# KVAR and Capacitance



## eng787 (Sep 30, 2010)

Problem 1:

You're given a 3 phase induction motor that is running at 480 volts and it's reactive component is 57,790 vars. What is the PER PHASE capacitance (farads) needed to correct the power factor to 1.0? (if you need it, current total power output is 109,705 VA at a .85 power factor)

Is formula used in Kaplan's problem 2 for capaciatance=Q/2*PI*f* V-phase^2 is wrong. I used this formula and found the same result given by PE_Flyer as follow.

Problem 1:

VPhase = 480/sqrt(3) = 277 V

VAR/Phase = 57.79 kVAR/3 = 19.26 kVAR

Z=V2/S = 2772/19.26k = 3.98 ohms

Z = 1/(377*C)

C = 1/(377*Z) = 1/(377*3.98) = 666 x 10-6 Farads.

Also in privious thread :

Problem 2:

You want to correct a 1406 kVA .99 power factor load. I calculated ~180 kVAR oughta do it. Voltage is 6900 Y. What is the PER PHASE capacitance (farads) to correct it to 1.0?

I think 180 KVAR are wrong. Is not it should be 198.34 if we are calculating from 1406 KVA and .99 P.F.

Any help is appreciated. Thanks


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## cocky (Sep 30, 2010)

problem 1) Thats the right solution you got there. The new power factor is 1.00, so capacitor should provide all the necessary KVAR.

problem 2) This is same as the other one, the per hase KVAR that I noted down is 66.11.



Baljit Gill said:


> Problem 1:
> You're given a 3 phase induction motor that is running at 480 volts and it's reactive component is 57,790 vars. What is the PER PHASE capacitance (farads) needed to correct the power factor to 1.0? (if you need it, current total power output is 109,705 VA at a .85 power factor)
> 
> Is formula used in Kaplan's problem 2 for capaciatance=Q/2*PI*f* V-phase^2 is wrong. I used this formula and found the same result given by PE_Flyer as follow.
> ...


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## eng787 (Sep 30, 2010)

do the way kaplan problem is solved is correct. It follows like this:

Capcitance= Q_Per_phase/ 377* V_phase^2

= 19.26 kvar/377*277^2

=666 x 10-6 Farads

It hase same result


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## cableguy (Sep 30, 2010)

I get the same result (666) if I calculate for a Y connection, but it's 222 uF if connected in a Delta. That's something that annoys me about these questions; this one doesn't state whether you're connecting the caps in a Y or delta. On a motor, don't you normally connect them in a Delta? So if you use the Kaplan formula, you get the answer wrong. I think this is a bad question, and could be thrown out by NCEES were it on an exam. They need to specifically state Y or Delta. The multiple choice actually includes both 666 and 222 as options; both answers are technically correct - both can set the PF to 1.0. So it's a bad question.

I had this same problem in my Testmasters book, and yes, it threw me for a loop too. They did not state Y or delta.

My thread:

http://engineerboards.com/index.php?showtopic=13362

Has the procedures for both Y and delta calculation at the bottom of the thread.


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## Flyer_PE (Sep 30, 2010)

^I would tend to agree that they should tell you whether you are connecting wye or delta. I think delta is more common, but either is possible.


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