Category: WASTE WATER Engineering Objective Questions

250+ TOP MCQs on Characteristics of Wastewater – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Characteristics of Wastewater – 1”.

1. The surface water quality is affected by _______ and infiltration from rainfall.
a) Precipitation
b) Run off
c) Wetlands
d) Farming
Answer: b
Clarification: Even when the catchment area is preserved in its natural condition, the surface water quality is affected by run off and infiltration resulting from rainfall. The impact of these is dependant on the contact of the water with the impurities.

2. Function of the intended uses of water can achieve the desired water quality.
a) True
b) False
Answer: a
Clarification: The function of the land used in catchment area shows the existing water quality. The function of the intended uses for the water shows the desired water quality.

3. There is a direct relation between water use and its required quality.
a) True
b) False
Answer: a
Clarification: There is a direct relation between water use and its required quality. The most demanding use can be considered as domestic water supply, which requires the satisfaction of the various quality criteria.

4. _________ are constructed for water supply, electricity generation, recreation, irrigation and others.
a) Swimming pools
b) Ponds
c) Reservoirs
d) Tanks
Answer: c
Clarification: There are multiple uses assigned to water bodies, resulting in the necessity of satisfying diverse quality criteria. Reservoirs are constructed for water supply, electricity generation, etc.

5. What is the next step after water abstraction in wastewater treatment?
a) Supply to consumers
b) Treatment
c) Combine with stormwater
d) Supply to receiving body
Answer: b
Clarification: Initially, water is abstracted from the river, lake or water table and has a certain quality. Hence water is sent for treatment, where it undergoes transformation to be able to comply with intended uses.

6. Which of the following requires aesthetically pleasant water?
a) Domestic use
b) Industrial use
c) Irrigation
d) Aquaculture
Answer: b
Clarification: Industries need aesthetically pleasant water with low turbidity, colour, taste and odour, absence of macro-organisms. The water should be free from organisms harmful to health.

7. The addition of substances or energy forms that directly or indirectly alter the nature of the water body is called?
a) Water contamination
b) Water pollution
c) Sanitation
d) Water treatment
Answer: b
Clarification: Water pollution is the addition of substances or energy forms that directly or indirectly alter the nature of the water body in such a manner that negatively affects its legitimate uses.

8. What is the type of pollution where the pollutants reach the water body in points called?
a) Point-source pollution
b) Diffuse pollution
c) Point-source contamination
d) Diffuse contamination
Answer: a
Clarification: In point-source pollution, the pollutants reach the water body in points concentrated in space. Usually, the discharge of domestic and industrial wastewater generates point-source pollution.

9. What percentage of solids does wastewater contain?
a) 0.5%
b) 5%
c) 0.1%
d) 1%
Answer: c
Clarification: Wastewater contains 99.9% water and 0.1% solids. The goal is to remove this 0.1% of solids. Wastewater is the water generated by industries and communities.

10. What is the growth of huge amounts of algae and other aquatic plants leading to the deterioration of the water quality called?
a) Eutrophication
b) Algae growth
c) Nitrification
d) Denitrification
Answer: a
Clarification: Eutrophication is the growth of huge amounts of algae and other aquatic plants leading to the deterioration of water quality. The untreated wastewater is rich in organic matter which demands oxygen.

11. At higher temperatures, the amount of Dissolved Oxygen decreases.
a) True
b) False
Answer: a
Clarification: At higher temperatures, the amount of Dissolved Oxygen in the water decreases. This is because the rate of biochemical reactions increases as the temperature increases. This results in the depletion of Dissolved Oxygen.

12. The optimum temperature for bacterial activity is in what range?
a) 25-35 degree Celsius
b) 50 degree Celsius
c) 20 degree Celsius
d) 40 degree Celsius
Answer: a
Clarification: The optimum temperature for bacterial activity is 25-35 degree Celsius. Aerobic digestion and nitrification stop when the temperature increases up to 50 degree Celsius. When the temperature drops to 15 degree Celsius methane producing bacteria become inactive.

250+ TOP MCQs on Screens and Comminutors – 2 and Answers

Waste Water Engineering Questions and Answers for Experienced people on “Screens and Comminutors – 2”.

