SOLUTIONS TO CHAPTER 14: HEALTH EFFECTS OF FINE POWDERS EXERCISE 14.1 Determine, by calculation, the likely fate of a 20 μm particle of density 2000 kg/m3 suspended in the air inhaled by a human at a rate giving rise to the following velocities in parts of the respiratory system:
Part
Number
Diameter (mm)
Length (mm)
Typical air Typical velocity (m/s) residence time (s)
Mouth
1
20
70
3.2
0.022
Pharynx
1
30
30
1.4
0.021
Trachea
1
18
120
4.4
0.027
2 main bronchi
2
13
37
3.7
0.01
EXERCISE 14.1 SOLUTION: We will assess the likelihood of the 20 μm particles depositing due to inertial impaction, diffusion and sedimentation.
Inertial Impaction: Firstly we calculate the dimensionless ratio known as the Stokes Number. Stk
=
x 2 ρ pUp
(Text Equation.14.4)
18μD
The greater the value of this ratio is above unity, the greater will be the tendency for particles to impact with the airway walls and so deposit. The farther the value is below unity, the greater will be the tendency for the particles to follow the gas. We will assume that the relevant particle velocity U p is equal to the air velocity. For 20 μm particles this will be a reasonable assumption.
(20 × 10 − )
6 2
Thus, Stk =
× 2000 × Up − 3 Up 2 . 415 10 = × D 18 × 18.4 × 10 −6 × D
Applying this equation to the different areas of the respiratory tract: Part
Stokes Number
Mouth
0.386
Pharynx
0.113
Trachea
0.59
2 main bronchi
0.69
The vales of Stokes number in the trachea and main bronchi are near enough to unity to expect deposition of a substantial proportion of the 20 μm particles by inertial impaction.
SOLUTIONS TO CHAPTER 14: HEALTH EFFECTS OF FINE POWDERS
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Brownian Motion: To assess the importance of diffusion for a 20 μm particle, we calculate its likely random displacement using Text-Equation 14.5: Root mean square displacement in time t, L =
6 αt
(Text-Equation.14.5)
where α is the diffusion coefficient given by:
α=
kT 3πμx
for a particle of diameter x in a fluid of viscosity μ at a temperature T. -23 k is the Boltzmann constant, which has the value 1.3805 x 10 J/K. Assuming the air temperature is 30°C, 1.3805 × 10 −23 × 303 α= = 1.21× 10 −12 −6 −6 3π × (18.4 × 10 )× (20 × 10 ) And so, from Text-Equation 14.5, the displacement in 1 second, L
= 6(1.21× 10 −12 )× 1 = 2.7 × 10 −6 m
The typical diffusion velocity of the 3 μm particle is therefore 2.7 x 10-6 m/s Sedimentation: To determine the importance of sedimentation, we need to calculate the particle’s terminal velocity. This is the maximum velocity that the particle would reach due to sedimentation. For particles of the size of relevance here, falling in air, the drag force will be given by Stokes’ Law (Text-Equation 2.13) and the terminal velocity will be given by Text-Equation 2.13, reproduced below:
UT
=⎜ ⎝
x2 (ρ p
− ρf )g ⎟ 18μ
(Text-Equation 2.13)
Assuming the air temperature is 30°C,
(20 × 10 − ) (2000 − 1.2) × 9.81 = 0.0237 m/s = 18 × (18.4 × 10 − ) 6 2
UT
6
We can estimate the residence time that would be required in each part of the respiratory system for motion by sedimentation to result in deposition. This will be the time required for the particle, moving at its terminal velocity plus diffusion velocity, to travel a distance equivalent to the part diameter. (In this case, the diffusion component is actually negligible compared to the sedimentation component). In the table below the required residence times are compared with the actual residence time.
