MINISTRY OF SCIENCE AND TECHNOLOGY by b6owQ81Z

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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.

								
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