Wt Chapter 5 Sed
description
Transcript of Wt Chapter 5 Sed
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SEDIMENTATION
AAiT Water TreatmentBy Zerihun Alemayehu
INTRODUCTIONSedimentation is removal of particulate materials suspended in
water by quiescent settling due to gravity
Commonly used unit operation in water and wastewatertreatment plants
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SEDIMENTATION
Particles
FlocculatingDiscrete
Suspension
ConcentratedDilute
Size
Shape
weight
displacement of water
velocity field interference
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TYPES OF SEDIMENTATION
No interaction between particles
Settling velocity is constant for individual particles
Dilute solids concentration
Examples: presedimentation in water treatment, grit removal in wastewater
Particles collide and adhere to each other resulting in particle growth
Dilute solids concentration
Examples: coagulation/flocculation settling in water treatment and primary sedimentation in wastewater treatment
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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
Size and shape
Specific gravity
Properties of the water
Specific gravity
Viscosity
Physical environment of the particle
Velocity of the water
Inlet and outlet arrangements of the structure
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DISCRETE PARTICLES SETTLING (TYPE 1)
Fd
mg
Fb
Fnet = mg - Fd - Fb
D
ss
C
gdv
w
w
)(
3
4
Re < 1, Cd = 24/Re (Laminar flow)1
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EXAMPLEEstimate the terminal settling velocity in water at a temperature of15oC of spherical silicon particles with specific gravity 2.40 andaverage diameter of (a) 0.05 mm and (b) 1.0 mm
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SOLUTIONStep 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 Reynolds number
Since R > 2, the Stokes law does not apply. Use Eq. 1 to calculate v
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SOLUTIONStep 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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GENERALLYAll particles with d>= do, such that v >= vo, will arrive at or pass the
sampling port in time to.
A particle with dp < do will have a terminal settling velocity vp < voand will arrive at or pass the sampling port in time to, with original position at, or below a point Zp.
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 to 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.
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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 v1 < Z0/t1 is
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PROCEDURE SETTLEABLITY ANALYSIS
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PROCEDURE SETTLEABLITY ANALYSIS
For a given detention time to, an overall percent removal can be obtained.
All particles with settling velocities greater than v0=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
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
Vt x 102, m/min 3.0 2.5 2.0 1.55 1.0 0.83 0.48
x = Ci
=189
= 0.63Co 300
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SOLUTIONStep 2. Plot mass fraction vs. settling velocity
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0 0.5 1 1.5 2 2.5 3 3.5
X, W
t. f
ra
cti
on
rem
ain
ing
Velocity, m/min X 10-2
vo = 1.74 x 10-2 m/min
xo
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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
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0 0.5 1 1.5 2 2.5 3 3.5
X, W
t. f
ra
cti
on
rem
ain
ing
Velocity, m/min x 10-2
x = 0.06
x = 0.1
x = 0.1
x = 0.1
x = 0.06
x = 0.06
x = 0.06
vo=
1.7
4 x
10
-2m
/min
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SOLUTION
Step 5. x.vt by graphical integration
x 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
%72
74.1
46.046.0
.1
o
to
v
vxxX Step 6. Determine overall
removal efficiency
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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
Depth, m
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 62
Step 1. Determine the removal rate at each depth and time
xij = (1- Ci/Co)x100
x11 = (1 133/250) x 100 = 47
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SOLUTION
Step 2. Plot the isoconcentration lines
Step 3. Construct vertical line at to = 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 flowlongitudinal, radial flow
Vertical flowcircular (upward flow)
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LONG-RECTANGULAR BASINSLong 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 horizontally and vertically
across the tankthe outlet zone in which water flows
upward and over the outlet weir
settling zone, which occupies the
remaining volume of the tanksludge zone, which extends from the
bottom of the tank to just above the
scraper mechanism
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LONG-RECTANGULAR TANKDetention time: is the theoretical time that the water is detained
in a settling basin. it is calculated as the volume of the tankdivided by the rate of flow, and is denoted as = V/Q.
vo or Q/As is called overflow rate or surface loading rate or surface overflow rate (SOR).
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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.
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CIRCULAR BASINS
Simple sludge removal mechanisms
require less maintenanceno excessive weir overflow
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SHORT-CIRCUITING AND REDUCTION OF
EFFICIENCY
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INLET ZONE Inlets should be designed to dissipate the momentum and accurately distribute the
incoming flow
Multiple opening across width
Baffle
Multiple opening across width
Baffle
Flow distribution 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/m2
horizontal flow circular tank 30,000 to 40,000 1/d/m2
Slopes: Rectangular 1 % towards inlet and circular 8% Weir loading rate, m3/m/d < 248
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EXAMPLEFind 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
AAiT Water TreatmentBy Zerihun Alemayehu
EXAMPLE
Design a circular basin. 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.
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SOLUTIONV=6000x4/24=1000
Depth=20x4/24=3.33=3.5m
As=V/d=1000/3.5=285.7m2
As=xD2/4
Diameter=19m
Hence provide 5 m deep (1 m for sludge and 0.5 m free
board) by 19 m diameter tank
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EXAMPLE
Design a long-rectangular settling basin for
type-2 settling. 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.
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SOLUTIONQ = 15,000 m3/d, SOR (V) = 20 m/d and d = 2.5 m
Calculate surface area
A = Q/V = 15,000 / 20 = 750 m2
Use two tanks A = 375 m2 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
=
sin
=
( cos + 2)u
v
w
H
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EXAMPLE
A water treatment work treats 1.0 m3/s andremoves flocs larger than 0.02 mm. Thesettling velocity of the 0.02mm flocs ismeasured in the laboratory as 0.22 mm/s at 15oC. Tube settlers of 50 mm square honeycombsare inclined at a 50o angle and its verticalheight is 1.22 m. Determine the basin arerequired for the settler module.
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SOLUTIONStep 1. Determine the area needed for the settler modules
Q = (1m3 /s)/2 = 0.5 m3/s = 30 m3 /min
w = 50 mm = 0.05 m
H = 1.22 m
= 50o
=
( cos cos2 )
=0.5(0.0508)
(1.22 0.643 + 0.0508 0.6432)
=0.312
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SOLUTION A safety factor of 0.6 may be applied to determine the designed
settling velocity. Thus,
Step 2. find surface loading rate Q/A
= 0.6 0.00022/ =0.0312
= 2362 ( 240 2)
=
0.5 24 60 603
2402= 1803/(2)
Step 3. Compute flow velocity in the settlers
v = Q/A sin = 180/0.766
= 235 m/d
= 0.163 m/min
= 0.0027 m/s
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SOLUTION
30 4
3= 40
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SOLUTIONStep 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
=0.05082
4 0.0508= 0.0127
R =
=(0.0027/)(0.0127)
0.0000011312
30 < 2000, !
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SOLUTIONStep 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)
16 m
Launder
Baffle wall
Diffuser wall
Inlet channel
Outlet channel
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SOLUTION
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QUESTIONS?