kW, kVAR and kVA
Real power (kW) does work. Reactive power (kVAR) sustains the magnetic fields motors need. Apparent power (kVA) is their vector sum: kVA = √(kW² + kVAR²), and PF = kW ÷ kVA. At 0.75 PF, the same 75 kW load demands 100 kVA. Correcting to 0.95 lowers that to about 79 kVA.
Why low power factor matters
- Utility charges: many commercial tariffs bill on kVA demand or add penalties below a PF threshold, often 0.90 or 0.95. Check your own rate schedule.
- Lost capacity: transformers, generators and feeders are rated in kVA. Reactive current uses up that rating.
- Higher losses and voltage drop: extra current heats conductors (I²R) and depresses voltage at the far end of feeders.
Correction methods
Fixed capacitors at the motor
Capacitors connected at individual large motors that run continuously. They are simple, but must be sized carefully: oversizing can cause self-excitation when the motor is disconnected.
Automatic capacitor banks
A controller switches capacitor steps in and out at the service or a main distribution board as load changes. This is the most common approach for whole-facility correction.
Detuned banks and active filters
Where VFDs and other non-linear loads create harmonics, plain capacitors can resonate and fail. Detuned (reactor-equipped) banks or active harmonic filters correct PF while managing harmonics. See Common Three-Phase Problems.
Sizing a correction project
Required kVAR ≈ kW × (tan θ₁ − tan θ₂), where θ₁ and θ₂ are the angles for the existing and target PF. A proper study uses interval metering data, utility bills and a harmonic survey before equipment is specified.