as pf not giving ,taking pf 0.8
let pf of load =0.8
kw of load= 65 hp=65*0.746=48.5kw
if we want to improve pf to 0.95 then
improved kva= 48.5/0.95=51.052
improved kvar= sqrt(51.052^2-48.5^2)=15.9411
requard capacitor range = 36.375-15.94=20.43kvar.
first we should calculate the load(inductive or capacitive
in case of resistive load-----
in case of resistive load the, actual power factor will be
01(maximum),in this case the voltage will linior with
respecte to load current.
so tatal resistive load is(in example 65 hp),
so total load in kw = 65*0.746
As we know
Power factor = KW/KVA
for 65 Hp load(if pure resistive)
Considering Preent P.F=1
Target pf =0.99
Multiplying Factor by standard table= 1.56
Kvar= 48.49 KW * 1.56= 75.64
hence for the load of 65hp, capacitor range will be 75.64
Kvar.(in pure resistor).