Question

Given the soil cylinder and the test setup shown in Figure 2 10 cm 5 cm 10 cm SOIL 10 cm Datum Filter screen Figure 2

Answer 1

Head loss Across soln = T_0 -T_B = 35 - 25 = 10 cm therefore hydrotropic gradient = i = 10/L = 10/20 = 0.5 therefore total head at c = t_d -i times 10 t_c 35 - 0.5 times 10 t_c = 30 cm pressure head at c = p_c = t_c = d_c p_c = 30 - 10 p_c = 20 cm Here, vol = 42.5 cm^3 Time t = 5 min = 300 seconds i = 0.5 D = 6cm Therefore A = pi/4 times D^2 = pi/4 times 6^2 = 28.27 cm^2 Using Darcy's formula Q = k times l times A vol/t = k times 0.5 times 28.27 42.5/300 = k times 0.5 times 28.27 k = 0.01002 cm/s Now, Discharge velocity v_d = Q/A v_d = (val/t) /A = (42.5/300)/28.27 v_d = 0.00501 cm/s Dry weight of soil w_d = 9.33 N g_s = 2.7 Using following formula, r_d = w_d/vol. of soil = G_s. r_w \r\n