CE3303 UNIT 2 WATER TREATMENT PPT.pptx

Unit 2 Water Treatment
CE3303 water supply and waste
water engineering
Department of civil engineering
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PREPARED BY M.Vinitha AP/Civil

Water Treatment :-
1) It is the removal impurities from the water, which are harmful the human health
(Environment).
2) It is the process for enhancing the quality of water so that it meets water quality
criteria for its fitness for particularpurpose.
3) Water treatment is processes that makes water more acceptable for an end-use,
which may be used for drinking, industry, or medicine.
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Water Treatments :-
 Water treatment is a combination of physical, chemical & biological processes.
 These are three types of unit operation and unit processes.
1) Physical Unit Operation: in which physical forces are predominate for removal
of impurities. (screening, equalization, mixing, flocculation, sedimentation,
floatation, etc.)
2) Chemical Unit Processes: in which removal of contamination are brought about
by chemical activity. (chemical precipitation, disinfection, gas transfer,
adsorption, etc.)
3) Biological Unit Processes: : in which removal of contamination are brought
about by biological activity. (Aerobic, Anaerobic & Aerobic-anaerobic
Biological Unit processes)
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Biological Unit Processes:
 in which removal of contamination are brought about by biological activity.
(Aerobic , Anaerobic & Aerobic-anaerobic Biological Unit processes)
1) Aerobic Biological Unit Processes: removal of impurities or decomposition of
organic matter in presence of oxygen (RSF, Trickling filter, aerated lagoons, etc)
2) Anaerobic Biological Unit Processes: removal impurities or decomposition of
organic matter in absence of oxygen (anaerobic lagoons, etc)
3) Aerobic-Anaeerobic Biological Unit Processes: combination of above two
(stabilization pond etc)
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PREPARED BY M.Vinitha

Methods of Water Treatment :-
1) Screening
2) Aeration
3) Plain Sedimentation or Primary Sedimentation.
4) Coagulation & Flocculation
5) Secondary Sedimentation or Sedimentation with Coagulant (Clarifier,
Clariflocculator)
6) Filtration
7) Disinfection
8) Softening etc.
5