1. With respect to the operation which of these screens is the most expensive?
a) Chain driven
b) Continuous belt
c) Catenary
d) Reciprocating
Answer: b
Clarification: For a continuous belt replacement of screen is a time consuming process. Also this is a very expensive process. Hence the Continuous belt screen is considered the most expensive type of screen.

2. Which of these type of screen can be used both as a fine and coarse screen?
a) Continuous belt
b) Catenary
c) Reciprocating
d) Chain driven
Answer: a
Clarification: A continuous belt type screen can be used both as a fine and coarse screen. This is because the screen openings range from 0.5-30 mm. This is a continuous self- cleaning screen.

3. What is the maximum allowable approach velocity for a bar screen at peak flow rates?
a) >0.9 m/s
b) <0.9m/s
c) 1-2 m/s
d) 2.5m/s
Answer: b
Clarification: The approach velocity at peak flow rates for a bar screen should not exceed 0.9m/s. This is to prevent the pass-through of debris. These screens should be designed in such a way to prevent accumulation of debris or grit.

4. What should be installed in order to control the velocity of the bar screens?
a) Sluice gate
b) Grit chamber
c) Parshall flume
d) Hoppers
Answer: c
Clarification: Parshall flumes are installed at the downstream of a bar screen. These are installed in order to control the approach velocity of a bar screen. The approach velocity is a very important factor while designing the screens to prevent the accumulation of grit and debris.

5. What is the mathematical representation of hydraulic loses through a screen?
a) hL= 1/C(V2/2g)
b) hL= 1/C(V2-v2/2g)
c) hL= 1/C(V2-v2/g)
d) hL= 1/C(V2/g)
Answer: b
Clarification: The hydraulic loss/ head loss is represented as hL= 1/C(V2-v2/2g) .Where hL is the head loss of the screen and C is the empirical discharge coefficient. V is the velocity of the bar screen through the openings, v is the approach velocity in the upstream and g is the acceleration due to gravity (9.81m/s).

6. Which of these screens is used to remove secondary suspended solids?
a) Medium drum rotary screen
b) Fine drum rotary screen
c) Tangential fine screen
d) Horizontal reciprocating screen
Answer: b
Clarification: Fine drum rotary screens are used to remove secondary suspended solids. The mesh is generally made up of stainless steel. The screen cloth would be polyester material.

7. What is the headloss for fine screens?
a) 0.8-1.4m
b) 2-2.5m
c) 3-3.5m
d) <0.8 m
Answer: a
Clarification: The head loss for fine screens would be 0.8-1.4m. In many applications fine screens is limited to plants where head loss through screens is not a problem. Water sprays are supplied in order to keep the screens clean continuously.

8. What is the amount of BOD removed by a fixed parabolic fine screen?
a) 25-40 %
b) 40-50%
c) 5-20%
d) 30-40%
Answer: c
Clarification: The amount of BOD removed by a fixed parabolic fine screen is 5-20%. The amount of TSS is removed 5-30%. Compared to a rotary drum fine screen, the efficiency of these screens is low.

9. How is the headloss for a fine screen calculated?
a) hL= 1/2g(Q/CA)2
b) hL= 1/g(Q/CA)2
c) hL= 1/2g(Q/C)2
d) hL= 1/2g(Q/CA)
Answer: a
Clarification: The headloss for a fine screen is calculated as hL= 1/2g (Q/CA)2. Where hL is the headloss through the fine screen. Q is the flow rate, A is the effective open are of the submerged screen.

10. What is the size of the openings for a microscreen?
a) 35-50µm
b) 10-35µm
c) 50-60µm
d) 60-6µm
Answer: b
Clarification: The size of the openings for a microscreen range from 10-35µm. Microscreens utilize the use of variable low speed. This speed is up to 4r/min.

11. What is the removal efficiency of TSS in case of microscreens?
a) 80-85%
b) 85-90%
c) <10%
d) 10-80%
Answer: d
Clarification: The removal efficiency of the TSS by microscreens is around 10-80%. Usually it is around 55 % in most of the cases. Microscreens can’t handle solid fluctuations.

12. What is the typical headloss through the microscreens?
a) 150-200mm
b) 75-150mm
c) <75mm
d) >200mm
Answer: b
Clarification: The typical headloss through the microscreens is 75-150mm. In case the headloss exceeds 200 mm then bypasss should be provided. The mesh is generally made up of stainless steel.

