Shown below is a schematic diagram of a wastewater treatment process. The key information of the...

 

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Shown below is a schematic diagram of a wastewater treatment process. The key information of the process is given on the diagram. A unique feature of the process is having an activated carbon column at the end to remove soluble BOD before discharging the treated wastewater to a river. The recycle ratio (R) of the secondary treatment process is 0.5. Based on experimental data, the VSS/TSS ratio of the all the solids in the process is 0.80. The secondary clarifier can completely remove all solids. The removal of soluble BOD only occurs in the aeration tank and the activated carbon column. There is no nitrogen in the influent and the system is at steady state. Influent wastewater Q = 20,000 m³/d S. = 200 mg/L nbVSS, = 25 mg/L Q₁ = 1,000 m³/d Q₂ = 1,000 m³/d X = 3,500 mg VSS/L S = 100 mg/L Qs = 1,000 m³/d settling tank NGUL Secondary settling tank Aeration tank Return activated sludge Neutralization tank Q Coagulation & Neutralization flocculation. tank Q₁ = 3,000 m³/d Primary nbVSS, = 25 mg/L Sludge production Sludge Production Activated carbon column Treated wastewater Q = 16,000 m³/d S6 = 20 mg/L Calculate the total amount of solids that needs to disposed of from the entire system (in kg TSS/d). Shown below is a schematic diagram of a wastewater treatment process. The key information of the process is given on the diagram. A unique feature of the process is having an activated carbon column at the end to remove soluble BOD before discharging the treated wastewater to a river. The recycle ratio (R) of the secondary treatment process is 0.5. Based on experimental data, the VSS/TSS ratio of the all the solids in the process is 0.80. The secondary clarifier can completely remove all solids. The removal of soluble BOD only occurs in the aeration tank and the activated carbon column. There is no nitrogen in the influent and the system is at steady state. Influent wastewater Q = 20,000 m³/d S. = 200 mg/L nbVSS, = 25 mg/L Q₁ = 1,000 m³/d Q₂ = 1,000 m³/d X = 3,500 mg VSS/L S = 100 mg/L Qs = 1,000 m³/d settling tank NGUL Secondary settling tank Aeration tank Return activated sludge Neutralization tank Q Coagulation & Neutralization flocculation. tank Q₁ = 3,000 m³/d Primary nbVSS, = 25 mg/L Sludge production Sludge Production Activated carbon column Treated wastewater Q = 16,000 m³/d S6 = 20 mg/L Calculate the total amount of solids that needs to disposed of from the entire system (in kg TSS/d).

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Answer Xu 10500mg dis Amount of solids that needs to posed from Primary ... View the full answer