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TRANSMISSION LINE FORMULAS
FOR
ELECTRICAL ENGINEERS
AND
ENGINEERING STUDENTS
BY
HERBERT BRISTOL DWIGHT, B. Sc.
ASSOCIATE OF THE AMERICAN INSTITUTE OF
ELECTRICAL ENGINEERS
NEW YORK
D. VAN NOSTRAND COMPANY
25 PARK PLACE
1913
Engineering
Library
V
COPYRIGHT, 1913, BY
D. VAN NOSTRAND COMPANY
Stanbopc jpress
F. H. GILSON COMPANY
BOSTON, U.S.A.
PREFACE.
THE object of this book is to compile a set of instructions
for engineers, which will enable them to make electrical
calculations for transmission lines with the least possible
amount of work.
The chart and working formulas have for the most part
been developed independently by the author. Where the
same or similar methods have been previously published,
the fact is generally stated in the footnotes, but it has not
been found possible to make these references absolutely
complete.
The second part of the book is for reference and contains
the derivation of the principal formulas used in connec-
tion with transmission lines. As many recent articles on
transmission lines make use of formulas which are only
roughly approximate, or are even incorrect, a reliable col-
lection of formulas, with the method of obtaining them,
should be found valuable.
It should not be presumed, because the second part of
the book requires the use of the integral calculus, that
the working formulas will require a knowledge of higher
mathematics. The first five or six chapters are complete
in themselves, and are planned for the use of those who
have an ordinary acquaintance with alternating-current
calculations.
H. B. DWIGHT.
HAMILTON, CANADA,
August, i pi 2.
in
266994
CONTENTS.
PART I. WORKING FORMULAS.
CHAPTER PAGE
Frontispiece REGULATION CHART
PREFACE iii
I. INTRODUCTION i
II. ELEMENTS or A TRANSMISSION LINE 4
III. REGULATION CHART 8
IV. FORMULAS FOR SHORT LINES 16
V. K FORMULAS 25
VI. CONVERGENT SERIES 41
PART II. THEORY.
VII. CONDUCTORS 53
VIII. TRANSMISSION LINE PROBLEMS 57
IX. REACTANCE OF WIRE, SINGLE-PHASE 62
X. SKIN EFFECT 67
XI. REACTANCE OF CABLE, SINGLE-PHASE 76
XII. REACTANCE OF TWO-PHASE AND THREE-PHASE LINES 79
XIII. CAPACITY OF SINGLE-PHASE LINE 84
XIV. CAPACITY OF TWO-PHASE AND THREE-PHASE LINES 93
XV. THEORY OF CONVERGENT SERIES 98
PART III. TABLES.
TABLE
REGULATION CHART
I. FORMULAS FOR SHORT LINES 104
Conditions given at Receiver End.
II. FORMULAS FOR SHORT LINES 106
Conditions given at Supply End.
III. K FORMULAS 108
Conditions given at Receiver End.
IV. K FORMULAS 112
Conditions given at Supply End.
V
VI CONTENTS
TABLE PAGE
V. CONVERGENT SERIES 1 16
Conditions given at Receiver End.
VI. CONVERGENT SERIES 117
Conditions given at Supply End.
VII. RESISTANCE OF COPPER WIRE AND CABLE 118
VIII. RESISTANCE OF ALUMINUM CABLE 120
IX. REACTANCE OF WIRE, 25 CYCLES 121
X. REACTANCE OF CABLE, 25 CYCLES 122
XI. REACTANCE OF WIRE, 60 CYCLES 124
XII. REACTANCE OF CABLE, 60 CYCLES 125
XIII. CAPACITY SUSCEPTANCE OF WIRE, 25 CYCLES 127
XIV. CAPACITY SUSCEPTANCE OF CABLE, 25 CYCLES 128
XV. CAPACITY SUSCEPTANCE OF WIRE, 60 CYCLES 130
XVI. CAPACITY SUSCEPTANCE OF CABLE, 60 CYCLES 131
XVII. POWER FACTOR TABLE 133
ALPHABETICAL INDEX 135
Transmission Line Formulas
PART I.
WORKING FORMULAS.
CHAPTER I.
INTRODUCTION.
THE determination of the electrical characteristics of
transmission lines is a problem of considerable practical
importance to engineers. It occurs frequently in electri-
cal engineering work, and various methods have been pro-
posed for carrying out the calculations with greater or less
degrees of accuracy. Unfortunately, most of the methods
so far presented have been such as to require special mathe-
matical skill in using unfamiliar forms such as hyperbolic
sines, etc. Even the approximate methods, whose results
are not intended to be reliable for lines of considerable
length, are often too cumbersome to be used by an engi-
neer who has not time to make himself thoroughly familiar
with them. The result has been that engineers have often
been satisfied with calculations for practical cases which
were not nearly as correct as they might easily have been.
Working methods have been developed, and are presented
in the following chapters, for solving problems in connec-
tion with actual transmission lines. As these working
methods involve comparatively simple operations in algebra
2 , TRANSMISSION .LINE FORMULAS
.>\ .,,- -i {..*;/.\
and arithmetic only, they should be found useful by all
electrical engineering students and engineers. Groups of
problems with answers are added to provide practice in the
use of the formulas. The working formulas can be imme-
diately used by electrical engineers without the delay
caused by working out the true meaning and correct oper-
ation of long and intricate systems of calculation. They
are also arranged to require the minimum amount of labor
for routine work where many lines are to be calculated.
The first method, which is described in Chapter III, is in
the form of a chart which gives the regulation or voltage
drop of a line, and which also shows directly the required
size of conductor for given conditions.
In Chapter IV are given formulas for distribution lines
and transmission lines only a few miles long. These are
extended in Chapter V by means of the constant K to
apply to transmission lines of any length in ordinary
practice.
For purposes of checking different formulas, and for
the calculation of extremely long and unusual lines, the
fundamental formulas of transmission lines are expressed
by rapidly converging series in Chapter VI. While these
series require much more arithmetical work than the K
formulas, they will give the exact results to any degree of
accuracy desired. The method of convergent series in-
volves the use of complex numbers, that is, numbers in
which "j" terms appear. They are easier to handle,
however, than logarithms, sines and cosines of angles, or
hyperbolic functions, and therefore the use of these other
mathematical functions has been avoided.
Each of the above groups of working formulas is printed
in a table, ready for practical use. The tables will be
INTRODUCTION 3
found in the collection at the back of the book, as well as
in the separate chapters describing them.
When any formula is given which uses approximations,
the limits of its accuracy should be clearly stated so that
one can tell at a glance whether the method is sufficiently
accurate for the purpose in hand, or whether a longer
method giving greater accuracy is desirable. This is espe-
cially necessary in the calculation of transmission lines,
because approximate formulas are quite permissible for
lines only a few miles long, but become very untrust-
worthy when the length is increased to one hundred miles
or more. For this reason, each table of formulas has its
percentage and range of accuracy printed in a prominent
position, so that the most suitable method for any case
may be instantly chosen.
CHAPTER II.
ELEMENTS OF A TRANSMISSION LINE.
THE essential elements of a transmission line have been
described many times, but a short discussion of them,
with an explanation of some of the terms used in connec-
tion with the subject, may be useful before proceeding
with the actual calculations.
A transmission line consists of two or more conductors
insulated from each other so that they can carry energy by
electric currents to some more or less distant point.
The conductors may be solid copper wires, copper cables,
or aluminum cables. The diameters and resistances of
various standard conductors are given in Tables VII and
VIII, pages 118 to 120. It will be noted that the exact
value of the resistance of a conductor differs slightly when
a direct current, and an alternating current of 25 or 60
cycles, is flowing. This is due to the "skin effect," by
which an alternating current tends to flow near the surface
of a conductor, as explained in Chapter X. The drop in
voltage due to resistance is proportional to the current
and is in phase with it when the current is alternating.
Only overhead lines, carrying alternating currents, will
be considered in this book. Such lines are supported by
poles or steel towers at a considerable height above the
ground. The conductors are separated from each other by
a distance which may be several inches or several feet.
This distance is called the " spacing" of the conductors
4
ELEMENTS OF A TRANSMISSION LINE 5
and it has an important bearing on the electrical charac-
teristics of the line.
An alternating magnetic field is formed around, and in-
side of, conductors carrying alternating currents. This
field generates a voltage along the conductors which is
proportional to the current, like the voltage drop due to
resistance, but which is 90 out of phase with the current.
This voltage is called the reactance drop. Tables of re-
actance of transmission lines will be found in Part III.
Since the voltage drop in a transmission line is due to
resistance and reactance, a simple line may be considered
to be made up of the elements shown in Fig. i. If R is
Resistance
Receiv-
Suppty Eg E er0 r
Resistance ^^Sr i Load.
Fig. i.
the total resistance, the voltage drop in phase with the
current / will be IR, and if X is the total reactance, the
voltage drop in quad-
rature with the current
will be IX.
The vector diagram of
the above quantities will
be as in Fig. 2. The cur-
rent is in general not in
phase with the voltage Fig - 2-
E, but lags behind it by an angle 6, according to the
power factor, cos 6, of the load. The resistance drop IR
will therefore not be added directly to E, but must be
added vectorially, along with the reactance drop IX, as in
6 TRANSMISSION LINE FORMULAS
Fig. 2. It is evident that the voltages E and E 8 , and the
power factors cos 6 and cos 0, at the two ends of the line,
are not the same in value.
A long transmission line acts as a condenser and this
fact also must be taken into account. A condenser con-
sists of two electrical conductors placed close together but
insulated from each other so that a direct current cannot
pass between them. However, if an alternating voltage
be applied between them, a charge of electricity propor-
tional to the electrostatic capacity of the condenser will
flow into and out of the conductors. The result is that an
alternating current will appear to flow between them, pro-
portional to the t capacity susceptance of the condenser.
This current, called the charging current, will be 90
out of phase with the voltage, and, unlike most currents
in ordinary practice, it will lead the voltage in phase, in-
stead of lagging behind it. The amount of the charging
current may be determined by means of the tables of
capacity susceptance of transmission lines, in Part III.
A current in phase with the voltage will flow between
the conductors, but it is only noticeable at very high
voltages. Part of it is a leakage current flowing over the
insulators, and part is a discharge through the air, and
produces the glow called corona, on high-voltage con-
ductors.
The elements of a transmission line accounting for the
leakage current and charging current are shown in Fig. 3,
in which resistances and condensers are shunted across the
line all along its length.
Considering for the present that the voltage of the line
is the same at all parts and is equal to , the current in
phase with E flowing across from one conductor to tjie
ELEMENTS OF A TRANSMISSION LINE
other will be EG, where G is the total conductance between
the wires. So also, if B is the capacity susceptance of the
Supply
P Receiv-
E er.
Fig. 3.
line considered as a condenser, EB will be the value of the
shunted current in quadrature
with E.
The vector diagram for the
line indicated in Fig. 3 (neg-
lecting the voltage drop in the
conductors) is shown in Fig. 4.
It is seen that the current I 8 at
the supply end is different in
magnitude and phase from the current / at the receiver.
In order to calculate the combined effect of the above
phenomena, formulas must be used which will take into
account the fact that the resistance, capacity, etc., are
uniformly distributed along the line, and that the line cur-
rent and voltage are different at all parts of the line.
Fig. 4.
CHAPTER III.
REGULATION CHART.
THE characteristic of a transmission line which limits
the load it may carry is its regulation, or the variation in
voltage which occurs when the load is thrown on and off.
This is especially true when the load has a low power
factor, which is the case in most instances at the present
time.
For estimating the regulation of a line, or the size of
conductor required, the regulation chart which forms the
frontispiece of the book may be used, and it will give the
required result much quicker than any method of calcula-
tion. An extra copy of the chart is inserted at the back
of the book; it may be found useful for cutting out and
mounting on cardboard.
The chart is accurate within approximately \ of i% of
full line voltage, when the regulation is less than 10% and
the line is not more than 100 miles long.
In using the chart,* one places a straightedge across it
from the point on the left corresponding to the spacing of
the transmission line, to the point on the right, correspond-
ing to the resistance of the conductor per mile. The reg-
ulation voltage, V, per total ampere per mile of line, is
then read directly from the chart for the power factor of
load considered. The regulation is taken as the change in
* The process of using the chart is similar to that used with the trans-
former regulation and efficiency charts published by J. F. Peters, Electric
Journal, December, 1911.
8
REGULATION CHART 9
load voltage when the load is thrown on or off, assuming
constant supply voltage.
The total regulation is quickly figured on the slide rule
from the following formula for two-phase (four wire) or
three-phase lines:
i . w u looo K.V.A. X IV
Regulation Volts = -
where K.V.A. = Kilovolt-amperes of load, at the receiver
end.
E = Line voltage at the load, or receiver end.
/ = Length of line in miles.
For single-phase lines use 2V instead of F, making the
formula as follows:
_> . . _, u looo K.V.A. X I X 2 F
Regulation Volts = -
The regulation volts may be expressed as a percentage
of E to give the per cent regulation, and a formula is given
on the chart for obtaining this result directly.
The line drop, or difference in voltage between the supply
end and the receiver end of the line, is the same as the
regulation for lines less than about 20 miles long, but for
longer lines the effect of the charging current must be taken
into account by the formula
Line Drop = Regulation Volts
aooo/
where K = 2.16 for 60 cycles,
and K = .375 for 25 cycles.
It is seen that the voltage due to the charging current
is proportional to the line voltage E, and to the square
of the number of miles, but is independent of the size or
spacing of the conductors, within the assigned limit of ac-
curacy. The constant K does not need to be used in the
10 TRANSMISSION LINE FORMULAS
v
formula for regulation, since the charging current is present
at both no load and full load.
In selecting the spacing point on the chart, one notes
whether the frequency is 25 or 60 cycles, and whether the
conductor is of copper or aluminum. The spacing points
are the same for both solid wire and cable. When the
wires of a three-phase line are not spaced at the corners of
an equilateral triangle, but are at irregular distances a, b,
and c from each other, as in Figs. 5 and 6, the equivalent
spacing _
5 = v abc
should be used.
........... c ......... ~
Fig. 6. Irregular Flat Spacing.
K -- a ..... -><- ...... a ----->j
I ............ 2a ............ j
Fig. 5~ Irregular Triangular Spacing. Fig. 7. Regular Flat Spacing.
With regular flat spacing, as in Figs. 7 and 8, the equa-
tion for the equivalent spacing becomes simply
s = i. 26 a.
It makes no difference whether the plane of the wires
with flat spacing is horizontal, vertical or inclined.
The spacing of a two-phase line is the average distance
between wires of the same phase. The distance between
wires of different phases is not considered.
The points marked on the resistance scale at the right
of the chart are for cables at 20 C., assuming hard-drawn
copper of a conductivity equal to 97.3% of the Annealed
REGULATION CHART II
Copper Standard, and hard-drawn aluminum of 60.86%
conductivity, and allowing an increase of i%
in resistance for the effect of spiralling of the
wires in the cable. However, these resistance
points are placed on the chart for convenience
only, and are not essential. If other assump-
tions are made, or if other sizes of conductor
are used, all that is needed is to find the re-
sistance of the conductor per mile, and use the
corresponding resistance point on the chart Fi s- 8 - Regular
to find "V." FtatSpactag -
One of the most common problems in estimating new
projects is to determine the size of wire needed for any
given value of regulation, and the chart will be found
especially applicable to this work. "V" is first found from
the equation,
v = % Reg'n X E 2
~ 100,000 K.V.A. X I '
Then lay a straightedge through "V" and the point for
the spacing to be used, and the nearest size of conductor
can be seen, at a glance, on the resistance scale at the right.
The chart is quite as useful for finding the voltage drop,
or required size of conductor, for distribution lines a few
hundred feet long as it is for transmission lines many miles
long.
PROBLEM A.
Find, by means of the chart, the regulation and line drop for the
following set of conditions:
Length of line 100 miles.
Spacing 8 feet.
Conductor No. 3 copper cable.
Load (measured at receiver end) , 3000 K. V. A.,
66,000 volts, 90% P.F., three phase, 60 cycles.
12 TRANSMISSION LINE FORMULAS
Lay a straightedge from the 8-foot spacing point (60 cycles, copper
conductor) to the point on the resistance scale for No. 3 copper
cable. It is found to cross the 90% P.F. line at the reading 1.344.
Then, by the formula on the chart,
^ T. , ,. 100,000 X 3000 x 100 x 1.344
Per cent Regulation = z 223
66,000 X 66,000
= 9.26%, or approximately 9.3%.
The calculated value of the regulation of this line is 9.40% (Chap.
VI, Prob. 2), so that the error involved in using the chart is less than
I of i% of line voltage.
The per cent line drop, according to the chart, is
9.26 100 x 2.16 x T<T = 7-io%.
As the calculated value is 7.08% (Chap. VI, Prob. 2), the error
from the chart is less than rV of i% of the line voltage.
PROBLEM B.
1 To find the size of copper required to give approximately 10%
voltage drop in the following case:
Length of line 3 miles.
Flat spacing as in Fig. 7. Wires 2 feet apart.
Load (measured at receiver end), 250 K.V.A.,
2200 volts, 85% P.F., three phase, 60 cycles.
First, find V from the formula on the chart.
V = 10 X 2200 X 2200 =
100,000 X 250 X 3
The equivalent spacing is 1.26 X2, or 2.52 feet. The proper
spacing point will therefore be just below the spacing point for 2^
feet, copper conductor, 60 cycles. Lay a straightedge from this
point to the reading 0.64 on the line for 85% P.F. and it cuts the
resistance scale at 0.36 ohm per mile. The nearest size of copper is
seen to be No. ooo.
