Category: WASTE WATER Engineering Objective Questions

250+ TOP MCQs on Hydrogen Sulphide in Sewers and Answers

Waste Water Engineering Multiple Choice Questions on “Hydrogen Sulphide in Sewers”.

1. H2S is intoxicant in nature.
a) True
b) False
Answer: b
Clarification: H2S gas can be toxic even at low concentrations. Under certain conditions H2S gas which can corrode the internal walls of sewers, manholes, pump stations and other concrete and steel structures by getting converted to sulphuric acid.

2. _______ corrodes internal walls of sewers.
a) Hydrochloric acid
b) Sulphuric acid
c) Sulphur dioxide
d) Carbon dioxide
Answer: b
Clarification: Under certain conditions H2S gas which can corrode the internal walls of sewers, manholes, pump stations and other concrete and steel structures by getting converted to sulphuric acid.

3. H2S gas is denser than air.
a) True
b) False
Answer: a
Clarification: H2S gas is denser than air so it may sit at the bottom of maintenance structures such as tanks, wells, enclosures, pits, buildings, storage areas etc.

4. H2S gas is _______ to generate sulphuric acid.
a) Oxidised
b) Reduced
c) Oxidised and reduced
d) Sulphonated
Answer: a
Clarification: H2S gas can be oxidised and converted into sulphuric acid within the sewer headspace on the sewer pipe wall.

5. Hydrogen sulphide is formed under ______ conditions.
a) Aerobic
b) Anaerobic
c) Anaerobic and aerobic
d) Any
Answer: b
Clarification: Hydrogen sulphide is formed under anaerobic conditions at low flow velocities and warm temperatures. The rate of release is increased at points of high turbulence and at the outlets of inverted syphons and pressure mains.

6. What is the flow velocity condition required for the formation of hydrogen sulphide?
a) Low
b) Medium
c) High
d) Rapid
Answer: a
Clarification: Low flow velocity condition is required for the formation of hydrogen sulphide. Sulphuric acid can corrode the internal walls of sewers, manholes, pump stations and other concrete and steel structures.

7. What is the temperature condition required for hydrogen sulphide formation?
a) Warm
b) Cold
c) Extreme cold
d) Intermediate
Answer: a
Clarification: Warm temperature conditions are required for hydrogen sulphide formation. Hydrogen sulphide is formed under anaerobic conditions at low flow velocities and warm temperatures.

8. The rate of release of H2S is high at the points of ______
a) Low velocity
b) Low flow
c) Low turbulence
d) High turbulence
Answer: d
Clarification: The rate of release of H2S is high at the points of high turbulence. Warm temperature conditions are required for hydrogen sulphide formation.

9. What is the exposure limit of H2S in a day?
a) 2 ppm
b) 1 ppm
c) 0.5 ppm
d) 0.1 ppm
Answer: d
Clarification: The exposure limit of H2S in a day is 0.1 ppm. The exposure limit of H2S is 2 ppm for 30 minutes. The exposure limit of H2S is 0.014 for 90 days.

10. What is the limit of H2S exposure for 30 minutes?
a) 0.5 ppm
b) 1 ppm
c) 1.5 ppm
d) 2 ppm
Answer: d
Clarification: The exposure limit of H2S is 2 ppm for 30 minutes. The exposure limit of H2S in a day is 0.1 ppm. The exposure limit of H2S is 0.014 for 90 days.

11. What is the odour threshold limit of H2S gas?
a) 0.0008 ppm
b) 0.008 ppm
c) 0.08 ppm
d) 0.8 ppm
Answer: b
Clarification: The odour threshold limit of H2S gas is 0.008 ppm. 4 ppm causes increased eye problems.

12. What amount of H2S causes increased eye problems?
a) 1 ppm
b) 2 ppm
c) 3 ppm
d) 4 ppm
Answer: d
Clarification: 4 ppm causes increased eye problems. 20 ppm causes neurological effects including memory loss and dizziness.

