Category: WASTE WATER Engineering Objective Questions

250+ TOP MCQs on Eco-Friendly Toilets and Answers

Waste Water Engineering Multiple Choice Questions on “Eco-Friendly Toilets”.

1. What is the amount required to build a concrete septic tank?
a) Rs 30,000
b) Rs 60,000
c) Rs 80,000
d) Rs 1,00,000
Answer: a
Clarification: Low water table means pit toilets, constructed under the government’s sanitation programme, do not work. Water collected in the pit cannot percolate down because the soil is already saturated with water. Rs 30,000-40,000 is needed to build a concrete septic tank, where water flows out into a drain instead of percolating down.

2. What is the population that lacks access to sanitation in India?
a) 100 million
b) 200 million
c) 400 million
d) 600 million
Answer: d
Clarification: In India, nearly 600 million people lack access to adequate sanitation, increasing the risk of groundwater contamination. According to the World Bank, diseases like diarrhea can kill approximately 800,000 children, under age five, every year and leaves millions malnourished and stunted.

3. What is the expected year to achieve 100% sanitation?
a) 2010
b) 2015
c) 2020
d) 2022
Answer: d
Clarification: In an effort to address this absence of sanitation, which is especially acute in rural areas that lack access to water and sewage infrastructure, the Indian government has set the target of 100% sanitation by 2022 and is working to boost cleanliness and end open defecation in the country.

4. What is the power of solar panels used in glaciers?
a) 200 watt
b) 240 watt
c) 280 watt
d) 350 watt
Answer: b
Clarification: In glaciers, temperature is around -40°C. Toilets are fitted with solar panels of 240 watt to keep the excreta warm for processing.

5. What is the average amount of water required for normal flushing?
a) 6 litres
b) 9 litres
c) 12 litres
d) 15 litres
Answer: b
Clarification: The toilet flushes itself before and after every use, using a minimum amount of water that is determined through sensors: On an average, each flush uses 1.5 litres of water, compared to the 8-10 litres used by a normal flush.

6. What is the number of chambers in UDDT toilets?
a) 2
b) 4
c) 6
d) 8
Answer: a
Clarification: EcoSan or UDDT toilets make great individual toilets for areas where digging the ground is highly complicated. UDDT is built above the ground level using conventional bricks or hollow blocks. It has two chambers – the urine, faeces and cleansing water go into separate holes.

7. What is the duration of use of each UDDT chamber?
a) 10 months
b) 12 months
c) 14 months
d) 16 months
Answer: b
Clarification: UDDT has two chambers – the urine, faeces and cleansing water go into separate holes. The floor of the chamber is also paved with concrete to prevent water or soil coming into contact with the faeces. Each chamber will be used for about 12 months alternatively.

8. What is the maximum amount of water consumption per flush to certify toilets in North America?
a) 1.5 gallons
b) 1.6 gallons
c) 1.8 gallons
d) 2.0 gallons
Answer: b
Clarification: In Canada, there is the Canadian Standards Association (CSA) and the US has the American National Standards Institute (ANSI) as well as the American Society of Mechanical Engineers (ASME). These associations set the standard for water consumption in North America and will only certify toilets which meet their requirements of 6 Litres or 1.6 gallons of water per flush.

9. Flush toilets work well because of a part called “the S bend” or “the siphon”.
a) True
b) False
Answer: a
Clarification: There is a part in flush toilets called “the S bend” or “the siphon”. This part is located within the hole at the bottom of the toilet bowl. This tube is shaped like an S and connects to the sewer plumbing under the toilet.

10. Flush toilets are said to be more sanitary than squat toilets.
a) True
b) False
Answer: b
Clarification: Squat toilets are said to be more sanitary than flush toilets as there is no contact with the toilet and it is also said to be less expensive and easier to clean and maintain.

11. What is the percentage of the world lacking access to toilets?
a) 23 %
b) 35 %
c) 42 %
d) 56 %
Answer: b
Clarification: Infrastructure and cost of conventional toilets are high. Conventional toilets are not suitable for developing countries. 35% of the world’s population lacks access to toilets.

