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area per unit mass of coarse particles and fine particles, respectively. (c) Determine the surface equivalent sphere diameter ds of coarse particles and fine particles, respectively. (d) Calculate the number weighted mean of ds and the volume weighted mean of ds for the sample. SHOW STEPS ON HOW TO REACH: answers = W2/T5 (a) 0.832 ; 0.99 (b) 287 m2 /kg ; 314 m2 /kg (c) 1 x 10-4 m ; 2.35 x 10-5 m (d) d1,0 = 8.71 x 10-5 m ; d4,3 = 9.92 x 10-5 m

area per unit mass of coarse particles and fine particles, respectively. (c) Determine the surface equivalent sphere diameter ds of coarse particles and fine particles, respectively. (d) Calculate the number weighted mean of ds and the volume weighted mean of ds for the sample. SHOW STEPS ON HOW TO REACH: answers = W2/T5 (a) 0.832 ; 0.99 (b) 287 m2 /kg ; 314 m2 /kg (c) 1 x 10-4 m ; 2.35 x 10-5 m (d) d1,0 = 8.71 x 10-5 m ; d4,3 = 9.92 x 10-5 m

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Consider a sample of cylindrical particles with circular cross-section of diameter D and with varying length L. The sample contains 99% (by weight) of coarse particles with length of 1 mm and 1% (by weight) of fine particles with length of 50 um. The value of diameter D is equal to 10um, independent of the particle length. The density of particles is 1400 kg/m3 . (a) Calculate the number fraction and the volume fraction of coarse particles in the sample. (b) Calculate the surface area per unit mass of coarse particles and fine particles, respectively. (c) Determine the surface equivalent sphere diameter ds of coarse particles and fine particles, respectively. (d) Calculate the number weighted mean of ds and the volume weighted mean of ds for the sample.

SHOW STEPS ON HOW TO REACH:

answers = W2/T5 (a) 0.832 ; 0.99 (b) 287 m2 /kg ; 314 m2 /kg (c) 1 x 10-4 m ; 2.35 x 10-5 m (d) d1,0 = 8.71 x 10-5 m ; d4,3 = 9.92 x 10-5 m

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Relative frequency: ƒ,= f(x) <-²₁ = Q(x₁) - Q(x₁) = q (x)dx i- XH-1 Cumulative frequency distribution: Q(x) Differential frequency distribution: q (x) Weighted mean size: x j=1 Zw, J Total volume of all particles: == = w₁x, N,w.x, = Amount of particles with size ≤ x Amount of all particles or q(x₁) ≈ Nbins S₂ = Σ №₁ ( ²nd² + nd₁L); i=1 dQ(x) dx n-th moment of particle size distribution: 4 = Number weighted mean size: μ₁/Mo Length weighted mean size: μ₂/M1 Surface weighted mean size: μ3/μ₂ Volume weighted mean size: μ4/μ3 = ΣN,w, i=1 fw,x, Σf,w, i=1 i=1 i=1 =[fix"=√x²q₁(x) dx Nbins V₁ = Σ N₁Bx²³ = BN x³ qn(x) dx i=1 [w(x) xq, (x) dx Tw(x) q₁ (x)dx Total surface area of all particles (for cube and sphere shapes): Nbins St = Σ N₁ax² = aN √ x² qn(x) dx i=1 Q(xi+1)-Q(xi) Xi+1-Xi Total surface area of all particles (for cylinder shapes with constant L): -N (ES 7x² qn(x) dx +nL il ( x Qan (x) dx) qn Total surface area of all particles (for cylinder shapes with constant d): Nbins S₁ = - Σ N. (2nd² + ndl.,) = N( ½ d² + xd ( x 9, (x) dx) i=1
Transcribed Image Text:Relative frequency: ƒ,= f(x) <-²₁ = Q(x₁) - Q(x₁) = q (x)dx i- XH-1 Cumulative frequency distribution: Q(x) Differential frequency distribution: q (x) Weighted mean size: x j=1 Zw, J Total volume of all particles: == = w₁x, N,w.x, = Amount of particles with size ≤ x Amount of all particles or q(x₁) ≈ Nbins S₂ = Σ №₁ ( ²nd² + nd₁L); i=1 dQ(x) dx n-th moment of particle size distribution: 4 = Number weighted mean size: μ₁/Mo Length weighted mean size: μ₂/M1 Surface weighted mean size: μ3/μ₂ Volume weighted mean size: μ4/μ3 = ΣN,w, i=1 fw,x, Σf,w, i=1 i=1 i=1 =[fix"=√x²q₁(x) dx Nbins V₁ = Σ N₁Bx²³ = BN x³ qn(x) dx i=1 [w(x) xq, (x) dx Tw(x) q₁ (x)dx Total surface area of all particles (for cube and sphere shapes): Nbins St = Σ N₁ax² = aN √ x² qn(x) dx i=1 Q(xi+1)-Q(xi) Xi+1-Xi Total surface area of all particles (for cylinder shapes with constant L): -N (ES 7x² qn(x) dx +nL il ( x Qan (x) dx) qn Total surface area of all particles (for cylinder shapes with constant d): Nbins S₁ = - Σ N. (2nd² + ndl.,) = N( ½ d² + xd ( x 9, (x) dx) i=1
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