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LECTURE NOTES MG – 4111 HYDRO-ELECTROMETALLURGY Semester I, 2010/2011 DR. M. Zaki Mubarok Department of Metallurgical Engineering, Faculty of Mining and Petroleum Engineering (FTTM)-ITB Course Outline I. II. Introduction to Hydrometallurgy Thermodynamic and Kinetic Aspects in Hydrometallurgy III. Leaching and Solid-Liquid Separation IV. Solution Purification and Metals Recovery Methods from Pregnant Leach Solution Course Outline V. Leaching and Recovery of Metals and Oxides Ores (Au, Ag, Zn, Al, Cu, Ni) VI. Leaching and Recovery of Sulphide Ores (Zn, Ni, Cu) VII. Introduction to Electrometallurgy VIII. Metals Production by Electrolysis in Aqueous Solution IX. Fused Salt Electrolysis Literatures 1. 2. 3. 4. 5. Havlik,T., ”Hydrometallurgy: Principles and Applications,” CRC publisher, 2008. Habashi, F. ”A Textbook of Hydrometallurgy”, Metallurgie Extractive, Quebec,1993 Norman L. Weiss, “SME Mineral Processing Handbook“, Volume II, SME, 1985 Unit Processes in Extractive Metallurgy: Hydrometallurgy, A Modular Tutorial Course of Montana College of Mineral Science and Technology Biswas, A.K. And Davenport, W.G., “Extractive Metallurgy of Copper”, Pergamon, Oxford, fourth edition, 2002 Literatures 6. Unit Processes in Extractive Metallurgy: Electrometallurgy, A modular tutorial course of Montana College of Mineral Science and Technology 7. Yannopoulus, J.C,”The Extractive Metallurgy of Gold”, Von Nostrand Reinhold, New York, 1991 Course Structure and Mark Distribution • Course Structure – Lecture – Tutorial – Assignment and Lab Work • Mark Distribution – – – – 45% Midterm Exam 45% Final Exam 5% Assignment 5% Lab Work • Attendance: 70% minimum CHAPTER I INTRODUCTION TO HYDROMETALLURGY Hydrometallurgy Extraction, recovery and purification of metals, through processes in aqueous solutions. Metals are also recovered in the other forms such as oxides, hydroxides. Electrometallurgy Recovery and purification of metals through electrolytic processes by using electrical energy. Hydrometallurgy Scope • Traditionally, hydrometallurgy is emphasized for metals extraction from ores. • Hydrometallurgical processing may be used for the following purposes: Production of pure solutions from which high purity metals can be produced by electrolysis, e.g., copper, zinc, nickel, gold, and silver. Production of pure compounds which can be subsequently used for producing the pure metals by other methods. For example, pure alumina to produce smelter grade aluminium. • However, hydrometallurgy principles can be applied to a variety of areas such as metals recycling from scrap, slag, sludge, anode slime, waste processing, etc. Unit Processes in Hydrometallurgy • In general, hydrometallurgy involves 2 (two) main steps: 1. Leaching Selective dissolution of valuable metals from ore. 2. Recovery Selective precipitation of the desired metals from a pregnant-leach solution. General outline of hydrometallurgical processes Ore/concentrate Leaching agent Oxidant leaching Solid-liquid separation Solid residu to waste Pregnant Solution Precipitant or electric current Solution purification Precipitation Pure compound Metals • Commonly, solution purification is conducted prior to metals recovery from the solution. • Solution purification is aimed at obtaining a concentrated solution from which valuable metals can be precipitated in the next processes effectively • Solution purification methods which are commonly used are as follows: – Adsorption by activated carbon – Adsorption by ion exchange resins – Solvent extraction (using organic solvents) – Precipitation with metals (cementation) Solution purification • Solution purifications