Liquid-liquid extraction

Liquid-liquid extraction

Basic principles In liquid-liquid extraction, a soluble component (the solute) moves from one liquid phase to another. The two liquid phases must be either immiscible, or partially miscible. usually isothermal and isobaric can be done at low temperature (good for thermally fragile solutes, such as large organic molecules or biomolecules) can be very difficult to achieve good contact between poorly miscible liquids (low stage efficiency) extracting solvent is usually recycled, often by distillation (expensive and energy-intensive) can be single stage (mixer-settler) or multistage (cascade)

Extraction equipment Batch: Continuous: single-stage: column: separatory funnel mixer-settler rotating-disk contacter a. agitator; b. stator disk

Design Mixer-settlers, both as stand-alone and as in- Mixer-settler column type, are offered for special column applications. As implied by the name, the mixer-settler-column is a series of mixersettlers in the form of a column. It consists of a Mixer-settlers operate with number a purely of stagewise contact. After every mixer other, there each is hydraulically a separated, and each stages installed one on top of the ettler. Mixer-settlers can be with operated a mixing in and a settling zone (see below). multistage, co- or countercurrent This design fashion. enables the elimination of some of the main disadvantages of conventional mixersettlers, whilst maintaining stage-wise phase contact. The mechanical design of the mixer-settlercolumn is comparable to the agitated ECR Kühni column. Key characteristics For long residence times: >5 min Extraction controlled by residence time Reactive extraction systems Long phase separation For extraction controlled by ph (stage-wise ph adjustment) For batch extraction

points Separation of high-boiling products or pollutants Agitated that are present column in only low concentrations Separation of components with similar boiling points or components forming ction with high mass transfer and/or azeotropes ng physical properties, this is the Separation of mixtures with of choice. The geometry of the thermally sensitive components d compartments can be adapted for Selective separation of single ng hydrodynamic conditions. Other components out of a complex mixture atures are the special mixing turbines Our portfolio includes a complete range of perforated partition plates. liquid-liquid-extraction equipment, enabling us to provide you with the most appropriate solution for your requirements. In addition to agitated columns, it includes mixer-settlers and packed columns. Design The agitated Kühni column has a simple and robust design. The drive unit and the shaft are supported at the top of the column, allowing you to use all common types of shaft seals (stuffing box, mechanical seals). In special cases, the seal can be replaced by a magnetic drive. Only radial slide-bearings are necessary

Packed extraction column The ECP packed column is based on c Packing The special Sulzer extraction packing Liquid distributors In order to create an even liquid flow Main benefits High specific throughput fa Small column diameters Revamp of existing columns Use in cases of difficult phy Low density difference < 50 Low interfacial tension: < 2 Tendency to form emulsion Reliable scale-up

Stream labeling feed solvent mixture raffinate extract E, y A, F, x A,0 mixer settler Feed (F) contains solute A (x A ) dissolved in diluent D (x D = x A ). Solvent (S) extracts A (y A ), creating the product extract stream (E). The depleted feed becomes the product raffinate stream (R). Equilibrium (no longer VLE!) is defined by the distribution ratio, K d : K d = y A /x A S E solvent flow rate = F S = constant S, y A,N+ N F R diluent flow rate = F D = constant R, x A,N Note that y A does not refer to gas composition. Usually specified: y A,N+, x A,0, F D /F S and x A,N.

McCabe-Thiele analysis: Counter-current extraction with immiscible liquids Equation of the operating line: Y = F D F S X + (Y F D F S X 0 ) (analogous to operating line for stripper column). X 0 (F D /F S ) max gives F S,min for N =. (X 0,Y ) Y For dilute solutions, y = R E x + (y R E x 0) 2 Can also use Kremser eqns, if solutions are dilute and equil. line is straight. N = ln me R y N+ y 0 y y 0 ( ) ln R me + me R 3 (X N,Y N+ ) X N = 3

Cross-flow cascade From mass balance around stage j: y j = R E j x i + (y j,in + R E j x j ) Figure 3-8 Cross-flow cascade From Separation Process Engineering, Third Edition by Phillip C. Wankat (ISBN: 03382276) Copyright 202 Pearson Education, Inc. All rights reserved. Increase overall efficiency by introducing fresh extracting solvent at each stage. Each stage has its own mass balance and operating line Uses much more solvent than countercurrent cascade (requires much more solvent recovery) A mixer-settler is just one cross-flow stage. Y (x 3,y 3 ) N = 3 (x 2,y 2 ) x 3 (x,y ) (x 2,y 3,in )(x,y 2,in ) X (x 0,y,in )

