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