by adsorption with activated carbon, ion exchange resins (IX) and solvent extraction (SX) have the same unit operations, namely: – Loading, and – Elution • In the elution step, the adsorbers are usually regenerated for another process cycle. Hydrometallurgy development Hydrometallurgy is developed after pyrometallurgy. Metals smelting has been practiced since thousands years ago. Hydrometallurgy was developed after the people discovered acid and base solutions. However, modern hydrometallurgy development is commonly associated with the invention of Bayer Process for bauxite leaching and cyanidation for gold extraction at the end of 19th century (1887). One of important highlights of hydrometallurgy development is uranium extraction (Manhattan Project) aimed at nuclear weapon production in second world war (1940‘s). Important milestones in the development of hydro-electrometallurgy • Cementation & Aqua Regia Use - 8th Century • Cyanidation - 1887 • Bayer Process - 1887 • Hall-Heroult Process - 1886, 1888 • Copper Electrowinning - 1912 • Zinc Electrolytic Process - 1916 • Manhattan Project (IX/SX) - 1940’s • Biooxidation of Sulphide Concentrates - 1960’s • Pressure Leaching – Sherrit Gordon Nickel Process - 1954 – Pressure Acid Leaching of Ni Laterites - 1955 • Large Scale Copper SX/EW - 1960’s Important milestones in the development of hydro-electrometallurgy • Carbon in Pulp (CIP)/Carbon in Leach (CIL) for Gold Recovery - 1980’s • Pressure Oxidation of Zinc Sulphides - 1981 • Two-Stage Zinc Pressure Leach - 1993 • Atmospheric Leaching of Zinc Sulphides – Albion (1993), Outokumpu (1999) Recent Developments: • Skorpion Project (Anglo American) – 2003 (Zn from ZnS) • Hydrozinc (TeckCominco) - 2004 • Inco’s Goro and Voisey Bay Projects - 2007 • Leaching of Chalcopyrite (CuFeS2) Ores Hydrocopper (Outokumpu) Cu from sulfidic ores • Atmospheric leaching of nickel laterite ore: 2008? Hydrometallurgy vs. Pyrometallurgy Hydrometalurgy Pyrometallurgy Treat high grade ore? Less economic More economic Treat low grade ore? Possible with selective leaching Unsuitable Treat sulphide ore No SO2; otherwise So or SO42- are generated SO2 generated (can be converted to H2SO4) Separate similar metal, such as Ni and Co Possible with certain method Not possible Pollutant Waste water, solid/slurry residues Gases and dust Reaction rates Slower Rapid Hydrometallurgy vs. Pyrometallurgy Hydrometalurgy Pyrometallurgy Scale of operation? Possibly economic to Unconomic at smale be done at small scale operation scale operation and expansion is easier Capital cost Generally lower than pyrometallurgy Higher Energy cost Lower Higher Materials Handling Slurry Easy to be Pumped and Transported Handle Molten Metal, Slag, Matte Residues Residues – Fine and Less Stable Slags – Coarse and Stable CHAPTER II Thermodynamic and Kinetic Aspects in Hydrometallurgy Spontaneous Reaction, Equilibrium State • As has been learned in basic engineering courses, chemical reaction will spontaneously occur when the Gibbs free (G) < 0. G = Go + RT ln K • G = 0 process is in equilibrium state – – – – Go = standard Gibbs free energy R = ideal gas constant = 8,314 J/K.mol T = absolute temperature of the system (K) K = equilibrium constant • Standard Gibbs free energy is determined at: – Gaseous components partial pressure = 1 atm – Temperature = 25 oC (298 K) – Ions activity = 1 Equilibrium Constant • For reaction: aA + bB cC + dD C D K A B a a c a a a d b a A = activity of A = γ[ A] →[ A] = A concentration = activity coefficient of component A For gaseous component of X→ a X = γ p X, in which p X = parsial pressure of X Nernst Equation • Hydro-electrometallurgical processes often involve electrochemical