If the discharge in the river is 160 lit/sec and its BOD is 5 mg/lit, find out the BOD of the diluted water.

Sewage discharge= $Q_s= \frac{7 \times 10^6}{24 \times 60 \times 60 } = 81.01 \ l/s$

Discharge of river = $Q_r= 160 \ l/s$

BOD of sewage = $C_s= 260 \ mg/l$

BOD of river = $C_r=5 \ mg/l$

BOD of diluted mixture = $C= \frac{C_s.Q_s + C_r.Q_r}{Q_s + Q_r }$

$ C = \frac{260 \times 81.01 + 5 \times 160}{81.01 + 160 }$

$C = 90.71 \ mg/l$