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POWER FACTOR CALCULATION

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bgl2010

Electrical
May 26, 2010
6
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.
 
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It looks like you're measuring on the low side and being metered on the high side. I assume your calculations are based on the load on the low side only.

Transformers have two contributions to reactive load. The first is no load reactive which is present as long as the transformer is energized and is fairly constant. This will be higher than normal because the transformer is oversized.

In addition, the load current produces an I^2 X through the transformer leakage reactance, which is also reactive or lagging kvar. Since the load current varies, the I^2 X also varies. This should be fairly low because the transformer is oversized.

I'm not straight on your sign conventions. I'm thinking the negative kvar means that the total load reactive load plus capacitors is lagging. Some one else will have to chime in on this.

You ought to get the utility to tell you what the high side PF readings are or you will have to get a test report on the transformer to see what the no load excitation loss is.

 
"I'm thinking the negative kvar means that the total load reactive load plus capacitors is lagging."
op should clarify, but I assume the opposite. Since a positive real number represents a real load here, a positive imaginary number would be inductive/lagging (S = V I*).

In that case, during work hours when you are seeing lagging load on the low side, it is even more lagging (lower pf) on the high side due to transformer reactive losses mentioned by magoo. Additionally, if for some reason there is high harmonic content, that lowers your power factor as well.



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(2B)+(2B)' ?
 
Your measurements indicate very low load compared to the size of the transformer. It's inductance will be a major portion of the reactive power seen by the utility, but not by your meter, in this situation.

First you need to determine exactly how the utility is measuring and billing the power factor. It may be power factor - demand or average, but commonly it is measured as kVAR demand, KVAR usage or kVAR average. Once you know the parameter that has to be optimized, estimate or measure the reactance of the transformer and then make adjustments to your low side capacitance. It may take some trial and error to get it right. The utility may be able to help you with the measurements.

Alan
“The engineer's first problem in any design situation is to discover what the problem really is.” Unk.
 
The utility may not be giving you credit for a leading power factor at night.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
I have doubts the bill uses an average power factor. The bill likely uses a 15 minute demand for the kW and kVAR. Then, those demands are used to calculate what is called the demand power factor. In other words, you pay penalties for the 15 minutes of power demanded when your power draw is the most and your power factor is the worst.

I'll assume the 220kW demand is the number which was billed, or multiplied by the different charges for the bill. In this case, correcting the power factor would reduce the demand charges from 220kW to 215kW, or take 5kW off the demand charges.

Mentioning capacitors, negative PF at night and positive PF during the day leads me to think negative is reactive. However, that is still a guess.

 
A nit-picky note on terminology.

Assuming we are dealing with a load (not generation) and real power is always positive:

In lagging case, the reactive power Q is positive, the angle of S = VI* is positive, the cos of that angle (power factor) is positive.

In leading case, the reactive power Q is negative, the angle of S = VI* is negative, the cos of that angle (power factor) is still positive.

We distinguish between the two types by using the word leading or lagging after power factor. We don’t get negative power factor unless real power reverses direction.

Sorry for being picky. Lionel, you are much more knowledgeable than me on many of these things and I know this is nothing new to you.

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(2B)+(2B)' ?
 
And if it's not clear, I was reacting to the word "negative pf" which Lionel used but OP didn't use. Maybe it was just a typo.

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(2B)+(2B)' ?
 
Yes, I shouldn't say negative power factor. Which brings up the question - is it correct to say power factor could be negative if generating?

I had it wrong anyways. It appears to me the OP is referring to negative reactive power at night which leads me to believe that his negative is capacitive, as in the power factor caps are left connected and they are over correcting at night when the plant is idle.
 
Generating adds further confusion: a leading generator is a sink for reactive power, where a leading (capacitive) load is a source. Conversely a lagging generator is a source of reactive power, while a lagging (inductive) load is a sink.


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If we learn from our mistakes I'm getting a great education!
 
Hi,

Thank you all for your answer. I still don't get the difference between PF value the Fluke 1735 is giving me (column one in my earlier post), and the value a calculate with the kW and kVar (column 4). Because I only get this difference when the kVar are negatives, I'm using the equation kW/(kW^2+kVar^2)^0.5 in excel and when the kVar are negatives the difference is big.

This system doesn't have a lot of harmonics, then the real power factor and the displacement power factor should be almos the same.

