The distribution of a solute between the raffinate and the extract can be expressed in terms of the partition coefficient K:

Transcription

1 Extraction Introduction An extraction process makes use of the partitioning of a solute between two immiscible 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 antibiotics 2. Purification of alkaloids 3. Protein purification using aqueous two-phase systems 4. Purification of peptides and small proteins 5. Purification of lipids 6. Purification of DNA Solvent systems Solvent extraction as used in the bio-industry can be classified into three types depending on the solvent systems used: 1. Aqueous/non-aqueous extraction 2. Aqueous two-phase extraction 3. 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 be used. 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 this region will split up into two phases in equilibrium with each other. For a mixture 1

2 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 the binodal solubility curve represents the single-phase region where all three components in the system are mutually miscible. Theory of extraction The different components involved in an extraction processes are summarized in Fig 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: 2

3 Where C E = equilibrium solute concentration in extracting solvent (kg/m 3 ) C R = equilibrium solute concentration in raffinate (kg/m 3 ) Extraction Factor The extraction factor, λ, is defined as: λ = amount of solute in extracting solvent/amount of solute in raffinate λ = EC E /RC R λ = 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 C R and C E 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 requires intimate 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 settling Once 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 two phases (i.e. raffinate and the extract) should have sufficient density difference to facilitate segregation of phases. Collection of phases After 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 solvent and raffinate 3

4 p = EC E /(EC E + RC R ) Dividing both nominator and denominator by RC R p = (E/R)(C E /C R )/{ (E/R)(C E /C R ) + 1} p = λ/(λ + 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/(λ + 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: ε = 100 K / [K + (R/E)] where R and E are the volumes of the aqueous and organic phases respectively, or ε = 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 = q n Therefore, the extracted fraction, E, is: E = 1 - q n The percent extracted after n extractions is: %E = (1 - q n )*100 Taking into account that q = R / (KE + R) 4

5 If K is large, i.e. > 10 2, 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/( ) = 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: C Rn = C R0 [R/(KE + R)] n C Rn = C R0 [1/(λ + 1)] n Or simply: C R =C R0 * q n where C Rn is the amount of solute remaining in the aqueous phase, volume R, after n extractions each with volume E, of organic phase, and C R0 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 I 2 in a carbon tetrachloride/water system is 85, then for a single extraction with 25 ml of CCl 4 q = R / (KE + R) q = %E = (1 - q n )*100 %E = (1 {[ R / (KE + R)]} 1 *100 %E = (1 {[ 50 / (85* )]} 1 *100 %E = 97.7% Amount remaining can also be calculated: C R1 = C R0 [R/(KE + R)] 1 C R1 = 0.1[50/(85* )] 1 C R1 = 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. 5

6 99 = (1 {[ 50 / (85*E + 50)]} 1 * = {[ 50 / (85*E + 50)]} 0.017E = 1 E = 58.2 ml Since, 97.7% of the I 2 is extracted in one extraction with 25 ml CCl 4, for three extractions with 8.33 ml of CCl 4, %E of the I 2 extracted can be calculated as follows: C Rn = C R0 [R/(KE + R)] n C R3 = 0.1[50/(85* )] 3 C R3 = 0.1[0.066] 3 = 0.1 * 2.87*10-4 = 2.87*10-5 g q = R / (KE + R) q = q 3 = 2.87*10-4 %E = (1 {[ 50 / (85* )]} 3 *100 %E = 99.97% One can also calculate amount remaining: C Rn = C R0 [R/(KE + R)] n C R3 = 0.1[50/(85* )] 3 C R3 = 0.1[0.066] 3 = 0.1 * 2.87*10-4 = 2.87*10-5 g If we are to use 6 ml of CCl 4 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 = = (1 (0.089) n )* = 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 extraction to 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 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 K A and K B, the separation factor β is defined as the ratio K A /K B where K A >K B. Assuming that K A 6

7 = 10 2, for different values of K B and β. For an essentially quantitative separation β should be at least 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/ K A K B ) 1/2 Successive extractions, whilst increasing the efficiency of extraction of both solutes, may lead to a poorer separation. For example, if K A = 10 2 and K B = 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 1 st extraction, assuming equal volumes of E and R q B = 1/(K + 1) q B = 1/( ) = 0.91 %E B = (1 - q n )*100 %E B = (1 0.91)*100 = 9.1% q A = 1/(K + 1) q A = 1/( ) = %E A = (1 - q n )*100 %E A = ( )*100 = 99% After the 2 nd extraction %E B = (1 (0.91) 2 )*100 = 17% %E A = (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 if K A = 10 2 and K B = 10 1, then we have: (E/R) = (1/ K A K B ) 1/2 (E/100) = (1/10 2 *0.1) 1/2 (E/100) = 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. 7

