Question

6. The aeration tank for a completely mixed aeration process is being sized for a design wastewater flow of 7500 m²/day. The influent BOD is 130 mg/l with a soluble BOD of 90 mg/1. The design effluent BOD is 20 mg/l with a soluble BOD of 7.0 mg/l. Recommended design parameters are a sludge age of 10 days and volatile MLSS of 1400 mg/l. The kinetic constants from a bench-scale treatability study are Y=0.60 mg VSS/mg soluble BOD and Ka = 0.06 day!. Calculate the volume of the aeration tank, aeration period, food/microorganism ratio, sludge production, and efficiency of BOD removal.

Answer 1

Solution : Given daka ©. calculate the volume of aeration kank.: - vo Oc YQ (S.-Se) X (1+ Kd Oc). Here Q = mean cell residence line = 90 days. @ influent coaslecaker flow = 7500 n°/day. So - influent Soluble BOD = 90 mglt. Se = effluenk Soluble Bop = 7 mg/l Y = growth yield = 0.60 mg. Kd = 0.06 day! X -- volatile MLSS - 1400 mgla. va lo x 0.60 x 7500 (90-7) 1400 *(1+0.06 x 10).

Vz 3735000 2240 V = 1667.4107 mi? 6. Aeration period. :- we know, o 5 V=1667.41070? 1 | Q = 7500m?/d. 0 1667.4107 7500 10 2 0-1923 days "70-5.398 heurs. @ food JMicroorganism ratio : So = 90 mgle. o - 0.2223 days

X21400 mgle. E 90 0.2223x1400 1 :0.2892 mg per day Ing of mivss. W. Shoge production : - Q.X. = 7500 milday x 1400 mgle = 10500 kg/day ©. efficiency of Bod removal. w in e change in Bob influent: BoD. (40.7) math 90 mgte. 0.92922 292.222 %.