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
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
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

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?

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
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