Assign_2_2018_flowsheets_min_proc_SOLUTIONS.doc

DEPARTMENT OF MATERIALS ENGINEERING MTRL 358 - Hydrometallurgy I NOTE: Partial or complete copying of another person's work is academic misconduct. See policy on academic misconduct in the syllabus or on the course Vista website. Assignment 2, 2018 - Flowsheets and Mineral Processing SOLUTIONS 1. A simplified flowsheet for the Galvanox process is shown below. This process is being developed at UBC. Chaclcopyrite is the most abundant mineral of copper in nature. However, it resists leaching by conventional hydrometallurgical methods; the mineral rapidly passivates (after a short while it leaches very slowly) when leached in sulfate solution. In this process a copper concentrate is produced which contains pyrite and copper minerals, particularly chalcopyrite. If pyrite (FeS 2 ) is present in the concentrate then chalcopyrite is readily leached. Pyrite, being an iron mineral, often occurs in sulfide mineral deposits. During leaching chalcopyrite is oxidized to form a solution of Cu +2 , Fe +2 and solid elemental sulfur. Pyrite itself undergoes little reaction. (Mainly it seems to act as a catalyst for oxygen reduction.) After leaching the slurry is subjected to solid-liquid separation. The solids may be recycled to leaching to reutilize the pyrite. Much of the solution proceeds to solvent extraction, which is used to obtain a much purer copper solution. The details are not important here. Some of the solution also proceeds to an oxyhydrolysis step (this is done in an autoclave) in which ferrous ion is oxidized to form hematite (Fe 2 O 3 ) and sulfuric acid. This acts as an outlet for iron and prevents its build-up in solution. Hematite is a very suitable iron product for disposal. The solvent extraction process also generates acid, which together with that formed by hematite formation can be reused in the leaching step. The concentrated copper sulfate solution from solvent extraction proceeds to an electrowinning step. Here very pure copper metal is produced by electrolysis. Answer the following questions related to the flowsheet. 1

(a) Write the balanced chemical reaction in ionic and neutral forms for the leaching of chalocopyrite. CuFeS 2 + O 2 = Cu +2 + Fe +2 + S Oxygen is the oxidant, as seen from the flowsheet. The rest is evident from the text. (a) CuFeS 2 = Cu +2 + Fe +2 + S O 2 = ? (b) CuFeS 2 = Cu +2 + Fe +2 + 2S O 2 = ? (c) CuFeS 2 = Cu +2 + Fe +2 + 2S O 2 = + 2H 2 O (d) CuFeS 2 = Cu +2 + Fe +2 + 2S O 2 + 4H + = + 2H 2 O (e) CuFeS 2 = Cu +2 + Fe +2 + 2S + 4e - O 2 + 4H + + 4e - = + 2H 2 O (f) CuFeS 2 = Cu +2 + Fe +2 + 2S + 4e - O 2 + 4H + + 4e - = + 2H 2 O CuFeS 2 O 2 + 4H + = Cu +2 + Fe +2 + 2S + 2H 2 O ionic form Balance + charges with sulfate: CuFeS 2 s + O 2 g + 2H 2 SO 4 aq = CuSO 4 aq + FeSO 4 aq + 2S s + 2H 2 O l neutral form (g) CHECK: Left side Right side 1Cu 1Cu 1Fe 1Fe 4S 4S 10O 10O 4H 4H 0 charge 0 charge OK (b) Referring to the generalized hydrometallurgical flowsheet on p. 30 of the Introduction to Extractive Metallurgy course notes, identify the parts of the flowsheet that correspond to mineral separation, leaching, solution purification and metal production. Use the flowsheet on the page below, circle and label the appropriate parts of the flowsheet. Hand this in with your completed assignment. See flowsheet below (c) Referring to the generalized metallurgical flowsheet that follows, indicate which route corresponds most closely to this process. You may indicate just the number in your assignment paper. See flowsheet below 2

Generalized extractive metallurgy flowsheet 1 (c). Route 3 makes the most sense. A concentrate is made and then leached directly. The solution is purified and then metal is precipitated by electrowinning. 4

