Machine Lesson Final1

Courses In Electrical Engineering Volume IV COURSE ON ELECTRICAL MACHINES FOR TECHNICAL COLLEGE

By Jean-Paul NGOUNE DIPET I (Electrotechnics), DIPET II (Electrotechnics) M.Sc. (Electrical Engineering) Teacher in the Electrical Department, GTHS KUMBO, Cameroon.

Electrical Machines_Jean-Paul NGOUNE

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To Jesus-Christ, My Lord and Saviour.


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FOREWORD

This is the first chapter of a Course on Electrical Machines for Technical College that I am writing. The concern of this chapter is the study of Direct Current Machines. That is, DC Generators and DC Motors. The chapter begins with the study of DC generators and end with that of DC motors. Many examples are treated in order to enable the reader to assimilate easily the course. My aim is to bring my humble contribution for the improvement of Technical Education in my country (Cameroon, Central Africa) and to help anyone to whom this document may be useful. This Document and many other pedagogical resources produced by me are available and freely downloadable at the following link: www.scribd.com/jngoune. Any suggestion or critic is warmly received; send me a mail at the following address: [email protected]. Stay blessed.

NGOUNE Jean-Paul, 03 August 2012, 12:18 Kumbo, Cameroon.


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CONTENTS

1.0

Specific objectives…………………………………………………………. 6

1.1

Introduction…………………………………………………………………. 6

1.2

Structure of Direct Current machines……………………………………. 8

1.3

DC generators………………………………………………………………

11

1.3.1

Working theory of DC generators………………………………………...

11

1.3.2

Equation of induced emf…………………………………………………..

13

1.3.3

Classification of DC machines……………………………………………

14

1.3.4

Equations of Separately Excited DC generators……………………….

16

1.3.5

Equations of Self Excited DC generators……………………………….

17

1.3.6

Armature Reaction…………………………………………………………

22

1.3.7

Compensating windings…………………………………………………...

23

1.3.8

Commutating poles (Interpoles windings)……………………………….

23

1.3.9

Losses in DC generators………………………………………………….

24

1.3.10 Powers and efficiencies…………………………………………………… 26 1.3.11 DC generators characteristics…………………………………………….

29

1.4

DC motors…………………………………………………………………..

47

1.4.1

Principle of DC motors…………………………………………………….

47

1.4.2

Back emf…………………………………………………………………….

48

1.4.3

Power relationship in DC motors…………………………………………

49

1.4.4

Types of DC motors……………………………………………………….

49

1.4.5

Conditions for maximum power…………………………………………..

54

1.4.6

Torque……………………………………………………………………….

55

1.4.7

Power flow and efficiency…………………………………………………. 56

1.4.8

Speed control……………………………………………………………….

1.4.9

Speed regulation…………………………………………………………… 67

66

1.4.10 DC motors characteristics…………………………………………………

68

1.4.11 Starting and breaking methods of DC motors…………………………..

75

REVIEW QUESTIONS…………………………………………………….

84

References/ Acknowledgements…………………………………………

93

About the author……………………………………………………………

94


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Courses In Electrical Engineering Volume IV ELECTRICAL MACHINES FOR TECHNICAL COLLEGE CHAPTER ONE: DIRECT CURRENT MACHINES

By Jean-Paul NGOUNE DIPET I (Electrotechnics), DIPET II (Electrotechnics) M.Sc. (Electrical Engineering) Teacher in the Electrical Department, GTHS KUMBO, Cameroon.


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Chapter One

DIRECT CURRENT MACHINES 1.0 Specific objectives: At the end of this chapter, the student will be able to: -

Define electrical machine;

-

Describe the structure of a direct current machine;

-

Differentiate a direct current motor from a direct current generator;

-

Establish the fundamental equations for the various types of direct current motors an generators;

-

Plot interpret the characteristic curves of direct current motors and generators for different types of test (open circuit characteristics, external characteristics);

-

Give specific applications for each type of direct current motors and generators;

1.1 Introduction: Electrical machines are converters that are used to continuously translate electrical input to mechanical output or vice versa. This process of translation is known as an electromechanical energy conversion. If the conversion is from mechanical energy to electrical energy, the electrical machine is said to be a generator; if it is from electrical energy to mechanical energy, the electrical machine is a motor. The electromechanical conversion results basically from the two following electromagnetic phenomena: 

When a conductor moves in a magnetic field, voltage is induced across the conductor (Generator action);


6



When a current carrying conductor is placed in a magnetic field, the conductor experiences a mechanical force (Motor action). In p u t

O u tp u t M o to r

E le c tric a l e n e rg y

M e h a n ic a l e n e rg y

In p u t

O u tp u t G e n e ra to r

M e c h a n ic a l e n e rg y

E le c tric a l e n e rg y

Figure 1.1: Energy conversion for motor and generator Field

M

Armature

Figure 1.2: Symbol of a DC machine

The two electromagnetic phenomena mentioned above occur simultaneously whenever energy conversion takes place from electrical to mechanical or vice versa. In motoring action, the electrical system makes current flow through conductors that are placed in the magnetic field. A force is produced on each conductor. If the conductors are placed on a structure free to rotate, an electromagnetic torque will be produced, making the rotating structure (rotor) to rotate at a given speed. If the conductors rotate in a magnetic field, a voltage will also be induced across each of them. In generating action, the rotating structure is driven by a prime mover such as a steam turbine or a diesel engine. A voltage is induced in the conductors that are rotating with the rotor. If an electrical load is connected to the windings formed by those conductors, and electrical power will be produced to supply it. Moreover, the current flowing in the conductors will interact with the magnetic field to produce a reaction torque, which will tend to oppose the torque applied by the prime mover.


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1.2 Structure of direct current machine:

The following figure presents the main parts of a direct current machine.

Figure 1.3: Structure of a direct current machine.

Figure 1.4: Photographic view of a DC machine


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The essential parts of a DC generator consist of: 

Magnetic frame or yoke,



Pole coils or field coils,



Armature windings,



Brushes and bearings,



Pole cores and pole shoes,



Armature core,



Commutator.

Of these, the pole core, the armature core, the yoke and the air gaps between the poles and the armature cores form the magnetic circuit whereas the rest form the electrical circuit. 

Yoke: The yoke is the outer frame of the machine. It provides mechanical support for

the poles and also canalizes the magnetic flux produced by the poles. It acts as a protective covering for the whole machine. The pole is made of cast iron or cast steel. 

Pole cores and pole shoes: The pole shoes help to support the field coils and also to spread out the

magnetic flux in the air gap. The pole cores and pole shoes are made up of steel plates laminated and retrieved together. They are bolted to the yoke. The purpose of laminating the core is to reduce Eddy current losses. 

Pole coils or field coils: The field coils consist of insulated copper wires wound round the pole cores.

When current is flowing through these wires, they magnetise the poles which produce the necessary flux that is cut by the rotating armature.

Figure 1.5: Field coil of a DC machine Electrical Machines_Jean-Paul NGOUNE

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

Armature core: The armature core houses the armature conductors in the slots. It is built up of

laminated steel disc mounted on a shaft.

Figure 1.6: Photographic view of an armature of DC machine 

Armature windings: Two types of windings are mostly employed for the armatures of DC

machines; they are: Lap Winding and Wave Winding. The difference between the two is merely due to the different arrangement of the connections at the front or commutator end of the armature. The following rules apply to the types of windings: -

In a lap winding, the number of parallel paths is always equal to the number of poles and also the number of brushes.

-

In wave windings, the number of parallel paths is always two and there may be two or more brush positions.

The lap winding is suitable for high current, low voltage machines like welding plants. The wave winding is suitable for high voltage, low current machines, like DC generators used for lighting. 

Brushes: The brushes are used to collect current from the commutator in the case of

generator. In the motor, they lead current into the armature windings through the commutator. The brushes are made of carbon. Electrical Machines_Jean-Paul NGOUNE

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

Commutator: The function of the commutator in a generator is to convert the alternating

current induced in the armature conductor into direct current in the external load circuit. The commutator has a cylindrical structure and is built up of insulated copper segments.

Figure 1.7: Photographic view of commutator.

1.3 DC generators. When a conductor is moving in a magnetic field, an emf e is induced across its terminals. That emf is given by the Faraday’s law as follows: e

d  BLv dt

e = induced emf in volts (V) Where

B = magnetic flux density (T) L = Length of conductor cut by the flux (m) v = Speed of the conductor (m/s)

DC generators are built using this basic principle.

1.31 Working theory of DC generators: DC generator is actually an ac machine which is furnished with a special device: the commutator. The commutator is therefore a rectifying unit in a dc generator as it converts the alternating current generated in the armature to a direct current at brushes.


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Generators are driven by some mechanical means like petrol or diesel engine, hydropower, steam engine etc. The brushes always keep the same polarity despite the fact that the emf induced in the armature windings changes his polarity twice per period.

Figure 1.8: Simple loop generator.

Ua

Emf induced in the armature windings 0

t

Ub

Emf received on the brushes

0

t

Uc

Direct voltage after filtering and stabilisation

0

t

Figure 1.9: Voltages at armature and at brushes


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1.3.2 Equation of induced emf. Let us assume that:   Flux per pole;

Z = Total number of armature conductors; P = Number of pairs of poles; N = Armature speed in rps; E = emf induced in one of the parallel path; A = Number of parallel path (A = 2 for wave winding, A = 2P for lap winding). According to the Faraday’s law of electromagnetic induction, we have: e

d . dt

For one revolution, each conductor is cut by a flux under the north pole and the and the south pole (2P). Therefore, the change in flux is d    2 P

The average emf induced per conductor is e  For one revolution, the time taken is dt  Hence

d   2 P  dt dt

1 n

d   2 P     2P  n . 1 dt n This is the emf generated across each armature conductor. Per path, the

number of conductors is given as the total number of armature conductors divided by the number of parallel paths, that is

Z . Therefore, the emf generated per path is A

given by: E    2P  n 

E

 E

Z  N  Z    2P       A  60   A 

ZN 2 P  60 A

For wave wound machine, A = 2, hence ZNP  ZnP 60


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

For lap wound machine, A = 2P, hence

E



ZN  Zn 60

Application exercise.

A 6 pole DC generator has 1530 armature conductors. The machine is nun at 2000 rpm and the useful flux per pole is 10 mWb. Calculate the emf generated in the following cases: -

The machine is wave wound;

-

The machine is lap wound.



Solution to the application exercise.

Data   10mWb; P 

6  3; N  2000; Z  1530 2

Case 1: wave winding E

ZNP 1530  2000  10  103  3   1530V 60 60

Case 2: Lap winding E

ZN 1530  2000  10  103   510V 60 60

1.3.3 Classification of DC machines. The field circuit and the armature circuit of a DC machine can be interconnected in various ways to provide a wide variety of performance characteristics and outstanding advantage of DC machines. Also, the field poles can be excited by two field windings, a shunt field winding and a series field winding as shown in the following figure.

Figure 1.10: Shunt and series field windings for a DC generator.


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The shunt winding has a large number of turn and takes only a small current (less than 5% of the rated armature current). The resistance of the shunt winding is far greater than that of series winding. The shunt winding is connected in parallel with the armature. The series winding has fewer turns but carries a large current. It is connected in series with the armature In short we can say that there are two main groups of DC machine: separately excited DC machines and self-excited DC machines. 

Separately excited DC machines. In the separately excited DC machine, the field winding is excited from a

separate source (Figure 1.11a). 

Self-excited DC machines. For this type of DC machine, the field winding can be connected in three

different ways: -

In series with the armature resulting in a series DC machine;

-

In parallel with the armature resulting in shunt DC machine;

-

Both shunt and series windings may be used, resulting in compound machine. There are two types of compound DC machines: long shunt and short shunt. The figure 1.11 below show the classification of DC machines.

