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Leaching Liquid Extraction 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) concentrated solution (A+C) overflow phase (liquid) solvent (C) extracted solid (A+B+C) underflow phase (slurry) Liquid extraction 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 Ideal stages in countercurrent leaching assumption : solute–free solid is insoluble in the solvent, solid flow rate is constant solvent solid ( solute, L : flow rate of retained liquid on the solid contaminated solvent ) V : flow rate of overflow solution FIGURE 23.2 Countercurrent leaching cascade6 . Solution retained by entering solid: xa Solution retained by leaving solid: xb Fresh solvent entering the system: yb Concentrated solution leaving the system: ya Assumption : concentration of liquid retained by the solid leaving any stage ( xn ) = ( yn ) the liquid over flow from the stage. ( xe = ye ) 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 : Vn+1 + La = Va + Ln --- (23.1) Solute : Vn+1 yn+1 + La xa = Ln xn + Va ya --- (23.2) Solving for yn+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 V y L x n a a a a yn1 xn --- (23.3) Vn1 Vn1 OP line : passes through ( xa , ya ) and ( xb , yb ) L slope 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 ( L1 > L2 > … ; Ln–1 > Ln > Ln+1 ) slope of OP line varies from unit to unit Number of ideal stages for constant underflow (1) in leaching , xe = ye , EQ line is straight ( y = mx , m = 1 ) (2) constant underflow , OP line is straight Eq (20.24) can be used * * n [( y b y b ) /( y a y a )] --- (20.24) N * * n[(yb ya ) /( yb ya )] 9 EQ : y = x * * ya = xa , yb = xb = ( yN ) Eq (20.24) can not be used if La ≠ L ( solution in feed differs from solution in underflow ). In such case , the performance of 1st 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. ( xa , ya ) ; ( xb , yb ) 2. find intermediate point L is function of x Overall mass balance mass balance to solute choose xn Ln Vn+1 yn+1 Eq (23.1) Eq (23.2) 3. plot the OP curve : ( xa , ya ) , ( xn , yn+1 )1 , ( xn , yn+1 )2 , ( xb , yb ) ( 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 ) 10 ya xb L b Va 11 oil benzene Vb = 10 + 655 Va yb = (0.015) ya N 1 solid : 1000 kg La oil : 400 kg (425) oil 60 kg xa ben. : 25 kg Lb = ? xb = ? (0.941) 12 Solution : x , y : mass fraction of oil , solvent : Benzene Vb = 10 + 655 = 655 kg yb = 10 ⁄ 655 = 0.015 determine xb by trial Lb × xb = 60 kg if xb = 0.1 , solution retained is 0.505 kg solid 60 then L = 1000 × 0.505 = 505 , and x = = 0.119 ( ≠ 0.1 try value ) b b 505 assume xb = 0.119 , solution retained is 0.507 60 L = 0.507 × 1000 = 507 , x = = 0.118 ( close enough ) b b 507 Benzene at Lb = 507 – 60 = 447 , Lb = 507 13 oil : 10 b a Ben : 655 V oil : b Ben : N 1 L b oil : 400 oil : 60 L Ben : 447 a Ben : 25 At solid inlet La = 400 + 25 = 425 xa = 400 ⁄ 425 = 0.941 Overall balance to get Va , ya oil in extract 400 + 10 – 60 = 350 Va = 350 + 233 = 583 benzene in extract 655 + 25 – 447 = 233 ya = 350 ⁄ 583 = 0.60 Ans : (a) ya = 0.6 , (b) xb = 0.118 (c) L = 507 kg / h , (d) V = 583 kg / h b a 14 y2 (0.408) Vb = 665 V2 y = ? Va = 583 yb = 0.015 n+1 ya = 0.6 n 1 L = 425 x = 0.3 a Lb = 507 n x = 0.941 (L = 530) a xb = 0.118 n x1 = 0.6 kgL = 595 (e) 1kg solid determine the inlet and exit concentrations for the first stage : x1 = ya = 0.6 solution retained : 0.595 L1 = 0.595 × 1000 = 595 kg kg V2 = L1 + Va – La = 595 + 583 – 425 = 753 h oil balance V2 y2 + La xa = Va ya + L1 x1 753 y2 + 425 × 0.941 = 583 × 0.6 + 595 × 0.6 y2 = 0.408 15 terminal point ( x1 , y2 ) = ( 0.6 , 0.408 ) ( xb , yb ) = ( 0.118 , 0.015 ) determine an intermediate point , choose xn = 0.3 solution retained : 0.53 , Ln = 0.53 × 1000 = 530 Vn+1 = Va + Ln – La = 583 + 530 –425 = 688 Vn+1 yn+1 = Va ya + Ln xn – La xa , yn+1 = 0.158 intermediate point ( 0.3 , 0.158 ) kg h 16 ( xa , ya ) = ( 0.941 , 0.6 ) ( x1 , y2 ) = ( 0.6 , 0.408 ) ( xb , yb ) = ( 0.118 , 0.015 ) ( xn , yn+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) N 2 1 x 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 18 Liquid Extraction Applied when distillation is ineffective close – boiling mixtures , substances that cannot withstand the temperature of distillation. utilizes chemical differences 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 extract ( solvent plus extracted solute ) feed raffinate mix and settle 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 9 empirical equation 1/3 0.3 0.15 a 1 2 g 0.5 2 v 0.5 1.54 0.41 c V s,c (1 R ) ( ) C1 ( ) 2 0.5 (23.4) g a ( a ) d v c v 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 3 ρd = density of dispersed phase, kg/m Δρ = density difference between phases, kg/m3 2 3 av = specific surface area of packing, m /m ε = fraction voids or porosity of packed section α = 1.0 for continuous-phase wetting, 1.2 for dispersed-phase wetting C1 = function of packing type The function C1 is 0.28 for Raschig rings of Intalox saddles.
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