reactions. • For electrochemical reaction G = -nFE, Go = -nFEo, therefore RT E E ln K nF o Nernst Equation In which, E = potential for reduction-oxidation reaction Eo =standard potential for reduction-oxidation reaction n = number of electron involved in the electrochemical reaction, F = Faraday constant = 96485 Coulomb/mole of electron • Spontaneous process E > 0 G < 0 Chemical reactions usually perform in leaching processes • Dissolution by acid – Example: ZnO(s) + 2H+ → Zn2+(aq) + H2O(l) • Dissolution by base – Example: Al2O3(s) + 2OH- → 2AlO2-(aq) + H2O(l) • Dissolution by complex ion formation Example: CuO(s) + 2NH4+(aq) + 2NH3(aq) → Cu(NH3)42+(aq) + H2O(l) Chemical reactions usually perform in leaching processes • Dissolution by oxidation – Ex: CuS(s)+ 2Fe3+ → Cu2+(aq) + 2Fe2+ + So(s) Other oxidators: O2, ClO-, ClO3-, MnO4-, HNO3, H2O2, Cl2 • Dissolution by reduction mechanism – Ex: MnO2(s) + SO2(aq) → Mn2+(aq) + SO42-(aq) Correlation of free energy (G) and heat (enthalphy = H) G = H - TS Go = Ho - TSo ∆Ho = Standard enthalpy (kJ/mol) ∆Go = Standard entropy (kJ/mol) ∆Go (reaction) = ∆Go (products) - ∆Go (reactants) ∆Ho (reaction) = ∆Ho (products) - ∆Ho (reactants) ∆So (reaction) = ∆So (products) - ∆So (reactants) ΔHT = ΔH298 + T ∫Cp dT 298 Cp = heat capacity at constant pressure (J/molK) Where possible, processes are designed to be autothermal → maintain constant temperature by the heat given by the reaction Calc. example 1 • Find K for each reaction using a) Standard free energy data b) Standard electrode potential data Calc. Example 2 a) What is the electrode potential of the Ni2+/Ni reaction in sulphate solution at 25°C at a Ni2+ concentration of 0.005 M (assumption: activity of Ni2+ is equal to its molar concentration) b) At what pH is H2 at 10 atm at equilibrium with this solution and pure nickel? Ni2+ + 2 e = Ni E° = -0.26 V 2H+ + 2 e = H2 E° = 0.00 V Pourbaix Diagram • Pourbaix Diagram = Potensial (Eh) – pH Diagram. • The diagram represents thermodynamic equilibrium of metal, ions, hydroxides (or, oxides) in aqueous solution at certain temperature (isothermal). • The boundary of stability regions of metal, ion, hydroxides (or oxides) are equilibrium lines. • Does not reflect reaction kinetics. Pourbaix Diagram • Three possible types of equilibrium lines: – Horizontal – Vertical – Slope • Variations in ion activities are plotted as contours/dashed lines • Horizontal Line: for equilibrium reactions that are independent of pH. Horizontal Line • Example: Fe3+ + e = Fe2+ Eo = 0.77 V R = 8.314 J/Kmol, T = 298 K, F = 96500 C/mol e-, n = 1 mol eRT aFe 2 + o E =E ln nF a Fe3 + If all ion concentrations are assumed to be equal to their molar concentrations 10-6 M. 3+ [ ] RT Fe o E =E ln nF [Fe 2+ ] 8,314 x 298 [Fe 2+ ] E = 0,77 ln [Fe3+ ] 2x96500 [Fe2+ ] E = 0,77 - 0,0592 log [Fe3+ ] E = 0,77 Vertical Line • Reactions do not involve electron → n = 0, no potensial , the equilibrium depends only on pH. • Example: Fe2O3 + 6H+ = 2Fe3+ + 3H2O K = [Fe3+]2/[H-]6 pH = - lo ga H+ ≈ - log[H+ ] = 14 - pOH - pOH = - lo ga (OH)- ≈ - log[OH] For certain Fe3+ concentration we can determine the equilibrium pH for the above reaction. Slope Line • For reactions that depend both on potensial (Eh) dan pH. • Example: If all ion concentrations are assumed to be equal to their molar concentrations 10-6 M. Water stability region (dotted lines) • Upper boundary line At pO2 = 1 atm • Lower boundary line At pH2 = 1 atm Eh-pH diagram of Fe-H2O system at 25°C Eh-pH Diagram of Zn-H2O System at 25 oC. Eh-pH Diagram of Cu-H2O System at 25 oC. Application of Eh – pH diagram in hydrometallurgy • Predicting potential leaching behaviour for certain mineral system • Predicting