Practical Transmission Line Calculation

ENGIN.

LIBRARY UC-NRLF

Meeh, dept.

PRACTICAL CALCULATION TRANSMISSION LINES

OF

PEACTICAL CALCULATION OF

TBANSMISSION LINES FOR DISTRIBUTION OF DIRECT AND ALTERNATING CURRENTS BY MEANS OF OVERHEAD, UNDERGROUND, AND INTERIOR WIRES FOR PURPOSES OF LIGHT, POWER, AND TRACTION

BY L.

W. ROSENTHAL, o ASSOCIATE MEMBER,

E.E.

A.I.E.E.

NEW YORK MCGRAW PUBLISHING COMPANY 239

WEST

39TH

1909

STREET

Engineering library

COPYRIGHT,

1909,

BY THE

McGKAW PUBLISHING COMPANY

NEW YORK

Stanbopc Jpresa F.

H.G1LSON COMPANY BOSTON.

U.S.A.

PREFACE.

THIS little book is offered to the engineering profession with the hope that it may be of practical help in the rapid and accurate Its existence is the outcome of calculation of transmission lines. Its chief mission is to is in part barren. substitute a direct solution for the trial method which was formerly

the belief that this field

a necessary evil. The arrangement of the formulas, tables and text has been dictated solely by the needs of the rapid worker. All sections except the last include the important effects of

The section relating to temperature and specific conductivity,, direct-current railways is novel in the form of its tables, and the have been found rapid and comprehensive., alternating-current division presents a new and original method

methods outlined in

The

it

It is the only method known to for the solution of these problems. the author which determines the size of wire directly from the volt

and it also possesses unique features of scope, accuThe chapter on single-phase railways is in and simplicity. racy accord with most of the consistent data published on the subject, loss in the line,

although further accurate investigations of installed lines modify to some extent the present accepted values.

may

The author desires to call particular attention to the fallacies of some familiar methods of calculating alternating-current transmisIt will be sion lines which heretofore have been in common use. evident from Tables 11, 28, and 36 that their results are wholly erroneous under certain practical conditions, and indicate wires which may be entirely at variance with the specified requirements. The scope of this book has been confined to methods of calculation. Hence, the most desirable limits of line losses are not discussed, but the tables are sufficiently extended to cover all cases There is no discussion of the characterlikely to arise in practice. istics of alternating-current

transmission lines, although the tables

of wire factors render apparent

many of their important features. does not either the determination of the book include Furthermore, iii

254556

PREFACE

iv

the size of conductors for conditions of

maximum economy

or the

consideration of alternating-current circuits in series. The preparation of the tables has involved thousands of calcula-

but thorough checks by the methods of differences and curve plotting have probably eliminated almost all errors of material

tions,

influence.

However, some discrepancies may have crept

in,

and the

author would be glad to learn of them. L.

NEW YORK

CITY. December, 1908.

W.

R.

CONTENTS. CHAPTER

DIRECT-CURRENT DISTRIBUTION FOR LIGHT AND POWER.

I.

PAGES

PAR. 1.

2. 3. 4. 5. 6. 7.

8.

INTRODUCTION PROPERTIES OF CONDUCTORS CURRENT-CARRYING CAPACITY PARALLEL RESISTANCE OF WIRES GIVEN ITEMS

6^ :

9 10 10

FORMULAS AMPERE-FEET EXAMPLES.

10 11

.

CHAPTER

DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS.

II.

15.

INTRODUCTION RESISTANCE OF RAILS PARALLEL RESISTANCE OF RAILS AND FEEDERS NEGATIVE CONDUCTORS POSITIVE CONDUCTORS RESISTANCE OF CIRCUIT GIVEN ITEMS

16.

EXAMPLES

9.

10.

11. 12.

13. 14.

CHAPTER

.

5

6

17 18

18 19

19

20

.

III.

16

16

ALTERNATING-CURRENT TRANSMISSION BY

OVERHEAD WIRES. 17. 18.

19.

20.

INTRODUCTION OUTLINE OF METHOD

RANGE OF APPLICATION MAXIMUM ERROR

22.

TRANSMISSION SYSTEMS BALANCED LOADS

23.

TEMPERATURE

21.

CONDUCTIVITY SOLID AND STRANDED CONDUCTORS 26. SKIN EFFECT 27. WIRE SPACING 24. SPECIFIC

25.

v

29 30 31 31 32 33 33 33 33 33

34

CONTENTS

vi PAR. 28. 29.

ARRANGEMENT OF WIRES FREQUENCY

34.

MULTIPLE CIRCUITS CURRENT-CARRYING CAPACITY TRANSMISSION VOLTAGE VOLT LOSS POWER TRANSMITTED

35.

POWER

36.

POWER-FACTOR WIRE FACTOR GIVEN ITEMS SIZE OF WIRE PER CENT VOLT LOSS CHARGING CURRENT

30.

31. 32. 33.

37. 38. 39. 40. 41.

LOSS

42.

CAPACITY EFFECTS

43.

EXAMPLES

.

CHAPTER

PAGE 34 34 34 34 35 35 35 36 36 36 37 37 37 38 38 39

.


IV.

UNDERGROUND CABLES. 45.

INTRODUCTION MAXIMUM ERROR

46.

TEMPERATURE

47.

50.

PROPERTIES OF CONDUCTORS THICKNESS OF INSULATION CURRENT-CARRYING CAPACITY CAPACITY EFFECTS

51

EXAMPLES

44.

48. 49.

.

CHAPTER

V.

63 63 64 64 64 64 65 65

INTERIOR WIRES FOR ALTERNATING-CURRENT DISTRIBUTION.

55.

INTRODUCTION PROPERTIES OF CONDUCTORS SPACING OF WIRES AMPERE-FEET

76

56.

EXAMPLES

76

52. 53. 54.

CHAPTER

VI.

DISTRIBUTION FOR SINGLE PHASE RAILWAYS.

57.

INTRODUCTION

58.

METHOD OF CALCULATION

59.

IMPEDANCE OF RAIL PERMEABILITY OF RAIL IMPEDANCE AND WEIGHT OF RAIL IMPEDANCE OF RAIL AND FREQUENCY

60. 61.

62.

75 75 75

85 85 85 86 86 86

CONTENTS

vii

PAGE

PAR. 63.

FORMULA FOR RAIL IMPEDANCE

64.

POWER-FACTOR OF TRACK, HEIGHT OF TROLLEY EFFECT OF CATENARY CONSTRUCTION IMPEDANCE OF COMPLETE CIRCUIT MULTIPLE TRACKS .. .,.. EXAMPLES ,.....-..*

65.

66. 67. 68. 69.

,

.

86 87 87 87

,

.

=

.

.

,

_

87 88 88

TABLES. CHAPTER

DIRECT-CURRENT DISTRIBUTION FOR LIGHT AND POWER.

I.

PAGE 6 6

No.

T

and aluminum and aluminum F Properties of copper and aluminum Ampere-feet per volt drop and current-carrying capacity Values of H for copper or aluminum. Formulas for direct-current wiring Values of a for copper and aluminum.

Values of 2. Values of 1.

3.

4. 5.

6.

for copper for copper

8. 9.

7

,

8 8

,

CHAPTER 7.

.

II.

14

15

DISTRIBUTION FOR DIRECT CURRENT RAILWAYS.

Values of T l for steel Equivalents of copper of 100 per cent conductivity. Resistance to direct current of one steel rail

Formulas for direct-current railway A for wires and rails

circuits

17 .

20 24 25

,

,

10. Values of

CHAPTER

III.

26


OVERHEAD WIRES. Error in per cent of volt loss , 12. Maximum error in per cent of true values at. 20 cent 13. Values of c for overhead wires 14. Reactance factors ,...., Formulas for a. c. transmission by overhead wires 15. Values of volt loss factors 16. Values of A for balanced loads 17. Values of B for balanced loads. for overhead copper wires at 15 cycles per second 18. Values of for overhead copper wires at 25 cycles per second 19. Values of for overhead copper wires at 40 cycles per second 20. Values of 21. Values of for overhead copper wires at 60 cycles per second. ..... 22. Values of for overhead copper wires at 125 cycles per second 23. Values of for overhead aluminum wires at 15 cycles per second. 24. Values of for overhead aluminum wires at 25 cycles per second. 25. Values of for overhead aluminum wires at 40 cycles per second. 26. Values of for overhead aluminum wires at 60 cycles per second. 27. Values of for overhead aluminum wires at 125 cycles per second. 11.

.

.

,

M M M M M M M M M M

.

.

.

.

.

.

.

.

.

ix

31

32 39 39 50 51

52 52 53 54 55

56 57

58 59

60 61 62

X

TABLES

CHAPTER


IV.

UNDERGROUND CABLES. PAGE

No. '

28. Error in per cent of true volt loss 29. Maximum error in per cent of true values at 20 30. Values of c for

Formulas for

33. Values

63 64 65

cables

71

underground cables

a. c.

transmission

by underground

VQ '" = F (1 - 0.01 F of A for balanced loads of B for balanced loads

31. Values of 32. Values

cent

72 72

)

72 73 74

M for multiple conductor copper cables Values of M for multiple conductor copper cables

34. Values of 35.

CHAPTER

INTERIOR WIRES FOR ALTERNATING-CURRENT

V.

DISTRIBUTION. 36.

Error in per cent of true volt

Formulas

for a.

37. Values of a 38. Values of 39. Values of 40. Values of

CHAPTER

and

B

...

82 82 83 84

DISTRIBUTION FOR SINGLE-PHASE RAILWAYS.

and calculated values of impedance per mile Formulas for single-phase railway circuits

41. Test

42. Values of

81

for balanced loads

M for copper wires in interior conduits M for copper wires in molding or on cleats

VI.

76

loss

interior wiring b for balanced loads

c.

M for single-phase railway circuits

88 92 93

PART

I.

DIRECT-CURRENT DISTRIBUTION BY MEANS OF OVERHEAD, UNDERGROUND AND INTERIOR WIRES FOR PURPOSES OF LIGHT, POWER AND TRACTION

DIRECT-CURRENT DISTRIBUTION CHAPTER

I.

DIRECT-CURRENT DISTRIBUTION FOR LIGHT AND POWER. 1.

Introduction.

Problems

in

transmission

direct-current

The same formula covers

and distribution are

relatively simple. conditions of installation and operation whether by overhead, underground or interior wires. The formulas give accurate all

results,

and

all

items of influence are easily included. The tables and will cover almost all the

are concise but comprehensive

usual and unusual requirements of varied practice. The resistance of stranded and 2. Properties of Conductors.

and length is practically the 3, page 8, gives properties of wires at 20 fahr. for copper of 100 per cent and aluminum of

solid conductors of the

the same.

same

cross section

Table

cent, or 68

62 per cent conductivity in Matthiessen's standard scale. The resistance at any other temperature and conductivity may be found for copper or aluminum from formulas (1) and (2).

v *

Ohms

resistance per 1000 feet

=

Ohms

resistance per mile

= 54/700 X T

S =

>

.

.

S o

^

^

Cross section of metal in circular mils.

T = Temperature

factor.

Table

1,

page

6.

Thus, at 40 cent, the resistance per mile of No. of 98 per cent conductivity

is

1

copper wire

TRANSMISSION CALCULATIONS

6

Table

i.

Values of

Temperature

T for

Copper and Aluminum.

DIRECT-CURRENT DISTRIBUTION 3. Properties of Copper and Aluminum at 20 Cent, or 68 Fahr. Conductivity in Matthiessen's Standard Scale ; Copper 100 Per Cent, Aluminum 62 Per Cent.

Table


8 Table

4.

