Liquid-Liquid Extractions Last lecture covered: • Using a simple spreadsheet to apply the Rachford Rice Procedure • Bubble point and Dew Point temperature and pressure calculations • An example of a dew point temperature calculation This lecture will cover: • Ternary Liquid-Liquid extractions. • Ternary phase diagrams. • A procedure to determine the product compositions and flow rates of a liquid-liquid extraction separation. Lecture 7: Liquid-Liquid Ternary Single Stage 1 Liquid-Liquid Ternary Single Equilibrium Stage We last covered the Flash Calculations where a liquid phase and a vapor phase were in equilibrium. For that system we needed equilibrium data, for example, from a Depriester chart to determine the distribution of components between the two phases. Now we’ll analyze the following system: Ternary Liquid-Liquid Extraction In this case: • We create two liquid phases by introducing a solvent (C) (MSA) to a liquid mixture of a carrier (A) and a solute (B) • Solvent (C) and carrier (A) have very little solubility in each other Liquid-Liquid Extraction Solvent Feed S, C Solvent Rich Liquid Out E, A, B, C Liquid Feed F, A, B Carrier Rich Liquid out R, A, B, C Define the raffinate as the exiting phase rich in carrier. Define the extract as the exiting phase rich in solvent. Lecture 7: Liquid-Liquid Ternary Single Stage 2 Liquid-Liquid Ternary Single Equilibrium Stage If the solvent and carrier have some solubility in each other, then the raffinate will have a small amount of solvent in it and the extract will have a small concentration of carrier: Liquid-Liquid Extraction Solvent Feed Extract out S, XC(S),T, P E, XA (E), XB (E), XC (E), T, P Liquid Feed F, XA(F), XB, T, P Raffinate out R, XA (R), XB (R), XC (R), T, P The raffinate is the exiting phase rich in carrier. The extract is the exiting phase rich in solvent. Lecture 7: Liquid-Liquid Ternary Single Stage 3 Liquid-Liquid Ternary Single Equilibrium Stage If the solvent and carrier have no solubility in each other, then the raffinate will have no solvent in it and the extract will have no carrier in it: All of solvent exits in the extract Liquid-Liquid Extraction Solvent Feed S, XC(S),T, P Extract out E, XB (E), XC (E), T, P Liquid Feed F, XA(F), XB, T, P Raffinate out R, XA (R), XB (R), T, P All of carrier exits in the raffinate The raffinate is the exiting phase rich in carrier. The extract is the exiting phase rich in solvent. Lecture 7: Liquid-Liquid Ternary Single Stage 4 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 XB: The ratio of mass of component B to another component of the stream. Mole ratio XB: 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: Solvent Feed Extract out S, XC(S),T, P E, XB (E), XC (E), T, P Liquid Feed Raffinate out F, XA(F), XB, T, P R, XA (R), XB (R), T, P F FB XB FA E E EB X B EC XB S R R RB XB RA XB FA S S B X B SC 0 Lecture 7: Liquid-Liquid Ternary Single Stage 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. 5 Material Balances Solute Material Balance: Solvent Feed Extract out S, XC(S) E, XB (E), XC (E) Liquid Feed Raffinate out F, XA(F), XB R, XA (R), XB (R) F FB XB FA Rate of B in the feed is the ratio of B to A, times feed rate of A. E E EB X B EC XB S R R RB XB RA XB FA S S B X B SC 0 Solute Material Balance: 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. FB SB RB EB F S R E X B FA XB SC XB RA X B EC F R E X B FA XB FA X B S Lecture 7: Liquid-Liquid Ternary Single Stage 6 Equilibrium Distribution The way the solute will distribute itself between the extract and raffinate at equilibrium is given by the K-Value: E R X B K'DB XB Solvent Feed Extract out S, XC(S) Liquid Feed F, XA(F), XB 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. B E, XB (E), XC (E) Raffinate out R, XA (R), XB (R) Note that concentrations of exiting streams from an equilibrium stage are always related by equilibrium. Lecture 7: Liquid-Liquid Ternary Single Stage 7 The Extraction Factor The degree of separation of the solute between the exiting streams is expressed as the extraction factor: Extraction Factor: The ratio of solute flow in the extract to solute flow in the raffinate. E E EB X B EC XB S R R RB XB RA XB FA XB E S EB B R RB X FA B Combining this definition with the equilibrium relationship: E R X B K'DB XB results in another expression for the extraction factor: B Lecture 7: Liquid-Liquid Ternary Single Stage K'DB S FA The larger the equilibrium driving force to separate B, and the larger the ratio of solvent to feed, the larger the extraction factor. 