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
.
ALTERNATING-CURRENT TRANSMISSION BY
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
ALTERNATING-CURRENT TRANSMISSION BY
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
ALTERNATING-CURRENT TRANSMISSION BY
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
TRANSMISSION CALCULATIONS
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
resistance per 1000 feet
=
-
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
resistance per 1000 feet
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.
TRANSMISSION CALCULATIONS
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.
TRANSMISSION CALCULATIONS
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.
REQUIRED ITEMS. From (8),
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.
REQUIRED ITEMS. From (8),
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
resistance per 1000 feet
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.
TRANSMISSION CALCULATIONS
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.
DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS
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
TRANSMISSION CALCULATIONS
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
DISTRIBUTION FOR DIRECT-CURRENT RAILWAYS
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'
oooo oooo
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o
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t-s
slis
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om
n
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coo's
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oooo
oooo
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SSS8
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iT\rr\
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o' o'
Ills ills
To * ST oo
o'
o'
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o 2 o oo rs
d o' o'
"^ ^" O O o^ O OOOO rr,
rr,
\oo
111! llli o' o' o' o' d o' d )'
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a
dd
fs,-
poo oooo
1
I
888S
o m r^
oo oo
I
ISIi
.S8.
g5
!
>o o
oooo OOjQ
I
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ITS
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liSs oo'oo
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ill! tn
sss;
0000 0000 r^^^t'ir* oooo oooo
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tr\
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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
TRANSMISSION CALCULATIONS
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.
TRANSMISSION CALCULATIONS
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
TRANSMISSION CALCULATIONS
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
ALTERNATING-CURRENT TRANSMISSION
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
TRANSMISSION CALCULATIONS
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
ALTERNATING-CURRENT TRANSMISSION
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
TRANSMISSION CALCULATIONS
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.
ALTERNATING-CURRENT TRANSMISSION
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.
TRANSMISSION CALCULATIONS
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,
REQUIRED ITEMS. From (47),
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.
TRANSMISSION CALCULATIONS
42
From
(55),
From
Kilowatts at source.
W Amperes per
From
=
50
^_ =
=
/
2
X
0.95
From
Power-factor at source.
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.
From Table From Table
Per
page
16,
page
21,
A =
0.0379.
56,
M=
1.38
(c
-
d).
REQUIRED ITEMS. From (50),
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.
Kilovolts at source.
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
ALTERNATING-CURRENT TRANSMISSION
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.;
From Table From Table
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
.
TRANSMISSION CALCULATIONS
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
Kilowatts at source.
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
Power-factor at source.
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.
ALTERNATING-CURRENT TRANSMISSION
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.
TRANSMISSION CALCULATIONS
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).
REQUIRED ITEMS. From (50),
"" 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
Kilovolts at source.
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
Kilowatts at source.
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
Power-factor at source.
(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.
From Table From Table
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.
TRANSMISSION CALCULATIONS
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
Power-factor at source.
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
Kilowatts at source.
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.]
TRANSMISSION CALCULATIONS
50
Formulas for A. Required Items.
C. Transmission
by Overhead Wires.
ALTERNATING-CURRENT TRANSMISSION Table 15.
V
Values of Volt Loss Factors.
51
52
TRANSMISSION CALCULATIONS
Table 16
Values of
A
for
Balanced Loads.
ALTERNATING-CURRENT TRANSMISSION
-S
i
a
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TRANSMISSION CALCULATIONS
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TRANSMISSION CALCULATIONS
56
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TRANSMISSION CALCULATIONS
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SSS-K CM r^
I .
o
o
^CMiTiS
ALTERNATING-CURRENT TRANSMISSION
NO O S 8
U^
8
8
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2
S
1
5
S S
1
5
1-
S
S
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rN
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a
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TRANSMISSION CALCULATIONS
60
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ALTERNATING-CURRENT TRANSMISSION
61
62
TRANSMISSION CALCULATIONS
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.
TRANSMISSION CALCULATIONS
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
page 74, use No. 00, for which
35,
From
volt loss.
v
X
10 per cent; 0.98 lag.
///
-
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),
Kilowatts at source.
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),
Power-factor at source.
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.
REQUIRED ITEMS. Size of each wire.
5 per cent;
6
TRANSMISSION CALCULATIONS
68
From Table Per cent
page 73, use No. 0000, for which
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
Power-factor at source.
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
REQUIRED ITEMS. From (69),
volt loss.
v _
A =
72,
K= 0.95 lead.
L=
M=
X
X
0.11
1500
X
5
_
cent
1Q 9
2
(6)
Kilovolts at source.
# =
6
(1
Per cent power
+
From
0.109)
loss.
=
(71),
6.65 kv.
From Table
3,
page
7,
R=
0.518 ohm.
TRANSMISSION BY UNDERGROUND CABLES
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
Power-factor at source.
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.
TRANSMISSION CALCULATIONS
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
Kilowatts at source.
(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
TRANSMISSION CALCULATIONS
72 Table
FO
31.
Values of
F '" = F
(i-o.oi
F
).
TRANSMISSION BY UNDERGROUND CABLES
73
74
TRANSMISSION CALCULATIONS
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
TRANSMISSION CALCULATIONS
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.
REQUIRED ITEMS. From (84),
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;
From Table From Table
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.
TRANSMISSION CALCULATIONS
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
Power-factor at source.
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=
5 per cent; 0.90 lag.
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
REQUIRED ITEMS. From (84),
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
WIRES FOR ALTERNATING-CURRENT DISTRIBUTION
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),
REQUIRED ITEMS. Size of each wire.
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.
TRANSMISSION CALCULATIONS
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.
REQUIRED ITEMS. Size of each wire.
From
(84),
,.1100 X 3.75_ 100 X 225 '
From Table
39,
Volt toss.
From
Per cent
page 77, use No. 00, for which
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
TRANSMISSION CALCULATIONS
82 Table 37.
Temperature in
Degrees.
Values of a and b for Balanced Loads.
WIRES FOR ALTERNATING-CURRENT DISTRIBUTION
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.
2^E
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
oor^o^ -o ix
SS3
M
T*^GO
3S s
ssss
83
84
TRANSMISSION CALCULATIONS
tApr,
o^CMf^cs
oo-
a
OO
s S
5?
S
r^O^ S
".
O^r^\t> -.
.
.
".
S S S
--
--
o
IT
ssss
1
O ^I^O
OOO
OO
"r
fs,
oo
SSS in ^O
*
o fei
s so
=3.8.-.
1
I
I
r^r^o
oo
a a s
552
4
s
=
r-rtn
228
a
S
55
TT
& S
-.
.
s O^OO^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
TRANSMISSION CALCULATIONS
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.
TRANSMISSION CALCULATIONS
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.
DISTRIBUTION FOR SINGLE-PHASE RAILWAYS
From
(107)
obtained:
and Table
42,
page
89
93, the following results are
TRANSMISSION CALCULATIONS
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
REQUIRED ITEMS. From (104),
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.
DISTRIBUTION FOR SINGLE-PHASE RAILWAYS
From
Kilowatts at source.
W
=
500
(1
+
K =
+
(1
(111),
=
0.072)
From
Power-factor at source.
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.
REQUIRED ITEMS. From (104),
_ 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.
DISTRIBUTION FOR SINGLE-PHASE RAILWAYS
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