LIQUID-LIQUID EXTRACTION (LLE)

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1 LIQUID-LIQUID EXTRACTION (LLE) Expected Outcomes Students should be able to: 1. Explain the basic theory and principle of liquid-liquid extraction. 2. Demonstrate the liquid-liquid extraction under batch and continuous mode. 3. Calculate the number of theoretical stages and height equivalent to a theoretical stage using theory as well as graphical methods. 4. Discuss industrial liquid extraction equipment.

2 Easy Difficulty of separation Difficult

3 EXAMPLES OF LLE COLUMN IN PLANT WHAT DO YOU SEE?? DOES IT LOOK EASY OR COMPLICATED?? Source from: 3

4 Figure 8.1 TYPICAL LLE SYSTEM OF ACETIC ACID IN WATER o If mixture has purity < 50%, then extraction is considered ousually will followed by 2 distillation columns to separate Extract stream (D1) and Raffinate stream (D2) oaddition of solvent to solute soluble species in Feed (to be 3 components in liquid phase) 4

5 Table major components in LLE Carrier =A ; Solute = B ; Solvent = C Justification of solvent selection is based on solubility of the solute with the solvent (due to chemical structure difference in the components) Choice of distillation/extraction for separation depends on cost Examples of LLE in industrial chemical plants: Acetic acid in water using Ethyl acetate Penicillin in broth in pharmaceutical High MW of fatty acids separated from vegetable oils with liquid propane Lecture 7: Liquid-Liquid Ternary Single Stage 5

6 BASIC PRINCIPLE FOR LIQUID-LIQUID EXTRACTION

7 Liquid-Liquid Ternary Single Equilibrium Stage Ternary Liquid-Liquid Extraction (LLE) In this case: We create two liquid phases by introducing a solvent (C) to a liquid mixture of a carrier (A) and a solute (B) Solvent (C) and carrier (A) have very little solubility in each other while solvent (C) and solute (B) is highly soluble or partly soluble in each other Liquid-Liquid Extraction Solvent Feed, S Solvent Rich Liquid Out (Extract), E C A, B, C Liquid Feed, F Carrier Rich Liquid out (Raffinate), R A, B A, B, C Define the raffinate as the exiting phase rich in carrier ; the extract as the exiting phase rich in solvent. 7

8 Liquid-Liquid Ternary Single Equilibrium Stage Case 1 (carrier and solvent have some solubility in each other) then the raffinate will have a small amount of solvent in the stream and the extract will have a small concentration of carrier: Solvent Feed S : X C (S) [T, P] Liquid Feed F : X A (F), X B (F), [T, P] Liquid-Liquid Extraction Extract out E : X A (E), X B (E), X C (E) [T, P] Raffinate out R : X A (R), X B (R), X C (R), [T, P] A= carrier, B= solute, C= solvent The raffinate is the exiting phase rich in carrier. The extract is the exiting phase rich in solvent. All A, B, C components are in the Extract and Raffinate 8

9 Liquid-Liquid Ternary Single Equilibrium Stage Case 2 (carrier and solvent have no solubility in each other) then the raffinate will have no solvent in then stream and the extract will have no carrier in it: Solvent Feed S : X C (S) [T, P] Liquid Feed F : X A (F), X B (F) [T, P] Liquid-Liquid Extraction Extract out E : X B (E), X C (E) [T, P] Raffinate out R : X A (R), X B (R) [T, P] All of solvent exits in the extract A= carrier, B= solute, C= solvent The raffinate is the exiting phase rich in carrier. The extract is the exiting phase rich in solvent. All of carrier exits in the raffinate 9

