9/12/2018. Course Objectives MSE 353 PYROMETALLURGY. Prerequisite. Course Outcomes. Forms of Assessment. Course Outline

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1 Kwame Nkrumah University of Science & Technology, Kumasi, Ghana MSE 353 PYROMETALLURGY Course Objectives Understand the fundamental concepts of pyrometallurgy Understand the concepts of materials and energy balance and its application to industrial units Provide methodologies for producing metals (iron oxide to steel, bauxite to aluminium and chalcopyrite to copper) Ing. Anthony Andrews (PhD) Department of Materials Engineering Faculty of Mechanical and Chemical Engineering College of Engineering Website: Understand the physico-chemical reactions in pyrometallurgical extraction of metals from ores Select appropriate process routes for the refining of metals 2 Course Outcomes Use Ellingham diagram in the extraction of metals Gain general knowledge of the various methods available for the production iron, copper and aluminium Prerequisite Knowledge of thermodynamics of materials (phase diagrams, Gibbs free energy, physical chemistry) Principles of material and energy balance across process units Design a system, component, or process to meet the desired needs Understand principles underlying processing-structure-propertiesperformance 3 4 Forms of Assessment Quizzes 5 Assignments 10 Mid-Sem Exam 15 Final Exam 70 Total 100 Introduction Course Outline Drying and Calcination, Roasting Principles of pyrometallurgy; reduction of metal oxides (use of Ellingham diagram) Beneficiation of iron ores Blast furnace iron making, alternative iron-making processes, slag-metal reactions, steel making processes, refining processes Pyrometallurgy of nonferrous metals Alumimium, Copper 5 6 1

2 Recommended Books Steel: processing, structure and performance: Kraus Physical metallurgy of steels: Leslie Mechanical metallurgy: Dieter Fundamental of physical metallurgy: Verhoeven Recovery of metals from ores Ore: deposit that contains enough metal that we can extract economically. Most metals are found in minerals. Most important ores are oxide, sulfide and carbonates. Why? Introduction to the thermodynamics of materials: Gaskel The chemistry of gold extraction, Ellis Horwood: Marsden, J. and House, I. The extractive metallurgy of gold, Chapman and Hall, London: Yannopoulos, J.C. Reducing agent converts the oxide to metal In pyrometallurgy, thermal energy is one of the important inputs 7 8 Principal mineral sources of some common metals Occurrence and Distribution of Metals Metallurgy is the science and technology of extracting metals from minerals. Five important steps are involved: o Mining (getting the ore out of the ground) o Concentrating (preparing it for further treatment) o Reduction (to obtain the free metal in the zero oxidation state) o Refining (to obtain the pure metal) o Mixing with other metals (to form an alloy). 10 Principles of Pyrometallurgy Pyrometallurgy: branch of extractive metallurgy It involves the thermal treatment of minerals/ores resulting in physical and chemical transformations in the materials to enable recovery of valuable metals. Pyrometallurgy requires energy input Sources of energy include fossil fuel combustion, exothermic reactions, electrical heat. Elements that can be extracted by pyrometallurgical processes include the oxides of less reactive elements like Fe, Cu, Zn, Cr, Sn, Mn. 11 Pyrometallurgy Pyrometallurgy: using high temperatures to obtain the free metal. Four major steps are employed, all requiring high amount of thermal energy: 1. Calcination is heating of ore to cause decomposition and elimination of a volatile product (CO 2 or H 2 O): PbCO 3 (s) PbO(s) + CO 2 (g) 12 2

3 Pyrometallurgy Pyrometallurgy 2. Roasting: heating which causes chemical reactions between the ore (solid) and the furnace atmosphere (gas). That is solid-gas reactions at elevated temperatures 2ZnS(s) + 3O 2 (g) 2ZnO(s) + 2SO 2 (g) 2MoS 2 (s) + 7O 2 (g) 2MoO 3 (s) + 4SO 2 (g) 3. Smelting: melting process which separates chemical reaction products into 2 or more layers. Slag consists mostly of molten silicates in addition to aluminates, phosphates, fluorides, and other inorganic materials. 4. Refining: the treatment of crude metal product to improve its purity Principles of Calcination Calcination: a thermal treatment process and applied to ores and other solid materials for the following reasons: 1. thermal decomposition 2. phase transition 3. to remove volatile fractions such as CO 2 and H 2 O Material is heated below the melting point in rotary kiln or fluidized bed reactor. Rate of calcination is governed by the supply of necessary heat of decomposition. Calcination is done in the solid state in the absence or limited supply of air or oxygen 15 Principles of Calcination Calcination is more endothermic than drying. Calcination involves the removal of: free water adsorbed water loosely bound water and strongly bound water The last two are chemical in nature and represent phase transformations. Furnaces for calcination processes: shaft furnaces, rotary kilns, and fluidized bed reactors. 16 Applications of Calcination To produce lime from CaCO 3 in cement production CaCO 3 = CaO + CO 2 To cause decomposition of hydrated minerals 2Al(OH) 3 = Al 2 O 3 + 3H 2 O Heat/Energy Balance in Calcination Calcination requires thermal energy. For calculation, we need several thermo-chemical values like heat of formation, specific heat, and heat content. To cause decomposition of volatile matter contained in petroleum coke Heat treatment of minerals to effect phase transformation Devitrification of glass materials

