Could someone explain power factor for me please (and I guess a few others)
?

The reason being is that I have a lot of CFLs and so the rated power
consumption is somewhat less than if these were ordinary filament bulbs.

However the other day I plugged in a few to my gizmo that tells power being
drawn in watts and also power factor.

The power factor was around 60% on the CFLs.

I believe the elec company doesn't like devices with a poor (low?) power
factor but I don't know why.

The other concern is how does the electricity meter measure power drawn by a
device with a low power factor ? - does it over or under read ? Also does
this vary between the older eddy current spinning discs and the newer
flashing LED meters with some other method of measuring power.

Basically my elec bill is still way higher than I think it should be and I
am trying to get to the bottom of it.

Thanks,

Nick

Hi,

Nick wrote:

Could someone explain power factor for me please (and I guess a few
others) ?

Power factor = 1 means that voltages and current are in phase
Power factor = 0 means they are 90 deg out of phase.
In all cases, PF = cos(V/A phase angle)

Try this for a fuller description:

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

The reason being is that I have a lot of CFLs and so the rated power
consumption is somewhat less than if these were ordinary filament bulbs.

However the other day I plugged in a few to my gizmo that tells power
being drawn in watts and also power factor.

The power factor was around 60% on the CFLs.

I believe the elec company doesn't like devices with a poor (low?) power
factor but I don't know why.

Because it means that while you are paying for a certain amount of energy
used (in phase), there are disproportionatly larger currents flowing in the
suppliers cabling, which means that if everyone ran with a PF much 1, the
supplier would need to beef up their distribution (= more dosh for no more
sales). There might be a further impact on the generating stations too, but
the engineering of that is beyond me.

The other concern is how does the electricity meter measure power drawn by
a
device with a low power factor ? - does it over or under read ?

It reads the correct amount of in phase *energy* used - they are carefully
designed to do exactly this. IIRC, some industrial sites may actually pay
for the kVAh used, rather than the kWh as in residential supplies. However
big industrial sites usually employ PF correction measures - eg capacitors
to compensate for highly inductive loads.

Also does
this vary between the older eddy current spinning discs and the newer
flashing LED meters with some other method of measuring power.

In principle, no. The spinning disc meters used a principle of physics to
measure the true power throughput and integrate over time. I have no idea
how the electronic meters work, but the end result will be the same.

Basically my elec bill is still way higher than I think it should be and I
am trying to get to the bottom of it.

Can't think it would be related to the PF. Your CFLs constitute a smaller
fraction of your overall demand, so your mean PF will hopefully be much
nearer to 1.

I'm not an engineer, so there may be ******** above, but it is my
understanding.

Cheers

Tim

Nick wrote:

Could someone explain power factor for me please (and I guess a few others)

It is a measure of how far out of phase the current being drawn from the
supply is with the voltage. With straight resistive loads the current
will be in phase with the voltage. The more capacitive or inductive
(i.e. the more reactive) the load appears, then the more out of phase
the current drawn will be (up to 90 degrees shift for a perfect inductor
or capacitor).

So a "unity" unity power factor of 1 (or 100%) is a resitive load, a
power factor of 0 is a fully reactive one.

The reason being is that I have a lot of CFLs and so the rated power
consumption is somewhat less than if these were ordinary filament bulbs.

This is typically true of all fluorescent lights unless corrected. They
normally have an inductive choke in series with the

However the other day I plugged in a few to my gizmo that tells power being
drawn in watts and also power factor.

The power factor was around 60% on the CFLs.

I believe the elec company doesn't like devices with a poor (low?) power
factor but I don't know why.

Imagine you are riding a bike up a hill. How hard you have to push on
the pedals will depend on a number of things - the steepness of the
hill, and how fast you want to get up it for example. Now say some evil
sod came along and modified your bike by attaching a large spring to the
right hand pedal, and fixing the other end to the seat post. Now every
time you push the right pedal down, you are not only overcoming the
resistance to move the bike, but you are stretching the spring. This
would make pushing the pedal harder. However when you go to push the
left pedal, the spring on the other side is now pulling and hence
helping. All the energy you stored in the spring on one cycle you now
get back on the other. The result is a bike that consumes the same
amount of real energy to do the same task, however it appears harder to
ride because the total peak push required is higher.

Poor power factors work the same way. The actual power dissipated will
remain unchanged, but the peak currents that flow around the circuit
will be higher.

Hence the generators need higher current capacity, and you suffer more
transmission losses in the distribution network. There is also a
additional problem that because reactive loads are drawing current that
is not in phase with the voltage waveform they can tend to distort the
waveform - taking lumps out of it. This has the effect of adding noise
to the mains supply for other users (which in turn can cause them some
power wastage since this harmonic noise is often dissipated as extra
heat in their appliances)

The other concern is how does the electricity meter measure power drawn by a
device with a low power factor ? - does it over or under read ? Also does
this vary between the older eddy current spinning discs and the newer
flashing LED meters with some other method of measuring power.

For domestic customers you are billed only for real power consumed - the
meter should be unaffected by non unity power factors. As to how good
the various models are in this respect I don't know.

