education

1. Chapter five Conventional
methods of water treatment
Mizan Tepi University
Daniel Kebede

2. COAGULATION AND FLOCCULATION
Purpose
 The primary purpose of the coagulation/flocculation
process is the removal of turbidity from the water.
 Turbidity is a cloudy appearance of water caused by
small particles suspended therein.
 Water with little or no turbidity will be clear.

3. Coagulation is a chemical process in which charged particles
(colloids) are destabilized
The addition and rapid mixing
of coagulants
The destabilization of colloidal
and fine particles
The initial aggregation of
destabilized particles

4. • Coagulation is commonly achieved by adding different types of chemicals
(coagulants) to the waste water to promote destabilization of the colloid
dispersion and agglomeration of the resulting individual colloidal particles.
words "coagulation" and "flocculation" are often used interchangeably they
refer to two distinct processes
 Coagulation indicates the process through which colloidal particles and
very fine solid suspensions are destabilized so that they can begin to
agglomerate if the conditions are appropriate
Flocculation refers to the process by which destabilized particles actually
conglomerate into larger aggregates so that they can be separated from the
wastewater.

5. • Principle of coagulation
• when certain chemicals are added to water an insoluble, gelatinous, flocculent is
formed .
• First the coagulant are mixed in the water to produce the required precipitate ,then
the water is sent in the sedimentation.
• The principle of coagulation can be explained from the following two conditions:
• I. Floc formation
• When coagulants (chemicals) are dissolved in water and thoroughly mixed produce
a gelatinous precipitate. This precipitate is known as floc.
• The gelatinous precipitate has therefore, the property of removing fine and colloidal
particles quickly.
• II. Electric charges
• Most particles dissolved in water have a negative charge, so they tend to repel each
other. As a result, they stay dispersed and dissolved or colloidal in the water.
• The purpose of most coagulant chemicals is to neutralize the negative charges on the
turbidity particles to prevent those particles from repelling each other.
• The amount of coagulant which should be added to the water will depend on the
zeta potential. If the zeta potential is large, then more coagulants will be needed.
• .

6. COLLOID STABILITY
When two colloids come in close proximity there are two
forces acting on them.
Electrostatic potential (Zeta potential) – a repelling force
and,
Van der Waals force – an attraction force
Colloids - Particle size between 0.001 to 1.0 micron
Can be hydrophilic- readily dispersed in water (organic colloids) or hydrophobic- no
affinity for water; “stable in water because of electric charge (inorganic colloids

7. COLLOID DESTABILIZATION
Most colloids found in water are negatively charged the
addition of positively charged ions (Na+, Mg2+, Al3+, Fe3+ etc.)
can neutralize the colloidal negative charges.
Coagulant salts dissociate when added to water and produce
positively charged hydroxo-metallic ion complexes
When complexes adsorb to the surface of the colloid, the zeta
potential is reduced and particle is destabilized
Destabilized particles aggregate by attraction due to van der
Waals forces.

8. MECHANISMS OF DESTABILIZATION
Ionic layer compression. A high ionic concentration compresses the
layers composed predominantly of counter ions toward the surface of
the colloid.
Adsorption and charge neutralization. surface charges are
decreased when charged species (particularly trivalent) attach to the
surface of the colloid.
Sweep coagulation. The Al(OH)3 forms in amorphous, gelatinous
flocs that are heavier than water and settle by gravity.
Interparticle bridge. Synthetic polymers linear or branched and are
highly surface reactive.
•IMPURITIES REMOVED BY COAGULATION
 Miscellaneous fragments of animals and vegetables
 Plankton, mainly phtoplankton
 Colloids including clay
 Organic colouring matter

9. • FACTOR AFFECTING COAGULATION
• 1 Type of coagulant
• 2. Dose of coagulant
• 3. Characteristic of water
 Type and quantity of suspended matter
 Temperature of water
 pH of water
• 4. Time and method of mixing
• Common Coagulants
• Coagulant chemicals come in two main types - primary coagulants and
coagulant aids.
• Primary coagulants neutralize the electrical charges of particles in the
water which causes the particles to clump together.
• Coagulant aids add density to slow-settling flocs and add toughness to the
flocs.

