What is power factor, anyhow?

[trader4]

Ouch!

[Mike Granby]

It isn't complicated. Like I said, certain loads put a strain on the trasnmission system without actually using power. That's all that has to be said. All this show-boating about phase angles, reactance and moving the PF back to unity is what is getting in the way.

[George]

So your opinion trumps everyone then? Suppose someone wanted to actually understand what was happening rather than the trivialized don't know any more than you did before you read it explanation?

[HeyBub]

"Those who say a job is too difficult simply haven't tried hard enough." [Ronald Reagan]

[HeyBub]

Well, if one can't understand the problem, he just has to take the word of his betters. I guess.

[Twayne]

$File/Qpole%20Pole%20Mount%20Brochure.pdf OUTstanding, heybub! Thanks for the education, really. I have never seen them and if I had I'd be sure to have remembered them! At first I thought they might be all Pac Rim and EU located, but no, there are a few for N.A. too, so obviously they are used here, IMO.

I sincerely apologize to you; hope you'll accept it. I was wrong and made suppositions I shouldn't have. I was surprised to see polyprop as the material, too. Guess I shoulda known better; next time I'll be a lot more careful.

Twayne

[David Combs]

Now, a prior piece of this thread said it'd generate heat, the "missing" power, and the next one agreeded with that.

How does that assertion jibe with it's "using" less?

(Sorry for the delayed followup.)

David

[David Combs]

In article , The Daring Dufas wrote: ...

Sorry to butt in, but isn't that the same guy who had an election OPENLY stolen from him, and not only didn't complain but snuck (slithered?) off, tail between legs?

(And then the next guy, who swore he'd fight it to the death if they stole it from him -- didn't he do exactly the same thing?)

Sickening. Am sickened. Especially so because in each case, NO ONE ELSE complained either! Or seemed to care.

Just what is this country coming to?

David

PS: again, sorry for the intrusion into a really interesting thread, but when I saw that name, I just couldn't help myself!

[David Combs]

In article , hr(bob) [e-mail] wrote: ...

A not only much delayed but probably stupid question:

If you phase-shifted volt-amp pair, and multiplied them together at each time-instant, and addeded them up (ie integrated the product dt)?, would that give the same result as multiplying the peak voltage and peak current by the pf?

Thanks,

David

(I gotta go read that wikipedia article!)

[Steve N.]

No, you're remembering it all wrong. He was the presidential candidate whose supporters were so dumb that they couldn't vote properly enough that their votes could be counted. It doesn't take any intelligence to have the right to vote, but it does take a smidgen of intelligence to know how to mark your ballot, so that your vote can be tallied.

[Steve N.]

No, you're remembering it all wrong. He was the presidential candidate whose supporters were so dumb that they couldn't vote properly enough that their votes could be counted. It doesn't take any intelligence to have the right to vote, but it does take a smidgen of intelligence to know how to mark your ballot, so that your vote can be tallied.

[Steve N.]

No, you're remembering it all wrong. He was the presidential candidate whose supporters were so dumb that they couldn't vote properly enough that their votes could be counted. It doesn't take any intelligence to have the right to vote, but it does take a smidgen of intelligence to know how to mark your ballot, so that your vote can be tallied.

[Steve N.]

No, you're remembering it all wrong. He was the presidential candidate whose supporters were so dumb that they couldn't vote properly enough that their votes could be counted. It doesn't take any intelligence to have the right to vote, but it does take a smidgen of intelligence to know how to mark your ballot, so that your vote can be tallied.

[trader4]

They are two seperate but related issues. If the voltage and current are not perfectly in phase you are in fact delivering less power into the loads in the house than you would if they were in phase. That means to get the same amount of useful power into the house, it takes more current. Since wires, switches, etc all have some small but finite amount of resistance, using more current to deliver the same amount of power results in more loss that doesn't do any useful work. If you were an electric utility with miles of wire, this becomes a real issue. However for residential users, the effect is so small that it's insignificant, hence power factor correction isn't an issue.

[Wayne Whitney]

I assume you mean voltage and current waveforms that may be out of phase.

You have the right idea but the details are a little off. In the numerator, you need to integrate over 1 second. And in the denominator, you need to use the root-mean-square values of voltage and current (which are the commonly referenced values anyway). Then you'll have power factor. Otherwise, you'd be off by a constant factor, i.e. you won't get a value of 1 when the voltage and current are actually in phase.

Cheers, Wayne

[trader4]

That's OK, we'll give you a pass since you're obviously suffering from BDS.

[trader4]

Actually he's right as far as using the instantaneous voltage and current and doing the integration. That gives the true power. But as you say, the comparison value calculated the normal value would be to use the RMS voltage and current and power factor, not peak values.

[Wayne Whitney]

I think we are in agreement, I guess I just wasn't clear. If you integrate V*A over a time interval, you get energy. So to get true power, you need to divide by the length of the time interval. Which is why I suggested integrating over an interval of 1 second.

Cheers, Wayne

[trader4]

Yes, I agree. We're thinking the same thing, but I was a little off there too :) After doing the integral, you do have to divide by the time interval, whether it's one cycle or one sec, etc.

[David Combs]

Got it.

Thanks,

David