KVAR needed to correct PF to 0.95

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To correct the power factor (PF) from 0.89 to 0.95 for a 9000 kW 3-phase load, the total reactive power supplied by a capacitor bank is calculated. The initial reactive power at 0.89 PF is found to be 4610 kVAR, while at 0.95 PF it is 2958 kVAR. Therefore, the capacitor bank needs to supply 1652 kVAR to achieve the desired PF improvement. This process involves the capacitor bank producing reactive power to offset the inductive load. The calculations confirm the effectiveness of the capacitor bank in raising the power factor.
Derill03
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Hello All,

I need to know if i have done this correctly (its been some time):

Question:

The 3-phase load at a bus is 9000 kW at .89 lagging PF. What is the total reactive power supplied by a 3-phase capacitor bank that will increase the PF to .95 lagging?

Work:

9000kW/0.95 = 9473.68kVA

kVAR = sqrt( (9473.68kVA)^2 - (9000kW)^2 )
kVAR = 2958.14

Thanks in advance.
 
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Yes until now it's correct, but:
Derill03 said:
What is the total reactive power supplied by a 3-phase capacitor bank that will increase the PF to .95 lagging?
( from PF = 0.89 ).
 
I think the way i did this is correct because it is asking for the total kVAR needed to achieve 95% PF, i think you are implying it says what is the additional kVAR needed to achieve 95% which would be difference of kVAR at 95% and kVAR at 89%.

Thoughts?
 
Derill03 said:
The 3-phase load at a bus is 9000 kW at .89 lagging PF. What is the total reactive power supplied by a 3-phase capacitor bank that will increase the PF to .95 lagging?
You must calculate the amount of KVAr's consumed by PF = 0.89. Now the capacitor will compensate/produce an amount of kVAr's, so that the new PF becomes 0.95.
A capacitor is consuming negative reactive power, thus producing reactive power, and thereby increasing an inductive PF.

Thoughts: I'm not in doubt about that. :wink:
 
Ok so then if i understand correctly the following should be true:

kVAR's at 89% = 4610
kVAR's at 95% = 2958

so kVAR's supplied by capacitor bank should be 4610-2958 = 1652 kVAR

This would be true because the capacitor bank will effectively cancel out some kVAR to raise the power factor.
 
Derill03 said:
so kVAR's supplied by capacitor bank should be 4610-2958 = 1652 kVAR

This would be true because the capacitor bank will effectively cancel out some kVAR to raise the power factor.
Yes, that's absolutely correct.
 
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