I have a Lidl unit that measures how much electricity something is using -
it also measures the 'Power Factor'

My reading is that a Power Factor of 1 is good and 0.5 means it is using
twice as much electricity as it needs to. Am I correct?

I also read (I'm planning to give up reading) that you can adjust the Power
Factor. Is this feasible and/or worthwhile?

I'm wondering as two of my fridges have a power factor of about 0.58 so
adjusting them could produce a reasonable saving in electricity.

--
Mark BR

On Nov 22, 3:33*am, "Mark BR" wrote:
I have a Lidl unit that measures how much electricity something is using -
it also measures the 'Power Factor'

My reading is that a Power Factor of 1 is good and 0.5 means it is using
twice as much electricity as it needs to. Am I correct?

no

I also read (I'm planning to give up reading) that you can adjust the Power
Factor. Is this feasible and/or worthwhile?

no

I'm wondering as two of my fridges have a power factor of about 0.58 so
adjusting them could produce a reasonable saving in electricity.

absolutely none


NT

On Sun, 22 Nov 2009 04:39:55 +0000, John Rumm
wrote:

Sort of. In a domestic situation a poor power factor will not result in
you being charged for more electricity, although the load with a lower
power factor will draw more current. Poor PFs are bad for distribution
efficiency though, and can result in the mains supply waveform getting
misshapen and noisy - so power supply companies tend to penalise big
industrial users if they don't control their PFs.

snip

Nice answer. ;-)

Andy C

On Sun, 22 Nov 2009 04:39:55 +0000, John Rumm wrote:

Mark BR wrote:

I have a Lidl unit that measures how much electricity something is using -
it also measures the 'Power Factor'

This is quite a "deep" subject when you get into all the detail, but I
will attempt to give some basics here. The wakypedia article is not a
bad starting place on this one:

http://en.wikipedia.org/wiki/Power_factor

My reading is that a Power Factor of 1 is good and 0.5 means it is using
twice as much electricity as it needs to. Am I correct?

Sort of. In a domestic situation a poor power factor will not result in
you being charged for more electricity,

This is *the* relevant bit of the answer, and needs re-iterating. For
domestic customers just ignore power factor. It genuinely has no relevance
except in industrial processes where very large motors are used.



--
The Wanderer

I may be omniscient, but don't expect me to know everything.

On Sun, 22 Nov 2009 07:29:55 +0000
The Wanderer wrote:

On Sun, 22 Nov 2009 04:39:55 +0000, John Rumm wrote:

Mark BR wrote:
snip
Sort of. In a domestic situation a poor power factor will not result in
you being charged for more electricity,

This is *the* relevant bit of the answer, and needs re-iterating. For
domestic customers just ignore power factor. It genuinely has no relevance
except in industrial processes where very large motors are used.

Is this also true of my little workshop? I have several machines that
have induction motors, up to about 3 hp. Not, of course, that they are
significant in terms of overall electricity usage when compared to the
tumble dryer and stupid washing machine that insists on heating cold
water to 40 (both for next years recycling!).

R.

P.S. I loved John's bicycle analogy, first lucid explanation of PF
I've heard.

"NT" wrote in message
...
On Nov 22, 3:33 am, "Mark BR" wrote:
I have a Lidl unit that measures how much electricity something is using -
it also measures the 'Power Factor'

My reading is that a Power Factor of 1 is good and 0.5 means it is using
twice as much electricity as it needs to. Am I correct?

no

eh? The power company has to supply V * I watts of power. Your meter
measures and charges for V * I * Cos phi where phi is phase angle between V
and I.
If you stick a capacitor across your mains supply and draw 10A the power
company has to generate 230V * 10A = 2300W. The current will be 90 deg out
of phase with the volts so you meter will not register (V * I * cos phi =
0). The 10A is real and will cause heating in the power distribution system.
This is the worst case where your PF is zero. Unfortunatly that 2300W is of
zero use to you.

On Sun, 22 Nov 2009 04:39:55 +0000, John Rumm
wrote:

Mark BR wrote:

I have a Lidl unit that measures how much electricity something is using -
it also measures the 'Power Factor'

This is quite a "deep" subject when you get into all the detail, but I
will attempt to give some basics here. The wakypedia article is not a
bad starting place on this one:

http://en.wikipedia.org/wiki/Power_factor

My reading is that a Power Factor of 1 is good and 0.5 means it is using
twice as much electricity as it needs to. Am I correct?

Sort of. In a domestic situation a poor power factor will not result in
you being charged for more electricity, although the load with a lower
power factor will draw more current. Poor PFs are bad for distribution
efficiency though, and can result in the mains supply waveform getting
misshapen and noisy - so power supply companies tend to penalise big
industrial users if they don't control their PFs.

The effect of the non unity (i.e. 1) PF is to cause the peak
alternating current drawn by a load to not line up with the peak
alternating voltage (there are other causes of poor PFs but we can skip
those for the moment). This misalignment shows up as a lead or a lag in
the respective waveforms. It comes about when the load contains
components with an ability to store energy (i.e. a capacitors and
inductors). Capacitive loads cause the current to lead the voltage, and
inductive ones cause a lag.

If you want an analogy, imagine riding a bike up hill. You stick a
certain about of push into the pedals to keep it moving overcoming
resistance, and more to add the energy you are acquiring by climbing the
hill. A poor PF is like someone attaching a big spring to one pedal and
the seat post, such that every time you push the right pedal down you
also need to stretch the spring. As you can imagine this will take more
"push" from you to keep riding. However that extra push is only required
on the right pedal. When you push the left pedal you have the energy
stored in the spring pulling up on the right pedal and hence working for
you. So the result is the bike is harder is harder to ride, but the
total energy required to get up the hill is actually the same.


