KVAR needed to correct PF to 0.95

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Discussion Overview

The discussion revolves around calculating the reactive power (kVAR) needed from a capacitor bank to correct the power factor (PF) of a 3-phase load from 0.89 lagging to 0.95 lagging. The scope includes mathematical reasoning and technical explanation related to power factor correction in electrical engineering.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the total kVAR needed to achieve a PF of 0.95, resulting in 2958.14 kVAR.
  • Another participant questions the interpretation of the problem, suggesting that it may require the additional kVAR needed rather than the total kVAR.
  • A different participant clarifies that the capacitor bank produces reactive power, which compensates for the inductive load, thus improving the power factor.
  • One participant calculates the kVAR at 0.89 PF as 4610 and at 0.95 PF as 2958, proposing that the difference (1652 kVAR) represents the kVAR supplied by the capacitor bank.
  • Another participant agrees with the calculation of the kVAR supplied by the capacitor bank as 1652 kVAR, reinforcing the idea that it effectively cancels out some kVAR to raise the power factor.

Areas of Agreement / Disagreement

There is some agreement on the calculations presented, particularly regarding the kVAR values at different power factors. However, there is disagreement on whether the question asks for total kVAR or additional kVAR needed for correction.

Contextual Notes

Participants have not fully resolved the interpretation of the original question regarding whether it seeks total kVAR or the additional kVAR required for correction. The calculations depend on the assumptions made about the nature of the power factor correction.

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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