Treatment Units & Function :-
1) Screens : To remove large size floating matter like debris, fish, vegetation,
garbage, etc. from the raw water by protective bar screen.
2) Aeration: To add dissolved oxygen in water.
3) Plain sedimentation: To remove settleable suspended solids & other matter by
passing water through large volume tanks where the flow speed slowsdown.
4) Coagulation: To Add coagulant in water.
5) Flocculation: To form floc in water.
6) Secondary sedimentation (Clarifier, Clariflocculator): water is applied to
large volume tanks where the flow speed slows down and the suspended solids
or floc are removed.
7) Filtration: To trap remaining impurities by physical straining.
8) Disinfection: To kill pathogens.
9) Softening: To remove hardness of water.
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Screens :-
1) Screening devices are the 1st unit operation in treatmentplant
2) To remove large size floating particles such as debris, animals, bushes, pieces of
woods, plastics, leaves, aquatic plants, rags, solids, lumber, tree, branches, roots,
etc.
3) It consists of parallel bars, rods or wire, grating, wire mesh or perforated plate,
having opening of any shape but circular & rectangular arecommon.
4) The materials removed by these devices are termed as screening.
5) The screenings are usually no biodegradable materials & can be used for land
filling purposes.
6) To prevent the pump, pipe, other equipment from getting damage or clogging.
7) To increase the efficiency of other treatment process.
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Screens :-
1) To prevent the pump, pipe, other equipment from getting damage or clogging.
2) To increase the efficiency of other treatment process.
 Types of Screens:-
1) Fine Screens (Opening < 25 mm)
2) Medium Screens (Opening = 25 to 50 mm)
3) Coarse Screens (Opening = 50 to 150 mm)
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Coarse Screens :-
1) Are classified as either bar racks (trash racks) or bar screen.
2) It consists of parallel bars or rods of steel or stainlesssteel.
3) It can be used to entrap the coarse materials to protect the damaging of pumping
units.
4) Bar racks are used as pretreatment applications. These bar racks may be hand
cleaned if small in size or mechanical cleaned if large sizes.
5) To prevent the settling of coarse materials in the channel, the velocity inthe
approach channel must be in range of 0.3 m/sec to 0.6 m/sec.
6) Width of the bar 5 mm to 15 mm.
7) Allowable headloss 150 mm.
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Aeration :-
1) Aeration is the process of exchange of gases by creating good interfacebetween
liquid phase and gas phase.
2) It is the process of bringing water in intimate contact with air, while doing sothe
water absorbs oxygen from the air.
3) CO2 can remove upto 60%.
4) Iron, Manganese.
5) H2S & other gases are also removed up-to certain extent from thewater.
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Objectives of Aeration :-
1) To add the oxygen, to impart freshness to water.
2) To remove or decrease the CO2 content of water & thereby raises its pH value.
3) To remove H2S content of water & thereby removal ofodor.
4) To remove CH4 & taste caused due to organicdecomposition.
5) To convert iron & manganese from their soluble states to their insoluble states,
so that these can be precipitated & removed and thereby we can remove color,
taste & odor.
6) To add oxygen, to oxidize organic matter & thereby removal of volatile
substances.
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Factors Affecting Aeration Process :-
1) If the partial pressure is more, more will be the solubility & rate of exchange of gas &
thereby more will be aeration.
2) If the temperature of water is less, less will be the solubility & less will be aeration.
3) If the concentration of impurities is more then solubility will be less & thereby aeration
will be less.
4) If the surface area of interface is more, more will be the rate of exchange of gases &
thereby aeration will be more.
5) If the thickness of interface is more, less will be the rate of exchange of gas & thereby
aeration will be less.
6) If time of contact is more then more will be exchange of gas & thereby aeration will be
more.
7) If degree of under saturation is more then fast will be rate of exchange & thereby
aeration will be more.
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Limitation of Aerations :-
1) It is not an efficient process for removal of tastes & odors caused by relatively non-
volatile substances such as oils, algae & industrial wastewater.
2) Odor removal is 50% when symura bacteria is present.
3) Aeration may add more oxygen in water & making it more corrosive.
4) Fe & Mn can not be precipitated by aeration when organic matter is present.
5) Possibility of air borne contamination is there.
6) Additional lime may be required to neutralize the CO2 that would be removed byaeration
7) Aeration is not economical in colder months or state or area.
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Types of Aerators :-
1) Spray Aerators.
2) Diffused Air Aerators.
3) Gravity Aerators:
a) Cascade Aerator
b) Multiple Tray Aerator
c) Inclined ApronAerator
d) Slopping TrayAerator.
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Cascade Aerators :-
1) Simplest free fall aerator.
2) Waterfalls & weirs of any kind are cascade aerator.
3) Water allowed to fall through a certain height of 1 to 3 m due to this water comes into
close contact with air.
4) It consist of 3 to 9 steps of concrete or metal.
5) Rise of steps should be 20 to 50 cm.
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Inclined Apron Aerators :-
1) Water allowed to fall along an inclined apron which is usually studded with riffle plates.
2) Due to the riffle plates water jumps into the atmosphere causing more aeration.
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Multiple Tray Aerators :-
1) Consists of closely stack perforated trays or panel.
2) Water is allowed to fall freely from tray to tray due to which water comes in contact with
air.
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Spray Aerators :-
1) Spray aerator divide the water flow into fine streams & small droplets which come into
intimates contact with the air in their trajectory.
2) Water is sprinkled into atmosphere in the form of fine spray or jets or droplets through
nozzles.
3) It requires considerable head (0.75 to 1.5 Kg/cm2) 2 to 9m.
4) Reduces CO2 by 70 % or more & H2S by 90% or more.
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Diffused Air Aerator:-
1) Perforated pipe network is installed at the bottom of tank & compressed air is blown
through these pipes.
2) The air bubbles travel upward through water, thus causing aeration.
3) Air diffuser basins have a retention period of about 10 to 30 minutes, depth of 3 to 5m &
width of 3 to 9m..
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Design Criteria for Design of Cascade Aerator:-
 Area required for Design of Aerator = 0.015 – 0.045m2/m3/hr.
 No. of cascade = 3 to 9.
 Height of aerator = 1 to 3m.
 Rise of each step = 20 to 50 cm.
 Velocity of inlet pipe = 0.3 to 0.9 m/s
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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.
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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
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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.
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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. 30

