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Chapter 2. Particle Size Estimation and Distributions
2. INTRODUCTION
The term mineral particle is loosely used in mineral processing. Particles in a mineralprocessing plant are never a single size but consist of many different sizes. The particles arealso many different shapes which makes characterisation of the quantity, size, very difficult.
Unless a particle is spherical or cubic its size determination is never an absolute process.Particles such as (a) and (b) in Fig. 2.1 can uniquely be described by the diameter of a sphere,ds or the length of the side of a cube, do However, the dimension of particle (c) is difficult tocharacterise as either the maximum, dviAx or the minimum dMiN or some dimension in betweencould be used to represent the particle size.
Jmax
mm
a b
Fig. 2.1. Characterisation of particle size.
2.1. Methods of size estimationThere are many ways the size of an irregular particle could be characterised, with none
being absolutely right. For instance, the particle size can be characterized by determining thesize of hole or aperture the particle will just pass through (sieve size), or the time the particletakes to settle in a fluid such as water and express the particle size as the size of a sphere thathas the same settling rate (Stokes diameter). Particle size could also be expressed as thediameter of a circle that has the same cross sectional area or projected area as the particle,when placed either on its most stable position or randomly oriented positions. In threedimensions, the equivalent volume or surface area of a sphere have also been used to describea particle size. Visual examination under the optical microscope or electron microscope or thediffraction of a laser beam have been used to derive statistical dimensions of particles. All ofthese methods do not necessarily give the same measure of size for the same sample ofparticles. Hence it is necessary to mention the measurement method when quoting particlesize.
The particle dimensions measured by optical methods are limited to sizes greater than 100(im. With an electron microscope, particle sizes down to 0.001 urn can be measured. Forgravity sedimentation methods, the particle size should be greater than 1 micron.
33
Fixed direction
equivalent perimeter diameter
equivalent area diameter
Martin’s diameter (divides profile into 2 equal areas)
Feret’s diameter
33
Where individual measurements are not required approximate size ranges can be obtainedby sieving. The closer the successive sieve sizes are, the closer is the approximation to the realsize of the particles held between successive sieves. This information is usually sufficient formost metallurgists and sieving is the most common method of determining size.
2.1.1. Microscopic MethodThe use of a microscope to measure the size of irregularly shaped mineral particles can be atedious and difficult process. A video camera linked to a computer can speed up the process ofsize recognition by the projected outline of a particle. Video systems are used to size largelumps of rock on conveyor belts and even on the back of dump trucks as the load is tippedinto crusher feed bins. The obvious technique is to measure the average distance between twoextremities of the particle and the arithmetic or geometric mean of a number of measurementstaken. The extremities of the distance to be measured is standardised [1] and taken as themean cord length of the projected outline as defined by Martin's or Feret's diameter (Fig. 2.2).Martin's diameter is the length of the line bisecting the image of the particle. The bisectingline is taken parallel to a fixed direction, irrespective of the orientation of the particle (Fig.2.2). Feret's diameter is the mean distance between two tangents on the opposite sides of theapparent outline of the particle. The tangents are perpendicular to an arbitrarily fixeddirection, irrespective of the orientation of the particle.
The relative size of particles may also be indicated by reference circles having an equalprojected area or perimeter, (Fig. 2.2). The diameter of the reference circle is the arithmetic orgeometric mean value of the distances by Martin's or Feret's method. The arithmetic meandiameter measured by either method may be approximated as:
• + < kH _ "MAX + UM1Na A M ~ (2.1)
where duAX anddiameters and dAgiven by:
are the averages of several measurements of Martin's or Feret'sis the arithmetic mean diameter. The geometric mean diameter likewise is
Martin's diameter(divides profile into 2 equal areas)
equivalent areadiameter
equivalent perimeterdiameter
Feret's diameter Fixed direction
Fig. 2.2. Particle size equivalents.
3434
dGM = (di d2 d3 d4... dN)1/N (2.2)
or logdoM = £^ l 0 g d N^ (2.3)
where N is the number of particles measured.
In some cases the average estimations of diameters is expressed as the Mellor's meandiameter (diyuj which is given by the expression:
dML= 0.632 [ (dMAx + dMiN XdwAx2 + dMiN2 )f33 (2.4)
This method gives a slightly higher average than the arithmetic mean size of a particle.Therefore when reporting particle size it is necessary to quote the method used.
2.7.2 Particle Size in terms of Volume and Surface AreaDue to the difficulty in measuring length, sometimes the particle size is described in terms ofits volume or surface area, as a volume mean diameter or a surface mean diameter. In asample consisting of several particles, this concept can only be used if it is assumed that allthe particles have the same shape. Thus if d is the dimension of the particle and pp thedensity, then the mass would be proportional to d3pp.
In a mass of sample, Mp, containing N number of particles of characteristic size dp :
MP = kNdpp s (2.5)
where k is a constant, a function of the shape factor, and ps the density of the solid particles.
Eq. (1.48) may be re-written as:
4 = - M P _ (2.6)kN P s
If all the particles are assumed to have equal dimensions, which is expressed as the meanvolume diameter, dy, then the sum of the volumes of all the particles of size dy is equal to thevolume of particles in the sample mix. Then;
_ _ (2.7)
Hence d v = ^ ^ - (2.8)
Using Eq. (2.6), N may be eliminated from Eq. (2.8). Then on simplifying, the volumemean diameter may be written as:
3535
dv3 = ^ p = '— (2.9)
•yMP V M p
Similarly, the surface mean diameter, ds, has been derived by Coulson and Richardson [2]as:
- 2
Example 2.1.Gold nuggets were examined under a microscope and the following frequency distributionwas obtained:
IntervalSize, mmNumber
10-13250
21-21800
32-4400
44-8120
58-1612
616-324
Assuming the nuggets are of similar shape, determine the size distribution on a mass fractionbasis.
