Bài tập chương 2: Kỹ Thuật Phân Tán Pha 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) Number in range, fi Relative number, Pi Cumulative no Undersize,Wi Relative no per m, Pp,i 50 0 ∑= 525 ∑= 2804559,375 Formula: Relative number: Pi fi f Mid point: xi i Cumulative no undersize: Wi Wi 1 Pi P Relative no per mm: Pp ,i i xi xi 1 Cumulative mass undersize: mi f i x i f x i Relative mass: Sv 6 mi xi i 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: fx f i i i Where fi 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): a: 12.4 b: 13.5 c: 25.0 d: 28.2 Solution: xmean 50.6 150.8,5 200.12,5 55.17,5 45.25 20.35 5.45 13,5 m f 525 i The mean size by mass, calculated in a similar way, is (µm): a: 12.4 b: 13.5 c: 25.0 d: 28.2 f i xi Solution: Size range ( m) Mass fraction 50 Mass fraction: mi f i x i f i x i ; xmean mi xi m ; mi i The complete distribution specific surface area per unit volume (Sv) is the total surface area divided by the total volume It may be calculated by several routes, one is shown left For this distribution Sv is ( m): a: 0.213 b: 0.240 c: 0.264 d: 0.444 Solution: Sv 6 Size range ( mm) Mass fraction mi/xi mi 0,264173 xi 50 Now convert your answer into one with SI units You will need the equations relating specific surface (Sv) 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 µm and 100% by number at 101 µm The equation for N0(x) is: N0(x)= a: x-1 b: x/100-1 c: x/100-0.01 d: x-100 N o (1) Solution: no ( x) 0,01 0,01x N (101) o Note, work in fractional terms and NOT percentages (ii) The equation for n0(x) is: n0(x)= a: b: 1/100 c: 100 d: x/100-1 dN o ( x) 0,01 dx What is the shape of the n0(x) graph given by your answer to part (ii)? Solution: no ( x) (iii) (iv) Solution: no ( x) 0,01 đường thẳng cắt trục no ( x) song song với x The best equation to use for calculation of specific surface is: 101 101 x dx x dx 1 100 a Sv 101 b Sv 101 c Sv 101 x dx 1 xdx 1 x dx 1 100 101 Solution: Sv 101 x no ( x)dx 101 x n ( x)dx o (v) x dx 100 x3dx 100 101 chọn A The specific surface area per unit volume is ( m1 ): a 0,013 b 0,045 c 0,059 d 0,079 (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 Solution: S ssphere c 0,066 d 0,053 0,079 0,088 m1 0,9 (i) The spercific surface of another size distribution having the same limits as that given in question but on a mass distribution N3(x) basis is ( m1 ) a 0,013 b 0,138 c 0,277 d 0,544 Solution: n3 ( x) dN3 ( x) d 0,01x 0,01 0,01 dx dx 101 101 n ( x) 0,01 Sv dx dx 0,277 x x 1 (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: xSv 21,7 m Sv 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 For 50 x 15 m; N o ( x) 0,83 0,043ln( x) 15 x m; N o ( x) 0,30 0,293ln( x) x 0,5 m; N o ( x) 0,21 0,304ln( x) where N0(x) is the cumulative fraction by number less than the particle size in µm (i) Write down below the equations,for n0(x) for: dN o 0,043 50 x 15 µm : no ( x) dx x dN o 0,239 dx x dN o 0,304 x 0.5 µm : no ( x) dx x 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): 15 x µm : no ( x) (ii) 15 50 x n ( x)dx x n ( x)dx x n ( x)dx 2 o Sv 0,5 o 15 o 15 50 x n ( x)dx x n ( x)dx x n ( x)dx 3 o 0,5 o o 15 Just considering the size grade 50≥ x ≥15 µm the area integral in the above equation (i.e ∫ x 2n0 (x)dx ) is (µm2): a 48.91 b: 12.35 50 Solution: 50 x n ( x)dx x 2 o 15 15 c: 293.5 0,043 dx 48,91 x d: 1.286 (iii) The volume integral for the same grade is (µm3): a: 5895 b: 48.91 c: 1743 d: 102.1 0,043 Solution: x3no ( x)dx x dx 1743 x 15 15 The area and volume integrals for the other sizr ranges are as follows: 50 (iv) 50 for: 15≥ x ≥4 µm area is 24.98 µm2 volume is 263.8 µm3 4≥ x ≥0.5 µm area is 2.39 µm2 volume is 6.473 µm3 The specific surface area per unit volume of the fog is (µm-1): a: 0.0379 b: 0.227 c: 2.96 d: 0.493 Solution: 15 50 x n ( x)dx x n ( x)dx x n ( x)dx 2 o Sv 0,5 4 15 o 15 50 x n ( x)dx x n ( x)dx x n ( x)dx 3 o 0,5 (v) (vi) o o 2,39 24,98 48,91 0,227 6,473 263,8 1743 o 15 Now convert your answer into one with SI units: 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 I ln ln Io I 100 1,6.106 Solution: exp SvCL C 1 Io Sv L 0,227.106.50 4 ... the mean size by number is (µm): a: 12. 4 b: 13.5 c: 25 .0 d: 28 .2 Solution: xmean 50.6 150.8,5 20 0. 12, 5 55.17,5 45 .25 20 .35 5.45 13,5 m f 525 i The mean size by mass, calculated... is (µm): a: 12. 4 b: 13.5 c: 25 .0 d: 28 .2 f i xi Solution: Size range ( m) Mass fraction 50 Mass fraction:... m): a: 0 .21 3 b: 0 .24 0 c: 0 .26 4 d: 0.444 Solution: Sv 6 Size range ( mm) Mass fraction mi/xi mi 0 ,26 4173 xi