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SEDIMENTATION
INTRODUCTION
emoval of par particulate materials suspended in Sedimentation is remo
wate waterr by quie quiesc scen entt sett settliling ng due due to grav gravit ityy unit operation in water and wastewater Commonly used treatm treatment ent plants plants
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SEDIMENTATION Particles
Discrete
Flocculating
Size Shape weight
Suspension displacement of water velocity field interference
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Dilute
Concentrated
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TYPES OF SEDIMENTATION
No interaction between particles Settling velocity is constant for individual particles p articles
Dilute solid’s concentration
Examples: presedimentation presedimentation in water treatment, treatment, grit removal removal in wastewater
Particles collide and adhere to each other resulting in particle growth
Dilute solid’s concentration
Examples: coagulation/flocculation settling in water treatment and primary sedimentation in wastewater treatment AAiT
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SEDIMENTATION Particles
Discrete
Flocculating
Size Shape weight
Suspension displacement of water velocity field interference
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Dilute
Concentrated
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TYPES OF SEDIMENTATION
No interaction between particles Settling velocity is constant for individual particles p articles
Dilute solid’s concentration
Examples: presedimentation presedimentation in water treatment, treatment, grit removal removal in wastewater
Particles collide and adhere to each other resulting in particle growth
Dilute solid’s concentration
Examples: coagulation/flocculation settling in water treatment and primary sedimentation in wastewater treatment AAiT
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TYPES OF SEDIMENTATION
Particles are so close together movement is restricted Intermediate solids concentration Solids move as a block rather than individual particles Fluidic interference causes a reduction in settling velocity Distinguishable solids liquid interface Intermediate solids concentration Example: settling of secondary effluents
Particles physically in contact Water is squeezed out of interstitial spaces Volume of solids may decrease High concentration of solids (sludges)
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DISCRETE PARTICLES SETTLING (TYPE 1)
Characteristics of the particles
Properties of the water
Size and shape Specific gravity Specific gravity Viscosity
Physical environment of the particle
Velocity of the water Inlet and outlet arrangements of the structure AAiT
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DISCRETE PARTICLES SETTLING (TYPE 1) Fb
Fd
Fnet = mg - Fd - Fb Fnet = 0
mg v
v s
4 ( s
3
w
w
) gd C D
Re < 1, Cd = 24/Re (Laminar flow) 1
• • •
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STOKE’S LAW For Re < 1 (laminar flow) and Re = wvsd/ (for perfect sphere),
Stoke’s law:
2
v s
gd ( s
18 2
v s
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w )
gd (S g
1)
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EXAMPLE Estimate the terminal settling velocity in water at a temperature of 15oC of spherical silicon particles with specific gravity 2.40 and average diameter of (a) 0.05 mm and (b) 1.0 mm
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SOLUTION
Step 1. Using Stokes equation for (a) at T= 15oC
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SOLUTION…
Step 2. Check with the Reynolds number
Step 3. Using Stokes’ law for (b)
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SOLUTION…
Step 4. Check the Reynold’s number
Since R > 2, the Stokes’ law does not apply. Use Eq. 1 to calculate v
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SOLUTION…
Step 5. Re-calculate Cd and v
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SOLUTION…
Step 6. Re-check Re
Step 7. Repeat step 5 with new R
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SOLUTION…
Step 8. Re-check Re
Step 9. Repeat step 7
(b) The estimated velocity is around 0.16 m/s
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SETTLING COLUMN
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GENERALLY
All particles with d>= do, such that v >= vo, will arrive at or pass the sampling port in time t o.
A particle with dp < do will have a terminal settling velocity v p < vo and will arrive at or pass the sampling port in time t o, with original position at, or below a point Z p.
If the suspension is mixed uniformly then the fraction of particles of size dp with settling velocity vp which will arrive at or pass the sampling port in time t o will be Zp/Zo = vp/vo.
