Electricity · Industrial electricity

Power factor correction and reactive compensation

From active power P and the initial and target power factors, calculate Q before and after correction, required compensation Qc, and the reduction in apparent power.

Created v1.0

By Thibaut Grzelak, Mechanical Analysis Engineer

Industrial electricity
01Calculation inputsQc = Q₁ − Q₂
cos φ₁Q₁Qccos φ₂Q₂
From active power P and the initial and target power factors, calculate Q before and after correction, required compensation Qc, and the reduction in apparent power.
02

Results

Target reactive power Q₂
—

Reactive power corresponding to the target power factor.

Required compensation Qc
—

Capacitive reactive power to compensate to reach the target.

Initial apparent power S₁
—

Apparent power before correction.

Target apparent power S₂
—

Theoretical apparent power after correction.

Relation and numerical substitutionQc = P · [tan(arccos(cosφ₁)) − tan(arccos(cosφ₂))]

—

Sinusoidal steady state with active power P assumed unchanged by the correction. Power factors must satisfy 0 < cos φ ≤ 1 and the target must be greater than the initial value. The result sizes reactive power in kvar; it does not directly size capacitors in µF. Harmonics, resonance, capacitor tolerances and utility requirements are not included.

01Formulas and symbols

Formulas used

Q₁P · tan(arccos(cosφ₁))Reactive power before correction
Q₂P · tan(arccos(cosφ₂))Reactive power at the target
QcQ₁ − Q₂Reactive power to compensate
S₁P / cosφ₁Apparent power before correction
S₂P / cosφ₂Apparent power after correction
02Assumptions and limits

Scope of validity

  • Sinusoidal steady state with active power P assumed unchanged by the correction.
  • Power factors must satisfy 0 < cos φ ≤ 1 and the target must be greater than the initial value.
  • The result sizes reactive power in kvar; it does not directly size capacitors in µF.
  • Harmonics, resonance, capacitor tolerances and utility requirements are not included.
03Validation example

Reference numerical case

  1. P = 10 kW, cos φ₁ = 0.80 and cos φ₂ = 0.95.
  2. Q₁ = 10 × tan(arccos 0.80) = 7.50 kvar.
  3. Q₂ = 10 × tan(arccos 0.95) ≈ 3.29 kvar.
  4. Qc = Q₁ − Q₂ ≈ 4.21 kvar; S falls from 12.50 to 10.53 kVA.
04References

Technical references

  1. IEC 60050-131 — International Electrotechnical Vocabulary: circuit theory.
  2. BIPM — The International System of Units (SI), 9th edition.