noticed something odd about CFLs

[pete]

Thank you. I suppose this is about average for usenet:

1 question asked, 38 replies. 37 talking complete bollocks and one that's informative. Keep up the good work!

[<me9]

They are not. VAr is from the out of phase component. W is the in phase component. THe 'power' (VA) seen by the transmission medium is the vector sum of the W and VAr.

At some places we had metering for all three. Usually we were charged for KW, but also had a maximum demand charge for the maximum KVA over any half hour within the charging period.

[Kevin]

yup bolloxs rules OK

[Mike Clarke]

Because that's what happens, the two waveforms may well be out of phase but at any instant the voltage and current at that time are used to determine the power. There's no extra adjustment needed for the difference of phase angle it's already accounted for by using the instantaneous values.

That's right, it's already determined the real power, there's no need to do anything with power factors.

I suspect the term "allowing for power factor" has led to much confusion here. Some people might think in terms of multiplying RMS volts x RMS amps to get VA and then making an allowance for the power factor to arrive at watts. This is not relevant when considering electricity meters which integrate the products of instantaneous values over time to measure true power consumption.

[Kevin]

ahhhhhhhhhhhhhhhhhhhhhhhhh a peak voltage of 240v and a peak amps of 10A is 2400VA if the current leads the voltage by any ammount,at the time the voltage reaches its peak then the current is all ready falling so it might be say 9 amp then the meter will read 240X9=2160VA

[Andy Burns]

The 230V (or 240V if you prefer) figure is RMS, so the peak is 1.4x that at 320V.

[The Natural Philosopher]

The two statements are entirely consistent and factual.

If you cat se that, go and read up

They do.

But they don't know that they do.

The time integral of an instantaneous VI product gives true power, as the instantaneous value of I with respect to V includes all the information necessary to calculate a putative phase angle. But the meters don't bother to do that because they don't need to to get the right answer.

Phase angles and power factors are for people who cant do calculus. Fortunately, due to being designed by cleverer people than yu, and electricity meter is a crude form of analogue computer that can.

[The. Wanderer]

They're not. Think vectors, google for vector analysis. The link I posted earlier illustrated the point adequately well.

[John Rumm]

[The Natural Philosopher]

sausage roll withdrawal symptoms.

[John Rumm]

Yup, I think you have it...

Imagine you sample the voltage and current 50 times (or whatever) a second. Each reading you multiply together to get the instantaneous power reading. The result you can sum over time (in effect integrating the area under the power curve). So the meter gets the right answer for the real power consumed regardless of any phase shift between current and voltage waveforms (or any non sinusoidal components in either). The thing it can't do easily is tell you what the power factor[1] is.

[1] And even if it could, knowing the power factor does not tell you the full story, since it assumes sinusoidal voltage and current waveforms.

[The Natural Philosopher]

Hmm. Strictly I THOUGHT it was the ratio of the actual power to the RMS volts times RMS amps...

which has some sort of meaning off sinusoid.

But hey, lets not split hairs.

The saasage rolls are almost cooked...

[John Rumm]

Hmmm, you just reminded me what I forgot to buy the other day!

Oh well never mind, Happy Christmas everyone!

[Michael Shergold]

For goodness sake, why doesn't one of you switch everything off but one CFL lamp and take a look at their meter reading for a few minutes; or an hour if you are not very good at maths. Michael

[Andy Champ]

Ah! All is clear. Thanks!

I don't suppose you feel like fixing the Wikipedia article for the rest of us...

Andy