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 

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

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?

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

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

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.

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

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!

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.

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

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!