Category: WASTE WATER Engineering Objective Questions

250+ TOP MCQs on Activated Sludge Process – 2 and Answers

Waste Water Engineering online test on “Activated Sludge Process – 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) High sludge volume index
b) Low sludge volume index
c) Low oxygen demand
d) Floc formation is less
Answer: a
Clarification: Readily biodegradable COD if it is high in the influent, then the sludge volume index will be very high. Also this high level of COD would contribute to the floc formation. This also results in higher oxygen demand.

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 the 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 considered.

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 m3d.

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.

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250+ TOP MCQs on Anaerobic Digestion and Answers

Waste Water Engineering Multiple Choice Questions on “Anaerobic Digestion”.

1. Anaerobic digestion occurs in the presence of oxygen.
a) True
b) False
Answer: b
Clarification: Anaerobic digestion is a collection of processes by which micro-organisms break down biodegradable material in the absence of oxygen. The process is used for industrial or domestic purposes to manage waste or to produce fuels.

2. Anaerobic digestion can be used to produce fuel.
a) True
b) False
Answer: a
Clarification: Anaerobic digestion process is used for industrial or domestic purposes to manage waste or to produce fuels. Much of the fermentation used industrially to produce food and drink products, as well as home fermentation uses anaerobic digestion.

3. _________ uses anaerobic digestion.
a) Incineration
b) Combustion
c) Fermentation
d) Oxygenation
Answer: a
Clarification: In industries, fermentation is used to produce food and drink products. Anaerobic digestion occurs naturally in some soils and in a lake and oceanic basin sediments, where it is usually referred to as “anaerobic activity”.

4. What is biogas composed of?
a) O2 and CO2
b) CO2 and NO2
c) CH4 and O2
d) CH4 and CO2
Answer: d
Clarification: Anaerobic digestion is a biological process. A gas is produced that is composed of methane and carbon-dioxide which is called as biogas. These gases are produced from organic wastes such as livestock manure, food processing waste, etc.

5. What is the percentage of methane in the biogas?
a) 10-20 %
b) 20-40 %
c) 45-65 %
d) 55-75 %
Answer: d
Clarification: Anaerobic processes could either occur naturally or in a controlled environment such as a biogas plant. Depending on the waste feedstock and the system design, biogas is typically 55 to 75 percent pure methane.

6. How many steps are present in anaerobic digestion?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: There are three steps present in the anaerobic digestion process. The first step is the decomposition of plant or animal matter and the second step is the conversion of decomposed matter to organic acids. In the final step, the acids are converted to methane.

7. What is the temperature that needs to be maintained during the process?
a) 90˚F
b) 95˚F
c) 100˚F
d) 105˚F
Answer: c
Clarification: Process temperature affects the rate of digestion and should be maintained in the mesophilic range (95 to 105 degrees Fahrenheit) with an optimum of 100 degrees Fahrenheit.

8. The pathogens in digestate are _________
a) Highly active
b) Inactive
c) Partly inactive
d) Neutral
Answer: c
Clarification: The pressure exerted by the rising gas can be used to transport the gas to the collection vessel or directly to where it is going to be sued. The digestate is rich in organics and nutrients, almost odourless and pathogens are partly inactivated.

9. What is the range of retention time during anaerobic digestion?
a) 10-20 days
b) 20-30 days
c) 40-80 days
d) 40-100 days
Answer: d
Clarification: The range of retention time during anaerobic digestion is 40-100 days. Biogas reactors are often installed at household or community level in rural areas for the co-digestion of animal manure and toilet products.

10. How much is the cow yield from anaerobic digestion that can be obtained?
a) 0.1 m3/Kg dung
b) 0.2 m3/Kg dung
c) 0.3 m3/Kg dung
d) 0.4 m3/Kg dung
Answer: d
Clarification: The cow yield from anaerobic digestion is 0.4 m3/Kg dung. Animal manure and kitchen waste contain a lot of organic matter and generally, the process produces enough biogas for the family to cover cooking requirements.

11. In anaerobic digestion system which of the following statement holds good?
a) Acids are necessary to be added to make the pH neutral
b) Alkalines are necessary to be added to make the pH neutral
c) A flocculant is to be added to bring about flocculation
d) A coagulant is necessary to be added in order to bring about coagulation
Answer: b
Clarification: Alkalines are necessary to be added to make the pH neutral. The pH in these systems drops very low. Hence the addition of such alkaline to increase the pH becomes necessary.

