Question

A 72 KVA transformer supplies a heating and lighting load of 12 kW at unity power factor and a motor load of 70 kVA at 0.766 (lagging) power factor. Calculate the minimum rating of the power-factor improvement capacitors which must be connected in the circuit to ensure that the transformer does not become overloaded

Answer 1

A 72 kVA transformer supplies a heating and lighting load of 12 kW at unity p.f. and a motor load of 70 kVA at 0.766 p.f. lagging. Calculate the minimum capacitor rating so the transformer is not overloaded.

Motor: kW = 70 x 0.766 = 53.62 kW. sin phi = sqrt(1 - 0.766^2) = sqrt(0.4132) = 0.6428. Motor kVAr = 70 x 0.6428 = 45.0 kVAr (lagging).
Total kW = 12 + 53.62 = 65.62 kW. Total kVAr = 45.0 kVAr.
Present total kVA = sqrt(65.62^2 + 45^2) = sqrt(4306 + 2025) = sqrt(6331) = 79.6 kVA. This exceeds 72 kVA, so the transformer is overloaded.
To avoid overloading, the total kVA must be reduced to 72 kVA. The kW (65.62 kW) is fixed, so the allowable kVAr is:

kVAr = sqrt(72^2 - 65.62^2) = sqrt(5184 - 4306) = sqrt(878) = 29.6 kVAr.

Capacitor kVAr required = 45.0 - 29.6 = 15.4 kVAr.

So the minimum rating of the power-factor improvement capacitors is about 15.4 kVAr (say 16 kVAr).

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