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If r is the resistance of each three-phase conductor, then the loss per wire is 12, 3, and the total loss is 12, while in the singlephase system it is 2/2R. Hence, to get the same loss, 12, 212R, or r = 2R; that is, each three-phase wire has twice the resistance and half the copper of each single-phase conductor, or in other words, the three-phase, three-wire system requires 75 per cent as much copper as the single-phase at the same maximum voltage. In the three-phase, three-wire system, with the lamps connected between the neutral point and the three outer wires, that is, in Y fashion (p. 144),— the voltage between the outer wires, or ▲ potential, will be √3, if V is the lamp or Y voltage. In other words, the potential between lines is raised from to V√3; and since it has been shown previously that the weight of copper is inversely as the square of the voltage, the weight of copper for the three wires will be one-third as great as in the preceding case, and 94 = as much as for the two-phase. The addition of a fourth or neutral conductor (Fig. 166) will increase this to, so that the three-phase, four-wire requires 33.3 per cent as much copper as the single-phase, two-wire system at the same minimum voltage.

The monocyclic system, when supplying lamps, is practically the same as a single-phase circuit (p. 208); and most of the energy for motors is carried by the two main conductors, except in starting, when the auxiliary wire furnishes a certain amount of current. If the extra wire is omitted on certain circuits, then the copper required is the same as for the single-phase, two-wire system. If the third wire is one-half as large as each of the other two, then the monocyclic calls for 125 per cent; and if the three conductors are all of the same size, then the copper demanded is 150 per cent of that used by the single-phase circuit. Hence the monocyclic system requires as much copper as an equal voltage single-phase, two-wire system, plus the copper in the auxiliary conductor.

This table is issued by the General Electric Company, and gives in convenient form the constants to be used in calculating overhead electric transmission lines, the various quantities having the following significance : —

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L = Inductance in milhenrys per 1000 feet of conductor.

C= Capacity in microfarads per 1000 feet of conductor.

iCharging current at 100 cycles and 10,000 volts to neutral, that is, in a

20,000 volt single-phase, and a 17,300 volt three-phase line.

SIZE OF WIRE

A.W.G.

io= 2 X X frequency X CX EX 10-6; where E is the E.M.F. between a line and neutral.

x = reactance = 2 X X frequency X L X 10—3.

The E.M.F. consumed by resistance r, of the line, is = Ir, and in phase with the current /

The E.M.F. consumed by the reactance x, of the line, is = Ix, and in quadrature with the current I.

The E.M.F. consumed in the line is neither Ir nor Ix, but depends upon the phase relation of current in the receiving circuit.

The loss of energy in the line is = 12r, hence does not depend upon the reactance, but only upon the resistance.

Two wires in parallel have the same resistance and about half the reactance (if strung on separate insulators and intermixed) of a single wire of double cross-section. Thus replacing one No. 0000 wire by two No. 0 wires, the resistance, weight of copper, etc., will remain the same, but the reactance will be reduced practically to half, so where lower reactance is desired, the use of several conductors, strung on independent insulators and intermixed, is advisable.

The values given for L, C, 。, and x are calculated for sine waves of current and

E.M.F.

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52633 .197
41742 .249
33102 .314
26250 .395
20816 .499 .352 .00306
16509 .629 .360 .00300
13094 .792 .366 .00294 .0185 .0575.1380 .288
10382 .999 .373 .00288 .0181 .0585.1405 .293

.00351.0220 .0486.1166
.317 .00342 .0215|| .0498 .1194
.324 .00334 .0210|| .0509.1220
.332.00326 .0205|| .0521.1248

.339 .00320.0201 .0532.1277
.346 .00313 .0197|| .0543 .1301
.0193 .0553.1327
.0189.0565.1355

.243

.249

.254

.261

.266

.271

.276

.283

The above figures apply to parallel wires that are 18 inches apart, but are not much modified by moderate changes in interaxial distance. For example, the inductance L is decreased about 10

per cent when the wires are 12 inches apart, and is increased about 10 per cent when they are put 30 inches apart. A similar change would be produced in the reactance 2fL. The capacity C and charging current i are increased about 10 per cent, if the interaxial distance is reduced to 12 inches; and is decreased about 10 per cent when it is raised to 30 inches. For wires placed close together in cables, the inductance and reactance would be greatly reduced, and the capacity and charging current greatly augmented, so that their values in the table would not be even approximately true. In such cases, or whenever there may be any doubt, the general formulæ for inductance and capacity, given on pages 129 and 137 respectively, should be used. The table on page 130 shows the inductance of wires from No. .0000 to No. 12, and for interaxial distances from 3 to 96 inches.