Figure 1.11: Classification of DC machines Electrical Machines_Jean-Paul NGOUNE

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A rheostat is normally included in the circuit of the shunt winding to control the field current and thereby to vary the field mmf. Field excitation may also be provided by permanents. This maybe considered as a form of separately excited machine, the permanent magnet providing a separate but constant excitation.

1.3.4 Equations of separately excited DC generators. Let us consider the figure below presenting the electrical diagram of a separately excited DC generator.

If

I U

Uf Rh

RA

Rf E

Let us assume the following notations: If = Field current; Rf = Resistance of the field winding; Uf = Field circuit voltage; E = Emf generated; U = Terminal voltage; Ra = resistance of the armature winding; Ia = Armature current; I = line or load current; Rh = Rheostat.

We can deduce the following formula:

If 

Uf R f  Rh

; U  E  Ra.Ia  E  Ra.I


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1.3.5 Equations of self-excited DC generators.

The field winding of a self excited generator is energized by the current produced by the generator itself. In this machine, residual magnetism must be present in the machine iron to get the self excitation process started. When the armature is rotated, some emf is produced which goes through the field coils and strengthens the residual magnetism.

a. Shunt excited DC generator. Ish

I U

Ia

U Rsh Ia  I  Ish U  E  RaIa Ish 

RA

Rh E




A 4 pole lap wound DC shunt generator has a useful flux per pole of 0.07Wb. The armature winding consists of 440 conductors having a total resistance of 0.055Ω. Calculate the terminal voltage when running at 900rpm if the armature current is 50A. 

Solution to the application exercise

Data: P 

4  2 ; Z  440 ; Ra  0.055 ; I a  50 A ;   0.07Wb ; lap winding (A =2P). 2 Is h

I U

Ia RA

Rh E

We know that U  E  RaIa Let us first determine the emf E of the generator. Since the machine is lap wound, we have: E 

ZN 60


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E

440  900  0.07  462V 60

Then we have U  E  RaIa  462  0.055  50  459.25V

Exercise 1.1 An 8 pole DC shunt generator with 778 wave connected armature conductors and running at 1000 rpm supplies a load of 12.5Ω resistance at a terminal voltage of 250V. The armature resistance is 0.24Ω and the field resistance is 250Ω. Determine: a. the armature current; b. The induced emf; c. The flux per pole.

b. Series excited DC generator. I

Rs U

Ia

I  Ia U  E  Ra  Rs I

RA

E




An 8pole DC series generator with 778 wave wound armature conductors running at 1000 rpm supplies a load of 20Ω.The useful flux per pole amount to 4mWb. The armature and series field resistances are 0.08Ω and 0.12Ω respectively. Determine: a. The emf generated; b. The armature current; c. The terminal voltage. 

Solution to the application exercise.

Data: P 

8  4 ; Z  778 ; N = 1000rpm; R = 20; Ra = 0.08Ω; Rs = 012Ω; Wave 2

winding: A = 2.   4mWb


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I

Rs U

Ia RA

E

a. Emf generated. E

ZNP 778  1000  0.004  4   207.46V 60 60

b. Armature current E  Ra  Rs  R Ia  0  Ia 

E 207.46   10.27 A Ra  Rs  R 0.08  0.12  20

c. Terminal voltage.

U  E  Ra  Rs Ia  R.Ia  205.4V

Exercise 1.2 A series generator supplies a load of 10Ω and the line current is 25A. The armature and field resistances are 0.08Ω and 0.12Ω respectively. Determine: a. The terminal voltage; b. The emf generated.

c. Compound excited generator. We will give the equations for long shunt connected, and then for short shunt connected DC compound generators.

i)

Long shunt DC compound generator. Is h

I Ia

Rs

Ia RA

Rh E


U

I sh 

U Rh

I a  I  I sh

U  E  I a Ra  Rs 

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ii)

Short shunt DC compound generator.

Ish

Rs

I

Ia

U

I sh 

I a  I sh  I

Ra

Rsh

U  E  Ra I a  IRs

E



U  IRs Rsh

Remark

In DC machines, there is usually a voltage drop Ub at the brushes. If mentioned, it should be taken into account in the total drop.

Ub Ish

Rs

I

Ia

U U  E  Ra I a  Rs I  U b 

Ra

Rsh

E



Application exercise

A 4 pole, long shunt, compound generator supplies 100A at a terminal voltage of 500V. If armature resistance is 0.02Ω, series field resistance is 0.04Ω and shunt field resistance 100Ω, find the generated emf. Take drop per brush as 1V. 

Solution Is h

I Ia

Rs

Ia RA

Rh E

U

I  100 A Ra  0.02 Rs  0.04 U  500V Ub  1V Rh  100

E  Ra  Rs Ia  2Ub  U (We have two brushes, so the drop due to the brushes is 2Ub) Electrical Machines_Jean-Paul NGOUNE

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 E  U  Ra  Rs Ia  2Ub But Ia  I  Ish  I 

U 500  100   105 A Rh 100

Finally,

E  500  0.02  0.04105  2  508.3V

Exercise 1.3 A long shunt compound generator supplies a load of 10kW at 250V. The armature, series and shunt resistances are 0.08Ω, 0.03Ω and 100Ω respectively. Calculate the emf generated.

Exercise 1.4 Let a short shunt compound generator supplying 100A at a terminal voltage of 230V. The armature, series and shunt resistances are 0.02Ω, 0.05Ω and 200Ω respectively. Calculate the field current and the emf generated. 

Remark: Parallel operation of shunt generators.

The connection of generators in parallel means that they are connected to the same load, usually through a common busbar system. To connect two or more shunt generators to the same busbar, two conditions must be observed: -

The emf of the incoming generator should be practically equal to the busbar voltage V.

-

Like polarities should be connected together.

Exercise 1.5 A 4 pole, DC shunt generator with a shunt field resistance of 100Ω and an armature resistance of 1Ω has 378 wave-connected conductors in its armature. The flux per pole is 0.01Wb. If a load resistance of 10Ω is connected across the armature terminals and the generator is driven at 1000rpm, calculate the power absorbed by the load.


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1.3.6 Armature reaction (AR) By armature reaction is meant the effect of magnetic field set up by armature current on the distribution of flux under main poles. With no current flowing in the armature, the flux in the machine is established by the mmf produced by the field current. However, if the current flows in the armature circuit, it produces its own mmf (hence flux) that opposes the flux produces by the field current under main poles. The armature magnetic field has two effects: -

It demagnetises or weakens the main flux;

-

It cross-magnetises or distorts the main flux.

a)

With no armature current, the flux under pole is not distorted.

d) With current flowing in its conductors, the armature produces its own flux that distorts the flux under main poles.

Figure 1.12: Armature reaction.


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1.3.7 Compensating windings These are used for large direct current machines which are subjected to large fluctuations in load, such as rolling mill motors and turbo-generators. Their function is to neutralise the cross magnetizing effect of armature reaction. In the absence of compensating windings, the segments of the commutator may be short-circuited during rapid changes in the load of the machines.

1.3.8 Commutating poles (Interpoles windings) To counter the effect or armature reaction in medium and large-power DC machines, a set of commutating poles (sometimes called interpoles windings) is always placed between the main poles as shown by the figure below.

Figure 1.13: Commutating poles

These narrow poles carry windings that are connected in series with the armature. The number of turns on the windings is designed so that the poles develop a magnetomotive force mmfc equal and opposite to the magnetomotive force mmf a of the armature. As the load current varies the two magnetomotive forces rise and fall together exactly bucking each other at all time. By nullifying the armature mmf in this way, the flux in the space between the main poles is always zero, solving then the problem of armature reaction.


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Exercise 1.6 The following information is given for a 300kW, 600V, long shunt compound generator: shunt field resistance = 75Ω, armature resistance including brush resistance = 0.03Ω, commutating field winding resistance = 0.011Ω, series field resistance = 0.012Ω, divertor resistance ( in parallel with series field resistance) = 0.036Ω. When the machine is delivering full load, calculate the voltage and the power generated by the armature.

Exercise 1.7 A 4 pole, lap wound DC shunt generator has a useful flux per pole of 0.07Wb. The armature winding consists of 220 turns each of 0.004Ω resistance. Calculate the terminal voltage when running at 900 rpm if the armature current is 50A.

Exercise 1.8 A 4 pole, DC shunt generator with a shunt field resistance of 100Ω and an armature resistance of 1Ω has 378 wave-connected conductors in its armature. The flux per pole is 0.01Wb. If a load resistance of 10Ω is connected across the armature terminals and the generator is driven at 1000 rpm, calculate the power absorbed by the load.

1.3.9 Losses in DC generators The three main losses in DC generators are copper losses, iron losses and mechanical losses. 

Iron losses

These losses are due to the rotation of the iron core of the armature in the magnetic field produced by the poles. Iron losses consist of hysteresis and Eddy Current losses. i)

Hysteresis losses

These losses are due to the reversal of magnetism of the armature core. That is the energy required to magnetize and demagnetize the armature core as it passes through the magnetic flux of the north and the south pole. ii)

Eddy current losses

As the armature core rotates, its conductors cut the magnetic lines of flux produced by the main poles. An emf is therefore induced in those conductors and a current Electrical Machines_Jean-Paul NGOUNE

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known as Eddy current (In French it is called ‘courrants de Foucault’) is circulating through them. These currents cause losses called Eddy current losses. To minimize theses losses, the core should be laminated. 

Copper losses

They cause heat in the machine. They are due the resistance of the windings found in the machine. Hence, we have armature copper losses (RaIa2), shunt copper losses (RshIsh2), series copper losses (RsIs2) 

Mechanical losses

Mechanical losses consist of friction losses at the bearings of the rotating armature and also of windage losses. Mechanical losses and iron losses are known as constant losses.

The following chart summarises the losses mentioned above. Armature copper loss Copper losses Field copper loss Useful output Input Hysterisis loss Total losses

Iron losses Eddy current loss

Friction loss Mechanical losses Windage loss

Figure 1.14: Losses in DC generators


25

1.3.10 Powers and efficiencies

a) Powers Let us assume: Po = Output power (also called useful power or electrical power); Pin = Input power (power absorbed from the prime mover); Pm = Mechanical power developed in the armature; PJ = Total copper losses; Pc = Constant losses (iron and mechanical losses). The power distribution can be sketched as follows.

Pm

Pin

Pc

Po

Pj

 2N  Pin  T .   T  60 

Where T = Shaft torque of the prime mover in (N.m). N = Rotating speed in rpm.

 = Angular speed in (rad/s). Pm  Pin  Pc  E.Ia

Where E = emf generated in (V) Ia = armature current (A) Po  Pm  PJ  UI Where U = terminal voltage in (V) I = line current in (A).


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b) Efficiencies

m 

e 





Mechanical efficiency



Electrical efficiency



Overall efficiency

Pm Pin

Po Pm

Po P P  m  o   m  e Pin pin Pm



Remark

Efficiency is maximal when Joule losses equal constant losses (PJ = PC). 

Application exercise

A 220 DC generator runs at 1500rpm and supplies a load of 15A. If the input torque is 28N.m, calculate the efficiency. 

Solution

By definition,  

Po Pin

Po  UI  220  15  3300W But

2  3.14  1500  2N  Pin  T .  4396W   28  60  60 

Hence,  

3300  0.75068 pu  75.068% 4396

Exercise 1.10 A long shunt compound wound generator running at 16.67rps supplies 11kW at a terminal voltage of 220V. The resistances of the armature, shunt and series field windings are 0.05ΩΩ, 110Ω, and 0.06Ω respectively. The overall efficiency is 81.5%. Determine: a) The total copper loss; b) The iron and friction losses; Electrical Machines_Jean-Paul NGOUNE

27

c) The torque exerted by the prime mover; d) The electrical and mechanical efficiencies.

