Power Factor Correction

The use of inductive loads in domestic premise is a big subject. what
can we use to reduce power factor at home, and what are the benefits.
Reply to
GDD
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The benefits are saving energy, but are not nearly as significant as some would have you believe. Take for example a 1/2 hp motor running 12 hours a day with a power factor of 0.4 (pretty darn low, even for single phase equipment). That motor draws about 7.8 amps at 120V. Let us *assume* it is being fed from 50 ft (100 ft round-trip) of 14 AWG wire. The power dissipated in the supply wiring as heat is ~15.7 watts when the motor is running.
If we could correct the power factor at the motor terminals to 1.0, then the motor would only draw about 3.1 amps, and the power dissipated in the supply wiring would be 2.5 watts when the motor is running.
So the savings of ~13.2 watts in power, over 12 hours a day for 365 days a year would be ~58 kwh. Even in an expensive market of $0.30 /kwh, that's just $17.40 a year.
Can we correct the power factor to 1.0 with a device that costs under $17.40?
If the motor initially has a pf of 0.6, and we can correct it to 0.95, (somewhat more realistic goal) then the savings would be only about $5.53 per year.
Some folks mistakenly believe that by correcting the power factor, and reducing the motor running current from say 7.8 amps to 3.2 amps, that they will cut the operating costs by this simple ratio (3.2/7.8 => 41% less electricity???). Scam artists hope you don't know the real savings and use this kind of calculation so 'justify' buying their product.
But that is not the case. Electric meters not only measure the current, but also the power factor and measure only the *energy* used.
Notice also, that the savings is in the wiring inside your home. So 'whole house power factor correction' does even less for you. If the power factor is corrected at the service panel instead of at the motor terminals, then the wiring inside your home is still carrying the same current, and dissipating the same heat. The reduced current in the relatively short, low-resistance conductors from the service panel back to the electric meter is the only savings in this situation.
There are situations when the utility charges an industrial customer extra fees for poor power factor loads. In those cases it is worthwhile for the customer to correct their power factor on the premises. But those types of situations are not all that common, and I've never heard of a residential customer being charged for poor power factor conditions.
For folks living 'off-grid', a poor power factor can cause trouble with loading of static inverters or small generators they may be using. In these cases, power factor correction may help avoid buying a larger unit and so can be of benefit.
And finally, if you look at the number of inductive loads in a typical household (number of electric motors is a good estimate) and how often they run, you start to realize that this isn't really that much of an issue.
Note I said 'inductive loads', I'm not considering the poor power factor caused by electronic power supplies. That requires a different form of power factor correction.
daestrom
Reply to
daestrom
One of the best explanations of the true "savings" of power factor correction that I have seen on Usenet. Good job.
I deal with calls from clients being promised the sun, moon, and stars with some new-fangled power factor correction technology (usually simple caps in a different looking can with some lights).
Charles Perry P.E.
Reply to
Charles Perry
Around here the Utility tells me that residential PFs are normally 0.95 and up. Do you have a link to a source that describes this as a big subject? I'm curious to see that opinion. And, I assume you mean 'What can we use to _increase_ power factor at home...'?
j
Reply to
operator jay
EXCELLENT response.
If the original poster would provide the name of the utility, I will BET that the utility meter rate is not measuring PF on residences. Since they do not measure it. No worries mate.
I knew of someone that spent a plethora of time and effort installing cap banks on motor loads at his house trying to reduce his electric bill. 3 a/c compressors in 4 years tells ya something. And 2 washing machines in the sa me time frame.
Reply to
SQLit
Great post! Ed
Reply to
ehsjr
ditto
i remember a TV expose' a few years back where one user swore he had to turn the thermostats up 3 times after installing the "magic gizmo" on his coolers.
Reply to
TimPerry
This is true. But then it depends on how one defines 'power factor'. If the harmonic currents drawn by nonlinear loads are considered, then the power factor calculated may be lower.
I think that this might be a bigger issue than residential inductive loads. Furthermore, the steps one would take to improve power factors differ depending on whether they are inductive and linear or nonlinear.
