Solution Manual Vol 7

# 2 in the flow net there are 37 flow channels and 11

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Unformatted text preview: is calculated by using Equation 2.16: Nf 3:7 1:3 106 m3 =s per m q kh 106 4:00 11 Nd al h0 log At1 h1 Figure Q2.2 Seepage 7 2.3 The flow net is drawn in Figure Q2.3, from which Nf 3.5 and Nd 9. The overall loss in total head is 3.00 m. Then, q kh Nf 3:5 5:8 105 m3 =s per m 5 105 3:00 9 Nd The pore water pressure is determined at the points of intersection of the equipotentials with the base of the structure. The total head (h) at each point is obtained from the flow net. The elevation head (z) at each point on the base of the structure is 2.50 m. The calculations are tabulated below and the distribution of pressure (u) is plotted to scale in the figure. Point 1 2 3 4 5 6 h (m) 2.33 2.00 1.67 1.33 1.00 0.67 h z (m) 4.83 4.50 4.17 3.83 3.50 3.17 u w (h z) (kN/m2) 47 44 41 37 34 31 e.g. for Point 1: 7 3:00 2:33 m 9 h1 z1 2:33 2:50 4:83 m h1 Figure Q2.3 8 Seepage u1 9:8 4:83 47 kN=m2 The uplift force on the base of the structure is equal to the area of the pressure diagram and is 316 kN per unit length. 2.4 The flow net is drawn in Figure Q2.4, from which Nf 10.0 and Nd 11. The overall loss in total head is 5.50 m. Then, q kh Nf 10 2:0 106 m3 =s per m 4:0 107 5:50 11 Nd Figure Q2.4 2.5 The flow net is drawn in Figure Q2.5, from which Nf 4.2 and Nd 9. The overall loss in total head is 5.00 m. Then, q kh Nf 4:2 4:7 106 m3 =s per m 2:0 106 5:00 9 Nd Seepage 9 Figure Q2.5 2.6 The scale transformation factor in the x direction is given by Equation 2.21, i.e. pffiffiffiffiffiffiffi pffiffiffiffiffi kz 1:8 xt x pffiffiffiffiffi x pffiffiffiffiffiffiffi 0:60x kx 5:0 Thus in the transformed section the horizontal dimension 33.00 m becomes (33.00 0.60), i.e. 19.80 m, and the slope 1:5 becomes 1:3. All dimensions in the vertical direction are unchanged. The transformed section is shown in Figure Q2.6 and the flow net is drawn as for the isotropic case. From the flow net, Nf 3.25 and Nd 12. The overall loss in total head is 14.00 m. The equivalent isotropic permeability applying to the transformed section is given by Equation 2.23, i.e. k0 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kx kz 5:0 1:8 107 3:0 107 m=s Thus the quantity of seepage is given by q k0 h Nf 3:25 1:1 106 m3 =s per m 3:0 107 14:00 12 Nd 10 Seepage Figure Q2.6 2.7 The scale transformation factor in the x direction is pffiffiffiffiffiffiffi pffiffiffiffiffi kz 2:7 xt x pffiffiffiffiffi x pffiffiffiffiffiffiffi 0:60x kx 7:5 Thus all dimensions in the x direction are multipled by 0.60. All dimensions in the z direction are unchanged. The transformed section is shown in Figure Q2.7. The equivalent isotropic permeability is pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k0 kx kz 7:5 2:7 106 4:5 106 m=s The focus of the basic parabola is at point A. The parabola passes through point G such that GC 0:3 HC 0:3 30 9:0 m Thus the coordinates of G are x 48:0 and z 20:0 Substituting these coordinates in Equation 2.34 48:0 x0 20:02 4x0 Seepage 11 Figure Q2.7 Hence, x0 2:0 m Using Equation 2.34, with x0 2.0 m, the c...
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## This note was uploaded on 09/21/2009 for the course CVEN 3718 taught by Professor Dobroslavznidarcic during the Spring '08 term at Colorado.

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