13. What is the chemical used to prevent H2S gas formation from Magnesium hydroxide?
a) Calcium nitrate
b) Ferric nitrate
c) Oxygen injection
d) Carbon injection
Answer: a
Clarification: Calcium nitrate is the chemical used to prevent H2S gas formation from Magnesium hydroxide. Ferric chloride is the chemical used to prevent H2S gas formation from Sodium hydroxide.

14. What is the chemical used to prevent H2S gas formation from Sodium hydroxide?
a) Ferric chloride
b) Magnesium hydroxide
c) Ferric nitrate
d) Calcium nitrate
Answer: c
Clarification: Ferric chloride is the chemical used to prevent H2S gas formation from Sodium hydroxide. Calcium nitrate is the chemical used to prevent H2S gas formation from Magnesium hydroxide.

15. What is the chemical used to prevent H2S gas formation from Ferric chloride?
a) Magnesium hydroxide
b) Sodium hydroxide
c) Oxygen injection
d) Ferric nitrate
Answer: c
Clarification: Oxygen injection is done to prevent H2S gas formation from Ferric chloride. Calcium nitrate is the chemical used to prevent H2S gas formation from Magnesium hydroxide.

250+ TOP MCQs on Characteristics of Wastewater – 2 and Answers

Waste Water Engineering Questions and Answers for Freshers on “Characteristics of Wastewater – 2”.

1. Turbidity in water is caused by which of these following?
a) Total dissolved solids
b) Suspended solids
c) Ions
d) Heavy metals
Answer: b
Clarification: Suspended solids majorly contribute to the turbidity present in water. This is defined as an expression of optical property that causes light to be scattered and absorbed with no change in the flux level. This is measured in NTU.

2. What is the relationship between turbidity and TSS For a settled secondary effluent?
a) TSS = (2.3) x Turbidity
b) TSS = (2.0) x Turbidity
c) TSS = (2.5) x Turbidity
d) TSS = (1.3) x Turbidity
Answer: a
Clarification: TSS = (2.3) x Turbidity. This relationship holds good for a settled secondary effluent. In case of effluent from a granular filter bed TSS = (1.3 to 1.6) x Turbidity.

3. Which characteristic of water does Calcium contribute to?
a) Hardness
b) Suspended solids
c) pH
d) Colour
Answer: a
Clarification: Hardness is caused by the salts of Calcium and magnesium. The sulphate salts contribute to permanent hardness whereas the carbonate and bicarbonate salts contribute to temporary hardness. The hardness is measured in mg/L.

4. Sodium ions contribute to which characteristic of the water?
a) pH
b) Total Dissolved Solids
c) Colour
d) Suspended solids
Answer: b
Clarification: Sodium ions contribute to Total Dissolved Solids. This is removed by the ion exchange method. In sea water the Sodium chloride is > 25000ppm.

5. What is the role of chlorine in water treatment?
a) To remove hardness
b) To remove ions
c) Coagulant agent
d) To remove bacteria
Answer: d
Clarification: Chlorine is used to remove bacteria from the water. The free chlorine in water should not exceed 2 mg/L. In surface water, the chloride level is around 10 mg/L.

6. Which heavy metal is found in cooling water treatment?
a) Zinc
b) Arsenic
c) Lead
d) Copper
Answer: a
Clarification: Zinc is the heavy metal found in cooling water. Other heavy metals such as arsenic are found in the waste from the pesticide industry. Zinc is also found in wastes dissolved from galvanized pipes.

7. How is hardness represented?
a) Hardness = 2 ( Ca2+) +2( Mg2+)
b) Hardness = Ca 2+ + 2 (Mg 2+)
c) Hardness = 3(Ca 2+) + Mg 2+
d) Hardness = 3(Ca 2+) +2( Mg2+)
Answer: a
Clarification: Hardness is 2 (Ca 2+) +2( Mg 2+). Hardness is contributed by the Magnesium and Calcium salts. It is of two types; permanent and temporary hardness.

