Monroe L. Weber-Shirk School of Civil and

Environmental Engineering

Flocculation

Overview

Flocculation definition Types of flocculatorsMechanical DesignFractal Flocculation Theory applied to

hydraulic flocculatorsCFD analysis of hydraulic flocculatorsHydraulic Flocculator Design

0.001 0.01 0.1 11 10

4

0.001

0.01

0.1

1

10

Floc diameter (mm)

Ter

min

al v

eloc

ity (

mm

/s)

What is Flocculation?

A process that transforms a turbid suspension of tiny particles into a turbid suspension of big particles!

RequiresSticky particles (splattered with adhesive nanoglobs)Successful collisions between particles

Flocs are fractals (“the same from near as from far”)The goal of flocculation is to reduce the number of

colloids (that haven’t been flocculated)One goal is to understand why some colloids are

always left behind (turbidity after sedimentation) Another goal is to learn how to design a flocculator

Mechanical Flocculation

Shear provided by turbulence created by gentle stirring

Turbulence also keeps large flocs from settling so they can grow even larger!

Presumed advantage is that energy dissipation rate can be varied independent of flow rate

Disadvantage is the reactors have potential short circuiting (some fluid moves quickly from inlet to outlet)

Recommended G and Gq values: Turbidity or Color Removal (Mechanical flocculators)

Type

“Velocity gradient” (G) (1/s)

Energy Dissipation Rate

(e)+ Gqq*

(minute)

Low turbidity, color removal

20-70 0.4 – 4.9 50,000-250,000

11 - 210

High turbidity, solids removal

70-180 4.9 - 32 80,000-190,000

7 - 45

Sincero and Sincero, 1996 Environmental Engineering: A Design Approach

mW

kg

* Calculated based on G and Gq guidelines+ average value assuming viscosity is 1 mm2/s

G is the wrong parameter… (Cleasby, 1984)

Mechanical Design: mixing with paddles

3 3

2D P paC A a V

GV

“v

elocit

y grad

ient”

Drag co

effici

ent

Projec

ted ar

ea of

padd

les

Ratio o

f rela

tive t

o

abso

lute v

elocit

y of p

addle

s

Reactor volume1.9DC

extra

Mechanical Flocculators?

Waste (a small amount of) electricityRequire unnecessary mechanical

componentsHave a wide distribution of energy

dissipation rates (highest near the paddles) that may break flocs

Have a wide distribution of particle residence times (completely mixed flow reactors)

extra

Ten State Standards

The detention time for floc formation should be at least 30 minutes with consideration to using tapered (i.e., diminishing velocity gradient) flocculation. The flow‑through velocity should be not less than 0.5 nor greater than 1.5 feet per minute.

Agitators shall be driven by variable speed drives with the peripheral speed of paddles ranging from 0.5 to 3.0 feet per second.  External, non-submerged motors are preferred.

Flocculation and sedimentation basins shall be as close together as possible.  The velocity of flocculated water through pipes or conduits to settling basins shall be not less than 0.5 nor greater than 1.5 feet per second.  Allowances must be made to minimize turbulence at bends and changes in direction.

Baffling may be used to provide for flocculation in small plants only after consultation with the reviewing authority.  The design should be such that the velocities and flows noted above will be maintained.

http://10statesstandards.com/waterrev2012.pdf

Hydraulic flocculators allowed only by special permission!

Hydraulic Flocculators

Horizontal baffle

Vertical baffle

Pipe flow

Gravel bedVery low flows and pilot plants

A bad idea (cleaning would require a lot of work)

Flocculator Geometry

2 Channels

Upper Baffles

Port between channels

Entrance

Lower Baffles

Exit

extra

Why aren’t hydraulic flocculators used more often?

Simple construction means that there aren’t any items that private companies (venders) can sell as specialized components

Consulting firms want to be able to pass the design responsibility off to a vender

The presumed operation flexibility of mechanical flocculators (variable speed motor driving a slow mixing unit)

Poor documentation of design approach for hydraulic flocculators (special permission required to use in the US!)

