Category: WASTE WATER Engineering Objective Questions

250+ TOP MCQs on Characteristics of Industrial Wastewater and Answers

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

1. Which of the following is not a physical characteristic?
a) Colour
b) pH
c) Odour
d) Temperature
Answer: b
Clarification: The principal physical characteristics of wastewater include solids content, colour, odour and temperature. The total solids in a wastewater consist of the insoluble or suspended solids and the soluble compounds dissolved in water.

2. Total solids in a wastewater consist of insoluble solids alone.
a) True
b) False
Answer: b
Clarification: The total solids in a wastewater consist of the insoluble or suspended solids and the soluble compounds dissolved in water. The suspended solids content is found by drying and weighing the residue removed by the filtering of the sample.

3. What is the minimum percentage of solids in wastewater?
a) 30 %
b) 40 %
c) 50 %
d) 60 %
Answer: b
Clarification: The organic matter consists mainly of proteins, carbohydrates and fats. Between 40 and 65 % of the solids in an average wastewater are suspended. Usually, about 60 % of the suspended solids in a municipal wastewater are 2 settleable (Ron & George, 1998).

4. What is the percentage of settleable solids in municipal wastewater?
a) 60 %
b) 70 %
c) 80 %
d) 90 %
Answer: a
Clarification: Settleable solids, expressed as millilitres per litre, are those that can be removed by sedimentation. Usually, about 60 % of the suspended solids in a municipal wastewater are 2 settleable (Ron & George, 1998).

5. What is the minimum required the temperature to categorize as volatile solids?
a) 500˚C
b) 600˚C
c) 700˚C
d) 800˚C
Answer: b
Clarification: The minimum temperature to categorize volatile solids is 600˚C. Solids may be classified on various other factors. Volatile solids are usually organic in nature.

6. What is the colour of the wastewater after 6 hours of a generation?
a) Grey
b) Light brown
c) Dark brown
d) Dark grey
Answer: b
Clarification: After six hours of duration the colour of wastewater is light in brown. If the wastewater undergoes some degree of decomposition the colour of the water is light grey.

7. What is the colour of septic wastewater?
a) Grey
b) Black
c) Light brown
d) Dark brown
Answer: b
Clarification: The colour of the septic water is black. If the wastewater undergoes some degree of decomposition the colour of water is light grey.

8. The formation of ferrous sulphide leads to the blackening of water.
a) True
b) False
Answer: a
Clarification: The blackening of wastewater is often due to the formation of various sulphides, particularly, ferrous sulphide. This results when hydrogen sulphide produced under anaerobic conditions combines with divalent metal, such as iron, which may be present.

9. Which of the following is not a volatile organic compound?
a) Hydrochloric acid
b) Acetaldehyde
c) Formaldehyde
d) Dichloromethane
Answer: a
Clarification: Many of organic substances are classified as priority pollutants such as polychlorinated biphenyls (PCBs), polycyclic aromatic, acetaldehyde, formaldehyde, 1, 3-butadiene, 1, 2-dichloroethane, dichloromethane, hexachlorobenzene (HCB), etc.

10. What is the maximum pH that the bacteria can sustain in terms of alkalinity?
a) 7
b) 7.5
c) 8
d) 8.5
Answer: d
Clarification: Operation outside of the pH range of 7.0 to 8.5 can be toxic to bacteria; however, if the change is gradual the micro-organisms can become acclimated to pH levels slightly beyond this range.

11. Which of the following is not correct with respect to the effect of oil on treatment systems?
a) Foam formation
b) Toxic to anaerobic bacteria
c) Interferes with settling
d) Explosive with pure oxygen
Answer: a
Clarification: The oil in treatment systems can be toxic to anaerobic bacteria; it might interfere with settling and be explosive with pure oxygen.

12. Which of the following is not correct with respect to the effects of acids on treatment systems?
a) Destroy microbes
b) Upset anaerobic digester
c) Corrode structures
d) Interferes with settling
Answer: d
Clarification: The acids in treatment systems may destroy all the essential microbes, upset anaerobic digester and corrode the structures in the treatment plant.

