Question

two loads are connected in parralel acrros a 100 Vrms, 60Hz source. one load absorbs 500kwat 0.6pf log, the secound load absorbs 400kw and 600 kvar. determine the value of the capacitor that should be added in parralel with the 2 losds to improve the overall pf to 0.9 log

Answer 1

Load 1: - 500 kw, 0.6 p.f Load 2: - 400 kw, 600 KVAR. Load 1: - P = 500 kw, cos phi = 0.6, phi = 53.13 degree P = integral cos phi integral = P/cos phi = 500/0.6 integral = 833.333 KVA phi_1 = integral sin phi = 833.333 times sin 53.13 phi_1 = 666.6666 KVAR I*_1 = S/V = K 833.3333 k/100 I_1 = 8.3333 angle -53.13 K Amps \r\n Load 2: - 400 kw, 600 KVAR P = 400 kw, Q = 600 KVAR cos phi = cos [tan^-1 (Q/p)] = cos [tan^-1 (600/400)] cos phi = 0.5547 phi = 56.309 degree I_2 = P/V times cos phi = 400 k/120 times 0.5547 I_2 = 6 angle -56.309 K amps I = I_1 + I_2 = (8.333 angle -53.13 + 6 angle -56.309) k I = 14327.631 angle -54.46 Amps Q_total = Q_1 + Q_2 = 666.666 + 600 Q_total = 1266.666 KVAR \r\n Overall P. f = cos (-54.46)