting

Figure 21.40a Delay angle: zero.

line 3 to

The

K

resulting

and

A

ripple-free,

is

still

conducts

segment has a du-

60 degrees. Furthermore, the current

mains constant and

un-

does not shorten

the conduction period; each thyristor

given by

Q5

occurs,

takes place (but at

later times) for the other thyristors.

1.35 E.

a

instead of switch-

/t „

in

is

ensure

triggering pulses by an angle

choppy waveshape between terminals shown in Fig. 21.40b.

and the average

is

delay

no appreciable dc voltage drop

Because there

to

mode

gate pulse gl triggers Ql.

line

any given time.

acts as a rectifier

or dc voltage between

practice, the

at the desired instant.

and the potential of point

on. Thus,

tell at

now

ing over to

are labelled so as to

which they are

Q1-Q2, Q2-Q3, Q3-Q4, and so

by referring

is fired. In

of 15° (Fig. 21.40b). Current

a time; the conduction pairs are,

at

the other

(along with Q6) will

made wide enough

commutation occurs

Q2.

to

follows the peaks of the

The thyristors

indicate the sequence in

Q5

,

On

if

start

21.28 Delayed triggering—

This switching process continues indefinitely and,

K

gl fires after 0 ()

and the resulting com-

mutation transfers the load current from

as in Fig. 21.21, point

()

Thus,

in Fig. 21.40a.

conduction cannot

Commutation oc-

conducting, taking over from Q5.

starts

if

triggering pulses are

0° point (0 O ), thyristor

at the

(21.12)

then negative.

is

1

,

continue to conduct until gl

are conducting, carrying load current /d

(Fig. 21.40a).

shown

gl fires slightly ahead of 0

conditions are stable. Initially, suppose thyristors

Q5 and Q6

-E0 )/R

(Ed

triggering time has to be fairly precise to obtain

the rectified voltage

are triggered in succes-

operation for

in

=

fd

We assume that the con-

sion at 60-degree intervals.

been

re-

The

fed from a 3-phase transformer.

gates of thyristors

verter has

E0

with a smoothing inductor. The

in series

505

due

to the

re-

presence

Figure 21 -40a

Delay angle: zero.

506


of the big inductor. The level of point

K follows the

E=

tops of the individual sine waves, but the average

voltage

£d

,

than before.

between

K and A,

is

We can

prove that

it

Ed =

age

a =

obviously smaller is

effective value of the ac line-to-line volt-

|V

given by

According 1.35

1

firing angle [°]

Ecosa

smaller as

(21.13)

to Eq. 21.13,

to or less than £"0

where

Ed

,

dc voltage produced by the 3-phase,

is

6-pulse converter [V]

cause the

smaller than

E0

SCRs

.

However,

Delay angle: 15°.

Figure 21.40c Delay angle: 45°.

Q2

Q3

Q4

Q5

Q6

Ed becomes equal

would reverse when

Q1

Ed

this is impossible, be-

can only conduct

Figure 21.40b

Q1

if

the load current flows intermit-

tently. Ordinarily, the current

Ed =

becomes smaller and

a increases. However,

in the

forward


507

Figure 21.40d Delay angle: 75°.

direction.

We

will not study the condition

of

The dc load current

inter-

mittent current flow. Figs. 21.40c

tween

K

and

A

for

Note

that the ac

large,

compared

a = 45° and

component

to the dc

in

now

is

very

500

2

H Calculate the power supplied

b.

voltage

£d = =

to the load for

= Because

The dc output voltage of the converter

£d =

£ cos a - 1.35 X 480 cos = 626 V

is

is

15°

negligible, the IR drop across the 2

nal resistance

£d

is

63

=

39.4

A

is

kW

1.35

£cos a

1.35

X 480 X

167.7

less than

in reverse.

This

is

is

cos 75°

V

£0

,

the current tends to

impossible and, conse-

simply zero and

so, too,

the power.

inverter

O interIf triggering is

is

E — £d — £0 = 626 - 500 =

63

21.29 Delayed triggering

Because the dc voltage drop across the inductor is

flow

quently, the current

1.35

=

is

triggering delays of (a) 15° and (b) 75°.

a.

126/2

With a phase angle delay of 75°, the converter

dc source having an internal resistance of

Solution

therefore

to the load

P = Ed Id = 626 X

component.

a

= EIR =

The power supplied

75°, respectively.

E KA

Example 21-7 The 3-phase converter of Fig. 21.39 is connected to a 3-phase 480 V, 60 Hz source. The load consists of

V

/d

and 21.40d show the waveform be-

is

mode

delayed by more than 90°, the voltage

£d developed by the converter becomes negative, according to Eq. 21.13. This does not produce a nega-

126

V

tive current because, as

we

said,

SCRs

conduct

in


508

—

\

Q3

Q1 'a

1

1

3-phase

3-phase

/b

Transline

(

former /c

-
Q4

Q6

4

Figure 21.41 Three-phase, 6-pulse converter

in

the inverter mode.

only one direction. Consequently, the load current

simply zero. However,

Fig. 2

is

we can force a current to flow

1

shows

.42

by connecting a dc voltage of proper magnitude and

ated by the inverter

polarity across the converter terminals. This external

reaches a

voltage

E0

must be

slightly greater than

for current to flow (Fig. 21.41).

Ed

in

order

The load current

= (E0 - Ed )/R

the load

,

is

actually a source, delivering a

power output

P = EJd

pated as heat

in

mainder

is

.

Part of this

power

the circuit resistance

If

we

R

is

dissi-

and the

re-

secondaries of the

delivered to the

3-phase transformer.

SCR

power P ac

that is

we

are left with a net active

original rectifier has

now become

an

in-

verter,

converting dc power into ac power. The tran-

sition

from

rectifier to inverter

is

smooth, and

re-

quires no change in the converter connections. In the rectifier

mode,

the firing angle lies

and 90°, and the load the inverter 1

Ed = —

1

It

E at a fir-

.35

80°.

1

mode, the

may

between 0°

be active or passive. In

firing angle lies

between 90°

80°, and a dc source of proper polarity

provided.

triggering angle of a given thyristor

rectifier

must be

The

between 15° and 90° and

is

usually

thyristor acts as a as an inverter be-

tween 90° and 165°. Under these conditions, the dc voltage developed reaches 15° and 165°;

The

it

is

zero

triggering angle

rectifier

its

maximum

value

at

at 90°. is

seldom

mode. The reason

is

that

less than 8° in the

sudden

line volt-

age changes might cause a thyristor to misfire, thus

producing a discontinuity

delivered to the 3-phase line.

and

angles

given by Eq. 21.13.

is still

value of

kept between 15° and 165°.

subtract the small trans-

former losses and the virtually negligible

The

ing angle of

The

Because current flows out of the positive termi-

losses,

maximum

at firing

The dc voltage Ed gener-

21,30 Triggering range /

E0

waveshapes

is

given by

nal of

the

of 105°, 135°, and 165°.

in the

dc output current.

mode we seldom permit the firing angle to exceed 165°. If we go beyond this point, the inverter may lose its ability to switch from one thyristor to the next. As a result, the currents build In the inverter

up very quickly

some 1

until the circuit breakers trip. In

cases, the firing angle

is

not allowed to exceed

50°, to ensure an adequate safety margin. Fig. 2

firing

1

.43

shows

the allowed

and forbidden gate

zones for a particular thyristor

in a 3-phase,


Triggering

Figure 21 .42a sequence and waveforms with a delay angle

of

105°

Figure 21.42b sequence and waveforms with a delay angle

of

135°

Triggering

6-pulse converter. Specifically, Fig. 21.39.

The other

thyristors

it

refers to

have similar

Ql

in

firing

zones, but they occur at different times.

changes both

upon

tor has 1

21.31 Equivalent circuit of a converter ^

We may

think of a converter as being a static ac/dc

motor-generator set whose dc output voltage £j

.

2.

in

magnitude and

the gate pulse delay.

some

polarity,

However,

509

depending

the dc genera-

special properties:

only one direction.

It

can carry current

It

produces an increasingly large ac ripple

in

volt-

age as the dc voltage decreases.

The analogy may be represented by Fig. 21 .44, in which

the circuit of


510

one

cycle-

Figure 21.43 Permitted gate

•

firing

E. xc represents the

•

Ed

•

ec

is

is

zones

for thyristor

Q1

3-phase line voltage.

•

the dc voltage generated by the converter. the ac voltage generated

by the converter

on the dc side (mainly the 6th and monics).

1

2th har-

•

D

is

in

only one direction.

a diode to remind us that current can flow

The dotted that active

line

between

£ac and Ed

power can flow

indicates

in either direction be-

tween the ac and dc systems.


5

1

I

oA

I

Figure 21.46

Figure 21.44

Equivalent circuit of a 3-phase thyristor converter

Equivalent circuit of a thyristor converter.

in

the inverter mode. •

Unlike a motor/generator

set,

the dc and ac sys-

tems are not electrically isolated from each

other.

angle of 45°.

The

thyristors flow for

When the converter is operating as a rectifier, the equivalent circuit is shown in Fig. 21.45. When operating as an inverter, the circuit

21.46.

The to

given by Fig.

ac voltage generated by the converter

appears across inductor

sumed

is

L.

Its

inductance

is

as-

be sufficiently large to ensure an almost

ripple-free dc current. It

will also

rents, as

be recognized that the currents flow-

we have

dc current

harmonic cur-

already seen (Section 21.14).

.

/

2,

/

when

4

i

,

/

5,

6 in the

equal

is

This holds true for any firing an-

The only

a plain 3-phase diode rectifier (Fig. 2 .20). 1

difference

is

that they flow later in the cycle.

The waveshapes of the corresponding ac

line cur-

found because they are equal to the

difference between the respective thyristor currents.

Thus, referring to Fig. 2

1

These

have a peak value

flow

line currents also

in positive

.39, line current

/.,

=

/,

—

i

4

.

/d but they ,

and negative blocks of 20°. 1

effect of the ac line currents

is

im-

portant because they usually flow in the windings of a converter transformer.

The I~R

shows the voltage and current waveshapes

we know

depends upon

loss

the effective value / of the current. Fig. 2 1 .47

/

3,

rents in a thyristor converter are identical to those in

The heating

21.32 Currents in a 3-phase, 6-pulse converter

Id

/,,

20°, and their peak value

between zero and 180°. Consequently, the cur-

gle

rents are easily

ing in the 3-phase lines are not sinusoidal. Thus, on the ac side, the converter generates

to the

currents 1

From

Eq. 21.6

that

the converter functions as a rectifier at a firing /

The

=

0.816/d

(21.6)

effective value of the ac line current

is,

there-

dc output current and

is

unaffected by the firing angle. Clearly, the same

is

fore, directly related to the

true

when

the converter operates as an inverter.

21.33 Power factor

We recall

that in the 3-phase, 3-pulse diode rectifier

(Figs. 21. 16

Figure 21.45 Equivalent circuit of a 3-phase converter fier

mode.

and 21.17), the currents

and 3 are symmetrical with respect in

the recti-

in lines

neutral voltages. Thus, rectangular current actly

in

the

1,

2,

to the line-to-

middle of the positive

i

£2 n

2 is

ex-

wave.


512

In essence (and in actual fact),

£9 N This .

two

is

i

2 is in

phase with

also true for the currents in the other

lines, as regards their respective voltages.

This

Referring

now

to Fig. 21.47,

has been delayed by 45°, currents have

all

we

where triggering

note that the thyristor

been shifted (displaced) by 45°

to

reflected back into the primary of the

the right. Consequently, the line currents lag the re-

transformer, and from there to the 3 -phase feeder.

spective line-to-line voltages by 45°; the displace-

Because the currents are

ment power factor

condition

is

in

phase with the voltages,

the displacement power factor the rectifier

is

1

00%. As

a result,

draws no reactive power from the

line.

(cos 45°

=

is

no longer unity but only 0.707

means that a converter abpower from the ac system to which it

0.707). This

sorbs reactive


is

connected. This

is

1

whether the converter op-

true

erates as a rectifier or inverter. is

5

The

reactive

power

given by

£=

/>tana

(21.14)

where

Q=

power absorbed by

reactive

the converter

[var]

P =

dc power of the converter (positive for a rectifier,

a =

negative for an inverter) [W]

triggering angle

The reader

f°]

waveshapes of the cur-

will note that the

rents in Fig. 21.47 are the

same

as those in a con-

ventional 6-pulse rectifier (Fig. 21.20).

Figure 21.48

See Example

1

b.

for a triggering angle of 15°,


b. c.

c.

d. e.

The displacement power factor The reactive power absorbed by The total power factor

1.

Calculate a.

Example 21-8 In Example 2 -7, and

21-1

the converter

The dc current carried by each SCR The dc voltage generated by the inverter The required firing angle a The effective value of the ac line currents The reactive power absorbed by the inverter

Solution a.

Each

SCR

carries the load current for one-third

of the time. The dc current

is,

therefore,

Solution a.

The displacement between

current and the line-to-neutral voltage


/

the fundamental line

factor

is

a =

b.

The

a =

=

cos 15°

0.966, or

power supplied

active

96.6%

to the converter

P = Ed Id = 39 .4

E0

less the

Q = P tan a

c.

Knowing 12

-

10.6kvar

the inverter

is

IR drop. Thus,

is

Hence

39.4 tan 15

900/3

The voltage Ed generated by equal to

kW

=

=

7/3

= 300 A

15°.

is

b.

cos

=

000

=

16 000

- 900 X

=

15 100

V

1

that the effective ac line voltage

V, the firing angle can be

is

found from

Eq. 21.13:

c.

The

total

factor

X

power

factor

distortion

=

displacement power

power

15

Ed = 100 =

factor

cos

=

0.966

-

0.923

X 0.955 = 92.3% This

Example 21-9

A 1

16

kV

*

if

-

12 supplies

900

A

to a 12 kV, 3-phase, 6-pulse,

(Fig. 21.48).

£ cos a X 12 000

.35

1.35

a =

0.932

a -

21.2°

the firing angle that

cos

a

would be required

the converter operated as a rectifier.

,

dc source having an internal resistance of

60 Hz inverter

is

1

However, because

it

is in

the actual firing angle

a = 180 -

the inverter mode,

is

21.2

=

158.8°


514

The

d.

effective value of the ac line current

is

45° /

* \

=

0.816

d

=

0.816

X 900

(21.6)

(a)

= 734 A -

The dc power absorbed by

e.

the inverter

is

—

=

15 100

=

13.6

-

120°

^

Q3

—

^conducting

^

conducting

X 900 commutation

MW

overlap

actually negative because the inverter ab-

is

— —

conducting

P = EJa

P

120°

45°

sorbs dc power; hence,

P = -13.6MW

(b)

-

The

reactive

power absorbed by

the inverter

120°

— Q5

is

conducting

Q = PVdna = -13.6 =

(21.14)

"conducting

Figure 21.49 a.

In practice, the actual reactive

the calculated value,

due

to

power is higher than b.

commutation overlap.

Instantaneous commutation 45° (see Fig. 21.58).

in

Section 21

.9,

that the current in a

three-phase rectifier cannot switch instantaneously

from one diode

to

the

process takes time and this

Thus,

The commutation

next.

also true for thyristors.

is

in a six-pulse converter,

from Ql

to

Q3

Q3

followed by

taneous (as assumed

the

to

Q5

in Fig. 21.47),

shown in Fig. 21.49b. The transfer of /d from one

commutation is

but

a

rectifier

The commutation overlap delays build-up by angle

mentioned

in

when a =

Same conditions with commutation overlap of 30°, showing current waveshapes in Q1, Q3, Q5.

21.34 Commutation overlap

We

-

tan 158.8

Mvar

5.27

_^

conducting

Q3

u. It

by the same angle. Owing fective firing angle

is

to these delays, the ef-

somewhat

greater than the

power

triggering angle a. This reduces the

the converter in both

modes.

It

the current

also delays the current cutoff

the

rectifier

factor of

and inverter

Edc

also reduces the average dc voltage

.

not instan-

is

more

like

21.35 Extinction angle

that

is

thyristor to the next

effected during the so-called commutation over-

lap period, defined by angle

u.

The amount of over-

lap varies with the current /d At full-load, u lies typically between 20° and 30°. At light load it can be .

as small as 5°.

On

account of commutation overlap,

the current in each thyristor flows for a period of

120

+

u degrees instead of 120°, as

we have

as-

sumed so far. The commutation overlap modifies the waveshape of £AK but we will not examine this ,

aspect of converter behavior.

We

have seen

inverter

that

mode,

when

it is

be initiated prior to

a converter operates in the

very important that conduction

a =

1

80°.

an ideal inverter flows for

1

Because the current

also cease before the angle of (180 is

reached.

tation

The

and 300°

21 .50).

recover

interval is

+

120)

=

300°

between the end of commu-

called the extinction angle

The extinction angle permits its

in

20°, the conduction must

blocking ability before

y

thyristor its

(Fig.

Ql

anode

to

(1)

again becomes positive with respect to the cathode K. The value of y lies typically between 5° and 20°. 1


515

Figure 21.50

Waveshape ity

of ^ in thyristor

before the

ode.

The

critical

figure also

Q1

shows

we

From

between angles

often define the

by the angle of advance

the angle of delay a.

shown

a delay angle a. The extinction angle 7 permits Q1 to establish its blocking abilreached. At 300° the anode of Q1 becomes positive with respect to its cath-

is

the relationship

In the case of an inverter,

ing instant

for

angle of 300°

(3,

fir-

rather than

Fig. 21.50

by

can be

it

that the following relationships exist be-

tween the commutation angle the angle of

advance

(3,

0 = p =

u,

the delay angle a,

and the extinction angle 180 w

- a

+ 7

7:

(21.15)

we have

tronic switching

accomplished by

thyristors.

and

that

u.

one of

its

tion only stopped

naturally to zero.

shortcomings was that conduc-

when

We

the

Although

anode current dropped

it is

possible to force the

anode current

to zero

mentioned

Section 21.18, the additional circuit

in

components make

by special techniques, such

this solution

expensive. Another problem is

limited to a

To overcome

maximum

this

is

as

cumbersome and

that thyristor switch-

of about 2 kHz.

problem, special semiconductor

switches have been developed whereby conduction

can be

initiated or

blocked by controlling the gate cur-

These devices are constantly be-

ing improved upon, but

studied circuits in which the elecis

saw

7,

rent or gate voltage.

21.36 Semiconductor switches far

(3,

ing

(21.16)

DC-TO-DC SWITCHING CONVERTERS

So

a,

we

will limit our attention to

those that are most frequently used.

As mentioned

the introduction to this chapter, they are

GTOs,

in

bipo-


516

power MOSFETs, and IGBTs. These

available.

Most

controllable on/off switching devices enable us to de-

far within

me

lar transistors,

Fig. 21.5 lb

sign dc-to-dc and dc-to-ac converters of extraordinary

The basic

versatility.

principles of these switch-type

converters are explained

Thyristor

from

and

GTO

in

Basic Characteristics Apart

GTOs are

very similar to ordinary thyristors. The characteristics

of both these devices


the current

in the

Thus,

on and off states are

in the off state,

the

maximum

bands

(Fig. 2

bounded by

limits

is

anode current

E AK

,

up

to

the cross-hatched

During the on

1. 5 la).

thyristor conducts, the figure

voltage drop

when

zero the thyristor can withstand both

is

forward and reverse blocking voltages

when

state,

shows

the

£ AK

that the

about 2 V, and the upper limit of the / AK is

again indicated by the cross-

hatched band. These bands merely indicate the

maximum

broad-brush

shows

that

GTOs

values that are currently

3

V compared to 2 V for thyristors. As in

GTO

a thyristor, conduction in a

below

holding

the

GTO

However, the

is

tive current into the base for a

To ensure

off

thyristor

GTO.

which the anode

few microseconds.

pulse has to be about one third the value of the an-

GTOs

ode current.

are high-power switches,

some

of which can handle currents of several thousand

amperes

BJT

at

voltages of up to

4000

terminals

base.

V.

Basic Characteristics The bipolar junction or BJT, is

due

is

designated bipolar because

to the migration of both electrons

named

When

transistor has three

collector C, emitter E, and base

The

to emitter is initiated

and maintained

/ B to

flow into the

operated as a switch, the base current

must be large enough

to drive the

BJT

into satura-

Under these conditions, the voltage ECE between the collector and emitter is about 2 to 3 volts, tion.

(a)

at rated collector current.

as the base current

is

Conduction ceases as soon

suppressed.

The characteristics

J3 V 3 kA

GTO

|/AK

/max « 500 Hz

C

,|/c

holding current

/G

GTO

£AK 4kV (b)

Figure 21.51 Typical properties

and

and approximate limits on and off states.

thyristors in the

B

collector current /c that flows

by causing a sustained current

E AK 4kV

4kV

in-

extinction, the amplitude of the gate

from collector

G_

current

by

current can be blocked by injecting a strong nega-

(Fig. 21.52). holding

of the

current

a device in

and holes within the device. The |/AK

initiated

is

about

der to keep conducting, the anode current must not fall

conduction

3kA

is

the case of

jecting a positive current pulse into the gate. In or-

transistor,

thyristor

are able to with-

stand forward voltages but not reverse blocking voltages. Furthermore, the voltage drop

the ensuing sections.

their important gate turn-off feature,

thyristors are designed to operate

Hmits shown.

of

GTOs


and approximate

limits of

BJTs.


B JT

of the

in the

on and off

shown

states are

in Fig.

21.52, together with the approximate limits of the collector-emitter voltage Ic

.

Note

ECF

that the transistor

ECF Power

values of

and collector current

cannot tolerate negative

transistors can carry currents

.

ECE volt-

of several hundred amperes and withstand

ages of about

1

kV.

To establish

collector currents of

100 A, the corresponding base current about

typically

is

MOSFET

Basic

The

Characteristics

de-

vice having an anode and cathode, respectively called drain D, source S, and gate

The drain current /D maintaining a voltage

EGS

is

initiated

£GS

G

(Fig. 21.53).

by applying and

of about

1

V

2

between

and the source. Conduction stops whenever

below a threshold

falls

limit (about

1

V).

The

gate currents are extremely small; consequently,

power is needed to drive this electronic The characteristics in the on and off states shown in Fig. 21.53, together with typical max-

very

little

switch. are

imum /D

.

limits of drain voltage

The

E os To within

£DS

and drain current

MOSFET cannot tolerate negative values of meet

this requirement,

it

has incorporated

a reverse-biased diode, as

it

identified the

namely istics

same way

the

on and off

also a

is

are

terminals

as those in a transistor,

and base. The character-

collector, emitter,

in

IGBT

whose

switch

shown

states are

in Fig.

21.54, together with the limiting voltages and cur-

The

rent.

collector current in an

higher than

in a

symbol for the device. Power

shown

MOSFETs

in the

can carry

drain currents of about a hundred amperes and with-

stand

ECE

when

driven into saturation, the

voltages of about 500 V. At rated current,

ranges from about 2

V

E DS

Compared

power

MOSFET is a voltage-controlled three-terminal

the gate

Basic Characteristics The

voltage-controlled

MOSFET.

IGBT

is

Consequently, the

much

IGBT

can handle more power.

A.

1

IGBT

517

voltage drop

to 5 V.

BJTs,

to

GTOs, an important

MOSFETs, and IGBTs

is

feature of

their fast turn-on

and turn-off times. This enables these switches used

at

much

As

higher frequencies.

to

be

a result, the as-

sociated transformers, inductors, and capacitors are

smaller and cheaper. Typical are

shown

in Figs.

maximum

frequencies

21.51 to 21.54. Another advan-

tage of high-speed switching

semicon-

that the

is

ductor switches can generate lower-frequency volt-

ages and currents whose waveshapes and phase can

be tailored to meet almost any requirement.

21.37 DC-to-DC switching converter In

some power systems

there

is

a need to transform

dc power from one dc voltage level to either a higher or lower dc level. For example,

in

a public

4000 V dc overhead drive a 300 V dc motor

transportation system, a

line

may

in

be the source to

bus. In other cases, a 12

power

V

a device rated at 120

battery

V

may have

a to

dc. In alternating-

current systems the voltage step-up or step-down

can easily be done with a transformer. But

in

dc sys-

100 A

D| I'D

MOSFET / max - 200 kHz

—#m

E DS

1

MOSFETs.

eg

kV


f

2

///// b \

/max ~ 50 kHz

E GS

,

and approximate

limits of


and approximate

limits of

IGBTs.


518

terns,

an entirely different approach

is

required.

When

It

the switch opens (Fig. 21.57) the current

involves the use of a dc-to-dc switching converter,

collapses and ^11 the stored energy

sometimes called a chopper.

the arc across thaiwitch. At the

Suppose

that

power has

high-voltage dc source load

E One {)

.

solution

is

to be transferred

Es to

from a

to a lower-voltage dc

this

open and close

rent

to

the circuit periodically with a switch (Fig. 21.55). In

order to follow the transfer of energy, assume the

T During this inductor is £ s — E0

voltage across the ties as

shown

{

.

,

lates volt-seconds,

is

rapidly discharging the volt-seconds

with polari-

decreases very quickly.

/

in-

Although some energy source there

is

£s

to the load

,

and the switch

(when is

/

has reached a value

/a

about to open), the current

=

(£ s

- E0 )T

}

/L

shown

effect

to the

because

Figure 21.56 Energy is stored

in

inductor.

the inductor.

loss

by adding a diode

in Fig. 21.58.

its

cathode

When the switch

in the

(2 8)

ceeds

is

The diode has

positive with respect

anode and so the diode does not conduct.

the switch opens, current

high value

Figure 21.55 Energy transfer using an

the

closed,

is

.

x

is

W= 2 U

from is

therefore poor.

closes, the current rises to 7 a as before.

ducing a voltage e inductor

is

,

(21.17)

The corresponding magnetic energy stored

transferred

is

while the switch

We can prevent this energy

no /a

E0

it

result, the current

a great loss of energy every time the switch

opens. The efficiency

(E s - E 0 )t

to the circuit as

1

it

was increasing (compare Figs. 21.56 and The high negative voltage indicates that the

inductor

L say,

opposite to what

had previously accumulated. As a

creases at a constant rate given by

After a time 7

The polarity of was when the cur-

collapsing so quickly.

is

interval, the

and the resulting current

=

is

voltage

time, a high

induced across the inductor because

21.57).

The inductor accumu-

in Fig. 21.56.

is x

the current

connect an inductor be-

tween the source and the load and

switch closes for a time

voltage e

dissipated in

is

same

it

/

again begins to

When

fall, in-

However, e L cannot jump

reached before because as soon as

the

to the it

ex-

anode of the diode becomes positive

'

*

Figure 21.57 Energy is dissipated

in

the arc. Note the polarity of e L

Figure 21 .58 Energy transferred without

loss.

.


and so the diode begins diode voltage drop

E0

.

Because

ductor

is

E

Assuming

follows that e

it

the

=

x

constant, the voltage across the in-

is

()

to conduct.

negligible,

is

from

also constant. Starting

current

/a ,

/

therefore falls at a uniform rate given by

=

i

l

a

can calculate

.

T2 because

V

s

.

T must x

zero and riod

we have

1

=

as a

Instead of letting the load current swing between

the volt-seconds

Referring to Fig. 2 .59,

during charge period

known

21.38 Rapid switching

T

equal the volt-seconds released during the dis-

T2

is

after a time

accumulated during the charging period charge interval

what

inductor, and diode constitutes

(2.28)

The current eventually becomes zero

T2 We

The switch is actually a GTO, MOSFET, or 1GBT, whose on/off state is controlled by a signal applied to the gate. The combination of the electronic switch, step-down dc-to-dc converter, or buck chopper.

^

~

519

x

/a

,

we can

(as in Fig. 2

desired value

/a

.

close the switch for a short pe.59) until the current reaches the

1

We

then open and close the switch

rapidly so that the current increases and decreases

V-s during discharge

by small increments. Referring

period

switch

to Fig. 21.60a, the

T and open

closed for an interval

is

during

x

A (+) = A — EJ

(£ s

T\

_ (

an interval

During

When delivered

- EJ T, ^—^

open, the load cur-

lower value

to a

/.,

flows

/b

When

same

rate as

did in Fig. 21.59.

it

the current has fallen to a value / b

,

the

switch recloses. Because the cathode of the diode the current all its

is

zero, the inductor will

stored energy to load

E

have

(

{)

.

taneously, the diode will cease to conduct.

Simul-

We

and repeat the cycle indefinitely. Consequently,

),

flowing, and the source

T

now

current then builds up and ]

is

the current in the diode immediately stops

can

therefore reclose the switch for another interval

circuit enables us to transfer

now +

.

in the inductor,

and the freewheeling diode. The current de-

creases at the

(21.18)

peak value

its

this interval, current

the load,

^

from

rent falls

()

Consequently,

T2 =

is

.

)

= E T2

(E*

Th When the switch

supplies current

when

it

7b

.

The

reaches the value

(after a time T.J, the switch reopens.

The

free-

this

energy from a high-

wheeling diode again comes into play and the cycle repeats.

The

current supplied to the load fluctuates

voltage dc source to a lower-voltage dc load without therefore between

/.,

(Fig. 21.60b). Its

average or dc value

and the slightly lower value

incurring any losses. In effect, the inductor absorbs

energy livers

at

it

a relatively high voltage (E s lower voltage E0

at a

The diode diode because

— E0

)

and deequal to

7a

,

but the exact value

is

/b

nearly

7() is

given by

.

is

sometimes called a freewheeling

it

automatically starts conducting as

4,

soon as the switch opens and stops conducting

Whereas

when

stant,

the switch closes.

=

(/ a

+

the current in the load

is

is

essentially con-

having only a small ripple (Fig. 21.60b), the

current supplied by the source ries

(21.19)

b )/2

of sharp pulses, as shown

is

composed of a seWhat

in Fig. 21 .60c.

the average value of these pulses?

It is

found by

noting that the average current during each pulse o ;

(duration

A-::.:.,k»-

t

repeat

TJ

is (/ a

average current

7S

+

/ b )/2

=

70

.

Consequently, the

during one cycle (time T)

is

the cytte

+ T +t

T2

Is

=

70

(TJT)

that is

Figure 21.59

E and

/

in

the inductor of Fig. 21 .58.

7S

=

70

D

(21.20)


520

If

we

above equation, we

substitute Eq. 2 1 .20 in the

obtain

•EL

Tin

source

i ^

|

/D

load -=

which gives the important relationship

= DES

£„

1

(21.21)

where Figure 21.60a Currents in a chopper

E0 = £s =

circuit.

dc output voltage of the converter [V]

dc voltage of the source [V]

D-

duty cycle

Equation 21.21 signifies that the dc output voltage

EQ can be controlled simply by varying the duty cycle D. Thus, the converter behaves like a highly efficient

dc transformer

Figure 21.60b Current

in

as

which the "turns

ratio"

this ratio

electronic switch, such as an

— 7b—

and off If

D. For a

is

can be changed

needed by varying the on time of the switch. In practice, the mechanical switch

......

la

in

given switching frequency,

the load.

at a

frequency that

more power

is

may be

required, a

is

replaced by an

IGBT It can be turned on as high as

50 kHz.

GTO is used, wherein the

frequency could be of the order of 300 Hz.

T

L,

Figure 21.60c

Example 21-10 The switch in Fig. 2

Current pulses provided by the source.

quency of 20 Hz and remains closed for

where

cycle.

load

= = o T.A = T= D= s

The

a.

dc current absorbed by the load A]

dc ammeter connected

3

in series

ms

per

with the

70 A,

indicates a current of

If

a dc

ammeter

is

connected

in series

[

on time of the switch period of one cycle

duty cycle

circuit

waveshapes tinuous, they

by

E0

.60a opens and closes at a fre-

dc current drawn from the source f AJ

=

source, what current will

[s]

b.

[s]

T.JT

of Fig. 21. 60a shows the current

in the source, the load,

still

What

is

it

indicate?

the average current per pulse?

Solution a.

Using Eq. 21

.20,

we have

and the diode.

Although the waveshapes are choppy and disconstant

A

1

obey Kirchhoff s current law,

period

T

_J_

"

in-

duty cycle

instant.

=

Ta

T

Turning our attention to the power aspects, the

50 ms

20 3 •

~ 50

dc power drawn from the source must equal the dc

power absorbed by is no power loss in freewheel in 2 diode.

the load because, ideally, there the switch, the inductor, or the

We

can, therefore, write

=

70

X

=

4.2

A

0.06

0.06

with the


The average

b.

T.J is

is

current during each pulse (duration

70 A. Considering

that the

/c

average current

only 4.2 A, the source has to be specially de-

signed to supply such a high 70 cases a large capacitor

A pulse.

In

most

2400/600

=

4

A

To

calculate the average current in the diode,

we

refer to Fig. 21 .61a. Current /()

/s

connected across the

is

= PIE,

521

was found

is

20

A and

4 A. By applying

to be

Kirch hoff's current law to the diode/inductor terminals of the source.

It

the high current pulses as

can readily furnish it

junction, the average diode current /D

discharges.

=

/n

Example 21-11

We

wish to charge a

1

20

V battery from a 600 V dc The average is

16A

battery cur-

20 A, with a peak-to-peak ripple of

2 A. If the chopper frequency

c.

The duty cycle

200 Hz, calculate

D=

the following:

b. c.

d.

The The The The

is

=

Eq/Es

T= a.

*s

= 20-4

source using a dc chopper. rent should be

/.0

120/600

=

-

ms

1//'= 1/200

5

Consequently, the on time

dc current in the diode duty cycle

Ta = DT =

X

0.2

Ta

5

is

ms =

1

ms

inductance of the inductor

The waveshapes of /s and / D 21.61c and 2

circuit

diagram

is

shown

desired battery current fluctuates

between 19

average of 20 a.

0.2

dc current drawn from the source

Solution

The

A with

120

is

in Fig.

21.61a and the

given in Fig. 21.61b.

A and

It

21 A, thus yielding an

to the battery

V X

20

A-

The power supplied by the source 2400 W. The dc current from the source is

.6

1

are

d, respectively.

shown

interval T.A the average voltage across is

(600

-

120)

= 480

V.

1

(

during the interval

is

)

2 A; consequently,

M = A + /L

therefore,

i

2

=

(2.28)

)

0.48/L

L = 0.24 H

21

19

(0

(a)

21

19

Figure 21.61 Circuit of Example 21-1

a.

Current

in

the load.

1

in Figs.

Note the sharp

The volt-seconds accumulated by the inductor ms = during this interval is A + = 480 V X = 480 mVs 0.48 V s. The change in current

W

is,

During

the inductor

is

2400

1

pulses delivered by the source. d.

a peak-to-peak ripple of 2 A.

The power supplied

P =

b.

is

Figure 21.61 Current drawn from the source.

c.

d.

Current

in

the freewheeling diode.

A


522

The

Thus, the inductor should have an inductance of 0.24 H.

If

a larger inductance were used, the

would be smaller, but the dc ages and currents would remain the same. current ripple

converter, therefore, has the ability to trans-

form the resistance of a fixed

volt-

Thus, the chopper again behaves like a dc trans-

cle.

former

21 .39

resistor to a higher

value whose magnitude depends upon the duty cy-

which the turns

in

ratio

is

D.

Impedance transformation Example 21-12 The chopper in Fig. 2 .62 operates at a frequency of 4 kHz and the on time is 20 |jls. Calculate the apparent resistance across the source, knowing that R 0 .

So far, we have assumed that the converter feeds power to an active load E However, it may also be {) .

used to connect a higher-voltage dc source lower- voltage

load

Equations 21.20 and 21.21

have

the

R0

resistor

additional

apply, but

still

relationship

Furthermore, the apparent resistance terminals of the source

is

E0 Rs

Es

to a

21.62).

(Fig.

we now = I0 R 0 .

across the

1

=

a

12

Solution

The duty cycle

given by

is

D = TJT = TJ=

20 X 10

6

X 4000 =

0.08

Applying Eq. 21.22, we have

We

can therefore write

R s = R^D 1 = 12/(0.08) 2

Rs = /y/s

= /{)

D

D

per.

2

where

R0 =

D=

[fl]

is

two.

The reason

flow

in

is

that a transformer permits

both directions

— from

down chopper we have

is

power

the high-voltage side

low-voltage side or vice versa. The step-

often required,

just studied can transfer

the high-voltage side to the low-

voltage side. Because

duty cycle

to a trans-

an important difference between the

power only from

actual load resistance [fl]

re-

using a chop-

Although a chopper can be compared

to the

apparent resistance across the source

many times by

can be increased

former, there

Rs =

875 il

This example shows that the actual value of a sistor

£(>

I{)

1

power flow

in

we now examine

both directions a dc-to-dc con-

verter that achieves this result.

L

(D

1

21.40 Basic 2-quadrant dc-to-dc converter 1. 63a in which two mechanical S2 are connected across a dc voltage

Consider Fig. 2 switches S

©

J

Figure 21.62 A chopper can make a fixed resistor R0 appear as a variable resistance between terminals 1-2.

source in

I

to

E u The .

switches open and close alternately

way that when S is closed, S2 is open and versa. The time of one cycle is T, and SI is

such a

vice

l

closed for a period

of S SI

l

is

is

D=

T. v It

follows that the duty cycle

TJT, while that of S2

closed, terminal

l

is at

is (

l

-

D).

When

the level of point 3 and


523

S1

Eh

/

S2

Figure 21.63a Two-quadrant dc-to-dc converter.

Figure 21.64 Power can flow from

to be. This

-12

converter.

Terminal Ta

Tb

voltage that

Figure 21.63b of

E12

dc

always

and average value

EL

+

(

to transfer

fluctuating while

is

EL

dc power from

Then, when SI for a period fore,

open, S2

is

Tb The .

between

Eu E

erage dc value

E ]2 = £ H

whose dc

(Fig. 21.64).

Knowing

EQ

is

for a period

E ]2

varying

(Fig.

x

D from zero to

circuit

21.63b) and

between points

flow into terminal

1

,

it

minal 2 either via S2 source is

circulate,

av-

it is

S2

2

I

We

vary the mag-

if

is

L

as

Suppose duty cycle

shown to dc.

2

1,

is

current / happens to its

way back

to ter-

closed) or via S

1

is

EL

that

evident that current its

/

can

al-

to use

It

in-

an

in-

has the ad-

We

EL

,

no dc power exchange

1

EL

,

IL

EH

and 2

is

constant. will

will take place. / L will

magnitude

and

the

dc comIf

E0

flow and

But

if

E0

is

flow from terminal is

given by

= (E^-E0 )/R

(21.23)

to £ L / L will, therefore, flow from ter2 toward the battery. This dc power can

Power equal 1,

1

no dc current

a dc current

into terminal 5. Its

minals

that the load has a

/?.

between points

exactly equal to

less than

assume

fixed. Consequently, the

and

direction happens

is

in Fig. 21.64.

would

the efficiency

both the voltage source

D are

Because one of the switches

no matter what

which would reduce

losses

vantage of opposing ac current flow while offering

ponent apparent that the

is

es-

is

it

could place a

5, but that

of the converter. The best solution

ductance

(21.21)

.

it

R

and

1

small internal resistance

we can

,

can find

(if

E H if S2 is open.

always closed,

its

volve

no opposition

on the left-hand side of terminals

never open. For example,

ways

1

£L from zero to Eu examining Fig. 21.63a,

nitude of In

.

given by Eq. 21.21

is

DE„ By

7a =0

output voltage oscillates, there-

and zero

constant,

.

closed and so

is

2.

ter-

as a battery,

short-circuit currents will result.

is

fixed:

between the two, otherwise

sential to place a buffer

resistor

so the output voltage

remains


)

()

E l2

but the

in either direction,

voltage

E [2 to a load such E52 has a value E

minals

Waveshape

can flow

Suppose we want

T

<

is

1

vice versa.

called a two-quadrant converter be-

/

of the

polarity

FQ and

to

a crucially important feature of the

is

It is

cause current

EH


524

only this

come from the higher voltage source E H mode of operation, with E0 less than E L

In

.

,

the

d=

converter acts like the step-down (buck) chopper

we covered in Section 2 .37. On the other hand, if E0 is

~=20kHz

0.2

1

greater than

£

L

,

a dc

current /L will flow out of terminal 5 and into ter-

minal

1. Its

magnitude

— E L )/R. Power

(E0

the low- voltage battery side

the higher voltage side tion,

=

is / L

E0 to E H In this mode of operawith E greater than E L the converter acts like

now flows from

.

,

{)

a step-up (boost) chopper.

The

mechanical

21.63a

system

switching

therefore able to transfer dc

is

both directions

—from

of

Fig.

power

in

high-voltage side to low-

Figure 21.65 Example 21-13.

Circuit of

voltage side or vice versa. Again, because the cur-

EL

rent can reverse while the polarity of

remains

the same, this buck/boost converter operates in

two quadrants.

now examine

Let us

more

verter

closely,

the behavior of the con-

To determine Assuming

the current

is

given on a buck/boost con-

EH =

switching frequency

30 V R = 2il L = = 20 kHz with a duty

mH

10

A X

=

2 II


|jls.

Let us

and

The value and direction of the dc current / L The peak-to-peak ripple superposed on the dc

/

20

V.

1, it

(jls

20

totals

that

S2

V X

is

40

is

V .jxs/IO mH = now

X

100

V =

20

A

again momentarily 5

Its

/]

average value IL

=

open (40 ns)

is

greater than

£L

,

/ 10

is

100

20 V)/2

closed

O =5A

10

|jls

1

is

1/20

is

000 = 50|xs

closed for a time

and S2

is

closed for 40

T. x

=

(jls.

0.2

X 50

5

|

(40 us)

(/' j

The duration of one cycle

Thus, S

A

2Q

V S2

(30 V -

7 = \/f=

mH

4 1.

(jls

=

=

Figure 21.66

See Example 21-13.

is

800

closed

(Fig. 21.67).

/

cur-

flows out of terminal 5 and into terminal

|jls

when SI

S1

E h = DE H —

V.

Because the battery voltage rent

volt-

closed, the

0.08 A.

see what happens

Solution

0.2

in

Knowing

follows that I L

current

Referring to Fig. 21.65, the value of

its

Therefore the current increases by an amount

A/ = 800

Determine the following:

b.

(

5

seconds and during the 40

V

to

across the inductor

flowing into terminal 4 and that terminal 4

is

+)

V -

magnetic "charge"

cycle

D of 0.2 for SI. a.

E4I

must be increasing. The inductor accumulates

V E0 =

100

closed (Fig. 21.66).

is

momentarily equal

is

equal to the battery voltage minus the IR drop

that / L is

verter (Fig. 21.65):

/

dc value of 5 A, the voltage

by means of an example.

is

when S2

situation

the resistor: 30

Example 21-13 The following data

the peak-to-peak ripple, let us ex-

amine the

A

increasing)

The


The

current flowing into the battery

and

its

value would (45

Direct-current

525

would reverse

become

V-

30V)/2

ft

=

7.5

A

power now flows from

the 100

source toward the battery, causing the

latter

charge up. Under these conditions, the converter said to operate in the buck

from boost

tion

to

mode. Thus, the

V to is

transi-

buck can be effected very

smoothly

by

Fig. 21.65

can be considered to be the mechanical

simply

varying

the

duty

cycle.

equivalent of a buck/boost chopper.

Figure 21.67

21.41 Two-quadrant electronic converter

See Example 21-13.

Figs. 21.66

flow

in

and 21.67 show the direction of current

the case of a boost converter. If the converter

operated in the buck mode, the currents would follow the

same paths but

in the

opposite direction. With me-

chanical switches this creates no problem because

they can carry current real

world

we have

But

in either direction.

in the

to deal with electronic switches,

which inherently carry current

in

only one direction.

Therefore, in order to get bidirectional ity, diodes have to

be placed

in antiparallel

with the respective semi-

conductor switches Ql and switch contacts are

ple,

voltage across the inductor

V=

80

V, but terminal

4

is is

now

100

V —

(30

—

when

tinue

on

source

current flows into terminal

to terminal 2 either

EH

or

negative with respect

1

from the lower voltage battery toward the higher voltage source.

Note

21.69).

The converter

is

said to function in

mode.

that if the duty cycle

were raised

value of £, would increase to 100

X

to 0.45, the

0.45

=

45

V.

1,

it

The

to indi-

exam-

can con-

by way of diode Dl and

by way of Q2, provided

to terminal l. The current / is therefore decreasing. The volt-seconds discharged during the 10 |xs interval is 80 V X 10 |jls = 800 V (jls. The change in current is A/ = 800 V (jls/10 mH = 0.08 A. We observe that the decrease in current when SI is closed is the same as the previous increase when S was open. Consequently, the peak-to-peak ripple is 0.08 A. The dc current fluctuates between 5.04 A and 4.96 A (Fig. 21 .68). Direct-current power flows

the boost

(Fig.

cate the allowed direction of current flow. For

Figure 21.68

10)

Q2

shown with an arrowhead

Figure 21.69 Two-quadrant electronic converter.

Q2

is

closed.


526

Similarly

if

current flows out of terminal

£H

,

provided that Ql

same way

Q2

,

it

can take

arm

arm

A

B

closed.

together perform the

together perform the

same way

as

/ Q1

D

as the mechanical switch SI. Similarly,

D2

and

is

Ql and Dl

In this figure,

1

D2 or the path through Ql and

the path through diode

(1-D)

me-

Ell

A

B

EA

EB

chanical switch S2. Fig. 21.69 therefore represents the essence of a 2-quadrant electronic converter. If

a dc voltage

EH

applied between terminals 3 and

is

(1_D

£ L between is again E L =

the converter generates a dc voltage

2,

terminals

and 2 and the relationship

1

£>£ H where ,

It is

closed

4 ;

\

Ql and Q2 cannot be

important to note that at the

period, called

The

/Q

I

D is the duty cycle of Ql. same

time, otherwise a short-circuit

will result across source

open.

V 02^

dead

current

is

£H

Thus, for a very brief

.

time,

both switches must be

carried

by one of the two diodes

Figure 21.70 Four-quadrant dc-to-dc converter.

E A = DE H

during this instant.

Power can be made

to

flow from the higher volt-

age side to the lower voltage side or vice versa.

power transported in one direction or the other depends upon the respective voltages and the duty cycle. The 2-quadrant converter of Fig. 2 .69 is the ba1

sic

The dc voltage £ B between terminals B, 2

The

is

E B = (\- D)E H

A

The dc voltage E u between terminals the difference

between

building block for most switch-mode converters.

^ll

E A and E B

= EA

and B

is

:

"~

= DE H -(\ ~D)E H

21.42 Four-quadrant dc-to-dc converter

thus

E XJL = E H

The 2-quadrant converter we have studied can only be used with a load whose voltage has a specific poThus,

larity.

terminal 2.

We

1

in Fig.

21 .69, given the polarity of

can only be

can overcome

(4-

)

It

by means of a

consists of

two

identical

2-quadrant converters arranged as shown 21.70. Switches Ql,

Q2

in

,


this restriction

4-quadrant converter.

EH

converter arm

in Fig.

A

open

alternately, as do switches Q3, Q4 in conarm B. The switching frequency (assumed to be 100 kHz) is the same for both. The switching sequence is such that Ql and Q4 open and close simultaneously. Similarly, Q2 and Q3 open and close

aD-

Equation 21.24 indicates that the dc voltage

D =

when

0.5.

D

1

D

,

voltage can therefore be either positive or negative.

Moreover,

if

a device

is

connected between termi-

nals A, B, the direction of dc current flow can be ei-

A to B

B

and close

ther

converter of Fig. 21.70 can function

Ql

is

A

it

will also be

duty cycle for

Q2

and

if

D for Q4. Q3

is (

1

-

the duty cycle for It

follows that the

D).

The dc voltage E A appearing between terminals A, 2

is

given by

is

Furthermore, the voltage = changes linearly with D, becoming +£" H when = and —E H when 0. The polarity of the output zero

verter

simultaneously. Consequently,

(21.24)

l)

from

or from

to

A. Consequently, the in

all

four

quadrants.

The instantaneous voltages E A2 and E B2 late

constantly between zero and

+£H

-

oscil-

Fig. 21.71

shows the respective waveshapes when D = 0.5. Similarly, Fig. 21.72 shows the waveshapes when D = 0.8. Note that the instantaneous voltage £ AB between the output terminals A, B tween

+ EH

and

~E H

.

oscillates be-

In practice, the alternating


f

527

E A2

0

4-quadrant dc-to-dc

converter

"B2

+ £H 0

'

Figure 21.73

E AB

Four-quadrant dc-to-dc converter feeding a passive dc

+ EH

load R. 0

-L U 10

0

15

20

ws

that

Consequently, only the dc component

connected

ternal device

when D =

0.5.

The average

voltage

zero.

appear between terminals A,

are

£ LL remains as the active driving emf across the ex-

Figure 21.71 Voltage output

B

components filtered out.

is

to terminals

A, B.

Consider, for example, the block diagram of a

R (Fig. E H As we the magnitude and polarity of £ LL can be

converter feeding dc power to a passive load 21 .73).

The power

have seen,

is

provided by source

varied by changing the duty cycle D.

frequency

E A2

'

/

of several kilohertz

constant. Inductor

L and

that the dc current

flowing

is

capacitor

.

The switching assumed to be

C act as filters so

in the resistance

has neg-

Because the switching frequency

ligible ripple.

is

high, the inductance and capacitance can be small,

making for inexpensive filter components. The dc currents and voltages are related by

thus + £H

power-balance equation 0

E H I H = E U J L We .

the switching losses and the small control

sociated with the

£ LL = 0.6E H

-mi

Fig. 21.74

active device

the

neglect

power as-

D and /input signals.

shows the converter connected to an which could be either a source or

E0

,

a load. If need be, the polarity of

E0 could be the re-

verse of that shown. In all these applications

0

5

10

15

"20

25

flow from

EH

to

E0

,

we can

force

power

to

or vice versa, by simply ad-

justing the duty cycle D. This 4-quadrant dc-to-dc

converter

Figure 21.72 Voltage output Ell. is

0.6

EH

.

when D =

0.8.

The average voltage

is

therefore an extremely versatile device.

The inductor L is converter.

It

alone

a crucially important part of the

is

able to absorb energy at one


528

snubber

= turn-on time

T]

T2 = on-state time 73 = 4-quadrant

A

K

dc-to-dc

turn-off time

Ta

= off-state time

D

= duty cycle =

converter

Ti2

-

switching

semiconductor

A

Figure 21.75a

A

Switching semiconductor and snubber.

D

f

During each

Figure 21.74 Four-quadrant dc-to-dc converter feeding an active dc

EQ

source/sink

GTO

neous current voltage level (high or low) and release

And

voltage level (low or high).

it

it

at

another

performs

this

ing one cycle because

energy dissipated

in the

cycle, divided by

T

21.43 Switching losses

temperature switches

all

a

function essentially the

GTO. The

GTOs,

as

losses that affect their

The

and switching efficiency.

rise

focus our analysis is

It

such

switches

we assume

of the

same way, but to

Turn-on time T, The current :

in the

GTO

it is

rapidly decreasing.

ble value / T , is

3.

T2 The

On-state time

:

VT

across the

GTO

:

current in the

GTO

rapidly falling while the voltage across

T4 The

Off-state time

:

current in the is

Fig. 2

is

{

where/,

is

.75a shows a

1

GTO

with

its

loss

anode, cath-

ode, and gate. In addition to the circuit that

GTO. A snubber R, L,

is

is

is

be-

connected

an auxiliary circuit com-

C components

(usually including

semiconductor devices) that control the magniwell as the anode current

snubber

is

Fig. 2

E AK as The purpose of a

to aid

1

/.

commutation and

to

reduce the

GTO.

.75b indicates the voltage, current, power,

and energy associated with each of the four

intervals

it is

GTO

stage is

zero

relatively high.

The sum of T + T2 + T3 + T4

T of one

power

of the switching operation. For example, during

while the forward voltage

riod

a result, the instantaneous

is

rapidly increasing. 4.

As

during these intervals can be very high.

losses in the

volts.

r3 The

Turn-off time

GTO is

and turn-off periods,

tude and rate of rise of the anode voltage current has reached a sta-

and the voltage

about 2 to 3

equal to the

being far greater than the on-state voltage drop of 2 to 3 volts.

posed of

is

rapidly increasing while the voltage across

2.

is

during one complete

so happens that the voltage across the

to the .

GTO

ing switched (not shown), a snubber

brief intervals: 1

average power

substantial during the turn-on

the switching device

switching operation involves four

GTO. The

However, we power loss dur-

it.

determines the temperature

it

switches and their duty cycle.

semiconductor

times the instanta-

are mainly interested in the average

rise

MOSFETs, and IGBTs have

it

through

that flows

duty automatically, in response to the electronic

All

equal to the product of the

is

instantaneous voltage across

.

power

interval the instantaneous

dissipated in the

cycle which, in turn,

the switching frequency.

equal to the peis

equal to \lfc

T

the instantaneous voltage across the {

Vj, the instantaneous current

power

dissipated

is

P

}

.

is /,,

GTO is

and the average

Consequently, the energy

dissipated during this stage

is

P{T

joules. {

On

other hand, the energy dissipated during interval is

zero because the current

is nil.

the

T4


permit them to operate frequencies than

"

ra

\*

much

anode-cathode

not overlook the losses (albeit

n

higher switching

GTOs.

In addition to the

T2

T\

at

529

we must

losses,

much

smaller) asso-

They

ciated with the gate voltages and currents.

are

not covered here. instant.

y

{

voltage

DC-TO-AC SWITCHING CONVERTERS

instant.

/ {

current

We have

average

p

power

studied the 2-quadrant and 4-quadrant dc-

to-dc switching converters. In this section

T'T

j

examine the 4-quadrant converter p t

energy

x

VT I T T2

x

Figure 21.75b Four stages of a

The

GTO

form of heat

in the

therefore given by

is

=

energy

21.44 Dc-to-ac rectangular wave converter

switching operation.

energy dissipated

total

during one cycle

PJ

Referring back to the 4-quadrant converter shown in Figs.

+ VT IT T2 + P 3 T3

}

will

converter.

P\T\

0

we

as a dc-to-ac

(21.25)

21.70 and 21.71,

duty cycle

D has

it

is

seen that

when

the

a value of 0.5, the dc output volt-

E lA is zero. However, the instantaneous value ELL oscillates symmetrically between +E U and

age

The on-time is

T. r

mentioned

T by

also related to

where

earlier, is

equal to

the expression

D is the duty cycle. We can, T2 =

7V

It

Ta = DT,

therefore, write

DT

of

—E H at a rate determined by the switching frequency. Consequently, the converter is able to transform the dc voltage into a rectangular ac voltage.

we

Substituting in Eq. 21.25,

energy dissipated

The

total

{

in the

(21.26)

GTO is this energy

T

power

PJ, + VT A, D T + P 3 T3

loss

Recognizing

power

X

that

loss

The rectangular wave can have any frequency,

ranging from a few cycles per hour to several hun-

= P T + VTIT DT + P 3 T3

power dissipated

divided by

obtain

dred kilohertz.

EH

.

vice \/fc

,

we

get

E LL oscillates

(21.27)

is

is

1

.27

power dissipated

quent temperature

in the

rise. It

GTO

and

its

conse-

can be reduced

and turn-off times are shorter. advantages offered by

whose

ThaJ:

MOSFETs

if is

fc

the turn-on

one of the

-100-

and IGBTs,

brief turn-on and turn-off times (T and ]

T3

)

+ 00 V 1

is

100 V, the

and

1

0.90

Figure21.76a

£H

.

If

an external de-

the output terminals A,

power can flow from

— 127 V + 100-

and the duty cycle D. The equation also indicates that the dissipation

to

the ac side and vice versa.

can be seen that the dis-

sipation increases with the switching frequency

E H N2 =

connected

Equation 21.27 reveals the factors that determine the

between

Consequently, the effective value of the funda-

(Fig. 21.76b),

= P\T\fc + VT fT D + P 3 T3fc

dc supply voltage

(Fig. 21 .76a).

mental

T=

If the

— 00 V The wave contains a fundamental sinusoidal component whose peak amplitude is 1.27 output

B

the dc side to


530

A -o-

4-quadrant

external

dc-to-ac

device

converter

f

-OB A

A

D

= 0.5

ELL =

%(2D-1)

(a)

Figure 21.76b Single-phase dc-to-ac switching converter

D=

0.5 and

f

can be

which

in

varied.

+ EH

Although the frequency can be varied over a

wide range,

rectangular-wave converter can

this

only generate an output that fluctuates between

+ E H and — E H

.

Furthermore, the wave contains

rather large 3rd, 5th,

D=

(b)

0.8

E LL = + Q.6E H

may

and 7th harmonics which

be objectionable.

21.45 Dc-to-ac converter with pulse width modulation When we verter,

we

0.5

+

EH

studied the 4-quadrant dc-to-dc con-

discovered that

it

produced an average

output voltage given by

L

E LL = E H (2D-

(21.24)

l)

(C)

/J

= 0.5

_£H

E LL = Q

Consider the 4-quadrant dc-to-dc converter of Fig. 21.77a, which

frequency

fc

is

operating

soon become apparent,

will

at

a constant switching

of several kilohertz. For reasons that it is

called carrier fre-

quency. Suppose that the duty cycle

average value of

E LL

E LL = E H

(2

This average dc value

is,

H

L

it

is

set at 0.8.

X 0.8-1) = 0.6E H is

+ EH

The

therefore,

buried

in the

£0 which continually fluctuates — E (Fig. 21 .77b). However, by ,

is

E LL

= -0.6E H

L

output voltage

between

+£H

(d)

D

= 0.2

and

using a small

filter

possible to eliminate the high frequency

com-

ponent /c and thereby obtain the desired dc voltage

ELL =

-EH

-Q.6E H

Figure 21.77 Four-quadrant dc-to-ac switching converter using carrier

frequency

fc

and three

fixed values of D.


across the output terminals LI, L2. Note that the

0.8

on for a considerably

positive voltage pulses are

0.2

longer period than the negative pulses. If

0'

D is set to 0.5, the average output voltage £ LL

becomes

531

zero, buried again within the fluctuating

E0

ac voltage

Note

(Fig. 21.77c).

of the positive voltage pulses

is

that the duration

now

equal to that of

the negative pulses.

Next,

is

is

ELL

D=

if

0.2,

we

find that the average value

—0.6 EH this condition is seen in Fig. 21.77d. The duration of the positive voltage pulses of

now why

is

;

less than that

of the negative pulses and that

the average (dc) output

Suppose now,

that

D

switching suddenly between at a

frequency

(Fig.

/,.

E lA

will

negative.

D=

0.8 and

much lower than 21.78). As a result,

frequency /that

voltage

is

varied periodically,

is

fluctuate

D=

0.2

the carrier

is

the output

continually

between

+ 0.6 E H and —0.6 E H The

filtered output voltage

between terminals LI L2

therefore a rectangular

.

,

wave having

frequency/ The big advantage over

a

the rectangular

nitude of E

,

x

,

trolled at will.

duty cycle at

is

is

wave of

as well as

Fig. 21 .76 its

is

that the

mag-

frequency/ can be con-

For example,

switched from

if

EH =

D=

100

0.65 to

V

will fluctuate

between +24

0.35

V and —24 V at a

frequency of 73 Hz. The only requirement

the carrier frequency /, be at least ten times the desired output

is

that

frequency/

Fig. 2 .79 gives the 1

result as Fig. 2

the right by an angle of 9 degrees.

1

.78,

ex-

phase-shift

is

signal.

Next, consider Fig. 2 .80 wherein the duty cycle 1

is

varied gradually between 0.8 and 0.2, following

a triangular pattern. This causes the filtered output

voltage

E LL to vary between +0.6 EH

faithfully reproducing the triangular

0.8

0.2

0.2

0

0

Figure 21.80 D.

The

achieved by simply delaying the duty cycle

0.8

Figure 21.78 Frequency and amplitude control by varying

same

cept that the rectangular voltage has been shifted to

and the

D=

a frequency of 73 Hz, the resulting rectangular

wave

Figure 21.79 Frequency, amplitude, and phase control by varying D.

Waveshape

control by varying D.

and -0.6 £ H wave. Clearly,

,


532

the 4-quadrant converter

cause

it

is

a versatile device be-

can generate an ac output voltage of almost

EH (whose value is fixed) and knowing the desired value of ELl (t) as a func-

Consequently, knowing

any shape. The frequency, phase angle, amplitude,

tion of time, the pattern of

and waveshape can

For example, suppose

all

be adjusted as needed by

simply modifying the duty cycle pattern.

Another important property

power can

that

is

E given

put voltage

under

all

conditions.

The reason

no matter

that

is

what the polarity of the output voltage happens can always flow

be, the current

According

impedance

very low because the output terminals A, fectively connected to the dc supply

£H

,

B

D = The

low

internal

ratio

it-

.28,

+

(360 ft

D

6)

given by

is

0.5

+

1

sin

+

(360//

Em /EH

is

impedance.

wave converter

expressed

which

0.5

+ m

[1

+

sin (360//

6)]

(21.30)

determine the duty cycle pattern to genIn this equation, /is in hertz,

erate a desired output voltage, consider Fig. 21.81 in

wave can be

as:

D= In order to

(21.29)

called amplitude modulation ra-

duty cycle pattern to generate a sine

21.46 Dc-to-ac sine

0)

designated by the symbol m. Consequently, the

tio,

self has a very

sin

is

are ef-

which

1

to generate an out-

to

one direction or

in

the other. Furthermore, the ac output

to Eq. 2

can be programmed.

by

E = Em

flow from the dc side to the ac side and vice versa,

D

we want

E UM)

is

a voltage having an arbitrary

in

waveshape and frequency. Applying Eq. 21.24, we can write

The tio,

EH (2D-

1)

seconds, and 0

ratio c //is called frequency


m

f

modulation

ra-

.

Example 21-14

A

from which we immediately deduce

200

V

dc source

connected to a 4-quadrant

is

switching converter operating

D =

in

t

degrees.

0.5

(21.28)

quency of

kHz.

8

a carrier fre-

at

desired to generate a sinu-

It is

soidal voltage having an effective value of 120 at a

frequency of 97

lagging.

Hz and phase

V

angle of 35°

Calculate the value of the amplitude

modulation

ratio

and derive an expression for the

duty cycle. Solution 4-quadrant

The peak value

switching

device

converter

Em

of the output voltage

Em =

120 <2

The amplitude modulation

fc

m = EJEH =

L2

A

A

0=0.5

( 1

V

£LL(/)\ +

"*f

Figure 21 .81

170

V

ratio is

170/200

The frequency modulation

E" )

=

is

=

0.85

ratio is

=/
The expression

for

D is

Four-quadrant switching converter producing an arbitrary

waveshape

ELL{t)

.

D=

0.5[1

+

0.85 sin(360

X

97

t

+

35)]

(21.31)

FUNDAMENTAL ELEMENTS OF POWER ELECTRONICS 100 V

wave

21.47 Generating a sine

533

V-

86.6

To understand the switching process, let us determine the switching sequence to generate a sine wave between terminals A and B of Fig. 2 .70. Here 1

50 V-

are the specifications:

Peak ac voltage required: 100

V

Frequency required: 83.33 Hz Carrier frequency: 1000

Dc

Hz

E H 200 V

supply voltage

1

ms Type of switching

is

We reason

Q4

standard: Ql,

do Q2 and Q3

open and

A

(see Fig. 2 1 .70).

60

30

\

close simultaneously, as

I

I

!

I

/o

90

Hz

Duration of one cycle of the desired 83.33

150

180

_

\

C

B

D

—

6000

\*

as follows:

120

E jus

G

F

—

>\

fun-

Figure 21.82

damental frequency

Positive half-cycle of the fundamental 83.33

T=

l//= 1/83.33

=

0.012

s

=

12 000

age comprises

jjls

six carrier

periods of

1

ms

Hz

volt-

each.

This period corresponds to 360 electrical degrees.

Proceeding

Period of the carrier frequency:

Tc = Number 12

000

=

\lfc

1/1000

=

s

ms = 1000

1

|xs

=

|jls

Angular interval

(at

fundamental frequency) cov-

ered by one carrier period

360°/ 2 1

=

calculate the value of is

30°. In

one

cycle of the fundamental voltage there are 12 such

D

reached. Table

lists

the items of

interest.

Knowing

12.

=

we

way,

2 IE organizes the information and

=

of carrier cycles per fundamental cycle

ijls/1000

this

end of the cycle (360°)

until the

the duty cycle for each interval, the cor-

responding time that Q

For example, during

l

,

Q4 are on can be determined.

interval B, they

for 0.625

X 000 |xs = 625

TABLE

E

1

|uls. It

must be closed

follows that Q2,

Q3

intervals.

Fig. 21.82

shows

the positive half-cycle of the

sine wave, with the 30° intervals labelled

The average voltage

for interval

cycle during this interval

D =

0.5

(

+

1

~'LL(t) \

A

to

A is zero. The

duty

=

therefore,

+

0.5(.

4,

=

0.5

The average voltage for interval B is 100 sin 30° = 50 V. The duty cycle during this interval is, therefore,

D -

0.5(1

+

LL(Q

-0.5

1

50

+

0.625

200 The average voltage 86.6 V.

for interval

C

is

100

The duty cycle during this -interval

D =

0.5

1

+

86A -— 200

=

is,

0.716.

sin

60°

GENERATING A SINE WAVE

=

therefore,

Ql,Q4on

angle [deg]

is,

21

G. [VI

Q2,

Ljjls J

[

Q3 on [jls

]

interval

A

0

0

0.5

500

500

30

50

0.625

625

375

B

60

86.6

0.716

716

284

C

0.75

750

250

D

120

86.6

0.716

716

284

E

150

50

0.625

625

375

F

180

0

0.5

500

500

G

210

-50

0.375

375

625

H

240

-86.6

0.284

284

716

I

0.250

250

750

J

K

90

270

100

-100

300

-86.6

0.284

284

716

330

-50

0.375

375

625

L

360

0

0.5

500

500

M


5 34

r

200

200

100

210 340 270 300 330 360

7f

200

60

30

0

120 150

90

18(K

a|b|c|d|e|f|g|h!i!j|k|l|a Figure 21.84

Figure 21.83

Sequential (+) and

and (-) pulses contain the sinusoidal

Alternative (+)

i

-)

pulses contain the sinusoidal

components.

component. here. In

must be closed

for the remaining part of the interval,

- 625 = 375 Note that when Ql, is momentarily positive, and when £ Q4 AB E is momentarily negative. are closed Q2, Q3 AH Thus, during interval B, £ AB is +200 V for 625 fxs namely

1

000

jjls.

are closed

and

and

-200 (

—

It is

)

V for 375

fxs. Fig. 21.83

shows these (+)

seen that although the carrier period

(— 1000

[is),

the on/off pulse widths are

continually. That

is

why

this

fixed

type of switching

called pulse width modulation, abbreviated It is

is

changing

PWM.

worth noting that during each switching

terval (A, B, C, etc.), the area of the

V

200

is

in-

V positive

cases the switching sequence

control. In the simple

the intention

to

is

show

The

is

under

method used above,

the fundamental principle

upon which pulse width modulation

based.

is

sinusoidal voltage buried in the pulse train

appears grossly distorted

in

21.84. However, once the

quency

polarities.

many

computer

is filtered out,

both Figs. 21.83 and

Hz

1000

carrier fre-

the resulting voltage will be

very sinusoidal. Indeed, the lowest harmonics are clustered around 1000 Hz, which

fundamental 83.33

A

Hz

12 times the

is

frequency.

higher carrier frequency would yield a better

waveshape because more points could be established along the sinusoidal curve and the filtering would be

is

easier.

But a higher frequency would increase the

equal to the volt-second area of the sinusoidal seg-

power

losses in the

pulse minus the area of the 200

ment during

that interval.

It

negative pulse

under the positive sine wave, shown dotted 21.83,

is

equal to the

IGBTs

that are being switched.

follows that the area

sum of the

in Fig.

areas of the seven

21.48 Creating the

PWM

positive pulses less the areas of the six negative

We

The same remarks apply to the negative half-cycle. Thus, whenever a continuous voltage is transformed into a chopped PWM form, the volt-

waveshape chopped up

each interval

seconds during any given interval are the same.

frequency. In the previous section

pulses.

generates the

shows another switching pattern that same sinusoidal voltage. However, the

pulses are

positive during the positive half-cycle

Fig. 21.84

all

and negative during the negative half-cycle. Again, in this

case the

sum of the

areas of the five positive

pulses included in the 180° interval area of the dotted sine

wave during

is

equal to the

that interval.

There are many other ways of programming the switching sequence, which will not be discussed

that

have seen

that to transform a desired voltage

into

PWM

form,

waveshape is The duration of

the

into small intervals. is

pulse train

T of the carrier we showed how

equal to the period

can be done by actual calculation. In practice, a

very ingenious method

is

employed

to create the

PWM pulses. Consider one arm of a converter as Fig. 21 .63a.

illustrated in

Suppose the desired voltage

undulating shape

shown

in Fig.

E L has

the

21.85 and that the

fixed dc voltage at the input to the converter

is

EH

.

We want to transform the continuous E L into a series of pulses having a fixed amplitude

EH

,

a fixed period


535

both the desired voltage £L and the fixed dc voltage E H The successive crossing points of the two signals .

establish the firing pattern of the

IGBTs or GTOs.

21.49 Dc-to-ac 3-phase converter shows a dc-to-ac 3-phase PWM concomposed of three switching arms. In this schematic diagram the duty cycle for each arm is programmed so that the output voltages EAN E BU Fig. 21.86a

verter

,

,

ECN

are equal and mutually out of phase by 120°.

This

is

achieved by having the respective duty cy-

cles follow the expressions:

D,

=

D2 = D3 = By

0.5

[1

0.5

[

1

0.5

[1

+ m

sin

+ m

sin (360//

-

+ m

sin (360//

-

assigning values to

(360 ft)] 1

20)]

(2

1

.32)

(2

1

.33)

240)] (21.34)

m and/,' the amplitude and fre-

quency of the output voltages between terminals A, Figure 21.85 Transforming a desired continuous voltage

PWM

T and

an appropriate conduction period

pulse.

To achieve

y

.

into

X~7

a

Ta for each we first draw a series of having a base T and height

It is

Eh

this result,

isosceles triangles, each

EH

EL

voltage.

/

/

clear that the sides of the triangles will in-

waveshape £L at various points. The following simple rule applies: Conduction Ta takes place whenever £L lies above the triangular wave, and conduction ceases whenever it lies tercept the

below. Thus, conduction 0-1, 2-3, 4-5, 6-7,

the remaining intervals.

shown

occurs during intervals

forth,

The

and ceases during

resulting pulse train

in gray contains the original signal

carrier frequency

smooth

signal

were

£L would

In practice, the period that

7a

and so

shown

(a)

filtered out,

£L

.

If

the

the original

immediately appear.

T is made much

shorter than

in the figure. Consequently, the

£L during one carrier period that indicated in Fig. 21.85.

is

change

in

considerably less than

A higher carrier frequency

automatically improves the faithful reproduction of the original signal buried in the

PWM pulse train.

In an electronic converter such as that illustrated in Fig. 21.69, the gates are fired at appropriate in-

stants

by using proportionally reduced signals of

Figure 21 .86 Three-phase dc-to-ac switching converter.

B,


536

and

0

C can be set to any desired value. The peak line-toAN £ BN £CN are given by

neutral voltages

,

,

+

L1 0.5/77

The 3-phase

£H

(V3/V2)

The phasor diagram

converter

in Fig.

21.86b.

=

0.612

Fig.

D3

21.87

is

a

in Fig. 21 .69.

L

filter

out the

wanted ac voltages

between terminals LI, L2, L3. Power can again flow

from the dc

side to the 3-phase side

Fig. 21.88 of

shown

is

block diagram of the dc-to-

3-phase converter. The inductances

m, 0


(21.36)

.

21.86a each represents

shown

carrier frequency, leaving only the ,

m EH

for these voltages

in Fig.

the electronic switches

LI i /

therefore,

understood that the three dual

It is

mechanical switches

L3

D2

0.5/?7

is,

device

dc-to-ac

D1

(21.35)

effective line-to-line voltage

£rms =

L2

EH

a dc-to-ac 3-phase switching

shows

the

such a 3-phase converter

converter. 240°

120°

o°

and vice versa.

PWM voltages generated by when connected

to a

500

V

360°

^ ;nnnr^JU±MiiTTiriJiJJLi 0

500 V 0

V

1

500 V

500 V

phase

AB 0

V -500 V

phase

I

500 V

BC 0

V -500 V

500 V

phase

CA 0

V -500 V

Figure 21.88 Three-phase PWM voltages produced by a dc-to-ac switching converter operating at 540 Hz with a 500 V dc input. Top: ân> ^bn> ^cn outputs, peak 60 Hz sinusoidal component — 200 V. Bottom: Eabi ^bc> ^ca outputs, peak 60 Hz sinusoidal component = 346.4 V, rms value = 245 V


dc source. The carrier frequency

is

540 Hz and the fun-

anode

damental wanted frequency /'is 60 Hz. The amplitude modulation

Em = =

peak

ratio is 0.8; consequently, the

neutral sinusoidal voltage

m'EH

0.5

X

0.5

537

cathode gate

line-to-

choke

is

filter

(21.35)

0.8

chopper

X 500 = 200V

peak inverse voltage rectifier

The

effective 3-phase line-to-line voltage

is

inverter

Erms =

=

0.612/77

EH

X

0.8

0.612

(21.36)

X

500

=

245

harmonic

commutation

V

line

should be noted that the converter illustrated

It

Figs. 21.86

and 21.87

is

in

a 4-quadrant converter that

can deliver or receive 3-phase active power (watts) the fundamental frequency.

power

ceive reactive

It

can also deliver or

fundamental

(vars) at the

commutation

converter

cycloconverter

at

bridge rectifier

re-

self

fre-

commutation

displacement power factor

quency. Consequently, as far as the output terminals LI, L2,

L3

behaves ex-

are concerned, the converter

actly like the


we

Chapter

16. Its

''synchronous reactance"

to 2ir/L,

where

/'is

2 -6 1

Xs

is

1

1

.

in Fig.

6 produces a secondary line voltage

of 2.4 kV. The dc load current

equal

the fundamental frequency and

The 3-phase transformer shown 2

studied in

600 A.

/ tl is

L Calculate

The

the inductance of the carrier frequency filters.

is

The dc voltage across the load The average current carried by each diode The maximum current carried by each diode

a.

latter are

used to diminish the carrier frequency cur-

rents in the 3-phase line.

The device

in Fig.

2 .87 1

b.

may

c.

be a passive or active load or even a 3-phase source. 2 -7 1

The 3-phase transformer shown

21.50 Conclusion This chapter has given some of the basic concepts of


1

1

.

of 2.4 kV. electronic devices

work

and converters.

It

lays the ground-

a.

dc and ac

b.

for the study of electronic drives for

motors and the control of large blocks of power electric utilities. In

subsequent chapters

we

in

2

1

-8

will see

If

the dc load current

600 A,

is

The dc voltage across the load The average current carried by each diode

An

ac source having an effective voltage of

600

further applications of switching converters.

in Fig.

9 produces a secondary line voltage

2

V,

60 Hz

is

connected to a single-phase

bridge rectifier as

The load

shown

in Fig. 2

resistor has a value of

1

30

.

1

4a.

II.

Questions and Problems Calculate Practical level

a.

2

b.

1

-

1

2 -2 1

State the basic properties of a diode.

c.

State the basic properties of a thyristor.

d.

2 -3 1

What

is

the approximate voltage drop

across a diode or

SCR when

it

conducts?

e.

2 -9 1

maximum

2 -4

What

2 -5

Explain the meaning of the following terms:

1

is

the approximate

permis-

sible operating temperature of a thyristor? 1

The The The The The

dc voltage dc voltage

£ 34 E54

dc load current

/

average current carried by each diode active

power supplied by

the ac source

The chopper shown in Fig. 2 .62 is connected to a 3000 V dc source. The chopper frequency is 50 Hz and the on-time 1

is

l

ms.

538


21-15

Calculate a.

b.

21-10

In

The voltage across resistor R The value of /s if R 0 -

Problem 2

calculate the

-9, if

1

()

The chopper shown in Fig. 2 .62 is connected to a 2000 V dc source, and the load resistor R & has a value of 0. 5 fl. The on1

1

we double

time

the on-time

new power absorbed by

across the resistor

the

load.

a.

b. a.

In


pated by the six diodes

if

power

the

b.

What

a.

The

b.

The

dissic.

the average voltage

drop during the conduction period

is

d.

0.6 V. e.

the efficiency of the rectifier alone?

is

f.

21-12

current in Fig. 21.6 has a value of

-6 A. What

the polarity of

is

is

negative.

Is


peak value of / s apparent resistance across the dc

£34 ?

g.

21-16

Draw

the

waveshapes of 7 S

Fig. 21.14a

is

connected

to a

shown

120 V, 60 is

is

e.

21-14

The inductance of the choke The peak-to-peak ripple across

The

line voltage

240

is

V,

b.

in Fig.

current of

21,19.

c.

is

21-17 the

b.

d.

choke

The dc load draws

needed

to generate

60

V

needed

to generate

60

V

mode)

firing angle

mode)

The converter shown line voltage

lay angle

a

in Fig.

2 1 .36

is

con-

of 40 kV, 60 Hz. The

is

450 A.

If the de-

75°, calculate the following:

b.

The dc output voltage The active power drawn from

c.

The

d.

The

a.

the ac line

effective value of the secondary line

current

The dc voltage produced by the rectifier The active power supplied by the 3-phase in

each diode

The duration of current flow

in

The The

reactive

power absorbed by

the con-

verter

21-18

The peak current

In a.

each diode

b. c.

Problem 21-17, calculate the following: The peak positive value of £ KA The peak negative value of E KA The peak-to-peak ripple across the inductor

effective value of the secondary line

21-19

reactive

power absorbed by

the con-

The peak-to-peak inductor

The

electronic contactor in Fig. 21 .33 con-

trols the

power

to a 15 11 heater.

The

sinu-

soidal supply voltage has an effective value

verter g.

firing angle

load draws a dc current of

750 A.

current f.

The

ondary

60 Hz on the

[msl e.

for a firing angle of

nected to a transformer that produces a sec-

source c.

The

(inverter

Calculate a.

line.

The dc output voltage

(rectifier

in the

secondary side of the converter trans-

former

V

Calculate

about 5 percent of the dc current d.

in

directly connected to a

90°

peak-to-peak ripple

that the

.

Hz

3 ft, calcu-

a.

The dc load current The PIV across the diodes The energy that must be stored choke so

and I D

in

following:

late the

/() ,

the current in-

rectifier

source. If the load resistance

,

The 3-phase, 6-pulse converter shown 3-phase, 208

c.

V.

chopper frequency

creasing or decreasing?

a.

60

power supplied to the load power drawn from the source dc current drawn from the source

Fig. 21.36

The single-phase bridge

b.

fxs

source

current in Fig. 21.7 has a value of

+ 6 A and E65 21-13

is

and the dc voltage

Calculate


fixed at 100

is

ripple across the

of 600

V at a frequency of 60 Hz. If the fira = 0°, calculate the following:

ing angle


a.

b. c.

d. e.

The effective current The power supplied to the heater The effective value of the harmonic currents The displacement power factor The reactive power furnished by the ac line

g and g4 if we wish to generate an output frequency of 12 Hz. Draw the waveshape 1

of the voltage across the load resistance.

21-24

Referring to Fig. 21 15, and recognizing

a.

.

that area

Advanced 2 1 -20

In

b.

component

is

known

to

2 -25 1

d. e. f.

g.

21-21

The rms value of all the harmonic currents The power dissipated in the heater [W] The distortion power factor The apparent power supplied by the source The total power factor The displacement power factor The reactive power supplied by the line [var]

The

rectifier

shown

purpose

L.

1

Inductor

breakers to

comes too rent

is

trip

Assuming

1

The maximum

c.

The average power delivered

d.

Draw

2

minimum

1

used

in a battery

shown 120

in Fig.

V

Calculate the

is

a.

After 2 s?

b.

After 4 s?

c.

After 6 s?

=

10

The

A 3-phase, dc side

battery voltage

and

ft.

maximum

1

it

what

is

the diode will not break

how many

diode conduct is

c.

2 -23 1

the

is

2

1

.59.

on for

the value of the

is

R =

shown

in

connected 10 mft.

The

The

20

V

battery

ac side

is

connected

to a

120 V, 60

1

Hz

line. If the bat-

500 A,

calculate

that

the following:

down.

a.

b.

rms secondary voltage

c.

300 V?

What

[W|

effective sec-

electrical degrees will the if

is

to be used as an inverter.

tery delivers a current of

For

in Fig.

chopper

6-pulse converter is

to a 3-phase,

ondary voltage of the transformer so

b.

to the load

as a function of

with that

-25, if the s,

/

Will anything prevent the current from

Fig. 21.41

charger similar to the one

21.11a.

and/?

V

waveshape of

building up indefinitely?

50 ms.

having a PIV rating of 600

the

d. After 8 s?

2 -27

A diode

current in the inductor

current

R become

A after

and off for

transferred to the load, per

and off for 2

s

s

[J]

Problem 2

In

so that the short-circuit current

does not exceed 3000

on for 2

The energy

e.

L

is

a.

time and compare

2 -26

the initial cur-

1000 A, calculate the

value of

chopper


b.

L

before the dc current be-

large.

s,

cycle

to limit the rapid build-

is

up of dc current should load

a.

peak-to-peak current ripple must not

The chopper shown in Fig. 2 .58 transfers power from a 400 V source to a 00 V load. The inductor has an inductance of 10

short-circuited. This enables the circuit

is

If the

5 H. If the

21.19 produces

in Fig.

A at 250 V.

reduces the current ripple, but an additional

1

is

[V-s].

1

a dc output of 1000

2 -22

60 Hz

V,

choke

Calculate

c.

the effec-

if

,

exceed 7 A, calculate the inductance of

be/F - 12.34 A.

a.

A (+

2000 /

found to be 17.68 A. The rms value of

the fundamental

b.

almost triangular, calculate

produced by the source

tive voltage

firing angle is in-

creased to 120° and the effective current is

is

the approximate value of

level

Problem 21-19 the

539

the peak current in the diode?

The cycloconverter in Fig. 2 .34 is connected to a 60 Hz source. Calculate the 1

time interval between the firing of gates

The firing angle required The active power delivered to the ac line The reactive power absorbed by the converter

Industrial applica turn

21-28

A 24 V a 12

V

battery (rms),

resistor.

is

connected

in series

60 Hz generator

with

to a 10 ft


540

Anode

Calculate

The maximum and minimum voltage

a.

from zero

current: rises linearly

2. On-state time:

to

260

A

to

550

V

250 ms

across the resistor

The

b.

On-state voltage:


c.

in the resistor

3.

2 -29 1

A distorted

A and a THD of 26

percent.

Calculate the peak value of the fundamental

component and

all

the harmonics taken together.

One of the diodes

voltage: rises linearly

Anode

current: falls linearly

the effective value of

in a

21.13 fails and becomes The source voltage has an effective value of 208 V and, be-

a.

b.

and determine

21-31

the resistor

c.

d.

One of the most

thyristors

(Q2)

in Fig.

2 .34

short-circuited. Describe

1

A GTO

in the

GTO dur-

dissipated in the

GTO dur-

The peak power dissipated The peak power

The approximate average power during The approximate average power during

likely

happen

in this

unfused

The approximate energy

GTO f.

operates under the following con-

the

g.

h.

ms

dissipated in the

The approximate energy The

in the

during the on-state period

GTO during

Turn-on time: 2

total

dissipated in the

the turn-off period

power dissipated

in the

GTO,

glecting gate losses

Anode

the

during the turn-on period

The approximate energy dissipated

GTO

ditions: 1.

mA

1

what

circuit.

2 -32

V

on-state period e.

will

ms

turn-on period

it.

becomes

zero

ing the on-state period

and the effective value of the voltage across

to

ing the turn-on period

sorbed a power of 1400 W. Describe what

new power absorbed by

A

Calculate

fore the diode failed, the resistor ab-

the

V

from 420

Off-state reverse current: 6

lar to that in Fig.

after the failure

Off-state time: 375

from 3.2

Off-state voltage: 1300

bridge circuit simi-

open-circuited.

happens

ms

Anode

4.

1

Turn-off time: 6

60 Hz current has an effective

value of 547

2 -30

A 3.2 V

On-state current: 420

effective value of the voltage across

the resistor

voltage: drops linearly from 1200 to 3.2

V

V

i.

The frequency and duty cycle of the switching operation

ne-

Chapter 22 Electronic Control of Direct-

Current Motors

22.0 Introduction

quadrant

stricted to

and the speed

High-speed, ductor

change

reliable,

in the

and inexpensive semicon-

have

devices

produced

a

voltage.

dramatic

we examine some

22.

of the basic principles of such

The

inverters already covered in Chapter 2

1

.

The reader

armature

Consequently, the

many

switch S

methods of control, we

only study the behavior of power

circuits.

The reason is that they constitute, by themcomplex subject involving sophisticated

connected

field current I r rectifier.

is

External inductor

smooth armature

La

is

(Fig.

provided by a single-

current.

L ensures

is is

a

The armature

usually large enough, and so the

initially at rest

The

and the disconnecting

open. in-

power

In addi-

processor can be set for any desired motor

speed and torque. The actual values are compared

and

with the desired values, and the processor automatically generates gate pulses to bring the

important subject does not detract from the thrust of to explain the

is

by means of suitable transducers.

tion, the

microprocessors. Nevertheless, the omission of this

is

6-pulse converter

torque, etc. These inputs are picked off the circuit

selves, a

this chapter, which

fixed,

is

puts such as actual speed, actual current, actual

ingenious ways of shaping

electronics, logic circuits, integrated circuits,

field excitation

A gate triggering processor receives external

and controlling triggering pulses are not covered here.

The

inductance

drives should also be consulted.

will

The

external inductor can often be dispensed with.

ceeding further. Sections 20.16 and 20.17 on elec-

In describing the various

1).

relatively

should, therefore, review this chapter before pro-

tric

A 3-phase,

phase bridge

and

circuits involve rectifiers

.

between the armature and a 3-phase source

control of dc motors. In this chapter

electronic drives.

l

varied by changing the armature

is

two

as close

together as possible. Limit settings are also incor-

fundamentals of

porated so that the motor never operates beyond ac-

electronic dc drives.

ceptable values of current, voltage, and speed.

22.1

We

First

Gate pulses are

quadrant speed control

Switch S

begin our study with a variable speed drive for

a dc shunt motor.

initially

delayed by an angle

90° so that converter output voltage

We assume that its operation is re-

that

54

Ed

is

then closed and

a

is

Ed

is

a = zero.

gradually reduced so

begins to build up. Armature current

/d starts

542


limit settings

/

»

v

max min/max

max current

speed

etc.

_L_J__i_

desired current

jr

-

n

actual armature current

gate

control

external inputs

|

I

desired speed

actual speed

triggering

settings

from the system

processor

desired etc.

i

actual etc.

TTTTTT

a

G,

G3 G 3 G4 G5 G6

I

3-phase line

6 single

phase

source

Figure 22.1 Armature torque and speed control

of

a dc motor using a thyristor converter.

flowing and the motor gradually accelerates. During the starting period, the current ically.

is

The dc voltage

4.

Furthermore, the gate-triggering processor

excess

motor runs

speed.

As

mature

The power source

tR losses except those

the

its

power is

apply during the start-up phase.

all

is

loss in the thyristors

the active

is

small; conse-

power drawn from

the ac

Even if an inexperienced operator tried to start the motor too quickly, the current-limit setting would override the manual command. In effect, lowable preset value.

exceed the

al-

speed, the firing an-

usually between 15° and 20°. Converter volt-

gle

is

age

Ed is slightly

greater than induced voltage

E0 by

an amount equal to the armature circuit /d /? a drop.

The converter voltage

available to drive the load.

the armature current can never

When the motor reaches full

in the ar-

itself.

quently,

3.

when

low speed while developing

armature resistors are needed; consequently,

there are no

2.

at

a result, power-factor correction

difficult to

No

much lower

diminishes continually as the motor picks up

Four features deserve our attention as regards

.

is

rated torque. Furthermore, the reactive

of, say, 1.6 pu.

the start-up period: 1

start-up

absorbs a great deal of reactive power

is

preset so that the pulses can never produce a current in

Ed during

than rated voltage. Consequently, the converter

monitored automat-

is

Ed =

given by the basic equation 1.35

£ cos- a

(21.17)

To reduce the speed, we increase the firing angle becomes less than E0 In a Ward-Leonard system this would immediately cause the armature

a

so that Ed

.

current to reverse (see Section 5.5). Unfortunately,

ELECTRONIC CONTROL OF DIRECT-CURRENT MOTORS

the current cannot reverse in Fig. 22.

because the

1

SCRs conduct in only one direction. As a result, when we increase a, the current is simply cut off and motor coasts

the

to the

terval,

E

comes

less than the

()

lower speed. During

gradually falls and,

new

setting of

£d

is

b.

SCR

The

a under

firing angle

rated full-

firing angle required so that the

velops

its

rated torque at

motor de-

400 r/min

the armature

Solution builds

equal to the load torque, the motor

At full-load the converter must develop a dc output of 250 V:

lower speed.

efficiency at the lower speed

cause the

,

The required

load conditions

a. it

will continue to run at the

The

this in-

Calculate a.

eventually be-

it

The torque quickly

current again starts to flow.

up and, when

when

543

losses are small.

is still

However,

voltage generated by the converter

is

high be-

the ripple

Ed = 250 -

1.35

E cos a

1.35

X 208

a =

0.89

a =

27°

greater than

under rated full-load conditions because a

cos is

(21.14)

cos

a

greater

(Section 2 .28). Consequently, the armature current 1

is

not as smooth as before, which tends to increase

the armature copper losses and iron losses.

equally serious problem

is

the large reactive

Armature IR drop IR

power

absorbed by the converter as the firing angle

is

at rated current:

An in-

= 2500 A X

Counter-emf (cemf)

when the firing angle is 45°, absorbs as much reactive power from

0.004 1)

at

1

200

=

10

V

r/min:

creased. For example, the converter

the 3-phase line as

To stop

it

the motor,

does active power.

we

b.

delay the pulses by 90° so

that

Ed =

that

depends on the mechanical load and the

0

V.

The motor will coast

E0 = 250 -

to a stop at a rate

To develop

rated torque at

ture current

must

400 r/min ()

to a

208

V, 3-phase,

60 Hz

line using a

is

2500

A and the armature resistance

desired values

-

gate

J[

-

Ed =

3-phase bridge

is

4 mil.

j

^[processor

is

+

80

10

= 90 V

The converter must, therefore, generate 90 To determine the firing angle, we have

Ed — 90 =

~j

triggering U limits

1

connected

converter (Fig. 22.2a). The full-load armature current is

= 0V

Armature terminal voltage 1200 r/min dc motor

external

1

.35

1.35

£ cos a X 208

a = 71°

208

V

3-phase

Figure 22.2a 22-1.

a

(see Fig. 22.2b) ,10 V,

Figure 22.2b Rated torque at 400

r/min.

V.

(21. 14)

cos

inputs

Jgates

See Example

at

X 240 = 80 V

(400/1200)

Armature IR drop V,

the arma-

is

£ =

Example 22-1 250

400 r/min,

be 2500 A. The cemf

still

inertia

of the revolving parts.

A 750 hp,

= 240 V

10


544

Example 22-2 Referring to Example

we please. We 22-1, calculate the reactive

power absorbed by

the converter

develops

at

full

torque

400

when

the

motor

r/min.

dynamic braking, using

often resort to

a resistor connected across the armature. However, the converter can ajso be verter,

made

to operate as an in-

feeding power back into the 3-phase

regenerative braking

is

Such

line.

preferred for large motors be-

Solution

The load condition is given in Fig. 22.2b. The dc power absorbed by the motor is

P = EJ = 90 X 2500 - 225

kW

lX

If

we

cause the kinetic energy is

the generator output can be precisely controlled to obtain the desired rate

To make neglect the relatively small converter losses,

of

larity

the active

power supplied by

the ac source

is

also

225 kW. reactive

power drawn from

the ac source

Ed

of change

in

speed.

the converter act as an inverter, the po-

must be reversed,

as

shown

in Fig. 22.3.

E0 Ed must be adjusted to be slightly less than £0

This means Finally,

The

during the deceleration

lost

converted to useful electrical energy. Furthermore,

we must

also reverse the polarity of

.

,

is

to obtain the desired braking current / d (Fig. 22.3).

given by

Q = Plana -

(21.14)

225 tan 71°

= 653

kvar (compare

the active

with

this

power of 225 kW)

This example shows that a large amount of reac-

power

tive It

is

required as the firing angle

is

even exceeds the active power needed

increased.

at full-load.

Figure 22.3 Motor control by

field reversal.

Capacitors could be installed on the ac side of the converter to reduce the burden on the 3-phase feeder line.

Alternatively, a variable-tap transformer could

be placed between the 3-phase source and the converter.

By reducing

speeds, the reactive erably.

The reason

is

the ac

voltage

the lower

at

power can be reduced considthat the firing angle

can then be

kept between 15° and 20°. However, this

be a feasible solution

if

the speed of the

may

the reactive

other

ways

to

than 90°. However, to change the polarity of

reduce

time. Furthermore, after the generator (braking)

phase

is

over,

when field 1:

verter.

again reverse the armature or

machine runs

circuit so that the

mo-

controlling the

we can make the speed fall

reversal

as a motor. Bearing

we now is

list

the steps to be

employed.

Delay the gate pulses by nearly 180° so

Ed becomes

quite large and negative. This

operation prepares the converter to act as an

simply coasts to a lower speed. To obtain a quicker re-

generator output,

we must

field so that the

Step

cannot always tolerate a situation where a motor

By

and

equipment. Reversing the

We

temporarily as a generator.

,

this requires additional

that

tor acts

£0 we

field or the armature,

22.2 Two-quadrant controlfield reversal

we have

in-

must reverse either the

these conditions in mind,

sponse,

as they first ap-

can be changed almost

stantaneously by delaying the gate pulses by more

taken

modify the

Ed

armature or field also takes a significant length of

power demand.

to

polarity of

to

be continually varied. The tap-changing becomes too

show

The

not

motor has

frequent to be practical. Later sections in this chapter (Sections 22.8 and 22.9)

These changes are not as simple pear.

as fast as

It

takes a

which current Step

2:

/d

is

zero.

Reverse the

quicklyas

field current as

possible so as to reverse the polarity of total

in-

few milliseconds, following

reversing time

may

last

from

1

E0

.

The

to 5 sec-

ELECTRONIC CONTROL OE DIRECT-CURRENT MOTORS

545

onds, owing to the high inductance of the shunt

The armature

field.

current

zero during

is still

this interval.

Step

3:

Reduce a so

less than

Eih

that

Ed becomes

slightly

enabling the desired armature cur-

The motor now

rent to flow.

acts as a generator,

feeding power back into the ac line by the inverter.

lower

way

of

Figure 22.4

speed drops rapidly toward the

Its

Motor control by armature reversal.

setting.

What do we do when reached?

We

the lower speed

reduce speed, the same steps are followed as

is

case of field reversal, except that the armature

quickly rearrange the circuit so

the dc machine again runs as a motor. This

versed instead of the

in-

in the is re-

field.

volves the following steps:

22.4 Two-quadrant control-

Step 4: Delay the gate pulse by nearly 180° so that

Ed becomes

two converters

quite large and negative. This

operation takes a few milliseconds, after which

again zero.

current

I d is

Step

Reverse the

5:

field current as quickly as

make E

possible so as to

()

ing time again lasts from this interval, the

Step

6:

1

is

During

zero.

acts as a motor,

is

Both are connected

tifier

at

connected

we

use

reverse paral-

to the armature, but only

a given time, acting either as a rec-

The other converter is whenever power to

or inverter (Fig. 22.5).

no need

is

faster,

in

to take over

the armature has to be reversed. there

to reverse the

Consequently,

armature or

field.

The

time to switch from one converter to the other

The machine

and the converter

lel.

on standby, ready

Ed becomes positive greater than Eih enabling the de-

to the rectifier

speed control has to be even

identical converters

one operates

that

sired armature current to flow.

now

revers-

to 2 seconds.

armature current

Reduce a so

and slightly

The

positive.

When two

10 ms. Reliability

typically

back

and maintenance

proved,

mode.

is

is

is

considerably im-

reduced.

Balanced

against these advantages are the higher cost and in-

creased complexity of the triggering source.

22.3 Two-quadrant control armature reversal In

some

Because one converter

industrial drives, the long delay associated

with field reversal

is

unacceptable. In such cases

we

reverse the armature instead of the field. This requires a high-speed reversing switch designed to

carry the full armature current.

The

control system

arranged so that switching occurs only

mature current

is

zero.

Although

wear and arcing, the switch

still

this

Due

to its

when

the ar-

reduces contact

thousand amperes.

low inductance, the armature can be

reversed in about 150 ms, which

is at

least 10 times

faster than reversing the field. Fig. 22.4

fied circuit

is

has to be fairly large

to carry a current, say, of several

is

a simpli-

showing a dc shunt motor connected

the converter by

means of a reversing

always ready to take

is

over from the other, the respective converter volt-

contactor.

to

To

ages are close to the existing armature voltage, both value and polarity. Thus,

in 1

in Fig. 22.6a,

converter

power to the motor at higher than the cemf E During

acts as a rectifier, supplying

a voltage slightly

()

this period, gate pulses are

2 so that

it is

inactive. Nevertheless, the control cir-

cuit continues to generate pulses

so that

Ed2 would be

allowed

equal to

to reach the gates

To reduce

.

withheld from converter

the

EcU

(G7

to

having a delay a 2 if

the pulses were

G12,

Fig. 22.5).

motor speed, gate pulses a are det

layed and, as soon as the armature current has fallen to zero, the control circuit

converter

1

withholds the pulses

to

and simultaneously unblocks the pulses

to converter 2.

Converter

1

becomes

inactive and

546


converter

converter 2

1

Figure 22.5 Two-quadrant control using two converters without circulating currents. the delay angle

L

comes

a2

is

Ed2

then reduced so that

slightly less than

E0

,

be-

thus permitting reverse

current Id2 to flow (Fig. 22.6b).

This current reverses the torque, and the motor

speed decreases rapidly. During the deceleration

a2

phase,

is

varied automatically so that

the rapidly decreasing value of is

a

gle

Figure 22.6a Converter

in

1

this

i

tracks

the pulses operation; converter 2 blocked.

In

Ed2

follows

some cases a 2

period the control circuit continues to

generate gate pulses for converter

converter 2

1

.

varied to maintain a constant braking current.

During converter

E0

If the

a2

so that

were allowed

Ed]

1 ,

and the delay an-

would be equal

to reach the gates

motor only operates

in

quadrants

(G 1

to 1

Ed2

if

to G6).

and 4, the

direction of rotation never reverses. Consequently, L

converters

and

1

and 2 always

act respectively as rectifier

inverter.

22.5 Four-quadrant control -two converters with circulating current Some converter

1

in

converter 2

torque. Unfortunately, the converter current operation; converter

and

down to zero speed. This means that the torque may at times be much less than rated

Figure 22.6b Converter 2

industrial drives require precise speed

torque control right

1

blocked.

is dis-

continuous under these circumstances. In other


words, the current for

1

20°. Thus, at

erratic,

To

in

each thyristor no longer flows

low torques the speed tends

and precise control

is


that function

simultaneously.

we

use two converters

They

Example 22-3 The dc motor in

are connected

When

Converter

and vice

versa.

The armature

current /

the

is

b.

difference between currents /d] and /d2 flowing in the

two converters. With

this

arrangement, the currents

when

both converters flow during 120°, even zero. Obviously, with

operation, there

one

to the other.

is

no delay

at all in

The armature

c.

/ is

d.

in

most sophisticated control system

most expensive. The reason

verters operate simultaneously,

is

ac circulating currents.

V, calculate the following:

1

The The

firing angles for converters

reactive

power drawn from

is

available.

The dc power delivered by converter

A

1

is

It is

when con-

that

= 450 X

L2

)

1

800

= 810kW

to limit the

The power absorbed by converter 2

fed

as an inverter)

the isolated secondary

shown

incoming

Solution

(operating

is

typical circuit

composed of a delta-connected primary and two wyeconnected secondaries

and 2

this represents

The converters may be

windings of a 3-phase transformer.

1

the

line

each must be pro-

vided with a large series inductor (L,,

from separate sources, such as

360

The dc power associated with converters and 2 The active power drawn from the incoming

3-phase

a.

the

is

of 1800 A, and

,

ac line voltage for

If the

switching from

current can be reversed

almost instantaneously; consequently,

also the

delivers a current / d

3-phase line

in

two converters continuously

1

each converter a.

inverter,

Fig. 22.8 has an armature voltage

while drawing a current of 1500 A.

converter 2 absorbs 300 A.

an

rectifier the other functions as

V

of 450

difficult to achieve.

back-to-back across the armature (Fig. 22.8).

one functions as a

be

to

547

in Fig. 22.8.

= 450 X 300

Other

=

transformer circuits are sometimes used to optimize

performance, to reduce cost, to enhance

reliability,

or

Note

to limit short-circuit currents.

kW

135

that the converter voltages

£tn

and

Ed2

are

essentially identical because the dc voltage

drops

in L,

and L 2 are negligible. This means

that the respective triggering of converters

and 2 cannot be voltage-controlled (by example). In practice, the triggering controlled and

is

made

converter currents b.

The

active

/ dI

to

and

is

£"

1

0 for ,

current-

depend on the desired /d2

.

power drawn from

the incoming ac

line is

P = P - P2 {

= 810 -

135

= 675 kW Secondary winding

Figure 22.7 Precise 4-quadrant electronic speed control

vided

in

a modern steel


mill.

is

pro-

1,

2, 3 delivers

while secondary winding

kW.

It

7, 8,

810

kW

9 receives 135

follows that the net active power drawn

from the

line (neglecting losses)

is

675 kW.

548


3-phase line

/d.

G3 -

/

= /d.-/d-

G4 -

converter

Figure 22.8 Two-quadrant control

c.

The approximate

of

1

a dc motor using two converters with

firing angle for converter

can be found from Eq. 21.13: 1.35

E cos

1.35

a

0.926

cos

circulating currents.

Note

assuming Id2

that

would take hence

in

is

a very small

£dJ

)

to

make

held constant,

change

in cti

it

(and

a big change in /d]

The

.

oti

X 360

450

converter 2

^2

cos

a

|

reason

is

tween

£dl

that 7dl

is

E0

and

mature resistance

,


divided by the very low ar-

R a Such .

a big change in /d]

|

=

would produce

22.2°

a corresponding big change in

the current / feeding the armature.

Because

£d2

is

nearly equal to

angle for converter 2

same approximate because angle

it

is

found

value.

£dl to

,

the firing

have the

However,

operates as an inverter, the

same way, a very small change

In the

a2

in

(with /dj held constant) produces a big change in /d2 and, therefore, in the

armature current.

Thus, although the approximate values of

a

{

and a 2 are mainly determined by the magnitude

is

-

a2 =

180

-

180

-

157.8°

-

of the armature voltage a,

upon the speed), 22.2

E0

(which depends

their precise values are set

the desired value of the armature current is

why

7d

|

and

Id2

have

to

/.

by

That

be current-controlled.


d.

The

power drawn by converter

reactive

x

=

33

kvar

1

power drawn by converter 2

reactive

is

= P tan a, - 810 tan 22.2°

Qi

The

1

549

Ql = F*2 tan a 2 - -135 tan

=

is

Figure 22.9 157.8°

Hoist raising a load.

55 kvar

Consequently, the reactive power drawn from

incoming 3-phase

the

Q=

line

is

+ Qi = 331 + 55 Qi

= It is

interesting to note that

whereas the active

= P v — P 2 the reactive = (Q Q + Q 2 The reason is that

powers subtract (P

powers add:

lower

386 kvar

)t

Figure 22.10 Hoist lowering a load.

).

x

a line-commutated converter always absorbs re-

active power, whether

it

functions as a rectifier

or inverter.

22.6 Two-quadrant control with positive torque

power to the motor, becomes a generator. We can feed the resulting electric power into the ac line by making the converter act as an inverter. The gate pulses are simply delayed by more than 90°, and Ed is adjusted to the descending weight delivers

and so

it

obtain the desired current flow (Fig. 22. 10).

Hoisting and lowering can, therefore, be done in

So

far,

we have

discussed various ways to obtain

when

torque-speed control

However, many that

always act

the

torque reverses.

industrial drives involve torques in

one

even when the

direction,

speed reverses. Hoists and elevators

fall into this

whether the load moves up or down. Operation 1

and

to the

The armature

is

is

connected

output of a 3-phase, 6-pulse converter.

the load

is

When

being raised, the motor absorbs power

from the converter. Consequently, the converter acts as a rectifier (Fig. 22.9). directly

When verses,

The

is

armature has to be reversed.

22.7 Four-quadrant drive

We can

readily achieve 4-quadrant drive of a dc

ma-

chine by using a single converter combined with ther field or armature reversal.

of switching is

may be

ei-

However, a great deal

required. Four-quadrant control

possible without field or armature reversal by using

speed depends

two converters operating back-to-back. They may function either alternately or simultaneously, as pre-

<

upon the weight of the

the load

field or

Ed The armature

lifting

upon the converter voltage

current depends

required.

2.

Consider a hoist driven by a shunt motor having constant field excitation.

is

downward

category because gravity always acts

therefore in quadrants

manner and no field or armature reversal However, the empty hook may not descend by itself. The downward motion must then be assisted by the motor, which means that either the a stepless

viously described in Sections 22.4 and 22.5.

load.

being lowered, the motor

re-

which changes the polarity of E0 However, .

The following example control of an industrial drive.

illustrates

4-quadrant


550

Note: Clockwise speed and torque are positive.

torque

2

3

1

& speed

!

t

Figure 22.11 Torque-speed characteristic

of

an

industrial drive.

FOURTH

THIRD

SECOND

generator

motor

generator

inverter

rectifier

inverter

(+)

(-)

H

r/min

Nm

3 cr

9

4

3

CD cd

&

1000 800 600 400 200

— — — —

quadrant

machine mode converter

mode

converter polarity

speed

0

X-i 14

200 400 600 800 1000

—

r-?

'

16

18

20'

\

ry—\

i

22

24

—

26 S

/

time

Figure 22.12

See Example

22-4.

Example 22-4

An

industrial drive has to

four quadrants. In doing so,

develop the torque-speed

characteristic given in Fig. 22.1

1

.

Adc

shunt motor

is

zero.

from one quadrant

powered by two converters operating back-to-

sition

back.

The converters function

Fig. 22.1

alternately (only

one

at

state

of each converter over the 26-second operating

period, and indicate the polarity at the terminals of the

dc machine. The speed and torque are considered positive

when

acting clockwise.

The

analysis of such a drive

1,

at 2, 8, 15,

simplified by subdi-

viding the torque-speed curves into the respective

tran-

Referring to

and 25

s.

We draw vertical lines through these points (Fig. 22.12). We then examine whether the torque and speed are positive or negative during each subdi-

Knowing

can immediately state is

to another.

the speed or torque passes through zero 21,

vided interval.

Solution

mo-

for those

These moments always coincide with the

used,

a time) as explained in Section 22.4. Determine the

we look

ments when either the torque or speed pass through

in

the respective signs,

we

which quadrant the motor

is

operating. For example, during the interval from

2

s

to 8

s,

both the torque and speed are positive.


Consequently, the machine 1.

On

25

s,

is

operating in quadrant

the other hand, in the interval from 21

the speed

is

the quadrant,

machine functions assuming

to

negative and the torque positive,

indicating operation in quadrant

Knowing

s

and whether

it

1

TABLE 22A

which converter

is in

Operating

machine

acts as a generator.

Consequently, one of the two converters must function as an inverter. first

But which one? To answer the

look

at the polarity

Because the speed

polarity

is

negative,

is

machine

can carry

Converter 2

1

2-8

as

shown

in

acts as a generator.

Fig.

22.13b.