Ampere-Feet per Volt Drop and Current-Carrying Capacity.

DIRECT-CURRENT DISTRIBUTION

9

current per wire for lead-covered underground cables is based on tests for a temperature rise from 70 fahr. initial to 150 fahr.

Formula

final.* rise

(4)

may

be solved for

Amperes per wire d

C

to find the temperature

under given conditions.

=

.....

\~r

(4)f

Diameter of wire in inches.

C = Temperature

H=

=H

Heat

factor.

rise in

degrees centigrade.

Table

5,

page

8.

T = Temperature factor at final temperature.

Table

1,

page

6.

The size of wires is sometimes determined by their currentcarrying capacity, especially in interior work where the runs are short. For longer lines it is usually advisable to calculate the loss and then note that the wire has sufficient carrying capacity, while for shorter stretches

it is

often better to select a wire of

proper current capacity and then find by calculation whether the loss is within the specified limit.

The parallel resistance at any temperature for a number of wires of any conductivity is given by the following formulas: 4.

Parallel Resistance of Wires.

Ohms


=

-

i

n

u>

+ T + 12 1

T

Ohms

=

resistance per mile

-

"/

T 1

T2

(5)

*

.

.

.

*

i

Where all the wires have the same temperature and conductivity, formulas (5) reduce to those below.

Ohms


X T S + + S, = 54 700 X T S + S 2 410 350

-

'

.

-

2

Ohms

resistance per mile

r

^

l

In formulas Sit

T,

page

(5)

and

(6)

S 2 = Circular mils in T T 2 = Temperature l}

respective wires. factors of respective wires. Table

1,

6.

* Standard

Underground Cable

Co.'s

Handbook No. XVII, page

192.

" t Based on formulas in Foster's Electrical Engineers' Pocketbook," 1908, page 208.


10

Thus, the resistance per mile at 30 cent, of one No. 00 trolley wire of 97 per cent conductivity in parallel with one 1,000,000cir. mil aluminum feeder of 62 per cent conductivity is

=

54,700 133,100

1,000,000

1.069

1.674

0.0758 ohm.

Similarly, the parallel resistance per 1000 feet at 40 cent, of one No. and one No. 4 copper wires of 97 per cent conductivity 18

X +

10,350

105,500 5.

Given Items.

1.110

=

Q 078Q

ohm

41,740

The problem may

tion of the size of wire

the given size of wire.

from the volt

require the determinaloss, or the volt loss from

In either case the other required items

and temperature, current and distance of If power in watts is specified, then the voltage transmission. at load or source must also be given. The term " source " is used in this book to designate the point from which the circuit, or the part of circuit under consideration, starts. Thus in direct-current distribution it may signify the are conductor metal

generator, rotary converter, storage battery, connection to main, to feeder or to sub-feeder, or simply a certain point of the circuit. 6. Formulas. Formulas for the complete solution of direct-

current problems are given on page 14, and the method of procedure will be clear from the arrangement of the table. The

may be determined from the volt drop, or from the cent volt per drop, which in direct-current systems is always equal to the per cent power loss. The value of a is found in Table 6,

size of wire

and the

size of each wire is noted from Table 3, page 7, corresponding to the required circular mils. The per cent volt drop is expressed in terms of power as well as current, so that problems involving voltage at source and watts at load may be solved without preliminary approximation. It should be carefully noted that percent loss, V or V is expressed as a whole number, and that the length of transmission, I, is the dis-

page

15,

,

tance from source to load, which is the same as the length of one wire. In formula (14), r is the resistance per foot of one wire, equal to the values in Table 3, columns 4 or 8, divided by 1000.

The size 7. Ampere-Feet. mined from the ampere-feet per

may be readily deterdrop as given by formula (7).

of wires volt

11

DIRECT-CURRENT DISTRIBUTION The wire having Table

4,

a corresponding

page

Ampere-feet per volt drop

= Amperes. = Distance from = Drop in volts.

/ I

v

value

then noted from

is

8.

=

(7)

source to load, in feet.

The values in Table 4 are for copper and aluminum of 100 per cent and 62 per cent conductivity, respectively, at 20 cent. For other conductivities or temperatures, formula is

(8),

page

14,

more convenient.

In case of distributed loads, as in interior lighting work, II is sum of the products given by each load, /, multiplied by Thus if 6 amperes are to be transits respective distance, I. the

mitted 25 II

3

ft.,

=

X

6

amperes 50 25

+

X

3

ft.,

50

+

and 2 amperes 100 2

X

100

=

ft.,

500 ampere-feet.

Examples. Examples in practice may take innumerable of procedure in any case will be clear but the method forms, from the table of formulas on page 14. The following problems 8.

are typical and serve to illustrate the simplicity of the calculations for direct-current distribution. .

Example

A

i.

to deliver 200

copper circuit of 97 per cent conductivity is for a distance of 1000 feet with a loss of

amperes

10 volts at 40

cent.

GIVEN ITEMS. I

=

From

v = 10 volts; 200 amperes; Table 6, page 15, a = 23.0.

I

REQUIRED ITEMS. From (8),

Size of each wire.

= 23 X

200

X

1000

=

46Q 000

ci]

10

Use 500,000

From

Volt drop.

_

23

X

cir.

mil wires.

(9),

200

X

500,000

1000

=

9.2 volts.

=

1000

feet.


12 Watt

From p

=

From Tables 1 and 3, T = 1.110, and r = 0.0000207 ohm.

loss.

2

(14),

X

1.11

X

X

0.0000207

2

(200)

X

=

1000

1840 watts.

the voltage at the load at any value, say 100, cent the per drop and per cent power loss are found to be the

By assuming

same, 9.2 per cent.

Example feet with a circuit

is

A

2.

motor

is

to take 25 kw. at a distance of 240

per cent of the 110 volts generated, while the to consist of copper wires of 98 per cent conductivity loss of 5

with a temperature of 30 cent.

GIVEN ITEMS.

w=

From

v=

(19),

From Table

6,

=

e

25,000 watts;

X 110=

0.05

page

21.9

a

15,

7 =5per

cent;

Z

= 240

feet.

5.5 volts.

=

21.9.


Size of each wire.

s=

110 volts;

X

239

X

240

=

5.5

Use No. 0000 Per

wires, for

From 2L9 X

cent volt drop.

V

"'

=

0.01

X

which

S=

211,600

cir.

mils.

(11),

25,000

211,600

X 240 X (HO)

=

.

-

2

7 = 5.4 per cent. v = 0.054 X 110 = 5.9 volts. Volt drop. From (19), From (13), e= 110 - 5.9 = 104.1 volts. Volts at load. 054 X 25 QOQ = From (14), p = Wattloss. 1430 watts.

From Table

31,

page

72,

-

'

1

Watts

at generator.

w =

From 25,000

~~

U.Uo4

(15),

+ 1430=

26,430 watts.

From Table 4, page 8, it is seen that the wire should have weatherproof insulation or else be increased to 250,000 cir, mils.

DIRECT-CURRENT DISTRIBUTION Example

The load on

3.

a feeder

is

13

to consist of 20

amperes

at 50 feet, 25 amperes at 100 feet, and 40 amperes at 150 feet, from the source. Calculate the required size of a uniform circuit of 100 per cent conductivity for a total loss of 2 volts at 20 From (7), Ampere-feet per volt drop

X

20

+

50

X

25

X

100

X

40

cent.

150

2

From Table

4, page 8, the required size of each wire is No. copper and No. 000 for aluminum. Example 4. Copper mains of 98 per cent conductivity are to deliver 500 amperes to a point 550 feet from a rotary converter with a loss of 3 per cent of the voltage at load. Calculate the

for

98 per cent conductivity and at 50

size of wire of

cent,

if

220

volts are generated.

GIVEN ITEMS. 7

= 500 From

=220

e

amperes;

v

(18),

=

V=3

volts;

X

03

-

per cent;

Z

= 550

feet.

6.4 volts.

^

From Table

6,

page

23.6

a

=

23.6.


Size of each wire.

a=

15,

X

500

X

550

_

Use 1,000,000 Volt drop.

Per cent

- From

(9),

volt drop.

t;

=

From

23 6 '

1;010>000 cir cir.

mils.

X

*

^ 1,UUU,UUU

(19),

7 = -5A = 2^

Volts at load.

Watts

at load.

55

=

6.5 volts.

2.95 per cent.

,2\j

From (13), e = 220 - 6:5= 213.5 volts. From (16), w = 213.5 X 500 = 106,750 watts.

Find the combined resistance of 1500 feet of Example 5. 500,000 cir. mils of 98 per cent copper wire at 20 cent, in parallel with the same length of 1,000,000 cir. mils of aluminum wire of 62 per cent conductivity at 30 cent.

From

(5)

and Table

1,

1.02

1.674

TRANSMISSION CALCULATIONS Formulas For Direct-Current Wiring. Required Items.

DIRECT-CURRENT DISTRIBUTION Table

6.

Values of a For Copper and Aluminum.

Temperature in

degrees.

15

CHAPTER

II.

DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS. 9.

Introduction.

track

rails for

Electric railways almost always use the the return of current. Besides this, other differ-

ences between circuits for railways and those for power and lighting purposes are the higher voltage employed on the trolley, the greater per cent loss allowed in the line and the variable nature of the loads.

Due

moving and

to the track rails the calculations of railway feeders

and

working conductors are somewhat complex and uncertain. However, the recent character of bonding has made the calculations more reliable by giving a higher and more permanent value to the conductivity of the grounded side. In electric railways such as the open conduit and the double trolley systems, the track rails are not used for the return current, and the calculations for their circuits are therefore simpler and more definite. Where the feeders and working conductors form a complete

copper 10.

circuit,

the formulas on page 14

Resistance of Rails.

may

be used.

Table

9, page 24, gives the equivalent copper section and the resistance of third rails or track rails at 20 cent., for various values of relative resistance of steel to

The corresponding value for any number of rails is copper. found by multiplying the equivalent copper section, or by dividThe resistance at any other ing the resistance, by the number. temperature is found by multiplying the value in the table by the temperature factor of resistance for steel, T1} Table 7, page 17, while the equivalent section of copper is equal to the tabular value divided by 7\; or, the values may be found for one rail for any other condition of relative resistance and temperature by

means

of the following formulas: cir. mils of 100 per cent copper

Equivalent

=

125,000

T X l

X Pounds per yard Relative resistance 10

DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS

Ohms


T X

Relative resistance

t

12.1

Ohms

17

X Pounds

(21)

per yard

resistance per mile

T\ X Relative resistance X Pounds per yard

2.28

T = Temperature

(22)

Table 7, page 17. factor of steel. As an example, the total resistance per mile at 30 cent, of two 65-lb. track rails having a relative resistance of 13.2 is l

1.05

2.28

Table

Temperature, deg. cent

X 7.

X

13.2

65

X

=

0.0468 ohm.

2

Values of

T

for Steel.


18

copper feeder of 97 per cent conductivity and one 500,000-cir. mil aluminum feeder of 62 per cent conductivity is

=

X

833^000

2

500,000

500,000

1.069

1.674

1.05 12.

Negative

Conductors.

The track

rails

and negative

feeders carry the return current. The size of rails is fixed by conditions other than electrical conductivity and usually give a Electrototal resistance much below that of the positive side.

may require that negative feeders be connected to the rails at certain points, but otherwise the rails generally lytic conditions

have ample conductivity without any copper reinforcement. The section of negative conductors may be based on a maxi-

mum feeder

The size of the negative allowable drop in the return. is found by subtracting the equivalent copper section of

the rails from the total circular mils required. The additional resistance of bonds rnay be included by increasing the true relative resistance of the rail to

an apparent value.