10 Mass and Mole Ratios Often the concentrations are as mass or mole ratios, rather than mass or mole fractions. This is generally done to simplify the expressions used in the analysis. Mass ratio X B : The ratio of mass of component B to another component of the stream. Mole ratio X B : The ratio of moles of component B to another component of the stream. Note that the basis (choice of component) for the mass or mole ratio must be chosen. Mass Ratio Example: F F B X B FA Solvent Feed S : X C (S) [T, P] Liquid Feed F : X A (F), X B (F) [T, P] E E E B X B EC X B S R R R B X B RA X B FA S S B X B SC 0 No solubility of solvent with carrier Extract out E : X B (E), X C (E) [T, P] Raffinate out R : X A (R), X B (R) [T, P] Rate of B in the feed is the ratio of B to A, times feed rate of A. Rate of B in the extract is the ratio of B to C, times rate of C. Rate of B in the raffinate is the ratio of B to A, times rate of A. Rate of B in the solvent is the ratio of B to C, times feed rate of C. 10

11 Material Balances Solute (B) Material Balance: Solvent Feed S ; X C (S) Liquid Feed F : X A (F), X B F F B X B FA E E E B X B EC X B S R R R B X B RA X B FA Extract out E : X B (E), X C (E) Raffinate out R : X A (R), X B (R) Rate of B in the feed is the ratio of B to A, times feed rate of A. Rate of B in the extract is the ratio of B to C, times rate of C. Rate of B in the raffinate is the ratio of B to A, times rate of A. S B X S B S C 0 Rate of B in the solvent is the ratio of B to C, times feed rate of C. Solute, B, Material Balance: F B S B R B E B F S R E X B FA X B SC X B RA X B EC F R E X B FA X B FA X B S 11

12 Equilibrium Distribution The way the solute (B) will distribute itself between the extract and raffinate at equilibrium is given by the K-Value: E X B ' R KDB X B Note that the K-value is primed to signify that this is a ratio of mass or mole ratios, not a ratio of mole fractions. Solvent Feed Extract out S: X C (S) B E: X B (E), X C (E) Liquid Feed Raffinate out F: X A (F), X B (F) R: X A (R), X B (R) B will be transferred from pink to blue part in the LLE column Note that concentrations of exiting streams from an equilibrium stage are always related by equilibrium. 12

13 The Extraction Factor The degree of separation of the solute (B) between the exiting streams (extract and raffinate) is expressed as the extraction factor: B Extraction Factor, : The ratio of solute flow in the extract to solute flow in the raffinate. E E E B X B EC X B S R R R B X B RA X B FA B E R B B X X E B R B S F A E C = S (all solvent in Extract), R A = F A (all carrier in Raffinate) Combining this definition with the equilibrium relationship: E X B ' R KDB X B results in another expression for the extraction factor: B ' KD B F A S The larger the equilibrium ' driving force to separate B, K DB, and the larger the ratio of solvent to feed (S/F A ), the larger the extraction factor B. 13

14 Extraction Efficiency We can determine the amount not extracted (ratio of R B /F B ) starting with the material balance of the solute (B): F E X B FA X B S R XB FA We substitute in the K-value ratio: F ' R X B FA K DB X B S R XB FA E E E B X B EC X B S R R R B X B RA X B FA And simplify the above equation: This ratio gives the amount of solute (B) left in the raffinate to the amount originally in the feed stream, R X B F X B ' K DB F A S F A R X B 1 ' K DB S 1 F A X B F 1 B 1 The larger the extraction factor, the smaller fraction of solute (B) not extracted 14

15 Ternary Phase Diagrams It is convenient to construct ternary phase diagrams on a Gibbs Triangle (shown at right). Note that the variables for these diagrams are only composition and that pressure and temperature are held constant (means that these diagrams are slices through a four dimensional space with constant T and P). C At each point, x A + x B + x C = points located on the diagram: C B Point 1: x A =? ; x B =? ; x C =? Point 2: x A =? ; x B =? ; x C =? Point 3: x A =? ; x B =? ; x C =? Point 4: x A =? ; x B =? ; x C =? A B Lecture 7: Liquid-Liquid Ternary Single Stage A 15