13. What is the temperature limit of the plastic pipe?
a) 30˚C
b) 40˚C
c) 50˚C
d) 60˚C
Answer: b
Clarification: Plastic pipe (PVC) has temperature limitations of around (40 °C) and can fail if used for hot water transport (I.W.T, 1999). If the O-rings fail, exfiltration or infiltration of the collection system may occur.

14. What is the neutral pH?
a) 6.0
b) 7.0
c) 7.5
d) 8.0
Answer: b
Clarification: The acceptable pH range for the discharge of industrial wastewater to the POTW collection system, as regulated in much industrial waste or sewer-use ordinances, is 6.0 to 9.0. In some ordinances, the pH range may be widened.

15. What is the minimum excess amount of chlorine required to corrode equipment?
a) 30mg/L
b) 40mg/L
c) 50mg/L
d) 60mg/L
Answer: b
Clarification: Many platers will over-chlorinate their cyanide wastewater to ensure they meet the requirements for cyanide concentrations. However, 40 to 50 mg/L excess chlorine can be corrosive to equipment and dangerous to personnel servicing a pump station.

250+ TOP MCQs on Pumps & Pumping Stations – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Pumps & Pumping Stations – 1”.

1. An apparatus for raising, driving or compressing fluids or gases is called _________
a) Piston
b) Pump
c) Compressor
d) Force drive
Answer: b
Clarification: A pump is a device used to transfer or force the liquid or gas against gravity. There are different types of pumps based on the requirements and the pumps are designed for different loads.

2. ________ pumps produce a head and a flow by increasing the velocity of the liquid with the help of the rotating vane impeller.
a) Displacement pumps
b) Positive pumps
c) Centrifugal pumps
d) Rotating pumps
Answer: c
Clarification: Centrifugal pumps produce a head and a flow by increasing the velocity of the liquid with the help of the rotating vane impeller. Centrifugal pumps include radial, axial and mixed flow units.

3. What are the pumps that operate by alternately filling a cavity and then displacing a given volume of liquid called?
a) Centrifugal pump
b) Rotating pump
c) Positive displacement pump
d) Roto-dynamic pump
Answer: c
Clarification: A positive displacement pump alternately fills a cavity and then displaces a given volume of liquid. For each cycle, this pump delivers a constant volume of liquid independent of discharge pressure or head.

4. The two types of pumps behave very differently regarding pressure head and flow rate.
a) True
b) False
Answer: a
Clarification: There are two types of basic pumps. One is the centrifugal pump and the other one is positive displacement pump. Centrifugal pump is also called as a roto-dynamic pump. These two pumps behave very differently with respect to flow rates and pressure head.

5. A steam pump is a type of centrifugal pump.
a) True
b) False
Answer: b
Clarification: There are different types of centrifugal pumps such as end suction pumps, in-line pumps, axial-flow pumps, regenerative pumps, etc. Steam pump is a type of positive displacement pump.

6. In a centrifugal pump, the flow is _______ when the viscosity is increased.
a) Reduced
b) Increased
c) Greater
d) Same
Answer: a
Clarification: In a centrifugal pump, the flow is reduced when the viscosity is increased. This is because the viscosity of any fluid restricts its capacity to flow easily. On thickening the solution or liquid, the flow is reduced.

7. What are the pumps with one or more impellers called?
a) ANSI process pumps
b) API process pumps
c) Centrifugal pumps
d) Positive displacement pumps
Answer: c
Clarification: The general name for pumps with one or more impellers is called centrifugal pumps. Many types and configurations of centrifugal pumps are used for different applications.

8. Why are booster pumps used?
a) Higher flow
b) Boost pressure
c) Chop solids
d) Maintain flow
Answer: b
Clarification: Booster pumps are used to further boost the pressure in a system. It may be in-line circulator, horizontal split case, or vertical turbine in a can type of pump. Chopper Pumps are designed to chop up solids and stringy material as it pumps.

9. Why are cryogenic pumps used?
a) Boost pressure
b) Handle low temperature liquids
c) Handle high temperature liquids
d) Pump small quantities of liquids
Answer: b
Clarification: Cryogenic pumps are used to handle very low temperature liquids. Booster pumps are used to further boost the pressure in a system. Drum pumps are used to pump out small quantities of liquid of out of drums and carboys.