250+ TOP MCQs on Dry Weather Flow & Variations in Rate of Sewage and Answers

Waste Water Engineering Multiple Choice Questions on “Dry Weather Flow & Variations in Rate of Sewage”.

1. What is the wastewater flow in a sewer system during the periods of dry weather with minimum infiltration called?
a) Wet weather flow
b) Dry weather flow
c) Wet flow
d) Dry flow
Answer: a
Clarification: Summers are usually dry. Hence the rainfall is minimum and is called as dry weather flow in sewers. During the rainy season or periods of high rainfall or storm, the sewers will be full and are called wet weather flow.

2. In order to determine the section of the sewer, it is essential to know the total quantity of the sewage that would flow through the sewer.
a) True
b) False
Answer: a
Clarification: The quantity of sewage that would flow in dry weather is different from the quantity of the sewer that would flow in the wet weather. Hence peak conditions must be considered and sewer size should be determined.

3. The dry weather flow is also called as sanitary sewage.
a) True
b) False
Answer: a
Clarification: Dry weather flow mainly consists of domestic sewage and industrial wastewater as the rainfall or storm is absent or less in the times of summer of dry weather. Hence dry weather flow is also called as sanitary sewage.

4. Which of the following factors does the dry weather does not depend on?
a) Rate of water supply
b) Population growth
c) Infiltration of groundwater
d) Design of sewer system
Answer: d
Clarification: Dry weather flow depends on the type of area served, rate of water supply, population growth, infiltration of groundwater. Design of sewer system is to determine the size of a sewer system in peak conditions. It has nothing to do with the factors responsible for dry weather flow.

5. Which of the following common unit is used to express the rate of water supply to a city or town?
a) Litres/capita/second
b) Litres/capita/day
c) Millilitres/capita/day
d) Millilitres/capita/hour
Answer: b
Clarification: The common unit used to express the rate of water supply to a city or town is in terms of litres per person for a day. Capita refers to each person or an individual. Hence the unit is litres/capita/day.

6. The quantity of sanitary sewage entering the sewers would be ___________ the total quantity of water supplied.
a) Less than
b) Equal to
c) More than
d) Greater than
Answer: a
Clarification: The quantity of sanitary sewerage entering the sewers would be less than the total quantity of water supplied because the water is lost in domestic consumption, evaporation, lawn sprinkling, fire fighting and industrial consumption.

7. The quantity of sanitary sewage directly depends on ______
a) Rate of water supply
b) Area
c) Population
d) Precipitation
Answer: c
Clarification: As the population increases the rate of water consumption also increases which leads to the direct increase in sanitary sewage. This also increases water demand. Hence the quantity of sanitary sewage directly depends on the population.

8. The quantity of water supply is equal to the rate of supply multiplied by the ______
a) Population
b) Quantity of sewage
c) Area
d) Net usage
Answer: a
Clarification: The rate of usage of water varies from individual to individual and different sectors. Hence the total quantity of water supplied is equal to the rate of supply multiplied by population.

9. Wet weather flow is _________ than the dry weather flow.
a) Lower
b) Higher
c) Equal to
d) Not equal to
Answer: a
Clarification: Wet weather flow is higher than the dry weather flow because of the inflow and infiltration in the sewer system. Wet weather flow is the combination of stormwater and sanitary sewage but dry weather flow consists of only sanitary sewage.

10. Dry weather flow+ storm water=_________
a) Dry weather flow
b) Stormwater
c) Wet weather flow
d) Sanitary sewage
Answer: c
Clarification: Dry weather flow is the wastewater flow in sewers during dry weather. Dry weather flow in combination with stormwater is called as wet weather flow.

11. Sewage flow rates vary by source and with a time of the day, the season of the year and weather conditions.
a) True
b) False
Answer: a
Clarification: The rate of usage of water depends on the population of a city of a country. Sewage generation in terms depends on water usage. Different cities use the quantity of water based on their needs. Hence the sewage flow rates vary by source and with a time of the day, season of the year and so on.

