ERT 320/ 3 BIO-SEPARATION ENGINEERING PRODUCT ISOLATION & CONCENTRATION (PART I) ERT 320/ 3 BIO-SEPARATION ENGINEERING MISS WAN KHAIRUNNISA WAN RAMLI

OUTLINES INTRODUCTION BASIC PRINCIPLES EQUIPMENTS OPERATING MODES BASIC CALCULATION

Solvent/ liquid-liquid extraction

INTRODUCTION LLE A mass transfer operation in which a liquid solution (the feed) is contacted with an immiscible/ nearly immiscible liquid, called the solvent that exhibits preferential affinity or selectivity towards one/ more of the components in the feed. ADVANTAGES LIMITATIONS BASIC PRINCIPLES & PARAMETERS SCALE UP DESIGN OF EXTRACTOR COLUMN

Extraction is preferred for several applications: Dissolved/ complexed inorganic substances in organic/ aqeous solution Removal of a component present in small concentrations. Eg: hormones in animal oil A high-boiling component present in small concentration in an aqeous waste stream. Eg: Recovery of acetic acid from CA Recovery of heat-sensitive materials. Eg: Antibiotics Separation of a mixture according to chemical type rather than relative volatility Separation of close-melting or close-boiling liquids; solubility differences exploited Mixtures that form azeotrope

The simplest LLE involves only a ternary system Feed2 miscible components: carrier, C + Solute, A. Solvent, S pure compound C & S are at most PARTIALLY soluble in each other. A is soluble in C & completely/ partially soluble in S During the extraction process, mass transfer of A from Feed to Solvent occur, with LESS transfer of C to the solvent or S to the feed. However, A is NOT completely transferred to the solvent & seldom achieved in 1 stage only Need multi stages. Rafinatte Solvent, S Feed, F Extract

BASIC PRINCIPLES When LLE is carried out in a test tube/ flask, the two immiscible phases are stirred/ mixed together to allow molecules to dissolve into the preferred solvent phase

The solute originally present in the aqueous phase gets distributed in both phases If solute has preferential solubility in the organic solvent, more solute would be present in the organic phase at equilibrium The extraction is said to be more efficient Extract = the layer of solvent + extracted solute Raffinate = the layer from which solute has been removed The distribution of solute between two phases is express quantitatively by distribution coefficient, KD Eq. 1 Higher value of KD indicates higher extraction efficiency

EXAMPLES OF KD VALUES

MIXING BY MECHNICAL AGITATION THE TWO PHASES MUST BE BROUGHT INTO INTIMATE CONTACT WITH A HIGH DEGREE OF TURBULENCE IN ORDER TO OBTAIN HIGH MASS TRANSFER RATES. EXTRACTION EQUIPMENT MIXING BY MECHNICAL AGITATION MIXING BY FLUID FLOW

MIXED-SETTLERS For Batchwise Extraction: The mixer and settler may be the same unit. A tank containing a turbine or propeller agitator is most common. At the end of mixing cycle the agitator is shut off, the layers are allowed to separate by gravity. Extract and raffinate are drawn off to separate receivers through a bottom drain line carrying a sight glass. The mixing and settling times required for a given extraction can be determined only by experiment. (e.g: 5 min for mixing and 10 min for settling are typical) - both shorter and much longer times are common.

For Continuous Extraction: The mixer and settler are usually separate pieces of equipment. The mixer; small agitated tank provided with a drawoff line and baffles to prevent short-circuiting, or it may be motionless mixer or other flow mixer. The settler; is often a simple continuous gravity decanter. In common used; several contact stages are required, a train of mixer-settlers is operated with countercurrent flow.

PACKED EXTRACTOR COLUMN Tower extractors give differential contacts, not stage contacts, and mixing and settling proceed continuously and simultaneously. Extraction; can be carried out in an open tower, with drops of heavy liquid falling through the rising light liquid or vice versa. The tower is filled with packings such as rings or saddles, which causes the drops to coalesce and reform, and tends to limit axial dispersion. In an extraction tower there is continuous transfer of material between phases, and the composition of each phase changes as it flows through the tower. The design procedure ; is similar to packed absorption towers.