10. End suction pumps are the common type of _________ pumps.
a) Drum pumps
b) Centrifugal pumps
c) Positive displacement pumps
d) Grinder pumps
Answer: b
Clarification: End suction pumps are the common type of centrifugal pump. It has a horizontal shaft with an overhung impeller. The flow goes in the end of the casing, and out the top.

11. Which pump among the ones mentioned below can be located above the suction reservoir without an external priming system?
a) Slurry pumps
b) Self-priming pumps
c) Submersible pumps
d) Trash pumps
Answer: b
Clarification: Self-Priming pumps are a type of centrifugal that can be located above the suction reservoir without an external priming system. It has an end suction configuration but an enlarged case to support priming.

12. Which type of pump is designed to handle rocks and other solids?
a) Trash pumps
b) Submersible pumps
c) Slurry pumps
d) Self-priming pumps
Answer: a
Clarification: Trash pumps are a type of submersible centrifugal pump designed to handle rocks and other solids while dewatering. It is used in dewatering construction sites, mines, and utility pits.

13. Which type of pump should be used in order to handle low viscosity fluids?
a) Centrifugal Pump
b) Displacement Pump
c) Submersible Pump
d) End Suction Pump
Answer: a
Clarification: Centrifugal pumps can pump liquids which are of low viscosity. It cannot handle liquids such as oil. The liquid that is pumped by centrifugal pumps should be free from air.

14. Which type of pump should be selected in order to pump the sewage from a septic tank to the water treatment system?
a) Vertical Sump Pump
b) Progressive Cavity Pump
c) Submersible Pump
d) Screw Pump
Answer: c
Clarification: Submersible pumps are used to handle liquids with solids. Also, these are non-clog pumps which are fully or partially submerged in the tanks. In cases where the pump is partially submerged the motor is above the water level and the motor is connected to the pump by an extended shaft.

15. In an activated sludge process which type of pump is used to recirculate the sludge?
a) Booster Pump
b) Centrifugal Pump
c) Vane Pump
d) Vertical Turbine Pump
Answer: b
Clarification: The activated sludge does not contain any solids. Also, it is not very viscous. Hence a centrifugal pump can be used for this application.

16. Which type of pump is used while handling the sludge disposal system?
a) Screw Pump
b) Multistage Pump
c) Self-priming Pump
d) Vertical Pump
Answer: a
Clarification: Screw Pumps utilize intermeshing screws driven by timing gears in order to move the viscous liquids. These pumps are used to pump thick liquids. Hence it is suitable to handle sludge.

250+ TOP MCQs on Methods for Treatment of Wastewater – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Methods for Treatment of Wastewater – 1”.

1. Screening and comminution are preliminary treatment processes.
a) True
b) False
Answer: a
Clarification: Screening and comminution are preliminary treatment processes utilized to protect mechanical equipment in the treatment works, to aid downstream treatment processes by intercepting unacceptable solids and to alter the physical form of solids so they are acceptable for treatment.

2. _____ devices remove materials which would damage equipment or interfere with a process.
a) Grit
b) Screening
c) Oxidation
d) Reduction
Answer: b
Clarification: Screening devices remove materials which would damage equipment or interfere with a process or piece of equipment. Screening devices have a varied application of wastewater treatment facilities.

3. __________ represents the heavier inert matter in wastewater.
a) Debris
b) Waste
c) Screens
d) Grit
Answer: d
Clarification: Grit represents the heavier inert matter in wastewater which will not decompose in treatment processes. It is identified with matter having a specific gravity of about 2.65 and design of grit chambers is based on the removal of all particles of about 0.011 inch or larger (65 mesh).

4. Supplemental means of aeration are often employed with an equalization basin to provide better mixing.
a) True
b) False
Answer: a
Clarification: supplemental means of aeration are often employed with an equalization basin to provide better mixing, chemical oxidation of reduced compounds, some degree of biological oxidation, agitation to prevent suspended solids from settling.

5. Which of the following should be provided in the case where aeration is absent?
a) Screening devices
b) Mechanical mixers
c) Grit removers
d) Sedimentation tank
Answer: b
Clarification: If aeration is not provided, baffles or mechanical mixers must be provided to avoid stratification and short circuiting in equalization basins. The size and shape of an equalization facility will vary with the quantity of waste and the patterns of waste discharge.

