Practical Transmission Line Calculation

ENGIN.

LIBRARY

PRACTICAL

CALCULATION

OF

PEACTICAL

CALCULATION

OF

TBANSMISSION

LINES

FOR

DISTRIBUTION

OF DIRECT

AND

ALTERNATING

CURRENTS BY

MEANS

OF OVERHEAD,

UNDER-GROUND,

AND

INTERIOR WIRES

FOR

PURPOSES

OF

LIGHT,

POWER,

AND

TRACTION

BY

L.

W.

ROSENTHAL,

E.E.

o

ASSOCIATE MEMBER, A.I.E.E.

NEW

YORK

MCGRAW

PUBLISHING

COMPANY

239

WEST

39TH

STREET

Engineering

library

COPYRIGHT, 1909,

BY THE

McGKAW

PUBLISHING

COMPANY

NEW

YORK

StanbopcJpresa F. H.G1LSON COMPANY

PREFACE.

THIS little book is offered to the _{engineering profession} with

the_hope thatit

_may

_{be of practical help}inthe _rapid and accurate calculation of transmission lines. Its existenceis the outcome of the belief that this field is _{in part barren.} Itschief mission is to substitute a direct solution for the trial method which was_formerly a _necessaryevil.

The

arrangement of the formulas, tables andtext

has been dictated _{solely by}the needs ofthe_rapid worker.

All sections _except the last include the _important effects of

temperature and specific conductivity,,

The

section relating to direct-current _railways is novelin the form of its _tables, and the

methods outlined in ithave been found _rapid and comprehensive.,

The

alternating-current division presents a

new

andoriginalmethod for_{the solution of these problems.} It is_{the only}method

known

to

the author which determines the sizeof wire_directlyfrom thevolt

loss inthe_line,and it also_{possessesunique} features of scope, accu-racy and simplicity.

The

chapter on single-phase railways is in

accord with mostof the consistent _{data published} onthe _subject,

although further accurate _{investigations} of installed lines

_may

modifyto some extent_{the presentaccepted} values.

The

authordesires to call_{particular attention to}the fallaciesof

some familiar methods of calculating alternating-current transmis-sion lineswhich heretofore have been in

common

use. It will be evident from Tables 11, 28, and 36 that their results arewholly 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 confinedto methodsof

calcula-tion. _Hence, the most desirable limits of line losses are not discussed,but the tables are sufficiently extendedto cover all cases likely to arise in practice. There is no discussion of the

character-istics of _{alternating-current} transmission _{lines, although} the tables of wirefactors _{render apparent}

_many

of their _important features. Furthermore,the book does not include either thedeterminationof

iii

iv

PREFACE

the size of conductors for conditions of

maximum

_economy or the consideration of alternating-currentcircuits in series.

The

_{preparation of the tables} has involved thousands of calcula-tions, but thorough checksby themethods of differences and curve

plotting have probably eliminated almost all errors of material

influence. _{However, somediscrepancies}

_may

have _creptin,and the

author would be _gladto learn ofthem.

L.

W.

R.

CONTENTS.

CHAPTER

I.

DIRECT-CURRENT

DISTRIBUTION

FOR

LIGHT

AND POWER.

PAR. PAGES 1. INTRODUCTION 6^ 2. PROPERTIES OF CONDUCTORS : 5 3. CURRENT-CARRYING CAPACITY 6

4. PARALLEL RESISTANCE OF WIRES 9

5. GIVEN ITEMS 10

6. FORMULAS 10

7. AMPERE-FEET 10

8. EXAMPLES. . 11

CHAPTER

II.

DISTRIBUTION

FOR DIRECT-CURRENT

RAILWAYS.

9. INTRODUCTION 16

10. RESISTANCE OF RAILS 16

11. PARALLEL RESISTANCE OF RAILS AND FEEDERS 17 12. NEGATIVE CONDUCTORS 18

13. POSITIVE CONDUCTORS 18

14. RESISTANCE OF CIRCUIT 19

15. GIVEN ITEMS 19

16. EXAMPLES.. 20

CHAPTER

III.

ALTERNATING-CURRENT

TRANSMISSION

BY

OVERHEAD

WIRES.

17. INTRODUCTION 29 18. OUTLINE OF METHOD 30 19. RANGE OF APPLICATION 31 20. MAXIMUM ERROR 31 21. TRANSMISSION SYSTEMS 32 22. BALANCED LOADS 33 23. TEMPERATURE 33 24. SPECIFIC CONDUCTIVITY 33

25. SOLID AND STRANDED CONDUCTORS 33

26. SKIN EFFECT 33

27. WIRE SPACING 34

vi

CONTENTS

PAR. PAGE 28. ARRANGEMENT OF WIRES 34 29. FREQUENCY 34 30. MULTIPLE CIRCUITS 34 31. CURRENT-CARRYING CAPACITY 34 32. TRANSMISSION VOLTAGE 35 33. VOLT LOSS 35 34.

POWER

TRANSMITTED 35 35.

POWER

LOSS 36 36. POWER-FACTOR 36 37. WIRE FACTOR 36 38. GIVEN ITEMS 37 39. SIZE OF WIRE 37

40. PER CENT VOLT LOSS 37

41. CHARGING CURRENT 38

42. CAPACITY EFFECTS 38

43. EXAMPLES. . 39

CHAPTER

IV.

ALTERNATING-CURRENT

TRANSMISSION

BY

UNDERGROUND

CABLES. 44. INTRODUCTION 63 45. MAXIMUM ERROR 63 46. TEMPERATURE 64 47. PROPERTIES OF CONDUCTORS 64 48. THICKNESS OF INSULATION 64 49. CURRENT-CARRYING CAPACITY 64 50. CAPACITY EFFECTS 65 51. EXAMPLES 65

CHAPTER

V.

