Waste Water

Treatment

Chapter – 2
Preliminary and primary waste water
treatment methods
1
Course content
2.1. Preliminary Treatment
 Screening

 Comminutors

 Grit Removal Basins

2.2. Primary Treatment (sedimentation)

 Primary clarifier

 Skimming tanks

2
 Introduction
• Water treatment typically consists a range of unit processes which are usually

operated in series.

• The main unit processes in water treatment are (in order of operation):
 Screening, - clarification or sedimentation
 storage -filtration
 Sedimentation -pH adjustment
 Aeration -disinfection
 coagulation -softening and other tertiary process
 flocculation -sludge removal.

• The selection of the necessary unit processes depends on

 the source & quality of the raw water and

 the desired quality of the finished water.
3
 Recommended treatment for specific impurities

4
 Conventional Water Treatment

Raw water
Screening Filtration
sludge

Alum Coagulation Cl2 Disinfection

Polymers

Flocculation Storage

Sedimentation Distribution
sludge
5
2.1 Preliminary treatment processes

• Any physical, chemical or mechanical processes used before main water

treatment processes.

• May include the following processes

 Removal of debris

 Control of algal growth

 Pre-sedimentation
 Aeration to remove dissolved gases or to enhance oxidation

 Pre‐disinfection

6
Cont…
• Preliminary treatment consists in separating the floating materials and heavy
settleable inorganic solids.
• Reduces the BOD of the WW by about 15 to 30%. The process used are:

1. Screening
Are under Preliminary Treatment
2. Comminutors
3. Grit champers

1. primary clarifier. Are under Primary treatment

2. Skimming Tanks.

7
 1.Screening
• Screens are located at the intake structure, to remove large floating or suspended

matters before they enter the pump station and It does not change the chemical
& bacteriological quality of water.

• Floating maters include pieces of clothes, papers, wood, hair, fiber , kitchen

refuse, fecal solids, etc.

• What if floating materials not removed? It will choke the pipes or adversely

affect the working of the treatment plant.

• The main idea of providing screens is to protect the pumps and other equipment

from the possible damages due to the floating matter of the sewage.

8
Classification of Screens
• Based on the dimension of the opening (Spacing) screens can be classified as:-

1. Course screening, for spacing of over 40mm

2. Medium screening, for spacing of 10mm to 40mm.
3. Fine screening, for a spacing under 10 mm.

• Screens are usually installed at the entrance of the wastewater treatment plant

 to protect mechanical equipment and avoid interference with plant

operations

 prevent objectionable floating materials

9
A. Coarse screens (bar screen)
• Removes large floating objects like: pieces of clothes, papers, wood, hair, fiber

aquatic plants, leaves, or other floating debris

• If not removed, this material can

o blockage equipment
o damage pumps and piping

• It generally consists of a number of parallel bars, with a width of 40mm or above

• bar stand is inclined to facilitate cleaning. Collect about 6 litters of solids per

million litters of WW.

10
Manual and mechanical bar screen

11
B. Medium screens
• The spacing between the bars is about 10 to 40mm. These screens ordinarily collect

30 to 90 litters of material per million liter of sewage. The screenings usually

contain some quantity of organic material, which may decompose and become
offensive.
c. fine screens
 The spacing between the bars is under 10mm. Fine screens are usually used for

pre-treatment of industrial wastes to remove materials which tend to produce

excessive layer or foam.

 They remove as much as 20% of the suspended solids from sewage. These

screens, however, get clogged very often, and need frequent cleaning. This
screens will considerably reduce the load on further treatment units.

12
2, Comminutors /shredder:-
• A devices which breaks the larger sewages solid to about 6mm size

• Typically located after the screening process, which further reduce the size of
the remaining materials for easier handling and recycling.

13
Disposal of screenings

• Are maters separated by screens and disposed off. It contains 85%-90% moisture
and other floating matter.

• It may also contain some organic load which may putrefy (decompose),

causing bad smell. To avoid such possibilities, the screenings are disposed
of either by: - Burning

- Burial

- Dumping

14
3. Grit Removal Basins
• The grit chamber remove the inorganic particles [specific gravity about 2.65 or
larger) such as sand, gravel, egg shells and other non-putresible (non degradable)
material.

• Grit may clog channels or damage pumps due to abration and to prevent their
accumulation in sludge digesters.

