3 phase power factor correction -> Capacitance

[R2KBA]

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?

 

[EEVA PE]

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?

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

 

[R2KBA]

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?

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

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) -> substitute V/X for I and angle is 90degrees which then turns into Q=root3(V^2)/X

 

[snerts50]

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?

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

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) -> substitute V/X for I and angle is 90degrees which then turns into Q=root3(V^2)/X

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

 

[Insaf]

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?

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

---------------------------------------------------------

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).

 

[ElecPwrPEOct11]

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.

 

[R2KBA]

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.

 

[EEVA PE]

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.

 

[mudpuppy]

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.

 

[DK PE]

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

 

[snerts50]

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.

 

[EEVA PE]

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?

 

[DK PE]

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?

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.

 

[R2KBA]

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.

 

[ElecPwrPEOct11]

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.

 

[R2KBA]

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.

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.

 

[mudpuppy]

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

 

[ElecPwrPEOct11]

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?

 

[R2KBA]

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.

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?

 

[DK PE]

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.