6. Which process is employed to gain sufficient head for the wastewater?
a) Screening
b) Pumping
c) Oxidation
d) Fermentation
Answer: b
Clarification: Pumping facilities may be employed to gain sufficient head for the wastewater to flow through the treatment works to the point of final disposal. Pumping is also generally required for recirculation of all or part of the flow around certain units within the plant. Pumping facilities are classified as influent, effluent, or recirculation stations and perform a critical function.

7. What is the most common used coagulant?
a) Alum
b) Ferric sulphate
c) Limestone
d) Coal
Answer: a
Clarification: Sedimentation using chemical coagulation has been implied mainly to pre-treatment of industrial or process wastewaters and removal of phosphorus from domestic wastewaters. Alum is mostly used as it is cheap and easily available.

8. What is the intermediate zone composed of in aerobic-anaerobic ponds?
a) Algae
b) Aerobic bacteria
c) Facultative bacteria
d) Organic solids
Answer: c
Clarification: Aerobic-anaerobic ponds consist of three zones: a surface zone of algae and aerobic bacteria in a symbiotic association; an intermediate zone populated with facultative bacteria (aerobic or anaerobic); and an anaerobic bottom zone where settled organic solids are decomposed by anaerobic bacteria.

9. Nitrification efficiency is significantly suppressed as the temperature is _________
a) Increased
b) Decreased
c) Neutral
d) Maintained
Answer: b
Clarification: Two important considerations in nitrification are the maintenance of a proper pH and temperature. Nitrification is a very temperature-sensitive system and the efficiency is significantly suppressed as the temperature decreases.

10. __________ is a process which involves further removal of the nitrogen.
a) Nitrification
b) Denitrification
c) Ammonification
d) Reduction
Answer: b
Clarification: As with nitrification, denitrification is a process which involves further removal of the nitrogen by conversion of the nitrate to nitrogen gas. This represents a process for the ultimate removal of nitrogen from wastewater.

11. In rotating biological contractors, what percent of corrugated plastic discs are submerged?
a) 20
b) 50
c) 80
d) 40
Answer: d
Clarification: In the process of rotating biological contractors, the large diameter corrugated plastic discs are mounted on a horizontal shaft and placed in a tank. The medium is slowly rotated with about 40 percent of the surface area always submerged in the flowing wastewater.

250+ TOP MCQs on Sand Filters – 2 and Answers

Waste Water Engineering Quiz on “Sand Filters – 2”.

1. Calculate the diameter of a dual media filter with the following data.
Flow: 2200 m3/hr
Velocity: 15 m/hr
a) 15.5 m
b) 13.6 m
c) 13.9 m
d) 16.5 m
Answer: b
Clarification: The calculated diameter is 13.6 m. Area is first calculated by dividing the flowrate with velocity. That is 2200/15 =147 m2. Then square root (147 x (4/3.14)) = 13.6 m.

2. Calculate the diameter of a multimedia filter with the following data.
Flow: 2200 m3/hr
Velocity: 18 m/hr
a) 13.3 m
b) 14.5 m
c) 12.5 m
d) 12.9 m
Answer: c
Clarification: The calculated diameter is 12.5 m. Area is first calculated by dividing the flowrate with velocity. That is 2200/18 =122 m2. Then square root (122 x (4/3.14)) = 12.5 m.

3. While designing a blower for backwash for sand filters without air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: a
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for sand filters without air scouring is considered as 36m/hr. In case air scouring is there the velocity considered would be different.

4. While designing a blower for backwash for sand filters with air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: c
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for sand filters with air scouring is considered as 24 m/hr. In case air scouring is not there the velocity considered would be different.

5. While designing a blower for backwash for dual media filters without air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: b
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for dual media filters without air scouring is considered as 30m/hr. In case air scouring is there the velocity considered would be different.

6. While designing a blower for backwash for multimedia filters without air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: c
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for multimedia filters without air scouring is considered as 24 m/hr. In case air scouring is there the velocity considered would be different.