PROBLEM C.
Find the voltage drop of the following two-phase line:
Length of line 80 miles.
Spacing 10 feet.
Conductor No. oo aluminum cable.
Load (measured at receiver end), 15,000 K.V.A.,
100,000 volts, 95% P.F., two phase, 25 cycles.
REGULATION CHART 13
Laying a straightedge across the chart from the lo-foot spacing
point for 25 cycles and aluminum conductor, to the resistance point
for No. oo aluminum, the value of V for 95% P.F. is found to be
0.750. Then the line drop, in volts, is equal to
1000x15.000x80x0.750 _ o o o8 o o8
100,000
= 9000 240
= 8760 volts.
The calculated value is 8810 volts (Chap. VI, Prob. i). The
error from the chart is 0.05% of line voltage. .
PROBLEM D.
Find the regulation of the following single-phase line:
Length of line 15 miles.
Spacing 3 feet.
Conductor No. o copper wire.
Load (at receiver end), 300 K.V.A., 50% P.F.,
11,000 volts, single phase, 60 cycles.
From the chart, V = 0.851.
Therefore Regulation = io.oo Xoo X i S X a Xo.8 S i _ 6 %
11,000 X 11,000
[Calculated value, 6.40% (Chap. IV, Prob. A). Error from chart,
0.07% of line voltage.]
PROBLEM E.
Find the K.V.A. which can be delivered at the end of the following
line, with 8% regulation:
Length of line 75 miles.
Spacing 8 feet, regular flat spacing.
Conductor No. oo aluminum cable.
Character of load (at receiver end),
88,000 volts, 85% P.F., three phase, 25 cycles.
Equivalent spacing 5 = 8 X 1.26 = 10.08 feet.
V = 0.755.
KVA = 8 X 88,000 X 88,000
100,000 X 0.755 X 75
= 10,900.
14 TRANSMISSION LINE FORMULAS
PROBLEMS, CHAPTER III.
(REGULATION CHART.)
1. Find the size of copper cable which is needed to deliver 200
K.V.A. at a distance of 3 miles with 10% drop or less.
Spacing of line 2 feet.
Character of load (at receiver end), 2200 volts,
80% P.F., three phase, 60 cycles.
[Ans. No. o.]
2. Assuming No. o copper cable for the previous problem, find
the volts drop in the line.
[Ans. 219 volts. Calculated, 221 volts (Chap. IV., Prob. i).
Error 0.1% of line voltage.]
3. Find the size of copper required for a drop of 6% or less in the
following case:
Length of line 5000 feet.
Spacing 18 inches.
Load (at receiver end), 75 Kw. (79 K.V.A.),
95% P.F., 2000 volts, single phase, 60 cycles.
[Ans. No. 4 copper.]
4. Assuming No. 4 copper wire of 1.312 ohms per mile, find the
per cent line drop and the supply voltage. (See Elec. Journal, Apr.,
1907, p. 231.)
[Ans. 543%, 2110 volts, calc. 5-43%, 2109 volts (Chap. IV.,
Prob. 3)-]
5. Find the required size of copper for 9% regulation in the fol-
lowing case :
Length of line 25 miles.
Spacing 3 feet.
Load (at receiver end), 2500 K.V.A.,
20,000 volts, 60% P.F., three phase, 25 cycles.
[Ans. No. o copper.]
6. Assuming No. o copper wire of 0.520 ohm per mile, find the
per cent regulation. (See Elec. Journal, Apr., 1907, p. 231.)
[Ans. 8.42%, calc. 8.51% (Chap. IV, Prob. 5). Error 0.09%
of line voltage.]
REGULATION CHART 15
7. Find the per cent regulation and the voltage drop of the fol-
lowing line:
Length of line 75 miles.
Spacing, 8 feet, regular flat spacing.
Conductor No. oo aluminum cable.
Load (at receiver end), 10,000 K.V.A.,
88,000 volts, 85% P.F., three phase, 25 cycles.
[Ans. 7-36% Reg'n., 7.15% line drop, calc. 7-35% Reg'n., 7.13%
line drop (Chap. V, Prob. 3).]
8. Find the K.V.A. which can be delivered at the end of the fol-
lowing line, with 10% regulation, at 75% and at 90% P.F.:
Length of line 100 miles.
Spacing 10 feet.
Conductor No. oooo aluminum cable.
Receiver voltage 110,000 volts, three phase,
60 cycles.
[Ans. 14,300 K.V.A., 16,300 K.V.A.]
CHAPTER IV.
FORMULAS FOR SHORT LINES.
THE effect of capacity is inappreciable with short lines
as it amounts to only y 1 ^ of i% for a line about 20 miles
long. Thus distribution lines and many short transmission
lines can be quite accurately calculated without considering '
the line capacity at all. The formulas in Tables I and II,
pp. 1 8 and 20, enable one to solve many problems con-
nected with such lines.
The formulas are divided into two groups, those in
Table I being used when all the particulars describing the
load, such as K.V.A., voltage and power factor, are speci-
fied at the receiver end. Table II is used when these par-
ticulars are specified at the supply end.
One first finds the quantities P and <2'or P s and Q s . These
are the values of in-phase current, and reactive or quadra-
ture current, at the point where the conditions are speci-
fied. It is to be noted that only values of current
expressed in total amperes are to be used in connection
with the formulas in this book. The number of amperes
per wire is never used in the calculations (except with
single phase lines), but if it is desired to be known, it
may be determined from the formulas at the bottom of
the tables.
The next step is to find the quantities A and B, or F and
G. One is then ready to find the value of any of the ten
quantities, whose formulas are given in the tables. It
16
FORMULAS FOR SHORT LINES 17
should be remembered that each of these ten quantities
may be determined independently of all the others. Thus
it is not necessary to work out the first six equations in
order to obtain the value of the seventh, since the seventh,
like any of the others, may be calculated directly.
18 TRANSMISSION LINE FORMULAS
TABLE I. FORMULAS FOR SHORT LINES.
CONDITIONS GIVEN AT RECEIVER END.
These formulas are exact when the line is short. When the line is 20
miles long, they are correct within approximately T ^ of i% of line voltage.
Conditions given:
K.V.A. = K.V.A. at receiver end.
E = Full load voltage at receiver end.
cos 6 = Power factor at receiver end.
K.W. = K.V.A. cos 6.
r = Resistance of conductor per mile. (From Tables VII-VIII.)
x = Reactance of conductor per mile. (From Tables IX-XII.)
I = Length of line in miles.
1000 K.V.A. cos 6
Then P= = = In-phase current at receiver end (m
total amps.).
_. looo K.V.A. sin 6 _ f ..
Q = - = Reactive current at receiver end (in
total amps.) when current is lagging
1000 K.V.A. sin 6 , .
= -- = - when current is leading.
hi
Find the following quantities:
Three phase or two phase. Single phase.
A=E + Prl + Qxl. A = E + 2 Prl + 2 Qxl.
B = Pxl-Qrl. . B = 2 Pxl-2Qrl.
Formulas (capacity neglected) :
(1) Voltage at supply end = A -\ -- -.
2 A.
B z
(2) Regulation of line = A H -- j E. (Same as line drop.)
2 A.
(3) Per cent regulation of line = - - - - '- per cent.
(Same as per cent line drop.)
(4) K.V.A. at supply end = rr^- X K.V.A.
L
(5) K.W. at supply end = (AP - BQ).
IOOO
(6) Power factor at supply end = ( Ap ~ ^Q) E
(in decimals).
FORMULAS FOR SHORT LINES
AP-B
(7) In-phase current at supply end =
(8) Reactive or quadrature current at supply end
in total amperes.*
When this quantity is positive, the current is lagging.
When this quantity is negative, the current is leading.
(9) K.W. loss in line = - (AP - BQ - EP).
(10) Per cent efficiency of line = . _ ^ per cent.
Total amps.
* Amperes per wire, three phase = -- -j= -
V 3
Total amps.
Amperes per wire, two phase = -
20 TRANSMISSION LINE FORMULAS
TABLE II. FORMULAS FOR SHORT LINES.
CONDITIONS GIVEN AT SUPPLY END.
These formulas are exact when the line is short. When the line is 20
miles long, they are correct within approximately tV f z % f nne voltage.
Conditions given:
K.V.A. = K.V.A. at supply end.
E s = Full load voltage at supply end.
cos 6 = Power factor at supply end.
K.W. = K.V.A. cos 6.
r = Resistance of conductor per mile. (From Tables VII-VIII.)
x = Reactance of conductor per mile. (From Tables IX-XII.)
I = Length of line in miles.
loco K.V.A. cos 9
Then P a = - ^ = In-phase current at supply end
s
(in total amps.).
_. 1000 K.V.A. sin 9 _
Q 8 = - = - = Reactive current at supply .end
c-s
(in total amperes) when current is
lagging.
1000 K.V.A. sin 9 , . . , ,.
= -- - when current is leading.
E 8
Find the following quantities:
Three Phase or Two Phase. Single Phase.
F = E 8 - P 8 rl - Q 8 xl F = E 8 -2 P 8 rl - 2 &*'
G = Q 8 rl - P 8 xl G = 2 Q 8 rl - 2 P 8 xl.
Formulas (capacity neglected):
G 2
(1) Voltage at receiver end = F +
(2) Regulation of line = E 8 F p. (Same as line drop.)
2
(3) Per cent regulation of line = - per cent.
(Same as per cent line drop.)
F+-
(4) K.V.A. at receiver end = = X K.V.A.
L 8
FORMULAS FOR SHORT LINES 21
(5) K.W. at receiver end = ^^ (FP 8 GQ 8 ).
i (FP 8 -GQ g )Es
(6) Power factor at receiver end = y ^ \ x K.V.A.
(in decimals).
(7) In-phase current at receiver end = FPs " Qa in total amperes.*
, GPs + FQ S
(8) Reactive or quadrature current at receiver end = ^r
in total amperes.*
When this quantity is positive, the current is lagging.
When this quantity is negative, the current is leading.
(9) K.W. loss in line = ^ (E 8 P S - FP a + GQs).
(10) Per cent efficiency of line = ^ p ^ P er cent -
Total amps.
* Amperes per wire, three phase, = ^=
_ Total amps.
Amperes per wire, two phase, -
22 TRANSMISSION LINE FORMULAS
PROBLEM A.
Find the regulation of the following single-phase line:
Length of line ....................... 15 miles.
Spacing ............................. 3 feet.
Conductor .......................... No. o copper wire.
Load (at receiver end), 300 K.V.A., 11,000 volts, 50% P.F., single
phase, 60 cycles. (Same as Prob. D, Chap. III.)
From the tables in Part III,
r = 0.5331.
* = 0.686.
P = looo X 300 X 0.50 = 6
11,000
= looo X 300X0.866 =
11,000
A = 11,000 + 2 X 13-64 X 0.5331 X 15
+ 2 X 23.62 X 0.686 Xi5
= 11,704.
B = 2 X 13.64 X 0.686 X 15 - 2 X 23.62 X 0.5331 X 15
= " 97 '
Per cent regulation = I0 ( 11,704 + 97 X 97 -- 11,000 )
11,000 \ 2 X 11,704 /
= 6.40%.
PROBLEM B.
Calculate the volts drop for the following case, where all the
conditions are specified at the supply, or generator, end of the line:
Length of line ........................ 10 miles.
Spacing .............................. 3 feet.
Conductor ........................... No. 2 copper wire.
Quantities measured at supply end: 600 K.V.A., 6600 volts,
80% P.F., three phase, 60 cycles.
From the tables hi Part III,
r = 0.8469.
* = 0.714.
FORMULAS FOR SHORT LINES 23
_ looo X 600 X 0.80 _
Pe ~ 6600 ~ 72>73 '
n _ looo X 600 X 0.60 _
V* ~ ~ ~ 54-55-
6600
F = 6600 - 72.73 X 8.47 - 54-55 X 7-14
= 5594-5-
G = 54-55 X 8.47 - 72.73 X 7-i4
= - 57-
Voltage at receiver end
= 5594-5 + 0-3
= 5595 approximately.
Drop in volts = 6600 5595
= 1-005 volts.
Per cent drop = I0 X I0 *
5595
= 17.96% of receiver voltage.
PROBLEMS, CHAP. IV.
(FORMULAS FOR SHORT LINES.)
1. Find the voltage drop in the following case:
Length of line ......................... 3 miles.
Spacing ............................... 2 feet.
Conductor ............................ No. o copper cable.
Load (at receiver end), 200 K.V.A., 2200 volts, 80% P.F., three
phase, 60 cycles. (Prob. 2, Chap. III.)
[Ans. 221 volts.]
2. Find (a) the P.F. at the supply end,
(b) the per cent efficiency of the line, for the case in Prob. i.
[Ans. (a} 78.7% P-F. (b) 92.3% efficiency.]
3. Find the supply voltage in the following case:
Length of line ........................ ...... 5000 feet.
Spacing .................................... 18 inches.
Conductor, No. 4 copper wire of 1.312 ohms per mile.
Load (at receiver end), 75 Kw., 2000 volts, 95% P.F., single phase,
60 cycles. (Prob. 4, Chap. III.)
[Ans. 2109 volts.]
24 TRANSMISSION LINE FORMULAS
4. Find the volts drop and the watts loss, in the following line:
Length of line . . 20 miles.
Spacing 5 feet.
Conductor No. i aluminum cable.
Two phase, 25 cycles.
K.V.A. at supply end 10,000.
Volts at supply end 35,000.
P.F. at supply end 80%.
[Ans. 5950 volts, 1770 Kw.j
5. Find the per cent regulation of the following line:
Length of line 25 miles.
Spacing 3 feet.
Conductor, No. o copper wire of 0.520 ohm per mile.
Load (at receiver end), 2500 K.V.A., 20,000 volts, 60% P.F.,
three phase, 25 cycles. (Prob. 6, Chap III.)
{Am. 8.51%.]
CHAPTER V.
K FORMULAS.
WHEN a transmission line is more than 20 miles long,
the formulas for short lines given in Chapter IV are no
longer accurate, and other formulas must be used, which
will take into account the capacity of the line. Such
formulas, called K formulas, will be found in Tables III
and IV, pp. 28 to 35. The same tables will be found in
Part III at the end of the book.
The K formulas will be found very similar to those of
the last chapter, and while they require more arithmet-
ical work, they should not be found any more difficult to
understand. No more values of line constants need to be
looked up for the K formulas than for the " Short Line"
formulas. The capacity of the line does not enter into the
calculations, since its effect is allowed for by means of the
constant K which is the same, at any one frequency, for
all values of line capacity.
The formulas of this chapter assume that the leakage
current is zero; that is, that no power is lost from leakage
over the insulators or from corona. This is a correct
assumption to make for all voltages except the very highest
in use. If it is desired to make allowance for corona loss,
the formulas of Chapter VI should be used.
The accuracy of the K formulas is given as approxi-
mately YO of i% of line voltage for lines up to 100 miles
long and with regulation up to 20%, and as \ of i% for
lines up to 200 miles long, and with the same regulation.
25
26 TRANSMISSION LINE FORMULAS
These limits are close enough for commercial work, so that
the K formulas can be recommended for all ordinary
engineering calculations of the performance of long power
transmission lines under steady conditions, where the
corona loss is small. The accuracy of the electrical calcu-
lations will be better than the accuracy with which the
resistance and the physical dimensions of the line are
generally known.
The K formulas are well adapted to the solution of
long transmission lines which have substations at inter-
mediate points between the ends. In such cases each
section of the line between substations must be calculated
separately, beginning with the end where conditions are
known. The first step is to find the voltage, in-phase
current and quadrature current at the first substation.
The load taken by the substation, expressed as in-phase
current and quadrature current, must be added to, or sub-
tracted from, the above values of current. When condi-
tions are given at the receiver end and one is proceeding
toward the supply end, the substation load must be added
to the line load. When conditions are given at the supply
end, the substation load must be subtracted from the line
load, since one is proceeding away from the supply. Hav-
ing thus found complete conditions at one end of the
second section of the line, the calculation of this section
may be taken up in the same way as for the first section.
In this manner the entire line may be calculated and the
voltage and current at the unknown end may be deter-
mined.
Examples are worked out, which will give a clear idea of
the manner in which the K formulas are used. Many
other such examples have been calculated and carefully
K FORMULAS 27
compared with the fundamental formulas. As these ex-
amples have covered the range of practicable transmission
lines, a sound basis is afforded for the estimate of the
accuracy of the K formulas and for the statement that
they are sufficiently reliable for all ordinary engineering
purposes in the calculation of electric power transmission
lines.
28 TRANSMISSION LINE FORMULAS
TABLE III. ^ FORMULAS FOR TRANSMISSION LINES.
CONDITIONS GIVEN AT RECEIVER END.
Accurate within approximately T V of i% of line voltage up to 100 miles,
and ^ of i% up to 200 miles, for lines with regulation up to 20%.
^ = 6 (cycles) 2 # =2>l6 for 60 cycles. K = 0.375 for 25 cycles.
10,000
Conditions given:
K.V.A. = K.V.A. at receiver end.
E = Full load voltage at receiver end.
cos 6 = Power factor at receiver end.
K.W. =K.V.A. cos0.
r = Resistance of conductor per mile. (From Tables VII-VIII.)
x = Reactance of conductor per mile. (From Tables IX-XII.)
