extraction chapter

7/27/2019 Extraction Chapter

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Extraction

Introduction

An extraction process makes use of the partitioning of a solute between twoimmiscible or partially miscible phases. For example the antibiotic penicillin is more

soluble in the organic solvent methyl isobutyl ketone (MIBK) than in its aqueous

fermentation medium at acidic pH values. This phenomenon is utilized for penicillin

purification. When the extraction takes place from one liquid medium to another, the

process is referred to as liquid-liquid extraction. When a liquid is used to extract

solutes from a solid material, the process is referred to as solid-liquid extraction or

leaching. In this chapter we will mainly discuss liquid-liquid extraction. When a

supercritical fluid is used as an extracting solvent, the process is referred to as

supercritical fluid extraction (SFE). SFE will be briefly discussed at the end of this

chapter.

Typical applications of extraction in bioprocessing include:1. Purification of antibiotics2. Purification of alkaloids3. Protein purification using aqueous two-phase systems4. Purification of peptides and small proteins5. Purification of lipids6. Purification of DNA

Solvent systemsSolvent extraction as used in the bio-industry can be classified into three types

depending on the solvent systems used:

1. Aqueous/non-aqueous extraction2. Aqueous two-phase extraction3. Supercritical fluid extraction

Low and intermediate molecular weight compounds such as antibiotics, alkaloids,

steroids and small peptides are generally extracted using aqueous/non-aqueous solvent

systems. Biological macromolecules such as proteins and nucleic acids can be

extracted by aqueous two-phase systems. Supercritical fluids are used as extracting

solvents where organic or aqueous solvents cannot be used satisfactorily.

Ideally, the two solvents involved in an extraction process should be immiscible.

However, in some extraction processes partially miscible solvent systems have to beused. For partially miscible solvent systems, particularly where the solute

concentration in the system is high, triangular or ternary phase diagrams such as

shown in Fig. 7.1 are used. In such diagrams the concentration of the components are

usually expressed in mole fraction or mass fraction. Fig. 7.1 shows the phase diagram

for a solute A, its initial solvent B and its extracting solvent C. Such phase diagrams

rely on the fact that all possible composition of the three components can be

represented by the area within the triangle. The composition of the mixture

represented by point H on the diagram is such that content of A is proportional to HL,

content of B is proportional to HJ and content of C is proportional to HK. The curve

shown in Fig. 7.1 is called the binodal solubility curve. The area under the curve

represents the two-phase region. Any mixture represented by a point within thisregion will split up into two phases in equilibrium with each other. For a mixture


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having an overall composition H, the composition of the two phases are represented

by points P and Q which are obtained from the points of intersection of the binodal

solubility curve with the tie line PQ which passes through H. The tie lines are straight

lines which connect together the compositions of the two phases, which are in

equilibrium with one another. These tie lines are experimentally determined. The

point F on the binodal solubility curve is called the plait point. The area above thebinodal solubility curve represents the single-phase region where all three components

in the system are mutually miscible.

Theory of extractionThe different components involved in an extraction processes are summarized in Fig.

7.2. The feed consists of the solute to be extracted in its original solvent e.g. penicillin

in fermentation media. The extracting solvent (e.g. MIBK) is the phase to which the

solute (i.e. penicillin in this case) is to be transferred and device within which this

transfer of solute takes place is called an extractor. The raffinate is the spent feed

while the extract is the enriched extracting solvent.

The distribution of a solute between the raffinate and the extract can be expressed in

terms of the partition coefficient K:


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Where

CE = equilibrium solute concentration in extracting solvent (kg/m3)

CR= equilibrium solute concentration in raffinate (kg/m3)

Extraction Factor

The extraction factor, , is defined as:

= amount of solute in extracting solvent/amount of solute in raffinate

= ECE/RCR

= K(E/R)

Where: E and R are the volume of extracting solvent and initial solvent, respectively.

The value of K is often independent of the solute concentration, particularly at low

solute concentrations. However, at higher solute concentrations, deviation from the

linearity between CR and CE may be observed in some systems. The partition

coefficient of a solute between two phases is usually determined by experimental

methods.

Liquid-liquid extraction involves transfer of solute from one liquid phase to another.

The three basic steps common to all liquid-liquid extraction processes are:

Mixing or contacting

Transfer of solute between two partially or completely immiscible liquids requiresintimate contact of the two. This is usually achieved by dispersing one liquid (the

dispersed phase) as tiny droplets in the other liquid (i.e. the continuous phase). Solute

transport rate depends on the interfacial mass-transfer coefficient which depends on

the hydrodynamic conditions in the system and on the available contact area.

Vigorous mixing can achieve both these requirements.

Phase separation or settlingOnce the desired extent of solute transport has been achieved (i.e. equilibrium has

been reached), the next step is to allow the droplets of the dispersed phase to coalesce.

This eventually leads to the separation of the two liquids into distinct layers due to

density difference. In a liquid-liquid extraction process, it is important that the twophases (i.e. raffinate and the extract) should have sufficient density difference to

facilitate segregation of phases.

Collection of phasesAfter settling (or phase separation), the extract and raffinate phases are collected as

separate streams by appropriate means.

Efficiency of Extraction

The fraction extracted, p, can be defined as:

p = amount of solute in extracting solvent/total amount of solute in extracting solventand raffinate


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p = ECE/(ECE + RCR)

Dividing both nominator and denominator by RCR

p = (E/R)(CE/CR)/{ (E/R)(CE/CR) + 1}

p = /( + 1)Or:

p = K/ [K + (R/E)]

p = KE/ [KE + R]

Using the same approach, the fraction of solute remaining in raffinate, q, can be

shown to be equal to:

q = 1/( + 1) = 1/ [K + (R/E)]

q = R/ (KE + R)

Evidently,

p + q = 1

The efficiency of a single extraction depends on the magnitude of K and on the

relative volumes of the liquid phases. The percentage extraction is given by:

= 100 K/ [K + (R/E)]

where R and E are the volumes of the aqueous and organic phases respectively, or

= 100 K/ [K + 1]

when the two phases are of equal volume.

