232
AC POWER
[CHAP. 10
EXAMPLE 10.13 How much capacitive Q must be provided by the capacitor bank in Fig. 10-10 to improve the power factor to 0.95 lagging?
Fig. 10-10 Before addition of the capacitor bank, pf ¼ cos 258C ¼ 0:906 lagging, and 240 08 ¼ 68:6 258 A 3:5 258 240 68:6 pffiffiffi þ258 ¼ 8232 258 ¼ 7461 þ j3479 S ¼ Veff I eff ¼ pffiffiffi 08 2 2
I1 ¼
After the improvement, the triangle has the same P, but its angle is cos 1 0:95 ¼ 18:198. 3479 Qc ¼ tan 18:198 7461
or
Then (see Fig. 10-11),
Qc ¼ 1027 var (capacitive)
The new value of apparent power is S 0 ¼ 7854 VA, as compared to the original S ¼ 8232 VA. The decrease, 378 VA, amounts to 4.6 percent.
Fig. 10-11
The transformers, the distribution systems, and the utility company alternators are all rated in kVA or MVA. Consequently, an improvement in the power factor, with its corresponding reduction in kVA, releases some of this generation and transmission capability so that it can be used to serve other customers. This is the reason behind the rate structures which, in one way or another, make it more costly for an industrial customer to operate with a lower power factor. An economic study comparing the cost of the capacitor bank to the savings realized is frequently made. The results of such a study will show whether the improvement should be made and also what final power factor should be attained. EXAMPLE 10.14 A load of P ¼ 1000 kW with pf ¼ 0:5 lagging is fed by a 5-kV source. A capacitor is added in parallel such that the power factor is improved to 0.8. Find the reduction in current drawn from the generator. Before improvement: P ¼ 1000 kW; cos ¼ 0:5; S ¼ P= cos ¼ 2000 kVA; I ¼ 400 A After improvement: P ¼ 1000 kW; cos ¼ 0:8; S ¼ P= cos ¼ 1250 kVA; I ¼ 250 A