Chain Drive Calculation for Belt Conveyors 1

[Vish S]

I'm trying to design chain & sprockets for a chain drive powering a belt conveyor.
Things known to me:

Drive Pulley Dia– 36"
Motor – 150 HP 1780 RPM
Gearbox – M1170DH2-A-11.090:1 Ratio Falk
Backstop - TA30 Morse
Snubber Pulley Dia - 18"
Conveyor Rise – 16 Degrees
Conveyor Length – 350 Ft.
Conveyor Speed – 437 FPM

Is this enough information for me to make chain & sprocket selection?

Thanks for any information you may be able to provide me

Vish

 

[mfgenggear]

OP
What the output torque and RPM required.

 

[JohnRBaker]

If your figures are correct, you'll need to seriously overdrive the sprocket and chain set-up. I mean, in order to achieve 437 FPM, the drive pulley has to be driven at something over 46 RPM, but the output from your gearbox is less than 20 RPM, that's a better than a 2:1 overdrive, something I would never attempt with a chain drive.

I would go with 20:1 gear box, or 25:1, if that was more readily available, which would give you a much more reasonable chain drive setup and one that was not being over-driven. If you had to, you go go as high as 30:1 but that would be about as far as I would go.

John R. Baker, P.E. (ret)
Irvine, CA
Siemens PLM:

The secret of life is not finding someone to live with
It's finding someone you can't live without

 

[3DDave]

Per

https://www.rexnord.com/contentitems/techlibrary/documents/m161-110_manual

pg 20.

M Standard metric input and output shafts (diameter/bore, length, key & keyway)
1 1000 Series
170 Drive Size
DH D = Drive One designation, H = Parallel, horizontal L.S. shaft
2- Number of reductions/Stages in gear drive
A- Model Code 1, Initial Model A
11.090:1 Ratio Code
Exact ratio expressed as (5) characters including decimal point Examples: 1.321:1, 14.95:1, 155.7:1, 1196.:1.

(no ":1" in the description.)

There seem to be a few characteristics missing, but this is enough for the calculations.

The output from this attached to the motor should be 160 RPM.

 

[JohnRBaker]

Sorry, I was looking at the ratio as being 90:1

John R. Baker, P.E. (ret)
Irvine, CA
Siemens PLM:

The secret of life is not finding someone to live with
It's finding someone you can't live without

 

[dvd]

It will be a little off by just using the drive pulley diameter, instead of a neutral axis through the belt, but depending on belt thickness, should be fairly close.

You will need to choose the appropriate chain size based on chain tables from a source like this

(source -

https://www.martinsprocket.com/docs...log/sprocket-engineering-data-horsepower.pdf)

Chain drives are pretty poor, you will need lubrication at that speed. A synchronous belt would be a lower maintenance choice. Any belt or chain drive will result in high bending moments on your driver and driven shafts, and be more subject to fatigue failure, or bearing failure. Choosing the correct gearbox ratio and directly coupling would be much better.

 

[MintJulep]

It's been a very long time since I selected drives for a conveyor.

But, don't you need to know the mass flow of the product being conveyed to figure out the power?

Or are you trusting that whoever decided on the 150 HP motor did that job correctly?

 

[3DDave]

I figured something like that. It's not an ideal encoding system. If it was the European separator it's worse - 11,090 in American.

It took far more digging to find that than expected. I'd nearly given up and ended up typing tiny fragments of the part number into Rexnord's site.

96 vertical feet in 48 seconds; 150hp -> 82500 ft-lbf/sec; so nominally 40,000 pounds of material on the belt? A bit over 100 pounds per foot. Which seems like gravel, maybe?

 

[mfgenggear]

150 hp at 1500 rpm = 525 torque approx.