It is also easy to recognize that the fraction remaining in the original solution after n

extractions will be :

Fraction remaining after n extraction = qn

Therefore, the extracted fraction, E, is:

E = 1 - qn

The percent extracted after n extractions is:

%E = (1 - qn)*100

Taking into account that q = R/ (KE + R)


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If K is large, i.e. > 102, a single extraction may effect virtually quantitative transfer of

the solute, whereas with smaller values of K several extractions will be required.

p = K/ [K + (R/E)]

For R = E

P = 100/(100 + 1) = 0.99

%E = 0.99 * 100 = 99%

The amount of solute remaining in the aqueous phase is readily calculated for any

number of extractions with equal volumes of organic solvent from the equation:

CRn = CR0 [R/(KE + R)]n

CRn = CR0 [1/( + 1)]n

Or simply:

CR =CR0 * qn

where CRn is the amount of solute remaining in the aqueous phase, volume R , after n

extractions each with volume E, of organic phase, and CR0 is the amount of solute

originally present in the aqueous phase.

Large volume extraction versus small volume multi extractions

If the value of K is known, the equation above is useful for determining the optimum

conditions for quantitative transfer. Suppose, for example, that the complete removal

of 0.1 g of iodine from 50 mL of an aqueous solution of iodine and sodium chloride is

required. Assuming the value of K for I2 in a carbon tetrachloride/water system is 85,

then for a single extraction with 25 mL of CCl4

q = R/ (KE + R)

q = 0.023

%E = (1 - qn)*100

%E = (1 {[ R/ (KE + R)]}1

*100%E = (1 {[ 50/ (85*25 + 50)]}

1 *100

%E = 97.7%

Amount remaining can also be calculated:

CR1 = CR0 [R/(KE + R)]1

CR1 = 0.1[50/(85*25 + 50)]1

CR1 = 0.0023 g in 50 mL

Assume we are looking for a 99% extraction using one extraction, calculate the

volume of extracting solvent to be used with the 50 mL of the aqueous iodine solution

described above.


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99 = (1 {[ 50/ (85*E + 50)]}1 *100

0.01 = {[ 50/ (85*E + 50)]}

0.017E + 0.01 = 1

E = 58.2 mL

Since, 97.7% of the I2 is extracted in one extraction with 25 mL CCl4, for threeextractions with 8.33 mL of CCl4, %E of the I2 extracted can be calculated as follows:


CR3 = 0.1[50/(85*8.33 + 50)]3

CR3 = 0.1[0.066]3 = 0.1 * 2.87*10-4 = 2.87*10-5 g

q = R/ (KE + R)

q = 0.066

q3 = 2.87*10-4

%E = (1 {[ 50/ (85*8.33 + 50)]}3 *100

%E = 99.97%

One can also calculate amount remaining:


CR3 = 0.1[50/(85*8.33 + 50)]3

CR3 = 0.1[0.066]3 = 0.1 * 2.87*10-4 = 2.87*10-5 g

If we are to use 6 mL of CCl4 in each of multiple extractions so that 99% is extracted,

how many extractions are necessary?

q = R/ (KE + R)q = {[ 50/ (85*6 + 50)]}

q = 0.089

99 = (1 (0.089)n)*100

0.99 = 1 (0.089)n

(0.089)n = 0.01

n = 1.91

This means that two extractions with 6 mL each are enough to obtain 99% of iodine

from the aqueous solution. This is less than half the volume needed for one extractionto obtain the same percentage extraction of the iodine.

It is clear therefore that extracting several times with small volumes of organic

solvent is more efficient than one extraction with a large volume. This is of particular

significance when the value of K is less than 102.

Selectivity of Extraction

Often, it is not possible to extract one solute quantitatively without partial extraction

of another. The ability to separate two solutes depends on the relative magnitudes of

their distribution ratios. For solutes A and B, whose distribution ratios are KA and KB,the separation factor is defined as the ratio KA/KB where KA>KB. Assuming that KA


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= 102, for different values of KB and . For an essentially quantitative separation

should be at least 105.

A separation can be made more efficient by adjustment of the proportions of organic

and aqueous phases. The optimum ratio for the best separation is given by the Bush-

Densen equation

(E/R) = (1/ KAKB)1/2

Successive extractions, whilst increasing the efficiency of extraction of both solutes,

may lead to a poorer separation. For example, if KA = 102 and KB = 10

1, one

extraction will remove 99.0% of A and 9.1% of B whereas two extractions will

remove 99.99% of A but 17% of B.

After 1stextraction, assuming equal volumes of E and R

qB = 1/(K + 1)

qB = 1/(0.1 + 1) = 0.91

%EB = (1 - qn)*100%EB = (1 0.91)*100 = 9.1%

qA = 1/(K + 1)

qA = 1/(100 + 1) = 0.0099

%EA = (1 - qn)*100

%EA = (1 0.0099)*100 = 99%

After the 2nd extraction

%EB

= (1 (0.91)2)*100 = 17%

%EA = (1 (0.0099)2)*100 = 99.99%

Although two extractions removed 99.99% of A, the extracting solvent contains also

17% of B.

This means that increasing the number of extractions will affect the purity of the

better extracted solute negatively. This is where the concept of reverse extraction may

prove very efficient.

A good approach is to choose the volume ratio which should be used. For example, if

we have 100 mL of aqueous feed solution containing compounds A and B, where ifKA = 102 and KB = 10

1, then we have:

(E/R) = (1/ KAKB)1/2

(E/100) = (1/102*0.1)1/2

(E/100) = 0.316

E = 31.6 mL

In practice, a compromise must frequently be sought between completeness of

extraction and efficiency of separation. It is often possible to enhance or suppress the

extraction of a particular solute by adjustment of pH or by complexation. This

introduces the added complication of several interrelated chemical equilibria which

makes a complete theoretical treatment more difficult.


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pH Effect

Consider the extraction of a carboxylic acid from water into ether. At low pH, where

the acid is undissociated, CHA ~ [HA] and the acid is extracted with greatest

efficiency. At high pH, where dissociation of the acid is virtually complete, K

approaches zero since almost all the acid is dissociated and will not partition in theorganic phase, and extraction of the acid is thus negligible.

Effect of Complex Formation

Returning to the extraction of iodine from an aqueous solution of iodine and sodium

chloride, the effect of adding iodide to the system is to involve the iodine in formation

of the triiodide ion. Thus, the presence of iodide affects K in such a way that at very

low concentrations CI2 ~ [I2]and iodine is extracted with greatest efficiency. At high

iodide concentrations, Kf[I]aq >> 1, which means that most I2 is converted to I3

-, and

K is reduced with a consequent reduction in the extraction of iodine.

