WATER TREATMENT TECHNOLOGY (TAS 3010) LECTURE NOTES 9b - Flocculation

Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT

© SHAHRUL ISMAIL, DESc.University College of Science and Technology Malaysia

CHAPTER 3:Environmental Microbiology

CHAPTER 9 : CHAPTER 9 : CHAPTER 9 : CHAPTER 9 :

TAS 3101 : WATER TREATMENT TECHNOLOGY

Water Water

Treatment Treatment

Process :Process :

FlocculationFlocculation

Water Water

Treatment Treatment

Process :Process :

FlocculationFlocculation


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1) Introduction

2) Flocculation Theory

3) Flocculation Factors

4) Velocity Gradient

5) Flocculation Types


Introduction

- Slow

- Slow mixing process in which particles are brought into contact in

order to promote their agglomeration.

- Conversion factors simplify changing from one unit of measurement to

another.

- There are two conditions under which water is measured—water at

rest and water in motion. Water at rest is measured in units of volume.

Water in motion is measured in units of flow— unit of volume for a

convenient time unit. It is important that the difference between a unit of

volume and a unit of flow be kept in mind.


Agglomeration of destabilized particles into microfloc – second stage of floc growth

Floc : Bulky floccules which can be settled

Flocculant / Flocculant aid : Addition of another reagent promote the formation of the floc.

Relies on turbulence

Accomplished by gentle stirring

Enhance sedimentation

Flocculation


Flocculation


Flocculation Theory


Factors affecting performance of flocculation process

1) pH

i. Prominent factor

ii. Optimum pH range

iii. The type of colloid

iv. Chemicals to adjust the pH of the water

v. Trial and error testing

Flocculation Factors


2) Tank volume

The volume of the tank is calculated using the following formula:

V = Q t

Where:

V = volume

Q = flow

t = detention time

Flocculation Factors – Con’d


3) Detention time

Time required for a small amount of water to pass through a tank at a given flow rate. Mathematically, detention time is given by the following formula:

t = V Q

Where:

t = detention time

V = tank volume

Q = flow

Flocculation Factors – Con’d


Velocity Gradient

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

the floc.

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.

Where

G = Velocity gradient

P = Power input

u = dynamic viscocity

V = Volume

Velocity Gradient


- Design parameter for flocculation is Gt (dimensionless

number)

- Where;

G = velocity gradient, s-1 (20 – 75 s-1)

t = time, s (10 – 60 min)

Gt = 12 000 – 270 000 (unitless)

- High G values with Low t : Small and Dense flocs

- Low G values with High t : Larger and Lighter flocs

Velocity Gradient


- Vary from cylindrical to rectangular or cubical.

- Mixed simply by water flowing around baffles, or with a variety of

types of paddles, turbines, and propellers.

Vertical-shaft, turbine-type impeller Horizontal-shaft

paddle

Flocculation - Mixers


Baffled Chamber Flocculator


Flocculation - Types

TYPESTYPES

MECHANICAL MECHANICAL DEVICESDEVICES

HYDRAULIC HYDRAULIC METHODSMETHODS


1) MECHANICAL DEVICES

- Most often in small plants

Advantages :

i) Prevents water from rotating continuously in the same direction

around the shaft

ii) Low head loss

Disadvantages :

i) Low velocity around the shaft

ii) High operation and maintenance cost

Mechanical Devices


DESIGN DETAIL - POWER

Power = Force x Velocity

P = FD x Vp

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

Mechanical Design - Power


DESIGN DETAIL - FORCE

FD = ½ CD Ap Vp2 p

Where;

CD = coefficient of drag, 1.8 for flat blades

Ap = area of paddle blades, m2

p = density of water, kg/m3 ( 998.2 kg/m3)

Mechanical Design - Force


DESIGN DETAIL – POWER

P = G2 µ

Where;

G = velocity gradient, s-1

µ = dynamic viscocity, Ns/m2

= volume of mixing tank, m3

P = power input, Watt or Nm/s

THEREFORE;

Overall Equation :

P = CD AP (Vp)3 p

2

Mechanical Design - Power


Measure of the reduction in the total head (sum of elevation head,

velocity head and pressure head) of the fluid as it moves through a

fluid system

Present :

i. Friction between the fluid and the walls of the pipe; the

ii. Friction between adjacent fluid particles as they move

relative to one another;

iii. Turbulence caused whenever the flow is redirect or

affected

Proportional to the length of pipe, the square of the fluid velocity

Head Loss


Resistance of water to flow due to internal

molecular forces.