AGITATED EXTRACTOR COLUMN It depends on gravity flow for mixing and for separation. Mechanical energy is provided by internal turbines or other agitators, mounted on a central rotating shaft. Flat disks disperse the liquids and impel them outward toward the tower wall, where stator rings create quite zones in which the two phases can separate. In other designs, set of impellers are separated by calming sections to give, in effect, a stack of mixer-settlers one above the other. In the York-Scheibel extractor, the region surrounding the agitators are packed with wire mesh to encounter coalescence and separation of the phases. Most of the extraction takes place in the mixing sections, but some also occurs in the calming sections. The efficiency of each mixer-settler unit is sometimes greater than 100 percent.

OPERATING MODES Batch Extraction: The aqueous feed is mixed with the organic solvent After equilibrium, the extract phase containing the desired solute is separated out for further processing iii. Routinely utilized in laboratory procedures iv. This can be carried out for example in separating funnel or in agitated vessel Schematic representations of BATCH extraction: Single Multiple stages (crosscurrent)

PENICILIN

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

CALCULATION METHODS EXTRACTION OF DILUTES SOLUTION 1. Extraction factor, E is defined as: From Eq. 1

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

EXAMPLE 1 Penicillin F is recovered from a dilute aqueous fermentation broth by extraction with amyl acetate, using 6 volumes of solvent (V) per 100 volumes of the aqueous phase (L). At Ph 3.2 the distribution coefficient KD is 80. (a) What fraction of the penicillin would be recovered in a single ideal stage? (b) What would be the recovery with two-stage extraction using fresh solvent in both stages?

Draw the material balance diagrams SOLUTION 1: Draw the material balance diagrams Material Balance: L(x0) + V(y0) = L (x1) + V(y1) L(x0) – L(x1) = V(y1) – V(y0) Since y0=0 (at initial no penicillin in solvent phase) So, L(x0)-L(x1) = V(y1) L(x0-x1)= V(y1) Since KD = y1/x1, y1=KDx1  Refer to EQ. 1 So, L(x0-x1)=V(KDx1) x1[(VKD/L )+ 1)]= x0, where VKD/L = E  Refer to EQ. 2 E = VKD/L = (6)(80)/100 = 4.8

Material Balance (cont.): x1/x0 = 1/ (1+E)  Refer to EQ. 3  (frac. of penicillin in raffinate phase = frac. remaining) = 1/ (1+ 4.8) = 0.1724 Fraction of penicillin recovered = Fraction of penicillin in extract phase = 1- 0.1724 = 0.828 = 82.8% Or calculated using EQ. 4, E/(1+E)= 4.8/ (1+4.8) =0.828; 82.8% recovery

b) Draw the material balance diagrams SOLUTION 1: b) Draw the material balance diagrams Material Balance: L(x1) + V(y0) = L (x2) + V(y2) L(x1) – L(x2) = V(y1) – V(y0) Since y0=0 (at initial no penicillin in solvent phase) So, L(x1)-L(x2) = V(y2) L(x1-x2)= V(y2)

Material Balance: Since KD = y2/x2, y2=KDx2  Refer to EQ. 1 So, L(x1-x2)=V(KDx2) x2[(VKD/L )+ 1)]= x1, where VKD/L = E  Refer to EQ. 2 E= (6)(80)/100 = 4.8 x2/x1 = 1/ (1+E)  Refer to EQ. 3  (frac. of penicillin in final raffinate phase from raffinate phase in 1st stage) = frac. remaining) = 1/ (1+ 4.8) = 0.1724 x2/x0 = (x2/x1) * (x1/x0) = (0.1724) * (0.1724) = 0.0297  (frac. of penicillin in final raffinate phase from feed phase = frac. remaining from ) Fraction of penicillin recovered = Fraction of penicillin in extract phase from feed phase = 1- 0.0297 = 0.9703 = 97.0%

