Yes, that's 400V +/-5%. Can't be all that recent if it shows HP!
As I'm sure you know, that's getting somewhat off the subject. At turndown, the speed and volts are reduced by the electronic controls.
What amp fuse for a fridge freezer
- Thread starter Ihavenojob
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Some still do!
I'll happily replace the 10HP with 7.5KW
What electronic controls?
That's obviously true. However, this all started with the statement (variants of which hjave been repeated) ...
If "for a given power output" does not mean "for a given [i.e. a particular] power output", then what on earth does it mean?
However, I confess that I'm getting rather confused, presumably due to my ignorance of how motors work', when looking at the figures in the catalogues he's been posting (figures which I need to study and play with more carefully).
If power output (hence also power input, if we assume that efficiency remains unchanged) remains unchanged, then if an increase in supply voltage results in current remaining unchanged or falling, then the only possible explanation is that PF has reduced. That does happen (in the figures in the catalogue), but it doesn't look (on 'eyeballing') as if PF falls by anything like enough to explain the behaviour of current (I need to do some sums to check - by that's the impression of my eyeballs).
If, as seems to be inevitable, the (somewhat non-intuitive) behaviour of current is the consequence of a reduction in PF, by what mechanism does the PF reduce when voltage is increased? As you say, the increased voltage will result in a very small increase in speed, but that looks (in catalogue figures) as if it's only a tiny increase in speed (generally no more than 0.2%, other than for the smallest motors, when it can be as high as ~2%) nowhere near enough to 'cancel', let alone 'reverse', the effect of increasing voltage by about 9% (from 380V to 415V).
As said, I'm going to play a bit with the numbers, since my first impression from eyeballing the figures doesn't seem to 'add up'.
Kind Regards, John
I assume the speed control of a heating pump uses electronics, but correct me if that's wrong.
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You are suitably corrected.
I don't see the problem. That's exactly what it means. Change it to rated power if it helps.
plugwash explained it pretty well in #43, but to recap. If the voltage increases, the speed increases (slightly, as you say below) and the load rises. How much it rises depends on the type of load. With a constant-torque load the rise is lower than a squared-torque load. (They're the most common types of load, but if it were important enough I'm sure somebody could design a load where the power input did not vary with speed, at least over a certain speed range). But in either case the power increase is lower than the voltage rise, so there is a reduction in current. If the load is then adjusted to bring the power back to rated, the current drops a little more. The motor data sheets are on that basis, same power at each voltage.
The current only remains unchanged if the power increase is the same as the voltage increase, but it isn't, it's less. There's no need to bring PF into it, but a reduction in PF causes the current to rise (for a given power).
No, it's not inevitable! You seem to have explained it to yourself there.
That speed, depends upon the type of motor - a synchronous motor, will remain in sync to the mains frequency, so as voltage rises, the current demand for the same load, will decrease proportionately.
OK, so how does it work? I'm curious. Not that it has anything much to do with the current discussion.
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I said in #11 - I'm talking about an electric motor, 3-phase induction to be specific. But the result is the same for synchronous - higher volts = lower amps.
I'm not going too get heavily involved in this as I simply don't know enough, The majority of my experience is controls and based around the trip setting on the overload device, any mechanical experience is effectively incidental.
An 'ideal' induction motor running free with no mechanical load SHOULD present a constant power electrical load within its designed voltage. Reason being the back EMF SHOULD balance out as Fixitflav mentions.
In practice changing the voltage on a loaded motor (Let's think about a water pump) increasing the voltage on the same 'ideal' motor the current is likely to drop and reducing voltage, current is likely to increase.
But my experience is the actual amount of change is not likely to retain the same power as predicted by the charts for a loaded motor.
Now of course not all motors are equal and some present a fairly constant impedance (my explanation), increasing voltage will result in increased current and increased heat, some motors run very much hotter when approaching their lower voltage capability and of course all will struggle and heat as they approach their stalling load - think about putting a load on an electric drill.
Now suggesting a drill or any other brush type motor puts a different slant on things, increasing voltage will result result in increased speed, current and hence heat for a given load.
Yes, exactly, I was agreeing with you. The point is, that is only true of a fixed speed synchronous motor.
I've not been following this very closely, but you both seem to be saying the same thing - voltage increase, current reduction - assuming the both speed (synchronous) and load remain constant.
Increase the load and the current will increase, because the motor has to struggle more, to maintain sync. Which is why at startup, such motors can draw quite large currents, until they gain sync..
Another problem with electric drill motors is, the higher the load, the higher the current, more heat generated, and as the speed falls - the motor cooling fan is less able to supply cooling air. The major cause of armature burnout.
No, it's true for asynchronous motor also. I've gone through the explanation several times
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