13. What is the hydraulic loading rate for a microscreen?
a) 1-3 m3/m2.min
b) 7-9 m3/m2.min
c) 3-6 m3/m2.min
d) 10-12 m3/m2.min
Answer: c
Clarification: The hydraulic loading rate for a microscreen is 3-6 m3/m2.min. The hydraulic loading rate is based on the submerged surface area of the drum. The drum diameter is 2-2.5 m.

14. Comminutors are usually installed in which size of water treatment plants?
a) 0.2 m3/sec
b) 3/sec
c) 0.5 m3/sec
d) 1 m3/sec
Answer: b
Clarification: Comminutors are usually installed in plants which are <0.2 m3/sec. These are installed in order to shred material. These require high operating and maintenance cost.

15. What is the headloss through a communitor?
a) 0.3-0.4 m
b) 0.1-0.3 m
c) 0.5-0.6 m
d) 0.6-0.7 m
Answer: b
Clarification: The headloss through the communitor is 0.1-0.3m. In large units it can approach 0.9m. These may be preceded by grit chambers.

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250+ TOP MCQs on Secondary Clarifiers – 2 and Answers

Waste Water Engineering Multiple Choice Questions & Answers on “Secondary Clarifiers – 2”.

1. Which of these require a flash mixer and flocculation prior to it?
a) High rate solid contact clarifier
b) Clariflocculator
c) Secondary clarifier
d) Dissolved aeration flotation
Answer: c
Clarification: The secondary clarifier requires a flashmixer prior to it. It also requires flocculation prior to it. In this type of clarifier only clarification takes place.

2. Which of these require a flash mixer prior to it?
a) High rate solid contact clarifier
b) Clariflocculator
c) Secondary clarifier
d) Dissolved aeration flotation
Answer: b
Clarification: A clariflocculator requires a flash mixer prior to it. A clarifier requires both a flash mixer and flocculator prior to it. Clarifiers are used to remove suspended solids.

3. Which of these don’t require a flash mixer and a flocculator prior to it?
a) High rate solid contact clarifier
b) Clariflocculator
c) Secondary clarifier
d) Dissolved aeration flotation
Answer: a
Clarification: High rate solid contact clarifiers don’t require a flash mixer or a flocculator prior to it. This has both the flocculation zone and flash mixing zone inside. It also has a draft tube separating the flocculation zone and the settling zone.

4. Which of these is the most efficient?
a) High rate solid contact clarifier
b) Primary clarifier
c) Clariflocculator
d) Tube settler
Answer: a
Clarification: High rate solid contact clarifier is the most efficient. It removes around 97% of the suspended solids present in the waste water. The capital cost for the same is very high.

5. While designing a clarifier what is the maximum limit velocity that is assumed?
a) 1.5 m/h
b) 2.5 m/h
c) 3 m/h
d) 3.5 m/h
Answer: a
Clarification: While designing a clarifier, the velocity assumed is 1.5 m/h. Based on this the diameter has arrived. Also the diameter for a clarifier would be greater compared to that of a clariflocculator.

6. While designing a clariflocculator what is the maximum limit velocity that is assumed?
a) 1.5 m/h
b) 2.0 m/h
c) 3 m/h
d) 3.5 m/h
Answer: b
Clarification: While designing a clariflocculator, the velocity assumed is 2.0 m/h. Based on this the diameter arrives. Also, the diameter for a clariflocculator would be greater compared to that of a HRSCC.

7. While designing an HRSCC what is the velocity assumed?
a) 1.5 m/h
b) 2.5 m/h
c) 3 m/h
d) 3.5 m/h
Answer: c
Clarification: While designing an HRSCC, the velocity assumed is 3.0 m/h. Based on this the diameter has arrived. Also the diameter for an HRSCC would be smaller than the other clarifier or clariflocculator.

8. While designing a clariflocculator what is the maximum detention time assumed for the flocculation zone?
a) 40 mins
b) 60 mins
c) 90 mins
d) 120 mins
Answer: a
Clarification: While designing a clariflocculator, the time assumed for the flocculation zone is 40 mins. This usually ranges from 10-40 mins. For a better and efficient process generally, the time is chosen as 40 mins.