INTERIOR

WIRES FOR ALTERNATING-CURRENT

DISTRIBUTION. 52. INTRODUCTION 75 53. PROPERTIES OF CONDUCTORS 75 54. SPACING OF WIRES 75 55. AMPERE-FEET 76 56. EXAMPLES 76

CHAPTER

VI.

DISTRIBUTION

FOR

SINGLE

PHASE

RAILWAYS.

57. INTRODUCTION 85

58. METHOD OF CALCULATION 85

59. IMPEDANCE OF RAIL 85

60. PERMEABILITY OF RAIL 86

61. IMPEDANCE AND WEIGHT OF RAIL 86

CONTENTS

vii

PAR. PAGE

63. FORMULA FOR RAIL IMPEDANCE 86

64. POWER-FACTOR OF TRACK, , 87

65. HEIGHT OF TROLLEY 87

66. EFFECT OF CATENARY CONSTRUCTION,.. 87

67. IMPEDANCE OF COMPLETE CIRCUIT 87

68. MULTIPLE TRACKS..=.,.. , 88

TABLES.

CHAPTER

I.

DIRECT-CURRENT

DISTRIBUTION

FOR

LIGHT

AND

POWER.

No. PAGE

1. ValuesofTforcopperand aluminum 6

2. Valuesof

F

forcopperandaluminum. 6

3. Propertiesof_copperand aluminum , 7

4. _{Ampere-feetper}voltdropandcurrent-carrying capacity 8

5. Valuesof

H

forcopperoraluminum., 8

Formulasfordirect-current wiring 14

6. Valuesofaforcopperand aluminum. 15

CHAPTER

II.

DISTRIBUTION

FOR

DIRECT

CURRENT

RAILWAYS.

7. Valuesof Tlforsteel 17

8. _Equivalentsof_copperof100 per cent conductivity.., 20

9. Resistanceto directcurrentofonesteelrail 24 Formulasfordirect-current railwaycircuits , 25

10. Valuesof

A

forwiresandrails 26

CHAPTER

III.

ALTERNATING-CURRENT

TRANSMISSION

BY

OVERHEAD

WIRES.

11. Errorin_percentofvolt loss , 31

12.

Maximum

error in_{per cent}of true valuesat.20 cent 32

13. Valuesof c foroverheadwires 39

14. Reactancefactors ,...., 39

Formulasfora. c.transmission_byoverheadwires 50

15. Valuesofvolt loss factors .. 51

16. Valuesof

A

forbalancedloads 52

17. Valuesof

B

forbalancedloads. , 52

18. Valuesof

M

for_{overhead copper}wiresat15_cyclespersecond 53

19. Valuesof

M

foroverhead copperwiresat25cyclespersecond 54

20. Valuesof

M

foroverhead copperwires at40cyclespersecond 55

21. Valuesof

M

for_{overhead copper}wires at60cyclesper second. ... 56

22. Valuesof

M

foroverhead copperwiresat125cyclespersecond 57

23. Valuesof

M

foroverheadaluminumwiresat15_{cyclesper second.}.. 58

24. Valuesof

M

foroverheadaluminumwiresat25cyclespersecond... 59

25. Valuesof

M

foroverheadaluminumwiresat40_cyclespersecond... 60

26. Valuesof

M

foroverheadaluminumwiresat60_cyclespersecond... 61

27. Valuesof

M

foroverheadaluminumwires at_{125 cycles per second.}. 62

X

TABLES

CHAPTER

IV.

ALTERNATING-CURRENT

TRANSMISSION

BY

UNDERGROUND

CABLES.

No. PAGE

28. Errorin_{per cent}oftruevoltloss

'

63

29.

Maximum

error in_{per cent}oftrue valuesat20 cent 64

30. Valuesofcfor_undergroundcables 65

Formulasfora.c. transmissionbyundergroundcables 71

31. Valuesof

V

_Q'"

= F

(1

-

0.01

F

) 72

32. Valuesof

A

forbalancedloads 72

33. Valuesof

B

forbalancedloads 72

34. Valuesof

M

formultipleconductor coppercables 73

35. Valuesof

M

formultipleconductor coppercables 74

CHAPTER

V.

INTERIOR

WIRES FOR ALTERNATING-CURRENT

DISTRIBUTION.

36. Errorinper centoftruevoltloss 76

Formulasfora.c.interior_wiring 81

37. Valuesofaandbforbalancedloads 82

38. Valuesof

B

forbalancedloads 82

39. Valuesof

M

forcopperwiresin interiorconduits 83

40. Valuesof

M

forcopperwiresin_moldingoroncleats ... 84

CHAPTER

VI.

DISTRIBUTION

FOR

SINGLE-PHASE RAILWAYS.

41. Testandcalculated valuesofimpedanceper mile 88

Formulasforsingle-phaserailwaycircuits 92

PART

I.

DIRECT-CURRENT

DISTRIBUTION

BY MEANS

OF

OVERHEAD,

UNDERGROUND

AND

INTERIOR

WIRES

FOR PURPOSES OF

DIRECT-CURRENT

DISTRIBUTION

CHAPTER

I.

DIRECT-CURRENT

DISTRIBUTION

FOR

LIGHT

AND POWER.

1. Introduction. Problems in direct-current transmission

and

_{distribution are relatively simple.}

The same

formulacovers

all conditionsofinstallation

and

_operation whether

_by

_overhead, underground or interior wires.

The

formulas _give accurate results,

and

all itemsofinfluence are easily included.

The

tables

are concise but comprehensive

and

will cover almost all the usual

and

unusual requirements of varied _practice.

2. _{Properties of}Conductors.

The

resistance of stranded

and

solidconductorsofthe

same

crosssection

and

_lengthis_practically the same. Table3, page 8, gives the properties of wires at 20

cent, or 68 fahr. for _copper of _{100 per cent}

and aluminum

of

62 per cent conductivity in Matthiessen's standard scale.