Basically my elec bill is still way higher than I think it should be and I
am trying to get to the bottom of it.

It would seem unlikely that poor power factor is your problem. It does
become an issue for some industrial consumers since they are billed
based on the apparent power consumed.

--
Cheers,

John.

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|-----------------------------------------------------------------|
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\================================================= ================/

Thanks guys - I wasn't too far out in my assumptions but the replies have
served to confirm
and remove doubts that the meters are too far out in what they are
measuring...

So I will continue to fine where the consumption and quiescent drain ( some
400 watts) is going.

I'll start a new thread to see if this is reasonable.
Thanks,

Nick

Nick wrote:

Thanks guys - I wasn't too far out in my assumptions but the replies have
served to confirm
and remove doubts that the meters are too far out in what they are
measuring...

So I will continue to fine where the consumption and quiescent drain (
some 400 watts) is going.

I'll start a new thread to see if this is reasonable.
Thanks,

Nick

You might try shutting the house down, then switching on circuits one at a
time whilst watching the meter. When you find the circuit that adds 100W,
try going around and turning all your video/dvd/tv/computers etc off. I bet
it's the combined drain of all of those things - maybe not much each
(except the PC which could account for a couple of hundred watts by
itself), but they may add up.

The blinking led on the meter will give a reasonable measurement of
instantaneous load.

HTH

Tim

In message , Tim
writes
Hi,

Nick wrote:

Could someone explain power factor for me please (and I guess a few
others) ?

Power factor = 1 means that voltages and current are in phase
Power factor = 0 means they are 90 deg out of phase.
In all cases, PF = cos(V/A phase angle)

Try this for a fuller description:

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

Which is all true for linear loads (be that resistive, inductive or
capacitive).
However the input to CFLs is not linear. Its diodes into a capacitor
(followed by a high frequency inverter to drive the lamp).
A diode capacitor input arrangement only draws current from the mains to
"top up" the capacitor near mains voltage peak, i.e. the current only
flows for 2-3ms of every 10ms half cycle, and has a higher rms
(root-mean-square) value than you might expect.

PF can be defined as real power divided by apparent power,
where apparent power is Vrms * Irms.
Remember Irms is higher for non-linear loads because the current is
narrow and peaky (i.e. the squared term is much bigger before you
average (mean) it then square-root it). Therefore you get low
power-factors.

Real / Apparent power works just as well for linear loads, where as
cos(phase angle) doesn't work for non-linear loads.


The reason being is that I have a lot of CFLs and so the rated power
consumption is somewhat less than if these were ordinary filament bulbs.

However the other day I plugged in a few to my gizmo that tells power
being drawn in watts and also power factor.

The power factor was around 60% on the CFLs.

True, and typical of diode-cap inputs.

I believe the elec company doesn't like devices with a poor (low?) power
factor but I don't know why.

Because it means that while you are paying for a certain amount of energy
used (in phase), there are disproportionatly larger currents flowing in the
suppliers cabling, which means that if everyone ran with a PF much 1, the
supplier would need to beef up their distribution (= more dosh for no more
sales). There might be a further impact on the generating stations too, but
the engineering of that is beyond me.

The same applies to the non-linear loading version of PF, more rms
current flowing, bigger cables/transformers, more losses, but the energy
co's aren't selling any more power.

The other concern is how does the electricity meter measure power drawn by
a
device with a low power factor ? - does it over or under read ?


The same non-linear, high rms input current, factors apply to (small)
switch mode power supplies. Depending upon equipment class, larger
switch mode power units (75W for most) must have some form of power
factor correction, either simple passive to slug off the current input
peak, or clever electronics to force a near-sinusoidal input current.
Good designs will often achieve a PF of 0.99.



--
steve

In article ,
Tim writes:
Hi,

Nick wrote:

Could someone explain power factor for me please (and I guess a few
others) ?

Power factor = 1 means that voltages and current are in phase
Power factor = 0 means they are 90 deg out of phase.
In all cases, PF = cos(V/A phase angle)

No, that's only in the special case of a current phase shift,
which actually doesn't apply to CFLs (or at least, isn't the
main contributor of low PF in that case).

In all cases, PF=W/VA

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

Unfortunately, that's really not a good article. The author doesn't
seem to understand the difference between power and energy.

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

In message , Steven Briggs
writes
The same non-linear, high rms input current, factors apply to (small)
switch mode power supplies. Depending upon equipment class, larger
switch mode power units (75W for most) must have some form of power
factor correction, either simple passive to slug off the current input
peak, or clever electronics to force a near-sinusoidal input current.
Good designs will often achieve a PF of 0.99.



For anyone who wants to know more,
http://www.reo.co.uk/files/kbase/handbook%20en%2061000-3-2.pdf
is a readable introduction to the subject, pages 4-5 for the basic
explanation.

--
steve

Steven Briggs wrote:

In message , Tim
writes
Hi,

Nick wrote:

Could someone explain power factor for me please (and I guess a few
others) ?

