Power Calculation for Rotating Cylinder

[edisonf]

I need help, please.

I have an application with a rotating cylinder in which material is tumbled for processing purposes. The following quantities are known although I have need of generic formulas in the future as these are variables between system applications.

Known Quantities:
Vessel mass (empty):146,000 lbs.
Inside diameter:11’-9”
Outside diameter:11’-10”
Weight of material:80,000 tons based on 32 lbs/ft3 bulk material density
Drive arrangement:1800 rpm electric motor coupled to gear reducer (55.2:1 gear reduction) driving steel wheel, 24” in diameter.
Number of drives4) each. The (4) steel wheels drive and cradle the vessel through (2) circumferentially mounted, 14’-1” diameter driven rings.
Time to accelerate:approx. 10 seconds
Full speed rpm: 6 rpm

My Methodology:

1. Calculate moment of inertia I of vessel (neglecting everything else) using:

m•(di2 + do2)/8 = 5,052,310 lbs-ft2

2. Calculate average angular ? acceleration using:

6 rpm • 2•?/(60 sec/min•10 sec)= 0.0628 rad/sec2

3. Calculate total torque required to accelerate from:

T = I•? = 317,285 lbs-ft

4. Calculate torque required from gear reducer output. If 317,285 lbs-ft is required at 7.04’ (radius of driven ring), 45,058 lbs. of force must be developed at the drive wheel. Since the radius of the drive wheel is 1’, the torque developed at the output of the gear reducer must be:

45,058 lbs. x 1’ lever arm/4 drives =11,264.5 lbs-ft per drive

5. Determine torque input to the gear reducer. Since the gear reduction ratio is 55.2:1:

11,264 lbs-ft/55.2 = 204 lbs-ft motor torque

6. Determine motor HP rating from the relationship:

T•rpm/5252 = 69.9 HP per drive

Conclusion:

Based on my calculations, each motor will need to develop approx. 204 lb-ft of torque to accelerate the vessel. I realize that the motor could be sized based on breakdown torque, not using the above simple HP=T•rpm/5252 relationship which yields approx. 69.9 HP to be developed by each motor. If breakdown torque is approximately 230% of full speed torque, (4) 30 HP motors would suffice accounting for a 1.15 service factor. I just need verification that my methodology is valid and that my figures are reasonable. Also, I realize the above does not account for the material to be processed in the vessel. I must also take this into account and would very much appreciate any help you can provide along these lines as well.

 

[JStephen]

For material in the cylinder, figure it's in there sloped at angle of repose, and you can calculate the torque required to rotate- I'd add a fudge factor to allow for uncertainty in the slope.

I assume that is 80,000 lbs, not 80,000 tons up there.

 

[zekeman]

First of all, you get all the moments of inertia of the drive wheels,output gear of gearbox and motor inertia, each reflected to the motor input shaft-i.e for each inerta Ii, the reflected inertia is:
Ii,n^2 where n is the transmission ratio of each inertia in question. Then the sum of the reflected inertias is
SUM of Ii/ni^2. If you call this I, then the dynamics, ignoring friction, are
T=Idw/dt from which
dt=I/T*dw
where w= angular velocity and T= torque delivered. From the Torque-speed curve of the motor the torque is represented as a function of omega or
T=T(w)
and the equation is
dt=I/T(w)*dw
If you know T(w), then you integrate (area under the I/T(w) curve) both sides to get
t=Integral of I/T(w)dw , integrated from 0 to wf or the rated speed of the motor (1750?). This will yield the time,t it takes to get to the final speed. If the time is significantly more (or less) than the 10 seconds required, you go to the next size motor and redo the integral allowing for a safety factor (t<10 sec).

As far as the additional material, I would forget the angle of repose since the material is being constantly rotated. Just let the volume fill the bottom of the vessel and get its moment of inertia to add to the vessel's. This would be reasonably conservative.

 

[tygerdawg]

Go to seweurodrive.com , dodge.com, nord.com, Eaton Power Transmission, or other gearmotor company websites and download their free catalogs and engineering manuals. This will give you basic tutorials on gearmotor sizing methods. The SEW stuff is especially rigorous if you can figure out how to navigate through to the German website's "dokumentation" area and get the English-language engineering manuals. They are metric, but the math's the same.

In the gearmotor books you will learn the concept of "Service Factor" which is basically oversizing the gearmotor with a fudge factor based upon the type of usage the gearmotor will see. The Dodge books are especially fat in this respect. Really, for what you get in a gearmotor, torque is cheap. So be prepared to purchase plenty of it. Your motor will create only the torque necessary to do the job asked of it anyway. Oversizing will give you more life to your application.

Account for all possible loadings and torque-eaters, like the force on the steel wheels required to overcome the microscopic indentation caused by loading. See the Machinery's Handbook for more formulas.

You must also upsize the motor for the number of starts per hour. Starting a motor is essentially a short circuit until the magnetic field is developed, so things get hot momentarily. If you start frequently, then you need more thermal mass so things don't melt. SEW sizing methods are very persnikety about this.

TygerDawg