Water Supply Engineering Environmental Engg I Prof Rajesh

Water Supply Engineering (Environmental Engg. -I) Prof. Rajesh Bhagat Asst. Professor, CED, YCCE, Nagpur B. E. (Civil Engg. ) GCOE, Amravati M. Tech. (Enviro. Engg. ) VNIT, Nagpur Achievement v Selected Scientist, NEERI-CSIR, Govt. of India. v GATE Qualified Three Times. v UGC - NET Qualified in First Attempt. v Selected Junior Engineer, ZP Washim. v Three Times Selected as UGC Approved Assistant Professor. v Assistant Professor, PCE, Nagpur. v Assistant Professor, Cummins College of Engg. For Women, Nagpur v Topper of Ph. D Course Work at UGC-HRDC, RTMNU Nagpur Mobile No. : - 8483002277 / 8483003474 Email ID : - rajeysh 7 bhagat@gmail. com Website: - www. rajeysh 7 bhagat. wordpress. com

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) The other type of particles are those which change their size, shape & weight and thus loose their identity. 3) Settling- process by which particulates settle to the bottom of a liquid and form a sediment. 4) Sedimentation is process by which the suspended particles that are heavier than water are removed by gravitational settling. 5) 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. 6) The average time theoretically for which the water is detained in the tank is called detention period. 7) Sedimentation process can remove 60% of S. S. & 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: Ø Simplest technologies Ø Little energy input Ø Relatively inexpensive to install and operate Ø No specialized operational skills Ø Easily incorporated into new or existing facilities Disadvantages of Sedimentation: Ø Low hydraulic loading rates Ø Poor removal of small suspended solids Ø Large floor space requirements Ø 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: Raw water continuously admitted into the tank and allowed to flow slowly to settle down particles which is suspension. 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: (Quiescent Type or Fill & Draw Type) In this tank, water is completely brought to rest. This type of tank works intermittently.

1) Horizontal Flow Tanks: a) Rectangular Tanks with Longitudinal Flow b) Circular Tanks with Radial Flow 2) Vertical Flow Tanks: ( Tube Settlers or Plate Settlers) Ø Series of very small tubes is known ass tube settlers. Tubes may be square, circular, rectangular, triangular, hexagonal, etc. Ø Doubling the surface area by inserting tray in a settling tank would double the settling capacity of the tank. Tubes provides laminar flow condition.

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

Typical Dimensions of Sedimentation Tanks Description Dimensions Range Rectangular Depth, m Length, m Width, m 6‑ 10 Typical 3‑ 5 15‑ 90 3 -5 25‑ 40 3‑ 24 Circular Diameter, m Depth, m 4‑ 60 3‑ 5 Bottom Slope, mm/m 60‑ 160 12‑ 45 4. 5 80

SEDIMENTATION Size Shape weight Particles Discrete Flocculating Suspension Dilute Concentrated 16

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: 1) This corresponds to the sedimentation of discrete particle in a suspension of low solids concentration. 2) No interaction between particles 3) Settling velocity is constant for individual particles Type II: Hindered Settling: 1) This type of settling refers to dilute suspension of particles that flocculate during sedimentation process. 2) Highest settling velocity occurs in this zone. 17

Type III: Zone of Settling: 1) Refers to flocculent suspension of Intermediate solids concentration 2) Solids move as a block rather than individual particles 3) Mass of particles subside as whole Type IV: Compression settling 1) Flocculent suspension of high concentration. 2) Volume of solids may decrease

Inlet & Outlet Arrangement for Sedimentation Tank

Design Concept of Sedimentation: 1) Flow Velocity, v. H = (Q/(B x H)) 2) Over Flow Rate = Surface Loading, v. O = (Q/(B x L)) 3) Settling Velocity, vs : - (v. H / v. S ) = (L / H) 4) Particles with settling velocity ‘v. S’ equal to or greater than v. O will settle down. 5) Over flow rate = 500 to 750 lit/Hr/m 2 ( Plain Sedimentation) 6) Over flow rate = 1000 to 1250 lit/Hr/m 2 ( Sedimentation with coagulation) 7) Length of Tank (L) greater than 4 B (Width of tank, B should not exceed 12 m)

