L/O/G/O 單元操作 ( 三 ) Chapter 23 Leaching and Extraction 化學工程學系李玉郎

Leaching Solid extraction, dissolve soluble matter from its mixture with an insoluble solid solid (solute A+ inert B) solvent (C) concentrated solution (A+C) overflow phase (liquid) Liquid extraction extracted solid (A+B+C) underflow phase (slurry) Separate two miscible liquids by use of a solvent 2

Leaching Percolation of solvent through stationary solid beds done in a tank with a perforated false bottom extraction battery a serious of such tank Shank s process extraction battery in countercurrent operation diffusion battery a series of pressure tank operated with countercurrent solvent flow. ( solvent is volatile, force pressure is required ) 3

Moving-bed leaching (a) Bollman extractor 4

(b) Rotocel extractor countercurrent extraction fresh solvent feed, last compartment before the discharge point 5

Dispersed solid leaching Solid is dispersed in the solvent by mechanical agitation in a tank or flow mixer. The leached residue is then separated by settling or filtration. Principles of Continuous Countercurrent Leaching solvent Ideal stages in countercurrent leaching assumption : solute free solid is insoluble in the solvent, solid flow rate is constant L : flow rate of retained liquid on the solid V : flow rate of overflow solution FIGURE 23.2 solid ( solute, contaminated solvent ) 6 Countercurrent leaching cascade.

Solution retained by entering solid: x a Solution retained by leaving solid: x b Fresh solvent entering the system: y b Concentrated solution leaving the system: y a Assumption : concentration of liquid retained by the solid leaving any stage ( x n ) = ( y n ) the liquid over flow from the stage. ( x e = y e ) attained simply or with difficulty depending on the structure of the solid. 7

Operation Line material balance to the control surface shown upward Total solution : V n+1 + L a = V a + L n --- (23.1) Solute : V n+1 y n+1 + L a x a = L n x n + V a y a --- (23.2) Solving for y n+1 gives to operating line equation, which is the same as that derived earlier for the general case of an equilibrium stage cascade [Eq. (20.7)] : L n Va ya La xa yn 1 xn --- (23.3) Vn1 Vn 1 OP line : passes through ( x a, y a ) and ( x b, y b ) slope L V ( if flow rate constant ) 8

Variable underflow : if the density and viscosity of solution change with x, solid from lower number of stage retain more liquid ( L 1 > L 2 > ; L n 1 > L n > L n+1 ) slope of OP line varies from unit to unit Number of ideal stages for constant underflow (1) in leaching, x e = y e, EQ line is straight ( y = mx, m = 1 ) (2) constant underflow, OP line is straight Eq (20.24) can be used N n[( y n[( y b b y y * b a ) /( y ) /( y a * b y y * a * a )] )] --- (20.24) 9

EQ : y = x y a * = x a, y b * = x b = ( y N ) Eq (20.24) can not be used if L a L ( solution in feed differs from solution in underflow ). In such case, the performance of 1 st stage is calculated by material balance, ad then apply Eq (20.24) to the remaining stages. Number of ideal stages for variable underflow 1. determine terminal points by material balance. ( x a, y a ) ; ( x b, y b ) 2. find intermediate point L is function of x choose x n L n V n+1 y n+1 3. plot the OP curve : ( x a, y a ), ( x n, y n+1 ) 1, ( x n, y n+1 ) 2, ( x b, y b ) ( only one intermediate point is enough usually to plot OP curve, unless change of L and V are large or two curves are very close ) Overall mass balance mass balance to solute Eq (23.1) Eq (23.2) 10

L b V a x b y a 11

oil benzene V b = 10 + 655 y b = (0.015) oil 60 kg L b =? x b =? N 1 V a y a solid : 1000 kg L a oil : 400 kg x (425) a ben. : 25 kg (0.941) 12

Solution : x, y : mass fraction of oil, solvent : Benzene V b = 10 + 655 = 655 kg y b = 10 655 = 0.015 determine x b by trial L b x b = 60 if x b = 0.1, solution retained is 0.505 kg kgsolid 60 then L b = 1000 0.505 = 505, and x b = = 0.119 ( 0.1 try value ) 505 assume x b = 0.119, solution retained is 0.507 60 L b = 0.507 1000 = 507, x b = = 0.118 ( close enough ) 507 Benzene at L b = 507 60 = 447, L b = 507 13

