Long-term estimation of soil heat flux by single layer soil temperature

Abstract

Soil heat flux is one of the important components of surface energy balance. In this study, long-term estimation of soil heat flux from single layer soil temperature was carried out by the traditional sinusoidal analytical method and the half-order time derivative method of Wang and Bras [Wang and Bras (1999) J Hydrol 216:214–226]. In order to understand the characteristics of soil heat flux and to examine the performances of the two methods, a field experiment was conducted at a temperate and humid grassland in Cork, Ireland. Our results show that the soil heat flux had the same magnitude as the sensible heat flux at this grassland site. It was also demonstrated that the analytical method did not predict the soil heat flux well because the sinusoidal assumption for the temporal variation in soil heat flux was invalid. In contrast, good agreement was found between the soil heat flux measurements and predictions made by the half-order time derivative method. This success suggests that this method could be used to estimate soil heat flux from long-term remotely sensed surface temperature.

Appendix: Derivation of the half-order time derivative method

The algorithm and derivation of the half-order time derivative method was first proposed by Wang and Bras (1999). Here, details of the derivation are provided.

First of all, define two new variables: $\tilde z = {z \mathord{\left/ {\vphantom {z {\sqrt {D_0 } }}} \right. \kern-\nulldelimiterspace} {\sqrt {D_0 } }}$ and $\Theta = T - T_0$, and then transfer Eqs. 7, 9, and 10 to:

$$\frac{{\partial \Theta }}{{\partial t}} = \frac{{\partial ^2 \Theta }}{{\partial \tilde z^2 }}$$

(A1)

$$\Theta = 0,\,{\text{for t = 0,}}\,\tilde z <0$$

(A2)

$$\Theta = 0,\,{\text{for t}}\,{\text{ $>$ }}\,{\text{0,}}\,\tilde z \to - \infty $$

(A3)

Now, take the Laplace transform of both sides of Eq. A1. According to the definition of Laplace transform, $L\left\{ {\frac{\partial }{{\partial t}}f\left( t \right)} \right\}\left( s \right) = sF\left( s \right) - f\left( 0 \right)$, the left hand side of A1 then becomes $s\tilde \Theta \left( {\tilde z,s} \right) - \Theta \left( {\tilde z,0} \right) = s\tilde \Theta \left( {\tilde z,s} \right)$ since A2 requires $\Theta \left( {\tilde z,0} \right) = 0$. The Laplace transform of the right hand side remains the same since the Laplace transform is based on the variable t. Hence, Eq. A1 is transformed to

$$s\tilde \Theta \left( {\tilde z,s} \right) = \frac{{\partial ^2 \tilde \Theta }}{{\partial \tilde z^2 }}$$

(A4)

where $\tilde \Theta$ is the Laplace transform of $\Theta$ and defined as:

$$\tilde \Theta \left( {\tilde z,s} \right) = \int\limits_0^\infty {\exp \left( { - st} \right)\,} \Theta \left( {\tilde z,t} \right)dt$$

(A5)

The general solution of Eq. A4 is

$$\tilde \Theta \left( {\tilde z,s} \right) = A\left( s \right)\exp \left( {\tilde z\sqrt s } \right) + B\left( s \right)\exp \left( { - \tilde z\sqrt s } \right)$$

(A6)

where A(s) and B(s) are arbitrary functions of s and can be determined by the boundary conditions. By Eq. A3, B(s) is found to be zero and Eq. A6 then becomes:

$$\tilde \Theta \left( {\tilde z,s} \right) = A\left( s \right)\exp \left( {\tilde z\sqrt s } \right)$$

(A7)

Differentiating both sides of Eq. A7 with respect to $\tilde z$ gives:

$$\frac{\partial }{{\partial \tilde z}}\tilde \Theta \left( {\tilde z,s} \right) = \sqrt s A\left( s \right)\exp \left( {\tilde z\sqrt s } \right)$$

(A8)

Now, substituting A7 into A8 yields

$$\frac{\partial }{{\partial \tilde z}}\tilde \Theta \left( {\tilde z,s} \right) = \sqrt s \tilde \Theta \left( {\tilde z,s} \right)$$

(A9)

