d Floc - Cornell

Report
Flocculation
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Overview
Flocculation definition
Types of flocculators
Mechanical Design
Fractal Flocculation Theory applied to
hydraulic flocculators
CFD analysis of hydraulic flocculators
Hydraulic Flocculator Design
Terminal velocity (mm/s)
10
1
0.1
0.01
0.001
4
110
0.001
0.01
0.1
Floc diameter (mm)
1
What is Flocculation?
 A process that transforms a turbid suspension of tiny
particles into a turbid suspension of big particles!
 Requires
 Sticky 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
Low turbidity,
color removal
High turbidity,
solids removal
“Velocity
gradient”
(G) (1/s)
20-70
70-180
Energy
q*
Dissipation Rate
 mW 
+
(e)  kg 
Gq
(minute)
0.4 – 4.9
50,000- 11 - 210
250,000
4.9 - 32
80,000- 7 - 45
190,000
* Calculated based on G and Gq guidelines
+ average value assuming viscosity is 1 mm2/s
Sincero and Sincero, 1996 Environmental Engineering: A Design Approach
G is the wrong parameter… (Cleasby, 1984)
extra
Mechanical Design: mixing with
paddles
3
G
CD  1.9
CD AP a V
2V 
3
pa
extra
Mechanical Flocculators?
Waste (a small amount of) electricity
Require unnecessary mechanical
components
Have 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)
Ten State Standards
http://10statesstandards.com/waterrev2012.pdf
 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.
Hydraulic flocculators allowed only by special permission!
Hydraulic Flocculators
Horizontal baffle
Vertical baffle
Pipe flow
Very low flows and pilot plants
Gravel bed
A bad idea (cleaning would
require a lot of work)
extra
Flocculator Geometry
Entrance
2 Channels
Port
between
channels
Exit
Lower Baffles
Upper Baffles
Why aren’t hydraulic flocculators
used more often?
extra
 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
Schulz and Okun (on
Hydraulic flocculators)
Surface Water Treatment for
Communities in Developing
Countries, (1984) by Christopher R.
Schulz and Daniel A. Okun.
Intermediate Technology Publications
Recommend velocity between 0.1 and 0.3 m/s
Distance between baffles at least 45 cm
Water must be at least 1 m deep
Q 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 correctly
What length scale could make a dimensionless parameter?
extra
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
The Challenge of Flocculation
We would like to know
How 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 scale
The Al(OH)3 nanoglobs begin at a scale of 100
nanometers
Flocs 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 collision
 Colloids 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
*hypothesis from 2012!
http://dx.doi.org/10.1016/j.watres.2010.06.026
,
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=2
2 collisions 2+2=4
3 collisions 4+4=8
4 collisions 8+8=16
n
Number of original particles in the floc = 2
n
What is n to obtain 1,000,000,000 = 2 ?
log(1000000000)  n log(2)
n=30
This assumes volume is conserved!
n
log(1000000000)
log(2)
Fractal Dimensions
What happens to the density of a floc as it grows
larger? Floc density approaches the density of
water because the floc includes water
1
d  d0i DFractal
3
V  V0i DFractal
Number of primary particles in the floc
If volume is conserved, what is DFractal? ____
3
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
V  V0i DFractal
Floc Volume Fraction
Floc
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   VFloc  CFloc
Floc
VSuspension
 Floc
dominates
Use conservation of volume to estimate
initial Floc
C AlOH   CClay CClay C Al OH 
Floc  Clay   Al (OH )
0
3
3
 Floc
0

