Mineral Processing 345 Past Paper PDF

Mineral Processing 345 Basic metallurgical concepts and mass balancing 1 Introduction Natural raw materials, along with waste (biological, municipal, and industrial waste), form the building blocks of all material and chemicals that we deal with daily. Normally, the furt...

Mineral Processing 345 Basic metallurgical concepts and mass balancing 1 Introduction Natural raw materials, along with waste (biological, municipal, and industrial waste), form the building blocks of all material and chemicals that we deal with daily. Normally, the further we progress along the production chain, the purer the raw materials that are processed into products. The most complex raw materials that occur are often those that occur naturally. Each has its own physical-chemical properties, but occurs integrated in solid form. Both natural raw materials (e.g. ores and mineral associates, coal) and solid matter waste (e.g. municipal waste, electronic waste, ash) share important similarities: Both have complex chemical compositions (numerous chemical species, each with a potentially complex structure itself). Both tend to have a multi-phase origin (numerous solid phases can be present). Neither natural raw materials nor waste have much economic value or usefulness in their natural or unmanufactured form. Many of the natural raw materials and wastes that are economically viable tend to present in the solid form. The solid form adds some further complications to the workability, namely: Solid phases (e.g. various minerals, metals, ceramics, polymers and biological materials) can be physically coalesced. The solid phases have varying physical natures and quantities. The phase crystal structure, density, chemistry, magnetic susceptibility, electrical and thermal conductivity, transformability and breaking properties can differ. The intergrowths can form complex geometric patterns. The solid materials have a particular nature. The particles thus take a variety of shapes, forms and surface behaviour. The economically reclaimable species often appear in lower concentrations, often as low as parts per million (e.g. gold from electrical current boards, platinum in ores). Beyond the precious metals, few metals occur in elemental form and most occur in various minerals whose chemical compositions vary. Each compound appears different chemically and physically. To say that a mineral deposit contains 30 ppm gold, for example, does not necessarily imply that it is an economically exploitable ore body, or that the extraction process is known. It is the role of the chemical engineer to develop new and innovative processes to extract minerals and metals from these complex ores. Macro- and microscopic ore properties are looked at in order to determine the economic viability of extraction/processing of an ore body. These properties and analytical techniques for characterisation have been discussed as part of the geology section of the course. The macroscopic properties influence the way in which the ore is taken from the ground in order that it can be processed further. The design of the mine, which is the task of the mine construction engineer, is an important factor that determines the costs of the raw materials for the processing plant. Natural raw Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 1 of 18 materials are thus not without their costs; the ore itself can become a considerable portion of the industrial costs of the plant. The microscopic properties of ores determine to a large extent the economic and technical feasibility of extraction routes towards refining the minerals and extracting the metals out of them. Important micro- properties include, amongst others: the mineral phases that occur; the composition of the various phases; the volumes and mass fractions of each mineral; the mineral and ore grain sizes and their diffusion; the physical-chemical properties of the mineral phases; and, the shape and micro-structure of the grains. The micro-structure influences the physical- and chemical processing method of the ore. A combination of physical techniques and chemical techniques (hydrometallurgy and pyrometallurgy) is used to extract/upgrade/refine valuable species, as illustrated in Figure 1. In this course the focus is more on natural raw material processing, although the principles are also valid to waste processing. The motivation for this is the large number of chemical and metallurgical engineers that find themselves in the mineral processing industry, which remains of importance in South Africa. The reclamation of metals and minerals from secondary sources like waste is rapidly rising in importance in Europe as a result of the drive towards a circular economy. Figure 1. High-level overview of raw material processing flowsheet. The course will be structured around the high-level process overview presented in Figure 1. Physical processing can be divided further based on the various techniques and processes commonly used, as shown in Figure 2. Agglomeration and solid-liquid separation techniques were discussed in Particle Technology 316. In this course, the focus will be on comminution and size classification (approximately 3 weeks’ lectures) as well as physical separation (beneficiation) techniques based