The

resistance at

_any

other _temperature

and

_conductivity

_may

be found for_copperor

aluminum

fromformulas ₍₁₎

and

_(2).

v

Ohms

resistance_{per 1000}feet

=

>

*

. .

S

Ohms

resistance_{per mile}

=

54/700

X

T

^

^

o

S

=

Cross section of metal in circular mils.

T

=

Temperature factor. Table 1, page 6.

Thus, at 40 cent, the resistance _{per mile} of No. 1 _copperwire

6

TRANSMISSION CALCULATIONS

Table i. Values of

T

for_{Copper and}

Aluminum.

DIRECT-CURRENT

DISTRIBUTION

Table 3. Properties ofCopper and

Aluminum

at20 Cent, or 68 _{Fahr. Conductivity}inMatthiessen's Standard Scale;

8

TRANSMISSION CALCULATIONS

Table 4.

Ampere-Feet

per Volt

Drop

and Current-Carrying

DIRECT-CURRENT

DISTRIBUTION

9

current per wire for lead-covered _underground cablesis based on tests for a _temperature rise from 70 fahr. initial to 150 fahr.

final.*

Formula

₍₄₎

_may

be solvedfor

C

to findthe_temperature rise under _given conditions.

Amperes

per wire

=

H

\~r

...

(4)f d

=

Diameterofwire in inches.

C

=

_Temperature risein _{degrees centigrade.}

H=

Heat

factor. Table 5, page 8.

T

=

_Temperaturefactor atfinal_temperature. Table1, page6.

The

size of wires is sometimes determined

_by

their current-carrying 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 hassufficient _{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.

4. Parallel Resistance of Wires.

The

parallel resistance at

any

temperature for a

number

of wires of

_any

_conductivity is

given

by

the following formulas:

in

Ohms

resistance_{per 1000} feet

=

-

u>

+

1

₊

T

12

T

*

Ohms

resistance _{per mile}

=

-"/

(5)

. . .

T

1

T

i * 2

Where

allthewireshave the

same

_temperature

and

_{conductivity,}

formulas ₍₅₎ reduce to those below.

Ohms

resistance _{per 1000} feet

-

10'350

X T

.

S,

+

S

2

+

-Ohms

resistance _{per mile}

=

54r700

X

T

^

S

l

+

S

2

4-In formulas (5)

and

(6)

Sit

S

2

=

Circular mils in respective wires.

T,

T

l}

T

2

=

Temperature factors of respective wires. Table 1,

page 6. *

Standard Underground CableCo.'sHandbookNo._{XVII, page}192.

t Based onformulasinFoster's"Electrical_{Engineers'Pocketbook,"1908,}

10

TRANSMISSION CALCULATIONS

Thus, theresistance _{per mile} at 30 cent, ofone No. 00 _trolley wire of _{97 per cent conductivityin parallel} with one

1,000,000-cir. mil

aluminum

feederof_{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 _copperwiresof_{97 per cent conductivity}

18

10,350

X

1.110

₌

Q078Q

ohm

105,500

+

41,740

5. Given Items.

The

problem

may

require the determina-tion of the size of wire from the volt loss, or the volt loss from the given size of wire. In either case the _{other required items}

are conductor metal

and

_{temperature, current}

and

distance of

transmission. If

_power

in watts is _specified, then the voltage

atload or source

must

also be _given.

The

term "source" is used inthis

book

to _{designate the point}

from which thecircuit,orthe part ofcircuit under _{consideration,} starts.

Thus

in direct-current distribution it

_may

_signify the generator, rotary converter, storagebattery, connectionto _main, tofeeder orto sub-feeder, orsimplyacertain _pointofthe circuit.

6. Formulas. Formulas for the _complete solution of

direct-current _problems are _{given on page} 14,

and

the

method

of

pro-cedure will be clear from the _arrangement of the table.

The

size of wire

_may

be determined from the _{volt drop, or} from the per centvoltdrop, whichin direct-currentsystems is_alwaysequal

to _{the per cent}

_power

loss.

The

value of a is found in Table 6,

page 15,

and

the size ofeach wire is noted from Table _{3, page7,} corresponding to _{the required} circular mils.

The

_{per cent} volt _drop is_expressed in termsof

_power

aswell as current, so that problems involving voltage at source

and

watts

at load

_may

be solved without preliminary approximation. It should becarefullynotedthat percentloss,

V

or

V

,isexpressedas

a_{whole number,}

and

_{that the length}of _{transmission,I,} isthe dis-tance fromsource 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.

7. Ampere-Feet.

The

size of wires

_may

be readily

DIRECT-CURRENT

DISTRIBUTION

11

The

wire _having a _{corresponding} value is then noted from

Table 4, page 8.

Ampere-feet per volt _drop

=

(7)

/

=

_Amperes.

I

=

Distance fromsource to load, in feet.

v

=

_Drop

in volts.

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 (8), page 14, is

more

convenient.

In _{case of distributed loads, as in interior lighting work,} II is the

sum

of _{the products given}

_by

each load, /, multiplied

by

its _{respective distance,} I.

Thus

if 6 amperes are to be

trans-mitted 25 _ft., 3 _amperes 50 ft.,

and

2 amperes 100ft.,

II

=

6

X

25

+

3

X

50

₊

2

X

100

=

500 _ampere-feet.

8. _Examples.

_Examples

_{in practice}

_may

take innumerable forms, but the

method

of _procedure in

_any

case will be clear

from thetable of formulas _{on page} 14.

The

_{following problems}

are typical

and

serve to illustrate the _{simplicity of} the calcu-lations fordirect-current distribution. .

Example

i.

A

_copper circuit of _{97 per cent conductivity} is

to deliver _{200 amperes} for a distance of 1000 feet with a loss of 10 volts at 40 cent.

