A playful child left alone has run a vacuum cleaner in reverse, creating a dust cloud. If the cloud consists of 0.001, 0.01, and 0.1 mm diameter particles, and the particle density is 700 kg/m³, will the child be able to clean up the mess by dusting the furniture before her mother returns an hour

Question

A playful child left alone has run a vacuum cleaner in reverse, creating a dust cloud. If the cloud consists of 0.001, 0.01, and 0.1 mm diameter particles, and the particle density is 700 kg/m³, will the child be able to clean up the mess by dusting the furniture before her mother returns an hour later? Assume that the particles near the ceiling must settle 2.5 m before depositing on various surfaces, and that the air temperature is 20°C.

Accepted Answer

For a particle settling at terminal velocity, a vertical force balance shows that
W = F_air
where W = ρPg[latex]\sout{V}[/latex] is the weight of a particle of density ρ_P, and F_air is the total force applied by the air to the particle. The force applied by the air consists of a drag force F_D that accounts for the relative motion of the particle through the stationary air and a buoyancy force F_B that accounts for the effects of the hydrostatic pressure variation in the air. Because the buoyancy force is not included when the drag force is calculated by using a drag coefficient, we must write

W = F_D + F_B (A)

(Another way to think about this problem is revealed by rearranging this equation as W − F_B = F_D. Since W − F_B is the weight of the particle as measured in air, we see that the force balance equates this weight, which causes the particle to settle, to the drag force that resists the settling motion.) Substituting for each term in the force balance (A) gives
[latex]ρ_Pg\sout{V} =C_D\frac{1}{2} ρ_{air} U^2A+ρ_{air} g\sout{V}[/latex]
Solving for the terminal velocity we obtain

[latex]U=\sqrt{\frac{2(ρ_P −ρ_{air} )g\sout{V} }{C_Dρ_{air}A}}[/latex] (B)

We will assume a spherical particle and a creeping ﬂow drag coefficient given by C_D = 24/Re. Note that since C_D = 24/Re = 24µ/(ρ_airU D) in this case, the terminal velocity also occurs in the drag coefficient in (B). The area and volume are A = [latex]\pi[/latex]D²/4 and [latex]\sout{V}[/latex] = [latex]\pi[/latex]D³/6, respectively; hence [latex]\sout{V}[/latex]/A =2D/3. Inserting these values into (B) shows that the terminal velocity is given by

[latex]U=\frac{(ρ_P −ρ_{air} )gD^2 }{18\mu} [/latex] (C)

The time needed for a particle to settle from a height H is t = H/U. Thus the settling time is

[latex]t=\frac{18\mu H}{(ρ_P −ρ_{air} )gD^2} [/latex] (D)

Since the diameter occurs in the denominator, the smallest particles take the longest time to settle. The maximum settling time t_max is found using H = 2.5 m. Inserting data for air ρ_air = 1.2 kg/m³, µ = 1.81 × 10⁻⁵ (N-s)/m², and other values into (B)–(D), we can construct the following table showing t_max for each particle size:

D(mm)	U(m/s)	t_max (minutes)	Re
0.001	2.1 × 10⁻⁵	2000	1.4 × 10⁻⁶
0.01	2.1 × 10⁻³	20	1.4× 10⁻³
0.1	2.1 × 10⁻¹	0.2	1.4

The calculated Reynolds numbers conﬁrm the validity of the creeping ﬂow assumption. It is evident that the smallest particles, which take ∼33 h to settle, will pose a problem. Note that since ρ_air = 0.7% ρ_P, buoyancy is negligible in this example.

Question 14.3: A playful child left alone has run a vacuum cleaner in rever...

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