mineral processing design and operation

particle size, diameterSauter mean diameter discharge mass ratio liquid/solid displacement, distance, diameter partition coefficient of size i = recovery of size i in the U/F corrected p

Mineral Processing Design and Operation

PREFACE

In nature minerals of interest exist physically and chemically combined with the host rock.Removal of the unwanted gangue to increase the concentration of mineral in an economicallyviable manner is the basis of mineral processing operations This book treats the strategy ofbeneficiation as a combination of unit operations Each unit process and its operation istherefore treated separately Integration of these units leading to the development of viableflow sheets that meets the final objective, is then indicated

The greatest challenge to a mineral processor is to produce high grade concentratesconsistently at maximum recovery from the ore body To quantify recovery a reasonable idea

of the initial concentration of mineral in a lode is required Proper sampling representing theore body is therefore essential The book therefore commences with the techniques ofsampling of ore followed by the design and operation of unit processes of comminution thathelp to release the mineral from the associated rocks Separation and concentration processesusing techniques involving screening, classification, solid-liquid separations, gravityseparation and flotation then follow In the book some early methods of operation have beenincluded and the modern methods highlighted

The design and operation of each unit process is a study by itself Over the years,improvements in the understanding of the complexities of these processes have resulted inincreased efficiency, sustained higher productivity and grades Mathematical modeling hashelped in this direction and hence its use is emphasized However, the models at best serve asguides to most processes operations that invariably involve complex interdependent variableswhich are not always easily assessed or manipulated To solve the dilemma, plants areincreasingly being equipped with instruments and gadgets that respond to changes muchfaster than humans can detect Dynamic mathematical models are the basis of operations ofthese gadgets which are usually well developed, sophisticated, electronic equipment In thisbook therefore, the basics of instrumental process control is introduced the details of whichbelong to the province of instrument engineers

This book is written after several years on plant operation and teaching The book is biasedtowards practical aspects of mineral processing It is expected to be of use to plantmetallurgist, mineral processors, chemical engineers and electronics engineers who areengaged in the beneficiation of minerals It is pitched at a level that serves as an introduction

to the subject to graduate students taking a course in mineral processing and extractivemetallurgy For a better understanding of the subject solved examples are cited and typicalproblems are set Most problems may be solved by hand-held calculators However mostplants are now equipped with reasonable numbers of computers hence solution to problemsare relatively simple with the help of spreadsheets

The authors are grateful for the help received from numerous friends active in the field ofmineral processing who have discussed the book from time to time Particular thanks are due

to Dr Lutz Elber and Dr H Eren who painfully went through the chapter on process control.Authors are also grateful for permission received from various publishers who own materialthat we have used and acknowledged in the text And lastly and more importantly to ourrespective families who have helped in various ways and being patient and co-operative.

A.Gupta and D.S.Yan

Perth, Australia, January 2006

A general convention used in this text is to use a subscript to describe the state of the quantity, e.g Sfor solid, L for liquid, A for air, SL or P for slurry or pulp, M for mass and V for volume A subscript

in brackets generally refers to the stream, e.g (O) for overflow, (U) for underflow, (F) for feed, (C) forconcentrate and (T) for tailing There are a number of additions to this convention which are listedbelow

b Rosin-Rammler distribution parameter

By breakage distribution function

c a constant

C a constant

C circulation ratio or load

C concentration, (mass solid/volume of slurry) kg/m3

Ct concentration at time t (mass of solid/volume of slurry) kg/m3

CS(c) concentration of solid (C = concentrate, F = feed, T = tail

f = froth, P = pulp)

Cu, CF concentrations of the underflow and feed respectively, kg/m3

(mass of solid/volume of slurry)

CF correction factor

CI confidence interval

CR confidence range

CS(u) solids concentration in the underflow (O = overflow,F = feed) %

concentration by mass of solids in the feed %

particle size, diameter

Sauter mean diameter

discharge mass ratio (liquid/solid)

displacement, distance, diameter

partition coefficient of size i = recovery of size i in the U/F

corrected partition coefficient

energy

corrected partition coefficient

energy of rebound

specific grinding energy

efficiency based on oversize

Ecart probability, probable error of separation

efficiency based on undersize

total energy

a constant

ball wear rate

ball load-power function

suspensoid factor

function relating to the order of kinetics for pulp and froth

mass fraction of size i in the circuit feed

feed size

floats at SG

froth stability factor

feed mass ratio (liquid/solid)

settling factor

% kg/m3-mmmmicrons

cm, mmmmmmmmmm-m-mmmm -kWh-WhkWh/t -kW-kg/h cm, -

80% passing size of feed

Rowland ball size factor

correction factor for extra fineness of grind

correction factor for oversized feed

correction factor for low reduction ratio

mass flow rate

Bond slurry or slump factor

gravitational constant (9.81)

grade (assay)

net grams of undersize per revolution

grinding parameter of circulating load

height of ball charge

height of the start of the critical zone in sedimentation

height of the clarification zone (overflow)

height of rest

hindered settling factor

mudline height at the underflow concentration

height after infinite time

impact crushing strength

imperfection

fraction of mill volume occupied by bulk ball charge

fraction of mill volume in cylindrical section occupied by balls

and coarse ore

superficial gas velocity

fraction of mill volume occupied by bulk rock charge

fraction of mill volume filled by the pulp/slurry

constant

rate constant for air removal via froth and tailings respectively

rate constant for fast and slow component respectively

comminution coefficient of fraction coarser that ith screen

screening rate constant, crowded condition, normal and half size

screening rate constant, separated condition, normal and half size

-kg/s,t/h

-m/s2

%, g/t, ppmg/rev

J-m-

-m, cm

mm

mmmmm-mm

kg.m/mm

m/s

-min"1

-t/h/m2

m1 kW

length of cylindrical and cone sections

Nordberg loading factor

minimum and maximum crusher set

crusher throw

length of vortex finder

length from end of vortex finder to apex of a cyclone

moisture (wet mass/dry mass)