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)
 Series of very small tubes is known ass tube settlers. Tubes may be square,
circular, rectangular, triangular, hexagonal, etc.
 Doubling the surface area by inserting tray in a settling tank would double the
settling capacity of the tank. Tubes provides laminar flow condition.
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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.
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Circular Basin Rectangular Basin
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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
Flocculating
Discrete
Suspension
Concentrated
Dilute
Size Shape
weight
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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.
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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
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Inlet & Outlet Arrangement for Sedimentation Tank
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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)
4) Particles with settling velocity ‘vS’ equal to or greater than vO will settle down.
5) Over flow rate = 500 to 750 lit/Hr/m2 ( Plain Sedimentation)
6) Over flow rate = 1000 to 1250 lit/Hr/m2 ( Sedimentation with coagulation)
7) Length of Tank (L) greater than 4B (Width of tank, B should not exceed 12 m)
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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 = vH x DT
7) Volume of Tank, V = Q x DT
8) Depth of Tank = 3 to 4.5 m
9) Flow Velocity, vH = (Q/(B x H))
10) C/S Area of Tank = V / L
11)C/S Area = Q / vH vH = (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 0.5m 46

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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)
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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
50

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 = 395.83 m3/Hr
Capacity of Sedimentation tank required = V = Q x DT = 395.83 x 8
= 3167 m3
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= 3167 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
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= 3167 m3
The c/s area of the tank is required = ( Capacity of tank / Length of Tank )
= 3167 / 96 = 32.99 m2
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= 3167 m3
= 3167 / 96 = 32.99 m2
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 (< 12m)
54

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 m3/Hr
= 3167 m3
= 3167 / 96 = 32.99 m2
The width of tank , B = C/s area / Depth = 32.99/4 = 8.25 m (< 12m)
FB = 0.3m
Depth for sludge = 0.3 m
Overall Depth of Tank = 4 + 0.3 + 0.3 = 4.6 m
Dimension of Tank = 96 x 8.25 x 4.6m 55

3) Design a plain sedimentation tank for a population of 100000 with water
supply rate 135 lpcd.
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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 m3/ Hr
Assuming Detention Time = 6 Hours
= 6075 m3
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3) Design a plain sedimentation tank for a population of 100000 with water
supply rate 135 LPCD.
= 6075 m3
Assuming Velocity of flow through the tank, v = 0.6 m / min.
58

3) Design a plain sedimentation tank for a population of 100000 with water supply
rate 135 LPCD.
= 6075 m3
Assuming Velocity of flow through the tank, v = 0.6 m / min.
= 6025 / 216 = 28.125 m2
The width of tank , B = C/s area / Depth = 28.125/4 = 7.03 m (< 12m)
FB = 0.3m
Depth for sludge = 0.3m
Overall Depth of Tank = 4 + 0.3 + 0.3 = 4.6 m
Dimension of Tank = 216 x 7.03 x 4.6m 59

Over Flow Rate = Surface Loading, vO = (Q / (B x L))
Over flow rate = 500 to 750 lit / Hr /m2 ( Plain Sedimentation)
Check for Que. 02 :-
vO = (Q / (B x L))
vO = (395834 / (8.25 x 96))
vO = 500 Lit / Hr /m2
Check for Que. 03 :-
vO = (Q / (B x L))
vO = (1012.5 / (7.03 x 216))
vO = 666.78 Lit / Hr /m2
60

Coagulation:-
1) Process of mixing or adding certain chemicals to form insoluble precipitate
for absorbing suspended & colloidal matter
2) Certain chemicals is called Coagulant. Ex. Alum
3) Finely divided impurities not removed in plain sedimentation without
longer detention periods
4) To Remove turbidity up to 20 NTU & bacterial load by 70%
5) Adopted when turbidity more than 40 NTU
6) Coagulant produces thick gelatinous precipitate & has the property of
arresting & attracting the suspended matter
7) Floc posses +ve charge & impurities has – ve charge
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Why Coagulation :-
1) Even after sedimentation, some finely divided particles such as clay and
coloring matter are held in suspension.
2) These are removed only if their size is increased by using some chemical
compounds.
3) In coagulation, coagulants are added and particles are made to settle down by
neutralization.
62

