UGPA3173 UNIT OPERATION II
May Trimester 2014
Tutorial I
1)
Table 1 shows the screen analysis of crushed coal which segregated into masses of
particles with different density, sphericity and shape factor i.e. particle A, B and C. Given
that the density, sphericity, s and shape factor, respectively of A: 1970 kg/m3, 0.375,
0.4; B: 2010 kg/m3, 0.498, 0.5 and C: 2110kg/m3, 0.528. 0.6. The screen analysis of
particles A, B and C are given in Table 1 based on the mass retained in each mesh.
Based on this screen analysis, calculate for mixture of A, B and C:
(i)
Specific surface area of mixture (Aw) in mm2/g
(ii)
Total population of the mixture (Nw) in particles/g
(iii)
Volume-surface mean diameter, DS
(iv)
Mass mean diameter, DW
(v)
Volume mean diameter, DV
(vi)
Surface mean diameter, DS'
Table 1: Screen analysis of crushed coal
Particles
Particle size
(mm)
Mass retained
(g)
0.025
0.050
0.075
0.100
0.125
0.150
0.175
0.200
0.225
6.35
11.13
19.82
29.19
33.75
27.66
20.15
17.61
10.99
2)
Mixture of particle containing 40% cubes and 60% spherical particles with diameter of 4
mm. Find the equivalent sphere diameter based on:
(i)
Volume of mixture
(ii) Surface of mixture
(iii) Surface-volume ratio of mixture
3)
A crusher reduces coal from coarse particles to finer particles with the following mass
distribution as in Table 2. Given the shape factor is 0.5, sphericity is 1 and density of
particle is 2700 kg/m3. Calculate:
Table 2: Mass distribution of crushed coal
(i)
Particle size (mm)
Mass (percent)
0.05
0.25
0.50
0.75
1.00
10
15
35
20
10
Volume mean diameter,
(ii) The number of particles distribution by fraction.
(iii) Total volume of particles in mixture.
(iv) Total surface of particles in mixture.
UGPA3173 UNIT OPERATION II
May Trimester 2014
Tutorial I
(Solution)
Question 1
Particles
Particle
size
(mm), Dpi
Mass
retained
(g)
mass
fraction,
Xi
Dpi
0.025
6.35
0.036
0.050
11.13
0.075
B
Xi/Dpi
Xi/Dpi
1.563E-05
0.0009
1.440
2304.000
0.063
1.250E-04
0.0032
1.260
504.000
19.82
0.112
4.219E-04
0.0084
1.493
265.480
0.100
29.19
0.165
1.000E-03
0.0165
1.650
165.000
0.125
33.75
0.191
1.953E-03
0.0239
1.528
97.790
0.150
27.66
0.157
3.375E-03
0.0236
1.047
46.520
0.175
20.15
0.114
5.359E-03
0.0200
0.651
21.270
0.200
17.61
0.100
8.000E-03
0.0200
0.500
12.500
0.225
10.99
0.062
1.139E-02
0.0140
0.276
5.443
176.65
1.000
3.164E-02
0.1305
9.845
3422.003
Specific surface area of mixture (Aw) in mm2/g
i)
XiDpi
Given:
-
For particle A, the A = 1970 kg/m3 = 0.00197g/mm3, sA = 0.375 and A = 0.4.
For particle B, the B = 2010 kg/m3 = 0.00201g/mm3, sB = 0.498 and B = 0.5.
For particle C, the C = 2110 kg/m3 = 0.00211g/mm3, sC = 0.528 and C = 0.6.
Therefore Total surface area of the mixture = AwA + AwB + AwC
6
6
(1.440 1.260 1.493)] [
(1.650 1.528 1.047)]
(0.375)(0.00197)
(0.498)(0.00201)
6
[
(0.651 0.500 0.276)]
(0.528)(0.00211)
AW [
AW 34054.82 25325.18 7685.26
AW 67065.26 mm2/g
ii)
Total population of the mixture (Nw) in particles/g
Given:
-
For particle A, the A = 1970 kg/m3 = 0.00197g/mm3, sA = 0.375 and A = 0.4.
For particle B, the B = 2010 kg/m3 = 0.00201g/mm3, sB = 0.498 and B = 0.5.
For particle C, the C = 2110 kg/m3 = 0.00211g/mm3, sC = 0.528 and C = 0.6.
