Wye start, Delta run...

Luketrician

Senior Member
Was reading a closed thread that had this discussion and I would like to bring it back up again.


Some had mentioned that in a wye configuration the torque of the motor would be less due to the fact that the windings are seeing less voltage than in a Delta config ( ie soft start equip)

My understanding is that In a Wye config..the torque would be greater, since in this configuration the current would be greater due to the Wye config, then when motor is up to speed it is then changed to a delta since less power is now needed during running conditions, would this be correct?

Thanks
Luke
 

don_resqcapt19

Moderator
Staff member
The purpose of a wye/delta start/run system is to reduce the starting current. The current is less when you have the motor connected in wye.
 

rcwilson

Senior Member
Luke- You're thinking of a motor as a constant power device: V x I = constant. If volts go down; amps go up.
That is true for a running motor. But during starting the motor is not a constant power device. It follows ohms law (more or less). If voltage is lower, current is lower. Lower current = less torque.
 

Besoeker

Senior Member
Was reading a closed thread that had this discussion and I would like to bring it back up again.


Some had mentioned that in a wye configuration the torque of the motor would be less due to the fact that the windings are seeing less voltage than in a Delta config ( ie soft start equip)

My understanding is that In a Wye config..the torque would be greater, since in this configuration the current would be greater due to the Wye config, then when motor is up to speed it is then changed to a delta since less power is now needed during running conditions, would this be correct?

Thanks
Luke
For a cage induction motor, torque is approximately proportional to the square of the applied voltage.
If you connect the three windings in WYE they will each see 1/SQRT(3) of the delta voltage - about 277V if the line to line is 480V.
It will thus produce about (1/SQRT(3))^2 or 1/3 of the torque in delta.
 
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