Related to the liquid's temperature.

Viscosity


A mechanical flocculator is used to treat 38000m3/day water with

detention time 20 minutes.

a) Design the dimension of the tank if L : W : d = 1 : 4 : 2

b) Find the power required when velocity gradient is 55s-1 and dynamic

viscosity 1.002 x 10-3 N.s/m2

c) If the tank have 3 paddles and every paddle have 4 plate with

relative velocity of paddles is 0.38 m/s and coefficient drag is 1.8,

find the area of 1 plate.

(Water density = 998.2 kg/m3)

Mechanical - Problems


Solution :

For a) Use volume tank formula to find each

dimension of L, W, d

For b) Find power, use formula that can fits all the

information given

For c) Use formula Power that fits all the information

given



Solution :

a) = Qt

= 38,000 m3/day x 20 mins

= 527.8 m3.

= L x W x d

= 527.8m3 = 8L3

L3 = 527.8 /8

L = 4.04m W = 4 x 4.04 = 16.16m d = 8.08 m



Solution :

b) P = G2

=(55)2 (1.002 x 10-3)(527.80) p

= 1599.8 Watt.

c) P = CD Ap (Vp)3 p

2

Ap = 1599.8 (2)

(1.8) (0.38)3 (998.2)

= 32.45 m2 So, area of 1 plate = 2.7 m2



- Baffle types

- Induce required velocity gradients for achieving floc

formation

Advantages :

i) Simple to construct and operate

ii) Less chance of short circuiting

Disadvantages :

i) Cannot be easily adjusted

ii) Increase head loss

Hydraulic Flocculation


1. Velocity in the channel : 10 – 30cm/s

2. Width of the channel : 45 cm

3. Depth of flow : > 1.0m

4. Detention time : 10 – 20 minutes

5. Loss of head : 15 – 60 cm

Hydraulic – Design Detail


To design;

1) Tank volume,

2) Distance of flow, D

3) Cross Sectional Area between baffle, Ax

4) Depth of channel, d

5) Clear distance between baffle and the end of wall, Y

6) Effective length of each channel, Pe

7) No. of channel required for each compartment

8) Length and width basin, L & W

Hydraulic – Design Detail


1) Total volume,

= Q x t


D = t x v


Ax = / D


d = A / Distance between baffles

Hydraulic – Design Formulas



Y = 1.5 x Distance between baffles


Pe = Width of each compartment - Y

7) No. of channel required for each compartment,


Hydraulic – Design Formula

No. of channel for total distance flow = D/ Pe

So, for each compartment = No. of channel/ of total distance flow / total compartments


1) Total volume,

= Q x t

= 20 x 106 L/day x 25 min x1 m3/1000L x day/1440 min

= 347 m3


D = t x v

= 25min x 0.17 m/s x 60s /1 min


Ax = / D


d = A / Distance between baffles




Ax = / D

= 347m3 /255 m

= 1.36 m2


d = Ax / Distance between baffles

= 1.36 m2 / 0.75 m

= 1.81m




Y = 1.5 x Distance between baffles

= 1.5 x 0.75

= 1.125 m


Pe = Width of each compartment – Y

= 3m – 1.125m

= 1.875m




No. of channel for total distance flow = D/ Pe

So, for each compartment = No. of channel/ of total distance flow / total compartments



No. of channel for a total distance of flow = D/Pe

= 255

m/1.875

= 136

channel

No. of channel for EACH compartment = 136 / 3= 45.3





Length = (135 x 0.75 m) + ( 135 x 0.075m)

= 111.375m

Width = 1.875 m

Check Volume = 111.375 x 1.875 x 1.81 = 380 m.



Design a baffle type basin of round the end type with the

following data :

Daily quantity of water to be treated = 20 MLD

Detention time = 25 min

Average velocity of flow = 0.17 m/s

Distance between baffles = 0.75 m.

Thickness of each baffle = 0.075 m

3 compartment,

Hydraulic – Problems


Hydraulic – Problem Solution

1) Calculate the tank volume,

= Qt

= 20 x 106 L/day x 25 min

= 260 m3


D = tv

= 25 min x 0.17 m/s

=


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WATER TREATMENT TECHNOLOGY (TAS 3010) LECTURE NOTES 9b - Flocculation

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