Reflected Power

Yep, with a difference. For industrial size loads, it would be rather impractical to employ the same methods found in switching power supplies and such. For example, one such PFC standard: covers only devices running up to 220VAC 16Amps draw and exempts: Equipment with rated power less than 75W, except class C equipment. Professional equipment with power > 1 kW. Symmetrically controlled heating elements with power =< 200W. Independent dimmers for incandescent luminaries with power =< 1 kW. In other words, it doesn't apply to high power systems. For example, it would be impractical to use the same load tracking PFC circuitry, on a factory full of sewing machines, where the load might vary from 1 sewing machine to hundreds over very short time periods. However, applying it to individual sewing machine motors would be appropriate. Compared to PFC IC techniques, the simple bank of PF correcting caps used across power lines to the entire factory are crude and imperfect, but good enough for todays applications.

--
Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
Santa Cruz CA 95060 http://802.11junk.com 
Skype: JeffLiebermann     AE6KS    831-336-2558

Oh yes, I will go away, because you tell me to. Who do you think you are... Bill Sloman?

joe

I don't define it that way. IEEE does. But you should be able to convince yourself that it is true if you apply first principles and some education.

Yes, it could be if the reactive load is 0. But your problem is that not all reactive loads are 0. So, suppose you have an inductor of 1 ohm reactance across a 100V source. It has no resistance, just inductance.

What is the current? What is the power used? What is the power factor?

Note that the current is *not* infinite.

This is not a trick question. Show your analysis.

I can't find any online explanation of knitbitting, so I can't understand that sentence.

If your assertion is true, it doesn't matter what theoretical case I dream up. You must explain why your assertion works in *all* situations, including mine, because that is what theory is all about. Do you have an example of your assertion that you can prove?

No, you learned about the very special case of pure sine waves. That's not the real world. John's formula is the general case.

Yes, and that includes harmonics (usually caused by nonlinear loads).

They do the same thing; cause power loss.

Only if the voltage is zero. No power, no heat. So what?

I'll take your word for it. Oh wait! Now I remember from some time ago that the book about power electronics was talking about Power Quality Factor, if, at least, I got

*that* right. Which makes me concede to your PF definition, but wondering how the PQFactor ended up on the wikipedia page as being PF.

Yes it *is* a trick question. :) If you are allowed a zero resistance, then I am allowed an infinite current. Currents aren't infinite because there is resistance.

Then I don't understand why my US-english spell-checker didn't go off! By the way, it does go off on the word 'english', strange enough.

Ok, you really insist on it, don't you? So let's re-assert it:

If you want to talk about the theoretical case of PF = 0, then theoretically the current could be infinite for a kW != 0.

Do you agree that this is the statement to prove?

Then take a load that has an adjustable (down to zero) power factor PF, and a resistance that's proportional to PF, let's say: Z = r*PF. Then for all PF the active power is (V^2/Z)*PF = [V^2/(r*PF)]*PF and in the limit for PF-->0 this is V^2/r != 0. The current is I = V/R and as PF --> 0, I --> inf.

I think I just have proven _some_ case in which the current _could_ theoretically be infinite while kWh != 0. ;)

Not that it has any practical meaning whatsoever... But none the less I accepted your challenge, was fun. Or did I introduce some dirty math trick in there... I really don't know. When you prove me wrong I'll know it. :)

joe

thank you

Mark

Yep, the same thing happened to me.

In EE school, they taught us about PF relative to pure sine waves.

Then in the mid 80's harmonic current correction in switching power supplies became popular and in the real world we all had to re-learn the definition of PF.

I went through the exact same scenerio.

Mark

Joe, here is one that will blow your mind...

Assume a 1V source that is switched at 50% duty cycle. It has a 1 ohm load.

What is the power in the resistor?

What is the source Vrms*Irms?

Now, what is the power factor seen by the source?

Conclusion: Power factor can be calculated no matter what the conditions. DC, AC, arbitrary stuff.

