Froth Flotation_1


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Froth Flotation_1 presentation includes flotation fundamentals, performance calculation, hydrophobicity or hydrophilicity, and also particle or bubble contact. Next, Front Flotation_2 will brightly discuss about collection in the froth layer, reagents and flotation's equipment.

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  • Selective attachment of air bubbles to hydrophobic particles. The buoyancy of the bubbles then carries these particles to the surface, leaving the hydrophilic particles behind.
  • Froth Flotation_1

    1. 1. FrothFlotation_1 Flotation Fundamentals Performance Calculation Hydrophobicity/Hydrophilicity Particle or BubbleContactBy Pambudi Pajar Pratama BEng, MSc
    2. 2. Flotation Fundamentals Froth flotation is a highly versatile method forphysically separating particles based on differencesin the ability of air bubbles to selectively adhere tospecific mineral surfaces in a mineral/water slurry. The particles with attached air bubbles are thencarried to the surface and removed, while theparticles that remain completely wetted stay in theliquid phase. Froth flotation can be adapted to a broad range ofmineral separations, as it is possible to use chemicaltreatments to selectively alter mineral surfaces sothat they have the necessary properties for theseparation.
    3. 3. Flotation Fundamentals It is currently in use for many diverse applications,with a few examples being:o Separating sulfide minerals from silica gangue(and from other sulfide minerals);o Separating potassium chloride (sylvite) fromsodium chloride (halite);o Separating coal from ash-forming minerals;o Removing silicate minerals from iron ores;o Separating phosphate minerals from silicates;o And even non-mineral applications such as de-inking recycled newsprint. It is particularly useful for processing fine-grainedores that are not amenable to conventionalgravity concentration.
    4. 4. The Flotation SystemChemistry ComponentsCollectorsFrothersActivatorsDepressantspHOperation ComponentsFeed RateMineralogyParticle SizePulp DensityTemperatureEquipment ComponentsCell Design AgitationAir FlowCell Bank ConfigurationCell Bank ControlThe flotation system includes many interrelated components, and changes inone area will produce compensating effects in other areas (Klimpel, 1995).
    5. 5. Flotation Fundamentals Changes in the settings of one factor (such asfeed rate) will automatically cause or demandchanges in other parts of the system (such asflotation rate, particle size recovery, air flow, pulpdensity, etc.) As a result, it is difficult to study theeffects of any single factor in isolation, andcompensation effects within the system can keepprocess changes from producing the expectedeffects (Klimpel, 1995). This makes it difficult to develop predictive modelsfor froth flotation, although work is being done todevelop simple models that can predict theperformance of the circuit from easily-measurableparameters such as solids recovery and tailingssolid content (Rao et al., 1995).
    6. 6. Performance Calculations There is no one universal method for expressing the effectiveness of aseparation, but there are several methods that are useful forexamining froth flotation processes:a) Ratio of Concentration, the weight of the feed relative to the weight ofthe concentrate. Starting with the mass balance equations, and thedefinition of the ratio of concentration:F = C + TFf = Cc + TtRatio of Concentration = F/Co The Ratio of Concentration is F/C, where F is the total weight of the feedand C is the total weight of the concentrate. One limitation with thiscalculation is that it uses the weights of the feed and concentrate.o While this data is available in laboratory experiments, in the plant it is likelythat the ore is not weighed and only assays will be available. However, it ispossible to express the ratio of concentration in terms of ore assays.o F, C, and T are the % weights of the feed, concentrate, and tailings,respectively; and f, c, and t are the assays of the feed, concentrate, andtailings.
    7. 7. Performance CalculationsIt needs to eliminate T from these equations so thatit can solve for F/C:o Ff = Cc + Tt, And multiplying (F = C + T) by t gives:o Ft = Ct + Tt, So subtracting this equation from the previouseliminates T and gives:o F(f - t) = C(c - t), And rearranging produces the equation for theratio of concentration:o F/C = (c – t)/(f – t)
    8. 8. Performance Calculationsb) % Metal Recovery, or percentage of the metal inthe original feed that is recovered in theconcentrate. This can be calculated using weightsand assays, as (Cc)/(Ff) * 100. Or, since C/F =(f –t)/(c – t), the % Metal Recovery can be calculatedfrom assays alone using 100(c/f)(f – t)/(c –t).c) % Metal Loss is the opposite of the % MetalRecovery, and represents the material lost to thetailings. It can be calculated simply by subtractingthe % Metal Recovery from 100%.
    9. 9. Performance Calculationsd) % Weight Recovery is essentially the inverse of theratio of concentration, and equals 100 * C/F = 100 *f – t)/(c – t).e) Enrichment Ratio is calculated directly from assaysas c/f, weights are not involved in the calculation.Example Calculations:Problem: A copper ore initially contains 2.09% Cu. Aftercarrying out a froth flotation separation, the productsare as shown in Table 1. Using this data, calculate:a) Ratio of concentrationb) % Metal Recoveryc) % Metal Lossd) % Weight Recovery, or % Yielde) Enrichment Ratio
    10. 10. Performance CalculationsFeedf = 2.09% CuF = 100% WtConcentratec = 20% CuC= 10% WtTailingsT = 0.1% CuT = 900% WtProduct % Weight % Cu AssayFeed 100 2.09Concentrate 10 20.0Tailings 90 0.1Table 1
