Question

Dust particles from a blasting operation rise up in the atmosphere to a height of 500 m. The viscosity of air is \( 1.80 \times 10^{-5}\ \text{kg}\,\text{m}^{-1}\,\text{s}^{-1} \). Assume the particles have quiescent settling (no turbulence) in atmosphere. Neglecting air density, the settling time, in days, of a dust particle of 2.5 \(\mu\)m diameter and 3600 \( \text{kg}\,\text{m}^{-3} \) density is . (rounded off to two decimal places)

Hint:

A dust-sized particle in still air settles at a constant Stokes' law terminal velocity; divide the fall height by that velocity to get the time.

Answer 1

Correct Answer: 8.49

Step 1: Pick the governing law for a tiny settling particle.
A very small particle falling through still (quiescent) air settles at a constant terminal velocity given by Stokes' law, valid for the low Reynolds number of dust-sized particles: \[ v_t = \dfrac{(\rho_p - \rho_a)\,g\,d^2}{18\mu} \] where \( \rho_p \) is particle density, \( \rho_a \) is air density, \( g \) is gravity, \( d \) is particle diameter and \( \mu \) is the air viscosity.

Step 2: Simplify using the problem's assumption.
The question tells us to neglect air density, so \( \rho_a = 0 \) and the formula reduces to \[ v_t = \dfrac{\rho_p\,g\,d^2}{18\mu} \]

Step 3: Substitute the numbers.
Convert the diameter to metres: \( d = 2.5\ \mu\text{m} = 2.5 \times 10^{-6} \) m, so \( d^2 = 6.25 \times 10^{-12}\ \text{m}^2 \). With \( \rho_p = 3600\ \text{kg}\,\text{m}^{-3} \), \( g = 9.81\ \text{m}\,\text{s}^{-2} \) and \( \mu = 1.80 \times 10^{-5}\ \text{kg}\,\text{m}^{-1}\,\text{s}^{-1} \): \[ v_t = \dfrac{3600 \times 9.81 \times 6.25 \times 10^{-12}}{18 \times 1.80 \times 10^{-5}} = \dfrac{2.207 \times 10^{-7}}{3.24 \times 10^{-4}} \approx 6.81 \times 10^{-4}\ \text{m/s} \]

Step 4: Find the time to fall 500 m.
Since the particle settles at this constant speed, time is height divided by speed: \[ t = \dfrac{500}{6.81 \times 10^{-4}} \approx 733945\ \text{s} \] Converting to days by dividing by \( 86400\ \text{s/day} \): \[ t = \dfrac{733945}{86400} \approx 8.49\ \text{days} \]

Final Answer:
The dust particle takes about 8.49 days to settle back down through 500 m of still air. \[ \boxed{8.49\ \text{days}} \]