10. A coagulant is the substance (chemical) that is added to the
• water to accomplish coagulation.
Three key properties of a coagulant:
Trivalent cation. most efficient cation
Nontoxic. For production of safe water
Insoluble in neutral pH range can precipitate without
leaving high concentration of ions in water.
•In water treatment plants, the following are the
coagulants most commonly used:
• Alum: Al2(SO4)3.14H2O
Ferric chloride: FeCl3
Ferric sulfate: FeSO4
Polyelectrolytes (Polymers)
Aluminum salts are cheaper but iron salts are more effective

11. ALUMINUM

12. EXAMPLE: ALKALINITY CALCULATION
• If 200 mg/L of alum to be added to achieve complete
coagulation. How much alkalinity is consumed in mg/L as
CaCO3?
• Solution:
• 594 mg 366 mg
• Al2(SO4)3.14 H2O + 6HCO3
-  2Al(OH)3+ 6CO2 + 14H2O +
3SO4
-2
 594 mg alum consumes 366 mg HCO3
-
 200 mg alum will consume (366/594) x 200 mg HCO3
-
• = 123 mg HCO3
-
Alk as CaCO3  123
Equivalent weight of CaCO3

13. JAR TEST
Jar test Is used to
• determine:
Proper coagulant
Proper coagulant aid
Proper coagulant dose
Procedure
Add reagents (4 to 6 beakers )
Rapid mix (100 to 150 rpm for 1 minute)
Slow mix (25 to 30 rpm for15 to 20 mins )
Allow Settling (30 to 45 mins)
Turbidity and color removal (%)
pH of supernatant
Plot the values and determine the optimum

14. Figure b
Jar test apparatus with turbid water (a) and three samples during
flocculation (b).
Figure a

16. • The results of the two jar tests are plotted in Figure a .
In the first test, the optimal pH was chosen as 6.0, and
this pH was used for the second jar test.
• From the second jar test, the optimal alum dose was
estimated to be about 12.5 mg/L.
• In actual practice, the laboratory technician would
probably try to repeat the test using a pH of 6.25 and
varying the alum dose between 10 and 14 to pinpoint
the optimal conditions.
• From Figure b , the optimum pH was estimated to be
6.0 and the optimum dose was estimated to be 12.5
mg/L.

17. JAR TEST
– OPTIMUM PH AND DOSE

18. PH ADJUSTMENT
Is used if pH of water to be treated is not within the
• optimum pH of the coagulant
pH is increased using lime
pH is reduced using sulfuric acid
• ALKALINITY ADDITION
Is used when natural alkalinity is not enough to produce
• good floc
Hydrated or slaked lime is used
Soda ash (Na2CO3) is also used (expensive)

19. TURBIDITY ADDITION
Is used to provide sufficient particulate concentration to achieve
rapid coagulation through sufficient interparticle collision
Is done by recycling chemically precipitated sludge
Clays are also used for that purpose
•COAGULANT AIDS
Insoluble particulate materials such as clay, sodium silicate, pure
precipitated calcium carbonate, and activated carbon
They are used in waters with low concentrations of particles
Because their density is higher than most floc particles, floc
settling velocity is increased
The dosage must be carefully controlled to avoid lowering the
water quality.

20. COAGULATION-FLOCCULATION PROCESS
• Coagulation/flocculation process is a two step process - the water first
flows into the Flash mix (Rapid mix) chamber, and then enters the
flocculation basin.
RAPID MIXING AND FLOCCULATION
Rapid mixing is used to:
Disperse chemicals uniformly throughout the mixing basin
Allow adequate contact between the coagulant and particles
Microflocs are produced
Lasts only for about 45 secs
Flocculation is used to:
Agglomerate microflocs to larger ones

21. RAPID MIXING AND FLOCCULATION
• RAPID MIXING DEVICES
RAPID MIXING DEVICES
A tank utilizing a vertical shaft mixer
Hydraulic jump

22. RAPID MIX TANK IMPELLERS
• Mixing equipment include an electric motor, gear-type speed
reducer, and either a radial-flow or axial-flow impeller
The power imparted to the liquid in a baffled
tank by an impeller
P = power, W
N p = impeller constant (power
number)
n = rotational speed, revolutions/s
Di = impeller diameter, m
 = density of liquid, kg/m 3
impeller constant =0.31

23. •
• RAPID MIXING DESIGN
 Design parameters for rapid-mix units are mixing time t and
velocity gradient G.
 The velocity gradient is a measure of the relative velocity of
two particles of fluid and the distance between.
G 
W