Only converter

of current flow, and so

this direction

the one in operation.

s

8-15

rectifier

off

off

inverter

s

15-21

s

off

rectifier

21-25

s

inverter

off

of the arma-

negative, the armature

Current flows out of the positive terminal because the

Converter

interval

operation,

acts as a rectifier or inverter.

clear that the

we

mode

Time

Thus, taking the interval from 21 to 25 seconds,

ture.

encourage the reader to verify them.

3a),

the required direction of current

flow. This tells us

question,

We

for the

results are tabulated in Table

the

Finally,

sponds to a positive armature voltage (Fig. 22.

is

The

other intervals.

that a positive (clockwise) speed corre-

we can deduce

it

similar line of reasoning enables us to deter-

mine the operating mode of each converter 22A.

2.

we know whether

motor or generator.

as a

A

551

it

1

is

22.8 Six-pulse converter with freewheeling diode

When

a dc

motor

is

started up,

we can

signifi-

cantly reduce the reactive power absorbed by the

converter by placing a diode across the converter output (Fig. 22.15). The usefulness of this free-

wheeling diode

best illustrated by a numerical

is

example.

Suppose a dc motor has the following characteristics:

hp

rated power: 100

converter

1

converter 2

rated armature voltage:

240

rated armature current:

320

armature resistance: 25 mil

armature inductance:

1.7

Figure 22.13a Polarities

when

the speed

is

positive.

V A

We

mH

begin our analysis using a conventional

6-pulse converter to drive the motor. (rectifier) is fed

by a 3-phase,

will analyze the voltages tor is at standstill

1

The converter

84 V, 60 Hz

line.

We

and currents when the mo-

with rated current flowing

in the

As a result, the motor will develop rated The circuit of the motor and converter is

armature. torque.

shown in Fig. The motor converter

1

converter 2

from 21 s to 25

needed

to

is

s.

stalled

and the value of

Ed

supply the armature IR drop.

Ed =

Figure 22.13b Interval

that

22. 14a.

=

IR 8

= 320 A X

V

25 mfl

is

only


552

Figure 22.14a Conventional

To develop is

supplying rated current to a stalled dc armature. Firing angle

rectifier

this

dc voltage, the required firing angle

£d = 8

-

l

.35

power

E cos a

is

so large.

£ 1N

l

.35

X

1

84 cos a

The

a = 0.0322

a =

power P supplied

to the converter

is

necessarily equal to that absorbed by the armature.

Thus, 8

= = =

reactive

power

X 320 = 2560

W

Q absorbed by

Pima =

the converter

is,

2560

would be useful

261

+

is

A

if

137

wave

we

E KA The

will then increase.

and

tan 88.15°

£ KA wave. The peak

.

1

V

and

— 123

V, but

only 8 V.

is

could remove the nega-

reason

is

that for a given

5°) the average voltage

Ed

The negative voltages can be

A (Fig.

22. 15a). In effect, as soon as

K K becomes

negative with respect to A, the diode starts con-

79.25 kvar

that the reactive

power

is

3

1

times greater than

the active power.

sponding line current

Q

1

is

/a

are

£ 1N

shown

and the correin Fig.

22.14b.

As

a result,

triggered 88.15° after B (V

the center of the positive current pulse lags 88.15°

all

currents

cease to flow in the converter, and so the line currents


that

/c ,

(21.6)

ducting. During the conduction period

Note

,

suppressed by placing a diode between terminals

2= =

The

to /b

X 320

0.816

the average value of the

tive voltages in

therefore,

Note

/ a lags

the reactive

/d

a sawtooth

is

firing angle (such as 88.

The

0.816

value oscillates between

It

tl

Thus,

Figure 22.14b also shows that voltage

88.15°

across the armature

P = EJ =

.

why

effective value of the line current /

active ac

is


and so

The

and that

and their respective line-to-neutral voltages.

whence cos

£ ]N

behind the positive voltage peak almost 90° behind

given by

88.15°.

is

/

a , / b , /c

are also zero.

Note

that / a

is

now com-

posed of a double positive pulse of current followed

by a double negative pulse.

The presence of the diode produces the E KA wavein Fig. 22. 5a. The negative voltages are

shape shown

1


— Q5

(

Q1

Q2

Q3

Q4

Q5

Q6

Q1

fires

fires

fires

fires

fires

fires

fires

Q6— — Q6,

Q1

— -Q1, 02-4-02, Q3

-Q3,

Q4+ *-Q4,

553

Q5-4-Q5, Q6-4

conducting conducting conducting conducting conducting conducting conducting

1.414£ 1.225

E

\\

m

/vv\ /o\

120

\

—V

/*** '

? •*****/

V /

\

/

X

\

/

/

-

\

240

180

/\

^300^

/\ /

480

540

\/ \

/

g5^^\^g6^^\_g1

94

320

420

360

A

-320 A Figure 22.14b Voltage and current waveshapes

when armature

is

stalled while developing rated torque.

clipped off, and so the dc voltage across the armature

becomes much ture current also

order to

larger than 8 V.

It

result, the

arma-

becomes much greater than 320 A.

make Ed =

angle a.

As a

8 V,

we must

can be proved

that the

In

£d =

is

1.35

E(\ - cos |120 -

1.35

X

184(1

-

cos 1120

=

1.35

X

184(1

-

cos 31.85°)

=

37.4

V

= =

dc voltage of a 3-

88.15])

The

resulting armature current would, therefore, be

given by

1.35

£(l - cos [120 -

/=

a]) (22.1)

This

where

£d = E=

a])

increase the firing

phase 6-pulse converter equipped with such a free-

wheeling diode

£j

is

dc voltage [VJ

is

37.4/0.025

=

1496

nearly five times the rated current, which

clearly unacceptable. In order to obtain the desired

effective value of line-to-line voltage

A

£d =

8 V,

we must

increase the firing angle such that

[V]

a = 1.35

=

firing angle

(between 60° and 120°)

a constant that applies only

when a

between 60° and 120° [exact value 3 \ 2/77

£ 8 -

[°]

tl

lies

= 0.9678

=

1.35

1.35

£(1 - cos [120 -

X

cos (120

184

(1

-

a)

-

a]) (22.1)

cos [120

I

therefore In our

88.15°

example, the dc voltage with a firing angle of is

120

- a =

arcos 0.9678

=

14.6°

-

a])

X

—

Q1

4

k

Q3

'a

1

1

3- phase

3-phase Trans-

line

/b 1

former

-©1

'4

,Q4

L

Q6

88.15°

^

Figure 22.15a Conventional

rectifier

and freewheeling diode supplying current

554

to

a stalled dc armature. Firing angle

is

88.15°


consequently,

£"i

a = 120 With

14.6°

this firing angle, the

-

105.4°

waveshapes of £ KA and

.

power

1

The

a is

is

active

again 320

power

Q

therefore 82.7° and

is

factor

is

PF (displacement) = cos

/a

shown in Fig. 22. 5b. The positive half-cycle composed of two current pulses. Each has an amplitude of 320 A and a duration of (120° — 105.4°) = 14.6°. The effective value of a therefore, are as

of

N The displacement angle

the displacement

555

82.7°

-

0.127

power P drawn from the 3-phase line 8 V = 2.56 kW. The reactive

A X

is

,

= 320V (4 X

/ a (eff)

Thus, the 3-phase line currents are in Fig. 22. 14a,

mature

is

=

14.6)/ 360

128.9

Q = P tan

A

=

much lower than

even though the dc current

Thus, with the freewheeling diode the converter

1

1

=

(105.4

+

=

only draws 20 kvar from the

180)/2

= 142.7°. On E lN occurs at

The reader should note

1.414 1.225

<5 d

=

(142.7°

-

60°)

=

that

82.7° behind

E KA

starts to

become

the diode to exert

its

It is

The

Q1

Q2

Q3

Q4

Q5

Q6

Q1

fires

fires

fires

fires

fires

fires

— Q5, Q6^ — Q6, Q1-* *-Q1. Q2^ -Q2, Q3 + -Q3, Q4—

Q6—

conducting conducting conducting conducting conducting

inducting EQ5,

line current is

Figure 22.15b rectifier

and freewheeling diode supplying current

fir-

only then

negative, thus permitting

influence.

fires

E£

Conventional

to

that the freewheeling

ing angle lies between 60° and 120°.

60°.

by angle

compared

diode only begins to produce an effect when the

Consequently, the fundamental component of lags

line,

79.26 kvar with no diode.

1

the other hand, the positive peak of 4> e

kvar

in the ar-

the same.

<&;

2.56 tan 82.7°

- 20

is centered The positive half-cycle of current midway between 05.4° and 80° (see Fig. 22. 5b).

Thus, angle

(displacement angle)

to a dc armature. Firing angle

is

105.4°,


556

a =

not sinusoidal and possesses a strong harmonic con-

can be shown that the fundamental compo-

tent. It

firing angle

L8o°)

•

nent of the line current lags behind the correspond-

=

0.675

30°

d

a constant that only applies for

(1.5

a/2

4-

(222)

needed

firing angle

a =

displacement phase angle firing angle

and 120°) 30°

The

=

(must

lie

[°]

Ed = 8 =

between 60°

[°1

/

=

/

d

is

given by

V(120 - a)/ 90


factor

(displacement)

The displacement angle the reactive power Q:

=

The

cos 4> d

(22.4)

O d can be used to calculate

0.675 £(1

X

0.675

+

Section 22.8, the

in

Ed =

produce

8

V is given by

cos a)

184

cos a)

4-

(1

a = 159.32°

(22.3)

line-current pulses have an amplitude equal

armature current, namely 320 A. The du-

to the dc

PF

to

whence

given by

is

beis

—0.936 = cos a

a constant for this type of converter


a

V 2)/ttJ

Using the same example as

where

Od =

between 60° and

lie

tween 60° and 180° [exact value

ing line-to-neutral voltage by an angle <£ d given by

* =

(must

n

ration of the current pulses

is

-

(180°

159.32°)

=

20.68°. Based upon the example of Fig. 22.16, the positive current pulse starts at 159.32° and ends at 180°.

Q-Ptan^

(22.5)

The negative

and ends

at

through zero

22.9 Half-bridge converter

The half-bridge converter is another way whereby the reactive power can be reduced when the dc output voltage is low. Fig. 22.16 shows a 3-phase, 6-pulse

-

c>

current pulse starts at 279.32°

300°. Thus, current at

an angle

/.,

essentially passes

0 0 given

by

+

=

1/2(159.32°

300°)

229.66°

Consequently, the positive peak of sentially at an angle of (229.66°

Thus, the displacement angle

/a

occurs es-

-

90°)

=

79.66°

=

139.66°.

is

converter in which three thyristors have been re4> t

placed by three diodes D2, D4, D6. This half-bridge converter has properties similar to those of a conven-

,

=

-

(139.66°


60)

factor

is

tional thyristor bridge with a freewheeling diode.

The 180°.

firing angle

a can be

However, the freewheeling diode

effect only

begins for angles greater than 60°. In Fig. 22. 6, the 1

firing angle is is

assumed

to

be

1

PF

varied from zero to

35°.

As

a result,

positive during successive 45° periods.

Note

As

(displacement)

before, the active

cos

d

=

0.179

power supplied by

the 3-phase

line is

E KA

P=

that

the positive and negative current pulses in each

=

The

reactive

2.56

kW

power absorbed

is

3-phase line also flow for 45°.

The average value of £ KA

£d = in

is

0.675 £(1

+

2.56 tan 79.66°

14kvar cos a)

(22.6)

The reader will note that even power is needed with the half-bridge

which

£d = E=

Q = P tan $ d =


dc voltage across the load [V] effective value of line voltage [V]

less

reactive

rectifier than

with the freewheeling diode circuit of Fig. 22. 15b.

The

effective value / of the line current

is

given by

557


558

/c f

X

20.68°

X

2

=

2

I

X

improvement

the

360°

some of the whence /

=

A 0.339

-

/d

0.339

X 320 =

A

108.4

that has taken place, let us

equipped with,

train

started with both "motors

two dc motors,

say,

connected

As the speed picks The motors are then

external resistor.

Table

22B sums up

the basic properties of the three

we have

types of rectifier converters the case of converters its

are those during

takes place.

B and

discussed. In

C, the firing angle lim-

which freewheeling operation

The values of Ed

,

Od

,

/,

and so

forth, are

only valid within the stated limits. Furthermore, the load

is

assumed

to be resistive.

shorted out.

is

connected

produces small

modified to make use of the ad-

vantages offered by thyristors and

GTOs. Existing some cases,

operate on dc and, in still

which, of course, are repeated

jolts,

tracks, with

it

also produces slippage

consequent loss of traction. The it

and speed.

used.

To modify such

sys-

We now

study

some simple chopper

cir-

cuits used in conjunction with series motors.

of the special properties of the dc series motor.

dc series motors are

its

during the electric braking process. Although a jolt

dc chopper overcomes these problems because

and buses had for years been de-

still

paralleled and

permits smooth and continuous control of torque

signed to run on direct current, principally because

trolley lines

up, the resistor

shorted out, as the train reaches

is

is

an

nominal torque and speed. The switching sequence

affects passenger comfort,

22.10 Detraction

Many have been

in series to

with another resistor. Finally,

in series

the last resistor

on the

Electric trains

review

features of the older systems.

Fig. 22.17

shows the armature and

motor connected

ries

Supply voltage

head

E

s

is

to the output

field

of a se-

of a chopper.

picked off from two bare over-

trolley wires. Capacitor

C

{

furnishes the high-

whose amplitudes are equal to the dc armature current drawn by the motor. The

current pulses

tems, high-power electronic choppers are installed

large

on board the vehicle (see Section 21.37). Such

inductor L, has a smoothing effect so that current /

choppers can drive motors rated

horsepower with outstanding

hundred

at several

results.

To appreciate

drawn from

the trolley line (or catenary) has a rela-

tively small ripple.

PROPERTIES OF SOME RECTIFIER CONVERTERS (RESISTIVE LOAD)

TABLE 22B

Converter

A

Converter

B

Converter

C

Items 3-phase. 6-pulse

0

firing angle (a) limits

90°

to

dc output voltage (£d )

1.35

displacement angle (O d )

a

PF

(displacement)

=

effective line current

Total apparent

Total active

(total)

a

cos 3> d

cos

0.816

(S)

EI

power (Q)

cos

60°

a

\

freewheeling diode

E{\ -

+

/

d \

+

(120

El V3

3

cos [120

-

a])

0.675 E(i

a/2)

- a)/ 90

tan

PIS

<£>

d

P

tan

PIS

/

d

\

/:/ \

d

cos a)

cos a/2 (

1

SO

3

Ed Id P

+

a/2

a/2

cos (30 IL[

60° to 180°

to 120°

1.35

30

(/)

power (P)

Total reactive

PF

power

E

Half-bridge

3-phase, 6-pulse

+

P

tan <£ d

PIS

-

a)

/

180


trolley wires

559

trolley wires

Figure 22.17 Direct-current series motor driven by a chopper.

Figure 22.18a

The chopper is not a switch as shown, but a commutated SCR.

See Example

As

motor

far as the

a.

concerned, the

is

force-

total

tance of the armature and series field

enough

induc-

and release the energy needed dur-

to store

ing the chopper cycle. Consequently, no external in-

ductor

is

required.

When

chopper frequency

is

corresponding on time

T.

some older systems, Ta

is

switching frequency (about 2000 Hz)

motor

the

starts up, a

typically

is

500

[is.

In

}

kept constant while the

The top frequency

varies.

limited by the switching and

is

Most choppers function

at

Tlv

such cases,

T.d

may

tor delivers

still

a.

T.d

range from 20

to

240

A X

0.

Consequently,

is

related to the input voltage

E0 =

a

s

= DE

(21.21)

s

T.

is

the

x

E0 =

is

V

and

s

10

6

/ = 57.14 Hz \/f= 1/57.14 17

500

|jls

(Fig. 22.18b)

The dc current drawn from /

= PIE =

=

/s

=

8.23

S

24

the catenary

is

X 240/700

A 7*.

trolley-bus

+ 7b

17 500 .

is

driven by a 150 hp, 1500 r/min,

V series motor. The nominal full-load current is 200 A and the total resistance of the armature and field is 0. H. The bus is fed from a 700 V dc line. A chopper controls the torque and speed. The 600

—

I

I

1

I

1

chopper frequency varies from 50 Hz but the on time

Ta

is

fixed at 600

|jls.

to

1

600 Hz,

Figure 22.18b / s drawn by the chopper from the 700 V source when the motor is stalled.

Current pulses

zero

E = 700 V.

from

on time.

Example 22-5

A

24

700/ X 600 X

24

800

by the equation

and the cemf

E0 = E*fTa

D is the duty cycle, /is the chopper frequency

where and

EJT

£

V,

is at standstill.

find the frequency

times are varied. Thus, the chopper output voltage {)

= 24

il

1

because the motor

mains the same, no matter how the on-off switching

E

mo-

1

is

are var-

jjls

the

rated output.

its

Nevertheless, the basic chopper operation re-

(jls.

when

stand-

Referring to Fig. 22. 8a, the armature IR drop

more sophis-

and

is at

Solution

constant frequency, In

ticated controls, both the frequency ied. In

Calculate the chopper frequency

We can

turn-off time of the thyristors.

but with a variable on time

b.

low

50 Hz. The

used, typically

Calculate the chopper frequency and the current

drawn from the line when the motor still and drawing a current of 240 A.

large

is

22-5.

560


during start-up)

Example 22-6 Referring to Example 22-5 and

At rated output the voltage across the motor

late the

(Note the very low current drawn from the

b.

terminals

600

is

frequency

is

V

(Fig. 22. 19a).

line

The required

tor

peak valueôf currents

/s

Fig. 22.18a, calcu-

and

/

when

mo-

the

standstill.

is at

therefore given by:

Solution

E0 = 600

EJT

a. X

= 700/ X 600 X

10

Although the average value of / s peak value

6

ries

/=

1429

Hz

Ta + Tb = 1//=

= 700

(Fig.

b.

22.19b)

/

/s

=

its

in a se-

the other hand,

at 240 A. The average value of line current / is 8.23 A. The voltage across the capacitor fluctuates and is

steady

have a ripple because inductor

/ will

Consequently, the peak value of

PIE,

= 600 X =

8.23 A,

L, does not have infinite inductance.

/;

=

On

of brief, sharp pulses.

so current

Line current

is

240 A. The current flows

the armature current / 0

1/1429

ijls

is

171

slightly greater than the

/ will

be

average value.

200/700

22.11 Motor drive using a dc-to-dc

A

switching converter In Section 2

trolley wires

200 A

converter.

1

.42

It is

we

studied the 4-quadrant dc-to-dc

eminently suited for dc motor drives.

Consider Fig. 22.20

in

which a 3-phase source

is

converted to dc by means of a 6-pulse uncontrolled rectifier.

The dc output

is

is

applied to a 4-quadrant

L C. The converter composed of IGBT switches Ql Q2, Q3, Q4 and

switching converter via a

filter

cl ,

,

their associated diodes. Its output terminals

B

A,

are

connected to the armature of a dc motor. The armature

Conditions

when

is

composed of

the armature resistance

/?.„

and the counter emf

EiV

armature inductance L a

Figure 22.19a the motor

is

running at rated torque

and speed.

The shunt

field

is

,

excited from a separate source

(not shown).

Currents

/h

values. Current

sumed

700 nt

ceed,

to

we

/2

,

/d

and

/.,

represent instantaneous

supplied by the rectifier

as-

is

be constant and ripple-free. As we prowill

make

distinctions

between

instanta-

neous and average, or dc values.

We could

—

analyze the drive

in

terms of symbolic

equations, but such an algebraic approach lacks the

—100 ms

interest

and impact of dealing with numerical

ues. Consequently,

we

will use

two examples

val-

to

il-

(b)

lustrate the factors that

Figure 22.19b Corresponding current pulses from the 700 V source.

The /

s

drawn by the chopper

first

come

into play in a dc drive.

examines the system when the motor op-

erates at full-load.

The second covers

under dynamic braking conditions.

the behavior


/J

= 61.7 A

"

£ 12 = 324V

£ CB = 250V

1. -

-t-

Q3

Q1

+

~

C

recti-

500

238 V

80 A

_-

fier

< Q4*

/a

Q2

Figure 22.20 Dc motor controlled by a 4-quadrant dc-to-dc

L^y-J

150

240 V 3-ph

56

Figure 22.21a converter.

See Example

22-7.

Example 22-7

A 25

250 V, 900 r/min dc motor

hp,

connected to

is

thus,

a dc-to-dc converter that operates at a switching fre-

quency of 2 kHz. The converter rectifier

Ld

A 500

(jlF

C

capacitor

line

and

D=

is

80 A.

We

wish

to deter-

The 250

V

appears between terminals A,

The required duty cycle when its

B

(Fig.

22.21a).

Because the motor develops rated torque, the

the following:

ops

c.

0.886

1

mature draws

b.

1)

so,

and an induc-

The armature resistance and inrespectively 50 mfl and 4 mH. The

rated dc armature current

a.

(2D -

act as filters.

ductance are

mine

Hz

connected to a 240 V, 3-phase, 60

(Fig. 22.2 la).

tor

250 = 324

fed by a 6-pulse

is

the

motor devel-

its

voltage drop in the armature resistance

rated torque at rated speed

The waveshape of currents The waveshape of voltages

/,, / 2 ,

E i2

and

80

The induced armature 900 r/min is, therefore,

.

Solution

The 3-phase

rectifier

0.15(1

=

12

is

V

/.,

£AB

and

A X

ar-

namely 80 A. The

rated current,

produces a dc voltage

Ed

voltage, or counter emf, at

E = 250 t ,

12

= 238 V

given by

The dc power input

Ed =

1.35

E

=

1.35

X 240 V = 324 V

P = 250 V X

This voltage appears between the input terminals

motor

Neglecting the losses

80

A=

produce the rated 250

V across the ar-

that current /d

is

in the converter,

relationship

is

is

and recalling

324

V,

it

follows

given by

mature, the duty cycle of the converter has to be ad-

The

W

20 000

that the dc output of the rectifier

justed accordingly.

is

1

2 of the converter. In order to

to the

(21.4)

324

given by Eq.

/d

= 20 000

/d

=

61.7

A

the converter

is

21.24:

output voltage

=

input voltage X.(2

E LL = E H (2D -

1)

D—

The frequency of 1)

(21.24)

period of one cycle

T=

2

kHz and

is

\/f= 1/2000

= 500

fxs

so the


562

/

d =

7

61.7A

/

1

1 1

d

A

= 61.7

£ AC = 324 - 250 = 74 V

--80A

—

= -80

l\

A

A

80

1

£ CB = +250 V

h h

Q1

18.3

500

141.7

A C:

^

A

C m"

B

\

C 500

|iF /

80 A

a

80 A

are "on." Current /a

is

increas-

times of

Ql (and Q4)

are, respec-

(and Q3) are 57

juls

and 443

ously, followed by

when D2 and D3

Q2

When Ql and Q4

(= 80 A) now flows toward terminal

in Fig.

this

in the positive direction.

—

61.7)

=

80 A.

18.3

It

(= 80 A)

/,

Note however,

the rectifier only furnishes 61. 7 A, is

22.2 lb. This

time

whereas the

come from

ar-

the capaci-

The capacitor discharges, causing the voltage across it to drop by an amount AE given by

tor.

AE = QIC =

18.3

A X

Ql and Q4 then open

Q2 and Q3

443 u.s/500 |xF for

57

|xs.

,

which

op-

is

furnished by the rectifier

/d

a result, by Kirchhoff 's cur-

its

value

=

During

16

V

this in-

are closed (Fig. 22.21c), but they

cannot carry the armature current because

it is

is

.

+

(80

6

=

.7)

1

lights the absolute necessity of

the circuit. Without

it,

141 .7 A. This high-

having a capacitor

inhibited during this 57

jjls

interval.

The

capacitor charges up and the increase in voltage is

in

the flow of armature current

AE

given by

AE = QIC=

141.7

A X

57 |xs/500

|jlF

=

16

V

that

follows that the difference

A must

1

22.21b.

in Fig.

must flow into the capacitor

would be

shown

had

it

Ld As

and

operate simultane-

are conducting, the armature

and during

|jls

a is

/

rent law, the current I 2

ence of inductor of

and Q3, which also open and

current follows the path

mature current

£CB = +250V

are conducting. Current

Meanwhile, current

close simultaneously.

443

£ AC = -574V

jus)

continues to flow unchanged because of the pres-

|xs.

Ql and Q4

recall that

lasts for

80 A

jjls

primes

follows that the corresponding on and

terval

Circuit

posite to the direction

flows

a

decreasing.

Tu = DT = 0.886 X 500 = 443 Th = 500 - 443 - 57 jjls

(80

/

ID2

(57

tively,

We

B

jliF

Q2;

D3

Figure 22.21c

when Q1 and Q4 ECA = -74 V.

The on and

Q2

Q3:

238 V

^

Q4;

Figure 22.21b

It

15 °

mH

(443 us)

Circuit ing.

A

flow-

Note

that the increase in voltage across the capacitor

during the 57

|jls

interval

crease during the 443

|jls

is

exactly equal to the de-

interval.

ripple across the capacitor

is,

The peak-to-peak

therefore, 16 V. Thus,

the voltage between points 1 and 2 fluctuates be-

tween (324

+

8)

=

332

V and

(324

-

8)

=

3 16

V

This 2.5 percent fluctuation does not affect the operation of the motor.

Let us

now

look more closely

at the

armature

current, particularly as regards the ripple. In Fig.

ing opposite to the direction permitted by these

22.21b the voltage across the armature inductance

IGBTs. However, the current must continue

can be found by applying

to

flow

because of the armature inductance. Fortunately, a offered by the diodes

path

is

with

Q2 and Q3,

as

shown

D2

and

D3

in the figure.

associated

Note

that

/,

KVL:

£ac + £cb + £b2 +

E AC +

250

+

0

+ £ia = - 324 + 0 = Ei\

0 0


Hence

563

£ CB = 250 V

74

-AC

+ 324

V

Therefore, the volt seconds accumulated during this

443

interval

|jls

is

74

X 443 -

32 782

jxs-

sulting increase in armature current A/., 6 A/ = A/L a = 32 782 X 10~ /0.004

=

tl

V.

The

re-

is

A

8

(2.28)

Next, consider Fig. 22.21c. The voltage across the

armature inductance can again be found by applying

KVL: ÂC + £CB + £b1 + ^12 + ^2A = 0

E AC + 250 + EAC = —574 V.

Hence,

0

+ 324 +

0

=

0

This negative voltage causes a

very rapid decrease in the armature current. The decrease during the 57

A/a The

8

=

given by

jxs interval is

X

574

A decrease

57

10~ 6/0.004

X

during the 57

=

8

A

(2.28)

jxs interval is pre-

443 The peak-to-peak ripple is, therefore, A, which means that the armature current fluctu-

cisely equal to the increase during the previous jxs interval.

8

ates

between (80

+

4)

=

84

A

-

and (80

4)

=

61.7A

76 A. Figure 22.2 Id shows the waveshapes of the various voltages and currents.

Example 22-8

-76 A

We now

-84 A

consider the question of dynamic braking.

The same motor is used as in Example 22-7, and we assume it is running at 900 r/min at the moment that braking is applied. We further assume that the inertia of the motor and its load is very large. As a result, the speed cannot change quickly. The connection is

between the converter and the 6-pulse

removed and

a braking resistance of

nected between terminals

500

1

and

capacitor (Fig. 22.22).

|jlF

2,

20

V

332 A

316A

rectifier

fl is con-

along with the

We assume

443 us

that a

500 us

braking torque equal to 75 percent of nominal torque

is

sufficient.

mature current

is

Consequently, the required ar-

0.75

X

ing frequency remains

wish

to

V

80

A=

60 A. The switch-

unchanged

determine the following:

at

2 kHz.

We

Figure 22.21 d

Waveshapes

of currents

and voltages

in

Example 22-7.


564

other hand, the voltage should not be too high,

£h

1

otherwise 4

150

mH

ity

X

LJ*

20 n

500

b.

238 V

1$0A

Q4

IGBT devices.

The average current in the resistor is 524 V/20 ft = 26 A. Knowing the input and output voltages of the

we can determine

converter,

Q2

could exceed the withstand capabil-

it

of the switching

£ LL = £ H (2D- 1) 229 = 524(2 D - 1)

Figure 22.22 Dynamic braking. See Example 22-8.

b. c.

The voltage across the resistor The duty cycle required The braking behavior of the system

The on and

Because the motor

moment

E0 remains

voltage tor

60

is

at

induced

238 V. However, the mo-

must now operate as a generator and so the

A braking current flows out of the + (

minal, as

shown

tance

0.15 ft

is

X 60 A =

To

the armature resis-

9 V.

1

,

E LL of the

converter.

we

2 of the converter,

£H

Due

Q2

Ql and Q4

Q2

period

is

229 hence

V X

60

This

lasts for

900 r/min

is

The capacitor

age across

to

it

are closed, the armature

|xs

shown

in Fig. 22.23.

and during

this

resistor.

power

AE given

86

A X

140 ^s/500 jxF

26 A

1

20 a

*-

4

A

150

mH

LJ

C"

V 500

60 A

Q2

(140

238 V

\is)

higher than the previactually an

from continu-

to the drive system.

24

by

V

60 A

+

^F

It is

=

h

1

Thus,

(£ H ) /20ft

On

the

=

discharges, causing the volt-

advantage because the higher voltage automati-

ing to feed

(

x

re-

2

cally prevents the input rectifier

time I

.

drop by an amount

86 A

ous operating voltage of 324 V.

|xs.

The current in 60 A) flows out of terminal the resistor is still 26 A. It follows that a current (60 + 26) = 86 A must come from the capaci-

equal to the power ab-

much

and 360

for, say,

E H = E V2 = 524 V

This voltage

140

—

A=

Q3

and

the generator during

sorbed by the 20 ft braking

jjls

operate simultaneously, as do

and Q3.

When Q2

10 cycles of the converter switching frequency.

this 10-cycle

|±s

reason as

speed will

to the large inertia, the

The power output of

0.72

(and Q3) are 140

still

A£ = QIC =

essentially constant at

are, therefore,

follows that the corresponding on and off

times of

tor.

between

follows:

main

DT =

1

calculate the dc input voltage

terminals

Ql (and Q4)

current follows the path

The dc voltage between terminals A, B is (238 — 9) = 229 V, which is the required average output voltage

It

ter-

)

in Fig. 22.22.

The voltage drop across

0.72

X 500 = 360 Tb = 500 - 360 = 140 jjis

turning at 900 r/min at the

that braking is applied, the

off times of

Ta =

Solution a.

(21.24)

Therefore

D= a.

the value of the

duty cycle:

Figure 22.23 Current flows through IGBTs

Q2 and

Q3.

Q3


1

565

rotation

2

Figure 22.24 Current flows through diodes D1 and D4.

Next,

when Q2, Q3 open and Ql, Q4 close, the Dl and D4 22.24). Applying KCL, a current of (60

current has to circulate via diodes (Fig.

=

26)

360

34

|jls.

A must flow

The

into the capacitor during

resulting increase in voltage

AE = QIC =

34

A X

360

jjls/500 jjlF

=

/

is

24

V

Figure 22.25 Special current-fed dc motor.

Thus, the increase in voltage during 360

|jls

is

exactly equal to the decrease during the remaining 140

|jls

of the switching cycle. The voltage

across the resistor fluctuates between 524 12

= 536 V

and 524

-

12

=

512

+

power

down and

voltage between terminals A,

B

By

the speed

is

60

With the armature flows

is

it is

possi-

A braking current until

only a fraction of

This adjustment

its

N

rated value.

in coil

broken with

this coil,

it is

immediately established

next coil. Consequently, conductors facing the

pole always carry currents that flow into the page,

therefore, continuous and

may

be expressed by

T = kIB

(22 J)

where

electronic drives involve direct-current motors at all like

dc machines. The reason

is

T = motor

= — B /

that the usual rotating

shown, current

out of the page (toward the reader). The motor torque

circuit.

motors do not look

in the position

A

field.

A and the resulting torque causes the ar-

22.12 Introduction to brushless dc

Some

brushes are con-

while those facing the S pole carry currents that flow

is,

that

con-

mature to turn counterclockwise. As soon as contact is

in the

of course done automatically

by means of an electronic control

Two narrow

coil are

segments of a

opposite

permanent magnet N, S creates the magnetic

the

will decrease

continually adjusting the duty

ble to maintain the

The two ends of each

diametrically

feeds current into the coils as the armature rotates.

to the passive braking resistor. In

cycle during the deceleration period,

to

nected to a constant-current source that successively

so doing, the motor will slow

progressively.

nected

6-segment commutator.

V.

This example shows that the converter can transfer

other (Fig. 22.25).

commutator is replaced by a sta-

tionary electronic converter.

We now

discuss the the-

current

in

the conductors [A]

average flux density surrounding the currentcarrying conductors [T]

ory behind these so-called "brushless" dc machines.

Consider a 2-pole dc motor having three independent armature coils A, B, and C spaced at 1 20° to each

torque [N-m]

k

=

a constant, dependent upon the turns per coil

and the

number of

size of the armature


566

If the current

ing torque

is

and flux density are fixed, the

also fixed, independent of

result-

motor speed.

are 60° wide; conse-

The commutator segments

quently, the current in each coil flows in 60° pulses.

Furthermore, the current

in

each

every

coil reverses

time the coil makes half a turn (Fig. 22.26). The ternating nature of the current

is

al-

of crucial impor-

tance. If the current did not alternate, the torque de-

veloped by

each

would

coil

act

in

first

is the number of poles and n is the speed The frequency in the coils is automatically related to the speed of the motor because the faster the machine rotates, the faster the commutator

where p (r/min).

switches from one coil to the next. In other words,

commutator generates an ac current in the coils whose frequency is such that a positive torque is de-

the

veloped

As

one

speeds.

at all

the coils rotate, they cut across the magnetic

An

by the N, S poles.

direction, then in the opposite direction, as the ar-

field created

The net torque would be zero, and so the motor would not develop any power. Fig. 22.26 shows that the ac currents in the three

therefore, induced in each coil, and

coils are out of phase by 120°. Consequently, the ar-

coils are

mature

rotates.

behaves

mature

as

if

it

were

by

soidal. Basically, the

commutator

are mutually displaced at 120° due to the

a

voltages brushes.

waveshapes are rectangular instead of sinu-

rent

ical

is

acts as a

mechan-

converter, changing the dc current from the dc

is

given by

dc

a

is

the

between the

voltage

that the

way

The induced ac

brushes are always

moving

in the

same

in

di-

is

E

}

the armature

duced voltage (22.8)

always the same (see Section

4.2).

brushes were connected to a dc voltage

If the

is

180°

is

rection through the magnetic field; consequently,

5.2).

—

the armature.

appear as

The reason

the polarity

source

/= pnt\2Q -

mounted on

contact with coils that are

source into ac current in the coils. The frequency of the current

is,

frequency

also given by Eq. 22.8. Furthermore, the voltages

that the cur-

excited

3-phase source. The only difference

ac voltage

its

would accelerate

E0 was

What determines

the speed

fed from a current source, as

speed will increase

until the in-

about equal to

when

it

until the load

is in

E

(Section

the armature

our case? The

torque

is

equal to

-]

the torque developed by the motor. Thus, while the

speed of a voltage-fed armature depends upon equilibrium between induced voltage and applied voltage, the speed of a current-fed armature depends

upon equilibrium between motor torque and load The torque of a mechanical load always

torque.

rises with increasing speed.

h

Consequently, for a

given motor torque, a state of torque equilibrium

always reached, provided the speed

is

is

high enough.

Care must be taken so that current-fed motors do not run

22.13

away when

the load torque

is

removed.

Commutator replaced by reversing switches

Recognizing that each alternating current, 1

1

I

0

1

1

1

we

coil in Fig. 22.25 carries an

can eliminate the

tator

by connecting each

coil to a pair of slip-rings

and bringing the leads out

current

to a set

of mechanical

reversing switches (Fig. 22.27). Each switch has

Figure 22.26

The dc

commu-

I

60 120 180 240 300 360 0 60 120 180 240

changes

to

ac current

in

the coils.

four normally open contacts. Considering coil A,


coil

A

Coil

4

B

h

2

coil

slip rings

h

3

5

C

TV < ^/8> < > < > 14^^12 10\

o

it?\^/>ie

current source

Figure 22.27 The commutator can be replaced by an array chanical switches and a set of slip-rings.

of

me-

example, switch contacts 7 and 8 are closed during the 60° interval when coil side faces the for

10

pole (Fig. 22.28). The contacts are then open for

120° until coil side 4 faces the

N

pole,

whereupon

contacts 9 and 10 close for 60°. Consequently, by

Figure 22.28 showing how current

The dc motor from the one

in Fig.

we

obtain the

same

result as if

we

B and C

1

.

energized

at different times.

If

used a com-

22.29 looks so different

we

increase the dc current

/

or the field

consequently, the speed will increase.

same way, but they are Fig. 22.27 shows how

If

we

brushes against the direction of

shift the

rotation in Fig. 22.25, current will start flowing in

each coil a

little

earlier than before.

Consequently, the ac current

nected to the current source. The reversing switches

in

each

lead the ac voltage induced across ing dc

power

mechanical inverter, chang-

into ac power.

The

slip-rings

We can

merely

coil will

its

terminals.

produce exactly the same effect by

ing the thyristors a

provide electrical contact between the revolving

A.

22.25 that

the array of 12 contacts and 6 slip-rings are con-

really act as a 3-phase

in coil

strength of poles N, S, the torque increases, and

2.

operate the

controlled

For example:

mutator.

Coils

is

we would never sussame properties. And yet they do.

in Fig.

pect they have the

synchronizing the switch with the position of coil

A,

8

Circuit

1

N

567

little

fir-

earlier in Fig. 22.29.

ar-

Under these circumstances, the machine furpower to the three thyristor bridges, at the same time as it absorbs active power from them.

mature and the stationary switches and the dc power

nishes reactive

supply. Clearly, the switching arrangement of Fig. 22.27 is

more complex than

the original

commutator.

However, we can simplify matters by making the

ar-

3.

If

we

shift the

brushes by

1

80°, the current in

mature stationary and letting the permanent magnets

each coil flows

rotate.

By thus literally turning the machine inside we can eliminate 6 slip-rings. Then, as a final step, we can replace each contact by a.GTO thyris-

shown

out,

voltage in each coil remains unchanged be-

tor (Fig. 22.29).

The

12 thyristors are triggered

by

gate signals that depend upon the instantaneous position

of the revolving

rotor.

cause

in the

in Fig. 22.25.

it

opposite direction to that

However, the induced

depends only on the speed and

tion of rotation. Consequently, the

comes a

generator, feeding dc

the current source.

direc-

machine be-

power back

into


568

The same

result occurs if

we

fire the thyris-

The thyristors then feeding power back to the

tors 180° later in Fig. 22.29.

behave as

rectifiers

dc current source. It is

now

its

performance.

First, the

"synchronous motor" can never

of step because the stator frequency

ence between them

is

that

one

is

the gates of the

a stationary electronic

By

we produce

pull out

not fixed, but

SCRs

are triggered

is

that

by a signal

that

differ-

depends upon the instantaneous position of the

equipped with a ro-

commutator composed of 12

rotor.

For the same reason, the machine has no tendency

mechanical commutator, while the other has

thyristors.

is

changes automatically with speed. The reason clear that the machines in Figs. 22.25

and 22.29 behave the same way. The only tating

"brushless" dc machine. This has a profound effect

upon

oscillate or hunt

Second, the phase angle between the ac current

firing the thyristors earlier or later,

the

same

in a

winding and the ac voltage across

it

can be

effect as shifting the brushes.

modified by altering the timing of the gate pulses. This enables the synchronous motor to operate leading, lagging, or unity

22.14 Synchronous motor as a

The revolving-field motor

in Fig.

22.29

is

it

receives

its

ac power,

it

at

factor.

spective voltages and currents can be fully conbuilt like

a 3-phase synchronous motor. However, because of

way

power

Third, because the phase angle between the re-

brushless dc machine

the

to

under sudden load changes.

behaves

like a

trolled, the

machine can even function

as a genera-

tor,

feeding power back to the dc current source.

The

thyristor bridges then operate as rectifiers.


569

"synchronous"

motor

gates G, to

G6

1

e-

rotor position

gate I

triggering {

-

processor

!

other inputs

Figure 22.31

Figure 22.30 Brushless dc motor being driven by a converter.

This elementary dc motor circuit of Fig.

Currents

60 degree machine.

/

h

i

i

2,

3

in Fig.

doubled

to

exciting

them by

1

20°,

equivalent to the entire

these conditions, the input to the gate triggering

dc

processor no longer depends on rotor position or ro-

conduction period can be

tor speed.

wye and

Obviously then, the behavior of the machine as a

a 3-phase, 6-pulse converter (Fig.

commutatorless dc motor or synchronous motor de-

by connecting the

22.30). This reduces the half.

is

22.30.

22.29 flow only during

intervals, as they did in the original In practice, the

1

Furthermore,

coils in

number of

thyristors

by

improves the current-carrying

it

pends upon the way the gates are gering frequency

constant, the

is

On

fired. If the trig-

machine

acts as a

capacity of the windings because the duration of

synchronous motor.

current flow

gering frequency depends on the speed of the rotor,

is

doubled. Gate triggering

pendent upon the position of the angle between line voltage

E

s

and

power

again de-

The phase

later. In

factor of the

power absorbed by the converter. As a matter of interest, the converter and motor of Fig. 22.30 could be replaced by the dc motor in Fig. 22.31.

nected in

wye and

The armature

coils are con-

the 3 leads are soldered to a

3-segment commutator. The respective voltages

and currents are identical

in the

behaves

like a

commutatorless dc motor*.

the

motor

has to be slightly leading to provide the reactive

shown

it

line current / is

modified by firing the gates earlier or circuit of Fig. 22.30, the

is

rotor.

the other hand, if the trig-

two

22.16 Practical application of a brushless dc motor One is

at

practical application of the brushless dc


12

V

dc has an output of only

miniature construction,

cepts that reflect the theory

The motor

figures.

machine

in

two

sets

is

1

W. For

all its

represents hi-tech con-

we have just

studied.

a permanent magnet synchronous

which the armature

field rotates.

22,15 Standard synchronous motor

it

motor

This small blower, rated

is

stationary and the

The armature has four

of identical

coils,

A,

A and

salient poles

and

B, B, Coils A,

and brushless dc machine The machine shown

in Fig.

22.30 can be made to

Readers familiar with feedback theory the basic distinction

function as a conventional synchronous motor by

applying a fixed frequency to the

SCR

gates.

Under

will recognize that

between the two machines

is

that

one

functions on open loop while the other operates on closed loop.

A


570

Figure 22.32 This miniature blower, rated at the

left

on the

1

W, 12 V

dc,

2500

r/min,

right consists of four coils that

are connected so as to produce are excited, as

shown

like

a pair of brushes

two N poles when they

in Fig. 22.33.

As

consequent south poles are created (S c

a result,

two

in the figure).

The same remarks apply to coils B, B; when they are excited, they create two N poles where the two consequent S c poles existed before. The stationary coil A,

A and

B

B,

are excited sequentially for equal

lengths of time by

two

electronic switches.

therefore, dealing with a brushless dc tually a 2-phase

The

we assume

it

is

pickup device

H

the

N

We

motor that

rotating clockwise.

A

stationary

detects the successive presence of

N pole, as shown in Fig. 22.33,

nal that causes

ac-

it

riding

field.

by. If the rotat-

produces a

one of the electronic switches

A, A. This produces a

sig-

to ex-

cw magnetic

is

timed by a position-

on a 4-segment commutator.

torque between the rotor and stator poles, thereby sustaining the

cw rotation. The flux pattern between

the respective poles at this

moment

is

shown

in the

figure.

On

the other hand,

der a S pole,

it

set

A,

A to

torque.

As

ing.

when

H

lies

momentarily un-

causes the other switch to close, which

excites coil set B, B.

from one

and S poles as the rotor sweeps

cite coil set

is

permanent magnet poles and

pickup happens to be under the influence of a ing

are,

synchronous motor.

rotor has four

The 7-blade impeller on The stationary armature

driven by a brushless dc motor.

are commutated by an electronic switch. The switch

sensing detector; together they behave

sets

is

contains a circular 4-pole permanent magnet that constitutes the revolving

At the same time,

it

causes coil

be de-energized. This also produces a

set

Thanks

of coils to the other keeps the rotor goto the

presence of the pickup device

(which acts as a position detector), the switching

quency

is

cw

a result, the successive switching action

always related

Fig. 22.34

shows

to the

fre-

speed of rotation.

the switching converter that

generates the 2-phase power.

It

consists of

two

Ql and Q2, which behave like switches. The base of Ql receives the signals from the pickup device H. The latter is actually a Hall effect transistors


571

+ 12V

0

12

6

ms

18

£3 g02V)

BB

off

Figure 22.33 Construction of a 12

V,

blower application. The this instant, the Hall

1

W brushless dc motor for

coil

structure

detector

H

is

0

12

6

ms

stationary. At

triggers coils

A on and

Figure 22.34 Switching

Boff.

circuit of

brushless dc motor and wave-

shapes.

detector,

which produces

when under voltage

is

zero

in the

+2V

a voltage of about

the influence of a

N

pole.

The

signal

presence of a S pole. The

2.2 |xF capacitors absorb the inductive energy re-

shows

rents in the coil sets.

the

One

is,

speed

2500

is

therefore,

it

is

computer components

will

In the next chapter

we

will

encounter brushless

machines are always connected to a large 3-phase

cycle lasts for about

1/0.012

=

The

fre-

83.3

Hz and

the

Its

chapter on ac motor drives.

in


brushless motor offers several adit

requires absolutely

no mainte-

nance, even after thousands of hours of service.

Second,

vital

for lack of adequate cooling.

r/min.

vantages. First,

cles

damaged

ac source. For this reason, they are discussed in the

This small blower serves to cool components a computer.

ensures that

the cur-

waveshape of

12 ms, which corresponds to half a turn.

quency

not be

dc motors of several thousand horsepower. These

leased every time the coils are de-energized. Fig. 22.34 also

bility

Practical level

22-1

pollution-free because no dust partisur-

a.

much quieter than

b.

from worn-out brushes can contaminate

rounding components. Third,

it

is

State in

a conventional dc motor because brushes are noisy,

which quadrants a dc machine op-

erates

22-2

As As

A dc

a

motor

a generator

machine

is

turning clockwise in

Does

develop a clockwise or

both mechanically and electrically, on account of

quadrant

brush friction and sparking. Finally,

counter-clockwise torque?

its

high relia-

3.

it

572

22-3


A 2-pole What duced

22-4

dc motor runs

5460

at

voltage

r/min.

the frequency of the voltage in-

is


1

,

the converter

nected to a 3-phase 480 V, 60

Hz

is

a.

con-

is set at 15°.

and

line

Switch S

is

c.

closed and the armature current

is

270 A.

d.

22-10

Calculate

c.

The dc voltage across the armature The power supplied to the motor The average current in each diode

d.

The power output

a.

Explain

why

be reversed

in

[hp] if the armature cir-

b.

a.

What

is

meant by

is

60 A.

The delay angle required The reactive power absorbed by

the con-

Does inductor L absorb

reactive

power

from the ac line? 1

power

Referring to Fig. 22.5, an ac ammeter inserted in series with line

gives a reading

1

of 280 A. Furthermore, a 3-phase power

the term commutatorless its

started at

is

line.

the basic behavior of the

dc machine? Describe

Problem 22-9

in

the rotat-

drive of Fig. 22.5 with that of Fig. 22.8.

22-7

in

verter

22-1

Compare

motor

Calculate

c.

power from

ing armature back into the ac

22-6

The motor

a.

order that the converter feed

that the

rated voltage

The reactive power absorbed by the converter The effective value of the line currents The induced voltage E0 at 900 r/min

limited to

the field or armature has to

may

Fig. 22.1

its

reduced voltage, and the starting current

cuit has a resistance of 0.07 fl

22-5

60 Hz and the motor

The delay angle required so operates at

b.

b.

V,

following:

in the coils?

the delay angle

208

is

operates at full-load, calculate the

factor meter indicates a lagging displace-

construction and

ment power

factor of 0.83.

principle of operation, b.

What is meant by a half-bridge converter? What advantages does it have over a con-

Calculate a.

ventional 3-phase bridge converter?

b.

The motor shown

1

has a shunt

22- 2 1

Calculate the effective value of the 60

Hz

b.

What

is

a.

Does

b.

the peak-to-peak voltage ripple

Draw

e.

What

ac

the

22- 1 3

in the

In

a.

b.

A

10 hp, 240 V, 800 r/min permanent magnet dc motor has an armature resis-

c.

1

tance of 0.4

H and

rent of 35 A.

verter

22.9

is lift-

speed

if

{)

the armature current Id

=

that R.

0.

1

fl,

is

150 A.

calculate the

x

Ed

Problem 22- 1 2 (and referring

same mass

is

to Fig.

lowered

at a

following: the effective value of the ac line

current?

22-9

in Fig.

lb at a constant

constant speed of 100 ft/min, calculate the

line. is

E

Knowing

22.9), if the

waveshape of the current

5000

value of converter voltage

the field current contain a substantial

ripple? Explain. d.

motor shown

Neglecting gear losses, calculate the value of

circuit.

across the field terminals? c.

hoist

of400ft/min.

ac

voltage that should be applied to the single-

phase bridge

The

ing a mass of

180 V, 2 A.

field rated at a.

in Fig. 22.

1

operating alone as an inverter

is

Intermediate level

22-8

The value of the dc load current Id The approximate delay angle if converter

shown

It is

d.

flow

a rated armature cur-

energized by the con-

in Fig. 22.1. If the ac line

The armature current and its direction The value of E0 and its polarity The value of Ea and its polarity In which direction does the active power

22- 14

In

in the ac line?

Problem 22- 1 2,

held

still

if

the

mass

is

simply

in midair, calculate the following:


The value of E The armature current Id The value and polarity of Ed

a.

c.

22- 5 1

If

armature resistance: 12 ml)

()

b.

the 3-phase line voltage

is

armature inductance: 350

240

V,

22- 6 1

In

b.

In

During current

is

a. Referring to Fig. 22. 8a, calculate the aver1

b.

1

is

the

PIV across

electronic chopper

600

V

the diode?

is

a. b. c.

If

(uls.

The switching frequency

The armature voltage The armature current Draw the waveshape of

Advanced 22- 8 1

negligible, calculate the duty

A

in the

d.

level

If the

wise,

it

cemf E has

the

{)

when is

whether converters

1

rectifiers or inverters

is

cycle and the

were operated as

The following specifications are given the motor shown in Fig. 22.25: armature diameter: 100

the arma-

a 4-quadrant unit?

for

mm mm

armature axial length: 50

200

turning clock-

quadrant

in

new duty

During the start-up phase, would the cur-

turns per coil:

operates

is stalled.

rent ripple be seriously affected if the con-

the armature turns

machine

it

voltage drop across the switches

ture current actually flows in the direction

the

armature when

peak-to-peak current ripple. the current in the

Referring to Fig. 22.8, the

shown and

a result, the con-

Calculate the peak-to-peak current ripple

verter

clockwise. Furthermore,

As

the voltage drop across the

is

2 V, calculate the

high.

shown when

Q4 GTO

under these conditions. c.

22-20

polarity

Assuming

of 620

freewheeling diode, assuming the armature is

kept open.

is

lasts for

the dc current in the trolley

inductance

In order to

cycle needed to establish an average current


is

is

switches

b.

wire

620 A.

ing the start-up phase. a.

placed between a

800 Hz, and each power pulse

400

average armature

at

verter acts as a 2-quadrant converter dur-

dc trolley wire and the armature of

a series motor.

start-up, the

maintained

always maintained closed and the Q3

GTO

by the freewheeling diode,

What

An

is

limit the current fluctuations, the

age current and also the peak current car-

22- 7

A

60

Problem 22-13 Problem 22-14

ried

|ulH

620

rated armature current:

Hz, calculate the delay angle required a.

573

1.

rotational speed:

State

840 r/min

and 2 are acting as

flux density in air gap: 0.5

when

armature current: 5

the

machine

T

A

operates a.

In

b.

In

quadrant 3

c.

In

quadrant 4

d.

Using

quadrant 2

Make

a.

b.

a sketch of the actual direction of

current flow and the actual polarity of

£

c. ()

in

d.

each case.

e.

22- 1 9

A 200

hp,

250

V,

600 r/min dc motor

is

driven by a 4-quadrant converter similar to the

one shown

in Fig, 22.20.

Hz, and the voltage of

280

V.

The motor has

the'

f.

22-2

1

a.

GTOs

are used operating at a frequency of 125

characteristics:

this information, calculate the fol-

lowing:

dc source

the following

is


voltage induced in each coil

dc voltage between the brushes

frequency of the voltage

in

each

coil

frequency of the current

in

each

coil

power developed by

motor

Referring to Fig. 22. effect if capacitor C,

b.

the

torque exerted by the motor 1

7,

what would be

the

were removed?

In Fig. 22.18a, calculate the

approximate

value of capacitor C, \\xF\ so that the volt-

age across

50

it

V during

does not drop by more than the time of a current pulse.

574


22-22

A 3-phase,

6-pulse rectifier

equipped

is

22-24

with a freewheeling diode. The 3-phase feeder has a line voltage of 240

V and the

interval of 3 ms,

the 4

22-25

mH

Calculate

b.

The The

cause 60

to

through the armature when

c.

firing angle

reactive

needed

it

and

]

it is

53 A,

C=

known /,

7000

that at a

= + 140 A, \xF.

standstill

which the voltage across the capacitor

power absorbed by

changing.

the con-

Is

is

the capacitor charging or dis-

charging?

22-26

armature current

is

1

,

suppose the 80

is

decreas-

the polarity of terminal

respect to terminal

A

actually flowing in the

direction shown. If the current is

In Fig. 22.20, the following information

is

given:

Referring to Fig. 22.2

what

= +

to attain this current

Industria I appl ica tion

ing,

/d

Calculate the value of I2 and the rate at

verter

22-23

and the voltage across


E {2 = +300 V,

A to flow

is at

and

inductance during the interval.

given instant

The voltage needed

+ 80 A, A/s.

Calculate the value of the current after an

is

a.

is

6000

increasing at the rate of

composed of an armature having a resistance of 0.4 fi. The rated armature current is 40 A. dc load

Referring to Fig. 22.2 1 a, /a

C?

A with

L a = 20 mH, R a =

1

= -5

E0 = +65 V, E AB = +60 V. Is

.2 fl,

IA

= +5

/A

increasing or decreasing and at what

rate

is it

A,

/,

changing?

A,

Chapter 23 Electronic Control of Alternating-

Current Motors

23.0 Introduction

We

AC

saw

trol

is

in

Chapter 22

that the electronic con-

of dc motors enables us to obtain high efall

6.

torques and speeds. Full 4-quadrant


electronic control of ac motors. Thus,

we

AC

built in

DC

machines can run

much

larger sizes:

machines are limited

at

to

speeds up to 100 000

about 2000 r/min.

to the

find that

In this chapter

squirrel-cage and wound-rotor induction motors, as

in

well as synchronous motors, lend themselves well

Whereas dc machines

to

r/min, whereas large dc machines are limited to

possible to meet precise high-speed in-

dustrial standards.

machines can be

50 000 kW. about 2000 kW. up

ficiency at control

5.

we

cover 3-phase motor controls

keeping with the power emphasis of the book.

However, the reader should

first

review the basic

are con-

principles of electronic drives covered in Chapters

by varying the voltage and current, ac ma-

21 and 22. Furthermore, to understand the basic

chines are controlled by varying the voltage and fre-

principles of ac motor control, the reader should

to electronic control.

trolled

we may ask, if dc machines do such outstanding job, why do we also use ac ma-

quency. Now,

also review Sections 20.18

an

plain

chines? There are several reasons: 1

.

AC AC

machines have no commutators and brushes;

machines cost

less (and

weigh

23.1

less) than

dc

Types of ac drives

Although there are many kinds of electronic ac

machines.

drives, the majority can be 3.

AC

4.

AC

and 20.20, which ex-

variable frequency affects the behavior

of a squirrel-cage induction motor.

consequently, they require less maintenance. 2.

how

machines are more rugged and work better

in hostile

ages: up to 25 kV. 1

the fol-

environments.

machines can operate

about

grouped under

lowing broad classes:

000

DC

at

mueh

1

.

Static

frequency changers

higher volt-

machines are limited

to

V.

575

2.

Static voltage controllers

3.

Rectifier-inverter systems with line

commutation


576

4.

5.

Rectifier-inverter systems with self-commutation

Rectifier-inverter systems with self-commutation rectify the

Pulse- width modulation systems

is

changers convert the incoming

Static frequency line

frequency directly into the desired load

quency. Cycloconverters they

used

are

synchronous and

both

drive

to

fre-

category, and

fall into this

verter

control by varying the ac voltage.

They

used to

are used with

soft- start induction

motors

(Fig. 23.2).

is

incoming

line

is

commutation

line-commutated by the very motor

Such systems are mainly used

self-commutated, generating

own

its

new development

it

in

widespread industrial

as far as

They enable

applications are concerned.

variable

speed induction motor drives ranging from zero speed and up. Their appearance

due

directly

to the

marketplace

in the

availability of high-speed

IGBTs (Fig. 23.7). The seven block diagrams shown in Figs. 23-

switching devices such as

through 23-7 are examples of these ac drives.

drives.

to control synchro-

'f.U

nous motors (Fig. 23.3). Similar systems are used to control the speed of wound-rotor induction motors

3-phase source

controlled

dc

inverter

rectifier

link

(line-comm)

_

(Fig. 23.4).

appropriate current

upper/lower^

i

3-phase source

I

—I

in-

fre-

and 23.6).

frequency to dc, and the dc

reconverted to ac by an inverter. The inverter,

turn,

is

control squirrel-cage induction motors (Figs. 23.5

is

Rectifier-inverter systems with line rectify the

frequency to dc, and the dc

Pulse-width modulation systems are a relatively

and torque

squirrel-cage induction motors. Static voltage controllers are also

line

quency. Such rectifier-inverter systems are used to

squirrel-cage induction motors (Fig. 23.1). Static voltage controllers enable speed

incoming

reconverted to aeby an inverter. However, the

limits

cycloconverter

(

M

Jit

t-n-

control

appropriate frequency

—

and

firing unit

"

desired speed

real

E,

)

3®

/.

values

n, T, etc.

Figure 23.3 appropriate voltage

U

upper/lower^

At

desired speed

and

control

limits


Variable-speed synchronous motor drive using a controlled rectifier

from a dc

firing unit

"

real

values

E.

n, etc.

I.

3-phase source

Figure 23.1

link

and a line-commutated

3OT

;

\

Variable-speed drive system using a cycloconverter

inverter fed

current source (see Section 23.2).

diode rectifier

(see Sections 23.3 and 23.5).

dc link

3-phase source

_

static

inverter

switches

(line-comm)

appropriate voltage

appropriate voltage Jit

upper/lower^ limits

control

and

fji

desired speed upper/lower^ limits

desired speed

control

and

firing unit

real

firing unit E,

Figure 23.2

Figure 23.4

Variable-speed drive using a static switch (see

Variable-speed drive

Section 23.6).

tor

for

(see Section 23.12).

/.

ii,

values T,

etc.

a wound-rotor induction mo-

ELECTRONIC CONTROL OF ALTERNATING-CURRENT MOTORS

3-phase source

_

cause large synchronous motors are generally con-

current

controlled

d.c.

rectifier

(self-comm)

appropriate current

reader


upper/lower

control

limits

desired speed

and

firing unit

to

be alternating-current machines. The

may

therefore consider that in electronic dri-

sidered

inverter

link

577

ves the synchronous motor

a sort of hybrid ani-

is

mal that can be treated either as an ac machine or

as

a brushless dc machine, depending upon the point

*

values

real

£,

/.

of view. In this chapter,

T, etc.

/?,

we

consider

to be an ac

it

machine when fed from a 3-phase source. Figure 23.5 rectifier

self-commutated inverter fed from a dc

link

and a

current

_

rectifier

voltage

dcTlr link

J

inverter (self-corn m)

3®

limits

appropriate frequency control

consists of

and

firing unit

"

real

£,

/,

n,

1

acts as a controlled rectifier, feeding dc

power to converter 2. The latter behaves as a linecommutated inverter whose ac voltage and frequency are established by the motor.

A appropriate voltage

upper/lower

It

Converter

~/ controlled

drive.

tween a 3-phase source and the synchronous motor.

source (see Section 23.9).

3-phase source

shows a typical synchronous motor two converters connected be-

Fig. 23.8

Variable-speed drive using a controlled

smoothing inductor L maintains a ripple-free

current in the dc link between the

desired speed

Current

values

a current source.

T

controlled by converter

/ is

A

Variable-speed drive using a controlled

rectifier

self-commutated inverter fed from a dc

link

and a

Es

is

naturally-commutated by voltage

induced across the terminals of the motor. This

voltage

voltage

source (see Section 23.10).

is

created by the revolving magnetic flux in

the air gap.

The

flux depends

rents and the exciting current

inverter

Gate triggering of converter

appropriate

quency (60 Hz) while motor frequency. The

frequency desired speed real £,

/,

stator cur-

flux

is

usually

Es

is

to the

1

is

done

at line fre-

that of converter 2

is

done

at

latter is directly

proportional

to controls, information

picked off at

motor speed.

values

With regard

n, T, etc.

various points

Figure 23.7

is

assimilated

in the

gate-triggering

processors, which then send out appropriate gate-

Variable-speed drive using a diode

commutated

The

proportional to the motor speed.

(self-comm)

limits

upon the

/,-.

kept fixed; consequently, the induced voltage

voltage

upper/lower

,

verter 3) supplies the field excitation for the rotor.

Converter 2

—

1

smaller bridge rectifier (con-

etc.

Figure 23.6

3-phase source

two converters. which acts as

PWM

inverter fed from

rectifier

a dc

and a

link

self-

voltage

firing pulses to converters

1

and

2.

Thus, the proces-

sors receive information as to desired speed of rota-

source (see Section 23.13). tion, actual speed,

instantaneous rotor position, stator

voltage, stator current, field current, etc.

23,2

In

pret

Synchronous motor drive using current-source dc link combination of a synchronous motor and

position-commutated inverter behaved less

dc motor. This presents a

bit

inter-

mal conditions and emit appropriate gate pulses correct the situation or to meet a specific

Chapter 22, Sections 22.l4.and 22.15, we saw

that the

They

whether these inputs represent normal or abnor-

its

like a brush-

of a dilemma, be-

The gate pulses of converter the position of the rotor. This set

is

to

command.

2 are controlled by

accomplished by a

of transducers that sense the revolving magnetic

field.

They

are

mounted on

the stator next to the air


578

Figure 23.8 Synchronous motor driven by a converter with a dc 60 Hz, thus permitting high speeds.

employ

gap. Other methods

mounted on

the

end of the

link.

position transducers

shaft.

Due to

this

method

The output frequency can be considerably greater than

produce a revolving

MMF that moves in jerks around

the armature. This produces torque pulsations, but

damped out (except at low The shaft there-

of gate control, the synchronous motor acts the

they are almost completely

same way

speeds), due to the inertia of the rotor.

as a brushless dc machine.

may be

speed

current

/

The motor

increased by raising either the dc link

or the field current

E

Stator voltage

s

emf E2

produces a dc

E2 =

E cosa 2

1.35

fore turns smoothly

The motor

/,.

s

given by

voltage

are essentially sinusoidal.

rent, line current,

speed.

£ LN The

and

line

field cur-

and triggering are adjusted so

that

line current /s leads the line-to-neutral voltage (Fig.

23.9).

s

at rated

(21.13)

where

E2 — £ = a2 —

Es

when running


The reason

must operate dc voltage generated by converter 2 fV] reactive

at

is

that the

synchronous motor

power

factor to supply the

leading

power absorbed by converter

2.

effective line-to-line stator voltage [V]

Converter

1

is

designed so that under full-load

firing angle of converter 2 [°]

conditions, firing angle

Similarly, the voltage

produced by converter

1

is

the reactive

a

,

is

about

power drawn from

the

minimize 60 Hz ac line.

1

5°, to

given by E,

Link voltages

E

= and

x

1

.35

E2

E

v

cos

a

|

are almost equal, differing

only by the negligible IR drop in the inductor. Firing angle aj

is

automatically controlled so that the

magnitude of the

link current is just sufficient to de-

velop the required torque.

The

stator line current /s

pulses, as

shown

flows

in Fig. 23.9.

in

These

120° rectangular step-like currents

Figure 23.9 Typical voltage

and current waveshapes

in Fig.

23.8.


Regenerative braking

accomplished by

shift-

d.

the firing angle

ing the gate-firing pulses so that converter 2 acts as

e.

the voltage

is

a rectifier while converter

The

polarity of

E2

and

E

x

operates as an inverter.

1

g.

the reactive

Power

h.

the

same

pumped back

therefore,

direction.

into the 3-phase,

and the motor slows down. During

line

this

Starting the

no voltage

is

zero

at standstill.

To

Consequently,

way

it

f=

The

successive pulses create N, S poles in the stator that

Like a dog chasing

535 r/min

that short current

are always just ahead of their opposite poles

when

to the

motor

propor-

is

speed. Because the rated speed

its

450 r/min at a frequency of 60 Hz, quency at 535 r/min is:

commuta-

pulses flow successively in phases ab, be, and ca.

and,

tional to

is

the fre-

get around this difficulty, the

converters are fired in such a

rotor.

The frequency applied

a.

available to produce the line

is

tion of converter 2.

1

Solution

motor creates a problem, because the

E2

stator voltage

1

power supplied to converter mechanical power developed by the motor

60 Hz period

motor functions as an ac generator.

the

a 2 of converter 2

of the dc link

the firing angle a, of converter

f.

reverses, but the link cur-

rent continues to flow in the is,

E2

579

71.3

Fundamental component of the

b.

Hz

stator current:

on the /p

the rotor accelerates

its tail,

X 60 Hz =

450 r/min

reaches about 10 percent of rated speed,

= =

0.955

/s

0.955

X 239 = 228 A

(21.8)

converter 2 takes over and commutation takes place normally. This pulse

mode

motor as

to brake the

The speed

it

1

is

also used

Current

c.

approaches zero speed.

is

I,

pumps

228

Firing angle

d.

= e.

3-phase synchronous motor rated

similar to that

a2 = —

line.

Example 23-1 60 Hz, 450 r/min,

shown

600

200 kW,

A

:

FP = - arccos - 18.2° = 161.8°

arccos

180°

0.95

link:

(21.4)

S

"

=

1.35

X

511

X

655

cos!61.8°

The dc voltage drop across inductor L

60 Hz.

V,

a2

Voltage E2 of the dc E2 = \35E cos a 2

The three-phase

in Fig. 23.8.

electric utility voltage is

at

connected to a drive

is

293

Converter 2 acts as an inverter, consequently:

are

be smoothly synchronized with the

V,

=

0.78

two examples. Pumped-storage hydropower plants also use this method to bring the huge synchronous machines up to speed so they circulating

480

0.78

industry and

brushless synchronous motors for nuclear reactor

A

(21.7)

d

applied to motors ranging

textile

link:

I,

kW to several megawatts. Permanent magnet

synchronous motors for the

may

dc

in the

control of synchronous motors using a

current-source dc link

from

of operation

gible, consequently,

E = E2 =

at

a speed of 535 r/min.

fective terminal voltage

draws an effective factor of 95

is

5

1

1

V

line current / s of

The

ef-

and the motor

f.

Firing angle

a

Converter

acts as a rectifier, hence:

239 A at a power

%. The motor has an efficiency of 93 %.

Neglecting the losses in the converters, calculate: a.

the frequency applied to the stator

b.

the fundamental

component of the

c.

1

:

{

£,

=

1.35£ L cosa,

655

=

1.35

a,

=

arccos 0.808

X 600 X

(21.4)

cos a,

-

36.0°

stator g.

current

negli-

V

655

l

The motor runs

is

V

Active power supplied to converter

1

/s

the current / flowing in the dc link

/>

=

£,/

=

655

X 293 =

191 915

W

-

192

kW


580

Displacement power factor of converter

PF =

=

cos a,

=

cos 36.0°

-

0.809

80.9

Apparent power absorbed by converter

=

S = 192kW/0.809

237

made

%

23.1

kVA

139 kvar

0.93

=

179

rectifier

supplying the

stator currents

The

and of /, are

kW ~

is adjusted so that the motor operates at power factor at low frequency. However, even unity power factor (/.„ / b 7C respectively in phase

excitation

240 hp

,

with

£aN £ bN £cN ,

active

,

),

the cycloconverter absorbs re-

power from the 60 Hz

delayed triggering

is

line.

The reason

needed on the 60 Hz

is

that

line to

Hz

The power factor is typically 85 percent when the motor runs at rated power and speed. Fig. 23.12 shows a large low-speed synchronous motor that is driven by a cycloconverter. The speed

used, the cycloconverter output frequency

can be continuously varied from zero to 15 r/min.

sometimes used

a higher frequency to a lower

to drive

slow-speed synchronous

motors rated up to several megawatts.

typically variable

from zero

to 10 Hz.

If

a 60

Such a low

generate the sinusoidal low-frequency voltage. input

The low speed permits

direct drive of the ball-mill

frequency permits excellent control of the wave-

without using a gear reducer.

shape of the output voltage by computer-controlled

by altering the gate

firing of the thyristor gates.

commutated, with the

The

thyristors are line-

result that the

the electronics surrounding each

Fig. 23. 10

generator, feeding

SCR

is

consider-

The motor

firing so that the

power back

is

stopped

motor

acts as a

into the ac line.

Similar high-power, low-speed cycloconverter

complexity of

drives are being used with propulsion motors on

board ships. For example, a popular 70 000 ton

ably reduced.

to the

is

controlled so as to keep a constant flux in the air

frequency (Section 21.24). These converters are

is

output voltage

unity

power from

is

line, the

gap. Furthermore, the gate pulses are timed and the

We have seen that cycloconverters can directly con-

source

Hz

magnitudes of the three

Synchronous motor and cycloconverter

vert ac

(Fig.

reduce the reactive power ab-

The 3-phase controlled

at

23,3

fo

waveshape. 1

Mechanical power developed by the motor:

kW X

However,

field current /f functions as a current source.

= V237 2 -192 2 =

192

1).

sorbed from the 60

1

Q = VF^P 2

Pm -

approach a sine wave quite closely

to

usually designed to have a trapezoidal, flat-topped

Reactive power absorbed by converter

h.

low-frequency voltage can be

ate gate firing, the

1

shows

three cycloconverters connected

wye-connected

stator

of a 3-phase synchro-

cruise liner (Fig. 23.59),

is

propelled by two 14

MW

synchronous motors. The motors are directly cou-

nous motor. Each cycloconverter produces a single-

pled to propeller shafts that are driven at speeds of

phase output, based upon the principle explained

zero to 140 r/min.

in

Section 21.24. Referring to phase A, the associated

cycloconverter

+A

and

- A,

is

composed of two 3-phase

bridges,

each fed by the same 3-phase 60

Hz

23.4 Cycloconverter voltage and frequency control

line.

Bridge

+ A generates

voltage for line

a,

the positive half-cycle of

while bridge

—A generates

the

Returning to Fig. 23.1

1

we can

frequency output voltage

is

see that the low-

composed of

segments of the 3-phase 60 Hz

selected

The two bridges are prevented from operating at the same time so as to prevent circulating currents between them. The resulting lowfrequency wave is composed of segments of the

ventional 6-pulse rectifier, except that the firing an-

60 Hz voltage between

gle

negative half.

lines

I,

2, 3.

By

appropri-

line voltage.

segments are determined by the gate

SCRs. The triggering is

is

The

firing of the

identical to that of a con-

continually varied during each low-frequency

3-phase Converter

I

+ -

A A

Q1 to Q6 Q7 toQ12

+ B -

B

+

C

60 Hz

line

3-phase

60 Hz

line

Thyristors

-C

Q13 Q19

to

Q25 Q31

to

to

to

I

I

/V

s '

7

7

7

Q18 Q24

Q30 Q36

primary

limit

control

settings settings i

secondary

!_

gate triggering I

I

processor

'-rnexternal inputs

Typical inputs to gate triggering processor f 1

a.

rotor position

gate |

C

b.

motor speed

c.

cycloconverter input current

d.

cycloconverter output current

e.

input and output voltages

f.

desired speed

|

I

triggering |

processor |

~H~" inputs

Figure 23.10

A is a slowly chang10 times less than the supply frequency. Thyristors Q1 to Q12 are triggered so as to track the desired sine wave as closely as possible. This produces the sawtooth output voltCycloconverter driving a large synchronous motor. The output voltage associated with phase ing sine

wave having a frequency

©f 6 Hz,

which

is

age shown in Fig. 23.1 1 The power factor at the input to the motor is assumed to be unity. The corresponding power factor at the input to the cycloconverter is less than unity, due to the delayed firing angles. .

581

I


582

,-dead zone

rdead zone

-A<+Hn operation-

-

-r=

At-)in operation-

1/6 s-

Figure 23.11 Voltage between lines a and N of

Fig. 23.10.

period so as to obtain an output voltage that ap-

stead of 20 Hz, and the amplitude of the output volt-

proaches a sine wave. During the positive half-

age

Q6 are triggered in sequence, thyristors Q7 to Ql2 for the negative

cycle, thyristors

followed by

Ql

to

half-cycle. In Fig. 23. II the low-frequency output

voltage

has

the

same peak amplitude

3-phase line voltage; consequently,

it

has the same

effective value. In this figure, the frequency

of the line frequency, or 6

We

Hz on

a

the

as

I/IO

is

case the output frequency

The 60 Hz

the

as

firing

is

trig-

20 Hz on a 60 Hz sys-

line voltages are indicated, as well

sequence for the various SCRs.

Although the resulting waveshape

is

The

see,

a very

produced. Nevertheless, the cur-

rent flowing in the windings will soidal.

gate pulses are

we can

A low-output

still

be quite sinu-

voltage requires a large firing

angle delay, which in turn produces a very low

power

factor

on the 60 Hz

line.

Although we have only discussed the behavior

gering process by referring to Fig. 23.13. In this

tem.

also reduced by one-half.

jagged voltage

60 Hz system.

can gain a better understanding of the

is

altered accordingly and, as

B and

of phase A, the same remarks apply to phases

C

(Fig. 23.10).

The

gate firing

low-frequency line currents

/a

timed so that the

is ,

/b

/c are

,

mutually

out of phase by 120°.

The cycloconverter

drive

is

excellent

when

high-

starting torque

and relatively low speeds are needed.

does follow the general shape of the desired sine

However,

not suitable

wave (shown

one-half the system frequency are required.

is

The

as a dash line).

very jagged,

it

gate-triggering

it

is

if

frequencies exceeding

times are quite irregular (not evenly spaced) to obtain the desired output voltage.

That

is

why

the

fir-

ing program has to be under computer control. If this

20 Hz voltage

is

applied to the motor of Fig.

23. 0, the resulting current will 1

sine

wave. In

effect, the

be a reasonably good

inductance of the windings

smooths out the ragged edges

that

would otherwise

be produced by the sawtooth voltage wave.

To reduce

the speed, both the frequency

and

same proportion.

Thus,

is

23.14 the frequency

now

10

Fig. 23.

1

5

Hz

in-

shows a 3-phase squirrel-cage induction

motor connected converter.

of Fig. 23.

voltage have to be reduced in the in Fig.

23.5 Squirrel-cage induction motor with cycloconverter

The 1

0,

to the output of a

circuit

arrangement

3-phase cyclo-

is

similar to that

except that the windings are directly

fed from a 3-phase line. Consequently, the windings cannot be connected in

wye

of delta but must

be isolated from each other. Motor speed

is

varied

Figure 23.12a

6400 kW (8576 hp), 15 r/min, 5.5 Hz, 80°C used to drive a ball-mill in connected to a 50 Hz cycloconverter, whose output frequency is variable from zero stator: 8000. mm; active length of stacking: 950 mm; slots: 456.

Stator of a 3-phase synchronous motor rated

a cement

factory.

The

to 5.5 Hz. Internal

stator

is

diameter of

(Courtesy of ABB)

583

Figure 23.12b

The 44

rotor poles are directly

on the right-hand side

mounted on the

of the poles bring the

ball-mill

dc current

so as

to eliminate the

need

of

a gearbox. The two slip-rings

into the windings.

Figure 23.12c End view of the ball-mill showing the enclosed stator frame in the background. The mill contains 470 tons of steel 3 balls and 80 tons of crushable material. The motor is cooled by blowing 40 000 m of fresh air over the windings, per hour.

(Courtesy of ABB)

584

Q5 6123

4

5

tint

1

1

E 12

E]2 £l3 &21 ^2]

£.11

6

1

t

t

E 32 E \2

1112

789

tint

10 11

12

7

nit

5 6

1

2 -

thyristor triggering sequence

tin 60 Hz

^1.1

Figure 23.13

Waveshape of the output voltage EaN in Fig. 23.10 at a frequency of 20 same value as the effective input voltage between the 3-phase lines.

056

1

23456

1

2

34

5

6

1

2

3

4

89

Hz.

The

10 11 12

effective output voltage

7

8

—

has the

thyristor triggering sequence

Figure 23.14

Waveshape

of

EaN

in Fig.

23.10 at a frequency of 10 Hz. The thyristors are triggered

later in the cycle

so that the '

effective value of the output voltage

the

same

in this

figure

as

it

is in

is

only half that between the 3-phase lines.

Fig. 23.13.

585

As a

result, the flux in the air

gap


586

by applying appropriate gate pulses

to the thyris-

and frequency. For

tors to vary the output voltage

example, the speed of a 2-pole induction motor can be varied from zero to 1500 r/min on a 60

Hz

line

by varying the output frequency of the cyclocon-

from

verter

Good

0.

Hz

1

25 Hz.

to

torque-speed characteristics

in

four

all

and decelerated

started, stopped, reversed,

by regenerative braking. Standard 60 can be used. The stator voltage

Hz

motors

automatically ad-

is

justed in relation to the frequency to maintain a con-

machine. Consequently, the torque-

stant flux in the

speed curves follow the same pattern and have the

same properties

positive, converter 2 transfers

is

L[

It is

important

to.

erates at a time. Thus, ation, converter 2

to the

3-phase

line.

note that only one converter op-

when

converter

1

shown

240°

later.

At 15 Hz an angle of 120° corresponds

X

a delay of (1/15)

(120/360)

=

The smooth current and voltage waveshapes shown in Fig. 23. 7 are actually jagged sine waves, 1

switching between output and

example, to obtain regenerative braking, the

fre-

input. Consequently, cycloconverter-fed

frequency corresponding to

the speed of the motor. Thus,

motor turns

Hz

in

a 4-pole induction

495 r/min, the cycloconverter

at

quency must be 16.5

if

slightly less than (495

consider phase

across

through

winding

the it is

I.

. x

tively positive

fre-

4)/ 120

order to feed power back into the

To understand the operation of verter,

X

=

line.

the cyclocon-

A in Fig. 23.15. The voltage is E a and the ac current

The current

is,

£a

and negative. Because of the

in-

by an angle of about 30° (Fig. 23.17).

Suppose the cycloconverter generates

a frequency

of 15 Hz.

now

can only be furnished by converter

tion.

The

thyristors

Q5

current obviously returns

by way of

Q2, Q4, Q6. This converter can

as a rectifier

when

ways

is

less than

A 3-phase

when Ea

is

may

be

resistive loads.

to the

it

A

power from

is

V, 1760 r/min,

connected as shown

The cycloconverter line

60 Hz. The

is

connected

in Fig. 23.15.

to a 3-phase,

60 Hz

and generates a frequency of 8 Hz. Calculate the

approximate value of the following:

d.

The effective voltage across each winding The no-load speed The speed at rated torque The effective current in the windings at rated

e.

The

acts

phase

480

20 A.

This motor

act either

positive,

squirrel-cage induction motor has a full-

three independent windings each carry a rated cur-

b.

the

torque the 3-phase line.

Similarly, a negative current

nished by converter

separate blower

84 percent, even for

load rating of 25 hp,

negative, the con-

verter acts as an inverter, delivering

A winding to

A

low speeds.

Example 23-2

c. is

at

The cycloconverter can furnish the reactive power absorbed by the induction motor. However, a lot of reactive power is drawn from the 60 Hz line; the power factor is therefore poor. Indeed, with sinusoidal outputs the displacement power factor is al-

a.

When Ea

and delivers power

winding. Conversely,

tifier

1,

because only

"point" in the proper direc-

as a rectifier or inverter.

phase

ing must be provided.

needed

to Fig. 23.15, a positive current/,,

Ql, Q3,

motors run

about 10°C hotter than normal and adequate cool-

rent of

Referring

thyristors

to the constant

therefore, alterna-

ductive nature of a squirrel-cage motor, 7a lags be-

hind

to

22.2 ms.

due

slightly less than the

The

is illus-

The converters in phases B and C function the same way, except that the thyristors (similar to those in phase A) are fired respectively 120° and

For

quency produced by the cycloconverter must be

oper-

trated in Fig. 23. 17.

in Fig. 23.16.

as those

is in

blocked, and vice versa.

is

rectifier/inverter behavior of the converters

quadrants can be obtained. Consequently, the motor

can be

when E

versely,

power from. the, winding

2.

/a

This converter acts as a rec-

negative, and during this period

the converter supplies

power

effective voltage of the

60 Hz

line

can only be fur-

to the winding.

Con-

Solution a.

The all

flux in the

motor should remain the same

frequencies. Consequently, for a frequency

at

control

limit

settings settings

J_J_

positive

V

current

converter

gate triggering

processor

negative

Vtt

y current

converter

external inputs a.

actual speed

b.

absolute

c.

stator volts

d.

stator current

slip

e. etc.

Figure 23.15 Squirrel-cage induction motor fed from a 3-phase cycloconverter.

587


588

of 8 Hz, the voltage across the windings must

be reduced

in

proportion. Thus, the voltage 8

E =

b.

The

full-load speed at

Consequently,

is

60 Hz

m

X 480 V

230 V, 15 Hz

-160,

i

is

synchronous speed

Hz

64

V

speed

l^ t- sj

y

Hz

at 8

n

N

this

y

c.

the

speed

at rated

motor whose

1800 r/min. The no-load

8

When

1760 r/min.

is

therefore,

is"

Hz —X — 60 Hz

—

460 V, 30 Hz

|

is

60 Hz

a 4-pole

motor

is

1800

= 240rmin

operating on 60 Hz, the slip

torque

(1800

is

40 r/min. Consequently,

-

1760)

the slip speed

= is

again

40 r/min when the motor develops rated torque at 8 Hz. The speed at rated torque is, therefore,

= 240 - 40 = 200

n d.

Because the flux 8

Hz

as

it

was

motor

in the

60 Hz,

at

it

torque will be developed

r/min is

the

same

at

follows that rated

when

the stator windings reaches

its

the current in rated value,

namely 20 A. e.

Ideally, the

the

-200

peak

line-to-line voltage applied to

motor should be equal

to that of the

supply. In other words, the

60 Hz

60 Hz

rms value of the

should be the same as the rms

line voltage

Figure 23.16 Typical torque-speed curves of a 2-pole induction tor driven

by a cycloconverter. The cycloconverter

connected

a 460

to

V,

3-phase, 60 Hz

voltage output of the cycloconverter.

mo-

Consequently, the line voltage should be about

is

64

line.

V.

A higher line

voltage could be used, but

/

—

Z*r

...ul

0

3D

s

M 10

0

2

°\

2 0

3e>0

i

> J ) } >

3! 10

J-

/

4

converter

^/blocked

rectifier

1

rectifier

> ? ) > }

converter 2

,

blocked^

rectifier

,

Figure 23.17 Operating

mode

of converter

1

and converter 2 when current

/

a

lags 30° behind

Ea

blocked

y//


would require a greater

this

589

lag in the firing an-

gle to obtain the desired output voltage.

The

converter would draw more reactive power from

power

the line and the

would be

factor

poorer.

23.6 Squirrel-cage motor and static voltage controller

autotransformer

Figure 23.18

The speed of a 3-phase

mo-


Variable-speed blower motor.

by simply varying the stator voltmethod of speed control is particularly a motor driving a blower or centrifugal

tor can be varied

age. This

useful for

pump. To understand why, suppose connected

the stator

is

18 4%

3-phase autotrans-

to a variable-voltage

former (Fig. 23.18).

At rated voltage, the torque-speed characteristic of the motor

is

given by curve

apply half the rated voltage,

Because torque

is

1

9. If

we

obtain curve

2.

of Fig. 23.

we

1

§ 125 cr

o 100

proportional to the square of the

(3),

applied voltage, the torques in curve 2 are only 1/4 J2).

of the corresponding torques in curve

1

.

For exam-

46%

breakdown torque drops from 184% to 46%. Similarly, the torque at 60 percent speed drops from 175% to 43.75%. The load torque of a blower varies nearly as ple,

the

much

60

40

—

*~

100

80

speed

%

t

synchronous speed

as the square of the speed. This typical char-

acteristic,

the

I

20

0

shown by curve

3, is

superimposed on

motor torque-speed curves. Thus,

voltage, the intersection of curves

blower runs

that the

speed.

On

at

rated

and 3 shows

90 percent of synchronous

the other hand, at half rated voltage, the

blower rotates speed.

at

1

Figure 23.19 Torque-speed curve of blower motor at rated voltage (1) and 50 percent rated voltage (2). Curve 3 is the torque-speed characteristic of the fan.

at

To reduce

only 60 percent of synchronous

By varying

the voltage this way,

we

can

is

the voltage across the motor, the firing

delayed

still

more. For example, to obtain

50 percent rated voltage,

control the speed.

The variable-voltage autotransformer can be

re-

placed by three sets of thyristors connected back-to-

shown in To produce

angle 0

The

all

the pulses are delayed

by about 100°. The resulting distorted voltage and current

waveshape

for phase

A are pictured very ap-

the respective thyristors are fired with a delay 0

The distortion increases the losses in the motor compared to the autotransformer method. Furthermore, the power factor is

equal to the phase angle lag that would exist

considerably lower because of the large phase angle

back, as

Fig. 23.20.

valves.

rated voltage across the motor,

motor were directly connected

to

sets are called

the

line.

if

the

Fig.

23.21 shows the resulting current and line-to-neutral

C

The valves i,n. phases B and same way, except for an addi-

voltage for phase A.

are triggered the

tional delay of 120° and 240°, respectively.

proximately

lag

in Fig.

23.22.

Nevertheless, to a

0.

first

approximation, the

torque-speed characteristics shown still

in

Fig.

23.19

apply.

Due power

to the considerable 1

2

R

losses

factor, this type of electronic

and lower

speed control

is


590

only feasible for motors rated below 20 hp. Small

creases

linearly

hoists are also suited to this type of control, because

reached.

Some

with time

until

voltage

full

is

of these ramp-up schedules incorpo-

overcome

they operate intermittently. Consequently, they can

rate a short pulse of full voltage to

cool off during the idle and light-load periods.

tic friction

of machinery that has not operated for

some time

or that

23.7 Soft-starting cage motors

schedules the starting current can be limited automatically

In

many

applications an induction motor must not

when switched across the For example, some loads, such as con-

accelerate too quickly

power

line.

veyor

belts,

have

to

be started slowly to prevent

tip-

ping or spilling the goods. In other cases, a centrifugal

a

pump must

damaging water-hammer

pipes. In

a

not start too quickly, otherwise

power

effect could burst the

still

other instances, the voltage drop along

line

may be

induction motor

is

excessive when, say, a 500 hp

slammed

to, say,

is

covered with

frost.

the sta-

In other

Some

four times the rated current.

of the starting and stopping features offered are

illus-

trated in Fig. 23.23a.

Once tactor

is

the

motor reaches rated speed, a bypass con-

sometimes used

to short-circuit the thyristors

to eliminate the heat loss.

Due

to the voltage

tween the anode and cathode, the to

600 hp motor drawing a

500 A,

will dissipate

drop be-

3-phase loss

W per ampere. Thus, the

amounts tors of a

about 3.5

total

about 3.5

thyris-

line current of, say

X 500 =

1750 W.

across the line.

In all these applications the back-to-back static

switch of Fig. 23.20 can be used to soft-start or softstop a squirrel-cage motor by applying reduced volt-

age across the

stator.

The

starting controls are set so

that, initially, the

voltage builds up rapidly until the

motor just begins

to turn, after

which the voltage

in-

Figure 23.21

Waveshapes

control

.

settings

|

limit

settings

9 e triggering

I

processor

externa

\

L

,n P uts

1

i

Figure 23.22

Figure 23.20 Variable-voltage

at rated voltage.

speed

control of a squirrel-cage

duction motor using back-to-back thyristors.

in-

Waveshapes age.

(very approximately) at

50%

rated volt-

In


591

the absence of a bypass contactor, forced cooling

would have

be used to get

to

When power

is

rid

of the heat.

shut off, the motor

may

coast to

a stop too quickly. In such cases the slow ramp(a)

down tage.

feature of the electronic starter

During

terminals

comes

this

reduced gradually

is

to rest.

an advan-

is

phase the voltage across the motor

The

starter

field to generate the

until

the

motor

can be programmed

in the

ramp-up and ramp-down

fea-

tures that are best suited to the load (Fig. 23.23b).

Another feature of soft and absolutely tacts clap

(b)

starters is their reliability

silent operation.

when they open and

ing coil to worry about, and

No mechanical con-

close,

no noisy hold-

—most importantly

no worn contacts to replace. Soft starting of induction motors

from

1

hp up

to several

fers an excellent alternative

1.0

is

available

thousand horsepower. choice to series

It

of-

resis-

tance and autotransformer starters as well as wyeacross-the-line

E

delta

and part-winding

starters. Retrofitting older

start (p.u.;

starters is

motion

an important application of soft

starters.

(c)

starts

(d)

start

run

soft stop

4.0

current limit start

/

(P-u.) (e)

1.0

Figure 23.23b

460 V, 60 Hz. Startup time adjustable 5-50 s; initial torque adjustment 0-75%; current limit schedule 75%-400%; kick-start time adjustable 0-1 .5 s. In background is 40 hp, 460 V Solid-state soft starter rated 5 hp,

0start

Figure 23.23a Five typical options to control the soft starting

and

soft

stopping of a cage induction motor (per-unit values).

soft starter.

(Courtesy of Bafdor Electric Company)


592

SELF-COMMUTATED INVERTERS

•

The

inverter

power

loss

assumed to be neglipower is equal

is

gible; consequently, the dc input to the active ac output

23.8 Self -com mutated inverters for

cage motors

•

The

reactive

power.

power generated by

the inverter

is

not produced by the commutating capacitors inIn

we saw

Section 23.2

that a

synchronous motor

can be driven by a line-commutated inverter. This

cluded

is

possible because the synchronous motor can pro-

The

power

reactive

due

is

of the load.

•

The reactive power output power input.

•

The IGBTs,

power needed by the inverter. Unfortunately, if the synchronous motor is replaced vide the reactive

by an induction motor, the frequency conversion

in the circuitry.

to the nature

thyristors, or

requires no net dc

GTOs

connect the dc

system breaks down, because an induction motor

input terminals to the ac output terminals in a

cannot deliver reactive power. Worse

controlled sequence, with negligible voltage

ally

still, it

actu-

drop.

absorbs reactive power.

Nevertheless,

we can drive an

induction motor us-

It

a.

follows that

In a voltage source inverter, the ac line

ing a self-commutated inverter (also called force-

voltages are successively equal to

commutated inverter). It operates quite differently from a line-commutated inverter. First, it can gener-

dc input voltage, or zero

ate

own

its

b.

frequency, determined by the frequency

of the pulses applied to the gates. Second,

it

can

In the latter case, the thyristors are

thyristor

is

arranged

in

surrounded by an array of capacitors,

To control the speed of a squirrel-cage motor, we and inverter are connected by a dc is

connected

power would

thyristors to not,

is

to

is

connected

dc links are used

link.

The

60 Hz supply

to the stator.

—constant

Two

rectifier

line

and

types of

current and constant

in-

The purforce some

conduct when normally they

and to force other thyristors to stop con-

voltage-source inverters* mentioned above.

By

virtue of an inductor L, the constant current

a constant current to the inverter,

link supplies

which

is

then fed sequentially into the three phases

of the motor (Fig. 23.24a). Similarly, by virtue of

ducting before their "natural" time.

It is

precisely

capacitor this forced

switching action that enables these con-

verters to generate

we show

C (Fig.

23.26a), the constant-voltage link

furnishes a constant voltage to the inverter, which

and absorb reactive power.

Because of the variety of the switching used,

to the 3-phase,

voltage. This gives rise to the current-source and

ductors, diodes, and auxiliary thyristors.

pose of these auxiliary components

the

use a rectifier-inverter system in which the rectifier

a

However, each

circuit.

±

dc current, or zero

the inverter

conventional 3-phase bridge

the

In a current source inverter, the ac line

currents are successively equal to

ei-

ther absorb or deliver reactive power. The reactive power generated or absorbed depends upon the nature of the load and the switching action of the power semiconductors. The switches may be IGBTs, power MOSFETs, GTOs, or ordinary thyristors.

±

is

switched sequentially from one phase to the next of

circuits

the self-commutated inverter as a

simple 5-terminal device having two dc input

the induction motor.

Many

switching methods have been devised. In

ter-

minals and three ac output terminals to provide

the following Sections 23.9 to 23.12,

we

first

de-

3-phase power to the motor. There are two basic types of inverters: current-source inverters (Fig.

23.24a)

and

voltage-source

inverters

(Fig.

In 2-quadrant is

from

23.26a). This simple representation helps us un-

derstand the basic features of inverters:

all

self-commutated

and 4-quadrant motor drives, the term inverter

somewhat misleading because power can flow the dc side to the ac side (inverter

not only

mode), but also from

the ac side to the dc side (rectifier mode). For this reason

prefer the term converter rather than inverter

power can flow

in

both directions.

we

whenever ac/dc


scribe the

methods

that generate a rectangular

current or a rectangular

on

to describe

wave

voltage.

We

wave

voltage

then go

EAN The .

voltage finds

own

its

593

place, so to

speak, depending upon the particular speed, torque,

and direction of rotation the motor happens

pulse-width modulation methods.

to

have.

We can obtain regenerative braking (generator ac23.9 Current-source self-

tion)

commutated frequency converter (rectangular wave)

in

The current-source frequency converter shown Fig. 23.24a

is

such that the current pulse that flows for

1

in

each phase

link.

C

verter,

that the 3-phase rectangular current

is

is

almost sinusoidal

the line current

/.,

in

in shape. Fig.

EAN

sponds

power

to the operating

.

factor of the motor. itself

although instant

t {

The

is

we

also reverse the polarity 1

shown

now

acts as an in-

Note

in Fig. 23.24c.

direction of rotation

is

quency, the torque-speed

changed by

in all

four quad-

and

shown

forth, as

in Fig. 23.

1

6.

High-inertia loads

L

'

self-

3-phase 2

£2

'

commutated

3

inverter

o-

squirrel cage

induction motor

gate

control settings

external

triggering limit settings

Figure 23.24a Current-fed frequency converter.

al-

By changing the frecurve can be moved back

converter 2

line

that con-

can be quickly brought up to speed by designing the

1

Ce-

The

rants with high efficiency.

not determined by the zero crossing point of

.

line.

power to the mo-

easily

frequency converter can operate

coincides with the firing of

converter

dc

the gates of converter 2. Consequently, this static

It

and

the thyristor connected to phase A, the timing of the

pulse

into the

tering the phase sequence of the pulses that trigger

not upon the switching action of the inverter. In effect,

.

tor during this regenerative braking period.

Phase angle 6 corre-

depends upon the properties of the motor

1,

verter 2 continues to supply reactive

23.24b shows

E2

same

relationship between stator voltage and

stator current is

one phase, and the associated


power

in the

feeding power back into the 3-phase

new phase

pulses together produce a revolving magnetic field

flow

retarding the triggering of the

of £]. Consequently, converter

nearly sinusoidal.

is

By simultaneously

thyristors in converter

is

a rectangular

20°. Nevertheless, the resulting

voltage between lines A, B,

The reason

is

to

direction and so converter 2 feeds

in

was explained

Section 20.20. This reverses the polarity of

However, the dc current continues

used to control the speed of individual

cage motors. The switching action of the inverter

that

by changing the firing angle and reducing the

gate pulse frequency of converter 2, as

L

processor

-

j

inputs

motor speed dc current etc.


594

verter.

The

efficiency of the motor

is

88% and

that

of the inverter js 99.4%. Referring to Fig. 23.24, calculate the approximate value of the following:

The dc power input to converter 2 The current in the dc link The dc voltage E produced by converter

a.

b. c.

1

{

Figure 23.24b

Solution

Motor voltage and current.

a.

The

active

power absorbed by 40 X 746

P

=

the

motor

is

kW

33.9

0.88

The

power absorbed by converter 2

active

verter)

The

b.

33 9

=

P?2

=

kW

34.1

0.994

effective value of the full-load current

stamped on the nameplate. This

Figure 23.24c Asynchronous generator voltage and

(the in-

is

is

is

52 A,

the funda-

mental component of current /a (Fig. 23.24b).

current.

Consequently, the dc current has an approxicontrol system so that full-load torque as the

is

mate value of

developed

motor accelerates.

In practice, the output

tangular

wave

inverter using thyristors

may be

in

P,

£ = /;=

X

34.1

«

changed

(21.7)

is

}

than 200 Hz. At

rated torque the ac voltage has to be

A

66.7

The dc value of E (and of E2 )

c.

commercial applica-

tions, frequencies are usually less

=

0.78

var-

ied over a range of 10:1, with top frequencies of

about 400 Hz. However,

52

h =

frequency of such a rec-

10

3

^6.7-

=

5,1

v

in pro-

portion to the frequency so as to maintain a constant stator flux.

Consequently, the dc link voltage

must be reduced as the speed

is

speed. This voltage reduction

E

x

23.10 Voltage-source self-

reduced below base is

achieved by

commutated frequency

in-

converter (rectangular wave)

creasing the firing angle of the thyristors in converter

1

.

Unfortunately, this tends to increase the re-

power drawn from the 3-phase line. The dc link voltage is held constant when the motor operates above base speed. The motor then

active

develops less than rated torque because ning

in the

it

is

run-

constant horsepower mode. Fig. 23.25

shows the physical

some

In

mills, the

speeds of several motors have to rise and

together.

fall

These motors must be connected

to a

common bus in order to function at the same voltage and frequency. Under these circumstances we use a voltage-source frequency converter (Fig. 23.26a).

A 3-phase

size of a current- source vari-

able-frequency drive.

industrial applications, such as in textile

E\

.

bridge rectifier produces a dc voltage

The capacitor ensures

a

stiff

dc voltage

at

the in-

put to the inverter, while the inductor tends to

Example 23-3

smooth out the current

A 40

The

hp,

motor

is

1

165 r/min, 460 V, 52 A, 60 Hz, 3-phase

driven by a current-source frequency con-

/d supplied

by the

inverter successively switches voltage

the lines of the 3-phase motor.

rectifier.

E 2 across

The switching

pro-

Figure 23.25 Current-source variable-frequency electronic drive tion motor.

The output frequency can

The design

is

for

a conventional 500 hp, 460

be- varied from zero to

72 Hz and the

a current-source, 6-step output that can operate

(Courtesy of Robicon Corporation)

595

in all

V,

1780

r/min,

3-phase, 60 Hz induc-

efficiency at rated load

4 quadrants.

and speed

is

95%.


596

Figure 23.26a Voltage-fed frequency converter.

K120°H t

u

duces positive and negative rectangular voltage pulses of 20° duration (Fig. 23.26b). The frequency 1

ranges typically from about 10

Hz

to

The fundamental component of voltage

200 Hz.

the line-to-line

directly related to the dc voltage

is

by the

expression

£ line = 0.78£d

(23.1)

Figure 23.26b Motor line-to-line voltages.

where

=

£iine

effective fundamental

component of

Ed = 0.78 = Up

dc voltage

at

input to converter [V]


=

output voltage

£ ]ine

frequency so as to maintain a constant flux (or motors).

age

equal to the dc voltage

is

that rectifier voltage £,

quency fore be

minute

varies.

in the

Because the flat-topped ac

motor

Ed (= E 2

),

must be varied

The speed of

the

it

volt-

follows

as the fre-

motor can there-

controlled from a few revolutions per to

maximum

the voltages are held constant this

constant horsepower mode, the torque decreases as

Regenerative braking

varied in proportion to the

is

all

the speed rises.

V6/tt]

base speed, the amplitude of the inverter

to

Above base speed

while the frequency continues to increase. In

line-to-line voltage [V]

while developing

full torque.

when

current /d reverses ator.

Voltage

E2

possible, but the link

is

the

motor

acts as a gener-

does not change polarity as

it

does

in a current-source inverter. Because converter

l

cannot accept reverse current flow, a third converter (not

shown) has

with converter

The

to 1

be installed

in reverse parallel

to permit regenerative braking.

third converter functions as an inverter and,

while

it

operates, converter

1

is

blocked.

As

a result,

voltage-source drives that actually return power to


Time T3

T1

30

20

6

1

contacts

T2

1

X

X

2 X

4

X

5

X

T7

X X

X

X X

X

X

X X

X X

X X

X

6

X

X X

X

T11

T10

T8

X

X

X

T9

T6

X

3

intervals

T5 T4

597

X

X

X

X

X

X

(b)

r Figure 23.27 Three mechanical switches could produce the same voltage pulses as a voltage-fed inverter. b. Table showing the switching sequence of the

a.

switches. c.

Voltages produced across the motor terminals. T3

T1

T2

the ac line tend to be

source drives. In

more expensive than

many

cases a resistor

is

T4

T7

T6

T9 T8

T1

T10

current-

used to ab-

I

120

power delivered during the braking process. Unless the power is large, such dynamic braking is much cheaper than feeding power back sorb

T5

240

360

480

600

the

into the line. In other installations a dc current

is in-

(C)

23.11

jected into the stator windings (see Section 14.9).

The switching

of a

wound-rotor induction motor

action of converter 2 can be rep-

shown The opening and closing sequence

Chopper speed control

We have already seen that the speed of a wound-rotor

resented by three mechanical switches, as

in

Fig. 23.27a.

is

induction motor can be controlled by placing three

with the resulting

variable resistors in the rotor circuit (Section 13.16).

given

in the chart (b), together

rectangular line voltages.

switch contact

is

lustrates that thyristors in

An X

indicates that a

closed. This mechanical

model

il-

and other electronic devices

converter 2 really act as high-speed switches.

The switching

action

is

called~6-step because the

Another way to control the speed phase bridge

feed the rectified

The

is

to connect a 3-

rectifier across the rotor terminals and

power into a single

variable

resulting torque-speed characteristic

to that obtained with

resistor.

is identical

a 3-phase rheostat. Unfor-

has to be varied me-

switching sequence repeats after every 6th step, as

tunately, the single rheostat

can be seen from the chart.

chanically in order to change the speed.

still


598

We

can make an all-electronic control

the secondary circuit. In this circuit, capacitor plies the high current pulses

drawn by

R

{)

The

relationship

where

A" d is

A

1

,

Ed = -

either

(21.22)

between termi-

D is the chopper duty cycle.

Knowing

P

r

the torque,

T

at

30

kW

(40 hp),

in Fig.

23.28

is

200

rated

170 r/min, 460 V, 60 Hz. The

1

resistor

R0

is

is

400

V,

torque of 200

N m

at

900

135

is

200

Part of

P

r

is

The rated syn1200 r/min. The slip at

clearly

.

(n s

-

=

(1200

=

0.25

n)/n s

-

The power of 6282 sistor

R0

,

but

EJd

it

130

W (13.7)

T

W

is

0.25

X

6282

W

actually dissipated in re-

(13.2)

135 7d

6282 46.5

Jju

—q

/ 1

—(chopper

i

Figure 23.28 a wound-rotor induction motor using a load

resistor

25 130

obviously equal to the rectifier

is

JIM

control of

power

(13.19)

s

Thus,

.

900)/ 1200

rectifier

Speed

PJn

PjT = sP

output

is

s

calculate the

r/min.

ples covered in Chapters 13 and 21 is

V

dissipated as heat in the rotor circuit:

This problem can be solved by applying the princi-

900 r/min

100

9.55 /\71200

Solution

chronous speed

(21.4)

X

we can

P = 25

so that the motor develops a

T. x

is

and the load

0.5 il. If the chopper frequency

Hz, calculate time

the bridge rectifier

1.35£ 1.35

9.55

=

r

open-circuit rotor line voltage

(13.4)

delivered to the rotor:

Example 23-4 The wound-rotor motor shown

is

= sEoc = 0.25 X 400 - 100 V

The dc voltage developed by

given by

the apparent resistance

A2, and

made

- R 0 /D~

7? d

nals

is

D

By

900 r/min

the apparent resis-

tance across the bridge rectifier can be

high or low.

^.E

the chopper.

has already been explained in Section 21.37. T, v

rotor line voltage at

to

C sup-

The purpose of inductor L and freewheeling diode varying the chopper on-time

The

(Fig.

23.28) by adding a chopper and a fixed resistor

and chopper.

A

ELECTRONIC CONTROL OF ALTERNATING -CURRENT MOTORS

The apparent per circuit

Given

is,

that

resistance at the input to the chop-

therefore,

23.12 Recovering power in a wound-rotor induction motor

= Ed /Id = 135/46.5

Instead of dissipating the rotor

=

23.29).

we

2.911

could use

0.5 fl, and applying Eq. 21.22,

=

X D2

it

/

Example 23-5 In Example 23-3,

2.08

ms

where

s is the slip

therefore 2.08 ms.

is

£oc

is

On

the other hand, rectified output voltage

Ed = \35E

calculate the magnitude of the the capacitor.

Because the IR drop

flowing

R0

in

is

a steady current

we

Ed = E2

.

in the

is

(21. 4)

1

3.4 and 2

1

is .4,

obtain

E, 1.35 2

In

X

0.5 /()

=

Recognizing that

6282

=

112

tion

A

1

12 A.

the repetition rate

The pulse width

is

is

2.08

200 pulses per second.

On the other hand, the rectifier continuously charges the capacitor with a current Id of 46.5 A.

shows

control using a variable-voltage battery.

a fixed quantity, the equa-

is

E2

F

depends exclusively upon Consequently,

-

we

could

vary the speed from essentially synchronous speed (s

=

0) to zero (s

=

1)

age from zero to 1.35

by varying the battery

£oc

volt-

.

In practice, instead of charging a battery to ab-

sorb the rotor power,

rectifier

Figure 23.29

Eoc

that the slip

the battery voltage

therefore delivers current pulses hav-

ing an amplitude of

Speed

Ed

smoothing inductor

Combining Eqs.

given by

ms and

the open-circuit rotor

given by

negligible,

The capacitor

(13.4)

voltage at standstill (Section 13.10).

drawn from

I0

and

200

Solution

The current

is

0.415

The chopper on-time

current pulses

E = sEw:

= —— =

D —

the rotor terminals

given by

0.415

7d = 1

a resistor,

Assuming the battery voltage E2 can be varto some arbitrary maximum value, let

The ac voltage E across 2.9

D=

in

dc battery (Fig.

us analyze the behavior of the circuit.

we have 0.5

power

to charge a large

from zero

ied

R0 =

599

we

use a 3-phase inverter that


600

returns the

power

mutated inverter

value of

T

ET

is

lies

The line-com-

connected to the same feeder

power

that supplies

transformer

to the ac source. is

to the stator (Fig. 23.30).

A

usually added, so that the effective

between 80 and 90 percent of

This ensures that the firing angle

is

E2

.

reasonably close

to equal

power absorbed by

the inverter.

As

usual,

the voltages are related by the equation

E2 =

1.35

£T

cos

The

(21.13)

is

that for

two

[V]

is

very efficient be-

not dissipated in a group of

returned to the

line.

from no-load

to full-load.

.

