×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Power Factor - Power Consumption
2

Power Factor - Power Consumption

Power Factor - Power Consumption

(OP)
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.

RE: Power Factor - Power Consumption

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.

RE: Power Factor - Power Consumption

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.

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

RE: Power Factor - Power Consumption

(OP)
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...

RE: Power Factor - Power Consumption

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

RE: Power Factor - Power Consumption

It's all about the suds!

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

RE: Power Factor - Power Consumption

(OP)
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.

RE: Power Factor - Power Consumption

(OP)
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?

RE: Power Factor - Power Consumption

Quote:

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.

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

RE: Power Factor - Power Consumption

(OP)
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.

RE: Power Factor - Power Consumption

Quote:

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)

Quote:

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).

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

RE: Power Factor - Power Consumption

(OP)
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.?

RE: Power Factor - Power Consumption

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.





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

RE: Power Factor - Power Consumption

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.

RE: Power Factor - Power Consumption

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.

RE: Power Factor - Power Consumption

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.

RE: Power Factor - Power Consumption

(OP)
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.

RE: Power Factor - Power Consumption

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....

RE: Power Factor - Power Consumption

(OP)
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?

RE: Power Factor - Power Consumption

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.

RE: Power Factor - Power Consumption

said another way, say u put your pf caps 10 feet away from this motor. nameplate is 100amps, no load say it shows 40 amps on clamp on ammeter, friction to rotate is 1 amp say. you will have 40 amps on those 10 feet of wires out to the pf caps, then the rest of the wires from there on back to the power company will have only the real current of 1 amp on them. this imaginary current is only circulating back and forth.

RE: Power Factor - Power Consumption

(OP)
Ok yes that is exactly what I was trying to get at, thank you. So, if we know that the reactive current is 40 amps and full load current is 100, and we have a capacitor at the motor, then the wiring from the the motor back to the power supply don't need to be spec'd for any larger than 60 amps... right? Anything else would be overkill.

I looked up the circuit for a PF correction capacitor and a motor. It shows the capacitor in parallel with the motor. So I guess the power supply wires will only carry 1 amp of current and the wires connecting the capacitor will carry 39 amps (for the no load case).

RE: Power Factor - Power Consumption

Quote:

yes that is exactly what I was trying to get at, thank you. So, if we know that the reactive current is 40 amps and full load current is 100, and we have a capacitor at the motor, then the wiring from the the motor back to the power supply don't need to be spec'd for any larger than 60 amps... right?
Maybe a nitpick, but quantitatively important:

If motor was drawing FLA and you had 100% correction (you shoudln't have that much if caps are switched with motor) then the real component supplied upstream of capacitors would be sqrt(100^2 - 40^2) = 91.6A

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

RE: Power Factor - Power Consumption

(OP)
So in this case, at full load, the 100% PF correction only reduces the real supply current by 8.4%. BUT, this doesn't actually change the PF (it just corrects it) because the apparent current is still 100 amps and the real current is still 91.6. The power factor is 0.916 in both cases. If there were no correction, the real current would still be 91.6 and the apparent 40; however, the current in the supply lines would then be 100 amps.

Please god tell me this is right...

RE: Power Factor - Power Consumption

A simple example of why power companies are justified in charging penalties for poor power factor:
A 1 kW motor at 100% power factor will require 1 KVA transformer capacity.
A 1 kW motor at 50% power factor will require 2 KVA transformer capacity.
If you don't like it, correct your power factor. ROIs are often less than one year.

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

RE: Power Factor - Power Consumption

Jdogg05 you about got it! couple last comments since you are ready :)

remember that the reactive current is exactly 90 degrees out of phase with the real current (ELI the ICEman). so the math is like finding the lengths of the sides of a triangle not algebraic addition and subtraction; hence the sum of the squares.

when a motor is loaded, the reactive current goes up due to leakage inductance inside so the 40 amps no load in my example is probably closer to 50 or 55amps at full load; so the math shows probably sqrt[100^2-55^2] or 83amps real for a more realistic .83 pf....

RE: Power Factor - Power Consumption

Quote (jdogg05)

then the wiring from the the motor back to the power supply don't need to be spec'd for any larger than 60 amps... right?

Just to be clear something up, Your local electrical code will tell you that you can never size the feeder wire to a motor by using the corrected current. There are good reasons why sizing the feeder cable to match the motor rated current is not overkill, such as the capacitor failing and being able to start the motor without an excessive voltage drop.

RE: Power Factor - Power Consumption

(OP)
Ok I finally understand this now. Thanks everyone for all the input.

RE: Power Factor - Power Consumption

FWIW, real world power factors are typically 80% plus. Generators are sold like that for ex., 100 KVA @ 80% pf = 80 KW. For large consumers, if the electric company is able to bill by load PF (most are not, my utility for ex., does not) they would love to make you pay more when your PF falls below 80 or 90%, whatever they can get the regulating agency to approve. So examples of 0% or 10% PF are just theoretical examples, not normally seen in the real world.

RE: Power Factor - Power Consumption

1capybara just wrote:

So examples of 0% or 10% PF are just theoretical examples, not normally seen in the real world.

To which I respond: 'It depends on your world...'

In my working world, we use capacitors to 'supply' the lagging vars drawn by the predominantly inductive loads connected to the grid. What we call 'low-voltage' caps [meaning with a phase-to-phase voltage of 14, 28 or 44 kV] typically have a rating of between 10 and 30 MX. Our directly-connected 'high-voltage' caps [either 115 or 230 kV] range from 96 up to a whopping 410 MX output.

Most of the HV caps and a number of the LV ones have smoothing reactors installed in series with their connection to the grid to dampen that initial 'kick' when they're placed in service. Alternatively, depending on the location of the installation and the electrical characteristics prevailing there, independent pole operation breakers may be utilized to very precisely time the individual contact closure to co-ordinate with the null or zero-crossing point in the waveform so as to facilitate smooth insertion.

On the opposite side of the fence, we also use reactors to absorb VARs from the system, particularly in association with 500 kV circuits which generate great amounts of 'lagging' VARs when lightly loaded...

But I digress.

Capacitors have almost a perfectly leading power factor; reactor have very close to a lagging power factor. The switching duty is severe, particularly for cap switching, and sulphur-hexafluoride-insulated [SF6] breakers are gradually replacing the older oil circuit breakers in LV applications.

A factor that often bears on power system operation is the need to transport large amounts of power from where it's generated to where it's used; and as an earlier poster alluded to, peak times are the worst. During these periods especially, proper deployment of reactive resources can be used to optimize the delivery limit of a circuit which might otherwise be constrained due to either thermal or stability limits. [We have also recently commissioned series capacitor installations to improve the ratings of some of our 500 kV circuits.]

A final note: capacitor output is all or nothing, and varies as the square of the applied voltage. Reactors have a similar characteristic, but since it's based on frequency and since there is typically far less variation in frequency than in voltage profile, VAR consumption by reactors is almost pegged...

The latest thing is to use Static VAR Compensators [SVCs] which consist of a combination of reactors and thyristor-switched capacitors; SVC's can thus have a soothly adjustable 'lagging' reactive buck / boost function, for example, from 40 MX 'in' to 60 MX 'out.' As a power system operator, I consider these the cat's miaow; they're almost as good as a synchronous condenser...

Autotransformer tertiary windings often prove a very handy place to hook these gizmos up.

Hope this was at least interesting, even if it didn't help...

CR

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources