Or bearings {mechanical ) or iron or eddy or I squared R (eletrical)

Edited By jason udall on 16/09/2012 23:21:42

Steve,

Yes, I did read it.

Perhaps you misunderstood me. A motor that is running with no external mechanical load must have zero efficiency because you are putting power in and getting none out. The power to drive the fan, and the losses in the bearings the iron losses and the losses due to the winding resistance are not part of the output power.

Efficiency is defined as (useful power output)/(total power input} and thus, if a motor is running lightly loaded such as driving a free running grinding wheel, it's efficiency must be very low.

I know it's now 45 years since I studied electrical machines at university but the principles remain the same.

Regards,

Russell

Well I agree with Jason. In the case of the fan, not just a mechanical load, but a useful one – it's keeping the damn thing cool! I went through all this stuff as a student back in the dark ages as well – and I recall full well being told that it's not possible to have a completely unloaded motor, although you could get pretty close with air bearings, and a lot of care with balancing, etc.

The other thing, of course, is what effect this has on the end result of applying Brook Crompton's formula – you can muck about with the efficiency figure quite a bit without making a huge amount of difference to the end result. So, it's still a 1hp motor!

As a pragmatic physicist , I've always found it best not to ignore awkward things that get in the way of a definition, especially a rather restricted one.

And as such I prefer the definition in the Collins dictionary, which says " the ratio of the useful work done by a machine, engine, device, etc., to the energy supplied to it, often expressed as a percentage ". This implies that because some of the work done services the motor itself, the efficiency percentage will never reach 100% in terms of the power available at the output shaft. I would respectfully suggest that as a definition, it's of more practical use because it will indicate what's left for other use. IOW, this isn't a theoretical motor we're dealing with, but a real one.

It's all very well to talk about the motor having zero efficiency when it's stalled, but you can't do a lot of grinding like that, can you?

As for the question of adequacy for the task – then yes of course it's fine. My Exe grinder seems to work remarkably well with a motor rated at 3/4hp.

The original question was "What is the most accurate method of calculating it's KW or HP?", and I can't agree with the 2kW figure, because that would make it at least a 2hp motor. The snag with this is simply that Brook Crompton didn't make a 2.5+hp motor with a frame that small.

The input PD across each winding is an average of 240V, and the input current is 3.3A. That's 792W, not 2kW. How do you arrive at 2kW?

Edited By Steve Garnett on 19/09/2012 18:27:33

Ah, sorry – misunderstood your meaning.

Regds, Steve