Another Motor Theory Question

[philly]

First of all I want to thank everyone who has replied to my threads. I have learned more about induction motors on this site than I have anywhere else.

I'm using the attached motor speed vs torque curve as a reference. I have reports of this motor spiking on a trend up to 1000A. I found this to be a very high value. Looking at the attached curve I see that the 1000 A corrolates to a position that is to the left of the torque breakdown point. After the motor has been up to speed and running, wouldn't any point to the left of the breakdown point be considered a stalled motor? As the motor is running and loaded up, it moves to the left along the torque curve until it gets to the peak of the torque curve which beyond this peak is a motor stall. Is this correct. So tracing this current value of 1000A it would appear that the motor was in a stalled condition.

The motor overloads did trip this motor. If this motor stayed in this stalled condition would the current value move all the way back to a S=1 thus drawing LRC in a stalled condition?

 

[Besoeker]

Philly

I see nothing attached.

 

[philly]

Philly

I see nothing attached.

Sorry I forgot attachment

 

[Jraef]

... After the motor has been up to speed and running, wouldn't any point to the left of the breakdown point be considered a stalled motor?

Not necessarily, but for sure you are flirting with a stall condition at that point.

As the motor is running and loaded up, it moves to the left along the torque curve until it gets to the peak of the torque curve which beyond this peak is a motor stall. Is this correct. So tracing this current value of 1000A it would appear that the motor was in a stalled condition.

Once you drop behind the breakdown torque point, now you are drawing more and more current with less and less effect; that is certainly the beginnings of a stall. But if it recovers, then it isn't a stall.

The motor overloads did trip this motor. If this motor stayed in this stalled condition would the current value move all the way back to a S=1 thus drawing LRC in a stalled condition?

That's why Overload Relays are an I2t curve; it gives you time to recover if possible, but the higher the current gets, the less time you have to effect that recovery. And yes, the current would move all the way back to LRC if the motor comes to a complete standstill.

 

[philly]

Not necessarily, but for sure you are flirting with a stall condition at that point.

Once you drop behind the breakdown torque point, now you are drawing more and more current with less and less effect; that is certainly the beginnings of a stall. But if it recovers, then it isn't a stall.

O.K. I had thought that anything to the left of the breakdown point was a stall but I guess not. Esentially what you are saying, that since the breakdown toruqe is the max torque that the motor can produce, then anything to the left of the breakdown torque is drawing more current without producing any additional toqure.

I guess technically the motor is not in a stall condition until it stops and S=1 all the way to the left of the curve.

When you move to the left of the breakdown torque are you still producing the max torque but just drawing additional current, or does the torque output drop to the left of the breakdown torque as seen on the plot?

Obviously as the motor is becoming overloaded the slip is increasing and the motor is slowing down as you approach the breakdown point. The motor starts slowing down even more as is goes to the left of the breakdown point and the slip incerases thus causing current to increase as well.

Will you see a large jump in current as you move across the breakdown point or will the current continue to increase linerally?

 

[Mayimbe]

O.K. I had thought that anything to the left of the breakdown point was a stall but I guess not. Esentially what you are saying, that since the breakdown toruqe is the max torque that the motor can produce, then anything to the left of the breakdown torque is drawing more current without producing any additional toqure.

I guess technically the motor is not in a stall condition until it stops and S=1 all the way to the left of the curve.

This is true.

When you move to the left of the breakdown torque are you still producing the max torque but just drawing additional current, or does the torque output drop to the left of the breakdown torque as seen on the plot?

the latter.

I believe it depends on the load and the inertia of the machine (which is given).

Obviously as the motor is becoming overloaded the slip is increasing and the motor is slowing down as you approach the breakdown point. The motor starts slowing down even more as is goes to the left of the breakdown point and the slip incerases thus causing current to increase as well.

Will you see a large jump in current as you move across the breakdown point or will the current continue to increase linerally?

Yes. the current will increase in a lineal way. That large jump in current can happen also if the machine has an abrupt change in the load.

 

[gar]

091112-0831 EST

philly:

You need to find some good books on AC machinery. The one I have from many years ago is Alternating-Current Machinery by Bailey and Gault, 1951, McGraw-Hill. Chapters 12 and 13 cover induction motors.

If you have a low resistance rotor, as is the case for the curve you included, then you have the negative resistance type of effect in the curve. Your motor is a low resistance rotor. A high resistance rotor has a torque curve that does not fold back. See plot in figure 12-10 on page 209 of Bailey and Gault.

In the fold-back type of curve if you have a load with a nominal torque that is constant independent of speed and just below the breakdown torque, then a slight overload for a short time that takes you beyond the breakdown point may cause the motor to recover at a lower speed. Note this bi-stable possibility by drawing a straight line at constant torque and and see the two intersection points with the motor speed-torque curve.

Learn how to draw load speed-torque curves overlaid on your motor speed-torque curve to determine the operating point of the motor and load.

.

 

[Besoeker]

The motor starts slowing down even more as is goes to the left of the breakdown point and the slip incerases thus causing current to increase as well.

Will you see a large jump in current as you move across the breakdown point or will the current continue to increase linerally?

The increase in current will follow the curve you have shown. It isn't linear and there is no discontinuity or jump. Sure, the changes are likely to happen quickly if you exceed the maximum torque but the current will still follow the curve.

Maybe here's a different way of looking at it.
The current the motor takes depends on slip. This is true regardless of loading. From your curves, the motor will always take about 1230A at 600 rpm or s = 0.5 no matter how it got to that point. If the motor torque is greater than the required load torque (and, in normal circumstances, it would be) it will accelerate. And vice versa.

Over the normal operating range, increasing load torque results in increased slip and increased current until the motor output torque is in equilibrium with the required load torque.