EXTRACTION OF CONCENTRATED SOLUTION Equilibrium relationship are more complicated3 or more components present in each phase. Equilibrium data are often presented on a triangular diagram such as Fig 23.7 and 23.8. TERNARY SYSTEM TYPE I 1 immiscible pair Solute, A Solvent, S Carrier, C

TRIANGULAR DIAGRAM Point E  PLAIT POINT [The composition of extract & raffinate phases approach each other] Line ACE  Extract phase Line BDE  Raffinate phase TIE LINE TRIANGULAR DIAGRAM

Point M: 0.2 Acetone, 0.3 water, 0.5 MIK How to obtain the phase composition using the triangular diagram? #Example: If a mixture with 40 % acetone and 60 percent water is contacted with equal mass of MIK, the overall mixture is represented by point M in Figure 23.7: Assume mass of mixture and solvent Determine the F/S ratio to pinpoint the mixing, M point. Point M: 0.2 Acetone, 0.3 water, 0.5 MIK Draw a new tie line, Point M to Extract & Raffinate phase Extract phase: 0.232 acetone, 0.043 water, 0.725 MIK Raffinate phase: 0.132 acetone, 0.845 water, 0.023 MIK Ratio of acetone to water in the product = 0.232/0.043 = 5.4 7. Ratio of acetone to water in the raffinate = 0.132/0.845 = 0.156

0.3 Water F 0.2 Acetone S 0.5 MIK

Solute, A Tie line in Fig 23.8 slope up to the right – extraction would still be possible But more solvent would have to use. The final extract would not be as rich in desired component (MCH) Raffinate phase Solvent, S Carrier, C Extract phase TERNARY SYSTEM TYPE II 2 immiscible pairs

COORDINATE SCALE • Refer to Treybal, Mass Transfer Operation, 3rd ed., McGraw Hill • The book use different triangular system • The location of solvent (B) is on the right of the triangular diagram (McCabe use on the left) • Coordinate scales of equilateral triangles can be plotted as y versus x as shown in Fig 10.9 • Y axis = wt fraction of component A (acetic acid) • X axis = wt fraction of solvent B (ethyl acetate)

TERNARY SYSTEM TYPE I SOLUTE, A SOLVENT, S CARRIER, C

SOLUTE, A SOLVENT, S CARRIER, C TERNARY SYSTEM TYPE II

SINGLE-STAGE EXTRACTION

• The triangular diagram in Fig 10 • The triangular diagram in Fig 10.12 (Treybal) is a bit different as compared to Fig. 23.7 (McCabe) • Extract phase  on the RIGHT • Raffinate phase  on the LEFT • Fig 10.12 shows that we want to extract Solute, A from C by using solvent, S • Total material balance: • Material balance on A:

Amount of solvent to provide a given location for M1 on the line FS: • The quantities of extract and raffinate: • Minimum amount of solvent is found by locating M1 at D (Raffinate phase) • Maximum amount of solvent is found by locating M1 at K (Extract Phase)

MULTISTAGE CROSSCURRENT EXTRACTION

Continuous or batch processes Refer to Fig 10.14 Raffinate from the previous stage will be the feed for the next stage The raffinate is contacted with fresh solvent The extract can be combined to provide the composited extract The total balance for any stage n: • Material balance on A:

DESIGN CALCULATION

COUNTER-CURRENT EXTRACTION N-EQUILIBRIUM STAGES Consider a countercurrent-flow, N-equilibrium-stage contactor for solvent extraction of a ternary system under isothermal, continuous & steady state flow Feed, F contains the carrier, C & solute, A, & can slso contain solvent, S. The entering solvent, S can contain C & A F ialah mass flow rate of feed S ialah mass flow rate of solvent En ialah mass flow of extract leaving stage n Rn ialah mass flow of raffinate leaving stage n yn ialah mass fraction of solute in extract leaving stage n xn ialah mass fraction of solute in raffinate leaving stage n Feed 1 2 n N-1 N Extract Solvent Raffinate

