©
SHAHRUL ISMAIL, DESc.
University
College
of
Science
and
Technology
Malaysia
CHAPTER 3:
Environmental Microbiology
Water
Treatment
Process :
Flocculation
Flocculation theory
TYPES
MECHANICAL
DEVICES
HYDRAULIC
METHODS
FLOCCULATION

TYPES
Vertical

shaft,
turbine

type
impeller
Horizontal

shaft
paddle
In

line jet rapid mixing
device
Rotating blade
flocculators
Horizontal paddle
flocculator
Vertical paddle
flocculator
Baffled chamber
flocculator
CLARI

FLOCCULATOR
Temp (
°
C)
Density (kg/m
3
)
100
958.4
80
971.8
60
983.2
40
992.2
30
995.6502
25
997.0479
22
997.7735
20
998.2071
15
999.1026
10
999.7026
4
999.9720
0
999.8395
−10
998.117
−20
993.547
−30
983.854
The density of water in kilograms per cubic
metre
(SI unit)
at various temperatures in degrees Celsius.
The values below 0
°
C refer to
supercooled
water.
Density of water (at 1
atm
)
Drag force and
Coefft
. Of Drag
1.
In
fluid dynamics
,
drag
(fluid resistance) refers to forces
that oppose the relative motion of an object through
a
fluid
.
2.
The
drag equation
calculates the force experienced by an
object moving through a
fluid
at relatively large velocity
(i.e. high
Reynolds
number
,R
e
>
~1000)
3.
The
power
required to overcome the aerodynamic drag is given by:
4.
Power = Drag force X
Velovity
5.
In
fluid dynamics
, the
drag coefficient
s a
dimensionless
quantity
that is used to quantify the
drag
or resistance of
an object in a fluid environment such as air or water.
6.
It is used in the
drag equation
, where a lower drag
coefficient indicates the object will have
less
aerodynamic
or
hydrodynamic
drag. The drag
coefficient is always associated with a particular surface
area.
[1]
S.
N
o
Parameters
Flash Mixer
Flocculator
1.
Detention time
20

60 s
10

40 min
2.
Tank depth
3
–
4.5 m
3
–
4.5 m
3.
Peripheral
velocity of
paddles (paddle tip vel.)
1.75
–
2 m/s
0.2
–
0.6 m/s
4.
Velocity
gradient, G
> 300 (1/s)
10
–
75 (1/s)
DESIGN CRITERIA FOR FLASH MIXER & FLOCCULATOR
S.No
Parameters
Value
1.
Detention time
10

40 min
2.
Tank depth
3
–
4.5 m
3.
Total area of paddles
10
–
25% of CS area
of tank
4.
Peripheral
velocity of paddles
0.2
–
0.6 m/s
5.
Velocity
gradient, G
In tapered flocculation,
I stage
II stage
III stage
10
–
75 (1/s)
100 (1/s)
50
–
60 (1/s)
< 20 (1/s)
6.
Dimensionless
parameter,
Gt
For Alum coagulants
For Ferric coagulants
10
4

