- posted
16 years ago

can we use to reduce power factor at home, and what are the benefits.

- posted
16 years ago

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.

can we use to reduce power factor at home, and what are the benefits.

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- posted
16 years ago

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

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

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

- posted
16 years ago

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.

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.

- posted
16 years ago

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

j

- posted
16 years ago

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.

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.

- posted
16 years ago

Great post!
Ed

- posted
16 years ago

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.

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.

- posted
16 years ago

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.

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.

- posted
16 years ago

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.

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.

- posted
16 years ago

----------------------------

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

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

- posted
16 years ago

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.

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.

- posted
16 years ago

If by "distortion power factor" you mean something to do with current spikes
or such, it seems to me that capacitors CAN "help."

- posted
16 years ago

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.

Charles Perry P.E.

- posted
16 years ago

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

--Dale

- posted
16 years ago

Sure, so anyone selling "watt-savers", or the like, to residential
customers is a crook! ...which is the point here.

- posted
16 years ago

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.

- posted
16 years ago

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.

Basically the PF is the (multiplicitave) difference between the real power over the total of the real plus "imaginary" power. In

- posted
16 years ago

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.

Later,

j

Vrms

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

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

- posted
16 years ago

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

Bill

-- Ferme le Bush

- posted
16 years ago

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

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

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 *

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

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