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 yn1   xn  --- (23.3) Vn1  Vn1

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 (e) L1 = 595 determine the inlet and exit concentrations for the first stage :

x1 = ya = 0.6 kg solution retained : 0.595 kg solid

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 kg solution retained : 0.53 , Ln = 0.53 × 1000 = 530 h

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 )

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. Values for other 9 0.5 2 packings range 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 Side–to–side ( with or without scraper )

 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)

Figure 23.6 (a) perforations in horizontal plates. (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 phase,  Extraction of dilute solutions y

change in flow rate can be neglected , Solvent (s)  y  K  i  is constant  distribution coefficient. D  x   i  Diluent (b) E : extraction factor (equivalent to stripping factor ) + solute (a) x K V Raffinate E  D phase L

30 V  If (1) one single stage , (2) pure solvent yo = 0 L  x  1  fraction of solute remaining    x x0  x0  1 E y  x  E  fraction of recovery 1    x  1 E V y = L ( x0 – x )  0  V y x0 x V y V  1 E 1  0 Cascade   KD  E L x x L x L x Kremser equation [ Eqs. (20.24) , (20.25) , (20.26) , (20.28) ] can be used.

V L

31 V (a) y = 0 , L = 0.06 , KD = 80 V y = 0 x 1 KDV   0.172 E   800.06  4.8 x 1 E L L 0 E x x0 fraction of recovery  0.828 1 E  x  y 1 1   x   0 

material balance V ( y1 – y0 ) = L ( x0 – x1 )

0

V y1 = L ( x0 – x1 ) , y1 = KD x1

VKD  VKD   x1 1  x0   E   L   L  x 1 1   0.172 x 1 E 0 32 (b) two stages extraction , ( same value of E )

x1 1 x2 1 V V  ,  x0 1 E x1 1 E y = 0 y = 0

x 1 2 L x0 x1 x2   2  0.0297 x0 1 E

x recovery : 1  2  0 . 9703 or 97% x 0

(c) using Eq (20.28) the same recovery as (b)

* * n[(xa  xa ) /(xb  xb )] N  n E

33 n[(x  x* ) /(x  x* )] N  a a b b n E xb  0.03 , let x = 100 , x = 3 xa ya x a b a 100 V ( ya – 0 ) = L ( 100 – 3 ) L 10097 ya  (97)  1617 V 6 V = 6 * ya 1617 L = 100 xa    20.2 KD 80 * xb = 0 n[(100  20.2) /(3 0)] N   2.09 n 4.8 3 Compare : based on 100 volume aqueous xb yb = 0 (b) (c) 2 stage 2.09 …countercurrent flow V : 62 V : 6 1 concentration lower higher of extract 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

Extracted solute curve ACE : extract layer (MIK) curve BDE : raffinate layer (water)

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 diluent 36 in Extract :

H2O < 2%, (for Csolute = 0)  as acetone concentration increases, water content also increases. as shown 

y A maximum at y = 0.27 y A H2O

37 Type II  The solvent ( aniline ) is only partially miscible with both The other components.

FIGURE 23.9 System aniline-n-heptane- MCH at 25oC: a, solute, MCH: b, diluent, n-heptane; s, solvent, aniline. (After 16 Varteressian and Fenske. ) methyl cyclo hexane

38 Use of McCabe – Thiele method

yA : mass fraction of solute in extract phase , V

xA : mass fraction of solute in raffinate phase , L equilibrium data : yA , xA values at two ends of tie lines.

OP line : Eq (23.3)

 L  V y  L x  n  a a a a yn1   xn  Vn1  Vn1

 two terminal points ( xa , ya ) , ( xb , yb ) one ( or more ) intermediate points.

39 L = 100 V = ? Solution : a a xA = 0.4 A = 39.6 H O : n EQ curve : from Fig 23.8 A = 40 2 MIK : 100 – m determine terminal point H2O = 60

Base : feed 100 mass per hour , La = 100 , xa = 0.4

acetone : A = 40 ; H2O = 60 A in extract ( 99% recovery ) : 40  0.99 = 39.6 A in raffinate (99% recovery) : 40  0.01 = 0.4 Lb= ? let n : mass of H2O in extract A = 0.4 H O : 60 – n 2 Vb = 100 (MIK) m : mass of MIK in raffinate MIK : m yb = 0 Va = 39.6 + n + ( 100 – m ) = 139.6 + n – m ... (1) 40

