7 Sedimentation F11

8/12/2019 7 Sedimentation F11

1/22

7-Sedimentation_F11

1

SEDIMENTATION (3rd

DC 211; 4thDC 266-282)

a. Definition- Sedimentation is the removal of solid particles from suspension by gravity.

b. Objectives:

1) Clarification - separation of particles from water (clarifier)2) Settling - collection of sediment (settling tank)3) Thickening - concentrating the removed particles (thickener)

Primary sedimentation tank, Pocatello wastewaer treatment plant, Pocatello, Idaho

Secondary sedimentation tank, Pocatello wastewater treatment plant, Pocatello, Idaho


2/22

2

c. Sedimentation Types depend on:a) Concentration of the suspension (i.e., dilute or concentrated)b) Characteristics of the particles (i.e., discrete or flocculent).

Type I Sedimentation

a) deals with dilute suspensions of discrete(nonflocculating) particles- gravity separation of nonflocculating discrete particles in a dilute suspension.

b) Each particle falls independently of all other particles near by.- settling is unhindered

c) Settle at constant velocitySettling velocity = f (fluid properties, particle characteristics)

d) Example: grit particles (e.g., sand) in wastewater (WWTP)-Concentrations (up to 10,000 mg/L)

e) Applications:CGrit chamber (grit removal tank)

Design criteria: settling velocity vs= 0.075 fps (for 0.2-mm particle, SG = 2.65)

C Pre-sedimentation tank in WTP

Type II Sedimentation

a) Dilute suspensions of flocculentparticles (~500 mg/L)

b) Deals with hindered settling of flocculent materials

c) Settling velocity is changing because particle size is constantly changing

d) Examples:Silt, ground toilet paper, Al(OH)3, CaCO3, Mg(OH)2

e) Applications:

C Flocculator in WTP (Alum or iron coagulation)C Primary settling tank in WWTP


3/22

7-Sedimentation_F11

3

Type III Sedimentation (Zone Settling)

a) Concentrated suspension of flocculent particles(1000 mg/L)

-flocculent materials in higher concentrations than type II suspension.-settle as a mass, there is a distinct clear zone and sludge zone (sludge blanket).

b) Hindered settling- particles stay in the same position relative to other particles.

c) Examples:C Biological sludge (e.g., activated sludge)C Chemical sludge (e.g., lime sludge in water softening)

d) Applications:C

Settling tank for biological sludge (activated sludge in a secondary clarifier) inWWTP.C Settling tank for chemical flocculent (e.g., lime sludge, coagulant sludge) in

WTP, lime-softening sedimentation.

(Type IV) Compression

a) Very concentrated suspension of flocculent particles.b) The particles are in physical contact with one another and supported partially by

the compacting mass.

- compression results when the concentration increases to the point where theparticles are in physical contact with one another and supported partially by thecompacting mass.

- Settling is extremely slow- Floc forms a structure and water is squeezed out through pores

c) Example: thickened activated sludged) Applications: Thickener in WWTP


4/22

4

Summary

Type II

Type I

Type III

Zone

Settling

Compression

Time or Distance

Column

Levelofparticleorblanket

Depth

Grit chamber Primary Sed. Aeration Tank Second. Sed.

Tank Tank

(Type I) (Type II) (Zone settling)

Thickner

(Compression)


5/22

7-Sedimentation_F11

5

Type I Sedimentation

- Gravity separation of discrete (non-flocculating) particles in dilute suspensions.- Settling velocity = f (fluid properties, particle characteristics )

Isaac Newton (1687)- showed that a particle falling in a quiescent fluid accelerates until the frictionalresistance (or drag) on the particle is equal to the gravitational force of the particle.

Fig. 3-35 (3rdDC 221); Fig. 4-38 (4

thDC 276)

Free-body diagram for Type I Particle- Forces acting on a free-falling particle in a fluid

FG= gravitational force; the force due to gravityFB= buoyancy force; the buoyant force due to fluidFD= drag force; the frictional force

- FDis a function of: roughness, size, shape, velocity of the particle, density,velocity of the fluid.

The downward acceleration of the particle

From the Newtons second law: F = m a

m (dvs/dt) = FG- FB - FD (1)

wherem = mass of the particlevs= linear settling velocity of the particle

t = time

Note: Newtons second law - the acceleration of an object is directly proportional to the resultant force acting on it and inverselyproportional to its mass.

FG= sg Vp (2)

FB= g Vp (3)

FD= CDApvs2/ 2 (4)

where

FD FD

FB

Fg


6/22

6

s= density of particle, kg/m3

= density of fluid, kg/m3

g = acceleration of gravity, m/s2

Vp= volume of particle, m3

CD= drag coefficient (Newtons dimensionless drag coefficient)Ap = cross sectional area of particle (projected particle area in the direction of

flow), m

2

vs= velocity of particle, m/s

Substituting (2), (3), and (4) into (1) yields

m (dvs/dt) = sg Vp - g Vp - CDApvs2/ 2 (5)

This is the eqn for the dynamic behavior of the particle.

