Sedimentation MCQ Quiz - Objective Question with Answer for Sedimentation - Download Free PDF
Last updated on May 15, 2025
Latest Sedimentation MCQ Objective Questions
Sedimentation Question 1:
A soil is composed of solid spherical grains of identical specific gravity and diameter between 0.075 mm and 0.0075 mm. If the terminal velocity of largest particle falling through water without flocculation is 0.5 mm/sec then, that for the smallest particle would be
Answer (Detailed Solution Below)
Sedimentation Question 1 Detailed Solution
Concept:
The settling velocity of sand particles is given by Stoke's Law,
\({{\rm{V}}_{\rm{s}}} = \frac{{g \times \left( {G - 1} \right) \times {d^2}}}{{18\;ν }}\)...... (1)
G -Specific gravity of sand particles
ν - Kinematic viscosity
d - Diameter of sand particles
Calculation:
Given:
Let V1 and V2 be settling velocities of larger and smaller grain respectively
V1 = 0.5 mm/sec
Diameter of larger grain d1 = 0.075 mm
Diameter of smaller grain d2 = 0.0075 mm
g is constant, specific gravity identical for both grains, both the particles are falling through water hence ν remains same. Hence equation 1 becomes
⇒ \(\frac{{{V_1}}}{{{V_2}}}\; = \;{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}\)
⇒ \(\frac{{{0.5}}}{{{V_2}}}\; = \;{\left( {\frac{{{0.075}}}{{{0.0075}}}} \right)^2}\)
⇒ V2 = 0.005 mm/sec
Sedimentation Question 2:
In a sewage treatment system, a grit chamber of dimensions 10 m × 1.5 m × 0.75 m liquid depth has a flow of 750 m3 / h. The surface loading rate and detention time of the grit chamber are respectively:
Answer (Detailed Solution Below)
Sedimentation Question 2 Detailed Solution
Explanation:
Grit Chamber Dimensions 10 m × 1.5 m × 0.75 m
i.e.Surface area = 10 m × 1.5 m
Flow Rate (Q) = 750 m3/h
We know,
a) \(\rm Surface \ loading \ rate = \dfrac{Q}{{Surface\;Area}}\)
\(= \dfrac{{750}}{{10 \times 1.5}}\)
= 50 m3/hr/m2 or 50,000 l/hr/m2
b) \(\rm Detention \ time= \dfrac{{Volume}}{Q}\)
\(= \dfrac{{10 \times 1.5 \times 0.75}}{{750}} \times 60 = 0.9\; \rm min\)
Additional Information
Grit chambers:
(i) These are long narrow rectangular channels with an increased cross-sectional dimension that decelerate flow and allow grit to settle.
(ii) These units are provided in order to carry out the removal of the inorganic suspended solid and to pass forward the suspended solid in PST.
(iii) These chambers are designed in the form of a long channel, which may either be rectangular or parabolic in cross-section.
(iv) The grit chamber is a settling tank with a provision to control flow velocity.
(v) A velocity control device known as proportional weir flow and Parshall flume is provided at the outlet of the grit chamber to maintain constant flow velocity irrespective of the flow rate.
(vi) If the rectangular section is adopted, the proportional weir is provided in front of it and if the parabolic section is adopted, a parshall flume is provided in front of it.
Sedimentation Question 3:
Wastewater is treated in the sedimentation stage which is divided into four types as shown in the figure.
Match the following process of sedimentation corresponding to type I, II, III and IV against A, B, C, and D.
Hindered settling, discrete setting, compression settling and Flocculant settling.
Based on this, select the correct option
Answer (Detailed Solution Below)
Sedimentation Question 3 Detailed Solution
Explanation:
Type I: Discrete particle settling(A)
- Particles settle individually without interaction with neighboring particles.
Type II: Flocculant particle settling(B)
- Flocculation causes the particles to increase in mass and settle at a faster rate.
Type III: Hindered or Zone settling(C)
- The mass of a particle tends to settle as a unit with individual particles remaining in fixed positions with respect to each other.