Power factor = 1 means that voltages and current are in phase
Power factor = 0 means they are 90 deg out of phase.
In all cases, PF = cos(V/A phase angle)

Try this for a fuller description:

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

Which is all true for linear loads (be that resistive, inductive or
capacitive).
However the input to CFLs is not linear. Its diodes into a capacitor
(followed by a high frequency inverter to drive the lamp).
A diode capacitor input arrangement only draws current from the mains to
"top up" the capacitor near mains voltage peak, i.e. the current only
flows for 2-3ms of every 10ms half cycle, and has a higher rms
(root-mean-square) value than you might expect.

PF can be defined as real power divided by apparent power,
where apparent power is Vrms * Irms.
Remember Irms is higher for non-linear loads because the current is
narrow and peaky (i.e. the squared term is much bigger before you
average (mean) it then square-root it). Therefore you get low
power-factors.

Real / Apparent power works just as well for linear loads, where as
cos(phase angle) doesn't work for non-linear loads.


The reason being is that I have a lot of CFLs and so the rated power
consumption is somewhat less than if these were ordinary filament bulbs.

However the other day I plugged in a few to my gizmo that tells power
being drawn in watts and also power factor.

The power factor was around 60% on the CFLs.

True, and typical of diode-cap inputs.

I believe the elec company doesn't like devices with a poor (low?) power
factor but I don't know why.

Because it means that while you are paying for a certain amount of energy
used (in phase), there are disproportionatly larger currents flowing in
the suppliers cabling, which means that if everyone ran with a PF much
1, the supplier would need to beef up their distribution (= more dosh for
no more sales). There might be a further impact on the generating stations
too, but the engineering of that is beyond me.

The same applies to the non-linear loading version of PF, more rms
current flowing, bigger cables/transformers, more losses, but the energy
co's aren't selling any more power.

The other concern is how does the electricity meter measure power drawn
by a
device with a low power factor ? - does it over or under read ?


The same non-linear, high rms input current, factors apply to (small)
switch mode power supplies. Depending upon equipment class, larger
switch mode power units (75W for most) must have some form of power
factor correction, either simple passive to slug off the current input
peak, or clever electronics to force a near-sinusoidal input current.
Good designs will often achieve a PF of 0.99.


Thank you to both Steven and Andrew - I was only considering linear loads
and a phase shifted sinusoidal current and voltage curve. Probably because
my old man taught me about this stuff and they didn't so many electronic
gismos in his day(!).

Interesting stuff.

Cheers

Tim

Tim wrote:

Try this for a fuller description:
http://en.wikipedia.org/wiki/Electric_power

Try this for something more relevant:
http://en.wikipedia.org/wiki/Power_factor

--
Andy

Nick wrote:

The other concern is how does the electricity meter measure power drawn by a
device with a low power factor ? - does it over or under read ? Also does
this vary between the older eddy current spinning discs and the newer
flashing LED meters with some other method of measuring power.

Domestic meters are designed to read true energy/power only. They are
carefully designed to perform this calculation.

Power = Voltage x Current. The meters are designed in such a way as to
measure the instantaneous voltage and instantaneous current, multiply
them, and then record the result.

In the case of mechanical meters, there are some cleverly arranged
electromagnets - one produces a field proportional to voltage, and one
proportional to the current. They are arranged so that one induces a
current in the disk, such that the current flow is at right angles to
the other magnet. The result is a torque on the disk which is
proportional to the instantaneous voltage and instantaneous current.
Finally, there is a permanent magnet which causes eddy current braking
on the disk, making the disk rotation speed proportional to the torque
acting on it.

Electronic meters use analogue-to-digital converter chips to measure
voltage and current at several 10s of kHz (using appropriate
shunts/potential dividers). An integrated microprocessor performs the
multiplications digitally and adds up the total energy. It also triggers
an LED every time a certain quantity is totted up.

The mechanical ones tend to lose accuracy outside of their nominal
range. Friction leads to inaccuracy at low and high ends, and extremely
poor power factor can also degrade accuracy.

The electronic meters design is such that they should accurately record
at all loads, and all power factors, including the nasty waveforms
produced by electronic equipment. I'd expect them to be better than the
mechanical meters in this regard.

Andy Wade wrote:
Tim wrote:

Try this for a fuller description:
http://en.wikipedia.org/wiki/Electric_power

Try this for something more relevant:
http://en.wikipedia.org/wiki/Power_factor

and:

http://www.allaboutcircuits.com/vol_..._11/index.html


--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

On Wed, 28 Feb 2007 01:39:27 UTC, John Rumm
wrote:

http://www.allaboutcircuits.com/vol_..._11/index.html

Excellent. Thanks, John.

--
The information contained in this post is copyright the
poster, and specifically may not be published in, or used by
http://www.diybanter.com

On 27/02/2007 23:32, Mark wrote:

The mechanical ones tend to lose accuracy outside of their nominal
range. Friction leads to inaccuracy at low and high ends.

What is the usual electricity meter replacement policy? In 17 years at
this house, my gas meter has been routinely replaced twice, the
electricity meter not at all.