K = CEI2/CRI2

In presence of iodide:CRI2 = [I2] + [I3

-]

Most iodine is converted to triiodide which make the value of CRI2 very large and

facilitates transfer of I2 from extracting solvent to aqueous phase, in order to form the

complex at high iodide concentration. This very much lowers K.

Effect of Association

The distribution ratio is increased if association occurs in the organic phase.

Carboxylic acids form dimers in solvents of low polarity such as benzene and carbon

tetrachloride. Dimerization is only slight in oxygenated solvents and extraction into

them is therefore less efficient than into benzene or carbon tetrachloride.

Craig Counter Current Extraction

Discontinuous counter-current distribution is a method devised by Craig, which

enables substances with close distribution ratios to be separated. The method involves

a series of individual extractions performed automatically in a specially designedapparatus. This consists of a large number (50 or more) of identical interlocking glass

extraction units (see figure below) mounted in a frame which is rocked and tilted

mechanically to mix and separate the phases during each extraction step.

A Craig apparatus consists of a series of glass tubes (r: 0, 1, 2..) that are designed and

arranged such that the lighter liquid phase can be transferred from one distribution

tube to the next. The liquid-liquid extractions are taking place simultaneously in all

tubes of the apparatus.


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The lower (heavier) phase of the two-phase solvent system (e.g. water, blue layer in

the picture) is the "stationary phase", whereas the upper (lighter) phase (e.g. hexane,

red layer in the picture) is the "mobile phase". In the beginning, tube # 0 contains themixture of substances to be separated in the heavier solvent and all the other tubes

contain equal volumes of the pure heavier solvent phase. Next, the lighter solvent is

added to tube # 0, extraction (equilibration) takes place, and the phases are allowed to

separate. The upper phase of tube # 0 is then transferred to tube # 1, fresh solvent is

added to tube # 0, and the phases are equilibrated again. The upper layers of tubes # 0

and # 1 are simultaneously transferred to tubes # 1 and # 2 respectively. This cycle is

repeated to carry on the process through the other CCD elements.

In this system, substances with a higher distribution ratio move faster than those with

a lower distribution ratio. In order to understand the whole process, it is helpful to

examine the distribution of a substance A in each tube after a given number of

equilibration/transfer cycles. Assuming that the volumes of each solvent phase are

equal (V), p and q represent the fraction of A with distribution ratio of D.

The fractions of solute in successive tubes after each extraction step are shown in thefollowing figure:


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Tube No. 0 1 2 3

0Before

equilibration 1

PAfterequilibration q

E

xtraction

First transfer, n = 1

Tube No. 0 1 2 3

0 pBefore

equilibration q

pq p2After

equilibration q

2

pq

Extraction

The resulting fractions are: q2 + 2pq + p2 = (p + q)2

After the second transfer, n = 2

Tube No. 0 1 2 3

0 pq p2Before

equilibration q2 pq

pq2 2p2q p3After

equilibration q3 2pq2 p2qExtraction

After the 3rd transfer, n = 3

Tube No. 0 1 2 3

0 pq2 2p2q p3Before

equilibration q3 2pq2 p2q

Fractions q3 3pq2 3p2q p3 Extraction

The resulting fractions are: q3 + 3pq2 + 3p2q + p3 = (p + q)3

After the nth transfer we have:

The fractions distributed according to the binomial (p + q)n

We observe that, after n transfers/equilibration cycles, and since the ratio D = p/q

must be maintained for each tube after the equilibration step, the total fraction of A in

each tube corresponds to the terms of the binomial expansion (p+q)n. Therefore, the

total fraction of a solute in tube r after n transfers is given by:


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By combining with the previous expressions of p and q, we finally obtain

A general rule of CCD is that the greater the difference of the distribution ratio of

various substances, the better the separation between each other. A much largernumber of tubes is required to separate mixtures of substances with close distribution

ratios.

Today, the Craig apparatus is only very rarely used (as instruments are hardly

available any more), mostly because of the efficiency and convenient handling of

modern chromatographic instruments. However, the principle of countercurrent

extraction provides a very useful educational and scientific example, as it introduces

the fundamental concepts of equilibration between mobile and stationary phases. In

CCD, each tube in which a full equilibration can take place corresponds to one

theoretical plate of a chromatographic column.

Aqueous two-phase extractionAqueous two-phase extraction which is a special case of liquid-liquid extraction

involves transfer of solute from one aqueous phase to another. The two immiscible

aqueous phases are generated in-situ by addition of substances such as polymers and

salts to an aqueous solution. Historically, gelatin-agar and gelatin-soluble starch were

used. Two types of aqueous two-phase systems are commonly used:

1. Polymer-polymer two-phase system2. Polymer-salt two-phase system

A polymer-polymer two phase system can for instance be obtained by mixing dextranand PEG at a certain composition. By adding specific amounts of these polymers to an

aqueous feed phase (which contains the solute), two aqueous phases, one rich in PEG

and the other rich in dextran can be obtained. Aqueous two-phase systems can also be

generated using a polymer (e.g. PEG or dextran) and a salt such as sodium or

potassium phosphate. Aqueous two-phase separations take place at certain

compositions only. The figure below shows a PEG-dextran phase diagram where a

solubility curve separates the two-phase region (above the curve) from the single

phase region (below the curve). Such "binary" phase diagrams which are based on the

compositions of the two polymers (or polymer and salt) are used for determining the

concentrations needed for an extraction process. These phase diagrams also predict

the polymer/salt content of the raffinate and the extract phases. The composition of

the individual phases generated can be obtained using tie-lines as shown in the figure.


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Factors Affecting Protein Partitioning in Two-Phase Aqueous Polymer Systems

1. Protein molecular weight2. Protein charge, surface properties3. Polymer molecular weight4. Phase composition5. Salt effects6. Affinity ligands attached to polymers7. pH8. Temperature

In PEG/dextran aqueous two-phase extraction of proteins, the partition behavior

depends to a great extent on the relative polymer composition. It also depends on the

solution pH and the molecular weight of the protein. Generally speaking, proteinpartitioning into the PEG rich phase is favored. When a polymer-salt combination is

used to generate the aqueous two-phase system, a protein partitions favorably into the

polymer rich phase. The general scheme for aqueous two-phase extraction is shown in

Fig. 7.4.

Extraction by an ATPS offers advantages for processing on a large scale, such as the

possibility of obtaining a high yield, the possibility of continuous processing and a

reduction in operational cost in relation to the costs of conventional processes.