Dilute fractional extraction A common situation: the feed contains two important solutes (A, B), and we want to separate them from each other. Choose two solvents: A prefers solvent ( extract ) B prefers solvent 2 ( raffinate ) K d,a = y A /x A > K d,b = y B /x B < F z A z B extract y A, y B, E solvent y A,N+ = 0 y B,N+ = 0 N solvent 2 x A,0 = 0 x B,0 = 0 R E R raffinate x A,N x B,N absorbing section stripping section

McCabe-Thiele analysis: dilute fractional extraction One operating line for each solute i, in each section of the column (i.e., 4 total). Top operating lines (absorbing section): y i = R E x i + (y i, R E x i,0) Bottom operating lines (stripping section): y i = R E x i + (y i, R E x i,0 ) Equilibrium data is different for each solute (use separate McCabe-Thiele diagrams!) Y y A, Operating lines intersect at feed composition (not shown, may be very large). 2 6 5 3 NF = 4, feed stage If y A, and x A,N are specified, and N F is known, use M-T diagram to obtain N, then use trial-and-error to find x B,0 and x B,N+ x A,N If y A, and x B,N are specified, vary N F (trial-and-error) until N is the same for both solutes. X

Center-cut extraction When there are 3 solutes: A, B and C, and B is desired solvent + A solvent 2 solvent 3 + B solvent 2 (A and C may be > component each) Requires two columns: column separates A from B+C column 2 separates B from C F z A, z B, z C Requires three extracting solvents: A prefers solvent over solvent 2 B, C prefer solvent 2 over solvent B prefers solvent 3 over solvent 2 C prefers solvent 2 over solvent 3 solvent solvent 2 + B + C solvent 3 solvent 2 + C

There are two liquid phases Partially miscible solvents Each phase is a ternary (3-component) mixture of solute A, diluent D and solvent S Ternary equilibrium diagrams have 3 axes: usually, mole or mass fractions of A, D, and S Literature data is commonly presently on an equilateral triangle diagram (note NO origin) Each axis is bounded 0 x Miscibility boundary = equilibrium line (depends on T, P) Figure 3-4 Effect of temperature on equilibrium of methylcyclohexane-toluene-ammonia system from Fenske et al., AIChE Journal,,335 (955), 955, AIChE From Separation Process Engineering, Third Edition by Phi l lip C. Wa nkat (ISBN: 03382276) 202 Pearson Education, Inc. All rights reserved

Reading ternary phase diagrams Consider the point M: water content (x A ) is? ethylene glycol content (x B ) is? furfural content (x C ) is? 0.9 0.20 0.6 check: x A + x B + x C = Read the mole/mass fraction of each component on the axis for that component, using the lines parallel to the edge opposite the corner corresponding to the pure component. The mixture M lies inside the miscibility boundary, and will spontaneously separate into two phases. Their compositions (E and R) are given by the tie-line through M. The compositions of E and R converge at the plait point, P (i.e., no separation). region of partial miscibility A-C A 2-component mixture of furfural and water is partially miscible over the composition range from about 8 % furfural to 95 % furfural. Separation by extraction requires a furfural/water ratio in this range (otherwise single phase).

Right-triangle phase diagrams Raffinate (diluent-rich): x A + x B + x C = Extract (solvent-rich): y A + y B + y C = We need to specify only two of the compositions in order to describe each liquid phase completely. This can be shown on a right-triangle phase diagram, which is easy to plot and read. raffinate compositions are represented by coordinates (x A, x B ) extract compositions are represented by coordinates (y A, y B ) Figure 3-2 Equilibrium for water-chloroform-acetone at 25 C and atm From Separation Process Engineering, Third Edition by Phillip C. Wankat (ISBN: 03382276) Copyright 202 Pearson Education, Inc. All rights reserved. Vertical axis corresponds to both x A and y A. Horizontal axis corresponds to both x B and y B More tie-lines can be obtained by trial-anderror, using the conjugate line. Q: Where does pure C appear on this diagram? Ex.: find the tie-line that passes through M.