9. While designing an HRSCC what is the detention time assumed for the draft tube?
a) 60 secs
b) 90 secs
c) 120 secs
d) 150 secs
Answer: a
Clarification: While designing an HRSCC, the detention time assumed for the draft tube is around 60 secs. Based on the detention time and the flow rate of the water, the volume of the draft tube is calculated. From this and the height, the area is calculated.

10. While designing an HRSCC what is the detention time assumed for the detention hood?
a) 60 secs
b) 90 secs
c) 120 secs
d) 150 secs
Answer: c
Clarification: While designing an HRSCC, the detention time assumed for the detention hood is around 120 secs. Based on the detention time and the flow rate of the water, the volume of the detention hood is calculated. From this and the height, the area is calculated.

11. While designing an HRSCC what is the Solid loading rate assumed for the Clarification zone?
a) 1 m/h
b) 2 m/h
c) 0.75 m/h
d) 0.9 m/h
Answer: d
Clarification: While designing an HRSCC, the solid loading rate for the clarification zone is assumed as 0.9 m/h. Area of the clarification zone = Flow/Solid loading rate. From the area, the diameter for the clarification zone is calculated.

12. Calculate the diameter of the draft tube for the following data.
Flow: 1100 m3/h.
Height of the draft tube: 1.5 m
a) 4 m
b) 5 m
c) 6 m
d) 7 m
Answer: a
Clarification: The volume is calculated by multiplying flow x time. (1100 x 60/3600=18.33). The volume is divided by the height to obtain the area. Square root (area x 1.27) =3.98 (rounded off to 4m).

13. Calculate the clarifier volume for a HRSCC for the following data.
Flow: 500 m3/h.
a) 40 m2
b) 90 m2
c) 75 m2
d) 100 m2
Answer: c
Clarification: The retention time is considered as 90 mins for the flocculation zone. The volume is calculated by multiplying retention time and flow. The area is then calculated by dividing the volume by the height. From the area the diameter can be calculated.

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

Waste Water Engineering Question Bank on “Sludge Treatment Process – 2”.

1. Out of the following methods which provide uniform dewatering?
a) Belt filter press
b) Filter press
c) Sludge bed
d) Centrifuge
Answer: a
Clarification: Belt filter press results in uniform dewatering. This prevents clogging of the solids. This also prevents warping of rollers.

2. The usage of which type of pump results in less wear and tear while pumping the sludge?
a) Vertical turbine pumps
b) Submersible pump
c) Progressive cavity pump
d) Non-clog pump
Answer: c
Clarification: By using progressive cavity pumps to pump the sludge wear and tear can be reduced. This is also one method to reduce the clogging of solids. Thus mostly progressive cavity pumps are preferred over other pumps in order to pump sludge.

3. The cyclone gritter removes the grit particles based on which of the below mentioned phenomenon?
a) Centrifugal force
b) Gravity
c) Brownian motion
d) Sedimentation
Answer: a
Clarification: The cyclone gritter removes grit particles based on centrifugal force. The heavier particles move to the outside of the cylinder section. These are then discharged through the conical section.

4. Primary sludge constitutes which of the following?
a) Chemical sludge
b) Settlable solids
c) Biological solids
d) Biological and settlable solids
Answer: b
Clarification: Primary Sludge constitutes of settleable solids. This is mostly returned to the primary clarifier. This is done in order to carry out mixing to provide a uniform mixture.

5. Secondary sludge constitutes which of the following?
a) Chemical sludge
b) Settlable solids
c) Biological solids
d) Biological and settlable solids
Answer: d
Clarification: The secondary sludge constitutes biological sludge. This also constitutes settleable solids. The secondary sludge is sent to the primary clarifier, where mixing is carried out in order to provide an uniformity.

6. Advanced waste water sludge constitutes which of the following?
a) Chemical sludge
b) Settlable solids
c) Biological solids
d) Chemical and biological solids
Answer: d
Clarification: The advanced waste water sludge constitutes biological sludge. This also constitutes chemical sludge. The advanced waste water sludge is sent to the primary clarifier, where mixing is carried out in order to provide an uniformity to the mixture.