12. What is the optimum temperature at which anaerobic digestion is carried out?
a) 25-35 degree Celsius
b) 55 degree Celsius
c) 45 degree Celsius
d) 40-50 degree Celsius
Answer: a
Clarification: Anaerobic digestion is carried out at 25-35 degree Celsius. At lower temperatures, higher Solid retention time is expected. Attached or suspended growth anaerobic digestion is carried out at temperatures as low as 10-20 degree Celsius.

13. For what range of COD level anaerobic digestion is carried out?
a) 400-800
b) 900-1200
c) 1500-2000
d) 300-400
Answer: c
Clarification: The COD level would be around 1500-2000 mg/L if it is subjected to anaerobic digestion. In case if it is around 1300 mg/L then aerobic digestion is carried out. In anaerobic digestion, methane is produced.

14. What is the percentage of Carbon dioxide produced during anaerobic digestion?
a) 70-80%
b) 80-90%
c) 30-50%
d) 50-60%
Answer: c
Clarification: Around 30-50% of gas is produced. This carbon dioxide produced reduces the pH. Hence alkaline is required to be added.

15. What is the rate limiting step in anaerobic digestion?
a) Methanogenesis
b) Hydrolysis
c) Acidification
d) Biogas production
Answer: a
Clarification: In the case of anaerobic digestion, methanogenesis is the rate limiting step. This results in longer solid retention time. In the case of acid fermentation, hydrolysis is the rate limiting step.

16. What is the volumetric organic loading in Kg COD/ m3 day for a completely mixed anaerobic digestion process?
a) 1-5
b) 5-7
c) 7-9
d) 9-12
Answer: a
Clarification: The volumetric organic loading in Kg COD/ m3 day for a completely mixed anaerobic digestion process is 1-5. This is the loading rate obtained at 30 degree Celsius. This value holds good for suspended growth anaerobic digestion.

17. What is the volumetric organic loading in Kg COD/ m3 day for an anaerobic contact digestion process?
a) 1-8
b) 8-10
c) 10-12
d) 12-15
Answer: a
Clarification: The volumetric organic loading in Kg COD/ m3 day for an anaerobic contact digestion process is 1-8. This is the loading rate obtained at 30 degree Celsius. This value holds good for suspended growth anaerobic digestion.

18. What is the volumetric organic loading in Kg COD/ m3 day for an anaerobic sequence batch reactor?
a) 4-6
b) 1.2-2.4
c) 2.6-3.4
d) 6.4-7.6
Answer: b
Clarification: The volumetric organic loading in Kg COD/ m3 day for an anaerobic sequence batch reactor digestion process is 1.2-2.4. This is the loading rate obtained at 30 degree Celsius. This value holds good for suspended growth anaerobic digestion.

19. What is the hydraulic retention time for a completely mixed anaerobic digestion process?
a) 15-30 days
b) 40-50 days
c) 50-60 days
d) 60-80 days
Answer: a
Clarification: The hydraulic retention time for a completely mixed anaerobic digestion process is 15-30 days. This is dependent on the influent COD. This is mostly assumed as 20 days in most cases.

20. What is the hydraulic retention time for a completely mixed anaerobic contact type process?
a) 8-10 days
b) 5-8 days
c) 0.5-5 days
d) 10-12 days
Answer: c
Clarification: The hydraulic retention time for a completely mixed anaerobic digestion process is 0.5 -5days. This is dependent on the influent COD. This is mostly assumed as 3 days in most cases.

250+ TOP MCQs on Biological Nitrification and Denitrification and Answers

Waste Water Engineering Multiple Choice Questions on “Biological Nitrification and Denitrification”.

1. Nitrification is performed by a small group of _______
a) Autotrophic bacteria
b) Eutrophic bacteria
c) Fungi
d) Viruses
Answer: a
Clarification: Nitrification is an aerobic process performed by small groups of autotrophic bacteria and archaea. This process was discovered by the Russian microbiologist Sergei Winogradsky.