The point is that everything you have learned can be applied in every situation *if applied properly*.

I was just arguing that I was the one incorporating the harmonics in the PF formula by adding the multiplicative term I1/Irms. Oh, I also explained it:

I1, rms is the RMS value of the base harmonic current component (50 Hz or 60 Hz depending on your location) and Irms the RMS value of the total current.

As I argued indeed. :)

Yes, we agree. :)

The crucial word here is: 'theoretical'. Expect some math involving limits into infinity or zero.

cheers :)

joe

I have never heard of Power Quality Factor. I am at a disadvantage here.

Okay, Joe, what inductance with zero resistance does it take to have infinite current? Do you not understand R+jX?

Sorry, I used Google to try to find it. But, not a problem. That is minor.

No, I do not. Worse, I do not understand kW != 0.

Sorry, Joe, the above has no meaning to me. Perhaps it is my inability to understand. After all, I'm no spring chicken.

The only way that current from a source can be infinite is that the load on the perfect source is a perfect 0+j0 ohms.

If you can show otherwise, please do.

I don't think you *intentionally* introduced some trick because I don't think that was you intention at all. This is an interesting discussion and I'll be happy to supply what little knowledge I have to further understanding, if you wish.

Let me know what you need/want for proof.

I'd figured as much, that's all your good for.

Jamie

To follow up on this, how would you calculate the power into a load with arbitrary waveforms for current and voltage? The obvious way, don't you think, is to multiply the instantaneous voltage times the instantaneous current to get instantaneous power and then average the result over the period of your interest? Is there something wrong with that?

Then, at the same time, over the same period of interest, you calculate (or measure) the Vrms and Irms. Now you can calculate the power factor with power/(Vrms*Irms).

(By the way, Vrms and Irms are also integrals. But, let's not get into that)

Power factor is usually only applicable to power line situations even though it can be applied most anywhere. It just makes sense only in special situations. You have to decide when that situation is applicable.

Please tell us how a linear load can distort? If not, what do they do?

And you're (not your) good for what?

The definition that you presented differs from the one I was taught 'back in the days' that I was learning electronics.

I really don't feel to argue in such a negative way about it. I have explained it extensively and am not prepared to discuss this with you while being attacked in such a tone as you are presenting here.

Thank you!

joe

I think we already agreed that the definition you have been taught is newer and more general than the one I was using.

joe

Did you see the wikipedia page on power factor?

There is no inductance with zero resistance and there is no infinite current. Your question is an extreme theoretical case and can not be answered with a practical situational answer. It is a limit case. I'll try to expand my apparently too cryptic earlier answer below.

Well, maybe you can ask Bill Sloman. I'm sure he doesn't mind me mentioning him here, as he has the habit attacking other people who are totally non-involved in threads in which I've been discussing with him.

!= is short for 'not equal'. I think I used it in bash before, or php or whatever. After explaining this please don't get upset if I continue using this definition of unequal (!=), ok?

There is no 0+j0 ohms. So we need to take mathematically the limit case.

I define active power (measured in kW) in a harmonic current and voltage system, 1 frequency, no harmonics or other higher frequencies, as P = V * I * PF, in which PF = cos(phi), in which phi is the phase angle between V and I.

We know that I = V / Z with the impedance of the load equal to Z = r*PF.

Then P = V * I * PF and its value as we bring PF to zero is

P = lim [V^2/(r*PF)]*PF = V^2/r != 0 PF->0

I think I indeed did trick you, not sure though if that influenced the outcome. But indeed it wasn't intentional.

Now it's your turn, where did I trick you? :)

joe

** Trick question.

The PF seen by the source is of course 1, due to the resistive load. That the source goes on and off makes no difference, cos a dead source has no PF.

However, if the load were switched on and off instead the PF is then 0.71.

PF is all about the efficient use of equipment and cable and the most efficient use is when the load is steady DC.

.... Phil

I'm not attacking you.

What is V*I*cos(90)?