Category: WASTE WATER Engineering Objective Questions

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.

250+ TOP MCQs on Correlation of Mass Transfer with Water Treatment and Answers

Waste Water Engineering Multiple Choice Questions on “Correlation of Mass Transfer with Water Treatment”.

1. Which among the following reactors has important advantages?
a) Bubble column bioreactors
b) Fluidized bed bioreactors
c) Cylindrical orbitally shaken bioreactors
d) Continuous stirred tank bioreactors
Answer: c
Clarification: Among disposable bioreactor systems, cylindrical orbitally shaken bioreactors show important advantages. They provide a well-defined hydrodynamic flow combined with excellent mixing and oxygen transfer for mammalian and plant cell cultivations.

2. __________ is the net movement of mass from one location or component to another.
a) Osmosis
b) Mass transfer
c) Dilution
d) Reverse osmosis
Answer: b
Clarification: Mass transfer is the net movement of mass from one location, usually meaning stream, phase, fraction or component, to another. In processes like precipitation, membrane filtration, and distillation mass transfer takes place.

3. Which of the following do not involve mass transfer in terms of water treatment?
a) Clarification
b) Aeration
c) Air Stripping
d) Adsorption
Answer: a
Clarification: Clarification is the process in which the suspended solids are removed. This is done by sedimentation. Flocculants such as alum is added to obtain flocs. Then this settled down. The clear supernatant flows into the next treatment channel.

4. What is the flux not dependent on?
a) Pressure
b) Mass solute
c) Area
d) Time
Answer: a
Clarification: The flux is not dependent on the pressure. Flux is defined as the mass of solute transported through an area per unit time. Mass flow is the product of flux multiplied by area.

5. In which of the following cases the flux is considered as volumetric flux?
a) Desalination/Reverse Osmosis
b) Ion exchange
c) Adsorption
d) Aeration
Answer: a
Clarification: In desalination/ reverse osmosis, the flux is termed as volumetric flux. The material moving across the membrane is measured in terms of volume. The corresponding flux is termed as volumetric flux.

6. In water treatment what is the driving force due to which mass transfer occurs?
a) Gibbs Energy
b) Electric potential
c) Gravity
d) Pressure
Answer: a
Clarification: Gibbs Energy is the cause due to which the mass transfer occurs. The Gibbs Energy is also known as the concentration gradient. When there is a concentration gradient present between the phases there will be mass transfer from a higher concentration gradient to a lower one.

7. In case of aqueous solution which of the following mathematical formula holds good?
a) J = Na
b) J = Na x Xa
c) J = (1-1/xa) X Na
d) J = 1/xa X Na
Answer: a
Clarification: In aqueous solution, the solute is considered to be negligible. Hence J = Na holds well. In general cases J = Na(1/1-Xa).

8. How is the value for displacement value for solute calculated?
a) Vb= Dt/ x
b) Vb= 2 Dt/ x
c) Vb= Dt2/ x
d) Vb= 2 Dt2/ x
Answer: b
Clarification: Vb = 2 Dt/ x. Where Vb= Brownian velocity. Dt is dffussion coefficient in liquid phase. X is the displacement value for solute.

9. Which of these do not relate the mass transfer coefficient and diffusion coefficient?
a) Reynolds number
b) Schmidt number
c) Sherwood number
d) Boltzamn constant
Answer: d
Clarification: Sherwood number relates the mass transfer coefficient and diffusion coefficient. kf(avg)L /Df =0.664Re1/2 Sc1/3. Where kf(avg) is the mass transfer coefficient, Re is the Reynolds number and Sc is the Schmidt number.

10. Which of these is used to analyse the minimum amount of extraction phase for treatment?
a) Operating Diagram
b) Freundlich Isotherm
c) Langmuir Isotherm
d) Isobar graphs
Answer: a
Clarification: Operating diagrams are used to analyse the amount of extraction phase for treatment. For example, in case of adsorption the activated carbon amount is calculated from the V/M slope. It was found that at greater V/M slope value treatment is feasible.

11. What is the mass transfer formula for a plug flow reactor?
a) q = Ms(Co)
b) q = Q (Co-C)
c) q = Q Ms(Co-C)
d) q = Q MsC
Answer: c
Clarification: q = Q Ms(Co-C).Where q is concentration of solute adsorbed. Q is the flow rate. Ms is the feed rate of the solute.

12. The liquid phase controls the rate of mass transfer for compounds which have an H value of which of the following?
a) H>0.05
b) H=0.02
c) H<0.002
d) H=1
Answer: a
Clarification: The liquid phase controls the rate of mass transfer for compounds which have an H value >0.05. The gas phase controls the rate of mass transfer for compounds which have an H value <0.002. For compounds with H value between 0.002 and 0.05, both liquid and gas control the rate of mass transfer.

250+ TOP MCQs on Sedimentation Tanks – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Sedimentation Tanks – 1”.