Obtaining the conjugate line Each point on the conjugate line is composed of - one coordinate from the extract side of the equilibrium line - one coordinate from the raffinate side of the equilibrium line On this graph, which component is the diluent? which is the solute?

Hunter-Nash analysis of mixer-settler F S M R E mixer settler Why does F appear on or near the hypotenuse? Flow rates of E and R are related by mass balance. Compositions of E and R are also related by equilibrium. coord.: (y D,y A ) Why does S appear at or near the origin? Overview of solution using RT diagram:. Plot F and S and join with a line. 2. Find mixing point, M, which is co-linear with F and S. 3. Find tie-line through M; find E and R at either end (colinear with M). 4. Find flow rates of E and R. E S mixing line M tie-line F R coord.: (x D,x A )

F S mixer M TMB: F + S = M Co-linearity CMB A : Fx A,F + Sx A,S = Mx A,M = (F + S)x A,M Why are F, S and M co-linear on the Hunter-Nash diagram? solve for coordinates of M: (x A,M, x D,M ) x A,M = Fx A,F + Sx A,S F + S CMB D : Fx D,F + Sx D,S = Mx D,M = (F + S)x D,M x D,M = Fx D,F + Sx D,S F + S F S = x A,M x A,S x A,F x A,M = x D,M x D,S x D,F x D,M rearrange x A,M x A,S x D,M x D,S = x A,F x A,M x D,F x D,M CMB A S (x D,S,x A,S ) CMB D M (x D,M, x A,M ) F (x D,F, x A,F ) slope from M to S slope from F to M Therefore F, S and M are co-linear. To locate M on the FS line: calculate either x A,M or x D,M.

The lever-arm rule Another way to locate M: To calculate flow rates E and R: S MS MF M F similar triangles E EM similar triangles M MR R Fx A,F + Sx A,S = Mx A,M M = R + E Fx A,F + (M F)x A,S = Mx A,M F(x A,F - x A,S ) = M(x A,M - x A,S ) R M = x A,M y A,E x A,R y A,E = ME RE F M = x A,M x A,S x A,F x A,S = MS FS Your choice! Use mass balances, or measure distances and use lever-arm rule.

Hunter-Nash analysis of cross-flow cascade S S 2 F = R 0 R 2 R 2 E E Treat each stage as a mixer-settler. each R i, S i pair creates a mixing line find each E i, R i pair using a tie-line E M F R E 2 M 2 R 2 S

Hunter-Nash analysis of counter-current cascade F S M R mixer separator (column) E and R are both points on the equilibrium line. But they are not related by the same tie-line. Overview of solution using RT diagram:. Plot F and S and join with a line. 2. Find mixing point, M, which is colinear with F and S. 3. Plot specified x A, on raffinate side of equilibrium line to find R. 4. Extrapolate R M line to find. 5. Find flow rates of E and R. x A, S mixing line M NOT a tie-line F R

Stage-by-stage analysis R x A, S = E 0 y A,0 stage TMB: E 0 + R 2 = E + R E 0 R = E R 2 = E 2 R 3 etc. R 2 E constant difference in flow rates of passing streams = E j R j+ = constant stage CMB A : E 0 y A,0 + R 2 x A,2 = E y A, + R x A, N E 0 y A,0 R x A, = E y A, R 2 x A,2 = etc. F = R N+ x A,N+ y A,N+ constant difference in compositions of passing streams net flow of A: x A, = E j y A,j R j+ x A,j+ net flow of D: x D, = E j y D,j R j+ x D,j+

The difference point Define a difference point,, with coordinates (x A,, x D, ): x A, = E y R x 0 A,0 A, x = E y R x 0 D,0 D, D, does not necessarily lie inside the RT graph. All pairs of passing streams E j, R j+ are co-linear with. Using the -point to step off stages on Hunter-Nash diagram: using the specified location of R (as x A, ), can find E (use tie-line); given the location of E, can find R 2 (use ); given the location of R 2, can find E 2 (use tie-line); given the location of E 2, can find R 3 (use ); and so on, until desired separation is achieved. First, need to locate. It may be on either side of the Hunter-Nash diagram.

Finding the -point Procedure:. Plot F (= R N+ ), S = (E 0 ). Locate M. 2. Plot R and locate. M F = R N+ last mixing line R 3. Extend the lines joining E 0 -R, and -R N+, to find at the intersection point. first mixing line S = E 0 4. All intermediate mixing lines must pass through.