2. In nitrification, ammonia is converted to _____
a) Nitrogen
b) Nitrate
c) Nitrite
d) Amide
Answer: b
Clarification: The first step is nitrification, which is the conversion of ammonia to nitrate through the action of nitrifying bacteria. The second step is the nitrate conversion (denitrification), which is carried out by facultative heterotrophic bacteria under anoxic conditions.

3. Nitrate conversion takes place only through assimilatory cellular functions.
a) True
b) False
Answer: b
Clarification: The first step is nitrification, which is the conversion of ammonia to nitrate through the action of nitrifying bacteria. The second step is the nitrate conversion (denitrification), which is carried out by facultative heterotrophic bacteria under anoxic conditions.

4. In dissimilatory denitrification, ______ serves as the electron acceptor in energy metabolism.
a) Nitrate
b) Nitrite
c) Nitrogen
d) Ammonia
Answer: a
Clarification: In dissimilatory denitrification, nitrate serves as the electron acceptor in energy metabolism and is converted to various gaseous end products but principally molecular nitrogen, N2, which is then stripped from the liquid stream.

5. Microbial yield under aerobic conditions is lower than anoxic conditions.
a) True
b) False
Answer: b
Clarification: The microbial yield under anoxic conditions is considerably lower than under aerobic conditions. Hence a relatively small fraction of the nitrogen is removed through assimilation.

6. Denitrification releases _______
a) Nitrogen
b) Oxygen and nitrogen
c) Carbon-dioxide
d) Nitrogen and carbon-monoxide
Answer: a
Clarification: Denitrification releases nitrogen which escapes as an inert gas to the atmosphere while oxygen released stays dissolved in the liquid and thus reduces the oxygen input needed into the system.

7. How many molecules of oxygen are required for one molecule of nitrogen?
a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: Four molecules of oxygen are required for one molecule of nitrogen during nitrification but releases back 2.5 molecules in denitrification. Thus, theoretically, 62.5% of the oxygen used is released back in denitrification.

8. Which of the below is used to denitrify nitrates?
a) Carbon
b) Aluminium
c) Iron
d) Copper
Answer: a
Clarification: The use of artificial carbon source should be avoided as far as possible since it adds to the cost of treatment and also some operating difficulties may arise from dosing rate of methanol.

9. What is the process for converting organic-nitrogen to ammonia?
a) Hydrolysis
b) Nitrification
c) Denitrification
d) Oxidation
Answer: a
Clarification: Maximum percentage of nitrogen in sewage is converted from nitrogen to ammonia through a process called hydrolysis.

10. What is the minimum amount of dissolved oxygen required for nitrification?
a) 0.5 mg/L
b) 1.0 mg/L
c) 1.5 mg/L
d) 2.0 mg/L
Answer: b
Clarification: Nitrification occurs only under aerobic conditions at dissolved oxygen levels of 1.0 mg/L or more. When DO concentrations are less than 0.5 mg/L, the growth rate is minimal.

11. What is the optimum pH for nitrosomonas?
a) 6.5-7.5
b) 7.5-8.5
c) 9.5-10.5
d) 13.5-14
Answer: b
Clarification: For Nitrosomonas optimum pH is between 7.5 and 8.5 and as well as for Nitrobacter. Treatment plants effectively nitrify with a pH of 6.5 to 7.0. Nitrification stops at a pH below 6.0.

12. What is the minimum temperature required for denitrification process?
a) 3˚C
b) 5˚C
c) 15˚C
d) 30˚C
Answer: b
Clarification: The minimum temperature required for denitrification process is 5˚C. The rates increase with temperature and type of organic source present. When methanol or acetic acid used, the highest growth rate can be found.

13. What is the denitrification rate for the following data?
Temperature: 10 degree Celsius
Dissolved oxygen: 10mg/L
a) 0.038 per day
b) 0.0038 per day
c) 0.38 per day
d) 3.8 per day
Answer: a
Clarification: Overall denitrification rate (ODR)= specific denitrification rate (SDNR)x 0.42 x(1-Dissolved oxygen). SDNR is assumed as 0.1 Kg NO3-N/kg MLVSS per day. ODR = 0.1x 0.42 x (1-0.1) = 0.038 per day.