16 Ternary Phase Diagrams Compositions are read as follows: Draw three lines from the Composition point parallel to the composition lines. Read the compositions off of the three axes. Note: Only two mole fractions are needed. Meaning that the total percentage is 100% at each point. So the third fraction as a check only. Compositions can be in mole fractions or mass fractions. C [94% C, 3% B, 3% A] [33% C, 33% B, 33% A] [30% C, 70% B] [100% A] A B Lecture 7: Liquid-Liquid Ternary Single Stage [20% C, 20% B, 60% A] 16

17 Partially Soluble Ternary Systems If the two phases both have a partial solubility of the other component, then the analysis is somewhat more complicated. The difficulty is that now equilibrium data must be obtained for the ternary which relates the partial solubilities. Equilibrium data can be obtained graphically, or from tables. The ternary phase diagram is a typical way of representing the equilibrium compositions of the two phases. Ethylene Glycol Solute 66% EthGly 7% Furfural 27 % Water A composition where only a single liquid exists. Lecture 7: Liquid-Liquid Ternary Single Stage 50 % EthGly 50% Furfural 100% Furfural Furfural Solvent 17% EthGly 27% Furfural 56 % Water A composition where two liquid phases coexist. Water Carrier 21

18 Specification of Liquid-Liquid Equilibrium For two phase equilibrium (either complete insolubility, or partially solubility): the equilibrium is between two liquids phases ( = 2) three components (ternary) distribute between the two phases (C = 3) For the static equilibrium case we can specify 3 variables: If we specify T and P we are left with one additional variable: Thus if we specify the concentration of one component in either of the phases this completely defines the state of the system. Ethylene Glycol Water Equilibrium-Line: Show the compositions of the equilibrium phases. Tie-Lines: Show the compositions of the equilibrium phases. Lecture 7: Liquid-Liquid Ternary Single Stage Furfural 22

19 Partially Soluble Ternary Systems Example: Consider a feed of 200 kg of 30% ethylene glycol in water. Add 300kg of pure furfural solvent. Solvent Feed Extract out S, X C (S) E, X B (E), X C (E) Liquid Feed Raffinate out F, X A (F), X B R, X A (R), X B (R) Ethylene Glycol Water Furfural Lecture 7: Liquid-Liquid Ternary Single Stage 23

20 Determination of Solvent Free Point In order to know the highest percent of separation (Separation efficiency) that we might get in certain feed (mixtures of solute and carrier) solvent flows, these steps need to be done in a specific ternary diagram: Step 1: Locate the Solvent and Feed points Step 2: Locate the mixing point M Step 3: Use the tie-line to get the raffinate and extract compositions Step 4: Determine the amount of extract and raffinate (can use lever rule) Step 5: Determine the solvent free extract: Mixtures of A and B. Extend line from S through E to solvent free point at H. Lecture 7: Liquid-Liquid Ternary Single Stage 24

21 Partially Soluble Ternary Systems Example: Consider a feed of 200 kg of 30% ethylene glycol in water. Add 300kg of solvent which is pure furfural. Step 1: Locate the Solvent and Feed points Ethylene Glycol Water F 60 kg EG 140 kg water S 300 Kg Furfural Lecture 7: Liquid-Liquid Ternary Single Stage 25

22 Partially Soluble Ternary Systems Step 2: Locate the mixing point M: FA X B S X B F F S S kg 0 300kg 500kg 0.12 From Solute (B) Ethylene Glycol Water F 60 kg EG 140 kg water M S 300 Kg Furfural Lecture 7: Liquid-Liquid Ternary Single Stage 26

23 Partially Soluble Ternary Systems Step 3: Use the tie-line to get the raffinate and extract compositions. Extract (4% water, 14%EG, 82% furfural) Raffinate (87% water, 8%EG, 5% furfural) Get the most accurate ratio of both extract and raffinate between the two tie lines that locate M Ethylene Glycol Water E M F 60 kg EG 140 kg water R S 300 Kg Furfural Lecture 7: Liquid-Liquid Ternary Single Stage 27