Using electricity is cool, design innovation is suspect… Prior to AguaClara we didn’t have a design algorithm based on

the fundamental physics

extra

Schulz and Okun (on Hydraulic flocculators)

Recommend velocity between 0.1 and 0.3 m/sDistance between baffles at least 45 cmWater must be at least 1 m deepQ must be greater than 10,000 m3/day (115 L/s)

These aren’t universal constants!

We need to understand the real constraints so we can scale the designs correctlyWhat length scale could make a dimensionless parameter?

Surface Water Treatment for Communities in Developing Countries, (1984) by Christopher R. Schulz and Daniel A. Okun. Intermediate Technology Publications

AguaClara Flocculator Design Evolution

We began using conventional guidelines based on “velocity gradient” but we were aware that this system was fundamentally incorrect (it doesn’t capture the physics of turbulence)

We were concerned that by using a defective model we could potentially produce defective designs

If the model doesn’t capture the physics, then it won’t scale correctly (and we are designing plants for scales that hadn’t been tested)

extra

Edge of Knowledge Alert

Why would we ever think that the baffled flocculators invented over 100 years ago were the optimal design for flocculation?

We have better coagulants, shouldn’t that influence flocculator design?

We are only now beginning to understand the physics of fractal flocculation

We are improving the design of hydraulic flocculators based on our evolving understanding of the physics of flocculation

extra

The Challenge of Flocculation

We would like to knowHow do particles make contact to aggregate?What determines the time required for two flocs

to collide?How strong are flocs?

The challenge of the large changes in scaleThe Al(OH)3 nanoglobs begin at a scale of 100

nanometersFlocs end at a scale of 100 micrometers

Hydraulic Flocculation Theory

Turbulence is caused by the expansion that results when the water changes direction as it flows around each baffle

Colloids and flocs are transported to collide with each other by turbulent eddies and over short distances by viscous shear.

Flocs grow in size with each successful collisionColloids have a hard time attaching to large flocs*

(maybe because the surface shear is too high???)Flocculation model and collision potential for reactors with flows characterized by high Peclet numbers Monroe L. Weber-Shirk

, a, and Leonard W. Liona

http://dx.doi.org/10.1016/j.watres.2010.06.026*hypothesis from 2012!

How do the flocs grow?

Exponential growth

How many sequential collisions are required to make a 1 mm particle starting from 1 mm particles?

How much larger in volume is the 1 mm diameter particle?

____________1,000,000,000 !

Doubling Collisions

1 collision 1+1=22 collisions 2+2=43 collisions 4+4=84 collisions 8+8=16Number of original particles in the floc = 2

n

What is n to obtain 1,000,000,000 = 2n?

n=30This assumes volume is conserved!

log(1000000000) log(2)

log(1000000000)

log(2)

n

n

Fractal Dimensions

What happens to the density of a floc as it grows larger?

1

0FractalDd d i

diam

eter o

f prim

ary

parti

cle

diam

eter o

f flo

c

Number of primary particles in the floc

Frac

tal d

imen

sion

If volume is conserved, what is DFractal? ____3

3

0FractalDV V i

Floc density approaches the density of water because the floc includes water

Fractal Flocculation

Fractal geometry explains the changes in floc density, floc volume fraction, and, ultimately, sedimentation velocity as a function of floc size

The fractal dimension of flocs is approximately 2.3 (based on floc measurements)

3

0FractalDV V i

Floc Volume Fraction

The fraction of the reactor volume that is occupied by flocs

For fractal dimensions less than 3 the floc volume fraction increases as floc size increases

