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MINISTRY OF SCIENCE AND TECHNOLOGY DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION GOVERNMENT TECHNOLOGICAL COLLEGES/INSTITUTES AGTI Year II (Metallurgy) Date : 28-9-2006 (Thursday) Time : 8:30 – 11:30 AM MET-02012 MINERAL PROCESSING ANSWER ANY FIVE. I. (a) What are the steps involved in dressing an ore? Explain them. Ans:The steps involved in dressing an ore are (1) Communication (2) Sizing (3) Concentration (4) Dewatering (1) Communication Communication means reduction of large ore pieces to a smaller size. (2) Sizing Sizing is the separation of the material into products of different sizes. (3) Concentration Concentration is done after liberation and is based on differences in physical and properties of the various minerals. (4) Dewatering Dewatering is the removal of water from concentrate to produce a dry product. I. (b) Define crushing and draw a rough sketch of a jaw crusher labeling important parts. Also discuss the important characteristics. Solution Crushing is the first size reduction process conducted on run-of- mine ore to produce suitable size material for feeding to ball mill for grinding or to heavy media separation. Fly wheel Feed eccentric pivot G pitman Fixed jaw rare (back) toggle rare toggle seat Movable S jaw Product Front toggle G = gape S = set Jaw Crusher Characteristics - The disadvantages of jaw crushers are that they work on half cycle only, because of their reciprocating motion of the movable jaw (2). - Owing to alternate loading and release, it must be rugged and require strong foundations to dampen vibrations. - Reduction ratio varies from 4:1 to 7:1. II. (a) Define critical speed of a ball mill. If the actual speed of a 6 ft diameter ball mill is 25 rpm, calculate its critical speed and percent critical speed of the mill. Solution Critical speed of ball mill Critical speed of ball mill is defined as theoretical speed m revolutions per minute at which the mill charge in contact with the shell will just centrifuge assuming no slippage between it and the shell liner. actual speed = 25 rpm d = 6 ft 54 .18 Nc = R 54 .18 = d 2 54 .18 = 6ft 2 = 31.281 % critical speed = ? actual speed % critical speed = x 100 theoretical speed 25 rpm = 100 31 .281 rpm = 79.92% II. (b) Define screening efficiency and drive its formula. On what factors does it depend? Solution Screening efficiency Screening efficiency can be defined as the percentage ratio of the weight of the undersize actually passed through the screen to that of the undersize that is contained in the feed. The percentage screening efficiency E can be expressed as follows: Utons E = 100 x % u F tons 100 Utons E = 10000 % uFton But E can be computed as follows without weighting U and F. F=U+O u b F =U+O 100 100 u b F = U + (F – U) 100 100 F (100 b) (u - b) = U 100 100 U ( u b) = F (100 b) U Can be substituted , F ( u b) E = 10,000 % u (100 b) Factors it depend, (a) Feed rate of ore (b) Mesh size of screen (c) Moisture in ore feed (d) Type of screen surface (e) Percentage of screen opening area (f) Shape of particles (g) Type of screen motion (h) Percentage of “near-mesh” particles (i) Blinding of screen openings (j) Stickiness of ore particles (k) The degree of separation of feed (l) Slope of the screen III. (a) Define gravity concentration. What methods are available to separate minerals with large differences in specific gravity? Solution Gravity concentration Gravity concentration is the process of separating valuable minerals from gangue after comminution. The following methods are available to separate minerals with large differences in specific gravity. (a) Heavy medium separation (b) Jigging (c) Sluicing (d) Tabling (e) Humphrey Spiral concentration (f) Pinched Sluice concentration (g) Tilting Table concentration III. (b) 2 cc of