Mass transfer and separation technology Massöverföring och separationsteknik ( MÖF-ST ) , 7 sp

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1 Mass transfer and separation technology Massöverföring och separationsteknik ( MÖF-ST ) , 7 sp 14. Extraction and leaching Ron Zevenhoven Åbo Akademi University Thermal and Flow Engineering Laboratory tel ; ron.zevenhoven@abo.fi 14.1 Liquid-liquid extraction mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik 2

2 Liquid-liquid extraction /1 Liquid-liquid extraction involves three stages: Contacting / extraction Separation, and Solvent recovery Important examples are Separation of aromatics from paraffins Metallurgical processes (acid solid-liquid extraction leachates) see section 9.2 Picture: 3 Liquid-liquid extraction /2 L-L extraction replaces or complements to distillation if distillation would take too much heat, or too many stages azeotropes are formed, heating must be avoided, the components to be separated are very different Pictures CRBH83 4

3 Liquid-liquid extraction /3 Often, cascaded processes are used with several sections and/or different solvents Terminology: the component of main interest is transferred from the feed to the extract, the residue of the feed is the raffinate Picture 5 Liquid-liquid extraction /4 Typical equipment used are Mixer-settlers see section 9.2 Tray or packed columns, spray columns (a) Columns with mechanical agitators or centrifugal effects, for example rotating disc contactors (RDC) (b), Kühni columns (c) a b c Pictures 6/34

4 14.2 Equilibrium mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik 7 Liquid-liquid equilibrium /1 Assuming that the carrier liquid (C) and the solvent (S) are immiscible the component of interest solute (A) will distribute over extract (E, phase II or 2) and raffinate (R, phase I or 1). This is determined by an equilibrium condition for the concentrations with equilibrium constant K: c A,II /c A,I = K, or x A,E /x A,R = x A,II /x A,I = x A,2 /x A,1 = K K = K (ρ mol,e / ρ mol,r ) Note: x A + x S + x C = 1 8/34

5 Liquid-liquid equilibrium /2 Similar to relative volatility for G/L systems, a relative selectivity β AC of the solute with respect to the carrier can be defined as β AC = (x A,II / x A,I ) / (x C,II / x C,I ) = (x A /x C ) II / (x A /x C ) I Adding the solvent S as a third component to the binary system A + C should not produce a homogeneous (one phase) solution Most suitable are two partly miscible liquids, or two immiscible liquids. 9/34 Liquid-liquid equilibrium /3 Picture see p. 22 If the solvent and carrier liquids are immiscible, a plot of x II against x I can be set up; for partial miscibility a ternary diagram is used for the components A, S and C at given T and p. Preferable is a system with one pair of immiscible liquids (left), less preferable is a system of two immiscible pairs (right) 10/34

6 Liquid-liquid equilibrium /4 Typical effect of temperature on a ternary liquid system Picture The phase behaviour is hardly affected by pressure (except near critical conditions) but depends strongly on temperature. Changing temperature can change the type of ternary diagram. 11/34 Liquid-liquid equilibrium /5 Picture The type of ternary system determines the possible separation possible: in the diagrams given above, only C/A mixtures with compositions ranging from C to F can be separated 12/34

7 14.3 Equilibrium stages mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik 13 Graphical method using ternary diagram (Hunter-Nash) /1 A counterflow L-L extraction process (iso-thermal, continuous, steady-state) can be presented as a series of N equilibrium-stage contactors Using the equilibrium constant K, an extraction factor can be defined as E = K S/R, for solvent and raffinate streams S and R Mass balance for stage 1: F + E 2 = E 1 + R 1, for stage n : R n-1 + E n+1 = E n + R n, etc. Picture 14/34

8 Graphical method using ternary diagram (Hunter-Nash) /2 Similar to the McCabe- Thiele method for binary systems, a graphical procedure can be used to calculate the number of equilibrium stages (and reflux ratios) Picture The first step is to determine the so - called mixing point M (from an overall mass balance) for which the ratio of the lenght of line sections M-F/M-S = S/F, with feeds S and F (kg/s or kmol/s) Then, for given R N mass balance gives product E 1 15 Graphical method using ternary diagram (Hunter-Nash) /3 The next step is to find the so-called operation point P, = the point (outside the diagram) where the line through E 1 and F crosses the line through S and R N After this, produce the so-called tie-line to R 1 based on equilibrium with E 1 (the diagram should give the eq. lines!) Picture 16/34

9 Graphical method using ternary diagram (Hunter-Nash) /4 Next, point E 2 is found from line through P raffinate R 1 R 2 follows from equilibrium with E 2, et c. Continue until R n reaches (or passes) final raffinate R N Values for minimum and maximum solvent-to-feed ratio s may also be determined graphically Picture 17/34 Example Hunter-Nash procedure old exam question 382, /34

10 Note Tie lines Example Hunter-Nash procedure old exam question 382, 386 M a. Total mass balance: E 1 + R N = F + S = 0,22 kg/s R N = 0,22 - E 1 p-balance: E 1 x P,E1 + R N x P,RN = F x P,F + S x P,S diagram, point M follows from feed-m : M-benzene = S : F = 1:1 line R N E 1 through R N and M x P,E1 = 0,32 E 1 0,32 + (0,22-E 1 ) 0,05 = 0,11 0,5 + 0,11 0 E 1 = 0,163 kg/s E 1 R N P 19/34 Note Tie lines Example Hunter-Nash procedure old exam question 382, 386 M (a). Total mass balance : E 1 + R N = F + S = 0,22 kg/s R N = 0,22 - E 1 b-balance: E 1 x b,e1 + R N x b,rn = F x b,f + E 2 x b,s E 1 0, = 0 + 0,11 1 E 1 = 0,170 kg/s w-balance: E 1 x w,e1 + R N x w,rn = F x w,f + S x w,s E 1 0,02 + 0,95 R N = 0,5 0,11 + 0,11 0 E 1 = 0,165 kg/s E 1 3 values for E 1 E 1 ~ 0,165 kg/s R N ~ 0,055 kg/s R N P 20/34

