If a dust particle has a physical density significantly greater than 1 g/cm3, how will its aerodynamic diameter (da) compare to its actual geometric diameter (dg)?
- Ada will be smaller than dg
- Bda will be equal to dg
- Cda will be larger than dgCorrect
- Dda is independent of density
Solution
Concept
Aerodynamic diameter and particle density
Formula
da = dg × √(ρp / ρ0)
Step-by-step solution
1. Aerodynamic diameter (da) is the diameter of an equivalent unit-density sphere (ρ0 = 1 g/cm3) having the same terminal settling velocity as the actual particle. 2. The relationship is governed by da = dg × √(ρp / ρ0), where dg is geometric diameter and ρp is particle density. 3. Given ρp > 1 g/cm3, the density ratio term √(ρp / ρ0) is greater than 1. 4. Consequently, da is larger than dg. Answer: C
Explanation
The aerodynamic diameter of a particle is defined as the diameter of a standard unit-density sphere (density 1 g/cm3) that exhibits the exact same terminal settling velocity in air. When a particle has a physical density significantly greater than 1 g/cm3, a larger unit-density sphere is required to match its higher settling rate, meaning the aerodynamic diameter exceeds its geometric diameter.
Why the other options are wrong
A) da is smaller than dg only when physical particle density is less than 1 g/cm3 (such as porous soot or hollow cenospheres). B) da equals dg only for spherical particles with a density exactly equal to 1 g/cm3. D) Aerodynamic diameter is fundamentally dependent on particle density and shape factor.
Shortcut
Density > 1 -> Aerodynamic diameter > Geometric diameter (heavy particles settle faster, behaving like larger unit-density spheres).
Common mistake
Inverting the ratio and assuming higher density compacts the aerodynamic diameter.
Correct answer: C) da will be larger than dg