the possibility of metals ion precipitation at the purification of pregnant-leach solution Application of Eh – pH diagram in hydrometallurgy Fe(OH)3 or Fe2O3 can be precipitated from Fe3+ at lower pH than the precipitation of Zn2+ to Zn(OH)2 or ZnO. Fe2+ have to be oxidized to Fe3+ to gain lower pH value for Fe(OH)3 precipitation. Pourbaix Diagram can be constructed at various temperature for more than two systems Eh-pH diagram of Zn-S-H2O system at 25oC Diagram Pourbaix in Presence of Complex Ion • Example: Au-H2O system with the presence of cyanide (CN-) ion (case of gold cyanidation leaching) • Equilibrium of Au3+/Au Au 3 3e Au G 0 G 0 f (Au ) G 0 f (Au 3 ) 433 KJ / mol Standard reduction potential for Reaction 1: 0 3 G 433 10 E0 nF 3 96500 1.5 V (1) • Equilibrium reaction of O2/H2O O 2 + 4H + + 4e = 2H 2 O (2) Eo = 1.23 V. Au 3+ / Au = 1.5 O 2 / H2O = 1.23 E Au 3 + / Au > EO2 / H2O Therefore, Au3+ ions are not stable in water and readily reduced to Au by oxidation of H2O to O2 (the opposite of Reaction 2). In the other word, gold can not be oxidized (dissolved) in water only with the presence O2. Potensial – pH diagram of Au–H2O system without the presence of complexing agent With the presence of CN-, Au3+ forms STABLE COMPLEX of “aurocyanide“ (Au(CN)2-) and the potential-pH diagram for Au changes significantly as follow: Eh-pH Diagram of Au-CNH2O system at 25 oC for [Au] = 10-4 M and [CN-] = 10-3 M • By the presence of cyanide ions, Au+ + e = Au E = 1.69 – 0.0591 log [Au+] Au+ + 2CN- = Au(CN)2(K = 2 x 1038) Au(CN)2- + e = Au + 2CN- ........................... (3) E 1.69 0.0591log K 0.0591log CN a 2 / a AuCN 2 In comparison to the first reaction that has Eo of 1.69 V, Reaction (3) has much lower Eo at -0.57 V. Dissolution of Au is limited by the following equilibrium of Reaction (3). • During cyanidation leaching, dissolved oxygen is required to oxidize Au prior to the formation of stable complex of Au(CN)2-. Interactions in Electrolyte Solution Two types of interactions in electrolyte: - Ion-ion interaction, and - ion-solvent interaction Knowledge of interaction in electrolyte solution is important because the interactions affect solvation effects, diffusion, conductivity, ionic strength and activity coefficients of ions in solution. Interactions in electrolyte solution influence the transport properties of ions in solution. Ionic Strength and Activity Coefficient - Ionic strength (I), expresses the ionic concentration that includes the effects of ionic charge. - Ionic strength (I) is defined as follow: 1 2 I ci z i 2 i in which ci = concentration of ion i in molar (mol/L) and zi = the charge of ion i. - It is found that activity coefficient, electrical conductivity and the rates of ionic reactions are all the functions of ionic strength. Ionic Strength for unit concentration in molal - Remember, molality = moles of solute in 1 kg solvent. Molality can be converted to molality by the following correlation: ci mi = 0.001ρ - ∑c iMi in which Mi = the molar mass of each solute in kg/mol (not in g/mol), ci = molarity of solute i, and is the density of the solution in kg/m3 (=g/L) - In dilute solutions, ci 0.001mio (in which o = density of pure solvent). Ionic Strength for unit concentration in molal - Therefore for dilute solution, 1 1 2 ∑ I= c i z i = ∑0.001m i ρ o z i 2 2 i 2 0.001ρo ∑m z 2 I= i i 2 If the solvent is water at 25oC (density 1000 kg/m3), then: 1 Similar form with ionic 2 ∑ I ≈ mi zi strength in molarity 2i - Molar activity coefficient can be converted to molal activity coefficient by the following correlation:) f 1 M s m for salt, or i fi 1 M s m for single ion. in which = total moles of ion formed during complete