1. Flocculation

3. Overview Flocculation definition Types of flocculators
Mechanical Design
Fractal Flocculation Theory applied to hydraulic flocculators
CFD analysis of hydraulic flocculators
Hydraulic Flocculator Design

4. 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

5. 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)

6. Recommended G and Gq values: Turbidity or Color Removal (Mechanical flocculators)
Type
“Velocity gradient” (G) (1/s)
Energy Dissipation Rate (e)+ Gq q*
(minute)
Low turbidity, color removal
20-70
0.4 – 4.9
50, ,000
High turbidity, solids removal
70-180
80, ,000
7 - 45
Cleasby, J. L., "Is Velocity Gradient a Valid Turbulent Flocculation Parameter?"
ASCE, Journal of Environmental Engineering, Vol. 110, No. 5, Oct.,
1984
* 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)

7. Mechanical Design: mixing with paddles
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Mechanical Design: mixing with paddles
Ratio of relative to absolute velocity of paddles
Projected area of paddles
Drag coefficient
“velocity gradient”
Reactor volume

8. Mechanical Flocculators?
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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)

9. 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.
Hydraulic flocculators allowed only by special permission!

10. Hydraulic Flocculators
Horizontal baffle
Vertical baffle
Pipe flow
Gravel bed
Very low flows and pilot plants
A bad idea (cleaning would require a lot of work)

11. Flocculator Geometry Entrance 2 Channels Port between channels Exit
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Flocculator Geometry
Entrance
2 Channels
Port between channels
planta de 50 Lps con 2 cámaras de sedimentación
Exit
Lower Baffles
Upper Baffles

12. Why aren’t hydraulic flocculators used more often?
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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

13. 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
Schulz and Okun (on Hydraulic flocculators)
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!
The velocity might be related to creating collisions between particles or preventing breakup of flocs from too much shear. In any case, velocity is not the right parameter because it isn’t dimensionless and thus it might not scale properly across a wide range of plant flow rates.
Surface Water Treatment for Communities in Developing Countries, (1984) by Christopher R. Schulz and Daniel A. Okun. Intermediate Technology Publications
We need to understand the real constraints so we can scale the designs correctly
What length scale could make a dimensionless parameter?

14. AguaClara Flocculator Design Evolution
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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)

15. Edge of Knowledge Alert
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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

16. 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 16

17. 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!

18. How do the flocs grow?

19. 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 !

20. Doubling Collisions 1 collision 1+1=2 2 collisions 2+2=4
Number of original particles in the floc = 2n
What is n to obtain 1,000,000,000 = 2n?
n=30
This assumes volume is conserved!

21. 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
diameter of primary particle
diameter of floc
Fractal dimension
Number of primary particles in the floc
If volume is conserved, what is DFractal? ____ 3

22. 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)
Adachi Wat. Res. Vol. 31, No. 3, pp. 449~54, 1997

23. 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
We are ignoring the initial diffusion controlled aggregation of aluminum hydroxide molecules with each other and with clay. We are assuming that this aggregation process produces particles that are relatively dense (high fractal dimension) and that this process occurs quickly relative to the subsequent shear and turbulent eddy mediated collisions (in less than a minute).
dominates

24. Floc Volume Fraction Primary particle diameter (clay + coagulant)
“super fluffy” flocs
Fractal dimension of 1 would suggest that the flocs would occupy the entire flocculator by the time they reached 0.2 mm (for this example)
Fractal dimension of 3 gives volume conservation.
Dense flocs

25. Buoyant Density of Flocs
Will these flocs settle faster than the primary particles?
Here we assume that the primary particle (clay plus coagulant) is 1 micrometer in diameter.
The assumption is that we are starting with clay at 2650 kg/m^3.

26. Floc Terminal Velocity
Laminar flow
Floc Terminal Velocity
Upflow velocity for floc blankets
Capture velocity for AguaClara plate settlers
Why flocculation is necessary!
The shape factor adjusts the coefficient of drag for non spherical geometry and according to Tambo has a value of 45/24 for flocs (Tambo et al., 1979). This value is probably not yet well characterized. In any case it is expected to be order 1 because it is a correction to the drag coefficient.
All fractal dimensions greater than one are expected to increase the terminal velocity as the floc diameter increases.
Both floc blanket upflow velocity and plate settler capture velocity are AguaClara current design values, but are not optimized.
1 mm
DFractal = 2.3 and d0 = 1 mm
shape factor for drag on flocs
The model takes into account the changing density of flocs

27. 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
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.
Are the two flocs in different eddies?

