Calculation of Torque for Design of Drive Gear

Is this for a liquid or a solid?

It is for sludge. Solids shall be 2% by volume.

Viscosity? Temperature?

What is this agitator intended to do?

Is the tank baffled?

What shape is the sweeper?

What is the diameter of the sweeper?

Viscosity shall be same as water. It is sludge which needs to be swept from the tank floor. Temperature shall be 35 degree centigrade max.

Agitator shall be required to sweep the settled sludge from base to a central pit. The floor is slightly tilted, say 1 in 12 slope towards centre. Paddles located on the sweeping arm force the sludge to centre. There shall be about 22 paddles, each of 1m length x 0.1m ht. tilted at about 30 degrees tilted to centre upon the sweeping arm.

The tank is not baffled. It is a circular tank with a slight conical bottom with 2 arms sweeping the floor from a centre shaft taking drive from the gear. Inlet motor shall be 1440 rpm which will be reduced to 57600 times as under:

Inlet rpm 1440, reduced to 2:1 by chain sprocket, reduced to 8:1 by gearbox and again to 30:1 by gearbox. Final drive gearbox has a reduction ratio of 120:1 (pinion versus drive gear) to which torque shall be transferred for sweeping the arm. The earlier ratios are set. What i need to know is how much torque is needed so i can work backwards towards the motor.

Without doing field trials or comparing the installation with similar plant, 1/4hp (0.2kW) might be enough for such a duty.

What similar installations are available to make comparisons?

You are right. What i need is the torque needed to design the gear. The motor i have chosen is 1 hp or 0.75 kw standard motor of 1440 rpm. The final drive gear of 120 ratio is the one which needs to be designed since it is the final gearbox giving drive to shaft which holds the arms.

In order to design the gear, i need to know the torque. I think it is about 38.8 kg-cm. Please inform.

Shaft power = shaft torque multiplied by angular velocity.

Your gear ratios give a rake speed 1.5, not 1 rev/h.

As PWSlack says, you need to base torque on existing installtions. Thickener torques vary widely between water treatment and secondary sewage sludge at the low end to metallurgical slurries at the top end. Also torque varies hugely depending on the solids particle size distribution.

Having said that, your description sounds like a light duty. But your torque figure 38.8 kg.cm is completely wrong, this is close to the motor full-load torque at 0.75kW, 1440 rpm. The torque increase as the speed reduces, and each step must be rated accordingly. Assuming typical overall drive train efficiency 50%, and the motor is at rated power, torque at the rakes is > 2 million kg.cm.

It's some years since I worked with thickeners, but that does sound high for the description given. Maybe PWS's 0.2 kW might be nearer the mark, but it's really up to you. Best to base it on a similar application!

Codey

You are right, of course. The calculation or 38.8 kg-cm is wrong. I need the correct one. The sludge volume being scrapped shall be about 9 cum or 9 tons. What is the torque needed if the arm is 27m long?

Which textbook did this question come from?

When in doubt, make it stout. Design the gear to suit the motor.

Does two arms on a central shaft= 4.4 meters² of paddle area?

I agree that the first question looks like it was coppied from a test.

I agree this reads like a question from the P.E. exam.

But given your clarifications, I would use the following formula found in Machinery's Handbook:

T = 63025 x HP/N

where T = transmitted torque in lbf-in,

N = speed in rpm and

HP = horsepower

Given a 1 hp motor rotating at 1 rev/hr = 1/60 rpm we get:

T = 63025 x 1/ (1/60)

= 3,781,500 lbf-in

= 4,356,759 kg-cm.

Granted this is a worst-case, very conservative approach, but if you design for this, your application will be very safe and highly unlikely to fail.

Finally, the right answer. I think the answer is the closest to the calculated torque. Thanks friend.

How do you know it's the right answer? It's right for your arbitrarily chosen 1 hp motor, assuming all the power gets to the rake mech. It might bear no relationship to the torque needed to move the sludge.

That wasn't the question, the question was what load should the gears be designed for.

Re-read my answer.....I did NOT say this was the torque needed to move the load, but rather than maximum torque that could be transmitted with a 1 HP motor and a 1 RPM load rotation. That means the number is the maximum torque that can be applied the shaft of the rotor. From that all the loads and torques applied to the gear can be calculated and the gears designed accordingly.

It does not take into account shock loading, locked rotor torque, motor and gear efficiency, bearing and support considerations and more. But it's a quick answer that can be refined if necessary.

I didn't say you did say this was the torque needed to move the load. I was querying TRV's assumption that 1 hp is the appropriate motor power. With that power at the rake I agree with your torque.

But usual procedure for designing a thickener is first to decide maximum torque needed for the duty. Design the rake mechanically for that. Max torque with rake speed gives power at the rake. Allow an efficiency to get motor power. Used to use 50%, that includes some safety, but with several reduction stages in series it won't be high.

The system is protected by torque measurement somewhere along the drive train, set to trip at 95 - 100% rake torque. Usual also to have an alarm at 85 - 90%.

With your torque 4,356,759 kg-cm, if it actually ran at that the motor would be overloaded by about 2 x. Would need to set the trip at ~ 50%, or use a bigger motor.

Also TRV needs to check the drive train ratios. As I said in #8, comes to 1.5 rev/h. At that speed, he'll either get less torque or need more power.

Codey

You are right, of course. What i need to know is the amount of torque needed to push/scrape sludge of volume 17 cum at a density of 1.02 through a scrapper lenght of 28m and ht. 0.1m to a circular distance of 180 metres. Once i know the necessary torque, i can reverse calculate.

Impossible to give you the torque from the information provided. As several posters have said, it's best to base it on a similar installation.

From your data density 1.02 (kg/litre?) and concn by volume 2%, I back-calculate particle density 2 kg/litre. Is this about right? What is the material and application? You haven't told us yet, and that would help give a clue to the duty. Also the particle size distribution has a big effect on torque.

Codey

I have a problem with a motor 1440 RPM that is connected to a gear by means of a chain and tooth wheel.

"Inlet rpm 1440, reduced to 2:1 by chain sprocket, reduced to 8:1 by gearbox and again to 30:1 by gearbox."

A chain is used for slow turning shafts …

Wouldn't it be better to use a gear motor with output speed 1440 : (2 x 8) = 90 RPM ?

An interesting sidelight to this question is whether you really want to design the gear and shaft for the torque required under working conditions, instead of, for example, start-up? It is not unheard of to meet power outages from unexpected sources. Then there are the expected outages (maintenance, repair, etc). A recurrent theme in this thread is reference to actual working data as - at the very least - a guide to check the theoretical output. It has even been known to use the actual data instead of theoretical as the primary design input. (!)

The torque for which the system shall be designed shall be 100,000 lb.ft. The factor of safety shall be about 1.25. The volume of sludge scrapped shall be 9 cum per arm or 18 cum in all. According to this value, the structural members and gear shall be designed.

Kindly comment or confirm.