8 ph Effect Consider the extraction of a carboxylic acid from water into ether. At low ph, where the acid is undissociated, C HA ~ [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 the organic 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 C I2 ~ [I 2 ] and iodine is extracted with greatest efficiency. At high iodide concentrations, K f [I ] aq >> 1, which means that most I 2 is converted to I 3 -, and K is reduced with a consequent reduction in the extraction of iodine. K = C EI2 /C RI2 In presence of iodide: C RI2 = [I 2 ] + [I 3 - ] Most iodine is converted to triiodide which make the value of C RI2 very large and facilitates transfer of I 2 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 designed apparatus. 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. 8

9 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 the mixture 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 the following figure: 9

10 Tube No Before 0 equilibration 1 After equilibration P q Extraction First transfer, n = 1 Tube No Before 0 p equilibration q After pq p 2 equilibration q 2 pq Extraction The resulting fractions are: q 2 + 2pq + p 2 = (p + q) 2 After the second transfer, n = 2 Tube No Before 0 pq p 2 equilibration q 2 pq After pq 2 2p 2 q p 3 equilibration q 3 2pq 2 p 2 q Extraction After the 3 rd transfer, n = 3 Tube No Before 0 pq 2 2p 2 q p 3 equilibration q 3 2pq 2 p 2 q Fractions q 3 3pq 2 3p 2 q p 3 Extraction The resulting fractions are: q 3 + 3pq 2 + 3p 2 q + p 3 = (p + q) 3 After the n th 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: 10

11 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 larger number 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 extraction Aqueous 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 system 2. Polymer-salt two-phase system A polymer-polymer two phase system can for instance be obtained by mixing dextran and 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. 11

12 The partition of a solute between the two aqueous phases depends on its physicochemical properties as well as those of the two polymers (or polymer and salt). 12

13 Factors Affecting Protein Partitioning in Two-Phase Aqueous Polymer Systems 1. Protein molecular weight 2. Protein charge, surface properties 3. Polymer molecular weight 4. Phase composition 5. Salt effects 6. Affinity ligands attached to polymers 7. ph 8. 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, protein partitioning 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 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 decreased as the PEG MM increased, and for cytochrome C the effect was the smallest. This was 13

14 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 better separation 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 citrate partitioned 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 14

15 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 successfully recycled. More recently PEG-sulphate systems have begun to be used where separation of some biomaterial was achieved with PEG 4000 and (NH 4 ) 2 SO 4 at ph 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. 15

16 The degree of affinity partitioning, K aff, can be described by the ratio between the partition 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 Schemes Extraction 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 two phases depending on its partition coefficient. The rate at which the transfer of solute 16

17 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 The mixer unit must be able to generate high interfacial 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. 17

18 If a batch of feed containing R volume of initial solvent and an initial solute concentration of C0 is mixed with S volume of pure extracting solvent, the concentration distribution in the extract and the raffinate at equilibrium is given by: C E = KC R Where C E solute concentration in extract (kg/m 3 ) C R solute concentration in raffinate (kg/m 3 ) By performing a solute material balance, we get: RC R0 = RC R + EC E 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: C R0 = C R + (E/R) C E K = C E /C R, or C R = C E /K Therefore we have C R0 = C E /K + (E/R) C E C R0 = C E {(1/K) + (E/R)} Multiply both sides by K KC R0 = C E {1 + K(E/R)} The extraction factor (λ) is defined as: λ = K(E/R) KC R0 = C E {1 + λ} We then have : The fraction extracted is given by: Also, fraction remaining in raffinate can be calculated from the relation: q = 1/(1+λ) 18

19 Always, we have: p + q = 1 Example 100 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 = 100 C R 2, where C R and C E 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: RC R0 + SC S = RC R + EC E However, C s = 0 since the initial organic solvent is solute free.therefore we have: RC R0 = RC R + EC E 100 * 1 = 100C R +10C E The equilibrium relationship is: C E = 100 C R 2 By substitution we get: C E = g/1 C R = g/1 The fraction of citric acid extracted in the first batch extraction is: P = (7.298*10)/(100*1) = 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: 19

