# power factor correction



## Rei (Apr 11, 2010)

The question is "find the per-phase capacitance needed to change the power factor to 1.0."




I'm too lazy to post up the entire solution, but basically the total 3-phase reactive power needed to correct power factor to 1 is 57.8kVAR. The solution then divide it by 3 since we are looking for per phase capacitance value.

I understand the solution, but I assume the capacitance for the power factor correction is placed on the transformer secondary (between the transformer and the motor loads) which mean it's a wye connection. Therefore, the capacitance calculation on the last line of the solution should taking into account the voltage divided by square root of 3, right? The solution just used V^2 and not (V/1.73)^2.


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## pelaw (Apr 11, 2010)

1. the capacitance is in parallel with the 2 motors and is connected to the same bus motors are connected to.

2. From the summation of motor powers, the Q = 180 kVARs.

3. Since the capacitor is in parrellel with motors it has the same voltage applie, V = 6900 V.

4. The capacitor's impedance Z equals to jXc because there is no R.

5. The power in the capacitor is: S = all Q = V^2/Z = V^2/jXc

6. The power supplied to each phase: S ph = S total / 3; similarly, Q ph = Q total / 3; So Q ph = 180 kVARs / 3 = 60 kVARs

6. From there, Z = jXc = V2/S = 6900^2/60000 = j793

7. Xc = 1 / (2 pi 60Hz * C)

8. C = 1 / 2 pi 60 * 793.5 = 0.00 00 03 343 = 3.34 * 10^-6


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## pelaw (Apr 11, 2010)

I think you are right. The C should be 10 uF. They divided by 3 unnecessarily.


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## DK PE (Apr 12, 2010)

I agree with the 3.34 uF answer you calculated. As stated, you need a total of 180 kVAR of reactace or 180/3 per phase. Without any additional information, I would assume the capacitors are connected line-line and use the 6900 volts on that side. Then you need 793 ohms of reactance PER PHASE and your math is correct.

Just because the transformer secondary is connected in wye doesn't mean the loads are connected in that manner.


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## pelaw (Apr 12, 2010)

Yes, I have to back off of my earlier assertion.

Total = 180,000 kVARs = 6900^2 / Z phase

Per phase: (1/3) * 180,000 kVARs = (1/3) 6900 ^2 / Z phase

60,000 kVARs = (6900/sqrt 3)^2/ Z phase

Z phase = (6900/sqrt 3)^2/60,000

Z phase = 3984^/60,000 = j264

C = 1/2 pi 60 * 264 = 1 uF =C

(BTW: S total = sqrt 3 * V line * I line = sqrt 3 * V line * [(V line/sqrt 3) / Z phase] = V line ^2/ Z phase)


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## DK PE (Apr 12, 2010)

Sorry PElaw, but I don't agree with either your 1uF or 10 uF answers. The TOTAL reactive power you require is 180kVAR and this has to be supplied by three capacitors or 60kVAR per capacitor. PF correction is typically corrected in delta so the voltage across each capacitor is 6900 volts. From there you have the solution you wrote earlier in stpes 6/7, i.e. what value of capacitor with 6900 volts across it creates a capacitive reactance of 793 ohms --&gt; which can be provided by a 3.34 uF capacitor. You need three of them @ this value.


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## Bluekayak (Apr 12, 2010)

DK PE said:


> Sorry PElaw, but I don't agree with either your 1uF or 10 uF answers. The TOTAL reactive power you require is 180kVAR and this has to be supplied by three capacitors or 60kVAR per capacitor. PF correction is typically corrected in delta so the voltage across each capacitor is 6900 volts. From there you have the solution you wrote earlier in stpes 6/7, i.e. what value of capacitor with 6900 volts across it creates a capacitive reactance of 793 ohms --&gt; which can be provided by a 3.34 uF capacitor. You need three of them @ this value.


I must be missing part of the problem statement because without efficiency data, I would assume 1kVA = 1hp (see IEEE std 399-1997, the "Brown Book", Table 7-3) for the 1,200hp induction motor. Obviously kW indicates input power (at least in the US) for the synchronous motor. Thus the complex powers for the motors would be:

Sim = 1200kVA&lt;31.8°

Ssm = 500/0.8kVA&lt;-36.9° = 625kVA&lt;-36.9°

Real components:

Pim = 1200x0.85 = 1020kW

Psm = 500kW

Reactive components:

Qim = 1200sin(31.8) = 632kVAR

Qsm = 625sin(-36.9) = -375kVAR

Qt = 257kVAR, thus 257kVAR of reactive power required to correct pf to unity.

