Answer Posted / deepakindi1

Boss what if u dont have the table?

let Cos(@1) be the present PF

@1=cos'(@1) (cos inverse)
let Cos(@2) be the desired PF
@2=cos'(@2) (cos inverse)


KVAR required= KW[ tan(@1)-tan(@2)]

for 65 Hp load
KW= 65HP*0.746 = 48.49KW
Considering Preent P.F=0.50
desired pf =0.99
ie cos(@1)=0.5
@1=cos'(0.5)=60
@2=cos'(.99)=8.1

kvar=48.49*[tan(60)-tan(8.1)]
=48.49*[1.732-.142]
=77.08

kvar=77.08

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