Therefore Total surface area of the mixture = AwA + AwB + AwC
1
1
(2307 504 265.48)] [
(165 97.79 46.52)]
(0.4)(0.00197)
(0.5)(0.00201)
1
[
(21.27 12.50 5.44)]
(0.6)(0.00211)
NW [
NW 3,900,369.059 307,769.636 30,971.587
NW 4,239,110.28 particles/g
(iii)
Volume-surface mean diameter
DS
DS
1
xi
i 1
pi
n
D
1
9.845
DS 0.1016 mm
(iv)
Mass mean diameter
n
DW xi D pi
i 1
DW 0.1305 mm
(v)
Volume mean diameter
1
DV
n x
i3
i 1 D pi
(vi)
1/ 3
DV
3422.003
1/ 3
DV 0.0664 mm
Surface mean diameter
x
i
i 1 D pi
DS'
n
xi
i 1 D pi3
1
2
1
9.845 2
DS'
3422.003
DS' 0.0536 mm
Question 2
Volume of cubes = 4 X 4 X 4 = 64 mm3
Surface area of cubes = 6 (4 X 4) = 96 mm2
Volume of sphere = D3 / 6 = (4)3 / 6 = 33.5 mm3
Surface area of sphere = D2 = (4)3 = 50.27 mm2
Therefore
total volume of mixture = (40% x 64) + (60% x 33.5) = 45.7 mm3
total surface area of mixture = (40% x 96) + (60% x 50.27) = 68.56 mm2
(i)
To find the equivalent sphere diameter based on volume,
Volume of mixture = volume of sphere
45.7 = D3 / 6
therefore D = 4.44 mm
(ii)
To find the equivalent sphere diameter based on surface area,
surface of mixture = surface of sphere
68.56 = D2
therefore D = 4.67 mm
(iii)
To find the equivalent sphere diameter based surface-volume ratio
surface-volume of mixture = surface-volume of sphere
68.56/45.7 = (D2)/(D3 / 6)
6/D = 1.500
therefore D = 3.999, D~4.00 mm
Question 3
Particle
Mass
size, Dpi (percent)
(mm)
Mass
fraction,
xi
Xi/Dpi
(mm-1)
Xi/Dpi3
(mm-3)
Number of Particles
particles in number
mix
fraction,
Ni
Volume
particle
(mm3)
0.05
10
0.1111
2.222
888.800
658370.601
0.9840
6.545 x 10-5
43.090
7.854 x 10-3
5170.843
0.25
15
0.1667
0.667
10.669
7902.966
0.0118
8.181 x 10-3
64.654
0.196
1548.981
0.50
35
0.3889
0.778
3.111
2303.705
0.0034
0.065
149.741
0.785
1808.408
0.75
20
0.2222
0.296
0.527
390.371
0.0006
0.221
86.272
1.767
689.786
1.00
10
0.1111
0.111
0.111
82.222
0.0001
0.524
43.084
3.142
258.342
Sum
90
1.00
4.074
903.218
669049.865
of Volume
Surface
of
particle
particle/g (mm2)
mix
(mm3)
386.841
of Surface of
particle/g
mix (mm2)
9476.36
i.
Volume mean diameter,
1
DV
n xi
3
i 1 D pi
ii.
1/ 3
1
DV
903.218
1/ 3
DV 0.1035 mm
The number of particles distribution by fraction.
To find the number of particle distribution:
First need to find the number of particles in every particle size range, (Nw).
Then the total number of particles and fractions in each particle range.
For example Nw0.05,
NW
1
a p
NW 0.05
xi
D
i 1
pi
1
(888.800)
(0.5)(0.00270)
NW 0.05 658,370.601
So the final number of particle in each particle size range and number of particle distribution
is as in column 6 & 7 of Table 2-1.
Number of particle distribution
Particle size 0.05
(mm)
Number
0.9840
fraction, Ni
iii.
0.25
0.50
0.75
1.00
0.0118
0.0034
0.0006
0.0001
Total volume of particles in mixture.
To find the total volume of particle:
First need to find the volume of single particle (for each size range) which based on
the shpericity = 1 is a sphere. Hence volume of single particle = D3 / 6
For example D = 0.05, V = (0.05)3 / 6 , V = 6.545 x 10-5 mm3
Then multiplying volume of single particle (column 8) with number of particle in each
range (column 6) results in the volume of total particle in each particle range.
For example, total volume V0.05 = (6.545 x 10-5)(658370.601) = 43.090 mm3
Finally the total volume of particles in mixture = sum of particle volume in each size
range.
Total volume of particles in mixture = 386.841 mm3 as in column 9 of Table 2-1
iv.
Total surface of particles in mixture.
To find the total surface of particle:
First need to find the surface area of single particle (for each size range) which based
on the shpericity=1 is a sphere. Hence surface area of single particle = D2
For example D = 0.05, S0.05 = (0.05)2 , S0.05 = 7.854 x 10-3 mm2
Then multiplying surface area of single particle (column 10) with number of particle in
each range (column 6) results in the surface area of total particle in each particle
range.
For example, total surface area S0.05 = (= 7.854 x 10-3)(658370.601) = 5170.843 mm2
Finally the total surface area of particles in mixture = sum of particle surface area in
each size range.
Total surface of particles in mixture = 9476.36 mm2 as in column 11 of Table 2-1
