Capacitor range is actually given in Kvar, so to find that
range in kvar

As we know
Power factor = KW/KVA
for 65 Hp load
KW= 65HP*0.746 = 48.49KW
Considering Preent P.F=0.50
Target pf =0.99
Multiplying Factor by standard table= 1.59
Kvar= 48.49 KW * 1.59 = 77.09 Kvar

hence for the load of 65hp, capacitor range will be 77.09
Kvar.

as pf not giving ,taking pf 0.8
let pf of load =0.8
kw of load= 65 hp=65*0.746=48.5kw
kva =48.5/0.8=60.625kva
kvar =sqrt(60.625^2-48.5^2)=36.375kvar
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
or resistive).
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
=48.49.
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).

hi can you give any body fault level calculation, generator
fault level calculation,transformer fault level calculation
for system fault level with equation? i have 3.3kv
generation voltage three gas gen sets, one is 1.9MW,2nd is
2.5MW,& 3rd is 3.8MW & i have 11 kv generation voltage one
gas gen sets is 3.8MW, we are step up to 22kv by transformer

What is main cause of cable's terminal point burned in Lug
on high temperature although lug and cable selection is ok
as per stantard and also tightness is ok as per situation.
so please take in deep and tell correct answer thanku.