Solution to exercise 1.10 Is h

n  16.67 rps P  11kW

I Rs

Ia

U

U  220V Ra  0.05 Rsh  110

Ia RA

Rs  0.06

Rh E

  0.815

a) Total copper losses PJ  Ra  Rs I a2  RshI sh2

But I a  I sh  I 

U P 220 11000     2  50  52 A Rsh U 110 220

Hence, PJ  0.05  0.06  52 2  110  22  737.44W b) Iron and friction losses Pm

Pin

Pc

Po

Pj

PC  Pin  Pm Po 11000   13496.93252W  0.815 But Pm  Po  PJ  11000  737.44  11737.44 Pin 

Hence, Pc  13496.93  11737.44  1759.43W c) Torque exerted by the prime mover Pin  T  2n  T  T 

Pin 2N

13496.9325  128.925 N .m 2  3.14  16.67

d) Electrical and mechanical efficiencies 

Electrical efficiency


28

e 

Po 11000   0.9371 pu  93.71% Pm 11737.44

 m 

Mechanical efficiency Pm 11737.44   0.8696 pu  86.96% Pin 13496.9325

Exercise 1.11: (Probatoire F3 2010) An asynchronous three phase motor drives a shunt generator which supplies in full load a current of 40A under a voltage of 320V. The useful power of the driving motor is equal to 20.614kW at full load; its armature resistance is 1.25Ω and its field resistance is 200Ω. Determine: a) The useful power of the generator; b) The current in the field circuit and in the armature; c) The emf of the generator; d) The constant losses.

Exercise 1.12: Compound wound generator (Probatoire 2008) A shunt generator is to be transformed into a short shunt compound generator by addition of series field windings. A test carried out only on the shunt field gave the following results: 

A current of 4.5A produces a voltage of 250V at no load;



A current of 5.5A produces the same voltage at full load of 40A.

The shunt windings have 1200 turns. a) calculate the number of series turns necessary to maintain this voltage ; b) Armature, shunt field and series field resistances are respectively 0.5Ω, 70Ω, and 0.3Ω. The generator supplies a load of 5kW at a voltage of 250V. Calculate: i)

The emf of the generator;

ii)

The full load efficiency, if constant losses are estimated to 120W;

iii)

The efficiency at 3/4full load.

1.3.11 DC generator characteristics The important characteristics of DC generators are: 

The no load characteristics;



The load characteristics;


29



Emf vs. speed characteristics.

a) No load characteristics (open circuit characteristics): E  f ( I f ) The connection diagram used to record data for the plot of no load characteristics is the following, no matter the type of generator.

I=0 A

If

E

Eo Rh

Rf

G

V

As the value of rheostat is varying, so the field current is varying and its value is recorded from the ammeter. The speed is kept constant and the generated emf on no load (I = 0) is measured by the voltmeter V. The corresponding values of E and If are recorded and the graph E  f ( I f ) can be plotted. Mathematically the relationship between the emf E and the field current If can be shown. E  Zn

But the flux is function of the field current:   kI f Hence, E  ZnkI f  f I f . The general appearance of the open circuit characteristics for DC generators is as follows. E(v)

Eo 0

If(A)

Figure 1.15: No load characteristics of DC generators. Electrical Machines_Jean-Paul NGOUNE

30

The difference OA represents the emf generated due to the residual magnetism in the poles. Exercise 1.13: (Probatoire 2011). The no load test of a DC generator functioning at separated field has given the following results at the constant speed N = 1500rpm. I(A)

0

0.05

0.1

0.15

0.2

0.25

0.3

0.4

0.5

0.6

0.7

E(V)

10

40

70

100

135

165

195

220

230

235

237

The field resistance is r = 100Ω; the armature resistance R = 0.5Ω. The machine is used as a shunt generator. 1. Establish the no load characteristics E = f(i). 2. Determine the emf due to the residual magnetism in the poles; 3. Calculate the value of field rheostat in order to have an emf at no load of E = 220V. 4. The field current is and has its value of question 3. The generator supplies a load with a constant current of I = 49.6A. Calculate. a) The total current supplied by the generator; b) The voltage across the generator; c) The total copper losses; d) The useful power of the generator; e) The power absorbed by the generator; f) The efficiency of the generator knowing that constant losses are equal to 100W. Solution of exercise 1.13 1. Plot of the no load characteristics Scale: 1cm  0.1A (Abscissa); 1cm  10V (Ordinate). See the following page. 2. Emf due to the residual magnetism in the poles It is the emf when the field current is equal to zero. According to the curve below, we have Eo = 10V. 3. Value of the field rheostat in order to have an emf at no load of 220V. Rh I f  R f I f  E  Rh 

E  Rf If


31

For E = 220V, If = 0.4A Hence, Rh 

220  100  450 0 .4

E(V)

250

200

100

10 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

1.0

I(A)

4. a) Total current supplied by the generator. I a  I  I f  49.6  0.4  50 A

b) Voltage across the generator. VG  E  Ra I a  220  0.5  50  195V c) Total Copper losses Pm

Pin

Pc

Po

Pj

PJ  Pm  P0  EI a  VG I  220  50  195  49.6  1328W d) Useful power of the generator.


32

Po  VG I  195  49.6  9672W e) Power absorbed by the generator. Pa  Pm  EI a  220  50  11000W f) Efficiency of the generator. 

p0 Po 9672    0.87135 Pu  87.135% Pin Pm  PC 11000  100

Exercise 1.14: Shunt generator The open circuit characteristics at 1000rpm of a 4 pole, 220V shunt generator with 72 slots and 8 conductors per slot with armature conductors lap connected is as follows: Field current (A) 0.25 0.5 emf (V)

25

50

1

2

3

4

5

100 175 220 245 255

The field circuit resistance is 75Ω. 1. Explain how this test was carried out 2. Plot the curve and determine: a) The emf induced due to residual magnetism; b) The emf generated at the given field resistance when the generator is under normal operation. 3. The useful flux per pole. 4. The residual flux. Solution of exercise 1.14 Data: N = 1000rpm; p 

4  2 ; U = 220V; Z = 72x8 = 576; lap connection; 2

Rsh = 75Ω; shunt generator. 1. Test procedure for the plotting of open circuit characteristics. The connection of the DC generators for the determination of the open circuit characteristics is as follows. 0A

If A

E + Rh

G

V

-


33

The field current If is varied rheostatically and its value measured by an ammeter. The speed si kept constant and the generated emf in the load is measured by the voltmeter V. The corresponding values are recorded and the graph of E = f(If) is plotted. 2. Plot of the curve. The curve is sketched on the page below (the curve normally should be plotted son a graph paper. Scale: 1A = 1cm (abscissa) 10V = 1cm (ordinate) E(V) 260

T

S

150

R

A 10 0

If (A) 1

2

3

4

5

6

7

a) To know the value of the emf induced due to the residual magnetism, we just have to project the curve back ward to cut the ordinate axis (point A). We obtain E0 = 10V b) The emf for a field resistance of 75Ω To know the maximum emf the generator will generate on normal operation, we should draw the shunt resistance line.


34

To draw the shunt resistance line, take any value of If (for example, let us take 2A), multiply this value by the shunt resistance Rsh = 75Ω. Mark the corresponding point on the ordinate axis. Let that point be R. 75x2 = 150V, hence, R(2A, 150V). Draw the line joining the origin O and the point R, it cuts the open circuit characteristics at the point S. Draw a horizontal line from S to T. OT gives the maximum emf generated with 75Ω as shunt resistance. From the curve we can read: OT = E =210V (almost). 3. Useful flux per pole. E

ZN 2 p ; Lap winding  2 p  A  60 A

ZN 60 E  60 ZN Hence, 60  210   21.875mWb 576  1000 E

4. Residual flux. E  E0  10V    1.04mWb .

b) Load (or external) characteristics U = f(I)

i)

Separately excited generator.

Figure 1.16: Separately excited generator under load.

We assume that the separately excited generator is driven at constant speed and that the field winding is excited by a battery. The exciting current is constant and so is the resultant flux. The induced voltage Eo is therefore fixed. When the machine Electrical Machines_Jean-Paul NGOUNE

35

operates at no load, the terminal voltage E12 is equal to the induced voltage Eo because the voltage drop in the armature resistance is zero. However, if we connect a load across the armature, the resulting load current I produces a voltage drop across the armature resistance Ra and causes the terminal voltage to drop. As we increase the load, the terminal voltage also decreases progressively. The load characteristic of a separately excited generator is presented in the following figure.

Figure 1.17: Load characteristics of a separately excited generator

ii)

Shunt generator

When a shunt generator is loaded, after voltage build up, the terminal voltage U drops with increase in load current I. The decrease depends on the armature drop IaRa and it is however small except the machine is overloaded. Ish

I U

Ia RA

U  E  Ra I a

Rh E


36

The sketch of the load characteristics of a shunt generator is as follows. U(V) E Un

0 I(A)

Figure 1.18: load characteristics of a shunt generator.

The terminal voltage of a self-excited shunt generator falls off more sharply with increasing load than that of a separately excited generator. The reason is that the field current in a separately excited machine remains constant, whereas in a self-excited generator, the exciting current falls as the terminal voltage drops ( I ex 

U ). For a self-excited generator, the drop in voltage from no load to full load Rsh

is about 15 percent of the full load voltage, whereas for a separately excited generator, it is usually less than 10%. The voltage regulation is said to be 10 percent and 15 percent respectively. iii)

Series generator. RS

RA

I U LOAD

E

When the switch S is closed with the load resistance R comparatively large, the machine does not excite; but as R is reduced, a value is reached when a slight reduction of R is accompanied by a relatively large increase of terminal voltage. When the machine is on open circuit, the terminal voltage is very small (voltage due to the residual magnetic field in the poles). The sketch of the load characteristics of a series generator is presented in the following figure. Electrical Machines_Jean-Paul NGOUNE

37

U(V)

0 I(A)

Figure 1.19: Load characteristics of a series generator.

Because of the rising voltage characteristics of DC generators, they are mostly used as boosters on DC power lines to compensate for voltage drop. Series wound generator is quite unsuitable when voltage is to be maintained constant or even approximately constant over a wide range of load current.

iv)

Compound generators

The compound generator was developed to prevent the terminal voltage of a DC generator from decreasing with increasing load (as it is the case for shunt generator. Thus, although 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 distribution system of a ship supplies power to both DC machinery and incandescent lamps. The current delivered by the generator fluctuates continually, in response to the varying loads. These current variations produce corresponding changes in the generator terminal voltage, causing the light flicker. Compound generators eliminate this problem A compound generator is similar to a shunt generator, except that it has additional field coils connected in series with the armature. These series field coils are composed of a few turns of heavy wire, big enough to carry the armature current. The total resistance of the series coils is therefore small. The figure bellow is a schematic diagram showing the series and shunt field connections


38

Figure 1.20: Shunt and series field connections of a compound generator.

