Hello everyone,
I've been trying for a few days to size a roller chain pallet conveyor.
It involves transferring 6 pallets (1200mmx800mm) of 500KG each, on 2 double roller chains type B (Europe) (on stainless steel rail).
For the moment I have these results:
Friction coefficient: 0.2
and
CP = 9.81 x μc [(2.05 x Wc x L) + W]
Uc: Friction coefficient
Toilet: Mass of the chain per metre (in kg/m).
L: The length in metres
W: Total mass on the conveyor (in kg).
Cp: Chain pull in newton.
I then take a coef of 12 to size the chain (for the breaking load).
To begin with, does this method seem correct to you?
Then I'd like to calculate the inertia of the conveyor, and for the time being I'm drying a little!
If anyone specializes in conveyors....! Thank you very much!
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Hello
For the calculation of the effort this seems logical (unless the movement is not horizontal). Just the coeff 12 seems very high to me. For handling, it seems to me that a coeff 5 is sufficient.
For inertia, what exactly is the need? The inertia of a linear system can be summarized by the fundamental principle of dynamics: sum(forces (N)) = m(kg)*a(m/s2). Be careful, it's in vector.
But in general, we rather resonate with rotational inertia at the level of the engine for example. It then depends more or less on the engine torque and acceleration.
Do you have a spec of downtime or motion for example?
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Hello, thank you very much for your answer, unfortunately I don't have acceleration, I just know that I must be able to change a palette in 10 seconds.
After I guess I can go and see pallet conveyor catalogs to get an idea?
In fact I am trying to size the power of the motor in relation to the load.
Indeed, often in the catalogues, there are forms to help with the determination.
I imagine that the 6 paddles are moving simultaneously, positioned at a regular pace and that the 10 seconds are to go from one pallet to the next, that the movement is not continuous (so start, move, brake, stop).
The problem of acceleration is THE sizing criterion: 3 tons to move 800mm (in the best case) in 10 seconds, the engine will be beautiful baby.
Moreover, the sellers of geared motors have good tools, this may be the best way to go.
You may need to provide a flywheel and a coupling system, depending on the cadence.
Hello
We probably need to dig a little deeper into the question but here may be a first clue:
sum(F) = motor force - resistance of the charged chain (CP) = m*a, so a=F/m
integrating with respect to time we have velocity = v(t) = F*t/m + C. If v(0)=0 (start stopped) then C=0.
then position x(t) = F*t^2/2*m +C'. ditto if x(0) = 0 then C'=0.
Considering linear acceleration and braking, the maximum speed should be reached after 5 seconds. This allows you to go back to the maximum strength and speed.
Once you have the diameter of the drive ring, you can deduce a torque and a rotational speed. And therefore the power of the motor since P (in W) = C (in N.m) *w (in rad/s).
Obviously, you have to adjust all this to the real conditions of your conveyor belt (distance to be covered, time, mass, ...) and don't forget to take a small safety factor.
Hello
I worked for the company CERMEX specializing in packaging lines, but not for this type of machine. Try with Gérald (Lynkoa member), he also worked in the same kind of company, he can help you.
http://www.lynkoa.com/users/sepm
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