3 Phase Motor HP?

I have a 3 phase motor looks like an older squirrel cage Brooks/Crompton sadly it's missing it's information plate.

What is the most accurate method of calculating it's KW or HP?

The motor is wired Delta and has an across the terminals resistance of 8.2 Ω dividing this by 1.5x gives a single winding resistance of 12.3 Ω.

Using Ohms law my calculations have given me this:

V/R = I. therefore 230v ÷ 12.3 Ω gives 18.6992 Amps (I)

230v x 18.6992 Amps = 4.3 Kw ÷746 = 5.765 HP

Looking at the size of the motor it dosen't seem big enough to be this powerful, but that said it won't run for long on my 2HP RPC before the switch trips out.

Any & all help appreciated.

Paul

you have measured the "DC" resistance of the COLD windings..

I say cold since that value varies with temperature and rises as the motor heats up.

Simpler way to power( watts or horse) would be ( 230*230/R) ( just algerbra)

1) your calculated current would be closer to the stall current (0 rpm) of the motor ( again cold but would drop as the windings heat up ..eeek!)

2 FLA is a term oftern used with motors ( full load amps ) this figure would be closer for that calculation of horse power..( and is rule of thumb about 1/10 of stall current)

3 You say delta...so lets assume three phase ... thus each "phase " is doing the (230 *230)/R thing so hp would be triple what you calculate... ( but see 1 and 2)

so my guess would be (3*230*230/12.3 )/10(ish !) =1290 W or1.7 hp

stall/fla guess --------kW------ hp

5------------------------- 2.5 ------3.4

8 ------------------------ 1.6------ 2.2

10 ---------------------- 1.3 ----- 1.7

12 ---------------------- 1.0 ----- 1.4

hope this helps and not just fogs the matter

Indeed edited by Jason..to fix the formating decisions made for me by this PXXXXY editor 

V
V
V

Edited By jason udall on 11/09/2012 09:34:20

Hi Paul,
There are a few errors in your reasoning. The resistance part of your calculation is correct. The current calculation is wrong as the voltage between phases is 398 (sqrt 3 x 230) Also you have only considered the power in one winding so you also need to multiply by 3 This is not the main flaw in your reasoning. Your reasoning treats each winding a a pure resistor. If this was true all the power would be converted to heat. You cannot work out the power consumed by a motor by just considering the winding resistance. I think you would be better off trying to find a cataloge of motors of a similar age and rotational speed and compare physical sizes. Do not compare it with modern motors which are smaller in size. These tend to run hotter as modern insulation can withstand higher temperatures.

Les.

There is a little thing called inductance that is missing from all of this, we are dealing with AC. What diameter is the main casing and the shaft diameter?

Indeed Kwil.. inductance and back emf and various others.. but the OP started simply and a rule of thumb type approach appeared usefull.

A motor is not even a pure inductor , motors are more complex and theres phase angle to include in the V.I product.( vector vs scalar arithmatic ) I know or at least have forgotten much of my AC theory.

But ..but the OP has made some sensible tests.. might suggest test motor off load measure I and work from there...

The only real way to measure the power is to power it up and load it with a brake until the speed drops to its rated speed, usually 1440 rpm., then measure the output power. Dificult unless you have or can improvise a dynamometer. Instead you could measure the electrical power input which will be perhaps 10% higher. Again that's difficult without a power meter, amps times volts doesn't work.

I really think your best approach is to see if it will do the job. If it runs slow or overheats it's not powerful enough.

Russell

KWIL is asking the sensible questions, I think - give us some physical dimensions and we can make a much better guess at what the rating plate might have said... I have 2 or 3 different Brookes/Comptons to check against, and I'm sure that others do as well.

DC resistance will only tell you the approximate stall current of the motor, which will, in practice, still be a bit less due to inductive effects on AC.

It is, indeed, the physical size of the motor which is the best guide to its power as this limits how much heat it can dissipate. Also, the diameter of the copper wire in the windings should give a clue as to the maximum working current.

Follow some of these links: **LINK**

Neil

Thank you gentlemen some very good food for thought.

Some more information:

Dimensions:

Length bell to bell 10"

Diameter 6-3/8"

Shaft Diameter 5/8"

Running (unloaded) RPM 2970 (difficult to load up as it's a T&C grinder)

Voltages:

L1 - L2 = 250v

L2 - L3 = 230v

L1 - L3 = 240v

Voltage mismatch is down to the homebrew 2HP RPC it is running from.

Amperages:

L1 = 1.5

L2 = 1.1

L3 = 0.7

Above measurements taken with motor running @ 28 degrees C

And as they say a picture paints a thousand words....

Thats a 6" wheel in the pics.

Paul

Quick and dirty way of working it out from the under ower figures posted:

Multiply highest voltage by highest amperage = approx watts that winding, mutiply that by 3, multiply that by 0,8 .That should be accurate enough for most purposes. I figure it at about 1.1 to 1.3 Kw.

Nathan.

You could of course ignore my previous post and look at the obvious, it,s running on your 2hp(1100w) converter !! Why didn,t I do that?

Nathan.

Hmm... not quite sure how you arrived at that exactly...