Thus, the removal efficiency of any size particle from suspension is the ratio of the settling velocity of that particle to the settling velocity vo defined by Zo/to. AAiT
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PROCEDURE – SETTLEABLITY ANALYSIS
Usually 2m high column
Mix the suspension thoroughly
Measure initial SS concentration, Co
Measure concentrations at certain intervals, Ci
All particles comprising C1 must have settling velocities less than Z0/t1. Thus the mass fraction of particles with v 1 < Z0/t1 is
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PROCEDURE – SETTLEABLITY ANALYSIS
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PROCEDURE – SETTLEABLITY ANALYSIS For a given detention time t o, an overall percent removal can be obtained. All particles with settling velocities greater than v 0=Z0/t0 will be 100 percent removed.
Thus, 1 – xo fraction of particles will be removed completely in time t0. The remaining will be removed to the ratio vi/vo, corresponding to the shaded area in Fig. 4.2. If the equation relating v and x is known the area can be found by integration:
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EXAMPLE : SETTLING COLUMN ANALYSIS OF TYPE-1 SUSPENSION
A settling analysis is run on a type-1 suspension. The column is 2 m deep and data are shown below.
What will be the theoretical removal efficiency in a settling basin with a loading rate of 25 m3/m2-d (25m/d)?
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SOLUTION
Step 1. Calculate mass fraction remaining and corresponding settling
rates =
Ci
0.63
Time, min
60
80
100
130
200
240
420
Mass fraction remaining
0.63
0.60
0.56
0.52
0.37
0.26
0.09
3.0
2.5
2.0
1.55
1.0
0.83
0.48
Co
=
189
=
x
V t x 102, m/min
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SOLUTION
Step 2. Plot mass fraction vs. settling velocity 0.7
g0.6 n i n i a0.5 m e r0.4 n o i t0.3 c a r f . 0.2 t W , X0.1
xo
vo = 1.74 x 10-2 m/min
0 0
0.5
1
1.5
2
2.5
3
3.5
-2
Velocity, m/min X 10
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SOLUTION…
Step 3. Determine vo
vo = 25m3/m2.d = 1.74 x 10-2 m/min
Step 4. Determine xo = 54 %
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SOLUTION
Step 5. x.vt by graphical integration 0.7
g0.6 n i n i a0.5 m e r0.4 n o i t0.3 c a r f . 0.2 t W , 0.1
x = 0.06
n i m / m
x = 0.06 x = 0.1
2 -
0 1 x 4 7 . 1 = o v
x = 0.1 x = 0.1
X
x = 0.06 x = 0.06
0 0
0.5
1
1.5
2
2.5
3
3.5
-2
Velocity, m/min x 10
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SOLUTION… x
Step 5. x.vt by graphical
integration
vt
x.vt
0.06
1.50
0.09
0.06
1.22
0.07
0.1
1.00
0.10
0.1
0.85
0.09
0.1
0.70
0.07
0.06
0.48
0.03
0.06
0.16
0.01
x.vt=0.46
Step 6. Determine overall removal efficiency
X 1 xo
0.46
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x.vt vo
0.46 1.74
72%
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TYPE-2 SETTLING
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TYPE-2 SETTLING…
Mass fraction removed is calculated as:
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EXAMPLE: SETTLING COLUMN ANALYSIS OF FLOCCULATING PARTICLES
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SOLUTION Step 1. Determine the removal rate at each depth and time xij = (1- Ci/Co)x100 x11 = (1 – 133/250) x 100 = 47 Depth, m
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Time of sampling, min 30
60
90
120
150
180
0.5
47
67
80
85
88
91
1
28
51
63
74
78
83
1.5
19
40
53
63
72
77
2
15
33
46
56
64
72
2.5
12
28
42
51
59
68
3
10
25
38
47
55
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SOLUTION… Step 2. Plot the isoconcentration lines Step 3. Construct vertical line at t o = 105 min
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SETTLING TANKS /SEDIMENTATION TANKS
The principle involved in these tanks is reduction of velocity of flow so that the particles settle during the detention period.