8. What is the level of Calcium carbonate for a moderately hard water?
a) Hardness >150 ppm
b) Hardness= 50 ppm
c) Hardness is in between 50-100 ppm
d) Hardness <50 ppm
Answer: c
Clarification: Hardness is in between 50-100 ppm in the case of moderately hard water. In the case of soft water it is less than 50 ppm. In the case of hard water, it is in between 100-150 ppm. In case of very hard water it is greater than 150 ppm.

9. How is TDS and conductivity related?
a) TDS = 0.5 conductivity
b) Conductivity= 0.5 TDS
c) TDS=3 Conductivity
d) TDS=Conductivity
Answer: a
Clarification: TDS = 0.5 Conductivity. This relationship holds good at low TDS. At high TDS values, TDS = 0.9 (slope of line) × electrical conductivity.

10. Colour is contributed to which of the following?
a) Natural organic matter
b) Inorganic compounds
c) Total dissolved solids
d) Suspended solids
Answer: a
Clarification: Natural organic compound contributes to the colour of the water. This is secreted by the metabolic activity of algae, protozoa etc. This imparts a yellowish colour to the water.

11. What will be the Total Organic Carbon (TOC) of coloured water?
a) 100-200ppm
b) 100ppm
c) 200ppm
d) 200-300 ppm
Answer: a
Clarification: The TOC of coloured water would be in the range of 100-200 ppm. The TOC of groundwater is 0.2-2 ppm. The TOC of surface water is 1-20 ppm.

Waste Water Engineering for Freshers,

250+ TOP MCQs on Aerated Grit Chambers and Answers

Waste Water Engineering Multiple Choice Questions on “Aerated Grit Chambers”.

1. In aerated grit chambers, there are no moving parts below the water surface.
a) True
b) False
Answer: a
Clarification: Aerated grit chambers consist of extremely simple design. It consists of no moving parts and air lifting pumps can be used as well as air blowers to pump air. Maintenance is significantly reduced.

2. A grit chamber is usually installed ______ primary sedimentation tanks.
a) Before
b) After
c) In between
d) In
Answer: a
Clarification: A grit chamber is usually installed before primary sedimentation tanks and it is sensible to say that it should be placed well before wastewater pumps. There are basically three different types of grit chambers.

3. At what velocity will the particles settle down?
a) 0.5 m/s
b) 0.3 m/s
c) 0.1 m/s
d) 0.25 m/s
Answer: b
Clarification: The design of a horizontal flow type is such that as wastewater flows through in horizontal direction at a certain velocity (0.3m/s) particles will start to settle at the channel before reaching the outlet point.

4. Grit chamber is constructed to protect the further mechanical equipment of the water treatment plant.
a) True
b) False
Answer: a
Clarification: Grit chamber is needed to protect major mechanical equipment from wear and tear including damage. Other than that, it also ensures that formation of deposit especially in pipelines can be avoided and overall this can save the time that otherwise would be spent to clean the accumulation of waste.

5. Grit chambers are constructed to prevent their accumulation in _______
a) Sedimentation tank
b) Sludge digesters
c) Coagulation tank
d) Sand filter
Answer: b
Clarification: Grit chamber is needed to protect major mechanical equipment from wear and tear including and remove the inorganic particles to prevent damage to the pumps and to prevent their accumulation in sludge digesters.

6. What is the cleaning period for manual grit chambers?
a) 3 days
b) 5 days
c) 48 hours
d) 1 week
Answer: d
Clarification: Manually cleaned grit chambers should be cleaned at least once a week. The simplest method of cleaning is by means of shovel. The grit washing mechanisms are also of several designs most of which are agitation devices using either water or air to produce washing action.

7. Grit chamber is similar to which of the following?
a) Screening unit
b) Sedimentation tank
c) Coagulation tank
d) Sand bed
Answer: b
Clarification: Sedimentation tanks and grit chambers are similar, designed to separate the intended heavier inorganic materials (specific gravity about 2.65) and to pass forward the lighter organic materials.

