1. SETTLING
AND
SEDIMENTATION

2.  Filtration versus settling and sedimentation: Filtration
Introduction (1/4)
 Filtration versus settling and sedimentation:
Filtration
 The solid particles are removed from the slurry by forcing the fluid through a filter medium, which blocks the passage of the solid particles and allows the filtrate to pass through.
Settling and sedimentation
 The particles are separated from the fluid by forces acting on the particles.

3.  Applications of settling and sedimentation:
Introduction (2/4)
 Applications of settling and sedimentation:
* Removal of solids from liquid sewage wastes
* Settling of crystals from the mother liquor
* Separation of liquid-liquid mixture from a solvent-extraction stage in a settler
* Settling of solid food particles from a liquid food

4.  Free settling versus hindered settling: Free settling
Introduction (3/4)
 Free settling versus hindered settling:
Free settling
 A particle is at a sufficient distance from the walls of the container and from other particles so that the fall is not affected.
 Interference is less than 1% if the ratio of the particle diameter to the container diameter is less than 1:200 or if the particle concentration is less than 0.2 vol% in the solution.
Hindered settling
 Occurred when the particles are crowded so that they settle at a lower rate.

5.  What is sedimentation?
Introduction (4/4)
 What is sedimentation?
 The separation of a dilute slurry or suspension by gravity settling into a clear fluid and a slurry of higher solid content.

6. THEORY OF PARTICLE MOVEMENT THROUGH A FLUID
For a rigid particle of mass m moving in a fluid, there are three forces acting on the body:
Gravity force, Fg, acting downward
(2) Buoyant force, Fb, acting upward
where r = density of the liquid
rs = density of the solid particle
Vs = volume of the particle

7. where CD = the drag coefficient A = the projected area of the particle
THEORY OF PARTICLE MOVEMENT THROUGH A FLUID (2/8)
For a rigid particle of mass m moving in a fluid, there are three forces acting on the body:
(3) Drag force, FD, acting in opposite direction to the particle motion
where CD = the drag coefficient
A = the projected area of the particle
The resultant force equals the force due to acceleration.

8. The falling of the body consists of two periods:
THEORY OF PARTICLE MOVEMENT THROUGH A FLUID (3/8)
The falling of the body consists of two periods:
(1) The period of accelerated fall
 The initial acceleration period is usually very short, of the order of a tenth of a second or so.
(2) The period of constant velocity fall
and solve the above equation for v.
Set 
* vg is called the free settling velocity or terminal velocity.

9. For spherical particles of diameter d,
THEORY OF PARTICLE MOVEMENT THROUGH A FLUID (4/8)
For spherical particles of diameter d, 

10. THEORY OF PARTICLE MOVEMENT THROUGH A FLUID (5/8)
The drag coefficient for rigid spheres has been shown to be a function of the Reynolds number.

11. In the Stokes' law region (NRe < 1),
THEORY OF PARTICLE MOVEMENT THROUGH A FLUID (6/8)
In the Stokes' law region (NRe < 1), 
* The Stokes’ law region is almost always satisfied for biological solutes.

12. * If the particles are quite small, Brownian motion is present.
THEORY OF PARTICLE MOVEMENT THROUGH A FLUID (7/8)
 Brownian motion: the random motion imparted to the particle by collisions between the molecules of the fluid surrounding the particle and the particle.
* If the particles are quite small, Brownian motion is present.
 This movement of the particles in random directions tends to suppress the effect of gravity.
 Settling of the particles may occur more slowly or not at all.

13. THEORY OF PARTICLE MOVEMENT THROUGH A FLUID (8/8)
 Brownian motion (continued)
* At particle sizes of a few micrometers, the Brownian effect becomes appreciable and at sizes of less than 0.1 mm, the effect predominates.
 In very small particles, application of centrifugal force helps reduce the effect of Brownian motion.