10
5
2
–
6 x
10
4
1
–
1.5 x
10
4
DESIGN CRITERIA FOR MECHANICAL FLOCCULATOR
S.No
Parameters
Value
7.
Power consumption
10
–
36 KW/MLD
8.
Velocity
of flow:
i)
From rapid mixing unit
to
flocculator
ii)
Through flocculation basin
iii)
From flocculation basin to
settling tank through pipe or
channel (to prevent settling
or breaking of
flocs
)
0.45
–
0.9 m/s
0.2
–
0.8 m/s
0.15
–
0.25 m/s
Step 1 Design of Influent Pipe
Assume velocity of flow
through pipe and use continuity
eqn. to find Diameter of pipe
V =
1 m/s (assume)
Q =
A V
255.1 (m
3
/h)=
(
π
D
2
/4) (1 x 3600)
D =
0.24 m
Provide
an influent pipe of 300 mm diameter.
Step 2 Design of
flocculator
1) Assume G & t,
calculate Gt.
2) If
Gt
is ok, calculate volume of
flocculator
(V =
QxDT
)
3) Assume water depth & from volume, obtain plan area
of
flocculator
(V = A x Depth)
4) From area of
flocculator
, obtain tank diameter.
Step 2 Design of
flocculator
Provide
a tank diameter of 6.6 m.
G =
40 (1/s) (assume)
t =
20 min (assume)
G t =
=
40 (1/s) (20 x 60) (s)
4.8 x 10
4
(Hence
Ok)
V =
=
255.1 (m
3
/h) (20/60) (h)
8.5 m
3
Water depth =
2.5 m (assume)
Plan area
of
floccualator
=
8.5/2.5 = 34 m
2
Let D =
Diameter
of
flocculator
D
p
=
Diameter
of inlet pipe
π
/4 (D
2
–
D
p
2
) =
34 m
2
D =
6.58 m
Step 3 Dimensions of Paddles
Approach:
1)
Calculate power input to
flocculator
(P = µG
2
V)
2)
Calculate area of paddles (P = ½ C
D
ρ
A
p
V
r
3
)
3)
Check total area of paddles = 10
–
25% of CS area of tank.
4)
Find out no. of paddles & paddle dimension (
HeightxWidth
)
5)
Find out no. of shafts to support paddles.
6)
Calculate the distance of shaft from the center line of
flocculator
7)
Calculate the distance of paddle edge from the center line of
vertical shaft (V = 2
π
rn
/60), assume no. of revolutions of
paddles.
Step 3 Dimensions of Paddles
8) Assume velocity of water below partition wall between
Flocculator
& clarifier and use continuity
eqn
(Q = A V)
for calculating Area of opening required below the
partition wall. Finally calculate depth below partition
wall.
9) Calculate depth to be provided for sludge storage (as
25% additional depth)
10)Calculate total depth of tank at partition wall,
assuming free board.
Total Area of Paddles (A
p
)
Provide
a tank diameter of 6.6 m.
P =
=
P =
µG
2
V
(0.89x10

3
) (40)
2
(
π
x6.6
2
x2.5/4)
122 W
P =
½ C
D
ρ
A
p
V
r
3
C
D
=
1.8 (assume)
ρ
@ 25
°
C
=
997 Kg/m
3
V
r
=
(0.75 x Vel. Of tip of blades)
Vel. Of tip of blades
=
0.4 m/s (assume)
V
r
=
=
0.75 x o.4
0.3 m/s
122
W
=
½ x 1.8 x
997 x
A
p
x (0.3)
3
A
p
=
5.04 m
2
Total Area of Paddles (A
p
)
CS Area of
flocculator
=
Π
⠶⸶
–
0.3) 2.5
(Total area of paddles /
CS Area of tank) =
=
5.04 /
Π
(6.6
–
0.3) 2.5
10.2% (Hence ok)
Provide 8 no
. of paddles of height 2 m & width
0.32 m
Total Area of Paddles (A
p
)
No. of shafts =
2 (each shafts will
support 4 paddles)
Dis. Of shaft from centre
line of
flocculator
=
=
(6.6
–
0.3) / 4
1.58 m
n (paddles) =
4 rpm
Dis. Of paddle edge
(r)
from center
line of vertical shaft, V =
0.4 =
r =
2
π
rn
/ 60
2
π
r4 / 60
1 m
Total depth of tank @ partition wall
Vel. Of water below partition
wall
bt.
Flocculator
& clarifier
=
0.3 (m/min) (assume)
Area of opening required below
partition wall, Q =
250 (m
3
/h)=
A =
A V
A x (0.3x60) (m/h)
13.9 m
2
Depth below partition
wall
=
=
13.9
/ (
π
x6.6)
0.67 m
Depth provided for sludge
storage =
=
Total
depth of tank assuming a free
board of 0.3 m =
=
0.25 x 2.5 (25%
extra)
0.63 m
0.3+2.5+0.67+0.63
4.1 m
Design of Clarifier
1)
Assume
SLR, calculate Dia. Of
clariflocculator
2)
Calculate length of weir & check weir loading rate.
SLR =
40 m
3
/m
2
/day
(assume)
SA of clarifier
=
=
(255.1 (m
3
/day) x 24) / 40 (m
3
/m
2
/day)
153.06 m
2
π
/4 (D
cf
2
–
6.6
2
) =
153.06 m
2
D
cf
=
15.44 m
Length of
weir =
=
Π
砠ㄵ⸴㐠
㐸⸵㌠4
t敩爠汯慤楮朠㴠
=
<
255.1 x 24 (m
3
/day) (1/48.53 (m) )
126 .2
(
m
3
/
day.m
)
300
(
m
3
/
day.m
) (OK)
SEDIMENTATION
•
It is the separation from water by gravitational settling
of SS that are heavier than water.
•
The most commonly used in water treatment.
•
Factors influencing settling are:
1) Size, shape, density & nature of particles
2) Viscosity, density & temp. of water
3) SLR
4) Velocity of flow
5) Inlet and outlet arrangements
5) Detention time
6) Effective depth of settling zone
TYPES OF SS
1.
Finely divided silt, silica & clay having specific gravity
ranging from 2.65 for sand, 1.03 for flocculated mud
particles containing 95% water.
2.
Alum & Iron
flocs
, specific gravities range from 1.18