Use conservation of volume to estimate initial

Floc FlocFloc

Suspension Floc

V C

V

0 3( )Floc Clay Al OH

3 3

0 3

ClayAl OH Al OHClay

Floc Clay Al OH

C C CC

Floc

Floc

dominates

Floc Volume Fraction

0.001 0.01 0.1 1 101 10

5

1 104

0.001

0.01

0.1

1

Fractal dimension = 1Fractal dimension = 2.3Fractal dimension = 3

Floc diameter (mm)

Flo

c V

olum

e F

ract

ion

0

3

0

FractalD

Floc Floc

d

d

Primary particle diameter (clay + coagulant)

“super fluffy” flocs

Dense flocs

Buoyant Density of Flocs

0.001 0.01 0.1 1 100.1

1

10

100

1000

1 104

Fractal dimension = 1Fractal dimension = 2.3Fractal dimension = 3

Floc diameter (mm)

Flo

c de

nsity

- w

ater

den

sity

(kg

/m^3

)

2 0 2

3

0FractalD

Floc H O Floc H O

d

d

Will these flocs settle faster than the primary particles?

0.001 0.01 0.1 11 10

4

0.001

0.01

0.1

1

10

Floc diameter (mm)

Ter

min

al v

eloc

ity (

mm

/s)

Floc Terminal Velocity

1 mm

Upflow velocity for floc blankets

Capture velocity for AguaClara plate settlers

Why flocculation is necessary!

The model takes into account the changing density of flocs

0 2

2 2

120

018

FractalD

Floc H Ot

H O H O

gd dV

d

DFractal = 2.3 and d0 = 1 mm

2

2 2

2

18

Floc H Ot

H O H O

gdV

shape factor for drag on flocs

Laminar flow

Analytical Model of the Flocculation Process

The floc porosity increases with floc diameterThe velocity between flocs is a function of

whether the separation distance is less or greater than the Kolmogorov scale

The time required per collision is a function of the relative velocity between flocs, the average separation distance between flocs, and the floc size In the next slides we will explore how to characterize

collision time for flocsWe will assume that collisions occur between similar

sized flocs. That assumption will need to be evaluated, but it is probably a good assumption for the initial growth of flocs.

Are the two flocs in different eddies?

How much water is cleared (filtered) from a floc’s perspective?

Volume cleared is proportional to a collision area defined by a circle with diameter = sum of the floc diameters

Volume cleared is proportional to timeVolume cleared is proportional to the

relative velocity between flocs

2Cleared Floc rV d v t

2Flocd

trv

rv t 2Flocd

How much volume must be cleared before a collision occurs?

What is the average volume ofwater “occupied” by a floc?

Need to know floc diameter (dFloc)

And floc volume fraction (fFloc)

3

6 Floc

FlocOccupied

d

V

Floc volumeSuspension volume

3

6 Floc

OccupiedFloc

dV

13

6Separation FlocFloc

L d

Use dimensional analysis to get a relative velocity given a length scale

2

3

L

T

,rv f L

1

3rv L

Viscous range Inertial range

, ,rv f L

rv L

2L

T

,rv Lf Assume linear

The origin of the G notation

13 4

KL

Re=

1

laminar turbulent

If the flocculator has laminar flow, then this side doesn’t apply and the G, Gq approach applies.

L is separation distance

Summary for Particle Collisions

1

3r Separationv Lr Separationv L

3

6 Floc

OccupiedFloc

dV

tc is average time per collision

Floc separation distance

13

6Separation FlocFloc

L d

viscous inertia

13 4

KL

2Cleared Floc r cV d v t

113 2

23

1 6 1

6c

Floc

t

11 2 39

89

1 6 1

6Floc

c

Floc

dt

0

3

0

FractalD

Floc Floc

d

d

6Floc

cFloc r

dt

v

Is tc a function of d? Yes!