The equations on page 138 give results for capacity that are twice as great as those found in the above table, the reason being that each conductor is considered separately with respect to zero potential, which would ordinarily exist at a point midway between two wires forming a circuit; whereas the equations (p. 138) give the capacity with respect to the other wire, which is one-half as much. The calculated charging current would be the same in both cases, because the total voltage would be used in one instance, and the potential with respect to zero is taken in the table. The latter plan is generally better. Other convenient tables issued by the General Electric Company are given below. They apply particularly to overhead circuits with wires 18 inches apart, and are sufficiently accurate for most practical purposes, when the distance between wires is approximately that amount. If the conductors are less than 18 inches apart, the loss in voltage is lower than that given by the formulæ, and if they are close together, as in cables or interior wiring, the loss will be only that due to resistance.

The following general formula may be used to determine the size of copper conductors, current per conductor, volts loss in lines, and weight of copper per circuit for various systems of electrical distribution.

Area of conductor, Circular Mils =

Current in main conductors =

D× W × K

Px E2

W× T

E

(77) (78)

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W = Total watts delivered.

D = Distance of transmission (one way) in feet.

Р = Loss in line in per cent of power delivered, that is of W.
E=
Voltage between main conductors at receiving or consumer's
end of circuit.

For continuous current K = 2160, T=1, M = 1, and A = 6.04.

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0000 000

1.23 1.29 1.33 1.34 1.62 1.84 1.99 2.09 2.35 2.86 3.24 8.49 0000 1.18 1.22 1.24 1.24 1.49 1.66 1.77 1.95 2.08 2.48 2.77 2.94

000

00 1.14 1.16 1.16 1.16 1.34 1.52 1.60 1.66 1.86 2.18 2.40 2.57 0 1.10 1.11 1.10 1.09 1.31 1.40 1.46 1.49 1.71 1.96 2.13 2.25

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1

12

84

1.07 1.07 1.05 1.03 1.24 1.30 1.34 1.36 156 1.75 1.88 1.97 2 1.05 1.04 1.02 1.00 1.18 1.23 1.25 1.26 1.45 1.60 1.70 1.77 3 1.03 1.02 1.00 1.00 1.14 1.17 1.18 1.17 1.35 1.46 1.53 1.57 4 1.02 1.00 1.00 1.00 1.11 1.12 1.11 1.10 1.27 1.35 1.40 1.43 51.00 1.00 1.00 1.00 1.08 1.08 1.06 1.04 1.21 1.27 1.30 1.31 6 1.00 1.00 1.00 1.00 1.05 1.04 1.02 1.00 1.16 1.20 1.21 1.21 6 7 1.00 1.00 1.00 1.00 1.03 1.02 1.00 1.00 1.12 1.14 1.14 1.13 8 1.00 1.00 1.34 1.00 1.02 1.00 1.00 1.00 1.09 1.10 1.09 1.07

5

13

78

10

9 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.06 1.06 1.04 1.02 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.04 1.03 1.00 1.00

9

10

The value of K for any particular power factor is obtained by dividing 2160, the value for continuous current, by the square of that power factor for single-phase, and by twice the square of that power factor for three-wire, three-phase, or four-wire, two-phase.

The value of M depends on the size of wire, frequency, and power factor. It is equal to 1 for continuous current, and for alternating current with 100 per cent power factor and sizes of wire given in the preceding table of wiring constants.

The value of T depends upon the system and power factor. It is equal to 1 for direct current and for single-phase current of 100 per cent power factor.

The value of A and the weights of wire in the table are based upon .00000302 lb. as the weight of one mil foot of copper.

It should be observed that P stands for the per cent loss of the delivered power, and not the per cent loss of the power at the generator; and that E is the potential at the end of the line, and not at the generator.

When the power factor cannot be more accurately determined, it may be assumed to be as follows for any alternating current system operating under average conditions: Lighting with no motors, 95 per cent; lighting and motors together, 85 per cent; motors alone, 80 per cent. The size of wire in (77) is for each of the main or outside conductors of a given system, for example, the three wires of a three-phase, the four wires of a two-phase, or the main two wires of a monocyclic system. The neutral wire in the threewire system for direct (p. 72) or alternating currents (p. 192) may in the case of feeders be made one-half or one-third size, or omitted entirely (p. 76), depending upon how well the system may be balanced. For local or secondary circuits it should generally be the same size as each of the main wires. These statements also apply to the middle wire of the four-wire, three-phase arrangement (Fig. 166). The size of the auxiliary conductor in the monocyclic system should be in the same proportion to either main wire as the motor load in amperes is to the total load in amperes.

A simple method of calculating the drop and other data of alternating current circuits is represented in Fig. 181, being derived from the little book on Alternating Current Wiring and Distribution, by W. L. R. Enmet. Assume a case where 500 incandescent lamps, of 57.5 watts each, are to be fed from the sec

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