Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Find differential and cumulative undersize and

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Unformatted text preview: 0.063 0.0 8.8 21.3 138.2 211.6 161.7 81.6 44.1 28.7 13.2 9.6 8.8 7.4 0.00 1.20 2.90 18.80 28.79 22.00 11.10 6.00 3.90 1.80 1.31 1.20 1.01 2.855 2.030 1.440 1.015 0.725 0.513 0.363 0.256 0.181 0.128 0.098 0.077 0.0120 0.0290 0.1880 0.2879 0.2200 0.1110 0.0600 0.0390 0.0180 0.0131 0.0120 0.0101 3.350 2.360 1.700 1.180 0.850 0.600 0.425 0.300 0.212 0.150 0.106 0.090 0.063 100.00 98.80 95.90 77.10 48.31 26.31 15.21 9.21 5.31 3.51 2.20 1.01 0.00 735.0 100.00 Total 1.0000 Exercise 17.3 (continued) 0.35 0.30 Differential Plot 0.25 0.20 Mass Fraction 0.15 0.10 0.05 0.00 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Average Particle Size, mm 100 90 80 70 60 Cumulative Mass, % Cumulative Undersize Plot 50 40 30 20 10 0 0.0 0.5 1.0 1.5 2.0 Aperture Size, mm 2.5 3.0 3.5 4.0 Exercise 17.4 Subject: Mean particle diameter from a particle-size analysis Given: Particle-size analysis in terms of the number of particles in each size fraction Assumptions: All particles have the same shape Find: Expressions for the surface-mean and mass-mean diameters based on numbers of particles, Ni, rather than mass fractions, xi, for each size range. Analysis: It is important to note that regardless of whether the expression is in terms of mass fractions or numbers of particles, the numerical result for the mean diameter is the same. Surface-mean diameter: From (17-7), the surface-mean diameter in terms of mass fractions is: 1 DS = n xi i =1 D pi (1) From (17-10), mass fraction is related to number of particles by, xi = ( )ρ N i f v D pi p (2) Mt Substituting (2) into (1) gives, DS = 3 1 fvρ p n Mt i =1 Ni D 2 pi Summing (2) over all mass fractions gives, fvρ p 1 =n 3 Mt N i D pi i =1 Substituting (4) into (3) gives the desired expression for surface-mean diameter: n DS = i =1 n i =1 3 N i D pi 2 N i D pi (3) (4) Exercise 17.4 (continued) Mass-mean diameter: From (17-8), the mass-mean diameter in terms of mass fractions is: n DW = i =1 Substitution of (2) into (5) gives, DW = (5) xi D pi fvρ p n Mt i =1 () N i D pi 4 Substitution of (4) into (6) gives the desired expression for mass-mean diameter: n DW = i =1 n i =1 4 N i D pi 3 N i D pi It may be noted that expressions for the other two mean diameters in terms of numbers of particles are given as (17-9) for arithmetic mean and as (17-14) for the volume mean. (6) Exercise 17.5 Subject: Calculation of mean particle diameters from a screen analysis. Given: Screen analysis of sodium thiosulfate crystals from Exercise 17.3. Assumptions: All particles have the same sphericity and volume shape factor. Find: Surface-mean, mass-mean, arithmetic-mean, and volume-mean diameters Analysis: The following equations apply for n screen fractions, i: Surface-mean diameter, (17-7): DS = 1 n i =1 Mass-mean diameter, (17-8): DW = n i =1 where D pi is average aperture xi D pi xi D pi n Volume-mean diameter, (17-15): i =1 D pi n xi i =1 Arithmetic-mean diameter, (17-12): D N = xi D pi 2 3 1 DV = n xi i =1 1/ 3 D pi 3 The calculations are made with a spreadsheet on the following page where mesh apertures...
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