Question

An industrial firm has two electrical loads connected in parallel across the power source. Power is supplied to the firm at 4000 V rms. One load is 30 kW of heating use, and the other load is a set of motors that together operate as a load at 0.6 lagging power factor and at 150 kVA. Determine the total current and the plant power factor.

Answer 1

Answer:

I = 27.65A < 40.59°

PowerFactor = 0.76

Explanation:

Current on the heating load is:

I1 = 30KW / 4KV = 7.5A < 0°

Current on the inductive load:

I2 = (150KVA*0.6) /4KV = 22.5A  with an angle of acos(0.6)=53.1°

The sum of both currents is:

It = I1 + I2 = 7.5<0° + 22.5<53.1° = 27.65<40.59°

Now, the power factor will be:

pf = cos (40.59°) = 0.76