AC WOUND ROTOR INDUCTION MOTOR STARTER
AC WOUND ROTOR INDUCTION MOTOR STARTER
(OP)
Please,is it possible to short out the slip ring-set of a 3-ph wound rotor induction motor,and start the same with auto-transformer by the motor-stator instead of rotor,if yes,what are the implications?





RE: AC WOUND ROTOR INDUCTION MOTOR STARTER
Usually this results in basically a simple induction motor with inrush currents and everything else that comes with an induction motor. However, occasionally, shorting the rotor results in a motor that is too tight, ie, it has too little slip. When that happens, the inrush current goes way up and the starting torque goes way down.
Exactly how a little impedance is added to the rotor to replace the shorts is unknown to me. Some of the other folks on this board can surely describe that or other fixes in detail.
Stand by.
RE: AC WOUND ROTOR INDUCTION MOTOR STARTER
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RE: AC WOUND ROTOR INDUCTION MOTOR STARTER
Rext + Rr = |Xls + Xlr|
i.e. the total rotor resistance will match the impedance magnitude of the source (neglecting stator resistance and magnetizing reactance which are small contributors) to give maximum power transfer and maximum torque.
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RE: AC WOUND ROTOR INDUCTION MOTOR STARTER
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Eng-tips forums: The best place on the web for engineering discussions.
RE: AC WOUND ROTOR INDUCTION MOTOR STARTER
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RE: AC WOUND ROTOR INDUCTION MOTOR STARTER
Changing the resistance in the rotor circuit of an induction motor alters the torque that the motor will develop for a given start voltage and a given slip.
If you have zero rotor resistance, the motor will not produce torque under high slip conditions, and the start current will e limited by the reactance of the rotor circuit.
The effect of shorting out the rotor on a wound rotor motor, is that under high slip conditions, the torque will be low and the current will be high.
If you then apply a reduced voltage starter to the stator, the current will be reduced and the torque will reduce by the square of the current reduction.
If we consider a standard cage motor which may have a LRC in the order of 650% and LRT in the order of 180%, then we can determine the start current to develop 40% torque by multiplying 650% by the square root of 40/180.(= 306%)
If we consider a wound rotor motor with the rotor shorted, we may have a LRC of 1200% and a LRT of 60%. The start current to develop 40% torque will be 979% which is worse than a standard cage motor started across the line.
In cases where the start torque requirements are very low, you will get away with starting with a shorted rotor, but the start current will be very high.
If the application requires a high start torque, then the only option is to use rotor resistors.
Best regards,
Mark.
Mark Empson
L M Photonics Ltd
RE: AC WOUND ROTOR INDUCTION MOTOR STARTER
There are 3 reasons to use a wound rotor motor.
-The load is very difficult to start and requires the extra torque and acceleration time the wound rotor motor allows.
-The motor was just installed as a means of "soft starting" the application.
-The load requires speed control.
The first 2 basically are the same but with different loads. In the first case, using a standard replacement induction motor could get you in trouble unless you use a VFD. In the second case, any standard induction motor could be used to replace the wound rotor motor. Since you are asking about shorting the slip rings then you must not want speed control and the third case is out.
In general, I find that adding a small amount of resistance to the rotor circuit will allow a very easy reduced voltage start - we use this in combination with SCR based soft-starters. Once the motor reaches full speed then the resistor is shorted out. This is where the trick lies, too much resistance and there is a large torque and current transient when the resistor is shorted. I've been involved with many applications such as ball mills and conveyers where this worked well.