As an

of the determination of the size of

example

negative feeder, suppose that S in formula (25) on page 26 should come out 2,000,000 cir. mils for the negative side of a circuit for which the track is to consist of two 70-lb. rails of an apparent relative

resistance

feeders

in

625,000

X

parallel 2,

cir.

or 750,000

cir.

Based on Table

ductivity.

773,000

14 (including bonding). The required with the track should have 2,000,000

of

mils of copper of 97 per cent conductivity, or 1,210,000 of 62 per cent conductivity.

cir.

mils of

mils of copper of 100 per cent con8, page 20, this is equivalent to

aluminum

The positive conductors consist of 13. Positive Conductors. auxiliary feeders in parallel with trolleys or third rails. Trolley wires vary in size from Nos. to 0000, while third rails usually range from 60 to 100 pounds per yard. The drop in the positive conductors is found by subtracting the calculated negative drop from the total that is allowed. Then for a third-rail system, the method

exactly like the determination of negative feeders paragraph 12; while for systems using trolley wires, the total section of positive conductors is calculated from formula is

in

(25),

page

difference

25.

The

size of the auxiliary feeder is

given by the

between the total circular mils required and that of

the trolley wire.


19

The temperature is included in the calculation of the circular by means of A in Table 10, page 26. For conductivities other than 100 per cent, the value of S in formula (25) is divided by the given conductivity. Thus if S comes out 1,000,000 cir. mils mils

of 100 per cent copper, the required section of 1

cent conductivity

noted from Table 14.

'

is

1

,

0.97 8,

copper of 97 per

ooo 000

page

or 1,030,000

cir.

mils, as

may

be

20.

Resistance of Circuit.

The

total resistance of the circuit

obtained by adding the resistance of the grounded side and the overhead. Where there are no negative feeders, the resist-

is

ance of the rails may be taken directly from Table 9, page 24, and if no positive feeders are used, the resistance of the trolleys is

given in Table

3,

page

7.

Thus the

resistance per mile at

cent, of a single-track road having two 60-lb. rails of an apparent relative resistance of 14 (including bonding) and a

20

No. 00 trolley of 100 per cent conductivity

2^2 +

o 411

=

is

o.462 ohm.

The determination of the proper loading on 15. Given Items. which to base the feeder systems is usually more difficult than In general the requirement will the calculation of the feeders. be for a maximum drop with a very severe condition of loading or for a much smaller drop with a distributed loading of an average In either case the loads and their positions are first value. settled and then the formulas are applied successively to the separate loads; or, where the feeder system is uniform throughout, one determination is made, using the total of the loads at their center of gravity. In case the feeder system

is

known

the total loss

is

easily

calculated by considering the loads separately, or by considering their combined effect where the feeder system is uniform

throughout the area of loading. For convenience in calculation the formulas have been expressed in terms of, current. The per cent volt loss has also been given in terms of power, so that no preliminary approximation is necessary when the voltage is known only at the source. 16. Examples. Although^ in practice, these problems occur in a great many different forms, the application of the formulas .will be clear from the following typical examples.

TRANSMISSION CALCULA TIONS

20 Table

8.

Equivalents of Copper of 100 Per Cent Conductivity. Conductivity in Matthiessen's Standard Scale.

DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS in the negative feeder from 500,000 to 1,000,000 to an increase of 100 per cent.

cir.

21

mils, equal

For a temperature of 40 cent., determine the which take 100 amperes each at respective locations of 500, 2000, 3000, and 5000 feet from a power station generating 550 volts, if the circuit consists of four

Example

7.

line voltage at successive cars

having a relative resistance of 14 (including bonding), and two No. 00 trolleys of 97 per cent conductivity in parallel with one 500,000 cir. mil aluminum feeder of 62 80-lb. track rails

per cent conductivity. From Tables 7 and 9, the resistance of the tracks Resistance per 1000

From

(23)

feet =

and Tables

1

-

and

0145

3,

X

i- 09

=

is,

0.00395 ohm.

the resistance of the overhead

is,

Resistance per 1000 feet

10350 133,100

X

=

2

total resistance per 1000 feet of road

0.00395 1st

2d

Car

Car

3d Car 4th Car

ohm<

1.738

1.11

Hence the

Q 01962

500,000

+

0.01962

=

is

0.02357 ohm.

= 0.02357 X 400 X0.5 = 550 - 4.7 Additional drop = 0.02357 X 300 X 1 .500 - 545.3 - 10.6 Line volts Additional drop = 0.02357 X 200 X 1 .000 = 534.7 - 4.7 Line volts Additional drop = 0.02357 X 100X2.000 = 530 - 4.7 Line volts Total drop Line volts

= 4.7 volts. = 545.3. = 10.6 volts. - 534.7. = 4.7 volts. = 530. = 4.7 volts. = 525.3.

In the above typical problem, the resistance of the grounded is but 20 per cent of the resistance of the overhead. A single-track road with two 75-lb. rails having Example 8.

side

a relative resistance of 13 plus 10 per cent for bonding (apparent resistance = 14.3), is to supply four cars with 150

relative

amperes each when located at

0.8, 1.0, 1.5

and

2.5 miles

from a

If the minimum line e.m.f. substation generating 550 volts. is to be 400 volts at 50 cent., what size copper feeder of 97 per cent conductivity should be in parallel with a No. 00 trolley

of 96 per cent conductivity?

22


From

and Table

(22)

7,

page

Resistance per mile of two

Total drop in track

41.5

=

17,

X

150

X 14 3 = 0.0477 ohm. X 75 X 2 (0.8 + 1.0 + 1.5 + 2.5) -

0.0477

=

150

- 41.5=

1<14

=

rails

'

2.28

volts.

Allowable drop in overhead

108.5 volts.

From

(25), in which IL is the sum of products of each load multiplied by its distance from source, and from Table 10, page 26,

s=

61,100

X

+

150(0.8

1.0

+

+W =

1.5

lOo.O

490>000

ci

,

mils

of 100 per cent copper.

Based on Table copper

in trolley

8,

page 20, the equivalent section of 100 per cent

is

X 0.96=

133,100

128,000

cir.

mils.

Required section of 100 per cent copper in feeder

-

-

490,000

128,000

=

362,000

Required section of 97 per cent copper

= 5^2,220 ,

373,000

0.97

Example

Find the

9.

cir.

mils.

in feeder

mils.

cir.

size of a third rail of relative resistance

7.5 (including bonding) required to start two cars taking 1000 amperes at a point midway between substations 8 miles apart,

the drop at 20 cent, is to be 25 per cent of the 600 volts at Each track rail is to weigh 80 Ib. per yard and to have a relative resistance of 13 (including bonding).

if

rotaries.

Distance from either substation to cars = f = 4 miles. Current from each substation = ^-^ = 500 amperes.

Total allowable drop = 600 X 0.25 = 150 volts. From Table 9, page 24, resistance per mile of two-track

= Drop

in track

=

0.0713

0.03565

Allowable drop in third

From

(25), in

s=

= 71.3 volts. 150 71.3= 78.7 volts. Table 10, page 26 = 54,700,

500

rail =

A from x 500 XJ =

which

54,700

X

= 003565 ohm X

4

rails


From Table

9,

page

24, this section

23

corresponds to an 85-lb.

Then from above and Table 9, Total resistance per mile= R = 0.03565 + 0.0387

rail.

From From From

=

=

drop

(40),

V= From

in per cent of volts at load is

From

X

148.7^-0.01

451.3

=

33.0 per cent.

(41), drop in per cent of volts at substation is 24.8 per cent. 148.7-r-0.01 600

7 =

Watt

= 0.07435 ohm.

=

0.07435 X 500 X 4 148.7 volts. (28), total drop at cars 148.7= 451.3 600 volts. (37), voltage

From

loss.

above,

= X R = 0.07435 ohm. p = 0.07435 X (500) X = 148,700 watts. 2

(35),

Example

8

Calculate the size of positive feeders at 30 cent, 600 amperes per mile

10.

required for a uniformly distributed load of

between two substations 6 miles apart with an average

loss of

5 per cent of the 650 volts generated, if there are to be eight 100-lb. track rails and four 70-lb. third rails having relative resistances (including bonding) of 12.5 and 8.0, respectively.

Since the average loss is to be 5 per cent, the maximum is 10 per cent or 65 volts. The result is equivalent to the total load concentrated at a point one-quarter the distance between substations from either one.

Total load per substation

Center of

^ = fiOO V

load

gravity of

=

f

=

fi

=

1800 amperes.

2 miles

from

either sub-

station.

From Tables

7

and

9,

resistance per mile of four tracks

= 0.0548 X 1.05-5-8 =0.00719 ohm. = 0.00719 X 1800 X 2 = 25.9 volts. in track Drop Allowable positive drop = 65 - 25.9= 39.1 volts. From (25), 5 = 57,400 X 1800 X 2-4-39.1 = 5,280,000 of 100 per cent copper. From Tables 7 and 9, third-rail section =1,090,000 X 4,150,000 cir. mils of 100 per cent copper.

cir.

4-*-

=

Required section of 100 per cent copper feeder

=

-

5,280,000 4,150,000 = 1,130,000 of section 97 Required per cent copper feeder

cir.

= 1,130,000-4-0.97 = 1,165,000 cir. mils. of 62 per cent aluminum feeder section Required

=

l,130,000-s-0.62

=

1,823,000

cir.

mils.

mils.

mils

1.05

TRANSMISSION CALCULATIONS oooo oooo o o' o' o o' o o o' oooo oooo d o' o' o' o' o' o o'

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ISIi

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DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS Formulas for D. Required Items.

C.

Railway

Circuits.

25

26

TRANSMISSION CALCULATIONS Table 10. Temperature degrees.

Values of in

A

for

Wires and Rails.

PART

II.

ALTERNATING-CURRENT TRANSMISSION BY MEANS OF OVERHEAD, UNDERGROUND AND INTERIOR WIRES FOR PURPOSES OF LIGHT, POWER AND TRACTION

27

ALTERNATING-CURRENT TRANSMISSION CHAPTER

III.

ALTERNATING-CURRENT TRANSMISSION BY OVERHEAD WIRES Introduction.

17.

The methods

in

common

use for calcu-

lating alternating-current transmission lines from the volt loss This condition results from are either indirect or inaccurate. it is impossible to devise an exact formula for any alternating-current system which will directly indicate the size of wire required for the transmission of a given amount of power

the fact that

with a given volt loss. Methods of the indirect class require that the size wire be known, and hence are trial methods. Approximations of the second class may be sufficiently close

under certain conditions but give wholly erroneous results for other practical cases.

The

method is that it should any system and frequency, the size of

essential feature^pLthe desirable

directly determine for

and power-factor at generator when given the line, power received, load power-factor, voltage at generator or load and distance between wires. The following method accomplishes this result with commercial wire,

power

volt

loss,

loss

length of

accuracy over a sufficiently wide range of conditions to cover

almost air practical cases.

