Prof. Rajesh BhagatAsst. Professor

Civil Engineering Department

Yeshwantrao Chavan College Of Engineering

Nagpur

B. E. (Civil Engg.) M. Tech. (Enviro. Engg.)

GCOE, Amravati VNIT, Nagpur

Mobile No.:- 8483003474 / 8483002277

Email ID:- rajeysh7bhagat@gmail.com

Website:- www.rajeysh7bhagat.wordpress.com

ENVIRONMENTAL ENGINEERING-I

UNIT-IV

1) Sedimentation: Principles types of setting basins, inlet and outlet

arrangements.

2) Clariflocculators: Principles and operation.

3) Filtration: Mechanism of filtration, types of filters RSF, SSF, pressure

filters, elements of filters, UDS, design aspects of filter and operational

problems in filtration.

Sedimentation

1) Discrete or granular particle are those which do not change their size, shape & weight.

2) Settling- process by which particulates settle to the bottom of a liquid and form a

sediment.

3) The other type of particles are those which change their size, shape & weight and thus

loose their identity.

4) The basin in which the flow of water is retarded or storage is offered is called the settling

tank or sedimentation tank or basin or clarifier.

5) The average time theoretically for which the water is detained in the tank is called

detention period.

6) Sedimentation is process by which the suspended particles that are heavier than water are

removed by gravitational settling.

7) Sedimentation process can remove 60% of SS & 75% of bacteria.

8) Floatation is a process by which lighter particles can remove.

Factors Affecting Sedimentation:

1) Flow velocity

2) Viscosity of water

3) Temperature

4) Size of particle

5) Shape of particle

6) Specific gravity of particle

Characteristics of the particles

Size and shape

Specific gravity

Properties of the water

Specific gravity

Viscosity

Physical environment of the particle

Velocity of the water

Inlet and outlet arrangements of the structure

Advantages of Sedimentation:-

Simplest technologies

Little energy input

Relatively inexpensive to install and operate

No specialized operational skills

Easily incorporated into new or existing facilities

Disadvantages of Sedimentation:-

Low hydraulic loading rates

Poor removal of small suspended solids

Large floor space requirements

Re-suspension of solids.

Types of Sedimentation Tank

A. Depending upon shape:

1) Circular

2) Rectangular

3) Square

B. Depending on process of operation:

1) Continuous Tank:

Flow velocity of water is reduced by providing sufficient length of travel. This tank is

designed such that the time taken by the water particle to travel from one end to another end

is kept slightly more than the time required for settling of suspended particles in water.

2) Intermittent Tank:

In this tank, water is completely brought to rest. This type of tank works intermittently.

SEDIMENTATION BASIN ZONES

1) Inlet Zone: The inlet or influent zone should distribute flow uniformly across the inlet to the

tank. The normal design includes baffles that gently spread the flow across the total inlet of the

tank and prevent short circuiting in the tank.

2) Settling Zone: The settling zone is the largest portion of the sedimentation basin. This zone

provides the calm area necessary for the suspended particles to settle.

3) Sludge Zone: The sludge zone, located at the bottom of the tank, provides a storage area for the

sludge before it is removed for additional treatment or disposal. Sludge is removed for further

treatment from the sludge zone by scraper or vacuum devices which move along the bottom.

4) Outlet Zone : The basin outlet zone (or launder) should provide a smooth transition from the

sedimentation zone to the outlet from the tank. This area of the tank also controls the depth of

water in the basin.

Circular Basin Rectangular Basin

Critical Settling Velocity and Overflow Rate

Critical settling velocity (Vo)is the settling velocity of particles which are

100% removed in the basin

Typical Dimensions of Sedimentation Tanks

______________________________________________________Description Dimensions

Range Typical

______________________________________________________

Rectangular

Depth, m 3-5 3.5

Length, m 15-90 25-40

Width, m 3-24 6-10

Circular

Diameter, m 4-60 12-45

Depth, m 3-5 4.5

Bottom Slope, mm/m 60-160 80

______________________________________________________

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SEDIMENTATION

Particles

FlocculatingDiscrete

Suspension

ConcentratedDilute

Size

Shape

weight

15

Types of Settling:-

Depending upon the concentration of solids and the tendency of a particle to interact the

following types of settling may occur

Type I: Discrete Settling:

This corresponds to the sedimentation of discrete particle in a suspension of low

solids concentration.

No interaction between particles

Settling velocity is constant for individual particles

Type II: Hindered Settling:

This type of settling refers to dilute suspension of particles that flocculate during

sedimentation process.

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Type III: Zone of Settling:

Refers to flocculent suspension of Intermediate solids concentration

Solids move as a block rather than individual particles

Mass of particles subside as whole

Type IV: Compression settling

Flocculent suspension of high concentration.

Volume of solids may decrease

Inlet & Outlet Arrangement for Sedimentation Tank

Design Concept of Sedimentation:

1) Flow Velocity, vH = (Q/(B x H))

2) Over Flow Rate = Surface Loading, vO = (Q/(B x L))

3) Settling Velocity, vs :- (vH / vS ) = (L / H)

Particles with settling velocity ‘vS’ equal to or greater than vO will settle down.

