1. Water
Treatment
Process:
Coagulation &
Flocculation
CHAPTER 5

2. Coagulation aided with sedimentation

3. Coagulation and flocculation
 Objective:
 In this chapter we will answer the following questions:
 How do coagulation and flocculation fit into the water
treatment process?
 Which chemical principles influence coagulation and
flocculation?
 Which chemicals are used in coagulation?
 How the optimal dose of coagulant can be determined?
 What factors influence coagulation and flocculation?
 Purpose:
 To remove turbidity, color, and some bacteria from
water.

4. Definitions
 Coagulation is
 The addition and rapid mixing of coagulants
 The destabilization of colloidal and fine particles
 The initial aggregation of destabilized particles
 Flocculation is
 The gentle agitation to aggregate destabilized
particles to form rapid-settling floc

5. Coagulation Chemistry
 The hydraulic settling values of small size particles in water
are very small
 they require longer time to settle in plain sedimentation tanks
 In a 3m depth sedimentation tank:
 a slit particle of size 0.05mm will require about 11hrs to settle
down
 clay particle of size 0.002mm will require about 4 days time to
settle
 Due to electrical charge, colloidal impurities remain
continuously in motion and never settle down by gravity in
water.

6. Coagulation Chemistry
 When water is turbid due to presence of such fine size
and colloidal impurities, plain sedimentation is of no use.
 Coagulation:
 A chemical process which removes all fine size and
colloidal impurities within reasonable period of 2 – 3hrs.
 Coagulant:
 the chemical used in the process
 The objective of coagulation is :
 to unit several colloidal particles together to form
bigger sized settable flocs which may settle down in
the tank.

7. Coagulation Chemistry
 The principle of coagulation can be explained from the
following two conditions:
1. Floc formation
 When coagulants are dissolved in water and thoroughly
mixed, they produce a think gelatinous precipitate = floc.
• Floc
 has the property of arresting suspended impurities in water
during downward travel towards the bottom of tank.
 Hence, removing fine and colloidal particles quickly.

8. Coagulation Chemistry
2. Electric charges
 Most particles dissolved in water have a negative charge, so
they tend to repel each other.
 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
 The amount of coagulant will depend on:
 Zeta potential
 a measurement of the magnitude of electrical charge surrounding
the colloidal particles.
 the amount of repulsive force which keeps the particles in the
water.

9. Coagulation Chemistry
 Coagulants tend to be positively charged.
 Due to their positive charge, they are attracted to the
negative particles in the water.
Positively charged coagulants attract to
negatively charged particles due to electricity
Negatively charged particles
repel each other due to electricity

10. Coagulation Chemistry
 The combination of positive and negative charge results in
a neutral.
 The next force which will affect the particles is known as
Van der Waal's forces.
 Van der Waal's forces refer to the tendency of particles in
nature to attract each other weakly if they have no charge.
Neutrally charged particles attract
due to Van der Waal's forces
When enough particles have joined together, they become floc and will settle out of the water.
When enough particles have
joined together, they become
floc and will settle out of the
water.
Particles and coagulants join
together into floc

11. Destabilization of Colloidal Dispersion
 Non-settleable Solids
 Have particle size between 0.0001 and 0.1mm
 Colloids
 Have particle size between 0.000001 to 0.001mm
 Most of non-settleable solids are colloidal particulates
 Colloids do not settle by the force of gravity
 Are stable in suspensions
 Colloid removal requires that they should be agglomerated
in to large particles.
 This requires surface charge to be destabilized.

12. Destabilization of Colloidal Dispersion
 Particle destabilization can be achieved by four
mechanisms:
1. Double layer compression
 Addition of electrolyte/coagulant to water shrinks the layer
of charged ions around the particle.
 If reduced enough, the attractive Van der Waals force (which
acts close to particle) can overcome Zeta potential.
 Diffused double layer is created by cations attaching to
negatively charged particles (fixed layer) and cations and
anions loosely attaching in outer diffused layer.