G = velocity gradient, sec-1
W = power imparted per
unit volume of basin, N-
m/s-m3
P = power imparted, N-m/s
V = basin volume, m3
 = absolute viscosity of
water ( =0.00131 N-
s/m2)

24. • RAPID MIXING DESIGN
Rotary mixing devices can be
Turbines
Paddle impellers
propellers
Basins are either circular or square in plan
Depth of basin is 1 to 1.25 of the basin diameter or width
Impeller diameter should be 0.3 to 0.5 of the tank width
Baffled tanks are recommended since they minimize vortexing and
ROTARY MIXING

25. • EXAMPLE – RAPID MIXING
• A square rapid-mixing basin, with a depth of water equal to 1.25 times the width, is
to be designed for a flow of 7570 m3/d. The velocity gradient is to be 790 s-1, the
detention time is 40 seconds, the operating temperature is 10 C, and the turbine
shaft speed is 100 rpm. Determine:
• The basin dimensions
• The power required
• µ=0.0013 N-s/m2 at 10oC
• SOLUTION
• Find the volume of the basin,
 The dimensions are (W)(W)(1.25W) = 3.50 m3 W = 1.41 m
• The depth of the basin, H = (1.25)(1.41 m) = 1.76 m Use W
= 1.41 m; H = 1.76
• Using the velocity gradient equation
• P  G2
V  (0.0013N  s / m2
)(790 / sec)2(1.411.411.76m3
)
• P  2863N  m / s
V (7570m3
/1440 min)X (min /60sec) 40sec  3.5m3

26. • FLOCCULATION
 Flocculation is stimulation by mechanical means to agglomerate
destabilized particles into compact, fast settleable particles (or flocs).
•
• Gt values for flocculation
•Values of Gt from 104 to 105 are commonly used, with t ranging from
10 to 30 min.

27. • FLOCCULATION BASINS
Flocculation is normally accomplished with
an axial-flow impeller,
a paddle flocculator, or
a baffled chamber
Flocculation basins are composed of minimum 3
compartments to:
Minimize short circuiting
Facilitate tapered flocculation
BAFFLED CHAMBER FLOCCULATOR

28. • IMPELLER FLOCCULATOR
PADDLE FLOCCCULATOR

29. • FLOCCULATION TANK DESIGN DATA

30. • EXAMPLE ON FLOCCULATION
• A cross-flow, horizontal shaft, paddle wheel flocculation basin is to be
designed for a flow of 25,000m3/d, a mean velocity gradient of
26.7/sec (at 10 C), and a detention time of 45 minutes. The GT value
should be from 50,000 to 100,000. Tapered flocculation is to be
provided, and the three compartments of equal depth in series are to
be used. The G values determined from laboratory tests for the three
compartments are G1 = 50 s-1, G2 = 20 s-1 , and G3 = 10 s-1 . These
give an average G value of 26.7 s-1 . The compartments are to be
separated by slotted, redwood baffle fences, and the floor of the
basin is level. The basin should be 15 m in width to adjoin the
settling tank. Determine:
1. The GT value
2. The basin dimensions
3. The power to be imparted to the water in each compartment

31. • SOLUTION
• The GT value = (26.7/sec)(45 min)(60 sec/min) = 72,100
• Since GT value is between 50,000 and 100,000, the detention time is satisfactory.
• Basin volume, V = (flow)  (detention time) = (25,000 m3/d)(45 min)(hr/60 min) = 781 m3
Profile area = (volume / width) = (781 m3 / 15 m) = 52.1 m2
• Assume compartments are square in profile, and x is the compartment width and depth.
Thus, (3x)(x) = 52.1 m2 = 17.37  x = 4.17 m and 3x = 3(4.17) = 12.51m
• Then, width = depth = 4.17 m and length = 12.51 m
• volume = (4.17)(12.51)(15.0) = 783 m3
• The Power, P = G2V (at 10 C,  = 0.00131 N-s/m2)
• P (for 1st compartment) = (0.00131 N-s/m2)(502/s2)(783 m3/3) = 855 N-m/s = 855 J/s = 855 W
P (for 2nd compartment) = (0.00131)(202)(783/3) = 137 W
• P (for 3rd compartment) = (0.00131)(102)(783/3) = 34.2 W
• PADDLE FLOCCULATOR

32. • D ESIGN
• GUIDELINES
• PADDLE WHEEL FLOCCULATOR
EXAMPLE
A water treatment plant is being designed to process 50,000 m3/d of water. Jar testing
and pilot-plant analysis indicate that an alum dosage of 40 mg/L with flocculation at a Gt
value of 4.0 x 104 produces optimal results at the expected water temperatures of 15oC.

