ERT 320/ 3 BIO-SEPARATION ENGINEERING
MISS WAN KHAIRUNNISA WAN RAMLI
OUTLINES
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INTRODUCTION
BASIC PRINCIPLES
EQUIPMENTS
OPERATING MODES
BASIC CALCULATION
LLE A mass transfer operation in which a liquid
solution (the feed) is contacted with an
immiscible/ nearly immiscible liquid, called the
solvent that exhibits preferential affinity or
selectivity towards one/ more of the components in
the feed.
ADVANTAGES
LIMITATIONS
BASIC PRINCIPLES & PARAMETERS
SCALE UP
DESIGN OF EXTRACTOR COLUMN
Extraction is preferred for several applications:
a. Dissolved/ complexed inorganic substances in organic/
aqeous solution
b. Removal of a component present in small concentrations.
Eg: hormones in animal oil
c. A high-boiling component present in small concentration in
an aqeous waste stream.
Eg: Recovery of acetic acid from CA
a. Recovery of heat-sensitive materials.
Eg: Antibiotics
a. Separation of a mixture according to chemical type rather
than relative volatility
b. Separation of close-melting or close-boiling liquids;
solubility differences exploited
c. Mixtures that form azeotrope
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The simplest LLE involves only a ternary
Rafinatte
system
Feed2 miscible components: carrier, C + Solvent, S
Solute, A.
Solvent, S pure compound
C & S are at most PARTIALLY soluble in
each other.
A is soluble in C & completely/ partially
soluble in S
During the extraction process, mass transfer
of A from Feed to Solvent occur, with LESS
transfer of C to the solvent or S to the feed.
However, A is NOT completely transferred
to the solvent & seldom achieved in 1 stage
only
Feed, F
Need multi stages.
Extract
When LLE is carried out in a test tube/ flask, the
two immiscible phases are stirred/ mixed
together to allow molecules to dissolve into the
preferred solvent phase
• The solute originally present in the aqueous phase gets
distributed in both phases
• If solute has preferential solubility in the organic solvent,
more solute would be present in the organic phase at
equilibrium
• The extraction is said to be more efficient
• Extract = the layer of solvent + extracted solute
• Raffinate = the layer from which solute has been removed
• The distribution of solute between two phases is express
quantitatively by distribution coefficient, KD
•
Eq. 1
• Higher value of KD indicates higher extraction efficiency
THE TWO PHASES MUST BE BROUGHT INTO
INTIMATE CONTACT WITH A HIGH DEGREE OF
TURBULENCE IN ORDER TO OBTAIN HIGH MASS
TRANSFER RATES.
MIXING BY
MECHNICAL
AGITATION
MIXING BY
FLUID FLOW
MIXED-SETTLERS
For Batchwise Extraction:
The mixer and settler may be the same unit.
A tank containing a turbine or propeller agitator is most common.
At the end of mixing cycle the agitator is shut off, the layers are allowed
to separate by gravity.
Extract and raffinate are drawn off to separate receivers through a
bottom drain line carrying a sight glass.
The mixing and settling times required for a given extraction can be
determined only by experiment. (e.g: 5 min for mixing and 10 min for
settling are typical) - both shorter and much longer times are common.
For Continuous Extraction:
The mixer and settler are usually separate pieces of equipment.
The mixer; small agitated tank provided with a drawoff line and
baffles to prevent short-circuiting, or it may be motionless mixer or
other flow mixer.
The settler; is often a simple continuous gravity decanter.
In common used; several contact stages are required, a train of
mixer-settlers is operated with countercurrent flow.
PACKED EXTRACTOR COLUMN
Tower extractors give differential
contacts, not stage contacts, and mixing
and settling proceed continuously and
simultaneously.
Extraction; can be carried out in an
open tower, with drops of heavy liquid
falling through the rising light liquid or
vice versa.
The tower is filled with packings such
as rings or saddles, which causes the
drops to coalesce and reform, and tends
to limit axial dispersion.
In an extraction tower there is
continuous transfer of material between
phases, and the composition of each
phase changes as it flows through the
tower.