mineralogical factor

mass fraction of undersize in the feed

mass fraction of makeup balls of size k

mass fraction of undersize in the oversize

cumulative mass fraction of balls less than size r

mass rate of ball replacement per unit mass of balls

mass fraction of undersize in the undersize

mass of size i in the underflow (F = feed)

mass

mass

mass/mass fraction of i* increment

cumulative mass fraction retained on i* screen at zero time

mass percent of the i* size fraction and j * density fraction

Nordberg mill factor

minimum mass of sample required

mass of rock

mass fraction of rock to total charge (rock + water)

cumulative mass fraction of balls of size r in the charge

mass of striking pendulum

mass of solid

mmmmmm-mmmm-kg/m3 -kg/h.t-kg

g

kg,tkg,t

%kgkgt/hkgtkgkg,t-kg,tkg-kgke,t sec), SOD n ^5 of solid feed, concentrate and tailing respectively kg, t

mass of solid in froth

MSK mass of sinks kg, t

AM(t) mass of top size particle kg, t

Mj mass of new feed g

Mw mass of water kg, t

n number of increments, measurements

order of rate equation

cumulative number fraction of balls of size less than r

mass fraction of size i in the overflow

binomial probability of being selected in a sample

mass fraction of size i in the new feed

probability of adherence, collision, emergence, froth recovery

power of the conical part of a mill

power for the cylindrical part of a mill

particle distribution factor

proportion of gangue particles

proportion of particles in the i size and j * density fractions

relative mill power

particle shape factor

power at the mill shaft

potential energy

pressure drop

alternate binomial probability = 1 — p

capacity

makeup ball addition rate

basic feed rate (capacity)

tonnage of oversize material

capacity of the underflow

flowrate of solids by mass in the overflow (U = U/F, F = feed)

mass flow of solid in concentrate

capacity, of feed slurry by mass

flowrate by volume in concentrate, tailing and feed respectively

capacity (flowrate) of liquid by volume in the overflow

-min'1-

-nf1g"1 -micronsmicrons-Pa-W kWkW -kWkWkWs kWkWkPa-t/hkg/dayt/h/mt/ht/ht/ht/ht/h

m3/hnrVh(U=underflow, F=feed)

QVOP(U) flowrate by volume of entrained overflow pulp in the U/F

QVOL(U) flowrate by volume of entrained overflow liquid in the U/F m3/h

Qvos(u) flowrate by volume of entrained overflow solids in the U/F m3/h

Qvs(O) flowrate by volume of solids in the overflow (U = U/F, F = feed)

Qv(f) flowrate by volume in the froth

Qv(O) flowrate by volume of overflow (pulp) (U = underflow)

Qw ball wear rate

r0 fraction of test screen oversize

r ball radius

r ratio of rate constants = ICA /(kA+kA)

ri, r2 radius within the conical section of a mill

R radius

R recovery

R reduction ratio

Ri,R2,R3 Dietrich coefficients

R the mean radial position of the active part of the charge

R' fractional recovery, with respect to the feed to the first cell

R ' mass of test screen oversize after grinding

Re radius of cone at a distance Lj from cylindrical section

ReA, Rec Reynolds number in the apex and cone section respectively

R eP particle Reynolds number

R F froth recovery factor

Ri radial distance to the inner surface of the active charge

Ro mass of test screen oversize before grinding

R P radial distance of particle from the centre of a mill

RRO optimum reduction ratio

R T radius at the mill trunnion

Rv recovery of feed volume to the underflow

Roo recovery at infinite time

S speed

S sinks at SG

S surface area

SB surface area of ball

SB bubble surface area flux

S; breakage rate function

SF Nordberg speed factor

to detention or residence time

tR effective residence time

tu time for all solids to settle past a layer of concentration C

tio size that is one tenth the size of original particle

t A mean time taken for active part or charge to travel from

the toe to the shoulder

m

m

mgmm

m/s

m2

m2s"1min"1m

h, min, sh

s

hmms

mean time for free fall from the shoulder to the toe s

mass fraction of size i in the underflow

fraction of void space between balls at rest, filled by rock

fraction of the interstitial voids between the balls and rock charge

-in a SAG mill occupied by slurry of smaller particles

volume fraction of solids in the overflow, (U=underflow, F=feed)

-volume fraction of solids finer than the d50 in the feed (Vd5o/VS(F))

dimensionless parameter

volume of solids finer than the dso in the feed

percent of mill volume occupied by balls %volume dilution in the feed = VL(F/VS(F)

volume of liquid in the feed, (U=underflow, F=feed) m3

volume dilution in the overflow = VL(O/VS(O> or QVL(O/QVS(O)

percent of mill volume occupied by rock %volume of solids in the feed, (U = underflow, O = overflow) mterminal velocity m/sunknown true value

velocity of block pendulum before impact m/svelocity of block pendulum after impact m/svelocity of striking pendulum before impact m/svelocity of striking pendulum after impact m/sdistribution variance

width mdimensionless parameter

effective width mBond Work Index kWh/tBond Work Index, laboratory test kWh/toperating work index kWh/tcorrected operating work index kWh/twater split = QML(O)/QML<F)

deviation from the true assay

geometric mean of size interval micronsRosin-Rammler size parameter micronsSample mean

