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All questions of Treatment of Water for Civil Engineering (CE) Exam
What is the dimension formula of mean velocity gradient?- a)1/T
- b)1/T2
- c)T
- d)T2
Correct answer is option 'A'. Can you explain this answer?
What is the dimension formula of mean velocity gradient?
a)
1/T
b)
1/T2
c)
T
d)
T2
|
Surbhi Kulkarni answered |
Explanation: Mean velocity gradient is expressed in metre per second/m or Sec-1, so it has dimension formula of 1/T.
The period of cleaning of a slow sand filter is ______ than rapid sand filter.- a)Smaller
- b)Larger
- c)Slightly smaller
- d)Slightly larger
Correct answer is option 'B'. Can you explain this answer?
The period of cleaning of a slow sand filter is ______ than rapid sand filter.
a)
Smaller
b)
Larger
c)
Slightly smaller
d)
Slightly larger
|
Shanaya Das answered |
Explanation: The period of cleaning of a slow sand filter is 1-2 months and that of the rapid sand filter is 2-3 days.
The quantity of water flowing per hour per unit horizontal area is called ____- a)Detention time
- b)Flowing through period
- c)Displacement time
- d)Overflow rate
Correct answer is option 'D'. Can you explain this answer?
The quantity of water flowing per hour per unit horizontal area is called ____
a)
Detention time
b)
Flowing through period
c)
Displacement time
d)
Overflow rate
|
Milan Sen answered |
Explanation: Overflow rate = Q/A where Q is discharge rate in litre per day, A is the horizontal area in m2.
Which of the following statement is wrong regarding Anthracite?- a)It requires less water
- b)It is less inert to caustic solutions than sand
- c)It is more costly per tonne than sand
- d)It is used in industrial filters.
Correct answer is option 'B'. Can you explain this answer?
Which of the following statement is wrong regarding Anthracite?
a)
It requires less water
b)
It is less inert to caustic solutions than sand
c)
It is more costly per tonne than sand
d)
It is used in industrial filters.
|
Aman Kapoor answered |
Explanation: Anthracite is a costlier filter material used in industrial filter which is more inert to hot caustic solutions than sand.
Multistage flash evaporator produces _____ kg of water per kg of prime steam.- a)5
- b)10
- c)15
- d)20
Correct answer is option 'D'. Can you explain this answer?
Multistage flash evaporator produces _____ kg of water per kg of prime steam.
a)
5
b)
10
c)
15
d)
20
|
Pioneer Academy answered |
Multistage flash evaporator produces 20 kg of water per kg of prime steam.
They operate on the brine recycling basis.
They operate on the brine recycling basis.
The maximum daily demand of a city is 24*105 and the rate of filtration is 100 per litre per hour per m2. What is the area of the filter? - a)10
- b)100
- c)1000
- d)1500
Correct answer is option 'C'. Can you explain this answer?
The maximum daily demand of a city is 24*105 and the rate of filtration is 100 per litre per hour per m2. What is the area of the filter?
a)
10
b)
100
c)
1000
d)
1500
|
|
Puja Desai answered |
Explanation: Maximum daily demand = 24*105
Rate of filtration = 100
Area of filter=Maximum daily demand/Rate of filtration=24*105/(100*24)
=1000m2.
Rate of filtration = 100
Area of filter=Maximum daily demand/Rate of filtration=24*105/(100*24)
=1000m2.
Which of the following is the best evaporative desalinating method?- a)Vertical tube evaporator
- b)Multistage flash evaporator
- c)Tube settler
- d)Multi-effect multistage flash evaporator
Correct answer is option 'B'. Can you explain this answer?
Which of the following is the best evaporative desalinating method?
a)
Vertical tube evaporator
b)
Multistage flash evaporator
c)
Tube settler
d)
Multi-effect multistage flash evaporator
|
Sharmila Gupta answered |
Introduction:
Evaporative desalination is a process that involves the removal of salt and other impurities from seawater or brackish water. It utilizes the principle of evaporation to separate fresh water from the saline solution. Among the given options, the best evaporative desalinating method is the Multistage Flash Evaporator.
Explanation:
The Multistage Flash (MSF) evaporator is a widely used method for desalination due to its high efficiency and low energy consumption. It operates on the principle of thermal vapor compression, where seawater is heated under high pressure and then rapidly flashed into multiple stages.
Working Principle:
The MSF evaporator consists of multiple stages, each containing a series of flash chambers. In each stage, seawater is heated using steam, which raises its temperature and pressure. As the high-pressure seawater enters the flash chamber, it is suddenly exposed to a low-pressure environment. This causes the water to evaporate instantly, leaving behind the concentrated brine.
Key Advantages:
The Multistage Flash evaporator offers several advantages over other evaporative desalination methods, making it the best choice:
1. High Efficiency: The MSF evaporator has a high thermal efficiency, allowing it to produce a large quantity of fresh water from seawater or brackish water.
2. Low Energy Consumption: Compared to other methods, the MSF evaporator requires less energy for operation, reducing the overall cost of desalination.
3. Scalability: The MSF evaporator can be easily scaled up or down to meet the desired water production capacity.
4. Reliability: The MSF evaporator has a simple design and is known for its reliability and long operational life.
5. Flexibility: The MSF evaporator can handle a wide range of feedwater salinities, making it suitable for various applications.
Conclusion:
In conclusion, the Multistage Flash (MSF) evaporator is the best evaporative desalinating method among the given options. Its high efficiency, low energy consumption, scalability, reliability, and flexibility make it an ideal choice for large-scale desalination projects.
Evaporative desalination is a process that involves the removal of salt and other impurities from seawater or brackish water. It utilizes the principle of evaporation to separate fresh water from the saline solution. Among the given options, the best evaporative desalinating method is the Multistage Flash Evaporator.
Explanation:
The Multistage Flash (MSF) evaporator is a widely used method for desalination due to its high efficiency and low energy consumption. It operates on the principle of thermal vapor compression, where seawater is heated under high pressure and then rapidly flashed into multiple stages.
Working Principle:
The MSF evaporator consists of multiple stages, each containing a series of flash chambers. In each stage, seawater is heated using steam, which raises its temperature and pressure. As the high-pressure seawater enters the flash chamber, it is suddenly exposed to a low-pressure environment. This causes the water to evaporate instantly, leaving behind the concentrated brine.
Key Advantages:
The Multistage Flash evaporator offers several advantages over other evaporative desalination methods, making it the best choice:
1. High Efficiency: The MSF evaporator has a high thermal efficiency, allowing it to produce a large quantity of fresh water from seawater or brackish water.
2. Low Energy Consumption: Compared to other methods, the MSF evaporator requires less energy for operation, reducing the overall cost of desalination.
3. Scalability: The MSF evaporator can be easily scaled up or down to meet the desired water production capacity.
4. Reliability: The MSF evaporator has a simple design and is known for its reliability and long operational life.
5. Flexibility: The MSF evaporator can handle a wide range of feedwater salinities, making it suitable for various applications.
Conclusion:
In conclusion, the Multistage Flash (MSF) evaporator is the best evaporative desalinating method among the given options. Its high efficiency, low energy consumption, scalability, reliability, and flexibility make it an ideal choice for large-scale desalination projects.
Solids are removed from the water by which of the following unit operation?- a)Inter facial contact
- b)Solid stabilization
- c)Ion transfer
- d)Solids transfer
Correct answer is option 'D'. Can you explain this answer?
Solids are removed from the water by which of the following unit operation?
a)
Inter facial contact
b)
Solid stabilization
c)
Ion transfer
d)
Solids transfer
|
Pranab Joshi answered |
Explanation: During solids transfer, solids are removed from water by straining, sedimentation, flotation and filtration.
In Electro-dialysis, the electric energy which required is _____- a)Directly proportional to cation permeable membrane
- b)Directly proportional to anion permeable membrane
- c)Directly proportional to salt concentration
- d)Inversely proportional to salt concentration
Correct answer is option 'C'. Can you explain this answer?
In Electro-dialysis, the electric energy which required is _____
a)
Directly proportional to cation permeable membrane
b)
Directly proportional to anion permeable membrane
c)
Directly proportional to salt concentration
d)
Inversely proportional to salt concentration
|
Lavanya Menon answered |
In Electro-dialysis, the passage of electric current helps the diffusion of cation and anion permeable membrane and the energy required is directly proportional to the concentration of salt in the saline water.
The tanks built with mechanical means for continuous removal of solids being deposited by sedimentation are called ____- a)Clarifiers
- b)Settling basins
- c)Sedimentation tanks
- d)Eco-pons
Correct answer is option 'A'. Can you explain this answer?
