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HW-10-08-08s-Chs16&amp;18

# HW-10-08-08s-Chs16&amp;18 - IE 215 Solutions for...

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IE 215 Solutions for Problems due Oct 8, 2008 (Chs 16 and 18) 16.1 A screen with 325 mesh count has wires with a diameter of 0.001377 in. Using Eq. (16.1), determine: (a) the maximum particle size that will pass through the wire mesh, and (b) the proportion of open space in the screen. Solution : (a) By Eq. (16.1), particle size PS = 1/ MC - t w = 1/325 - 0.001377 = 0.003077 - 0.001377 = 0.00170 in (b) There are 325 x 325 = 105,625 openings in one square inch of the mesh. By inference from part (a), each opening is 0.00170 inch on a side, thus each opening is (0.0017) 2 = 0.000002889 in 2 . The total open area in one square inch of mesh = 105,625(0.000002889 in 2 ) = 0.30523 in 2 . This is total open space. Therefore, the percent open space in one square inch of mesh = 30.523%. 16.5 A pile of iron powder weighs 2 lb. The particles are spherical in shape and all have the same diameter of 0.002 in. (a) Determine the total surface area of all the particles in the pile. (b) If the packing factor = 0.6, determine the volume taken by the pile. Note: the density of iron = 0.284 lb/in 3 . Solution : (a) For a spherical particle of D = 0.002 in, V = π D 3 /6 = π (0.002) 3 /6 = 0.00000000418 = 4.18 x 10 -9 in 3 /particle Weight per particle W = ρ V = 0.284(4.18 x 10 -9 in 3 ) = 1.19 x 10 -9

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HW-10-08-08s-Chs16&amp;18 - IE 215 Solutions for...

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