i* measurement

deviation from standard unit

fractional average mineral content

Lynch efficiency parameter

angle

toe and shoulder angles of the charge

the slurry toe angle

function of charge position and mill speed

volume fraction of active part of the charge to the total charge

surface energy, surface tension, interfacial tension

coefficient of restitution

void fraction

a ball wear parameter

a ball wear parameter, wear distance per unit time

porosity of a ball bed

ratio of experimental critical speed to theoretical critical speed

fraction with the slow rate constant

fraction of critical speed

settling or sedimentation flux

density of the mineral and the gangue respectively

standard deviation (where o2 = var(x))

statistical error in assay

standard deviation of a primary increment

standard deviation on a mass basis

standard deviation of the proportion of particles in a sample

standard deviation of preparation and assay

statistical error during sampling

total error

nominal residence time

angle

-radiansradiansradians N/m -kg/m2/skg/m2/skg/m2/s-kg/m3, t/m3kg/m3,-t/m3kg/m3kg/m3kg/m3kg/m3kg/m3kg/m3kg/m3kg/m3kg/m3, t/m3-

-sradians,degree

-viscosity

velocity

critical speed

mNm, Pa.sm/srpm

velocity across a screen

normalised tangential velocity = VR/VT

overflow rate

ideal overflow rate

tangential velocity at distance Rp

rise velocity

settling velocity

initial settling velocity

settling velocity at time t

tangential velocity at the inside liner surface

terminal velocity

rotational speed, angular velocity

mean rotational speed

rotational speed of a particle at distance Rp

a milling parameter = function of volumetric filling of mill

percent solids

m/minrpmm/sm/srpmm/sm/sm/sm/srpmm/ss"1, rpm, Hzrpms'1, rpm-

%

1 INTRODUCTION

A processing plant costs many millions of dollars to build and operate The success of thisexpenditure relies on the assays of a few small samples Decisions affecting millions ofdollars are made on the basis of a small fraction of the bulk of the ore body It is thereforevery important that this small fraction is as representative as possible of the bulk material.Special care needs to be taken in any sampling regime and a considerable effort in statisticalanalysis and sampling theory has gone into quantifying the procedures and precautions to betaken

The final sampling regime adopted however is a compromise between what theory tells usshould be done and the cost and difficulty of achieving this in practice

1.1 Statistical Terminology

A measurement is considered to be accurate if the difference between the measured valueand the true value falls within an acceptable margin In most cases however the true value ofthe assay is unknown so the confidence we have in the accuracy of the measured value is alsounknown We have to rely on statistical theory to minimise the systematic errors to increaseour confidence in the measured value

Checks can be put in place to differentiate between random variations and systematic errors

as the cause of potential differences A random error (or variation) on average, over a period

of time, tend to zero whereas integrated systematic errors result in a net positive or negative

value (see Fig 1.1)

The bias is the difference between the true value and the average of a number of

experimental values and hence is the same as the systematic error The variance betweenrepeated samples is a measure of precision or reproducibility The difference between themean of a series of repeat samples and the true value is a measure of accuracy (Fig 1.2)

A series of measurements can be precise but may not adequately represent the true value.Calibration procedures and check programs determine accuracy and repeat orreplicate/duplicate measurements determine precision If there is no bias in the samplingregime, the precision will be the same as the accuracy Normal test results show that assaysdiffer from sample to sample For unbiased sampling procedures, these assay differences arenot due to any procedural errors Rather, the term "random variations" more suitablydescribes the variability between primary sample increments within each sampling campaign.Random variations are an intrinsic characteristic of a random process whereas a systematicerror or bias is a statistically significant difference between a measurement, or the mean of aseries of measurements, and the unknown true value (Fig 1.1) Applied statistics plays animportant role in defining the difference between random variations and systematic errors and

in quantifying both

True value, v

Systematic error

SAMPLE

Fig 1.1 Representation of a random and systematic error

1.1.1 Mean

The most important parameter for a population is its average value In sampling and weighing

the arithmetic mean and the weighted mean are most often used Other measures for the average value of a series of measurements are the harmonic mean, and the geometric mean.

Mode and median are measures of the central value of a distribution The mode forms the peak

of the frequency distribution, while the median divides the total number of measurements into

two equal sets of data If the frequency distribution is symmetrical, then its mean, mode andmedian coincide as shown in Figs 1.3 and 1.4

0 0.1 0.2 0.3 0.4 0.5 0.6

ASSAY

0 0.1 0.2 0.3 0.4 0.5 0.6

ASSAY

mean, mode, median

-0

mean, mode, median

10

Fig 1.3 Normal distribution

For a binomial sampling unit of mixed particles the average percentage of mineral A iscalculated by adding up all measurements, and by dividing their sum by the number ofmeasurements in each series

-0

Fig 1.4 Asymmetrical distribution

The weighted percentage is calculated, either from the total number of particles in each series,

or by multiplying each incremental percentage with the mass in each correspondingincrement, and by dividing the sum of all products by the total mass for each series However,the small error that is introduced by calculating the arithmetic mean rather than the weightedaverage, is well within the precision of this sampling regime The following formula is used tocalculate the weighted average for a sample that consists of n primary increments:

( 2

M

where AM; = mass of Ith increment

M = mass of gross sample

Due to random variations in the mass of each primary increment the weighted average is abetter estimate of v, the unknown true value, than the arithmetic mean

1,1.2 Variance

The variance, and its derived parameters such as the standard deviation and the coefficient of

variation, are the most important measures for variability between test results.

The term range may be used as a measure of variability.

Example 1.1

Consider a binary mixture of quartz and hematite particles with approximately 10% hematite.Samples are taken and the number of hematite particles are counted to obtain the percentage

of hematite in the sample Table 1.1 gives the result often samples For a binomial sampling

unit the range is (maximum value - minimum value) = 12.6 - 5.7 = 6.9%.

