Electrical Power Capacitor Question

[jakesaround]

I am taking the Electrical Power exam in October and I have been studying a bit.

Here is a question/concept that I need some help on.

If I have a 3 phase capacitor rated at 13,800 volts and 600kVar and I apply it to a 3 phase system rated 12,470 volts how many vars will be applied to this system?

I saw this concept in the Electrical Power practice questions from NEECES but I didn't follow their logic on the explanation.

Can someone give me a common sense approach/solution to this problem?

Thanks.

 

[cdcengineer]

Let me try...

Divide 12470 by 13800. That's the % you than multiply time 600kvar.

Does that match the answer?

 

[cableguy]

If I recall correctly, it varies according to the square of the voltage ratio. (V1/V2)^2, instead of just V1/V2.

I have a formula here from an industrial power book, it gives:

KVAR = 2*pi*f*C*(kV^2)/1000 where f=freq, C=capacitance in uF

also KVAR = 1000 (kV^2) / Xc where Xc is reactance in ohms

So from those formulas, you can surmise that the KVAR varies according to the square of the voltage - hence, the square of the voltage ratio.

 

[jakesaround]

The solution according to the study guide is as follows;

A: The Vars will vary as the square of the applied voltage.

System Voltage

KVAR = (-----------------------)^2 * (kVAR)

Capacitor Voltage

12,470 volts

KVAR = (-----------------)^2 * (600kVAR)

13,800 volts

KVAR = 489.92 kVAR

The math is simple enough I was just wondering if anyone could provide some insight into how to derive this relationship.

Thanks for your help.

 

[Flyer_PE]

No matter what the ratings are, a capacitor bank is simply an impedance. Rather than giving you the capacitor size in Farads or as an impedance value at 60 Hz, they give you the numbers in kVAR at a given voltage. If you consider any fixed impedance, the power dissipated varies by the square of the voltage (S=V2/Z).

Hopefully, I didn't just make it worse.

 

[jakesaround]

Thanks all.

That helps alot.