Particle Dispersion in the Neutral Atmospheric Surface Layer

Abstract

We address theoretically the longstanding problem of particle dispersion in the lower atmosphere. The evolution of particle concentration under an absorbing boundary condition at the ground is described. We derive a close-form solution for the downwind surface density of deposited particles and find how the number of airborne particles decreases with time. The problem of the plume formation above the extended surface source is also solved analytically. At the end, we show how turbophoresis modifies the mean settling velocity of particles.

Appendices

Appendix 1

Let us consider the dispersion of $N_0$ particles released initially at the height $z_0$ above the ground level. Our goal is to derive the number $N(t)=\int _{0}^{\infty }\tilde{\theta }(\theta ,t)\text {d}z$ of particles in the air as a function of time. To be able to impose the boundary condition at the surface, we regularize the problem by having a non-zero diffusivity at $z=0$. This give the transport equation

$$\begin{aligned} \partial _t \tilde{\theta }=\mu \partial _z [(z+r)\partial _z \tilde{\theta }]+g\tau \partial _z \tilde{\theta }, \end{aligned}$$

(39)

subject to the conditions $[\mu (z+r)\partial _z \tilde{\theta }+g\tau \tilde{\theta }]_{z=0}=[v_\mathrm{d} \theta ]_{z=0}$ and $\tilde{\theta }(z,0)=N_0\delta (z-z_0)$. The spatial scale r is a regularization parameter. At the end of the calculations, the limit $r\rightarrow 0$ will be taken.

Performing the Laplace transform

$$\begin{aligned} \tilde{\theta }_\mathrm{s}(z)=\int \limits _{0}^{+\infty }\text {d}te^{-st}\tilde{\theta }(z,t), \end{aligned}$$

(40)

one obtains an ordinary second-order differential equation

$$\begin{aligned} \mu \partial _z [(z+r)\partial _z \tilde{\theta }_\mathrm{s}] +g\tau \partial _z \tilde{\theta }_\mathrm{s}-s\tilde{\theta }_s=-N_0\delta (z-z_0), \end{aligned}$$

(41)

which should be supplemented by the condition $[D(z)\partial _z \tilde{\theta }_\mathrm{s} +g\tau \tilde{\theta }_\mathrm{s}]_{z=0}=[v_\mathrm{d} \theta _\mathrm{s}]_{z=0}$.

We pass to the new variable $\xi =\sqrt{z+r}$ and substitute $\tilde{\theta }_\mathrm{s}=\xi ^{-\gamma }f$. At $\xi \ne \sqrt{z_0+r}$ the function f obeys the modified Bessel’s equation

$$\begin{aligned} \mu \xi ^2\frac{\text {d}^2f}{\text {d}\xi ^2}+\mu \xi \frac{\text {d}f}{\text {d}\xi }-\left( \gamma ^2+\frac{4s}{\mu }\xi ^2\right) f=0. \end{aligned}$$

(42)

Therefore, two linearly independent solutions of Eq. 41 for $z\ne z_0$ can be chosen as

$$\begin{aligned} \tilde{\theta }_{s1}(z)= & {} (z+r)^{-\gamma /2}\mathcal{I}_\gamma (2\sqrt{s(z+r)/\mu }), \end{aligned}$$

(43)

$$\begin{aligned} \tilde{\theta }_{s2}(z)= & {} (z+r)^{-\gamma /2}\mathcal{K}_\gamma (2\sqrt{s(z+r)/\mu }), \end{aligned}$$

(44)

where $\mathcal{I}_\gamma$ and $\mathcal{K}_\gamma$ denote the modified Bessel functions of the first and second kind respectively Abramowitz and Stegun (1964).