• These inorganic materials are removed by the process of sedimentation due to

gravitational forces.

• The important point in the design of the grit basins is that the flow velocity should
neither be too slow nor should it be so high b/c.

- if too slow lighter organic matter will settle down

- if too high, even silt and grit not settle down.

Design of Bar Screens
Approach Velocity

• Optimum Velocity : 0.6 m/s (through the screen opening)

• Maximum Velocity : 0.9 m/s (to prevent entrapped materials being forced

through the bars).

• Minimum Velocity : 0.4 m/s to prevent deposition of solids.

Angle of Inclination (60 - 80 degree)

• This angle is important, particularly in manually cleaned bar screens.

The opening between the bars is 40-100 mm

The head loss through the screen is generally limited to about 150 mm.

16
Head loss in Bar screen
• Head loss represents the amount of energy lost as fluid flows through a system

and can have a significant impact on system performance and efficiency.

• The head loss can be easily calculated:

17
Example 1 (bar screen design)
Determine the head loss (hL) across the bar screen if:
a) The screen is clean
b) If 50 % of the screen is blocked

Parameter symbol Unit Value

Incoming flow velocity Va m/s 0.6
Pass through velocity Vb m/s 0.9
Screen cross sectional area A open m2 0.19
Friction coefficient for clean C 0.7
screen
Friction coefficient for C 0.6
clogged screen

18
solution
a) Head loss for clean screen

0.9

b) Head loss for 50% clogged

Q = V*A

Vb = 2*Vb = 2*0.9 m/s

19
= 1.8 m/s
Higher resistance due to blockage requires more energy
Example 2 (bar screen design)

Dimension the following pre-treatment unit for a WTP for a maximum flow of
48,000 m3/day. A bar screen is used to remove the coarse waste. The velocity in
the screen channel (Va) is 0.5 m/s and flow-through velocity of the clean screen
(Vb) is 0.8 m/s.

a) Calculate the required open screen area for reaching this screen velocity
during treatment

b) Calculate the head loss of this clean screen

c) Calculate the maximum clogging allowed before the screen needs to be

cleaned. (hL max = 0.15m)

20
Solution
a) Q max-day = 48,000m3/d = 0.56 m3/s

b) Head loss for clean screen

21
Solution
C) % of Area clogged when hL (max) = 0.15m

= 1.5788 m/s

0.354m2
1.5788

From a A total = 0.69 m2

56%
22
Design of a rectangular grit chamber
There are two design parameters called detention time (dt) and settling velocity
(Vs).

• Detention time: is a time required by WW to reach outlet from the inlet and or the
time taken to settle to the bottom of chamber.

• Settling velocity: is the velocity required by a grit to reach the bottom of

chamber.

Length(L)=flow velocity(Vt)*dt

Height(H)= settling velocity*dt

After fixing the depth and the detention time we can design the dimension of rectangular
chamber.
Two to three separate chambers in parallel should provided. In
manual cleaning as one unit can work, the other will be shout down
for cleaning.
 Grit removal basin

• The grit champers can be cleaned periodically at about 3 weeks interval either manually,
mechanically or hydraulically.
 EXAMPLE 3 (GRIT CHAMBER DESIGN)
A rectangular grit chamber is designed to remove particles with a specific gravity
of 2.65. settling velocity of this particles has been found from 0.016 to 0.022m/s,
depending on their shape factor. A flow through velocity of 0.3m/s will be
maintained by proportioning weir. Determine channel dimension for a maximum
waste water flow of 10,000m3/day.

Given Required
Cont…

34.5 s Length= flow velocity*detention time

0.022 Height= settling velocity*detention time

Length= flow velocity*detention time

Length= 0.3m/s*34.5 sec L= 10.37m

W= 0.507m H= 0.761m L=10.37m

26
EXAMPLE 4 (GRIT CHAMBER DESIGN)
Design a suitable grit chamber tank for a sewage treatment plant having
400l/s discharge passing through it. Assume the flow velocity through the Tank
as 0.2 m/s and detention period of 2 min. Assume free board= 0.3 and dead
space depth=0.45 at the bottom of the tank.

