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Feeders Q&A

  •   The feed rate is 70 tn/hr, the bulk density of this product is 120 lb/cu ft, the particle size of the product is less than 3/4 in., and the maximum moisture content of this product is less than 2 %. My question is how small of a vibratory feeder, both in size and weight, would be calculated to handle this specific process? Also, can the drive be either pneumatic or electromagnetic to reduce overall size and total weight?

    Answered March 10th, 2010 by Expert: Bob Kolatac

    There are many factors that must be considered when selecting a feeder, such as whether the feeder is vibratory, screw, belt, rotary, tray, en masse, etc. Design factors, which should be considered but not be limited to, include: functions to be accomplished, properties of the material, conveying speeds and distances, and economics.
    If, after reviewing of all details such as those above, it appears that a vibratory pan (or tube) is the best selection, then some basic guidelines would apply. Remember all feeding and conveying applications require a well engineered and designed storage/discharge supply capability. Poor flow to the feeder will result in poor feeder performance. You did not indicate how you intend to supply the aggregate to the feeder. What sort of feed accuracy is required, and is control feedback information is required?
    A pan (or tubular) feeder/conveyor is simply a trough (or tube) with vibration (force) applied in a manner so as to convey or feed bulk materials. The vibration or force is applied to the feeder at a very specific angle, so as to lift and move the individual particle along the feeder’s surface.
    A particle sitting on the surface is moved upward and forward with the feeder. The feeder retracts on its downward stroke, with the particle continuing forward, landing further along the feeder’s surface. The more dense the particle, the more efficiently it is conveyed. The cycle continues. This ‘throw and catch’ motion conveys material along the feeder’s surface at velocities variable from 0 – 100 ft / min (or more!), depending upon the combination of drive frequency, amplitude, drive angle, and material properties. A major difference between a feeder and a conveyor is use of high amplitude and low frequency for a conveyor to high frequency and low amplitude for a feeder. For feeder calculations, a pan (or tube) velocity of 25 ft/min is generally used. (Remember this may vary based on material properties, etc.)
    For vibratory feeder and conveyors, many drive configurations can be used including a.c. unbalanced ‘vibrators’, electromagnetic, and others. The resultant displacement must be linear, at an angle to the trough (or tube). The feeder/conveyor will be supported/mounted on suitable isolator mountings. (Some conveyors have other means of support from ‘leaf springs’, air mounts/springs, cables or wire rope, etc., depending upon the type of unit selected.)
    Vibratory feeders generally operate in the range from 900 cpm with displacements (stroke) of ¼ in., to a high frequency of 7,200 cpm with displacements of 0.035 in., with the norm from 3,600 to 5,700 cpm.
    Vibratory conveyors seldom exceed 1000 cpm (more like 300 – 600 cpm with displacements from ¼  to 4 in.
    Pan Feeder capacity (tn/hr) = [W] [D] [R] [d] /33.3, where:

    W = width of pan, in ft
    D = material depth, in ft
    R = linear velocity, in ft/min
    d= bulk density, in lb/ cu ft.

    Your application, 70 tn/hr with a bulk density of 120 lb/cu ft., results in a volumetric ‘rate  of 1,167 cu ft/hr, or 19.44 cu ft/min.
    For sizing pan width and an appropriate nozzle we are going to use a ‘typical’ horizontal velocity of 25 ft/min, with the equation: 19.44 cfm/25fpm = 0.7776 sq ft. x 144 = 111.9 sq in., which represents bed width and height. (Here we are using inches since most pans are sized in inches). Therefore, pan width and product depth must be equal to some combination of 111.9 sq in.
    When considering nozzle or gate openings, chute diameters, etc. with granular, aggregate particulates, a rule of thumb is to use six times the largest particle size, in this case ¾ in. Using this ratio we have: (6)(0.75) = 4 ½ in. Dividing 111.9 by 4.5 results in a nozzle width of 24.8 in. We can be creative here to minimize pan configuration, etc., by shortening the width and increasing bed depth of the infeed nozzle. Therefore a 20 in. wide pan, with a bed depth of +6 in., using a convey velocity of 25 ft/min will safely obtain your desired convey rate.
    A quick check: WDRd/33.3 = (1.66)(0.5)(25)(120) /33.3 = 74.77 tn/hr!
    For this configuration electromechanical, inverter-controlled drives or electromagnetic drive versions are available. I do not know of any pneumatic drive which would be suitable for such an application.

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