28. 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 time
Volume cleared is proportional to the relative velocity between flocs

29. How much volume must be cleared before a collision occurs?
What is the average volume of water “occupied” by a floc?
Need to know floc diameter (dFloc)
And floc volume fraction (fFloc)
Floc volume Suspension volume

30. Use dimensional analysis to get a relative velocity given a length scale
laminar
turbulent
Viscous range
Inertial range
L is separation distance
Assume linear
Re=1
The length scale is the average separation distance of the flocs
The origin of the G notation
If the flocculator has laminar flow, then this side doesn’t apply and the G, Gq approach applies.

31. Summary for Particle Collisions
viscous
inertia
tc is average time per collision
Floc separation distance
Yes! Because the phi.floc is a function of d!
Find the time to clear the volume occupied by one floc.
Then we need to find the average relative velocity of the two flocs based on their average separation distance.
Is tc a function of d?
Yes!

32. Successful Collision Models
Number of collisions is equal to time over collision time
Successful Collision Models
viscous
inertia
Time for one collision
Number of successful collisions
G 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

33. Relationship between laminar and turbulent
Set the collisions equal to get the relationship
The ratio of Gq to y 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 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.
This means that we can find the relationship between Gtheta calculated the traditional way from the head loss and compare that with Gtheta calculated correctly.
The

34. Minimum time to grow from colloid to large floc
Viscous
DFractal = 2.3
d0 = 1 mm
e = 6 mW/kg
Inertial
400 s
120 s
20 collisions to grow from 1 mm to 0.4 mm
40 s
Take the low NTU case because that is the most difficult to flocculate. Takes about 20 seconds per collision. About 20 collisions. Roughly 400 seconds or 7 minutes. Note that this doesn’t take into account any reactions that occur in the nm to micrometer scale. This assumes a perfectly uniform energy dissipation rate throughout the reactor.
At this energy dissipation rate flocs would only grow to 0.4 mm
This is with relatively high coagulant dosages of 50 mg/L for 100 NTU, 20 mg/L for 10 NTU, and 8 mg/L for 1 NTU. We now know that with well designed plants the coagulant dose can be reduced.
How much time is required to produce a 0.4 mm floc?

35. 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.
Or maybe phase 1 ends when a sign

36. 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 _______________
Performance is not based on the production of big flocs!
residual turbidity

37. 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?

38. 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

39. Flocculation Model: Integrating and tracking residual turbidity
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)
Separate variables
Integrate from initial colloid concentration to current colloid concentration
Would it make sense for design purposes to integrate from C.Colloid to C.ColloidFinal? That way the target is a final colloid concentrate as it really is in practice. The question is what would we use for the y axis with a goal of being able to plot experimental data in such a way that all of the data nicely collapses and the y axis represents progress toward an effluent turbidity goal? pCtarget=-log(Ctarget/C) so that the goal is to reach zero.
Rapid growth phase produces large number of flocs at maximum size and leaves significant number of notsettleable flocs.
Residual turbidity reduction phase – max size flocs are useless. Less than max size flocs continue to collide with colloids and reduce nonsettleable floc concentration.
We must have a distribution of floc sizes at the end of the initial growth phase that could be predicted by statistics. We will have many big flocs, few colloids. Most colloids are colliding with flocs that are already large enough to be captured by the sed tank. Thus most collisions result in effective removal of colloids. A few collisions are between two colloids that form a floc that is still too small to be captured.
Classic first order reaction with number of successful collisions replacing time

40. Laminar Flow Cases
Only colloids can collide effectively (big flocs are useless)
Floc volume fraction that matters for successful collisions is the colloid (small floc) fraction
Colloids can attach to all flocs
k has a value of approximately 0.7 (based on Karen Swetland’s data).
Floc volume fraction is a function of floc size (d) (which is a function of the reaction progression and shear conditions in the reactor

41. Comparison of Flocculation Hypotheses
If colloids could aggregate with all of the flocs then colloids would aggregate VERY rapidly
100 NTU clay suspension
Gamma of 0.1
The detention time for floc formation should be at least 30 minutes (ten state standards)
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

42. Coiled Tube Flocculation Residual Turbidity Analyzer
Here we plotted the residual turbidity, the turbidity of the water after settling, against the coagulant dose. We noticed that when we used a log-log plot, there was a linear trend in the data. So we explored further.
Dr. Karen Swetland Dissertation research