13. Destabilization of Colloidal Dispersion
1. Double layer compression
A negatively colloidal particle with its electrostatic field
Two major forces acting on colloids:
1. Electrostatic repulsion, Zeta potential
2. Intermolecular, Van der Waals, attraction

14. Destabilization of Colloidal Dispersion
2. Adsorption and charge neutralization
 Adding positively charged ions that adsorb to particle
surface can reduce surface charge and repulsion
3. Entrapment in precipitate
 Al and Fe salts added at right pH will precipitate as flocs
with colloids as nuclei.
4. Particle bridging
 Large organic molecules attach to multiple particles
‘bridging’ them (often used in addition to metal salts)

15. Factors affecting coagulation
1. Characteristic of water
a) Type and quantity of suspended matter
b) Temperature of water
c) pH of water
2. Type of coagulant
3. Dose of coagulant
4. Time and method of mixing

16. Factors affecting coagulation
 Low temperature affects coagulation and flocculation
processes by altering:
• coagulant solubility
• increasing water viscosity
• retarding the kinetics of hydrolysis reactions and particle
flocculation
 Higher alkalinity waters have higher pH.
 Metal coagulants are acidic, and coagulant addition
consumes alkalinity.
 For low alkalinity (pH) waters, coagulant addition may
consume all of the available alkalinity, depressing the
pH to values too low for effective treatment.

17. Types of coagulant
 Coagulant chemicals come in two main types:
 primary coagulants
 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
 add toughness to the flocs so that they will not break up during
the mixing and settling processes
 generally used to reduce flocculation time
 Lime, calcium carbonate and bentonite are examples

18. Coagulants
 Mainly aluminum and iron salts
 Aluminum sulfate – Alum - Al2(SO4)3.18H2O
 Ferrous sulfate
 Ferric sulfate
 Ferric chloride
 Sodium Almunate
 Polyelectrolytes
Al3OH2+
Al3(OH)2
+
[Al3(OH)4]4+
[Al3(OH)4]5+
[Al3O4(OH)24]7+
Intermediates in the
formation of Al(OH)3
precipitate
These species may adsorb very strongly
onto the surface of most negative colloids

19. 1. Aluminum Sulfate - Alum
 To produce the hydroxide floc, enough alkalinity should
present in the water.
floc
 If alkalinity is not enough, then it should be added. Usually
hydrated lime is used for that purpose.
 Optimum pH is 6.5 – 8.5
 Its dose may vary from 5 - 30mg/l (14mg/l)
 Dose of coagulant depends on various factors such as
turbidity, color, taste, pH value, temperature
Al 2(SO4)3.18H2O + 3Ca (HCO3) 2  2Al(OH) 3 + 3CaSO4 + 6CO2 +18H2O
Al2(SO4)3.18H2O + 3Ca(OH)2  2Al(OH)3 + 3CaSO4 + 18H2O
Al2 (SO4)3.18H2O + 3Na2CO3  2Al(OH)3 + 3Na2SO4 + 3CO2 + 18H2O

20. 1. Aluminum Sulfate
 Alum is the most widely used chemical coagulant, with
the following reason:
1. It is very cheap
2. It removes taste and color in addition to turbidity
3. It is very efficient
4. Flocs formed are more stable and heavy
5. It is not harmful to health
6. It is simple in working, doesn’t require skilled
supervision for dosing

21. 2. Sodium Aluminate -- Na2Al2O4
 It can remove carbonate and non-carbonate hardness
 React with calcium and magnesium salts to form
flocculent aluminates of these elements
 The pH should be with in the range of 6 and 8.5
 More expensive than alum
Na2Al2O4 + Ca (HCO3) 2  CaAl2O4 + Na2CO3 + CO2 + H2O
Na2Al2O4 + CaSO4  CaAL2O4 + Na2SO4
Na2Al2O4 + CaCl2  CaAl2O4 + 2NaCl

22. 3. Chlorinated Copperas
 Combination of Ferric sulphate and Ferric chloride
 When solution of Ferrous Sulphate is mixed with chlorine,
both Ferric sulphate and Ferric chloride are produced.
floc floc
 Both Ferric sulphate and Ferric chloride can be used
independently with lime as a coagulant
 If alkalinity is insufficient, lime is added
 Ferric chloride: effective pH range 3.5 – 6.5 or above 8.5
 Ferric sulphate : 4 – 7 or above 9.
6FeSO4.7H2O + 3Cl2 2Fe3(SO4)2 + 2FeCl3 + 42H2O
2FeCl3 + 3Ca(OH)2  2Fe(OH)3 + CaCl2
Fe2(SO4)3 + 3Ca(OH)2  2Fe(OH)3 + 3CaSO4
FeCl3
Fe2(SO4)3

