1. Particulate Solids

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SECTION 2-1Particulate SolidsPROBLEM 1.1

The size analysis of a powdered material on a mass basis is represented by a straight linefrom 0 per cent at 1 hm particle size to 100 per cent by mass at 101 pm particle size.Calculate the surface mean diameter of the particles constituting the system.

SolutionSee Volume 2, Example 1.1.

PROBLEM 1.2The equations giving the number distribution curve for a powdered materialaredn/dd = dfor the size range 0-10 pm, and dn/dd = lo0,000/d4 for the size range 10-100 pmwhere d is in pm. ketch the number, surface and mass distribution curves and calculatethe surface mean diameter for the powder. Explain briefly how the data for the constructionof these curves m a y be obtained experimentally.


PROBLEM 1.3The fineness characteristicof a powder on a cumulative basis is represented by a straighttine from the origin to 100 per cent undersize at a particle size of 50 pm. f the powderis initially dispersed uniformly in a column of liquid, calculate the proportion by masswhich remains in suspension in the time from commencement of settling to that at whicha 40 pm particle falls the total height of the column. It may be assumed that Stokes lawis applicable to the settling of the particles over the whole size range.

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SolutionFor settling in the Stokes law region, the velocity is proportional to the diameter squaredand hence the time taken for a 40 Fm particle to fall a height h m is:

t = h/402kwhere k a constant.During this time, a particle of diameter d w rn has fallen a distance equal to:

kd2h/402k= hd2/402The proportion of particles of size d which are still in suspension is:

= 1 - d2/402)and the fraction by mass of particles which are still in suspension is:

= l m [ l - (d2/402)]dwSince dw/dd = 1/50, the mass fraction is:

= (1/50) l [1- d2/402)1dd= ( 1 /50 ) [d- d3/4800)]p= 0.533 or 53.3 per cent of the particles remain in suspension.

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PROBLEM 1.4In a mixture of quartz of density 2650 kg/m3 and galena of density 7500 kg/m3, the sizesof the particles range from 0.0052 to 0.025 mm.

On separation in a hydraulic classifier under free settling conditions, three fractionsare obtained, one consisting of quartz only, one a mixture of quartz and galena, and oneof galena only. What are the ranges of sizes of particles of the two substances in theoriginal mixture?

SolutionUse is made of equation 3.24, Stokes law, which may be written as:

uo = kd2(ps - P I .where k (= g/18p) s a constant.For large galena: uo = k(25 x 10-6)2(7500 - l0o0) = 4.06 x 10-6k m / sFor small galena: uo = k ( 5 . 2 x 10-6)2(7500- 1OOO) = 0.176 x 10-6k m/sFor large quartz: uo = k(25 x 10-6)2(26J0- O00) = 1.03 x 10% m / sFor small quartz: uo = k(5 .2 x 10-6)2(2650- OOO) = 0.046 x 10-6km/s

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If the time of settling was such that particles with a velocity equal to 1.03 x lo-% m /ssettled, then the bottom product would contain quartz. This is not so and hence themaximum size of galena particles still in suspension is given by:

1.03 x 10% = kd2(7500- 1OOO) or d = O.oooO126m or 0.0126mm.Similarly if the time of settling was such that particles with a velocity equal to 0.176 x

m /s did not start to settle, then the top product would contain galena. This is notthe case and hence the minimum size of quartz in suspension is given by:

0.176 x 10-6k= kd2(2650- 1OOO) or d = O.oooO103 m or 0.0103 mm.It may therefore be concluded that, assuming streamline conditions, the fraction of

material in suspension, that is containing quartz and galena, is made up of particles ofsizes in the range 0.0103-0.0126 mm

PROBLEM 1.5A mixture of quartz and galena of a size range from 0.015 mm to 0.065 mm is to beseparated into two pure fractions using a hindered settling process. What is the minimumapparent density of the fluid that will give this separation? How will the viscosity of thebed affect the minimum required density?

The density of galena is 7500 kg/m3 and the density of quartz is 2650 kg/m3.

SolutionSee Volume 2,Example 1.4.

PROBLEM1.6The size distribution of a dust as measured by a microscope is as follows. Convert thesedata to obtain the distribution on a mass basis, and calculate the specific surface, assumingspherical particles of density 2650 kg/m3.

