Power Factor Correction

IEC 60831-1/2 · IEEE 18-2012 · IEEE 1036-2010 · IEC 60871-1 · Qc sizing · Harmonic resonance · Detuned reactor · Energy savings

IEC 60831-1/2 IEEE 18-2012 IEEE 1036-2010 IEC 60871-1 EN 50160 IEEE 1459 IEEE 519-2014
Presets
System Parameters
Capacitor Bank Configuration
Voltage Rating (IEC 60831-1 §5.2)
Add Short-Circuit MVA above to enable resonance check
Specify if you know which harmonic dominates the load
Methodology
Qc = P × (tan(arccos PF₁) − tan(arccos PF₂))   [IEC 60831-1 / IEEE 18-2012 §5]

C_delta (µF) = Qc / (3 × 2πf × V_LL²) × 10⁶
C_wye   (µF) = Qc / (3 × 2πf × V_LN²) × 10⁶

f_r = f_sys × √(Ssc / Qc)   [IEEE 519-2014 §5]
  where Ssc = Short-Circuit MVA

Harmonic risk: |h_r − h| / h ≤ 10%  (h ∈ {5, 7, 11, 13})
Detuned reactor: p% = (f_sys / f_p)² × 100   [IEC 60831-2]
Enter system parameters and press Calculate.

FAQ

What is power factor correction?

PF correction adds capacitive reactive power (kVAR) to counteract inductive reactive demand from motors, transformers and reactors. This raises the ratio of useful power (kW) to total power (kVA), reducing current, losses, and utility demand charges.

How is Qc calculated?

Qc (kVAR) = P × [tan(arccos PF₁) − tan(arccos PF₂)] per IEEE 18-2012 §5 and IEC 60831-1. For example, 500 kW at PF 0.75→0.95: tan(arccos 0.75)=0.881, tan(arccos 0.95)=0.329, Δtan=0.552, Qc=276 kVAR.

Why check harmonic resonance?

Capacitors and system inductance form a parallel LC circuit. The resonant frequency f_r = fsys × √(Ssc/Qc) must not coincide with characteristic harmonics (5th=250 Hz, 7th=350 Hz on 50 Hz systems). IEEE 519-2014 §5 requires this check. If risk exists, a detuned reactor (typically 7%) shifts the resonance below the problematic harmonic.

What voltage rating is required?

IEC 60831-1 §5.2 requires capacitor banks to be rated at ≥ 1.10 × V_system to accommodate steady-state voltage rise and transient overvoltages. For a 415 V system, select 440 V rated capacitors; for 11 kV, select 12 kV rated.

Fixed vs. step-switched capacitor banks?

Fixed banks are simpler and less expensive — suitable for loads with low variation. Step-switched (automatic) banks adjust reactive output in steps as the load changes, avoiding over-correction during light-load periods and resonance during switched operation.

What energy savings can be expected?

By raising PF, I²R losses in conductors and transformer reduce proportionally to the square of the current ratio. A typical 500 kW industrial plant raising PF from 0.75 to 0.95 saves roughly 276 kVAR × 3000 hrs/yr ≈ 828,000 kWh/yr in losses, worth ~€99,000/yr, avoiding ~330 tonnes CO₂.