34. • SOLUTION…
• The tank will contain three cross-flow paddles, so its length will be
divided into three compartments. For equal distribution of velocity
gradients, the end area of each compartment should be square, i.e.,
depth equals 1/3 length. Assuming maximum depth of 5 m, length is
SOLUTION
Power requirements
Assume G value tapered as follows First compartment, G = 40 s-1
Second compartment, G = 30 s-1 Third compartment, G = 20 s-1
The Power, P = G2V (at 15 C,  = 0.001139 N-s/m2)
P (for 1st compartment) = (0.001139 N-s/m2)(402/s2)(771.5 m3/3) = 470 W P
(for 2nd compartment) = (0.001139)(302)(771.5/3) = 260 W
P (for 3rd compartment) = (0.00139)(202)(771.5/3) = 120 W

35. SEDIMENTATION
Water treatment

36. •
• SEDIMENTATION
• Particles
•
• Discrete Flocculating
•
• Suspension
•
• Dilute Concentrated
Size
Shape
weight
displacement of water
velocity field
interference

37. INTRODUCTION
 Sedimentation is removal of particulate materials suspended in water by quiescent
settling due to gravity.
 Commonly used unit operation in water and wastewater treatment plants

38. • TYPES OF SEDIMENTATION
No interaction between particles
Settling velocity is constant for individual particles
Dilute solid’s concentration
Examples: pre sedimentation in water treatment, grit removal in
• wastewater
Particles collide and adhere to each other resulting in particle growth
Dilute solid’s concentration
Examples: coagulation/flocculation settling in water treatment and
primary sedimentation in wastewater treatment

39. particles are so close together movement is restricted
intermediate solids concentration
solids move as a block rather than individual particles
distinguishable solids liquid interface
example: settling of secondary effluents
physically in contact
water is squeezed out of interstitial spaces
volume of solids may decrease
high concentration of solids (sludges)

40. • Discrete particles settling(TYPE1)
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

42. • DISCRETE PARTICLES SETTLING (TYPE 1
mg
Fnet = mg - Fd -
Fb
Fnet = 0

46. • EXAMPLE
• Estimate the terminal settling velocity in water at a temperature of 15 oc
• Of spherical silicon particles with specific gravity 2.4 0 and average diameter of (a) 0.05
mm and (b)1.0 mm

47. • SOLUTION
Step 1. Using Stokes equation for (a) at T= 15oC
Step 2. Check with the Reynolds number

48. o Step 3. Using Stokes’ law for (b)
Step 5. Re-calculate Cd and v
Step 4. Check the Reynold’s number
Since R > 2, the Stokes’ law does not apply. Use Eq. 1 to
calculate v

49. • Step 6. Re-check Re
Step 7. Repeat step 5 with new R
Step 8. Re-check Re

50. • Step 9. Repeat step 7
SETTLING COLUMN

51. • procedure – settleability analysis
 usually 2m high column
 Mix the suspension thoroughly
 Measure initial ss concentration, co
 Measure concentrations at certain intervals, ci
 All particles comprising c1 must have settling velocities less than z0/t1.
thus the mass fraction of particles with v1 < z0/t1 is

52. • PROCEDURE – SETTLEABLITY ANALYSIS
 For a given detention time to, an overall percent removal can be
obtained.
 All particles with settling velocities greater than v0=z0/t0 will be 100
percent removed.
• Thus, 1 – xo fraction of particles will be removed completely in time t0.
the remaining will be removed to the ratio vi/vo. if the equation relating v
and x is known the area can be found by integration:

53. • Example :settling column analysis of type-1 suspension
• a settling analysis is run on a type-1 suspension. the column is 2 m deep and
data are shown below.
• SOLUTION
What will be the theoretical removal efficiency in a settling basin
with a loading rate of 25 m3/m2-d (25m/d)?

54. Step 1. Calculate mass fraction remaining and corresponding settling
rates

55. • SOLUTION
• Step 2. Plot mass fraction vs. settling velocity
Step 3. Determine vo
vo = 25m3/m2.d = 1.74 x 10-2 m/min
Step 4. Determine xo = 54 %

56. • SOLUTION
• Step 5. x.vt by graphical integration

57. • Step 5. x.vt by graphical integration
• Step 6. Determine overall removal efficiency

58. • SETTLING TANKS /SEDIMENTATION TANKS
 The principle involved in these tanks is reduction of velocity of flow so
that the particles settle during the detention period.
 Such tanks are classified into,
Fill and draw types tanks (batch-process)
Continuous flow tanks.

