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.
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:
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The reason being is that I have a lot of CFLs and so the rated power
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.
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.
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.
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 bollocks above, but it is my understanding.
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.
This is typically true of all fluorescent lights unless corrected. They normally have an inductive choke in series with the
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)
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.
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.
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,
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.
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.
True, and typical of diode-cap inputs.
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 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.
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
seem to understand the difference between power and energy.
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(!).
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.
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.
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