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Is The Size Of 3 Phase Im Dependent On Its Nominal Rpm?


bumeshrai

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I have a Inverter driven variable speed, 3600 nominal rpm 3 phase Induction Motor of around 250KW. It runs a wheel on track through gear train. I wanted to change the Induction motor to 3000 nominal rpm with the same wattage. I would change the gear ratio to get the same linear speed at the wheel. My designer says that the weight of Induction Motor would increase due to lowering of nominal speed.

 

Is he correct? The torque remain the same at the driven wheel, then why should the weight of the motor increase?

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Yes

 

You have made an incorrect assumption; " ...that the torque remains the same at the driven wheel."

 

To obtain the same linear speed at the wheel, the Gear Ratio would have to DECREASE ......

 

The motor would have to supply more torque into the gear box to make up for the loss of torque delivered to the wheel by the reduction of the gear ratio.

 

If you increase the torque rating of the motor, the motor will need more iron to support the higher torque value.

 

To put more iron into the motor, the size of the motor must increase.

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  • 2 months later...

I have a Inverter driven variable speed, 3600 nominal rpm 3 phase Induction Motor of around 250KW. It runs a wheel on track through gear train. I wanted to change the Induction motor to 3000 nominal rpm with the same wattage. I would change the gear ratio to get the same linear speed at the wheel. My designer says that the weight of Induction Motor would increase due to lowering of nominal speed.

 

Is he correct? The torque remain the same at the driven wheel, then why should the weight of the motor increase?

I know it's kind of late now, but I have been off of this forum for a while after changing jobs.

 

What JOhmega responded with is correct if you were changing speed via a different gear ratio as you had proposed.

 

What your designer compatriot said would be "inversely" correct if you were changing speed by changing the number of motor poles in the stator, however it goes the other way (hence "inversely correct"). If you wanted to maintain the same torque from a motor by changing pole count to slow it down, the motor would get physically SMALLER, not larger and of course your kW rating would be cut in half as well. But you cannot change the number of poles to affect such a small speed change, you can only change them in pairs. So at 3600RPM (assuming nominal), that is 2 poles. Your next option is 4 poles, which would be 1800RPM. There are no other values in between, so his point is essentially moot for this application.

 

But you said "inverter driven", so let's explore that option as well. If you use your inverter to change the speed, you will in fact retain the SAME torque at the lower speed. that is what an inverter drive does. In that effort then, it remains the SAME motor, so there is of course no change in the physical size of the motor. But there are a few caveats:

  1. The motor mechanical (shaft) power will be lower than original, because the mechanical power is a function of torque and speed. So even though your torque remains the same, you are reducing the speed by 17%, so you will reduce the mechanical power (kW) by the same amount and you will have a 208kW motor.
  2. The motor cooling capacity will also be reduced by a similar amount. This is for the most part offset by the fact that the motor will be drawing less current (presumably), but is worthy of consideration depending on the type of motor you have and the ambient conditions.
  3. I would highly suggest that the inverter drive be of the "Vector" type for the best performance on traction systems like you have described.

"He's not dead, he's just pinin' for the fjords!"
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