8. What is the specific gravity of settleable particles in the grit chamber?
a) 2.98
b) 2.56
c) 2.65
d) 2.89
Answer: c
Clarification: The specific gravity of heavier particles is 2.65. Grit chambers are nothing but like sedimentation tanks, designed to separate the intended heavier inorganic materials (specific gravity about 2.65) and to pass forward the lighter organic materials.

9. Stoke’s law holds good for Reynolds’s number ______
a) Less than 1
b) Equal to 1
c) Greater than 1
d) Nearer to 1
Answer: a
Clarification: The settling velocity of discrete particles can be determined using an appropriate equation depending upon Reynolds number. Stoke’s law holds good for Reynolds number, Re below 1.

10. What is the horizontal velocity considered for grit chamber?
a) 0.25-0.4m/sec
b) 0.6-0.9 m/sec
c) 1-2 m/sec
d) 3-4m/sec
Answer: a
Clarification: The horizontal velocity considered while designing a grit chamber is 0.25-0.4m/sec. In most cases, it is considered as 0.3 m/sec. The setting velocity for the removal of 21mm material is 1-1.3m/ sec.

11. What is the detention time considered for aerated grit chamber?
a) 5-6 mins
b) 2-5 mins
c) 1-2 mins
d) 3-4 mins
Answer: b
Clarification: The detention time for aerated grit chambers is 2-5 mins. Usually, in most cases, a detention time of 3 mins is considered. The release of VOC is one application of these aerated grit chambers.

12. Which of these is not a grit removal equipment installed in the aerated grit chamber?
a) Screw conveyers
b) Tubular conveyers
c) Jet pumps
d) Communitors
Answer: d
Clarification: Communitors are used to grind the materials. These are prior to a screen. Screw conveyers, tubular conveyers and jet pumps are used to remove the grit.

13. What is the amount of air supplied for these grit chambers?
a) 1-2 m3/m.min
b) 0.2-0.5 m3/m.min
c) 0.5-0.8 m3/m.min
d) 0.8-1 m3/m.min
Answer: b
Clarification: The air supplied for an aerated grit chamber ranges from0.2-0.5 m3/m.min. There are air diffusers installed in these chambers. These aid in diffusing the air supplied to the chamber.

14. At what depth are the air diffusers placed in the grit chamber?
a) 0.6-0.9 m
b) 1-2 m
c) 0.1-0.4m
d) 0.45-0.6 m
Answer: d
Clarification: The air diffusers are placed at 0.45-0.6m at the chamber. Influent and effluent baffles are also placed at the chamber. These improve the grit removal efficiency.

15. What is the amount of grit quantities captured by the aerated grit chamber?
a) 0.004-0.20 m3/103.m3
b) 0.25 m3/103.m3
c) <0.004 m3/103.m3
d) 0.5 m3/103.m3
Answer: a
Clarification: The amount of grit quantities captured by the aerated grit chamber ranges from 0.004-0.20 m3/103.m3.Usually it is in the range 0.015 m3/103.m3. In terms of US unit the typical value is 2 ft3/Mgal.

16. Usually while designing what is the range of the width of the aerated grit chamber assumed?
a) 0.5-2 m
b) 8-10 m
c) 2.5-7.0m
d) 12m
Answer: c
Clarification: While designing an aerated grit chamber the width is assumed in the range 2.5-7.0m. The ratio of the width: depth is 1:1 to 5:1. Typical ratio is generally assumed as 1.5:1.

17. Usually while designing what is the range of length of the aerated grit chamber assumed?
a) 7.5-20
b) 4-6 m
c) 25 m
d) 50 m
Answer: a
Clarification: While designing an aerated grit chamber the width is assumed in the range 7.5-20 m. The ratio of the length: width is 3:1 to 5:1. Typical ratio is generally assumed as 4:1.