Given Required

Discharge- 400l/s= 0.4m3/s

Flow velocity= 0.2m/s

Dt= 2min= 120sec

27
Cont…
Solution
Length= flow velocity*detention time

0.2m/s*120s= L= 24 m

Q= V*A

= A= Q/V= 0.4m3/s/0.2m/s

= 2m2

Let assume H= 1.2m

A= H*W W= A/H

= 2m2/1.2m = W= 1.67m

At the top there is free board of 0.3m and dead space depth=0.45

At the bottom
28 L=there
24mis dead
H=space
1.95mof 0.45W=1.67m
EXAMPLE 5 (GRIT CHAMBER DESIGN)
Design a grit chamber for treatment of municipal wastewater. The following design criteria
are applicable.
 Maximum flow rate- 5m3/s
 flow through velocity= 0.25m/s
 Dt= 1min (60 sec)
 Length to width ratio- 2:1

Answer
Step 1- calculate Area….Ax= 20m2

Step 2- Calculate length= Velocity*dt= L=15m

From L/W= 2….w=15/2…. W= 7.5m

Step 3- calculate height from, Volume= Q*dt

L*W*H= 5m3/s *60s….H= 2.7m

29
2.2 Primary treatment processes
• Primary waste water treatment is the removal of settle-able organic and inorganic
solids coming out from the grit chambers by sedimentation in primary clarifier and
the removal of suspended organic solids and FOG (fat, oil & grease) that will float
by skimming tank.

• Approximately 25 to 50% of the incoming

biochemical oxygen demand (BOD5),
50 to 70% of the total suspended solids (SS),
and 65% of the oil and grease are removed
during primary treatment.
 Primary clarifier
• Is a rectangular or circular tank in which wastewater is held for a period of time

to allow heavier solids to settle to the bottom as sludge and lighter materials to
float to the water surface as scum.

• Is a pre-sedimentation which involves settling of suspended solids without the

aid of a coagulant. It removes most of the sediment in the water during the
pretreatment stage.

• So pre-sedimentation will reduce the load on the coagulation/flocculation

basin and on the sedimentation chamber, as well as reducing the volume of

coagulant chemicals required to treat the water.

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Skimming tank
• Employed for removing fat, oils and grease and place before sedimentation tanks.

These oil and grease materials may be removed by an air which is blown by an
aerating device through the bottom.

• The rising air tends to coagulate and congeal(solidify) the grease and cause it to

rise to the surface.

• A detention period of about 3-5 minutes is sufficient.

Surface area required for the tank can be found out using the formula

where q=flow rate

Vr =rising velocity of greasy material to be removed
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33
The principle behind sedimentation
• The very fundamental principle is that: the organic matter present in sewage

have a specific gravity greater than that of water.

 In still sewage: these particles will, tend to settle down by gravity.

 In flowing sewage: particles are kept in suspension, because of the turbulence

of water.

• Hence, as soon as the turbulence is retarded by offering storage to sewage,

these impurities tend to settle down at the bottom of the tank.

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Types of settling
• Depending on the particles concentration and the interaction between particles,

there are 4 types of settling.

1. Type I: Discrete particle settling

The particles settle without particle interaction and occur under low solids
concentration.

Settling velocity is constant for individual particles

Examples: grit removal in preliminary treatment

35
2. Type II: Flocculent settling

• Particles initially settle independently, but flocculate in the depth of the

clarification unit. The velocity of settling particles is usually increasing as the
particles aggregates.

• Particles collide and adhere to each

other resulting in particle growth
• Examples: coagulation/flocculation
settling in water treatment

Type I settling and Type II settling are the common

sedimentation processes found in water treatment.
36
 3. Type III: Hindered or zone settling

Inter-particle forces are sufficient to hinder the settling of neighboring particles.

The particles tend to remain in fixed positions with respect to each others.

• Particles are so close together movement is restricted

• Solids move as a block rather than individual particles
• Fluidic interference causes a reduction in settling
velocity

37
4. Type IV: Compression settling
• After hindered settling continues for a long period of time, a layer of particles

with a definite structure begins to form at the bottom of the basin or pond.

• Occurs when the particle concentration is so high that particles at one level are

mechanically influenced by particles on lower levels. The settling velocity then

drastically reduces.
• Particles physically in contact
• High concentration of solids (sludge)

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Settling region for concentrated suspensions
Factors Affecting Sedimentation
1. Particle size
 The size and type of particles to be removed (i.e. density)

e.g. sand vs. colloidal material

2. Particle shape
 for example, round particle will settle much more readily than a particle that

has irregular edges.

