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...
View
Full Document
 Fall '11
 StuartChalk
 Analytical Chemistry, Erosion, main text, mm/s

Click to edit the document details