 

[philly]

I went ahead and took the advice above and overlaid some load torque curves on this motor torque curve to see what happens. I might have gonnen carried away, but heres the scenarios and questions I came up with.

Case I

For the case I plot below I plotted (3) different constant torque loads. I belive this is what a constant torque load curve looks like due to the fact that it has a constant torque throughout the speed range. As the load increases the horizontal curve is shifted upwards and intersects the motor torque curves at higher points. As the load is increased more torque is required from the motor and thus the current increases as well. I can see that curves #1 and #2 in this plow will operate within the motor torque range, however I'm not sure about curve #3? It intersects the motor torque curve in two places both to the left and right of the breakdown point. Which point will this load operate at? Will it ever reach the breakdown point since it hits the motor torque curve to the left of the breakdown point?

Case II

For the Case II plot I have shown a variable torque load. This varialbe torque load increases with speed as I have plotted (3) different load curves. I'm assuming that as the load increases on a variable torque load the curve will also shift upwards as shown in curves #1-3. Curves 1 and 2 will operate within the torque range of the motor although at #2 curve point there will be excessive current as it reaches the breakdown point. What about curve #3? This crosses the motor torque curve during the starting of the motor to the left of the breakdown point. Will the motor try to start and only reach a speed given by this intersection point of the load and toruqe curve? Or will the motor attempt to start and stall due to the fact the load torque goes above the motor torque?

Case III

In case III I have drawn a load torque curve that intersects the motor torque curve again both to the left and right of the motor torque curve breakdown point. What happens in this case with a variable torque load? During starting will the motor stall or only reach a speed determined by the intersection on the left of the breakdown point and draw excessive current? If the motor was already running at speed and then is loaded will it operate at the intersection point on the right of the breakdown point, or will it somehow sift over to the intersection points on the left?

Case IV

The final case in case IV shows a load torque curve that goes above the motor torque curve very shortly after zero speed. In this case, I would expect that the motor will not start at all and will barely move. It may even stay in a locked rotor condition. Is this correct?

I appreciate everyones responses to my questions, for they have helped me a great deal.

 

[Mayimbe]

Case I

For the case I plot below I plotted (3) different constant torque loads. I belive this is what a constant torque load curve looks like due to the fact that it has a constant torque throughout the speed range. As the load increases the horizontal curve is shifted upwards and intersects the motor torque curves at higher points. As the load is increased more torque is required from the motor and thus the current increases as well. I can see that curves #1 and #2 in this plow will operate within the motor torque range, however I'm not sure about curve #3? It intersects the motor torque curve in two places both to the left and right of the breakdown point. Which point will this load operate at? Will it ever reach the breakdown point since it hits the motor torque curve to the left of the breakdown point?

Its very difficult to know with your point of view. What I would do is see where curve#3 intercepts with the current curve, then make a projection from that point to the torque curve and see where is at. From your curves I estimate that the torque value will be decreasing and near the value of 1800inlb/ft

Case II

What about curve #3? This crosses the motor torque curve during the starting of the motor to the left of the breakdown point. Will the motor try to start and only reach a speed given by this intersection point of the load and toruqe curve? Or will the motor attempt to start and stall due to the fact the load torque goes above the motor torque?

Same as before, make the projection, the motor will work in a high current low torque status. It will damage the motor eventually.

Case III

In case III I have drawn a load torque curve that intersects the motor torque curve again both to the left and right of the motor torque curve breakdown point. What happens in this case with a variable torque load? During starting will the motor stall or only reach a speed determined by the intersection on the left of the breakdown point and draw excessive current? If the motor was already running at speed and then is loaded will it operate at the intersection point on the right of the breakdown point, or will it somehow sift over to the intersection points on the left?

breakdown point. in that scenario.

Case IV

The final case in case IV shows a load torque curve that goes above the motor torque curve very shortly after zero speed. In this case, I would expect that the motor will not start at all and will barely move. It may even stay in a locked rotor condition. Is this correct?

Correct. its like you try to move an elefant. Although the current will be very very high in this condition

 

[philly]

I found an acutal motor speed vs torque curve with the load torque already overlaid on it.

I am confused by this because it appears to show the load torque curve intersects the current cuve at about 1.8 P.U. This would mean that the motor would be running over its rated current.

Or am I wrong and you ignore where the load torque crosses the current cuve and only look where it crosses the motor torque cuve and then extrapelate a line up or down to see where it intersects the current curve.

On second thought that looks exactly like what is going on, and the current ponit is the point marked just below .5 P.U. current.

 

[Besoeker]

I found an acutal motor speed vs torque curve with the load torque already overlaid on it.

I am confused by this because it appears to show the load torque curve intersects the current cuve at about 1.8 P.U. This would mean that the motor would be running over its rated current.

Or am I wrong and you ignore where the load torque crosses the current cuve and only look where it crosses the motor torque cuve and then extrapelate a line up or down to see where it intersects the current curve.

On second thought that looks exactly like what is going on, and the current ponit is the point marked just below .5 P.U. current.

Current and torque are shown on different axes with different scaling.

What the curves don't tell you are the actual values of torque. 1.0 pu load torque might be less than 1.0 pu motor torque thus occur at less than 1.0 motor rated current.

 

[philly]

Yes but I would think that 1 P.U. torque will corrospond to 1 P.U. torque output. Like we stated earlier for any torque point on the curve we can find the corrosponding current by drawing a vertical line, and seeing where this vertical line crosses the current curve.

I see now what I was missing in regard to the load torque. If you look at where the load torque intersects with the motor torque its a little less than 1 P.U. torque and looks to be about .8 P.U. torque. If you draw a vertical line at this point you will see that this vertical line intersects with the current curve at about .8 P.U. current as well. This point is marked on the current curve although it may be difficult to see on this pdf.