14. [Example] Many animal cells can be cultivated on the external surface of dextran beads. These cell-laden beads or “microcarriers” have a density of 1.02 g/cm3 and a diameter of 150 mm. A 50-liter stirred tank is used to cultivate cells grown on microcarriers to produce a viral vaccine. After growth, the stirring is stopped and the microcarriers are allowed to settle. The microcarrier-free fluid is then withdrawn to isolate the vaccine. The tank has a liquid height to diameter ratio of 1.5; the carrier-free fluid has a density of 1.00 g/cm3 and a viscosity of 1.1 cP. (a) Estimate the settling time by assuming that these beads quickly reach their maximum terminal velocity. (b) Estimate the time to reach this velocity.
Hint:
(To be continued)

15. Solution:  vg = 0.022 cm/s Check: Liquid volume, V
Example: settling of dextran beads
Data: d = 150 mm = cm; m = 1.1 cP = g/cm-s; rs = 1.02 g/cm3; r = 1.00 g/cm3; g = 980 cm/s2
(a) Estimate the settling time by assuming that these beads quickly reach their maximum terminal velocity.
Solution:
 vg = cm/s
Check:
Liquid volume, V
 h = 52.3 cm  Settling time
(To be continued)

16. Solution (cont’d): Force balance: ;   (I.C.: t = 0, v = 0) 
Example: settling of dextran beads
(b) Estimate the time to reach the terminal velocity.
Solution (cont’d):
Force balance: ;  
(I.C.: t = 0, v = 0) 
(To be continued)

17.  When t >> 1.16  10-3 s, v = vg
Example: settling of dextran beads
(b) Estimate the time to reach the terminal velocity.
Solution (cont’d):
At steady state (t  ),
 When
 When t >> 1.16  10-3 s, v = vg
 For v = 0.99vg, t = 5.34  10-3 s #

18. ISOPYCNIC (SAME-DENSITY) SEDIMENTATION
 To capture particles in a solution having density gradient.
 Application: determining the density of the solute or suspended particle.
* There are three methods for establishing conditions for isopycnic sedimentation:
(1) Layer solutions of decreasing density, starting at the bottom of the tube.
(2) Centrifuge the solution containing a density-forming solute (such as CsCl) at extremely high speed.
(3) Use the gradient mixing method.

19.  Produce an outflow with a linear solute gradient.
ISOPYCNIC (SAME-DENSITY) SEDIMENTATION (2/3)
* Methods for establishing conditions for isopycnic sedimentation (cont’d)
(3) Use the gradient mixing method.
 Produce an outflow with a linear solute gradient.
 The most widely used method.
Q/2 Q Q

20. where C1,0 = initial solute concentration in the mixed chamber
ISOPYCNIC (SAME-DENSITY) SEDIMENTATION (3/3) C2
C, Q
where C1,0 = initial solute concentration in the mixed chamber
C2 = solute concentration in the non-mixed chamber (constant)
Q = outflow rate from the mixed chamber
V0 = initial volume in each vessel

21. ISOPYCNIC (SAME-DENSITY) SEDIMENTATION

22. [Example] You wish to capture 3 mm particles in a linear density gradient having a density of 1.12 g/cm3 at the bottom and 1.00 g/cm3 at the top. You layer a thin particle suspension on the top of the 6 cm column of fluid with a viscosity of 1.0 cp and allow particles to settle at 1 g. How long must you wait for the particles you want (density = 1.07 g/cm3) to sediment to within 0.1 cm of their isopycnic level? Is it possible to determine the time required for particles to sediment to exactly their isopycnic level?
Solution:
(a) 
(To be continued)

23. The dependence of liquid density r on the distance x is:
Example: isopycnic sedimentation (cont’d)
The dependence of liquid density r on the distance x is:
The isopycnic level of r = 1.07 g/cm3 is:
The time needed for the particle to sediment to 3.4 cm can be obtained from:
(To be continued)

24. 
(b) It is not possible to determine the time required for particles to sediment to exactly their isopycnic level (3.5 cm). #

25. DIFFERENTIAL SETTLING (or CLASSIFICATION)
 Separation of solid particles into several size fractions based upon the settling velocities in a medium.