3. Electrical charge of particles

 the same charge tend to repel each other i.e keeps the particles from

collecting into flocs and settling.

4. Water temperature
 When the temperature decreases, the rate of settling becomes slower.
40
Cont…
5. Current
 wind currents, thermal and other currents

i.e. Tend to distribute the floc unevenly throughout the tank; as a result, it does not
settle out at an even rate.

6. Detention Time
 Time that the water is in the system

In addition to the above listed factors sedimentation efficiency can also be influenced by:
short- circuiting
gases in the water
algal growth on tank walls and

41
 Sedimentation Tank
 Are designed for effecting settlement of particles by reducing the flow velocity

or by detaining the sewage in them.

 These particles may settle at the bottom of the tank and are removed by using

scrapers.

Types of sedimentation tanks

Based on methods of Based on shape

Based on location
operation a. Circular tank
a. Primary tank
a. Fill and draw type tank b. Rectangular tank
b. Secondary tank
b. Continuous flow type c. Hopper bottom
42tank tank
Classification based on methods of operation

a. Fill and Draw Type Sedimentation Tank

• Store sewage for a certain period and keep it in complete rest. Water from inlet

is stored for up to 24 hour. In that time, the suspended particles are settled at the
bottom of the tank.

• After 24 hours, the water is discharged through outlet. Then settled particle are

removed. This removal action requires 6-12 hours. So, one complete action of
sedimentation requires 30-36 hours.

43
Cont…
b. Continuous Flow Type Sedimentation Tank
• In this case sewage is not brought to complete rest. Flow always takes place

but with a very small velocity. During this flow, suspended particles are settle
at the bottom of the tank. The flow may be either in horizontal direction or
vertical direction.

44
Cont…
Horizontal flow
• Tanks are generally in rectangular shape and flow takes place in horizontal
direction. They have more length twice its width. Because they need to flow
more distance to settle all suspended particles.

Vertical flow
• Tanks are generally in circular or hopper bottom shape and flow takes place in
vertical direction. When the liquid enters near the bottom of the tank and is
withdrawn at the surface. Hopper bottom is provided at the bottom of the tank
to dispose the collected sludge.

45
classification based on shape

a. Circular tanks

• Are preferred for continuous vertical flow type sedimentation tanks. In this

case influent is sent through central pipe of the tank and radial flow takes
place. Mechanical sludge scrappers are provided to collect the sludge and
collected sludge is carried through sludge pipe provided at the bottom.

• Are uneconomical as compared to rectangular tanks but they have high

clarification efficiency.

46
 Cont…
b. Rectangular basins

• Are mostly preferred tanks which are widely used. Have the simplest design and

low maintenance cost allowing water to flow horizontally through a long tank.

• This type of basin is usually found in large-scale water treatment plants.

• In addition, rectangular basins are the

least likely to short-circuit, especially if
the length is at least twice the width.
Baffle walls are provided for rectangular
tank to prevent short circuiting.

47
Cont…
c. Hopper Tank

• In case of hopper bottom tank, a deflector box is located at the top which

deflects the influent coming from central pipe to downwards. Sludge is

collected at the bottom and it is disposed through sludge pump.

• The water enters from the top inlet channel. Are vertical flow tanks, because

water flows upward and downward in these tanks.

48
classification based on LOCATION
a. Primary tank

• These tanks are the first step in the sedimentation process, where wastewater is

held to allow larger suspended solids to settle out due to gravity. This stage
significantly reduces the organic load before the water moves on to biological
treatment processes. The settled sludge is removed periodically for further
processing.

b. Secondary Tank

• Secondary sedimentation tanks are used to separate the activated sludge

(biological solids) from the treated water. This step ensures that the effluent
meets discharge standards by removing remaining suspended solids and
pathogens that were not captured during primary treatment.

49
Zones of sedimentation tanks

• All sedimentation basins have four zones - the inlet zone, the settling zone, the

sludge zone, and the outlet zone

• Each zone should provide a smooth transition between the zones. In addition,

each zone has its own unique purpose.

• Water can also enters the basin from the center rather than from one end and

flows out to outlets located around the edges of the basin.

50
A. the inlet/influent should distribute flow
uniformly across the inlet to the tank.
The normal design includes baffles that
gently spread the flow across the total
C. the outlet zone to provide a smooth
inlet of the tank.
transition from the settling zone to the
effluent flow.

