Capacitor Bank Sizing Calculator

Estimate the kvar needed to improve power factor from the present value to the desired target.

This calculator estimates the required correction kvar from a steady kW load. Final capacitor-bank design should also consider harmonic conditions, switching steps, and detuning requirements.

What This Calculator Solves

Poor power factor increases current for the same real power and can make a system look larger than it needs to be. In practical terms, that can mean higher losses, more voltage drop, less spare capacity in transformers and feeders, and utility penalties in some installations. A capacitor bank reduces reactive demand so the same kW load can operate with lower current.

This capacitor bank sizing calculator estimates the kvar needed to move from an existing power factor to a better target value. It is useful for motors, mixed industrial loads, workshop feeders, commercial services, and planning studies where the user wants to know whether a 25 kvar, 50 kvar, or 75 kvar correction bank is in the right range.

The result is based on the standard reactive-power relationship using the load in kilowatts and the angle difference between the present and target power factors. The calculator also shows the approximate line current before and after correction so the effect is easier to understand.

Correction Formula

Required capacitor kvar: Qc = P x (tan phi1 - tan phi2)
Line current before correction: I1 = P / (sqrt(3) x V x PF1)
Line current after correction: I2 = P / (sqrt(3) x V x PF2)

Inputs and Meanings

InputMeaningUnit
LoadReal power demand to be correctedkW
Existing PFPresent operating power factor0 to 1
Target PFDesired improved power factor0 to 1
Line VoltageSupply voltage used for the current comparisonV

Units and Practical Interpretation

The correction result is shown in kvar because capacitor banks are normally selected by reactive power, not by kW. The current comparison is shown in amperes so the user can see how feeder or transformer loading changes when the same real power is delivered at a better power factor. In other words, the kW load stays the same, but the current demand usually falls.

Worked Examples

Example 1: Workshop load correction

Load = 180 kW, existing PF = 0.72, target PF = 0.95

tan phi1 = 0.9647 and tan phi2 = 0.3287

Qc = 180 x (0.9647 - 0.3287) = 114.48 kvar

A designer would then compare that result with practical bank steps such as 100 kvar or 125 kvar depending on switching strategy.

Example 2: Mixed motor feeder

Load = 75 kW, existing PF = 0.8, target PF = 0.95

tan phi1 = 0.75 and tan phi2 = 0.3287

Qc = 75 x (0.75 - 0.3287) = 31.60 kvar

A 30 kvar or 37.5 kvar correction stage could be examined next.

Design Notes

This page estimates how much reactive compensation is needed, but it does not decide whether the bank should be fixed, automatic, detuned, or split into stages. That decision depends on how the load varies over time, whether harmonics are present, and whether the bank will operate close to non-linear equipment such as VFDs or rectifiers.

For that reason, this calculator is best used as a first design step. It tells you the correction range, helps you understand current reduction, and gives a sound base for the next engineering decision. After that, capacitor switching steps, protection, detuning reactors, and harmonics should be checked before final selection.

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