Yes, I just meant in that particular case.
Do you mean varying the supply voltage, or the voltage applied to the motor, determined by the control circuit. In the latter case I would guess the electronics would give the right voltage to the motor, even if the supply varies somewhat, but I make no claim to much knowledge of the subject.
As I said, even that 'particular case' said that it used PWM
The latter would presumably be the more relevant.
I'm not sure what you mean by 'the right voltage'. I was talking about the changes in current if/when that voltage varied.
Well, the current does go up as the voltage goes up - which I suppose would please Mr Ohm
It is perhaps a case of it not 'mattering' all that much? I imagine there was a day in the past when a domestic CH pump was a domestic CH pump, and that was about the only basis one had on which to make a choice (i.e. no choice!)
This actually makes things 'worse' (more confusing), rather than better, but ...
You and I are familiar with that, and should understand that not only should R = V²/W become Z = V²/W when we're talking about AC (Z being impedance), but also that, even Z = V²/W is only correct when PF=1, otherwise it needs to be Z = (V²/W) x PF.FWIW I used V = W/I & R= W/I² and then rounded so yes there will be 'errors'.This actually makes things 'worse' (more confusing), rather than better, but ...
[ for a start, assuming you were using R = V²/W for your calculations, with your W and V figures .I get fractionally different values of 'R' from you - 941.78Ω, 889.35Ω & 829.72Ω respectively - but those differences from yours are trivial and of no importance ]
That would, indeed, confuse Mr Ohm - or, rather, the Law named after him, since, as has been said, I don't think Mr Ohm himself talked at all about 'resistance' or knew anything of 'his equation'.
Mr Ohm would probably have become even more confused when he discovered that we were talking about something called 'alternating current'
However, the tabulated data we've been shown appears to indicate that, with a motor, PF falls (a little) with increasing voltage, which means that the impedances you've calculated for 231V and 241V probably should be even more lower than the impedance at 217V than they are in your calculations.
Don't ask me
Kind Regards, John
OK - but if you substitute the first of those equations into the second, you get the R = V²/W that I mentioned, so (thankfully) we did the same!
Fair enough.
True. However if one 'assumes' (always pretty dangerous, even more so in this situation, since it's little more than a guess) that higher voltage means higher speed then, on the basis of the figures we've been shown that would suggest that PF will fall as voltage increases, which would mean that impedance would fall even more with increasing voltage than your calculations suggested.
But that applies to a free-standing motor as well. The only difference is in the latter case the rated power is shaft power, so need to also divide by eff and PF (and root 3 for 3ph), whereas for a domestic appliance it's the input watts. And of course it gives instantaneous watts, no guide to consumption over a period.
It applies to anything, Stated ('running') power and voltage provide a very good estimate of ('running') current, for anything. As you go on to say, in the case of things like motors, the rating plates rarely, if ever, give any indication of 'start-up' power/current, despite the fact that the start-up current can be considerably greater than the running current. That's why, in post #4 back on 'page 1', I wrote...
Kind Regards, JohnI don't know what the start-up current for a domestic fridge-freezer is likely to be, and it probably doesn't persist for long enough to blow a 3A fuse, but it could well be considerably greater than 3A, so the OP's question seems perfectly reasonable.
I would probably be inclined to use a 10A or 13A fuse, just to be sure' - after all, one doesn't want the freezer losing power whilst one is on a Spanish beach
You would think not, but I don't know for sure. I don't suppose the heater has much of an inrush current, so 10 amp seems to allow for a substantial inrush on the motor. Maybe 10 amp is just to play safe (though not so safe in terms of protection).
Same here, although I suppose that it's not impossible that there needs to be some flow of the refrigerant even when the heater ison.
Indeed. I don't think that 1.3A for the heater (even if on simultaneously with the compressor), with no 'inrush', makes that much difference. As I said at the start, and recently re-quoted, I would probably personally use a 10A or 13A fuse, 'just to be sure'. Although I suppose you're strictly right in saying that a 10/13A fuse would be 'less safe in terms of protection' than would be a 3A one, I'm not really sure what relevant 'protection a 3A one would be providing (over and above the protection afforded by a larger fuse).
Are you talking about a 'frost-free' one? If so, my understanding (which may well not be correct!) is that, in at least some of them, all of the 'cooling' actually happens outsdie of the cavity of the appliance (hence 'frost' arises primarily on ezternal coils, and that the fan is then used to blow cold air (over those tubes into the cavity. With such a system, the fan only runs when the door is closed (but probably not continuously) - the fan then certainly stops when one opens the door (and re-starts when one closes it), presumable to avoid all the cooled air being 'blow out' of the cavity through the door.
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