A city discharges 1500 liter per second of waste water into a river, whose minimum rate of flow is 3500 lit per second.

The temperature of waste water as well as river water is 20°C. The 5day BOD of waste water at that temperature is 300mg/lit and that of river water is 1 mg/lit.

The DO content of waste water is zero and that of the stream is 90% of the saturation D.O.

If the minimum D.O.to be maintained in the stream is 4.0mg/lit. Find out the degree of waste water treatment required. Assume the coefficient of de-oxygenation ($K_D$) as 0.1 and coefficient of re-oxygenation ($K_R$) as 04.

D.O at $20^0 C$ = $9.17 \ mg/l$ (assume fire value)

$\begin{aligned} \text{D.O. content of the stream} &= \text{90 % of saturation D.O.} \\ &= \frac{90}{100} \times 9.17 \\ &= 8.25 \ mg/l \end{aligned}$

$\begin{aligned} \text{Minimum D.O. to be maintained in the stream} \\ \text{Max permissible saturation deficit} \space (D_c) &= 9.17 - 4.0 \\ &= 5.17 \ mg/l \end{aligned}$

from equations given below

$\bigg( \frac{L}{D_C.f} \bigg)^{f-1} = f \bigg( 1 - (f - 1) \frac{D_0}{L} \bigg)$

$D_0 = 3.39 \ mg/l, \space \space D_c = 5.17 \ mg/l$

$f = \frac{K_R}{K_D} = \frac{0.4}{0.1} = 4$

we get,

$\begin{aligned} \bigg( \frac{L}{5.17 \times 4} \bigg)^{4-1} &= 4 \bigg( 1 - (4 - 1) \frac{3.39}{L} \bigg) \\ L &= 28.28 \ mg/l \end{aligned}$

$\begin{aligned} y_1 &= L \big( 1 - 10^{-k_O.l} \big) \\ &= 28.29 ( 1 -10^{-0.1 \times 5} \big) \\ &= 19.34 \ mg/l \end{aligned}$

now,

$\begin{aligned} BOD_{mix} \space 19.34 &= \frac{C_5 \times 1500 + 1 \times 3500}{1500 \times 3500} \\ C_s &= 62.13 \ mg/l \end{aligned}$

$\begin{aligned} \text{Degree of treatment required (%)} &= \frac{Original_{BOD} - Permissible_{BOD}}{Original_{BOD}} \times 100 \\ &= \frac{300 - 62.13}{300} \times 100 \\ &= 79.29 \% \end{aligned}$