14. Calculate the residence time for the following data.
Influent nitrate: 22 mg/L
Effluent nitrate: 3 mg/L
Overall denitrification rate: 0.038 per day.
MLVSS: 2000 mg/L
a) 1.5 hrs
b) 5 hrs
c) 6 hrs
d) 9 hrs
Answer: c
Clarification: Residence time: (Influent nitrate – Effluent nitrate)/ ODR x MLVSS. Residence time= (22-3)/(0.038 x 2000)= 19/(0.038 x2000). Residence time= 6 hrs.

15. Which of these chemicals is added in the post anoxic step?
a) Acetate
b) Propanol
c) Butanol
d) Chloroform
Answer: a
Clarification: Acetate is added in the post anoxic step. Sometimes even methanol is added. This is done in order to increase the rate of denitrification.

16. Ammonia is converted into nitrogen by which type of bacteria in anaerobic conditions?
a) Autotrophic
b) Heterotrophic
c) Organotrophs
d) Lithotrophs
Answer: a
Clarification: Ammonia is converted into nitrogen by autotrophic bacteria in anaerobic conditions. Under aerobic conditions ammonia is converted to nitrogen by aerobic bacteria. Organotrophs and lithotrophs are a type of heterotrophic bacteria.

17. What is the temperature at which the denitrification process takes place when it is carried out in a fluidised bed reactor?
a) 30-35 degree
b) 35 degree
c) 40 degree
d) 25 degree
Answer: a
Clarification: The temperature at which the denitrification process takes place is 30-35 degree. This can be done aerobically or anaerobically. Post anoxic tanks have a slower rate of denitrification when compared to post anoxic tanks.

250+ TOP MCQs on Softners and Answers

Waste Water Engineering Multiple Choice Questions on “Softners”.

1. ________ is the removal of calcium, magnesium and certain other metal cations in hard water.
a) Disinfection
b) Sedimentation
c) Softening
d) Cleaning
Answer: c
Clarification: Softening is the removal of calcium, magnesium and certain other metal cations in hard water. The product water that is soft water extends the life of plumbing.

2. Soft water can foul plumbing.
a) True
b) False
Answer: b
Clarification: Hard water leads to fouling of plumbs by building up lime scales and cause galvanic corrosion. In industries, effluent flow from the regeneration process can lead to scale precipitation.

3. ________ form insoluble salts.
a) Aluminium ions
b) Sodium ions
c) Calcium ions
d) Manganese ions
Answer: c
Clarification: The rinse water contains calcium or magnesium ions which form insoluble salts and leaves a coating of insoluble stearates on tub and shower surfaces, commonly called soap scum, in hard water areas.

4. Which of these methods does not remove hardness?
a) Reverse osmosis
b) Ion exchange resins
c) Chelating agents
d) Sedimentation
Answer: d
Clarification: The methods such as reverse osmosis, ion exchange resins and chelating agents are used to remove hardness. Sedimentation can be used to remove suspended solids.

5. Ion exchange resins are ______
a) Inorganic polymers
b) Organic polymers
c) Sandy materials
d) Liquids
Answer: b
Clarification: Ion exchange resins are usually organic polymers. It contains anionic functional groups to which divalent cations bind strongly than monovalent cations.

6. Ion exchange resins contain cation functional groups.
a) True
b) False
Answer: b
Clarification: Inorganic materials called zeolites also exhibit ion-exchange properties. They are the organic polymers and contain anionic functional groups to which divalent cations bind more strongly.

7. _____ is added to hard water to make it softer.
a) Lime
b) Chlorine
c) Hydraulic acid
d) Sulphuric acid
Answer: a
Clarification: Lime softening is the process in which lime is added to hard water to make it softer. It has several advantages over the ion-exchange method but requires full-time, trained personnel to run the equipment.

8. ________ is used in detergents to soften water.
a) Hydraulic acid
b) Sulphuric acid
c) Citric acid
d) Chlorine
Answer: c
Clarification: Citric acid is used to soften water in soaps and laundry detergents. A commonly used synthetic chelator is ethylenediaminetetraacetic acid (EDTA).