V b oil : 10 Ben : 655 b N a 1 oil : Ben : At solid inlet L b oil : 60 Ben : 447 L a = 400 + 25 = 425 x a = 400 425 = 0.941 L a oil : 400 Ben : 25 Overall balance to get V a, y a oil in extract 400 + 10 60 = 350 V a = 350 + 233 = 583 benzene in extract 655 + 25 447 = 233 y a = 350 583 = 0.60 Ans : (a) y a = 0.6, (b) x b = 0.118 (c) L b = 507 kg / h, (d) V a = 583 kg / h 14

V b = 665 y b = 0.015 L b = 507 x b = 0.118 y n+1 =? x n = 0.3 (L n = 530) n y 2 (0.408) V 2 1 x 1 = 0.6 L 1 = 595 (e) determine the inlet and exit concentrations for the first stage : x 1 = y a = 0.6 solution retained : 0.595 L 1 = 0.595 1000 = 595 kg kg kgsolid V 2 = L 1 + V a L a = 595 + 583 425 = 753 kg h 15 V a = 583 y a = 0.6 L a = 425 x a = 0.941 oil balance V 2 y 2 + L a x a = V a y a + L 1 x 1 753 y 2 + 425 0.941 = 583 0.6 + 595 0.6 y 2 = 0.408

terminal point ( x 1, y 2 ) = ( 0.6, 0.408 ) ( x b, y b ) = ( 0.118, 0.015 ) determine an intermediate point, choose x n = 0.3 solution retained : 0.53, L n = 0.53 1000 = 530 V n+1 = V a + L n L a = 583 + 530 425 = 688 kg h kg h V n+1 y n+1 = V a y a + L n x n L a x a, y n+1 = 0.158 intermediate point ( 0.3, 0.158 ) 16

( x a, y a ) = ( 0.941, 0.6 ) ( x 1, y 2 ) = ( 0.6, 0.408 ) ( x b, y b ) = ( 0.118, 0.015 ) ( x n, y n+1 ) = ( 0.3, 0.158 ) y 4 ideal stages are required FIGURE 23.3 McCabe Thiele diagram for leaching (Example 23.1). 17

concentrated solution If solute is of limited solubility and the concentrated solution reaches saturation solvent input to stage N should be large enough that all liquid, except that adhering to the underflow from stage 1 should be unsaturated. V y (saturate) Stage efficiency Solid of permeable rate of leaching is largely governed by the rate of diffusion through the solid. Solid of impermeable a strong solution is confined on the solid surface, the approach to equilibrium is rapid. η 1 N 2 1 x 18

Applied when distillation is ineffective close boiling mixtures, substances that cannot withstand the temperature of distillation. utilizes chemical differences Liquid Extraction Example : separate petroleum products that have different chemical structures but about the same boiling range. When either distillation or extraction may be used, the choice is usually distillation. extraction offer greater flexibility in choice of operation conditions ( choice of solvent and temperature) 19

Extraction Equipment two phases must be brought into good contact. because two liquid phases have comparable densities, energy for mixing and separation is usually small. That is, hard to mix and harder to separate (energy is supplied ) can be operated batchwise or continuously ( if gravity flow is used) solvent feed mix and settle extract ( solvent plus extracted solute ) raffinate 20

Mixer settlers Batchwise may be the same unit Continuous process typical : mixing 5 min settling 10 min screen or pad of glass fiber promote the coalescence of droplets 21

Spray and packed extraction tower differential contacts, mixing and settling proceed simultaneously and continuously. phases : continuous, dispersed phases, extracted phases, raffinate phases choice of dispersed phase : 1) phase with higher flow rate may be dispersed to give a higher interfacial area. 2) phase with more viscosity give higher settling velocity Wettability continuous phase wet packing FIGURE 23.5 Spray tower; A, nozzle to distribute light liquid. 22

Rate of mass transfer is relatively low compared to distillation or absorption. contact is more effective in the region where the drop are formed ( due to higher mass transfer rate in newly formed drops, or back mixing of the continuous phase ) So, effective method to increase the mass transfer stages : 1. redisperse the drops at frequent intervals throughout the tower. 2. fill the tower with packing: ring, saddles,, the packing causes the drops to coalescence and reform, thus, increase the number of stages. Flooding velocity in packed towers flow rate of one phase is held constant, and that of the other increases gradually. A point is reached where dispersed phase coalesces, both phases leave : together through the continuous phase outlet. The larger the flow rate of one phase at flooding, the smaller is that of the other. 23

