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.
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:
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.
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.
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.
no
no
NT
Nice answer. ;-)
Andy C
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.
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.
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
And office blocks and warehouses where there are large numbers of fluorescent lights which may need to be power factor corrected.
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!
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.
They're nearly all electronic, and near-as-damn-it all PF=1, if installed in the last 10 years.
Yes, unless you're a very large industrial customer.
WM heating is unlikely to be significant. Tumble dryer could well be though. As for the machinary, depends how long you use it for.
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).
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.
Its all down to how you are metered. The domestic electric meters basically sum the instantaneous product of current and voltage over time. Hence they only "see" the real power consumed and not the reactive.
Thankyou kind sir! ;-)
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.
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