Maximum Torque of Induction Motor

Why it is said Maximum torque of induction motor is inversely proportional to frequency square ? Maximum torque formula for induction motor is below. I can't understand why max torque is derated according to frequency square ? K constant in formula is inverse proportional to frequency but not square.

@boodo23:
This is what I learned from the very beginning.

.
T= HP X 63.025
___________
S

T = TORQUE IN LB. IN.

S = SPEED IN RPM

T = Hp * 5252/RPM


RPM = 120 * f/# of motor poles


VFD will maintain volt/hz ratio from 0 - 60hz. Above that current output begins to drop to maintain max voltage, which means torque will drop as well.


Base f/operating f = % torque loss

Here’s a few things for you to think about
Early on I was told
An electric motor below base speed runs in constant Torque
and above base speed runs in constant HP

But here’s a question of you genius
How do you calculate the torque at zero speed
Jet a motor will develop torque at zero speed it could even be above the rated torque
I have had motors develop 100% torque at zero or near zero speed
Check a suspended load on a hoist

GaryS said:

HereÂ’s a few things for you to think about
Early on I was told
An electric motor below base speed runs in constant Torque
and above base speed runs in constant HP

But hereÂ’s a question of you genius
How do you calculate the torque at zero speed
Jet a motor will develop torque at zero speed it could even be above the rated torque
I have had motors develop 100% torque at zero or near zero speed
Check a suspended load on a hoist

Click to expand...

If you take HP * 5252 (constant) / rated speed, that gives you your motor output torque. At rated speed (which implies rated slip), rated current.

If your motor is rated at 1790 rpm (for example) at 60 Hz (4 pole motor, US) then it is rated for 10 rpm slip.

If your VFD is putting out 0.16667 Hz (* 60 = 10 rpm) then your motor should give rated torque, and may be a bit over rated current if the VFD has boost programmed in.
If your load does not move at this torque, and the VFD increases the output frequency, to say 0.333 Hz (20 rpm) then you climb up the torque curve for the motor. It can't do this output torque continuously, but to begin moving the load it can.
If your load still does not move at this torque, and the VFD increases the output frequency, to say 1 Hz (60 rpm) then you climb up the torque curve some more and get .. likely .. close to where the motor peak torque (also called breakdown torque) happens. None of the motor manufacturers talk about this value, but I have gotten over 250% torque on small motors when testing this out. I can't test big motors

https://www.engineeringtoolbox.com/electrical-motors-torques-d_651.html


This is how I understand it. My depth of understanding is not perfect, of course. YMMV

I would welcome this discussion in classroom
Let me say this most people miss the details
When you as a 1790 rpm motor you tell me a lot with just that much information
1790 means it was manufactures for use in the US. It’s a 4 pole motor rated to run on 60HZ system and as you say it has a rated slip of 10 rpms. And if it is truly running at 1790 it producing it’s rated output torque. It also tells me that on 50hz system it would be running at approximately 1490 rpm’s
But it would still have the same volts to hertz ratio 8 volts per hertz 60hz / 480V 50hz / 400V
You can take it down to 10hz or up to 120hz or even higher it you want
Having said that you still did not answer the original question
How do you calculate the output torque at zero speed / RPM
In your reply you used 10 rpm’s to calculate the torque while there is no question that a squirrel cage motor will develop 100% torque at as slow as 1 rpm but all of the formulas require an rpm number to calculate the torque and number 0 just does not work.
And to add another question, what is output torque of a 4 pole 60hz squirrel cage motor running at exactly 1800 RPM ( Remember exactly 1800 rpm not 1790) assuming the power input frequency is 60hz

GaryS said:

I would welcome this discussion in classroom
Let me say this most people miss the details
When you as a 1790 rpm motor you tell me a lot with just that much information
1790 means it was manufactures for use in the US. It’s a 4 pole motor rated to run on 60HZ system and as you say it has a rated slip of 10 rpms. And if it is truly running at 1790 it producing it’s rated output torque. It also tells me that on 50hz system it would be running at approximately 1490 rpm’s
But it would still have the same volts to hertz ratio 8 volts per hertz 60hz / 480V 50hz / 400V
You can take it down to 10hz or up to 120hz or even higher it you want
Having said that you still did not answer the original question
How do you calculate the output torque at zero speed / RPM
In your reply you used 10 rpm’s to calculate the torque while there is no question that a squirrel cage motor will develop 100% torque at as slow as 1 rpm but all of the formulas require an rpm number to calculate the torque and number 0 just does not work.
And to add another question, what is output torque of a 4 pole 60hz squirrel cage motor running at exactly 1800 RPM ( Remember exactly 1800 rpm not 1790) assuming the power input frequency is 60hz

Click to expand...

I had hoped to side-step the 0 rpm debate. I guess it didn't work

For a motor, the torque (according to my limited understanding) has to do with the induced current in the rotor, current-carrying conductors crossing magnetic lines of flux, etc.

So I guess I'll re-phrase my answer. The torque at 0 rpm is dependant on the current into the motor, and the slip that the motor is rated for.
No math here - my reasoning is that the math is worked out for a motor at rated rpm, slip, poles, driven at 60 Hz or 50 Hz.
Since the motor torque is known at rated speed, with rated current, and 50 or 60 Hz input ... that is what it generates at 0 hz. The rotor current, flux, et al is not the same, but it should be close enough if you are not using good test instruments
If the VFD drives the motor at rated current, rated slip, you will get about rated torque


As for torque over the rated speed, I need more information. Does the VFD have more voltage to apply?

