Part of the magic of sine waves is you can add any number of them (at the
same frequency) with different phases and still end up with a sine wave.
This is how QAM works (add two 90 degree out of phase signals - they're in
quadrature - and the result is a sine wave whose amplitude you can vary -
square root of the sum of the squares - and whose phase angle you can vary
- arctan of the ratio of the two amplitudes). Kinda magic! So, adding two
sine waves that are 120 degrees out of phase (as across two hots of a
three phase Y) results in a sine wave.

As to power factor variations with line voltage, I'm GUESSING that with a
higher line voltage, the rectifier needs to conduct for a smaller portion
of the cycle since, even if the same current were drawn at the higher
voltage, you'd have more power, so you'd need to deliver that power from
the AC line for a shorter period of time, resulting in a lower power
factor.

In your power factor formula, where did the 1.73 come from? Imagine a
resistive load, which should have a power factor of 1, and has W=VA. It
would not come out to be 1 in that formula.

Finally, the watt hour meters used by the electric companies indeed
measure watt hours, not VAH. So you actually pay for the energy you get.
The advantage to high power factor is largely, I believe, in efficiency.
Losses tend to be proportional to the square of the current (I squared R
losses), so a higher current that does not actually deliver more power
DOES result in increased losses. These losses are in transmission lines
and transformers. These losses can cause both of these to overheat. To
avoid overheating, they must be oversized, which costs money.

For info on these losses (especially excess neutral current) in light
dimming systems, see
http://www.dovesystems.com/pages/apnotes/LDI2001/img0.htm

Harold


> Al,
>
> On Wed, 10 Mar 2004 07:11:19 -0800, al smith wrote:
>
>> I did some measurements, where I found peak current for 122VAC single
>> phase
>> to be 1.554A being pulled on the AC, and then did the same test using
>> 212VAC
>> (two legs of a 208 3 phase) and found the AC current draw to be 0.935A
>
> This has always perplexed me: surely if you're connecting across 2 of 3
> phases (something we don't do in the
> UK) then the voltages are at 120 degrees phase angle to each other, so you
> will be getting a distorted
> waveform, not a sine wave?
>
>> Thus wattage is 189.6 for single phase, and 198W for the '3phase'.
>> Pretty
>> close in value, although I might expect them to be pretty much the same,
>> since the same load was applied.  So thats the first curious item.
>
> Unless you're using "True RMS" meters, I think the distortion I mention
> would mean that the readings would be
> skewed.
>
>> Then, using the formula  A=W/(V*pf*1.73)  I get a pf of 0.58, much lower
>> than I might expect. So is there a relationship of the pf related to
>> voltage?  I would tend to think not, as its the comparison of apparent
>> vs
>> real power, or the efficency of the supply.  I wonder if running it
>> using
>> two of the 3 legs of a 3 phase power has anything to do with it?  I do
>> have
>> a variac, so maybe I will run the same test, using a single phase 212VAC
>> and
>> see if there are any differences.
>
> I'd be interested to hear the results - and to see a 'scope trace of the
> waveform, so I can see if I was
> right! :-)
>
> Incidentally, is power factor correction (PFC) a Good Thing for the user
> (ie. does the meter go round less for
> a given power used) or for the supplier (the meter goes round more)?
>
> Cheers,
>
> Howard Winter
> St.Albans, England


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