FLOTATION

ChE 154 - X

Cayabyab, Jamel P.

Ferrer, Joseph Rafael T.

Lontoc, Janella A.

Mantes, Redel Erap

Maravive, Angelica C.

Porcalla, Jamaeca M.

Santos, Quennie Mae

I. Introduction

One of the goals of many unit operations is to separate the components of raw materials to the desired products. In solid-solid separation, one of the most widely used operations is flotation. This is used to separate and concentrate ores by altering their surfaces to a hydrophobic or hydrophilic condition and is developed in the mineral industries. The material floated off is called the concentrate as it contains the desired mineral. The other material which sinks in the water and is removed from the bottom is called tailings,

II. Principles

According to Brown (1950), flotation includes any operation in which one solid is separated from another by floating one of them at or on the surface of a fluid. Perry defines flotation as separation of mixed liberated particles based on the difference in their wettability. The diagram below shows the typical setup of flotation system.

Flotation is used in the mineral process industry to concentrate mineral values (i.e. copper, lead, zinc, molybdenum and nickel) and for the recovery of fine coal and for the concentration of a wide range of mineral commodities. It is also applicable in wastewater treatment to remove particulate, organic and biological contaminants, extraction of metallic values, removal of heavy metal compounds from hydrometallurgical streams (precipitation flotation), and recovery of bitumen from tar sands.

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The above pictures show some of the real application of flotation: the froth flotation used for concentrating sulfide ores (upper left), froth flotation for separating individual plastics for recycling ( upper right) and froth flotation used in mining industry to separate valuable minerals from waste rocks (lower).

The table below shows the classification of flotation.

Types Description

Ore Flotation involves the separation of ore particles from gangue particles by

Macroflotation the removal of macroscopic particles

Microflotation the removal of microscopic particles such as microorganisms and colloids

Precipitate flotation

in which a precipitate is removed and the precipitating agent is other than a surfactant

Ion flotation the removal of surface-active ions through the use of a surfactant which

Molecular flotation

in which surface-inactive molecules are removed through the use of a surfactant which yields an insoluble product

Adsorbing colloid flotation

the "piggy-back" removal of dissolved material which is first adsorbed on colloidal particles

Figure19-63 (Perry, 1997) illustrates the two major steps of unit operation of flotation.

In conditioning step, the slurry or the pulp, consisting of particles to be separated is fed in the conditioning to which the necessary flotation reagents are added. The particles are finely divided material reduced by the ball mill or other fine crusher, usually varying in size from 20 mesh to under 200 mesh. This step aims to create physical-chemical conditions for achieving appropriate selectivity between particle species that are to be separated and to cause complete “filming” of the solid by the reagent.

Flotation reagents can be collectors, frothers, and modifiers. Flotation reagents Description

Collectors surface-active agents that are added to the flotation pulp, where they adsorb selectively on the surface of the

particles and render them hydrophobicFrothers surface-active agents added to the flotation pulp primarily

to stabilize the air bubbles for effective particle-bubble attachment, carryover of particle-laden bubbles to the

froth, and removal of the froth*similar to ionizing collectors except that they concentrate

primarily at the air-liquid interface

Modifiers Activators- used to make a mineral surface amenable to collector coating

pH regulators- used to control or adjust pH, a very critical factor in many flotation separations

Depressants- assists in selectivity (sharpness of separation) or stop unwanted minerals from floating

Dispersants and flocculants- control slimes that sometimes interfere with the selectivity and increase reagent consumption

Table 1. Descriptions of flotation reagents.

Flotation involves relative interaction of three phases: solid, water and air to contact solid

particles suspended in water with air bubbles and cause a stable bubble-particle attachment.

Recall that flotation depends upon the relative adsorption or “wetting” of the solid surfaces by the fluid. This is controlled by surface or interfacial energy. The best equation that describes the interfacial energies is the Young -Dupre Equation.

Generally, the sum of the components of the interfacial tensions equals to zero.

γSG = γSL + γLG (cos θ) --------> Young-Dupre Equation

where γ = interfacial tension of the solid-gas (SG), solid-liquid (SL) and liquid-gas (LG),respectively

θ = contact angle

The importance of the young equation is the value of the contact angle can affect the relationship of the phases. If the contact angle is very small, then the bubble (gas) does not attach to the solid surface, while a very large contact angle results in very strong bubble attachment. A contact angle near 90 degrees is sufficient for effective froth flotation.

