# Induction Motor - Torque Speed Characteristics

Discussion in 'General Electronics Chat' started by shespuzzling, Apr 22, 2010.

1. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
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Can someone please explain the torque/speed characteristics of an induction motor? In particular, I am confused about the following graph:

http://www.ecmweb.com/mag/405ecm08fig2.jpg

I think maybe my understanding of torque is not accurate...I thought it was just a force that was exerted on an object causing it to spin (because there is a fixed axis of rotation). In the case of an induction motor, the torque should be dependent on the current through the rotor, which is in turn dependent on the speed of the rotor...so torque and speed should be dependent on one another. The book (Wildi - Electrical Machines) I'm reading doesn't seem to agree with me there. Also, I'm not sure why the torque starts out high (assuming it has something to do with momentum & starting from rest?), and then gets lower, and reaches the breakdown torque at a speed lower than the full load speed....if it breaks down before it gets to full load speed, how can it ever get there?

As the rotor starts to turn, the torque would be at its highest (I think) because the most amount of current is flowing through the rotor bars and thus the force on it is strongest. But then as the rotor speeds up, the current in the rotor decreases because the changing flux decreases so shouldn't the torque decrease as well?

2. ### Markd77 Senior Member

Sep 7, 2009
2,803
594
Looks like this motor is being driven with a fixed frequency so it has a peak in torque there (right side). Pretty much all motors have good torque at low speeds so that's why the peak at zero speed.
Imagine a normal DC motor curve which starts high and then drops off to zero added to this peak at the frequency the motor is driven at.

3. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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Hm, I just started learning about motors so I'm not sure I understand...my biggest problems have to do with breakdown torque (why does it happen at a lower speed than full load speed? how is that possible?) and the fact that a rotor can keep picking up speed even though the torque that corresponds to a faster speed should be smaller because the current in teh rotor is less.

4. ### Markd77 Senior Member

Sep 7, 2009
2,803
594
This is the curve for a normal DC motor. The induction motor is being driven at the frequency at the far right side of your graph. The Full load speed is for maximum power output and maximum power does not occur at maximum torque because power is proportional to speed X torque.
For the DC motor maximum torque is at zero speed and although the motor is using a lot of power, the mechanical power is zero, because it is not doing any work.

5. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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I'm sorry, but I'm not following. Where does frequency come into play in the graph I posted? I am assuming that the induction motor is being fed by some constant frequency generator. My understanding was that the graph I posted was a generic example of torque/speed characteristics for all induction motors (depending on resistance of the rotor, etc...I've actually only learned about squirrel cage & wound rotors so maybe not ALL IM's, but at least those).

So, according to that graph, as a motor picks up speed, the torque starts out higher than full load torque. That I understand...when the rotor is initially at rest the flux changes fastest w.r.t. the rotor so maximum current is induced and there is a large torque. So why doesn't the torque (and speed) decrease as the rotor starts to rotate? Is there momentum that is keeping the rotor moving and less torque is required to turn it?

Then, on the graph, the torque goes down to a minimum point (pull-up point I think), and picks up again. But this all happens as the rotor is picking up speed, so that would mean that the current is getting smaller, so how is the torque getting bigger? And, in order for the rotor to be at full-load speed, the graph implies that it would have to eventually pass the speed associated with break-down torque, so why doesn't it stop before it gets to full-load speed?

Thanks.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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An induction motor is normally sized for the anticipated full load torque. Load torque will depend on the type of load - be it a pump or fan etc. Different loads will have different torque-speed characteristics. As long as the load torque-speed characteristic lies inside the motor torque-speed envelope then normal operation is possible. The designer must be careful to ensure that the mechanical load never approaches the pull-out torque value. That result is highly undesirable.

The torque-speed graph simply shows the operating envelope for the motor over a full range of torque-speed conditions. Normally, the motor would not traverse the curve when starting from zero speed and accelerating to the particular torque-speed demand from the load. If you overlay a load torque-speed characteristic on the motor torque speed characteristic you will come up with a resultant operating point.