D. sludge zone, provides a storage

area for the sludge before it is
removed for additional treatment or
disposal.
B. Settling zone is the largest portion of the
sedimentation basin. Provides the calm area
51 necessary for the suspended particles to settle.
Determining the capacity of the settling zone

• The capacity of the settling zone can be determined on the basis of over flow

rate.

• It is assumed that the settlement of a particle at the bottom of the tank does not

depend on the depth, but on the surface area of the tank. This assumption can
be proved theoretically as follow;

Let L=Length of the settling zone

W=Width of the tank H=depth of the tank C=Capacity of the tank

T=time of horizontal flow(detention time)

Vh(Vd)=Horizontal flow velocity (velocity in the direction of flow)

Vs= vertical settling velocity

52 Q=Discharge of flow
53
 •shows that the velocity of
settlement of the particle is
independent on the depth
of the tank.

54
Design of horizontal flow Tanks
• The design of continuous flow sedimentation tank depends on the following

assumptions:

1. Particles settles in settling zone of the tank

2. the flow is horizontal and steady and the velocity is uniform in all parts of settling
zone for time equal to td.

3. A particle is removed when it reach the bottom of settling zone

• The design of sedimentation basins is governed by three basic criteria:

1. The quantity of water to be treated

volume of wastewater flowing through a
2. The selected surface loading rate (overflow rate)
unit area of the treatment tank per day

3. The selected detention period

• To reduce the likelihood of short-circuiting a length to width ratio (L/W) of 3 or

55
more is recommended.
 Cont…

 The two characteristics are important in designing a sedimentation basin.

1. The overflow rate (also known as surface loading or the surface overflow rate)
is defined as the ratio between the influent flow rate and the surface area of the
tank.

2. The weir loading is also known as weir overflow rate is another important factor
in sedimentation basin efficiency. Is the number of gallons of water passing over a
foot of weir per day.

 Is the flow rate of clarifier effluent, divided by the length of the overflow weir.

56
Cont…
• When the sedimentation basin specifications are given the following steps are

followed to determine the surface area, dimensions, and volume of the

sedimentation tank as well as the weir length.
Spec

1. Divide flow in to at least two tanks. Depth – 7-16 ft

L:W= 4:1
2. Calculate the required surface area.
Velocity- 0.5ft/min
3. Calculate the required volume. weir loading- 15,000gal/ft/day

4. Calculate the tank depth.

5. Calculate the tank width and length.

6. Check flow through velocity.

- If velocity is too high, repeat calculations with more tanks.

7. Calculate the weir length

57
1. Divide the flow in to two tanks
The flow should be dividing in to at least two tanks and the flow through each tank
should be calculated using the formula shown below:

Where:
Qc = flow in one tank
Q = total flow
n = number of tanks
We will consider a treatment plant with a flow of 1.5MG/D. We will divide the
flow in to three tanks, so the flow in one tank will be:
QC = 1.5MG/D/3
QC = 0.5MG/D
58
2. Surface area
• We will base this surface area on an overflow rate of 500gal/ft2/day in order to

design the most efficient sedimentation basin.

• The surface area is calculated using the following formula:

….derived from Q=V*A

Where: `A=surface area,ft2

Qc=flow, gal/day

O.R. = overflow rate(settling velocity), gal/day-ft2

In our example, the surface area of one tank is calculated as follows:

A = (500,000gal/day)/ (500gal/ft2/day )

A=1,000ft2

(Notice that we convert the flow from 0.5MG/D to 500,000gal/day before

59
beginning our calculations.)
3. Volume
• The tank volume is calculated by multiplying flow by detention time.

• We will consider a tank with automatic sludge removal, so the detention time

should 4 hours.

• The volume of one our tanks is calculated as follows:

• V=(500,000gal/day)(4hr) 1gal= 0.133681ft3

V=11,141ft3 4hr= 0.1666day

• (Notice the conversions between days and hours between cubic feet and

gallons.)

60
4. DEPTH
The tank depth is calculated as follows:

Where: d=depth, ft V=volume,ft3 A= surface area, ft2

For our example, the depth is calculated to be:
d=(11,141ft3 )/(1,000ft2 ) d = 11.1 ft
• The specifications note that the depth should be between 7 and 16 feet. Our
calculated depth is within the recommended range.
• If the depth was too large, we would begin our calculations again, using a larger
number of tanks. If the depth is too shallow, we would use a smaller number of
tanks.