The design procedure ; is similar to
packed absorption towers.
AGITATED EXTRACTOR COLUMN
It depends on gravity flow for mixing and for
separation.
Mechanical energy is provided by internal turbines
or other agitators, mounted on a central rotating
shaft.
Flat disks disperse the liquids and impel them
outward toward the tower wall, where stator rings
create quite
zones in which the two phases can separate.
In other designs, set of impellers are separated by
calming sections to give, in effect, a stack of mixersettlers one above the other.
In the York-Scheibel extractor, the region
surrounding the agitators are packed with wire mesh
to encounter coalescence and separation of the
phases.
Most of the extraction takes place in the mixing
sections, but some also occurs in the calming
sections.
The efficiency of each mixer-settler unit is
sometimes greater than 100 percent.
Batch Extraction:
i. The aqueous feed is mixed
with the organic solvent
ii. After equilibrium, the extract
phase containing the desired
solute is separated out for
further processing
iii. Routinely utilized in laboratory
procedures
iv. This can be carried out for
example in separating funnel
or in agitated vessel
Schematic representations of BATCH
extraction:
(a) Single
(b) Multiple stages (crosscurrent)
Continuous Extraction (Co-current & Counter-current):
Schematic representations of
(a) co-current
(b) Counter-current operations:
EXTRACTION OF DILUTES SOLUTION
1. Extraction factor, E is defined as:
From Eq. 1
2. For a single-stage extraction with pure solvent:
Penicillin F is recovered from a dilute aqueous fermentation
broth by extraction with amyl acetate, using 6 volumes
of solvent (V) per 100 volumes of the aqueous phase (L). At
Ph 3.2 the distribution coefficient KD is 80.
(a) What fraction of the penicillin would be recovered in a
single ideal stage?
(b) What would be the recovery with two-stage extraction
using fresh solvent in both stages?
SOLUTION 1:
a) Draw the material balance diagrams
Material Balance:
L(x0) + V(y0) = L (x1) + V(y1)
L(x0) – L(x1) = V(y1) – V(y0)
Since y0=0 (at initial no penicillin in solvent phase)
So, L(x0)-L(x1) = V(y1)
L(x0-x1)= V(y1)
Since KD = y1/x1, y1=KDx1  Refer to EQ. 1
So, L(x0-x1)=V(KDx1)
x1[(VKD/L )+ 1)]= x0, where VKD/L = E  Refer to EQ. 2
E = VKD/L = (6)(80)/100 = 4.8
Material Balance (cont.):
x1/x0 = 1/ (1+E)  Refer to EQ. 3
 (frac. of penicillin in raffinate phase = frac. remaining)
= 1/ (1+ 4.8)
= 0.1724
Fraction of penicillin recovered = Fraction of penicillin in extract phase
= 1- 0.1724
= 0.828
= 82.8%
Or calculated using EQ. 4, E/(1+E)= 4.8/ (1+4.8) =0.828; 82.8% recovery
SOLUTION 1:
b) Draw the material balance diagrams
Material Balance:
L(x1) + V(y0) = L (x2) + V(y2)
L(x1) – L(x2) = V(y1) – V(y0) 0
Since y0=0 (at initial no penicillin in solvent phase)
So, L(x1)-L(x2) = V(y2)
L(x1-x2)= V(y2)
Material Balance:
Since KD = y2/x2, y2=KDx2  Refer to EQ. 1
So, L(x1-x2)=V(KDx2)
x2[(VKD/L )+ 1)]= x1, where VKD/L = E  Refer to EQ. 2
E= (6)(80)/100 = 4.8
x2/x1 = 1/ (1+E)  Refer to EQ. 3
 (frac. of penicillin in final raffinate phase from raffinate
phase in 1st stage) = frac. remaining)
= 1/ (1+ 4.8)
= 0.1724
x2/x0 = (x2/x1) * (x1/x0)
= (0.1724) * (0.1724)
= 0.0297  (frac. of penicillin in final raffinate phase from feed phase
= frac. remaining from )
Fraction of penicillin recovered
= Fraction of penicillin in extract phase from feed phase
= 1- 0.0297
= 0.9703
= 97.0%
EXTRACTION OF CONCENTRATED SOLUTION
1. Equilibrium relationship are more complicated3 or
more components present in each phase.