When ET =

0,

both

E 2 and Ed are

power of

ple, if the

control using a rectifier

and

naturally

commutated

symmet-

is

economical be-

the rotor,

power

which

considerably

is

to the stator.

lowest desired motor speed

For examis

80%

of

synchronous speed, the power handled by the con-

are

to the stator.

much

have

inverter.

It

smaller than

stator circuit

Figure 23.30

Speed

It is

cause the rectifier and inverter only have to carry

verters (at rated torque)

23.31a shows the torque-speed curves for

settings of E T

rectangular and flows

unity.

power

,

is

.

ET

is

shifted to the

£ AN Consequently, the displacement power factor of the load across the rotor is

Another advanthe speed

is

with respect to the respective line-to-neutral

less than the input is

.

speed decreases only slightly with

°]

any given setting of

practically constant Fig.

T

1

This method of speed control

[

This method of speed control

is

that the

rotor current / R

the slip

resistors, but

,

rotor voltage

a

secondary line voltage of transformer

cause the rotor power

When ET is adjusted *he torque-speed characteristic

during 120° intervals (Fig. 23.31b).

dc voltage developed by the inverter [V]

firing angle

£uc

increasing torque.

always

a =

tage

Note

left.

rical

where

E2 = ET =

0.4

(curve 2) has the same shape but

to the permissible limit of 165° while reducing the

reactive

Consequently, the slip-rings are short-cir-

zero.

cuited, thus yielding curve

if

where the

to be controlled.

is

only

20%

of the input

follows that the converters they were placed in the full

stator

power would


601

Solution

Synchronous speed

a.

ns

is

=

120,///?

=

120

X

- 900 5

-0.4

s

|

E v = 0.4E(K

!

!

,

\

,

1

1

(

20

0

40

60

Slip

\E V =0

100

r/min

is

=

%

(n s

—

n)/n s

-

=

(900

=

0.222

r

80

60/8

-0

s

q_|

(13.1)

(13.2)

700)/900

* speed

Mechanical power Figure 23.31a Torque-speed characteristics for

two settings

of voltage

ET

of

is

P m = 800kW(=

a wound-rotor motor

1072 hp)

but

.

(13.8)

= P

800

Power supplied

r

Electric

90

60

30

120 150 18(T

-

0.222)

to rotor is

P =

0

\ r (

kW

1028

power output of

rotor:

P = sP

degrees

angle

(13.7)

0.222

Consequently, 228 in Fig.

b.

23.30.

kW

is

1028

kW

228 Figure 23.31b Rotor voltage and current

X

fed back to the ac

E = sEoc

£-

Example 23-6

A 3-phase,

3000 hp, 4000 V, 60 Hz, 8-pole woundmotor drives a variable-speed cen-

rotor induction trifugal

4160

V

1800 V.

pump. When

motor

is

3-phase 4160 V/480

V

transformer

connected between the inverter and the 23.32). If the motor has to develop 800

a.

The power output of the

Rotor voltage and link voltage

c.

Link current

d.

Firing angle of the inverter

e.

Current

/c

in the

transformer

T

,

DC

1800

link voltage:

£d = =

is is

line (Fig.

kW

c.

at

DC

1.35

E

1.35

X 400 = 540 V

link current:

a /d

rotor

and rotor current

(13.4)

X

= 400 V

speed of 700 r/min, calculate the following:

b.

0.222

connected to a

open-circuit rotor line voltage

line, the

A

the

line.

Rotor line-to-line voltage:

The

- /y£ = 228 = 422 A t

,

000/540

effective value of the rotor current

is

•

primary and secondary lines of

/R

=

0.816

/d

-

0.816

X 422

-

344

A

(21.6)


602

3 phase

4160

40 A

2240 kW, 4000

V, 8-pole

E CQ = 1800 V

Figure 23.32

See Example

23-5.

d.

E2 — 540

=

a =

1

.35

1.35

ET cos a X 480 cos a

and currents. in the

firing angle

is

tions that are

actually (l

80

-

33.5)

=

146.5°

The current

in

each phase of the 480

The

effective value

V

is

line

rectangular

given by

is

Eq. 21.6:

wave

= 344 A

/

=

(480/4160)

4160

X

60

is

6

=

360 Hz.

Hz

synchronous speed

V

torque vibration side

is

the frequency of the

the synchronous speed

is

1.5

is

is

On

the other hand,

when

applied to the stator, the

45 r/min and the associated

X

6

=

9 Hz.

Torque pulsations, such as 360 Hz, are damped

X 344 - 40 A

out at moderate and high speeds, due to mechanical inertia.

PULSE-WIDTH MODULATION DRIVES

a 9

However,

at

vibration

is

Hz

speed

modulation

low speeds (such as 45

is

where

r/min),

very noticeable. Such torque

fluctuations are unacceptable in plications,

23.13 Review of pulse-width

The self-commutated frequency converters cussed so far generate rectangular waveshapes

When

60 Hz,

is

a frequency of 1.5

Effective line current on the

six

1800 r/min and the corresponding torque pulsa-

tion

/

is

For example, suppose the drive involves a 4pole induction motor.

flows during 120° intervals and has a peak value of 422 A.

superimposed on the main driving

The frequency of the torque pulsation

times that of the fundamental frequency.

because the converter acts as an inverter. e.

these harmonic currents flow

33.5° torque.

The

When

motor windings, they produce torque pulsa-

some

fine speed control

industrial ap-

down

to zero

Under these circumstances,

required.

in-

stead of using rectangular waveshapes, the motor dis-

that

contain substantial 5th and 7th harmonic voltages

driven

by pulse-width

Converters that generate ered in Chapter 2

1

,

modulation

PWM

outputs were cov-

Sections 2 .45 to 2 1 .48. 1

is

techniques.


To

briefly review the basic principles, consider

voltage-source

the

shown

frequency

in Fig. 23.33.

A

3-phase bridge rectifier

which appears

produces a fixed dc voltage sentially

undiminished as

commutated converter

way

2.

system

converter

E2

at

1

is

triggered

the

same

time, the pulse spacings are rearranged so that their

weighted average again approaches a sine wave. In

es-

the input to the self-

The converter

of pulses per half-cycle from 5 to 50. At

603

some cases

the output voltage has to be re-

duced while maintaining the same fundamental quency. This

is

done by reducing

all

fre-

the pulse widths

com-

in proportion to the desired reduction in output volt-

posed of a series of short positive pulses of constant

age. Thus, in Fig. 23.34b, the pulses are half as wide

in a special

so that the output voltage

is

amplitude followed by an equal number of short

as those in Fig. 23.34a, yielding an output voltage

negative pulses (Fig. 23.34a). The pulse widths and

half as great, but having the

We can therefore

pulse spacings are arranged so that their weighted

average approaches a sine wave, as shown

This wanted sine wave

figure.

is

in the

called the funda-

same frequency.

vary both the output frequency

and output voltage using a fixed dc input voltage.

As

a result, a simple diode bridge rectifier can be

frequency ranges typically from

used to supply the fixed dc link voltage. The

dis-

400 Hz. The frequency of the pulses, called carrier frequency, can range from 200 Hz to 20 kHz, depending on the application and the type of switch employed (GTO, IGBT, etc.). The pulses in the figure all have the same width,

placement power factor of the 3-phase supply

line

mental and 0.1

Hz

its

to

is

therefore close to unity.

The presence of the

carrier frequency eliminates

the low-frequency harmonics of the

embedded

fun-

damental frequency. The only harmonics present

but in practice, the ones near the middle of the sine

are the carrier frequency itself and close multiples

made broader than those near the edges. By increasing the number of pulses per half-cycle, we can make the fundamental output frequency as low as we please. Thus, to reduce the output frequency of Fig. 23.34a by a factor of 10, we increase the number

thereof. Thus, a

wave

are

Figure 23.33

Speed

control by pulse width modulation.

PWM drive that generates a funda-

mental frequency of 2 Hz, using a carrier frequency of 2500

Hz,

would have harmonics

clustered

around 2500 Hz and multiples of 2500 Hz. The har-

monics of 2 Hz do not show up. Consequently,


604

formed almost instantaneously

h—

12°—H

any arbitrary shape. As a

cies of

can

tor servo drives

into higher frequenresult, induction

now respond

mo-

commands

to

as

quickly as the best dc drives. Fig. 23.35 illustrates a typical application of In order to

motor

to induction

Figure 23.34a

this

Voltage waveform across one phase.

motor

a

in

PWM drives.

understand the application of drives,

somewhat

that induction

fixed frequencies and

was needed,

different light.

at

The reason

whenever a variable-speed was to

the immediate solution

much

use a dc drive. Consequently, not

was directed

PWM

necessary to look

motors have traditionally operated

is

at

drive

is

it

to variable-speed induction

havior. In the sections that follow,

attention

motor be-

we examine

this

aspect of induction motors.

Figure 23.34b Waveform yielding the same frequency

TORQUE AND SPEED CONTROL OF INDUCTION MOTORS

but half the

voltage.

torque vibrations at low speeds (and even zero

speed) are imperceptible.

However, the rier

distortion of current

due

We begin our discussion of speed and torque control by referring

In addition, the carrier frequency

tors run

As

a result, standard induction

by a 60 Hz

PWM

voltage source as

Hz

to several

is

dc motor (Fig. 23.36a). The field <£>

that

stationary in space

is

which can be varied by means of the

When

and

field current.

the brushes are in the neutral position, the ar-

mature current

way

such a

/

flows

in the

that every

armature conductors

conductor

is

in

subjected to a

to a

force tending to turn the motor ccw. In this diagram is

control of the gate triggering. it

compared

produces a flux

sinusoidal source.

Pulse-width modulation

IGBTs,

mo-

about 10°C to 20°C hotter when supplied

conventional 60

to a

in the

voltage that appears across the windings increases the iron losses.

flux orientation

to the car-

frequency increases the copper losses

motor windings.

DC motor and

23.15

effected by computer

is at

right angles to

The current axis is therefore in line axis. The important feature is that the

the brush axis.

By

firing the gates of

possible to control induction motors up

hundred horsepower.

the axis of the armature current

If

GTOs

with the field

resulting torque

is

then

are used,

motors of several thousand horsepower can be driv-

en electronically.

maximum and

portional to the product

and

/

OA

directly pro-

Because the quantities

can be varied independently,

it

is

very easy

to control the torque.

23.14 Pulse-width modulation and induction motors The important

feature of

PWM

is

that

it

enables the

The speed can

s

.

by raising and low-

Thus, a very low speed

ing a low voltage accompanied by a large armature

production of very low -frequency sinusoidal voltages

and currents, using a relatively high-frequency car-

It is

A further advantage is that the waveshapes can be

E

can be obtained with high torque by simply apply-

current

rier.

also be varied

ering the applied voltage

/,

while keeping the flux

at its rated value.

important to note that the orientation of the

flux axis with respect to the armature current axis

altered in a fraction of a millisecond. Consequently,

has a direct impact on the torque. For example,

even low-frequency sinusoidal voltages can be trans-

the brushes are shifted as

shown

in Fig.

if

23.36b, the


605

angle between the flux axis and current axis

is al-

tered and this will produce a smaller torque. Indeed, it

if the brushes were shifted off by 90°, the angle between the flux and cur-

can be seen that

neutral

rent axes

would

also shift by 90° and the resulting

The reason

torque would be zero.

that the forces

is

on the armature conductors now cancel each

other.

Thus, flux orientation relative to the armature current axis

is

just as important as are the

and

/ val-

ues themselves.

23.16 Slip speed, flux orientation,

and torque Figure 23.35 Food processing is one of countless industrial applications where PWM drives are used. A good example is the production line at the Wortz® Company in Poteau, Oklahoma, pictured here. This 700-foot long baking and packaging line, which produces 1000 boxes of saltine crackers every 10 minutes, is equipped with 44 PWM drives plus 35 other individual motors. The drives allow precise control over the entire production process. Downtime and maintenance of the drive system are much less compared to older baking lines that use mechanical drives. (Courtesy of Baldor Electric Company)

a

—~r

The fundamental behavior of an induction motor having p poles can be understood by reference to Fig. 23.37. It shows two successive N, S poles, created by the stator (not shown), sweeping to the right at

synchronous speed n s The flux per pole .

tributed sinusoidally, with a

=

0.8 T.

The

flux in question

is

rotor and

its

follows that the flux at a slip

is

//,

speed given by

S =

Figure 23.36a control of

a dc motor.

1

1

).

Figure 23.36b DC motor with brushes

is

moving

less than n s

to .

It

cutting across the rotor bars

(n s

-

n)

/

Speed and torque

3.

where n

R H==I=h

1

rotor bars are also

the right, but at a speed

dis-

B peak

the mutual flux that

crosses the air gap (see Section

The

is

peak flux density

off neutral.


606

N

14.4

-4.8

N

Figure 23.37 Air-gap flux and resultant voltages, currents, and forces produced

As

where

S =

A

voltage

duced slip

=

synchronous speed [r/min]

n

=

rotor speed [r/min]

which

is

There are

maximum

and

The rotor frequency/? is quency/by the expression

proportional to the slip in

in time.

We

in bars

4 and

fi

=

related to the stator fre-

sf

(13.3)

which the

bar happens to be immersed. Thus, the voltage

in bars 1, 7,

each will vary sinusoidally

apply to an induction motor:

therefore, induced in each rotor bar,

speed S multiplied by the flux density

momentarily

in

speed [r/min]

ns

is,

the flux cuts across the bars, the voltage in-

learned in Chapter 13 that the following equations

the amplitude of

trical

in rotor.

10,

is

Furthermore, the

and zero

slip s is

s

=

(n s

given by

-n)h h

(13.2)

13.

six rotor bars per pole

angle separating them

is

and so the elec-

18076 =

30°.

and, finally, n s

is

related to the stator frequency

and the number of poles p by the equation

/

ELECTRONIC CONTROL OE ALTERNATING-CURRENT MOTORS

=

ns

From

(13.1)

120.///;

we deduce

these equations

from the

axis of the flux

tween zero and

stator for all loads be-

full-load. Clearly, under these con-

the following ex-

motor compares very favor-

ditions the induction pression:

607

ably with a dc motor as far as flux orientation

VI 20

f2 =

(23.3)

However, when

where

the rotor frequency

high (say,

is

30 Hz or more), the reactance of the rotor bar

f2 =

rotor frequency [HzJ

siderably greater than

— slip speed [r/min] p = number of poles

Thus,

rotor frequency

Frequency plays an important

role

of the induction motor. At

full-

in the flux orientation

load and rated torque, the rotor frequency of conventional

motors

is

2

Hz

or less.

Rotor Current The voltage induced

in a particular

rotor bar will cause a current to flow in the bar equal

by the bar impedance. The

to the voltage divided latter, in turn,

and

its

depends upon the resistance of the bar

reactance. Because of the reactance, the cur-

rent in a particular bar will lag behind the voltage,

which means

that

reaches

it

maximum

its

brief instant after the voltage has reached

mum. For example, rent

is

assumed this bar

sition bar 2

is

value a

maxi-

its

referring to Fig. 23,37, the cur-

to lag

60° behind the voltage; con-

sequently, the current reaches

because

its

maximum

60° behind bar 4, which

occupied

at the instant its

in

is

axis of the currents in the rotor

voltage

is

at all

that to

ensure good flux orientation

speeds, the frequency

Hz

low: typically, 2

in the rotor

must be kept

or less.

Torque. The force exerted on each rotor bar to

through

and the flux density

is

it

is

pro-

the product of the current flowing

portional

to be.

is

lags 60° behind the voltage.

For example,

in Fig.

in

which

happens

it

23.37, the force on bar 3

proportional to the current (208 A) times the flux

F = BLI

density (0.693 T). Using yields a force of 14.4

length of 10 cm.

N

for a rotor bar having a

The sum of all

the rotor bars, multiplied

(Eq. 2.26), this

the forces acting on

by the radius arm,


to the torque

is

equal

Fig. 23.37

the voltage, current, flux density, and

illustrates

force for the individual rotor bars.

bar 2

the po-

was

momentarily maximum.

The

We conclude

therefore directly proportional

is

con-

23.37, where the rotor frequency

in Fig.

40 Hz, the current to the slip speed.

is

resistance. Consequently,

its

the current will lag significantly behind the voltage.

S

The

is

concerned.

therefore

23.17 Features of variable-speed control-constant torque mode

displaced from the axis of the air gap flux by an an-

Before going into a detailed analysis of speed and

gle of 60°. This

torque control, which involves using the equivalent

is

equivalent to shifting the brushes

of a 2-pole dc motor by 60° from the neutral. The flux orientation in this figure

However, given load

is

therefore poor.

that the rotor

typically less than 2 Hz,

reactance of a rotor bar

is

frequency it

at full-

follows that the

very low compared to

the current and the voltage

The current

is

fore, essentially the

same

is,

there-

as tbe voltage distribu-

This means that the rotor bar in which the

duced voltage

maximum

its

typically less than 5°.

distribution in the rotor bars

is

current.

rent in the rotor

is

maximum As

will

in-

also carry the

a result, the axis of the cur-

almost directly

in line

circuit to

At full-load the phase angle between

resistance.

tion.

is

with the

look

diagram of the induction motor,

main

at the

operating

modes of

duction motor. 3-phase, 60

Hz

It

a

I

kW,

4-pole,

1

it

is

useful

shows

six

740 r/min

in-

features. Fig. 23,38

has a nominal rating of 4 16 V,

but

is

designed to run over a broad

range of speeds, including zero speed, by varying the stator frequency. In these six

held constant at in the air

sity

fl^k

gap

=

is

°- 8

modes we assume its

that the torque

is

rated value. Furthermore, the flux

held constant with a peak flux den-

T Tne -

f"irs t

example

(Fig. 23.38a) ex-

hibits rated operating conditions, together with in-

formation on rotor voltages and currents. These

Figure 23.38 Features of an induction motor: constant torque mode.

608


used to analyze

details are then

erating

modes

(Fig.

23.38b

flux

2338a: Rated Operating Mode, The

Fig.

quency/applied

to the stator

synchronous speed ops rated torque

at

is

1

800

fre-

r/min.

= 60

corresponding rotor frequency

r/min.

The

=

8 V; frequency 2

Hz

Motor Operating As a Brake. The

Fig. 23.38d:

stator frequency

dropped

is

still

further to 0.5 Hz,

with the result that the synchronous speed becomes

therefore given by

is

800

60/1

A>

The motor devel-

1740)

by the mutual

in the stator

is

E = 240 X

60 Hz, and so the

is

1740 r/min, which corresponds

speed of (1800

to a slip

The voltage induced

the subsequent op-

in

to 23.38f).

609

However, the motor can

15 r/min.

develop

still

Eq. 23.3: rated torque provided the slip speed

f2 =

(23.3)

60 r/min. This

at

achieved

is

if

opposite direction to the flux

and so

= —45

60)

f2 =

(60

X

-

4)/ 120

2

r/min.

However,

The peak voltage induced is

known

in

corresponding peak current

known

each rotor bar under 100

to be

lies

stator windings,

of 240 V,

(h

at

it

induces

in

was before

—

is

/?)

is

in

realized

is

183

as in

must be

magnitude, frequency, the slip speed (n s

again 60 r/min. Thus, full-load torque

produced when the rotor turns

at

(183

—

60)

is

=

123 r/min.

The voltage E^ induced

in the stator is

than before, because the flux r/min. ratio

The value of E (b

is

is

now

only turning

readily calculated

at

less

Ai

183/1800

=

flowing

=

24.4 V; frequency

by the

6.

1

Fig.

2338c: Motor Stalled. The

now reduced is

to 2

Hz and

Hz.

stator

=

0 and so the motor

is

is

speed

torque^- the slip

must again be 60 r/min. This means 60)

frequency

so thasynchronous speed

60 r/min. To produce rated

in the stator

windings. Nevertheless, the

that n

not turning at

=

(60

all.

—

is 60 r/min. This happens when the 60 r/min in either direction. Again,

The

as a brake.

flux

is

ro-

the

not rotating and so

0.

2338f: Operation Above Base Speed. As

stant torque

an-

mode, the

stator

frequency

is

raised to

150 Hz, giving a synchronous speed of 4500 r/min. Rated torque

speed at

is

4440

is

again obtained

60 r/min. The rotor r/min.

now quite = 600 V.

Note

that

high, reaching

is,

when

the slip

therefore, turning

induced

in the stator is

E$ = 240 X 4500/1800

modes examined The peak voltage induced in the rotor bars remains at 100 mV, the peak current remains at 250 A, and the freis

revealing that

in all six

above the rotor conditions are

quency

is

frequency Fig.

mo-

in the sta-

other example of motor performance in the con-

It

183

method:

E = 240 X

is

The frequency

same torque

when

The voltage induced

— the

2338e: Stator Excited by DC. The stator freis now zero, which means that a dc current

motor acts

Fig. 23.38a, the current in the rotor bars it

mode when

quency

each phase a voltage

with the result that the synchronous speed

and phase. This

Fig.

tor turns at

r/min. In order to produce the

speed of (15

2 V.

is

ative speed

applied to the stator has been reduced to 6.1 Hz,

exactly as

tor

the stationary

at 6,1 Hz.

in this

rated torque can again be obtained provided the rel-

a frequency of 60 Hz.

2338b: Operation

Fig.

also

is

It

4° behind the flux axis.

The flux orientation is excellent. As the mutual flux sweeps across

the rotor turns in the at a

and the

that the current lags 4° behind the voltage

and so the current axis

E

mV

250 A.

is

maintained

flux and rotor turn in opposite directions, the

Hz

tor acts as a brake.

these conditions

is

tor

unchanged is

at 2

identical.

Hz. In effect, as the stator

varied, the entire behavior of the

seen to depend upon the slip speed

is

portant to recall that the flux per pole fixed.

To meet

this

5. It is

moim-

was kept

requirement, the magnetizing

current that produces the mutual flux must some-

how

be held constant.


610

23.18 Features of variable-speed control-constant

^ X t « 3 CD

o°

1

horsepower mode

o ~

V -f

10

"~

1

50

= 4500

Returning to the motor described by Fig. 23.38a,

suppose the electronic power supply can only deliver the rated

maximum

of 240 V, but that the frequency

can be raised to 400 Hz, the

motor speed

need

if

wish

were held

at its

1

50 Hz.

normal

E2

2(peak)

250 A

(peak)

mV

100

to raise

about 4500 r/min, which means

to

raising the stator frequency to

flux density

We

be.

y

peak

If the

Figure 23.39 Features of an induction motor: constant horsepower mode.

level, the stator

voltage would have to be 600 V, as previously seen

But because the stator voltage

in Fig. 23.38f.

ited to

240

fall in

proportion to the increase

V, the

B peak

(60/ 50) 1

This flux density

X

lim-

in

0.8

frequency. Thus,

T=

0.32

is

depicted

T

in Fig.

.

23.39.

see what happens to the torque and speed,

In order to

we

develop the

maximum

means

ing the thermal limits. This

current should again be

To produce 250 A,

3.

were equal

If tfpeak

B pc

times

.

is lk

less.

now

mV

is

is

is

X 60 =

flux per pole

is

only 1/2.5 of

is

also 1/2.5 of

motor speed

rated base speed and so the

is

is

its

mV.

X

few extra degrees behind

larger than in Fig. 23.38, but the reduc-

due

to this

change

in flux orienta-

minimal.

be reduced by a factor of 20. To generate

to

mV in the rotor bars,

the 100 to

sponds

the slip speed would 200 r/min. This correa rotor frequency of 20 X 2 Hz = 40 Hz.

be 20

to

X 60 =

1

flux orientation

sustained.

2.5

23.19 Feature of variable-speed control-generator mode

slip

150

4500 —

In variable-speed induction

rated value, rated value.

nearly 2.5 times

is

2.5

would be poor, and the resultwould be so serious that the constant horsepower mode could probably not be

its

horsepower remains

The motor constant horsepower mode.

is

[f)

have

erator

its

rotor frequency

angle between the flux axis and current

tion in torque is

Hz

make E2

to

The required

and so the torque

at its rated value.

axis

no longer 2

2.5 times greater than

ing drop in torque

for the slip speed to in-

The resulting motor speed 150 = 4350 r/min.

the

.

The

speed of

way

r/min.

However,

induced

generate 100

to

speed must therefore be 2.5

The

peak

mV.

only 0.32 T, which

Therefore, the only

equal to 100

E2

The

is

5 Hz. This higher frequency will cause

E2 The

have

to 0.8 T, a slip

crease by a factor of 2.5.

4.

that the

the peak voltage

60 r/min would suffice But

Hz ~

speed

tor of

250 A.

bars must again be 100

23.38a.

in Fig.

in the rotor is

slip

However, if the speed had to be raised by a fac20 (i.e., to 20 X 1800 = 36 000 r/min), while limiting £ to 240 V, the flux density would

possible

torque, the current in the rotor bars should be

in the rotor

because the

tion

as large as possible, without, however, exceed-

2.

The frequency

the current to lag a

reason as follows:

1

5.

2

2.5 times less than the rated

is

peak value. The situation

To

is

peak flux density will automatically

operating in the

mode

motor

drives, the gen-

of operation comes into play very fre-

We

recall that

an induction motor becomes

a generator

whenever

the stator flux turns in the

quently.

same

direction as the rotor, but at a slower speed.

We

will study

running

two

cases,

at close to rated

the speed

is

much

lower.

one where the motor

is

speed and another where

These conditions are rep-

ELECTRONIC CONTROL OF ALTER NA TING- CURRENT MOTORS

Fig. 23.40b:

1

I

Generator Mode, Low Speed. The

frequency

stator

6

case

in this

synchronous speed of

243 r/min, and the

slip

is

Hz, producing a

6.1

83 r/min. The rotor turns

1

speed

same

rotor voltages and currents are, therefore, the

Power

as in Fig. 23.40a.

is

at

again 60 r/min. The

is

again being fed from the

rotor to the stator.

23.20 Induction motor and equivalent circuit

its

Torque and speed control of a squirrel-cage induction

motor

is

more

difficult to

motor because the rotor

is

achieve than

rotor current cannot be controlled directly. tor current

is

in

a dc

not accessible, and so the

The

induced by the current flowing

ro-

in the

Furthermore, the stator current also produces

stator.

the very flux that

needed

is

to

produce the torque.

This complex situation can best be resolved by

re-

ferring to the equivalent circuit of a 3-phase induction motor.

The complete circuit for one phase (drawn from 15), is shown in Fig. 23.41 It is very simi-

Chapter

.

lar to the circuit

diagram of a transformer. The pa-

rameters of the motor are listed as follows:

Figure 23.40

r,

=

stator resistance

X\

=


The generator

x2

=

rotor leakage reactance referred to the stator

equal to the rated torque, and

r2

=

rotor resistance referred to the stator

Features of an induction motor: generator mode.

resented in Figs. 23.40a and 23.40b.

torque in both cases the

same data

Fig. 23.40a:

is

is

used as

in Figs.

Generator Mode, Rated Speed. The

stator

frequency

speed

is

1800 r/min. The rotor turns

previous examples,

by

4°.

is

2

Hz and

it

is

axis it

by

is

was

in Fig.

now

slightly

4°.

1860 r/min; in the

follows that the rotor fre-

the current again lags behind

However, there

it

at

60 r/min. As

is

E2

an important difference be-

cause the direction of current flow

what

xm = s =

stator


slip (not slip

speed)

60 Hz and so the synchronous

is

consequently, the slip speed

quency

23.38 and 23.39.

is

the reverse of

23.38a. Furthermore, the current

ahead of

The important point

versal of the rotor current.

the flux axis, leading

to

It is

remember

is

the re-

the reversal that pro-

duces the generator action of the motor.

we have

In this figure, in the interest of simplicity,

not

included

the

branch representing the iron

losses.

E between

The applied voltage tral

N

produces a stator current

of two parts,

lm

and

/2

.

/

Current

line

/

m

is

tizing current that produces the flux

gap. Current actually

1 2 is

flows

in

and neu-

1

which consists

[,

the in

magnethe air

a reflection of the current that

the

rotor;

it

is

the

torque-

producing component of stator current. Flux the mutual flux that links the stator and rotor.

is

It is


6 2

1.5

Q

?fi cv 265V

3Q

<

60 Hz

1-2x3600

r

a * i 97 %

£l30O V

\

90 = 48 Q

,

w s = 3600 n = 3510

T=

10.3

Nm

Figure 23.42a Equivalent circuit of a 5 hp, 2-pole, 460 60 Hz induction motor at full-load.

Figure 23.41

V,

3-phase,

Equivalent circuit for one phase of a 3-phase induction

motor (see Chapter

When

15).

whose

precisely the flux illustrated in Figs. 23.38,

peak flux density

B (pcilk) and which

is

induces

voltage

The

in the circuit

fluxes <£| and

02

spectively.

The sum of

linking the stator.

tal

It

O

sum of 4> and

Similarly, the

The

and

resistance

R 2 /s

senting the active

<£,

is

is

It

2 is

induces

that

r

re-

in

me

=

3600/90

way of

=

R 2 nJS =

1.2

X

ft.

After solving the circuit, /,

in the rotor.

The

5.6

A =

A

it

found that

is

5.2

A

1.9

/„.

A

full-load torque developed

by

all

3 phases

is

given by:

repre-

9.55

T =

transmitted

is

therefore, equal to

is,

48

sta-

equal to the to-

E4N

a simulated

power P

and rotor

,

the total flux

E2N

induces voltage

flux linking the rotor.

and x 2

associated with

,

are the leakage fluxes for the stator

tor.

the motor operates at full-load, the speed is 3510 r/min, which corresponds to a slip speed of (3600 - 3510) = 90 r/min. The line-to-neutral voltage E is 460/V3 = 265 V. The power resistance

P 1

X

„

3

(13.9)

across the air gap, from stator to rotor, by induction.

_

This "power resistance" can be expressed in terms

X

9.55

of the slip speed S:

power

2

X 48

X

3

=

10.3

N-m

3600

The magnetizing

resistance

(23.4,

S

s

which

When

Equivalent circuit of a practical motor

machine, 23.42b.

What information can such a circuit diagram

yield in

current / m produces the flux

value

air gap. Its

value,

23.21

5 .2

is

at full-load is

97

%

of

the reason for designating

its

$ in the no-load as

97 %.

these currents are observed in the actual

we

The

obtain the picture flux

is

rotating

dragging the rotor along with

ccw it.

shown in Fig. at 3600 r/min,

We

discover that

the case of a practical motor? Consider Fig. 23.42a,

the axis of the rotor current lags only 2.4° behind

which shows

the axis of the flux. Consequently, the flux orienta-

the equivalent circuit of a

commercial

460 V, 3-phase, -60 Hz, 35 0 r/min cage motor. The parameters at 60 Hz are listed as follows: 5 hp,

1

r,

-

I

.5

tion is

is

excellent.

The reader

will note that the 2.4°

equal to the phase angle determined by the power

resistance (48 ft) and the leakage reactance

11

x2

(2 ft)

of the rotor.

r2

=

1.2 ft

xm

=

130 ft

Consider now the situation when the rotor

3600 r/min

is

locked, with full voltage applied to the stator (Fig.


613

Figure 23.42b At

full-load, flux in

the air gap

is

oriented at 2.4° with

respect to rotor current. Flux rotating ccw.

23.43a).

The power

to 7? 2 or

1

,

.2 ft.

that the currents are

and

resistance

is

now simply

After solving the circuit,

much

larger,

it

is

equal

found

and the orientation

have changed. Indeed, the rotor current axis lags 59° degrees behind the flux axis; conseof

/2

quently, there

is

a large reduction in the torque that

would otherwise be

available. Furthermore, the air

gap flux has dropped

to

witnessed by the drop

which

now

is

42%

in the

of

its

rated value, as

magnetizing current,

only 0.83 A. Clearly, this

is

an unsat-

isfactory condition as far as torque production

concerned.

speed (S tor

It is

is

directly attributable to the high slip

= 3600 r/min)

and the consequent high

this

analyze the behavior of the motor

when

it

will is

now

driven

by a variable frequency source.

hertz rule.

23.22 Volts per hertz of a practical 1

motor

gap

is

oriented at 59°

rule

motor remains

let

1

If

is

we

is

is

held con-

essentially the

However,

works out when our 5 hp, 35 0

which

a

problem

us see

how this

r/min,

driven by a 6

460

Hz

V,

60

source,

one-tenth of the base frequency.

apply the volts per hertz

rule, the line volt-

The line-to-neuThe reactances are

one-tenth of 460 V, or 46 V.

age

is

speed curve retain the same shape provided that the

tral

voltage

all

range.

low speeds. For example,

Hz commercial motor

8 and 204 1 9 that in variable-

varied in proportion to the frequency. This

the volts/hertz ratio

same over a broad speed

speed drives, the torque-speed curve, and current-

is

When

stant, the flux in the

arises at

voltage

the air

has given rise to the so-called constant volts per

background information, we

We saw in Sections 20.

rotor, flux in

with respect to rotor current. Flux rotating ccw.

ro-

frequency (60 Hz).

Using

Figure 23.43b At locked

is,

therefore, 26.5 V.

diminished by a factor of

10, as

can be seen

in


614

Fig. 23.44a.

Assuming

same

the

r/min in order to get rated torque,

power

resistance

pedances

is

equal to 4.8

it

il.

1

.5

Q

Q

0.3

follows that the

Thus,

the im-

all

23.44a are ten times less than those

in Fig.

23.42a-except for the

in Fig.

speed of 90

slip

1.5 il stator resistance

which remains unchanged. After solving the circuit of Fig. 23.44a,

we

disrt

cover that the torque due to the three phases

s

is /?

/

P

9.55

X

3

4-

X

= 360 = 270 = 6.1 N-m

(13.9)

Figure 23.44a

x

9.55

5 hp induction motor operating at one-tenth of rated

4.8

X

3

=

6.1

voltage and frequency.

N-m

360 This torque

much

is

value of 10.3

less than the rated

N-m. What has happened? The calculations indicate that the magnetizing current is only .5 A compared 1

to

1

gap

.9

A in

Fig. 23.42a. Therefore, the flux in the air

much

is

That

less than before.

the reason for

is

the large drop in torque.

Thus, the constant volts/hertz rule leads to a big

drop

in

torque

at

lower speeds. The culprit

tor resistance. If

volts/hertz rule

The drop

is

the sta-

were not present, the constant

it

would work

perfectly.

torque can be remedied by system-

in

atically raising the stator voltage, to

compensate for

= 76

compenmotor is con-

the IR drop in the stator. This torque boost sation can be introduced

by a

trolled electronically

when

the

PWM drive.

Figure 23.44b Motor rotating

23.23 Speed and torque control of induction motors The problem

in controlling


current

I2

the current

/,

I2

in the stator. In

components. Furthermore,

is

its

it is

that

order to conits / ni

advantageous

rated value, to ensure the

as great as possible without excessive satura-

track of

all

these variables, the circuit

parameters of the motor must be known. Toward this end,

some

r/min.

time

when

When

the drive

the

motor

is

is

done once,

at the

installed.

is in

operation, the stator volt-

age, stator current, frequency, and speed are sensed

by transducers and compared with the wanted ues.

The computer

mines the / m and

in the control

/2

val-

system then deter-

components and automatically

sets the required voltage, frequency,

and current

that are required.

Sensing the speed creates a problem, because

tion of the iron.

To keep

270

parameters. This measurement is

namely

current must be split into

at

feature that actually measures the rotor and stator

the torque-producing

flowing

that / m be held close to

flux

m and

into a single current,

trol the torque, this

and

/

torque and speed

merged

are

%

= 90 r/min

PWM control

systems incorporate a

shaft encoders

have

to be added. This

is

not an easy

matter for motors that are already installed and

whose

shaft extensions are not accessible. For this


reason other algorithms exist whereby the computer

can estimate the speed, without feedback from the This gives good results,

shaft.

low. But

minute, or is

the speed

if

if

mandatory

is

is

it

in

required,

is

many

forth, as part

located

more than 50

ni

of the control

If a

PWM drive

is

used

vary the speed of a

to

pressor or fan, the change in speed rather slowly and once set, the tially at

tor, to

computer by the ultimate

trol

the high-speed calculations

of induction motors

many

user.

mo-

to the converter driving the

achieve the desired is

made

devices,

sophisticated

Thus,

result.

PWM con-

possible thanks to addition

in

to

the

motor behaves

ranging from

kHz

1

about

to

16 kHz. These frequencies can often be changed

One

the field to satisfy particular needs.

cerns

noise which,

is

in

quiet environments,

ticularly noticeable in the

1

0

kHz

to 2

is,

say,

47 Hz instead of

60 Hz. The fundamental voltages and currents are sinusoidal and the equivalent circuit diagram ficient to describe the is

kHz

is

par-

range. For

is

suf-

motor behavior, even while

changing. in

some machine-tool

applications, the

change without warning,

all in

stop,

and

start

may suddenly

a matter of millisec-

onds. Under such conditions, the behavior of the mo-

can only be described by special equations

tor

are far

more complex than those covered here by

that

the

equivalent circuit. During such transient conditions, the voltages and currents are

no longer sinusoidal,

frequencies have been raised to

and the computer-generated waveshapes change

kHz and more because they are beyond the range human audibility. Unfortunately, the higher fre-

ods, the flux must be maintained both in value and

power-

orientation so as to instantaneously develop the re-

this reason, carrier

of

1

in

of the con-

commade

an ordinary induction motor, ex-

while responding to torques that

drives for induction motors use various car-

usually

motor runs essen-

motor must rapidly accelerate, reverse,

23.24 Carrier frequencies frequencies,

like

cept that the frequency

However,

PWM

is

constant speed. Under these conditions, the

the speed

switching converter.

rier

the

extra features

system. Furthermore, special features can be pro-

The computer makes

when

converter.

23.25 Dynamic control of induction motors

grammed

and sends signals

from the

it

charge of speed and

possible to include

into the

have been

filters

is

an encoder.

and so

line

motor

such as rate of acceleration, deceleration, overcurrent protection,

cases,

only a few revolutions per

Having a computer torque,

some

In

to attenuate the effect, particularly

a servo position control to use

insulation.

added

is

1

than normal, causing an eventual breakdown of the

not too

the speed

if

6

quencies always require a decrease

in the

handling capability of the semiconductors.

The sharp sulation.

The

owing

some motors,

fast rise

to the

instant to instant.

During these

as regards the in-

time produces two effects.

mismatch between the high-fre-

It is

rier

dynamic control

possible.

that of

tor

motor

Obviously, vector control

PWM

volt-

age. Second, during these sharp peaks, the distrib-

uted

inductance

and capacitance

of

motor

the

windings causes the doubled voltage to appear across the first

few

turns of the

the dielectric stress

is

much

control

but

other

It is

this

type of

often called flux vec-

names is

at car-

conjunction

are

also

used.

not needed to drive a fan

or compressor, where fast changes of speed are not required.

Nor

is

vector control necessary to drive

high-inertia loads that inherently take considerable

time to change speed. Indeed, inertia plays an important role in the setting of

all

PWM drives.

result,

There are many ways of designing high-response

greater

induction motor drives, but so far no single best

motor windings. As a

on these turns

in

with high-speed computers, that makes

the connecting cable, the voltage across the

terminals tends to double for a fraction of a mi-

IGBTs operating

the fast switching of

frequencies of several kilohertz,

quency impedance of the motor windings and

crosecond during every impulse of the

transition peri-

quired torque.

rise-time of the carrier voltage has also

created problems in

First,

from


616

method has evolved. However, a few basic ples are common scribe

them

to all

vector drives, and

we

princi-

32 A

30 A

18

A

20 A

14

A

10

will de-

briefly in the next section.

3 O

23.26 Principle of flux vector control

stator

When

an induction motor runs

In this

way we

arrive at a simple circuit

rotor

we

at steady-state,

mechanical angle between

can use any one of the three phases as a model for all.

A

rotor

and

stator

= 20°

diagram Figure 23.45

and a few simple equations

that

adequately describe

The respective phase

the behavior of the motor.

Instantaneous currents

and

rotor

in

stator windings of

a 2-pole, 3-phase induction motor.

voltages and currents have sinusoidal waveshapes,

whose frequency

is

constant,

all

neatly separated by

This reassuring situation

when

motor

the

subjected to rapidly changing

is

if

behavior

particularly

the drive

is

is

completely upset

is

suddenly has to change speed. The

torques, or

it

complex when

so small that

its

the inertia of

mechanical time con-

is

trical

time constant of the drive.

When

such high

re-

when commands, the

sponse drives are subjected to disturbances, or they must follow rapidly changing

voltages and currents are no longer sinusoidal, and the term phase angle loses

Under these voltages in

all

its

three phases

and

must be considered on

explanation that follows,

wound-rotor motor so the currents

that

it

is

we assume

easier to visualize

and rotor position. Nevertheless, a mo-

is

in

a schematic diagram of a 2-pole,

the stator

and rotor windings are as

shown. The stator currents are generated by an appropriate current source, which establishes both their instantaneous rates of change.

magnitudes and instantaneous

The instantaneous magnitudes

duce speed voltages

in the rotor

windings

proportional to the speed of rotation.

On

in-

that are

the other

hand, the instantaneous rates of change of the stator currents induce voltages in the rotor windings duction.

The sum of

rise

the

to

rotor

The delay

is

currents.

by

in-

the speed voltages and induc-

in-

due to the inductance

of the rotor windings. Consequently, the

L 2 /R 2

tnTle

constant of the rotor (per phase) plays an important

high-performance drives.

role in the response of

The

current source

changes

in

is

designed to produce

fast

selected rotor currents by modifying

both the magnitude and rate of change of specific stator currents. Therefore, tor drive

rents

is

one of the duties of a vec-

to target those stator currents that will in specific rotor cur-

while developing the required torque and

maintaining the rated flux and in the air

gap. Evidently, this

its

is

proper orientation

no small

feat.

In explaining the principle of a vector drive,

much

it

is

easier to use a numerical example. Thus, the

magnitudes and directions of the instantaneous currents are

3-phase motor. Suppose the instantaneous currents

flowing

types of voltages.

a

having a cage rotor performs the same way. Fig. 23.45

give

produce the desired changes

meaning.

special conditions, the currents

an instantaneous basis, both for the stator and the rotor. In the

voltages

stantaneously in response to changes in these two

of the same order of magnitude as the elec-

stant

tor

tion

Unfortunately, the rotor currents cannot change

phase angles of 120°.

shown schematically

in Fig. 23.45.

physical location of the stator windings A, B,

C

The and

instantaneous position of the rotor windings X, Y, are

shown

in Fig.

Z

The distributed stator and shown as single coils. The in-

23.46.

rotor windings are

stantaneous current flows

in the

respective coils are

The individual coils are assumed to have 10 turns. The rotor is rotating clockwise. At

also shown.

this instant, its position

Z

is

such that windings X, Y,

are displaced by 20° clockwise

from the corre-

sponding stator windings A, B, C.

The currents

in

the three stator coils produce

magnetomotive forces

that are oriented at rieht an-


617

gles to the plane of the individual coils. Thus, by

applying the right-hand rule, phase

mmf of cally

32

A X

upward

duces an

mmf

of 18 left

vertical. In turn,

10 turns

from the

A

=

A X

at

phase

produces an

A, directed

phase

(Fig. 23.47). Similarly,

rected to the

X

= 320

10 turns

10 turns

=

verti-

B

pro-

180 A,

di-

an angle of 60° from the

C

produces an

mmf of

140 A, slanted to the right

at

14

A

60°

vertical.

458

Figure 23.48 Instantaneous magnitude and orientation

space

in

of

the rotor mmfs.

32

A

The vector sum of This

mmfs

placed

as

It is

at this particular

if

the entire stator were re-

moment by

coil carrying a current of 48. fictitious coil is tilted at

Now

let

1

a single 10-turn

A. The plane of

coils

Instantaneous position of the rotor and stator windings.

case of the

us consider the rotor.

mmfs

stator.

this

4° to the horizontal.

The

individual

produce oriented mmfs the same way as

Figure 23.46

duce

gives a resultant

s

three phases.

all

these

mmf / of 481 A, slanted at 4° to the vertical. single mmf represents the combined effect of

stator

in

the

Thus, the respective currents pro-

of 300 A, 200 A, and 100 A, which are

oriented at 60° to each other. However, on account

of the position of the rotor relative to the

481

entire

®

group of

mmfs

is

shifted

by 20°

stator,

the

(Fig. 23.48).

resultant of these mmfs is a rotor mmf / R of 458 A directed downward at an angle of 9° clockwise to

The

320

the vertical. Therefore, at this instant, the three rotor

windings can be replaced by a single

turns, carrying a current of 45.8 A.

coil of 10

The plane of this

fictitious coil is tilted at 9° to the horizontal.

Let us

now combine the

23.48 into a single It

the stator mmfs.

how

in

space

of

However, /R

it

23.49).

the stator and rotor vectors

bine to produce the resultant

Figure 23.47 Instantaneous magnitude and orientation

again shows

vectors in Figs. 23.47 and

mmf vector diagram (Fig. mmf vectors

also reveals that the vector

produces a vector

M

/

.

It is

/s

sum

com-

and / R

of

this important net

s

.

and

mmf


6 8 1

Tn observing Fig. 23.49,

it

is

obvious that the

six

vectors generated by the six windings will change

by

instant

instant, both in

magnitude and direction.

The challenge of vector

control

changing torques and speeds, /

is

keep vector

the dotted circle, (2) to keep angle

M on

90°, and (3) to keep

/s

from exceeding

face of

the

in

(1) to

its

\\i

close to

maximum

permissible limits.

The reader will note trol

of current

that the accent

in the three stator

is

on the con-

windings. To pro-

duce the required currents, the voltages generated

PWM

by the

converter must,

in turn,

have appro-

priate waveshapes. In addition, the instantaneous

must be inferred from the

rotor currents

neous

stator voltages

and

feedback from an encoder

mine fed

which

Figure 23.49

The sum sultant

of the stator

mmf M which /

,

and

rotor

mmf's produces a

creates the flux

in

all

necessary) to deter-

These readings are

mathematical model of the motor,

in

the drive parameters are stored.

most vector drives, the 3-phase readings are

In

re-

(if

the position of the rotor.

into a

instanta-

currents, together with

converted into equivalent 2-phase values because

the air gap.

they are easier to manipulate. Tn effect, for purposes that is

produces the mutual tlux

the

same

where of

1

flux that appears in Fig. 23.41

is

The

in this chapter.

mmf /M

and

This else-

has a magnitude

mmf

R and the mutual flux

.

The

relationship

T= maximize

close to 90°, and that

PWM

It

that

all

values in terms of direct and

a tribute to the designers of vector drives

is

such sophisticated control systems have been

in Fig. / R 4> sin

(23.5)

i|j

the torque, angle is

i[i

should be

one of the objectives of the

vector control unit. Another objective

is

adjust the magnitude of / M to produce rated flux in the air

converted into

invented. But a glance at the ever-changing vectors

given by

In order to

mits expressing

created by the interaction of the ro-

is

is

an equivalent 2-phase machine. This approach per-

quadrature axes.

10 A, inclined at 14° to the horizontal.

The torque tor

in the air gap.

of computation, the 3-phase motor

gap. Thus, the actual magnitude of

23.49, and their underlying concept, reveals

that fast

PWM switching and nanosecond computer

response

in real time, is

what makes such remark-

able drives possible.

to

/

23.27 Variable-speed drive

M

and

should be adjusted to equal the desired /* M Tn Fig.

electric traction

.

23.49 the value of / M ing a radius /* M

We

is

a

should

ally the tip of / M

little

lie

too large because ide-

on the dotted

circle hav-

one of the objectives of vector

to decompose the stator mmf /s into two component that produces the tlux and the component that produces the torque. These components are made evident in Fig. 23.49; they

control

speed of an induction motor have to be controlled.

We

.

recall that

There are many applications where the torque and

is

parts: the

correspond respectively to vectors

/

M and / R

.

will

examine

electric traction

because

in this

category the torque and speed cover a particularly

broad range, including generator action. Electric traction

power

phase power,

power

is

also interesting because

lines for electric trains furnish at

voltages from 5

kV

60 Hz, to

many

single-

25 kV. The

factor of the transportation vehicle should be


619

Figure 23.50 This blower-cooled, flux vector-controlled motor with

an

optical

encoder (not

visible) that

is

equipped

permits accurate

sensing of the shaft position at any instant of time. Standard encoders produce 1024 pulses per revolution. Nominal rating of motor: 10 hp, 230 V, 3-phase, 60 Hz, base speed 1800 r/min. The speed is variable from zero to 4500 r/min. PWM carrier frequency is nominally 2.5 kHz or 8 kHz. {Courtesy of Baldor Electric Company)

Figure 23.51 Internal view of

and a host

a

PWM flux vector control

of other

components under the

with information provided by the trol.

keypad

unit

showing the complex

circuitry.

control of a microprocessor.

instructions, permits

It

comprises IGBTs,

The feedback from

an extremely wide range

amplifiers,

filters,

the encoder, together

of position

and speed con-

Thus, torque can be programmed according to position and speed, ranging from zero to several thousand rev-

olutions per minute. Typical features are 0.01 percent free motoring

down

to

speed

regulation,

1

ms

torque reversal, and smooth vibration-

zero speed.

(Courtesy of Baldor Electric Company)

as close as possible to unity so as to

minimize the

line voltage drop. In addition, the current

from the

line

it

draws

should be sinusoidal and tree from

harmonics, so as to prevent interference with adjacent telephone lines.

As we

will

see, the

PWM

switching converter offers an elegant

way of

meet-

ing these requirements without having to resort to

huge

filters

and power factor correcting

capacitors.

Traction motors are relatively large and hence

GTOs

are often used.

The switching frequency of


620

GTOs

mum

variable but

is

is

typically limited to a maxi-

of about 250 Hz. This relatively low carrier

frequency demands particular attention

when

PWM

of the motor windings. The 5th and 7th harmonics are dominant: on a

spond

However,

methods are applied. In the

range from 60

Hz and

up, the switching

60 Hz fundamental, they corre-

to frequencies of if

300 Hz and 420 Hz.

a rectangular

wave

is

used

at

low

fre-

quencies the corresponding current harmonics be-

PWM

converters are arranged to deliver rectangular wave-

come

shapes, as already discussed in Section 23.10. There

Pulse-width modulation can be synchronized or un-

is

an important advantage

in

doing

so,

because for a

given dc link voltage, the fundamental rms voltage is

a

higher

PWM

when using

a rectangular

wave

rather than

wave. Although the voltage harmonics are

damped out by

the reactance

methods must be used.

synchronized. Synchronization means that the carrier

frequency

is

arranged to be an exact integral multiple

of the wanted fundamental frequency. The multiple

number such

preferably an odd

important at these frequencies, the corresponding current harmonics are

too great, and

It

as 3, 5, 7,

and so

has been found that synchronization

tageous whenever the carrier frequency

is

is

forth.

advan-

is less

than

Figure 23.52 This 3 hp, 460

V

PWM drive and keypad

integrates the control unit

and motor

into

torque the speed can be varied from 180 r/min to 1800 r/min. Above 1800 r/min the motor operates

in

a single package. At constant

— up

to

a

maximum

of

3600 r/min—

the constant horsepower mode. For optimal performance the carrier frequency can be set to

any value between 1 125 Hz and 18 kHz, [Courtesy ofBaldor Electric Company)

albeit with progressive

power

derating.


ten times the

wanted frequency. Thus,

frequency

variable but limited to a

250 Hz,

say, of

it

if

the

fed into a

should be synchronized whenever

wanted frequency happens

to

this

GTO.

43.67

=

If the

Thus, the carrier should be

218.35

Hz

PWM

wherein the carrier limit,

is

is

In this

at the

When line,

power at

(3)

1

The catenary 5 kV, 60 Hz to

single

Hz

PWM

and thereby helps reduce the ripple

the system

is first

connected to the power

switch (11) and resistor (12) limit the inrush

(13)

is

immediately closed thereafter.

(3).

of

a traction drive.

At the same time,

it

by

(8) traction

motor

@ @ (Tj)

pantograph switch resistor

(Q) switch

filter for

the

the

converter

establishes the nature of

power flow between conIn

of inductive reactance.

and chopper

PWM converter

generated

verter (3) and transformer (1).

(9) trolley line

©filter

frequency

a transformer

converter

it

series-tuned filter

leakage reactance of the transformer

phase transformer

(6) capacitor


120

line supplies

inductor

(4) braking resistor

(7)

(3)

the active and reactive

the basic elements of a drive for a

© @

A

current due to the presence of capacitor (6). Switch in the

components

transportation vehicle.

by capaci-

(8).

Inductive reactance (2) acts as a

23.28 Principal

single-phase

stiff

(7) furnishes the

dc link voltage.

carrier

shows

held

absorbs the double frequency current generated

following example.

Fig. 23.53

is

(5)

GTO

250 Hz.

These points are brought out more clearly

link voltage

during a fast stop.

in the

mode, the harmonics

are clustered around multiples of

is

all

by converter

method can be employed,

simply held fixed

namely 250 Hz.

voltage

event that the catenary line cannot absorb

in the

X

25 Hz, the

less than

The

V.

A braking resistor and chopper (4) absorb power

exactly.

wanted frequency

unsynchronized

set at 5

The

and switching converter

3-phase power to traction motor

be 43.67 Hz, the

odd multiple of does not exceed the 250 Hz limit

frequency that

530

to

PWM converter (3) that delivers 700 V to

the dc link. tor (6),

carrier should be set at the highest

of the

down

that steps the voltage

the carrier

maximum,

wanted frequency exceeds 25 Hz. For example,

the if

is

621

practice, the fills

the role


622

It

must be understood

that the circuit of Fig.

23.53 has been highly simplified to illustrate only those points

we want

to

The 3-phase, 4-pole

motor

is

described

minimum j/s

a

maximum

of 300

a question of choice regarding

the level of current harmonics that are permissible.


power:

When

kW

60

1

0

speed range:

to

V

motor runs between 1800 r/min (base

temperature class: forced ventilation:

waveshape

6-step

V dc

used

is

1

00 Hz.

A

rectangular

range; thus, the

in this

full

The motor operates in the constant horsepower mode, and the fundamental rms voltage across the motor terminals is 700

H 12

between 60 Hz and

cies lie

3000 r/min

1800 r/min

base speed:

the

speed) and 3000 r/min, the corresponding frequen-

545

rated voltage:

mass:

the carrier frequency will be held be-

The maximum depends upon the allowable switching losses in the GTOs, plus safety margins.

as follows: type:

mode

Hz. The

emphasize. traction

PWM

tween a minimum of 200 Hz and

nrVmin

link voltage

available.

is

given by

520 kg

=

£iine

=

°- 78

0-78

X 700 = 546 V

(23.1)

At frequencies immediately below 60 Hz, where

23.29 Operating mode of the 3-phase converter

more important,

the harmonics begin to be

synchronized

The operating mode of converter (7) can be followed by referring to Fig. 23.54. The GTOs have a frequency limit of, say, 300 Hz. We assume that in the

PWM

mode

is

the

Bearing

initiated.

Hz and bewanted frequency is lowered, the carfrequency must be decreased in proportion,

be used to generate frequencies of 60 low. rier

As

the

both frequencies following the sloped line constant torque

When

constant hp.

nized

320 300 280

rectangular

wave

PWM

PWM

:

-

c

CD

g-

quency

is

the

200

limit

On

this

quency

^

new

5

= 40

operating

starts at 7

low the 300 Hz quency

falls

reached,

is

to a

GTO

carrier fre-

Hz, which

is

just be-

Then, as the fundamental

gradually from 40

Hz

—

30 Hz

30 Hz, the

it is

always ex-

30

40

this point, a transition

chronized 60

is

running

800

is

at

When

about 900

210 Hz

—

just

above the 200 Hz minimum. At

20

reached, the motor

r/min, and the carrier frequency "I

1

0

is

fre-

to

actly 7 times the fundamental frequency.

120 100 80 60 40

fre-

frequency ratio of

line, the

X 40 = 280

limit.

5.

evi-

Hz.

carrier frequency keeps track so that

160

it is

corresponding fundamental

-

The schedule then jumps 7.

210

o 200

200 Hz

the lower

dent that synchronized

unsynchro-

in

mind that the GTO frequency must not exceed 300 Hz, it follows that a frequency ratio of 5 must

100

PWM

is

made

mode, using the

frequency (300 Hz), which

is

to the unsyn-

maximum GTO

10 times the wanted

fundamental frequency (Hz)

frequency (30 Hz). Thus, for fundamental frequen-

below 30 Hz, the

Figure 23.54

cies

Generating variable fundamental frequencies with

stant at

moderate

line (Fig. 23.54).

carrier frequencies.

GTO

frequency

is

held con-

300 Hz, as shown by the bold horizontal


Fig. 23.55 illustrates the typical

motor voltage

the line-to-line ating

We recall that EAB is the difference between the voltages EAN and EBN that are produced by the two arms

waveshapes of

EAB for the four oper-

modes of the converter displayed

of the 3-phase converter (see Fig. 21.86, Chapter 21).

in Fig. 23.54.

+700 V

+700 V

142 V

-700 V (a)

fundamental: 100 Hz, 546 carrier:

V

rms,

3000

-700 V r/min (d)

100 Hz, 700 V peak


V

rms, 330 r/min

300 Hz, 700 V peak

+700 V

-AB 82 V

-700 V

(b)


V

rms, 1410 r/min

235 Hz, 700 V peak (e) line-to-neutral

fundamental:

+700 V

carrier:

voltages 1 1

-ab

-700 V fundamental: 35 Hz, 318 carrier:

V

rms, 1050 r/min

245 Hz, 700 V peak

Figure 23.55

Waveshapes a. b.

of line-to-line voltage for operating

Constant horsepower mode: 100 Hz, 3000 Synchronized PWM at 47 Hz; rfl = 5.

modes shown

f

d.

Synchronized PWM at 35 Hz; m f = 7: Unsynchronized PWM at 1 1 Hz; m = 27.27.

e.

Line-to-neutral voltages of individual switching

c.

in Fig.

r/min.

f

arms

EAN

and

EBN

.

23.54.

£ AN and

Hz, 58.3

V rms

300 Hz, 700 V peak

dc component: 350 V

(c)

623

i

'BN


624

For speeds above base speed (1800 r/min), implying frequencies above 60 Hz, the converter operates in the rectangular

age

120°

during

applied

is

wave mode. The

effective line-to-line voltage V. Figure

when

is

and

intervals

the

held constant at 546

EAB waveshape

23.55a shows the typical

the frequency

step volt-

slightly

is

700 V dc supply voltage. 23.55b shows the £ AB waveshape

greater than the

47 Hz,

corresponding to a synchronous speed of 1410

The

r/min.

=

carrier frequency

235 Hz, synchro-

is

dc supply. Consequently, the

Vs

when

shorter is

700

However,

GTOs

the

momentary fundamental voltage

142 V.

less than

does not mean that individual

this

are being switched on,

switched

off.

We

and a fleeting 0.68 ms

must remember

that the

£ AN

£ BN

and

when

on/off switching intervals. In effect,

the amplitude of the fundamental (11 Hz) out-

£AN

on interval

nized to be exactly 5 times the fundamental fre-

put voltage

equal to the ^'interval, namely about 3.33

also synchronized with respect to the

order to produce the symmetrical pulses

in

tal,

fundamen-

shown. The peak value of the fundamental voltage

is

now 428 V2 = 605

V

V,

which

700

less than the

is

23.55c

Fig.

The

£ AB

waveshape of

the

is

—

1

is

low, the

The same

.67 ms.

is

true for

ing rate presents no problem for the

and

£ BN

shown

carrier frequency

in Fig.

now 245

is

35 Hz.

at

Hz, which

ex-

is

in

contains a dc component of 350

£AB

equal

to

V

in Fig.

illustrated in

23.55a the volt-

the positive half of the sine wave.

tained is

23.55d

when

is

this

is

in the

is

is

pulse durations

width varies

In the is

1

Hz.

It

in the

In

fixed at 300 Hz.

very

are

Nevertheless, although the figure cannot their

1

ob-

unsynchronized mode.

the carrier frequency

individual

short.

show

it,

course of each half-cycle.

middle of the cycle, where the peak voltage

142 V, the pulse width

culation

is

made

is

Chapter 21, Section 21.46 we saw that a switch-

is

equal to the pe-

riod of the carrier frequency, which

is

s

=

3.33 ms.

the fundamental

is

momentarily 142

When

ms amount

T=

1/300

the voltage of V, the volt-

to 142

X

3.33

=

wave of any

fre-

quency, amplitude, and phase angle. This feature has a direct application

in the

power exchange

tween the ac and dc side of the converter

The

23.53.

redrawn

be-

(3) in Fig.

converter, dc link, and transformer are

in Fig. 23.56.

The reactance x

is

the leak-

age reactance of the transformer, including the

re-

actance of the catenary line referred to the sec-

ondary side of the transformer. Let us Voltage

as follows:

seconds during 3.33

latter is related to

-82 V

ing converter can generate a sine

only 0.68 ms. The cal-

The duration of each pulse

Hz = 0.00333

M3

V and a superposed

The

23.30 Operating mode of the single-phase converter In

£ AB

waveshape

waveshape obtained when the con-

operating

mode,

The

the multipulse

the fundamental frequency

typical of the

verter

V

by 142

V.

the

seconds of the positive pulse are equal to those of

Fig.

+700 V

pulses. This general

waveshapes

For example,

of

volt-seconds

the

rule holds true for all the Fig. 23.55.

GTOs. The £ AN

both cases. However, the average output voltage

seconds under the dotted sine wave during one halfcorresponding seven 700

2

This switch-

23.55e. Note that the instantaneous

peak ac component of 82

are

.

ms

switching intervals and waveshapes are

actly 7 times 35 Hz. Note, in passing, that the volt-

cycle

£ BN

almost

is

output voltage fluctuates between zero and

dc supply voltage.

£ AB

pulses are produced by the difference between the

quency. Furthermore, the phase angle of the carrier is

V

0.68 ms. Clearly, the pulse widths are even

later at

V

vided by the .700

duration of the ompu\se must be 473

100 Hz. Note that the peak

is

amplitude of the fundamental voltage

Fig.

473 mVs. However, these volt-seconds are pro-

former

now

£ 12

look

at the

fixed because

is

voltages in Fig. 23.56.

on the secondary side of the its

are directly related to the catenary

work. of

£34

On is

trans-

frequency and amplitude

power

line net-

the other hand, the magnitude and phase

determined by the switching action of the

converter

(3).

By

controlling the magnitude and


phase angle of tive

£ 34

is

it

,

possible to control the ac-

and reactive power flow between the secondary

side of the transformer

and the converter.

phase angle of

by an angle

6,.

E34

it

)

is

E ]2

lags behind

Furthermore, suppose

£34

rms value of

adjusted so

adjusted so

£ 34(rms)

it

is

,

(the

equal to

it

E l2

leads

As

23.57b).

as far as the

the converter; consequently, the catenary supplies

say,

.

power is

at unity

power

transferred

is

factor.

The

active

power P

given by

P =

M

^ is

sin 9,

1

(16.8)

is

obtained

to the

dc link by con-

£34

so that the

80° out of phase with

E [2

(Fig.

power factor is again unity catenary power line is concerned.

a result, the

£34 generated by the converter

pulse-width modulated

at

a carrier frequency

of,

4 kHz. The voltage between terminals 3 and 4

composed of 700 60 Hz.

V

impulses that are modulated

can be shown that the principal har-

It

at

these terminals are

given by the equation: ./,.

coasting downhill.

Instead of applying the brakes, the drag

The power furnished

is

at

by having the motor run as an asynchronous generator.

is

monic frequencies generated

E]2

Next, suppose the vehicle

.

The 60 Hz voltage is

is

E 34 so that

by an angle 6 2 At the same time, the

resulting current

.

that

to be fed into the

This power reversal

converter adjusts the magnitude of

Under these conditions the resulting / is forced to be in phase with E V2 As a result, only active power is delivered from the transformer to £i2(rms/ cos 9]

now 3.

obtained by shifting the phase of voltage

Consider, for example, Fig. 23.57a wherein the is

verter 7 (Fig. 23.53) has

transformer by converter

625

wherein fH rier

the

is

=

2fc ±f

harmonic frequency, fc

is

the car-

frequency and /is the wanted frequency. In our

=

case,/H

2

X 4000 ± 60 = 8060 Hz

These frequencies are so high

and 7040 Hz.

that the correspond-

ing ac currents are almost completely filtered out by the reactance is

x.

As

a result, the current

The 60 Hz

sinusoidal current

tified current Id

Thus,

consists of

Id

ponent

is

age

£56

chopped

(Fig. 23.56)

appears as a rec-

120

Hz component.

120 Hz. As a

LC

Hz

dc link volt-

ripple.

the ac motor does not ex-

perience the same problem because

in

balanced

3-phase systems the instantaneous power stant, so there is

ripple

is

that

converter

waves

This com-

series filter (5)

result, the

contains only a small 120

The converter feeding

the converter.

rectified sine

short-circuited by the

that is tuned to

/

on the dc side of

that contain a strong

Figure 23.56 Power transfer between the transformer and the ac side of a converter.

/

nearly a pure sine wave.

no ripple on

this account.

is

con-

The only

caused by the switching action of the

at carrier

frequency.

/

Figure 23.57a Active power delivered from transformer to converter.

Figure 23.57b power delivered from converter

Active

to transformer.

Figure 23.58 its power from an 1 1 kV, 25 Hz single-phase catenary. The ac voltage is rectified to produce a fixed dc voltage of 2400 V for the dc link. The GTO thyristors in the PWM converters generate a variable frequency (0-120 Hz) and variable 3-phase voltage (0-1870 V).They drive four 815 kW, 3-phase induction motors. The fully loaded trainset is 140 m long and has a mass of 343 t. It has reached speeds of 277 km/h (172 mi/h). The tilt-body feature enables the train to move through curves at higher speeds, thereby saving time while ensuring passenger comfort. Vehicle dynamic tests, carried out in the U.S. Northeast Corridor on this ABB X2000 tilt-body trainset were sponsored by Amtrak, by SJ (Swedish State Railways), and by ABB, and supported by the Federal

This tilt-body electric train draws

Railroad Administration.

{Courtesy of ABB Traction

Inc.)

626

Figure 23.59

MS Fascination contains a power plant composed

of six diesel-electric synchronous 6820 kVA. The 6600 V, 3-phase, 60 Hz, 75% power factor generators are respectively driven by four 12-cylinder and two 8-cylinder, 512 r/min diesel engines. The number of diesel engines in service at a given time depends upon the electrical load. Generators that are brought on

This 70 367-ton Superliner

generators, four of which are rated at 10

line

260 kVA and two

at

are synchronized automatically.

The propulsion system comprises two 14 MW, 1000

V,

3-phase, 14-pole synchronous motors that are directly

r/min to 140 r/min. The motors each have two 3-phase windings which can be operated independently or in tandem. Each winding has a rating of 7 MVA. The rotating field is supplied by the excitation unit which provides the excitation current for the Propulsion Motor. The excitation unit is supplied from the 450 V network through a 450 V/400 V, 400 kVA excitation transformer. A control unit controls both the excitation current of the Propulsion Motor and the output currents of the cycloconverters, i.e., the stator currents of the Propulsion Motor. In addition, the control unit takes care of the speed control of the motor, overload protection of the supply network, and the synchronization to the propellers. The 6.6 kV main bus voltage is stepped down to 1500 V by means of transformers. The secondary sides are connected to the input of four cycloconverters, two of which are used for each propulsion motor. The cycloconverters each employ 36 thyristors. This ultramodern Superliner has a length of 260.6 m, a beam of 31 .5 m and operates at a service speed of 19.5 knots. The passenger capacity is 2040, complemented by a captain and crew of 920. (Courtesy of Carnival Cruise Lines)

coupled

to

two propeller shafts driven

at

speeds ranging from 50

627

Figure 23.60 an important step toward the creation of new jobs. The inroad of computers industry has now expanded into the electric power sector as well. Thus, electric power technology is rapidly changing to embrace these new devices and concepts. Top left: This modular educational console employs 200 machines (dc, synchronous, induction) for direct handson training of electronic drives. Physical connections between machines and electronic converters are made, using

The

training of technical personnel

and electronics

into

every sector of

is

commerce and

W

Measurements are taken using standard instruments. The student observes the physical reality of sudden changes in speed, and so forth. Top right: In a more advanced program, data acquisition is used to display waveshapes, voltages, currents, active and reactive power, as well as harmonics. A computer makes the necessary calculations in real time, thus permitting torque-speed characteristics and other drive features to be observed. Printouts of observations save valuable time for students and instructors alike. By becoming accessible and visible, harmonics lose their mystery and become a source of interest and even fascination. Lower right: Simulation is becoming a popular way of performing experiments without using any hardware at all. The special program pictured here permits simulated modules to be taken from "inventory," pulled into the "console," connected with flexible "wires," and mechanically coupled to "loads." The "modules" are exact replicas of those shown in the picture at left. The interesting feature of this simulation program is that the static and dynamic properties of the individual machines (and loads) are stored in the computer. As a result, the steady-state and transient state behavior of a drive can be observed as the real machines and converters were present. protected leads.

torque, inertia, overloads,

if

(Courtesy of Lab-Volt Ltd.)

628


23-2

23.31 Conclusion

Why

are

two converters needed

phase winding This chapter has covered several types of ac drives.

The converters making up the control part of the drive can be classed into two main groups: linecommutated and self-commutated. Line-commu-

23-3

We

because the current

is

Line-commutated con-

verters are used in large ac

motor drives.

to the fore

availability of

high-power switches such as ability of

In addition, the ability of

23-4

in real time.

PWM switching convert-

perform as well as dc drives do. importance

b.

23-5

factor.

This

is

verter in the

damental

made

power

while

filtering

out the

filters.

The

fed back

permitting real

power

to

PWM

dc side and vice versa has opened

where power has

line.

What

is

is

is

driv-

fed from a

the approximate maxi-

speed that can be attained with

this

What

is

the basic difference

between a

inverter?

23-7

mo-

A line-commutated

inverter can be used to

why. 23-8

switching

In

comparing the physical arrangement of

the bridge rectifiers in one phase of the

power while

cycloconverter of Fig. 23.15,

new frontiers

there any

arrangement of Fig. 23.33? 23-9

A

large squirrel-cage induction

motor has

to run at a very low, steady speed. If elec-

and rotating machines,

is required, what type of would be most appropriate?

tronic control

prominent power apparatus.

control

Intermediate level


23-10 Practical level

Name

is

difference with the bridge rectifier

to be controlled. Thus,

switching converter deserves a special

in the hierarchy of

1

6-pole induction motor

line-commutated and a self-commutated

flow freely from the ac side

place, along with transformers

23-

speed [r/min]

full- load rated

not a 3-phase induction motor. Explain

for all induction

converters to generate or absorb reactive

PWM

in Fig. 23.19.

the line at high-power factor and low-

Indeed, the inherent property of

the

23-6

high-

tor electronic drives.

in all areas

has the torque-

drive a 3-phase synchronous motor but

harmonic content, augurs well

to the

It

power can be drawn from, or

fact that

to,

60 Hz.

arrangement?

PWM mode and phase-shifting the fun-

voltage

8 has

full-load rated torque [N-m]

A 3-phase,

mum

to dc

frequency components with relatively inexpensive

The The

60 Hz

utility

possible by operating the con-

V, 3-phase,

en by a cycloconverter that

power and vice versa while operating sinusoidaUy at unity power feeders, can convert ac

1

Calculate

that

is

switching converters, connected to electric

in Fig. 23.

speed characteristic shown

a.

Another factor of great

flux in the air gap. Calculate

The blower motor shown 240

GTOs

complex waveshapes of any frequency and phase has opened the way to induction that

60 Hz.

4 poles and a full-load rating of 2 hp,

ers to generate

motor drives

same

ing the

high-speed computers

and microprocessors to process signals

V,

the voltage and frequency to be applied to

mainly because of two factors: the

and IGBTs, and the

460

to run at a no-load

the stator.

Drives using self-commutated converters have

come

rated at

is

want the motor

speed of about 225 r/min while maintain-

extinguished naturally

as the line voltages change.

for each

23.15?

3-phase, 16-pole squirrel-cage

induction motor

tated converters are particularly suited for thyristors

A standard

in Fig.

629

The induction motor

in Fig.

23.15 has six

poles and runs at a no-load speed of three types of drives used to

squirrel-cage induction motors.

power

160 r/min. The effective voltage across the windings

is

42

V.


630

23- 6

Calculate

1

a.

The frequency generated by

b.

The smallest possible

effective line voltage

of the 60 Hz source [V] 1

When the

23- 7

induction motor in Problem

23- 0 operates

at full-load

unchanged)

power factor

1

its

(with the frequency is

The blower motor inal rating

460

80 percent.

The time during which converter

1

acts as a

The time during which

it

acts as an inverter

23-18

The motor and blower shown in Fig. 23. have the properties shown in Fig. 23. 19. The synchronous speed is 1200 r/min, the rated torque

240

8

is

N

When When

c.

Is

In


the

The

b.

The

23- 19

m, and the rated voltage

torque, speed, and horsepower is

240 120

A 30

208

hp,

We

V.

diodes

V

V

240

a.

is

Calculate

wish

line voltage

of

to limit the locked-rotor

maximum

value of 40

N*m

so

A

connected to the three slip-rings. rheostat

is

connected

The approximate mechanical power devel-

b.

The The

The approximate dc output voltage The approximate resistance of the rheostat

In

Problem 23- 9 a dc chopper

and

23-20

1

power absorbed by converter

cir-

under locked-rotor conditions

d.

c.

a.

oped by the motor [hp]

1

3500 r/min,

The synchronous speed of the motor The power that is dissipated in the rotor cuit

reactive

at a

Calculate

b.

angle of converter

stator volt-

and

V,

82 percent.

firing

V

across the dc output of the rectifier.

60 Hz. The motor has an efficiency of

23- 5

V, 3-phase,

A single manual

line voltage is

voltage

reduced to 230

3-phase bridge rectifier composed of six

horsepower when

age and current are respectively 250

60 A. The 3-phase

is it

is

as to ensure a small starting current.

V

torque, speed, and is

when

Referring to Fig. 23.24a, the dc link volt-

Why

at rated

duces an open-circuit rotor

the voltage

23-14

losses

60 Hz, wound-rotor induction motor pro-

V.

the voltage

c.

R

the rotor hotter in (a) or (b)?

torque to a a.

motor runs

the stator voltage

Calculate

1

2

I

speed of 810 r/min.

250

23 - 3

respective torque-speed charac-

age required so that the blower runs

1

is

nom-

of the motor and blower are given

b.

a.

[ms|

1ms] 1

8 has a

(with the motor coupled to the blower)

rectifier

23- 2

1

23.19. Calculate the rotor

in Fig.

b.

in Fig. 23.

of 1/4 hp, 1620 r/min, 3-phase,

The

V.

teristics

Calculate a.

remains

varies linearly

it

Can you explain why?

2. 1

1

23.26

with the frequency generated by converter

verler [Hz]

23-

E2

23.33 the value of

In Fig.

fixed,..but in Fig.

the cyclocon-

its

power-handling capacity 1

is

con-

nected between the dc output of the recti-

1

easier to achieve regenerative

fier

and a 0.2 fl

resistor. If the

chopper

braking with a current-fed drive than with

operates

a voltage-fed drive?

calculate the duration of the on-time

Referring to Fig. 23.32, calculate the

under locked-rotor conditions.

fol-

at

a fixed frequency of 500 Hz,

7a

lowing: a. b. c.

d.

The The The The

average current

peak current

in

in

each diode

each diode

Advanced 23-21

a.

level

The squirrel-cage induction motor shown

peak inverse voltage across each diode

Fig. 23. 15 has a

frequency of rotor current / R

460

V

nominal rating of 50 hp,

per phase, 60 Hz,

1

100 r/min. The

in


3-phase line voltage

want the motor

208

is

run

to

at

V,

60 Hz.

we

If

200 r/min, while developing

The power factor of the 60 Hz The effective value of / s

g.

a speed of about

h.

full-load

i.

torque, calculate the approximate voltage

and frequency

be applied to the stator

to

23-25

windings. b.

current

If

has an effective value of 60 A,

line

Show

the flow of active and reactive power

in the

converters.

The

solid-state starter illustrated in Fig.

23.23b /a

631

is

used with a 5 hp, 460 V,

3-

phase 1760 r/min motor that drives a

calculate the approximate value of the peak

current carried by each thyristor.

torque of 2 pu.The kick-start voltage

23-22

The self-commutated

inverter in Fig.

23.24a furnishes a motor current having

What

is

A standard 200 tor

50 hp,

Calculate the value of the kick-start torque

b.

Knowing

shown

in Fig.

that the full-load current

tangential force exerted

applied to the stator so that the motor de-

per pole.

its

Assume

rated torque at

that the flux in

on the rotor

bar currents were

If the rotor

b.

400 r/min. the machine

the rotor bar voltages,

on the rotor

The peak current is assumed 240 A and the length of the ro-

bars, per pole? to


remain

tor bars

A 50 hp dc

motor

is

000 and 30 000

23-27

at

10 cm.

r/min.

Owing

to

at these

speeds,

com-

it is

in Fig.

23.37 has 8 poles

connected to a 60

Hz

rotor has a diameter of 140 tor

de-

it is

The motor shown and

mutation problems associated with a standard commutator

is

required to drive a

centrifuge at a speed ranging between 8

phase with

in

what would be the

net tangential force exerted

1

bars,

is

constant.

23-24

6.2 A,

Referring to Fig. 23.37 calculate the net

a.

Calculate the voltage and frequency to be

velops

is

power

dissipated in the starter.

23-26

23.24a.

torque.

initial

calculate the approximate thermal

60 Hz squirrel-cage induction modriven by a current-fed self-commu-

tated inverter

is

is set

0.4 pu.

V,

is

voltage

at

and the

750 r/min, 3-phase,

1

initial

a.

the

value of the dc link current?

23-23

pu and the

set at 0.8

an effective value of 26 A.

belt

conveyor. The motor has a starting

a.

cided to use a commutatorless dc motor b.

frequency

is

source.

The

mm and the ro-

40 Hz. Calculate

The speed of the rotor r/min] The torque developed by the rotor [N-m] |

driven by two converters with a dc link

A 2-pole

(Fig. 23.8).

motor

selected

is

23-28

having a nominal rating of 50 hp, 30 000 r/min,

460

V,

tor leading.

60 A, 90 percent power

When

the

motor delivers

2

is

1

55°. If the available

60 Hz

line volt-

gates of converter 2

e. f.

The fundamental

c.

d.

23-29

The is

E2

its

in

order that the motor will

N-m.

rated torque of 10.1

full-load efficiency of the

97.3% when

it

1

4

M

in Fig.

23.59

operates at unity power

factor.

Taking field,

into

account the losses of the dc

which amount

to

nominal stator current

1

ripple frequency in

needed

propulsion motor described

a.

The dc link voltage The delay angle of converter The dc link current if the motor draws an input power of 41 .5 kW The fundamental ripple frequency in £,

is

develop

its

age is 575 V, calculate the following: a. The triggering frequency applied to the b.

that

fac-

rated output, the delay angle for converter

In the equivalent circuit of Fig. 23.44a,

calculate the line-to-line stator voltage

b. Calculate the

84 kW, calculate the at full-load.

nominal stator current per

winding. c.

Calculate the peak current carried by the thyristors.


632

23-30

a.

The frequency range of the cycloconverters The rated active and reactive power supplied by each of the 10

a.

b.

MVA diesel-

260

c.

electric generators c.

The 15% power tric

most

factor that the diesel-elec-

90%

is

23-33

much lower

The magnitude and phase of £ 34 The amplitude modulation ratio

A single-phase, in Fig.

60 Hz converter such

as

23.56, operates at a syn-

chronous carrier frequency of 300 Hz. The

Can you give a plausible low power factor rating?

680

V

and the effective

dc link voltage

is

value of the 60

Hz fundamental

voltage

The time required to cover a distance of 500 miles when this Superliner runs at its rated

£34

service speed.

and the on/off time when the ac voltage

In Fig.

23.56

420/137°;

/

it is

=

E l2 = = 0.2 Q.

given that

330/142°; x

a.

b.

23-34 to,

or received by, the converter

V. Calculate the duty cycle


phase when the motor runs

of 12

92

V. Calculate the following:

a.

The value of the power

terminals 3 and 4 can be adjusted by con-

b.

The

c.

The magnetizing component of

car-

60 Hz power

line.

d.

e.

and the

is

E l2 = 400/162°. The magnitude EM are adjusted so that the

and phase of

converter supplies 161

kW to the dc

while ensuring that the current

The torque-producing component of

the sta-

The

total

torque

volt-

age on the secondary side of the trans-

former

the stator

tor current

The reactance x has a £2,

resistance

stator current

current

rier frequency of fc is 800 Hz and the fundamental frequency is equal to that of the

60 Hz impedance of 0.8

cir-

at

600 r/min while operating at a frequency Hz and a line-to-line voltage of

V

The magnitude

The

is

The equivalent circuit of a 2-pole, 5 hp, 480 V, 3-phase, 60 Hz motor is shown in

and phase angle of the ac voltage between tinually varying the duty cycle.

D

+500 V; -30 V.

cuit per

Fig. 23.56, the 4-quadrant single-phase

converter produces a dc voltage of 805 6.

430

Fig. 23.42a.

Referring to the electric vehicle drive of

between terminals 5 and

is

momentarily:

Determine the following: a. The magnitude and phase of E 34 b. The active and reactive power delivered

23-32

in

The magnitude of d The magnitude and phase of /

shown

lagging power factor rating of

.

alternators.

reason for this d.

d.

generators can supply

than the

1

Neglecting the losses

phase with

the converter, calculate the following:

b.

23-3

E ]2

In reference to Fig. 23.59, calculate the

following:

/ is in

link,

23-35

Referring to Fig. 23.44a, but assuming a greater load, calculate the line current,

magnetizing current, and torque tor turns at

if

the

mo-

90 r/min instead of 270 r/min.

Part Four Electric Utility

Power Systems

Chapter 24 Generation of Electrical Energy

shows how the system demand (power)

24.0 Introduction

ing a typical day in the

Now

that

we

we

mapower devices,

are familiar with the principal

chines, transformers, and other are in a position to see

large electrical system.

how

the winter.

varies dur-

a typical day in

pattern of the daily

demand

is re-

markably similar for the two seasons. During the

they are used in a

Such a system comprises

The

summer and

winter the peak

demand of

1

5

GW = (

15

000

M W)

all

the apparatus used in the generation, transmission,

and distribution of electric energy, starting from the generating station and ending up in the most remote

summer home

in the country.

This chapter and the

next three chapters are, therefore, devoted to the

following major topics: •

the generation of electrical energy

•

the transmission of electrical energy

•

the distribution of electrical energy

•

the cost of electricity

24.1

Demand

of

3

an

electrical

°0

system The

total

utility

power drawn by

4

12

8 *-

the 'customers of a large

system fluctuates between wide

16

20

24 h

time

Figure 24.1

limits, de-

Demand curve of a larger system during a summer day and a winter day.

pending on the seasons and time of day. Fig. 24. 635

ELECTRIC UTILITY POWER SYSTEMS

636

is

higher

than

the

summer peak of

Nevertheless, both peaks occur about

because increased domestic activity

1

GW.

10

These power blocks give

7:00 (5 p.m.)

at this

generating stations:

time coa.

incides with industrial and

commercial centers

Base-power stations

still

operating

minimum demand

throughout the year never

falls

the base load of the system.

nual

summer

during the

examining the curve, we note

In

peak load is

1

5

below 6

We also

GW. The

for only 0.

1

at

tions are particularly well adapted to furnish

base demand. b.

relatively quickly to

GW). demand

GW.

This

Intermediate-power stations ally

changes

that

in

can respond

demand, usu-

by adding or removing one or more gener-

ating units:

Hydropower

adapted for

this purpose.

stations are well

is

see that the anc.

Peak-generating stations

that deliver

base load has to be fed

percent of the time.

we have

(6

that the

100 percent of the time, but the peak load

extremes,

power

times: Nuclear stations and coal-fired sta-

at full capacity.

The load curve of Fig. 24.2 shows the seasonal variations for the same system. Note that the peak demand during the winter (15 GW) is more than twice the

that deliver full

that all

are

rise to three types of

for brief intervals during the day:

may occur

must be put

Between these two

Such

power stations

into service very quickly.

Consequently, they are equipped with prime

intermediate loads that have to

movers such as

diesel engines, gas turbines,

be fed for less than 100 percent of the time. If

we

plot the duration of each

nual base,

we

demand on an

compressed-air motors, or pumped-storage

bines that can be started up in a few minutes. In

obtain the load duration curve of Fig. this regard,

demand of 9 GW lasts 70 percent of the time, while a demand of 12 GW lasts for only 15 percent of the time. The 24.3. For example, the curve

graph

is

tur-

an-

shows

it is

worth mentioning that thermal

that a

generating stations using gas or coal take from

4

to 8 hours to start up, while nuclear stations

may

divided into base, intermediate, and peak-

take several days. Obviously, such gener-

ating stations cannot be used to supply short-

The peak-load portion usually indemands that last for less than 5 percent of

load sections.

cludes

the time.

term peak power.

1

On

this basis the

system has to deliver

GW of base power, another 6 GW of intermediate power, and 3 GW of peak power. 6

Returning to Fig. 24.3,

it

so happens that the ar-

eas of the dotted and cross-hatched parts are proportional to the relative

amount of energy (kW

GW 15

y M day-H r*-1

-1

day -winter

Figure 24.2

Demand curve

of

a large electric

utility

system during one year.

-

-spring-

day-H

h)

GENERATION OF ELECTRICAL ENERGY

Figure 24.3 Load duration curve

of

a large electric

utility

system.

associated with the base, intermediate, and peak loads. Thus, the

637

base-power stations supply 58 per-

ergy. Fig. 24.4 also

shows some of the obstacles from following the

that prevent transmission lines

Due

cent of the total annual energy requirements, while

shortest route.

the peak-load stations contribute only 1.3 percent.

cal

The peak-load

zigzag path between the generating station and the

stations are in service for an average

hour per day. Consequently, peak power

is

very expensive because the stations that produce

it

of only

are idle

1

most of the time.

and

to these obstacles, both physi-

legal, transmission lines often

follow a

ultimate user.

24.3 Types of generating stations There are three main types of generating

24.2 Location of the generating station

1

In planning an electric utility system, the physical

.

Thermal generating

stations:

stations

2.

Hydropower generating

3.

Nuclear generating stations

stations

location of the generating station, transmission lines,

and substations must be carefully planned

arrive at an acceptable,

economic

sometimes locate a generating

solution.

to

We can

station next to the

Thermal generating electrical

energy

in the

stations

produce most of the

United States. Nevertheless,

primary source of energy (such as a coal mine) and

important hydropower stations and nuclear generat-

use transmission lines to carry the electrical en-

ing stations produce about 20 percent of the total re-

ergy to where

it

is

needed.

practical or economical,

When

we have

primary energy (coal, gas,

oil ^

this is neither

by ship,

train, or

pipeline to the generating station (pig. 24.4).

generating station may, therefore, be near

away from,

quirements.

Some

to transport the

to,

Although we can harness the wind,

The

or far

the ultimate user of the electrical en-

of the largest hydropower stations are lo-

cated in Quebec and British Columbia, in Canada.

lar

tides,

and so-

energy, these energy sources represent a tiny part

of the

total

energy

we

need.


638

refinery

Figure 24.4 Extracting, hauling,

and transforming the primary sources of energy is done in different ways. The dotted transmisG with the consumers must go around various obstacles. G T thermal

sion lines connecting the generating stations station;

GH

:

hydro station;

GN

:

:

nuclear station.

power balance between generator and load

24.4 Controlling the

The

electrical

energy consumed by the thousands of

customers must immediately be supplied by the ac generators stored.

because electrical energy cannot be

How

do we maintain

this

almost instanta-

neous balance between customer requirements and generated power? To answer the question,

let

us

bine and the increased

R

(Fig. 24.5). i

Water behind the dam

is

immediately trans-

On

the other hand, the electric

power P L drawn

from the generator depends exclusively on the

When

load.

power Pr supplied to the rois tor equal to the electrical power P consumed by the load, the generator is in dynamic equilibrium and its speed remains constant. The electrical system is said to be stable. the mechanical

x

However, we have

consider a single hydropower station supplying a regional load

power

mitted to the generator.

mand

just seen that the

fluctuates continually, so

P

PT

{

is

P

system de-

sometimes

flows through the turbine, causing the turbine and

greater and sometimes less than

generator to rotate.

than P x the generating unit (turbine and generator) the tur-

bine depends exclusively on the opening of the

wicket gates that control the water flow. The greater

more water

If

is x

greater

,

The mechanical power PT developed by

the opening, the

.

is

admitted to the

tur-

begins to slow down. Conversely,

PT

,

if

P

{

is

less than

the generating unit speeds up.

The speed

variation of the generator

is,

therefore,

an excellent indicator of the state of equilibrium be-


639

burner heat

dam

wicket gates

)

steam valve

steam valve

hydraulic

turbine

q

synchronous

^

A

0


generator

tween

transmission

line

line

to three

if

rises they

it

ous

we

consumers

independent regions.

P L and PT and, hence, of the stability of the sys-

tem, if the speed

falls the

wicket gates must open, and

must close so as

to maintain a continu-

PT and P L Although could adjust the gates manually by observing the state

of equilibrium between

.

speed, an automatic speed regulator

Speed

They can

small as 0.02 percent. Thus, ator increases

from

governor begins nism. will

If

is

always used.

regulators, or governors, are extremely

sensitive devices.

1

detect speed changes as

the speed of a gener-

if

800 r/min

to act

to

1

The governors of thermal and nuclear stations opsame way, except that they regulate the steam valves, allowing more or less steam to flow through the turbines (Fig. 24.5). The resulting

erate the

change

in

change

in the rate

it

back

when

sponding change

quency bility

is

The same

in the

reduce combustion

boiler pressure will quickly

exceed the safety

limits.

24.5 Advantage of interconnected

systems

corrective ac-

Consider the three generating stations of

Fig. 24.5,

system freguency. The

each can operate

of a system. The system is

to

a

case of

change produces a corre-

is

suddenly removed.

fre-

therefore an excellent indicator of the sta-

the frequ e n cy

we have

in the

connected to their respective regional loads R h R 2 and R 3 Because the three systems are not connected,

the load

Clearly, any speed

be accompanied by

800.36 r/min, the

on the wicket gate mecha-

to rated speed.

to

of combustion. Thus,

as soon as the valves are closed off, otherwise the

drop momentarily, but the governor will quickly

bring

steam flow has

a coal-burning boiler,

the load should suddenly increase, the speed

tion takes place

generator

transmission

consumers

Figure 24.5 Power supplied

synchronous

eon s tan t.

is

stable so long as

,

.

at its

own

frequency, and a distur-

bance on one does not affect the others. However, is it

it

preferable to interconnect the systems because (1)

improves the overall

stability, (2)

it

provides better


640

Figure 24.6 Three networks connected by four

continuity of service, and (3)

it

is

tie-lines.

more economical.

for annual inspection

shows four interconnecting transmission

Fig. 24.6

lines, tying

together both the generating stations and

High-speed

the regions being serviced.

and

remaining

circuit break-

lines

is

stations.

Energy flowing over the

the station that supplies

power

charges.

electric

case of a fault and to reroute the flow of

power.*

We now

discuss the advantages of

Stability.

lines are

Systems

greater reserve

have

that are interconnected

power than a system working

alone. In effect, a large system

is

better able to

withstand a large disturbance and, consequently, it

is

inherently

more

stable.

For example,

if

the

load suddenly increases in region R,, energy im-

mediately flows from stations

over the interconnecting load

is,

G2

tie-lines.

and

G3

and

The heavy

therefore, shared by all three stations in-

stead of being carried by one alone. 2.

Continuity of Service. should break down, or

If if

it

less

it,

any wheeling

utility

when

transmission

its

used to deliver power to a third party.

Economy. When

several regions are intercon-

nected, the load can be shared

among

the various

generating stations so that the overall operating cost

is

ating

minimized. For example, instead of oper-

all

three stations at reduced capacity dur-

ing the night

down one

when demand

is

station completely

carry the load. In this

way we

low,

and

we can

shut

the others

let

greatly reduce the

operating cost of one station while improving the efficiency of the other stations, because they

a generating station has to be shut

3.

tie-

A wheeling charge is the amount paid to

another electric

such a network. 1.

it

automatically metered and credited to

ers d, to d, () are installed to automatically interrupt in

customers

repair, the

serves can temporarily be supplied by the two

down

now

run closer to their rated capacity.

Electric utility

companies

are, therefore,

inter-

ested in grouping their resources by a grid of inter-

connecting transmission The IEEE Std 100-1992

dition that causes a device, a fail to

perform

circuit, a

in

lines.

A central

dispatching

states that a fault is a physical con-

component, or an element

to

a required manner, for example, a short-

broken wire, or an intermittent connection.

office (control center) distributes the load

among the

various companies and generating stations so as to

minimize the costs

(Fig. 24.7).

Due

to the

complex-


641

Figure 24.7 Technicians patching

in

rooms

the control

office,

of

two generating stations communicate with each

of some systems, control decisions are invariably made with the aid of a computer. The dispatching ofity

has to predict daily and seasonal load

fice also

other regions.

generating units so as to maintain good stability of

immense and complicated network. New England Power Exchange

For example, the

(NEPEX)

coordinates the resources of

utility

companies

Island,

Maine, and

vises

New

1

3 electrical

Connecticut,

serving

New

Hampshire.

power flow between

the state of

this

Rhode

also super-

It

huge network and

York and Canada.

necessarily operate at the still

be allocated

same frequency,

among

the load

the individual generat-

ing units, according to a specific program. Thus,

a generating unit has to deliver more power,

ernor setting is

is

changed

cal output

decrease

from

this unit

in the total

its

if

gov-

more power The increased electri-

slightly so that

delivered to the generator.

The sudden

loss of an important load

constitutes a major contingency. If a

to

big load

is

suddenly

the turbines begin

lost, all

speed up and the frequency increases everywhere

on the system.

On the other hand,

if

a generator

dis-

is

connected, the speed of the remaining generators decreases because they suddenly have to carry the entire load.

The frequency

the rate of 5

Hz

no time must be

starts to

decrease

—sometimes

at

per second. Under these conditions, lost and, if conventional

methods

are

unable to bring the frequency back to normal, one or

Although such interconnected systems must can

dis-

or a permanent short-circuit on a transmission line

changes and to direct the start-up and shut-down of

the

a central

other, or with

while supervising the operation of their respective generating units.

produces a corresponding

power supplied by

all

the other

generating units of the interconnected system.

more loads must be dropped. Such load shedding done by frequency-sensitive relays circuit breakers as the

on a 60

Hz system

percent of the

frequency

the relays

that

falls.

may

be

is

open selected For example, set to

shed 15

system load when the frequency

reaches 59.3 Hz, another 15 percent

58.9 Hz, and a final 30 percent

when

58 Hz. Load shedding must be done

when

it

reaches

the frequency in less than

is

one

second to save the loads judged to be of prime importance.

As

far as the disconnected

customers are con-

cerned, such an outage creates serious problems.

24.6 Conditions during an outage

A

major disturbance on a system

r

(.cal

must be taken

to prevent

it

led contin-

from spreading

start to

moves through

a paper

cool down, paper tears as

gency) creates a state of emergency and immediate steps

Elevators stop between floors, arc furnaces

to

mill,

traffic

Clearly,

it is

lights in

rupted service.

it

stop functioning, and so forth.

everyone's interest to provide uninter-


642

Experience over many years has shown

most system short-circuits are very

brief.

that

They may

pends, therefore, on these two factors. The available

hydro power can be calculated by the equation

be caused by lightning, by polluted insulators, by

by overvoltages created when

falling trees, or cuit breakers

open and

close.

"> =

Such disturbances

where

usually produce a short-circuit between two phases

P — q =

or between one phase and ground. Three-phase short-circuits are very rare.

Because

h

line short-circuits are, in general, very

9.8

major outage can usually be prevented by

brief, a

simply opening a short-circuited line and reclosing it

very quickly. Naturally, such fast switching of cir-

cuit breakers

happens

in

done automatically because

is

(24.1)

9.8
cir-

all

it

a matter of a few cycles.

Owing

power [kW]

available water

water rate of flow [m

=

coefficient to take care of units

to friction losses in the

power output of

1

water conduits,

itself,

somewhat

less effi-

is

.

ciency of large hydraulic turbines

is

between 90 and

94 percent. The generator efficiency

The frequency of

a system fluctuates as the load

governors always bring

varies, but the turbine

back

to

60 Hz. Owing

1

When

to these fluctuations, the

the accumulated loss or gain

80 cycles, the error

is

is

corrected by making

about all

the

generators turn either faster or slower for a brief period.

The frequency correction

is

ranging from 97 to

affected ac-

Example 24-1

A large hydropower station has a head of 324 m and 3 an average flow of 1370 m /s. The reservoir of water

b.

5

1

84 000 cycles

in a

24-hour period. Electric clocks

connected to the network indicate the correct time to within 3 seconds,

second hand

is

composed of a

km 2

se-

.

The available hydraulic power The number of days this power could be sustained if the level of the impounded water were allowed to drop by m (assume no precipita1

because the position of the

directly related to the

is

of lakes covering an area of 6400

Calculate

60 Hz network generates exactly

a

behind the dams and dikes

ries

a.

way

even higher,

size of the generator.

cording to instructions from the dispatching center. In this

is

99 percent, depending on the

it

system gains or loses a few cycles throughout the day.

the mechani-

However, the

the turbine

than that calculated by Eq. 24.

24.7 Frequency and electric clocks

/s]

head of water [m]

turbine casing, and the turbine cal

3

tion or evaporation

number of

in

by surrounding

and neglect water brought

rivers

and streams)

elapsed cycles. Solution

HYDROPOWER GENERATING

a.

The

available

STATIONS Hydropower generating stations convert the energy of moving water into electrical energy by means of a hydraulic turbine coupled to a synchro-

power

The power that can be extracted from a waterfall depends upon its height and rate of flow. The size and physical location of a hydropower station de-

9.8

is

qh

X 324 350 000 kW = 4350 X

1370

MW

A drop of m in the water level corresponds to 6 3 6400 X 10 m of water. Because the flow is 1

1

24.8 Available hydro

9.8

=

4 b.

nous generator.

P =

hydropower

370

m

3

/s,

the time for

through the turbines t

= 6400 X = 4.67 X

=

all this

water to flow

is

1298 h

6

10 /1370

10

6 s

= 54

days

GENERATION OE ELECTRICAL ENERG Y

As

a matter of interest, a flow of

10 times the

New

1370 nrVs

amount of water used by

York and

its

is

about

the city of

suburbs,

24.9 Types of hydropower stations Hydropower

stations are divided into three groups

depending on the head of water: 1.

High-head development

2.

Medium-head development

3.

Low-head development High-head developments have heads

in

excess of

300 m, and high-speed Pelton turbines are used. Such generating stations are found in the Alps and

The amount of impounded water is usually small. Medium-head developments have heads between 30 m and 300 m, and medium-speed Francis turbines are used. The generating station is fed by a huge reservoir of water retained by dikes and a dam. The dam is usually built across a river bed in a relatively mountainous region. A great deal of water is impounded behind the dam (Fig. 24.8). Low-head developments have heads under 30 m, and low-speed Kaplan or Francis turbines are used. These generating stations often extract the energy from flowing rivers. The turbines are designed to handle large volumes of water at low pressure. No other mountainous regions.

reservoir

is

provided (Fig. 24.9).

Figure 24.8 Grand Coulee Dam on the Columbia River in the state of Washington is 108 m high and 1270 m wide. It is the largest hydropower plant in the world, having 18 generating units of 125 MW each and 12 generating units of 600 MW each, for a total of 9450 MW of installed capacity. The spillway can be seen in the middle of the dam. (Courtesy of General

Electric)

643

644


the basic features

and components of a hydropower

plant (see Fig, 24.10).

Dams. Dams

7.

rrtade

of earth or concrete are built

across river beds to create storage reservoirs.

Reservoirs can compensate for the reduced precipitation during dry seasons

and for the abnor-

accompany heavy

mal flows

that

ing snow.

Dams

rains

and melt-

permit us to regulate the water

flow throughout the year, so that the powerhouse

may

run

at close to full capacity.

Spillways adjacent to the

dam

are provided to dis-

charge water whenever the reservoir level

Figure 24.9 The Beauharnois generating station on the St. Lawrence River contains 26 3-phase alternators rated 50 MVA, 13.2 kV, 75 r/min, 60 Hz at a power factor of 0.8 lagging. An additional 10 units rated 65 MVA, 95,7 r/min make up the complete installation. The output ranges between 1000 and 1575 depending upon the seasonal water flow. (Courtesy of Hydro-Quebec)

MW

24.10

Makeup

A hydropower

MW

We

have seen

that the

demand

is

too high.

for electricity varies

considerably throughout the day, and from season to season. Consequently, the available water cannot

ways be used

to

water reservoir

supply energy to the system.

is

small or almost nonexistent (such

as in run-of-river stations), let

al-

If the

we

unfortunately have to

the water through the spillway without using

it.

Dams

irri-

often serve a dual purpose, providing

gation and navigation facilities, in addition to their

of a

hydropower plant

installation consists of

dams, water-

ways, and conduits that form a reservoir and channel the water toward the turbines. These, and other

items described, enable us to understand

some of

power-generating

2.

Figure 24.10

medium-head hydropower

plant.

The

Conduits, Penstocks,

integrated system of the is

a good example.

and Scroll-Case.

installations, conduits lead the

dam

site to the

draft

Cross-section view of a

role.

Tennessee Valley Authority

In large

water from the

generating plant. They

may

be


open canals or tunnels carved through rock. The

3.

which bring the water

pipes),

meters

diameter, enable the water supply to be

in

valves,

sometimes several

The penstocks channel

the water into a scroll-case

evenly distributed around

its

is

circumference. Guide

vanes and wicket gates control the water so that

it

flows smoothly into the runner blades (see Figs. 24.

1

1 ,

24. 2, 1

and 24.

1

3).

The wicket

gates open and

close in response to a powerful hydraulic that

is

mechanism

controlled by the respective turbine governors.

Figure 24.11 case feeds water around the circumference turbine. 483 (Courtesy of Marine Industrie) Spiral

draft tube

improves the hydraulic efficiency of the

turbine.

leads out to the tailrace, which channels

It

the water into the

shut off in the conduits.

that surrounds the runner (turbine) so that water

of

has passed

designed vertical channel, called draft tube. The

to the individual

Enormous

that

through the runner moves next through a carefully

conduits feed one or more penstocks (huge steel

turbines.

Draft Tube and Tailrace. Water

645

4.

downstream

river bed.

Powerhouse. The powerhouse contains the synchronous

transformers,

generators,

circuit

breakers, etc., and associated control apparatus.

Instruments, relays, and meters are contained in a central

room where

the entire station can be

monitored and controlled. Finally, many other devices (too numerous to mention here) the complete

hydropower

make up

station.

a

MW

Figure 24.13 Runner of a Francis-type turbine being lowered into position at the Grand Coulee Dam. The turbine is rated at 620 MW, 72 r/min and operates on a nominal head of 87 m. Other details: runner diameter: 10 m; runner mass: 500 t; maximum head: 108 m; minimum head: 67 m; turbine efficiency: 93 percent; number of

Figure 24.12 Inside the spiral case, a set of adjustable wicket gates control the

amount

of

water flowing into the turbine.


wicket gates: 32;

mass per wicket gate: mass of shaft: 175

shaft length: 6.7 m;

6.3

t;

turbine

t.

(Courtesy of Les Ateliers d'ingenierie. Dominion)

ELECTRIC UTILITY POWER S YSTEMS

646

Pumped-storage

24.11

installations

(shown by a energy

We

have already seen

needed

to

that

peak-power

stations are

it

plus), the

peaking station returns the

had.previously stored.

This second solution has two advantages:

meet the variable system demand. To un1

.

derstand the different types of peaking systems used,

The base-power station more efficient.

is

larger and, conse-

is

much

quently,

consider a network (electric system)

which the

in

MW and 160 MW,

demand varies between 100 as shown in Fig. 24.14. One obvious variable demand is to install a 100 daily

solution to this

MW base-power

station

and a peak-power unit of 60

termittently

by a gas

During

is

to install a larger

MW and a smaller peaking

MW. The peaking

to both deliver

in-

and absorb 30

must be able

station

MW of electric power.

lightly loaded periods (indicated

by a minus

sign in Fig. 24.15), the peaking station receives and stores energy provided

The peak-power

station

by the base-power generat-

ing plant. Then, during periods of heavy

demand

smaller and,

therefore, less costly.

Large blocks of energy can only be stored mechanically, and that

is

why we

often resort to a hy-

draulic pumped-storage station.

However, another solution 30

driven

turbine.

base-power unit of 130 station of

MW,

2.

Such a peak-power

generating station consists of an upper and a lower reservoir of water connected by a penstock and an

associated generating/pumping unit. During system

peaks the station acts

like

an ordinary hydropower

generating station, delivering electrical energy as

water flows from the upper to the lower reservoir.

However, during reversed.

light load periods the process

The generator then operates

is

as a synchro-

nous motor, driving the turbine as an enormous

pump. Water now flows from

MW 160

/V

reservoir, thereby storing

the next system •

^peak power unit 60 MW

0

100

is

repeated

Peak-power generators have ratand 500 MW. They are re-

MW

versible because the direction of rotation has to be

1

changed when the turbine operates as a pump. Starting such big synchronous motors puts a

time

Figure 24.14 MW base power station and a 60 peak power station can supply the network demand.

MW

A 100

preparation for

The generating/pumping cycle

ings between 50

MW

in

(Fig. 24.16).

once or twice per day, depending on the nature of the system load.

base power

station

peak

the lower to the upper

energy

heavy load on the transmission

line,

and special

methods must be used to bring them up to speed. Pony motors are often used, but static electronic frequency converters are also gaining ground. (A pony motor is a machine that brings a much larger machine up to speed.) Pumped-storage

installations operating in con-

junction with nuclear plants

make

a very attractive

combination because nuclear plants give best ciency

when operating

at

effi-

constant load.

THERMAL GENERATING STATIONS time

The hydraulic resources of most modern countries

Figure 24.15 base power station and a 30 pumped storage unit can also supply the network demand.

A 130

MW

MW

are already fully developed. Consequently, to rely

on thermal and nuclear

growing need for

we have

stations to supply the

electrical energy.


647

Figure 24.16

pumped storage

Tennessee pumps water from Lake Nickajack to the top of Raccoon Mountain, where m head. The four alternator/pump units can each deliver 425 MVA during the system peaks. The units can be changed over from generators to pumps in a few minutes. (Courtesy of Tennessee Valley Authority)

This it

is

stored

in

a2

station

km 2 (^500

Thermal generating

in

acres) reservoir, giving a 316

stations

produce

electricity

from the heat released by the combustion of or natural gas.

oil,

MW and

tween 200

high efficiency and

Such a

Most

1500

have ratings be-

MW so

economy of a

station has to

as to attain the

its

to

it

exhausts from

sult,

is

always low because of the inherent low efficiency

is


=

(1

we

T = x

T2 = In

20°

+

273°

= 293 K

- r2 /r,)100

that steel in It

turns out that the highest feasible temperature 7,

is

about 550°C. As a

result,

T3 = 550° + 273° = 823 K It

(24.2)

is

the corresponding high steam pressures.

follows that the

maximum

possible efficiency

of a turbine driven by steam that enters

293 T|

efficiency of the

a re-

and other metals can safely withstand, bearing

K

—

(

at

823

K and

is

where

=

As

cannot use temperatures above those that

exists at

r\

be lower than the ambient

usually about 20°C.

should be as high as possible. The problem

The maximum efficiency of any ma-

r\

is

be less than

T2 =

chine that converts heat energy into mechanical en-

ergy

T2 cannot

T2 cannot

mind

efficiency of thermal generating stations

of the turbines.

should be as small as possible. However,

This means that to obtain high efficiency, T,

the turbines.

