Answer / s. kotiyal

Sir u need 6 capacitor pannel
of 25 kvar.

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

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.

A TRANSFORMER 440/400V,DELTA/STAR CONNECTED.THE NEUTRAL CONNECTED WITH GROUND CABLE THAT IS DIRECTLY CONNECTED TO THE PANEL,FROM PANEL IT IS CONNECTED EARTH PIT.IS IT RIGHT? IF IT IS RIGHT LEAKAGE CURRENT AFFECTED.SECONDATY IS PROTECTED WITH EARTH LEAKAGE RELAY.

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