One

familiar

method

of calculating alternating-current lines

assumes that the impedance volts equals the line loss, or in Figs. 1 and 2 AB = AC. Several other methods in common use assume that the line loss equals the projection of the impedance volts on the vector of delivered voltage. The errors of these approximations in terms of the actual volt loss are shown in Table 11, for copper wires on 36-inch centers and a true volt loss of 10 per cent of the generated volts. The errors for leading power-factors, larger wires or greater spacings exceed those shown in Table 11, and it is evident

then, that both methods

may

lead to erroneous results. 29

The


30

assumption gives wires which are too large, while the other assumption gives wires too small. By stating the errors of calculated the 11 in terms of Table values, the percentages for the first assumption and decreased would be shown greatly case. in second the similarly increased All items which depend on the 18. Outline of Method. load of wire, power-factor and wire spacing were frequency, size first

grouped into one term, which wire factor.

is

denoted by

M

and

called the

All other variables that determine the size of wire

Fig. 2

were grouped into the second member of the equation of M. after introducing the proper transmission factors, the equation took the form shown in the table of formulas opposite the size of wire. The remaining formulas are simple derivations or from well known relations. either from the equation of The vector diagram from which the fundamental equation was

Then

M

derived

is

shown

in Fig.

leading current, wherein,

1

for lagging current

and

in Fig. 2 for

ALTERNATING-CURRENT TRANSMISSION 01 = OA =

Current at load.

OB

Volts at load.

AC= OB - OA = Volt loss.

AD=

Resistance volts.

DB =

Reactance

volts.

AB= Impedance

volts.

(f> Q

31

Volts at source.

= = =

Power-factor angle of load. Power-factor angle of line. Power-factor angle at source.

of Application. The formulas apply for all volt than 20 per cent of the generated voltage or 25 per cent of the voltage at load, and for a sufficient range of wire At high sizes to cover the usual and unusual cases in practice. standard frequencies and for power-factors near 100 per cent for the largest wires were omitted in order to the values of 19.

Range

losses less

M

confine the in

Table

12,

maximum

possible error within the limits prescribed

page 32.

Table

Size of

Wire.

A.W.G.

n.

Error in Per Cent of True Volt Loss.


32

reduces to a zero error near 10 per cent loss. The point where the should be correct was arbitrarily fixed at 10 per cent volt loss, since this is about the mean value in practice,

formula for

M

After the wire

is determined, the actual per cent drop is calcuIn order to minimize its error and that of the remaining has been divided into a items, each column in the tables of maximum of three sections. It will be observed that any value of is found between the section letters a and b, b and c,

lated.

M

M

and d, depending upon the and ing power-factor. or c

size of wire, frequency, spac-

The maximum

errors in the calculations are shown in Table These errors occur near per cent and 20 per cent volt losses but gradually reduce to zero near 10 per cent loss. In practical problems the calculated value of is most often between the section letters c and d, and therefore the errors of

12 below.

M

the remaining calculations are negligible. It should be observed that the values in Table 12 are expressed in per cent of the true result. Thus for an error of 2 per cent with a true volt loss of 5 per cent, the calculated value would

be no greater than 5.1 and no

Table 12.

Maximum

less

than 4.9 per cent.

Errors in Per Cent of True Values at 20

Calculated Items.

Cent.

ALTERNATING-CURRENT TRANSMISSION

33

Accurate formulas cannot be given for ,the two-phase three-wire system where the size of the common wire is different from the others. However, the result may be .approximated by calculating for three equal wires and then making the proper allowance for the larger cross section of the common lead. (See In the three-phase four- wire system Example 16, page 44.) the neutral wire carries no current when the system is balanced,

and hence the

results are exactly similar to those for the three-

phase system with three wires. 22. Balanced Loads. A balanced load is assumed in all the In practice such is often not the case, although formulas. the variation from a balanced condition is usually small. The effect of unbalancing is to alter the voltage in proportion to the amount of unbalancing. The other items are also affected, but seldom is the discrepancy of any practical importance. All results have been calculated for a 23. Temperature. varies temperature of 20 cent, or 68 fahr. The value of directly with the temperature, but it also depends to a lesser degree on other conditions. However, as the error of commission is much less than the error of omission, the values of A in Table 16, page 52, have been calculated for various temperatures. The calculations have been made 24. Specific Conductivity. for a specific conductivity of 100 per cent for copper and 62 per The value of varies inversely with the cent for aluminum. specific conductivity but also depends somewhat upon other conditions. However, sufficiently accurate results are obtained by including the proper conductivity of the conductor by the

M

M

method shown in Table 16. 25. Solid and Stranded Conductors.

Wires smaller than No. have been considered solid, while the larger sizes have been taken stranded. However, since the inductance of a stranded wire

is

between that

of a solid wire of the

same

cross section of

metal and one of the same diameter, the discrepancy in the results for

final

any ordinary variation from the assumed conditions

not appreciable. 26. Skin Effect.

Owing

to

a

decreasing current

is

density

toward the center of wires carrying alternating currents, the effective

resistance

sizes of

is

increased

in

direct

proportion

to

the

and the frequency. In the usual transmission wire the effect is negligible, and even in the

product of

its

cross section


34

largest sizes shown for. the higher frequencies the maximum additional resistance is less than 5 per cent, which value decreases very rapidly with the size of wire.

Wire Spacing. - Inductance increases directly with the distance from center to center of the wires. However, the effect on the impedance of the line for large variations in the spacing -

27.

not great, even at the higher commercial frequencies. In all cases in practice, the wire factor has been In calculated for three different spacings at each frequency.

is

order to cover

general the results for wires on 18 and 60-inch centers will be than 27 inches and greater than 48 inches, respectively, while the values for 36 inches will

sufficiently accurate for spacings less

But, where greater accuracy desired for spacings other than shown, the values of may be readily interpolated from the tables.

cover spacings from 27 to 48 inches.

M

is

28.

Arrangement

of Wires.

It is

assumed

in the

two and

three-phase systems with three wires that the conductors are For other placed at the three corners of an equilateral triangle. arrangements, with the wires properly transposed, little error is

introduced by taking the distance from center to center of the wires as the average of the distances between wires 1 and 2, In the two-phase four-wire system the 2 and 3, and 1 and 3.

same phase should be used. tables have been calculated for The 29. Frequency. standard frequencies. The values for 15 cycles per second distance between wires of the

all

are

necessary for the design of single-phase railway systems, while 25, 40, 60, and 125 cycles per second cover the remaining systems Howof transmission for purposes of light, power, and traction.

where an odd frequency is to be employed, the required may be interpolated from the tables. Where circuits are in parallel from 30. Multiple Circuits. the source to receiver, the load should be proportioned between them and each line calculated separately. ever,

value of

31.

city in

M

The current-carrying capaCurrent-carrying Capacity. in Table 4, page 8, for both is shown wire amperes per

copper and aluminum. The values in the table are based on a temperature rise of 40 cent, or 72 fahr., but the currentcarrying capacity for any temperature elevation may be found from formula (4), page 9. In general, overhead transmission lines are of sufficient length


35

insure the necessary current-carrying capacity when the conditions of volt loss are met. However, it is desirable to note from Table 4 that such is the case, after each determination of to

wire.

Transmission Voltage. The voltage is taken between either at or the load source. The term "source" designates wires, 32.

the generator, the secondary terminals of step-up or step-down transformers, or a certain point of the circuit from which the calculation is made. In general the voltage at the source is given, although in some cases of transmission to a single center of distribution, the voltage at the load is specified. The formulas have been stated in terms of both voltages in order to cover all cases without any preliminary approximation. For convenience the voltage has been expressed in kilovolts, thousands of volts. In a two-phase four-wire system the voltage between the wires

same phase is specified. With lagging current the voltage

of the

than the voltage at the generator. voltage at load

is less

when the sum

at the load

is

always

less

With leading current the of the power-factor angles

and line is less than 90 degrees, but it gradually becomes greater than the voltage at the generator as their sum increases above 90 degrees. In an alternating-current system the volt 33. Volt Loss. loss in the line is the difference between the voltages at the generator and at the load. The line loss is always less than the impedance volts and almost always greater than the projection of the impedance volts on the vector of delivered volts. It may be greater or less than the resistance volts. (See Figs. 1 and 2, of load

page 30.) The per cent volt

be expressed in terms of the volts However, either may be obtained from the other by means of the simple equations shown below the table of formulas on page 50. For convenience in calculations the cent loss is a whole number. as per expressed loss

may

at load or source.

Power Transmitted. The power transmitted is always at the and is specified load, usually expressed as true power in kilowatts or as apparent power in kilo volt-amperes. Where 34.

the load

is given in amperes the equivalent value of kilowatts be obtained from equations (59) or (60) solved for W. The may values of B in Table 17, page 52, serve to make the formula


36 for

amperes per wire

to

applicable

all

systems

of

trans-

mission.

For balanced loads in one-phase, two-phase four-wire or threephase circuits the current is the same in each wire, while in the two-phase three-wire system, the current in the common lead is In the two-phase 1.41 times that in each of the other wires. three-wire system / represents the amperes in each of the outer

same as the current per phase. The power loss in any system depends only upon the current and resistance. The per cent power loss may wires, this being the 35. Power Loss.

be greater or less than the per cent volt loss. It is less than the per cent volt loss, V, when the power-factor at the source is less than the power-factor of the load, and is greater when the reverse An exception to this statement may occur when capacity is true. (See Example 20, page 48, and Example 25, page 69.) The power at the source is the sum of the power at load and the power loss in line. The power-factor of the load is always 36. Power-Factor. have been calculated from 95 per cent The values of given.

effects in the line are included.

M

lead to 75 per cent lag, for all frequencies except 125 cycles per second, giving a range sufficient to cover almost all practical

The power-factor at the load should be known accurately, as the value of varies greatly with it. The power-factor at the source depends upon the power-factor

cases of transmission.

M

at the load, the per cent power loss and the per cent volt loss. For current leading at the load, the power-factor is less leading at the source than at the load, and even may become lagging at

the source.

(See Example 24, page 68.) Lagging current at the always accompanied by lagging current at the source. The power-factor at 'the source is lower than the powerfactor of the load when the volt loss, V, is a larger percentage than the power loss, P, and it is higher at the source when the reverse is true. This statement may be modified by capacity

load

is

effects in the line.

page

(See

Example

20,

page

48,

and Example

25,

69.)

Wire Factor.

The wire factor, M, depends upon the and on the power-factor angles of the impedance load and line. The values have been calculated for copper and aluminum at all standard frequencies, for a range of sizes and 37.

of the wires,

load power-factors sufficient to cover

all

practical cases likely


37

Tables 18 to 22 give the results to arise in transmission design. for copper, and Tables 23 to 27 for aluminum. Those values which, under certain conditions, might lead to errors greater than previously specified have been omitted. 38.

The given problem may be of two kinds; and the size of wire is required, or In both specified and the line loss is required.

Given Items.

either the line loss

the size of wire

is

is

specified

cases the additional items to be given are system of transmission,

conductor metal, temperature, frequency, wire spacing, distance power delivered, power-factor of load, and the

of transmission,

voltage between wires at the source of load. The size of wire is generally determined 39. Size of Wire.

from the per cent volt loss, although in some instances it is fixed by the condition of per cent power loss. Formulas are given for both methods of procedure. When the per cent volt loss is given, the value of A is taken from Table 16, page 52, for the given system of transmission, temperature and specific conductivity. From Table 15, page 51, or VQ is found, depending upon whether the voltage is given is then calculated from The value of at the load or source. of or and the size wire is noted from Tables 18 formula (46) (47),

V

M

to 27, opposite the required value of

M.

given the per cent power loss, R is calculated from formula (48) or (49), and the size of wire is found in Table 3, page 7,

When

The wire found by opposite the required resistance per mile. of each conductor. either method is the required size It is desirable to calculate the 40. Per Cent Volt Loss. per cent volt loss after the size of wire is determined. The value

of

M whether between the section letters a and

d, is

noted from the proper table.

The

b,

b

value of

and c, or

V"

or

c

F

and " is

calculated from formula (50) or (51), and then located in Table 15, page 51, in the column headed by the same section letters as

noted above for M. The corresponding value of per cent volt loss in terms of the volts at load or source, respectively, is then Where the calcuobtained in the column headed by V or V lated value does not appear in the table, the corresponding value of V or V is easily found by interpolation. .

