Chapt5_Theory of Sedimentation.pdf

cmutsvangwa: Water Quality & Treatment, Dept of Civil & Water Eng. 10/10/2006 5-1

Chapter 5 Theory of Sedimentation

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Chapter 5

THEORY OF SEDIMENTATION Downward movement of small suspended particles by gravity. Sedimentation is classified upon the characteristics and concentration of suspended materials:

• discrete particles • flocculent • dilute suspension

Discrete particles (Type 1) Particle whose size, shape and specific gravity do not change with time i.e. non-interactive settling of particles from a dilute suspension. Examples are grit and sand, and their mass is constant. Flocculant particles (Type 2) Particles which agglomerate (coalesce/flocculate) during settling i.e. no constant characteristics. Their mass varies during the process of settling and an increase in mass causes a faster rate of settlement.

Fig. 1: Settling paths of discrete and flocculent particles

Dilute suspension Concentration of particles is not sufficient to cause significant displacement of water as they settle.

Time

Dep

th Flocculent particle

path

Discrete particle path



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Hindered particles Or called zonal settlement, the particles interact and the concentration of particles is high. Settlement is slow because as particles move down (large in numbers), water is displaced upwards hindering downward settlement. Theory of sedimentation Discrete particles Discrete particles will accelerate until a limiting terminal velocity is reached when placed in a liquid of lower density. Gravitational force=frictional drag force

( )gVFforceGravity wsg ρρ −== (1) Where: sρ =density of particle wρ =density of fluid V =volume of particle

Fdrag

Fg

Fig. 2: A settling particle in water

2

2s

CddragvACF ωρ= (2)

Where: Cd =Newton’s drag coefficient Ac =cross-sectional area of particle perpendicular to the direction of motion vs -settling velocity of particle



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The coefficient of drag varies with shape and the regime of flow which is defined by the Reynolds number, Re:

For laminar flow, Re≤1 Re24

=dC

Transitional flow, 1<Re≤103 34.0Re3

Re24

++=dC

Turbulent flow, Re>103 Cd =0.4

and µρφ dv ws=Re or

γφ dvs=Re

Where; φ =shape factor

d =diameter of particle At equilibrium; dragg FF =

Therefore; ( )2

2s

wcdwsvACgV ρρρ =−

⇒ ( )wcd

wss AC

gVv

ρρρ −

=2

Particles are assumed spherical and for perfect spheres, the shape factor, Ø=1. The shape factor accounts for irregularities of particles.

⇒ 6

3dV π= Or

3

234

⎟⎠⎞

⎜⎝⎛=

dV π

4

2dA π=

Hence ( )wd

wss C

gdv

ρρρ

34 −

= (Newton’s law

Or ( )134

−= sd

s SCgdv Specific gravity

w

ssS

ρρ

=



4

Therefore for laminar flow: ( )γ18

12 −= s

sSgd

v (Stokes Law)

Or ( )

µρρ

18

2ws

sgd

v−

= ( )

γ18

2ws

sSSgd

v−

=

Where; Ss =specific gravity of particle Sw =specific gravity of liquid To use the above equations for nonspherical particle, the diameter d must be the diameter of equivalent spherical particle. The volume of the equivalent spherical

particle: 23

234

sphernonsphere ddV −=⎟⎠⎞

⎜⎝⎛= φπ

⇒ spherenondd −= 33.024.1 φ Estiamte values for the shape factors (Sincero, 1996) are in Table 1: Table 1; Estimated values of shape factors

Materila shape factor (φ ) Angular sand 0.64 Sharp sand 0.77 Worn sand 0.86 Perfect sphere 1.0

In water treatment flow is usually laminar and transitional, but the sphericity is not always 1, i.e. particles not always spherical. The effects of irregular shape are not pronounced in low settling velocities. This suite most sedimentation processes because they are designed to remove small particles which settle slowly.

ρµ

ρ

µγ ==,

,cos,cosdensitymass

ityvisabsoluteityvisKinematic

Therefore dynamic viscosity, γρµ = Example 1 Determine the terminal settling velocity for a sand particle with an average diameter of 0.5mm and a density of 2600kg/m3 settling in water at 20oC.



