Which capacitor banks would provide desired power factor?

Certification Forum NETA Level 4 Exam
1
  1. To increase the power factor of a substation 750 kVA load that has a 0.76 PF to a PF of 0.95, which of the following capacitor banks would provide desired power factor?**
    Your Answer: 250 kVAR
    Correct Answer: 300 kVAR

PLease explain

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2

I think your answer of 250kVAR is correct.

Key formula for this question is
Q = S x sin(cos-1(pf))

750kVA at 0.76PF
Q = 750 x sin(cos-1(0.76))
Q = 750 x sin(40.54)
Q = 750 x 0.65
Q = 487.44

750kVA at 0.95PF
Q = 750 x sin(cos-1(0.95))
Q = 750 x sin(18.19)
Q = 750 x 0.312
Q = 234.18

Correction = 487.44 - 234.18
Correction = 253.26

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3

Your solution assumes that the apparent power (S) is held constant. For power factor correction, true power (P) is held constant.

Let:
PF1 = original power factor
PF2 = desired power factor
P = True Power
Q1 = Original Reactive Power
Q2 = Desired Reactive Power
Qc = Capacitor Bank Reactive Power
S1 = Original Apparent Power
S2 = Desired Apparent Power

For this problem,

P = S1 * PF1 = 750kVA * 0.76 = 570 kW

S2 = P / PF2 = 570kW / 570 = 600kVA

Q1 = SQRT(S1^2 - P^2) = 487 kVAR

Q2 = SQRT(S2^2 - P^2) = 187 kVAR

Qc = Q1 - Q2 = 487 - 187 = 300 kVAR

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4

I noticed a slight mistake on the line calculating S2. It should be 570kW / 0.95 = 600kVA

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