Extraction. - separation of a liquid mixture by adding a liquid solvent forming a second phase enriched by some component(s) Notation:

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Tobias Shelton
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1 Extraction - separation of a liquid mixture by adding a liquid solvent forming a second phase enriched by some component(s) Notation: A extracted component B added solvent C original solvent F feed R rafinate S solvent E extract Two cases: 1) immiscible solvents B and C (only A is transported) 2) partially miscible solvents (A is transported, but B and C also a little) Example: extraction of acetic acid from water solution by adding an ether; after extraction, ether is regenerated by distillation; this is cheaper than distilling water solution of acetic acid directly Extractors 1) staged a. cascade of mixers and separators (settlers) b. column with plates (vibration is used to facilitate separation) 2) continuous a. packed or spray columns (slow) b. centrifuge (fast) General extractor is described as a general equilibrium stage (see earlier chapter), only enthalpy balance is often neglected (no heat effects). For correction to non-equilibrium efficiency - only A crosses the interface - F and R contain only A+C (phase x) - S and E contain only A+B (phase y) It is convenient to use relative mass fractions: Special case immiscible solvents X A = m A = x A ; x 1 x A = X A ; m A 1 + X A = x A (m A + ) A

2 Y A = m A = y A ; y m B 1 y A = Y A ; m A 1 + Y A = y A (m A + m B ) A m B and are constant. Single stage For mass balances we use finite time intervals no flows only masses - mass balance of A at steady state: X AF + m B Y AS = X A + m B Y A or - equilibrium relation for A: y A = ψ A x A or Y A = φ A X A By using the definition of relative fractions we obtain: (X AF X A ) = m B (Y A Y AS ) Y A = ψ A 1 (ψ A 1)X A X A Remark: for low concentrations x A = X A and y A = Y A and thus ψ A = φ A = const To find X A, Y A we solve the balance and equilibrium equations simultaneously either graphically or algebraically. Graphical solution: We rewrite the balance of A: Y A = m B (X A X AF ) + Y AS which is an operational line with slope equal to m B passing through point [X AF, Y AS ].

3 Algebraic solution: By combining balance of A and equilibrium relation X AF + m B Y AS = X A ( + ψ A m B ) / 1 X AF + m B Y AS = X A (1 + ψ Am B ) we introduce an extraction factor ζ x = ψ A m B and ζ y = ψ A m B = 1 ζ x Thus: X A = (1 + ζ x ) 1 (X AF + Y AS m B ) In general, solution of this equation is by iterations, because ζ x may depend X A on through ψ A = ψ A (X A ). That is, we estimate X A (0), then calculate XA (1) from the equation, and then use XA (1) to calculate XA (2), etc. until X A (n+1) XA (n) < ε where ε is small. For a real stage we have to combine mass balance, equilibrium relation and efficiency. We denote equilibrium composition by stars: Y A = φ A X A And efficiency is: E A = Y A Y AS Y A Y AS = X AF X A X AF X A

4 Graphical solution Algebraic solution 1) solution for equilibrium case: 2) from efficiency we have: X A = (1 + ζ x ) 1 (X AF + Y AS m B ) X A = X AF E A (X AF X A ) Repeated extraction (or cross-flow extraction) - mass balance of A in the entire cascade: or X AF + m B,n Y AS = X A + m B,n Y A (X AF X A ) = m B,n ( Y A Y AS ) = m A where m A is the mass of A being transferred from phase x to phase y. This equation serves for determining m A or Y A

5 - mass balance of A in the stage n X A,n 1 + m Bn Y AS = X A,n + m B,n Y A,n ; n = 1 N or Y A,n = m B,n (X A,n X A,n 1 ) + Y AS which are operational lines with slopes m B,n passing through points [X A,n 1, Y AS ]; n = 1 N Equilibrium condition (in equilibrium stages): Y A,n = φ A X A,n ; n = 1 N Solution of mass balance and equilibrium conditions leads to all N unknown pairs X A,n, Y A,n. Graphical solution - intersection of operational lines and equilibrium curve - either repeat N times (if N is given) or do until a given is reached Algebraic solution - the use of combined balance and equilibrium equations with extraction factor ζ x,n - in analogy to one extractor: where X A,n = (1 + ζ x,n ) 1 (X A,n 1 + Y AS m B,n ) ζ x,n = ζ A,n m B,n simulation (or control) calculation

