Flocculator, ClariFlocculator, Clarifiers

©
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.