Clay

3
 AlOH 
3
Floc Volume
Fraction
Floc
d 
 Floc0  
 d0 
“super fluffy” flocs
1
Floc Volume Fraction
Primary particle
diameter (clay +
coagulant)
3 DFractal
Fractal dimension = 1
Fractal dimension = 2.3
Fractal dimension = 3
0.1
0.01
0.001
4
110
5
110
0.001
0.01
0.1
Floc diamet er (mm)
1
10
Dense flocs
Buoyant Density of Flocs
 Floc   H O    Floc   H O 
2
0
2
 d0 
 
d 
3 DFractal
Will these flocs settle faster than the primary particles?
Floc density - water density (kg/m^3)
4
110
1000
100
10
Fractal dimension = 1
Fractal dimension = 2.3
Fractal dimension = 3
1
0.1
0.001
0.01
0.1
Floc diameter (mm)
1
10
Laminar flow
Floc Terminal Velocity
Upflow velocity for floc
blankets
Terminal velocity (mm/s)
10
1
Capture velocity for
AguaClara plate settlers
0.1
0.01
0.001
4
110
gd 2  Floc   H 2O
Vt 
18 H 2O
 H 2O
0.001
1 mm
2
0
gd
Vt 
18 H 2O
0.01
0.1
Floc diameter (mm)
 Floc   H O  d 
 
H O
 d0 
0
2
1
Why flocculation is
necessary!
DFractal 1
DFractal = 2.3 and d0 = 1 mm
2
shape factor for drag on flocs
The model takes into account the changing density of flocs
Analytical Model of the Flocculation
Process
 The floc porosity increases with floc diameter
 The velocity between flocs is a function of
whether the separation distance is less or greater
than the Kolmogorov scale Are the two flocs in different eddies?
 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 flocs
 We 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.
How much water is cleared (filtered)
from a floc’s perspective?
Volume cleared is proportional to a
2


d
Floc
collision area defined by a circle with
diameter = sum of the floc diameters
Volume cleared is proportional to time  t
 vr
Volume cleared is proportional to the
relative velocity between flocs
vr t
2
VCleared   d Floc
vr t
d
2
Floc
How much volume must be cleared
before a collision occurs?
What is the average volume of
water “occupied” by a floc?
1
Need to know floc diameter
   3
LSeparation  d Floc 
(dFloc)

6

 Floc 
And floc volume fraction (Floc)

Floc  6
3
d Floc
VOccupied

Floc volume
Suspension volume
VOccupied  6
3
d Floc
Floc
Use dimensional analysis to get a
relative velocity given a length scale
laminar
Viscous range
vr  f e , , L 
 
LK   
e 
L is separation distance
Assume linear
e
vr  L

The origin of
the G notation
L 
e  3
T 
 L2 
  
T
Re=1
vr  Lf e , 
2
3
1
4
turbulent
Inertial range
vr  f e , L 
vr  e L 
1
3
If the flocculator has
laminar flow, then this
side doesn’t apply and
the G, Gq approach
applies.
Summary for
Particle Collisions
 
LK   
e 
3
1
4
inertia
viscous
2
VCleared   d Floc
vr tc

VOccupied  6
3
d Floc
vr  LSeparation
Floc
Floc
d Floc
tc 
6Floc vr
d 
 Floc0  
 d0 
e

tc is average time
per collision
vr   e LSeparation 
1
3
3 DFractal
LSeparation
Is tc a function of d? Yes!
1
1
2
  
 d Floc 

6

 Floc 
1  6    1
tc      2
6  e   3
Floc
3
1
1
3
Floc
separation
distance
1
1 6   d  3 1
tc    

6     e   89
Floc
9
2
Floc
Successful
Collision Models
Number of collisions is equal
to time over collision time
inertia
viscous
Nc 
t
tc
1
2
1
3
Time for one tc  1  6     1
2
3
6

e
    Floc
collision
Number of
successful
collisions
N cviscous
1
1
2
e
3 
 4.8Floc   t N c
 
2
1
1 6  9  d  3 1
tc    

6     e   89
Floc
inertial
2
Floc
1
3


e
9
 5.6Floc
 2  t
 d Floc 
8
2
3
Ncviscous  4.8Floc
Gt
 is fractional surface coverage of colloids with adhesive nanoglobs
dFloc 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
Relationship between laminar
and turbulent
1
3


1
3
9
4.8Floc
Gq  5.6Floc
 2  
 d Floc 
8
2
2
Gq
3
4.8Floc
 1 
9
5.6Floc  2 
 d Floc 
8
 6
 3
   dClay
Gq



2
9
 CClay

  VClay Clay
Gq
2
9
1

3
Set the collisions equal to get
the relationship
The ratio of Gq to  is the
number of clay particles raised
to the 2/9th power!
Use the target settled water turbidity as the best guess for a
relationship between conventional Gq and . 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   65 m2/3. But this isn’t calculated
correctly. Should
take their Gq and energy dissipation rate
2
2
9
tomake
C theconversion.

