How can a three-phase motor have 20 Nm of torque at 1432 RPM and 1.8 kW?

Question

I am trying to work out the maximum torque a motor puts out to do some stress analysis on a structure, and I have just been given the motor details from the manufacturer. They said it produces 20 Nm of torque, rotates at 1432 RPM and has a power output of 1.8 kW. I didn't pick up on this at first but later found some errors in my calcs, which pointed back to the motor specs not being in line with each other. From a mechanical standpoint, I've always gone with the Power[W] = Torque[Nm] X Angular Velocity[rad/s] to do any rotational calcs. In this case, if the power output and speed are correct, the torque would be 12 Nm not 20 Nm.

The manufacturer has come back and told me he confirmed the original figures he provided me and they are correct. I was wondering if anyone can tell me how this would work?

Answer 1

20 Nm of torque does not coincide with maximum RPM so you can't just multiply that torque and maximum (unloaded) radians per second because that would be nonesense.

The manufacturer is most likely telling you that the motor can produce a maximum torque of 20 Nm at a significantly lower speed.

Maximum torque will be when the motor is significantly loaded and turning at 859.4 RPM or 60.0% of full (unloaded) speed.

Answer 2

Induction motors need to run very close to their “synchronous speed” which in this case is 1500 rpm. This is a 4 pole motor designed to run on 50 Hz.

As the motor slips due to losses and loads, torque goes up roughly proportional to the slip, up until the rated torque of the motor.

So your calculation of 12 N.m is correct, assuming the RPM they gave you is the rated speed and the 1.8kw is the rated or continuous power of the motor. This is the rated or continuous torque. There are other torque specs of a motor like Peak Torque or Breakdown torque. They typically run 175-300% of the rated torque. This is most probably the 20 N.m the mfr. mentions. Run continuously at that point, the motor will fail from overheating.

So mfrs rate their motor thermally, or continuous and they will give you the maximum torque you can pull continuously. They will give you an ambient temperature maximum for this rating. It is at that point the motor has slipped to 1432 rpm. If you run the motor in a hotter environment you will have to de-rate the motor and not load it past that point as the motor´s cooling ability will be reduced.

If you load it more, you can get it to produce a higher peak torque, but you may only be able to run it for seconds or minutes until the motor overheats. Run more load past this point and the motor torque will decrease and still thermally be in peril.

Answer 3

Torque (N-M) = (kW x 9550) / RPM

1.8 x 9550 = 17190, divided by 1432 = 12 N-m Full Load Torque at rated speed.

Starting / Locked Rotor Torque in a standard Design B AC induction motor is 160% of Full Load Torque, +-10%

20 N-m / 12 N-m = 1.66

So I'd say they were giving you the Locked Rotor Torque value. Many people mistakenly assume this is the maximum torque the motor can develop, but that maximum value is actually called Break Down Torque (BDT, as mentioned earlier) and occurs at roughly 80% of rated speed. So the maximum torque that motor would deliver (again, assuming standard Design B torque-speed curve, the most common type) would be approximately 220% FLT or 26.4 N-m, but would take place at approximately 1145 RPM (assuming the motor is getting full voltage at 50Hz).