Next it is straightforward to find that the function

$$\begin{aligned} \tilde{\theta }_{s}(z)=\left\{ \begin{array}{ll} A_1\tilde{\theta }_{s1}(z)+A_2\tilde{\theta }_{s2}(z), &{} 0\le z\le z_0,\\ \\ A_3\tilde{\theta }_{s2}(z), &{} z\ge z_0, \end{array} \right. \end{aligned}$$

(45)

with

$$\begin{aligned} A_1= & {} \frac{2N_0}{\mu }(z_0+r)^{\gamma /2}\mathcal{K}_\gamma \left( 2\sqrt{\frac{s(z_0+r)}{\mu }}\right) , \end{aligned}$$

(46)

$$\begin{aligned} A_2= & {} \frac{\sqrt{\frac{sr}{\mu }}\mathcal{I}_\gamma '\left( 2\sqrt{\frac{sr}{\mu }}\right) -\left( \frac{v_\mathrm{d}}{\mu }-\frac{\gamma }{2}\right) \mathcal{I}_\gamma \left( 2\sqrt{\frac{sr}{\mu }}\right) }{\left( \frac{v_\mathrm{d}}{\mu }-\frac{\gamma }{2}\right) \mathcal{K}_\gamma \left( 2\sqrt{\frac{sr}{\mu }}\right) -\sqrt{\frac{sr}{\mu }}\mathcal{K}_\gamma '\left( 2\sqrt{\frac{sr}{\mu }}\right) }A_1, \end{aligned}$$

(47)

$$\begin{aligned} A_3= & {} \frac{\mathcal{I}_\gamma \left( 2\sqrt{\frac{s(z_0+r)}{\mu }}\right) }{\mathcal{K}_\gamma \left( 2\sqrt{\frac{s(z_0+r)}{\mu }}\right) }A_1+A_2, \end{aligned}$$

(48)

satisfies Eq. 41 together with the boundary condition $[D(z)\partial _z \tilde{\theta }_\mathrm{s} +g\tau \tilde{\theta }_\mathrm{s}]_{z=0}=[v_\mathrm{d} \tilde{\theta }_\mathrm{s}]_{z=0}$.

The ground deposition flux $\tilde{j}_z(t) = -[v_\mathrm{d} \tilde{\theta }]_{z=0}$ is given by the following contour integral

$$\begin{aligned} \tilde{j}_z(t)= & {} -\frac{v_\mathrm{d}}{2\pi i}\int _C \text {d}s\ e^{st}\left( A_1\tilde{\theta }_{s1}(0)+A_2\tilde{\theta }_{s2}(0)\right) \nonumber \\= & {} -\frac{N_0}{2\pi i}\frac{v_\mathrm{d}}{\mu }\int _C\text {d}s\ e^{st} \left( 1+\frac{z_0}{r}\right) ^{\gamma /2}\frac{\mathcal{K}_\gamma (2\sqrt{s(z_0+r)/\mu })}{\left( \frac{v_\mathrm{d}}{\mu }-\frac{\gamma }{2}\right) \mathcal{K}_\gamma (2\sqrt{sr/\mu })-\sqrt{\frac{sr}{\mu }}\mathcal{K}_\gamma '(2\sqrt{sr/\mu })}.\nonumber \\ \end{aligned}$$

(49)

Now we put $r\rightarrow 0$ and obtain

$$\begin{aligned} \tilde{j}_z(t)=-\frac{N_0}{i\pi \varGamma (\gamma )}\int _C \text {d}s\ e^{st} \left( \frac{sz_0}{\mu }\right) ^{\gamma /2} \mathcal{K}_\gamma \left( 2\sqrt{\frac{sz_0}{\mu }}\right) . \end{aligned}$$

(50)

