A sewage is discharged in a river flowing at a velocity of 0.25 $m / s e c$ . The D.O content and $B O D_{5}$ of mixture at the point of disposal is 2.50 $m g / l$ and 15 $m g / l$ . The deoxygenation constant and reaeration constant on base are 0.20 $d a y^{- 1}$ and 0.40 $d a y^{- 1}$ respectively. The distance (in km, rounded off to nearest integer) from point of disposal where critical D.O. deficit will occur is

62.5

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Solution

The correct option is A 62.5
Given
$k_{d} = 0.20 d a y^{- 1}$

$k_{d} = 0.40 d a y^{- 1}$

$D_{0} = 2.50 \frac{m g}{l}$

$B O D_{5} = B O D_{u} (1 - e^{- k_{d} \times 5})$
$15 = L (1 - e^{- 0.20 \times 5})$
L=23.73 $m g / l$
Time $(t_{c})$ after which critical D.O. deficit occur is
$t_{c} = \frac{1}{k_{d} (f - 1)} l o g_{e} [(1 - (f - 1) \frac{D_{0}}{L}) f]$
Self purification constant $(f) = \frac{K_{r}}{K_{d}} = \frac{0.40}{0.20} = 2$
$t_{c} = \frac{1}{0.20 \times 1} l o g_{e} [(1 - \frac{2.5}{23.73}) \times 2]$
=2.91 days
Distance= v $\times t_{c}$
$= 0.25 \times 2.91 \times 24 \times 60 \times 60 \times 10^{- 3} k m s$
=62.856 km

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