Tool
kVAR calculator
Work out how much reactive compensation a plant needs to move from its present power factor to a target.
The compensation required is Q = P × (tan φ₁ − tan φ₂), where P is real power in kW and φ₁, φ₂ are the present and target power factor angles. For a 500 kW load moving from 0.75 to 0.95, that gives 277 kVAR.
Reactive compensation required
Use the maximum demand that will actually occur, not the annual average — the penalty is applied at peak.
Capacitor output falls with voltage and is reduced by roughly 7% when a 7% detuning reactor is fitted. If harmonics are present, add the reactor and size on detuned voltage.
# φ₁ = acos(present PF), φ₂ = acos(target PF)
# 500 kW, 0.75 → 0.95
tan(acos 0.75) = 0.8819
tan(acos 0.95) = 0.3287
Qc = 500 × 0.5532 = 277 kVAR
How to use the result
Three decisions follow from the kVAR figure, and they are not independent.
1. Capacitors or an active device
Capacitors are the cheapest way to supply reactive power — typically $20–45 per kVAR installed. They are also a low-impedance path for harmonic current, so on a distorted network they can enter resonance and fail. If voltage THD is above roughly 3%, a detuned reactor is required; above 8%, capacitors may not be the right answer at all.
2. Fixed or automatic
A plant whose load varies between 20% and 100% cannot be corrected by a single fixed bank without going leading at light load. Automatic step control with 6–12 stages, or a static var generator with continuous output, is needed. The kVAR figure sizes the total; the step size is set by the smallest load increment.
3. One bank or distributed
Central compensation is cheaper per kVAR and easier to maintain. Distributed compensation at each board avoids long reactive current paths and reduces losses in the distribution cables. Above roughly 500 kVAR, the loss saving often justifies distributing.
Quick reference
| Present PF | → 0.90 | → 0.95 | → 0.98 |
|---|---|---|---|
| 0.60 | 53 | 75 | 95 |
| 0.70 | 31 | 53 | 73 |
| 0.75 | 23 | 45 | 65 |
| 0.80 | 15 | 37 | 57 |
| 0.85 | 8 | 29 | 49 |
| 0.90 | 0 | 21 | 41 |
Values in the table are kVAR per 100 kW. Multiply by your load in hundreds of kW. The 0.98 column exists because some tariffs pay a credit above 0.95 — but correcting past 0.98 risks leading power factor at light load.
Two things the formula does not tell you
It does not account for harmonics. If the load has significant distortion, part of the apparent power is harmonic rather than reactive, and capacitors cannot supply it. The plant will still be penalised after the correct kVAR of capacitance is installed. In that case the measurement has to separate fundamental reactive power from distortion before sizing anything.
It assumes the load stays at P. Adding compensation does not reduce real power, so the saving comes entirely from the demand charge and any reactive charge — not from energy consumption. Payback calculations that include a kWh saving are usually wrong.
Verifying a compensation proposal?
Send the load list and a recent power quality recording. We will check the kVAR figure, the detuning choice, and whether capacitors are appropriate for the distortion level.