I = Length of transmission line in miles.
Then P = ^- . . . cos _ ^ j^gg curren t a t receiver end
(in total amps.) .
Q = 1000 K.V.A. sing = Reactiye current afc rece i ve r end
(in total amps.), when current is
lagging.
1000 K.V.A. sin & . L . , ,.
= -- = - , when current is leading.
K FORMULAS
29
Find the following quantities:
Full Load.
. 3
Load.
i-*(J-Vi.
\ioooj
. The above are for two- and three-phase lines. For single-phase
lines use 2 r and 2 x in place of r and #.
30 TRANSMISSION LINE FORMULAS
TABLE III. (Continued.}
CONDITIONS GIVEN AT RECEIVER END.
Formulas:
Full Load. No Load.
Voltage at receiver end.
:* (,)&--_
(for constant supply voltage).
Regulation at receiver end in volts, for constant supply voltage.
N.B. The regulation at receiver end may be expressed as a percentage
of E.
Voltage at supply end.
(for constant receiver voltage) .
Regulation at supply end in volts, for constant receiver voltage.
f*\ L. B * B *
(6) A H -- j AQ -- j-
2 A 2 Ao
N.B. The regulation at supply end may be expressed as a percentage
K FORMULAS 31
Current at supply end in total amperes*
(7) VC 2 + D 2 . (8) Co 2 + A 2 .
K. V.A . at supply end.
( 9 ) -
1000 2A
.JF". a/ supply end.
(n) (4C + 3D). (12) ^
. (10) -^-(^o+^j- )VCo 2 +A> 2 .
Power factor at supply end, in decimals.
AC + BD A Q C +B D
(13) 7 -
A +
In- phase current at supply end in total amperes.*
AC+BD , ,, A C +B D
Reactive current at supply end in total amperes*
, . BC-AD , Q x ^o
(I7) "'
When this quantity is positive, the current is lagging.
When this quantity is negative, the current is leading.
K.W. loss in line.
(19) (AC + BD -EP). (20) (A C + B D )
IOOO IOOO
[same as No. 12]
Per cent efficiency of line.
, . 100 EP
AC+BD
Total amps.
* Amperes per wire, three-phase, = -p
Total amps.
Amperes per wire, two phase, =
32 TRANSMISSION LINE FORMULAS
'TABLE IV. K FORMULAS FOR TRANSMISSION LINES.
CONDITIONS GIVEN AT SUPPLY END.
Accurate within approximately T V of i% of line voltage up to 100 miles
and \ of i% up to 200 miles, for lines with regulation up to 20%.
T ^ 6 (cycles) 2 , , , ..
K = . K 2.16 for 60 cycles. K 0.375 for 25 cycles.
10,000
Conditions given:
K.V.A. = K.V.A. at supply end.
E 8 = Full load voltage at supply end.
cos 9 = Power factor at supply end.
K.W. = K.V.A. cos0.
r = Resistance of conductor per mile. (From Tables VII-VIII.)
x = Reactance of conductor per mile. (From Tables IX-XII.)
/ = Length of transmission line in miles.
Then P 8 = -^~~ = In-phase current at supply end (in
c-s
total amps.).
Q 8 = ' ' = Reactive current at supply end (in
c-s
total amps.), when current is lagging
1000 K.V.A. sin 9 . . , ,.
= = , when current is leading.
K FORMULAS 33
Find the following quantities:
Full Load.
/
Vh-i, _LY.
3 \ioooy
= ES ' - I - I *
X \IOOOj
. The above are for two- and three-phase lines. For single-
phase lines use 2 r and 2 x in place of r and #.
34 TRANSMISSION LINE FORMULAS
TABLE IV. (Continued.)
CONDITIONS GIVEN AT SUPPLY END.
Formulas:
Full Load. No Load.
Voltage at receiver end.
(for constant supply voltage).
Regulation at receiver end in volts, for constant supply voltage.
N.B. The regulation at receiver end may be expressed as a percentage
of E.
Voltage at supply end.
F+
(4) E.. (5) E 08 = 2 ^E a
F +^L
2F
(for constant receiver voltage).
Regulation at supply end, in volts, for constant receiver voltage.
N.B. The regulation at supply end may be expressed as a percentage
ofE a .
' K FORMULAS 35
Current in total amperes.* _
( 7 ) VM* + N* at receiver end. (8) Vjjf 2 + N Q 2 at supply end.
K.V.A.
( ) __L_ {p+**.} VM* + N* at receiver end.
w ' 1000 \ 2 Fj
(10) - E 8 Vlf 2 + No 2 at supply end.
tf.TF.
(ll ) _J_(FM + Gtf) at receiver end. (12) -^- E 8 M at supply end.
7 1000
Power factor, in decimals.
FM + GN - at receiver end.
* *' / fT 2 \
f F + ^ )
2V 2
In-phase current in total amperes*
( IS ) FM + ^ N at receiver end. (16) Jfo at supply end.
^ + 5
Reactive current in total amperes*
( I7 ) GM ~ - V at receiver end. (18) ]V at supply end.
When this quantity is positive, the current is lagging.
When this quantity is negative, the current is leading.
K.W. loss in line.
( IQ ) _I_ ( Es p 8 - FM - ON). (20)
Per cent efficiency of line.
( 2I ) f J?p P er cent -
Total amps.
* Amperes per wire, three phase, =
Total amps.
Amperes per wire, two phase, = - -
36 TRANSMISSION LINE FORMULAS
PROBLEM A.
Find by the K formulas, the line drop in the following case:
Length of line 100 miles.
Spacing 8 feet.
Conductor No. 3 copper cable.
Load (at receiver end), 3000 K.V.A., 66,000 volts, 90% P.F.,
three phase, 60 cycles. (Prob. A, Chap. III.)
From the tables, r = 1.078, x = 0.840.
1000 X 3000 X 0.90
P = -f *= = 40.91.
66,000
O = looo X 3000 X Q.4359 = o
66,000
2 l6
A = 66,000 66,000 X :
100
+40.91 X 107.8(1 --X )
\ 3 loo/
+ 19.81 X8 4 (i-^X
\ 6 100
= 70,580 volts.
. 1.078 .. 2.16
B = 66,000 X e- X
0.840 100
+ 40.91 X 84(1 -\ X^-J 19-81 X 107.8 ( i- - X )
\ 6 100 / V 3 I00 /
= 3150 volts.
Supply voltage = 70,580 + = 7o,6 5 o.
Line drop = 4650 volts = 7.05%.
By the fundamental formulas, using the same line constants, the
line drop is 7.08% (Prob. 2, Chap. VI). The discrepancy is 0.03%
of line voltage.
PROBLEM B.
Find by the K formulas the voltage at the supply end of the
following line:
Total length of line ......................... 300 miles.
Spacing ................................... 12 feet.
K FORMULAS 37
Conductor, 266,800 c.m. aluminum cable.
Load at receiver end of line, 9000 K.V.A., 80% P.F. (lagging),
100,000 volts, three phase, 60 cycles.
Load taken by a substation at the middle of the line, 150 miles
from either end, 2000 K.V.A. at the line voltage and at 70%
P.F. (lagging).
Solution of first section of line.
r = 0.3410 x = 0.791 / = 150,
P = 72 Q = 54.
From the K formulas,
A = 100,000 4860 + 3560 + 6360
= 105,060 volts.
B = 2100 -f 8470 2670
= 7900 volts.
C= 72 -3-50+ 1.13 -0.57
= 69.06 amperes.
D = 1.51 - 54 + 2.63 -f 80.59
= 30.73 amperes.
A H = Ei = 105,360 volts.
AC + BD
In-phase current = -
4 I -^
Reactive current
= 71.15 amps
BC - AD
' 2A
= 25.46 amps
Solution of second section of line.
Conditions at middle of line :
Ei = 105,360.
In-phase current of substation load
looo X 2000 X 0.70
= = 13.29 amps.
105,360
Reactive current of substation load
looo X 2000 X 0.7 141
L - a - = 13.56 amps.
105,360
38 TRANSMISSION LINE FORMULAS
s
PI = 71.15 -|- 13.29 = 84.44 amps.
Qi = 25.46 + 13.56 = 11.90 amps.
Then by the K formulas,
Ai = 105,360 5120 + 4180 1410
= 103,010 volts.
BI = 2200 + 9940 + 590
= 12,730 volts.
T? 2
AI -\ j- = 103,800 volts
2 AI
= voltage at the supply end of the line.
[By the fundamental formulas, supply voltage = 103,900 volts
(Prob. B, Chap. VI).]
PROBLEMS, CHAP. V.
(K FORMULAS.)
1. Find, by means of the K formulas, the voltage drop from
the supply end to the receiver end (the line drop) of the following
line:
Length of line 200 miles.
Spacing 9 feet.
Conductor No. ooo aluminum cable.
Load (at receiver end), 4500 K.V.A., 66,000 volts, 80% P.F.,
three phase, 60 cycles.
Ans. 6650 volts.
[By the fundamental formulas, 6700 volts. (Chap. VI, Prob. A).]
2. Find the regulation of the line in Prob. A, Chap. V. [See
Prob. A, Chap. III.]
Ans. 9.37%.
[By the fundamental formulas 9.40%. (Chap. VI, Prob. 2.)
Error 0.03% of line voltage.]
3. Find the per cent regulation and voltage drop of the following
line:
Length of line 75 miles.
Spacing, 8 feet, regular flat spacing.
Conductor No. oo aluminum cable.
Load (at receiver end), 10,000 K.V.A., 88,000 volts, 85% P.F.,
three phase, 25 cycles. (Prob. 7, Chap. III.)
Ans. 7-35% reg'n, 7.13% drop.
'K FORMULAS 39
4. Find, by the K formulas, the per cent voltage drop, the per
cent loss, and the power factor at the supply end of the following
line:
Length of line 100 miles.
Spacing 6 feet.
Conductor No. oooo copper wire.
Take r = 0.267, * -7 2 7> & = 6.03 X lo" 6 .
Load (at receiver end), 100 amperes per wire, 60,000 volts,
95% P.P., three phase, 60 cycles. [Problem of Pender and
Thomson, Proc. A. /..., July, 1911.]
Ans. 13-09% drop, 7.61% loss, 96.58% P.F.
[Calc. by series, 13.03% drop, 7.60% loss, 96.66% P.F. (Prob. 4,
Chap. VI).]
5. Find, by the K formulas, the K.V.A. and voltage at the supply
end, and the efficiency of the following line:
Length of line 250 Km. = 155.34 miles.
Spacing 6 feet.
Conductor No. ooo copper wire.
Total resistance of one conductor .... 51.5 ohms.
Total reactance of one conductor .... 48 . o ohms.
Load (at receiver end), 15,000 K.V.A., 86,600 volts, 80% P.F.,
three phase, 25 cycles.
[See page 91, "Application of Hyperbolic Functions," by A. E.
Kennelly, University of London Press, 1912.]
Ans. 15,130 K.V.A., 97,920 volts, 89.68%.
[By series, 15,153 K.V.A., 97,934 volts, 89.71%.]
6. Find (a) star voltage at supply end at full load,
(6) star voltage at supply end at no load,
(c) regulation volts (star) at the supply end,
(d) amperes per wire at supply end at full load,
(e) power factor at supply end at full load,
(/) loss in line at full load,
(g) efficiency of the transmission line,
(h) amperes per wire at supply end at no load (i.e., the
"charging current"),
(i) power factor at supply end at no load,
(7) loss in line at no load,
40 TRANSMISSION LINE FORMULAS
for the following line:
Length of line 300 miles.
Spacing 10 feet.
Conductor, No. ooo copper cable of 0.330 ohms per mile.
Load (at receiver end), 18,000 K.V.A., 104,000 line volts,
90% P.F., three phase, 60 cycles.
[Prob. A, page 2, G. E. Review Supplement, May, 1910.]
Answers:
By K Formulas.
(a) 69,820
(b) 48,610
(c) 21,210
(d) 97.0
(e) 92.2
(/) 2530
(2) 86.5
( 91-3
W 7.1
0') 940
7. Find, by the K formulas, the voltage at the supply end of
the following line:
Total length of line 400 miles.
Spacing 15 feet.
Conductor No. oooo copper cable.
Load at receiver end of line, 5000 K.V.A., 85% P.F. (lagging)
110,000 volts, three phase, 60 cycles.
Load taken by a substation at the middle of the line, 200 miles
from either end, 2500 K.V.A., at the line voltage and at 90% P.F.
(lagging).
Ans. 89,720 volts.
[By the convergent series, 90,190 volts (Prob. 6, Chap. VI). Error
0.5%.]
A method of operation of transmission lines is coming
into prominence, by which the voltage of the line is kept
constant at all points, and the inconveniences due to poor
By Fundamental
Formulas.
Error in per cent
of full voltage
or current.
69,670 volts
0-3%
48,950 volts
0.6%
20,720 volts
o.9%
96 . 59 amps.
o.5%
92.35%
0-2%
2440 Kw.
0.6%
86.90%
o.5%
90.97 amps.
o-5%
6-47%
o.7%
860 Kw.
0.5%
K FORMULAS 4OA
regulation are obviated. Synchronous machinery, con-
sisting of either synchronous motors or generators, is
installed in the stations throughout the transmission
system in sufficient quantity to hold the voltage at a
constant value by controlling the amount of leading or
lagging current supplied to the line. This method will
probably come into considerable favor, for there seems to
be practically no limit to the extent of a transmission
system operated at constant voltage.
The following problem outlines the method of calcula-
tion for such cases.
A line 400 miles long has a substation at the middle,
200 miles from either end. A load of 10,000 Kw. is
taken at the receiver end of the line, and 8000 Kw. at
the substation. Find the power factor which is required
for these loads, in order that the voltage at the generator,
substation, and receiver end may be 110,000 volts, the
following data being given:
Conductor, 250,000 c.m. copper cable, 1 4-foot spacing,
3-phase, 60 cycles,
r = 0.2284, x = 0.813.
ist section of line, / = 200 miles.
By the K formulas,
n 1000 X 10.000
P = - -=90.91;
Q is unknown;
A = 110,000 9500 + 3910 + 160.3 Q
= 104,410 + 160.3 Qj
40B TRANSMISSION LINE FORMULAS
B = 2670 + 14,570 - 43-o Q
= 17.240 - 43-o Q-
Now the voltage at the substation end of the ist section
of the line is 110,000; that is,
.
+-^ = no,ooo;
squaring both sides, A 2 + B 2 = 121 X io 8 .
This gives a quadratic equation in Q,
2.754 Q 2 + 3198 Q - 90,150 = o,
from which ()= 27.53 total amps.;
Power factor = = 95.7%,
94.99
and, as Q is positive, this is a lagging power factor.
The power factor obtained by the hyperbolic formulas
is 95-9%-
Using the above value of Q, we obtain
C = +82.78,
D = + 90.59.
In-phase current at substation end of ist section
= + 95- 11 '-
Reactive current at substation end of ist section
= ~ 77-53-
2nd section of line, / = 200 miles.
K FORMULAS 400
In-phase current in ist section = +95.11,
In-phase current of substation load
1000 X 8000
- = + 72.73
110,000
In-phase current at substation end of 2nd section
= 167.84.
Reactive current at substation end of 2nd section
= Q*- 77-53>
where Q x is the unknown reactive current of the sub-
station load.
By the K formulas, as before,
A I = 95>3 + J 6o.3 Q x ,
B l = 32,910 - 43.0 Q x ,
and voltage at generator end of line
AI-\ l = 110,000.
24l
From the above we obtain as before a quadratic equa-
tion in Q x , which gives
&=+ 65.59.
40D TRANSMISSION LINE FORMULAS
Power factor of substation load = 74.3%, lagging.
The power factor obtained by the hyperbolic formulas
is 74.6%,
CHAPTER VI.
CONVERGENT SERIES.
THE mathematical expression for finding the operating
characteristics of a transmission line, in which exact ac-
count is taken of all the electrical properties of the line,
has been published many times. It involves the use of
hyperbolic sines and cosines, as well as of complex quan-
tities,* and, without some special arrangement, cannot be
directly applied to the calculation of a particular case.
For this reason, most of the systems so far published for
calculating transmission lines have used approximate for-
mulas which have been based on the hyperbolic formulas.
In a few cases, an attempt has been made to devise a
system pi working which would give the exact results of the
fundamental hyperbolic formulas, but generally the labor
required in using the systems is so great as almost to pro-
hibit obtaining the exact result, or else the accuracy of
the work is seriously impaired by the necessity of inter-
polating values, from tables of hyperbolic functions which
have been recently prepared for this purpose and are not
as large and complete as they should be for good working.
The original hyperbolic formulas can be expressed in
the form of convergent series.f In this form they do not
* The hyperbolic formulas are given in Chap. XV.
f Prof. T. R. Rosebrugh, Applied Science Magazine, University of To-
ronto, March, 1909; Prof. T. R. Rosebrugh, Proc. A. I.E. E., Nov., 1909,
p. 1460; J. F. H. Douglas, Electrical World, April 28, 1910; Dr. C. P.
Steinmetz, Electrical World, June 23, 1910; Dr. C. P. Steinmetz, "Engi-
neering Mathematics," Chap. V, 1911.
41
42 TRANSMISSION LINE FORMULAS
involve hyperbolic or trigonometrical functions, and so do
not require any mathematical tables, the only operations
being multiplication and addition. The series can be car-
ried to any accuracy desired by merely using enough of the
terms, which diminish very rapidly when commercial fre-
quencies are involved.
