Torque & Motor H.P Required for Drum Rotation of 3T Weight

[cosyindia]

We have designed a system that rotates a horizontal drum of 3 tons weight (drum wt 400 kgs + liquid load 2600 kgs). I am unable to find the torque required to rotate this drum or for that matter the H.P of the motor required. The details are as follows:

Drum empty wt: 400 kgs
Drum (water) loaded wt: 3000 kgs
Drum diameter: 1500mm
Drum length 2400 mm

The drum is horizontally placed on four rubber rollers of 150 mm diameter that are in turn rotated by a reduction gear drive. The Drum needs to rotate @ 0 to 15 RPM and the Motors RPM is 1400 so the reduction gear box has to be of ratio 1:93

 

[Mech_Engineer]

It takes more than weight to determine the torque you'll need. For a start you need to specify how quickly you want the cylinder to accelerate from 0-15 rpm. Also, you need to look at the drum's center of gravity and moment of inertia which is going to be tricky considering the fluid in the drum will nonlinearly affect it.

 

[cosyindia]

The drum needs to reach 15 RPM in say around 10 seconds (the gradual increase in RPM for the drum is required as it would jump due to sudden rotation of the rollers). And to the CG we can consider the drum filled with water so if a value of torque is arrived we would be in a position to know the Horse Power. But a close value and formula is sufficient for us to check.

 

[janik.mech]

Dear cosyindia:

This is a problem that I have encountered recently while designing a coater drum for the food industry. I am assuming the following:

-that the water remains quasi-stationary at the bottom of the drum and that it is not lifted by some sort of baffle to a higher point in the drum.
-if not I assume that the drum is completely filled with water and needs only be rotated.

If this is the case then the only thing you require to establish the size if the motor is the rolling friction between the rubber wheels and the drum and some losses that you can estimate through driving the shaft with a sprocket and chain.

Lets assume that the drum lies on the four rollers that are spaced 90degrees appart and 45degrees with the vertical axis.

Then the vertical force per wheel will be 7354.5N per wheel. The normal force is then the 10400 N per wheel.

According to: http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm
the rolling friction between a rubber roller and steel roller is 0.0077m.

The formula to calculate the rolling friction is:

F = f x W/R

where R is the radius of the wheel.

Thus the rolling friction force is:

F_rf = 106.7N

Thus the torque required to drive one wheel is:

T = 106.7 * (0.15/2)
= 8 N.m

And in total 4 wheels is 32N.m.

This is only valid for steady state rotation. To accelerate the drum in a specific time the mass moment of inertia is required and you will have to calculate that using, I recommend, a software program such as solidworks.

Regards
Janik Bessinger