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Transformers & PF

Power Factor Correction Capacitor Calculator

Size a capacitor bank to raise power factor from its present value to a target. The calculator gives the kVAr needed, the capacitance per phase for star- and delta-connected banks at 50 or 60 Hz, and the drop in supply current and kVA.

For estimation and education. Not a substitute for a licensed engineer, the applicable codes (IEC 60364, NEC/NFPA 70, BS 7671, local regulations) or manufacturer data. Read the disclaimer.

System

Load

Power unit
kW

Often 0.95; check your tariff

Supply

V
Frequency

Result

Power factor correction — Engineerly. Estimate for education only; verify against applicable codes and manufacturer data.

Capacitor bank
69.15kVAr
C per phase, delta
458.6µF
C per phase, star
1,376µF
Current reduction
26.3%
kVAr before → after
102 → 32.87kVAr
kVA before → after
142.9 → 105.3kVA
Line current before → after
206.2 → 151.9A
Capacitor bank line current
99.81A
Multiplier (tan φ1 − tan φ2)
0.6915

Working, with your numbers

  1. 1. tan φ before and after
    tan φ = √(1 − PF²) / PF
    tan φ1 = 1.0202, tan φ2 = 0.328684
    = Δ = 0.69152
  2. 2. Capacitor reactive power
    Qc = P × (tan φ1 − tan φ2)
    Qc = 100 kW × (1.0202 − 0.328684)
    = 69.152 kVAr
  3. 3. Capacitance per phase, delta
    C_Δ = Qc / (3 × 2πf × V_LL²)
    C_Δ = 69,152 var / (3 × 2π × 50 Hz × 400² V²)
    = 458.58 µF
  4. 4. Capacitance per phase, star
    C_Y = Qc / (3 × 2πf × (V_LL/√3)²)
    C_Y = 69,152 var / (2π × 50 Hz × 400² V²)
    = 1,375.7 µF
  5. 5. Line current before → after
    I = P / (√3 × V × PF)
    I1 = 206.2 A, I2 = 151.93 A
    = −26.32 %

Assumptions

  • Real power stays constant; only reactive power changes.
  • Sinusoidal voltage at the frequency selected; harmonics are ignored.
  • Balanced three-phase bank at the line-to-line voltage entered. Delta capacitors see V_LL, star capacitors see V_LL/√3.
  • Capacitance is the value needed at the stated voltage and frequency, not a catalogue rating.
System: three
Real power: 100 kW
PF: 0.7 → 0.95
Voltage: 400 V
Frequency: 50 Hz

Check before you rely on this

  • Capacitors can resonate with the supply inductance at a harmonic frequency. Where VFDs, UPS or rectifiers are present, use detuned (reactor) banks and a harmonic study (IEEE 519, IEC 61000 series).
  • Avoid over-correction to a leading PF at light load; automatic stepped banks follow the load.
  • A capacitor rated for a higher voltage than the system delivers less: Q scales with (V / V_rated)².
  • Estimation and education only. Not a substitute for a licensed engineer, the applicable code (IEC 60364, NEC/NFPA 70, BS 7671 or local regulations) or manufacturer data.

How to use this calculator

  1. Choose single- or three-phase and enter the load's real power (kW, or hp as real power).
  2. Enter the existing power factor (from a meter or bill) and the target PF. The target must be higher than the existing value.
  3. Enter the supply voltage (line-to-line for three-phase) and select 50 Hz or 60 Hz.
  4. Read the kVAr, the capacitance per phase and the current reduction.

How it works: the formula

Qc = P × (tan φ1 − tan φ2), tan φ = √(1 − PF²) / PF Per capacitor: Q = V² × 2πf × C Delta (each sees V_LL): C_Δ = Qc / (3 × 2πf × V_LL²) Star (each sees V_LL / √3): C_Y = Qc / (2πf × V_LL²) = 3 × C_Δ Single-phase: C = Qc / (2πf × V²) Current: I = P / (√3 × V_LL × PF) → reduction = 1 − PF1 / PF2

A capacitor supplies reactive power locally, so less of it has to come through the supply. Real power does not change. Only the reactive side of the power triangle shrinks. The factor tan φ1 − tan φ2 is the “kVAr per kW” multiplier printed in many capacitor tables.

Variables and units

SymbolMeaningUnit
PReal power of the loadkW
PF1, PF2Existing and target power factor—
QcCapacitor reactive power requiredkVAr
fSupply frequencyHz
CCapacitance per phaseµF

Worked example

100 kW at PF 0.70, corrected to 0.95 on a 400 V, 50 Hz three-phase supply:

tan φ1 = √(1 − 0.49) / 0.70 = 1.0202 tan φ2 = √(1 − 0.9025) / 0.95 = 0.3287 Qc = 100 × (1.0202 − 0.3287) = 69.15 kVAr C_Δ = 69 152 / (3 × 314.16 × 400²) = 458.6 µF per phase C_Y = 3 × 458.6 = 1375.7 µF per phase I: 206.2 A → 151.9 A (−26.3 %)

A delta bank needs one third of the capacitance of a star bank for the same kVAr, because each capacitor sees √3 times the voltage and Q rises with V². In practice you would choose a standard bank such as 70 or 75 kVAr, preferably in automatic steps.

Assumptions

  • Load real power is constant.
  • Sinusoidal supply at the stated frequency.
  • Balanced bank on a balanced three-phase supply.

Limitations

  • Capacitors combined with supply inductance form a resonant circuit. With harmonic-producing loads this can amplify harmonics and damage capacitors. Detuned (reactor) banks and a harmonic assessment are then needed (see IEEE 519 and the IEC 61000 series).
  • Fixed capacitors sized for full load can over-correct at light load. Automatic stepped banks avoid a leading power factor.
  • A capacitor's rated kVAr applies at its rated voltage and frequency. At a lower voltage it delivers less, in proportion to (V/V_rated)².
  • Switching transients, inrush current and discharge requirements are not covered here.

References

Standards and texts that cover this topic. Use the edition adopted in your jurisdiction.

  • IEEE Std 1036, Guide for the Application of Shunt Power Capacitors.
  • IEC 60831-1, Shunt power capacitors of the self-healing type for AC systems up to 1000 V.
  • IEEE Std 141 ("Red Book") — power factor and its improvement in industrial plants.
  • IEEE Std 519, Standard for Harmonic Control in Electric Power Systems.

Frequently asked questions

Why not correct to exactly 1.0?

The last few hundredths of PF need disproportionately large banks, and any load drop then pushes the system leading. Tariffs usually reward 0.95 or so. That is a common target.

Should the bank be star or delta?

Low-voltage banks are usually delta-connected because they need a third of the capacitance. Medium-voltage banks are often star-connected, frequently ungrounded, for protection reasons. The kVAr required is the same either way.

Where should the capacitors go?

Central banks at the main switchboard are simplest. Capacitors at individual large motors also cut current in the feeder cables. Motor-connected capacitors must not over-excite the motor when it is switched off; maker guidance limits their size.

Will power factor correction reduce my energy bill?

It reduces reactive-power charges and maximum demand charges where kVA is billed. It also cuts I²R losses in cables a little. It does not reduce the kWh your equipment actually uses.

Related calculators

How these formulas were checked: methodology and testing.