Assignment 1 (2006) Solutions

CHEE 2940 Particle Processing 2006 – Assignment #1 Due 5 PM Tuesday 7 March - Chemical Engineering Office 1. For a cube with dimensions 5.00 x 5.00 x 5.00 mm, calculate the following parameters: a) Equivalent volume diameter (2 marks), b) Equivalent surface diameter (2 marks), c) Sauter diameter (2 marks), and d) Corresponding sphericities (8 marks). 2. Below is the result of a particle size analysis: Aperture size (mm or µm)

amount retained (g)

-10 mm

+9.5

0.36

-9.5

+6.8

5.16

-6.8

+4.75

24.36

-4.75

+3.4

50.04

-3.4

+2.36

83.64

-2.36

+1.7

115.08

-1.7

+1.18

131.28

-1.18

+850 µm

129.6

-850

+600

118.92

-600

+425

101.04

-425

+300

85.44

-300

+212

73.2

-212

+150

62.76

-150

+106

52.56

-106

+75

46.2

-75

+53

34.92

-53

+38

25.68

-38

0

59.76

a) Determine the mass fractions of the amounts retained, the (arithmetic) mean diameter (using the mid-point diameter for each size range), and the standard deviation. Show your table. (16 marks) b) Determine the cumulative oversized and undersized, and complete the following table. (15 marks)

Assignment #1-1

Size range Mid-point Mass Mass retained fraction γi (micron) di (micron) (g) 0 0 - 38 59.760 38 - 53 25.680 53 - 75 34.920 75 - 106 46.200 106 - 150 52.560 150 - 212 62.760 212 - 300 73.200 300 - 425 85.440 425 - 600 101.040 600 - 850 118.920 850 - 1180 129.600 1180 - 1700 131.280 1700 - 2360 115.080 2360 - 3400 83.640 3400 - 4750 50.040 4750 - 6800 24.360 6800 - 9500 5.160 9500 - 10000 0.360 sum 1200

Cumulative Cumulative Undersized Oversized P Q

c) Plot the particle size distribution as a histogram, normalised histogram and continuous distribution (normal axes and lognormal axes), cumulative oversized and undersized products (normal axes and lognormal axes). (25 marks) d) Determine the most frequent (mode) diameter and median diameter of the particles. You can use the available tables and diagrams in the answers to the above questions. Show your answers on the diagrams. (10 marks) e) Determine the volume-equivalent, surface-equivalent, and Sauter-equivalent diameters of the particles (using the mid-point diameter for each size range). Show your table. (20 marks)

Assignment #1-2

Solutions 1a.

Equivalent volume diameter (2 marks): d volume =

1b.

6V

=

π

3

6 × 53

= 6.2 mm

π

Equivalent surface diameter (2 marks): d surface =

1c.

A

π

=

6 × 52

π

= 6.9 mm

Sauter diameter (2 marks): d Sauter

1d.

3

3 d sphere 6V = = 2 = 5.0 mm A d surface

Volume sphericity (2 marks): ψ V = π ( dV ) / A = 0.81 2

Surface sphericity (2 marks): ψ A = π ( d A ) / ( 6V ) = 1.38 3

Sauter-diameter sphericities:

ψ VA = π ( d32 ) / A = 0.52 (2 marks) 2

ψ AV = π ( d32 ) / ( 6V ) = 0.52 (2 marks) 3

2a.

The table is shown below. (4 marks) Size range Mid-point

Mass Mass fraction retained γi (micron) di (micron) (g)

00 - 38 38 - 53 53 - 75 75 - 106 106 - 150 150 - 212 212 - 300 300 - 425 425 - 600 600 - 850 850 - 1180 1180 - 1700 1700 - 2360 2360 - 3400 3400 - 4750 4750 - 6800 6800 - 9500 9500 - 10000 sum

19 45.5 64 90.5 128 181 256 362.5 512.5 725 1015 1440 2030 2880 4075 5775 8150 9500

59.76 25.68 34.92 46.2 52.56 62.76 73.2 85.44 101.04 118.92 129.6 131.28 115.08 83.64 50.04 24.36 5.16 0.36 1200