When the generator runs at no load, the current in the series coils is zero. The shunt coils, however, carry exciting current Ix which produces the field flux, just as in a standard self excited shunt generator. As the generator is loaded, the terminal voltage tends to drop, but load current Ic now flows through the series field coils. The mmf developed by these coils acts in the same direction as the mmf of the shunt field. Consequently, the field flux under load rises above its original no-load value, which raises the value of Eo. If the series coils are properly designed, the terminal voltage remains practically constant from no-load to full load. The rise in the induced voltage compensates for the armature drop. In some cases we have to compensate not only for the armature voltage drop, but also for the IR drop in the feeder line between the generator and the load. The generator manufacturer then adds one or two extra turns on the series winding so that the terminal voltage increases as the load current rises. Such machines are called over compound generators. If the compounding is too strong, a low resistance can be placed in series with the series field (The name of this resistance is diverter resistance). This reduces the current in the series field and has the same effect as reducing the number of turns. For example, if the value of the diverter resistance is equal to that of series field, the current in the latter is reduced by half. In a differential compound generator, the mmf of the series field acts opposite to the shunt field. As a result, the terminal voltage falls drastically with increasing load. We can make such a generator by simply reversing the series field of a standard compound generator. Differential compound generators were formerly used Electrical Machines_Jean-Paul NGOUNE

39

in DC arc welders, because they tended to limit the short circuit current and to stabilize the arc during the welding process.

Figure 1.21: Load characteristics of compound generators

c) Saturation curve of a DC generator. The saturation curve is the plot of the field flux under poles  against the exciting current Ix. When the exciting current is relatively small, the flux is small and the iron in the machine is unsaturated. Very little mmf is needed to establish the flux through the iron. Because the permeability of the air is constant, the flux increases in direct proportion to the exciting current, as shown by the linear portion Oa of the saturation curve.

Figure 1.22: Saturation curve of a DC generator. However, as we continue to raise the exciting current, the iron in the poles and the armature begins to saturate. A large increase in the mmf is now required to Electrical Machines_Jean-Paul NGOUNE

40

produce a small increase in flux, as shown by the portion bc of the curve. The machine is now said to be saturated. Saturation of the iron begins to be important when we reach the so-called “knee” ab of the saturation curve.

Figure 1.23: Saturation curve of a DC generator. Since the emf is directly proportional to the flux under the poles, the saturation curve can also be seen as the plot of the emf Eo at no load against the exciting current Ix.

Figure 1.24: Saturation curve of a DC generator.

d) Controlling the voltage of a shunt generator.

It is easy to control the induced voltage of a shunt excited generator. We simply vary the exciting current Ix by means of a rheostat connected in series with the shunt field. The following circuit presents such a connection.


41

a

x

n p R m Ix

Eo

G

b y

Figure 1.25: Voltage control of a shunt generator.

To understand how the output voltage varies, suppose that Eo is 120V when the movable contact p is in the centre of the rheostat. If we move the contact towards the extremity m, the resistance Rt between p and b decreases, which causes the exciting current to increase. This increases the flux and, consequently, the induced voltage Eo. On the other hand, if we move the contact towards the extremity n, Rt increases, the exiting current decreases, the flux decreases, and so Eo will fall. We can determine the no load value of Eo if we know the saturation curve of the generator and the total resistance Rt of the shunt field circuit between p and b. We draw a straight line corresponding to the slope of Rt and superimpose it on the saturation curve. This dotted line passes through the origin, and the point where it intersects the curve yields the induced voltage. For example, if the shunt field has a resistance of 50Ω and the rheostat is set at the extremity m, then Rt = 50Ω. The line corresponding to Rt must pass through the coordinate point E = 50V, I = 1A. This line intersects the saturation curve where the voltage is 150V (see figure of the following page). That is the maximum voltage the shunt generator can produce. By changing the setting of the rheostat, the total resistance of the field circuit increases, causing Eo to decreases progressively. For example, if Rt is increased to 120Ω, the resistance line cuts the saturation curve at a voltage Eo = 120V.


42

Figure 1.26: The no load voltage depends upon the resistance of the shunt field circuit.

If we continue to raise Rt, a critical value will be reached where the slope of the resistance line is equal to that of the saturation curve in its unsaturated region. When this resistance is attained, the induced voltage suddenly drops to zero and will remain so for any Rt greater than this critical value. The critical resistance corresponds to 200Ω.


43

Exercise 1.15: (Probatoire 2009) A DC generator has the following magnetisation characteristic at 1200rpm Field aimed

1

2

4

6

8

10

Emf (V)

192

312

468

566

626

660

1. Draw the magnetisation curves: a) at 1200rpm; at 1000rpm. Take scale 20mm = 100V; 10mm = 1A. 2. If the generator is shunt excited and driven at 1000rpm, determine: a) The voltage at which it will build up on open circuit. b) The value of the critical resistance of the shunt field circuit at 1000rpm. c) The terminal potential difference and the load current for a load resistance of 30Ω. Note: Field resistance = 60Ω and armature resistance = 0.5Ω

Solution of exercise 1.15: 1. Plot of the magnetisation curves. For the same value of the field current, the variation of the emf is proportional to that of the speed. Hence, emfs at 1000rpm can be deduced from those at 1200rpm using the following reasoning. E2 N 2 N   E2  2 .E1 E1 N1 N1

Where: E2 = emf at 1000rpm, N2 = 1000rpm, N1 = 1200rpm, E1 = emf at 1200rpm. We obtain the following table of values Iex(A)

1

2

4

6

8

10

E(V) at

192

312

468

566

626

660

160

260

390

471.67

521.67

550

1200rpm E(V) at 1000rpm

The magnetisation curves are then plotted on the following figure.


44

E(V) 700V N = 1200rpm 600V Eo

N = 1000rpm

500V

400V Eo = 60If 300V

200V

100V If(A) 0

1

2

3

4

5

6

7

8

9

10

2. a) The open circuit build up voltage is situated at the intersection of the curves

E  f (I f ) E  60 I f  R f .I f (resis tan celine) So, if we draw a line passing between the origin of the axis and the point having the coordinates E = 60V, I = 1A; that line cut the magnetisation curve at a point whose ordinate is the open loop build up voltage. We find Eo = 550V b) The critical resistance is the resistance for which the resistance line has the same slope with the linear part of the saturation curve. For Rf = 160Ω, the resistance line E  R f I f has the same slope with the linear part of the saturation curve. Hence, the

critical resistance is Rf =160Ω. Electrical Machines_Jean-Paul NGOUNE

45

c) Determination of load current and load voltage. These quantities are such that: Ish

I U

Ia RA

Rh E

U  Rf I f U  RI U  E  R I  I f 

U  60 I f U  30 I U  550  0.5I  I f 



And we find after calculations the following results U = 536.58V; I = 17.89A.

f) Emf vs speed characteristics. The emf generated is directly proportional to the speed (provided the flux is maintained constant). Any increase in speed is accompanied by an increase in generated emf. E

ZN 2 P  60 A

Since P and A are constant quantities, we deduce that: E  N 

E1 E2 E   ...  n N1 N 2 Nn

So, the emf vs speed characteristics can be sketched as follows. E(V)

0

Nc

N(rpm)

Figure 1.27: Emf vs speed characteristics


46

Below a certain speed (Known as the critical speed Nc), no emf is generated. So the characteristic is not supposed to start at the origin of the axis.

1.4 DC motors. 1.4.1 Principle of DC motor A DC motor is a machine which converts electrical energy into mechanical energy. Its action is based on the principle that when a current carrying conductor is placed in a magnetic field, it experiences a mechanical force (called Laplace’s force).

Current carrying conductor.

B

I

F

Figure 1.28: Current carrying conductor in a magnetic field The magnitude of the force F is given by the following formula F  BIL

Where B =Flux density of the magnetic field; I = Current flowing in the conductor; L = Length of the conductor.

There is no basic difference in construction between DC generator and DC motor. In fact, the same machine can theoretically be used interchangeably as a generator and as a motor. The main difference between the two machines is that


47

when the machine is motorizing, the generated emf is less than the terminal voltage meanwhile in a generator, the generated emf is greater than the terminal voltage.

1.4.2 Back emf When the motor’s armature rotates, the conductors wound on it cut the magnetic flux under the poles. An emf is therefore induced in them. The direction of this induced emf, known as back or counter emf is such that it opposes the applied voltage. Since the back emf is induced due to the generator action, the magnitude of it is therefore given by the same expression as for DC generators.

Eb 

Where:

ZN 2 P  60 A

Eb = Back emf in volts; Z = Number of armature conductors; N = Rotational speed of the armature in rpm; Ф = Flux per pole in Webers; P = Number of pairs of poles; A = Number of parallel paths in the armature (A = 2 for wave winding; A = 2P for lap winding).

The armature of a DC motor is hence equivalent to a source of emf Eb in series with a resistance Ra. The supply voltage across the armature should therefore be large enough to balance both the voltage drop in the armature and the back emf all the time. Ia U Ra

U  Eb  Ra I a

Eb


48

1.4.3 Power relationship in a DC motor The voltage equation of a DC motor is: U  Eb  Ra I a Multiplying each term of the voltage equation by Ia, we get: UI a  Eb I a  Ra I a2

The equation above is known as the power equation of the DC motor. The term UIa represents the power supplied to the motor armature, Ra I a2 represents the power lost in the armature, EbIa represents the mechanical power developed in the armature causing the rotation of the armature. The power developed EbIa is not all available at the shaft since some of it is used to overcome the mechanical power losses of the motor.

1.4.4 Types of DC motors Different types of Dc motors are the following: 

Permanent magnet DC motors;



Separately excited DC motor;



Series wound DC motors;



Shunt wound DC motors;



Compound wound DC motors.

a) Permanent magnet DC motors It consists of an armature and one or several permanent magnet encircling the armature. Field coils are not usually required. However, some of these motors do have coils wound of the poles. If they exist, these coils are intended only for recharging the magnets in the event that they loose their strength. Permanent magnet DC motors work like separately excited motors.

b) Separately excited DC motors For these motors, field coils and armature conductors are supplied by different sources


49

Ia

Rh

U Ra Field E

Eb

Figure 1.29: Separately excited DC motor. U  Eb  Ra I a  Eb  U  Ra I a Pm  P  PJ  UI a  Ra I a2  U  Ra I a I a  Eb I a

c) Series wound DC motors In series motor, field and armature circuits are connected in series, as shown in the figure 1.30 below; so Ia = If. Rs

I= Ia =If U

Ra Eb

Figure 1.30: DC series motor.

The field coils consist of a few turns of thick wires. The cross sectional area for the wire of the coils has to be fairly to carry the armature, but owing to the large current, the number of turns of wire in them need not to be large. U  Eb  RS  Ra I Power drawn from the main: P  UI

Mechanical power developed: Pm  P  Ploss Electrical Machines_Jean-Paul NGOUNE

50

Pm  UI  I 2 Ra  RS   I U  Ra  RS   Eb .I

The speed of DC series motor is very high at no load and very low at heavy load. The torque is very high at low speed. Therefore DC series motor is suitable for duties requiring a large starting torque and also frequent starting. Applications for DC series motor are: railway traction, hoist, crane…

d) shunt wound DC motor The armature circuit and the shunt field circuit are connected across a DC source of fixed voltage U. An external field rheostat is sometimes used in the field circuit to control the speed of the motor.

I Ia

Ish

U

Ra Rsh Eb

Figure 1.31: DC shunt motor.

I sh 

U Rsh

I a  I  I sh U  Eb  Ra I a Power drawn from the main: P  UI

Mechanical power developed:

Pm  P  Ploss

Pm  UI  UI sh  Ra I a2  U I  I sh  Ra I a2  UI a  Ra I a2  I a U  Ra I a   I a Eb An important characteristic of shunt motor is that it has a fairly constant speed for a fairly wide range of loads. Shunt motor applications are found in driving shafts, machine tools, blowers…


51

e) Compound wound DC motor. Compound wound DC motor has both series winding and shunt field winding. It is usually connected long shunt. Rs Ish

U

Ia Ra

Rsh Eb

Figure 1.32: Compound wound DC motor.