The tanks built with mechanical means for continuous removal of solids being deposited by sedimentation are called ____
a)
Clarifiers
b)
Settling basins
c)
Sedimentation tanks
d)
Eco-pons
|
Anjana Mukherjee answered |
Clarifiers
Clarifiers are tanks built with mechanical means for the continuous removal of solids deposited by sedimentation. They are commonly used in wastewater treatment plants and other industrial processes to separate solid particles from liquids.
Function of Clarifiers
The main function of clarifiers is to remove suspended solids from a liquid stream through the process of sedimentation. Sedimentation is the process by which particles settle to the bottom of a liquid due to gravity. Clarifiers provide a large surface area for the settling of particles and use mechanical mechanisms to facilitate the removal of settled solids.
Working Principle
The working principle of clarifiers involves the following steps:
1. Inflow of wastewater: The wastewater containing suspended solids enters the clarifier tank.
2. Settling zone: The wastewater is directed to the settling zone, where the velocity of the flow is reduced. This allows the solid particles to settle down under the influence of gravity.
3. Mechanical mechanisms: Clarifiers are equipped with mechanical mechanisms such as scrapers or rakes that continuously remove the settled solids from the bottom of the tank. These mechanisms help in preventing the accumulation of solids and maintain the capacity of the clarifier.
4. Effluent discharge: The clarified liquid, free from settled solids, is discharged from the top of the clarifier tank.
5. Sludge removal: The removed solids, known as sludge, are collected in a hopper or sludge pit at the bottom of the clarifier. The sludge is periodically removed from the clarifier for further treatment or disposal.
Advantages of Clarifiers
- Efficient removal of suspended solids: Clarifiers are designed to provide a large settling area, ensuring effective removal of suspended solids.
- Continuous operation: The mechanical mechanisms in clarifiers allow for continuous removal of settled solids, enabling uninterrupted operation.
- Flexibility: Clarifiers can be designed to accommodate different flow rates and handle a wide range of solid concentrations.
- Space-saving design: Clarifiers have a compact design, making them suitable for installations with limited space.
In conclusion, clarifiers are tanks built with mechanical means for the continuous removal of solids deposited by sedimentation. They play a crucial role in wastewater treatment and other industrial processes by effectively separating suspended solids from liquids.
Clarifiers are tanks built with mechanical means for the continuous removal of solids deposited by sedimentation. They are commonly used in wastewater treatment plants and other industrial processes to separate solid particles from liquids.
Function of Clarifiers
The main function of clarifiers is to remove suspended solids from a liquid stream through the process of sedimentation. Sedimentation is the process by which particles settle to the bottom of a liquid due to gravity. Clarifiers provide a large surface area for the settling of particles and use mechanical mechanisms to facilitate the removal of settled solids.
Working Principle
The working principle of clarifiers involves the following steps:
1. Inflow of wastewater: The wastewater containing suspended solids enters the clarifier tank.
2. Settling zone: The wastewater is directed to the settling zone, where the velocity of the flow is reduced. This allows the solid particles to settle down under the influence of gravity.
3. Mechanical mechanisms: Clarifiers are equipped with mechanical mechanisms such as scrapers or rakes that continuously remove the settled solids from the bottom of the tank. These mechanisms help in preventing the accumulation of solids and maintain the capacity of the clarifier.
4. Effluent discharge: The clarified liquid, free from settled solids, is discharged from the top of the clarifier tank.
5. Sludge removal: The removed solids, known as sludge, are collected in a hopper or sludge pit at the bottom of the clarifier. The sludge is periodically removed from the clarifier for further treatment or disposal.
Advantages of Clarifiers
- Efficient removal of suspended solids: Clarifiers are designed to provide a large settling area, ensuring effective removal of suspended solids.
- Continuous operation: The mechanical mechanisms in clarifiers allow for continuous removal of settled solids, enabling uninterrupted operation.
- Flexibility: Clarifiers can be designed to accommodate different flow rates and handle a wide range of solid concentrations.
- Space-saving design: Clarifiers have a compact design, making them suitable for installations with limited space.
In conclusion, clarifiers are tanks built with mechanical means for the continuous removal of solids deposited by sedimentation. They play a crucial role in wastewater treatment and other industrial processes by effectively separating suspended solids from liquids.
Which of the following is the expression of the Carmen equation where symbols have their usual meanings? - a)hf = f* (L/(s *d)) * ((1-e)/e3) * (v2/g)
- b)hf = f* (L/(s *d)) * ((1-e)/e3) * (v/g)
- c)hf = f* (L/(s *d)) * ((1-e)/e) * (v2/g)
- d)hf = f* (L/(s *d)) * ((1-e)/e2) * (v2/g)
Correct answer is option 'A'. Can you explain this answer?
Which of the following is the expression of the Carmen equation where symbols have their usual meanings?
a)
hf = f* (L/(s *d)) * ((1-e)/e3) * (v2/g)
b)
hf = f* (L/(s *d)) * ((1-e)/e3) * (v/g)
c)
hf = f* (L/(s *d)) * ((1-e)/e) * (v2/g)
d)
hf = f* (L/(s *d)) * ((1-e)/e2) * (v2/g)
|
Nishtha Tiwari answered |
Explanation: The expression of the Carmen equation is given by hf = f* (L/(s *d)) * ((1-e) /e3) * (v2/g) where hf =head loss, d= diameter of pipe. L= length of pipe and v=approach velocity.
What is the value of specific gravity of sand filter?- a)2
- b)2.6
- c)3.1
- d)3.5
Correct answer is option 'B'. Can you explain this answer?
What is the value of specific gravity of sand filter?
a)
2
b)
2.6
c)
3.1
d)
3.5
|
Bhargavi Mukherjee answered |
Explanation: The specific gravity of sand filter lies in the range of 2.55-2.65. So, the value 2.6 lies in the range of specific gravity of sand filter.
Which of the following statement is wrong regarding filtration?- a)It removes fine particle
- b)It removes suspended solids not removed by sedimentation
- c)It does not remove turbidity
- d)It removes color
Correct answer is option 'C'. Can you explain this answer?
Which of the following statement is wrong regarding filtration?
a)
It removes fine particle
b)
It removes suspended solids not removed by sedimentation
c)
It does not remove turbidity
d)
It removes color
|
|
Kiran Choudhary answered |
Filtration is a process that involves the separation of solid particles from a fluid or gas by passing it through a porous medium. It is commonly used in various industries such as water treatment, pharmaceuticals, food processing, and many others. Filtration is an effective method for removing suspended solids and other impurities from a liquid or gas stream. However, there are certain misconceptions about filtration.
c) It does not remove turbidity
This statement is incorrect. Filtration is actually a widely used method for removing turbidity from water or other liquids. Turbidity refers to the cloudiness or haziness of a fluid caused by the presence of suspended particles. These particles can be clay, silt, organic matter, or other insoluble materials.
Filtration works by passing the turbid fluid through a porous medium, such as a filter media or a filter cartridge. The porous medium acts as a physical barrier, trapping and removing the suspended particles from the fluid stream. As the fluid passes through the filter, the particles get trapped in the filter media, while the clarified fluid passes through. This results in the removal of turbidity and the production of a clear, filtered liquid.
Filtration is particularly effective in removing fine particles and suspended solids that are not easily removed by sedimentation. Sedimentation is a process in which solid particles settle down under the influence of gravity. While sedimentation can remove larger particles, it is less efficient for smaller particles and colloidal suspensions. Filtration, on the other hand, can effectively remove particles of various sizes, including fine particles and colloids.
Filtration can also remove color from a liquid. Color in water or other liquids can be due to the presence of dissolved organic compounds, dyes, or other substances. These colored compounds can be adsorbed or trapped on the surface of the filter media, resulting in the removal of color from the filtered liquid. However, it is important to note that the effectiveness of filtration in removing color depends on the nature of the color-causing compounds and the type of filter media used.
In summary, filtration is an effective method for removing fine particles, suspended solids, turbidity, and even color from a liquid or gas stream. It is a widely used process in various industries and plays a crucial role in ensuring the quality and purity of the final product.
c) It does not remove turbidity
This statement is incorrect. Filtration is actually a widely used method for removing turbidity from water or other liquids. Turbidity refers to the cloudiness or haziness of a fluid caused by the presence of suspended particles. These particles can be clay, silt, organic matter, or other insoluble materials.
Filtration works by passing the turbid fluid through a porous medium, such as a filter media or a filter cartridge. The porous medium acts as a physical barrier, trapping and removing the suspended particles from the fluid stream. As the fluid passes through the filter, the particles get trapped in the filter media, while the clarified fluid passes through. This results in the removal of turbidity and the production of a clear, filtered liquid.