1 (d). Leaching requires 4 moles of H + /mole of CuFeS 2 . What fraction of the required acid is produced by the oxyhydrolysis step? (Consider the balanced reaction that forms Fe 2 O 3 .) Here a chemical reaction is needed: Fe +2 + O 2 = Fe 2 O 3 (a) Fe +2 = Fe 2 O 3 O 2 = ? (b) 2Fe +2 = Fe 2 O 3 (c) 2Fe +2 + 3H 2 O = Fe 2 O 3 (d) 2Fe +2 + 3H 2 O = Fe 2 O 3 + 6H + (e) 2Fe +2 + 3H 2 O = Fe 2 O 3 + 6H + + 2e - O 2 + 4H + + 4e - = 2H 2 O (f) 2Fe +2 + 3H 2 O = Fe 2 O 3 + 6H + + 2e - 0.5O 2 + 2H + + 2e - = H 2 O 2Fe +2 + 2H 2 O + 0.5O 2 = Fe 2 O 3 + 4H + (Required to estimate stoichiommetric acid formed.) or: Fe +2 + H 2 O + 0.25O 2 = 0.5Fe 2 O 3 + 2H + FeSO 4 + H 2 O + 0.25O 2 = 0.5Fe 2 O 3 + H 2 SO 4 (g) Check Left Right Left Right 1Fe 1Fe 0 Charge 0 Charge 1S 1S 5.5O 5.5O 2H 2H OK This shows that for every one Fe leached then precipitated, 1 mole of H 2 SO 4 is formed = 2 mole H + /mol CuFeS 2 . Hence 50% of the required acid comes from this step . 2. In the zinc Roast-Leach-Electrowin (RLE) process a zinc concentrate is roasted in air to form ZnO. Iron minerals combine with zinc to form zinc ferrite. See the reactions below. ZnS s + 3/2O 2 g = ZnO s + SO 2 g ZnS s + 2FeS 2 s + 7O 2 g = ZnFe 2 O 4 s + 5SO 2 g Zinc ferrite is difficult to leach, whereas ZnO is easily dissolved in diluted acid. Thus high-iron ZnS concentrates are more challenging to treat. A hypothetical sphalerite concentrate contains 48% Zn as sphalerite and 6% Fe as pyrite plus other minerals. What proportion of the total zinc in the calcine will be present as zinc ferrite? 5

Tyler mesh number Opening  m Mass retained g 1" 25400 10.9 0.75" 19050 61.3 0.5" 12700 280.9 0.375" 9525 114.7 0.25" 6350 128.6 6 3360 80.1 20 841 140.5 48 297 101.9 200 74 64.3 325 44 4.0 -325 2.0 (a) Calculate the cumulative mass passing each screen size. For instance, for the 1 inch sieve all but 10.9 g passes through. For the 0.75 inch sieve size all but 10.9 + 61.3 g passes, etc. Data are shown below: Mesh no. Opening  m log(Opening) Mass retained g Fraction retained % Cumulative mass passing % 1 25400 4.404834 10.9 1.1 98.9 3/4 19050 4.279895 61.3 6.2 92.7 1/2 12700 4.103804 280.9 28.4 64.3 3/8 9525 3.978865 114.7 11.6 52.7 1/4 6350 3.802774 128.6 13.0 39.7 6 3360 3.526339 80.1 8.1 31.6 20 841 2.924796 140.5 14.2 17.4 48 297 2.472756 101.9 10.3 7.1 200 74 1.869232 64.3 6.5 0.6 325 44 1.643453 4.0 0.4 0.2 -325 2.0 0.2 e.g. The total mass is the sum on each screen plus what passed through the finest screen (325 mesh) = 989.2 g. The fraction retained for the no. screen, for instance is 10.9/989.2 = 1.1%. (b) Plot the cumulative percent passing each screen versus the log 10 of the opening size and determine the P 80 of the crushed ore. The cumulative mass passing any given screen is everything below that size. For example, the cumulative mass passing in % through the no. 3/8 screen is 13.0 + 8.1 + 14.2 + 10.3 + 6.5 + 0.4 + 0.2 = 52.7%. This is plotted on the figure below: 7 Plot of cumulative fraction passing versus log of sieve opening size after screening of a copper ore.