When a solid particle attaches itself to a bubble, there is a loss in surface energy –ΔE per unit area of surface δ, equal to the loss in the surface tension

Since,γSG = γSL + γLG (cos θ)

γSL - γSG = -γLG (cos θ)

then

Next to conditioning step is the separation. In this step, air bubbles is generated and introduced in to the process vessel. Particles attached the air bubbles are in most application removed from the process vessel (flotation cell) as froth. Thus knowledge about process vessel is important.

III. Equipments

Equipments used in flotation are designed according to methods of generation and introduction of air bubbles into the tank. The diagram below shows most commonly used in flotation.

These equipments can be classified into electrolytic, dissolved-air and dispersed-air flotation units. Electroflotation (see Fig. 19-70 of Perry et al, 1997) uses electrodes to generate air bubbles. The electrodes used can be aluminum, platinized titanium, titanium coated with lead dioxide and stainless steel. Power is supplied to the electrode at a low voltage potential of 5 – 10 volts and these consume about 0.5 to 0.7 kW/ sqm of flotation tank surface area depending on the conductivity of the liquid and the distance between the electrodes. Electrolytic flotation unit creates smallest bubble size (10-50 micrometer) and very little turbulence created by bubble formation. It is attractive for the separation of small particles and fragile flocs and for small installations in the flowrate range of 10 – 20 cu.m/h. However, the electrodes used are prone to fouling that may require mechanical cleaning devices. It is not also suitable for potable water treatment due to heavy metal contamination.

Dissolved-air flotation (see Fig. 19-71 of Perry et al, 1997) is the saturation of processed stream with air and generation of air bubbles by releasing pressure. In this flotation system, no addition of frother-type chemical reagents is needed to treat process streams with low solid concentration (0.01- 2% by vol). The size of the bubble ranges from 20 – 100 micrometers. It is applicable for sewage and potable water treatment and also for treatment of slaughterhouse, poultry processing, seafood processing, soap, and food processing wastes.

Vacuum flotation and pressure flotation are the two main types of dissolved-air flotation. In vacuum flotation, process stream is saturated with air at atmospheric pressure and is introduced to the flotation tank on which vacuum is applied. This can be run only as a batch process. On the other hand, pressure flotation consists of pressurizing and aerating the process stream and introducing it into the flotation vessel that is maintained at atmospheric pressure. The reduction of pressure results in formation of fine air bubbles. Compared to vacuum flotation, this can be operated on a continuous basis. Pressurization can be carried out through full-flow or split-flow pressure flotation.

The last equipment is the dispersed-air flotation unit. This unit generates relatively large air bubbles (at least 1 mm in size) by pneumatic or mechanical means. Frothers are added to control the size and stability of air bubbles.

Pneumatic cell and mechanical cell are generally called flotation cell. Flotation cell is the equipment in which the material is actually separated or floated from the residual tailings. It consists of a vessel provided with a feed at one end, an overflow for froth removal, and a discharge for tailing at the opposite end. In mechanical cell (see Fig. 19-73 of Perry et al, 1997) a mechanical agitator draws in air and beats it into the pulp by means of a rotation impeller on an upright shaft. Because of more violent agitation, more thorough flotation is achieved and tailings become more nearly free from material desired in the concentrate. Pneumatic cells on the contrary depend upon compressed air for mild agitation. These produce a clean froth relatively free from gangue. Compared to mechanical cells, pneumatic cells (see Fig. 19-78 of Perry et al, 1997) require 50% longer contact time and full conditioning of the pulp before flotation. An example of pneumatic devices is the flotation columns where air-bubble generation is accomplished by a gas-sparging system.

IV. Sample Problem

Problem 1:

It is desired to recover lead from an ore containing 10 percent lead sulfide (PbS) and the balance assumed to be silica, 500 tons of ore being treated per 24-hr day.

It is assumed that the concentrate from a single cell is of acceptable purity but the tailings are to be retreated in scavenger cells with return of scavenger concentrate to the rougher.

