Dear All,
We need to install 125uF at 380Volts, three phase delta
connected capacitor bank parallel to 37KW Motor,
1. The
capacitor bank has to produce 125uF at 380Volts, the equivalent KVAr rating
from the formula KVar=2*pi*f*C*V2 is 5.6 KVar.
2. Due
to the unavailability of the capacitor bank rated at 380Volts,we proposed the
Kvar rating of 7.5Kvar at 440 Volts.
3. On
applying 380Volts to the capacitor rating of (7.5Kvar,440Volts), the capacitor
has to produce the Kvar rating of 5.6Kvar by the equation , Actual
Kvar=Nameplate Kvar x (Applied Voltage)2/(Rated Voltage)2.
What will be the
capacitance output of the capacitor bank
at this condition? Will the capacitance also reduce with the reduction in Kvar?
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