dissociation, m = ionic molality and Ms = molecular weight of solvent (kg/mol). Activity and Activity Coefficient, DEBYE-HUCKEL LAW - Debye Huckel Law correlates the activity coefficient (fi , i) with ionic strength (I). - Forms of Debye-Huckel equations depend on concentration of solution and the unit concentration used. - For dilute solution at 25 oC and I given in molar (M), log f i 0.51159zi 2 I log f ± = 0.51159 z + z - I for single ion, and for salt. - The above equations are known as LIMITING DEBYE HUCKEL LAW. The limitation of LIMITING Debye-Huckel Equation • The D-H Limiting Law is called a ”limiting” law because it becomes increasingly accurate as the limit of infinite dilution is approached. • Up to concentrations of about 0.01m THE LIMITING D-H LAW gives reasonable values, but at higher concentrations the calculated activity coefficient become inaccurate (high %error compared to the values determined experimentally). Debye-Huckel Law for Concentrated Solution - For concentrated solution (> 0.01 molal), Limiting Law D-H is modified by considering the ionic size parameter: log f A z z I 1 Ba I - in which A and B are constants that depend on the kind of solvent and temperature, a = ion size parameter. - For aqueous solution at 25 oC, A = 0.51159 and B = 3.2914 x 109 meter. ACTIVITY AND MEAN ACTIVITY - Molar activity and molar activity of a single ion i is determined as follow: a i i mi a i fi Ci and - For 1 mole of M+A- salt that dissociates to + mol of Mz+ and - mole of AzM+A- + Mz+ + - Az- = + + - ACTIVITY AND MEAN ACTIVITY Mean molal activity coefficient can be determined by the following correlation: 1/ Mean molality, m m m 1 / Note that m m Thus, mean molal activity, a m m 1 / m Exercise: 1 1. Determine the molar activity coefficient of Ca2+ at 25oC using relevant Debye Huckel Equation in the following solution: a. 0.0004 mole of HCl and 0.0002 mole of CaCl2 in one liter solution b. 0.004 mole of HCl and 0.002 mole of CaCl2 in one liter solution c. 0.4 mole of HCl and 0.2 mole of CaCl2 in one liter solution Ion size parameter for Ca2+ = 0.4 nm. Exercise 2: 2. The stoichiometric mean activity coefficient at 25 oC of the sulphuric acid in a mixture of 1.5 molal sodium sulphate (Na2SO4) + 2 molal H2SO4 is 0.1041. If the second dissociation constant, K2, for sulphuric acid is 0.0102 and the pH of the solution is –0.671, calculate: a) the molal activity of H2SO4 b) the molal activity of SO42c) the molal activity of HSO4d) the mean activity of H2SO4 Exercise:3 1 gram FeCl2, 1 gram NiCl2 and 1 gram of HCl are added to 200 ml of deaerated water. Platinum electrodes are used to deliver electrical current so that the electrolysis performs. The anodic and cathodic current density are 1000 A/m2. The following are the reactions and Eo (in the reduction direction) that may occur: Fe2+ + 2e = Fe Eo = -0,277 V Ni2+ + 2e = Ni Eo = -0,250 V 2H+ + 2e = H2 Eo = 0 V Cl2 + 2e = 2ClEo = 1,359 V a) b) c) d) Calculate molar activity coefficients of the cations and anion contained in the solution (use the Finite Size of Debye Huckel Limiting Law) Calculate the activity of the cations and anion contained in the solution Determine the half cell potential of the above reactions Which pair of redox (reduction –oxidation reaction) that would occur (based on the calculation of c) Exercise: 3 (cont.) e) What would be the cell voltage of the reaction d Data: Atomic weight Fe = 55.8, Ni = 58.7, Cl = 35.5, H =1 Ion size parameter in nm : Fe2+ = Ni2+ = 0.6, H+ = 0.9, Cl- = 0.3 H2 overpotential = 0.28 V Cl2 overpotential = 0.03 V Ohmic overpotential = 0.25 V. Kinetics in Hydrometallurgy • Kinetics in hydrometallurgy deals with the kinetics of leaching, adsorption and