4)Over flow rate = 500 to 750 lit/Hr/m2 ( Plain Sedimentation)

5) Over flow rate = 1000 to 1250 lit/Hr/m2 ( Sedimentation with coagulation)

6) Length of Tank, L greater than or equal to 4B (Width of tank, B should not exceed 12 m)

7) Flow velocity = 0.15 to 0.9 m/min.

8) Length of Tank = L = vH x DT (Detention Time = 4 to 8 Hrs)

9) Depth of Tank = 3 to 4.5 m

10)Provision for Sludge = 0.3 to 1.2 m & FB = 0.3 to 0.5m

Design Concept of Sedimentation:

1) Stokes Law :- Vs = ((g/18)(Gs-1)(d2/ u))

g = acceleration due to gravity

Vs = settling velocity

D = diameter of particle, cm

Gs = specific gravity

u = kinematic viscosity = 0.01 cm2 /s

Find the settling velocity of a particle in water for following condition:

Diameter of particle = 0.06 mm, specific gravity of particle = 2.65,

temperature of water = 200c, Kinematic Viscosity of water at 200c = 1.007

centistokes.

G = 9.81 m/s2 = 9810 mm/s2, u = 1.007 centistokes = 1.007mm2/ s

Design Concept of Sedimentation:

1) Stokes Law :- Vs = ((g/18)(Gs-1)(d2/ u))

g = acceleration due to gravity

Vs = settling velocity

D = diameter of particle, cm

Gs = specific gravity

u = kinematic viscosity = 0.01 cm2 /s

Find the settling velocity of a particle in water for following condition:

Diameter of particle = 0.06 mm, specific gravity of particle = 2.65, temperature of water = 200c, Kinematic Viscosity of water at 200c = 1.007 centistokes.

G = 9.81 m/s2 = 9810 mm/s2, u = 1.007 centistokes = 1.007mm2/ s

Vs = 3.21 mm/s

Re = (Vs. Dp) / u = (3.21 x 0.06 )/1.007 = 0.195 < 1

Therefore stokes law is applicable.

sedimentation

A circular sedimentation tank is generally provided with its bottom cone

shaped with a slope of 1 vertical to 12 horizontal. Under this condition, its

diameter is given by:

V = D2 (0.011 D + 0.785 H)

Design a plain sedimentation tank for max. daily demand of water 9.5 x 106 lit /

day. Assume the velocity of flow to be 20 cm / minute and detention time 8 hours.

Ans: Quantity of water to be treated = 9.5 x 106 lit / day = 395833.33 lit / Hr

Capacity of Sedimentation tank required = V = Q x DT = 395833.33 x 8

= 3.17 x 106 Lit

= 3170 m3

Velocity of flow to be maintained through the tank, v = 20 cm / min = 0.2 m / min.

The length of tank required, L = v x DT = 0.2 x 8 x 60 = 96 m

The c/s area of the tank is required = ( Capacity of tank / Length of Tank )

= 3170 / 96 = 33 m2

Assuming the water depth in the tank is 4 m.

The width of tank , B = C/s area / Depth = 33/4 = 8.25 m (< 12m)

Design a plain sedimentation tank for max. daily demand of water 9.5 x 106 lit / day.

Assume the velocity of flow to be 20 cm / minute and detention time 8 hours.

Ans: Quantity of water to be treated = 9.5 x 106 lit / day = 395833.33 lit / Hr

Capacity of Sedimentation tank required = V = Q x DT = 395833.33 x 8

= 3.17 x 106 Lit

= 3170 m3

Velocity of flow to be maintained through the tank, v = 20 cm / min = 0.2 m / min.

The length of tank required, L = v x DT = 0.2 x 8 x 60 = 96 m

The c/s area of the tank is required = ( Capacity of tank / Length of Tank )

= 3170 / 96 = 33 m2

Assuming the water depth in the tank is 4 m.

The width of tank , B = C/s area / Depth = 33/4 = 8.25 m (< 12m)

FB = 0.5m

Depth for sludge = 1m

Overall Depth of Tank = 4 + 1 + 0.5 = 5.5 m

Dimension of Tank = 96 x 8.25 x 5.5m

Design a plain sedimentation tank for a population of 100000 with water supply rate

135 lpcd.

Ans: Quantity of water to be treated = 100000 x 135 x 1.8 = 1012.5 m3/ hr

Assuming Detention Time = 6 Hours

Capacity of Sedimentation tank required = V = Q x DT = 1012.5 x 6

= 6075 m3

Assuming Velocity of flow through the tank, v = 0.6 m / min.

The length of tank required, L = v x DT = 0.6 x 6 x 60 = 216 m

The c/s area of the tank is required = ( Capacity of tank / Length of Tank )

= 6025 / 216 = 28.125 m2

Assuming the water depth in the tank is 4 m.

The width of tank , B = C/s area / Depth = 28.125/4 = 7.03 m (< 12m)

FB = 0.5m

Depth for sludge = 1m

Overall Depth of Tank = 4 + 1 + 0.5 = 5.5 m

Dimension of Tank = 216 x 7.02 x 5.5m

Filtration

1) Process of removal of impurities like very fine suspended particles,

bacteria, color, taste, etc. by passing the water thorough bed of

granular materials.