1. Sedimentation is a process using gravity to remove suspended solids from water.
a) True
b) False
Answer: a
Clarification: Sedimentation is a physical water treatment process using gravity to remove suspended solids from water. Solid particles entrained by the turbulence of moving water may be removed naturally by sedimentation in the still water of lakes and oceans.

2. What are the ponds constructed for removing entrained solids called?
a) Clarifiers
b) Settling basins
c) Eco-ponds
d) Ditches
Answer: b
Clarification: Settling basins are ponds constructed for the purpose of removing entrained solids by sedimentation. Clarifiers are tanks built with mechanical means for continuous removal of solids being deposited by sedimentation.

3. The tanks built with mechanical means for continuous removal of solids being deposited by sedimentation are called _________
a) Clarifiers
b) Settling basins
c) Sedimentation tanks
d) Eco-pons
Answer: a
Clarification: Clarifiers are tanks built with mechanical means for continuous removal of solids being deposited by sedimentation. Settling basins are ponds constructed for the purpose of removing entrained solids by sedimentation.

4. Sedimentation tank is also called as settling tank.
a) True
b) False
Answer: a
Clarification: Sedimentation tank, also called settling tank or clarifier is a component of a modern system of water supply or wastewater treatment. The suspended particles settle down thereby providing some degree of purification.

5. What is the accumulated layer at the bottom of the tank called as?
a) Flocs
b) Sediment
c) Sludge
d) Sewage
Answer: c
Clarification: A layer of accumulated solids called sludge forms at the bottom of the tank and is periodically removed. Prior to sedimentation, coagulants are added to the water in order to facilitate the settling process, which is followed by filtration and other treatment steps.

6. How many types of sedimentation tanks are there based on the method of operation?
a) 2
b) 4
c) 5
d) 7
Answer: a
Clarification: Based on the method of operation, there are two types of sedimentation tanks. The first one is filled and draw type tank and the second one is the continuous flow type tank.

7. How many types of sedimentation tanks are present based on the shape of the tank?
a) 2
b) 3
c) 4
d) 5
Answer: a
Clarification: Based on the shape of the sedimentation tank, there are two types of sedimentation tank. They are circular sedimentation tank and rectangular sedimentation tank.

8. What is the time period for which the water is stored in fill and draw type of sedimentation tank?
a) 48 hours
b) 24 hours
c) 52 hours
d) 76 hours
Answer: b
Clarification: In case of fill and draw type sedimentation tank, water from the inlet is stored for some time. The time may be 24 hours. The suspended particles settle down. After 24 hours, the water is discharged through the outlet.

9. How many hours does the removing of Settleable particles require in fill and draw type of sedimentation tanks?
a) 3-6
b) 6-9
c) 6-12
d) 12-15
Answer: c
Clarification: Then settled particle is removed. This removal action requires 6-12 hours. 30-40 hours is required for one complete action in case of fill and draw type sedimentation tank.

10. In which of the following sedimentation tanks, the flow may be either horizontal or vertical?
a) Circular
b) Rectangular
c) Fill and draw type
d) Continuous flow
Answer: d
Clarification: In the case of continuous flow type, water is not allowed to rest. Flow always takes place but with a very small velocity. During this flow, suspended particles are settling at the bottom of the tank. The flow may be either in a horizontal direction or vertical direction.

11. What is the maximum permissible velocity in horizontal flow type sedimentation tank?
a) 0.2 m/s
b) 0.3 m/s
c) 0.5 m/s
d) 1 m/s
Answer: b
Clarification: Horizontal flow type sedimentation tanks generally in rectangular shape. They have more length twice its width. Because they need to flow more distance to settle all suspended particles. The maximum permissible velocity in this case is 0.3m/sec.

12. The vertical type of sedimentation tanks are usually in _________ shape.
a) Circular
b) Rectangular
c) Semi-circular
d) Cylindrical
Answer: a
Clarification: The vertical flow type sedimentations tanks are generally in circular shape and flow takes place in vertical direction. The flow is usually horizontal generally in rectangular tanks.

13. Why are the baffle walls provided?
a) Prevent short circuit
b) Abstruct flow
c) Reduce velocity
d) Collect sediment
Answer: a
Clarification: The sedimentation tanks preferred are rectangular and are used widely. The flow takes place in horizontal direction that is length wise in rectangular tanks. Sometimes baffle walls are provided for rectangular tank to prevent short circuiting.

14. Sludge is disposed through _______
a) Hopper bottom
b) Sludge pump
c) Deflector
d) Launder
Answer: b
Clarification: A deflector box located at the top. This deflects the influent from central pipe downwards in case of hopper bottom tanks. Sludge is collected at the bottom and it is disposed of through sludge pump.

15. ________ is provided after activated sludge process.
a) Primary sedimentation tank
b) Flocculation tank
c) Secondary sedimentation tank
d) Sand filter
Answer: c
Clarification: The suspended particles present in the wastewater contains microbes and are removed and reflected towards aerobic filter to maintain high microbe concentration in an aerobic filter.