—312.5 apparent K.W.
.8 The current, therefore, at 2,000 volts will be 312,500 =156.25 amperes.
2,000 From the table of reactances under the heading "18 inches," and corresponding to No. 0 wire, is obtained the constant .228. Bearing the instructions of the table in mind, the reactance-volts of this line are, 156.25 (amperes) X 10 (thousands of feet) X .228=356.3 volts, which is 17.8 per cent of the 2,000 volts to be delivered.
From the column headed "Resistance-Volts" and corresponding to No. 0 wire, is obtained the constant .197. The resistance-volts of the line are, therefore, 156.25 (amperes) X 10 (thousands of feet) X .197=307.8 volts, which is 15.4 per cent of the 2,000 volts to be delivered.
Starting, in accordance with the instructions of the table, from the point where the vertical line (which at the bottom of the chart is marked "Load Power Factor" .8) intersects the inner or smallest circle, lay off horizontally and to the right the resistance-E. M. F. in per cent (15.4); and from the point thus obtained, lay off vertically the reactance-E. M. F. in per cent (17.8). The last point falls at about 23 per cent, as given by the circular arcs. This, then, is the drop, in per cent, of the E. Al. F. delivered. The drop, in per cent, of the genera tor E. M. F. is, of course, 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 R for the line, or, what amounts to the same thing, by multiplying the resistance-volts by the current.
The resistance-volts in this case are 307.8, and the current 156.25 amperes. The loss i3 307.8 X 156.25=48.1 K. W. The percentage loss is 48.1 16.1 per cent.
250 + 48.1 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. If 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 12.4 = 11 per cent, instead of 18.7 per cent. 112.4 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 inter action. For instance, suppose that in the example worked out above, two No. 3 wires were used instead of one No. 0 wire. The resistance volts would be practically the same, but the reactance-volts would be less in the ratio X _244 = 535, since each circuit would bear half the .22S 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 if 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.