Can anybody teach me how to do calculation for Power Factor Correction in Electical engineering?

Answer 1

Just remember that

(Watts)2 + (VAr)2 = (VA)2

This formula links the actual power in watts to the volt-amps which is the volts times the amps, and the bit left over is the VAr or volt-amps reactive.

For example you have a 250 v load of 400 watts that draws 2 amps. The

power factor is 0.8 because

VA = 250 x 2 = 500, and 0.8 x 500 gives the power of 400 watts.

The VAr formula is:

(VAr)2 = (VA)2 - (Watts)2

so calculate 5002 - 4002 and the answer is 90000 and take the square root. That means that there are 300 VArs that need to be got rid of.

So if we put a capacitor of 300 VAr across the load, that will take out the VArs and the overall current taken drops to 1.6 amps, which will make you one of the supply company's star customers. Actually they don't mind with a 400 watt load but if it was 400 kW they would be pleased.

300 VArs means that the capacitor must draw 300/250 amps or 1.2 amps and the formula for current in a capacitor is 2.pi x freq x C x V so that the capacitance must be 1.2 / (2.pi x 50 x 250) for a 50 Hz system so the capacance is 0.0000153 Farads or 15.3 microfarads. On a 60 Hz system it would be less, 12.7 microfarads.

The above assumes that the load has a 'lagging' power factor, because the phase of the current lags behind the voltage, which is usual for induction motors and many other types of load. If the load already had a leading power factor you would have to put an inductor on to correct it, but that would be exceptional and the supply company might even be happy to let you continue without PF correction because you would be putting 'leading' VArs into the system.

Answer

Power-factor capacitor banks are expressed in reactive volt amperes (var), not in terms of their capacitance (farads). The capacitance of power-factor capacitor banks are of academic interest only, because they are always labelled in reactive volt amperes. To determine the necessary reactive volt amperes (of the capacitor bank) you do the following.

Step 1: Using the existing power factor, determine the existing reactive power of the load.

Step 2: Using the desired power factor, determine the required reactive power of the load.

Step 3: Subtract the required reactive power (step 2) from the existing reactive power (step 1). and the answer will give you the reactive power of the necessary capacitor bank.