    11. 11. Performance Calculationsa) From Table 1, the Ratio of Concentration can becalculated as F/C = 100/10 = 10. If only assays areavailable, the ratio of concentration equals (20 –0.1)/(2.09 – 0.1) = 10.So, for each 10 tons of feed, the plant would produce 1 ton ofconcentrate.b) Using the example data from Table 1,o The % Cu recovery calculated from weights and assays is:% Cu Recovery = [(10*20)/(2.09*100)]*100 = 95.7%o The calculation using assays alone is% Cu Recovery = 100*(20/2.09)*(2.09 – 0.1)/(20 – 0.1) = 95.7%o This means that 95.7% of the copper present in the ore wasrecovered in the concentrate, while the rest was lost in thetailings.
    12. 12. Performance Calculationsc) The % Cu Loss can be calculated by subtracting the% Cu Recovery from 100%:% Cu Loss = 100 – 95.7 = 4.3%o This means that 4.3% of the copper present in the ore waslost in the tailings.d) The % Weight Recovery is equal to the % Weight ofthe concentrate in Table 1.o It can also be calculated from the assay values given inthe table, as follows:% Weight Recovery = 100*(2.09 - 0.1)*(20 – 0.1) = 10%e) The Enrichment Ratio is calculated by dividing theconcentrate assay in Table 1 by the feed assay:Enrichment Ratio = 20.0/2.09 = 9.57o This tells that the concentrate has 9.57 times the copperconcentration of the feed.
    13. 13. Performance CalculationsGrade/Recovery Curves While each of these single calculated values areuseful for comparing flotation performance fordifferent conditions, it is most useful to considerboth the grade and the recovery simultaneously,using a “Grade/Recovery Curve”. This is a graph of the recovery of the valuablemetal achieved versus the product grade at thatrecovery, and is particularly useful for comparingseparations where both the grade and therecovery are varying.
    14. 14. Performance CalculationsGrade/Recovery Curves  If 100% of the feed is recovered tothe product, then the product willobviously have the samecomposition as the feed, and sothe curve starts at the feedcomposition with 100% recovery. Similarly, if the purest mineral grainthat contains the metal of interestis removed, this will be themaximum grade that can beproduced by a physicalseparation, and so the 0%recovery end of the curveterminates at an assay less than orequal to the assay of the purestgrains available in the ore.Figure 1. Typical form of Grade orRecovery Curves for frothflotation.
    15. 15. Hydrophobicity/hydrophilicityThe basis of froth flotation is the difference inwettabilities of different minerals. Particles range from those that are easily wettable bywater (hydrophilic) to those that are water-repellent(hydrophobic). If a mixture of hydrophobic and hydrophilic particles aresuspended in water, and air is bubbled through thesuspension, then the hydrophobic particles will tend toattach to the air bubbles and float to the surface. The froth layer that forms on the surface will then beheavily loaded with they hydrophobic mineral, and canbe removed as a separated product. The hydrophilicparticles will have much less tendency to attach to airbubbles, and so it will remain in suspension and beflushed away (Whelan and Brown, 1956).
    16. 16. Hydrophobicity/hydrophilicity Particles can either be naturally hydrophobic, or thehydrophobicity can be induced by chemical treatments. Naturally hydrophobic materials include hydrocarbons,and non-polar solids such as elemental sulfur. Coal is agood example of a material that is typically naturallyhydrophobic, because it is mostly composed ofhydrocarbons. Chemical treatments to render a surface hydrophobicare essentially methods for selectively coating a particlesurface with a monolayer of non-polar oil.Figure 2. Selective attachmentof air bubbles to hydrophobicparticles.
    17. 17. Hydrophobicity/hydrophilicityFigure 3. Contact angle betweenand air bubble and a solidsurface immersed in liquid.
    18. 18. Hydrophobicity/hydrophilicityFigure 3. Contact angle between and air bubble and a solidsurface immersed in liquid.If the contact angle is very small, then the bubble does notattach to the surface, while a very large contact angleresults in very strong bubble attachment. A contact anglenear 90° is sufficient for effective froth flotation in most cases.
    19. 19. Particle or Bubble ContactFigure 4. Simplified schematicof a conventional flotationcell. The rotor draws slurrythrough the stator andexpels it to the sides,creating a suction thatdraws air down the shaftof the stator. The air is then dispersed asbubbles through the slurry,and comes in contactwith particles in the slurrythat is drawn through thestator.
    20. 20. Particle or Bubble ContactFigure 4. Simplified schematicof a conventional flotationcell. Once the particles arerendered hydrophobic, theymust be brought in contactwith gas bubbles so that thebubbles can attach to thesurface. If the bubbles and surfacesnever come in contact, thenno flotation can occur. Contact between particlesand bubbles can beaccomplished in a flotationcell such as the one shownschematically in Figure 4.
    21. 21. Particle or Bubble ContactFigure 4. Simplified schematicof a conventional flotationcell. Particle/bubble collision isaffected by the relative sizesof the particles. If the bubbles are largerelative to the particles, thenfluid flowing around thebubbles can sweep theparticles past without comingin contact. It is therefore best if thebubble diameter iscomparable to the particlediameter in order to ensuregood particle/bubblecontact.
    22. 22. FrothFlotation_2 Collection in the FrothLayer ReagentsBy Pambudi Pajar Pratama BEng, MScContinue to nextpresentation