This set of Waste Water Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Diffused and Mechanical Aeration Systems”.
1. Aeration consumes less energy in activated sludge process compared to other processes.
a) True
b) False
View Answer
Explanation: Aeration occurs in many water and wastewater treatment processes, but in the activated sludge process and its variants, it consumes more energy than other processes, by far.
2. What does OTR stand for?
a) Oxygen transfer rate in standard conditions
b) Oxygen transfer rate in clean water
c) Oxygen transfer efficiency in clean water
d) Oxygen transfer efficiency in standard conditions in the clean water
View Answer
Explanation: OTR stands for oxygen transfer rate in clean water. SOTR stands for oxygen transfer rate in clean water. OTE stands for Oxygen transfer efficiency in the clean water. SOTE stands for oxygen transfer efficiency in standard conditions in clean water.
3. Which of the following is not a standard condition for aeration?
a) 40˚C
b) 1 atm
c) Zero salinity
d) Zero DO
View Answer
Explanation: Various standard conditions are mentioned for the aeration process. The standard temperature is considered to be 20˚C along with 1 atmospheric pressure, zero salinity and zero dissolved oxygen.
4. What is the symbol used to define the ratio of the process to clean water mass transfer?
a) Alpha
b) Beta
c) Gamma
d) Sigma
View Answer
Explanation: The ratio of the process to clean water mass transfer is called alpha factor and is symbolized by alpha.
5. _________ is the actual mass of oxygen transferred per unit time.
a) OTR
b) SOTR
c) OTE
d) SOTE
View Answer
Explanation: The OTR is the actual mass of oxygen transferred per unit time and it is the key process variable for design. The DO saturation concentration is the concentration of dissolved oxygen at saturation with no reactions in the liquid.
6. High solubility of CO2 ______ pH.
a) Maintains
b) Reduces
c) Increases
d) Nullifies
View Answer
Explanation: High solubility of CO2 reduces the pH of water which causes excessive consumption of lime or other neutralizing agents in coagulation and softening process. The corrosiveness of water is also higher at lower pH values.
7. What is the range of pH above which aeration is required?
a) 2 mg/L
b) 5 mg/L
c) 8 mg/L
d) 10 mg/L
View Answer
Explanation: Exposure of water droplets to air for 2 seconds will lower the CO2 by 70-90%. If CO2 > 10mg/L aeration is recommended. Otherwise, the lime addition should be used to neutralize the CO2.
8. H2S is highly soluble in water.
a) Tue
b) False
View Answer
Explanation: H2S is highly soluble in water. Hydrogen sulphide poisoning is one of the leading causes of accidents. If its concentration is above 5 ppm moderate odour is present. If its concentration increases above 300 ppm, it leads to unconsciousness and death.
9. Which of the following is not used for iron removal?
a) Oxidation
b) Reduction
c) Gasification
d) Absorption
View Answer
Explanation: Oxygenation of water is done by gasification and absorption. The removal of iron from water or wastewater can be done by various methods such as oxidation, gasification and absorption.
10. How many types of aerators are present?
a) 2
b) 3
c) 4
d) 5
View Answer
Explanation: There are four types of aerators. They are gravity aerators, spray aerators, diffused air aeration systems and mechanical aerators. Air stripping is done to remove toxic volatile organics and volatile compounds.
11. What is the value for Ɵ for mechanical and diffused aeration for the following equation.
KLA(T) = KLA(20 Degree Celsius)ƟT-20
a) 1.015
b) 1.1
c) 1.024
d) 1.04
View Answer
Explanation: The value for Ɵ is usually between 1.015 to 1.04. In case of mechanical and diffused aeration this is 1.024. KLA is Oxygen mass transfer coefficient.
12. What is the value for α for diffused aeration for the following equation.
α= KLA(wastewater) /KLA(Tap water)
a) 0.4-0.8
b) 0.2-0.3
c) 0.8-1
d) >1
View Answer
Explanation: Usually the value for α varies between 0.3-1.2. The value for α is usually between 0.4-0.8 for diffused aeration. KLA is Oxygen mass transfer coefficient.α is the correction factor.
13. What is the value for α for mechanical aeration aeration for the following equation.
α= KLA(wastewater) /KLA(Tap water)
a) 0.6-1.2
b) 0.2-0.3
c) 0.3-0.5
d) >1.2
View Answer
Explanation: Usually the value for α varies between 0.3-1.2. The value for α is usually between 0.6-1.2 for mechanical aeration. KLA is Oxygen mass transfer coefficient.α is the correction factor.α depends on the geometry of the basin, degree of mixing and wastewater characteristics.