4 Assignment 1 (Due 17/09/18) 1. Calculate the heat energy required to calcine 1000 kg limestone of composition 84% CaCO 3, 8% MgCO 3 and 8% H 2 O charged at 298K. Lime is discharged at 1178K and gases leave at 473K. 2. It is desired to produce 10 kg.mol lime from calcinations of CaCO 3 (pure) in a rotary kiln. Producer gas of composition 7.2% CO 2, 1.6% O 2, 16.6% CO and 74.6% N 2 is combusted with 20% excess air to obtain the desired temperature in the kiln. The limestone and air are supplied at 298K, whereas producer gas is heated to 900K. Lime is discharged at 1200K and at 500K. Calculate the amount of the producer gas (1atm and 273K). Roasting Roasting consist of thermal gas-solid reactions. It includes oxidation of metal sulphides to give metal oxides and sulphur dioxide. It is strongly exothermic process Roasting is carried out below melting points of sulphides and oxides involved (i.e o C) Note: Use thermochemical values from literature 19 Typical examples are: Roasting Types of Roasting The solid product from roasting is often called calcine ZnS + 1.5O 2 = ZnO + SO 2 ΔH 298 = kj/mol When all sulphides are converted into oxides - dead roasting Example: extraction of Zn from its ore 2FeS O 2 = Fe 2 O 3 + 4SO 2 ΔH 298 = kj/mol Cu 2 S + 1.5O 2 = Cu 2 O + SO 2 ΔH 298 = kj/mol In addition, other reactions may take place: formation of SO 3 and metal sulphates and formation of complex oxides such as ZnO.Fe 2 O 3. Both reactants and products are in solid states. When less than required amount of oxygen is supplied to fully oxidize the feed (Sulphur is partially removed) partial roasting When sulphide is converted into sulphate as sulphate can be dissolved easily into an aqueous solution sulphating roasting Typically used for hydrometallurgical extraction of lead sulphide ores Roasting can be carried out in a number of furnaces, including multiple hearth furnace, fluid bed roasting. 1. Heat of reaction: S SO kcal/kg.mol S SO kcal/kg.mol Several oxidation reactions possible (release heat): Fe Fe 2 O 3 FeS FeO ZnS ZnO Heat of formation of some compounds Calculate the heat released for the following reactions: ZnS + 1.5O 2 = ZnO + SO 2 Calculated from heat of formation. 4

5 Ore concentrate is mixed with solid fuel Solid fuels contain combustible mass (mainly carbon, hydrogen and Sulphur) and non-combustible mass (water and ash) Fuel are divided according to the physical state: Solids (coal and coke) Liquid (fuel oil) Gaseous (CO, H, hydrocarbons, hydrogen sulphide, etc) Fuel characterized by calorific value (CV kj/kg) The Dulong s formula gives the heat of combustion as: where NCP is the net calorific power, and C, H, O, S and W represent the mass percent of the four elements and moisture in the fuel. For gross calorific power (Dulong s formula) GCP Determination of CV of gaseous fuels Combustion components include CO, H 2, NH 3, etc O 2, CO 2, N 2 are diluents Question Calculate the gross calorific value for a coal with the following analyses: 74%C; 6%H; 1%N; 9%O; 0.8%S; 2.2% moisture and 8% ash Question Write combustion equations for CH 4, C 2 H 6 and C 3 H 8 and calculate the heat of combustion values. In roasting, air is used for the oxidation of sulphides as well as combustion of coal Calculation of amount of air (theoretical air) is important Consider the following reactions: C + O 2 = CO 2 H O 2 = H 2 O ZnS + 1.5O 2 = ZnO + SO 2 PbS + 1.5O 2 = PbO + SO 2 5

6 For the purposes of combustion calculations the composition of air is approximated as a simple mixture of oxygen and nitrogen: oxygen = 21% nitrogen = 79% Ultimate analysis gives the percentage by mass of each element present in the fuel. An example of an ultimate analysis of a liquid fuel (oil) might be : Component % by mass Carbon (C) 86 Hydrogen (H 2 ) Each constituent is considered separately via its own combustion equation. For the carbon: C + O 2 CO 2 12kg 32kg 44kg or for 1 kg of fuel (kg) So each kg of oil requires 2.29 kg oxygen for combustion of its carbon and produces 3.15 kg CO 2 as product. Similarly H 2 + ½ O 2 H 2 O 2kg 16kg 18kg or per kg of fuel (kg) 2 2 In order to burn the hydrogen content of the oil 1.12 kg oxygen are needed and 1.26 kg water is formed. The total oxygen requirement is thus ( ) or 3.41 kg. A given quantity of air consists of 21% by volume of oxygen. We can simply transform to a mass basis thus: Component vol fraction(vf) vf MW Mass fraction Oxygen Nitrogen We can now establish that 3.41 kg oxygen, which is the stoichiometric requirement, will be associated with: kg nitrogen The stoichiometric air-to-fuel ratio is thus = 14.6 :

7 Assignment 2 (Due 17/09/18) 1. A fuel has the following composition by mass; carbon 86% and hydrogen 14%. Calculate the theoretical air supply per kg of fuel. 2. A fuel has the following composition by mass; carbon 86%, hydrogen 11.75%, Oxygen 2.25%. Calculate the theoretical air supply per kg of fuel and the mass of products of combustion per kg of fuel. 3. A slurry of chalcopyrite is continuously fed into a fluid bed roaster. Enough air is supplied to the roaster. The slurry containing 10% water is continuously fed into the roaster at a rate of 12,000kg/h. a) Write a balanced equation for the reactor. b) Calculate the amount of air required. c) Calculate the amount of gas produced. d) Amount of solid products formed. 7