GIVEN

ITEMS.

I

=

_{200 amperes;} v

=

10 volts; I

=

1000 feet.

From

Table 6, page 15, a

=

23.0.

REQUIRED

ITEMS.

Size of each wire.

From

(8),

=

23

X

200

X

1000

₌

46Q 000 ci]

10

Use

500,000 cir. mil wires. Volt drop.

From

(9),

_

23

X

200

X

1000 500,000

12

TRANSMISSION

CALCULATIONS

Wattloss.

From

Tables 1

and

_3,

T

=

1.110,

and

r

=

0.0000207

ohm.

From

(14),

p

=

2

X

1.11

X

0.0000207

X

(200)2

X

1000

=

1840 watts.

By

assuming the voltage at the load at

_any

_{value, say 100,}

the per cent drop

and

per cent

power

loss are found to be the same, 9.2 _per cent.

Example

2.

A

motor

is to take 25 kw. at a distance of 240

feet withaloss of_{5 per cent} ofthe 110 volts _{generated, while the}

circuit is to consist of _copper wires of _{98 per cent conductivity}

with a _temperature of 30 cent.

GIVEN

ITEMS.

w=

25,000 watts; e

=

110 _volts;

7 =5per

cent; Z

=

240 feet.

From

(19),

v=

0.05

X

110=

5.5 volts.

From

Table6,page 15, a

=

21.9.

REQUIRED

ITEMS.

Size _{of each} wire.

From

(8),

s

=

21.9

X

239

X

240

=

5.5

Use

No. 0000 wires, forwhich

S=

211,600 cir. mils. Percent volt_drop.

From

_(11),

V

"'

=

2L9

X

25,000

X

240 = . -0.01

X

211,600

X

(HO)

2

From

Table 31, page 72,

7 =

5.4 per cent.

Volt drop.

From

(19), v

=

0.054

X

110

=

5.9volts.

Voltsat load.

From

_(13),

e=

110

-

5.9

=

104.1 volts.

Wattloss.

From

(14),

p

=

-054

X

25

'QOQ

=

1430watts.

1 ~~ U.Uo4 Watts at _generator.

From

_(15),

w =

25,000

+

1430=

26,430watts.

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

13

Example

3.

The

load on a feeder is to consist of 20 _amperes

at 50 feet, 25 amperes at 100 feet,

and

40 amperes at 150 feet, from thesource. _{Calculate the required}sizeofauniform circuit of _{100 per cent conductivity} fora totallossof2volts at20 cent.

From

(7), Ampere-feet per volt _drop

20

X

50

+

25

X

100

X

40

X

150

2

From

Table 4, page 8, the requiredsize of each wire is No. for _copper

and

No. 000 for aluminum.

Example

4. Copper mainsof _{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 size of wire of _{98 per cent conductivity}

and

at 50 cent, if 220

volts are _generated.

GIVEN

ITEMS.

7

=

_{500 amperes;} e

=220

_volts;

V =

3 _{per cent;} Z

=

550 feet.

From

_(18), v

=

^

03

X

-6.4 volts.

From

Table 6, page 15, a

=

23.6.

REQUIRED

ITEMS.

Size of each wire.

From

(8),

a

=

23.6

X

500

X

550

_

1;010>000 cir

Use

1,000,000 cir. mils.

Volt drop.

-

From

(9),t;

=

23'6

X

^

*

55

=

6.5 volts. 1,UUU,UUU

Percent volt _drop.

From

_(19),

7 =

-5A

=

2.95 _per cent.

2^,2\j

Volts at load.

From

(13), e

=

220

-

6:5=

213.5 volts. Wattsatload.

From

(16),

w

=

213.5

X

500

=

106,750 watts.

Example

5. Find the

combined

resistance of 1500 feet of

500,000cir. mils of _{98 per cent copper}wireat20 _{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,

TRANSMISSION

CALCULATIONS

Formulas

For Direct-Current Wiring. RequiredItems.

DIRECT-CURRENT

DISTRIBUTION

15

Table 6. Values of

a

_{For Copper}

and Aluminum.

Temperature

in

CHAPTER

II.

DISTRIBUTION

FOR DIRECT-CURRENT

RAILWAYS.

9. Introduction. Electric railways almost always use the track rails for the return of current. Besides this, other differ-encesbetweencircuitsfor _railwaysand thosefor

_{power and}

light-ingpurposes are _{the higher voltage}

_employed

on the _trolley, the

greater per cent loss allowed in the line

and

the

_{moving and}

variable nature of the loads.

Due

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 calcula-tions

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 calcula-tions for their circuits are therefore _simpler

and more

definite.

Where

the feeders

and

_working conductors form a _complete

coppercircuit, theformulas on page 14

_may

be used.

10. Resistance of Rails. Table 9, page 24, gives the equiva-lent _coppersection

and

the resistance ofthird rails or track rails

at 20 _cent., for various values of relative resistance of steel to

copper.

The

corresponding value for

_any

number

of rails is

found

_by

_{multiplying the equivalent coppersection, or}

_by

divid-ing the resistance,

by

the number.

The

resistance at

_any

other temperature isfound

_by

_{multiplying the value}inthetable

by

the temperature factor of resistance for _steel,

T

1} 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:}

Equivalent cir. mils of _{100 per cent copper}

=

125,000

X

Pounds

per yard

T

l

X

Relative resistance

DISTRIBUTION

FOR DIRECT-CURRENT

RAILWAYS

17

Ohms

resistance _{per 1000}feet

T

t

X

Relativeresistance

12.1

X

Pounds

_peryard

Ohms

resistance _{per mile}

T

_\

X

Relativeresistance 2.28

X

Pounds

_{per yard}

(21)

(22)

T

l

=

Temperature factor of steel. Table 7, page 17.

