Day A. Mining chemicals hand book

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b. Recovery, %

Represents the ratio of the weight of metal or mineral value recov-

ered in the concentrate to 100% of the same constituent in the

heads or feed to the process, expressed as a percentage. It may be

calculated in several different ways, depending on the data available.

By assays f, c and t only:

R =

c(f – t)

x 100 = recovery % (27)

f(c – t)

By K plus assays f and c

R =

x 100 = recovery % (28)

By weights F and C, plus assays c and t

R =

x 100 = recovery, % (29)

Cc+t(F–C)

EExxaammppllee::

A copper concentrator is milling 15,000 tons/day of a

chalcopyrite ore assaying 1.15% copper. The concentrate and tailings

produced average 32.7% and 0.18% copper, respectively. Calculate:

by (25) K =

32.7 – 0.18

= 33.53

1.15 – 0.18

by (26) C =

15,000

(15,000)(0.97)

= 447.4 tons

33.53 32.52

by (27) R =

(32.7)(1.15 – 0.18)

X 100 = 84.8%

1.15(32.7 – 0.18)

by (28) R =

32.7

X 100 = 84.8%

(33.53)(1.15)

Metallurgical computations

239

11.6.2 Three product (bi-metallic) formulas

Frequently, a concentrator will mill a complex ore requiring the

production of two separate concentrates, each of which is enriched

in a different metal or valuable mineral, plus a final tailing accept-

ably low in both constituents. Formulas have been developed which

use the feed tonnage and assays of the two recovered values to obtain

the ratios of concentration, the weights of the three products of

separation, and the recoveries of the values in their respective

concentrates. For illustrative purposes data from a copper-zinc

separation is assumed.

Definition and notation

Product Weight % Cu % Zn Calculated

or Wt.% Assay Assay

Feed F c

Cu concentrate C c

Zn concentrate Z c

Tailing T c

Ratios of concentration K

and K

Recovery, % R

and R

The ratios of concentration, K

and K

are those for the copper and

zinc concentrates, respectively, with R

and R

the percentage

recoveries of the metals in their corresponding concentrates.

As follows:

C = F x

– c

)(z

– z

) – (z

– z

)(c

– c

)

= tons Cu concentrate (30)

– c

)(z

– z

) – (z

– z

)(c

– c

)

Z = F x

– c

)(z

– z

) – (c

– c

)(z

– z

)

= tons Zn concentrate

(31)

– c

)(z

– z

) – (z

– z

)(c

– c

)

C x c

x 100 copper recovery, % (32)

F x c

Z x z

x 100 zinc recovery, % (33)

F x z

and K

= ratio of concentration (34, 35)

C Z

Mining Chemicals Handbook

240

EExxaammppllee::

Product

Assay %

Tons Copper Zinc

Feed 1000 2.7 19.3

Cu concentrate C 25.3 5.1

Zn concentrate Z 1.2 52.7

Tailing T 0.15 0.95

Then,

C = 1000 x

(2.7 – 0.15)(52.7 – 0.95) – (19.3 – 0.95)(1.2 – 0.15)

(25.3 – 0.15)(52.7 – 0.95)(5.1 – 0.95)(1.2 – 0.15)

C = 1000 x

131.96 – 19.27

112,690

= 86.9 tons Cu concentrate

1301.51 – 4.36 1297.15

Z = 1000 x

(25.3 – 0.15)(19.3 – 0.95) – (2.7 – 0.15)(5.1 – 0.95)

(25.3 – 0.15)(52.7 – 0.95) – (5.1 – 0.95)(1.2 – 0.15)

C = 1000 x

461.50 – 10.58

450,920

= 347.6 tons Zn concentrate

1301.51 – 4.36 1297.15

(89.9)(25.3)

x 100 =

2198.6

x 100 = 81.4%

(1000)(2.7) 2700

(347.6)(52.7)

x 100 =

18,318.5

x 100 = 94.9%

(1000)(19.3) 19,300

(1000)

= 11.51, K

(1000)

