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Given voltage, measured current draw, and RPMs of an AC motor, and an estimated efficiency rating, is it possible to calculate the actual instantaneous output torque? Or is other information required?

I can also measure its physical dimensions; shaft diameter, etc. With some effort I should be able to measure the time it takes to spin up to full speed as well.

Here is the motor I have and the only label on it. I don't know where it came from:

enter image description here

enter image description here

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  • \$\begingroup\$ Is this a theoretical question, or do you actually have a motor and you want to know how much torque it can put out? If you have the HP rating and rated RPM's, you can calculate exactly how much torque it can put out. You should take a picture of the motor, especially the name plate if it has one, and edit your question to include the picture. Also, are you after rated torque, or actual torque under no-load? \$\endgroup\$ – mkeith Nov 21 '18 at 5:23
  • \$\begingroup\$ I actually have a motor. I'm after actual torque under no load. I'll add a picture now. \$\endgroup\$ – Jason C Nov 21 '18 at 5:29
  • \$\begingroup\$ @mkeith It says 180W on that label; which I guess is something like 1/4 hp, assuming that's a measure of output power although I only barely know what I'm talking about. \$\endgroup\$ – Jason C Nov 21 '18 at 5:40
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    \$\begingroup\$ Well, don't be offended, but I have to make sure you know what that means. The no-load torque is the amount of torque required to overcome whatever drag forces are acting on the motor. I would not say that this information is useless, but normally I don't worry much about it. So are you sure that is what you want? \$\endgroup\$ – mkeith Nov 21 '18 at 6:54
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    \$\begingroup\$ The motor torque curves are somewhat standardized. industrialelectricalco.com/wp-content/uploads/2014/01/… \$\endgroup\$ – mkeith Nov 22 '18 at 0:40
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\$cos(\phi)\$ is also required, The phase difference between voltage and current. The efficiency in that load is available in the catalog. \$ V * I * cos(\phi)*efficiency= rotational speed(rad/s)*Torque(NM)\$

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  • \$\begingroup\$ Great, thanks. I can measure that with a scope. What do you mean by "catalog", though? \$\endgroup\$ – Jason C Nov 21 '18 at 5:37
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    \$\begingroup\$ @JasonC, The formula is for single phase motor. \$\endgroup\$ – M KS Nov 21 '18 at 5:43
  • \$\begingroup\$ With what is given, we can solve for the product of cos(phi) and efficiency. But unfortunately, we can only solve for the product at full load. We have no way to know efficiency or cos(phi) under no-load conditions. \$\endgroup\$ – mkeith Nov 21 '18 at 7:20
  • \$\begingroup\$ cos(phi) * efficiency = 180W / (2.8A * 120V) = 0.54. But this only applies at full load. At no-load, the efficiency and cos(phi) would likely change, so 0.54 cannot be used. \$\endgroup\$ – mkeith Nov 21 '18 at 7:23
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    \$\begingroup\$ @JasonC, Can you explain more? why do you need torque? when in no load the external torque on the shaft is zero. \$\endgroup\$ – M KS Nov 21 '18 at 9:10
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I think that is an induction motor from a washing machine. 180W is the rated output power (mechanical power). 1600 rpm is the rated speed. So that means that when you power it from 120V at 60Hz, it will deliver 180W when it is spinning at 1600 RPM. This also implies that the synchronous speed of the motor is 1800 rpm. Conceptually, synchronous speed is the speed at which the torque is zero. As you load the motor, the motor slows down (this is called slip) and the torque increases linearly with slip.

So 1800-1600 = 200 rpm. That 200 rpm is the slip required for the motor to put out its rated power and torque. Let's calculate torque.

Power = speed * torque

If we use rad/sec and N-m, power will be in Watts. So let's convert. 1600 rpm = 26.7 rev/sec. 26.7 * 2pi = 168 rad/sec.

If Power = speed * torque, then torque = power/speed. So 180W / 168 rad/sec = 1.07 N-m at rated power. So now we know that when slip = 200 rpm, torque will be 1.07 Newton meters. Likewise, if slip = 100 rpm, then torque must be 0.5035 Newton meters.

So in general, torque is as follows:

T = 1.07 * (slip / 200)

It will be in Newton meters. Slip is in rpm. You calculate slip as follows: 1800 - rpm, where "rpm" is the actual rpm.

So, breaking it all down, measure the actual no-load rpm. Take 1800 rpm, and subtract the actual no-load rpm. This is the no-load slip.

Divide by 200, and multiply by 1.07. Now you have the no-load torque. Approximately. You can also calculate the no load output power, since all you need is torque and speed (just multiply rad/sec * Newton-meters to get power in Watts).

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  • \$\begingroup\$ I think I understand; this makes sense. But I am driving home for the holidays (at a rest stop) and won't be able to consider carefully until Thursday or Friday. May I ask how you obtained the synchronous speed? I see that it is a multiple (30x) of the AC frequency but I am not clear where that came from. \$\endgroup\$ – Jason C Nov 21 '18 at 22:52
  • \$\begingroup\$ I think I found it. 120 * 60 Hz / 4 poles per phase? en.m.wikipedia.org/wiki/Synchronous_motor#Synchronous_speed . I will read more about slip. Ok enough procrastinating, I've finished my Red Bull, time to get back on the road 😂 \$\endgroup\$ – Jason C Nov 21 '18 at 23:06
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    \$\begingroup\$ Yeah. Synchronous speed is always line frequency divided by the number of pole pairs. But you have to convert frequency to RPM first. So, a two pole (one pair) motor is 60 Hz * 60 sec/min = 3600 RPM. For a four pole (two pairs) it is 3600/2. For six poles (three pairs) it is 3600/3, etc. \$\endgroup\$ – mkeith Nov 22 '18 at 0:20
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    \$\begingroup\$ For a motor driven at line frequency (not driven by a variable frequency control) current is a good approximation of torque. Measuring slip is also a good approximation of torque (see my answer for the math). Motor power output is speed in rad/sec * torque in Newton-meters. If you give a motor an intertial load (heavy spinning wheel) you can infer torque by noting the rate of change of speed as well as actual speed. \$\endgroup\$ – mkeith Nov 22 '18 at 0:26

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