Is Upgrading to a Wider Conveyor Belt Worth the Investment?

[Volantis92]

Hey guys,

some insight to the following problem I have here would be helpful.
I need to see if a conveyor is worth upgraded (i.e. see what the output Tonne/per hour is) from a belt 745mm wide, idler pulley 800mm wide... to a 800mm belt and 850mm wide idler pulley.
Lets assume the length of the conveyor is 17m

I have a conveyor with 3 idlers on an angle of 45 deg, and assuming an angle of repose of 25 deg.
currently there is a 18KW motor on there, and let's assume the idler pulley barrel is 300mm. I want to upgrade to a 30KW motor.

How do I go about working out the max possible T p/hr?

i assume i need the speed of the motor, get velocity. Find the volume of the material over the conveyor length?

Any help is appreciated.

Cheers

 

[Baluncore]

The rate of material mass flow will be determined by height difference? and conveyor velocity.

Material falling onto the conveyor must be accelerated to the conveyor velocity. KE = ½∙m∙v²

The change in height will require (or release) potential energy proportional to; PE = m∙g∙h

 

[Volantis92]

Thanks for that mate,

So assuming I have incline angle of 15 deg. my equation would be:
T p/hr = 3.6 * (cross sectional area of trough) * (belt speed m/s) * (material density) * Cos(15)?

Only problem I have is getting an accurate calc for the cross sectional area of the trough.
How would I do that with idler trough angle of 45 deg
Belt 745 mm
angle of repose for material = 25 deg

cheers

 

[Baluncore]

You are confusing yourself by your fixation on the geometry of the conveyor and the density of the material.

So assuming I have incline angle of 15 deg. my equation would be:
T p/hr = 3.6 * (cross sectional area of trough) * (belt speed m/s) * (material density) * Cos(15)?

What is the coefficient 3.6 ?
Where is the acceleration g = 9.8 needed to work out the change in potential energy?

Product has a change of height, the potential energy must be provided. That height change is a fixed critical parameter.
Product must be accelerated to belt speed, that kinetic energy is not recovered. Belt speed is a critical variable.

 

[Baluncore]

Length = 17 m. Incline = 15 deg.
Height change, h = 17 * Sin( 15 ) = 4.4 m.

Material flow of 1 kg will require energy, E = KE + PE.
KE = 1/2 * 1kg * v²
PE = 1kg * 9.8 * h
E per kg = ( 1/2 * v² ) + ( 9.8 * h )

But belt speed, v, is unknown, we still need to know motor RPM.