Plant InSites: Come Up With a Solid Estimate

Use geometry to figure out the amount of material in storage or equipment.

By Andrew Sloley, Contributing Editor

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Large-scale solids-handling operations often rely on storing material in piles. The pile layout and solid's properties set the volume storable in a given area. Piles most often approximate one of four common shapes: a cone: a frustum of a cone; a wedge; or a frustum of a wedge (Figure 1) — a frustum essentially is the underlying shape with its top cut off.

Angle of repose, Θ, which is the stable angle between the surface of the pile and the horizontal, is the major bulk material property affecting volume. Angle of repose varies with base material, particle size, particle size distribution (or fines content), particle shape and other factors such as moisture content. Table 1 (at the bottom of this page) lists reasonable values of Θ for some common materials; many references also provide such information. Values for specific cases can vary considerably. So, always try to get data from industrial applications similar to yours; in critical cases, you may need to test your actual materials.

For a cone, the volume is:
v = πhr2/3        (1)

where h is the height and r is the radius.
For a frustum of a cone, the volume is:

v = πh(r12 + r22 + r1r2)/3           (2)
where r2 = r1h/tan Θ.

For a wedge, the volume is:

v = hb(2a1 + a2)/6 = hb[3a1 – 2h/tan Θ]/6   (3)
where a2 = a1 – 2h/tan Θ1.

For a frustum of a wedge, the volume is:

v = h[a1b1 + (a1 + a2)(b1 + b2) + a2b2]/6      (4)
where b2 = b1 – 2h/tan Θ2.

You can use these equations, which apply both to unconstrained piles and bulk solids flowing in constrained spaces such as hoppers, with those for other standard shapes to determine a variety of volumes.

For example, let's find the volume of grain remaining in a 10-ft.-dia. 50-ft.-high feed hopper when it's nominally half full (Figure 2). We break the hopper into three zones: a frustum of a cone on the bottom, a cylinder in the middle, and a cylinder minus a cone on the top.

The first section has 1 ft. of height cut from the cone. We determine r2 as 5 - 4/tan 45° or 1 ft. This gives a volume of the bottom section of π4[52 + 12 + (5 x 1)]/3 or 130 ft3.

The second section has a cylinder height of 25 - 5 tan 28° or 22.3 ft. This gives a cylinder volume of 1,753 ft3.

The third section has the volume of a cylinder 25 - 22.3 or 2.7 ft. tall minus the frustum height. Volume is π52 × 2.7 - π52 × 2.7/3 or 141 ft3.

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