1.34
3.
Precipitated crystals of calcium carbonate, obtained
from lime soda softening. Their specific gr. is 2.7 with
particle size 15
–
20 µm
SETTLING VEL. OF DISCRETE PARTICLES
The following equations may be used in arriving at settling
vel. Of discrete spherical particles:
1)
Stoke’s
law (Laminar):
V
s
= g/18 ((
ρ
s
–
ρ
)/µ) d
2
) , N
R
= 1, 0.1 mm particle size
2)
Hazen’s (Transition):
V
s
= [4/3 (g/C
D
((
ρ
s
–
ρ
)/
ρ
)]
0.5
, N
R
= 1
–
1000, 0.1
–
1mm
particle size
3)
Newton’s (Turbulent):
V
s
= [3.3 g ((
ρ
s
–
ρ
)/
ρ
) d ]
0.5
, N
R
= 10
3
–
10
4
, greater
than 1 mm particle size
SETTLING VEL. OF DISCRETE PARTICLES
C
D

> dimensionless drag coefficient
C
D
= (24/N
R
) + (3/√N
R
) + 0.34
N
R
= Reynolds No. = V
s
d
ρ
/µ
(dimensionless)
Types of tanks
•
Horizontal flow tanks or vertical flow tanks on the basis
of direction of flow of water in the tanks.
•
Rectangular, Square, Circular in plan.
Horizontal flow tanks
1)
Radial flow circular tank with central feed:

> Water enters at the centre of the tank

> Through openings in the circular well in the centre of
the tank, it flows
radially
outwards in all directions
equally.

> Horizontal vel. decreases as water flows towards
periphery

> sludge is taken to central sump mechanically
2)
Radial flow circular tanks with peripheral feed:

> Water enters the tank from periphery
3)
Rectangular tanks with longitudinal flow
4)
Rectangular tanks with longitudinal flow where sludge is
mechanically scrapped to sludge pit located near influent
end.
Vertical flow tanks
•
They combine sedimentation with flocculation.
•
They are square or circular in plan
•
Influent enters bottom of the unit, where flocculation
takes place
•
Upflow
vel. decreases with increased CS area of the
tank.
•
There is a formation of blanket of
floc
through which
rising
floc
must pass.
•
It is also known as ‘Sludge Blanket Clarifier’.
•
The clarified water is withdrawn through circumferential
weir.
CLARI

FLOCCULATORS
•
2 Or 4 flocculating paddles placed.
•
The paddles rotate on their vertical axis.
•
Paddles may be rotor

stator type, rotating in opposite
direction around this vertical axis.
•
Clarification unit is served by inwardly raking rotating
blades.
•
Flocculated water passes out from bottom of flocculation
tank to clarifying zone through a wide opening, the area of
opening being large enough to maintain a very low velocity.
•
Clear effluent overflows into the peripheral launders.
Tank Dimensions
•
Rectangular tank
upto
30 m length.
•
L to W ration = 3:1 to 5:1
•
Circular tanks
upto
60 m in
dia
are in use, but
upto
30 m
better to reduce wind effects.
•
Square tanks
upto
20 m
•
Depths = 2.5
–
5 m (preferably 3 m)
•
Bottom slopes vary from 1% in rectangular tanks to 8%
in circular tanks.
•
The slopes of
sludge hoppers range from 1.2:1 to 2:1
SLR & DT
TANK
TYPE
SLR (m
3
/m
2
/d)
DT
(h)
Particles
normally
removed
Range
Typical
value
for
design
Range
Typical
value
for
design
Plain
sedimentation
Upto
6000
15
–
30
0.01
–
15
3
–
4
Sand, slit &
clay
Horizontal flow
(circular)
25
–
75
30
–
40
2
–
8
2
–
2.5
Alum & Iron
floc
Vertical flow
(
Upflow
)
clarifiers