on the difference in physical properties between minerals in the ore, or the differences in the minerals’ surface chemistry (approximately 3 weeks’ lectures). In all the above cases, the minerals are not altered chemically, but are merely separated from each other. Once the raw material has undergone comminution and the minerals of interest have been concentrated by means of physical processing, chemical processing techniques are required to recover the metals of interest from the minerals in a suitable, saleable form. A summary of the most common chemical processing techniques is presented in Figure 3. Pyrometallurgy involves the high-temperature treatment of materials to facilitate separation of species; pyrometallurgy as a subject field is dealt with in Mineral Processing 415. In this course, the focus will be on hydrometallurgy which involves separation and recovery Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 2 of 18 of metals and metal compounds from raw material through chemical transformation in aqueous media (approximately 3 weeks’ lectures). Figure 2. Summary of common physical processing techniques. Figure 3. Summary of common chemical processing techniques. Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 3 of 18 Ores consist of more than one valuable species and the procedures necessary to separate chemical species and metals are very complex. To treat any particular ore body and produce a saleable commodity from it will require a unique process route. No two orebodies are the same from a metallurgy point of view, and each requires careful testwork and design to produce the optimum flowsheet (combination of any number of physical separation and chemical processing units) for that particular orebody. Stable process control of the mineral processing plant is also challenging given that the physical and chemical character of the supply stream changes constantly as different parts of the reef are mined. The overall aim of this module is to provide students with the knowledge and skills to develop conceptual flowsheets as well as to analyse and assess the performance of these mineral processing / extractive metallurgy processes. 2 Comminution and liberation Liberation is introduced in Week 4 to provide some context for the role of ore properties in mineral processing, but will be discussed in further detail in Week 5. Raw materials are not pure and different chemical phases are physically intergrown in particles. The objective of the physical processing steps is to separate mineral types to produce concentrates having a greater relative abundance of the desired mineral. To achieve separation of different phases, these must be first loosened or liberated from each other. An ore supply stream to a plant contains large particles consisting of a distribution of smaller grains of different phases. As illustrated in Figure 1, the first physical processing step involves crushing and milling (comminution) of ore in order to reduce the ore particle size. As the size of the large ore particles becomes smaller, so the particle size approaches the mineral grain size, resulting in the release (i.e. liberation) of valuable minerals from gangue minerals (gangue minerals refer to minerals that are of no economic value to the particular process). As the particle size post-breaking becomes smaller, so the level of liberation increases. Particles are usually classed based on particle grades. Classification varies from source to source; in some contexts, for example, particles containing 80% of more of the mineral of interest can be classified as being liberated. Particles with a mineral of interest content of between 30% and 80% are then referred to as the middlings fraction (i.e. these are multi-phase grains with intergrowth of significant residual unliberated mineral of interest), while the mineral of interest is considered locked in particles containing less than 30% of the specific mineral. Particles that contain no mineral of interest can also be thought of as free or liberated gangue. Complete liberation only occurs once the ore particle is milled much smaller than the original mineral grains. Where the valuable mineral occurs finely spaced, for example, in the ore matrix, it could be highly uneconomical to mill the ore until the mineral is completely liberated (i.e. there is no middlings or locked mineral of interest). The valuable mineral can also occur as a shell structure around the ore matrix, or as coarse grains. In such cases, mineral liberation by size reduction is easier and more economical. The different texture types behave differently during refinement processes, even if the number and types of mineral phases are the same. Figures 1.2a to 1.2i in the prescribed textbook (pp. 8-11) give good examples of microscopic analysis of typical ores, and the associated implications for comminution and liberation efficiency. Methods that quantify the mineral texture and liberation properties are necessary to determine how divisible different minerals are. Finely spaced minerals in a gangue matrix are separated from each other Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 4 of 18 and the matrix with difficulty. Intergrowths lead to multi-phase (multi-mineral) grains after crushing and milling. The percentage