Design Concept of Sedimentation: 4) Flow velocity = 0. 15 to 0. 9 m/min. 5) Detention Time = 4 to 8 Hrs 6) Length of Tank = L = v. H x DT 7) Volume of Tank, V = Q x DT 8) Depth of Tank = 3 to 4. 5 m 9) Flow Velocity, v. H = (Q/(B x H)) 10) C/S Area of Tank = V / L 11)C/S Area = Q / v. H = (Q / (B x H)) 12) C/S Area of Tank = B x H 13)Provision for Sludge = 0. 3 to 1. 2 m & F. B. = 0. 3 to 0. 5 m

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 = D 2 (0. 011 D + 0. 785 H)

2) 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

2) Design a plain sedimentation tank for 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 = 395. 83 x 8 = 3167 m 3 = 395. 83 m 3/Hr

2) 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 = 395. 83 m 3/Hr Capacity of Sedimentation tank required = V = Q x DT = 395. 83 x 8 = 3167 m 3 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

2) 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 = 395. 83 m 3/Hr Capacity of Sedimentation tank required = V = Q x DT = 395. 83 x 8 = 3167 m 3 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 ) = 3167 / 96 = 32. 99 m 2

2) 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 = 395. 83 m 3/Hr Capacity of Sedimentation tank required = V = Q x DT = 395. 83 x 8 = 3167 m 3 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 ) = 3167 / 96 = 32. 99 m 2 Assuming the water depth in the tank is 4 m. The width of tank , B = C/S Area / Depth = 32. 99 / 4 = 8. 25 m (< 12 m)

2) 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 = 395. 83 m 3/Hr Capacity of Sedimentation tank required = V = Q x DT = 395. 83 x 8 = 3167 m 3 Velocity of flow to be maintained through the tank, v = 20 cm / min The length of tank required, L = v x DT = 0. 2 x 8 x 60 = 0. 2 m / min. = 96 m The c/s area of the tank is required = ( Capacity of tank / Length of Tank ) = 3167 / 96 = 32. 99 m 2 Assuming the water depth in the tank is 4 m. The width of tank , B = C/s area / Depth = 32. 99/4 = 8. 25 m (< 12 m) FB = 0. 3 m Depth for sludge = 0. 3 m Overall Depth of Tank = 4 + 0. 3 = 4. 6 m Dimension of Tank = 96 x 8. 25 x 4. 6 m

3) Design a plain sedimentation tank for a population of 100000 with water supply rate 135 lpcd.

3) Design a plain sedimentation tank for a population of 100000 with average water supply rate 135 LPCD. Ans: Quantity of water to be treated = 100000 x 135 x 1. 8 = 1012. 5 m 3/ Hr Assuming Detention Time = 6 Hours Capacity of Sedimentation tank required = V = Q x DT = 1012. 5 x 6 = 6075 m 3

3) 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 m 3/ Hr Assuming Detention Time = 6 Hours Capacity of Sedimentation tank required = V = Q x DT = 1012. 5 x 6 = 6075 m 3 Assuming Velocity of flow through the tank, v = 0. 6 m / min. The length of tank required, L = v x DT = 0. 6 x 60 = 216 m

3) 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 m 3/ Hr Assuming Detention Time = 6 Hours Capacity of Sedimentation tank required = V = Q x DT = 1012. 5 x 6 = 6075 m 3 Assuming Velocity of flow through the tank, v = 0. 6 m / min. The length of tank required, L = v x DT = 0. 6 x 60 = 216 m The c/s area of the tank is required = ( Capacity of tank / Length of Tank ) = 6025 / 216 = 28. 125 m 2 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 (< 12 m) FB = 0. 3 m Depth for sludge = 0. 3 m Overall Depth of Tank = 4 + 0. 3 = 4. 6 m Dimension of Tank = 216 x 7. 03 x 4. 6 m

Over Flow Rate = Surface Loading, v. O = (Q / (B x L)) Over flow rate = 500 to 750 lit / Hr /m 2 ( Plain Sedimentation) Check for Que. 02 : v. O = (Q / (B x L)) v. O = (395834 / (8. 25 x 96)) v. O = 500 Lit / Hr /m 2 Check for Que. 03 : v. O = (Q / (B x L)) v. O = (1012. 5 / (7. 03 x 216)) v. O = 666. 78 Lit / Hr /m 2

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

Slow Sand Filter

Slow Sand Filter

Slow Sand Filter v Construction: 1) Open basin rectangular in size 2) Watertight shallow tank constructed in stone or brick masonry 3) Depth of tank 2. 5 to 4 m 4) Surface area 100 to 2000 sq. m 5) Filtration rate 100 to 200 liters per hour per sq. m. 6) Total depth of media 90 to 110 cm. 7) Sand depth 60 to 90 cm and Gravel bed 30 to 60 cm 8) Bed slope 1 in 100 to 1 in 200 towards central drain 9) Finer sand better bacterial removal efficiency but slower filtration rate 10) Effective size of sand (D 10) 0. 2 to 0. 35 11) Uniformity coefficient (D 60/D 10) 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.