SampleHematite11191075111281214

Total116151109105889810394110127

%Hematite9.512.69.26.75.711.211.78.510.911.0

Fig 1.5 Range of experimental values.

If each series of measurements is placed in ascending order, then the range is numerically

equal to xn - xi so that the range does not include information in increments X2, X3, , xn_i

For a series of three or more measurements the range becomes progressively less efficient as a

measure for variability as indicated in Fig 1.5

For two samples, the range is the only measure for precision but this is not sufficient to

estimate the precision of a measurement process The precision of a measurement process

requires the mean of absolute values of a set of ranges calculated from a series of four or more

simultaneous duplicates This is the variance

The classical formula for the calculation of the variance is:

11 19 10 7 5 11 12 8 12 14

Arithmetic mean Variance Standard deviation Coeff of variation

Total 116 151 109 105 88 98 103 94 110 127 SUM

% Hematite 9.48 12.58 9.17 6.67 5.68 11.22 11.65 8.51 10.91 11.02 96.89 9.69 5.0141 2.2392 23.1

Quartz 504 350 597 394 428 438 508 533 438 490

<

(

Large increment Hematite

53 45 56 52 43 52 55 56 50 49

Total 557 395 653 446 471 490 563 589 488 539 SUM Arithmetic mean Variance Standard deviation Coeff of variation

%Hematite 9.52 11.39 8.58 11.66 9.13 10.61 9.77 9.51 10.25 9.09 99.51 9.95 1.1264 1.0613 10.7

The physical appearance of a sample that consists of fifty primary increments of 5 kg each

is similar to that of a sample containing five increments of 50 kg, or to that of 250 kg of a bulksolid However, the difference in intrinsic precision (as indicated by the variance) may bedramatic, particularly if the variability within the sampling unit is high

In practical applications of sampling bulk solids we compromise by collecting andmeasuring unknown parameters on gross samples, and by reporting x, the sample mean as thebest estimate for v, the unknown true value If all increments are contained in a single grosssample, we have no information to estimate the precision of this sampling regime If we want

to know more about the precision of samples, systems and procedures, it is essential thatduplicate or replicate measurements be made, from time to time, to determine the coefficient

of variation for each step in the chain of measurement procedures

1.1.3 Confidence Intervals

Other convenient measures for precision are confidence intervals (CI) and confidence ranges(CR) 95 % confidence intervals and 95 % confidence ranges may be used, although ifconcern over sampling precision is high, then 99% or 99.9% confidence limits must beconsidered

That is, if we repeat a particular experiment 100 times, then 95 times out of 100 the resultswould fall within a certain bound about the mean and this bound is the 95 % confidenceinterval Similarly, confidence limits of 99 % and 99.9 % mean that 99 times out of 100, or

999 times out of 1000 measurements would fall within a specified or known range

In the draft Australian Standard, DR00223, for estimating the sampling precision in sampling

of particulate materials, a confidence interval of 68% is chosen [1]

Fortunately, we don't need to repeat a measurement one-hundred times if we want todetermine its 95% confidence interval, either for individual measurements or for their mean.Applied statistics provides techniques for the calculation of confidence intervals from a

limited number of experiments The variance between increments, or between measurements,

is the essential parameter

The most reliable estimate of a2 is var(x), the variance of the sample The reliability of thisestimate for a2 can be improved by collecting, preparing and measuring more primaryincrements, or by repeating a series of limited experiments on the same sampling unit

The 95 % confidence interval for a normal distribution is equal to ±1.96 a from thedistribution mean In practice, we often use the factor 2 instead of 1.96, to simplifycalculations and precision statements The 68 % confidence interval is equal to ±0.99 a, forinfinite degrees of freedom and if the number of replicate results exceeds 8 then a factor of 1.0

is an acceptable approximation

1.2 Mineral particles differing in size - Gy's method

Representing large bodies of minerals truly and accurately by a small sample that can behandled in a laboratory is a difficult task The difficulties arise chiefly in ascertaining a propersample size and in determining the degree of accuracy with which the sample represents thebulk sample

In each case the accuracy of the final sample would depend on the mathematical probabilitywith which the sample represents the bulk material The probability of true representationincreases when incremental samples are taken while collecting from a stream, like a conveyorbelt for solids and off pipes for liquids or slurries

Several methods have been put forward to increase the probability of adequatelyrepresenting the bulk minerals [2-5] One such method involving both the size and accuracy of

a sample taken for assay has been developed by Gy and is widely used [6, 7] Gy introduced amodel based on equiprobable sample spaces and proposed that if:

= dimension of the largest particle

MMIN ~ minimum mass of sample required

o2 = variance of allowable sampling error in an assay (in the case of a normal

distribution this equals the standard deviation)

then:

MM I N = 2 U-5)

a

K is usually referred to as the sampling constant (kg/m3)

In mineralogical sampling the dimension of the largest piece (dMAx) can be taken as the

screen aperture through which 90-95 % of the material passes As ± 2a represents the

probability of events when 95 out of 100 assays would be within the true assay value, 2a isthe acceptable probability value of the sample The sampling constant K is considered to be afunction of the material characteristics and is expressed by:

K = P S P D P L H I (1.6)

where Ps = particle shape factor (usually taken as 0.5 for spherical particles, 0.2 for gold

ores)

PD = particle distribution factor (usually in the range 0.20 - 0.75 with higher valuesfor narrower size distributions, usually taken as 0.25 and 0.50 when thematerial is closely sized)