precipitation • Studying of leaching kinetics is done for the establishment of the rate expression that can be used in design, optimization and control of metallurgical operations. • The parameters that need to be estabished: – Numerical value of the rate constant – Order of reaction – Rate determining step – Activation energy Leaching Kinetics • Consider the dissolution of a metal oxide, MO, with an acid by the following reaction: MO(s) + 2H+(aq) M2+(aq) + H20(aq) • The reaction rates for this leaching system can be given by dC MO 1 dC H+ rR = = dt 2 dt or dC rP = M2+ dt = dC H2O dt Leaching Kinetics • For general example if a chemical reaction involves A and B as reactants and C and D as products, the stoichiometric reaction can be written as follows: k1 aA + bB ⇔cC + dD (1) k2 where a, b, c, and d = stoichiometric coefficients of species A, B, C, and D, respectively k1, k2 = reaction coefficients in the forward and reverse directions, respectively Leaching Kinetics • The rate expression of this stoichiometric reaction can be written in a more general way: 1 dC A 1 dC B 1 dC C 1 dC D = = = = k1CnA CBm - k 2CpCCDq a dt b dt c dt d dt (2) where CA, CB, Cc, and CD are concentrations of species A, B, C, and D, respectively and m, n, p, q are orders of reaction. Leaching Kinetics • However, if the reaction given in Eq. 1 is irreversible, as in most leaching systems, Eq. 2 is reduced to the following form: 1 dC A = k 1CnA CBm a dt or dC A = k1' CnA CBm dt where k1’= k1 x a. • For this system, the rate constant, k1', and the orders of reaction, n and m, should be determined with the aid of leaching experimental data. • The rate expression given in the above equations can be further reduced if the reaction is carried out in such a way that the concentration of A is kept constant. • For such situations, the rate expression is reduced to: dC A = k 1″ CBm dt where k1”= k1’ x CAn. It should be noted that the rate constant and the order of reaction are constant as long as the temperature of the system is maintained constant. • Consider the dissolution of zinc in acidic medium: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) • For the above reaction, the rate of disappearance of H+ ion is directly related to the rate of appearance of Zn2+ ion; thus, dC Zn2 + dt = 1 dCH+ = k′ CmZnCHn = kCHn 2 dt • If concentration of zinc metals is assumed to be constant and CH is further abbreviated generally as CA, then the equation can be written as follow: dC A = kC nA dt • The order of reaction, n, can be any real number (0, 1, 2, 1.3, etc.). • When n = 0, the reaction is referred to as “zero order” with respect to the concentration of A. dC A = kC A 0 dt CA ∫dC A Co A or = CA - C XA = o A t = k ∫dt 0 = -kt k t CoA where CAo represents the concentration of A at t = 0, and XA represents the fractional conversion, i.e., XA = [ (CAo — CA)/ CAo]. • If the plot of XA versus t gives a straight line, the zeroorder assumption is consistent with experimental observations and the k value can be obtained from the slope of the plot. XA k/CAo time • When n = 1, the reaction is first order with respect to the concentration of A: dC A = kC A dt t dC A ∫ = -k ∫dt = -kt o 0 C CA A CA CA ln o = -kt CA ln(1 - X A ) = - kt or X A = 1 - e -kt ln (1 - XA) k time For second order reaction, t dC A ∫ 2 = - k ∫dt = -kt 0 CA Co A CA 1 1 = -kt C A CoA XA = CoA kt 1 - XA XA (1 - XA) k CAok time If the second-order assumption is valid, we obtain a straight line from a plot of XA/(1 - XA) versus t, and the rate constant can be determined from the slope of the plot. Temperature Effect on the Reaction Rate (Arrhenius Law) Reaction rate increases markedly with increasing temperature. It has