2. Equilibrium data are often presented on a triangular
diagram such as Fig 23.7 and 23.8.
Solute, A
Solvent, S
Carrier, C
TERNARY SYSTEM TYPE I
1 immiscible pair
Point E  PLAIT POINT
[The composition of extract &
raffinate phases approach each other]
Line ACE  Extract phase
Line BDE  Raffinate phase
TIE LINE
How to obtain the phase composition using the triangular
diagram?
#Example:
If a mixture with 40 % acetone and 60 percent water
is contacted with equal mass of MIK, the overall mixture is
represented by point M in Figure 23.7:
1. Assume mass of mixture and solvent
2. Determine the F/S ratio to pinpoint the mixing, M point.
Point M: 0.2 Acetone, 0.3 water, 0.5 MIK
3. Draw a new tie line, Point M to Extract & Raffinate
phase
4. Extract phase: 0.232 acetone, 0.043 water, 0.725 MIK
5. Raffinate phase: 0.132 acetone, 0.845 water, 0.023 MIK
6. Ratio of acetone to water in the product = 0.232/0.043
= 5.4
7. Ratio of acetone to water in the raffinate = 0.132/0.845
= 0.156
F
0.2 Acetone
S
Solute, A
•Tie line in Fig 23.8 slope up to
the right – extraction
would still be possible
•But more solvent would have
to use.
•The final extract would not
be as rich in desired
component (MCH)
Raffinate phase
Solvent, S
Carrier, C
Extract phase
TERNARY SYSTEM TYPE II
2 immiscible pairs
COORDINATE SCALE
• Refer to Treybal, Mass Transfer Operation, 3rd ed., McGraw
Hill
• The book use different triangular system
• The location of solvent (B) is on the right of the triangular
diagram (McCabe use on the left)
• Coordinate scales of equilateral triangles can be plotted as y
versus x as shown in Fig 10.9
• Y axis = wt fraction of component A (acetic acid)
• X axis = wt fraction of solvent B (ethyl acetate)
Bahan larut B
SOLUTE, A
0.9
0.1
0.8
0.2
0.7
0.3
0.6
Pecahan jisim B
0.4
0.5
0.3
Pecahan jisim S
0.5
0.4
Rafinat
0.6
Ekstrak
0.7
0.2
0.8
0.1
Pembawa C
CARRIER,
C
0.9
0.9
0.8
0.7
0.6
0.5
0.4
Pecahan jisim C
0.3
TERNARY SYSTEM TYPE I
0.2
0.1
SOLVENT,
Pelarut
S
S
Bahan larut B
SOLUTE, A
0.9
0.1
0.8
0.2
0.7
0.3
0.6
Pecahan jisim B
0.4
0.5
Rafinat
Pecahan jisim S
0.5
0.4
0.6
0.3
0.7
0.2
0.8
Ekstrak
0.1
Pembawa C
CARRIER,
C
0.9
0.9
0.8
0.7
0.6
0.5
0.4
Pecahan jisim C
0.3
0.2
TERNARY SYSTEM TYPE II
0.1
SOLVENT,
S
Pelarut S
SINGLE-STAGE EXTRACTION
• The triangular diagram in Fig 10.12 (Treybal) is a bit
different as compared to Fig. 23.7 (McCabe)
• Extract phase  on the RIGHT
• Raffinate phase  on the LEFT
• Fig 10.12 shows that we want to extract Solute, A from C by
using solvent, S
• Total material balance:
• Material balance on A:
Amount of solvent to provide a given location for M1 on the
line FS:
• The quantities of extract and raffinate:
• Minimum amount of solvent is found by locating M1 at D
(Raffinate phase)
• Maximum amount of solvent is found by locating M1 at K
(Extract Phase)
MULTISTAGE CROSSCURRENT EXTRACTION
• Continuous or batch processes
• Refer to Fig 10.14
• Raffinate from the previous stage will be the feed for the
next stage
• The raffinate is contacted with fresh solvent
• The extract can be combined to provide the composited
extract
• The total balance for any stage n:
• Material balance on A:
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Consider a countercurrent-flow, N-equilibrium-stage contactor for solvent
extraction of a ternary system under isothermal, continuous & steady state
flow
E1
E2
E3
En
En 1 EN 1
EN
S
Extract
y1
y2
1
Feed
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y3
yn
2
yn 1
n
F
R1
R2
xF
x1
x2
Rn 1
yN 1
yN
N-1
Rn
RN  2
yN 1 Solvent
N
RN 1
RN
Raffinate
xn 1
xn
xN  2
xN 1
xN
Feed, F contains the carrier, C & solute, A, & can slso contain solvent, S.
