Question 2.7: Avoiding a New Transformer by Improving the Power Factor. A ......

Avoiding a New Transformer by Improving the Power Factor. A factory with a nearly fully loaded transformer delivers 600 kVA at a power factor of 0.75. Anticipated growth in power demand in the near future is 20%. How many kVAR of capacitance should be added to accommodate this growth so they don’t have to purchase a larger transformer?

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At PF = 0.75, the real power delivered at present is 0.75 × 600 kVA = 450 kW. And the phase angle is \theta \ = \ \cos ^{-1}\left(0.75\right) \ = \ 41.4º. If demand grows by 20%, then an additional 90 kW of real power will need to be supplied. At that point, if nothing is done, the new power triangle would show

\begin{matrix} \quad \quad \text{Real power} \ P & = & 450 \ + \ 90 \ = \ 540 \ kW \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \text{Apparent power} \ S & = & 540 \ {kW}/{0.75} \ 720 \ kVA \ \text{(too big for this transformer)} \\ \ \text{Reactive power} \ Q & = & VI \ \sin \theta \ = \ 720 \ kVA \ \sin \left(41.4º\right) \ = \ 476 \ kVAR \quad \quad \end{matrix}

For this transformer to still supply only 600 kVA, the power factor will have to be improved to at least

PF \ = \ {540 \ kW}/{600 \ kVA} \ = \ 0.90

The phase angle now will be \theta \ = \ \cos ^{-1} \left(0.90\right) \ = \ 25.8º. The reactive power will need to be reduced to

Q \ = \ 600 \ kVA \ \sin 25.8º \ = \ 261 \ kVAR

The difference in reactive power between the 476 kVAR needed without power factor correction and the desired 261 kVAR must be provided by the capacitor. Hence

PF \ \text{correcting capacitor} \ = \ 476 \ − \ 261 \ = \ 215 \ kVAR

The power triangles before and after PF correction are shown below:

Example 2.7

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