SOLUTIONS TO CHAPTER 14: HEALTH EFFECTS OF FINE POWDERS
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Part
Required Actual residence residence time (s) time (s)
Mouth
0.84
0.022
Pharynx
1.27
0.021
Trachea
0.76
0.027
2 main bronchi
0.55
0.01
We conclude that deposition of the 20 μm particles by sedimentation is unlikely to be significant in these parts of the respiratory system. So, on balance, the likely fate of the 20 μm particles is that they will be deposited by inertial impaction in the trachea and main bronchi. EXERCISE 14.2 Given the following information, in which region of the respiratory tract is a 3 μm particle of density 1500 kg/m3 most likely to be deposited and by which mechanism? Support your conclusion by calculation. Part
Diameter (mm)
Length (mm)
Typical air Typical velocity (m/s) residence time (s)
Trachea
18
120
4.4
0.027
Bronchioles
2
20
0.6
0.032
Terminal bronchioles
0.7
5
0.2
0.023
Alveolar ducts
0.8
1
0.0023
0.44
Alveoli
0.15
0.15
0.00004
4
EXERCISE 14.2 SOLUTION
We will assess the likelihood of the 3 μm particles depositing due to inertial impaction, diffusion and sedimentation. Inertial Impaction: Firstly we calculate the dimensionless ratio known as the Stokes Number. Stk
=
x 2 ρ pUp 18μD
(Text Equation.14.4)
The greater the value of this ratio is above unity, the greater will be the tendency for particles to impact with the airway walls and so deposit. The farther the value is below unity, the greater will be the tendency for the particles to follow the gas. We will assume that the relevant particle velocity U p is equal to the air velocity. For 3 μm particles this will be a reasonable assumption.
SOLUTIONS TO CHAPTER 14: HEALTH EFFECTS OF FINE POWDERS
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(3 × 10 − )
6 2
Thus, Stk =
× 1500 × Up − 5 Up 4 . 08 10 = × D 18 × 18.4 × 10 −6 × D
Applying this equation to the different areas of the respiratory tract: Part
Stokes Number
Trachea
9.9 x 10-3
Bronchioles
1.2 x 10-2
Terminal bronchioles Alveolar ducts
1.16 x 10 -2
Alveoli
1.08 x 10-5
1.17 x 10-4
The values of Stokes number in all parts is substantially less than unity and so the 3 μm particles are most unlikely to be deposited by inertial impaction. Brownian Motion: To assess the importance of diffusion for a 3 μm particle, we calculate its likely random displacement using Text-Equation 14.5: Root mean square displacement in time t, L =
6 αt
(Text-Equation.14.5)
where α is the diffusion coefficient given by:
α=
kT 3πμx
for a particle of diameter x in a fluid of viscosity μ at a temperature T. -23 k is the Boltzmann constant, which has the value 1.3805 x 10 J/K. Assuming the air temperature is 30°C, ( 1.3805 × 10 −23 )× 303 α= = 8.04 × 10 −12 −6 −6 3π × (18.4 × 10 )× (3 × 10 ) And so, from Text-Equation 14.5, the displacement in 1 second, L
= 6(8.04 × 10 −12 )× 1 = 6.94 × 10 −6 m
-6 The typical diffusion velocity of the 3 μm particle is therefore 6.9 x 10 m/s
Sedimentation: To determine the importance of sedimentation, we need to calculate the particle’s terminal velocity. This is the maximum velocity that the particle would reach due to sedimentation. For particles of the size of relevance here, falling in air, the drag force will be given by Stokes’ Law (Text-Equation 2.13) and the terminal velocity will be given by Text-Equation 2.13, reproduced below:
SOLUTIONS TO CHAPTER 14: HEALTH EFFECTS OF FINE POWDERS
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⎛ x2 (ρ p − ρf )g UT = ⎜ ⎟ 18μ ⎝
(Text-Equation 2.13)
Assuming the air temperature is 30°C,
( 3 × 10 − ) (1500 − 1.2) × 9.81 = = 4 × 10 − − 18 × (18.4 × 10 ) 6 2
UT
4
6
m/s
We can estimate the residence time that would be required in each part of the respiratory system for motion by a combination of sedimentation and diffusion to result in deposition (assuming, in the worst case, that diffusion moves the particle in the same direction as sedimentation). This will be the time required for the particle, moving at its terminal velocity plus diffusion velocity, to travel a distance equivalent to the part diameter. (In this case, the diffusion component is actually negligible compared to the sedimentation component). In the table below the required residence times are compared with the actual residence time.