Q3-phase = (2pi60)(Cp-n)(VLL^2)

I agree with everything else DK just said.


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## Flyer_PE (Apr 12, 2010)

Bluekayak said:


> I must be missing part of the problem statement because without efficiency data, I would assume 1kVA = 1hp (see IEEE std 399-1997, the "Brown Book", Table 7-3) for the 1,200hp induction motor. Obviously kW indicates input power (at least in the US) for the synchronous motor. Thus the complex powers for the motors would be:


The 1kVA/HP assumption works if you don't have efficiency or power factor. However, for this problem, you have the power factor.

1HP = 0.746 kW

The kW value for the motor is 1200 HP * 0.746 kW/1HP = 895 kW

The kVA value is then 895 kW / 0.85 = 1053 kVA


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## Flyer_PE (Apr 12, 2010)

^Time to fall on my sword on that one. The way the answer key answers that question, they are applying the 1200 hp number as PIN rather than POut. They didn't apply an efficiency value. They probably should have.


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## DK PE (Apr 12, 2010)

I was trying to help the original poster out with the "per phase and sqrt(3)" issues that seemed to be coming up. Thinking about it in more detail now, I agree an assumption really needs to be made regarding the motor's efficiency since the real power input is greater than 746W/HP and one has to assume 1200 HP output on the shaft. I think the 1kVA/HP incorporates both efficiency and power factor as Flyer stated and so you really can't apply that and use the given power factor. To really get the correct answer, it seems like you would have to assume an efficiency of (maybe 90% on this size machine), then use the power factor given to get kVA. You arrive then in the neighborhood of 1kVA/HP and that would give a slightly different answer than the 3.34 uF I provided earlier.

Edit: crossed posts with Flyer's... I agree poorly worded problem and given solution seems to apply Power input which is unconventional


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## pelaw (Apr 12, 2010)

DK PE said:


> Sorry PElaw, but I don't agree with either your 1uF or 10 uF answers. The TOTAL reactive power you require is 180kVAR


True. correction is required for the total 180 kVARs. The point is that whether you pursue 3phase or single phase formula, both use Z phase, and Z phase is what is needed to get the capacitor value.

In the 3 phase formula, S total = V line ^ 2 / Z phase; in the per phase formula S phase = (V line/sqrt 3) ^ 2 / Z phase.

In either case Z phase = [6900 ^ 2 / 180,000 VARs] = (6900/1.732)^2 / 60,000 VARs = j264 = Z phase.

As far as efficiency, I don't see any need to know the efficiency to solve this problem correctly.


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## DK PE (Apr 12, 2010)

pelaw said:


> DK PE said:
> 
> 
> > Sorry PElaw, but I don't agree with either your 1uF or 10 uF answers. The TOTAL reactive power you require is 180kVAR
> ...


Not trying to beat a dead horse and if you understand your terminology I'm fine but my point is you need 793 ohms per capacitor, not 264. If you are stating Zphase of 264 ohms, I believe that is incorrect.


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## pelaw (Apr 12, 2010)

I see. No problem.

So the answer to the question is: since motors do not have a neutral, capacitors are connected accross the lines of wye connected motor load. Therefore capacitors are connected in delta. Therefore, there is no sqrt 3. Got it.


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## DK PE (Apr 12, 2010)

You got it ... but again, there is no statement that the motor is connected in wye... in fact the motor windings are likely connected in delta. Don't rush to the conclusion the motor is a wye load just because the secondary of the Txfmr is connected wye.


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## DK PE (Apr 13, 2010)

pelaw said:


> As far as efficiency, I don't see any need to know the efficiency to solve this problem correctly.


Again, not beating a dead horse but if this problem was stated differently such that 1200HP was the shaft output horsepower (which is "normal" and is what you get when you purchase a 1200HP motor), then you would need to compute the input real power using the efficiency. Once you have the input real power and power factor, you can determine the reactive power you need to correct.

From the solution given, they assumed the 1200HP is real input power which is not normal, in my opinion.


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