Stepping off stages on the H-N diagram Procedure: N = 3. Use R and conjugate line to find E 2. Use E and -point to find R 2 3. Use R 2 and conjugate line to find E 2 4. Use E 2 and -point to find R 3 E 2 E S = E 0 M F = R N+ R 3 R 2 R 3. Use R 3 and conjugate line to find E 3 Stop when you reach or pass.

Using McCabe-Thiele diagram instead of Hunter-Nash M-T diagram can be used with much greater accuracy than H-N diagram Need to transfer ternary equilibrium data from RT diagram Need to obtain the operating line Transferring equilibrium data from RT diagram A extract compositions P y A x A 0 0 D raffinate compositions Each tie-line represents a pair of equilibrium streams extract composition represented by y A raffinate composition represented by x A Each (x A, y A ) pair is a point on the M-T equilibrium line y A P 0 0 x A equilibrium line ends at P

Obtaining the M-T operating line R x A, S = E 0 y A,0 A Mixing lines represent passing streams. All mixing lines lie between the limits: (x, y 0 ) and (x N+, y N ) y N R N+ x M R y 0 E 0 D x N+ WAIT! In general, operating line is not straight. Plot arbitrary intermediate mixing lines to obtain more points. F = R N+ x A,N+ N y A,N+ Note: passing streams are (x j+, y j ) instead of (x j, y j+ ) as in distillation, simply due to our labeling convention (feed enters at stage N). y A P (x N+, y N ) 0 0 (x, y 0 ) x A

Choice of extracting solvent flow rate As S increases, separation improves, but extract becomes more dilute As S decreases, N must increase to maintain desired separation S min achieves the desired separation with N = A S F M M min D M max as M moves towards S, (S/F) increases (lever-arm rule) when M reaches the equilibrium line, all feed dissolves in extracting solvent (M max ) as M moves towards F, (S/F) decreases before reaching the equilibrium line, there is usually a pinch point (M min ) It is not easy to locate this pinch point on a McCabe-Thiele diagram, since the operating line curvature changes as S changes. On a Hunter-Nash diagram, min (corresponding to M min ) occurs when a mixing line and a tie-line coincide.

weight fraction acetone 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0. Acetone-water-trichloroethane at 25 C and atm E N,min Minimum solvent flow rate On H-N diagram whose tie-lines have negative slopes: M min 0 S 0 0. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 F R. Plot S = E 0, F = R N+, R 2. Join S and F 3. Extend SR mixing line 4. Locate several tie-lines 5. Extend tie-lines to the SR mixing line 6. Tie-line with furthest intersection from S locates min 7. Mixing line from min through F locates,min 8. Connecting R and,min completes the mass balance 9. M min is located at the intersection of SF and R,min min 0. (S/F) min = (FM min )/(SM min ) weight fraction water Rule-of-thumb: (S/F) act ~.5 (S/F) min

Minimum solvent flow rate 0.5 Portion of right triangle phase diagram for water-acetic acid-isopropyl ether at 20 C, atm On H-N diagram whose tie-lines have positive slopes: Strategy:. Plot S = E 0, F = R N+, R 2. Join S and F 3. Extend SR mixing line 4. Locate several tie-lines 5. Extend tie-lines to SR mixing line 6. Find tie-line which gives closest intersection to S; this locates min 7. Draw mixing line from min through F to locate,min 8. Connect R and,min to complete mass balance weight fraction acetic acid 0.4 0.3 0.2,min 0. M min F R 0 0 0. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 S weight fraction water min 9. M min is at the intersection of SF and R,min 0. (S/F) min = (FM min )/(SM min )

Two feed counter-current column R E 0 = S F S M R F 2 F T mixer mixer 2 separator R E F 2 Feed balance: F + F 2 = F T R E Overall balance: hypothetical mixed feedstream FT is co-linear with F, F 2 F = R N+ N Stage-by-stage analysis: mass balance changes where F 2 enters the column upper and lower sections have different sets of operating lines different -points

Hunter-Nash analysis of 2-feed column Overall balance:. Plot F and F 2. Locate F T (colinear with F and F 2 ). M F F T F 2 R 2. Plot S. Locate M (co-linear with S and F T ). 3. Plot R. Locate (co-linear with R and M).. Calculate flow rates R and. S = E 0

