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Conversion from Physical to
Aerodynamic Diameters for
Radioactive Aerosols
Jeffrey J. Whicker
Los Alamos National Laboratory
Health Physics Meeting 2007
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Problem
• Many articles provide data on the physical diameter of
particles
• Most inhalation dose models require information on
aerodynamic diameters
Physical
Diameter
=
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Aerodynamic
Diameter
Slide 1
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Additional complications
• Ranges of particle sizes for which conversion is needed
can be huge
– Reynolds numbers spanning 6 orders-of –magnitude
– 3 different flow regions
– The Stokes region
– The transition region
– Newton’s law region
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Slide 2
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Definition of Aerodynamic Diameter
• Aerodynamic diameter is the diameter of a unit density
particle (1 gm/cm3) that has the same settling velocity as
the particle.
Physical diameter
(Pu)
Physical diameter
(water)
Equal settling
velocities means
equal aerodynamic
diameters
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Slide 3
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Calculation of terminal settling velocity in the Stokes
region
Formula for terminal settling velocity:
 p d p2 gC p
VTS 
18
Where:
 = particle density (g cm-3)
dp = physical diameter (cm)
g = gravitational acceleration
Cp = Cunningham correction
factor
 = viscosity
 = shape correction factor
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Slide 4
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Setting Vts equations equal and solving
Physical diameter
(1 g/cm3)
Physical diameter
(Pu)
 p d p2 gC p  ae d ae2 gC ae
VTS 

18 p
18 ae
Equal settling
velocities means
equal aerodynamic
diameters
d ae  d p
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 pC p
 ae C ae
Slide 5
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Cunningham Slip Correction problem
Slip correction is needed because the particles are small enough
to “slip” between air molecules without collision. Cp gets larger
as the particle sizes decrease.
 p d p2 gC p
VTS 
18
Where: Cp  1 

dp 
2
.
514

0
.
8
exp(

0
.
55
)

dp 
 
 is the mean free path between collisions with air molecules
(0.066 m at 1 atm and 20oC)
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Slide 6
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Equations show interdependency of particle diameter
and Cunningham Slip Correction
d ae  d p
 pC p
 ae C ae
d ae C ae  d p
 pC p
 ae 
Solution: pick particle size (dp), solve for Cp, then solve right
side of equation (set p=11.46 g cm-3, ae = 1 g cm-3,  = 1.5
(ICRP 66), then iteratively solve for dae
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Slide 7
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Conversion in the transition regions (Re >1 but <1000)
Include for larger particles greater than about 50 m
VTS 
4 p d p g
3C D  g
Re

6.6d p 
Where: CD is the coefficient of drag
Unfortunately, to calculate the Re you need VTS , and
you need the VTS to calculate Re
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Slide 8
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Independence of Re and Cd from Settling Velocity
CD Re 2 
4  g pgd 3
3 2
Solution: Re was determined using the above equation*
then substituted into the equation below to calculate VTS
Re
VTS 
6.6d p 
*Using
table 3.5 in Hinds (1985) Aerosol Technology: Properties, Behavior, and Measurement
of Airborne Particles. John Wiley & Sons. New York, New York
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Slide 9
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Conversion of dp to dae
VTS  p  VTS  ae
Re p
Re ae

6.6  d p 6.6    d ae
 dp
d ae

Re ae
Re p
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Slide 10
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Conversion for really big particles (Re >1000, particle >
350 m)
VTS 
4 p d p g
3C D  g

4 ae d ae g
3C D  g
Coefficient of drag is relatively constant in the Newton’s region,
so taking a ratio of the two terms above this reduces to:
p
d ae  d p
 ae 
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Slide 11
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Useful relationship spanning all three Reynolds
regions:
Relationship for Plutonium Particles
up to 10,000 um
30
6
Log aerodynamic diameter (um)
Aerodynamic Diameter (um)
Relationship For Plutonium Particles up to 10 um
25
y = 2.7661x + 0.128
20
R2 = 1
15
10
5
0
0
1
2
3
4
5
6
Physical Diameter (um)
7
8
9
10
5
y = 1.0724x + 0.505
R2 = 0.9935
4
3
2
1
0
-1
-2
-2
-1
0
1
2
3
4
5
Log physical diameter (um)
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Slide 12
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Conclusions:
• Equations were developed to convert physical diameters
to aerodynamic diameters
• Examples were provided for plutonium particles
• BUT, this approach is valid for any particle with known
density
• The 2.8 (okay 3) rule for quick conversion of respirable
particle sizes of plutonium
• Simple relationships were developed for conversions of
particle sizes that span over 6 orders of magnitude in
Reynolds numbers (0.1 um up to 10,000 um diameters)
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Slide 13