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TASK 6: Consider a sample of disc-shaped (cylindrical) particles with circular cross-section of varyin diameter D. All particles have the same thickness L equal to 1 µm. The sample contains 90 (by volume) of particles with diameter 100 µm, 9% (by volume) of particles with diameter 1 um and 1% (by volume) of particles with diameter 1 µm. (a) Calculate the number fraction of particles of each diameter in the sample. (b) Determine the volume equivalent sphere diameter dy of particles of each diameter. (c) Calculate the number weighted mean, dv,N, and the volume weighted mean, dv,v, of the volume equivalent sphere diameter of particles in the sample. (d) What volume fractions of each of the three sizes would you choose in order to get a sample with a number weighted mean, dv, N, equal to 10 µm?

TASK 6: Consider a sample of disc-shaped (cylindrical) particles with circular cross-section of varyin diameter D. All particles have the same thickness L equal to 1 µm. The sample contains 90 (by volume) of particles with diameter 100 µm, 9% (by volume) of particles with diameter 1 um and 1% (by volume) of particles with diameter 1 µm. (a) Calculate the number fraction of particles of each diameter in the sample. (b) Determine the volume equivalent sphere diameter dy of particles of each diameter. (c) Calculate the number weighted mean, dv,N, and the volume weighted mean, dv,v, of the volume equivalent sphere diameter of particles in the sample. (d) What volume fractions of each of the three sizes would you choose in order to get a sample with a number weighted mean, dv, N, equal to 10 µm?

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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(a) 0.00819; 0.0819; 0.9099

(b) dv = 24.7, 5.31 and 1.14 um

(c) d1,0 = 1.67 um ; d4,3 = 22.7 um

(d) There are many different solutions (for example, volfrac1 = 0.9696, volfrac2 = 0.0304, volfrac3 = 0)

um = 1x10-6

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Relative frequency: f₁ = ƒ(x)x₁₁<x²x₁ = Q(x;) - Q(x₁) = ↑ q(x)dx XHI Cumulative frequency distribution: Q(x) Differential frequency distribution: q (x) Weighted mean size: x N j=1 Number weighted mean size: M₁/Mo Length weighted mean size: μ₂/M₁ Surface weighted mean size: μ3/μ₂ Volume weighted mean size: μ4/μ3 Total volume of all particles: St = W ; x ; Σω; St= N₁ i=1 Nins St= = i=1 Amount of particles with size ≤ x Amount of all particles or q (xi) ≈ dQ(x) dx Nwx i=1 Nbin ΣN,w, i=1 Nins n-th moment of particle size distribution: μ = = Σf₁x² = √x"q₂(x) dx i=1 = (=d² +nd L₁) Nins [ƒ,wx, [w(x)xq„(x)dx i=1 Ni Total surface area of all particles (for cube and sphere shapes): Nbins Σfiw, i=1 x= Nbins V₁ = Σ N₁³x³ = BN [ x³ qn(x) dx i=1 x Tw(x) q, (x) dx Total surface area of all particles (for cylinder shapes with constant L): Nbins • Σ N₁ (²nd² + xd₁L) = N( N₁ax² = aN x² qn(x) dx Q(xi+1)-Q(xi) Xi+1-Xi Total surface area of all particles (for cylinder shapes with constant d): Nbins x² qn(x) dx + лL - l . [ ) x qn(x) dx = + N (17 d² + md [ x 9, (x) dx ) qn x
Transcribed Image Text:Relative frequency: f₁ = ƒ(x)x₁₁<x²x₁ = Q(x;) - Q(x₁) = ↑ q(x)dx XHI Cumulative frequency distribution: Q(x) Differential frequency distribution: q (x) Weighted mean size: x N j=1 Number weighted mean size: M₁/Mo Length weighted mean size: μ₂/M₁ Surface weighted mean size: μ3/μ₂ Volume weighted mean size: μ4/μ3 Total volume of all particles: St = W ; x ; Σω; St= N₁ i=1 Nins St= = i=1 Amount of particles with size ≤ x Amount of all particles or q (xi) ≈ dQ(x) dx Nwx i=1 Nbin ΣN,w, i=1 Nins n-th moment of particle size distribution: μ = = Σf₁x² = √x"q₂(x) dx i=1 = (=d² +nd L₁) Nins [ƒ,wx, [w(x)xq„(x)dx i=1 Ni Total surface area of all particles (for cube and sphere shapes): Nbins Σfiw, i=1 x= Nbins V₁ = Σ N₁³x³ = BN [ x³ qn(x) dx i=1 x Tw(x) q, (x) dx Total surface area of all particles (for cylinder shapes with constant L): Nbins • Σ N₁ (²nd² + xd₁L) = N( N₁ax² = aN x² qn(x) dx Q(xi+1)-Q(xi) Xi+1-Xi Total surface area of all particles (for cylinder shapes with constant d): Nbins x² qn(x) dx + лL - l . [ ) x qn(x) dx = + N (17 d² + md [ x 9, (x) dx ) qn x
TASK 6: Consider a sample of disc-shaped (cylindrical) particles with circular cross-section of varyin diameter D. All particles have the same thickness L equal to 1 µm. The sample contains 90 (by volume) of particles with diameter 100 µm, 9% (by volume) of particles with diameter 10 um and 1% (by volume) of particles with diameter 1 µm. (a) Calculate the number fraction of particles of each diameter in the sample. (b) Determine the volume equivalent sphere diameter dy of particles of each diameter. (c) Calculate the number weighted mean, dv,N, and the volume weighted mean, dv,v, of the volume equivalent sphere diameter of particles in the sample. (d) What volume fractions of each of the three sizes would you choose in order to get a sample with a number weighted mean, dv,Ñ, equal to 10 μm?
Transcribed Image Text:TASK 6: Consider a sample of disc-shaped (cylindrical) particles with circular cross-section of varyin diameter D. All particles have the same thickness L equal to 1 µm. The sample contains 90 (by volume) of particles with diameter 100 µm, 9% (by volume) of particles with diameter 10 um and 1% (by volume) of particles with diameter 1 µm. (a) Calculate the number fraction of particles of each diameter in the sample. (b) Determine the volume equivalent sphere diameter dy of particles of each diameter. (c) Calculate the number weighted mean, dv,N, and the volume weighted mean, dv,v, of the volume equivalent sphere diameter of particles in the sample. (d) What volume fractions of each of the three sizes would you choose in order to get a sample with a number weighted mean, dv,Ñ, equal to 10 μm?
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