Sedimentation
Downstream Processing
Short Course
May 2007
Kevin Street
Gavin Duffy
Bioprocess Overview
Solid

liquid
Separation
Concentration
Purification
Formulation
Intra

Cellular
Product
Final Product
Extra

Cellular
Product
Cell Disruption
Upstream Processing
Centrifugation/Sedimentation,
Extraction, Filtration
Evaporation, Ultrafiltration,
Adsorption, Precipitation
Chromatography
Crystallisation, freeze drying,
Spray drying, sterile filtration
Chemical/Enzymatic/
Mechanical/Physical
Basic Biotechnology, 2
nd
Ed, Ch 9
Learning Outcomes
After this lecture you should be able to…
Describe the sedimentation process and equipment
Describe the motion of particles in free fall
Calculate the terminal velocity of a particle
Sedimentation
This is the separation of a liquid from particles
suspended in the liquid
A particle, falling from rest, accelerates under the force
of gravity
The drag force increases so the acceleration decreases
(liquid viscosity is important here)
Acceleration eventually becomes zero
–
the terminal
velocity is reached
Terminal velocity is reached quickly, e.g. a 100
m
particle in water reaches 2 mm/s in 1.5 ms
Upward velocity of liquid must be less than terminal
velocity for sedimentation to work
We must know the terminal velocity!
Sedimentation Tank
Single Particle Terminal velocity
For low Particle Reynolds number:
Creeping flow
Drag coefficient increases with velocity
Stokes law region
For high Particle Reynolds number:
Inertial flow (fluid must accelerate out of path)
Drag coefficient constant
18
2
g
d
u
f
p
T
2
1
74
.
1
f
f
p
T
g
d
u
Drag coefficient
The drag coefficient is defined as:
R’ is the drag force per unit projected area (N)
u is the velocity (m/s)
ρ
f
is the fluid density (kg/m
3
)
(What are the units of C
D
?)
Stokes’ law region:
Intermediate region:
Newton’s law region:
2
2
u
R
C
f
D
p
D
C
Re
24
44
.
0
Re
24
p
D
C
44
.
0
D
C
Drag curve for motion of a particle in fluid
Introduction to Particle Technology, Martin Rhodes, Ch 1
Stokes’
Newton’s
BL separation
Sphericity
Sphericity = surface area of equivalent sphere
surface area of particle
Equivalent sphere = sphere of same volume as particle
Deviation from sphere does not matter in Stokes’ law
region as much as in Newton’s law region
Particles fall with their small surface pointed
downwards in Stokes’ law region
The largest surface is pointed downwards in
Newton’s law region
Activity
–
Calculate Terminal Velocity
What are the particle Reynolds number and terminal
velocity for the following system?
Diameter 3
m
Density of solid phase 1090 kg/m
3
Cell free liquid density 1025 kg/m
3
Cell free liquid viscosity 0.005 Pa.s
Data taken from a case study of r

HSA production with recombinant
Pichia Pastoris
prepared by L Van der Wielen, European Federation on
Biotechnology
If you don’t know which region…
Calculate C
D
Re
2
from the following eqn:
Use result to draw a line on the drag curve
For example, suppose C
D
Re
2
= 8
Then,
for Re = 10
Re
2
= 100
C
D
= 0.08
for Re = 1
Re
2
= 1
C
D
= 8
for Re = 0.1
Re
2
= 0.01
C
D
= 800
Use these points to draw the line and read the Particle
Reynolds number. The velocity is then obtained
2
3
2
3
4
Re
g
d
C
f
p
f
p
D
…..use the Re v Drag coefficient chart
x
x
x
The Thickener
Feed added gently just below surface
Upward velocity of liquid must be less than u
T
Capacity depends on area: big area = low velocity (Q = va)
Degree of thickening depends on residence time which depends
on height
Can heat tank to reduce viscosity and increase u
T
Limit to solids flux
http://www.filtration

and

separation.com/thickener/sld004.htm
20/4/07
Batch Settling Test
Thickener Area Calculation
where
A = area (m
2
)
Q
0
= feed rate of suspension (m
3
/s)
Y = mass ratio liquid to solid in feed
U = mass ratio liquid to solid in underflow
C = particle volume fraction (1

ε
)
ρ
s
= density of solid (kg/m
3
)
u
T
= terminal velocity at conc. C (m/s)
ρ
f
= density of liquid (kg/m
3
)
f
T
s
u
C
U
Y
Q
A
0
Activity
–
Calculate Terminal Velocity
based on worked example 2.1 from Rhodes.
Comments 0
Log in to post a comment