COUNTER-CURRENT EXTRACTION DEGREE OF FREEDOM The degree of freedom for liquid-liquid extraction for n-equilibrium stages is 2C + 3 For ternary system, C = 3, thus the degree of freedom = 9 In most cases, specifications for Feed, F, xF, yS dan T will be provided There is 6 more specifications for LLE S & xN given S & y1 given y1 & xN given Number of equilibrium stages, N & xN given Number of equilibrium stages, N & y1 given Number of equilibrium stages, N & S given

COUNTER-CURRENT EXTRACTION MATERIAL BALANCE Overall total mass balance for control volume 1 (7.1) Overall mass balance of the component for control volume 1 (7.2) Solvent-to-Feed (S/F) and Extract-to-Raffinate (E/R) Flow rate ratios (7.3) (7.4) Feed 1 2 n N-1 N Extract Solvent Raffinate

OVERALL MATERIAL BALANCE OVERALL MATERIAL BALANCE HUNTER-NASH METHOD OVERALL MATERIAL BALANCE E1 F RN M OVERALL MATERIAL BALANCE

OPERATING POINTS & LINES HUNTER-NASH METHOD OPERATING POINTS & LINES OPERATING LINELocus of passing streams in a cascade It is given by the difference between inlet and outlet streams of each stage (7.5) This difference point, P  Operating point This point is established graphically by finding the intersection between TWO extrapolated operating lines which connect the Feed, F with Extract, E1 point & the Solvent, S with Raffinate, RN point. (7.6) Suapan 1 2 n N-1 N Ekstrak Pelarut Rafinat

OPERATING POINTS & LINES Pembinaan Titik Kendalian P HUNTER-NASH METHOD OPERATING POINTS & LINES Operation point Operation line [F, E1] E1 F M P Operating line [S, Rn] RN Pembinaan Titik Kendalian P

COUNTER-CURRENT EXTRACTION NUMBER OF EQUILIBRIUM STAGES Extract and raffinate streams which exited each stage was at equilibrium. This concludes that Exract and raffinate point at each stage are connected by the tie line. (7.7) The operating line then connects the extract and raffinate points at the next stage with the operating point, P according to Eq. (7.5) This construction of lines are repeated until the final required extract point is reached. Suapan 1 2 n N-1 N Ekstrak Pelarut Rafinat

COUNTER-CURRENT EXTRACTION NUMBER OF EQUILIBRIUM STAGES Operating line Operating point, P E1 F M Operating line RN Determination of number of equilibrium stages (2)

COUNTER-CURRENT EXTRACTION MINIMUM SOLVENT-TO-FEED FLOWRATE RATIOS The actual solvent-to-feed ratio is determine from the multiples of the minimum solvent-to-feed ratio. Minimum solvent-to-feed flowrate ratios usually happen when the number of stages needed is close to infinity. This happen when the operating line coincident with a tie line. The line connecting Solvent point,S and Raffinate point, RN is extended in both direction of raffinate and extract. The farthest point which intersects with any extended tie line on SRN line determine the operating line of N-stages. Suapan 1 2 n N-1 N Ekstrak Pelarut Rafinat 5

COUNTER-CURRENT EXTRACTION MINIMUM SOLVENT-TO-FEED FLOWRATE RATIOS E1Smin F MSmin’ P1 P2 P3 = P min Operating line Smin RN Determination of minimum solvent-to-feed flow rate ratio

