Power Factor vs service capacity

[sgunsel]

A refrigerated facility is considering increasing service capacity for a possible expansion, which would require larger utility transformers. There are several compressors and other motors throughout the facility, and I believe they are being charged a penalty every month for poor power factor. The service is 208Y/120, 800amps, about 288KVA. KVAR ranges from 50-90, demand from 140-195.

If the PF is increased at the service entrance, will that effectively increase usable capacity from the utility (in addition to reducing PF penalties)?

Thanks.

 

[charlie b]

No, it will not. From the point of connection of the pf correction device, the upstream distribution system will see the benefit. From that point downstream, the KVA you had before is the same as the KVA you will have afterwards. So if you connect the pf device at the service, the utility will see a lower load, and you will get a lower bill. But the facility itself will still be running at the same KVA, and thus you will not have increased the capacity of the system. In general, you want to connect pf correction devices as close to the load as you can.

 

[sgunsel]

Charlie, Are you saying that the KVA after the service entrance will be greater than the KVA coming into the service entrance?

 

[mivey]

Charlie, Are you saying that the KVA after the service entrance will be greater than the KVA coming into the service entrance?

He is saying the KVA after the capacitors is the same. The kVA ahead of the capacitors is less. That is because the capacitor is a stored energy system. It stores energy for 1/2 of a cycle and releases it to inductors on the other 1/2 cycle (i.e., the inductors store energy on the opposite 1/2 cycle).

Since this energy exchange takes place after the service, the only thing the service has to supply (after the initial storage charge) is the amount of stored energy that gets converted to heat. This heat loss is because the stored energy gets pushed up and down a system with resistance (sometimes called var losses in the power industry).

 

[gar]

110513-0826 EDT

P = VA*PF

If power is the same on two sides of a point, but the PFs are different, then see if you can answer your own question.

.

 

[sgunsel]

Isn't the PF the same on both sides after correction?

 

[Besoeker]

Isn't the PF the same on both sides after correction?

Think about the simple case of power factor correction for a single motor. Adding PFC doesn't alter the power factor of the motor itself. It will still take the same current.
But the supply current is reduced.

 

[charlie b]

Isn't the PF the same on both sides after correction?

Absolutely not. That is the point of pf correction. The essence of reactive power is that the magnetic field of the generator is exchanging energy with the magnetic field of the motor. That exchange is in addition to the energy that the motor is using to perform its physical work. That is also the reason that the total current is higher than it would be, if there were no inductive loads (such as motors).

When you add pf correction, what happens is that the magnetic field of motor will now be exchanging energy with the electric field of the capacitor. The power factor of the motor will be the same, but the source of the "extra energy being exchanged" is no longer the generator, but instead is the capacitor. So as far as the generator is concerned, there is no "extra energy being exchanged," and the generator only has to supply the energy needed to get the motor to do its physical work (i.e., the KW, or "real power"). The pf upstream of the pf correction is therefore higher than the pf downstream.

 

[gar]

110513-0950 EDT

sgunsel:

If you understand that Preal = VA*PF, and you also understand that the real power from the source is equal to the real power consumed by the load, then the following results:

(VA source) * (PF source) = (VA load) * (PF load)

and if the two PFs are different, then the two VAs have to be different. If the two voltages are the same, then the currents have to be different.

(P real source) = (P real load)

Comes from the theory of the conservation of energy. You can not create or destroy energy, you can only change its form or state.

.

 

[antenna2001]

Because of the high cost of a capacitor system, It might be worth the effort to trace the source of the increased power factor (pf). If the culprit is a single device, you can achieve the same results by correcting the problem at its source with a smaller set of capacitors.

Another way of correcting pf is by using a synchronous motor. Once you know how large the problem is, you can consult with an engineer to find the least costly alternate.

When you take the measurements, compare kW to KVA. Where they are different, there is a problem.

 

[antenna2001]

I forgot to say that once you get your readings, using a power meter, divide KVA by KW. The result is the power factor expressed as a decimal. Consult with the utility to find out what they want the power factor to be. I think they require somewhere between 0.8 and 0.9 and they don't want it higher because that results in a different set of ploblems.