PEG-DEXTRAN SYSTEMS

Effect of Polymer Molecular Mass (MM)

An increase in the molecular mass of dextran or of PEG will lower the concentration

required for phase separation. The polymer molecular mass influences protein

partitioning as a direct result of interactions between the two polymers. It has been

found that the partitioned protein behaves as if it were more attracted by smaller

polymer sizes and more repelled by larger polymers, provided all other factors such as

polymer concentrations, salt composition, temperature and pH are kept constant. It

was observed that smaller protein molecules and amino acids were not affected as

much as larger ones. For some proteins the partition coefficients increased as the MM

of dextran increased if all other conditions were kept constant, but little effect was

found for low MM proteins (Cytochrome C, 16,000). When the same proteins were

partitioned in systems with different PEG MM, their partition coefficients decreasedas the PEG MM increased, and for cytochrome C the effect was the smallest. This was


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attributed to the fact that when the PEG MM is increased, a weaker repulsion energy

is required to cause phase separation. Repulsive interactions between the polymer and

the protein become stronger as the polymer MM is increased, resulting in a

distribution of the protein towards the phase containing the polymer with an

unchanged MM. A weak net repulsion between the proteins and the polymer is

sufficient to change the distribution when the polymer MM is changed.

Effect of Polymer Concentration

It was found that proteins with MM less than 20,000 showed a linear relationship

between the ln K in PEG-dextran systems and a difference in PEG concentration

between the phases, for any particular system. They found that it was possible to

predict the partitioning of a protein at any concentration in that particular system if

one partition coefficient in the system were known.

However, others found that for some proteins the partition coefficient was inversely

correlated to phase concentration in a PEG-dextran system, showing that betterseparation could be achieved at high polymer concentrations. This, however, may also

affect the concentration of proteins that can be manipulated in the system as polymer

concentration has a directly inverse effect on protein solubility.

Effect of Salts

Salts can affect protein partitioning in different ways in PEG-dextran systems: one is

by altering the physical properties of the systems the hydrophobic difference between

the phases and the other is by the partitioning of ions between the phases, which

affects the partitioning of proteins according to their molecular charge.

Salts have been added to PEG-dextran systems to increase the selectivity of protein

partitioning in the aqueous two-phase methodology application for biological

separations.

It was observed that salt ions partition differently between the phases, causing an

uneven distribution in the system that generates a difference in electrical potential

between the phases. This difference in electrical potential would be independent of

salt concentration, but linearly dependent on the partition behaviour of the ions.

It was also observed that polyvalent anions such as phosphate, sulphate and citratepartitioned preferentially into dextran-rich phases, while halides partitioned nearly

equally. As an example, negatively charged materials have higher partition

coefficients in phases containing sodium sulphate rather than sodium chloride, while

the reverse holds for positively charged materials. Partition coefficients of negatively

charged materials decrease when the cationic series is changed from lithium to

sodium to potassium. The ratio between the phosphate ions, rather than the

concentration, was decisive for the difference in electrical potential. This applies to

multivalent ions, which show a series of pH-dependent dissociations and was clearly

the reason for the potential difference found between the two phases (Kula et al.,

1982).

PEG-SALT SYSTEMS


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The formation of PEG-salt systems was first observed in the 1950s, but the theoretical

fundamentals have not been well explained. It was found that for PEG solutions the

addition of some inorganic salts (sulphates and carbonates) is more effective than the

addition of others in reducing the critical concentration of cloud point curves. These

inorganic salts dramatically reduced the PEG cloud point at high temperatures.

PEG-salts systems have been introduced for the practical application of large-scale

protein separation because of the larger droplet size, greater difference in density

between the phases, lower viscosity and lower costs, leading to a much faster

separation than in PEG-dextran systems. Industrial application of PEG-salt systems

was improved by the availability of commercial separators, which allowed faster

continuous protein.

Initially PEG-phosphate systems were widely used where scientists have studied ways

of recycling the phosphate phase of the systems to minimize environmental pollution.

The recycling of the phosphate phase was achieved by its separation from the solids

by the use of alcohols. PEG from the top PEG-rich phase can also be successfullyrecycled.

More recently PEG-sulphate systems have begun to be used where separation of some

biomaterial was achieved with PEG 4000 and (NH4)2SO4 at pH 7-7.5. The presence of

2% NaCl (0.17 M) made the separation much worse. With 4% NaCl (0.34 M), a poor

separation was obtained (a tenfold decrease in K for aspartase). Since a pH or phase

ratio change was not observed, the dramatic change in K was considered to be due to

a change in hydrophobicity between the phases.

AFFINITY PARTITIONING

In the last 30 years, several groups have studied methods to increase partitioning by

the use of biospecific interactions in ATPSs.

The initial works on affinity partitioning in ATPSs were to purify trypsin by using

PEG-bound ligand p-aminobenzamidine and S-23 myeloma protein by using

dinitrophenol as ligand.


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The degree of affinity partitioning, Kaff, can be described by the ratio between thepartition coefficients of proteins with and without a ligand:

This equation describes the increase in the partition coefficient of a protein by the

binding of a specific ligand to the PEG-rich phase.

Affinity partitioning results in specific extractions of proteins, nucleic acids,

membranes, organelles and even cells, mainly when biospecific ligands are used.

Large Scale Extraction SchemesExtraction processes can be divided into two general schemes:

Batch extractions Continuous extractions,

Continuous extractions can also be further divided into the following schemes:

Single stage continuous extraction

Multi stage continuous extraction

In turn, multi stage continuous extraction can be divided into two general modes as:

Crosscurrent continuous extraction Counter current continuous extraction

These will be studied in the following sections.

Batch extraction

In a batch extraction process a batch of feed solution is mixed with a batch of

extracting solvent in an appropriate vessel. The solute distributes between the twophases depending on its partition coefficient. The rate at which the transfer of solute


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takes place from the feed to the extracting solvent depends on the mixing rate. Once

equilibrium is attained, the mixing is stopped and the extract and raffinate phases are

allowed to separate. The separation funnel commonly seen in chemistry laboratories is

the simplest small-scale batch extraction device. Mixer-settler units are usually used

for large-scale batch extraction. The basic principle of batch extraction using a mixer

settler unit is shown in Fig. 7.5. The mixer unit must be able to generate highinterfacial area, must provide high solute mass transfer coefficient and cause low

entrainment of air bubbles. The settler unit must have a low aspect ratio (L/D), i.e. be

of flat geometry, must allow easy coalescence and phase separation, and must allow

for easy collection of raffinate and extract as separate streams. The antibiotic

penicillin partitions favorably in an organic solvent from an aqueous fermentation

media at acidic conditions. However, at a neutral pH, the partitioning from organic

phase to aqueous phase is favored. Thus the antibiotic could be purified by sequential

reversed batch extraction, where the antibiotic is moved from aqueous to organic

phase and back again (as shown in Fig. 7.6). This sequence is usually repeated a few

times in order to obtain highly pure antibiotic.