I sh 

U Rsh

I a  I  I sh

U  Eb  Ra  Rs I a Power drawn from the main: P  UI

Mechanical power developed by the motor Pm  P  Ploss

Pm  UI  UI sh  Ra  RS I a2

 U I  I sh  Ra  RS I a2  UI a  Ra  RS I a2

 I a U  Ra  Rs I a  Pm  I a Eb Compound motor combines the characteristics of shunt and series motor, and finds applications in elevators, hoists, frequent starting duties such as refrigerators and air compressors.

Exercise 1.16 A 200V series DC motor with armature and field resistance of 0.5Ω and 0.3Ω respectively draws a current of 45A. Calculate the emf generated.


52

Solution of exercise 1.16 Rs

I= Ia =If

U  200V Ra  0.5

U

RS  0.3

Ra

I  45 A Eb

Eb  U  Ra  RS I

 200  0.5  0.3 45  164 Eb  164V

Exercise 1.17 An 11.19 kW shunt motor draws 51A from the supply. The armature and field resistances are 0.1Ω and 240Ω respectively. The motor efficiency is 91.4%. Calculate: a) The terminal voltage; b) The emf generated.

Solution of exercise 1.17

I Ia

Ish

Ra Rsh

U Po  11.19kW ; I  51A; Ra  0.1A; Rsh  240;  91.4%

Eb

a) Terminal voltage: Pin  UI  U 

But  

Pin I

Po P  Pin  o Pin 


53

Hence, U 

Po 11190   240.056V I 51  0.914 b) Emf generated.

U  Eb  Ra I a  Eb  U  Ra I a But I a  I  I sh  51 

240  50 A 240

Hence, Eb  240  0.1  50  235V

Exercise 1.18 A compound motor running from a 400V DC supply draws a current of 102A. The armature, series and shunt field resistances are 0.08Ω, 0.04Ω and 200Ω respectively. Calculate the generated emf.

Solution of exercise 1.18 Rs Ish

Ia Ra

U

U  400V ; I  102 A; Ra  0.08; RS  0.04 Rsh  200

Rsh Eb

U  Eb  Ra  RS I a  Eb  U  Ra  RS I a I a  I  I sh  I 

U 400  102   100 A Rsh 200

Hence, Eb  400  008  0.04  100  400  12  388V

1.4.5 Condition for maximum power. The mechanical power developed by the motor is: Pm  UI a  I a2 Ra

Differentiating both sides with respect to Ia we have: dPm  U  2 I a Ra  0 dI a Electrical Machines_Jean-Paul NGOUNE

54

Then, I a Ra 

U 2

As U  Eb  I a Ra and I a Ra  Then U  Eb 

U 2

U U  Eb  2 2

Thus mechanical power developed by a motor is maximum when back emf is equal to half the applied voltage. This condition is, however, not realised in practice, because in that case current will be much beyond the normal current of the motor. Moreover, half the input would be wasted in the form of heat and taking other losses into consideration (mechanical and magnetic), the motor efficiency will be well below 50 percent.

1.4.6 Torque. By the term torque is meant the turning or twisting moment of a force about an axis. It is measured by the product of the force and the radius at which this force acts. Consider a pulley of radius r meters acted upon by a circumferential force F Newton which causes it to rotate at n rps.

r

F Then, torque T  F  r Work done by this force in one revolution = Force x distance = F * 2r Power developed in one revolution: P

W F  2r T 1sec ond , with F  and t   t t r n

T  2r 2N r  2n  T   T    T With N in rpm and  in rad/s Then, P  1 60 n


55

We the finally deduce that: T 

60  P 2N

The torque is expressed in Newton metre (N.m)

1.4.7 Power flow and efficiency. DC motors, like DC generators, experience three main types of losses which influence the power flow. These losses are: 

Mechanical losses (friction and windage losses);



Copper losses;



Iron losses (Eddy current losses)

The power flow chart of a DC motor can be drawn as follows:

Pm Pin

Po

Pj

Pc

Figure 1.33: Power flow chart of a DC motor. a) Powers

Pin  UI

Where: U = Terminal voltage (V); I = line current (A); Pin = Input power (W). Pm  Eb I a  2n  Tm Where: Eb = Back emf (V); Electrical Machines_Jean-Paul NGOUNE

56

Ia = armature current (A); n = motor speed in rps Tm = armature torque, also called total torque or gross torque (N.m) Pm = Mechanical power developed in the armature (W). Also: Pm  Pin  Pj

With Pj = Total copper losses. Po  2n  T  Pm  Pc  BHP  735.5 Where: Pc = Iron and mechanical losses; BHP = Brake horse power; 1BHP = 746W (British) Po = Motor output power (W); T = Shaft torque. The torque which is available for doing the useful work is known as shaft torque T. It is so called because it is available at the shaft. The horsepower obtained by using the shaft torque is called Brake Horse Power (BHP) because it is the horse power available at the brake.

2n  T 735.5 735.5  B.H .P T  2n B.H .P 

The difference Tm  T is known as lost torque. The lost torque is absorbed by the mechanical losses: Pc  2n  TL  TL 

Pc 2n

Where TL is the loss torque in (N.m).


57

b) Efficiencies 

Mechanical efficiency:

m 



e 

 

Po Pm

Electrical efficiency: Pm Pin

Overall efficiency: Po   m  e Pin

Remark: Calculation of armature torque Pm  2nTm  Eb I a ZN 2 p   Ia 60 A 2p  2nTm  Zn   Ia A Finally ,  2nTm 

Tm    I a 

Zp A

Exercise 1.19: A 4 pole DC shunt motor has a lap connected armature with 60 slots, each slot containing 20 conductors. The useful flux per pole is 23 mWb and the armature current is 50A. Calculate the torque developed in the armature.

Solution of exercise 1.19: 2p  4  p  2 Lap  A  2 p  4 Z  60  20  1200   23mWb; I a  50 A Electrical Machines_Jean-Paul NGOUNE

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I Ia

Ish

U

Ra Rsh Eb

Tm    I a 

Zp A

Tm  23  10 3  50 

1200  2  219.7452  220 N .m 3.14  4

Exercise 1.20: In a brake test on a DC motor, the following information was recorded: 

Tachometer reading:1000rpm;



Electrodynamometer reading:50N.m;



Voltmeter reading: 220V



Ammeter reading (line current): 27.27A.

Determine for the motor the following: a) The output power; b) The efficiency.

Solution of exercise 1.20: a) Output power: 2N T 60 2  3.14  1000  50 Po  60 Po  5233.33W Po  2n  T 

b) Efficiency: 

Po P 5233.33  o   0.8723 pu  87.23% Pin UI 220  27.27


59

Exercise 1.21: A no load test on a DC shunt motor produces the following results: 

Supply voltage: 400V;



Line current:6.83A;



Speed: 12.5rps.

The armature and field resistances are 0.8Ω and 120Ω respectively. Calculate the motor efficiency and speed when the armature current is 30A.

Solution of exercise 1.21 U  400V I 0  6.83 A n0  12.5rps Ra  0.8 Rsh  120 I a  30 A

I Ia

Ish

U

Ra Rsh Eb

1. Motor efficiency By definition, the efficiency is found as follows: 

P0 Pin

Po and Pin being respectively the input and the output power of the motor. Pin  UI

 U I a  I sh   U  400    400 30   U  I a   Rsh  120     13333.33W Pin  13333.33W Electrical Machines_Jean-Paul NGOUNE

60

Po  Pin  PL , PL being the total loss of the machine. Let us calculate PL. 

Constant losses ( iron and mechanical losses)

The armature input at no load is equal to the constant losses. I ao  I o  I sh  I o 

U 400  6.83   3 .5 A Rsh 120

Where Io and Iao are respectively the line current and the armature current of the motor at no load. Hence the constant losses can be deduced. Pc  UI ao  400  3.5  1400W 

Copper losses

Shunt copper losses: PJsh  Rsh I sh2  120  3.332  1330.66W

Armature copper losses PJa  Ra I a2  0.8  30 2  720W

Total copper losses PJ  PJsh  PJa  1330.66  720  2050.66W The total loss can therefore be deduced: PL  PC  PJ  1400  2050.66  3450.66W The output power is then: Po  Pin  PL  13333.33  3450.66  9882.67W We can therefore deduce the efficiency 

Po 9882.66   0.7411 pu  74.11% Pin 13333.33

2. Speed when the armature current is 30A. For shunt motor, the back emf is proportional to the speed: E

ZN 2 P   KN  N ,  being a constant, since the flux  is constant for a 60 A

shunt motor. 

Back emf at no load:

E0  U  Ra I ao  400  0.8  3.5  397.2V 

Back emf at 30A:

E1  U  Ra I a  400  0.8  30  376V Electrical Machines_Jean-Paul NGOUNE

61

Then, due to the proportionality relationship between speed and back emf for shunt motor, we can write: E0 n0 E n   n1  1 0  11.75rps E1 n1 E0

As the armature current increases, the speed decreases (due to the decreasing of the back emf).

Exercise 1.22: A 500V, 50b.h.p. (37.3kW), 1000rpm DC shunt motor has on full load an efficiency of 90 percent. The armature circuit resistance is 0.24Ω and there is total voltage drop of 2V at the brushes. The field current is 1.8A. Determine: a) Full load current; b) Full load shaft torque in N.m; c) Total resistance in the motor starter to limit the starting current to 1.5 times the full load current.

Solution of exercise 1.22:

I Ia

Ish

U

Ra Rsh Eb P0  37.3kW  50 BHP; U  500V ; N  1000rpm;  90%; Ra  0.24;U drop  2V ; I sh  1.8 A

a) Full load current. Let us first determine the motor input: Pin 

Po 37300   41444W  0.9

Then the full load current can be deduced:


62

I

Pin 41444   82.9 A U 500

b) Full load shaft torque Po 

2N 60  Po 60  37300 T  T    356 N .m 60 2N 2  3.14  1000

c) Starter resistance. The starting line current is 1.5I, that is: I st  1.5  82.9  124.3 A Then the armature current at starting is: I ast  I st  I sh  124.3  1.8  122.5 A At starting, the back emf is equal to zero (because no speed), then we have: U  I ast Rst  Ra  U drop

Rst being the starter resistance in series with the armature. We deduce: Rst 

U  U drop I ast

 Ra

500  2  0.24 122.5 Rst  3.825



Exercise 1.23: A 4 pole, 220V shunt motor has 340 lap-wound conductors. It takes 32A from the supply mains and develops 7.5h.p. (5.595kW).The field winding takes 1A. The armature resistance is 0.09Ω and the flux per pole is 30mWb. Calculate: (i) The speed and (ii) the torque developed in N.m.

Solution of exercise 1.23: i)

Determination of the speed:

Data: 2 P  4; Z  340; I  32 A; Po  5.595kW , I sh  1A;  30mWb; Ra  0.09 Eb 

ZN 2 P ZN , lap winding  2 P  A  Eb   60 A 60

Hence, N 

60  Eb Z

But Eb  U  Ra I a  U  Ra I  I sh   220  0.0932  1  217.2V Then we have: Electrical Machines_Jean-Paul NGOUNE

63

N

60  Eb 60  217.2   804.4rpm  13.4rps Z 540  0.03

ii)

Torque developed in rpm:

Po  2n  T  T 

Po 5595   66.5 N .m 2n 2  13.4

Exercise 1.24: A DC series motor takes 40A at 220V and runs at 800rpm. If the armature and series field resistances are 0.2Ω and 0.1Ω respectively and the iron and friction losses are 0.5kW, find the torque developed in the armature. What will be the output of the motor? Solution of exercise 1.24: Rs

I= Ia =If U

Ra Eb

Data: I  I a  40 A;U  220V ; N  800rpm; Ra  0.2; Rs  0.1; Pc  0.5kW i)

Determination of the armature torque (or gross torque):

Pm  2n  Ta  Ta 

Pm EI  a 2n 2n

Eb  U  I a Ra  RS   220  400.2  0.1  208V

But,

n

800 40  rps 60 3

Then, Ta  ii)

208  40  99.3 N .m 2  40 / 3

Motor output:

Po  Pin  Ploss 

Armature and series field joule losses.