0.050 0.021 0.029 0.039 0.044 0.052 0.061 0.071 0.084 0.099 0.108 0.109 0.096 0.070 0.042 0.020 0.004 0.000

Standard deviation

Mean diam di * γi

(di - d) * γi

di * γi

0.946 0.974 1.862 3.484 5.606 9.466 15.616 25.810 43.153 71.848 109.620 157.536 194.677 200.736 169.928 117.233 35.045 2.85

65561.804 26886.804 35364.126 44565.201 47227.448 50782.880 50557.324 46012.140 36001.491 19307.105 2475.220 8189.995 71524.476 204671.388 352782.688 431157.649 209714.522 20834.720

17.978 44.303 119.194 315.325 717.619 1713.400 3997.696 9356.125 22115.656 52089.438 111264.300 226851.840 395194.310 578119.680 692454.563 677017.688 285616.750 27075.000

2

2

1.000 1166.389 1723616.981 3084080.864

Assignment #1-3

The mass fractions of the amounts retained are shown in the fourth column. (4 marks) The calculation of the mean diameter is shown by the firth column, which gives m

d = ∑ γ i di = 1166.389 µm. (4 marks) i =1

The standard deviation can be calculated in two ways which are shown by the last two columns. 4 marks - The first way gives

σ=

m

∑γ (d i =1

i

− d ) = 1723616.981 = 1312.866 µ m 2

i

- The second way gives 

m



σ = ∑ γ i ( di )  − ( d ) = 3084080.864 − 1166.3892 = 1312.866 µ m  i =1

2

2



The two calculations give the same result. 2b. The completed table is shown below. Size range Mid-point

Mass Mass fraction Cumulative Cumulative retained Undersized Oversized γi (micron) di (micron) (g) P Q 0 0 0 0 1.000 0 - 38 19 59.76 0.050 0.050 0.950

38 - 53

45.5

25.68

0.021

0.071

0.929

53 - 75

64

34.92

0.029

0.100

0.900

75 - 106

90.5

46.2

0.039

0.139

0.861

106 - 150

128

52.56

0.044

0.183

0.817

150 - 212

181

62.76

0.052

0.235

0.765

212 - 300

256

73.2

0.061

0.296

0.704

300 - 425

362.5

85.44

0.071

0.367

0.633

425 - 600

512.5

101.04

0.084

0.451

0.549

600 - 850

725

118.92

0.099

0.550

0.450

850 - 1180

1015

129.6

0.108

0.658

0.342

1180 - 1700

1440

131.28

0.109

0.768

0.232

1700 - 2360

2030

115.08

0.096

0.864

0.136

2360 - 3400

2880

83.64

0.070

0.933

0.067

3400 - 4750

4075

50.04

0.042

0.975

0.025

4750 - 6800

5775

24.36

0.020

0.995

0.005

6800 - 9500

8150

5.16

0.004

1.000

0.000

9500 - 10000 sum

9500

0.36 1200

0.000

1.000

0.000

Cumulative undersized = (5 marks); Cumulative oversized = (5 marks); Table = (5 marks)

Assignment #1-4

2c. The histogram for the particle size distribution is obtained by plotting the third versus first column in the table shown in the answer to Q. 2b. (5 marks)

Mass retained (g)

140 120 100 80 60 40 20

-1 21 50 2 -3 42 00 5 85 60 0 0 -1 17 1 00 80 34 23 00 60 68 47 00 50 -9 50 0

-7

10 6

53

0

-3 8

5

0

Size range (micron) The normalised histogram for the particle size distribution is obtained by plotting the fourth versus first column in the table shown in the answer to Q. 2b. (5 marks)

0.120

0.080 0.060 0.040 0.020

10 - 75 6 21 150 2 42 300 5 85 - 60 0 0 17 11 00 80 34 - 23 0 0 60 68 - 47 00 5 0 -9 50 0

53

-3

8

0.000

0

Mass fraction

0.100

Size range (micron)

Assignment #1-5

The continuous distribution is obtained by plotting the fourth versus second column in the table shown in the answer to Q. 2b. The horizontal axe can be in either the normal or lognormal scale. (7 marks)

0.12

f(d)

0.08

0.04

0 0

2000

4000 6000 d (microns)

8000 10000

0.12

f(d)