After an initial transient period, the velocity becomes constant and acceleration becomes zero:

Vs> vt (vt= terminal settling velocity, constant)

dvs/dt = 0

g ( s - ) Vp = CDAp vs2/ 2

Solve for vs

vg V

C As

s p

D p

=

(( )

) /

21 2

(6)

For spherical particle,

V

A

d

d d

p

p

p

pp= =

4

3 2

2

2

3

3

2

( )

( )

(7)

where dp= particle diameter

Substituting (7) into (6) yields

v g

Cd

g d

Cs

s

Dp

s p

D

=

=

( ( )

) (( )

)/ /2 2

3

4

3

1 2 1 2

In the flow regime for which Re < 1, CD= 24 / Re (9)

where laminar flow prevails,

Re = the Reynolds number

Re = dp vs / (10)

where = dynamic viscosity, pa.s (kg/m .s)

Substituting (10) into (9) yields

CD= 24 / dpvs (11)

Substituting (11) into (8) gives Stokess law

( )v

g ds p=

2

18


7/22

7-Sedimentation_F11

7

Stokes law

- If the particle is spherical, the settling velocity can be described by Stokess law under laminar flow condition.

( ) 2sg - dv =

18 (3

rdDC 222; 4

thDC 277)

wherev = settling velocity, fps (m/s)g = acceleration of gravity, 32.174 ft/s

2(9.81 m/s

2)

s = density of particles, lb.s2/ ft

4(kg/m

3)

= absolute (dynamic) viscosity of water, lb.sec/ ft2(kg/m.s)

d = diameter of the spherical particles, ft (m)= density of water,lb.s

2/ ft

4(kg/m

3)

since =

where = kinematic viscosity, ft2

/sec (m2

/s)

2( )

18

sg d

=

2( 1)

18

sg d

=

2( 1)18

gSG d

=

where SG = s/= specific gravity

Unit: =

kg

m s

m

s

kg

m

=

2

3

= pas

[ ]

kg

m s s

kg

m s2

=

Other unit:

Pa N

m

kg

m s= =

2 2


8/22

8

Example: Stokes Law

Given:Water temp, T = 9.5C;Particle diameter, dp= 6.56 x 10

-4ft (0.2 mm or 0.2 x 10

-3m)

S.G. of the particle = 2.65;

Acceleration of gravity, g = 32.2 ft/s

2

(9.81 m/s

2

)

Determine the settling velocity of the particle.

(Solution)

Using Storkes law,Kinematic viscosity, v= 1.41 x 10

-5ft

2/s (1.36 x 10

-6m

2/s) at T = 9.5C,

Note: Table A-1 (4th

DC 977),

Kinematic viscosity, v= 1.307 (m

2

/s) at 10C

m2/s x 10

-6= (m

2/s)

SI unit:

( )( )

( )( )

2

22

3

6 2

11

18

9.8 / 12.65 1 0.2 10

18 1.36 10 /

0.0264 /

o p

gv SG d

m sx m

x m s

m s

=

=

=

English Unit

( )( )

( )( )

2

22

4

5 2

11

18

32.2 / 1

2.65 1 6.56 1018 1.41 10 /

0.09 /

s p

gv SG d

ft s

x ftx ft s

ft s

=

=

=


9/22

7-Sedimentation_F11

9

Example Type I Sedimentation: Ten State Standards (GLUMRB)

Particle diameter, dp= 0.2 mm (6.56 x 10-4

ft), S.G. = 2.65Flow velocity, V = 1.0 ft/s, Q = 4.0 MGD = 6.1 ft

3/s

Temp = 9.5 C, kinematic viscosity, = 1.41 x 10-5

ft2/s

Acceleration of gravity, g = 32.2 ft/s2= 9.81 m/s

2

H = depth; (let H = 2W)

Make 2 tanks (H = depth, W = width, L = length)

(Solution)

design Q = (6.1 ft3/s) = 3.1 ft

3/s

Using Storks law

( )( )

( )( )

v g

SG d

xx

ft s

o p=

=

=

18

11

32 2

18

1

141 102 6 5 1 6 5 6 10

0 09

2

5

4 2

.

.. .

. /

This is voof the particles which will be removed 100%.

Ax Q

V

ft s

ft sft= = =

31

1 031

32. /

. /.

Ax = WH = W (2W) = 2W2= 3.1 ft

2

W = 1.25 ft

H = Ax/W = 3.1 ft2/1.25 ft = 2.5 ft

t H

v

ft

ft ssd

s

= = =2 5

0 0927 8

.

