Localized Point Mixing Rate Potential in Heterogeneous Velocity Fields

Abstract

Mixing is driven by an interplay between diffusion and flow-induced concentration gradients. Accurately describing the effect of flow heterogeneity on the mixing state of a plume is a challenging problem that has important repercussions on the modeling of plume dilution and chemical reactions. In this technical note, we propose a simple, semi-analytic measure to quantify the local mixing potential at a point based on the local properties of the flow-induced strain. Specifically, it is the trace of the local strain matrix squared, ${\text {Tr}}(\varvec{s}^2)$, which we demonstrate controls mixing from a point source over small times. Due to its mathematically similar influence to a shear flow on mixing, we propose an ansatz to attempt to use this metric as a predictor for mixing. We test its performance via random walk particle tracking simulations in heterogeneous Darcy flow through lognormal permeability fields. The ansatz appears to work better and better the more heterogeneous the flow, unlike more traditional approaches that rely on weak heterogeneity assumptions. While we cannot rigorously demonstrate why this is, we find it sufficiently promising that it may guide future model development.

Appendix: Full Solution for Dilution Index with Arbitrary Constant Deformation

For an arbitrary deformation tensor for which strain and rotation do not necessarily commute, we will calculate the exponentials of $\varvec{\varepsilon }$ and $\varvec{\varepsilon }^T$ via a result that follows from the Cayley–Hamilton theorem. Because the exponential function is analytic in the whole complex plane, we can write:

$$\begin{aligned} \begin{aligned} e^{\varvec{\varepsilon }u}&= a_\mu (u) \varvec{\varepsilon }^\mu ,\\ e^{\lambda _i u}&= a_\mu (u) \lambda ^\mu _i , \end{aligned} \end{aligned}$$

(18)

where $\mu$ runs over 0 to $d-1$ (and summation over repeated indices is implied), and the $\lambda _i$, $i=1,\ldots ,d$, are the eigenvalues of $\varvec{\varepsilon }$. Since $\varvec{\varepsilon }$ and $\varvec{\varepsilon }^T$ have the same eigenvalues, the same expressions hold if $\varvec{\varepsilon }$ is replaced by $\varvec{\varepsilon }^T$. The time-dependent $a_\mu$ can be determined by solving the linear system encoded in the second line, and they are found to be:

$$\begin{aligned} a_\mu (u) = b_{\mu j} \, e^{\lambda _j u} , \end{aligned}$$

(19)

where:

$$\begin{aligned} b_{\mu j} = \frac{1}{\prod _{i \ne j} (\lambda _j - \lambda _i)} \times {\left\{ \begin{array}{ll} 1 , &{} \quad \mu = d - 1 \\ -\sum _{i \ne j} \lambda _i , &{} \quad \mu = d - 2 \\ \prod _{i \ne j} \lambda _i , &{} \quad \mu = d - 3 \end{array}\right. }. \end{aligned}$$

(20)

These formulas are valid in 2- and 3-d ($a_{-1}$ in 2d should be ignored). Then:

$$\begin{aligned} \begin{aligned} \varvec{\sigma }( t )&= c_{ij}(t) b_{\mu i} b_{\nu j} \varvec{\varepsilon }^\mu \varvec{\varepsilon }^{T\nu }, \\ c_{ij}(t)&= \frac{e^{(\lambda _i + \lambda _j)t} - 1}{\lambda _i + \lambda _j}. \end{aligned} \end{aligned}$$

(21)

Note there is no sum implied in the definition of the $c_{ij}$. Also, these formulas are still valid if a pair of eigenvalues sums to zero, in which case they are to be understood in the limit. For such a pair, $c_{ij} = t$.