of ore material that still occurs as multi-phase particles (middlings) has a great impact on the refinement processes that follow size reduction processes. High quantities of multi-phase particles (middlings) lead to the following problems: Multi-phase particles that consist of mainly valuable mineral will end up in the concentrate stream of a divider, and thus dilute the concentrate with worthless gangue associated with the particles. Multi-phase particles that consist of mainly gangue will end up in the refuse stream of a divider, and the associated valuable minerals will be lost with the refuse. The above-mentioned problems can occur even if the piece of equipment is optimally designed for implementation. A combination of perfect mineral liberation and perfect mechanical operation of the machine is necessary for perfect mineral separation. The quality of mineral separation is thus determined by both the fraction of liberated particles (dependent on mineralogy) and the machine operating capacity (dependent on machine design). 3 Process efficiency The objective of mineral processing is to separate mineral types. These processes will produce at least two product streams, namely: (1) the concentrate, which has a greater relative abundance of the desired mineral than the feed stream, and (2) the tailings, which has a greater relative abundance of the gangue mineral than the feed stream. Several performance criteria are defined below to quantify the quality or efficiency of mineral separation processes. 3.1 Grade The grade (or assay) of a stream or ore usually refers to the content of the marketable end product in the material. Thus, in metallic ores, the percentage metal is often quoted as the ore grade; in non-metallic operations, grade usually refers to the valuable mineral content of the material. 3.2 Recovery The recovery, in the case of the concentration of a metallic ore, is the percentage of the total metal contained in the ore (process feed) that is recovered from the concentrate. In other words, a recovery of 90% means that 90% of the metal in the ore is recovered in the concentrate and 10% is lost to the tailings. The recovery, when dealing with non-metallic ores, refers to the percentage of the total mineral contained in the ore that is recovered into the concentrate. With reference to the basic, generic separation process shown in Figure 4, the mass of metal or mineral of interest (species i) in any stream can be calculated as the product of the total ore flow rate and the ore grade: 𝑓𝑓𝑖𝑖 Mass of species i in feed = 𝐹𝐹 100 𝑐𝑐𝑖𝑖 Mass of species i in concentrate = 𝐶𝐶 100 Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 5 of 18 where: fi is the grade of the feed material (% species i in the feed ore) F is the mass (or mass flow rate) of ore in the feed ci is the grade of the concentrate (% species i in the concentrate material) C is the mass (or mass flow rate) of the concentrate The recovery (R) of species i for the process can be calculated as the mass (or mass flow rate) of the mineral or metal of interest (species i) in the concentrate divided by the mass (or mass flow rate) of the mineral or metal of interest in the feed, and is therefore given by Equation 3: 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 i 𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑖𝑖 𝐶𝐶 % Recovery, 𝑅𝑅 = × 100 = × 100 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 i 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑓𝑓𝑖𝑖 𝐹𝐹 Figure 4. Diagram of basic separation process. Grade is specified for species i, where species i may refer to the metal or mineral of interest. It is also possible to calculate the recovery using only the grades of the feed, concentrate and tailings by means of the so-called two product formula. To derive the two product formula, consider the total mass balance as well as a species i mass balance for the process shown in Figure 4. Total mass balance: 𝑇𝑇 = 𝐹𝐹 − 𝐶𝐶 Species i mass balance: 𝑡𝑡𝑖𝑖 𝑇𝑇 = 𝑓𝑓𝑖𝑖 𝐹𝐹 − 𝑐𝑐𝑖𝑖 𝐶𝐶 By substituting Equation 4 in Equation 5, the following expression is obtained for the ratio of concentrate mass to feed mass: 𝐶𝐶 𝑓𝑓𝑖𝑖 − 𝑡𝑡𝑖𝑖 = 𝐹𝐹 𝑐𝑐𝑖𝑖 − 𝑡𝑡𝑖𝑖 Substituting Equation 6 into Equation 3 yields the two product formula for recovery, which shows that the recovery can be determined by measuring only the grade (mass concentration of the mineral or metal of interest) of the respective streams: 𝑐𝑐𝑖𝑖 𝐶𝐶 𝑐𝑐𝑖𝑖 (𝑓𝑓𝑖𝑖 − 𝑡𝑡𝑖𝑖 ) % Recovery, 𝑅𝑅 = × 100 = × 100 𝑓𝑓𝑖𝑖 𝐹𝐹 𝑓𝑓𝑖𝑖 (𝑐𝑐𝑖𝑖 − 𝑡𝑡𝑖𝑖 ) Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 6 of 18 Example 1 An ore stockpile consists of 1000 t ore containing 1% chalcopyrite. Chalcopyrite is the only copper bearing mineral in the ore. Preliminary processing produces 250 t of concentrate with 2% chalcopyrite content. a) Calculate the copper grade of the concentrate. b) Calculate the copper recovery for the process. a) The copper grade of the concentrate can be determined from the specified mineral grade. The copper grade of the concentrate is equal to the mass of copper in the concentrate (CCu) divided by the mass of concentrate (C): 𝐶𝐶𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = × 100 𝐶𝐶 The mass fraction of copper in the chalcopyrite can be calculated using the ratio of molecular weights, so that the mass of copper in the