Gq
306
Clay
9

 
  N Clay  2

  M Clay 

m3
 
LK   
e 
3
1
4
Minimum time to grow from
colloid to large floc
400 s
DFractal = 2.3
 dn 
d0 = 1 mm n  DFractal log 2  d 
 0
e = 6 mW/kg
120 s
40 s
20 collisions to grow
from 1 mm to 0.4 mm
Collision time (s)
30
20
10
0
0.001
0.01
0.1
1
10
Floc diameter (mm)
1 NT U suspension
10 NT U suspensio n
100 NT U suspension
tcinertial
tcviscous
1
9
8
d 
1  6  31 32
9
 
  e d Floc0
6 
 d0 
1
3
1
2
2
d 
1  6   
3
    Floc0  
6  e 
 d0 
How much time is required to produce a 0.4 mm floc?
8 DFractal 3
9
2 DFractal
2
3
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!
residual turbidity
 Performance is based on _______________
Remember our Goal?
Introduce pC*
Sloppy parlance… log removal
What 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
dCColloid
  kCColloid
dN c
dCColloid
  kdN c
CColloid

CColloid
CColloid0
dCColloid
  k  dN c
CColloid
 CColloid
ln 
 CColloid
0


  kN 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)
e 
2
Separate variables
1
2
3
N cviscous  4.8Floc
  t
 
1
N cinertial
 e 3
9
 5.6Floc
 2  t
 d Floc 
8
Integrate from initial colloid concentration to
current colloid concentration
Classic first order reaction with number
of successful collisions replacing time
 CColloid
ln 
 CColloid
0

Laminar Flow Cases

  kN c

1
2
2
e
3 
N cviscous  4.8Floc
  t
 
Only colloids can collide effectively (big flocs are useless)
1
2
e
3 
N cviscous  4.8Floc
  t
 
2
2
3
Floc volume fraction that
matters for successful collisions
is the colloid (small floc)
fraction
1
2
 CColloid0 
1  Colloid    
t


   ln 
4.8k   CColloid   e 
C
 Colloid 
Colloids can attach to all flocs
2
3
N1cViscous
 CColloid0   d 
 4.8t 
  

 Colloid   d0 
1  Colloid
t

4.8k   CColloid0
2
3
  d0 
  
 d 
2 3 DFractal 
2 3 DFractal 
3
3
Floc
e 
 
 
1
2
1
2
 CColloid0 
 

  ln 
C
e 
 Colloid 
d 
 Floc0  
 d0 
3 DFractal
Floc volume fraction is a
function of floc size (d)
(which is a function of the
reaction progression and shear
conditions in the reactor
Comparison of Flocculation
Hypotheses
If colloids could aggregate
with all of the flocs then
colloids would aggregate
VERY rapidly
100
Turbidity (NTU)
colloids aggregate
everything aggregates
10
1
2
3
0.1
10
100
1000
10000
1
2
 CColloid0 
1  Colloid    
t


   ln 
4.8k   CColloid   e 
 CColloid 
Time (s)
Ten state standards 30
minute flocculation time.
The graph is laminar flow
case, full scale flocculators
are turbulent flow.
t
1  Colloid

4.8k   CColloid0
2
3
  d0 
  
 d 
2 3 DFractal 
3
1
2
 CColloid0 
 
ln


 
e 
 CColloid 
100 NTU clay suspension,  = 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
pC * 
2

