In article ,
John Rumm writes:
Since my answer to a question on PF a while ago seemed well received, I
thought I may as well hack it about a bit for a wiki article. Comments,
edits etc as you see fit:

http://wiki.diyfaq.org.uk/index.php?title=Power_factor

John, it's discussing power factor only in the context of phase
shift (which was the context of the original thread IIRC).
Nowadays, lots of low power factor loads have little or no phase
shift contribution to their low power factor.

It should probably include the definition of PF, i.e. PF = W / VA
and also the special case of power factor due to phase shift only,
where PF = cos(Ø).

--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]

Mark wrote:

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

I learnt about power factor back in the early 60s and I too do not
remember anything other than phase shift. Memory fades of course but the
goal posts may have moved since if switched mode power supplies either
weren't around then or hadn't established sufficient presence to matter.

I don't recall anything about the 'useless' power just being temporarily
stored and then returned every cycle either which again could just be
loss of memory but it leaves the nagging question which someone might
care to answer - If the useless power is returned in full why is it that
commercial users get charged for kva?

On Tue, 15 Dec 2009 15:04:46 +0000, Roger Chapman wrote:

Mark wrote:

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

I learnt about power factor back in the early 60s and I too do not
remember anything other than phase shift. Memory fades of course but the
goal posts may have moved since if switched mode power supplies either
weren't around then or hadn't established sufficient presence to matter.

I don't recall anything about the 'useless' power just being temporarily
stored and then returned every cycle either which again could just be
loss of memory but it leaves the nagging question which someone might
care to answer - If the useless power is returned in full why is it that
commercial users get charged for kva?

Yes, SMPS weren't exactly ten a penny when I studied Electrical
Engineering!

This link gives a fairly comprehensive explanation in pretty much
straightforward language

http://www.cip.ukcentre.com/Power%20Factor.htm

Might also be worth adding to the wiki.


--
The Wanderer

Whenever I look for something, it's always in the last place I look.

In article ,
Mark writes:
On Sat, 12 Dec 2009 10:24:28 +0000 (UTC),
(Andrew Gabriel) wrote:

In article ,
Andy Dee writes:
The Wanderer wrote:


Alright, it's early yet, and I haven't fully got my brain in gear, but what
sort of load will have a poor pf but little or no phase shift contribution?

I would like to know this too...

All power supplies which rectify mains into a DC storage capacitor
(except those with power factor correction). This is all switched
mode power supplies 25W*, which includes nearly all compact fluorescent
retrofits. The low power factor is because they draw current only on
the +ve and -ve peaks of the sine wave. However, there's negligable
phase shift.

In the more general case, anything which doesn't draw a sinusoidal
current waveform is going to have a low power factor component which
isn't due to phase shift (and may also have a component which is due
to phase shift).

* Switched mode power supplies above 25W now have to be power factor
corrected in the EU, although many older ones exist which aren't.

As pf = COS(Ø) which is the phase angle, no other term is involved.

That is the definition of power factor when it's only due to phase
shift, as I said. It's a simplification (suitable only for reactive
loads) of the proper power factor definition PF = W / VA.

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

OK, I put together some demos so you can try this for yourself, and
save me trying to explain it only in words...

http://www.cucumber.demon.co.uk/PowerFactor/

What happens in the case of a SMPSU (the last demo) is that the energy
draw is concentrated into some small areas, which requires large current
peaks. The RMS value of the current actually _increases_, and it's this
that represents heating effects in wiring, size of supply transformers,
etc. This is why the power factor drops.

--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]

On Tue, 15 Dec 2009 16:14:39 +0000, The Wanderer
wrote:

On Tue, 15 Dec 2009 15:04:46 +0000, Roger Chapman wrote:

Mark wrote:

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

I learnt about power factor back in the early 60s and I too do not
remember anything other than phase shift. Memory fades of course but the
goal posts may have moved since if switched mode power supplies either
weren't around then or hadn't established sufficient presence to matter.

I don't recall anything about the 'useless' power just being temporarily
stored and then returned every cycle either which again could just be
loss of memory but it leaves the nagging question which someone might
care to answer - If the useless power is returned in full why is it that
commercial users get charged for kva?

Yes, SMPS weren't exactly ten a penny when I studied Electrical
Engineering!

This link gives a fairly comprehensive explanation in pretty much
straightforward language

http://www.cip.ukcentre.com/Power%20Factor.htm

Might also be worth adding to the wiki.

Isn't there an error in this article? Surely if peak power (in the
example) is 2W the average power is not 1W?