Reply to
Paul Hovnanian P.E.
The only meaningful definition of power-factor is watts/(VARS+watts). That is, the usefull power vs. volts*amps.
Sure. ...and quite obviously. However homeowners (at least in the US) pay for watts. PF isn't important. Were I running an aluminum smelter, things would be way different. I rather doubt that even the nonlinear loads in a typical residence do much the PF of the system either. Could be though, the worry-wart Europeons seem to think PF correction on PC power supplies is a big deal.
Reply to
Keith
----------------------------
Sure, residential customers pay for watts (rather kilowatt-hours) but the rate/KWH charged does include costs which are affected by power factor. Your supply voltage is also affected by poor power factor. The reason that residential customers do not face demand charges is that the typical power factor is close enough to unity to make the costs of measuring KVA demand not worth the effort. Throw in a bugger factor to account for the average PF or (better yet) peak KVA demand - and go with only energy costs biased accordingly. --
Don Kelly @shawcross.ca remove the X to answer
Reply to
Don Kelly
That depends on what you plan on doing with the power factor data. If you applying capacitors, you want displacement power factor since capacitors cannot correct for distortion power factor.
Kind of like saying the only thing you care about paint is the color. Well, let someone paint your car with latex and you will be annoyed. You have to know the application before you know what data you need.
Charles Perry P.E.
Reply to
Charles Perry
If by "distortion power factor" you mean something to do with current spikes or such, it seems to me that capacitors CAN "help."
Reply to
John Gilmer
No. Distortion power factor is that portion of power factor directly related to harmonic current. Caps do not help harmonics, as I am sure you know. They can lead to resonance problems causing high harmonic voltages. They can also, in some cases, push transformers into interesting saturation conditions that can lead to high harmonic currents.
Charles Perry P.E.
Reply to
Charles Perry
Can someone recommend a decent primer that explains these different kinds of power factor you are yakking about? I've always thought it was just one kind of thing.
--Dale
Reply to
Dale Farmer
Sure, so anyone selling "watt-savers", or the like, to residential customers is a crook! ...which is the point here.
Reply to
Keith
Your first sentence says it all. Adding capacitors does nothing if the problem is crap from badly=behaved switchign power supplies. However, PF alone doesn't tell you this.
Hell, even black paint isn't "black". There are thousands of shades of "black". ...your only stating the obvious.
Reply to
Keith
I can't recommend a text, though there was an article here in the last day or so that had a pretty good explanation of the result of a crappy PF on a residential load.
Basically the PF is the (multiplicitave) difference between the real power over the total of the real plus "imaginary" power. In *all* cases we can calculate the real power by measuring the instantaneous power (volts times amps) and averaging over at least a cycle. If we can only measure current and voltage independantly, we can calculate VA but need more information before we can calculate the real power. The relationship between voltage and current is needed. If both are sinusoids the PF is the cosine of the phase difference. If not, as is the case with harmonics (e.g. switching power supplies and fluroscent lamps) the phase difference isn't meaningful but the average of the instantaneous V*A still works.
Reply to
Keith
Here is a nutshell version to look at:
Vrms * Irms = S = Apparent power [measured in Volt Amps or VA] Actual power delivered [measured in Watts] = P P = S * pf = S * pf(Displacement) * pf(Distortion)
Resistive load, e.g. incandescent light bulb: - Fundamental current is close to in-phase with fundamental voltage, pf(Displacement) ~= 1 - Very little distortion, pf(Distortion) ~= 1 - Overall power factor pf ~= 1
Inductive load, e.g. motor: - Fundamental current is significantly out of phase with fundamental voltage, pf(Displacement) ~= 0.8 - Very little distortion, pf(Distortion) ~= 1 - Overall power factor pf ~= 0.8
Certain harmonic loads, e.g. 3-phase UPS input (w/ no input filters): - Fundamental current is fairly close to in-phase with fundamental voltage, pf(Displacement) ~= 0.95 - Significant distortion, pf(Distortion) ~= 0.7 - Overall power factor pf ~= 0.66
So, say you measure Vrms to the UPS and find it to be 600V, and you measure Irms and find it to be 40A, the apparent power is S=600*sqrt(3)*Irms=41.6kVA. The actual power flowing in is only around P=41.6*0.95*0.7~=27.7kW. Though it has low pf, capacitors will help only very little ... if they improved the pf(Displacement) to 1.0 but didn't improve the pf(Distortion) you still have an overall pf=0.7.