Data: density of gold nugget = 17580 kg/m3
Solution
Step 1Let Mp equal the mass fraction of particles of size dn and specific gravity ps and let Nn equalthe corresponding number of particles in each fraction.
n is the number of size intervals, equal to 6, then:
Mp = kN nd n3p s for n= l -6
k is a constant equivalent to the shape factor. In this example, k is taken as 1
Therefore, the mass fraction for each size interval would be:
N H3 rrn
Mp = — -2J Nn d3 p s
The calculations can now be made as follows to estimate Mp values for different values of n:
3636
n123456
dn, mm0.51.53.06.0
12.024.0
N325018004001201204
0.1253.375
27216172813824
406.256075108002592020736055296
Ps175801758017580175801758017580
4 3 N p s
7.14x10*1.07 xlO8
1.90 x 10s
4.56 xlO8
3.65 x 109
9.72 x 10s
MP
0.00130.01990.03530.08470.67790.1808
2.1.3. Sedimentation Method - Gravity SedimentationIn metallurgical practice where sedimentation or classification of particles is required, theStokes diameter is often used to indicate the size of the particle. Stokes' diameter is computedby observing the rate of fall of particles through a stationary fluid medium. The terminalvelocity (VT) of the particle is given by the relation:
VT = g (PS-PF) m/s (2.11)
where d = diameter of a spherical particle, mg = acceleration due to gravity, m/s2
ps pF = density of solid particle and fluid respectively, kg/m3
\i = viscosity of the fluid, Ns/m2 or Pa s (For water, n = 0.001 Pa s)Vj = terminal velocity, m/s
Eq. (2.11) is strictly applicable for spherical particles falling through a perfectly stillmedium and not being "hindered" by other particles or reacting with the medium. Toprevent hindrance by other particles, the concentration of the suspension should preferablycontain less than 1 % solids. Any hindrance by the walls of the vessel is generally neglectedespecially when the diameter of the vessel is about 100 times greater than the diameter of theparticle being measured.
The expression is not applicable for very small particles or conditions that lead toBrownian movement of particles.
When applying Eq. (2.11) to non-spherical particles the ratio of the maximum to minimumdiameter of the particles should be equal to or less than 4. For irregular particles, therefore,the relation does not give the true diameter of a particle, but describes the equivalent sphericaldiameter or the mean projected diameter dp, of the particle. The mean projected area would beit dp
2/4 and the mean volume expressed as, v = k dp . The constant k is a function of the shapeof the particle.
For Spherical particles,Angular particles,Mineral particles,
k = 3t/6k = 0.4 (approximately)k = 0.2-0.5
Note: Stokes diameter x 0.94 = sieve size (approximately).
3737
Eq. (2.11) can therefore be used to determine the size of individual particles less than 50um and preferably greater than 1 um.
In practice, difficulty arises in determining the terminal velocity (VT) as the free fall ofparticles in a medium depends on it's friction against the fluid and buoyancy. The frictionalforce and the buoyancy acting upwards opposes the downward forces on the falling particle.The combined opposing force known as the "drag" has to be determined to estimate the truevalue Of VT. For a spherical particle under streamline conditions (also known as laminar orNewtonian conditions), Stokes has determined the drag force, FD, as:
F D = 6 j t r n V T (2.12)
where r = radius of spherical particle.
The expression Eq. (2.12) is known as Stokes' Law for laminar flow.It has been found that FD varies with the particle diameter, density, velocity of particle and
also the viscosity of the medium. The ratio of (d PF V/U) is a dimensionless number known asthe Reynolds number (Re). When the character of fluid flowing past the particle is streamline,then the value of Re is less than 2100. When the flow is turbulent, Re is greater than 4000.Between a Re of 2100 and 4000 is the intermediate or transition stage.
The relation between the drag force and Reynolds number for streamline flow conditionsis:
F D = — (2.13)Re
For turbulent conditions, Newton derived the forces due to drag as:
FDN= 0.055 7id2VT2pF (2.14)
During the settling of a particle under streamline conditions, the force acting on the particledue to gravitational acceleration is given by:
mg = -nr3(ps - p F ; g (2.15)5
where m = effective particle mass, andr = particle radius
This would be equal to the force at the terminal velocity of the particle, given by Eq. (2.12),
4-7t r 3 (p s - p F ) g = 6nry.VT or
- ^ - r 2 ( p s - p F ) g (2.16)9u,
3838
This equation is Stokes' equation as given in Eq. (2.11). Under turbulent conditions,similarly:
-7 i r 3 (p s - p F ) g = 0.227tr2pFVTr:i (2.17)
where VTT is the terminal velocity under turbulent conditions, or
L P F
Even though the terminal velocities can be calculated from Eqs. (2.16) and (2.18), inpractice it is difficult to ascertain whether the flow past a falling particle is streamline orturbulent. To solve this predicament, FD is recalculated by considering the dimensionlessquantity [(FD/PFAVT2) Re2 ] to yield an expression which does not include VT. NOW if weconsider the drag force/unit projected area, A, of a particle falling freely as FD', then FD' =FD/A. We can now write:
4FD-"r2 = j i r 3 ( p s - p F ) g or
FD = | r ( P s - p F ) g (2.19)
By multiplying both sides of Eq. (2.19) by the dimensionless factor, [(FD'/PFVT2) Re2] wehave:
] 2 2dg(P s-pF)rdpFyT l
J 3 p F V T2 L M- J
2
PFVT2L V- J 3pFVT
2 L M-
_ ^ _ [ R e ] » = 2d 3 p F g(p s -p F )
Similarly we can arrive at an expression, which is independent of d by using thedimensionless factor [(FD'/PFVT2) Re"1];
Rfi =
P F V T2 3 P F
2 V T3
The right hand side of Eq. (2.20) deals with the fluid properties only and therefore can beeasily evaluated. The diameter d of the particle falling under any condition can now becomputed by equating this value to the left side of Eq. (2.20).Similarly, the terminal velocity of the particle can be computed from knowledge of the lefthand side of Eq. (2.21).
3939
Tables and charts are available, plotting log [(FD/PFAVT 2)Re2 ] against log [Re] and log
[(FD/PFAVJ2 ) Re"1] against Re. Tables 2.1 and 2.2 give the relationship over a limited rangefor spherical particles [3]. This data is also presented in graphical form in Fig. 2.3, which canbe derived from the empirical correlation [4]:
= [l.84ReO31 + 0.293 Re006]345 (2.22)pFVT
As the tables are applicable for spherical particles, Heywood [3] introduced a correctionfactor for non-spherical particles. To use the correction factor Heywood considered thediameter of the projected area, dp (of a particle sitting on its most stable position). Hecalculated the corresponding volume as k.dp3, where k was a constant and a function of shapeand the factor (FD/pF AV2) Re2.