= 2.88

(86.9) 347.6

The three product solution illustrated above can be somewhat

simplified by taking an intermediate tailings sample between the

two stages of concentration; i.e., a copper tail (zinc feed) sample in

the previous example. Then, adding the notations:

Copper tail (zinc feed) = CT

with copper and zinc assays = c

and z

Metallurgical computations

241

Assume mill feed, F, as Unity 1

Then, C + CT = 1 (a)

(

C x c

) + (CT x c

) = c

(b)

(

C x c

) + (CT x c

) = c

(c)

Subtracting (c) from (b),

C(c

– c

) = (c

– c

)

Then, C = F

– c

)

= tons copper concentrate

(36)

– c

)

and similarly, Z = (F – C)

– z

)

= tons zinc concentrate

(37)

– z

)

EExxaammppllee::

It is decided to take a copper tail (zinc feed) sample in

order to provide a check on the results calculated in the previous

example. The sample (CT) assayed 0.55% Cu (c

) and 20.9% Zn (z

respectively. The check weights of the copper and zinc concentrates

are computed as follows:

Copper concentrate,

C = 1000 x

(2.7 – 0.55)

= 1000 x

(2.15)

= 86.9 tons

(25.3 – 0.55) (24.75)

Zinc concentrate,

Z = (1000 – 86.9) x

(20.9 – 0.95)

= 913.1 x

(19.95)

= 352.0 tons

52.7 – 0.95 (51.75)

As can be seen, the calculated weights of the copper concentrate

check exactly, while the zinc concentrate checks within 1.3%.

An average of the zinc concentrate weights, obtained using both

methods, could be used if desired.

It should be understood that there are certain limitations to the

use of three-product formulas, since it is required by definition that

two of the three products involved must be concentrates of essentially

different metals or mineral components. The formulas will only give

reliable results when the assays indicate that a differential concentra-

tion of the two components into separate concentrates has occurred.

Mining Chemicals Handbook

242

11.7 Flotation reagent usage formula

The consumption or usage rate of the chemicals employed in flota-

tion is generally expressed in terms of grams per metric ton of ore

treated. Depending upon the particular reagent, it may be fed as a

dry solid, as a water solution or dispersion, or in the undiluted

"as-is" liquid form. The normal procedure when checking or setting

reagent feed rates is to measure the amount being fed to the circuit

per unit time, usually per minute. Liquid or reagents in solution or

dispersion are measured in ml and dry solids in grams. When

feeding liquids, the specific gravity and weight percent strength of

the reagent must also be known. With this information, along with

the known ore tonnage being treated per unit time, the reagent

measurements can then he translated into grams/metric ton

consumption rates, as follows:

11.7.1 For dry reagents

(g reagent / min.)(1440 min. / day)

g reagent

(38)

tons ore / day ton ore

11.7.2 For liquid reagents

(ml reagent / min.)(reagent sp. gravity)(1440 min. / day)

g reagent

(39)

tons ore / day ton ore

11.7.3 For reagents in solution

(ml solution / min.)(g reagent / liter solution)(1440 min. / day)

g reagent

tons ore / day x 1000 ton ore

NNoottee::

0.0020lb

metric ton per short ton

EExxaammppllee::

At a 10,000 tons/day milling rate, a plant is using 590

ml/min. of a 200 g/L xanthate solution. Calculate the dosage rate.

(590)(200)(1440)

= 17g/t

10,000 x 1000

Metallurgical computations

243

(40)

11.8 Material balance software

In the past few years several software programs have been introduced

to perform aforementioned computations as well as to provide

material balances in operating circuits using several sophisticated

statistical tools. Examples of commercially available software packages

include MATBAL* and JKSimMet**. Excel Solver can also be used.

* MATBAL is a proprietary program of Algosys Inc.

** JKSimMet is a proprietary program of JK Tech/Contract Support Services

11.9 Bibliography

1. The Denver Equipment Co., Handbook, 1954 Edition.

2. Mineral Processing Flowsheets: Denver Equipment Company,

Denver, CO, 1962.

