Designing a Capacitor Bank For 125kVAR

Why reinvent the wheel (rhetorical question)? It might be better to pick up the telephone and get on to the people that do this sort of thing all the time, perhaps?

KVAR=VI sinθ

I = KAVR/Vsinθ

For capacitive load sinθ=1

I = KAVR/V=125000/V - eq-1

I =V/2*Π*Frequency*Capacitance - eq2

By eq 1 and 2

125000/V =V/2*Π*Frequency*Capacitance

required capacitance= (125000*2*Π*Frequency) /V²

So you need capacitor of (125000*2*Π*Frequency) /V² farads and of V volts.

the eq-2 given should be like this

I = V/Xc = V/(1/2*3.14(pie)*f*C) i.e Xc= (1/2*pie*f*C)

so I = V*2*pie*f*C

Now Go ahead.

thank u

yes thanks for correction. it should be like this-

KVAR=VI sinθ

I = KAVR/Vsinθ

For capacitive load sinθ=1

I = KAVR/V=125000/V - eq-1

I =V*2*Π*Frequency*Capacitance - eq2

By eq 1 and 2

125000/V =V*2*Π*Frequency*Capacitance

required capacitance= (125000*2*Π*Frequency)

So you need capacitor of (125000*2*Π*Frequency) farads and of V volts.

dear Sir,

Please note Xc= V x V/kvar

Hence V xV /Kvar = 1/2phi*f * C

Hence C= kvar *1000/(V X VX 2 Phi Xf) where C in Fd and V in Volts and f in frequency in Hz

Hence the Value of C =Kvar*1000/(v*v*2 phi*f)

Regards

V.Ambarani

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Rakesh you seem to know your stuff about capacitors. Could I liase with you about a special use I have for a capacitor. Maybe you might email me at michael@roof2000.com and I will explain in detail what I am trying to do.