Power Factor - Power Consumption 2

[jdogg05]

Magnetizing current consumes no power. Reactive power is not real. However, the current in the wires is real. As I understand it, the reactive power DOES actually consume energy each cycle but it is like a spring. It compresses/decompresses/compresses, etc. providing the energy for the next cycle. The reason a low PF is bad is because the cables that supply the motor, or the cables that supply your residential area, have a curernt capacity associated with them, and thus when a low PF exists, the current in the wires is higher than it needs to be... which results in heat dissipation losses. THEREFORE, the magnetizing current DOES actually result in energy consumption, just not in the intended load. Heat dissipation is real. Energy cannot be created or destroyed. It can just change forms... This is why you get charged a lot for buildings that have low PFs because the utility company doesn't care and is simply charging you for the total energy consumed, both in the distribution lines and in the loads in your building. Is this correct logic!? Please, I am going crazy trying to comprehend this.

 

[ScottyUK]

You're on the right track.

The conductors do indeed have a power loss which limits their capacity to carry current, but the value of that power loss isn't the main concern. The problem for the utility is that they would ideally like to use those conductors to deliver energy to loads. A low power factor load means that they have to use a significant portion of their conductor's capability carrying reactive current, leaving less capacity for delivering real power. Phrased differently, to to deliver a given amount of power to a low power factor load they need a much bigger and more expensive conductor than they would at a high power factor. In the case of a transformer, which is rated in VA rather than watts, they would have to pay for a much larger transformer, containing more core steel, more copper, and more oil, requiring a bigger and more expensive substation equipped with larger circuit breakers. The same reasoning applies to generating plant.

The quick answer is that low power factor loads make inefficient use of the transmission and distribution system, so the operators naturally pass on some of this cost to those who have low power factor loads through penalties.

 

[electricpete]

I agree with op and Scotty. Reactive power is not itself associated with energy useage, but transmitting reactive power along with real power requires that the supply network carry higher current which causes higher I^2*R losses within the supply network.

Scotty brings other important points... effect on supply system. The vars have to be "generated" and transmitted. This uses equipment and capacity and also alters the voltage profile.

I think scotty left out a crucial and central technical point, though: Suds in your glass leave less room for beer.

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(2B)+(2B)' ?

 

[jdogg05]

Thank you for the explanation. So can the utility companies measure end users PFs? Is this how they penalize them? I mean if all they can measure is energy consumption (and average peak power by differentiating with respect to time), then it seems that they would not be able to determine the PF of the users load...

 

[DRWeig]

Yes, the utility can measure power factor and charge a penalty. They can also meter VARs (that's volt-ampere reactive) along with kW for demand, and charge more based on the ratio. I've seen one utility charge for consumption of VAR-hours as a penalty.

So in essence, a utility can meter anything. It can also charge fees in any manner to give customers an incentive to correct their power factor.

Best to you,

Goober Dave

Haven't see the forum policies? Do so now: Forum Policies

 

[electricpete]

It's all about the suds!

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(2B)+(2B)' ?

 

[jdogg05]

Ok this is starting to make way more sense now. Thanks everyone. So basically, if the utility provider decided not to build their distribution system with adequate capacity (if they just spec'd it for everyone having a power factor of 1), then there would LITERALLY be huge losses from heat dissipation from this extra current. But the companies are not stupid so they build there distribution systems properly, and accounting for reactive power, size everything larger, hence increasing capital costs, which are then absorbed by the end user.

 

[jdogg05]

Also, as per the power loss equation, P = I^2*R, if the ACTUAL current is made much larger because of the reactive current, how is there no power loss...? How can this be explained from a numerical/quantitative perspective?

 

[electricpete]

Also, as per the power loss equation, P = I^2*R, if the ACTUAL current is made much larger because of the reactive current, how is there no power loss...?

There is increased Itotal^2*R power loss in the supply path associated with increased current to carry reactive power (Itotal = sqrt(Ireal^2 +Ireactive^2)). No-one said otherwise. Suds.


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(2B)+(2B)' ?

 

[jdogg05]

As per ScottyUK's comment, the power loss is not the main concern. I don't understand why. If the load has a PF of 0.1 then almost all the power is being consumed by the conductors. The capital costs of the bigger distribution system can't really be that much bigger than the operating costs of the energy production... if that is the case.

 

[electricpete]

If the load has a PF of 0.1 then almost all the power is being consumed by the conductors.

