Talk:Power factor
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The article states "The significance of power factor lies in the fact that utility companies supply customers with volt-amperes, but bill them for watts." At least in Australia, this is not correct. The electricity meters measure volt-ampere hours. The bill incorrectly says kWh, but really you are charged for kVAh. This means that it is in the customer's interest to have unity power factor, since this reduces the amount they pay for electricity.
The article is confusing. What do the various values of the power factor mean? Are they the same (just magnitude) for both sources and sinks? A diagram with at least -1, 0, and +1 would be helpful. How does leading and trailing play into this? Do only sinks have leading or trailing characteristics or do sources too? Does the utilization depend on matching them, or is +1 always desirable for sinks no matter what the value of the source?
If this isn't the right place for comments about the entry I apologise in advance. Please direct me to the proper place.
I'm a rather bored power engineer, so i had a hack about, hopefully it makes a bit more sense now. let me know if it needs clearer explanation
Much better. It might be nice to illustrate the power triangle with something like the graphic at http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html.
This article only talks about "displacement" power factor. What about the effects of non-sinusoidal currents? --Wtshymanski 22:35, 16 Dec 2004 (UTC)
Non-sinusoidal (non linear)analysis is very difficult, and a bit too tricky to try and explain here. I have included some details about harmonics. --Euripides 15:53, 3rd Jan 2005
I think that this sentence could do with a bit more explanation: "Since this stored energy returns to the source and is not available to do work at the load" - why is this exactly? --TimSmall 13:16, 1 September 2006 (UTC)
- The energy does not go into the load; it is reflected back (down the power lines) to the source, and so is wasted. — Omegatron 13:53, 1 September 2006 (UTC)
- Current that is not in step with the voltage does not transfer energy from the source to the load but continually circulates energy back and forth between the source and the load. This energy circulation is not 100% efficient. During each "trip" from source to load or load to source, some energy is lost as heat in the wires and other parts of the power generation, transmission and distribution equipment. Does that help? --C J Cowie 14:24, 1 September 2006 (UTC)
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- The energy doesn't necessarily get bounced back and forth, either. It can disappear into the source. — Omegatron 14:42, 1 September 2006 (UTC)
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- Most of the reactive power flows back and forth between the source and load such that "On one half-cycle, the source supplies energy to the energy-storage element, and on the next half-cycle the energy-storage element returns energy to the source....currents required to supply the stored energy produce losses in the generating and transmission system..." Scott, Ronald E. (1960). Linear Circuits. Reading, MA: Addison-Wesley. “Low power factor means more current and greater I2R losses in the generating and transmitting equipment.” Fitzgerald, A. E.; Kingsley, Charles Jr. and Umans, Stephen D. (1983). Electric Machinery, 4th ed., Mc-Graw-Hill, Inc.. ISBN 0-07-021145-0.
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[edit] Explanation of Apparent and Reactive power
There does not appear to be any proper explanation of what apparent and reactive power are, or how to calculate them.
Mrwooster 02:42, 25 February 2007 (UTC)
[edit] Form Factor
In the case of non-sinusoidal currents and voltages, the term "form factor" is sometimes used to refer to the ratio between the mean value and the RMS (root mean square) value. The form factor can become large in the case of pulsed waveforms with a high mark to space ratio.
- True, but I didn't think it clarified power factor at all. --Wtshymanski 03:04, 16 November 2005 (UTC)
[edit] Question re load sharing generators
dear sirs,
i've just ran into your discussion about <power factor>. would you participate in following:
two equal generators running in parallel load sharing mode, - both of them loaded equal due (as sample 2 x 150kW) but one has too less current than another one (difference is about 150 Amps).
if to rely on the fact that P = square root 3 x U x I x Cos Phi we face reducing power factor for one of generators ... it is clear.
while peaks of load, as a result, one with higher current is tripping due to overcurrent protection.
the practical question is where the the influence for <power factor> appeared from (?) if measurements taken around both generators' excitation systems shows no difference.
with other words: could it be that consumers' =low state of insulation= causes such a behaviour ?
thanks for yr participation brgds Alex
- This looks like a question for http://eng-tips.com electric power engineering section --C J Cowie 15:24, 1 March 2006 (UTC)
[edit] Advertising links
The last four of the five external links seem to various extents to be advertising links. However each one seems to be a direct link to useful information related to the subject of this article. Should they all be removed? --C J Cowie 01:30, 23 August 2006 (UTC)
I was wrong about that. Learn about Power Factor provides forms to be filled out to request information by email. Very little information is directly available from the linked page. I am removing that one now. --C J Cowie 17:43, 23 August 2006 (UTC)
User:Wtshymanski acted while was still talking. --C J Cowie 17:46, 23 August 2006 (UTC)
[edit] Physical Explanation of Leading Power Factor
I was just wondering if anyone can explain how a leading power factor makes sense physically. Current is a result of voltage, isn't it? So how can an effect possibly precede its cause? Unless it is really lagging by >=3/4 of a period...? Anyone? GBMorris 22:00, 9 December 2006 (UTC)
[edit] Amateur explanation of power lead/lag
Having never studied this stuff to Uni level but having been interested since childhood, here is my take on the logic of current lead/lag:
A capacitor is like a short-term rechargeable battery, only a rechargeable needs to be charged carefully. A capacitor will act effectively like a short circuit until it is storing the same voltage as is being applied. At the other end of the scale, a capacitor that is sitting across a 12v DC line and is charged to 12v presents an almost infinite (minus leakage) resistance across the 12v line. Should that 12v line waiver, the capacitor's charge will try to either boost the deficiency or grab the increase.
