"Sieve size (cm) x Mass retained on the %retain on sieve %finer than Ysieve(g)25.4 20 10 9012.7 50 26 645.08 90 47 172.54 18 9 81.27 10 5 3< 1.27 5 3 x(cm) Y = log[ln1/(1-y)] X= log x25.4 0.3622 1.40412.7 0.0093027 1.1045.08 -0.73 0.7062.54 -1.0789 0.4041.27 -1.5163 0.104 0.5 y = 1.474x - 1.686 R² = 0.993 0 0 0.5 1 1.5 -0.5 Y-Values Linear (Y-Values) -1 -1.5 -2 Axis Title n = slope = 1.4742X = 10* 1.686/1.4742o= 11.44 cmb) Yes, in order to allow less number of retain mass in the sieves1c) E = 10E [1/ ? 1.61 /? - 1/ ? ? ]i 4 1/1.4742 1/2 1/2= 10 *200/[11.44 *10 (1.6) ] – 10*200/[250000]1/1.4742 1/2 1/2 = 10 *200/[114400(1.6) ] – 10*200/[250000] = 1.0418 kWh/tonnes per unit µmQ2) Hydraulic Parameter = area/wetted parameter = (0.5*2)/(2*0.2 +2*2)Axis Title = 0.2 m-5 Re = (1.2) * (0.15) * (0.2)/2.0 * 10= 1800Re<2000 gas is laminarTerminal velocity Vt -6 2 -5 V = (1500)(9.8)(8 *10 ) /18 * 2.0 * 10t -3 = 2.6133 * 10 m/sDetermine collection efficiency = -3= (2.6133 * 10 )(10 m)/(0.15 * 0.5) = 0.34844 OR 34.844%Q3) component Mass(Kg) % of massMoisture energycontent% Food 80 32 70 4700Paper 25 10 6 16700Plastics 20 8 2 32500Textile 50 20 10 17400Yard 15 6 60 6500Metals 40 16 3 700Miscellaneous 20 8 6 7000 a)0.32*0.3*4700 + 0.1*0.94*16700 + 0.08*0.98*32500 + 0.2*0.9*17400 + 0.06*0.4*6500 +0.16*0.97*700 + 0.08*0.94*7000 = 8492.04 kJ/kgwet MSW Overall energy = 0.32*4700 + 0.1*16700 + 0.08*32500 + 0.2*17400 + 0.06*6500 + 0.16*700+0.08*7000= 10316 kJ/KgQ4) solid recovery = (Q C )/C Q * 100x x o o 3 Q = 200 + 40 = 240 m /hrout Q *C = Q C + Q *Cout out x x in in 240 * C = 200 * 20 + 40 * 800o -3C = 150 gmo b) rate of sludge m + m1 2= 800 *40 + 200 * 20= 36000 g/hrQ5) C= q/k = 0.0005/0.02 = 0.025Tana = s = 0.02H = 20cmmax2 2 20cm = L*0.025/2[0.02 /0.025 +1 – 0.02/0.025 0.02 +0.025 ] L = 1799.75 cm "