Question: YAB 4 2 4 3 YAB 4 2 4 3 2 . A schematic system to demonstrate a complete - mixed activated sludge process is

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2. A schematic system to demonstrate a complete-mixed activated sludge process is shown in FIGURE Q2.
FIGURE Q2: Schematic diagram of complete-mixed activated sludge process.
A wastewater stream containing biochemical oxygen demand (BOD5) of 200mgL is to be treated via an activated sludge process. The wastewater stream also has a flow rate of 3000m3d in addition to the following characteristics for aeration tank and secondary clarifier in order to achieve the desired treatment efficiency during the commencement of treatment system.
Aeration Tank:
Food/Microorganism (F/M) ratio =0.25-5.00kgBOD5? MLVSS.d
Mean Cell Residence Time (MCRT)=5-10d
Hydraulic retention time =5-7h
Mixed liquor suspended solids (MLSS) concentration =4000-6000mgL with 75% consisted of volatile matter
Recirculation ratio =0.4-1.0
Required effluent BOD5=20-25mgL
Excess sludge wastage rate =0.5-0.8kgkgBOD5 removed
Secondary Clarifier:
Hydraulic loading =35-50m3m2.d
Depth of clarifier =3-6m
Solids loading =150kg VSS/m ?2.d
NOTE: All the ASSUMPTIONS made must be explicitly stipulated along the designing steps.
a. Design a suitable volume of aeration tank by considering the F/M ratio.
[10 marks]
b. Evaluate an appropriate BOD5 removal efficiency.
[10 marks]
c. Identify the MCRT value that fits the aeration tank.
[10 marks]
d. Predict the diameter of secondary clarifier as calculated from recirculation ratio.
[10 marks]
e. Formulate a suitable volume of secondary clarifier that satisfies the depth required.
[10 marks]
END OF PAPER -
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YAB 4 2 4 3 YAB 4 2 4 3 2 . A schematic system to

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