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If the efficiencies of BOD removal of first stage and second stage trickling filters are each 65%, then what is the overall BOD removal efficiency of these filters?
1. 87.75%
2. 77.25%
3. 92.6%
4. 65%

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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%

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