A Method for Estimating the Threshold Wind Speed for Dust Emissions as a Function of Soil Moisture

Abstract

A method is proposed to estimate the threshold wind speed for dust emissions as a function of soil moisture in arid regions. This method, which is applicable at the local scale, employs a model of the surface heat budget to estimate the spatial distribution of thermal inertia-derived soil moisture (TISM) and an analytical footprint model to estimate dust source areas. It incorporates readily available satellite and meteorological data. The soil moisture inside the dust source area is estimated for individual dust phenomena observed at a synoptic surface observation site, and then, together with the corresponding observed wind speed, the threshold wind speed as a function of soil moisture is found by employing the parametrization of Fécan et al. (Ann Geophys 17:149–157, 1999). This relationship represents the local dust climatology at the observation site, although the data include some outliers. The results show that readily available data for the natural environment can be used, as an alternative to wind-tunnel data, with the parametrization used to derive the threshold wind-speed relation. The derived relation indicates the TISM and wind speed at which the probability of a dust phenomenon is 10%. Moreover, the threshold wind-speed results are not sensitive to the principal footprint-model parameters. The TISM standard error of approximately 0.04 m³ m⁻³ could significantly affect the results under dry and weak-wind conditions, but this sensitivity can be avoided if the soil clay content or the soil type at the observation point is known.

Appendix 1: Derivation of the Concentration Footprint Parameters in the Kormann and Meixner (2001) Model

The specific formulations of the concentration footprint model that do not appear in Kormann and Meixner (2001) (hereafter abbreviated as KM2001) are given below. First, the maximum crosswind integrated footprint for concentration $c_{\hbox{max} }$ and the $x_{d}$ value $x_{{_{\hbox{max} } }}^{(c)}$ are derived.

The crosswind integrated concentration footprint $c(x,z)$(see Eq. 20 in KM2001) is given by

$$ c(x,z) = \frac{1}{\Gamma (\mu )}\frac{r}{{Uz^{1 + m} }}\frac{{\xi^{\mu } }}{{x^{\mu } }}\exp \left( { - \frac{\xi }{x}} \right), $$

(5)

where differentiation yields

$$ \frac{\partial c}{\partial x} = \frac{1}{\Gamma (\mu )}\frac{{r\xi^{\mu } }}{{Uz^{1 + m} }}\left[ { - \frac{\mu }{{x^{\mu + 1} }}\exp \left( { - \frac{\xi }{x}} \right) + \frac{1}{{x^{\mu } }}\exp \left( { - \frac{\xi }{x}} \right) \cdot \frac{\xi }{{x^{2} }}} \right] $$

$$ = \frac{1}{\Gamma (\mu )}\frac{{r\xi^{\mu } }}{{Uz^{1 + m} }}\frac{1}{{x^{\mu + 1} }}\exp \left( { - \frac{\xi }{x}} \right)\left[ { - \mu + \frac{\xi }{x}} \right]. $$

(6)

Then, $c(x,z)$ attains a maximum value when

$$ \frac{\partial c}{\partial x} = \frac{1}{\Gamma (\mu )}\frac{{r\xi^{\mu } }}{{Uz^{1 + m} }}\frac{1}{{x^{\mu + 1} }}\exp \left( { - \frac{\xi }{x}} \right)\left[ { - \mu + \frac{\xi }{x}} \right] = 0. $$

(7)

By solving this equation with respect to $x$, the along-wind position of the maximum crosswind integrated concentration footprint $x_{\hbox{max} }^{(c)}$ is determined as

$$ x_{\hbox{max} }^{(c)} = \frac{\xi }{\mu }. $$

(8)

Then, by substituting Eq. 8 into Eq. 5, the maximum value of $c(x,z)$ is derived as

$$ c_{\hbox{max} } = \frac{1}{\Gamma (\mu )}\frac{r}{{Uz^{1 + m} }}\mu^{\mu } \exp \left( { - \mu } \right). $$

(9)

Next, the specific form of the function for concentration with respect to a unit point source $\gamma (x,y,z)$ is derived. In Eq. 8 of KM2001, $\gamma (x,y,z)$ is given as

$$ \gamma (x,y,z) = D_{y} (x,y) \cdot c(x,z). $$

(10)

The specific forms of $c(x,z)$ is given by Eq. 5 above and $D_{y} (x,y)$ is given as

$$ D_{y} (x,y) = \frac{1}{{\sqrt {2\pi } \sigma (x)}}\exp \left( { - \frac{{y^{2} }}{{2\sigma (x)^{2} }}} \right), $$