Factors Affecting Coagulation:-
1) Dose of coagulant
2) Type of coagulant
3) Form of coagulant
4) Type of mixing
5) Type of feeding
6) pH value
7) Temperature
8) Impurities
9) Time
63

Types of coagulant:-
1) Alum, Al2(SO4)3.18H2O, Aluminium Sulphate
2) Copperas FeSO47H2O, Ferrous Sulphate
3) Chlorinated Copperas, Fe(SO4 ) 3 + FeCl3
4) Sodium Aluminate, Na2Al2O4
5) Magnesium Carbonate, MgCO3
64

Alum:-
1) Widely used due to easy availability and minimum cost
2) Recovery is possible and recovery cost is just ¼ th of cost of alum
3) Dose of alum = 10 to 30mg/l
4) Effective when pH = 6.5 to 8.5
5) Requires presence of alkalinity in water
6) Many water have bicarbonate alkalinity naturally
Al2(SO4)3.18H2O + 3Ca(HCO3)2 = 2Al(OH)3 + 3 CaSO4 + 6CO2 + 18H2O
Above reaction indicate that 1 mg of Al3+, reacting with 5.56 mg of alkalinity as
CaCO3, Form 2.89 mg of aluminium hydroxide flocs.
If insufficient alkalinity in water then lime will be added in water then reaction will be
Al2(SO4)3.18H2O + 3Ca(OH)2 = 2Al(OH)3 + CaSO4 + 18H2O
Sometimes soda also added into water
Al2(SO4)3.18H2O + 3NaCO2 = 2Al(OH)3 + 3 NaSO4 + 3CO2 + 18H2O
Soda ash doesn’t cause hardness but expansive than lime
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Advantages of Alum:-
1) Reduces taste and odor in addition to turbidity
2) Cheap and easily available
3) Simple in working
4) Doesn’t required skilled labor
5) Crystal clear water
6) Heavy and better floc
7) Quite tough to broken down
66

Disadvantages of Alum:-
1) Difficult to dewater the sludge
2) Sludge is not easy to dispose off
3) Imparts permanent hardness CaSO4
4) Produces CO2 and can causecorrosion
5) Effective pH range is very small (6.5 to 8.5)
67

Iron compounds as a coagulant:-
1) Ferrous Sulphate, Ferric Chloride and Ferric Sulphate
2) Ferrous sulphate is known as Copperas used as coagulant in conjunction with lime.
FeSO4. 7H2O + Ca (HCO3)2 = Fe (HCO3)2 + CaSO4 + 7 H2O
Fe (HCO3)2 + 2 Ca (OH)2 = Fe (OH)2 + 2 CaCO3 + 2 H2O
1) Further Fe (OH)2 Ferrous Hydroxide is converted to Fe (OH)3 Ferric Hydroxide.
2) Dose of Iron salt depend upon on turbidity, natural alkalinity & CO2 in water.
3) Because of addition of lime, color of water may be permanently set.
4) Moreover iron salt are less expensive than alum but skilled supervision is required.
68

Advantages of Iron compounds as a coagulant:-
1) Floc produced is heavy and readily settles under favorable conditions.
2) They are more efficient & requires less time for reaction.
3) Active over a pH range of 3.8 to 10.
4) Cost of treatment is usually low.
5) Widely used for industrial purpose but less often forpotable.
6) Removes H2S, taste, odor etc
69

Disadvantages of Iron salt as coagulant:-
1) Iron salt causes staining.
2) Iron salt are difficult to store and handle.
3) Lime is required to be added which increases cost of treatment.
4) Promotes growth of iron bacteria & adversely affect the RSFprocess.
5) Imparts more corrosiveness than alum.
6) To maintain proper dosage skilled labor is required.
7) Ferrous fluoride introduces fluorine in supply which can cause ‘mottledteeth.
8) By Addition of lime, the color of water may be permanently set.
70