The entering solvent, S can contain C & A
F ialah mass flow rate of feed
S ialah mass flow rate of solvent
En ialah mass flow of extract leaving stage n
Rn ialah mass flow of raffinate leaving stage n
yn ialah mass fraction of solute in extract leaving stage n
xn ialah mass fraction of solute in raffinate leaving stage n
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The degree of freedom for liquid-liquid extraction for nequilibrium stages is 2C + 3
For ternary system, C = 3, thus the degree of freedom = 9
In most cases, specifications for Feed, F, xF, yS dan T will be
provided
There is 6 more specifications for LLE
1.
2.
3.
4.
5.
6.
S & xN given
S & y1 given
y1 & xN given
Number of equilibrium stages, N & xN given
Number of equilibrium stages, N & y1 given
Number of equilibrium stages, N & S given
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Overall total mass balance for control volume 1
F  S  M  E1  RN
(7.1)
Overall mass balance of the component for control volume 1
FxF  SyN 1  MxM  E1 y1  RN xN
(7.2)
Solvent-to-Feed (S/F) and Extract-to-Raffinate (E/R) Flow rate ratios
S
x  xM
 F
F xM  y N 1
1
Extract
(7.4)
E1
E2
E3
En
En 1
EN 1
EN
S
y1
y2
y3
yn
yn 1
yN 1
yN
yN 1
1
Feed
E1 xN  xM 1

RN xM 1  y1
(7.3)
2
n
N-1
Solvent
N
F
R1
R2
Rn 1
Rn
RN  2
RN 1
xF
x1
x2
xn 1
xn
xN  2
xN 1
RN Raffinate
xN
Bahan larut B
0.9
0.1
0.8
0.2
0.7
0.3
0.6
Pecahan jisim B
0.4
0.5
0.3
Pecahan jisim S
0.5
0.4
Rafinat
0.6
Ekstrak
0.7
E1
F
0.2
0.1
Pembawa C
0.8
M
RN
0.9
0.9
0.8
0.7
0.6
0.5
0.4
Pecahan jisim C
0.3
0.2
OVERALL MATERIAL BALANCE
0.1
Pelarut S
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OPERATING LINELocus of passing streams in a cascade
It is given by the difference between inlet and outlet streams of each
stage
(7.5)
F  E  R  E  ...  R  E  ...  R  S  P
1
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1
n 1
2
n
N
This difference point, P  Operating point
This point is established graphically by finding the intersection between TWO
extrapolated operating lines which connect the Feed, F with Extract, E1 point &
the Solvent, S with Raffinate, RN point.
E1 P E1  P F


E1
E1
FP
Ekstrak
E1
E2
E3
En
En 1
EN 1
EN
S
y1
y2
y3
yn
yn 1
yN 1
yN
yN 1
1
Suapan
(7.6)
2
n
N-1
Pelarut
N
F
R1
R2
Rn 1
Rn
RN  2
RN 1
xF
x1
x2
xn 1
xn
xN  2
xN 1
RN Rafinat
xN
Bahan larut B
0.9
0.1
0.8
0.2
0.7
0.3
0.6
Pecahan jisim B
0.4
0.5
Operation
point
Operation line
[F, E1]
0.3
0.2
P
0.5
0.4
Rafinat
0.6
Ekstrak
0.7
E1
F
0.8
M
0.1
Operating line
[S, Rn]
Pembawa C
Pecahan jisim S
0.9
RN0.9
0.8
0.7
0.6
0.5
0.4
Pecahan jisim C
Pembinaan Titik Kendalian P
0.3
0.2
0.1
Pelarut S
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Extract and raffinate streams which exited each stage was at equilibrium.