Other shape factors have been proposed, such as the sphericity factor, S, definedmathematically as [5, 6]:
No min al volume diameter dvo —
No min al surface diameter d
where dy = diameter of a sphere of equal volumeds = diameter of a sphere of equal surface area
The values of log [(FD/PFAV2) Re"1 ] and log [(FD/PFAVT2)RC2 ] corresponding to differentvalues of shape factor, k, are given in Table 2.3. To use this table, log [(FD /PFAV2) Re"1] iscalculated from a measured settling velocity, Vj, using Eq. (2.21) for a spherical particle andthe equivalent Re value found from Table 2.2. The correction value from Table 2.3 is thenapplied to the log Re value to account for the non-spherical shape of the particle. The particlesize can be calculated from this corrected value of Re. A similar procedure follows for valuesof log [(Fi)/pFAVT2)Re2 ] calculated from values of particle size d.
Example 2.2Specular haematite particles from Labrador were sampled to determine the size of individualparticles. The average terminal velocity of the particles was determined to be 15 mm/s byallowing single particles to fall through a tall cylinder of water. The density of the oreparticles was 5400 kg/m3 and the water at a temperature of 25 ° C. Estimate the approximatesize of the particles.
Data: n (H20) at 25 ° C - 0.8904 mPa.s
40
1.E-06
1.E-05
1.E-04
1.E-03
1.E-02
1.E-01
1.E+00
1.E+01
1.E+02
1.E+03
1.E+04
1.E+05
1.E+06
1.E-03 1.E-02 1.E-01 1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05
Re
F(d
Vr/2
eR)
2F( ,
dVr/
2e
R/1)sphericity = 0.1 0.6 1.0
0.1
0.6
1.0
log [(FD/ρFAV T2)Re -1 ]
Log[(FD/ρFAV T2)Re2]
40
CDDC
CD
ou
1.B06
1.E+05
1.B04
1.B03
1.B02
1.B01
1.B00
1.E01
1.E02
1.E08
1.E04
1.E05
sphericity= 0.1 0.6 1.0
1.E06
f
A
==^
/ '
M
-Sk
= = ^
i- -V
: : ^\— A -
^ ^
m >
- Log[(FD/pFAVT^)Re1
:
\
\
£
In
>\' v—NS
J
f /
/1
-4-J-l' J
1
losrrFn/o AV ^Re"1! : : :
x
v
5
s.
\
s —\
\
— 5(
s
S
\
s
\
— 5 ( - -
— s - - -
0.1
06
1.0
1.E08 1.E02 1.E01 1.E-+00 1.E-+01 1.E-+02 1.E-+03 1.E-+04 1.E+05
Fte
Fig. 2.3. Relationship between log [(FD/pFAVT2)Re2 ] and log Re and log [(FD/pFAVT
2)Re"' ] and logRe for different particle shape factors, S.
4141
Table 2.1Relation between log [(FD/pFAVT
2)Re2 ] and log Re [2]./Re 2 )
1
0
1
2
3
4
5
0.0
3.919
2.908
T.8740.738
1.491
2.171
2.783
0.1
2.018
1.007
1.9670.817
1.562
2.236
2.841
0.2
2.117
1.1050.008
0.8951.632
2.300
2.899
0.3
2.216
1.203
0.1480.972
1.702
2.363
2.956
0.4
2.315
1.301
0.236
1.048
1.771
2.425
3.013
0.5
2.414
1.3980.324
1.124
1.839
2.487
3.070
0.6
2.513
1.495
0.410
1.199
1.907
2.548
3.127
0.7
2.612
1.591
0.495
1.273
1.974
2.608
3.183
0.8
2.711
1.686
0.577
1.346
2.040
2.667
3.239
0.9
2.810
1.781
0.569
1.419
2.106
2.725
3.295
Where ./(Re2) = Log[(FD/pAVT2)]Re2
Table 2.2Relation between log [(Fp/ppAVT2) Re"1 ] and log Re |ARe"')
4
3
tol
1
01
2
3
4
0.0
3.316
2.517
1.829
1.200
0.616
0.072
T.554
T.047
2.544
0.1
3.231
2.443
1.763
1.140
0.560
0.019
T.503
2.996
2.493
0.2
3.148
2.372
1.699
1.081
0.505
T.969
T.452
2.946
2.443
0.3
3.065
2.300
1.634
1.022
0.449
T.919
T.401
2.895
2.393
0.4
2.984
2.231
1.571
0.963
0.394
T.865
T.350
2.845
2.343
0.5
2.903
2.162
1.508
0.904
0.339
T.811
T.299
2.794
2.292
0.6
2.824
2.095
1.496
0.846
0.286
T.760
T.249
2.744
0.7
2.745
2.027
1.383
0.788
0.232
T.708
T.198
2.694
0.8
2.668
1.961
1.322
0.730
0.178
T.656
T.148
2.644
0.9
2.591
1.894
1.260
0.672
0.125
T.605
T.097
2.594
Where/Re1) = log[(FD/pAVT2)Re'
Table 2.3Correction for shape,factor k,and log [(FD/pFAV2) Re'1 ] and log [(Fp/pFAVT
2)Re2 ] [2].log[(FD/pFAV2)Re']
4.04.53.03.52.02.51.01.5
0.01.02.03.04.05.0
k=0.2
+0.289+0.231+0.173+0.119+0.072+0.033+0.007-0.003-0.007-0.008-0.010-0.012-0.013-0.014
k=0.3
+0.217+0.175+0.133+0.095+0.061+0.034+0.018+0.013+0.011+0.009+0.007+0.005+0.003+0.002
k=0.4
+0.185+0.149+0.114+0.082+0.056+0.038+0.028+0.024+0.022+0.019+0.017+0.015+0.013+0.012
log[(FD/pFAVT
2)Re2]21
012
2.53
3.54
4.55
5.56
k=0.2
+0.032+0.030+0.026+0.021+0.012
0.000-0.022-0.056-0.089-0.114-0.135-0.154-0.172
k=0.3
-0.002-0.003-0.005-0.010-0.016-0.020-0.032-0.052-0.074-0.093-0.110-0.125-0.134
k=0.4
-0.022-0.023-0.025-0.027-0.031-0.033-0.038-0.051-0.068-0.083-0.097-0.109-0.120
4242
Solution
For simplicity we may assume that the iron ore particles are spherical in shape, thereforeEq. (2.11) may be considered applicable.