concentrate is given by the following equation for the case where chalcopyrite is the only copper bearing mineral: 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 The mass of chalcopyrite in the concentrate can be calculated from the specified grade: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝑔𝑔𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 = 𝐶𝐶 100 Combining the equations above yields the expression for copper grade: 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶 𝐶𝐶 100 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = × 100 𝐶𝐶 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶 63.55 ∴ 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 2 = 0.69 % 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 183.53 b) The copper recovery for the process will be the same as the chalcopyrite recovery for the process: 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝐶𝐶 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝑅𝑅𝐶𝐶𝐶𝐶 = = = = 𝑅𝑅𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝐹𝐹𝐶𝐶𝐶𝐶 𝑀𝑀𝑀𝑀 𝐹𝐹𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝐶𝐶𝐶𝐶 𝐹𝐹𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝑀𝑀𝑀𝑀𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 The recovery of chalcopyrite can be calculated from the specified feed and concentrate masses and grades: 𝑐𝑐𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝐶𝐶 2(250) 𝑅𝑅𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 = × 100 = × 100 = 50 % 𝑓𝑓𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶2 𝐹𝐹 1(1000) Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 7 of 18 3.3 Metallurgical efficiency 3.3.1 Recovery-grade curve Since concentrate grade and recovery are metallurgical factors, the metallurgical efficiency of any concentration operation can be expressed by a curve showing the recovery attainable for any value of concentrate grade. The goal of obtaining maximum recovery and maximum grade simultaneously conflicts in principle. The grade-recovery curve for a specific feed material, for a specific separating unit or process, illustrates that there is a significant interplay between grade and reclamation. Figure 5 shows a typical grade-recovery curve, with the marked points on the solid curve corresponding to the three scenarios presented in Figure 6. Figure 5. Grade-recovery curve. (a) (b) (c) Figure 6. Schematic illustration of recovery-grade trade-off in concentration operations: (a) high grade, low recovery; (b) intermediate grade and recovery; (c) low grade, high recovery. Consider the schematic diagrams in Figure 6. Concentrator plants separate minerals based on the differences in their physical or surface properties. These differences in properties are the largest between fully liberated gangue and fully liberated mineral of interest, with the properties of middling particles being between that of the gangue and the mineral of interest depending on the relative proportion of the respective minerals. If a concentrator aims to produce a high grade concentrate, it will typically target recovery of the liberated mineral of interest as shown in scenario (a). As a result, any mineral of interest located in the middling particles report to the tailings and this results in a low recovery. On the other hand, Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 8 of 18 concentrator operations that aim to achieve high recoveries will also recover middling particles to the concentrate in order to avoid losses of the mineral of interest located in the middling particles to the tailings (scenario (c)). However, the effect of this is that middlings dilute the concentrate and the gangue associated with the middling particles that are recovered reduces the grade of the concentrate. In practice, the optimum grade and recovery targets for a concentrator operation will depend on economic factors and can be at intermediate targets (scenario (b)). In order to move the grade-recovery curve towards the dotted curve in Figure 5 (i.e. in order to achieve higher recovery at a fixed target grade or a higher grade at a fixed target recovery), improved liberation of the feed material is required so that a larger proportion of the mineral of interest can be recovered to the concentrate without diluting the concentrate with gangue. 3.3.2 Separation efficiency Several indicators are used to quantify separation efficiency of unit operations or processes. These include: Ratio of concentration: this is the ratio of the mass of the feed (F) to the mass of the concentrate (C). Enrichment ratio: this is the ratio of the grade of the concentrate (ci) to the grade of the feed (fi). Separation efficiency: this is a single index that combines recovery and grade to define metallurgical efficiency of the separation. It is defined as the recovery of the mineral of interest (Rmineral) minus the recovery of gangue (Rgangue) to the concentrate: Separation efficiency, 𝑆𝑆𝑆𝑆 = 𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑅𝑅𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 The expression for the separation efficiency in terms of flow rates and grades is derived on page 18 of the prescribed textbook. Take note that the symbol C is defined differently in the prescribed textbook (refer to the discussion of Equation 1.2 on