AC Motor Selection 1

[creighbm]

So I am working a design to rotate a large drum at 5 deg/min (very large inertia) sitting upright on a bearing. I am able to take care of the gear ratios and motor HP calcs but what I am stuck on is how to determine the required torque to move this drum. I know that T = I*alpha but I can input any range of alphas to get a range of torques. Alpha can be infinitely small in this equation which yields a low torque requirement but the motor will most certainly stall. Shouldn't alpha be dictated by the motor spec sheet then one could compare this to the stall torque? I missing something? Thanks in advance.

 

[electricpete]

Someone please clarify for me... do we have:

A: motor---belt---gearset---drum
or
B: motor---belt---drum---gearset

B doesn't make much sense to me.

At any rate I think the equation I2 = I1*(N1/N2)^2 is a correct way to convert inertia from one speed to another for either scenario.

The equation Ieffective = Ibearing/(50/d)^2 (where 50/d is diameter ratio associated with both ends of the belt) wouldn't make sense to me if we're talking scenario A and if Ibearing represents the drum... this expression doesn't include both speed transformation stages.

I am probably missing something. Start with scenario A or B please.
Thanks.

=====================================
(2B)+(2B)' ?

 

[zekeman]

"The equation Ieffective = Ibearing/(50/d)^2 (where 50/d is diameter ratio associated with both ends of the belt) wouldn't make sense to me if we're talking scenario A and if Ibearing represents the drum... this expression doesn't include both speed transformation stages."

Please read or more carefully reread the dialogue closely.

1)If you read my suggestion of using the dryerlike belt configuration for the first stage the "gear ratio" would be the large bearing dia/motor shaft dia, then you would see that it is correct, since all other reflected inertias would be negligible in comparison..

2)I2 = I1*(N1/N2)^2 is incorrect since it does not include the motor inertia, only one shaft inertia, probably the first.

 

[creighbm]

I think I am good now...thanks for all of your help. The reason I was so hung up on the angular acceleration was you need this to figure out the load on the gear teeth. Using the equation "t=V/a=V*I/(T-Tf)" solved this problem as I can compute how fast things really do get up to speed. Thanks again.

 

[DesignerGuy16]

Ps, a vfd and low rpm 6 pole motor may be in your future. An inverter duty 1200 rpm motor could be run way down below 20hz insted of 60hz. 19000:1 reducer would be nutty, but 5000:1 is much easier to come by. Also with that kind of ratio the output torque will be very high.

I'm not in a position to run any numbers for you, but that's my perspective as a machine designer. Just get your torque requirements and look into motors and vfd drives. If sized correctly heat, etc will not be an issue.

James Spisich
Design Engineer, CSWP