This concludes that Exract and raffinate point at each stage are
connected by the tie line.
yn *  f  xn 
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(7.7)
The operating line then connects the extract and raffinate points at the next
stage with the operating point, P according to Eq. (7.5)
This construction of lines are repeated until the final required extract point is
reached.
Ekstrak
E1
E2
E3
En
En 1
EN 1
EN
S
y1
y2
y3
yn
yn 1
yN 1
yN
yN 1
1
Suapan
2
n
N-1
Pelarut
N
F
R1
R2
Rn 1
Rn
RN  2
RN 1
xF
x1
x2
xn 1
xn
xN  2
xN 1
RN Rafinat
xN
Bahan larut B
0.9
0.1
0.8
0.2
0.7
0.3
0.6
Pecahan jisim B
0.4
0.5
Operating
point, P
Operating
line
0.3
Pecahan jisim S
0.5
0.4
Rafinat
0.6
Ekstrak
0.7
E1
F
0.2
0.8
M
0.1
Operating
line
Pembawa C RN
0.9
0.9
0.8
0.7
0.6
0.5
0.4
Pecahan jisim C
0.3
0.2
Determination of number of equilibrium stages (2)
0.1
Pelarut S
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The actual solvent-to-feed ratio is determine from the multiples of the
minimum solvent-to-feed ratio.
Minimum solvent-to-feed flowrate ratios usually happen when the
number of stages needed is close to infinity.
This happen when the operating line coincident with a tie line.
The line connecting Solvent point,S and Raffinate point, RN is extended
in both direction of raffinate and extract.
The farthest point which intersects with any extended tie line on SRN line
determine the operating line of N-stages.
Ekstrak
E1
E2
E3
En
En 1
EN 1
EN 5
S
y1
y2
y3
yn
yn 1
yN 1
yN
yN 1
1
Suapan
2
n
N-1
Pelarut
N
F
R1
R2
Rn 1
Rn
RN  2
RN 1
xF
x1
x2
xn 1
xn
xN  2
xN 1
RN Rafinat
xN
Bahan larut B
0.9
0.1
0.8
0.2
0.7
0.3
E1Smin
0.6
Pecahan jisim B
0.4
0.5
F
0.3
Pecahan jisim S
0.5
0.4
Rafinat
0.6
Ekstrak
0.7
MSmin’
P2 P1 0.2
0.8
P3 = P min
0.1
Operating
line
Pembawa C
RN
0.9
0.9
0.8
0.7
0.6
0.5
0.4
Pecahan jisim C
0.3
0.2
0.1
Determination of minimum solvent-to-feed flow rate ratio
Pelarut S
Smin
•
The difference in total outflow for Smin/F ratio
F  E1S min  RN  Smin  Pmin
•
•
(7.8)
Point MSmin can be determine by intercepting both line: FSmin dan RNE1Smin
Therefore, by using the inverse-lever-arm rule, we can determine the minimum
solvent-to-feed ratio, Smin/F
M S min F
Smin

F
Smin M S min
•
(7.9)
The actual S/F is 1.2 – 2.0 times the SSmin/F
Ekstrak
E1
E2
E3
En
En 1
EN 1
EN
S
y1
y2
y3
yn
yn 1
yN 1
yN
yN 1
1
Suapan
2
n
N-1
Pelarut
N
F
R1
R2
Rn 1
Rn
RN  2
RN 1
xF
x1
x2
xn 1
xn
xN  2
xN 1
RN Rafinat
xN
•Use widely in separation of proteins, enzymes, viruses, cells
and cell organels
•Not denature the biological entities as they might be in
organic solvents.