43. Two Phase Floc Model: PACl
These are the results! For a capture velocity of 0.12 mm/s the data collapses to a line with an R^2 around 0.9 for both. Keep in mind that this graph encompasses 244,000 data points…
Initial floc growth

44. Two Phase Floc Model: Alum
Initial floc growth

45. Solving for pC*: Lambert W or ProductLog Function (Laminar flow)
Use Wolframalpha to solve for C*
Get a solution from wolframalpha by entering -ln(y)/(y^(2/3))=x
W is the Lambert W function
Log is base 10

46. Laminar Flow Floc Model
Sedimentation velocity of ???
Collisions to make first big flocs = 0.4
Empirical
Floc volume fraction
Fractional surface coverage of colloid by coagulant
Lambert W Function
Flocculation time
Sedimentation tank capture velocity
Velocity gradient

47. Turbulent Flow Case Only colloids can collide and attach effectively
Why are almost all of these collisions in the inertial range?
Colloids can attach to all flocs
Collisions are in inertial range because these are the few colloids that are left after phase 1 and thus the spacing is larger than Komogorov length scale.

48. Turbulent Flow Flocculator: “Big Flocs are useless” hypothesis
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.
The predicted performance is quite similar given a wide range of influent turbidities
The predictions seem reasonable
Gamma of 0.1
= 0.1 for these plots
e = 2.6 mW/kg

49. Turbulent Flow Flocculator: Colloids can attach to all flocs hypothesis
Here the separation distance between reactive flocs might be less than the Kolmogorov length scale
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!
For this analysis I assumed that d.floc was 48 micrometers with a sedimentation velocity of 0.12 mm/s.
Average energy dissipation rate average was 2.6 mW/kg
Conclusion is that small flocs CAN’T attach to flocs that have reached their maximum size.
For these plots
G= 0.1
e= 2.6 mW/kg
pC* proportional to time

50. Solving for pC*: Lambert W or ProductLog (Turbulent Flow)
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Solving for pC*: Lambert W or ProductLog (Turbulent Flow)
Use Wolframalpha to solve for C*
Get a solution from wolframalpha by entering -ln(y)/(y^(8/9))=x
W is the Lambert W function
Log is base 10

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Lambert W Function
Performance varies very little over wide range of inputs.
Diminishing returns on investment
To increase from 90% removal to 99% removal by holding all parameters constant except residence time, you would have to increase residence time by more than a factor of 10.

52. Proposed Turbulent Flocculation Sedimentation Model (missing phase 1)
Lambert W Function
Flocculation time
Energy dissipation rate
Sedimentation velocity of ???
Sedimentation tank capture velocity
-log(fraction remaining)
Characteristic colloid size
The term with floc time and energy dissipation rate should really be the integral of those terms over the time in the flocculator.
Here eta.coag/Vcapture replaces the constants 5.6k
Eta.coag has a value of approximately 0.6 for the experimental conditions of Karen Swetlands dissertation.
Initial floc volume fraction
Fractional surface coverage of colloid by coagulant G
What does the plant operator control? ________
What does the engineer control? __________ 𝜙0
What changes with the raw water? __________

53. 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

54. 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?

55. 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.

56. 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

57. Required Collision Potential*
Agalteca, Alauca, Marcala 2 were designed to have y=100 m2/3
We are currently using y=75 m2/3
Atima???
*The model has not been validated for turbulent hydraulic flocculator. Thus this is only a rough estimate.

58. 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?

59. Surface coverage (G) of clay by coagulant precipitate
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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.

60. Floc Model Equations for G
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Floc Model Equations for G
Surface area of clay divided by total surface area of clay + reactor walls
Surface area to volume ratio for clay normalized by surface area to volume ratio for a sphere

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

62. extra
Poisson distribution and coagulant loss to walls combined to get surface coverage
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

63. Clay platelet geometry
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Clay platelet geometry
Model clay as cylinder with height and diameter
D.Clay is the diameter of a sphere with equal volume as the clay platelet
D.ClayPlatelet is the diameter of a cylinder given ratio of height to diameter and equal volume spherical diameter

64. Clay platelet geometry
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Clay platelet geometry

65. Surface coverage of clay
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Surface coverage of clay
If you neglect wall loses

66. Solving for Coagulant Dose
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Solving for Coagulant Dose
Coagulant dose is inside coagulant volume fraction
Solve for dose…
Given a target residual turbidity, solve for required G, then solve for coagulant dose

67. Loss of clay to reactor walls
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Loss of clay to reactor walls
FractalFlocModel.xmcd (Mathcad 15 files)
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.

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