Question 2.10: Correcting the Power Factor in a Three-Phase Circuit. Suppos......
Correcting the Power Factor in a Three-Phase Circuit. Suppose that a shop is served with a three-phase, 208-V transformer. The real power demand of 80 kW is mostly single-phase motors, which cause the power factor to be a rather poor 0.5. Find the total apparent power, the individual line currents, and real power before and after the power factor is corrected to 0.9. If power losses before PF correction is 5% (4 kW), what will they be after power factor improvement?
Before correction, with 80 kW of real power being drawn, the apparent power before PF correction can be found from
P_{3\phi } \ = \ 80 \ kW \ = \ \sqrt{3} V_{line} I_{line} \ \cos \theta \ = \ S_{3\phi } \ \cos \theta \ = \ S_{3\phi } \ \cdot \ 0.5so, the total apparent power is
S_{3\phi } \ = \ \frac{80 \ kW}{0.5} \ = \ 160 \ kVAFrom (2.70), the current flowing in each leg of the three-phase system is
S_{3\phi } \ = \ 3V_{phase}I_{line} \ = \ 3 \ \cdot \ \frac{V_{line}}{\sqrt{3} } \ \cdot \ I_{line} \ = \ \sqrt{3} V_{line}I_{line} (2.70)
I_{line} \ = \ \frac{S_{3\phi }}{\sqrt{3} · V_{line}} \ = \frac{160 \ kVA}{\sqrt{3} \ \cdot \ 208 \ V} \ = \ 0.444 \ kA \ = \ 444 \ AAfter correcting the power factor to 0.9 the resulting apparent power is now
S_{3\phi } \ = \ \frac{P_{3\phi }}{\cos \theta } \ = \ \frac{80 \ kW}{0.9} \ = \ 88.9 \ kVAThe current in each line is now,
I_{line} \ = \ \frac{S_{3\phi }}{\sqrt{3} \ \cdot \ V_{line}} \ = \ \frac{88.9 \ kVA}{\sqrt{3} \ \cdot \ 208 \ V} \ = \ 0.247 \ kA \ = \ 247 \ ABefore correction, line losses are 5% of 80 kW, which is 4 kW. Since line losses are proportional to the square of current, the losses after power factor correction will be reduced to
\text{Losses after correction,} \ P_{line} \ = \ 4 \ kW \ \cdot \ \frac{\left(247\right) ^{2}}{\left(444\right) ^{2}} \ = \ 1.24 \ kWSo, line losses in the factory have been cut from 5 kW to 1.24 kW—a decrease of almost 70%.