The remaining

calculations are clearly outlined in the table of When the voltages at the load and source formulas on page 50. become known, either formula for any required item may be used.

38


The total charging current for a 41. Charging Current. single-phase circuit or for each phase of a two-phase four-wire circuit

is

EJL

0.000122

D

2 log,,

-j-

The charging current per wire circuit

is

EJL

0.000141

E = / = L = D = d =

of a three-phase three-wire

Kilovolts between wires at source.

Frequency in cycles per second. Distance from source to load, in miles. Distance between center of wires. Diameter of wire in same unit as D.

42. Capacity Effects. Capacity influences the voltage loss, In all systems the power loss and the power-factor at generator. or commercial high frequency, except those of unusual length capacity may be entirely neglected without any disturbing error. However, the effect of capacity may be easily included in the

calculations

by assuming that the same

result

is

produced by

substituting for the distributed capacity of the line one-half its total at each end.* The true power-factor of the load is then

replaced by an apparent value determined as follows:

.'....

.

(44)

After if has been calculated, the corresponding apparent power-factor is taken from Table 14, page 39-, and this value is for the determination of the size of wire, used in the table of

M

volt loss

and power

loss.

The true power-factor at the source is obtained by the follow' using the apparent ing method: Formula (61) is solved for ' power-factor of load for K. The value of t corresponding to

K

KQ

is

noted from Table

14,

39, and substituted in formula The true power-factor at source

page

(45) for the reactance factor.

then obtained from Table

is t

.

(See

Example

20,

page

48,

by finding and Example

14,

WQ *

Method

of

',

K

corresponding to

25,

page 69.) (45)

H. Fender, Proceedings of A.I.E.E., June, 1908, page 772.


39

In formulas (44) and (45) above,

= E = EQ =s / = L = = c

= =

=

W= W=

Capacity factor. Table 13, page 39. Kilovolts between wires at load. Kilovolts between wires at source.

Frequency in cycles .per second. Distance from source to load, in miles. Reactance factor for true power-factor of load. Table 14. Reactance factor for apparent power factor of load. Reactance factor for true power-factor at source. Reactance factor for apparent power-factor at source. Kilowatts at load.

Kilowatts at source. Table 13.

Cir. Mils

or A.W.G.

Values of c for Overhead Wires.


40

has been inclosed in brackets after

of

many

the

calculated

values.

A

Example n.

transformer substation of an electric railway 3200 kw. at 15 cycles per second and 80 per cent lagging power-factor over 40 miles of single-phase copper The loss is to be 15 per line with its wires on 60-inch centers. cent of the 33,000 volts generated, the temperature 20 cent., and is to be supplied with

the wires are to be of 100 per cent conductivity.

GIVEN ITEMS.

W

=

3200 kw.;

From Table From Table

16,

EQ = 33 kv.; 7 = L = 40 miles; K = page 52, A = 0.200. page 51, V = 10.9. '

15,


Size of each wire.

2

M= .

From Table (o

-

15 per cent;

0.80 lag.

(33) X10.9 2 X 3200 X 40

page 53, use No. 00, for which

18,

M

=

0.494

d).

Per cent

From

volt loss. __

T/

From Table

15,

page

V = E = From v

Per cent power

=

From 33

=

3200

X

40

=

n6

f

c

_^\

[15.3 per cent.]

(53),

-

(1

0.153)

=

28.0 kv.

-

5000

[28.0 kv.]

(54),

1000 (33

-

28)

From Table

3,

volts.

page

[5000 volts.]

7,

0.411 ohm.

Kilowatts at source.

W

X

51,

loss.

R =

(51),

0.2

15.3 per cent.

Kttovolts at load.

VoUloss.

X

0.494

3200

From (1

+

[21.0 per cent.] (57),

0.21)

=

3870 kw.

[3870 kw.]

ALTERNATING-CURRENT TRANSMISSION Watt

From

loss.

(58),

P= 1000 (3870-3200) = 670,000 From Table

wire.

Amperes per

From

41

17,

watts.

page

52,

[670,000 watts.]

B=

1.000.

(60),

33

(1

From Table

4,

page

8,

No. 00 conductors will safely carry this

current.

From

Power-factor at source.

=

(1

+

(61),

-

0.21) (1

0.153) 0.80

=

0.820.

[0.820.]

A lighting transformer 10,000 feet distant is Example 12. to receive 50 kw. at 95 per cent lagging power-factor and 125 cycles per second with a loss of 5 per cent of the 2000 volts at load at a temperature of 30 cent. The line is to be single-phase and of copper conductors of 100 per cent conductivity on 18inch centers.

GIVEN ITEMS.

#=2kv.; L = 10,000

T7=50kw.; From Table From Table

16, 15,

page page

feet;

7=

5 per cent;

K=

0.95 lag.

A = 0.0393. V = 4.8.

52,

51,

REQUIRED ITEMS.

From

Size of each wire.

^

(2)

0.0393

From Table Per cent

22, use

= 0.905 X

From Table

15,

E = Per cent power

loss.

X

X

7=

From

+

X

10

which

M

=

0.905

(b

-

c).

(50),

0.0393

2 (1

4.8

50

0, for

page 51,

Kilovolts at source.

(46),

X

From

volt loss.

v/ ,

No.

2

50

X

10

= 445/5

4.75 per cent.

\

[4.75 per cent.]

(52),

0.0475)

From Table

= 3,

2.10 kv.

page

7,

[2.10 kv.]

R=

0.518 ohm.


42

From

(55),

From


W Amperes per

From

=

50

^_ =

=

/

2

X

0.95

From


Example

(From

13.

=

0.0282)

From Table

wire.

(59),

(57),

+

(1

51.4 kw.

page

17,

[51.4 kw.]

A=

52,

26.3 amperes.

1.000.

[26.3 amperes.]

(61),

Electric Journal, 1907,

page 231, Ex.

2.)

GIVEN ITEMS.

#=2kv.; L = 5000 feet;

TF=75kw.;

Wires

=

K=

No. 4; 0.95 lag.

cycles, 18-inch spacing. wires of 100 per cent copper and temperature of 20

System: 1-phase, 60

Assume cent.


Per

page

16,

page

21,

A =

0.0379.

56,

M=

1.38

(c

-

d).


cent volt loss.

r/ ,

52,

_ 1.38X0.0379 X

75

X 5_

4

9(Hc

_

2

^

(2)

From Table

page

15,

V=

5.40 per cent.


E= v= Example

2 (1

From

Volt loss.

14.

51,

From

+

[5.40 per cent.]

(52),

0.054)

=

2.11 kv.

[2.11 kv.]

(54),

1000 (2.11

-

2.00)

=

110 volts.

(From "Alternating Currents

Esty, 1907, page 321.)

"

[110 volts.]

by Franklin and


43

GIVEN ITEMS.

# =23kv.; L = 30 miles;

TF=1000kw.;

E =

20 kv.;

K =

0.85 lag.

System: 1-phase, 60 cycles, 18-inch spacing. Assume temperature of 20 cent, and copper of 100 per cent conductivity.

From Table

page

16,

7= -^=15 per cent.

(20)

X

21,

page

_

From Table Example

15,

page

15,

51,

V=

13.1.

X13.1

X

1000

(c

30

56, use No.

-

1,

for

which

d}.

(50),

X 1000 X 30 __+ Anff ^ ~7207" [15.5 per cent.] page 51, V = 15.75 per cent. A load of 750 kw. of 2-phase power at 25 cycles

0.936

15.

From Table

From

volt loss.

T/"

2

0.2

M= 0.936 Per cent

0.200.

REQUIRED ITEMS From (46),

Size of each wire.

From Table

A=

52,

X

0.2

per second and 100 per cent power-factor is to be delivered over 5 miles of aluminum line of 4 wires with 36 inches between conductors of the same phase, with a loss of 7.5 per cent of the 6600 volts generated, at 30 cent.

GIVEN ITEMS.

W= 750

kw.;


16, 15,

E = L =

6.6 kv.;

7 =

5 miles;

K=

page 52, page 51,

A = F '=

7.5 per cent; 1.00.

0.104. 6.4.

REQUIRED ITEMS. Size of each wire.

From

(47),

()' xa.4-- 0716 M- 0.104 X 750 X 5

.


44

From Table

M Per cent

0.755

-

(c

0, for

(51),

0.755X0.104X75X5 =

y

which

d).

From

volt loss.

No.

58, use

page

23,

=

2

6 ?6

(c

_

^

(6.6)

From Table

page

15,

F =

From

Kilovolts at load.

E =

p =

-

6.6 (1

Per cent power

From

51,

8.16 per cent.

[8.10 per cent.]

(53),

=

0.0816)

6.06 kv.

From Table

loss.

3,

[6.07 kv.]

page

7,

R=

0.837 ohm.

(55),

0.837

X

X

0.104

X

750

5

=88Qpercent

[8

.

82 per cent>]

(6.06)

From


W = 750 Amperes per

From

7

From Table

(1

0.0889)

From Table

wire.

(59),

(57),

+

=

'

X

5

75Q

= 61.9

page

8,

817 kw.

page

17,

6.06 4,

=

52,

amperes.

[816 kw.]

B =

[61.8 amperes.]

aluminum conductors

No.

0.500.

will easily

carry this current.

From


K = Example

(1

+

0.0889). (1

A two-phase

16.

(61),

-

0.0816)

=

1.00.

[1.00.]

three-wire line 30 miles long

is

to deliver 2000 kw. at 40 cycles per second and 98 per cent lagging power-factor with a loss of 10 per cent of the 20,000 volts at load.

Calculate the size of copper wires of 98 per cent conducfor a temperature of 50 cent.

tivity on 36-inch centers

GIVEN ITEMS.

W= 2000 From Table From Table

16, 15,

kw.;

E= 20 L= 30

page

52, 'A

page

51,

=

kv.;

miles;

^~ = U. C/O

V=

9.1.

V= K = 0.138.

10 per cent; 0.98 lag.


45

REQUIRED ITEMS.

From

Size of each wire.

M= 0.138 From Table the

common

Per cent T///

(46),

X

X

2000

30

page 55, use three No. 00 wires, or increase = 0.466 (c - d). For No. 00,

20,

M

lead to No. 000.

volt loss.

_

0.466

X

From

(50),

0.138

X

2000

X

30

_

Q R4

^

/

2

(20)

From Table

15,

page

51,

Since the volt loss in the

7=

10.84 per cent.

common lead is

of the outer wires,

Per cent volt

loss in

outer wire

=

1.41 times that in each

-:

=

4.50 per cent.

With a common lead one number

larger in the A. than cent each outer wire, larger) per Approximate per cent volt loss in common lead

= 10.84-

W. Gauge

(26

4.5

1.26

Therefore, in the above case with No. 00 outer wires

000 for the

common

total per cent volt loss

Approximate

and No.

lead,

=

+

5.03

= 9.53

52,

B =

0.500.

4.50

per

cent.

Amperes per

From

In

wire.

(59), in

common

From Table

17,

page

each outer wire,

wire,

/=

51

X

1.41

=

71.9 amperes.

Two substations located respectively at 40 Example 17. and 50 miles away are supplied from a No. 00, 25-cycle threephase aluminum line with wires on 60-inch centers. The total load at the nearer substation (No. 1) is 2000 kw. at 95 per cent lagging power-factor, while the other (No. 2) takes 1000 kw. at 98 per cent lagging power-factor, the generator voltage being 33,000 and the temperature 30

cent.


46

GIVEN ITEMS. For No.

1

W= 2000 kw.

:

K= For No.

