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CHE4162 – TUTORIAL 2 - SOLUTIONS PARTICLE SIZE ANALYSIS (Chapter 1) EXERCISE 1.1: For a regular cuboid particle of dimensions 1.00 x 2.00 x 6.00 mm, calculate the following diameters: (a) equivalent volume sphere diameter (b) equivalent surface sphere diameter (c) surface-volume diameter (the diameter of a sphere having the same external surface to volume ratio as the particle). (d) sieve diameter (the width of the minimum aperture through which the particle will pass) (e) projected area diameters (the diameter of a circle having the same area as the projected area of the particle resting in a stable position). SOLUTION TO EXERCISE 1.1: (a) volume of cuboid = 12 mm3 If xv is the equivalent volume sphere diameter, then

π 3 x = 12 6 v

Hence, xv = 2.840 mm. (b) surface area of cuboid = (6 × 2 × 2) + (6 × 1 × 2) + (1 × 2 × 2 ) = 40 mm2

If xs is the equivalent surface sphere diameter, then πx2s = 40 . Hence, xs = 3.568mm.

40 = 3.333 mm2 / mm3 12 6 = 3.333 (surface-volume ratio If xsv is the surface-volume sphere diameter, then x sv (c) Surface to volume ratio of the cuboid =

for a sphere of diameter x is 6/x). Hence, xsv = 1.8 mm. (d) Sieve diameter is the second largest dimension, i.e. 2 mm. (e) The cuboid has three stable resting positions and so has three projected areas: projected area 1 = 6 mm2 projected area 2 = 2 mm2 projected area 3 = 12 mm2 If xp is the projected area diameter, then

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.1

π 2 π 2 π 2 xp1 = 6; xp2 = 2 x = 12 4 4 p3 4 Giving three projected area diameters: xp1 = 2.76 mm; xp2 = 1.60 mm; xp3 = 3.91 mm. EXERCISE 1.2: Repeat exercise 1.1 for a regular cylinder of diameter 0.100 mm and length 1.00 mm. SOLUTION TO EXERCISE 1.2: πx2 h π(0.1) 2 × 1.0 −3 3 (a) volume of cylinder = = 7.854 × 10 mm 4 4 π 3 −3 If xv is the equivalent volume sphere diameter, then xv = 7.854 × 10 6 Hence, xv = 0.247 mm.

(b) surface area of cylinder = ⎛ πx2 ⎛ π(0.1)2 ⎞ ⎞ 2 ⎜ ⎜ ⎟ × 2 + πxh = × 2⎟ + π × 0.1× 1.0 = 0.3299 mm ⎝ 4 ⎠ ⎝ 4 ⎠ If xs is the equivalent surface sphere diameter, then πx2s = 0.3299 Hence, xs = 0.324 mm. (c) Surface to volume ratio of the cylinder =

0.3299

2 3 −3 = 42.0 mm / mm

7.854 × 10 6 If xsv is the surface-volume sphere diameter, then = 42.0 (surface-volume ratio x sv

for a sphere of diameter x is 6/x) Hence, xsv = 0.143 mm. (d) Sieve diameter is the second largest dimension, i.e. 0.1 mm. (e) The cylinder has two resting positions: projected area 1 = 0.1 x 1.0 = 0.1 mm2 πx2 −3 2 projected area 2 = = 7.854 × 10 mm 4 If xp is the projected area diameter, then π 2 π 2 −3 xp1 = 0.1; xp 2 = 7.854 × 10 4 4 Giving two projected area diameters: xp1 = 0.357 mm; xp2 = 0.1 mm. (The cylinder is unlikely to be stable in position 2) SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.2

EXERCISE 1.4: 1.28 grammes or a powder of particle density 2500 kg/m3 are charged into the cell of an apparatus for measurement of particle size and specific surface area by permeametry. The cylindrical cell has a diameter of 1.14 cm and the powder forms a bed of depth 1 cm. Dry air of density 1.2 kg/m3 and viscosity 18.4 x 10-6 Pas flows at

a rate of 36 cm3/min through the powder (in a direction parallel to the axis of the cylindrical cell) and producing a pressure difference of 100 mm of water across the bed. Determine the surface-volume mean diameter and the specific surface of the powder sample. SOLUTION TO EXERCISE 1.4:

We will assume that the pressure drop - flow relationship for this packed bed of powder is described by the Carman-Kozeny equation:

(− Δp) H

= 180

(1 − ε) 2 μU ε3

x2

(Chapter 4: Equation 4.9)