If you mean that for load pf of 0.1, then almost all the I^2*R losses in the conductors is associate with supplying reactive current, I agree. (How that compares with actual power consumed by the load is unknowne without more info)

As per ScottyUK's comment, the power loss is not the main concern.

I agree with you, which concern is largest will depend on the situation. I think Scotty was just directing your attention to another aspect which is potentially more important than the cost of generation associated with supplying that I^2*R if the local network is already taxed to capacity (remembering that generation cost tends to be a lot less than consumer cost).

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(2B)+(2B)' ?

 

[jdogg05]

OK. This is starting to become clear now. Thanks everyone for helping me understand this. I have struggled with this concept for way too long.

electricpete, when you say consumer cost, are you meaning distribution networks? All the power lines, transformers, substations, etc.?

 

[electricpete]

My terminology

consumer cost - how much does energy cost the user. Almost 20 cents per kw-hr in my neck of the woods.

generating cost - how much does it cost the utility to generate the energy. In my neck of the woods it is around 2 cents per kw-hr, excluding generation capital costs. A factor of 10 lower. Because utility also has to build and maintain the T&D infrastructure to deliver the generated power to the user.

How much does a given kw-hr of I^2*R loss cost the utility?
Convert it using the low generating cost, not the high consumer cost.







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(2B)+(2B)' ?

 

[ScottyUK]

Don't get too focused on losses in conductors. The power loss in the conductors isn't normally a problem due to its monetary value, it is a problem because it causes heating in the conductor. Conductors have a maximum safe operating temperature, which imposes a limit to how much current can be carried by any given conductor. Cables and lines are expensive assets, so utility companies don't want to make them significantly larger than they need to be. By keeping the power factor near unity the cable or line is being used to transmit the greatest possible amount of active power. Transmitting the same active power at a lower power factor would require larger cables and lines, which cost more.

 

[Compositepro]

One major factor that has not been mentioned yet is that generation and distribution assets must be sized for peak loads, not average loads. VARs add to peak loads but do not get billed as kwh. Capital equipment is a major cost of power, as others have said.

 

[ScottyUK]

Hmm, most transformers and generators have such a long time thermal constant that short-term peaks are of little consequence. In contrast AVRs have to be sized to support reactive peaks otherwise the generator terminal voltage will sag or collapse.

 

[jdogg05]

So there are a lot of negative things associated with low PF: capital costs of larger capacity generation and distribution infrastructure, lost energy due to I^2R losses, and reactive peak demands that can result in terminal voltage collapse.

Just to ratify my understanding. Say, for instance, a small genset was running a little induction motor. If all the cabling and everything in the system was sized so as to account for all possible reactive power (view everything as sunk costs), there would be no issues with a low PF. Please tell me this is correct.

 

[mikekilroy]

So there are a lot of negative things associated with low PF: capital costs of larger capacity generation and distribution infrastructure, lost energy due to I^2R losses, and reactive peak demands that can result in terminal voltage collapse.

Just to ratify my understanding. Say, for instance, a small genset was running a little induction motor. If all the cabling and everything in the system was sized so as to account for all possible reactive power (view everything as sunk costs), there would be no issues with a low PF. Please tell me this is correct.

Keep in mind as others have said, the I^2R losses are the power companies problem mostly, not yours, and are a small precentage of the total so don't get it blown out of proportion....

with that thought, even if you have low PF, if you size YOUR generator to YOUR loads nameplate data, it will COVER you. Don't get caught up thinking low PF will make YOUR loads bigger than nameplate data; ie., if you have a 100amp nameplate rated motor, you will likely size your wiring for 100 amps..... if that motor is unloaded, it may draw only 1 amp of REAL power (that u pay for) but still 40 amps of imaginary, so PF will be real low, but it is SSTILL way less than the 100amps YOU sized for....

 

[jdogg05]

Ok, that being said, if your motor nameplate says 100 amps, does the namufacturer assume it is being connected across the line without capacitors or whatever else would lower the PF? i.e. doesn't the manufacturer have to specify this rated nameplate current for the worst possible case (no PF correction in the circuit)? I mean, they have to account for all possible reactive current in this specification don't they?

 

[mikekilroy]

capacitors or any other pf correction scheme has no effect on the motor current! to be a motor, it has reactive current. pf caps only give it a more local closer spot to cycle back and forth to rather than going all the way back to the pocos generator 20 miles away. 100 amps is 100 amps.