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PARTICLE SIZE ANALYSIS The size analysis of a powdered material on a dry basis is represented by a straight line from 0% at 1 micrometer particle size to 100% by mass at 101 micrometer particle size. Calculate the surface mean diameter of the particles constituting the system. Surface mean diameter = 21.7 micrometer The fineness characteristic of a powder on a cumulative basis is represented by a straight line from the origin to 100% undersize at particle size of 50 micrometer. If the powder is initially dispersed uniformly in a column of liquid, calculate the proportion by mass which remains in suspension in the time from the commencement of settling to that at which a 40 micrometer particle falls the total height of the column. It may be assumed that Stokes' law is approachable to the settling of the particles over the whole size range. 0.533 by mass remains in suspension In a mixture of quartz of density 2650 kg/cu. m. and galena of density 7500 kg/ cu. m., the sizes of the particle range from 0.0052 to 0.025 mm. On separation in a hydraulic classifier under free settling conditions, three fractions are obtained, one consisting quartz only, one a mixture of quartz and galena, and one of galena only. What are the ranges of sizes of particles of the two substances in the original mixture? Size range for galena and quartz, originally, is 0.0103 to 0.0126 mm. A mixture of quartz and galena of size range from 0.015 mm to 0.065 mm is to be separated into two pure fractions using hindered settling process. What is the minimum apparent density of the fluid that will give this separation? How will the viscosity of the bed affect the minimum required density? The density of galena is 7500 kg/cu.m. and the density of quartz is 2650 kg/cu.m. Apparent density = 1196 kg/ cu. m. for Newtons Range = 2377 kg/ cu. m. for Stokes Range The size distribution of a dust is measured by a microscope is as follows. Convert these data to obtain the distribution on a mass basis, and calculate the specific surface, assuming spherical particles of density 2650 kg/cu.m. Size range, micrometer No. of particles in range 0 -2 2000 2-4 600 4-8 140 8-12 40 12-16 15 16-20 5 20-24 2 Surface mean diameter = 8.2 micrometer Volume of this particle = 288.7 cubic micrometer Surface are of this particle = 211.2 square micrometer Specifice surface = 211.2/288.7 = 0.731 square micrometer/cubic micrometer The performance of a solids mixer was assessed by calculating the variance occurring in the mass fraction of a component amongst a selection of samples withdrawn from the mixture. The quality was tested at intervals of 30s and the data obtained are:

mixing time, s 30 50 90 120 150 sample variance 0.025 0.006 0.015 0.018 0.019 If the component analysed represents 20% of the mixture by mass and each of the samples removed contains approximately 100 particles, comment on the quality of the mixture produced and present the date in graphical form showing the variation of mixing index with time. Answer: time = 60s The size distribution by mass of the dust carried in a gas, together with the efficiency of collection over each size range is as follows: Size range, (µm) 0–5 5–10 10–20 20–40 40–80 80–160 Mass (per cent) 10 15 35 20 10 10 Efficiency (per cent) 20 40 80 90 95 100 Calculate the overall efficiency of the collector and the percentage by mass of the emitted dust that is smaller than 20 µm in diameter. If the dust burden is 18 g/m3 at entry and the gas flow us 0.3 m3/s, calculate the mass flow of dust obtained. Over-all efficiency = 73.5% Dust that is smaller than 20 µm in diameter emitted = 90.1% Mass emitted = 1.43 x 10-3 kg/s (0.12 tonne/day) The collection efficiency of a cyclone is 45 per cent over the size range 0–5 µm, 80 per cent over the size range 5–10 µm, and 96 per cent for particles exceeding 10 µm. Calculate the efficiency of collection for a dust with a mass distribution of 50 per cent 0–5 µm, 30 per cent 5–10 µm and 20 per cent above 10 µm. Efficiency = 65.7% A sample of dust from the air in a factory is collected on a glass slide. If dust on the slide was deposited from one cubic centimeter of air, estimate the mass of dust in g/m3 of air in the factory, given the number of particles in the various size ranges to be as follows: Size range (µm) 0–1 1–2 2–4 4–6 6–10 10–14 Number of particles (-) 2000 1000 500 200 100 40 It may be assumed that the density of the dust is 2600 kg/m3, and an appropriate allowance should be made for particle shape. Mass of dust in g/m3 = 0.25 g/m3 A cyclone separator 0.3 m in diameter and 1.2 m long, has a circular inlet 75 mm in diameter and an outlet of the same size. If the gas enters at a velocity of 1.5 m/s, at what particle 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 density of the particles is 2700 kg/m3. Theoretical cut size = 2.17 micrometer Theoretical cut velocity = 3.83 x 10-4 m/s SIZE REDUCTION A material is crushed in a Blake jaw crusher such that the average size of particle is reduced from 50 mm to 10 mm, with the consumption of energy of 13.0 kW/(kg/s). What will be the consumption of energy needed to crush the same material of average size 75 mm to average size of 25 mm: (a) assuming Rittinger’s Law applies,