I also read (I'm planning to give up reading) that you can adjust the Power
Factor. Is this feasible and/or worthwhile?

You can do what it called power factor correction. Worthwhile in an
industrial setting where customers are usually charged based on their VA
loading rather than their real power loading in watts, but less so in a
domestic one where the meter will give a reasonable indication of the
actual power consumption regardless of PF.

I'm wondering as two of my fridges have a power factor of about 0.58 so
adjusting them could produce a reasonable saving in electricity.

alas no. Trading up to a modern low energy consumption one may help, but
there it not much you can do for an old one.



I have a lot of admiration for people who can take a very complex
subject, such as this, and express it in simple, easily understandable
English without losing any of its meaningful content, as here.

Very well done. ;-)

The Wanderer wrote:
.. It [power factor] genuinely has no relevance
except in industrial processes where very large motors are used.


And office blocks and warehouses where there are large numbers of
fluorescent lights which may need to be power factor corrected.

On Sun, 22 Nov 2009 04:39:55 +0000, John Rumm wrote:

snip

If you want an analogy, imagine riding a bike up hill. You stick a
certain about of push into the pedals to keep it moving overcoming
resistance, and more to add the energy you are acquiring by climbing the
hill. A poor PF is like someone attaching a big spring to one pedal and
the seat post, such that every time you push the right pedal down you
also need to stretch the spring. As you can imagine this will take more
snip


Beautifully put! I wish I'd thought of that... :-)

Just as a matter of interest, we're just installing some power factor
correction gear for a local company who use large, slow AC motors to
drive mixers. The PF is absolutely horrid and consequently they are also
charged for a larger kW supply (more amps) than they would need with a
better PF. The PF correction suppliers have told us that, using their
equipment to get the load down the company may also be able to get a
reduction in their kW supply from the distribution company. If so the
savings should be in the region of 1000 quid per year on the electricity
bill. That's a fair bit!

--
Mick (Working in a M$-free zone!)
Web: http://www.nascom.info
Filtering everything posted from googlegroups to kill spam.

Phil Jessop wrote:

eh? The power company has to supply V * I watts of power. Your meter
measures and charges for V * I * Cos phi where phi is phase angle between
V and I.
If you stick a capacitor across your mains supply and draw 10A the power
company has to generate 230V * 10A = 2300W. The current will be 90 deg out
of phase with the volts so you meter will not register (V * I * cos phi =
0). The 10A is real and will cause heating in the power distribution
system. This is the worst case where your PF is zero. Unfortunatly that
2300W is of zero use to you.

Although they have to generate that 10 amps for this very hypothetical load
the power lost is only 10*10*R where R is the resistance of the
transmission lines. This loss would be very much less than 2300W.

Basically the power station only supplies the full 2300W for every alternate
half cycle, your (exceedingly large!) capacitor accepts this and then gives
it all back to the power company for a refund in the next half cycle.

The overall power factor for domestic consumers is sufficiently close to
unity for it to be not worth the extra cost of monitoring peak current as
well as power.

--
Mike Clarke

In article ,
Dave Osborne writes:
The Wanderer wrote:
. It [power factor] genuinely has no relevance
except in industrial processes where very large motors are used.


And office blocks and warehouses where there are large numbers of
fluorescent lights which may need to be power factor corrected.

They're nearly all electronic, and near-as-damn-it all PF=1,
if installed in the last 10 years.

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

In article ,
TheOldFellow writes:
On Sun, 22 Nov 2009 07:29:55 +0000
The Wanderer wrote:

On Sun, 22 Nov 2009 04:39:55 +0000, John Rumm wrote:

Mark BR wrote:
snip
Sort of. In a domestic situation a poor power factor will not result in
you being charged for more electricity,

This is *the* relevant bit of the answer, and needs re-iterating. For
domestic customers just ignore power factor. It genuinely has no relevance
except in industrial processes where very large motors are used.

Is this also true of my little workshop? I have several machines that

Yes, unless you're a very large industrial customer.

have induction motors, up to about 3 hp. Not, of course, that they are
significant in terms of overall electricity usage when compared to the
tumble dryer and stupid washing machine that insists on heating cold
water to 40 (both for next years recycling!).

WM heating is unlikely to be significant. Tumble dryer could well be though.
As for the machinary, depends how long you use it for.

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

In article ,
"Phil Jessop" writes:

"NT" wrote in message
...
On Nov 22, 3:33 am, "Mark BR" wrote:
I have a Lidl unit that measures how much electricity something is using -
it also measures the 'Power Factor'

My reading is that a Power Factor of 1 is good and 0.5 means it is using
twice as much electricity as it needs to. Am I correct?

no

eh? The power company has to supply V * I watts of power. Your meter
measures and charges for V * I * Cos phi where phi is phase angle between V
and I.

It actually measures energy, which it does by
integrating instantaneous value of V * I.

This is only equal to V * I * Cos phi in the
case of phase shifts, but the meter also correctly
works with low power factor which isn't due to
phase shifts (where V * I * Cos phi does not hold).

If you stick a capacitor across your mains supply and draw 10A the power
company has to generate 230V * 10A = 2300W. The current will be 90 deg out
of phase with the volts so you meter will not register (V * I * cos phi =
0). The 10A is real and will cause heating in the power distribution system.
This is the worst case where your PF is zero. Unfortunatly that 2300W is of
zero use to you.

You effectively take it from the power company during
part of the mains cycle, and give it back during another
part of the mains cycle.

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

"John Rumm" wrote in message
o.uk...
Mark BR wrote:

I have a Lidl unit that measures how much electricity something is
using -
it also measures the 'Power Factor'


Many thanks. I tried the Wikipedea article but was not much wiser, now I
understand a bit more and at least I now know I do not have to do anything
about it.

One less thing to worry about.

--
Mark BR