PL = liberation factor (0 for homogeneous (unliberated) materials, 1 forheterogeneous (liberated) and see Table 1.3 for intermediate material)

m = mineralogical factor

The mineralogical factor, m, has been defined as:

(1-7)

where a is the fractional average mineral content and

PM and po the specific gravity of the mineral and the gangue respectively

The liberation factor, PL, is related to the top size, dMAX and to the liberation size, dL of themineral in the sample space It can be determined using Table 1.3 In practice, PL is seldomless than 0.1 and if the liberation size is unknown then it is safe to take PL as 1.0

Table 1.3

Liberation factor as a function of liberation size [9]

Top Size/Liberation Size, <1 U4 4^10 10-40 40-100 100-400 >400

Lib Factor (PL)) 1.0 0.8 0.4 0.2 0.1 0.05 0.02

When a large amount of sample has been collected it has to be split by a suitable methodsuch as riffling At each stage of subdivision, samples have to be collected, assayed andstatistical errors determined, hi such cases the statistical error for the total sample will be thesum of the statistical errors during sampling (as) and the statistical error in assay (aA), SO thatthe total variance (ax2) will be:

When the sample is almost an infinite lot and where the proportion of mineral particles hasbeen mixed with gangue and the particles are large enough to be counted, it may be easier toadopt the following procedure for determining ap

Let PM = proportion of mineral particles

PG = proportion of gangue particles

N = number of particles

Then the standard deviation of the proportion of mineral particles in the sample, Gp; willbe:

aP = (1.9)

The standard deviation on a mass basis (CTM) can be written in terms of the percent mineral

in the whole sample provided the densities (p) are known Thus if pM and PG are the densities

of the mineral and gangue, then the mass percent of mineral in the entire sample, consisting ofmineral and gangue (the assay), will be:

5600 kg/m3 and the density of the gangue is 2500 kg/m3

Solution

From the data, dMAx = 2.5 cm

Since the confidence interval required is equal to ± 0.1 % of a 9% assay,

2a = 0.1/9 or a = 0.011/2 = 0.00555

Again from the data dMAx/dL = 25 / 0.075 = 333.33, hence from Table 1.3, PL = 0.05

As the ore contains chalcocite (Cu2S) assaying about 9% Cu, it can be considered to contain[159.2/(63.56 x 2)] x 9 = 11.3% of Cu2S Thus a = 0.113

i.e The copper content of CU2S is given by the ratio of atomic masses;

(63.56 x 2) x 100 = 127.1 x 100 = 79.8% Cu (atomic mass; Cu = 63.56,(63.56x2)+ 32.1 159.2 S = 32.1)

The chalcocite content of the ore is then given by:

Thus the minimum sample size should be 131 kg

Note the importance of not rounding off the numbers until the final result If cr is rounded

to 0.0055 then MMIN = 133 kg and if a is rounded to 0.0056 then MMIN = 128 kg

Example 1.4

A composite sample of galena and quartz was to be sampled such that the assay would bewithin 0.20% of the true assay, of say 5.5 %, with a probability of 0.99, ie the sample assay

would be 5.5 % + 0.20 %, 99 times out of 100 Given that the densities of galena and quartz

were 7400 kg/m3 and 2600 kg/m3 respectively and the average particle size was 12.5 mm with

a mass of 3.07g, determine the size of the sample that would represent the composite

To determine N it is necessary to find aj

Deviation fromStandard Unit, X

X=X/CT

2.1702.3262.5763.2913.8904.4174.892

Hence the mass of sample necessary to give an assay within the range 5.5 ± 0.20 %, 99 timesout of 100 would be =228,201.9x3.07 = 700,579 g * 701 kg

13 Mineral particles of different density

Where variations in density of individual particles and their composition occur, thefollowing considerations may be adopted to provide the sample size [9]:

1 Divide the material into n density fractions, pi, p2, P3, pn varying from purely onemineral to the other, e.g copper and quartz in a copper ore, and consider n size fractions,

di, d2, d3 dn

2 Consider the mass percent of the i size fraction and j density fraction as MM and

3 Consider Ay and Py as the assay and the proportion of particles in the i size and j *density fractions

For sampling a mixture of two components (mineral and gangue) the proportion ofparticles in the ij fraction would be;

For a multi-component system the principles developed in the above Eqs 1.13 to 1.15 may

be extended Their solution can be achieved easily by a computer The general equation foroverall sample assay is:

100.058.90.0

4.8

3.62.65

The lot has to be sampled so that it would assay ± 0.15 % of the true assay having 5% with aprobability of 99%

Solution

Stepl

As the stipulated probability is 0.99 Table 1.4 may be used

(4)100.058.90.0

SG

(5)4.83.62.65

Av

Diad;

(6)111

0)

111

d / x p

(7)x (5)(8)4.83.62.65

Mass %

di3 p(3)/(8)(9)0.522.76.0329.23

Proportion

by No.(9)/£9(10)0.0170.7760.2071.0Step 3

Withkri

Now follow Example 1.4 to determine the sample size

With known P and 0, N can be estimated using the equation a 2 =———

1.4 Incremental Sampling

Theoretically, unbiased gross samples can be obtained by collecting a sufficiently largenumber of single particles from a sampling unit If each particle has a finite chance of beingselected for a gross sample, and if this probability is only a function of its size, then thiscollection of particles will constitute an unbiased probability sample

In the practice of sampling bulk solids, such a sample collection scheme is highlyimpractical Therefore, we collect groups of particles for primary increments

Sampling experiments that are based on the collection of a series of small and largeincrements from a large set of particles, demonstrates that the precision of a single incrementsampling regime is a function of the number of particles in a primary increment As a matter

of fact, the formula for the variance of a binomial sampling unit, var(x) = N.p.q, shows thefundamental relationship between probabilities (p, q), the total number of particles in anincrement (N), and thus its mass, and the precision for the parameter of interest Hence, theprecision of a single increment is essentially determined by its mass