The fundamental formulas as expressed by convergent
series have been rearranged, and some new convergent
series have been added, to make the formulas as tabulated
in this chapter directly applicable to the exact solution of
all the problems treated by the K formulas. Exactly
the same final formulas in A, B, C, D, etc., are used with
the convergent series as with the K formulas.
Unlike the K formulas, which are expressed in .the
simplest algebraical form, the convergent series involve
the use of complex numbers, that is, numbers containing
the well-known "j" terms. No difficulty should be expe-
rienced on this account, however, as the rules for using
complex quantities are quite straightforward, and even
one who has never worked with them should be able to
make use of the formulas described in this chapter by
closely following the instructions.
Each of the complex quantities, (A +JB), (P JQ),
Z = (r + jx) I* Y = (g +jb) lj etc., is composed of two
parts, the first, a so-called "real" term, and the second, a
j term. In adding complex numbers, the j terms must
be kept separate from the others. Thus
4 +j 5 added to 7 +73 = n +78.
In multiplying two complex quantities, the simple rules
* The notation Z = (r + jx) I, etc., is used in accordance with the
resolution adopted by the International Electrotechnical Commission in
Turin, Sept., 191 1.
CONVERGENT SERIES 43
of ordinary algebra are followed, and it must be remem-
bered that
j-xj-'f
= -i,
and, therefore,
-jXJ=+i
j* =+J, etc.
Thus (4 +j 5) X (7 +y 3) is worked out as follows:
4+75
7+73
+ 28+jf35
- 15 +y 12
+ 13+747-
In using the convergent series, E, P, and Q are the same
as used with the K formulas, E being expressed as a real
number without any j term. Z is equal to (r -\-jx) I,
where r and x are taken from Tables VII to XII, Part III,
for resistance and reactance per mile. F is equal to
(g -\-jb) I. The leakage conductance, g, per mile, should
be estimated from the most suitable test data available,
giving insulator leakage and corona loss under conditions
similar to those of the line considered. The capacity sus-
ceptance, 6, per mile, will be found in Tables XIII to
XVI, Part III.
After F and Z have been written down in the form of
complex numbers, the product FZ should be found, as
described above for the multiplication of complex quan-
tities. From this is obtained
FZ FZ FZ
- , - , and ,
24 6
44 TRANSMISSION LINE FORMULAS
each expressed as a complex number of a single real term
and a single j term. Multiplying the last two together
F 2 Z 2 F 2 Z 2
gives -, from which may be written
2-3-4 2.3.4.5
down. In most cases no more terms need to be calcu-
lated, even for very accurate work, but this is to be de-
termined while doing the work, as one usually figures out
the terms of these series until they become too small to be
FZ
considered when added to
2
By addition of terms obtained above, the values of
FZ , F 2 Z 2 FZ F 2 Z 2
1 h etc. and 1 + etc.
2-3-4 2-3 2-3.4.5
are obtained, each as a complex number of two terms.
/FZ \
Multiply E by the value found for ( + etc.) and add
it to E. Multiply (P - jQ) by Z, or (r +jx) /, and by the
/Y7 \
value of I + etc.) and add it to (P - jQ) Z. The
above quantities are added together to give A -\-jB, the
sum of all the real parts being equal to A , and the sum of all
the j terms being equal to B.
Similarly, C +JD is found by adding
(P -y, (P -JQ) (^ + etc.), EY, and EY (~ + etc.).
These values of A, B, etc., are inserted in equations i to
21 given with the K formulas, in exactly the same way
as the values of A, B, etc., found according to the second
page of Table III. Each step of the above procedure is
shown in the examples in this chapter.
The use of Table VI is the same as that of Table V de-
scribed above.
CONVERGENT SERIES 45
TABLE V. CONVERGENT SERIES FOR TRANSMISSION LINES.
CONDITIONS GIVEN AT RECEIVER END.
The convergent series give the results of the fundamental formulas as
accurately as desired, if a sufficient number of terms is used.
When conditions are given at the receiver end, the same as with the K
formulas, find the quantities:
Full Load.
1 ( P if)} 7\^
2-3-4 2-3.4.5.6 }
YZ Y 2 Z 2 Y 3 Z 3 ^
~T \f JV) ^ 1 I
i n ( P iD"> ( T
' 2 3 ' 2-3.4.5 ' 2.3.4-5-6-7'
-r JMJ \Jr JV) \ i -
I 1 1 ..- t~ CtC. 1
\
II 1 1 etc
1
1 YZ
2-3 2-3.4-5 2.3.4-5-6.7
No Load.
V%72 V373 \
1 1 r+r 1
r
u I ]DV> -c-l I | 1
/ y
,_l_ in n F.V T 4-
2.3.4 2.3.4.5-6 y'
Z Y^Z 2 Y Z Z 3 \
f .
V 2-3 2-3.4.5 2. 3. 4'5'6. T I*
where Z = (r +y) /.
r = resistance of conductor per mile.
x = reactance of conductor per mile.
I = length of transmission line in miles.
Y=(g+jb)l.
g = leakage conductance of conductor per mile.
b = capacity susceptance of conductor per mile.
Use A, B, C, D, etc., with the equations in the third and fourth pages
of Table HI to solve transmission line problems.
j2
Note i. In the formulas, A -\ r is used instead of VA 2 + B 2 .
2 A
This approximation may be used for very accurate work, as it is correct
within approximately T ^ of i% when the regulation is not more than 20%.
Note 2. The above are for two- and three-phase lines. For single-
phase lines use 2 r and 2 * in place of r and x, and use | g and | b in place
of g and b.
46 TRANSMISSION LINE FORMULAS
TABLE VI. CONVERGENT SERIES FOR TRANSMISSION LINES.
CONDITIONS GIVEN AT THE SUPPLY END.
The convergent series give the results of the fundamental formulas as
accurately as desired, if a sufficient number of terms is used.
When conditions are given at the supply end, the same as with the
K formulas, find the quantities:
Full Load.
(
2-3 2-3.4.5 2.3-4-S
No Load.
+ JG, = E. (i - h YZ + Y>Z* - g Y'Z* + ^ YW - etc.),
, +jlf, = E,Y(I -$YZ + Y*2?-^- YW + -,- Y<Z> - etc.),
\ -"-5 o-'S 2035 /
where Z = (r +jx) I.
r = resistance of conductor per mile.
x = reactance of conductor per mile.
I = length of transmission line in miles.
Y=(g+jb)l.
g = leakage conductance of conductor per mrle.
b = capacity susceptance of conductor per mile.
Use F, G, M, N, etc., with the equations in the third and fourth pages
of Table IV to solve transmission line problems.
Note i. In the formulas, F H -- - is used instead of Vp* + G 2 . This
2 F
approximation may be used for very accurate work, as it is correct within
approximately yf ^ of i% when the regulation is not more than 20%.
Note 2. The above are for two- and three-phase lines. For single-
phase lines use 2 r and 2 x in place of r and x, and use \ g and \ b in place of
g and b.
CONVERGENT SERIES 47
PROBLEM A.
Find the line drop, by means of the convergent series, for the
following line:
Length of line 200 miles.
Spacing 9 f ee t-
Conductor No. ooo aluminum cable.
Load (at receiver end), 4500 K.V.A., 66,000 volts, 80% P.F.,
three phase, 60 cycles.
From the tables,
r = 0.5412, * = 0.784, 6=549Xio- 6 .
Then p = looo X 4500 X 0.8 ^^^
06,000
looo X 45QO X 0.6 =
66,000
Z = 108.24+.; 156.8.
Y = -j-j 0.001098.
YZ = -0.17216 +j 0.11885.
YZ
= 0.08608 +j 0.05942.
YZ
= - 0.04304 +y 0.02971.
4
YZ
= 0.02869 +.70.01981.
6
0.00059 .70.00085.
+ 0.00124 j 0.00085.
F 2 Z 2
= + 0.00065 j 0.00170.
2-3.4
YZ
= 0.08608 +j 0.05942.
IY7 \
h etc. = - 0.08543 +^'0.05772.
\ 2 /
F 2 Z 2
= + 0.00013 j 0.00034.
2.3.4-5
YZ
= 0.02869 +y 0.01981.
2 '3
( +etc.)= 0.02856+70.01947.
\2 3 /
48 TRANSMISSION LINE FORMULAS
Now E = 66,000.
(YZ \
-- h etc. J = 0.08543 +.;' 0.05772.
~ + etc.) = -5640 +.7-3810.
P-JQ= 54-55 -3 40.91-
708.24 +y 156.8.
+ 5900 -j 4430.
+ 6420+^8550,
(P - JQ) Z= + 12320 + j 4120.
YZ \
etc. ] = 0.02856 + 7*0.01947.
*
\23
80+^*240.
-350 -./i 20.
(V7 \
+ etc. ) = - 430 +j 120.
2*3 /
E = 66,000.
E( h etc.J = 5640+^3810.
( p -jQ)Z= 12320+^4120.
(P - JQ) Z>(^- + etc.) = - 430 +j 120.
A+jB= 78,320 - 6070+^*8050
= 72,250+^8050.
B z
A H = 72,700 volts.
2 ^1
Line drop = 6700 volts.
PROBLEM B.
Find, by the convergent series, the voltage at the supply end of
the following line :
Total length of line 300 miles.
Spacing 12 feet.
Conductor, 266,800 c.m. aluminum cable.
Load at receiver end of line, 9000 K.V.A., 80% P.F. (lag-
ging), 100,000 volts, three phase, 60 cycles.
CONVERGENT SERIES 49
Load taken by a substation at the middle of the line, 1 50 miles
from either end, 2000 K.V.A., at the line voltage and at 70%
P.F. (lagging). (Prob. B, Chap. V.)
Solution of first section of line:
r = 0.3410, x = 0.791, b = 5.44 X io~ 6 , / = 150.
Z= 51.15-!-.; 118.65.
y = +j 0.000816.
YZ = 0.09682 +,70.04174.
( + -^- + etc. ) = - 0.04809 +j 0.02053.
\ 2 2 3 '4 I
F__, -- F 2 Z 2 -I- etc.) = - 0.01607 +j 0.00689.
2-3 2-3.4-5 /
E = + 100,000.
E ~ + etc = - 4810 +j 2050.
(P-jQ)Z= 10,090 +./5780.
(P - jQ) Z ( + etc.
- 200 - 20.
A+jB= 105,080 +77810.
B z
A + = 1 = 105,370 volts.
In a similar manner, it is found that
C+jD = 69.08 + j 30.36 amps.
In-phase current, - = 71.15 amps.
A + JL ;
2 A
Reactive current, - - = 25.15 amps.
A+-
2 A
Solution of second section of line:
Conditions at middle of line,
Ei = 105,370 volts.
In-phase current of substation load
1000 X 2000 X 0.70
= - - L - = 13.29 amps.
105,370
Reactive current of substation load
1000 X 2000 X 0.7141
_ - i * . = I3>55 amps<
105,370
50 TRANSMISSION LINE FORMULAS
Current of substation load = 13.29 j 13. 55.
Current of first section = 71.15 -\-j 25.15.
Pi JQi = 84.44 +j 1 1. 60.
1 = 4- 105,370.
(YZ \
4- etc. J - 5070 4- J 2, 160.
(Pi ~~ JQi) Z = 2 >94 4- J 1 0,6 10.
(Pi JQi) Z i- '- 4- etc.] = - 120 -y 150.
* * 3 /
^i -}-jBi = 103,120 4~J 12,620 volts,
yli -!> j- = 103,900 volts
= voltage at the supply end of the line.
PROBLEMS, CHAP. VI.
(CONVERGENT SERIES.)
1. Find, by the convergent series, the voltage drop of the follow-
ing line:
Length of line 80 miles.
Spacing 10 feet.
Conductor No. oo aluminum cable.
Load (at receiver end), 15,000 K.V.A., 100,000 volts, 95%
P.F., two phase, 25 cycles. (Prob. C, Chap. III.)
Ans. 8810 volts.
2. Find, by the convergent series, the per cent line drop and the
per cent regulation of the following line:
Length of line 100 miles.
Spacing 8 feet.
Conductor No. 3 copper cable.
Load (at receiver end), 3000 K.V.A., 66,000 volts, 90% P.F.,
three phase, 60 cycles. (See'Prob. A., Chap. Ill, and Prob. A,
Chap. V.)
Ans. 7.08% drop, 9.40% reg'n.
3. Find, by the convergent series, the K.V.A. and voltage at the
supply end, and the efficiency of the following line:
CONVERGENT SERIES 51
Length of line 250 Km. = 155.34 miles.
Spacing 6 feet.
Conductor No. ooo copper wire.
Total resistance of one conductor, 51.5 ohms.
Total reactance of one conductor, 48.0 ohms.
Total susceptance of one conductor, 3.724 X icf 4 mhos.
Load (at receiver end), 15,000 K.V.A., 86,600 volts, 80%
P.F., three phase, 25 cycles. (Prob. 5, Chap. V.)
Ans. 15,153 K.V.A., 97,934 volts, line; 56,542 volts, star;
89.71%-
4. Find, by the convergent series, the per cent voltage drop, the
per cent loss, and the power factor at the supply end of the following
line:
Length of line 100 miles.
Spacing 6 feet.
Conductor No. oooo copper wire.
Take r 0.267, x -7 2 7> ^ = 6.03 X io~ 6 .
Load (at receiver end), 100 amperes per wire, 60,000 volts,
95% P.F., three phase, 60 cycles. [Prob. 4, Chap. V.]
Ans. 13.03% drop, 7.60% loss, 96.66% P.F.
5. Find, by the convergent series,
(a) star voltage at supply end at full load,
(b) star voltage at supply end at no load,
(c~) regulation volts (star), at the supply end, .
(d) amperes per wire at supply end at full load,
(e) power factor at supply end at full load,
(/) loss in line at full load,
(g) efficiency of the transmission line,
(ti) amperes per wire at supply end at no load (i.e., the
"charging current"),
({) power factor at supply end at no load,
(/) loss in line at no load, for the following line:
Length of line 300 miles.
Spacing 10 feet.
Conductor, No. ooo copper cable of 0.330 ohm per mile.
Load (at receiver end), 18,000 K.V.A., 104,000 volts, 90% P.F.,
three phase, 60 cycles. (Prob. 6, Chap. V.)
52 TRANSMISSION LINE FORMULAS
Ans. (a) 69,670 volts, (b) 48,950 volts, (c) 20,720 volts, (d)
96.59 amps., (g) 92.35%, (/) 2440 Kw., (g) 86.90%, (A)
90.97 amps., (i) 6.47%, (;) 860 Kw.
6. Find, by the convergent series, the voltage at the supply end
of the following line:
Total length of line 400 miles.
Spacing 15 feet.
Conductor No. oooo copper cable.
Load at receiver end of line, 5000 K.V.A., 85% P.P. (lagging),
110,000 volts, three phase, 60 cycles.
Load taken by a substation at the middle of the line, 200 miles
from either end, 2500 K.V.A. at the line voltage and at 90% P.F.
(lagging). (Prob. 7, Chap. V.)
Ans. 90,190 volts.
PART II.
THEORY.
CHAPTER VII.
CONDUCTORS.
THREE main classes of conductors are used for overhead
lines for the transmission of electric power; namely, copper
wires, copper cables and aluminum cables. The cables used
are generally strands of seven wires; that is, they consist
of a central straight wire with six
wires wound spirally around it, as
indicated by the cross section in
Fig. 9.
From this figure it is seen that ..
the maximum diameter of a y-wire
strand is equal to 3 times the di-
ameter of one of the wires.
The outside wires do not follow
a straight path parallel to the
central wire and the axis of the cable, but lie in a spiral
around it, as mentioned above. As there is always a
slight insulating film of oxide on any wire, the current
flowing in the cable tends to stay in the individual wires,
and so follows the longer path. Thus, the resistance of a
cable is greater than that of a solid wire of the same area
of cross section. The amount of the difference depends on
the number of wires in the cable and the pitch of the
53
2s
Fig. 9. 7-Wire Strand.
54 TRANSMISSION LINE FORMULAS
spiralling, but an average value of i% is assumed in mak-
ing up the tables in Part III. The cross section of the
cable is assumed to be equal to the sum of the cross sections
of the individual wires. The weight per unit length of the
cable calculated from this cross section must be increased
by the same percentage as the above increase in resistance,
due to the extra length of the outside wires. Since the
cross section in Fig. 9 does not cut the outside wires ex-
actly at right angles, their sections as shown in the figure
are really ellipses, and the di-
ameter of the cable is slightly
greater than 6 pi. However,
this difference is small and has
been neglected in the figures
for diameter of cable tabulated
in Part III.
The number of wires in a
strand varies in practice ac-
j<_ 2$ J cording to the degree of flexi-
bility and mechanical strength
Fig. 10. 19- Wire Strand. J
desired by the user. The num-
ber of wires per strand in the tables represents average
practice for overhead lines. The larger cables often have
19 or even 37 wires.
The section of a i9-wire strand is shown in Fig. 10, and
it is seen that the maximum diameter is 5 times the diam-
eter of one of the individual wires. The same increase of
i% in resistance is allowed as with a 7-wire strand.
There is only a very slight difference in the reactance
and capacity of a 7-wire and a i9-wire strand of the same
sectional area, so that values listed for 7 wires may be
used for 19 wires, and vice versa, without very much error.