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If a batch of feed containing R volume of initial solvent and an initial solute concentration ofC0 is mixed with S volume of pure extracting solvent, the concentration distribution in the

extract and the raffinate at equilibrium is given by:

CE = KCR

WhereCE solute concentration in extract (kg/m3)

CRsolute concentration in raffinate (kg/m3)

By performing a solute material balance, we get:

RCR0 = RCR+ ECE

E and R are the volumes of extracting solvent and raffinate, respectively.

Therefore, from the two equations above, dividing the first equation by R gives:

CR0 = CR+ (E/R) CE

K = CE/CR , or

CR= CE/K

Therefore we have

CR0 = CE/K + (E/R) CE

CR0 = CE {(1/K) + (E/R)}

Multiply both sides by K

KCR0 = CE {1 + K(E/R)}

The extraction factor() is defined as:

= K(E/R)

KCR0 = CE {1 + }

We then have:

The fraction extracted is given by:

Also, fraction remaining in raffinate can be calculated from the relation:

q = 1/(1+)


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Always, we have:

p + q = 1

Example100 litres of an aqueous solution of citric acid (concentration = 1 g/1) is contacted

with 10 litres of an organic solvent. The equilibrium relationship is given by C E = 100CR

2, where CR and CE are the citric acid concentrations in the raffinate and extract

respectively and are expressed in g/1. Calculate:

a) The concentration of citric acid in the raffinate and the extract.

b) The fraction of citric acid extracted.

Solution

A citric acid mass balance for the first batch extraction gives:

RCR0 + SCS = RCR+ ECE

However, Cs = 0 since the initial organic solvent is solute free.Therefore we have:

RCR0 = RCR+ ECE

100 * 1 = 100CR+10CEThe equilibrium relationship is:

CE = 100 CR2

By substitution we get:

CE = 7.298 g/1

CR= 0.270 g/1

The fraction of citric acid extracted in the first batch extraction is:

P= (7.298*10)/(100*1) = 0.7298

Example, Continued

If the extract thus obtained is then contacted with a further 100 litres of aqueous

solution of citric acid (concentration = 1 g/1) calculate:

c) The concentration of citric acid in the raffinate and extract phases of the second

extraction. Comment on these results.

A citric acid mass balance for the second batch extraction gives:


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RCR0 + ECE1 = RCR+ ECE

(100 x l) + (10 x 7.298) = 100CR+ 10CE

Since we have CE = 100 CR2

Then by substitution we get:

CE= 13.609 g/1

CR= 0.369 g/1

The amount of citric acid extracted in the second extraction was lower than in the

first. This is due to the fact that extracting solvent already had some citric acid in it

and consequently the concentration driving force was lower.

The approach discussed above is not suitable when the equilibrium relationship

between the extract and raffinate concentration is non-linear. For non-linear

equilibrium functions, a graphical method is preferred. This graphical method is based

on plotting the equilibrium function and the material balance on a graph. Theequilibrium function which is also called the equilibrium line is of the form shown

below:

CE =f(CR)

The solute material balance equation can be written as shown below:

RCR0 = RCR+ ECE

Dividing by R, and rearrangement we get:

CE = (R/E)[CR0-CR]

CE = (R/E)CR0 - (R/E)CR

(R/E)CR0 is constant for a specific E, R, and CR0

CE = b - (R/E)CR

Graphical solution for batch extraction using immiscible solvents

This method is more useful when the equilibrium relationship between extract and raffinateconcentration is non-linear. The graphical method is based on drawing the equilibrium lineand the operating line as shown in the figure below.

Equilibrium line is given by:

CE =f(CR)

The operating line is obtained from material balance:

CE = b - (R/E)CR


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Graphical solution for batch extraction

The point of intersection of these two lines gives the equilibrium concentrations in the extractand raffinate phases.

Graphical solution of the first part of the example above, step by step:

1. Draw the equilibrium line from

CE = 100 CR2


2. Locate b on the graph (intercept) where b = (R/E)CR0

b = (100/10) * 1 = 10

E uilibrium line

Operating line

CR

CE

RCR0/E = b

Slope = -R/E

E uilibrium line

CR

CE


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3. From the slope (-R/E) and the point at the intercept draw the operational line:Slope = -(100/10) = -10

4. The point of intersection of these two lines gives the equilibrium concentrations in theextract and raffinate phases.

Continuous Extractions

1. Single-stage continuous extraction using immiscible solventsThe single-stage continuous extraction process is merely an extension of the batch

extraction process. Continuous feed and extracting solvent streams enter the mixer

unit. The transfer of solutes takes place within the mixer and residence time within

this device should be sufficient for complete extraction. The emulsion produced in themixer unit is fed into a settler unit where phase separation takes place and continuous

E uilibrium line

CR

CE

b = (R/E)CR0

CR

E uilibrium line

Operating line

CE

RCR0/E = b

Slope = -R/E


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raffinate and extract streams are thus obtained. The analytical solution of the single

stage continuous extraction using completely immiscible solvent systems is based of

the following assumptions:

1. The exit streams from the extractor i.e. the raffinate and extract are inequilibrium

2. There is negligible entrainment of the other phase with each exitstreamComposing material balance relationship, we get:

FCF + SCS = RCR+ ECE

Assume F = R, and S = E

Also, if a fresh extracting solvent is used, we have CS = 0,

therefore, we have:

RCF = RCR+ ECE

Where, in this case, E and R represent flow rates of extracting solvent and feed,respectively.