The utility here uses the kWh (monthly active energy) and the kVarh (monthly reactive energy) to determine an average PF, with this equation> kWh/(kWh^2+kVah^2)^0.5. Therefore, since I'm having readings of negative kVar during idle hours, that means the capacitors are compensating all the reactive power of the load during that time, but they can't compensate the trafo reactive power. So adding more capacitors in the low voltage side won't avoid the penalty, because the utility meter is not bi-directional.

What about changing the tap position in the trafo, would that reduce it's reactive power consumption?

Regards
 
You original readings posted all came from the same instrument at the same time?

Have you measured the harmonic content?

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(2B)+(2B)' ?
 
Were the kW and kVAR values measured with the Fluke also?

This meter can display both real power factor and displacement power factor. Also, it can show individual phase values and totals. Which you are looking at?

Your calculated values are correct based on the kW and kVAR. Of course the pf may be thought of as leading or lagging depending on whether you think about it from source point of view or load point of view. Read the instruction manual carefully, especially the section called "measurement theory", to be sure you understand what it is showing you.

If the kVARh meter won't run backwards, you may not be able to offset daytime usage with leading pf at night (which would make sense from the utility's point of view). Changing the tap position probably won't make much difference.

Alan
“The engineer's first problem in any design situation is to discover what the problem really is.” Unk.
 
Sorry epete. I started composing my post and came back to it later and missed yours.

Alan
“The engineer's first problem in any design situation is to discover what the problem really is.” Unk.
 
The advantage of over correcting KVARs at night and on the weekend is that you do get pure Watts-Hours to add to your monthly KWHRs and this helps to improve your average PF.
The other side of that coin is that you may not get many KWHrs when the plant is not running, still 128 hours a week does help.
KVARs added to the transformer secondary will offset the transformer primary KVAR consumption.


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Why do you think you can't compensate for the transformer reactance? Capacitive kVAR on the low side of the transformer will be reflected to the high side of the transformer.

How does the utility use that calculated power factor number to penalize you?

It actually sounds like the utility sums the negative and positive kVAR's together and you are being penalized for leaving the excessive capacitors connected at night. Try switching the capacitors on and off to match the plant load.
 
Hi Lionel.
Back in the days of electro-mechanical KWHr meters when phase shifting transformers were added to a standard KWHr meter to meter KVARHrs, a common PF penalty formula was 0.9 minus the average monthly PF = % penalty.
If the monthly average PF was 85% the penalty would be 0.9 - 0.85 = 0.05 and a 5% penalty would be added to the bill.
Some meters were allowed to run bi-directionally and it didn't take much capacity to bring the monthly average above the penalty point.
Some meters had ratchets fitted that prevented the metering of leading KVARHrs.
Some utilities used a more punitive formula to determine penalties.
There are so many ways to measure PF and asses penalties that you must call the local utility to determine local practices.
In support of your post, Lionel, many new electronic meters measure all four quadrants and positive and negative energy are summed. At one time, some customers would invert the KWHr meter for a number of days each month. This would cause the meter to run backwards and subtract the usage from the monthly total.
With the electronic meters the default setting sums all power through the meter, forward or backwards.
This could easily be the default KVARHr setting for some electronic meters.
Bottom line, you have to check locally.


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Hi,

All readings are from the same moment and the same instrument. The harmonics distortion is very low, according to the equipment. All the values are total, I'm not sure if the PF data in logging mode is displacement or real, but the harmonics are low.

If I could compensate the trafo reactance with capacitors in the low voltage side, is there a limit, wouldn't be any reasonance or over voltage in the low side.

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

The penalty calculation is done the way waross explained "If the monthly average PF was 85% the penalty would be 0.9 - 0.85 = 0.05 and a 5% penalty would be added to the bill."

Regards.
 
If the transformer is over corrected the voltage will rise. Small amounts of over correction are usually acceptable. Two major factors in regards to the magnitude of the over-voltage are the amount of over-correction and the impedance of the supply lines. The over-correction will cause a voltage rise in the transformer and in the distribution feeder conductors. This voltage rise is at 90 degrees to the real power component of the current and must be quite severe before it causes an unacceptable terminal voltage at the transformer secondary.
With a lot of capacity and a soft distribution system it is possible to drive the voltage high enough to cause rapid and frequent failure of lamps.
Measure your secondary voltage during the weekend.


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
If the data coming out of your test equipment is not self-consistent, you really need to study it that to understand that aspect. The data you posted in your first post does not seem self-consistent unless there are harmonics present (more likely in current than voltage). If you are positive about the harmonics, break out the manual and study it or call up tech support.

=====================================
(2B)+(2B)' ?
 
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