Generally, we assume that 1hp is about 750W. And with that frame size, and it being a 2-pole motor (we can tell that from the unloaded speed), Brooke Compton typically made 1 and 1.5hp motors. The 2hp ones are a little bit larger - much easier to keep them cool. And they have a larger diameter shaft, as well.

But from the numbers given, I'd say that by the simple expedient of multiplying the average voltage by the total current (which is the correct product!), you get a power consumtion of 792W - which, taken along with the frame dimensions, pretty firmly makes it a 1hp motor, I'd say.

Edited By Steve Garnett on 12/09/2012 10:46:57

If it's running at 2970 rpm it is very lightly loaded indeed. Nominal no load speed will be 3000 rpm and fully loaded speed is 2880 rpm. For small degrees of slip ( the speed drop) slip is proportional to load power. Thus if you are taking about 1 hp at the moment you can take nearly 4 hp from it before the speed drops below 2880 rpm (provide of course that it doesn't overheat which would be unlikely for ittermittent loading). More than adequate for your grinder I should think!

Russell

Not from an old Brookes Crompton he won't... These are pretty lossy, and will lose power rapidly. You can get a much more sensible drive for them from a modern inverter, and with care compensate for the torque/speed characteristics somewhat. I still remember the first time I did this - something of a revelation!

Gentlemen again thank you for some very positive advice.

It seems I have by popular consensus a 1HP (or thereabouts) motor.

I can appreciate the difficulties of having a mixed bag of parameters to work from with differing motor types/sizes but is there no definitive equation?, how do the motor repair workshops establish a "rogue" motors HP or do they use some kind of Dyno?

Regards

Paul

Both the torque and the current are directly proportional to the slip up to normal working levels. Motors under 5 hp or so are designed for a slip rating of about 4% at their designed output power. Bigger motors have a lower slip rating. Without slip there is no torque developed. As I said you can load the motor until it's speed drops to about 2880 rpm and get the rated power output.

I do however suspect that his input power measurement is a little off. The bottom line however is that he is currently running at a slip of only 1% so is running at about 1/4 rated power. The motor is more than adequate for the use it is put to.

I must agree with you that an electronic inverter drive is the way to go. My lathe and mill are both fitted with them.

Russell.

You've also assumed that the rpm measured is accurate - generally it's easier to measure current accurately than speed (unless you are specifically set up to do this). And if you go back to the original post, you'll see that this motor, with its low loading, already won't run for long without tripping out the rpc that's driving it - supposedly rated at 2hp.

So whilst I also suspect the current consumption, it's for a slightly different reason; there's something altogether not right about it. We know that those motors are rated, at that size, at 1.5hp max, and this one's drawing, to a first approximation, the sort of current that it should be when loaded and running at its rated speed for a 1hp. And running at 28 degrees with that light a load? Seems to be a tad warm to me, especially as this is likely to be a case reading - how warm is it inside? I think I'd be wanting to have a careful look at this before relying on it.

Edited By Steve Garnett on 13/09/2012 21:17:38

As an edit to my 2nd post I,ll change the wattage figure for the converter from 1100W to 1500W, apologies.

The very quick calc is one I was given by an instructor at Maddon generators more years ago than I care to remember and is the one I,ve used at work to calculate the generated output required to start a motor of any size (3PH) from a mobile/portable Genset--- works for me--- for the previous 35 years. Just use Ohms law to apply it to motor Wattage.

I do agree that the L1 figure is "iffy" but is probably correct given that the inverter is driving L2/L3 through capacitors. Small changes of capacitor values serving L2/L3 would give more balance to the circuit.

Doing the calc L1+L2+L3 in running Watts= 776 W, so given the motor/winding temp 1100/1300 W is somewhere very close.

The motor/winding temp will fall with better phase balance.

Do the calc, Total V(L1+L2+L3) divided by Total A (L1+L2+L3) This also puts you somewhere close.

There will always be variation due to temp and stopping/starting a motor during testing will always make it worse as the motor does not dissipate start up heat until it,s been run for a while.

Nathan.

Edited By Nathan Sharpe on 13/09/2012 23:39:34

Agreed. With the measured 1% slip an error of 0.5% in the measured speed would give a 50% error in the torque or power calculations!

With an ordinary multimeter the current measurement is likely to be within about 5% but the input power is not given by I x V. Power factor must be taken into account, which is likely to be quite low at such low loading.

I think the only valid conclusion from all the above posts is that the motor is adequate for the job but there is a problem with the converter.

Russell.

FWIW, I looked up what Brook Crompton have to say about iron-frame motors and their overall performance, which entirely confirms what I suspect. They give a perfectly reasonable forumula which says:

output hp= (V * A * Efficiency * Power Factor * 1.732)/746

If we assume that the efficiency and power factor figures are in the normal range for an unloaded three-phase motor, then the product of these * 1.732 is going to be approaching unity (I've guessed a value of around 0.9), then the end figures look like:

(240 * 3.3 * 0.9)/746 = 0.955hp.

That's as near to 1 as makes no odds. All the stuff about temperature will make a difference to the efficiency figure, certainly - but not a huge amount of difference to the end result.

It is a 1hp motor!