my company total load is 1000kw, 415volt, all load are inductive load. incomin

my company total load is 1000kw, 415volt, all load are
inductive load. incoming powerfactor is .8. we want to
have0.95 pf connecting capacitors. so now how much rating of
capacitors we have to connect? show how?

Question Posted / jay

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my company total load is 1000kw, 415volt, all load are inductive load. incoming powerfactor is .8...

Answer / s. kotiyal

Sir u need 6 capacitor pannel
of 25 kvar.

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my company total load is 1000kw, 415volt, all load are inductive load. incoming powerfactor is .8...

Answer / rman

currently, the load is 0.75MVAR operating at 0.8pf. to enhance it to 0.95pf, the MVAR must be decreased to 0.329MVAR.
Q=0.75-0.329=0.36MVAR
0.36MVAR must be your operating reactive power. therefore your capacitance must be {0.36M/[0.36M/(415*0.95*sqrt(3))]^2*sqrt(3)} which will be your reactance at 0.0969 per phase. also to find the capacitor to be inserted per phase,assuming your frequency is 60Hz, C=(2*pi*60*0.0969)^(-1)
therefore the C=27.37mF

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my company total load is 1000kw, 415volt, all load are inductive load. incoming powerfactor is .8...

Answer / rsundareee

your original p.f is 0.8(Tan 0.8) and you required p.f is
0.95(Tan 0.95).Required p.f = kW(Tan0.8 - Tan 0.95),when
you select the breaker rating of Capacitor bank you have to
conside the Hormonics (33%),For exambe breaker rating will
be 100A it shall be (100+33=133) 160A MCCB.

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