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.

Rapid Sand Filter

Rapid Sand Filter

Rapid Sand Filter v 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. 5 m 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 (D 10) 0. 35 to 0. 6 10) Uniformity coefficient (D 60/D 10) 1. 2 to 1. 7

Rapid Sand Filter v 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 (30 cm to 3 m) v 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

Mechanism involved 1) Mechanical Straining: 2) Sedimentation: 3) Adsorption: 4) Biological Metabolism: 5) Electrolytic changes: in Filtration:

4) Design a slow sand filter for 40000 Population & 200 lpcd max. water supply rate: Water to be treated per day = max. daily demand x population = 200 x 40000 = 8000000 lit/day

4) Design a slow sand filter for 40000 Population & 200 lpcd max. water supply rate: Water to be treated per day = max. daily demand x population = 200 x 40000 = 8000000 lit/day Assuming rate of filtration = 200 lit/hr/m 2 = 200 x 24 = 4800 lit/day/m 2

4) Design a slow sand filter for 40000 Population & 200 lpcd max. water supply rate: Water to be treated per day = max. daily demand x population = 200 x 40000 = 8000000 lit/day Assuming rate of filtration = 200 lit/hr/m 2 = 200 x 24 = 4800 lit/day/m 2 Total surface area of filter reqd = (water reqd to be treated per day / rate of filtration) = (8000000/ 4800) = 1667 m 2

4) Design a slow sand filter for 40000 Population & 200 lpcd max. water supply rate: Water to be treated per day = max. daily demand x population = 200 x 40000 = 8000000 lit/day Assuming rate of filtration = 200 lit/hr/m 2 = 200 x 24 = 4800 lit/day/m 2 Total surface area of filter reqd = (water reqd to be treated per day / rate of filtration) = (8000000/ 4800) = 1667 m 2 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 m 2

4) Design a SSF for 40000 Population & 200 lpcd max. water supply rate. Water to be treated per day = max. daily demand x population = 200 x 40000 = 8000000 lit/day Assuming rate of filtration = 200 lit/hr/m 2 = 200 x 24 = 4800 lit/day/m 2 Total surface area of filter reqd = (water reqd to be treated per day / rate of filtration) = (8000000/ 4800) = 1667 m 2 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 m 2 Assuming L= 2 B 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/m 2 3) Total surface area of filter required 4) No. Of units 5) Area of each unit (10 to 100 m 2) 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. 6 m 10) FB = 0. 5 m 11) UDS = 0. 6 m (1. 25 to 1. 35)

Pressure Filter

Pressure Filter 1) Vertical and horizontal 2) Similar to RSF 3) Filtration rate = 6000 to 15000 lit/hr/m 2 4) Pressure = 3 to 7 kg/cm 2 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: Mud & other arrested impurities sticks to sand grain and formation of mud balls. Improper washing of bed leads to formation more mud balls. Compress air along with backwashing for 4 minutes, also supported by manual surface racking or caustic soda. 2) Cracking of filter beds: Fine sand in top layers shrinks & forming cracks 3) Air Binding: Negative pressure making bubbles to stick to sand reducing filtration rate. 4) Sand Incrustation: It is due to accumulation of sticky gelatinous material. Sand grains enlarge in size and effective size changes. carbonization of water can prevent this problem.

Filter Troubles or Problems 5) Jetting & Sand Boils: It occurs when backwashing water follows path of least resistance and break through to the scattered points due to small differences in porosity and permeability. Use of 8 cm thick layer of coarse garnet is also recommended. 6) Sand Leakage: It results when smallest gravels are displaced during backwashing. Water will enter the under-drainage system unfiltered. It can be reduced by properly proportioning of sand gravel layer. In between sand gravel garnet layer can be used to tackle this type of problem.