9. Which is the commonly used chelator?
a) Rhodamine B
b) EDTA
c) Chlorine
d) Bromine
Answer: b
Clarification: A commonly used synthetic chelator is ethylenediaminetetraacetic acid (EDTA). Chelators are used in chemical analysis, as water softeners and are ingredients in many commercial products such as shampoos and food preservatives.

10. Calculate the amount of softener resin required for the following information.
Magnesium: 80 ppm as CaCO3
Capacity: 47 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs
a) 4.5 m3
b) 3.5 m3
c) 5.1 m3
d) 2.5 m3
Answer: c
Clarification: Magnesium load= Magnesium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Magnesium load/ (Exchange capacity x Correction factor). Resin Volume =[(80 x 135 x 20)/1000]/(47 x 0.9)= 5.1 m3.

11. Calculate the theoretical softener resin bed depth for the following data:
Magnesium Load: 130 ppm as CaCO3
Capacity: 47 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs
Linear velocity: 16.6 m3/h/ m3
Space velocity: 20 m3/h/ m3
a) 1.5 m
b) 1.2 m3
c) 2.6 m
d) 2.5 m
Answer: b
Clarification: Magnesium load= Magnesium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Magnesium load/ (Exchange capacity x Correction factor). Resin Volume = [(130 x 135 x 20)/1000]/(47 x 0.9)= 1.8 m3. Bed depth = Linear Velocity/Space velocity. Bed depth = 20 /16= 1.2 m.

12. Calculate the area of the softener vessel required for the following information.
Magnesium Load: 180 ppm as CaCO3
Capacity: 47 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs
Linear velocity: 16.6 m3/h/ m3
Space velocity: 20 m3/h/ m3
a) 8.5 m2
b) 11.6 m2
c) 9.7 m2
d) 8.5 m3
Answer: b
Clarification: Magnesium load= Magnesium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Magnesium load/ (Exchange capacity x Correction factor). Resin Volume = [(180 x 135 x 20)/1000]/(47 x 0.9)= 11.6 m3. Bed depth = Linear Velocity/Space velocity. Bed depth = 20 /16= 1.2 m. Area = 11.6/1.2= 9.7 m2.

13. Calculate the diameter of the softener vessel required for the below given data.
Magnesium Load: 500 ppm as CaCO3
Capacity: 57 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs
Linear velocity: 16.6 m3/h/ m3
Space velocity: 20 m3/h/ m3
a) 5.2 m
b) 5.0 m
c) 7.6 m
d) 9.8 m
Answer: a
Clarification: Magnesium load= Magnesium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Magnesium load/ (Exchange capacity x Correction factor). Resin Volume =[(500 x 135 x 20)/1000]/(57 x 0.9)= 26 m3 Bed depth = Linear Velocity/Space velocity. Bed depth = 20 /16= 1.2 m. Area = 26 /1.2= 22 m2. The diameter for the vessel is 5.2 m.

14. Calculate the fast rinse flow of the regenerant liquid for a softener with the below given data.
Magnesium Load: 10 ppm as CaCO3
Capacity: 47 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs
a) 4.5 m3 /hr
b) 7.2 m3/hr
c) 1.7 m3 /hr
d) 7.5 m3/hr
Answer: b
Clarification: Magnesium load= Magnesium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Magnesium load/ (Exchange capacity x Correction factor). Resin Volume =[(10 x 135 x 20)/1000]/(47 x 0.9)= 0.6 m3 Velocity is considered as 12-16 m3/hr/m3. In this case we consider 12 m3/hr/m3. Thus the fast rinse flow = Resin volume x velocity = 0.6 x 12 = 7.2 m3/hr.

15. Calculate the regenerant flow required for the softener system with the given below data.
Calcium Load: 85 ppm as CaCO3
Capacity: 37 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs
a) 25 m3/hr
b) 28 m3/hr
c) 12 m3/hr
d) 21 m3/hr
Answer: d
Clarification: Calcium load= Calcium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Calcium load/ (Exchange capacity x Correction factor). Resin Volume =[(85 x 135 x 20)/1000]/(37 x 0.9)= 7.0 m3 Velocity is considered as 3 m3/hr/m3. Thus the regenerant flow = Resin volume x velocity = 7 x 3 = 21 m3/hr.