23. 4. Polyelectrolytes (Polymeric Coagulants)
 Anionic—ionize in solution to form negative sites along
the polymer molecule.
 Cationic—ionize to form positive sites.
 Non-ionic—very slight ionization.
 Synthetic polymers (more common in coagulation)
 Uses:
 Broaden the pH range over which satisfactory flocculation
could occur
 Reduce the quantity of primary coagulant required
 Dosage: usually 1mg/l
Polymer: a substance which has a molecular structure built up chiefly or completely
from a large number of similar units bonded together

24. Dose of the coagulant
 Dosage is the required concentration of the chemical
within the water (mg/l).
 Chemical feed is the amount of chemical you add to
the water (mg/day).
 High coagulant dose destabilization of floc
 Low coagulant dose insufficient
 Optimal coagulant dose is requires
 Fixing of the optimal coagulant dose requires
laboratory experiment --- Jar Test

25. Jar Test
 It is used to predict the functioning of a large scale
treatment operation.
 Is used to determine:
 Proper coagulant and coagulant aid
 Proper coagulant dose
 Procedure:
 Rapid mixing – 1min (80 RPM)
 Slow mixing - 15 – 20min (20 RPM)
 Settling - 30min
 Then measure the final turbidity in each container.
 The final turbidity can be evaluate roughly by sight or more
accurately using a nephelometer.
 The optimum coagulant dose is the dose which meets the
specified turbidity required on the regulatory permit.

26. Jar Test

27. Example 1
Find out the quantity of alum required to treat 18 million
liters of water per day. The dosage of alum is 14mg/lit. Also
work out the amount of CO2 released per liter of treated
water.
Solution:
Ans: 252 kg of alum per day
100 kg of CO2 per day
The chemical reaction of alum is given by:
Al2(SO4)3.18H2O + 3Ca(HCO3)2  2Al(OH)3 + 3CaSO4 + 6CO2 +18H2O

28. Feeding of coagulant
 In order to fed chemicals to the water regularly and
accurately, some type of feeding equipment must be
used.
 Coagulants may be put in raw water either in
i. powder form (Dry-feeding equipment)
ii. solution form (Solution-feeding equipment)

29. Feeding of coagulant
1. Dry-feed Type
• Dry powder of coagulant is filled in the conical hopper.
• Agitating plates are used to prevent arching of chemicals
• Feeding is regulated by the speed of toothed wheel or helical
spring
• Activated carbon and lime are also added in powder form
 Control of dose is difficult
Chemical feed rate of a dry chemical is
given by:
Chemical feed (mg/day)
= Flow (l/day)*Dosage (mg/l)

30. Feeding of coagulant
2. Solution-feed Type
 Can be easily controlled with automatic devices
 Not desirable (more labor)
 There might be corrosion
 Procedure:
 First, solution of required strength of coagulant is
prepared.
 The solution is filled in the tank and allowed to mix
in the mixing channel in required proportion to the
quantity of water.

31. Mixing devices
 The process of floc formation greatly depends upon:
 the effective mixing (rapid mixing) of coagulant with the raw
water.
 Rapid mixing is used:
 To disperse chemicals uniformly throughout the mixing basin
 To allow adequate contact between the coagulant & particles
 To the formation of microflocs
 Mixing is done by mixing device.
1. Hydraulic jump
 flume with considerable slope is developed
2. Pump
 centrifugal pump is used to raise raw water

32. Mixing devices
3. Compressed air method
 compressed air is diffused from bottom of the mixing tank
4. Mixing channels
 Mixing of raw water and coagulant is made to pass through
the channel in which flume has been done.
 Vertical baffles are also fixed at the end of the flumed part
on both sides of the channel.
Mixing channel

33. Cont…
5. Mechanical mixing basins
 Mechanical means are used to agitate the mixture to achieve
the objective of thorough mixing.
 Flash mixers and deflector plate mixers are used
A. Flash mixer
 The mixing of coagulant in water
is achieved by rotating vigorously
fans fixed in the mixing basin.
 The deflecting wall avoids
short circuiting and deflects
the water flow towards the fan
blades (paddles).
 Chemical pipe discharges the coagulant just near the rotating
fan.