 
3log(e)  2 
W  Gt 03  N C1  Coag 
3

2
 VCapture 
 
2
pC*
1.5
q=1200s,
5 NTU
q=1200s, 15 NTU
q=1200s, 50 NTU
q=1200s, 150 NTU
q=1200s, 500 NTU
q=800s, 5 NTU
q=800s, 15 NTU
q=800s, 50 NTU
q=800s, 150 NTU
q=800s, 500 NTU
Fitted Model
1
0.5
0
0.1
Initial floc growth N C1
1
10
*G*q*^2/3
100
Two Phase Floc Model: Alum
pC * 
2

 
3log(e)  2 
W  Gt 03  N C1  Coag 
3

2
 VCapture 
 
2
pC*
1.5
q=1200s,
5 NTU
q=1200s, 15 NTU
q=1200s, 50 NTU
q=1200s, 150 NTU
q=1200s, 500 NTU
q=800s, 5 NTU
q=800s, 15 NTU
q=800s, 50 NTU
q=800s, 150 NTU
q=800s, 500 NTU
Fitted Model
1
0.5
0
0.1
Initial floc growth N C1
1
10
*G*q*^2/3
100
Solving for pC*: Lambert W or
ProductLog Function (Laminar flow)
2
3
 CColloid0 
 CColloid0

 ln 
 CColloid 
 CColloid

ln  C * 
C
2
*3
C*  e
2
3
0
 Gt
Coag
VCapture
2
Coag
3  2
 W Gt03
2 3
VCapture





2
Coag

3
  Gt0
VCapture

Use Wolframalpha to solve for C*
W is the Lambert W function
2

Coag 
3log(
e
)
2
*
3
pC 
W  Gt0


2
VCapture 
3
Log is base 10
Laminar Flow Floc Model
Coag  0.4
mm Sedimentation
s
velocity of ???
Empirical
Collisions to make
first big flocs = 0.4
2




3log(
e
)
2
Coag
*
3
pC 
W
 Gt 0  N C1  
2
 3 VCapture 
 
Floc volume
fraction
Lambert W Function
VCapture
mm Sedimentation tank
 0.12
capture velocity
s
Fractional surface
coverage of colloid by
coagulant
Flocculation time
e 
Velocity gradient  
 
1
2
Turbulent Flow Case
N cInertial
 CColloid 
ln 
 kN c
 CColloid 
0 

1
3
8  e

9
 5.6Floc
 2  t
 d Floc 
Only colloids can collide and attach effectively
8
9
N cInertial
C
  e 
 5.6t   Colloid   2


d
Colloid
Colloid

 

1
3
Why are almost all of these
collisions in the inertial range?
Colloids can attach to all flocs
8
9
1
2
 3  CColloid0 
1  Colloid   dColloid
t


 
 ln 
5.6k   CColloid   e 
 CColloid 
8
9
NcInertial
 CColloid0   d 
 5.6t 
  

 Colloid   d0 
1  Colloid
t

5.6k   CColloid0
8
9
  d0 
 

d
  Floc 
2 3 DFractal 
3
2 3 DFractal 
3
 e 
 2 
 d Floc 
1
1
3
 d  3  CColloid0 


 ln 
e
C


 Colloid 
2
Floc
Floc
d 
 Floc0  
 d0 
3 DFractal
Turbulent Flow Flocculator:
“Big Flocs are useless” hypothesis
8
9
1
2
 3  CColloid0 
1  Colloid   dColloid
t


 
 ln 
5.6k   CColloid   e 
 CColloid 
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.
Ten state standards 30 minute
flocculation time.
Turbidity (NTU)
100
3 NTU
10 NTU
30 NTU
100 NTU
300 NTU
10
The predicted performance
is quite similar given a wide
range of influent turbidities
The predictions seem
reasonable
1
0.1
100
1000
time (s)
10000
 = 0.1 for these plots
e = 2.6 mW/kg
Turbulent Flow Flocculator:
Colloids can attach to all flocs hypothesis
1  Colloid
t