61
 5. Width and Length
• You will remember that the volume of the rectangular solid is calculated as

follows: V= LWd

Where: V=volume d=depth L= length W=width

• For our tank, the length has been defined as follows: L= 4W or L:W= 4:1

• Combined these two formulas, we get the following formula used to calculate

the width of our tank: V=4w2d

w = √(V/4d)

w=15.8ft

The length is calculated as: L = 4(15.8ft) L= 63.2ft

62
 6. Flow through velocity
• First, the cross sectional area of the tank is calculated:

Ax = Wd
Ax = (15.8ft) (11.1ft)
Ax = 175.4ft2

• Then the flow through velocity of the tank is calculated (with a conversion from

gallons to cubic feet and from days to minutes):

• Convert gal/day to ft3/min……1gal/d= 9.28337*10-5ft3/min

=46.41ft3/min

• = 46.41ft3/min/175ft2 V=0.26ft/min

The velocity for our example is less than 0.5ft/min, so it is acceptable.

63
As a result, we do not need to repeat our calculations.
7. WEIR LENGTH
• Weir length represents the perimeter of the top of the tank.
• Weir loading, also known as weir overflow rate, is the number of gallons of
water passing over a foot of weir per day.

• Where: Lw: weir length, ft

• Qc: flow in one tank, gal/day
• W.L: weir loading, gal/day-ft
We have a weir loading of 15,000gal/ft/day
Lw = (500,000gal/day)/(15,000gal/ft/day) Lw = 33.3ft

64
Example- 1
Design a rectangular horizontal flow sedimentation tank for the following data.
Design flow- 3ML/day
Detention period- 4hr
Velocity of flow- 10cm/min
Solution
Length= flow velocity*dt
=10cm/min*4h
=0.1m/min*(4*60)hr
=24m
Convert 3ML/d to m3/hr………125m3/hr
Volume= Q*dt
Volume= 125m3/hr*4h= volume= 500m3
From the volume formula we can find for Area……Area= Volume/Length
= 500m3/24m = 20.23m2
Assume H= 3m ……from A= H*B……B= 7m
65
Provide 1m depth for sludge and 0.5m as free board H= 3m+1m+0.5m= 4.5m
Example 2
Design a sedimentation basin for the following data
i. Flow rate
maximum flow rate =10,000m3 /day
flow through velocity= 0.5ft/min

ii. Design parameters

td = 4hr
L/W=2
SOR=10m3/m2/d
Design flow rate to be 25% of maximum flow rate

66
 1. Determine the number of tanks
Qc= 25%(Qtotal)
Qc= 25/100(10,000m3/day) Qc= 2,500m3/day
Qc= 2,500m3/day

n=Q/Qc
n=10,000m3/day/2,500m3/day= 4

2. Calculate surface area

From Q=V*A….. Surface Area = flow rate/SOR (settling velocity)
= 2500m3/day/10m3/m2/d
= 250m2

67
3. Calculate volume
Volume=Qc*dt
= 2,500m3/d * (4/24)day
= 417m3

4. Calculate tank depth(effective depth)

V=L*W*H or H= Vs*dt
V= As*H =10m3/m2/d / (4/24)day
H=V/As = 1.67m
= 417m3/250m2 = 1.67m Provide the sludge and the free board zone of 0.5m
Ht=2+0.5+0.5= 3m

5. Calculate tank width and length

As=L*W…it is given that L/W=2
W=√As/2= 11.18m L= 2*11.18m= 22.3m
(Note that: if it is requested to provide inlet and out-let zone, it is assumed to be to be equal to th
68
effective depth of the tank. (L= 2+2+22= 26m)
 6. Check flow through velocity
Q=V*A
Vh= Qc/Ax-section
= 2,500m3/d / 11.8m*1.67m
= 133.9m/day…….0.001549m/s
Convert m/s to ft/min…………..1m/s= 196.85ft/min.
0.001549m/s= 0.305ft/min……………..OK.