Note that the deposition velocity $v_\mathrm{d}$ drops out in this limit. The branch cut for the analytic continuation of the integrand in (50) is defined on the negative real axis of the complex plane. Then, we use so-called Hankel integration contour which extends from the point $-\infty -0\cdot i$, around the origin counter-clockwise and back to the point $-\infty +0\cdot i$. This leads us to the following result

$$\begin{aligned} \tilde{j}_z(t)= & {} -\frac{N_0}{i\pi \varGamma (\gamma )}\left( \frac{z_0}{\mu }\right) ^{\gamma /2} \int \limits _0^\infty s^{\gamma /2}e^{-st}\left[ e^{i\pi \gamma /2}\mathcal{K}_\gamma (2i\sqrt{sz_0/\mu })\nonumber \right. \\&\left. -\,e^{-i\pi \gamma /2}\mathcal{K}_\gamma (-2i\sqrt{sz_0/\mu })\right] \text {d}s =-\frac{N_0}{\varGamma (\gamma )}\frac{z_0^\gamma }{\mu ^\gamma t^{\gamma +1}} \exp \left( -\frac{z_0}{\mu t}\right) . \end{aligned}$$

(51)

The deposition flux $\tilde{j}_z$ and the number of airborne particles N are related by identity $\tilde{j}_z=\text {d}N/\text {d}t$. Performing integration of (51) over t we obtain the survival probability $p(t)\equiv N(t)/N_0$ in the explicit form (17).

Appendix 2

Here we derive the exact expression (27) for the surface distribution of deposited particles $\sigma (x)$. For this aim we turn to Eq. 3 and add the constant correction $\mu r$ to diffusivity in order to avoid the vanishing of coefficient near highest spatial derivative. The resulting transport equation is as follows,

$$\begin{aligned} \partial _t \theta = \mu \partial _z [(z+r)\partial _z \theta ] n+g\tau \partial _z \theta -\beta z^m\partial _x \theta . \end{aligned}$$

(52)

The initial and boundary conditions are chosen to be $\theta (x,z,0)=N_0\delta (x)\delta (z-z_0)$ and $[\mu (z+r)\partial _z\theta +g\tau \theta ]_{z=0}=[v_\mathrm{d}\theta ]_{z=0}$.

The surface density $\sigma (x)$ is given by the total number of settled particles per unit length in the downwind direction, i.e.

$$\begin{aligned} \sigma (x)=-\int \limits _{0}^{+\infty } j_z(x,z=0,t)\text {d}t, \end{aligned}$$

(53)

where $j_z=-\mu (z+r)\partial _z\theta -g\tau \theta$ is the vertical component of particle flux. Let us rewrite this relation as

$$\begin{aligned} \sigma (x)=- J_z(x,z=0), \end{aligned}$$

(54)

where $J_z=-\mu (z+r)\partial _z\varTheta -g \tau \varTheta$ is the flux for integrated concentration $\varTheta (x,z)=\int _0^\infty \theta (x,z,t)\text {d}t$ which obeys the equation

$$\begin{aligned} \mu \partial _z [(z+r)\partial _z \varTheta ] n+g\tau \partial _z \varTheta -\beta z^m\partial _x \varTheta =-N_0\delta (x)\delta (z-z_0), \end{aligned}$$

(55)

with the boundary condition $[\mu (z+r)\partial _z\varTheta +g\tau \varTheta ]_{z=0}=[v_\mathrm{d}\varTheta ]_{z=0}$.

Next, we perform the Laplace transform of $\varTheta (x,z)$ with respect to x

$$\begin{aligned} \varTheta _\mathrm{s}(z)=\int \limits _{0}^{+\infty }e^{-sx}\varTheta (x,z)\text {d}x. \end{aligned}$$

(56)

This transformation leads to the following inhomogeneous differential equation for $\varTheta _\mathrm{s}(z)$

$$\begin{aligned} \mu \partial _z [(z+r)\partial _z \varTheta _\mathrm{s}] n+g\tau \partial _z \varTheta _\mathrm{s}-s\beta z^m\varTheta _\mathrm{s}=-N_0\delta (z-z_0). \end{aligned}$$

(57)

For $0\le z\le z_0$ the solution of this equation under the condition $[\mu (z+r)\partial _z\varTheta +g\tau \varTheta ]_{z=0}=[v_\mathrm{d}\varTheta ]_{z=0}$ reads

$$\begin{aligned} \tilde{\theta }_{s}(z)=B_1\varTheta _{s1}(z)+B_2 \varTheta _{s2}(z), \end{aligned}$$