The

{

temperature, which

size.

stations are usually located near a river

condense the steam as

T2 IT

temperature

enor-

or lake because large quantities of cooling water are

needed

tient

large installation.

be seen to appreciate

mous complexity and Thermal

stations

coal,

1

—

293/823)100 = 64.4%

machine [%]

temperature of the gas entering the turbine [K]

Due

to other losses,

some of

the most efficient

steam turbines have efficiencies of 45%. This temperature of the gas leaving the turbine [K]

most thermal generating stations the gas

is

steam. In order to obtain a high efficiency, the quo-

means

that 65%; of the thermal energy

is

lost

during

the thermal-to-mechanical conversion process.

enormous

loss of heat

and

how

to dispose

The of

it


648

represents one of the major aspects of a thermal

pressure turbine

generating station.

superheater *S*>.

HP after having passed

The

superheater,

through

composed of a

series of tubes surrounding the flames, raises the

Makeup

24.12

steam temperature by about 200°C. This increase

of a thermal

in

generating station

temperature ensures that the steam

absolutely

is

dry and raises the overall efficiency of the station.

The

basic structure and principal

components of a

thermal generating station are shown

They •

24.17.

are itemized and described below.

A huge

A high-pressure

row upon row of

(HP) turbine

(3) converts ther-

mal energy into mechanical energy by steam expand as

boiler (1) acts as a furnace, transferring

heat from the burning fuel to

blades.

it

moves through

letting the

the turbine

The temperature and pressure

at the out-

put of the turbine are, therefore, less than at the

water tubes S,, which entirely surround the

input. In order to raise the thermal efficiency

flames. Water

and to prevent premature condensation, the

tubes by a •

in Fig.

A drum

is

pump

kept circulating through the

steam passes through a reheater S 3 composed of

P,.

,

(2) containing water and steam under high

pressure produces the steam required by the turbines.

It

also receives the water delivered

boiler- feed

pump P 3 Steam .

by

races toward the high-

a third set of heated tubes.

The medium-pressure (MP)

turbine (4)

to the high-pressure turbine,

except that

ger so that the steam

may expand

still

in

-cooling water out

Figure 24.17 Principal components

of

a thermal power

plant.

similar

it is

more.

cooling

—water

is

big-


•

The low-pressure (LP)

turbine (5)

two

and right-hand sections.

identical left-hand

The turbine

remove

sections

649

composed of

is

the remaining avail-

able energy from the steam (Fig. 24.

1

The

8).

steam flowing out of LP expands into an almost perfect •

vacuum

Condenser ting

it

(6)

created by the condenser (6).

causes the steam to condense by

let-

flow over cooling pipes S 4 Cold water from .

an outside source, such as a river or lake, flows

through the pipes, thus carrying away the heat. the

A condensate pump P 2

removes the lukewarm

condensed steam and drives (7)

It is

condensing steam that creates the vacuum.

toward a feedwater

through a reheater

it

pump

(8).

Figure 24.18 Low-pressure section

•

The

reheater (7)

a heat exchanger.

is

It

re-

ceives hot steam, bled off from high-pressure

of

a 375

MW, 3600

r/min steam-

turbine generator set, showing the radial blades.


Electric)

turbine HP, to raise the temperature of the

Thermodynamic

feedwater.

studies

the overall thermal efficiency

when some steam than letting all •

of gas,

bled off this way, rather its

normal course through

supply and control the amount

(9)

or coal injected into the boiler. Coal

oil,

pulverized before

heavy bunker atomized jet

oil is

to

it is

injected. Similarly,

preheated and injected as an

improve surface contact (and

combustion) with the surrounding •

that

three turbines.

The burners is

is

follow

it

show

improved

is

A forced-draft

air.

fan (10) furnishes the

quantities of air needed for

enormous

combustion

(Fig.

24.19). •

An

induced-draft fan (11) carries the gases and

other products of combustion toward cleansing

apparatus and from there to the stack and the outside •

air.

Generator G, directly coupled to

all

three tur-

bines, converts the mechanical energy into electrical

Figure 24.19 This forced-draft fan provides 455

energy.

m 3/s of air at a pres-

sure difference of 5.8 kPa for a thermal power station. In practice, a

steam station has hundreds of other

components and accessories ciency, safety, and

to ensure high effi-

economy. For example, control

is

driven by a 3-phase induction motor rated 12

(8955 kW), 60 Hz, 890 r/min. (Courtesy of Novenco Inc.)

It

000 hp


650

amount of steam flowing to complex water purifiers maintain the

valves regulate the

the

turbines;

re-

quired cleanliness and chemical composition of the feedwater; lubricated.

pumps keep

oil

the bearings properly

However, the basic components we

have just described enable us to understand the operation and

mal

some of

the basic problems of a ther-

station.

24.13 Turbines The low-, medium-, and high-pressure turbines possess a series of blades mounted on the drive shaft (Fig. 24.18). The steam is deflected by the blades, producing a powerful torque. The blades are

made of special

steel to

withstand the high temper-

ature and intense centrifugal forces.

The HP, MP, and LP gether to drive a

some

generator.

generator while the

MP

other one having the

However,

in

HP

turbine drives one

and LP

turbines drive an-

large installations the

same

at

rating.

The condenser

is

known

have seen that about one-half the energy proin the boiler

when

steam

it

has to be removed from the

exhausts

into

ture of the to

1

0°C

as

away

The temperacooling water increases typically by 5°C it flows through the condenser tubes. The

needed

to carry

the heat.

condensed steam (condensate) usually has a tem-

27°C and 33°C and the corresponding absolute pressure is a near-vacuum of about 5 kPa. The cooling water temperature is only a few degrees below the condensate temperature perature between

even

that for every

at

air.

MJ

far

away from

the condenser,

located in a dry region, or

a river or lake,

it is

of heat. Consequently,

Consider

now

a tub containing 100 kg of water

at a certain temperature. If l

kg of water

we

can

to evaporate, the

remaining 99 kg

will inevitably

conclude that whenever ter evaporates, the

l

somehow

cause

temperature of the

drop by 5.8°C.

We

percent of a body of wa-

temperature of the remaining

water drops by 5.8°C. Evaporation

therefore, a

is,

very effective cooling process.

But how can we produce evaporation? Surprisingly, all that

is

needed

to

is

expose a large

to

air.

sur-

The simplest

do this is to break up the water into small and blow air through this artificial rain.

In the case of a thermal station, the is

lake evapo-

droplets,

24.15 Cooling towers thermal station

A

kilogram of water that evapo-

rates, the lake loses 2.4

way

If the

thermal

low temperatures, and

face of water to the surrounding

(see Fig. 24.20).

in

evaporation causes the lake to cool down.

condenser.

the

Consequently, enormous quantities of cooling water are

as important as the boiler

a large surface to the surrounding

24.14 Condenser duced

large pipes

and nuclear power stations. (Courtesy of Foster-Wheeler Energy Corporation)

rates continually,

We

220 MW. Note the

feeding cooling water into and out of the condenser.

turbines are coupled to-

common

Figure 24.20 Condenser rated

we

still

one way or another.

have

We

to cool

often use

water flowing out of the condenser of a cooling tower (Fig. 24.21 into small droplets.

As

)

is

where

the droplets

warm

cooling

piped to the top it

is

fall

broken up

toward the

evaporation to produce the cooling effect. To un-

open reservoir below, evaporation takes place and the

derstand the principle, consider a lake that exposes

droplets are chilled.

The cool water

is

pumped from


the reservoir

where steam.

it

and recirculated through the condenser,

again removes heat from the condensing

The cycle then

repeats.

is

lost

by evaporation. This

up by a stream, or small

except for the small portion consumed by the

lost,

motor and pump.

losses in the

Approximately 2 per-

cent of the cooling water that flows through the con-

denser

loss can

be made

24.17 Energy flow diagram for a steam plant

lake.

Modern thermal generating

boiler- feed

pump

drives the feedwater into the

This means

that materials are strained to the limits

of safety as

far as temperature, pressure,

gether with the

The high back pressure tolarge volume of water flowing

are concerned.

requires a very powerful motor to

available to

pump

it. In modern steam stations power represents about percent of

drive

l

output.

Although

this

the

pumping

the generator

appears to be a significant

we must remember that the energy expended in pump is later recovered when the high-pressure

steam flows through the turbines. Consequently, the energy supplied to the feed

pump motor

is

not really

all,

and centrifugal forces

Because the same materials are the resulting steam plants are nec-

essarily similar. Fig. 24.22

turbine-generator

set,

and

shows a

typical

Fig. 24.23

is

a

540

MW

view of the

control room.

loss,

the

designers

all

strive for high efficiency at lowest cost.

high-pressure drum.

through the

stations are very simi-

throughout the world because

lar

24.16 Boiler-feedpump The

651

Most modern boilers furnish steam at a temper550°C and a pressure of 6.5 MPa. The

ature of

1

overall efficiency (electrical output/thermal input) is

then about 40 percent. The relative amounts of

energy, steam flow, losses, and so forth, do not

change very much, provided the temperature and pressure have the approximate values indicated

above. This enables us to draw a diagram showing the energy flow, steam flow, water flow, and so on, in a

reduced-scale model of a typical thermal gen-

Figure 24.21 Cooling tower installed

in

a nuclear power station

Oregon. The generator output

is

a

1

280

MVA

at

in

a

power factor of 0.88. Tower characteristics: height: 152 m; diameter at the base: 117 m; diameter at the 3 top: 76 m; cooling water: 27 m /s; water loss by evaporation: 0.7

m 3 /s.The temperature

drops from 44.5° to 24° as

it

of the -cooling

water

passes through the tower.

(Courtesy of Portland General Electric Company)

Figure 24.22 This 540

3600

MW steam-turbine generator set

r/min, generating

pressure turbine and alternator are


runs at

a frequency of 60 Hz. The low-

Electric)

in

the background.


652

400 kg/s

Steam output

40 X 10 kg/s 40 X 30 40 X 8 kg/s

Cooling water

40 X 360 kg/s

14

Air intake

MW

Boiler thermaj power

1

200

MW

320 kg/s

400

kg/s

Heat carried away

40

by the cooling water If

a large river or lake

ing tower

is

required,

q

= 2% X

it

is

X

15

MW

MW

not available and a cool-

would have

14

600

to evaporate

400 = 288 kg/s

of cooling water. This loss by evaporation has to be

made up by

Figure 24.23 Control room of the 540 generator {Courtesy of General Electric)

MW

erating station. Fig. 24.24

ducing 12 Using tics

24.18 Thermal stations and the environment

set.

The products of combustion of thermal generating

shows such a model pro-

model,

we can

stations are an increasing subject of concern, their

MW of electrical power.

this

a local source of water.

of any thermal power station. For example, a 480

to

Carbon dioxide (C0 2 ), sulfur dioxide (S0 2 ), and

estimate the characteris-

MW station (40 times more powerful than the model)

due

impact on the environment.

water are the main products of combustion when oil, coal,

or gas are burned.

Carbon dioxide and wa-

produce no immediate environmental

ter

has the following approximate characteristics:

effects,

but sulfur dioxide creates substances that give rise Electric

power output

40 40

Coal consumption

X X

12 1

MW

kg/s

480

MW

40 kg/s

to acid rain. that

may

Dust and

fly

ash are other pollutants

reach the atmosphere. Natural gas pro-

stack ty)

floss 3

A

MW

steam: 550°C, 16.5

MPz

(absolute)

10 kg/s

electric

alternator

power

—fH

boiler

30

MW

27

MW

12

MW

0MW

360

burners

kg/s

cooling water

t

2

in

* 10°C

-t l

8 kg/s

ESvffgy heat coal 1

kg/s

air

10 kg/s

Figure 24.24 Scale model of a typical thermal generating station.

condensate: 30°C;

0.005

MPa

(absolute)

15

MW

lost

GENERATION OE ELECTRICAL ENERGY

duces only water and

C0 2

.

This explains

used (rather than coal or

oil),

pollution must be reduced to a

A good

example of pollution control

Eraring generating station Australia, about 100 is

It

why gas

is

when atmospheric minimum.

located

in

is

the large

Newcastle,

km north of Sydney (Fig. 24.25).

equipped with a special fabric

cleaning system (Fig. 24.26).

huge vacuum cleaners

to

filter flue

The fabric filters

remove

particles

act like

from the

The fabric filter for each boiler composed of 48 000 filter bags, each 5 m long and

16

cm

in

diameter (Fig. 24.27).

When

a boiler oper-

ates at full capacity, they capture dust particles at the rate of

28

rial is later

jects.

kg/s.

A substantial

proportion of this mate-

mixed with concrete

The following technical

for road-building pro-

specifications enable us

to appreciate the size of this station.

Electrical data

number of generators: 4 power per generator: 660 speed: 3000 r/min voltage: 23 000 V

MW

frequency: 50 Hz, phases: 3

gas

boiler-gas flue stream. is

653

Thermal and mechanical data number of steam turbines: 4 number of condensers: 4 number of boilers: 4 steam flow per turbine: 560 kg/s steam temperature: 540°C

steam pressure: 16.55

MPa

cooling water per condenser: 21 000 kg/s coal consumption per boiler: 51.5 kg/s

dust captured by cleaning system: 28 kg/s

Figure 24.25 View of the Eraring Power Station in Newcastle, Australia. The large building on the left is the turbine-generator hall: 27 m wide x 38 m high x 41 8 m long. Te the right can be seen the four structures that house the steam boilers. A portion of the flue gas cleaning system can be seen between the emission stack in the foreground and the boiler structures. (Courtesy of the Electricity Commission of New South Wales)


654

Figure 24.26 General construction of the flue gas cleaning system showing the filter bags that capture the dust, which then

falls into

the hoppers below.

(Courtesy of Electricity Commission of New South Wales)

m

length of one turbine-generator unit: 50

weight of one turbine-generator

number of emission

stacks:

unit:

1342 tons

2

Figure 24.27

m 20 m

height of emission stack: 200

Installation of the fabric filter bags.

outside diameter at bottom:

long

outside diameter at top:

and 16 cm

in

Each bag

is

15

m

diameter.

(Courtesy of Electricity Commission of New South

11.6m

Wales)

Another interesting feature station

mines

is

brought

that are only

the station

and near

its

is

that coal for the

by conveyor

in 1

is

.5

km

and 4.5

ideally located near

belts

km

its

from two

away. Thus,

A nuclear station

source of fuel

source of cooling water, on the shore of

tor.

Lake Macquarie.

The

NUCLEAR GENERATING STATIONS Nuclear stations produce electricity from the heat released by a nuclear reaction.

an atom

splits in

two

sion), a considerable

Note

that a

When the nucleus of

(a process called

amount of energy

atomic is

fis-

released.

chemical reaction, such as the combus-

tion of coal,

produces only a rearrangement of the

atoms, without in any

way

affecting their nuclei.

is

replaced by a nuclear reac-

reactor contains the fissile material that

generates the heat. tains a

identical to a thermal station,

is

except that the boiler

A nuclear station, therefore, con-

synchronous generator, steam turbine, con-

denser, and so on, similar to those found in a con-

ventional thermal station. also similar (between

The

overall efficiency

is

30 and 40 percent), and a

cooling system must be provided

for.

Consequently,

nuclear stations are also located close to rivers and lakes. In dry areas, cooling towers are installed.

Owing

to these similarities,

we

will only

the operating principle of the reactor

examine

itself.


24.19 Composition of an atomic nucleus; isotopes The nucleus of an atom contains two types of partiprotons and neutrons. The proton carries a

—

electron.

The

neutron, as

its

Neutrons

tracted to nor repelled

name

implies, has

no

are, therefore, neither at-

in nature: 235

840 times as much as electrons do. The mass of an atom is concentrated in its nucleus. in the nu-

cleus depends upon the element. Furthermore, be-

electrons

24A

is

is

electrically neutral, the

number of

equal to the

number of

protons. Table

gives the atomic structure of a few important

elements used

in

two types of hydrogen atoms that can be distinguished from each other only by the makeup First, there is

whose nucleus contains Next, there

is

I

ordinary hydrogen (H),

proton and no neutrons.

a rare form, deuterium (D),

clei

contain

ton.

This rare form

When two

tion

called an isotope of hydrogen.

is

atoms of ordinary hydrogen unite

we obtain ordinary water (H 2 0) called light water. On the other hand, if 2 atoms of deuterium unite with atom of oxygen, we 1

a molecule

of heavy water (D 2 0).

oceans contain about

7000 kg of sea

TABLE 24A

we

are about to discuss.

24.20 The source of uranium Where does uranium come from? from the ore found tains the

8

The

kg of heavy water for every

1

tually

composed of

(3

atoms of uranium and

so happens that

It

238

tively precise ratio of

U0 8

1

235

and

less interesting 235

tope

from

U

is

1

398 parts of the

for every 10 parts of the iso-

is

It

very difficult to separate

238

cal properties.

In order to use these substances in nuclear reactors,

they

(U0 2 ). The 235

U0 2

are

processed

natural

in the ratio

into

of 1398:10.

Some nuclear reactors require of the isotope

235

U

uranium dioxide

U0 2 again contains 238 U0 2 and

than natural

U0 2 that has more U0 2 does. This is

produced by an enrichment process whereby the tio

of

235

U0 2 to 238 U0 2

than the natural ratio of lot

of

238

U0 2

is

0:

1

is

1

398. In this enrichment

obtained as a byproduct

ATOMIC STRUCTURE OF SOME ELEMENTS

hydrogen deuterium tritium

helium

Symbol

Protons

H D 3

H

He

Electrons

Neutrons

(neutrons

+

1

1

0

1

1

1

1

2

1

1

2

3

2

2

2

4

carbon

C

6

6

6

12

iron

Fe

26

26

30

56

92

92

143

235

92

92

146

238

uranium 235 uranium 238

235

\J

23K

IJ

"

ra-

raised to 50: 1398 rather

Mass number Element

ac-

U0 8 U0 8 because they possess identical chemi-

U.

235

238

U 3O s

U0 8 in the rela-

398: 1 0.

In other words, the ore contains

process a

water.

U308

compound

obtained

is

It

uranium mines. This ore con-

in

atoms of oxygen).

with one atom of oxygen,

obtain

because both are essential to the operation of

the nuclear reactors

whose nu-

neutron, in addition to the usual pro-

1

235

common, but the isotope U is rare. Uranium 235 and heavy water deserve our atten-

very

nuclear reactors. For example,

there are

of the nucleus.

U).

1

The number of protons and neutrons cause an atom

(

by protons and electrons.

Protons and neutrons have about the same mass,

and both weigh

238

U) and uranium 238 Each contains protons, but U has 92 ( 235 146 neutrons and U has 143. Uranium 238 is

positive charge, equal to the negative charge on an

electric charge.

isotopes of uranium are

uranium 238

found

235 cles

same way, two

In the

655

protons)


656

10 parts

natural

235

U 30 8

U 30 8

10 parts

natural (fuel for

235

U0

50

2

U0 2

enriched

235

parts

U0 2

U0

2

(« 3.6%)

(fuel for light

heavy

water reactors)

water reactors)

Figure 24.28 in the manufacture of nuclear fuel for heavy-water and light-water reactors. This extremely simplified 238 diagram shows that in the process of enriching uranium dioxide, it is inevitable that large amounts of U0 2 remain as a byproduct.

Various steps

that

must be

stored.

As we

shall see, this

byproduct

alent to the heat given off

also has useful applications.

of coal. Uranium

The process of converting uranium ore into these uranium derivatives is shown in highly simplified

mass when

form

it

is

by burning 3 thousand tons

one of those elements

fissions.

that loses

However, uranium 235

sionable, whereas uranium

238

not,

is

is fis-

and so large

separating plants have been built to isolate molecules

in Fig. 24.28.

containing

235

U from those containing 238 U.

24.21 Energy released

24.22 Chain reaction

by atomic fission When

The

way

is

there

atom fissions, it splits in mass of the two atoms formed in this

the nucleus of an

two.

is

total

usually less than that of the original atom. If a loss in mass, energy

is

released according

How can we provoke the fission of a uranium atom? is to bombard its nucleus with neutrons. A

One way

neutron makes an excellent projectile because not repelled as

speed

to Einstein's equation;

ing a

E = mc 2

(24.3)

where

is

loss of

mass

it

has a good chance of scor-

is

strong enough, the nucleus

23
X

[3

of energy

X

releases

form of

The

MeV

218

heat. Fission

reaction on an atomic scale, and

1

1

J,

ond important

10* m/sj is

move

released be-

cause, according to this formula, a loss in

only one gram produces 9

in the

U

it

is

fission of

of energy,

a very violent

produces a sec-

[kg]

speed of light

An enormous amount

[JJ

it is

if its

the impact

will split in two, releasing energy.

mainly energy released

approaches the nucleus and,

not too great,

hit. If

one atom of

E= m= c =

it

which

mass of is

equiv-

at

effect:

high speed

It

ejects 2 or 3 neutrons that

away from

the broken nucleus.

These neutrons collide with other uranium atoms, breaking them up, and a chain reaction quickly takes place, releasing a tremendous

amount of heat.


This

the principle that causes atomic

is

bombs

to

explode. Although a uranium mine also releases neutrons, the concentration of to

235

we have

to

nuclei.

Toward

fissionable masses of uranium fuel

mersed

in a

this end,

(U0 2 ) may

moderator. The moderator

slow

can

that

absorbing them.

By

small

erator, the

are im-

1.

down

neutrons

uranium

sure that

mod-

2.

it

at

away

may

the heat. This coolant

be used, or

reactors.

ide 3.

may be

cir-

However,

moves

in

a closed circuit

The

must be used containing about 3 percent

High- Temperature Gas Reactor

(HTGR

).

23S

or carbon dioxide.

heat to a steam gen-

CANDU:

Due

Th

Canada Deuterium Uranium, developed by

coolant

heat exchanger

aling water

condenser


of

a nuclear power station.

i

to the high operating

Atomic Energy Commission of Canada.

erator that drives the turbines (Fig. 24.29). Thus,

coolant circulating

U.

reactor uses an inert gas coolant such as helium

which includes a heat ex-

latter transfers the

as

enriched uranium diox-

helium or carbon dioxide. The hot coolant

changer.

in

ordinary water boiling under high

in all light-water reactors,

heavy water, ordinary water, liquid sodium, or a gas like

CANDU*

need for a heat exchanger, because the steam

an acceptable level,

a liquid or gas has to flow rapidly through the reactor to carry

in

culates directly through the turbines.

as the chain reaction starts, the temper-

To keep

used

pressure and releasing steam. This eliminates the

will a chain reaction take place,

ature rises rapidly.

is

kept under such high pres-

Boiling-Water Reactors (BWR). The coolant this reactor is

causing the reactor to go critical

As soon

is

it

cannot boil off into steam. Ordinary

heavy water, as

speed of the neutrons can be reduced so

Only then

it

and

water, as in light-water reactors

without

fuel within the

Pressure-Water Reactor (PWR). Water as a coolant

be ordi-

they have the required velocity to initiate other fusions.

lead us to believe,

ing are the most important:

using an appropriate geometri-

cal distribution of the

name would

There are several types of reactors, but the follow-

nary water, heavy water, graphite, or any other material

its

not cool but searingly hot.

slow

the neutrons to increase their chance of strik-

uranium

is

24.23 Types of nuclear reactors

In the case of a nuclear reactor,

ing other

the coolant

U atoms is too low

produce a chain reaction.

down

contrary to what

657

pump

feedwater

pump

the


65 8

Figure 24.30 Aerial view of a light-water nuclear generating station.

The

large rectangular building

in

the

foreground houses a 667 MVA, 90-percent

power

factor,

generator

19

set;

60 Hz, 1800 r/min turbo-

kV,

the circular building surrounds

the reactor.

(Courtesy of Connecticut Yankee Atomic

Power Company; photo by Georges Betancourt)

temperature (typically 750°C), graphite as a moderator.

changer

is

The steam created by

as hot as that

tional coal-fired

steam

overall efficiency of

produced

boiler.

HTGR

used

is

the heat ex-

in a

conven-

Consequently, the stations

is

about

40 percent. 4.

Fast Breeder Reactor (FBR). This reactor has the

remarkable

ability to

both generate heat and create

additional nuclear fuel while

it

is in

operation.

24.24 Example of a lightwater reactor Figure 24.31 Reactors that use ordinary water as a moderator are

Looking down

similar to those using heavy water, but the uranium-

the reactor.

dioxide fuel has to be enriched. Enrichment means

(Courtesy of Connecticut Yankee Atomic Power

that the fuel 235

bundles contain between 2 and 4 percent

U, the remainder being

of

23S

U. This enables us

On

the other hand, the reactor has to be shut

down about once

a year to replace the

expended

Company; photo by Georges Betancourt)

to

reduce the size of the reactor for a given power output.

into the water-filled refueling cavity of

fuel.

posed of a massive vertical ternal diameter of 4.5

steel

tank having an ex-

m and a height of 12.5 m. The

tank contains 157 vertical tubes, which can lodge 3

m

total

of

reaction

is

157 large fuel assemblies. Each assembly

The generated heat, created mainly by the fission of uranium 235, is carried away by a coolant such

long and groups 204 fuel rods containing a

as ordinary water, liquid sodium, or a gas such as

477 kg of enriched

C0 2

kept under control by 45 special-alloy control rods.

.

As

it

flows through the heat exchanger, the

coolant creates the steam that drives the turbine.

A typical

nuclear power station (Figs. 24.30 and

24.31) possesses a light-water reactor that

is

corn-

When

U0 2

.

The nuclear

is

these rods are gradually lowered into the

moderator, they absorb more and more neutrons.

Consequently, they control the rate of the nuclear


reaction and, hence, the

amount of heat released by

659

izontal tubes, each housing 12 fuel bundles con-

kg of U0 2 Each bundle

the reactor.

taining 22.2

The nuclear station drives a 3-phase, 667 MVA, 90 percent power factor, 19 kV, 60 Hz, 1800 r/min

372.5

a total of

synchronous generator.

MW of thermal power.

.

kW while

it is

in

releases about

operation. Because there

4680 bundles,

Twelve pumps, each driven by an 1100

CANDU

actors in that fuel.

One of

the

differs

from

all

the biggest installations of

Toronto, Canada.

provide

It

is

other re-

uses natural uranium dioxide as a

it

located at Pickering,

Each reactor

reactor and the heat exchangers in a closed loop.

reactor uses heavy water, both as

moderator and coolant.

a

its

few kilometers

The nuclear

kind east

interface

is

of

4 reactors.

station has

coupled to 12 heat exchangers

that

between the heavy-water

The heat exchangers produce the steam to drive the four turbines. The steam exhausts into a condenser that is cooled by water drawn from Lake Ontario.

Each turbine drives a 3-phase, 635 MVA, factor, 24 kV, 1800 r/min, 60 Hz

85 percent power alternator.

The

one end of

fuel bundles are inserted at

they are withdrawn from the other end.

bines (Fig. 24.32).

are inserted

is

enclosed in a large horizontal

vessel (calandria) having a diameter of 8

length of 8.25 m.

The

m

and a

calandria possesses 390 hor-

814kg/s 252°C

4.1

MPa

^ —

The bundles

and removed on a continuous basis

— an

average of nine bundles per day. Table 24B compares the typical characteristics of light-water and heavy-water reactors.

635 steam

MVA

cos $ = 0.85

safety valve

1800 r/min 24 kV 60 Hz

coolant

moderator

I

I

steam cooling water

Figure 24.32 Simplified schematic diagram of a

one

CANDU

alternator.

{Courtesy of Atomic Energy of Canada)

the

calandria and, after a 19-month stay in the tubes,

coolant and the ordinary steam that drives the tur-

Each reactor

kW

motor, push the heavy-water coolant through the

24.25 Example of a heavy-water reactor The

is

the reactor develops 1740

nuclear generating unit

composed

of

one heavy-water reactor

driving


660

TYPICAL LIGHT-WATER AND HEAVY-WATER REACTORS

TABLE 24B

Light- Water Reactor

Heavy-Water Reactor

b

Reactor Vessel

external diameter

4.5

length

12.5

416

weight empty position

total

604

t

mass of

75

fuel

390

U0 2 (3.3%)

enriched

t

horizontal

57

J

type of fuel

mm

25.4

vertical

number of fuel canals

m

8.25

mm

274

vessel thickness

8m

m m

U0 2

natural

104

t

t

Moderator

heavy-water

light-water

type

volume

13.3

m

3

m3

242

Reactor Cooling

heat produced in reactor

128

rate

coolant temperature leaving the reactor

total

pumps 12

Hz synchronous

a.

b.

it

containing

fissionable

The core

surrounded by a blanket composed of

is

238

238 (

U).

containing

No

plutonium 239 nonfissionable

moderator

is

239 (

atoms of

238

239

Pu

bombard

nonfissionable

U. This nuclear reaction produces two

important results:

the fissioning core can be

Some atoms

of

238

U

in the

surrounding blanket

239

Pu. In other words, the passive fis-

sionable atoms of plutonium 239.

As time goes

uranium

used; consequently,

the

The heat released by

MW

atoms of uranium 238 are transmuted into

Pu).

the high-speed (fast) neutrons generated by the

fissioning

540

fissionable

possesses a central core

nuclear fuel.

It

MW

capture the flying neutrons, thereby becoming

can extract more of the available en-

in the

substances

12

I4MW

used to drive a steam turbine.

breeder reactor differs from other reactors

because

t/s

249°C 294°C

MW

600

generator

24.26 Principle of the fast breeder reactor

ergy

7.73

t/s

3

Output

3-phase, 1800 r/min, 60

fast

m

130

4

pump power Electrical

A

3

285°C 306°C

coolant temperature entering the reactor

coolant

m

249

MW

1661

heavy-water

light-water

volume flow

MW

1825

coolant

238

U

is

by, the blanket of nonfissionable

gradually transmuted to fissionable

and waste products. The blanket

moved and is

Pu

periodically re-

the materials are processed to recover

the substances containing

covered

is

239

239

Pu.

The nuclear fuel

re-

placed in the central core to generate heat


and

produce

to

still

more

fuel in a

newly relined

of the energy shortage because hydrogen

common

blanket of substances containing uranium 238.

is

661

the most

element on earth.

This process can be repeated until nearly 80 percent of the available energy in the uranium tracted. This

to

much more

is

ex-

being extracted by conventional reactors.

The breeder reactor is particularly well adapted complement existing light-water reactors. The

reason

that a great deal of

is

byproduct

238

U

is

available as a

manufacture of enriched

in the

235

it

load and a peak-load generating plant.

Why

24-2

By

capturing fast

ing plant next to the

in the

factors

splitting the nucleus of a

We

What

in a

heavy

decrease in

light ele-

when they The required

ap-

proach each other

ve-

to a

is

Give two reasons why

at

high speed.

The

24-7

are high enough, a self-sustaining chain reaction

24-8

5000 mYs

at

mal

hydropower

plant, a

plant,

and a nu-

Name two

24-9

basic differences between a

light-water reactor and a heavy-water reactor.

24-10

Explain what

is

meant by moderator, fisand heavy water

sion, fusion, neutron,

We can, therefore, produce heat by the fusion of two elements, and the hydrogen bomb is a good example of this principle. Unfortunately, we run into almost insurmountable problems when control the fusion reaction, as

we

we must do

in

at the

a nu-

same time showing them down.

voted to solve this problem.

is

it

1

1

The Zaire

River, in Africa, discharges at a

being de-

It

1

300

km 3

of water per year.

has been proposed to build a series of

dams

in the

region of Inga, where the river

drops by 100 m. Calculate

in

b.

The water flow [nrVs] The power that could be harnessed [MW]

could mean the end

c.

The discharge

If scientists

domesticating nuclear fusion,

24-

try to

confining and controlling high-speed par-

major worldwide research effort

Intermediate level

constant rate

clear reactor. Basically, scientists have not yet suc-

A

is

Explain the operating principle of a ther-

will result.

in

river flow in Fig. 24.9

clear plant.

close to the speed of light and corresponds

without

electric utility sys-

hydraulic power.

thermodynamic temperature of several million

ticles

meant by the term network?

a height of 24 m. Calculate the available

degrees. If both the atomic concentration and speed

ceeded

is

system?

tems are interconnected.

between the two nuclei (both

are positive), they only unite (fuse)

locity

24-6

released by the fusion of an atom of

deuterium with an atom of tritium. However, owing to the strong repulsion

the best indicator of stability (or

What

process called nuclear fusion. For examis

is

24-5

can also produce

energy by combining the nuclei of two

energy

mine mouth. What

into play in determining the

instability) of an electric utility

mass and a release of energy.

ple,

come

best solution?

element such as uranium results

in a

we

to a

generating plant or installing the generat-

24.27 Nuclear fusion

ments

stations not suited

have the choice of hauling the coal

used up.

have seen that

power

peak loads?

Referring to the coal mine in Fig. 24.4,

24-3

24-4

We

are nuclear

to supply

could be rejuvenated, as explained

above, until most of the potential energy is

Explain the difference between a base-

1

stored) could be used to surround the

core of a fast breeder reactor. neutrons,

24-

U

(see Fig. 24.28). This otherwise useless material

(now being

uranium


efficient than the 2 per-

Practical level

now

cent

is

succeed

a.

in

cubic miles per year


662

24- 2 1

For how long does a

have

500

1

MW generator

A fuel

24- 1 8

produce the same quantity

to run to

has

of energy as that released by a 20 kiloton

bomb? (See conversion

atomic

1

MW and

between 60

1

10

MW

ergy,

MW.

we have

to the

unit

and a

Advanced 24-19

6000

280

is

[TW

Assuming

ficiency

is

a. b.

The The

ft

24-20

Grand

active

an ab-

IGW]?

Referring to Fig. 24.32, the temperature of

294°C

to

249°C

in

heat exchangers.

the average turbine ef-

is

cooled

ter,

power output [MW]

power supplied

at

what would the peak

the heavy-water coolant drops from

and the generators

0.98, calculate the following:

reactive

system having

energy were consumed

solutely uniform rate,

0.92 and the average generator is

[g]

h] of the electric utility

If this

b.

M VA at a power factor of 0.9

lagging.

efficiency

weight of the bundle, due

in

energy released

load be

a particular day, the head of

deliver

and

[J]

Calculate the annual energy consumption

each case?

On

amount of heat released

the load duration curve given in Fig. 24.3.

unit.

are the respective capacities of the

Coulee dam

energy

in the reactor,

level

a.

base power and peaking power plants in

24-14

releases an

the following options:

pumped-storage

What

total

The reduction

b.

base-power generating unit and a

Install a

it

LBtu]

power

diesel-engine peaking plant. b.

The

a.

in

the required en-

base-power generating

Install a

a.

To produce

19-month stay

its

If

kW of thermal


regularly

the course of one day, the average

being 80

heavy-water reactor.

into a

during

The demand of a municipality varies

bundle of natural uranium dioxide mass of 22.2 kg when first inserted

average of 372.5

charts in

Appendix.) 24- 3

a.

passing through the

Knowing

at the rate

of 7.7

calculate the heat

that the reactor t/s

[MW]

of heavy wa-

transmitted to

the heat exchangers (specific heat of

to the

heavy water

system

4560

is

J/kg).

[Mvar] e.

Industrial Application

The amount of water flowing through

the

24-2

turbines [ydVsJ

24- 5 1

1

On November

Explain the principle of operation of a

1

A modern

coal-burning thermal station

1

992,

at

1

0:09

am

a large

produces an electrical output of 720

8 823

MW to suddenly lose 1050 MW of

generating power. In a matter of seconds,

MW.

Calculate the approximate value of the

the system frequency

fell

from 60 Hz

to

59.97 Hz. The power output of the other

following: a.

2,

causing the interconnected power pool of

cooling tower.

24-16

1

generator on the East Coast tripped out,

generators on the system was selectively in-

The amount of coal consumed

[tons (not

creased and the rated 60

Hz

frequency was

tonnes) per day] b.

The amount of smoke,

gas,

and

fly

ash re-

utes.

leased [tons per day] c.

The cooling water flowing through

the con-

10°C |m 1

In

3

1

6, if a

cooling tower

quired,

how much

from a

local stream [rrrVs]?

be recycled?

raised

above

for a time to recover the cycles lost,

The

behavior of the frequency before and after

/s]

Problem 24-

The frequency was then

60 Hz

thereby correcting the electric clocks.

denser, assuming a temperature rise of

24- 7

restored after an interval of about 7.5 min-

is

re-

water must be drawn

Can

this

water

the incident

is

shown

in Fig. 24.33.

Calculate a.

The average frequency during minute restoration period

the 7.5-

GENERA TfON OF ELECTRICA L EN ERG Y

exactly one minute

60.02

exactly 60 Hz.

^ I ô

60.01

when

663

the frequency

How many

is

turns did the

minute hand make during the 7.5-minute

,

ii

interval?

60.00

What

is

the error in the minute-

hand reading, expressed

in

milliseconds?

§ 59.99 24-22

£

59.98

J—

59.97

1

8

823

tors.

09.50

10.00

10:10

10:20

10:30

AM

AM

AM

AM

AM

the

c.

that

had not occurred and the at

varies as the cube

falls

from 60 Hz

A summer camp is a

in

when

power the fre-

to 59.97 Hz.

located near a 55

ft

wa-

show that the stream delivers minimum of 270 cubic feet per minute in Tests

It is

proposed to

in-

would have been

generated during the 6-minute interval the accident

power

the course of a year.

The number of cycles

quency had stayed

drive fans and similar

of a 10 000 hp blower motor

terfall.

7.5-minute period

that half the

of the speed. Calculate the drop

24-23

The number of cycles generated during

we assume

Some motors

quency

cell.

b.

1

MW load consists of induction mo-

loads wherein the

Figure 24.33 Dry

In Problem 24-2

stall

motor and drive

it

if

fre-

as a generator. Calculate the approximate

horsepower of the motor

60 Hz

that could har-

ness 80 percent of the capacity of the

d. Electric clocks are designed so that the

minute hand makes one complete turn

a 3-phase induction

in

falls.

Chapter 25 Transmission of Electrical Energy

tem must

25.0 Introduction

satisfy

some

basic requirements. Thus,

the system must

The

transmission of electrical energy does not

usually raise as

much

interest as

ation and utilization; consequently,

does

its

tend to neglect this important subject. This

involved in transmission are those employed

in

is

is

much

3.

Maintain a stable frequency that does not vary

4.

Supply energy

5.

Meet standards of safety

6.

Respect environmental standards

We

two generating

show some of the power flow are con-

then

the voltage and

an acceptable price

tions,

G, and

G2

,

a

It

consists of

few substa-

an interconnecting substation and several

is

carried over lines designated extra-high

voltage (EHV), high voltage (HV),

components power distribution system

Principal

In order to provide electrical

stations

commercial, residential, and industrial loads. The energy

in

at

shows an elementary diagram of a

Fig. 25.1

trolled in an electric utility system.

of a

Hz

transmission and distribution system.

low-voltage, high-power, low-power, aerial lines,

25.1

1

study these properties for

several types of transmission lines: high-voltage,

lines.

more than ± 10%

by more than ±0.

that greatly affect the transmission of electrical en-

ways whereby

consumers

greater than

nary, they possess important electrical properties

and underground

that

Maintain a stable, nominal voltage that does not vary by

carried by conductors such

we

power

unfor-

Although these conductors appear very ordi-

ergy. In this chapter,

at all times, the

material resources

overhead transmission lines and underground ca-

ble.

Provide,

need

generation.

Electrical energy as

1.

2.

human and

tunate because the

gener-

we sometimes

medium

volt-

age (MV), and low voltage (LV). This voltage classification

energy to consumers

is

made according to a scale of standardwhose nominal values are given in

ized voltages

Table 25A.

usable form, a transmission and distribution sys-

664

TRANSMISSION OF ELECTRICAL ENERGY

PRODUCTION h I

medium

1

i 1

-TRANSMISSION,.

^

.

.

,

extra high voltage

665

-DISTRIBUTIONmedium

,

high voltage

low voltage

voltage

tie-line

115 -

t

t-r

kV

2.4

|

kV

120/240

V

single-phase

to

to

230 kV

69 kV

f-

to

1

V

600

-

I

three-phase

medium

heavy

industry industry^ generating transmission

interconnection

transmission

distribution

small industry

substation

substations

substations

commerce

substation

stations

residences

Figure 25.1 diagram

Single-line

of

a generation, transmission, and

Transmission substations line voltage

(Fig. 25.

1

)

change the

by means of step-up and step-down

transformers and regulate

it

by means of

static

var

compensators, synchronous condensers, or trans-

distribution

faulted lines

apparatus,

medium

may have

1

.

low voltage. The low

2.

V, 3-phase.

It

serves to

power

to increase the stability of the

overall network.*

distribution systems in which the voltage gener-

between 120

25,2 Types of The design of lowing

circuit

breakers,

arresters, to protect

expensive

1

and lightning

line voltage

V

and 69 kV. Distribution

kV

69 kV) and low-

to (

1

20

V to 600 V)

power

tie different

These substations also contain

dis-

kV and 800 kV

voltage distribution systems

systems together, to enable power exchanges be-

fuses,

which the

in

15

distribution systems (2.4

ments, and small industry.

tween them, and

1

power

categories:

systems, in turn, are divided into medium-voltage

dences, commercial and institutional establish-

Interconnecting substations

roughly between

ally lies

private resi-

divide their

two major

transmission systems is

voltage ranges from 120/240V single phase to

600

utilities

volt-

automatic tap-changing

capabilities to regulate the

power

tribution systems into

age to low voltage by means of step-down trans-

In addition, control

part of a substation.

Electrical

Distribution substations change the

from the system.

power measuring devices, disconnect

switches, capacitors, inductors, and other devices

may be

formers with variable taps.

formers, which

system.

apparatus, and to provide for quick isolation of

.

2.

a

power

power

line

lines

depends upon the

fol-

criteria:

The amount of active power

it

The distance over which

power must be

the

has to transmit

carried

A network

is

3.

The

4.

Esthetic considerations, urban congestion, ease

cost of the

power

line

"an aggregation of interconnected conductors

consisting of feeders, mains and services"

(ref.

IEEE

Standard Dictionary of Electrical and Electronics Terms).

of installation, and expected load growth

666


TABLE 25A

VOLTAGE CLASSES AS APPLIED TO INDUSTRIAL AND COMMERCIAL POWER Nominal system voltage

Voltage class

Two-wire

low voltage

120

Three- wire

Four- wire

—

120/240 single phase

single phase

LV medium

j

voltage

2

120/208

480

V

277/480

600

V

347/600

400

4 160 4 800

MV

6 900

800

7 200/12 470

23 000

7 620/13 200

1

3

34 500

high voltage

1

7 970/13 800 I

A

A (\{\

A

f\ A / \

940

46 000

14 400/24

69 000

19 920/34 500

1

J

UUU

1

1

|

|

138 000 161 000

HV

230 000 345 000

extra-high voltage

500 000

EHV All

735 000-765 000

voltages are 3-phase unless indicated otherwise.

Voltages designated by the symbol

are preferred voltages.

Note: Voltage class designations were approved for use by

We distinguish four types

of power

lines,

IEEE Standards Board (September 4,

accord-

In

Low-voltage (LV)

lines provide

power

buildings, factories, and houses to drive tors, electric stoves,

conditioners. tors,

usually

mercial buildings consists of a grid of underto

lines

lamps, heaters, and air

may be overhead

or under-

at

600

V or less.

Such

a network provides dependable service, be-

cause even the outage of one or several cables

The lines are insulated conducmade of aluminum, often extend-

ing from a local pole-mounted distribution

The

ground cables operating

mo-

transformer to the service entrance of the consumer.

areas, the distribution

system feeding the factories, homes, and com-

ing to their voltage class: 1.

some metropolitan

1975).

will not interrupt 2.

customer service.

Medium-voltage (MV) to

lines tie the load centers

one of the many substations of the

company. The voltage

is

utility

usually between 2.4

kV

ground, and the transformer behaves like a

and 69 kV. Such medium-voltage

miniature substation.

tion systems are preferred in the larger cities. In

radial distribu-

TRANSMISSION OF ELECTRICAL ENERG Y

like fingers

power

from one or more substations

to feed

such as high-rise

to various load centers,

and campuses.

buildings, shopping centers,

The

substations to the generating stations. are

composed of aerial conductors

ground cables operating kV. In this category

we

at

lines

Extra-high-voltage

when generating load centers.

We

must

of-

sufficiently thick to prevent

breakdown un-

power systems,

To

increase the leakage path (and hence the

leakage resistance), the insulators are molded with

From

a mechanical standpoint,

they must be strong enough to withstand the dy-

namic

also find lines that

(EHV)

must be

wave-like folds.

or under-

pull

and weight of the conductors.

There are two main types of

to

insulators: pin-type

insulators and suspension-type insulators (Figs.

increase the stability of the network. 4.

electric standpoint, insulators

high resistance to surface leakage currents and

stand.

voltages below 230

transmit energy between two

fer a

der the high-voltage stresses they have to with-

High-voltage (HV) lines connect the main

3.

From an

systems the transmission lines spread out

radial

667

25.2 and 25.3). The pin-type insulator has several lines are used

stations are very far

from the

put these lines in a separate

class because of their special electrical proper-

Such

ties.

lines operate at voltages

up

to

800

kV

and may be as long as 1000 km.

25.3 Standard voltages To reduce

the cost of distribution apparatus and to

facilitate its protection, standards-setting organiza-

have established a number of standard

tions

ages for transmission in

lines.

volt-

These standards, given

Table 25A, reflect the various voltages presently

used

North America. Voltages that bear the sym-

in

are preferred voltages. Unless otherwise in-

bol

dicated,

all

voltages are 3-phase.

Figure 25.2 Sectional view of a 69 kV pin-type insulator. BIL: kV;

125

25.4

Components

of a

transmission

line

A transmission sulators, /.

line

is

HV

kV.

{Courtesy of Canadian Ohio Brass Co.

composed of conductors,

and supporting

270

60 Hz flash-over voltage under wet conditions: Ltd.)

metal cap

in-

structures.

Conductors. Conductors for high- voltage lines

are always bare. Stranded copper conductors, or

steel-reinforced

ACSR

aluminum cable (ACSR)

porcelain

are used.

conductors are usually preferred because

they result in a lighter and more economical line.

Conductors have

to

long. Special care

be spliced when a line

must be taken so

is

very

that the joints

have low resistance and great mechanical strength. Figure 25.3 2.

Insulators, Insulators serve to support and anchor

the conductors

and to insulate them from ground.

Insulators are usually

made of porcelain,

but glass and

other synthetic insulating materials are also used.

Sectional view of a suspension insulator. Diameter:

254 mm;

BIL: 125 kV, 60 Hz flash-over voltage, under wet conditions: 50 kV. (Courtesy of Canadian Ohio Brass Co. Ltd.)


668

porcelain skirts (folds) and the conductor the top.

A

is

fixed at

steep pin screws into the insulator so

it

can be bolted to a support.

salts to

lators are used, strung together

metallic parts.

4

to 7; for

by

The number of 1

10 kV,

230 kV, from

1

3 to

we 1

their

cap and pin

insulators

depends

generally use from

6. Fig.

insulator arrangement for a

735 kV

of 35 insulators each, to

in parallel

line. It is

com-

Supporting structures. The supporting structure

at

at

a safe height from the

an adequate distance from each other.

For voltages below 70 kV, we can use single wooden poles equipped with cross-arms, but for higher voltages,

wood

two poles is

For very high-voltage

The spacing between conductors must be suffiwind conditions.

The spacing has

to

be increased as the distance be-

tween towers and as the

line voltages

become

higher.

25.5 Construction of a line

poles,

ground and

rotting.

cient to prevent arc-over under gusty

Once we know

provide both mechanical and electrical strength.

must keep the conductors

from

25.4 shows an

posed of 4 strings

3.

it

towers are used, made of galvanized an-

gle-iron pieces thatâre bolted together.

For voltages above 70 kV, suspension-type insu-

upon the voltage: for

prevent

lines, steel

are used to create an H-frame.

The

treated with creosote or special metallic

the conductor size, the height of the

and the distance between the poles (span), we

can direct our attention

A

to stringing the

conductors.

wire supported between two points (Fig. 25.5)

does not remain horizontal, but loops middle.

The

vertical distance

line joining the points

point of the conductor

down

between the

at the

straight

of support and the lowest

is

called sag.

The

tighter the

wire, the smaller the sag will be.

Figure 25.4 Lineman working bare-handed on a 735 kV line. He is wearing a special conductive suit so that his body is not subjected to high differences of potential. In the position shown, his potential with respect to ground is about 200 kV. (Courtesy of Hydro-Quebec)


669

Figure 25.5

Span and sag

of

a

line.

Before undertaking the actual construction of a line,

it is

important to calculate the permissible sag

and the corresponding mechanical things, the

pull.

Among other

summer to winter temperature range must

be taken into account because the length of the conductor varies with temperature. Thus,

if

the line

is

strung in the winter, the sag must not be too great,

otherwise the wire will stretch even more during the

summer heat, with the result that the clearance to ground may no longer be safe. On the other hand, if the line

is

KM

summer, the sag must not

installed in the

be too small otherwise the wire, contracting

in

win-

may become so dangerously tight as to snap. Wind and sleet add even more to the tractive force, which may also cause the wire to break (Fig. 25.6). ter,

Figure 25.6 During winter, steel towers must carry the combined weight of conductors and accumulated

ice.

(Courtesy of Hydro-Quebec)

length of the transmission line. In addition, corona

25.6 Galloping lines

emits high-frequency noise that interferes with If

a coating of sleet

is

deposited on a line during

nearby radio receivers and

TV

windy conditions, the line may begin to oscillate. Under certain conditions, the oscillations may be-

corona,

come

diameter or by arranging them

so large that the line

is

seen to actually gallop.

we must

sets.

To diminish

reduce the electric field (V/m)

around the conductors, either by increasing in sets

their

of two, three,

Galloping lines can produce short-circuits between

or

phases or snap the conductors. To eliminate the

25.7a and 25.7b). This bundling arrangement also

problem, the cial

line is

sometimes equipped with spe-

mechanical weights,

or to prevent

to

dampen

them from building

the oscillations

up.

more bundled conductors per phase

reduces the inductive reactance of the abling

it

to carry

more power. This

25.8 Pollution

The very high voltages

phere

tinual electrical discharge

around Jhe conductors,

due to local ionization of the corona

effect,

Dust, acids,

use tqday produce a con-

air.

This discharge, or

produces losses over the entire

line,

en-

constitutes an

important additional benefit.

25.7 Corona effectradio interference in

(see Figs.

settle

salts,

and other pollutants

on insulators and reduce

properties. Insulator pollution cuits during storms or

in the

atmos-

their insulating

may produce short-cir-

momentary

possibility of service interruption

overvoltages.

The

and the necessity

to


670

clean insulators periodically

concern

to the. utility

therefore a constant

is

company.

In addition to pollution, there

problem of

the

is

lightning, discussed in the following sections.

25.9 Lightning strokes During stormy weather, by a process not yet

fully

understood, a charge separation takes place inside clouds, so that positive charges

move

upper

to the

part of the cloud while negative charges stay be-

low

(Fig. 25.8). This transfer of electric charge

up

sets

an

electric

within

field

Furthermore, the negative charge

the

cloud.

base of the

at the

cloud repels the free electrons on the ground below. Consequently, region

charged, by induction. field

T becomes

positively

follows that an electric

It

and difference of potential

will be established

between the base of the cloud and the

earth.

Furthermore, another electric field exists between

T

the electrons repelled from region

Figure 25.7a Four bundled conductors 3-phase, 735 kV

make up

this

phase

of

As more and more positive charges move upward within the cloud, the electric field below the

a

line.


cloud becomes more and more intense. Ultimately,

where

air be-

gins to break down. Ionization takes place

first at

it

-457

mm -35

mm

reaches the

critical ionization level

the tips of church spires

and the top of high

and may sometimes give s

7 steel strands (2.5

mm)

42 aluminum strands (4.6

mm)

Mariners of old observed

this

When

the electric field

light St.

it

becomes

tense, lightning will suddenly strike earth.

mm

trees,

bluish light.

rise to a

masts of their ships and called

457

and the posi-

tive charge at the top of the cloud.

around the

Elmo's

fire.

sufficiently in-

from cloud

to

A single stroke may involve a charge transfer

of from 0.2 to 20 coulombs, under a difference of potential of several

hundred million

rent per stroke rises to a

croseconds and

50 is

|xs.

What

often

is

peak

falls to half its

in

volts.

The

cur-

one or two mi-

peak value

in

about

visually observed as a single stroke

composed of several strokes following each The total discharge time

other in rapid succession.

may

last as

long as 200 ms.

Discharges also occur between positive and neg-

Figure 25.7b

ative charges within the cloud, rather than

Details of the bundled conductors.

the base of the cloud and ground.

between


671

Figure 25.8 Electric fields created

by a thundercloud.

The thunder we hear pressure wave.

It is

is

produced by a supersonic

created by the sudden expansion

of air surrounding the intensely hot lightning stroke.

enormous overvoltage between The dielectric strength of air

rise

less than

lightning arresters are metallic rods that

the lightning toward a

ground electrode by means of

a conducting wire. This prevents the high current

from passing through the building cause a rester

fire

or endanger

its

itself,

occupants.

and anything connected to

it

50

which might

A lightning ar-

can be dangerous

line discharges in typically

|jls.

Unfortunately, the

above the highest point of a building, channeling

immediately ex-

and the overvoltage disappears

ground

The simplest

is

ceeded and a flashover occurs. The itself

25.10 Lightning arresters on buildings

the line and ground.

(initiated

between the

arc

by the lightning

a highly ionized path

stroke),

which behaves

and

line

produces con-

like a

ducting short-circuit. Consequently, the normal ac line voltage

immediately delivers a large ac current

that follows the ionized path.

current

open

may

at the

This follow-through

sustain the arc until the circuit breakers

end of the

line.

The

fastest circuit break-

ers will trip in about l/l5th of a second,

which

is al-

to the touch; during a discharge, a very high voltage

most 1000 times longer than the duration of the

can exist between the protective system and ground.

lightning stroke

The reason

is

that the resistance

rod and the ground

itself is

between the ground

seldom

ohms. Thus, a discharge current of

1

less than 0.5

0

kA

duce a momentary touch voltage of 5000

Much more

can pro-

V.

sophisticated lightning arresters are

used on electrical

utility

systems.

They

divert light-

ning and high-voltage switching surges to ground before they

damage

costly and critical electrical

equipment.

lightning will strike the overhead ground

often,

wire that shields the

charge very

still

high local

lines

When

lightning

line,

deposits a large electric charge, producing an

it

makes a

direct hit

on a transmission

producing a

line,

overvoltage. This

swiftly

move

in

opposite directions

speed of light (300

concentrated

m/(jis).

at

close to the

The height of the impulse

represents the magnitude of the surge voltage

that exists

and transmission

line. In the latter case, a local

accumulates on the

charge immediately divides into two waves that

wave

25.11 Lightning

itself.

Direct hits on a transmission line are rare; more

from point

to point

between the

line

and

The peak voltage (corresponding to the crest of the wave) may attain one or two million volts. Wave front ab is concentrated over a distance of about 300 m, while tail be may stretch ground

(Fig. 25.9).

out over several kilometers.


672

Figure 25.9 Flow

of electric

The wave

charge along a transmission represents

also

the

value of the current flowing in the

point-to-point line.

For most

surge current corresponds to a resistance of about

400 is

A surge voltage of 800 000 V at a given point

(2.

therefore

accompanied by a

800 000/400

As

the

local surge current of

= 2000 A.

wave

it

Should the wave encounter a

takes

in

about

is

(jls,

1

In turn, the apparatus within the substation

that

it

does not

trip.

fail,

the line until is

the it

On

to several

will flash over,

if

and the

the insulator

will continue to travel along

wave can produce

It

real

havoc. The windings of transformers, synchronous etc., are seriously damaged when they flash over to ground. Expensive repairs and even more costly shut-downs are incurred while the apparatus is out of service. The overvoltage may also damage circuit breakers, switches, insulators, and relays, that make up a substation. To reduce the

condensers, reactors,

a 1000

kV

kV

surge

station ar-

to ground.

beyond

The

wave

residual impulse

that travels

the arrester then has a peak of only

400 kV.

This impulse will not damage station apparatus built to

withstand an impulse of 550 kV.

25.12 Basic impulse insulation level (BIL)

re-

eventually encounters a substation.

here that the impulse

if

hun-

corresponding to the

the other hand,

wave

considerably higher than the arrester

rester diverts a substantial part of the surge energy

line insulator, the

sulting follow-through current will cause the circuit

breakers to

is

voltage enters a substation, the 400

length of wavefront ab. If the insulator cannot with-

stand this overvoltage,

all

designed to withstand an impulse voltage, say

and

The overvoltage period is equal to the time for the wave to sweep past the insulator. The

dred kilovolts

400 kV.

to flatten out,

nominal value

ar-

lines.

Lightning arresters are designed to clip off

clipping voltage. Consequently,

be briefly subjected to a violent over-

its

incoming

and

voltage. it

all

voltage peaks that exceed a specified level, say,

550 kV, 2

the peak of the surge voltage decreases.

voltage rises from

must be installed on

resters

R

travels along the line, the I

corona losses gradually cause

latter will

impulse voltage on station apparatus, lightning

between surge voltage and

lines the ratio

aerial

line.

How

do insulating materials

react to impulse volt-

ages? Tests have shown that the withstand capability

increases substantially

when

plied for very brief periods.

we wish

To

voltages are ap-

illustrate,

to carry out an insulation test

former, by applying a 60

Hz

tween the windings and ground. As occurs. Let us

voltage

On

is

assume

we

we

slowly raise

where break-

that the

46 kV (RMS) or 65 kV

the other hand, if

trans-

sinusoidal voltage be-

the voltage, a point will be reached

down

suppose

on a

breakdown

crest.

apply a dc impulse volt-

age of extremely short duration between the windings


X 50

1.2

microsecond wave of 900 kV,

673

said to

is

possess a basic impulse insulation level (or BIL) of

900 kV.

shows an insulator BIL impulse test.

Fig. 25.1

subjected to a

The BIL of higher than

its

1

a device

usually several times

is

nominal ac operating voltage. For

example, the standards require that a 69

°0

However, there

BIL and nominal voltage. As the BIL rises, we must amount of insulation which, in turn, in-

impulse voltage used to determine

of

creases the size and cost of equipment. In conclusion, the

rester begins to

we

discover that

peak voltage (or

1

it

takes about twice the

BIL of

the

peak voltage

at

which an

ar-

conduct must always be lower than

the apparatus

it

is

intended to protect.

30 kV) before the insulation breaks

down. The same phenomenon

is

observed

in the

case

of suspension insulators, bushings, spark gaps, and so on, except that the ratio crest ac voltage

kV.

no special relationship between

is

increase the

the BIL rating of electrical apparatus.

and ground,

distrib-

/is

time Figure 25.10 Standard shape

kV

BIL of 350

ution transformer must have a

50

1.2

being

string

between impulse voltage and

closer to

is

l

.5.

25.13 Ground wires we can

In Fig. 25.6

supported

at the

discern two bare conductors

very top of the transmission-line

towers. These conductors, called ground wires, are

of standardization, and to enable a

intended to shield the line and intercept lightning

comparison between the impulse withstand capa-

strokes so they do not hit the current-carrying con-

of similar devices, standards organizations

ductors below. Grounding wires normally do not

In the interest

bility

have defined the shape and crest values of several impulse waves. Fig. 25. 0 shows one of these stan1

dard impulse waves.

and

falls to

attains

its

one-half the peak

in

It

peak

50

to

1550

kV

steel.

They

ground

are connected to

Transmission-line towers are always solidly con-

(see Table 25B).

the

TYPICAL PEAK VOLTAGES FOR 1.2 x 50 |xs BIL TESTS Values are

in

made of

each tower.

25.14 Tower grounding nected to ground. Great care

TABLE 25B

at

jjls

The peak range from

jjls.

voltage has a defined set of values that

30 kV

after 1.2

carry current; consequently, they are often

ground resistance

ning hits a

line,

it

is

is

taken to ensure that

low. In effect,

when

across the insulators as the lightning current dis-

charges to ground. Such a voltage

kilovolts

rise

may

duce a flash-over across the insulators and 1550

825

250

1425

750

200

650 550

150

450 350

90

1300 1175

1050

900

The peak voltage

is

pro-

con-

sequent line outage, as shown by the following

110

30

used to specify the basic im-

Example 25-1

A 3-phase

69

300 kV

supported on steel towers and protected

by a

is

kV

transmission line having a BIL of

circuit breaker (Fig. 25.12). is

20

line

is

tance at each tower

piece of equipment (transformer, insulator, capaci-

the transmission

bushing, etc.) that can withstand a

transformer just

resistor,

a

example.

pulse insulation level (BIL) of'equipment. Thus, a

tor,

light-

creates a sudden voltage rise

ahead

O

The ground

resis-

whereas the neutral of

solidly

of

the

grounded

at

the

circuit-breaker.

Figure 25.11 4 000 000 V impulse causes a flashover across an insulator string rated at 500 kV, 60 Hz. Such impulse tests crease the reliability of equipment in the field. The powerful impulse generator in the center of the photo is 24 m high and can deliver 400 kJ of energy at a potential of 6.5 MV. {Courtesy of IREQ)

A

674

in-


During an

electric storm,

one of the towers

is

hit

by

615

3-phase short-circuit initiated by the lightning stroke will continue to be fed and sustained by a

a lightning stroke of 20 kA.

heavy follow-through current from the 3-phase a.

Calculate the voltage across each insulator source. This short-circuit current string

under normal conditions circuit breaker,

b.

Describe the sequence of events during and

af-

In

producing a

a load interruption,

Under normal conditions, the line-to-neutral voltage is 69 kV/V3 = 40 kV and the current flowing

tial

in the

steel

tower ground resistance

tower

is

as the ground.

therefore at the It

is40V2 = 57

When

is

zero.

same poten-

follows that the peak volt-

age across each insulator string (line

b.

to

tower)

kV

try to limit the

number of

kA X 20

= 400

kV.

The voltage between is therefore 400 kV,

the tower and solid ground

and so the potential difference across insulator strings

Because

this

jumps

to the

same

all

this

example,

of 300 kV, a flashover immediately occurs across the insulators, short-circuiting lines to the steel cross-arm.

lightning

The

as

it

all

if

the tower re-

would have

voltage across the insulators

200

kV

Note

risen to

and no flashover would have occurred. that lightning currents of

frequent, and they last only a

20

kA

are quite

few microseconds. line

outage

is

to use

a circuit breaker that recloses automatically, a

cycles after

it

trips.

to lightning will

By

that

few

time the disturbance due

have disappeared and normal op-

eration of the system can resume.

three

value.

impulse exceeds the insulator BIL

Figure 25.12 Flash-over produced by

towers and ground. In

sistance had been 10 ft instead of 20 17, the impulse

Another way of avoiding a

lightning strikes the tower, the voltage

across the ground resistance suddenly leaps to

20

we

outages by ensuring a low resistance between the

Solution

The

the

view of the many customers affected by such

ter the lightning stroke

a.

/ sc will trip

line outage.

three

resulting

flows to ground.

25,15 Fundamental objectives of a transmission line The fundamental purpose of a transmission or distribution line is to carry active power (kilowatts)


676

from one point

to another. If

also has to carry re-

it

dividual reactances x

j

.

Similarly, the total capacitive

Xc is equal to the sum of the xc reactances,

active power, the latter should be kept as small as

reactance

possible. In addition, a transmission line should

except they are connected

possess the following basic characteristics:

nient to

1.

The voltage should remain

of the line as constant as possi-

ble over the entire length of the line,

source to load, and for

all

value 2

from

loads between zero

line losses

must be small so

as to attain a

high transmission efficiency 3.

2

The

I

R

losses

XL

must not overheat

Hz

a

Hz power line, pro250 km. Note that R and

or 60

less than

is

The regood ap-

line. is

Xc decreases

to

with increasing length.

circuit of Fig.

25.14 can also be

represent one phase of a 3-phase line.

one conductor and

same conductor and

E

is

the voltage

between the

neutral.

until the require-

ments are met.

25.17 Typical impedance values

25.16 Equivalent circuit of a line In spite of their great differences in

power

Table

25C

gives typical values of the inductive and

capacitive reactances per kilometer for practical rating,

voltage levels, lengths, and mechanical construc-

transmission lines operating the respective

tion,

Xc each having a

Current / corresponds to the actual current flowing in

and inductors, must be added

length

at

60 Hz. Surprisingly,

impedances per unit length are

rea-

transmission lines possess similar electrical

sonably constant for

all

aerial lines.

properties. In effect, an ac line possesses a resis-

tance R, an inductive reactance

reactance

Xc

,

XL

,

and a capacitive

These impedances are uniformly

about 0.5 fl/km and xc

is

Thus, x L

about 300 000

dis-

tributed over the entire length of the line; conse-

quently,

we can

represent the line by a series of

identical sections, as

shown

ample), and elements

We

xc represent the imped-

r,

ances corresponding to

Each seckm, for ex-

in Fig. 25.13.

tion represents a portion of the line (I

this unit length.

can simplify the circuit of Fig. 25.13 by

lumping the individual resistances a total resistance R. In the tal

conve-

increase as the length of the line increases,

used alone cannot satisfy the above require-

parts,

located at each end of the

The equivalent

the conduc-

ments, supplementary equipment, such as capacitors

its

whereas

tors If the line

Xc

composed of two

sulting equivalent circuit of Fig. 25. 14

vided

The

is

in parallel. It is

that the total capacitive reactance

proximation of any 50

and rated load 2.

assume

inductive reactance

Figure 25.13 Distributed impedance

r together to yield

same way, we obtain a

to-

XL equal to the sum of the in-

of

a transmission

line.

Figure 25.14 Equivalent lumped

circuit of

a transmission

line.

is

H-km


whether the transmission or whether the

power

line voltage

is

high or low,

great or small.

is

a.


b.

Draw

Type of line

A-

L ini

aerial line

0.5

underground cable

0.1

The same can be

a-c

Solution a.

smaller. Thus,

one hundred times smaller than whereas x

x

is

line

xc

is

However, the

such a wide range that

it

typical value to n Table

and ampacity for several

is

0.5

xc =

300 000/50

The

2XC =

aerial conductors.

the

_

A 3-phase 230 kV transmission line having a length km is composed of three ACSR conductors

having a cross-section of 1000 kcmil. The voltage

load

is

230 kV

(line-to-line)

and

that at the

220 kV.

TABLE 25D Conductor

a

- 6000 a

X 6000 = is

12

ka

23(W3 =

source and 220/V3 = 127 kV circuit per

phase

is

133

kV

at

at the load. is

shown

in

Fig. 25.15. b.

The complete equivalent circuit of the 3-phase line is shown in Fig. 25.16. Note that the line capacitance acts as if it were composed of six capacitors connected between the lines and

ground. This circuit arrangement holds true

RESISTANCE AND AMPACITY OF SOME BARE AERIAL CONDUCTORS Resistance per conductor

size

AWG

2

The voltage per phase

gives the resistance

of 50

X 50 = 3.25 X 50 = 25 a

0.065

The equivalent

is

-km

capacitive reactance at each end of the line

size varies over

Example 25-2

of the source

left

R = XL =

impossible to assign a

25D

300

impedances, per phase, are

line

distance that ac

power can be transmitted by cable. The resistance r per unit length depends upon the size of the conductor.

a/km

0.5

about

that of aerial lines,

maximum

0.065 (I/km

XL Xc =

000

about five times smaller. This fact has

a direct bearing on the

=

r

The

the approxi-

impedances per unit length are

said of underground cables, ex-

much

25C and 25D,

Referring to Tables

mate

inj

cept that the inductive and capacitive reactances of

3-phase cables are

line.

300 000 3

per phase.

circuit,

the complete equivalent circuit of the

3-phase

TABLE 25C TYPICAL IMPEDANCE VALUES PER KILOMETER FOR 3-PHASE, 60 HZ LINES

677

Cross-section

[mm

2 |

Copper [ft/km J

at

ACSR

75°C [n/km]

Ampacity

in free air*

ACSR

Copper A] |

10

5.3

3.9

6.7

70

7

10.6

2.0

3.3

110

[A]

4

21.1

0.91

1.7

180

140

1

42.4

0.50

0.90

270

200

3/0

85

0.25

0.47

420

300

300 kcmil

152

0.14

0.22

600

500

600 kcmil

304

0.072

0.11

1000 kcmil

507

0.045

0.065

*The ampacity indicated

is

the

practice, the actual line current

maximum that may be used without weakening may be only 25 percent of the indicated value.

750

950 1300

'

the conductor by overheating.

1050 In


678

3.25

H

elements shown

a

25

in Fig. 25.14.

The

validity of this

simplification depends upon the relative magnitude

of the active and reactive powers

^12

ciated with the line,

12 kft zt:

133 kV

127

kV

that

compared

Ph g L Qc ,

to the active

asso-

power P

delivers to the load. Referring to Fig. 25.17,

it

these powers are 7// //////J//////////////////////////////'

P =

Aground

power absorbed by

active

the load

P = I2 R, active power dissipated in the line reactive power absorbed by the line <2l — 2 = E /Xc reactive power generated by the <2c }

Figure 25.15 Equivalent circuit of one phase (Example 25-2).

,

even when the neutrals of the source and load

The

are not grounded.

result

age acrosêach capacitor

is

(We assume

line.

the source and load volt-

ages have the same magnitude)

that the volt-

given by

With the exception of P these powers

the respective line

portional to the length of the line. If one of

is

}

Ec =

E/V3, where

E

is

voltage at the source or load. Thus, knowing that the line voltage is

and 220

kV

230 kV

at the load, the

at the

source

voltages across

<2 L ,

tive

or

Q c — is

power

P,

negligible

we

pro-

the ac-

it.

For example, low-voltage lines are always short

and because the voltage

!33kV

ligible.

Low-voltage

is

low,

lines

E2 /Xc is always neg-

can therefore be repre-

sented by the circuit of Fig. 25.

Ecl = 220/V3 = !27kV

happens

to

be small, such as

cuits, the resistance

25.18 Simplifying the equivalent

1

in

8. If

the conductor

house-wiring

cir-

predominates and the inductive

portion of Fig. 25.18 can also be neglected.

On

circuit

We

compared with

all

them

can neglect the corresponding

element that produces

circuit

the respective capacitors are

230/V3

P j,

are

the other hand, extra-high-voltage lines are

always long, and so the reactive powers associated can often simplify the equivalent circuit of a

transmission line by eliminating one, two, or

3.25

n

25

all

the

with the line capacitance and line inductance be-

come ciency

Q

small.

important. Furthermore, because the is

high,

it

follows that the I

The equivalent

circuit

2

R

can therefore be

resented by the circuit of Fig. 25.19.

source

133

12

kn

12

Figure 25.16 Equivalent circuit of a 3-phase

ka:

Figure 25.17 line.

effi-

losses are

Active and reactive powers of a transmission

line.

rep-


a.

— R

The

679

and reactive powers associated with

active

the line: b.

The approximate equivalent

circuit, per phase.

is Solution

III LjU

—

o

Referring to Fig. 25.20a, a.

-»

The

we

have:

line-to-neutral voltage at the load

E = 230/Y3 -

1

is

33 kfl

Figure 25.18 Equivalent circuit of a short LV

The

line.

active

phase

power transmitted

to the load per

is

P = 300MW/3 = 100 The load

I

2XC

2XC

I

CD

I

—

|

CD

I

|

>

I

I

/

current

MW

is

= 100MW/133 kV - 750 A

I

If

we

temporarily neglect the presence of the

12 kfl capacitor in parallel with the load, the line

6

—

current 2

f

Figure 25.19 Equivalent circuit of a long

HV

Pj

In

general, medium-voltage and high-voltage

can be represented by a simple inductive reac-

We

will

make

=

is

equal to the load current (Fig. 25.20b).

losses in the line are 2

I

line.

lines

tance.

R

R= =

3.25

X 750 2

1.83

MW(1.8

Reactive power absorbed by the line

QL -

use of this fact in developing

=

the properties of transmission lines.

Example 25-3 The transmission 300

MW

to the

is

2

/

XL =

14.1

25

Mvar

line

shown

3-phase load.

in Fig. If

230 kV, determine

(

1

4 percent of P) at

is

25.16 delivers

Q = E2/Xc =

the line voltage at

=

1.47

2

133 000 /12 000

Mvar

the following:

25

Figure 25.20 Progressive simplification of a 735 kV

is

X 750 2

Reactive power generated by the line

end

both the sending end (source) and receiving end (load)

percent of P)

line

(Example 25-3).

n

3.25 O,

25

n

each


680

Total reactive

b.

Qc =

2

X

=

3

Mvar

Comparing and

P,

power generated by

it

is

1

the line

voltage

is

.47 (3 percent

of P)

the relative values of

clear that

Es

is

fixed, but the receiver voltage

ER

de-

pends upon the power drawn by the load.

we can

25.20 Resistive line

P QL Qc }

,

,

,

The transmission

sistance R. Starting

The

tance and the capacitance of the line. is

of Fig. 25.2 a possesses a 1

from an open

circuit,

we

re-

gradu-

re-

ally

sulting equivalent circuit

line

neglect the resis-

reduce the load resistance

a simple inductive

During reactance of 25 il (Fig. 25.20c).

ER

process

this

we

until

it

becomes

zero.

observe the receiver voltage

across the load, as well as the active

power P

it

absorbs. If numerical values were given, a few sim-

25.19 Voltage regulation

would enable us to draw a graph of P However, we prefer to use a generalized curve that shows the relationship between E R and P for any transmission line having an ple calculations

and power-transmission

ER

capability of transmission lines

as a function of

arbitrary resistance R.

Voltage regulation and power-handling capacity are

two important features of a transmission

line.

Thus,

The generalized shape of Fig. 25.21b.

It

this

graph

is

given

reveals the following information:

the voltage of a transmission line should remain as

constant as possible even under variable load conditions. Ordinarily, the voltage regulation to full-load

should not exceed

voltage (though tion as

As

±5%

we can sometimes

R

from zero

of the nominal

accept a regula-

high as ±10%).

regards power-handling capacity,

it

may come

as a surprise that a transmission line can deliver only

so

much power and no more. The power

can be

that

transported from source to load depends upon the

impedance of the

line.

transmitting active

We

are mainly interested in

power because only

it

can do

useful work. In order to determine the voltage regulation

and to establish

pability,

1

.

power transmission

their

we now examine

ca-

four types of lines:

Resistive line

2.

Inductive line

3.

Inductive line with compensation

4.

Inductive line connecting two large systems In

our analysis the lines connect a load (or

ceiver)

have

all

R

to a source (or sender) S.

re-

The load can

0

19

„

possible impedance values, ranging from

no-load to a short-circuit. However,

we

are only in(b)

terested in the active

power

the line can transmit.

Consequently, the load can be represented by a variable resistance absorbing a

power

P.

The sender

Figure 25.21 Characteristics of a resistive

line.

100%

in


a.

There

an upper limit to the power the line

is

can transmit to the load. In

maximum

this

voltage

is

kW/25

kW =

0.19, or 19%,

the percentage predicted by the curve

of Fig. 25.21b.

(25.1)

reached when the receiver

is

that 4.75

which

P n ^ = Es 2 /4R and

Note

effect,

25.21 Inductive line

is

ER =

Let us

Es

0.5

now

consider a line having negligible

tance but possessing an inductive reactance b.

The power delivered to the load when the impedance of the load

is

maximum

is

equal to the

25.22a).

resistance of the line. c.

If

we

maximum

permit a

cent (E R line

=

0.95

Es ),

.

regulation of 5 per-

the graph

can carry a load that

P max The

is

shows

that the

only 19 percent of

ER would

factor,

2

I

R

it

can be represented by a variable

power P As

sistive line, voltage

ER

in the

case of a

diminishes as the load

(Fig. 25.22b). In effect, the generalized

as a function of

P reveals

graph of

P

The

line

maximum power

can transmit a

to the

losses in the line. (25.2)

single-phase transmission line having a resis-

0

1

fi is

connected

to a fixed

sender voltage

of 1000 V. Calculate a.

The maximum power

the line can transmit to

the load b.

The receiver power 950 V

for a receiver voltage of (a)

Solution a.

The maximum power

that

can be transmitted

P niilx = E s 2 /4* = 1000 2 /(4 X - 25 kW b.

When ER = 950

V, the voltage

Es - E R = The

line current /

1000

is,

=

(£s

=

5

drop

0.95

E

0.707

E

is

t

10)

in the line is

- 950 = 50 V

therefore,

- ER )/R =

50/10 100

A

The receiver power

*

is

P = ES I = 950 X = 4.75 kW

5

ER

load given by

£ s 72 X

A

in-

the following information:

Example 25-4 tance of

re-

re-

then be too low.

sender must furnish the power

absorbed by the load, plus the

receiver again operates at unity power

sistance absorbing a

a.

that the

The

and so

resis-

X (Fig.

creases, but the regulation curve has a different shape

could transmit more power, but

line

the customer voltage

Note

681

= 4750

(b)

W

Figure 25.22 Characteristics of an inductive

line.

%


682

The corresponding

receiver voltage

ER =

/. Taking E R = 950 V as the reference phasor, we-draw the phasor diagram for the cir-

current

is

Es

0.707

cuit of Fig. 25.23.

Thus, for a given line impedance and sender voltage, the reactive line can deliver twice as

much power

as a resistive line can

P = E s 2 /2 X
/ is in

If

we

(compare

E s = ER +jIX = 950 + 10y7

R).

This equation corresponds to the phasor dia-

maximum

shows

percent, the graph

given line

load

line.

again allow a

a load that

E R because the we can write

phase with

Furthermore,

is resistive.

The power delivered to the load is maximum when the resistance of the load is equal to the reactance of the

c.

Current

regulation of 5

that the line

can carry

gram of write,

2

60 percent of P max Thus, for a impedance and a regulation of 5 peris

From

Fig. 25.24.

this

diagram we can

by inspection

£s

.

I000

2

= £r 2 + (10 if = 950 2 + 100 2

cent, the inductive line can transmit six times as

much

active power as a resistive line can. The sender has to supply the active power P consumed by the load plus the reactive power 2

I

X absorbed

by the

from which /

The power

line.

A

single-phase transmission line having an induc-

connected

is

to a fixed

sender

voltage of 1000 V.

c.

=

31.22

to the receiver

is,

kW

A

therefore,

31.22

Note

that

P miiX

(50 kW), as predicted by the curve of Fig.

29.66

The maximum

active

power

is

equal to 60 percent of

the line can deliver 10

to a resistive load b.

975

25.22.

Calculate a.

\

P = ES I = 950 X = 29.66 kW

Example 25-5 tive reactance of 10 fl

=

12

The corresponding receiver voltage The receiver power when the receiver voltage is 950 V

Solution a.

The maximum power the load

that

can be transmitted to

is

Figure 25.23

P max = E

2

r

=

1000

50 b.

According at

0.707

707

X ^ 2

/2

See Example

X

V,

we

10

kW

is

Es =

10/

0.707

X

1000

V

/

In order to calculate the receiver

ER = 950

25-5.

25.22b the receiver voltage

maximum power

ER = c.

to Fig.

12fZ

first

power when

calculate the value of the

Figure 25.24

See Example

25-5.

£R

=

950V


25.22

We

Compensated inductive

line

Beyond

ER gradually decreases to zero

this limit,

shown by

a diagonal line, as

can improve the regulation and power-handling

683

in

the graph of Fig.

25.25b. Note the following:

capacity of an inductive line by adding a variable capacitive reactance

Indeed,

we can

the value of

Xc

Xc

across the load (Fig. 25.25a).

so that the reactive

supplied by the capacitor

is at all

power

age

£s

,

£R

will

Es 2 /Xc

2

the line. For

sated line can. Moreover,

of maintaining a constant load voltage.

always be equal to the sender

irrespective of the active

volt-

power P absorbed

(E R

still

is

an upper limit to the

the line can transmit. that

we can

= Es ) up to

a

A

detailed analysis

maintain a constant load voltage

maximum

P max

of

2

one-half the line; the

Es

supplied by the sender

has the advantage

it

reactive

remaining half

If necessary,

.

add a second capacitor Xc (dash

we

can

line in Fig. 25.25a)

The source has then only to P, while the reactive power

at the input to the line.

supply the active power is

= Es 2 /X

Xc supplies XL absorbed by the

Capacitor

power I is

However, there

The voltage regulation is perfect until the load 2 power reaches the limiting value P nViiX = Es /X. The compensated inductive line can deliver twice as much power (f max ) as an uncompen-

the receiver

by the load.

power shows

b.

times equal to one-

power I X absorbed by such a compensated line the value of

half'the reactive

voltage

a.

get perfect regulation by adjusting

supplied by the capacitors

at

both ends.

(25.3)

Example 25-6

A

single-phase line possesses an inductive reac-

tance

X of

10 (1 and

is

voltage of 1000 V. If late the

a.

it

connected to a fixed sender fully

is

compensated, calcu-

following:

The maximum

active

power

that the line can

deliver to a resistive load b.

The capacitive reactance on the receiver

c.

The

that

must be

installed

side, in (a)

capacitive reactance that must be installed on

the receiver side

when

the active

power

is

40

kW

Solution a.

The maximum power the receiver

that

can be transmitted to

is

P nrdX = Es 2 /X= 1000 2/10 - 100 kW b.

Fig. 25.26

shows

the

compensated

line with the

capacitive reactance carrying a current Ic

.

The

phasor diagram of Fig. 25.27 gives us the key to solving the value of

Using

ER =

1000

V

Xc that

is

required.

as the reference phasor

reason as follows:

The

IR

Figure 25.25 Characteristics of a

current in the resistive load

compensated

inductive

line.

= P m JER = = 100 A

is

100 000/1000

we


684

x 10

Consequently,

= tt

2

100O7c = 5(/c + 100 2

5/c - f000/c + Solving

this quadratic /c

The value of Xc

)

X

5

10

we

equation

100

4

=

0

find

A

given by

is


=

2

Xc = E R /IC - 1000/100 = 10 ft = 100 A means E R by 45°. Consequently, £ s must be 90° ahead of E R Thus, the power transfer is maximum when the phase angle between E s and £ R is 90°. This is compatible with In Fig. 25.27, the fact that /c

that current / leads

,

Eq. 16.8 in Section 16.23. c.

When

the load

is

40 kW,

the load current

is

/c i

/R

- P/E R = - 40A

40 000/1000

Letting the current in the capacitor be /c the resulting line current is ,

=

/B

£R

100A

= 1000

V

= V/2 + 40 2


The The

The

current /c leads reactive

£R

reactive

Xc

is

Qc = Er?c

The

reactive

= iooo/c The

current

the line

/ in

/

=

= The

reactive

Full

\% +

2

I

X=

2

(IC

compensated

Qc =

+

line

1000 Ic

2

100

the line 2 )

Solving

0.5

=

)10

that

Gl

5 (40

this equation,

2

/C

the line

we

2

+

2 /c

)

find

is

Ic

10

we have

0.5 <2 L

X=(40 2 +

Qc =

/R

100

2

/

compensation requires

2

power absorbed by

QL = In a fully

+

power absorbed by

GL =

is

V£

is

Qc = /c^r = 1000/c

by 90°.

power generated by

Xc

power generated by

The value of Xc

8.35

A

is

= ER =

^ ^

=

1000

F35

=,,9

-

8il

is


10

685

11

REGION

REGION R

S

Figure 25.29

Es

leads

ER

.


The

circuit

sated line

We

is

and phasor diagram for given

in Fig.

E R and Es Es

note that although the values of

the same,

ER

compen-

this

25.28.

lags considerably behind

are

Es

and E R are fixed, each possessing the same magnitude E. Regarding the exchange of active power

between the two regional consumers, we examine three distinct possibilities:

.

1.

Es

and

2.

Es

leading

ER

by an angle 8

3.

Es

lagging

ER

by an angle 8

/.

E s and

25.23 Inductive line connecting

two systems Large

cities

and other regional users of

transmission lines. Such a network improves the

to better

system and enables

it

is

S supplies power

disturbances. Interconnecting lines also enable en-

nies.

utility

The 60 Hz frequency of such

a

sor diagram,

compa-

network

R in Phase In

zero and no power

active

sentially independent of

and

in

each other, both

phase. In effect, because of their

when an

infinite

buses.

if

is

is

Region

and, from the pha1

6.23) that the

given by

Er

sin

8

(25.4)

where

value

P =

E= X=

they

What happens

additional transmission line

(Fig. 25.29).

R

we can prove (Section

P = A

enormous

power, the regional consumers appear as

were independent,

in

transmitted.

to region

is

large regional users remain es-

this case, the line current

power transmitted

everywhere the same.

The voltages of

is

E s Leads E R by an Angle 8

2.

endure momentary short-circuits and other

ergy exchanges between electrical

phase

in

electrical

energy are always interconnected by a network of

stability of the electric utility

ER

put in ser-

two such regions? ^ shows the equivalent circuit of such an inductive line connecting two regional consumers S and R. We assume the terminal voltages

8

vice between

=

active

power transmitted per phase [MW'|*

line-to-neutral voltage [kV]

inductive reactance of the line, per phase

phase angle between the voltages

end of the

Fig. 25.29

*

at

[ill

each

line [°]

In this equation, if

E

power transmitted by

represents the line voltage, the three phases.

P

is

the total


686

power increases

(6 =

0° a 90°)

power: decreases (0 = 90° a 180°)

Figure 25.30a Power versus angle

Fig. 25.30a

from region S

100%

180

degrees

Figure 25.30b Voltage versus power characteristic.

characteristic.

shows

the active

to region

R as a

power transmitted

function of the phase

angle between the two regions. Note that the power

maximum

increases progressively and attains a

value of

E2 IX when

the phase angle

just as in the other transmission lines

mit only so

line.

the

is

transmit

still

we

power when

the

mode

8 approaches 90°, the

two regions

are

Fig.

In summary, there is always a limit to the amount of power a line can transmit. The maximum power is

proportional to the square of the sender voltage and

trip.

E R as

25.30b shows the load voltage

tion of the active

transmission

of op-

point of pulling apart, and the line circuit

at the

breakers will

power

transmitted.

horizontal line that stretches to a

P 11UIX = E~/X before

It is

maximum

back again

falling

a func-

simply a value

to zero (dot-

ted line). This voltage regulation curve should be

compared with

25.25b for a compen-

that of Fig.

sated line.

Note

REGION

Ex is voltages E s


even though the terminal equal

in

magnitude

clear that

ER

Ex

drop

3.

Es Lags Behind E R by an Angle it

ER

are it is

increases.

active

now

region

and

increases as the phase angle between

and

but

quite large,

(Fig. 25.29). Furthermore,

Es

The

R

side.

25.24 Review of power

avoid this condition

corresponds to an unstable

When

eration.

flows from the leading to the lagging voltage

compensated inductive

as that of a

Although we can

it

stud-

much power and no more. The power

same

phase angle exceeds 90°, because

we have

connecting two power centers can trans-

ied, a line

limit

90°. In effect,

is

power versus phase angle is identical to that shown in Fig. 25.30a. If we compare Figs. 25.29 and 25.31, we note that the direction of power flow does not depend upon the relative magnitudes of £ s and E K (they are equal), but only upon the phase angle between them. On inductive lines, active power always

power has flows

the

in the

8 (Fig. 25.3

same value

1

).

as before,

opposite direction, from

toward region

S.

The graph of

active

Figure 25.31

ER

leads

Es

.

S

REGION R

TRANSMISSION OF ELECTRICA L ENERG Y

inversely proportional to the impedance of the line. Fig. 25.32 enables us to

power and voltage

studied.

impedance of 10

Cl

Es

curves

of 1000 V.

become

Each model possesses an

and the sender furnishes a It is

flatter

ER

clear that the

and

flatter as

25.25 Choosing the line voltage

actual values of

for the four transmission line

models we have age

compare

volt-

versus

P

we progress from

We

have seen that for a given transmission

P max

that

E

where

can be transmitted

kW, whereas kW.

Because also

show

the inductive line can

some

resistance,

the voltage-power curve of a

transmit drops to 80 line possessing

no

maximum power

kW, compared

than

can

kW for a

resistance.

those

Nevertheless, the

100

given

method of

E — ^

E2

oc

P

Because Z is proportional

7 to the length of the line,


can be expressed by

E = kVpi

in

powers are much these

analysis

is

examples. the same.

(25.5)

where

E = 3-phase line voltage [kV] P = power to be transmitted [kW = length of the transmission line ]

/

In practice, the voltages and'

higher

to

it

oc

and so

compen-

O (curve 5). This fifth line also has an im-

pedance of 10 H, but the

imped-

we

sated line having a reactance of 9.8 Ct and a resis-

tance of 2

its

2

we deduce

lines possess

all

and

ance. Thus,

of 5 percent or better. Thus, the resistive line can transmit 4.75

is

the voltage of the line and Z,

is

P max

The table next to the graph shows the maximum power that can be transmitted assuming a regulation

line

maximum power 2 proportional to £ /Z,

for a given voltage regulation the

a resistive to an inductive to a compensated line.

transmit 30

687

k

=

coefficient that depends line tion.

[kmj

on the type of

and the allowable voltage regulaTypical values are


688

k

=

:

0.

for an

1

uncompensated

regulation of

=

k

From Table 25C we

having a

line

5%

XL =

H/km X 20 km =

0.5

10 ft

0.06 for a compensated line c.

Equation 25.5

very approximate, but

is

it

The !R drop

in the line is

does give 1R

an idea of the order of magnitude of the line voltage E.

find

The value finally chosen depends upon economic

The

=

drop

167

X

= 3006 V

18

in the line is

factors as well as technical considerations; in gen-

voltage selected will

eral, the actual

0.6

E and

1

.5 E.

Example 25-7 Power has to be feed a

1

0

carried over a distance of

MW unity power factor load.

km

20

to

Tf the line is

=

IX L

between

lie

X

167

=

10

1670

V

The line-to-neutral voltage across the load is 19 900 V. The complete circuit diagram per phase is given in Fig. 25.33a. The corresponding phasor diagram is given in Fig. 25.33b. The sender voltage

may be

calculated as follows:

uncompensated, a.

Determine the

Es = V(19 900 + = 22 967 V

line voltage

b.

Select an appropriate conductor size

c.

Calculate the voltage regulation

If the

3006)

load were removed,

22 967

V.

The voltage

2

+

1670

ER would

regulation

is,

2

rise to

therefore,

Solution a.

Because the

line is not 1

1

.

E =

-

0.1

=

44.7

as-

regulation

equal to

is

(25.5)

V10 000 X 20

Note

and

1.5

that this

(22 967

=

3967/19 000

=

0.209 or 20.9%

44.7

kV (= 67 kV)

is

19 000)/ 19 000

medium-voltage

line is

more

re-

3006 V

.1670 V.

X

-

=

sistive than inductive.

kV

voltage between 0.6

44.7

MW

V P/

k

Any

X

we

compensated,

sume k = 0. A power of 0 10 000 kW. Consequently,

kV (= 27 kV) feasible.

We

shall use a standard line voltage of 34.5 kV.

The


E = b.

The conductor

34.5/ V3 size

/

S/(V3E)

=

10

=

167

According

ACSR

X

The

MW

19 900

line current is

Figure 25.33a Transmission

line

under load.

(8.9) 6

10 /(1.73

X

34 500)

96^

A

to Table

V

= 19.9kV

depends mainly upon the

current to be carried.

=

3.33

is

1670

25D, we can use a No.

= 200 A R = 0.9 ft/km X 20 km =

1

167

conductor:

ampacity

A

19 900

V

Figure 25.33b 18

a


3006

V

V

TRA NSMISSION OF ELECTRICA L ENERG Y

689

25.26 Methods of increasing the power capacity High-voltage lines are mainly inductive, possessing a reactance of about 0.5 fl/km. This creates prob-

lems when we have to transmit large blocks of power over great distances. Suppose, for example, that 4000 has to be transmitted over a distance of 400 km. The reactance of the line is 400 km X 0.5 il/km — 200 H, per phase. Since the highest

MW

practical line-to-line voltage

about 800 kV, the

is

3-phase line can transmit no more than 2 P max - E IX ^ l/X 1

= 800 2 /200 = 3200 MW

MW,

To transmit 4000 two

lines in parallel,

(25.3)

Figure 25.34

the only solution

one beside the

other.

to use

is

Note

that

doubling the size of the conductors would not help,

because for such a

line

it

is

the reactance and not

Two 735 kV transmission electrical

posed

of

lines

in

parallel carrying

energy to a large city. Each phase is com4 bundled conductors (see Fig. 25.7).


the resistance of the conductors that determines the

maximum power that

can be transmitted.

With

Additional lines are also useful to provide sys-

tem security out,

due

in the

the scheduled

maining

power can

Thus, still

if

one

line

arrangement, the

maximum power

be carried by the re-

P max = E2 /(XL - Xcs where

Xcs

is

we sometimes

and even four transmission

erect two, three,

lines in

which follow the same corridor across

(25.6)

)

the reactance of the series capacitors

per phase. Such series compensation

large blocks of power,

is

trips

is lost,

line.

To carry parallel,

event that a parallel line

to a disturbance.

this

given by

is

also used to

regulate the voltage of medium-voltage lines

when

the load fluctuates rapidly.

the

countryside (Fig. 25.34). In addition to high cost, the use of parallel lines often creates serious prob-

25.27 Extra-high-voltage lines

lems of land expropriation. Consequently, special

When

methods

voltages, special problems arise that require the in-

mum

are

sometimes used

power of a

to increase the

X

when we can no we try to reduce the

line. In effect,

longer increase the line voltage, line reactance

maxi-

by greatly increasing the effective

x

is done by using two or more conductors per phase, kept apart by spacers. Such bundled conductors can reduce the

diameter of the conductors. This

much

reactance by as

as

40 percent, permitting an

increase of 67 percent in the power-handling capability

of the

in series

with

the three lines to artificially reduce the value of

XL

.

energy

stallation of large

is

transmitted

at

extra-high

compensating devices to regulate

the voltage and to guarantee stability.

Among

these

devices are synchronous capacitors, inductive reactors, static

var compensators, and shunt and series

capacitors.

To understand

the need for such devices, and to

appreciate the magnitude of the powers involved,

consider a 3-phase, 735 kV, 60 length of

line.

Another method uses capacitors

electrical

600 km. The

Hz

line,

line operates at

having a

727

kV and

the inductive and capacitive reactances are respectively 0.5 fl

and 300

kH

for each kilometer of

690


length.

We

first

determine the equivalent circuit of

the transmission line per phase:

Mvar)

Sender voltage per phase (line-to-neutral)

is

XL -

.

have a capacity of 176 Mvar, per phase.

Despite this inductive compensation,

we

still

have a reactive power of 176 Mvar, generated by

Xcl

is

X 600 = 300

0.5

XC2 (420 /1000 = 176 XL2 The latter must,

by

entirely absorbed

is

therefore,

E s = 727/V3 = 420 kV Inductive reactance per phase

power generated by

reactive

which has

,

be absorbed by the synchronous

to

generator G. However, a capacitive load

17

at the ter-

minals of a generator creates overvoltages, unless Capacitive reactance per phase

xc =

300 kH/6oo

we

is

n

= 500

reduce the alternator exciting current (Section

16.13).

But underexcitation

because Equivalent capacitive reactance the line

each end of

at

is

= XC2 = 2 X 500 = 100012

*ci

The equivalent

circuit,

now

Fig. 25.35. Let us

it

install a

Mvar

close to the generating station. In the case of

second inductive reactance of 176

line to distribute the inductive

per phase,

is

shown

in

over

its

Inductive reactors (fixed or variable) are com-

study the behavior of the line

No-Load Operation. At no-load with

XC2

produces a

nance and the terminal voltage In effect,

we

the circuit

ER

formed

partial reso-

rises to

600 kV.

split

coil

placed inside a tank and im-

in oil (Fig. 25.37).

up into a

magnetic

A

laminated steel core

series of short air

flux. Intense

oped across the

obtain:

compensation evenly

length.

mersed

in series

1500 km),

to

several inductive reactances are installed along the

posed of a large

XL

km

very long transmission lines (500

Ct

recommended we

not

must

under no-load and full-load conditions.

by

is

leads to instability. Consequently,

gaps carries the

magnetic forces are devel-

air gaps.

On

a 60

Hz

system, these

forces continuously oscillate between zero and sev-

Xq 2

eral tons, at a

X

Xqi

mechanical frequency of 20 Hz. The 1

core laminations and

\ j

iooo

= 420 kV X 1000

^-

a

fore,

300

Q

- 600 kV

metallic parts must, there-

all

be firmly secured

to

reduce vibration and

to

limit noise to an acceptable level.

Operation Under Load, Characteristic Impedance. This represents an increase of 43 percent above the

kV

nominal voltage of 420 normally high voltage

way

sible

to reduce

XL2

actance

X L2

at the

it

(Fig. 25.35).

Such an ab-

The only

is

unacceptable.

is

to connect an inductive re-

end of the

fea-

line (Fig. 25.36). If

we

XC2 the resulting parallel resonance brings voltage E R back to 420 kV. In effect, the make

equal to

,

300

Returning again to the uncompensated line on opencircuit (Fig. 25.35), let us

power

we

connect a variable unity

factor load across the receiver terminals. If

progressively increase the megawatt load, re-

ceiver voltage

ER

will gradually decrease

open-circuit value of

load

it

will

become

600 kV and

for

from

its

one particular

exactly equal to the sender volt-

a

300

n

L l

*C2

420 kV

1

Tooon

transmission

420 kV*

100012

1000O

Figure 25.36

Figure 25.35

EHV

Tooon

line at

no-load.

EHV

l

*C2

Ci

£r=600i
reactor compensation.

o / o OV /

L2 =

420 kV


691

Figure 25.37 Three large 1 10 Mvar, single-phase reactors installed in line

a substation

to

compensate

the

capacitance of a very long 3-phase 735

kV transmission line. (Courtesy of Hydro-Quebec) age

Es

(Fig. 25.38).

This particular load

is

called the

active

power generated by

surge-impedance load. For most

aerial lines the im-

line is equal to that

pedance of this load corresponds

to a line-to-neutral

line, in effect,

load resistance of about

400

II per phase. This par-

ticular load resistance (called surge

impedance)

is

independent of the system frequency or the length of the line.

The surge impedance load (SIL) of

3-phase transmission line

therefore, given

is,

by

a

= £2 /400

compensates

its

still

limited to the value given

the

2

P max =

3

X

(E IX)

=

3

X

(420 /300)

2

(25.7)

= 1764MW

= E=

surge impedance load

[MWJ

3-phase line voltage [kV]

In Fig. 25.38, the total surge-impedance load 2

approximately 727 /400

When

-

1320

its

V>/420kV

power

surge-impedance load, the

300

1( 1000 ft

Figure 25.38 Surge impedance loading

is

MW.

a transmission line delivers active

corresponding to

re-

n

1000f2

420 kV,

Figure 25.39 compensator for HV [Courtesy of General Electric) Static var

of

a

line.

The

the load ex-

by Eq. 25.3, namely

where

SIL

inductance.

itself. If

ceeds the surge-impedance load, we can keep £ R at 420 kV by adding extra capacitors at the receiver end of the line. However, the maximum power is

approximate equation:

SIL

the capacitance of the

absorbed by

line.


692

If the

load

is less

than the surge- impedance load,

must add inductive reactance

at the receiver

we

end of the

line to maintain a constant voltage. Conversely, if the

load

is

greater than the surge-impedance load, a ca-

pacitive reactance

must be added. Because the load

transmission lines (not shown) (Fig. 25.40). spective voltages

Eb

by an angle

Eu

Eb

and

we

8. If

are equal, but

The

ZTa

re-

leads

decide to connect the two

centers by an extra transmission line having a reac-

power P

tance X, the active

A

will automatically flow

B because £ a

Eb

continually changes throughout the day, the magni-

from

tude of the capacitive and inductive reactance must

25.23). Furthermore, phase angle 8 and reactance

continually be varied to keep a steady voltage. This

will completely dictate the

done by means of static var compensators or rotating synchronous machines. liver or

The

is

(Fig. 25.39),

latter

1

7.

1

transmitted because

5).

we wish

if

We

sometimes have line

to install

tightly interconnected. to

Such a

that

line

may

are

Ed

be required

meet the energy needs of a rapidly growing area

or to improve the overall stability of the network. In

such cases,

we

we want

Eb cm force

leads

However, we

already

(E /X) sin

8.

to achieve.

is

by

a

power exchange

artificially

to

introduce

a

phase-shift

in

one

modifying the

we have

to

autotransformer

(Section 12.11) at one end of the line; by varying the

phase angle of

this transformer,

control the active

we can

completely

power flow between the two centers.

use special methods so that the ad-

ditional line will transmit the required power.

Consider, for example, two major power centers

A and B that are already

interconnected by a grid of

Example 25-8 Fig.

25.41a shows the voltages and phase angle be-

tween two regions

A

and B

that are already inter-

connected by a network (not shown). Voltage

known

to lead

£a

value of 100 kV.

X=

20

by

11 °,

A new

fl connects the

Eb

tie line

having a reactance

two regions.

The power transmitted by the line and the dipower flow, if no phase-shift transformer is employed

rection of

Figure 25.40 Power flow between two regions.

x = 20n

Figure 25.41a ordinary transmission line causes

power

to flow in the

wrong

direction.

is

and both voltages have a

Calculate a.

An

B

.

phase angle between the two regions. All

do

For ex-

energy from region

A, the installation of a simple line will not

do, because

an additional trans-

between regions

X

magnitude of the power

to transmit

direction or the other

mission

(see Section

2

P=

not correspond to what

ample,

to region

25.28 Power exchange between power centers

leads

However, the magnitude and direction of P may

can de-

absorb reactive power, according to whether

they are over- or under-excited (Section

to


100 kV + 19°

100 kVL2?

693

100 kV I+H°

t

autotransformer

Figure 25.41b

A b.

power

phase-shift autotransformer can force

The required

to flow

in

the desired direction (Example 25-8).

from which

phase-shift of the transformer so

that the line will transmit

70

MW from A to B

Solution a.

Consequently, voltage

The power transmitted

in Fig.

P = (E2/X) sin 5 = (!00 2/20) sin

25.4 la

is

given by

order that 70

(25.4)

=

ll°

95.4

B b.

Eb

leads

£a

the 95.4

,

MW

MW will flow from

lows that

E b already Ed must be

ends of the transmission transmit 70

MW. We P= 70 =

sin 8,

=

so that

it

° 4-

(Fig.

19°

it

fol-

ahead of

therefore, proits

primary

and secondary windings, and the secondary

will

voltage

Ed must

lead the primary voltage E. v

We

2

the line or even in the middle. sin 8,

=

8°

have

(E /X)

8° in

B

x

1

duce a phase-shift of 19° between

between opposite

line,

to

leads E. by 11°, 1

The autotransformer T must,

E. v lb, let us first calculate

the phase angle 8, required

Eb by

25.41b). Referring to the phasor diagram and

to A.


produced by the

MW may flow from A

noting that

Because

Ed

phase-shift transformer must lead

can put the autotransformer

at either

end of

(25.4)

2

(l00 /20) sin

8,

25.29 Practical example

0.14

of

'^CONNECTICUT Norwalk,,

We now

power exchange

consider a practical application of a phase-

shift transformer.

tween the

state

To

facilitate

power exchange

of Connecticut and Long Island,

York, six single-phase

1

38

be-

New

kV submarine cables (two

per phase) were installed between Norwalk and Northport, 25.42).

at

the bottom of

Long

connected by a grid of transmission ground,

it

was decided

transformer

at

imum

lines

MW.

phase-shift of

Norwalk, a 300

(Fig. inter-

above

to install a phase-shift auto-

Northport to control a

power flow of 300 at

Sound

Island

Because the two regions were already

maximum

Variable taps enable a max-

±25

degrees.

On

the other side,

MVA variable- voltage

autotrans-

Figure 25.42

former was installed to provide voltage control of up

Six single-phase submarine cables join the State of

to 10 percent, without phase-shift.

Connecticut to Long Island.

phase-shift and the voltage at either end of the 19

By

varying the

km

694


line,

it is

two

the

possible to control the regions, in

power flow between

one direction or the

other, de-

pending on the need. The following technical

show 750

a

the magnitude of the

powers involved

details in

Each of

the six single-phase cables (Fig. 25.44)

possesses a resistance of 1.3

Figure 25.43

tance of

Equivalent circuit of each submarine cable.

The

1

latter

.

fl,

an inductive reac-

H, and a capacitive reactance of 375 (1. can be represented by two reactances of 1

Figure 25.44 is one of seven cables submerged and Norwalk (Connecticut).

Cross-section view of submarine cable (138 kV, 630 A). This cable Island

Sound between Northport (Long

(Courtesy of

Pirelli

Cables Limited)

Island)

such

a cable transmission system.

in

Long


750 H

each end of the cable. The full-load current

at

per cable

is

630 A, and

the line-to-neutral voltage

80 kV. Referring to the equivalent for one cable (Fig. 25.43),

we can

circuit

it.

and

These powers

,

,

25-3 }

d.

Sag of a transmission

e.

Reactance of a

Why

IN

A SUBMARINE CABLE

must transmission

long, transmits a

Power

Total

installation

per cable

power

Is

line

towers be

ground?

A 735 kV transmission a.

Submarine cable

line

line

solidly connected to

25-4

POWERS

Suspension-type insulator

Ground wire Corona effect

resistive

g L Qc P

are listed in Table 25 E.

TABLE 25E

a. b. c.

diagram

readily calculate

the value of the inductive, capacitive,

powers associated with

is

695

line,

745 miles

power of 800

MW.

there an appreciable voltage difference

between the two ends of the

line,

measured

line-to-neutral? 2

R

I

b.

losses:

P = 630 2 X .3 = 0.516 MW s

0.5I6MW

3.1

MW

25-5

Reactive power generated:

Qc = =

2

17

areas,

two

identical 3-phase

Mvar

102

Mvar

these two lines by a single line, by simply doubling the size of the conductors?

= 630 2 X = 0.436 Mvar 1

some

In

by separate towers. Could we replace

Mvar

Reactive power absorbed: <2 L

there an appreciable phase angle between

lines are installed side-by-side, supported

(80 000) /375 17.06

Is

corresponding line-to-neutral voltages?

1

Explain.

.

0.44

Mvar

2.6

Mvar 25-6

Why

do we seldom

install

underground

Active power transmitted:

cable (instead of aerial transmission lines)

P = 630 X 80 000 - 50 MW

50

MW

300

between generating

MW

stations

and distant

load centers? In

comparing them with the active power

mitted (300

MW), we

of the cable

is

far

more important than

tance or inductance.

mous

A cable

behaves

are like

capacitor, contrary to an aerial line

mainly inductive.

trans-

25-7

In Problem 25-4 the average line span is 480 m. How many towers are needed between the source and the load?

25-8

A 20 km

can see that the capacitance its

resis-

an enor-

which

is

1

It is

precisely the inherent large

shock

transmit energy over long distances. This restriction is

in

to

used because 25-9

capacitance then has no effect.

kV

transmission line operating

A lineman

he does not

if

could receive a

first

connect the

ground before touching

What

is

High-voltage dc transmission lines are covered

ACSR

Chapter 28.

can

a

the ampacity of a

it.

600 kcmil

cable suspended in free air?

1

classes.

Name them

and

state the ap-

proximate voltage range of each. 25-2

Explain what terms:

is

con-

Intermediate level

Standard voltages are grouped into four

main

Why

copper conductor having the same

siderably greater?

Practical level

25-

fatal

line

Explain.

cross-section carry a current that


at

has just been disconnected from

the source.

capacitance that prevents us from using cables to

does not apply when direct current

3.2

is

meant by the following

25- 1 0

Each phase of the two 735 kV

lines shown composed of 4 bundled subconductors. The current per phase is 2000 A and the resistance of each subcon-

in Fig.

25.34

ductor

is

is

0.045 (I/km.


696

Calculate a.

The

power transmitted by both

total

unity b.

power 2

The

total I

lines at

loss,

The I~R

knowing

25-17

the lines are

1

loss as a percent of the total

A

power

single-phase transmission line possesses

R

a resistance

Es

of

1

5 II (Fig. 25.2

1

a. Calculate the terminal voltage

power P absorbed by

b.