Knowing the pressure drop (-Δp), the packed bed depth, H, the bed voidage ε, the gas density μ and the superficial gas velocity U, we can determine the particle size x, which will be the mean surface volume diameter of the particles in the bed (see Chapter 4). π ⎛ 1.14 ⎞ 2 Cross-sectional area of cell = ⎝ = 1.0207 × 10 −4 m 2 ⎠ 4 100 Cell volume = area x depth = 1.0207 × 10 −4 × 0.01 = 1.0207 × 10−6 m 3 Mass of powder sample, M = (1− ε )ρ p (cell volume ) Knowing M = 1.28 x 10-3 kg and particle density = 2500 kg/m3, 1.28 × 10 −3 voidage of bed, ε = 1 − = 0.4984 2500 × 1.0207 × 10− 6 Superficial gas velocity, U = gas volume flow 36 × 10− 6 1 = × = 5.878 × 10 −3 m / s − 4 cross sectional area of cell 60 1.0207 × 10

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.3

Pressure drop, equivalent to 100 mm of water, 100 × 1000 × 9.81 = 981 Pa. Δp = hw ρw g = 1000 (assuming water has a density of 1000 kg/m3) Substituting into the Carman-Kozeny equation: 981 (1 − 0.4984)2 18.4 × 10− 6 × 5.878 × 10 −3 = 180 2 0.01 0.49843 xsv Which gives x sv = 20.08 μm Checking the packed bed Reynolds number: xsvρf U 20.08 × 10 −6 × 1.2 × 5.878 × 10 −3 Re ′ = = = 0.0153 μ(1 − ε ) 18.4 × 10− 6 × (1 − 0.4984) The Reynolds number is much less than 10, the limit for laminar flow, and so the assumption that the Carman-Kozeny equation can be used is justified. For spheres of diameter 20.08 μm, the surface area per unit volume, S is given by: 6 6 S= = = 2.988 × 10 5 m 2 / m 3 xsv 20.08 × 10−6

And so the specific surface or surface area per unit mass of particles is given by: S 2.988 × 10 5 = = 119.5 m 2 / kg. 2500 ρp (If we use the laminar flow part of the Ergun equation, xsv = 18.3 μm and specific surface = 131 m2/kg.) EXERCISE 1.5: 1.1 grammes or a powder of particle density 1800 kg/m3 are charged into the cell of an apparatus for measurement of particle size and specific surface area by permeametry. The cylindrical cell has a diameter of 1.14 cm and the powder forms a bed of depth 1 cm. Dry air of density 1.2 kg/m3 and viscosity 18.4 x 10-6 Pas flows at

a rate of 36 cm3/min through the powder (in a direction parallel to the axis of the cylindrical cell). The measured variation in pressure difference across the bed with changing air flow rate is given below: Air flow (cm3/min)

20

30

40

50

60

Pressure difference across 56 the bed (mm of water)

82

112

136

167

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.4

Determine the surface-volume mean diameter and the specific surface of the powder sample.

EXERCISE 1.6: Estimate the (a) arithmetic mean, (b) quadratic mean, (c) cubic mean, (d) geometric mean and (e) harmonic mean of the following distribution. size cumulative % undersize

2

2.8

4

5.6

8

11.2

16

22.4

32

44.8

64

89.6

0.1

0.5

2.7

9.6

23

47.9

73.8

89.8.

97.1

99.2

99.8

100

SOLUTION TO EXERCISE 1.6: In general, we apply Equation 1.5 to calculate the different means

g(x ) =

1

∫0 g(x)dF

(1.5)

where x is the mean and g is the weighting function, which is different for each mean definition. Examples are given in Table 1.6.1 below: Table 1.6.1 Definitions of means g(x)

x x2 x3 logx 1/x

Mean and notation arithmetic mean, xa quadratic mean, xq

cubic mean, xc geometric mean, xg harmonic mean, xh

The calculations according to Equation 1.5 are summarised in Table 1.6.2 (Page 1.11). Each distribution [g(x) versus F] is calculated from the original arithmetic distribution. Each distribution is then numerically integrated to give the right hand side of Equation 1.5. The means are then determined. For example: For the arithmetic mean, xa =

1

∫0 g(x)dF = 13.59

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.5

1 For the quadratic mean, x2q = g(x2 )dF = 259.86

hence, xq = 259.86 = 16.12

∫0

1 For the cubic mean, x3c = g(x3 )dF = 7165.30

∫0

3

hence, xc = 7165.30 = 19.28 1

∫

For the geometric mean, ln xg = g(ln x)dF = 2.4435 0 hence, xg = exp(2.4435) = 11.51 For the harmonic mean, hence, xh =

1 ⎛1 1 = g ⎞ dF = 0.1025 0 ⎝ x⎠ xh

∫

1 = 9.76 0.1025

EXERCISE 1.7 The following volume distribution was derived from a sieve analysis: size

37 -45

45 - 53 53 - 63 63 - 75 75 - 90 90 -

(μm) volume

0.4

3.1

11

21.8

27.3

106 -

126 -

150 -

180-

106

126

150

180

212

22

10.1

3.9

0.4

0

percent in range

(a) Estimate the arithmetic mean of the volume distribution. From the volume distribution derive the number distribution and the surface distribution, giving assumptions made. Estimate: (b) the mode of the surface distribution (c) the harmonic mean of the surface distribution Show that the arithmetic mean of the surface distribution conserves the surface to volume ratio of the population of particles. SOLUTION TO EXERCISE 1.7: (a) The arithmetic mean of the volume distribution is defined as (from equation 1.5): SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.6

xav =

1

∫0 xdF v

This is calculated numerically in column 21 of Table 1.7.1 (page 1.12-1.13), which summarises the calculation for this exercise. The result is arithmetic mean of volume distribution = 86 μm. The number distribution The volume distribution is now converted into a number distribution and a surface distribution.

