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# Average power factor

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#### alphacat

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I want to know how inefficient a house is.

I'm using an IC that calculates the power factor of the load connected to it.
(It measures instantaneous values of current and voltage, then calculates active and apparent powers, and finally calculates the power factor).

I spread these ICs around the house and have each electrical home appliance connected to such IC, and I'd like to calculate the "average power factor" of the house, in order to know how much inefficient my house is.

Once I have the power factor of each appliance, what should be the correct way to calculate the average power factor of the house?

I thought of:
[1 / ∑P(i)] * ∑[P(i)*PF(i)]

The reasons behind this are:
1. I give a larger significance to larger consumers.
2. When an appliance is off, then its calculated PF is 0, and I dont want to take this into account.

Thank you.

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Industry is charged extra for having a poor power factor. Homes are not because homes are charged only for the amount of real power they use. The electricity company loses power when homes have a poor power factor.

Industry uses huge electric motors but homes do not. It is the motors that cause a poor power factor. Your refrigerator motor and the fan motors in your furnace and air conditioner have a poor power factor but their power is low.

Thanks!

So, you're saying that an energy provider wouldnt be interested in the average power factor of an house since even if its bad (relatively to other houses - not factories), the losses spent on this house are not major at all?

I think the loss of power to an energy supplier of a home that has a high power factor is very low. But there are many homes and the losses all add together. Maybe in the future homes will be charged extra for a high power factor.

Hello,

If you still want to calculate this, you could add up all the apparent powers
and add up all the real powers and then do one calculation to determine
actual power factor, not 'average' power factor.
If you had a resistor and capacitor and you calculated the power factor,
then you had another resistor and another capacitor in the same circuit,
how would you calculate the power factor for both? Same way. Calculate
the real power and the apparent power (for both R's and C's) and go from
there, keeping them separate until all of them are known.

TO get a more accurate PF measurement all you need to do is take all the measurements off of your mains that supply the power into your house. Then do a basic data logging system that records it over a regular time period.

Most modern equipment is already power factor corrected to some degree so your actual reading are likely going to be fairly consistent with being around .8 to .9 which is all ready a good number.

Its the very high load users like factories and large manufacturing sites that get fines for having PF numbers around .4 to .6 that have to monitor their numbers and make corrections.

Most modern residential power supply systems on the HV side of the power transformers already have automatic PF correction devices built in to help stabilize the average PF for an area of homes. You may have seen them on the overhead power line poles. They are typically a cluster of 3 or more gray rectangular boxes that have two or three wires going in and out of them from the lines that carry the HV. They are often mistaken for HV circuit breakers which are connected in series with the HV lines and not aross them.

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My electricity comes out of the ground like water and natural gas. The telephone and cable TV/internet wires too.

My sewer goes into the ground. Maybe the ground converts my sewer waste into electricity and all the other utilities.

My city now has "smart elecricity meters". They have an LCD screen and know the time of day. Soon we will be charged the exact hourly cost of electricity so we can use the washer and dryer at night when electricity costs less.

Thank you a lot guys!

Hi again,

Here is something else to think about...

For a circuit consisting of a 1 ohm resistor in series with a 1000uf cap
connected to a 120vac 60Hz power source, it has a certain power
factor lets call it PF1.
For another circuit consisting of a 1 ohm resistor in series with a 7.036mH
inductor connected to the same source, it has a power factor the same
as the cap and resistor above, so we could call it PF1 too.

Now say we have the two circuits connected to the same source which is the
120vac power line...now the power line sees a power factor of 1 because
the cap and inductor together draw a current that is in phase with the
supply source.

The question is, how can we find an equation that allows us to add the
two arbitrary power factors?

The answer is: if we dont know whether they are inductive or capacitive

Thank you very much for this.

A great example!

Motors in a home draw a fairly high inductive current. Capacitor loads in a home are low power like radios, TVs and a computer or two. I think the load is inductive and the capacitive load is almost negligible.

But since homes are not charged extra for the poor power factor then it doesn't matter to us, but it costs the energy provider extra.

Thank you very much for this.

A great example!

