Discussion in 'General Electronics Chat' started by strantor, Sep 21, 2011.

1. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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for example, if you took a 3phase 230V 5hp AC induction motor rated at 3450rpm @ 60Hz, and you fed it 3phase 36VAC or 38VAC 60Hz, what would happen? would it just lack torque?

which leads me to another question, what happens when you overload an AC induction motor? I've never seen it happen. In a dc motor, it will draw more amps and the speed will droop, but since the speed of the AC motor is determined by frequency, I'm assuming as you inrease the load higher and higher, it would just get to a point where the rotating magnetic field outruns the rotor, and then what happens?

2. ### #12 Expert

Nov 30, 2010
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I can help with the second part. Working with fan motors, I find that excessive load increases current. I am constantly on the watch for a mismatched fan blade or incorrect (not enough) resistance to the air flow causing an overload condition.

One of the basic principles of an induction motor is that if the motor ran at exactly the speed that the number of poles dictate, no current would be drawn because the back EMF would exactly match the applied EMF. The operation is dependent on the "slippage". 4 instance, the 3450 RPM rating of your motor says, "this motor must slip by (3600-3450 = 150) rpm in order to draw enough current to supply 5 horsepower to the load".

I hope this point of view helps illustrate how induction motors work.

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3. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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amazing! I've done that math before while blindly following the instruction manual setting up VFDs but never made the connection that slip is causing you to actaully lose RPM. I always thought of slip as being the phase angle difference between the rotating magnetic field in the stator and the rotor. So this means that with 150rpm of slip, after running for an hour, the rotor is 9000 revolutions behind where it's "supposed" to be.
will additional load cause it to slip further? if so, how far can it slip?

if so, does this mean as you can slow an induction motor down by loading it? that would rock my understanding of AC motors

if not, and it's a fixed slip of 150rpm, what happens as you load it higher and higher? does the current just increase and increase until the motor catches fire? what if it's current limited?

I conceive in my mind that if you loaded the induction motor too high, eventually the rotor wouldn't be able to keep up with the magnetic field and would stall and then oscillate, but not sure if that's right

4. ### VoodooMojo Active Member

Nov 28, 2009
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if a piece of wood is fed into a saw run by an AC motor too fast, what happens?
the motor stops and if the saw is not turned off pronto, the fuse or circuit breaker trips....thats the current limiter

5. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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right, because if not, the motor would burn up. ok, got it. thanks!

6. ### #12 Expert

Nov 30, 2010
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There is a thing called a synchronous motor, but you asked about induction motors.

The amount of slippage is directly proportional to the load. At no load, the motor runs pretty close to the 3600 RPM dictated by the 2 poles.

7. ### steveb Senior Member

Jul 3, 2008
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Perhaps another piece of the puzzle to consider is that voltage and frequency should ideally be changed together. So you might not want to use 36 V at 60 Hz, but instead a lower frequency would be appropriate for that voltage level.

Look up Volts-per-Hertz control of induction motors to get more information. For example, see the following.

http://www.ece.umn.edu/users/riaz/animations/imvfmovie.html

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9. ### GetDeviceInfo Senior Member

Jun 7, 2009
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I look at it a bit different;

remember your basic magnetic induction where voltage is proportional to the speed that you cut the lines of flux. In an induction motor, the rotor windings (bars) will be cut by the rotating stator field at a max frequency when the rotor is stalled. This induces a maximum voltage into the rotor, resulting in a high current field strength in the rotor. This field then reacts with the stator field to give high torque. At or near synchronous speed, there is little or no frequency difference between stator and rotor, resulting in low voltage/current in the rotor and subeqently a weak field interaction.

So what creates the stator current?

remember that your induction motor consists of both resistance and inductance components. At synchronous speed, your reactive component is near nill while at stall it is max. As motor slip increases, so does the out of phase current. When in phase and out of phase currents equal, this is know at the 'torque breakdown' point, beyond which the motor stalls. At start up or stall, high current is due to the reactive component.

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10. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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aaah! back to the books

from wikipedia:
In addition to my misunderstanding of the slip, I also thought that "synchronous" meant that the rotor was always aligned to the poles in the stator. I thought that "asynchronous" meant that there was a phase offset between the rotor and stator (which I thought to be "slip"). so, now 2 big misconceptions I've had for a long time have been cleared up within 30 min. Thanks again #12!

11. ### GetDeviceInfo Senior Member

Jun 7, 2009
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you would be correct if you said always aligned to the stators rotating magentic field. The phase offset between the two is your 'torque angle' and has implications in resulting voltage/current phasing. Typically though, when 'pullout' torque is exceeded, this motor will stall as well.

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12. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
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So, is it a stretch to say that the normal means of using "volts/herts" from 0 to 60hz is considered "constant torque" and using an alternate method of a fixed voltage and variable frequency would be considered "constant hp"?

if for example 48Vac was the maximum voltage available, and you employed volts/hertz all the way up to 48V (12Hz) and then 48vac constant after that, then would you just derate the motor by the fraction of available voltage? (48/230)*5hp = 1.04hp?
wow, that sucks if that's the case.

13. ### steveb Senior Member

Jul 3, 2008
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No, I don't think it's a stretch to say that at all, but I would be careful how you word it. Instead of "constant torque" and "constant power", I'd say you are providing control that allows you to be near the same torque and power operating points. The term "constant" might be interpreted as "control", which an open-loop system can't really provide. The mechanical loading will dictate the actual torque or power.

I think that you can consider the use of volts/hertz as a type of open-loop rotor flux control for steady state operation. Establishing the flux to the proper value allows the motor to provide the needed torque and power, with a reasonable value for the slip.

14. ### #12 Expert

Nov 30, 2010
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Here's a cute experiment with a synchronous motor...pull the front off the ice maker in a refrigerator/freezer, start an eject cycle by turning the gears using the slot provided. After the motor starts, you can feel a BIG difference when you break the synchronization. (This works a lot better when there is no water or ice in the ice maker.) Must have been a smart designer to use that kind of motor so it pretty much quits struggling when the ice stalls the motor. In a minute or so, the mold heater loosens the ice, the motor resumes turning, and there is lots of torque to shove the ice out.

It's just so interesting to actually feel the synchronization "break" when you exceed the torque limit of the motor.

Last edited: Sep 22, 2011
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15. ### GetDeviceInfo Senior Member

Jun 7, 2009
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I did a Millwright apprenticeship in a steelmill many years ago where they drove the rolling mill rolls with two synchronous motors. I don't know what thier ratings were, but the motor frames where about 8' in diameter. I didn't understand it back then, but they must surely have been used for power factor correction as well.