Calculating capacitance from KVAR value

[cableguy]

I'm working some problems and am confused about 2 very similar problems, and can't quite figure out why. My issue is that the solution methods are identical, yet in one they use a mysteriously appearing square root of 3, and the other has no square root of 3. Which one is right, and/or why is that root 3 in there?

Here we go.

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)

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?

Answers:

#1, I got 128x10^-6, their solution worked out like this:

Q=57.8 kVARS, PF=1.0

Qphase=Q/3=19.26 kVARs

Cphase=Qphase=Qc: :wv: ::^2)=[19260]/[377*(480)^2/sqrt(3)]=222x10^-6 Farads

Where did that sqrt 3 come from? I worked it a different way. I did:

1/(2*pi*f*C)=V^2 * sqrt(3)/Q and solved for C, then divided by 3. Maybe I needed to divide by square root of 3 instead of 3, because we're going line to neutral with the caps. I can understand that.

But, #2 solution has no square root of 3.

#2, I got 9.8x10^-6, their solution:

Q=180 KVARs

Qphase=Q/3=60 kVARs

Cphase=Qphase=Qc/: :wv: ::^2)=[60000]/[377*(6900)^2]=3.34x10^-6

What I did was to divide Q/3 first to get 60, and then used my formula above, and instead of line to line voltages, I used line to neutrals in my formula. If you divide my answer by 3, you get their answer (which could be associated with my line to phase conversion).

Since neither of my answers matched, and they are using different formulas to solve pretty much the same exact problem, I don't know who is right. lol.

Thanks for reading. Any insight is appreciated.

Or if anyone has a quick and known good formula for this I can just scribble down in my book, that'd work for me too, lol.

Edit: and what's with my omega-V's becoming the West Virginia symbol? Bwah ha.

 

[Flyer_PE]

Dunno if this helps, but here's what I get:

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.

Problem 2:

VPhase = 6.9kV/sqrt(3) = 3.98 kV

VAR/Phase = 180 kVAR/3 = 60 kVAR

Z=V2/S = 39802/60k = 264.5 ohms

Z = 1/(377*C)

C = 1/(377*Z) = 1/(377*264.5) = 10.03 x 10-6 Farads.

 

[cableguy]

Thanks for the reply. These questions are confusing me big-time. The stupid part is, they shouldn't be that hard! It's my references that are contradicting each other. I need to figure out what the right method is.

I'm working the Kaplan sample exam, and Morning #2 is one of these problems as well. Of course I tried the Testmasters solution method, and got the wrong answer. Kaplan has their own method, and it's straightforward, but it doesn't work for the above Testmasters questions.

The formula they use does not divide the Qtotal by 3; it simply goes:

C = Qtotal/(2*pi*f*Vln^2), where Vln = Voltage line-to-neutral

They do specify a balanced Y format, but they wanted the capacitance value per phase as well.

Basically, you need to convert 1.67 MVAR to a capacitance when you're working with a 13.6kV line voltage. I got 8uF by the Testmasters method, which was nowhere near any of the answer choices.

The Kaplan method doesn't get near any of the Testmaster choices.

I guess I'm going to dig through my other references and see if any of them can explain it.

 

[GabeM]

Answers:#1, I got 128x10^-6, their solution worked out like this:

Q=57.8 kVARS, PF=1.0

Qphase=Q/3=19.26 kVARs

Cphase=Qphase=Qc:: :wv: :::^2)=[19260]/[377*(480)^2/sqrt(3)]=222x10^-6 Farads

Where did that sqrt 3 come from?

I think the square root of three should be in paranthesis with the 480: (480/sqrt(3))^2 not 480^2/sqrt(3). This would make this solution the same as the solution to problem #2. It also makes sense because then the KVArs and volts are both per a single phase.

 

[cableguy]

Well, a coworker loaned me a book that had a very good section on capacitor calcluation in it (Electric Power Distribution System Engineering by Gonen). The 2 Testmaster answers "agree" with the method in this textbook. The textbook provides a method for both Delta and Y capacitor connections; the Testmasters questions result in a delta configuration, though it was not specified in the question. Unfortunately, the Kaplan Practice Exam Morning #2 question looks to be "bad", because it asks for the capacitance per phase with a Y connection - and applying the Gonen method, you get 24 uF (which would be per phase) - but the Kaplan answer is 72 uF (which looks to be 3 phase). I don't trust the answers a lot in the Kaplan practice exam any more, too many errors. The problems are good, but the answers are bad, lol.

Anyhow, if anyone is interested in the method, here it is:

Find the KVAR you want to absorb - Qc

Line current is I

Delta phase current is Ic

For a delta format, each capacitor:

I=Qc / (sqrt3 * Vline)

Ic=I / sqrt3

Xc=Vline/Ic

Capacitance then = 10^6 / (377 * Xc)

For a Y format, each capacitor:

Ic = I = Qc / (sqrt3 * Vline)

Xc = Vline / (sqrt3 * I) [note that we're converting Vline-line to Vline-neutral with the sqrt3 here]

Capacitance = 10^6 / (377 * Xc)

 

[cableguy]

Forgot to mention above, the result given in the above calculations is already in microfarads. So your answer is "8" above, that's 8 microfarads.