COUNTER-CURRENT EXTRACTION MINIMUM SOLVENT-TO-FEED FLOWRATE RATIOS The difference in total outflow for Smin/F ratio (7.8) Point MSmin can be determine by intercepting both line: FSmin dan RNE1Smin Therefore, by using the inverse-lever-arm rule, we can determine the minimum solvent-to-feed ratio, Smin/F (7.9) The actual S/F is 1.2 – 2.0 times the SSmin/F Suapan 1 2 n N-1 N Ekstrak Pelarut Rafinat

AQUEOUS TWO PHASE EXTRACTION Use widely in separation of proteins, enzymes, viruses, cells and cell organels Not denature the biological entities as they might be in organic solvents. The proteins are partitioning between two aqueous phases which contains mutually incompatible polymers or other solutes. EXAMPLE Light phase is water + 10% polyethylene glycol (PEG) and 0.5% dextran Heavy phase is water + 1% glycol and 15% dextran Proteins are partitioned between phases with distribution coefficient (KD) that depends on the pH. KD can vary from 0.01 to more than 100.

Factors that affect protein partitioning in Aqueous Two Phase System: 1. Protein molecular weight 2. Protein charge, surface properties 3. Polymer(s) molecular weight 4. Phase composition, tie-line length 5. Salt effects 6. Affinity ligands attached to polymers

COUNTER-CURRENT EXTRACTION EXAMPLE 1 Acetone is to be extracted from a feed mixture of 30 wt.% acetone (A) and 70 wt.% ethyl acetate (C) at 30°C by using the same weight of pure water (S) as the solvent. The final raffinate is to contain 5 wt.% of acetone. Determine the weight percent of acetone that can be extracted using a single stage Determine the number of equilibrium stages required Suapan 1 2 n N-1 N Ekstrak Pelarut Rafinat

SOLUTION Assume F = 50 g, so S = 50g E F y1 Xf = 0.3 S Yn = 0 R Xn = 0.05 Material balance: F + S = M 50 + 50 = 100 Material balance on A: Fxf + Syn = Mxm (50x0.3) + (50x0) = 100xm Xm = 0.15 M = E1 + R1, R1 = M – E1 Mxm = Ey1 + Rxn Mxm = E1y1 + (M-E1)xn E1 = M (xm-xn) / (y1 – xn)

Find point S (100 wt. % S) & F (30 wt. % A, 70 wt Find point S (100 wt.% S) & F (30 wt.% A, 70 wt.% C), draw a line connecting point F & S From material balance, we have found xm & M point MUST lie in FS line Find point M using xm on FS line Then, we already have the fraction of A we need in raffinate stream, xn = 0.05 Find point R1 using xn which needs to be on the equilibrium curve Draw a line connecting point R1 & M, extrapolate until reached the Extract side. The point which intercept both line R1,M and equilibrium curve are point E1 & y1. Read the value for y1, settle the material balance to find E1 & R1 Equilibrium stages Draw line connecting F with E1, R1 with S, extrapolates until both lines intercept each other at point P.

S F E1 Xm = 0.15 M R1

COUNTER-CURRENT EXTRACTION EXAMPLE 1 Solution of Acetic Acid and water containing 30 wt.% of Acetic Acid is to be extracted by contacting to twice weight of isopropyl ether at 40°C to give a raffinate containing 5 wt.% of Acetic Acid. Determine the quantities & composition of each streams for a single stage extraction process Determine the number of equilibrium stages required

STEPS IN SOLVING MATERIAL BALANCE Construct the equilibrium curve in triangular diagram & connects the dot to form tie line Find operating point of the diagram by extrapolating tie lines until they intercepts at P Draw the material balance diagram (based on the q) Solve the material balance at each stage using eq. 7.1 & 7.2 to find mixing point, M Find the F & S point in the diagram, connects the points to form line FS Since we only know the fraction of solute, A in M, find M(?, xm) on line FS Draw a tie line passing M & extended to point P Intercepting points at equilibrium curve are the solute concentration of E & R point of the extraction process, respectively at both extract & raffinate phase Solve the material balance for R & E

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