Analytical Chem Soil HW Solutions 33

Analytical Chem Soil HW Solutions 33 - k 1(0.0504 – 5k 1...

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33 q = A.k.i k = q/Ai, where A = 112 2 mm 2 . (With A in mm 2 and the flowrate q in mm 3 /sec, the permeability k is calculated in mm/s) Hydraulic gradient, i 1 2 5 10 k A , mm/s (from test 1) 6.3 × 10 -3 6.3 10 -3 6.3 10 -3 - k B , mm/s (from test 2) - - 2.55 10 -4 2.63 10 -4 Table Q3.4: Processed permeability test data (The sample in test 1 has flow parallel to the laminations as the measured permeability is the greater) Taking k H = 6.3 10 -3 mm/s, k V = 2.6 10 -4 mm/s, d 1 = 5 mm for the sand and d 2 = 3 mm for the silt and substituting these values into main text Equations 3.21 and 3.22 with permeabilities k 1 and k 2 for the sand and silt respectively, k H = 6.3 10 -3 mm/s = {(k 1 5 mm + k 2 3 mm)} ÷ 8 mm k V = 2.6 10 -4 mm/s = 8 mm {(5 mm/k 1 ) + (3 mm/k 2 )} Rearranging the equation for k H and working with all permeabilities in mm/s, k 2 = (0.0504 – 5k 1 )/3 Substituting this into the equation for k V , {5 k 1 } = {9 (0.0504 – 5k 1 )} = {8 2.6 10 -4 } Multiplying both sides by
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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 2-1566.77k 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 10-4 mm/s The first of these gives k 2 = 10-4 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) = 10-5 m/s; k 2 (silt) = 10-7 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.

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