Dividing the first equation by R gives:

CF = CR+ (E/R) CE

K = CE/CR , or

CR= CE/K

Therefore we have

CF = CE/K + (E/R) CE

CF = CE {(1/K) + (E/R)}Multiply both sides by K

KCF = CE {1 + K(E/R)}

The extraction factor for single stage continuous extraction is defined as:

= K(E/R)

KCF = CE {1 + }

CE = KCF/{1 + }

CR= CF/{1 + }


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When the equilibrium function is non-linear, a graphical method is used, the approach

being similar to that shown in the previous example. The operating line is of the form

shown below:

CF = CR+ (E/R) CE , multiply by (R/E)

(R/E)CF = (R/E)CR+ CE

CE = (R/E)CF - (R/E)CR

CE = b - (R/E)CR

b = (R/E)CF

CE = (R/E)CF - (R/E)CR

CE = (R/E){CF CR}

ExampleFermentation broth enters a continuous mixer settler extraction unit at a flow rate of

100 1/min. This contains 20 g/1 antibiotic and its pH has been adjusted to 3.0. Butyl

acetate which is used as the extracting solvent enters the extractor at a flow rate of 10

1/min. At pH 3.0 the equilibrium relationship is given by CE = 40CR, where CRand CE

are the antibiotic concentrations in the raffinate and extract respectively and are

expressed in g/1. Calculate:

a) The antibiotic concentration in the extract and the raffinate.

b) The fraction of antibiotic extracted.

Solution

The separation process is summarized in the Figure below:


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An antibiotic mass balance for the first stage gives us:


Cs = 0

(100*20) = 100CR+ 10CE

The equilibrium relationship is:

CE = 40CR(100*20) = 100CR+ 10*40CR

Solving for CRequation we get:

CE =160g/1

CR= 4g/1

The fraction of antibiotic extracted in the first stage is:

P= ECE/RCF

Graphical solution:1) Draw the equilibrium line where CE = 40CR2) Locate the intercept point at b = (R/E)CF = (100/10)*20 = 2003) Find the slope = -(R/E) = - (100/10) = -104) Draw the operational line using the point at the intercept and the slope5) From the intersection between the equilibrium and operational lines find CE

and CR. Same results are obtained

Continue Example

The organic extract from the first extraction step enters a second continuous mixer

settler extraction unit. An aqueous extracting solvent phase at pH 7.0 is fed into this

extractor at a flow rate of 5 1/min. At pH 7.0 the equilibrium relationship is given by

CE = 37CR, where CE is the antibiotic concentration in the aqueous extract phase while

CRis the concentration in the organic raffinate phase. Calculate:


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c) The concentration of the antibiotic in the extract and raffinate from second

extraction.

d) The overall fraction of antibiotic extracted.

From an antibiotic mass balance for the second stage we get:

ECE + SCS = R2CR2 + E2CE2

Assume E = R2 , S = E2 , and CS = 0 due to using a fresh solvent

ECE = R2CR2 + E2CE2

(10*160) = 10 CR2 + 5 CE2

CE = 37CR

(10*160) = 10 CR2 + 5 *37 CR2

WhereCE2 = antibiotic concentration in aqueous extract

CR2 = antibiotic concentration in organic raffinate

Solving for CR2 we get:

CE2 = 303.6 g/1 CR2 = 8.205 g/1

The overall fraction of antibody extracted is:

P= amount extracted/total amount

Graphical solution:

1) Draw the equilibrium line where CE = 37CR2) Locate the intercept point at b = (R/E)CF = (10/5)*160 = 3203) Find the slope = -(R/E) = - (10/5) = - 24) Draw the operational line using the point at the intercept and the slope5) From the intersection between the equilibrium and operational lines find CE

and CR. Same results are obtained

2. Cross-current continuous extraction using immiscible solvents

Any extraction process is limited by the equilibrium constant. In order to recover

residual solute from the raffinate stream this could be sent to one or several

subsequent stages for further extraction. The multi-stage extraction scheme shown in

Fig. 7.10 is referred to as cross-current extraction. Fresh extracting solvent is

generally fed into each stage and a corresponding extract is obtained. All extract

streams are combined together to obtain the overall extract while the overall raffinate

is obtained from the last stage. A cross-current extraction system is usually solved in

stages, i.e. starting with the first stage, obtaining the raffinate and extract


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concentrations and forward substituting these values. Some of the basic assumptions

are:

1. The raffinate and extract streams from each stage are in equilibrium2. There is negligible entrainment of the other phases3. The same equilibrium relationship holds good for all the stages,

i.e. the value of K is independent of solute concentration For i = 1 to n:

Material Balance

ECE = E1CE1 + E2CE2 + E3CE3The final volume of extracting solvent is:

S1 + S2 + S3 = E(Input 1) + (input 2) + (input 3) = (output 1) + (output 2) + (output 3)

(FCF + S1CS) + (R1CR1 + S2CS) + (R2CR2 + S3CS) = (E1CE1 + R1CR1) + (E2CE2 +

R2CR2) + (E3CE3 + R3CR3)

Since a fresh extracting solvent is used in each step, CS = 0, omitting similar terms on

both sides gives:

FCF = E1CE1 + E2CE2 + E3CE3 + R3CR3

ECE = E1CE1 + E2CE2 + E3CE3FCF = ECE + R3CR3


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It should be clear that:

F = R1 = R2 = R3 = R, and

E is the total volume of extracting solvent and CE is the total extracted concentration,

and CR3 is the concentration of solute remaining in the feed, which can be generally

taken as CR. Therefore, we have:

RCF = ECE + RCR

RCF = RCR+ ECE

Where, E and R represent flow rates of extracting solvent and feed, respectively.

Dividing the first equation by R gives:

CF = CR+ (E/R) CE

K = CE/CR , or

CR= CE/KTherefore we have

CF = CE/K + (E/R) CE

CF = CE {(1/K) + (E/R)}

Multiply both sides by K

KCF = CE {1 + K(E/R)}

Since = K(E/R)

KCF = CE {1 + }

CE = KCF/{1 + }

CR= CF/{1 + }

The major disadvantage of using cross-current extraction is the diminishing

concentration driving force.

3. Staged counter-current extraction

The drawback of diminishing concentration driving force observed in cross-current

extraction can be eliminated using staged counter-current extraction. This refers to

extraction in the form of a chain of counter-current cascades. As with cross-current

extraction the individual elements of the cascade are referred to as stages. Fig. 7.11

shows an idealized staged counter-current scheme. The feed stream enters the nth

stage and the extracting solvent stream enters the 1st stage. The raffinate phase is

collected from the 1st stage while the extract phase is collected from the nth stage.

Using this scheme the concentration driving force is maintained more or less uniform

in all the stages comprising the cascade. Equipment used for staged counter-current

extraction varies widely. Fig. 7.12 shows a staged counter-current extraction set-uphaving three stages made up of mixer-settler units.