been found empirically that temperature affects the rate constant in the manner shown in the following equation: k = k oe-Ea / RT ln k = ln k o - Ea 1 R T log k = log k o - Ea 1 2.303R T where Ea is the activation energy and k° is a constant known as the frequency factor, frequently assumed to be independent of temperature. Modeling of heterogenous reaction kinetics • Heterogenous reaction between solid and fluid in hydrometallurgical processes is frequently modelled with “shrinking core“ model. • If we select a model we must accept its rate equation, and vice versa. • If a model corresponds closely to what really takes place, then its rate expression will closely predict and describe the actual kinetics; • If a model differ widely from reality, then its kinetic expressions will be useless. • Detailed of modeling and relevant kinetics equations for various rate determining steps can be found in previous course (Metallurgical Kinetics). • For determination of Ea, number of experiments, at least at three or four different temperatures are needed, with all other variables being kept constant. The next step is to calculate the rate constant for each temperature as discussed previously. • A plot of In k versus 1/T yields a straight line from which the activation energy, Ea, can be determined • Activation energy value can be used to predict the rate determining step of the reaction: • Ea = 40 – 80 kJ/mol: process is controlled by surface chemical reaction • Ea = 8 – 20 kJ/mol: process is controlled by diffusion to and from the surface Mass Transfer in Solution • For hydrometallurgical system, mass transfer of component i in solution frequently consists of a molecular diffusion term, migration term, and convective diffusion term, as indicated in the following expression: Ni = - Di∇Ci - ziμiFCi∇Φ + Ci V where Ni= flux of i, Ci = concentration of i, Di = diffusion coefficient of i Ci = concentration gradient of i, zi = valence of the specified ion, µi = ionic mobility, F = the Faraday constant, Ф = electrical potential gradient, and V = net velocity of the fluid of the system First and Second Fick’s Law of Diffusion • If Ni consists of the molecular diffusion term only, Ni = -Di∇Ci (Fick's first law) ∂Ci + Di∇ 2Ci = 0 ∂t (Fick' s second law ) • Dimensionless Parameter for Convection Calculation LV LV Di Di where µ = the viscosity of the fluid, ρ = the density of the fluid Dimensionless Parameter for Convection Calculation • The parameter LV/Di is known as the Peclet number and can be separated into two other parameters: Lvρ/µ that known as the Reynolds number, and µ/ρDi is the Schmidt number. • Peclet number is regarded as a measure of the role of convection against diffusion, LV ∇Ci V convective diffusion = = D i D i (∇Ci / L) molecular diffusion • For most hydrometallurgical systems, the Schmidt number is on the order of 1,000 because the diffusivity of ions and kinematic viscosity of water are, respectively, on the order of 10-5 cm2/s and 10-2 cm2/s. • Therefore, if the Reynolds number is greater than 10-3, the Peclet number is greater than 1, and consequently, convective diffusion is more dominating than molecular diffusion in such systems. Mass Transfer Coefficients for Convective Diffusion • For systems with large Peclet numbers, it is frequently assumed that there is a diffusion boundary layer at some distance from the solid surface. For such systems, it is quite common to write the mass flux from the bulk solution to the solid surface as follows: Ni = km (Cb - Cs) where Ni = mass flux of species i km = mass transfer coefficient, in cm/s Cb = concentration of species i in the bulk solution, in mol/cm3 Cs = concentration of species i at the