Power factor correction and energy

[don_resqcapt19]

The following was in an e-mail newsletter that I get. I don't think the last sentence is correct. It is my understanding the the only savings by putting the power factor correction at the load it the reduction in the I^2R losses on the feeder conductors. Would that amount to "far more kW per hour"?

Why are we talking about PF at the feeder level? If you correct it at the entrance, then you avoid the PF penalties from the electric utility. But that?s all you do.
You may have PF of 99% at the service and only 77% at the load. This means the load has to draw significantly more power to do the same amount of work. You aren?t paying the higher rate per kWH, but you?re using far more kW each hour.

 

[Besoeker]

The following was in an e-mail newsletter that I get. I don't think the last sentence is correct. It is my understanding the the only savings by putting the power factor correction at the load it the reduction in the I^2R losses on the feeder conductors. Would that amount to "far more kW per hour"?

I agree. It is nonsense.

 

[Jraef]

I agree. It is nonsense.

But if they say it with enough authority*, it becomes truth does it not?


* Think Cartman on the adult cartoon show South Park;
"You will respect my authori-TAH'"

 

[charlie b]

Let’s talk motor loads, for that is what we deal with most often. A motor is an inductive load because it is built with wires wrapped in coils. A generator is also built with wires wrapped in coils. Current passing through each coil creates a magnetic field. The essence of poor power factor, the essence of inductive loads, the essence of reactive power, is that the magnetic field of the generator is exchanging energy with the magnetic field of the motor. That energy exchange is in addition to the energy that causes the motor to do its job. The motor could not spin, if it were not for the exchange of energy, but the motor draws more power (i.e., more KVA) because of it.

What happens when you insert a capacitor into the circuit, at any point between the generator and the motor, is that (1) From that point to the motor, the electric field of the capacitor will be exchanging energy with the magnetic field of the motor, and (2) The generator does not see that energy exchange. Thus, from that point upstream, you will have a higher power factor, but from that point to the load the power factor is the same as it was before.

So the statement that the pf can be .99 at the service and .77 at the motor is a valid statement. But the statement that the motor is doing more work, or drawing more power, is not true. If the pf was .77 before you connected the capacitor, then no matter where you put the capacitor, the pf is going to be that same .77 between that point of connection to the motor.

Don, you are right in saying that the advantage of connecting the capacitor close to the motor is that the feeder (or branch circuit) conductors between the service and the capacitor connection point will have a higher pf, and thus a lower overall current, and thus a lower amount of I2R losses along the wire. It will also have a lower voltage drop along those conductors.

 

[gar]

090310-1836 EST

The statement "but you?re using far more kW each hour" is nonsense. You can not get more output energy than you put in. So if the input energy is 1 KWH at the main panel you can not have more than 1 KWH at the load, and it actually can not be that large at the load.

.

 

[don_resqcapt19]

...
Don, you are right in saying that the advantage of connecting the capacitor close to the motor is that the feeder (or branch circuit) conductors between the service and the capacitor connection point will have a higher pf, and thus a lower overall current, and thus a lower amount of I2R losses along the wire. It will also have a lower voltage drop along those conductors.

And my understanding that the only energy savings you get by putting the PF correction at the load over putting it at the main service is that gained by the reduction in voltage drop and the lower I^2R losses. This will result in a small savings in kWH. The newsletter implies a much larger savings. The newsletter was not an add from a seller, but the MRO Insider published by EC&M magazine. I didn't find a link to this issue, but the back issues are here.

 

[charlie b]

And my understanding that the only energy savings you get by putting the PF correction at the load over putting it at the main service is that gained by the reduction in voltage drop and the lower I^2R losses. This will result in a small savings in kWH. The newsletter implies a much larger savings.

Your understanding is correct, and the newsletter would be wrong in making such a claim.

 

[Besoeker]

But if they say it with enough authority* it becomes truth does it not?

Perhaps in the minds of the gullible.
:wink:

 

[ronmath]

I believe they are talking about KVA not KW. The savings of putting it at the source of the bad power factor might be in wire and conduit sizes as they have to be sized to carry the full load amperage, not just the KW load. It does seem to make some sense this way. IMHO

 

[realolman]

This means the load has to draw significantly more power to do the same amount of work. You aren’t paying the higher rate per kWH, but you’re using far more kW each hour.

Would the pink highlighted words have any bearing on the subject?

You would not get the Horsepower with a bad PF?

 

[charlie b]

I took that to mean that the writer was asserting that a significantly higher value of apparent power (KVA) would be needed to do the same work (KW). My response: Higher? Yes. Significantly higher? No.

 

[richxtlc]

And my understanding that the only energy savings you get by putting the PF correction at the load over putting it at the main service is that gained by the reduction in voltage drop and the lower I^2R losses. This will result in a small savings in kWH. The newsletter implies a much larger savings. The newsletter was not an add from a seller, but the MRO Insider published by EC&M magazine. I didn't find a link to this issue, but the back issues are here.

I wrote EC&M an email concerning that statement, I hope that they print a correction in the next issue.

 

[Besoeker]

I took that to mean that the writer was asserting that a significantly higher value of apparent power (KVA) would be needed to do the same work (KW). My response: Higher? Yes. Significantly higher? No.

This means the load has to draw significantly more power to do the same amount of work

It's just plaing wrong at every level.
Leaving aside the point that power and work are two different things, the inference is that poor power factor of a load means poor efficiency and that PFC external to the load would be a fix. Taurean turds come to mind.
The very best it could do would be to reduce I^2R losses in the supply cabling to the poor PF load. If that dissipates significant power, the poor power factor might be the least of your problems....

 

[LarryFine]

There's little we can do about the load's power factor. Reactive power is a necessary evil of reactive loads. What we try to do is minimize the effects of that reactive power on the electrical supply system, as well as other loads.

As I understand it, PF correction merely "shunts" the reactive power out of the rest of the supply system. Normally, the reactive power has to be carried from the load all the way back to the source by the entire electrical system.

Instead, that reactive power only has to be carried by the parts of the supply system between the load and the corrective devices. It relieves the rest of the building wiring, service equipment, and POCO equipment from having to carry it.

The reason standard KwH meters don't respond to the reactive power is that the voltage and current peaks don't coincide, so the two peaks don't combine to spin the meter as fast as the same peaks would if they were simultaneous.

 

[SIMROX]

It all depends on the metering of the current! Most metering is in kWH and the power factor has no impact on the kWH (real power) used. If the load is inductive (or capacitive) the current is not in phase with the voltage and the real power KW and apparent power differ. The larger the difference in the phase angle between the current and voltage the smaller the power factor is (Cosine of angle between Apparent Power VA and Real Power W in power triangle).



If you are being billed on current or VA?s ( Volt Amps) only (unlikely) the worse the power factor is, current is out of phase with the voltage the higher the costs. If you are being billed on Watts (V x A x pf) you are not being billed of the reactive power component but supply authorities define the minimum required power factor of the installation as they still have to produce the current being consumed.

Where power factor is a burden is when you have poor power factor the line current is higher than when you are running low power factor. You may not be paying for the current but the volt drop caused by the high current can cause problems. This is very common on small sub kW motors where the power factor at low load my only be 0.4. ie a 0.37 kW motor on 400V 3 phase supply at 100% load my be 0.85 but at no load the motor power factor my drop to 0.45 and in simple terms the motor line current increases when the motor load reduces as the phase angle changes. On the other hand the motor load reduces the input power also reduces, as the meter dealing in KWH.