You have not written which motor you are using
You decide your load as per your requirement.You said you know the voltage rating now depending on your load findout current rating.Then you can have its KW rating according to its power factor.convet KW rating to HP rating
1 HP = 746 KW------------British ratings
1 HP = 736 KW------------Asian ratings

Reguards

I guess, you're trying to find the rating of a motor you have.

The first thing one would ask for is: is it an AC or DC motor (or some special type of motors like Universal Motor). I guess it is a DC- for which 24 V is a standard voltage rating. (I'm not aware of standard AC motor ratings of 24 V). But, in that case, I don't think the power rating  will go above say, 100 W (0.13 hp).

Please give some more info. If you have an idea on  where it is being used/ what type of load it is driving/ the Amps of the fuse in its supply line, we can arrive at a better answer. But first of all, please clarify if it's AC or DC motor.

Regds

Umn. (umn@ieee.org)

Here's a little back round that might help.  i just got this project dumped on me...so bare with me.  the motor is a 24 V DC motor.  it is being manufactured for us in china, we need to rate these motor's over here for UL purposes.  in our application we use thes motors to power a scrub brush, by increaseing the down force of the sctub deck we increase the load, thus the current. Now, we also want to sell the motor to anyone else that wishes to buy them, so we need a generic type testing proceedure to determine the horsepower rating.....thanks for all the help so far

Here's a rough way to do it- assuming that the motor is Shunt wound or permanent magnet type.

This method assumes that the motor operates at around 80 % of its maximum efficiency on rated load. Which is a good approximation. I repeat, it is not 80% efficiency- it is 80% of the MAXIMUM efficiency. For example, if you find (how to find- we'll see down) the max efficiency as say, 75 %, at full load operating point, approximate efficiency  is: 80 % of 75% = 60%.

Load test to find max efficiency:

Keep the supply vultage constant at 24 VDC. You'll have to measure
i) the input current and
ii) the mechanical load applied on the motor shaft by standard shaft loading methods.

1. Run the motor on no-load. P_out = Load = 0 hp. Find the input power as
P_in = input amps x supply power / 746  -> hp
Efficiency= P_out * 100/ P_in = 0%.

2. Increase the load in small steps. Find the mechanical load on the motor as P_in. The input current would increase. Find the new P_in and then, the new efficiency. You'll get non-zero values now.

3. Continue step 2. The efficiency increases with load upto a point. Then it starts decreasing. Get a couple of points on the decreasing side also. Make sure that you don't run the motor for long in this load.

Plot the variation of efficiency with load. Find the rated load as indicated in pargraph 2 above.

NOTES:
i) If the motor heats up excessively when you increase the load, this  trick may not work. The method assumes a normal shunt/ permanant magnet DC motor.

ii) In  "80% of maximum efficiency ", you could refer to standard motor manuals of similar capacity to come up with a better value for the factor. Still, 80% is a fairly good value, I guess.

Please keep posted, if you have some other ideas/ you tried out something.

Regards

umn@ieee.org