Successful Collision Models

113 2

23

1 6 1

6c

Floc

t

11 2 39

89

1 6 1

6Floc

c

Floc

dt

1

2234.8

viscousc FlocN t

13

89

25.6

inertialc FlocFloc

N td

viscous inertia

Time for one collision

Number of successful collisions

G is fractional surface coverage of colloids with adhesive nanoglobsdFloc is perhaps mean floc size of flocs that are capturing colloids

For completeness we should probably include a correction for hydrodynamic effects that make it difficult for non porous particles to approach closely. This may increase the time for the first few collisions when flocs aren’t very porous

234.8

viscousc FlocN Gt

cc

tN

t Number of collisions is equal

to time over collision time

Relationship between laminar and turbulent

13

823 9

2

23

138

92

2

9

3

2 229 99

2

3

14.8 5.6

4.8

15.6

6

306

Floc FlocFloc

Floc

FlocFloc

Clay

Clay ClayClay

Clay Clay Clay

Gd

G

d

G

d

C CG GN

V Mm

The ratio of Gq to y is the number of clay particles raised to the 2/9th power!

Set the collisions equal to get the relationship

Use the target settled water turbidity as the best guess for a relationship between conventional Gq and y. For a turbidity of 1 NTU the number of clay particles is 150,000,000 per liter.Schulz and Okun suggest 20,000 as a minimum value of Gq equivalent to = y 65 m2/3. But this isn’t calculated correctly. Should take their Gq and energy dissipation rate to make the conversion.

0.001 0.01 0.1 1 100

10

20

30

1 NTU suspension10 NTU suspension100 NTU suspension

Floc diameter (mm)

Col

lisio

n tim

e (s

)Minimum time to grow from

colloid to large floc

20 collisions to grow from 1 mm to 0.4 mm

How much time is required to produce a 0.4 mm floc?

DFractal = 2.3d0 = 1 mme = 6 mW/kg

0

21 1 22 33 23

0

1 6

6

Fractal

viscous

D

c Floc

dt

d

0

8 311 2 8 99

3 3 9

0

1 6

6

Fractal

inertial

D

c Floc

dt d

d

InertialViscous

20

log nFractal

dn D

d

13 4

KL

400 s

120 s40 s

Initial Floc Growth: Phase 1(the 50% solution)

Initial growth phase of flocculation can be modeled with the equations on the previous slide (summing the collision times until the maximum floc size is reached)

The end of the initial growth phase is reached when a significant fraction of the flocs reach a size that can be removed by sedimentation.

Observations: Collision time model

The previous analysis was tracking the time required to make the first big flocs

For a highly turbid suspension it may only take a few seconds to produce visible flocs

This is why successful flocculation can be observed very early in a flocculator

This doesn’t mean that a flocculator with a residence time of 100 s will perform well

We need to track the colloids that are left behind! Performance is based on _______________residual turbidity

Remember our Goal?

Introduce pC*Sloppy parlance… log removalWhat is the target effluent turbidity for a

water treatment plant?What is pC* for a water treatment plant

treating 300 NTU water?

Tracking the residual turbidity: Phase 2 of Flocculation

After the initial production of flocs at their maximum size the interaction of the colloids change

Most of the collisions are ineffective because collisions with flocs that are maximum size apparently are useless and most flocs are their maximum size

Flocculation Model: Integrating and tracking residual turbidity

ColloidColloid

c

dCkC

dN

Colloidc

Colloid

dCkdN

C

0

Colloid

Colloid

CColloid

cCColloid

dCk dN

C

0

ln Colloidc

Colloid

CkN

C

The change in colloid concentration with respect to the potential for a successful collision is proportional to the colloid concentration (the fraction of the colloids swept up is constant for a given number of collisions)

Integrate from initial colloid concentration to current colloid concentration

Separate variables

Classic first order reaction with number of successful collisions replacing time

1

2234.8

viscousc FlocN t

13

89

25.6

inertialc FlocFloc

N td

Laminar Flow Cases

0

2 32 13 3 2

10

4.8

Fractal

Viscous

D

Colloidc

Colloid

C dN t

d

0

3

0

FractalD

Floc Floc

d

d

Only colloids can collide effectively (big flocs are useless)