- The concentration of particles is so high that sedimentation can only occur through compaction of the structure.
- In the primary treatment stage of a wastewater treatment plant, the removal of large suspended organic solids takes place in sedimentation tanks.
- These tanks are designed to allow the wastewater to flow slowly, which promotes the settling of heavier solids.
- As the wastewater enters the sedimentation tanks, the flow velocity decreases, allowing the larger and denser organic solids to settle to the bottom of the tank under the force of gravity. This process is called sedimentation or settling.
Sedimentation Question 4:
In continuous flow sedimentation tank the quantity of water passing per hour (or per day) per unit horizontal area is known as:
Answer (Detailed Solution Below)
Sedimentation Question 4 Detailed Solution
Concept:
Surface Loading Rate (SLR) or Surface Overflow Rate (SOR)
Volume of water applied in unit time per unit surface area of the settling tank is known as the Surface Loading Rate. It is also denoted by Vo.
\({\rm{SOR\;or\;}}{{\rm{V}}_{\rm{o}}} = \frac{{\frac{{{\rm{Volume\;of\;water\;}}}}{{{\rm{Time}}}}}}{{{\rm{Surface\;area}}}} = \frac{{{\rm{Discharge\;}}}}{{{\rm{Surface\;area}}}}\)
It is expressed in m3/d/m2
Additional Information
O ther design parameters of Primary sedimentation tank:
1) Detention Time (DT)
The average theoretical time taken by water to travel from inlet to outlet of settling tank is known as detention time. Detention time should be greater than the settling time of the particle to capture the particle in the settling tank.
\({\rm{DT}} = \frac{{{\rm{Volume\;of\;settling\;tank}}}}{{{\rm{Discharge}}}} = \frac{{\rm{V}}}{{\rm{Q}}}\)
It is expressed in hours
2) Flow through velocity
The speed with which the water travels from inlet to outlet of the settling tank.
\({\rm{V}} = \frac{{{\rm{Discharge\;}}}}{{{\rm{Cross}} - {\rm{section\;area\;of\;tank}}}} = \frac{{\rm{Q}}}{{\rm{A}}}\)
It is expressed in m/min
3) Weir Loading Rate (WLR)
Discharge per unit length of the weir is known as the weir loading rate.
\({\rm{WLR}} = \frac{{{\rm{Discharge}}}}{{{\rm{Length\;of\;weir}}}} = \frac{{\rm{Q}}}{{\rm{L}}}\)
It is expressed in m3/day/m
Sedimentation Question 5:
Stokes' law is applicable to soil particles of size between
Answer (Detailed Solution Below)
Sedimentation Question 5 Detailed Solution
Explanation:
Stokes Law
Assumptions
- It is assumed that particles show discrete settling (i.e. grains of different sizes fall through a liquid at different velocities).
- Particles assumed spherical in shape.
- Medium is assumed infinite.
- Particles size range 0.2 mm to 0.0002 mm.
Settling velocity,
\({V_s} = {\frac{{(G - 1){γ _w}d}}{{18μ }}^2}\)
where,
G = specific gravity, γw = unit weight of water, d = diameter of particle, μ = dynamic viscosity
Additional Information
Limitations in the use of Stoke's Law in Sedimentation analysis:
- Stoke's law is applicable for spherical particles only. The fine clay particles are not spherical in shape. While applying Stoke's law, the concept of equivalent diameter is used. The equivalent diameter of a soil particle is defined as the diameter of an imaginary sphere that has the same specific gravity as the soil particle and settles with the same terminal velocity as that of the soil particle.
- It is assumed that every particle settles independently without interference from other particles as well as from the sides of the jar. To minimize error due to this assumption, it is recommended that not more than 50 gms of soil particles be taken in 1000 ml of soil suspension.
- The soil particles in the soil suspension may have different values of specific gravity. But in the computations, an average value of G is used.