The P 80 size is indicated on the diagram: log(opening) = 4.2; opening = 10 4.2 = 15,850  m = 1.59 cm . This means that 80% of the sample is less than 1.59 cm in size. (c) Assume the ore is mainly a silicate rock with a Bond work index about the same as that of silica (see Introduction to Mineral Processing notes). Calculate the energy requirement for crushing in kWh/ton. For the purposes of the Bond work index equation, assume that the mined ore is very large. The ore will be reduced in size from large (nominally infinite size) to 15,900 microns. The Bond equation uses size in microns! W = 10W i (1/d O 1/2 - 1/d I 1/2 ) = 10 x 14 kWh/ton x (1/15900 1/2 - 1/  ) = 1.11 kWh/ton (d) The as mined ore will be reduced in size to a P 80 of 2.5 cm. Suppose the copper can be 80% leached. Estimate how much power in MW (megawatts) is needed for size reduction if the production rate is to be 50,000 tonnes of copper per year? (1 tonne = 2200 pounds). Multiplying the feed rate in tons/h times the energy in kWh/ton gives the power needed in kW. The ore mass flow rate is determined from the copper production rate: 50,000 t Cu prod . x 1 t ore x 1 t Cu in ore x 2200 lb x 1 ton y 0.0070 t Cu in ore 0.80 t Cu prod. t ore 2000 lb = 9.8214 x 10 6 tons/year = 1121.17 tons/hour The energy requirement is: 8

Roughers-copper roughers and copper roughers-copper cleaners . In either case most of the feed (ore in the first case; bulk concentrate in the second) passes through the first stage. Two concentrates are obtained from the second stage, one a flotation concentrate and the other the tailings from that operation . For the purposes of this discussion, the copper concentrate product the copper roughers is a flotation concentrate product. It then becomes the feed to the cleaners, from which two concentrates are obtained, one a flotation product and the other the tailings. (e) Which of the following statements are true as concerns how the objectives of good grade and good recovery are achieved in the flotation of the copper concentrate. ( i ) Roughers and scavengers employ aggressive flotation conditions as indicated by the extra added xanthate at the scavengers to ensure good recovery of CuFeS 2 , while a subsequent cleaners operation employing mild flotation conditions ensures high grade. (Not true - roughers employ moderate conditions for moderate recovery and grade.) ( ii ) Sodium cyanide is used to promote aggressive flotation conditions ensuring good recovery. (Not true - cyanide is a depressant, not an activator.) ( iii ) Moderate flotation conditions are used in the roughers to provide moderate grade and recovery. Scavengers uses aggressive flotation conditions to ensure good recovery. True ( iv ) Ball mills are used in the flotation circuit to effect better liberation of minerals. The ball mill products are recycled through flotation to ensure better recovery and grade. True ( v ) The copper cleaners flotation uses mild flotation conditions with NaCN added as a depressant so that the flotation product from this step has a high grade of CuFeS 2 . True ( vi ) Sodium cyanide is used as a depressant in copper cleaners flotation to ensure that mainly the high-grade CuFeS 2 mineral particles are floated. True ( vii ) Ball mills are used in the flotation circuit to effect better exposure of mineral grains, allowing for better attachment of collector and better recovery. (Not true - liberation is the goal, not exposure.) (f) A pH of 9 is used in rougher flotation to obtain a Ni-Cu concentrate. (i) How will this affect pyrrhotite in rougher flotation? (ii) Would the critical pH curve for pentlandite lie to the right or the left of pH 9 and the collector concentration used in this process? (iii) In the course notes no critical pH curve for pentlandite is provided. Do you think the critical pH for pentlandite would lie to the left or the right of the curve for chalcopyrite? From the critical pH curves, a pH of 9 is at the right of the pyrrhotite curve for most xanthate concentrations. Hence pyrrhotite will not float . Right of the curve = does not float. (This can be understood by considering the adsorption equilibria. High pH = increased [OH - ]: 10

