|May19-09, 01:33 PM||#1|
Mechanization of a Tank Rolling Process
I am currently doing a mechanical engineering internship for a chemical manufacturing and engineering company. Many of the chemicals that we deal with are extremely volitile. The 700kg tanks that these chemicals are shipped in must be cleaned after use. Currently, the empty tanks are loaded by an overhead hoist onto 4 dolly wheels that are fixed to the ground, and are then filled with HCL, Caustic, and H20. Once filled, and allowed appropriate time to sit, the tank is then manually rolled (rotated about the horizontal axis) on the dollies until it is upside down, and can be drained.
Because the tanks weigh approx 3255kg when fully loaded, I have been charged with motorizing the rolling procedure in order to make the process easier on the workers.
My first step has been trying to figure out exactly what force would be required to roll the tank.
I have done the calculations several times, consulted my notes from last semester, looked on google, and am still getting a number that seems rather unrealistic. Any knowledgable insight would be GREATLY appreciated.
Here is what I have so far.
I want the tank to be able to make a full rotation in 30seconds. Therefore:
2*Pi radians = 1 revolution
Omega(angular velocity) is therefore = 2*Pi radians/ 30 seconds =>
Omega= Pi/15 radians/second
Using the relation: Omega = Omega initial + Alpha(angular acceleration)*Time
(Pi/15) = 0 + Alpha*30seconds
Alpha = Pi/450 radians/second2
Now, for the physical constants of the tank:
The radius of the tank = 104.15cm => 1.0415m
The mass of the tank and the liquid inside = 3255.92kg
As far as I can tell, all the above calculations are correct
Here is where I have made some assumptions that simplify the process, and my not be quite correct.
When you look at the tank from the round end, it looks like a circle. I approximated it as a such, with mass M found above.
*Also, because it rotates on the supporting wheels, about a contant centerpoint, I approximated this as essentially being fixed at the center.
Essentially it looks likeoOo
Doing this, my equillibrium equations are: Sum of moments about the centerpoint := I*Alpha= F*Radius
Where F is the force applied at the radius that is necissary to turn the cylinder.
I am using (1/2)(M*Radius2) for I and the values above for Alpha and Radius.
Doing this calculation yeilds F= 11.84newtons.
This is not valid. I have turned this tank. It takes considerably more effort than that.
I think the error may be in my oversimplification of the FBD.
However, if I treat both the wheels as reactions instead of the tank being fixed at the center, I end up with two vertical forces(which can be solved for) and two CO-LINEAR horizontal reactions that oppose my force F. These can't be solved for without utilizing deformation equations...which seems far too complex for a simple dynamics problem.
I guess what I am really looking for is: whichsimplifications can be made validly, and which ones cause problems in the real world.
Sorry for the super lengthy post, but I wanted to be thorough.
Much Thanks -Ian
P.S. Once I find the force required to turn the cylinder, I plan on running keyed shafts through the axles of the wheels, and using sprockets and chain to connect them to a motor. I just need to know how much force to apply, and hence what size motor, sprockets, chain, and gearbox to use.
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