40
–
50

1
–
1.5
Flocculent
Inlets & Outlets
•
To obtain uniform vel. of flow, water is passed through
dispersion perforated by holes/slots.
•
Vel. Of flow through slots = 0.2
–
0.3 m/s
•
Headloss
= 1.7 x velocity head
•
Outlet structures:
Weir, Notches, Orifices, Effluent trough, Effluent
launder, Outlet pipe.
•
V

notches attach to sides of troughs and are placed 150
–
300 mm c/c
•
Weir length relative to surface area determines strength
of outlet current.
•
Normal weir loadings are
upto
300 m
3
/d/m
Sludge removal
•
Sludge is normally removed under hydrostatic pr. through
pipes.
•
Pipe
dia
= 200 mm or more (for non

mechanized units)
= 100
–
200 mm (for mechanized units)
•
Floor slope = shall not be flatter than 1 in 12 (for circular
tanks with scrapper)
= 1 in 10 (for manual cleaning)
•
Scrapping mechanism is rotated slowly to complete 1
revolution in 30
–
40 min
•
Tip vel. Of scraper should be 0.3 m/min or below
•
Power requirements are 0.75 W/m
2
of tank area
Settling tanks are capable of giving settled water having
turbidity not exceeding 20 NTU,
preferably < 10 NTU
Presedimentation
& Storage
DT = 0.5
–
3 h
High SLR = 20
–
80 m
3
/m
2
/d
Tube Settlers
•
Settling
η
of basin is dependent on SA, independent of
depth.
•
Very small
dia
tubes inserted in the basin provide
laminar flow conditions.
•
Such tube settling devices provide excellent clarification
with DT equal to or less than 10 min
•
Tubes cab be horizontal or steeply inclined.
•
In tubes inclined @ 60
°
, sludge will not slide down the
floors.
•
Tubes may be : Square, Rectangular, Triangular, Circular,
Hexagonal, Diamond shaped
•
Thin plastic sheet (1.5 mm) black in color
Settling Tank Efficiency
The
η
of basins is reduced by currents induced by inertia
of incoming water, wind, turbulent flow, density & temp
gradients. Such currents short circuit the flow.
The
η
of real basin,
Y/
Y
o
= 1
–
[1+(
nV
o
/(Q/A))]

1/n
Y/Y
o

>
η
of removal of suspended particles
η

>
Coeff
. that identifies basin performance
V
o

> Surface overflow rate for ideal settling basin
Q/A

> Required surface overflow rate for real basin
to achieve an
η
of Y/Y
o
for given basin
performance
Settling Tank Efficiency
Values of n:
n = 0 for best performance
= 1/8 for very good performance
= ¼ for good performance
= ½ for
ave
. performance
= 1 for very poor performance
A well designed tank should be capable of having a
volumetric
η
of
atleast
70%
To achieve better clarification, flow regime in settling
basin should be as close as possible to ideal plug flow.
A narrow and long rectangular tank approximates plug
flow conditions than peripheral feed circular tank & centre
feed radial flow tank.
Design of Radial Circular Settling Tank
Design a secondary circular settling tank to remove alum
floc
with following data.
1) Ave. output from settling tank = 250 m
3
/h
2) Amt. of water lost in
desludging
= 2%
3) Ave. design flow = (250/98%)x100% = 255.1 m
3
/h
4) Min. size of alum
floc
to be removed = 0.8 mm
5) Sp.gr. Of alum
floc
= 1.002
6) Expected removal
η
of alum
floc
= 80%
7) Assumed performance of settling tank = very
good (n = 1/8)
8) Kinematic viscosity of water @ 20
°
C = 1.01x10

6
m
2
/s
Design of Radial Circular Settling Tank
Approach:
1)
Using
Stoke’s
law calculate settling vel. Of particles
(V
s
) & check N
R
<1. If N
R
exceeds 1, Hazen’s formula
could be used.
2)
Calculate decreased SLR (Q/A)
3)
Using decreased SLR, calculate SA & Tank dia.
Assume DT & calculate depth.
4)
Check for weir loading rate
Design of Radial Circular Settling Tank
Step 1: Calculate Settling vel. Of particles:
Vs = (g(Ss