69
Example 3 (circular sedimentation tank)
A circular sedimentation tank is to have a minimum detention time of 4hour and an
overflow rate of 20m/d. Determine the required diameter of the tank and the depth if
the average flow rate through the tank is 6ML/d.
Solution
Volume= Q*dt depth=Vs*dt
= 6,000m3/d*(4/24)day = 20m/day *(4/24)day
= 1000m3 = 3.33m
From V= ∏D2/4 *h………..
D= 19.5m

Hence provide 5m deep (1m for sludge and 0.5m free board) by 19.5m diameter tank)

70
Example 4 (circular sedimentation tank)
Design a circular sedimentation tank if the average flow rate of tank is 10,000m3/d.
Water will stay for 2 hours in tank (detention period).
Solution
= 10,000m3/d*(2/24)d = 833.34m3

Volume= As*H
= ∏D2/4*H…….Assume height of the tank is 3m
833.34m3= ∏D2/4*3m
= D=18.8m….say 19m
Provide 1m for sludge and 0.5m as a free board, so total height/ depth
= 3m+1m+0.5= 4.5m

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Example 5 (rectangular sedimentation tank)
Design a rectangular sedimentation tank for a town having population of 50,000
with the water supply rate of 180l/c/d.
Solution
Given
Population= 50,000
Water supply rate =180l/c/d
Find flow rate or (Q) = population*supply rate/p*conversion factor
Q= 50,000*180l/c/d*0.8
Q= 7,200m3/d
Assume detention period 2hr

Volume= 7,200m3/d*(2/24)d = 600m3

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 Cont…
Assume depth(h)= 3m
Volume= As *h
As= V/h = 600m3/3
As = 200m
Assume L/W= 4
As= L*W
200m= 4W2
W= 7m and L= 4* 7m = 28m
 Provide 3m for inlet and outlet arrangement
Total length= 28m+3m+3m = 34m
 Provide free board of 0.5m and sludge depth of 1m
Overall depth= 3m+0.5m+1m= 4.5m
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Example 6 (rectangular sedimentation tank)
The maximum daily demand of water purification plant has been estimated as
12Ml/day. Design the dimension of a suitable sedimentation tank (fitted with
mechanical sludge removal arrangement). Assume a detention period of 6hour and
the velocity of flow is 20cm/min.
Solution

12Ml/d*(6/24)d= V= 3,000,000l or 3,000m3

L= Vh*dt
= 20cm/min*(6*60)min = 7,200cm= 72m
V=l*w*h
V= l*Acrossection
74 Acrossection= V/l = 3000m3/72m = 42m2
 Cont…
Assume depth(d)= 4m
Acrossection=w*h
W= A/h
42m2/4m = 10.5m
 Provide free board and sludge depth of 0.5m.
Overall depth= 4m+0.5m+0.5m= 5m

Dimensions of the tank= 72m*10.5m*5m

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Settling of discrete particles (type 1 settling)
• When discrete particles are placed in dormant fluid, it will accelerate until the friction
resistance (drag force, FD) is equal to the impelling force (driving force) acting on the
particles.
• At this stage, the particles attain a uniform or terminal velocity and settles down with
this constant velocity called settling velocity, Vs. 3 forces which apply on particle are:-

The 3 forces which apply on particle are:

• For laminar settling (d < 0.1mm) here Re<0.1, Stoke’s formula can be applied

• Where ƥ p= density of particle, ƥ= density of water and g= gravitational

constant(9.81m/s2).
Cont…

Where CD is the function of Re

Reynolds number is a dimensionless parameter that facilitates the prediction of

flow behavior. A low Reynolds number indicates laminar flow while a high
Reynolds number indicates turbulent flow.

 For transitional settling (d between 0.1mm and 1mm) here 1<Re<103

 For turbulent settling (d >1.0mm) here Re>103

Where CD is the function of Re

Example 7
Find the terminal settling Velocity of a spherical Particle with diameter of 0.5mm
and a specific gravity of 2.65 settling through water (at 200C).

Solution
Given
D=0.5mm 0.5*10-3m
Sg=2.65

Assume the flow is Laminar

= 0.22m/sec…… Check the flow is laminar or not

 Cont…
……..not laminar

The flow is transitional flow

Check Re

………….OK

Therefore Vp = 0.11m/sec

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 e

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Quiz-1
1. What is short circuiting?

2. Design rectangular sedimentation tank treating of 2ML/d of raw water with

detention time of 4hours. The settling velocity= 0.55m/hr and L/W= 4.

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