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Chapter 8

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.

Step-by-Step

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.