12. A proper assessment should be made before sewers are sized or treatment plant capacities are set.
a) True
b) False
Answer: a
Clarification: Different treatment plants produce different quantities of wastewater depending upon the water needs and usage within the industry. The quantity of wastewater in sewer depends upon the population in that area and water usage. Hence proper assessment should be made considering both peak and normal conditions.

13. What is the lowest wastewater flow in hospitals?
a) 50 gpcd
b) 700 lpcd
c) 900 lpcd
d) 40 lpcd
Answer: b
Clarification: The unit lpcd stands for litres per capita per day and gpcd stands for gallons per capita per day. 40 lpcd indicates the minimum water usage for places like kids play home and schools. 700 lpcd is for hospitals.

14. __________ in residential areas resemble water usage patterns in those areas.
a) Hydrographs
b) Sewage flow patterns
c) Dry period patterns
d) Daily flow
Answer: b
Clarification: The water usage depends upon the population, number oh houses and the type of water usage in that area. Sewage flow in turn depends upon these activities. Hence sewage flow patterns in residential areas resemble water usage patterns in those areas.

15. A graph showing discharge versus time past a specific point in a conduit carrying flow is called?
a) Water flow graph
b) Precipitation graph
c) Hydrograph
d) Variation graph
Answer: c
Clarification: A graph showing discharge versus time at a specific point in a conduit carrying flow or river is called ‘hydrograph’. Hydrograph helps in understanding the flow characteristics of water at different time intervals.

250+ TOP MCQs on Self Purification of Natural Streams and Answers

Waste Water Engineering Multiple Choice Questions on “Self Purification of Natural Streams”.

1. Which one of the following is the basic indicator of river health?
a) BOD
b) COD
c) DO
d) ThOD
Answer: c
Clarification: The amount of dissolved Oxygen (DO) in water is one of the most commonly used indicators of a river health. Below 4 or 5 mg/l, the life forms that survive are reduced.

2. What is the minimum amount of DO required for the life survival of aquatic animals?
a) 10 mg/l
b) 5 mg/l
c) 2 mg/l
d) 1 mg/l
Answer: c
Clarification: As DO drops below 4 or 5 mg/L the forms of life that can survive begin to be reduced. A minimum of about 2.0 mg/L of dissolved oxygen is required to maintain higher life forms.

3. Oxygen demanding wastes improves DO.
a) True
b) False
Answer: b
Clarification: Oxygen demanding wastes remove DO. Plants add DO during the day but remove it at night. Respiration of organisms removes oxygen. Temperature is reduced in summer which inturn reduces flow.

4. In the concept of self purification of natural streams, complete the following phrase. Solution to pollution is _________
a) Control
b) Dilution
c) Reuse
d) Recycle
Answer: b
Clarification: Waste water disposal practices are based on the premise that “the solution to pollution is dilution”. In this method relatively small quantities of waste are discharged into large bodies of water.

5. _________ is accomplished by the replenishment of oxygen lost to bacterial degradation of organic waste.
a) Gas transfer
b) Dilution
c) Filtration
d) Re-suspension
Answer: a
Clarification: The transfer of gases into and out of water is an important part of the natural purification process. Gas transfer is accomplished by the replenishment of oxygen lost to bacterial degradation of organic waste.

6. On which of the following does the self purification process does not depend?
a) Volume
b) Flow rate
c) Temperature
d) Aquatic species
Answer: d
Clarification: The speed and completeness of the self purification process depend upon many factors like volume, flow rate, the turbulence of flow, variation in sunlight, etc. Aquatic species are affected by polluted streams.

7. Flowing water bodies recover rapidly.
a) True
b) False
Answer: a
Clarification: Flowing water bodies like streams, canals and rivers can recover rapidly from degradable, oxygen demanding wastes and excess heat through a combination of dilution and bacterial decay.

8. In a flowing stream, the breakdown of degradable wastes by bacteria ________ dissolved oxygen.
a) Increases
b) Depletes
c) Maintains
d) Improves
Answer: b
Clarification: In a flowing stream, the breakdown of degradable wastes by bacteria depletes dissolved oxygen. This eliminates the populations with high oxygen requirements until the stream is cleansed of wastes.