For a 208V VFD driving a 208V motor - the torque goes down. Back-of-the-napkin numbers would be rated torque * 1790/1800


For a 575V VFD driving a 208V VFD rated motor with the properly rated motor insulation, torque should be constant up to (575/208*1800)-10 rpm. Rated current will give you rated output torque at rated slip.


If the motor is not 'pushing' rated load, it has less slip, will turn slightly faster, and generate less torque. If it is trying to 'push' over rated load, it will have more slip ... if it is supplied proportionally more current then it will drive the load. Up to breakdown torque. If the VFD is limiting motor current below what the motor needs to generate the required torque, the motor will stall.


IMHO

I think there is some confusion between the torque demand of the load and the torque capability of the motor. They aren't always equal.

For a centrifugal pump, for example, the torque demand of the pump drops as the square of the speed. That's because the power demand drops with the cube of the speed, and per the formulas others provided the torque will drop with the square.

The motor should be capable of providing more torque than the load demands. If not the motor will decelerate. If the torque demand from the load exceeds the breakdown torque the motor will stall.

At starting the motor will put out the locked rotor torque and accelerate the load. It will continue to put out maximum torque during acceleration. If that isn't enough to get the load to speed before the overloads trip you have a failure. At steady-state the load torque matches the output torque. That may not be the rated torque of the motor. Think about putting the fan blade from your bathroom ventilator onto a 50 hp motor. The torque, and power, will be much less than the motor capability, won't it?

https://www.engineeringtoolbox.com/electrical-motors-torques-d_651.html

Tom Jenkins said:

I think there is some confusion between the torque demand of the load and the torque capability of the motor. They aren't always equal.

For a centrifugal pump, for example, the torque demand of the pump drops as the square of the speed. That's because the power demand drops with the cube of the speed, and per the formulas others provided the torque will drop with the square.

The motor should be capable of providing more torque than the load demands. If not the motor will decelerate. If the torque demand from the load exceeds the breakdown torque the motor will stall.

At starting the motor will put out the locked rotor torque and accelerate the load. It will continue to put out maximum torque during acceleration. If that isn't enough to get the load to speed before the overloads trip you have a failure. At steady-state the load torque matches the output torque. That may not be the rated torque of the motor. Think about putting the fan blade from your bathroom ventilator onto a 50 hp motor. The torque, and power, will be much less than the motor capability, won't it?

https://www.engineeringtoolbox.com/electrical-motors-torques-d_651.html

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Good catch!


I guess I did not list my assumptions. I assumed that we were discussing a motor, at 0 rpm, driven by a VFD.


@Tom Jenkins has supplied the across-the-line answer. Which is very much industry-accepted/empirically determined/proven in the field.

Need Slip to produce Torque

To Answer Gary's Question, motor will not produce torque with out slip so 1800 rpm motor at 1800 rpm no torque.

I thought I would take minute and reply
Some of you have done a great job of explaining how the torque is developed in a motor at zero speed
And that’s great to help some of us understand how things work To first get the motor moving

But the original question was how do you calculate the torque at zero speed. The simple answer is you can’t, all the formulas require some rpm value if you use 0 (zero ) the calculation does not work
While we all know and understand that the motor will develop a large amount of torque at start up but the formula just doesn’t work with 0 rpms

The 2nd question on a 4 pole squirrel cage motor running exactly at 1800 rpm’s what is the torque
Thank you AkaHammer you got it right and supplied the explanation
A squirrel cage motor will never run at base speed because at base speed they develop zero torque and rpm’s will slip back, the higher the torque requirements the larger the slip
The motor will develop its rated torque at its rated slip
The base speed of a 4 pole motor on a 60hz system is 1800 rpms the same motor on 50hz system will have a base speed of 1500 rpms
This only applies to squirrel cage motor PM motors and Sync motor are a different thing altogether

Instead of putting 0 for rpm, just put .00001; would this not work? So that way, it won't cancel everything out? Maybe even the constant 1 would work?

bodoo23 said:

Why it is said Maximum torque of induction motor is inversely proportional to frequency square ? Maximum torque formula for induction motor is below. I can't understand why max torque is derated according to frequency square ? K constant in formula is inverse proportional to frequency but not square.

Click to expand...

In an AC induction motor, if you keep frequency the same, maximum torque is proportional to the VOLTAGE squared...

So if you apply 1/2 voltage to the motor, the maximum torque (Break Down Torque) drops to 1/4 of the normal value.
This is the design basis of all Reduced Voltage Starters.

alive15 said:

Instead of putting 0 for rpm, just put .00001; would this not work? So that way, it won't cancel everything out? Maybe even the constant 1 would work?

Click to expand...

No.

The torque capability of the motor is a function of geometry, electromagnetics, and all kinds of stuff that appear to me to be pure magic. The motor manufacturer can supply locked rotor torque, breakdown torque, etc.

The load torque is a function of, well, the load. For a conveyor at start, it is the load plus breakaway for friction. For a fan at start it is essentially zero.