Power to turn a horizontal cylinder

[rory pringle]

Trying to work out geared electric motor required for turning a rotary drier:
18 m 1.1m diameter pipe weighing 4 tons lying horizontally on frictionless bearings.
Weight of material being turned is 500kg wood shavings.
Pipe has a chain driven ring gear outside diameter 1.2 and 83 tooth chain drive.
Rotational speed variable 2 cycles per minute to 6 cycles per minute
Drive sprocket 30 tooth.
Can someone help?

 

[mfgenggear]

Please make a rough sketch. With all units with correct data.
Goal or objectove
what is the torque required and rom from motor.

What is the requirements output of gear box.
Input torque and rpm into chain drive
What is yhe rpm of the pipe and torque required to rotate

 

[3DDave]

Consider the problem as one of continuously pumping 500 kg of material about 0.5 meters up at whatever rate. The chips won't get much above the centerline of the pipe before sliding.

Power = weight * height / time. The weight is 500kg * 9.81 m/sec^2 = whatever Newtons, the height is 0.5 meters, and the time is whatever a 1/4 turn of the pipe takes.

2 cycles per minute is 1 cycle in 30 seconds, so 1/4th of that is about 8 seconds.

Or you can just go with a 10-15 Hp motor and realize that if it is over-sized, there is a lot of inertia to overcome at startup and you don't want to pop a breaker getting the thing running from a standstill.

I'm too lazy to try to back-drive what the wall thickness of the cylinder is from the OD and the mass; without that I don't know what the inertia is and therefore what a reasonable torque to get it up to speed fast enough not to cause an electrical failure. I guess it could be like a thin shell at 0.55 m Radius, so mass * radius^2? But then I need to know how long the motor can stand being at too low an RPM for continuous operation.

I think the 300% range on speed is the larger problem but I don't do anything with VFD (variable frequency drive) and electric motor calcs so I cannot recommend one.

I do recall that a lot of the time these things are driven with a hydraulic motor and pump. A variable displacement pump can easily manage the speed range and won't be bothered by the initial starting torque as long as the motor can handle it. The electric motor to drive the pump can be much smaller; it only needs to be large enough for the hydraulic losses and to operate at the desired wood chip turning speed. It will take whatever time it takes to get up to speed; shortening the time means a bigger electric pump motor and a higher allowable pressure setting on the variable displacement pump.

 

[mfgenggear]

Op
Waiting for sketch, Dave pointed out good advice. I don't memorize formulas, I just google them.


+2
To calculate torque and RPM from weight capacity, you need to first convert the weight into a force, then use the formula for torque (τ = F * r) where "F" is the force (weight) and "r" is the radius of the lifting mechanism (lever arm), which will give you the torque required to lift the weight; RPM is not directly obtainable from weight capacity alone, and you'd need additional information like the motor's power or the desired lifting speed to calculate it.
Key points to remember:
Force calculation: Convert the weight capacity (in pounds or kilograms) to force by multiplying it by the acceleration due to gravity (9.8 m/s² or 32.2 ft/s²).
Torque formula: Torque (τ) = Force (F) * Radius (r).
RPM and weight: Weight capacity alone cannot directly give you RPM; you need to consider the motor's power or the desired lifting speed to calculate RPM.
Example calculation:
Problem:
A lifting mechanism has a weight capacity of 100 pounds and a lever arm of 2 feet. Calculate the required torque.
Steps:
Convert weight to force: Force = Weight * Gravity = 100 lbs * 32.2 ft/s² = 3220 lb-ft/s²
Calculate torque: Torque = Force * Radius = 3220 lb-ft/s² * 2 ft = 6440 lb-ft.
Important considerations:
Lever arm length: The length of the lever arm significantly affects the required torque.
Efficiency: The calculation assumes 100% efficiency, which is not always the case in real applications.
Motor selection: When designing a system, you need to choose a motor with sufficient torque and RPM to handle the desired weight capacity and lifting speed.
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[rory pringle]

W

Consider the problem as one of continuously pumping 500 kg of material about 0.5 meters up at whatever rate. The chips won't get much above the centerline of the pipe before sliding.

Power = weight * height / time. The weight is 500kg * 9.81 m/sec^2 = whatever Newtons, the height is 0.5 meters, and the time is whatever a 1/4 turn of the pipe takes.

2 cycles per minute is 1 cycle in 30 seconds, so 1/4th of that is about 8 seconds.

Or you can just go with a 10-15 Hp motor and realize that if it is over-sized, there is a lot of inertia to overcome at startup and you don't want to pop a breaker getting the thing running from a standstill.

I'm too lazy to try to back-drive what the wall thickness of the cylinder is from the OD and the mass; without that I don't know what the inertia is and therefore what a reasonable torque to get it up to speed fast enough not to cause an electrical failure. I guess it could be like a thin shell at 0.55 m Radius, so mass * radius^2? But then I need to know how long the motor can stand being at too low an RPM for continuous operation.

I think the 300% range on speed is the larger problem but I don't do anything with VFD (variable frequency drive) and electric motor calcs so I cannot recommend one.

I do recall that a lot of the time these things are driven with a hydraulic motor and pump. A variable displacement pump can easily manage the speed range and won't be bothered by the initial starting torque as long as the motor can handle it. The electric motor to drive the pump can be much smaller; it only needs to be large enough for the hydraulic losses and to operate at the desired wood chip turning speed. It will take whatever time it takes to get up to speed; shortening the time means a bigger electric pump motor and a higher allowable pressure setting on the variable displacement pump.

Wow Thanks guys. What a great help so far. I will get back on this shortly