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Solution: 1 Determine the terminal settling velocity using Stokes Law:

2 Check the Reynolds number and assuming Φ=0.85 for sand.

NR =93.2 3 Because the Reynolds number is greater than 1, then it should be computed

from the equation: 34.0Re3

Re24

++=dC



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Example 2 Determine the terminal settling velocity of a discrete spherical particle having a diameter of 0.6mm and specific gravity of 2.65. T=22oC.



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Example 3 Determine the terminal settling velocity of a discrete worn sand particle having a measured diameter of 0.6mm and specific gravity 2.65. T=22oC. Solution



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SETTLING IN AN IDEAL SETTLING BASIN FOR TYPE 1 PARTICLES An ideal horizontal settling zone is free from inlet and outlet disturbance, in which particles settle freely at terminal settling velocities in quiescent conditions without any disturbances and flocculation is absent (Fig. 3). The particles are distributed uniformly In the design of sedimentation basins, the usual procedure is to select a particle with a terminal velocity vs and to design the basin so that all the particles that have a terminal velocity equal to or greater than vs will be removed.

sAvQ = (1) Where: A =surface area of sedimentation basin

vs =settling velocity or surface loading, m3/m2.day (AQvs = )

H

Outlet zone Inlet

zone vp

Particle trajectory vs

vpSettling zone

h

Fig. 3 Type I settling in a horizontal basin



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Design velocity for a continuous flow sedimentation:

TH

timeention

depthvs ==det

he length of basin and the time a unit of water spend in the basin (detention time)

• effects of inlet and outlet

g

articles with velocity less than vs will not be removed during the detention time, but

Tshould be such that all particles with velocity vs will settle at the bottom of basin, but adjustments must be made for:

• turbulence • short circuitin• sludge storage

Psome particles with velocity less than vs which enter the tank at distance from the bottom not greater than H will be removed e.g. at h. Assuming that particles of various sizes are uniformly distributed on the entire depth H, at inlet, then particles with settling velocity vp less than vs will be removed in the ratio:

s

pr v

vX =

here: Xr =fraction of particle with settling velocity vp that are removed

e. particle with settling velocity vp less than vs which enter the tank at a distance

o determine the efficiency of removal for a given settling time, t it is necessary to

etermination of settling velocities

• sieve analysis and hydrometer test combined with Stokes Law:

W i.from the bottom not greater than H that are removed. Tconsider the entire range of settling velocities present in the tank. D

( )

λ181 2dSgv s

s−

=

• settling column

ettling column analysis S



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• a settling column 2 to 3m deep, and diameter at least 100x largest particle size to prevent wall effects is used (Fig. 4).

• the initial suspended solid concentration of the suspension is noted, Co in mg/l • sample is placed in a jar and mixed completely to ensure uniform distribution

of particles. • suspension is allowed to settle quiescently • samples are drawn at time intervals at a point h (one point) discrete settling

particles, the depth of sampling will not affect the resultant distribution curves

of the settling velocities: i

i th

v =

• all particle with velocity vp>vs will pass the sampling point and settle, and particles remaining must have settling velocity less than vs (vp<vs). However, there us a partial removal of some particles with velocity vp<vs and will be removed in the ratio:

s

pr v

vX =

• the procedure is repeated for time intervals t2, t3; t4; t5……..tn, and these values of settling velocities are plotted against mass fraction remaining to give the settling velocity characteristic distribution curve for the suspension (Fig. 5).

Fig. 4 Column analysis for discrete particles

Sampling point

vp<vs

h

vp>vs



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1.0

1-Xs

Fig. 5: Settling velocity distribution curve for the mass fraction remaining

The total removal is given as: ( ) ∫+−=sx

s

ps dx

vv

xR0

1

Where; Xs =particles with vp=vs 1=Xs =fraction of particles with vp≥vs removed

∫sx

s

p dxvv

0

=fraction of particles with vp<vs removed

Example Determine the total removal efficiency given the following data:

• settling analysis results Table 2 • column is 1.6m deep • surface loading is 30m/day • Co=200mg/l

Table 2

Time, min 0 40 80 120 160 200 240 280 Conc, Ci, mg/l 200 175 170 160 155 110 80 35

Solution

Xs

Xp

vsvp

Pro

porti

on o

f par

ticle

s w

ith le

ss

than

sta

ted

settl

ing

velo

city

Suspension settling velocity distribution curve for the mass fraction remaining

Settling velocities

Removed particles



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1. Compute mass fraction remaining and corresponding velocities (Table 3) Table 3 Time (min)