- The lower limit of particle size for the validity of Stoke's Law is 0.0002 mm. However, the upper limit for the same is 0.2 mm. For particles of size less than 0.0002 mm, Brownian movement affects their settlement and in the case of particles larger than 0.2 mm, turbulence affects the settlement.
- Stokes's law is valid upto a maximum Reynolds number of 1.
Top Sedimentation MCQ Objective Questions
For a given discharge, the efficiency of sedimentation tank can be increased by:
Answer (Detailed Solution Below)
Sedimentation Question 6 Detailed Solution
Download Solution PDFExplanation:
- The efficiency of the sedimentation tank indicates the overall percentage removal of suspended matter at a given overflow rate or surface loading rate.
- The efficiency of the sedimentation tank increases if the overflow rate reduces (more time available to particles for settle).
overflow rate, \({V_s} = \frac{Q}{{BL}}\)
From the equation, it is clear that if the surface area (B × L) of the tank increases, the overflow rate reduces, and efficiency increases for a given discharge.
Important Points
Efficiency depends on the following parameter during sedimentation:
- The velocity of flow: Efficiency increases if the velocity of flow reduces
- Viscosity: Efficiency increases if viscosity reduces (Viscosity changes but we can not do)
- Size of particle: Efficiency increases if the size of particle increases.
The maximum daily demand at a water purification plant has been estimated as 12 million litres per day.Find the cross-sectional area of the sedimentation tank required if a detention period of 6 hours and the velocity of flow of 0.2 m/minute is assumed.
Answer (Detailed Solution Below)
Sedimentation Question 7 Detailed Solution
Download Solution PDFExplanation:
The Cross-sectional area required for the sedimentation tank is calculated as:
\(\text {Cross-Sectional Area} = \frac{\text{Design Flow}}{\text{Velocity of flow}}\)
Calculation:
The Design flow will be the maximum daily demand which is given as 12 million liters per day.
Q = 12 × 106 × 10-3 m3/day
Or
Q = 12000/(24× 60) = 8.33 cum/minute
Velocity of flow, V = 0.2 m/minute
Therefore, the cross-sectional area required for the sedimentation tank is calculated as:
A = 8.33/0.2
∴ A = 41.7 m2
24,00,000 litres of water passes through a sedimentation tank which has rate of flow 300 cubic meter per day. The detention time for the tank is:
Answer (Detailed Solution Below)
Sedimentation Question 8 Detailed Solution
Download Solution PDFConcept:
Detention Period: Detention period of settling tank is the average theoretical time required for water to flow through the tank length and is given by the ratio of the volume of the tank to rate of flow.
Detention period of a rectangular tank \({\rm{\;}}\left( {{{\rm{t}}_{\rm{d}}}} \right) = \frac{{{\rm{Volume\;of\;the\;tank}}}}{{{\rm{Rate\;of\;Flow}}}} = \frac{{{\rm{LBH}}}}{{\rm{Q}}}\)
Calculation:
Given,
Volume of tank = 2400 m3
Rate of flow = 300 m3/day
∴Detention period of the rectangular tank \({\rm{\;}}\left( {{{\rm{t}}_{\rm{d}}}} \right) = \frac{{{\rm{LBH}}}}{{\rm{Q}}} = \frac{{2400}}{{300}} \)
Detention time = 8 days
Important Points
Detention period of a circular tank, \({\rm{\;}}{{\rm{t}}_{\rm{d}}} = {\rm{\;}}\frac{{{\rm{Volume\;of\;the\;tank}}}}{{{\rm{Rate\;of\;Flow}}}} = \frac{{{{\rm{d}}^2} \times \left( {0.011{\rm{d}} + \frac{{\rm{\pi }}}{4}{\rm{H}}} \right)}}{{\rm{Q}}}\)
Where, d is the diameter of the circular tank, H is the vertical wall depth or side water depth and Q is the rate of flow.
In a sedimentation tank design, surface overflow rate (S. O. R) is calculated as
Answer (Detailed Solution Below)
Sedimentation Question 9 Detailed Solution
Download Solution PDFConcept:
Surface Loading Rate (SLR) or Surface Overflow Rate (SOR)
Volume of water applied in unit time per unit surface area of the settling tank is known as the Surface Loading Rate. It is also denoted by Vo.