9. What is the objective of water quality management?
a) Control the discharge of pollutants
b) Pollutants are discharged into flowing streams
c) Selective pollutants are released
d) Only highly toxic pollutants are released
Answer: a
Clarification: To control the discharge of pollutants so that the water quality is not degraded to an unacceptable extent below the natural background level is the objective of water quality management.

10. The impact of pollution depends upon nature of the pollutants and the ___________
a) Toxic contaminants
b) Season
c) Contaminants
d) Characteristics of river
Answer: d
Clarification: The impact of pollution depends upon nature of the pollutants and the characteristics of the river like discharge and speed of flowing water, depth of the river, type of bottom, surrounding vegetation, etc.

250+ TOP MCQs on Chemical Clarification – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Chemical Clarification – 1”.

1. Clarification removes most of the turbidity.
a) True
b) False
Answer: a
Clarification: Most of the turbidity is removed by clarification, making the water crystal clear. Disinfection is usually the final step in the treatment of drinking water which destroys pathogenic microbes.

2. What is the first step in clarification?
a) Sedimentation
b) Coagulation
c) Flocculation
d) Screening
Answer: b
Clarification: Finely divided particles suspended in surface water repel each other because most of the surfaces are negatively charged. Coagulation is the first step to neutralize the charged particles and form flocs.

3. The agglomeration of destabilized particles into large particles is called _________
a) Sedimentation
b) Coagulation
c) Flocculation
d) Disinfection
Answer: c
Clarification: Flocculation is the agglomeration of destabilized particles into large particles and can be enhanced by the addition of high-molecular-weight, water-soluble organic polymers. These polymers increase floc size by charged site binding and by molecular bridging.

4. What is the density of alum content used in water treatment?
a) 9.8 lb/gal
b) 10.1 lb/gal
c) 11.1 lb/gal
d) 9.2 lb/gal
Answer: c
Clarification: Alum is the primary coagulant. Alum is widely used because it is cheap and easily available. The chemical formula for alum is AlCl3. It is used in the form of a liquid and density is 11.1 lb/gal.

5. What is the speed of the impeller in a flash mixer?
a) 90 rpm
b) 100 rpm
c) 110 rpm
d) 120 rpm
Answer: c
Clarification: The impeller used in the flash mixer is a propeller type. The average speed of the impeller used in the flash mixer is 110 rpm. Rpm refers to the number of rotations per minute.

6. Microflocculation is brought about by which phenomenon?
a) Velocity gradient
b) Differential settling
c) Brownian movement
d) Differential mixing
Answer: c
Clarification: Microflocculation is brought about by Brownian movement. This is also known as perikinetic flocculation. In this type of flocculation, particle aggregation is brought about by random thermal motion.

7. Macro flocculation is brought about by which phenomenon?
a) Velocity gradient
b) Differential settling
c) Brownian movement
d) Differential mixing
Answer: a
Clarification: Macroflocculation is brought about by velocity gradient. This is also known as orthokinetic flocculation. In this type of flocculation, particles to be flocculated are mixed.

8. Alum precipitation occurs at which pH?
a) 1-3
b) 3-5
c) 5-7
d) 7-9
Answer: c
Clarification: Alum precipitation occurs at 5-7. Minimum solubility occurs at 6. Alum is a coagulating agent.

9. Iron precipitation occurs at which pH?
a) 1-3
b) 3-5
c) 5-7
d) 7-9
Answer: d
Clarification: Iron precipitation occurs at 7-9. Minimum solubility occurs at 8. Water containing iron usually is brown in colour.

10. Out of these which is not used as a chemical coagulant?
a) Alum
b) Calcium chloride
c) Ferric Chloride
d) Poly Iron Chloride
Answer: d
Clarification: Poly Iron chloride is not a chemical coagulant. It is a flocculant. It is used in bringing about flocculation.

11. What is the one disadvantage of adding chemicals to bring about precipitation?
a) The TSS increases
b) The TDS increases
c) BOD increases
d) It imparts colour
Answer: b
Clarification: The disadvantage of adding chemicals to bring about precipitation is that this increases the TDS. There is an increase in the dissolved constituents. Addition of chlorine increases the TDS in the effluent.

