# 3 phase power factor correction -> Capacitance



## R2KBA (Oct 14, 2011)

Ok, so p.f. correction is supposed to be a gimmie question, and I thought I had this nailed down, but apparently not. From what I can tell, the NCEES problems stop once you determine the VAR necessary to raise the p.f. I found a problem that wants the total capacitance. It is Camara sample exam 1 problem 10. I won't post it b/c I'm not sure about copywrite issues and such.

Basically, Camara and I agree up to the point where we decide we need a total of -150VAR from a 3-phase capacitor. He asks for the total capacitance. The way I have always done it was Q=-2*pi*f*C*V^2. That is "Vars equals 2pi time frequency times capacitance times RMS voltage squared). Simply move everything on the right under the Q and solve for C. Done. But apparently I may be wrong.

Camara says Q = root3*V^2/Xc, solves for Xc and then converts reactance to capacitance. That root3 is what I have a problem with. Ive never seen it before in that formula. Also, the problem didn't specify whether the capacitors would be connected in delta or wye.

I've always used the formula: C = (Qadd)/(2*pi*f*V^2) and always got the problems right. What's the issue here?


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## EEVA PE (Oct 14, 2011)

R2KBA said:


> Ok, so p.f. correction is supposed to be a gimmie question, and I thought I had this nailed down, but apparently not. From what I can tell, the NCEES problems stop once you determine the VAR necessary to raise the p.f. I found a problem that wants the total capacitance. It is Camara sample exam 1 problem 10. I won't post it b/c I'm not sure about copywrite issues and such.
> 
> Basically, Camara and I agree up to the point where we decide we need a total of -150VAR from a 3-phase capacitor. He asks for the total capacitance. The way I have always done it was Q=-2*pi*f*C*V^2. That is "Vars equals 2pi time frequency times capacitance times RMS voltage squared). Simply move everything on the right under the Q and solve for C. Done. But apparently I may be wrong.
> 
> ...



I think the root 3 is due to this being a 3 phase problem. You are thinking single phase.

I have for 3 phase Q = (sq3) V^2 / Xc

I have for single phase Q = V^2 / Xc


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## R2KBA (Oct 14, 2011)

EEVA said:


> R2KBA said:
> 
> 
> > Ok, so p.f. correction is supposed to be a gimmie question, and I thought I had this nailed down, but apparently not. From what I can tell, the NCEES problems stop once you determine the VAR necessary to raise the p.f. I found a problem that wants the total capacitance. It is Camara sample exam 1 problem 10. I won't post it b/c I'm not sure about copywrite issues and such.
> ...


Yeah, I guess you are right. I guess it just looks strange to me. I guess I wanted to see it in a book somewhere. I suppose it can be derived from the formula we are all familiar with: Q=root3(VLL)(I)sin(angle) -&gt; substitute V/X for I and angle is 90degrees which then turns into Q=root3(V^2)/X


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## snerts50 (Oct 14, 2011)

R2KBA said:


> EEVA said:
> 
> 
> > R2KBA said:
> ...


This was great. I'm glad I saw this. I went through my notes and noticed I hadnt accounted for this.


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## Insaf (Oct 14, 2011)

EEVA said:


> R2KBA said:
> 
> 
> > Ok, so p.f. correction is supposed to be a gimmie question, and I thought I had this nailed down, but apparently not. From what I can tell, the NCEES problems stop once you determine the VAR necessary to raise the p.f. I found a problem that wants the total capacitance. It is Camara sample exam 1 problem 10. I won't post it b/c I'm not sure about copywrite issues and such.
> ...


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The voltage you used line to line or line to Neutral? I believe Qph=Q/3 and Vln=V/sqrt(3) can be used to calculate Xc=Vln^2/Qph and finally,C=1/(2*pi*Xc).


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## ElecPwrPEOct11 (Oct 15, 2011)

Is this covered in Camara's Power Reference Manual? I don't think I've seen it before but want to take a good read through it.


