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Fundamentos De Flotation


Enviado por   •  4 de Mayo de 2013  •  8.586 Palabras (35 Páginas)  •  178 Visitas

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1 Froth Flotation – Fundamental Principles

Froth flotation is a highly versatile method for physically separating particles based on

differences in the ability of air bubbles to selectively adhere to specific mineral surfaces in a

mineral/water slurry. The particles with attached air bubbles are then carried to the surface and

removed, while the particles that remain completely wetted stay in the liquid phase. Froth

flotation can be adapted to a broad range of mineral separations, as it is possible to use chemical

treatments to selectively alter mineral surfaces so that they have the necessary properties for the

separation. It is currently in use for many diverse applications, with a few examples being:

separating sulfide minerals from silica gangue (and from other sulfide minerals); separating

potassium chloride (sylvite) from sodium chloride (halite); separating coal from ash-forming

minerals; removing silicate minerals from iron ores; separating phosphate minerals from

silicates; and even non-mineral applications such as de-inking recycled newsprint. It is

particularly useful for processing fine-grained ores that are not amenable to conventional gravity

concentration.

Figure 1: The flotation system includes many interrelated components, and changes in one area

will produce compensating effects in other areas (Klimpel, 1995)

Froth flotation is a good example of an engineering “system”, in that the various important

parameters are highly inter-related, as shown in Figure 1. It is therefore important to take all of

Chemistry Components

Collectors

Frothers

Activators

Depressants

pH

Equipment Components

Cell Design

Agitation

Air Flow

Cell Bank Configuration

Cell Bank Control

Operation Components

Feed Rate

Mineralogy

Particle Size

Pulp Density

Temperature

Flotation

System

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these factors into account in froth flotation operations. Changes in the settings of one factor (such

as feed rate) will automatically cause or demand changes in other parts of the system (such as

flotation rate, particle size recovery, air flow, pulp density, etc.) As a result, it is difficult to

study the effects of any single factor in isolation, and compensation effects within the system can

keep process changes from producing the expected effects (Klimpel, 1995). This makes it

difficult to develop predictive models for froth flotation, although work is being done to develop

simple models that can predict the performance of the circuit from easily-measurable parameters

such as solids recovery and tailings solid content (Rao et al., 1995).

1.1 Performance Calculations

There is no one universal method for expressing the effectiveness of a separation, but there are

several methods that are useful for examining froth flotation processes:

(a) Ratio of Concentration, the weight of the feed relative to the weight of the concentrate, The

Ratio of Concentration is F/C, where F is the total weight of the feed and C is the total weight of

the concentrate. One limitation with this calculation is that it uses the weights of the feed and

concentrate. While this data is available in laboratory experiments, in the plant it is likely that

the ore is not weighed and only assays will be available. However, it is possible to express the

ratio of concentration in terms of ore assays. Starting with the mass balance equations, and the

definition of the ratio of concentration:

F = C + T, Ff = Cc + Tt, Ratio of Concentration = F/C

where 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, and tailings. We now need to eliminate T from these

equations so that we can solve for F/C:

Ff = Cc + Tt, and multiplying (F = C + T) by t gives us:

Ft = Ct + Tt, so subtracting this equation from the previous eliminates T and gives:

F(f - t) = C(c - t), and rearranging produces the equation for the ratio of concentration:

F/C = (c – t)/(f – t)

(b) % Metal Recovery, or percentage of the metal in the original feed that is recovered in the

concentrate. This can be calculated using weights and assays, as (Cc)/(Ff)·100. Or, since C/F =

(f – t)/(c – t), the % Metal Recovery can be calculated from assays alone using 100(c/f)(f – t)/(c –

t).

(c) % Metal Loss is the opposite of the % Metal Recovery, and represents the material lost to the

tailings. It can be calculated simply by subtracting the % Metal Recovery from 100%.

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(d) % Weight Recovery is essentially the inverse of the ratio of concentration, and equals

100·C/F = 100·(f – t)/(c – t).

(e) Enrichment Ratio is calculated directly from assays as c/f, weights are not involved in the

calculation.

Example Calculations:

Problem: A copper ore initially contains 2.09% Cu. After carrying out a froth flotation

separation, the products are as 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 process.

Product % Weight % Cu Assay

Feed 100 2.09

Concentrate 10 20.0

Tailings 90 0.1

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

Feed

f = 2.09% Cu

F = 100% Wt

Concentrate

c = 20% Cu

C = 10% Wt

Tailings

t = 0.1% Cu

T = 900% Wt

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The calculation using assays alone is

% Cu Recovery = 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 – 95.7 = 4.3%

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