(58)

where

$$\begin{aligned} \varTheta _{s1}(z)= & {} (z+r)^{-\gamma /2}\mathcal{I}_{\gamma _m}\left( 2\sqrt{\frac{\beta s}{\mu }} \frac{(z+r)^{\frac{m+1}{2}}}{m+1}\right) , \end{aligned}$$

(59)

$$\begin{aligned} \varTheta _{s2}(z)= & {} (z+r)^{-\gamma /2}\mathcal{K}_{\gamma _m}\left( 2\sqrt{\frac{\beta s}{\mu }} \frac{(z+r)^{\frac{m+1}{2}}}{m+1}\right) , \end{aligned}$$

(60)

$$\begin{aligned} B_1= & {} \frac{2N_0}{(m+1)\mu }(z_0+r)^{\gamma /2}\mathcal{K}_{\gamma _m} \left( 2\sqrt{\frac{\beta s}{\mu }}\frac{(z_0+r)^{\frac{m+1}{2}}}{m+1}\right) , \end{aligned}$$

(61)

$$\begin{aligned} B_2= & {} \frac{r^{\frac{m+1}{2}}\sqrt{\frac{\beta s}{\mu }}\mathcal{I}_{\gamma _m}'\left( 2\sqrt{\frac{\beta s}{\mu }} \frac{r^{\frac{m+1}{2}}}{m+1}\right) -\left( \frac{v_\mathrm{d}}{\mu }-\frac{\gamma }{2}\right) \mathcal{I}_{\gamma _m}\left( 2\sqrt{\frac{\beta s}{\mu }} \frac{r^{\frac{m+1}{2}}}{m+1}\right) }{\left( \frac{v_\mathrm{d}}{\mu }-\frac{\gamma }{2}\right) \mathcal{K}_{\gamma _m}\left( 2\sqrt{\frac{\beta s}{\mu }} \frac{r^{\frac{m+1}{2}}}{m+1}\right) -r^{\frac{m+1}{2}}\sqrt{\frac{\beta s}{\mu }}\mathcal{K}_{\gamma _m}'\left( 2\sqrt{\frac{\beta s}{\mu }} \frac{r^{\frac{m+1}{2}}}{m+1}\right) }B_1, \end{aligned}$$

(62)

and $\gamma _m=\gamma /(m+1)$.

Now we apply the inverse Laplace transform to the solution (58) and substitute result into (54). This gives the surface density of particles in the form of an integral in the complex plane

$$\begin{aligned} \sigma (x)=\frac{v_\mathrm{d}N_0}{2\pi i}\int _C \frac{e^{sx}\left( 1+\frac{z_0}{r}\right) ^{\gamma /2}\mathcal{K}_{\gamma _m}\left( 2\sqrt{\frac{\beta s}{\mu }}\frac{(z_0+r)^{\frac{m+1}{2}}}{m+1}\right) }{\left( v_\mathrm{d}-\frac{\mu \gamma }{2}\right) \mathcal{K}_{\gamma _m}\left( 2\sqrt{\frac{\beta s}{\mu }}\frac{r^{\frac{m+1}{2}}}{m+1}\right) -r^{\frac{m+1}{2}}\sqrt{\mu \beta s}\mathcal{K}_{\gamma _m}'\left( 2\sqrt{\frac{\beta s}{\mu }}\frac{r^{\frac{m+1}{2}}}{m+1}\right) }\text {d}s, \end{aligned}$$

(63)

which in the limit $r\rightarrow 0$ becomes

$$\begin{aligned} \sigma (x)=\frac{N_0z_0^{\gamma /2}\beta ^{\gamma _m/2}}{\mu ^{\gamma _m/2}(m+1)^{\gamma _m}\varGamma (\gamma _m)\pi i} \int _C e^{sx}s^{\gamma _m/2}\mathcal{K}_{\gamma _m} \left( 2\sqrt{\frac{\beta s}{\mu }}\frac{z_0^{\frac{m+1}{2}}}{m+1}\right) \text {d}s. \end{aligned}$$

(64)

To calculate this integral we define contour C as the Hankel path extending from the point $-\infty -0\cdot i$, circling the origin counter-clockwise, and returning to the point $-\infty +0\cdot i$. After some algebra the closed form expression (27) can be derived.

