Homework Help is Here – Start Your Trial Now!
Engineering
Chemical Engineering
TASK 6: Consider a sample of disc-shaped (cylindrical) particles with circular cross-section of varying 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, d,, equal to 10 μm?

TASK 6: Consider a sample of disc-shaped (cylindrical) particles with circular cross-section of varying 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, d,, 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
See similar textbooks
icon
Related questions
Question

answers are

(a) 0.00819; 0.0819; 0.9099 (b) dv = 24.7, 5.31 and 1.14 m (c) d1,0 = 1.67 um ; d4,3 = 22.7 um

show wokring on how to reach these

 

PLEASE DONT COPY AND PASTE FROM OTHER ANSWERED QUESTIONS AS THEY ARE WRONG 

WILL LIKE IF RIGHT 

 

 

TASK 6: Consider a sample of disc-shaped (cylindrical) particles with circular cross-section of varying 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 μm 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,№, 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?
Transcribed Image Text:TASK 6: Consider a sample of disc-shaped (cylindrical) particles with circular cross-section of varying 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 μm 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,№, 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?
Relative frequency: f₁ = f(x)}¸₂¸‹‹«x, = Q(x,) — Q(x₁) = † q(x)dx Cumulative frequency distribution: Q(x) Differential frequency distribution: q (x) = Weighted mean size: x j=1 w; x ; N Σω, j=1 Number weighted mean size: μ₁/μo Length weighted mean size: μ₂/M₁ Surface weighted mean size: μ3/μ₂ Volume weighted mean size: μ4/μ3 Total volume of all particles: St= N = i=1 i=1 Amount of particles with size ≤ x Amount of all particles N,w.x, ΣN,w, i=1 dQ(x) or q(xi) ~ dx Nhi n-th moment of particle size distribution: μ = Σfx" = [x"q„(x)dx i=1 Nbins = Σ N₁ (2nd²7 + nd₂l) = N( Ni i=1 Nins Σfwx i=1 Nin fiwi [w(x) xq (x)dx Tw(x)q, (x) dx = x Nbins V₁ = Σ N₁Bx³ = µN √ x³ qn(x) dx i=1 Total surface area of all particles (for cube and sphere shapes): Nbins S₁ = N₁ax² =aN x² qn(x) dx Nfx² Total surface area of all particles (for cylinder shapes with constant L): (√x Q(xi+1)-Q(xi) Xi+1-Xi x² qn(x) dx + πL + L ( x Q₁ (x) dx ) qn x Total surface area of all particles (for cylinder shapes with constant d): Nbins St = • Σ N₁ (²nd² + nadly) = N( 7d² +nd [ x 4₁ x xqn (x) dx i=1
Transcribed Image Text:Relative frequency: f₁ = f(x)}¸₂¸‹‹«x, = Q(x,) — Q(x₁) = † q(x)dx Cumulative frequency distribution: Q(x) Differential frequency distribution: q (x) = Weighted mean size: x j=1 w; x ; N Σω, j=1 Number weighted mean size: μ₁/μo Length weighted mean size: μ₂/M₁ Surface weighted mean size: μ3/μ₂ Volume weighted mean size: μ4/μ3 Total volume of all particles: St= N = i=1 i=1 Amount of particles with size ≤ x Amount of all particles N,w.x, ΣN,w, i=1 dQ(x) or q(xi) ~ dx Nhi n-th moment of particle size distribution: μ = Σfx" = [x"q„(x)dx i=1 Nbins = Σ N₁ (2nd²7 + nd₂l) = N( Ni i=1 Nins Σfwx i=1 Nin fiwi [w(x) xq (x)dx Tw(x)q, (x) dx = x Nbins V₁ = Σ N₁Bx³ = µN √ x³ qn(x) dx i=1 Total surface area of all particles (for cube and sphere shapes): Nbins S₁ = N₁ax² =aN x² qn(x) dx Nfx² Total surface area of all particles (for cylinder shapes with constant L): (√x Q(xi+1)-Q(xi) Xi+1-Xi x² qn(x) dx + πL + L ( x Q₁ (x) dx ) qn x Total surface area of all particles (for cylinder shapes with constant d): Nbins St = • Σ N₁ (²nd² + nadly) = N( 7d² +nd [ x 4₁ x xqn (x) dx i=1
Expert Solution
steps

Step by step

Solved in 5 steps with 27 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Introduction to Chemical Engineering Thermodynami…
Introduction to Chemical Engineering Thermodynami…
Chemical Engineering
ISBN:
9781259696527
Author:
J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:
McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemical Engineering
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
Elements of Chemical Reaction Engineering (5th Ed…
Elements of Chemical Reaction Engineering (5th Ed…
Chemical Engineering
ISBN:
9780133887518
Author:
H. Scott Fogler
Publisher:
Prentice Hall
Process Dynamics and Control, 4e
Process Dynamics and Control, 4e
Chemical Engineering
ISBN:
9781119285915
Author:
Seborg
Publisher:
WILEY
Industrial Plastics: Theory and Applications
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:
9781285061238
Author:
Lokensgard, Erik
Publisher:
Delmar Cengage Learning
Unit Operations of Chemical Engineering
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:
9780072848236
Author:
Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:
McGraw-Hill Companies, The
SEE MORE TEXTBOOKS