mineral·xanthate adsorbed + OH - aq = mineral·OH - adsorbed + xanthate aq (ii) In the course notes no critical pH curve is provided for pentlandite. Would the critical pH curve for pentlandite lie to the right or the left of pH 9 and the collector concentration used in this process? The pentlandite curve should lie to the right of pH 9 ; a pH below (to the left) of the curve = floats. (iii) Do you think the critical pH for pentlandite would lie to the left or the right of the curve for chalcopyrite? Based on the flowsheet, penatlandite reports to tailings at the pH where chalcopyrite floats. Hence the pentlandite critical pH should be to the left of the chalcopyrite curve , qualitatively as per the diagram below: (g) If the collector concentration was increased, would you expect this shift the critical pH curve for chalcopyrite in the presence of cyanide to lower pH, to higher pH or not at all? Critical pH curves curve up and to higher pH as collector concentration is increased. This shifts the curve to higher pH . This would occur with or without cyanide present. This makes sense in terms of LeChatelier's principle and the equilibria: mineral·xanthate adsorbed + OH - aq = mineral·OH - adsorbed + xanthate aq or mineral·xanthate adsorbed + CN - aq = mineral·CN - adsorbed + xanthate aq (h) What are the purposes of the collector and frother chemicals? Briefly explain. Collector (ethyl xanthate) attaches to a mineral surface and renders the surface hydrophobic . The particle then may attach to an air bubble . Frother is used to stabilize the mineralized froth. It keeps the bubbles from bursting prematurely and keeps the mineral particle-bubble aggregates at the top of the cell long enough to be skimmed off. 11

Calculate the following quantities: Mass flow of tailings; Bulk concentrate production rate; the copper and nickel grades in the tailings; the nickel grade of the bulk concentrate, the recoveries of nickel and copper from the ore. You will need to write 6 mass balance equations and solve for the unknowns. A simplified diagram of the process showing the flows and only the main operations might help. Consider the total mass flow as well. A simplified diagram of the flows appears below: There is one input stream (ore). Recycle streams are in and out, so they don't affect the mass balance on a stage. For the roughers-scavengers alone there are two outlet streams: bulk concentrate and scavenger concentrate. For the overall process there are three outlets: scavengers concentrate, Ni con and Cu con; the bulk concentrate is an in and an out within the overall process. There is data for the ore, bulk concentrate, nickel concentrate and copper concentrate. Mass balance equations can be written for the whole ore, copper and nickel through the process. Note that we can write mass balance equations for these over the roughers and scavengers alone (ore, tailings and bulk concentrate) and for the whole of the process (ore, tailings, Ni con and Cu con.) This will give 2 x 3 = 6 equations. Let O stand for ore, t for tailings, Cu con and Ni con for the respective concentrates and B for the bulk concentrate. Let C Cu Co con stand for the grade of copper in the copper concentrate, C Cu Ni con stand for grade of copper in the nickel concentrate, etc. Finally, let M Cu con stand for the mass flow rate of the copper concentrate, and so on. The mass balance equations are: 1 M O = M t + M Cu con + M Ni con Ore Ni con 3 M O x C Cu O = M t x C Cu t + M Cu con x C Cu Cu con + M Ni con x C Cu Ni con Cu overall 13 Bulk Con

4 M B x C Cu B = M Cu con x C Cu Cu con + M Ni con x C Cu Ni con Cu bulk con 5 M O x C Ni O = M t x C Ni t + M Cu con x C Ni Cu con + M Ni con x C Ni Ni con Ni overall 6 M B x C Ni B = M Cu con x C Ni Cu con + M Ni con x C Ni Ni con Ni bulk con The unknowns are highlighted in yellow. From equation 2 it is apparent that: M B = 1586.2 + 3413.8 = 5000 tpd From equation 2: M O = M t + M B = 50,000 = M t + 5000 M t = 45,000 tpd For equation three there are two unknowns still: C Cu t and C Cu Ni con . For equation 4 there is one unknown still: C Cu Ni con . we can solve for this, then for C Cu t . 5000 x 10% = 1586.2 x 29% + 3413.8 x C Cu Ni con C Cu Ni con = 1.17177% = 1.2% Then from equation 3: 50,000 x 1.2% = 45,000 x C Cu t + 1586.2 x 29% + 3413.8 x 1.17177% C Cu t = 0.2222% = 0.22% From equation 5: 50,000 x 1.2% = 45,000 x C Ni t + 1586.2 x 0.90% + 3413.8 x 13.2% C Ni t = 0.3002% = 0.30% From equation 6: 5000 x C Ni B = 1586.2 x 0.90% + 3413.8 x 13.2% C Ni B = 9.298% = 9.3% tpd Ni grade % Cu grade % Bulk concentrate 5000 9.3 10.0 Copper concentrate 1586.2 0.90 29.0 Nickel concentrate 3413.8 13.2 1.2 14