A Note on Upscaling Retardation Factor in Hierarchical Porous Media with Multimodal Reactive Mineral Facies

Abstract

We present a model for upscaling the time-dependent effective retardation factor, ${\widetilde{R}}(t)$, in hierarchical porous media with multimodal reactive mineral facies. The model extends the approach by Deng et al. (Chemosphere 91(3):248–257, 2013) in which they expanded a Lagrangian-based stochastic theory presented by Rajaram (Adv Water Resour 20(4):217–230, 1997) in order to describe the scaling effect of $\widetilde{R}(t)$. They used a first-order linear approximation in deriving their model to make the derivation tractable. Importantly, the linear approximation is known to be valid only to variances of 0.2. In this note, we show that the model can be derived with a higher-order approximation, which allows for representing variances from 0.2 to 1.0. We present the derivation and use the resulting model to recalculate $\widetilde{R}(t)$ for the scenario examined by Deng et al. (2013).

Appendix: Derivation of Cross-Covariance of \(v_1\) and \(R\)

The derivation of $C_{v_1 R} (\xi )$ by Deng et al. (2013) is presented here for ease of reference. The $C_{v_1 R} (\xi )$ is found by taking the Fourier transform of the Eq. (18). Here we consider a unimodal porous media with the spectral density function $S_{ff} (k)$ as in Eq. (19). Note that the same integration method is used three times for the three exponential terms in Eq. (20). The $C_{v_1 R} (\xi )$ is found by:

(23)

We set $I$ equal to the integration part in (23). Therefore,

(24)

One can use a spherical coordinate system and define the following:

$$\begin{aligned} k_1= & {} k \cos \,\beta \end{aligned}$$

(25)

$$\begin{aligned} k_1/k= & {} \cos \,\beta \cos \,\chi +\sin \,\theta \sin \,\chi \cos \,\alpha \end{aligned}$$

(26)

$$\begin{aligned} \hbox {d}k_1 \hbox {d}k_2 \hbox {d}k_3= & {} k^{2}\sin \,\theta \,\hbox {d}k\,\hbox {d}\alpha \,\hbox {d}\theta \end{aligned}$$

(27)

where $\chi$ is the angle between the separation vector $\xi$ and the direction of mean flow $k_1$, and $\theta$ is the angle between $\xi$ and $\kappa$. The $\chi$ and $\xi$ are coordinates of the covariance function. The $k,\theta$, and $\alpha$ are spherical coordinates in wave number space. Substituting (25), (26), and (27) into (24) gives:

$$\begin{aligned} I= & {} \frac{\sigma _f^2 }{\lambda \pi ^{2}}\int \limits _{k=0}^\infty \int \limits _{\alpha =0}^{2\pi } \int \limits _{\theta =0}^\pi \left\{ \left[ 1-\cos ^{2}\theta \cos ^{2}\chi -\sin ^{2}\theta \sin ^{2}\chi \cos ^{2}\theta \right. \right. \nonumber \\&\left. \left. \quad -\,2\cos \theta \cos \chi \sin \theta \sin \chi \cos \alpha \right] \frac{k^{2}\lambda ^{4}}{\left( 1+k^{2}\lambda ^{2}\right) ^{2}}\hbox {e}^{ik\xi \cos \theta }\sin \theta \,\hbox {d}k\,\hbox {d}\theta \,\hbox {d}\alpha \right\} \end{aligned}$$

(28)

One can let $\cos \theta =y$ and use the relationship of $\hbox {e}^{ik\xi y}=\cos k\xi y+i \sin k\xi y$ to change the (28) to the following expression:

$$\begin{aligned} I= & {} \frac{\sigma _f^2 }{\lambda \pi ^{2}}\int \limits _{k=0}^\infty {\int \limits _{\alpha =0}^{2\pi } } \nonumber \\&\quad \times \,\int \limits _{y=-1}^1 \left\{ \left[ \left( 1\!-\!y^{2}\cos ^{2}\chi \right) \!-\!\left( 1\!-\!y^{2}\right) \sin ^{2}\chi \sin ^{2}\alpha \right] \frac{k^{2}\lambda ^{4}}{\left( 1\!+\!k^{2}\lambda ^{2}\right) ^{2}}\cos k\xi y\right\} \hbox {d}k\,\hbox {d}y\,\hbox {d}\alpha \quad \quad \nonumber \\ \end{aligned}$$

