PRODUCT ISOLATION & CONCENTRATION

(PART I)ERT 320/ 3 BIO-SEPARATION ENGINEERING

MISS WAN KHAIRUNNISA WAN RAMLI

OUTLINES

• INTRODUCTION• BASIC PRINCIPLES• EQUIPMENTS• OPERATING MODES • BASIC CALCULATION

SOLVENT/ LIQUID-LIQUID EXTRACTION

INTRODUCTION

ADVANTAGESLIMITATIONSBASIC PRINCIPLES & PARAMETERSSCALE UPDESIGN OF EXTRACTOR COLUMN

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.

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

d. Recovery of heat-sensitive materials. Eg: Antibiotics

e. Separation of a mixture according to chemical type rather than relative volatility

f. Separation of close-melting or close-boiling liquids; solubility differences exploited

g. Mixtures that form azeotrope

• The simplest LLE involves only a ternary system

• Feed2 miscible components: carrier, C + 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

• Need multi stages.Feed, F

Extract

Solvent, S

Rafinatte

BASIC PRINCIPLES

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

EXAMPLES OF KD VALUES

EXTRACTION EQUIPMENT

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 COLUMNIt 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 mixer-settlers 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.

OPERATING MODESBatch Extraction: i. The aqueous feed is mixed

with the organic solventii. 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)

PENICILIN

Continuous Extraction (Co-current & Counter-current): Schematic representations of (a) co-current (b) Counter-current operations:

CALCULATION METHODS

EXTRACTION OF DILUTES SOLUTION

1. Extraction factor, E is defined as:

From Eq. 1

2. For a single-stage extraction with pure solvent:

EXAMPLE 1

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) Since y0=0 (at initial no penicillin in solvent phase) So, L(x1)-L(x2) = V(y2) L(x1-x2)= V(y2)

0

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 complicated3 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.

TERNARY SYSTEM TYPE I1 immiscible pair

Solute, A

Solvent, S Carrier, C

Point E PLAIT POINT[The composition of extract & raffinate phases approach each other]

TIE LINE

Line ACE Extract phase Line BDE Raffinate phase

TRIANGULAR DIAGRAM

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 solvent2. 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

phase4. Extract phase: 0.232 acetone, 0.043 water, 0.725 MIK5. Raffinate phase: 0.132 acetone, 0.845 water, 0.023 MIK6. Ratio of acetone to water in the product = 0.232/0.043

= 5.47. Ratio of acetone to water in the raffinate = 0.132/0.845 = 0.156

0.2 Acetone

0.3

Wat

er

0.5 MIK

S

F

TERNARY SYSTEM TYPE II2 immiscible pairs

Extract phase

Raffinate phase

• 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)

Solute, A

Carrier, CSolvent, S

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)

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Rafinat Ekstrak

TERNARY SYSTEM TYPE I

SOLUTE, A

CARRIER, CSOLVENT, S

TERNARY SYSTEM TYPE II

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Rafinat

Ekstrak

SOLUTE, A

CARRIER, CSOLVENT, 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:

DESIGN CALCULATION

• Consider a countercurrent-flow, N-equilibrium-stage contactor for solvent extraction of a ternary system under isothermal, continuous & steady state flow

• 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

COUNTER-CURRENT EXTRACTIONN-EQUILIBRIUM STAGES

1 2

1 2F

F R R

x x xFeed

1 2 3

1 2 3

E E E

y y y

1

1

n n

n n

R R

x x

1

1

n n

n n

E E

y y

2 1

2 1

N N N

N N N

R R R

x x x

1

1 1

N N

N N N

E E S

y y y

1 2 n N-1 N

Extract Solvent

Raffinate

• The degree of freedom for liquid-liquid extraction for n-equilibrium 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. S & xN given2. S & y1 given3. y1 & xN given4. Number of equilibrium stages, N & xN given5. Number of equilibrium stages, N & y1 given6. Number of equilibrium stages, N & S given

COUNTER-CURRENT EXTRACTIONDEGREE OF FREEDOM

1 1 1F N M N NFx Sy Mx E y R x

1

F M

M N

S x x

F x y

• Solvent-to-Feed (S/F) and Extract-to-Raffinate (E/R) Flow rate ratios

(7.1)

(7.2)

(7.3)

• Overall total mass balance for control volume 1

1 NF S M E R • Overall mass balance of the component for control volume 1

11

1 1

N M

N M

x xE

R x y

1 2

1 2F

F R R

x x xFeed

1 2 3

1 2 3

E E E

y y y

1

1

n n

n n

R R

x x

1

1

n n

n n

E E

y y

2 1

2 1

N N N

N N N

R R R

x x x

1

1 1

N N

N N N

E E S

y y y

1 2 n N-1 N

Extract Solvent

Raffinate

1

(7.4)

COUNTER-CURRENT EXTRACTIONMATERIAL BALANCE

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Rafinat Ekstrak

OVERALL MATERIAL BALANCE

F

RN

E1

M

HUNTER-NASH METHODOVERALL MATERIAL BALANCE

HUNTER-NASH METHODOPERATING POINTS & LINES

1 1

1 1

E P E P F

E EFP

(7.5)

• OPERATING LINELocus of passing streams in a cascade• It is given by the difference between inlet and outlet streams of each

stage1 1 2 1... ...n n NF E R E R E R S P

1 2

1 2F

F R R

x x xSuapan

1 2 3

1 2 3

E E E

y y y

1

1

n n

n n

R R

x x

1

1

n n

n n

E E

y y

2 1

2 1

N N N

N N N

R R R

x x x

1

1 1

N N

N N N

E E S

y y y

1 2 n N-1 N

Ekstrak Pelarut

Rafinat

(7.6)

• 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.