26. The terminal settling velocities of components A and B are:
DIFFERENTIAL SETTLING (2/6)
If the light and heavy materials both have a range of particle sizes, the smaller, heavy particles settle at the same terminal velocity as the larger, light particles.
The terminal settling velocities of components A and B are:
For particles of equal settling velocities, vgA = vgB.

27. In the turbulent Newton's law region, CD is constant.
DIFFERENTIAL SETTLING (3/6)
In the turbulent Newton's law region, CD is constant. 
For laminar Stokes’ law settling, 

28. For transition flow between laminar and turbulent flow,
DIFFERENTIAL SETTLING (4/6)
In the turbulent Newton's law region, CD is constant,
For laminar Stokes’ law settling,
For transition flow between laminar and turbulent flow,

29. * Size range dA3 to dA4: pure fraction of A
DIFFERENTIAL SETTLING (5/6)
 Settling a mixture of particles of materials A (the heavier) and B (the lighter) with a size range of d1 to d4 for both types of material:
* Size range dA3 to dA4: pure fraction of A
 No B particles settle as fast as the A particles in this size range.
* Size range dB1 to dB2: pure fraction of B
 No particles of A settle as slowly.

30. * Increasing the density r of the medium.
DIFFERENTIAL SETTLING (6/6)
* Size range of A particles from dA1 to dA3 and size range of B particles from dB2 to dB4: form a mixed fraction of A and B
* Increasing the density r of the medium.
 The spread between dA and dB is increased.

31. Solution: A particles: galena; B particles: silica
[Example] A mixture of silica and galena (方鉛礦; PbS) solid particles having a size range of 5.21  10-6 m to 2.50  10-5 m is to be separated by hydraulic classification using free settling conditions in water at 20C. The specific gravity of silica is 2.65 and that of galena is Calculate the size range of the various fractions obtained in the settling. The water viscosity at 20C is  10-3 Pa-s.
Solution: A particles: galena; B particles: silica
Assuming Stokes’ law settling,
 Check the validity of the Stokes’ law region.
(To be continued)

32. For the largest particle and the biggest density,
Example: hydraulic classification
Solution (cont’d):
For the largest particle and the biggest density,
dA = 2.50  10-5 m and rsA = 7.5 g/cm3 = 7500 kg/m3
Check:
= < 1
 O.K. with the Stokes’ law region.
(To be continued)

33. For particles of equal settling velocities,
Example: hydraulic classification
Solution (cont’d):
For particles of equal settling velocities, 
 dA3 =  10-5 m
The size range of pure A (galena) is:
dA3 =  10-5 m to dA4 = 2.50  10-5 m
(To be continued)

34. The size range of pure B (silica) is:
Example: hydraulic classification
Solution (cont’d): 
 dB2 =  10-5 m
The size range of pure B (silica) is:
dB1 = 5.21  10-6 m to dB2 =  10-5 m
The mixed-fraction size range is:
dA1 =  10-6 m to dA3 =  10-5 m
dB2 =  10-5 m to dB4 = 2.50  10-5 m #

35. INCLINED SEDIMENTATION
 Particle-free overflow exits the upper end, and particle-rich suspension leaves in the underflow.
 Rapid removal of high density solids can be achieved.
 The particles need to settle only a distance of order b (compared with a distance of order L in a vertical settler).

36. (2) Concentrate (batchwise) the particulate fraction.
INCLINED SEDIMENTATION (2/2)
 Applications:
(1) Continuously (or batchwise) harvest supernatant from particle (cell)-laden broth.
(2) Concentrate (batchwise) the particulate fraction.
(3) Perform a binary particle classification by size.
 Scaleup of inclined settlers: increasing the area for settling.

37. The End of
SETTLING AND
SEDIMENTATION