E =

;

0.95 lag;

W= 1000 kw.; M= 0.655 -

2:

Per cent T/

volt

16,

page

drop

0.104.

X

0.104

X

2000

40

_

:

VQ

(33)

From Table

15,

KilovoltsatNo.

Per cent

page

V =

51,

From

1.

-

A

.

9ft

2

6.26 per cent.

#=

(53),

drop from No.

volt

;

.

REQUIRED ITEMS. From (51), No. 1.

to

_ 0.690 X

//

A=

52,

;

d).

(c

From Table

L = 40 miles 0.690 (c d) 10 miles; K = 0.98 lag; 33 kv.

M= L=

to

1

33

No.

(1

2.

- 0.0626) = 30.9 kv. E =30.9 kv. from

above.

From T/

VQ

(51),

_ 0.655 X

X

0.104

1000

X

_

10

:

n 71

^

/

2

(30.9)

From Table

15,

page

Kilovolts at No. 2.

V =

51,

From

(53),

0.81 per cent.

E=

30.9 (1 -0.0081)

= 30.6

kv.

A three-phase load of 5000 kw. and 25 cycles at is to be delivered 30,000 volts to a receiver having 95 per cent lagging power-factor, over 40 miles of No. 00 copper wires of Example

18.*

100 per cent conductivity on 48-inch centers, at a temperature of 20

cent.

GIVEN ITEMS.

W= 5000

kw.;

E= 30

kv.;

L=

Per cent

volt loss.

T///

_ 0.457 X

K= 0.95

40 miles;

By interpolation from Table 19, page From Table 16, page 52, A = 0.100.

54, M

=

0.457

lag. (c

d).


"" 0.1

X

5000

Problem of H. Fender, Proceedings

X

40

_

n

A

n

,

,.

}

of A.I.E.E., June, 1908,

*

page 771.

ALTERNATING-CURRENT TRANSMISSION From Table

page

15,

From


EQ =

30

From

Volt loss.

=

v

+

(54),

-

11.3 per cent.

33.4 kv.

=

30.0)

From Table

loss.

[11.3 per cent.]

(52),

=

0.113)

1000 (33.4

Per cent power

From

(1

V =

51,

47

[33.4 kv.]

3400 3,

volts.

page

7,

[3400 volts.]

R =

0.411 ohm.

(55),

From


WQ = Amperes per

From

5000

+

(1

(57),

=

0.101)

5505 kw.

From Table

wire.

17,

page

[5505 kw.] 52,

B =

0.578.

(59),

7

=

From Table

'

578

X

500

page

4,

=

101.4 amperes.

[101.4 amperes.]

No. 00 wires have ample current-

8,

carrying capacity.

From


(61),

=

o.94 lag.

[0.94 lag.]

A three-phase load of 10,000 kw. at 60 cycles 19. and second 98 per cent leading power-factor is to be delivered per over 65 miles of three aluminum wires on 60-inch centers, with Example

a loss of 15 per cent of the 44,000 volts generated; temperature

40

cent.

GIVEN ITEMS.

W=

10,000 kw.


16,

15,

;

E = L =

page page

44 kv.

;

65 miles;

F =

A = 0.108. TV = 10.9.

52, 51,

REQUIRED ITEMS. .

Size of each wire.

M==

From

(47),

(44)

0.108

X

2

X

10.9

10,000

15 per cent;

K= 0.98

X

65

lead.


48

From Table (a

-

No. 0000, for which

61, use

page

26,

M = 0.316

6).

Per cent T,

volt loss.

_

X

0.316

From

(51),

0.108

X

X

10,000

65

_

1 1

A

M

,

(44)*

From Table

15,

Per cent power

From

F =

page 51,

17.3 per cent.

From Table

loss.

3,

page

[17.4 per cent.]

7,

R=

0.417 ohm.

(56),

From


K =

(1

+

-

0.23) (1

(61),

0.173)

0.98=

1.00.

[0.994 lead.]

A three-phase load of 7500 kw. at 90 per Example 20.* cent lagging power-factor and 60 cycles per second is to be delivered over 140 miles of three copper wires of 100 per cent conductivity on 96-inch centers, with a voltage loss of 18.7 per cent of the volts at load, the voltage at the source being 71,200.

Include the effect of capacity and assume a temperature of

20 cent.

GIVEN ITEMS.

W= 7500

kw.;

J

=

L= From Table From

(63),

From Table

16,

page

V = 15,

page

71.2 kv.;

V=

140 miles;

K= 0.90

52,

A-

|yj^ 51,

T

lag.

0.100.

-

V=

18.7 per cent;

15-8 per cent. 11.2.

REQUIRED ITEMS. Apparent power-factor of load. Assuming that the required will be between Nos. 0000 and 8, c is approximately 0.00005. (From Table 13, page 39.) From Table 14, page 39, the reactance factor corresponding

size of wire

to 90 per cent lagging power-factor

=

0.49.

* Based on Problem of H. Fender, Proceedings of A.I.E.E., June, 1908,

page 774.

ALTERNATING-CURRENT TRANSMISSION From ,

=

(44),

page 38,

Q 4Q

_

Q.QOQQ5[71.2

-

(1

0.158)]

2

60

X

140

49

=

Q 2g

7500

By interpolation in Table 14, page 39, the apparent powerfactor of load, K', = 0.96 lag From

Size of each wire.

X

0.1

(47),

X

7500

140

From Table

21, page 56, use No. 00, for to be 0.600 (c d). interpolation

Per cent

volt loss.

T

X

0.600

T7

From

(51),

X

7500

0.1

X

140

:

_

M

which

10

A

is

found by

,,

(

(71.2)'

From Table

From

Kilovolts at load.

E= Per cent power

From

=

Page 51,F

15,

=

59.3 kv.

3,

page

0.167)

From Table

loss.

[17.2 per cent.]

(53),

-

71.2 (1

16.7 per cent.

7,

[59.0 kv.]

R=

0.411 ohm.

(55),

From


W

=

7500

(57),

+

(1

0.133)

=

8500 kw.

[8500 kw.]

From

Power-factor at generator (not including capacity).

K '=

+

(1

0.133) (1

From Table 14, page From (45), page 38,

'

39,

?

-

m

0.167) 0.96

From Table

14, Z

(61),

0.91.

0.46.

0.00005 X(71.2) 2 8500

(corresponding to

=

X

60

X

140

n 91

page 39, the true power-factor at load, ),

=

0.98 lag.

K

[0.98 lag.]


50

Formulas for A. Required Items.

C. Transmission

by Overhead Wires.

ALTERNATING-CURRENT TRANSMISSION Table 15.

V

Values of Volt Loss Factors.

51

52


Table 16

Values of

A

for

Balanced Loads.


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i

9

9

9

<

S

S

*

S

S

9

S

1

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s

i

i

NO m

-

r* r-^

S'S-i

a O^ m

>O

OO

S.

-

O

0> :

s

5

CO

O

1

*

^T

S 2 I

(N

5Q

*

ir%

? O

gj

-

S ON

m

-a

T NO

?5

CN!

O m

O a

?;

pg

CN|

OO

i

S

*

E

t>t

rv

mo*N~* S? -^

S

in

o

S

S

NO

t^

oo

SS ..

I

<

81

;.s

2

s

g.g

1 a

lolS A f u

in to

*
O?

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r^

f

O

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fO m

i

S

S-S

S

S

I

? "A

O NO

S NO

O ON

O

SinooS

^

t^

__

O

rôoONON

m o ....

NO

ro

oo

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S 1

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s

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.*

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g


61

62


CHAPTER

IV.

ALTERNATING-CURRENT TRANSMISSION BY UNDERGROUND CABLES. 44.

Introduction.

The

calculations for

underground cables

are very similar to those for overhead lines. The main difference in results is due to the proximity of the wires, in consequence of which inductance effects are considerably reduced. However,

with commercial thickness of insulation such effects are not entirely eliminated for the larger wires at the higher frequencies.

The

difference may be ascertained by comparing the values of M- at 100 per cent power-factor in Tables 34 and 35 with the resistance per mile of the same wires in Table 3. As the conductors in underground cables are almost always copper, the results have been calculated for that metal alone. The calculations for 15 cycles per second have been omitted.

The

error of the

common assumption

that the resistance drop

equals the volt loss is indicated in Table 28 below for insulation inch thickness. of

Table 28.

Error in Per Cent of True Volt Loss,

Assuming Resistance Drop = Volt Loss.

TRANSMISSION CALCULATIONS than 20 per cent of the generated volts or 25 per cent of the As in the case of overhead lines the formulas voltage at load. are approximate, but the error in general is much reduced, even less

without subdivision of the wire factors into sections. The maximum error in the calculation for the per cent volt loss is about 5 per cent near per cent and 20 per cent volt losses, but it gradually reduces to a zero error near 10 per cent loss. Thus for a true volt loss of 5 per cent the extreme range of the calculated result is about 4.85 to 5.15 per cent. Under ordinary

conditions the error

is

immeasurably small.

The maximum errors of the formulas under the most severe conditions are shown in the following table. Table

29.

Maximum

Error in Per Cent of True Values, at 20 Cent.

TRANSMISSION BY UNDERGROUND CABLES and three-conductor

cables.

determined only by the volt

65

In most cases the size of cable is loss, but in some problems the

Hence current-carrying capacity may be the ruling factor. advisable to note from Table 4 that the calculated wire

is

it is

capable of carrying the required current without undue heating.

Although underground lines are of Capacity Effects. short length, yet the closeness of the conductors comparatively 50.

and the high values insulation sometimes

The city values. tion of volt loss,

of the specific inductive capacity of the it advisable to determine the capa-

make

method power

of including capacity in the calculaand the power-factor at source is

loss

exactly like that given for overhead lines in paragraph 42, page 38.

The capacity

factor, c, for underground cables is to be taken from Table 30 below. The specific inductive capacity was assumed to be 2.5, but for any other value, c in Table 30 should be multiThus c for a threeplied by the ratio of the given value to 2.5. No. 00 and a conductors cable having specific inductive phase

capacity of

3, is

0.0007X3

=

0.00084.

2.5

Table 30.

Values of c for Underground Cables,

Circular Mils, or

A.W.G.


66

The meaning

term "source" is specified are Following typical examples illusparagraph 32, page on of formulas the the page 71 trating application 21. A one-phase underground cable 5 miles long is Example

source

is

specified.

of the

35.

in

.

800 kw. at 98 per cent lagging power-factor and 60 second with a loss of 10 per cent of the 6600 generated cycles per with volts, copper of 100 per cent conductivity and a temperature of 40 cent. GIVEN ITEMS. to deliver

# =6.6kv.; 7 = L = 5 miles; K = = 0.215. page 72, A = 9.0. page 72, 1Y"

TF=800kw.; From Table From Table

32,

31,

REQUIRED ITEMS. From (66), (6.6)^X9 =

Size of each wire.

M= 0.215 From Table Per cent

X

800

Q.456.

5


35,

From

volt loss.

v

X


///

-

M = 0.448.

(70),

X

0.448

0.215

X

.8005 _

2

(6.6)

From Table

page 72,

31,

From

Kilovolts at load.

TF

=

800

From

+

(1

From

loss.

Amperes

From

=

650

5.95 kv. volts.

0.411 ohm.

(75),


Watt

-

= 1000 (6.60 (73), 5.95) From Table 3, page 7, R = loss.

Per cent power

From

9.84 per cent. = 6.6 (1 - 0.0984) (72), E

v=

From

Volt loss.

VQ =

(76),

= 883 kw. p = 1000 (883 -

0.104)

(77),

From Table

per wire.

