Air-Parcel Residence Times Within Forest Canopies

Abstract

We present a theoretical model, based on a simple model of turbulent diffusion and first-order chemical kinetics, to determine air-parcel residence times and the out-of-canopy export of reactive gases emitted within forest canopies under neutral conditions. Theoretical predictions of the air-parcel residence time are compared to values derived from large-eddy simulation for a range of canopy architectures and turbulence levels under neutral stratification. Median air-parcel residence times range from a few sec in the upper canopy to approximately 30 min near the ground and the distribution of residence times is skewed towards longer times in the lower canopy. While the predicted probability density functions from the theoretical model and large-eddy simulation are in good agreement with each other, the theoretical model requires only information on canopy height and eddy diffusivities inside the canopy. The eddy-diffusivity model developed additionally requires the friction velocity at canopy top and a parametrized profile of the standard deviation of vertical velocity. The theoretical model of air-parcel residence times is extended to include first-order chemical reactions over a range of of Damköhler numbers (Da) characteristic of plant-emitted hydrocarbons. The resulting out-of-canopy export fractions range from near 1 for $Da =10^{-3}$ to less than 0.3 at $Da = 10$. These results highlight the necessity for dense and tall forests to include the impacts of air-parcel residence times when calculating the out-of-canopy export fraction for reactive trace gases.

Appendix 1: The First Passage Solution

If the turbulent transport of air parcels is assumed to be diffusive with a constant eddy diffusivity $K_\mathrm{eq}$, then the time evolution of the particle position can be modelled by a Wiener process without mean drift

$$\begin{aligned} \mathrm{d}z=\sqrt{2K_\mathrm{eq}}\mathrm{d}W, \end{aligned}$$

(24)

where W is a Wiener process with independent Gaussian increments. Under these conditions, the time evolution of the probability density function (PDF) of the particle position satisfies the Fokker–Planck equation (Thomson 1987; Rodean 1996)

$$\begin{aligned} \frac{\partial P(z,t;z_\mathrm{rel})}{\partial t}=K_\mathrm{eq}\frac{\partial ^2 P(z,t;z_\mathrm{rel})}{\partial z^2}, \end{aligned}$$

(25)

where $P(z,t;z_\mathrm{rel})$ is the probability of a particle released at $z=z_\mathrm{rel}$ at $t=0$ to be found at z at a time t. Considering a semi-infinite domain (i.e., $z<h_c$), the first time a parcel crosses the boundary can be modelled by placing an absorbing boundary at $z=h_c$. Thus, the two boundary conditions for Eq. 25 are given by $P(z=h_c,t;z_\mathrm{rel})=0$ and $P(z\rightarrow -\infty ,t;z_\mathrm{rel})=0$. Together with the initial condition $P(z,t=0;z_\mathrm{rel})=\delta (z-z_\mathrm{rel})$, the solution is the well-known Gaussian function

$$\begin{aligned} P(z,t;z_\mathrm{rel})=\frac{1}{\sqrt{4\pi K_\mathrm{eq}}}t^{-1/2}\left\{ \exp {\left[ -\frac{(z-z_\mathrm{rel})^2}{4K_\mathrm{eq}t}\right] }-\exp {\left[ -\frac{(z-\left( 2h_c-z_\mathrm{rel}\right) )^2}{4K_\mathrm{eq}t}\right] }\right\} \end{aligned}$$

(26)

In the present case, the probability of an air parcel still being in the domain $z<h_c$ (i.e., it has not crossed the boundary yet) is given by

$$\begin{aligned} F(t;z_\mathrm{rel})=\int _{z=0}^{\infty }{P(z,t;z_\mathrm{rel})\mathrm{d}z}=\mathrm{erf}\left( \frac{h_c-z_\mathrm{rel}}{\sqrt{4K_\mathrm{eq}t}}\right) \end{aligned}$$

(27)

The probability $F(t;z_\mathrm{rel})$ is usually referred to as the survival probability. The rate of decrease in $F(t;z_\mathrm{rel})$ at a given time $\tau$ is equal to the probability of particles crossing the boundary at that time. Thus, the first-passage distribution for the present scenario is given by

$$\begin{aligned} p(\tau ;z_\mathrm{rel})=-\left. \frac{\partial F(t;z_\mathrm{rel})}{\partial t}\right| _{z=h_c,t=\tau }=\frac{(h_c-z_\mathrm{rel})}{\sqrt{4\pi K_\mathrm{eq}}}\tau ^{-3/2}\exp {\left[ -\frac{(h_c-z_\mathrm{rel})^2}{4K_\mathrm{eq}\tau }\right] }, \end{aligned}$$

(28)

which is identical to Eq. 1.