7. While designing a blower for backwash for multimedia filters with air scouring what is the velocity considered?
a) 36 m/h
b) 30 m/h
c) 24 m/h
d) 20 m/h
Answer: d
Clarification: For backwashing of filters blowers are required. The velocity of these blowers while designing them for multimedia filters with air scouring is considered as 20 m/hr. In case air scouring is not there the velocity considered would be different.

8. While designing a pump for backwash for a filter what will be the velocity assumed?
a) 24 m/h
b) 36m/h
c) 30 m/h
d) 20 m/h
Answer: a
Clarification: The velocity of the backwash pump is considered as 24m/h. This is in case of Dual media filter or Pressure sand filter. A horizontal centrifugal pump would be used as a backwash pump.

9. Calculate the flow rate of the backwash pump to be used for a dual media filter with the following data.
Flowrate: 2200 m3/h
Rise velocity of DMF: 15 m/h
a) 4500 m3/ h
b) 3520 m3/ h
c) 3750 m3/ h
d) 4920 m3/ h
Answer: b
Clarification: First the area of the filter is found out. Area = 2200/15=147 m2. Then by multiplying area x the velocity of the backwash pump assumed we derive at the flowrate. Flowrate= 147 m2 x 24 m/h = 3520 m3/h.

10. Calculate the flow rate of the backwash pump to be used for a pressure sand filter with the following data.
Flowrate: 2200 m3/h
Rise velocity of PSF (Pressure Sand filter): 10 m/h
a) 4500 m3/ h
b) 5280 m3/ h
c) 6480 m3/ h
d) 7580 m3/ h
Answer: b
Clarification: First the area of the filter is found out. Area = 2200/10=220 m2. Then by multiplying area x the velocity of the backwash pump assumed we derive at the flowrate. Flowrate= 220 m2 x 24 m/h = 5280 m3/h.

11. Calculate the flow rate of the backwash pump to be used for a multimedia filter with the following data.
Flowrate: 2200 m3/h
Rise velocity of multimedia filter: 18 m/h
a) 3700 m3/ h
b) 2900 m3/ h
c) 5300 m3/ h
d) 3800 m3/ h
Answer: b
Clarification: First the area of the filter is found out. Area = 2200/18=122 m2. Then by multiplying area x the velocity of the backwash pump assumed we derive at the flowrate. Flowrate= 122 m2 x 24 m/h = 2900 m3/h.

12. What is the bulk density of pebbles assumed while designing a Pressure sand filter?
a) 1650 kg/m3
b) 1600 kg/m3
c) 650 kg/m3
d) 550 kg/m3
Answer: b
Clarification: The bulk density of the pebble used is generally assumed as 1600 kg/m3. This value is used to calculate the amount of pebble required. Generally the amount of pebble is calculated by number of bags required. Each bag constitutes around 140 Kg of pebble.

13. What is the bulk density of sand assumed while designing the amount required for a sand filter?
a) 1750 Kg/m3
b) 1300 Kg/m3
c) 1000Kg/m3
d) 790 Kg/m3
Answer: a
Clarification: While designing a pressure sand filter the bulk density of the sand is considered as 1750 kg/ m3. The sand is generally measured in bags. Each bag weighs around 140 Kg.

14. Calculate the amount of pebbles required for the following data.
Flow: 200 m3/h
Velocity of dual media filter: 15m/h
Height: 0.6 m
a) 12668 Kg
b) 13792 Kg
c) 35000 Kg
d) 15792 Kg
Answer: a
Clarification: First the diameter is calculated by finding the area. Area= 200/15. Diameter= square root (1.27 × 13.33) = 4.11. Amount of pebbles equals πd2/4. This equals 0.785 x 4.11 x 4.11 x 0.6 x 1600 = 12668 Kg.

15. Calculate the amount of sand required for the following data.
Flow: 200 m3/h
Velocity of dual media filter: 15m/h
Height: 0.4 m
a) 12668 Kg
b) 9237 Kg
c) 1020 Kg
d) 12752 Kg
Answer: b
Clarification: First the diameter is calculated by finding the area. Area= 200/15. Diameter= square root (1.27 × 13.33) = 4.11. Amount of sand equals πd2/4. This equals 0.785 x 4.11 x 4.11 x 0.4 x 1750 = 9237 Kg.

250+ TOP MCQs on Facultative Ponds and Answers

Waste Water Engineering Multiple Choice Questions on “Facultative Ponds”.

