Answer Posted / karuppiah

Let total power is 1500kw

assume
cos Q1 =power factor before improvement =0.75
cos Q2= poer factor after improvement =0.98
tan1=0.88{cos Q1=0.75,Cos-0.75=46.01degree,Now angle is
46.01,so tan(46.01)=0.88}

tan2=0.2

and
kvar 1= power*tan1=1500*0.88=1320kvar
kvar 2=power*tan2=1500*0.2=300kvar

now, required rating of capacitor bank =kvar1-kvar2
=1320-300
=1020kvar
Capacitor Bank should connect parallel with the load.
Thanks
Karups

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