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Subject: Estimation of sphericities
Given: (a) a cylindrical needle with height = 5 times the diameter.
(b) a rectangular prism of sides a, 2a, and 3a.
Find: The sphericities
Analysis: The sphericity, ψ, is defined by (171) and (172):
ψ= surface area of a sphere of same volume as the particle
6 Vparticle
=
surface area of particle
D p S particle (1) (a) Let the diameter of the needle be D and the height be H = 5D
2
πD 2 H πD ( 5 D ) 5πD 3
=
=
The volume of the needle = Vneedle =
4
4
4
2
πD
πD 2 11 2
The surface area of the needle = S needle = πDH + 2
= πD ( 5 D ) +
= πD
4
2
2 Substitution into (1), gives: ψ = 6
Dp 5πD 3
4
15 D
=
11 2
11 D p
πD
2 From the definition of the sphericity, Vneedle = Vsphere
Therefore, Dp
D = 30
4 (2) 3
5πD 3 πD p
or
=
4
6 1/ 3 = 1.96 and from (2), ψ= 15 1
= 0.696
11 1.96 (b) Let the sides of the prism be a, 2a, and 3a.
The volume of the prism = Vprism = a(2a)(3a) = 6a3
The surface area of the prism = S prism = 2 ( a ) ( 2a ) + ( a ) ( 3a ) + ( 2a ) ( 3a ) = 22a 2
Substitution into (1) gives: ψ = 6 6a 3
36 a
=
2
D p 22a
22 D p From the definition of the sphericity, Vprism = Vsphere or 6a 3 =
Dp 36
Therefore,
=
a
3.14 1/ 3 = 2.55 and from (2), ψ= πD 3
p
6 36 1
22 2.25 = 0.727 Exercise 17.2
Subject: Sphericity of a thin circular plate
Given: Circular plate of thickness, t, and diameter, D, with sphericity, ψ, equal to 0.594
Find: The ratio of t to D.
Analysis: The sphericity, ψ, is defined by (171) and (172):
ψ= surface area of a sphere of same volume as the particle
6 Vparticle
=
surface area of particle
D p S particle The volume of the plate = Vplate = πD 2t
4 The surface area of the plate = S plate = 2 Substitution into (1), gives: ψ = πD 2
πD 2
+ πDt =
+ πDt
4
2
πD 2t
4 6
6
tD
=
2
D p πD
D p 2 D + 4t
+ πDt
2 From the definition of the sphericity, Vsphere = Vplate or 3
Therefore, D 3 = D 2t
p
2 (1) and from (2), πD 3
p
6 = ψ = 0.594 = (2) πD 2t
4
6 32
Dt
2 1/ 3 tD
2 D + 4t (3) Let R = t/D Therefore, t = RD
Substitution into (3) gives: 0.594 = 6
33
DR
2 1/ 3 RD 2
R
= 5.24 1/ 3
R ( 2 + 4R )
( 2 D + 4 RD ) which can be rewritten as, 0.297 R1/ 3 + 0.594 R 4 / 3 − 1.31 = 0
Solving (4), a nonlinear equation, with Polymath, for an initial guess of 0.3, R = t/D = 0.166 (4) Exercise 17.3
Subject: Differential and cumulative undersize plots of a screen analysis for crystals of sodium
thiosulfate.
Given: U.S. Screen analysis.
Assumptions: particles approximate spheres
Find: Differential and cumulative undersize plots of the data
Analysis: Use a spreadsheet with apertures for U.S. Mesh from Table 17.4. Results are as
follows with the two plots on the next page. Arithmetic plots are preferred here.
Differential and Cumulative Undersize Screen Analyses
Differential Analysis Cumulative Undersize
Analysis U.S.
Mesh
Number Aperture,
Dp, mm Mass
retained,
grams % Mass
retained Average
Particle
Size, mm Mass
Fraction,
xi Aperture,
Dp, mm Cumulative
Undersize
wt% 6
8
12
16
20
30
40
50
70
100
140
170
230 3.350
2.360
1.700
1.180
0.850
0.600
0.425
0.300
0.212
0.150
0.106
0.090...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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