Hi again,

Oh you are welcome

You know another comparative example would be
to compare two cap circuits like the one in my previous post to
the cap and inductor circuit working together. The two cap
circuit would have some power factor PF1 less than 1 while the
cap and inductor circuit would have a power factor of 1,
yet all of the individual power factors are the same. Thus
in one case we would get an addition of power factors equal
to some number less than 1, and in the other case we would
get the addition of power factors (the same as previous) equal
to exactly 1. We cant have it both ways

PF1+PF1<1
and
PF1+PF2=1

or with functions:

f(PF1,PF1)=1
and
f(PF1,PF1)<1

it is fairly logical to see that the function f(x,y) can not both equal 1 and
also be less than 1.
Alternately, the two power factors might function out to greater than 1, but
along the same reasoning line the function can not be both greater than 1
and also equal to 1 either.

Numerical:
With the two circuit examples given (1 ohm and 1000uf, and 1 ohm and 7.036mH)
the power factors both come out to about 0.36, so adding we get:
PF=0.36+0.36=0.72

which clearly is not correct as the true power factor for the two circuits in
parallel is 1 as the inductor was chosen to perfectly cancel the effect of the
phase of the capacitive circuit.

If we took the average:
PF=(0.36+0.36)/2=0.36

it is still way off. Same story taking the sqrt of the sum of squares.

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You've got to take into account the sign of the individual power factors; lagging or leading?

See: Power factor - Wikipedia, the free encyclopedia

Hello,

Ok, so what is the resulting power factor when we have two power
factors 0.3605 and 0.03768, both leading ???

If you can answer that this would be quite amazing.
Perhaps *you* should take another look at Wikipedia...

We can combine power factors but only if we know both the power
current for each load (and yeah, also if they are leading or lagging).

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Hello,

Ok, so what is the resulting power factor when we have two power
factors 0.3605 and 0.03768, both leading ???

If you can answer that this would be quite amazing.
Perhaps *you* should take another look at Wikipedia...

We can combine power factors but only if we know both the power
current for each load (and yeah, also if they are leading or lagging).

You seem to be offended by my suggestion. I certainly wasn't trying to offend, and I'm sorry if you were in fact offended.

I didn't suggest that one could calculate a resultant PF with only the knowledge of the individual power factors, so I don't think having another look at Wikipedia is going to tell me anything I don't already know.

It isn't possible to calculate a resulting total power factor if you don't know the magnitude of the individual currents as well as the individual PF's, but you do need to take into account the signs (if they're different), which you weren't doing.

You seem to be offended by my suggestion. I certainly wasn't trying to offend, and I'm sorry if you were in fact offended.

I didn't suggest that one could calculate a resultant PF with only the knowledge of the individual power factors, so I don't think having another look at Wikipedia is going to tell me anything I don't already know.

It isn't possible to calculate a resulting total power factor if you don't know the magnitude of the individual currents as well as the individual PF's, but you do need to take into account the signs (if they're different), which you weren't doing.

Hi there Electrician,

Well, the original poster seemed to want a way to compute power factors
based on having the power factors alone, with no additional information.
My reply was to show that it would not be possible given that little
information. Then you replied to my post saying that i excluded something,
when i was only responding to the desires of the original poster, and to
top it off you ALSO did not include enough information to show how we
might go about calculating the power factors for some loads, so i figured
i might as well reply again with the correct information.

As long as you wanted to get into how it would be possible to calculate
the new power factor we might as well post a method for doing this, just
in case anyone wants to do this in the future. That would mean your
post, although incomplete as well, would at least get us talking about
the possibilities given all the information we need to do this.

For example, if we are given the following information about two loads:
1. The power factors for both loads (and their leading or lagging status)
2. The current measurement for both loads

we can then say that the resulting power factor can be calculated as the
cosine of the angle of the vector sum of the two, with current as magnitude
and of course power factor as the cosine of the angle for each load.

As a worked example, say we have two loads, one with PF=0.36052 and
current of 42.33 amps rms, and the other with PF=0.037681 and current of
4.52 amps rms, both leading. The resulting power factor comes out to
approximately 0.331 also leading. The resulting current is the square root
of the sum of the squares of the individual currents of course.

other with PF=0.037

Hi there Electrician,

Well, the original poster seemed to want a way to compute power factors
based on having the power factors alone, with no additional information.
My reply was to show that it would not be possible given that little
information. Then you replied to my post saying that i excluded something,
when i was only responding to the desires of the original poster, and to
top it off you ALSO did not include enough information to show how we
might go about calculating the power factors for some loads, so i figured
i might as well reply again with the correct information.

In post #5 you said:

"If you still want to calculate this, you could add up all the apparent powers
and add up all the real powers and then do one calculation to determine
actual power factor, not 'average' power factor.
If you had a resistor and capacitor and you calculated the power factor,
then you had another resistor and another capacitor in the same circuit,
how would you calculate the power factor for both? Same way. Calculate
the real power and the apparent power (for both R's and C's) and go from
there, keeping them separate until all of them are known."

