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:-

http://electrialstandards.blogspot.in/2015/05/capacitor-in-series-and-parallel.html

Visit for Capacitors in Series and Parallel:-

http://electrialstandards.blogspot.in/2015/05/capacitor-in-series-and-parallel.html