At the bottom Vb = 100 Va = 39.6 + n + ( 100 – m ) = 139.6 + n – m

Lb = 0.4 + ( 60 – n ) + m = 60.4 + m – n ... (2) assume : n , m are small and tend to cancel

Va  140 , Lb  60  ( will be adjusted after calculating n , m )

39.6 n: H O  y   0 . 283  y = 0.049 ... ( n ) 2 (extracted) Aa H2O m: MIK 140 0.4  0 Fig 23.8 (raffinate)  xAb   0 . 0067  xMIK = 0.02 ... ( m ) 60 139.6  m  from (1)  n     0 . 049 ... (3) 1 0.049   V a n if m is small , n ≒ 7.2  0.049 139.6  n  m m = 1.09

 60.4  n  m n = 7.14 (2)  m     0 . 02 ( n = 7.2 )  0.02  0.98  60.4  m  n Lb  m ≒ 1.1 used to revise n by Eq (3)  n 41= 7.1 n: H2O m: MIK calculate Va = 139.6 + n – m (1)  ya

Lb = 60.4 + m – n ... (2)  xb

39.6 V = 139.6 + 7.1 – 1.1 = 145.6 ; y   0 . 272 ( 1st : 0.283 ) a a 145.6

0.4 Lb = 60.4 + 1.1 – 7.1 = 54.4 ; x b   0 . 0074 ( 1st : 0.0067 ) 54.4 terminal points : ( 0.4 , 0.272 ) ; ( 0.0074 , 0 )

Find an intermediate point on the operation line

yH2O = 0.03 pick yA = 0.12  V  L  x Fig 23.8 yMIK = 0.85  used to revise A

assume amount of MIK in VMIK  100 ( MIK in raffinate is small , (1.1) ) 100 V yMIK  100 V  117.6 0.85 43 100 V yMIK  100, V  117.6 0.85 xA = 0.4 ya = 0.272 overall balance :

L  V + Lb – Vb  117.6 + 54.4 – 100 = 72.0 L V balance to A : xA = ? yMIK = 0.85 yA = 0.12 L xA  V yA + Lb xb – Vb yb = 117.6  0.12 + 54.4  0.0074 – 0 = 14.5 0.4

14.5 x   0.201  rough value A 72 A = 0.4 MIK = 1.1 MIK

Lb = 54.4 Vb = 100 xb = 0.0074 yb = 0

44 xA = 0.4 ya = 0.272 Calculate corrected values of V , L , xA L xA = 0.201  xMIK  0.03 V Fig 23.8 xMIK yMIK = = 0.03 0.85 balance to MIK : yA = 0.12 L xMIK + Vb = Lb xb,MIK + V yMIK A = 0.4 MIK = 1.1 MIK 72  0.03 + 100 = 1.1 + V (0.85) Vb = 100 Lb = 54.4 yb = 0 xb = 0.0074 V = 101.1/0.85 = 118.9 ( revised ) 1st assume : 100 mole MIK (V=117.6) revised L = 118.9 + 54.4 – 100 = 73.3 ( overall balance )

Vy  0.4  0 118.90.12  0.4 revised x  A   0.2 A 73.3 73.3 intermediate point ( 0.2 , 0.12 ) From Fig 23.11 , N = 3.4 stage 45 46 Countercurrent extraction of type II systems using reflux

Solvent plays the same part in extraction that heat does in distillation

47 Limiting reflux ratios RD  L  L       D min D

RD → ∞ (total reflux) , min. No. of plates

R =  L  , No. of plate is ∞ D D min Practical examples of extraction with reflux aniline – heptane – methylcyclohexane ( MCH ) system ( Fig23.9 )

  MCH  in extract only modestly greater than that in raffinate  heptane  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.

(五環硫氧烷)

Extract : nearly all aromatic and few % distillation : ( azetropic distillation ) paraffine and nophthenes (suefolane and hydrocarbons form light hydrocarbon + water  type I system with plait point) azetropic ( low boiling Temp. ).

boiling points overlap Backwash : low boiling hydrocarbons ( countersolvent or reflux ) 49 50 Special Extraction Techniques

(1) recovery of sensitive biological products  using only aqueous phases.

Example : separation of proteins

two aqueous phase : (a) polyethylene glycol

( 80 ~ 90% water ) (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  o Tc = 31.06 C useful solvent for supercritical extraction : CO2 Pc = 73.8 bar In supercritical region : 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 54 V , y y , V V , y Vb , yb n+1 n+1 n n a a N n 1 La xn L , x L xa b b Ln 1 x1