20 RC R0 + EC E1 = RC R + EC E (100 x l) + (10 x 7.298) = 100C R + 10C E Since we have C E = 100 C R 2 Then by substitution we get: C E = g/1 C R = 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. The equilibrium function which is also called the equilibrium line is of the form shown below: C E = f(c R ) The solute material balance equation can be written as shown below: RC R0 = RC R + EC E Dividing by R, and rearrangement we get: C E = (R/E)[C R0 -C R ] C E = (R/E)C R0 - (R/E)C R (R/E)C R0 is constant for a specific E, R, and C R0 C E = b - (R/E)C R Graphical solution for batch extraction using immiscible solvents This method is more useful when the equilibrium relationship between extract and raffinate concentration is non-linear. The graphical method is based on drawing the equilibrium line and the operating line as shown in the figure below. Equilibrium line is given by: C E = f(c R ) The operating line is obtained from material balance: C E = b - (R/E)C R 20

21 RC R0 /E = b Equilibrium line C E Slope = -R/E Operating line C R Graphical solution for batch extraction The point of intersection of these two lines gives the equilibrium concentrations in the extract and raffinate phases. Graphical solution of the first part of the example above, step by step: 1. Draw the equilibrium line from C E = 100 C R 2 Equilibrium line C E C R Graphical solution for batch extraction 2. Locate b on the graph (intercept) where b = (R/E)C R0 b = (100/10) * 1 = 10 21

22 b = (R/E)C R0 Equilibrium line C E C R Graphical solution for batch extraction 3. From the slope (-R/E) and the point at the intercept draw the operational line: Slope = -(100/10) = -10 RC R0 /E = b Equilibrium line C E Slope = -R/E Operating line 4. The point of intersection of these two lines gives the equilibrium concentrations in the extract and raffinate phases. Continuous Extractions 1. Single-stage continuous extraction using immiscible solvents The 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 the mixer unit is fed into a settler unit where phase separation takes place and continuous C R 22

23 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 in equilibrium 2. There is negligible entrainment of the other phase with each exit stream Composing material balance relationship, we get: FC F + SC S = RC R + EC E Assume F = R, and S = E Also, if a fresh extracting solvent is used, we have C S = 0, therefore, we have: RC F = RC R + EC E Where, in this case, E and R represent flow rates of extracting solvent and feed, respectively. Dividing the first equation by R gives: C F = C R + (E/R) C E K = C E /C R, or C R = C E /K Therefore we have C F = C E /K + (E/R) C E C F = C E {(1/K) + (E/R)} Multiply both sides by K KC F = C E {1 + K(E/R)} The extraction factor for single stage continuous extraction is defined as: λ = K(E/R) KC F = C E {1 + λ} C E = KC F /{1 + λ} C R = C F /{1 + λ} 23

24 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: C F = C R + (E/R) C E, multiply by (R/E) (R/E)C F = (R/E)C R + C E C E = (R/E)C F - (R/E)C R C E = b - (R/E)C R b = (R/E)C F C E = (R/E)C F - (R/E)C R C E = (R/E){C F C R } Example Fermentation 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 C E = 40C R, where C R and C E 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: 24

25 FC F + SC S = RC R + EC E An antibiotic mass balance for the first stage gives us: FC F + SC S = RC R + EC E C s = 0 (100*20) = 100C R + 10C E The equilibrium relationship is: C E = 40C R (100*20) = 100C R + 10*40C R Solving for C R equation we get: C E =160g/1 C R = 4g/1 The fraction of antibiotic extracted in the first stage is: P = EC E /RC F Graphical solution: 1) Draw the equilibrium line where C E = 40C R 2) Locate the intercept point at b = (R/E)C F = (100/10)*20 = 200 3) Find the slope = -(R/E) = - (100/10) = -10 4) Draw the operational line using the point at the intercept and the slope 5) From the intersection between the equilibrium and operational lines find C E and C R. 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 C E = 37C R, where C E is the antibiotic concentration in the aqueous extract phase while C R is the concentration in the organic raffinate phase. Calculate: 25

26 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: EC E + SC S = R 2 C R2 + E 2 C E2 Assume E = R 2, S = E 2, and C S = 0 due to using a fresh solvent EC E = R 2 C R2 + E 2 C E2 (10*160) = 10 C R2 + 5 C E2 C E = 37C R (10*160) = 10 C R2 + 5 *37 C R2 Where C E2 = antibiotic concentration in aqueous extract C R2 = antibiotic concentration in organic raffinate Solving for C R2 we get: C E2 = g/1 C R2 = g/1 The overall fraction of antibody extracted is: P = amount extracted/total amount Graphical solution: 1) Draw the equilibrium line where C E = 37C R 2) Locate the intercept point at b = (R/E)C F = (10/5)*160 = 320 3) Find the slope = -(R/E) = - (10/5) = - 2 4) Draw the operational line using the point at the intercept and the slope 5) From the intersection between the equilibrium and operational lines find C E and C R. 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 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 26