33,

800)

page 72,

= 83,000 watts. B = 1.000.

(79),


K = Q

(1

+

0.104)

From (1

-

(80),

0.0984)

0.98=

0.975.

TRANSMISSION BY UNDERGROUND CABLES

67

A one-phase underground cable with No. 1 Example 22. copper conductors of 98 per cent conductivity and 10,000 feet long, delivers 150 kw. at 85 per cent lagging power-factor and 125 cycles per second. The generator voltage is 2200 and the temperature is assumed 50 cent. GIVEN ITEMS.

W=

# =

150 kw.;

L=

2.2 kv.;

From Table

35,

page 74,

From Table

32,

page

10,000 feet;

M=

0.896.

A=

Q 0423

72,

'

=

K= 0.85

lag.

0.0432.

yo

\)

REQUIRED ITEMS. Per cent

volt loss. // ,

From Table

31,

From

X

0.896

(70),

V =

page 72,

From

Kilovolts at load.

X

0.0432

150

X

10

1on

14.0 per cent.

E=

(72),

2.2 (1

-

0.14)

=

1.89 kv.

A three-phase load of 3000 kw. and 25 cycles Example 23. per second is to be delivered to a receiver having a lagging powerfactor of 95 per cent, over 6 miles of underground cable of 97 per cent conductivity with a loss of 5 per cent of the 11,000 volts

generated; temperature 20

cent.

GIVEN ITEMS.

W = 3000

kw.

# =11

;

L=

kv.

;

6 miles;

From Table

31,

page 72, VJ"

From Table

32,

page 72,

=

7 = K =

A = 5L122 =

Q.103.

0.97

Jf-

From (ID*

0.103

X

X

(66),

4.8

3000

X

0.95 lag.

4.8.


5 per cent;

6


68

From Table Per cent


34,

From

volt loss.

V '" = 0-280 X

(70),

X

0.103

(H)

From Table

31,

Volt loss.

From

Per cent power

From

=

(73), v

X

X

4.5

=

4 30

page

3,

= 495 volts. = 0.258 ohm. 7, R

11

(75),

X 3000 X p _ 0.258 X 0.103 [11 (1 0.045) 0.95]

^

6

2

Amperes per

From

6

4.50 per cent.

10

From Table

loss.

X

3000

2

VQ =

page 72,

M = 0.280.

From Table

wire.

33,

cent

B =

page 72,

0.578.

(79),

I

=

578

'

-

11 (1

From Table

><

page

4,

3QQQ

=

0.045) 0.95

174 amperes.

seen that the cable will not over-

8, it is

heat.

From


K =

+

(1

0.048) (1

(80),

-

0.045)

0.95=

0.95.

A

conductors of three-phase cable with No. kw. at 6000 delivers 2500 98 per cent conductivity volts, 95 and 60 cycles per second, to a per cent leading power-factor

Example

24.

receiver 3 miles from the generator, the temperature of the wires being 40 cent.

GIVEN ITEMS.

IF-

1500 kw.; #=6kv.; From Table 35, page 74,

From Table

Per cent

page

32,

0.475

5 miles; 0.475.

-^ =

0.110.

(J.9o


volt loss.

v _

A =

72,

K= 0.95 lead.

L=

M=

X

X

0.11

1500

X

5

_

cent

1Q 9

2

(6)


# =

6

(1

Per cent power

+

From

0.109)

loss.

=

(71),

6.65 kv.

From Table

3,

page

7,

R=

0.518 ohm.


From

(74),

= 0.518 X .

0.11 .

From

1500

X

From Table

wire.

-5

^

=

cent

(0.95)

(6)

Amperes per

X

2

2

69

33,

page 72,

B=

0.578.

(78), 7

=

X

0.578 6

X

1500

0.95

From Table 4, page 8, this current is slightly in excess of the safe current-carrying capacity of the cable.

From


d+0.132) 0.95 a Q97a 1 + 0.109

g There

is

(80),

no indication from the above result whether the curIt then becomes is leading or lagging.

rent at the source

necessary to construct a vector diagram similar to Fig. 2, page 30, it will be found that the current in the above case is

from which

leading at the source.

A

three-phase load of 400 kw. at 95 per cent lagging power-factor and 125 cycles per second is to be transmitted 8 miles with a loss of 7.5 per cent of the generated voltage,

Example

25.

Include capacity effects and assume wires of 97 per cent conductivity and a temperature of 30 cent.

6600.

GIVEN ITEMS.

W= 400

EQ =

kw.;

L=8 From Table

31,

page 72,

From Table

32,

page 72,

7 =

6.6 kv.;

K

miles;

7 "'=

7.5 per cent;

=0.95

lag.

6.95.

A - ^~- =

0.107

0.97

REQUIRED ITEMS. Assuming that the required size Apparent power-factor of load. will be between Nos. 1 and 6, c= 0.0006 (from Table 30, page 65). From Table 14, page 39, the reactance factor for 95 per cent of conductors

lagging power-factor

=

0.33.


70

From

(44),

0.0006 [6.6

-

(1

X

2 0.075)] 125

8

97

~400~~ interpolation in Table 14, page 39, the apparent factor of load, K', = 0.97 lag.

By

From

Size of each wire.

From Table

35,

(66),

X

0.107

v

-

X

8

approximately.

From

volt loss.

"'

400

page 74, use No. 2 cable, for which

M = 0.925, Per cent

power

X

0-925

(70),

X

0.107

400

X

8

_7

97

2

(6.6)

From Table

Per cent power

From

7 =

page 72,

31,

7.87 per cent.

From Table

loss.

page

3,

R=

7,

0.824 ohm.

(75),

From


(76),

W

=

400

(1

+0.081)

Power-factor at source (not including capacity).

K '=

+

(1

Q

0.081)

From Table 14, page From (45), page 38, t

~

From Table

14,

page ),

'

39,

t

0.0727) 0.97

=

=

~~

39,

=

From

(80),

0.97 lag.

0.27.

0.0006 (6.6) 2 125

n 97

(corresponding to

(1

-

= 432 kw.

X

8 _

ft

91

the 'true power-factor at load,

0.98 lag.

K

TRANSMISSION BY UNDERGROUND CABLES Formulas

for A. C. Transmission

Required Items.

by Underground Cables.

71


72 Table

FO

31.

Values of

F '" = F

(i-o.oi

F

).


73

74


CHAPTER

V.

INTERIOR WIRES FOR ALTERNATING-CURRENT DISTRIBUTION. Interior wiring involves short runs of con52. Introduction. ductors; therefore the conductors are often determined by their current-carrying capacity rather than by conditions of maximum

drop.

It is generally advisable to note the required size of wire

both conditions. Table 4, page 8, gives the National Electric Code Standard for current-carrying capacity of interior wires. The formulas for interior wiring calculations are similar to for

The the preceding ones for alternating-current transmission. units of power, voltage and distance have been changed in order All the required items are expressed terms of the current per wire and the per cent volt loss is also given in terms of the power at load so that problems involving watts at load and voltage at source may be solved without

to facilitate calculation.

in

preliminary approximation. The ordinary error of the calculation

is considerably less than stated in paragraph 45 for underground conductors, on account of the smaller wires usually employed. The error at 20 cent,

common assumption

that the resistance drop equals the indicated in Table 36, page 76, for wires in conduits and on 3-inch centers.

of the

volt loss

is

It is apparent that results based on Table 36 may be much greater or much less even at power-factors near unity. The power-factor angles of the load and line. page 30.)

the assumptions of than the true values, error is due to the (See Figs.

1

and

2,

M

have been The values of 53. Properties of Conductors. calculated for wires of 100 per cent Matthiessen's Standard at a temperature of 20 cent, or 68 fahr. However, the effects of temperature and conductivity are introduced by means of a and b page 82. Table 39, page 83, gives the values of 54. Spacing of Wires. for wires with a thickness of insulation of inch, while

in Table 37,

M

75


76

Table 40, page 84,

is

calculated for conductors on 3-inch centers,

Any slight variation from either of these spacings will not appreciably alter the results, even for the largest wires at the highest commercial frequency. Table 39 is to be used for wires in conduit, duplex cables, multi-conductor cables or twisted wires. Table 40 is for wires in molding, or for open work such as wires on cleats or knobs. A simple method of dealing with dis55. Ampere-Feet. tributed lamps is by use of formula (84) on page 8L The term is equal to the sum of the products / each load, given by multiplied by its distance I.

II is

the ampere-feet and

Error in Per Cent of True Volt Loss, Assuming Resistance Drop = Volt Loss.

Table 36.

Size

of

Wire.

A.W.G.

WIRES FOR ALTERNATING-CURRENT DISTRIBUTION

77

A group of lamps having a power-factor of Example 26. 95 per cent lag is to be supplied with 20 amperes at 60 cycles per second over a single-phase copper cable 150 feet long. Calculate the size of wire for 2 volts drop at 30

cent, for copper

of 97 per cent conductivity.

GIVEN ITEMS. /=

v=

2 volts; I 20 amperes; From Table 37, page 82, a = 963

= X

K = 0.95 lag.

150 feet; 0.97 =

934.


Size of each wire,

934 x 2 20 X 150

0.623.

M

From Table Example

=

0.623. 39, page 83, use No. 5, for which a cable with At 40 cent, No. 8 27. single-phase

conductors of 97 per cent conductivity and 200 feet long, delivers 25 amperes at 98 per cent lagging power-factor from a trans-

former which gives 100 volts at 125 cycles per second.

GIVEN ITEMS. /

=

25 amperes;


e

=

100 volts;

39, page 83, M 37,

=

I

=

200

K = 0.98

feet;

1.27,

page 82,

a = 928

X

0.97

=

b

900;

=

= 222. -^ y \j

/

*

REQUIRED ITEMS. Volt loss.

From

(85),

1.27

Volts at load.

Per cent

xsx 200 = 70voltg

From (89), e = From (86),

100

-

7

=

93 volts.

volt loss.

Per cent power

loss.

From Table

3,

page

7, r

=

0.000627.

lag.


78

From

(90),

p _ 0.000627 X Watts

at load.

From

(94),

From Table

w=

25

From

loss.

X

93

A

28.

=

?

38,

X

page 82,

B=

1.000.

98

-

2280 watts.

0.076

X

2280== 173 watts.

'

.00

p=

(92),

From


Example

200

X^25

1

Watt

X

222

60-cycle

(96),

step-down transformer with 120

volts at its secondary terminals is connected to a load of 2000 watts at 90 per cent lagging power-factor over 200 feet of singlephase circuit on cleats. Determine the size of wire for 5 per cent

drop at 20 cent, and copper of 98 per cent conductivity.

GIVEN ITEMS.

w=

2000 watts;

=

e

120 volts;

1= 200

feet;

7 = K=


From Table 31, page 72, 7 ///= 4 -8v= 0.05 X 120 = 6 volts. From (98), From Table 37, page 82, a = 1000 X 0.98 - 980. B from Table 38, page 82, = From (95), (in which

1.000),

2000 120

Size of each wire.

-

(1

0.05) 0.90


98QX6 =1.51. M= 19.5 X 200 From Table Per cent

page

40,

volt loss.

84, use

From

No.

8, for

which

M= 1.20.

(87),

v '" = L2Q X 0.01 X

200 980

X 20 X (120)

m 3 40 2


From Table

31,

Volts at load.

Amperes

V = From (89), e= page

72,

From

per wire.

3.50 per cent.

120

(1

= From Table

4,

by No. 8

page

-

0.035)

=

116 volts.

(94),

X

116

carried

79

8, it is

0.90

19.2 amperes.

seen that the current will be safely

wires.