1)d
2
)/18
√
= (9.81 (1.002

1)(0.8x10

3
)
2
)/(18x1.01x10

6
)
= 6.91 x 10

4
(m/s)
N
R
=
V
s
d
/
√
= ((6.91x10

4
)(0.8x10

3
)) / (1.01x10

6
)
= 0.55 < 1 (Hence
Stoke’s
law applicable)
Design of Radial Circular Settling Tank
Step 2: Calculate SLR:
V
s
= V
o
(for ideal basin for complete removal)
=
6.91 x 10

4
(m/s)
= 59.7 (m/d)
Due to short

circuiting, basin
η
is reduced & decreased
SLR (
i.e
Q/A) is calculated as,
Y/Y
o
= [1
–
[1+(nV
o
/Q/A)]

1/n
Y/Y
o
= 0.8
n = 1/8
V
o
= 59.7 m/d
Q/A = 33.49 m/d (OK, since it is within design range
of 30
–
40 m
3
/m
2
/d)
Design of Radial Circular Settling Tank
Step 3: Determine Dimensions of Tank:
SA of tank, A = Q/(Q/A)
= (255.1x24)/33.49
= 182.8 m
2
Tank dia.
= 15.26 m
DT
= 2.5 h (assume)
Depth
= (255.1 x 2.5) / 182.8
= 3.5 m
Design of Radial Circular Settling Tank
Step 4: Check for Weir loading:
Weir length = Periphery of tank
=
π
x 15.25
= 47.94 m
Weir loading = (255.1 x 24) / 47.94
= 127.7 m
3
/d/m
< 300 m
3
/d/m , hence OK
Design of Rectangular Sedimentation Tank
1.
Desired
ave
. outflow settling tank = 250 m
3
/h
2.
Water lost in
desludging
= 2%
3.
Design
ave
. flow = (250/98%) x 100% = 255.1 m
3
/h
4.
Min. size of particle to be removed = 0.02 mm
5.
Expected removal
η
of min. particle size = 75%
6.
Nature of particles = Discrete & Non

flocculatig
7.
Sp. Gravity of particles = 2.65
8.
Assumed performance of settling tank = good (n = ¼)
9.
Kinematic viscosity of water @ 20
°
C = 1.01 x 10

6
(m
2
/s)
Design of Rectangular Sedimentation Tank
1)
Calculate settling vel. Of min. size particles:
Vs = (g(S
s
–
1) d
2
) / 18
√
= (9.81(2.65

1)(0.02x10

3
)
2
)/(18x1.01x10

6
)
= 3.56 x 10

4
(m/s)
N
R
= V
s
d /
√
= (3.65x10

4
)(0.02x10

3
)/(1.01x10

6
)
= 704 x 10

3
< 1 (OK)
Design of Rectangular Sedimentation Tank
2) Calculate decreased SLR:
V
s
= V
o
(Theoretical SLR for 100% removal)
= 3.65 x 10

4
m/s
= 30.76 m/d
Y/Y
o
= 0.75
n = ¼
Q/A = 18.53 (m/d)
= OK (since within design range of 15
–
30
m
3
/m
2
/d)
Design of Rectangular Sedimentation Tank
3) Determine Dimensions of Tank:
SA, A = Q/(Q/A)
= (255.1x24) / 18.53
= 330.4 m
2
L/W = 4 (assume)
W = 9.09 m
L = 36.36 m
DT = 4 h (assume)
Depth = (
QxT
) / A
= (255.1 x 4)/(36.36 x 9.09)
= 3.09 m
Design of Rectangular Sedimentation Tank
4) Influent Structure:
•
It consists of an influent channel, submerged orifices &
baffles in front of orifices.
•
Provide 0.6 m wide & 0.6 m deep influent channel that
runs across the width of the tank.
•
Provide 4 submerged orifices (0.2m x 0.2m) in the inside
wall of an influent channel to distribute the flow uniformly
into basin.
•
A baffle of 1 m deep is provided at a distance of 1 m away
from orifices to reduce turbulence.
Design of Rectangular Sedimentation Tank
Effluent Structure:
•
It consists of effluent weir, effluent launder, outlet box &
outlet pipe.
•
Weir loading rate = 200 m
3
/d/m (assume)
Weir length = (250 x 24 ) / 200 = 30 m
•
Provide 30 m length of effluent launder with V

notches
fixed on one side of launder.
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