Filtration is particularly effective in removing fine particles and suspended solids that are not easily removed by sedimentation. Sedimentation is a process in which solid particles settle down under the influence of gravity. While sedimentation can remove larger particles, it is less efficient for smaller particles and colloidal suspensions. Filtration, on the other hand, can effectively remove particles of various sizes, including fine particles and colloids.
Filtration can also remove color from a liquid. Color in water or other liquids can be due to the presence of dissolved organic compounds, dyes, or other substances. These colored compounds can be adsorbed or trapped on the surface of the filter media, resulting in the removal of color from the filtered liquid. However, it is important to note that the effectiveness of filtration in removing color depends on the nature of the color-causing compounds and the type of filter media used.
In summary, filtration is an effective method for removing fine particles, suspended solids, turbidity, and even color from a liquid or gas stream. It is a widely used process in various industries and plays a crucial role in ensuring the quality and purity of the final product.
In which settling type, dilute suspension of particles takes place?- a)Zone settling
- b)Compression settling
- c)Hindered settling
- d)Discrete settling
Correct answer is option 'C'. Can you explain this answer?
In which settling type, dilute suspension of particles takes place?
a)
Zone settling
b)
Compression settling
c)
Hindered settling
d)
Discrete settling
|
|
Gitanjali Iyer answered |
Explanation: In this type of settling, particle mass increase and it settle down due to flocculation.
Which of the following cannot be the velocity of flow in the horizontal flow sedimentation tank?- a)12cm/min
- b)39cm/min
- c)5cm/min
- d)20cm/min
Correct answer is option 'B'. Can you explain this answer?
Which of the following cannot be the velocity of flow in the horizontal flow sedimentation tank?
a)
12cm/min
b)
39cm/min
c)
5cm/min
d)
20cm/min
|
|
Puja Desai answered |
Understanding Velocity in Horizontal Flow Sedimentation Tanks
In the context of horizontal flow sedimentation tanks, the velocity of flow is a crucial factor that affects the sedimentation process. The velocity must be optimized to allow sufficient time for particles to settle without causing re-suspension.
Typical Velocity Ranges
- Horizontal flow sedimentation tanks generally operate within a specific range of flow velocities.
- Typical design velocities are often between 1 to 20 cm/min, depending on the type of particles being treated and the tank design.
Analysis of the Options
- 12 cm/min: This falls within the acceptable range for sedimentation, allowing particles to settle effectively.
- 39 cm/min: This velocity is significantly higher than the typical range. Excessive velocity can lead to turbulence, preventing effective sedimentation and causing re-suspension of settled particles. This is why option 'B' is not feasible.
- 5 cm/min: This is a low but acceptable velocity that can promote effective sedimentation.
- 20 cm/min: This is at the upper limit of the normal operational range, but still allows for effective particle settling.
Conclusion
The velocity of 39 cm/min cannot be used in a horizontal flow sedimentation tank as it exceeds the typical operational limits, leading to ineffective sedimentation processes. Proper design and operational practices dictate that values above this threshold can disrupt the sedimentation process significantly.
In the context of horizontal flow sedimentation tanks, the velocity of flow is a crucial factor that affects the sedimentation process. The velocity must be optimized to allow sufficient time for particles to settle without causing re-suspension.
Typical Velocity Ranges
- Horizontal flow sedimentation tanks generally operate within a specific range of flow velocities.
- Typical design velocities are often between 1 to 20 cm/min, depending on the type of particles being treated and the tank design.
Analysis of the Options
- 12 cm/min: This falls within the acceptable range for sedimentation, allowing particles to settle effectively.
- 39 cm/min: This velocity is significantly higher than the typical range. Excessive velocity can lead to turbulence, preventing effective sedimentation and causing re-suspension of settled particles. This is why option 'B' is not feasible.
- 5 cm/min: This is a low but acceptable velocity that can promote effective sedimentation.
- 20 cm/min: This is at the upper limit of the normal operational range, but still allows for effective particle settling.
Conclusion
The velocity of 39 cm/min cannot be used in a horizontal flow sedimentation tank as it exceeds the typical operational limits, leading to ineffective sedimentation processes. Proper design and operational practices dictate that values above this threshold can disrupt the sedimentation process significantly.
When the pH is between 5 and 10, the chlorine in the water acts as ______- a)Hypochlorous acid
- b)Hypochlorite ions
- c)Molecular chlorine
- d)Hypochlorous acid hypochlorite ions
Correct answer is option 'D'. Can you explain this answer?
When the pH is between 5 and 10, the chlorine in the water acts as ______
a)
Hypochlorous acid
b)
Hypochlorite ions
c)
Molecular chlorine
d)
Hypochlorous acid hypochlorite ions
|
Diya Patel answered |
Explanation:
pH Range:
- When the pH of water is between 5 and 10, the chlorine in the water acts as a combination of hypochlorous acid and hypochlorite ions.
Role of Hypochlorous Acid (HOCl):
- Hypochlorous acid (HOCl) is a weak acid that acts as a powerful disinfectant and sanitizer.
- It is effective in killing bacteria, viruses, and other harmful microorganisms in water.
Role of Hypochlorite Ions (OCl-):
- Hypochlorite ions (OCl-) are the conjugate base of hypochlorous acid.
- They also contribute to the disinfection process by oxidizing and destroying organic and inorganic contaminants in water.
Equilibrium:
- In the pH range of 5 to 10, there is an equilibrium between hypochlorous acid and hypochlorite ions.
- The proportions of HOCl and OCl- depend on the pH of the water.
Importance of pH Range:
- Maintaining the pH of water within the range of 5 to 10 ensures that the chlorine remains in the form of hypochlorous acid and hypochlorite ions, which are effective disinfectants.
- If the pH strays outside this range, the efficiency of chlorine as a disinfectant may decrease.
In conclusion, in the pH range of 5 to 10, chlorine in water primarily acts as a combination of hypochlorous acid and hypochlorite ions, which help in disinfecting and sanitizing the water effectively.
The thickness of the base material on which filter media are supported is- a)10-100cm
- b)20-80cm
- c)30-75cm
- d)10-30cm
Correct answer is option 'C'. Can you explain this answer?
The thickness of the base material on which filter media are supported is
a)
10-100cm
b)
20-80cm
c)
30-75cm
d)
10-30cm
|
|
Nandini Mishra answered |
The thickness of the base material on which filter media are supported is 30-75cm.
Introduction:
In the field of chemical engineering, filter media are commonly used in various processes to separate solids from liquids or gases. These filter media are typically supported on a base material, which provides structural integrity and acts as a support for the filter media. The thickness of this base material is an important parameter that affects the performance and efficiency of the filtration process.
Importance of the base material thickness:
The thickness of the base material plays a crucial role in determining the overall effectiveness of the filtration process. It affects the capacity of the filter to hold the filter media firmly in place, prevents media migration, and allows for the proper flow of the fluid through the filter.
Optimal thickness range:
The correct answer to the given question is option 'C' - 30-75cm. This range of thickness is considered to be optimal for supporting the filter media. Let's understand why:
1. Structural integrity: The base material should have sufficient thickness to provide structural integrity to the filter. Thicker base material ensures better stability and prevents deformation or collapse of the filter during operation.
2. Preventing media migration: The base material should be thick enough to prevent the filter media from migrating or being washed away by the fluid flow. This ensures that the filter media stay in place and effectively capture the desired particles.
3. Uniform flow distribution: The thickness of the base material also affects the flow distribution of the fluid through the filter. A thicker base material allows for better distribution of the fluid flow, ensuring uniform filtration and preventing channeling or bypassing of the fluid.
4. Optimal pressure drop: The thickness of the base material influences the pressure drop across the filter. A thicker base material can accommodate a larger volume of fluid, resulting in a lower pressure drop. This is important to maintain efficient filtration without excessive energy consumption.
5. Handling and maintenance: The thickness of the base material should be practical for handling and maintenance purposes. A thickness range of 30-75cm strikes a balance between providing sufficient support and ease of handling during installation, replacement, and cleaning of the filter media.
Conclusion:
In conclusion, the thickness of the base material on which filter media are supported is typically in the range of 30-75cm. This range ensures structural integrity, prevents media migration, allows for uniform flow distribution, and provides optimal pressure drop. It also considers practical aspects of handling and maintenance.
Introduction:
In the field of chemical engineering, filter media are commonly used in various processes to separate solids from liquids or gases. These filter media are typically supported on a base material, which provides structural integrity and acts as a support for the filter media. The thickness of this base material is an important parameter that affects the performance and efficiency of the filtration process.