12. When alum is added to the waste water containing calcium ions which compound is formed as a precipitate?
a) Al(OH)3
b) Al2 (SO4)3
c) Ca(OH)2
d) CaCO3
Answer: a
Clarification: When alum is added to a waste water containing calcium salts Al (OH)3 is precipitated. This sis a gelatinous floc. This reaction sweeps out suspended particles.

13. Along with ferrous sulphate what other chemical is added in order to bring about precipitation?
a) Alum
b) PAC
c) Lime
d) Polyelectrolyte
Answer: c
Clarification: Along with ferrous sulfate, lime is generally added to bring about precipitation. This is done in order to increase the pH. Ferric hydroxide precipitates are formed only at very high pH.

14. Why is Ferric ammonium sulfate generally not used as a coagulant?
a) It is very expensive
b) It is dependent on the Dissolved oxygen present in the waste water
c) It is a very slow process
d) It works only at high pH
Answer: b
Clarification: Ferric ammonium sulfate is not a preferred coagulant. This is because only if oxygen is present ferrous hydroxide forms ferric hydroxide which is a precipitate. Thus this is dependent on the dissolved oxygen of the waste water.

15. What is the recommended surface loading rate in case of alum floc suspension?
a) 30-70 m3/m2.d
b) 10-30 m3/m2.d
c) 70-100 m3/m2.d
d) <10 m3/m2.d
Answer: a
Clarification: The recommended surface loading rate for alum floc suspension is 30-70 m3/m2.d. The typical value is 70 m3/m2.d. In case of iron floc suspension the recommended surface loading rate is 30-70 m3/m2.d

250+ TOP MCQs on Activated Sludge Treatment Systems – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Activated Sludge Treatment Systems – 1”.

1. Why is the sludge aerated?
a) To avoid bacterial growth
b) To increase bacterial growth
c) To maintain pH
d) To maintain temperature
Answer: b
Clarification: An activated-sludge reactor is a system in which pre-treated sewage (i.e. having passed through primary treatment) is aerated to promote the growth of bacteria (cells) that gradually consume the organics in the sewage.

2. ________ is treated in activated sludge reactor.
a) Pre-treated sludge
b) Treated sludge
c) Macronutrients
d) Micro-organisms
Answer: a
Clarification: Pre-treated sludge is treated in an activated sludge reactor. This helps to reduce a load of treatment on the activated sludge process.

3. After the treatment, the BOD demand ______
a) Remains constant
b) Decreases
c) Increases
d) Alters
Answer: b
Clarification: After the treatment, the BOD demand decreases. BOD stands for biological oxygen demand and COD stands for chemical oxygen demand.

4. Which is the next reactor after activated sludge reactor in the treatment process?
a) Flocculation unit
b) Aeration unit
c) Clarifier
d) Disinfection unit
Answer: c
Clarification: Clarifier (settling tank) is the next reactor where the solids (mostly cells, called sludge at this stage) are separated from the water. The system is commonly operated in continuous mode (as opposed to batch mode).

5. Where is the sludge at the bottom of the clarifier processed to?
a) Settling unit
b) Aerator
c) Flocculation unit
d) Disinfection unit
Answer: b
Clarification: The sludge coming from the bottom of the clarifier is processes to aerator unit and this clearly indicates activated sludge process.

6. Which of these is not an alternative to activated sludge treatment systems?
a) Stabilization ponds
b) Rotating biological reactors
c) Trickling filter
d) Screening units
Answer: d
Clarification: In the trickling filter, pre-treated sewage is sprayed on the substrate. Cells can grow by the available oxygen. Stabilization ponds are similar to aerated lagoons and the process is natural.

7. An activated sludge system consists of two components, an aerator and ________
a) Screening units
b) Disinfection unit
c) Flocculation unit
d) Clarifier
Answer: d
Clarification: Aerator and clarifier are the two components present in activated sludge systems. An activated sludge system consists of two components such as aerator and clarifier.

