# Homework Help: KVAR needed to correct PF to 0.95

1. Oct 5, 2015

### Derill03

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

2. Oct 5, 2015

### Hesch

Yes until now it's correct, but:
( from PF = 0.89 ).

3. Oct 7, 2015

### Derill03

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. Oct 7, 2015

### Hesch

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.

5. Oct 7, 2015

### Derill03

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. Oct 7, 2015

### Hesch

Yes, that's absolutely correct.