Importance of the base material thickness:
The thickness of the base material plays a crucial role in determining the overall effectiveness of the filtration process. It affects the capacity of the filter to hold the filter media firmly in place, prevents media migration, and allows for the proper flow of the fluid through the filter.
Optimal thickness range:
The correct answer to the given question is option 'C' - 30-75cm. This range of thickness is considered to be optimal for supporting the filter media. Let's understand why:
1. Structural integrity: The base material should have sufficient thickness to provide structural integrity to the filter. Thicker base material ensures better stability and prevents deformation or collapse of the filter during operation.
2. Preventing media migration: The base material should be thick enough to prevent the filter media from migrating or being washed away by the fluid flow. This ensures that the filter media stay in place and effectively capture the desired particles.
3. Uniform flow distribution: The thickness of the base material also affects the flow distribution of the fluid through the filter. A thicker base material allows for better distribution of the fluid flow, ensuring uniform filtration and preventing channeling or bypassing of the fluid.
4. Optimal pressure drop: The thickness of the base material influences the pressure drop across the filter. A thicker base material can accommodate a larger volume of fluid, resulting in a lower pressure drop. This is important to maintain efficient filtration without excessive energy consumption.
5. Handling and maintenance: The thickness of the base material should be practical for handling and maintenance purposes. A thickness range of 30-75cm strikes a balance between providing sufficient support and ease of handling during installation, replacement, and cleaning of the filter media.
Conclusion:
In conclusion, the thickness of the base material on which filter media are supported is typically in the range of 30-75cm. This range ensures structural integrity, prevents media migration, allows for uniform flow distribution, and provides optimal pressure drop. It also considers practical aspects of handling and maintenance.
The amount of chlorine used for plain chlorination of water is about - a)0.2 ppm
- b)0.3 ppm
- c)0.4 ppm
- d)0.5 ppm
Correct answer is option 'D'. Can you explain this answer?
The amount of chlorine used for plain chlorination of water is about
a)
0.2 ppm
b)
0.3 ppm
c)
0.4 ppm
d)
0.5 ppm
|
|
Sankar Dasgupta answered |
Chlorination of Water
Chlorination of water is the process of adding chlorine or its compounds to water to disinfect it and kill any disease-causing microorganisms present in it.
Amount of Chlorine Used
The amount of chlorine used for plain chlorination of water is about 0.5 ppm (parts per million). This means that for every million parts of water, 0.5 parts of chlorine are added.
Importance of Chlorine Concentration
The concentration of chlorine in water is very important for effective disinfection. If the concentration is too low, it may not kill all the microorganisms present in the water. On the other hand, if the concentration is too high, it may lead to the formation of harmful byproducts that can have adverse effects on human health.
Factors Affecting Chlorine Concentration
Several factors can affect the chlorine concentration in water, including:
1. Water temperature: Higher water temperatures require more chlorine to be effective.
2. pH level: The pH level of water can affect the stability of chlorine and its ability to disinfect water.
3. Organic matter: The presence of organic matter in water can react with chlorine, reducing its effectiveness.
4. Contact time: The amount of time that chlorine is in contact with water can affect its effectiveness.
Conclusion
In conclusion, the amount of chlorine used for plain chlorination of water is about 0.5 ppm. This concentration is important for effective disinfection of water and several factors can affect the stability and effectiveness of chlorine in water.
Chlorination of water is the process of adding chlorine or its compounds to water to disinfect it and kill any disease-causing microorganisms present in it.
Amount of Chlorine Used
The amount of chlorine used for plain chlorination of water is about 0.5 ppm (parts per million). This means that for every million parts of water, 0.5 parts of chlorine are added.
Importance of Chlorine Concentration
The concentration of chlorine in water is very important for effective disinfection. If the concentration is too low, it may not kill all the microorganisms present in the water. On the other hand, if the concentration is too high, it may lead to the formation of harmful byproducts that can have adverse effects on human health.
Factors Affecting Chlorine Concentration
Several factors can affect the chlorine concentration in water, including:
1. Water temperature: Higher water temperatures require more chlorine to be effective.
2. pH level: The pH level of water can affect the stability of chlorine and its ability to disinfect water.
3. Organic matter: The presence of organic matter in water can react with chlorine, reducing its effectiveness.
4. Contact time: The amount of time that chlorine is in contact with water can affect its effectiveness.
Conclusion
In conclusion, the amount of chlorine used for plain chlorination of water is about 0.5 ppm. This concentration is important for effective disinfection of water and several factors can affect the stability and effectiveness of chlorine in water.
When the impurities are separated by the gravitation of settling particles, the operation is called ______- a)Sedimentation with coagulant
- b)Plain sedimentation
- c)Secondary sedimentation
- d)Disinfection
Correct answer is option 'B'. Can you explain this answer?
When the impurities are separated by the gravitation of settling particles, the operation is called ______
a)
Sedimentation with coagulant
b)
Plain sedimentation
c)
Secondary sedimentation
d)
Disinfection
|
Advait Chatterjee answered |
Explanation: In plain sedimentation, the impurities are separated from a suspending fluid by use of natural forces like gravitational forces.
Why Alum is preferred over other coagulants?- a)It is easy to dewater the sludge formed
- b)It imparts corrosiveness to water
- c)It reduces taste and odor in addition to turbidity
- d)The time required for floc formation is less
Correct answer is option 'C'. Can you explain this answer?
Why Alum is preferred over other coagulants?
a)
It is easy to dewater the sludge formed
b)
It imparts corrosiveness to water
c)
It reduces taste and odor in addition to turbidity
d)
The time required for floc formation is less
|
|
Lalit Yadav answered |
Alum is preferred over other coagulants because it reduces the taste and odor in addition to turbidity and produce lighter flocs that can be broken easily.
The effective size of a rapid sand filter is ______- a)0.45mm
- b)0.7mm
- c)0.8mm
- d)0.9mm
Correct answer is option 'A'. Can you explain this answer?
The effective size of a rapid sand filter is ______
a)
0.45mm
b)
0.7mm
c)
0.8mm
d)
0.9mm
|
Mehul Nambiar answered |
The effective size of a rapid sand filter refers to the size of the sand particles that are most effective in removing impurities from the water being filtered. It is an important parameter in the design and operation of sand filters used in water treatment processes. The correct answer to the question is option 'A', which is 0.45mm.
- Importance of Effective Size in Rapid Sand Filters:
The effective size of the sand particles in a rapid sand filter determines the filter's ability to remove suspended solids and other impurities from the water. It affects the filter's efficiency, flow rate, and overall performance. The smaller the effective size, the greater the surface area available for filtration, leading to better removal of impurities.
- Definition of Effective Size:
The effective size of a sand filter refers to the diameter of the sand particles that are larger than 10% of the total sand particles by weight. In other words, it is the size of the sand particles through which 10% of the total sand volume passes. It is typically expressed in millimeters (mm).
- Options and Selection:
In the given options, the effective size of 0.45mm is the correct answer. This means that 10% of the sand particles in the filter are smaller than 0.45mm in diameter. This size is commonly used in rapid sand filters for water treatment due to its effectiveness in removing impurities.
- Other Options:
The other options provided, such as 0.7mm, 0.8mm, and 0.9mm, are larger than 0.45mm. This means that a higher percentage of the sand particles in the filter would be larger than these sizes. As a result, the filter's efficiency in removing impurities would be reduced compared to using sand with an effective size of 0.45mm.
- Conclusion:
The effective size of a rapid sand filter plays a crucial role in its performance. The correct answer to the question is option 'A', which represents a sand size of 0.45mm. This size allows for efficient removal of impurities and is commonly used in water treatment processes.
- Importance of Effective Size in Rapid Sand Filters:
The effective size of the sand particles in a rapid sand filter determines the filter's ability to remove suspended solids and other impurities from the water. It affects the filter's efficiency, flow rate, and overall performance. The smaller the effective size, the greater the surface area available for filtration, leading to better removal of impurities.
- Definition of Effective Size:
The effective size of a sand filter refers to the diameter of the sand particles that are larger than 10% of the total sand particles by weight. In other words, it is the size of the sand particles through which 10% of the total sand volume passes. It is typically expressed in millimeters (mm).
- Options and Selection:
In the given options, the effective size of 0.45mm is the correct answer. This means that 10% of the sand particles in the filter are smaller than 0.45mm in diameter. This size is commonly used in rapid sand filters for water treatment due to its effectiveness in removing impurities.
- Other Options:
The other options provided, such as 0.7mm, 0.8mm, and 0.9mm, are larger than 0.45mm. This means that a higher percentage of the sand particles in the filter would be larger than these sizes. As a result, the filter's efficiency in removing impurities would be reduced compared to using sand with an effective size of 0.45mm.