A load of 1500 watts at 110 volts, 85 per cent Example 29. lagging power-factor and 60 cycles per second is to be delivered 250 feet over a two-phase four- wire circuit on cleats, with a loss of 2 per cent of the generated volts. Calculate the size of wire of 98 per cent conductivity for a temperature of 30 cent.

GIVEN ITEMS.

W=

1500 watts;

1= 250 feet; From

v

(98),

=

= 110 volts; K=0.85 lag.

V =

e

-

X

02 ~

JL

11Q

-

2 per cent;

2.25 volts.

U.U^

^

From Table 37, page 82, a = 963 X 0.98 = 944. From (94) (in which B from Table 38, page 82, =

0.500),


From

(84),

X 2.25 M = 944 8 X 250 From Table Per cent

40,

page

volt loss.

Volts at source.

84, use

From

From

No.

8, for

which

M = 1.18.

(86),

(88), e

=

110

(1

+

0.0227)

=

113 volts.


80

A 125 volt, 125 cycle generator is to supply a of load 100 amperes per wire at 100 per cent powerthree-phase factor over 225 feet of cable, with a loss of 3 per cent of the generated volts. The wires are to have a conductivity of 99 per cent Example

30.

and a temperature

of 30

cent.

GIVEN ITEMS. 7

=

100 amperes;

=

e

125 volts;

F =

K=

/= 225 feet; From (98), v=0.03 X 125=3.75 From Table 37, page 82, a

=

1110

X 0.99=

1100;

3 per cent

;

LOO.

volts.

&

=

-=181.


From

(84),

,.1100 X 3.75_ 100 X 225 '

From Table

39,

Volt toss.

From

Per cent


volt loss.

Volts at toad

ft100

M = 0.163.

(85),

From

From

(98),

(89),

e=

125

- 3.3=

121.7 volts.

WIRES FOR ALTERNATING-CURRENT DISTRIBUTION Formulas for A.C. Interior Wiring. Required Items.

81


82 Table 37.

Temperature in

Degrees.

Values of a and b for Balanced Loads.


S S - _

S 3 9 N V o 2

08 5r **

S -

S

288:

So fn ?R
S o 3 ? in >o

cs

1

to t-

3

I

<

.

T

""

S :<=>

!n

S8S5 SS.R .s CA

i^2

SSSi 58 5.8.

2Ê

8

5

2 5-2-2

s 2.s u-v

----

O

as i

!z

2 5

s

i

S 1

s 1

S

.

i orsu^^ ONOr^

l

TOOOOO fsAmT

oorô^ -o ix

SS3

M

T*^GO

3S s

ssss

83

84


tApr,

o^CMf^cs

oo-

a

OO

s S

5?

S

rÔ^ S

".

O^r^\t> -.

.

.

".

S S S

--

--

o

IT

ssss

1

O ÎÔ

OOO
OO

"r

fs,

oo

SSS in Ô

*

o fei

s so

=3.8.-.

1

I
I

r^rô

oo
a a s

552

4

s

=

r-rtn

228

a

S

55

TT
& S

-.

.

s OÔO^T

"

a

02

O

^

tOCOOOO

^fo^ *

IM

CHAPTER

VI.

DISTRIBUTION FOR SINGLE-PHASE RAILWAYS. 57.

Introduction.

A

an double or multiple

trolley circuit consists primarily of

overhead trolley wire suspended from

single,

catenary construction, and the track rails. The trolley wire may be in parallel with an auxiliary feeder or " by-pass." However, the installation of auxiliary feeders is the exception rather than the rule, and since their position with respect to the trolley wire

and

much importance but made to include them in

rails is of

has been

of wide variation, no attempt The table of

M

the results.

has been calculated for 15 and 25-cycle systems with one and two bonded rails per track. The rails have a relative resistance of

copper of 12, while the trolley wires are taken at 100 per cent conductivity and 20 cent. The value of is based on the 58. Method of Calculation. " " and on isolated Report of Electric Railway Test Commission

steel to

M

calculations published in various places. General laws have been deduced for the grounded portion of the circuit, and they have been found to agree sufficiently well with the experimental results. The properties of the trolley and catenary construction have been calculated and the results have been combined with

those for the

rails to

obtain the values of the complete circuit. are similar to those for overhead trans-

The formulas on page 92

mission lines except that the former have been expressed in terms amperes in order to facilitate computation. The per cent volt loss is also given in terms of the kilowatts at load in order to

of

avoid approximation when the voltage

is

known only

at

the

source.

Due to skin effect, an alternating cur59. Impedance of Rail. rent flowing through a steel rail has a diminishing density toward the center, thereby increasing the effective resistance of the rail.

Eddy currents and hysteresis losses produce a further increase in the effective resistance, resulting in an impedance to alternating current considerably greater than the resistance of the 85


86

The

impedance divided by the impedance ratio. The impedance of a complete track depends upon the frequency of the system, on the height of the trolley wire, on the shape, size, permeability and specific resistance of the rail, and on the character of bonds and roadbed. It is evidently a complex quantity and can be determined only by tests of installed track under normal conditions of operation.

rail to direct

current.

resistance to direct current

ratio of

is

called the

The permeability of a given rail Permeability of Rail. of the current. The method of variation depends upon density 60.

irregular but an average value of the permeability can be assumed without causing much error in the final result. 61. Impedance and Weight of Rail. The impedance depends on the size and shape of rail and differs in the standard bullhead It section of foreign practice from the T-rail of this country.

is

changes very slightly with the weight of the rail commonly In any case the impedance has-been found by experiment to vary inversely with the perimeter of the section. 62. Impedance of Rail and Frequency. Experiment shows, at least at lower frequencies, that the impedance of rails varies

installed.

directly as the square root of the cycles per second.

Thus

at

25 cycles per second the impedance is about 1.3 times the corresponding value at a frequency of 15. It has been concluded 63. Formula for Rail Impedance. from experiments that the impedance of rails varies directly as

the square root of

the frequency and

inversely

as

the

However, perimeter of standard T-rails is approximately proportional to the square root of their weight

perimeter.

the

per yard. The author has developed the following formula and has found that it agrees sufficiently well with tests to serve for purposes of practical calculation. Impedance per mile of one T-rail,

bonded and

installed,

'Cycles per second

-"v/i Pounds per yard

The constant 0.8 is based on an impedance ratio of 6.6 at 25 cycles per second for a 75-pound bonded rail of a relative resistance of steel to copper of about 12, with the trolley wire about 20 feet above the track.* * General Electric Co.'s Bulletin, No. 4392, 1904.

DISTRIBUTION FOR SINGLE-PHASE RAILWAYS

87

The power-factor increases with 64. Power-Factor of Track. the current, but at the lower frequencies this variation is small for normal current densities. Tests of the Electric Railway Commission show a

slight increase of power-factor with the but other experiments indicate practically no change.* The power-factor of the track has been assumed constant for all rail weights, at 65 per cent for 15 cycles and 55 per cent for 25 cycles per second.

weight of rail

Height of Trolley. Trolley wires for single-phase in from about 18 to 22 feet above rails. height systems range The height assumed in the calculations of is 22 feet, but any variation within standard limits will have small effect on the impedance of the trolley, and almost no effect on the impedance 65.

M

of the track. of Catenary Construction. The trolley wire is from a or from two wires arranged supported single catenary wire, 66.

Effect

either horizontally or vertically. In almost every case the trolley and catenary wires are electrically connected. However, since

the catenary wires are steel, their skin effect prevents them from carrying more than a small part of the current, which for single catenary construction has been determined by calculation to

be equivalent to about 10 per cent.f In double or multiple catenary construction the proportional part of the current carried

by the

trolley

is

somewhat reduced, but

in order to be safe in

calculation for either type of overhead construction, the above result has been used.

The ohmic resistance 67. Impedance of Complete Circuit. and the reactance of the track and trolley having been calculated, the two were combined to obtain the total impedance. Table 42, page 93, gives the values of M. It will be observed that the effect of varying the rail weight is less pronounced than of varying the size of trolley, or, in other words, that the greater part of the total impedance is due to the trolley. Since the impedance of the overhead system is susceptible of fairly accurate calculation, it follows that any error and variation in the assumed value of the track impedance will have a small effect on the final result.

This conclusion

Following

is

is

corroborated by

tests.

a comparison between the calculated impedance

* Foster's "Electrical Engineer's Pocketbook," 1908, page 795. t J, B. Whitehead, Proceedings of A.I.E.E., May, 1908, page 627.


88

per mile and the published results of tests on installed lines. The values of the Electric Railway Test Commission are not included, as their test track is not similar to modern construction.

Table 41.

Rails.

Test and Calculated Values of Impedance Per Mile at 25 Cycles Per Second.


From

(107)

obtained:

and Table

42,

page

89

93, the following results are


90

Example bonded

A

33.

25-cycle single-phase road with four 80-

and two No. 000

trolleys supplies each of three cars with 50 amperes at 85 per cent power-factor when located at 3, 5, and 10 miles away, respectively, the power station Ib.

rails

voltage being 11,000. Owing to the power-factor at the solution

two

first

cars, the following

slightly in error.

is

GIVEN ITEMS.

IL = 50

From Table

Per cent

(3

+

+

5

10)

volt loss to

FQ=

900 ampere-miles.

M=

page 93,

42,

=

=

0.0315.

REQUIRED ITEMS. last car. From (103),

0.0315

X

900

From

Kilovolts at last car.

=26pe

(106),

#=11

(1-0.026) =10.7 kv.

A 25-cycle single-phase car starting 8 miles Example 34. from a transformer station generating 6600 volts takes 500 kw. at 80 per cent power-factor from a circuit consisting of two 70Ib. rails and one No. 000 trolley. GIVEN ITEMS.

TF=500kw.; # =6.6kv.; L= 8 miles; From Table 42, page 93, M = 0.066.

Per cent


volt loss.

v

,

K= 0.80 lag.

0.066

=

X

500

X

8

_

fi

2

(6.6) 0.80

From Table

page

31,

From

Kilovolts at load.

Amperes.

From

72,

8.3 per cent.

(106),

E=

6.6 (1

-

0.083)

=

6.05 kv.

(112),

6.05

Per cent power

7 =

loss.

X

0.80

From Table

103 amperes

42,

page 93,

R = 0.425 ohm.


From


W

=

500

(1

+

K =

+

(1

(111),

=

0.072)

From


91

-

0.072) (1

536 kw.

(114),

0.083) 0.80

=

0.79 lag.

Two 15-cycle single-phase locomotives start miles from a power station generating 11,000 10 simultaneously Determine the line voltage at the locomotives if each volts. takes 3000 kv.-amp. (kilo volt-amperes) at 75 per cent powerExample

35.

factor, over a circuit consisting of eight 100-lb. rails

and four

No. 0000 trolley.

GIVEN ITEMS. IF'

= 6000

From Table

page

42,

M= W Per cent

-

volt loss.

/

T/

L= 10 miles; K = 0.75 lag.

# =llkv.;

kv.-amp.;

93,

46

=

0.0115.

4

6000

X

=

0.75

4500 kw.


_ 0.0115 X

4500

X

10

_

c

7

0.75

From Table

31,

page 72,

7 =

Kilovolts at locomotives.

tf=ll(l -

6.1 per cent.

From

(106),

0.061)=

10.3 kv.

92

TRANSMISSION CALCULATIONS Formulas Required Items.

for Single-Phase

Railway

Circuits.


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s

UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return This book

is

to desk

DUE

from which borrowed.

n-rfielMt data Damed below.

1 ? 19497

I

LD

21-100m-9,'48(B399sl6)476

tea-

Engineering library

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tJL

Practical Transmission Line Calculation

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