Saturday, March 7, 2015

Power factor Improvement formula; Additional KVAR required for Power factor improvment.


Every industry monitor its power factor so that they will not pay for reactive power at lower factor. It become very important to know what additional capacitor bank required for improvement of power factor. Every industry consists of inductive load such as lights and motors which usually have power factor in the range to 0.8 to 0.85. So power factor improvement is of prime importance as if they kept running their system at same power they suffer loss of 20 to 15%. As if they keep power factor at 1 they will not have pay any additional charges at lower power factor.


So let’s calculate what additional capacitor bank needed for power factor improvement. Let’s take an example that system is running at 0.8 power factor and inductive load is 200 KW what additional capacitor bank required for improvement of power factor from 0.8 to 1.


KVAR Required = KW X [tan (cos-1(running P.F)- tan (cos-1(Desired P.F)]


Now KW is 200 KW and running P.F. is 0.8 and desired PF is 1 putting these values into the formula we get:-


KVAR Required= 200 [tan (cos-1(0.8)- tan (cos-1(1)]

KVAR Required= 200 [tan 36.86- tan 0]

KVAR Required= 200X 0.75 = 150 KVAR

150 KVAR is additional capacitor bank required for increasing power factor from 0.8 to 1.

This is simpler way of calculating additional KVAR required for power factor improvement.
Visit for Capacitors in Series and Parallel:-

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