As an example, thetotalresistance _{per mile}at30 cent, of

two

65-lb. trackrails _havinga relative resistance of 13.2is

1.05

X

13.2 2.28

X

65

X

2

=

0.0468

ohm.

Table 7. Values of

T

for Steel. Temperature, deg.cent

18

TRANSMISSION

CALCULATIONS

copper feeder of 97 per cent conductivity

and

one 500,000-cir.

mil

aluminum

feederof_{62 per cent conductivity}is 833^000

X

2 _{500,000 500,000}

=

1.05 1.069 1.674

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 total resistance

much

belowthat of the positive side.

Electro-lytic conditions

may

require that negative feeders be connected

to the rails at _{certain points,} but otherwise the rails _generally

have

_ample

_conductivity without

_any

_copper reinforcement.

The

section of _{negative conductors}

_may

be based on a

maxi-mum

allowable _drop in the return.

The

size of the negative feeder 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

example

of the determination of the size of _{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 of 14 _{(including bonding).}

The

required

feeders in _parallel with the track should have _2,000,000 625,000

X

2, or 750,000 cir. mils of copper of _{100 per cent} con-ductivity.

Based

on Table 8, page 20, this is _equivalent to

773,000cir.mils of_copperof_{97 per cent conductivity,}or_1,210,000

cir. mils of

aluminum

of _{62 per cent conductivity.}

13. Positive Conductors.

The

positive conductors consist of

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

is _exactly like the determination of _negative feeders in _{paragraph 12;} while for _{systems using trolley wires,} the

total section of positive conductors is calculated from formula (25), page 25.

The

size of the auxiliary feeder is _given

_by

the

difference between the total circular mils _required

and

that of

DISTRIBUTION

FOR DIRECT-CURRENT

RAILWAYS

19

The

_temperature is included in the calculation of the circular mils

_{by means}

of

A

inTable10,page26. Forconductivitiesother 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 of _{100 per cent copper, the required} section of _copperof _{97 per}

1 ooo 000

cent conductivity is ' 1 , or 1,030,000 cir. mils, as

may

be

0.97

noted from Table 8, page20.

14. Resistance of Circuit.

The

total resistance of thecircuit is obtained

_by

_adding the resistance of the _grounded side

and

the overhead.

Where

there are no negative feeders, the resist-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

20 cent, of a _{single-track road having}

two

60-lb. rails of an apparent relative resistance of 14 _{(including bonding)}

and

a No. 00 trolley of 100 per cent conductivity is

2^2

₊

o411

=

o.462

ohm.

15. GivenItems.

The

determinationof the proper loadingon which to base the feeder _systems is _usually

more

difficult than the calculation of the feeders. _{In general the requirement} will befora

maximum

_{dropwith a very severe condition}of_loadingor fora

much

smaller_drop with adistributed loading of an_average

value. In either case the loads

and

their _positions are first

settled

and

then the formulas are _{applied successively} to the separate loads; or, where the feeder system is uniform

through-out, 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 loadsseparately, 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 ex-pressed in terms of, current.

The

_{per cent} volt loss has also

been _given in terms of _power,so that no _preliminary approxi-mationis _necessary

when

_{the voltage}is

known

_only atthesource.

16. _{Examples. Although^} in _practice, these _problems occur

in _{a great}

_many

different forms, the application ofthe formulas

20

TRANSMISSION

CALCULA

TIONS

Table 8. _Equivalents of _Copper of 100 Per Cent _{Conductivity.}

DISTRIBUTION

FOR DIRECT-CURRENT

RAILWAYS

21

in _{the negative feeder} from 500,000 to _1,000,000 cir. _{mils, equal}

to anincrease of 100 percent.

Example

7. For a temperature of 40 _cent., determine the line voltage at successive cars 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 offour 80-lb. track rails _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 per cent conductivity.

From

Tables 7

and

_9, theresistance ofthe tracksis, Resistance per 1000 feet

=

-0145

X

i-09

=

0.00395

ohm.

From

₍₂₃₎

and

Tables 1

and

3, theresistance oftheoverheadis, Resistance _{per 1000} feet

10350

133,100

X

2 _500,000 1.11 1.738

=

Q01962

ohm<

Hence

thetotal resistance _{per 1000}feet ofroadis 0.00395

+

0.01962

=

0.02357

ohm.

1st Car Total _drop

=

0.02357

X

400

X0.5

=

4.7 volts.

Line volts

=

550

-

4.7

=

545.3.

2d Car Additional_drop

=

0.02357

X

300

X

1.500

=

10.6volts.

Line volts

-

545.3

-

10.6

-

534.7.

3d Car Additional_drop

=

0.02357

X

200

X

1.000

=

4.7volts.

Line volts

=

534.7

-

4.7

=

530.

4th Car Additional_drop

=

0.02357

X

100X2.000

=

4.7 volts.

Line volts

=

530

-

4.7

=

525.3.

In the above _{typical problem, the} resistance of the _grounded

sideisbut 20 per centoftheresistance ofthe overhead.

Example

8.

A

single-trackroad with

two

75-lb. rails _having a relative resistance of 13 plus 10 per cent for _{bonding (apparent} relative resistance

=

_14.3), is to _supply four cars with 150 amperes each

when

located at 0.8, 1.0, 1.5

and

2.5 miles from a

substation _generating 550 volts. If the

minimum

line e.m.f.

is to be 400 volts at50 _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

_{TRANSMISSION CALCULATIONS}

From

₍₂₂₎

and

Table 7, page 17,

Resistance per mileof

two

rails

=

1<14

X

14

'3

=

0.0477

ohm.

2.28

X

75

X

2

Total_drop in track

=

0.0477

X

150 _(0.8

₊

1.0

+

1.5

+

_2.5)

-41.5 volts.

Allowable _drop in overhead

=

150

-

41.5=

108.5 volts.

