Question

O The town of Gangnam in Seoul, Korea discharges 17360 m3/day of treated wastewater in the Han River. The treated wastewater has a BOD5 of 12 mg/L and a BOD decay constant, k, of 0.12 day at 20 °C. The river has a flow rate of 0.43 m/s and an ultimate BOD, Lo, of 5 mg/L. The DO of the river is 6.5 mg/L and the D0 of the wastewater is 1.0 mg/L (a) Calculate the ultimate BOD of the wastewater. Ans: 26.6 mg/L (b) Calculate the initial DO (DOo) and initial ultimate BOD (Lo) after mixing. Ans: DO. 4.75 mg/L,Lo11.86 mg/L

Answer 1

BOD5, Y5 = 12 mg/l

BOD decay constant, k = 0.12/day at 20 C

t = 5 days

Ultimate BOD, L0 =

L5 = Lw⁰ (1-e^-kt) = 12 mg/l

Lw⁰ (1-e-^{(0.12 x 5)}) = 12 mg/l

Lw⁰ (1-e^-0.6) = 12 mg/l

Lw⁰ (1-0.54) = 12 mg/l

Lw⁰ = 12 mg/l/0.46 = 26.6 mg/l

b) Initial DO :

Flow rate of wastewater , Qw = 17360 m³/day

per day = 86400 secs

per sec = 17360/86400 = 0.2 m³/s

Flow rate of river, Qr = 0.43 m³/s

DO of river, DOr = 6.5 mg/l

DO of wastewater, DOw = 1 mg/l

DO = QwDOw + QrDOr/(Qw+Qr) = (0.2 m³/s x 1 mg/l + 0.43 m³/s x 6.5 mg/l)/(0.2+0.43 m³/s) = (0.2 + 2.795 mg/l)/0.63 = 2.995/0.63 = 4.75 mg/l

Initial ultimate BOD after mixing : La = QwLw + Qr Lr/(Qw+Qr) = ( 0.2 m³/s x 26.6 mg/ml + 0.43 m³/s x 5 mg/l)/(0.2+0.43) = 5.32 + 2.15)/0.63 = 7.47 mg/ml/0.63 = 11.857 = 11.86 mg/ml