We will now focus on the two-dimensional case. The three-dimensional case can be tackled using similar techniques, but the algebra is substantially more complicated, and fully characterizing the system requires knowledge of the eigenvalues of strain, the absolute value of the vorticity and also the orientation of the vorticity vector, for example with respect to the principal axes of strain. The final result, while it can be explicitly written, yields little physical insight and/or utility. However, for two dimensions we have for the eigenvalues of strain, rotation and deformation $\lambda _{\varvec{s},i} = \pm \lambda _{\varvec{s}}$, $\lambda _{\varvec{s}} \in \mathbb {R}$, $\lambda _{\varvec{\xi }, i} = \pm \lambda _{\varvec{\xi }}$, $\lambda _{\varvec{\xi }} \in \imath \mathbb {R}$, and $\lambda _i = \pm \lambda$, $\lambda = \sqrt{|\lambda _{\varvec{s}}|^2 - |\lambda _{\varvec{\xi }}|^2} \in \mathbb {R} \cup \imath \mathbb {R}$, which are all easily seen to be true for 2-d incompressible flow ($\imath$ is the imaginary unit). We can now use the results above to find:

$$\begin{aligned} \varvec{\sigma }(t) = \frac{2D}{4 \lambda } \left[ [\sin \,h(2\lambda t) + 2\lambda t] \mathbbm {1} + [ \cos \,h(2\lambda t) - 1] \frac{\varvec{\varepsilon }+ \varvec{\varepsilon }^T}{\lambda } + [\sin \,h(2\lambda t) - 2\lambda t] \frac{\varvec{\varepsilon }\varvec{\varepsilon }^T}{\lambda ^2} \right] .\!\!\nonumber \\ \end{aligned}$$

(22)

Using $\varvec{\varepsilon }= \varvec{s} + \varvec{\xi }$ we find:

$$\begin{aligned} \begin{aligned}{} \varvec{\sigma }(t)&= \frac{2D}{4 \lambda } \bigg [ [\sin \,h(2\lambda t) + 2\lambda t] \mathbbm {1} + [ \cos \,h(2\lambda t) - 1] \frac{2\varvec{s}}{\lambda } \\&\quad + [\sin \,h(2\lambda t) - 2\lambda t] \frac{\varvec{s}^2 - \varvec{\xi }^2 + [\varvec{\xi },\varvec{s}]}{\lambda ^2} \bigg ]. \end{aligned} \end{aligned}$$

(23)

Now from the results for commuting strain and rotation we expect that if $[\varvec{\xi },\varvec{s}] = 0$, rotation should play no part, so the $\varvec{\xi }^2$ term is troublesome. However, the application of the Cayley–Hamilton theorem above means that there is a relation between $\varvec{\varepsilon }$ and $\varvec{\varepsilon }^T$ and their corresponding powers. Specifically, using Eq. (18) for $\varvec{\varepsilon }$ and $\varvec{\varepsilon }^T$ and looking at the coefficients of $t^2$ we find the relation:

$$\begin{aligned} \varvec{\xi }^2 = \lambda ^2 {\mathbbm {1}} - \varvec{s}^2. \end{aligned}$$

(24)

Using this relation, we can rewrite $\varvec{\sigma }$ as:

$$\begin{aligned} \varvec{\sigma }(t) = \frac{2D}{\lambda } \left\{ \lambda t \mathbbm {1} + \frac{ \cos \,h(2\lambda t) - 1 }{2\lambda } s +\frac{\sin \,h(2\lambda t) - 2\lambda t }{4\lambda ^2}(2\varvec{s}^2 + [\varvec{\xi },\varvec{s}]) \right\} . \end{aligned}$$

(25)

In this form, we can clearly identify the effect on spreading of pure diffusion, and the coupling between diffusion and strain, and diffusion, strain and rotation.

Let us now consider the (direct-oriented) basis where $\varvec{s}$ diagonalizes. This basis always exists, because $\varvec{s}$ is symmetric. Note that if $\varvec{s} = 0$ it diagonalizes in any basis and our expressions will still hold. We will use hatted indices (e.g., $\hat{i}$) to denote components in this basis. The matrices s and $s^2$ diagonalize trivially in this basis, but one can also show without too much effort that:

$$\begin{aligned}{}[\varvec{\xi }, \varvec{s}]_{\hat{i} \hat{j}} = -\lambda _{\varvec{s}} Re (\lambda _{\varvec{\xi }}) \begin{pmatrix} 0 &{}\quad 1 \\ 1 &{}\quad 0 \end{pmatrix}, \end{aligned}$$

