ERT 313 BIOSEPARATION ENGINEERING EXTRACTION Prepared by: Miss Hairul Nazirah Abdul Halim
Definition of Extraction Liquid-Liquid extraction is a mass transfer operation in which a liquid solution (the feed) is contacted with an immiscible or nearly immiscible liquid (solvent) that exhibits preferential affinity or selectivity towards one or more of the components in the feed. Purpose of Extraction To separate closed-boiling point mixture Mixture that cannot withstand high temperature of distillation Example: - recovery of penicillin from fermentation broth solvent: butyl acetate - recovery of acetic acid from dilute aqueous solutions solvent: ethyl-acetate
Extraction Equipment Batchwise or continuous operation Feed liquid + solvent (put in agitated vessel) = layers (to be settled and separated) Extract the layer of solvent + extracted solute Raffinate the layer from which solute has been removed Extract may be lighter or heavier than raffinate. Continuous flow more economical for more than one contact process
Extraction Equipments: - Mixer settlers - Packed extraction towers - Perforated plate towers - Baffle towers - Agitated tower extractors Auxiliary equipment: - stills, evaporators, heaters and condenser
Extraction of Dilute Solution Extraction factor is defined as: Where: E = extraction factor K D = distribution coefficient V = volume of solvent L = volume of aqueous
For a single-stage extraction with pure solvent; 1 - the fraction of solute remaining is 1 E - the fraction recovered is E 1 E
Extraction of Concentrated Solution Equilibrium relationship are more complicated 3 or more components present in each phase Equilibrium data are often presented on a triangular diagram such as Fig 23.7 and 23.8
Consider Fig 23.7 Line ACE shows extract phase Line BDE shows raffinate phase Point E is the plait point the composition of extract & raffinate phases approach each other Tie line a straight line joining the composition of extract & raffinate phases. Tie line in Fig 23.7 slope up to the left extract phase is richer in acetone than the raffinate phase. This suggest that most of the acetone could be extract from water phase using moderate amount of solvent.
Consider Fig 23.8 Line AD shows extract phase Line BC shows raffinate phase Tie line in Fig 23.8 slope up to the right extraction would still be possible But more solvent would have to use. The final extract would not be as rich in desired component (MCH)
How to obtain the phase composition using the triangular diagram? - Point M: 0.2 Acetone, 0.3 water, 0.5 MIK - Draw a new tie line - Extract phase: 0.232 acetone, 0.043 water, 0.725 MIK - Raffinate phase: 0.132 acetone, 0.845 water, 0.023 MIK - Ratio of acetone to water in the product = 0.232/0.043 = 5.4 - Ratio of acetone to water in the raffinate = 0.132/0.845 =0.156 Let s compare with Fig 23.8. Which system is more effective?
Coordinates Scale Refer to Treybal, Mass Transfer Operation, 3 rd ed., McGraw Hill The book use different triangular system The location of solvent (B) is on the right of the triangular diagram (McCabe use on the left) Coordinate scales of equilateral triangles can be plotted as y versus x as shown in Fig 10.9 Y axis = wt fraction of component C (acetic acid) X axis = wt fraction of solvent B (ethyl acetate)
Single-Stage Extraction
The triangular diagram in Fig 10.12 (Treybal) is a bit different as compared to Fig. 23.7 (McCabe) Extract phase on the left Raffinate phase - on the right Fig 10.12 shows that we want to extract component C from A by using solvent B. Total material balance: Material balance on C:
Amount of solvent to provide a given location for M 1 on the line FS: The quantities of extract and raffinate: Minimum amount of solvent is found by locating M 1 at D Maximum amount of solvent is found by locating M 1 at K
Multistage Crosscurrent Extraction
Multistage Crosscurrent Extraction Continuous or batch processes Refer to Fig 10.14 Raffinate from the previous stage will be the feed for the next stage The raffinate is contacted with fresh solvent The extract can be combined to provide the composited extract The total balance for any stage n: Material balance on C:
Tutorial 4 Batch Extraction EXAMPLE 23.2. Penicillin F is recovered from a dilute aqueous fermentation broth by extraction with amyl acetate, using 6 volumes of solvent per 100 volumes of the aqueous phase. At ph 3.2 the distribution coefficient K D is 80. (a) What fraction of the penicillin would be recovered in a single ideal stage? (b) What would be the recovery with two-stage extraction using fresh solvent in both stages?
Continuous Single Stage Extraction An inlet water solution of 100 kg/h containing 0.010 wt fraction nicotine in water is stripped with a kerosene stream of 200 kg/h containing 0.0005 wt fraction nicotine in a single stage extraction unit. It is desired to reduce the concentration of the exit water to 0.0010 wt fraction nicotine. Calculate the flow rate of the nicotine in both of the exit streams.
SOLUTION Nicotine in the feed solution = 100 (0.01) = 1 kg/h nicotine Water in feed = 100 (1-0.01) = 99 kg/h water 2. Nicotine in solvent = 200 (0.0005) = 0.1 kg/h nicotine Kerosene = 200 (1 0.0005) = 199.9 kg/h kerosene 3. Exit stream of aqueous phase, L 1 Water = 99 kg/h = (1 0.0010) L 1 L 1 = 99.099 kg/h (nicotine + water) Nicotine = 99.099 99 = 0.099 kg/h nicotine in exit stream 4. Exit stream of solvent phase, V 1 Solvent = 199.9 kg/h Nicotine in solvent = 0.1 + (1 0.099) = 1.001 kg/h in exit stream Solvent + Nicotine = 199.9 + 1.001 = 200.9 kg/h
Multistage Crosscurrent Extraction If 100 kg of a solution of acetic acid (C) and water (A) containing 30% acid is to be extracted three times with isopropyl ether (B) at 20 C, using 40 kg of solvent in each stage, determine the quantities and compositions of the various streams. How much solvent would be required if the same final raffinate concentration were to be obtained with one stage? The equilibrium data at 20 C are listed below [Trans. AIChE, 36, 628 (1940), with permission].
SOLUTION The horizontal rows give the concentrations in equilibrium solutions. The system is of the type shown in Fig. 10.9, except that the tie lines slope downward toward the B apex. The rectangular coordinates of Fig. l0.9a will be used, but only for acid concentrations up to x = 0.30. These are plotted in Fig. 10.15.
Point M 1 is located on line FB. With the help of a distribution curve, the tie line passing through M 1 is located as shown, and x 1 = 0.258, y 1 = 0.117 wt fraction acetic acid. Eq. (10.8):
Stage 3 In a similar manner, B 3 = 40, M 3 = 130.1, x M3 = 0.1572, x 3 = 0.20, y 3 = 0.078, E 3 = 45.7, and R 3 = 84.4. The acid acetic content of the final raffinate is = x 3 R 3 = 0.20(84.4) = 16.88 kg. The composited extract is E 1 + E 2 + E 3 = 43.6 + 46.3 + 45.7 = 135.6 kg, The acid content in the composited extract: E 1 y 1 + E 2 y 2 + E 3 y 3 = 13.12 kg.
If an extraction to give the same final raffinate concentration, x = 0.20. were to be done in one stage, the point M would be at the intersection of tie line R 3 E 3 and line BF of Fig. 10.15, X M = 0.12. The solvent required would then be, by Eq. (10.6), S 1 = 100(0.30-0.12)/(0.12-0) = 150 kg, Hence, 150 kg of solvent is required for single stage extraction 120 kg of solvent is required in the three-stage extraction.