a 10% collector solution is added to a pulp weighing 2 kg in a conditioner. If the pulp contains 500 g of ore with sp-gr. 2.7, calculate (i) collector consumption in lb/ton of ore (ii) pulp density in conditioner (iii) collector concentration in mg/l (iv) sp gr of pulp Solution 100 ml of water 10g (ωt of collector) 2 10 2 ml of water = 0.2 g of collector 100 (i) 500 g of ore 0.2 g of collector 2000 0.2 2000 lb of ore 500 = 0.8 lb/ton ( collector consumption) ωt of ore (ii) Pulp density = x 100 ωt of pulp 500 g = x 100 = 25 % 2000 g (iii) collector concentration in mg/l = ? 1000 mg ωt of collector = 0.2 g x = 200 mg 1g ωt of water = ωt of pulp - ωt of ore = 2000 – 500 = 1500 g 1500 g Volume of water = = 1500 cc 1g cc = 1500 x 10-3 l 0.2 10 3 mg collector concentration in mg/l = 1500 10 3 l = 133.33 mg/l IV. (a) A conditioner fro a 800 tdp mill has as installed volume of 300 cu.ft. when 6 minutes conditioning time a allowed. Calculate the percentage change in pulp density if the conditioning time is to be increased 50% for the same daily feed. Compute also the daily water consumption under the new set of conditions if the sp.gr. of ore is 2.8. Solution % change in pulp density = ? ωt of ore Pulp density = x 100 ωt of pulp ωt of pulp = ωt of ore + ωt of water 1day 1hr ωt of ore (for 6 min) = 800 ton day x x 6 min 24hours 60 min = 6666 lbs Volume of pulp = 300 ft3 ωt of ore Volume of ore = sp.gr of ore 6666 lbs = (2.8 62 .4 lb ) ft 3 = 38.152 ft3 volume of water = volume of pulp - volume of ore = 300 ft3 – 38.16 ft3 = 261.84 ft3 ωt of water = volume of water x density of water = 261.84 ft3 x 62.4 lb/ft3 = 16338.816 lbs ωt of ore Pulp density = x 100 ωt of pulp 6666 lbs = 100 (6666 lbs 16338 .816 lbs ) = 28.98 % Conditioning time is increased 50%. 50 New condition time = 6 x = 3 min 100 Total time = 3 + 6 = 9 min 1day 1hr ωt of ore = 800 ton day x x 9 min 24hours 60 min = 5 ton 2000lb = 5 ton x = 10,000 lb 1ton 1 Volume of ore = 10000 lb x (62 .4 2.8) lb ft 3 = 57.234 ft3 volume of water = volume of pulp - volume of ore = 300 ft3 – 57.234 ft3 = 242.766 ft3 ωt of water = volume of water x density of water = 242.766 ft3 x 62.4 lb/ft3 = 15148.5984 lbs ωt of ore Pulp density = x 100 ωt of pulp 10000 lbs = 100 (10000 lbs 15148 .5984 lbs ) = 39.76 % Change in pulp density = 39.76 – 28.98 = 10.78 % 10.78% % change in pulp density = 100 28.98% = 37.198 % Daily water consumption =? Solid liquid = 10000 lb: 15148.5984 lb = 1:1.52 1.52 lb 1ton 32 ft 3 800 ton 2000 lb 1day 1hr = 1lb 2000 lb 1ton 1day 1ton 24 hr 60 min = 27.02 ft3/min ft 3 60 min 24 hr = 27.02 min day = 38908.8 ft3/day IV. (b) Write notes on the preparation of xanthate and oxidation when exposed to moisture and air. Solution Preparation of Xanthate The basic minerals for the preparation of technical xanthates are alcohols, hydroxides and carbon disulphide. Potassium or sodium xanthates are generally used. Therefore either potassium hydroxide or sodium hydroxide can be the starting materials. The simplest method of preparation consists of dissolving hydroxide in alcohol. KOH + ROH = ROK + H2O In this way, alcoholate is obtained which reacts with carbon disulphide as follows: ROK + CS2 = ROCS2K+ One method of purifying xanthate is by double re-crystallization from alcohol. The crystals are thoroughly washed with ether, dissolved in acetones and saited out by ether. All, solvents used should be of high purity such purified xanthates are used in the pH range 7 to 12. V. (a) A 1000 tpd mill treats ore assaying 10% PbS using roughing and scavenging operations with the scavenger concentrate assaying 8% PbS with pulp dilution 1:1 and the final tailing assays 1%PbS. Also tailing from rougher analyzes 4% PbS. Rougher flotation time of 8 minutes with 30% pulp density and scavenging operation of 14 minutes with 25% pulp density are to