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Question: secondary treatment method numerical
0

Determine the size of high rate trikling filter for following data

1) sewage flow:4MLD

2) Recirculation ratio:1:5

3) $BOD_5$ of raw sewage:280 mg/l

4) BOD removal in PST=25%

5) Final effluent $BOD_5$ desired=30mg/l

Subject:-Environmental Engineering-II

Topic:-Secondary treatment method

Difficulty:High

ee2(77) • 73 views
 modified 27 days ago by Sanket Shingote ♦♦ 220 written 9 weeks ago by
0

### Given:-

1) sewage flow: 4MLD

2) Recirculation ratio: 1:5

3) $BOD_5$ of raw sewage: 280 mg/l

4) BOD removal in primary clarifier=25%

5) Final effluent $BOD_5$ desired=30mg/l

### 1) To find efficiency of filter(n):-

Total BOD present in raw sewage=4 ml *280 mg/lit=1120kg

BOd removed in primary tank=25%

∴ BOD left in the sewage entering per day in the filter unit= 1120*0.75=840kg

BOD concentration desired in final affluent = 30mg/lit

$\therefore$ total BOD left in the effluent per day = 4ml X 30 = 120 kg

∴BOD removed by the filter=840-120=720 kg

Efficiency of the filter(η)=$\frac{BOd removed}{total BOD}*100$

=$\frac{720*100}{840}=85.71%$

### 2) Volume of filter in ha-m:-

Efficiency of the filter is given by:-

$η=\frac{100}{1+0.0044*\frac{Y}{V*F}^\frac{1}{2}}$

where Y=total BOD applied to the filter in kg=840kg

V=Volume of filter in ha-m

F=Recirculation Factor

$=\frac{1+R}{(1+0.1*R)^2}$

$F=\frac{1+1.5}{(1+0.1*1.5)^2}$

$=\frac{2.5}{1.15}$

F=1.89

∴$85.71=\frac{100}{1+0.0044\frac{840}{V*1.89}^\frac{1}{2}}$

$85.71=\frac{100}{1+0.0044\frac{444.44}{V}^\frac{1}{2}}$

$1+0.0044\frac{444.44}{V}^\frac{1}{2}=\frac{100}{85.71}$

$1+0.0044\frac{444.44}{V}^\frac{1}{2}=1.166$

$0.0044\frac{444.44}{V}^\frac{1}{2}=1.166-1$

$0.0044\frac{444.44}{V}^\frac{1}{2}=0.166$

$\frac{444.44}{V}^\frac{1}{2}=\frac{0.166}{0.0044}$

$\frac{444.44}{V}^\frac{1}{2}=37.22$

Squaring on both the sides

$\frac{444.44}{V}=37.22^2$

$V=\frac{444.44}{37.22^2}=0.3/hectre-m$

$V=3100m^3$

### 3) Surface area required:-

Depth of filter according to design aspect should be from 1.8 to 2.4 m and diameter of filter should be 30 m to 60 m

Assuming the depth of filter as 2.4 m, we have

$Surface \space area=\frac{3100}{2.4}m^2=1291.66m^2=1292m^2$

### 4) Diameter(D) of the circular high rate trikling filter

$\frac{πD^2}{4}=1292$

$D^2=\frac{1292*4}{π}$

$D=\frac{1292*4}{π}^\frac{1}{2}=40.558 sap 41 m$

D=41m

∴So diameter is between 30 to 60 m for the trickling filter