18. Usually while designing what is the range of depth of the aerated grit chamber assumed?
a) 0.5-2 m
b) 8-10 m
c) 2-5 m
d) 12 m
Answer: c
Clarification: While designing an aerated grit chamber the width is assumed in the range 2-5 m. When expressing the same in US unit it is 7-16 ft. The release of VOC by theses aerated grit chambers is one thing which we have to consider and take proper measures to reduce the same.

19. Aerated grit chambers are usually designed to remove particles in which range of size?
a) 0.1mm diameter
b) 0.35 mm diameter
c) 0.21 mm diameter
d) 0.41 mm diameter
Answer: c
Clarification: The aerated grit chambers are designed to remove particles in the range 0.21mm diameter. The cross section of these tanks is similar to that of an activated sludge process tank. The only difference is that these chambers have hoppers of about 0.9m in depth.

250+ TOP MCQs on Tube Settlers and Answers

Waste Water Engineering Multiple Choice Questions on “Tube Settlers”.

1. Tube settlers are used to increase the settling capacity of ______
a) Flocculation unit
b) Clarifier
c) Sand filter bed
d) Flash mixing unit
Answer: b
Clarification: Tube settlers and parallel plates increase the settling capacity of circular clarifiers and/or rectangular sedimentation basins by reducing the vertical distance a floc particle must settle before agglomerating to form larger particles.

2. A floc particle must settle after agglomerating to form larger particles.
a) True
b) False
Answer: b
Clarification: Tube settlers decrease the vertical distance of settling. Hence the floc particles settle faster. A floc particle must settle before agglomerating to form larger particles.

3. What is the angle of slope in tube settlers?
a) 30˚
b) 60˚
c) 90˚
d) 180˚
Answer: b
Clarification: Tube settlers use multiple tubular channels sloped at an angle of 60° and adjacent to each other, which combine to form an increased effective settling area. This provides for a particle settling depth that is significantly less than the settling depth of a conventional clarifier, reducing settling times.

4. The fine flocs escape and cannot b captured by tube settlers.
a) True
b) False
Answer: b
Clarification: Tube settlers capture the settleable fine floc that escapes the clarification zone beneath the tube settlers and allows the larger floc to travel to the tank bottom in a more settleable form.

5. The settling depth provided by tube settlers is _________ than the actual settling depth.
a) Less
b) Greater
c) Equal
d) Nearer
Answer: a
Clarification: The multiple tubular sections combine to form an effective settling area. This provides for a particle settling depth that is significantly less than the settling depth of a conventional clarifier, reducing settling times.

6. What is the angle at which a tube settler is usually inclined at?
a) 45-60 degree
b) 45 degree
c) 60 degree
d) >60 degree
Answer: a
Clarification: The tube settler removes suspended solids. It is classified as a secondary treatment. This is inclined at a position at 45-60 degree.

7. What would happen if the tube settler is inclined greater than 60 degrees?
a) The efficiency will increase
b) The efficiency will decrease
c) Colour will be removed
d) Odour will be removed
Answer: b
Clarification: The tube settler is usually after a flash mixer and flocculator. It is usually inclined at 45-60 degrees. In case it is inclined more than 60 degrees than the efficiency reduces.

8. What would happen if the tube settler is inclined at an angle lesser than 45 degrees?
a) Solids will accumulate in the plates
b) Efficiency Increases
c) It becomes 100 percent efficient
d) You don’t require a flocculator prior to it
Answer: a
Clarification: Usually the tube settlers are inclined at an angle 45-60 degree. This is usually classified as secondary treatment. In case the tube settlers are placed at an angle lesser than 45 degrees then this results in accumulation of solids between the plates.

9. What is the nominal spacing between the plates?
a) 1 inch
b) 2 inch
c) 3 inch
d) 4 inch
Answer: b
Clarification: The nominal spacing between the plates are 2 inches. These are inclined at a length of 2-3 m. To control biological growth, the accumulated solids must be flushed periodically.