CONDUCTORS 55
The resistances for direct current tabulated in Part III
have been calculated in accordance with the recommenda-
tions of the Bureau of Standards for the preparation of wire
tables.* The Standardization Rules of the* American In-
stitute of Electrical Engineers are in agreement with these
recommendations. According to the Bureau of Standards
and the A. I. E. E., the "Annealed Copper Standard,"
which is of 100% conductivity, is represented by a resis-
tivity of 0.153022 ohm per meter-gram at 20 C. This is
equivalent to 1.72128 micro-ohms per centimeter cube at
20 C., assuming a density of 8.89. This is the same as
the resistivity of Matthiessen's Standard at 20 C., for-
merly used by the A. I. E. E. The conductivity of hard
drawn copper recommended for wire tables by the Bureau
of Standards is 97.3%, this value representing an average
for good commercial copper. The average conductivity
given by the Bureau of Standards for hard drawn alu-
minum on the centimeter cube basis, assuming a density of
2.699, is 60.86%. The above values have been used in
preparing the tables in Part III, i% being added to the
resistance for the effect of spiralling, as already noted.
If it is desired to calculate the resistance of copper con-
ductors for other temperatures than 20 C., the tem-
perature coefficient, a 2 o, for hard drawn copper of 97.3%
conductivity should be used in connection with the formula
R t = RW { i + 2o (t - 20) }
where / is the temperature in degrees Centigrade lor which
the resistance R t is desired and where
2o = 0.00383.
* Bulletin of the Bureau of Standards, Vol. VII, pp. 71-126, Washington,
1911; Proc. A. I. E. ., Dec., 1910.
56 TRANSMISSION LINE FORMULAS
For other initial temperatures and other conductivities,
temperature coefficients should be used as given in the
table of temperature coefficients in Part III, which is
taken from Appendix E of the Standardization Rules of the
A. I. E. E.
For the temperature coefficient of hard drawn aluminum,
a value of
2o = 0.0039,
which is recommended by the Bureau of Standards, may
be used.
CHAPTER VIII.
TRANSMISSION LINE PROBLEMS.
WHEN conditions are given at the receiver, or load, end of
a transmission line, the convergent series of Table V give
at once the voltage, A +JB, and the current, C +JD, at
the other end of the line. By putting the load current
equal to zero, we obtain the following expression for the
no-load voltage at the supply end:
CV7 17272 \
i+ + JL ^ L - + etc.)
2 2-3.4 /
Thus the ratio of the voltages at the two ends of the line at
no load is
E \ 22-3-4
which is independent of the voltage E, and depends only
on the constants of the line.
The absolute value of a complex quantity like the volt-
age AQ -\-jBo, is its total numerical value independent of its
phase relation. This is the same, in the case of the voltage
A Q +JB , as its measured value, and is equal to
or + -
2 AQ
to a very close approximation when BQ is smaller than A .
Since the two complex quantities making up equation (i)
are equal in all respects, their absolute values are eaual,
and hence
CV7 V%7 2 \
i + + J - jL - + etc.) (2)
2 2-3-4 /
57
58 TRANSMISSION LINE FORMULAS
When the line is carrying full load, the measured value
of the receiver voltage is E, and of the supply voltage,
B 2
A H -- j- If the load be thrown off and the supply volt-
2 A.
B 2
age be kept constant at A -\ -- - > then the receiver voltage
2 A.
will rise to a value E Q . The line is now at no load, and the
ratio of the voltages at the two ends is, by equation (2),
I YZ Y 2 Z 2 \
= absolute value off i + H ---- (- etc. I
V 2 2 3 4 /
A +
Thus Eo =
(equation 2, Table III).
We are now in a position to obtain the regulation of the
line, since by the definition in the A. I. E. E. Standardiza-
tion Rules,
En E
Per cent regulation = *-= X 100.
rL
Thus, the regulation volts at the receiver end which are
to be expressed as a percentage of E, are
as in equation 3, Table III.
TRANSMISSION LINE PROBLEMS
59
It is often desirable to find the regulation of a line at the
supply end, that is, the per cent change in supply voltage
from full-load conditions to no-load conditions, when the
receiver voltage is kept constant. If the receiver voltage
is E, we have seen in the preceding paragraph that the full-
load supply voltage is equal to
2 A
and the no-load supply voltage is
E 0a =A + -
The per cent regulation at the supply end is
E s E 0a
E 8
Xioo,
and the regulation volts at the supply end are, thus
77
E a
2 A
A
AQ
2A
as in equation 6, Table III.
In the expression C +JD for current at the supply end,
the quantity C denotes the
component of current which is
in phase with the voltage E at
the other end of the line. We
can, however, find the com-
ponent of supply current which
is in phase with the supply
voltage, by first finding the k
watts at the supply end.
Let the supply voltage be
E 9 = A +JB
Fig. n.
60 TRANSMISSION LINE FORMULAS
and let its phase be denoted by the angle 0, Fig. n, where
7?
tan = >
r>
and, therefore, sin =
^4
and cos =
Similarly, let the current at the supply end be C + JD,
at a phase angle 0, where
and, therefore
and
VC 2 + D 2
The watts at the supply end are equal to the current,
multiplied by the voltage, multiplied by the power factor;
that is,
Watts = absolute value of I a X absolute value of E s
X cos (0 - 0)
= \/C 2 + D 2 X VA 2 + B 2 X (cos cos 0+sin sin 0)
= VC 2 + > 2 X
BD
as in equation n, Table III.
The quadrature volt-amperes, or reactive power, are
given by the following equation :
TRANSMISSION LINE PROBLEMS 6l
Reactive power
= absolute value of I 8 X absolute value of 8 Xsin (0 0).
= VC 2 + D 2 X VA 2 + B 2 (sin 6 cos - cos 6 sin 0)
j x c__ ^ x z> . )
= C - ylZ).
When the expression BC AD has a positive value, the
current at the supply end is lagging behind the supply
voltage, and when the expression has a negative value, the
current leads the voltage in phase.
We can now obtain the in-phase component of current,
which is equal to watts divided by voltage (equations 15
and 1 6), and in the same way the quadrature component
of current, which is equal to reactive power divided by
voltage (equations 17 and 18). The power factor at the
supply end is equal to watts divided by volt-amperes
(equations 13 and 14). Since the power supplied is known,
being AC + BD, and the power delivered at the receiver is
also known, being equal to EP, their difference represents
the loss of power in the line due to resistance of the con-
ductors, leakage over the insulators and corona loss.
The equations in F, G, M and N are quite similar to the
above equations in their derivation, and they give the solu-
tions of similar problems when conditions are given at the
supply end of the line.
CHAPTER IX.
REACTANCE OF WIRE, SINGLE-PHASE.
Effect of Flux in Air. Let there be an alternating
current, /, in the transmission line wire, A, indicated in
Fig. 12.
The magnetic field set up
by the current at P, a dis-
tance x away from the wire,
will be at right angles to
the wire. The intensity of
the field will be equal to the
force on a unit magnetic pole
at P due to the current in the wire. The force due to the
current in a short length, dl, of the wire will be
cos 6 = - cos 6 d8,
r 2 x
since dl cos = r dd
, x
and
cos0
The total force at P due to the current in the wire A is
equal to
1C
C~ 2 I cosOdd , , . .v
(where re is a constant)
J <* ^
TT
I sin 0"P
2/
X
62
REACTANCE OF WIRE, SINGLE-PHASE 63
where / is measured in absolute electromagnetic units.
When I is in amperes, the field at distance x is
- lines per sq. cm. (i)
10 x
r s-AV ---!
_A 41 $5.
_ r-r -J~. ^^
Fig. 13.
In Fig. 13 is shown the cross section of a single-phase
transmission line. The lines of force in the path of thick-
ness dx surrounding the wire A are
21 j
dx
IOX
per centimeter of the transmission line. These lines cut
the wire A and produce an alternating voltage in it which
is 90 out of phase with the current and is equal to
2/
ju dx X icr 9 volts,
x
where co = 2 TT X number of cycles per second, and where
/ is in amperes.
The voltage drop between the wires A and B, due to
flux in the air produced by the current in A, is obtained by
integrating the above expression from x = p to x = s.
The integration is not carried beyond x = s, since flux
which cuts both A and the return wire B does not produce
any voltage between them. The voltage drop is equal to
* 2 7 s
jw dx X io~ 9 = y 2 / loge - X io~ 9 .
TRANSMISSION LINE FORMULAS
There will be an equal drop due to the flux produced by
the current in the wire B, so that the total drop due to
flux in air is
jco4/log e - X lo- 9
p
volts per centimeter of line,
2jw X 741-1 logio- X
p
(2)
volts per ampere per mile of single-phase line.
Effect of Flux in the Conductor. Let i be the current
per unit area of section at any point in the wire shown in
Fig. 14. (For the present
assume that i is the same at
all points of the section.)
The total area of section of
the wire is irp 2 and therefore
the total current in the wire
is
/ = Trip 2 .
Fig. 14. Section of Wire.
The total current inside the
circle of radius x is
/l = irix 2 .
This is the only current forcing flux around the circular
path of width dx t since currents flowing nearer the surface
of the wire do not tend to produce magnetic lines in a
path which does not surround them. Thus the flux density
at the radius x is
n I _ /> /w/f/v*'"
10 X
IOX
2irix
IO
REACTANCE OF WIRE, SINGLE-PHASE 65
The total flux in the outer ring of the section is
dx _ TTJ (p 2 x 2 )
10 10
This cuts the element dx of the wire and produces a voltage
along it equal to
juTri (p 2 x 2 ) io~ 9 volts per cm. (3)
This voltage leads the current by 90 in phase at all sec-
tions. It is greatest at the center and zero at the surface
and so is an unbalanced voltage; it therefore causes a local
quadrature current to flow along the center of the wire and
return near the surface.
Let the local current at the element dx be i( X ) per unit
area of section. Then the average voltage drop along the
wire due to the flux inside it and the resulting local bal-
ancing current, is equal to
juILi = juiri (p 2 - x 2 ) io~ 9 + i (x ) r,
where L\ is the self -inductance of the wire due to the
above-mentioned flux inside it, and where r is the specific
resistance of the metal in centimeter units, that is, the re-
sistance of a centimeter cube of the metal. The current
i (X ) adjusts itself so that the drop is the same at all parts
of the section. From the last equation, we have
jwrip^Li juiri (p 2 - X 2 ) 9 , ,
*(*) = j j ~ X 10 y , (4)
since / = Trip 2 .
As i( X ) is a local current in the wire, and does not in-
crease or decrease the main current /, its sum when added
up all over the section must be zero, and thus
r. _
2 1^x^( X ) ax = o,
.
66 TRANSMISSION LINE FORMULAS
that is, j ^^ fVLi* - P 2 x icr 9 + x 3 icT 9 ) dx = o
r Jo
Now LI is a constant, independent of x, and so
P 4 Z,! - p 4 io~ 9 + - io~ 9 = o.
2
Therefore Li = \ X icr 9 (5)
The voltage drop between the wires A and B due to the
flux inside both wires is
2JuILi = 2/w/ X | X lo" 9 volts per cm.
= 270) X 80.47 X icT 6
volts per ampere per mile. The total reactive drop be-
tween the wires is thus
2./ (80.47 + 741.1 logio-) X io~ 6 (6)
volts per ampere per mile of single phase line.
This may be written in the following form which is more
convenient for computation by means of logarithm tables:
Reactance drop = 2Ju X 741.1 Iogi X io~ 5 (7)
0.779 p
volts per ampere per mile of single-phase line.
The above is the usual formula for reactance of a single-
phase line. The proof is longer than that generally given,
but it has the advantage of giving a correct idea of
the distribution of current and magnetic flux inside the
wire. As the irregular distribution of current produces the
"skin effect" described in the next chapter, and necessi-
tates slight corrections in the above formula for reactance
and in the resistance, the importance of calculating the
correct current distribution is evident. The above formula
is sufficiently accurate, however, for calculating the tables
of reactance of wire in Part III.
CHAPTER X.
SKIN EFFECT.
IN the last chapter a local quadrature current i( X ) was
assumed, whose resistance drop balances up the unequal
voltages produced at the center and near the surface by
the flux inside the wire. This local current, i (X ), when
added up over all parts of the section of the' wire, amounts
to zero, and so cannot produce magnetic lines in the air
outside the wire. But it can produce lines inside the wire,
and the effect of these will now be calculated.
The reactive drop in one wire due to the flux inside it
produced by the main current i is
juirip 2 Li = j(mif^ X \ X io~ 9 volts,
where i is in amperes.
Then at a distance x from the center we have, from equa-
tion (4), Chap. IX,
X X TO-' - - p _ * x
2 r
10"
This is a lagging current at the center and a leading
current at the surface, and it equals zero when integrated
over the entire section.
The current i( X ) , integrated over the circle of radius x } is
68 TRANSMISSION LINE FORMULAS
The flux density at the element dx, due to the above cur-
rent, is
2 Ii x \ jwirH _ Q / 9 ~x
- ^ = * - X 10 9 (- p 2 X + X s ).
iox 10 r
The flux in the ring outside of the circle of radius x, due
to I (x) , is
( x) - ^ x 10-9 r (-^4. *s) ^
ior J x
jW^io- 9 / P 4 . p 2 * 2 . p 4 * 4 \
= - r- H --- 1 ---- J
ior \ 2 2 4 47
/COTT 2 ^' IO~ 9 / 9 9 4\
= < - (- p 4 + 2 p 2 X 2 - ^ 4 ).
40 r
This flux produces a voltage at the element dx, equal to
U>Vi IO~ 18 / 4 , 9 o 4 \ / \
- (- p 4 + 2 p 2 * 2 - ^ 4 ). (i)
4 P
A local current, f (2x ), will flow in order to keep the volt-
age drop uniform over the section. Let the average drop
due to 4>( X ) be
then
" Vilo ~ 18 ( P 4 - 2 p
Integrate ^ (2 x) over the entire surface and as it is a local
current
2irxi(2x)dx = o
r Jo
2
SKIN EFFECT 69
18 / 6 6 "
Therefore, /Z^P 4 = - ( - - - +
2 T \2 2
, . I COTTp 2 I0~ 18 f N
and 2 = j (3)
Thus i ( 2x) = - C 7r * I (2 p 4 - 6 p 2 * 2 + 3 x 4 ) (4)
This current is in phase with the main current and, as it
is negative at the center and positive at the surface, it
produces a stronger resultant current near the surface of
the wire. This is the well-known "skin effect." The
effect of the quadrature current i^ is to increase the re-
sultant current both at the center and near the surface,
but its effect is not as large as that of the in-phase current
i (2x ) and so the net result is a crowding of current toward
the surface.
The above process may be continued indefinitely, each
step adding a smaller correction than the one before to
the current at radius x and to the average drop in the wire.
Thus the expression for i($ x) , equation (4) , may be inte-
grated over the circle of radius #, and will give the value
of 1(2 x )' This current produces a flux density at the radius
x, and by integrating this over the outer ring of the section,
the value of 0@ X ) is obtained. The flux <j> ( 2 X ) produces an
unbalanced voltage which must be corrected by a local
current i(3 X ), so as to give a uniform drop over the section,
due to the inductance Z, 3 . Equating the total local current
to zero, as before, gives
i coVV io~ 27
LZ = -
In the same way it is found that
. i coVp 6 io~ 36
70 TRANSMISSION LINE FORMULAS
1 C0 4 7T 48 IO- 45
and
8640 r 4
Let the resistance of the wire per centimeter be R, where
R = - ohms per cm.,
COTTp 2 IO~ 9
and let m = - - -
r
- ^ IO ~ 9
R
Then the total drop in the wire is
IR
( 2 Iog 6 - H --- j
\ p 2 J 12
= IR + ;W io~ 9 2 Io 6 - H --- m
- * + - i2 - ^ 4 ---- ^ (5)
48 180 8640 /
volts per centimeter.
The total drop in phase with the current is
-Y (6)
/
12 180
The total copper loss due to all the currents in the wire is
therefore equal to
PR(I+ -Lm 2 --Lm 4 + ...V
V 12 180 /
This can be checked by integrating the losses due to the
total in-phase and quadrature currents in all parts of the
section of the wire, the above result being obtained by
this method also. Thus, in every respect, both as to volt-
age drop and watts loss, the resistance of the wire to the
alternating current is
12
SKIN EFFECT 71
Values of R' for both 25 and 60 cycles are tabulated in
Part III. When taking the resistance of a conductor from
the tables, R' should always be used for alternating current,
and R should be used only when the conductor carries
direct current.
The total drop in quadrature with the current is
= /co/io" 9 ) 2log - + -(i -- m 2 -\ -- 3-m* - ) (7)
( p 2\ 24 4320 I)
The series i - m 2 + -^-w 4 -
24 4320
is thus a correction factor for the term J or 80.47 in the
ordinary formula for reactance. Its effect is too small,
however, to make any appreciable change in the tabulated
values of reactance.
Proof by Infinite Series. The above formulas for the
resistance and inductance of a wire carrying alternating
current are sufficiently accurate for transmission line calcu-
lations with ordinary frequencies. They may .also be ex-
tended to include more terms without undue labor. How-
ever, as skin effect formulas are generally obtained and
expressed by means of infinite series which can be carried
out to any degree of accuracy for high-frequency work, a
short outline of the derivation of the infinite series will be
given. It will prove a check upon the correctness of the
formulas given above, but it will probably not give as
clear an idea as they do of the actual distribution of
current in the wire.