p. 18) than in these notes. The derivation is repeated here based on the current definition of C as the mass flow rate of the concentrate. From the definition of recovery, Rmineral and Rgangue can be written in terms of flow rates and grades: 𝐶𝐶 × 𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑅𝑅𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = × 100 𝐹𝐹 × 𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶 × 𝑐𝑐𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑅𝑅𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = × 100 𝐹𝐹 × 𝑓𝑓𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 Because gangue refers to all the minerals that are of no economic value to the particular process, the gangue content in any stream is equal to the total content minus the mineral content. In the case of the concentrate stream, for example: 𝑐𝑐𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 100 − 𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 By substituting Equations 9 to 11 into Equation 8, the expression for separation efficiency in terms of flow rates and grades is given by Equation 12, which simplifies to Equation 13: 𝐶𝐶 × 𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶 × (100 − 𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ) 𝑆𝑆𝑆𝑆 = 100 − 𝐹𝐹 × 𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹 × (100 − 𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ) Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 9 of 18  100𝐶𝐶 (𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ) 𝑆𝑆𝑆𝑆 = 100 × 𝐹𝐹𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 (100 − 𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ) In the case of metallic ores where the ore grade is specified in terms of the metal content, the mineral content can be calculated from the metal content using the fractional metal content of the valuable mineral (zmetal) calculated based on the molecular weights (MW) of the respective elements constituting the mineral (assuming there is only a single metal-bearing mineral): 𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑧𝑧𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 where, for the general mineral AaBb consisting of metal A and non-metal B: 𝑎𝑎𝑀𝑀𝑀𝑀𝐴𝐴 𝑎𝑎𝑀𝑀𝑀𝑀𝐴𝐴 𝑧𝑧𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = = 𝑀𝑀𝑀𝑀𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑎𝑎𝑀𝑀𝑀𝑀𝐴𝐴 + 𝑏𝑏𝑀𝑀𝑀𝑀𝐵𝐵 Substituting Equation 14 in Equation 13 yields the separation in terms of the metal grade of the material (note that the term 100zmetal corresponds with symbol m, the percentage metal content of the mineral, as used in equation 1.3 in the prescribed textbook): 𝐶𝐶 (100𝑧𝑧𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 )(𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ) 𝑆𝑆𝑆𝑆 = 100 × 𝐹𝐹𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 (100𝑧𝑧𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ) Example 2 (Example 1.1 in the prescribed textbook) A tin concentrator treats a feed containing 1% tin, and three possible combinations of concentrate grade and recovery are: High grade 63% tin at 62% recovery Medium grade 42% tin at 72% recovery Low grade 21% tin at 78% recovery Determine which of these combinations of grade and recovery produce the highest separation efficiency. Assume tin is totally contained in SnO2 (MW = 150.71 g/mol). This example appears as Example 1.1 in the prescribed textbook. Only the calculations for the first scenario (63% tin at 62% recovery) are shown here to highlight key aspects. The same calculations can be repeated for the other scenarios. There are two approaches that could be followed to solve the problem: (1) substituting values into Equation 16, or (2) calculate the mineral and gangue recoveries from a mass balance to substitute into Equation 8. Approach 1 From Equation 16: 𝐶𝐶 (100𝑧𝑧𝑆𝑆𝑆𝑆 )(𝑐𝑐𝑆𝑆𝑆𝑆 − 𝑓𝑓𝑆𝑆𝑆𝑆 ) 𝑆𝑆𝑆𝑆 = 100 × × 𝐹𝐹 𝑓𝑓𝑆𝑆𝑆𝑆 (100𝑧𝑧𝑆𝑆𝑆𝑆 − 𝑓𝑓𝑆𝑆𝑆𝑆 ) Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 10 of 18 The metal grades of the feed (fSn = 1%) and concentrate (cSn = 63%) are specified. The metal fraction in the mineral can be calculated using Equation 15: 𝑀𝑀𝑀𝑀𝑆𝑆𝑆𝑆 118.71 𝑧𝑧𝑆𝑆𝑆𝑆 = = = 0.788 𝑀𝑀𝑀𝑀𝑆𝑆𝑆𝑆𝑆𝑆2 150.71 Make use of the specified recovery to calculate the ratio of the concentrate flow rate to feed flow rate: 𝐶𝐶 × 𝑐𝑐𝑆𝑆𝑆𝑆 𝑅𝑅𝑆𝑆𝑆𝑆 = × 100 𝐹𝐹 × 𝑓𝑓𝑆𝑆𝑆𝑆 𝐶𝐶 𝑅𝑅𝑆𝑆𝑆𝑆 𝑓𝑓𝑆𝑆𝑆𝑆 62 1 ∴ = = = 9.841 × 10−3 𝐹𝐹 100 𝑐𝑐𝑆𝑆𝑆𝑆 100 63 Now the separation efficiency can be calculated: 𝐶𝐶 (100𝑧𝑧𝑆𝑆𝑆𝑆 )(𝑐𝑐𝑆𝑆𝑆𝑆 − 𝑓𝑓𝑆𝑆𝑆𝑆 ) (78.8)(63 − 1) 𝑆𝑆𝑆𝑆 = 100 × × = 100 × (9.841 × 10−3 ) × = 61.8% 𝐹𝐹 𝑓𝑓𝑆𝑆𝑆𝑆 (100𝑧𝑧𝑆𝑆𝑆𝑆 − 𝑓𝑓𝑆𝑆𝑆𝑆 ) 1(78.8 − 1) Approach 2 In this case, the separation efficiency is calculated using Equation 8: 𝑆𝑆𝑆𝑆 = 𝑅𝑅𝑆𝑆𝑆𝑆𝑆𝑆2 − 𝑅𝑅𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 Because SnO2 is the only mineral containing tin, the SnO2 recovery will be equal to the recovery of tin: 𝑐𝑐 𝐶𝐶 × 𝑐𝑐𝑆𝑆𝑆𝑆𝑆𝑆2 𝐶𝐶 × 𝑆𝑆𝑛𝑛 𝑧𝑧𝑆𝑆𝑆𝑆 𝐶𝐶 × 𝑐𝑐𝑆𝑆𝑆𝑆 𝑅𝑅𝑆𝑆𝑆𝑆𝑆𝑆2 = × 100 = × 100 = × 100 = 𝑅𝑅𝑆𝑆𝑆𝑆 𝐹𝐹 × 𝑓𝑓𝑆𝑆𝑆𝑆𝑆𝑆2 𝑓𝑓 𝐹𝐹 × 𝑓𝑓𝑆𝑆𝑆𝑆 𝐹𝐹 × 𝑆𝑆𝑆𝑆 𝑧𝑧𝑆𝑆𝑆𝑆 The recovery of gangue to the concentrate can be calculated using standard mass balances and the definition of recovery. For the mass balances, assume a basis of 100 ton feed material per hour (because the recovery is a ratio of mass flow rates, the value selected as the basis for calculation will not affect the final answer): 