5.6k   CColloid0
8
9
  d0 
 

d
Floc



2 3 DFractal 
3
1
2
 d Floc
 3  CColloid0 


 ln 
e


 CColloid 
Turbidity (NTU)
100
3 NTU
10 NTU
30 NTU
100 NTU
300 NTU
10
100
1000
time (s)
Ten state standards 30
minute flocculation time.
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!
1
0.1
10
Here the separation distance between
reactive flocs might be less than the
Kolmogorov length scale
10000
For these plots
= 0.1
e= 2.6 mW/kg
pC* proportional to time
extra
Solving for pC*: Lambert W or
ProductLog (Turbulent Flow)
8
9
 CColloid0 
 CColloid0

 ln 
 CColloid 
 CColloid
 ln  C
*
8
* 9
C 
C*  e
  
te
8
9
0
d
1
3
2
3
Colloid
1
3
Coag

te
   2
VCapture

3
dColloid
8
9
0
Coag
VCapture
1

8

9
8
te 3 Coag
 W  09 2
8 9
VCapture
3

d
Colloid






C* 
CColloid
CColloid0
C
pC *   log  Colloid
 CColloid
0




Use Wolframalpha to solve for C*
W is the Lambert W function
1


8
3

9 log(e)  8 9 te
Coag 
pC * 
W
0 2
9
8
VCapture 

3
dColloid


Log is base 10
extra


8

9 log(e)  8 9 te
Coag 
pC * 
W 0 2
9
8
VCapture 

3
dColloid


Lambert W
Function
1
3
3
pC*
2
1
0
0.1
1
10
100
1


8
3

Coag 
 8   9 te
9 0 2
VCapture 

3
dColloid


1000
Performance
varies very
little over
wide range of
inputs.
Diminishing
returns on
investment
Proposed Turbulent Flocculation
Sedimentation Model (missing phase 1)
Flocculation time
Energy dissipation rate
Sedimentation
velocity of ???

Lambert W
Function
1

8
9log(e)  8 9 te 3 Coag 
*
pC 
W 0 2
9
8
VCapture 

3
dColloid


-log(fraction
remaining)
Sedimentation tank
capture velocity
Characteristic colloid size
Initial floc volume fraction
Fractional surface coverage
of colloid by coagulant

What does the plant operator control? ________
VCapture
te
What does the engineer control? __________
0
What changes with the raw water? __________
1
3
Turbulent Flocculation
Sedimentation Model Questions
Coag  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
2
3
dColloid
 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
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 flocs
Perhaps 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.
q
Flocculator Collision
Potential
1
3
   te dt
0
 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  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
Required Collision Potential*

VCapture d
Coag
2
3
Colloid
8
9

8
9
 CColloid0 
 CColloid0 
ln




C
C
 Colloid 
 Colloid 
q
1
3
   te dt
0
Turbidity (NTU)
100
Agalteca, Alauca,
Marcala 2 were
designed to have
=100 m2/3
3 NTU
10 NTU
30 NTU
100 NTU
300 NTU
10
We are currently
using =75 m2/3
1
0
20
40
60
80
100
ψ (m^2/3)
*The model has not been validated for turbulent hydraulic
flocculator. Thus this is only a rough estimate.
Review
1


8
3
Coag 
9 log(e)  8 9 te
*
pC 
W 0 2
9
8
VCapture 

3
dColloid


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?
extra
Surface coverage () 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.
extra
Floc Model Equations for 
 AClay ATotal 
SAClayTotal
SAClayTotal  SAWall
SAClayTotal 

4
2
Tube Tube
D
L

 ClaySphere
 AClay ATotal
SAClay
VClay
1

SAWall
1
SAClayTotal
DClay 3
2
3 DTube
LTube  ClaySphere
2 DClay
Surface area of clay divided by
total surface area of clay +
reactor walls
Clay
SAClay 6
SAClay DClay


2
VClay  DClay
VClay 6
SAClayTotal 
SAWall   DTube LTube
Clay
SAClay
VClay

6 ClaySphere
DClay
Surface area to volume ratio for clay
normalized by surface area to volume ratio
for a sphere
extra
Ratio of clay surface area to total
surface area (including reactor walls)
 AClay ATotal
1

