Question

Suppose a gust of wind has carried a 52-μm-diameter dust particle to a height of 340 m. If the wind suddenly stops, how long will it take the dust particle to settle back to the ground? Dust has a density of 2700 kg/m^3, the viscosity of 25∘C air is 2.0×10−5N⋅s/m^2, and you can assume that the falling dust particle reaches terminal speed almost instantly.

Answer 1

First, let's find the terminal speed of the dust particle. Terminal speed is reached when the force of gravity on a particle equals the drag force acting on it.

For a spherical particle falling through a fluid, we can use Stokes' Law to find the drag force. Stokes' Law states that the force of gravity on the particle is given by:

F = 6πηrv

Where F is the drag force, η is the fluid viscosity, r is the radius of the sphere, and v is the terminal speed of the particle.

We also know the force of gravity on the particle, given by:

F_gravity = mg

Where m is the mass of the particle and g is the acceleration due to gravity.

We can find the mass of the particle using its density:

m = Vρ

Where V is the volume of the particle and ρ is its density. For a sphere, the volume is given by:

V = (4/3)πr^3

So, the mass is:

m = (4/3)πr^3ρ

Now, substituting the mass back into the force of gravity equation, we get:

F_gravity = (4/3)πr^3ρg

Now, we set the force of gravity equal to the drag force and solve for the terminal speed:

6πηrv = (4/3)πr^3ρg

v = (2/9)(r^2ρg/η)

Now we can plug in the given values:
r = 52 μm/2 = 26 μm (convert diameter to radius)
ρ = 2700 kg/m³ (density of dust)
g = 9.81 m/s² (acceleration due to gravity)
η = 2.0 × 10⁻⁵ N⋅s/m² (viscosity of air)

v = (2/9)((26 × 10⁻⁶ m)² × (2700 kg/m³)(9.81 m/s²)/(2.0 × 10⁻⁵ N⋅s/m²))
v ≈ 0.0339 m/s

Now that we have the terminal speed, we can use the equation for speed to find the time it takes for the particle to fall:

distance = speed × time
time = distance/speed

Plug in the given distance (height) and calculated terminal speed:

time = (340 m)/(0.0339 m/s)
time ≈ 10028 s

It will take about 10028 seconds, or about 2.8 hours, for the dust particle to settle back to the ground.