solid surface, in mol/cm3 • Because the units of measure of km are the same as those of (D/), where is the diffusion boundary layer thickness, km, is often substituted by this ratio. Therefore, D N i = (Cb - Cs ) δ • The diffusion boundary layer thickness is often estimated by the relationship km = D/δ, provided km is known. Mass Transfer from or to a Flat Plate. • The mass transfer coefficient for a flat plate where fluid is flowing over the plate at a velocity V0 has been well documented. • The mass transfer coefficient for such a system can be estimated from first principles and has the following form: k m = 0.664D2 / 3ν -1/ 6 L-1/ 2 V01/ 2 For Re < 106 where D = the diffusivity of the diffusing species v = the kinematic viscosity of the fluid L = the length of the plate Rotating Disk. • Although it is not a practical geometry, because the mathematical representation of the system is exact and follows very closely to the experimental data, a rotating disk is frequently used to determine the mass flux and the mass transfer coefficient. • The mass transfer coefficient for this system is as follow k m = 0.62D2 / 3ν -1/ 6ω1/ 2 • The equation is valid for the Reynolds number, r2ω/ν is less than 105, where r and ω are, respectively, the radius and the angular velocity of the disk. Particulate System • It has been demonstrated that the mass transfer coefficient for particulate systems can be given by the following equation: 2D km = + 0.6Vt1/ 2 d -1/ 2 ν -1/ 6 D 2 / 3 d where d = the diameter of the particle, Vt = the slip velocity, which is often assumed to be the terminal velocity of the specified particle. • The terminal velocity of a particle can be calculated using the following equation depending on the Reynolds number of the system, which is defined by dVtρ/μ, where ρ is the density of the fluid: 2r 2 (ρs - ρ)g Vt = 9μ where ρs is the density of the particle. The preceding equation is often referred to as the Stokes' equation and is valid as long as the Reynolds number is less than 1. When the Reynolds number is between 1 and 700, the following equations are used: μ A Vt = 10 dρ 1/ 2 ( ) A = 5.0 0.66 + 0.4 log K - 5.55 4gd 3ρ(ρs - ρ) K= 3μ2 where g is the gravitational coefficient. Example 1: • A cementation reaction, Zn + Cu2+ → Cu + Zn2+, is taking place at the surface of a zinc plate of 10 cm x 10 cm area. • Feed flowing parallel to the plate at a velocity of 1 m/s contains copper at 1 mo/dm3. • Suppose we want to estimate the rate of deposition assuming that the mass transfer of Cu2+ to the zinc plate is rate determining step. The diffusivity of Cu2+ is 7.2x10-6 cm2/s, and the kinematic viscosity of water is 0.01 cm2/s. 1 dN Cu 2+ = k m (Cu 2b+ - Cu s2+ ) = k m Cu 2b+ S dt where S Ncu2+ Cub2+ Cus2+ = the surface area of the plate = the number of moles of Cu 2+ ion = the concentration of Cu2+ in the bulk = the concentration of CU2+ at the interface km = 0.664 (7.2 x 10-6)2/3 (0.01)-1/6 (10)-1/2 (100)1/2 = 0.664 x 3.7 x 10-4 x 2.15 x 0.316 x 10 = 1.7 x 10-3 cm/s. Therefore, 1 dN Cu 2 1.7 10 3 1,000 1.7 mol / cm 2 .s S dt and 100 10 Re 105 0.01 Example 2: • Consider the situation from the previous example, except that instead of a zinc plate, zinc particles 100 µm in diameter are suspended in a 1 mol/dm3 Cu2+ solution. Suppose we want to estimate the rate of deposition of Cu2+ (Note that the density of Zn is 7.14 g/cm3.) For particulate, 4 981 0.01 1 7.14 1 K 80.3 4 3 10 1/ 2 A 50.66 0.4 log 80.3 5.55 0.412 3 Therefore, Vt 100.412 2.58 cm / s Re 10 2 2.58 / 10 2 2.58 As a result , 2 7.2 10 6 km 0.6 2.581/ 2 0.011/ 2 0.011/ 6 7.2 10 6 0.01 1.44 10 3 7.75 10 3 9.19 10 3 cm / s Finally , 1 dN Cu 2 km 9.19 10 3 1,000 9.19 mol / cm 2 .s S dt 2/3