# Capacitor Banks - calculating series/parallel kVAR



## stinkycheese (Apr 11, 2012)

There are a few questions from Complex Imaginary that ask about capacitor banks. To pick one (4-70), a 3P motor needs its PF improved by 3360 kVARs, using 140 kVAR units. The answer seems to be a straightforward 1120 kVAR/phase, or 8 units per phase. What I'm confused about is that CI claims that for three-phase banks, whether the units are in series or parallel, you just add the kVARs. Is this correct? AFAIK a capacitor's kVAR rating is for a spec'd supply voltage, and that kVAR rating varies with the square of voltage. So if you put 8 of those units in series, each unit will have its kVAR rating cut by a factor of 64-- and 8 140kVAR units become a 17.5 kVAR equivalent. Whereas 8 in parallel at rated voltage would give you the 1120kVAR/phase you're looking for. Is CI assuming that if you're putting 8 in series, you'd pick units rated for 140kVAR at Vline/8?

Thanks!


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## stbtigerr (Apr 11, 2012)

I don't have this volume. Can you give a little more info about their solution? Is it 8 per phase? Do they give you the operating voltage and rated voltage?


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## stinkycheese (Apr 11, 2012)

The question asks- for a three-phase motor requiring 3360kVAR in PF correction, you have 140 kVAR units. How many would you need per phase, if the units are to be connected in series?

CI's answer is for 8 per phase. They say in the solution that series/parallel combinations don't matter, you just add the total kVAR. No mention of the individual units' kVAR.

If you have the other books, 1-57, 2-17, and 3-72 are similar.


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## stbtigerr (Apr 11, 2012)

I don't have a good answer. The only thing I can think of is that they're specifiying kVar vaules instead of farad values of the capacitors, so if you have 140kVar units connected in parallel, you get that kVar value at full voltage. If you connect 140kVar units in series, you get that kVar value at the divided voltage amongst the other units but either way, you get the same total kVar. The solution explantion may be misleading, because I don't see how the exact same units could be connected in series or parallel and still have the same total kVar value. Hopefully someone can confirm/explain better.


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## cos90 (Aug 1, 2017)

Anybody have a good proof for this? Adding kvar of a capacitor bank in series / parallel?


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## Messi (Mar 16, 2019)

Does anyone have any idea why we are adding KVARs with both series / parallel cap bank connections ?


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## roy167 (May 16, 2019)

stinkycheese said:


> The question asks- for a three-phase motor requiring 3360kVAR in PF correction, you have 140 kVAR units. How many would you need per phase, if the units are to be connected in series?
> 
> CI's answer is for 8 per phase. They say in the solution that series/parallel combinations don't matter, you just add the total kVAR. No mention of the individual units' kVAR.
> 
> If you have the other books, 1-57, 2-17, and 3-72 are similar.


I Can see how it can be confusing. You are correct about series/parallel capacitors are not the same. KVAR = V^2/Z(c), voltage is not the same if connected in Y or Delta. 

The problem has already given you that 3360  total KVAR is required for PF correction. That means per phase would be 3360/3 whether series or parallel. Then divide by 140 KVAR = 8 units

There is difference in Capacitance(ohms) and Capacitor (micro farad)

It is like they have given 3 phase KVA or KW and asking to find out what is per phase power in Y or Delta. It's going to be three phase quantity/3 whether Y or Delta.


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