Let an alternating current, 7, of sine wave form and of
steady value, flow in a round wire of radius p. (See Fig. 14,
72 TRANSMISSION LINE FORMULAS
Chap. IX.) Let it take up such a distribution that the
drop at all parts of the section of the wire, due to re-
sistance and to magnetic flux, is the same. Then if i' be
the current density at radius x, we may assume
*' = flo + <*i* 2 + 2* 4 + + a n x 2n +:.-- (8)
where OQ, a\, . . . a nj etc., are constants, independent of x.
(As the same value of i f would be obtained for both +x
and x, only even powers of x need be assumed for the
series.)
The total current in the part of the section inside a
circle of radius x will be
/
Jo
= / 2irxi f dx
i . n ..i ,.
= 2 TT [ -- 1 --- r H ---- r r (9;
\ 2 4 2n /
The flux density at the radius x is
iox io \ 2 n
and the total flux in the outer ring of the section, outside
the circle of radius x, is
2 I f
r
= I
J x
,
dx
iox
io 2 3 n
The drop at radius x due to the flux 0' is
jot' X io~ 8
and the resistance drop due to the current at the same
SKIN EFFECT 73
part is i'r. Thus the total drop per centimeter of wire,
which is the same at all parts of the section, is
V = jwtf icr 8 + i'r
2' 3 * /
+ r (oo + aix 2 + a*x* + - - + a n # 2n +)" (n)
The above expression for V is the same for all values of x,
and we may therefore equate each coefficient of x 2 , # 4 , etc.,
to zero. Thus, putting
corrp 2 ICT 9
= m,
r
we have a\ =
2p
np
... ... etc.
and
V = fl^+WTo
Substituting the values of ai, 02, etc., we obtain
V OQT -\-jwu lo" 9
(jm) 2 a p 2 (jm} n ~ l a^ .
+ '
(12)
74
TRANSMISSION LINE FORMULAS
Now by putting x = p in the expression for /', equation
(9), we obtain the value of the total current in the wire,
2 3
fgi + ^
n
Therefore,
^
n(jm} n ~ l
Substituting this value of OQ in equation (12), and putting
the resistance per centimeter of the wire, we obtain
(J m Y i , (J m Y i
+ ... + _ + ...
r j. j.v
.
"
i 6 \J'"'l i
(W
+
_
V*Oy
+
This expression can evidently be carried to any accuracy
desired. It will give the same results as were previously
obtained in equation (5), by expanding the denominator as
a binomial of the form (i + x)~ l and multiplying by the
numerator. This gives
V =
or,
io
~ 9 -
This is the voltage drop, omitting the effect of the flux
SKIN EFFECT 75
outside the wire, and is the same as the value previously
obtained. (See equations 6 and 7.)
REFERENCES.
Maxwell, Elec. and Magn., Vol. II, Para. 689-690.
Rayleigh, Phil. Mag., 1886, Vol. 21, page 381.
Kelvin, Math. Papers, 1889, Vol. 3, page 491- *
Rosa and Grover, Bulletin of Bureau of Standards, Washington, 1911,
Vol. 8, No. i, pages 173-181.
CHAPTER XL
REACTANCE OF CABLE, SINGLE-PHASE.
As stranded cables are very commonly used for trans-
mission lines, it is desirable to have a special formula for
the reactance of cables. An outline will be given of the
method of obtaining the formula for a seven-wire cable.
This formula was used in preparing the reactance tables
in Part III.
A seven-wire strand consists of a central straight wire,
with six wires of the same size laid around it in a spiral.
The spiralling of the wires increases the resistance of the
cable by an amount which is taken as i%. The spiralling
also increases the outside diameter of the cable by about
T V of i%. (See Chapter VII.)
In calculating the reactance of the cable, the first step is
to plot the flux density at various distances from the center
of the cable. (See Fig. 15.) For points entirely outside
the cable, the flux density obeys the law
27
where x is the distance from the center. The total voltage
due to these lines which cut the entire cable is
2log e - X io~ 9
p
volts per ampere per centimeter. When x is less than the
radius of the cable, the flux at the distance x is
'
76
REACTANCE OF CABLE, SINGLE-PHASE
77
I (X ) is proportional to the area of conductor inside the
circle of radius x, and this must be measured from the
diagram of the section of the conductor (Fig. 15).,
/'
~
x
^^
^
"-*
-,
s'
'^
*
^^
-^
/
/-
x ^
X
x~
' -*
^
^>
-^
^
^
^x
^
.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 I.I
X
p
Fig. 15.
Plot a curve of flux density, / (x) , for various values of #.
Let (x) be the area of the curve of / (x) between the values
# and p, where p is the outside radius of the cable. Also,
let C( X ) be the length of that part of the circle 2 TX which
lies in the section of the wires of the cable.
As shown in Chapter IX, page 65, the voltage drop
along different parts of the cable is not uniform, and must
be balanced by a local quadrature current. Thus we have
the conditions
io
~ 8
at any section,
78 TRANSMISSION LINE FORMULAS
and 2X'z)(z) dx = o.
o
Substituting the value of i( X ) from the first equation, we have
dx ->zCz dx io~ 8 = o.
<r
Now I LI is a constant, and
2.c (X )dx = A,
(T
the area of the section of the cable.
By dividing p into a number of equal parts and calling
each part dx, the value of 4>( X )C (X ) dx for each successive
value of x is found. Adding these together, a close estimate
of the value of
dx
is obtained. This, divided by I A gives the value
Li = 0.633 X icr 9 .
If the conductor were a solid wire of radius p, with a
straight line curve of / (x) , 1 would be J X lo" 9 , which
would become 80.5 X io~ 6 when reduced to the ordinary
formula for reactance per mile. Thus the reactance per
mile of cable is
x = (80.5 X ^3 + 74I .! lo glo -) io- X 2 irf
\ 0.500 p/
= ( 102 + 74I.I logio-j I0~ 6 X 2 7T/
ohms per mile.
The same process applied to a ig-wire strand gives the
formula
x = ($9 + 741.1 logio-j icr 6 X2irJ
ohms per mile.
CHAPTER XII.
REACTANCE OF TWO-PHASE AND THREE-PHASE LINES.
Reactance, Two-phase. The reactive drop in a single-
phase line, in which round wire is used, is
2JI X 2 7r/f 80.5 -f 741.1 log-) io~ 9
volts per mile of line. In a two-phase four-wire line, the
drop will be the same as the above, when / is the current
in one phase.
AT T K.V.A.
Now 2 / = - = the total amperes.
hi
Therefore the reactive drop per mile of line, in absolute
value, is
^|^ X 2 T/(8o.s + 741.1 log-) io-'
\ p/
K.V.A.,
~E~~ Xx >
wnere x is the tabulated value of reactance.
Reactance, Three-phase, Irregular Spacing. When the
conductors of a three-phase line are spaced so that they are
not exactly equidistant, the voltage drop due to reactance
is not the same in the different phases. It is the practice
with such lines tp interchange, or transpose, the conductors
at intervals along the line, so that the different reactive
voltages are applied to an equal extent to all three con-
ductors. Such a line, when carrying a balanced load (equal
currents in each conductor, at 120 in phase from each
79
8o
TRANSMISSION LINE FORMULAS
other), will have the voltages of the three phases equal at
the end of the line.
The average reactance of an irregularly spaced three-
phase line, in which the conductors are transposed at regu-
lar intervals, may be calculated as follows:
Fig. i 6.
17.
Let Fig. 1 6 represent the spacing of the conductors A, B,
and C of a transmission line. Let the currents in the con-
ductors be represented by the vectors OP, OQ, and OR
(Fig. 17). If the power factor is 100%, these vectors may
also represent the star voltages, and the line voltages will
be represented by the vectors PQ, QR, and RP.
Let the current in conductor A be
and let
and
I A = OP = i .00 / amperes,
IB = OQ = ( 0.50 + o.S66j) I amperes,
In = OR = ( 0.50 0.866 j) /amperes.
Let also
Voltage from neutral to A = i.oo V.
Voltage from neutral to B = (- 0.50 + 0.866.7) V.
And voltage from neutral to C = ( 0.50 0.866.7") V.
Then the measured, or absolute, value of voltage between
B and C is 1.732 V, where V is the star voltage of the line.
REACTANCE OF TWO-PHASE AND THREE-PHASE LINES 8 1
The reactive voltage on conductor A per mile is
1,00/80.5 + 74I.I log \Ij 2 7T/ X IQ- 6
(due to flux from 7^)
+ (-0.50 +0.8667) (741-1 logyJ//27r/X io~ 6
(due to flux from IB)
+ (- 0.50 - 0.866 j) (741.1 logy JIj 2 TT/ X io~ 6
(due to flux from I c )j
where p is the radius of the wire.
The reactive voltage on conductor B per mile is
1.00(741.1 log J7; 2 irf X io~ 6
(due to flux from I A)
(due to flux from IB)
+ ( 0.50 0.866 j) (741.1 log J Ij 2 irf X I
(due to flux from /<?).
The reactive voltage between A and B is therefore
(1.50 - o.
+ (0.50 + 0.866 j) 741.1 (log- - log-)// 27T/ X iQ- 6 .
\ p pi
Suppose at the end of one mile the line is transposed
so that the above two conductors occupy the positions B
82 TRANSMISSION LINE FORMULAS
and C. Then the reactive voltage between these conductors
for the next mile is
(1.50 - 0.8667) (80.5 + 74i.i log-) Ij 2 TT/ X iQ- 6
+ (0.50 + 0.8667) 74LI (log- - log-) Ij 2 7T/ X I0~ 6 .
\ p p/
Let the line be transposed again. Then for the third
mile the reactance voltage between these same two con-
ductors is
(1.50 - 0.8667) (80.5 + 741.1 log -} Ij 2 TT/ X lo- 6
+ (0.50 + 0.8667) 74I.I (log- - log-) Ij 27T/X IO- 6 .
\ p P/
The total reactive voltage for the three miles is
(1.50-0.8667) j 80.5X3 + 741.1 (log- + log- + log-) I
( \ P p P/ )
X Ij 2 7T/ X XT*.
/Thus the average reactive voltage per mile, which is the
same for all phases, is, in absolute value,
1.732(80.5 + 741.1- log- -J/27T/X IO- 6 .
\ P /
Now 1.732 7 = the total amperes.
_ K.V.A.
E
Therefore the drop in line voltage, in absolute value, is
(8o.5 + 7 4 i.ilog^27T/Xio- 6
in volts per total ampere per mile of transmission line,
where
s = ^ dbc.
REACTANCE OF TWO-PHASE AND THREE-PHASE LINES 83
Thus the total reactive drop is equal to
K ' VA - (80.5 + 741.1 log 2 TT/ X io- 6
E
K.V.A.
Xx,
E
where x is the tabulated value of reactance.
Reactance, Three-phase, Regular Spacing. When the
conductors are spaced at the corners of an equilateral tri-
angle of side 5, then the expression for reactance is the
same as the usual formula:
x = (80.5 + 741.1 logio-J 27T/
X
Reactance, Three-phase, Flat Spacing. When the three
conductors lie in one plane (either
horizontal or vertical), the center j - ? - j
one being equidistant from the k-- a -4- ..... a -!
other two, as in Fig. 18, the react- Fig. 18.
ance per mile is
i a "^i X i X 2
80.5 + 741.1 log - -
P
, 1.260
= 80.5 + 741. i log -- .
P
This is approximately 4% higher in an ordinary case
than the reactance for spacing on an equilateral triangle
of side a.
The formulas of this chapter will apply to cable as well
as to wire, if the term 80.5 is changed as per Chapter XI.
CHAPTER XIII.
CAPACITY OF SINGLE-PHASE LINE.
Capacity of Two Round Wires. The conductors of a
transmission line form a condenser, the electrostatic capac-
ity of which can be calculated from the dimensions of the
line. The simplest line to calculate is a single-phase line
consisting of two round wires, and this case will be in-
vestigated first.
Fig. 19.
Fig. 20.
Suppose that A and B (Fig. 19) are two long parallel
wires of infinitesimal section and that they are spaced a
distance t centimeters apart. Let A carry a charge of + q
electrostatic units of electricity per centimeter, and let B
carry q units per centimeter.
First, find the force exerted on a unit charge near the
wire A .
From the symmetry of the arrangement it is evident that
the resultant force on a unit charge at P (Fig. 20) will be a
repulsion away from the wire at right angles to it, since the
total effect of the right half of the wire must be equal to
the total effect of the left half. The force at right angles
84
CAPACITY OF SINGLE-PHASE LINE 85
to the wire exerted by the charge on the element dl will be
= -^ cos dd,
r\
since dl cos = r dB
V,
and r =
cos
The total force exerted by the wire will be
The potential of the point (Fig. 19) midway between
the wires, will be zero, since the effect of the positive charge
on A will be equal to the effect of the negative charge on B.
The potential difference between P and O is the work done
in moving a unit charge from one point to the other. The
2 Q
force due to the wire A on a unit charge at any point is * ,
acting directly away from A. Therefore the work done
in moving a distance dr toward A is
and thus the total work in moving from P to against the
force due to A is
i_
r~2Q, t
-* dr = 20 loge
r 2ri
Similarly, the work against the force due to B is equal to
2r 2
86 TRANSMISSION LINE FORMULAS
Therefore the potential difference between P and is
equal to the total work, and is
2 q log e -
Fig. 21.
At P (Fig. 21) make the angle APD equal to the angle
PBD. Then the triangle PBD is similar to the triangle
APD, and therefore
PB ^r*
AP " n
p
If we draw a circle of radius p about the fixed point D,
then at any point on this circle similar triangles are formed
by p, r\j and r^ as in Fig. 21, and therefore
r 2 DB
= = constant.
ri p
where r\ and r 2 are the distances of the point on the circle
from A and B respectively. Therefore, the potential
2#log -
fi
will be the same at all points on this circle.
Now let a solid cylindrical conductor fill all the space
inside the circle of radius p. All points on its surface will
be at the same potential. The distribution of potential
outside of the cylinder will not be altered from the previous
condition when all points on the circle of radius p were
CAPACITY OF SINGLE-PHASE LINE 87
also at the same potential. The potential of the cylinder
will be
, W , DB
2qlog e - = 2qlog e
r\ P
In the same way, let the wire B be replaced by a solid
cylinder of radius p and center E, as in Fig. 22.
s
... t ..
Fig. 22.
The potential of this cylinder, which carries q units
per centimeter, will be
AE , DB
P P
since the second cylinder is symmetrical with the first.
Thus the potential difference between the two cylinders is
DB a
4?k)ge = *,
P v
where C is the capacity per centimeter of line.
ThuS If \ C= ^DB-
4log
P
DB FB
Now - = ,
p FA
since F is a point on the circle.
DB DB- p
rpr, (
Therefore
p DB + p - s
where 5 is the interaxial distance between the two solid
cylinders, and is equal to DE, Fig. 22.
88 TRANSMISSION LINE FORMULAS
Therefore,
DB 2 + DB -p - DB -s = DB -p - p 2
or DB 2 - DB s + p 2 = o.
Solving this quadratic equation in DB, we have
2
[The negative value of the radical must not be used,
since it would give the value of DA instead of DB.]
Therefore,
i
$4V-'
\ 2 p * 4 p
which may be expressed as
4 cosh" 1 ( )
\2p/
or it may be expanded by the Binomial Theorem to give
the approximate value
T
per centimeter,
or, very nearly,
4log e -
P
* It is evident that the expression
i
C =
P
which is sometimes published, is less accurate than the simpler expression
4 log,-
CAPACITY OF SINGLE-PHASE LINE 89
Transferring to other units,
r = T x * 4343 X 2 ' 54 X I2 X 528
" 4X9XXO"
x
2
loglo (_)
V s/
38.83 x io- 9
, (S p\
logio ---)
Vp s/
farads per mile of single-phase line.
The capacity susceptance is
38.83 X
fr f v N/
2 7T/ C = 2 7T/ X - X
mhos per mile of single-phase line. The charging current
in this line will be
amperes,
where b is the tabulated value of capacity susceptance.
REFERENCE. "A Treatise on the Theory of Alternating Currents," by
Alexander Russell, 1904, Vol. I, page 99.
Capacity of Cable. The formula for capacity of a line
using stranded cables will be the same as the above formula
for solid wires, p being taken as the maximum radius of
the cable. All the electrostatic charge on the cable does
not lie at the maximum radius from the center, but as
actual cables are generally slightly larger than the calcu-
lated diameters in the tables, it will be sufficiently close to
take p from the tables and use it in the regular formula for
capacity.
90 TRANSMISSION LINE FORMULAS
Effect of the Earth on Capacity of Line. The effect of
bringing a conducting plane, such as the earth, near to two
charged wires is to change their electrostatic field and in-
crease their capacity.
Consider two long parallel wires, A and A l (Fig. 23), of
infinitesimal section and carrying + q and q units of
electricity per centimeter
respectively. As in Fig.
19, the point O midway
M __ / o N between the two wires
will be at zero potential.
All points having the same
potential must have equal
to a constant. It is evi-
dent that all points at the same potential as lie in the
plane MON, perpendicular to AA 1} since for all such
points
Therefore, the wire A may be replaced by a solid con-
ducting plane MN, which will be at zero potential. Thus,
when the conducting plane is the earth, its effect is the same
as that of a charged wire at a depth below the surface
equal to the height of the original wire.* The assumed
wire is called an image wire, since it occupies the same
position as the image of the real wire, considering the
surface of the ground as a mirror.
