Terminal Velocity of a Particle from a Volcano
A volcano has erupted, spewing stones, steam, and ash several thousand meters into the atmosphere (Fig. 10–14). After some time, the particles begin to settle to the ground. Consider a nearly spherical ash particle of diameter 50 µm, falling in air whose temperature is -50°C and whose pressure is 55 kPa. The density of the particle is 1240 kg/m³. Estimate the terminal velocity of this particle at this altitude.
SOLUTION We are to estimate the terminal velocity of a falling ash particle.
Assumptions 1 The Reynolds number is very small (we will need to verify this assumption after we obtain the solution). 2 The particle is spherical.
Properties At the given temperature and pressure, the ideal gas law gives 𝜌 = 0.8588 kg/m³. Since viscosity is a very weak function of pressure, we use the value at -50°C and atmospheric pressure, 𝜇 = 1.474 × 10^{-5} kg / m \cdot s.
Analysis We treat the problem as quasi-steady. Once the falling particle has reached its terminal settling velocity, the net downward force (weight) balances the net upward force (aerodynamic drag + buoyancy), as illustrated in Fig. 10–15.
Downward force: F_{\text {down }}=W=\pi \frac{D^{3}}{6} \rho_{\text {particle }} g (1)
The aerodynamic drag force acting on the particle is obtained from Eq. 10–12, and the buoyancy force is the weight of the displaced air. Thus,
Drag force on a sphere in creeping flow: F_{D}=3 \pi \mu V D (Eq. 10-12)
Upward force: F_{\text {up }}=F_{D}+F_{\text {buoyancy }}=3 \pi \mu V D+\pi \frac{D^{3}}{6} \rho_{\text {air }} g (2)
We equate Eqs. 1 and 2, and solve for terminal velocity V,
\begin{aligned}V &=\frac{D^{2}}{18 \mu}=\left(\rho_{\text {particle }}-\rho_{\text {air }}\right) g \\&=\frac{\left(50 \times 10^{-6} m \right)^{2}}{18\left(1.474 \times 10^{-5} kg / m \cdot s \right)}\left[(1240-0.8588) kg / m ^{3}\right]\left(9.81 m / s ^{2}\right) \\&= 0 . 1 1 5 m / s\end{aligned}
Finally, we verify that the Reynolds number is small enough that creeping flow is an appropriate approximation,
\operatorname{Re}=\frac{\rho_{\text {air }} V D}{\mu}=\frac{\left(0.8588 kg / m ^{3}\right)(0.115 m / s )\left(50 \times 10^{-6} m \right)}{1.474 \times 10^{-5} kg / m \cdot s }=0.335
Thus the Reynolds number is less than 1, but certainly not much less than 1.
Discussion Although the equation for creeping flow drag on a sphere (Eq. 10–12) was derived for a case with Re ≪ 1, it turns out that the approximation is reasonable up to Re ≅ 1. A more involved calculation, including a Reynolds number correction and a correction based on the mean free path of air molecules, yields a terminal velocity of 0.110 m/s (Heinsohn and Cimbala, 2003); the error of the creeping flow approximation is less than 5 percent.