2) Types of filter:

1) SSF

2) RSF

3) Pressure Filter

Mechanism involved in Filtration

1) Mechanical straining

2) Sedimentation

3) Adsorption

4) Biological metabolism

5) Electrolytic changes

Slow Sand Filter

1) Removes about 98 to 99 % of bacterial load

2) Removes turbidity.

3) Less efficient in removal of color about 20 to 25 %

4) Not highly efficient in the removal of colloidal matter.

Slow Sand Filter

Slow Sand Filter

Slow Sand Filter

Construction:-

Open basin rectangular in size

Watertight shallow tank constructed in stone or brick masonry

Depth of tank 2.5 to 4m

Surface area 100 to 2000 sq. m

Filtration rate 100 to 200 liters per hour per sq. m.

Total depth of media 90 to 110 cm.

Sand depth 60 to 90 cm and Gravel bed 30 to 60 cm

Bed slope 1 in 100 to 1 in 200 towards central drain

Finer sand better bacterial removal efficiency but slower filtration rate

Effective size of sand (D10) 0.2 to 0.35

Uniformity coefficient (D60/D10) 2 to 3

Slow Sand Filter

Operation:-

1) Waters should not be coagulated.

2) Filter is filled with a raw water to a depth of 1 to 1.5 m above the

surface of sand.

3) Water is passed with filtration rate 100 to 200 liters/hr/ sq. m.

4) Loss of head 60 cm

5) For cleaning 20 to 30 cm top sand is scraped & replaced

6) First filter water is collected by under drainage system

Rapid Sand Filter

Rapid Sand Filter

Rapid Sand Filter

Construction:-

1) Open basin rectangular in size

2) Watertight shallow tank constructed in stone or brick or concrete masonry

3) Surface area 10 to 100 sq. m

4) Length to width ratio 1.25 to 1.35

5) Filtration rate 3000 to 6000 liters/hr/sq. m.

6) Depth of tank 2.5 to 3.5m

7) Total depth of media 90 to 130 cm.

8) Sand depth 60 to 70 cm & Gravel bed 60 to 70 cm

9) Effective size of sand (D10) 0.35 to 0.6

10) Uniformity coefficient (D60/D10) 1.2 to 1.7

Rapid Sand Filter

Operation:-

1) Enters filter through inlet pipe

2) Uniformly distributed on the sand bed

3) After passing through bed collected by UDS

4) Initially head loss very small as increases beyond limit bed requires

washing (30cm to 3m)

Back washing:-

1) Back flow of water

2) First water drained out form filter

3) All valves are closed & compressed air is passed for 2-3 minutes

4) Water passes thorough drain, gravel & sand.

5) Sand expands and all impurities are carried away with water drain.

6) After 24 hr & requires 10 minutes

Design of slow sand filter

40000 Population & 200 lpcd max. water supply rate:-

Water to be treated per day = max. daily demand x population

= 200x40000 = 8000000 lit/day

Assuming rate of filtration = 200 lit/hr/m2 = 200x24 = 4800 lit/day/m2

Total surface area of filter reqd = (water reqd to be treated per day / rate of filtration)

= (8000000/ 4800) = 1667 m2

Provide four units, out of four keep one stand by while repairing and hence only three

units provide the necessary filter area reqd.

The area of each filter unit = (1667/3) = 556 m2

Assuming L= 2B

Area of each filter unit = 556 = L x B = 2 B x B

B = 16.7 m & L = 33.4 m

Assuming thickness of sand filter media = 0.9 m

Assuming thickness of gravel filter media = 0.5 m

Depth of water = 1.2 m

Free board = 0.3 m

Under drainage system = 0.6 m

Overall Depth of Filter = 0.9 + 0.5 + 1.2 + 0.3 + 0.6 = 3.5 m

Dimension of filter = 33.4 x 16.7 x 3.5 m

Rapid Sand Filter

1) Water reqd to treat per day in liter

2) Rate of filtration = 3000 to 6000 lit/hr/m2

3) Total surface area of filter reqd

4) No. Of units

5) Area of each unit (10 to 100 m2)

6) L=1.5 B

7) Thickness of sand bed = 60 to 70 cm

8) Thickness of gravel bed = 60 to 70 cm

9) Depth of water = 0.9 to 1.6m

10) FB = 0.5m

11) UDS = 0.6m

Pressure Filter

Pressure Filter

1) Vertical and horizontal

2) Similar to RSF

3) Filtration rate = 6000 to 15000 lit/hr/m2

4) Pressure = 3 to 7 kg/cm2

5) Few houses, private estate, industries, swimming pools, railway station,

etc.

6) Least efficient in removal of bacteria & turbidity

7) Difficult to inspect, clean & replace

8) Effectiveness of backwashing is not visible.

9) Good quality of water

Pressure Filter

Filter troubles or Problems

1) Formation of Mud Balls

2) Clogging and Cracking of filter beds

3) Air binding

4) Sand Incrustation

5) Jetting & Sand Boils

6) Sand Leakage

Filters during backwash