6 o N is given; X A,n is calculated consecutively, n = 1 N design calculation o X A,n is given, N is calculated either for n = 1,2 until X A,N is reached or for n = N, N 1 until X A,0 is reached (more convenient if φ A,n depends on X A,n ) For a real cascade efficiency must be used in addition to mass balance and equilibrium condition. - for equilibrium we use asterisks: - the efficiency is Graphical solution (analogous to one stage): Y A,n = φ A,n X A,n E A = Y A,n Y AS = X A,n 1 X A,n Y A,n Y AS X A,n 1 X A,n Algebraic solution 1) solution for the equilibrium stage n: 2) the use of efficiency: X A,n We can introduce overall efficiency: and use it to calculate N real. Special case: E C = = (1 + ζ x,n ) 1 (X A,n 1 + Y AS m B,n ) X A,n = X A,n 1 E A,n (X A,n 1 X A,n ) N number of equilibrium stages = N real number of real stages a) m B,n = const = m B b) φ A,n = const = φ A linear equilibrium line then also ζ x,n = const = ζ x = φ A m B

7 Combined mass balance and equilibrium equation reads: X A,n = (1 + ζ x ) 1 (X A,n 1 + Y AS m B ) let us assume for simplicity that the solvent B is pure Y AS = 0. Then X A,n X A,n 1 = (1 + ζ x ) 1 or X A,n 1 X A,n = 1 + ζ x For all stages together: X A,0 X A,1 X A,N 1 = (1 + ζ X A,1 X A,2 X x ) N A,N X A,0 X A,N A more general form with Y AS > 0 is: X A,0 Y AS φa X A,N Y AS φa = (1 + ζ x ) N Remark: A similar formula with efficiency for real stages can be derived. Counter-current cascade (for flow systems) Mass balance for A in the entire cascade or m CX AF + m BY AS = m CX A,N + m BY A,1 m C(X AF X A,N ) = m B(Y A,1 Y AS ) = m A where m A is flow of A from phase x to phase y. Mass balance of A in stage n: m B(Y A,n Y A,n+1 ) = m C(X A,n 1 X A,n ); n = 1 N or

8 (Y A,n Y A,n+1 ) (X A,n 1 X A,n ) = m C m B operational line with slope m C passing through the points [X m A,n 1, Y A,n ]; n = 1 N B Equilibrium condition: Y A,n = φ A,n X A,n ; n = 1 N Graphical solution - slope is constant, all points [X A,n 1, Y A,n ] lie on one line - stages are found as rectangular steps Minimal consumption of the solvent (B) - two cases o no tangent o tangent

9 Minimal consumption can be calculated from: m C m B,min = (Y A,max Y AS ) (X AF X A,N ) m (X AF X A,N ) B,min = m C (Y A,max Y AS ) Remark: if m B = m B,min then N ; in practice m B = γm B,min where γ = Algebraic solution The mass balance in stage n can be rewritten as m CX A,n 1 + m BY A,n = m CX A,n + m BY A,n+1 = m A where m A is constant for all stages and can be determined from balance of all stages. By using equilibrium condition Y A,n = φ A,n X A,n we get: X A,n 1 = φ A,nm B m A + m C m C ζ x,n This formula can be used repeatedly to calculate N if is X A,N given or X A,N if N is given. For real stages a new efficiency must be defined Murphree efficiency where X A,n E X,n = X A,n 1 X A,n X A,n 1 X A,n = Y A,n φa,n (i.e. Y A,n is for real stage, not Y A,n for equilibrium stage). E Y,n = Y A,n Y A,n+1 Y A,n+1 Y A,n

10 where Y A,n = φ A,n X A,n (i.e. X A,n is for real stage, not X A,n for equilibrium stage). The two efficiencies E X,n and E Y,n are not equal. Graphical representation Graphical solution of the cascade Special case - φ A,n = const = φ A linear equilibrium then ζ X = φ Am B m C = const we introduce effectiveness factors Graphical representation η X = X AF X AN X AF Y AS φ A ; η Y = Y A1 Y AS φ A X AF Y AS