Stage-by-stage analysis R E 0 = S Balance around top of column: R E 0 = R j+ E j = R, E 0, are colinear Balance around bottom of column: R N+ = E k R k+ = 2 R N+,, 2 are co-linear F 2 j R k R N F = R N+ E E Overall balance: F 2 + R N+ + E 0 = + R F 2 = ( R N+ ) + (R E 0 ) = + 2 F 2,, 2 are co-linear feed-line 2 is located at the intersection of two mixing lines: R N+,, 2 and F 2,, 2 Need another line to locate : TMB: F T = F + F 2 = + (R E 0 ) = + F T,, are co-linear is located at the intersection of two mixing lines: R, E 0, and F T,, 2 Note: and 2 may be on different sides of the phase diagram.

Using the feed-line Acetone-water-trichloroethane at 25 C and atm. Locate at intersection of R E 0 and F T weight fraction acetone 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0. E 2 E 0 S 0 0. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 weight fraction water F M F T R 2 R F 2 2. Locate 2 at intersection of F 2 and R N+ 3. Step off stages, initially using to generate the first mixing lines 4. Identify the optimum feed stage when the mixing line crosses the feed line, F 2 2 feed line 2 5. When the tie-line crosses the feed line, the next mixing line will be generated using 2

Countercurrent liquid-liquid extraction with reflux R E 0 = S How to increase y A,N? need to increase x A,N+ make R N+ an reflux stream R makeup solvent E 0 R E extract reflux (no benefit to raffinate reflux) F R E F = R N+ x A,N+ N y A,N Turning extract into raffinate : extract is mostly solvent raffinate is mostly diluent We need to remove solvent, e.g., distillation, stripping R N+ reflux In a conventional liquid-liquid extraction column: y A,N is related by equilibrium to x A,N x A,N depends on x A,N+ dilute feed gives dilute extract highest y A,N obtained with S S min, but this requires very large N R N E recovered solvent S R solvent separator Q P E product extract

Analogy to distillation reflux V L 0 D N recovered solvent S R Saturated liquid reflux stream is obtained by condensing V (vapor stream rich in A) to give L 0 (liquid stream rich in A) R N+ reflux solvent separator Q P E product extract External reflux ratio = L 0 /D Internal reflux ratio = L/V Extract reflux stream is obtained by removing solvent from (extract stream rich in A and solvent) to give R N+ (raffinate stream rich in A and depleted in solvent) External reflux ratio = R N+ /P E Internal reflux ratio = R N+ /

Stage-by-stage balances Similar to 2-feed liq-liq extraction column: - two -points (mass balance above and below feed stage) - if F, E 0, R and R N+ are specified, same stage-by-stage analysis But R N+ is an internal stream, usually not specified. Usually specified: F, x A,F, x D,F plot F y A,0, y D,0 plot E 0 x A, plot R on sat d raffinate curve x A,PE, x D,PE plot P E (same location as R N+ and Q, different flow rates) y A,SR, y D,SR plot S R R N+ /P E F = F + R can t locate F (or E ) because we don t know R

Mass balance: solvent separator R N+ Q S R solvent separator P E = Q + S R = R N+ + P E + S R is co-linear with Q and S R. also lies on sat d extract line. Obtain /S R from lever-arm rule. We will also need R N+ /S R : A S R E 0 P E, Q, R N+ F D R S R = R N+ S R S R S R P E = R N+ P E + P E + S R don t know ++ S R P E S R P E = R N+PE + SR R N+ S R = S R P E S R

Finding the -points 2 = - R N+ 2 x A, 2 = y A,N - R N+ x A,N+ x A, 2 = y A,N R N+ x A,N+ R N+ 0.9 0.8 Acetone-water-trichloroethane at 25 C and atm We don t know the individual flow rates, R N+, but we know /S R and R N+ /S R. We can calculate x A, 2 and thereby locate 2 on the R N+ line. Locate at the intersection of two mixing lines: 2 = E 0 - R F = + 2 Proceed to step off stages. weight fraction acetone 0.7 0.6 0.5 0.4 0.3S R 0.2 E 0 0. P E, Q, R N+ 0 0 0. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 F weight fraction water R