10. Tube settlers are operated in which mode?
a) Concurrent
b) Countercurrent
c) Cross flow
d) Hindered settling
Answer: b
Clarification: Tube settlers are operated in a countercurrent mode. The water is passed from the bottom and flows upwards and exists from the basin above the modules. The solids that settle within the tube settle down due to gravity.

11. What is the mathematical representation when many plates are used?
a) V=Q/Nwb
b) V= Q x Nwb
c) V= Q/Nw
d) V=Q x Nw
Answer: a
Clarification: When a number of plates are used V= Q/Nwb. Where V = settling velocity. Q is the flowrate. N is the number of plates. w is the perpendicular distance between the surfaces. b is the dimensions of the surface at right angles to w.

12. Which of these represent the settling velocity of tubesettlers?
a) 11-15 m2/mm3
b) 20-25 m2/mm2
c) 40-45 m2/mm2
d) 30-35 m2/mm2
Answer: a
Clarification: The surface loading rate of tube settlers is 11-15 m2/mm2. Based on this the area required is calculated. Also the number of tube settlers is then calculated.

13. Calculate the area of the tube settlers for the given data:
Flow: 1500 mm3/h
Surface loading rate: 7 m2/m3/h
a) 215 m2
b) 300 m2
c) 400 m2
d) 500 m2
Answer: a
Clarification: Area= Flow rate/ Surface loading rate. This is represented as mm2. Area = 1500/7 = 215 mm2.

250+ TOP MCQs on Sludge Characteristics, Thickening and Concentration – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Sludge Characteristics, Thickening and Concentration – 1”.

1. Sludge is composed of liquid components alone.
a) True
b) False
Answer: b
Clarification: Sewage sludge treatment describes the processes used to manage and dispose of sewage sludge produced during sewage treatment. Sludge is mostly water with lesser amounts of solid material removed from liquid sewage.

2. How many types of sludge are present?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: There are two types of sludge present, primary and secondary sludge. Primary sludge includes settleable solids removed during primary treatment in primary clarifiers.

3. _______ is a term used for reuse of sewage sludge.
a) Solids
b) Biosolids
c) Potential solids
d) Manure
Answer: b
Clarification: “Biosolids” is a term often used in wastewater engineering publications and public relations efforts by local water authorities when they want to put the focus on reuse of sewage sludge after the sludge has undergone suitable treatment processes.

4. Biosolids are the inorganic wastewater solids.
a) True
b) False
Answer: b
Clarification: Biosolids are defined as organic wastewater solids that can be reused after stabilization processes such as anaerobic digestion and composting. The term “biosolids” was introduced by the Water Environment Federation in the U.S. in 1998.

5. Who introduced the term biosolids?
a) UNEPA
b) USEPA
c) EPA
d) WEF
Answer: d
Clarification: The term “biosolids” was introduced by the Water Environment Federation (WEF). Some people argue that a term is a form of “propaganda” with the aim to hide the fact that sewage sludge may also contain substances that could be harmful to the environment when the treated sludge is applied to land.

6. Which year was the term biosolids introduced?
a) 1990
b) 1993
c) 1995
d) 1998
Answer: d
Clarification: The term biosolids was introduced in the year 1998 by the Water Environment Federation (WEF).

7. Which of the following processes is not used for water reduction?
a) Centrifugation
b) Filtration
c) Heating
d) Evaporation
Answer: c
Clarification: Water content of sludge may be reduced by centrifugation, filtration and by evaporation to reduce transportation costs of disposal or to improve suitability for composting. Centrifugation may be a preliminary step to reduce sludge volume for subsequent filtration or evaporation.

8. Sand drying bed is used for filtration.
a) True
b) False
Answer: a
Clarification: Centrifugation may be a preliminary step to reduce sludge volume for subsequent filtration or evaporation. In a sand drying bed or as a separate mechanical process filtration may occur through underdrain in a belt filter press.

