1. U N C L A S S I F I E D Conversion from Physical to Aerodynamic Diameters for Radioactive Aerosols Jeffrey J. Whicker Los Alamos National Laboratory Health Physics Meeting 2007

2. U N C L A S S I F I E D Slide 1 Problem Many articles provide data on the physical diameter of particles Most inhalation dose models require information on aerodynamic diameters Physical Diameter = Aerodynamic Diameter

3. U N C L A S S I F I E D Slide 2 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

4. U N C L A S S I F I E D Slide 3 Definition of Aerodynamic Diameter Aerodynamic diameter is the diameter of a unit density particle (1 gm/cm 3 ) that has the same settling velocity as the particle. Physical diameter (water) Physical diameter (Pu) Equal settling velocities means equal aerodynamic diameters

5. U N C L A S S I F I E D Slide 4 Calculation of terminal settling velocity in the Stokes region Formula for terminal settling velocity:  = particle density (g cm -3 ) d p = physical diameter (cm) g = gravitational acceleration C p = Cunningham correction factor  = viscosity  = shape correction factor Where:

6. U N C L A S S I F I E D Slide 5 Setting V ts equations equal and solving Physical diameter (1 g/cm 3 ) Physical diameter (Pu) Equal settling velocities means equal aerodynamic diameters

7. U N C L A S S I F I E D Slide 6 Cunningham Slip Correction problem Slip correction is needed because the particles are small enough to “slip” between air molecules without collision. C p gets larger as the particle sizes decrease. Where: is the mean free path between collisions with air molecules (0.066  m at 1 atm and 20 o C)

8. U N C L A S S I F I E D Slide 7 Equations show interdependency of particle diameter and Cunningham Slip Correction Solution: pick particle size (d p ), solve for C p, 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 d ae

9. U N C L A S S I F I E D Slide 8 Conversion in the transition regions (Re >1 but <1000) Include for larger particles greater than about 50  m Where: C D is the coefficient of drag Unfortunately, to calculate the Re you need V TS, and you need the V TS to calculate Re

10. U N C L A S S I F I E D Slide 9 Independence of Re and C d from Settling Velocity Solution: Re was determined using the above equation * then substituted into the equation below to calculate V TS * Using table 3.5 in Hinds (1985) Aerosol Technology: Properties, Behavior, and Measurement of Airborne Particles. John Wiley & Sons. New York, New York

11. U N C L A S S I F I E D Slide 10 Conversion of d p to d ae

12. U N C L A S S I F I E D Slide 11 Conversion for really big particles (Re >1000, particle > 350  m) Coefficient of drag is relatively constant in the Newton’s region, so taking a ratio of the two terms above this reduces to:

13. U N C L A S S I F I E D Slide 12 Useful relationship spanning all three Reynolds regions:

14. U N C L A S S I F I E D Slide 13 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)