Question

[3 pts] A water treatment plant consists of coagulation, flocculation, sedimentation, filtration and disinfection. The suspended solids concentration in the raw water coming into the plant is 500 mg/L and the plant treats 8 MGD. Alum (Al2(S04)3 14H20) is used as a coagulant with a dose of 50 mg/L. Compute the total solid sludge produced daily if a complete reaction of alum to aluminum hydroxide (Al(OH)3] occurs and 98% total solids are removed by the sedimentation and filtration steps. The total solid sludge produced per day (kg/d) is the suspended solids removed in the sludge plus the aluminum hydroxide precipitate. (*Hint: you'll need to do a mass balance for the treatment process) Al2(SOJ3°14H20+ 6 ???? . > 2 Al(OH)3 (s) + 6 CO2 + 14 H2O + 3 SO-24

Answer 1

C_1 = 500mg/L, theta = 0.42 m^3/s Iorated water (Co) = 0.02 times C_1 =0.02 times 500 times 10 mg/L Material balance for ss: - Accumulation Input Output Generation 28304332 = Consumption with in system theta = Input - Output + 0 Input , Output QC_1, QC_0 + qC_S or qC_s = Q(C_1 - C_0) 0.42(500 - 10) qC_s = 0.42 times 4 g_0 qC_s = 205.8g/sec Material balance for Aluminium Hydroxide: - Al_2(SO_4)_3 14H_2Ot ? 2Al(OH)_3 + ? +? i.e. 594 gm of alum produces 156 gm of Alum Hydroxide (Sludge) 1 gm ---= 1.56/594 - for (theta times C) 50 mg/L or (2rho gm/s) = 156/594 times 21.5 46 gm/s Total solid, SS Removed + Alum Hydroxyde (Sludge) =205.8 + 5.46 = 211.26 gm/s or 18252.86 kg/day