8. How is air pumped in the aerator unit?
a) Bubbled from bottom
b) Sides
c) Bubbled from top
d) Sprayed
Answer: a
Clarification: As cells need oxygen for their metabolism, air is injected from the bottom of the aerator. Rising bubbles agitate the water well and create good contact between the three ingredients: cells, sewage and oxygen.

9. Which are the three ingredients in activated sludge systems?
a) Cells, sewage and oxygen
b) Cells, sewage and nitrogen
c) Solids, sewage and oxygen
d) Solids, water and oxygen
Answer: a
Clarification: The cells need oxygen for their metabolism, air is injected from the bottom of the aerator. The water is well agitated by the rising bubbles and creates good contact between the three ingredients: cells, sewage and oxygen.

10. Mechanical stirring can be done instead of injection of air from the bottom.
a) True
b) False
Answer: a
Clarification: Activated-sludge aerators are well agitated by mechanical stirring from the top or injection of air from the bottom.

11. The sludge particles concentration is increased by the growth of the organism in aeration tanks.
a) True
b) False
Answer: b
Clarification: The wastewater is first screened and then it is mixed with different amounts of the recycled liquid containing high proportion of organisms and sludge concentration decreases as organism growth increases.

12. The organisms feed on _____ in aeration tanks.
a) Water
b) Air
c) Sludge particles
d) Bacteria
Answer: c
Clarification: The organisms in the reactor multiply by feeding of the organic solids present in the reactor. The reactor consists of wastewater containing organic solids.

250+ TOP MCQs on Oil and Grease Removal Methods and Answers

Waste Water Engineering Multiple Choice Questions on “Oil and Grease Removal Methods”.

1. Oil and grease is the presence of inorganics in wastewater.
a) True
b) False
Answer: b
Clarification: Oil and Grease (O&G) are a common occurrence in wastewater. An EPA commissioned study recently concluded that O&G is an indicator of the presence of numerous other organics in a wastewater, the types that partition into oil.

2. The surfactants have ______ chains.
a) Linear
b) Hydrocarbon
c) Complex
d) Carbon
Answer: b
Clarification: Oil is chemically emulsified in water when emulsifiers such as surfactants or soaps are present. The surfactants have hydrocarbon chains. The simplest ones are sodium laurel sulphate or stearic acid which has a hydrophilic (water loving) and a lipophilic (oil loving) end.

3. Lipophilic end is water liking.
a) True
b) False
Answer: b
Clarification: The surfactants have hydrocarbon chains. The simplest ones are sodium laurel sulphate or stearic acid which has a hydrophilic (water loving) and a lipophilic (oil loving) end. The lipophilic end enters the oil droplet, while the hydrophilic end remains in the water.

4. The lipophilic end ______
a) Remains in water
b) Enters oil droplet
c) Dissolves
d) Stays on top of water surface
Answer: b
Clarification: The lipophilic end enters the oil droplet, while the hydrophilic end remains in the water. Since this creates a charge on the otherwise neutral oil droplet, the droplets will repel each other and disperse.

5. What is the size of the oil droplets?
a) Less than 50 microns
b) Less than 40 microns
c) Less than 30 microns
d) Less than 20 microns
Answer: d
Clarification: The droplets are less than 20 microns, while the colour of the water is white. The white colour is an indicator that the emulsion must be split to allow removal of the oil. The source of such oils is metal working fluids, coolants, lubricants, motor oil, hydraulic fluids, etc.

6. What is the colour of the emulsion?
a) White
b) Grey
c) Black
d) Yellow
Answer: a
Clarification: The white colour is an indicator that the emulsion must be split to allow removal of the oil. The source of such oils is metal working fluids, coolants, lubricants, motor oil, hydraulic fluids, etc.

7. What is the size of dissolved oil droplets?
a) Less than 10 microns
b) Less than 8 microns
c) Less than 7 microns
d) Less than 5 microns
Answer: d
Clarification: These are oils from the light end of the oil spectrum such as benzene, toluene and xylene. The molecules are less than five microns in size. They are removed very effectively by activated carbon.