14. Calculate the theoretical oxygen requirement for the following details.
Flow: 200 m3/h
BOD load: 800 mg/L
a) 45.5 Kg/h of O2/hr
b) 23.05 Kg/h of O2/hr
c) 25.05 Kg/h of O2/hr
d) 50.5 Kg/h of O2/hr
View Answer
Explanation: BOD load in Kg/day = Inlet BOD in mg/l x Flow in m3/day /1000 =800 x 200 / 1000=160 Kg/day. Theoretical oxygen requirement = BOD load x Kg of O2/Kg of BOD= 160 x 1.8 = 288 Kg of O2/day. Standard O2 Requirement = Theoretical O2 requirement/ Alpha x [Beta x Csw – D.O.] x Theta ^ (Tww-20°C)/ Css= 288/24/[0.7 x [0.95 x 8.64 – 2] x 1.024^ (30-20)]/9.86=23.05 Kg of O2/hr.
15. Calculate the amount of air required for the following details.
Flow: 300 m3/h
BOD load: 800 mg/L
a) 550 m3/hr
b) 518 m3/hr
c) 700m3/hr
d) 618 m3/hr
View Answer
Explanation: BOD load in Kg/day = Inlet BOD in mg/l x Flow in m3/day /1000 =800 x 300 / 1000=240 Kg/day. Theoretical oxygen requirement = BOD load x Kg of O2/Kg of BOD= 240 x 1.8 = 432 Kg of O2/day. Standard O2 Requirement = Theoretical O2 requirement/ Alpha x [Beta x Csw – D.O.] x Theta(Tww-20°C)/ Css= 432/24/[0.7 x [0.95 x 8.64 – 2] x 1.024(30-20)]/9.86=38.29 Kg of O2/hr. Air requirement = Actual Oxygen requirement/ (Density of air x % of O2 in air by weight x SOTE%) = 38.29 /( 1.201 x 0.232 x 0.1599)= 696 m3/hr. (Rounded off to 700 m3/hr). Factors assumed: SOTE=17.27%; Alpha=0.7; Beta=0.95; Theta=1.024; Surface saturation C=7.97 (Calculated); Site basin saturation Csw=8.64 (Calculated); Standard basin saturation Css=9.86 (Calculated).
16. Calculate the number of diffusers for the following.
Blower Flow rate: 535 m3/hr
Coarse diffusers are used.
a) 54
b) 60
c) 65
d) 70
View Answer
Explanation: Number of diffusers equals =flow/ flow rate per diffuser. Flow rate considered for coarse diffusers are 10 m3/hr. Therefor number of diffusers= 535/10= 54.
17. Calculate the number of fine diffusers required for the following data.
Flow: 300 m3/h
BOD load: 1000 mg/L
a) 150
b) 146
c) 170
d) 186
View Answer
Explanation: BOD load in Kg/day = Inlet BOD in mg/l x Flow in m3/day /1000 =1000 x 300 / 1000=300 Kg/day. Theoretical oxygen requirement = BOD load x Kg of O2/Kg of BOD= 300 x 1.8 = 540 Kg of O2/day.
Standard O2 Requirement = Theoretical O2 requirement/ Alpha x [Beta x Csw – D.O.] x Theta ^ (Tww-20°C)/ Css= 540/24/[0.7 x [0.95 x 8.64 – 2] x 1.024^ (30-20)]/9.86=48 Kg of O2/hr. Air requirement = Actual Oxygen requirement/ (Density of air x % of O2 in air by weight x SOTE%) = 48/( 1.201 x 0.232 x 0.1599)= 875 m3/hr. Factors assumed: SOTE=17.27%; Alpha=0.7; Beta=0.95; Theta=1.024; Surface saturation C=7.97 (Calculated); Site basin saturation Csw=8.64 (Calculated); Standard basin saturation Css=9.86 (Calculated).
Number of fine diffusers required = 875/6=146. (flow rate considered for fine diffusers is 6 m3/hr).
18. Calculate the amount of air required for the equalization tank with the following details
Flow: 500 m3/h
BOD load: 800 mg/L
a) 950 m3/hr
b) 800 m3/hr
c) 2000 m3/hr
d) 785 m3/hr
View Answer
Explanation: Volume of tank = flow x retention time considered = 500 x 8 =4000 m3. The air requirement is 0.5 m3/hr/ m3 tank volume. Therefor the air requirement considered is 4000 x 0.5 = 2000 m3/hr.
Sanfoundry Global Education & Learning Series – Waste Water Engineering.
To practice all areas of Waste Water Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.
