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reactors for slipring motor starting

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tony11

Electrical
Apr 4, 2003
2
Can anyone recommend the best way to calculate a starting reactor? Can it be as simple as X = Rotor volts/root 3.Rotor amps for full load torque(assuming negligable resistance)
 
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Hello Tony
If you are looking to build a secondary starter, you will need to add resistance to the rotor circuit in order to develop torque. If you add pure inductance only, you wil reduce the start current, but you will also reduce the start torque. i.e. this will behave in a similar fashion to a standard reduced voltage starter with a slip ring motor and a shorted rotor.
Under shorted rotor conditions, the slip ring motor exhibits a very high Locked Rotor Current and a low locked rotor torque. Reducing the voltage reduces the start current, but also reduces the start torque by the square of the current reduction.

Best regards,

Mark Empson
 
Suggestion: The wound rotor induction motors normally do not use starting reactors for starting on the rotor side. They only use resistors to start the motor and in some instances to control motor speed. Where have you found that application?
 
The application is a compressor with an old resistor starter where the resistors burnt out. They tried shorting out the slipprings and starting via a soft starter. The motor stalled and they blew up the soft starter! A rotor reactor looks an elegant solution as the impedence falls with the rotor frequency. However it sounds like you need some resistance to move the pull-out torque point towards zero speed.
 
Hello tony
You have it in a nutshell. You need rotor resistancde to develop torque. You would probably find that with the rings shorted, you would have a locked rotor current of 1400% or thereabouts, and a locked rotor torque that could be less than 50%. That will not start much at all. Primary side starting will reduce the voltage and current, and the torque by the square of the reduction. Secondary side starting with reactors will have a similar effect. Resistors in the rotor will shift the torque curve and will enable the machine to start.
You could slect resistors equal to the final stage, or with a maximum torque at about 90% speed, and then use a primary side starter. At this point, the motor would behave similar to a standard cage motor. Once up to speed, short out the resistors.
The start current will be higher than could be obtained with a properly designed secondary resistance starter.
Best regards,

Mark Empson
 
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