Series Capacitor Application

[kartracer087]

Hi,

Say that my system voltage is 13,800Y/7,970V.

If one was to connect a capacitor that is nameplate rated 100kVAR 7,970V in series with a 50kVAR unit with nameplate rating of 7,970V, would these two series capacitors be able to provide me with roughly 33.3kVAR per the series capacitor rule? Will the actual kVAR of the series assembly be reduced from the calculated series rating of 33.33kVAR because the voltage across each individual unit is less than nameplate of 7,970V? Not sure how that works.

In this case would it be required to spec one unit at roughly 5,340V and the second unit at roughly 2,630V (this would be the approximate voltage split across each capacitor)? Or can the capacitors both be specified as 7,970V rated and you'll still get roughly the nameplate kVAR out of the series combined units?

Thanks,

 

[kartracer087]

I think this is correct, that specifying the 7,970V units is proper.

So basically the equivalent capacitance of 33kVAR at 7,970V is 1.392uF.

If I take my base voltage of 7,970V and 50 kVAR I get 2.09uF, and similarly if I use a 100kVAR I get 4.18uF.

Series capacitor ratings are like two parallel resistors, so the formula is the same except (C1*C2)/(C1+C2) instead of R1 and R2, thus

(2.09*4.18)/(2.09+4.18) = 1.392uF.

So this is exactly 33kVAR. The total voltage drop across both capacitances amounts to the total kVAR. To maintain the same relationship, this means the voltage across each capacitor is less than rated, hence the reason why the series connection is less. The actual capactance "seen" across each would be:

5,312V across the 50kVAR/7,970V rated capacitor = 22.22kVAR
2,656V across the 100kVAR/7,970V rated capacitor = 11.11kVAR

33.33kVAR total (reactive power across each capacitor sums in series capacitors).

Obviously this entire discussion is considering only one phase connected in wye. The other two phases could be identical for a combined bank rating of 100kVAR in this configuration.