𝐹𝐹 = 100 𝑡𝑡/ℎ From the tin feed grade, the mass flow rate of tin is known: 𝑓𝑓𝑆𝑆𝑆𝑆 1 𝐹𝐹𝑆𝑆𝑆𝑆 = 𝐹𝐹 × = 100 × = 1 𝑡𝑡/ℎ 100 100 The mass flow of SnO2 in the feed can be calculated based on the Sn flow rate because SnO2 is the only mineral containing tin: 𝐹𝐹𝑆𝑆𝑆𝑆 1 𝐹𝐹𝑆𝑆𝑆𝑆𝑆𝑆2 = = = 1.27 𝑡𝑡/ℎ 𝑧𝑧𝑆𝑆𝑆𝑆 0.788 The balance of the feed is gangue material: 𝐹𝐹𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 𝐹𝐹 − 𝐹𝐹𝑆𝑆𝑆𝑆𝑆𝑆2 = 100 − 1.27 = 98.73 𝑡𝑡/ℎ Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 11 of 18  In approach 1 it has been shown that: 𝐶𝐶 𝑅𝑅𝑆𝑆𝑆𝑆 𝑓𝑓𝑆𝑆𝑆𝑆 62 1 = = = 9.841 × 10−3 𝐹𝐹 100 𝑐𝑐𝑆𝑆𝑆𝑆 100 63 The total flow rate of the concentrate can therefore be calculated ∴ 𝐶𝐶 = (9.841 × 10−3 )𝐹𝐹 = 0.984 𝑡𝑡/ℎ The flow rate of gangue in the concentrate can be calculated using the same approach followed in the case of the feed: 𝑐𝑐𝑆𝑆𝑆𝑆 63 𝐶𝐶𝑆𝑆𝑆𝑆 = 𝐶𝐶 × = 0.984 × = 0.62 𝑡𝑡/ℎ 100 100 𝐶𝐶𝑆𝑆𝑆𝑆 0.62 𝐶𝐶𝑆𝑆𝑆𝑆𝑆𝑆2 = = = 0.787 𝑡𝑡/ℎ 𝑧𝑧𝑆𝑆𝑆𝑆 0.788 𝐶𝐶𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 𝐶𝐶 − 𝐶𝐶𝑆𝑆𝑆𝑆𝑆𝑆2 = 0.984 − 0.787 = 0.197 𝑡𝑡/ℎ The recovery of gangue can now be calculated because the flow rates of gangue in the feed and in the concentrate are known: 𝐶𝐶𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 0.197 𝑅𝑅𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = × 100 = × 100 = 0.2% 𝐹𝐹𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 98.73 To calculate the separation efficiency: 𝑆𝑆𝑆𝑆 = 𝑅𝑅𝑆𝑆𝑆𝑆𝑂𝑂2 − 𝑅𝑅𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 𝑅𝑅𝑆𝑆𝑆𝑆 − 𝑅𝑅𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 62 − 0.2 = 61.8% 3.4 Economic efficiency Since the purpose of mineral processing is to increase the economic value of the ore, the importance of the recovery-grade relationship is in determining the most economic combination of recovery and grade which will produce the greatest financial return per tonne of ore treated in the plant. Important cost considerations in determining the most profitable operation are discussed on pp. 19-20 of the prescribed textbook. Read this section to understand the concept of net smelter return and how to calculate the mine or concentrator profit. The smelter operation mentioned in this discussion in the textbook refers to the extractive metallurgical process that treats the concentrate, produced by physical concentration, to produce the final product (as illustrated in Figure 1). The case studies / examples provided on pp. 20-26 of the textbook will provide further insight into the assessment of economic performance; pay particular attention to the calculation of the contained metal value using the concentrator recovery and concentrate grade. 4 Flowsheets and mass balancing 4.1 Pulp density From the grinding (milling) stage onwards, most mineral processing operations are carried out in slurry streams, the water and solids mixture being transported through the circuit via pumps and pipelines. For the purposes of metallurgical accounting, the weight of dry solids contained within the slurry is important. Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 12 of 18 For large-scale operations, it is usually necessary to measure the slurry density by online instrumentation. For mass balancing purposes, the percentage solids by weight in a slurry stream is usually required. Knowing the densities of the slurry and dry solids, the % solids by weight, x, can be calculated if volume additivity is assumed: 𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 + 𝑉𝑉𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 The volume of each phase (Vi) can be written in terms of phase mass (mi) and density (ρi): 𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜 𝑚𝑚𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜 𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = + = + 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 Considering the definition of x as being the percentage solids by weight in the slurry,: 𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜 𝑥𝑥 = × 100 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 it is possible to write the slurry volume as a function of the slurry mass and the respective phases’ densities (by substituting Equation 19 in Equation 18): 𝑥𝑥 100 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 1 − 𝑥𝑥 100 𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = + 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 1 𝑥𝑥 100 1 − 𝑥𝑥 100 ∴ = = + 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 Rearranging Equation 21 yields the expression for x as a function of slurry and phase densities: 100𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 − 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑥𝑥 = 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 − 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 Because the liquid is usually water, it is generally acceptable to assume a density of 1000 kg/m3 for the liquid phase. The percentage solids by volume is related to the percentage solids by weight (x) as follows: 𝑚𝑚 100𝑉𝑉𝑜𝑜𝑜𝑜𝑜𝑜 100 𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 100𝑚𝑚𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 % 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑏𝑏𝑏𝑏 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = = 𝑚𝑚 = 𝑉𝑉𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝑥𝑥 100 100 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∴ % 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑏𝑏𝑏𝑏 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = = 𝑥𝑥 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 Examples 3.1 to 3.4 in the prescribed textbook (pp. 52-54) provide good examples of where the equations derived above can be used. Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 13 of 18 4.2 Mass balances The same mass balancing principles and techniques that have been taught in Chemical Engineering 254 apply in mineral processing. The physical processing units that will be discussed from Week 4 to Week 8 in this module, where there is no chemical transformation of species, can be treated as non-reactive systems. It is important to account for or distinguish between balances per phase (e.g. liquid vs ore vs slurry) or per property class (e.g. different classes of liberation, or different particle size classes). Three examples are presented here, but these skills will be developed further as we consider different unit operations and processes throughout this course. Example 3 Run of mine ore contains 23% CuFeS2 in quartz. Following crushing and milling, 68% of the CuFeS2 and 84% of the quartz are completely (100%) liberated. If particles with less than 100% liberation are classified as middlings, calculate the middlings fraction and grade in the mill product. This is an example of a mass balance performed on the basis of liberation classes; the different classes are considered, namely liberated mineral of interest (CuFeS2), liberated gangue, and middlings. Because the problem statement requires the middlings fraction and grade (both fractions) to be calculated, any basis for calculation can be used or the problem can be solved in terms of the total feed mass to the mill (F). Although the mill product leaves as a single stream, visualising the liberation classes as separate product streams, as shown below, can assist when solving the problem. Start by calculating the mass of the liberated CuFeS2 and liberated gangue using the liberation percentages and feed composition specified in the problem statement: 68 68 23 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑆𝑆2 = 𝐹𝐹 = 𝐹𝐹 = 0.1564𝐹𝐹 100 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑆𝑆2 100 100 84 84 (100 − 23) 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 = 𝐹𝐹 = 𝐹𝐹 = 0.6468𝐹𝐹 100 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 100 100 Perform a total mass balance to calculate the mass of middlings: 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐹𝐹 − 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑆𝑆2 − 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑙𝑙𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞𝑞 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐹𝐹 − 0.1564𝐹𝐹 − 0.6468𝐹𝐹 = 0.1968𝐹𝐹 This shows that 19.7% of the feed that enters the mill leave the mill as middlings. Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 14 of 18 To calculate the middlings grade, the mass of CuFeS2 in the middlings fraction needs to be calculated. Because the mass of CuFeS2 in the feed and in the liberated CuFeS2 streams are now known (and the liberated quartz does not contain CuFeS2), a CuFeS2 species balance around the mill will yield the mass of CuFeS2 in the middlings: 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑆𝑆2 𝑖𝑖𝑖𝑖 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝐹𝐹𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑆𝑆2 − 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑆𝑆2 = 0.23𝐹𝐹 − 0.1564𝐹𝐹 = 0.0736𝐹𝐹 Calculate the grade of the middlings: 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑆𝑆2 𝑖𝑖𝑖𝑖 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 0.0736𝐹𝐹 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = × 100 = × 100 = 37.4% 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑚𝑚𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 0.1968𝐹𝐹 Example 4 Consider the closed milling circuit in the figure below. For the recycle stream, R, the density of the solids (dS) is 3.7 kg/L and that of the slurry (dSL) is 1.9 kg/L. a) Calculate the recycle mass flow rate, RSL, given that F2S = 25 t/h. b) Calculate the circulating load ratio (ratio of solids recycled to solids fed, RS/F1S) c) Calculate the rate of water addition, W. In this example, no information about the mineral composition, size classes or liberation of the ore is provided so the ore is treated as a single phase in the mass balance. Here it is important to distinguish between the phases (solids, water or slurry) when performing the mass balances. Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 15 of 18  a) Because the solids and slurry densities for the recycle stream are specified, it is possible to calculate the percentage solids in the recycle stream using Equation 22 (assuming water has a density of 1 kg/L): 100𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 − 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 100(3.7)(1 − 1.9) 𝑥𝑥𝑅𝑅 = = = 64.9% 𝜌𝜌𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜌𝜌𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 − 𝜌𝜌𝑜𝑜𝑜𝑜𝑜𝑜 1.9(1 − 3.7) Because no material is added or removed in the conditioning tank, the mass flow rate of solids in O must equal the mass flow rate of solids in F2 (solids balance around the conditioning tank): 𝑂𝑂𝑠𝑠 = 𝐹𝐹2𝑆𝑆 = 25 𝑡𝑡/ℎ To calculate the recycle mass flow rate, system boundaries for the mass balances must be defined such that the recycle stream crosses the system boundaries. The percentage solids in M and O are specified, while the same has been calculated for the recycle stream; the solids flow rate in O is also known. Considering mass balances around the cyclones, there are two unknowns (solids flow rates in M and R, respectively) which can be solved through simultaneous solution of two independent mass balances (total or slurry flow rate as well as solids flow rate). The respective mass balances are: Slurry (total flow) mass balance: 𝑀𝑀𝑆𝑆𝑆𝑆 = 𝑅𝑅𝑆𝑆𝑆𝑆 + 𝑂𝑂𝑆𝑆𝑆𝑆 Solids mass balance: 𝑀𝑀𝑆𝑆 = 𝑅𝑅𝑆𝑆 + 𝑂𝑂𝑆𝑆 where: 𝑥𝑥𝑀𝑀 𝑀𝑀𝑆𝑆 = 𝑀𝑀 = 0.33 𝑀𝑀𝑆𝑆𝑆𝑆 , 𝑅𝑅𝑆𝑆 = 0.649 𝑅𝑅𝑆𝑆𝑆𝑆 , 𝑂𝑂𝑆𝑆 = 0.15 𝑂𝑂𝑆𝑆𝑆𝑆 = 25 𝑡𝑡/ℎ 100 𝑆𝑆𝑆𝑆 Solving the two mass balances simultaneously yields a recycle flow rate, RSL of 94 t/h. b) Because the recycle flow rate (RSL) and percentage solids in the recycle (xR) have been calculated in part (a), this part of the question requires only the solids flow rate in the feed stream (F1S) to be calculated. This can be done if the entire process is considered as the system for mass balance purposes: the recycle stream falls inside the system boundaries and the water stream contains no solids. So the solids mass balance for the overall process reduces to: 𝐹𝐹1𝑆𝑆 = 𝐹𝐹2𝑆𝑆 = 25 𝑡𝑡/ℎ The recirculating load ratio is therefore: 𝑥𝑥 𝑅𝑅𝑆𝑆 𝑅𝑅 100 𝑅𝑅𝑆𝑆𝑆𝑆 0.649 × 94 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = = = = 2.44 𝐹𝐹1𝑆𝑆 𝐹𝐹1𝑆𝑆 25 c) To calculate the water addition rate, the system boundaries can be defined in a number of different ways. If the overall process is defined as the system, for example, the solids flow rates and solids contents of streams F1 and F2 are already known; a water mass balance will include the water addition as the only unknown: Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 16 of 18  𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏: 𝑊𝑊 + 𝐹𝐹1𝑊𝑊 = 𝐹𝐹2𝑊𝑊 Where: 𝑥𝑥𝐹𝐹1 𝑥𝑥𝐹𝐹1 𝐹𝐹1𝑆𝑆 𝐹𝐹1𝑊𝑊 = 1 − 𝐹𝐹1 = 1 − 𝑥𝑥 100 100 𝐹𝐹1 100 The flow rate of water in stream F2 can be calculated from the solids content and solids flow rate in the same manner. From the water mass balance, the water addition rate is then: 𝑥𝑥𝐹𝐹2 𝐹𝐹2𝑆𝑆 𝑥𝑥𝐹𝐹1 𝐹𝐹1𝑆𝑆 𝑊𝑊 = 𝐹𝐹2𝑊𝑊 − 𝐹𝐹1𝑊𝑊 = 1 − 𝑥𝑥 − 1 − 𝑥𝑥 100 𝐹𝐹2 100 𝐹𝐹1 100 100 15 25 95 25 𝑊𝑊 = 1 − − 1 − = 140.4 𝑡𝑡/ℎ 100 15 100 95 100 100 Example 5 The feed to a processing plant assays 0.8% copper. The concentrate produced assays 25% Cu, and the tailings 0.15% Cu. Calculate the recovery of copper to the concentrate, the ratio of concentration, and the enrichment ratio. To solve problems like these, it is important to know and understand the definition of the different efficiency indicators. Because all involve the ratio of masses or flow rates, any basis can be selected for calculation and mass balances performed accordingly. However, with the derivation of the so called two product formula it has been shown that the recovery can be calculated from the streams’ grades only, as specified in this problem. Starting with the recovery calculation, Equation 7 can be used: 𝑐𝑐𝑖𝑖 (𝑓𝑓𝑖𝑖 − 𝑡𝑡𝑖𝑖 ) 25(0.8 − 0.15) % Recovery, 𝑅𝑅 = × 100 = × 100 = 81.7% 𝑓𝑓𝑖𝑖 (𝑐𝑐𝑖𝑖 − 𝑡𝑡𝑖𝑖 ) 0.8(25 − 0.15) The ratio of concentration is defined as the ratio of the feed ore mass to the concentrate mass. From Equation 6: 𝐹𝐹 𝑐𝑐𝑖𝑖 − 𝑡𝑡𝑖𝑖 25 − 0.15 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = = = = 38.2 𝐶𝐶 𝑓𝑓𝑖𝑖 − 𝑡𝑡𝑖𝑖 0.8 − 0.15 The enrichment ratio is calculated directly from the specified grades: 𝑐𝑐𝑖𝑖 25 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = = = 31.25 𝑓𝑓𝑖𝑖 0.8 Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 17 of 18 5 Supplementary reading Topic Prescribed textbook reference* Comments Read discussion for additional background information. Study figures 1.2a to 1.2i as Introduction “Mineral processing methods”, pp. 7-11 examples of how ore texture could impact liberation. Study carefully to ensure understanding of the Comminution concept of liberation. Pay special attention to and “Liberation”, pp. 14-15 the discussion of figures 1.5 and 1.6. (The liberation discussion of liberation models will be explored further in Week 5.) Most of the concepts on metallurgical efficiency “Concentration”, pp. 16-18 are described in the notes. Read the discussion in the textbook for additional insight. Study carefully to ensure understanding of the Process “Concentration”, pp. 19-20 difference between and calculation of NSR and efficiency mine profit. “Economics of tin processing” & Work through these case studies thoroughly to “Economics of copper processing”, pp. understand how the costs, revenues and profits 20-26 are calculated. Mass “Slurry streams”, pp. 51-54 Work through examples 3.1 to 3.4. (Information balances about the standard density test procedure is not important at this stage.) *https://www-sciencedirect-com.ez.sun.ac.za/book/9780750644501/wills-mineral-processing-technology 6 Acknowledgements Parts of these notes have been taken from course material prepared by Professors AJ Burger and JJ Eksteen. Notes prepared by C Dorfling Basic metallurgical concepts and mass balancing Page 18 of 18