The following information

is

(2.

Draw

the

the load

successively 285

ER

when

11,

c.

Inductive reactance X, of the

power

In

Problem 25-

angle between

1

what

,

1

ER

and

II,

is

1

reactive

line current /

c.

The

line in

25- 8

Problem 25-

1

1

is

25- 9

Problem 25- 17, what

In

the terminals of

the load and the

absorbs for the

same impedance b.

25-14

Draw

25- 5 1

ER

Problem 25-13, what is the phase angle between E R and £ s when the load impedance is 45 II? Does E R lead or lag

A

single-phase transmission line possesses

by a source

line for the

285

£s

25-16

It is

The

sup-

of 6000 V.

ER

at the

xL

=

xc

= 300

the

0.5 I) r

impedances are

=

25-20

What

meant by

is

0.25

O

kfl

reduced

Advanced

X

to that

v

,

R,

and

shown

Xc if the

in Fig. 25.14.

the term surge imped-

25-21

A

level

3-phase 230

nects

kV

transmission line hav-

two regions

II,

per phase, con-

that are

50 miles

apart.

following capacitive loads:


the load

is

45

ER

and

II.

re-

15 a).

ER

at the

end of the

line for a capacitive load

of 45

II.

Calculate the voltage

the

two ends of

the line

is

the voltages at 20°.

Calculate

Problem 25-15 possesses a

sistance of 15 II (instead of a reactance of

a.

which

ing a reactance of 43

a 45 n.

line in

in

circuit is

end of the

Calculate the phase angle between

Es when

the phase angle

each section of the

3,

1

Calculate the values of

£s ?

a. Calculate the voltage

b.

is

£s ?

ance load?

an inductive reactance of 15 H. plied

km

as a function of P.

In

behind

1

values.

the graph of

and


of

Calculate the voltage

ER

circuit represents a transmission line length

H (Fig. 25.22a). E R at power P it

the capacitor

6kV

1

reactance of 15

with the

the load

replaced by another having an inductive

a.

15 (2

reactive power absorbed by the line The reactive power supplied by the sender The apparent power supplied by the sender The voltage E s of the sender The voltage E R and the power P when E s is

the phase

45 II?

The transmission

power supplied by

The

The

between 25- 3

line:

in parallel

II

a.

f.

1

is

50

b.

e.

£R

P.

£ s when

1

Xc

Xc in paralwith the source, calculate the following:

lel

g. 1

:

Neglecting the dotted reactance

the im-

45 H, 15

E R 6000 V

Terminal voltage

Equivalent load resistance: 45 il

load:

and the

graph of the terminal voltage

as a function of the

25- 2

raise the

given for

a.

d.

and 5

is

b.

is

pedance

we

by connecting

it?

d. Capacitive reactance

The

a).

6000 V, and the impedance of the unity power factor load varies between 285 I) and 5 II. source

line

Fig. 25.25a:

transmitted

25-1

end of the

a capacitor across

factor load

R

purely resistive, can

vol.ta.ge at the

350 miles long c.

If a line is

b.

power transmitted by

c.

The The The

total reactive

d.

The

reactive

a. b.

active

the line

current in each conductor

each region

power absorbed by the line power supplied to the line by

TRANSMISSION OF ELECTRICAL EN ERG Y

25-22

A 3-phase aerial

line

connected

to a 115

3-phase source has a length of 200 km.

composed of

three

600 kcmil-type

kV

25C and 25D,

if

there

is

ACSR

e.

25-23

XL

Xc

b.

The voltage between conductors

,

and

,

The current drawn from phase

power received by

2

total I

R

at the

the source, per

load

the

loss in the line

the source

if a

Compare

drawn

line,

this current

the conductors.

the current

3-phase short occurs

across the end of the b.

per phase

(open end) c.

total reactive

Problem 25-22, calculate

from

no load on

The value of R,

The

a. Tn

the line, calculate the following: a.

The

source

It is

conductors. Referring to Fig. 25. 17 and

Tables

d.

697

with the ampacity of

Chapter 26 Energy

Distribution of Electrical

consuming

26.0 Introduction

centers.

The

distribution substations

we mentioned

Chapter 25

In power system

that

an electrical

equipment

in

such

similar to that found in sub-

stations associated with generating plants.

composed of high-voltage transmission lines that feed power to a medium-voltage (MV) network by means of substations. In North America these MV networks generally operate at is

voltages between 2.4

kV

and 69 kV. In

26.1 Substation

A

that function

In this chapter,

between 120

V

turn, they

and 600

V.

we cover the following main topics:

1.

Substations

2.

Protection of medium-voltage distribution

equipment

medium-voltage substation usually contains the

following major components:

supply millions of independent low-voltage sys-

tems

electrical

is

Transformers

Surge arresters

Circuit breakers

Current-limiting reactors

Horn-gap switches

Instrument transformers

Disconnect switches

Relays and protective devices

Grounding switches In the description that follows,

systems

sic principles

3.

Low-voltage distribution

4.

Electrical installation in buildings

understand

we

study the ba-

of this equipment. Furthermore, to

how

it

all fits

together,

we conclude

our

study with a typical substation that provides power to a large suburb.

SUBSTATIONS 26.2 Circuit breakers

Substations are used throughout an electrical sys-

tem. Starting with the generating station, a substation raises the

Circuit breakers are designed to interrupt either nor-

medium-voltage generated by the

mal or

synchronous generators to the high-voltage needed to transmit the

energy economically.

The high transmission-line voltage in

short-circuit currents.

switches that

They behave

may be opened

like big

or closed by local

pushbuttons or by distant telecommunication is

then reduced

sig-

nals emitted by the system of protection. Thus, cir-

those substations located close to the power-

cuit

698

breakers will

automatically

open a

circuit

DISTRIBUTION OF ELECTRICAL ENERGY

whenever

the line current, line voltage, frequency,

and so on, departs from a preset

The most important types of circuit breakers

are

.

that has to

carries the line current of the

be protected.

If

Oil circuit breakers

lay contacts

(OCBs)

C b C2

the tripping coil

to close.

is

As soon

Air-blast circuit breakers

source. This causes the three

3.

SF 6

open, thus interrupting the circuit.

4.

Vacuum

circuit breakers

/.

circuit breakers

cates (1) the

maximum in cycles.

a circuit breaker usually indi-

maximum steady-state current it can maximum interrupting current, (3) the

line voltage,

The

and

time

(4) the interrupting

interrupting time

may

last

from 3

In

Hz system. To interrupt large currents we have to ensure rapid deionization of

combined with rapid cooling. High-speed damage to transmission lines and equipment and, equally important, it helps to maintain the stability of the system whenever a con-

set

of fixed

movable contacts, actuated simulta-

neously by an insulated rod, open and close the

When the circuit breaker is closed,

porcelain bushing, flows through the contact, the

interruption limits the

tact,

movable

cir-

the line cur-

rent for each phase penetrates the tank by

the arc,

oil.

(Fig. 26.2), three porcelain bushings

channel the 3-phase line currents to a contacts. Three

cuit.

this quickly,

line contacts to

a steel tank filled with insulating

one version

to 8

cycles on a 60

main

Oil Circuit Breakers. Oil circuit breakers are

composed of The nameplate on

as they close,

energized by an auxiliary dc

2.

carry, (2) the

phase

the line current exceeds

a preset limit, the secondary current will cause re-

the following: 1

nected to the secondary of a current transformer.

The primary

limit.

699

way

first

of a

fixed

contact, the second fixed con-

and then on out by a second bushing.

tingency occurs.

The

triggering

breaker to open

is

action

that

causes

usually produced by

a

circuit

means of an

overload relay that can detect abnormal line conditions.

For example, the relay

coil in Fig. 26.

1

is

con-

transmission line

Figure 26.2 Cross-section of an

shows Figure 26.1 Elementary tripping

a

circuit breaker.

circuit breaker.

The diagram

keeps the a satisfactory temperature during cold weather. (Courtesy of Canadian General Electric)

oil

circuit for

oil

at

four of the six bushings; the heater


700

If

an overload occurs, the tripping coil releases a

powerful spring that pulls on the insulated rod, causing the contacts to open.

As soon

arate, a violent arc is created,

surrounding

oil.

as the contacts sep-

which

volatilizes the

The pressure of the hot gases

creates

turbulence around the contacts. This causes cool to swirl

around the

arc, thus

extinguishing

oil

modern high-power breakers, the arc is conchamber so that the pressure of the hot gases produces a powerful jet of oil. The jet is made to flow across the path of the arc, to exIn

it.

Other types of

signed so that the arc

is

circuit breakers are de-

deflected and lengthened by

a self-created magnetic field.

The

arc

is

blown

against a series of insulating plates that break up the arc

and cool

it

down.

Figs. 26.3

Air-Blast

and 26.4 show the

appearance of two typical OCBs.

Breakers. These

Circuit

breakers interrupt the circuit by

pressed air

at

circuit

blowing com-

supersonic speed across the opening

Compressed

contacts.

air is stored in reservoirs at a

MPa (—435

pressure of about 3

and

psi)

is

replen-

The

ished by a compressor located in the substation.

most powerful

it.

fined to an explosion

tinguish

2.

circuit breakers

40

short-circuit currents of

765 kV

in

kA

can typically open at

a line voltage of

a matter of 3 to 6 cycles on a 60

The noise accompanying

the air blast

is

Hz

line.

so loud that

noise-suppression methods must be used

when

the

circuit breakers are installed near residential areas.

Fig. 26.5

breaker.

shows a

typical 3-phase air-blast circuit

Each phase

modules connected

is

composed of

in

series. Fig.

three contact

26,6 shows a

cross-section of the contact module. 3.

SF 6

Circuit Breakers. These totally enclosed

circuit-breakers, insulated with

SF 6

gas*, are used

* Sulfur hexafluoride.

Figure 26.3 Three-phase oil circuit breaker rated 1200 A and 1 15 kV. It can interrupt a current of 50 kA in 3 cycles on a 60 Hz system. Other characteristics: height: 3660 mm; diameter: 3050 mm; mass: 21 t; BIL: 550 kV. {Courtesy of General Electric)

Figure 26.4 Minimum oil circuit breaker installed in a 420 kV, 50 Hz substation. Rated current: 2000 A; rupturing capacity: 25 kA; height (less support): 5400 mm; length: 6200 mm; 4 circuit-breaking modules in series per circuit breaker. (Courtesy of ABB)


movable contact

fixed contact

701

actuating rod

Figure 26.6 Cross-section of one module of an air-blast breaker.

When

the circuit breaker

circuit

the rod

trips,

is driv-

en upward, separating the fixed and movable contacts. The intense arc is immediately blown out by a jet of

compressed

sistor

dampens

air

coming from the

The

re-

when

the

orifice.

the overvoltages that occur

breaker opens.


Electric)

Figure 26.5 Air blast circuit breaker rated

a current of 40 kA

interrupt

2000 A

in

at

362

kV.

It

can

3 cycles on a 60 Hz

consists of 3 identical modules connected each rated for a nominal voltage of 121 kV. The compressed-air reservoir can be seen at the left. Other characteristics: height: 5640 mm; overall length: 9150 mm; BIL 1300 kV. {Courtesy of General Electric)

system. in

It

series,

whenever space

town substations are

much

is at

a premium, such as in

(Fig. 26.7).

These

down-

circuit breakers

smaller than any other type of circuit

breaker of equivalent power and are far less noisy than air circuit breakers. 4.

ers

Vacuum

Circuit Breakers. These circuit break-

operate on a different principle from other

breakers because there contacts open.

They

is

no gas

to ionize

when

the

are hermetically sealed; conse-

quently, they are silent and never

become polluted

(Fig. 26.8). Their interrupting capacity

is

limited to

about 30 kV. For higher voltages, several circuit

Figure 26.7 Group of 15 totally enclosed SF 6 stalled

Vacuum

circuit breakers are often

ground distribution systems.

in

under-

of

breakers

a large

in-

city.

Rated current: 1600 A; rupturing current: 34 kA; nor-

breakers take up only 1/16 of the volume of

conventional

used

circuit

an underground substation

mal operating pressure: 265 kPa (38 psi); pressure during arc extinction: 1250 kPa (180 psi). These SF 6 circuit

breakers are connected in series.

in

circuit

rupting capacity.

(Courtesy of ABB)

breakers having the

same

inter-


702

26.4 Disconnecting switches Unlike air-break switches, disconnecting switches are unable to interrupt any current at

when

only be opened and closed

They

They must

all.

the current

is

zero.

are basically isolating switches, enabling us to

isolate oil circuit breakers, transformers, transmis-

sion

and so

lines,

forth,

from a

network.

live

Disconnecting switches are essential to carry out

maintenance work and Fig. 26.12

switch.

It

to reroute

shows a 2000 A,

power

15

kV

equipped with a latch

is

flow.

disconnecting

to prevent the

switch from opening under the strong electromagnetic forces that

accompany

engaged by pulling the

short-circuits.

The

latch

is dis-

movable blade out of the fixed

Fig. 26. 13

shows another "disconnect"

a larger current, but at a

contact. that carries

much lower voltage.

opened by means of a manual hookstick.

It,

too,

shows how

the fixed

is

Fig. 26.14

shows another type of disconnecting switch and 26. 15

and

inserting a hookstick into the ring

Fig.

and movable contacts en-

gage. Fig. 26.16 shows maintenance personnel work-

Figure 26.8 Three-phase vacuum circuit breaker having a rating of 1200 A at 25.8 kV. It can interrupt a current of 25 kA in 3 cycles on a 60 Hz system. Other characteristics: height: 2515 mm; mass: 645 kg; BIL: 125 kV. (Courtesy of General Electric)

ing on a large disconnecting switch in a

26.5 Grounding switches Grounding switches a transmission line

are safety switches that ensure

is

grounded while

definitely

pairs are being carried out. Fig. 26.

3-phase switch with the blades

26.3 Air-break switches

zontal) position.

Air-break switches can interrupt the exciting currents of transformers, or the

moderate capacitive

currents of unloaded transmission lines.

They can-

not interrupt normal load currents.

HV substation.

all

To

1

re-

7 shows such a

open

in the

(hori-

short-circuit the line to ground,

three grounding blades swing up to engage the

connected

contact

stationary

Grounding switches

when

are

to

each

phase.

opened and closed only

the lines are de-energized.

Air-break switches are composed of a movable blade that engages a fixed contact; both are mounted

on insulating supports

(Figs. 26.9, 26.10).

Two

ing horns are attached to the fixed and movable contacts.

When

the

main contact

is

broken, an arc

up between the arcing horns. The

arc

produces and the magnetic

becomes longer 26.1

1

).

until

it

field.

As

is

set

moves upward

due to the combined action of the hot

26.6 Surge arresters

arc-

air currents

the arc rises,

it

it

eventually blows out (Fig.

The purpose of a surge arrester* voltages that

electrical apparatus

ing surges.

due either

to limit the over-

to lightning or switch-

The upper end of the

arrester

to the line or terminal that has to

the lower

is

may occur across transformers and other

end

is

is

solidly connected to ground.

Although the arcing horns become pitted and

gradually wear out, they can easily be replaced.

*

connected

be protected, while

Also called lightning

arrester, or surge diverter.

Figure 26.9

One

pole of a horn-gap disconnecting switch rated 600 A, 27 kV, 60 Hz;

(left) in

the open position,

(right) in

the

closed position.

(Courtesy of Dominion Cutout)

Figure 26.10 pole of a 3-phase 3000

One

60 Hz horn-gap disconnecting switch in the open position (left); in the be operated manually by turning a hand wheel or remotely by means of a motorized drive located immediately below the hand wheel. Other characteristics: height when closed: 12 400 mm; length: 7560 mm; mass: 3 1; maximum current-carrying capacity for 10 cycles: 120 kA; BIL: 2200 kV. (Courtesy of Kearney)

closed position

(right).

The

A,

735

kV,

switch* can

703


704

Figure 26.11 arc produced between the horns

The

as

ing switch

it

Figure 26.13 of

a disconnect-

cuts the exciting current of a

former provides the

light to

HV

trans-

Disconnecting switch rated 10 kA,

1

kV

for

indoor use.

(Courtesy of Monteli Sprecher and Schuh)

take this night picture.

{Courtesy of Hydro-Quebec)

Figure 26.14

Figure 26.12 This hookstick-operated disconnecting switch

2000

A, 15

kV and has a BIL

of

95

is

rated


current,

way

any voltage

if

need be,

maximum, by to

be diverted

the arrester absorbs energy

surge.

The E-I

for

sidewise

(Courtesy of Kearney)

Ideally, a surge arrester clips

cess of a specified

Disconnecting switch rated 600 A, 46 kV operation.

kV.

in

ex-

permitting a large to

ground. In

this

from the incoming

characteristic of an ideal surge ar-

rester

is,

therefore, a horizontal line

corresponds to the

maximum

whose

level

permissible surge

voltage. In practice, the E-I characteristic slopes up-

ward

(Fig. 26.18) but

sonably

flat.

is still

considered to be rea-


current

limited by the resistance of the valve

is

blocks and the arc

cooled

705

is

simultaneously stretched and

a series of arc chambers.

in

quickly snuffed out and the arrester

is

The

arc

The

protect the line against the next voltage surge.

discharge period

is

is

then ready to

very short, rarely lasting more

than a fraction of a millisecond.

A more modern type of arrester has valve blocks made of any

air

stacked zinc-oxide discs without using

gaps or other auxiliary devices.

acteristic rester,

more

is

Its

E-I char-

similar to that of a silicon carbide ar-

except that

it

is

much

and therefore

flatter

effective in diverting surge currents. These

metal-oxide varistor

(MOV)

arresters are largely

used today. Lightning arresters with very also enable us to reduce the

paratus installed

HV

and

EHV

in

flat

characteristics

BIL requirements of ap-

On

substations (Section 25.12).

systems, the reduction in

BIL

signif-

icantly reduces the cost of the installed apparatus. Fig.

26.19 shows a lightning arrester installed

EHV

in

an

substation.

Figure 26.15

The blade in tight

of

a vertical motion disconnecting switch

is

26.7 Current-limiting reactors

contact with two fixed contacts, due to the

pressure exerted by two powerful springs. switch opens, the blade twists on

its

When

axis as

it

the

MV

bus

eral feeders,

upward. During switch closure the reverse rotary

movement

The

moves

a substation usually energizes sev-

in

which carry power

to regional load

centers surrounding the substation.

exerts a wiping action against the fixed

impedance of the

contacts, thus ensuring excellent contact.

that the output

(Courtesy of Kearney)

very low. Consequently,

if

It

so happens

MV bus

is

usually

a short-circuit should oc-

cur on one of the feeders, the resulting short-circuit current could be disastrous.

Some lain

arresters are

composed of an

external porce-

tube containing an ingenious arrangement of

stacked discs, air gaps, ionizers, and coils. The discs (or valve blocks) are material.

The

composed of a

silicon carbide

resistance of this material decreases dra-

matically with increasing voltage.

Under normal voltage conditions, spark gaps column. Consequently, the resistance of the

rester

is infinite.

However,

if

ar-

down and the surge The 60 Hz follow-through

220

and a nominal secondary current of 1600 A.

supplies the

MVA,

transformer having an impedance of

power

common

to eight

MV

200

A feeders A

ing an interrupting capacity of

transformer impedance

is

8%,

is

circuit breaker hav-

4000 A. Because it

the

can deliver a sec-

ondary short-circuit current of /

It

connected to

bus (Fig. 26.20). Each feeder

a serious overvoltage

occurs, the spark gaps break

discharges to ground.

8%

kV

protected by a 24.9 kV, 200

prevent any current from flowing through the tubular

Consider, for example, a 3-phase 69

kV/24.9

- 1600 X (1/0.08) = 20 000 A


706

Figure 26.16 Like all electric equipment, disconnecting switches have to be overhauled and inspected at regular intervals. During such operations the current has to be diverted by way of auxiliary tie lines within the substation. The picture shows one pole of a 3-phase disconnect rated 2000 A, 345 kV. (Courtesy of Hydro-Quebec)

This creates a problem because

comes

if

a feeder be-

short-circuited, the resulting current flow

could be as high as 20 000 A, which

the feeder (Fig. 26.21).

enough

to

The reactance must be high

keep the current below the interrupting

five times

capacity of the circuit breaker but not so high as to

greater than the interrupting capacity of the circuit

produce a large voltage drop under normal full-load

breaker protecting the feeder.

could be destroyed cuit.

over

in

The

is

circuit breaker

attempting to interrupt the

cir-

Furthermore, the feeder might be damaged its

entire length,

from the circuit-breaker

a violent explosion

fault. Finally,

at the fault itself,

due

to the

ing reactor

is

this

would take place

tremendous amount of

from happening, a current-limit-

connected

in series

shows another application

wherein three current-limiting reactors are connected

in series

with a

HV

line.

to the

thermal energy released by the burning arc.

To prevent

conditions. Fig. 26.22

with each phase of

26.8 Grounding transformer

We sometimes have to create a neutral on a 3-phase, 3-wire system to change

it

into a 3-phase, 4-wire

system. This can be done by means of a grounding


707

Figure 26.17

Combined disconnecting

switch and grounding switch rated at 115 kV.The grounding switch blades are

the open, horizontal position.

These blades

pivot

upward

to

engage three

fixed contacts at the

same

shown

time the

in

line is

opened. (Courtesy of Kearney)

transformer.

former

in

It

is

which

basically a 3-phase autotrans-

identical primary


in series

and secondary

but in zigzag fash-

ion on a 3-legged core (Fig. 26.23). If

line

we connect

a single-phase load between one

and neutral, load current

/

divides into three

equal currents 1/3 in each winding. Because the currents are equal, the neutral point stays fixed

line-to-neutral voltages

and the

remain balanced as they

would be on a regular 4-wire system.

In practice,

the single-phase loads are distributed as evenly as

possible between the respective three phases and neutral so that the unbalanced load current / re-

mains

26.9 8

10 12 14 16 18 20 peak current

relatively small.

Example

of a substation

kA

Figure 26.18 Voltage-current characteristic of a surge- arrester having a nominal rating of 30 kV (42.4 kV peak), used on a 34.5 kV line (28.5 kV peak, line-to-neutral).

Fig. 26.24 shows the principal elements of a typical modern substation providing power to a large suburb. Power is fed into the substation at 220 kV and is

distributed at 24.9

within about a 5

km

kV

to various load centers

radius.

220 kV

69

MVA

24.9

MV circuit

/

I16OOA

j

breaker

\]

/

< ,c] *-

[]

E]

cli

c]

eIi

kV

bus bar reactor

o^

) (fault 8 feeders rated 200

A

Figure 26.21 Current-limiting reactors reduce the short-circuit current.

Figure 26.19

MOV

surge arresters protect


this

EHV transformer.

Electric)

220 kV

69MVA 24.9 kV bus bar

.__ c,rcu,t ||

breaker

2o

MV

kA

^

£^ CO

[]

CO

C]

[]

) (fault 8 feeders rated 200

Figure 26.20 MV busbar feeding eight

lines,

Figure 26.22 Three 2.2 il reactors rated 500 A are connected in series with a 120 kV, 3-phase, 60 Hz line. They are insulated from ground by four insulating columns, and each is protected by a surge arrester. (Courtesy of Hydro-Quebec)

A

each protected by a

circuit breaker.

708


709

26.10 Medium-voltage distribution Thirty-six 3-phase feeders (30 active and 6 spares),

400 A lead outward from Each feeder is equipped with three

rated at 24.9 kV,

the sub-

station.

current-

limiting reactors that limit the line to ground short-

maximum

circuit currents to a

of 12 kA.

Some

feeders are underground, others overhead, and

still

others are underground/overhead.

Underground feeders are composed of three

sin-

gle-phase stranded aluminum cables insulated with polyethylene.

by a

spiral

The

insulation

is in

turn surrounded

wrapping of tinned copper conductors

which act as the ground. The cable is pulled through underground concrete duct (Fig. 26.27) or simply Figure 26.23 Grounding transformer

The

substation

sion lines,

3-phase

all

is

buried in the ground. Spare cables are invariably to create a

3-phase

neutral.

buried along with active cables to provide alternate

fed by three separate transmis-

operating

transformers

at

220 kV.

rated

at

220 kV/24.9 kV. The windings

Tt

contains six

MVA,

36/48/60

are connected in

wye-delta and automatic tap-changers regulate the

neutral

is

established on the

MV

side

by

means of 3-phase grounding transformers. Consequently, single-phase power can be provided at

=

24.9

oil circuit

breakers having an inter-

HV

kA

Conventional

breakers having an inter-

oil circuit

rupting capacity of 25

by

all

protect the

kA are

used on the

For nearby areas the 24.9

MV side.

having an interrupting capacity

station covers an area of

235

m X

170 m. However,

switching and other operations can be carried out

by telecommunications from a dispatching

center.

substation provides service to hundreds of

single-family homes, dozens of apartment buildings, several business

tricts,

at

the substation. Tn

more remote

special measures have to be taken to

keep

dis-

the

self-regulating autotransformers (Fig. 26.29) are

26.11 Low-voltage distribution At the consumer end of the

down by

transformers from 24.9

We now

study the distribution system

kV

is

stepped-

to the

lower voltages needed by the consumers.

much

Two

low-

voltage systems are provided on this typical subur-

ban network:

and shopping centers, a large

branches out from the substation.

MV feeders that spread

out from the substation, the voltage

.

Single-phase 120/240

and some industries Figs. 26.25 and

26.26 show the basic layout and components of the substation.

reg-

often installed.

1

that

is

side.

This completely automatic and unattended sub-

university,

line voltage

the outgoing feeders are protected

circuit breakers

The

kV

ulated within acceptable limits by the tap-changing

of 12 kA.

line

commercial establish-

to residences,

ments, and recreation centers (Fig. 26.28).

voltage reasonably stable with changing load. Thus,

rupting capacity of 32

Furthermore,

off at various points to supply 3-phase and single-

transformers

14.4 kV.

Minimum

The 24.9 kV aerial lines are supported on wooden poles. The latter also carry the LV circuits and telephone cable. The 24.9 kV lines are tapped phase power

secondary voltage.

A

service in case of a fault.

2.

Three-phase 600/347

The

first

system

is

V with grounded neutral V with grounded neutral

mainly used

in

individual

dwellings and for single-phase power ranging up to

Figure 26.24 Aerial view of a substation serving a large suburb.

connecting switches

(2)

and

are connected to an

MV

bus

power through

circuit (5)

breakers

The 220 kV

lines (1) enter the substation

and move through dis(4). The secondaries

energize the primaries of the transformers

operating at 24.9 kV. Grounding transformers (6) and

current-limiting reactors (8).

ergize the suburb

(3) to

The power

is

carried

away by 36

aerial

MV circuit breakers

(7)

and underground feeders

feed to en-

(9).

Figure 26.25 This sequence of 12 photos on the right lines until 1.

2.

3.

4.

it

shows how energy flows through the

substation, starting from the

220 kV

leaves by the 24.9 kV feeders.

220 kV incoming line. The line passes through three CT's (left) and the substation apparatus is protected by 3 Three H V disconnecting switches are placed ahead of the circuit breakers. Minimum volume oil circuit breakers composed of three modules in series permit the

lightning arresters (right).

line to

be opened and

closed under load. 5.

Three-phase transformer bank steps down the voltage from 220 kV protect the

6.

MV

7.

Grounding transformer and

3.

Current-limiting reactors.

9.

10,

line

.

circuit

its

associated

due

right

oil

circuit

breaker having an interrupting capacity of 25 kA.

breaker having an interrupting capacity of 12 kA.

MV underground feeder rated 400 A,

12. All steel

on the

from the transformer feeds the 24.9 kV bus.

Three-phase 1 1

to 24.9 kV. Lightning arresters

HV windings.

24.9 kV/14.4 kV leads into the ground toward a load center

in

the suburb,

supports are solidly grounded by bare copper conductors to prevent overvoltages across equipment

to lightning strokes

and other disturbances. Typical 710

station

ground resistance:

0.1 O.

10

HV LINE

1

220 kV lightning arrester

HV

HV

^ ^4

X>

>*6

f

32 kA

HV

disconnect

motorized disconnect

36/48/60

HV

{ C

tie line

.

J"

V8

10

MVA, 60 Hz Y±

220kV/24.9kV grounding transformer

MV

grounding switch

disconnect

circuit breaker

lightning arrester

220 kV

^
disconnect

HV

HVLINE

X current transformer

'

circuit breaker

MV

A Ml

25 kA*£l4

disconnect

T

line

breaker

reactor

17

r

feeder 18* -t'Nyiv^b—c/\*A>.

j

19*

20*

21

*

22*

23* 24.9


kV FEEDERS LEADING TO SUBURB

of the substation in Fig. 26.24.

712

2

DISTRIBUTION OF ELECTRICA L ENERG Y

Figure 26.27 underground feeders

MV

in

713

concrete duct.

I

Figure 26.29 Automatic tap-changing autotransformer maintains steady voltages on long


rural lines.

Electric)

150 kVA. The second

used

is

in

industry, large

buildings, and commercial centers where the

requirement

is

power

under 2000 kVA.

For single-phase service, the transformers are

kVA

usually rated between 10

and 167

kVA

and

The voltage rating is typi400 V/240-120 V. The transformers pos-

they are pole-mounted. cally 14

sess a single high-voltage bushing connected to one side of the

ing

turn,

to

HV winding. The other side of the wind-

connected

is is

to the steel enclosure

ground

of 3-phase installations, 3 single-

phase transformers rated

The

neutral

in

(Fig. 26.30).

In the case

used.

which,

connected to the neutral conductor and also

at

14

units are connected in

on the primary side

is

400 V/347 V

are

wye-wye and

the

solidly grounded.

The secondary side provides a line voltage of 600 V, and it may or may not be grounded. Such standard distribution transformers have no taps,

Figure 26.28

MV

aerial feeder serving

a residential

district.

and no

circuit breakers or fuses are used

on the

714


4

Figure 26.31

Figure 26.30

A

fused cutout (top

left)

and

Expulsion type fused cutout rated 7.5 kV, 300 A.

lightning arrester (top


a single-phase transformer rated 25 kVA, 14.4 kV/240 V-120 V, 60 Hz. right) protect

secondary

side.

ever, protected

cessive

damage

Figs. 26.30

The primary by a cutout to

HV terminal

in

equipment

is,

how-

order to prevent exin

case of a fault (see

studies reveal that

one ing

line all

70 percent take place between

and ground. Finally, short-circuits involv-

three phases of a transmission line are rare.

The methods of protection

and 26.31).

tistics

and on the necessity

are based

upon these

sta-

to provide continuity of

service to the customers.

PROTECTION OF MEDIUMVOLTAGE DISTRIBUTION SYSTEMS Medium-voltage

lines

must be adequately pro-

tected against short-circuits so as to limit

equipment and

to restrict the

damage

26.12 Coordination of the protective devices

to

outage to as small an

When

a fault occurs, the current increases sharply,

area as possible. Such line faults can occur in vari-

not only on the faulted line, but on

ous ways: falling branches, icing, defective equip-

rectly or indirectly lead to the short-circuit.

To

pre-

ment, lines that touch, and so forth. According to

vent the overload current from simultaneously

trip-

statistics,

85 percent of the short-circuits are tem-

porary, lasting only a fraction of a second.

The same

ping

all

all lines that di-

the associated protective devices,

design the system so that the devices

we must

trip selectively.

DISTRIBUTION OF ELECTRICAL ENERG Y

715

substation

Figure 26.32 Protective devices

must be coordinated.

A well-coordinated system will cause only those devices next to the short-circuit to open, leaving the others intact.

To achieve

this, the

and tripping time of each device line

is

1.

Fused cutouts

2.

Reclosers

3.

Sectionalizers

tripping current set to protect the

and associated apparatus, while

restricting the

outage to the smallest number of customers. Consider, for example, the simple distribution

26.13 Fused cutouts

A fused

cutout

is

system of Fig. 26.32 composed of a main feeder F-

switch.

The fuse

F from a substation, supplying a group of subfeeders. The subfeeders deliver power to loads A, B, C,

switch.

It

D, and E. to

A protective device

each subfeeder so that

if

is

installed at the input

a short-circuit occurs,

alone will be disconnected from the system. For

it

in-

stance, a short-circuit at point 1 will trip device

but not

but not

P 2 Similarly, a fault at point 2 will open P 3 P4 and so on. A short-circuit must be cleared .

,

essentially a fused disconnecting

constitutes the

movable arm of the

pivots about one end and the circuit can

be opened by pulling on the other end of the fuse with a hookstick (Fig. 26.3

1

).

Cutouts are relatively

inexpensive and are used to protect transformers

and small single-phase feeders against overloads.

They

are designed so that

when

the fuse blows,

has occurred on the

line.

matter of a few cycles. Consequently, the coor-

Fused cutouts possess a fuse link

dination between the protective devices involves de-

by a spring. The fuse link assembly

in a

lays that are

therefore

measured

know

in milliseconds.

We

must

the tripping characteristics of the

fuses and circuit breakers throughout the system.

The most important

protective devices used on

lines are the following:

MV

it

automatically swings down, indicating that a fault

that is

is

kept taut

placed inside

a porcelain or glass tube filled with boracic acid, oil,

or carbon tetrachloride.

replaced every time

it

a relatively long outage. tion,

The

fuse link must be

blows, which often results

in

To ensure good coordina-

the current/time characteristics are selected


716

very carefully for each cutout. link

A

burned-out fuse

must always be replaced by another having ex-

actly the

same

the internal control setting of the recloser. If the short-circuit dQes not clear itself after

attempts to reclose/the

rating.

circuit permanently.

26.14 Reclosers

A recloser is circuit

the fault,

remove

it,

Reclosers rated a circuit breaker that opens on short-

and automatically recloses

after a brief time

The delay may range from a fraction of a second to several seconds. The open/close sequence may be repeated two or three times, depending on

delay.

rents

up

to 12

A repair crew

must then locate

kV can

24.9

at

are

interrupt fault cur-

made for either single-

and are usually pole-mounted

lines,

(Fig. 26.33). Reclosers are self-powered, their

energy from the

actuating springs by

three

opens the

and reset the recloser.

000 A, They

phase or 3-phase

two or

line, the recloser

line

and storing

in

it

drawing powerful

means of electromagnets.

26.15 Sectionalizers When

a main feeder

is

protected by several fuses

spotted over the length of the line,

it is

to obtain satisfactory coordination

often difficult

between them,

based on fuse-blowing time alone. Under these

we

cumstances, izer

is

resort to sectionalizers.

a special circuit breaker that trips depending

on the number of times a recloser has tripped up the

cir-

A sectional-

line. In

further

other words, a sectionalizer works ac-

cording to the "instructions" of a recloser.

For example, consider a recloser tionalizer

R

and a

sec-

S protecting an important main feeder

(Fig. 26.34). If a fault occurs at the point

shown, the

recloser will automatically open and reclose the circuit,

according to the predetermined program.

A

recorder inside the sectionalizer counts the number

of times the recloser has tripped and, just before

it

trips for the last time, the sectionalizer itself trips

but permanently. In so doing,

C

Figure 26.33 Automatic recloser protecting a 3-phase feeder.

and

D

of power but

Consequently,

when

it

it

deprives customers

also isolates the fault.

the recloser closes for the last

circuit

breaker

sectionalizer

recloser

fault

is l :

substation

Figure 26.34 Recloser/sectionalizer protective system.

0 0

0

0


A and B

LV

111

we

briefly

continue to receive service. Unlike reclosers, sec-

cover the organization of a low- voltage (LV)

distrib-

tionalizers are not designed to interrupt line cur-

ution system.

time,

it

rents.

will stay closed

Consequently, they must

when

val

and customers

the line current

when

with the time

trip

zero,

is

will

which coincides is

open.

Sectionalizers are available for single-phase and

They offer several advantages over fused cutouts. They can be reclosed lines.

on a dead short-circuit without fear of exploding, and there

no delay

is

having the correct

caliber.

26.16 Review of If

we examine

searching for a fuse link

in

MV

we

find

it

protection

station

The

may

The most common LV systems used America are Single-phase, 2-wire, 120

Single-phase, 3- wire, 240/120

3.

Three-phase, 4-wire,

4.

Three-phase, 3-wire, Three-phase, 4-wire, Three-phase, 3-wire,

contains dozens of auto-

7.

Three-phase, 4-wire,

it

North

V 208/120 V 480 V 480/277 V 600 V 600/347 V

and hundreds of fused

be coordinated with reclosers and sec-

makes

in

V

1.

2.

5.

tionalizers elsewhere in the system.

protection of

26.17 LV distribution system

In

reclosing of circuit breakers at the sub-

vices available

circuits. In this section

6.

matic reclosers, sectionalizers, cutouts.

by

diagram of a typical dis-

the single-line

tribution system,

finally

during the inter-

the recloser itself

3-phase transmission

and

The

variety of de-

possible to provide adequate

MV lines by using combinations such as

Europe and other

380/220

V,

parts of the world, 3-phase

50 Hz systems are widely used. Despite

the different voltages employed, the basic princi-

ples of

LV

everywhere the same.

distribution are

V System.

Single-Phase, 2-Wire, 120 distribution system

This simple

only used for very small loads.

is

Circuit breaker-fuse

When

2.

Circuit breaker-fuse-fuse

system

3.

Circuit breaker-recloser-fuse

are required. Furthermore, the line voltage drop under

4.

Circuit breaker-recloser-sectionalizer

1.

is

not satisfactory because large conductors

load becomes significant even over short distances.

Single-Phase, 3 -Wire, 240/120 5.

V

heavier loads have to be serviced, the 120

V System.

In order

Circuit breaker-sectionalizer-recloser-sectionto

reduce the current and hence conductor

alizer-fuse, etc

voltage In

urban areas the lines are relatively short and

the possibility of faults

is

rather small.

Such

lines

are subdivided along their length into three or four

sections,

cutout.

each

protected

single-pole

fused

Reclosers and sectionalizers are not re-

On the other kV line may be

quired.

24.9

by

more exposed

hand,

in

outlying districts, a

quite long and consequently

to faults. In

such cases, the

line is

subdivided into sections and protected by reclosers

and sectionalizers

to

provide satisfactory service.

LOW-VOLTAGE DISTRIBUTION We have seen that electrical energy is delivered to the consumer via

HV

substations through

MV networks

120

V

240

raised to

is

level

is still

size, the

However, because

V.

the

very useful, the 240 V/120

V

3-wire system was developed. This type of distribution system

is

widely used. Fig. 26.35

is

a highly

schematic diagram, showing the essential elements

The

of such a system.

dual voltage

is

produced by a

distribution transformer having a double secondary

winding (Section neutral,

11.1).

"live" lines

A and B

wire, called

When

the

are equally loaded, the current

When

in the neutral is zero.

the neutral current

tween the

The common

solidly connected to ground.

is

is

line currents / A

V

the loading

is

unequal,


-

to distribute the

120

between the two

live wires

/ B (Fig.

26.35).

We

try

loads as equally as possible

and the

neutral.


7 8 1

Figure 26.35 Single-phase 240 V/120

V

distribution system.

3-phase loads

Figure 26.36 Three-phase, 4-wire, 208 V/120

What 1

.

V

distribution system.

are the advantages of such a 3-wire system?

The line-to-ground voltage which

is

is

only 120 V,

reasonably safe for use in a

and, moreover, the intensity will vary as refrigerator motors, electric stove elements,

home

switched on and

off.

the neutral conductor 2.

Lighting and small motor loads can be energized

at

120 V, while larger loads such as elec-

tric

stoves and large motors can be fed from the

240

V

Both

line.

by fuses or

circuit

However, such protective devices must

never be placed

The reason

is

in series with the neutral conductor.

that if the device trips, the line-to-

neutral voltages

become unbalanced. The voltage

across the lightly loaded 120

V

line

goes up, while

more heavily loaded side goes down. This means that some lights are dimmer than others

that across the

is

is

forth, are

that

when

open, the line fuses are ren-

dered useless.

Three-Phase, 4-Wire, 208/120

V System, We

can

create a 3-phase, 4-wire system by using three

single-phase transformers connected in delta-wye.

live lines are protected

breakers.

and so

Another reason

The in

neutral of the secondary

is

grounded, as shown

Figure 26.36.

This distribution system

is

used

in

commercial

buildings and small industries because the 208 line voltage

large loads, while the 120 lighting

V

can be used for electric motors or other

circuits

V

lines

can be used for

and convenience

outlets.

The

single-phase loads between the three respective


24.9

719

kV 600 V 600 V 600 V

r lighting

1

load

120

V

Figure 26.37 Three-phase, 3-wire, 600

V

distribution

system.

"live" lines and neutral are arranged to be about

When

equal.

the loads are perfectly balanced, the

current in the neutral wire

is

Three-Phase, 3- Wire, 600 3-phase, 3-wire system

is

metal housing of electrical equipment cally

motors are

V

A

System,

used

installed,

(Fig. 26.37). Separate

systemati-

is

systems.

zero.

in factories

600

V,

26.19 Electric shock

where It is

fairly large

LV

grounded on HV, MV, and

ranging up to 500 hp

V

600 V/240-120

difficult to specify

whether a voltage

ous or not because electric shock

is

step-down the current that flows through the

transformers, spotted throughout the premises, are

used to service the lighting loads and convenience

is

danger-

actually caused by

human

body.

upon the skin contact

current depends mainly

The

resis-

tance because, by comparison, the resistance of the outlets.

Three-Phase, 4-Wire, 480/277

V

body System. In large

buildings and commercial centers, a

480

V,

itself is negligible.

4-wire It is

distribution system

is

used because

it

enables motors

V while fluorescent lights operate at V convenience outlets, separate transformers are required, usually fed from the 480 V line.

to run at

277

V.

480

For 20


to

600/347

resistance varies

current

is

Between 10

20

may

not be able to

50 mA, the consequences can be

go; above

The

mA and

potentially dangerous because the victim

loses muscular control and

V, 4-wire

mA mA the

generally claimed that currents below 5

are not dangerous.

1

systems.

human

resistance of the

ranges from 500 11 to 50 kO. dry hand

installations The grounding of electrical systems

is

probably one

of the less well understood aspects of electricity. Nevertheless,

it

a very important subject and an

is

way of preventing accidents. As we have seen, most electrical

distribution

buildings are grounded, usually by con-

necting the neutral to a water pipe or the massive steel structure.

On

pose of grounding

is

=

12

much

low-voltage' systerns, the pur-

mainly to reduce the danger of

electric shock. In addition, for reasons of safety, the

say,

body, taken be-

(/

= 600 V/50 kfl

clammy hand

is

lower, so that an ac voltage as low as 25

V

the resistance of a

could be dangerous

if

letting go. it

the person

is

unable to

let

go.

ac current flows through the body, the

muscular contractions

when

leg,

the resistance of a

50 kD, then momentary contact

V line may not be fatal

mA). But

When

effective

in

is,

with a 600

If

let

fatal.

tween two hands, or between one hand and a

26.18 Grounding electrical

systems

The contact

with the thickness, wetness, and resistivity of the skin.

may

The current

flows

in the

may

particularly dangerous

region of the heart.

temporary paralysis and, fibrillation

prevent the victim from

is

if

it

It

result. Fibrillation is a rapid

coordinated heart beat that

is

induces

flows long enough,

and un-

not synchronized with


720

14.4

14.4

kV

kV *

neutral

neutral

J

1

cutout

— H—

V'

c

(a)

C

.

"

-

5

(b)

V

120

2

cutout

2

5000 V

i

I

ground

ground

Figure 26.38

may produce

Transformer capacitance

high voltages on the LV side of a transformer.

the pulse beat. In such cases, the person can be re-

could touch either one of them without harm because

vived by applying

there

artificial respiration.

have shown that there

Statistical investigations* is

one chance

in ten that

a current

may

cause death

/

^

1

1

6/

return.

However,

Depending upon the

V7

(26.1)

where

may be

as high as

relative

20

to

magnitude of Cj and

40 percent of

voltage. If a person touches either

=

current flow through the

= =

time of current flow

body [mA]

116

[sj

person

outcome

t

is

if /c is

longer be able to

one of the only 20

go

let

/c

sec-

could

mA,

V

conductor. This in the trans-

restricted to

fatal.

the

(Fig. 26.38b).

could be caused by an internal fault

former, or by a branch or tree falling across the

LV

,

serious, suppose that a high-voltage

wire accidentally touches a 120

between 8 ms and 5 s. For example, a current of 58 mA flowing for 4 s

The time

could be

Even more

an empirical constant, expressing the probability of a fatal

may no

C2

the primary

ondary wires, the resulting capacitive current be dangerous. For example,

t

between the

primary, secondary, and ground can produce a high

it

/

this is not true.

C h C2

capacitive coupling

voltage between the secondary lines and ground.

the following equation:

if it satisfies

no ground

is

First, the

MV

Under these circumstances, the lowvoltage system would be subjected to 4.4 kV. This and

lines.

1

high voltage between the secondary conductors and

26.20 Grounding of 120

V

ground would immediately produce a massive

and 240 V/1 20 V systems

over.

secondary network, possibly inside a

Suppose the primary winding of a former 1

4.4

is

kV

tors are

*

connected between the

line (Fig. 26.38a). If the

ungrounded,

it

and neutral of a

secondary conduc-

would appear

DazieLCF, 968 "Reevaluation 1

distribution trans-

line

that a

and General

Application, Vol. IGA-4. No. 5. pages 467-475.

home

the

or fac-

Consequently, an undergrounded secondary

system

is

a potential fire hazard and

may produce

se-

rious accidents under abnormal conditions.

person

of Lethal Electric

Currents," IEEE, Transactions on Industry

tory.

flash-

The flash-over could occur anywhere on

On

the other hand, if

one of the secondary

lines

is

firmly grounded, the accidental contact between

a

HV and a LV conductor produces a short-circuit.

The

short-circuit current follows the path

Fig. 26.39.

The high current

will

blow

shown

in

the fuse on


Figure 26.40 Undergrounded

721

metallic enclosures are potentially

dangerous.

Suppose the motor Figure 26.39 A HV to LV fault

if

the secondary

part of an appliance, such as a

that the

motor frame

is

connected

is

to the

grounded.

solidly

and

refrigerator,

not dangerous

is

is

ungrounded metal enclosure.

If a

touches the enclosure, nothing will happen

equipment the

MV

side, thus effectively

transformer and

from the

MV line.

V system

the 120

disconnecting the

secondary distribution system In conclusion,

is

if

the neutral on

solidly grounded, the potential

difference between the ground and live conductor 1

will only slightly

exceed 120 V. However,

ground electrode has a high resistance the voltage on conductor short-circuit current tially

produced by an

1

may

(say,

if

50

the fl),

MV-LV

be large and poten-

still

dangerous.

is

megohms

drop from several

The consumer of electrical

electricity

equipment of

all

is

to only a

may

few hundred

or less. A person having a body resistance R h would complete the current path to ground as shown in Fig. 26.40. If R G is small, the leakage current / L flowing through the person's body could be

ohms

dangerously high. This system

As tion

Equipment grounding

the

faulty, resistance

the windings and the motor frame

a first approach,

we

is

unsafe.

could remedy the situa-

by bonding the metal enclosure

neutral wire (Fig. 26.41).

26.21

if

functioning properly. But should the

motor winding insulation become

R e between

person

to the

The leakage

grounded

current

now

flows from the motor windings, through the motor

constantly touching

kinds, ranging from do-

mestic appliances and hand-held tools to industrial motors, switchgear, and heating equipment.

As we

have seen, the voltages and currents associated with this

equipment far exceeds those the human body can

tolerate.

In order to

modern

equipment

safe to the touch.

source connected to a motor solidly

grounded

bond

entrance panel board

understand the safety features of

distribution systems,

is

.

service

is

lej;

M

at the

—

(Fig. 26.40).

metal enclosure

,

service

entrance

us begin with a

simple single-phase circuit composed of a 120

neutral

^

Consequently, special precautions are taken

to ensure that the

ground

V

The

service entrance.

Figure 26.41 Bonding the enclosure

to the neutral wire

appears safe.


722

frame and the enclosure, and straight back neutral

ground potential a person touching would not experience any shock.

The wire

to the

Because the enclosure remains

wire.

trouble with this solution

may become

is

at

the enclosure

that the neutral

open, either accidentally or due

to a faulty installation.

For example,

controlling the motor

inadvertently connected in

is

if

the switch

enclosure and

still

be turned on and

off.

However, a per-

son touching the enclosure while the motor

is off,

could receive a bad shock (Fig. 26.42). The reason is

that

is off,

when

the

motor with the defective windings

the potential of the

motor frame and the en-

led to the system

ground

at the ser-

ment, the enclosute

forced to remain at ground po-

is

tential.

A faulty connection such as that in Fig. 26.42

would

result in a short-circuit, causing the fuse to

blow. In grounded systems the neutral wire must

never be connected to the enclosure, except for special

cases permitted by the National Electrical Code.

The ground wire may be

series with the neutral rather than the live wire, the

motor can

is

vice entrance panel (Fig. 26.43). With this arrange-

is

bare, or,

if

insulated,

colored green. In armored-cable and conduit

stallations,

the

armor and conduit serve as

it

in-

the

ground conductor. The locknuts, squeeze-connectors, threads,

so as to

and bushings must be screwed on

make

a

good

tight

connection between

electrical

closure rises to that of the live conductor.

the service entrance ground and the hundreds of

To get around this problem, we install a third wire, called ground wire. It is bonded (connected) to the

outlets that

Most

sometimes make up a large

electrical outlets are

now

ceptacles having three contacts switch

tral,

"off

and one ground

electrical appliances

installation.

provided with

—one

live,

re-

one neu-

(Fig. 26.44). Consequently,

and portable hand tools such

as electric drills are equipped with a 3-conductor

120

V

receptacle

i

-

service

entrance panel board

u

T

metal enclosure

.

service

entrance

ground

Figure 26.42 Enclosure-to-neutral bonding can

ground wire

still

be dangerous.

q

service

entrance panel board r service

entrance

ground

Figure 26.43

A separate ground conductor bonded is

safe.

to the

enclosure

Figure 26.44

The metal housing

of

hand

tools

must be grounded.


cord and a 3-prong plug.

One

exception

is

the

double-insulated devices that are completely en-

closed in plastic enclosures. the

They

are

exempt from

ground wire requirement and so they are

will

produce a hazardous leakage current throughout

the pool, even

the source of

in

26.22 Ground-fault circuit breaker

25

far are

usually adequate, but further safety measures are in

some

cases.

Suppose

for

example, that a

person sticks his finger into a lamp socket (Fig. 26.45). Although the metal enclosure

grounded, the person will

still

is

securely

receive a painful shock.

Or suppose a 120 V electric toaster tumbles into a swimming pool. The heating elements and contacts

power

as soon as such accidents occur.

ondary winding

shown

as

in Fig. 26.46.

The

sec-

connected to a sensitive elec-

is

tronic detector that can trigger a circuit breaker that

is in

series with the

conditions the current

/

V

120

line.

w—

the net current (/

in the toroidal core

from the

is

w in the line conductor is ex-

zero. Consequently,

does not

live

and so

flowing through the hole

/N )

in the core, the

CB

CB

Under normal

actly equal to the current /N in the neutral,

Suppose now

induced voltage

no

flux

is

E F is zero,

trip.

that a fault current I h leaks directly

wire to ground. This could happen

if

someone touched a live terminal (Fig. 26.45). A fault current / L would also be produced if the insulation

neutral

down between

broke ground

sure.

a

motor and

flowing through the hole of the

is

.

A flux

induced, which trips

only 5

mA has

to

is

grounded enclo-

Supermalloy™

is

CT is

set

no longer zero

up and a voltage

EF

CB. Because an imbalance of

be detected, the core of the trans-

former must be very permeable afford

its

Under any of these conditions, the net current

but equal to /F or / L

protection.

will typically trip

small current transformer surrounds the live

and neutral wires

and breaker

Figure 26.45 Special case where a grounding wire does not

securely

ms if the leakage current exceeds 5 mA. How do

produced

line

is

these protective devices operate?

A The grounding methods we have covered so needed

the frame of the toaster

if

grounded. Devices have been developed that will cut

These ground-fault circuit breakers

equipped with 2-prong plugs.

723

at

low flux

densities.

often used for this purpose because


724

has a relative permeability of typically 70

it

flux density of only

000

2

a

at

4 mT.

In general, the I

t

know-

factor can be calculated

ing (a) the cross section of the conductor, (b)

composition (copper or aluminum), and

maximum

26.23 Rapid conductor heating: 2 the / f factor

temperature

can tolerate. The

it

(c) 2

I

t

its

the

factor

copper and aluminum conductors are given by

for

the following equations:

sometimes happens

It

that a current far greater than

normal flows for a brief period

in a

conductor.

for

copper conductors,

The

.

I

I~R losses are then very large and the temperature of

hundred degrees

the conductor can rise several

can flow

short-circuit, intense currents

and cables before the

circuit is

in

for

I

conductor increases very rapidly. What

and

that is typically less than

erated in the conductor

is

1

that

it

5 seconds.

The

rise

we

I Rt

(26.2)

At for a given value of Q:

f-Rt

= mcAt

— mc

depends upon the /"/

known

4

A 2 Io So &

(^-^) +

(26.5)

%J

V234

duration of the short-circuit

[s]

net cross-section of conductor without

90

=

initial

0m

=

final

[mm 2

]

temperature of conductor [°CJ

temperature of conductor [°C]

line

made of aluminum conductor No. Under nor-

AWG has a cross-section of 26.6 mm 2

.

conductor can continuously

this

maximum

Calculate the

permissible f

2

0

(26.3)

factor.

t

factor initial

is

80°C and

that the

maximum

A maximum short-circuit current of 2000 A is can

it

line.

For

how

circulate without exceeding the

long

250°C

temperature limit? Solution

that high temperatures

therefore, very important because

2

during a short-circuit, knowing that the

foreseen on this overhead (/

the insulation that covers a conductor. is,

io

temperature should not exceed 250°C.

follows that for a given conductor the temperature

well

(26.4)

\

short-circuit current [A]

temperature

b.

=

=

mal conditions

a.

from which

It is

X

5.2

An overhead

hence

It

i

carry a current of 160 A.

Q = mcAt

rise

em \

Example 26-1

can calculate the temperature

At

+

log l0

heat gen-

given by

3 3.17,

/234

2

/l

counting the empty spaces

flows for a period

2

Q= From Eq.

=

~ — A

Suppose the conductor has a mass m, a resistance R, and a thermal heat capacity c. Moreover,

t

t

t

is

the temperature rise under these conditions?

/

2

/

dissipated to the surroundings and so the tempera-

is

4

where

Furthermore, the heat does not have time to be

suppose the current

10

the

fuse or circuit breaker.

ture of the

X

1.5

aluminum conductors,

conduc-

opened by

1

in a

fraction of a second. For example, during a severe

tors

t

-

.

=

2

The I it

damage 2 t

a.

Using Eq. 26.5 we find

factor

determines

f-t

=

5.2

X

\

5.2

X

10

0 *A

2

logJ—^L^) [

the temperature rise under short-circuit conditions.

For example, a No. 2 tially at

AWG

copper conductor,

ini-

a temperature of 90°C, cannot endure an I

factor in excess of 22 to be limited to

X

10

6

A

250°C during

2

s if its

temperature

a short-circuit.

2 t

4

X

26.6

2

X

/ 234

log m Bn) \

is

=

7

X

10

6

A

2 s

+ 250\

234 + 80

/


b.

The 2000 A current can flow 2

l t

2000

2 f

=

1

X

10

=

7

X

10

26.25 Electrical installation in buildings

6

The

t

.75

1

proposed to use a No. 30

a temporary fuse.

source of electrical energy. All such in-house

If its initial

AWG copper wire as

2

f

t

temperature

50°C,

is

Safety

1.

needed

to melt the wire (copper melts at

Protection against electric shock

damage

The time needed

to melt the wire if the short

circuit current is

30

A

Solution

From

a.

Protection against overloads

d.

Protection against hostile environments

Conductor voltage drop

2.

Eq. 26.4

c.

we have It

2

l

t

=

11.5

Protection of conductors against physical

b.

1083°C) b.

X

10

4

X

+ 1083 ^4 + g

should not exceed

0.0507 log 10

The

distribution system should last a mini-

mum of 50 years.

For a current of 30

30

A we

obtain

h=

197

2

=

197

t

=

0.22

in

Economy

4.

The

cost of the installation should be mini-

mized while observing the pertinent standards.

r

Thus, the fuse will blow

Standards are set by the National Electrical Code* s

and every

approximately 220 ms.

electrical installation

an inspector before

of

with the conductor. The fuse must 2

be selected so that will

its

I

t

rating

is

less than that

produce an excessive temperature

the conductor. In effect,

we want

rise

the fuse to

of

blow

before the conductor attains a dangerous temperature, usually

taken to be 250°C. In practice, the

rating of the fuse

temperatures

below

important element

2

I

components

electrical installation

2

I t

Many components

makeup of an The block diagrams of Figs.

are used in the

electrical installation.

26.47 and 26.48, together with the following definitions

',

will help the reader

understand the purpose

of

some of the more important

1.

Service Conductors. These are the conductors

that extend

from the

street

items.

main feeder or from

such as to ^produce conductor

is

far

Nevertheless, the

an

must be approved by

can be put into service.

conductor from excessive tem-

In order to protect a

peratures during a short-circuit, a fuse must be in series

it

26.26 Principal

26.24 The role of fuses

which

or 2 percent.

Life expectancy

3.

= 197A 2 s

placed

1

/234

2

y

b.

distri-

certain basic requirements:

a.

The

is

original

bution systems, be they large or small, must meet


in a building

between the consumer and the

s

Example 26-2 It is

system

electrical distribution

the final link

=

t

given by

6

for a time

725

t

this

maximum

rating of the conductor

in the

choice of the fuse.

limit. is

an

*

In

Canada, by the Canadian Electrical Code.

Some

taken from the National Electrical Code.

a

726


service

service

conductors

service

conductors

equipment

service e(}uipment

feeder

feeder

metering equipment

switchboard

feeder

branch circuits

panel board

feeder

outlets

branch circuits

@ 0 E

outlets

panelboard

outlets

receptacle utilization

u

u

u

equipment

branch circuits

Figure 26.47 Block diagram of the electrical system in a residence. In many cases the meter is installed upstream from the service equipment.

Figure 26.48 Block diagram of the electrical system

in

an

industrial

or commercial establishment.

Equipment Various

transformer to the service equipment on the con-

3.

sumer premises.

recorders to indicate the electrical energy

2.

Service Equipment. The necessary equipment,

usually consisting of a circuit breaker or switch and fuses,

and

their accessories, located near the point

of entrance of service conductors to a building or

Metering

meters

and

consumed

on the premises. 4.

Panel Board.

units designed for

A single

panel or group of panel

assembly

in the

form of a single

panel; including buses, automatic overcurrent de-

other structure, or an otherwise defined area, and in-

vices,

tended to constitute the main control and means of

of

cutoff of the supply.

placed

and with or without switches for the control

light, heat, or in

power

circuits;

a cabinet or cutout

designed to be

box placed

in or


against a wall or partition and accessible only

the steel structure if the station

from the

50 kA lightning

5.

front.

A large single panel,

Switchboard.

frame, or as-

sembly of panels on which are mounted on the

What

26-7

face,

devices buses,

from the front and are not intended

to

and the

final branch-circuit

istics

overcur-

8.

final

and the

Outlet

having the character-

in Fig. 26.

1

8

connected

is

to

34.5 kV.

Branch Circuit The

circuit

arrester

shown

a line having a line-to-neutral voltage of

rent device.

tween the

surge arrester

A lightning

26-8

vice equipment, or the generator switchboard of an

7.

ground wire

current-limiting reactor

Intermediate level

Feeder. All circuit conductors between the ser-

isolated plant,

receptacle

recloser

be

installed in cabinets. 6.

by a

the purpose of the following

fuse cutout

and usually instruments.

Switchboards are generally accessible from the rear as well as

is hit

stroke.

equipment:

back, or both, switches, over-current, and other protective

is

727

A

circuit

conductors

be-

overcurrent device protecting the

Calculate a.

outlet(s).

b.

point on the wiring circuit at which

The peak voltage between The current flowing in the

line

and neutral

arrester

under

these conditions

current 9.

is

taken to supply utilization equipment.

A

Receptacle.

contact device installed

26-9

In

the

at

outlet for the connection of a single attachment plug.

Utilization

10.

Equipment. Equipment which

uti-

a.

heating, lighting, or similar services.

The

b.

greatly simplified diagrams of Figs. 26.47

c.

and 26.48 indicate the type of distribution systems used respectively

in a

home and

in

an industrial or

1

Fig. 26.26

(jls

shows

the ac-

equipment. Can you correlate the

symbols of the schematic diagram with

is

the difference

between a

circuit

26-

1

1

Repairs have to be carried out on cuit breaker

breaker and a disconnecting switch?

the three

four types of circuit breakers.

vice,

26-3

lasts for 5

a schematic diagram of a

the equipment?

Practical level

Name

is

substation and Fig. 26.25


26-2

neutral.

The peak current in the arrester The peak power dissipated in the arrester The energy dissipated in the arrester if the

tual

What

and

surge effectively

26- 0

commercial establishment.

26-1

line

Calculate

energy for mechanical, chemical,

lizes electrical

Problem 26-8 an 80 kV surge appears

between

The disconnecting switch shown

in Fig.

maximum of maximum allowable

No. 6 shown

220 kV

lines

HV cir-

in Fig. 26.26. If

must be kept

in ser-

which disconnecting switches must

be kept open?

26.13 can dissipate a rated

200 W. Calculate the

26-12

The

current-limiting reactors (8)

in Fig.

value of the contact resistance.

rent to 12

26-4

What

26-5

Name some

is

the purpose of a grounding switch?

each 26- 3 1

26-6

The ground 0.35

f2.

resistance of a substation

is

Calculate the rise in potential of

kA on

the 24.9

kV

feeders.

Calculate the reactance and inductance of

of the main components of a

substation.

shown

26.25 limit the short-circuit cur-

coil.

In Fig. 26.35 resistive loads,

respectively, absorb 1200

3600 W. Calculate

1 ,

2,

and 3

W, 2400 W, and

the current:


728

26-14

A and B

a.

In lines

b.

In the neutral

c.

In the

In Fig.

HV

a.

conductors

line

b.

26.37 the lighting circuit

is

26-15

Draw

26- 9 1

300

1

for

mA and 2 A. 26-20

The

oil in

The following loads are connected to the 240 V/120 V line shown in Fig. 26.35.

load

=

6 kW, cos 6

load 2: 4.8

kW, cos

0

kVA, cos

3: 18

6

1

=

the 3-phase, 24.9

What What

A and

loads have the following ratings:

M

1

:

the

feeder (14.4

kV

line-

have a cross-section of

The cable possesses

the fol-

inductive reactance: 0.1 (2

MV line? the power factor on the MV side?

capacitive reactance:

Referring to Fig. 26.36, the connected

motor

1,

resistance: 0.13 11

the current in the

single phase loads:

MCM.

kV

1

make up

that

B, and

a. 1

aluminum conductors

0.7 lagging

the neutral.

26- 8

in-

kilometer of length:

=

b.

is

power transformer shown

lowing characteristics, per phase, and per

0.8 lagging

Calculate the currents in lines

c.

the big

to-neutral) each

500

.0

a.

is

this

Referring to Fig. 26.25, items 10, three

:

system

Can you explain

not grounded.

power flow from the three 220 kV incoming lines, state which circuit breakers and which disconnecting switches have to be opened, and in what order? 26-21

level

1

V

terrupting the

Explain the operation of a ground fault

load

and

line

has to be filtered and cleaned. Without

mA for 10 ms 30 mA for 2 min 300

circuit breaker.

26-17

V

on the extreme left-hand side of Fig. 26.26

dangerous:

Advanced

one 600

phenomenon?

State whether the following conditions are

26-16

V between

a graph (/ versus

currents between 10

b.

In Fig. 26.37 a sensitive voltmeter reads

is

of Eq. 26.

factor

side.

ground, even though the 600

Crosshatch the potentially mortal regions.

a.

V

MV lines. t)

power

Calculate the line currents and

on the 2400

off and

two motors together draw 420 kVA from the 600 V line. Calculate the current the

in the

Calculate the currents in the secondary

windings.

30

b.

=

motor M2: 160 kVA, cos 0

0.5 lagging

-

0.80 lagging

If

fl

the equivalent circuit of

the line length

is

one phase

no current-limiting

c.

occurs

at the

Given

the 12

is

if

5 km. line reactors are used,

calculate the short-circuit current

kW each

50 k VA, cos 6

Draw

3000

end of the

kA rating

if

a fault

line.

of the circuit breakers,

a line reactor needed in this special case?

Chapter 27 The Cost of Electricity

27.0 Introduction In

1999 the electric power

utilities in the

States supplied approximately 3 trical

1

.

1

tinually at the service of every

24 hours

a day.

and distribution of costs that

gories

in-

in

30

hundreds of billions of dollars.

TW h of elec-

Operating costs include

may be

—fixed

Bearing

kW of power con-

in

two types of

man, woman, and

mind

weekly expense.

the relative importance of these

costs, utility

companies have estab-

lished rate structures that attempt to be as equitable

The production, transmission,

this

salaries, fuel costs, ad-

ministration, and any other daily or

customers (Table 27A). This enormous

amount of energy represents child,

1

United

energy to their industrial, commercial, and

residential

and distribution of electrical energy. These

vestments represent enormous sums measured

sion,

as possible for their customers.

energy involves important

The

rates are based

upon the following guidelines:

divided into two main cate-

costs and operating costs. 1

.

The amount of energy consumed [kW

h]

Fixed costs comprise the depreciation charges against buildings, dams, turbines, generators, circuit breakers, transformers, transmission lines,

CONSUMPTION OF ELECTRICAL ENERGY Number

Type of customer

"

residential

!

These

It is

to

statistics for

IN

is

con-

factor of the load

THE UNITED STATES Total

which energy

(1999)*

amount

Monthly consumption

consumption [TW-h]

per

consumer [kW-h]

1018 971

5 792

109 817 000

1141

866

161

000

1999 were drawn from information supplied by the Energy Information Administration (www.eia.doe.gov).

expected that the yearly increase

3525 TW-h.

The power

at

527 000 13

1

3.

or rate

964 000

industrial

commercial

of

consumers

The demand, sumed fkW]

and

other equipment used in the production, transmis-

TABLE 27 A

2.

in

energy consumption will be about 2 percent. Thus,

in

2005

the total energy will

amount


730

27 A

Tariff

based upon energy

night and day, including Saturdays and Sundays,

constantly drawing 1000

The

of electricity depends,

cost

upon the

first,

amount of energy (kW-h) consumed. However, even if

a customer uses no energy at

minimum

all,

service charge, because

keep him connected

As consumption

may

he has to pay a

money

costs

to

to the line.

increases, the cost per kilowatt-

hour drops, usually on a sliding mestic tariff

it

scale.

Thus, the do-

20 cents per kilowatt-hour

start at

kW h, fall to 10 cents/kWh for the next two hundred kW h, and bottom out at 8

for the first

one hundred

cents/kWh for the rest of the energy consumed. The same general principle applies to medium-power and large-power users of electrical energy.

27.2 Tariff The monthly

kW of active power. At the

it has consumed a total of 720 h = 720 000 kWh of energy. Factory B consumes the same amount of energy, but its load is continually changing. Thus, power fluctuates between 50 kW and 3000 kW, as shown

end of the month (720

h),

kW X

1000

in the figure.

Obviously, the capacity of the trans-

former and the transmission line supplying factory

B must be The ital

greater than that supplying factory A.

electric utility must, therefore, invest

to service factory B; consequently,

able that factory It is

B

it

should pay more for

more capis

its

reasonenergy.

advantageous, both to the customer and the

company that energy be consumed rate. The steadier the power, the less

a con-

utility

at

stant

the cost.

based upon demand

cost of electricity supplied to a large

customer depends not only upon the energy con-

sumed, but also upon the

which

27.3

Demand meter

The graphs of

Fig. 27.1

show a number of power

used. In

spikes that last for a very short time. These spikes

other words, the cost also depends upon the active

correspond to the high power drawn by motors

power (kW) drawn from

when

rate at

the line.

it

is

To understand

the

reason for this dual rate structure, consider the fol-

lowing example.

Two

factories

TB

and

B

(Fig. 27.1). Factory

TA

last

are respectively con-

A operates

factory

at full-load,

maximum demand: 1000 kW

start-up

I

I.

|

i

ZL

i

0 14

28 30 d

21

•-time

energy: 720

000 kW-h

jmaximum demand: 3000 kW

U

factory B

Figure 27.1 Comparison between two

28 30 d

factories

arises:

last, in

How

order to

be considered significant? The answer depends

P 1000

equipment by

company. The question then

long does the power surge have to

kW

"\aaaaa

trr mm 1 A

TA

the utility

start-

long enough to warrant the

installation of correspondingly large

A

nected to a high-voltage line by transformers

and

they are started up. However, the high

up power does not

consuming the same energy but having

different

demands.

THE COST OF ELECTRICITY

upon several be

to

users,

factors, but the period

such as municipalities, the averaging period

may be as long mand interval. To monitor meter

usually taken

is

30 minutes. For very large power

15, or

10,

is

as

the

60 minutes.

It is

power drawn by

cessive

demand

a plant, a special

intervals (15 minutes, say).

the

demand every

by, the

meter

ries a

is

maximum demand,

second pointer that

is

which

to ity

it

is

the

used for

Example 27-1 The graph in Fig. 27.4 drawn by a

represents the active

large factory

power

between 7:00 and 9:00

in

aver-

called the

pointer reads 2 JVIW while the second (pushed)

The

the morning.

moves up In

the meter car-

pushed upscale by the first. sits at

the highest position

pushed. At the end of the month, a

employee takes

is

both diagnostic and billing purposes.

faithfully records

15 minutes and a pointer

The second pointer simply

printout constitutes a

The demand meter has a 30 min demand interval. Let us assume that at 7:00 the first

and down a calibrated scale as the demand changes. order to record the

It

power during suc-

age power measured during each interval

demand. As time goes

The

permanent record of the demand and

called the de-

installed at the customer's service entrance.

automatically measures the average

ing large industrial loads.

731

util-

maximum demand reading and

resets the pointer to zero.

This special meter it

is

tries

is

called a

demand meter, and

installed at the service entrance of

most indus-

and commercial establishments

(Fig. 27.2).

Fig. 27.3

shows a

printing

demand meter for meter-

Figure 27.3 Printing

demand

meter.


Electric)

10

p

s

0

15

7:00

45

60 8:00

75

90

105

120

mm

900

Figure 27.4

Figure 27.2

Combined energy and demand (Courtesy of Sangamo)

30

meter.

Instantaneous record of the active power absorbed by

a

plant.


732

MW. What

pointer indicates 3

The demand between 8:00 and 8:30

are the meter read-

ings at the following times? a.

7:30

b.

8:00

c.

8:30

d.

9:00

Pd =

5X5+1 Thus, 4.67

Solution a.

to the graph, the

Consequently, pointer

MW

b.

1

is

2

continues to indicate

7:30 and pointer 2 stays where

at

The average power and 8:00

is

it

was

During

demand) between 7:30

(or

During

this

MW

min + 2

X 20

moves from 2

MW.

at

Note

5

min

MW

(at

1

7:30) to 4. 7 1

4.

1

7

siderably less than the 7

MW

power

to

and

produce

X

20)/30

interval, pointer

MW,

unity and so, too,

where they are

con-

MW peak which oc-

P.S

factory

X

transformers,

magnetic

their

is

the

A

absorb fields.

factor of the factory

low power factor

example shows.

kW,kVA

kW

1000

kVA

1000 0 0

kW,

IT factory

ps *

7

14

21

kVA

28 30

2000 kVA

2000

1000 kW

1000 0

Y

14 time

*-

Figure 27.5 low plant power factor requires larger

A

utility

reactive

The power

1000 kW 2000 kVA

/VWW\

factor

therefore, less than

is,

power

installed.

1000

a

sits at

in-

creases the cost of electrical energy, as the following

T

drops

maximum demand.

kW kVA

0

1

but pointer 2

upon power

machines

factor of these

is

+

is

alternating-current machines, such as induc-

tion

curred during this interval.

1000 1000

to 2.83

the previous

motors

gradu-

MW.

demand reading

to

+

8:00 both pointers indicate

that the

MW,

30-minute

MW

27.4 Tariff based

interval, pointer

pushing pointer 2 to

(at 8:00),

X

min)/30 min

Many

30-minute

Consequently, 4. 17

5

(7

this

from 4.67

equal to the energy divided by time:

MW X 4 MW = 4.17 MW

ally

moved up

MW

2.83

4.67

Pd =

8:30 both pointers have

(1X5+12X5

MW.

MW.

+

MW.

average power (or

demand) between 7:00 and 7:30 2

3X5

MW

4.67 at

+ 4X5 + X 5)/30

The demand between 8:30 and 9:00

According

at 3

X5 + 8X5

(7

is

company

lines

and equipment.

21

28 30 d

THE COST OF ELECTRICITY Consider two factories X and Y that consume the same amount of energy (kW-h) and, in addition, have the same maximum demand (kW). However, the power factor of X is unity while that of Y is 50

733

27.5 Typical rate structures Electrical

power

utility rates

vary greatly from one

we can

area to another, and so

only give a general

overview of the subject. Most companies divide

percent (Fig. 27.5).

The energy and demand being

the same, the

watthourmeters and demand meters will show the

same reading at the end of the month. At first glance would appear that both users should pay the same bill. However, we must not overlook the apparent power drawn by each plant. Apparent power drawn by factory X: it

their

1

—power corresponding

Domestic power

.

=

—power of Medium power— power of

Small power

2.

=

1000

more, but

(7.7)

1000/1.0

their

to the

needs of houses and rented apartments

3.

S = P/cos 0

customers into categories, according to

power demand. For example, one utility company distinguishes the following four power categories:

less

Large power

4.

than 100 kilowatts

less 1

00 kilowatts and

than 5000 kilowatts

—power

in

excess of 5000 kilo-

watts

kVA

Table 27B shows, on a comparative basis, the

Apparent power drawn by factor Y:

type of rate structures that apply to each of these

S

=

/7cos 6

=

1000/0.5

categories. In addition, a contract

- 2000 kVA

up between the

electrical utility

is

usually

drawn

and the medium- or

may stipulate a minimum power fac-

large-power customer. The contract

Because the

line current

parent power, factory rent as factory X.

Y

The

is

proportional to ap-

minimum monthly demand,

draws twice as much curconductors feeding

line

must, therefore, be twice as big. Worse

still,

Y

the

transformers, circuit breakers, disconnect switches,

Y must

have

therefore, invest

more

and other devices supplying energy

to

tor,

the voltage regulation,

a

and various other clauses

concerning firm power, growth

rate, liability, off-

peak energy, seasonal energy, price increases, and so forth. However, in the case of residences, the rate

schedule

is

quite straightforward (see Table 27B).

twice the rating of those supplying X.

The

utility

company must,

capital to service factory Y; consequently, that

it

should pay more for

consumes the same amount. ture

is

its

it

is

logical

energy, even though

designed to automatically increase the billing

whenever the power factor is low. Most electrical ities

require that the

clients be

the it

is

power

minimum

rate.

When

the

in

order to benefit from

power

factor

is

too low,

usually to the customer's advantage to improve

done by

pay the higher monthly

bill.

This

is

it,

usually

Demand

controllers

For industrial and commercial

maximum demand

consumers the

plays an important role

piling the electricity

bill.

be made by keeping the as possible. Thus, an

in

com-

Substantial savings can

maximum demand

as low

alarm can be installed

sound a warning whenever the demand

is

to

about to

exceed a pre-established maximum. Loads

that are

not absolutely essential can then be temporarily

switched off

until the

peak has passed. This proce-

installing capacitors at the service entrance to

dure can be carried out automatically by a demand

on the load side of the metering equipment.

controller that connects and disconnects individ-

the plant,

These capacitors may supply for

util-

factor of their industrial

90 percent or more,

rather than

tive

it

In practice, the rate struc-

27.6

patf, or all,

of the reac-

power required by the plant.' Industrial capacitors power factor correction are made in single-phase

and 3-phase

units rated

from 5 kvar

to

200

kvar.

ual loads so as to stay within the prescribed

mum demand (Fig.

27.6).

Such

maxi-

a device can easily

save thousands of dollars per year for a medium-

power customer.


734

TYPICAL RATE STRUCTURES

TABLE 27B

**

Residential rate structure Typical clauses: "... This rate shall apply to electric service in a single private dwelling

1.

2.

.

.

This rate applies to single-phase alternating current

.

at

60 Hz

.

.

.

." .

.

Rate schedule

Minimum monthly 100

first

next 200

kW h

charge: $5.00 plus

per month

kW-h per month

kW h

excess over 200

at 5 at

3

cents/kWh cents/kWh

per month

2 cents/kW h

at

General power rate structure (medium power) Typical contract clauses: ".

1.

.

The customer's maximum demand

.

5000 2.

.

3.

...

4.

.

.

.

.

.

kW

.

.

for the month, or

its

contract

demand,

is at

least

kW but

50

not

more than

.

Company shall make available to the customer 1000 kW of firm power during the term of this contract The power shall be delivered at a nominal 3-phase line voltage of 480 V, 60 Hz The power taken under this contract shall not be used to cause unusual disturbances on the Utility Company's

Utility

.

.

.

system. In the event that unreasonable disturbances, including harmonic currents, produce interference with nication systems, the customer shall at his expense correct such disturbances 5.

6.

7.

.

.

.

Voltage variations shall not exceed 7 percent up or

Company

down from

the

.

.

nominal

commu-

.

line voltage

.

.

.

make periodic tests of its metering equipment so as to maintain a high standard of accuracy Customer shall use power so that current is reasonably balanced on the three phases. Customer agrees to take corrective measures if the current on the more heavily loaded phase exceeds the current in either of the two other phases by more than 20 percent. If said unbalance is not corrected, Utility Company may meter the load on individual phases and compute the billing demand as being equal to three times the maximum demand on any phase The maximum demand for any month shall be the greatest of the demands measured in kilowatts during any 30-minute period of the month .

.

.

.

.

.

.

.

.

Utility

shall

.

.

8.

.

9.

.

.

.

.

.

.

.

.

90 percent of the highest average kVA measured during any 30-minute period ." mand, such amount will be used as the billing demand ... If

.

is

higher than the

maximum

de-

.

Rate schedule

Demand

charge: $3.00 per

Energy charge:

month per

4 cents/kWh for the

kW of billing demand first

100 hours of billing

2 cents/kWh for the next 50 000 1

.2

demand

kWh per month

cents/kW h for the remaining energy General Power Rate Structure (large power)

Typical contract clauses: 1.

"... Customer's

2.

Clauses similar

3.

.

.

.

4.

.

.

.

maximum demand

to clauses

.

.

.

demand is greater than 5000 kW General Power (medium power) contract given above.

month or

2 to 8 listed in the

Contract shall be for a duration of 10 years

The minimum Utility

for

bill

previous 36 months 5.

for the

Company

.

.

any one month

.

.

its

contract

.

." .

.

shall equal

70 percent of the highest

maximum demand

charge during the

.

shall not

be obligated to furnish power

in greater

amounts than the customer's contract demand

." .

.


735

TYPICAL RATE STRUCTURES

TABLE 27B

Rate Schedule

Demand

charge: First 75

kW of demand per month at $2.50 per kW kW of demand per month at $2.00 per kW

000

Excess over 75 000

Extra charge for any

Energy charge:

First

demand

in

excess of customer's contract

20 million kW-h per month

at 6.

Next 30 million kW-h per month Additional energy *

1

mill

= one

at

at

5.9 mills per

1

demand

at

$2.20 per month per

kW

kW-h* per kW-h

mills per

6.0 mills

kW-h

thousandth of a dollar, or one tenth of a cent.

Example 27-2

apparent power (kVA). Calculate the electricity

using the

medium-power

rate

bill

schedule (Table 27B).

Billing of a domestic customer

A homeowner consumes 900 kW-h month of August. Calculate

during the

the electricity

using

bill

the residential rate schedule given in Table

Solution

Clause 9

demand

27B.

is

Solution

Minimum First

100

Next 200

charge

kW h @ 5 cents/kW-h kWh @ 3 cents/kWh

is

is

90% of the kVA 90% X 700 = 530 kVA, which the maximum demand of 1200 kW.

important here because

equal to

greater than

1

1

Consequently, the demand for billing purposes

= = =

$5.00

=

12.00

1

530

kW

and not

1

200 kW. The power

is

factor of the

5.00

6.00

Remaining energy consumed (900

-

300)

600 kW-h

TOTAL bill

@

= 600 kW

h

2 cents/kW-h

for the

month

$28.00

This represents an average cost of 2800/900

=

3.11

cent/kWh.

Demand

meters

are

not

usually

installed

in

homes because the maximum demand seldom exceeds lOkW. Table 27C shows the energy consumed by various electrical appliances found is

in

an example of an all-electric

a

home.

home

Fig. 27.7

heated by

baseboard heaters.

Example 27-3 Billing for a

A

medium-power customer

small industry operating night and day, 7 days a

week, consumes 260 000 kW-h per month. The max-

imum demand demand factory

is

is

is

1200 kW, and the-maximum

1700 kVA. Note

that the

demand

kVA

in this

measured for both active power (kW) and

Figure 27.6 Automatic

demand

loads whenever the

controller that

sheds nonessential

demand reaches

a preset

level or

according to a set schedule. This model can control

up to 96 loads. (Courtesy of Gentec

Inc.)


736

AVERAGE MONTHLY CONSUMPTION OF HOUSEHOLD APPLIANCES

TABLE 27C _

t

•-

Average monthly consumption and a dishwasher

of a family of five

kW

Appliance

people

in

a modern house equipped with an automatic washing machine

consumed

h

kW-h consumed

Appliance

Hot water heater (2000 gallons/month)

500

Automatic washing machine

Freezer

100

Coffee maker

9

Stove

100

Stereo system

9

Lighting

100

Radio

7

4

100

Dryer

70

Dishwasher

30

Lawn mower Vacuum cleaner

Electric kettle

20

Toaster

4

Electric skillet

15

Clock

2

Electric iron

12

plant

kW)

is

is

530

demand (1530 demand (1200 kW). the demand charge is

low; consequently, the billing

higher than the metered

Applying the 1

7

kW @

rate schedule,

$3.00/kW

The energy charge

for the first 100 h

=

$ 4 590

=

6 120

is

kW X 100 hours = 153 000 kW h @ 4 cents/kW-h 1530

=

153 000

X

0.04

The energy charge for the next 50 000 kW-h is = 000 50 000 kWh @ 2 cents/kWh 1

The remainder of the energy is (260 000 - 153 000 - 50 000) = 57 000

kWh

Figure 27.7

The energy charge for the remainder of the energy is 57 000 kW-h @ .2 cents/kWh 684 1

TOTAL bill

for the

The average unit cost

Fig. 27.8

mand

month

12

bill

a plant equipped with both a de-

that

consumes a maximum of 9400 minimum of 2100 kW h in July.

a.

b.

_

Example 27-4

using the large-power rate schedule given

in

Solution

pacitors.

customer

A paper mill consumes 28

home

January, and a

Table 27B.

controller and power-factor correcting ca-

Billing of a large-power

in

peak demand of 43 000 kW. Calculate the monthly

is

394/260 000

4.77 cents/kW h

shows

kW-h

$12 394

cost of energy

= =

All-electric

million kilowatt-hours of

energy per month. The demand meter registers a

The demand charge is 43 000 kW @ $2.50/kW The energy charge for the

first

20 million kW h is 20 million kW h

mill/kW h

20 X 10

6

X

@

6.

1

=

6.1/1000

The energy charge

for the next

=

$ 107 500

122 000


737

Power factor correction

27.7

Power factor correction ically feasible

(or

improvement)

whenever the decrease

is

econom-

in the

annual

cost of electricity exceeds the amortized cost of installing the required capacitors. In

some cases the cus-

tomer has no choice but must comply with the mini-

mum power factor specified by the utility company. The power

factor

may be improved by

installing

capacitors at the service entrance to the factory or

commercial enterprise.

In other cases

it

may be

power factor of an device, or machine, if its power factor sirable to correct the

de-

individual particu-

is

larly low.

Example 27-5

A factory power

Figure 27.8 All-electric industry

covering an area of 1300

m2

.

It

is

heated by passing current through the reinforcing wire

mesh embedded

in

the concrete

floor.

A

load con-

connects and disconnects the heating sections depending on the level of produc(priority loads). The demand is thereby kept below

kVA

draws an apparent power of 300

factor of

65%

at a

(lagging). Calculate the kvar

capacity of the capacitor bank that must be installed at the service

entrance to bring the overall power

factor to

troller

(nonpriority loads)

a.

Unity

tion

b.

85 percent lagging

the desired preset level. Annual energy consumption:

kW h; maximum demand in winter: 92 kW; maximum demand during the summer: 87 kW.

375 000

Solution a.

Apparent power absorbed by the plant

is

{Courtesy of Lab-Volt)

S

=

300

kVA

Active power absorbed by the plant

(28

-

20)

X

10

8 million

kW h @

8 million 8

=

6

X

TOTAL bill

for the

A

monthly

high, but

=

6

month

$ 277 500

=

= V300 2 X 10~ 3 =

9.9

kW

sents less than 5 percent of

that

the.,

it

may appear

therefore,

selling price of the

Fig. 27.9 gives an idea of the

b.

power and energy

all

(7.4)

195-

power

=

228 kvar

factor to unity,

the reactive

load (228 kvar).

probably repre-

finished product.

a large city.

raise the

supply

of nearly $300 000

is

2

9.9 mill

To

we must remember

consumed by

Q = Vs - P 2

cent/kW h bill

195

Reactive power absorbed by the factory

48 000

kW h =

277 500/28 X 10 1

kWh is

(7.7)

6.0 mill/kW h

6.0/1000

Average cost per

or about

P = Scosd = 300 X 0.65 =

is

we have

power absorbed by

The 3-phase capacitors

have a rating 228 kvar.

Fig.

to

the

must,

27.10a

shows the active and reactive power flow. The factory continues to draw the same amount of active power (195 kW) because the mechanical

and thermal loads remain unchanged.


738

Figure 27.9 In

1998, the

mand

city of

Montreal with 908 343 customers

during the winter, 6695

stitutional

MW;

MW.

Consequently, because the to

is

new

overall

la

C.I.D.E.M., Ville

power

230

195/0.85 reactive

Because the plant

kVA

power supplied by

Q = V230 2 -

in-

195

still

2

=

is

121 kvar

if

we

common

is

-

cusre-

large customers, an automatic controller switches

lies slightly

power

factor

above 95 percent.

draws 228 kvar and the

Example 27-6

=

121)

107 kvar

A 600 kW induction furnace connected to an 800 V lagging.

down

can accept a power factor of 0.85

we can

install a

bank and, hence, reduce the

shows the new power flow factory.

industrial

practice to install a variable ca-

single-phase line operates

(228

(instead of unity),

and the

it

always

is

Q=

installation.

capacitor units in and out so that the

must come from the capacitors. The rating of these units

active and reactive power, irrespective

pacitor unit at the service entrance. In the case of the line

line furnishes only 121 kvar, the difference

line

de-

tomers varies greatly throughout the day. As a sult,

The new

same

of the size of the capacitor

The demand of commercial and

must be

the line

P/cos 6

itor

Maximum

834 935; general and

de Montreal)

the

be 0.85 lagging, the apparent power

drawn from

Thus,

of electrical energy.

Residential customers:

customers: 67 234; industrial customers: 6174.

(Courtesy of Hydro-Quebec and of Service de

factor

consumed 26 335 GW.h

during the summer, 3591

Note

is

at

a

supplied by a 4

power

kV

a.

Calculate the current in the 4000

b.

If

that the factory

draws

and a

step-

transformer (Fig. 27.11).

10b

transmission

factor of 0.6

line

smaller capac-

cost. Fig. 27.

in the

It

a

500 kvar capacitor

is

V

installed

line.

on the

side of the transformer, calculate the

power

factor and the

new

new

line current.

HV

'0

195

kW 1 300 kVA

195

kW

228 kvar

MM EL

(a)

228 kvar

Q

/>[^195 />[^195

kW

kW

5

I

300 kVA

I 230 kVA

(b)

Q [^121

Q [^228

kvar

kvar

Q. 107 kvar-

O Q

Figure 27.10 Overall power b. Overall power a.

factor corrected to unity

(Example

27-5).

factor corrected to 0.85.

4kV 600

kW

P

S

0

800 V

^671 kVA

induction

^S=

300 kvar Q

1

600 kW 1000

kVA

800 kvar

250

67 A.

500 kvar

<2>

-

molten iron

A 600

kW

800 kvar transformer

eddy currents induction furnace

Figure 27.11 Individual load

coil

power

factor correction

(Example 27-6). 739

i o o o o


740

The new

Solution

This

is

line current is

an interesting example where individual /

power factor correction must be is

that the induction furnace

whereas the plant line.

We

is

applied.

The reason

a single-phase device

is

certainly energized

by a 3-phase

cannot correct the power factor of single-

By

= SJE = = 168 A

671 000/4000

installing a single-phase capacitor bank, the line

250

current drops from

A

to 168

phase equipment by adding balanced 3-phase ca-

sents a decrease of 33 percent.

pacitors at the service entrance.

loss

a.

Active power absorbed by the furnace

and voltage drop on the supply

greatly

is

rises

P = 600

kW

=

0

to

line currents are

89% which

more

at the service

4

line

The reader may want

is

= S/E= = 250 A

l

b.

kV

be reasonably

it

refer to

to

is

2

600

27.8 Measuring electrical energy, the watthourmeter

We

2

have already seen that the SI unit of energy

the joule.

However,

for

many

years,

have been using the kilowatt-hour Reactive power supplied by the capacitor

is

(kW

h)

is

is

ally

kvar

Active power drawn from the line

is

kW

power by

time.

The electricity

line is

=

New power factor

671 is

employed

Fig. 27.12

shows

the principal parts of such a

B p wound with many turns of B c an aluminum disc D ;

supported by a vertical spindle; a permanent mag-

2

net A;

kVA

and a gear mechanism that registers the num-

ber of turns

made by

the disc.

connected to a single-phase

of the line

= PJS L =

600/671

=

0.89

precision motor.

When

the meter

line, the disc is

jected to a torque which causes

cos 6

for residen-

metering.

fine wire; a current coil

= V600 + 300 2

usu-

sumed during one month. Watthourmeters must,

meter: a potential coil

Qc

bill is

based upon the number of kilowatt hours con-

therefore, be very precise. Induction watthourmeters

tial

Apparent power drawn from the

+

One

exactly equal to 3.6 MJ.

are practically the only types

2

measure the

to

energy are called watthourmeters; they are designed to multiply

Gl = G - Qc

P L = 600

is

utilities

Meters which measure industrial and residential

Reactive power that the line must supply

= 800 - 500 = 300

power

energy supplied to industry and private homes. kilowatt hour

Q c = 500 kvar

V/> L

8,

a single-

and inductors.

= VToOO 2 = 800 kvar

-

Chapter

shown how

is

unity power-factor 3-phase load by using capacitors

1000/4

VS^^P

Sl

bal-

phase load can be made to appear as a balanced,

Reactive power absorbed by the furnace

Q

Finally, the 3-phase

likely to

entrance despite the presence of

Section 8.22, wherein in the

will significantly re-

bill.

this large single-phase load.

600/0.6

= lOOOkVA Current

R

be

is

anced

S = P/cos

60%

line will

2

Furthermore, the power factor

reduced.

from

duce the monthly power

Apparent power absorbed by the furnace

A, which repre-

follows that the f

It

it

is

sub-

to turn like a high-


74

27.9 Operation of the

watthourmeter The operation of a watthourmeter can be understood by referring to Fig. 27.13. Load current / produces an alternating flux O c which crosses the aluminum ,

disc,

inducing

eddy currents

in

/,

.

a voltage and, consequently,

it

On

the other hand, potential coil

produces an alternating flux current

I

.

t

The

disc

is

p

,

therefore subjected to a force

(Section 2.22), and the resultant torque causes rotate. It

can be shown that the torque

tional to flux

<£>

p

and current

tively proportional to voltage

E

follows that the motor torque 0,

which

is

the active

load. This, however,

rotating disc


the meter.

Electric)

D

is

is

it

to

propor-

and the cosine of the <£>

El cos

Figure 27.12b Components making up

lt

angle between them. Because

it

Bp

which intercepts

and

/,

are respec-

and load current is

/,

proportional to

power delivered

only part of the story.

to the


742

Figure 27.13 Principle of operation of watthourmeter.

As

moves between

the disc

the poles of

perma-

nent magnet A, a second whirlpool of eddy-currents is

induced

in the disc.

The

interaction of the flux

magnet and these eddy currents pro-

the permanent

duces a braking torque whose value the

speed of the

ways equal it

from

disc.

is

proportional to

Because the motor torque

Example 27-7 The nameplate of a watthourmeter shows K h = makes 17 turns in 2 minutes, energy consumed by the load during and the average power of the load. If the disc

3.0.

calculate the this interval

is al-

Solution to the braking torque (see Section 3.1

follows that the speed

is

1),

proportional to the motor

The latter, we have seen, is proportional to the active power supplied to the load. Consequently, the number of turns per second is proportional to the number of joules per second. It follows that the number of turns of the disc is proportional to the number

Each turn corresponds to an energy consumption of 3.0 W-h. Energy consumed during the 2-minute in-

torque.

terval

is

£h

=

=

3.0

-

51

x number of turns X 17 W-h

of joules (energy) supplied to the load.

Average power absorbed by the load during

27.10 Meter readout In addition to other details, the

watthourmeter

lists

the

amount of energy,

through

the

Consequently,

Kh

.

in watt hours,

Constant

we can

calculate the

Kh

= =

which flows

meter for each turn of the

disc.

amount of en-

ergy that flows through a meter by counting the

number of turns. Then, dividing energy by

time,

can calculate the average value of the active supplied to the load during the interval.

we

power

this in-

is

P = Eh/t = 51/2Wh/min = 51/(1/30) W-h/h

nameplate of a

the rated voltage, current and

frequency, and a metering constant is

terval

51

X

1530

30

W

W

Most watthourmeters have four dials to indicate the amount of energy consumed. The dials are read from left to right and the number so obtained is the number of kilowatthours consumed since the meter was first

put in service. In reading the individual dials,


Figure 27.14 Reading the dials

we always over

last.

of

take the

743

a watthourmeter.

number which

For example,

the pointer swept

in Fig. 27.14, the

reading

is

1-5-9-0, or 1590 kW h. To measure the energy consumed during one month, we simply subtract the

dle and using a single register (Fig. 27.15). rent

the

and potential coils are connected

same way

as those of

Fig. 27.16

is

The

cur-

to the line in

two wattmeters.

a 3-phase, solid-state watthour-

readings at the beginning and end of the month.

meter having a precision exceeding that of induc-

Some modern

tion-type

watthourmeters.

meters are

now

watthourmeters give a digital readout

which, of course,

is

much

easier to read.

watthour-

Electronic

being developed that will monitor

harmonic content and phase unbalance as

27.11 Measuring 3-phase energy

part of the

metering process to meet special contractual

re-

quirements.

and power The energy consumed by a 3-phase load (3-wire system) can be measured with two single-phase watthourmeters. The two meters are often combined into one by mounting the two discs on the same spin-

Questions and Problems Practical level 27-1

Explain what

is

meant by the following

terms:

27-2

demand

billing

maximum demand

fixed cost

mill

demand

Using the

rate

calculate the

schedule given

power

who consumes 920 27-3

Explain

why

bill

27-5

Table 27B,

one month.

low power factor

a

tory results in a higher

27-4

in

interval

homeowner

for a

kW h

in

demand

power

in a fac-

bill.

Explain the behavior of a demand meter.

Using the

27B

rate

schedule given

in

Table

medium-power customer, calcumonthly power bill under the fol-

for a

late the

lowing conditions:

demand meter reading Figure 27.15 Watthourmeter for a 3-phase, 3-wire (Courtesy of General Electric)

billing

demand

= 20 kW 1

= l5()kW

circuit.

energy consumed

= 36 000 kW-h

744


demand interval of 15 new value of the

other one having a

minutes, calculate the

maximum demand 27-8

a.

Give an estimate of the energy consumed one year by

If

the average rate

late the

27-9

modern

a

city of

North America

itants in b.

at 8:00.

is

in

300 000 inhab-

(refer to Fig. 27.9).

40 mill/kW

h,

annual cost of servicing the

A motor draws 75 kW from a cos 8 = 0.72 lagging.

calcucity.

a 3-phase line

at

a. Calculate the value

of

Q and

S absorbed by

the motor. b.

If

20 kvar, 3-phase capacitor

a

P

value of c.

is

with the motor, what

in parallel

Q

and

connected

is

the

new

supplied by the line?

Calculate the percent drop in line current after the capacitor is installed.

27-10

A plant

draws

1

60

kW at a lagging power

factor of 0.55. a. Calculate the capacitors [kvar] required to

power

raise the b.

If

the

power

factor to unity.

factor

how much

ging,

bank cost

(in

only raised to 0.9 lag-

is

less

would

the capacitor

percent)?

Figure 27.16 This high-precision electronic watthourmeter gives a

27-

1

a.

Assuming

that

power

in a large industry

mill/kW

numerical readout of the energy delivered by a 3-phase

be purchased

line. It has an accuracy of 0.2 percent, which compares favorably with the 0.5 percent accu-

having an efficiency of 96 percent.

some

watthourmeused on high-power lines where the monthly consumption exceeds 10 GWh. (Courtesy of Siemens) racy of

of the best induction-type

b.

a.

If

a.

power

factor to 0.9,

maximum demand b.

Would

the billing

would

demand have been

27-

1

to

Example 27-

1

,

the

42

maximum

in

demand registered at 8:00 is 4. 7 MW. If demand meter were replaced by an1

the

in

kW h

if

it

A barrel

af-

1

According

have an

if

effi-

Referring to the residential rate schedule

given

ning

fected?

27-7

to

Table 27B, calculate the cost per only 20

kW h are consumed during

The heating element on an electric stove is rated at 1200 W. Using the same rate schedule, what is the least possible cost of run-

b.

the

have been affected?

motor were redesigned

a given month.

capacitors had been installed so as to

raise the

motor runs night and day, 365 days

ciency of 97%.

The demand meter in a factory registers a maximum demand of 4300 kW during the month of May. The power factor is known be less than 70 percent.

If the

the

27-12

to

can

estimate the

per year, what would the annual saving be

is

Intermediate level

27-6

h,

hourly cost of running a 4000 hp motor

transmission

ters.This meter

at 15

for

of

one hour ?

oil

costing 32 dollars contains

gal (U.S.) having a heating value of

15

000

Btu/gal.

When

the fuel

is

burned

a thermal generating station to produce

electricity, the overall efficiency is typi-

cally 35 percent. Calculate the

minimum


27- 4 1

cost per kilowatt-hour, considering only


the price of the fuel.

27-19

Describe the construction of a watthourmeter.

27- 5 1

Explain

The

why

1

for a load of 10 is

connected

what 27- 6 1

the

is

We want to

kW.

If a 5

in parallel

new

0 r/min

1

a.

The

home by means

found that the disc makes

b.

in

1

minute. If

power of the

open

of

in

10

on the speed of

ro-

and the precision of the meter?

the effect

resistance of coil

A domestic

Does

B c changes

this affect the

800

with tem-

if

the

at

proposed

to use a variable

speed

wide

high- and low-water flows. the

.

Valve fully open

2.

Valve throttled

3.

Inverter drive with valve fully open.

energy

speed of rota-

is

6 cents/kWh and the throttled

condition

day of the

watthourmeter has a precision

possible error

less

partly closed

The cost of a variable speed inverter for the 200 hp motor is $32 000. The cost of

remains fixed?

of 0.7 percent. Calculate the

is

magnet

tation

is

1

1

27.13 decreases by 0.5 percent

tion if the active load 1

is

The following information regarding three modes of operation is given in Table 27D:

heater.

What

The

needed, the valve

1760

3.0,

years.

perature.

27- 8

Kh =

flux created by the permanent

in Fig.

is

It is

determine the power of an elec-

calculate the 1

When

inverter drive and leave the valve

complete turns

27- 7

at

so as to throttle the flow.

with the load,

a watthourmeter. All other loads are shut it is

000 gallons of

r/min with the valve fully open.

rate of rotation?

tric heater installed in a

off and

delivers 10

water per minute when running water

kvar capacitor

induction motor driving a cen-

pump

tifugal

the disc rotates.

disc in Fig. 27. 3 turns at

A 200 hp

745

is

on 17 hours per day, every

year. Referring to Table

27D,

answer the following questions: a. How much energy is saved per day by

maximum

monthly consumption

using

an inverter instead of throttling the valve?

kW h. b.

How much money

is

saved per year by

in-

stalling the inverter?

TABLE 27D Motor input [kW]

Motor losses [kW]

Pump

losses [kW]

Valve losses [kW]

Flow [gal/min]

000

(1)

135

10

25

0

(2)

133

10

27

32

8000

(3)

89

9

16

0

8000

10

Chapter 28 Direct-Current Transmission

28.0 Introduction

teract

and dampen out the power

Quick power control also means development of high-power, high-voltage

The

circuit currents

made it possible to blocks of power using di-

electronic converters has

transmit and control large current.

rect

transmission

Direct-current

can be limited to

oscillations. that dc short-

much lower

values than those encountered on ac networks. 2.

offers

DC

power can be transmitted

We

in

have seen

cables over

that the capaci-

unique features that complement the characteristics

great distances.

of existing ac networks.

We cover here some of the ways it is being adapted and used, both in North America and throughout the world. However,

tance of a cable limits ac

various

a

before undertaking this chapter, the reader should

by cable capacitance exceeds the rating of the

first

ered

Beyond cable

review the principles of power electronics covin

Chapter 2

is

What

are the advantages of transmitting

dc rather than by ac? They I.

this limit, the reactive

itself.

into play

1

may

power by

be listed as follows:

power generated

Because capacitance does not come

under steady-state dc conditions, there

theoretically

no

limit to the distance that

unthinkable. Furthermore, underground dc cable

DC

may be used

to deliver

power

into large ur-

power can be controlled much more quickly. For example, power in the megawatt

ban centers. Unlike overhead

lines,

range can be reversed

ground cable

from

in a

dc line

one second. This feature makes

it

in less

than

When

instability

is

is

invisible, free

of securing rights of way.

about to oc-

3.

We

have seen that ac power can only be trans-

cur (due to a disturbance on the ac system), the

mitted between centers operating

dc power can be changed

frequency. Furthermore, the

in

under-

atmospheric pollution, and solves the problem

useful to op-

erate dc transmission lines in parallel with exist-

ing ac networks.

to

power may be carried this way. As a result, power can be transmitted by cable under large bodies of water, where the use of ac cables is

Features of dc transmission

28.1

power transmission few tens of kilometers (Section 25.29).

amplitude to coun-

746

at the same power transmitted

DIRECT- CURRENT TRANSMISSION

depends upon

line reactance

between the voltages

gle

(Section 25.23). But

by

when power and

into the picture,

not limit the steady-state thing,

it

is

tem. Furthermore, the power flow between the

and the phase an-

each end of the line is

systems can be modified and even reversed matter of milliseconds

transmitted

line reactance

power

does Unlike ac transmission

flow. If any-

power

means

that

fect,

power can

lines,

not easy to tap

it is

off at different points along a dc line. In ef-

However,

this is a

large blocks of ac

dc lines are usually point-to-point systems, ty-

ing one large generating station to one large power-

be transmitted over greater distances by using

consuming

converters

Electronic

center.

are

marginal benefit because

power

installed at each

are already being car-

ried over distances exceeding

none

in

1000 km,

end of the transmission

line,

but

between. However, a multiterminal dc

line

originating at Radisson, near 4.

in a

faster than could

only the resistance of the line that

limits the flow. This also

dc.

— much

be achieved on an ac system.

and phase angles do not

dc, frequencies

come

at

747

Overhead dc transmission

lines

become eco-

built to

nomically competitive with ac lines when the

supply power

to

James Bay, has been

New

England and three

other points in Quebec.

length of the line exceeds several hundred kilometers.

The width of

and experience due

the

to date

to lightning are

power corridor

is less,

28.2 Basic dc transmission system

has shown that outages

somewhat reduced.

A dc

Consequently, dc transmission lines are being

used

to carry

bulk power directly from a gener-

ating station located near a coal fall, to

5.

transmission system consists basically of a dc

transmission line connecting two ac systems.

converter

mine or water-

into dc

the load center.

at

one end of the

power while a

converter acts therefore as a

back-to-back converters, which interconnect

inverter.

only a few meters long. Back-

the transmission system

to-back converters enable the two systems to operate

at their

result,

system do not tend

disturbances on one

line

^Q3

be represented by is

1

1

into dc power.

The dc power

is

carried over a

converter 2

1

Q2

05

Q6

04:

1

O

£d2 *1

-© Q4

Q6

Q5j Q3

Q2

Qi:

Firing

a 3-phase,

2-conductor transmission line and reconverted to

-CZDAQ1

One

other as an

6-pulse rectifier that converts the ac power of line

to destabilize the other sys-

converter

may

the circuit of Fig. 28.1. Converter

respective frequencies and

phase angles. As a

into ac power.

rectifier, the

Stripped of everything but the bare essentials,

large adjacent ac systems with a dc transmisis

A

power

similar converter at the other

end reconverts the dc power

At the opposite extreme of great distance are

sion line that

line converts ac

angle

Figure 28.1 Elementary dc transmission system connecting 3-phase

Firing

line

1

to

3-phase

line 2.

angle

a2

line

2

E2


748

ac power by means of converter

Both the

verter.

rectifier

acting as an in-

2,

and inverter are

line-

and power

28.3 Voltage, current, relationships

commutated by the respective line voltages to which they are connected (Sections 21.9, 21.20,

In a practical transmission line,

21.28, and 21.29). Consequently, the networks

tors L,

can

between the ripple-free dc voltage

function

entirely

at

frequencies

different

without affecting the power transmission between

and L 2

(Fig. 28.2)

2 .26, and 21.31

Power flow may be reversed by changing the firing angles a, and a 2 so that converter becomes an inverter and converter 2 a rectifier. Changing the

tween

pears across

angles reverses the polarity of the conductors, but

harmonic currents flowing

the direction of current flow remains the same. This

to an acceptable level.

1

1

mode

of operation

can only conduct current

in

The dc voltages Ed[ and Ed2

at

The drop

glect

is

each converter

shown

JR drop

E 2G

usually so small that

except insofar as

it,

it

affects

in the

Due

to the high voltages

actually in series.

lines,

composed of

is

typically be

nected

in series.

in

composed of 50

Each converter

300

converter

Figure 28.2 Smoothing inductors

shown

cr-i

is

1

28.3b.

in Fig.

thyristors

inverter.

.35

E

We

150°, respectively,

Consequently, the dc

cos a]

x

=

1.304

E

x

Similarly,

Ed = In the case

angle a,

is

1

E2

cos

of dc transmission

a2 lines, the rectifier

simply designated a. Furthermore, the

inverter firing angle (ot 2 )

.35

is

not considered as a delay

with respect to the rectifier zero firing point.

converter 2

1

(or a)

and L 2 are required between

rectifier is as

waveshape of have assumed fir-

given by

1.35 E, cos 15°

converter

each bridge arm are triggered simultaneously, so

angle

as

Ed = -

28.1 would,

The 50

transmission line

on the dc side of the

thyristors con-

in Fig.

thyristors.

A

together they act like a super-thyristor.

firing

E ]C

losses, effi-

often called a

would

in the

and

is

and

inductors also reduce the ac

.

for the rectifier

in trans-

E 2G

in Fig. 28.3a. Similarly, the

several thyristors connected

Such a group of thyristors

therefore, contain

is

L 2 The

ne-

we can

in Fig. 28.1

Thus, a valve for a 50 kV, 1000

ap-

= 15°anda 2 =

encountered

valve.