Now put this across rectified AC, empty capacitor, 0v line. Line voltage increases, capacitor absorbs. Load also starts to absorb. Although the load will not reach full current until the voltage peaks, the capacitor will grab every electron it can get its layers on and therefore the current through the capacitor will be hugely higher than that through the load. Therefore, the current the line supplies will be far more to do with what the capacitor wants than what the real load wants. By the time the AC peaks, the capacitor is fully charged and presents no load so the real load gets everything.
Now, the AC wave starts to diminish. The line starts to become less charged than the capacitor, so the capacitor starts to discharge into the line. The load is supplied by both the line and the capacitor. The current drawn by the line is therefore much less than would be expected, due to the capacitor's help.
This is why the current curve precedes the voltage curve by 0.25 of a full sinusoidal cycle with a capacitor across the line.
Coils are the opposite of capacitors in that they fight, or kick against, any change in voltage. Put a coil across the same rectified AC line and it will appear as a near infinite resistance whilst the voltage is rising. As soon as the voltage has peaked and started to decrease, however, the coil then becomes less and less resistant. The nearer the voltage gets to zero, the less resistant the coil is. Near zero it is almost a short circuit and so the line current is at its highest. Having said that, the strange part is that the coil is now fully charged. Once the voltage starts to increase again, the coil then starts to discharge. So although the load drawn by the load is xAmps, the large part of this will be supplied by the coil and only a small amount by the line. This is why the current curve lags the voltage curve.
Read up on loudspeaker crossovers and passive graphic equalizer units. Also look at power supplies which used to have a diode valve (American:Tube) followed by an inline choke (coil), which then became bridge rectifier diodes followed by capacitor across the line.
Then look at rechargeable batteries: put 4 in parallel and charge to 1.2v. Once charged, switch them to series and they will return 1.2v x 4 = 4.8v. Now apply this to capacitors and you have a switched mode power supply. —The preceding unsigned comment was added by APNorth (talk • contribs) 23:41, 21 December 2006 (UTC).
[edit] ELI the CIVIL ICE man
Having gone straight from ham radio in high school to physics in college I missed out on all the nice EE mnemonics like "Little star up in the sky, power equals R squared I" or whatever it is, so ELI the CIVIL ICE man is news to me. Mulling it over, I concluded there must be four types of EE.
Type 1. Depends on mnemonics to remember whether current leads or lags voltage in capacitors and inductors.
Type 2. Intuits that current lags voltage in an inductor by thinking of inductors as flywheels that respond sluggishly to an applied force. Doesn't intuit current lead for capacitors (how could a mere capacitor predict the future?) but does remember that capacitors are the opposite of inductors.
Type 3. Is equally at home with voltage causing current and current causing voltage. An AC current applied to a capacitor causes the voltage to rise from zero when the current is at its peak and causes the voltage to reach its peak just when the current has dropped to zero and is starting to go negative. Hence in a capacitor voltage lags current by 90 degrees.
Type 4. Understands flywheels by analogy with inductors. --Vaughan Pratt (talk) 08:21, 26 January 2008 (UTC)
- I don't know if there are four types of EE; after a few years in the business you don't think about this any more than you think about how to tie a tie. The mnemonic is just there to backstop you on an exam. Analogies are tricky - best to learn the physics ( then you don't have to remember which property is which - is speed more like current or voltage? )--Wtshymanski (talk) 04:30, 27 January 2008 (UTC)
[edit] Distribution vs. Generation Losses
If a poor power factor is balanced by compensating complementary elements, the distribution losses only occur within that sub-system. Low power factors do not lead to generation losses unless they are not compensated somewhere, within the system as a whole. So, in general, small users with some equipment with poor power factor need not feel guilty. But if poorly designed computer power supplies are commonly used, then there could be significant system-wide issues.-69.87.193.242 12:49, 4 April 2007 (UTC)
[edit] Power factor for AC
I know to calculate power in an AC circuit you have to multiply the current times the voltage times the power factor, but what if you are actually measuring current with a meter? the initial equation for power factor is based on ohm's law. If I stick a current meter in a hot AC circuit does it measure a current that can be multiplied by the actual voltage to get power? using cosine theta would seem to me to throw this off. Then again I'm not an expert by any means. Expert advise please, and also perhaps if I'm right and the power factor only accounts for inductance as a means of correcting ohm's law for an inductive load that it should stipulate that you can use ohm's law directly with measured variables on a physical circuit.
[edit] Graphs
The graphs are pretty nice, but I think they should show RMS (root mean square) power rather than average. RMS voltage, power and current is the quantity typically discussed with AC signals.
Showing that the average AC power is zero in the power factor 0 graph is true but misleading. The load does not see 0 watts, and this is what a casual reader may conclude. —Preceding unsigned comment added by 24.68.59.219 (talk) 21:22, 10 April 2008 (UTC)
- I believe that average power is what the original illustrator intended to demonstrate; that is what I did when I reimplemented the graphs in Gnuplot/SVG. There is some justification in doing so; an ordinary integrating electric meter will also show zero energy used when presented with a zero-PF load, which is why industrial consumers are nearly always demand-metered. 121a0012 (talk) 06:24, 17 April 2008 (UTC)