(26)

where Re denotes the real part. Incidentally, this shows that in 2-d we have $[\varvec{\xi }, \varvec{s}] = 0$ iff there is no rotation or no strain. We thus find:

$$\begin{aligned} \begin{aligned}{} \sigma _{\hat{i} \hat{j}}(t)&= \frac{2D}{\lambda } \\&\quad \times \left( \begin{array}{l@{\quad }l} \lambda t + \frac{ \cos \,h(2\lambda t) - 1 }{2\lambda } \lambda _{\varvec{s}} + \frac{\sin \,h(2\lambda t) - 2\lambda t }{4\lambda ^2} 2\lambda _{\varvec{s}}^2 &{} - \frac{\sin \,h(2\lambda t) - 2\lambda t }{4\lambda ^2} 2 \lambda _{\varvec{s}} Re({\lambda _{\varvec{\xi }}}) \\ \quad - \frac{\sin \,h(2\lambda t) - 2\lambda t }{4\lambda ^2} 2 \lambda _{\varvec{s}} Re (\lambda _{\varvec{\xi }}) &{} \lambda t - \frac{ \cos \,h(2\lambda t) - 1 }{2\lambda } \lambda _{\varvec{s}} + \frac{\sin \,h(2\lambda t) - 2\lambda t }{4\lambda ^2} 2\lambda _{\varvec{s}}^2 \end{array}\right) . \end{aligned}\nonumber \\ \end{aligned}$$

(27)

We can now find the determinant, which is frame independent as long as direct orientation is kept, directly using this basis. Some manipulation gives:

$$\begin{aligned} \begin{aligned} \det \varvec{\sigma }(t) = 4 D^2 \left[ t^2 + \lambda _{\varvec{s}}^2 \frac{ \sin \,h^2(\lambda t) - \lambda ^2 t^2 }{\lambda ^4} \right] , \end{aligned} \end{aligned}$$

(28)

and thus the dilution index is:

$$\begin{aligned} \begin{aligned} E(t)&= 4 \pi e D \sqrt{ t^2 + \lambda _{\varvec{s}}^2 \frac{ \sin \,h^2(\lambda t) - \lambda ^2 t^2 }{\lambda ^4} }. \end{aligned} \end{aligned}$$

(29)

This expression recovers the solutions for pure rotation and pure strain found under the assumption of commuting strain and rotation. Note also that $\sin \,h^2(x) - x^2$ is positive for all $x \in {\mathbb {R}} \cup \imath {\mathbb {R}}$, which implies that strain always has an enhancing effect on mixing.

The case $\lambda = 0$, or equivalently $|\lambda _{\varvec{s}}| = |\lambda _{\varvec{\xi }}|$, represents shear flow and must be understood as the limit of the formulas above. It leads to:

$$\begin{aligned} \varvec{\sigma }(t) = 2Dt \left\{ \mathbbm {1} + t \varvec{s} + \frac{t^2}{3}(2\varvec{s}^2 + [\varvec{\xi },\varvec{s}]) \right\} , \end{aligned}$$

(30)

$$\begin{aligned} \sigma _{\hat{i} \hat{j}}(t) = 2Dt \begin{pmatrix} 1 + \lambda _{\varvec{s}} t + \frac{2}{3} \lambda _{\varvec{s}}^2 t^2 &{}\quad - \frac{2}{3} \lambda _{\varvec{s}} Re ({\lambda _{\varvec{\xi }}}) t^2 \\ - \frac{2}{3} \lambda _{\varvec{s}} Re({\lambda _{\varvec{\xi }}}) t^2 &{}\quad 1 - \lambda _{\varvec{s}} t + \frac{2}{3}\lambda _{\varvec{s}}^2 t^2 \end{pmatrix}, \end{aligned}$$

(31)

$$\begin{aligned} E(t) = 4 \pi e D t \sqrt{ 1 + \frac{1}{3} \lambda _{\varvec{s}}^2 t^2 }. \end{aligned}$$

(32)