What the torque and rpm be for a conveyor
And mass load it can carry

 

[Vish S]

All, thanks for active participation.
After digging through some archives and asking around, I have following information:

Chain – 22 ft. 200-2 Cottered Chain
Drive Sprocket – D200B17H x 5.1181 {130mm} (17 teeth)
Driven Sprocket – D200M60 (60 Teeth)

Also the GB ratio is slightly different from original : M1170DHC2-A-11.2 where 11.2:1 is Ratio as stated by 3DDave

The chain drive is existing and maintainence wise as most of you stated already is a pain. AN option to change to direct drive was given bu8t its expensive

Now, I am tasked with checking if this chain and sprocket spec correct for the conveyor system we have.
Material is Aggregates & approx. 1100 TPH.

Is there a way I can calculate chain length (total or in pitches) & Sprocket PCD, # of tooth etc. with the information I have (except for the existing chain & sprocket)?

 

[3DDave]

The chain length depends on the distance between the sprockets. Being that chain comes in inconvenient integers and is sensitive to tiny individual link length variations plus the inevitable effects of wear, you may need an adjuster or a tensioner. That is necessary.

Otherwise you can get close with using half the circumference of each sprocket added to twice the center distance. The half-circumference of a 60 tooth sprocket with a 200-2 appears to be 2.5 inch pitch, so 30 teeth * 2.5 inches per pitch is the length of the half circumference. 17 teeth is an odd number, that will change from 8 to 9 teeth engaged, but 8.5 teeth average will get close enough.

If the chain you have is 20 ft, then the sprockets are taking, roughly 8 feet of chain leaving, roughly, enough for 6 feet of center distance. This is not exact because the wrap on the big sprocket will be more and on the smaller one will be less, but not in 1:1 ratio. Check my math. I am unpaid and therefore only enthusiastic.

1100 TPH at 96 feet is 58666 ft-lbf/sec = 106 HP, which leaves margin for losses in the drive and belt system.

 

[Vish S]

3DDave

In your calculations what is 96 feet? Is this conveyor lift / height of Drive pulley from ground? How did you come up with 106 HP?

 

[3DDave]

You said it was a 16 degree slope that is 350 feet long. sin(theta) = opposite/hypotenuse.

 

[Vish S]

Correct, wanted to double check if that's it is

 

[dvd]

fyi on chain length -

from p. 17 - Martin Sprocket

 

[Vish S]

Is there a some way to calculate sprocket PCD & # of tooth?

 

[dvd]

PCD = chain pitch in inches / sin(180°/ no. of teeth in sprocket)

example

RC 200 chain = 2.5 inch pitch

60 tooth sprocket

PCD = 2.5 inch / sin (180/60) = 2.5"/ sin 3° = 47.768"

see page 31 of previously linked document for PCD of RC sprockets.

CONTACT A POWER TRANSMISSION COMPONENTS REPRESENTATIVE FOR SOME ASSISTANCE YOU SEEM TO BE STRUGGLING WITH THIS.

 

[3DDave]

I detect there may be a problem - Transmission components reps like to talk before the customer has bought the parts. This seems to be "This stuff has been bought, what do I do with it?" and it would take a really really understanding rep to help on someone else's sale.

 

[Vish S]

dvd
I already contacted a transmission component rep and he was of no help. He still owes me some info so lets see what he comes back with.

The issue here is, the conveyor has chain drive, Gear Box & Motor (all specs are mentioned in previous threads) which is a pain maintenence wise. SO I am tasked to see if the chain, sprocket & lubrication spec currently installed is correct or we need to change some thing.
So trying to do everything from scratch as far as Chain & sprocket design and came here for help on calc.

 

[3DDave]

If it has been moving material at the required rate, there are a near infinite number of combinations of reduction to get exactly that result. Greater reduction lowers the stress on the individual parts at the expense of increased bearing wear/fatigue. I expect there is a wide range of solutions where stress is traded off for cycles if there is some concern over fatigue.

What is the nature of the maintenance that is a problem? Perhaps someone has a belt drive to replace the chain drive to remove a notable wear item if there is abrasive dust that builds up. That is beyond my experience to advise on, but belt drives to handle 150 HP certainly exist.