Based on the equation derived by the fractional calculus (Miller and Ross 1993),

$$L\left\{ {D^\nu f\left( t \right)} \right\} = s^\nu F\left( s \right) - \sum\limits_{k = 0}^{m - 1} {s^{m - k - 1} D^{k - m + \nu } f\left( 0 \right)} $$

(A10)

the right hand side of Eq. A9 becomes

$$\sqrt s \tilde \Theta \left( {\tilde z,s} \right) = L\left\{ {\frac{{\partial ^{\frac{1}{2}} }}{{\partial t^{\frac{1}{2}} }}\tilde \Theta \left( {\tilde z,t} \right)} \right\}$$

(A11)

By the initial condition Eq. A2, the last term of Eq. A10, f(0), is found to be zero. Replacing the right hand side of Eq. A9 by Eq. A11, we have

$$\frac{\partial }{{\partial \tilde z}}\tilde \Theta \left( {\tilde z,s} \right) = L\left\{ {\frac{{\partial ^{\frac{1}{2}} }}{{\partial t^{\frac{1}{2}} }}\tilde \Theta \left( {\tilde z,t} \right)} \right\}$$

(A12)

Inverting the Laplace transform of Eq. A12 leads to

$$\frac{\partial }{{\partial \tilde z}}\Theta \left( {\tilde z,t} \right) = \frac{{\partial ^{\frac{1}{2}} }}{{\partial t^{\frac{1}{2}} }}\Theta \left( {\tilde z,t} \right)$$

(A13)

With the definition of Θ, Eq. A13 becomes

$$\frac{\partial }{{\partial z}}T\left( {z,t} \right) = \frac{1}{{\sqrt {D_o } }}\frac{{\partial ^{\frac{1}{2}} }}{{\partial t^{\frac{1}{2}} }}\left[ {T\left( {z,t} \right) - T\left( 0 \right)} \right].$$

(A14)

Now, applying the fractional calculus (Miller and Ross 1993)

$$\frac{{d^\alpha f\left( t \right)}}{{dt^\alpha }} = \frac{1}{{\Gamma \left( {1 - \alpha } \right)}}\frac{d}{{dt}}\int\limits_0^t {\frac{{f\left( s \right)}}{{\left( {t - s} \right)^\alpha }}ds} $$

(A15)

to the right hand side of Eq. A14, we have

$$ \begin{array}{*{20}c} {\frac{\partial } {{\partial z}}T{\left( {z,t} \right)} = \frac{1} {{{\sqrt {\pi D_{0} } }}}{\int\limits_0^t {\frac{{\partial T{\left( {z,s} \right)}}} {{\partial s}}\frac{{ds}} {{{\sqrt {t - s} }}}} }} \\ { = \frac{1} {{{\sqrt {\pi D_{0} } }}}{\int\limits_0^t {\frac{{dT{\left( {z,s} \right)}}} {{{\sqrt {t - s} }}}} }} \\ \end{array} $$

(A16)

Finally, the prognostic result can be evaluated by applying Eq. A16 to Fourier’s law, and then we have

$$G\left( {z,t} \right) = \sqrt {\frac{{k\rho _s c_s }}{\pi }} \int\limits_0^t {\frac{{dT\left( {z,s} \right)}}{{\sqrt {t - s} }}} $$

(A17)

Equation A17 can also be written as:

$$G\left( {z,t} \right) = \sqrt {\frac{{k\rho _s c_s }}{\pi }} \int\limits_0^t {\frac{{\partial T\left( {z,s} \right)}}{{\partial s}}\frac{{ds}}{{\sqrt {t - s} }}} = 2\sqrt {\frac{{k\rho _s c_s }}{\pi }} \int\limits_0^t {\frac{{\partial T\left( {z,s} \right)}}{{\partial s}}} d\sqrt {t - s} $$

(A18)

And its discrete form is

$$G = 2\sqrt {\frac{{k\rho _s c_s }}{\pi }} \sum\limits_{i = 0}^N {\frac{{T_{i + 1} - T_i }}{{t_{i + 1} - t_i }}\left[ {\sqrt {t_N - t_{i + 1} } - \sqrt {t_N - t_i } } \right]} $$

(A19)

where N is the number of intervals.