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## R2KBA (Oct 20, 2011)

snerts50, I wouldn't be so quick to accept that formula with root3 in it. I am still unsure about it, and I am leaning towards what Insaf has proposed, which is use V^2/X for either case, with V either being line-to-line or line-to-neutral, depending on how it is set up. I have not been able to find any clear examples or derevations to support that formula with root3 in it.


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## EEVA PE (Oct 20, 2011)

I got this formula Q = sqrt3 (V^2)/Xc from the answer solution in Camara's Sample exam#1 prob 10. I am assuming it is correct because I have not seen a problem with errors in Camara's stuff that is not on the errata. I have tried to validate that this formula is correct but have not seen it used in any other text and even searched online for it, so I can prove it is wrong or right. I am just hoping Camara knows his stuff or it does not get asked on the exam.


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## mudpuppy (Oct 21, 2011)

This problem is on my whiteboard at work because the same question came up there. A coworker and I (both PEs) agree with R2KBA if that helps muddy the waters.


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## DK PE (Oct 21, 2011)

I'm throwing my vote with R2KBA/mudpuppy.... the voltage used is whatever voltage is across the capacitor.


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## snerts50 (Oct 21, 2011)

R2KBA said:


> snerts50, I wouldn't be so quick to accept that formula with root3 in it. I am still unsure about it, and I am leaning towards what Insaf has proposed, which is use V^2/X for either case, with V either being line-to-line or line-to-neutral, depending on how it is set up. I have not been able to find any clear examples or derevations to support that formula with root3 in it.


Since the comment by Insaf I've been doing some research myself and I think I agree with him. It makes sense.


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## EEVA PE (Oct 21, 2011)

I gave this some thought, this is how i see it.

If I needed to find the total capacitance to correct the power factor in a 3 phase system i would then:

1. Divide the total QVARS by 3 to get Q per phase

2. Find Vln

3. Calculate the Xc using Xc = Vln^2/Qphase

4. Calculate C=1/wXc per phase.

5. Then I will get 3 capacitors of the value calculated in item 4 above and put it on each phase.

6. Then to answer the question on what my total capacitance is I just add up the 3 capacitor values.

If I had used the other formula with the sqrt3 I would get a value of one capacitor for all three phases, but then how would I hang the capacitor on the 3 phase system? Which phase do i put the capacitor on? So therefore using the sqrt3 leads to a non practical solution?

Is my logic correct? Are there any holes in my thought process above?


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## DK PE (Oct 21, 2011)

EEVA said:


> I gave this some thought, this is how i see it.
> 
> If I needed to find the total capacitance to correct the power factor in a 3 phase system i would then:
> 
> ...


If I'm following your process (at the point I underlined), you appear to be assuming the caps are connected line-neutral when normally, power factor correction is connected in delta, I believe. The capacitor can only provide the Q that results from the voltage impressed across it.


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## R2KBA (Oct 21, 2011)

Not wishing to trash Camara (although I do have to say he LOVES his Wheatstone Bridges), but I believe there may be errors that have not been reported. For example, on exam 1 problem 38 the solution says that option B is true, which says "per unit impedances on T1 (a transformer) differ from primary to secondary side", but then in the solution it states (correctly) that p.u.impedance is the same on both sides. In this case he is conceptually right but the correct answer would be "B and D" since both are false."

I am by no means the ultimate power guru, but I did spend at least an hour yesterday drawing capacitor connections, converting from line-to-line to line-to-neutral, playing around with the commonly known power formulas, and I absolutely could not find a way to get that formula in problem 10. But either way, the NCEES sample exam shows that they tend to ask for VARs and not capacitance, so we should be OK.


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## ElecPwrPEOct11 (Oct 21, 2011)

Hopefully we don't need this for the test but thought I'd contribute with an example I made up.

Say you need a total of -1000 VAR on a 480V/3phase 60Hz system.

For delta:

1) 1000=Q = sqrt(3) * Vline^2 / XC, total

2) Xc= sqrt(3) * Vline^2 / 1000 = 399 across all 3 phases, or = 133 per phase.

3) Use Xc = 1/(w* C) to find the value of C.

4) C = 1.99 * 10^-5 Farads.

