The following size distribution was obtained for glass spheres by microscope examination.. Complete the table rows showing the relative number of particles, the cumulative number undersi
Trang 1Bài tập chương 2: Kỹ Thuật Phân Tán Pha
1 The following size distribution was obtained for glass spheres by microscope examination Complete the table rows showing the relative number of particles, the cumulative number undersize and the relative (i.e fractional) number per micron range (Câu 1+2)
Solution:
Size range ( m) <5 5-7 7-10 10-15 15-20 20-30 30-40 40-50 >50
Relative number, Pi 0 0,0952 0,2857 0,3810 0,1048 0,0857 0,0381 0,0095 0
Cumulative no Undersize,Wi 0 0,0952 0,3810 0,7619 0,8667 0,9524 0,9905 1,0000
Relative no per m, Pp,i 0 0,0476 0,1429 0,1905 0,0524 0,0429 0,0190 0,0048 0
(mid point)3.frequency,
3
i i
f x 10800 92118,75 390625 294765,625 703125 857500 455625 ∑= 2804559,375 Cumulative mass undersize ,mi 0 0,0039 0,0367 0,1760 0,2811 0,5318 0,8375 1,0000
Relative mass, Sv 0 0,0039 0,0259 0,0845 0,0964 0,1276 0,1436 0,1333
Formula:
i
i
f P
f
Cumulative no undersize: W i W i1P i
2
i
p i
P
2
i i i
Cumulative mass undersize:
3 3
i i i
i i
f x m
f x
v
i
m S
x
Trang 23 The mean particle size by number is best calculated by determining the fractional contribution to the mean from each size range, or grade This is the product of the relative number and the mid-point size in the grade Summing all the contributions together gives the mean size Note that this is mathematically equivalent to the more conventional:
i i i
f x f
Where f i is the frequency (or number) of occasions that xi occurs, eg You may have used this formula before to calculate
an average mark
Select the correct answer from the following:
Thus the mean size by number is (µm):
Solution: 50.6 150.8,5 200.12,5 55.17,5 45.25 20.35 5.45 13,5
525
i i mean
i
f x
4 The mean size by mass, calculated in a similar way, is (µm):
Solution:
Trang 3
Mass fraction:
3 3
i i i
i i
f x m
f x
i i
i
m x
m
5 The complete distribution specific surface area per unit volume (S v ) is the total surface area divided by the total volume
It may be calculated by several routes, one is shown left For this distribution S v is (m):
a: 0.213 b: 0.240 c: 0.264 d: 0.444
Solution: 6 i 0, 264173
v
i
m S
x
mi/xi 0,0006418 0,0038642 0,0111426 0,0060058 0,0100283 0,0087358 0,0036102
Now convert your answer into one with SI units
6 You will need the equations relating specific surface (S v ) to number frequency and also to mass fracion (both per micron) See section 2.3 for the derivation of the last relation starting from the former For spherical particles the two equations are given in the box on the right These equations are based on an idealised cumulative size distribution that can be represented as a single straight line on the upper figure give to the right
(i) The cumulative size distribution of a powdered material may be represented as
a straight line on a % number undersize versus particle diameter (x) graph
passing through the points 0% by number at 1 µm and 100% by number at 101
µm The equation for N 0(x) is: N0(x)=
Solution: (1) 0 ( ) 0,01 0,01
(101) 1
o
o o
N
N
Trang 4Note, work in fractional terms and NOT percentages
(ii) The equation for n 0(x) is: n0(x)=
Solution: ( ) o( ) 0,01
o
dN x
n x
dx
(iii) What is the shape of the n 0(x) graph given by your answer to part (ii)?
Solution: n x o( )0,01 đường thẳng cắt trục ( )n x song song với x o
(iv) The best equation to use for calculation of specific surface is:
a
101 2
1
101 3
1
100 6
100
v
x dx S
x dx
1
6
v
S
xdx
c
101 2 1 101 3 1
6
v
x dx S
x dx
Solution:
2
3
( )
100
( )
100
o v
o
x dx
x n x dx S
x dx
x n x dx
chọn A
(v) The specific surface area per unit volume is (m1):
a 0,013 b 0,045 c 0,059 d 0,079
Trang 5(vi) Now convert your answer into one with SI units
(vii) If sphericity of material is really 0,9 the specific surface is (m1
):
a 0,071 b 0,088 c 0,066 d 0,053
0,088 0,9
sphere
s
7
(i) The spercific surface of another size distribution having the same limits as that given in question 6 but on
a mass distribution N 3 (x) basis is (m1)
a 0,013 b 0,138 c 0,277 d 0,544
Solution:
3 3
( )
3
v
n x
(ii) The Sauter mean diameter of the distribution in part (i) is (m)
a 50 b 43,5 c 21,7 d.10,8
Solution:
6 21,7
v
S
v
8 The attenuation of a parallel beam of light is used to monitor visibility on an airport approach during
fog conditions It is found, experimentally, that the droplet size distribution of the fog may be
represented by the equations
Trang 6For
o o o
where N 0(x) is the cumulative fraction by number less than the particle size in µm
(i) Write down below the equations,for n 0(x)
for:
0,043 ( )
o
dN
n x
0, 239 ( )
o
dN
n x
x
d
0,304 (
o
n x
dx
x
(ii) The specific surface area per unit volume (Sv ) of the fog is the total surface area divided by the
total volume, in number distribution terms (assuming spherical droplets):
6
v
S
Just considering the size grade 50≥ x ≥15 µm the area integral in the above equation (i.e ∫ x 2
n0 (x)dx ) is
(µm 2 ):
Solution:
o
x
Trang 7(iii) The volume integral for the same grade is (µm 3 ):
Solution:
0,043
o
x
(iv) The area and volume integrals for the other sizr ranges are as follows:
for:
15≥ x ≥4 µm area is 24.98 µm2 volume is 263.8 µm 3
4≥ x ≥0.5 µm area is 2.39 µm2
volume is 6.473 µm 3 The specific surface area per unit volume of the fog is (µm -1 ):
Solution:
6, 473 263,8 1743
v
S
(v) Now convert your answer into one with SI units:
(vi) If the path length is 50 m, the concentration of fog that will reduce the intensity of the beam by a
factor of 100 is (-):
a: 1.6x10-6 b: 1.6x10-5 c: 1.6x10-4 d: 1.6x10-3
6
1
o v
o
v
I I I