In the case of a single-phase transmission line, image
wires A' and B' must be assumed for both wires A and B
* See " Elements of Electricity and Magnetism," by J. J. Thomson, page
138.
CAPACITY OF SINGLE-PHASE LINE
A'
---->
-, (
(Fig. 24) and the capacity of the entire system of four wires
is then calculated as follows:
Let h be the distance of the
wires from the ground and 5
their distance apart from center
to center. Let A carry a charge
of + q units per centimeter; A', 7
q units; B, q units; and
B' , + q units. Let a unit charge
be carried from the surface of
A to that of B. Assuming that Fig 24
the charges are concentrated at
the centers of the wires, the total work done is equal to
/
J p
-I
2 q (s - x)
dx.
The sum of the first two integrals has been shown to be
approximately
The sum of the last two integrals is approximately
4/r>-f s 2
2 * log iT7-
Therefore, the total work is equal to
, , zh
P
92 TRANSMISSION LINE FORMULAS
Therefore, C is approximately
! S . 2 k
4log e - +
P
Taking as an average case,
h = 360 inches (30 feet),
s = 120 inches (10 feet),
p = 0.25 inch,
we have - = 480,
p
2 h .
and = 6.
s
Therefore, - =
\/4 h 2 + s 2
Now, Iogio48o = 2.681,
while logio r^ = 0.006.
6
Thus the capacity is changed by the nearness of the
ground by less than } of i%, even with the comparatively
wide spacing of 10 feet.
Tests have shown that the effect of the ground in increas-
ing the capacity is even less than the above amount, due
partly to the fact that the ground is a poor conductor.
As the effect of the ground is so slight, it has been neglected
entirely in the calculations in this book.
REFERENCE. For an alternative proof, see "A Treatise on the Theory
of Alternating Currents," by Alexander Russell, 1904, Vol. I.
See also " The Calculation of Capacity Coefficients for Parallel Suspended
Wires," by Frank F. Fowle, Elec. World, Aug. 19, 1911.
CHAPTER XIV.
CAPACITY OF TWO-PHASE AND THREE-PHASE LINES.
Capacity, Two-phase. The charging current of a single-
phase line was shown in Chapter XIII to be
*HH)
amperes per mile of line
-i*
where b is the tabulated value of capacity susceptance per
mile.
In a two-phase, four-wire line, each phase is quite similar
to a single-phase line, and so the charging current per wire
is
\Eb.
As this amount of charging current flows in each phase,
the total amperes of charging current are
Eb,
for a two-phase, four-wire line.
Capacity, Three-phase, Irregular Spacing. When the
wires of a three-phase transmission line are not spaced at
the corners of an equilateral triangle, but the transposi-
tion of the conductors is carried out at regular intervals,
the charging current in the wires will be a balanced, three-
phase current, since each wire will have passed through the
same average conditions. This is shown in an approxi-
mate manner as follows:
93
94
TRANSMISSION LINE FORMULAS
As when calculating the self-induction of an irregularly
spaced line, consider a line three
miles long which is transposed
at the end of each mile.
The work in carrying a unit
charge from C to B (Fig. 25)
assuming the charges concen-
trated at the centers of the
wires, is approximately
Fig . 25 .
E a = q B 2 loge- - q c 2 loge- + q A 2 loge--
p p
Now q B is a periodic quantity, which alternates in value
at the same frequency as the voltage or current.
We have q B = CiE
= _,-v
27T/
where IB' is the charging current flowing into the capacity
Ci of the wire B.
Thus
/ 2 loge - - /</ 2 loge - + // 2 loge \
p p tv
2 7T/
In the second mile the conductors are transposed into
new positions. Let the currents in them remain the same.
Therefore,
E a = =l(l B ' 2 loge - - I C ' 2 log, - + // 2 log -),
27T\ p CI
T/
and in the third mile
E a
~ - I C ' 2
p
" +
2 loge -)
al
CAPACITY OF TWO-PHASE AND THREE-PHASE LINES 95
Let
and
and
Then
Adding together and dividing by 3, we obtain the ap-
proximate average value per mile,
that is,
where
Similarly,
and
E a =^(V-v) 2 logA
5= ^abc.
E b = 4( V - I A) 2 log -
2 7T/ \ / p
Fig. 26.
E a = i.ooEj as in Fig. 26,
^b (-5o o.866j) ,
E c = (o.5o+o.866j)E.
j 2 irfE (0.50 + 0.8667)
5
P
(l)
(0.50 - 0.8667), (3)
96 TRANSMISSION LINE FORMULAS
also I A + IB + Ic = o, (4)
since they are currents flowing in a three-phase line.
From equations (i) and (3)
2log e -
P
Adding (4) and (5), we have
7 '_ _L_ y 2 irfE X 1 .00
3 2loge~
P
-r, ,,. T , i . . 2 TT/E ( 0.50 -f 0.866 /)
From this, I B = j=. X - =
3 2 log -
P
and I ' * x **fE(--SQ-<>.*66fl.
^ 2 loge ~
P
The vectors for IA, IB and /c' may now be plotted as
in Fig. 27, and it is seen that the vectors are the same
, length and are at 120 to each other.
Thus the charging current is a bal-
. anced three-phase current.
The power factor of the charging
current is zero, since the current in
any wire I A (Fig. 27) is at right angles
to the direction OP (Fig. 26) of the
Fig. 27. corresponding star voltage or in-phase
current.
The total amperes of charging current are
T / 27T/E
I A - per centimeter
2 loge-
P
CAPACITY OF TWO-PHASE AND THREE-PHASE LINES 97
, 5
logio -
P
= Eb,
where b is the tabulated value of capacity susceptance per
mile.
Capacity, Three-phase, Regular Spacing. When the
conductors are placed an equal distance, s, from each other,
the formula for b is
, 38.83 X 27T/ Vy .,
_ o o j_ x 10 a mhos per mile.
logio ~
P
Capacity, Three-phase, Flat Spacing. When the wires
lie in one plane (either horizontal or vertical), the certer
one being at a distance a from the other two (see Fig. 18),
and the wires being transposed at regular intervals, the
formula for susceptance is
b= 38.83 X 2 rf _ yjn _.
, .
logio
P
mhos per mile of transmission line.
CHAPTER XV.
THEORY OF CONVERGENT SERIES.
THE well-known fundamental formulas for a transmis-
sion line without branches, in which the load is delivered
only at the end of the line, are as follows:
E 8 = E cosh VYZ + 7/b| / sinh VYZ,
and i a = / cosh VYZ + ~^ sinh VYZ~
where E 8 and I 8 are the voltage and current at the
supply end,
E and I are the voltage and current at the re-
ceiver end,
Y is the line admittance
and Z is the line impedance.
The above .equations have been published at various
times. They are obtained as follows:
Let r = resistance of conductor per unit length,
and x = reactance of conductor per unit length.
Then z = r +jx
= impedance of conductor per unit length.
Let g = leakage conductance from conductor per
unit length,
and b = capacity susceptance of conductor per unit
length.
Then y = g +jb
= admittance of conductor per unit length.
98
THEORY OF CONVERGENT SERIES 99
Let EI = voltage of line at a distance / from the re-
ceiver end,
and let I l = P l -jQ l
= current in the line at a distance / from the
receiver end.
(Since Ii is usually not in phase with the voltage, it
must be expressed as a complex quantity.)
Now in an element of length, dl, of the line, the voltage
consumed by impedance is
dE l = zl t dl.
The current consumed by admittance is
dl t = yE t dl.
Thus 1f' = Z/ " (l)
and ' ^ = yEl : (2)
Differentiating (i)
d 2 Et _ dlj
dl 2 Z dl '
Substituting (2) in this gives
This is a differential equation of the second order and
may be expressed in the form
(D* - yz) E, = o,
and we have
and from (i),
II = z~dT
(5)
100 TRANSMISSION LINE FORMULAS
Now at the supply end,
yl = Y,
and zl = Z
Therefore, E a = A l v + A 2 e ~ ^ (6)
(7)
At the receiver end,
|o
and E = A 1 +A 2 (8)
and /=Vf <Xi-^ 2 ). (9)
From (6)
(10)
which, by the definition of cosh 6 and sinh 6, is
(A 1 + A 2 ) cosh VYZ + (Ai- At) sinh
Therefore, from (8) and (9),
E 3 = E cosh VYZ + = 7Z sinh VFZ: (i i)
Similarly, from (7),
= 7 cosh VFZ + = EY sinh VFZ. (13)
Equations (n) and (13) are the fundamental formulas of
transmission lines, as generally written.
THEORY OF CONVERGENT SERIES
101
Now
I . YZ . F 2 Z 2 \
= 21 H 1 hetc.l.
\ 2 2-3 -4 /
Similarly c VT1 -
/=7^ / FZ F 2 Z 2 \
= 2 v FZ i + H h etc. 1.
V 2-3 2-3.4.5 /
Substituting these results in (10) and (12), we can ex-
press the fundamental equations as follows:
and
/. =
T
2-3-4
r 2-3-4-5 '6
r cue. i
FZ
F 2 Z 2
F 3 Z 3
1 r*tr
r
-f-etc.
2-3
2-3-4-
5 2-3.4.5-6
7
FZ ,
F 2 Z 2
I
etc.)
T
H
2-3-4
2.3.4.5-6
FZ
F 2 Z 2
F 3 Z 3
_ O-Atr 1
.) (14)
/
.)*(i5)
2-3 2.3-4-5 2.3.4.5.6-7
Equations (14) and (15) are the same as those tabulated
in Chapter VI for obtaining E a or A + jB, and I 8 or C + JD.
For no-load values, all that is necessary is to put the load
current, /, equal to zero.
When conditions are given at the supply end, the same
equations for full-load conditions are obtained, except that
* See references to T. R. Rosebrugh, J. F. H. Douglas, and C. P. Stein-
metz, page 41.
102 TRANSMISSION LINE FORMULAS
the second half of the expressions for voltage and current
is negative, since power is now flowing away from the point
where the voltage is specified, instead of toward it. Thus
we have the series fox F +jG and M +JN.
At no load, the conditions are really not all specified at
the supply end, but the current is specified to be zero at
the receiver end, and this necessitates the use of special
series. From Table V we have the ratio of the voltage at
the two ends of the line,
E Qs _ A Q +JB<>
E ~ E
, YZ , F 2 Z 2
H 1 hetc
2 2.3.4
This ratio is independent of the voltage E, and depends
only on the constants of the line. Thus, if E s , the voltage
at the supply end at no load, is given, we can obtain the
no-load voltage at the receiver end from the equation,
etc
2 2.3-4
V7 V 2 7 2 X- 1
or
2 2-3-4
which, when expanded by the binomial theorem, gives
E Q =
- etc]
/
2 24 720 8064
(16.)
as in Table VI.
The no-load current at the supply end is
CV7 1727-2 \
i+ +- -+etcl
2-3 2-3.4.5 /
as in the equation for C Q +JD .
r ' ' .
THEORY OF CONVERGENT SERIES 103
Substituting the value of E from equation (16), we have
-YZ+^-Y 2 Z 2 - Y*Z 3 +-m-Y*Z*-e
2 24 720 8064
2.3 2-3.4.5 2.3.4.5.6.7
F 4 7 4 \
+ - ~ - 5 + etc. ).
2.3.4.5.6.7.8.9 /
Multiplying the two series together by the ordinary
algebraical method, we obtain
F 4 Z 4 -
3 i5 3i5 2835
as given in Table VI of convergent series.
PART III
TABLES.
TABLE I. FORMULAS FOR SHORT LINES.
CONDITIONS GIVEN AT RECEIVER END.
These formulas are exact when the line is short. When the line is 20
miles long, they are correct within approximately ^ of i% of line voltage.
Conditions given:
K.V.A. = K.V.A. at receiver end.
E = Full load voltage at receiver end.
cos 6 = Power factor at receiver end.
K.W. = K.V.A. cos 0.
r = Resistance of conductor per mile. (From Tables VII- VIII.)
x = Reactance of conductor per mile. (From Tables IX-XII.)
/ = Length of line in miles.
1000 K.V.A. cos 8 , ,.
Then P= - =; - = In-phase current at receiver end (in
c,
total amps.).
_ 1000 K.V.A. sin _, . , ,.
Q = - - = Reactive current at receiver end (in
total amps.) when current is lagging.
1000 K.V.A. sin ,
= -- - when current is leading.
hi
Find the following quantities:
Three phase or two phase. Single phase.
A = E + Prl+QxL A = E + 2 Prl + 2 Qxl.
B = Pxl- Qrl. B = 2 Pxl- 2Qrl.
Formulas (capacity neglected) :
B* *
(1) Voltage at supply end = A -\ -\
R2
(2) Regulation of line = A + - E. (Same as line drop.)
(3) Per cent regulation of line = - - - -^ - '- per cent.
(Same as per cent line drop.)
,. r 104
: ' ' r"" r ' r ' "'
rrrrffr
,,,M,|,,,,,,,,",M, ,,,,,,,,,,,,,,,,,,
in > i* _
T"T" '"J""!"* 1 ! 1
4k M M
T-JL,^ M1
i/
P
]i' , ,
. i . i . 1 1 1 1 1 1
? Uj
,,,.,, ,,, ,..., j, . .. i
ro
., 1 ,,.. | ... .j ... i ,,., .j .,-,.,,.
cd
Ohms Resistano
r ro j-
T L L
I I
.ro
les.
FeetSpacing
Copper
Aluminum
jo.ro ro
wNoaxnot^oj K 5J FeetSpacing
- FeetSpacing Ml I MM I I i r^^r
T^rnTTTTTT^ Alumlnum
Swo 03 ^ 01 -^ 04 ^oi ~" FeetSpacing
25 Cycles.
Q -<Q
" JT^
ct 1 |
mn
fo-
o
Z
i Jo i i ii i i
I"".!""'"".!"
CD 00
T" 1 '".!""''"'! 1 ' J J
i0 00 "*l . 0* W|
= Regulation Factor.
pmr
|* ||
fe Ii
st ^
1
8
"\
^ ^*5
f^
^
/
; / .^ ?
^
\
y ii'
I
. , . . . IJ . . . . 1
1 1 1 J 1
rrrf njrrrrrfn
'1
1 1 ' 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
O)
In
!^
m
fo o
Conductor per Mile.
TABLES 105
TABLE I. (Continued.)
B 2
(4) K.V.A. at supply end = -j^- X K.V.A.
(5) K.W. at supply end = ^ (AP - BQ).
(6) Power factor at supply end ""
1000. . .
(in decimals).
(7) In-phase current at supply end = ~ in total amperes.*
(8) Reactive or quadrature current at supply end =
in total amperes.*
When this quantity is positive, the current is lagging.
When this quantity is negative, the current is leading.
(9) K.W. loss in line = -- (AP - BQ - EP).
(10) Per cent efficiency of line = -^= - ^r per cent.
AJc n\)
. Total amps.
* Amperes per wire, three phase, = - _
V 3
. . . Total amps.
Amperes per wire, two phase, = -- *
106 TRANSMISSION LINE FORMULAS
TABLE II. FORMULAS FOR SHORT LINES.
CONDITIONS GIVEN AT SUPPLY END.
These formulas are exact when the line is short. When the line is 20
miles long, they are correct within approximately T V of i% of line voltage.
Conditions given:
K.V.A. = K.V.A. at supply end.
E 8 = Full load voltage at supply end.
cos 6 = Power factor at supply end.
K.W. = K.V.A. cos 6.
r = Resistance of conductor per mile. (From Tables VII- VIII.)
x = Reactance of conductor per mile. (From Tables IX-XII.)
/ = Length of line in miles.
_. 1000 K.V.A. cos T .
Then JP = - ~ - = In-phase current at supply end
/i
(hi total amps.).
_ 1000 K.V.A. sin
Qt = - 5 - = Reactive current at supply end
>
(in total amperes) when current is
lagging.
1000 K.V.A. sin 9
= -- = - when current is leading.
-c-a
Find the following quantities:
Three Phase or Two Phase. Single Phase.
F = E 8 - P 8 rl - Q 8 xl F = E 8 -2 P 8 rl - 2 Q a xl
G = Q a rl - P 8 xl G = 2 Q a rl - 2 P a xl.
Formulas (capacity neglected):
(1) Voltage at receiver end = F -\ ^
(2) Regulation of line = E 8 F -- . (Same as line drop.)
(3) Per cent regulation of line = - -^ - - per cent.
(Same as per cent line drop.)
(4) K.V.A. at receiver end = ^ X K.V.A.
TABLES 107
TABLE II. (Continued.)
(5) K.W. at receiver end ^^ (FP S GQ a ).
i (FP 8 - GQ 8 ) E a _
(6) Power factor at receiver end :
(in decimals).
(7) In-phase current at receiver end = - -^- - in total amperes.*
GPs + FQ
(8) Reactive or quadrature current at receiver end = ^
F +TF
in total amperes.*
When this quantity is positive, the current is lagging.
When this quantity is negative, the current is leading.
(9) K.W. loss in line = (E S P S - FP S + GQ 8 ).
(10) Per cent efficiency of line = - ^ per cent.
Total amps.
* Amperes per wire, three phase, = ?=-
Total amps.
Amperes per wire, two phase, =
108 TRANSMISSION LINE FORMULAS
TABLE in. K FORMULAS FOR TRANSMISSION LINES.
CONDITIONS GIVEN AT RECEIVER END.