Coagulants Aids:-
 Certain chemicals added to increase rate of reaction, to reduce the coagulant
dose, to extend the optimum pH range, to produce faster and tougherflocs.
1) Activated silica.
2) Bentonite clay.
3) Lime.
4) Soda ash.
5) Polyelectrolyte.
71

Jar Test :-
 Consists of jar test assembly.
 Used to determine optimum coagulant dose.
 Sample (1L) has to placed into number of beaker.
 Various amounts of dose in each beaker is added.
 Paddle speed = 100 rpm for 1 minute then make it slow for next 10 to25
minutes with 40 rpm.
 After fast & slow mixing allow it to settle down for 30 minutes.
 Smallest dose that produces a more floc is optimum coagulant dose.
 This is trial and error method.
 One factor is varying and others are constant. 72

J a r t e s t
73

Feeding Methods for Coagulant:-
1) There are two methods: Dry feeding and Wet feeding.
2) Choice between dry & wet feeding depends on characterization of coagulants,
convenience of its application, dosage of coagulant and size of plant.
3) Alum being uniform in grain size and not affected by atmospheric moisture can
be fed by dry feeding.
4) Iron salt can not be fed by dry feeding due to hygroscopic property.
5) If dose of coagulant is small and accuracy is needed then coagulant must be fed
in a solution form.
6) Generally for large treatment plant wet feeding is adopted and for small treatment
plant dry feeding is adopted because of more cost of wet feeding device.
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1) Dry feeding:-
 Simple in operation
 Requires less space
 Feeding machine is cheaper
 Control of dose is difficult
 Old method
 Powder form
75

2) Wet feeding:-
 Equipment can be adjusted more readily and easily controlled. Solution is prepared and
stored in tank. Corrosive chemicals creates problems in wet feeding
 Water level in mixing channel increases, water level in float chamber also increases,
thereby float rises and pinion rotates pulley which lifts the plug and more quantity of
coagulant solution is supplied.
76

Mixing Devices:-
 Floc formation depends on the mixing of coagulant.
 Point of turbulence.
 Methods of mixing:-
1. Centrifugal pump
2. Compressed air agitation
3. Narrow mixing channel with flume
4. Mixing basin with baffle wall
5. Mixing with mechanical devices
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1) Centrifugal pump
 Used to raise the water to settling tank
 Dose can be added to the suction pipe
 Water with coagulant passes through the impeller of the pump, mixing by
agitation
2) Compressed air agitation:-
 Water with coagulant is agitated by compressed air frombottom
78

3) Narrow mixing channel with flume (Hydraulic jump)
Coagulant fed from feeding tank.
Turbulence caused by vertical baffle mixes the chemical.
Flume provided to develop the hydraulic jump for turbulence.
79

4) Mixing basin with baffle wall:-
Water flows horizontally for short distance complete turns & continuous back&
forth around the ends of baffle. This cause turbulence &mixing
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5) Mixing with mechanical device:-
1) Most of WTP have mechanical devices ie Flash Mixer
2) Coagulant agitated vigorously by paddle
3) Propeller type impeller in flash mixer with speed of 100 to 1400 rpm
4) Detention time , t = 30 to 60 sec.
5) Velocity gradient, G = 300 to 600 sec-1
6) Power requirements = 1 to 3 watts per m3/hr
81

Flash Mixer:-
 Typical mixing with mechanical driven impeller or paddle.
 Impeller provided at lower end of vertical shaft driven by motor in a deep circular
or rectangular tank.
 Impeller is rotated more than 100 RPM.
82

Flocculation:
 The process of floc formation. After mixing, slow and gentle stirring is permitted to
agglomerate the floc.
84

Clariflocculator :-
1) Consists of feeding devices, mixing devices, flocculator & settlingbasin.
2) Coagulant fed through feeding devices
3) Mixed and agitated in the mixing basin
4) Allowed to agglomerate in flocculation tank
5) Flocculated water passed into sedimentation tank
6) Clarified water taken out through outlet to RSF
7) Removes turbidity up to 20 NTU and 70 % of bacterial load
86

