3 Phase Motor HP?

Steve,

That formula is correct but the efficiency of an unloaded motor must, by definition be zero as it provides no output power.

His motor is loaded but only very lightly so the efficiency will be much lower than 0.9. What exactly it is we cannot know.

Russell.

Gentlemen, thank you very much for the inclusive edification.

My homebrew RPC had some issues so after rewiring it and making quite a few replace & test (poke & hope) capacitance changes I now get the following:

The grinder running (unloaded) at a consistent 2970 RPM (RPM measured using a Chinese LED/Laser counter)

Voltages:

L1 - L2 = 239v

L2 - L3 = 239v

L1 - L3 = 230v

Voltage mismatch is down to what I think is an acceptable difference but once I have some alternate sizes of capacitors delivered I will try to tune if further.

Amperages:

L1 = 1.1

L2 = 1.1

L3 = 0.9

As a test I also ran my Boxford Lathe(3/4hp Gryphon) off the same RPC at the same time this gave little variation in voltage but the amperage readings then developed into this:

L1 = 1.9

L2 = 1.9

L3 = 2.4

The changes to the RPC have made the grinder motor very smooth with no trip outs, bizarrely the Boxford motor now runs smoother on the RPC than when it is running from my Teco inverter.

Steve, the formula from Brook Crompton was exactly what I was looking for many thanks.

Regards

Paul

Glad to hear you have it sorted.

Russell.

Russel my thanks to you also, your prognosis about the RPC being "off" was quite correct.

It;s still not 100% but a hopefully a bit more capacitor tuning will get it better, what I have noticed though is as more motors are added the values dont climb exponentially as you might expect I am putting this down to the different types/styles of motor.

I am beginning to wonder if it may also be a factor that all the motors currently tested are not "natively" 230v and have been converted from Star (415v) only to Delta (230v) by me converting (picking out the star point in) each device.

I now find myself with more questions than answers.

Regards

Paul

You didn't read what I said, did you? I said that the product of the efficiency, power factor and 1.732 would be approaching unity, and that I guessed at 0.9 which may be just a tad on the high side, but not much. The numbers I used to arrive at that figure already took account of a low efficiency value - low, because there's no such concept as an unloaded motor of this type. For a start, it's powering its own fan...

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

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

Quite!

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

Motor eff. is zero at stall ( max torque min rpm) and also zero at "no load rpm". ( max rpm min torque)

max eff. is where the product of these two is max. ( for pm dc motors half max rpm serves as first order approximation) for induction motors ( from memory ) synchronisation equates to max rpm (unloaded) where no power is absorbed into mechanical load..so heres the point.. fan and bearings constitute mechaical loads and cannot be met with zero slip... I forget how to convert slip into watts but there will be a small but non zero load on the motor thus the rpm is not the off load rpm..

where all that leads the OP I don't know. apart from if the wheel cuts with no significant rpm drop then there is adequate power.

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!

O.K. we seem to be arguing over a minor point; what is the definition of efficiency?

In my book the efficiency of any system is the ratio of the energy extracted from the system to that put in. The energy drawn from the system does not include the losses or energy used within that system. So, as Jason said, the efficiency is zero at no load.

Your definition seems to be the efficiency of the conversion from electrical to mechanical energy and includes the friction and cooling fan as output energy. Thus this efficiency will be non zero.

Still we have wandered way off topic. His motor is adequate for the task.

Regards,

Russell.

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.

Steve. eff. in this case only helps us determine guess the available power to power the grinding wheel wrt. input power ( again "estimated" from V and I   ie 2 kW in 70 % eff means 1400 Watts say

conversly the question " will my converter power this under load? " which was probably the original point ( I forget now) is still mostly un answered by this disscusion..any one care to start a thread for this ? .

intresting though this is I think its done to death in this thread .

OVER

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

Steve Garnett....How do you arrive at 2kW?

2kW ...just an (unhelpful it seams) example.

Ah, sorry - misunderstood your meaning.

Regds, Steve

No worries