Unformatted text preview: k 1 .(0.0504 – 5k 1 ), 0.252 – 25k 1 + 9k 1 = 30769.231 k 1 .(0.0504 – 5k 1 ) 153846.16k 1 21566.77k 1 + 0.252 = 0 k 1 = [1566.77 ± √ (1566.77 2 – 4 153846.16 0.252)] [2 153846.16] k 1 = 0.01002 mm/s or 1.63 104 mm/s The first of these gives k 2 = 104 mm/s for the silt; the second gives k 2 = 0.0165 mm/s. As the sand must have a greater permeability than the silt, the solution is k 1 (sand) = 105 m/s; k 2 (silt) = 107 m/s To estimate the flowrates at fluidization in upward flow, you will need (a) to derive main text Equation 3.33, and (b) to assume a unit weight for the soil. Main text Equation 3.33 is derived by considering a plug of soil on the verge of uplift (main text Figure 3.24). Neglecting side friction, uplift will just occur when the upward force due to...
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This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk
 Analytical Chemistry

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