From

_{(25), in} which

IL

is the

sum

of _products of each load multiplied

by

itsdistancefrom_source,

and

from Table_{10,page 26,}

s

=

61,100

X

150(0.8

+

1.0

+

1.5

+W

₌

490>000 ci

,

mils lOo.O

of _{100 per cent copper.}

Based

on Table8,page 20,the equivalentsection of100 per cent copperin _trolleyis

133,100

X

0.96=

128,000 cir. mils. Required section of _{100 per cent copper}in feeder

-

_490,000

-

_128,000

=

_362,000 cir. mils. Required section of97 per cent copperin feeder

=

5^2,220

_,

373,000 cir. mils.

0.97

Example

9. Findthe sizeof athirdrail of relative resistance 7.5 _{(including bonding) required} to start

two

cars _taking 1000 amperes at a _point

_midway

between substations 8 _{miles apart,} if the _drop at 20 cent, is to be 25 per cent of the 600 volts at

rotaries.

Each

track rail is to _weigh 80 Ib. _{per yard}

and

to

have a_{relative resistance of 13 (including bonding).} Distance from either substation to cars

=

_f

=

4 miles.

Currentfrom each substation

= ^-^ =

500 amperes. Total allowable _drop

=

600

X

0.25

=

150 volts.

From

Table 9, page 24, resistance per mile of two-track rails

=

0.0713

=

₀₀₃₅₆₅

_ohm

Drop

in track

=

0.03565

X

500

X

4

=

71.3 volts. Allowable _drop in third rail

=

150

-

71.3=

78.7 volts.

From

(25), in which

A

from Table 10, page 26

=

54,700,

DISTRIBUTION

FOR DIRECT-CURRENT

RAILWAYS

23

From

Table 9, page24, thissection corresponds to an 85-lb.

rail.

Then

from above

and

Table 9,

Totalresistance_per

mile=

R

=

0.03565

+

0.0387

=

0.07435

ohm.

From

_(28), total _drop

=

0.07435

X

500

X

4

=

148.7 volts.

From

_{(37), voltage} at cars

=

600

-

148.7=

451.3 volts.

From

(40), drop in_{per cent} of volts atload is

V =

148.7^-0.01

X

451.3

=

33.0_percent.

From

(41), drop in _{per cent}of volts at substationis

7 =

148.7-r-0.01

X

600

=

24.8 _per cent.

Wattloss.

From

_above,

R =

0.07435

ohm.

From

_{(35), p}

=

0.07435

X

(500)2

X

8

=

_148,700 watts.

Example

10. Calculate thesizeof positive feeders at30 cent,

required fora _uniformlydistributedloadof_{600 amperesper mile}

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

sub-stations from either one.

fiOO

V

fi

Total load per substation

=

^

=

1800 amperes.

Center of _gravity of load

=

_f

=

2 miles from either

sub-station.

From

Tables 7 and 9, resistanceper mile of fourtracks

=

0.0548

X

1.05-5-8

=0.00719 ohm.

Drop

in track

=

0.00719

X

1800

X

2

=

25.9 volts. Allowable _{positive drop}

=

65

-

25.9=

39.1 volts.

From

_(25),

5

=

57,400

X

1800

X

2-4-39.1

=

_5,280,000 cir. mils of _{100 per cent copper.}

From

Tables 7

and

_9, third-rail section_=1,090,000

X

4-*-1.05

=

_4,150,000cir. mils of _{100 per cent copper.}

Required section of _{100 per cent copper} feeder

=

_5,280,000

-

_4,150,000

=

_1,130,000cir. mils. Required section of _{97 per cent copper} feeder

=

1,130,000-4-0.97

=

1,165,000 cir. mils. Required section of _{62 per cent}

aluminum

feeder

TRANSMISSION

CALCULATIONS

o

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22

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oooo

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odd

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llli

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oooo

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8888

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omMrq

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SSS8

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o'o'd d

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DISTRIBUTION

FOR DIRECT-CURRENT

RAILWAYS

25

Formulas for D. C. _RailwayCircuits.

26

TRANSMISSION

CALCULATIONS

Table 10. Values of

A

for Wires

and

Rails.

Temperaturein

PART

II.

ALTERNATING-CURRENT

TRANSMISSION

BY MEANS

OF

OVERHEAD,

UNDERGROUND

AND

INTERIOR

WIRES

FOR PURPOSES

OF

LIGHT,

POWER

AND

TRACTION

ALTERNATING-CURRENT

TRANSMISSION

CHAPTER

III.

ALTERNATING-CURRENT

TRANSMISSION

BY

OVERHEAD

WIRES

17. Introduction.

The methods

in

common

use for

calcu-lating alternating-current transmission lines from the volt loss

are either indirect or inaccurate. This condition results from the fact that it is _impossible to devise an exact formula for

_any

alternating-current system which will _{directly indicate} the size

of_{wire required}forthe transmission of_{a given}

amount

of

_power

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

essential _{feature^pLthe} desirable

method

is thatit should

directly determine for

any

system

and

frequency, the size of wire,

power

loss

and

_power-factor at _generator

when

_{given the} volt _{loss, length} of _line,

_power

_received, load _{power-factor,} voltage at _{generator or load}

and

distance between wires.

The

following

method

accomplishes this result with commercial 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 theline 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 lossof 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.

The

30

TRANSMISSION

CALCULATIONS

first _assumption gives wires which are too large, while the other _assumption gives wires too small.

By

stating the errors of Table 11 in terms of the calculated values,the percentages

shown

would be _greatly decreased for the first _assumption

and

similarly increased in thesecond case.

18. Outline of Method. All items which

_depend

on the frequency, size of wire, load power-factor

and

_{wire spacing} were grouped into one term, which is denoted

_by

M

and

called the wire factor. All other variables that determine the size of wire

Fig. 2

were grouped into the second

member

of the equation of

M.