34. Cont…
 Design criteria of flash mixer
1. Detention period – 30 to 60 sec
2. Velocity of flow – 0.9m/sec
3. Depth – 1 to 3m
4. Rotation per minute of blade – 80 to 100
5. Power required – 0.041kW/1000m3/day

35. Cont…
B. Deflector plate mixer
 Mixing is achieved by diffusing water through a
deflection plate.
 Chemical pipe discharges the coagulant just near the
deflector plate.

36. Cont…
 The formula used to determine the chemical feed rate of
liquid chemicals is shown below:
Chemical feeder setting 𝑚𝑙/𝑚𝑖𝑛
=
Flow (l/min) ∗ Dosage (mg/l)
Liquid chemical concentration (mg/ml)
Liquid chemical concentration 𝑚𝑔/𝑚𝑙
=
Dry chemical weight (mg)
Volume of water (ml)

37. Cont…
Example 1: If you added 230 mg of polymer to 100 ml of water,
you would have a liquid polymer solution with a concentration
of 2.3 mg/ml.
Example 2: Consider a situation in which the flow of water into
the water treatment plant is 18.9MLD and the alum dosage is 10
mg/l. The liquid alum has a concentration of 520 mg/ml.
The chemical feeder setting would be determined as follows:
𝐶ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑓𝑒𝑒𝑑𝑒𝑟 𝑠𝑒𝑡𝑡𝑖𝑛𝑔 (𝑚𝑙/𝑚𝑖𝑛) =
𝐹𝑙𝑜𝑤 (𝑙/𝑚𝑖𝑛) ∗ 𝐷𝑜𝑠𝑎𝑔𝑒 (𝑚𝑔/𝑙)
𝐿𝑖𝑞𝑢𝑖𝑑 𝑐ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 (𝑚𝑔/𝑚𝑙)
𝐶ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑓𝑒𝑒𝑑𝑒𝑟 𝑠𝑒𝑡𝑡𝑖𝑛𝑔 (𝑚𝑙/𝑚𝑖𝑛) =
13143 𝑙/𝑚𝑖𝑛 ∗ 10 𝑚𝑔/𝑙
520 𝑚𝑔/𝑚𝑙
= 252𝑚𝑙/𝑚𝑖𝑛
The setting on the liquid alum feeder should be 252 ml/min.

38. Flocculation
 After adding the coagulant to the raw water:
 rapid agitation is developed in the mixture to obtain a
thorough mixing. (for about 1 to 3 min)
 next to rapid mixing, mixture is kept slowly agitated for
about 30 to 45min
 Slow mixing process in which particles are brought into
contact in order to promote their agglomeration is
called flocculation.
 The tank/basin in which flocculation process is carried
out is called
flocculation
chamber

39. Flocculation
 Flocculation occurs by:
1. Brownian motion
 the random motion of particles suspended in a fluid
resulting from their collision
 important for small particles (< 0.5m) --- Perikinetic
flocculation
2. Stirring
 mechanical stirring strong enough to cause particle
collisions but not so strong as to break up particles. ---
Orthokinetic flocculation
3. Differential settlement
 larger, faster particles catch up with smaller, slower
particles

40. Flocculation
 The velocity of flow in the flocculation chamber is kept
between 0.12 – 0.18m/sec.
 Activated carbon in powder form can be used to speed
up the flocculation.
 The rate of agglomeration or flocculation is dependent
upon:
 Type and concentration of turbidity
 Type of coagulant and its dose
 Temporal mean velocity gradient caused by mixing – G in
the basin

41. Flocculation
 Mean Velocity Gradient (G)
√ The velocity of water flowing through the flocculation
basin must be within a very specific range, designed to
gently mix the water without breaking apart of the floc.
√ G is the rate of change of velocity per unit distance
normal to the section - (T-1).
√ Measurement of the intensity of mixing in the chamber.
√ Determines how much the water is agitated in the tank.
√ Determines how much energy is used to operate the
flocculator.

