Electrical Design guide Understand a concept

Why a low power factor costs money, and how much correction is needed

Converting kW and power factor into the reactive power to remove, and choosing a defensible target.

The engineering question

The meter records fewer kWh than the equipment consumes. What is the missing current, and how do you remove it?

Worked example

The figures below were produced by running the Power Factor Correction Calculator with the same inputs — during the build, from its authoritative engine.

Results
Capacitor bank required	138.3082 kVAr
Reactive power now	220.4793 kVAr
Reactive power at target	82.171 kVAr
What to install	Install ≈ 138 kVAr of shunt capacitance (a three-phase bank of 46 kVAr per phase) to lift power factor from 0.75 to 0.95.

Reactive power at a given power factor is P·tan(acos(pf)).
The correction required is the difference between the existing and target reactive power.
Install that in kVAr, and consider detuning reactors where harmonics are present.

Sanity check

Correcting to 0.95 then 0.98 should add only a small further bank. If the second step doubles the required kVAr, the target has been misread.

Tools for this problem

These are the EngiMetric calculators that compute the numbers in this topic. Each one shows its formula, its units and its limits rather than returning a bare result.

Power Factor Correction Calculator Calculate the reactive power in kVAr that must be removed with capacitors to reach a target power factor.

Engineering review required

This topic supports professional engineering work; it does not replace a sealed design. The result depends on site data, the governing standard and engineering judgement, and should be confirmed by a competent engineer before it is built.

References & standards

IEEE 141 — reactive power compensation
IEC 60831 — shunt capacitors