Design criteria for flash mixer:-
1) Temporal mean velocity gradient, G = 300 to 600 sec.-1
2) Power requirements = 1 to 3 watts per m3 per hour
3) RPM of Impeller = 100 to 1400 rpm
4) Ratio of tank dia. To impeller dia. = 5:1 to 3:1
5) Tangential velocity of impeller = 3 m/s at blade tip
6) Ratio of tank ht to tank dia. = 1:1 to 3:1
7) Product, G.t =18000
8) G = (P/uV)1/2
91

Design Criteria For Flocculator :-
1) Temporal mean velocity gradient, G = 10 to 75 sec.-1
2) Power requirements = 10 to 36 Killo watts per MLD
3) RPM of Impeller = 4.5 rpm
4) Velocity of blades = 0.2 to 0.6 m/s
5) Velocity of flow = 0.2 to 0.8 m/s
6) Outlet flow velocity = 0.15 to 0.25 m/s
7) Ratio of tank ht to tank dia. = 1:1 to 3:1
8) Product, G.t = 10000 to 100000
9) G = (P/uV)1/2
10) Area of paddles = 10 to 25 % of c/s area of tank
11) Velocity ratio, k = 0.25
12) Kinematic viscosity, u= 1.0087x10-3 at200c
13) Density, ρ = 998 kg/m3 at 200c
14) Dept of tank = 3 to 4.5m 92

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
93

Slow Sand Filter
94

Slow Sand Filter
95

Slow Sand Filter
 Construction:-
1) Open basin rectangular in size
2) Watertight shallow tank constructed in stone or brick masonry
3) Depth of tank 2.5 to 4m
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 (D10) 0.2 to 0.35
11) Uniformity coefficient (D60/D10) 2 to 3
96

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.
97

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.
98

Rapid Sand Filter
99

Rapid Sand Filter
100

101

102

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
103

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 (30 cm 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
104

105

106

Mechanism involved in Filtration:
1) Mechanical Straining:
2) Sedimentation:
3) Adsorption:
4) Biological Metabolism:
5) Electrolytic changes:
107

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
= 200x40000 = 8000000 lit/day
108

= 200x40000 = 8000000 lit/day
Assuming rate of filtration = 200 lit/hr/m2 = 200x24 = 4800 lit/day/m2
109

= 200x40000 = 8000000 lit/day
Total surface area of filter reqd = (water reqd to be treated per day / rate of filtration)
= (8000000/ 4800) = 1667 m2
110

= 200x40000 = 8000000 lit/day
= (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
111

4) Design a SSF for 40000 Population & 200 lpcd max. water supply rate.
= 200x40000 = 8000000 lit/day
= (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
112

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
113

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 required
4) No. Of units
5) Area of each unit (10 to 100 m2)
6) L=1.5 B (1.25 to 1.35)
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
114

Pressure Filter
115

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
116

117

Pressure Filter 118

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 and 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. 119

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 and gravel layer. In between sand and
gravel garnet layer can be used to tackle this type of problem.
120

121

Disinfection:-
1) Killing of pathogens
2) Sterilization is the process of killing of all bacteria
3) Materials or agent used are called disinfectant
4) Disinfectant destroys microorganism by damaging cell wall of and
inactivation of critical enzyme system responsible for metabolic activity.
Necessity:-
 Water may contain bacteria & other micro-organism which may be pathogenic. It
is therefore necessary to disinfect water to kill bacteria and other micro-organism
and thus prevent water borne diseases.
122

Disinfection:
123

Requirements of Good Disinfectant:-
1) Effective in killing all the harmful pathogens
2) Readily available at reasonable cost
3) Safe to handle
4) Simple method of application
5) Harmless, unobjectionable, economical & easily available
6) Not required skilled labor
7) Not required costly equipment for its application
8) Kill all pathogens in reasonable time at a normal temperature
9) Protect against recontamination
124