Flooding velocities in packed columns can be estimated from the empirical equation 9 1/3 0.3 0.15 2 0.5 2 a 0.5 1.54 0.41 1 v g c V s, c( 1 R ) ( ) C1 ( ) 2 0.5 (23.4) g d av ( ) c av where R V s, d V s, c V s, c, V s, d = superficial velocities of continuous and dispersed phases at flooding, respectively, m/s c = viscosity of continuous phase, Pa s σ = interfacial tension between phases, N/m ρ d = density of dispersed phase, kg/m 3 Δρ = density difference between phases, kg/m 3 a v = specific surface area of packing, m 2 /m 3 ε = fraction voids or porosity of packed section α = 1.0 for continuous-phase wetting, 1.2 for dispersed-phase wetting C 1 = function of packing type The function C 1 is 0.28 for Raschig rings of Intalox saddles. Values for other packings range 9 0.5 2 from 0.204 to 0.42. For a given system the term V s, c( 1 R ) is const. 24

Baffle towers ( 檔板塔 ) Disc and doughnut ( with or without scraper ) Side to side can handle dirty solutions containing suspended solids. flow liquid is smooth, it is valuable for liquids that emulsify easily. not effective mixers, each baffle equivalent to only 0.05 0.1 ideal stage. 25

Perforated plate towers ( 多孔塔板 ) (a) perforations in horizontal plates. Figure 23.6 (b) cascade weir tray with mixing and settling zones. dispersed phase light liquid, rising through perforated plate continuous phase heavy liquid, flow down the plate through downcomers. cascade weir tray reduces the quantity of oil carried downward by the solvent and increases the effectiveness of the extractor. Perforations : 1½ 4½ mm, plate spacing : 150 600 mm. 26

Pulse columns ( 脈衝塔 ) Agitation is provided by external means reciprocating pump: pulses the entire contents of the column at frequent intervals. Tower may contain ordinary packing pulsation disperses the liquids and eliminates channeling. Special sieve plates may also be used : holes are smaller, 1.5~3 mm diameter. used almost for processing highly corrosive radioactive liquids. No down comers are used. HETP 1/3 of unpulse column. 27

Agitated tower extractors 28

Agitated tower extractors Mechanical energy is provided by internal turbines or other agitators, mounted on a central rotating shaft. a) flat disks disperse the liquids and impel them outward toward the tower wall, stator tings create quiet zones in which the two phase can separate. b) the regions surrounding the agitators are packed with wire mesh to encourage coalescence and separation. Each mixer settler is 300 600 mm high. ( η > 100% sometimes ) The problem of maintaining the internal moving parts may be a serious disadvantage, particularly where the liquids are corrosive. 29

Principles of extraction Questions about ideal stages, stage efficiency, minimum ratio between the two streams, same important as in distillation. Extraction of dilute solutions Extraction phase, y change in flow rate can be neglected, K D y x i i is constant distribution coefficient. E : extraction factor (equivalent to stripping factor ) E K D V L Solvent (s) Diluent (b) + solute (a) x Raffinate phase 30

If (1) one single stage, (2) pure solvent fraction of solute remaining fraction of recovery Cascade Kremser equation [ Eqs. (20.24), (20.25), (20.26), (20.28) ] can be used. x x x E 1 x 0 1 E 0 1 1 E V L y x L V V K D E V y = L ( x 0 x ) V L y x x x x 0 1 V y o = 0 L x 0 y x0 E 1 x L 31

V (a) y = 0, = 0.06, K D = 80 E KDV L L 800.06 fraction of recovery x 1 1 x 0 4.8 0.828 material balance V ( y 1 y 0 ) = L ( x 0 x 1 ) V y 1 = L ( x 0 x 1 ), y 1 = K D x 1 1 E E 0 x x 0 1 1 E 0.172 x V y = 0 y x 0 L VK x L D 1 1 x0 x x 0 1 1 E 1 V D 0.172 K L E 32