Appendix 3

Here we consider the case where the surface source is switched on at $t=0$. For definiteness, we speak about the dust produced by the industrial area. Evolution of the dust concentration is described by (39). In this sub-section we put $\mu \rightarrow 1$ thus passing to

$$\begin{aligned} \partial _t \tilde{\theta }= \partial _z\left\{ [(z+r)\partial _z+\gamma ] \tilde{\theta }\right\} . \end{aligned}$$

(65)

with the initial condition $\tilde{\theta }(z,0)=0$ We consider two different boundary conditions corresponding to a fixed dust concentration at $z=0$ and to a fixed particle flux at $z=0$.

Let us produce the Laplace transform with respect to time

$$\begin{aligned} \tilde{\theta }_\mathrm{s}(z)=\int _0^\infty \text {d}t\ \exp (-st) \tilde{\theta }(z,t). \end{aligned}$$

(66)

The inverse Laplace transform reads

$$\begin{aligned} \tilde{\theta }(z,t)=\int _{C} \frac{\text {d}s}{2\pi i} \exp (st) \tilde{\theta }_\mathrm{s}(z), \end{aligned}$$

(67)

where integration is performed along a line parallel to the imaginary axis to the right of all singularities of $\tilde{\theta }_\mathrm{s}$. In terms of $\tilde{\theta }_\mathrm{s}$ the Eq. 65 becomes

$$\begin{aligned} s \tilde{\theta }_\mathrm{s}= \partial _z\left\{ [(z+r)\partial _z+\gamma ] \tilde{\theta }_\mathrm{s}\right\} . \end{aligned}$$

(68)

A solution of the Eq. 68 tending to zero as $z\rightarrow \infty$ is

$$\begin{aligned} \tilde{\theta }_\mathrm{s}(z)\propto (z+r)^{-\gamma /2} K_\gamma (2\sqrt{s(z+r)}). \end{aligned}$$

(69)

Let us pose the boundary condition $\tilde{\theta }=1$ at $z=0$. In terms of the Laplace transform the boundary condition is $\tilde{\theta }_\mathrm{s}(0)=s^{-1}$. The solution (69) with the boundary condition is

$$\begin{aligned} \tilde{\theta }_\mathrm{s}(z)= \frac{1}{s}\left( \frac{r}{z+r}\right) ^{\gamma /2} \frac{K_\gamma (2\sqrt{s(z+r)})}{K_\gamma (2\sqrt{sr})}. \end{aligned}$$

(70)

Now we should perform the inverse Laplace transform. When (70) is substituted into Eq. 67 one finds

$$\begin{aligned} \tilde{\theta }(z,t)=\int _{C} \frac{\text {d}s}{2\pi i s} \left( \frac{r}{z+r}\right) ^{\gamma /2} \frac{K_\gamma (2\sqrt{s(z+r)})}{K_\gamma (2\sqrt{sr})} \exp (st). \end{aligned}$$

(71)

For small values of argument $K_\gamma (x)=2^{-1}\varGamma (\gamma ) (2/x)^\gamma$. Substituting the expression into Eq. 71 one obtains

$$\begin{aligned} \tilde{\theta }(z,t)= & {} \frac{r^\gamma }{ \varGamma (\gamma )} \int _{C} \frac{\text {d}s}{\pi i s} \left( \frac{s}{z+r}\right) ^{\gamma /2} K_\gamma (2\sqrt{s(z+r)}) \exp (st)\nonumber \\= & {} \frac{1}{\varGamma (\gamma )} \frac{r^\gamma }{(z+r)^\gamma } \left( \frac{z+r}{t}\right) ^{\gamma /2} \int _{C} \frac{\text {d}\zeta }{\pi i \zeta } \zeta ^{\gamma /2} K_\gamma (2\sqrt{\zeta (z+r)/t}) \exp (\zeta ). \end{aligned}$$