- Conclusion:
The effective size of a rapid sand filter plays a crucial role in its performance. The correct answer to the question is option 'A', which represents a sand size of 0.45mm. This size allows for efficient removal of impurities and is commonly used in water treatment processes.
Which Chloramine is formed, when the pH range is less than 4.4?- a)Monochloramine
- b)Dichloramine
- c)Both Trichloramine and Dichloramine
- d)Trichloramine
Correct answer is option 'D'. Can you explain this answer?
Which Chloramine is formed, when the pH range is less than 4.4?
a)
Monochloramine
b)
Dichloramine
c)
Both Trichloramine and Dichloramine
d)
Trichloramine
|
Abhay Kapoor answered |
Formation of Chloramines at Different pH Ranges:
- Chloramines are formed when chlorine and ammonia are added to water.
- The pH of the water plays a crucial role in determining the type of chloramine that is formed.
pH Range Less than 4.4:
- When the pH range is less than 4.4, both dichloramine and trichloramine are formed, but trichloramine is the dominant species.
- This is because at low pH, ammonia is protonated to form ammonium ions (NH4+), which react with chlorine to form dichloramine (NH2Cl) and trichloramine (NCl3).
- However, the formation of trichloramine is favored over dichloramine because the latter is more susceptible to hydrolysis at low pH.
Significance of Trichloramine:
- Trichloramine is a potent disinfectant and can effectively kill bacteria, viruses, and other microorganisms.
- However, it is also a respiratory irritant and can cause respiratory problems, especially in indoor swimming pools and other enclosed spaces.
- Therefore, it is important to maintain the pH of the water in swimming pools and other water systems to prevent the formation of trichloramine.
Conclusion:
- When the pH range is less than 4.4, both dichloramine and trichloramine are formed, but trichloramine is the dominant species.
- Trichloramine is a potent disinfectant but can also cause respiratory problems, especially in indoor swimming pools and other enclosed spaces.
- Therefore, it is important to maintain the pH of the water in swimming pools and other water systems to prevent the formation of trichloramine.
- Chloramines are formed when chlorine and ammonia are added to water.
- The pH of the water plays a crucial role in determining the type of chloramine that is formed.
pH Range Less than 4.4:
- When the pH range is less than 4.4, both dichloramine and trichloramine are formed, but trichloramine is the dominant species.
- This is because at low pH, ammonia is protonated to form ammonium ions (NH4+), which react with chlorine to form dichloramine (NH2Cl) and trichloramine (NCl3).
- However, the formation of trichloramine is favored over dichloramine because the latter is more susceptible to hydrolysis at low pH.
Significance of Trichloramine:
- Trichloramine is a potent disinfectant and can effectively kill bacteria, viruses, and other microorganisms.
- However, it is also a respiratory irritant and can cause respiratory problems, especially in indoor swimming pools and other enclosed spaces.
- Therefore, it is important to maintain the pH of the water in swimming pools and other water systems to prevent the formation of trichloramine.
Conclusion:
- When the pH range is less than 4.4, both dichloramine and trichloramine are formed, but trichloramine is the dominant species.
- Trichloramine is a potent disinfectant but can also cause respiratory problems, especially in indoor swimming pools and other enclosed spaces.
- Therefore, it is important to maintain the pH of the water in swimming pools and other water systems to prevent the formation of trichloramine.
The head loss through the bed of solids of the filter can be determined by- a)Carmen-Kozney equation
- b)Rose equation
- c)Carmen-Kozney and Rose equation
- d)Charles equation
Correct answer is option 'C'. Can you explain this answer?
The head loss through the bed of solids of the filter can be determined by
a)
Carmen-Kozney equation
b)
Rose equation
c)
Carmen-Kozney and Rose equation
d)
Charles equation
|
|
Preethi Das answered |
Explanation: The head loss through the bed of solids of the filter can be determined by both Carmen-Kozney and Rose equation where two cases are considered, one for homogeneous mixed bed and other for stratified bed.
What indicates the permanent hardness when alum is added to water? - a)Al (OH)3
- b)CaSO4
- c)CO2
- d)Ca (OH)3
Correct answer is option 'B'. Can you explain this answer?
What indicates the permanent hardness when alum is added to water?
a)
Al (OH)3
b)
CaSO4
c)
CO2
d)
Ca (OH)3
|
|
Shounak Basu answered |
Explanation: The presence of calcium sulfate indicates the permanent hardness in water when alum is added to water.
1cm per second settling velocity corresponds to a surface loading of ______ litre per day per m2 - a)36000
- b)864000
- c)24000
- d)6000
Correct answer is option 'B'. Can you explain this answer?
1cm per second settling velocity corresponds to a surface loading of ______ litre per day per m2
a)
36000
b)
864000
c)
24000
d)
6000
|
Naveen Saha answered |
1. Introduction:
This question is related to settling velocity and surface loading in the context of chemical engineering. We are given the settling velocity of 1 cm per second and we need to determine the corresponding surface loading in liters per day per square meter.
2. Understanding the problem:
To solve this problem, we need to understand the definitions of settling velocity and surface loading.
- Settling velocity: Settling velocity is the speed at which particles in a fluid settle under the influence of gravity. It is a measure of the rate at which particles settle in a given medium.
- Surface loading: Surface loading refers to the rate at which a substance is applied to a specific surface area. In this case, it represents the rate at which particles settle on a specific surface area.
3. Calculation:
To calculate the surface loading, we need to convert the settling velocity from cm per second to liters per day per square meter.
- Conversion of settling velocity:
1 cm per second = 0.01 m per second (as 1 m = 100 cm)
= 0.01 * 60 * 60 seconds per hour (as there are 60 seconds in a minute and 60 minutes in an hour)
= 36 m per hour
- Conversion of surface loading:
Surface loading is calculated as the product of settling velocity and the density of the particles.
Let's assume the density of the particles is ρ kg/m^3.
Surface loading = settling velocity * density
= 36 m/hour * ρ kg/m^3
To convert this to liters per day per square meter, we need to multiply by appropriate conversion factors:
1 m^3 = 1000 liters
1 day = 24 hours
Surface loading = (36 m/hour * ρ kg/m^3) * (1000 liters/m^3) * (24 hours/day)
= 864,000 ρ liters per day per square meter
4. Conclusion:
Therefore, the surface loading corresponding to a settling velocity of 1 cm per second is 864,000 liters per day per square meter.
This question is related to settling velocity and surface loading in the context of chemical engineering. We are given the settling velocity of 1 cm per second and we need to determine the corresponding surface loading in liters per day per square meter.
2. Understanding the problem:
To solve this problem, we need to understand the definitions of settling velocity and surface loading.
- Settling velocity: Settling velocity is the speed at which particles in a fluid settle under the influence of gravity. It is a measure of the rate at which particles settle in a given medium.
- Surface loading: Surface loading refers to the rate at which a substance is applied to a specific surface area. In this case, it represents the rate at which particles settle on a specific surface area.
3. Calculation:
To calculate the surface loading, we need to convert the settling velocity from cm per second to liters per day per square meter.
- Conversion of settling velocity:
1 cm per second = 0.01 m per second (as 1 m = 100 cm)
= 0.01 * 60 * 60 seconds per hour (as there are 60 seconds in a minute and 60 minutes in an hour)
= 36 m per hour
- Conversion of surface loading:
Surface loading is calculated as the product of settling velocity and the density of the particles.
Let's assume the density of the particles is ρ kg/m^3.
Surface loading = settling velocity * density
= 36 m/hour * ρ kg/m^3
To convert this to liters per day per square meter, we need to multiply by appropriate conversion factors:
1 m^3 = 1000 liters
1 day = 24 hours
Surface loading = (36 m/hour * ρ kg/m^3) * (1000 liters/m^3) * (24 hours/day)
= 864,000 ρ liters per day per square meter
4. Conclusion:
Therefore, the surface loading corresponding to a settling velocity of 1 cm per second is 864,000 liters per day per square meter.
What is the settling velocity of the particle if its diameter is 2 * 10-3cm. Given G=2.65, viscosity v=8 * 10-3cm2/Sec? - a)0.01cm/Sec
- b)0.13cm/Sec
- c)0.24cm/Sec
- d)0.36cm/Sec
Correct answer is option 'B'. Can you explain this answer?
What is the settling velocity of the particle if its diameter is 2 * 10-3cm. Given G=2.65, viscosity v=8 * 10-3cm2/Sec?
a)
0.01cm/Sec
b)
0.13cm/Sec
c)
0.24cm/Sec
d)
0.36cm/Sec
|
|
Palak Sengupta answered |
Explanation: Settling velocity of the particle is given by vs=g (G-1) d2/18v
Specific gravity G=1, g=981cm2/Sec, viscosity v=8 * 10-3cm2/Sec
vs = 981* (2.65-1) *0.002*0.002/(18*.008) = 0.13cm/Sec..