(11)

which is the same as Eq. 9 in KM2001. Substituting Eqs. 11 and 5 into Eq. 10 yields

$$ \gamma (x,y,z) = \left( {\frac{1}{{\sqrt {2\pi } \sigma }}\exp \left[ { - \frac{{y^{2} }}{{2\sigma^{2} }}} \right]} \right)\left( {\frac{1}{\Gamma (\mu )}\frac{r}{{Uz^{1 + m} }}\frac{{\xi^{\mu } }}{{x^{\mu } }}\exp \left[ { - \frac{\xi }{x}} \right]} \right). $$

(12)

Next, we derived the maximum concentration footprint and its along-wind position, and as with Eq. 25 in KM2001, we differentiate $\gamma (x,y,z)$ with respect to $x$ at $y = 0$, leading to

$$ \frac{1}{\sigma }\frac{d\sigma }{dx} = \frac{\xi }{{x^{2} }} - \frac{\mu }{x}. $$

(13)

If crosswind diffusion is omitted, $\frac{d\sigma }{dx} = 0$, and so

$$ x_{\hbox{max} }^{(c)} = x_{\hbox{max} }^{(\gamma )} = \frac{\xi }{\mu }. $$

(14)

For $\sigma = \frac{{\sigma_{v} x}}{{\bar{u}}}$,

$$ \frac{1}{{\left( {\frac{{\sigma_{v} x}}{{\bar{u}}}} \right)}}\frac{{d\left( {\frac{{\sigma_{v} x}}{{\bar{u}}}} \right)}}{dx} = \frac{\xi }{{x^{2} }} - \frac{\mu }{x}. $$

(15)

Considering the effective plume velocity,

$$ \bar{u}(x) = \frac{\Gamma (\mu )}{\Gamma (1/r)}\left( {\frac{{r^{2} \kappa }}{U}} \right)^{{\frac{m}{r}}} Ux^{{\frac{m}{r}}} , $$

(16)

and the flux length scale,

$$ \xi (z) = \frac{{Uz^{r} }}{{r^{2} \kappa }}, $$

(17)

the left-hand side of Eq. 15 is rewritten as

$$ \begin{aligned} \frac{1}{{\left( {\frac{{\sigma_{v} x}}{{\bar{u}}}} \right)}}\frac{{d\left( {\frac{{\sigma_{v} x}}{{\bar{u}}}} \right)}}{dx} & = \frac{1}{{\left( {\frac{x}{{\bar{u}}}} \right)}}\frac{{d\left( {\frac{x}{{\bar{u}}}} \right)}}{dx} = \frac{\Gamma (\mu )}{\Gamma (1/r)}\left( {\frac{{r^{2} \kappa }}{U}} \right)^{{\frac{m}{r}}} \frac{1}{U}x^{{\frac{m}{r} - 1}} \frac{{d\left( {\frac{\Gamma (1/r)}{\Gamma (\mu )}\left( {\frac{{r^{2} \kappa }}{U}} \right)^{{ - \frac{m}{r}}} \frac{1}{U}x^{{1 - \frac{m}{r}}} } \right)}}{dx} \\ & = x^{{\frac{m}{r} - 1}} \frac{{d\left( {x^{{1 - \frac{m}{r}}} } \right)}}{dx} = x^{{\frac{m}{r} - 1}} \left( {1 - \frac{m}{r}} \right)x^{{ - \frac{m}{r}}} = \left( {1 - \frac{m}{r}} \right)\frac{1}{x} \\ \end{aligned} $$

(18)

Here, Eqs. 16 and 17 are the same as Eqs. 18 and 19 in KM2001, respectively.

Then, Eq. 15 is rewritten as

$$ \left( {1 - \frac{m}{r}} \right)\frac{1}{x} = \frac{\xi }{{x^{2} }} - \frac{\mu }{x}, $$

(19)

and solving with respect to $x$ yields

$$ \frac{1}{x}\left( {1 - \frac{m}{r} + \mu - \frac{\xi }{x}} \right) = 0. $$

(20)

Because $x$ is regarded as a finite value,

$$ \frac{\xi }{x} = 1 - \frac{m}{r} + \mu = 1 + \frac{1}{r}, $$

(21)

where $\mu = \frac{1 + m}{r}$.

Finally, we find

$$ x_{\hbox{max} }^{(\gamma )} = \frac{r\xi }{r + 1}. $$

(22)