From the data: ps = 5400 kg/m3
pL = 1000kg/m3
H (H20) at 25 ° C = 0.0008904 Pa.s
Substituting in Eq. (2.11):
0015= 9-81(5400-1000)di
18(0.0008904)
hence d = 0.0000746 m or 0.0745 mm.
Example 2.3
An irregular shaped mineral particle (shape factor, k = 0.3) had a projected area of 2.16 xIQ^m2. It was allowed to settle freely in water held at 20° C. The mean specific gravity ofthe particle was 2.65. Estimate the settling velocity of the particle of density, 1120kg/m3.
SolutionSteplProjected area of particle = it r2 = 2.16 x lO^m2
then r= 0.000829 m,d = 0.00166 m
From Appendix D-l PL H2O at 20 ° C = 998 kg/m3
And from Appendix D-2 U, H2O at 20 °C = 1.002 x 10"3 Ns/ m2
Step 2Let the projected diameter of particle be dp and rp the radius
The total drag force FD on particle = FD* * d p = (ps - PL) g k dP3
4
Next dividing both sides of the equation by the term p^V2 and multiplying by Re2
FD Re2 _ ^ dp g k (p s - p F ) Re2
p F V 2 7tpFV2
substituting Re = dp VDF/U, and simplifying:
4343
F D R e 2 _ 4 d p k g ( p s - p F ) p F
p F V 2 n\i2
Substituting the data gives:
F D Re 2 _ 4 x 0.00166 3 x 0.3 x 9.81 x (1120-998) x 998 = „ „_„ , .
p F V 2 3.14x (1.002x10 ~ 3 ) 2
F Re 2
log D • = log 2078.64 = 3.3178
P F V
Then from Table 2.1,
log Re =1.7143 or Re = 51.79From Table 2.3 the correction value for log Re corresponding to a shape factor of 0.3 equals -0.0447.
That is, the corrected value of log Re for the non-spherical particle = 1.7143 - 0.0447 =1.6696
and Re =46.73
Now Re = dpVpL/n =46.73
., . , 46.73 x (1.002xlO"3) „ . „ , .then V = = 0.025 m/s
0.00166 x 1120
Usually gravity segregation methods are used to determine particle sizes down to 0.02microns.
2.1.4. Sedimentation Method - Centrifugal SedimentationEq. (2.11) indicates that the terminal velocity of a particle can be increased and the settlingperiod reduced by increasing the value of the acceleration on the particle. This can easily beachieved both in the laboratory and in industry by the use of centrifuges (Fig. 2.4), where gcan be replaced by r co2, where r is the radius of the centrifuge and co is the angular velocity.For a spherical particle, obeying Stokes' Law, the motion of the particle in the centrifuge canbe expressed by the equation, for Re < 0.2;
d r I 7i dp p§ I dr 7i 3 / \ 2 I-I *)A\
^2 ^ 6 J + iH P d{ - P Ps PL
where dp = particle diameter.
44
R
r
ro rt
44
Fig. 2.4. Laboratory centrifuge.
Neglecting acceleration, Eq. (2.24) may be rewritten as:
3 ^ ^ = * d P P s " P L | r « o - or
dt 6 1 p s
(2.25)
dr _ dp ( p s - P L
Psco dt (2.26)
Assuming that the initial position of the particle in the centrifuge was r0 and the final positionwas rt, after time t, integration of Eq. (2.26) would give:
dp =
18up s
18 ji p s
t o r
0.5
(2.27)
Thus from a knowledge of r0 and rt the diameter d, of a particle, can be estimated. Whereturbulent conditions exist, the diameter of a particle can similarly be derived as:
Ps3dPco2(ps-pL) or
d P = - Ps3o>2(ps-pL)
(2.28)
In commercial centrifuges, the positions of a particle, rt, at times ti, t2, t3 ... tn is observedby a light source thus getting a profile of movement of particles when more than one particleis added for centrifuging.
4545
2.2. Particle size distribution
2.2.1. Sieve analysisWhichever method of measuring particle size is used, the characterisation of a sampleaccording to size is referred to as a size analysis. A size analysis in mineral processing isprimarily used to obtain quantitative data about the individual size and size distribution ofparticles in the process stream. This is important in assessing the quality of grinding,optimum size of feed to a process for maximum efficiency etc.
In most mineral processing applications an approximate size of particles is all that isrequired. Sieving through standard sieves of known apertures and determining the sieve sizeon which all the particles are retained can conveniently give the limiting size of a group ofparticles subjected to the operation. The closer the size of successive sieves the closer wouldbe the estimation to the true size.
Instead of individual measurements of sizes of particles, the mean size of a particle passingthrough and retained on sieves gives a good approximation of particle size within the sieveranges.
The sieve size analysis of a mineral sample giving the mass fraction of mineral retainedbetween successive sieve sizes is usually recorded as in Table 2.4. Points to note from thetable are:
1, There are no particles larger than 3350 um, (i.e. the fraction of particles retained on the3350 um screen is zero).
2. The mass of material on the 1680 um sieve equals 24.7 % of the total mass.3, The mass fraction retained on the 850 um screen and passing through the 1680 um
screen equals 26.2 %. Similarly, the mass of material between sieve fraction 420 um and210 um equals 9.7 % by mass.
4. The total material retained on the 420 um sieve equals 0 + 24.7 + 26.2 + 15.2 = 66.1 %by mass.
Similarly, the mass of mineral retained on the 210 \im sieve = 24.7 + 26.3 + 15.2 + 9.7= 75.8 %, and so on as seen in column 3 of Table 2.4.
Even though the subsequent sieve openings are related by a factor of iNl (Tyler's series), aplot of size against mass percent in most cases is far from a straight line. To obtain a specificscreen size that passes or retains a given mass, usually some interpolation of the tabled data isrequired. For example, in Table 2.4, to obtain the 70 % passing size, interpolation betweenthe 1680 and 850 micron screen size is required. This is best done when a linear relationshipexists between the size analysis data. A number of empirical relationships have beenformulated to linearise the data to make interpolation and extrapolation of the data moreaccurate. One such method is to rewrite the screen analysis by calculating the cumulativeamount retained or passing through a particular sieve as illustrated in Table 2.5 and Fig. 2.5,
4646
Table 2.4.Sieve analysis results
Size fraction, microns Mass, %(1)
-3350+1680-1680+850-850+420-420+210-210+105-105+75
-75Total
Table 2.5.