Colloids can attach to all flocs

0

2 13 21

ln4.8

ColloidColloid

Colloid Colloid

Ct

k C C

0

0

2 2 3 13 3 2

01ln

4.8

FractalD

ColloidColloid

Colloid Colloid

Cdt

k C d C

Floc volume fraction that matters for successful collisions is the colloid (small floc) fraction

0

ln Colloidc

Colloid

CkN

C

Floc volume fraction is a function of floc size (d) (which is a function of the reaction progression and shear conditions in the reactor

1

2234.8

viscousc FlocN t

1

2234.8

viscousc FlocN t

10 100 1000 100000.1

1

10

100

colloids aggregateeverything aggregates

Time (s)

Tur

bidi

ty (

NT

U)

Comparison of Flocculation Hypotheses

0

2 13 21

ln4.8

ColloidColloid

Colloid Colloid

Ct

k C C

0

0

2 2 3 13 3 2

01ln

4.8

FractalD

ColloidColloid

Colloid Colloid

Cdt

k C d C

If colloids could aggregate with all of the flocs then colloids would aggregate VERY rapidly

Ten state standards 30 minute flocculation time.The graph is laminar flow case, full scale flocculators are turbulent flow. 100 NTU clay suspension, G = 0.1

Show that pC* is proportional to time for “everything aggregates” model

Coiled Tube Flocculation Residual Turbidity Analyzer

Dr. Karen Swetland Dissertation research

Two Phase Floc Model: PACl

0.1 1 10 1000

0.5

1

1.5

2

=1200s, 5 NTU=1200s, 15 NTU=1200s, 50 NTU=1200s, 150 NTU=1200s, 500 NTU=800s, 5 NTU=800s, 15 NTU=800s, 50 NTU=800s, 150 NTU=800s, 500 NTUFitted Model

*G**̂ 2/3

pC*

Initial floc growth

1

2* 3

0

3log( ) 2

2 3Coag

CCapture

epC W Gt N

V

1CN

Two Phase Floc Model: Alum

0.1 1 10 1000

0.5

1

1.5

2

=1200s, 5 NTU=1200s, 15 NTU=1200s, 50 NTU=1200s, 150 NTU=1200s, 500 NTU=800s, 5 NTU=800s, 15 NTU=800s, 50 NTU=800s, 150 NTU=800s, 500 NTUFitted Model

*G**̂ 2/3

pC*

1

2* 3

0

3log( ) 2

2 3Coag

CCapture

epC W Gt N

V

Initial floc growth1C

N

Solving for pC*: Lambert W or ProductLog Function (Laminar flow)

Use Wolframalpha to solve for C*

W is the Lambert W function

Log is base 10

2* 3

0

3log( ) 2

2 3Coag

Capture

epC W Gt

V

* 2

302

*3

lnCoag

Capture

CGt

VC

2

30

3 2* 2 3

Coag

Capture

W GtVC e

0 0

2233

0lnColloid Colloid Coag

Colloid Colloid Capture

C CGt

C C V

Laminar Flow Floc Model

Velocity gradient

Flocculation time

Fractional surface coverage of colloid by

coagulant

Floc volume fraction

Sedimentation tank capture velocity

Sedimentation velocity of ???

1

2

1

2* 3

0

3log( ) 2

2 3Coag

CCapture

epC W Gt N

V

Lambert W Function

Collisions to make first big flocs = 0.4

Empirical

0.4Coag

mm

s

0.12Capture

mmV

s

Turbulent Flow Case

0

2 38 139 3

20

5.6

Fractal

Inertial

D

Colloidc

Colloid Floc

C dN t

d d

8 139

25.6

Inertial

Colloidc

Colloid Colloid

CN t

d

Only colloids can collide and attach effectively

Colloids can attach to all flocs

0

3

0

FractalD

Floc Floc

d

d

0

8 12 391

ln5.6

ColloidColloid Colloid

Colloid Colloid

Cdt

k C C

0

0

8 2 3 129 33

01ln

5.6

FractalD

ColloidColloid Floc

Colloid Floc Colloid

Cd dt

k C d C

0

ln Colloidc

Colloid

CkN

C

1

38

92

5.6Inertialc Floc

Floc

N td

Why are almost all of these collisions in the inertial range?