Specific gravity G=1, g=981cm2/Sec, viscosity v=8 * 10-3cm2/Sec
vs = 981* (2.65-1) *0.002*0.002/(18*.008) = 0.13cm/Sec..
What is formed when coagulant is added to water?- a)Scum
- b)Soap
- c)Bubbles
- d)Floc
Correct answer is option 'D'. Can you explain this answer?
What is formed when coagulant is added to water?
a)
Scum
b)
Soap
c)
Bubbles
d)
Floc
|
|
Parth Banerjee answered |
Explanation: When coagulant is added to water, then white gelatinous precipitate is formed called as floc.
Settling tank efficiency is reduced by
1. Eddy current
2. Surface current
3. Density current- a)1, 2
- b)1, 2, 3
- c)2, 3
- d)1, 3
Correct answer is option 'B'. Can you explain this answer?
Settling tank efficiency is reduced by
1. Eddy current
2. Surface current
3. Density current
1. Eddy current
2. Surface current
3. Density current
a)
1, 2
b)
1, 2, 3
c)
2, 3
d)
1, 3
|
|
Abhiram Choudhury answered |
Explanation: Eddy current is set up by the inertia of the incoming fluid, surface current is induced due to wind in open tank and density current cause cold water to under run the warm water to flow across its surface.
What will be the surface area of the basin for a flow of 2*106 litre per day having a surface loading rate 10,000 litre/day/m2? - a)100
- b)200
- c)300
- d)400
Correct answer is option 'B'. Can you explain this answer?
What will be the surface area of the basin for a flow of 2*106 litre per day having a surface loading rate 10,000 litre/day/m2?
a)
100
b)
200
c)
300
d)
400
|
|
Mihir Kumar answered |
Explanation: Surface area of basin= flow of water/surface loading rate = 2000000/10000=200m2.
_______ is the removal of calcium, magnesium and certain other metal cations in hard water.- a)Disinfection
- b)Sedimentation
- c)Softening
- d)Cleaning
Correct answer is option 'C'. Can you explain this answer?
_______ is the removal of calcium, magnesium and certain other metal cations in hard water.
a)
Disinfection
b)
Sedimentation
c)
Softening
d)
Cleaning
|
|
Lavanya Menon answered |
Softening is the removal of calcium, magnesium and certain other metal cations in hard water. The product water that is soft water extends the life of plumbing.
The effective size of sand of the slow sand filter is- a)0.1-0.2mm
- b)0.2-0.3mm
- c)0.4-0.7mm
- d)1-2mm
Correct answer is option 'B'. Can you explain this answer?
The effective size of sand of the slow sand filter is
a)
0.1-0.2mm
b)
0.2-0.3mm
c)
0.4-0.7mm
d)
1-2mm
|
|
Rohan Majumdar answered |
Explanation: The effective size of sand of the slow sand filter is 0.2-0.3mm and for rapid sand filter, the effective size is 0.4-0.7mm.
Boiling of water is a _____ method of disinfection.- a)Physical
- b)Chemical
- c)Mechanical
- d)Electrical
Correct answer is option 'A'. Can you explain this answer?
Boiling of water is a _____ method of disinfection.
a)
Physical
b)
Chemical
c)
Mechanical
d)
Electrical
|
|
Lavanya Menon answered |
Boiling of water is a physical method of disinfection which is also called as disinfection by heat.
The elapsed time in gravity filter is 1sec. What is the height of fall in a single descent?- a)4.4m
- b)4.9m
- c)8.8m
- d)9.8m
Correct answer is option 'B'. Can you explain this answer?
The elapsed time in gravity filter is 1sec. What is the height of fall in a single descent?
a)
4.4m
b)
4.9m
c)
8.8m
d)
9.8m
|
|
Niharika Kaur answered |
Answer: b
Explanation: t = 1sec, g = 9.8
Elapsed time, t = (2h/g) 1/2 in a single descent.
Height of fall, H = t2g/2 = 1*1*9.8/2= 4.9m.
Explanation: t = 1sec, g = 9.8
Elapsed time, t = (2h/g) 1/2 in a single descent.
Height of fall, H = t2g/2 = 1*1*9.8/2= 4.9m.
Consider the following statements regarding sand as filter material.
1. It should be free from clay or silt
2. Organic matter should be present in the sand
3. It should be nonuniform
Which of the above statement is/are correct?- a)1, 2
- b)1 only
- c)2, 3
- d)1, 2, 3
Correct answer is option 'B'. Can you explain this answer?
Consider the following statements regarding sand as filter material.
1. It should be free from clay or silt
2. Organic matter should be present in the sand
3. It should be nonuniform
Which of the above statement is/are correct?
1. It should be free from clay or silt
2. Organic matter should be present in the sand
3. It should be nonuniform
Which of the above statement is/are correct?
a)
1, 2
b)
1 only
c)
2, 3
d)
1, 2, 3
|
|
Lakshmi Das answered |
Explanation: Sand used as a filter material should be free from clay, silt, organic matter or other impurity. It should be uniform and of proper size.
Why Alum is preferred over other coagulants?- a)It is easy to dewater the sludge formed
- b)It imparts corrosiveness to water
- c)It reduces taste and odor in addition to turbidity
- d)The time required for floc formation is less
Correct answer is option 'C'. Can you explain this answer?
Why Alum is preferred over other coagulants?
a)
It is easy to dewater the sludge formed
b)
It imparts corrosiveness to water
c)
It reduces taste and odor in addition to turbidity
d)
The time required for floc formation is less
|
|
Sagarika Chopra answered |
Explanation: Alum is preferred over other coagulants because it reduces the taste and odor in addition to turbidity and produce lighter flocs that can be broken easily.
The head loss through screen depends on- a)Shape of screen elements
- b)Open area, block area
- c)Approach velocity
- d)Nature of their construction, approach velocity
Correct answer is option 'D'. Can you explain this answer?
The head loss through screen depends on
a)
Shape of screen elements
b)
Open area, block area
c)
Approach velocity
d)
Nature of their construction, approach velocity
|
|
Shraddha Nair answered |
Explanation: The head loss through the screen is given by, h=B (w/b) 4/3h1*sinx Where h=head loss, h1=approach velocity head, b=minimum width, x=angle of rack, B=shape factor. It also depends on the nature of their construction.
The bacterial removal efficiency of the rapid sand filter is ______ in comparison to slow sand filter.- a)More
- b)Less
- c)Equal
- d)Better
Correct answer is option 'B'. Can you explain this answer?
The bacterial removal efficiency of the rapid sand filter is ______ in comparison to slow sand filter.
a)
More
b)
Less
c)
Equal
d)
Better
|
|
Prasad Sengupta answered |
Explanation: The bacterial removal efficiency of the rapid sand filter is90-99%, whereas of a slow sand filter is 98-99%.
Which of the following process is used to remove the colloidal particles from water?- a)Chemical precipitation
- b)Chemical coagulation
- c)Ion exchange
- d)Adsorption
Correct answer is option 'B'. Can you explain this answer?
Which of the following process is used to remove the colloidal particles from water?
a)
Chemical precipitation
b)
Chemical coagulation
c)
Ion exchange
d)
Adsorption
|
|
Arjun Chawla answered |
Chemical coagulation is the process used to remove colloidal particles from water. It involves the addition of chemicals that destabilize the colloidal particles, causing them to clump together and form larger particles, which can then be easily removed through sedimentation or filtration.
Here is a detailed explanation of the process:
1. Introduction to Colloidal Particles in Water:
- Colloidal particles are small particles suspended in water, ranging in size from 1 to 100 nanometers.
- They are too small to settle under gravity and are resistant to filtration.
- Colloidal particles can cause turbidity in water and may carry impurities such as organic matter, bacteria, and viruses.
2. Chemical Coagulation:
- Chemical coagulation is a widely used method to remove colloidal particles from water.
- Coagulants, such as aluminum sulfate (alum) or ferric chloride, are added to the water.
- These coagulants neutralize the surface charge of colloidal particles, causing them to destabilize.
3. Destabilization of Colloidal Particles:
- Colloidal particles have a negative surface charge due to ionization of functional groups on their surface.
- The coagulant ions neutralize this charge, reducing the repulsive forces between particles.
- As a result, the particles come closer together, allowing attractive forces (van der Waals forces) to act between them.
4. Formation of Coagulation Flocs:
- The attraction between particles leads to the formation of larger aggregates called coagulation flocs.
- These flocs continue to grow as more particles are incorporated into them.