\({\rm{SOR\;or\;}}{{\rm{V}}_{\rm{o}}} = \frac{{\frac{{{\rm{Volume\;of\;water\;}}}}{{{\rm{Time}}}}}}{{{\rm{Surface\;area}}}} = \frac{{{\rm{Discharge\;}}}}{{{\rm{Surface\;area}}}}\)
It is expressed in m3/d/m2
Additional Information
Other design parameters of Primary sedimentation tank:
1) Detention Time (DT)
The average theoretical time taken by water to travel from inlet to outlet of settling tank is known as detention time. Detention time should be greater than the settling time of the particle to capture the particle in the settling tank.
\({\rm{DT}} = \frac{{{\rm{Volume\;of\;settling\;tank}}}}{{{\rm{Discharge}}}} = \frac{{\rm{V}}}{{\rm{Q}}}\)
It is expressed in hours
2) Flow through velocity
The speed with which the water travels from inlet to outlet of the settling tank.
\({\rm{V}} = \frac{{{\rm{Discharge\;}}}}{{{\rm{Cross}} - {\rm{section\;area\;of\;tank}}}} = \frac{{\rm{Q}}}{{\rm{A}}}\)
It is expressed in m/min
3) Weir Loading Rate (WLR)
Discharge per unit length of the weir is known as the weir loading rate.
\({\rm{WLR}} = \frac{{{\rm{Discharge}}}}{{{\rm{Length\;of\;weir}}}} = \frac{{\rm{Q}}}{{\rm{L}}}\)
It is expressed in m3/day/m
As per IS 10313:1982, which of the following factors does NOT influence the sedimentation process in a sedimentation tank ?
Answer (Detailed Solution Below)
Sedimentation Question 10 Detailed Solution
Download Solution PDFConcepts:
As per IS 10313: 1982, the factors that influence sedimentation are:
- Size, shape and weight of particle;
- Viscosity and temperature of water;
- Surface overflow;
- Surface area;
- Velocity of flow;
- Inlet and outlet arrangement;
- Detention periods; and
- Effective depth of basins.
Additional Information:
Sedimentation usually finds application in two principal ways in water treatment, that is, plain sedimentation and sedimentation following coagulation, flocculation or softening.
1. Plain sedimentation is used to remove settleable solids that occur naturally in surface waters. These solids settle without any previous treatment. Plain sedimentation is usually used as a preliminary process to reduce heavy sediments prior to subsequent treatment process such as coagulation.
2. Sedimentation following chemical coagulation and flocculation is used to remove suspended solids that have been rendered settleable by chemical treatment.
A sedimentation tank removes 210 kg of suspended solids per day. What will be the volume of sludge produced per day if the moisture content of sludge is 95% (by weight) and specific gravity of wet sludge is 1.05?
Answer (Detailed Solution Below)
Sedimentation Question 11 Detailed Solution
Download Solution PDFConcept:
'X' % of Moisture content is a sludge means that 'X; kg of water + (100 - x) kg of solids leads to formation of 100 Kg wet sludge.
Calculation:
Total solids = 210 kg/day (Given)
Moisture Content of sludge = 95% (Given)
95 kg of water + 5 kg of solids = 100 kg of wet sludge
5 kg of solids → 100 kg of wet sludge
1 kg of solids → \(\dfrac{100}{5}=20 \ kg\) of wet sludge
210 kg of solids → (20 × 210) = 4200 kg/day of wet sludge
Weight of wet sludge in one day = 4200 kg/day
Density of wet sludge = ρws = 1.05 × 1000 = 1050 kg/m3
The volume of wet sludge, \(V_{ws}=\dfrac{m}{\rho_{ws}}=\left(\dfrac{4200}{1050}\right)=4 \ m^3\)
Vws = 4 m3
Note: However, if in question, it is given that the moisture content of sludge is 98%. Instead of 95%, then we get the volume of wet sludge = 10 m3 which is option '2'.