Methods of Disinfection:-
 Physical Methods
1. Disinfection by Heat: Boiling of water
2. Disinfection by light: UV light
3. Sunlight is a natural disinfectant
 Chemical Methods
1. Chlorine
2. Bromine
3. Iodine
4. Ozone
5. Potassium Permanganate
6. Metal ions such as copper & silver
7. Treatment with excess lime 125

Boiling of Water
1) Bacteria killed by increasing temp. 800C
2) Most effective methods
3) Impracticable to boiled large quantity of water
4) In case of epidemic consumers may boiled water for drinking
5) Cant take care of future contaminations
126

Ultra Violet Ray Treatment
1) Very costly & effective method.
2) Sun is most powerful source of UV rays but it requires a large exposure area &
long time.
3) Hence generated by machines consisting of mercury vapour lamps enclosed in a
quartz globe.
4) Effective disinfection water should be flow in a thin clear layer.
5) Depth water should not exceeds 10 cm.
6) Colorless & turbidity should not more.
7) No residual effects against re-contamination.
8) Unsuitable for large scale treatment plant.
9) Private building, institutes, hospitals, swimming pools, etc.
10) No color & taste developed.
127

128

129

Treatment With Excess Lime:-
1) Enough lime to bring pH to 9.5 to kill bacteria
2) Removal up to 99 to 100 %
3) Re-corbonation is necessary
130

Iodine & Bromine Treatment
1) Dose = 8 mg/L.
2) Contact period = 5 minutes.
3) In the form of small pills.
4) More costly than chlorine hence used for small water supply.
5) Ex. Swimming pools, army, private plants etc.
6) Less irritating to the eyes than chlorine when used in swimming tank.
7) Provides longer lasting protection.
8) Reduced offensive taste and odors.
131

Halogen Group:
132

Ozone
1) More powerful as well as costly than chlorine.
2) Doesn't provide residual protection against recontamination.
3) Cant be stored.
4) Used on small scale.
5) Dose = 2 to 3 ppm.
6) Contact period = 10 minutes.
7) France and Russia on large scale.
8) Required skilled labor.
9) Gives pleasant and appealing appearance.
133

134

135

136

Potassium Permanganate
1) Good for well water supply or water which is less contaminated .
2) 98 % removal of bacteria.
3) 100 % removal of cholera causing bacteria.
4) Rural areas.
5) Oxidizes organic matter and iron.
6) Pink color means organic matter present.
7) Dose = 1 to 2 mg/L.
8) Contact period = 4 to 6 hrs.
9) Cheap and quite useful.
137

138

Silver & Copper Treatment
1) Removes algae.
2) Take care of future pollution.
3) Contact period = 15 minutes to 3 hrs.
4) No taste and no odor.
5) No harmful effect on human body.
6) Use of silver is very costly.
7) Generally not used for treating public water supplies.
139

Chlorination
1) Universally adopted.
2) Cheap, reliable & easy to handle.
3) Capable of providing residual disinfecting effects for long periods.
4) Ideal disinfectants.
5) Imparts bad taste.
6) Doesn’t affect pH.
7) Not effective when water is alkaline.
140

Purpose of Chlorination
1) To prevent the spread of diseases.
2) To remove bacteria, fungus, and other pathogens.
3) Oxidation of Iron, Fe and H2S.
4) Removal of taste and odor.
5) Oxidation of organic matter.
141

Factors Affecting Chlorination
1) Turbidity.
2) Metallic compound presence.
3) Ammonia compounds.
4) Temperature.
5) pH value.
6) Contact time.
7) Nature and concentration of bacteria.
142

Action of Chlorine
Cl2 + H2O HOCl + H+ + Cl-
HOCl H+ + OCl-
1) When chlorine added it forms hypochlorus acid or hypochlorite ions.
2) HOCL & OCl accomplish disinfection.
3) It has an effect on micro-organism.
4) pH = 10 only OCl ions are found.
5) pH = 7 to 5 HOCL exist.
6) Chlorine doesn’t reacts below pH = 5 & remains as elemental chlorine.
7) HOCl is 80 times more powerful than OCL.
8) Hence pH is maintained above 5 & near to or less than 7. 143

