What is power factor, anyhow?

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George wrote:

If power factor correction helped a piece of equipment produce less of a starting load, would that not help?
TDD
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Wayne Whitney wrote:

There are industrial units that automatically switch different capacitors in and out of the circuit to correct power factor. These units are quite expensive and not something a homeowner would buy unless you're Bill Gates and have a home that uses the power of a small factory. Oh yea, I forgot Al Gore, I have heard that the savior of humanity is quite the power hog. Any home can benefit from power factor correction but there is no one size fits all magic box as far as I know.
TDD
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Sounds like Bud is a middle-aged electrical engineer who knows what he is talking about!!
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...

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!
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wrote:

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.
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wrote:

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.
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wrote:

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.
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wrote:

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.
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On Jan 29, 11:21pm, snipped-for-privacy@panix.com (David Combs) wrote:

That's OK, we'll give you a pass since you're obviously suffering from BDS.
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I can just see... a back yard loaded up with oil drums chained together with jumper cables.
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I wonder what kind of IR losses are going to be in the pole pig transformers when you get a low enough PF to get free electricity. Ive seen what happens to an engine /generator set when it was feeding a UPS with a PF of .57. I couldnt convience the D... A.. enginneer that it OK to disconnect the bank of PF correction caps, that it wasnt written in stone that you had to use ALL of them. All the smoke started coming out, 1200 amp breaker tripped.
Jimmie
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I thought we dispensed with the idea that power factor has anything to do with a customer getting free electricity many posts back.

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In

lol, that is supremely ignorant and impossible! It's a silly myth and even in theory can't be made to work using today's physics.

--
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Often you'll find excellent advice on a newsgroup.
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For a power factor of one, the voltage and current are in exact agreement/phase for the peaks and nulls. If you apply a voltage to a pure resistor, they are in phase and the power used is the product of the voltage and current. If you apply a voltage to an inductor/motor/ transformer, the current lags behind the voltage somewhat and the power into the device is not the product of the voltage times the current, but the product of the voltage times the current times the power factor. If you apply a voltage to a capacitor/condensor, the current leads the voltage somewhat. Again, the actual power into the device is the product of the voltage times the curent times the power factor. If you have a very large inductance. the current will lag far behind the voltage, and the actual power used in the device is small. But, the power company still has to be able to provide the maximum current to the device, but does not get much $$ since the power atually consumed is much less. That is why the power companies do not like inductive or capacitive loads. Compact fluorescent lights tend to look like capacitive loads because they have a large inrush current that leads the voltage and so altho the power companies will save some power when these are adopted over a widespeard area, they will still have to be able to provide the peak current that these lamps draw. Fortunately, that peak curent is still well below what a resistive incandescent light that puts out the same lumens uses.
Hope this helps.
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hr(bob) snipped-for-privacy@att.net wrote:

If you study how a mechanical killowatt-hour meter works, you'll find that it depends on the voltage and current being in phase. Usually phase-shifting loads (transformers, motors, CHLs) are negligible (their resistive loads are much greater than their reactive loads) and their effect is ignored. Put a big enough reactive load (that shifts the phase between voltage and current 90) on the circuit and the old-fashioned, mechanical, killowatt-hour meter will do what?
It will stop. Or run backwards.
Specifically:
"Some combinations of capacitive and inductive load can interact with the coils and mass of a rotor and cause reduced or reverse motion. All of these effects can be detected by the electric company, and many modern meters can detect or compensate for them.[nudge-nudge, wink-wink]" ["Detection" is usually accomplished by looking at the WTF? box attached to the distribution panel.]
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hr(bob) snipped-for-privacy@att.net wrote:

I agree with other comments - particularly hr[bob], Wayne, George and trader.
Accuracy of watt-hour meters comes up at alt.engineering.electrical. Both mechanical and electronic meters accurately measure the real power (watts). There is no question. (If anyone is interested in how mechanical watt-hour meters work details are in a thread on alt.engineer.electrical starting 11-21-09 and titled "Balancing the Breaker Box". Much of the thread is forgettable, but look at posts by snipped-for-privacy@shawcross.ca and daestrom , particularly starting 11-30. This is only readable by techies.)
Adding to hr[bob]'s post - the current through an inductor (as in a motor) stores energy in a magnetic field. As the current is increasing, additional current is required to store the energy. As the current is decreasing the energy comes out as current that flows back into the grid. This shifts the peaks of the current flow away from the peaks of the voltage. For a motor, the overall current is increased over the current that does mechanical work.
In a capacitor, energy is stored in an electric field. As with an inductance, the energy flows into the field and back out to the grid, shifting the voltage and current peaks.
Watt-hour meters ignore reactive currents (inductance and capacitance) and measure only real power (watts). (They record energy - watt-hours.)
Energy storage in capacitors counter balances energy storage in inductances and can return power factor toward 1. Utilities have racks of PF correction capacitors to raise the PF toward 1, which reduces currents, which reduces utility resistance losses.
Utilities can meter several things. One is real power (energy), which is what watt-hour meters measure. For commercial and particularly industrial facilities they may also measure "demand" and "VARs".
"Demand", which is in Daring's posts, is the peak power consumed (max of watt-hours measured over a small time period). High peak power means the utility has to have capacity to supply that demand. If usage can be shifted, like cycling air conditioning when a facility is using a lot of power in other areas, the capacity the utility has to provide is reduced and the facility is rewarded by lower "demand" charges. IMHO motor starts are too short to change demand much. On a mechanical meter demand is a dial around the top of the face of the watt-hour meter.
"VARs" are volt-amps - reactive. This is reactive power, and peaks when voltage and current are at 90 degrees. If V and A are in phase (resistive load) the VARs are zero. It is measured with a separate watt-hour meter (could be combined in an electronic meter - don't know if they do). This reactive power is not "used" - it flows from the grid and back to the grid 120 times per second. But it increases the resistance losses for the utility, and the utility can charge a big penalty for VAR 'use'. That provides an incentive for facilities to reduce the VARs using PF correction capacitors which, as someone said, have to be switched in and out depending on motor load. PF correction at motors reduces the resistance losses within the facility.
Since residential users pay no VAR penalty "black boxes" to increase PF are a scam. Residential users pay only for power that is used.
Equipment that includes a DC supply distorts the current, with high current near the peak voltage and low or zero current the rest of the cycle. This includes switch mode power supplies (computers, CFLs) and variable speed drives. The V & A peaks are not so much shifted as the current wave form is non-sinusoidal. I think this is referred to displacement power factor. (Power factor is defined as power divided by volts times amps.) I believe the European Union has limits on this.
--
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...

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!)
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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
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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.
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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
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