Question

A waste water flow at the rate of 1 MLD having BOD of 120 mg/l enters a circular oxidation pond having diameter and depth 80 m and 1.5 m respectively. The decay constant is 0.1 per day. The efficiency of oxidation pond would be:

• A. 76.5%
• B. 78.6%
• C. 82.4%
• D. 88.6%

Answer 1

The correct option is C 82.4%

Sewage flow $=1MLD=1\times {10}^3{m}^3/\text{day}$

Influent BOD, L = 120 mg/l

Diameter of pond = 80 m

Depth of pound = 1.5 m

k = 0.1 per day

Detention time of oxidation pond, $t_d=\frac{1}{k}lo{g}_{10}\left(\frac{L}{L-y}\right)$

where y = BOD removed

$\frac{V}{Q}=\frac{1}{k}lo{g}_{10}\left(\frac{L}{L-y}\right),V$ = volume of oxidation pound
$\frac{\frac{\pi }{4}\times {80}^2\times 1.5\times 0.1}{1\times {10}^3}=log\left(\frac{120}{120-y}\right)$
$120=(120-y){10}^{0.754}$
$(120-y)-\frac{120}{5.675}$
$y=98.85mg/l$
Efficiency $=\frac{y}{L}\times 100$
$=\frac{98.85}{120}\times 100=82.4\%$