Then

after introducing the proper transmission factors, the

equation took the form

shown

in the table of formulas opposite thesize of wire.

The

_remaining formulas are _{simple derivations} either from the _equation of

M

or from well

known

relations.

The

vector_{diagram from which} the_{fundamental equation}

was

derived is

shown

_{in Fig.} 1 for _{lagging current}

and

in _Fig. 2 for

ALTERNATING-CURRENT

TRANSMISSION

31

01

=

Current at load.

OA

=

Volts at load.

AD=

Resistance volts.

DB

=

Reactance volts.

AB=

Impedance

volts.

OB

Volts atsource.

AC=

OB

-

OA =

Volt loss.

<p

=

Power-factor angle of load.

=

_{Power-factor angle} of line.

(f>Q

=

Power-factor angle at source.

19.

Range

of Application.

The

formulas _apply for all volt

losses less than 20 per cent of _{the generated voltage or 25 per}

cent of _{the voltage at load,}

and

for a sufficient _range of wire sizes to cover the usual

and

unusual cases in practice.

At

_high

standard _frequencies

and

for _{power-factors} near _{100 per cent} the values of

M

for the largest wires were omitted in order to confine the

maximum

possible errorwithin the limitsprescribed inTable 12, page 32.

Table

n.

Error in Per Cent ofTrue Volt Loss.

Size of

Wire.

32

TRANSMISSION CALCULATIONS

reducestoazero error_{near 10 per cent}loss.

The

pointwherethe formula for

M

should be correct

was

_{arbitrarily fixed at 10 per}

cent volt loss, since thisis about the

mean

value _{in practice,} After the wireis _{determined, the actual per cent drop}is calcu-lated. In orderto minimizeits error

and

that of the _remaining

items, each

column

in the tables of

M

has been divided into a

maximum

of three sections. It will be observed that

_any

value of

M

isfound between thesectionletters a

and

_b, b

and

c,

or c

and

_{d, depending}

_upon

the size of _{wire, frequency,}

spac-ing

and

power-factor.

The

maximum

errors in the calculations are

shown

in Table

12 below. 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

M

is most often between the section letters c

and

_d,

and

therefore the errors of

the _remaining calculations are _negligible.

Itshouldbe observedthat the valuesinTable 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 less than 4.9 _percent.

Table 12.

Maximum

Errors in Per Cent ofTrue Values

at 20 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

calcu-lating for three equal wires

and

then

making

the proper allow-ance for the _larger cross section of the

common

lead. (See

Example

16, page 44.) In the three-phase four-wire _system 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 formulas. In _practice such is often not the _{case, although} the variation from a balancedcondition is _usually small.

The

effect of _unbalancingis to alter_{the voltage} in _proportion to the

amount

of _unbalancing.

The

other items _{are also affected,} but seldomis _{the discrepancy} of

_any

_{practical importance.}

23. Temperature. All results have been calculated for a temperature of 20 cent, or 68 fahr.

The

value of

M

varies directly with the temperature, but it also _depends to a lesser degreeonotherconditions. _However, astheerror ofcommission

is

much

less thanthe error of_{omission, the values} of

A

in Table 16,page 52, have been calculated forvarious temperatures.

24. Specific Conductivity.

The

calculations have been

made

fora_{specific conductivity}of _{100 per cent} for _copper

and

_{62 per}

cent for aluminum.

The

value of

M

_{varies inversely} with the 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

method shown

in Table 16.

25. Solid

and

Stranded Conductors.

-

Wires smaller than

No. have beenconsideredsolid,while thelargersizes have been

taken stranded. However, since the inductance of a stranded wire is between that of a solid wire ofthe

same

cross section of

metal

and

oneof the

same

diameter, the discrepancy inthefinal

results for

_any

_ordinary variationfrom the

assumed

conditionsis

not _appreciable.

26. Skin Effect.

_Owing

to a _decreasing current _density toward the center of wires carrying alternating currents, the effective resistance is increased in direct _proportion to the product of its cross section

and

_{the frequency.} In the usual sizesoftransmission wire the effect is _negligible,

and

even in the

34

TRANSMISSION CALCULATIONS

largest sizes

shown

for.the higher frequencies the

maximum

additionalresistanceislessthan5_percent, which value decreases

very rapidly with the size of wire. 27.

Wire

Spacing. -- _Inductance

increases directly with the distancefrom center to centerof the wires. _However,theeffect on the _impedance of the line _{for large variations in the spacing}

is not _great, even at the _higher commercial frequencies. In

order to cover all cases in _practice, the wire factor has been calculated for three different _spacings at each _frequency. In general the results for wires on 18

and

60-inch centers will be

sufficiently accurate for spacings less than 27 inches

and

_greater

than 48 _{inches, respectively,} while the values for 36 inches will cover spacingsfrom27to48inches. _But,where_{greateraccuracy}

is desired for_{spacings other}than shown, the values of

M

_may

be

readily interpolatedfromthetables.

28.

_Arrangement

of Wires. It is

assumed

in the

two and

three-phase systems with three wires that the conductors are

placed atthe three corners of anequilateral triangle. Forother 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, 2

and

3,

and

1

and

3. In the _two-phase four-wire _system the

distance between wires ofthe

same

_phase should be used.

29. Frequency.

The

tables have been calculated for all

standard frequencies.

The

values for 15 cycles per second are

necessary for the design of single-phase railway systems, while

25, 40,60,

and

125 cyclespersecondcover the remaining systems

of transmissionfor_purposesoflight, power,

and

traction.

How-ever, where an

odd

frequency is to _{be employed, the required}

value of

M

_may

be interpolated from the tables.

30. Multiple Circuits.

Where

circuits are in parallel from

the source to receiver, the load should be proportioned between

them and

each linecalculated separately.

