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PF correction for 1250 kVA generator (+150 kW)

Short answer

Capacitor required is 353.06 kVAr. Generator then supplies 1150 kW at 0.92 pf, exactly 1250 kVA, not overloaded.

A 1250 kVA turbine generator set is operating at rated load with a power factor of 0.8. An additional load of 150 kW at 0.85 power factor is to be added. What kVAR value of capacitors is required so that the generator is not loaded?

The generator must not exceed 1250 kVA after the extra load is added, so capacitors must supply enough kVAR to keep the total kVA at 1250.

Existing load (rated, pf 0.8):

P1 = 1250 × 0.8          = 1000 kW
Q1 = 1250 × sin(cos⁻¹0.8) = 1250 × 0.6 = 750 kVAR

Additional load (150 kW, pf 0.85):

tan(cos⁻¹0.85) = 0.6197
Q2 = 150 × 0.6197        = 92.96 kVAR

Total load:

P  = 1000 + 150   = 1150 kW
Q  = 750 + 92.96  = 842.96 kVAR

Reactive power allowed on the generator at 1250 kVA:

Q_allowed = √(1250² − 1150²)
          = √(1 562 500 − 1 322 500)
          = √240 000 = 489.90 kVAR

Capacitor kVAR required:

Qc = 842.96 − 489.90 = 353.06 kVAR

New power factor of generator = 1150 / 1250 = 0.92 lagging.

Answer: About 353 kVAR of capacitors is required; the generator then supplies 1150 kW at 0.92 pf, i.e. exactly 1250 kVA (not overloaded).

✓ Answer checked by Claude Opus 5.5 (high effort)

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