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Thread: round belting belt conveyor

  1. #1
    luisss749 Guest

    round belting belt conveyor

    Hello

    I'm designing a belt conveyor/classificator for fruits/vegetables, you can see a video of this kind of conveyors in: https://www.youtube.com/watch?v=hAKtQbKkijM
    as you can see the conveying surface is provided by several round section belts, i already made a power calculation for: empty belt conveyor, move the charge horizontally and for accelerate the belt conveyor, now i have some questions i hope you guys can help me with:

    1. I'm using the same design approach that for trough belt conveyors, looking the formulas i can see that the calculation for the power for empty belt and for move the charge horizontally is the same: a) find the weight of the components (or conveyed load) W = w1+w2+wn, b) multiply this weight by the idler friction coeff. Ff= W*mu, c) multiply this weight by the lenght or the conveyor this give us the "work", Work= F*d, d)with the belt linear speed and the lenght of the belt we can obtain the time to empty the belt t=d/v, e) divide the work obtained in by the time to empty the belt, this give us the power P=Work/t. is this admissible for this kind of conveyors?
    2. For the power to accelerate of the belt first i've obtained the inertia moments of all the rotating components (i did that wit the help of cad software) to obtain the moment of inertia of the belting and the load (Wbl) first i've obtained the sum of both weights, i use the entire weight, i didn't multiply it by the idler friction coeff. then i use the formula: MomentInertia= Wbl(pi*D/2*pi) i know the sheave diameter based in the minimun required for a given belting diameter, then i make the asumption of get a starting time of 3 sec, as i know the sheaves diameter i can get the rotational speed of the sheaves, with all this datas i apply the formula Torque = (sum of moment of inertia*rpms)/(308*t) is this valid?
    3. for the power calculation empty belt/move charge horizontally, i'm using the friction coeff. of 0.033, the same value as for trough belt conveyors idlers but im using a 2 bolt flanged ball bearing with a diameter = 20mm (AST bearings). i've read from a textbook that ball bearings have a friction coefficient of 0.001 to 0.005, and also i found a document from ntn bearings (NTN 200tech) here there a table that gives values from 0.001 to 0.0015, now the value of 0.033 seem to high to me also the operating conditions never gonna be equal that for a trough belt conveyor, what you recommend me?
    4. For the revision of the tensions i have this formula: Te=33000HP/fpm, then i have the data of friction coeff. belt vs aluminium = 0.75, i also know the wrap angle = 90º, with this data i can get the value of T1= Te/ 1-(1/e*mu*k), this tension T1 is the value that must be below of the admisible tension of the belt, im right?

  2. #2

    Smile Unfortunately

    If there's a hard way to do something somebody will find it.

    The machine shown in the video is not a belt conveyor and most of the design information quoted is irrelevant.
    The bands distort laterally according to the dimensions of the burden & individual items of the burden rotate at will during travel.

    Your application is tailor made for DEA (discrete element analysis) and some members of the forum might be inclined to model the behaviour for a reasonable fee.
    Concerns about power consumption are very secondary indeed. All you really need is a light duty gearmotor with a variable speed drive.

    I, first of all, thought the video intended to show how to gently rip the stalks off cherries but when that process failed I moved on to the next video about the vacuum conveying of apricots. The fact that the fruit was already contained within the groundsheet seemed to have escaped the designers who went to great pains to then remove the fruit once more.

    I suggest that you turn over a new leaf. Ho Ho Ho.
    John Gateley
    johngateley@hotmail.com
    www.the-credible-bulk.com

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