How do you calculate capacitance to improve power factor in a industry?

Answer 1

A capacitor used to correct a lagging power factor has to take the right number of VAR (volt-amps reactive).

If you have a load of W watts with a power factor of P, the total VA (volt-amps) is W/P. For example a 400 watt load with a power factor of 0.8 draws 500 VA because 400/0.8=500.

The basic formula is that W2 + (VAR)2 = (VA)2

The VAR comes from this formula:

VAR = sqrt((VA)2 - W2)

So for the 400-watt load with a 0.8 power-factor, which draws 500 VA, the VAR is equal to 300.

On a 120 v supply the capacitor must draw 300/120 amps, or 2.5 amps, and its reactance is therefore 120/2.5 or 48 ohms.

The capacitance comes from this formula: 1 / (2pi times frequency times reactance).

On a 60 Hz system this is 1 / (377 times reactance)

So for a 48 ohm reactance the capacitance is 0.0005526 Farad or 55.3 microfarads.

This would completely correct the power factor in the example and the current drawn would come down from 4.16 amps to 3.33 amps, which would reduce the power lost in the supply wiring by 34%.

Answer 2

You don't normally care what the capacitance of a power-factor capacitor is, as power-factor improvement capacitors are rated according to their reactive power (reactive volt amperes). Their capacitance (in farads) are only of academic interest.

The basic procedure is to determine your load's existing reactive power, then determine the necessary reactive power to achieve the desired power factor. The difference between the two will give you the necessary reactive power of the p.f. improvement capacitor.