8 Extraction Efficiency We can determine the amount not extracted starting with the material balance of the solute: F E R X B FA XB S X B FA We substitute in the K-value ratio: F R R X B FA K'DB XB S XB FA And simplify: XB R This ratio gives the amount of solute left in the raffinate to the amount originally in the feed stream, FA F K' S F XB DB A XBR 1 F K' S XB DB FA Lecture 7: Liquid-Liquid Ternary Single Stage 1 1 B 1 The amount not extracted increases with the feed rate, and smaller ratio distribution between extract and raffinate and less solvent. 9 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 (that is that these diagrams are slices through a four dimensional space with constant T and P). A Lecture 7: Liquid-Liquid Ternary Single Stage C B 10 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 (use the third as a check). Compositions can be mole fractions or mass fractions. [20% C, 20% B, 60% A] C [94% C, 3% B, 3% A] [33% C, 33% B, 33% A] [30% C, 70% B] [100% A] A Lecture 7: Liquid-Liquid Ternary Single Stage B 11 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: Solute Ethylene Glycol Water 50 % EthGly 50% Furfural 17% EthGly 27% Furfural 56 % Water A composition where two liquid phases coexist. 100% Furfural Solvent Lecture 7: Liquid-Liquid Ternary Single Stage 66% EthGly 7% Furfural 27 % Water A composition where only a single liquid exists. Furfural Carrier 16 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 Tie-Lines: Show the compositions of the equilibrium phases. Lecture 7: Liquid-Liquid Ternary Single Stage Furfural 17 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, XC(S) E, XB (E), XC (E) Liquid Feed Raffinate out F, XA(F), XB R, XA (R), XB (R) Ethylene Glycol Water Furfural Lecture 7: Liquid-Liquid Ternary Single Stage 18 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 S 300 Kg Lecture 7: Liquid-Liquid Ternary Single Stage Water F 60 kg EG 140 kg water Furfural 19 Partially Soluble Ternary Systems Step 2: Locate the mixing point M: X BF FA XBS S 0.3 200kg 0 300kg 0.12 FS 500kg Ethylene Glycol Water F 60 kg EG 140 kg water M S 300 Kg Lecture 7: Liquid-Liquid Ternary Single Stage Furfural 20 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, 5%EG, 8% furfural) Ethylene Glycol Water F 60 kg EG 140 kg water M E R S 300 Kg Lecture 7: Liquid-Liquid Ternary Single Stage Furfural 21 Partially Soluble Ternary Systems Step 4: Determine the amount of extract and raffinate (can use lever rule) M E XC 0.63 0.82 R f R E 0.08 0.82 0.257 XC XC XC Extract (4% water, 14%EG, 82% furfural) Raffinate (87% water, 5%EG, 8% furfural) R 0.257 500kg 128.4kg E 1 0.257 500kg 371.6kg Ethylene Glycol Water F 60 kg EG 140 kg water M E R S 300 Kg Lecture 7: Liquid-Liquid Ternary Single Stage Furfural 22 Partially Soluble Ternary Systems Step 5: Determine the solvent free extract: Mixtures of E and S. Extend line from S through E to solvent free point at H. Solvent free extract H (20% water, 80% EG) R XB XB H Ethylene Glycol Water F 60 kg EG 140 kg water M E F R S 300 Kg Lecture 7: Liquid-Liquid Ternary Single Stage Furfural 23 Summary This lecture covered: • Ternary Liquid-Liquid extractions. • Ternary phase diagrams. • A procedure to determine the product compositions and flow rates of a liquid-liquid extraction separation. Next lecture will cover: • Leaching • Crystallization Lecture 7: Liquid-Liquid Ternary Single Stage 24