27 concentrations and forward substituting these values. Some of the basic assumptions are: 1. The raffinate and extract streams from each stage are in equilibrium 2. There is negligible entrainment of the other phases 3. 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 EC E = E 1 C E1 + E 2 C E2 + E 3 C E3 The final volume of extracting solvent is: S 1 + S 2 + S 3 = E (Input 1) + (input 2) + (input 3) = (output 1) + (output 2) + (output 3) (FC F + S 1 C S ) + (R 1 C R1 + S 2 C S ) + (R 2 C R2 + S 3 C S ) = (E 1 C E1 + R 1 C R1 ) + (E 2 C E2 + R 2 C R2 ) + (E 3 C E3 + R 3 C R3 ) Since a fresh extracting solvent is used in each step, C S = 0, omitting similar terms on both sides gives: FC F = E 1 C E1 + E 2 C E2 + E 3 C E3 + R 3 C R3 EC E = E 1 C E1 + E 2 C E2 + E 3 C E3 FC F = EC E + R 3 C R3 27

28 It should be clear that: F = R 1 = R 2 = R 3 = R, and E is the total volume of extracting solvent and C E is the total extracted concentration, and C R3 is the concentration of solute remaining in the feed, which can be generally taken as C R. Therefore, we have: RC F = EC E + RC R RC F = RC R + EC E Where, E and R represent flow rates of extracting solvent and feed, respectively. Dividing the first equation by R gives: C F = C R + (E/R) C E K = C E /C R, or C R = C E /K Therefore we have C F = C E /K + (E/R) C E C F = C E {(1/K) + (E/R)} Multiply both sides by K KC F = C E {1 + K(E/R)} Since λ = K(E/R) KC F = C E {1 + λ} C E = KC F /{1 + λ} C R = C F /{1 + λ} 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 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 shows a staged counter-current extraction set-up having three stages made up of mixer-settler units. 28

29 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 equilibrium 2. There is negligible entrainment of the other phases 3. The same equilibrium relationship holds good for all the stages, i.e. the value 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 FC F + SC S = E 1 C E1 + R Np C RNp F = R Np = R, and S = E 1 = E, therefore: RC F + EC S = EC E1 + RC RNp EC E1 EC S = RC F - RC RNp E(C E1 C S ) = R(C F - C RNp ) C E1 = (R/E)(C F C RNp ) + C s For any stage, n, we have: C En+1 = E 1 {(R/E)[C F C Rn ]} 29

30 C En+1 = E 1 {(R/E)[C F (C En /K)]} Figure 7.12: Analytical solution for staged counter current extraction using immiscible solvents For the n th stage, we get: Working stage by stage: starting with stage 1, we have: RC R2 + EC E0 = RC R1 + EC E1 If C E0 = 0, which is true for the fresh solvent, then RC R2 = RC R1 + EC E1 Dividing by R C R2 = C R1 + (E/R)C E1 K = C E1 /C R1 Substitution gives: C R2 = C R1 + (E/R) (KC R1 ) However, K(E/R) = λ 30

31 C R2 = C R1 + λ C R1 C R2 = (1 + λ) C R1 Working the second stage in the same manner, we have: RC R3 + EC E1 = RC R2 + EC E2 Dividing by R C R3 = C R2 + (E/R)C E2 (E/R)C E1 K = C E1 /C R1, K = C E2 /C R2 C R3 = C R2 + λc R2 λc R1 C R3 = C R2 (1 + λ) λc R1 Substituting for C R2 from above, where: C R2 = (1 + λ) C R1 We have: C R3 = (1 + λ) C R1 (1 + λ) λc R1 C R3 = (1 + λ) 2 C R1 λc R1 C R3 =C R1 {(1 + λ) 2 - λ} C R3 = (1 + λ + λ 2 ) C R1 Or we can generally state that for the (n-1) stage, we have: C Rn = (1 + λ + λ λ n-1 ) C R1 Now for the targeted nth stage RC Rn+1 + EC E0 = RC R1 + EC En C E0 = 0, dividing by R C Rn+1 = C R1 + (E/R)C En C Rn+1 = C R1 + (E/R)KC Rn C Rn+1 = C R1 + λ C Rn C Rn = (1 + λ + λ 2 + λ λ n-1 ) C R1 C Rn+1 = C R1 + λ(1 + λ + λ 2 + λ λ n-1 ) C R1 31