(b) two stages extraction, ( same value of E ) x x 1 0 1 1 E x x 1 2 2 0 1 E recovery : x 1 2 0.9703 or 97% x 0, x x 2 1 1 1 E 0.0297 x 0 V y = 0 L x 1 V y = 0 x 2 (c) using Eq (20.28) the same recovery as (b) N * * n[( xa xa) /( xb xb)] n E 33

x b 0.03 x a, let x a = 100, x b = 3 V ( y a 0 ) = L ( 100 3 ) N * * n[( xa xa) /( xb xb)] n E 100 x a y a x * a y a K y x * b = 0 n[(100 20.2) /(3 0)] N 2.09 n 4.8 Compare : based on 100 volume aqueous 3 x b y b = 0 (b) (c) 2 stage 2.09 countercurrent flow V : 62 V : 6 1 lower higher concentration of extract L 10097 ( 97) 1617 V 6 a D 1617 80 20.2 L = 100 V = 6 34

Extraction of concentrated solutions ; phase equilibria phase diagram : Type I partial miscibility of solvent (s) and diluent (b), complete miscibility of solvent and the component to be extracted (a). Solvent (s) Diluent (b) + solute (a) 35

Extraction of concentrated solutions ; phase equilibria curve ACE : extract layer (MIK) curve BDE : raffinate layer (water) Extracted solute E : plait point ( the composition of two phases approach each other ) tie lines : slope up to left extract phase is richer in acetone. a = 0.232 s = 0.725 b = 0.043 solvent 36 diluent

in Extract : H 2 O < 2%, (for C solute = 0) as acetone concentration increases, water content also increases. as shown y y H A 2 O maximum at y A = 0.27 37

Type II The solvent ( aniline ) is only partially miscible with both The other components. FIGURE 23.9 System aniline-n-heptane- MCH at 25 o C: a, solute, MCH: b, diluent, n-heptane; s, solvent, aniline. (After Varteressian and Fenske. 16 ) methyl cyclo hexane 38

Use of McCabe Thiele method y A : mass fraction of solute in extract phase, V x A : mass fraction of solute in raffinate phase, L equilibrium data : y A, x A values at two ends of tie lines. OP line : Eq (23.3) y n1 L V n n1 x n V a ya L V n1 two terminal points ( x a, y a ), ( x b, y b ) one ( or more ) intermediate points. a x a 39

Solution : EQ curve : from Fig 23.8 determine terminal point Base : feed 100 mass per hour, L a = 100, x a = 0.4 acetone : A = 40 ; H 2 O = 60 A in extract ( 99% recovery ) : 40 0.99 = 39.6 A in raffinate (99% recovery) : 40 0.01 = 0.4 let n : mass of H 2 O in extract m : mass of MIK in raffinate V a = 39.6 + n + ( 100 m ) = 139.6 + n m... (1) A = 40 H 2 O = 60 L a = 100 x A = 0.4 L b =? A = 0.4 H 2 O : 60 n MIK : m 40 V a =? A = 39.6 H 2 O : n MIK : 100 m V b = 100 (MIK) y b = 0

At the bottom V b = 100 V a = 39.6 + n + ( 100 m ) = 139.6 + n m L b = 0.4 + ( 60 n ) + m = 60.4 + m n... (2) assume : n, m are small and tend to cancel V a 140, L b 60 ( will be adjusted after calculating n, m ) 39.6 (extracted) y Aa 0.283 y H = 0.049... ( n ) 140 2O 0.4 0 Fig 23.8 (raffinate) x Ab 0.0067 x MIK = 0.02... ( m ) 60 139.6 m from (1) n 0.049... (3) 1 0.049 if m is small, n 7.2 V a 60.4 n (2) m 0.02 ( n = 7.2 ) 0.98 L b n 0.049 139.6 n m m 0.02 60.4 m n m 1.1 used to revise n by Eq (3) n = 7.1 41 n: H 2 O m: MIK m = 1.09 n = 7.14

n: H 2 O m: MIK

calculate V a = 139.6 + n m (1) y a L b = 60.4 + m n... (2) x b 39.6 V a = 139.6 + 7.1 1.1 = 145.6 ; y a 0.272 ( 1st : 0.283 ) 145.6 0.4 L b = 60.4 + 1.1 7.1 = 54.4 ; xb 0.0074 ( 1st : 0.0067 ) 54.4 terminal points : ( 0.4, 0.272 ) ; ( 0.0074, 0 ) Find an intermediate point on the operation line pick y A = 0.12 Fig 23.8 y H 2O = 0.03 y MIK = 0.85 used to revise V L x A assume amount of MIK in V MIK 100 ( MIK in raffinate is small, (1.1) ) V y MIK 100 V 100 117.6 0.85 43