1. The facultative pond functions like a ________
a) Primary clarifier
b) Secondary clarifier
c) Aerator
d) Sedimentation unit
Answer: a
Clarification: The facultative lagoon in the pond sequence functions like the primary clarifier of a conventional sewage treatment system. Heavy solids will settle to the bottom of the lagoon and lighter solids will float.

2. The surface area of facultative ponds should be small.
a) True
b) False
Answer: b
Clarification: The surface area provided for facultative ponds must be large and an atmospheric oxygen transfer rate adequate to prevent anaerobic conditions on the lagoon surface.

3. In facultative ponds, micro-organisms are able to oxidize only dissolved organics.
a) True
b) False
Answer: b
Clarification: The dissolved and suspended organics from the original wastewater and the products of anaerobic catabolism on the bottom of the lagoon are oxidized by the intermediate depths of the lagoon.

4. ______ occur when the rate of oxygen transfer from the lagoon surface is less than the rate of oxygen consumption in the lower levels of the lagoon.
a) Temperature rise
b) Objectionable odour
c) Raised pH
d) Temperature fall
Answer: b
Clarification: Objectionable odours are formed when the rate of oxygen consumption is more in the lower levels of the lagoon than the rate of oxygen transfer from the lagoon surface.

5. How much amount of oxygen should one acre of facultative lagoon must provide per day?
a) 10 pounds
b) 50 pounds
c) 75 pounds
d) 90 pounds
Answer: b
Clarification: A 4000 sqm area which is equal to 1 acre facultative lagoon might provide 50 pounds of oxygen per day (5 grams of oxygen per square meter per day) for biochemical catabolism.

6. ______ weather requires large oxygen transfer.
a) Cold
b) Warm
c) Rainy
d) Normal
Answer: b
Clarification: During cold weather, waste can get accumulated and cause short-term warm weather oxygen requirements to exceed long-term waste loading rates. Warm weather will require large oxygen transfer rates.

7. Ice mats _____ the oxygen transfer surface.
a) Alter
b) Increase
c) Reduce
d) Maintain
Answer: c
Clarification: Algal respiration can require additional oxygen during darkness but algae can provide surface oxygen during daylight hours. Ice or scum mats can reduce the oxygen transfer surface.

8. What is the depth range of facultative ponds?
a) 0.5-0.8 m
b) 0.8-1.0 m
c) 1-1.2 m
d) 1.2-1.5 m
Answer: d
Clarification: Secondary facultative ponds are 1.20 – 1.50 meters deep and are designed for BOD removal on the basis of a relatively low surface loading of 100 – 400 kg BOD/ha-d at the temperature between 20°C and 25°C.

9. What is the maximum allowable temperature in facultative ponds?
a) 15˚C
b) 20˚C
c) 25˚C
d) 30˚C
Answer: c
Clarification: Facultative ponds are designed for BOD removal on the basis of a relatively low surface loading of 100 – 400 kg BOD/ha-d at the temperature between 20°C and 25°C.

10. Which colour are the facultative ponds?
a) Blue
b) Green
c) Brown
d) Black
Answer: b
Clarification: Due to the presence of algae, facultative ponds are coloured dark green. These can optimize their vertical position in the pond water column in relation to incident light intensity and temperature more easily than non-motile forms (such as Chlorella, although this is also fairly common in facultative ponds).

250+ TOP MCQs on Rapid Filtration and Answers

Waste Water Engineering Multiple Choice Questions on “Rapid Filtration”.

1. When was the rapid sand filer first used?
a) 1930
b) 1920
c) 1950
d) 1960
Answer: b
Clarification: The first modern rapid sand filtration plant was designed and built by George W. Fuller in Little Falls, New Jersey. Fuller’s filtration plant went into operation in 1920 and its success was responsible for the change to this technology in the U.S.

2. Rapid sand filters require large area compared to slow sand filters.
a) True
b) False
Answer: b
Clarification: Rapid sand filters were widely used in large municipal water systems by the 1920s because they required smaller land areas compared to slow sand filters.

3. Pressure flow is considered in rapid filtration.
a) True
b) False
Answer: b
Clarification: The polluted or unfiltered water flows through the filter medium under gravity or under pumped pressure and the floc material is trapped in the sand matrix.

