PFC cost of operation.

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Theoretical question. Somewhat.
A customer with an 100 kvar pfc that is no longer needed. If they leave it connected to the line, what does it cost them to operate? Pick a rate.

I don't know if having a PF less than unity in the capacitive direction incurs a penalty or not.

I would be inclined to disconnect it. At some point it is going to fail and if it is not connected it will not hurt anything when it fails.
 
If the customer does not have enough inductive loads (lagging PF) for which this is compensating, then he will potentially be operating in a LEADING power factor, and the utility may end up penalizing him just the same as if it were lagging. In other words they WANT you to run at .95 lagging or better, but not leading either. You can have all kinds of problems with transients, resonnance etc. etc. and from the Utility standpoint, that may end up on THEIR side and make the problem theirs.

But to answer your specific question, there is a very small load, called "parasitic resistance" (a.k.a. ESR or Effective Series Resistance) in capacitors. The losses associated are usually figured at 1/2 watt per kVAR, so in your case, 50 watts.
 
If the customer does not have enough inductive loads (lagging PF) for which this is compensating, then he will potentially be operating in a LEADING power factor, and the utility may end up penalizing him just the same as if it were lagging. In other words they WANT you to run at .95 lagging or better, but not leading either. You can have all kinds of problems with transients, resonnance etc. etc. and from the Utility standpoint, that may end up on THEIR side and make the problem theirs.
Operating at leading power factor can result in increased supply voltage. I don't suppose it's very common. Mostly I deal with it for induction motor applications where there is a known correction required for the reactive part of the current so it isn't a situation I would be very likely to encounter. But I did once - it was about 32 years ago. It was a lightly loaded distribution transformer with excessive PFC connected. The old electrical guy in charge of the plant said that's what the problem was. At first, I didn't believe it. But when I did the calculations and played with a few vector diagrams, I had to concede that he was right.
 
If the customer does not have enough inductive loads (lagging PF) for which this is compensating, then he will potentially be operating in a LEADING power factor, and the utility may end up penalizing him just the same as if it were lagging. In other words they WANT you to run at .95 lagging or better, but not leading either. You can have all kinds of problems with transients, resonnance etc. etc. and from the Utility standpoint, that may end up on THEIR side and make the problem theirs.

But to answer your specific question, there is a very small load, called "parasitic resistance" (a.k.a. ESR or Effective Series Resistance) in capacitors. The losses associated are usually figured at 1/2 watt per kVAR, so in your case, 50 watts.

I was going to say any heat produced has to be from a resisitve component which will be a KW component and not a KVAR component in the circuit, Even the small amount of heat produced in the conductors supplying it.
 
The increased voltage due to PFC was pretty apparent when we checked the voltage. I did the math at the time and it was almost text book. This area seldom suffers from low voltage and generally it is on the high end. They were being dinged pretty hard for poor PF, and at the time one capacitor did the job with correction resulting in a credit on their billing. Paid for itself in months. No penalty for over correction. We did have the POCO change the taps and bring the voltage down. It is not used at all now, but I have always been curious as to operating costs when there is no PF or very little, to correct.
 
As indicated by others, there if very little loss in the capacitor itself. If the capacitor over compensates for reactive load, the current in cables ahead of the capacitor will be higher than it would be with unity power factor. If this is the case, there will be more I?R losses in the cables.
 
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