42. Flocculation
 The value of G can be computed in terms of power input
by the following equation:
Where P – power dissipated (watt)
µ - absolute viscosity (Ns/m2)
V - the volume to which P is applied (m3)
G - temporal mean velocity gradient caused by mixing (s-1)
 Power can also be calculated by:
Where: P = power input, Watt or Nm/s
FD = drag force on paddles, N
vp = velocity of paddles (velocity relative to the water), m/s
𝐆 =
𝑷
𝝁𝑽
𝟏
𝟐
𝐏 = 𝑭𝒐𝒓𝒄𝒆 ∗ 𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 = 𝑭𝑫
∗ 𝒗𝒑

43. Cont…
Where:
CD = coefficient of drag, 1.8 for flat blades
Ap = area of paddle blades, m2
𝑤 = density of water, kg/m3)
Therefore,
FD
=
1
2
CD
Ap
vp
2
ρ𝑤
P =
1
2
CD
Ap
vp
3
ρ𝑤
𝐆 =
CD
Ap
vp
3
ρ𝑤
𝟐𝝁𝑽
𝟏
𝟐

44. Cont…
 Rotating paddle wheels or vertical mounted turbines are
the most commonly used flocculation technique.
 The design criteria of a horizontal continuous flow rectangular
basin flocculator:
i. Depth of tank : 3 – 4.5m
ii. Detention time (t) : 30 – 45 minute
iii. Velocity of flow : 0.12 – 0.18m/s
iv. Total area of paddles : 10 – 25% of cross-
sectional area of the tank
v. Peripheral velocity of blades : 0.2 – 0.6m/s
vi. Velocity gradient (G) : 10 – 75 s-1
vii. Factor G*t : 104 - 105
viii. Power consumption : 49 – 196 kW/Mm3/d
ix. Outlet flow velocity : 0.15 – 0.25 m/s
Vertical-shaft,
turbine-type
impeller

45. Secondary sedimentation tank/ Clarifier
 After flocculation, water enters the settling tank which is
normally called a clarifier.
 Water is retained in the tank for a sufficient period to
permit the settlement of the floc to the bottom.
 Principles of clarifier design:
 the same as for plain sedimentation basin except that its
detention period is lower.
 The detention period: 2 - 2.5hrs
 Overflow rate: 1 - 1.2 m/hr (24 – 28.8) m3/m2/d
- Flash mixer - Flocculator
- Settling basin
Section of sedimentation unit consisting of

46. Example 1
1. Design a settling tank (coagulation–sedimentation)
with continuous flow for treating water for a
population of 48,000 persons with an average daily
consumption of 135 lit/head. Take detention period of
2.5 hrs. and maximum day factor of 1.8.
Solution:

47. Example 2
A mechanical flocculator is used to treat 38,000 m3/day
water with detention time 20 minutes.
a. Design the dimension of the tank if L : W : H = 4 : 1 : 2
b. Find the power required when velocity gradient is 55s-1
and dynamic viscosity 1.002*10-3 Ns/m2
c. If the tank have 3 paddles and every paddle have 4 plate
with relative velocity of paddles is 0.38m/s and
coefficient of drag is 1.8, find the area of one plate.

48. Example 2
Solution :
a) ∀ = 𝑄 ∗ 𝑡
= 38,000 m3/day * 20 min
= 527.8 m3
∀ = 𝐿 ∗ 𝑊 ∗ 𝐻
= 527.8m3 = 8W3
W3 = 527.8 /8
W = 4.04m
L = 4 * 4.04 = 16.16m
H = 8.08 m

49. Example 2
Solution :
b)
= 552 *(1.002 *10-3)*527.8
= 1599.8 Watt
c)
• So, area of one plate
=
32.4
3 ∗ 4
= 2.7𝑚2
𝐏 = 𝑮𝟐
𝐏 =
𝟏
𝟐
𝑪𝑫
𝑨𝒑
𝒗𝒑
𝟑
𝝆𝒘
𝑨𝒑
=
𝟐𝐏
𝑪𝑫
𝒗𝒑
𝟑 𝝆𝒘
𝐴𝑝
=
2 ∗ 1599.8
1.8 ∗ 0.383 1000
= 32.4𝑚2

50. Flocculation Basins