7. To remove 1% primary sludge which of the following size of mesh of a cyclone gritter is used?
a) 150
b) 100
c) 300
d) 50
Answer: a
Clarification: To remove the 1% primary sludge usually a 150 sized mesh of the cyclone degritter is used. In these cyclone degritters the sludge is fed tangentially. This is the most effective way to degrit sludge.

8. To remove 2% primary sludge which of the following size of mesh of a cyclone gritter is used?
a) 150
b) 100
c) 65
d) 28-35
Answer: b
Clarification: To remove the 2% primary sludge usually a 100 sized mesh of the cyclone degritter is used. Prior to thickening of sludge, degritting should be carried out. After degritting the sludge is stored before it is subjected to thickening.

9. To remove 3% primary sludge which of the following size of mesh of a cyclone gritter is used?
a) 100
b) 150
c) 65
d) 28-35
Answer: c
Clarification: To remove the 3% primary sludge usually a 65 sized mesh is used. After degritting, the sludge is stored for not more than 4-5 hours. Then the sludge is thickened using a centrifuge, gravity belt etc.

10. Which of the following methods is used for stabilization of the sludge?
a) Anaerobic digestion
b) Sludge drying bed
c) Centrifuge
d) Filter press
Answer: a
Clarification: The sludge is stabilized by anaerobic digestion. Lime conditioning and aerobic digestion are the other methods in order to bring about stabilization. The sludge is stabilized before it is subjected to conditioning.

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250+ TOP MCQs on Phosphorus Removal – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Phosphorus Removal – 1”.

1. Phosphorus is one of the responsible factors for eutrophication.
a) True
b) False
Answer: a
Clarification: Controlling phosphorous discharged from municipal and industrial wastewater treatment plants is a key factor in preventing eutrophication of surface waters.

2. What is the percentage of sludge in phosphorus removal by chemical precipitation?
a) 10%
b) 20%
c) 30%
d) 40%
Answer: d
Clarification: Phosphate removal is currently achieved largely by chemical precipitation, which is expensive and causes an increase in sludge volume by up to 40%.

3. What is the amount of phosphorus present in municipal wastewater?
a) 1-2 mg/L
b) 2-8 mg/L
c) 4-12 mg/L
d) 5-20 mg/L
Answer: d
Clarification: Municipal wastewaters may contain from 5 to 20 mg/l of total phosphorous, of which 1-5 mg/l is organic and the rest in inorganic. The individual contribution tends to increase because phosphorous is one of the main constituents of synthetic detergents.

4. Phosphorus is the main component of synthetic detergents.
a) True
b) False
Answer: a
Clarification: The individual contribution tends to increase because phosphorous is one of the main constituents of synthetic detergents. The individual phosphorous contribution varies between 0.65-4.80 g/inhabitant per day.

5. What is the average phosphorus contribution per person per day?
a) 1.2 g
b) 2.18 g
c) 3.26 g
d) 4.25 g
Answer: b
Clarification: The individual phosphorous contribution varies between 0.65 and 4.80 g/inhabitant per day with an average of about 2.18 g.

6. What is the phosphorus available for a biological process called?
a) Phosphate
b) Orthophosphate
c) Polyphosphate
d) Biophosphorus
Answer: b
Clarification: Orthophosphates are available for biological metabolism without a further breakdown. Molecules with two or more phosphorous atoms, oxygen and in some cases hydrogen atoms combine in a complex molecule to form polyphosphates.

7. ________ undergo hydrolysis and revert to the orthophosphate forms.
a) Biophosphorus
b) Orthophosphate
c) Phosphate
d) Polyphosphate
Answer: d
Clarification: Molecules with 2 or more phosphorous atoms, oxygen and in some cases hydrogen atoms combine in a complex molecule. Usually, polyphosphates undergo hydrolysis and revert to the orthophosphate forms. This process is usually quite slow.

8. How much amount of phosphorus is removed by secondary treatment?
a) 0.2-0.3 mg/l
b) 0.4-0.6 mg/l
c) 0.5-1 mg/l
d) 1-2 mg/l
Answer: d
Clarification: Normally secondary treatment can only remove 1-2 mg/l, so a large excess of phosphorous is discharged in the final effluent, causing eutrophication in surface waters.

