Behaviour of induction motor under varying load

Since this seemed to be a question causing some disagreement, I thought it might be fun to inject some actual data into the discussion...

So I plugged my Delta unisaw into my el cheapo power meter and took a few finger in the air readings. First with the saw running and no load, then while applying some load to the blade (by pressing the end of a MDF push block against the side of the blade to induce some frictional drag[1])

So these tests were carried out with a supply voltage at the time of about 235V (that will likely also fallen a little under load - but I have not measured the supply impedance in the workshop). There will be a significant amount of error in some of these numbers since my meter has poor LCD contrast, and I was reading it from a distance while applying a force to a spinning blade - my focus was very much on the blade and not the meter!

No load figures:

Current: 4A VA ~900 Power ~220W PF ~0.30

Suggesting just under 1A of that current was actual load, and 3A was reactive.

Applying load, I could push the figures up to :

Current: 5.25A

1300 VA 1100 W PF ~ 0.7 - 0.85

(note these were all metered separately since the meter needed to be switched top the appropriate range - so the load applied for each will not have been exactly the same each time. So they are not going to be mutually consistent - but ought to be in the ballpark)

So with increased load (i.e. slip) the real power consumption rose by around 5x, but the reactive current fell quite dramatically. i.e. the improving PF was not simply due to increasing real power transfer with a mostly static reactive component.

[1] judging by the motor sound this was certainly a larger load than say cutting any normal sheet material, or ripping pine boards etc. However still short of the load when say ripping a hardwood 8x2" or similar.
Reply to
John Rumm
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I would expect pf to improve with load.

If you do the calculations for reactive power: No load = sqrt ( 900^2 - 220^2) = 870W Part load = sqrt (1300^2 - 1100^2) = 1300W

Therefore the reactive power (and hence reactive current) changes less than real power but still increases 50% with your applied load.

Hopefully NP will see this real world example as well!

Reply to
Fredxxx

Is it just me or does this sound more like an industrial accident than research?!

:-)

Reply to
Scott M

Industrial accident waiting to happen, yes. I appreciate JR's efforts and I'm sure he means well, but it's not a sufficiently scientifically rigorous approach to draw any meaningful conclusions from.

Reply to
Cursitor Doom

Rigorous, no agreed. (Feel free to repeat a similar experiment to a better standard ;-)

However it does easily show show the significant impact of load on the reactive current drawn by the motor in this circumstance, which seems to be something some were claiming was not possible.

Reply to
John Rumm

No, if you read what was written, it was that the value of the reactive component does not change

Reply to
The Natural Philosopher

After initially making out that the reactive current didn't change, plus statements like "Load is by definition in phase current not 'wattless' current" and "What yiou have, with a motor, amounts to a fat inductor in parallel with a variable resistor, representing the variable mechanical load" it took you a long time to accept that the reactive current was indeed dependent on load.

Are you man enough to admit that?

Reply to
Fredxxx

No

Reply to
The Natural Philosopher

I'd love to. I've got all the metrology necessary here and more besides; but what I don't have is time. :(

Reply to
Cursitor Doom

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Reply to
harry

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