Free Available Chlorine:-
1) Together HOCl and OCl is known as Free Available Chlorine
2) Free Cl2 reacts with compounds such as ammonia, proteins, amino acids &
phenols and form chloramines & chloro-derivatives which are called Combined
Chlorine.
3) Combined chlorine also posses some disinfecting properties but less effective
compared to free available chlorine.
144

Break Point Chlorination
Two action:-
I. Killing of bacteria.
II. Oxidation of organic matter.
 Pure water has no chlorine demand.
 Raw water has chlorine demand.
1) First stage :- killing of bacteria, oxidation of organic matter and metals.
2) Second stage :- forming of combined residual chlorine will gradually increases.
Chloramines (Mono-chloramine, Di-chloramine & Tri-chloramine) have been
recorded as a combined residual chlorine.
3) Third Stage:- decrease in combined residual chlorine indicate the breakdown of
chloramines.
145

146

Break Point Chlorination
1) At point C, gives bad taste & odor (oxidation started or breakdown of
chloramines started).
2) At point D, bad smell & taste suddenly disappear. (oxidation completed)
3) Point D represent the Break Point Chlorination
4) After this point, addition of chlorine results in increase in residual chlorine
represented by line E with slope 450
5) At break point true residual free chlorine is revealed.
6) Addition of chlorine beyond point D is break point chlorination
7) Gives ideas to ensure the desired amount of residual chlorine.
147

148

Significance of Break Point Chlorination:
1) Removes taste and odor.
2) Have adequate effect on bacteria or pathogens.
3) Leave desired chlorine residual.
4) Removes Fe & Mn.
5) Oxidation of organic, ammonia & other compounds.
6) Removes color.
149

Forms of Chlorination
1) Plain chlorination
2) Pre-chlorination
3) Post chlorination
4) Double Chlorination
5) Multiple Chlorination
6) Super Chlorination
7) Dechlorination
150

Plain Chlorination
1) 0.5 to 1 mg/l.
2) Only chlorine treatment & no other treatment.
3) To remove bacteria, color, odor, etc.
4) Controls growth algae.
5) Clear water (turbidity less than 20 NTU)
6) In case of emergency.
151

Pre Chlorination
1) Application before treatment specially filtration.
2) Reduces the quantity of coagulants.
3) Reduces taste & odor.
4) Controls growth of algae.
5) Reduces bacterial load on filters.
6) Prevents decomposition of sludge in settling tanks.
152

153

Post Chlorination
1. Compulsory treatment.
2. Application of chlorine to water.
3. Standard forms of chlorination.
4. After filter and before it enters the distribution system.
5. Residual chlorine = 0.2 mg/l (1mg/L)
154

Multiple Chlorination
1) Application of chlorine at two or more points.
2) Consists of pre & post chlorination.
3) Highly contaminated water.
4) Large amount of bacteria.
155

Super Chlorination
1) Application of chlorine beyond the stage of break point.
2) At the end of filtration.
3) Whenever breakout of water borne diseases.
4) Dose = 10 to 15 mg/l.
5) Residual chlorine = 1 to 3 mg/L.
6) Necessary to remove the excess chlorine by dechlorination before supplied to the
consumers.
156

De-Chlorination
 Removal of excess chlorine from water before supplying to consumers.
 Some residual chlorine should remains in water.
I. Dechlorination by Aearation.
II. Addition of some chemicals such as:
1) Sodium thio-sulphate,
2) Sodium bio-sulphate,
3) Sodium sulphite,
4) Activated carbon,
5) Pottassium permanganate,
6) Ammonia,
7) Sulphur dioxide
157

Testing of Chlorine Residuals
1) Orthotolidine Test.
2) Starch Iodide Test.
3) Chlorotex Test.
4) DPD test.
158

DPD Test
(Diethylene-P-Phenylene Diamine)
1) Widely used in modern days.
2) DPD reagents used.
3) Comparators to compare the color.
4) 10 ml of sample in tube.
5) Add reagent and compare the color.
159

Orthotolidine Test
1) Commonly used to determine the residual chlorine.
2) 0.1 ml of orthotolidine & 0.2 ml Sodium Arsenite solution is added to 10 ml of
sample.
3) Compare the color with colored glass standard.
160

Thank you
161

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