Need Help on Chain Pull Calculations for Conveyor 1

[BreadTruck]

I'm not conveyor-wise and I'm sure most of you will get a kick out of this. I'm sure some of you have encountered this problem and could enlighten me on this.

The conveyor system is called a 2-strand roller flight floor conveyor. The conveyor track that supports the chain is fabricated from a structural channel / wear bar combination. The composition of the conveyor chain is such that each link of chain is composed of three (3) rollers -- (2) rolls that make contact with the conveyor track wear bar [at the bottom] and that have a coefficient of friction of 0.12 and (1) raised roll that protrudes above the link to make contact with the product being conveyed that has a coefficient of friction of 0.03. The raised roll is included in the design to allow accumulation of product on the length of the conveyor.

Chain pull force (F) is calculated as follows:

F = Wf

Where W = weight of the product being conveyed
f = coefficient of friction

Let's assume for simplicity that there is only one object [or product] being conveyed. My question is how would the distribution / interactions in the two different coefficient of frictions be determined in the above equation in order to finally determine the Chain Pull Force OR is there another equation that someone could offer for my review and comment.

Any help would be greatly appreciated.

Thank you in advance.

BreadTruck

 

[dulmant]

Hello BreadTruck,

First of all, it can be argued that F= Wxf is a too simple way to calculate the pull force of the chain. You must consider the weight of the chain itself, as well as its tensioning preload and the efficiency of the chain mounting system. Then you have to take into account the start-up acceleration of the system. Often these elements are more power-consuming than moving the load itself.

Simplifyed approach:

The force required to move a loaded roller on a given surface, without sliding, can be calculated as follows:

F= (Mr/R)+Fi, where
Mr= rolling moment = Lxs, where
L= the load on the roller,
s= the rolling friction coefficient, depending on the materials of the roller and of the track,
R= the outer radius of the roller,
Fi= the inertia of the load = (L/g)xa, where
g= 9,8m/s2,
a= the acceleration the roller will be pulled with.

In order to use as less power as possible, the rollers should not slide on the track but roll, therefore:
(s/R)<f, where
f= the sliding friction coefficient between the roller and the track.
On the limit (s/R)= f can be considered.

Therefore:

F= ((Lxs)/R)+((L/g)xa)= Lx((s/R)+(a/g))= Lx(f+(a/g))

Now, L comes from the weight of the load plus the weight of the chain system, divided by the number of supporting rollers.

I am not familiar with this type of chain but if I understand this right, the 3rd roller is in contact both with the object and the track, so the object will advance twice as fast as the chain with respect to, let's say, the track.

Before I take this further, maybe unnecessarily, please let me know if I understood right.

Hope this helped, regards,

dulmant