(2)28.522.813.28.45.02.4
19.6100.0
Size analysis with cumulative massesCum. % Passing,(100-Retained)
10071.548.635.427.022.019.6
0
Sieves, (im
3350168085042021010575
Pan
Cum. Mass, % retained(3)
28.551.464.673.078.080.4
100.0
Cum. Mass %retained0.0
28.551.464.673.078.080.4
100.0
Plotting the data in this form (Fig. 2.5) does not always result in a linear relationship.
2.2.2. Log-normal distributionThe distribution of sizes in a sample do not often follow a normal distribution, as in Fig. 1.3,but are skewed to the right. It may however match a log-normal distribution. That is, it willtake the shape of a normal distribution if the x-axis is a log scale. Then plotting thecumulative percent passing on a probability scale (y-axis) against the sieve sizes on a log scale(x-axis), should produce a straight line. The probability scale is an inverse normaldistribution, so plotting a normal distribution on this scale will give a straight line. Such alog-normal or log-probability plot is shown in Fig. 2.6 for the data in Table 2.5.
If this plot is linear (the size distribution is log-normal) then an approximation of the meanof the distribution is obtained from the dso, the 50% passing size. The geometric standarddeviation of the distribution is obtained from the 84% passing size since at this point, theparticle size is approximately equal to the mean size plus the standard deviation. That is:
Standard deviation = d»4 - ds. (2.29)
In some cases, curvature of the plot is observed in the coarse sizes because the maximumparticle size is not equal to infinity.
47
0102030405060708090
100
10 100 1000 10000
Particle Size (microns)
Cu
mu
lati
ve %
Pas
sin
g
47
C
'55(0
£°o.>JO
i3o
1009080
70
6050
4030
20
10
0
W
J/
,f<>*
/
/
P | i i i i
10 100 1000 10000
Particle Size (microns)
Fig. 2.5. Log-linear plot of cumulative % passing vs size.
2.2.3. Gaudin-Schuhmann Distributionhi some cases where the distribution is skewed, a linear plot can be obtained by plotting thelog of the cumulative undersize against the log of the screen aperture, preferably on log-logpaper (Fig. 2.7). Such plots are known as Gaudin-Schuhmann plots, hi most cases this wouldyield a straight line. The advantage of such plots is that a limited number of sieves may beused to measure the size of particles lying between any two sieve sizes. It also indicates thedistribution of sizes of particles that exist in a sample.
It is not necessary to plot both the cumulative oversize and undersize as each is the mirrorimage of the other as seen in Fig. 2.8.
The Gaudin-Schuhmann distribution is given as:
'=100 - (2.30)
where y = cumulative mass % passing size xx = screen aperture sizek = size parametera = distribution parameter
Taking the logs of both sides of this equation and rearranging it, then:
logy = log 100 + alogx - alogklog y = a log x + CONSTANT
or(2.31)
This is the equation of a straight line if x and y are plotted on a log-log scale. The slope ofthe straight line will be the distribution parameter, a, and the intercept of the straight line,when y = 100 %, will be the size parameter, k. These two parameters characterise the size ofthe sample. The distribution parameter is a measure of the range of different sizes in thesample and the size parameter is a measure of the top size.
48
10 100 1000 10000
Particle Size (microns)
Cu
mu
lati
ve %
Pas
sin
g
99.99
99.899.9
99
98
95
90
80706050403020
10
5
210.50.20.10.05
0.01
1
10
100
10 100 1000 10000
Particle Size (microns)
Cu
mu
lati
ve %
Pas
sin
g
48
D)
10
£Q)
|
g.gg
99
90
70
50403020
10521
0.50.20.1
0.05
0.01
•.—I r
_
10 100 1000 10000
Particle Size (microns)
Fig. 2.6. Log-normal plot of size distribution in Table 2,5.
100
D)
(0Q.
<D 10>
3
Eo
ti
*
'
10 100 1000
Particle Size (microns)
10000
Fig. 2.7. Log-log plot of cumulative % passing (undersize) vs size, (Gaudin-Schuhman plot).
49
0
10
20
30
40
50
60
70
80
90
100
10 100 1000 10000
Particle Size (microns)
Cu
mu
lati
ve %
Passing
Retained
49
ive
(03Eo
10090
80
70
60
50
40
30
20
10
0
1
1"
ks
\
/
J/
s /
\ \ \
\
—•— Passing
/
10 100 1000
Particle Size (microns)
10000
Fig. 2.8. Plot of cumulative % undersize (passing) and oversize (retained).
Calculation of the Slope of a Log-Log Plot
The slope of a graph is taken as the ratio (Y1-Y2)/(X1-X2) where (X1.Y1) and (X2,Y2) aretwo points on the graph (Fig. 2.9). On a log scale the points are plotted as (X,Y) but are inreality the log of these values. That is, the points correspond to (log X, log Y).
100
60
10
(X2,Y2)
10 100 1000 10000
Fig. 2.9. Log-log plot through the points (X1 ,Y 1) and (X2,Y2).
5050
In Fig. 2.9:
Slope * (60-81 but(1000-100)
Slope = (log 60-log 8)(log 1000-log 100)
Alternatively, if the distance in mm between one log cycle on the Y-axis (say 10-100) is thesame distance as one log cycle on the X-axis (10-100) then using a ruler, the slope can also bedetermined by;
Slope = (distance from Y2 to Yl in mm)(distance from X2 to XI in mm)
2.2.4 Rosin-Rammler DistributionAn alternative plot, especially suitable for finely ground particles, like that produced intumbling mills, is to plot log [log 100/(100-y)], or log [log 100/R], against the log of sievesize, where y is the cumulative % passing and R is the cumulative % retained. Such plots areknown as Rosin-Rammler plots, (Fig. 2.10). The double log scale expands the fine and coarseends of the size range (<25 % and > 75 %) and compresses the mid range (30-60 %). Underoperating plant conditions approximations in particle size computation and estimation aresufficient for most purposes.