31. Current-carrying Capacity.

The

current-carrying

capa-city in amperes per wire is

shown

in Table 4, page 8, for both

copper

and

aluminum.

The

values in the table are based on a temperature rise of 40 cent, or 72 fahr., but the current-carrying capacity for

_any

_temperature elevation

_may

be found from formula (4), page 9.

ALTERNATING-CURRENT

TRANSMISSION

35

to insure the _{necessary current-carrying capacity}

when

the conditionsof volt loss are met. _However, itis desirable to note from Table 4 that such is the case, after each determination of wire.

32. Transmission Voltage.

The

voltage is taken between

wires, either at the loador source.

The

term"source" designates 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

formu-las have beenstatedintermsofboth_voltagesinordertocoverall

cases without

_any

_{preliminaryapproximation.} For convenience the voltage has been _expressed in _kilovolts, thousands of volts. In a _two-phase four-wire _{system the voltage} between the wires of the

same

_phaseis _specified.

With

_{lagging current the voltage} at the load is _always less than the _voltage at the _generator.

With

leading current the voltage at load is less

when

the

sum

of _{the power-factor angles} of load

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.

33. Volt Loss. In an alternating-current system the volt

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

ofthe

_impedance

voltson the vectorofdelivered volts. It

_may

be greater or less than the resistance volts. _{(See Figs.} 1

and

_2,

page 30.)

The

per cent volt loss

_may

be expressed in terms of the volts at load or source. _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 per centlossis _expressedasawhole number.

34.

Power

Transmitted.

The power

transmitted is _always

specified at the load,

and

is _{usually expressed} as true

power

in kilowatts or as _apparent

_power

in kilo_{volt-amperes.}

Where

the load is _given in _{amperes the equivalent value} of kilowatts

may

be obtained from _equations (59) or (60) solvedfor

W.

The

36

TRANSMISSION CALCULATIONS

for _{amperes per} wire _applicable to all _systems of trans-mission.

For balancedloads in_{one-phase, two-phase} four-wire or

three-phase circuits the current is the

same

in each wire, while in the two-phase three-wire system, the current in the

common

lead is

1.41 times that in each of the other wires. In the _two-phase three-wire _{system/ represents the amperes} in each of the outer

wires, this _{being the}

same

as _{the current per phase.}

35.

Power

Loss.

The power

loss in

_any

_{system dependsonly}

upon

the current

and

resistance.

The

_{per cent}

_power

loss

_may

be _greater orless than_{the per cent} volt loss. Itis less than the per cent volt loss, V,

when

the power-factorat thesource is less than_{the power-factor}oftheload,

and

is _greater

when

the reverse

istrue.

An

_exceptiontothisstatement

_may

occur

when

_capacity

effects in the line areincluded. _(See

_Example

_{20, page48,}

and

Example

25, page 69.)

The power

at the source is the

sum

of

the

_power

at load

and

the

_power

loss in line.

36. Power-Factor.

The

power-factor of the load is _always

given.

The

values of

M

have been calculated from 95 per cent

lead to 75 _{per cent lag,} for all _{frequencies except 125 cycles per}

second, giving a range sufficient to cover almost all practical cases of transmission.

The

_power-factor at the load should be

known

_{accurately, as}the value of

M

_{varies greatly} withit.

The

_power-factor atthesource _depends

_upon

_{the power-factor}

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 thesource than at theload,

and

even

may

become

laggingat

thesource. _(See

_Example

24, page68.) Laggingcurrentat the load is _{always accompanied}

_by

_{lagging current} at the _source.

The

power-factor at 'the source is lower than the

power-factor 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 effects in the line. _(See

_Example

_{20, page 48,}

and Example

_25,

page 69.)

37.

Wire

Factor.

The

wire factor,

M,

depends

upon

the impedance of the wires,

and

on the power-factor angles of the load

and

line.

The

values have been calculated for _copper

and

aluminum

at all standard frequencies, for a range of sizes

and

load power-factors sufficient to cover all _practical cases

ALTERNATING-CURRENT

TRANSMISSION

37

to arisein transmission _design. Tables 18 to 22 _give the results 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. Given Items.

The

given problem

may

be of

two

_kinds;

either theline lossis _specified

and

thesize of wireis _{required, or}

thesizeofwireis _specified

and

theline loss is_required. In both

casesthe additional itemstobe _givenare_systemof _{transmission,}

conductor metal, temperature, frequency, wire spacing, distance of _{transmission,}

_power

_{delivered, power-factor of load,}

and

the voltage betweenwires atthe source ofload.

39. Size of Wire.

The

size of wire is _generally determined

from_{the per cent} volt_{loss, although}in

some

instancesit is fixed

by

the conditionof_{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, page51,

V

or _VQ is _{found, depending}

_upon

whether_{the voltage} is _given

at the load or source.

The

value of

M

is then calculated from formula(46) or (47),

and

thesize ofwire is noted from Tables 18

to _{27, opposite the required value} of

M.

When

given the per cent

power

loss,

R

is calculated from

for-mula

(48) or (49),

and

thesizeofwireisfoundinTable3, page7,

opposite the required resistance per mile.

The

wire found

_by

either

method

is _{the required} size of each conductor.

40. Per Cent Volt Loss. It is desirable to calculate the per cent voltloss afterthesizeof wireis determined.

The

value

of

M

whetherbetweenthesectionletters a

and

b,b

and

c,orc

and

d, is noted from the proper table.

The

value of

V"

or

F

"

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 obtained in the

column

headed

_by

V

or

V

.

Where

the

calcu-latedvalue does not _appearinthe_{table, the corresponding value}

of

V

or

V

is _easilyfound

_by

_{interpolation.}

The

_remaining calculations _{are clearly} outlinedin the table of formulas_{on page}50.

When

the voltages at the load

and

source