Q1. Particle size distribution of a sample of spherical particles is represented by the differential frequency distribution qn(d) which is a function of particle diameter d with the graph shown below. 9n(d) (µm-1) 0 0 (a) Calculate the value of qn(0). 250 d (μm) (b) Calculate values of the cumulative frequency distribution Qn(d) for the following values of d (in µm): 25, 125 and 225. (c) Calculate the length weighted mean diameter and the surface weighted mean diameter of particles in the sample.
Q1. Particle size distribution of a sample of spherical particles is represented by the differential frequency distribution qn(d) which is a function of particle diameter d with the graph shown below. 9n(d) (µm-1) 0 0 (a) Calculate the value of qn(0). 250 d (μm) (b) Calculate values of the cumulative frequency distribution Qn(d) for the following values of d (in µm): 25, 125 and 225. (c) Calculate the length weighted mean diameter and the surface weighted mean diameter of particles in the sample.
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
100%
Q1. (a) 8 x 10-3 um-1
(b) 0.19, 0.75 and 0.99
(c) 125 um and 150 um
(d) 1.22 x 106 particles and 0.04 m2
Transcribed Image Text:Relative frequency: f₁ = f(x)x₁,<x5x₁ = Q(x,) - Q(x₁) =
Cumulative frequency distribution: Q(x)
Differential frequency distribution: q(x):
Weighted mean size: x =
Number weighted mean size: M₁/Mo
Length weighted mean size: μ₂/M₁
Surface weighted mean size: μ3/μ₂
Volume weighted mean size: μ4/M3
Total volume of all particles:
W₁
St =
Nbins
S₁
=
Nins
-
n-th moment of particle size distribution: A4, =
Ni
Nin
Σw, x, [N, wx, [ƒ, wx, [w(x) xq (x) dx
i=1
Tw(x) q₁ (x)dx
i=1
Amount of particles with size ≤ x
Amount of all particles
dQ(x)
dx
ΣN,W
i=1
Nbins
• Σ N₁ (= ma² + xd₁L) = N(
i=1
=
Nbins
=[ N₁ (²nd² + 7dL₁) = N
=
лdL;
i=1
Ĵq(x)dx
i=1
~(ESX
x
or q (x₁)
i=1
Nhi
ΣΗ
i=1
Vs = ΣN₁Bx² = BN x³ qn(x) dx
fiw,
Total surface area of all particles (for cube and sphere shapes):
Nbins
S₁ = Σ N₁ax? = aN
i=1
f₁x" = √x"q, (x) dx
X
Total surface area of all particles (for cylinder shapes with constant L):
Anh da
x qn (x)
Total surface area of all particles (for cylinder shapes with constant d):
≈
[x² qn(x) dx
√ x²
Q(xi+1)-Q(xi)
Xi+1-Xi
x² qn(x) dx + L
+ N(57d² + md [ x 9₁ (x) dx)
S
qn
Transcribed Image Text:Q1. Particle size distribution of a sample of spherical particles is represented by
the differential frequency distribution qn(d) which is a function of particle diameter
d with the graph shown below.
an(d)
(μm-1)
0
0
(a) Calculate the value of qn(0).
250
d (μm)
(b) Calculate values of the cumulative frequency distribution Qn(d) for the
following values of d (in µm): 25, 125 and 225.
(c) Calculate the length weighted mean diameter and the surface weighted mean
diameter of particles in the sample.
(d) Calculate the total number and total surface area of particles in 1 cm³ of the
sample.
(e) Consider a gas laden with particles which exhibit the size distribution given in
this question. Describe how you could separate the particles from the gas.
Explain your reasoning.
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Follow-up Question
where does the 3.2x10-5 come from ?
Transcribed Image Text:Step 2: b) Cumulative frequency distribution
Cumulative frequency distribution Q₁,(x) = 9,(x)dx
For 0 ≤ x ≤ 250
= 0.008-3.2x 10-5x
9n(x):
Solution
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