Suppose a load has a linear torque-speed characteristic rising from zero torque at zero speed to 80% of the motor rated full load torque. The load torque-speed line will intersect the motor torque speed line at a speed slightly higher than the full load torque speed. That is, the motor would run slightly closer to synchronous speed at 80% of full load torque than it would at its rated full load torque. During startup the motor torque-speed values would simply follow the load torque-speed line. It would [by intentional design] not approach anywhere near the pull out torque.

Suppose while running at 80% of rated load an abnormality develops in the load - let's say it's a pump in which a bearing starts to seize up. Suppose the load torque now starts to rise dramatically above the rated value. It's now possible in extreme circumstance that the motor will start to traverse the torque-speed line up to the pull out torque. Once that occurs the motor supply protection provisions will come into play - probably by isolating the motor from the supply.

Last edited: Apr 22, 2010
7. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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Thanks, that makes things much clearer. So, are load torque-speed characteristics typically linear? Also, is the full-load torque and full-load speed exclusive to a particular load, but the general characteristics and the shape of the curve always pretty similar for a particular motor across all loads?

Does the fact that the pull-out torque occur at around 80% of the synchronous speed mean that if a load is applied with a torque equal to 2.5T (same as the value of the breakdown torque on the graph I attached), then it will accelerate to a speed that is 80% of the synchronous speed and at that point motor will stop?

Finally, for the starting torque and pull-up torque, if the load characteristics are linear from 0 speed to 80% of the full load torque, then the load torque is never equal to the starting torque or the pull-up torque. So, what do those values actually mean in terms of the load and the motor?

Thanks again, I think (hope) I have a better idea of what these graphs mean.

8. ### Markd77 Senior Member

Sep 7, 2009
2,803
594
No, at zero speed the motor cannot provide a torque of 2.5T so the motor will be stalled. 2.5T is only available at 80% synchronous speed, if it was running under those conditions and slowed down even slightly it would not be able to provide enough torque and would stop.
A normal load like a fan would have low torque at low speed and the torque would increase with speed.
A rotating drum which lifts a weight on a cable is an example of a load with constant torque.

9. ### retched AAC Fanatic!

Dec 5, 2009
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Right. Motors stall torque depends on the speed. This is usually linear in each motor, but not linear across motors in general.

10. ### GetDeviceInfo Senior Member

Jun 7, 2009
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Your graph represents a typical NEMA B torque curve. There are others.

Torque, as you originally surmissed, is related to current flowing within the rotor. However, to clarify, it's directly related to inphase current.

What bearing does frequency have?, in the classic formula, inductive reactance is determinent on frequency. Where does the specific frequency come from?, slip.

At full speed and down to approx 25% slip (B design), your resistive current is higher than your reactive current. At the 25% mark, you cross the threshold and reactive wins out over resistive causing a downward cascade to stall.

The curve itself comes from the addition of inphase and out of phase currents over the speed range of the motor.

11. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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Okay, so about the starting torque:

1. Why is the starting torque higher with higher rotor resistance? Wouldn't the lower current flowing in the rotor mean the torque acting on the rotor at start-up is actually lower?

2. Does the value for starting torque (1.5T for the NEMA B torque curve) just tell you the maximum available torque at start up that the motor can provide? If the load requires less than that amount of torque for start-up, then is the actual starting torque lower as well, but if the load requires more, then you get a stall?

12. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
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Just reviewed the torque section on this website and another question cropped up:

Why does the torque increase from the pull-up torque to the break-down torque even though current in the rotor is decreasing?

I got the impression from the info on this website that a motor does actually follow the torque-speed curve (NEMA B), as it starts up and eventually reaches equilibrium. Is the breakdown torque only detrimental if it is there for an extended period of time? Is it okay to momentarily reach the breakdown torque? Maybe I'm not reading this tutorial correclty, not sure. Anyway, answers to these questions would be most appreciated.

Thanks.

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Perhaps another way to view the torque-speed graph for the motor is that it represents the limiting difference between the actual load torque appearing at the motor shaft at any given speed and the 'reserve' accelerating torque available from the motor itself at that same speed.

So during acceleration from standstill the available accelerating torque varies from a minimum at the pull-up/pull-in torque point (B on the curve) to a maximum at the breakdown/pull-out torque point (C on the curve). The starting torque difference is somewhat more than the minimum available torque difference at the pull-up/pull-in point.