Lh

Ed

between

neglect commutation overlap, the wave-

line voltage

each thyristor shown

appears across inductor

ing angles of a,

ciency, and conductor heating.

mission

we

shape of

one direction.

station are identical, except for the line.

If

the undu-

Thus, the potential difference be-

).

E ]G and Ed

Similarly, the difference

required because thyristors

is

Ed and

output of the converters (Sections 21.10,

lating

them.

,

smoothing induc-

must be used as a buffer

firing

fluctuating

dc voltages

angle

E1G E2G and ,

,

a2

(or

(5)

ripple-free voltage

E&

Ol

Q2

Q3

Q4

Q5

Q6

Q1

fires

fires

fires

fires

fires

fires

fires

Figure 28.3a Rectifier voltage

waveshape E1G

for

a = 15° (neglecting commutation

overlap).

Figure 28.3b Inverter voltage

waveshape E2G

for

p

= 30°

(neglecting commutation overlap).

749


750

but as an angle of advance p. Thus, as regards the

a 2 = 150°

inverter, instead of stating that

have done

in

previous

all

inverter

(as

we

circuits

in

/,, I 2

is

related to

a 2 by

£"],

Ci>

Qi =

Note 90° and

The

80

- a2

a =

that the inverter voltage is zero

maximum when p —

when p

and 2 [AJ

effective values of the respective ac

p

is

=

powers absorbed by con-

reactive verters

1

1

line voltages [VJ

the simple equation

=

P

effective values of the rectangular

currents in ac lines

E2 =

Chapters 21, 22, and 23), we now refer to it as an angle of advance (3 = 30° (Fig. 28.3b). The value of

p

=

1

and 2

rectifier angle

[var]

of delay

[°]

inverter angle of advance

f°l

0°.

order to keep the reactive powers

In

power relationships of a dc transmission system are the same as those for any circuit containing ac/dc power converters. Referring to Fig. 28.2, and based upon equations we voltage, current, and

have already seen, the relationships

may

as low as possible,

proach

0°.

we

However,

attempt to

and

Q2

make a and p

ap-

Q

]

for practical reasons,

and

tak-

ing account of commutation overlap, the effective

value of

be stated

a

is

about 25°, while that of p

35°.** Using these values,

we

is

about

can calculate the

rel-

as follows:

magnitudes of the voltages and currents

ative

P = EJd Ed = 1.35

E, cos

Ed = .35 E 2 cos /, = U = 0.8 6 0, = /Man a 1

1

Q 2 = P tan

model of a transmission line. Tn we assume the line delivers 1000 A at a scale

a

(21.17)

100

Ed = /d =

active

dc

** The effective firing angles a* and

(21.18)

power transmitted [W]

a and

1.

The

2.

The commutation overlap angle

3.

The

actual firing angles

The approximate

line voltage [V]

(3

jul

relationship between these quantities

cos a*

=

1/2 (cos

cos

=

1/2 (cos

(3*

P

[

a + cos (a +

jul))

-

JUL))

|3 + cos = 7 + M-

(inverter) /d =

E = 82 kV o

£d

y

1000

= 100

A

kV

12

L!

/>

= 100

oE 2 *

Q2 ^

°

Figure 28.4 Scale model of a simple dc transmission system.

* 820 A

MW

^47 Mvar = 25

(0

is

converter 2

1

h % 820 A

"(effective)

the fol-

extinction angle 7

(rectifier)

<2i

depend upon

(3*

lowing:

dc line current A]

^=100 MW

potential of

(Fig. 28.4).

(21.6)

d

p

converter

a

model,

p

where

P =

kV

this

in

^(effective)

= 350

70 Mvar

P 2 = 100

MW

90 kV

DIRECT-CURRENT TRANSMISSION

Thus,

we have

Solution

£d = 100

1

kV =

.35 £, cos

a

1.35 £, cos 25

Using the scale model of

Fig. 28.4

by the appropriate

we

82

{

The ac

line voltages are

E = =

(15()kV/10()kV)

E2 = =

(150 kV/100 kV)

]

£2

1.35

kV - 1.35 E2 £2 = 90 kV

cos 35

c

The

b.

:

=

/2

=

0.816

X

0.816

820

123

X 82

kV

cos p

also =

and multiplying

find the following:

kV

Furthermore,

Ed =

ratios,

c

a.

E =

100

751

1

= U =

/,

The

active

power absorbed by

1000

60

The

reactive

is

150 kV X 400 A = 60 000 kW

A d.

the rectifier

=

P,

MW

X 820

(400/1000)

= 328 A c.

100

is

d

A

kV X

kV


1000

100

35

x 90

MW

power absorbed by each converter

is

Gi

:

=

P

tan

a

Qi

100 tan 25°

=

(60/100)

-

28

X 47

Mvar

47 Mvar

Q2 = P

e2

tan

(60/100)

X 70

= 42 Mvar

p

100 tan 35°

Figure 28.4

may

28.4

70 Mvar

:

thus be used as a scale model to

determine the order of magnitude of the voltages, currents,

and power

in

any dc transmission system.

Power fluctuations on a dc line

In order to ensure stability in transmitting dc power, the rectifier

and inverter must have special voltage-

current characteristics.

Example 28-1

A dc transmission current of

line operating at

400 A. Calculate

1

50

kV carries

a

the approximate value

of the following: a.

b. c.

d.

These characteristics

shaped by computer-controlled gate-firing

The ac line voltage at each converter station The ac line current The active power absorbed by the rectifier The reactive power absorbed by each converter

We

are

circuits.

can best understand the need for such controls

by studying the behavior of the system when

the

controls are absent. Fig. 28.5

resistance R.

and

£,12,

shows a dc transmission line having a The converters produce voltages Edl

and the resulting dc

line current

is

given by (28.1)


752

converter

network

converter 2

1

1

network 2

/d

Hh

Figure 28.5 small change

A

The

in

Ed1

either

resistance

line

Ed2

or

produces a very

can produce full-load current

Furthermore, a small variation

can produce a very big change

Ed]

increases by only a

Edl

in either in / d

.

few percent,

can easily double. Conversely,

if

Ed[

and

the

in

Ed]

E 2 may

in

/

d

.

Ed2

Ed2

The

fluctuate.

and

or

Ed2 if

the line current

increases by zero.

fall to

are subject to

sudden changes because the associated ac ages E\ and

change

line.

For example,

only a few percent, the line current can Unfortunately, both

big

^2

always small; conse-

is

quently, a very small difference between

Ed2

—W

£d2

£d1

line volt-

fluctuations

may

be due to sudden load changes on the ac networks or to

any number of other system disturbances

Owing

occur.

to the

of the converters and transmission rent could

that can

almost instantaneous response

Figure 28.6a Rectifier E-l characteristic.

dc cur-

line, the

swing wildly under these conditions, pro-

Eu

Edl

the dc output voltage

kept

ducing erratic power swings between the two net-

line voltage

works. Such power surges are unacceptable because

constant until the line current I d reaches a value

they tend to destabilize the ac networks

Beyond

at

each end,

is

true that the firing angles

modulated tions.

a and

|3

could be

to counteract the ac line voltage fluctua-

However,

it

is

preferable to design the system

so that large, unpredictable dc herently impossible.

power surges

We now show how this

keeping a constant desired value tomatically)

/,

at

are in-

to /,,

when Ed]

done.

were

to

is

Edl

.

As

28.5 Typical rectifier and inverter

=

0. In

controlled rectifier circuit I

curve shown

is

designed

in Fig. 28.6a.

to yield the

Assuming

E-

a fixed ac

other words,

would be equal

regards the inverter,

until the

E2

,

obtained by

equal

rate, so that / d is if

a short-circuit

occur across the dc side of the

line voltage

computer-

is

approaches the

firing angle then increases (au-

a very rapid

E-l curve shown

characteristics

until current ld

The

resulting dc current

In a practical dc transmission system, the

drops sharply, as can be seen

on the curve. This E-l characteristic

and because they produce misfiring of the SCRs. it

this point,

is

it

designed

is

in Fig.

28.6b.

voltage

Ed2

rectifier, the

to /,. to

Assuming

is

give the

a fixed ac

maintained

dc line current reaches a value

at

72

.

zero

This

means that from zero to 72 the firing angle |3 = 90°. As soon as the dc current approaches the desired ,


753

The

differ-

mined by

the inverter characteristic).

ence between A/.

/[

and

I 2 is

called the current margin

held constant and equal to about 10 percent

It is

of the rated line current. £"d2

If the line

has appreciable resistance, the IR drop

modifies the effective rectifier characteristic so that it

follows the dash line

in Fig. 28.7a. This,

however,

does not affect the operating point under normal conditions.

(b)

is,

28.6

/d

fier

Inverter E-i characteristic.

effective

the firing angle

/2 ,

decreases (automati-

(3

Under normal operating conditions the voltage level is kept slightly below the

the

power flow over

. {

The

and inverter characteristics

stant but

/|

and

I2

ElU

and

Ed2

are varied simultaneously while

rectifier

line current /d smaller than that in Fig. 28.7a.

I2

is

effect of these

(Fig. 28.7a).

The

for a transmission

we can cause

the dc

power

to vary over a

range. Note that the line voltage that

it

is

always determined by

rent correspond to the point of intersection of the

termined by the

tic)

/,

It is

obvious

that the line current /d

while the line voltage line

is

(determined by the rectifier characteris-

£d

is

equal to

Ed2

(deter-

At

wide

Ed2 is constant, and the inverter. On the

actual transmission-line voltage and the actual cur-

equal to

is

de-

may wonder why

the

rectifier.

this point, the reader

E-I characteristics have been given such odd shapes to attain

such a simple

result.

The reason

is

that the

IR drop

\

fdl

£di

operating point

operating point

Ed2

£"d2

1 inverter

inverter

£d

AI

A/

id

Figure 28.7b

Figure 28.7a Operating point when the transmission rated power.

line delivers

By

thus shifting the E-I characteristics back and forth,

other hand, the magnitude of the line current

two curves.

si-

are kept con-

shows the new E-I characteristics

constraints can best be seen by superposing the rectifier

the dc line, the recti-

inverter

voltage level. Furthermore, limiting current slightly smaller than l

. {

keeping the current margin fixed. Thus, Fig. 28.7b

cally) to a limiting value of about 30°.

made

to the inverter

Ed2 I

and inverter E-I characteristics are modified

multaneously. Voltages value

power input

Power control

To vary

Figure 28.6b

The

therefore, given by the product

Operating point rated power.

when

the

line delivers

20 percent

of


754

dc system must be able to accommodate serious ac voltage fluctuations at either end of the line without affecting the dc limit the

short-circuit

how

plain

power flow too much.

magnitude of the

occur on the dc

this

is

must also

It

fault currents, line.

We

will

Finally,

one of the worst conditions

tifier

maximum current /] while the inmaximum current /2 Consequently,

verter draws a

now

the fault current

.

achieved.

h=

28.7 Effect of voltage fluctuations

- h)

(A

only 10 percent of the normal line current (Fig.

is

28.9). Fault currents are, therefore,

Referring to Fig. 28.7a,

assume

us

let

that the

dc

line carries full-load current /,. If the ac voltage of line

but this has no effect on

power flow over

On

Ed2

Ed] rises in proportion, or E d2 Consequently, the

increases suddenly,

1

the other hand,

decreases

in

l x

the line

proportion.

Ed2

unaffected, but because

The dc is

However, the percent change ceed the percent change Next,

may

if

line current is

in

line

also

is

shown

This produces a

in Fig. 28.8.

creases suddenly from l to {

The dc /2 ,

less.

E2

new

1,

operating

line current de-

Ed2

not excessive. Consequently, the

bance

is

It is

is

zero, the

power delivered

to

small.

now

clear that the special shape of the E-l

power

fluctuations on

the line, and limits the short-circuit currents. In practice, the actual E-l characteristics differ slightly

from those shown

in Fig. 28.7.

However, the basic

principle remains the same.

Edx

to

£tU With .

current margin of 10 percent, the drop in current

again not affected too much.

Ed2 are close to

and

28.8 Bipolar transmission line

.

while the dc voltage

decreases equally suddenly from

smaller

power cannot ex-

ac voltage

a large disturbance occurs on line

fall drastically.

point,

in

decreases,

smaller than before,

dc power carried by the

the

EdX

the fault

characteristics prevents large

E2

voltage

if line

much

than on ac transmission lines. In addition, because

.

unaffected.

is

can arise

supplies a

should a ex-

that

a short-circuit on the dc line. Here again, the rec-

is

power flow

As soon

a is is

as the distur-

cleared, the E-l characteristics return to the

original curves uiven in Fis. 28.7a.

Most dc transmission sess a positive line

mon ground

return (Fig. 28. 10a).

end of each

stalled at the /d) 1

and

ld2

flow

and 3 act as

inverters.

is /d

line

They

pos-

and a com-

A converter is

and the

in-

line currents

shown. Converters

while converters 2 and 4 are

Power obviously flows over both lines to ac network 2. The ground cur-

— ,

line,

in the directions

rectifiers

from ac network rent

lines are bipolar.

and a negative

ld2

.

1

It is

usually small because the con-

verters automatically maintain equal currents in the

positive and negative lines.

The bipolar arrangement has First, the

ditions.

ground current

is

three advantages.

small, under normal con-

Consequently, corrosion of underground converter 2 (inverter)

If = /,-/;

Figure 28.9 The short-circuit

Figure 28.8

Change

in

operating point

when

falls drastically.

current at the fault cannot exceed 10

percent of the rated

line current.


pipes,

and

structures,

so

forth,

minimized.

is

/d2

Second, the same transmission-line towers can carry

two

doubling the power, with a

lines, thus

power flow on one

line

is

in the

same

direction as before

(Fig. 28.10b).

rel-

atively small increase in capital investment. Third, if

continue to flow

155

28.10

interrupted, the other

Components

of a

transmission

line

dc

can continue to function, delivering half the normal

power between

In order to function properly, a dc transmission sys-

the ac networks.

tem must have auxiliary components,

Power reversal

28.9

in addition to

the basic converters. Referring to Fig. 28.11, the

most important components are

To reverse power flow

in

the firing angles, so that verters

a bipolar line,

all

the rectifiers

we change become

1

in-

DC

.

line inductors (L)

and vice versa. This reverses the polarity of

2.

Harmonic

and

3.

Converter transformers (T,,

the transmission lines, but the line currents /d

converter

1

,

filters

on the dc side (F dc )

positive line

t

T2

)

converter 2

1 network

1

-m-

/d2

—

7 b

^ negative line

/J

4

= 35°

Figure 28.10a Bipolar line transmitting

converter

—m-

network

1

0,

power from network

1

to

network

2.

converter

1

= 35° network 2

converter 3

= 35°

Figure 28.10b Power reversal from network 2

a. = 25°

to

network

1

is

obtained by reversing the

line polarities.


756

microwave communications

.

link

/

p))—

Figure 28.11 Schematic diagram showing some

4.

Reactive power source

(Q^

Harmonic

6.

Microwave communications

7.

Ground electrodes (Gd)

more important components

of the

Q2

5.

(((1

HVDC

transmission system.

before the current becomes too large to handle elec-

on the ac side (F ac )

filters

an

This enables the thyristors to establish control

cur.

)

of

tronically.

link

between the

converter stations

The need

28.12 Converter transformers

components

for these

is

explained

in

the

The

following sections.

£L 28.11 Inductors

and harmonic

filters

verter.

tiary

converter)

3

1

).

They

2th harmonic currents, and such

1

allowed to flow over the dc

line,

could

produce serious noise on neighboring telephone lines.

Consequently,

the currents

filters are

to

sometimes added

ground

.

LC

(Fig. 28.

if

essentially

power

seen, the dc line voltage

constant

from

no-load

to

Ed

circuits,

1

1

is

a line fault should oc-

kept

Furthermore, to reduce the reactive power absorbed

by the converter,

firing angle

means

that the ratio

small. This

that £,

a

should be kept

between ac voltage

Because

Ed

is

constant,

it

is

es-

follows

must also be essentially constant.

Unfortunately, the network voltage

each

).

for

(Q\).

full-load.

input and dc voltage output of the converter

also prevent the dc line current

from increasing too rapidly

is

sentially fixed.

line.

tuned to respectively short-circuit the 6th and 12th

The inductors L

winding (Section 12.5)

in either

A lower-voltage ter-

or wye-delta, are used.

The filters shunt filter F dc The

two inductors L and a composed of two series

harmonic currents

by the con-

required to prevent

from flowing over the

consist of latter is

network voltage

Three-phase transformers, connected

As we have

Voltage harmonics are produced on the dc side of

if

to transform the ac

direct connection to a source of reactive

both the rectifier and inverter (Section 21 give rise to 6th and

is

to yield the ac voltage £, required

,

wye-wye

on the dc side (6-pulse

currents,

basic purpose of the converter transformer on the

rectifier side

£ u may

vary

significantly throughout the day. Consequently, the

converter transformers

on the

rectifier

side

are

equipped with taps so that the variable input voltage

£

L

,

will give a reasonably constant output voltage £,


The

taps are switched automatically

tap

changer whenever the network voltage

by a motorized

reasons, taps are needed on the converter transform-

on the inverter

28.16 Ground electrode

E L]

changes for a significant length of time. For the same

ers

757

Particular attention at

paid to the ground electrode

is

each end of the dc

side.

Direct currents in the

line.

ground have a corrosive

on pipes, cables and

effect

metallic structures. Consequently, the actual ground

electrode

28.13 Reactive power source The

reactive

power

Q

absorbed by the converters

must be supplied by the ac network or by

power

active

a local re-

Because the active power

source.

usually located several kilometers from

is

the converter station, to ensure that dc ground currents create

no

problem around the

local

The dc ground wire between tual

grounding

station.

and

the station

the ac-

pole-mounted or en-

site is either

a shielded cable. At the grounding

transmitted varies throughout the day, the reactive

closed

in

source must also be varied. Consequently, either

special

means

variable static capacitors or a synchronous capaci-

tance. This

tor are required (Section 17. 15).

system

is

minimize electrode

are used to

when

particularly important

operates

temporarily

in

a bipolar

monopolar

the

mode. Under these circumstances, the ground

28.14 Harmonic

on the ac side

filters

Three-phase, 6-pulse converters produce 5th, 7th, l

lth,

choppy current waveforms (Section 2

1

.

1

l

).

Again

cur-

rent

may exceed 1000 A, and

may

eventually dry out the grounding bed, causing

the heat generated

the ground resistance to increase.

3th (and higher) current harmonics on the

1

ac side. These harmonics are a direct result of the

site,

resis-

The

best grounds are obtained next

large bodies of water. But rate

even

to,

or

in,

in this case, elabo-

grounding methods must be used.

for reasons of telephone interference, these currents lines.

must not be allowed

to

flow over the ac

Consequently, the currents are bypassed

through low-impedance

tween the 3-phase

lines

filters

neutral point

each 3-phase

LC

resonant

is

grounded.

filter is

F ac connected be-

and ground. The

each frequency are connected

On

in

wye, and the

Hz

a 60

composed of

filters for

network,

a set of series-

circuits respectively tuned to 300,

LC circuits are almost entirely ca-

pacitive. Consequently, they also furnish part of the

reactive

power

Q

Fig.

28.12 shows the elementary circuit diagram of a

monopolar mercury-arc ing dc line operates at

essential.

For example,

to

the line ting

/,

is

maintain the current at

Two

is

fed into a

smoothing

in series

with the dc

line.

The two LC filters

and 12th harmonic

voltages generated on the dc side of the converter. The

Q and

wye

both ends of

communications link between them

margin A/ (Fig. 28.7), the inverter

The incom-

1

1

Q resistors make the filters

less sensitive

frequency changes on the outgoing ac

line.

Three single-phase transformers connected wye-

28.15 Communications link the line, a

line.

effectively short-circuit the 6th

to slight

In order to control the converters at

inverter station.

50 kV, and power

inductors, each having an inductance of 0.5 H, are

9

absorbed by each converter.

1

230 kV, 3-phase, 60 Hz power connected

420, 660, and 780 Hz.

At 60 Hz these

28.17 Example of a monopolar converter station

one end of

(with a tertiary winding) are connected to the

ac side of the converter.

A

160

Mvar synchronous

capacitor, connected to the tertiary winding, pro-

vides the reactive

power

for the converter.

Filters for the 5th, 7th,

1

1th,

and 13th harmonic

must "know" what the

rectifier current set-

currents are connected between the three ac lines

This information

continually relayed

and neutral of the 230

is.

is

kV

system.

As

previously ex-

by a high-speed communications link between the

plained, the filters shunt the ac harmonic currents so

two converters.

that they

do not enter

the

230 kV

line.

758


Figure 28.12 Simplified circuit of

a 150

kV,

1800

A,

60 Hz mercury-arc

inverter.

See components

illustrated in Figs.

28.13

to 28.17.

Figure 28.13 These 12 single-phase harmonic an

ters at

fil-

inverter station are tuned

300 Hz, 420 Hz, 660 Hz, and 780 They are connected between the three lines and neutral of the outgoing 230 kV, 60 Hz transmission line. The filter in the foreground is tuned to 720 Hz. It is a series circuit composed of a 2 (1 resistor, a group of for

Hz.

capacitors having a total capacitance of

0.938

of

44.4

jxF,

and an

oil-filled

mH. The 720 Hz

inductor

reactive

power associated with the LC circuit amounts to 18.8 Mvar. (Courtesy of GEC Power Engineering Limited, England)

Figs. 28. 3 through 28. 7 give us an idea of the 1

size of these various

1

components, and of the im-

mense switchyard needed

to

accommodate them.

28.18 Thyristor converter station

become almost standardized. Thus, each pole composed of two 6-pulse converters. Fig. 28.18a shows how two 200 kV converters are connected to produce a 400 kV dc output. The dc has is

sides are connected in series, while the ac sides are essentially connected in parallel across the

This means that converter 2

Mercury-arc converters have been supplanted by

230 kV, 3-phase

thyristor converters and the design of the latter

(and the secondary winding of transformer T2)

line.

Figure 28.16 Portion of the 3-phase, 6-pulse mercury-arc inverter rated

270 MW, 150

kV.

{Courtesy of Manitoba Hydro)

dc side of the inverter station. Theêcond inductor can be seen in the distance (lower right-tiand corner). The space between the two units permits installing the filters on the dc side. Courtesy of Manitoba Hydro) (

Figure 28.17 Partial

view of the refrigeration

inverters. (

159

Courtesy of Manitoba Hydro)

unit

needed

to cool the


760

functions

at a

dc potential of 200 kV. The wind-

ings must be especially well insulated to with-

stand these dc voltages,

in

addition to the

1

80

single-conductor cable operates mits 20

kV

connected

in

was

link

laid

in

Two

England and France.

former T2 are connected in wye-wye. This produces a 30° phase shift between the secondary

kV and

transsea.

+

100

kV and

English Channel between

the

are

while those of trans-

wye-delta,

100

at

ground current returns by the

English Channel. In 1961 a bipolar submarine

3.

ac voltage.

The 180 kV windings of transformer Tl

MW. Tbe

one operating

cables,

the other at

-

at

100 kV, laid side-by-

MW

60 of power in one direcThe power exchange between the was found to be economical because

side, together carry

1

tion or the other.

voltages of Tl and T2. Consequently, the thyristors in converter

same

the

and converter 2 do not

1

time. In effect, the

two converters

two countries

fire at

the time zones are different and, consequently, the act as

system peaks do not occur

same

the

at

time.

a 12-pulse converter.

One

important result of the 30° shift

Furthermore, France has excess hydro generating is

that the

two

5th and 7th harmonic currents generated by the

converters tend to cancel each other on the primary

Tl and T2 and do

side of

not, therefore,

appear

in

capacity during the spring, thus making the export

of power attractive. Pacific Intertie.

4.

1970 a bipolar link operat-

In

±400 kV was

230 kV line. Consequently, the filtering equipment for these frequencies is substantially reduced.

Oregon, and Los Angeles, California. The overhead

Furthermore, the 30° phase

line transmits a total

the

harmonic on the dc

equipment needed

for

shift eliminates the 6th

which reduces the

side,

Fdc

filtering

a 12-pulse converter, together with the ac and dc

The valves are called quadruple because each is composed of four bridge

connections. valves

spective

MW over a distance of

of 1440

to

flow

in either di-

depending upon the requirements of the

NW

and

SW

regions.

The dc

re-

link also

helps stabilize the 3-phase ac transmission system

connecting the two regions.

Nelson River. The hydropower generated by

5.

arms. Fig. 28.21b illustrates the impressive size of the

these valves.

between The Dalles,

installed

1370 km. Power can be made rection,

.

28.18b shows the three valves that make up

Fig.

ing at

Nelson

River,

km

890

situated

north

of

means of two bipolar lines operating at ±450 kV. Each bipolar line carries 1620 MW, which is converted and fed Winnipeg, Canada,

28.19 Typical installations Power transmission by direct current is being used in many parts of the world. The following installa-

into the ac

system near Winnipeg. According

studies made,

transmit

transmitted by

is

it

was

slightly

power by dc

rather

to

more economical to than by ac over this

tions give the reader an idea of the various types of

systems that have been built over the years, and the particular 1.

problem they were designed

Of

Schenectady.

mile, 5.25

between York,

in

MW,

30

to solve.

Eel River. The back-to-back

6.

historical

kV

interest

is

the

17 the

230 kV

electrical

and

Hz and 60 Hz

was

installed in

first

Schenectady,

New it

Brunswick. Although both systems operate

nominal frequency of 60 Hz,

Sweden,

in

km

it

was not

at

a

feasible to

tied

important dc transmission line

mainland by a 96

New

connect them

directly,

due

to stability considera-

system.

1954.

It

connected the

Island of Gotland (in the middle of the Baltic Sea) to the

between

transmission line installed

1936. Using mercury-arc converters,

Gotland. The

intertie

systems of Quebec and

tions. In this application, the 2.

station at Eel River,

Canada, provides an asynchronous

Mechanicville

together a 40

considerable distance.

submarine cable. The

is

dc ^transmission line"

only a few meters long, representing the length of

the conductors to connect the rectifiers and inverters.

Power may flow in of 320 (see

imum

MW

either direction, Figs. 28.

1

up

to a

9 and 28.20).

max-

primaries

+400 kV 0.5 H +

rwv

•

©

©

Q

dc

line

+300 kV

© +200

kV filters

400 kV

Fdc

(dc)

"T" I

I

Figure 28.18a Schematic diagram of one pole of a ±400 kV converter station. It consists of two 200 kV converters connected in series on the dc side. The converters are 6-pulse units, respectively connected to 3-phase voltages that are 30° out of phase. The dc filter is tuned to the 1 2th harmonic. The ac filters prevent the 1 1 th, 1 3th, and higher harmonics from entering the 230 kV system. The static var compensator supplies the reactive power needed by the converter.

Y1

Y2-

Y3-

converter

D1

D2-

D3-

converter 2

1

-® quadruple valve

1

quadruple valve 2

quadruple valve 3

Figure 28.18b Schematic diagram of the 12-pulse converter showing the two 6-pulse converters and quadruple valves constitute the main components in one pole of the valve hall. 761

line

connections. Three

762


Figure 28.19 This converter station and switchyard at Eel River connects the ac networks of

Quebec

and New Brunswick by means of a dc link. The rectifier and inverter are both housed in the large building

in

the center.

It

pioneered

the commercial use of solid-state thyristors

HVDC

in

applications.

(Courtesy of

New Brunswick

Electric

Power

Commission)

Figure 28.20 View of one 6-pulse its

thyristor valve

rectangular cubicle.

It

is

housed

in

fed by a 3-phase

converter transformer and yields an output of

2000 A

40

at

thyristors

it

kV.

The hundreds

of individual

contains are triggered by a

reliable,

interference-free fiber-optic control system.

Eight such cubicles, together with three synchronous capacitors and four converter trans-

formers,

make up

the entire converter terminal.


This pioneering station was the thyristors in a large 7.

CU

to

first

use

commercial application.

±400

kV.

A metallic

Electric)

ground return

is

provided, in

the event that one line should be out for a prolonged

Project The power output of a generating

mines near Underwood, North Dakota, is converted to dc and transmitted 436 miles eastward to a terminal near Minneapolis, Minnesota, where it is reconverted to

period (see Fig. 28.2

la).

Fig. 28.2 lb

three quadruple valves that

shows

make up one

the

12-pulse

station situated next to the lignite coal

ac.

The bipolar

line transmits

1

000

MW

at

1

250 A,

converter. 8.

Chdteauguay

Substation.

The

substation, located near Montreal,

converter station rated

at

1000

is

Chateauguay a back-to-back

MW (Fig. 28.22).

In


Coal Creek Terminal

-702

North Dakota

Dickinson Terminal

km

(436 miles)-

Minnesota Pole

TCC 500

1

TD

valve hall

Pole

1

"X +400kV ^

MW

763

1250

A

if

2x 1590 MCM ACSR

Of

Fdc

Fdc

T1

10.3

km

MCM

954

20 K km m MCM

Gd2 0.029

T2

n

ff 0.4 H

0.11

n

0.4

345 kV

H

system

~ Fdc

500

MW

2 x 1590

230 kV 308/339

MVA

Pole 2-

-400kV

The

Figure 28.21a

HVDC in

1000 MW over a distance of 702 km. The output from two 500 turboalternators is stepped up to 230 kV and transmitted to the Coal Creek Terminal where the ac power is converted to dc.The on-load tap-changing converter transformers TCC are connected wye-wye and wyeline delivers

MW

delta for 12-pulse converter operation.

The

0.4

H smoothing

inductors are

posed

of

a 48.8

mH

The dc

inductor

in

pacitor bank, prevent the 12th

reaching the dc

filters

series with

Fdc, each

series with a

1

MVA lines con-

20 km from the respective terminals. Under normal conditions the line currents are controlled automatically

so that the ground current

However,

if

one pole

is

ca-

is

20

A

or less.

out of service for short peri-

ods, the ground current can be as high as 1375 A.

The 12-pulse

inverter station (Dickinson Terminal)

feeds into a 345 kV, 60 Hz system and tap-changing

TD

are used to regulate the

in-

verter voltage level.

The

in-

com(jlF

292/321

and negative transmission

converter transformers in

the grounded lines, thus significantly reducing the sulation requirements.

positive

Pole 2 valve hall

two bundled conductors (2 x 1590 MCM, ACSR). The dc grounds are situated at 10.3 km and

North Dakota to the Dickinson Terminal

±400 kV

A

sist of

transmission system that links the Coal Creek in

1250

Pole 2 valve hall

Simplified schematic diagram of the bipolar

Minnesota. The

[|]

it

system

Terminal

MCM ACSR

control

terminal trol

system

is

arranged

to

operate each

unmanned from a telecommunications

center located

in

con-

Minnesota.

harmonic voltage from

lines.

Figure 28.21b View of three quadruple valves being installed in one of the Coal Creek Terminal valve halls. Together they constitute one pole of the 400 kV system. (Courtesy of United Power Association)

Figure 28.21c Aerial view of the Coal ing the

two valve

Creek Terminal show-

halls that respectively pro-

duce the +400 kV and -400 kV dc voltage. The switchyard contains circuit breakers, transformers, and filters. {Courtesy of United Power Association)

1 \

Figure 28.22a

A quadruple

1

140 kV, 1200 A IREQ, the Hydro-Quebec

valve, rated at

dc, being tested at

research center. The valve

m

wide, and 2.7

m

deep.

It

m

is

12

is

designed

high, 6.9 for

the Chateauguay substation and contains a total of

400

thyristors.

{Courtesy of Hydro-Quebec)

764


765

Figure 28.22b View of one 500 MW, 140

kV, 3600 A dc back-to-back converter at the Chateauguay substation. It is composed of 6 quadruple valves. The three valves on the right usually operate as rectifiers, and the three on the left usually function

as inverters. The valve

hall is

17.5

m

wide and 18

m

high.


order to ensure high reliability,

two independent valve

Power 735

kV

state of link,

halls,

usually flows from the

ac system to the 765

New

York.

Owing

composed of at 500 MW. Hydro-Quebec

is

it

each rated

kV

The

thyristors are water-cooled, using deionized

water and an elaborate water/glycol/air-heat exchanger.

ac system in the

to the rectifier/inverter


frequency changes on one system do not affect

the other system.

Furthermore, tbe direction of

power flow can be reversed, depending upon circumstances.

the

Practical level

28-

1

Give three examples where dc power transmission

is

particularly useful.

766

28-2


Name

the principal

components making

a. Calculate the

28-3

Which harmonics occur on

On

converter?

b.

the ac side of a

What

1

line current per pole

transmission line operating

50 kV,

at

estimate the total reactive power absorbed

600 A. The terminal con-

carries a current of

tains a single 3-phase, 6-pulse converter.

by each converter 28- 2 1

approximate value of the sec-

a. Calculate the

ondary ac

line voltage

of the converter

What

is

tance of 0.5

The bipolar

shown

line

erates at a potential of

in Fig. 28.

±

1

50 kV.

a.

A and

28- 3 1

28-7

The transmission

ground current

line

shown

possesses a resistance of 10 fier (converter

a.

kV

1

)

The

recti-

while the inverter generates 96 kV.

Id

it

is

of the inverter are fired a

dc

and

given that

= 800 A, and

the

1

Ed

two dc in-

If a short-circuit oc-

little

magnitude

ing the gate triggering remains unaltered.

28- 4 1

Each pole of the bipolar Nelson project (Fig. 28. 8) is composed of two conduc1

tors (2-conductor bundle) of

earlier in the cycle, will the

A

loss at the

of the rectifier current after 5 ms, assum-

produces a dc voltage of

Calculate the dc line current and the power

If the gates

700

1

rectifier station, calculate the

in Fig. 28.5

(I.

resis-

the line currents in

curs between line and ground close to the

transmitted to network 2. b.

kV,

ductance of 0.5 H. the

from the

smoothing inductors L each have an

transmitted between the two ac

The value of

102

fl. If


= 450

networks b.

km

5

electrode.

dc


The power

a bipolar con1

1400 A, calculate the power

10a op-

If the

600

line currents are respectively

400 A,

located

each pole are respectively

line current?

28-6

is

and possesses a ground

station,

the effective value of the secondary

station.

The ground electrode of verter station

transformer. b.

on the Pacific

and using the model of Fig. 28.4,

Intertie,

A dc

£he approximate peak inverse volt-

Referring to Section 28. 19, calculate the

ductors? 28-5

is

the dc side?

the purpose of the large dc line in-

is

What

age across each valve?

28-1

28-4

dc current per valve (bridge

arm).

up a dc transmission system.

line current

strands

ACSR

composed of 72 of aluminum (diameter 0. 6

Each conductor

cable.

is

1

in)

increase or decrease? Explain. c.

If the

inverter gates are fired so that the in-

verter generates

1

10 kV, will the

power

When

Each 2-conductor bundle 800 A over a distance of 550 miles. The voltage at the rectifier terminal is 450 kV. in) steel core.

carries a

flow reverse? Explain.

28-8

and a central 7-strand (diameter 0.1067

a short-circuit occurs on a dc line,

the current in the fault itself

is less

than

nominal current of

1

Neglecting the presence of the

steel core,

the full-load current. Explain.


28-9

Why

is

a

communications

link

needed be-

a.

tween the converter stations of a dc line?

The

effective cross section of the 2-

conductor bundle b.

Intermediate level

The

2

[in

]

line resistance of the

dle at a temperature of

28-10

The converters shown

in Fig.

28.2 are

identical 3-phase, 6-pulse units,

a voltage

1200 A.

Ed

of 50

kV

c.

producing

and a current

/d

of

d. e.

2-conductor bun-

20°C

The corresponding I~R loss The dc voltage at the inverter terminal The efficiency of the line (neglecting corona losses)


Advanced 28- 1 5

28- 16

level


1

a.

is

b.

The resonant frequency of the two dc filters The value of the respective series imped-

c.

What

Problem 28-

1

5, if the 6th

20 kV, calculate the approximate value of

b.

Calculate the value of the 6th harmonic voltage across the 0.5

the dc voltage across the capacitors?

volt-

the corresponding harmonic current.

ances is

harmonic

age generated on the dc side of the inverter

2, calculate the fol-

lowing; a.

In

767

c.

What

is

H

the value of the

inductor.

360 Hz voltage

the input to the second 0.5

H

at

line inductor?

Chapter 29 Transmission and Distribution Solid- State Controllers

without compromising reliability and

29.0 Introduction

idea

The successful development of thyristors, and other electronic switches changes

in controlling

power flow

is

GTOs,

utilize

promoting major

in the

ties

utilities.

On

multiplicity of transmission

and distribution

made

it

power flows

that

It is

lines has

created

demand

problem of

failures,

electric utilities

is

circuits.

the

power-handling capacity of their existing

power

lines that

now

handle currents of several

we have

seen

in

volts.

Chapter 28 that they are

HVDC

systems of several

These switching devices now being incorporated into equipment such as hundred

kilovolts.

ries capacitors, var ters,

compensators, harmonic

we

will

sefil-

cover some of the impor-

tant solid-state controllers that

lines

are

and ultra-high-speed switches.

In this chapter

ways whereby they can increase

are looking for

can

already being used in

that

electric utili-

ties

that

Indeed,

for electric

For these and other reasons,

possible to envisage

thousand amperes and several thousand

power continues to grow, while it is becoming more difficult to obtain rights of way to erect more transmission and distribution

the

now

possible thanks to the existence of thyristors and

GTOs

create stability problems that might get out of hand.

special

that instabili-

a potentially dangerous situation. This rapid action

switching surges, and sudden load shedding do not

The

is

instantaneously to any contingency and counteract

will take. Furthermore,

by equipment

the important problems

are ''active," in the sense that they can react almost

necessary to allow wide margins of safety so

instabilities

to

to carry the electrical load.

occur very quickly and can build up and spread

the other hand, the

is it

The

erwise exacerbate the situation.

the complexity of these transmission systems has

made

stability.

thermal limit and

play to disconnect devices and loads that might oth-

rel-

ever more difficult to predict the amount and

direction that

all

to their

onds. Thus, circuit breakers must be brought into

ing transformers, and static var compensators, trans-

atively passive elements.

them

them up

out throughout an entire system in a matter of sec-

With the exception of circuit breakers, tap-changmission and distribution systems have comprised

to load

One of

transmission

and distribution sectors of electric power

is

have been developed

recently and which have undergone tests in the field.

768

TRANSMISSION AND DISTRIBUTION SOLID-STATE CONTROLLERS

We

begin with power flow controllers for transmis-

sion,

which

FACTS

are

(Flexible

classified

AC

under

Transmission System),

lowed by power electronic controllers tion.

acronym

the

The FACTS program was pioneered by the Power Research Institute (EPRI) of Palo

Electric

manufacturers and electric power

equipment

in

The new TCSC approach is to vary the transmispower capability in accordance with imis

accomplished by

in series

with the line on an

mediate requirements. This varying the capacitance

when

instantaneous as and

required basis.

Consider, for example, Fig. 29.1a

FLOW CONTROLLERS we

of this equation was shown

sion line

utilities.

TRANSMISSION POWER In the transmission sector

derivation

Section 16.23.

fol-

for distribu-

Alto. California, in collaboration with

The

769

in

which two

capacitors having a reactance

a;,

series with a transmission line

having an inductive

are connected in

reactance X, per phase. Each capacitor can be con-

will look at the fol-

nected to an inductive reactance A a by means of

lowing equipment: B Thyristor controlled series capacitor

a.

synchronous compensator

b.

Static

c.

Unified power flow controller

d.

Static

(TCSC)

(STATCOM)

i-Mh

(UPFC)

frequency converter

Q2

Q1

29.1 Thyristor-controlled series Figure 29.1a

capacitor (TCSC)

One phase

Chapter 25, Section 25.26, we saw

In

the

that

power-handling capability of a 3-phase transmission line can be increased by introducing a fixed ca-

of the transmission line can be raised. the

power

We recall

tance of the line that

Ql valve

sin R o

.

are blocked, only the ca-

is

is

(X

—

2a;,).

On

the other hand,

if

triggered "on" so that uninterrupted

(25.4)

P =

active

E=

line-to-line voltage at

power transmitted [MW]

line

jx

where the voltages

Es E R power

8

7

-Vr-A

:

,

-J

-

vc

,

equation becomes in

(29.2)

at

line [°]

line are not equal, the

case,

A'J„

X +

[il]

phase angle between the voltages

more general

..

namely

x„

inductive reactance per phase

each end of the

with a l

each end of the

[kVJ

each end of the

falls in parallel

The resulting impedance between points 1, 2 becomes inductive (Fig. 29.1b), with value equal to /xcjf,/(A c — xa ). The effective reactance X clY of the line is then the sum of the impedances seen in the figure,

In the

Q2

conduction takes place, A a .

wherein

at

considerably smaller than a c

valves Ql,

given by

E2 X

X= 8 =

line.

pacitors are in the circuit and so the effective reac-

the is

x,d is

When

pacitor in series with each phase. This reduces the

power

series-compensated

back-to-back thyristor valves labeled Q. The reactance

effective series reactance and, therefore, the

of a

(29. 1)

Figure 29.1b Impedance when

Q1

Q2

closed

open

line is partially

compensated.

B -o


770

The

thyristor valves

Q L Q2 can

be switched on and

off independently; consequently, for a given phase

angle between points

A

and B, the active power

transported can be varied as required.

the

The switch-

done within one cycle, which means

ing can be

Fig. 29.3. In practice, for overvoltage protection,

metal oxide v.aristors are also included

(MOV)

and a

circuit breaker

They do not appear

the circuit.

in the figure.

We

that

power flow can be very quickly controlled. The following example will illustrate the switch-

in'

a.

The

wish to determine the following: effective

impedance of a single

capacitor/inductor unit ing process, and the results that can be achieved.

when

the thyristors con-

duct fully b.

Example 29- 1

A

ance of 54

km

(2,

R

per phase (Fig. 29.2).

long and comprises three

con-

line

c.

has an imped-

conductors

The voltages tween 215

in

is

line is

10

1

conductors

R

kV. Furthermore, the phase

tween 8° and

S always leading

17°, with region

As we learned

that region S will

in

is

is

a.

The inductive impedance x p of the capacitor parallel with the inductor is given

tempt

is

made

line

much power

To meet these

as possi-

1

.7

1

II

in parallel

b.

x -

=

2C

12

1.71 1.71

The maximum nominal power

is

is

determined by

transmission line:

each having

an impedance of 12 Q. Each capacitor can be connected

12

the nominal current and nominal voltage of the

objectives, the transmission line in series,

_ ~

(29.3)

at-

ble within the thermal limit of the conductors.

equipped with four capacitors

— xa

1

a

is

consequently, every

to transport as

xc

P

to re-

In addition to its role as a stabilizing link be-

facility;

in

by

re-

gion R.

tween the two regions, the transmission

237 kV, and the phase angle

15°

Section 25.23, this means

always deliver active power

revenue-producing

compensa-

Solution

angle between the two regions varies randomly be-

gion R.

region voltage

between them

1050 A.

both regions vary randomly be-

kV and 246

best configuration of the series

(number of valves Ql, Q2, etc., in action) when the S region voltage is 218 kV, the

having a cross-section of 1000 kcmil. The thermal limit of the

The

tion circuit

The

ACSR

that the trans-

mission line can carry

230 kV, 3-phase, 60 Hz transmission

necting two strong regions S and

The maximum nominal power

^nominal

EIX

3

= 230 000 X = 418MW

with an inductive reactance of

by means of thyristor valves 0. as shown

—

in

1050

x V3

region 12fl

R

12U

F <:

215 to 246 kV 110 km (5=8°

ACSR

to 19°

thermal

limit

1000 kcmil 1050

Q1

A

and R

Q3

Q4

Figure 29.3

Figure 29.2 Transmission

Q2

line

(receiver).

connecting two regions

S

(sender)

Transmission capacitors.

line with four thyristor-controlled series


c.

In order to carry the desired

418

MW, =

and 8

given

Es =

nominal power of

218 kV,

15°, the effective line

ER =

237 kV,

impedance

MW

The 393 is quite close to the maximum power of 4 8 MW. The phasor diagram for this condition is shown in Fig. 29.4b. This example shows that the thyristor-controlled rated

Xetr

can be calculated using the expression

1

series capacitors

EsEr

sin 8

(29.4)

ity

418

In

Xeff =

that

two

32

ft

given by Fig. 29.4a.

thyristor valves

ing while the remaining

It

can be seen

Q3, Q4 are conducttwo are not. The result-

ing net reactance of the transmission line

54 -

24

mitted

is,

+

4

to

of the transmission

= 34

ft.

The

some

line.

it is

useful to reduce the con-

actual

power

is

higher than

value. This vernier control enables the troller to vary the effective

mission line over a

is

much wider Example

trans-

the conduction period

is

(29.4)

e eff

218

X

range.

29-

1,

suppose that

The combination of

is

4 12 instead of

the 12 12 capacitor in

parallel with the 4 12 reactor gives an inductive re-

actance of 6

ft, as

follows:

23 7 sin 15°

(29.3)

34

P = 393

MW 2Q

12Q * 1= 1058 A P = 393

con-

trans-

shortened so that the effec-

tive reactance of the inductor

-^sin8 X

actual

its

TCSC

impedance of the

Referring back to

therefore,

_

applications

reactance x, of the inductors

1.71 ft.

218 kV

meet

duction period of the thyristors so that the effective

the closest configuration approaching

this result is

and out

in

29.2 Vernier control

whence

trial,

can be switched

any power requirement within the thermal capabil-

X 237

218

By

77

MW

2Q

_ 12X4 ~ 12-4 r 237 kV -15°

= +6 On

the other hand,

12 (inductive)

the conduction period

if

short-

is

ened even more, the effective reactor impedance can be raised to 36

(a)

sulting

LC parallel

reactance of

1

ft.

Under these conditions

the re-

combination yields a capacitive

8 ft, as follows:

1058 A

x

p

=

—

(29.3)

218/V3

_ ~

12 12

= —

237/V3

Thus, the

TCSC

X 36 - 36

18 ft (capacitive)

vernier technique

is

seen to offer

an additional advantage as compared to a conventional series

Figure 29.4 Circuit configuration for tion.

a given sender/receiver condi-

compensation arrangement.

However, care must be taken

to prevent the short-

ened conduction period from creating a condition of

772


parallel

resonance wherein the reactance of the

in-

that a forbidden

Nominal system voltage

2.

Nominal

3.

Nominal 3-phase compensation: 202 Mvar

its

4.

Nominal capacitive reactance, per phase

is

progressively increased

base value of 1.71

Vernier control stability

is

problems

f).

when between two regions. The

particularly advantageous

arise

5.

way as to counteract the osThe almost instantaneous action of the thyristors, supported by feedback signals and computer algorithms, makes such a maneuver feasible. Fig. 29.5 shows a large TCSC system installed in cillation.

6.

Hz transmission

TCSC

system.

modules

It is

7.

The

in

delayed conduction mode):

The

TCSC

conduction mode): 1.22 il is

designed to withstand the follow-

on 30-minute overload current: 4350

composed

10-second overload current: 5800

A A

maximum fault current through module: kA maximum crest fault current in thyristor valve: 60 kA

20.3

thyristor-

controlled series capacitor unit has the following rating and

effective capacitive reactance, per

ing overload conditions:

that are individually

protected by metal oxide varistors.

(in-

Effective inductive reactance, per phase (inductors in full

Bonneville Power Administration's 500 kV,

of six identical

A

24(2

transmission line in such a

the C.J. Slatt Substation in Northern Oregon,

Maximum

phase (inductors

be damped out by modulating the power flow over the

3-phase, 60

2900

ductors not in circuit): 8 il

low-frequency power oscillations that take place can

the

line current:

(line-to-line):

band must be skipped as the reac-

tance of the inductor

above

500 kV

1.

ductor approaches that of the capacitor. This means

components:

Figure 29.5a Overall view of the

TCSC

bypass breakers are isolate

are

in

at the C.J. Slatt Substation in

at the

left.

The

them from ground. On each

the boxes at the

right, and mounted on three platforms to reactors are in the middle, and thyristors

Northern Oregon. Disconnect switches are at the

capacitors, reactors,

and

thyristor valves are

platform, capacitors are at the right,

left.

The TCSC is part of EPRI's Flexible AC Transmission System (FACTS) program. The project was developed by Power Research Institute in collaboration with the Bonneville Power Administration and General Electric Company.

the Electric

{Courtesy of EPRI)


<

773

oypass

To Buckley

Disconnect

1

To

Slatt

Isolation

Disconnect lect

TCSC Series

Module

Capacitor Isolation

Varistor

Disconnect

/

-

\

Reactor

J

Reactor

Thyristor

Valve

O-

Bypass Breaker

Figure 29.5b Schematic circuit diagram of one phase (Courtesy General Electric Company)

It is

worth noting that

this installation

of the

is

TCSC

system, installed

the first of

kind in the world.

its

in

power

reactive

500 kV transmission

series with the

is

needed

line.

to prevent further voltage

collapse.

Switching converters were discussed

29.3 Static synchronous

21.44, and a 3-phase in

compensator

PWM

in

Section

version was described

Section 21.49. However, the converters

interested in

we

are

do not make use of high-frequency

that the voltage

PWM techniques because the megawatt powers in-

of a transmission line can be controlled by means of

volved require the use of GTOs, and these switch-

In Sections

25.22 and 25.27

we saw

a compensator located at the receiver end of the line.

The compensator in

power

delivers or draws reactive

order to stabilize the voltage. Traditionally, these

compensators have been rotating machines

(Fig.

17.24) or static var compensators that require large

capacitors and inductors (Fig. 25.39).

Today,

it

is

possible to replace these machines

and a group of transformers. This

chronous compensator, or

STATCOM,

static syn-

has numer-

ous advantages over previous compensators. it

acts

much

faster

and can respond to voltage

tuations in a matter of one cycle. Second, erate far

age

is

more

low

reactive

— which

is

power when

just the

hertz.

For

reason the converter oper-

this

wave mode

ates in the rectangular

on/off switching

The

basic

is

done

at the

60 Hz

in

it

First,

fluc-

can gen-

the system volt-

moment when

a lot of

duces.

is

It

which

the

line frequency.

STATCOM converter is represented in

Fig. 29.6 together with the rectangular

and devices by a switching converter, a dc capacitor,

ing devices can only operate at frequencies of a few

hundred

waves

it

pro-

essentially identical to the 6-step con-

verter described in Section 23.10.

The rectangular

line-to-line voltages contain a

fundamental component whose peak value equal to 1.10

£ H where E H ,

input to the converter. line-to-line voltage

and 0.78

the

Eu

I \

effective 3

0.45

1.10

E u I V2 =

line-to-neutral

EH

is

the dc voltage at the

follows that the effective

It

is

is

.

0.78

EH

voltage

is


774

XYZ

-T

Eh

EU

I

Eh

J

, 0

^C

)

t

t

gi

g2

t

1

Eh

£B 2

Figure 29.7 Principle of operation of a static

En

synchronous com-

pensator.

1.10 A

E

J

the

GTO gate pulses gl

,

g2, g3. Thus, the phase an-

gle of the converter voltage can be set to any value 1.10

£H

/-;

^

H

between zero and 360°, with respect

£H

mission line voltages U, V,

to the trans-

W on the secondary side

of the transformer.

To understand what happens

We select

the line-to-neutral voltages for terminals

A and

namely sator,

Figure 29.6 Converter and waveshapes

for static

var

EAn

we

and

compen-

line voltages are rectangular, they

contain the 5th, 7th, and higher odd multiples of the

Hz

frequency. Harmonics that are

multiples of three, called triplens, are absent. Fig. 29.7 installation.

a schematic diagram of a converter

It

comprises a 3-phase high-voltage

transformer T; three reactances x; a 3-phase

£H between termi-

converter; a capacitor C; and a dc voltage source

The magnitude of nals A, B, C,

phase angle

is

is

the ac voltage

controlled by varying

£H

,

Because

this is a var

U,

compen-

.

and the

controlled by appropriate timing of

arranged so

phase angle of converter voltage it

is in

£An

phase with the corresponding

transmission line voltage

E Vn We now .

examine

three cases. 1

is

transmission line X, Y, Z; an ideal 3-phase step-

down

.

are only interested in generating reactive

this result, the is

fundamental 60

E Un

power. Consequently, the line current /A must lag or lead the line-to-neutral voltages by 90°. To obtain

sator.

Because the

in the circuit, let us

consider one phase of the 3-phase system.

.

If £An = E Vn no current will flow in the reactance x and so the compensation is nil (Fig.

29.8a). 2.

If

£ An

is

less than

£ Un a current IA will flow E Un (Fig. 29.8b). Its mag,

that lags

90° behind

nitude

given by

is

— FUn

(29.5)


1

E Vn

A =0

115

tions-from the dc side to the ac side and vice versa. This feature can be put to remarkably good use.

£An

(a)

Suppose

phase angle of the converter

that the

delayed so that

it

sion line voltage by, say,

EAn

E \Jn

verter to receive active

-+>

line

and

power

this

1

°.

This will cause the con-

power from

will

have

to be

dc power supply, minus the losses

On the other hand, t

'a

(b)

voltage

if

1°,

from the converter

'a

only occur

By

at the

power

active

that

value of

EAn

of

EUn and depend upon /

is

be delivered

will

to the transmission line.

This can

expense of the dc power supply to the converter.

adjusting the phase control, the current

drawn from the power supply can be

(C)

Figure 29.8 Phasor relationships

absorbed by the in the converter.

the phase angle of the converter

which must now deliver dc power

>

I

the transmission

arranged to lead the transmission line

is

voltage by, say,

A

is

lags slightly behind the transmis-

required

is

to set the

verter voltages A, B,

C

/H

set to zero. All

phase angles of con-

so they lag slightly behind

the the corresponding transformer voltages U, V,

W

.

just

enough

to provide the losses in the converter.

The power supply E H can therefore be dispensed

As

a result, the compensator draws reactive

power from

with altogether, leaving only the capacitor

The compen-

maintain

the

required

dc

C

to

The voltage

voltage.

sator behaves like a large inductance even

across the capacitor can be increased or reduced by

though no coils are present and no magnetic

simply advancing and retarding the small phase an-

field is 3.

the transmission line.

If

£An

produced. is

gle

greater than

£un by 90°

E Un

(Fig. 29.8c).

,

current / A will lead

The magnitude of IA

again given by Eq. 29.5, except that

it is

mentioned above.

way

In this is

nega-

As a result, the converter delivers reactive power to the transmission line. The converter

tive.

voltage level

the capacitor

EH

across terminals A, B,

needed

to

Example 29-2 The converter in

field are present.

damental

if

it

were a large capacitor even

In practice, transformer

T

always has a certain

leakage reactance. Therefore, under real-life conditions,

it

is

the leakage reactance of the transformer

that constitutes the reactance

lows

x

charged up

to a dc

C

has precisely the value

produce the required var compensation.

though no electrostatic plates and no electric

then behaves as

is

so that the resulting ac voltage

Fig. 29.7

line voltage

is

rated to generate a fun-

ranging from 4

an effective current of 2000

kV

to

6

kV

at

A per phase.

The 230 kV transmission line voltage is stepped to 4.8 kV by means of a transformer. The

down

leakage reactance x of the transformer, referred to in Fig. 29.7.

It

fol-

the secondary side, has a value of 0.2 12.

The

ca-

that the transformer fills a dual role: trans-

pacitor

forming the voltage, while providing the reactance

needed to permit compensation to take place. Next,

let

EH and have already learned

us look at the dc power,snpply

the associated capacitor C.

We

that the converter can transfer

power

in

both direc-

bank on the dc side of the converter has

pacitance of 500 a.

(jlF.

Calculate the converter line voltage so as to deliver a total of 6.4

mission

line.

a ca-

Mvar

£ AB

needed

to the trans-


776

b.

Calculate the de voltage across the capacitor

by the leakage reactance of the transformers ensures

bank under these conditions.

correspondingly low harmonic currents. Fig. 29.9

Solution a.

The current needed

to

shows

produce 6.4 Mvar

Q - 770 /\ * ~ E UV V3 4800 V 3 6 400 000

-

The voltage drop across

E = x

/.v

the reactance

= 770 X

0.2

-

154

to a

kV

161

is

transmission /

a commercial installation of a

±100 Mvar STATCOM connected line.

contains

It

eight

converters,

whose voltages are phase-shifted to reduce the harmonic voltages and currents on the high-voltage side of the transformers. It was developed by EPR1

is

with

collaboration

in

V

TVA

Westinghouse

and

Electric Corporation.

Line-to-neutral voltage induced on the sec-

ondary side of the transformer

is

power flow

29.5 Unified

E Vn = 4800 /V3 = The converter exceed

£ Un

2771

V


E An

must

Consider two electric

utility

regions

by 154 V; hence

A

and B

that

are individually so strong that their voltages are es-

£ An =

2771

+

154

= 2925 V

EH =

assume

EB

and

EA]i = 2925 V3 = 5066 V Capacitor dc voltage

magnitude and

sentially fixed in

further

Converter line-to-line voltage

b.

controller

(UPFC)

in

phase. Let us

EA Under

that the line-to-neutral voltages

are equal and in phase (Fig. 29.10).

these conditions,

the regions are linked by a trans-

if

mission line having an impedance X, there can be

is

2925/0.45*

no active or reactive power exchange between

= 6500 V

them, because the line current

29.4 Eliminating the harmonics

Adding capacitors

The rectangular waves generated by a single 3-phase converter, such as that shown in Fig. 29.7, would produce large current harmonics in the trans-

tween the two ends of the

in series

/

would be

zero.

with the line would not

help because there would be no driving voltage be-

mission

For

line, a situation that

this reason, instead

could not be tolerated.

of only one converter, sev-

eral converters are used.

voltage, but the respective voltages are shifted

from

each other by definite, specified angles. These phaseshifted voltages are isolated

from each other and can,

be applied individually to low-voltage

transformer windings. The windings on the highvoltage side are connected in series in such a to cancel

way

out most of the harmonics (5th, 7th,

result

is

as

etc.),

while successively adding up the fundamental 60

components. The

may have excess used

in the

turbance

generating capacity that could be

neighboring region. Again, a sudden

in

Hz

a composite sinusoidal

bility. It

power in order to maintain stawould be most useful if the other region

could then help out to manage the contingency.

Speed

is

of the essence during such emergencies,

and so the flow of active and reactive power over the transmission line should be rapidly

and

selec-

tively controlled.

To meet these voltage source

be varied, line.

One

is

objectives, suppose that an ac

£c whose ,

magnitude and phase can

somehow connected

solution

is

voltage containing only the fundamental and high-

verter on the region

The

0.45 drawn from page 773.

dis-

one of the regions might require extra

frequency harmonics. The high impedance offered

*

line.

unfortunate because one of the regions

is

active and reactive

Each converter produces a rectangular output

therefore,

This

in series

with the

to use a dc/ac switching con-

A

side of the line (Fig. 29.1

la).

E T ahead of the line reactance is now the phasor sum of £ A and Ec rather than its former value E A If the phase angle between ET and resulting voltage

.


Liquid-to-air

111

Heat Exchangers

Figure 29.9a Physical layout of the components of a ±100 Mvar static synchronous compensator (STATCOM) installed in the Tennessee Valley Authority (TVA) Sullivan Substation near Johnson City, Tennessee. This project represents a joint collaboration between the Electric Power Research Institute (EPRI), TVA, and the Westinghouse Science and

Technology Center. (Courtesy of Westinghouse Electric Corporation)

EB

is 8, it

follows that active power will be trans-

mitted over the

line,

given by

P = The phasor diagram the phase angle

cj>

of

constant, the tip of

As a

result,

maximum

ET

sin 8

(29.6)

(Fig. 29.1 lb) is

shows

that if

varied while keeping

Ec

angle 8 will progressively change from a

it

will pass

maximum

through zero. Thus, the ac-

power transported over the line can be positive means that power can flow in ei-

as needed

its

when E A and E B /

magnitude can be varied

by varying the magnitude and phase of Ec

.

always

is

are equal and in

right angles to

at

Ec

.

power is delivered or absorbed by the converter. However, the converter delivers reactive power equal to Q c = EC I vars. This is precisely the reactive power absorbed by the line real

reactance X. Note also that so that

cj)

=

90°, active

Suppose now

that

£A

the phase of at

unity

(Fig. 29.12).

As

E H are E A being

and

have different values, with

EB

if

power

Ec

power

is

set

factor

be delivered by region A.

will

negative value,

or negative, which

ther direction. Moreover,

that

Consequently, no

will follow the dotted circle.

positive to a

during which tive

Ec

ET E M X

Note

phase, phasor

in

phase but

smaller than

Ec pivoting

before, the phase angle of

can be varied over a complete

around the end of phasor across the line impedance

is

circle,

E A The ET — E B

voltage drop

.

,

and current

/

Figure 29.9b is one of eight switching converters used in the TVA Sullivan Substation to control the reactive power of a 161 kV transmission line. The converter has a rating of 12.5 Mvar and operates from a nominal dc bus of 7600 V. The 3-phase, 60 Hz output voltage has a rating of 5.1 kV. Five GTOs rated at 4.5 KV and 4000 A turn-off current are connected in series for each switch arm. The entire converter station occupies an area 48 feet wide and 90 feet This

long.

(Courtesy of Westinghouse Electric Corporation)

Figure 29.10

778


780

of all switching converters. Converter 2

Ec and

nish any voltage

power

In addition to delivering real

able to fur-

is

phase angle that

is

filters also

to converter 3.

2 (via the dc link), converter

1

produce the reactive power ab-

sorbed by the converters.

required.

Series-tuned

filters to

ance path for the harmonic currents generated

absorb or deliver reactive power to region A, just

by the converters like a static var

compensator.

4.

and has indeed been named unified power flow con-

(UPFC).

formers. Furthermore, on account of

its

GTOs,

extremely

respond

to

the controller can be

made

Three-phase feeder

to

each 6-pulse converter. a 12-

current harmonics. 5.

Two

3-phase, 6-pulse ac/dc converter

bridges, connected in series, with grounded

any power flow contingency.

to

fre-

pulse output, which reduces the voltage and

rapid operation, which depends only on the switching speed of the

harmonic

The two converters together produce

can replace phase-shifting trans-

It

(5). Principal

quencies are 550 Hz, 650 Hz, and higher.

Thus, the two-converter arrangement of Fig. 29.13 constitutes a very versatile power controller

troller

provide a low imped-

can simultaneously

intermediate point.

The converters

are de-

signed for reversible power flow, hence the

29.6 Static frequency changer Frequency changers have been

back-to-back thyristors. This permits energy

in service for

to be returned to the 150

many

power for The low frequencies

years, mainly to provide low-frequency

railway transportation systems.

were needed

runs.

dc link operates

power lines. A furcommutation were the prime movers of

voltage drop, along the overhead

was

ther reason

locomotives

at the time.

shows a the

given

is

in

Chapter

17, Fig. 17.2.

power needed

electric utility to the 16 2/3

6.

7.

it

An

inductive

filter (6)

peak

on-state

reduces the harmonic

A dc circuit breaker offers when

Today, the availability of high-power switching

made

The

it.

protection in the

event of commutation failure of converters

for a railway system.

converters has

4400 V; mean

ripple of the dc current flowing through

It

Hz

1, 2.

a nominal voltage of

current: 1650 A.

rotary frequency converter that transforms

50 Hz power of an

are line-commutated

V. Thyristor rating: Repetitive

off-state voltage:

These frequency

converters always involved rotating machines, an ex-

ample of which

2650

at

to permit satisfactory

of ac series motors that electric

The converters

source when power on downhill

and feed the dc link between points

reduce the reactance and, hence, the

to

kV

trains regenerate braking

they are operating

in

the inverter

(5)

mode

(power being fed back toward the 150 kV

possible to effect the freline).

quency conversion without using any rotating machines at

all.

gram of

a

Fig. 29. 14

20

MW

shows the basic

circuit dia-

frequency converter.

static

8.

A harmonic

comprises the following numbered components:

duces the 33 1/3 age.

1

.

Transmission delivers

line,

power

Two

tuned

33 1/3 Hz, which

is

The single-phase power output of the

converter station obliges the dc link to deliver

A circuit HV line.

to the converter station.

pulsed power to the switching converters 9.

The

(10).

capacitor at the input to each single-phase

3-phase wye-delta-delta, and wye-deltaconverter acts as a

star

to

Hz output frequency. It reHz ripple of the dc link volt-

150 kV, 3 phase, 50 Hz, that

breaker permits disconnection of the 2.

filter

twice the 16 2/3

It

filter

and ensures

that the

transformer banks that reduce the 150 kV,

50 Hz

line voltage to

verter bridge. allel,

The

1

190

tertiary

V

converters operate in the voltage-source mode. for each con-

windings are

connected to harmonic

filters (3).

in par-

The

10.

The

16 2/3

Hz

switching converter modules are

single-phase water-cooled units.

The GTOs

in


gram

in

the

ABB

of

a 50 Hz

to

1

6

% Hz

static

frequency converter station. (Diagram adapted from a

Review, 5/95 edition)

78

circuit dia-


782

synchronous compensators (STATCOMs). These units are

connected

in parallel

with the

line.

The

possibility also exjsts of injecting a voltage in series

with the

trol

line,

using a

UPFC. The

latter

can con-

both the active and reactive power flowing

over the transmission

line, in

addition to providing

local var control.

Similar controllers are being developed for the distribution sector,

where individual consumer needs

are particularly important. In this sector, disturbances

such as voltage sags, voltage swells, harmonic dis-

power

tortion,

and power factor have

interruptions,

Table

29A

Figure 29.15

to

These water-cooled GTO converter modules are installed in the 50 Hz to 16 % Hz Giubiasco converter

the problems that have to be addressed.

station

in

be dealt with (Fig. 29.

bances originate on the consumer

Switzerland.

and

utility side,

(Courtesy of ABB)

150 Hz. Thus, the frequency modulation is 9.

GTO rating:

age:

4500 V; peak

ratio

Repetitive peak off-state voltturn-off current:

3000 A.

The outputs of the 12 converters are connected to the two primary windings of six transformers (11). The secondary windings are connected in series to produce the 66 kV, 16 2/3 Hz single-phase output (12). The converters are triggered sequentially so that the output voltages are out of phase. As a result,

most of the harmonics are eliminated and

the resulting

waveshape

is

almost a perfect sine

wave. Under full-load and unity power

harmonic distortion

The converter

is

less than

station

factor, the

0.35%.

designed to operate

is

network. In one special

mode

of operation, the sta-

tion serves as a single-phase static var

compen-

sator to stabilize the voltage of the network. In this

the converter station

the 150

kV

transmission

is

disconnected from

DISTRIBUTION CUSTOM

POWER PRODUCTS 29.7 Disturbances

on

distribution

systems We

have seen that the voltage on high-voltage

transmission lines can be regulated by using static

some of on the

side, others

others can be traced to both.

tree that falls

on a 24

utility side.

On

kV

feeder

the other hand, an arc furnace that

produces random and violent changes

in current

can

distort the voltage feeding the foundry, as well as

that of

neighboring consumers. Such voltage pollu-

tion

produced on the consumer

is

side, but a cus-

tomer on the same network plagued with flickering lights sees

it

as a utility disturbance. Thus, the link

between consumer and

mon

coupling

is

ble to distinguish

many

to

why

it

is

between the two

gin of a disturbance

Having said

utility at the

the reason

this,

is

point of

com-

often impossi-

as far as the ori-

concerned.

both the customer and the

utility

have distortion-free and reliable power.

In

instances consumers have installed uninter-

ruptible

power supplies (UPSs)

to prevent distur-

bances from reaching sensitive electronic equipment (Fig. 29.17). In hospitals, operating

port landing fields, where

power

kind cannot be tolerated, the electric generators

line.

lists

Some distur-

creates a disturbance that clearly originates on the

want

separately or in parallel with an existing 16 2/3

mode

still

For example, a

the converters operate at a carrier frequency of

16).

and dc

term emergency power utility

rooms, and

air-

interruption of any

UPS

includes diesel-

batteries to provide long-

in the

event of a prolonged

outage.

Owing

to the proliferation of nonlinear loads,

electronic drives, and other harmonic generating

some studies have shown that consumers would prefer to have the utility ensure the supply of quality power rather than doing so themselves. However, the reality of the situation is constantly devices,

random

power interruption

distortion

10000

5000

-5000 o

-10000 -1

5000

brownout

switching

transient

7th

15000

harmonic distortion, 14% THD

-i

> 10000 )ltag<

5000

-

•

> o ( neutn

-5000 -1

0000

-

-1

5000

•

ine-to-

Figure 29.16 Typical disturbances that occur

in

distribution

and transmission systems.

783

1

^

af6 0

5*Q

7/2

0


784

DISTRIBUTION DISTURBANCES

TABLE 29A

Duration

Nature of disturbance

consumer

low power factor

hours

voltage swells and sags

cycles

harmonics (current)

hours

harmonics (voltage)

hours

random voltage

hours

distortion

voltage transients

cycles

high short-circuit current

cycles

voltage regulation

hours

power

interruption

cycles

power power

interruption

seconds

side (C)

(U)

utility side

C C c c c c c

(shunt or series)

shunt

U

shunt or series shunt

u u u u u u u u

hours

interruption

'Solid-state breaker 2

Compensation

Origin

series

shunt or series

shunt or series series

shunt or series

+ SSB + SBB

shunt or series

shunt or series shunt

+ SSTS

1

1

2

(SSB)

Solid-state transfer switch

evolving, especially

(SSTS)

in the

context of the pending

deregulation of the electric utility industry. For instance, the lowest cost solution

may

be that Power

is

particularly attractive in distribution systems

because of the harmonics that are present

in

both

voltages and currents. These harmonics must be

minimum

when

Quality becomes a service provided by third parties

kept to a

to either the distributors of electricity or the indus-

must be taken either to eliminate them or to divert them into paths where they can do no harm. The

commercial, and residential consumers. The

trial,

changes that are occurring

institutional

fore, affect both the

will, there-

approach to the solution and the

magnitude diminishes with increasing

quency. Thus,

power quality. Toward this end, manufacturers, research

able

tutes, tric

and universities

utilities

—

are

—

in collaboration

lated converters in the kilowatt to the

range. These

nology used in

with elec-

developing pulse-width modu-

megawatt

PWM converters are based upon techin electric drives,

such as those covered

Chapters 21 to 23. The reader

may want

to refer

to these chapters to review the basic properties of

if all

in

many

cases

it is

deemed

fre-

accept-

harmonics below the 13th (780 Hz)

are

suppressed. If the

means

highest harmonic of interest

is

the 13th,

it

frequency should be about 10

that the carrier

times as great. The carrier frequency must therefore

be about

This

is

1

0

X

1

3

X 60 = 7800 Hz, or about 8

kHz.

within the capabilities of high-power IGBTs.

Converters operating troduce their

own

at

these carrier frequencies

in-

high-frequency distortion and

means must be taken

these converters.

they appear, means

harmonics are usually multiples of 60 Hz, and their

choice of products that are utilized to improve

insti-

and,

to limit the resulting carrier

current.

29.8

Why PWM

PWM

converters are extremely versatile because

converters?

they can generate a voltage of any shape, any fre-

Another reason

that favors

PWM

their ability to generate sinusoidal

rather than the rectangular

Consequently,

converters

is

60 Hz voltages

waves produced by GTOs.

PWM converters can be interfaced di-

quency, and any phase by simply applying an ap-

rectly with the distribution

propriate gating signal to the IGBTs. This feature

filter

network without having

or phase-shift the harmonics of the

60 Hz

to

rec-


static

785

bypass switch

output

input

T

rectifier

electric

circuit

breaker

utility

inverter critical

load

battery

Figure 29.17a This rudimentary single-line diagram shows the ba-

an on-line UPS. Power from the rectified and the output is connected to the terminals of a battery. The battery serves as a permanent standby source of energy and also ensures a ripple-free dc input to the inverter. The inverter generates the regulated, high-quality, 60 Hz

This 18 kVA, 120/208

voltage to power the

generates an output of 120/208

sic

elements

electric utility

If

a

utility

of

is

power

critical load.

interruption occurs, the inverter

continues to operate, typically for several minutes,

drawing

its

energy from the

battery.

The

static

bypass

switch serves to automatically connect the electric ity

to the load in the

event of a failure

in

util-

the conver-

Figure 29.17b V,

60 Hz

UPS

at output

power factors ranging from 0.7 lagging to 0.7 leading. The THD is less than 5 percent, even with nonlinear loads. The full-load efficiency is about 90 percent. The low noise level and small size are due to the high switching frequency (~16 kHz) of the IGBT inverter. If power

sion components.

3-phase, 60 Hz on-line

V,

interrupted, or

is

exceeds the input tolerance

window, the internal battery supplies power to the tangular pulses. However, converters,

A final

the case of large

where more than one converter

quired, phase-shift

many

in

methods can be used

reason for using

distorted

power

unit

is

re-

up to 10 minutes with no interruption (Courtesy Square D/Groupe Schneider) verter for

in-

to load.

to advantage.

PWM converters

waveshapes produced by

is

that 3-ph,

industrial

converter

processes contain voltage and current harmonics

60 Hz output + carrier

that bear

quency.

no relationship

to the

60 Hz

line fre-

PWM switching converters are able to gen-

erate voltages

and currents

in

error-correcting triggering signals

opposition to these

random distortions and thereby neutralize them. The signals driving the IGBT gates are derived

limit

triggering

other

from feedback circuits wherein the actual voltage or current

waveshape

is

compared with

the

the

two becomes

the correction signal that triggers

the gates (Fig. 29.18).

desired voltage or current

actual voltage or current

Figure 29.18 Generating a 3-phase voltage or current that

is

very

close to the desired voltage or current.

diagram of one phase of a 13.2 kV, 3-phase distribu-

29.9 Distribution system

tion In order to see the context in

processor

inputs

wanted

waveshape. The instantaneous difference between

gate

settings

which shunt and

series

compensators operate, Fig. 29. 19 shows a schematic

system composed of a

radial feeder

and

its

branches.

The feeder emanates from a

where

protected by a rec loser, such as described

it

is

substation,


786

residential

®©A

*6

MFG km

3

-

machine-tool center

high voltage transmission

6 km

line

T2

*5

SC2

neutral

~

C^-p

^112

x4

substation

transformer +

4 km

recloser

3

©

km

x2

5

*3

km

-

indus-

y ''®

8

km trial

:

SSB

xl

>

park

foundry

£Hl

tlLc

SC1

T1

shunt neutral

neutral

Figure 29.19

in Section 26. 14.

power

The feeder and

its

branches furnish

60 Hz component and any

tal

residual harmonics,

manufacturing area, a residential area, a

switching surges, and other minor disturbances that

precision machine-tool center, and an industrial park.

the compensator has not been able to eliminate or sup-

to a

foundry equipped with arc furnaces

In addition, a

to

be serviced

feeder and

its

in the

is

near future. Each section of the

branches

is

by jc h jc 2 x 6 We neglect the resistive component of the line impedance. .

A shunt

.

.

.

compensator

is

connected

The compensator

a transformer T 1 a

PWM converter SC

,

tery/capacitor

power

supply.

the input to

at

the industrial park.

is

composed of 1

,

and

its

bat-

The transformer has

a

The waveshape

critical

it

at

is

that

we are interested in. We assume that the

carrier frequency is adequately filtered so that

its

ages and currents can be ignored. The voltage e h entry to the industrial park consists of the

voltat the

fundamen-

permits the immediate

to the

later.

machine-tool center,

The compensator

is

com-

posed of three individual converters connected series with

in

each line and isolated therefrom by three

The diagram shows only one conThe leakage reactance xa of the transformer,

verter.

modulated by the much lower frequency com-

excellent,

protected by a series compensator SC2, located

the service entrance.

transformers T2.

carrier

therefore,

conditions that will be described

Turning our attention

leakage reactance x a referred to the secondary side.

ponent e a

is,

disconnection of the industrial park under certain

The converter generates

a voltage which consists of a

eh

A solid-state breaker SSB

several kilometers long and

possesses a certain inductive reactance, designated ,

press.

thanks to the presence of the compensator.

the voltage e h across the primary, and the voltage e a

generated by the converter bear the same symbols as in the case of the shunt compensator. is

However,

understood that their values differ from those

the shunt compensator.

it

in


The proposed foundry

787

a particularly disturb-

is

ing load because of the arc furnaces

They produce random changes

it

in current,

contains.

which

or-

dinarily produce corresponding fluctuations in the

As we will see, a shunt compenovercome this problem.

terminal voltage. sator

is

The tem

is

able to

electrical activity of

that occur,

Power

we

Among

many

the

disturbances

Figure 29.20

cite the following;

Equivalent

interruption

when

(b)

(a)

continually changing, and so the voltages and

currents fluctuate.

•

converter

such a distribution sys-

opens and recloses on a transient

circuit

diagram and symbol of a switching is represented by

converter. Optional energy storage

the recloser suddenly

battery.

fault

•

Sudden disconnection of an important load

•

Across-the-line start of a large induction motor

the base per-unit voltage corresponds to the rated

•

Switching surge when capacitors are turned on

compensator voltage.

•

Transient line-to-ground fault on one phase

agrams the transformers have a

The choice of shunt pends upon several

•

Major disturbance on

the high-voltage transmis-

sion line that supplies

power

in the

to the substation

shunt-type

condenser, or

The

PWM

converter used

compensators in

similar to the converter described

is

Sections 21.45 to 21.49.

We

assume a 3-phase

unit operating at a carrier frequency of 6

However,

to simplify the explanations

kHz.

we assume

a

single-phase converter applied to only one phase of the 3-phase system (Fig. 29.20a). is

simplified even further

shown

in Fig.

The compensator

29.20b, where

it is

as a simple voltage source e a associated with

an optional energy storage battery.

To understand

the impact of a

distribution system, analysis. This circuits

it

is

relatively easy,

is

and disturbances

make a circuit despite the many sub-

useful to

that

occur

such a system.

in

In a circuit analysis involving transformers,

it

is

which reduces

best to use the per-unit approach,

everything to a single voltage level. The transformers "disappear

'

and the resulting

circuit

easier to visualize and easier to solve. this result,

we simply assume

that in

is

much

To achieve

our circuit

di-

and

that

We

which

will be discussed

begin our study with the

referred to as distribution static

DSTATCON.

principle of operation The

PWM

tions.

It

shunt compensator has several applica-

can be used as a voltage regulator; short-

term power source; harmonic distortion suppressor; power factor corrector and active filter. This section illustrates these applications.

Consider the the shunt

circuit of Fig. 29.2 1 a,

compensator for the

which shows

industrial park

portion of the distribution circuit around identical to the

compensator on a

1

The shunt compensator:

29.11

both shunt and series

in

:

PWM distribution compensator (DSTAT-

COM), sometimes

29.10 Compensators and circuit analysis

1

or series compensation de-

factors,

ensuing sections.

ratio of

entire

network

system of Fig. 29.

placed by an equivalent reactance alent voltage

1

£eq

.

X

and

a

It

is

except that the

9,

to the left of point

it.

6 has been CC|

This simplification

re-

and an equivis

possible by

The equivalent voltfundamental 60 Hz compo-

virtue of Thevenin's theorem.

age

E

CL]

comprises the

nent e c and

all

the transient

tions e d that occur upstream Fig. 29.2 lb

is

and harmonic

from point

a replica of Fig. 29.2 la in which

the compensator has been replaced by

symbol and the

distor-

6.

industrial park

is

its

equivalent

represented by an


788

SSB

N

neutral

| (a)

z Figure 29.22 Voltage regulation by

The Figure 29.21 Deducing the equivalent circuit sator and the industrial park.

voltage

Ecq

shunt compen-

for the

-e a

We

.

can

now

ji a x a

Because e h

begin our study of voltage regulation, distortion,

it

and other matters of

xa

interest.

The purpose of

Voltage Regulation. is

to maintain a constant

the

compen-

60 Hz voltage

e h at

the service entrance of the industrial park in the face

of a varying voltage

For the

moment we

sator attempts to

.

L

and a varying industrial load.

will neglect e d

.

The compen-

keep e b constant by varying the

voltage e a (Fig. 29.22a). Varying

compensator current

/

a

,

which

e.d

will

change the

in turn will

modify

~ji.ax a

cast in the

closed.

nent e K and a disturbance voltage e d

sator

around the

circuit equation is taken

first

The equivalent has been replaced by a 60 Hz compois

shunt compensator.

of

right-hand loop of Fig. 29.22a:

which can be impedance Z. The SSB

means

+

eb

=

0

form

=

-e a

eh

(29.10)

regulated and hence constant,

is

as the reference phasor (Fig. 29.22b).

fixed and the magnitude of

is

zero to

/

a(max)

converter.

As

,

the

the

maximum

i.

d

we

take

The value of

can range from

rated current of the

magnitude and phase angle of

are varied with respect to e b

,

is

it

e.A

seen that the mag-

nitude and phase angle of / a will also change, subject to Eq. 29.10. In particular, if

value

*

a(max)

i

a is

kept

at its rated

while varying e a the phasor ,

will trace out the locus of a circle.

;7 a(maxyr a

Thus, by making

a relatively small change in the magnitude and phase /

c,

and hence e b In the following explanation we write

of e a

,

we can cause

to rotate

through 360°. This ob-

.

the pertinent equations to determine

what happens

under these changing conditions. The phasor dia-

grams

will

be of particular help.

servation leads us to a second circuit equation, this

time taken around the left-hand loop of Fig. 29.22a:

-e c + Aj:c +

eb

=

0

TRANSMISSION AND DISTRIBUTION SOLID STATE CONTROLLERS

xc 'a(max)

•

Figure 29.23a Relationship between the source voltage ec and the regulated voltage eb

vc

789

'a(max)

Figure 29.23b Voltage regulation

a given

limits for

load.

.

regulation should be effected without demanding

any

hence

power from the compensator. Referring means that / a must be at right

real

to Fig. 29.22a, this

=

ec

eh

+ ji cjc c

angles to

however,

In turn,

e. r

implies that

£?

pensator-must be

=

'c

We

+

Referring

between

eb

now

and

*

b is

determined by the power facConsequently, for a given

OM

load condition, the phasor

Phasor

fixed.

ec

plus phasor

phasor

i

a

is,

(=

therefore, the

/7 ajt c .

that the locus of 'a(max)» tne

Jt

// bjc c )

/

a(maxJ

.

It

/

as a

shown

in

29.23b.

Fig.

tip

of e c can

lie

anywhere

within the bounding circle without in any fecting the magnitude of e b far as the delivery of

We

compensator

way

af-

.

power

to load

Z is conis

unim-

knowing

the

value's of e 0 that will

still

are only interested in

maximum and minimum

to

phase with the voltage

in

eb

tion to

£?

b,

/

OP

minimum value of maximum

e c corresponds

and the

quite a broad range, but

is

rela-

seen that for the given load condi-

the

b,

phasor

This

It is

to

it

phasor OQ.

depends upon

the value of the line reactance x c If the line reac.

tance

is

small,

small, the diameter of the circle will be

which reduces the regulatable range of

Therefore,

if

ec

.

a large sag or swell occurs in e c the ,

compensator may be unable

from changing.

It

is,

to

keep the voltage e b

therefore, difficult to regu-

late the variable voltage

of a "stiff" feeder by

means of a shunt compensator whose kVA rating is small compared to that of the feeder. As we will see, this problem can be resolved by using a series compensator.

Power Interruption. The

industrial park represents

a load of several megawatts, and the service con-

cerned, the phase angle between e c and e b

enable the

follows

again a circle. At rated

can be varied from zero to

^ follows that the

portant.

can

it

extremity of phasor e c follows the out*

However, because

As

is

But we have just seen that

c / a{max) is

of the circle,

'a(max)'

+

eb

sum of phasor

can swing through 360° and that

have any value between zero and

line

(29.11)

phase angle 0

to Fig. 29.23a, the

tor of the industrial park.

this

the com-

29.23b shows the resulting phasor

Fig,

tionships.

K+ iW+M

=

Eq. 29.10,

to

across the load.

can, therefore, write fc

OM

'b

according

a — the voltage generated by

produce the constant

tract stipulates that

power

shall not

by transients lasting 10 seconds or

be interrupted

less.

Power can

be interrupted by either a sudden short-circuit or an

open

circuit

on the

feeder. In such cases the shunt

compensator can be equipped with

a battery to sup-

output voltage e b Furthermore, in order to elimi-

ply energy for the brief period that the feeder

nate the necessity for energy storage, the voltage

connected. However, before supplying power the

.

is

dis-


790

conductors feeding the park must be isolated from the

main feeder by means of a

breaker

SSB

(Figs. 29. 19

solid-state circuit

and 29.24).

It

sator,

A

consists of

thyristors

connected back-to-back, as previously

explained

in

Section 21.23. The reason for the iso-

which immediately converts

and delivers

it

few cycles

not that of other clients connected to the system.

Suppose

evident:

What might cause that a

a

power

interruption?

Suppose

snowstorm or hurricane has produced a mo-

mentary line-to-ground

from the substation recloser to

open

fault

on the main feeder, 4

(Fig. 29.19).

after

two or

km

This will cause the

three cycles.

However,

opens

and then recloses again, cleared. Sensing the

is

60 Hz power

to

after the start of the intense short-

circuit, the recloser

The compensator was designed to meet the emergency needs of the industrial park and lation

it

tp the industrial park.

for, say,

new

to

its

(

1/2 s)

fault

SSB

situation, the

and the compensator returns that the

30 cycles

which time the

at

has

recloses

normal

state.

MW.

park represents a load of 8

During the power interruption, the converter must supply 8

This

is

MW X

0.5

s

=

4MW

s

-

4

MJ of energy.

quite within the capability of such energy

storage systems, which can typically provide as

much

MJ.

as 100

lem, the solid-state breaker (SSB) opens the circuit so

The phasor relationships when only the compensator is supplying power to the park are illustrated in Fig. 29.24b. The compensator automatically keeps voltage e b at its rated value, which means that

quickly (within one-half cycle) that the voltage

it

before

it

can open, the short-circuit will cause the

voltage across the affected lines to collapse, which will

impact

all

consumers. To get around

this

prob-

at

point 6 has not had time to collapse. Simultaneously, the battery begins furnishing energy to the

compen-

must generate a voltage

e. v

Voltage Distortion. Looking tortion, consider Fig. 29.25,

voltage e

ci

appears

Hz

voltage.

at

in the distribution

direct our attention to

at

now

it

voltage dis-

wherein a harmonic system.

The impedances

are

now

higher than

60 Hz and are therefore labeled x cd and

compensator

We

alone, neglecting the 60

will react to eliminate the

x

K)

.

The

harmonic

voltage from appearing across the load, and so e M

=

0.

As

a result, the distortion current

lating in the load

is

monic current

will

/

d

also zero.

flow

in

It

/

bc

,

circu-

follows that a har-

both the source and

'b

(b)

Figure 29.24 Behavior of compensator during a power

interruption.

Figure 29.25 Behavior of compensator when source produces a voltage distortion ed .


compensator.

the

We

can write the following

*cd

pa

equations:

~e d +jidxcd +

e bd

=

0

(29.12)

~;Vad +

e bd

=

0

(29.13)

-*ad

From

79

these equations, and because e bd

is

zero,

we

obtain

*cd

In

most cases x ld

is

considerably larger than x cd

;

consequently, to prevent the harmonic voltage from

appearing across the load, the compensator must generate a greater harmonic voltage originating harmonic voltage e d It

than the

e.dd

.

should be noted that a similar analysis applies to

transient voltages, such as switching surges.

For ex-

ample, suppose that the manufacturing sector (Fig.

(b)

29.19) has a power-factor correction system that in-

When

volves the on/off switching of capacitors. capacitors are switched on,

it

generates for a few cy-

cles a surge of perhaps 5 kV, at approximately

Hz

This transient rides on top of the 60 Fig. 29. 1 6).

the

900 Hz.

voltage (see

As it travels along the feeder, its amplitude may still be substantial when

will diminish rapidly but it

Figure 29.26 Power factor correction.

reaches the input of the industrial park. Again the

compensator comes to the rescue, because

its

carrier frequency is considerably higher than

6 kHz

900 Hz

same time it furnishes the compensator with active power P = e.J.d cos Under these conditions, the battery supplying the dc input is not necessary. The cj>.

capacitor

and so the transient voltage can be suppressed.

e h as

Power Factor Correction. The shunt compensator can be used to correct the power factor at the input to the industrial park. To do so, the compensator voltage

converter.

ea

is

arranged so that current

/

a

,

lags 90° behind volt-

age e b (Fig. 29.26). Referring to the figure, suppose

compensator has losses represented by

that the

tance ra

.

We can

resis-

then write the following equation:

+

'a'a

+M +

=

=M + Va +

slightly

Note

behind

foundry

is

eh

it

slightly lags or leads

It

is

for the

c

STATCOM

examine the proposed fed by a line leading

Z (Fig.

(Fig. 29.19).

Q=

A

29.27a) because

draw a highly fluctuating current

during certain phases of their operation. As a rethe voltage drop along the distribution feeder

nonlinear, which in turn produces a distorted at

the service entrance to the foundry. it

(29.15) is is

shown

in

and lags

Consequently, the distribution

system receives reactive power

We now

a nonlinear load

the arc furnaces

sult,

that e a is greater than e b

it.

so that

Indeed, the current changes so erratically that

The corresponding phasor diagram Fig. 29.26b.

a

from point 2 on the main feeder

voltage e bd

was previously described

foundry installation.

is

hence

kept charged to the desired dc voltage

Nonlinear Load.

i

0

is

by controlling

level

e b i.

x

and

at the

impossible to express the voltage

in

terms of

harmonics and frequency-dependent reactances.

However, the instantaneous voltage drop along the line

is

the rate of

always equal

to

its

inductance

L times

change of current. The instantaneous


792

®

voltage c bd

L

at the

foundry

is,

therefore, given by

the equation,

e.

L

+ L

+ eM =

.

Ar

0

hence

For example,

in this

Hz reactance x 7 of

1

1

equation, fl, its

(a)

2irf~

(b)

E,

are taken.

fat I'll

11 1

A

'In

1 1

1

1

i

The jagged

Fig. 29.27c again

!

1

>

AA 1

flfl

\

on

1

y*

u J/

a

Ik *'

voltage reaches peaks of

shows

the current

tM

\ or\

M

/

c,

together

with the fundamental component of voltage e c point 2. Although c c

1

at this t

at the

current has an effective value

A while the

1

1*

K

[!

1

H

0.029

X 60

2tt

60

L is given by

l2kV.

!'

1

1

-'Vi

a feeder has a

inductance

29.27b shows the current and voltage

of about 2300

j

I/'

1

input to the foundry before any corrective measures

foundry prior to shunt compensation

at

I

if

(29. 6)

ll

•<7

Fig.

^.-L|^

=

^ bd

point of

is

at

sinusoidal, the actual voltage

common

coupling will be polluted to

!

The amount of pollution will depend upon the impedance upstream from point 2. If the impedance is suba certain extent because of the distorted current.

[f

stantial, the

as other

waveshape could be unacceptable

consumers are concerned. in the feeder, in

the current

addition to improving the

voltage waveshape at the input to the foundry (c)

12000

voltage and current

at

source

Fig. 29.28

shunt compensator SC3, together with transformer -

j

c

monitors the instanta-

in the feeder,

and

this signal is fed

A

into the gate triggering processor.

second input

provides the wanted instantaneous sinusoidal cur-

S

< >

A current transformer CT

neous current

4000-

g

itself.

the foundry, the feeder, and the

-

T3. 8000

shows

as far

therefore im-

waveshape of

portant to improve the

flowing

It is

rent. -4000

-

The processor compares

shape of

-8000

As -1

2000

the

two

signals and

generates the triggering pulses to correct the wave/

c

.

a result, the current in the feeder approaches a

-

sine wave. Consequently, the voltage drop along the

feeder

is

now

sinusoidal and so, too,

Figure 29.27

the entrance to the foundry.

Equivalent circuit of foundry, voltages, and currents

delivered to the foundry

prior to installation of

shunt compensator.

load

is

is

However,

is still

the voltage at the current

*'

h

distorted because the

inherently nonlinear. This

means

that the cur-


xl

793

C.T.

-M-

neutral

gate

wanted

actual

triggering

current

processor

current

foundry error-correcting

gate signals

SC3

T3

shunt neutral

Figure 29.28 Shunt compensator

rent

/

a

at

entrance to foundry.

supplied by the shunt compensator

is

the distorted portion of the factory current.

pensator Fig.

is

now

acting as an active

actually

The com-

it

still

at the factory

be

felt

by the

has been eliminated

park.

filter.

29.29a shows the voltage e b

will

A

residential sector at the

even though

entrance to the industrial

study of the related consequences of in-

stalling a

compensator

is

always advisable.

entrance and the sinusoidal current flowing in the feeder. Fig.

29.29b shows the same voltage, along

with the distorted current supplied to the foundry. Fig. 29.29c

shows

the current

compensator (about 800

A rms)

/

a

29.12 The series compensator: principle of operation

supplied by the

and the correspond-

The

series

compensator*

is

similar to a shunt

ing instantaneous power that it delivers. Note that the power of the compensator fluctuates continuously be-

pensator; the main difference

tween positive and negative values, reaching momen-

call that in Fig.

tary

peaks of 10

MW during the 80 ms interval cov-

ered by the graph. However, the net power, averaged

over a few 60

dom

Hz

cycles,

is

zero because of the ran-

it is

sator can respond to

seen that the shunt compen-

many

nected

different electrical distur-

at the input to

line reactance

con-

a machine-tool center.

The was

derived the same

way

as

is

represented by impedthe effective

by x c and the source by a 60 Hz sinu-

soidal voltage e c

sumer. However, correcting a power quality problem

,

and a distortion voltage e & compensator

the behavior of the

We now

—

first,

as

Fig. 29. 19

does not benefit everyone to the same degree. Thus, a disturbance created

We reis

ance Z, the compensator by voltage

examine

one point on the distribution system} of

is

com-

connected

for the shunt compensator. Thus, in Fig. 29.30,

bances and thereby ensure quality power to the con-

at

it is

29.19 a series compensator

equivalent circuit

done

that

with the feeder instead of in parallel.

the machine-tool factory

nature of the alternating current

In conclusion,

in series

is

by the manufacturing group

Sometimes called dynamic voltage restorer (DVR), tic

scries voltage regulator

(SSVR).

or sta-


794

voltage and current ahead of compensator

(a)

Figure 29.30 Series compensation.

a voltage regulator, and second, as a supply-side distortion neutral izer.

Voltage Regulation. Consider Fig. 29.3 1 a in which (b)

10000

-

8000

-

6000

-

E,

I

ac

/'

'

A

''\

1

\

'
M

fj

/

o 0

A

60 Hz source voltage

i

\

vl

|

V

1/

The load draws a

'

/

\

Ul /Hi!

1

/IU

i/\

H

r M

/

current

\

>

-4000

i

f

V

i

/

vJ

I

'

Up

"

\

U

'

^

i

V-f

'

b

We

.

can write the

fol-

+

=

+ji b xc +

+

ea

=

eb

0

hence

'

-6000 •

0000

•

'

\

1

-8000 -1

i

ji^c

-

-2000

is

lowing equation:

'a,

«

<

e c acts alone in the circuit

held constant by means of the series compensator.

>\

'

-I

2000

the

(no distortion), and the voltage e b across the load

I

4000

>«

compensation

Input to foundry after shunt

at

i

W

\' v

/

/

W

»'

\'

ec

v

Suppose

eb

that i b lags

(29 .17)

ea

behind e h and that the com-

pensator generates a constant voltage e a(max) equal ,

to

rated voltage. This yields the phasor diagram

its

of Fig. 29.31b. The phase of e ainrdX) can be varied as desired, (c)

power and current

In

compensator (duration 80 ms)

and so

its

locus describes a circle with

M.

center

Given

that e b

is

held constant, phasor e c of the

source can have any value and phase angle, pro-

vided that â(max)-

its

extremity

falls

within the circle for


portant; the only objective

is

to

ec

and e h

is

unim-

keep the magnitude

of e h constant. In achieving this result, the maxi-

mum possible value of e c is given by phasor OP and minimum by phasor OQ (Fig. 29.28c). Thus,

the

even

if

series

e Q sags

and swells over

compensator can

still

this

wide range,

the

hold the voltage con-

stant across the load.

However, the phasor diagram reveals

Figure 29.29 Voltages, currents,

shunt compensator

and converter power is

installed.

after the

compensator must supply

when

the

real

power

magnitude of phasor e c

is

that the

to the

system

equal to

OQ.


The reason

is

phasor —e. x(m

that

that is 180° out of

.

lx)

has a component

phase with current

when — £ a(max) is in quadrature with tive power of the converter is zero. compensator must absorb equal to OP.

If this

real

795

/

/

b

only

It is

.

b that the ac-

Similarly, the

power when

ec

happens to be a short-term

is

situ-

ation lasting, say, for 10 seconds, the battery will be

able to supply or receive the required power. Thus,

OQ to OP can be regulated.

the full range of e c from

But

cannot be a long-term solution.

it

Voltage regulation

is still

possible on a long-term

compensator drawing or deliver-

basis without the

ing real power. This implies that phasor

kept

at right

angles to phasor

/

in Fig.

29.3 Id. The magnitude of e c can, therefore,

vary from length

OP to

rower range than

that in Fig. 29.28c.

series

The

PQ shown

of phasor e c must then follow the line

tip

must be

e.d

b (Fig. 29.31a).

length

OQ, which

a nar-

is

Note

that the

compensator can regulate the voltage of even

a very stiff (low impedance) feeder. It

should be noted that the voltage disturbances on

the three phases

may be

quite different. For example,

a single-phase line-to-ground fault

generate

will

on the network

unbalanced voltages on the three

phases. Therefore, the three converters that

make up

compensator must operate independently of each

the

other in order to produce balanced line-to-line volt-

0

ages

ib

at

the entrance to the machine-tool center.

According

to statistics, voltage sags account for

about 90 percent of the disturbances affecting

criti-

(c)

cal loads

on a distribution system. The

series

com-

pensator can be specifically designed to resolve

problem because tire

power

it

this

does not have to provide the en-

(or energy) during such a disturbance.

Notably, the inverter power rating can be mini-

mized by

tailoring

its

output voltage to the expected

depth of sag, while the energy storage

by

its

is

determined

expected duration.

Consider, for example, a 480 V, 3-phase, 60

Q (d)

feeder that delivers 600

kVA

to a critical

Suppose the expected worst-case voltage sag percent of the line-to-neutral voltage and that

Figure 29.31 a. b.

c.

d.

pected duration

is

20 cycles. The

series

Hz

load. is

its

15

ex-

compen-

Voltage regulation with series compensator. Voltage regulation with series compensator.

Sag and

swell compensation. Quadrature series compensator.

sator must, therefore, boost the voltage by

cent during the sag while current. Consequently,

still

its

1

5 per-

carrying the same line

power

rating

need only


796

Figure 29.32 This

±2 MVA

converter assembly can be used

in

either a shunt or series

compensator (DSTATCON or DVR).

(Courtesy of Westinghouse Electric Corporation)

be

15% X 600 kVA = 90 kVA, and the energy storis 90 kW X 20 cycles X (1/60) s = 30

age required

kW

s

= 30

kJ.

circuit-breaker trips.

Both the power rating and energy

storage requirements are modest. In practice, the series

current flow to an arbitrary low value until the main

29.13 Conclusion

compensator can typically provide boost

We

voltages of 25, 50, 75, and 100 percent.

Thus, a series compensator effective solution to the sag solid-state breaker

is

often a

more

cost-

problem than using a

and a fully-rated shunt compen-

In

instantaneous control of the power flowing over

transmission

and distribution systems.

In

rendered possible by the rapid

re-

lines

every case this

sator plus energy storage.

Current Limiting.

have seen that series compensators, shunt com-

pensators, and static circuit breakers enable almost

is

sponse of switching converters.

some

Some

applications the series verters are also able to control the

compensator can be preset current in a stiff feeder.

Its

to limit the short-circuit

voltages and currents, thereby filling the role of acreaction time

is

so fast tive

that

it

of these con-

waveshape of

can immediately introduce a voltage

sition to the feeder voltage

and thereby

in

oppo-

harmonic

filters.

The converters can

also be used as high-power

limit the

frequency converters

in the

megawatt range.


191

Figure 29.33 This 13.8 kV, 3-phase solid-state

circuit

breaker (SSB) comprises both

GTOs and

thyristors.

The GTOs are

rated

600 A and the thyristors are rated for 8000 A. (Courtesy of Westinghouse Electric Corporation)

These new high-power devices

will

have a pro-

found impact on the management of power tric

utility

systems.

As

well, they enable

control of electric power, improve

29-4

The

solid-state switches in Fig. 29.1a

through one of the thyristors. 29-5

What

29-6

A 3-phase

is

meant by a

stiff feeder?

switching converter operates

from a dc bus of 2400

What

is

the

GTO and 29-2

Explain

main

distinction

between a

*

a thyristor?

why

a

high-frequency

GTO

cannot be used

PWM converter.

V. Calculate the ap-

proximate rms line-to-line voltage converter operates

in a

684 A.

Calculate the peak current that flows

enable currently unused capacity to be mobilized.

29-1

the fre-

is

carry an effective ac current of

power quality, and

Practical level

What

quency of the harmonic?

enhance the


A conductor carrying a 60 Hz current also bears a 23rd harmonic.

dynamic

control of system disturbances, thereby increasing the stability of the network. Finally, they

29-3

in elec-

a.

in the

rectangular

b.

in the

PWM

mode

wave mode with

m =

1

if

the


798

29-7

In

Problem 29-4 calculate the longest time

needed 29-8

What

is

60 Hz

to interrupt the

29- 1 7

A 6700 kW load

backed up by a shunt

is

compensator connected

current.

meant by vernier control of a

age

TCSC?

is

to a

group of

bat-

The energy storbe 40 MJ. Calculate the 240

teries operating at

designed to

V.

ampere-hour capacity of the battery pack. 29-9

Explain what

meant by

is

29- 8 1

1

29- 11

Problem 29-

1

7,

b.

Brownout

can the battery deliver power to the load

c.

Voltage swell

before

d.

UPS

begin to fall?

Advanced

A cable carries

Hz

a 60

A

current of 870

29- 9 1

its

terminal voltage will suddenly

level


we want to main24 kV across

,

tain a line-to-line voltage of

the effective value of the current.

the input of a conglomerate 3-phase load

In Problem 29- 1 0 calculate the

maximum

having a capacity of 6.8

power

factor.

Referring to Fig. 29.3, calculate the capacitance of the condensers

and the induc-

MW

The feeder has

of 5 II per phase. 1

1

and a 5th harmonic of 124 A. Calculate

possible value of the peak current.

29- 2

how many seconds

for

Switching surge

Intermediate level 29- 0

In

a.

source voltage

It is

may

at unity

a reactance

known

that the

vary between 25

kV

and 26.4 kV.

tance of the inductors. 29- 3 1

In Fig. 29.6 calculate the

peak 60 Hz

line-

to-neutral ac voltage if the voltage across

the capacitor

is

3400

Calculate a.

The maximum voltage

V,

standing that

29-14

Referring to the series compensator of Fig. 29.12,

and

EB =

it is

can develop a

and the

its

known

7.4 kV.

The

rated current

maximum

ing that

EA

and

EB

kV

compensator

voltage of

power

changed between regions

29- 1 5

6.9

1

.5

that can

A and

B,

The rated power of

c.

If a brief

be ex-

know-

d.

rent flowing in the 150 lines

when

kV

transmission

the converter delivers rated

single-phase

power

factor of the 150

at

kV

66 kV. The power

line

is

0.96 lagging.

the

minimum

unable to keep the output voltage

is

24 kV?

If a brief swell occurs,

age e.

4, calculate the cur-

is

can reach before the compen-

sator

voltage

pensator

1

compensator

it

mum

are in phase.

the

sag occurs, what

voltage

at

Explain the principle of operation of a


neither absorbs or delivers

b.

kV

UPFC. 29- 1 6

it

any long-term active power

800 A. Calculate

is

active

EA =

that

series

maximum

required, per phase,

for the series compensator, on the under-

at

If the

is

24

it

what

is

the maxi-

can reach before the com-

unable to keep the output volt-

kV?

compensator can deliver active and

reactive

power

for a brief period,

what

minimum voltage in part (c) and maximum voltage in part (d)?

the

is

the

Chapter 30 Harmonics

30.0 Introduction Voltages and The

torted.

currents in industry are often dis-

subject of distortion

Chapter

cussed

in

mend

the

reader

was

Section 2.12 and

2,

review

mentioned, the distortion

it

may

first dis-

we recom-

briefly.

As

was

be caused by mag-

netic saturation in the core of a transformer,

4

by the

switching action of thyristors, or by any other nonlinear load.

A distorted wave is

made up of a funda-

mental and one or more harmonics. For example, Fig. 30.1, reproduced

from

Fig. 2.23,

shows a

torted alternating current that contains a 5

lh

dis-

and 7

lh

harmonic and several other higher harmonics.

30.1

We

Harmonics and phasor diagrams

0

tude, nent.

and

initial

Waveshape It

indicates the frequency, ampli-

is

of

180

240

300

420

360

a distorted 60 Hz current having an

fective value of 62.5 A.

value of each^sinusoidal compo-

The amplitude

120

Figure 30.1

can represent a distorted wave by a composite

phasor diagram.

60

The

ef-

current contains the followth

components: fundamental (60 Hz): 59 A; 5 harth monic: 15.6 A; 7 harmonic: 10.3 A. Higher harmonics ing

equal to the peak value of the

voltage or current. For example, the phasor diagram

are also present but their amplitudes are small.

799


800

V

4

100 V

20 V, 300 Hz

EF

60 Hz

Figure 30.2

A

distorted voltage

can be represented by phasors

volving at different speeds. Their tion also affects the

re-

angular posi-

initial

waveshape.

of Fig. 30.2 represents a distorted voltage com-

posed of two components: 1

.

of

1

00

frequency of 60

V, a

angle of 0°. tates 2.

E F having

a fundamental voltage

ccw

at

We can

an amplitude

Hz and

an

Figure 30.3

Waveshape

initial

consider that the phasor ro-

in Fig.

of the voltage

generated by the phasors

30.2.

60 revolutions per second.

th

harmonic EH having an amplitude of 20 V, frequency a of 300 Hz, and an initial angle of

a 5

59°. This phasor also rotates

ccw but

5 times

faster than the fundamental.

The fundamental can be expressed by

E¥ =

Similarly, the

EH = The

20

+

100 sin (0

100 sin 360/r

harmonic can be expressed by:

sin (59

distorted

=

0°)

the equation:

+

59°)

wave can

=

20

sin (5

X

360ft

+

59°)

therefore be expressed by

the equation:

E=

+ 20

100 sin 6

sin (5 0

+

59°)

-120 1

where

Figure 30.4 0

= 360 ft = 360 X 60 X

Waveshape

t

5

The angles

are expressed in degrees.

shape of one complete cycle

is

shown

voltage

when

the

initial

angle of the

180°.

the fre-

30.2 Effective value of a distorted

wave

angular position with respect to the funda-

mental. For example,

if

the initial angle of the 5

harmonic

is

changed from 59°

30.2

is

The wave-

The waveshape depends not only on

in Fig.

of the

harmonic

in Fig. 30.3.

quency and amplitude of the harmonics but also on their

th

lh

to 180°,

The

effective value of a distorted voltage

is

given

by the equation:

the resulting voltage will have the flat-topped shape illustrated in Fig. 30.4.

E =

\jEl

+ El

(30.1)

HARMONICS

where

c.

E= E = EH = The

Using Eq. 30.1 we obtain:

effective value of the distorted voltage [V]

E = VE~ + El

effective value of the fundamental [VJ

[;

effective value of

£H

effective value

of

the harmonics

all

is

given

= V59.4 + El 2

66

the harmonics fV]

all

801

from which

E H = V66 2

59.4

=

2

28.8

V

by the equation;

The

E H = VEj + E] + - + El where the 2

E2

nd

3

,

E4

,

rd ,

4

.

th .