9. ______ is present in EBPR implementations.
a) Aeration tank
b) Anaerobic tank
c) Aerobic and anaerobic tank
d) No tanks
Answer: c
Clarification: Both aerobic and anaerobic is present in EBPR implementations. Enhanced biological phosphorus removal is applied to activated sludge systems for the removal of phosphate.

10. What does PAO stand for?
a) Poly-ammonium oxygenate
b) Polyphosphate-accumulating organisms
c) Poly-ammonium organisms
d) Phosphate-ammonium organisms
Answer: b
Clarification: A group of heterotrophic bacteria, called polyphosphate-accumulating organisms (PAO) are selectively enriched in the bacterial community within the activated sludge.

11. What is the percentage of phosphorus present in bacterial biomass?
a) 0.2%
b) 0.3%
c) 0.8%
d) 1%
Answer: d
Clarification: The percentage of phosphorous is 1-2 in all the bacterial biomass. A fraction of phosphorous is present in all the bacterial biomass.

250+ TOP MCQs on Ion Exchange Media Properties – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Ion Exchange Media Properties – 1”.

1. Amberjet 4400 has a total capacity of ________ in the chlorine form.
a) 0.5 eq/L
b) 1 eq/L
c) 1.5 eq/L
d) 2 eq/L
Answer: c
Clarification: Amberjet 4400 has a total capacity of about 1.5 eq/L in the Cl— form and 1.2 eq/L in the OH— form. The difference in the capacity is due to the swelling of the resin.

2. What is the percentage of swelling between the chlorine and the hydroxide forms?
a) 10%
b) 20%
c) 30%
d) 40%
Answer: c
Clarification: The difference in the total capacity of Amberjet 4400 is just due to resin swelling, it swells by up to 30 % between the Cl— and the OH— form.

3. The number of active groups in a resin sample fluctuates.
a) True
b) False
Answer: b
Clarification: The number of active groups in any of the resin is a constant. When the resin swells up, the density of the active groups decreases. Volume capacity is a measure of active group density.

4. ______ is the value of the sieve opening through which exactly 50 % of the resin volume passes.
a) Mean diameter
b) Effective size
c) Harmonic mean size
d) Uniformity coefficient
Answer: a
Clarification: The value for sieve opening is given by mean diameter. Exactly 50% of the resin volume passes through the mean diameter. It is abbreviated as “d50“.

5. Mean diameter is also called as ______
a) Actual diameter
b) Radius
c) Median diameter
d) Effective size
Answer: c
Clarification: Mean diameter is also called as median diameter. In the language of statistics, it is called the median diameter.

6. What is the percentage of resin volume considered for the mean diameter?
a) 20%
b) 30%
c) 40%
d) 50%
Answer: d
Clarification: The mean diameter is the value of the (theoretical) sieve opening through which exactly 50 % of the resin volume passes, i.e. that retains exactly 50 % of the resin sample.

7. What is the percentage of resin sample that is considered for effective size?
a) 10%
b) 20%
c) 30%
d) 40%
Answer: a
Clarification: The effective size is the value of the sieve opening through which exactly 10 % of the resin sample passes. It is abbreviated as “d10“.

8. Total capacity is the number of exchangeable bi-valent ions.
a) True
b) False
Answer: b
Clarification: Total capacity is the number of active groups i.e., the number of exchangeable monovalent ions. Both volume and dry weight capacity values must be reported.

9. ______ is the number of sites where exchange actually takes place during one cycle.
a) Total capacity
b) Operating capacity
c) Mean capacity
d) Resin capacity
Answer: b
Clarification: Operating capacity is the number of sites where exchange actually takes place during one cycle. Total capacities of new resins are measured for quality control.

10. Which of these is not a result of high moisture content?
a) Fast exchange
b) Good adsorption properties
c) Low total capacity
d) High total capacity
Answer: d
Clarification: The dry weight capacity indicates if a resin has been properly functionalised, irrespective of its water content. Whilst a high total capacity is generally desirable, not all exchange sites are used in a complete ion exchange cycle.

11. Which of these is not a result of low moisture content?
a) low total capacity
b) difficult to regenerate
c) no removal of big ions
d) tendency to fouling
Answer: a
Clarification: The low moisture content leads to the tendency of fouling, difficult to regenerate and there will be no removal of bigger ions.