Correct Answer - Option 1 : 87.75%
Concept:
Efficiency of First stage High rate trickling filter is:
\({\eta _I} = \frac{{100}}{{1 + 0.0044\sqrt {\frac{{{W_1}}}{{{V_1}{F_1}}}} }}\)
Where,
W1 = Amount of B.O.D applied to the first stage of trickling filter (kg/day)
V1 = Volume of filter media in first stage of trickling filter (hectare-meter)
F1 = Recirculation factor for First stage
\({F_1} = \frac{{1 + R}}{{{{\left( {1 + 0.1R} \right)}^2}}}\)
R = Recirculation ratio
Efficiency of Second stage High rate trickling filter:
\({\eta _{II}} = \frac{{100}}{{1 + \frac{{0.0044}}{{1 - {\eta _I}}}\sqrt {\frac{{{W_2}}}{{{V_2}{F_2}}}} }}\)
Where,
W2 = Amount of B.O.D applied to the second stage of trickling filter (kg/day)
W2 = WI(1-ηI)
V2 = Volume of filter media in second stage of trickling filter (hectare-meter)
F2 = Recirculation factor for second stage
F2 = F1 (If amount of recirculation in both the stages is same)
Overall efficiency of trickling filter:
\({\eta _o} = {\eta _I} + \left( {1 - {\eta _I}} \right){\eta _{II}}\)
Calculation:
Given,
Efficiency of BOD removal by first stage trickling filter = 65%
and ηI = 0.65
Efficiency of BOD removal by second stage trickling filter = 65%
ηII = 0.65
So, overall efficiency of trickling filter is
\({\eta _o} = {\eta _I} + \left( {1 - {\eta _I}} \right){\eta _{II}}\)
Putting the values of η1 and η2
η0 = 0.65 + (1 - 0.65) 0.65 = 0.65 + 0.2275 = 0.8775
η0 = 87.75%