The Rosin-Rammler (or Weibull) distribution is expressed as:
R = 100 exp (2.32)
where Rx1
b
cumulative mass % retained on size xsize parameter, anddistribution parameter
Rearranging and taking the logarithm of both sides of Eq. (2.32) gives:
l oehr- = -T lQge (2.33)
Taking logarithms a second time to remove the exponent gives:
, , f 100^ , . , . i , ,log log =blogx-blogx +logloge or
v R Jlog log] — ] = blogx +CONSTANT
v R i(2.34)
51
10 100 1000 10000
Size (microns)
Cu
mu
lati
ve %
Ret
ain
ed
0.10.1
510
20304050
60
70
80
9092
94
969798
98.5
9999.299.499.6
99.7
99.8
99.9
36.79
51
A plot of log log (100/R) versus log x should give a straight line. The parameters of theRosin-Rammler distribution, b and x1 are obtained from the slope of the straight line and theintercept at the horizontal line at R = 36.79, respectively. Together they completely describethe size distribution. To simplify the calculation of the double log, special graph paper isavailable (Rosin-Rammler/Weibull paper) where values of cumulative % retained (orcumulative % passing) can be plotted directly on the Y-axis. A line on the graph paper at acumulative % retained value of 36.79 is included to allow estimation of the parameter x1. Themethod of computing for a Rosin-Rammler plot is illustrated in Example 2.4.
Example 2.4A sieve analysis showed the following results (Table 2.6). Plot the data according to Gaudin-Schuhmann and Rosin-Rammler and determine the mass of the size fraction retained on 200\xm.
Fig. 2.11 is a plot of the cumulative % passing column in Table 2.6 versus size. TheGaudin-Schuhmann plot is a reasonable straight line that shows the fraction of materialpassing a 200 urn sieve was expected to be 10.1% by mass.
Rosin-Rammler plots can similarly be made either by calculations or using special logpaper known as Weibull graph paper that save labourious calculations.
Typical calculations using the data in Table 2.6 is illustrated in Table 2.7.This can now be plotted on normal graph paper (Fig. 2.12A) or on special Rosin-Rammler
paper (Fig. 2.12B). The plot is nearly a straight line, which indicates that the mass fractionretained on 200 microns is about 90.6 % by mass or 9.4 % passing.
I
0.10.
51020304050
60
70
<D 92
. > 94
w 96
— 97
|
99.2
99.4
99.6
10 100 1000
Size (microns)
10000
Fig. 2.10. Rosin Rammler plot of data from Table 2.4.
52
1
10
100
10 100 1000 10000
Particle Size (microns)
Cu
mu
lati
ve %
Pas
sin
g
52
Table 2.6Sieve analysis, Example 2.4.Aperture, Mass Cum. mass % Cum. mass Log Cum. mass Log
mm1330047502360118060030015075-75
020.921.120.910.315.7
4.32.34.5
020.942.062.973.288.993.295.5100
retained % passing % retained aperture100.079.158.037.126.811.16.84.50
1.321.621.801.861.691.971.982.00
4.123.683.373.072.782.482.181.88
IffIff
£s?a>>
10
1
i
y
i •
•
10 100 1000
Particle Size (microns)10000
Fig. 2.11. Gaudin-Schuhman plot, Example 2.4.
Table 2.7.Data for Rosin Rammler plot, Example 2.4.
logAperture
4.123.683.373.072.782.482.181.88
Cum. mass %Retained, R
020.942.062.973.288.993.295.5
1/R
0.04780.02380.01590.01370.01120.01070.0105
100/R
4.782.381.591.371.121.071.05
log 100/R
0.6800.3770.2010.1360.0510.0310.020
Log[logl00/R]
-0.168-0.424-0.696-0.868-1.292-1.515-1.699
53
10 100 1000 10000 100000
Size (microns)
Cu
mu
lati
ve %
Ret
ain
ed
0.10.5
1
510
20304050
60
70
80
90
92
94
96
97
98
98.5
9999.2
99.4
99.6
99.7
99.8
99.9
36.79
-1.8
-1.6
-1.4
-1.2
-1.0
-0.8
-0.6
-0.4
-0.2
0.0
1.5 2.0 2.5 3.0 3.5 4.0
log Size (microns)
log
[lo
g 1
00/R
]
200
AB
53
0.0
-0.2
-0.4
E -0.6
Z -0.8oi
£-1.0o>2 -1.2
-1.4
-1.6
-1.8
90.6 %/
/ :
/
200
>
!E3
o .
l
*
100 1000 10000 100000
Size (microns)1.5 2.0 2.5 3.0 3.5 4.0
log Size (microns)
A
Fig. 2.12. Rosin Rammler plot on: A normal graph paper according to Table 2.7.
B plot on special Rosin Rammler paper from Table 2.6.
2.3. Combining size distributionsSince sizing by different methods, (sieving, cyclosizer, laser sizing) covers different size
ranges, size distributions from different methods may need to be combined to give a completesize distribution.
Sizing of the same sample by different methods will generally produce different resultsbecause of the difference in measurement principle. The difference is related to particleshape. For spherical particles, the Stokes diameter calculated from the settling velocity isexact and the projected area of a sphere is that of a circle of the same diameter. Thus thecloser particle shape is to a sphere, the closer the measured sizes will be. For elongated or flatparticles the settling velocity will be lower than for spherical particles due to an increase inspecific surface area leading to a greater drag force and hence lower terminal velocity. Thus,the Stokes diameter will be lower for elongated particles. On the other hand, these particleswill be more stable when resting flatter on a viewing surface and hence will have a largerprojected area and hence a larger projected area diameter. In general, the Stokes diameter issmaller or equal to the same particle's sieve diameter.
Table 2.8 shows the results from a size analysis of a fine quartz sample sized by sieving,cyclosizing and with a laser sizer. The -212 micron quartz was dry screened at 75 micronsand the -75 micron fraction sized by screening and cyclosizing. The fraction from the lastcone of the cyclosizer was collected and allowed to settle before decanting the water anddrying the solids. This minus cone 5 material was laser sized for comparison.