.

.

En

are the effective values of

nth harmonics.

Combining equations

30.

we

et 30.2,

1

effective value of

all

the harmonics

is

28.8 V.

(30.2)

30.3 Crest factor and total harmonic distortion (THD)

obtain the

There are several ways of describing the degree of

expression:

distortion of a current or voltage.

E = V£ + E\ + Ej + - + El 2

(30.3)

Similar equations apply in the case of distorted

that are fre-

(THD).

distortion

By

currents.

Two

quently used are crest factor and total harmonic

definition, the crest factor of a voltage

is

equal to the peak value divided by the effective

Example 30-1

(rms) value.

.

Calculate the effective value of the distorted volt-

age

in Fig.

peak voltage

30.3. crest factor

(30.4) effective voltage

Solution Effective value of the fundamental:

In the case of a sinusoidal voltage (which evidently

EF =

Effective value of the 5

E3 =

=

100/V2 lh

70.7

V

has no distortion) the crest factor

harmonic:

20/V2

=

V

14.1

E =

Y'e

2

+ £2 =

= V5T97 -

10 J

\

72,1

+

flat-topped.

On

2

V

a.

84

tive value of all the

a crest

V.

the effective value of the fundamental

is:

(THD) =

In the case of a distorted voltage, the

the harmonics

b.

—

(30.5a)

THD is given by:

Solution

66

the effec-

The amTotal harmonic distortion


Effective value of the square

distortion

wave

b.

a.

factor

fundamental. In the case of a dis-

torted current, the equation

c.

all

A

harmonics divided by the effec-

V. Calculate:

the effective value of the square

1.41.

tends to be

By definition, the total harmonic (THD) of a current or voltage is equal to tive value of the

is

.4

pointy.

14,1

Example 30-2 A square wave has an amplitude of 66 plitude of the fundamental

other hand,

1

greater than 1.4 indicates a voltage that tends to be

Effective value of the distorted voltage: 2

a crest factor less than the

=

is a/2

wave having

wave

is

clearly

(THD) =

Total harmonic distortion

Cm —

(30.5b)

V.

Effective value of the fundamental:

Eu =

Em J<2 =

84A/2 = 59.4

From

V

these expressions,

it

is

voltages and currents have a

seen that sinusoidal

THD of zero.


802

Example 30-3 The distorted current shown in Fig. 30.1 has an effective value of 62.5 A. Knowing that the fundamental has an effective value of 59 A, calculate: the effective value

a.

harmonic

of

/H

the total

the effective value of all the harmonics

th

Q

source voltage

above 51 V 300 Hz

th

the amplitude of the 7

d.

24

distortion, in percent

b. c.

the 7

100 V 60 Hz

harmonics

the

all

harmonic

18.6

mH

Solution Effective (or rms) value of

a.

the harmonics:

all

Figure 30.5

2

V/ 2

(30.1)

'f

= V62.5 2 -

59

=

2

20.6

of the effective fundamental current and the effective values

H

THD

circuit.

A

Total distortion factor:

b.

RL

Distorted voltage applied to an

of the individual harmonic currents.

(30.5)

Example 30-4 Figure 30.5 shows a distorted voltage source com-

20.6

=

0.349

34.9

%

posed of a fundamental of 100 V, 60 Hz, and a 5

59

harmonic of 5 lh

Noting the values of the 5

c.

given

and 7

in Fig. 30.1, the effective

monics

>

7

/

th

harmonics

value of har-

1

V,

to a resistor of

24

18.6

300 Hz.* The source

v.,

is:

-

V/

2

-

r5 1

Y20.6

2

-

mH. At 60 Hz

the latter has a reactance:

1 1

X6Q =

2tt

1

15.6

8.66

=

2irfL

X 60 X 0.0186 =

However, 2

-

10.3

at

300 Hz

the reactance

th

5

300

X

=

10.3V2

=

14.6 A.

pendently of each other,

=

7 ft

A

harmonic

is

5 times greater:

2

Since the fundamental and 5

Amplitude of the 7

l

35 ft

harmonic act inde-

we can draw

separate

cuits to calculate the respective currents

and 30.7).

(Fig. 30.6

Harmonics and

7 ft

2

v

- V74^9 =

30.4

connected

ft in series with an inductance of

lh

=

d.

is

age source

circuits

is

cir-

and powers

We recall that whenever a volt-

ignored,

it

is

replaced by a short circuit.

Referring to Fig. 30.6, the impedance of the 60 It

is

important to

know how

a circuit responds to

circuit

V24 2 +

inductors, capacitors, and transformers, the various act

independently of each other. The fun-

damental and each harmonic behaves as ers

were not

there. In solving

particular harmonic,

the

if

Hz

is:

harmonics. In linear circuits composed of resistors,

harmonics

The fundamental current

7

2

=

25 ft

-

4

is:

the oth-

such a circuit for a

'60

=

voltage sources of the

100

—

A

25

other harmonics, including that of the fundamental, are replaced

th

total

rms

equal to the algebraic

sum

by a short-circuit. Then, the

current in each branch

is

In the

absence of information

to the contrary, the values

voltages and currents are rms values.

of

HARMONICS

803

300 Hz source was

(In these calculations, the

short-

circuited.)

A

4

Let us

W

384 24 Q

now

gards the 5

th

consider the circuit of Fig. 30.7 as re-

harmonic. Proceeding the same as be-

60 Hz source

fore but with the

in short-circuit,

we

obtain the following results:

100 V 60 Hz

Impedance of the 7

Z300 =

Q.

The 5

lh

fundamental current

resulting

V'24

2

+

35

2

300 Hz:

=

42.4 il

harmonic current:

/

Figure 30.6 Impedances and component.

circuit at

Em 300

——

Z

=

1

Active harmonic power dissipated 2

I m)

300

2

A

42 4

R =

.2

1

in the resistor:

X 24 =

2

34.6

W

Reactive powers and apparent powers are not de1,2

A

fined for harmonic voltages and currents. Conse34.6

51

W

Q

24

quently,

we

ignore them. Only active harmonic

powers are considered.

The harmonic current

V

300 Hz

lags behind the harmonic

voltage by an angle: 35

Q

We

can

=

arctan

6 ;nn

arctan

R

resulting 5

th

now combine

the fundamental and har-

P

6()

=

2

/(l0 tf

=

harmonic current.

in the resistor:

4

2

X 24

Apparent power ^60

Power

2

X

7

-

1

V4

+

£««,

= */«, =

= 400 VA

60 Hz: 384 0.96 or 96

%

400

/?/3(K)

()

lags /behind the funda-

arccos0.96

4.18

A

X 4 = 96 V

the resistor: 1.2

V'96

2

The current of 4.18

A

\

1

00

2

16.3°

-28.8 V

+

28.8

2

=

+

51

2

-

1

and the voltage of

(Fig. 30.8) are the values that

-

=

100.2

V

Total effective voltage of the source:

E = VE 2 + £ 200 =

mental voltage by an angle:

=

24

= 24X

= V£ R\, + El m =

,

6 60

2

Total effective voltage across the resistor:

£r

The fundamental current

1.2

the resistor:

The 300 Hz voltage across

ER3(X) =

00 X 4

= V4 2 +

4,o

112 var

60 Hz:

at

^6o4i

factor at

4

^60

=

The 60 Hz voltage across

= 384W

Reactive power absorbed by the reactance: Art)

as follows:

Total effective current in the circuit: /

Active power dissipated

55.6°

24

monic currents and voltages Figure 30.7 Impedances and

35 — =

by instruments designed

12.2

1

V

12.2

V

would be measured

to give

rms readouts.


804

Since the

A

4.18

418.6

24

we

W

total active

power in our example

is

4 8.6 W, 1

obtain:

r total

PF

n

^total

^

W

418.6

V

112.2

18.6

mH

However, the

0.893 ou 89.3

VA

469

power

traditional

the cosine of the angle

factor

Figure 30.8 to the

When

power dissipated

PF,displacement

in the resistor:

384

lolal

This

also the total active

is

source.

would

It

+

34.6

Example

30.4,

=

418.6

W

power supplied by

the

or 96 %.

When

and

if

it

were connected

in the circuit.

power factor power factor

total

to Fig. 30.6, the dis1

6.3°

=

0.96 dis-

power factor, and have the same value.

factor, the total

power

factor

all

30.6 Non-linear loads

linear load (Fig. 30.9).

Section 7.9, Chapter

factor 7.

Its

was first discussed in meaning must be en-

when distorted voltages and currents are present. The terms displacement power factor and total power factor are then used. larged

single-phase

circuit,

the

total

E connected to a

The load may be a

non-

saturable

reactance, a rectifier, or a set of mechanical switches that

The concept of power

is

(30.7)

4>

equal to cos

is

Consider a sinusoidal voltage

power

cos

corresponds to the power that a wattmeter

indicate

a

har-

called dis-

no harmonics are present, the

placement power the traditional

=

and referring

placement power factor

30.5 Displacement

In

is

fundamen-

In

Total

factor

placement power factor.

and currents due and harmonic components.

Effective voltages tal

power

this

given by

is

between the fundamental

voltage and the fundamental current.

monics are present,

%

open and close

On

periodically.

account of the non-linearity, the current will

not be sinusoidal.

component

/F

It

will contain a

fundamental

/ H The produced by the sinu-

and harmonic components

fundamental component soidal voltage E, but the

is

.

harmonic components are

apparent

equal to the product of the effective volt-

age times the effective current. In Example 30.4 and referring to Fig. 30.8,

MS

=

II

2.2

we have

of source

X

4.18

x RMS of source = 469 VA

non-linear

/,

The total power factor is equal to the total power divided by the total apparent power:

load

active

Figure 30.9

A PF,total

(30.6)

sinusoidal source connected to a non-linear load produces a fundamental current lF and the load generates harmonic currents /H .

HARMONICS

generated by the load. Clearly, the harmonic currents flow in the sinusoidal source

E

as well as in

assume the switch

concerned,

it

age E. Thus, reactive

switch were always closed, the voltage

If the

the fundamental

can

we

powers

component of current

lag, lead, or

be

in

is

phase with volt-

can attribute traditional active and

to this non-linear load.

producing no sparks

is ideal,

and having no losses.

the load.

As far as

805

across the resistor would be sinusoidal, and the current

would be 1000 V/10 fl = 100 A. The power form of heat would be:

dissipated in the

However, the

2

P=

1

R=

100

2

X

10

=

100

kW

product of the fundamental voltage and any one of the

However, because the switch

harmonic currents produces zero power.

is

open half the time,

power dissipated will be one-half of 100 kW, namely 50 kW. It follows that the chopped sinuthe

30.7 Generating harmonics The process whereby harmonics

soidal current has an effective value of 70.7 A, be-

are created

is

quite remarkable. Consider, for example, the cir-

30.10a

cuit of Fig.

nusoidal voltage

is

in

which a 1000

connected to a 10

series with a switch.

V,

60 Hz

si-

fl resistor in

The switch opens and

periodically in synchronism with the 60

closes

Hz

fre-

quency. Figure 30.10b shows that the switch closed during the

last half

of each half-cycle.

is

We

P=

2

ever

it is

1

R =

If

synchronous

= 70.7A

If

we decompose

the

chopped current

into

fundamental and harmonic components, we it

age (Fig. 30.

switch

A

A)

closed. Consequently, the switch doesn't

contains a fundamental 60

ponent of 59.3

= 59.3

(70.7

absorb any net active power.

cover that

/ rms

2

X 10 ft = 50 kW. The synchronous switch does not heat up because the current is zero when the switch is open and the voltage across its terminals is zero whencause

1

1

A ).

that lags 32.5°

Its

peak value


its

dis-

Hz com-

behind the volt59.3 V2

is

factor

= 84 A. —

cos 32.5°

is

1500

10Q 1000

84

A fundamental

500

\>"" component

^7~

H

ji

1

0 ,30 60 90 120150180 210 240 270300 330 360

-500

-1000

Figure 30.10 a.

-1500

A synchronous switch

sorbs reactive

in series

with a resistor ab-

power when current flow

by the switching action. b.

Waveshape of the chopped

current.

is

delayed

Figure 30.11

The chopped current contains a fundamental 60 Hz component whose amplitude is 84 A and that lags 32.5° behind the voltage of the source.


806

we

0.843 or 84.3 %. Consequently,

lowing

obtain the fol-

results: If

A

= 59.3

|-32:5'

Apparent fundamental power supplied by the source:

= EI = 1000 V X

S

A=

59.3

kVA

59.3

Active fundamental power supplied by the source:

P = 5 X PF =

59.3

0.843

power

fundamental

Reactive

X

= 50 kW by

supplied

the

source:

Q =

\f S

2

- P 2 = V59.3 2 - 50 2 =

31.9 kvar

Figure 30.12 showing the flow of active and reactive fundamental power and the harmonic power PH Circuit

.

This indicates that the load consisting of the syn-

chronous

switch

and

absorbs

resistor

reactive

power even though there is no magnetic field at all. The resistor cannot absorb reactive power and so the synchronous switch is

must be absorbing

That

it.

far as the 10 £1 resistor

now evident that the

active

power

is

concerned,

it

ab-

sorbs a fundamental active power: 2

I

R =

59.3

2

X

=

10 (I

kW

35.2

we know

kW of

it,

kW and

the resistor only absorbs 35.2

the remainder:

P swilch =

-

(50

kW and Q swilch = a resistance

35.2)

=

14.8

Furthermore, since the source delivers 3

1

kW. .9

R

resistor absorbs

also be absorbing

We

it:

none of

<2 sw tch

=

j

it,

the switch

As

31.9 kvar.

kW of fundamental active power. But if absorbs 14.8 kW must deliver an equal amount, otherwise it

it

it

would rapidly become very

ideal switch

hot.

does not heat up

But we know

at all.

this

So what has

to the fundamental power? The answer is that it is immediately converted into harmonic power of equal magnitude. The harmonic power P H of 14.8 kW generated by the

happened

switch

is

sipated by the resistor.

it

absorbs

P swilch =

we can

represent

=

14.8 it

by

with an inductive reactance

{

=

14.8

X

1000/59.3

2

=

31.9

X

1000/59.3

2

-

4.21 11

2

=

9.07 fl

fundamental (60 Hz) components are

far as the

concerned, the model of Fig. 30. 3a 1

is

a perfect rep-

resentation of the circuit. However, the "resistance"

R merely {

simulates the fundamental active power

absorbed which

monic power. We saw that the circuit

is

immediately converted into har-

is

the effective value of the current in

70.7

A and

that

its

fundamental com-

ponent has a value of 59.3 A. Consequently, the fective value of

then absorbed by the 10 fl resistor.

The fundamental and harmonic power flows are illustrated in Fig. 30.12. Note that the sum of the fundamental power of 35.2 kW plus the harmonic power of 14.8 kW does indeed equal the 50 kW dis-

that

31.9 kvar,

in series

must

have just found that the switch absorbs 14.8

composed of all odd-num-

2

PI I

kvar

Q/I

and the

Hz can

It

(Fig. 30.13a). Their values are respectively:

must be absorbing

the synchronous switch

is

harmonic powers.

the switch carries a fundamental

A and

current of 59.3

Since the source delivers a fundamental active

power of 50

into a series of

bered harmonics starting with the third harmonic/ H 3 X 60 Hz = 180 Hz.

Since

P=

switch behaves like a fre-

converts the fundamental 60

It

be shown that the series

a finding of great importance.

As

It is

quency converter.

V/

the harmonic currents

all 2

2

/

= V70.7 2 -

The harmonic voltage across therefore as far as

E=

IR

=

38.5

AX

59.3

2

=

38.5

A

the 10 i! resistor

10 ft

ef-

is:

= 385

V.

is

Thus,

harmonics are concerned, the synchronous

switch can be considered to be a distorted voltage

HARMONICS

63.1

switch (fundamental)

A

70.7

.

A

1000 V 60 Hz

1000 V 60 Hz

10

807

10Q

a /

H = 38.5

A

Figure 30.14

The capacitor can

Figure 30.13a Equivalent circuit

for

furnish the reactive

power ab-

sorbed by the synchronous switch.

the fundamental components.

switch (harmonics)

10Q

2

Figure 30.13b Equivalent circuit for

all

source of 385

V (Fig.

developed by

this

kW

and

power

it

that

the harmonic components.

30. 3b).

source

1

is

tive

VX

38.5

A=

1

4.8

comes from the fundamental 60 Hz was transformed by the switch.

30.8 Correcting the Given

The harmonic power

385

that the

changed. However, the harmonic component of

power factor

38.5

synchronous switch absorbs a reac-

power of 31.9

that a capacitor

kvar,

it is

reasonable to assume

could supply

this reactive

Let us connect a capacitor of 31.9 kvar

power.

in parallel

with the source (Fig. 30.14). The capacitor will

draw a current of 31.9 kvar/ 000 1

V =

3 1.9 A.

Figure 30.15

Waveshape of the current flowing in the 1 000 V source when the capacitor (Fig. 30.14) is in the circuit.

The

A continues

cause ics.

As

new /

in the

any way affect the voltage between terminals

1

2.

Consequently, the chopped current flowing

in the

switch and the 10 Vt resistor remains un-

and

in

the

1

000

V

source be-

all

harmon-

effective current flowing in the source

38.5

2 )

=

is

63.1 A.

Thus, the addition of the capacitor reduces the

power of 50 kW, which means a fundamental current of 50 A in phase with the 000 V source. in

flow

a result, with the capacitor installed, the

current in the 1000 to 63.

1

to

appears as a short-circuit for

= V(50 2 +

source will then only have to supply an active

But the presence of the cagacitor does not

it

1

A (Fig.

source

V source from 70.7 A (Fig. 30. 10)

30. 14).

is

The waveshape of the current sum of the chopped cur-

equal to the

rent flowing in the resistor

soidal current

drawn by

and switch, plus

the sinu-

the capacitor (Fig. 30.15).

The peak value of the latter is 3 .9 V2 = 45 A. The example we have studied demonstrates nature and origin of harmonics. It also shows 1

the

the

808


origin

of reactive power

These concepts permit a

in

non-linear circuits.

better understanding of the

come from

can only

the switch because the linear

resistor can certainly not provide

As

harmonics generated by electronic converters and

can be represented by the circuit of Fig. 30.

other non-linear devices.

switch behaves like a resistance

30.9 Generation of reactive

power

we saw

that a non-linear

load can absorb reactive power.

Depending upon

previous section

the relationship

between the fundamental voltage

and fundamental current, a non-linear load can also generate reactive power.

Consider the

circuit

first

stead of the last half.

of the current

of Fig. 30.16.

It is

identical

synchronous switch

The fundamental component again 59.3

A

but

it

leads the voltage by 32.5° (Fig. 30.17) instead of

again delivers 50

As

kW

The

field.

As

in the

case of Fig.

30.13a, the resistance of 4.21 II represents the ele-

ment that absorbs the fundamental power of

14.8

kW,

immediately converted into harmonic

which

is

power.

As

a result, the 1000

V

source

of active power but absorbs

31.9 kvar of reactive power. This reactive

(Fig. 30.

is

half of each half-cycle in-

in the circuit is

being 32.5° behind.

8a.

a matter of interest,

we show

the phasor

diagram of the fundamental voltages and currents

to Fig. 30. 10 except that the

closed during the

1

with a ca-

in series

even though the "capacitor" does not pro-

pacitor,

duce any electrostatic In the

it.

regards the^fundamental components, they

power

The

1

8b).

fact that a non-linear device

such as a syn-

chronous switch can absorb or deliver reactive power

opens many interesting

possibilities.

Such

artificial

capacitors and inductors, created by switching, are less

bulky than their

real-life counterparts

and store

only minimal amounts of energy (joules). Because the

energy stored

is

so small, the reactive powers can be

changed almost instantaneously. The nous compensator studied is

a practical

in

static

synchro-

Section 29.3, Chapter 29

example of how reactive power

is

gener-

ated and absorbed by using electronic switches.

synchronous switch

70.7

A

3

i

1000 V/

^

l:;:^> so

r-^

(

1500

kw R

1

31

.9

10

£2

kvar

r 2

-200

Figure 30.16 a. A synchronous switch in series with a resistor delivers reactive power when the switching action advances the current flow. b.

Waveshape

of the current in the circuit.

1500

Figure 30.17

The chopped current contains a fundamental 60 Hz component having a peak value of 84 A. The current leads the applied voltage by 32.5°.

HARMONICS

809

Example 30-5

switch A

A 442 when

600

u.F,

where

it

losses of 20

W

applied across

its

60 Hz capacitor has

The capacitor

terminals.

10Q

V,

rated sinusoidal voltage

is

installed in a factory

is

simultaneously carries a fundamental

rent of 100 A,

cur-

60 Hz and a harmonic current of 80 A,

300 Hz. The harmonic current

is

fundamental. Because the peak

is

25° ahead of

the

V2 times the

ef-

fective value, the distorted current can be described

by the following equation; /

We a.

=

+

100 V2 sin 0

80 V2

+

sin (50

25°)

wish to calculate:

the effective value and

THD of the current

in

the capacitor b.

the effective value and

across c.

its

THD of the voltage

terminals

the approximate value of the losses

(b)

Solution

Figure 30.18 a.

a.

effective value of the distorted current

is:

2

2

= V(l00 + 80 ) = 128A THD = / H // F = 80 A/ 00 A - 80

of Fig. 30.16. b.

The

Equivalent circuit for the fundamental components /

Phasor diagram

of the

fundamental voltages and

%

1

currents. b.

Reactance of the capacitor

X60 = EFFECT OF HARMONICS Now that we know the their effect

we

on

will study a

few

Reactance

nature of harmonics, what

equipment? To

electrical

1/2tt/C

is

X300 =

at

=

6

!0 /(2tt

When

(60 Hz/300 Hz)

its

same waveshape

EH =

terminals does not have the

the capacitive reactance

The reason not the same

is

harmonic

is

that

at the

fundamental

and

Harmonics also

affect the losses in the dielectric

separating the metal plates.

To

a first

approximafrequency

and to the square of the corresponding voltage across the capacitor.

=

6

fl

6 fl

=

1.

2

A

I

=

80

X

the capacitor:

=

6

600

V

the capacitor:

X L2 =

96

V

Total effective voltage across the capacitor:

E =

+ El = V600 2 + 96 2 = 608 V

THD of voltage:

frequencies.

tion, the losses are proportional to the

I00

Harmonic voltage across

a

as the current.

X

Fundamental voltage across

a capacitor carries a distorted current, the

voltage across

442)

300 Hz:

E b = lXm = in

X 60 X

illustrate,

practical examples.

30.10 Harmonic current capacitor

60 Hz:

at

THD = E H IE ¥ = Note

=

1

6

%

THD of voltage 6%) is much less THD of current (80%).

that the

than the

96/600 (

1

The equation of the terminal duced from current

/,

voltage

E

is

de-

while recalling that the

810


fundamental and harmonic voltages are respectively 90° behind the corresponding currents:

E = 600V2 96^2

sin (6

sin (56

+

25°

ence of the -harmonic current has increased the

by 13%.

losses

+ - 90°)

90°)

Harmonic currents conductor

30.11 from which

E = 60(W2 96V2

sin (6

sin (59

-

90°)

+

Whenever

65°)

it

Fig. 30. 19

shows

the

current and voltage.

It is

seen that the voltage

pected because the reactance

at

This

is

300 Hz

be ex-

to is

is

5 times

that

voltages can be treated separately,

we can

mate the losses associated with each. that the losses at

The

losses

600

due

V,

to the 5th

assume they are proportional V,

temperature.

We

harmonic

Hz

to the

300 Hz

Example 30-6

A No

AWG

4

m

cable having a length of 75

resistance of 25.7

mil when

it

is

has a

60 Hz cur-

carries a

then 70 °C in an

After the installation of an electronic drive, the

esti-

recall

cable was found to carry an additional 7

lh

harmonic

50 A, 420 Hz. We wish to calculate the new losses in the cable and the approximate tem-

current of will

be

We

perature

frequency

Solution

losses.

and the square of the voltage. Consequently, the

96

its

ambient of 25 °C.

60 Hz = 20 W.

calculated on the basis of the 60

losses at

a

a harmonic current flows in a conductor,

increases the losses and raises

rent of 100 A. Its temperature

60 Hz. the harmonic and fundamental

less than that at

Knowing

in

waveshape of the capacitor

less distorted than the current.

c.

W + 2.6 W = 22.6 W. The pres-

Total losses are 20

a.

are:

rise.

Original copper losses due to the current of

100 A: 2

PF =

Ir

R =

100

2

X

0.0257

= 257

W

Assuming the resistance of the cable remains the same when it carries the additional harmonic current, the losses caused by the harmonic component are

PH = The new an

2

IH

R = 50 2 X

losses are

257

of 25

increase

%

0.0257

= 64

W

64 = 321 W, which is compared to 257 W.

+

Consequently, the temperature

rise will increase

by 25 %. Since the previous temperature

was (70 will

—

25)

=

45 °C, the new temperature

be about 1.25

that the

X

45 °C

=

56

°C

It

rise rise

follows

temperature of the cable will increase

about (25 °C

+ 56

°C)

=

to

81 °C.

30.12 Distorted voltage and flux a coil

in

Figure 30.19 Distorted current

across

its

in

terminals.

a capacitor and resulting voltage

See Example 30-5.

In Section 9.2,

in a coil is

Chapter 9 we saw

that the

given by the equation:

peak flux

HARMONICS

811

E (9.2)

4.44//V

E

wherein

the effective sinusoidal voltage,

is

the frequency, and /V coil.

What happens To answer

torted?

the

is

if

number of

the applied voltage

we

the question,

/is

turns on the dis-

is

consider the

following example.

Example 30-7

A

distorted voltage

connected across the termi-

is

The voltage has

nals of a coil having 1200 turns.

fundamental component of 150 V, 60

harmonic of 20 V,

a 3

80 Hz. The harmonic lags

1

1

Hz and

1

a rd

35°

behind the fundamental. The respective voltages 495 nWb peak

can therefore be expressed by the equations:

E¥ = EH =

We a.

150V2

sin 6

=

120V2sin(36

212 sinO

- 135°)=

-

170 sin (36

135°)

wish to determine:

the

waveshape and

effective value of the dis-

torted voltage b.

waveshape of

the

the flux

and

its

peak value

Solution a.

The waveshape of the shown in Fig. 30.20a.

E= As

in

V(150

2

+

distorted voltage Its

is

effective value

120

2

=

)

192

is:

Figure 30.20

V

previous examples, the respective funda-

mental and harmonic voltages act independently of each other. Consequently, the flux created

we

a.

Distorted voltage applied to a

b.

Resulting magnetic flux and

coil.

its

components. See

Example 30-7.

can determine

by each, using Eq.

9.2.

The

fun-

damental flux has a peak value:

In the

same way,

monic

flux

is

the peak value of the 3

rd

har-

given by:

4.44/7V (9.2)

4>,

4.44/7V

150 4.44

The

flux

is

469 jjiWb

X 60 X

sinusoidal but

it

voltage (as seen in Fig. 9.1, Chapter

is

4.44

lags 90° behind the 9).

The

equation of the fundamental/flux, in microwebers,

120

1200

age and so 4> H

F

sin (0

-

90°)

180

X

-

125 uAVb

1200

This flux also lags 90° behind the harmonic volt-

therefore:

= 469

X

= =

its

equation

125 sin (36

125 sin (36

is:

-

135°

225°)

-

90°)


812

The

total flux inside the coil is

equal to the

the instantaneous fluxes (Fig. 30.20b).

peak value of 495

from

far different

Note

(JiWb.

that

sum of

waveshape

its

A 200

reaches a is

that of the applied voltage.

Furthermore, the flux waveshape torted.

It

The reason

amplitude of the flux

is

less dis-

is

given voltage, the

that for a

is

inversely proportional to the

frequency.

contains an iron core, the core will

If the coil

heat up on account of the hysteresis and eddy cur-

To

rent losses.

a

approximation, the

first

losses are equal to the

harmonic

losses.

To

sum of

total iron

the fundamental and

-200

1

calculate them, the loss curves

Figure 30.21

of the core material must be consulted.

Distorted current of

nent of 15

30.13 Harmonic currents in a 3-phase, 4-wire distribution

waveshape of

the

13 A and 3

in

each

rd

line of

harmonic compoa 3-phase, 4-wire

the line current,

it

was found

to

have the following components:

lighting systems of

commercial and

industrial

buildings often use fluorescent and halogen lamps

connected between the

and neutral of

line

The same

a 3-phase, 4-wire feeder.

is

com-

true of

A A 6.6 A 2.4 A

fundamental (60 Hz): 3

that are

1

flowing

feeder.

system The

A

rd lh

5

7

th

1

harmonic:

1

2

15

harmonic: harmonic:

puters and other single-phase electronic devices.

The problem

is

draw

that these devices usually

non-sinusoidal currents that contain a strong 3

harmonic.

When

ld

the loads on the three phases are

balanced, the fundamental components cancel out

conductor because their phasor sum

in the neutral is

zero (see Section 8.8, Chapter

8).

harmonics except those

true for all

The same

is

that are multi-

ples of three. Unfortunately, instead of cancelling out, the 3

neutral

rd

9

,

lh ,

I5

lh

conductor.

components

,

etc.,

harmonics add up

Consequently,

these

in the

triplen

are 3 times greater in the neutral con-

flowing that

in

shows

the

waveshape of the current

one phase of a 3-phase, 4-wire feeder

powers the lighting system

lights are

connected between

347 V/600 lines

1

V

line

in

a hangar.

The

and neutral of the

system. The currents

in the

other two

have the same waveshapes except they are dis-

placed by 120°.

The rms the neutral

confirm that the 3

rent in the neutral (45

A)

to

1

13

A and

the current in

be 45 A. After analyzing

ld

harmonic cur-

indeed 3 times that

is

in the

lines (15 A). It

is

easy to understand

line currents

add up

why

the 3rd

in the neutral.

harmonic

Figure 30.21

shows the waveshape of one of the line currents and ld its 3 harmonic component. Comparing the harmonic with the distorted wave, we see that it repeats every 120°. Since the distorted

themselves displaced by

harmonic currents

1

20°,

In

some cases

ld

the 3

neutral overheats

it

line currents are

follows that the 3

rd

in the three lines are all in phase. in the neutral.

harmonic

is

so great that the

and special measures must be

taken to reduce the current.

One way

to block

it

is

to

interpose a transformer connected in delta-star be-

tween the source and the tions, the 3

ld

neutral return

load.

Under these condi-

harmonic on the primary side

connection) cannot flow

line current is

was found

test results

Consequently, they add up

ductor than they are in the lines. Figure 30.2

The

is

absent.

in the lines

(delta

because the

However, the 3

,d

harmonic

current will continue to flow in the star-connected

HARMONICS

windings as well as

in the

delta-connected windings.

IMPEDANCE OF A THREE-PHASE FEEDER

In effect, the transformer acts as a filter as far as the

3

Rl

harmonic

harmonics

concerned. Unfortunately, the other

is th

(5

,

7

th

etc.)

,

813

flowing

in the

secondary

The impedance of a 3-phase feeder

feeder continue to flow in the primary feeder.

inferred

from

short-circuit

usually

is

short-circuit capability (or

its

MVA).

It is

given by the formula:

30.14 Harmonics and resonance eI

Harmonics created by non-linear loads such as

(30.8a)

^sc

elec-

converters can produce resonance condi-

tronic

tions. In

order to understand the problem,

we

or

begin

with a brief introduction, followed by an industrial application.

Figure 30.22 shows a factory that

medium-voltage 60 Hz feeder

Es The

The feeder former

is

is

T having

supplied by a

(30.8b)

Xs

short-circuit capability of the feeder

^sc

that provides a volt-

feeder has an impedance

.

/•s

where

Example 30-8

age

=

*s

1MVAJ

(see box).

line-to-line voltage of the feeder [kV]

connected to a step-down transa turns ratio of

a:

l

.

Its total

impedance of the feeder per phase \Q]

leakage

The impedance

is

considered to be entirely

For example, a 3-phase, 25

active.

kV

having a short-circuit capability of 52

re-

feeder

MVA has

an impedance given by:

a

:

E s2 _

load

T

J

(factory)

2

= 12Q

52

1

Figure 30.22

Medium

25

voltage source and transformer connected to

a factory that contains

some

XT is referred to the secondary side. The equipped with a variable capacitor bank Xc keep the power factor at an acceptable level. The

reactance plant

non-linear loads.

to

is

value of

The

Xc

corresponds to the 60-Hz reactance.

plant also contains

some non-linear

generate a harmonic current / H

Transposing load

all

loads that

.

elements on the primary side of

the transformer to the secondary side,

we

obtain the

equivalent circuit of Fig. 30.23. The procedure for shifting impedances was explained in Section 9.10. 2 Combining XT and Xs /a we obtain the single 60 Hz ,

inductive reactance

What

Figure 30.23 Circuit of Fig.

ferred to the

30.22

in

which the reactances are trans-

secondary side

of the transformer.

is

X

.

{

the nature of the circuit seen by the har-

monic current

Es /a becomes

/ H ? First,

the fundamental voltage

a simple short circuit (Fig. 30.24).


8 4 1

^s

'L

-<

Xj 4*

f

\

load

'l*s a

Jk\ a

v

J

I

Lr

T

load

i

i

a:1

Figure 30.24 as seen by the harmonic currents.

Figure 30.25 Harmonic voltages and currents on the primary and secondary side of the transformer.

Circuit

The value of

X LH

the harmonic inductive reactance

increases in proportion to the harmonic order h of

the harmonic,

we can

On

write

with the frequency. Consequently,

ie.

X

H

l

— hX L

ability to carry useful

.

the other hand, the

XCH

harmonic reactance

tomers connected to the same feeder. In order to appreciate the

of the capacitor bank decreases inversely with the frequency. Consequently,

Note

that

Because the

X

V

H

is

effectively in parallel with

monic frequency can have least

onance

XCH = Xc /h.

latter is variable,

that a parallel

XCH

different values, is

it is

at

one of the harmonics. Perfect resonance occurs

when

X, H

= XCHl but even partial resonance (X X CH can have important effects.

H

in the ca-

pacitor as well as current I L in the secondary of the

transformer can

become

several times greater than

the harmonic current / H

.

Consequently, there

is

a

/C // H

partial

Furthermore,

current

/H

XCH /XLH

and

/ L // H

multipliers.

many

ratio

is

it

is

well

high. Consequently, the

known

harmonic current

A

times.

For example, when the in

the ca-

10 times greater than I H The current in the transformer is 9 times greater. Thus, the multiis

.

plier effect

/ H of

30 A,

of 300 A is

no

means

that a

moderate harmonic current

can produce a capacitor current /c and a transformer current / L of 270 A. That say,

trivial matter.

TABLE 30A AMPLIFICATION OF CURRENT

harmonic voltage

IH

/

H

it

is

EH

(Fig.

V/ii

equal to

0.7

3.33

2.33

0.8

5

4

times the parallel imped0.9

10

9

0.95

20

19

>20

>2()

20

21

1.1

10

11

1.2

5

6

1.3

3.33

4.33

high harmonic voltage will distort the volt-

age throughout the plant. Finally, the

harmonic current

secondary of the transformer

is

/L

flowing

(Fig. 30.25). This

may

to the electric utility. First, the

the

in

reflected into the

primary and from there into the feeder and the

network

H

see that

that in a condition

30.24) can reach high values because

ity

We

equal to 0.9, the current

of resonance, the parallel impedance can be very

ance.

l

resonance can multiply the harmonic

of overheating the capacitor as well as the pri-

mary and secondary windings of the transformer.

the

XCH /X

resonance and the corre-

)

Under resonance conditions, current I c

risk

sponding

even

pacitors

l

importance of the res-

30A shows several

.

clear

possible for

Table

effect,

ratios in the vicinity of

and because the har-

resonance condition

nearly equal to

power. Second, the harmonic

may produce telephone interference; and third, it may upset sensitive electronic devices of other cus-

util-

be unacceptable

harmonic

1

1

.05

will in-

crease the losses in the feeder, thereby reducing

its

HARMONICS

The following example resonance

in

will

show

the importance of

an industrial context.

It

represents the

power of

480

6.

Total active

7.

Total reactive

8.

Harmonic currents I H generated by

the load:

power of the

815

kW

360 kvar

load:

potential situation of thousands of installations that

contain electronic drives or other loads that gener-

r

ate harmonics.

3 th

5

Example 30-9

A

7

factory, built in the

1000 kVA, 25 kV/600

1980s,

powered by

a

to

keep the plant power

9.

in

and

had been

installed, the fuses protecting the capacitors

began

blow occasionally. Furthermore, it was found that the 3-phase voltage was sometimes distorted. The distortion was particularly noticeable whenever the load drew 600 kVA at a power factor of 80 %. All 4 capacitors were then in service, raising the power factor of the plant to above 95 %. It was also found that the capacitor currents were sometimes considto

2.

3.

power

to the factory:

MVA

Analysis of the 60 Hz circuit Before considering the harmonics,

let

us

make

a

few calculations regarding the 60 Hz aspects in the plant. Fig. 30.27 shows the load conditions for one phase (line-to-neutral). 10.

Apparent power of the load: 600/3

1

Active power of the load: 480/3

1

.

= 200 kVA

=

160

=

12.

Reactive power of the load: 360/3

13.

Capacitive reactance of one capacitor:

kW

120 kvar

analysis of the situation yielded the follow-

ing information (Fig. 30.26): .

A

Short-circuit capability of the 25 kV, 3-phase

140

erably higher than normal.

1

25

harmonic:

out.

In 1995, after several electronic drives

An

harmonic:

feeder that supplies

%, four 3-phase 60 kvar capacitors,

located at the service entrance, are automatically

switched

th

negligible

90 A 60 A

th 1

V transformer that has an im-

pedance of 6.5 %. In order factor above 95

is

harmonic: harmonic:

the load:

Number

2tt/C

of capacitors in service: 4

Nominal 3-phase 60 kvar

rating of each capacitor:

14.

1

5.

4.

Line-to-neutral voltage (measured): 360

5.

Total apparent

power of the

600 V

T

load:

Z=

6.5

nil / £ i i

%

T

T T T 4 x 60 kvar

of

correcting capacitors.

=

1

all

.5

6(1

(2.11)

four capacitors:

a

Reactive power generated by the four capacitors:

<2c

2 = E

360

2

86.4 kvar

load

600 kVA 480 kW 360 kvar PF = 80 % 442

Figure 30.26 Single-line diagram

X 60 X 442

600 kVA

1000 kVA

3ph

V

6

Capacitive reactance of

|jlF

25 kV

2tt

6 11/4

Line-to-neutral capacitance of each capacitor:

442

10

1

a factory with its power See Example 30-9.

factor

/jF,

6

LI,

60 A, 21

.6

kvar

-

Figure 30.27 Equivalent circuit of the fundamental components.

816


16.

Reactive power supplied by the transformer:

QT = 17.

120 kvar

-

=

86.4 kvar

26.

Inductance of the feeder referred to the sec-

ondary side of the transformer:

33.6 kvar

X/a 2 _

Active power supplied by the transformer:

Pr = 160kW 18.

0.0026

Apparent power supplied by the transformer:

7 fxH

X 60

2tt

2tt/'

27. Total of the feeder inductance and leakage

inductance: 2 ST = V(160 +

19.

2

33.

=

)

kVA

163

L tolal = 62 +

7

=

69 fxH

Fundamental current on the secondary side of

shows

Fig. 30.28

the transformer:

that the

69

jjlH is

effectively in

parallel with the capacitors. Their capacitance /T

- ST/E =

= 454 A

63 000/360

1

vary from zero to

on the primary side of

20. Line-to-neutral voltage

768

jjlF in

4 steps of 442

1

.

=

(25 kV/600 V)

Fundamental current

15

F

in the

(30.9)

000 V) =

V/25

kV 25

1

0.9

A

2

=

4.5

L = inductance of the LC

C=

feeder:

a

(30.8b)

140

When L = 69 quency

Impedance of the feeder referred

24.

4.5 12 (600/25 000)

2

fxH and

v

2.6

C = 442

[F]

the resonant fre-

fiF,

is:

to the sec-

=

circuit [H]

LC circuit

capacitance of the

ondary side of the transformer:

Xs kr =

LC

2tt\

primary of the

where

= 454 A (600

22. Impedance of the 25

23.

Let

kV

transformer: /,

can

us calculate the resonant frequency for each step,

./resonance

2

(jlF.

using the formula:

the transformer:

ETP = 360 V X

1

_ ~

_

1

(30.9)

2tt\

LC

mft 1

Impedance of the transformer referred secondary side (see Sections

10.

1

to the

2ttV69 X

10

6

X 442 X

10"

2 to 10.14):

= 912 Hz

XT =

6.5

6.5

%

X Zn =

6.5

600

%X

%X

(10.9)

0.0234

This impedance

is

(2

=

C

1326 fxF

to

successively switches from 884

23.4

ma

reactance of the transformer.

now

lh

y

Hz analysis (see

is

2tt/

0.0234 2tt

X

to

very close

^—

90 A 60A-< 25 A *

— —

5 th

7 th 1

th

Fig. 30.27).

consider the impact of the harmonics.

—XT- =

julF

obtain the corresponding

harmonic frequency (420 Hz). Similarly,

phase

Analysis of the harmonic circuits 25. Leakage inductance of the transformtr.

L = ^

we

The resonant frequency of 456 Hz to the 7

69

Let us

fxF,

resonant frequencies of 644 Hz, 526 Hz, and 456 Hz.

effectively equal to the leakage

This terminates the 60

1768

2

000 000

=

Then, as

_

60

„

it

62 fxH

neutral

Figure 30.28 Equivalent circuit of the harmonic components.

HARMONICS

644 Hz

very close to the

is

(660 Hz). Thus, the 7 tentially

lh

lh

harmonic frequency

1

th

and

harmonics are po-

1

the four capacitors are in service as th

X

401

of 10.9

transformer and feeder inductances have a reac-

in the

X 420 X 69

=

jjlH

182 mfl. At

this

frequency, the capacitive reactance of the capacitors

1/(2tt

is

X 420 X 1768

= 214

fiF)

The impedance of the two reactances

The

than

in parallel

is:

_ ~

parallel

=

X 214 — 182

182

214

1217 mil

feeder)

will

th

AX

produce a voltage of 60

1.22 il

and so

=

73

it

in the

V

The harmonic current flowing four capacitors is therefore 73 V/214 mil = in series.

341 A. Also, the harmonic current flowing

secondary of the transformer 401 A. The 60

A harmonic

is

is

in the

73 V/182 mfl

=

therefore amplified in

both the capacitor bank and the transformer on ac-

by each capacitor

monic problem

the transformer car-

85 A. This

V(85

2

2

+

6()

)

=

104 A. The

after a certain time. This

when

A is

the har-

Hz harmonic

voltage across the

60 Hz

360

V,

line-to-neutral voltage in the factory.

The

resulting

line-to-neutral voltage will be V(36()

Compared

360

to the usual

increase and

V, this

2

is

+

73

2 )

=

367

V.

not a significant

would not be noticed by reading an

ordi-

nary voltmeter. However, an instrument that can mea-

is

73 V/360

would

V=

indicate that the voltage

20.3 %.

The

distortion will be

present throughout the plant and could affect sensitive

devices such as computers and electronic drives.

= xlfTl* = V454 + 40 1^ = 2

605

We saw that the 25 kV feeder carries a 420 Hz harA and a useful 60 Hz current of

monic current of 9.8

A

10.9 A. This

higher than the load current of 454 A.

The primary winding

=

capacitors will be superimposed on the

sure harmonic content

much

concerned, they

not immediately apparent.

is

Next, the 73 V, 420

THD

is

is

blow

The secondary winding of ries a total rms current:

This

14.5 A.

even greater than the nominal fundamen-

count of parallel resonance (Fig. 30.29).

r-

=

)

harmonic were absent.

far as the four capacitors are

is

2

and the losses greater

a worrisome situation, particularly

across the capacitors as well as across the two in-

ductances

9.6

capacitors will overheat. If they are protected by 75

1.22(1

A

+

V( 10.9

current of 60 A. Thus, the total rms current car-

ried

harmonic has a value of 60

2

=

will be hotter

fuses, the fuses will

The 7

is /

each carry a harmonic current of 34 1/4

tal

=

kV

if the

current

30.27).

effective current in the primary winding (and

25

As

unwanted

9.6 A. This

almost as large as the load current

is

A that was calculated previously (Fig.

The transformer

mil.

=

(600 V/25 000 V)

harmonic current

shown in Fig. 30.29. As expected, the 7 harmonic then creates a problem. At 420 Hz, the tance of 2tt

reflected into the primary in the ra-

is

of transformation. The primary harmonic current

tio is

dangerous.

Suppose

secondary

in the

817

also carries a current that

portionally greater. Thus, the 401

is

pro-

A harmonic current

is

electric utility

an undesirable situation as far as the is

concerned because the presence of

harmonic diminishes the loading capacity of the

the

feeder. Furthermore, the distortion could affect the

quality of the voltage supplied to other customers.

Note

that the

73

V distortion

across the secondary

**

/-v-v-s

/-v-\ .

18mi2

is

k

164 mil

not reflected into the primary according to the ra-

J

tio of turns.

To

calculate the harmonic voltage across

the primary

we

note that the voltage across the feeder

I i

!

7.2

341

V

ATt -

!

i

I

1768 liF 214 mil

73 V

load

impedance of 18 mil I

(Fig. 30.29).

age

The

Figure 30.29 Equivalent circuit for the 7 currents flowing

in

th

harmonic. Note the large

the transformer and the capacitors.

is

On

is

401

A X

18

the primary side, the

therefore 7.2

V X

resulting distortion

voltage of 25 kV/V3 barely acceptable.

=

(25 is

2

mil =

000 V/600 V)

%

7.2

harmonic

-

V

volt-

300

V.

of the line-to-neutral

14.4 kV,

which may be

just


818

We have only examined the effect of the 7 n harmonic. A fuller study would require analyzing the impact of

1

*'23.4

m£2

nant situation can occur in service. In this

when two

case, the

th

200 kVA 160 kW 120 kvar

kW

33 kvar 637 /jH, ^0.24 ft

1

the harmonics. For example, a reso-

all

63 kVA

•-160

capacitors are

harmonic creates a

1

resonance problem. In is

more complex

mandatory

situations a

to identify the

computer program

problems

can

that

As

reactances between the line and neutral.

442

arise.

For example, induction motors appear as inductive the

number of motors in service changes throughout the day, random resonances may occur which compli-

jjF,

6

60 A

ft,

Figure 30.30 Equivalent circuit of the fundamental components

tuned

cate the matter even further. For this reason, har1

.

monic filters are often used to provide a specific low-impedance path for the harmonic currents.

when

In effect,

where they can do no harm.

We

briefly

examine the

30.15 Harmonic

can create a resonance problem when harmon-

Inductance needed:

three-phase circuits the 5

th

harmonic that could

combined

to eliminate the

in series

4.

LC

circuit

problem

Under these conditions lh

to

add an

in-

C so that the th

harmonic.

the series circuit provides a

low-impedance path for the 5 a result, the 5

is

tuned to the 5

is

harmonic

th

harmonic current. As

will flow

by way of the

Xw =

As regards harmonics higher than

the 5

lh ,

the

always be inductive. Consequently,

/

this ''blocking filter" eliminates the res-

2tt

1

= 637

=

V/5.76

|xH

Hz

volt-

X 60 X 637 X

5.76

il

=

10

6

=

0.24 f)

LC circuit at 60 Hz

il.

is

The fundamental is

therefore

The 60 Hz

62.5 A.

line-

to-line voltage across the capacitor terminals

E60 = is

62.5

AX

about

6 il

10%

X V3 = 650

is

V.

higher than rated voltage of

addition of the coils does not significantly

power delivered by the transformer. we assume the secondary continues to deliver 163 kVA and 454 A per affect the


phase, as

it

somewhat

Example 30-10

.2 fl

60 Hz:

did in Fig. 30.27. However, the fun-

damental current flowing

onance problem.

in

the capacitors

is

is

the voltage

LC blocking filters now

in place, let

greater than before, as

across their terminals.

Figure 30.30 shows the inductances (coils) that are

With the

with each of the 442

us examine the

in series

at

the series

0.24 il

= 360

LC

impossible to produce a resonance condition. The

added

300)

current flowing in each capacitor

600 V. The

it is

=

Impedance of

This

following example, based upon Example 30-9,

shows how

5.

2 tt/L

LC

circuit rather than in other parts of the network.

circuit will

=

installed these coils, let us determine the

Reactance of the coils

6(1-

with each capacitor

th

5

300 Hz

at

har-

produce a resonance condition.

One way

X

1.2/(2tt

6 H;

Hz

ages and currents.

usually the lowest harmonic of interest.

ductance L

is

II.

3.

Having

the lowest

60 Hz

behavior of the system as regards the 60

tors

it is

I

Desired reactance of the coils

filters

have seen that power factor correcting capaci-

Consequently,

= 2

6 11/5

L = X,/2tt/=

We

is

at

2.

nature of these filters in the following section.

monic

—

impossible to suppress the

is

harmonics, they can be channeled through paths

ics are present. In

Reactance of each capacitor

consequently, the reactance at the 300

harmonic it

when

coils are installed in series with the capacitors.

jjiF

Their values are calculated as follows:

capacitors.

new flow of

erated by the load.

the harmonics gen-

HARMONICS

1

l

As regards

6.

we

pacitors in service,

shown

circuit

cause

tively nil.

As

1

at that

a result, the 5

carries a 5

th

300 Hz is effecharmonic current zero. Each capacitor

The harmonic voltage across the capacitors

As regards

,h

circuit

of Fig. 30.3

0.3 fl

X

=

the 7

lh

X V3 =

some of

-

0.42

x

,

1

impedance of the

the

=

0.42

X

0.3 (1

f1

The 7

l

1

206)

(300 Hz/420 Hz)

A from

the previous value of

banks of several thousand kvars,

the capacitor groups are tuned to the 5

lh

ll ,

1

to higher

th

and

\

1

cluding that of the electric

30.16 Harmonics

1

is

to

in the

network,

in-

utility.

the supply

in

may

A divides

manufacturing

that are generated in a

way

find their

into the electric utility

network. This produces a voltage distortion that fects the quality of service for all

harmonic current of 60

harmon-

The objective

.

af-

customers con-

nected to the same feeder. To understand the prob-

be-

lem, consider the one-line diagram of Fig. 30.33a.

tween the 182 mft and 206 mil impedances.

It

shows

the voltage source

Es

of an electric

utility

power to a number of customers A A2, A3, A4 by way of a main feeder having a 60 Hz reactance X s The feeder terminates at a point of common coupling (PCC)*. Fig. 30.33b shows one that provides

0.13

12

90 A, 5

th

th

network Harmonics

= 206 mft

0.206 ft

0.214

A. This

1

(see Fig. 30.29).

such as the 7

LC circuit to

therefore inductive:

is

28.

coils is

while that of

(Fig. 30.32). 8.

+

82

transformer has likewise

in the .9

In large capacitor

plant

X =

A

401

1

1

minimize them everywhere else

becomes

harmonic

a huge drop from the previous value of 341 A.

82/(

channel the harmonic flows to specific paths and to

46.8 V.


ft.

is

1

harmonic while others are tuned

harmonic, and based upon the

(420 Hz/300 Hz)

0.2 14

the terminals of

=

now 60 A X

fallen to 3

ics

0.3 fl

the 7

the capacitors

22.5 A.

is

E M){) = 90 A X 7.

=

harmonic current of 90/4

capacitor bank

in the

is

The current

th

transformer drops to

in the

A current LC branch be-

The 90

.

the resonant

in

impedance

its

obtain the equivalent

30.3

in Fig.

flows entirely


harmonic with the four ca-

the 5

819

H .voltage 5 th

H

I

.

7

= 0

62 |jH

f.tH

OA

90 A

1.2 £2/4 = 0.3 LI

I

1.2 12/4

= 0.3

V

phase of the 3-phase

load

on the

0

draws

circuit.

fundamental current

a

some non-linear

Among

the customers

manufacturing plant Al.

line is a large

loads,

/,

but,

also injects a harmonic

it

Figure 30.31 Equivalent circuit of the 5 the coils are installed

in

th

harmonic components when

current

into the distribution system.

/H

/ 2 , / 3 , /4 ,

but their harmonic contributions are neg-

The feeder therefore mental current / F where / = ligible.

.

F

60

A, 7 th

H ,

= 5.8

62 liH

7 juH

31.9

A

28.1

voltage 7 th

Ajj :

H

V

a

harmonic current / H The fundamental current

LI

0.214

Q

load

l

By Equivalent

circuit of

I } .X S

the coils are installed

in

/•>

+

/3

+

/4 .

and

produces

/,.

a voltage

along the length of the feeder. Similarly,

definition, a

PCC

is

the point of he public supply net-

work, electrically nearest

the 7

+

206 mLl

Figure 30.32 th

carries a total funda/,

.

drop 0.42

The other

customers respectively draw fundamental currents

series with the capacitors.

182 mi2

harmonic components when

series with the capacitors.

It

on account of

lation,

may

and

at

I

to a particular

which other consumers"

be. connected.

consumer's

instal-

installations are. or


820

Example 30-11 Factory Al .in Fig. 30.33a absorbs an apparent power of 4600 kYA from a 25 kV, 60 Hz network. Non-linear loads

in the plant

produce a 5

harmonic current. Compared current, the 5

lh

th

and 29

lh

fundamental

to the

harmonic has a value of 0.12 pu lh

2%) and the 29 harmonic has a value of 0.024 pu (2.4%). The feeder at the point of common coupling

(

1

has a short-circuit capacity of 97

MVA.

Calculate a.

b. c.

d.

The reactance Xs of the feeder th th The value of the 5 and 29 harmonics The value of the harmonic voltages at the PCC The relative values, with respect to the 25 kV feeder voltage, of the harmonic voltages at the

PCC

PCC e.

A2

A1

A3

A4

V3

The

THD at the PCC

Solution a.

Referring to Fig. 30.33, the 60 the feeder

Hz

reactance of

is: 2 El = 25 = 6.44

(b)

J

Q

Figure 30.33 a. Point of b.

common

b.

coupling PCC.

Fundamental current drawn by factory

Harmonic voltages created at the PCC on account harmonic currents generated by load A1

S

4600 X 10 3

£ S V3

25 000V3

106

of the

The current

/H

produces a harmonic voltage drop

in

5

I5

where h if / H is

is

a 7

lH

The 29

hXs

th

harmonic, h

=

=

7. c.

The

E H could voltage £ s con-

5

th

1

A

0.12

X

106

=

12.7

A

harmonic current:

=

I29

the order of the harmonic. For example, lh

A

harmonic current:

EH

given by

EH =

(30.8b)

97

sc

0.024

X

106

harmonic voltage,

=

2.5

A

line-to-line:

For a high value of h the voltage drop be quite large. Because the supply tains

no harmonics,

it

follows that

EH

must appear

E5 =

=

at

PCC. Thus, the harmonic current / H generated by factory Al affects the quality of the voltage supplied to all customers connected to the same PCC.

I5

hXs V3

12.7

X

5

X

6.44

X

V3 =

708

V

the

It is

therefore important to limit the magnitude of

*

ANSI/EEE Standard No. 519-1992 entitled "IEEE Recommended Practices and Requirements for Harmonic Control

in Electrical

cerning the

the harmonic currents

that flow into the electric

ity

network. In general, the

the

PCC

THD

should not exceed 3 %.*

util-

of the voltage

at

may

Power Systems" gives guidelines con-

maximum harmonic

inject into a

also gives a general in a distribution

currents that a customer

power system. This 100-page document overview of harmonic-related problems

system.

HARMONICS

The 29

lh

harmonic voltage,

E29 = =

I29 hXs

Relative value of

d.

E5

(pu)

=

=

(pu)

E5

V3 =

X

6.44

809 V

:

=

708/25 000

Relative value of

E29

F

V3

X 29 X

2.5

line-to-line:

E29

0.028

=

2.8

X

%

:

Figure 30.34

=

809/25 000

-

0.032

3.2

%

The leakage

flux at

eddy currents FS

in

/

Note

that the voltage distortion

harmonic

is

due

th

29

to the

greater than that due to the 5

monic even though the 29

th

fundamental frequency induces the copper windings.

th

har-

harmonic current

Suppose, for example, that a fundamental curis

rent of

th

5 times smaller than the 5

40

A

flowing

Total voltage distortion at the

THD =

V(0.028

2

+

0.032

2 )

PCC:

-

0.043

A 7 th

4.3

4

%

W

duce serious stray losses This

THD is considered to be high.

turn, affects their

Transformers and the K factor

a transformer winding.

Fig.

30,1 7

In a transformer,

some of the leakage

flux lines sur-

rounding the windings intercept the turns of the

mary and secondary windings. As

harmonic current of

same value would produce stray losses of 7" X = 196 W. Clearly, a distorted current can pro-

the

=

primary winding pro-

in the

.

duces stray losses of 4 W. e.

821

windings, which, in

in the

temperature

rise.

30.34 displays a short portion of one turn

torted current / F

produced by

/F

eddy current

/

pri-

a result, these

.

A

carries a

It

portion

$F

of the leakage flux

pierces the turn, inducing in

FS

The sum of these eddy

.

in

60 Hz undis-

it

an

currents

along the length of the winding produces the funda-

flux lines induce feeble voltages inside the copper

aluminum) conductors which,

(or

mental frequency stray losses. in turn,

produce

eddy currents.

These eddy currents produce additional losses

Suppose

that the dc resistance of the

R ohms. The {)

PR

losses.

above the usual Joule

when

the windings carry distorted currents.

sent a fraction g of these

upon the

2

losses.

/?<>/ F

Depending

size of the transformer, the value of g can

vary from 2

The

Distorted currents produce harmonic leakage

Let the stray losses repre-

/? () / h ".

effect

These additional losses are called stray

losses. Stray losses are particularly important

is

in

tance are equal to the windings over and

winding

Joule effect losses due to this resis-

% to

15

total losses

%.

PT at the

fundamental frequency

can therefore be expressed by the equation:

fluxes in addition to the fundamental leakage flux.

When

these harmonic fluxes pierce the copper con-

total losses

ductors, they induce harmonic voltages and therefore

=

+

Joule effect losses

stray losses

PT = P + P K }

harmonic eddy currents. Unfortunately, for a

=

given flux density, harmonic leakage fluxes of order

/? 0

/p

+ gR

{)

ll

h induce voltages that are h times greater than those

induced by the fundamental leakage

flux.

The

thus cor-

responding harmonic eddy currents are also h times greater. Consequently,

PT =

1}:(R ()

+ gR

() )

(30.10)

becausejosses increase as

the square of the current,

it

follows that stray losses

Example 30-12

increase as the square of the harmonic order of the

The primary of a transformer

leakage flux density.

current / F of 85 A.

The dc

carries an undistorted

resistance of the winding


822

0.04

is

The

11.

stray losses

losses. Calculate the value

the stray losses,

and the

amount

to

% of the PR

9

thus,

of the Joule effect losses,

"P v = Comparing Eq.

Solution

P

=

289

tf 0 /,r

}

=

0.04

X

W

of

9

PT =

Total losses:

c

/c

X 289 = 26

,

,

.

.

(30.11)

+

Il that

we

see that

the multiplier effect

on stray

with Eq. 30.10,

1

The value

by the equation: ll

K

same winding section when the winding carries a distorted current composed of a fundamental component / h and several harmonics /2 /4 / ]r The distorted current has an effecshows

Fig. 30.35

$>KR 0 )

to the presence of harmonics.

K is given

W 289 + 26 = 3 15 W

PK =

Stray losses:

due

losses

85"

30.1

K represents

the factor

Joule effect losses:

+

/?(/?„

total losses.

+

+

I\

+

2

+ »_ +

3 lj

+

1]

+

I]

••

Irl

+

/

is

the

+

K-

irll

(30.12)

.

tive value /[

This equation

Thus.

.

sometimes expressed

is

in

abbrevi-

ated form: /

Note tal

=V

1

il

+

1;

+

I]

that the leakage flux

flux lines

c|)

K

+

/

...

K-

comprises fundamen-

and harmonic flux

lines

.

individual stray losses contributed

As mentioned above,

the individual

nents of

/T

.

and the square of their harmonic orders

We

(frequencies). />,.

can therefore write:

— =

h

equal to the

losses are proportional to the square of the respective currents

K—

.

sum of the by the compo-

total stray losses are

(30.13)

where

$ H They

respectively induce eddy currents / FS et / HS

The

2

S/? /f1(pu)

Ai(pu)

due

stray loss factor

to

relative value of the

harmonic with

spect to the total effective current It is

seen that the

A' factor is a

re-

IT

property of the distorted

current and not of the transformer. Nevertheless, dicates the potential heating effect

= p + pK

harmonics

order of the harmonic

when

it

in-

the distorted

current flows in a transformer. For this reason,

some

3

=

/?„/-p

+

!>R 0 (ll

+

2-7;

+ 37= + - +

transformers are designed with a specific Irll)

K factor to

indicate the level of distortion they can tolerate with-

out overheating.

UR

(/,-;

+ 27 t + 37,2 +

»•

+

Irl j)

When

the current carried

0

undistorted,

K=

1

higher than the 31 ple, the is

Figure 30.35 The harmonic leakage fluxes induce eddy currents /HS in the copper windings.

sl

On are

by the windings

is

60 Hz system, harmonics usually ignored. For exama

K factor of a 60 Hz

square

wave

(Table

2A)

about 13. Figure 30.45 shows an instrument that

can measure the

S 7

.

K factor of a current.

Example 30-13 The primary winding of a transformer carries torted current having the following

fundamental current: rd

3

23

harmonic:

rd

harmonic:

a dis-

components:

A A 47 A

520 270

HARMONICS

The winding has

mfl and

a dc resistance of 3

stray losses are equal to

4

the

% of the Joule effect losses.

Calculate a.

b. c.

d.

e.

Joule effect loss f.

The component

3.380

140

W

6.06

example, the stray losses are 25

In this

%

0.24 or 24 called the f

2

W/

1

of the Joule effect losses,

R

former

one must a trans-

in

The

that carries a distorted current.

W=

037

1

commonly

losses. This illustrates that

be prudent when calculating the losses

situation

particularly serious because stray losses are not

is

produces the largest

in IT that

2>

X

251

The effective value /T of the current The A' factor The Joule effect losses in the primary winding The stray losses and the total losses in the primary winding The component in /x that produces the largest

K K

X

^K(h = 23)

823

uniformly distributed over the surface of the wind-

stray loss

Rather they are concentrated

ings.

in certain

re-

gions, often near the top and bottom of the wind-

Solution ings. a.

Effective value of /x

The

significantly reduce the service life of the trans-

K factor they

K factor: K=

2

+

3 lj

2 _ 520 _ +

23

^

3

2

X 270 2 +

0.782

+

1.898

(30.12)

2

23

X 47 2

30.18 Procedure for analyzing a periodic wave

We

2

have seen

The

+

3.380

=

6.06

effect losses:

0.003

=

1037

harmonics play an important role

to

Although special

measure them

its

%X

X

1037

6.06

=

25

1

W

1

.

total losses:

Given a in.

W + 251 W =

}

The the

largest Joule effect losses are

520

The procedure

A fundamental

distorted

wave you want

1288

For example,

H =

W 2.

produced by

7; if

it

is

H you the 7

H

10

th

harmonic,

H =

the fundamental,

Multiply the harmonic order

number

current:

it is

if

is

the

to analyze,

are interested

H

by

l.

10.

minimum number

The of read-

ings required per cycle to ensure that the ac(fundamental)

— =

^(/f 0.003

curacy will be of the order of

X 520

= 811W

ample,

2

=

The

largest stray losses are

harmonic:

10

in the

X

7

case of the 7

= 70

th

±

5

%. For ex-

harmonic,

10H

readings.

* *

f.

is

based on Fourier series analysis.

PT = P + P K = 1037 e.

we

wave

harmonic components

using a simple hand calculator.

decide which harmonic

The

instru-

in the field,

describe a method whereby any distorted

can be decomposed into

W

stray losses:

P K = gPjK = 4

that

installations.

ments are available

now

P3 = I$R 0 = 588 2 X d.

specified.*

/?3

in electrical

The Joule

is

2

588

c.

can tolerate

HARMONIC ANALYSIS +

I\

/T

=

considerably

former. Special transformers are built in which the

= 588 A b.

may become

These overheated regions can

hotter than others.

= V520 2 + 270 2 + 47 2

Ij

Thus, some regions

:

produced by the 23

,d

See ANSI/IEEE C57.

Recommended Capability

1

1

0-

1

986 Standard

entitled

"IEEE

Practice For Establishing Transformer

When

Supplying Nonsinusoidal Load Currents."


824

3.

Since one cycle comprises 360°, the approxi-

mate

D

interval

between readings

is

For example,

D =

we

if

=

0

235°,

A=

H =

49, and

7,

obtain for this row:

360710 H. In the case of a 7 harmonic, D = 360770 = 5.14°. This number can be rounded lh

to 5°, 4.

which simplifies the

Prepare a 4-column table similar to that

column

down

[

1

1

8.

the

will

list

360.

They

D

numbers

D =

5° the

0, 5, 10, 15

.

.

.

wave during one

write

0,

wave.

cos (7

column up

7,

to

9.

A cos

(HO),

the product of

S] of the values in

column

=

excluding the value corresponding to 0

360°.

cycle.

sum

Calculate the [3],

Then


X

according

to the equation:

X = S.D/180 10.

passes through zero.

it is

row

if 0 = 235°, A = 49, and we obtain A cos (HO) = 49 X X 235°) = 49 X cos 1645° = -44.4

choose the moment when the wave

However, no matter where the

1645°

235°)

= -20.7

headed by symbol

[41

usual (although not necessary)

It is

sin

X

cos (HO).

H =

Select a starting point on the distorted

to

sin (7

For example,

represent the angles, in degrees, of

the distorted

column

A X

headed by the symbol degrees. If

In

calculate for each

indicate the

the angles from zero to 360°, spaced by

intervals of

5.

= 49

listing in degrees.

shown in Table 30B. The crosses numbers you will be filling in. In

(H6f = 49 X

A sin

starting point

Calculate the [4],

sum S 2 of the

(30.14)

values in column

is,

360°.

always designated as the 0° angle.

=

excluding the value corresponding to 0

Then


Y according

to the equation: 6.

In

column

down

[2]

headed by symbol A, write

sponding to the angles 7.

In

wave correcolumn f 1].

S.D/180

(30.15)

the values of the distorted

column

|3]

listed in

headed by symbol

calculate for each row, the product

A sin

AX

1

1

.

(HO),

sin (HB).

AH 12.

The angular

position

a —

D =

harmonic H =

\

X + Y2 :

a

[1]

12]

13]

[41

6

A

A sin HO

A cos HB

0

X

X

X

X

X

X

X

The following

arc tan

13.

=

0°)

is

re-

given by:

Y/X

(30.17)

rule applies to Eq. 30. 17:

If the value of'X is negative,

added

(30.16)

of the harmonic with

spect to the starting point (0

HARMONIC ANALYSIS

TABLE 30B

The amplitude A H of the desired harmonic component is then given by:

180° must he

to the angle.

The harmonic component you

are seeking

is

given by the expression:

harmonic X

X

The

X

X

is

sum So

A0 -

S„D/36()

sum

X =

S]

S,D/180

effective or

equal to

sum S 2

Y=

14.

S 2 D/180

In

H = AH

some

ponent

,

(HO + a)

(30.18)

rms value of the component

A u Nl.

cases, the

A0

sin

in

wave may contain

a dc

comTo

addition to the ac components.

HARMONICS

determine values

=

in

360°.

value, calculate the

its

column

The value of A

The above

()

{)

is

H =

harmonic

given by

S„D/360

(30.19)

is

much

m

[2]

B

A 0

computer

easier using a

30

or spreadsheet.

Example 30- 14 wave

positive half of the

is

We

tive half is rectangular.

1

and 10

which

15

90

7.5

13

0

150

37.5

18.75

-32.5

180

0

to determine:

H =

10

X

36°. Let us

th

harmonic

1

=

Thus,

10.

360730° =

0

0

0

0

240

-20

17.3

10

270

-40

40

300

-40

34.6

slightly.

0

-20

330

0

0

0

360

0

0

0

=

S,

30°,

0

210

D is

make D =

improve the accuracy

must therefore take

6.5

wish

10 readings must be taken during one

will

0

3.75

triangular and the nega-

cycle of the distorted wave. The interval

360710 =

0

-15

analyze the fundamental; conse-

H =

0 7.5

60

Solution

therefore

HO

22.5

the amplitude and position of the 4

at least

cos

26

the amplitude and position of the fundamental

to

A

H9

sin

30

b.

want

14]

[3]

A

22.5

a.

quently,

30°

120

Figure 30.36 displays an unusual waveshape. The

We

D =

I

calculations can be done by hand,

but the task

a.

HARMONIC ANALYSIS

TABLE 30C

excluding the value for 0

[2],

A0 =

sum S of the

825

X -

=

S.D/180

Y =

= -43.5 = -7.3

S2

175.9

29.3

S 2 D/180

We

12 readings

during one cycle.

Note

that

value of the

the

wave changes

abruptly at 160°, 240°, and 320°. In such cases,

we

take the average of the

mum (40

(and

values. Thus, at

+ at

0)/2

= +

320°)

is

20.

-

maximum

and mini-

160° the average value

The average value

at

angle

determined and

columns

is

240°

is

listed in

necessary calculations are then [3]

The sum quently,

we

and

column

made

to

[2J.

The

complete

[4].

Sj in

column

[3]

gives

1

75.9.

Conse-

obtain:

20.

30C is the four-column table mentioned previously. Column shows the angles from 0° 360° of 30°. By tracking the in intervals D to

X=

Table

[

1

waveshape, the value

]

A

S,D/180

175.9

sum of

Similarly, the

yields S 2

=

= —

X 30/180 -

column

the values in

43.5. Consequently,

29.3

we

[4]

obtain:

corresponding to each

SoD/180

-43.5

X 30/180

-7.3

Applying Eq. 30.16 and 30.17, we obtain the amplitude A, and phase angle a of the funda-

40

mental 240° 0

320°

160°

360°

Vx

1 1

-40 Figure 30.36 Analysis of a distorted wave.

a =

=

2

2 2 2 + Y = V29.3 + (-73) =

'Y\ arctan

(

—

\X arctan

=

/

-

\

29.3

7.3>

arctan

(- 0.249) = -

14°

i

30.2


826

Figure 30.37 th Fundamental and 4 harmonic components

wave

of the

A

45

11.25

54

13.5

63 72

[41

L3J

A

sin

HB

A

cos

HB

0

-11.25

-7.94

-10.92

15.75

-14.98

-4.87

18.0

-17.12

5.56

#

81

20.25

-11.9

16.38

90

22.5

0

22.50

99

24.75

14.55

20.02

108

27.0

25.68

8.34

117

29.25

27.82

-9.04

126

31.5

18.52

-25.48

135

33.75

0

-33.75

144

36.0

-21.16

-29.12

153

38.25

-36.38

-11.82

162

0

0

0

the equation:

171

0

0

0

=

180

0

0

0

189

0

0

0

198

0

0

0

H,

The

A,

sin

effective

30.2/V2

=

(H8

therefore be expressed by

+

a)

value

=

-

30.2 sin (6

of the

14°)

fundamental

is

21.3.

Figure 30.37 shows the fundamental super-

posed on the original wave,

now consider we must select

th

Let us

the 4

case,

at least

harmonic. In 10

this

H = 10X4 =

40 readings on the distorted wave. The interval D must therefore be no greater than

360740 -

harmonic

H =

D =

4

11]

12]

131

e

A

A sin HO

9°

[41

A

cos

H9

0

0

0

0

9

2.25

1.32

1.82

18

4.5

4.28

1.39

27

6.75

6.42

-2.09

9.0

207

0

0

0

216

0

0

0

225

0

0

0

234

0

0

0

243

-40

38.04

12.36

252

-40

38.04

-12.36

261

-40

23.51

-32.36

270

-40

0

-40.0

279

-40

-23.51

-32.36

-12.36

9°

HARMONIC ANALYSIS

TABLE 30D

36

[2]

e


The fundamental can

b.

[1]

5.29

288

-40

-38.04

297

-40

-38.04

12.36

306

-40

-23.51

32.36

315

-40

0

40.0

324

0

0

0

333

0

0

0

342

0

0

0 0

351

0

0

360

0

0

-7.28 S,

(continued)

=

-29.11

0 S2

= -101.96

HARMONICS

30D again lists the 4 columns. In this case, sum S, of column [3] gives —29. and that

Table the

1

of column

L4J

= -

gives S 2

101.96.

It

827

60

1

follows that

X = S,D/180 - -29.11 X (9°)/180 = -1.46 Y = S 2 D/180 = -101.96 X (9°)/180 - -5.10

40

20

> CD

The amplitude of the 4

A 4 = VX + Y- = 2

harmonic 2

V(-I.46)

4-

is

therefore:

(-5.1)

B

0

40

0

2

80 120 160 200 240 280 320 360 400

5.30

l|

-20

and the phase angle

a =

Lh

/Y\ arctan

is:

=

/

-5.10

\

-1.46

-40

arctan

\XJ

-60

=

arctan 3.49

=

74°

Figure 30.38

However, since

X

is

we must add

negative,

to the calculated value of 74°.

therefore

The 4

a = 74° + Ul

180°

-

The

1

80°

true angle

is

Waveshape obtained by summing and the

first

12 harmonics of

254°.

harmonic can be expressed by the

The sum of the fundamental and is shown graphically in Fig. 30.38. The resulting waveshape is quite close to the original waveshape even though the num(see Table 30E).

equation:

the first twelve

H4 =

5.3 sin (46

Figure 30.37 shows the 4

th

the fundamental

Fig. 30.36.

+

254°)

harmonic superposed

harmonics

ber of harmonics

is

quite limited.

on both the original waveshape and the fundamental wave.

By

repeating the

same

we can

exercise,

expressions for harmonics from the 2

nd

Questions and Problems find

to the 12

th

Practical level 30-

1

A 60 Hz

distorted current contains a 5

harmonic of 20

TABLE 30E

A

HARMONIC COMPONENTS

a.

Fundamental 2

nd

3

4 5

6

-

30.2 sin (0

14°)

H

7.7 sin (26

+

100°)

rd

H

7.4 sin (36

+

119°)

lh

H

5.3 sin (48

+

254°)

lh

H

4.1 sin (50

+

110°)

th

H

4.6 sin (60

+

239°)

lh

H

2.1 sin

(70

269°)

H

3.3 sin (88

+ -

H

1.4 sin (90

+

184°)

30-2

A and

Ul

a fundamental of 30

(rms values). Calculate: the effective value of the distorted current

b.

the frequency of the fundamental

c.

the frequency of the

harmonic

A 60 Hz voltage having an effective value of 485 V contains several harmonics. The fundamental has an effective value of 481 Calculate the effective value of

all

V.

the

harmonics. 7

lll

8

9

,h

H)

lh

lh

ll th 1

H

.3.0 sin

(108

30-3

8°)

+

1°)

H

1.5 sin

(118

+

97°)

H

1.9 sin

(120

+

117°)

In

Problem 30-2, calculate the

THD

in

percent.

30-4

A current

has the following components:

fundamental: 960 A; 5 7

lh

lh

harmonic: 156 A;

harmonic: 235 A. Calculate:

a.

the effective value of the current

b.

the distortion factor, in percent


828

30-5

Problem 30-4, the current flows

In

in

dissipated and the

with the fundamental current

b.

with the 5 with the 7

c.

A sinusoidal

30-6

th lh

1

triangular

the exact

Table 2A, Chapter

2.

wave of Fig. 30.39 has

peak value of 100

harmonic

voltage of

A contains

The

30- 2

in

V.

a

Determine the ap-

proximate rms value of the fundamental.

harmonic

a non-linear load.

85

value given

power associated

a.

Compare your value with

30°.

a

2 12 resistor. Calculate the total power

The

480

V

is

applied to

resulting current of

a fundamental of 74

A that

lags 32° behind the voltage. Calculate:

power factor power supplied by the source power factor

a.

the displacement

b.

the active

c.

the total

A 3-phase, 4-wire cable

30-7

Figure 30.39

feeds a group of

halogen lamps that are connected between

and

line

neutral.

The

current in the lines has

an effective value of 320 is

due

to a 3

rd

See Problem 30-12.

A of which 47 A

harmonic. Calculate the

Advanced 30- 3

In Fig. 30.40, using the

1

value of the current flowing

level

in the neutral.

30.

1

8,

a.

the fundamental

b.

the 3

Intermediate level

A voltage

30-8

of 4300

V

has a

THD of 26

%.

rd lh

c.

method of Section

determine the peak values of

the 5

harmonic

harmonic

Calculate the rms values of: a.

the fundamental

b.

all

A

the harmonics

A 60 Hz source contains a fundamental of lh 730 V and a 5 harmonic of 108 V. The

30-9

source 5

mH

is

connected

in series

to

n

100 50

an inductance of

with a resistance of 10

(2.

Calculate the effective values of the fol-

-50

lowing currents and voltages:

H

a.

fundamental current

b.

5

c.


d.

voltage across the resistor

Figure 30.40

voltage across the inductor

See Problem

th

-100

harmonic current

e.

30* 0 1

A 3-phase cable carries a distorted current of 830 A that contains a 7 harmonic of 60 th

A. The resistance of the cable the

harmonic

is

is

suppressed, by

2 mCl.

30- 4 1

1

1

A square wave

how much

has an amplitude of

Using the method described 30.

1

8,

in

1

00

V.

Section

determine the peak value of the

fundamental component. Take intervals of

360

30-13.

In Fig, 30.4

1

,

determine the peak values

of

If

will the losses in the cable decrease?

30-

270

180

90

0

a.

the fundamental

b.

the 3

c.

30- 5 1

the 5

rd th

harmonic harmonic

In

Problem 30-

a.

the effective value of the current

b.

the effective value of the fundamental

1

3,

determine:

HARMONICS

A distorted voltage

30- 8 1

829

represented by the

is

following equation (angles expressed degrees,

E=

t

850

sin

340

sin

in

seconds):

in

1

000

8

+

r

(126 000

-

t

30°)

Calculate a.

the frequency of the fundamental and the har-

b.

the effective value of the fundamental and the

c.

the effective value of the distorted voltage

monic Figure 30.41

See Problem

30- 6 1

harmonic 30-14.

c.


d.

theTHD

all

to a

the instantaneous voltage

e.

Draw

torted voltage

commercial building has a

circuit capacity of

60

30-19

MVA,

A non-linear

5

a.

the reactance of the feeder, per phase

b.

the current per phase at the service

t

=

1

ms

waveshape.

its

load generates the following it is

connected to a 50 Hz

source:

Calculate

if

and sketch

harmonics when

short-

when

a phasor diagram that represents the dis-

the harmonics

A three-phase 24 kV feeder that supplies power

d.

The

a short-circuit occurred

entrance to the building upstream

th :

th

A

7

:

4

lb

A

1

l

:

9

A

th

13

:

8

A

currents flow in a circuit that contains an in-

ductance of

40

20

1

300

with a capacitor of

jjlH in parallel

jjlF.

from the step-down transformer Calculate 30- 7 1

In Fig. 30.42 the rectangular current has a a.

peak value of 100 A.

Its

amplitude

is

zero

across the terminals of the inductance

during successive intervals of 36°, as shown. b. a.

Calculate the effective value of the fun-

Show

that the 5

th

c.

harmonic

is


all

the

harmonic voltages

across the inductance

damental b.

harmonic voltages

the value of the respective

the effective value of the current flowing in the

essentially

capacitor

zero

30-20

In Fig. 30.

1

1

,

determine the amplitude and

phase angle of the fundamental compo-

A

nent of the chopped current. Use intervals

100

of 6°.

30-2 180° 18°

90°

162

360°

1

Figure 30.43 shows a periodic voltage consisting of a succession of sinusoidal

pulses having an amplitude of 100 V.

Using intervals a.

D of 6

°,

determine:

the amplitude and phase angle of the

fundamental

-100

component

b.

the dc

Figure 30.42

c.

the effective value of the voltage

See Problem 30-17.

d.


all

the harmonics.

830


Figure 30.45 This portable instrument, the "Power Harmonics

Figure 30.44

Analyzer" can measure the

an oscilloscope, a multimeter, and a recorder. It used to verify control systems including variablespeed electronic drives {courtesy of Fluke Electronics

total harmonic distortion st (THD) of a voltage or current up to the 31 harmonic. It can also measure the active, reactive and apparent power in a 3-phase circuit, and its power factor. It also gives a readout of the K factor of a current (courtesy

Canada

of Fluke Electronics

This portable instrument, the "Scopemeter" incorporates

is

Inc.).

Canada

Inc.).

Chapter

31

Programmable Logic Controllers

Today, the programmable logic controller (some-

31 .0 Introduction

times called programmable controller)

Of

manufacturing operations, the programmable logic controller (PLC) is one of the most important. The first

all

the devices that are used to control

PLCs were

introduced

in

Up

More

31.1 Capacity of industrial

until then

PLCs

PLCs were mainly used

equipment was achieved by using hundreds, and even thou-

hard-wired physical relays. This offered

sands, of relays enclosed in metal cabinets.

vantages because the

the automatic control of manufacturing

In the beginning,

The annual automobile-model changes required

control systems

took a

lot

less energy.

Because the

many

cators that

when

made

it

LED (light-emitting diode) indi-

control system and to diagnose problems.

computer technology, the performance of PLCs

system

pressive.

Today, thanks to the evolution

to replace the relay racks.

This presented a big challenge for

computers

that

many compare-

spond

the

to the

needs of industry. Little by

little,

control

found. However, a

full

are

still

and regulate

is

and im-

used to replace relays,

industrial

operations and

processes.

For

in-

stance, they can regulate temperature, pressure, flow rates,

techniques were improved and more users of the

new technology were

While they

in electronics

PLCs can now perform mathematical

had previously been

used to do accounting jobs were modified to

much

easy to check the operation of the

making connections. For these reasons, control engineers developed a computerized programmable

nies. In effect,

ad-

less space than

Furthermore, they were programmable

and equipped with

were complex, the modifications

of time, and errors often occurred

PLCs took up

to replace

conventional relay cabinets and consumed

frequent modifications to the production lines and their associated relay-control systems.

main

the

is

than 50 man-

ufacturers offer hundreds of different models.

1960s,

the early

mainly by the automobile industry.

control device used in industry.

can

decade

motor

drives,

and so

forth. In addition,

now communicate with each other as well

a central computer.

The

latter

can collect data, change

went by before the new concept was' systematically

the operating parameters, and even modify the

adopted by manufacturers.

programming.

831

PLCs

as with

PLC

832


Figure 20.24a (reproduction) Simplified schematic diagram of a starter with plugging control.

Some programmable date

controllers can

accommo-

more than 3000 inputs and outputs. These easily replace more than 10 000 con-

PLCs can

ventional relays. Consequently,

it

is

possible to

31.2 Elements of a control system During our study of control systems

we saw power

that a

PLC.

contactors

However,

is al-

Furthermore,

case of larger factories,

ways preferable

them by

a

processes,

the

to

following sections

sic principle

PLC.

Finally,

we

will

wired relays to

PLCs

is

made.

stop

motors.

electric

of Figs. 20.16b, 20.24a, and 20.25

in

Chapter 20,

we

number of control devices

used

case,

we

then followed by

of a

that are

ignore in

each

see that the fundamental difference be-

tween the three diagrams

examine how PLCs are apthe transition from hard-

how

and

examining the schematic diagrams

the

a closer look at the actual physical construction

plied in industry and

start

handle specific

will explain the ba-

is

in

If

of the programmable logic controller

using a very simple model. This

to

perceive that they are very similar.

PLCs can be made we

and low-

we

smaller, faster, and easier to program. In the

Chapter 20,

net-

communications

work. By using individual PLCs production

PLCs throughout

to install several

the plant and link

it

in

relays,

auxiliary contacts were able to control big

control a complete factory with a single in the

few pushbuttons,

lies in

the

way

the control

devices are interconnected.

Suppose

that

we have

1

a "black box' inside which

various connections can be trol

made between

the con-

devices (pushbuttons, auxiliary contacts) and

the controlled devices (holding coils of contactors, pilot lights). In the case

of Fig. 20.24a (reproduced

here for convenience), this approach results

in the

PROGRAMMABLE LOGIC CONTROLLERS

control

input terminals

controlled

output terminals

devices

833

devices j

AA

—o

D

—o

o—

'

o—

start

stop

Hh

Hh

connection box ("black box")

Axi

L c p

R

—

—

o

F

Figure 31.1

An automated system

same

is

composed

devices and controlled devices. These on/off devices are all of the automated system. The systems differ only as to the number of devices

of control

nature, irrespective of the type of

and the connections between them.

set-up

shown

in Fig. 3

1

.

1

.

The

control devices (start

power

and stop pushbuttons, auxiliary contacts, and zero-

supply

speed switch) are connected to the input terminals of the black box. In the

same way,

controlled (holding coils

A

the devices that are

and B) are connected

XL

to

the output terminals on the right-hand side of the

CPU

control

input

central

output

controlled

devices

module

processing

module

devices

black box. unit

Suppose the black box puter

is

is

a computer.

The com-

designed to simulate the relays, relay

coils,

between

relay contacts, as well as the connections

programming

them. This approach opens up enormous possibilities

unit

because the computer can simulate hundreds of

relays having thousands of contacts.

only limited by the computer's

The number

memory

is

capacity.

Figure 31.2

The

five parts of

a PLC.

The control system can therefore take the general form shown in Fig. 31.2. A programmable logic controller consists of five basic parts. 1

.

A central processing computer

that

unit (CPU),

which

2,

is

between them.

input module, which serves as an interface

a

can simulate the required relay

contacts and relay coils, as well as the connections

An

between the actual control devices and the CPU. 3.

An

output module, which serves as an interface

between the

CPU

and the actual devices

are being controlled.

that


834

output

input

CENTRAL PROCESSING UNIT

module (+)

RELAY COILS

rfo2i

-

RELAY CONTACTS

for output relays

(i) (ii)

(ii)

(conventional) (iii)

(i)

(4)

for internal relays

111

for output relays

112

H(

for internal relays

—<5h

01

-°

(

—V 02

<

03

(iii)

for internal relays

(iv)

for input relays

(time delay)

(10)

(time delay)

(-)

(conventional)

(50)

for internal relays

13

module

(CPU),

04

'

—

—

-°

oc

IC

Figure 31.3 The central processing unit (CPU) contains in its memory a "stock" of elements such as "counters " etc. The "relay coils" in the CPU are represented by truncated circles.

A programming unit consisting of a keyboard

4.

and monitor

to

program the CPU.

select different types of "relays" that the

way

It

and "contacts"

On

the other hand, "coil" 101

it

is

103 are in the

unit.

(i.e.,

CPU, where

Let us examine the construction and the role played

we choose

matters,

a very simple

3 input terminals and

The input module has 12,

and

13,

and a

PLC

4 output terminals

having only (Fig. 31 .3).

terminal IC.

The

ternal control devices are respectively

between the a

24-V

1 1

source.

,

12,

and

The

To simplify

three terminals labeled II

common

13 terminals

real ex-

connected

and one side of

other side of the source

is

con-

dress)

is

To understand it

is

associated with terminal 12.

the operation of the input module,

useful to imagine that each rectangle corre-

sponds

to a "relay coil" that

sociated

bers 111, 112, 113, and 114. These

external

"coil" 102

is

control

is

"excited" by the as-

device.

For example,

normally "excited" because the push-

button with which

it is

associated

is

normally closed

NO contacts are

sometimes the mechanical contacts of

relays. In

other cases, they are electronic switches (such as triacs) that will

nal received

Figure 3 1

is

not

1

and B and a

address 102

ple, the

is

,

A

in the figure.

small rectangle bearing a reference

they form part of the vir-

01 02, 03, and 04, as well as a common terminal OC. The four terminals are associated with four normally-open contacts that bear reference num-

number (adassociated with each terminal. For exam-

nected to terminal IC, as shown

NO. The

simulated) control circuit.

control circuit

four components listed above.

is

The simulated shown in Fig. 31.3. The output module (Fig. 3 .3) has four terminals

tual

first

it

"contacts" associated with "coils" 101, 102, and

power

modules, and by the programming

normally

normally "not excited" because the

is

auxiliary contact associated with

they are to be connected.

is

normally open (NO). In the same way,

"coil" 103

needed by the CPU, by the input/output (I/O)

by the

"relay coils," "contacts

"not excited" because the pushbutton associated

with

computer can simulate, as well as the

A power supply that furnishes the

5.

(NC).

enables us to

V 120 }

open or close depending on the

from a "relay coil" .3

in the

shows two external contactor

pilot

sig-

CPU. coils

A

lamp, respectively connected be-

tween three of these output module terminals and one side of a 120-V ac source. The other side of the ac source is

is

connected to terminal OC. Terminal

not connected to anything.

on the input and output modules are "addresses" tablished by the manufacturer of the

The tion

04

The reference numbers

central processing unit has a

and an operating function.

es-

PLC.

memory

func-

We can imagine that

PROGRAMMABLE LOGIC CONTROLLERS contains an enormous stock of "contacts" and "re-

it

lay coils." This inventory of parts

memory

of the

1

.

in its

bus bars

stored in the

is

CPU. The simple model of Fig.

contains the following virtual (simulated) nents

3

1

input

.3

that

correspond

number of "contacts" per

NC

condition

is

set

"coil"

and

by means of

namely 101, 102, and

NO

13

the pro-

four

NO contacts of the output module. These

and

3,

,114

04

OC

120 V

"coil" 112, and the connections between them and the "bus bars" are all programmed ,

from the keyboard. Rectangle 101 simulates the

"coil"

of relay 101.

111, 112,

Because there are four output terminals, the

The

03

14 as the contacts they activate.

number of "output 3.

113

IC

"Contact" 101

shown) would

same reference numbers 1

02

Figure 31.4

particular "coils" (not explicitly

1

H'l>24 V

103.

"coils" of the "relays" associated with the

1

112

CPU

The

carry the

-o

.

The

their

01

111

Hr

,

gramming unit. These particular contacts (not explicitly shown in Fig. 31.3) would bear the same reference numbers as the "relay coils,"

2.

112

-OH

to the relay

"coils" (rectangles) in the input module.

output

module 101

12

or

,_j

{+

module

PB1

compo-

memory:

The "contacts"

835

coils"

is

We

also four.

will

now

use five examples to illustrate the be-

havior of this simple PLC.

"relay coils" and "contacts" of the "inter-

nal relays."

The

CPU;

they do not appear

in the

Examples

31.3

"coils" and "contacts" of these

"internal relays" operate entirely inside the

of the

use of a PLC

Example 31-1 In Fig. 31.4, we want lamp L2 pushbutton PB is depressed.

input or output

modules.

up when

to light

1

The memory contains an almost unlimited number

Solution

of "contacts" that can be associated with any one of 1

the internal "relay coils."

The "contacts" bear

.

same reference numbers as the "relay coil" that activates them. Depending upon the requirements of the control system, we can program as many "con-

101 2.

CPU

and

(

—

).

Let us assume that the

CPU

(

+

3. )

inventory con-

tains the following internal items: 1.

ref-

erence numbers 701 to 750; and 1

1

By

is

the pushbutton

connected

is

1

1

,

"coil"

depressed.

to terminal

12 (relay contact or triac)

is

02,

closed

lights up.

virtue of

(

1

),

the operator

must

select a nor-

mally-open "contact" carrying reference number 101 from the stock of components

(2),

memory.

In the

same way, by

in

the

virtue of

he must select the output "relay coil" num-

bered

shown 2.

connected to terminal

when

Because lamp L2

CPU

50 "coils" of conventional "relays" bearing

is

1

excited

when L2

simulates a power

supply represented by two vertical "bus bars"

is

contact

tacts" per "relay" as needed. In order to "excite" the

various "relay coils," the

Because PB

the

1

12.

(The simulated relay coils are

as truncated circles.)

0 "coils" of "time-delay relays" bearing refer-

ence numbers 901

to 910.

The

tirrie.

delays as-

sociated with these "relays" can be set during the

programming

period.

The

selections are

Finally, using the

erator

made by using

the keyboard.

monitor and the keyboard, the op-

programs the connections between "contact"


836

bus bars

bus bars

\

/ input

(_)

(+)

PB1

<

12

&

120 V

is

similar to the

one

31 .4 except

in Fig.

has been proa normally-open (NO)

that a normally-closed "contact" 101

grammed

in

CPU

the

instead of

113

03

4fCPU

IC

Figure 31.6 By programming

Figure 31.5

02

04

-o

113

101

H'l> 24 V

01

112

114

^4-fT03l-

-o

OC

This diagram

101

112

111

I3

04

114

IC

111

.

,

03

113

CPU 24 V

101

02

112 12

13

output

(_)

module

01

111

o^-flQ2l-

j+)

module""

PB1

112

101

/

input

output

module

module

OC

additional "contacts"

and

120 V

"relay coils"

a more complex control system

into the ladder circuit,

can be developed.

contact.

a few keys on the keyboard the operator can I0l, "coil"

and the

ll 2,

+

(

)

and (-) "bus bars"

shown in Fig 3 1.4. As a result of this program, when pushbutton PB is

depressed, "coil" 101

is

activated by the external

24- V dc source. This causes "contact" 101 to close,

which

in turn excites "coil"

excited, contact

Note

that "contact"

ally exist.

lated

is

1

When

L2

"coil"

1

12

is

lights up.

0 and "coil"

1

1

1

When

"coil"

12 to close.

powered by

As

1

1

2

is

In Fig. 31.6,

V

1

1

to

1

14 in the output

.

that

when PB1

up and L3 goes

light

is

de-

out.

three lamps are connected to terminals

03 and

PLC

Ol

are therefore controlled by

contacts 111, 112, and

source. (As

triacs or

The

02, and

"excited"

module are usually

L2

re-

1

1

3.

Consequently, the

operator must select the three correspond-

ing "coils" 111, 112, and 113.

mentioned previously, the normally-open contacts 1

than a minute.

pushbutton PB1 must control three

pressed, LI and

a result, the real

the real 120

in less

Solution

2 do not

1

change

lamps (LI, L2, and L3) so

are merely virtual elements simu-

by the computer.

lamp L2

12.

this

Example 31-3

1

They

causes contact

it

1

12 closes and

1

make

as

2.

mechanical contacts.)

The input

"coil" 101

with three "contacts"

must now be equipped (all

labeled 101),

two of

which are normally open and one normally

Example 31-2

closed.

Referring to Fig. 31.5, pushbutton PB1 must again

tions as

lamp L2, but this time PB1 go out when is depressed.

the

activate pilot

the

lamp must

Note Solution 1

.

The setup

The operator must program the connecshown by the "ladder circuit" between ( + ) and (-) "bus bars."

that

we can add

"contacts," increase the

num-

ber of "relays," and change the "connections" by is

identical to the

one

in

Example 31-1

simply pushing a few keys on the keyboard.

except that the operator must select from the

never have to

memory a "contact" 101 that is normally closed. As a result, "coil" 12 is normally excited and so

As

1

real contact

1

12

is

normally closed.

By

pressing

strip

3 are real comThe 120-V source and the

before, contacts 111, 112, and

ponents (usually three

lamps

are,

We

wires or mount a relay on a rack.

triacs).

of course, also

1

1

real.


838

The

adding more "time-delay relays," "contacts," and so forth. In other words,

we can change

the on/off

performance of the lamps by simply punching a few

when we

keys. Quite a remarkable achievement

memory

volatile

of the

The CPU performs the following 1

.

During a brief input

31 .4 The central processing unit

components, starting

We

will

now

mode of operation with the CPU.

will briefly describe the

mil-

schedule, the on/off condition of the control de-

2). In

role

2.

CPU memory

The

CPU

runs the program of the user. During

and depending on the program and

describe

this period,

of these

the status of the inputs that have been recorded,

CPU

the

and

decides which outputs should be acti-

While

vated.

CPU

the

program

is

being carried

out, these decisions are stored in a

memory

sec-

tion reserved for this purpose.

how 3.

it

at

in-

The central processing unit is the brain of the PLC. It is a complex circuit composed of one or more microprocessors. Without going into details,

we

(The status must be checked

vices can change at any time.) 1.

examples we have shown the

the construction and

the

all

status of

consists of a central

put module, and an output module (Fig. 3

played by these components.

checks

lisecond intervals because during a production

processing unit (CPU), a programming unit, an

the five preceding

CPU

modules and records the on/off

to them.

PLC

interval, the

sequential tasks:

the external control devices that are connected

(CPU) that every

divided into

put modules, and the program of the user.

remain absolutely the same.

have seen

is

are used to record

the status of the input modules, the status of the out-

consider that the external hard-wired connections

We

PLC

them

several sections. Three of

The

CPU

transmits the

memorized decisions

to

used.

is

There are two types of memory. The

which means

volatile,

the output modules. During this interval, the outfirst is

nonput modules activate or deactivate the external de-

that its contents

cannot be vices that are connected to the output terminals.

The second is volatile, which memorized contents can be changed read-

erased or modified.

The operating cycle

means its ily. The non-volatile memory contains all the instructions needed for the management of the PLC. These instructions are used

to

(b) carrying out the user

and

vices.

It

to

program, and

(c) trans-

mitting the results to the output modules.

check the input mod-

operating cycle ules

consists of (a) taking

successive readings of the status of the inputs,

determine the status of the control de-

The

also transmits orders to the output modules.

is

The

named scanning.

sequential scanning process goes on

when the PLC is in operation. The make a complete scan depends upon the

continually Finally,

interprets

it

and executes instructions

that

time to are furnished by the keyboard, then carries out the

speed of the user program. In the

memory

section the

PLC

manufacturer

in-

10 stalls the

the

PLC.

range of functions that can be executed by In addition to relay-type

functions

—such —

as the coil function, the contact function, etc.

PLC

offers

the

hundreds of other functions. These may

be arithmetic

functions,

memory

ing parameters of the volatile

memory

and the size of the user pro-

4.

ms

to

make

rule,

it

takes from 2

During the scanning process, the atically

ms

to

a complete scan.

CPU

system-

checks the correct functioning of the

hardware by diagnostic analysis.

drum-switch functions,

timer functions, counters, and registers. In effect, the non-volatile

PLC

gram. As a general

establishes

all

PLC. The contents of the non-

are defined by the manufacturer

and cannot be erased or modified by the

31.5

Programming

unit

the operat-

user.

The programming and monitor,

is

unit,

used

to

composed of a keyboard program the PLC. But

usefulness does not end there.

It

its

also enables the

PROGRAMMABLE LOGIC CONTROLLERS bus bars

bus bars

module

module

PB1

HI—

112 02

111

715

,

11.3

c

112

715

fl03h

114

101

715

Hr-

<>

03

Hr

—(ssV

HI 907

114

-

°

01

^

113 03

13

04

111

112 02

715 RT 90 7

^

,

I3

module

01

111

output

(_)

module

PB1

715

101

12

input

output

(_)

(+)

\

/

\

/ input

837

04

nro3T

715

.

.

Hh—Q12}H'l«24 V

IC

Hl|H

113

715

+f-

oc

<

IC

24 V

120 V

OC

113V

120

V

715

Figure 31.7 The external devices connected to the input and output modules perform the same way as in Fig. 31 .6, but the CPU circuit is programmed in a different way. In this case, an internal auxiliary "relay" 715 has been added.

Figure 31.8 This

CPU

ladder circuit uses an internal "time-delay

relay" labeled

as are

RT907. The time delay

is

programmed,

the other elements between the (+) and (-)

all

"bus bars."

Example 31-4 In Fig. 3

the

1

.7,

lamps

PLC

will

the operation of pushbutton

PB and of 1

same as in Example 31-3, but now make use of an "internal relay." the

is

the

Example 31-5 In Fig.

31.8, the

Example 3 after the

1

-4,

same

control

is

required as

in

but lamp LI must light up 5 seconds

pushbutton

is

depressed.

Solution 1

.

Among 750

the

50

internal "relays" labeled 701 to

that the operator

memory

can select from the

CPU

Solution 1

.

"contacts,"

all

bearing the number 715.

Two

case the operator must add an internal .

715. In addition, he selects three associated

these contacts are normally open and one

In this

"time-delay relay" to activate lamp LI The re-

bank, he decides to choose "relay"

mainder of the

circuit stays the

same.

of 2.

The operator

is

from the

normally closed.

selects "time-delay relay"

CPU memory

bank.

He

RT 907

then adds nor-

mally-open "contact" 907 and makes the 2.

CPU

The operator then programs the tions" as shown in Fig. 3 .7.

"connec-

When PB1

is

depressed, "coil"

101

is

excited,

101. This excites "coil" 7 5, 1

When PB1

is

CPU

in Fig. 31.8. Finally,

programs the time delay of

1

which closes "contact"

"connections" as shown

RT 907

at 5

he

seconds.

depressed, "contact" 101 closes, which

excites the "coil" of relay 715.

The two normally-

causing "coils" 111 and 112 to become "excited".

open "contacts" 715 immediately close and nor-

As

mally-closed "contact" 715 opens. Consequently,

a result, contacts

1

1

1

and

ing lamps LI and L2. At the is

"deactivated,"

1

12 close, thus light-

same

time, "coil"

causing contact

113

to

1

13

open,

which thereby extinguishes L3.

The

external connections to the input

module and output module are the same.

in Figs.

lights

up

at

once and lamp L3 goes

RT907

is

only closes 5 seconds

We again note that this new circuit involves only the computer.

lamp L2

virtual "coil"

31.6 and 31.7

excited but

later.

its

out.

The

"contact" 907

Thus, lamp LI comes on

after a delay of 5 seconds.

be made in a minute. we could make the lamps any way we please by simply

These changes can Indeed,

it

is

clear that

blink on and off in

all


839

user to observe the status of the input and output to make changes to certain parameThe programming unit is also used as a tool to check and diagnose the PLC, Finally, it is used to

modules and

ters.

save the programs on tape or disc, and to retrieve these programs from the

same

supports.

However,

although the programming unit plays several roles,

PLC

needed when the

not

is

it

is

operating.

Consequently, the programming unit can be disconnected once the industrial process

is

in operation.

The programming unit can take the form of small hand-held unit composed of a keyboard and monitor (Figs. 31.9, 31.10).

tiny

It

a

a

can also be a

computer with a large screen and keyboard having special keys. use, the

Because

programming

it

is

unit

designed for industrial

must be portable and

ro-

bust.

Using the proper software, a personal com-

puter

is

31

.6

often used as a

The

I/O

As mentioned

PLC programming

modules

previously,

modules (designated by the interfaces

and the central processing unit

(CPU). The interface function the

CPU

input and output

the

the abbreviation I/O) are

between the external control and con-

devices,

trolled

crucial. In effect,

is

accepts and emits only low-voltage and

low-power dc signals

The

(0 to 5 V).

and could be damaged

sensitive

unit.

if

CPU

exposed

is

to sig-

nals exceeding this voltage range. Thus, all

munication links between the

very

com-

CPU and the external

devices must be done via the I/O modules.

Each input module and each output module can be connected to several devices. the

We

number of "points" of entry and

I/O modules possess 4,

16, or

8,

then speak of

exit. Individual

32 points; some

Figure 31.9 This industrial

PLC has been adapted

for

educational

purposes. The operator punches the keys of the handheld

programming

unit

and tracks the

effect

on the

The programming unit interacts with the the PLC that are mounted on the upright

small screen.

two parts of panel. The upper part contains the central processing unit (CPU), the power supply, and the input and output (I/O) modules.

The lower

part

is

simply an exten-

sion of the upper part, offering additional I/O modules.

This

PLC has

10 input points and 6 output points.

(Courtesy of Lab-Volt).

have as many as 96 points. The 16-point modules are the

most common. In the case of a large auto-

mated production process, several I/O modules may

section,

and a communication board. Apart from the

communication board,

configuration

this

is

re-

be required. peated for each entry point.

31.7 Structure of the input The

parts

module.

It

shown

in Fig. 3

1

.

1

1

modules

comprise the input

consists of a terminal board, a filter and

conversion section, a status indicator, an isolation

In order to prevent spurious signals

ing the

PLC,

from

activat-

the filter and conversion section sup-

presses electrical noise such as

may be caused by

in-

duced voltages or contact bounce. The conversion portion reduces the

input

voltage that appears

840


filter + conversion |

aLLEN-BRAIHjEY

ft

\

/

/

\

electric isolation i

status indicator

to

i

CPU

i

i

i

filter

4PTFJ-

<

i

>

it

*CTU>

u K *so^

\

/

/

\

+

conversion

electric isolation

Figure 31.11 Structure of the input module.

nected to the terminal board.

The from

isolation

is

and random voltage spikes. The

achieved by means of an optical coupler

which converts the incoming electric signal signal.

C—M

K

K

an

IT HCT0>

>r

The

light signal, in turn, is

electrical signal

On

*~

M

CPU

electric isolation section protects the

electric noise

into a light

converted back into

by means of a phototransistor.

the input side, the optical coupler can with-

stand peaks of up to the transmission

1

500

V.

Thus, while

it

permits

of signals, the optical coupler

completely isolates the sensitive

CPU

circuits

from

those connected to the input terminals.

As

regards the communication section,

ters all the status

CPU.

transmits them to the

CONSOLE OE PROGRAMMATION

regis-

it

conditions of the input circuits and

The impedance of

the input

module

is

one of

its

important properties. Depending upon the voltage

Figure 31.10

shows the keyboard

used for programming. The small screen above the keyboard enables the operator to see the "contacts," "relay coils," "time delays," etc. as they are programmed. These elements all bear a reference number or address. The programming unit can also be used to check the status of the external input and output devices. Consequently, it is a useful tool for both programming and diagnosing the control system. (

Co u rtesy of La b

which the module was designed, the input im-

for

This portable programming unit

-

Vol t)

pedance of each point

The 10

from 5 kfl

to

mA.

1

2 kf 1.

is

about

This low current permits a reduction

in the

size of the external control devices, as well as the size

of the cables linking them to the PLC.

Note

must furnish the power supply

that the user

for the external control devices. Several voltages

may be The

.

will range

current required to activate a circuit

used: 24

filter

V

to

120

V

ac or 10

V

to

100

V

dc.

and conversion section of the input mod-

ule drops these voltages to a level that

is

compati-

ble with the optical coupler.

across the input terminals and,

if

necessary, recti-

fies ac signals.

31.8 Structure of the output

The status indicator is a light emitting diode (LED) that is on or off depending on the signal re-

The output modules

ceived

displayed

at

each

external

terminal.

It

facilitates

checking the operation of the control devices con-

CPU,

modules

are built using the architecture

in Fig. 31.12.

Moving outward from

the constituent parts are the (a)

the

com mimic a-


Fuses are also recommended to protect the

electric isolation

c o "5

equipment controlled by the PLC. Their main pur-

i

pose

o

from

*E

CPU

is

to protect the wiring

and components

in the

i

3 E E o u

event of a catastrophic short

i

i

Voltage spikes can electric isolation

To eliminate

circuit.

damage

the problem,

it

the output modules.

is

recommended

Figure 31.12

external device tends to create transitory overvolt-

Structure of the output module.

ages across the output module.

PLCs

31.9 Modular construction of tion section, (b) electric isolation, (c) status indica-

and

(d)

that

whenever an

voltage-limiting devices be installed

tor,

84

power

circuit.

Except for the communi-

cation section, the construction

the

is

same

One

of the important features of a

lar construction.

for each

are

all

Thus, the

mounted

PLC

CPU and the

is its

individual slots (Fig. 3

in

modu-

I/O modules 1.

1

3a).

output terminal.

The communication section receives the commands transmitted by the CPU and memorizes them until it receives new commands. In effect, the CPU

Modularity offers a big advantage because ule

is

presumed

utes, the is

not constantly in touch with the output module.