. /.

td= detention time ;

L = V td= (1.0 ft/s)(27.8 s) = 27.8 ft

Thus,H = 2.5 ft, W = 1.25 ft, L = 27.8 ft

W = 1.25 ft

Q, V

H = 2.5 ft

Ax

L = 27.8 ft


10/22

10

Ideal Sedimentation Basin (4thDC 267)

A = W L Vol = W L H

W A

Q b

a c

d

L

a = Inlet zone; b = Settling zone;c = Outlet zone; d =Sludge zone

a. Inlet Zone- Incoming flow is uniformly distributed over the cross section of the tank.

b. Settling Zone- The concentration of each size particle is uniform throughout the cross section

(2D).

c. Outlet Zone- Clarified effluent is collected and discharged through an outlet weir.

d. Sludge Zone- Provides for the collection of particles removed from suspension.


11/22

7-Sedimentation_F11

11

An ideal rectanglar sedimentation tank

Outlet weir

V Settling zone

vo

Inflow, Q

H

V

v

h

L

Inletzone

Sludge zone

Outletzone

V = flow velocityvo= terminal settling velocity of a particle that is just removed when it enters at the

water surface (H).

Note:

1) All particles with terminal settling velocity vo are removed.2) Only part of particles with settling velocity < vo are removed.

If the area of the triangle having H and L represents 100% removal of particles, theremoval ratio of particles having a settling velocity v is h/H.

v

v

hL

HL

h

Ho =

/

/

2

2

v H

t

distance it falls

detention time of tanko

d

= =

since td= Vol / Q

o

H H Q Qv

Vol WLH WL A

Q Q

= = = =

3 3/ /

,2 2

2

Q ft s ft gpd m d mvo A s dft ft m

= = =


12/22

12

Q/A = over flow rate (gpd/ft

2) or (m

3/m

2d)

= surface hydraulic loading rate= clarification rate

Percentage of Particle Removal, P

P v

v

s

o

= ( )100 (3rd

DC 219; 4thDC 274)

wherevs= settling velocity of particlevo= terminal settling velocity (overflow rate).

Example 3-22 (3rd

DC 219); Example4-22 (4th

DC 274)The town of San Jose has an existing horizontal-flow sedimentation tank with an

overflow rate of 17 m3/d. m

2. The town wishes to remove particles that have settling

velocities of 0.1 mm/s, 0.2 mm/s, and 1 mm/s. What percentage of removal should beexpected for each particle in an ideal sedimentation tank?

(Solution)

1) Convert the overflow rate to compatible unit, e.g., mm/s

3

2

100017 ( )

17 / 17

86400

0.2o

mmm

m d m mOverflow Ratesm d

dd

mmv

s

= = =

= =

2) Compare the particle settling velocity with the overflow rate.

a. For the particles with vs= 0.1 mm/s

Since vs< vo, some fraction of the particles will be removed.

P = vs/vo(100) = (0.1 / 0.2) 100 = 50%


13/22

7-Sedimentation_F11

13

b. For the particles with vs= 0.2 mm/s

P = vs/vo(100) = (0.2 / 0.2) 100 = 100%

Since vs= vo, 100% of the particles will be removed (ideally).

c. For the particles with vs= 1 mm/s

Since vs> vo, 100% of the particles will be removed.

Note:Common overflow rates for sedimentation in water and wastewater treatment:

300 - 1000 gpd/ft

2

12 - 41 m3/m

2.day

Side-water depth 7 ft (2.1 m)

Typical Design Values for Primary Clarifier (VH 345)

Overflow Rate = 600 - 800 gpd/ft2

(24 32 m3/m

2d)

700 gpd / ft2= 0.33 mm / sec

700

7 48 1440

0065

2

3gal

day ft

ft

gal

day ft

. min

.

min=

40.065 1 min 3.30 10 / 0.33 / min 3.28 60

ft mx m s mm s

ft s

= =

1 gpd = 1.547 x 10-6

ft3/sec

1 gpd 1.547 x 106

ft3 1 m 10

3mm

--------- = ----------------------- ---------- ------------ft

2 sec ft

2 3.28 ft m

= 4.71646 x 104

mm / sec

Side-water depth 7 ft (2.1 m)

Weir loading:

For Q 1 MGD, 10,000 gpd/ft (125 m3/m d)

For Q > 1 MGD, 20,000 gpd/ft (250 m3/m d)


14/22

14

cf, Final clarifiers (for activated sludge)

Overflow rate based on the daily design flow:For Q 1 MGD, 600 gpd/ ft

2(24 m

3/m

2d)

For Q > 1 MGD, 800 gpd/ ft2(33 m

3/m

2d)

Overflow rate during the peak flow:For Q 1 MGD, 1 MGD, 1 MGD, 20,000 gpd/ft (250 m3/m d)

Final Settling TankMax. Overflow Rate 800 gpd/ ft

2 (33 m

3/m

2d)