100 1000 100000.1

1

10

100

3 NTU10 NTU30 NTU100 NTU300 NTU

time (s)

Tur

bidi

ty (

NT

U)

Turbulent Flow Flocculator:“Big Flocs are useless” hypothesis

0

8 12 391

ln5.6

ColloidColloid Colloid

Colloid Colloid

Cdt

k C C

The predicted performance is quite similar given a wide range of influent turbiditiesThe predictions seem reasonable

Ten state standards 30 minute flocculation time.

After the big flocs become non reactive, then the average separation distance between the remaining flocs is larger and thus the collisions between active particles is dominated by inertia.

G = 0.1 for these plots e = 2.6 mW/kg

Turbulent Flow Flocculator: Colloids can attach to all flocs hypothesis

The predicted performance is quite different from what we observe. High turbidity would be very easy to treat if this hypothesis were true!Flocculators would be tiny!

0

0

8 2 3 129 33

01ln

5.6

FractalD

ColloidColloid Floc

Colloid Floc Colloid

Cd dt

k C d C

Ten state standards 30 minute flocculation time.

10 100 1000 100000.1

1

10

100

3 NTU10 NTU30 NTU100 NTU300 NTU

time (s)

Tur

bidi

ty (

NT

U)

Here the separation distance between reactive flocs might be less than the Kolmogorov length scale

For these plots G= 0.1e= 2.6 mW/kgpC* proportional to time

Solving for pC*: Lambert W or ProductLog (Turbulent Flow)

0 0

8 189 39

0 2

3

lnColloid Colloid Coag

Colloid Colloid CaptureColloid

C C t

C C Vd

1* 8 3

908 2

* 9 3

lnCoag

CaptureColloid

C t

VC d

18

390 2

3

9 8* 8 9

Coag

Capture

Colloid

tW

Vd

C e

18 3

* 90 2

3

9log( ) 8

8 9Coag

CaptureColloid

e tpC W

Vd

Use Wolframalpha to solve for C*

W is the Lambert W function

Log is base 10

0

* Colloid

Colloid

CC

C

0

* log Colloid

Colloid

CpC

C

extra

Lambert W Function

Performance varies very little over wide range of inputs.

Diminishing returns on investment

18 3

* 90 2

3

9log( ) 8

8 9Coag

CaptureColloid

e tpC W

Vd

18 39

0 2

3

8

9Coag

CaptureColloid

t

Vd

0.1 1 10 100 10000

1

2

3

pC*

extra

8

3

*2

9

1

3

0

9log( ) 8

8 9Coll

C

C

oapture

id

oag

V

tC W

d

ep

Proposed Turbulent Flocculation Sedimentation Model (missing phase 1)

Energy dissipation rate

Flocculation time

Fractional surface coverage of colloid by coagulant

Initial floc volume fraction

Sedimentation tank capture velocity

Sedimentation velocity of ???

-log(fraction remaining)

Characteristic colloid size

Lambert W Function

What does the plant operator control? ________

What does the engineer control? __________

G1

3t CaptureV

What changes with the raw water? __________𝜙0

Turbulent Flocculation Sedimentation Model Questions

Must represent a characteristic sedimentation velocity of the floc suspension. It could be the average terminal velocity of the full size flocs in the flocculator effluent

Is a characteristic size of a colloid. Is a function of the characteristic size of the

coagulant precipitate, geometry of the colloids, and loss of coagulant to reactor walls

Coag

2

3Colloidd

Fractal Flocculation Conclusions

It is difficult to flocculate to a low residual turbidity because the time between effective collisions increases as the number of colloids and nonsettleable flocs decreases

Colloids don’t attach to full size flocsPerhaps because the surface shear on the big

floc is too high for a colloid to attach (surface shear increases with diameter)

If we could routinely break up full size flocs perhaps we could speed the aggregation process

Perhaps related to deformability?

