250+ TOP MCQs on Activated Sludge Treatment Systems – 2 and Answers

Waste Water Engineering online quiz on “Activated Sludge Treatment Systems – 2”.

1. Calculate the volume of the aeration tank required for the following data:
Flow: 800 m³/d
BOD: 1000 mg/L
F/M ratio: 0.15(extended aeration)
MLSS: 3500 mg/L
a) 1550 m³
b) 2550 m³
c) 3550 m³
d) 4550 m³
Answer: a
Clarification: The BOD load is calculated as Flow x BOD. Since the flow is in m³/d, the m³ is converted to L by multiplying by 1000. BOD load is expressed as Kg/L by dividing it by 1000000. BOD load = 800 x 1000 x 1000/ (1000×1000) = 800 Kg/d. Then Volume of tank = Biomass/MLSS. Biomass = BOD load/F/M. MLSS is converted to kg/ m³. Thus biomass= 800/ 0.15 =5333.3Kg. MLSS = 3.5 Kg/m³. Volume of tank = 5333.33/3.5 = 1523.81 m³. (Rounded off 1550 m³).

2. For the removal of BOD through Activated Sludge Process (ASP) what would be the Solid retention time considered?
a) 18 hours
b) 1-2 days
c) 10 hrs
d) 3 days
Answer: b
Clarification: For the removal of BOD through ASP, the solid retention time considered is 1-2 days. However, this depends on the temperature. In case the temperature is very low, then the solid retention time considered would be around 5-6 days.

3. For the conversion of particulate organics through Activated Sludge Process (ASP) what would be the Solid retention time considered?
a) 18 hours
b) 1-2 days
c) 2-4 days
d) 5 days
Answer: c
Clarification: For the conversion of particulate organics through ASP, the Solid retention time considered would be 2-4 days. Particulate organic carbon is defined as those materials that can pass through a filter which is of size 0.7 to 0.22 um. This usually constitutes soil organic matter that includes plant material, pollen etc.

4. For the development of flocculent biomass for treating domestic water through Activated Sludge Process (ASP) what would be the Solid retention time considered?
a) 18 hours
b) 4 days
c) 10 hrs
d) 1-3 days
Answer: d
Clarification: For the development of flocculent biomass for treating domestic water through Activated Sludge Process (ASP), the solid retention time considered is 1-3 days. In case it is industrial waste water, the retention time considered would be different. The difference is due to the difference in concentration of BOD in the influent.

5. For the development of flocculent biomass for treating industrial water through Activated Sludge Process (ASP) what would be the Solid retention time considered?
a) 18 hours
b) 3-5 days
c) 20 hrs
d) 1-3 days
Answer: d
Clarification: For the development of flocculent biomass for treating industrial water through Activated Sludge Process (ASP), the solid retention time considered is 3-5 days. This is a little higher than that of the Solid retention time required to treat domestic waste water. The difference is because the industrial waste water constitutes higher BOD values.

6. For the removal of nitrogen completely through Activated Sludge Process (ASP) what would be the Solid retention time considered?
a) 3-18 days
b) 1-2 days
c) 12 hours
d) 18 hours
Answer: b
Clarification: For the complete removal of Nitrogen through ASP, the solid retention time considered is 3-18 days. However this depends on the temperature. In case the temperature is very low, then the solid retention time considered would be around 15-18 days. For higher temperatures, lower solid retention time is considerd.

7. For the removal of phosphorous through Activated Sludge Process (ASP) what would be the Solid retention time considered?
a) 18 hours
b) 1-2 days
c) 10 hrs
d) 2-4 days
Answer: d
Clarification: For the removal of phosphorous through ASP, the solid retention time considered is 2-4 days. However, this depends on the temperature. In case the temperature is very low, then the solid retention time considered would be around 4 days.

8. What is the typical value of the F/M ratio considered for an ASP process?
a) 0.04 g/g.d
b) 0.1 g/g.d
c) 0.4 g/g.d
d) 0.01 g/g.d
Answer: a
Clarification: The typical F/M value considered for an ASP process is 0.04 g/g.d. This is a process parameter that is used to characterize the operating conditions. Based on this F/M value the sizing of the aeration tanks for the ASP is carried out.

9. What is the typical value of the F/M ratio considered for an extended ASP process?
a) 0.04 g/g.d
b) 0.1 g/g.d
c) 0.4 g/g.d
d) 0.01 g/g.d
Answer: b
Clarification: The typical F/M value considered for an extended ASP process is 0.1 g/g.d. This is a process parameter that is used to characterize the operating conditions. Based on this F/M value the sizing of the aeration tanks for the extended ASP is carried out.

10. While designing an aeration tank for the ASP what is the volumetric organic loading rate considered?
a) 0.3-3
b) 0.03-0.2
c) 3-5
d) 5-7
Answer: a
Clarification: While designing an aeration tank, the volumetric organic loading rate considered is 0.3-3. The volumetric organic loading rate is defined as the amount of COD/BOD applied to the aeration tank volume per day. It is expressed in kg BOD/COD m³.d.

11. Calculate the BOD load for the following data.
Flow: 800 m³/d
BOD load: 1000 mg/L
a) 1000 kg/d
b) 800 kg/d
c) 500 kg/d
d) 1500 kg/d
Answer: b
Clarification: The BOD load is calculated as Flow x BOD. Since the flow is in m³/d, the m³ is converted to L by multiplying by 1000. BOD load is expressed as Kg/L by dividing it by 1000000. BOD load = 800 x 1000 x 1000/ (1000×1000) = 800 Kg/d.

To practice all areas of Waste Water Engineering for online Quizzes,