Based on design daily flow 8.0 ft (2.4 m)Maximum Weir Loading

For Q 1 MGD, 10,000 gpd/ft (125 m3/m d)

For Q > 1 MGD, 20,000 gpd/ft (250 m3/m d)

Concept of Tube settlers/ Plate settlers

o

d

H H H Q Qv

Vol WLH t WL A

Q Q

= = = = =

P vv

s

o

= ( )100

ss

o

v AP v

v Q= =


15/22

7-Sedimentation_F11

15

Q/4

Q/4

Q/4

Q/4

Q

/ 2 22

2

vQ Q Asv P vo sA A v Q

o

= = = =

1


o

= = = =

Q

Q/2

Q/2

/ 4 44

4


o

= = = =


16/22

16


17/22

7-Sedimentation_F11

17


18/22

18

Type II Sedimentation(VH, p.345, Section 10.13)

- Dilute suspension of flocculating solids

Applications:a) Chemical softening sludge, CaCO3, Mg(OH)3b) Primary settling sludge in WWTP

Characteristics:a) % Removal = f (Q/A, H)

- Removal of type II SS depends not only on the clarification rate (overflow rate; Q/A, vo) but also the depth of the tank.

b) No mathematical relationship exists to describe the influence of flocculation.

c) Settling column analysis is required to evaluate the type II sedimentation.

3. The Standard Method for a settling column test.

a. Experimental Procedures

1) Place a suspension in a column.- Settling column has sampling ports at various depth.

2) Allow sedimentation under quiescent conditions.3) Samples are withdrawn at selected time intervals from different depth.4) Plot the data as shown in figure below.

- Each % removal is recorded at the proper coordinates of depth and time.- Lines representing percentage of removal are drawn through the data.

Settling Column


19/22

7-Sedimentation_F11

19

Overall SS Removal, R (%)

vo v1 v2RTa(%) = ---- (% Ra) + ---- (% increment) + ---- (% increment) + ...

vo vo vo

H h1 h2vo= ------ v1= ------ v2= ------ . . .

to to to

v1 h1/to h1 v2 h2/to h2--- = -------- = ---- ---- = --------- = ------ . . .vo H/to H vo H/to H

h1 h2RTa(%) = (% Ra) + ---- (% increment) + ---- (% increment) + . . .

H H

h1 h2RTa(%) = (% Ra) + ---- (% Rb - %Ra) + ---- (% Rc - %Rb) + . .

H H


20/22

20

Example A batch-settling test using a 7.0-ft column and coagulated water from theirexisting plant yielded the following data. Estimate the overall removal for a settling tankof 7.0-ft deep with an overflow rate of 3123 gpd/ft

2.

(Solution)

Note that 3123 gpd/ft2gives 0.2899 ft/min or 7.0 ft/24 min.

3123 gal ft3 1 d 0 2899 ft 7 ft

------------ ------------ ------------- = ------------- = ----------ft

2d 7.48 gal 1440 min min 24 min

Percent removal as a function of time and depth

Time (min)5 8 12 18 22 28 32 38 42 52

Depth (ft) 2 14 17 40 50 56 63 66 74 75 79

4 13 15 16 49 45 59 62 70 71 73

6 12 14 17 43 44 47 50 65 70 72


21/22

7-Sedimentation_F11

21

Procedure:

1) Develop graph with isoconcentration lines for given column depth.

2) Determine the % of the particles which has settling velocities greater than the givenoverflow rate (noting that Q/A = vs).

- all particle which has the settling velocity greater than or equal to vswill beremoved, say Ra %.

45 % of the particles (Ra) ==> vs> 7 ft/24 min

2) Obtain an average settling velocity for an additional % SS removal interval.

60 - 45 = 15 % of the particles ==> 3.4 ft/24 min

3) Obtain an average settling velocity for a second % SS removal interval.

75 - 60 = 15 % of the particles ==> 1.3 ft/24 min

4) The next increment has a negligible velocity.

5) Determine Overall Removal, RTa(%)

h1 h2

RTa(%) = (%Ra) + ---- (% Rb - %Ra) + ---- (%Rc - %Rb) + . .H H

3.4 ft 1.3 ft= 45% + -------- (60 - 45%) + ------- (75 - 60%) + . . .

7 ft 7 ft

= 45% + 7.3% + 2.8% + . . .

= 55%

The overall removal = 55%


22/22

Comparison of Type I and Type II equations:

Type I

( )% (100) 100%so

v hR P

v H

= = =

Type II

1 2 3% (% ) (% % ) (% % ) (% % ) ...T b a c b d ch h h

R Ra R R R R R RH H H

= + + + +

----------Note:

Type III (zone settling) and Type IV (compression) will be discussed in detail in ENVE 616 Biological

Treatment of Wastewater.

Documents

7 Sedimentation F11