This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.
The percentage loss of power in the line has not, as with direct current, the same value as the percentage drop. This is due to the fact that the line has reactance, and also that the apparent power delivered to the load is not identical with the true power - that is, the load power factor is less than unity. The loss must be obtained by calculating I2 R for the line, or, what amounts to the same thing, by multiplying the resistance-volts by the current.
LIVING ROOM IN DWELLING FOR MR. W. W. WILLITS. HIGHLAND PARK. ILL.
Frank Lloyd Wright, Architect, Oak Park, Ill.
The resistance-volts in this case are 307.8, and the current
156 25 amperes. The loss is 307.8 X 156.25 = 48.4 K. W. The percentage loss is
4 8 .1
250 + 48.1 = 16.l per cent.
Therefore, for the problem taken, the drop is 18.7 per cent, and the loss is 16.1 per cent, If the problem be to find the size wire for a given drop, it must be solved by trial. Assume a size of wire and calculate the drop; the result in connection with the table will show the direction and extent of the change necessary in the size of wire to give the required drop.
The effect of the line reactance in increasing the drop should be noted. It' there were no reactance, the drop in the above example would be given by the point obtained in laying off on the chart the resistance-E. M. F. (15.4) only. This point falls at 12.4 per cent, and the drop in terms of the generator E. M. F. would be
112.4 = 11 percent, instead of 18.7 percent.
Anything therefore which will reduce reactance is desirable.
Reactance can be reduced in two ways. One of these is to diminish the distance between wires. The extent to which this can be carried is limited, in the case of a pole line, to the least distance at which the wires are safe from swinging together in the middle of the span; in inside wiring, by the danger from fire. The other way of reducing reactance is to split the copper up into a greater number of circuits, and arrange these circuits so that there is no inductive interaction. For instance, suppose that in the example worked out above, two No. 3 wires were used instead of one No. 0 wire. The resistancevolts would be practically the same, but the reactance-volts would be less in the ratio 1/2 x .244 = .535,
.228 since each circuit would bear half the current the No. 0 circuit does, and the constant for No. 3 wire is .244, instead of .228 - that for No. 0. The effect of subdividing the copper is also shown in the example given it is desired to reduce the drop to, say, one-half. Increasing the copper from No. 0 to No. 0000 will not produce the required result, for, although the resistance-volts will be reduced one-half, the reactance-volts will be reduced only in the ratio .212 If, however, two inductively independent circuits of No. 0
.228 wire be used, the resistance- and reactance-volts will both be reduced one-half, and the drop will therefore be diminished the required amount.
The component of drop due to reactance is best diminished by subdividing the copper or by bringing the conductors closer together. It is little affected by change in size of conductors.
An idea of the manner in which changes of power factor affect drop is best gotten by an example. Assume distance of transmission, distance between conductors E. M. F., and frequency, the same as in the previous example. Assume the apparent power delivered the same as before, and let it be constant, but let the power factor be given several different values; the true power will therefore be a variable depending upon the value of the power factor. Let the size of wire be No. 0000. As the apparent power, and hence the current, is the same as before, and the line resistance is one-half, the resistance-
E. M. F. will in this case be 15.4,
2 or 7.7 per cent of the E. M. F. delivered.
Also, the reactance-E. M. F. will be
.212 X 17.8 = 16.5 per cent.
Combining these on the chart for a power factor of .4, and deducing the drop, in per cent, of the generator E. M. F., the value obtained is 15.3 per cent; with a power factor of .8, the drop is 14 per cent; with a power factor of unity, it is 8 per cent. If in this example the true power, instead of the apparent power, had been taken as constant, it is evident that the values of drop would have differed more widely, since the current, and hence the resistance- and reactance-volts, would have increased as the power factor diminished. The condition taken more nearly represents that of practice.
If the line had resistance and no reactance, the several values of drop, instead of 15.3, 14, and 8, would be 3.2, 5.7, and 7.2 per cent respectively, showing that for a load of lamps the drop will not be much increased by reactance; but that with a load, such as induction motors, whose power factor is less than unity, care should be taken to keep the reactance as low as practicable. In all cases it is advisable to place conductors as close together as good practice will permit.
When there is a transformer in circuit, and it is desired to obtain the combined drop of transformer and line, it is necessary to know the resistance- and reactance-volts of the transformer. The resistance-volts of the combination of line and transformer are the sum of the resistance-volts of the line and the resistance-volts of the transformer. Similarly, the reactance-volts of the line and transformer are the sum of their respective reactance-volts. The resistance- and reactance-E. M. F.s of transformers may usually be obtained from the makers, and are ordinarily given in per cent.* These percentages express the values of the resistance- and reactance-E. M. F.'s when the transformer delivers its normal full-load current; and they express these values in terms of the normal no-load E. M. F. of the transformer.