--
(\__/) M.
(='.'=) Due to the amount of spam posted via googlegroups and
(")_(") their inaction to the problem. I am blocking most articles
posted from there. If you wish your postings to be seen by
everyone you will need use a different method of posting.
[Reply-to address valid until it is spammed.]

In article ,
Mark writes:
On Tue, 15 Dec 2009 16:14:39 +0000, The Wanderer
wrote:

On Tue, 15 Dec 2009 15:04:46 +0000, Roger Chapman wrote:

Mark wrote:

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

I learnt about power factor back in the early 60s and I too do not
remember anything other than phase shift. Memory fades of course but the
goal posts may have moved since if switched mode power supplies either
weren't around then or hadn't established sufficient presence to matter.

I don't recall anything about the 'useless' power just being temporarily
stored and then returned every cycle either which again could just be
loss of memory but it leaves the nagging question which someone might
care to answer - If the useless power is returned in full why is it that
commercial users get charged for kva?

Yes, SMPS weren't exactly ten a penny when I studied Electrical
Engineering!

This link gives a fairly comprehensive explanation in pretty much
straightforward language

http://www.cip.ukcentre.com/Power%20Factor.htm

Might also be worth adding to the wiki.

Isn't there an error in this article? Surely if peak power (in the
example) is 2W the average power is not 1W?

It is correct. The average of that line will be a horizontal line
at 1W.

--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]

On Wed, 16 Dec 2009 11:32:21 +0000 (UTC),
(Andrew Gabriel) wrote:

In article ,
Mark writes:
On Tue, 15 Dec 2009 16:14:39 +0000, The Wanderer
wrote:

On Tue, 15 Dec 2009 15:04:46 +0000, Roger Chapman wrote:

Mark wrote:

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

I learnt about power factor back in the early 60s and I too do not
remember anything other than phase shift. Memory fades of course but the
goal posts may have moved since if switched mode power supplies either
weren't around then or hadn't established sufficient presence to matter.

I don't recall anything about the 'useless' power just being temporarily
stored and then returned every cycle either which again could just be
loss of memory but it leaves the nagging question which someone might
care to answer - If the useless power is returned in full why is it that
commercial users get charged for kva?

Yes, SMPS weren't exactly ten a penny when I studied Electrical
Engineering!

This link gives a fairly comprehensive explanation in pretty much
straightforward language

http://www.cip.ukcentre.com/Power%20Factor.htm

Might also be worth adding to the wiki.

Isn't there an error in this article? Surely if peak power (in the
example) is 2W the average power is not 1W?

It is correct. The average of that line will be a horizontal line
at 1W.

With a sinusoidal waveform the average is not 1/2 the peak. It's
1/sqrt(2) of the peak.

--
(\__/) M.
(='.'=) Due to the amount of spam posted via googlegroups and
(")_(") their inaction to the problem. I am blocking most articles
posted from there. If you wish your postings to be seen by
everyone you will need use a different method of posting.
[Reply-to address valid until it is spammed.]

Mark wrote:
On Wed, 16 Dec 2009 11:32:21 +0000 (UTC),
(Andrew Gabriel) wrote:

In article ,
Mark writes:
On Tue, 15 Dec 2009 16:14:39 +0000, The Wanderer
wrote:

On Tue, 15 Dec 2009 15:04:46 +0000, Roger Chapman wrote:

Mark wrote:

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

I learnt about power factor back in the early 60s and I too do not
remember anything other than phase shift. Memory fades of course but the
goal posts may have moved since if switched mode power supplies either
weren't around then or hadn't established sufficient presence to matter.

I don't recall anything about the 'useless' power just being temporarily
stored and then returned every cycle either which again could just be
loss of memory but it leaves the nagging question which someone might
care to answer - If the useless power is returned in full why is it that
commercial users get charged for kva?
Yes, SMPS weren't exactly ten a penny when I studied Electrical
Engineering!

This link gives a fairly comprehensive explanation in pretty much
straightforward language

http://www.cip.ukcentre.com/Power%20Factor.htm

Might also be worth adding to the wiki.
Isn't there an error in this article? Surely if peak power (in the
example) is 2W the average power is not 1W?
It is correct. The average of that line will be a horizontal line
at 1W.

With a sinusoidal waveform the average is not 1/2 the peak. It's
1/sqrt(2) of the peak.

The average of a half cycle.

But what you have in the diagram is a regular wave form with a maximum
of 2W and a minimum of 0W. The average has to be halfway between.

