Calculating Torque Needed for Compost Vessel

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SUMMARY

The discussion focuses on calculating the torque required to spin a 21 ft. long, 4 ft. diameter compost vessel weighing between 2000-3000 lbs, which will hold up to 2000 lbs of compost. The vessel is powered by a hydraulic pump driven by a 4-5 HP gas motor, coupled with a 60:1 gear reduction box. Key formulas provided include torque calculations based on total weight and radius, as well as horsepower equations for rotating objects. Efficiency concerns regarding the hydraulic system versus direct motor drive are also highlighted.

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compostguy
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I am designing an in-vessel composting system - a 21 ft. long 4 ft. diameter steel tube that is mounted and spun on wheels by a hydraulic pump/turbine assembly, coupled to a 60:1 gear reduction box - the hydraulic pump will likely be powered by a small 4-5 HP gas powered motor.

My question relates to calculating the necessary torque to spin this vessel, and how to determine how much speed is altered. The vessel weighs probably 2000-3000 lbs. and will hold probably up to 2000 lbs. of compost.
How can I calculate the necessary torque to spin this? I suspect it has something to do with the circumfrence, the radius from the axis of rotation, etc., but don't really know how it all fits together.

Thanks!
-Colin
 
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[tex]W=\vec \tau \dot \vec d[/tex]

Really, the torque is only necessary to get the vessel spinning, and to counteract friction. With really good bearings - for example an air bearing - and a well balanced solid cylinder there is essentially no torque needed. On the other hand, if the compost vessel is set in a rubber cradle, you'll have a hard time turning it.

The maximal torque is typically necessary when all of the compost is on the side of the cylinder that is going up. (This is really easier to deal with using drawings.)

It seems quite odd to me that you're taking the power from the engine, turning it into hydraulic pressure, feeding it into a turbine, and then taking power from the turbine. It will be more efficient to run the vessel directly off of the motor.

A first estimate for the necessary torque woud be total weight: 5000 lbs multiplied by the radius of the vessel. This overestimates the torque needed to turn the vessel, but ignores turning friction.
 
Horsepower for rotating objects:

HP = (TxN)/(5250)

where

HP = Horsepower
T = Torque (lb-ft)
N = Motor Base Speed (RPM)
----------------------------------
Torque for rotating objects:

T = (HPx5250)/N

where

T = Torque (lb-ft)
HP = Horsepower
N = Motor Base Speed (RPM)
----------------------------------
You posted this in more than one thread... hope this helps.. and correct me if this isn't what you are looking for.
 

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