Wt Chapter 5 Sed

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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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Water Treatment By Zerihun Alemayehu

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SEDIMENTATION Particles

Discrete

Flocculating

Size Shape weight

Suspension displacement of water velocity field interference

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Dilute

Concentrated


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






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


TYPES OF SEDIMENTATION 



No interaction between particles Settling velocity is constant for individual particles p articles








Examples: presedimentation presedimentation in water treatment, treatment, grit removal removal in wastewater

 

Particles collide and adhere to each other resulting in particle growth






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)



18


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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


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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300


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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

62


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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


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 



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 . 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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Cocurrent

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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 

Asafetyfactorof0.6maybeappliedtodeterminethedesigned settlingvelocity.Thus,  = 0.6 × 0.00022/ =





 = 236 ( 240   )

Step2.findsurfaceloadingrateQ/A  



0.0312

=

0.5 × 24 × 60 × 60   240 

= 180 /( )

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  



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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Wt Chapter 5 Sed

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