59. • SETTLING TANKS /SEDIMENTATION TANKS
 Depending on their shape, sedimentation tanks may be classified as,
circular,
rectangular, and
square.
 Depending on the direction of flow, as
horizontal flow—longitudinal, radial flow
vertical flow—circular (upward flow)

60. • LONG-RECTANGULAR BASINS
 Long rectangular basins are commonly used in treatment plants processing large flows.
• Hydraulically more stable, and flow control through large volumes is easier with this
configuration.

62. • LONG-RECTANGULAR TANK
The inlet zone in which baffles intercept
the oncoming water and spread the
flow
The outlet zone in which water flows
y upward and over the outlet weir
sludge zone, which extends from the
bottom of the tank to just above the
scraper mechanism
settling zone, which occupies
the main volume of the tank

63. • LONG-RECTANGULAR TANK
 Detention time: is the theoretical time that the water is detained in a settling basin.
• it is calculated as the volume of the tank divided by the rate of flow, and is denoted as θ
= V/Q.
vo or Q/As is called overflow rate or surface loading rate or surface
overflow rate (SOR).

64. • REMOVAL EFFICIENCY OF PARTICLES
 a particle initially at height h with settling velocity of VSH will just be removed by the time it
has traversed the settling zone.
 particles initially at heights less than h will also be removed and those at greater heights
will not reach the bottom before reaching the outlet zone.
 all particles with settling velocity VS < VSH are removed partly, depending on their
position at a height from the top of the sludge zone.
 the efficiency of removal of such particles is given by H/H.
The greater the surface area, the higher the efficiency

65. • CIRCULAR BASINS

66. • DESIGN DETAILS
 Detention period: for plain sedimentation: 3 to 4 h, and
• for coagulated sedimentation: 2 to 2 1/2 h
 Velocity of flow: < 18 m/h (horizontal flow tanks)
 Tank dimensions: l : b : 3 to 5 : 1
generally l = 30 m (common) maximum 100 m.
breadth: 6 m to 10 m.
circular: dia < 60 m, generally 20 to 40 m
 Depth 2.5 to 5.0 m (3 m)
 Surface loading (or) overflow rate or (sor)
for plain sedimentation- 12,000 to 18,000 1/d/m2
for thoroughly flocculated water 24,000 to 30, 000 1/d/m2
horizontal flow circular tank 30,000 to 40,000 1/d/m2
 Slopes: rectangular 1 % towards inlet and circular 8%
 Weir loading rate, m3/m/d < 248

67. • EXAMPLE
 Find the dimensions of a rectangular sedimentation basin for the
following data:
volume of water to be treated = 3 mld
detention period = 4 hrs
velocity of flow = 10 cm/min
• SOLUTION

68. • SOLUTION

69. • EXAMPLE
• Design a circular basin
• A circular sedimentation tank is to have a minimum detention time of 4 h and a maximum
overflow rate of 20 m3/m2.d. determine the required diameter of the tank and the depth if
the average flow rate through the tank is 6 ml/d.
• SOLUTION
 v=6000x4/24=1000
 depth=20x4/24=3.33=3.5m
 as=v/d=1000/3.5=285.7m2
 as=xd2/4
 diameter=19m
 hence provide 5 m deep (1 m for sludge and 0.5 m free board) by 19 m
diameter tank

70. • EXAMPLE
• Design a long-rectangular settling basin for type-2 settling. a city must treat about
15,000 m3/d of water. flocculating particles are produced by coagulation, and a
column analysis indicates that an overflow rate of 20 m/d will produce satisfactory
removal at a depth of 2.5 m. determine the size of the required settling tank.
• SOLUTION
Q = 15,000 m3/d, SOR(V) = 20 m/d and d = 2.5m
Calculate surface area
A = Q/V = 15,000 / 20 = 750 m2
 Use two tanks  A = 375 m2 for each
Take L: B = 3: 1
L = 33.50 m and B = 11.20 m

71. • HIGH-RATE SETTLING MODULES
 Small inclined tubes or tilted parallel plates which permit effective gravitational settling of
suspended particles within the modules.
 Surface loading  5 to 10 m/h