0.08

0.04

0 10

100 1000 log(d/microns)

10000

Assignment #1-6

The curves for cumulative oversized and undersized products can be obtained by plotting the firth and sixth versus second column in the table shown in the answer to Q. 2b. The horizontal axe can be in either the normal or lognormal scale. (8 marks)

1

Cumulative mass fraction

Undersized 0.8 0.6 0.4 0.2

Oversized

0 0

2000

4000 6000 d (microns)

Cumulative mass fraction

1

8000

10000

Undersized Oversized

0.8 0.6 0.4 0.2 0 10

100 1000 d (microns)

10000

Assignment #1-7

2d. The mode diameter can be best determined at the maximum frequency distribution which is about 1400 ± 50 mm (see the column for mass retained or mass fraction in the answer to Q. 2a or 2b). The diameter on the horizontal axis of the frequency distribution is shUhown by the red arrow in the following diagram. (5 marks)

0.12

f(d)

0.08

0.04

Mode 0 10

100

1000

10000

log(d/microns) The median diameter can be best determined at the 50% of cumulative oversized or undersized product distribution which is about 618 ± 5 mm (see the column for mass retained or mass fraction in the answer to Q. 2a or 2b). The diameter on the horizontal axis of the frequency distribution is shown by the red arrow in the following diagram. (5 marks)

Assignment #1-8

2e. The table showing the calculation is given below. (5 marks) Size range

Mid-point Mass Mass fraction Surface Equivalent Volume Equivalent retained 2 3 γi di * γi di * γi (micron) di (micron) (g) 0 0 0 0 0 0 - 38 59.76 19 0.050 17.978 341.578

38 - 53 53 - 75 75 - 106 106 - 150 150 - 212 212 - 300 300 - 425 425 - 600 600 - 850 850 - 1180 1180 - 1700 1700 - 2360 2360 - 3400 3400 - 4750 4750 - 6800 6800 - 9500 9500 - 10000

25.68 34.92 46.2 52.56 62.76 73.2 85.44 101.04 118.92 129.6 131.28 115.08 83.64 50.04 24.36 5.16 0.36

45.5 64 90.5 128 181 256 362.5 512.5 725 1015 1440 2030 2880 4075 5775 8150 9500

Sum

1200

0.021

44.303

2015.802

0.029

119.194

7628.390

0.039

315.325

28536.879

0.044

717.619

91855.258

0.052

1713.400

310125.454

0.061

3997.696

1023410.176

0.071

9356.125

3391595.313

0.084

22115.656

11334273.828

0.099

52089.438

37764842.188

0.108

111264.300

112933264.500

0.109

226851.840

326666649.600

0.096

395194.310

802244449.300

0.070

578119.680

1664984678.400

0.042

692454.563

2821752342.188

0.020

677017.688

3909777145.313

0.004

285616.750

2327776512.500

0.000

27075.000

257212500.000

3084080.864 12277302166.666

Volume equivalent diameter: (5 marks)

dV =

m

3

∑γ d i =1

i

= 3 12277302166.666 = 2306.929 µ m

3 i

Surface equivalent diameter: (5 marks)

dA =

m

2

∑γ d i =1

i

= 3 3084080.864 = 1756.155 µ m

2 i

Sauter diameter: (5 marks) m

d Sauter =

∑γ d

i

∑γ d

i

i =1 m

i =1

i

i

3

= 2

12277302166.666 = 3980.863 µ m 3084080.864

Assignment #1-9

Note: The surface area and volume for each fraction should be

π di 2 4

γ i and

π di 3 6

γ i , respectively. For

simplicity, the numerical factors have been dropped off in the above calculation. If the full expressions for the surface area and volume are used, the equivalent diameters should be calculated in the same way, i.e.,

π d A2 4

m

=∑ i =1

π di 2 4

γi ∴ dA =

4

π

m

∑ i =1

π di 2 4

γ i and

π dV 3 6

m

=∑ i =1

π di 3 6

γ i ∴ dV =

3

6

π

m

π di 3

i =1

6

∑

γi .

If the full expressions for the surface area and volume are used, the numbers in the last two columns should be different but the results for the equivalent diameters should be the same as those shown above.

Assignment #1-10