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Join Date: Sep 2008
Posts: 5

mechanical designer

09/17/2008 3:07 PM

When design steel belt conveyor, how to decide the hp of el motor? What formular should be used? What is the hp loss through reducer ( gear box)?

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Join Date: Aug 2007
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#1

Re: mechanical designer

09/17/2008 3:36 PM

Well after a while, its experience, but to set a reference point, Intralox has a rather good belt program even though it for plastic modular belting.

set the loading for metal also
I would use it. To set a reference point. But remember them, when you need a modular belt.

www.intralox.com

As far as losses through the gear box, your gearbox supplier should tell you the efficiencies,

phoenix911

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#2

Re: mechanical designer

09/19/2008 9:19 AM

Horsepower is a measure of work that a machine performs and loosely speaking is, Force (in lbs) multiplied by Speed (in feet per minute) divided by 33000. (Assuming this is a horizontal steel belt conveyor)

So, first you must add up all the weight of every component being driven, drive shaft(s), driven shaft, drums belt will ride on, sprockets, chain, couplings, mesh belt and of course the weight of the part(s) on the mesh belt.

Each belt will have a co-efficient of friction that the belt manufacturer can give you.

Total Weight x co-efficient of friction x belt speed (in feet/min.) = foot lb. of work

x 1.25 for inefficiencies of assemblies (e.g. sprocket mis-alignment, low on lube etc.)

x 1.25 for 'over-load' factor

now, divide this ft.lbs/min answer, by 33,000 to get horse power

It will never be a 'whole' number, so round it up to the next easiest hp rating. It is better to err on the safe side and account for unforeseen abuse by operators.

Should the application be other than horizontal, say inclined, then additional factoring may be needed.

Let us know if this answers you question, or, give us the needed numbers as indicated above and we can verify!

Good Luck!

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