8. _________ act as a coupling agent between oil oil/water phases.
a) Oil
b) Water
c) Emulsifier
d) Disinfectants
Answer: c
Clarification: Emulsifier act as a coupling agent between the oil/water phases. Because the emulsifier is polar on one end (i.e., it has a charge) and is non-polar at the other end, it prevents the oil from approaching and coalescing.

9. An emulsion is a _____ system.
a) Homogenous
b) Heterogenous
c) Natural
d) Oxidized
Answer: b
Clarification: An emulsion is a heterogeneous system that consists of at least one immiscible liquid intimately dispersed in another liquid in the form of droplets, whose diameter generally exceeds 0.1 microns.

10. Surfactants and finely divided solids _____the stability of the emulsion.
a) Increase
b) Decease
c) Neutralize
d) Nullify
Answer: a
Clarification: Surfactants and finely divided solids increase the stability of the emulsion. An emulsion is a heterogeneous system that consists of at least one immiscible liquid intimately dispersed in another liquid in the form of droplets, whose diameter generally exceeds 0.1 microns.

11. Organo clays are manufactured by modifying _____with quaternary amines.
a) Nitrogen
b) Bentonite
c) Potassium
d) Sodium
Answer: b
Clarification: Organo clays are manufactured by modifying bentonite with quaternary amines, a type of surfactant that contains a nitrogen ion. The nitrogen end of the quaternary amine (the hydrophilic end) is positively charged and ion exchanges onto the clay platelet for sodium or calcium.

12. What is the range of charge present on bentonite?
a) 20-30 meq/gram
b) 30-40 meq/gram
c) 40-60 meq/gram
d) 70-90 meq/gram
Answer: d
Clarification: The bentonite has a charge of 70-90 meq/gram. After it is treated with the quaternary amine, some 30-40 meq/gram remain, resulting in the organo clay also removing small amounts of the common heavy metals such as lead, copper, cadmium and nickel.

13. The design of oil/water separators is based on _________
a) Stoke’s law
b) Newton’s law
c) Boyle’s law
d) Charles’s law
Answer: a
Clarification: The design of oil/water separators is based on Stoke’s Law. The lighter oil droplets impact on the slant ribs of the media, coagulate and rise to the surface. The principle of air flotation is that oil droplets will adhere to air and gas bubbles and rise to the surface of the tank.

14. Compute the required pressure for the flotation thickener without recycle for the following information:
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 302 kPa
b) 640 kPa
c) 380 kPa
d) 680 kPa
Answer: a
Clarification: The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)/Sa. Then the P= p+101.35/101.35. p= 302 kPa.

15. Determine the surface area for the following information.
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 34.7 m2
b) 67.7 m2
c) 38 m2
d) 72 m2
Answer: a
Clarification: Area = Flow/ surface loading rate. Area = 400 m3/d/8 L/m2/min x 1440 min/d. Therefore area = 34.7 m2.

16. Compute the required recycled rate for the flotation thickener with recycle for the following information:
A/S = 0.008 mL/mg
p = 275 atm
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 462 m3/d
b) 642 m3/d
c) 380 m3/d
d) 680 m3/d
Answer: a
Clarification: P= P+101.35/101.35. p= 3.73 atm.The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)R/Sa. Q. R = 462 m3/d.

17. Determine the surface area with recycle for the following information.
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 40.1 m2
b) 47.7 m2
c) 38 m2
d) 72 m2
Answer: a
Clarification: P= P+101.35/101.35. p= 3.73 atm.The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)R/Sa. Q. R = 462 m3/d. Area =Recycle Flow/ surface loading rate. Area = 462 m3/d x 1000 L/m3/8 L/m2/min x 1440 min/d. Therefore area = 40.1 m2.

18. What is the velocity considered for a flotation thickener to remove oil?
a) 8 L/m2/min -160 L/m2/min
b) 180 L/m2/min
c) 200 L/m2/min-250 L/m2/min
d) 2 L/m2/min-8 L/m2/min
Answer: a
Clarification: Velocity to be considered for a floatation thickener to remove oil is 8 L/m2/min-160L/m2/min. With the velocity the surface area is found out. Mostly the velocity considered is 8 L/m2/min.