(29)

Deng et al. (2013) integrated (29) by presenting the following integrals:

$$\begin{aligned}&\displaystyle I_a = \int \limits _0^\infty {\frac{k^{2}\lambda ^{4}}{\left( 1+k^{2}\lambda ^{2}\right) ^{2}}\cos k\xi y}\hbox {d}k=\frac{\pi }{4}\lambda \left( 1-\frac{\left| {\xi y} \right| }{\lambda }\right) \hbox {e}^{-\frac{\left| {\xi y} \right| }{\lambda }} \end{aligned}$$

(30)

$$\begin{aligned}&\displaystyle I_b =\int \limits _0^{2\pi } {(1-y^{2})} \sin ^{2}\chi \cos ^{2}\alpha \hbox {d}\alpha =\pi \left( 1-y^{2}\right) \sin ^{2}\chi \end{aligned}$$

(31)

$$\begin{aligned}&\displaystyle I_c =\int \limits _0^{2\pi } {\left( 1-y^{2}\cos ^{2}\chi \right) }\hbox {d}\alpha =2\pi \left( 1-y^{2}\cos ^{2}\chi \right) = \pi \left( 2-2y^{2}\cos ^{2}\chi \right) \end{aligned}$$

(32)

Substituting (30), (31), and (32) into (29) gives:

$$\begin{aligned} I= \frac{\sigma _f^2 }{4}\int \limits _{y=-1}^1 \left\{ \left[ \left( 2-2y^{2}\cos ^{2}\chi \right) -\left( 1-y^{2}\right) \sin ^{2}\chi \right] \left( 1-\frac{\left| {\xi y} \right| }{\lambda }\right) \hbox {e}^{-\frac{\left| {\xi y} \right| }{\lambda }}\right\} \hbox {d}y \end{aligned}$$

(33)

Then, one can expand (33) as:

$$\begin{aligned} I= & {} \frac{\sigma _f^2 }{2}\left\{ \int \limits _0^1 {\left( 1+\cos ^{2}\chi \right) } \hbox {e}^{-\frac{\xi y}{\lambda }}\hbox {d}y+\int \limits _0^1 {\left( 1-3\cos ^{2}\chi \right) } y^{2}\hbox {e}^{-\frac{\xi y}{\lambda }}\hbox {d}y\right. \nonumber \\&\left. \quad -\,\int \limits _0^1 {\frac{\xi }{\lambda }\left( 1+\cos ^{2}\chi \right) } y \hbox {e}^{-\frac{\xi y}{\lambda }}\hbox {d}y-\int \limits _0^1 {\frac{\xi }{\lambda }\left( 1-3\cos ^{2}\chi \right) } y^{3}\hbox {e}^{-\frac{\xi y}{\lambda }}\hbox {d}y\right\} \nonumber \\= & {} \frac{\sigma _f^2 }{2}\left( I_1 +I_2 -I_3 -I_4\right) \end{aligned}$$

(34)

Next, one can find $I_1,I_2,I_3$, and $I_4$ as (Deng et al. 2013):

$$\begin{aligned} I_1= & {} \int \limits _0^1 {\left( 1+\cos ^{2}\chi \right) } \hbox {e}^{-\frac{\xi y}{\lambda }}\hbox {d}y=-\frac{\lambda }{\xi }\left( 1+\cos ^{2}\chi \right) \left( \hbox {e}^{-\frac{\xi }{\lambda }}-1\right) \nonumber \\ \end{aligned}$$