The short-circuiting occurring in a sedimentation tank is represented by
Answer (Detailed Solution Below)
Sedimentation Question 12 Detailed Solution
Download Solution PDFExplanation:
Short-circuiting
- The deviation at the actual flow of the tank from the pattern of the ideal tank is called short-circuiting.
- In short-circuiting conditions, top layers of water have detention time (t) which far less than the detention time of the bottom layer.
- When the water in the sedimentation basin short-circuits, the floc does not have enough time to settle out of the water, influencing the economy of the plant and the quality of the treated water.
- The degree of short-circuiting is the deviation of the actual flow pattern to the ideal flow pattern.
- Displacement efficiency, ηd = (Flow through period / Theoretical detention time) × 100
- Flow-through period is the period "during this period the particles in suspension settle down and clear water flows out continuously from the tank. ".
- Detention period - Time taken by water to pass between entry and exit.
What is the detention period and overflow rate for plain sedimentation tank compared to sedimentation with coagulation?
Answer (Detailed Solution Below)
Sedimentation Question 13 Detailed Solution
Download Solution PDFConcept: -
The detention period of the plain sedimentation tanks is more as the particles are smaller in size. Due to their smaller size, they will require a larger time to settle down. While for sedimentation with coagulation, the size of particles increases due to floc formation which results in a lesser detention period of time.
Overflow rate is nothing just another name of settling velocity. Since settling velocity is proportional to the size of the particle.
Greater is the size of particle greater will be the settling velocity.
Hence, sedimentation with a coagulation overflow rate is higher than plain sedimentation.
Note: -
Detention period is defined as the time for which the water was in the tank. And overflow rate is the velocity with which the tank fills.
What is the required plan size of a square sedimentation tank (as the primary sedimentation tank in sweage treatment), given that its effective depth is 3 m, and the flow rate is 40 MLD with admissible surface loading of 100,000 l/ m2 / day?
Answer (Detailed Solution Below)
Sedimentation Question 14 Detailed Solution
Download Solution PDFConcept:
Surface Overflow Rate or Surface Loading: Surface overflow rate (SOR) is parameter that describes the settling characteristics of solids in a specific water or wastewater. It is defined as the volume of water flow per unit of time divided by the surface area (or plan area) of the settling basin.
Hence, \({\rm{SOR}} = \frac{{{\rm{Discharge}}}}{{{\rm{Surface\;or\;Plan\;area\;of\;the\;basin}}}}\)
Calculations:
Given, discharge = 40 MLD = 40 × 106 litres/day and surface loading rate = SOR = 100,000 litre/ m2 / day.
∴ Surface or Plan area of the basin \(= \frac{{{\rm{Discharge}}}}{{{\rm{SOR}}}} = \;\frac{{40{\rm{\;}} \times {\rm{\;}}{{10}^6}{\rm{\;litres}}/{\rm{day\;}}}}{{100,000{\rm{\;litres}}/{\rm{day\;per\;}}{{\rm{m}}^2}}} = 400\;{m^2}\)
As the sedimentation tank is square in shape so plan dimension is \(\sqrt {400} \;m = 20m.\)
∴ The plan dimension of the square sedimentation tank 20 m × 20 m.
In the design of a continuous flow type sedimentation tank, if A= theoretical detention period and B= the actual average time is taken by a batch of water in passing through the settling tank, then what will be the displacement efficiency?
Answer (Detailed Solution Below)
Sedimentation Question 15 Detailed Solution
Download Solution PDFConcept:
Displacement Efficiency:
It is the ratio of flowing through the period to the detention period.
\(Displacement\space Efficiency = {Flowing \space through \space the\space period \over Detention\space period}\)
Explanation:
Given Data
Theoretical detention period = A
The actual average time is taken by a batch of water in passing through the settling tank = B
So, \(Displacement\space Efficiency = {Flowing \space through \space the\space period \over Detention\space period}= {B\over A}\)