(72)

If $z\ll t$ then the main contribution to the integral (72) is produced by the residue in the pole $\zeta =0$, therefore

$$\begin{aligned} \tilde{\theta }\approx (r/z)^\gamma , \end{aligned}$$

(73)

for $z\gg r$. If $z\gg t$ then the main contribution to the integral (72) stems from the saddle point $\zeta =(z+r)/t$, therefore

$$\begin{aligned} \tilde{\theta }\propto \left( \frac{h}{z}\right) ^\gamma \left( \frac{z}{t}\right) ^{\gamma -1} \exp \left( -\frac{z}{t}\right) , \end{aligned}$$

(74)

for $z\gg r$.

Let us now consider another boundary condition at $z=0$ that implies constant upward flux of particles:

$$\begin{aligned} - r\partial _z \tilde{\theta }- \gamma \tilde{\theta }=1. \end{aligned}$$

(75)

We focus on the the case $\gamma <1$. Then the main terms of the McDonald function expansion at small x are

$$\begin{aligned} K_\gamma (x)\approx \frac{1}{2} \left[ \varGamma (\gamma ) \left( \frac{x}{2}\right) ^{-\gamma } +\varGamma (-\gamma ) \left( \frac{x}{2}\right) ^{\gamma }\right] . \end{aligned}$$

Therefore the boundary condition (75) leads to

$$\begin{aligned} \tilde{\theta }_\mathrm{s}(z)=\frac{1}{s^{1+\gamma /2}(z+r)^{\gamma /2}} \frac{K_\gamma (2\sqrt{s(z+r)})}{\varGamma (1-\gamma )}. \end{aligned}$$

(76)

Substituting the expression into Eq. 67 we find

$$\begin{aligned} \tilde{\theta }(z,t)= & {} \int _C \frac{\text {d}s}{2\pi i} \exp (st) \frac{1}{s^{1+\gamma /2}(z+r)^{\gamma /2}} \frac{K_\gamma (2\sqrt{s(z+r)})}{\varGamma (1-\gamma )}\nonumber \\= & {} \left( \frac{t}{z+r}\right) ^{\gamma /2} \int _C \frac{\text {d}\zeta }{2\pi i \zeta ^{1+\gamma /2}} \exp (\zeta ) \frac{K_\gamma (2\sqrt{\zeta (z+r)/t})}{\varGamma (1-\gamma )}. \end{aligned}$$

(77)

At small z / t the integral is gained at $\zeta \sim 1$. Substituting the main asymptotic of the McDonald function and deforming the integration contour to the negative semi-axis, one obtains

$$\begin{aligned} \tilde{\theta }(z,t)= & {} \frac{\varGamma (\gamma )}{\varGamma (1-\gamma )} \left( \frac{t}{z}\right) ^{\gamma } \int _C \frac{\text {d}\zeta }{4\pi i \zeta ^{1+\gamma }} \exp (\zeta )\nonumber \\= & {} \frac{\varGamma (\gamma )}{2 \varGamma (1+\gamma )\varGamma (1-\gamma )} \left( \frac{t}{z}\right) ^{\gamma }, \end{aligned}$$

(78)

for $z\gg r$. At large z / t the integral is determined by the saddle point where the asymptotic expression for $K_\gamma$ can be exploited. As a result, one finds

$$\begin{aligned} \tilde{\theta }\propto \left( \frac{t}{z}\right) ^{1+\gamma } \exp \left( -\frac{z}{t}\right) , \end{aligned}$$

(79)

for $z\gg r$. The expressions (78,79) demonstrate the same self-similarity as the expressions (73,74) do, however, with other power prefactor.