4. Which of the following processes is not done before rapid filtration?
a) Mixing
b) Flocculation
c) Sedimentation
d) Chlorination
Answer: d
Clarification: Mixing, flocculation and sedimentation processes are typical treatment stages that precede filtration. Chemical additives, such as coagulants are often used in conjunction with the filtration system.

5. The direction of water in filter beds is reversed for ______
a) Filtration
b) Cleaning
c) Decreasing velocity
d) Decreasing flow
Answer: b
Clarification: Rapid sand filters must be cleaned frequently, often several times a day, by backwashing, which involves reversing the direction of the water and adding compressed air.

6. For a rapid sand filter, what is the ratio of length: breadth while designing it?
a) 1:2
b) 1:3
c) 2:3
d) 4:1
Answer: b
Clarification: For a rapid sand filter the ratio of length: breadth while designing it is 1:3. The area is calculated from the velocity and the flowrate of water. From this ratio and the area, the length and breadth of the filter are calculated.

7. Calculate the area of the rapid sand filter for the following details.
Flow: 2000 m3/h.
Velocity: 5m/h
Backwash time: 30 mins
a) 18 m2
b) 20 m2
c) 22 m2
d) 16 m2
Answer: a
Clarification: The total operation time equals 24 minus the back wash time. Area equals Flowrate × total hours of operation/ [velocity × total hours of operation × (total hours of operation-backwash time). Area= 2000 × 24/ (5 × 24 × 23.5) =18 m2.

8. Calculate the length of the rapid sand filter for the following details.
Flow: 1000 m3/h.
Velocity: 5m/h
Backwash time: 30 mins
a) 1.0 m
b) 1.9 m
c) 1.7 m
d) 1.2 m
Answer: b
Clarification: The total operation time equals 24 minus the back wash time. Area equals Flowrate × total hours of operation/ [velocity × total hours of operation × (total hours of operation-backwash time). Area= 1000 × 24/ (5 × 24 × 23.5) =9 m2. Length: Breadth is in the ratio 1:3. Length = 1.7 m.

9. Calculate the breadth of the rapid sand filter for the following details.
Flow: 500 m3/h.
Velocity: 5 m/h
Backwash time: 20 mins
a) 4.5 m
b) 5.7 m
c) 3.7 m
d) 7.5 m
Answer: c
Clarification: The total operation time equals 24 minus the back wash time. Area equals Flowrate × total hours of operation/ [velocity × total hours of operation × (total hours of operation-backwash time). Area= 500 × 24/ (5 × 24 × 23.7) =4.5 m2. Length: Breadth is in the ratio 1:3. Breadth = 3.7 m.

10. What is the turbidity achieved by using a rapid filtration system?
a) 0.1 NTU
b) 0.7 NTU
c) 1.5 NTU
d) 0.9 NTU
Answer: a
Clarification: The turbidity achieved by using a rapid filtration system is 0.1 NTU. This is used to operate systems which have rates greater than 100 times that of slow sand filtration. Coagulation is used to precondition the water.

11. What is the limit of turbidity which a 2 stage filtration a handle?
a) 180 NTU
b) <100 NTU
c) 50 NTU
d) 200 NTU
Answer: b
Clarification: Two stage filtration is a type of Rapid filtration. This can handle upto 100 NTU. This is followed by a rough filter after a coagulation system. The rough filter is followed by a filtration system.

12. What is the height of the anthracite maintained in a dual media filter?
a) 1.5-1.8 m
b) 2.2 -2.4 m
c) 2.5-2.8 m
d) 2.8-3.0 m
Answer: a
Clarification: The height of the anthracite maintained in dual media filter is 1.5-1.8 m. Initially, the height maintained used to be 0.45-0.6 m. GAC is sometimes used instead of anthracite in case of dual media filters.

13. What is the height maintained of a deep bed monomedia?
a) 1.5-1.8 m
b) 2.2 -2.4 m
c) 2.5-2.8 m
d) 2.8-3.0 m
Answer: a
Clarification: The height of the deep bed monomedia is maintained at 1.5-1.8 m. This is a type of rapid filtration. The media is usually GAC/Anthracite.