32 C Rn+1 = C R1 [1 + λ(1 + λ + λ 2 + λ λ n-1 )] C Rn+1 = C R1 [(1 + λ + λ 2 + λ λ n )] Mathematically, this can be written in the form: C Rn+1 = C R1 [(λ n+1-1)/(λ 1)] The fraction extracted is given by: p λ n ( λ 1) = n+ λ 1 1 The number of stages necessary to achieve a certain percentage extraction of a solute in multi stage counter current extraction process can be calculated from the relation: C R1 = C Rn+1 [(λ- 1)/(λ n+1 1)] From an overall solute material balance for n stages, we get: C En = (R/E){C Rn+1 C R1 ) It should be remembered that C Rn+1 represents the initial feed concentration of the solute, while C R1 is the final raffinate concentration of the solute. Example Penicillin is to be extracted from filtered fermentation media (concentration = 1 g/1) to MIBK using 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 C E = {(25C R ) /(1+ C R )}, where C R and C E 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 C E = {(25C R ) /(1+ C R )}. In an overall sense, the feed concentration (C Rn+1 ) is the independent variable and the extract concentration (C En ) is the dependent variable. Therefore: C En = (R/E){C Rn+1 C R1 ) This equation is used for calculation of extracted concentration in staged counter current extraction. 32

33 This equation gives us the operating line. In this problem we can draw this by locating C R1 on the graph. C Rl is 1/10 th of C R n+1 since there is 90% penicillin extraction. The operating line is drawn from this point and it has a slope equal to (R/E). Next C Rn+1 is located on the graph at 1 g/l penicillin, and work back to C R1 at 0.1g/L penicillin in the raffinate. This is done by drawing a line perpendicular to the C R axis from C Rn+1 point to the operating line. The point of intersection of this line with the operating line give the starting point for drawing staircase stages as shown in the figure. Two stages will be required for this extraction. Mathematical Solution K = C E1 /C R1 K(E/R) = λ λ = (80/550)( C E /C R ) C E = {(25C R ) /(1+ C R )} λ = (80/550)( {(25C R ) /(1+ C R )}/C R ) λ = (80/550)( {(25*0.1) /(1+ 0.1)}/0.1) λ = 3.31 C Rn+1 = C R1 [(λ n+1-1)/(λ 1)] 1 = 0.1[(3.31 n+1 1)/(3.31 1) 22.1 = 3.31 n+1 Log 24.1 = (n+1)log 3.31 n =

34 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. The equilibrium relationship is given by C E = 12 C R, where C R and C E 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. Solution In this problem n = wt% means that 20 kg solute present in each 100 kg feed, which further means that: R = 80 kg water/h E = 20 kg organic solvent/h Feed concentration = 20 (E/R)C E /C R = λ (20/80) (12C R /C R ) = λ λ = 3 From equation (7.26) we can write: /l-o _ ( 3-1 C R1 = C Rn+1 [(λ- 1)/(λ n+1 1)] C R1 = 20(3 1)/{( )} C R1 = From an overall material balance: C En = (R/E){C Rn+1 C R1 ) C En = (80/20){ ) C En = p = EC En /RC Rn+1 p = 20*79.43/80*20 = Example An antibiotic is to be extracted from an aqueous solution using pure amyl acetate. The equilibrium relationship is given by C E = 32 C R, where C R and C E 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? Solution The feed concentration = C R n+1 = 0.4 g/1 34

35 For 97% extraction, C R1 = g/1 (E/R)C E /C R = λ (30/500) (32 C R )/C R = λ λ = 1.92 C R1 = C Rn+1 [(λ- 1)/(λ n+1 1)] = 0.4 {(1.92 1)/(1.92 n+1 1)} Therefore: n = 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: C R1 = 0.4 {(1.92 1)/( )} C R1 = g/l C En = (R/E){C Rn+1 C R1 ) C En = (500/30){ ) C En = 6.18 The fraction extracted is therefore: p = EC En /RC Rn+1 p = 30*6.18/(500*0.4) p = Differential extraction In 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). 35

36 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 be in local equilibrium. Different types differential extractors are shown in Fig 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 extraction A 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 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 analytical applications. 36