To find the capacitance for a wye system you'd just use a different voltage

5) 1000 = 3 * Vphase^2 / Xc,phase

6) Xc,phase = 130

7) Cphase= 1/(w * Xc) = 2.04 * 10^-5 Farads.

Does this look right to everyone? The capacitance will depend on the arrangement of the capacitors.


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## R2KBA (Oct 21, 2011)

I believe the concensus here was that the formula you show in step 1 doesn't make any sense because we can't see how it is derived or where it came from. Let us know if you figure out how to derive it somehow.



ElecPwrPEOct11 said:


> Hopefully we don't need this for the test but thought I'd contribute with an example I made up.
> 
> Say you need a total of -1000 VAR on a 480V/3phase 60Hz system.
> 
> ...


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## mudpuppy (Oct 21, 2011)

FWIW, I've been working for a utility for 10 years and have never seen a delta-connected capacitor bank.


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## ElecPwrPEOct11 (Oct 21, 2011)

OK, so I'll forget I ever knew the formula in step 1. The consensus then is to use the formula in step 5 and solve for the per-phase capacitance?


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## R2KBA (Oct 21, 2011)

Use the formula Q=V^2/X where V is the voltage across one capacitor, which may be line-to-line or line-to-neutral, depending on how it is connected. I would personally assume line-to-line if they don't say how it is connected, but do so at your own risk. The only difference for 1 phase vs 3 phase is a factor of 3.



ElecPwrPEOct11 said:


> OK, so I'll forget I ever knew the formula in step 1. The consensus then is to use the formula in step 5 and solve for the per-phase capacitance?


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## DK PE (Oct 21, 2011)

I thought that for correction for motor power factor the units are usually connected in delta (but they're always spec'd in kVAR anyway, not uF).	For transmission, maybe they are not connected in delta as it would increase the withstand voltage required for the capacitor?

For example, look up the spec for an Eaton Cutler Hammer 1543PCMRA, it will say 3 X 58uF Δ at 480V 60 Hz and it is spec for 15kVAR correction.


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## DK PE (Oct 21, 2011)

here is pic


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## snerts50 (Oct 21, 2011)

DK PE said:


> I thought that for correction for motor power factor the units are usually connected in delta (but they're always spec'd in kVAR anyway, not uF).	For transmission, maybe they are not connected in delta as it would increase the withstand voltage required for the capacitor?
> 
> For example, look up the spec for an Eaton Cutler Hammer 1543PCMRA, it will say 3 X 58uF Δ at 480V 60 Hz and it is spec for 15kVAR correction.


I have seen motor PF correction connected delta.

I worked for a utility for a year and swore I saw it there but cannot remember...


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## snerts50 (Oct 21, 2011)

R2KBA said:


> I believe the concensus here was that the formula you show in step 1 doesn't make any sense because we can't see how it is derived or where it came from. Let us know if you figure out how to derive it somehow.
> 
> 
> 
> ...


I think the formula in step 1 is valid.

What about, Q=Vc * Ic (basic Power equation, usable for every type of power)

and Ic=Vc/Xc

So, Q=Vc^2/Xc

Using the root(3) would depend on if you are using per phase values or line values.


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## EEVA PE (Oct 22, 2011)

R2KBA said:


> Not wishing to trash Camara (although I do have to say he LOVES his Wheatstone Bridges), but I believe there may be errors that have not been reported. For example, on exam 1 problem 38 the solution says that option B is true, which says "per unit impedances on T1 (a transformer) differ from primary to secondary side", but then in the solution it states (correctly) that p.u.impedance is the same on both sides. In this case he is conceptually right but the correct answer would be "B and D" since both are false."
> 
> I am by no means the ultimate power guru, but I did spend at least an hour yesterday drawing capacitor connections, converting from line-to-line to line-to-neutral, playing around with the commonly known power formulas, and I absolutely could not find a way to get that formula in problem 10. But either way, the NCEES sample exam shows that they tend to ask for VARs and not capacitance, so we should be OK.



What is wrong with getting wheatstone bridge problems on the exam, they are easy. I know my blood pressure will drop 5 pts on Friday if the 1st question on the exam is a wheatstone bridge problem. My blood pressure will probably go up 10 pts if it is an op amp problem.


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