- The size and strength of the flocs depend on various factors such as coagulant dosage, pH, and mixing intensity.
5. Settling or Filtration:
- Once the colloidal particles have formed larger flocs, they can be removed from the water through sedimentation or filtration.
- Sedimentation involves allowing the flocs to settle under gravity in a settling tank, where they form a sludge layer at the bottom.
- Filtration can be used to remove smaller flocs that may not settle effectively.
- The filtered water is then further treated to remove any remaining impurities.
In conclusion, chemical coagulation is an effective method for removing colloidal particles from water. It involves the addition of coagulants to destabilize the particles, leading to the formation of larger flocs that can be easily removed through sedimentation or filtration.
Here is a detailed explanation of the process:
1. Introduction to Colloidal Particles in Water:
- Colloidal particles are small particles suspended in water, ranging in size from 1 to 100 nanometers.
- They are too small to settle under gravity and are resistant to filtration.
- Colloidal particles can cause turbidity in water and may carry impurities such as organic matter, bacteria, and viruses.
2. Chemical Coagulation:
- Chemical coagulation is a widely used method to remove colloidal particles from water.
- Coagulants, such as aluminum sulfate (alum) or ferric chloride, are added to the water.
- These coagulants neutralize the surface charge of colloidal particles, causing them to destabilize.
3. Destabilization of Colloidal Particles:
- Colloidal particles have a negative surface charge due to ionization of functional groups on their surface.
- The coagulant ions neutralize this charge, reducing the repulsive forces between particles.
- As a result, the particles come closer together, allowing attractive forces (van der Waals forces) to act between them.
4. Formation of Coagulation Flocs:
- The attraction between particles leads to the formation of larger aggregates called coagulation flocs.
- These flocs continue to grow as more particles are incorporated into them.
- The size and strength of the flocs depend on various factors such as coagulant dosage, pH, and mixing intensity.
5. Settling or Filtration:
- Once the colloidal particles have formed larger flocs, they can be removed from the water through sedimentation or filtration.
- Sedimentation involves allowing the flocs to settle under gravity in a settling tank, where they form a sludge layer at the bottom.
- Filtration can be used to remove smaller flocs that may not settle effectively.
- The filtered water is then further treated to remove any remaining impurities.
In conclusion, chemical coagulation is an effective method for removing colloidal particles from water. It involves the addition of coagulants to destabilize the particles, leading to the formation of larger flocs that can be easily removed through sedimentation or filtration.
Which gas is released when alum is added to water? - a)Al (OH)3
- b)CaSO4
- c)CO2
- d)Ca (OH)3
Correct answer is option 'C'. Can you explain this answer?
Which gas is released when alum is added to water?
a)
Al (OH)3
b)
CaSO4
c)
CO2
d)
Ca (OH)3
|
|
Prasad Menon answered |
Explanation: Carbon dioxide gas is released when alum is added to water, which is corrosive to metals.
Range of Displacement efficiency in the plain sedimentation tank is- a)0.25-0.5
- b)0.1-0.2
- c)0.5-0.8
- d)0.3-0.6
Correct answer is option 'A'. Can you explain this answer?
Range of Displacement efficiency in the plain sedimentation tank is
a)
0.25-0.5
b)
0.1-0.2
c)
0.5-0.8
d)
0.3-0.6
|
|
Kiran Choudhary answered |
Explanation: Displacement efficiency is defined as the ratio of flowing through period to the detention period and varies from 0.25 to 0.5 in the plain sedimentation tank.
Which material is provided for the up flow bed in dual media bi-flow filter?- a)Anthracite
- b)Sand
- c)Rice husk
- d)Graphite
Correct answer is option 'B'. Can you explain this answer?
Which material is provided for the up flow bed in dual media bi-flow filter?
a)
Anthracite
b)
Sand
c)
Rice husk
d)
Graphite
|
|
Surbhi Kulkarni answered |
Understanding Dual Media Bi-Flow Filters
Dual media filters utilize two different types of materials to enhance filtration efficiency. These filters are especially effective for removing suspended solids from water and other liquids.
Primary Materials in Dual Media Filters
- Anthracite: This is a type of coal known for its high carbon content and low ash content. It is often used as the top layer in filters due to its larger particle size, which allows for effective removal of larger particles.
- Sand: Sand is the primary material used in the upflow bed of a dual media bi-flow filter. Its smaller particle size creates a dense bed that effectively traps finer particles and provides a high surface area for biological growth.
- Rice Husk: While rice husks can be used in some filtration systems, they are not standard in dual media filters.
- Graphite: This material is not typically used in filtration applications due to its properties and lack of filtration advantages.
Why Sand is the Correct Answer
- Filtration Efficiency: Sand's fine granules allow for the capture of smaller particles that are not retained by larger materials like anthracite.
- Cost-Effectiveness: Sand is widely available and relatively inexpensive compared to other filtration materials.
- Biological Growth: Sand provides a suitable substrate for microbial growth, enhancing the biological filtration process.
In conclusion, sand is the correct material for the upflow bed in dual media bi-flow filters, as it significantly contributes to the overall filtration effectiveness and efficiency of the system.
Dual media filters utilize two different types of materials to enhance filtration efficiency. These filters are especially effective for removing suspended solids from water and other liquids.
Primary Materials in Dual Media Filters
- Anthracite: This is a type of coal known for its high carbon content and low ash content. It is often used as the top layer in filters due to its larger particle size, which allows for effective removal of larger particles.
- Sand: Sand is the primary material used in the upflow bed of a dual media bi-flow filter. Its smaller particle size creates a dense bed that effectively traps finer particles and provides a high surface area for biological growth.
- Rice Husk: While rice husks can be used in some filtration systems, they are not standard in dual media filters.
- Graphite: This material is not typically used in filtration applications due to its properties and lack of filtration advantages.
Why Sand is the Correct Answer
- Filtration Efficiency: Sand's fine granules allow for the capture of smaller particles that are not retained by larger materials like anthracite.
- Cost-Effectiveness: Sand is widely available and relatively inexpensive compared to other filtration materials.
- Biological Growth: Sand provides a suitable substrate for microbial growth, enhancing the biological filtration process.
In conclusion, sand is the correct material for the upflow bed in dual media bi-flow filters, as it significantly contributes to the overall filtration effectiveness and efficiency of the system.
The uniformity characteristics of sand expressed in terms of .- a)Effective size
- b)Effective size and uniformity coefficient
- c)Uniformity coefficient
- d)Mean velocity
Correct answer is option 'B'. Can you explain this answer?
The uniformity characteristics of sand expressed in terms of .
a)
Effective size
b)
Effective size and uniformity coefficient
c)
Uniformity coefficient
d)
Mean velocity
|
|
Nikhil Chawla answered |
Explanation: The uniformity characteristics of sand are expressed in terms of effective size and uniformity coefficient. Effective size is called as effective diameter and uniformity coefficient is a measure of particle range.
What is the diameter of drum of a Micro-Strainer?- a)1m
- b)2m
- c)3m
- d)4m
Correct answer is option 'C'. Can you explain this answer?
What is the diameter of drum of a Micro-Strainer?
a)
1m
b)
2m
c)
3m
d)
4m
|
|
Rishika Bajaj answered |
Micro-Strainer Drum Diameter:
The diameter of the drum of a Micro-Strainer is typically around 3 meters.
Explanation:
Micro-Strainer:
- A Micro-Strainer is a type of filtration system used for removing fine particles from liquids.
- It consists of a rotating drum with fine mesh screens that trap particles as the liquid passes through.
Drum Diameter:
- The diameter of the drum in a Micro-Strainer is an important factor in determining the filtration capacity of the system.
- A larger drum diameter allows for more surface area for filtration, increasing the efficiency of particle removal.
- The standard diameter of a Micro-Strainer drum is usually around 3 meters, although variations may exist depending on the specific application and manufacturer.
Conclusion:
- In conclusion, the diameter of the drum of a Micro-Strainer is typically around 3 meters, providing an optimal balance between filtration capacity and system size.
The diameter of the drum of a Micro-Strainer is typically around 3 meters.
Explanation:
Micro-Strainer:
- A Micro-Strainer is a type of filtration system used for removing fine particles from liquids.
- It consists of a rotating drum with fine mesh screens that trap particles as the liquid passes through.
Drum Diameter:
- The diameter of the drum in a Micro-Strainer is an important factor in determining the filtration capacity of the system.
- A larger drum diameter allows for more surface area for filtration, increasing the efficiency of particle removal.
- The standard diameter of a Micro-Strainer drum is usually around 3 meters, although variations may exist depending on the specific application and manufacturer.
