# NCEES #131



## danderson (Sep 11, 2012)

3 phase capacitor rated at 240V and 110 kvar is operated at 3 phase 208V. What reactive power provided at will be most nearly?

Ok, the answer is

kvar = (208V/240V)^2 * 110kvar = 82.6

I see what they did. It makes good sense, but where is this relationship come from? I know I have seen this, but I just can't put it together.


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## power62 (Sep 11, 2012)

KVAR = (motor Voltage/Capacitor Volt)^2 X (Capacitor kVAR)


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## danderson (Sep 12, 2012)

Thank you. Do you have a reference for that, or did you just know it? I understand it, but I'd like to read over some reference material for it.

THNX


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## Flyer_PE (Sep 12, 2012)

The capacitor rating is given in terms of kVAR but it's still just a passive component with an impedance of Z ohms. Power is proportional to V2.


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## Lielec11 (Sep 18, 2012)

I am also interested in the theory behind this. I can't seem to find anything on this topic, can someone please point me in the right direction?


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## Lielec11 (Sep 18, 2012)

I am also interested in the theory behind this and I can't seem to find anything on it. Can someone please point me in the right direction?


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## Flyer_PE (Sep 18, 2012)

It's a pretty basic equation: S=V2/Z


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## Lielec11 (Sep 18, 2012)

Right but there is nothing related to apparent power or impedance in this equation, that is what confused me.


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## danderson (Sep 19, 2012)

I think what you have to understand is the relationship. It's not a fundamental equation. You just have to understand that the KVAR proportional to the Voltage squared. Therefore, if the voltage goes down, the KVAR goes down proportional to the percentage change squared. If it goes up, it would go up in the same way. This is ignoring effects of increasing voltage past rating, of course.


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## Lielec11 (Sep 19, 2012)

Thanks for the clarifications, Flyer and danderson.


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## ruffryder (Oct 19, 2012)

Another way to do this problem, though longer, is to find out the actual Xc using Xc = (Vold^2)/Qold. Now that you have Xc you can calculate the new Q value based on the new voltage Qnew = (Vnew^2)/Xc.

The formula they show in the answer sheet is a combination of the two equations. The combination and solving for the new Q = [(Vnew^2)/(Vold^2)]*Qold.

I think it is very important to go back to basic equations to determine where the shorter direct equations that are in the solutions come from. That way you understand the theory that question is based on.

I hope this helps.


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## Insaf (Oct 20, 2012)

One way to understand: var1= V1^2/Xc and var2= V2^2/Xc, so new var2= (V2/V1)^2 x var1

Another way consider as base change problem: KVARnew= KVARgiven (Vnew/Vgiven)^2= (110)x(208/240)^2=82.62 KVAR

Thanks


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