Part
Required Actual residence residence time (s) time (s)
Trachea
44.3
0.027
Bronchioles
4.9
0.032
Terminal bronchioles Alveolar ducts
1.72
0.023
1.97
0.44
Alveoli
0.37
4.0
We conclude that deposition of the 3 μm particles by sedimentation is likely to be significant in the alveoli and may occur to some extent in the alveolar ducts.
EXERCISE 14.3 Compare the Stokes numbers for 2, 5, 10 and 40 μm particles of density 1200 kg/m3 in air passing through the nose. What conclusions do you draw regarding the likelihood of deposition of these particles in the nose? Data: Characteristic velocity in the nose: 9 m/s Characteristic diameter on the airway in the nose: 6 mm Viscosity of air: 1.81× 10 −5 Pas.
Density of air: 1.21 kg/m
3
EXERCISE 14.3 SOLUTION:
The Stokes Number is:
SOLUTIONS TO CHAPTER 14: HEALTH EFFECTS OF FINE POWDERS
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Stk
=
x 2 ρ pUp
(Text Equation.14.4)
18μD
We will assume that the relevant particle velocity U p is equal to the air velocity. The is a reasonable assumption for particles up to 40 μm . Thus, Stk =
x 2 × 1200 × 9 18 × 18.1× 10
−6
× 6 × 10
−3
= 5.52 × 10 9 x 2
Applying this equation to the different particle sizes: Particle size, x ( μm ) Stokes Number, Stk
2
5
10
40
0.022
0.138
0.55
8.84
The greater the value of this ratio is above unity, the greater will be the tendency for particles to impact with the airway walls and so deposit. The farther the value is below unity, the greater will be the tendency for the particles to follow the gas. So, the 40 μm particles are highly likely to be deposited in the nose, the 10 μm particles may be deposited and the 2 and 5 μm particles are unlikely to be deposited.
EXERCISE 14.4 Carrier particles are used in dry powder inhalers. What is a carrier particle? What is the role of a carrier particle? Why are carrier particles needed in these inhalers?
See text. EXERCISE 14.5 With reference to the control of dusts as a health hazard, explain what is meant by the hierarchy of controls.
See text. EXERCISE 14.6 The required dose of a particulate drug of particle size 3 μm and particle density 1000 3 kg/m is 10 micrograms. Estimate the number of particles in this dose and the volume occupied by the dose, assuming a voidage of 0.6. 3 5 [Answer: 0.25 mm ; 7x10 ] EXERCISE 14.6 SOLUTION: Assuming that 3 μm represents the equivalent sphere diameter, then the mass of one
particle =
π 6
x 3ρp
=
π
3 × 10 − ) 1000 = 1.414 × 10 − ( 6 6 3
14
kg
SOLUTIONS TO CHAPTER 14: HEALTH EFFECTS OF FINE POWDERS
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Number of particles in the does of 10 micrograms =
10 × 10 −9 1.414 × 10
−14
= 7.07 × 10 5
Mass of dose expressed in terms of volume, particle density and voidage: M = ρ p (1 − ε )V
Volume of dose, V =
(Text-Equation 7.24) M 10 × 10 −9 = = 2.5 × 10 −11 m3 (or 0.025 mm 3) ρ p (1 − ε ) 1000 × (1 − 0.6 )
SOLUTIONS TO CHAPTER 14: HEALTH EFFECTS OF FINE POWDERS
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