Such tanks are classified into,
Fill and draw types tanks (batch-process)
Continuous flow tanks.
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SETTLING TANKS /SEDIMENTATION TANKS
Depending on their shape, sedimentation tanks may be classified as,
circular,
rectangular, and
square.
Depending on the direction of flow, as
horizontal flow
Vertical flow
longitudinal, radial flow
—
circular (upward flow)
—
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LONG-RECTANGULAR BASINS
Long rectangular basins are commonly used in treatment plants processing large flows.
hydraulically more stable, and flow control through large volumes is easier with this configuration.
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LONG-RECTANGULAR TANK the inlet zone in which baffles intercept the oncoming water and spread the flow uniformity both vertically the outlet zone in horizontally which waterand flows across the upward andtank over the outlet weir settling zone, which occupies the sludge zone, which extends from the remaining volume of the tank bottom of the tank to just above the scraper mechanism
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LONG-RECTANGULAR TANK is the theoretical time that the water is detained in a settling basin. it is calculated as the volume of the tank divided by the rate of flow, and is denoted as θ = V/Q.
Detention
time:
vo or Q/As is called overflow rate or surface loading rate or surface overflow rate (SOR). AAiT
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REMOVAL EFFICIENCY OF PARTICLES A particle initially at height h with settling velocity of vsh will just be removed by the time it has traversed the settling zone. Particles initially at heights less than h will also be removed and those at greater heights will not reach the bottom before reaching the outlet zone. All particles with settling velocity vs < vsh are removed partly, depending on their position at a height from the top of the sludge zone. The efficiency of removal of such particles is given by h/H.
The greater the surface area, the higher the efficiency. AAiT
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CIRCULAR BASINS
Simple
sludge removal mechanisms require less maintenance no excessive weir overflow
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SHORT-CIRCUITING AND REDUCTION OF EFFICIENCY θ
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INLET ZONE
Inlets should be designed to dissipate the momentum and accurately distribute the
incoming flow
Baffle
Multiple opening across width
Flow distribution
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Baffle
Multiple opening across width
Typical baffle (diffuser) walls
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OUTLET STRUCTURE
consist of an overflow weir and a receiving channel or launder. The launder to the exit channel or pipe. re-suspension of settled solids must be prevented flow velocity in upward direction has to be limited Increase the length of the overflow overflow weir
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DESIGN DETAILS Detention period: for plain sedimentation: 3 to 4 h, and for coagulated sedimentation: 2 to 2 1/2 h Velocity of flow: < 18 m/h (horizontal flow tanks) Tank dimensions: L : B : 3 to 5 : 1
Generally L = 30 m (common) maximum 100 m. Breadth: 6 m to 10 m. Circular: Dia < 60 m, Generally 20 to 40 m
Depth 2.5 to 5.0 m (3 m) Surface loading (or) overflow rate or (SOR)
For plain sedimentation- 12,000 to 18,000 1/d/m2 for thoroughly flocculated water 24,000 to 30, 000 1/ d/m 2 horizontal flow circular tank 30,000 to 40,000 1/d/m 2
Slopes: Rectangular 1 % towards inlet and circular 8% 3 Weir loading rate, m /m/d < 248
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EXAMPLE
Find the dimensions of a rectangular sedimentation basin for the following data:
Volume of water to be treated = 3 MLD Detention period = 4 hrs Velocity of flow = 10 cm/min
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SOLUTION
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SOLUTION
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EXAMPLE A circular sedimentation tank is to have a minimum detention time of 4 h and a maximum overflow rate of 20 m3/m2.d. Determine the required diameter of the tank and the depth if the average flow rate through the tank is 6 ML/d. AAiT
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SOLUTION V=6000x4/24=1000 Depth=20x4/24=3.33=3.5m 2 As=V/d=1000/3.5=285.7m
As=xD2/4 Diameter=19m
Henceprovide5mdeep(1mforsludgeand0.5mfree board)by19mdiametertank
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EXAMPLE . A city must treat about 15,000 m3/d of water. Flocculating particles are produced by coagulation, and a column analysis indicates that an overflow rate of 20 m/d will produce satisfactory removal at a depth of 2.5 m. Determine the size of the required settling tank. AAiT
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SOLUTION Q = 15,000 m3/d, SOR (V) = 20 m/d and d = 2.5 m Calculate surface area 2 A = Q/V = 15,000 / 20 = 750 m 2 Use two tanks A = 375 m for each Take L: B = 3: 1 L = 33.50 m and B = 11.20 m
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HIGH-RATE SETTLING MODULES Small inclined tubes or tilted parallel plates which permit effective gravitational settling of suspended particles within the modules. Surface loading 5 to 10 m/h
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TYPICAL CONFIGURATIONS
Countercurrent
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TYPICAL CONFIGURATIONS
Cocurrent
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TYPICAL CONFIGURATIONS
Crosscurrent
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TUBE SETTLERS
Take advantage of the theory that surface overflow loading, which can also be defined as particle settling velocity, is the important design parameter.
