Question

A town having a population of 1.2 lakhs is producing sewage at a rate of 100 Ipcd having 200 $\frac{mg}{l}$ of BOD. A trickling filter having recirculation ratio 1.5 is design to produce effluent of BOD 20$\frac{mg}{l}$. The operating depth of filter is 2.5 . The diameter of the trickling filter is m.

• A. 38.2m
• B. 40 m
• C. 56.15 m
• D. 70 m

Answer 1

The correct option is C 56.15 m

Total BOD sewage=$(1.2\times {10}^5\times 100\times {10}^{-6})MLD\times 200\frac{mg}{l}$
=2400kg

BOD entering the trickling filter $=0.7\times 2400=1680kg$

Desired BOD in the effluent= $(1.2\times {10}^5\times 100\times {10}^{-6})\times 20$
=240 kg

Efficiency of the filter,$\eta =\frac{1680-240}{1680}=85.71$%
$\eta =\frac{100}{1+0.0044\sqrt{\frac{Y}{VF}}}$
Y=Total BOD m kg=1680 kg
V=Volume of thickling filter(in ha-m)
F=Recirculation factor=$\frac{1+R}{(1+0.1R{)}^2}=\frac{1+1.5}{(1+0.1\times 1.5{)}^2}=1.89$
$\therefore 85.71=\frac{100}{1+0.0044\sqrt{\frac{1680}{V\times 1.89}}}$
Surface area of filter =$\frac{0.6191\times {10}^4}{2.5}=2476.4m^2$
Also,$\frac{\pi d^2}{4}=2476.4$
$\Rightarrow d=56.15m<60$ m
Hence, okay.