Power Factor Calculator — kVAr Correction and Current Reduction

What power factor correction actually does

Inductive loads draw magnetising current that shifts current out of step with voltage. The current still has to be generated, transmitted and paid for in terms of cable and equipment capacity, but it does no useful work. A capacitor bank supplies that magnetising current locally, so the supply no longer has to.

Correcting the power factor does not reduce the real power your process uses. It reduces the apparent power the utility must deliver, which reduces line current, frees capacity in cables and transformers, and removes the reactive power penalty from the bill.

The correction formula

Where PF₁ is the existing power factor and PF₂ is the target. The tangent of the arccosine of a power factor gives the ratio of reactive power to real power at that operating point, so the difference between the two tangents is the reactive power that has to be supplied locally.

Current is inversely proportional to power factor at constant real power, so the line current falls in the ratio PF₁ ÷ PF₂. Improving from 0.75 to 0.95 reduces current by about 21% — and because losses rise with the square of current, resistive losses in the feeder fall by roughly 38%.

Two worked examples

1. A 100 kW motor load at 0.75 improving to 0.95

The reactive ratio at 0.75 is 0.8819 and at 0.95 is 0.3287, giving a difference of 0.5532. The required correction is 100 × 0.5532 = 55.3 kVAr, so a 60 kVAr bank is the nearest standard size above the calculation. On a 480 V three-phase supply the line current falls from about 160 A to 127 A, and the apparent power falls from 133 kVA to 105 kVA — releasing 28 kVA of transformer capacity that can be used for additional load.

2. A small workshop at 0.82 improving to 0.95

A 30 kW connected load. The reactive ratios are 0.6980 and 0.3287, a difference of 0.3693, so 11.1 kVAr is required — a 12.5 kVAr standard bank. The current falls by 13.7%. At this scale the saving is modest in absolute terms, but if the tariff includes a reactive charge the payback is usually measured in months rather than years.

Choosing a target, and when to stop

• 0.95 is the usual target. It satisfies most utility minimums and captures about three quarters of the available current reduction. Going to 0.99 captures only a little more and costs a great deal more in capacitance.
• Do not correct above 0.98. Over-correction makes the installation appear capacitive. At light load this can raise the voltage, and the combination of capacitors with variable frequency drives, rectifiers or transformers creates a risk of harmonic resonance — where a harmonic current is amplified far beyond its source magnitude.
• Detuned reactors are needed where harmonics are present. Installing plain capacitors on a supply feeding drives without detuning is a common and expensive mistake. If more than about 15% of the load is non-linear, specify detuned banks.
• Capacitors need their own protection. Fuses and a discharge resistor are required, and capacitors must discharge before being re-energised. Verify against the manufacturer's instructions and the local code.

Alternatives to capacitors

Capacitors are the cheapest correction for a stable load. Other options suit different cases:

• Synchronous condensers — a synchronous motor running unloaded can be over-excited to supply reactive power. Suits very large installations and is immune to harmonic resonance.
• Active power factor correction — power electronics that continuously adjust to hold the power factor near unity. Built into modern high-quality drive front ends, and the right answer when the load varies rapidly.
• Replacing oversized motors — a motor running at 25% load has a far worse power factor than one running at 75%. Sometimes the cheapest correction is a correctly sized motor.

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