(35)

$$\begin{aligned} I_2= & {} \int \limits _0^1 {\left( 1-3\cos ^{2}\chi \right) } y^{2}\hbox {e}^{-\frac{\xi y}{\lambda }}\hbox {d}y\nonumber \\= & {} \left( 3\cos ^{2}\chi -1\right) \left[ \left( \frac{\lambda }{\xi }\right) \hbox {e}^{-\frac{\xi }{\lambda }}+2\left( \frac{\lambda }{\xi }\right) ^{2}\hbox {e}^{-\frac{\xi }{\lambda }}+2\left( \frac{\lambda }{\xi }\right) ^{3}\left( \hbox {e}^{-\frac{\xi }{\lambda }}-1\right) \right] \end{aligned}$$

(36)

$$\begin{aligned} I_3= & {} \int \limits _0^1 {\frac{\xi }{\lambda }\left( 1+\cos ^{2}\chi \right) } y \hbox {e}^{-\frac{\xi y}{\lambda }}\hbox {d}y\nonumber \\= & {} -\left( 1+\cos ^{2}\chi \right) \left[ \hbox {e}^{-\frac{\xi }{\lambda }}+\left( \frac{\lambda }{\xi }\right) \left( \hbox {e}^{-\frac{\xi }{\lambda }}-1\right) \right] \end{aligned}$$

(37)

$$\begin{aligned} I_4= & {} \int \limits _0^1 {\frac{\xi }{\lambda }\left( 1-3\cos ^{2}\chi \right) } y^{3} \hbox {e}^{-\frac{\xi y}{\lambda }}\hbox {d}y\nonumber \\= & {} \left( 3\cos ^{2}\chi -1\right) \left[ \hbox {e}^{-\frac{\xi }{\lambda }}+3\left( \frac{\lambda }{\xi }\right) \hbox {e}^{-\frac{\xi }{\lambda }}+6\left( \frac{\lambda }{\xi }\right) ^{2}\hbox {e}^{-\frac{\xi }{\lambda }}+6\left( \frac{\lambda }{\xi }\right) ^{3}\left( \hbox {e}^{-\frac{\xi }{\lambda }}-1\right) \right] \qquad \end{aligned}$$

(38)

Substituting (35), (36), (37), and (38) into (34), which in turn goes into (23), finally gives:

$$\begin{aligned} C_{v_1 R} (\xi )= & {} \frac{\rho _\mathrm{b}}{n^{2}}K^\mathrm{G}K_\mathrm{d}^\mathrm{G}\textit{Ja} \frac{\sinh (\sigma _w)}{\sigma _w}\frac{\sigma _f^2 }{2}\left\{ \left( 1+\cos ^{2}\chi \right) \hbox {e}^{-\frac{\xi }{\lambda }}+\left( 1-3\cos ^{2}\chi \right) \right. \nonumber \\&\left. \times \left[ \hbox {e}^{-\frac{\xi }{\lambda }}+2\left( \frac{\lambda }{\xi }\right) \hbox {e}^{-\frac{\xi }{\lambda }}+4\left( \frac{\lambda }{\xi }\right) ^{2}\hbox {e}^{-\frac{\xi }{\lambda }}+4\left( \frac{\lambda }{\xi }\right) ^{3}\left( \hbox {e}^{-\frac{\xi }{\lambda }}-1\right) \right] \right\} \end{aligned}$$

(39)

when the separation vector is parallel to the flow direction, $\chi =0$ and $\cos ^{2}\chi =1$, then:

$$\begin{aligned} C_{v_1 R} (\xi )= & {} \frac{\rho _\mathrm{b}}{n^{2}}K^\mathrm{G}K_\mathrm{d}^\mathrm{G}\textit{Ja} \frac{\sinh (\sigma _w)}{\sigma _w}\sigma _f^2 \nonumber \\&\times \left[ 4\left( \frac{\lambda }{\xi }\right) ^{3}\left( 1-\hbox {e}^{-\frac{\xi }{\lambda }}\right) -4\left( \frac{\lambda }{\xi }\right) ^{2}\hbox {e}^{-\frac{\xi }{\lambda }}-2\left( \frac{\lambda }{\xi }\right) \hbox {e}^{-\frac{\xi }{\lambda }}\right] \quad \quad \end{aligned}$$

(40)