14. ### GetDeviceInfo Senior Member

Jun 7, 2009
1,571
230

It will follow the curve. The actual loading will determine the time that the curve executes. No load, rotor mass alone is relatively quick. With a load, the time frame extends and could trip overloads.

t_n_k has a good view point in that an uncoupled motor really is delivering no output torque.

Because your reactive component is decreasing along with slip.

any load that exceeds the motors torque range will cause it to slow, driving up the current. This induces heat into the motor which will cause it to burn out if not protected by overloads. Within allowed temperature rises, you can apply any load to the motor, even a locked rotor (stall) condition.

A major concern then becomes speed regulation.

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Also keep in mind that the torque-speed characteristic tells you nothing about the dynamic response of the motor+load combination.

It's worth conducting a "thought experiment" in which you consider a constant inertia flywheel attached to the motor shaft. How will the motor speed & torque evolve with respect to time during the startup interval - up to the point where the speed reaches a steady state value? One can use this "experiment" as a notional representation of direct-on-line [DOL] starting of an unloaded motor in which the effective inertia is that embodied within the rotor itself. The experiment might also add some losses to the mix to make it more realistic.

I can recommend a useful software tool which might help in gaining insights into these matters and which makes the "thought experiment" a lot easier to conduct.

16. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
0
I'm sorry but I do not understand this response, can you explain it a little more? My understanding regarding torque/speed is very black and white: more current in the rotor means more torque acting on the rotor, less current = less torque. Please let me know if that is not the case because that could be the root of many of my problems.

I'm having similar problems with this response, with the same reasoning as above. As the motor speeds up, slip is decreasing, which means less current is flowign through the rotor bars and there should be less torque acting on the rotor...as you can see it's this very fundamental issue that is making this very difficult for me to understand.

So, even if the load is very light, there will be a minimum torque (the pull-up torque) acting on the load? At no load, will the motor torque/speed curve look like a straight line? Finally, do all load torque/speed curves start at some non-zero torque at 0 speed? I think I maybe be misunderstanding something very fundamental about the nature of torque...but I just can't seem to figure out what it is.

t_n_k, yes, a software recommendation would be helpful!

Thanks.

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
It's sometime hard to get a handle on torque (pardon the pun).

How would one measure motor starting torque for instance? One method might be to clamp a bar to the motor shaft and anchor the other end of the bar to a spring balance mechanism of suitable load capacity. Turn on the AC supply and wait for equilibrium - the bar arm length multiplied by the balance indication gives the starting torque. Don't do this for too long or you'll burn out the motor.

Standstill torque certainly exists for the motor. In a sense load torque always exists if only for the physical requirement of overcoming mechanical inertia. This is only a transient phenomena. The steady state torque provided by the prime mover (motor) will depend on the work being done in / by the mechanical load. Neglecting inertia, one can model a physical system according to the load type. A conveyor may just have a linearly increasing torque-speed characteristic starting at zero torque at zero speed.

So consider such a load at standstill. The load presents zero torque but the motor can deliver standstill (starting) torque. What happens? The motor load combination accelerates in angular velocity. The instantaneous shaft mechanical torque (which might be measured with a dynamometer) would be the required load torque at the shaft instantaneous speed. One might say that the available motor accelerating torque is "hidden" - as part of the total electrical torque equivalent being supplied to the motor windings from the power supply. The measurable / observable mechanical shaft torque is what the load demands - not what the motor is capable of delivering at a particular speed. Since the motor + load system is changing dynamically the combined motor + load inertia will also play a role in the dynamic torque conditions. A simple relationship for instantaneous load torque might be ...

$T=K*\omega+J*\alpha$

where J is the load moment of inertia and α is the shaft angular acceleration at speed ω. K is a constant.

One can also imagine controlling the motor power supply so as to gently increase the load speed in a less stressful manner than occurs with a direct-on-line (DOL) start. It's quite conceivable that one can achieve a smooth linear increase of load speed or torque starting from standstill. The torque speed observations will be nowhere near those of the limiting (Nema) curve for the motor.

Download an evaluation copy of PSIM from "www.powersimtech.com". There's a learning curve but it's worth it.