Aerosol settling

In a gas, how fast does the magnetic particulate fall out — the gas-lattice question?

vs = \frac{ρp \cdot g \cdot \left( d \right)^2 \cdot Cc}{18 \cdot η}

Method

1. Look up the particulate's density and the gas's viscosity. A gas's viscosity is independent of its pressure — a result of kinetic theory that surprises everyone once — so the handbook value at one atmosphere is used at every pressure.
2. The mean free path scales inversely with pressure and directly with absolute temperature: λ = 70 nm × (760 ÷ P) × (T ÷ 293 K). Seventy nanometres is argon's at one atmosphere and 20 °C; the other gases are within a factor of two and the same figure is used for all of them.
3. The Knudsen number is 2λ ÷ d. Cunningham's slip correction is C_c = 1 + Kn·(1.257 + 0.4·e^(−1.1/Kn)), which is 1 when the particle is much bigger than the mean free path and grows without limit as it becomes smaller.
4. Stokes's terminal velocity, with the slip: v_s = ρ_p·g·d²·C_c ÷ (18·η). The gas's buoyancy is left out, being a thousandth of the particle's weight.
5. The relaxation time is the same expression without g: τ_p = ρ_p·d²·C_c ÷ (18·η). It is how long the particle takes to reach 63 % of a new gas velocity, and the settling velocity is g times it.
6. Divide the bore by the settling velocity for the time a particle takes to fall out of a horizontal tube.

Assumptions

• A rigid sphere in still gas, in the Stokes regime: the particle's own Reynolds number below about 1, which holds for iron up to roughly 50 µm in air. Above that the drag rises faster than Stokes and the particle falls slower than this says.
• The particles do not interact — no chains, no clumps, no magnetic attraction between them. A magnetised iron aerosol chains at once, and a chain of ten falls faster than a sphere of one.
• The mean free path is argon's at one atmosphere, scaled with pressure and temperature, and used for every carrier gas. Hydrogen's is longer, nitrogen's and oxygen's shorter, all within a factor of two; the slip correction is insensitive to this at ordinary pressures and sizes.
• A liquid carrier is outside this page. If one is chosen the calculation runs with no slip and a warning, and the Stokes settling calculation in the medium group, which includes the carrier's buoyancy, is the one to use.
• Nothing here is Meyer's except the 90 in/s the EPG #1 holding gives for a gaseous medium. Stokes (1851) and Cunningham (1910) are the physics; the estate does not say what the gas is, what the particles are, or at what pressure.