An industry discharges sewage into a river having flow rate of 8 $m^{3} / s e c$ . $B O D_{5}$ of sewage is 180 $m g / l$ . Dissolved oxygen of the river before the sewage is discharged was 8.6 $m g / l$ . The maximum permissible discharge(in lps rounded off to nearest integer) allowed into the river in order to maintain minimum D.O. for survival of aquatic life is
9200

Open in App

Solution

The correct option is A 9200
Minimum D.O. required for survival of aquatic life= $4 m g / l$
$(D . O)_{m i x} = \frac{C_{R} Q_{R} + C_{S} Q_{S}}{Q_{R} + Q_{S}}$
Given, $Q_{R} = 8 m^{3} / s e c, Q_{s} = ?, C_{R} = 8.6 m g / l C_{s} = 0$
$4 = \frac{(8.6 \times 8) + (0 \times Q_{s})}{8 + Q_{s}}$
$∴ Q_{s} = 9.2 c u m e c s = 9200 l p s$

Suggest Corrections

Similar questions

m / s e c

B O D_{5}

m g / l

m g / l

d a y^{- 1}

d a y^{- 1}

Untreated sewage is discharged into the rivers. How will it affect the population of aquatic animals in the river?

Join BYJU'S Learning Program