PJ  I 2 Ra  RS   40 2 0.3  480W


64



Iron and friction losses:

PC  500W Then the total loss can be deduced: Ploss  PJ  PC  480  500  980W 

Motor input:

Pin  UI  220  40  8800W Finally, the motor power input. Po  Pin  Ploss  8800  980  7820W

Exercise 1.25: A 4 pole, 240V, Wave connected shunt motor gives 11.19kW when running at 1000rpm and drawing armature and field currents of 50A and 1A respectively. It has 540 conductors. Its resistance armature is 0.1Ω. Assuming a drop of 1V per brush, find: (i) total toque (ii) useful toque (iii) useful flux per pole (iv) rotational losses (v) efficiency.

Solution of exercise 1.25: Data: i) Ta 

2 P  4;U  240V ; Po  11.19kW ; N  1000rpm; I a  50 A; I sh  1A; Z  540; Ra  0.1;U drop  1V  2. Total torque (armature or gross torque):

Pm E I  b a 2n 2n

But Ea  U  Ra I a  U drop  240  50  0.1  2  233V Then, Ta  ii) Tu 

Useful torque (or shaft torque):

Po 11190   106.9 N .m 2n 2  3.14  50 / 3

iii)

Useful flux per pole.

Eb  Zn 



233  50  111N .m 2  3.14  50 / 3

2p E A Eb , because A = 2 (wave winding)   b  A Zn  2 p Zn  2

233  12.9mWb 540  50 / 3  2 


65

iv)

Rotational losses

Let us consider the power flow chart of a DC motor:

Pm Pin

Po

Pj

Pc

According to the chart, PC  Pin  PJ  Po  Pdrop 



Armature copper losses:

PJa  Ra I a2  0.1  50 2  250W



Shunt copper losses

PJsh  UI sh  240  1  240W 

Brush contact loss

Pdrop  U drop  I  2  50  100W



Motor input

Pin  U I  I sh   24050  1  12240W 

Motor output

Po  11.19kW  11190W Then we can deduce the rotational losses Pc  12240  250  240  100  11190  460W v) 

Efficiency.

Po 11190   0.9142 pu  91.42% Pin 12240

1.4.8 Speed control: Speed control means intentional change of the drive speed to a value required for performing a specific work process. The concept of speed control or adjustment should not be taken to include the natural change in speed which occurs due to change in load on the drive shaft. Speed can be controlled manually or by some automatic control device.


66



Speed control by mechanicals:

Speed can be adjusted mechanically by means of: -

Stepped pulleys;

-

Set of change gears;

-

Variable speed friction clutch mechanism… 

Electrical speed control:

ZN 2 p  60 A  60 A  Eb E    N     b   2 pZ   Eb 

But Eb  U  Ra  R I a Hence, N  

U  Ra  R I a 

The above expression reveals that the speed can be controlled by adjusting any one of the three factors appearing on the right hand side of the expression: i)

The line voltage U;

ii)

The external resistance in the armature circuit R;

iii)

The flux per pole 

1.4.9 Speed regulation: The speed regulation is defined as the change in speed when the load on the motor is reduced from rated value to zero, expressed as percent of the rated load speed. Speed regulation and speed control should not be confounded.

% Speedregulation 

N .L.speed  F .L.speed  100 F .L.speed

Where N.L.speed = No load speed and F.L.speed = Full load speed.


67

1.4.10 DC motor characteristics: The performance and, therefore, suitability of a DC motor is determined from its

characteristics

known

as

performance

characteristics.

The

important

characteristics of a DC motor are: i)

Torque-Armature current characteristic: Tm  f I a 

This characteristic curve gives the relation between the torque developed in the armature and the armature current. It is also known as electrical characteristic. ii)

Speed-Armature current characteristic: N  f I a 

This characteristic curve gives the relation between the speed N and the armature current Ia. This is also known as speed characteristics. iii)

Speed-Torque characteristics: N  f Tm 

This characteristic curve gives the relation between the speed N and the armature torque Tm. This is also known as mechanical characteristics.

1.4.10.1 Characteristics of DC series motors: a) Magnetic characteristic: Flux

Saturation Knee

0

Ia

Figure 1.34: Saturation curve of a DC series motor. The flux  varies with the variation in the armature current as the field winding is in series with the armature. The flux first increases following a linear law with the increasing of armature current. At the “knee”, the variation stops being linear. It is the beginning of the saturation. Finally, the flux remains almost constant as the armature current is increasing.


68

b) Speed-Armature current characteristic. It is shown that: E

ZN 2 p  Z  2 p      N  N , where  is a constant. 60 A  60 A 

For DC series motor, we have: U  Eb  Ra I a  Eb  U  Ra I a  N  U  Ra I a N

U  Ra I a 

The armature resistance, added to the series field resistance causes an important voltage drop. On the other hand, the flux being directly proportional to armature current makes that the speed is roughly inversely proportional to the armature current. That is the speed decreases with the increasing of the armature current. The speed characteristic of DC series motor is as follows:

N

0

Ia

Figure 1.35: Speed characteristic of a DC series motor.

If the load falls to a very small value (the current is too small). The speed may become dangerously high. That is why a DC series motor should never be started on no-load.

c) Torque-Armature current characteristic. We have shown that: Pm  Eb I a  2n  Tm  Tm   .I a 

Z P  k . .I a A


69

In series motor, the flux is proportional to the armature current up to full load (before saturation). Hence, Tm  k k I a .I a  k .k I a2   .I a2  Tm  I a2

Therefore, on light load (before saturation), mechanical torque Tm is proportional to the square of the armature current and hence the curve drawn between torque and armature current is a parabola as shown in the figure below. After saturation point, the flux is almost independent of excitation current and so, the torque is directly proportional to the armature current. Hence the characteristic become a straight line. Tm

0 B e fo re s a tu ra tio n : P a ra b o la .

S a tu ra tio n : S tra ig h t lin e .

Ia

Figure 1.36: Electrical characteristic of DC series motor.

d) Speed-Torque characteristic. E.I a  2n  Tm  n 

E .I a 2Tm

From this relationship, we deduce that speed is inversely proportional to the torque. At low torque, speed is very high and at high torque, speed is very low.


70

n

0

Tm

Figure 1.37: Mechanical characteristic of DC series motor.

1.4.10.2 Characteristics of DC shunt motors: a) Speed-Armature current characteristic. N

U  Ra I a k

For shunt motor, the flux is not affected by the armature current and the value of the armature drop RaIa is too small (usually bellow 5% of U). So, the speed does not drop too much with the increasing of the armature current. N

0

Ia

Figure 1.38: Speed characteristic of DC shunt motor.

b) Torque-Armature current characteristic: Tm   .I a 

Zp  k . .I a A

The flux being almost constant for a shunt motor, we deduce that torque is directly proportional to armature current. Electrical Machines_Jean-Paul NGOUNE

71

Tm

0

Ia

Figure 1.39: Electrical characteristic of Shunt motor.

c) Speed-Torque characteristic. The torque being proportional to the armature current, we deduce the speedtorque characteristic will have the same aspect with the speed-armature current characteristic. n

0

Tm

Figure 1.40: Mechanical characteristic of DC shunt motor.

1.4.10.3 Characteristics of compound DC motors: Since the compound motor has a combination of shunt and series excitations, its characteristics are intermediate between those of shunt and series motors. We distinguish to main types of DC compound motors: differential or subtractive compound motors and cumulative or additive compound motors.


72

a) Differential or subtractive compound: In differential compound DC motors, the series field winding is connected in such a way that the series field opposes the shunt field. Since the flux decreases with the increase in load so the speed remains nearly constant as the load is increased and in some cases, the speed will increase even. The characteristics of differential compound DC motors are similar to those of shunt motor. Since the shunt motor develops a good torque and almost constant speed, therefore, differential motor is seldom used. b) Cumulative or additive compound DC motors: In these types of motors, the series field is made to assist the shunt field. These machines are used to drive heavy machine tools like punches, elevators, conveyors… The characteristics of compound motors are intermediate between those of series motors and those of shunt motors. N

Shunt m otor

Com pond m otor

Series m otor

0 Ia

Figure 1.41: Speed characteristic of DC cumulative compound motor. T

Compound motor Series motor

Shunt motor

0

Ia

Figure 1.42: Electrical characteristic of DC cumulative compound motor.


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Exercise 1.26: A series motor runs at 600rpm when taking 110A from a 230V supply. The resistance of the armature circuit is 0.12Ω and that of the series winding is 0.03Ω. Calculate the speed when the current has fallen to 50A, assuming the useful flux per pole for 110A to be 0.024Wb and that for 50A to be 0.0155Wb.

Solution of exercise 1.26: Data: N1  600rpm;U  230V ; Ra  0.12; Rs  0.03;1  0.024Wb;0  0.0155Wb. What should be noticed here is that for series motor, speed is no more proportional to the emf as it was the case for shunt motor. Rs

I= Ia =If U

Ra Eb

Let us first determine the emf when for the two values of the armature current: U  Eb  Ra  Rs I a  Eb  U  Rs  Ra I a 

For Ia = 110A

Eb1  230  0.12  0.03 110  213.5V 

For Ia = 50A

Eb 0  230  0.12  0.03 50  222.5V It is known that, Eb 

ZN 2 p  Z  2 p  E    N  kN  k  b 60 A  60  A  N

K is a constant, it means that it does not change when the flux or the current changes, provided that we are in the linear zone of the magnetization curve of the motor; hence: k

Eb1 213.5   14.82 N11 600  0.024

Then, for with Ia = 50A, the constant k does not change and we have:


74

Eb 0  kN 00  N 0 

Eb 0 222.5   969rpm k0 14.82  0.0155

1.4.11 Starting and breaking methods of DC motor:

1.4.11.1 Starting methods: The following methods of starting DC motors are in use: 

Direct on line starting (DOL starting);



Resistance starting;



Reduced voltage starting.

a) DOL starting: This method is also called full voltage starting. In this method, the motor is directly connected to the supply, and we have: Ia 

U E Ra

Because the armature resistance Ra is very low, usually less than 1Ω, the current will be excessively high at starting. This current can blow fuse and damage the motor. Therefore a DC motor must never be started on line. A starting resistor must always be inserted in series with the armature in order to reduce the starting current.

Remark: Calculation of the starting resistance Rst. Let us consider a shunt motor where we have inserted in series with the armature a starting resistor. Is h

I R st

U

Ra Rsh E

Rst is the starting resistance. Let us denote the armature current at start as I ast. At start, the speed is equal to zero, so the emf is also equal to zero, therefore, we have: Electrical Machines_Jean-Paul NGOUNE

75

E  0  U  I ast Ra  Rst  

U  Ra  Rst I ast

 Rst 

U  Ra Iast

Exercise 1.27: The starter of a Peugeot 504 (a series motor) takes 200A under a voltage of 12 V. Under these conditions, it runs at 1000rpm and delivers a useful power of 1500W. The constant losses are estimated at 100W. Calculate: a) The power absorbed by the starter. b) Its efficiency. c) Its useful torque. d) The joule losses. e) The total resistance (armature + field windings). f) The back emf of the motor. g) Determine the direct starting current. h) What is the value of the resistance to be connected in series with the motor to limit the starting current to 240A?

Solution of exercise 1.27: The stator of the Peugeot 504 is a series motor having the following characteristics: I = 200A; U = 12V; N = 1000rpm; Pu = 1500W; Pc= 100W. Rs

U Ra Eb

a) Power absorbed by the starter. P  UI  12  200  2400W

b) Efficiency. 