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Rafinat Ekstrak

Pembinaan Titik Kendalian P

F

RN

E1

MP

Operation point

Operating line[S, Rn]

Operation line[F, E1]

HUNTER-NASH METHODOPERATING POINTS & LINES

COUNTER-CURRENT EXTRACTIONNUMBER OF EQUILIBRIUM STAGES

• 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.

*n ny f x (7.7)

• 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.

1 2

1 2F

F R R

x x xSuapan

1 2 3

1 2 3

E E E

y y y

1

1

n n

n n

R R

x x

1

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n n

n n

E E

y y

2 1

2 1

N N N

N N N

R R R

x x x

1

1 1

N N

N N N

E E S

y y y

1 2 n N-1 N

Ekstrak Pelarut

Rafinat

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Rafinat Ekstrak

Determination of number of equilibrium stages (2)

F

RN

E1

M

Operating point, P

Operating line

Operating line

COUNTER-CURRENT EXTRACTIONNUMBER OF EQUILIBRIUM STAGES

COUNTER-CURRENT EXTRACTIONMINIMUM SOLVENT-TO-FEED FLOWRATE RATIOS

• 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.

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F R R

x x xSuapan

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1 2 3

E E E

y y y

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n n

R R

x x

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n n

n n

E E

y y

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N N N

N N N

R R R

x x x

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N N

N N N

E E S

y y y

1 2 n N-1 N

Ekstrak Pelarut

Rafinat

5

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Determination of minimum solvent-to-feed flow rate ratio

F

RN

E1Smin

Operating line

P3 = P min

P2P1

Smin

MSmin’

COUNTER-CURRENT EXTRACTIONMINIMUM SOLVENT-TO-FEED FLOWRATE RATIOS

minmin

min min

S

S

M FS

F S M

• 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

(7.8)

• The difference in total outflow for Smin/F ratio

1 min min minS NF E R S P

1 2

1 2F

F R R

x x xSuapan

1 2 3

1 2 3

E E E

y y y

1

1

n n

n n

R R

x x

1

1

n n

n n

E E

y y

2 1

2 1

N N N

N N N

R R R

x x x

1

1 1

N N

N N N

E E S

y y y

1 2 n N-1 N

Ekstrak Pelarut

Rafinat

(7.9)

• The actual S/F is 1.2 – 2.0 times the SSmin/F

COUNTER-CURRENT EXTRACTIONMINIMUM SOLVENT-TO-FEED FLOWRATE RATIOS

• 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.

AQUEOUS TWO PHASE EXTRACTION

EXAMPLELight phase is water + 10% polyethylene glycol (PEG)

and 0.5% dextranHeavy phase is water + 1% glycol and 15% dextranProteins 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 weight2. Protein charge, surface properties3. Polymer(s) molecular weight4. Phase composition, tie-line length5. Salt effects6. 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

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1 2F

F R R

x x xSuapan

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1 2 3

E E E

y y y

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n n

R R

x x

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n n

n n

E E

y y

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N N N

N N N

R R R

x x x

1

1 1

N N

N N N

E E S

y y y

1 2 n N-1 N

Ekstrak Pelarut

Rafinat

COUNTER-CURRENT EXTRACTIONEXAMPLE 1

F

R S

E

Xn = 0.05

Xf = 0.3 y1

Yn = 0

Assume F = 50 g, so S = 50g

Material balance:F + S = M50 + 50 = 100

Material balance on A:Fxf + Syn = Mxm(50x0.3) + (50x0) = 100xmXm = 0.15

M = E1 + R1, R1 = M – E1Mxm = Ey1 + RxnMxm = E1y1 + (M-E1)xnE1 = M (xm-xn) / (y1 – xn)

SOLUTION

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.

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Aseton A

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Rafinat Ekstrak

S

F

M

R1

E1Xm = 0.15

COUNTER-CURRENT EXTRACTIONEXAMPLE 1

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 BALANCE1. Construct the equilibrium curve in triangular diagram &

connects the dot to form tie line2. Find operating point of the diagram by extrapolating tie

lines until they intercepts at P3. 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, M5. Find the F & S point in the diagram, connects the points

to form line FS6. Since we only know the fraction of solute, A in M, find

M(?, xm) on line FS7. Draw a tie line passing M & extended to point P8. 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

THANK YOU