•The proteins are partitioning between two aqueous phases
which contains mutually incompatible polymers or other
solutes.
EXAMPLE
Light phase is water + 10% polyethylene glycol (PEG)
and 0.5% dextran
Heavy phase is water + 1% glycol and 15% dextran
Proteins are partitioned between phases with
distribution coefficient (KD) that depends on the pH.
KD can vary from 0.01 to more than 100.
• Factors that affect protein partitioning in Aqueous
Two Phase System:
1. Protein molecular weight
2. Protein charge, surface properties
3. Polymer(s) molecular weight
4. Phase composition, tie-line length
5. Salt effects
6. Affinity ligands attached to polymers
Acetone is to be extracted from a feed mixture of 30 wt.% acetone (A) and
70 wt.% ethyl acetate (C) at 30°C by using the same weight of pure water
(S) as the solvent. The final raffinate is to contain 5 wt.% of acetone.
•
•
Determine the weight percent of acetone that can be extracted using a
single stage
Determine the number of equilibrium stages required
Ekstrak
E1
E2
E3
En
En 1
EN 1
EN
S
y1
y2
y3
yn
yn 1
yN 1
yN
yN 1
1
Suapan
2
n
N-1
Pelarut
N
F
R1
R2
Rn 1
Rn
RN  2
RN 1
xF
x1
x2
xn 1
xn
xN  2
xN 1
RN Rafinat
xN
Assume F = 50 g, so S = 50g
E
F
y1
Xf = 0.3
S
Yn = 0
Material balance:
F+S=M
50 + 50 = 100
Material balance on A:
Fxf + Syn = Mxm
(50x0.3) + (50x0) = 100xm
Xm = 0.15
M = E1 + R1, R1 = M – E1
Mxm = Ey1 + Rxn
Mxm = E1y1 + (M-E1)xn
E1 = M (xm-xn) / (y1 – xn)
R
Xn = 0.05
Find point S (100 wt.% S) & F (30 wt.% A, 70 wt.% C), draw a line
connecting point F & S
From material balance, we have found xm & M point MUST lie in FS
line
Find point M using xm on FS line
Then, we already have the fraction of A we need in raffinate stream, xn
= 0.05
Find point R1 using xn which needs to be on the equilibrium curve
Draw a line connecting point R1 & M, extrapolate until reached the
Extract side.
The point which intercept both line R1,M and equilibrium curve are point
E1 & y1.
Read the value for y1, settle the material balance to find E1 & R1
Equilibrium stages
Draw line connecting F with E1, R1 with S, extrapolates until both lines
intercept each other at point P.
Aseton A
0.9
0.1
0.8
0.2
0.7
0.3
0.6
Pecahan jisim A
0.4
0.5
0.4
0.3
Xm = 0.15
Pecahan jisim C
0.5
Rafinat 0.6
Ekstrak
0.7
E1
M
0.2
0.8
0.1
Pelarut Air S
S
F
0.9
0.9
0.8
0.7
0.6
0.5
0.4
Pecahan jisim S
0.3
0.2
0.1
R1Asetat C
Solution of Acetic Acid and water containing 30 wt.% of Acetic Acid is to be
extracted by contacting to twice weight of isopropyl ether at 40°C to give a
raffinate containing 5 wt.% of Acetic Acid.
•
•
Determine the quantities & composition of each streams for a single stage
extraction process
Determine the number of equilibrium stages required
STEPS IN SOLVING MATERIAL BALANCE
1. Construct the equilibrium curve in triangular diagram &
connects the dot to form tie line
2. Find operating point of the diagram by extrapolating tie
lines until they intercepts at P
3. Draw the material balance diagram (based on the q)
4. Solve the material balance at each stage using eq. 7.1 &
7.2 to find mixing point, M
5. Find the F & S point in the diagram, connects the points
to form line FS
6. Since we only know the fraction of solute, A in M, find
M(?, xm) on line FS
7. Draw a tie line passing M & extended to point P
8. Intercepting points at equilibrium curve are the solute
concentration of E & R point of the extraction process,
respectively at both extract & raffinate phase
9. Solve the material balance for R & E