The sewage of a town is to be discharged into a river system. The quantity of sewage produced per day is 8 million liters and its B.O.D is 250 mg/liter. If the discharge in the river is 200 l/s and if its B.O.D is 6 mg/l. Find out the B.O.D of the diluted water.

Sewage discharge=$Q_s$

=$\frac{800*10^6}{24*60*60}$l/s

=92.59 l/s

Discharge of the river=$Q_r$=200 l/s

B.O.D of sewage=$C_s$=250 mg/l

B.O.D of water/river=$C_r$=6 mg/l

B.O.D of the diluted mixture

C=$\frac{C_S*Q_S+C_R*Q_R}{Q_S+Q_R}$

C=$\frac{250*90.59+6*200}{92.59+200}$

C=83.21 mg/s