The binomial sampling experiment in which p = 0.095, and q = 0.905 (q = alternatebinomial probability = 1 - p), can be used to set up a table in which the precision of a primaryincrement is given as a function of the average number of particles in an increment In Table

1.6, the 95 % confidence ranges (CR) that are calculated from corresponding 95 % confidence

intervals (CI) for the expected number of value particles in each increment, together with theircoefficients of variation, are listed

Since these ranges are based on properties of the binomial distribution, their values areobviously independent of particle size For a bulk solid with a certain density, its top sizedetermines the mass of a primary increment Top size is defined as the 95% passing sieve size

High29.015.011.4

9.89.6

9.53

CV in %98.031.0

9.83.11.00.3

The mass of an increment must be such that it is large enough to include the large particlesand particles present in the sample should be in the same proportions as in the lot beingsampled The minimum mass of the increment is therefore dependent on the size of theparticles being sampled

The top size of a bulk solid is a measure of length The mass of a particle is a function ofits volume and specific gravity and ultimately a function of its mean diameter or length Themass of a primary increment then can be defined in terms of the number and mass of particlesfrom the top size range If no other information is available, then an acceptable rule of thumb

is to collect primary increments with a mass equal to 1,000 times the mass of a top sizeparticle

AM = 1,000-AM(t) (1.17)where AM = mass of primary increment in kg

AM(t) = mass of top size particle in kg

Experience and theory are embodied in a number of national and international standards onsampling of particulate materials where the sampling regimes are defined in terms of the totalnumber of increments, and the average mass of a primary increment

It is generally accepted that a primary increment should contain no less than one-thousand(1,000) particles

In the standard on sampling of iron ore (ISO TCI02), the minimum mass for primaryincrements is specified in relation to the top size, and in Table 1.7 these recommendations aretabulated and compared to calculations for hard coal

The minimum mass for a primary increment should preferably be defined in terms of thevolume of a particle from the top size range

Once the mass for a primary increment is selected either in accordance with applicablestandards or on the basis of the previous guidelines, or determined by the critical designparameters of a mechanical sampling system, the required number of primary incrementsremains to be defined Too few increments will result in too low a precision while too manywould unnecessarily increase the costs for sampling and preparation

Most standards for bulk solids contain simple formulae to calculate the required number ofincrements for a consignment from a given number for the unit quantity (usually 1,000 t), or

from tables that list the minimum numbers of primary increments as a function of the mass of

a consignment ISO TC 102 for iron ore specifies this number on the basis of the three levels

of variability which presupposes some knowledge of the expected variability

0.30.84122040

ISO 1988Hard Coal Mass in kg

0.60.836915

Table 1.8 is extracted from ISO 1988 on sampling of hard coal, and lists the number ofprimary increments for a 1,000 t unit quantity

Table 1.8

Number of increments to attain a precision of ±0.1 of the true ash [11]

Condition of Conveyors Wagons Seagoing Stock Piles

Coal Falling Streams Barges Vessels

Cleaned 16 24 32 32

Uncleaned 32 48 64 64

Table 1.9 gives the mass and numbers of primary increments for hard coal, as specified inASTM D 2234, for consignments of up to 1,000 t, and for an estimated precision of ±10% ofthe ash content

For consignments larger than this unit mass of 1,000 tonnes, ASTM D 2234 and ISO 1988use the same formula to calculate the required number of primary increments:

(1.18)

where n = required number of increments

S M = mass of consignment in tonnes

n(t) = tabulated number of increments

1UncleanedMass in kg Number1

37

353535

The overall standard deviation of sampling, sample preparation and assay is a function ofthe variability of the particulate material, the number and mass of the increments and therandom errors associated with sample preparation and assay It can be expressed as:

^ (1.19)

where <7L = standard deviation of a primary increment (from Eq (1.3))

GPA = standard deviation of preparation and assay

In a well-balanced sampling regime the variance of sampling and the variance ofpreparation should be of the same order of magnitude

Dividing a consignment of bulk solids into lots, collecting, preparing and assaying samplesfrom each lot, and reporting composite assays for the consignment, impacts significantly onthe precision of the final result The variance of preparation and analysis may easily become alimiting factor for the precision of a sampling regime

If we solve n, the number of primary increments, from the simplified formula for asampling regime of only one gross sample per consignment and a single measurement in thefinal analysis sample, it follows that [10]:

n = var(d) + var(c)/AM (1.20)

var(t) - var(pa)where var(c) = composition variance

var(d) = distribution variance

var(pa) = preparation and analysis variance

var(t) = total variance

The denominator of this formula may easily become a limiting factor for the total precision

of sampling regimes since it shows that:

n —> oo for var(pa) —> var(t)

Logically, the total variance cannot be smaller than the variance of preparation andanalysis.

Example 1.6

Suppose that we have a consignment of 1,0001 of iron ore with an assay of 65.0% Fe, and that

we want to determine the assay with a precision of ±10%

A precision of ±10% of 65.0% Fe is equivalent to ±6.5% of this iron content, and results in

a standard deviation of, 6.5/2 = 3.25 for a total variance of 3.252 = 10.56 Substitution ofthis total variance of 10.56, composition and distribution variances of say 1.66 and 0.71, and

an average mass of 2 kg for primary increments, in the formula for n, Eq (1.20), results in:

If a greater iron assay precision of ±0.25 % is required then the denominator in the aboveequation becomes negative and it is not possible to obtain that degree of precision samplingfrom a single lot The consignment must be divided into sub-lots in order to determine theassay with this precision Table 1.10 shows a number of possible combinations

Lots11483264

Increments8903601551430860

These results show that the assay of this particular type of ore cannot be determined with aprecision of ±0.25% at affordable costs because the variance of preparation and analysisrapidly becomes a limiting factor for the total precision

In terms of standard deviations, the number of increments may be calculated from [1]:

When it is impossible to achieve the desired precision by testing a single gross sample from

a lot (eg The number of increments is impossibly large) then it is necessary to divide the lotinto a number of sub-lots, ns Then;

P = Jf e^dp (1.23)

where x = deviation from the true assay.

