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Power factor correction: capacitor bank rating

Reactive power of a capacitor bank that raises the power factor from cos φ_1 to cos φ_2, and the resulting reduction of the supply current and line losses.

Formula
Q_c = P · (tg φ_1 − tg φ_2)I_2 / I_1 = cos φ_1 / cos φ_2ΔP_2 / ΔP_1 = (cos φ_1 / cos φ_2)²

Capacitor output is proportional to the voltage squared: a 400 V bank at 380 V gives about 90 % of its rating. For individual motor correction, keep the capacitor rating below about 90 % of the motor's no-load reactive power, or the motor may self-excite after switching off. With variable frequency drives and other non-linear loads, use detuned banks with anti-resonance reactors.

Source: Power triangle: Q = P · tan φ

Inputs

You can change a field's unit: the value is converted to the formula's units automatically.

P
Calculate

Average power at peak hours, from the meter or the current.

cos φ_1

Plants with induction motors: usually 0.7–0.85.

cos φ_2

Usually 0.92–0.97; correcting to 1 does not pay off and risks overcompensation at light load.

Unit converter for this formulaPower · Reactive power · Apparent power · Fraction and percent
Metric
  • kW1
  • W1,000
  • MW0.001
  • PS (metric hp)1.35962
  • kcal/h859.845
US field units
  • hp1.34102
  • BTU/h3,412.14

PS is the metric horsepower (735.5 W), hp the mechanical horsepower (745.7 W).

All units
Result
Q_cCapacitor bank rating
Fill in all fields
  • tg φ_1tan φ before correction
  • S_2Apparent power after correction
  • ΔIReduction of the supply current
  • ΔP_lossReduction of load losses in the line and transformer

More in Motors and transformers

Results are engineering estimates from standard formulas; for design decisions check them against the codes, project documents and specialists' calculations. The formulas carried over from the original set are unchanged, and their errors are described in the notes.