Treatment Of Waste Water – GATE CE Environmental Engineering Practice Questions & PYQs – Page 2 | BattleExam | BattleExam

Q25GATE 2014MCQ

A waste water stream (flow=2 $m^{3}$ /s, ultimate BOD=90 mg/l) is joining a small river (flow=12 $m^{3}$ /s, ultimate BOD=5 mg/l). Both water streams get mixed up instantaneously. Cross sectional area of the river is 50 $m^{2}$ . Assuming the de-oxygenation rate constant, k'=0.25/day the BOD (in mg/l) of the river water, 10 km downstream of the mixing point is

🚀 Solve in Practice Mode

📖 Explanation

Flow of waste water stream, $Q_{w}=2 \mathrm{m}^{3} / \mathrm{sec}$ Ultimate BOD, $Y_{w}=90 \mathrm{mg} / l=90 \mathrm{gm} / \mathrm{m}^{3}$ Flow of river, $Q_{R}=12 \mathrm{m}^{3} / \mathrm{sec}$ Ultimate BOD of river, $Y_{R}=0.5 \mathrm{mg} / l=5 \mathrm{gm} / \mathrm{m}^{3}$ BOD of mixture,

\begin{aligned} Y_{0} &=\frac{(2 \times 90)+(12 \times 5)}{2+12} \\ &=\frac{180+60}{14}=\frac{240}{14} \\ &=17.143 \mathrm{gm} / \mathrm{m}^{3} \\ k_{D} &=0.434 \mathrm{K} \\ &=0.434 \times 0.25=0.1085 \\ Y_{t} &=Y_{0}\left[1-(10)^{-k_{0} t}\right]\$ &\ldots(i)\\ \text { Area of river } &=50 \mathrm{m}^{2} \\ \text { Flow of river } &=12 \mathrm{m}^{3} / \mathrm{sec} \\ \text { Stream velocity } &=\frac{14}{50}=0.28 \mathrm{m} / \mathrm{sec} \\ \text { Time taken, } & t=\frac{10 \mathrm{km}}{0.28 \mathrm{m} / \mathrm{s}}=0.4133 \text { days } \end{aligned}

$\therefore$ From eq. (i),

\begin{aligned} \mathrm{BOD}_{\text {left }} &=Y_{0}-Y_{t} \\ &=Y_{0} \times 10^{-k t} \\ &=17.143 \times 10^{-0.1085 \times 0.4133} \\ &=15.46 \mathrm{mg} / l \end{aligned}