Also, the reason you save by correcting at the source of the problem is that the correcting device needs to be rated only for the current that it supplies. If you correct at the service, the correcting device has to be rated for the total current that your system is designed for.

 

[antenna2001]

I forgot to say that once you get your readings, using a power meter, divide KVA by KW.

Sorry. It should be divide KW by KVA to obtain pf as a decimal.

 

[kingpb]

... I believe they are being charged a penalty every month for poor power factor. The service is 208Y/120, 800amps, about 288KVA. KVAR ranges from 50-90, demand from 140-195.

From above could someone define what is being called DEMAND, i.e. range 140-195, because DEMAND does not make sense in the context of the question. Is the 140-195 meant to be kW instead of DEMAND.

How are you determining that the pf is bad?

 

[sgunsel]

The original question (perhaps not very clear) was: if I have a fully loaded service with a PF of 0.6, and the PF is increased to 0.9, regardless of whether at the service entrance or at individual loads, does the usable KW increase from the utility?

For example, assume 100 amp service, single phase, 120 volts. From the utility:
P1 = 100A x 120V x 0.6 = 7.2KW (12KVA)
P2 = 100A x 120V x 0.9 = 10.8KW (12KVA)
Padditional = 10.8 / 7.2 = 1.5, a 50% increase in P for same KVA.

 

[charlie b]

The part you are missing is that it is not KVA that stays the same. KW stays the same.
If you start with 100 amps at 120 volts, with a 0.6 power factor, then,

• KVA is 100 times 120 (then divided by 1000), or 12 KVA
• KW is 12 KVA times 0.6 pf, or 7.2 KW.

Now if you use pf correction to improve pf from 0.6 to 0.9, here is what happens:

• KW stays at 7.2.
• KVA becomes 7.2 divided by 0.9, or 8.0 KVA.
• Total KVA dropped from 12 to 8.
• Current is now 8000 VA divided by 120 volts, or 67 amps.
• Current dropped from 100 amps to 67 amps.

 

[kingpb]

Further to CB's post, a simple formula to remember for the current drop of a capacitor installed on the motor side of the O/L relay is:

%Ired = 100-100*(Old pf/new pf)

In this example: %Ired = 100-100*(0.6/0.9) = 33% reduction

so, the 100A becomes 67A.

 

[sgunsel]

Charlie b, Looks like the end result is the same - with power factor correction
we end up using 33 fewer amps, so there is now additional capacity. Looks like my math wasn't so good, though. It is much clearer now. Thanks!

 

[charlie b]

we end up using 33 fewer amps, so there is now additional capacity.

Take care, here, for I think you are still missing something. In the original question, you said that the pf correction would be at the service. In that case, it is the utility, and not the owner, that gains a bit of extra capacity. Downstream of the pf correction device, the situation is not changed. The pf from that point downstream is still 0.6, the KVA is still 12, and the current is still 100 amps. So if you want to gain back some capacity that the owner can use, then you need to put the pf correction somewhere within the owner's facility, and the closer to the loads the better.

 

[sgunsel]

charlie b, Thanks for the clarification. My original intent was to verify that there might be some additional capacity to be had. As usual, the details are critical. The utility service is maxed out and even a small increase will be many $$$. I don't expect miracles, but sometimes you have to squeeze out all you can. It may not be feasible or cost effective, but it might be worth looking at. Thanks again.

 

[jghrist]

If the concern is the utility transformer capacity, then capacitors at the service might help. What is the existing transformer size? From the ranges of loads in the OP, the maximum existing kVA is sqrt(195? + 90?) = 215 kVA at a pf of 91%. The actual power factor might be lower, because the maximum demand may not coincide with the maximum kVAR. If you corrected the pf to 100%, the kVA would equal the kW (195 kVA), freeing at most 20 kVA of utility capacity. You won't actually be able to correct to exactly 100%, however, because you can add capacitors only in discrete amounts and having more kVAR of capacitors than kVAR of reactive load will result in kVA being higher than the kW.