9. How many products are obtained after dewatering?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Sludge treatment technologies that are used for thickening or dewatering of sludge have two products and those are the thickened or dewatered sludge and a liquid fraction.

10. The liquid has high contents of phosphorus and nitrogen, if the sludge has been______
a) Dewatered
b) Aerobically digested
c) Anaerobically digested
d) Heated
Answer: c
Clarification: Liquids require treatment as it is high in nitrogen and phosphorus, particularly if the sludge has been anaerobically digested. The treatment can take place in the sewage treatment plant itself or as a separate process.

11. Sludge treatment reduces the formation of struvite scales in pipes.
a) True
b) False
Answer: a
Clarification: For phosphorus recovery, sewage treatment plant operators of treating sludge dewatering streams is that it reduces the formation of obstructive struvite scale in pipes, pumps and valves.

12. Which of these is not a method of digestion?
a) Composting
b) Aerobic digestion
c) Evaporation
d) Anaerobic digestion
Answer: c
Clarification: Sludges are treated using a variety of digestion techniques, the purpose of which is to reduce the amount of organic matter and the number of disease-causing micro-organisms present in the solids. The most common treatment options include anaerobic digestion, aerobic digestion and composting.

13. Anaerobic digestion is carried out in ____
a) Presence of oxygen
b) Presence of carbon-dioxide
c) Absence of oxygen
d) Absence of carbon-dioxide
Answer: c
Clarification: Anaerobic digestion is a bacterial process that is carried out in the absence of oxygen. Sludge is fermented in tanks at a temperature of 55 °C in thermophilic digestion and in mesophilic digestion it is fermented at a temperature of around 36 °C.

14. What is the temperature to be maintained for thermophilic digestion?
a) 44˚C
b) 55˚C
c) 66˚C
d) 77˚C
Answer: b
Clarification: The temperature to be maintained for a thermophilic digestion is 55˚C and the temperature to be maintained for a mesophilic digestion is 36˚C. Thermophilic digestion is more expensive in terms of energy consumption for heating the sludge.

15. What is the duration of sludge in tanks for mesophilic anaerobic digestion?
a) 10 days
b) 12 days
c) 15 days
d) 18 days
Answer: b
Clarification: For treating sludge produced at sewage treatment plants mesophilic anaerobic digestion is a common method. The sludge is fed into large tanks and held for a minimum of 12 days to allow the digestion process to perform the four stages necessary to digest the sludge.

16. Which of the following is not a product of anaerobic digestion?
a) Water
b) Oxygen
c) Carbon dioxide
d) Methane
Answer: b
Clarification: These are hydrolysis, acidogenesis, acetogenesis and methanogenesis. In this process, the complex proteins and sugars are broken down to form more simple compounds such as water, carbon dioxide and methane.

17. Which of the following is an advantage of anaerobic digestion?
a) Capital cost
b) Time
c) Methane generation
d) Space
Answer: c
Clarification: Methane generation is a key advantage of the anaerobic process. Its key disadvantage is the long time required for the process (up to 30 days) and the high capital cost.

250+ TOP MCQs on Phosphorus Removal – 2 and Answers

Waste Water Engineering Assessment Questions and Answers on “Phosphorus Removal – 2”.

1. What is the retention time considered for the anaerobic zone for the AO process for removing phosphorus?
a) 0.5-1.5 hrs
b) 1.5-2 hrs
c) 2-4 hrs
d) 4-8 hrs
Answer: a
Clarification: The retention time considered for the anaerobic zone for the AO process for removing phosphorus is 0.5-1.5 hrs. This process involves aerobic oxidation followed by anaerobic digestion. This is a very effective method to remove phosphorus compounds.

