Power factor Improvement from 0.8 to 0.96 for 440 Volts, 500 Amps:
Calculate the reactive power consumption of exiting system:
Now consider power factor is 0.8 then lets we calculate the reactive power requirement:
cos〖∅=0.8〗
∅=cos^(-1)0.8
∅=36.6 deg
Reactive Power Drawing from the source=1.732 VI Sin∅
Q=1.732*440*500*sin36.6
Q=1.732*440*500*0.58
Q=220 KVAR
In this condition the load is drawn 220 KVR from the source.
Now consider the power factor is 0. 96.. then calculate the required reactive power
cos〖∅=0.96〗
∅=cos^(-1)0.96
∅=16.2 deg
Reactive Power requirement=1.732 VI Sin∅
Q=1.732*440*500*sin16.2
Q=1.732*440*500*0.27
Q=103 KVAR
Hence the net reactive power compensation to be done is 220-103= 117 KVAR
Hence to improve power factor we should contribute 117 KVAR.
We can add at least four number of 25 KVR capacitor bank to our system.
Read more: techdoct.com
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