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## Q. 8.6

Terminal Velocity of a Particle from a Volcano
A volcano has erupted, spewing stones, steam, and ash several thousand feet into the atmosphere. Consider a solid particle of diameter 200 μm, falling in air which is at -50. °C and 55. kPa. The density of the particle is 2200 kg/m³ (SG = 2.2).
To do: Estimate the terminal velocity of this particle at this altitude.

## Verified Solution

The density of the air is calculated from the ideal gas law,

$ρ=\frac{PM_{air}}{R_uT}=\frac{55,000\frac{N}{m^2}(28.97\frac{kg}{Kmol})}{8.3143\frac{kJ}{Kmol K}(223.15K)}(\frac{kJ}{1000Nm})=0.8588 \frac{kg}{m^3}$

The viscosity of the air is calculated from Sutherland’s law, Eq. (8-63); the result is μ = 1.452 × $10^{-5}$ kg/(m s). The mean free path of air molecules at these conditions is calculated from Eq. (8-65), which results in λ= 8.390 × $10^{-8}$ m (0.08390 μm). For this value of λ, the Knudsen number is Kn = 4.195 × $10^{-4}$ from Eq. (8-68), and from Eq. (8-69) the Cunningham slip factor is 1.001. The authors used Mathcad to solve the simultaneous equation set, Eq. (8-59) for $c_D$, Eq. (8-100) for $v_t$, and Eq. (8-101) for Re. The results are:

$μ=μ_0(\frac{T}{T_0})^{1.5}(\frac{T_0+S}{T+S})$    (8-63)

$λ=\frac{\frac{μ}{0.499}\sqrt{\frac{\pi}{8}}}{\sqrt{ρp}}$    (8-65)

$Kn=\frac{λ}{D_p}$    (8-68)

$C=1+Kn[2.514+0.80exp(-\frac{0.55}{Kn})]$    (8-69)

$c_D=0.4+\frac{24}{Re}+\frac{6}{1+\sqrt{Re}}$    (8-59)

$v_t=\sqrt{\frac{4}{3}\frac{(ρ_p-ρ)}{ρ}\frac{D_pgC}{c_D}}$   (8-100)

$Re=\frac{ρD_pv_t}{μ}$   (8-101)

Re = 18.1
$c_D$ = 2.87

and

$v_t=1.53\frac{m}{s}$

The Mathcad program that solves these equations is available on the book’s web site.

Discussion: This terminal velocity is valid only at the given altitude; $v_t$ changes as the particle falls to lower altitudes.