2. What is the retention time considered for the aerobic zone for the AO process for removing phosphorus?
a) 1-3 hrs
b) 3-5 hrs
c) 5-8 hrs
d) 8-11 hrs
Answer: a
Clarification: The retention time considered for the aerobic zone for the AO process for removing phosphorus is 1-3 hrs. This process involves aerobic oxidation followed by anaerobic digestion. This is a very effective method to remove phosphorus compounds.

3. What is the retention time considered for the anaerobic zone for the A2O process for removing phosphorus?
a) 0.5-1.5 hrs
b) 1.5-2 hrs
c) 5-7 hrs
d) 3-4 hrs
Answer: a
Clarification: The retention time considered for the anaerobic zone for the A2O process for removing phosphorus is 0.5-1.5 hrs. This process is a modification process of the AO process. This is a very effective method to remove phosphorus compounds.

4. What is the retention time considered for the aerobic zone for the A2O process for removing phosphorus?
a) 0.5-1.5 hrs
b) 1.5-3 hrs
c) 3-4 hrs
d) 4-8 hrs
Answer: d
Clarification: The retention time considered for the aerobic zone for the A2O process for removing phosphorus is 4-8 hrs. This process is a modification process of the AO process. This is a very effective method to remove phosphorus compounds.

5. What is the retention time considered for the anoxic zone for the A2O process for removing phosphorus?
a) 0.5-1 hrs
b) 1-2 hrs
c) 2-4 hrs
d) 4-6 hrs
Answer: a
Clarification: The retention time considered for the anoxic zone for the A2O process for removing phosphorus is 0.5-1 hrs. This process is a modification process of the AO process. This is a very effective method to remove phosphorus compounds.

6. What is the retention time considered for the anoxic zone for the University of Cape Town treatment process for removing phosphorus?
a) 2-4 hrs
b) 4-6 hrs
c) 6-8 hrs
d) 8-10 hrs
Answer: a
Clarification: The retention time considered for the anoxic zone for the University of Cape Town treatment process for removing phosphorus is 2-4 hrs. The MLSS level maintained is 3000-4000 mg/L. The internal recirculation considered is 200-400%.

7. What is the retention time considered for the anaerobic zone for the University of Cape Town treatment process for removing phosphorus?
a) 1-2 hrs
b) 3-4 hrs
c) 4-8 hrs
d) 8-12 hrs
Answer: a
Clarification: The retention time considered for the anaerobic zone for the University of Cape Town treatment process for removing phosphorus is 1-2 hrs. The MLSS level maintained is 3000-4000 mg/L. The internal recirculation considered is 200-400%.

8. What is the retention time considered for the aerobic zone for the University of Cape Town treatment process for removing phosphorus?
a) 4-12 hrs
b) 3-4 hrs
c) 1-2 hrs
d) 5-6 hrs
Answer: a
Clarification: The retention time considered for the aerobic zone for the University of Cape Town treatment process for removing phosphorus is 4-12 hrs. The MLSS level maintained is 3000-4000 mg/L. The internal recirculation considered is 200-400%.

9. What is the retention time considered for the aerobic zone for the Virginia Initiative Plant treatment process for removing phosphorus?
a) 4-6 hrs
b) 10-12 hrs
c) 2-4 hrs
d) 1-2 hrs
Answer: a
Clarification: The retention time considered for the aerobic zone for the Virginia Initiative Plant treatment process for removing phosphorus is 4-6 hrs. The MLSS level maintained is 2000-4000 mg/L. The internal recirculation considered is 100-200% for the anoxic tank.

10. What is the retention time considered for the aerobic zone for the Virginia Initiative Plant treatment process for removing phosphorus?
a) 3-5 hrs
b) 7-9 hrs
c) 1-2 hrs
d) 4-6 hrs
Answer: c
Clarification: The retention time considered for the anaerobic zone for the Virginia Initiative Plant treatment process for removing phosphorus is 1-2 hrs. The MLSS level maintained is 2000-4000 mg/L. The internal recirculation considered is 100-200% for the anoxic tank.

Waste Water Engineering Assessment Questions,