From Equation 1.2 the number distribution and volume distribution are related by: f V (x) = k vx3f N (x)

(1.2)

1 f v (x) Therefore, f N (x) = kv x3

(1.7.1)

The number distribution is calculated from the volume distribution (column 6) using Equation 1.7.1. However, kv is unknown, but can be found by noting that:

∫ f N (x)dx = 1 and so, assuming kv does not vary with particle size x,

∫

f v (x) x

3 dx = k v

f v (x) 3 dx is shown in column 9 of the table. The value x of the integral and hence the value of kv is shown at the bottom of this column (2.36 x

The calculation of the integral

∫

10-6). The number distribution is then calculated according to Equation 1.7.1 in column 10. Column 11 is a check that f N (x)dx = 1.

∫

Column 12 gives the cumulative number distribution FN .

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.7

The surface distribution From Equations 1.1 and 1.2:

f S (x) = k sx 2f N (x)

(1.1)

f V (x) = k vx3f N (x)

(1.2)

the relationship between the surface distribution and the volume distribution is: f S (x) =

k s f v (x) kv x

(1.7.2)

The ratio ks/kv may be found by noting that

∫ f S (x)dx = 1 and so, assuming kv and ks do not vary with particle size x,

∫

f v (x) dx = x

kv kS

f v (x) dx is shown in column 14 of the table. The x value of the integral, and hence the value of the ratio kv/ks, is shown at the bottom of

The calculation of the integral

∫

this column (0.0126). The surface distribution is then calculated according to Equation 1.7.2 in column 15 and is plotted in Figure 1.7.1 (Page 1.10).

∫

Column 16 is a check that f S (x)dx = 1. Column 17 gives the cumulative surface distribution FS. (b) The mode of the surface distribution can be seen from Figure 1.7.1 (page 1.10) to be around 70 μm. (c) The harmonic mean of the surface distribution (from Equation 1.5) is found from:

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.8

1 = xhS

1

1

∫ xdFS 0

1

The integral

1

∫ x dFS is calculated in column 19 of the table and its value is shown at 0

the bottom of the table (0.01331). The harmonic mean of the surface distribution in therefore the inverse of this value. Harmonic mean of the surface distribution, xhS = 75 μm Note: In early versions of the book, the wording of the final part of the question is incorrect. It should read "Show that the arithmetic mean of the surface distribution conserves the surface to volume ratio of the population of particles".

From Equation 1.5, the arithmetic mean of a surface distribution is: 1

∫ xdFS

xaS = 01

∫ dFS 0

From Equation 1.1, f S (x) = k sx 2f N (x) or in cumulative form dFS = ksx 2dF N 1

∫ xkSx2 dFN

Therefore xaS = 01

∫ k Sx dF N 0

2

1

∫x

3dF

Assuming ks is independent of particle size, xaS = 10

N

(1.7.3)

∫ x2 dFN 0

The surface-volume mean xSV , must conserve the surface and volume of the population; that is the mean must enable us to calculate the total volume from the total surface of the particles. i.e.

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.9

k xSV × (total surface) × v = (total volume) ks ∞ ∞ k f s (x) dx. v = f v (x)dx and therefore xSV 0 0 ks

∫

∫

and since f S (x) = k sx 2f N (x)

∫ ∫

∞

3

f V (x) = k vx3f N (x)

and

k v x f N (x)dx

k then, xSV = s 0∞ = kv ks x2 f N (x)dx 0

∞ 3 x f N (x)dx 0 ∞ 2 x f N (x)dx 0

∫ ∫

=

1 3 x dFN 0 1 2 x dF N 0

∫ ∫

This expression is same as that for the arithmetic mean of the surface distribution derived in Equation 1.7.3 above. Hence the arithmetic mean of the surface distribution conserves the surface to volume ratio of the population of particles. For interest, in columns 20 and 22 the calculations of xaS and xhV are shown.

Figure 1.7.1: Differential surface distribution for exercise 1.7.

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.10

Table 1.6.2: Summary of calculations of different means (Exercise 1.6).

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.11

% vol/(100*[xU – xL])

Table 1.7.1: Summary of calculation for exercise 1.7. SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.12

Table 1.7.1: Summary of calculation for exercise 1.7-continued.

SOLUTIONS TO Tute 2: PARTICLE SIZE ANALYSIS (Ch1)

Page 1.13