Conclusion:
- In conclusion, the diameter of the drum of a Micro-Strainer is typically around 3 meters, providing an optimal balance between filtration capacity and system size.
Organic contaminants are removed from the wastewater by- a)Water softening
- b)Demineralization
- c)Absorption
- d)Adsorption
Correct answer is option 'D'. Can you explain this answer?
Organic contaminants are removed from the wastewater by
a)
Water softening
b)
Demineralization
c)
Absorption
d)
Adsorption
|
|
Nikhil Choudhury answered |
Explanation: Organic contaminants are removed from the wastewater by the attraction and accumulation of one substance on the surface of another.
Calculate the amount of softener resin required for the following information.
Magnesium: 80 ppm as CaCO3
Capacity: 47 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs- a)4.5 m3
- b)3.5 m3
- c)5.1 m3
- d)2.5 m3
Correct answer is option 'C'. Can you explain this answer?
Calculate the amount of softener resin required for the following information.
Magnesium: 80 ppm as CaCO3
Capacity: 47 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs
Magnesium: 80 ppm as CaCO3
Capacity: 47 Kg/m3
Correction factor: 0.9
Flow rate: 135 m3/hr
Operation hours: 20 hrs
a)
4.5 m3
b)
3.5 m3
c)
5.1 m3
d)
2.5 m3
|
|
Lalit Yadav answered |
Magnesium load= Magnesium as CaCO3 x Flow rate x Operating hours/1000. Resin Volume= Magnesium load/ (Exchange capacity x Correction factor). Resin Volume =[(80 x 135 x 20)/1000]/(47 x 0.9)= 5.1 m3.
The spacing between the bars in racks is- a)30mm
- b)50mm
- c)70mm
- d)90mm
Correct answer is option 'B'. Can you explain this answer?
The spacing between the bars in racks is
a)
30mm
b)
50mm
c)
70mm
d)
90mm
|
|
Nikhil Chawla answered |
Spacing between the bars in racks
The correct answer is option 'B' which states that the spacing between the bars in racks is 50mm. Let's understand why this answer is correct.
Introduction to racks
Racks are commonly used in various industries, including chemical engineering, to store and organize materials or equipment. They are designed with horizontal bars or shelves that provide support and allow for easy access to the stored items.
Importance of spacing in racks
The spacing between the bars in racks is an important consideration as it directly impacts the storage capacity and accessibility of the rack. The spacing should be chosen carefully to ensure optimal utilization of the available space while allowing for easy placement and retrieval of items.
Factors influencing spacing
Several factors are taken into account when determining the spacing between the bars in racks, including:
1. Size and shape of the stored items: The spacing should be suitable to accommodate the dimensions of the items being stored. If the items are larger or irregularly shaped, a wider spacing may be required to ensure they fit properly.
2. Weight-bearing capacity: The spacing should be chosen in such a way that it can support the weight of the stored items without causing any deformation or damage to the rack.
3. Accessibility and maneuverability: Sufficient space between the bars is necessary to allow for easy access to the stored items. This ensures that items can be placed or removed from the rack without any difficulty or risk of damage.
Choosing the ideal spacing
Considering the various factors mentioned above, a spacing of 50mm is often considered optimal for most applications. This spacing provides a good balance between storage capacity and accessibility. It allows for efficient utilization of the rack while ensuring easy placement and retrieval of items.
Other spacing options
While the correct answer is option 'B' (50mm), it is worth noting that other spacing options may be used in different scenarios. For example:
- Smaller spacing, such as 30mm, may be chosen when storing smaller items that require more support or when a higher storage capacity is desired.
- Larger spacing, such as 70mm or 90mm, may be chosen for storing larger or bulkier items that require more space or when the accessibility of items is not a primary concern.
Ultimately, the choice of spacing depends on the specific requirements of the application and the type of items being stored in the rack.
The correct answer is option 'B' which states that the spacing between the bars in racks is 50mm. Let's understand why this answer is correct.
Introduction to racks
Racks are commonly used in various industries, including chemical engineering, to store and organize materials or equipment. They are designed with horizontal bars or shelves that provide support and allow for easy access to the stored items.
Importance of spacing in racks
The spacing between the bars in racks is an important consideration as it directly impacts the storage capacity and accessibility of the rack. The spacing should be chosen carefully to ensure optimal utilization of the available space while allowing for easy placement and retrieval of items.
Factors influencing spacing
Several factors are taken into account when determining the spacing between the bars in racks, including:
1. Size and shape of the stored items: The spacing should be suitable to accommodate the dimensions of the items being stored. If the items are larger or irregularly shaped, a wider spacing may be required to ensure they fit properly.
2. Weight-bearing capacity: The spacing should be chosen in such a way that it can support the weight of the stored items without causing any deformation or damage to the rack.
3. Accessibility and maneuverability: Sufficient space between the bars is necessary to allow for easy access to the stored items. This ensures that items can be placed or removed from the rack without any difficulty or risk of damage.
Choosing the ideal spacing
Considering the various factors mentioned above, a spacing of 50mm is often considered optimal for most applications. This spacing provides a good balance between storage capacity and accessibility. It allows for efficient utilization of the rack while ensuring easy placement and retrieval of items.
Other spacing options
While the correct answer is option 'B' (50mm), it is worth noting that other spacing options may be used in different scenarios. For example:
- Smaller spacing, such as 30mm, may be chosen when storing smaller items that require more support or when a higher storage capacity is desired.
- Larger spacing, such as 70mm or 90mm, may be chosen for storing larger or bulkier items that require more space or when the accessibility of items is not a primary concern.
Ultimately, the choice of spacing depends on the specific requirements of the application and the type of items being stored in the rack.
Consider the following statement regarding depth of sand of the rapid gravity filter?
1. It should be checked against breakage of floc
2. The sand depth should lie between 100-200cm
3. The floc should be attached through the sand bed
Which of the above statement is/are correct?- a)Only 1
- b)1, 2
- c)2, 3
- d)Only 2
Correct answer is option 'A'. Can you explain this answer?
Consider the following statement regarding depth of sand of the rapid gravity filter?
1. It should be checked against breakage of floc
2. The sand depth should lie between 100-200cm
3. The floc should be attached through the sand bed
Which of the above statement is/are correct?
1. It should be checked against breakage of floc
2. The sand depth should lie between 100-200cm
3. The floc should be attached through the sand bed
Which of the above statement is/are correct?
a)
Only 1
b)
1, 2
c)
2, 3
d)
Only 2
|
Anirban Pillai answered |
Explanation: Sand depth should be such that floc do not break through the sand bed and the depth of sand should vary between 60-90 cm.
In the expression vS = 1.8 (g*d (G-1)) 1/2, the diameter of spherical particle is - a)0.1mm
- b)< 0.1mm
- c)1mm
- d)>1mm
Correct answer is option 'D'. Can you explain this answer?
In the expression vS = 1.8 (g*d (G-1)) 1/2, the diameter of spherical particle is
a)
0.1mm
b)
< 0.1mm
c)
1mm
d)
>1mm
|
|
Alok Chakraborty answered |
Explanation: The settling velocity of spherical particle is given by stokes law and when the diameter d of spherical particle is > 1mm, settling velocity is given by vS = 1.8 (g*d (G-1)) 1/2where G is the specific gravity of the particle.
Which of the following has highest shape factor as stated by Carmen?- a)Ottawa sand
- b)Pulverized coal
- c)Rounded coal
- d)Angular sand
Correct answer is option 'A'. Can you explain this answer?
Which of the following has highest shape factor as stated by Carmen?
a)
Ottawa sand
b)
Pulverized coal
c)
Rounded coal
d)
Angular sand
|
|
Nayanika Gupta answered |
Explanation: The shape factor as stated by Carmen is 0.95 for Ottawa sand, 0.73 for pulverized coal and angular sand and 0.82 for rounded sand.
Which of the following is correct regarding the cheapest filter material used in water treatment?- a)Sand>Anthracite>Garnet sand
- b)Sand<Anthracite>Garnet sand
- c)Sand<Anthracite<Garnet sand
- d)Sand=Anthracite>Garnet sand
Correct answer is option 'A'. Can you explain this answer?
Which of the following is correct regarding the cheapest filter material used in water treatment?
a)
Sand>Anthracite>Garnet sand
b)
Sand<Anthracite>Garnet sand
c)
Sand<Anthracite<Garnet sand
d)
Sand=Anthracite>Garnet sand
|
|
Charvi Gupta answered |
Explanation: Sand is the cheapest filter material followed by anthracite whereas Garnet sand is the costliest among them and is not used as a sole filter material but as a constituent in mixed media filter.
Chapter doubts & questions for Treatment of Water - Environmental Engineering 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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