V ymik 100, overall balance : V 100 117.6 0.85 x A = 0.4 y a = 0.272 L V + L b V b 117.6 + 54.4 100 = 72.0 balance to A : L x A V y A + L b x b V b y b = 117.6 0.12 + 54.4 0.0074 0 = 14.5 x A 14.5 72 0.201 0.4 rough value L A = 0.4 MIK = 1.1 L b = 54.4 x b = 0.0074 V x A =? y MIK = 0.85 y A = 0.12 44 MIK V b = 100 y b = 0

Calculate corrected values of V, L, x A x A = 0.201 x MIK 0.03 balance to MIK : Fig 23.8 L x MIK + V b = L b x b,mik + V y MIK 72 0.03 + 100 = 1.1 + V (0.85) V = 101.1/0.85 = 118.9 ( revised ) revised L = 118.9 + 54.4 100 = 73.3 ( overall balance ) V A 0.4 0 118.9 0.12 0.4 revised xa y 73.3 73.3 intermediate point ( 0.2, 0.12 ) From Fig 23.11, N = 3.4 stage 1st assume : 100 mole MIK (V=117.6) 0.2 x MIK = 0.03 A = 0.4 MIK = 1.1 45 x A = 0.4 y a = 0.272 L L b = 54.4 x b = 0.0074 V y MIK = 0.85 y A = 0.12 MIK V b = 100 y b = 0

Countercurrent extraction of type II systems using reflux Solvent plays the same part in extraction that heat does in distillation 47

Limiting reflux ratios R D (total reflux), min. No. of plates R D = L D min L D min L D, No. of plate is R D Practical examples of extraction with reflux aniline heptane methylcyclohexane ( MCH ) system ( Fig23.9 ) MCH heptane in extract only modestly greater than that in raffinate great many stage is needed. Low solubility of MCH and Heptane in aniline ( solvent ) large flow of solvent. 48

Sulfolane process Enrichment of the extract by countercurrent washing with another liquid ( dissolve in the extract, can be easily removed ) extraction of aromatics compounds. Extract : nearly all aromatic and few % paraffine and nophthenes (suefolane and hydrocarbons form type I system with plait point) Backwash : ( 五環硫氧烷 ) distillation : ( azetropic distillation ) light hydrocarbon + water azetropic ( low boiling Temp. ). boiling points overlap low boiling hydrocarbons ( countersolvent or reflux ) 49

Special Extraction Techniques (1) recovery of sensitive biological products using only aqueous phases. Example : separation of proteins two aqueous phase : ( 80 ~ 90% water ) (a) polyethylene glycol (b) dextran ( 葡萄聚醣 ) or phosphate salts partition coefficient : 0.01 ~ 100 ( dependent on ph ) 51

(2) Supercritical Fluid Extraction Extraction with a solvent held at pressure and temperature above the critical point of the solvent. Advantage : 1. selective dissolving power 2. low viscosity, low density, high diffusivity ( 100 times of ordinary liq. ) easily penetrates porous or fibrous solids. 3. solute can be recovered by changing T and P. 4. does not affect the characteristic flavor and aroma of extracted materials Disadvantage : high pressure. 52

Phase equilibria useful solvent for supercritical extraction : CO 2 In supercritical region : T c = 31.06 o C P c = 73.8 bar no distinction between liquid and gas, no phase transition, the supercritical fluid acts like a very dense gas or light mobile liquid. Solubility and selectivity : strong function of T and P. total extraction of solutes pressure is highest. selective removal of odor producing volatile compnents close to critical point (solubilities are smaller, but selectivity for the most volatile component is much higher ) Commercial process : Decaffeination of coffee Caffeine ( 0.7 to 3% ) 0.02% 53

V b, y b V n+1, y n+1 y n, V n V a, y a L b, x b N n 1 x n L n L 1 x 1 L a x a