It didn't appear to me that you were showing that it "wouldn't be possible given that little information". You didn't mention that just knowing the PFs wouldn't be enough. You may have felt that the OP would understand that from what you said, but you didn't come right out and say it. You expected the OP to infer it from what you did say. If you don't say it, the OP might think that his method would work, and that you were just suggesting an alternative method.

It looked like you were suggesting a method that you thought would work, and you thought the OP would understand from your suggested method that more information was needed. However, the method you suggested won't work, because it doesn't take into account the sign of the PFs.

In one of your later posts you considered the case where there were both capacitive and inductive currents, and you got a result which was obviously wrong.

I was simply suggesting that if you considered the sign of the PF you would get a better result. It was not my purpose to show a complete method for calculating the resultant PF in all cases. You were doing a good job of describing calculation of the resultant; you just left out an aspect which is needed for mixed capacitive and inductive loads.

For example, if we are given the following information about two loads:
1. The power factors for both loads (and their leading or lagging status)
2. The current measurement for both loads

we can then say that the resulting power factor can be calculated as the
cosine of the angle of the vector sum of the two, with current as magnitude
and of course power factor as the cosine of the angle for each load.

As a worked example, say we have two loads, one with PF=0.36052 and
current of 42.33 amps rms, and the other with PF=0.037681 and current of
4.52 amps rms, both leading. The resulting power factor comes out to
approximately 0.331 also leading. The resulting current is the square root
of the sum of the squares of the individual currents of course.

You're on the right track when you say:

"we can then say that the resulting power factor can be calculated as the
cosine of the angle of the vector sum of the two, with current as magnitude
and of course power factor as the cosine of the angle for each load."

But the magnitude of the resultant current is not the "square root
of the sum of the squares of the individual currents" unless the angle between the currents is 90 degrees. You have to use a slight variation of the law of cosines to get the correct vector sum for other angles: Law of Cosines -- from Wolfram MathWorld

The magnitude of the resultant current is given by:

SQRT(I1^2 + I2^2 + 2*I1*I2*cos(Φ)), where Φ is the angle between the two currents.

Φ = arccos(PF1) - arccos(PF2)

The correct formula for resultant PF is rather complicated.

You can see that the simple square root of the sum of squares isn't correct if you just consider the case of two resistive loads (a PF of 1 for both). Let's say the two currents are 3 A and 4 A. Obviously the total current is 7 A, but the square root of the sum of squares calculation would give a resultant current of 5 A, plainly incorrect.

For your example, the correct total current is 46.849 A with a PF of .362096

If you have more currents, it's probably easier to simply calculate the components of each current, add those up and the calculate the power factor of that vector sum. The best components to use are not the apparent powers and the real powers, but rather the real powers and the reactive powers.

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In post #5 you said:

"If you still want to calculate this, you could add up all the apparent powers
and add up all the real powers and then do one calculation to determine
actual power factor, not 'average' power factor.
If you had a resistor and capacitor and you calculated the power factor,
then you had another resistor and another capacitor in the same circuit,
how would you calculate the power factor for both? Same way. Calculate
the real power and the apparent power (for both R's and C's) and go from
there, keeping them separate until all of them are known."

It didn't appear to me that you were showing that it "wouldn't be possible given that little information". You didn't mention that just knowing the PFs wouldn't be enough. You may have felt that the OP would understand that from what you said, but you didn't come right out and say it. You expected the OP to infer it from what you did say. If you don't say it, the OP might think that his method would work, and that you were just suggesting an alternative method.

It looked like you were suggesting a method that you thought would work, and you thought the OP would understand from your suggested method that more information was needed. However, the method you suggested won't work, because it doesn't take into account the sign of the PFs.

In one of your later posts you considered the case where there were both capacitive and inductive currents, and you got a result which was obviously wrong.

I was simply suggesting that if you considered the sign of the PF you would get a better result. It was not my purpose to show a complete method for calculating the resultant PF in all cases. You were doing a good job of describing calculation of the resultant; you just left out an aspect which is needed for mixed capacitive and inductive loads.

You're on the right track when you say:

"we can then say that the resulting power factor can be calculated as the
cosine of the angle of the vector sum of the two, with current as magnitude
and of course power factor as the cosine of the angle for each load."

