Overhead Conveyor Load Torque 2

[reesechen15]

Hello,

Please help.

We have an overhead conveyor painting line, 145 meters in length. We hang metal piece there weighing 30Kgs each spaced at 1 meter apart, so a total load of 4,350Kgs.

Can you please help me compute the torque needed to rotate the sprocket of conveyor, 50 teeths, 315mm diameter?

If I halved the sprocket diameter to 25 teeths, to double the conveyor speed, would the torque needed be doubled too?

Please help. I'm checking if our existing motor and gearbox would suffice.

Thank you.

 

[3DDave]

It depends not only on the load but also on the rolling friction in bearings, sliding friction if there is any. Misalignment can add considerably to the torque required and so will any elevation change.

Find what it takes to move one piece, add about 10% for margin and multiply by the number of pieces.

Ideally to double the speed, doubling the torque would be required, but since some friction losses are speed dependent (such as moving lubricant in bearings) it will require a little more.

However, speed goes down as the sprocket size goes down if the motor RPM stays the same. To double the speed you need to double the sprocket diameter. Note that as the torque goes up so does the side-load on the motor or gearbox shaft that the sprocket is attached to while also increasing the torsion on that shaft. Depending on how much strength is excess in that shaft now it might work or it might fatigue and fracture.

Whoever sold the conveyor should have calculated the required motor size.

 

[mfgenggear]

reesechen15 (Electrical)(OP)8 May 21 05:19
Hello,

Please help.

We have an overhead conveyor painting line, 145 meters in length. We hang metal piece there weighing 30Kgs each spaced at 1 meter apart, so a total load of 4,350Kgs.

Can you please help me compute the torque needed to rotate the sprocket of conveyor, 50 teeths, 315mm diameter?

If I halved the sprocket diameter to 25 teeths, to double the conveyor speed, would the torque needed be doubled too?

Please help. I'm checking if our existing motor and gearbox would suffice.

Thank you.

what is the current Hp & rpm of the motor
what is the motor shaft have installed how many teeth, editor is direct connection with a coupling)
how many teeth on the input shaft of the gear box(omit if direct connection)
what is the gear ratio of the gear box (revolution of input vs the revolutions of the output shaft)
what is the out put shaft connected to
so many unknows
and that's not addressing 3DDave comments, if possible it would be better to consult the original manufacture
a diagram might be helpful revolutions input vs revolution on output connected to conveyor

 

[reesechen15]

Hi @3DDave

I just need an estimate so I can assume that the rolling frictions and other variables to be negligible. The allowable torque at the output of the gearbox is 49Kg-m. The ratio of the conveyor sprocket to the gearbox sprocket is 50 teeths/14 teeths, so I assume the allowable torque of our existing setup to be 49 x 50 / 14 = 175Kg-m.

I'm talking about the rear sprocket here. If the front sprocket (gearbox) size is the same, and the rear sprocket (conveyor) size is reduced to halved, the speed of the conveyor will be doubled. The allowable torque would then be reduced to 49 x 25 teeths / 14 teeths = 87.5Kg-m. I just want to know how to convert the 4,350Kg full load capacity of our conveyor to torque and if reducing the size of the conveyor sprocket will not create a problem.

(Please correct me if I am wrong as I don't have a mechanical background)

The conveyor was built 20 years ago and we can't contact anymore the contractor.

 

[reesechen15]

Hi @mfgenggear

There are so many unknowns,but I don't think the details about the motor and the gearbox are necessary.

I just want to know how to convert the 4,350Kg full load capacity of our conveyor to torque.

I have data that the allowable torque at the output of the gearbox is 49Kg-m. The ratio of the conveyor sprocket to the gearbox sprocket is 50 teeths/14 teeths, so I assume the allowable torque of our existing setup to be 49 x 50 / 14 = 175Kg-m. If I reduced the size of the conveyor sprocket from 50 teeths to 25 teeths, the allowable torque would then be reduced to 49 x 25 teeths / 14 teeths = 87.5Kg-m. I just want to know if reducing the size of the conveyor sprocket will not create a problem.

Anyway, if this data will help please feel free to refer below:

Motor: 2HP, 1720RPM, AC220V Delta
Output shaft: unknown

Gearbox
Ratio: 1/60
Input shaft: unknown (I assumed input speed to be 6RPM x 60 (gearbox ratio) = 360RPM
Output shaft: 14 teeths
Output RPM: 6RPM
Output Torque at 360RPM = 49Kg-m
Output OHL = 450Kg

Conveyor
Shaft: 50 teeths (to be reduced to 25 teeths)
RPM: 1.75RPM (to be doubled)

(Please correct me if I am wrong as I don't have a mechanical background)

The conveyor was built 20 years ago and we can't contact anymore the contractor.

 

[3DDave]

Find the amount of power the motor takes now to run the conveyor - current and voltage measurements. A motor service can do this. Then you can compare to the power the motor is intended to make - 2HP. Double the present power draw needs to be less than what the motor can produce.

 

[dvd]

Torque should be fairly constant if no other modifications besides changing speed, but power will double - power = torque x rpm / constant

 

[mfgenggear]

reesechen15
a little complex, but here is a link that can help it's online program calculate power

https://cyclo.shi.co.jp/apps/powerCalculationEn/chain_conveyer.htm

 

[mfgenggear]

here is a good white paper

https://www.mechatronicscenter.com/uploads/7/5/6/2/75629763/introduction_to_motor_sizing.pdf