3. Taggart, A. F., Handbook of Mineral Dressing: J. Wiley & Sons, Inc.,

New York, 1945.

4. Weinig, A. and Carpenter, C., “The Trends of Rotation”:

Colorado School of Mines Quarterly, Vol. 32, No. 4, October,

1937.

5. Williamson, D. R., “The Mathematics of Concentration

Processes”: Colorado School of Mines Mineral Industries Bulletin,

Vol. 3, No. 6, November, 1960.

6. Kelly, E. G., and Spottiswood, D. J., Introduction to Mineral

Processing: John Wiley & Sons Inc., New York, NY, 1982.

7. Weiss, N. L., SME Mineral Processing Handbook: Society of Mining

Engineers, New York, NY, 1985.

8. Wills, B. A., Mineral Processing Technology: Butterworth-

Heinemann, Oxford, UK, Sixth Edition, 1997.

Mining Chemicals Handbook

244

.12

STATISTICAL METHODS

IN MINERAL PROCESSING

Mining Chemicals Handbook

246

Section 12

Statistical methods in mineral processing

12.1 Statistics in laboratory work

The purpose of laboratory work is to screen potential products for

the customer’s application, to identify potential improvements

attainable using them, and to generate information to justify testing

at larger (plant) scale.

To be a sound basis for operating decisions, the data generated

in a program of tests, and the conclusions drawn from that data,

should meet accepted scientific standards. Statistical procedures are

accepted standard methods for drawing conclusions from data, and

using them will add credibility to conclusions. In addition, from

laboratory work it is often required that we characterize the per-

formance of a new proposed system as a function of several factors.

This will be the case when realizing improvements from a new

reagent requires, in addition, other changes in process conditions or

plant operations. Also part of the statistical approach, are methods

for modeling complex, multivariable systems over a range of

operations with a reasonable amount of effort.

12.1.1 Statistical distributions and summary statistics

How large a change in performance can be detected in a test pro-

gram depends on the magnitude of errors due to the test procedure

and to analyses, and on steps taken to minimize the impact of the

systematic sources of error. Error, as used in the statistical sense

referring to the numerical result obtained from an experiment, is

the difference between the actual result and its ideal or "true" value.

Just how large this is depends on (generally unperceived) variations

Figure 1.

Distribution of 200 observations from a theoretical normal

distribution with mean 91.0, standard deviation 1.0

Statistical methods in mineral processing

247

86 88 90 92 94

recovery

in materials and technique, and how sensitive the final result is to

those variations. A distribution function represents the variability of

a test result. Recovery of Cu for many tests with a standard reagent

might, for example, form a distribution like that illustrated in Figure

1. The horizontal axis gives values of the recovery; the vertical axis,

numbers or the fraction of observations in each category of recovery

value. If the total number of observations increased, the form of this

distribution would approach a limiting curve. It is unlikely you will

see so many observations in laboratory work; however, it is impor-

tant to understand that any particular experimental result is just one

observation from an ensemble of potential results described by such

a distribution curve.

Average (or mean) and standard deviation are the most common

statistics for summarizing either a distribution or a set of results.

The same terms (mean, standard deviation) are used to denote the

mean and to estimate its confidence limits from a small sample of

observations from the distribution. Standard formulae for calculat-

ing the mean and standard deviation for a small set of data are

below. Hand calculators with statistical functions can calculate these

directly, and spreadsheet software provides these as built-in functions.

x = –

∑

s =

√

∑

– x)

i=1

n –1

Confidence limits for the calculated mean are:

x +

√n

where t is the value obtained from Student’s t table with n-1

degrees of freedom and the chosen confidence level. 95% is the

most common confidence level for reporting.

Example: A flotation test is repeated 5 times on a substrate. The

recovery results are: 90.2, 90.5, 89.3, 90.0, 90.2. The average and

standard deviation are:

x =

90.2 + 90.5 + 89.3 + 90.0 + 90.2

= 90.04

√

(90.2 – 90.04)

+ (90.5 – 90.04)

+ ...

= 0.45

5 – 1

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248