The

CPU

communicates with the output module

module

fashion,

ms

successive scans varies from 10

ms 100

to 2

(

the

original

matter of min-

that contains the

memory

when

isolation sections

play the

program (previously saved) has

to be re-

programming unit. This operation, which takes only a few minutes, is equivalent to re-

to

same

and the status indicators

By

roles as in the input modules.

tical

coupling, they protect the sensitive

cuits

from voltage surges. The

LED

CPU

Another advantage of modularity op-

expand the capacity of the

The power

needs of the

to

is

the ability to

meet the growing

from the

user.

Thus, I/O modules can be added as

and when required. The memory capacity of the the only thing that limits the

1

so as to operate the devices connected to the seen, each output

each point point acts like a switch or contact (see Fig. 3 closes or opens the circuit that

is

A

close the circuit.

The user furnishes

is

1

often used to open and the

power sup-

Although the output modules are intended

to

control industrial devices, their current-carrying cais

limited.

Most of them can carry a maxi-

A

to

2

A

per output

terminal. If an external device requires a larger cur-

an auxiliary relay must be used. For example,

in Fig. 31. 14, auxiliary relay

vate the holding coil

A of a

B

is

31.10

We

employed

large contactor.

is

controlled on/off by a small relay.

to acti-

Remote inputs and outputs

have just seen

that users

can tailor the

PLC

to

The modular nature of the PLC also enables the installation of I/O modules at remote locations, far removed from the CPU itself. We then speak of remote I/O modules. Such modmeet

ply to drive the external devices.

current ranging from 0.5

1

.3). It

connected to the

external device.

triac

.

1

As we have

output terminals.

PLC

number of I/O modules that can be accommodated. Figure 3 3b shows the construction of a 6-point output module in which is

circuit amplifies the signal

PLC

cir-

indicators help

check the status of the controlled devices.

pacity

replaced

of the PLC. The

placing an entire relay rack.

The

rent,

can immediately

in a

that requires special attention

one

mod-

trieved using the

500 communications per second).

mum

it

a

can be up and running again. The only

according to the previously

mentioned scanning process. The time between two

CPU

PLC

in is

sequential

to be defective,

be replaced by another one. Thus,

if

their needs.

ules are placed close to the particular

automated

process or production line that has to be controlled.

These modules can be located as near as 3 meters to as far as 3

km

equipped with

a

from the CPU. Each I/O module is power supply and a communications

842


12

Figure 31.13a Modular PLC for the

3

456789

distributed control of I/O devices of

10

11

any type, including position control and the control

of in-

The PLC rack shown contains ten modules of various kinds. Racks having capacities to hold from two to ten modules are available. Seven such racks connected locally have a potential capacity of more than 5000 local I/O points (modules 1-4 above). However, most I/O modules are now installed close to the transducers and actuators and tied to the PLC by communication networks whose protocols are either proprietary or standarddustrial processes.

ized by industrial manufacturers' associations (module

5).

The remote modules can themselves be micro-PLCs. In general, rather than using a single PLC, several PLCs are installed in a factory, each controlling a machine or a section. The PLCs are often interconnected by means of a factory communication network (modules 6-7). Finally, the factory network can also be connected to the informanetwork of the business enterprise, making use of the embedded gateway capability of the PLC (module 8).

tion

Description of modules: 1

.

2. 3. 4. 5.

6.

7. 8.

9.

10. 1 1

.

12.

module with screw terminals module with screw terminals (see internal construction in Figure 31.13b) 96-point I/O module including pre-wired cable for external input terminal blocks and output modules 8-point configurable analogue input module DeviceNet® network module (mainly for the control of remote I/O modules, contactors, variable speed drives, etc.) Controller Link® network module using fiber optics (mainly for supervision and exchange of data at high speed) Controller Link® network module using twisted-pair cable. Network module for communication by Ethernet® protocol (mainly for supervision, data acquisition, and management of information) Optional CPU inner board with two programmable serial communication ports Central processing unit (CPU) of the PLC (two configurable serial ports can be seen on the front) Power supply for the PLC Optional memory card to save local files concerning the project. This may be the project development file itself, electrical wiring diagrams, or to hold rapid and intensive local data acquisition by the PLC. 16-point input

16-point output

{Courtesy

ofOMRON Canada Inc.)


843

Figure 31.13b the interior of a 16-point relay-type output module. The relays are mounted on plug-in bases be replaced by means of an extractor that is included in the module. On account of their robustness, versatility, and low cost, output modules with relays are often used instead of semi-conductor devices.

This picture

and can

shows

readily

However, a relay

is

an electromechanical device whose

useful

life is

limited

by the

electrical load

operations. Typical load capacity is*2 amperes. Note that the terminal board can be not have to be

{Courtesy

undone when the modufe'is removed from the PLC

ofOMRON Canada

Inc.)

rack.

lifted easily,

and the number

of

so the wiring does


844

Figure 31.13c to a PLC, such as shown in module 3 of Figure The connection is made by means of pre-wired cables of different lengths (up to a maximum of ten meters). The relay module can also be connected further away by adding a communication module of the desired choice. External output modules allow for a wider choice of output types. Also, modules having greater capacities

This "remote" I/O module can be used locally by connecting

it

31 .13a.

can be

installed, thus facilitating direct interfacing of the controlled

{Courtesy

A twisted

module. link,

3

1

.

1

devices connected to the output.

ofOMRON Canada Inc.) or coaxial cable, or a fiber-optic

connects the remote station to the

CPU. Figure

controls the start/run/stop operation of a motor.

Figure 31.14 uses a ations.

3c shows a remote output module.

and PLC

or

perform the same oper-

input to the input

module

of a relay. The "coir has one

like the coil

more "contacts," located

NO

spective

circuits

to

1

behaves

31.11 Conventional control circuits

PLC

We recall that each

or

NC

in the

conditions are

CPU, whose reprogrammed by

the user. It is

now

evident that

we can

use a

PLC

a conventional relay control circuit.

examples show how the change

is

in

Note

place of

The following

made.

that "contact" 102 associated with the stop

pushbutton

(NO). In

programmed

is

effect, since this

nected to terminal

Example 31-6

which causes

Consider the conventional control circuit of Fig.

be closed. This

20.16b (reproduced here for convenience) which

ton

is

the is

to

NC

102, "coil"

NO

be normally open

pushbutton 102

"contact" 102

is

is

con-

"excited,"

in the

CPU

to

true as long as the stop pushbut-

not depressed.

PROGRA MMA BLE LOGIC CONTROLLERS

contactor

disconnecting switch 1

0

i

j—oN>

c^-TT

thermal overload relay

M a

4

1

845

in

r

—c^^—Qr——c>— — 1

IT

1

control station

T

Figure 20.16b (reproduction) Schematic diagram of a 3-phase across-the-line magnetic

input (

—

reset

starter.

+)

(_)

output

module

module

start

pushbutton

102

101

i—

"l

Hh 103 103

112

1

11

V

CPU Figure 31.14 This

PLC

in Fig.

We added

virtual control

system produces the same machine performance as the hardware control system shown

20.16b.

also note that an auxiliary relay to the output

B had

module. The reason

is

to be

that ter-

that ''contact"

As

minal 01 cannot furnish the relatively large current

cited.

needed

cited."

A.

to excite to the holding coil of big contactor

By using

the small relay B,

whose contacts

are

robust enough to carry the exciting current of coil

A,

we

get around the problem.

Let us

now examine

contacts and coils of the start

pushbutton, "coil" 102

is

It

Prior to pushing the

excited which

means

rs

The "contacts"

closed.

a result, "coils"

module

put

A is

1

101 and

and 112 are not "ex-

1

1

1

1

and

1

12 in the out-

are open. Thus, auxiliary coil

excited and the pilot

coil

1

follows that contacts

a result, contact

the behavior of the virtual

PLC.

102

103 are open because "coils" 101 and 103 are not ex-

B

is

lamp connected

to

02

B

not

As

open, which means that holding

not excited and so the motor will not

Let us push the

is

is off.

start

and see what happens.

start.

pushbutton momentarily


846

input

output

CPU

module

start

module

control processing unit

101

102

104

ii

ii

1

ii

i* *i

01

111

(-)

(+)

111

112

02

.112 i> *i

(

-o

f

B

v

103

HI 102

Bx1

1Q4

103

111

II

#

W

OC

104

15

—o-

—

101

——

105

—w—iHf—

14

—o-

120 V

105

—

1 1

1 I

1

\r IC

I

24 V

Figure 31.15

PLC control system for starting and plugging a motor possesses same effect as the all-hardware system shown in Fig. 20.24a.

This

1

1.

"Coir 101

is

"excited," which closes "con-

same

security features

and produces the

which deenergizes coil B, thus deenergizing A, which stops the motor. It is readily

tact" 101. 2.

the

coil

Because "contact" 102

seen that the opening of is

already closed, "coil"

Ax

will extinguish

the pilot light. 1

"excited," which closes contact 111, thus

is

1

1

Note

exciting relay coil B. 3.

The

NO contact

B

closes,

which excites con-

that the wires

nals can be

As soon contact

as contactor

Ax

closes,

A closes,

the

NO

auxiliary

The simple

thus causing both "contacts" 103 to close.

A is

energized.

As

a result, contact

closes, causing the pilot 5.

When

the pressure

leased,

it

lamp

on the

1

12, 13,

,

1

is

lighter than in conventional relay

V power

at the

supply has

most.

control circuit of Fig. 20.16b

not justify using a

PLC. However,

it

would

illustrates the

principles that are involved.

12

to light up.

start

pushbutton

Example 31-7 is re-

causes "contact" 101 to open, but

since "contact" 103

much

control circuits. Also, the 24

only to deliver a few watts

which excites "coil" 103,

These contacts remain closed as long as contactor

1

Consequently, the cable connected to these termi-

tactor coil A, thus starting the motor. 4.

connected to terminals

and IC carry a current of only a few milliamperes.

closed, the

motor contin-

We

want

to use a

control system

PLC

shown

to replace the conventional

20.24a, displayed at

in Fig.

The resulting PLC cirand "ladder diagram" are shown in Fig. 31.15.

the beginning of this chapter.

ues to run. cuit 6.

When

the stop pushbutton

"coil" 102

is

depressed,

It

no longer excited, thus causing

"contact" 102 to open. is

is

As

a result, "coil"

deenergized, which opens contact 111,

1

1

(

is

—

)

called a ladder diagram because the

"bus bars" look

like the sidepieces

while the horizontal circuits the "contacts"

1, 2, 3, 4,

and "coils" look

(

+

)

and

of a ladder containing

like rungs.

PROGRAMMABLE LOGIC CONTROLLERS The

NC

NO contacts of the start pushbutton and the

associated with a device that serves to stop an

and

I]

12 of the

NC type.

action must he of the

contacts of the stop pushbutton are connected

to terminals

PLC. The cumbersome observed, a break

If this rule is not

mechanical interlock (for security purposes) of the

two pushbuttons of

NC

the

Fig.

20.24a

replaced by

NC

,

module. The former

NC

A

and B are respec-

T3

and T4 of the input

auxiliary contacts

(A X2

and B x2 ) are no longer required. The NO "contacts'" 103 and 104 of rungs 2 and 4 are the holding con-

The

tacts.

NC

"contacts" 104 and 103 of rungs

1

and 3 constitute a security interlock.

As 1

1

a further measure of security,

2 and

1

CPU

the

1

of rungs

1

ger of a break

Bx

. |

1

been realized

NC

"contacts"

the potential dan-

connection of contacts

in the

obvious

It is

that considerable

A XI

and

economies have

number of external devices con-

in the

it

cable con-

could produce

impossible to stop

an automated process.

31.13 Programming the

PLC

In order to

program a PLC, we must "write" the op-

erations

has to perform. These instructions are

it

typed on the programming unit keyboard, observed

on the monitor, and stored

the

in

CPU

memory. From

was devoted method of programming. The technical criteria

the very beginning, particular attention

and 3 were programmed into

memory. This eliminates

unexpected start-ups or make

in a

"con-

NO auxiliary contacts

The

and B x ,) of contactors tively connected to terminals

(A x

now

is

PLC

necting an input device to the

"contact" 102 of rung 3 and the

tact" 101, also of rung 3.

847

to the

stipulated that the system should be quickly

and eas-

programmable and reprogrammable by

the user.

ily

The PLC was

therefore carefully designed to

simple to use. However,

computer knowledge

to

is

it

make it some

useful to have

program a PLC.

nected to the I/O modules. The start and stop pushbuttons

now have only one contact and A and B have each only one namely A x and B X1

contactors contact,

the

main

The term programming language

.

|

The programming of ladder diagrams such as that of Fig. 31.15 takes considerable knowledge of logic circuits.

We

31.14 Programming languages

auxiliary

leave the analysis of the behavior

symbols

list

of

to

be

configured to program the PLC. The three principal

languages are: logic,

of this circuit to the reader.

refers to the

used and the way they have

that are

and

(

I

)

the ladder diagram, (2)

(3) the Sequential

Boolean

Flow Chart (SFC). The

European equivalent of the SFC language

is

called

Grafcet. Other languages have been developed in

31.12 Security rule

recent years, so that a total of about six options are

By

using a PLC,

it is

possible to invert the status of

the external contacts connected to the input ule.

Thus, a

NO contact

real

connected

to the input

module can be programmed by the user as an NC "contact" in the PLC's CPU (see Examples 31and 3

1

-2, section

state of a contact

3

1

.3).

This freedom to reverse the

must be used with discretion, par-

ticularly as regards the selection of the type of contact

(NO

or

NC)

put module.

of the devices connected to the in-

The following

rule

must always be

observed:

The ladder diagram

Among the several programming used, the ladder diagram

saying so explicitly, Figs.

31. 6,

31.7,

with a device that

serves to initiate an action of some kind must

NO

type. Conversely,

any "contact"

is

languages that are

the simplest. Without

we used

ladder diagrams

shows yet another ladder diagram. In programming a ladder diagram from board, the desired circuit

moved key

is

is

to create a

the key-

progressively displayed

this process, the cursor is

to the desired place

sired function

in

and 31.14. Figure 31.15

31.8,

on the monitor. During

Any "contact" associated be of the

available, at the discretion of the user.

mod-

on the screen and the de-

selected by pushing the appropriate

NO

or

NC

"contact," an "internal


848

and so

relay coil," an "output relay coil,"

When

it

is

number. The same procedure

is

followed for the

is

necting them. This

forth.

re-

risky because connection

is

errors can easily be

given a reference

chosen,

the function

made.

of the control circuit can be displayed on the screen.

The flexibility of PLCs is extraordinary. Thus, whenever a given control system is no longer required, it can readily be reprogrammed for a completely different system. With relay racks,

However, most PLCs

program-

such a changeover

a

would simply be scrapped, replaced, and

mainder of the control

•

circuit.

During the programming phase, only a portion that use the ladder

ming language can be connected

to

printer.

Consequently, the entire control circuit can be displayed, which enables the

programmer

diagram and the behavior of the system. Figure

gram displayed on

menu and

the monitor of a

ladder diaing a

was invented

One of the drawbacks one has

in

ladder diagram,

it

is

to is

in the

•

1

3

m

CPU

hav-

replaces hundreds of

of this language

reading is

it.

is

is

translate

it

in

more

is

and have

Thus, starting from a

hard to read a program

it

PLC consumes

coils.

far less energy.

reliable than a relay cabinet. is

moving

the absence of

Relay contacts wear out

older.

be replaced,

to

all

of which requires a

sustained maintenance program. ''Relay coils"

•

CPUs

never wear out.

and "contacts"

in

In addition, the

opening and closing of relay

contacts, while rapid, takes a certain time.

into the cor-

responding ladder diagram.

time interval

moreover,

The SFC language The sequential flow chart (SFC) language is a very effective tool in diagnosing PLCs and automated systems in general. It is a PLC programming method that enables the user to organize the indi-

it

is

not the

same

may change

for

all

with time. In

some

plications

where the opening and closing

quence

important, the time variations

is

The

relays and,

ap-

se-

may

in-

troduce control errors. Such errors are very difficult to

diagnose because of their random na-

ture. In the

case of PLCs, the "contact" opening

vidual machine operations of a process into a series

and closing times are

of steps and transitions. Ladder logic can then be

quence operations are never a problem.

used to implement the program.

One

parts.

Relays have moving parts that deteriorate as the

the dif-

dure

A PLC

equipment gets

easy to write a program using

Boolean language and then

connect the contacts and holding

important reason

middle of the 19th

However, the reverse proce-

difficult;

bulky than a conventional

less

0.

based on

used to solve prob-

the Boolean language. is

volume of

Furthermore, the

century by the British mathematician George Boole.

ficulty

much

is

control relays, as well as the hard wiring needed

Boolean algebra. This algebra It

The PLC

relay control system. For example, a

PLC.

Boolean language The Boolean programming language lems of logic.

not feasible and the racks

rewired.

to verify the •

31.16 shows a portion of the

is

•

The

fixed. Consequently, se-

relay cabinet has to be

assembled by hand.

Hundreds and even thousands of wires must be

31.15 Advantages of relay cabinets

PLCs over

which implies a big chance of making

There are many reasons for the universal popularity of PLCs. •

We

The PLC ble,

it

is

is

list

them as follows.

flexible.

Because

connected between the contacts and relay

it

is

programma-

easy to modify as the need arises. In the

These errors are with a PLC,

By

difficult to locate.

coils,

errors.

contrast,

draw a ladder diagram according to a plan. Here again, if an error is made, the hand-held programming unit (or the more sophisticated computer) conall that is

needed

case of control systems using physical relays,

tains utility functions that

any change means replacing relays and recon-

rect a mistake.

is

make

to

it

easy to cor-

PROGRAMMABLE LOGIC CONTROLLERS ™ £P

Ploi.tep

c*lh-c*1m[i221 V»ev/

File

V2

-Iptn

Insert

PLC

-

849

CX-Pjogra

Pl«r5*

joo!i

wydow Hrip_

Figure 31.16 Software programs

for

PLCs

are evolving rapidly. Small, dedicated programming consoles that permit only a very

limited interface with the user are

quick, minor modifications

in

the

now almost

only employed because of their portability and their

ability to

make

field.

is the portable computer whose software is based upon Microsoft Windows®. programs permit the development and configuration of the PLC, but they also enable the

But the most frequently used tool Evidently, these software

adequate documentation of the programs. Portable computers also enable an operator to communicate with more than one PLC at a time. Furthermore, changes to the program can be made without having to stop a machine or interrupt a process controlled by the PLC. They also constitute a powerful tool to diagnose or debug a problem. The reason is that the operator can visualize the situation in the form of chronograms and other graphic displays. Ladder diagrams can be "frozen" when specific conditions have been identified and registered in real time by the PLC. The most practical and frequently used programming language is still the ladder logic as displayed in this figure. However, more and more PLCs offer complementary programming tools such as sequential flow charts, functional block diagrams, literal (Boolean) language, and high-level languages such as Basic or C. In addition specialized software using dedicated languages can be tied in with the main development software. They are used for more specific tasks like positioning applications, process-manufacturing controls, or other tasks by integrated co-proces-

sors on the PLC. Finally other auxiliary software programs are used to perform simulations, supervision, data acquisition, control,

(Courtesy

and management.

ofOMRON Canada

Inc.)

•

•


850

•

As

regards cost, a

PLC

tional relay rack as

is

cheaper than a conven-

exceeds about ten or

fifteen.

The economy

is

obvi-

ously greater for more elaborate control systems.

Having

PLCs

said this,

are not perfect. Their

mode

lems. For example, the order in which the program is

from the

written can sometimes influence the behavior of

fact that

to another. This

ferent type of

programming manual

Thus, every time a it

is

ensure the proper opera-

to

programming

may

dif-

important to read the

tion of the various functions.

not big, but they

anomaly stems

manufacturers have not yet estab-

common standards. PLC is used,

lished

of operation sometimes causes sequencing prob-

program may not be the

Finally, the effect of a

same from one PLC

soon as the number of relays

The differences are when

require slight changes

the control circuit.

the controlled system.

MODERNIZATION OF AN INDUSTRY 31.16 Industrial application of

PLCs

year to complete.

It

required the input from con-

sulting engineering firms, visits to stevedoring op-

We

are

aware

all

gies.

have been greatly

that industries

transformed by the

new computer-based

technolo-

How does the transition from the old to the new

take place?

To answer

this question, let us

consider

a large stevedoring enterprise that modernized the

handling of one of its bulk materials by installing sophisticated

equipment and controls. During the

changeover, the

new equipment had

integrated with the old because

to be gradually

was unthinkable

it

to

make the changeover in one giant step. The stevedoring operation consists of unloading freighters carrying bulk products such as rock salt,

cement, pig

iron,

powdered

nickel, bauxite,

and alu-

mina. The modernization involved the handling of

alumina, a powder used to

chemical composition AI 2 0 3

.

some 700

1

given by the formula

tons per hour of alumina are sucked out

moved by conveyor

7) to a waiting train,

into cylindrical-shaped

loaded

Its

During a typical stevedoring operation,

of the ship's hold, 31.

is

make aluminum.

in a

belts (Fig.

and blown through ports

wagons.

A typical wagon

is

matter of 10 minutes.

erations that had already been modernized, cost calculations,

When management decided to modernize the alumina stevedoring procedure, a small team of experts was given the task of determining what methods should be used and how the entire process could be automated. This in-depth study took more than a

how

the modernization

the existing labor force.

After the study was made, the gradual transfor-

mation of the stevedoring operation began to take place. This involved the installation of

new

equip-

ment, driven by more than one hundred motors ranging in

power from

fractional size to

1

500 hp. The op-

had to be coordinated. But

eration of these motors

since the motors were scattered over a wide area, and

because they had to be monitored from a central control station,

communication of their

status

was of the

essence. Thus, the current, speed, phase unbalance,

temperature of the windings, temperature of the bearings, vibration, etc.,

uous

had

to

Furthermore,

basis.

be monitored on a contin-

if

anything ran out of

line,

an alarm had to be sounded to have the problem fixed.

For example,

tors (Fig. 3

1

.

1

8)

one of the 500-hp blower mo-

if

began

to vibrate, the situation

be corrected quickly because

damage

had

to

to the bearings

or to the driven equipment could result.

Motorized valves demanded special position controls.

31-17 Planning the change

and research on

would impact

Thus, as the alumina was rushing through 500-

mm pipes, the flow had to be kept at the desired level to ensure that

it

was

neither too high nor too low.

Clearly, such a complete control over the industrial

process could only be achieved by computers.

But during the transition, ment, together with

its

much

of the older equip-

relays, hard wiring, limit

switches, pushbuttons, and so forth, had to function


85

Figure 31.17

One

of the

many

bulk products, {

intersections of two conveyor belts that are

and

to load

and unload

more than Quebec.

1

km

The conveyors are used

long.

to store

freighters in the Port of

Courtesy of St. Lawrence Stevedoring)

as before. Consequently, the stevedoring operation

involved a progressive marriage of the old,

new with

the

wherein computers controlled one section

while

human

operators continued to control an-

other. Furthermore, the older hard wiring

integrated (temporarily) with

new

had

to

be

meant

that

many manual jobs would no

it

became necessary

to ensure that jobs that

would be replaced by tasks ing, less routine,

and more

that

were

were more

lost

satisfy-

interesting.

coaxial cable and

fiber-optic links. This required the installation of

31,18 Getting to

appropriate interfacing devices. All these alterations had to be made, not in a laboratory, but in the field,

longer be re-

quired. Because labor relations are always important,

where huge gantry cranes

How

was

the

know PLCs

new equipment

to look after it?

It is

installed

and who was

interesting to note that techni-

and powerful motors could easily wreck equipment

cians that had been doing their regular jobs for years

something went wrong. Thus, another important

new technology. The company representatives that were supplying the PLCs and fiber-optic devices enabled the older technicians to understand the new

if

aspect of the transition was the proper operation of the apparatus and the safety of personnel.

Another

vital

consideration was the impact of this

new technology on employment of ization

the working habits and continued

the workers. In effect, the

modern-

and automation of the stevedoring operation

were quickly able

to

absorb the

verbal exchange between

language of ladder logic diagrams, Ethernet®, net,

megabits per second,

etc.

And

so

it

inter-

became

evi-

dent that not only were the equipment and controls

852


Figure 31.18 to siphon alumina powder from freighter to special train wagons. The 3-phase motors have a rating 500 hp, 4160 V, 3600 r/min. The picture shows (1) a position-controlled pneumatic valve, (2) a thermal detector sense the temperature of the motor windings, and (3) a vibration detector.

Blower motors of to

(Courtesy of St. Lawrence Stevedoring)

became

became

gradually being changed, but that the same process

the lingo

human level. Valuable knowledge acquired over many years was gradually being enhanced by new information. Some individuals welcomed the challenge of learning something new, but others were more ap-

the

come part of the high-tech team. He now knows how to handle a keyboard, he understands what the computer screen reveals, he knows what to do if an

prehensive about their ability to cope with these

alarm sounds, and he can quickly communicate his

new-fangled concepts. But time and daily contact

findings to others (Fig. 31.19).

was going on

with son's

at the

new equipment inevitably transform a perway of thinking. As the transition took place,

familiar, the tasks

computers became

less intimidating,

routine,

and before

long the gray-haired veteran realized he had be-

The newly-hired technician

also had the advan-

tage of working closely with these experienced vet-


853

Figure 31.19 Strategic control

and surveillance center

networks, cameras, and status of

PLCs

of

all

bulk products handled

in

the Port of Quebec.

All

the communication

are centralized at this point.

{Courtesy of St. Lawrence Stevedoring)

erans.

As

the old

a result, the technical-human link

and the new

is

between

considerably reinforced.

staggering the use of large motors so they do not run at the

same

time.

Such peak-load control can save

thousands of dollars per month.

31.19 Linking the The

entire

PLCs

stevedoring facility

is

controlled by

about fifteen PLCs, two of which are assigned to the

alumina-handling operation.

Some

peak power

conjunction with manufacturers' representatives.

exam-

They programmed the PLCs using languages such as SFC and, more recently, block diagrams. For the

specific operations. Other

PLCs perform

to

that

PLCs. One reason is

drawn by

ple, the cost

is

order to coordinate isolated

be integrated with the sys-

tem. However, plans are under the

w#y

Naturally, the entire industrial process had to be

to centralize all

in

do not have

PLCs

programmed on computers. Towards this end, programming specialists were hired to implement the new control techniques and to incorporate them in the PLC memory blocks. This became the task of new graduates from technical institutes working in

are linked by a

communications network tasks that

31.20 Programming the

to control the

the stevedoring facility. For

of electric power can be reduced by


854

Figure 31.20 This 600-V, 3-phase motor control center is equipped with PowerLogic® devices that provide sophisticated monitoring capabilities such as active and reactive power, voltage, current, and waveform analysis. It includes Internet transmission capabilities and possesses embedded Web pages.

Figure 31.21 This motor starter with

it

motor against mechanical overloads, overand current unbalance, and phase

failure. In addition,

mits

all

more (

ground

monitors the power

it

faults,

and

this information to

protocol. latter are often

has a

In addition,

heating, voltage

make experienced programmer, the

on/off/jog pilot lights

includes a multifunction protection relay that pro-

tects the

ing time,

(Courtesy of Schneider Electric)

its

3-phase contactor and disconnect switch.

Copper

factor, start-

starting current.

PLC

a

using a

or fiber-optic cable

It

trans-

Modbus®

may be used

to

the connection.

Co u rtesy of Sch n e ide r

Ele ctric)

convenient to use than ladder diagrams.

up a new control sequence to handle, movement of a crane, the programmer must

In setting say, the

incorporate fail-safe features by simulating the operation of the crane to see

liminary simulation phase it

if is

any bugs

The

pre-

rupt a

As

arise.

integrity of the system. Finally,

when

put into actual operation for the

first

group watches closely

pushbuttons are spotted

very important because

enables the programmer to verify the safety and

nical

behaving as planned. Towards

a

program

is

time, the tech-

to see that everything is

3

1

wrongly-programmed regards

the

this end,

at strategic

emergency

points to inter-

activity.

motor control centers

(Fig.

.20), they contain the usual contactors to start

and

stop electric motors. However, they also include

monitoring devices that continually check the condition of the

motor

(Fig. 3

1

.2

1

).

This information

is


855

Figure 31.22 This fiber-optic junction box facilitates connections be-

tween cables. {

Courtesy of St. Lawrence Stevedoring)

relayed to the relevant

PLC

by fiber optics consist-

ing of a single strand of flexible cable (Fig. 31.22). In conclusion, the installation of this

modern

control system has significantly reduced the cost of

At the same

the alumina stevedoring operation.

time

it

has stimulated the operating personnel, and

Figures 31.23, 31.24, and 31.25 convey the It is

Quantum® PLC has module,

right: (1) I/O

At the far right

is (4)

a

three parts. From

CPU, and

(2)

Web

(3)

left

power

to

supply.

server which establishes

the 100-Mb/s Internet connection using either fiberoptic or wired cable.

has offered interesting job opportunities.

vor of this modernization process.

Figure 31.23 This

fla-

typical of

(

Co u rtesy of Sch neider

El ectric )

all

renovations and computer-modernizing procedures that are carried out

links, great efforts are

by industry today.

being

communication systems

31.21

activities.

The transparent enterprise

As an extension of

the automation process

just described, the

tendency today

atmosphere

we have

to integrate

it

with commercial transactions. In effect, the Internet

is

making volved

it

possible to

in the

is

communicate wjth everyone

in-

manufacturing, marketing, purchasing,

and end-use of products.

The end

In order to simplify these

result

made

that are is

is

in these

diverse

a "transparent" e-business

that not only facilitates

actions, but also

to standardize the

used

commercial

integrated with the actual

trans-

manu-

facturing and service activities of an enterprise.

Thus, the

PLC

is

no longer

lay-operating device;

it

has

just

an ingenious

become

re-

part of a so-

phisticated control system that reaches into every

echelon of business.

856



1

-

1

3 1-2

31-3

31-4

Name

the principal parts of a program-

mable

logic controller.

PLC, what module? In a PLC, what put module? In a

the purpose of the input

is

the purpose of the out-

In a control system,

of using a 31-5

is

A PLC

is

PLC

name

four advantages

instead of relay racks.

able to simulate relay coils,

contacts, and the connections

them. Explain it

how

this

between

simulation makes

easy to set up and modify an industrial

control system.

31-6

In a

PLC, describe

the purpose of the

central processing unit

3 1 -7

31-8

Name

(CPU).

three languages that are used in

programming a PLC. The PLCs in a large organization are often interconnected and tied to a main computer. Explain how this enables communication between the manufacturing process and the business and marketing activities

Figure 31.24 This programmable logic controller belongs to the

Momentum® family. has the usual I/O modules but embedded Web pages and possesses comIt

it

includes

munication

facilities that

enable

it

to receive

data by Internet protocol.

(Courtesy of Schneider Electric)

and send

of an enterprise.


Figure 31.25

The

four devices shown act as bridges between the Ethernet® and Modbus® protocols. The bridges enable older communication technologies to be transmitted over the Internet. Bridges are high-performance gateways. (

Co a rtesy of Schn e ide r Electric

857

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http://www.omron.com http://www.ge.com http://www.etacbe.com

R. D. 1971. Cooling towers. Scientific

American 224. No.

5:

http://www.schneiderelectric.com

70-78.

Motors, Single-Phase 74.

Zimmermann. fects of large

J.

A. 1969. Starting requirements and ef-

synchronous motors. IEEE Trans.

Gen. Appi. 1GA-5, No.

2:

Ind.

http://www.usmotors.com

and http://www.baldor.com

169-175.

http://www.pamensky.com

WEBSITES The following websites provide door

in this

Motors, Three-Phase http://www.usmotors.com

*

Topics and Websites

of topics covered

PLCs)

http://www.ab.com

437-441.

Woodson.

(fuses)

http://www.littlefuse.com

http://www.bussmann.com

38-42.

beam pumps. IEEE 5:

http://www.jenkins.com

1975. Effect of unbalanced voltage on opera-

Appi. lA-11.No. 72.

http://www.micromo.com

Motor Control

1964. High temperature insulation in ac

Westingliouse Engineer Mar. 1964: 1-5. F.

http://www.baldor.com

http://www.maxonmotor.com

26-35.

38: 387-390.

70.

http://www.usmotors.com

500 kV Substation. IEEE

Installation at Siatt

reactor. Scientific

http://www.magnetsales.com

Motors and Generators

Thyristor Controlled Series Compensation-Prototype

68.

Insulators

1978. Uranium: key to the nuclear fuel

P.

Power

Summers, Scientific

67.

cells

http://www.fetc.doe.gov

with "subharmonic" control in conjunction with re-

65.

Fuel

1468-1496. July 1995.

useful information on a variety

book. The websites usually open the

to other topics, in addition to the

ones

listed here.

http://www.baldor.com http://www.reliance.com

http://www.pamensky.com

864

REFERENCES

Power Electronics

Educational and General

http://www.irf.com

http://www.asee..oj"g

http://www.ixys.com

http://www.ieee.org

http://www.toshiba.com

http://www.abet.org

http://www.siliconix.com

http://www.asme.org

J

http://www.pwrx.com

http://www.astm.org

Power Systems

http://www.asm-intl.org/

http://www.tva.com

http://standards.ieee.org/catalog/sol

http://www.eei.org

http://www.abb.com/abbreview

http://www.ieee.org/power

http://www.eIectricityforum.com

http://www.ge.com

http://www.labvolt.com

http://www.hydroquebec.com

http://www.epri.com

http://www.alstom.com

http://www.prenhall.com

http://www.edt.com

Special information on using the internet

www.eia.doe.gov

http://bobalden.com

Resistors

http://www.ohmite.com

Bob Alden design.

http://www.bourns.com

uses his website for teaching three university-level

courses and consulting

He

in

both power engineering and

also writes a monthly

Information Highway with

column "Travelling

Bob Alden" which appears

http://www.vishay.com

IEEE's newspaper The INSTITUTE

Transformers

media.

in

web

the in

both print and on-line

http://www.ustransforiner.com

http://www.advancedcomponents.com http://www.ge.com

For further information on Electrical Machines, Drives, and Power Systems and other books by Theodore Wildi please contact:

http://www.alphatransfornier.com

http://www.wildi-theo.com

Variable Speed Drives

http://www.tbwoods.com http://www.rockwell.com

http://www.abb.com

US

Patents and Trademarks

http://patents.uspto.gov/

e-mail

[email protected]

We welcome your questions and

comments.

Appendix AXO

Conversion Charts

AX1

Properties of Insulating Materials

AX2

Properties of Conductors

AX3

Properties of

and

Insulators

Round Copper Conductors

AXO CONVERSION CHARTS The conversion charts listed in this Appendix make it very easy to convert from one unit to any other. Their use

is

explained in Chapter

Quantities such as

and so

l ,

Section

1

Examples tended to sion rule

0.

AREA, MASS, VOLUME,

the

bottom of each page are

which basically

in-

applying the conver-

states:

WITH THE ARROW — MULTIPLY

forth, are listed in alphabetical order for

quick reference. Multipliers between units are

at

assist the reader in

ei-

AGAINST THE ARROW — DIVIDE

ther exact, or accurate to ±0.1 percent.

Conversion charts adapted and reproduced with permission. Copyright Sperika Enterprises Ltd. All rights reserved.

865

©

1971. 1973 by Voita Inc.; copyright

©

1978. 1988. 1995 by

APPENDIX

866

SI

ELECTRIC CHARGE

AREA

MULTIPLES AND SUBMULTIPLES OF UNITS

[

ampere-hour

square mile

|

,2.59

|

[square kilometer

3600

km'

1000 1 100

10"

hectare

1000 I

2.47

aC

attacoulomb

1000

|

4047

L6.24

square meter

charge on an electron

1000

j 10.76 |

square foot

104

1000

square inch | 6.4516 |

square centimeter

100 |deca

square millimeter

da

[

1.97

,10

UNIT

|

ENERGY

MCM

1

,,0 |

TNT

kilotonne

106

square mil |

|

1000

deci

10 DO

I

,10

1

-

273

1.167 x 10 6

k cmil

|

centi

I

1 507 ,10

kWh

kilowatt hour

square micrometer

m

milli

|

1000

MJ

mega joule 1 277.8

1

W-h

watthour

CONDUCTANCE

1000

1

3.412

1000

British thermal unit

Btu

1000 1.055

mho [kilojoule

1000

1000

DENSITY |

L

|

1000

j 3.086

foot-pound force

pound per cubic inch t/m 3

tonne per cubic meter

joule

[gram per cubic centimeter g/cm*]

1000

|

ft-lbf

1.356

27.68

3L kilogram per

newton-meter

kg/L

liter

624 I 6.24 x 10 18

{pound per cubic foot 1 16.02 |

I

kg/m 3

kilogram per cubic meter

electronvolt

Example: Convert 1 590 MCM to square inches. 1590 MCM = 1590(- 1.97) (- 100) (- 6.4516)

Solution:


©

1

971, 1973 by Volta

Inc.;

in

2

= 1.25

copyright

©

in

2 .

1978, 1988, 1995 by

1

APPENDIX

MAGNETIC FIELD INTENSITY

FORCE |

kip

1 4.448

kilogram force

|

1121 ampere-turn per inch

kgf i

Ibf

1

I

79.6 ]

j 4.448 [

1000

|

ampere per meter

9.806

Mgl

,

139.37

pound force

[

MASS megagram

Oe

oersted

[

|

1.102

I

A/ml

|

ton (2000

lb)

907

|

|

kilogram

N]

newton

867

JcgJ ^ 2.205

"SI

pound |

MAGNETIC FLUX weber

FLOW |

m 3 /s|

cubic meter per second

10 6

MOMENT OF

,35.3 cubic foot per second

|

J

1000

4.55

|

,

cubic decimeter per second

kg-m 3

|

1.201

gallon (U.S.) per second

-

kilogram meter

23.73

gallon {Canadian) per second

|

I

100

,

r

|

MWb|

]

6.23

[pound mass foot 3

line of force

|

Ibft

3.785

liter

1

5.97

dm 3 /s

jounce force inch second 3

I

386.1

1

|

INERTIA

3

ounce mass inch 2

per second

MAGNETIC FLUX DENSITY

LENGTH

10

|mife

|

kilogausT"

|

1000 |

km

kilometer

PRESSURE AND STRESS

1000

\ 1.609

10

J

6

[gauss I

kgf/mm 2

| 6.45

|

m

meter J |

foot

|

inch

)

j

line per

J

square inch

|

9.81

MPa

megapascal

|

1 15.5

3.28 {

|

microtesla

K )0

| 1.02

1000

|

i

r

|kgf/cm 2

|

254 cm

(centimeter

|

]

pound per square

inch

(psi) j

j 6.89 |

mm

millimeter

10 6

|

kPa|

kilopascal |

j

MAGNETOMOTIVE FORCE

\ 39.37 mil (0.001 inch)

|

|

ampere

1000

millimeter of mercury (0°C) J

J

\ |

nanometer

nm

ampere- turn J

f

\

angstrom

Pal

pascal j

|

1.257

gilbert

|

|

newton per square meter |

Example: Convert 580 psi to megapascals. 580 psi = 580 (x 6.89) 1000) mpa = 4 mpa.

Solution:


©

1971,

1

973 by Volta

Inc.;

copyright

©

1

978,

1

988,

1

995 by

APPENDIX

868

POWER

TORQUE

SPEED

MW

megawatt

meter per second

I

m/s|

pound force foot

|

ftlb

|

'

| 2.237

,

1.356

:

Btu per second

1000

j

mile per hour

|

N-m|

newton-meter

I

J

,

1.055

1 1.467

kW

kilowatt

100

|

|

,8.851 |

foot per second

1000

pound force

I

\ 1.34

,

[kilometer per hour

hp

horsepower

km/h]

|

[

|

1000

inch

113

mN-m

millinewton-meter

|

27.8 [calorie per second

cm/s|

centimeter per second

|

| 4.184

1.97 |

foot per minute

VOLUME [cubic meter

joule per second I

1

IT

000

I

|

TEMPERATURE r

newton-meter per second

[degree Celsius

[

cubic f oot~

1000

j 22 6 |

cubic yard

milliwatt

|

1.308

3E

+273

xl.8

foot-pound force per minute

|

j 6.23 [gallon (Canadian)

C/F

| 1.201

.

4.546

gallon (U.S.)

+ 32

|

| 3.785

i degree Fahrenheit

[cubic decimeter

|

Examples: 100°C

=

100 + 273

=

373 kelvms

68° F = (68- 32) - 1.8 = 20°C

1000

02 cubic inch"

j

\ 16.4 |

cubic centimeter

milliliter

RESISTIVITY exaohm meter

10

|

1

Exfi-m

'

ohm-meter

J2-m|

10 9

|

ohm-centimeter

|

microhm-centimeter

FORCE OF GRAVITY

MASS

jioo |

] kilogram (force)

[kilogram (mass)

|

9.806

|e.oi5 |

ohm

circular mil per foot |

[pound (mass)

[

pound

(force)

| 4.448

j 1.662 |

nanohm-meter

nfim

|

|

newton

Example: Calculate the force of gravity (in newtons) that acts on a mass of 9 9 lb -> 9 (x 1) (x 4.448) N = 40 N.

lb.

Solution:


©

197

1,

1973 by Volta

Inc.;

copyright

©

1978, 1988, 1995 by

APPENDIX

869

TABLE AX1 PROPERTIES OF INSULATING MATERIALS Mechanical

Thermal Properties

Electrical Properties

Insulator dielectric

dielectric

strength

constant

max

operating

Properties

notes

thermal

temperature

conductivity

density

kV/mm

°C

W/(m-°C)

kg/m

3

2000

0.024

1.29

2.7

-

0.17

0.09

MV/m or

dry air

hydrogen

3

gas density is at

101

nitrogen

oxygen sulfur hexafluoride

(SF 6 )

30

3.5

-

0.024

1.25

3

-

0.025

1.43

-

0.014

6.6

2000

1

MV/m at 400

L-Pn

solid asbestos

1

1600

0.4

asbestos wool

1

1600

0.1

d.SK.d.1

C

12

1

0°C kPa

synthetic

1560

120

4.5

/inn

liquid (restricted)

glass

magnesium oxide 40

mica mineral

3.3

130

0.3

1600 to 2000

100

5 to 7

600

1.0

2500

3

4

1400

2.4

-

0.36

2800

20

epoxy

oil

mylar®

240

to

7

500

to

1000

(

10

2.2

110

0.16

860

400

3

150

-

1380

150

0.3

1140

120

0.17

1100

4.1

nylon

16

paper (treated)

14

polyamide

40

3.7

100 to 180

0.3

1100

polycarbonate

25

3.0

130

0.2

1200

polyethylene

40

2.3

90

0.4

930

polyimide

200

3.8

180 to 400

0.3

1100

35

3.6

90

0.35

1210

50

3.7

70

0.18

1390

4

6

1300

1.0

2400

65

0.14

950

polyurethane polyvinylchloride porcelain

(PVC)

rubber

1210,20

silicon

10

teflon

20

4

to 7

4 -

•

2

250

0.3

260

0.24

1800

to

2800

2200

powder)

a polyester a

polyamide

870

APPENDIX

TABLE AX2 ELECTRICAL, MECHANICAL AND THERMAL PROPERTIES OF SOME COMMON CONDUCTORS (AND INSULATORS) Thermal properties

Mechanical properties

Electrical properties

thermal

temp density

eoeff

resistivity p

ultimate

s pec i Tic-

conduc-

melting

strength

strength

heat

tivity

point

MPa

MPa

Chemical

Material

symbol

kg/m

0°C

or

composition

aluminum

Al

brass

- 70%

carbon/

C

Cu, Zn

0°C

2()°C

at

nll-m

(X

26.0

28.3

60.2

62.0

8000

3

10

)

ss .JJ

1

—

to

g/dm

3

2703

4.39 1

3

or

«s

£300

-

ZJUU

1

J/kg-°C

W/m-°C

°C

21

62

960

218

660

94

^70

^70

143

Q60

7 0

J.yJ

3600

1

30 000

graphite

constantan

yield

54%

Cu,

45%

Ni,

500

500

Cu

15.88

17.24

A 97

007U

gold

Au

22.7

24.4

3.65

19 300

iron

Fe

88.1

101

7 1A

7000

A Q H-. y

lead

Pb

84%

Cu,

12%

Mn

mercury

Hg

molybdenum

Mo

monel

30%

zz.O

^KO

3Q4

1

i

i

on

1% Mn

copper

manganin

M

A

<$y\)\)

4%

Ni,

Cu,

203

220

482

482

951

968

1

1

-+-

1

1

1

49.6

52.9

418

434

1080

1082

O

1

69%

Ni,

1% Fe

80%

Ni,

20% Cr

nickel

Ni

78.4

85.4

4.47

platinum

Pt

9.7

10.4

3.4

1

1

296

490

7Q

1

s

30

10

1

200

kwoo

sô

R400

8900 21

200

1

OKI

1063 i

jjj

397 jZ/

3S 1

1

090 uzu

O.H-

— j>y 3Q

38

2620

1

6QO

SÔ

6Q0

430

500

460

90

1455

131

71

1773

230

408

960

140

20

3410

380

110

420

994

0.024

400

silver

Ag

15.0

16.2

4.11

10 500

W

49.6

55.1

5.5

19 300

3376

zinc

Zn

55.3

59.7

4.0

7100

70

air

78% N 2

1.29

AC\

6Q0

tungsten

,

1

Zt

90 ZU

600 uuw

3 J

Q7

nichrome

130

9QO

i

i

1

3.3 1

69 j

k4 o

i

OQ

900 ZZU

ôo jWW

1

o o s

i

35

9S Z J

1 1

1 1

9 .z

1

3fSO

1

400

21% 0 2 hydrogen

H2

pure water

H20

0.09

1000

2.5

X

10

14

14

200

4180

0.17

0.58

0.0

TABLE AX3 PROPERTIES OF ROUND COPPER CONDUCTORS Gauge

Diameter of

number

bare conductor

Cross section

Resistance

mll/m

or

0/km

AWG/ D

OC

mm

o

MCM

mm

mils

25°C

cmils

105°C

Weight

Typical

g/m

diameter

or

of

kg/km

insulated

magnet 12.7

500

126.6

250 000

0.

138

0.181

1126

1.7

460

107.4

212 000

0.164

0.2 14

953

used

2/0

9.27

1cc 303

67.4

133 000

0.261

(\

1 A U.34

600

relays.

1/0

8.26

325

53.5

105 600

0.328

0.429

475

magnets.

250

4/0

1

1

1

7.35

289

42.4

87 700

0.415

0.542

377

2

6.54

258

33.6

66 400

0.522

0.683

300

3

5.83

229

26.6

52 600

0.659

0.862

237

4

5.

18

204

21.1

41

600

0.833

.09

187

wire in

motors,

formers,

1

5

A AO 4.DZ

1

16.8

33 120

1

A) J

1.3/

149

6

4.

162

13.30

26 240

1

.32

1

.73

118

7

3.66

144

10.5

20 740

1

1

80 oz

etc.

mm

93.4

Izo

8.30

16 380

.67 o

2.19

3.2 j

Z. 1Z

2.90

73.8

9

2.89

1

14

6.59

13

000

2.67

3.48

58.6

3.00

10

2.59

102

5.27

10

46.9

2.68

2.30

90.7

4.17

8

3.35 A TO 4.Z3

4.36

11

400 230

j. 34

37.1

2.39

12

2.05

80.8

3.31

6 530

5.31

6.95

29.5

2.14

72.0

2.63

5 180

6.69

8.76

25.4

1.91

04.

2.08

4 110

0.43

.u

18.5

1.71

8

i oc

1

O

1

C C

A

13

1.83

14

1

.03

15

J

.45

57.

1.65

3

260

10.6

13.9

14.7

1.53

16

1

.29

50.8

1.31

2 580

13.4

17.6

11.6

1.37

17

^ 1.13

4 3. 3

1.04

2 060

18

1.02

40.3

0.821

1

620

21.4

27.9

7.31

1.10

19

0.91

35.9

0.654

1

290

26.9

35.1

5.80

0.98

20

n o U.5

3Z.U

0.517

1

020

33.

A A 1 44.3

4.61

0.88

21

0.72

28.5

0.411

812

42.6

55.8

3.66

0.79

22

25.3

0.324

640

54.

70.9

2.89

0.70

23

0.64 n

ZZ.O

0.259

511

2.31

0.63

24

0.51

20.

0.205

404

25

17.9

0.162

320

26

0.45 n /in U.4U

n

0.128

253

i

1

i

1

i

1

1

1

1

A o D.V

1

00 ZZ.

86.0

9.24

1.22

12

1.81

0.57

142

1.44

0.5

70 /y

1.14

0.46

1

108 17/ 3

1

1

1

1

1

1

27

0.36

14.2

0.102

202

172

225

0.908

0.41

28

0.32

12.6

0.080

159

218

286

0.716

0.37

29

n on U.Z9

1

11.3

i

0.065

128

Z 11

334

0.576

0.33

30

0.25

10.0

0.0507

100

348

456

0.451

0.29

31

0.23

8.9

0.0401

79.2

440

0.27

n on u.zu

o.U

0.0324

64.0

J4

574 7no /uv

0.357

32

0.289

0.24

33

0. 18

7.

0.0255

50.4

689

902

0.228

0.21

34

0.16

6.3

0.0201

39.7

873

140

0.

79

0.19

35

0.14

5.6

0.0159

31.4

1110

0.141

0.17

1

1

1

1450

1

0.0127

25.0

1390

1810

0.1 13

0.15

0.0103

20.3

1710

2230

0.091

0.14

4.0

0.0081

16.0

2170

2840

0.072

0.12

3.5

0.0062

12.3

2820

3690

0.055

0.11

3.1

0.0049

9.6

3610

4720

0.043

0.1

36

0.13

5.0

37

0.11

4.5

38

0.10

39

0.09

40

0.08

,

Answers to Problems

Note: The following numerical answers are usually

Chapter 3

rounded off to an accuracy of ±1%. Answers that are preceded by the symbol ~ are accurate to ±5%.

1.

8.

Chapter 7.

22.

37.

8.TJ

mPa mg

9.

15.

mm Wb

38. kg/m' 39.

L

16.

r

24.

23.

2

mm

63. 126.9

MO

10.

kHz

17.

jjls

33.

W

71. 4.536 kgf 72. 0.93

A/m

75. 6049.6 1

76.

.34 kg 80.

mbar

83. 67.7

3.

6615

2940

393

0.33 91. 10.4 A, 23

1

2

GJ

mK

pAVb

13.

mrad

Hz

°C 41. kg 62.

or

65. 6.591

mm

69. 7.079

m 3 70.

2

18a.

mT

21.

34. L/s or nrVs 35.

K

9a. 1096

J

14.

209

36. rad

yd

13.56

Wb

81. 0.001

lb

X

K

10

6

82.

C

K

89. 120

X K

10

23. 1568

X

I0

kW, 336 hp

1

6. 10.8

kW,

W

9b.

J

15.

3288

209

W

9c.

J

16.

1

242 kJ lib. 50 N-m .89

kW, 2.53 hp

17b. 88.8 kJ

N-m, 47.8 r/min

Chapter 4 9a. 103

N

V

9b. 288

W

10b. polarity reverses

10% B =

N/nr

90. 1.67, 50,

20a.

5

5000

11a. 18

YC V

16.276.5

93a. 402 93b. 0.076

93e. 93.75 93f. 3.24

25

5.

2

km X 10 V-f

11 1.2

3

60 N-m

2

8304 kg/nr'

Pa 85.482.3

88. 366.3

78.

4.

17c.2195W 17d.2400W 18b. 1332 N-m 19a. 475 N-m 19b. 4455 N MJ 21. 7.68 MJ 22. 1568 N-m, 239 r/min

N-m 1222 N-m

20. 3.04

1.95

1

66. 2.59

107 miles 77. 288 000

12 92. 3.3 kil

16m 2 93d. 64m'

mA

12.

19.

J

73. 12 kilogauss 74. 1.417 kg

84. 1.378

86. 1.57 rad/s 87.

93c.

mL

kW

T

11. 18.

40.

64. 2.549 in

67. 76.77 Btu/s 68. 0.746

1

415.7

13. zero

1

cm

79.

2.

hp 7a. 100 kW, 134 hp 7b. 83.3% 7c. 68 256 Btu/h

17a. 13.1

MW

14.

N

392

14.5

=

A 18

9c. 10.4

N-m

A

E NL -

138

V

lib. 7800 A 12. 2 A 13. A = 18 V, V 14.at90°E A = 20 V, at 120°E A =

17a. 12 brush sets

23b. 83.25

18

150A 19.E XY is( + 0.436 T 21c. 530 (jls

17b.

E 34 is(-) 21a. 292 V 21b.

23a. 333

10a.

10c. voltage increases by less than

V

)

A

Chapter 2 2a.

0 8.

E 16 - -80 V

V,

- 30

3840

200

V

11. 70.7

1,

800

W

lags

W

1

2

13.

A

by 90° 15c.

17b.

A

12a. 12

23 Hz 14.

120V

18.

E

1

9c.

ms \ l

A

7.

E 52 - 0

Chapter 5 3.

+20

V,

l£a.

V 1,

1

.3

V, 14

12c. 1440 lags

1,

V 9b. 13 800 W 9c. 13.26 kW, 17.8 hp 10a. 1533 A 1.850 11a. 2975 A lib. 2400 A 12. 144 V, 9.6V 1.7 kW, 94% 13b. 10 A 13c. 8 mH. 223 V 13d. less

9a. 221

320, 160, 112

no 10b. 3

12b. 169.7

.67

lags

box

2c.

V

1.8

9a.

12d. 2880 15b.

E 25 = +80 V

+30 V 4. 20 V 5. 960 N 9b. 960 N

V,

A

2b.

10b.

V.

1

W

13a.

by 60°

by '150° ,16c. 2400 W,

19.no 20. -17.3 V, 8.13

1

than 4

A

15b. 45

V

16c.

873

14a. 0.48 12 14b. 30

kW, 8.43 kW, 0

70 kJ

kW

kW.

1.9

kW, 0

16a. 17.8 kg/m

kW 2

15a. 0.96 11

16b. 140 kJ

ANSWERS

874

Chapter 6

27.9mWb

99.1%

29a. 84.1 12 29b. 3.04% 29c. 23.36 m!2

29d.3.04% 29e.976W

kW, 547 A 10. 22.8 hp 11, 12a. 28°C 12b. 88°C too hot 14.47.2% \§h\ 16.
8.

18.

131.4

9.

A/mm 2

year 19a. 2.28

1

20b.985 cmil/A

W/kg

19b. 13.4

kW, 35hp

21. 26.2

e

jgiÔftECA

Chapter 5.

300 turns

W

9a. 1440

10b. 1440

source lid.

source

M

1 IF.

1440

10c.

the active source tive

VA

9b. 1440

VA

is

1

667 kVA, 291 kvar

7.

lb.

C

9c.

W

the reactive source

W

20b. 3512 21c. 769

18a. 2

VA

W

kW

17b.

L

is

and

3(X) var, both flowing

300

var

C

to

from B

to

is

A

A

O

16a. 320

17c. 1300

Q

23a. 5 19.6

W from D

23b. 5 1 9.6

6b.

to

W

C.

600 r/min 15b. 564 r/min 16a.

10

Hz 16c. N-m 18.

12b.

kVA

480

H

960 1

kN-m,

30.3

23b. 36.7° 24a. 9.6 (2 24b. 7.72

25b.

82%

26a. 21.3

kW

h 27a. 15.5% 27b. 88.4

W

19b.

A

17b. 882 r/min.

60

V 15

96%

V

23.56 m/s 22b. 3.3

Hz

kW

kW

23c. 34.5

26. 1500 A. 75

N

20

27.

22c. 196.35

A

21.

3.

mm

23d. 2251 kW,'

N

kW

28. 38

12,

Hz

kW 29e. 50 kW 30a. 8.6 M var 30b. 670 kW 30c. 1498 kW 30d. zero 30e. 23.8 kN-m 31a. 400 V 31b. 508 m!2, 314 kW 31c. 455 A 32.264.5 mm 29c. 1035 kvar 29d. 2.07

10.

1

mm, 120kW

5.79 12

26b. 18.1

78%

12a. 3 hp 12b. 54 or 56 13. 438, 280, 315 fMb; zero.

kW

15e.

A

16d.

14b. 586 r/min

14c. 222.4

kW

1509

15a. 140

000 Btu 19a. 456

20a. design

A lc. 1415 A Id. 0.9 mWb 2. 472 A, 360 V 4b. 30 A 4c. 18 A 4d. 10 800 W 5. 33.3 12 6. 400 V, 0.02 A

kW

14a. 3.8

Hz 15b.54kW 15c. 23.8 kvar 15d.65.6A 100 hp 16a. 1109 A 16b. 434 kW 16c. 90.2 kW 5212 A, 83.5 kN-m 16e. 43.3 kN-m 17.4.7°C 18a. 39.4 s

14d.5.3kW

22a. 4a.

Hz

240 Hz 20a. -8.66 A, 8.66 A, 0 20b. 86.6

675, 765 r/min

lb. 2.83

800

17a. 15 A. 90 A, 5.25

19a. 120 V, 30

A

V.

Chapter 14

Chapter 9

mWb

Hz

A

45 Hz 16b. 0.67

3 V,

29a. 4.49 ml2. 68.9 m!2 29b. 1067 V. 40 Hz: 16 V, 0.6

18b. 105

0.3

kV

.89

80.2%

12

4e. 10

1

14. 8 12

3.44

A

V,

97.2

to slot 8 22a.

34. NX) 13. 0.9 fl

15c. 37.2 kvar 15d.

21a.

la. 2

V. 15

1

23a. 87.9% 23b. 2298

kW, 7.78 kvar, 16.5 kVA 16b. 270 V 17a. XZY A 18b. 19.9 kvar 19. lamp connected to Z 20. 20 kW 18 A 21b. 130 A 21c. 72.1 A 22. 18.3 A 23a. 4.63 12.

kW

1

1

,

15b. 62.3

kVA

to-

mV

no 10a. 360 r/min 10b.no 10c. 20 12. 936 A. 156 A. 55

15a.

812

18a. 5

26c. 63.9

H2-X1

1

8a.

7.

4000 var

22b. 20.4

16a. 14.6

25a. 6.1

or

lib. 2.25

Chapter 13

D

12a. 160 12

89.5%

V

W

2a.

11c. 2.89 12

H1-X2

8.

11a. 0.45

kW

VA

- 100. 0. 50. 50, -86.6 V 2b. +,0.-. -, + 2c. 50, -86.6. - 100, 50. 86.6 V 3. yes 4a. 358 V 4b. 23.87 A 4c. 25.6 kW 5a. 694 A 5b. 13.2 kV 5c. 9.1 6 MW 5d. 27.5 MW b c 6b. yes 7. 26 kVA 8. 120 V 5e. 19 12 6a. a 9a. 13kW 9b.6.5kW 10. single-phase lla.41.6A llb.4l.6A 15a.

mA

A 2b. 20.8 A, 250 A 3. 506 A. 65 kA A 5a. 278 A 5b. yes 6a. delta-delta 577 A. 48. A 6c. 333 A. 27.8 A 7a. 236 kVA 7b. no 8b. 433 kVA 9a. 347, 600, 347 V 1

1

Chapter 8 V

300 kVA

1

A

!7.7kA, 2 75

4.

slot

4157

7.

Chapter 12

19c.

1.

10. 13

2a. 20.8 A, 433

the reac-

900 var 20a. 2765

V

V

kVA

is

the active

1

15c. 12. 16

12

500 var

22a. 102

F

11a.

s

A 13. 0.72, VA 14d. .9 A

2864

20c. 0.787 21a. 3845 var 21b.

VA

10a. 1440 var

12. 7.2 A, 9.6

18b. zero 19.

21d. 784.5

from

W

lc.

1

lie.

14b. 745 var 14c.

240 kvar 17a. 1200

s

lOe. 1/240

the active source

is

14e.96.5% 15a. 3.7111 15b. 6.32

17d. 0.923

W

W/kg

4.34 kvar

9d. 1/120

lOd. 1440

the reactive source

J is

2665 var 14a. 2765

16b.

W

2880

8.

6. 0.2

250 A/5

11c.

kVA 6.64.3%

100

2.

2.75% 30a. 1.912.36.612

1

gether 9. 92

Chapter 7

29f. 1.3%,

3
12

W/kg

feb. 156

20a. 58..'

22. 64 yeaJJ

453.6

H)b.

D

~ 1100 N-m 2

lb

ft

25a. 916

22b. Il.4kg-m

24b. 810 r/min. 260

N-m

25b.

mO

min 26g. 91

kW

2

270

23.

hp

12

795 kg,

kJ 24a.

24c. 702

ft-lb

-16

19b.

21b. 2.23

lb

ft

25c. 11.3 kJ 25d.6.7s

1.3 s

26c. 358

26a. 134 r/min 26b. 11:1 26f. 39

V. 1740 r/min

20b. same 21a. 54.5

A

26d. 50

l

26e. 728

MJ

h

Chapter 10 11.

120V 12.6()kV

14. 2.42

18a. 144

kW.

V

11

A, 22

18b. 2.25

13a.

A

1

15.

1

0.4

V

50 A, 1250

mWb

18c. 0.9

19c. additive 20a. short-circuit 23a.

24a. 126.4

V

13b. 0.353 A, 22.08

24b. 5.22 A, 95

A

A

mWb 72

16.

Chapter 15

A

1.88mWb

19a. zero 19b.

2a. 5.05

520

12 23b. 0.1 15 12

25a. 466.9

kW

25b. 466.9 kW,

V

12.

3a. yes. 366

82° 2b. 8

N-m

1

1

5. 0.4 12. 0.5 12, 1.48 12. 7. 13.9

r/min 2c. 45.4

3b. no. as a brake, 35

N-m, 35.5 N-m

268 N-m 8. 10.5

6.

A

2d. 332

N-m

4.

N-m

26.7

37.9 kN-m. 2.4

N-m. 42.6 N-m

1~2,

202

kN-m

12

ANSWERS

Chapter 16

Chapter 21

3a. 20 poles 3b.

360 r/min

4. increased

Hz

16c.

14411 16d. 10

16h. 36.9

kW

7500

kW

22a. 126

MW

V

16e. 1750

18b. 2.475°

19c. .916

MN m

V

92 12

kW

16g. 57.6

18c. 7.87 in

20a.8130kW 22.7°C 21. 0.409 T

MW

85.7% leading 23a. 228.5 0 W, delivers 63.3 Mvar to the bus

22c.

24.

14b. 0, J

1

1

15d.40()A 15e.30()Hz

15c. 12a

Mvar 18a.4()kV

19a.

40 A 19b.24kW

A

A

23. 1/24

25b.48kJ 25c. 4

Chapter 17

1

9c.

1

la.

A

333

lib. 3 12 12.

kVA

3457

17a.

500 r/min 14a. 36.9° 14b.

A

17b. 289

V

5889

17c.

2.

16

MW

17d. 1.44° 17e. 1569 kvar

9150 hp 18a. 4741V 18b. 32.8° 19a. 14.6

17f.

19b. 6807

19c. 142

lb

ft

A

kN-m

20b. 1704 kW, 36.2

V

1594

12,

kW. 45.3 kN-m kN-m 20d. -83 s

Chapter 18 1

A

4 A, 14

10b. probably not 12a. 50

13b.55%

13c. 0.28

14b. 4.44 pu

16b.

Hz

70.7% 10a. no

86.6% lagging

V

14a. 4.42

15a. 157.5

16c.

30

17c. 3.2

mhp

13a. 8

.06 pa

1

14d. 2.5 pu

17b. 21.2 A. 84

17a. 1.94 12

9d.

12c. 4.26.1

13d. 0.79 pu.

14c. zero

A

16a. 0.577

A

9b. 33.8° 9c. 26.8

W

V

ft-lb

16b. 3.6°

ms

18b. 13

16c. 1.8°

18c. 0.16

44%

A

25a. 120

26c. 200

A

27c. 57 kvar

1

1

would be much

+

20d. 14

Hz 20e.220W

22a. 24

V

2047

The

greater.

A and -807 A.

between

current

would

oscillate

44V 20b. 44 V 20c. N-m 21b. 2880 |jlH

14

20a.

20f. 2.5

Hz

22b. 97.8° 22c. 7340 var

16d.

37%

Chapter 23

N-m

10. 0.00

1

75

W

11a. 144

in

llb.78.5W 12a.30mN-m 12b.0.79mhp 12c. 1.77 14. 437.5 r/min 15a. 1.5 ms 15b. 4.5 ms 15c. 0.23 A 16a. 7.2°

kW

27a. 135.2° 27b. 57.5

230

V,

30 Hz 4a. 1620 r/min 4b. 8.78 N-m

Hz

10b. 42

1080 r/min,

9.240 ms

H

20e.

22b. 147°

15b. 93.2

Chapter 19 8. true

A

26b. 80

200

10a. 8

52 6.30°

A

26a. 120

V

V 4b. 169 kW 4c. 90 A 4d. 220 hp V 8b. 283 V 8e. 2 A 9a. 3 .3° 9b. 5. kvar 9c. 28.6 A 9d. 113 V 10a. 85.1° 10b. 16.8kvar 11a. 343 A lib. 146° 12a. 300 V 12b. 315 V 13a. 150 A 13b. 75 V 13c. 60 V 14a. zero 14b. 150 A 14c. 15 V 15a. 100.7° 15b. 87.3° 16a. 232 A, 240 A 16b. 700 V 17a. 192 V 17b. 250 A 19a. 0.02976 19b. 164 A 19c. 0.04576, 249 A 19d. The current

3.

4, ll

kVA

22a. 340

3.91 Hz 4a. 626

ripple

9a.

19e. zero

20d. 10.6

mH

kV

Chapter 22 8a.

20a. 4269 20c. 40.7

100%

A

367

17c.

18c. 54.64

24a. 3.29 V-s 24b. 0.47

s

kW

kV

956 kvar 9d. 32 poles

14d. zero 15a. 300 A, 0° 15b. 720 kvar delivered

100%

14c.

A

26d. 160

2300 hp 9a. 22 7 kVA 9b. 90.2%

70%

20f.63.5% 20g. 5.72 kvar 21. 6.25 22c. 30.4

7.

14.64

18b.

20c.

MW

17b. 6.3

19c. zero 19d.

kW

20b. 4.7

16712 16a. zero

15f.

kV

I6b.77.7° 16c. 102.3° 17a. 14

20a. 12.66

W

3565

12,

620

17d. 23.5

25a. 236 r/min 25b. 108°, 1944° 26a. 40 26b. 180 26c. 1.5 26d. 1.73 in 26e. 3.7

1

8a. 540

19d. 1.09

20d.

V 6b. 200 A 6c. 600 A 7a. 3240 V 7b. 200 A V 8b. 540 V 8c. 18 A 8d.9A 8e.972()W 9a. 150 V 9b. 3.75 A 10.45kW 11a. 0.72 kW lib. 99.96% 12a. negative 12b. increasing 13a. 36 A 13b. 170 V 13c. >65 J 13d. 0. H 13e. 70 V 14a. 324 V 14b. 243 kW 14c. 750 A 14d. 5.55 ms 14e.6l2A 14f.zero 14g. 45.4 V 15a.24kW 15b. 24 kW 6a.

50 Hz

16a. 145 12 16b.

16f. 3.031

18a. 44.56°

20c. 24.23

A

mm

V,

A 13.2400 V

50

12b.

807

15.

19b. 0.41212

20b. 720

23b. 7230

A

kV

17. 16

19a. 0.451)

A

12a. 150

353, 600, 750, 600, 0

poles 7. 4

6. 12

8a. decrease 8b. decrease 8c. increase 9a.

9b. 37.5 V, 0.25

875

.2

1

V

13b. 39.5° 13c. 12.4 kvar 15a. 141

19a.

3600 r/min 19b.

W

kW

15

18a. 3000 r/min, 0.93 hp

23.50

V, 15

A

17b. 36.3 19c.

Hz

20.325(jls 21a. 115 V, 15

ms

ms

lib. 13

5.

600 r/min

12a. 8

N-m,

hp 12b. 3.5 N-m, 720 r/min, 0.35 hp 13a. 16.5 hp

15d.46.7Hz 17a. 20.7

J

ms

11a. 50

15b. 422

W

337

21b. 85

A V

15c.

566

V

18. 194

V A

19d. 7.6 22. 31.8

12,

15

kW

A

Hz

Chapter 24 Chapter 20 7.

10a. quadrant

13a.

40

A

10b. quadrant

1

13b. 120

A

13c.

40

4

1 1.

A or

1

6.3 hp 12. 13

A

26

V, 7.5

Hz

14a. 2 or 4 14b.

3 respectively 14c. 4 or 2 respectively 15. clockwise 16a. 2 16b. 5

s

18.

23c. 30. 15

208 days 22. 28

kW

24.

25b. 633 V, 82.5 28c. 63.7

N

30b. 149

ft-lb

32b. 29.5

rn

ft-lb

Hz

1

s

23a. 18.85

.57, 16. 12.57

26.

28d. 56.4

ft-lb

V

29b. 10

Hz

23b.

1

25a. 345 V, 45

38 V. 5 Hz 28a? 30

31a. 307 V. 40 33. 282

kW

kW

Hz

A

1.3

or

1

min

kW

Hz

30a. -834

ft-lb.

358

30

ft-lb

312 mi

MW

16a.

3

14a.

lla.41.2X 12. 15.6 h

5400

MW

10'

13a.

mVs 60

14b. 2615

X 10 L\l, 1.76 X 90.6 TW-h 19b. 10.34

IO

,0

GW

lib.

40

4(X)

50

MW

Mvar

14c.

MW.

5715 tons 16b. -57 000 tons

18a. 1.86 19a.

28b. 4.29 hp

31b. 67.3 kJ 31c. 29.9 kj

1176MW

11c.

16c. 21.6

MW 13b.80MW. 9364 yd Vs

m Vs

Btu 18b. 0.207 g 20. 1580

MW

17.0.43m Vs

ANSWERS

876

Chapter 25 -2500

7.

600 kW;

750

9.

-3%

10c.

Chapter 29

la.

1

450

kW

V,

720 kW; 4243

V,

-240 kvar

kVA

17e. 802

a2

19.4

1

14

12. 0°

V,

17b. 139.2

17f.

5759

V

720

V,

15b. 0°

A

2.

450 kW; 3000

kW

1

26 kW; 5692

ER

14.

16a. 5692

V,

18.4°

is

1

V,

1

MW

kW

868

21b. 1074

The maximum- frequency of

Hz

about 2 kHz 3. 1380

V,

V 16b.no

17c. 290.6 kvar 17d. 50.6 kvar

17g. 625

37.5 kft 21a. 42

11,

V,

13a. 5992 V,

V

9000

MW

10b. 151.8

kW; 4500

1200 kW; 1897

laggings 15a. 6577 17a.

MW

10a. 5092 V,

500

1

A

5700

7.8.33ms

V

13.2159 19a. 5.57

879

10.

A

MW

19b. 2.7

19e. 14.4 kV, 33.7

A

V

6a. 1870 12. 4.5

16. 80.2

MVA

GTO

a high-speed

967

I406A

11.

14. 5.52

kV

4.

A

limited to

is

6b. 1470

V

mH, 221 pf

17.46.3A-h 18.6

kV

19c. 24

19d. 26.4

s

kV

kV

18. 20.3°

A

Mvar 21d. 74.3 Mvar 22a. 22 H, 100 H, 1500 il kV 22c. 45 A 22d. 9 Mvar 22e. 34.6 kW 23a. 648.5 A

Chapter 30

21c. 148.8 22b.

1

19

la. 36.0

A A

lb.

6a. 84.8

Chapter 26

60 Hz

lc.

4b. 29.4

%

6b. 30.1

kW

4a. 1000

%

300 Hz 2.62.2

5a. 1843

kW

%

6c. 73.8

V 3.12.9%

5b. 48.7

kW

5c.

A A

8a.

4163

7. 141

9b.

960

kV

6. 17.5

MW

8a. 48.8

9c.

4800

A

13b. 10

13a. 25 A, 35

J

kV

11. 5

A

and 7 12.

13c. 0.5

dangerous 15b. hazardous 17a. 17b.

2A

17c. 8 1.8% lagging

21b. 17.5

kA

8b. zero 9a. 12

A

1.2

000

n, 3.18

14. 9.74

A

A

mH

A

A

18b. 67.3 A, 86.8%

21c. yes

2.

$28.40

5.

$1470

lion dollars 9a.

9c.

13. 6.45 1

12a.

7.

4.33

MW

8a.

3975

GWh

8b. 159 mil-

kVA 9b. 75 kW, 52 kvar 10b. 32% 11a. 46.64 $/h

30cent/kW

eent/kW-h 15. lOr/min 16. 1800

MW

12.45kW

5b.

W

17a. meter reads

kW-h

-492 A 6a. 150MW 6b. 200 A 7a. 600 A, A 10b.64kV 11. 1800 A. 1000 Mvar

10a.

400

13.

4050

kV

A 17a. 85.6 A

14a. 110

15c. 21.1

50V

21d. 15.4V

21c.

A

14a. 2.9

2

in

14b. 13.38

16a. 17.7

A

16b. 20

B.auioTECA

kV

16c. 159

see section 31.2 2. see section 3

O

14c. 43.4

MW

V

1

.2

3. see section 3

31.15 5. see end of sections 3

6. see section 3

and 31.21

14d.426 kV 14e.94.7% 15a. 360 Hz, 720 Hz 151*911,110 15c. 150

A

15b. 67.5

4. see section

Chapter 28 57.6

V

13c. 18.7

Chapter 31 1.

h 12b. 2.4 cents

percent high 17b.no 18. ±5.6

5a.~45kV

A 1443 A

72 kvar. 104

12.5% 10a. 243 kvar

lib. $4205

-OA

15a. 70.7

21b. 31.8

Chapter 27

kW

A 14b. OA 14c. 2 1.6 A A 15d. 31.4% 16a. 9.6 Q 16b. 17b. / 5 = 0; / 3 = 24.7 A 18a. 50 Hz; 350 Hz 18b. 601 V; 240 V 18c. 647 V 18d. 601 V 19a. E 5 = = 107 V; £ l3 = 323 V 19b. 344 V 46.8 V; E 1 = 15.3 V; 19c. 55.1 A 20. 83.5 Apeak; -32.4° 21a. 35.36 V; 0° 13b.

15a. not

15.6 A, 124.7 A, 31 .6

1

18a. 777

9e.

10

V

V 9a. 71.7 A 9b. 7.86 A 9c. 72.1 9d.721 V 154V 10.21.6W 11.124.4V 12.58.6A 13a. 95.4 A

8b. 1082

pii

3. 2

1

1

.4

7. see section

31.14

1

.3

and 3

1

1

.2

.9

8. see sections

31.19

Answers to

Industrial Application

problems

Chapter

Chapter 4

1

94. 0.926 pu 95. 0.208 pu

l.4l6pu 97.6.0pu 0.39 pu

Chapter 2

24a. 480

A

26. 1385

N.m

W W

24b. 15 319 27. 261.6

24c. 5520

W

25. 203.5

W/kg

Chapter 5

-£,+//? = 0 21b. -£,-//? = 21c. £,+//? = 0 IR = 0 22a. / = 21d. - E2 + £, 3 A 22b. / = - 13 A 5/,= 0; 5 /, +2 /2 = 0; /, + / 3 = A 22c. / = 9 A 23a. -10 23b. - 98 - 7 /3 + 42 /, = 0; - 42 /, + 15 A = 0; /, + A + /, - 0 23c. -48+6 / 3 - 4 A = 0; 4 A +(7 + 12) /, = 0; /, + A ~ /, = 0 23d. 40 - 12 /4 +4/ 3 + 60 = 0; - 6 A +2 /, = 0; /, + A + /3 + /4 = 0 24a. 0.25 Hz 24b. 1000W 24c. 2000 J 24d.500W 24e. 70.7 V 24f. 70.7 V 24g. 50 V 25a. 0. 25 Hz 25b. 000 W 25c. 4000 J 25d. 500 W 25e. 70.7 V 25f. 70.7 V 25g. 0 V 26a. E 2i + 20 /, = 0: / 3 = /, + A; E 2i + A (50j) = 0 26b. E A + 20 /, = 0; EA + A 30 j) = 0; A = /, + A 26c. £ 21 + 20 /, = 0: £21 ~ '2 (~30j) = 0; /, = / 3 + A 26d. E b a — 20/, = 0; £ba +60 j A = 0; £b , - E A - / 3 (- 30 j) = 0; /4 + /, + /3 = A 26e. £ lb + (7 - 24j) / = 0: E ac + (- 24j) / = 0; Ebc - 7 / = 0 26f. E2l + E h - / (40j -45j) = 0; E l3 + 40j / - E B = 0; E 2} - I (- 45j) = 0 26g. - E 3 + 40j/ 3 + E 32 = 0; £" = 0; £ 23 - 30 A = 0; E + 40j /, = 0; A + /3 = /, 3 + E x2 21a.

()

1

19.2514r/min 20a. 400 conductors 20b. 21.

674

f'tVmin 22. 2

1

fl

W

425

85.

IV 20c.4.4mWb

23. 0.0273 pu 0.00346 pu

Chapter 6 24. 310 kg 25. 52.8

mH

26. 4.78 kg 27. 915

No 2 AWG 29. 0.81 V 223 W/m 30. A - /? 2 //f,(228 + /,) - 228 31. 1948 32a. 0.00614 O 32b. 0.092 V 32c. 1.29 V

W 224 V

28.

1

32e.

(

1

.33

N

W

32f. 0.528 J 32g. 26.4

ft/min 22 mi/h

32d. 38.7

32h. 5.8

W

%

Chapter 7

.

24. 345 microfarad 25a. 3.4

W 27a. 288 A

l.OMvar, 3 28b. 9.63

A

650

27b. 0.2

MVA

V

MW

27d. 10 300

28c. no 29a. 20.83

25b. 72.9 1

.0

V A

26. 68

J

Mvar

27c. 2.8

000

fl

MW,

28a. 1976 W, 1796 var 29b. 16.7

A

27t

Chapter 8

Chapter 3 24. 52

with

070

28.24A/C = 0

Btu, 7.63 h 25. Transformer temperature will rise

aluminum

paint 26. 52.9

kW

27.

~

31. 55.4 A, 48

47 °C

34b. 38.41

877

29. 16 A,

A

kVA

32.

16 A, 27.7

lOOkW

34c. 48.2

A

A

30.

156.6kW

33. 13591b/h 34a. 3 1.88

35a. 332

A

kW

35b.602kvar 35c. 25.8°

ANSWERS

878

Chapter 18

Chapter 9 7.

245

980

turns. 0

8c. 64.3

9.

turns 8a. 33.87 il 8b. 30.5 O,

=

<£>

80.9

= 0.516 mWb, The 330 V/I20V transformer

mWb,

0.657

<£>,,= 0.141

mWb

appropriate.

The 40 microfarad capacitor must £e

the

330

10.

mH

18a. 3.33 pu, 6 pu; 2.22 pu farad

is

35. 2 7 1

mm"

36. 85

W/kg, 225

mN.m

.5

1 1

34. slightly less than 60

1

36a. 348 13.

W

36b. 838

A 11. $177 13b. 465 V 13c. 12.9° 14. 121 A kVA 15. transformer 41.9 kVA. motor 33.8 kVA

32b.

125.8

32g. 231

1

J

32h. 1065

Chapter 22

35a. the speed will drop 35b. the starting torque will increase

23.

35c. the temperature will rise slightly 36a. 90 coils 36b. 30 36e.

1

08 100

mm

2

36f. 58.4

1

A

24f.

27. 4 times per year every 4.5 years 28. $17

ft.lbt

negative with respect to

is

857 V/s 26.

24a. 500

0.0547

31d.93%

32i. 0.395

/u is

C

O X

= 0.1620

900

$670

29.

25b. 2

A

31b. 1493

32. 0.39 pu 8.47 pu 2.88 pu

1

.7

27b. 16.1 1.36 pu

Hz 24b. -563 V

24c. 43.5° 24d. 73.7

3000 Hz 24g. 72.5% 24h. 60

30a. 5.83

W

A

N

26b. 57.6

N.m 28.57.2 V Hz to 16.33 Hz

29a. 8258

26a. 28.8

delivers 33.9 kvar 32a.

Chapter 15

32c. 513.5 V, angle

!5.9mH, 292

147

O

mH

25.

9b. 5X1,91.7

0

9c.

212

V

10a. 8

a

D=

30b. 7695

+

200

N.m

-

A

A

24e. 360

N A

max is 120 °C 28a. 220 MW at unity PF, 198 MW at 90% PF 28b. 96 Mvar 28c. 0.787 28d. 1.1 O, 28e. 2785 kW 29a. 194 370 hp 29b. 719 MJ 29c. 2279 MJ 29d. 5.67s 30. 28.6 °C 200011m

A

29c. 5839

kW, 6786 kvar 30d.

32b. 402.5

1

38

A angle +

62°

1.740 ms,

19° 34c.

1

.65

A

D=

0.478 34a. 7.2

34d. 5.95

35.8.41 A, 1.03 A, 8.83

A

N.m

Chapter 24 21d.

1

16

ms

Hz

21b. 26 993 cycles 21c. 27 000 cycles

22. 15 hp

23.-20 hp

Chapter 27 Chapter 17 22d. 260

22a. 17.57

kW

22h.0.4l3O

19a.748kWh t

22b. I23gal/min 22c. 32 500 lb.fr

22e. 97. c/c 22f. 4358 kvar 22g. 0.07 3 1

1

O

A

22.3 h

kW and

27a. 43.5 hp 27b.

21a. 59.985

Hz

27a. 300 r/min 29b. 4129

Chapter 16

A

3 A,

23.2° 32d. 0.902 33a. 2.892 ms, 0.441 ms.

0.868 33b. 1.593 ms,

34b. 6.26 A, angle

10b. 19.2 N.m, 6.3 hp

34e. 10.1

21. 96.8

1

25a. 25.8 N.m. 6.4 N.m,

31a. 430.8 V, angle 28° 31b. converter absorbs

9a.

V

24. 98 A, 24

decreasing 550 A/s

Chapter 23

Chapter 14 R=

W

mWb

N

37. Il96r/min 38. 14. 9

s

151 11

1

Chapter 13

31c. 1118

.32 pu, 35b. 14

V 28b. 26.8 V 28c. 72 W 29. 748.7 A. 37.6 A V 31. All thyristors will be destroyed 32a. 78 kW 1344W 32c.52kW 32d. 1344W 32e. 104 J 32f. 336 J

10. 1.3

31a.

1

yes 37. 1747 O,

28a. 4 IV, 7 30. 147.

30.90

35a. 4.39 pu,

s

W,

Chapter 21

Chapter 12

mm

8:25

1

S^J^

The LV winding is wound bn top-al V 14b. 5.31V 14c.~2300 V

winding 14a. 15. 15

36c. 5 36d. 266

mN.m

23. suggest a gear box having a ratio of

MGENiBM Chapter 20

DE

r

V

ms

21. 1.02

11

VWl^^f^

V ,67A

12. 13.88

N.m 20.56.5

0.5227 N.m, 0.07769

22d.

Chapter

18c. 11.7 micro-

22a. 200 pulses per second 22b. 22 mN.in 22c. 10.5

Chapter 10 37 72

N.m

Chapter 19 19.

O

18b. 12.2

19b. 24.8 microfarad

thejnost

V winding

34. 0.023

N.m

19a. 10.3

19b. $ 16 380

O

-1

Index

A

transient reactance,

Acceleration (of a drive system).

under load, 350

Back-to-back converters, 747, 760

voltage regulation, 352

57, 59.

Active power, 136, 140, 169, 675

Ambient temperature,

ACSR

Ampacity, 677

cable, 667.

677

Base speed,

ratio.

532

Air gap, 85

commutation, 514

Alternator, 71, 159. 335

delay,

three-phase {see Alternator,

two-phase, 160

"effective,"

firing,

335-364

Alternator, 3-phase,

brushless excitation of, 343

construction of,

505

336-340

of soft magnetic materials, 28. 29

BIL, 672, 705

750

extinction, 514,

3-phase)

B-H curve, 27 of vacuum. 27

Angle, 6

869

1

1

Basic insulation impulse level,

27

1

672, 705

Amplitude modulation

Aerial conductors. 677 Air,

B

359

Billing

750

750

demand, 734, 735 428

Bilevel drive,

Bipolar

line,

754

Bipolar winding, 424

of advance, 750

BJT,472, 516

phase, 22 torque, 359, 376.

377

Boiler,

648

Anode, 475, 492

efficiency,

elementary, 159

Apparent power, 141, 147

feed pump, 65

equivalent circuit, 346

Arc furnace, 791

Boolean language, 848

excitation of, 342

Arcing horns, 702

Boost chopper, 524

Armature

Braking, 463

cooling

of,

historical

339

example. 346

mechanical pole

power

shift,

357

of a dc generator, 73, 74, 76.

output, 358, 362

345

saturation curve,

short-circuit ratio,

synchronization

350

of.

353

synchronous reactance. 346 synchronous speed. 336. 340 torque angle, 358

86.

of a dc motor, 99, 100 1

1

Asynchronous generator, 311. 330 A u to ra n s for m er, 226-2 3 t

variable,

of a dc motor,

1

1

09

1

1

of an induction motor. 308, 309

90

reaction, 77.

647

of a synchronous motor, 383 regenerative, 312 time,

Bridge

1

1

rectifier.

brownout. 783

235

Auxiliary winding. 391. 396

879

Brush. 86, 87

480, 486-489

INDEX

880

Brushless dc motor, 569

forced,

Brushless excitation, 343, 371

line,

Brush

losses.

1

22

495

484

natural,

Buck chopper, 519

self,

1

^fcP

592

Cable, 677, 693

Compensation

i

QTECA

Coordination of protective devices, 714

1

submarine, 693

Cam switch, 44 L CANDU. 659

46

676. 720

Condenser, 649, 650

bundled, 669

Cathode, 475, 492

gauge number, 871

cemf

(see

Counter emf)

(see Central processing unit)

Current density, 121

Current (follow through), 671

Appendix AX3)

Catenary, 558

669

Crest factor, 801

Conductors, 677 {see also 1

effect,

Counter emf, 96

capacitor)

25

loss, 121

Corona

CPU

control, 831

synchronous, (see Synchronous

Carrier frequency. 530, 6

Copper

Cosine-sine conversion, 20

787

(shunt), 786,

Computer

Capacitor, 139. 145 in,

683, 688

786, 793

(series),

Capacitance (distributed), 230, 231

energy

(line),

Compensator

677

Current transformer, 23 1-234 Cutout, 715

Cycloconverter, 501, 580, 582

D

round copper, 87

Central processing unit, 833, 838

Conjugate (of vector), 151

Damper winding,

Centrifugal switch. 396. 397

Consequent poles, 304

DC DC

Celsius (degree)

Constant horsepower mode, 116

5, 7

Chain reaction, 656

Constant torque mode, 116

Characteristic impedance (see

Contact

Chopper, 5 8-52 1

1

558-560

,

Circuit

26

for,

(PLC), 844

three-phase, 158

two-phase, 160, 254 Circuit breakers air-blast,

700

minimum oil,

oil,

700

699

449

controller,

meter, 730

sulfur hexafluoride,

700

vacuum, 701, 702 Circular mil, 866

operation

Control diagram, 445, 832

properties of,

motors, 96

Conventional current flow, 15

generators, 71

Converter, 495

dc

535

Coil pitch, 89, 285, 288

four-quadrant, 526 1

13

PWM,

784

half bridge,

factor,

Distortion

529

dc-to-dc.517,522,560

dc machine. 91-93

Displacement power 490, 512

equivalent circuit of, 509

poles, 79,

476

Direct-current

Clock motor, 406, 642

Commutating

475

of,

Convection (heat loss by), 63

dc-to-ac, three phase,

84, 107

Diode

Coercive force, 32

Commutation

compound,

Contingency, 641

dc-to-ac,

169

Deuterium, 655 Differential

Appendix AX0)

797

in,

733

500 magnetic, 442

electronic,

Conversion charts, 8 (see

solid-state, 790,

HVDC

transmission)

(simulated), 833

Control system, 832

manual. 439

transmission (see

voltage and current

Contactor

solution of, 40-45, 148, 170

340, 370

576, 592, 596

Demand, 635, 730

normally open, 443 self-sealing,

equations

link,

Delta connection, 167, 169

normally closed, 443

surge impedance)

diode, 55

Cooling tower, 650

C of,

1

12-pulse, 760

Commutator, 72, 73j|6^99 Compensating winding, 14 impedance

factor of, 490, 5

two-quadrant, 525

^GENltRlA with freewheeling

overlap, 514,

Bushing, 231,232, 699

power

^3&^T

484

field, 78,

1

1

harmonic, 24, 799,810-812, 817, 823

power

factor,

490

Distribution systems

disturbances, 782

556

(mercury-arc), 757

low-voltage, 709, 717, 725

medium-voltage, 709

INDEX

three-phase, 3-wire, 719

transformation

three-phase, 4-wire, 718

unit of, 7

Disturbances on distribution


53

historical note,

Exciter, of), 57,

462

342

neutral zone,

343,371

brushless,

Drives, electronic

'-pilot,

rating,

336

shunt, 80

series motor,

559

converters with circulating current,

F

voltage,

Fahrenheit (degree), 61


Faraday, law of electromagnetic

546

560-565

electric traction,

Fast breeder reactor,

618-625

principles of, 57, 58,

462

first

Grand Coulee dam, 643 Ground

revolving, 335 Filter,

549

Fission,

592-5% wound

DSTATCON,

Ground fault Grounding

656

Flashover, 67

rotor motor,

resistance of,

597-602

1

Flux (see Magnetic flux)

of electrical systems, 7

Flux orientation, 604, 605

of equipment, 72

on

Dynamic Dynamo,

braking, 109

of gravity, 50

(see generator, dc)

unit of, 6

H

a conductor, 3

Half bridge converter, 556 Half-step drive, 425

Foucault currents, (see Eddy

E

currents)

E-business, 855

Harmonic

822

analysis)

823-827

Harmonic

analysis,

Harmonic

distortion (see

Distortion)

Fourier series analysis (see

currents, 34, 35,

1

Force, 50

DVR, 796

Eddy

Harmonics. 23, 783, 799

823-827

Effective value, 20

Freewheeling diode, 5

Efficiency, 53

Frequency, 19

and

Frequency converter, 309, 370,

and phasor diagrams, 799

ofdc machines, 123-125

Electromagnetic induction, 29

power circuits basic types, 496

Electronic

Electronvolt,

ratio,

G

in

measurement

of,

740

line,

669

Gate, 492

elimination

.

of,

square wave, 25

in

an alternator, 337

in

power systems, 812, 815, 819, 821

resonance, 813-816 Heat, 60

compound, 83

conduction

construction of, 84-90

convection, 63

differential

776, 790,818

756, 757,818

generation of, 805

Gear motor, 303 Generator, dc, 71-93

a thermal station, 65

three-phase circuits, 743

and transformers, 821

in a

Galloping

809-812

effective value of, 801

Fusion (nuclear), 661

866

consumed by a city, 738 consumed by appliances, 736 consumed in the U.S., 729

circuits, 802,

filter,

Energy, 53

in

532

analysis of,

Fuse, 715, 725

Enclosures, 299

flow

551

Frequency modulation

689

line,

19,

780, 806

of electrical machines, 362

EHV

723

GTO, 472,516

Flux vector control, 616

787, 796

Duty cycle, 520, 526

circuit breaker,

of dc terminals, 757

672, 674, 675

,

673

wire, 673, 722

481,486. 756, 757,818

Firing (see Triggering)

synchronous motor, 577-579

fault circuit

Grafcet, 847

of a dc machine, 85

induction motor, 582, 587,

Ground

breaker)

Feeder, 705, 727

quadrant control, 541

hoist control.

(see

660

Field

cycloconverter, 580, 582, 627

four-quadrant control, 549

76

Generator, ac. (see alternator)

GFCI,

induction, 29

current-fed dc link, 577 (dc motor),

76

84

separately-excited, 82

(types of ac), 575

chopper and

89

induced voltage, 75, 76, 80

eV, (see Electronvolt)

systems, 782

Drives (fundamentals

of,

compound, 84

of,

radiation of, 64

62

INDEX

882

Heat, continued (specific), (see

Table

Appendix,

AX2)

transmission

^

659

moment

to,

Historical machines, 89,

346

arresters, 671,

Hot spot temperature, 127 transmission,

life

746-765

Light water, 655

126-128

Line voltage, 166 Line commutated (see

expectancy, 126

Commutation,

Insulators

750

basic equations,

702

surge on a line, 672

Insulation classes,

pin-type,

667

Linear motion, 59

components

properties of, 869, 870

Limit switch, 442

755

harmonic rectifier

filter,

756, 757

Interpole (see

and inverter

characteristic,

scale model.

Commutating


750 757-765

self-commutated, 498, 529, 576, 592, 594

Hydrogen, 655, 869, 870 cooling, 339. 364, 386

I/O modules, 839

isotopes of, 655

Ionization,

Hydropower station, 642-646 power of, 642

reactive, 138, 154

Load duration curve, 637 Lorentz force, 31, 264 Losses machines, 120-125

in electrical

669

in

Iron losses, 33-36, 122 Isotope,

154

nonlinear, 791

1

line-commutated, 498, 503

of, 16,

active, 137, 154

poles)

Inverter (see also Converter)

752-754

typical converter stations,

Load, definition

suspension-type, 667

ground electrode, 757

natural)

Linear induction motor, 289

deterioration of, 126

bipolar line, 754 of,

276

Lightning, 670

56

353

Input module, 833, 834, 839

Horsepower, 52

flux, 199, 200,

Leakage reactance, 200, 322

56

of, 54,

Infinite bus,

Hertz, 6

1

of a synchronous machine, 336

Leakage

58,418

energy due

of electrical machines, 127-130 water, 655,

36, 183

Leading, 22, 143

effect of,

by induction, 237, 739

HVDC

of an induction motor, 285-288

in,

Inertia

Heating

Heavy

of a dc motor,

voltage induced

Inductor (see Inductance)

62

of,

00

smoothing, 748, 757

transmission lines, 676

in transformers,

655

(stray),

206, 821

821

Hysteresis, 33 loop, 33 loss,

33

motor, 405

J

M

Jogging, 450

Magnetic constant, 27

Joule, 6

field intensity, I

K

IGBT. 472, 517

kcmil, 866

flux density,

Impedance

K

levitation,

flux,

factor,

82

of ac circuits, 26, 41, 813

Kelvin, 5

per unit, 215, 216, 349

Kinetic energy, 54

ratio, 191

transformation, 192, 522

Impulse voltage, 672 Inching.

of rotary motion, 54

MCM,

141

voltage, 184

Faraday's law, 29

Inductance current

in,

36-40

energy

in,

25

force,

27

of, 5

838 (see kcmil)

Memory

(non

(volatile),

volatile),

838

838

Metals, properties of, 842

Induced voltage, 183 equality with applied

permeability, 28

Mass, unit

kVA,

450

27-29

293

Magnetomotive

of linear motion, 54

Kirchhoff (KVL, KCL), 40, 41

27

27

L

MeV,

Ladder diagram, 836, 847

Mil, 867

Lagging, 21,22, 143

Mil

Laminations, 36, 203, 265, 338

Mill,

Lap winding

Moment

of a dc generator, 75

(see

eV)

(circular),

866

735 of

inertia,

equations

for,

55

56

1

INDEX

MOSFET, 472,517

plugging

Motor, direct current, 96-1 braking, 109-111

sector type,

compound, 107

slip,

Penstock, 644 Permeability, 27, 28

and frequency, 275

rotor voltage

compound, 106 differential

607-610

265

rotating field,

569

brushless, 565, 568,

Peak inverse voltage, 476, 486

308

of,

principle of, 264,

1

of a vacuum, 27 Per unit impedance

288

of an alternator, 349

274, 291

of a transformer, 2 5-2

307, 605

mechanical power, 98

slip speed, 274,

permanent magnet,

standardization of, 299

Per unit system, 9-1

synchronous speed, 271

Phase

plugging, series,

1

1

1

10

104-105

torque, 279,

284

two speed, 303

starting of, 97, 108

98

volts per hertz rule,

wound

391-414 capacitor-run,

297

variable speed, 465

Motor, single-phase induction,

162

613

rotor, 264, 284,

336

Pilot exciter,

442 288

Pitch, 285, 286,

315

PIV,

476

PLC

(see

N

398

of,

sequence, 174-176

Pilot light,

402

capacitor-start,

meaning Phasor, 2

typical characteristics of, 276,

torque-speed curve, 116

1

angle, (see Angle)

283, 302, 308, 329, 330, 463

speed control. 100-103, 114

torque,

1

torque-speed characteristic, 281,

shunt, 103, 108

Programmable

logic

controller)

PLC

construction of, 39

National Electrical Code, 725

equivalent circuit, 409-413

Network, 665

advantages

mmf distribution,

Neutral

industrial application of,

394

of single-phase system, 225, 717

mmf, 410

of three-phase system, 164, 707,

principle of,

revolving

409

shaded-pole, 403 split-phase,

719, 812

396

zone, 76, 78, 89, 113


Newton,

Non-linear load, 804, 808

400

6,

vibration of, 401

154,

power

Synchronous motor) Motor, three-phase induction,

Plugging, 110, 308,453

17, 19, 15

1

472

of a voltage,

305

291,306 308, 309

stations,

654-661

level,

657

Power, 52

Ohm, 6

active, 136, 141, 169,

Oil (as coolant), 63

angle (see Torque angle)

Outage, 641

apparent, 141, 143

Outlet,

145,727

power

270

enclosures, 299

factor (see

45

Output module, 833, 840

in

3-phase

Over-compound

instantaneous, 134, 135, 160, 162

generator, 84

linear type,

289

mechanical power, 279

circuits,

162

of a motor, 52,58, 401

PAM

605

619

factor)

ac circuits, 146, 147

1

P

331,612 flux vector control, 616,

Power

806

in

of,

equivalent circuit, 323-325,

(flux orientation in),

472

transformer, (see Voltage

O

construction of, 263-267, 290 direction of rotation,

260

1

transformer)

basic equations, 273-275, 279,

of,

204

86, 204,

1

mark, 186

263-295

braking

coupling, 819

Potential

reactors (types of),

abnormal operating conditions, 310 as generator,

common

additive, subtractive,

P\

850

847

of a transformer,

Nuclear

Motor, synchronous (see

848

Polarity

50

Notation (E, I

of,

security rule,

Point of

torque-speed characteristic, 394,

883

„

of electrical machines, 362

measurement

motor, 305

PCC,

(see point of

of,

176

mechanical. 52

Pascal, 6

common

Peak load, 635, 636, 646

coupling)

reactive, 137, 138, triangle, 144, 148

806

INDEX

884

Power

factor, 143,

169,512.586

correction, 146, 737,

791,807

(displacement), 148,490,

490

Reactance

„

in a

three-phase machine, 265

in a

single-phase motor, 395

synchronous speed

732

without magnetic

490, 804

field, 148,

generation of, 808

base load, 637, 646

hydropower, 642-646 nuclear,

654-662

pumped

storage,

thermal model, 652

S

Salient pole,

Reactor

compensating, 691

nuclear,

of an alternator, 345 of a transformer, 206

657

smoothing, 748, 757

Scanning (of a PLC), 838

Real power (see Active power)

Scott connection, 255

Receptacle, 727

SCR

Recloser, 7

Scroll case,

Programmable logic 831-857

Rectifier (see also Converter)

controller,

1

bridge, 480,

485

controlled,

Prony brake, 53

power

Properties

three-phase, 3-pulse, 483

494 1

1

three-phase, 6-pulse, 485, 503

of insulators, 869, 870

Regenerative braking (see Braking)

407

Regulating transformer, 709

Pumped

storage,

646

Semiconductor switch, 515

controlled),

769

(control), 441, 831

overload, 44

1 ,

449

Series

motor

dc, 104

single-phase,

404

(simulated), 833

Service factor, 3 10

PWM (see pulse width modulation)

thermal, 441, 448

Servo (see Synchro)

time delay, 455, 837

SFC

Quadrant, 462

motor, 407

Quadrature component, 141

torque,

R Radiation, 64

378

Shock

(electric),

of an alternator, 360

Residual flux density, 32

of a transformer, 2

Residual magnetism, 103

protection (see Protective devices) ratio,

unit of, 6

SI units,

ground, 673

Siemens,

Resonance, 813, 816 plate,

84

236

4 6,

866

Sign notation hybrid, 45

Rheostat

nominal, 84

field, 81,

of a dc machine, 84

wound-rotor, 284

of an alternator, 342, 362

2,

350

Ramping, 422

name

1

Simulation (of relays), 833

Resistance

Rate structure, 733 Rating

719

Short-circuit

Remanent magnetism, 103

Resilient mounting, 401

emissivity, 65

(see sequential flow chart)

Shaded-pole, 403

Reluctance

Quadruple valve, 760, 761

592

Selsyn, (see Synchro drive)

Push button, 441

Q

72, 87

inverters,

Series compensation, 689, 769

Relay exciting current of, 443

602, 784

Self-commutated

Series capacitor (thyristor

714

Pulse width modulation, 530-536,

Segment (commutator),

Sequential flow chart, 847

Protective devices,

407

716

References, 859

Pull-in torque, 372,

Pull-out torque, 373,

645

Secondary winding, 185 Sector motor, 288

equivalent circuit, 5 factor of, 5

(see Thyristor)

Sectionalizer,

Programming language, 847 Programming unit, 834, 838

of conductors, 870

B-H curve)

of a dc generator, 80

Primary winding, 185

36

338

Saturation curve (see also

current limiting, 705 line

646 thermal, 646-654 transfer,

271

Sag, 669, 783

806, 808

Power generation

Power

of,

Reactive power, 137, 138

in rate structures, (total),

Rotating field

leakage, 200,217, 322

512, 804 (distortion),

of a transformer, 206

of a synchronous motor, 380

103

Ripple, 482, 486

positive and negative, 17, 19,

43,44 voltage, 17

INDEX

SIL {see Surge impedance load)

SVC

Single-phase to three-phase

Swell, 783

{see Static var compensator)

unit of, 5

Temperature

rise

by resistance method, 129

Switch

transformation, 178

of electrical machines, 125-130

702

Sine-cosine conversion, 20

air-break,

Single-phasing, 310

as non- linear load,

of insulation classes, 127, 128

805

397

866

Size of electrical machines, 130, 362

centrifugal, 396,

Slew speed, 42

disconnecting, 703

Tertiary winding, 248,

grounding, 702

Tesla, 6, 27,

Slip, 274,

291

267

Slip ring, 72, 264,

Tera,

THD

Switching losses, 528

harmonic distortion)

{see total

Snubber, 528

Synchro drive, 408

Source

Synchronization, 353

Thyristor,

(electrical diagram),

444

494

Synchronous capacitor, 385, 757

power

definition of, 16, 154

Synchronous generator, {see

principle of, 492,

61,870

Specific heat,

Synchronous motor (three-phase)

Speed, of a drive system, 57, 114

as brushless dc machine, 568,

Spillway, 644

braking

SSB, 790, 797

characteristics,

13

1

Star connection {see

Wye

Starter

446

autotransformer,

458

constant,

1

1

Torque, 51 angle, 358, 373,

376

breakdown, 282, 327

380

of a drive system, 57, 58

locked rotor, 282, 283

371,380 mechanical power, 376 power factor rating, 381 reactive power, 380

measurement

torque, 376,

pull-in, 372,

of,

pull-out, 373,

pull-up,

53

407

407

282

reluctance, 378

372


460

Time

494

493

equivalent circuit, 375

starting of,

dc motor, 108 part winding,

383

excitation of,

connection)

across-the-line,

569

construction of, 369-37

640

Stability, 359, 638,

Stabilized-shunt,

of,

gain,

properties of,

Alternator)

stations,

492

active, 137, 154

reactive, 138, 154

378

Total

harmonic

distortion, 491, 801

Transformers

primary resistance, 454

under load, 372

autotransformer, 226

reduced voltage, 454

V-curve, 382

classification of,

versus induction motor, 385

construction of, 203

solid-state,

wound

590

rotor,

284

wye-delta, 461 Start-stop stepping,

STATCOM, Static

Synchronous reactance, 346

(converter),

Synchronous speed

cooling

420

773

frequency changer, 780

Static switch,

500

757

285

Thermal generating 646-654

Symbols

307

Slip speed, 274,

of,

208

756

207

23

of single-phase motors, 393

(current),

of synchronous motors, 371

distribution,

of synchronous generators, 336

equivalent circuit, 187,202,

of 3-phase induction motors, 27

225

209, 217

Synchroscope, 354

exciting current, 197

T

grounding type, 706. 709

Stepper motor, 417-434

Taps, 205

(high-frequency), 238

Stray losses, 82

Tariff {see Rate structure)

high impedance type, 236

TCSC

ideal,

Static

synchronous compensator, 773

Static var

flux in, 185, 199

compensator, 237, 691

Stator, 263, 391

Substation, 665,

i

698,707,710

Submarine cable, 693

Surge impedance, 690, 691 Surge impedance load, 691

769

* '

-

183-195

impedance

Temperature, 60

Sulfur hexafluoride, 700, 869

Surge diverter, 702

885

of,

215-217

ambient, 127

impedance (measurement

hottest spot, 128

induced voltage, 185

rise {see

Temperature

scales, 5, 61

rise)

of),

leakage reactance, 200, 202 losses in,

206

212

INDEX

886

Transformers, continued

'

magnetizing current, 197 parallel operation of,

phase

243, 253, 256

shift,

polarity of, 186, rating of,

219

204

206

reflected impedance,

saturation curve,

" ;

191-195

rise,

V-curve, 382

surge impedance loading of, 691

Vector notation, 151

towers, 669, 673, 675

Volt, 7

types of, 665

Voltage ac, 18, 19

classes,

Traveling wave, 672

choice of transmission

Triplen, 812

effective value, 20,

209

Turbines, 639, 646, 650

induced, 30, 183

three-phase (see Transformers,

level,

Two-phase transformation, 254-256

peak, 18,20,21

peak inverse, 476, 486 Voltage transformer, 230


Transformers (three-phase)

Unified power flow controller, 776 Uninterruptible

autotransformer, 25

(see

244

delta-wye, 246

shift in,

Volt-second, 36-39, 519 Volts per hertz rule,

613

UPS)

W

Units

Ward-Leonard system, Watt,

base, 4,

conversion

243, 253, 256

of, 8,

9 (see

Appendix AX0)

phase shifting, 256 polarity of,

power supply

Unipolar winding, 425

open-delta, 248

phase

800

472-474

Two-speed motor, 303

230

delta-delta,

line,

666, 687

three-phase) (voltage),

666

'Transparent enterprise, 855

Triggering of gate, 492

234

(toroidal),

t?

Varmeter, 137, 143

206

taps, 5

temperature

T

687

submarine, 693

voltage classes, 666

r

ratio, 187, 188, 191

selection of line voltage,

260

Scott connection, 255

7,

101

52

Watthourmeter, 740 Wattmeter, 136

derived, 4, 5

Wave

in electricity, 8

Weber,

drive, 7,

424

30

in

magnetism, 8

Websites, 863

three-phase to two-phase, 254

in

mechanics, 7

Weight, (see Force of gravity)


in

thermodynamics, 7

Wheeling (charge), 640

wye-delta, 247

multiples, 7

Wire

wye-wye, 248

per-unit system, 9

Work, 51

tertiary

winding, 248, 757

Transient reactance, 359

SI,

Transmission lines dc,

746

choice of conductors, 677

components

of,

664, 667

4

Wound-rotor motor, 264, 284 as frequency converter, 315

UPS, 785 UPFC, 776 Uranium, 655

electronic control of,

enriched, 655

starting of,

interconnection of, 665

power and voltage

connection, 164, 169

voltage and current

in,

V

676 of,

680-685

597-601

284

torque-speed curve, 283

Wye

equivalent circuit, 676-678 of,

871

Universal motor, 404

dampers, 669

impedance

table,

Valve, 748,761

Z

Var, 137

Zero-speed switch, 454

169

Electrical Machines, Drives, and Power Systems 5E (Theodore Wildi)_text.pdf

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