Wye Start-Delta Run Motor Question

[nuckythompson]

I understand that you use a wye configuration to start the motor as it has reduced voltage and limits in-rush current. But I do not understand why you need to convert to the delta. Why couldn't you just leave it connected only as a wye? Thanks

 

[paulengr]

I understand that you use a wye configuration to start the motor as it has reduced voltage and limits in-rush current. But I do not understand why you need to convert to the delta. Why couldn't you just leave it connected only as a wye? Thanks

Torque is proportional to the square of the voltage: the motor sees 277 across each coil instead of 480.

If you don’t need it swap out for a motor 1/3rd of the size and just do straight ATL: Same LRC, FLA, and HP. Require to permanently delta and install smaller overloads.

If you just want simple remove all the contactors and the wye delta relay(s). Install a soft start with an integral bypass. Wire up motor as a 3 lead (permanently delta). You get variable starting current, no bump to delta run, and a lot of control options. Enjoy your extended motor life and smoother starting.

 

[don_resqcapt19]

I understand that you use a wye configuration to start the motor as it has reduced voltage and limits in-rush current. But I do not understand why you need to convert to the delta. Why couldn't you just leave it connected only as a wye? Thanks

When you use a 3 phase motor connected to a wye/delta starter, the windings see the line to line voltage divided by the square root of 3 when connected in wye, and they see the full line to line voltage when connected to delta. So assuming a motor rated for 480 volts when connected in delta, the windings only see 277 volts when connected to wye for starting.

 

[Jraef]

To put simply, when the voltage is reduced, so is the torque, in fact the peak torque is reduced by the SQUARE of the voltage reduction. So in Wye, the voltage seen by the motor coils is 58% of normal, so the peak torque is reduced to .58 x .58 or 33% of normal rated peak torque. Assuming that is still enough to accelerate the motor (otherwise Wye-Delta couldn’t be used), the problem becomes both the loss of that peak torque, which the motor uses to re-accelerate without stalling if there is a load change, but also that the full speed running torque is also reduced to 58% of normal.

Horsepower is a function of torque at a given speed and since speed is determined by frequency which isn’t changing, 58% torque = 58% HP. If the load could run with 58% HP, you were using too big of a motor!