What is the model missing?

The model doesn’t include any particle aggregation that occurs in the sedimentation tank. Residence time in the sed tank is long and energy dissipation rate is low. Flocculation in the sed tank (especially if there is a floc blanket) could be very important.

Thus real world performance is likely better than model predictions.

Flocculator Collision Potential

The majority of the collisions in a turbulent hydraulic flocculator are in the inertial range and the collision potential is determined by flocculator residence time, q, and energy dissipation rate, e.

The collision potential is given the symbol y and has the dimension of m2/3 and this length scale will be a property of the reactor geometry

The next set of notes provides guidance for designing the geometry of a flocculator to obtain a target collision potential

1

3

0

t dt

Required Collision Potential*

0 20 40 60 80 1001

10

100

3 NTU10 NTU30 NTU100 NTU300 NTU

ψ (m^2/3)

Tur

bidi

ty (

NT

U)

*The model has not been validated for turbulent hydraulic flocculator. Thus this is only a rough estimate.

Agalteca, Alauca, Marcala 2 were designed to have y=100 m2/3

We are currently using y=75 m2/3

0 0

2 83 9

8

9

lnColloid ColloidCapture Colloid

Coag Colloid Colloid

C CV d

C C

1

3

0

t dt

Review

Why is it that doubling the residence time in a flocculator doesn’t double pC* for the flocculator?

Why does increasing floc volume fraction decrease the time between collisions?

Which terms in the model are determined by the flocculator design?

8

3

*2

9

1

3

0

9log( ) 8

8 9Coll

C

C

oapture

id

oag

V

tC W

d

ep

Surface coverage (G) of clay by coagulant precipitate

In our modeling work we didn’t cover how to calculate what fraction of the clay surfaces are coated with coagulant. The following slides are the equation heavy derivation of that coverage. We assume coagulant precipitate has some characteristic diameter when it sticks to the clay and that the clay can be represented as a cylinder. We also assume that the coagulant sticks to everything including reactor walls. The coagulant approaches the clay surface randomly and thus accumulates in a Poisson distribution. The random bombardment means that some coagulant is wasted in double coverage of previously covered clay.

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Floc Model Equations for G

Clay Total

ClayTotalA A

ClayTotal Wall

SA

SA SA

Wall Tube TubeSA D L

2

4Clay

ClayTotal Tube Tube ClayClay

SASA D L

V

1

1Clay TotalA A

Wall

ClayTotal

SA

SA

3

26

6

ClayClay Clay Clay

ClaySphereClay Clay Clay

DSA SA D

V D V

6Clay ClaySphere

Clay Clay

SA

V D

23

2Tube Tube ClaySphere

ClayTotal ClayClay

D LSA

D

Surface area to volume ratio for clay normalized by surface area to volume ratio for a sphere

Surface area of clay divided by total surface area of clay + reactor walls

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Ratio of clay surface area to total surface area (including reactor walls)

2

1

13

2

Clay TotalA ATube Tube

Tube Tube ClaySphereClay

Clay

D LD L

D

12

13

Clay TotalA AClay

Tube ClaySphere Clay

D

D

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Poisson distribution and coagulant loss to walls combined to get surface coverage