Pu 1500   62.5% P 2400

c) Useful torque. Pu  2nT 

2N 60 Pu 60  1500 T T    14.33 N .m 60 2N 2  3.14  1000

d) Joule losses. Let us first draw the power stages chart. Electrical Machines_Jean-Paul NGOUNE

76

Pm Pin

Po = Pu

Pj

Pc

From the chart, we have:

Pj  Pin  Pm  Pin  Pc  Po   2400  1500  100  800W e) Total resistance (armature + field) Pj  ( Ra  Rs ) I 2  Rs  Ra 

Pj 800   0.02 2 I 200 2

f) Bock emf. U  Eb  Rs  Ra I  Eb  U  Ra  Rs I  12  0.02  200  8V g) Direct starting current. At starting, Eb = 0, hence, I st 

U 12   600 A Ra  Rs  0.02

h) Value of the resistance to be connected in series with the motor to limit the starting current to 240A.

Rst

Rs U

Ra Eb With Eb = 0 at starting, we have: I st 

U 0 U 12  Rst   Ra  Rs    0.02  0.03 . Ra  Rs  Rst I st 240

Exercise 1.28 A 30kW, 400V shunt motor has an armature resistance of 0.15Ω. The efficiency of the motor is 79%. Calculate the value of a series resistor which will limit the starting current to 1.5 of its full load value. Calculate the current the motor will take if started on line.


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Other starting methods for DC motors are the following:

b) Resistance starting: As studied above, the starting current will be reduced if a rheostat is inserted in series with the armature. That is resistance starting.

c) Reduced voltage starting: In high power motors, the starting rheostat becomes bulky and significant losses in energy take place in them. In such case, special systems like reduced voltage method of starting are used. This method consists in varying the voltage delivered in the motor in starting from some lower limit to the rated voltage. The Ward-Leonard System is one of the most used reduced voltage starting system.

1.4.11.2 Breaking methods: There are two main breaking methods of DC motors: 

Mechanical of frictional method;



Electric breaking.

a) Frictional breaking. This method is common but it has one inconvenient in that it is difficult to achieve smooth stop. b) Electric breaking. We have three main electrical method of breaking: 

Rheostatic or dynamic method of breaking;



Plugging of motors;



Regenerative breaking.

Exercise 1.29: (Probatoire 2012). A series excited DC motor is perfectly compensated at the rated working temperature. The resistance of the field winding is r = 0.03Ω and the resistance of the armature is R = 0.082  . The flux is proportional to the armature current. For the full load: The supply voltage U = 720V; the armature current I = 340A; the speed of revolution n = 1150rpm; the useful torque Tu = 1800Nm. Calculate: 1. The useful power and efficiency;


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2. The total joule losses; 3. The back electromotive force and the electromagnetic torque; 4. At starting, the motor is supplied under reduced voltage; the starting current is 600A. Determine the supply voltage at starting. 5. The machine is run at a high speed by reducing by half the rated field flux by shunting the shunt winding; the rated supply voltage remaining unchanged, the rotative speed moves to 2300rpm. a) Draw the corresponding diagram; b) Calculate the armature current.

Solution of exercise 1.29: Data: r  0.03; R  0.082;U  720V ; I  340 A; n  1150rpm; Tu  1800 Nm.

r

I U

R Eb

1. The useful torque: Pu 

2n 2  3.14  1150  Tu   1800  216660W  216.66kW 60 60

We can then deduce the motor efficiency: 

Pu P 216660 216660  u    0.885 pu  88.5% Pa UI 720  340 244800

2. The total Joule losses.

PJ  r  R I 2

PJ  0.03  0.082 3402  12947.2W  12.95kW 3. Back emf and electromagnetic torque: Eb  U  r  R I

Eb  720  0.03  0.082  340  681.92V Electrical Machines_Jean-Paul NGOUNE

79

The electromagnetic torque can therefore be deduced: 2n 60  Eb I Tm  Tm  60 2n 60  681.92  340 Tm   1925.24 Nm 2  1150 Eb I 

4. Supply voltage at starting: At starting, the emf is equal to zero, since the speed is also equal to zero, hence: U S  r  R I S  0.02  0.082 600  67.2V 5. The machine is run at high speed by reducing by reducing by half the rated flux. a) The corresponding diagram: Rsh r

I U

R Eb

b) The armature current I2:

Eb  kn I   I1  1; I 2  2  I1  1; I 2 

1 1 2

I 1 1 But, 2  1  1  2 2 I 2 1  1 2 I 340  I2  1   170 A 2 2

Exercise 1.30: A 460V series motor runs at 500rpm taking a current of 40A. Calculate the speed and percentage change in torque if the load is reduced so that the motor is taking 30A. Total resistance of the armature and field circuits is 0.8Ω. Assume flux is proportional to the field current. Electrical Machines_Jean-Paul NGOUNE

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Solution of exercise 1.30: We have shown that: Pm  Eb I a  2n  Tm  Tm   .I a 

Z P  k . .I a A

In series motor, the flux is proportional to the armature current up to full load (before saturation). Hence, Tm  k k I a .I a  k .k I a2   .I a2  Tm  I a2

And   I a Then,

T1  402 T2  302 

T2 900 9 9    T2  T1 T1 1600 16 16

Percentage in torque is: T1  T2  100  T1

Now

9 T1 16  100  7  100  43.75% T1 16

T1 

Eb1  460  40  0.8  428V

Eb 2  460  30  0.8  436V

But, we know that E  kn  n 

E k

  I a

Then, n 

E E E   k I a  k I a I a

Eb 2 n E I I So, 2  a 2  b 2  a1 Eb1 n1 Eb1 I a 2 I a1 Then,

n2 436 40    n2  679rpm 500 428 30


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Exercise 1.31: The input to a 220V, DC shunt motor is 11kW. Calculate (a) the output power (b) the efficiency (c) the speed at this load (d) the torque developed. The particulars of the motor are as follows: No load current = 5A; No load speed = 1150rpm; Armature resistance = 0.5Ω; Shunt field resistance = 110Ω. Solution of exercise 1.31:

I Ia

Ish

U

Ra Rsh Eb

a) Output power: No load input: P0  UI 0  220  5  1100W . I sh 

U  2A Rsh

I a 0  I 0  I sh  5  2  3 A

No load armature Joule loss: PJ 0  I a20 Ra  32  0.5  4.5W Then, constant losses are: Pc  P0  PJ 0  1100  4.5  1095.5W Now, when input is 11kW: Input current: I 

Pin 11000   50 A U 220

Armature current: I a  I  I sh  50  2  48 A Armature Joule loss: PJa  I a2 Ra  482  0.5  1152W Total losses: Ploss  PJa  PC  1152  1095.5  2247.5W Finally, the output power: Pout  Pin  Ploss  11000  2247.5  8752.5W b) Efficiency: 

Pout 8752.5   0.7956 pu  79.56% Pin 11000


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c) Speed at this load: Back emf at no load: Eb 0  U  I a 0 Ra  220  3  0.5  218.5V Back emf at given load: Eb1  U  I a1Ra  220  48  0.5  196V But we know that for shunt motor, E  kn  n , (because the flux  is constant.)

Then,

E1 n1 E 196   n1  n0  1  1150   1031.57 rpm E0 n0 E0 218.5

d) Torque developed: Ea I a  Tm 

2n 60  Ea I a  Tm  Tm  60 2n

60  196  48  87.1Nm 2  1031.57

Exercise 1.32: A 220V, series motor in which the total armature and field resistance is 0.1Ω is working with unsaturated field, taking 100A and running at 800rpm. Calculate at what speed the motor will run when developing half the torque. Solution of exercise 1.32: In the exercise 1.30 above, we have shown that, for series motor: n2 Eb 2 I a1   n1 Eb1 I a 2

Since the field is unsaturated, T  I a2 . Then,

Or,

T1  I a21 T2  I a22 T2  I a 2    T1  I a1 

2

1 1 I  When developing half torque, T2  T1    a 2  2 2  I a1 

Then, I a 2  I a1 

2

1 100   70.71A . 2 2

On the other hand, Eb1  U  Ra  RS I a1  220  100  0.1  210V

Eb 2  U  Ra  RS I a 2  220  70.71  0.1  212.9V Electrical Machines_Jean-Paul NGOUNE

83

We can then deduce the speed. n2  n1 

Eb 2 I a1 212.9 100   800    1147 rpm Eb1 I a 2 210 70.7

REVIEW QUESTIONS

Exercise 1: Separately excited DC generator (GCE A Level, 2010) The external characteristic of a separately excited generator, at rated speed is as follows. I(A)

0

20

40

60

80

100

U(V)

140

137

133

128

122

115

1. Plot the characteristic curve U = f(I). Scale: 1cm  10 A;1cm  10V . 2. This generator supplies a resistor R = 2.5  . Find the coordinates of the operating point. 3. The resistor is replaced by a storage battery of internal resistance r = 0.4Ω and an emf E’=100V. Find the coordinates of the operating point.

Exercise 2: Shunt motor (Probatoire F3, 2007) A perfectly compensated bipolar DC shunt motor has the following rated characteristics: 

Supply voltage of the armature: U = 120V;



Armature current: I = 100A;



Field current I = 2.3A;



Speed: N = 1500rpm;



Armature resistance: R = 0.05Ω

The motor is operating under constant voltage of 120V. The constant losses are Pc = 785W. 1. Calculate the back emf of the motor. 2. Compute: a) The useful power of the motor; Electrical Machines_Jean-Paul NGOUNE

84

b) The useful torque of the motor; c) The efficiency of the motor. 3. Find the value of the starting rheostat in order to limit the starting current at 2Ir. (Ir being the rated current of the motor)

Exercise 3: Shunt motor (GCE A Level, 2011). A shunt excited DC motor has an armature resistance of 0.1Ω and a field resistance of 120Ω. For a supply voltage V = 240V, the flux is proportional to the excitation current. This motor takes a current of 150A under normal operation and rotates at 1200rpm. 1. Calculate the armature and the field current then deduce the back emf of the motor. 2. Constant losses are 2/3 of Joule losses. Calculate the useful power and the efficiency of the motor and deduce the useful torque. 3. A field rheostat is added to the field circuit of the motor: a) Draw the electric circuit of this motor. b) Give the expression of the rotational speed (n) in function of the field current (i). c) Express the rotational speed (n) in function of the resistance (x) of the rheostat. d) Graphically represent the variations of the speed (n) in function of the rheostat resistance x. n = f(x).

Exercise 4: Explain clearly the effect of the back emf of a shunt motor. What precautions must be taken when starting a shunt motor? A four pole dc motor is connected to a 500V dc supply and takes an armature current of 80A. The resistance of the armature circuit 0.4Ω. The armature is wave wound with 522 conductors and the useful flux per pole is 0.025 Wb. Calculate: (a) the back emf of the motor; (b) the speed of the motor; (c) the torque developed by the armature.


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Exercise 5: A dc motor takes an armature current of 110A at 480V. The resistance of the armature circuit is 0.2Ω. The machine has 6 poles and the armature is lap connected with 864 conductors. The flux per pole is 0.05Wb. Calculate (a) the speed and (b) the gross torque developed by the armature.

Exercise 6: A six pole dc motor has a wave connected armature with 87 slots, each containing 6 conductors. The flux per pole is 20mWb and the armature has a resistance of 0.13Ω. Calculate the speed when the motor is running off a 240V supply and taking an armature current of 80 A. Calculate also the torque developed by the armature.

Exercise 7: Explain the necessity for using a starter with a dc motor. A 240V dc shunt motor has an armature of resistance of 0.2Ω. Calculate: (a) the value of resistance which most be introduced into the armature circuit to limit the starting current to 40A; (b) the emf generated when the motor is running at a constant speed with this additional resistance in circuit and with an armature current of 30A.