24 Partially Soluble Ternary Systems Step 4: Determine the amount of extract and raffinate (can use lever rule) R f X X M E C X C R E C X C R kg 128.4kg E kg 371.6kg Extract (4% water, 14%EG, 82% furfural) Raffinate (87% water, 5%EG, 8% furfural) Mixing (25%, 12%EG, 63% furfural) E=M(distance of RM/RE) Ethylene Glycol Water E M F 60 kg EG 140 kg water R S 300 Kg Furfural Lecture 7: Liquid-Liquid Ternary Single Stage 28

25 Partially Soluble Ternary Systems Step 5: Determine the solvent free extract: Mixtures of A and B only. Extend line from S through E to solvent free point at H. Solvent free extract H (20% water, 80% EG) (Separation efficiency) H Ethylene Glycol Water E M F 60 kg EG 140 kg water R S 300 Kg Furfural Lecture 7: Liquid-Liquid Ternary Single Stage 29

26 HUNTER NASH GRAPHICAL EQUILIBRIUM-STAGE METHOD FOR LIQUID-LIQUID EXTRACTION To find: Number of equilibrium stages needed for a countercurrent cascades LLE and Minimum@maximum solvent-to-feed (S/F) min@max Lecture 7: Liquid-Liquid Ternary Single Stage 30

27 Ternary Liquid-Liquid Extraction: Single-Stage Solute, B Ternary Phase Diagram Solvent: tetrachloroethane (TCE) Solute: Acetone Carrier: Water Plait Point P Extract E Mixing point Tie-lines F R Feed Raffinate S Lecture 7: Liquid-Liquid Ternary Single Stage Solvent C Carrier, A 31

28 Liquid-Liquid Extraction: Cascades What if we have a countercurrent cascade of Liquid-Liquid Contacting Stages? Extract F Carrier A (F A ) Solute B (F B ) E 1 E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R 2 R n-1 R n R N-2 R N-1 R N Solvent C S Raffinate Can we use a similar analysis to the one we used for countercurrent absorption or stripping? Yes, but also different since equilibrium here is given by a liquid-liquid ternary diagram rather than a vapor-liquid equilibrium. Lecture 7: Liquid-Liquid Ternary Single Stage Considerations: countercurrent flow; N-equilibrium-stage contactor for LLE under isothermal, continuous, steady state conditions at a pressure sufficient to prevent vaporization 32

29 Liquid-Liquid Extraction: Specifications Extract F Carrier A (F A ) Solute B (F B ) E 1 E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R 2 R n-1 R n R N-2 R N-1 R N Solvent C S Raffinate Specifications: F, (x i ) F, (y i ) S, T and one of: 1) S and (x i ) RN 2) S and (y i ) E1 3) (x i ) RN and (y i ) E1 4) N and (x i ) RN 5) N and (y i ) E1 6) S and N Lecture 7: Liquid-Liquid Ternary Single Stage 33

30 Liquid-Liquid Extraction: Hunter Nash Method Extract F Carrier A (F A ) Solute B (F B ) E 1 E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R 2 R n-1 R n R N-2 R N-1 R N Product Points: Step 1) Find the mixing point M=F+S Step 2) Determine mixing point compositions from component material balances or inverse lever rule Step 3) Since we know R N lies on the equilibrium curve and we know (x A ) RN we can determine (x B ) RN and (x C ) RN Step 4) Since we know R N, M and E 1 lie on a mixing line we can locate E 1 by extending a line from R N through M to the equilibrium curve where it intersects E 1. Extract E1 M Tie-lines Solute Plait Point P Solvent C S Raffinate Feed Raffinate R1 RN Lecture 7: Liquid-Liquid Ternary Single Stage Solvent C Carrier 34

