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PF correction for 1250 kVA generator (+170 kW)

Short answer

Existing load 1000 kW, 750 kVAr; extra 170 kW adds 105.36 kVAr, total 1170 kW, 855.36 kVAr overloads. Allowed Q at 1250 kVA is 440 kVAr, so 415.36 kVAr capacitors are required, resulting in 0.936 pf and exactly 1250 kVA.

Explain the need for power correction. A 1250 kVA generator set is operating at rated load with a pf of 0.8. An additional load of 170 kW at 0.85 pf is to be added. What value of capacitors is required so that the generator is not overloaded?

Need for power factor correction

Most industrial loads (induction motors, transformers, welding sets, fluorescent ballasts) draw lagging reactive power. Low pf means:

  • Higher current for the same kW → higher I²R losses in cables, transformers and generators.
  • Larger kVA rating of generators, transformers and switchgear is needed; less kW can be supplied from a given kVA.
  • Larger voltage drop and poor regulation.
  • Utility penalties / higher kVA demand charges.

Correcting pf with capacitors supplies the reactive power locally, reducing current, losses and kVA demand, and releasing capacity for extra load.

Numerical

Given: generator 1250 kVA at rated load, pf 0.8; extra load 170 kW at pf 0.85.

Existing load

P₁ = 1250 × 0.8 = 1000 kW
Q₁ = 1250 × 0.6 = 750 kVAr

Additional load

P₂ = 170 kW
Q₂ = 170 × tan(cos⁻¹0.85) = 170 × 0.6197
   = 105.36 kVAr

Total load without capacitor

P = 1000 + 170    = 1170 kW
Q = 750 + 105.36  = 855.36 kVAr
S = √(1170² + 855.36²) = 1449.3 kVA  (> 1250, overload)

Maximum kVAr allowed so that S = 1250 kVA

Q_allowed = √(1250² − 1170²) = √(193 600) = 440.0 kVAr
New pf    = 1170 / 1250 = 0.936 lag

Capacitor required

Qc = 855.36 − 440.0 = 415.36 kVAr

Answer: about 415.4 kVAr of capacitors is needed; the generator then supplies 1170 kW at 0.936 pf, i.e. exactly 1250 kVA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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