16. Calculate the rinse volume of the regenerant liquid for the softener system with the given below data.
Calcium Load: 60 ppm as CaCO3
Capacity: 37 Kg/m3
Correction factor: 0.9
Flow rate: 35 m3/hr
Operation hours: 10 hrs
a) 1.5 m3/hr
b) 3.5 m3/hr
c) 4.9 m3/hr
d) 2.5 m3/hr
Answer: b
Clarification: Calcium load= Calcium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Calcium load/ (Exchange capacity x Correction factor). Resin Volume = [(60 x 35 x 10)/1000]/(37 x 0.9)= 0.7 m3 Velocity is considered as 5 m3/hr/m3. Thus the rinse volume = Resin volume x velocity = 0.7 x 5 = 3.5 m3/hr.

17. Calculate the fast rinse period of the regenerant liquid for a softener system with the given below data.
Calcium Load: 40 ppm as CaCO3
Capacity: 37 Kg/m3
Correction factor: 0.9
Flow rate: 235 m3/hr
Operation hours: 10 hrs
a) 25 mins
b) 30 mins
c) 20 mins
d) 15 mins
Answer: a
Clarification: Calcium load= Calcium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Calcium load/ (Exchange capacity x Correction factor). Resin Volume =[(40 x 235 x 10)/1000]/(37 x 0.9)= 3.0 m3 Velocity for fast rinse is considered as 12-16 m3/hr/m3. In this case we consider 12 m3/hr/m3. Thus the fast rinse flow = Resin volume x velocity = 3 x 12 = 36 m3/hr. Velocity for rinse is considered as 5 m3/hr/m3. Thus the rinse volume = Resin volume x velocity = 3 x 5 = 15 m3/hr. Fast rinse period = Total rinse volume/ Fast rinse flow = 15/36 = 0.42 hrs= 25 mins.

250+ TOP MCQs on Collection and Disposal of Dry Refuse and Answers

Waste Water Engineering Multiple Choice Questions on “Collection and Disposal of Dry Refuse”.

1. _________ is the separation of biodegradable waste from non biodegradable waste for proper disposal and recycling.
a) Separation
b) Segregation
c) Removal
d) Composting
Answer: b
Clarification: Segregation is the separation of biodegradable waste from non-biodegradable waste for proper disposal and recycling. Improper segregation may cause mixing in landfills.

2. What is the amount of waste generated in urban India per day?
a) 155500
b) 188500
c) 175800
d) 168500
Answer: b
Clarification: Urban India produces up to 188,500 tonnes of waste per day. Segregation and disposal is a major part of disposing this waste.

3. Proper segregation can lead to mixing in landfills.
a) True
b) False
Answer: b
Clarification: Improper segregation may cause mixing in landfills. Methane gas is likely to be released in such circumstances, which is one of the most harmful greenhouse gases.

4. Methane gas is released due to ______
a) Proper segregation
b) Improper segregation
c) Disposal
d) Dumping
Answer: b
Clarification: Methane gas is likely to be released in such circumstances where improper mixing tales place, which is one of the most harmful greenhouse gases.

5. Garbage is a liquid waste.
a) True
b) False
Answer: b
Clarification: Garbage is not a liquid waste. It is a solid or semi-solid waste from preparing, cooking food and serving, etc. It does not include rubbish.

6. _________ is the solid or semi-solid waste incidental to preparing, cooking, and serving food and cleaning of food service items.
a) Rubbish
b) Waste
c) Garbage
d) Dirt
Answer: c
Clarification: Garbage is classified into two types. They are edible and non-edible. Edible garbage can be fed to animals, like scrap meat and vegetables.

7. _______ consists of wastes originating from various food facilities.
a) Compost
b) Manure
c) Rubbish
d) Garbage
Answer: c
Clarification: Rubbish consists of wastes which originate in food service facilities, barracks, wards, quarters and offices. Rubbish may be classified as combustible or non-combustible depending upon whether or not it can be burned.

8. How many types of rubbish waste are present?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Rubbish may be classified as combustible or non-combustible depending upon whether or not it can be burned.

9. How many types of garbage wastes are present?
a) 2
b) 3
c) 4
d) 5
Answer: a
Clarification: Garbage is classified into two types. They are edible and non-edible. Edible garbage can be fed to animals, like scrap meat and vegetables. Non-edible garbage cannot be fed to animals, like tea leaves, bones, etc.