40 80 120 160 200 240 280

Mass fraction remaining,

o

ii C

Cx =

0.88 0.85 0.8 0.78 0.55 0.4 0.175

thvs = ,

m/min

0.04 0.02 0.013 0.01 0.008 0.0067 0.0007

vs(m/min)

4x10-2 2x10-2 1.3x10-2 1x10-2 0.8x10-2 0.67x10-2 0.07x10-2



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TYPE II SETTLING (FLOCCULENT PARTICLES)

• Flocculation particles are in dilute suspension • settling is a result of inter-particle collisions • density of particles change because flocculating particles are continually

changing in size, shape and settling velocities • due to the above factors, Stokes law cannot be applies

Analysis of settlement for Type 2 particles

• analysis performed in column at least 300mm in diameter • depth equal to the proposed sedimentation tank • samples are withdrawn at regular time intervals from multiple ports or different

sampling heights and analysed to determine the reduction in suspended solids • the % removal is plotted as a numerical value against the depth and time • the concentrations obtained are used to compute mass fraction removal

instead of he mass fraction remaining • from the plot removal at various times, the theoretical efficiency is predicted

and a theoretical surface loading is established • the design surface loading should be 1/3 of that suggested by the settling tests

(theoretical), to get similar solids removal results to those obtained from a settling column

10010

×⎟⎟⎠

⎞⎜⎜⎝

⎛−=

CC

x ijij , %

Where: xij =mass fraction is % that is removed at the ith depth at jth time interval Co =initial solid concentration Cij =concentration at ith depth and jth depth time interval

• the values Cij and time are plotted to give isoremoval lines (Fig. 6), lines with the same concentration

• the slope at any point on any given isoremoval line is the instantaneous velocity of the fraction of particles represented by the line



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• velocity becomes greater at greater depth (the slope of the isoremoval lines becomes steeper), a common characteristic of flocculating suspensions, reflecting and increase in particle size and settling velocity because of continued collision and aggregation with other particles.

The % removal is given as:

⎟⎠⎞

⎜⎝⎛ +∆

++⎟⎠⎞

⎜⎝⎛ +∆

+⎟⎠⎞

⎜⎝⎛ +∆

= +

2.....

22% 1322211 nnn RR

hhRR

hhRR

hhremoval

Where: h =column height R1 =see diagram ∆h =see diagram Example Determine the overall removal efficiency of the sedimentation tank and surface loading given the following g data:

• initial solid concentration of sample Co =200mg/l • results of column analysis of flocculating suspension (Table 4) • height of sedimentation tank =2.4m • detention time =1 hr 20min

Table 4

Table 5



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Solution

• compute 10010

×⎟⎟⎠

⎞⎜⎜⎝

⎛−=

CC

x ijij (Table 5)

• plot iso-concentration lines (isoremoval lines), Fig. 6 • plot vertical line at t =1 hr 20 mins (80 mins, i.e. retention time) • from the graph at 80 mins, about 45% of the solids reach the floor i.e. 100%

removed • determine h∆• Overall removal, R



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Fig. 6 Plot of the iso-concentration curves

%76.73210090

4.24.0

29080

4.23.0

28070

4.25.0

27060

4.23.0

26050

4.26.0

25045

4.25.0%

=⎟⎠⎞

⎜⎝⎛ +

⎜⎝⎛ +

+⎟⎠⎞

⎜⎝⎛ +

+⎟⎠⎞

⎜⎝⎛ +

+⎟⎠⎞

⎜⎝⎛ +

+⎟⎠⎞

⎜⎝⎛ +

=removalOveral

Surface loading = ( )( ) 2

3

,/

mBLAreadaymQ

×

Detention time, Q

kofvolumet tan=

Q

AreaQ

heighArea 4.280 ×=

×=



18

Q

Aream

=4.2min80

Surface loading daymmdaymmareaQ ./2.43/2.43min/03.0

804.2 23====

Adjustment for full scale ( )daymmdaymSL ./8.28/8.285.12.43 23==

The surface loading for continuous flow tank should be 1/3 of that suggested by the settling column tests to get similar solids removal results to those obtained from a settling column. The optimum removal efficiency can be obtained by trying several detention times and then computing the surface loadings. The one which gives the maximum removal efficiency will be the one corresponding the maximum optimum surface loading. References

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Chapt5_Theory of Sedimentation.pdf