(b) assuming Kick’s Law applies? Which of these results would be regarded as being more reliable and why? (a) assuming Rittinger’s Law applies, 4.833 kW (b) assuming Kick’s Law applies, 8.88 kW – more reliable, uses bond work index, specific to material A crusher was used to crush a material with a compressive strength of 22.5MN/m2. The size of the feed was minus 50 mm, plus 40 mm and the power required was 13.0 kW/(kg/s). The screen analysis of the product was: Size of aperture (mm) Amount of product (per cent) through 6.0 all on 4.0 26 on 2.0 18 on 0.75 23 on 0.50 8 on 0.25 17 on 0.125 3 through 0.125 5 What power would be required to crush 1 kg/s of a material of compressive strength 45 MN/m2 from a feed of minus 45 mm, plus 40 mm to a product of 0.50 mm average size? For a feed of 1 kg.s, energy requirement = 47.8 kW A crusher reducing limestone of crushing strength 70 MN/m2 from 6 mm diameter average size to 0.1 mm diameter average size, requires 9 kW. The same machine is used to crush dolomite at the same output from 6 mm diameter average size to a product consisting of 20 per cent with an average diameter of 0.25 mm, 60 per cent with an average diameter of 0.125 mm and a balance having an average diameter of 0.085 mm. Estimate the power required, assuming that the crushing strength of the dolomite is 100 MN/m2 and that crushing follows Rittinger’s Law. Power required = 5.9 kW If crushing rolls 1 m diameter are set so that the crushing surfaces are 12.5 mm apart and the angle of nip is 31◦, what is the maximum size of particle which should be fed to the rolls? If the actual capacity of the machine is 12 per cent of the theoretical, calculate the throughput in kg/s when running at 2.0 Hz if the working face of the rolls is 0.4 m long and the feed density is 2500 kg/m3. Maximum size = 25.1 micrometer Throughput rate = 3 kg/s A crushing mill reduces limestone from a mean particle size of 45 mm to the following product: Size (mm) Amount of product (per cent) 12.5 0.5 7.5 7.5 5.0 45.0 2.5 19.0 1.5 16.0 0.75 8.0 0.40 3.0 0.20 1.0 It requires 21 kJ/kg of material crushed. Calculate the power required to crush the same material

at the same rate, from a feed having a mean size of 25 mm to a product with a mean size of 1 mm. Power required = 38.6 kJ/kg A ball-mill 1.2 m in diameter is run at 0.8 Hz and it is found that the mill is not working satisfactorily. Should any modification in the condition of operation be suggested? Speed should be halved,i.e., 0.38 Hz Power of 3 kW is supplied to a machine crushing material at the rate of 0.3 kg/s from 12.5 mm cubes to a product having the following sizes: 80 per cent 3.175 mm, 10 per cent 2.5 mm and 10 per cent 2.25 mm. What power should be supplied to this machine to crush 0.3 kg/s of the same material from 7.5 mm cube to 2.0 mm cube? Power required = 3.6 kW SEDIMENTATION Calculate the terminal velocity of a steel ball, 2 mm diameter and of density 7870 kg/m3 in an oil of density 900 kg/m3 and viscosity 50 mNs/m2. Terminal velocity = 0.189 m/s (If calculated, would give a 7% lower answer at 0.175 m/s) What is the terminal velocity of a spherical steel particle of 0.40 mm diameter, settling in an oil of density 820 kg/m3 and viscosity 10 mN s/m2? The density of steel is 7870 kg/m3. Terminal velocity = 0.051 m/s (If calculated, would give a 7% lower answer at 0.049 m/s) What are the settling velocities of mica plates, 1 mm thick and ranging in are from 6 to 600 mm , in an oil of density 820 kg/m3 and viscosity 10 mN s/m2? The density of mica is 3000 kg/m3. Settling velocities = 0.154 m/s for smallest and 0.159 m/s for largest, All mica plates will settle at approximately the same velocities. 2

A material of density 2500 kg/m3 is fed to a size separation plant where the separating fluid is water which rises with a velocity of 1.2 m/s. The upward vertical component of the velocity of the particles is 6 m/s. How far will an approximately spherical particle, 6 mm diameter, rise relative to the walls of the plant before it comes to rest in the fluid? Answer: 240 mm