11 For η X = 1 or η Y = 1 the cascade would have N. It can be shown that then for ζ Z = 1: 1 η X = ( 1 N ) = ζ N 1 η Y ζ Y and η X = η Y ; η X ζ Y = η X Y ζ X N = ln (1 η Z ζz 1 η ) Z ln ζ Z ; where Z is either X or Y η Z N = 1 η Z General case partially miscible solvents In addition to A, also B and C cross the interface. Therefore, all three mass balances must be used now relative mass fractions are not useful ordinary mass fractions should be used together with masses of the streams, not just. Equilibrium: - formally y k = ψ k x k ; k = A,B,C - ψ k depends on composition and temperature - graphical representation in a triangle diagram is more convenient x C = 1 x A x B ; y C = 1 y A y B

12 - data for the equilibrium curve are typically given in tables Mass balance of A, B, C is graphically represented in the triangle diagram by the lever rule. Extractor is thought of as a combination of a mixer and settler. - overall mass balance: m F + m S = m M = m R + m E - component k: m F x kf + m S y ks = m M z km = m R x k + m E y k Combination yields: m F (x kf y ks ) = m M (z km y ks ) in analogy the lever rule for the settler is or m F m M = y ks z km y ks x kf = SM SF lever rule for mixer m R = ME m M RE Other variants of the lever rule (less convenient for calculations)

13 - mixer: - settler: m F = MS m S FM m R = ME m E RM Repeated extraction (cross-flow) mass balance of stage n (assume constant mass of solvent) - overall: m R,n 1 + m S = m M,n = m R,n + m E,n - components: m R,n 1 x k,n 1 + m S y ks = m M,n z k,n = m R,n x k,n + m E,n y k,n This implies lever rules for points - mixer (R n 1, M n, S); n = 1 N - settler (R n, M n, E n ); n = 1 N Equilibrium in stage n: y k,n = ψ k,n x k,n ; n = 1 N or via graph Graphical solution a) lever ruler for stage 1: o mixer: m F = SM 1 determine M m M 1 SF 1 o then tie line and lever rule for settler: m R1 = M 1E 1 m E R 1 1 E 1 b) for stage 2 determine m R1

14 c) etc. o o mixer: settler: m R 1 m R 1 +m S m R 2 m R 1 +m S = SM 2 SR 1 determine M 2 = M 2E 2 R 2 E 2 determine m 2 Counter current cascade of extractors (or counter-current staged column) Mass balance of all stages summing form: - overall: m F + m S = m M = m R N + m E 1 - components: m Fx kf + m Sy ks = m Mz km = m R N x k,n + m E 1 y k,1 These equations imply lever rules connecting points (F, M, S) and (R N, M, E 1 ) Mass balance of stages 1 to n difference form: - overall: m F m E 1 = m R n m E n+1 = m P = const - components: m Fx kf m E 1 y k,1 = m R n x k,n m E n+1 y k,n+1 = m Pz kp = const Here the point P with composition z kp is hypothetical but useful; lever rule applies on the lines connecting (F, E 1, P) and (R n, E n+1, P) for n = 1 N in particular (R N, E N+1 =S, P ) Equilibrium in stage n - formally: y k,n = ψ k,n x k,n ; n = 1 N - in graph: equilibrium curve + tie lines Graphical solution 1) Summing form of mass balances serves to find the point M and then the point E 1 provided that the compositions of F, S and R N are given. Lever rule: m F m F+m S = SM SF determine M, then connect R N and M to find E 1 2) Difference form of mass balances serves first to find the point P (=pole) as intersection of line (F, E 1 ) and (R N, S). The tie line from E 1 determines R 1 and connection of P with R 1 determines E 2 (see the difference form of balances). This is repeated until R N is reached. In the figure below six step were needed to reach R N N = 6.

Liquid Liquid Extraction

Liquid Liquid Extraction By Dr. Salih Rushdi 1 Introduction Liquid-liquid extraction (sometimes abbreviated LLX) is a mass transfer operation in which a solution (called the feed which is a mixture of