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motor sizing


Subhashish

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Hi

 

I need to run a turn table weighing some 1000 kg within a speed range of 2 to 10 rpm using 3 PHASE INDUCTION MOTOR ,gear reducers and a variable speed drive .

Motor specs RPM - 1485, 0.37 KW , NOMINAL TORQUE 2.5 Nm

Gear reducers - There are two gear reducers ,first gear box of ratio 1/40 connected at motor output and the next one 1/10 connected to the output of the first gearbox . The output of the 2 nd gear bos is connected to the shaft of the turn table. That should provide 1000Nm torque at shaft turn table shaft .

 

I am not sure if the rating of the motor is adequate .CAN SOMEONE TELL ME? If its not sufficient how to decide a safe rating .

 

Also the turn table being a high inertia load do I need to consider breaking resistor for the drive.

Any specific thing to consider to decide the ramp up and down time .

Belive it should be considered constant torque load . Any advise or suggestion please.

Regards

 

Subhashish

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Hi Subhashish, Power [kW] = Torque [Nm] x Speed [rpm] / 9549 so at 10 rpm and 1000 N/m you will require around 1.05 kW at the turntable shaft, you must then consider the losses in the gearboxes, these can be as high as 50% depending on the gear box design. Once you have the loss factors for the gear box you will be able to calculate required input power to the gear box.

 

The load may be high inertia but it is turning slowly and due to the losses in the gear boxes you most likely will not need a braking resistor. This will depend on how quickly you require the turntable to stop.

 

You also need to consider the ability of the motor and drive to deliver full load torque at the low speed end in conjunction with a VSD.

 

Ken

An expert is one who knows more and more about less and less until he knows absolutely everything about nothing
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