11 R 1 Example Hunter-Nash procedure old exam question 382, 386 M c. Mass balance stage 1: E 1 + R 1 = F + E 2 = R 1 = F - E 1 + E 2 *) w-balance: E 1 x w,e1 + R 1 x w,r1 = F x w,f + E 2 x w,e2 E 1 = 0,165 kg/s, R 1 =?, x WE1 = 0,02 E 2 =?, x R1 = 0,79, x WE2 = 0,015 *) in w-balance E 2 = 0,128 kg/s x WE2 = 1,5 %; x BE2 = 88,5 %; x PE2 = 10 % R 1 = 0,128 0,055 = 0,073 kg/s E 1 E 2 b. N = 2 stages R N ~ R 2 Line E 1 -R 1 and E 2 -R 2 on tie line P 21/34 x i,x j diagrams The phase equilibrium information in a ternary diagram can also be used to produce an x 1,x 2 diagram for the two liquid phases (for the two-phase region!) Alternatively, the information from the ternary diagram can be processed into an x C, x A diagram x S x C Pictures 22/34

12 Example extraction: old exam question 422,378,362 Caffeine is to be removed from coffee in an extraction process that uses super-critical CO 2, i.e. scco 2 at T = 60 C, and pressure p = 272 atm (1 atm = 101,3 kpa). The equilibrium for caffeine in CO 2 (weight-% y) and in coffee (weight % x) describes with, x*, y or, x., y*,, x See the x,y phase diagrams A,B, C and D on the next page. The caffeine content in the coffee must be reduced from 1 weight-% to 0,05 weight-%. 23/34 Example extraction: old exam question 422,378,362 y* %-vikt A x %-vikt y* %-vikt 0.02 B x %-vikt 24/34

13 Example extraction: old exam question 422,378,362 y* %-vikt 0.05 C x %-vikt y* %-vikt 0.5 D x %-vikt 25/34 Example extraction: old exam question 422,378,362 a. How much pure scco 2 is needed (as kg scco 2 /kg coffee), for the process to operate with one (i.e. 1) equilibrium stage. (3 p.) In order to obtain a better result (i.e., to decrease the amount of scco 2 needed) the process will be operated as a counter-flow extraction see the Figure above. The scco 2 flow V (kg(h) that is used is 2 V min, where V min is the minimum amount that needed for doing the separation, with N = equilibrium stages. With coffee flow L (kg/h) it then follows that L/V = 0,5 (L/V) max. b. Draw the operating line in the phase diagram (diagram A for 0 < x 0,3 could be the most useful) that would give the maximum ratio (L/V) max, and determine the value for (L/V) max. (3 p.) c. Draw up the operation line for L/V = 0,5 (L/V) max, determine the concentration y N (weight-%) in the scco 2 product stream and determine how many equilibrium stages are needed to reduce the caffeine content in the coffee from from 1 weight-% to 0,05 weight-%. Use diagrams A and B or A and C, or A and D. (4 p.) Hand in the paper with the diagrams with your answer. Note that even if the intervals Δx and Δy are different in the diagrams (i.e. lenghts Δx Δy) this will not give errors when the number of stages is counted. 26/34

14 answer Example extraction: old exam question 422,378,362 a. 1 step: separation factor S = KV/L; not-transferred f = 0.05 = 1/S+1 S = 19 x = 0.05 % y* = % K = y*/x = V/L = S/K = kg scco 2 / kg coffee b. See diagram A, line through (x,y) = (0,05, 0) and tangent to eq. curve: (L/V) max = kg coffee / kg scco 2 = kg scco 2 / kg coffee (L/V) max = 0,0015/0,25 = 0,006 * ½ (L/V) = 0,003 27/34 answer Example extraction: old exam question 422,378,362 c. L/V = ½ (L/V) max = kg coffee / kg scco 2 = kg scco 2 / kg coffee diagram A and then further in diagram (B or) C: 4 equilibrium stages, and << kg scco 2 / kg coffee!!!! 28/34

15 14.4 Solid-liquid extraction (Leaching) mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik 29 Leaching Leaching (or washing) involves removal of a soluble fraction (the leachate) from a solid by a liquid solvent Important applications: Removal of metal from ore Extraction of sugar from sugar beets Removal of caffeine from coffee beans by supercritical CO 2 (sc- CO 2 ) Espresso machines Pictures 30/34

16 Leaching: Example /1 N units V N+1 L N S V 1 L L S L V L F S Source: 31/34 Leaching: Example /2 N units V N+1 V 1 L V L F S L N S L L S The composition of the liquids L i and V i is the same (with L also solids S) 32/34

17 Leaching: Example /3 33/34 Sources #14 CRBH83 J.M. Coulson, J.F. Richardson, J.R. Blackhurst, J.H. Harker Chemical engineering, vol. 2, 3rd ed. Pergamon Press (1983) Ch. 13 dhb07 A.B. De Haan, H. Bosch Fundamentals of industrial separations 2nd Ed., TU Eindhoven / U Twente, the Netherlands (2007) Chapter 5 WH92 J.A. Wesselingh, H.H. Kleizen Separation processes (in Dutch: Scheidingsprocessen) Delft University Press (1992) J.D. Seader, E.J Henley Separation process principles John Wiley, 2nd edition (2006) Chapters 8, 16 Z97 F. Zuiderweg Physical separation methods (in Dutch: Fysische Scheidingsmethoden) TU Delft 1987 (vol. 1, vol. 2) 34/34