AC Motor Sizing 1

[swertel]

I love finding deficiencies in my education. Can someone verify that my reasoning is correct?

We currently have a mixing maching that I prefer to call a rock tumbler (just like the ones at Toys-R-Us) because we have marbles in the slury as an agitator. It has begun to break down more and more often so we have decided to replace it. Since it was a custom build and all the information on it has been lost, the replace has also turned into a redesign.

To size an AC motor, the input torque and speed is needed. I can calculate all that based on gear reductions assuming I know the initial conditions. --Well, during my education, the initial conditions were always a part of the question.-- The RPM I can determine by measuring the existing setup. The torque I am not so sure about.

The rock tumbler has a cylindrical container with the mix in it. The container is supported on two rollers: one idler, one driver. I think I can determine the torque by multiplying the weight of the cylinder by the center to center distance of the driver roller and container. Is this correct or am I missing something? Friction?

Thanks in advance,
--Scott

 

[ecampos]

INERTIA!
Figure out how much Mass Moment of Inertia the system that is rotating has about its rotational axis.
Figure out the gear ratio from the drum to the motor.
Figure out how fast you want this system to rotate at maximum load (weight) and how fast you want to get it to steady state RPM.
Then T=I*Alpha (Alpha=angular acceleration)
This is the torque you need, but you still need to calculate your "Alpha"


Edson Campos
edsoncampos@earthlink.net

 

[swertel]

Thanks Edson,

Finding the inertia of the system is going to be the fun part!

Could you also check something else for me? I think I'm missing or adding a slug.

(1)Mass Moment of Inertia units are lb*in^2
(2)Alpha units are rad/sec^2
(3)Multiplying I get (lb*in^2*rad)/sec^2
(4)A slug is (lb*sec^2)/ft (converting without carrying coefficients is (lb*sec^2)/in)
(5)Multiplying (3)*(4)= lb^2*in.

I obviously would like torque to be in ft*lb or in*lb. Why do I get lb^2?

--Scott

 

[ecampos]

Mass moment of inertia has units of "ft*lbf*sec^2"
Multiplying should cancel the Sec^2 and you will get Ft*Lbf torque.
Why are you multiplying the Torque by Mass?
The mass part is taken care of in the "MASS" moment of Inertia part.
Let me know what happens.

Edson Campos
edsoncampos@earthlink.net

 

[swertel]

It wasn't a slug that got in the way, it was the graviational constant.

Here was my original thinking:
Since in my design, a cylinder is rotating; I can determine the Mass Moment of Inertia ("MMI" so as to not be confused with the pronoun "I&quot by computing MMI=(mr^2)/2. Units work out to be lb*ft^2 ~~ in^2 in my previous post. And, the previous post follows the rest of my reasoning. What the equation didn't tell me is that m is in units of lbm, not slugs or lbf.

But, as I now determined, m should be replaced with W/g. That yields MMI=lbf*ft*sec^2 as you mentioned above, where W is in lbf and g is in ft/sec^2. The units then clean up nicely to a Torque in units of ft*lbf.

Thank you very much for your help.

--Scott