Laboratory findings indicate that if water-to-solids ratio L/S = 2 and the contact time is 8 min in the rougher and L/s = 4 for 15 min in scavenger, with mechanically agitated machines of the Denver type, the following compositions will be found for the various products.

The density of PbS and SiO2 are 7.5 and 2.65 g/cc, respectively

Determine: a. Density for all solidsb. Mass of productsc. Volume of tanksd. Number of cells and power requirement using Denver No. 24 which has 50 cu. ft of volumee. Volume of delivered air and power requirement of air compressor when Air-Lift Machine by the Southwestern Engineering Co. is used (Assume 75 cfm air/ft in Rougher, 60 cfm air/ft in scavenger, and 25 percent longer contact time at 2 psi).

Table 19. POWER REQUIREMENTS OF LOW-PRESSURE BLOWERSAir delivered,

cfmApproximate Horsepower at Pressures of

2 psi 3 psi 4 psi 5 psi

500 6 9.5 13.5 17.5

1000 12 18.5 26 33

2000 24 38 54 70

3000 34 57 76 96

4000 46 76 105 135

5000 58 91 125 160

6000 70 120 168 220

7000 83 135 190 245

8000 96 165 225 310

Table 20. APPROXIMATE POWER REQUIREMENTS FOR MECHANICAL FLOTATION CELLS

Size, cubic ftHorsepower Consumed per Cell

Denver Fagergren

10 1

12 1.2 1.8-2.0

18 1.4

24 2.2 3.5-4.0

40 3.2 5

50 4.2 6

70 8

100 9

Solution:

Computation of density of all solidsIn feed, PbS: 10 g = 11.33 cm3

90 g = 34.00 cm 3 35.33 cm3

Average density

ρa = 176.5 lb/ft3

ρb

ρc

ρd

ρe

Feed, a

Rougher ScavengerFinal Tailings,

e

Concentrate, b

Scavenger concentrate, d

Rougher Tailings, c

Computation of the mass of productsBasis: 100 lb feedOMB: a = b + e = 100 b = 100 – e a = (100 – e) + e

Overall PbS balance: (0.1)(100) = (0.8)(100-e) + (0.005)(e) e = 88.1 lb b = 100 – e b = 11.9 lb

PbS balance around the scavenger: c = d + e (0.02)(d + 88.1) = (0.11)(d) + (0.005)(88.1) 0.09d = 1.3115 d = 14.57 lb c = d + e = 14.57 + 88.1 c = 102.67 lb

Yield

Computation of volume of tanks

Input to the rougher tank: a and d solids

a = 100 lb or

d = 14.57 lb or

114.57 lb 0.6468 ft3

Average density (ρad)

Since

Volume fraction of solids in pulp

Capacity per 1 ft3 of rougher tank volume:

Capacity required:

Therefore, required volume of rougher tank:

For scavenger cells:

c = 102.67 lb or

Capacity per 1 ft3 of rougher tank volume:

Capacity required:

Therefore, required volume of tank for scavenger cells:

Using Denver No. 24 machine50 ft3 of volume require 4.2 hp per cell to operate (From Table 20)

For rougher:

For scavenger:

Total: 21 cells and 88.2 hp

Using Air-Lift MachineStandard cross-sectional area of 9.85 ft2

25 % longer contact time2 psi

For rougher:

For scavenger:

Air for rougher:

Air for scavenger:

Total: 8,027.1 cfm

Power requirement for the air compressor 96 hp (From Table 19)

Problem 2:

A copper ore initially contains 2.09% Cu. after carrying out a froth flotation separation, the products are shown in Table 1.. Using this data, calculate:

(a) Ratio of concentration(b) % Metal Recovery(c) % Metal Loss(d) % Weight Recovery or % Yield(e) Enrichment Ratio

Table 1. Grade/recovery performance of a hypothetical copper ore flotation proces.

Products % Weight % Cu Assay

Feed F = 100 f = 2.09

Concentrate C = 10 c = 20

Tailings T = 90 t = 0.1

Solution: (a) From Table 1, the Ratio of concentration can be calculated as F/C = 100/10 = 10. if only

assays are available, the ratio of concentration equals

(c – t ) / (f - t) = (20-0.1) / (2.09-0.1) = 10.