be allowed. If the sp.gr. of the ore is 2.8. determine. (i) volume of rougher and scavenger cells in cu.ft., (ii) weight of concentrate produced/day (iii) percent of lead in ore lost in tailing Solution Step A- Flow sheet design a a c e Rougher Scavenger b d Overall weight balence, a = b + e = 1000 tpd 10 80 1 Overall PbS balence, xa= xb+ xe 100 100 800 1 x 1000 = 0.8 b + 0.01 e 100 100 = 0.8 b + 0.01 e Multiplying (1) by 0.8 and subtracting (2) 800 = 0.8 b + 0.8 e 100 = 0.8 b + 0.01 e 700 = 0.79 e e = 886.08 tpd b = 1000 – 886.08 = 113.92 tpd Weight balence around scavenger, c=d+e c = d + 886.08 PbS balence around scavenger, 4 8 1 c= d (886.08 ) 100 100 100 0.04 c = 0.08d + (886.08 x 0.01) Multiplying (3) by 0.04 and subtracting from (4) 0.04c = 0.08d + (886.08 x 0.01) 0.04c = 0.04d + (886.08 x 0.04) 0 0.04d - 26.5824 V. (b) Prepare a complete flow sheet for cyanidation or amalgamation. Solution Amalgamation Inhibitors: coatings, which inhibit amalgamation Undissolved substances Dissolved substances Antimony and arsenic sulfide minerals in pulp such as salts, alkali sulfide, oil and other organic contaminants which organic flotation reagent which become spread on the gold surface coated on gold surface Remedial Methods: 1. To use the alkali cyanides in pulp to remove oil coating on Au surface 2. Sodium or zinc amalgam instead of mercury 3. Electric current passing from pulp to amalgam Plate Amalgamation Amalgamation Practice Amalgam Barrel Amalgamation Treatment Retorting 20 to 45% Au Amalgam Sponge Au Flux Melting Slag Gold Bullion Flow sheet for Amalgamation VI. Define or explain the following. (i) critical pH (ii) dispersion agent (iii) flocculating agent (iv) sulphidizing agent (v) hydrophobic surface (vi) slime coating (vii) activation (viii) frothers Solution Ore Crushing Screening Grinding Classification 4Au+8NaCN+O2+2H2O Lime + Cyanide + Air Dissolution = 4NaAu(CN2)+4NaOH Solid liquid separation Solid Clear Pregnant solution Zn Dust Precipitation 2NaAu(CN)2+3Zn+4NaCN+2H2O = 2Au+2Na2Zn(CN)4+Na2ZnO2+2H2 Solid liquid separation Sulphuric Acid to Au precipitate Zn sulphate remove excess Zn Clacine Drying and calcining 500 to 550°C Flux Melting Slag Gold Bullion Flow Sheet for Cyanidation (i) Critical pH Critical pH is the pH of the pulp below which a given mineral will float and above which it will not float in the presence of controlled quantities of other reagents held at constant temperature. (ii) Dispersion agent Dispersion agent or dispersants are chemical reagents which when added to pulp remove slime coatings from the surface of coarse flotable mineral particles on which they have covered. Dispersants also tend to reduce adsorption of reagents on slimes. (iii) Flocculating agent Flocculating agents are mostly organic colloids which when added to pulp promote formation of very fine particles into flocs. (iv) Sulphidizing agent Sulphidizers or sulphidizing agents are chemical reagents, which can produce a coating of sulphide ions. (v) Hydrophobic surface This process of separating different minerals from each other is possible because of the differences in then surface reactivity. Most of the minerals have natural hydrophilic or water wettable surface. If means that they have more affinity for water than to air. (vi) Slime coating Slime coating may be defined as relatively adherent attachment of extremely fine minerals on the surfaces of granular coarse solids. (vii) Activation Activation is a process of forming a coating of heavy base metal ions on certain mineral surfaces to promote its flotation. (viii) Frothers Frothers are organic reagents which reduce the surface tension of water when added to pulp. They help to stabilized forth on the surface of the pulp. This allows the flotable mineral particles adhering to air bubbles to be removed from flotation machine.