31 Hunter Nash Method: Operating Lines Extract F Carrier A (F A ) Solute B (F B ) E 1 E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R n-1 R n R R N-1 R N R2 N-2 S Solvent C Raffinate Operating Points and Lines Mass Balance around entire cascade: We define the operating point P as the difference between passing streams: F S R N E 1 F E 1 R N S P Extract F Carrier A (F A ) Solute B (F B ) E 1 E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R n-1 R n R R N-1 R N R2 N-2 S Solvent C Raffinate Mass Balance around the first n stages: F E n 1 R n E 1 Lecture 7: Liquid-Liquid Ternary Single Stage We rearrange this equation to find that all passing streams are related by the same operating point P. F E 1 R n E n 1 P 35

32 Hunter Nash Method: Operating Point Extract E 1 F E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R n-1 R n R R N-1 R R N 2 N-2 S Solvent C Raffinate Operating Points and Lines Mass Balance around an internal stage (n): We can rearrange the above expression to find that R n is just a mixing point between P and E n+1. R n E n R n 1 E n 1 R n R n 1 E n E n 1 P E n 1 The following figure illustrates this concept: The stream R n is the mixing point between P and E n+1 because P is the net flow into stage n from passing streams R n-1, and E n. E n E n+1 n R n-1 R n Replace R n-1 and E n by P P n E n+1 R n The N mass balances around the N individual stages result in: F E 1 R n 1 E n R N S P Extract Lecture 7: Liquid-Liquid Ternary Single Stage E 1 F E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R R n-1 R n R R N-1 2 N-2 R N S Solvent C Raffinate 36

33 Hunter Nash Method: Operating Point Extract F Carrier A (F A ) Solute B (F B ) E 1 E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R n-1 R n R R N-1 R N R2 N-2 Solute Plait Point Operating Points and Lines Step 1) Locate the Operating Point by finding the intersection of operating lines for the leftmost and rightmost stages a) Draw a line through E 1 and F b) Draw a line through S and R N c) Locate the intersection P. This point is the operating point P. S Solvent C Raffinate E 1 M Feed Operating Point P Lecture 7: Liquid-Liquid Ternary Single Stage S R N Carrier 37

34 Hunter Nash Method: Stepping off Equilibrium Stages Operating Lines: The raffinate points are mixing points between P and corresponding extract points. This is shown graphically in the following diagram. Notice that to get the point P we need just F, S, E 1 and R N. Solute E6 E5 E4 E2 E3 E1 R1 F R N Operating Point P Lecture 7: Liquid-Liquid Ternary Single Stage S Carrier 38

35 Hunter Nash Method: Stepping off Stages Extract E6 E5 Lecture 7: Liquid-Liquid Ternary Single Stage E4 Solvent C F Carrier A (F A ) Solute B (F B ) E 1 E2 E3 E1 E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R n-1 R n R R N-1 R N R2 N-2 Solute Plait Point M R1 Feed R N Operating Lines and Tie Lines: Stepping Off Stages: Step 1) Locate point R 1 from the tie line intersecting E 1 Step 2) Draw a line from the operating point P through R 1 to the extract side of the equilibrium curve. The intersection locates E 2. Step 3) Locate point R 2 from a tie line. Step 4) Repeat Steps 2 and 3 until R N is obtained. Operating Point P In Carrier Step 3, Get the most accurate ratio of both E n and R n-1 between the two tie lines S Solvent C Raffinate 39