Exercise 8: Calculate the torque developed by a dc motor having an armature resistance 0.25Ω and running at 750rev/min when taking an armature current of 60A from a 480V supply.

Exercise 9: A six pole, lap wound, 220V, shunt excited dc machine takes an armature current of 2.5A when unloaded at 950rev/min. When loaded, it takes an armature current of 54A from the supply and runs at 950rev/min. The resistance of the armature circuit is 0.18Ω and there are 1044 armature conductors. For the loaded condition, calculate: (a) the generated emf; (b) the useful flux per pole; (c) the useful torque developed by the machine.


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Exercise 10: State Fleming’s left hand rule and explain how you would use it to determine the direction of a current induced in a conductor. Briefly describe and state the functions of each of the following: (a) armature; (b) commutator; (c) brushes; (d) interpoles. A series motor taking 120A is adjusted so that its field current is reduced to 90A. If the resistance of the field is 0.05Ω, calculate the value of the resistance to be connected in parallel with the field. Also calculate the value of the power being wasted in the additional resistance.

Exercise 11: A motor name plate shows the following rating: 10hp, 230V, 1350rpm, 37.5A, shunt. When the field current is 0.75A and the armature resistance is 0.38Ω, calculate the following: a) The full-load efficiency b) The terminal torque at rated load c) The internal torque and speed developed when the line current is 18A and the field current remains at 0.75A. Exercise 12: A 400V shunt connected dc motor has a full load output of 20kW at an efficiency of 85%. The shunt field resistance is 200Ω and the armature resistance is 0.15Ω. Calculate for this load: a) The input current b) The armature current c) The back emf d) The total copper losses e) The rotational losses.


87

Exercise 13: A perfectly compensated shunt motor is supplying an effective mechanical power of 15kW from a voltage 240V supply. The total joule losses represent 9% of the total power absorbed. The excitation current is 3.125A. The motor efficiency is 0.8. Calculate: a) The total power absorbed

e) The back emf of the motor

b) The joule losses in the stator

f) The useful torque

c) The joule losses in the armature

g) The constant losses

d) The current in the armature

h) The effective electrical power

Exercise 14: A separately excited dc motor has the following characteristics. 

Supply voltage U = 250V



Armature current Ia = 50A



Excitation current J = 1.5A



Armature resistance Ra = 0.3Ω

A no load test as a generator gave the following results: Uo = 308V for I = 1.5A and N = 1200rpm A no load test as a motor gave the following results: P 0 = 1200W and V = 250V. In the whole problem, U and I are constant. 1. What is the speed of the motor on no load? 2. Determine the no load torque T0. 3. Determine the speed of the motor at normal load and hence deduce the electromagnetic torque Te of the motor. 4. Show that the electromagnetic torque is proportional to the current I.

Exercise 15: A 4 pole, long shunt, compound dc generator supplies 100A at a terminal voltage of 500V. if armature resistance is 0.02Ω, series resistance is 0.04Ω and shunt field resistance 100Ω, find the generated emf. Take the drop per brush as 1V.


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Exercise 16: 1) Calculate the torque in Nm developed by a 440V dc motor having an armature resistance of 0.25Ω and running at 750rpm when taking a current of 60A. 2) A 4 pole lap connected dc motor has 576 armature conductors and draws an armature current of 10A. Calculate the flux per pole if the torque is 18.3N.m. 3) A shunt motor running on no load takes 5A at 200V. The resistance of the field and armature are 150Ω and 0.1Ω respectively. Determine the output and efficiency of the motor when the input current is 120A. 4) Calculate the value of a series resistance with the armature of the motor in question 3 if the starting current must not exceed the full load value. 5) The following information refers to a 4 pole, lap wound dc series motor with 200 armature conductors. Terminal voltage = 400V; Power output = 51kW; Efficiency = 85%; Speed = 500rpm; Field resistance = 0.06Ω; Armature resistance = 0.08Ω. Find: (a) the back emf, (b) the power developed in the armature, (c) the armature torque (d) the shaft torque (e) the useful flux per pole. 6) The following information was recorded on a brake test in a dc motor. Diameter of pulley = 0.3m; Net brake load = 350N; Supply voltage = 250V; Line current = 32A; Speed = 20rps. Calculate (a) the output power (b) the motor efficiency.

Exercise 17: Technology 1) For a dc motor, assume that: U = Armature supply voltage

N = number of armature conductors

r = Armature resistance

Ф = Flux under a pole

I = Armature current

N = speed of the armature

a) Show that n 

U  rI N

b) At no load, deduce that n 

U Where K is the constant of the field circuit NKI

and I is the field current. c) Give the role of the excitation resistance. d) Explain Why the DC motor turn in high speed when the field flux is cancel.


89

e) Indicate the two precautions to be taken to avoid the DC motor to turn in high speed. f) Explain why the DC series motor turns in high speed at no load 2) For series, shunt and compound dc motor, give the following characteristics: a) N = f(Ia) characteristics b) Tm = f(Ia) characteristics c) N = f (Tm) characteristics 3) Give 2 suitable applications of dc series motors, justify your answer. 4) Give 2 suitable applications of dc shunt motors, justify your answer. 5) Give 3 starting methods of dc motors. 6) Why is the full voltage starting dangerous for the motor? 7) Give 2 braking method of dc motor. 8) Give the 3 main electric braking methods. 9) What’s the drawback of the friction braking method? 10) What’s the Ward-Leonard system?

Exercise 18: A DC machine is connected across a 240 volts line. It rotates at 1200rpmand is generating 230 volts. The armature current is 40A. a) Is the machine functioning as a generator or as a motor? b) Determine the resistance of the armature circuit. c) Determine power loss in the armature circuit resistance and the electromagnetic power. d) Determine the electromagnetic torque in N.m. e) if the load is thrown off, what will the generated voltage and the rpm of the machine be assuming: (i) no armature reaction (ii) 10% reduction of flux due to armature reaction at 40A armature current.

Exercise 19: A DC shunt motor drives an elevator load which requires a constant torque of 300N.m. The motor is connected to a 600V DC supply and the motor rotates at 1500rpm. The armature resistance is 0.5Ω. a) Determine the armature current. b) If the shunt field flux is reduced by 10%, determine the armature current and the speed of the motor. Electrical Machines_Jean-Paul NGOUNE

90

Exercise 20 A DC shunt motor (50 hp, 250V) is connected to a 230V supply and delivers power to a load drawing an armature current of 200A and running at a speed of 1200rpm. Ra = 0.2Ω. a) Determine the value of the generated voltage at this load condition. b) Determine the value of the load torque. The rotational losses are 500W. c) Determine the efficiency of the motor if the field circuit resistance is 115Ω.

Exercise 21: Magnetic circuit, shunt generator. a) A magnetic flux of 0.4mWb exists in a magnetic circuit of a mean path length of 0.5m. The cross sectional area of this material is 500mm 2 and its relative permeability is 750. Calculate the value of magnetomotive force (mmf) required to produce this flux, knowing that the permeability of free space is µ0 = 4  10 7 . b) An eight pole shunt wound generator has 720 conductors and rotates at 12.5 rev/s. if the useful flux per pole is 0.45W, calculate: i)

The generated emf if the machine is lap wound.

ii)

The terminal voltage if the armature resistance is Ra = 0.35Ω ( Ish = 4.8 A and the load current is 54A).

Exercise 22: Long shunt compound wound dc generator. A long shunt compound wound generator running at 16.67r.p.s supplies 11kW at a terminal voltage of 220V. The resistance of the armature, shunt and series fields are 0.05Ω, 110Ω and 0.06Ω respectively. The overall efficiency is 81.5%. Find: a) The total copper loss; b) The iron and friction loss; c) The torque exerted by the prime mover; d) The electrical and mechanical efficiencies.

Exercise 23: Shunt wound generator. A shunt wound generator running at 16.67r.p.s supplies 10kW at a terminal voltage of 230V. The resistance of the armature and field coils are 0.12 Ω and 115 Ω respectively. The overall efficiency is 83%. Find: a) The total copper loss. Electrical Machines_Jean-Paul NGOUNE

91

b) The iron and friction loss; c) The torque exerted by the driving engine d) The electrical and mechanical efficiencies.

Exercise 24: Series generator. The following information refers to a series generator. 

Terminal voltage: 220V;



Power output: 10kW;



Efficiency: 80%;



Rated speed: 1000 r.p.m;



Field resistance: 0.06 Ω;



Armature resistance: 0.04 Ω

Find: a) The emf generated, b) The driving torque of the prime mover

Exercise 25: Two poles shunt generator. A 2 poles shunt generator runs at 1500rpm. The useful flux per pole is 9mWb. Armature resistance is 0.3Ω and field resistance is 125Ω. Three resistors (10, 20 and 25 Ω) are connected in parallel at the terminals of the generator. The voltage at the terminals of the machine is 220V. Calculate: a) The field current; b) The emf; c) The electrical efficiency; d) The number of active conductors.

Exercise 26: A DC shunt machine (23kW, 230V, 1500rpm) has an armature resistance R = 0.1Ω. 1. The DC machine is connected to 230V supply. It runs at 1500rpm at no load and 1480 rpm at full load armature current. a) Determine the generated voltage at full load. b) Determine the percentage reduction of flux in the machine due to armature reaction at full load condition.


92

2. The DC machine now operates as a separately excited generator and the field current is kept the same as in part 1. I delivers full load at rated voltage. a) Determine the generated voltage at full load. b) Determine the speed at which the machine is driven. c) Determine the terminal voltage if the load is thrown off.

Exercise 27: A DC shunt machine (10kW, 250V, 1200rpm) has Ra = 0.25Ω. The machine is connected to a 250V DC supply, draws rated armature current, and rotates at 1200rpm. a) Determine the generated voltage, the electromagnetic power developed, and the torque developed. b) The mechanical load on the motor shaft is thrown off, and the motor draws 4A armature current. (i) Determine the rotational loss. (ii) Determine the speed, assuming 10% change in flux due to armature reaction, for a change of armature current from rated value to 4A.

References: 1. Edward HUGHES, Electrical Technology, Fifth Edition, English Language Book Society (ELBS), 1977. 2. T.F. FOFANG, Electrical Principles for Technical Colleges, Volume three, Technician series, 2006. 3. Martin ANYANGWE, Principles of Electrotechnology for Technical Schools, Volume2, 2000. 4. Vikas MITTAL, A Presentation On Working Principle Of DC Motors.

Acknowledgements: 1. Many exercises treated in this course are taken from Probatoire Technique past questions proposed by the Cameroon General Certificate of Education Board (GCEB). 2. Part of the references used in this course is not specified because the documents in question were having no mention of their authors.


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ABOUT THE AUTHOR

NGOUNE Jean-Paul was born at Foreké-Dschang, Republic of Cameroon in 1984. He is a holder of a Master Degree in electrical engineering, obtained in 2010 in the Doctorate School of the University of Douala (UFD-PSI). He is also a holder of a DIPET II and a DIPET I respectively obtained in 2009 and 2007 in the Advanced Teacher Training College for Technical Education (ENSET de Douala). He is currently a permanent teacher of Electrical Engineering at the Government Technical High School of Kumbo, North-West region, Cameroon. His domain of research

concerns

the

improvement

of

energy

conversion techniques for an efficient generation of electrical energy from renewable sources (especially wind and solar energy, small and medium scale hydropower) and digital designing using FPDs.

NGOUNE Jean-Paul, M.Sc., PLET. P.O. Box: 102 NSO, Kumbo, Cameroon. Phone: (+237) 7506 2458. Email : [email protected] Web site: www.scribd.com/jngoune


94

Machine Lesson Final1

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