 DTube LTube
1
2
3 DTube
LTube  ClaySphere
2 DClay
 AClay ATotal 
1
1
2 DClay
3DTube  ClaySphereClay
Clay
extra
Poisson distribution and coagulant loss to
walls combined to get surface coverage
N CoagPerClay
SAClay

CCoag
1
VClay Clay
Coag  D 3 SAClay CClay
Coag
 AClay ATotal 
6
2
Coag
D
N CoagPerClay
SAClay
Coag
DClay

Clay  DCoag  ClaySphere
1
1
2 DClay
3DTube  ClaySphereClay
SAClay
VClay

6 ClaySphere
DClay
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.
  1 e

2
DCoag
NCoagPerClay
SAClay
 AClay ATotal
loss of coagulant to
the reactor walls
  1 e

Coag DClay  AClay ATotal
Clay DCoag  ClaySphere
extra
Clay platelet geometry
 HD 
VPlatelet 
VPlatelet 
H ClayPlatelet
DClayPlatelet

4

6
2
ClayPlatelet
D

4
VPlatelet   HD
H ClayPlatelet

4
3
DClayPlatelet
D.Clay is the diameter of a sphere with
equal volume as the clay platelet
3
DClay
VPlatelet   HD
DClayPlatelet
Model clay as cylinder with height and
diameter
3
ClayPlatelet
D
1
3


6
3
DClay
 2 
D.ClayPlatelet is the diameter of a

 DClay cylinder given ratio of height to diameter
 3 HD 
and equal volume spherical diameter
extra
Clay platelet geometry
1
3
SAClayPlatelet  2
SAClayPlatelet

4
DClayPlatelet
2
DClayPlatelet
  DClayPlatelet H ClayPlatelet

2
3
2
3
 2 

 DClay
 3 HD 
 HD 
 2  2
2  2
2 
 DClay   HD 
 DClay
4  3 HD 
 3 HD 
2
3
SAClayPlatelet
1
 2 
2
    HD  

D
Clay

2
3



HD 
 ClaySphere 
SAClayPlatelet
D
2
Clay
 ClaySphere
1
 2 
    HD  

2
3



HD 
2
3
H ClayPlatelet
DClayPlatelet
extra
Surface coverage of clay
  1 e
  1 e

Coag DClay  AClay ATotal
Clay DCoag ClaySphere
Coag DClay 1

Clay DCoag 
 AClay ATotal 
1
 ClaySphere 
  1 e
Coag DClay 1
Clay DCoag 
  1 e

3 DTubeClay
1
2
3
1
 2
 
  HD 
2
 3 HD  3 DTubeClay
Coag DClay 1
Clay DCoag 
3DTube  ClaySphereClay
2 DClay
 ClaySphere

1
1
2 DClay
2 DClay
1
 2 
    HD  

2
  3 HD 
 AClay ATotal
 ClaySphere

1
 ClaySphere 
2 DClay
3DTubeClay
1
2
1
 2  3


HD 

2
 3 HD 
2
3
If you neglect wall loses
extra
Solving for Coagulant Dose
  1 e
CCoag

Coag DClay 1
Clay DCoag 
1
2
3
2 DClay
1
 2




HD 

2
 3 HD  3 DTubeClay
Coagulant dose is inside
coagulant volume fraction
Solve for dose…
2


3
2 DClay   DCoagClay
1
 2 

  Coag    HD  

ln 1   

 2
  3 HD  3DTubeClay  DClay



d
2
3
Colloid
8
9

8
9
VCapture  CColloid0 
 CColloid0 

 ln 

Coag  CColloid   CColloid 
Given a target residual turbidity, solve for required , then solve for coagulant dose
extra
Fraction of coagulant on clay
Loss of clay to reactor walls
1
0.1
5 NTU
50 NTU
500 NTU
5000 NTU
0.01
0.001
0.001
0.01
0.1
1
10
Tube diameter (m)
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.
extra
Effect of stacking and wall loss in a 10
cm diameter flocculator with 20 NTU
Fractional clay surface coverage
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
Coagulant concentration (mg/L)
No Stacking and No Wall Loss
No Wall Loss
Stacking and Wall Losses
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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