In article ,
Mark writes:
On Wed, 16 Dec 2009 11:32:21 +0000 (UTC),
(Andrew Gabriel) wrote:

In article ,
Mark writes:
On Tue, 15 Dec 2009 16:14:39 +0000, The Wanderer
wrote:

On Tue, 15 Dec 2009 15:04:46 +0000, Roger Chapman wrote:

Mark wrote:

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

I learnt about power factor back in the early 60s and I too do not
remember anything other than phase shift. Memory fades of course but the
goal posts may have moved since if switched mode power supplies either
weren't around then or hadn't established sufficient presence to matter.

I don't recall anything about the 'useless' power just being temporarily
stored and then returned every cycle either which again could just be
loss of memory but it leaves the nagging question which someone might
care to answer - If the useless power is returned in full why is it that
commercial users get charged for kva?

Yes, SMPS weren't exactly ten a penny when I studied Electrical
Engineering!

This link gives a fairly comprehensive explanation in pretty much
straightforward language

http://www.cip.ukcentre.com/Power%20Factor.htm

Might also be worth adding to the wiki.

Isn't there an error in this article? Surely if peak power (in the
example) is 2W the average power is not 1W?

It is correct. The average of that line will be a horizontal line
at 1W.

With a sinusoidal waveform the average is not 1/2 the peak. It's
1/sqrt(2) of the peak.

Actually, the average of a sinusoidal waveform is where it's centred,
normally 0, but in this case 1.

1/sqrt(2) of the peak is the RMS value, not the average.

Have a play with the dynamic graphs I put up on
http://www.cucumber.demon.co.uk/PowerFactor/
As you change the phase angle, you will see the same graph morph
into the version that passes excess energy back and forth between
the load and the supply, and you'll see the current waveform grow
as a result of the increasing excess (reactive) energy being passed.

--
Andrew Gabriel
[email address is not usable -- followup in the newsgroup]

On Thu, 17 Dec 2009 01:27:58 +0000 (UTC),
(Andrew Gabriel) wrote:

In article ,
Mark writes:
On Wed, 16 Dec 2009 11:32:21 +0000 (UTC),
(Andrew Gabriel) wrote:

In article ,
Mark writes:
On Tue, 15 Dec 2009 16:14:39 +0000, The Wanderer
wrote:

On Tue, 15 Dec 2009 15:04:46 +0000, Roger Chapman wrote:

Mark wrote:

It's a very long time since I studied this but I don't recall any way
PF can be less than 1 unless there is a phase shift. In the example
you give you would just get a lower RMS current than a straight
sinewave so the input VA would be lower and the PF = 1. Can you
explain why I am wrong?

I learnt about power factor back in the early 60s and I too do not
remember anything other than phase shift. Memory fades of course but the
goal posts may have moved since if switched mode power supplies either
weren't around then or hadn't established sufficient presence to matter.

I don't recall anything about the 'useless' power just being temporarily
stored and then returned every cycle either which again could just be
loss of memory but it leaves the nagging question which someone might
care to answer - If the useless power is returned in full why is it that
commercial users get charged for kva?

Yes, SMPS weren't exactly ten a penny when I studied Electrical
Engineering!

This link gives a fairly comprehensive explanation in pretty much
straightforward language

http://www.cip.ukcentre.com/Power%20Factor.htm

Might also be worth adding to the wiki.

Isn't there an error in this article? Surely if peak power (in the
example) is 2W the average power is not 1W?

It is correct. The average of that line will be a horizontal line
at 1W.

With a sinusoidal waveform the average is not 1/2 the peak. It's
1/sqrt(2) of the peak.

Actually, the average of a sinusoidal waveform is where it's centred,
normally 0, but in this case 1.

1/sqrt(2) of the peak is the RMS value, not the average.

Have a play with the dynamic graphs I put up on
http://www.cucumber.demon.co.uk/PowerFactor/
As you change the phase angle, you will see the same graph morph
into the version that passes excess energy back and forth between
the load and the supply, and you'll see the current waveform grow
as a result of the increasing excess (reactive) energy being passed.

I was having a bad brain day yesterday. The RMS voltage or current is
1/sqrt(2) of the maximum. The power is 1/2 as in the article.

--
(\__/) M.
(='.'=) Due to the amount of spam posted via googlegroups and
(")_(") their inaction to the problem. I am blocking most articles
posted from there. If you wish your postings to be seen by
everyone you will need use a different method of posting.
[Reply-to address valid until it is spammed.]