For the screening samples, the +38 micron screen fractions were obtained using standard200 mm diameter BSS screens and the -38 micron fraction wet screened on a 10 micron nylonmesh screen. For comparison, a separate -212 micron sample was completely wet screened onall the above screens and also laser sized.
54
1
10
100
1 10 100 1000
Size (microns)
Cu
m %
Pas
sin
g
Dry screen
Cyclosizer
54
Ed
10
1 -
/
/
1
t
J
.. -o-
Dry screen
Cyclosizer
10 100
Size (microns)1000
Fig. 2.13, Screening and eyclosizing of-75 micron fraction of—212 micron quartz.
Each sizing method covered some of the same size range to make comparison possible. Fig.2.13 shows the results of the dry screening and the cyclosizer.
Since the sample that was cyclosized (-75 microns) represents only a portion of the wholesample (61.6%), the results have to be adjusted by multiplying the mass % in the cyclosizerfractions by the -75 micron proportion, ie. for the -45 +34 micron fraction, the scaledcumulative % passing is 0.802 x 61.6 = 49.4 %.
The same applies to the combination of the +75 micron screened results with the -75micron screened results.
When re-plotted, the data points line up with the +75 micron screen data as seen in Fig.2.14. The -75 micron dry screened data however deviates from the cyclosizer trend. When thewet screened results are plotted, as shown in Fig. 2.15, the results are aligned very well withthe cyclosizer results. This highlights the difference between wet screening and dry screening,particularly for fine material, in this case the -75 micron size range.
The dry screening results show a smaller percentage of material passing each screen below75 microns compared to the wet screening data. This results because the fine particles tend tostick to the coarser particles and do not pass the screens. The only way to get an accuratesizing by screening in these fine fractions is to use water to flush the fines through the screens.This is a more tedious process but as can be seen from the diagram, is the only way to getaccurate results for the fine fractions, hi the coarser fractions, fine particles attached to coarsefractions represent a smaller percentage of the mass in that fraction and hence wet screening isless critical. An acceptable procedure would be to wet screen the sample at 75 microns anddry screen the +75 micron fraction while wet screening the -75 micron fraction.
The laser sizer results of the cyclosizer minus cone 5 fraction, normalised to the totalsample mass using the procedure described above, is plotted in Fig. 2.16 and fits well with thetrend of the cyclosizer fractions and the wet screened results. The laser sized -212 micronsample results are plotted in Fig. 2.17 and are seen to overlap the cyclosizer and wet screeningdata. This represents a good result for this sample in that the different sizing methods give
55
Table 2.8Sizing data for a -212 micron ground quartz sample.
DRYSize
SCREEN
passing retained(um)
21215075
7563534538
WET
10
(um)
150750
6353453810
SCREEN
0
CYCLOSIZER (-75um)Sizepassing retained(um)
754534241612
(um)
45342416120
Mass
(g)
53148.7323.3525
13.148.765.914.329.4
9.8181.2
Mass
(g)
10.712.6106.62.611.6
54.1
Mass %
10.128.361.6100.0
7.226.936.47.916.2
5.4100.0
Mass %
19.823.318.512.24.821.4
100
Cum Mass %Passing
10089.961.6
100.092.865.929.521.6
5.4
Cum Mass %Passing
100.080.256.938.426.221.4
Combinedmass %
10.128.3
4.4516.5522.404.869.99
3.33100.0
Scaled towhole sample
61.649.435.123.716.213.2
Cum Mass%Passing
10089.9
61.657.140.618.213.3
3.3
WET SCREENSizepassing(um)
2121507553453810
retained(uni)
15075534538100
LASER SIZER (-212 um)
Sizepassing(um)
600404
27320313710275.650.941.828.219.011.67.083.922.39
retained(um)
404273
203137102
75.650.941.828.219.011.67.083.922.391.32
Mass
(g)
24.351.128.819.613.542.924.4204.6
Mass %Average
100.099.397.193.584.975.664.748.340.327.118.412.07.94.42.3
Mass %
11925.014.19.66.621.011.9100
Cum Mass%Passing
100.088.163.149.139.532.911.9
56
1
10
100
1 10 100 1000
Size (microns)
Cu
m %
Pas
sin
g
Dry screens
Cyclosizer
1
10
100
1 10 100 1000
Size (microns)
Cu
m %
Pas
sin
g
Dry screen
Cyclosizer
Wet screen
56
100
Q.
Eo
10
J/
/
if
JI \ *
A Dry screens
• —O—Cyclosizer
10 100
Size (microns)1000
Fig. 2.14. Adjustment of the -75 micron cyclosizer fraction to the total mass of the sample.
comparable results. This has been observed elsewhere for quartz and is an indication that theparticles of quartz are close to spherical [7].As there is no absolute measure of size for irregular shaped particles, different methods of sizemeasurement will give different size distributions. Generally, it is expected that irregularshaped particles will yield different sizing results when measured by different techniques.The differences can be quantified in terms of a shape factor. For example, a conversion shapefactor could be defined as:
D = di/d2
100
O>
Eo
10
(2.29)
//
/
-4-
{TI
•t
1—1
- o -
- • -
ury screen
Cyclosizer
Wet screen
10 100
Size (microns)1000
Fig. 2.15. Wet screen data for the-212 micron quartz sample.
57
1
10
100
1 10 100 1000
Size (microns)
Cu
m %
Pas
sin
g
Dry screen
Cyclosizer
Wet screen
Laser (-c5)
1
10
100
1 10 100 1000
Size (microns)
Cu
m %
Pas
sin
g
Cyclosizer
Wet screen
Laser (-212 um)
57
100
10
/
•J
fp
- 1s//k(
•
jj
Tl—Zi^\ *—I
—A—Dry screen
. —O— Cyclosizer
• Wet screen
- • — L a s e r (-C5)
10 100
Size (microns)
1000
Fig. 2.16. Laser sizing of-cone 5 from cyclosizer.
where di is the size assigned to particles in a narrow size interval and d2 is the size measuredby a second technique for the same interval.This factor is generally determined experimentally by selecting a number of cumulative %passing values and obtaining the ratio of di and d2 for each value and taking the average. Forexample, Fig. 2.18 shows a size distribution obtained by cyclosizer and screening where thedistributions don't match. The two plots are roughly parallel and where they overlap, takingthe cumulative % passing value of 75% and drawing a horizontal line gives the values of di
100
s i
|O
10
/
/
1
i/
f
t
f
ft
iTl—*|H F=#ipH >-|r<
—O— Cyclosizer
• Wet screen
—•—Laser (-212 um
> | I I I
)
10 100
Size (microns)1000
Fig. 2.17. Laser sizing of-212 micron quartz sample.