Theoretically, a shallow basin should be effective.
Use tubes of 25 to 50 mm diameter
At a 60o angle provide efficient settling
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TUBE SETTLERS
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PLATE /LAMELLA PLATE SETTLER
Taking advantage of the theory that settling depends on the settling area rather than detention time.
Distance between plates is designed to provide an upflow velocity lower than the settling velocity of the particles,
The effective settling area is the horizontal projected area
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PLATE /LAMELLA PLATE SETTLER
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DESIGN OF INCLINED SETTLERS
v
=
H
= u
sin
( cos + )
w
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EXAMPLE A water treatment work treats 1.0 m3/s and removes flocs larger than 0.02 mm. The settling velocity of the 0.02mm flocs is measured in the laboratory as 0.22 mm/s at 15 oC. Tube settlers of 50 mm square honeycombs are inclined at a 50o angle and its vertical height is 1.22 m. Determine the basin are required for the settler module. Water Treatment By Zerihun Alemayehu
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SOLUTION
Step 1. Determine the area needed for the settler modules Q = (1m3 /s)/2 = 0.5 m 3/s = 30 m3 /min w = 50 mm = 0.05 m H = 1.22 m = 50o = = =
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( cos − cos ) 0.5(0.0508)
(1.22 × 0.643 + 0.0508 × 0.643 ) 0.312
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SOLUTION
Asafetyfactorof0.6maybeappliedtodeterminethedesigned settlingvelocity.Thus, = 0.6 × 0.00022/ =
= 236 ( 240 )
Step2.findsurfaceloadingrateQ/A
0.0312
=
0.5 × 24 × 60 × 60 240
= 180 /( )
Step3.Computeflowvelocityinthesettlers v=Q/Asin =180/0.766 =235m/d =0.163m/min =0.0027m/s
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SOLUTION
4 30 × = 40 3
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SOLUTION
Step 5. Check horizontal velocity Q/A = (30 m3/min)/(4 m x 8 m) = 0.938 m/min
Step 6. Check Reynolds number (R) in the settler module
=
R=
0.0508 4 × 0.0508
=
= 0.0127
(0.0027/)(0.0127) 0.000001131
30 < 2000, ℎ !
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SOLUTION
Step 7. Launder dimension
Provide 3 launders for each basin. The launder must cover the entire length of the settler module; thus the length of the launder is 30 m. the flow rate in each launder trough is 0.5/3 = 0.167 m3/s For a rectangular trough section Q = 1.38 bh1.5
Select the width (b) as 0.5 m
Thus h = 0.39 m
Make the interior height of the launder 0.5 m (0.11 m freeboard)
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SOLUTION 10 m
30 m (covered with tubes)
Launder m 6 1
Baffle wall
Outlet channel Diffuser wall Inlet channel
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SOLUTION
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