36 Extract F Carrier A (F A ) Solute B (F B ) E 1 Hunter Nash Method: Stepping off Stages E 2 E 3 E n E n+1 E N-1 E N 1 2 n N 1 N R 1 R n-1 R n R R N-1 R N R2 N-2 S Solvent C Raffinate Product Points: Step 1) M=F+S Step 2) Determine mixing point compositions from component material balances or inverse lever rule Step 3) Since we know R N lies on a tie line and we know (x A ) RN we can determine (x B ) RN and (x C ) RN Step 4) Since we know R N, M and E 1 lie on a mixing line we can locate E 1 by extending a line from R N through M to the equilibrium curve where it intersects E 1. Operating Points and Lines Step 1) Locate the Operating Point by finding the intersection of operating lines for the leftmost and rightmost stage 1a) Draw a line through E 1 and F 1b) Draw a line through S and R N 1c) Locate the intersection P. This point is the operating point P. Lecture 7: Liquid-Liquid Ternary Single Stage Operating Lines and Tie Lines: Stepping Off Stages: Step 1) Locate point R 1 from the tie line intersecting E 1 Step 2) Draw a line from the operating point P through R 1 to the extract side of the equilibrium curve. The intersection locates E 2. Step 3) Locate point R 2 from a tie line. Step 4) Repeat Steps 2 and 3 until R N is obtained. 40

37 Hunter Nash Method: Minimum Solvent-to-Feed (S/F) min Solute Operating Points and Lines Step 1) Locate the raffinate Operating Line by extending a line from S through R N Step 2) Extend the tie lines to intersect the operating line Step 3) The tie line that intersects furthest from R N gives the minimum operating point, P min. Step 4) Extend a line from P min through F to the extract side of the equilibrium curve to find E 1. Step 5) Extend a line from E 1 to R N. The intersection with the line SF gives the minimum mixing point, M. E 1 Plait Point Note: If the tie lines slope down towards the solvent side of the diagram, then the minimum operating point will lie on the operating line at an intersection with a tie line nearest S. F M min S R N P min M max Lecture 7: Liquid-Liquid Ternary Single Stage Carrier 41

38 Hunter Nash Method: Minimum Solvent-to-Feed (S/F) min Solute Stepping off Stages for the minimum solvent case After locating the points P min, E 1, and M min the stages can be stepped off. If the minimum solvent is used then the separation will be pinched off and will require an infinite number of stages [means if we use (S/F) min, then N will be infinite] Plait Point E 1 F M min S R N P min Lecture 7: Liquid-Liquid Ternary Single Stage Carrier 42

39 Hunter Nash Method: Maximum Solvent-to-Feed (S/F) max Solute Stepping off Stages for the maximum solvent case After locating the points P min, E 1, and M min the stages can be stepped off. If the maximum solvent is used then the separation will require a minimum number of stages. A solvent ratio (S/F) must be selected such that (S/F) min < (S/F) < (S/F) max Plait Point F S R N P max M max Lecture 7: Liquid-Liquid Ternary Single Stage Carrier 43

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43 Use of Right-Triangle Diagram Ternary, countercurrent extraction calculations also can be made on a right-triangle diagram. Disadvantage: mass-percent compositions of only two of the components are plotted; the third being determined. Advantage: ordinary rectangular-coordinates graph paper can be used and either one of the coordinates can be expanded (to increase the accuracy of the constructions). 47

44 Wt fraction solute Use of Right-Triangle Diagram How to plot right- triangle diagram 1. Draw right-triangle with scale 2. Label x-axis and y-axis with solute, solvent or carrier 3. Construct equilibrium line by using equilibrium data 4. Draw tie-lines 5. Determine the location for Feed (F) point, Solvent (S) point, Extract (E) point and/or Raffinate (R) point and/or Mixing Point (M) point. Construction of Operating Line - The same method used as in ternary phase diagram Solute F Determination of composition in extract, raffinate and no. of equilibrium stages - The same method used as in ternary phase diagram (mass balance) 0.2 R S Solvent Carrier Wt fraction carrier 48

45 Use of Right-Triangle Diagram Examples of right-triangle diagrams Lecture 7: Liquid-Liquid Ternary Single Stage 49

46 Use of Kremser equation Ternary phase diagram & right-triangle diagram are used for partially miscible ternary system. While McCabe-Thiele or Kremser equation are used for completely immiscible ternary system. N = ln 1 me R ln y N+1 y 1 y1 y 1 R me + me R where me R = ye (Rx) Lecture 7: Liquid-Liquid Ternary Single Stage 50

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