KVAR needed to correct PF to 0.95

  • Thread starter Derill03
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In summary, the conversation discusses the calculation of total reactive power supplied by a 3-phase capacitor bank in order to increase the power factor from 0.89 to 0.95. The correct method is to calculate the kVARs at each power factor and then determine the difference, which in this case is 1652 kVAR. The capacitor bank will effectively cancel out some kVAR to raise the power factor.
  • #1
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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  • #2
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 ).
 
  • #3
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?
 
  • #4
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:
 
  • #5
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.
 
  • #6
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.
 

FAQ: KVAR needed to correct PF to 0.95

What is KVAR needed to correct power factor (PF) to 0.95?

KVAR is the unit of measurement for reactive power, which is the power required to maintain an electromagnetic field in an alternating current (AC) circuit. In order to correct the power factor to 0.95, a certain amount of KVAR is needed to counterbalance the reactive power in the circuit.

Why is it important to correct PF to 0.95?

Correcting the power factor to 0.95 is important because it ensures maximum efficiency in an AC circuit. A low power factor can result in increased energy consumption and higher electricity bills. Additionally, a low power factor can cause voltage drops and fluctuations, which can damage equipment and decrease its lifespan.

How is the amount of KVAR needed calculated?

The amount of KVAR needed to correct PF to 0.95 can be calculated using the following formula: KVAR = kW * tan(cos^-1(PF) - cos^-1(0.95)), where kW is the real power in kilowatts and PF is the current power factor.

Can KVAR be too high?

Yes, KVAR can be too high. This can occur when the reactive power in the circuit is overcompensated, resulting in a power factor that is higher than 0.95. In this case, the electricity bills may increase due to the excess reactive power being supplied.

How can KVAR be adjusted to correct PF to 0.95?

KVAR can be adjusted by installing power factor correction devices, such as capacitors, in the circuit. These devices absorb the excess reactive power and improve the power factor to 0.95. It is important to consult a professional to determine the correct amount of KVAR needed and the appropriate placement of the correction devices.

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