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Some of the basic assumptions needed for obtaining an analytical solution for staged

counter-current extraction are:

1. The raffinate and extract streams from each stage are in equilibrium2. There is negligible entrainment of the other phases3. The same equilibrium relationship holds good for all the stages, i.e. thevalue of K is independent of solute concentration

If the equilibrium relationship between the extract and raffinate leaving each stage is

assumed to be linear, we can write:

Material Balance

Input = Output

FCF + SCS = E1CE1 + RNpCRNp

F = RNp = R, and S = E1 = E, therefore:

RCF + ECS = ECE1 + RCRNp

ECE1 ECS = RCF - RCRNp

E(CE1 CS) = R(CF - CRNp)

CE1 = (R/E)(CF CRNp) + Cs

For any stage, n, we have:

CEn+1 = E1 {(R/E)[CF CRn]}


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CEn+1 = E1 {(R/E)[CF (CEn/K)]}

Figure 7.12: Analytical solution for staged counter current extraction using immiscible

solvents

For the nth

stage, we get:

Working stage by stage: starting with stage 1, we have:

RCR2 + ECE0 = RCR1 + ECE1

If CE0 = 0, which is true for the fresh solvent, then

RCR2 = RCR1 + ECE1

Dividing by R

CR2 = CR1 + (E/R)CE1

K = CE1/CR1

Substitution gives:

CR2 = CR1 + (E/R) (KCR1)

However, K(E/R) =


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CR2 = CR1 + CR1

CR2 = (1 + )))) CR1

Working the second stage in the same manner, we have:

RCR3 + ECE1 = RCR2 + ECE2

Dividing by R

CR3 = CR2 + (E/R)CE2 (E/R)CE1

K = CE1/CR1, K = CE2/CR2

CR3 = CR2 + CR2 CR1

CR3 = CR2 (1 + ) CR1

Substituting for CR2 from above, where:

CR2 = (1 + )))) CR1

We have:

CR3 = (1 + )))) CR1 (1 + ) CR1

CR3 = (1 + ))))2222CR1 CR1

CR3 =CR1{(1 + ))))2222 - }

CR3 = (1 + + 2) CR1

Or we can generally state that for the (n-1) stage, we have:

CRn = (1 + + 2 + .. n-1) CR1

Now for the targeted nth stage

RCRn+1 + ECE0 = RCR1 + ECEn

CE0 = 0, dividing by R

CRn+1 = CR1 + (E/R)CEn

CRn+1 = CR1 + (E/R)KCRn

CRn+1 = CR1 + CRn

CRn = (1 + + 2 + 3 + + n-1) CR1

CRn+1 = CR1 + (1 + + 2

+ 3

+ + n-1

) CR1


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CRn+1 = CR1[1 + (1 + + 2 + 3 + + n-1)]

CRn+1 = CR1[(1 + + 2 + 3 + + n)]

Mathematically, this can be written in the form:

CRn+1 = CR1[(n+1 - 1)/( 1)]

The fraction extracted is given by:

( )1

11

=

+n

n

p

The number of stages necessary to achieve a certain percentage extraction of a solute in multistage counter current extraction process can be calculated from the relation:

CR1 = CRn+1 [(- 1)/(n+1 1)]

From an overall solute material balance forn stages, we get:

CEn = (R/E){CRn+1 CR1)

It should be remembered that CRn+1 represents the initial feed concentration of the solute,

while CR1 is the final raffinate concentration of the solute.

Example

Penicillin is to be extracted from filtered fermentation media (concentration = 1 g/1) to MIBKusing staged counter-current extraction. The feed flow rate is 550 1/h while the extracting

solvent flow rate is 80 1/h. The equilibrium relationship is given by CE= {(25CR) /(1+ CR)},

where CR and CE are in g/1. Determine the number of theoretical stages needed for 90%

extraction of the antibiotic.

Solution

This problem has to be solved graphically. The equilibrium line is first plotted as

shown below.

The equilibrium line is based on the equation CE = {(25CR) /(1+ CR)}. In an overallsense, the feed concentration (CRn+1) is the independent variable and the extract

concentration (CEn) is the dependent variable. Therefore:


This equation is used for calculation of extracted concentration in staged counter

current extraction.


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This means that two stages are necessary

Example

A pharmaceutical compound is produced in an enzyme bioreactor as an aqueous

solution (concentration = 20 wt %). This compound is to be extracted with an organic

solvent using a staged counter-current extraction system having four stages. Theequilibrium relationship is given by CE = 12 CR, where CR and CE are the

concentrations in the raffinate and the extract respectively and are expressed in mass

ratio. The aqueous feed enters the extractor at a flow rate of 100 kg/h while the

organic solvent enters the extractor at a flow rate of 20 kg/h. Calculate:

a) The composition of the extract and the raffinate.

b) The fraction of pharmaceutical compound extracted.

SolutionIn this problem n = 4.

20 wt% means that 20 kg solute present in each 100 kg feed, which further meansthat:

R = 80 kg water/h

E= 20 kg organic solvent/h

Feed concentration = 20

(E/R)CE/CR=

(20/80) (12CR/CR) =

= 3

From equation (7.26) we can write: /l-O _ ( 3-1CR1 = CRn+1 [(- 1)/(n+1 1)]

CR1 = 20(3 1)/{(34+1 1)}

CR1 = 0.165

From an overall material balance:


CEn = (80/20){20 0.165)

CEn = 74.34

p = ECEn/RCRn+1

p = 20*79.43/80*20 = 0.9917

Example

An antibiotic is to be extracted from an aqueous solution using pure amyl acetate. The

equilibrium relationship is given by CE = 32 CR, where CR and CE are the

concentrations in the raffinate and the extract and are expressed in g/1. The antibiotic

concentration in the feed is 0.4 g/1 and the feed flow rate is 500 1/h. The solvent flow

rate is 30 1/h.

a) How many ideal counter-current stages are required to extract 97 % of the antibiotic?b) If 3 counter-current stages are used, what will be the fraction extracted?