72. • TUBE SETTLERS
 Take advantage of the theory that surface overflow loading, which can also be defined as
particle settling velocity, is the important design parameter.
 Theoretically, a shallow basin should be effective.
 Use tubes of 25 to 50 mm diameter
 At a 60o angle provide efficient settling

73. • TUBE SETTLERS

74. • DESIGN OF INCLINED SETTLERS

75. • EXAMPLE
• A water treatment work treats 1.0 m3/s and removes flocs larger than
0.02 mm. the settling velocity of the 0.02mm flocs is measured in the
laboratory as 0.22 mm/s at 15 oc. tube settlers of 50 mm square
honeycombs are inclined at a 50o angle and its vertical height is 1.22 m.
determine the basin are
• required for the settler module.

80. FIL
TRA
TION

81. FIL
TRA
TION
Filtration involves the removal of suspended and
colloidal particles from the water by passing it through
a layer or bed of a porous granular material, such as
sand.

82. CLASSIFICA
TIONOFFIL
TERS
MTU
 Based on the filter media
Sand filters, e.g. natural silica sand
Anthracite filters, e.g. crushed anthracitic coal
Diatomaceous earth filters, e.g. diatomaceous
earth
Metal fabric filters (microstrainers), e.g. stainless
steel fabric filter.

83. CLASSIFICA
TIONOFFIL
TERS
 Based on the depth of filter media
Deep granular filters, e.g. sand, dual-media
and multi-media (combination of two or more
media), granular activated carbon
Precoat filters, e.g. diatomaceous earth, and
powdered activated carbon, filters

84. CLASSIFICA
TIONOFFIL
TERS
 Based on the rate of filtration, sand filters can
be further classified as
Gravity filters
Slow sand filters
rapid sand filters
high-rate sand filters
Pressure filters

85. R
A
T
EOFFIL
TRA
TION
Rate of filtration (loading rate) is the flow rate of
water applied per unit area of the filter. It is the
velocity of the water approaching the face of the filter:
where va = face velocity, m/d = loading rate,
m3/d.m2 Q = flow rate onto filter surface,
m3/d
As = surface are of filter, m2
a
MTU Water
Treatment
A
v 
Q
s

86. E
X
A
M
P
L
E
A city is to install rapid sand filters downstream of the
clarifiers. The design loading rate is selected to be 160
m3/(m2 d). The design capacity of the water works is
0.35 m3/s. The maximum surface per filter is limited to
50 m2. Design the number and size of filters and
calculate the normal filtration rate.

87. • Example solution
• Step 1 determine the total surface area required

88. MECHANISM OF FIL
TRA
TION
 The theory of filtration basically involves,
transport mechanisms, and attachment
mechanisms.
 The transport mechanism brings small particles
from the bulk solution to the surface of the
media.
a) gravitational settling,
b) diffusion,
c) interception and
d) hydrodynamics.

89. MECHANISMOFFIL
TRA
TION
 They are affected by physical characteristics such
as size of the filter medium, filtration rate, fluid
temperature, size and density of suspended solids.
 As the particles reach the surface of the filter media,
an attachment mechanism is required to retain it.
This occurs due to
(i) electrostatic interactions
(ii) chemical bridging or specific adsorption.

90. S
L
O
WS
A
N
DF
IL
TERS
 In SSF water is allowed at a slow rate through a bed
of sand, so that coarse suspended solids are
retained on or near the surface of the bed.
 Loading rate of 2.9 to 7.6 m3/d.m2
 The raw water turbidity has to be < 50 NTU.
 The filtering action is a combination of straining,
adsorption, and biological flocculation.

91. S
L
O
WS
A
N
DF
IL
TERS
 Gelatinous slimes of bacterial growth called
‘schmutzdecke’ form on the surface and in the upper
sand layer, consists of bacteria, fungi, protozoa, rotifera
and a range of aquatic insect larvae.
 The underlying sand provides the support medium for
this biological treatment layer.
 Slow sand filters slowly lose their performance as the
Schmutzdecke grows and thereby reduces the rate of
flow through the filter. requires refurbishing

92. CL
E
A
N
I
N
G
S
L
O
WS
A
N
D
F
I
L
T
E
R
S
 Scrapping: the top few mm of sand is carefully
scraped off using mechanical plant and this
exposes a new layer of clean sand. Water is then
decanted back into the filter and re-circulated for a
few hours to allow a new Schmutzedecke to
develop. The filter is then filled to full depth and
brought back into service.
 wet harrowing: lower the water level to just above
the Schmutzdecke, stirring the sand and thereby
suspending any solids held in that layer and then
running the water to waste. The filter is then filled to
full depth and brought back into service.