Size range (Fm) Number of particles in range (-)0-2 20002-4 6004-8 1408- 12 4012-16 1516-20 520-24 2

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SolutionFrom equation 1.4, the mass fraction of particles of size dl is given by:

3XI = nIkld1Psr

where kl is a constant, n1 is the number of particles of size dl, and p, is the density ofthe particles = 2650 kg/m3.EX, 1 and hence the mass fraction is:

x1 = nlkld:ps/Xnkd3p,.In this case:

d n kd3np, X1 2003 6 0 06 140

14 1518 522 2

10 PO

5,300,000k 0.01 142,930,000k 0.09080,136,000k 0.168

106,000,000k 0.222109,074,000k 0.22977,274,000k 0.16256,434,400k 0.118

C = 477,148,400k X = 1.0The surface mean diameter is given by equation 1.14:

d, = Wd:) /Wld: )and hence:

d1361014

1822

n nd2 nd32000 2000 2000600 5400 16,200140 5040 30,24040 4Ooo 40,00015 2940 41,1605 1620 29,1602 968 21,296

C = 21,968 C = 180,056

Thus: d, = (180,056/21,968)= 8.20 V rnThis is the size of a particle with the same specific surface as the mixture.The volume of a particle 8.20 b m in diameter = (n/6)8.203= 288.7 wm3.

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The surface area of a particle 8.20 pm in diameter= (n x 8.202)= 211.2 pm2and hence: the specific surface = (21 .2/288.7)

= 0.731 pm2/pm3 or 0.731 x lo6 m2/m3

PROBLEM 1.7The performance of a solids mixer was assessed by calculating the variance occurring inthe mass fraction of a component amongst a selection of samples withdrawn from themixture. The quality was tested at intervals of 30 s and the data obtained are:

mixing time (s) 30 60 90 120 150sample variance (-) 0.025 0.006 0.015 0.018 0.019

If the component analysed represents 20 per cent of the mixture by mass and each of thesamples removed contains approximately 100 particles, comment on the quality of themixture produced and present the data in graphical form showing the variation of mixingindex with time.


PROBLEM 1.8The size distribution by mass of the dust carried in a gas, together with the efficiency ofcollection over each size range is as follows:Size range, (pm) 0-5 5-10 10-20 20-40 40-80 80-160Mass (per cent) 10 15 35 20 10 10Efficiency (per cent) 20 40 80 90 95 100Calculate the overall efficiencyof the collector and the percentage by mass of the emitteddust that is smaller than 20 pm in diameter. If the dust burden is 18 g/m3 at entry andthe gas flow is 0.3 m3/s, calculate the mass flow of dust emitted.


PROBLEM1.9The collection efficiency of a cyclone is 45 per cent over the size range 0-5 pm, 80per cent over the size range 5-10 pm, and 96 per cent for particles exceeding 10 pm.

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Calculate the efficiency of collection for a dust with a mass distribution of 50 per cent0-5 pm, 30 per cent 5-10 Fm and 20 per cent above 10 Fm.


PROBLEM1.10A sample of dust from the air in a factory is collected on a glass slide. If dust on the slidewas deposited from one cubic centimetre of air, estimate the mass of dust in g/m3 of airin the factory, given the number of particles in the various size ranges to be as follows:Size range (pm) 0- 1 1-2 2-4 4-6 6-10 10-14Number of particles (-) 2000 lo00 500 200 100 40It may be assumed that the density of the dust is 2600 kg/m3, and an appropriate allowanceshould be made for particle shape.

SolutionIf the particles are spherical, the particle diameter is d m and the density p = 2600 kg/m3,then the volume of 1 particle = ( n / 6 ) d 3m3, the mass of 1 particle = 2600(x/6)d3 kgand the following table may be producedSize (km) 0- 1 1-2 2-4 4-6Number of particles (-) 2000 1000 500 200Mean diameter (Km) 0.5 1.5 3.0 5.0

(m) 0.5 x 1.5 x 3.0 x 5.0 xVolume (m3) 6.54 x 3.38 x 1.41 x 6.54 x lo-Mass of one particle (kg) 1.70 x 8.78 x lo- 3.68 x 1.70 xMass of one particles insize range (kg) 3.40 x 8.78 x 1O-l2 1.83 x lo- 3.40 x lo-Size (pm) 6 - 10 10-14Number of particles (-) 100 40

(m) 8.0 x 12.0 xVolume (m3) 2.68 x 9.05 xMass of one particle (kg) 6.97 x 2.35 xMass of one particles in

Mean diameter (pm) 8.0 12.0

size range (kg) 6.97 x lo- 9.41 x lo-6


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Total mass of particles= 2.50 x lo-'' kg.As this mass is obtained from 1 cm3of air, the required dust concentration is given by:

(2.50 x lo-'') x lo3x lo6= 0.25 g/m3

PROBLEM 1.11A cyclone separator 0.3 m in diameter and 1.2 m long, has a circular inlet 75 mm indiameter and an outlet of the same size. If the gas enters at a velocity of 1.5ds, t whatparticle size will the theoretical cut occur?

The viscosity of air is 0.018 mNs/m2, the density of air is 1.3 kg/m3 and the densityof the particles is 2700 kg/m3.


1. Particulate Solids

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