KVAR needed to correct PF to 0.95

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SUMMARY

The discussion focuses on calculating the reactive power (kVAR) required to improve the power factor (PF) of a 3-phase load from 0.89 to 0.95. The initial load is 9000 kW, which translates to 9473.68 kVA at the desired PF. The calculations confirm that the total reactive power supplied by the capacitor bank is 1652 kVAR, derived from the difference between the kVAR at 0.89 PF (4610 kVAR) and at 0.95 PF (2958 kVAR). The methodology and final values are validated by participants in the discussion.

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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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