93. TYPICALS
L
O
WSANDFIL
TER
Sand filter bed
Grav
e
l
Schmutzecke
Supernatant water
System of
underdrains
Weir
Raw water
Finishe
d
water

94. AD
V
A
N
T
A
G
E
S
A
N
D
DI
S
A
D
V
A
N
T
A
G
E
S
 Advantages
Simple to construct and supervise
Suitable where sand is readily available
Effective in bacterial removal
Preferable for uniform quality of treated water
 Disadvantages
Large area is required
Unsuitable for treating highly turbid waters
Less flexibility in operation due to seasonal variations in
raw water quality

95. DESIGN CRITERIAFORS
S
F
Parameter Recommended level (UK experience)
Design life
Period of operation Filtration rate
Filter bed area Height of filter bed
Initial Minimum
Effective size Uniformity coefficient
Height of underdrains + gravel layer
Height of supernatant water
10-15 year
24 h/day
0.1 – 0.2 m/h
50-200 m2/filter (minimum of two filters)
0.8-0.9 m
0.5-0.6 m
0.15-0.3 mm
< 3
0.3-0.5 m
1 m

96. E
XAMPLE.S
S
FDESIGN
Design a slow sand filter to treat a flow of 800
m3/day.
Solution:
assuming a filtration rate of 0.15 m/h,
Required tank area = (800/24) x (1/0.15) = 222 m2
Use a tank 23 m long x 10 m wide.
From Table 6.1, the height of the tank require is:
 System underdrain + gravel ≈ 0.5 m
 Filter bed ≈ 0.9 m
 Supernatant water ≈ 1 m
Therefore, total tank height = 2.4 m and tank
dimension becomes 23 m long x 10 m wide x 2.4 m
high

97. RAPIDS
A
N
DF
IL
TERS
 The most common type of filter for treating municipal
water supplies.
 During filtration, the water flows downward through
the bed under the force of gravity.
 When the filter is washed, clean water is forced
upward, expanding the filter bed slightly and
carrying away the accumulated impurities. This
process is called backwashing.

98. A
D
V
A
N
T
A
G
E
S
A
N
D
DI
S
A
D
V
A
N
T
A
G
E
S
 Advantages
Turbid water maybe treated
Land required is lesscomparedto slow sandfilter
Operation iscontinuous.
 Disadvantages
Requiresskilled personnelfor operation and maintenance
Lesseffective in bacteriaremoval
Operational troubles

99. TYPICALGRADA
TION OFR
S
F
after backwashing, the larger
sand grains settle to the bottom
first, leaving the smaller sand
grains at the filter surface.
Allows in-depth filtration: provides
more storage space for the solids,
offer less resistance to flow, and
allows longer filter runs.

100. T
Y
P
E
S
OFR
S
F
 RSF based on filter material, three types:
Single-media filters: these have one type of
media, usually sand or crushed anthracite coal
Dual-media filters: these have two types of
media, usually crushed anthracite coal and
sand.
Multi-media filters: these have three types of
media, usually crushed anthracite coal, sand,
and garnet.

101. FIL
TERHYDRAULICS
The loss of pressure (head loss) through a clean stratified-sand
filter with uniform porosity was described by Rose:
where hL = frictional head loss through the filter, m
va = approach velocity, m/s D = depth of filter sand,
m CD = drag force coefficient
f = mass fraction of sand particles of diameter d
d = diameter of sand grains, m
ϕ = shape factor and = porosity

103. Example 2:
A dual media filter is composed of 0.30 m anthracite (mean particle size
0.20mm) that is placed over a 0.60 m layer of sand (mean particle size 0.70mm)
with a filtration rate of 9.78 m/h. Assume the grain sphericity ф = 0.75 and
porosity (ε) = 0.40 for both. Estimate the headless in the clean filter at 150C.

