Kvant Physics Problem 154
The physical system is a layer of fog consisting of identical spherical water droplets suspended in air.
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 7m50s
Source on kvant.digital
Problem
Estimate how many water droplets are contained in $1~\text{м}^3$ of fog if the visibility is 10 m and the fog settles out in 2 hours. The height of the fog layer is 200 m.
The air resistance force acting on a water droplet of radius $R~\text{(м)}$ moving with velocity $v~\text{(м/с)}$ is equal to $4{,}3Rv~\text{(Н)}$.
P. L. Kapitsa
Setup and Assumptions
The physical system is a layer of fog consisting of identical spherical water droplets suspended in air. The required quantity is the number density of droplets, denoted by $n$, measured in droplets per cubic meter.
The given data are:
$$L = 10,\text{m},$$
where $L$ is the visibility distance,
$$H = 200,\text{m},$$
where $H$ is the height of the fog layer,
$$t = 2,\text{h}=7200,\text{s},$$
where $t$ is the time during which the fog settles out,
$$F_{\rm d}=4.3Rv,$$
where $F_{\rm d}$ is the air resistance force in newtons, $R$ is the droplet radius in meters, and $v$ is the droplet speed in meters per second.
The density of water is taken as
$$\rho = 10^3,\text{kg/m}^3,$$
and the acceleration due to gravity as
$$g=9.8,\text{m/s}^2.$$
The droplets are assumed to be identical spheres. The settling process is assumed to occur at terminal velocity. Interactions between droplets are neglected. The visibility is interpreted as the mean free path of a light ray before encountering a droplet. Each droplet is treated as an opaque obstacle with geometric cross section
$$\sigma=\pi R^2.$$
The fog is assumed to be homogeneous throughout the layer.
Physical Principles
A droplet settling at terminal velocity has zero acceleration. The downward gravitational force is balanced by the upward air resistance force:
$$mg=F_{\rm d}.$$
The mass of a spherical droplet is
$$m=\frac43\pi R^3\rho .$$
The fog disappears because droplets fall through the entire layer height $H$ in time $t$. Hence the terminal velocity is
$$v=\frac{H}{t}.$$
For randomly distributed obstacles of number density $n$, the mean free path of a ray is
$$\lambda=\frac1{n\sigma}.$$
The visibility distance is identified with this mean free path:
$$L=\lambda.$$
Derivation
The terminal settling velocity follows from the observed settling time:
$$v=\frac{H}{t}.$$
Substituting the numerical values,
$$v=\frac{200,\text{m}}{7200,\text{s}} =2.78\times10^{-2},\text{m/s}.$$
The gravitational force on a droplet is
$$mg=\frac43\pi R^3\rho g.$$
At terminal velocity,
$$\frac43\pi R^3\rho g = 4.3Rv.$$
Dividing by $R$ gives
$$\frac43\pi \rho g,R^2 = 4.3v.$$
Hence
$$R^2=\frac{4.3v}{\frac43\pi\rho g}.$$
Substituting the numbers,
$$R^2= \frac{4.3,(2.78\times10^{-2},\text{m/s})} {\frac43\pi,(10^3,\text{kg/m}^3),(9.8,\text{m/s}^2)}.$$
The denominator equals
$$\frac43\pi(10^3)(9.8) \simeq 4.11\times10^4.$$
Thus
$$R^2\simeq \frac{0.119}{4.11\times10^4} \simeq 2.9\times10^{-6},\text{m}^2.$$
The geometric cross section of one droplet is
$$\sigma=\pi R^2 \simeq \pi(2.9\times10^{-6}) \simeq 9.1\times10^{-6},\text{m}^2.$$
The visibility condition gives
$$L=\frac1{n\sigma}.$$
Solving for the number density,
$$n=\frac1{L\sigma} =\frac1{L\pi R^2}.$$
Substituting $L=10,\text{m}$ and $\sigma=9.1\times10^{-6},\text{m}^2$,
$$n= \frac1{(10)(9.1\times10^{-6})} \simeq 1.1\times10^4,\text{m}^{-3}.$$
Result
The symbolic expression for the droplet number density is
$$n=\frac{1}{L\pi R^2}, \qquad R^2=\frac{4.3(H/t)} {\frac43\pi\rho g}.$$
Combining these relations,
$$n= \frac{\frac43\pi\rho g} {4.3,\pi,L,(H/t)} = \frac{4\rho g,t} {12.9,L,H}.$$
Substituting the numerical values,
$$n= \frac{1} {(10,\text{m})(9.1\times10^{-6},\text{m}^2)} \simeq 1.1\times10^4,\text{m}^{-3}.$$
$$\boxed{n \approx 1\times10^{4}\ \text{droplets/m}^{3}}$$
Thus one cubic meter of this fog contains approximately ten thousand water droplets.
Sanity Checks
The dimensions are consistent. The quantity $n\sigma L$ is dimensionless because
$$[\sigma]=\text{m}^2, \qquad [n]=\text{m}^{-3}, \qquad [L]=\text{m},$$
so
$$[n\sigma L]=1.$$
The computed radius is
$$R=\sqrt{2.9\times10^{-6}} \simeq1.7\times10^{-3},\text{m},$$
which corresponds to a droplet diameter of about
$$3.4\times10^{-3},\text{m}=3.4,\text{mm}.$$
A larger droplet settles faster, which agrees with the force balance relation $v\propto R^2$. Since the fog settles out in only two hours from a height of $200,\text{m}$, relatively large droplets are expected.
The most vulnerable step is the terminal velocity condition. Using $v=t/H$ instead of $v=H/t$, or forgetting to convert $2,\text{h}$ into $7200,\text{s}$, would change the answer by several orders of magnitude. Another common error is to use the droplet area $R^2$ instead of the geometric cross section $\pi R^2$ in the visibility relation.