So for each 10 tons of feed, the plant would produce 1 ton of concentrate.

(b) Using the example data from Table 1, the % Cu recovery calculated from weights and assays is:

% Cu Recovery = [(C*c) / (F*f)]*100

= [(10*20) / (2.09*100)]*100 = 95.7%

The calculation using assays alone is

% Cu Recovery = (c / f)*[(f - t)/(c - t)]*100

= 100*(20/2.09)*(2.09-0.1) / (20-0.1) = 95.7%

This means that 95.7% of the copper present in the ore was recovered in the concentrate, while the rest was lost in the tailings.

(c) The % Cu Loss can be calculated by subtracting the % Cu Recovery from 100%:

%Cu Loss = 100 - % Metal Recovery=

= 100 - 95.7 = 4.3%

This means that 4.3 % of the copper present in the ore was lost in the tailings.

(d) The % Weight Recovery is equal to the % Weight of the concentrate in the Table 1. It can also be calculated from the assay values given in the table, as follows:

%Weight Recovery = (f - t)*(c - t)*100

= 100*(2.09-0.1)*(20-0.1) = 10%

(e) The Enrichment Ratio is calculated by dividing the concentration assay in Table 1 by the feed assay:Enrichment ratio = c / f

= 20.0/2.09 = 9.57This tells that the concentrate has 9.57 times the copper concentration of the feed.

Problem 3: Sizing of Flotation CellConsider the following copper ore:

Specific gravity of dry ore: 2.7

Optimum Percent Solids in Laboratory Flotation MAchine = 30%

Optimum Laboratory Flotation Retention Time = 6.5 minutes

Assume that the flotation facility will need to process 14 500 metric tons of dry ore per day

**Note:

1.) Cells of 14.15 cubic meters (500 cubic feet) are a standard size.

2.) When the pulp is aerated, it will consist of approx. 15% air and 85% slurry by volume.

Solution:

From Table 2. Comparison of optimum flotation times in the laboratory and in the plant (Metso, 2006), the average laboratory flotation time for copper is 7 minutes and the average plant flotation time is 14.5 minutes.

The time needed for flotation in the plant for this particular ore is (14.5/7)*6.5 = 13.5 minutes. A feedrate of 14500 metric tons dry ore/day will require a flow of

(14,500 tons/day)(1day/24 hours)(1 hour/60minutes) = 10.07 metric tons/minuteand for a 13.5 minutes residence time there will need to be

(13.5 minutes)*(10.07 metric tons/minutes) = 136 metric tons of ore retained in the flotation bank.

Since the slurry must be 30% solids, 136 dry tons of ore will be equal to

(136 tons dry ore)*(1 tons of slurry/0.3 tons of dry ore) = 453 metric tons of slurry. Since the ore has a specific gravity of 2.7, the specific gravity of the slurry can be computed from the relation (100/density of slurry) = (%solid/density of solid) + ((100 - %solid)/density of water). Thus, the density of the slurry is 1.23. The total volume of the 453 tons of slurry that the cells must hold will be (453 metric tons)/1.23 = 368 cubic meters. however, it was noted that when the pulp is aerated it will consist of approximately 15% air and 85% slurry by volume, so the actual volume of pulp needed will be 368/0.85 = 433 cubic meters.

From the problem, Note 1 states that cells of 14.15 cubic meters are a standard size and so if these are used the number of cells needed is 433/14.145 = 30.6 cells or approximately 31 cells. Referring to Table 3, for copper 12 and 17 cells per bank must be used, with 14 to 17 cells being optimum. Thus, two banks of 16 cells would give 32 total cells, which would provide the 31 cells needed with some extra capacity. The cells would then be arranged in the individual banks as per the manufacturer’s recommendations.

V. References

Brown, George Granger. (1950). Unit Operations. Tokyo, Japan: Modern Asia Editions.

Metso. (2006). Basics in Minerals Processing. 5th ed. Section 4 - Separation, Metso Minerals, http://www.metso.com

Perry, Robert H. and Don W. Green. (1997). Perry’s Chemical Engineer’s Handbook. 7th ed. USA: McGraw-Hill, Inc.

http://www.chem.mtu.edu/chem_eng/faculty/kawatra/Flotation_Fundamentals.pdf