For decent primers, Wiki doesn't look great, but maybe OK if you read it over lightly.
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Or try a search along these lines
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Later,
j
Reply to
operator jay
On 1/31/06 7:36 PM, in article P0WDf.9794$ snipped-for-privacy@newsread3.news.atl.earthl>> No. Distortion power factor is that portion of power factor directly
It is not that complicated. Use the calculus of variations. Assume some fixed value of power transfer to a load and a sinusoidal applied voltage. it It is easy to show that the least conductive loss will occur if the load current is also sinusoidal and in phase with the voltage. No power transfer to the load results as a consequence of harmonic currents. Harmonic currents, however, will add to the I^2*R losses. Whether you want to lump effects of reactive currents and harmonic currents together to form some grand concept of power factor is a matter of choice rather of any deep significance.
Bill
-- Ferme le Bush
Reply to
Salmon Egg
Power factor is *always* equal to actual watts delivered divided by volt-amperes (often referred to as 'apparent power').
In a lot of applications, the voltage is a nice simple sine wave with constant frequency and voltage (such as 60 hz, 120V at your wall outlet). The type of load connected to this sort of voltage source will determine the current flow.
One kind of poor power factor can result from using a linear load that has a lot of reactance (either inductive, or capacitive). The current is also a nice sine waveform, but it doesn't peak at the same instant as the voltage. With inductive loads for example, the current peaks later in time than the voltage on each cycle. If you multiply the current times the voltage at many points throughout the sine wave, some of those products will be positive (both voltage and current happen to be positive at that moment, or both voltage and current happen to be negative at that moment). But some points throughout the sine wave, the product will be negative (either voltage or current happen to be negative while the other is positive). So the product (the instantaneous power flow) is sometimes positive, sometimes negative. The net power flow is the average value of all these separate 'instants'. Since some are negative, the average is lower than it would have been if the current was reaching its peak at the same instant as voltage. So the net power (watts) is lower than what just volts*amperes would suggest.
The other kind of poor power factor is a result of non-linear loads. Consider a full-wave rectifier feeding a capacitor and resistor in parallel. When the voltage waveform is less than the charge on the capacitor, the diodes in the rectifier are reverse biased and no current flows from the AC line. When the voltage rises above the capacitor charge, the diode is forward biased and a large current flows from the AC line to charge the capacitor and dissipate power in the resistor. Then as the line voltage drops again at the 'tail end' of the sine wave, the diode is reverse biased again and the current stops. So the current is *
not* a nice smooth sine wave, it is a chopped up thing with sharp upswings and downswings. This waveform has a lot of harmonics in it. These harmonics cause a poor power factor. The reason is pretty complicated and involves some Fourier transforms and calculus, I don't think I can explain it all hear.
Phase-shift power factor problems can be corrected by applying a reactance of the opposite type to that of the load. For an inductive load, installing a capacitor across the supply terminals can create a parallel L-C tank circuit and the capacitor supplies the imaginary current to feed the inductance while the line supplies just the real current to supply the load. But if the amount of inductance varies (as it does with a motor having a varying load), then ideally you would have varying capacitors. Variable caps that automatically adjust are pretty hard to find, so one usually settles for a fixed cap with a value that is a compromise.
Harmonic distortion power factor is harder to deal with. A harmonic trap that will prevent the high harmonic content current from propagating back to the supply is needed to remove the harmonic component. Simple capacitors across the line won't do it.
daestrom
Reply to
daestrom

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