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:A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter...

:A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter received flow 44L/Sec , the BOD concentration in raw sewage =440 mg/L , 30% BOD removed in the primary sedimentation tank, volume of filter (first stage )=826000L, volume of filter (second stage)=2563000L, recirculation rate =0.5 in second tank while no recirculation in first tank. Determine the following:
1-The effluent concentration of BOD in each unit.
2-Treatment efficiency of each unit.
3-Treatment efficiency of plant.
4-Diameters of each tanks if the depth is 5m.
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Answer #1

Solution:- the values given in the question are as follows:

flow rate(Q)=44 L/s

BOD concentration in raw sewage=440 mg/L

BOD removed by primary sedimentation tank=30%

filter volume in first stage=826000 L

filter volume in second stage=2563000 L

recirculation rate(second tank)=0.5

(1)

the effluent concentration of BOD in primary tank=30% of BOD concentration in raw sewage

the effluent concentration of BOD in primary tank=(30/100)*440

the effluent concentration of BOD in primary tank=132 mg/L

the effluent concentration of BOD in secondary tank=BOD concentration in raw sewage-the effluent concentration of BOD in primary tank

the effluent concentration of BOD in secondary tank=440-132

the effluent concentration of BOD in secondary tank=308 mg/L

(2)

treatment efficiency of primary tank(\etap)=[volume filter by primary tank/total volume]*100

treatment efficiency of primary tank(\etap)=[826000/(826000+2563000)]*100

treatment efficiency of primary tank(\etap)=24.3729%

treatment efficiency of secondary tank(\etas)=[volume filter by secondary tank/total volume]*100

treatment efficiency of secondary tank(\etas)=[2563000/(826000+2563000)]*100

treatment efficiency of secondary tank(\etas)=75.627%

(3)

treatment efficiency of plant(\eta)=[total filtered volume/total volume]*100

where, total filtered water=volume filter by primary tank+volume filter by secondary tank

total filtered water=826000+2563000

total filtered water=3389000 L

total volume=volume filter by primary tank+volume filter by secondary tank+recirculation volume

total volume=826000+2563000+0.5*2563000

total volume=4670500 L

treatment efficiency of plant(\eta)=[3389000/4670500]*100

treatment efficiency of plant(\eta)=72.5618%

(4)

depth of each tank=5 m

let diameter of primary tank is D1 and secondary tank is D2

volue(V)=area of tank*depth of tank

volue of primary tank(V)=area of primary tank*depth of primary tank

primary tank filter volume=826000 L

where, 1000 L=1 m^3

826000*10^-3=(3.14159/4)*D1^2*5

D1^2=210.3391

D1=14.5030 m

diameter of primary tank(D1)=14.5030 m

volue of secondary tank(V)=area of secondary tank*depth of secondary tank

secondary tank filter volume=2563000 L, or 2563 m^3

2563000*10^-3=(3.14159/4)*D2^2*5

D2^2=652.66259

D2=25.54726 m

diameter of secondary tank(D2)=25.54726 m

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