10. What is the minimum depth of the landfill required?
a) 1.5 m
b) 1.6 m
c) 1.7 m
d) 1.8 m
Answer: d
Clarification: Sites should be selected where the soil can be excavated to a minimum depth of 1.8 m. Clay-like soils or any soil that can form a seal over the compacted refuse can be acceptable.

11. What is the area of land required for 4000 people for disposal of waste?
a) 1.25 hectares
b) 1.5 hectares
c) 1.75 hectares
d) 2.0 hectares
Answer: a
Clarification: There should be at least 1.25 hectares per year for each 4000 personnel when the fill is to be 1.8 meters (m) deep.

250+ TOP MCQs on Sewer Sections and Answers

Waste Water Engineering Multiple Choice Questions on “Sewer Sections”.

1. In India which type of sewer systems are preferred?
a) Partial
b) Combined
c) Direct
d) Separate

Answer: b
Clarification: It becomes uneconomical to lay separate systems for both sanitary sewage and stormwater. Hence a combined system is used in countries like India which is a developing country and also because the rainfall is moderate.

2. Partial sewers carry a part of stormwater to sanitary sewers.
a) True
b) False

Answer: a
Clarification: In partial sewer systems a part of stormwater during rain is allowed to enter sanitary sewer while the remaining stormwater is carried through open drains to the point of disposal.

3. Two sets of sewers are laid in _________ system.
a) Partial
b) Separate
c) Combined
d) Partially combined

Answer: b
Clarification: In separate systems, two sets of sewers are laid. The sanitary sewage is carried through sanitary sewers while the storm sewage is carried through storm sewers. The sewage is carried to the treatment plant and stormwater is disposed of to the river.

4. Which of the following is an advantage of separate sewers?
a) Rivers are polluted
b) Size of the sewers are small
c) It proves economical
d) Nuisance potential is reduced

Answer: b
Clarification: In separate sewer systems, as two sets of sewers are used, the quantity of wastewater flow will also be less in sewers. Hence the size of the sewers will be small. As two sewers are used, it doesn’t prove to be economical.

5. When only one set of sewers are used to carry both sanitary sewage and storm water, it is called as __________ system.
a) Partial
b) Direct
c) Combined
d) Separate

Answer: c
Clarification: In separate systems, two sets of sewers are used to carry sanitary sewage and storm/ surface water separately. In the combined system, a single sewer with a larger size is used to carry both sanitary sewage and surface water together.

6. If sizes of the sewers are __________ choking problems are less.
a) Large
b) Small
c) Big
d) Minute

Answer: a
Clarification: Combined sewer systems consist of only one sewer to carry out both storm water and sanitary sewage. Hence the size of the sewers should be large. Large sewers are easy to clean and avoid choking problems.

7. Which of the following is a disadvantage with respect to the size of the sewers?
a) Cleaning
b) Choking problems
c) Handling and transportation
d) Economical

Answer: c
Clarification: As the combined systems consist of a single large sewer to carry both sanitary sewage and storm water, it is easy to clean and choking problems are avoided. As the size of the sewers is large, it becomes difficult to handle and transport the sewer pipes.

8. In which kind of cities are combined systems used?
a) Less population
b) High population
c) High residences
d) High storm

Answer: a
Clarification: If a city has less population, it consists of fewer residences and industries based on the city needs. Hence the quantity of sanitary sewage produced will also be less. It becomes uneconomical to lay separate sewers for both stormwater and sanitary sewage. Hence combined systems are preferred.

9. In partial systems, there is no over load on the treatment plants.
a) True
b) False

Answer: b
Clarification: In dry weather flow the stormwater is less or absent. During wet weather flow, the quantity of stormwater increases. As in partial systems, a part of stormwater is direct to sanitary sewers, the stormwater puts a load on treatment plants by increasing the quantity of water to be treated.

10. The treatment is costly in which of the following systems?
a) Combined
b) Separate
c) Direct
d) Partial

Answer: a
Clarification: Combined systems consist of both storm water and sanitary sewage. Hence it becomes costly for as the treatment for both are done. In separate systems, the required amount of treatment is done for stormwater and sanitary sewage separately.