But the magnitude of the resultant current is not the "square root
of the sum of the squares of the individual currents" unless the angle between the currents is 90 degrees. You have to use a slight variation of the law of cosines to get the correct vector sum for other angles: Law of Cosines -- from Wolfram MathWorld

The magnitude of the resultant current is given by:

SQRT(I1^2 + I2^2 + I1*I2*cos(Φ)), where Φ is the angle between the two currents.

Φ = arccos(PF1) - arccos(PF2)

The correct formula for resultant PF is rather complicated.

You can see that the simple square root of the sum of squares isn't correct if you just consider the case of two resistive loads (a PF of 1 for both). Let's say the two currents are 3 A and 4 A. Obviously the total current is 7 A, but the square root of the sum of squares calculation would give a resultant current of 5 A, plainly incorrect.

For your example, the correct total current is 46.849 A with a PF of .362096

If you have more currents, it's probably easier to simply calculate the components of each current, add those up and the calculate the power factor of that vector sum. The best components to use are not the apparent powers and the real powers, but rather the real powers and the reactive powers.

Hello again,

I think you caught a valid error when you noticed that i had worded one
response like this:
"The resulting current is the square root of the sum of the squares of the
individual currents of course"

but the correct wording should have been like this:
"The resulting current is the square root of the sum of the squares of the
individual component currents of course"

which is something that is well known. That is very good of you to
see that because that means now we will end up with a more word-for-word

There still seems to be a problem with your calculations however, and that
is that it is very hard to believe that you can have two currents, one
that is 42.33 and another that is 4.52 where they are out of phase with
each other and still end up with a current and i quote:
"46.849 A with a PF of .362096".

As a sanity check, you can see that 42.33 plus 4.52 equals 46.85, which is
0.001 amps different than your answer, and that could only happen if the
two where in phase with each other. This leads me to believe very strongly
that you made a mistake in your calculation and you should review that.

You were nice enough to take the time to reply and also to go through
the trouble of attempting a calculation of the resulting amplitude and
the PF too so i will take the time to go over my calculations and make
absolutely sure that there was no mistake in my calculation. I have
been dividing my time quite thin lately posting here and doing other things
so it may take me until tomorrow sometime but i promise i will get back to
this thread either sooner or by then. I'd like to also point out that
another way to do this is to simply look at the components for both
loads and that is quite simple. In the mean time, perhaps you can look
over your calculation of the resulting current and also the PF.
My goal now is to present the OP with a tried and proven method for
calculating the total load PF and current amplitude at the very least,
since we are now knee deep in discussing it

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"As a worked example, say we have two loads, one with PF=0.36052 and
current of 42.33 amps rms, and the other with PF=0.037681 and current of
4.52 amps rms, both leading. The resulting power factor comes out to
approximately 0.331 also leading. The resulting current is the square root
of the sum of the squares of the individual currents of course."

I copied one of the numbers incorrectly to a piece of paper. I copied the second PF as .37681, rather than .037681.

Let's go through the procedure to calculate the components of each current.

42.33 with PF=.36052 This PF corresponds to an angle of 68.8678 degrees, so the real component is 42.33*cos(68.8678) = 15.2608. The reactive component is 42.33*sin(68.8678) = 39.4834

4.52 with PF = .037681 This PF corresponds to an angle of 87.8405 degrees, so the real component is 4.52*cos(87.8405) = .17032, and the reactive component is 4.52*sin*(87.8405)= 4.51679.

The total current's components, real and reactive, are:

Itot = 15.43112 real and 44.00019 reactive

The angle of the resultant is arctan(44.00019/15.43112) = arctan(2.8514) = 70.674 degrees. The PF of the resultant is cos(70.674) = .33094, and the resultant magnitude is SQRT(15.43112^2 + 44.00019^2) = 46.6276.

This is slightly different from the result I got from the PF where I mistakenly dropped the zero in .037681.

If the RMS values of the various currents are denoted by I1, I2 and Itot, then the RMS value of Itot is not given by the square root of the sum of the squares of I1 and I2. In that case you have to use the formula Itot = SQRT(I1^2 + I2^2 + 2*I1*I2*cos(Φ)), where Φ is the angle between I1 and I2.

But, if you have decomposed I1 and I2 into their components and added the components to get Itot_real and Itot_reactive, then the RMS value of Itot is the square root of the sum of the squares of the two components of Itot.

These two currents aren't really very much out of phase. Their PFs are both leading, and the difference in their two phase angles is only about 19 degrees.

Hello again,

Ok, so you now got the same result i did several posts back?

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