Now if A is the average assay and A^ A2 A3 AN , are the individual assays of the nsample increments then:

n

and the standard deviation of each increment sample will be:

The standard deviation of the entire sample, o, is related to a L by the relation

22.5

32.2

43.2

52.5

62.5

72.3

82.9

93.3

102.1

The final sample should assay within ± 5% of the true value with a confidence of 99%.Estimate the number of increments

Stepl

Average assay = (Ai + A2 + A 3+ +A]0 )/10 = 2.65

Standard Deviation, cL = 0.400, from Eq (1.25)

Hence 61 increments have to be taken whose weight would be 61 x 2 = 122 kg

1.5 Continuous Sampling of Streams

When mechanical samplers are employed, the samplers are designed to cut into andwithdraw from a stream of travelling material at a predetermined frequency and speed Cutterscould operate linearly or rotate within the stream to be sampled

The rule of thumb for cutter openings is:

Normal Opening s 3 x largest particle size (dry stream)

Minimum Opening = 70 mm for fine particles

= 60 mm for fine slurry

SampleFig 1.6 Linear traversing sampler

It is expected that the opening sizes indicated here would prevent the formation of bridgesacross the openings in samplers Also, the wider the cutter opening the greater could be thespeed of traversing the cutter

The stroke length is adjusted to cover the width of the conveyer belt where the streamconsists of dry solids, or the width of the stream where liquids or slurries have to be sampled.The amount of sample (M) cut from any stream by a linear cutter is given by:

M = (Feed rate of stream') (Cutter opening')

(Cutter speed)

(1.29)

For a feed rate expressed as kg/s, a cutter speed in m/s and a cutter opening in m, the mass

M, is in kilograms Eq (1.29) can be expressed in terms of volume and can be written as:

V = (Vol rate of flow ) (Cutter opening)

where the units for volume rate of flow is m /s, cutter opening, m and cutter speed, m/s

1.5.2 Rotary Arc Cutter

The rotary type of cutters (Fig 1.7) allows samples to be collected or passed through asindicated in the segments For unbiased sampling, the edge of the cutters equals the radius ofthe circle forming the arc The effective radius, R, swept out by the cutter at any point, interms of the distance d from the centre of rotation of the cutter to the stream to be sampledalong the centreline is given by:

R = (1.31)cos 6

where 6 is the angle between the radius of the cutter and the centreline of the solids stream.The cutter opening changes as the cutter moves through the stream, then by simple geometry,the effective cutter opening, dcutter, at any point is approximately given by:

Cutter opening, dcutter = (1-32)

To determine the quantity of sample taken by the arc cutters, it is necessary to know the cutterangle, a This may be supplied by the manufacturer for a predetermined position of the cutter

or it can be calculated from the following relation:

^ » i Cutter Arc length „ „ ,, _„

Cutter Angle, a = - — x 360 (1.35)

27trThe mass of sample of solids, M, recovered per rotation can be computed from a known flowrate of mineral, Mp, by the expression:

M = ^ (1.36)

360 co

Mp is expressed in kg/s, and a> as rpm

Fig 1.7 Continuous rotary arc sampler

The volume, V, of liquid or slurry sampled can similarly be written as:

Mass of sample cut in 10 mins = 0.194 x 20 / 6 = 0.647 kg

1.6 Sampling Ores of Precious Metals

Precious metal deposits sometimes contain very low concentrations of discrete metallicparticles as in the case of gold deposits Extra care is therefore needed to collect arepresentative sample If it is assumed that the ore comprises of free particles of uniform sizethen, N, the number of particles required is given by :

( r

N = 0.45 - ^ (1.38)

)

where G = grade of the ore, expressed as volume fraction, and

as = probable error in sampling, expressed as volume fraction

The number of particles per gram, N', is related to the material density (ps) and particlesize by an empirical relation as [12]:

where d = limiting screen aperture This empirical equation covers a range of

particle sizes and shapes

The mass of sample, M to be taken can be given as:

143 kt1.14 kt

143 t

1.14 t

143 kg1.14 kg

143 g1.14g

H 4 g

14.3 g0.11 g

;/t10

1.43 kt

11.4 t

1.43 t11.4 kg1.43 kg

11.4g

1.43 g0.01 g

30

4771

3.81 t

477 kg 3.81 kg

477 g3.81 g0.48 g3.81 mg

In developing the table, a 90% confidence level and a relative 15% error has been assumed

NOTE: A usual practice is to crush 3.5 kg ore to 95% passing 75 micron sieve size and then taking a 200 gram sample for assay.