104. E
X
A
M
P
L
E4
Estimate the clean filter
headloss for a proposed new
sand filter using the sand.
Use the following
assumptions: loading rate is
216 m3/d.m2 , specific gravity
of sand is 2.65, the shape
factor is 0.82, the bed
porosity is 0.45, the water
temperature is 10oC, and the
depth of sand is 0.5 m.
Sieve No % retain d(mm)
8-12 5.3 2
12-16 17.1 1.42
16-20 14.6 1
20-30 20.4 0.714
30-40 17.6 0.0505
40-50 11.9 0.0357
50-70 5.9 0.0252
70-100 3.1 0.0178
100-140 0.7 0.0126

105. SOLUTION

107. SOLUTION

108. DISINFECTION

109. INTRODUCTION
•Disinfection is the process of killing all pathogens.
•Safe water means water
• (a) free from pathogenic bacteria,
• (b) aesthetically acceptable, and
• (c) free from excessive minerals, and poisonous matter.
•The substance used for disinfection is called
disinfectant.
•Disinfectants must effectively reduce all types of
pathogens without being toxic to humans or domestic
animals.
•Additionally, it must not drastically change the taste or
color of water and it must be persistent.

110. EFFICIENCY OF DISINFECTION
•Nature and concentration of organisms
•Nature and concentration of disinfectant
•Nature of water to be disinfected
(interfering substances like NH3, iron,
Mn, organic matter)
•Temperature of water
•Time of contact

111. DISINFECTION METHODS
• Heat
• Mechanical (ultrasonic vibration, membrane filtration)
• Radiation (Gamma, Ultraviolet radiation)
• Chemical
• Halogens (chlorine, bromine, iodine)
• Chlorine dioxide (ClO2)
• Chloramines (ClNH2, Cl2NH, Cl3N)
• Ozone

112. DISINFECTION KINETICS
•The kinetics of disinfection depends on the
following
•Time of contact
•Concentration of disinfectant
•Concentration of organisms
•Temperature of water

113. TIME OF CONTACT
• When a single unit of microorganisms is exposed to a single unit of
disinfectant, the reduction in microorganisms follows a first-order reaction.
• This equation is known as Chick’s Law, which states that the number of
organisms destroyed in a unit time is proportional to the number of organism
remaining:
• Where, N = number of microorganism (NO is initial number)
• k = disinfection constant
• t = time

114. DISINFECTANT CONCENTRATION
• The disinfection efficiency is generally estimated as,
• where C is the concentration of the disinfectant, t is
the time required to effect a constant % kill of the
organisms, and n is a coefficient of dilution.

115. CHLORINATION
•Chlorine is the most commonly
used disinfectant because it
• is readily available as gas, liquid, or powder
• is cheap
• is easy to apply due to relatively high solubility
(7000mg/l)
• leaves a residual in solution which, while not
harmful to humans, provides protection in the
distribution system
• is very toxic to most microorganisms, stopping
metabolic activities.

116. CHLORINATION
• The chlorine dosage required is a function of the water's organic content
(including the microorganisms) and the water's reduced inorganic content.
• Reduced inorganics include species such as Fe2+, Mn2+, NH3, etc. which will
be oxidized by chlorine.
• Chlorine is added to the water supply in two ways.
• as a gas, Cl2(g), from the vaporization liquid chlorine.
• As a salt, such as sodium hypochlorite (NaOCl) or bleach.

117. CHLORINATION
• Chlorine gas dissolves in water following Henry's Law
• When chlorine is added to water, a mixture of hypochlorous
acid (HOCl) and hydrochloric acid (HCl) is formed:
• Hypochlorous acid is a weak acid that dissociates to form
• hypochlorite ion (OCl-).

118. CHLORINATION
• Hypochlorous acid and hypochlorite ion compose what is called
free chlorine residual.
• These free chlorine compounds can react with many organic and
inorganic compounds to form chlorinated compounds.
• If the products of these reactions posses oxidizing potential, they
are considered as combined chlorine residual.
• In the presence of ammonia, chlorine form chloramines, which is
mainly monochloramine (NH2Cl), some dichloramine (NHCl2)
and trichloramine (NCl3).

119. BREAKPOINT REACTION
• Monochloramine Formation Reaction. This reaction occurs
rapidly when ammonia nitrogen is combined with free chlorine up
to a molar ratio of 1:1.
• Breakpoint Reaction. When excess free chlorine is added
beyond the 1:1 initial molar ratio, monochloramine is removed as
follows:

120. BREAKPOINT REACTION
• Chlorine demand: The difference between the amount of
chlorine added to water and the amount of residual chlorine after
a specified contact period.