2

1Coag CoagPerClay

A AClay TotalClay

D N

SAe

3

1

6

CoagPerClay Coag Clay Clay

Clay Coag Clay ClayCoag

N C V

SA SA CD

1

21

3

Clay TotalA AClay

Tube ClaySphere Clay

D

D

6Clay ClaySphere

Clay Clay

SA

V D

2

Coag CoagPerClay Coag Clay

Clay Clay Coag ClaySphere

D N D

SA D

1A ACoag Clay Clay Total

Clay Coag ClaySphere

D

De

The surface coverage is reduced due to stacking (which is handled by the Poisson distribution) and by the loss of coagulant to the reactor walls.

loss of coagulant to the reactor walls

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Clay platelet geometry

ClayPlateletHD

ClayPlatelet

H

D

2

4Platelet ClayPlatelet ClayPlateletV D H

3

6Platelet ClayV D

3

4Platelet HD ClayPlateletV D

3 3

4 6Platelet HD ClayPlatelet ClayV D D

1

32

3ClayPlatelet ClayHD

D D

D.Clay is the diameter of a sphere with equal volume as the clay platelet

Model clay as cylinder with height and diameter

D.ClayPlatelet is the diameter of a cylinder given ratio of height to diameter and equal volume spherical diameter

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Clay platelet geometry

2 2

3 32 22 2

24 3 3ClayPlatelet Clay HD Clay

HD HD

SA D D

2

ClayPlateletClaySphere

Clay

SA

D

2

31 2

2 3ClaySphere HDHD

224ClayPlatelet ClayPlatelet ClayPlatelet ClayPlateletSA D D H

1

32

3ClayPlatelet ClayHD

D D

ClayPlateletHD

ClayPlatelet

H

D

2

321 2

2 3ClayPlatelet HD ClayHD

SA D

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Surface coverage of clay

1A ACoag Clay Clay Total

Clay Coag ClaySphere

D

De

2

31 2

2 3ClaySphere HDHD

1 12

31

Coag Clay

ClayClay CoagClaySphere

Tube Clay

D

DD

De

12

13

Clay TotalA AClay

Tube ClaySphere Clay

D

D

12

3

Clay TotalA A

ClayClaySphereClaySphere

Tube Clay

D

D

2

3

1 1

21 2

2 3 3

1Coag Clay

Clay CoagClay

HDHD Tube Clay

D

DD

D

e

2

3

1 1

1 2

2 3

1Coag Clay

Clay Coag

HDHD

D

De

If you neglect wall loses

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Solving for Coagulant Dose

2

3

1 1

21 2

2 3 3

1Coag Clay

Clay CoagClay

HDHD Tube Clay

D

DD

D

e

2

3 21 2ln 1

2 3 3Clay Coag Clay

Coag Coag HDHD Tube Clay Clay

D DC

D D

0 0

2 8

3 9

8

9

lnColloid ColloidCaptureColloid

Coag Colloid Colloid

C CVd

C C

Coagulant dose is inside coagulant volume fractionSolve for dose…

Given a target residual turbidity, solve for required G, then solve for coagulant dose

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Loss of clay to reactor walls

0.001 0.01 0.1 1 100.001

0.01

0.1

1

5 NTU50 NTU500 NTU5000 NTU

Tube diameter (m)

Fra

ctio

n of

coa

gula

nt o

n cl

ay

Loss of coagulant to reactor walls can dominate for low turbidities and small reactors. For the LFSRSF in India treating 5 NTU water and injecting coagulant into 7.5 cm pipes the clay only gets 23% of the applied coagulant!There may be a way to design a larger contact chamber for the coagulant to reduce losses to the walls.

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Effect of stacking and wall loss in a 10 cm diameter flocculator with 20 NTU

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

No Stacking and No Wall LossNo Wall LossStacking and Wall Losses

Coagulant concentration (mg/L)

Fra

ctio

nal

clay

sur

face

cov

erag

e

Stacking due to random bombardment of the clay surface with coagulant nanoglobs. Stacking becomes significant for high surface coverage. Wall losses also cause a significant reduction in clay surface coverage.

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