1.7 Sampling Nomographs

When sampling to determine the grade of a large body of material, the sample of 94 kg (forexample) must be reduced to a few grams for chemical analysis To do this and still maintainthe sampling accuracy, Eq (1.5) suggests that the size of the particles should be reduced toallow a reduction in the sample mass To optimise a sampling regime, a sampling nomograph

Taking the logs of both sides of Eq (1.41) gives:

therefore, a nomograph plotted on a log-log scale will have a slope of-1, as shown in Fig 1.8.The use of the figure is illustrated in the following example A sample of 94 kg (94000g)

is crushed to minus 5 mm and a 30 g sub sample is split out for gold assay, using a riffle Theline on Fig 1.8 shows that as the sample mass is reduced, the error associated with a particlesize of 5 mm increases until it exceeds the minimum error specified for this sampling regimefor any sample mass below about 200g

To stay within the specified sampling accuracy, a sampling regime similar to Fig 1.9should be followed

line of specifiedminimum sampling accuracy

line ofspecifiedminimumsamplingaccuracy

Fig 1.9 A more appropriate sampling regime

C

σ2 = 10-2

Linesrepresentdecreasingvariance

The 94 kg of minus 5 mm material is sub divided to a sub sample mass of 1 OOOg in the firststage This 1 kg of sample is then pulverised to minus 0.5 mm and divided again to therequired sample mass of 30 g Following this sampling regime will maintain the samplingprocedure below the required sampling error

To derive a sampling nomograph, Eq (1.5) is used by calculating the sampling constant Kfrom known data

An alternative graphical method is to plot sample mass (M) vs top particle size (dMAx)showing the alternation of size and mass reductions required to maintain an acceptablesampling variance, governed by the Gy formula (Fig 1.10)

This can be achieved in the following manner:

From Gy's formula; Kd

and log MM I N = (log K - 2 log o) + 3 log dM A X (1.43)Thus for a given value of K and sampling accuracy, this relationship will yield a straightline of slope 3 when plotted on log-log graph paper (Fig 1.10) All points on this line, (lineBDF, Fig 1.10), have a constant fundamental variance, o The line represents a "safety line"and splits the graph into two areas

Top particle size (cm)

Fig 1.10 Alternative graphical sample regimes

10

To the left of this line, the sample mass is greater than the minimum required, MMINJ andthe fundamental sampling variance is less than c*2.

To the right of the line, the sample mass is less than the minimum and the samplingvariance is high and unacceptable To rectify the situation, either the mass has to be increased

or the particle size, d, has to be decreased to bring the sampling regime back to the left of thesafety line

A family of lines of equal variance can be constructed, each parallel to one another (Fig.1.10) From Fig 1.10, for a sample mass of 50 tonnes at a particle size of 9 mm, and a desiredvariance of 10 or better, the sample can be reduced in mass to 100 kg (A to B) and stillmaintain acceptable accuracy (to the left of the safety line) To reduce the mass further, thesample must first be reduced in particle size, eg to 3 mm (B to C) The sample mass can then

be reduced to 1 g through a sequence of mass reduction and size reduction (C to D to E to F)

1.8 Problems

1.1

Iron ore was sampled before stock piling with a stacker One hundred samples taken from the

stacker-conveyor showed a standard deviation in the iron assay of ± 0.5 % The ore assayed,

on average, 59 % Fe Sieve analysis of the samples showed that the largest size was less than

5 mm and the liberation of Fe was maximum in the size range, -400 + 300 \xm Given that the

specific gravity of the ore was 5.3 and the specific gravity of the gangue was 2.6, estimate theminimum mass of sample required representing the stack

1.2

Run-of-Mine iron ore was conveyed on a conveyor belt and sampled at regular intervals Thesieve analyses and Fe distribution obtained at different sampling frequencies gave thefollowing results:

2654.0

-5+43062.0

3062.2

-4+35064.5

4064.8

-31062.0

4565.5

3566.0

867.0The specific gravity of the ore was 5.5 and that of the gangue material was 2.55

Determine the minimum mass of sample required to represent the Run-of-Mine ore

1.3

A pile of gold tailings was augured to sample the dump A 5 cm diameter drill bit was usedand samples recovered from different depths The recovered samples were collected and aftermixing thoroughly, the composite was crushed, screened between 125 um and 75 jun andanalysed for gold The average of 10 gold analyses indicated a value of 200 +10 ppm gold.The liberation size of the gold was determined as -43 um The specific gravity of the gold

bearing minerals was 4.5 and the associated gangue minerals were 2.68 Estimate theminimum size of the sample

1.4

In a metallurgical test, the quality of feed was monitored Four operators and three similar testequipment were employed Material lost in the feed after each operation in each test was:Equipment No

1

2

3

112.512.812.012.411.813.212.612.213.0

Operator2

13.812.513.412.813.013.912.912.0712.7

312.212.812.113.513.012.213.213.02.8

411.513.112.8413.212.512.812.612.213.2Estimate the variance due to operator and experimental error

1.5

A mechanical sampler was used to sample a stream of iron ore conveyed on a 1.5 m widetravelling conveyor at a rate of 90 m/min and loaded to carry 12 mt/h of ore The samplecutter opening was 20 cm square and was operated at a frequency of 5 cuts per minute Therecovered sample was first crushed to -10 mm and then to -2.5 mm and analysed for Fecontent The liberation size of Fe was -65 urn and the standard deviation of the Fe contentwas ±0.15 after the first crushing and the same after the second crushing If the averageanalysis was 59% Fe, estimate;

1 The mass of sample cut per minute

2 The minimum mass of sample required to represent the Fe level of the ore

1.6

The average assay of a gold sample was 200 ppm Au that varied within 0.5 % of the trueassay.95% of the assays had a probability of 0.99 of the true assay Specific gravity of the goldore was 5.6 and the gangue was 2.54 Determine the size of a crushed gold ore sample thatwas taken

1.7

The mass fractions and distributions of a mixture of sphalerite, chert and middle fractionswere determined and the results tabulated below Compute the size of sample that should betaken such that its assay may be within 0.2% of the true assay, say 5% with a probability of0.99