SolutionThe feed concentration = CR n+1 = 0.4 g/1


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For 97% extraction, CR1= 0.012 g/1

(E/R)CE/CR=

(30/500) (32 CR)/CR=

= 1.92

CR1 = CRn+1 [(- 1)/(n+1 1)]

0.012 = 0.4 {(1.92 1)/(1.92n+1 1)}

Therefore: n = 4.296 The number of stages can only have integral values. Therefore

for 97% extraction, we will need 5 stages. For 3 counter-current stages n = 3. Using

equation (7.26), we can write:

CR1 = 0.4 {(1.92 1)/(1.923+1 1)}

CR1 = 0.029 g/L


CEn = (500/30){0.4 0.029)

CEn = 6.18

The fraction extracted is therefore:

p = ECEn/RCRn+1p = 30*6.18/(500*0.4)

p = 0.928

Differential extractionIn differential extraction the feed and the extracting solvent flow past one another

within the extractor. This is usually carried out in equipment having tubular geometry

in which the two phases enter from opposite directions (see Fig. 7.14).


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Within the equipment arrangements are made to ensure intimate mixing of the two

phases. The flow of one liquid past the other and their subsequent collection as

separate raffinate and extract streams is based on their density difference. When the

overall extraction process is considered, differential extraction is a "non-equilibrium"

process since the extract and raffinate stream are not in equilibrium with one another.

However, at any point within the extractor, the extract and raffinate streams could bein local equilibrium. Different types differential extractors are shown in Fig. 7.15.

Physically, towers designed for countercurrent contact can be open, but more usually

contain some form of packing or plates. The material of the packing is chosen so that

one phase wets it preferentially, thus increasing the surface area for mass transfer.

Similarly, the plates are designed to breakup droplets and increase the surface area. In

addition, the contents of the tower may be agitated either by an internal agitator or by

pulsing the fluids. The energy imparted by agitation or pulsation breaks up the

droplets of the dispersed phase.

The efficiency of solute transfer depends on the interfacial area generated within the

extractor as well as on the local solute mass transfer coefficients.

Supercritical fluid extractionA supercritical fluid (SF) is a material, which has properties of both liquid and gas.

Any substance can be obtained as a supercritical fluid above its critical temperature

and critical pressure as shown in Fig. 7.18. A supercritical fluid combines the gaseous

property of being able to penetrate substances easily with the liquid property of being

able to dissolve materials. In addition, its density can be changed in a continuous

manner by changing the pressure. The use of a supercritical fluid (e.g. supercritical

carbon dioxide or water) offers an alternative to organic solvents which are less

environment friendly. The dissolving power of a supercritical fluid offers a safe

solvent for pharmaceutical and food processing and the extraction of spirits and

flavors. It is also used for rapid removal of fouling components such as oil and grease

from devices. Supercritical fluid extraction has tremendous potential for analyticalapplications.


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Some of the main applications of supercritical fluid extraction are:

1. Separation and purification of essential oils and medicinal components2. Removal of grease and other fouling material from process equipment3. Food and beverage processing e.g. production of 'decaf coffee and

alcohol-free beer

As shown in Fig.7.18, a supercritical fluid can be obtained by increasing the

temperature and pressure of a pure substance above its critical temperature (Tc) and

critical pressure (Pc). The solubility of a substance in a supercritical fluid increases

with increase in pressure (see Fig. 7.19). This phenomenon is exploited to fine-tune an

extraction process. Increasing the pressure packs the supercritical fluid molecules

closer and facilitates the entrapment of more solute molecules. The density of a

supercritical fluid also increases quite significantly with increase in pressure. The

greater extracting power of a supercritical fluid when compared with a liquid is due to

the higher solute diffusivity in a supercritical fluid which is comparable to that of the

solute in the gaseous form

A supercritical fluid can be used for both liquid-SF extraction and solid-SF extraction.The SF is produced by pressurizing a gas such as carbon dioxide using a compressor.

In some SF generators an additional heat exchanger may also be used in series with

the compressor. Extractors used for supercritical fluid extraction are simple in design.

Minimal mechanical contacting arrangement is required in these devices due to the

high penetrability of the supercritical fluid. Solid-SF type extraction in the batch mode

may be efficiently achieved using a packed bed extractor.


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Liquid-SF extractionThe efficient dissolving power of a supercritical fluid means that the solute of interest

and the initial solvent are both likely to be soluble in it. Supercritical carbon dioxide is

not suitable for breaking water-ethanol azeotrope. However, is useful for removing

ethanol from ethanol water mixture.

Fig. 7.21 shows the set-up used for obtaining relatively concentrated ethanol from a

dilute ethanol-water mixture. Gaseous carbon dioxide is first converted to a

supercritical fluid by adiabatic compression, which is then fed into the extraction

vessel. The ethanol-water mixture is fed into this vessel at the operating pressure in a

counter-current direction. The supercritical fluid preferentially extracts the ethanol

and leaves the extractor from the top while the raffinate (predominantly water) exits

from the bottom of the extractor. The extract is sent through a pressure reduction

valve into an expansion chamber where gaseous carbon dioxide and liquid ethanol-

water mixture is obtained. The carbon dioxide is separated from the ethanol-water

mixture and sent back to the compressor.


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The use of a membrane as a porous barrier between the feed and the extracting solvent(i.e. SF) offers several advantages. Some of these are:

1. Efficient segregation of raffinate and extract2. Compactness of apparatus3. Large interfacial area4. Constant interfacial area which is independent of the fluid velocity

The use of membranes is more successful in liquid-SF extraction than in liquid-liquid

extraction since the SF can easily penetrate through the membrane into the feed phase,

extract the solute and diffuse back with it to the extract side. A new supercritical fluid

process called porocritical fluid extraction has been commercialized. The SF and thefeed liquid flow counter-currently through a module containing a porous membrane,

typically having 0.2-micron pores. Hollow fiber and spiral wound membranes are

preferred since they provide high interfacial area. Fig. 7.22 shows the principle of

porocritical fluid extraction.


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Solid-SF extractionThe high penetrability of a SF makes it ideally suited for leaching processes. With

conventional solid-liquid extraction, the solid particles need to the pulverized for

satisfactory solute recovery by extraction. Particle size reduction much less important

in solid-SF extraction. Such processes are usually carried out using packed beds. Fig.

7.23 shows the set-up used for extraction of caffeine from coffee beans. The extractor

is first packed with coffee beans and supercritical carbon dioxide which is used as the

extracting solvent is passed through this. The extract obtained consists mainly ofcaffeine dissolved in supercritical carbon dioxide. Caffeine and carbon dioxide are

then separated by pressure reduction and the gaseous carbon dioxide thus obtained is

recycled. The raffinate, i.e. caffeine free coffee beans, is suitable for producing 'decaf

coffee.

extraction chapter

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