POWER FACTOR CALCULATION

[bgl2010]

Hi,

I have a capacitors installed in the low voltage side of a 750 kVA transformer (12.5 kV/208-120 V). The utility meter is on the high voltage side of the transformer and the monthly average power factor on the bill is under the penalty limit of 0.9. Therefore I made some measurements with a Fluke 1735 to see the power factor over the day.

The data showed that during working hours the reactive power is positive and the total power factor is always between 0.95 and 0.99, but at night and in the early morning the total reactive power is negative and the total power factor is below 0.9. When I compare the total power factor of the measurments with the one I calculete with the active and reactive power, there is a big difference. Here are some value:

PF Total kW Total kVar Total PF Calculated
0.893 14.25 -4.65 0.950
0.89 13.65 -4.8 0.943
0.897 13.56 -4.47 0.949
0.894 14.34 -4.44 0.955
0.885 13.26 -4.83 0.939

I thought that since the reactive power was low during the working hours and negative during idle hours, then my monthly average power factor in the bill would be over 0.9, but it's around 0.88. Another factor I realized was that the maximum demand was only 220 kW and the transformer is 750 kVA, so it's underutilized.

Some people have told me that I should change the tap in the transformer so it would reduce the reactive power it´s using since the capacitors aren't compensating the reactive power of the trafo.

Regards.

 

[electricpete]

Is your load pulsating by any chance?

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(2B)+(2B)' ?

 

[electricpete]

The reason I ask that question is discussed in this thread:
thread237-249262


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(2B)+(2B)' ?

 

[electricpete]

I guess I would summarize that thread by saying that the concept of vars and harmonics apply to signals that a periodic at the line frequency (50 or 60hz) and multiples… and they don’t strictly apply when the signal is more complex such as pulsating load. In that case we need to consider very carefully the effect of the interval over which each item is computed, among other things.

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(2B)+(2B)' ?

 

[electricpete]

Last comment - The unique aspect of the load in the linked thread was not just that it was pulsating, but that the real power changed direction over the course of several cycles. That creates unique problems when we try to use watts and quasi-vars (I use the word quasi to excuse the fact that I refer to them in a non-periodic context where they are not strictly defined) to compute apparent power (or power factor). The reason is that the reversal of watts increases the apparent power, but we can’t tell that by looking at the total or average watts.

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(2B)+(2B)' ?

 

[LionelHutz]

The utility meter doesn't sum the negative and positive kVarh, it only record the positive.

Why are you trying to do negative reactive calculations then?

Why did you first post "I thought that since the reactive power was low during the working hours and negative during idle hours, then my monthly average power factor in the bill would be over 0.9" if you knew the meter didn't record any negative kVARh?

Right now, I highly doubt your measurements match, for lack of a better term. I suspect for example that you are quoting an average kW and an instanteous power factor. the most likely solution will be to RTFM for the meter.

The transformer isn't this massive inductive kVAR producing load. I'd suspect that if you use capacitors to adjust the power factor of your building loads to around 0.92 or 0.93 lagging then the transformer reactance would also be taken care of.

At this time, I think you need to take a step back and start over. Figure out exactly how you are being penalized. How the utility records the different billed values and how they do the calculations. How to use the Fluke meter to record the required data. Then, try to calculate how to correct your power factor again.