Water Characteristics and Sampling · 11/11/2013 9 AAiT Water Treatment By Zerihun Alemayehu...

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SEDIMENTATION

AAiT Water Treatment

By Zerihun Alemayehu

INTRODUCTION

Sedimentation is removal of particulate materials suspended inwater 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



TYPES OF SEDIMENTATION

No interaction between particles

Settling velocity is constant for individual particles

Dilute solid’s concentration

Examples: presedimentation in water treatment, grit 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

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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)



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<Re<104, Cd = 24/Re + 3/(Re)1/2 + 0.34 (Transition)•103 < Re < 104, Cd = 0.4 (Turbulent)

v

Fnet = 0



STOKE’S LAW

For Re < 1 (laminar flow) and

Re = wvsd/ (for perfect sphere),

Stoke’s law:

18

)(2

ws

s

gdv

18

)1(2

g

s

Sgdv

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EXAMPLE

Estimate 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



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)



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



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



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 to.

A particle with dp < do will have a terminal settling velocity vp < vo

and 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.



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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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)?



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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SOLUTION

Step 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



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



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



TYPE-2 SETTLING…

Mass fraction removed is calculated as:

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EXAMPLE: SETTLING COLUMN ANALYSIS OF

FLOCCULATING PARTICLES



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



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—longitudinal, radial flow

Vertical flow—circular (upward flow)



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 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



LONG-RECTANGULAR TANK

Detention time: is the theoretical time that the water is detainedin a settling basin. it is calculated as the volume of the tankdivided by the rate of flow, and is denoted as θ = 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.



CIRCULAR BASINS

Simple sludge removal

mechanisms

require less maintenance

no excessive weir overflow

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SHORT-CIRCUITING AND REDUCTION OF

EFFICIENCY

θ



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



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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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



SOLUTION

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SOLUTION



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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SOLUTION

V=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



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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SOLUTION

Q = 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



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




Cocurrent

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Crosscurrent



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



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



DESIGN OF INCLINED SETTLERS

𝑣 =𝑄

𝐴 sin 𝜃

𝑢 =𝑄𝑤

𝐴(𝐻 cos 𝜃 + 𝑤𝑐𝑜𝑠2𝜃)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.



SOLUTION

Step 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

𝑢 =𝑄𝑤

𝐴(𝐻 cos 𝜃 − 𝑤 cos2 𝜃)

=0.5(0.0508)

𝐴(1.22 × 0.643 + 0.0508 × 0.6432)

=0.312

𝐴

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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

𝑢 = 0.6 × 0.00022𝑚/𝑠 =0.0312

𝐴𝐴 = 236𝑚2 (𝑢𝑠𝑒 240 𝑚2)

𝑄

𝐴=

0.5 × 24 × 60 × 60𝑚3𝑑

240𝑚2= 180𝑚3/(𝑚2𝑑)

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



SOLUTION

30𝑚 ×4

3= 40𝑚

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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

𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐𝑠 𝑟𝑎𝑑𝑖𝑢𝑠 𝑟 =0.05082

4 × 0.0508= 0.0127𝑚

R =𝑣𝑟

𝜇=(0.0027𝑚/𝑠)(0.0127𝑚)

0.000001131𝑚2𝑠

30 < 2000, 𝑡ℎ𝑢𝑠 𝑖𝑡 𝑖𝑠 𝑎 𝑙𝑎𝑚𝑒𝑙𝑙𝑎 𝑓𝑙𝑜𝑤 𝑂𝐾!



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)

16 m

Launder

Baffle wall

Diffuser wall

Inlet channel

Outlet channel



SOLUTION

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QUESTIONS?

Water Characteristics and Sampling · 11/11/2013 9 AAiT Water Treatment By Zerihun Alemayehu...

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