58
1
10
100
1 10 100 1000
Size (microns)
Cu
m %
Pas
sin
g
Screen
Cyclosizer
d1=70 d2 = 110
1
10
100
1 10 100 1000
Size (microns)
Cu
m %
Pas
sin
g
Screen
Adjusted Cyclosizer
58
d1=70 d2 =
10IS
Q.
Eo
rnn-i
10
1
•
-1 ̂ Ht—1
—•—Screen
—•— Cyclcsizer
1 100010 100
Size (microns)
Fig. 2.18. Cyclosizer and screen results from irregular shaped particles
and d2 as 70 and 110 microns respectively. The conversion shape factor is then calculated asD =70/110 =0.636 for conversion of data from technique 2 to technique 1. To convert thedata from technique 1 to 2, each size point in distribution 1 is divided by the conversion shapefactor as shown in Table 2.9 and Fig. 2.19.
100
(BQ.
S?
E
10
A
/
A/
i f
—•—Screen
—A—Adjusted Cyclosizer
10 100
Size (microns)
1000
Fig. 2.19. Adjustment of cyclosizer data to form smooth equivalent screen sizings using theconversion shape factor D = 0.636.
5959
Table 2.9Conversion of cyclosizer data to equivalent screen data using the shape factor, D = 0.636
Passing(microns)
754534241612
SizeRetained(microns)
45342416120
Cum % Passing
79.059.649.933.723.018.8
Converted Size, (microns)
117.970.753.437.725.118.9
2.4. Problems
2.1Data obtained on analysing a tumbling mill product is given below. Plot the data using theGaudin-Schumann and Rosin-Rammler methods. From the plots estimate the mass fractionsretained on 200 um and 34 (am screens.
Size, urn 1200 600 300Feed, mass % retained 1.5 4.0 10.0
2.2Using the
Size, urnProduct,retained
15022.0
7533.0
-7529.5
Total100.0
following data, plot a Rosin-Rammler distribution curve and determine its
2000 1200 850 600mass % 4.0 9.5 12.5 15.5
50019.0
35016.0
25013.0
-25010.5
equation.
Total100.0
2.3Samples of the feed to a ball mill and the product after 15 mins of operation were analysedand the results are as follows:
Size, umFeed, mass % retainedProduct, mass % retained
12005.00.5
6001.02.0
30015.03.0
15020.020.5
7530.034.8
-7529.039.2
Total100.0100.0
Estimate the rate of reduction of the 1000 um feed size during the mill operation.
2.4The product from a grinding mill was sampled every 30 minutes for the first hour, then hourlyfor 4 hours. The samples were screened and analysed yielding the following average sizedistribution:
6060
ScreenSize| im
8004002501256835-35
Total
Feed
3.08.8
25.633.224.4
3.31.7
100.0
Product after Vih
1.83.0
26.025.026.0
5.03.2
100.0
Mass % Retained
Product after 1h
0.62.0
22.435.027.3
8.14.6100.0
Product after 2h
0.00.5
20.238.027.010.34.0100.0
Product after4 h0.00.2
13.032.031.212.26.4
100.0
Estimate the rate at which the 100 um feed size is reduced and its mass fraction after 2 hoursof operation
2.5Two sets of size analysis results are available for a given material. From a Gaudin-Schuhmatmplot determine the relations between size and mass distribution in these two cases. Using theequation for case B determine the fines fraction at -53 microns
Size, um+9510
-9510+6650-6650+1680-1680+850-850+420-420+210-210+105-105+75
-75Total
Case A, mass %0.0
55.525.010.55.02.41.60.00.0
100.0
Case B, mass %0.00.0
14.024.020.014.09.04.8
14.2100.0
2.6Gold nuggets were examined under a microscope and the following results were obtained.
Nugget size, mmnumber
(M U24500 2900
241100
4^8200
8̂ 16120
16-324
Determine: 1. the size distribution on a mass basis2. the size distribution on a surface area basis.
Data: density of gold nugget = 18500 kg/m3
6161
2.7A pond contains particles from a process tail stream. If the pond is 500 mm deep, calculatethe size of particles still suspended after 30 minutes from entering the pond. The particledensity is 2650 kg/m3.
2.8The diameter of 6 lots of particles from a sample were measured under an optical microscopeand a sedimentation method. The number of particles of each lot were counted and tabulatedas:
Lot number 1 2 3 4 5 6Size, mm -4 + 2 -2 + 1 -1+0.5 -0.5 + 0.25 -0.25 + 0.15 -0.15 + 0.075Number of 38 75 105 153 321 500particles
The particles were all of similar shape, having a shape factor of 0.3. Estimate:
1. the size of particles having the same specific surface as the mixture2. the surface area per unit volume of particles.
2.9A cylindrical glass tube of diameter 20 mm and capacity 22 cm3, is filled with a 12% watersuspension of mineral of specific gravity 2.4 and placed in a centrifuge. The centrifuge wasthen rotated at 3000 rpm for 2 seconds. Estimate:
1. The diameter of particles that would settle if the movement of particles was understreamline conditions.
2. The diameter of particles that would settle under turbulent conditions when therotational speed was increased to 4280 rpm.
Data: Temperature of water = 20°CViscosity of water at 20°C = 1.008 mPas
2.10A 1% suspension of hematite particles in methanol (1% solution in water) was filled in acylindrical centrifuge tube 3 cm long x 1.9 cm diameter and rotated at 5000 rpm. Estimate:
1. the size of particles that would settle every second for 5 seconds at the bottom of thetube.
2. If the number of particles settling every second were: 200,120, 80, 60, 20, determine themean specific surface of the hematite particles.
Data: Relative viscosity of methanol solution at 20°C = 1.3332Specific gravity of methanol at 20°C = 0.9964
6262
REFERENCES[1] B.H. Kay, Chemical Engineering, 73 (1966) 239.[2] J.M. Coulson and J.F. Richardson, with J.R. Backhurst and J.H. Harker, Chemical
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