Accurate within approximately T V of i% of line voltage up to 100 miles,
and \ of i% up to 200 miles, for lines with regulation up to 20%.
(cycles) 2
K = 6 '- . K= 2.16 for 60 cycles. K = 0.375 for 25 cycles.
Conditions given:
K.V.A. = K.V.A. at receiver end.
E = Full load voltage at receiver end.
cos 6 = Power factor at receiver end.
K.W. = K.V.A. cos 8.
r = Resistance of conductor per mile. (From Tables VII-VIII.)
x = Reactance of conductor per mile. (From Tables IX-XII.)
/ = Length of transmission line in miles.
,, 1000 K.V.A. cos 6
I hen P = - = In-phase current at receiver end
Hi
(in total amps.).
-. 1000 K.V.A. sin ^
Q = p; = Reactive current at receiver end
rL
(in total amps.), when current is
lagging.
1000 K.V.A. sin .
= -= when current is leading.
TABLES 109
TABLE III. (Continued.}
Find the following quantities:
Full Load.
_ The above are for two- and three-phase lines. For single-phase
lines use 2 r and 2 x in place of r and #.
110 TRANSMISSION LINE FORMULAS
TABLE in. (Continued.)
CONDITIONS GIVEN AT RECEIVER END.
Formulas:
Full Load. No Load.
Voltage at receiver end.
d)
(for constant supply voltage)
Regulation at receiver end in volts, for constant supply voltage.
A +TA
(3) - ng.X-A
N.B. The regulation at receiver end may be expressed as a percentage
ofE.
Voltage at supply end.
UE.-A + Z.- (5)*.-*+*!
(for constant receiver voltage).
Regulation at supply end in volts, for constant receiver voltage.
,,. * , & A B<?
(6) A H -- j AQ -- j-
2 A 2 AQ
N.B. The regulation at supply end may be expressed as a percentage
TABLES HI
TABLE III. (Continued.')
Current at supply end in total amperes.*
( 7 ) VcnTo 2 . (8)
K.V.A. at supply end.
fiT.PF. at supply end.
(n) (4C + 5Z>). (12)
Power factor at supply end, in decimals.
AC + BD , , A C +B D
(13)
In-phase current at supply end in total amperes*
AC+BD , ,
Reactive current at supply end in total amperes*
, . BC-AD , O s B
(I7) "' (
2
When this quantity is positive, the current is lagging.
When this quantity is negative, the current is leading.
K.W. loss in line.
(19) (AC + BD -EP). (20)
IOOO IOOO
[same as No. 12].
Per cent efficiency of line.
. 100 EP
percent '
Total amps.
* Amperes per wire, three-phase, = - -= -
^3
Total amps.
Amperes per wire, two phase, = - p
112 TRANSMISSION LINE FORMULAS
TABLE IV. K FORMULAS FOR TRANSMISSION LINES.
CONDITIONS GIVEN AT SUPPLY END.
Accurate within approximately fa of i% of line voltage up to 100 miles
and | of i% up to 200 miles, for lines with regulation up to 20%.
K = - . K = 2.16 for 60 cycles. K = 0.375 for 25 cycles.
10,000
Conditions given:
K.V.A. = K.V.A. at supply end.
E s = Full load voltage at supply end.
cos 6 = Power factor at supply end.
K.W. = K.V.A. cos0.
r = Resistance of conductor per mile. (From Tables VII-VIII.)
x = Reactance of conductor per mile. (From Tables IX-XII.)
I = Length of transmission line in miles.
Then P 8 = I0 K 'Y' A ' C S = In-phase current at supply end (in
&8
total amps.).
t supply end (in
total amps.), when current is lagging.
looo K.V.A. sin 9 , , . . ,.
= - when current is leading.
TABLES
TABLE IV. (Continued.)
Find the following quantities:
Full Load.
iooo
Kf I
No Load.
( Ti
Viooo/ )
Note. The above are for two- and three-phase lines. For single-
phase lines use 2 r and 2 * in place of r and .
114 TRANSMISSION LINE FORMULAS
TABLE IV. (Continued.}
CONDITIONS GIVEN AT SUPPLY END.
Formulas:
Full Load. No Load.
Voltage at receiver end.
(for constant supply voltage).
Regulation at receiver end in volts, for constant supply voltage.
Go 2 G 2
N.B. The regulation at receiver end may be expressed as a percentage
of E.
Voltage at supply end.
(for constant receiver voltage).
Regulation at supply end, in volts, for constant receiver voltage.
F + 1 21
(6) E s - - - if- E..
N.B. The regulation at supply end may be expressed as a percentage
of E,.
TABLES
TABLE IV. (Continued.)
Current in total amperes*
( 7 ) VM 2 + N* at receiver end. (8) ^M<? + N<? at supply end.-
K.V.A.
( Q \ _*_ IP+GL] VM 2 + # 2 at receiver end.
7 1000 \ 2V I
( I0 ) _JL_ a VM 2 + No 2 at supply end.
'
1000
()
^-(^M + G^) at receiver end. (12) 8 M at supply end.
IOOO
(13) / rm ^: GN - at receiver end -
Power factor, in decimals.
at receiver
(14) at supply end.
VM<? + No 2
In-phase current in total amperes*
(l ,\ FM + GN at recdver end> ( l6 ) MQ at supply end.
Reactive current in total amperes*
( I7 ) GM-FN at recdver end ( Ig ) jv at supply end.
When this quantity is positive, the current is lagging.
When this quantity is negative, the current is leading.
K.W. loss in line.
( IQ ) --(E 8 P 8 -FM-GN). (20) ^ s M (sameasNo.i2).
Per cent efficiency of line.
(21) =r-p per cent.
Total amps.
* Amperes per wire, three phase, = -
O
Total amps.
Amperes per wire, two phase, = -
Il6 TRANSMISSION LINE FORMULAS
TABLE V. CONVERGENT SERIES FOR TRANSMISSION LINES.
CONDITIONS GIVEN AT RECEIVER END.
The convergent series give the results of the fundamental formulas as
accurately as desired, if a sufficient number of terms is used.
When conditions are given at the receiver end, the same as with the K
formulas, find the quantities:
Full Load.
2-3-4-S 2-3.4.5.6.7
2.3 2.3.4.5 2.3.4.5..
where Z = (r +./*)*
r = resistance of conductor per mile.
x = reactance of conductor per mile.
I = length of transmission line in miles.
+ etc.
g = leakage conductance of conductor per mile.
b = capacity susceptance of conductor per mile.
Use A, B, C, D, etc., with the equations in the third and fourth pages
of Table III to solve transmission line problems.
2 _
Note i. In the formulas, A + 7- is used instead of V A 2 -f- .B 2 .
zA
This approximation may be used for very accurate work, as it is correct
within approximately T ^ of i% when the regulation is not more than 20%.
Note 2. The above are for two- and three-phase lines. For single-
phase lines use 2 r and 2 x in place of r and x, and use \ g and % b in place
of g and b.
TABLES 117
TABLE VI. CONVERGENT SERIES FOR TRANSMISSION LINES.
CONDITIONS GIVEN AT SUPPLY END.
The convergent series give the results of the fundamental formulas as
accurately as desired, if a sufficient number of terms is used.
When conditions are given at the supply end, the same as with the
K formulas, find the quantities:
Full Load.
Z3 etc
2-3-4 2-3- 4-5-
+ F2Z2 + - F3Z3 A + etc
2- 3^2- 3- 4. 5 2.3.4.5.6-7
{ , FZ , F 2 Z 2 PZ 3 u \
E 3 Y i 4 --- --- -- - -- r- etc. I .
V - 2-3 2-3.4.5 2-3-4-S-6-7 /
No Load.
iYZ+ ^-Y*Z*- F 3 Z 3 + ^ F 4 Z 4 - etc.Y
24 720 _ 8004 y
- I FZ -h F2Z 2 - -- F 3 Z 3 +
15 3 j s 2835
where Z = (r +jx) I.
r = resistance of conductor per mile.
x = reactance of conductor per mile.
I = length of transmission line in miles.
Y=(g+jb)l.
g = leakage conductance of conductor per mile.
b = capacity susceptance of conductor per mile.
Use F, G, M, N, etc., with the equations in the third and fourth pages
of Table IV to solve transmission line problems.
G 2 / -
Note i. In the formulas, F -\ -- - is used instead of V F 2 + G 3 . This
2r
approximation may be used for very accurate work, as it is correct within
approximately T f^ of *% when the regulation is not more than 20%.
Note 2. The above are for two- and three-phase lines. For single-
phase lines use 2 r and 2 x in place of r and x, and use | g and 6 in place of
g and b.
n8
TRANSMISSION LINE FORMULAS
TABLE VII. RESISTANCE OF COPPER WIRE AND CABLE.
Data assumed:
Temperature, 20 C. (68 F.).
Conductivity of hard drawn copper, 97.3% of the annealed copper
standard.
Increase of resistance of cables due to spiralling, i%.
Copper Wire.
Diam-
Resistance, ohms per mile.
B. & S.
Circular
ctcr
gauge.
mils.
(2P)
inches.
Direct
current.
25 cycles.
In-
crease.
60
cycles.
In-
crease.
Per cent
Per cent
0000
2ll,6oo
.4600
. . .
.2655
.2657
.08
.2667
44
000
167,800
.4096
3348
3350
05
.3358
.27
00
133,100
.3648
.4221
.4223
03
.4229
I?
o
105,500
3249
5326
.5327
.02
5331
. II
I
83,690
.2893
...
.6714
.6714
.OI
.6718
.07
2
66,370
.2576
.8466
.8466
.OI
.8469
.04
3
52,630
.2294
1.068
1.068
1. 068
03
4
41,740
.2043
1.346
1.346
I.346
.02
5
33,ioo
.1819
1.697
1.697
1.698
.01
6
26,250
.1620
2. I4O
2. I4O
2.140
.OI
7
20,820
1443
2.699
2.699
2.699
8
16,510
.1285
3-403
3-403
3.403
TABLES
119
TABLE VII. (Continued.)
Copper Cable.
Diam-
No. of
Resistance, ohms per mile.
B. & S.
gauge.
Circular
mils.
eter
(2p)
inches.
wires
as-
sumed.
Direct
current.
25 cycles.
In-
crease.
60
cycles.
In-
crease.
500,000
.8lII
19
1135
.1140
Per cent
41
.1162
Per cent
2-34
....
450,000
.7695
19
.1261
.1265
33
.1285
1.90
....
400,000
.7255
19
.1419
.1422
.26
.1440
1.50
350,000
.6786
J 9
.1621
.1625
.20
.1640
1.16
300,000
.6211
7
.1892
.1894
15
.1908
85
250,000
.5669
7
.2270
.2272
.10
.2284
.60
oooo
211,600
.5216
7
.2682
.2684
.07
.2693
43
ooo
167,800
4645
7
.3382
.3384
05
3391
27
oo
133,100
4137
7
.4264
4265
03
.4271
17
105,500
3683
7
5379
.5380
.02
5385
.11
I
83,690
.3280
7
.6781
.6782
.01
.6785
.07
2
66,370
.2921
7
8550
8551
.01
.8554
.04
3
52,630
.2601
7
1.078
1.078
1.078
-.03
4
41,740
.2317
7
1.360
1.360
1.360 .
.02
TABLE VII. (Continued.} TEMPERATURE COEFFICIENTS OF
COPPER.
For different initial temperatures and different conductivities.
Ohms per
meter-gram
at 20 deg.
cent.
Per cent
conduc-
tivity.
15
"20
25
"30
50
.16108
95
.00405
.00381
.00374
.00367
.00361
.00336
15940
96
. 00409
.00386
.00378
.00371
.00364
.00340
15776
97
.00414
.00390
.00382
00375
00368
00343
.15727
97-3
.00415
.00391
.00383
.00376
.00369
.00344
.15614
98
.00418
.00394
.00386
.00379
.00372
.00346
15457
99
.00423
.00398
.00390
.00383
00375
.00349
.153022
TOO
.00428
.OO4O2
00394
.00386
00379
00352
I5I5I
IOI
.00432
.OO4O6
.00398
.00390
.00383
00355
where Rt is the resistance at any temperature t deg. cent.
and Rti is the resistance at any "initial temperature" ti deg. cent.
From Appendix E, Standardization Rules of the A.I. E. E., June 27, 1911.
L20
TRANSMISSION LINE FORMULAS
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TABLES
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TABLE XVII. POWER FACTOR TABLE.
Cose.
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95
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.526783
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.866025
INDEX
PAGE
Absolute value of a complex quantity 57
Admittance 98
Aluminum cables, Table VIII 120
Annealed copper standard 55
Capacity 6 1
Capacity, derivation of formulas:
Cables 89
Effect of the earth 90
Two round wires 84
Two-phase 93
Three-phase, irregular spacing 93
Three-phase, regular flat spacing 97
Three-phase, regular spacing 97
Capacity susceptance 6
of cable, 25 cycles, Table XIV 128
of cable, 60 cycles, Table XVI 131
of wire, 25 cycles, Table XIII 127
of wire, 60 cycles, Table XV 130
Charging current 6
Chart, regulation, description of 8
Complex numbers 42
Condenser effect 6
Conductance 43
Conductivity of copper 55
of aluminum 55
Conductors 53
Convergent series 45, 1 16
derivation of 98
description of 41
Corona 6
Diameters of wires and cables 118
Elements of a transmission line 4
136 INDEX
PAGE
Fundamental transmission line formulas 41, 98
Hyperbolic transmission line formulas 41, 98
Image wire 90
Impedance 98
" J" terms 42
"K" formulas 28, 108
"K " formulas, description of 25
Leakage conductance 43
Leakage current 6
Line drop 9
Loss in line, derivation of formula 61
Magnetic field around wire 5, 62
Power factor table 133
Problem, 5000 feet 14, 23
3 miles, 80% P.F 14, 23
3 miles, 85% P.F 12
10 miles 22
15 miles 13, 22
20 miles 24
25 miles 14, 24
75 miles 13, 15, 3
80 miles 12, 50
100 miles, 60,000 volts 39, 51
100 miles, 66,000 volts n, 36, 38, 50
100 miles, 1 10,000 volts 15
155.34 miles 39> 5
200 miles 38, 47
300 miles 39*5!
300 miles, with substation 36, 48
400 miles, with substation 4. 5 2
Quadrature volt-amperes, derivation of formula 60
INDEX 137
.PAGE
Reactance of cable, 25 cycles, Table X 122
of cable, 60 cycles, Table XII 12$
of wire, 25 cycles, Table IX 121
of wire, 60 cycles, Table XI 124
Reactance, derivation of formulas:
single-phase, wire, effect of flux in air 62
single-phase, wire, effect of flux in the conductor 64
single-phase, cable 76
two-phase 79
three-phase, irregular spacing 79
three-phase, regular flat spacing 83
three-phase, regular spacing 83
Reactance drop $
Reactive power, derivation of formula 60
Regulation -. 8, 58
Regulation chart, description of 8
Resistance of aluminum cable, Table VIII 120
of copper wire and cable, Table VII 118
Resistance drop 4
Series, convergent 45, 116
derivation of 98
description of 41
Short line formulas 18, 104
description of 16
Skin effect . 4
Skin effect, derivation of formula 67
Skin effect formula, proof by infinite series 71
Spacing 4, 10
equivalent 10
flat 10
two-phase 10
Substation loads 26
Susceptance (see also " Capacity ") 6
Temperature coefficients 119, 120
description of 55
Watts, derivation of formula 59
Wire and cable tables n
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Alternating- Current
Machines
Being the Second Volume of
Dynamo Electric Machinery
Its Construction, Design
and Operation
By
Samuel Sheldon, A.M., Ph.D., D.Sc,
Prof, of Physics and Elec. Engineering, Polytechnic Inst., Brooklyn
AND
Hobart Mason, B.S.,E.E.
AND
Erich Hausmann, B.S.,-E.E.
THIS BOOK has been entirely re-written to bring it into accord
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It is so arranged that portions may be readily omitted in the case of the
latter without affecting the correlation of the remaining parts. The new
matter contains numerous problems, rigid derivations of the fundamental
laws of alternating current circuits, vector diagrams pertaining to the cir-
cuits of alternating current machines, methods of determining from tests
the behavior of machines and of predetermining their behaviour from
design data. There is also given a method of calculating all the leakage
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modern practice.
P. VAN NOSTRAND COMPANY
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THE NEW FOSTER
Fifth Edition, Completely Revised and Enlarged, with Four-
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Treating of the Latest and Best Practice
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CONTENTS
Symbols, Units, Instruments
Measurements
Magnetic Properties of Iron
Electro Magnets
Properties of Conductors
Relations and Dimensions of Con-
ductors
Underground Conduit Construction
Standard Symbols
Cable Testing
Dynamos and Motors
Tests of Dynamos and Motors
. The Static Transformer
Standardization Rules
Illuminating Engineering
Electric lighting (Arc)
Electric lighting (Incandescent)
Electric Street Railways
Electrolysis
Transmission of Power
Storage Batteries
Switchboards
I/ghtning Arresters
Electricity Meters
Wireless Telegraphy
Telegraphy
Telephony
Electricity in the U. S. Army
Electricity in the U. S. Navy
Resonance
Electric Automobiles X-Rays Lightning Conductors
Electro-chemistry and Electric Heating, Cooking Mechanical Section
Electro-metallurgy and Welding Index
D. VAN NOSTRAND COMPANY
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,
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YB 2419
^Engineering
Library
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