Can the coefficient of inertial resistivity ($\beta$-factor) be a 2nd rank tensor?

Question

Question: Can the Forchheimer $\beta$-factor be a 2nd rank tensor (like permeability, $k$)?

In most cases in literature, the quadratic Forchheimer equation is applied only to one-dimensional flow. However I have seen in some instances claims that the Forchheimer eqn is applicable in multiple dimensions as a vector equation. The proposed representation of the vector Forchheimer equation in Cartesian coordinates is (using Einstein's summation convention (or double index convention)): $$\tag{1} -\frac{\partial p}{\partial x_j} = \frac{\mu}{k_{ij}}q_i+\beta_{ij} \rho q_i|q_i|$$

In this case, the quadratic term is simply the magnitude of the specific discharge vector $|q_i|$ (the 'superficial velocity') times the directional specific discharge vector $q_i$.

If I underscore the variables in the equation with the letters 's', 'v', and 't' to represent the type of physical quantity they are, that is, scalar, vector, and 2nd rank tensor, respectively, I show:

$$\tag{2} \underbrace{-\frac{\partial p}{\partial x_j}}_{\text{v}} = \underbrace{\frac{\mu}{k_{ij}}}_{\text{t}}\underbrace{q_i}_{\text{v}}+\underbrace{\beta_{ij}}_{\text{t}} \underbrace{\rho}_{\text{s}} \underbrace{q_i}_{\text{v}}\underbrace{|q_i|}_{\text{s}}$$

I see that all terms on the righ-hand side of (2), when multiplied together, would result in a vector (consistent with the left-hand side).

But what it I wanted to rearrange (1) to solve for $q_i$?

One way I see how this may be down is as so:

Multiply through by the permeability tensor, $k_{ij}$ $$\tag{3} -k_{ij}\frac{\partial p}{\partial x_j} = \mu q_i+\beta_{ij}k_{ij} \rho q_i|q_i|$$

Extract the product $\mu q_i$ from the right-hand side of (3) $$\tag{4} -k_{ij}\frac{\partial p}{\partial x_j} = \mu q_i \left(1+\frac{\beta_{ij}k_{ij} \rho |q_i|}{\mu}\right)$$

Divide through by $\mu \left(1+\frac{\beta_{ij}k_{ij} \rho |q_i|}{\mu}\right)$ $$\tag{5} -\frac{k_{ij}}{\mu \left(1+\frac{\beta_{ij}k_{ij} \rho |q_i|}{\mu}\right)}\frac{\partial p}{\partial x_j} = q_i $$

And here is where I am unsure. In the first term on the LHS, I have one tensor in the numerator and two tensors in the denominator along with scalar quantities.

Questions: For those who are familiar with the theory of tensor analysis, is what I have done correct? If not, what have I done wrong? If I have done nothing wrong, does this mean that the Forchheimer $\beta$-factor cannot be a 2nd rank tensor?

Answer 1

Ultimately, the answer to the original question as posted is yes, the $\beta$-factor is a second-rank tensor. Per discussion in the comments, a new question had developed, which is: how would one write the x-components of the vector Forchheimer equation assuming a $x, y, z$ Cartesian coordinate system?

The solution to this question was given as $$k_{ij}J_j=\mu q_i+\rho k_{il}\beta_{lm}q_m\sqrt{q_nq_n}$$

which is the vector Forchheimer equation using Einstein's summation convention. In an effort to see the x-components as a result of the products of the terms in the equation above, summed over its range of values $(x,y,z)$, I found the following:

Writing the equation above as a single matrix equation, where the vectors and tensors are written in their compact matrix forms:

$$\begin{bmatrix}k_{xx}&k_{xy}&k_{xz}\\k_{yx}&k_{yy}&k_{yz}\\k_{zx}&k_{zy}&k_{zz} \end{bmatrix}\cdot\begin{bmatrix}J_x\\J_y\\J_z \end{bmatrix}=\mu \cdot \begin{bmatrix}q_x \\q_y\\q_z \end{bmatrix}+\rho \cdot \begin{bmatrix}k_{xx}&k_{xy}&k_{xz}\\k_{yx}&k_{yy}&k_{yz}\\k_{zx}&k_{zy}&k_{zz} \end{bmatrix}\cdot \begin{bmatrix}\beta_{xx}&\beta_{xy}&\beta_{xz}\\\beta_{yx}&\beta_{yy}&\beta_{yz}\\\beta_{zx}&\beta_{zy}&\beta_{zz} \end{bmatrix}\cdot \begin{bmatrix}q_x \\q_y\\q_z \end{bmatrix} \sqrt{\begin{bmatrix}q_x&q_y&q_z\end{bmatrix}\cdot \begin{bmatrix}q_x\\q_y\\q_z\end{bmatrix}}$$

The dot product of the tensor and vector on the LHS was found to be: $$\begin{bmatrix}k_{xx}J_x+k_{xy}J_y+k_{xz}J_z\\k_{yx}J_x+k_{yy}J_y+k_{yz}J_z\\k_{zx}J_x+k_{zy}J_y+k_{zz}J_z \end{bmatrix}$$

The dot product of the two tensors on the RHS was found to be: $$\begin{bmatrix}(\beta_{xx}k_{xx}+\beta_{yx}k_{xy}+\beta_{zx}k_{xz})&(\beta_{xy}k_{xx}+\beta_{yy}k_{xy}+\beta_{zy}k_{xz})&(\beta_{xz}k_{xx}+\beta_{yz}k_{xy}+\beta_{zz}k_{xz})\\(\beta_{xx}k_{yx}+\beta_{yx}k_{yy}+\beta_{zx}k_{yz})&(\beta_{xy}k_{yx}+\beta_{yy}k_{yy}+\beta_{zy}k_{yz})&(\beta_{xz}k_{yx}+\beta_{yz}k_{yy}+\beta_{zz}k_{yz})\\(\beta_{xx}k_{zx}+\beta_{yx}k_{zy}+\beta_{zx}k_{zz})&(\beta_{xy}k_{zx}+\beta_{yy}k_{zy}+\beta_{zy}k_{zz})&(\beta_{xz}k_{zx}+\beta_{yz}k_{zy}+\beta_{zz}k_{zz})\end{bmatrix}$$

The square root term on the RHS was found to be: $$(q_x+q_y+q_z)$$

If the coordinate system is chosen such that the $x,y,z$ coordinates coincide with the principal axes of the permeability anisotropy, the matrix equation becomes:

$$\begin{bmatrix}k_{xx}J_x\\k_{yy}J_y\\k_{zz}J_z \end{bmatrix}=\mu \cdot \begin{bmatrix}q_x \\q_y\\q_z \end{bmatrix}+\rho \cdot \begin{bmatrix}(\beta_{xx}k_{xx})&(\beta_{xy}k_{xx})&(\beta_{xz}k_{xx})\\(\beta_{yx}k_{yy})&(\beta_{yy}k_{yy})&(\beta_{yz}k_{yy})\\(\beta_{zx}k_{zz})&(\beta_{zy}k_{zz})&(\beta_{zz}k_{zz})\end{bmatrix}\cdot \begin{bmatrix}q_x \\q_y\\q_z \end{bmatrix}\cdot (q_x+q_y+q_z)$$

further multiplication on the RHS, we have: $$\begin{bmatrix}k_{xx}J_x\\k_{yy}J_y\\k_{zz}J_z \end{bmatrix}=\mu \cdot \begin{bmatrix}q_x \\q_y\\q_z \end{bmatrix}+\rho \cdot \begin{bmatrix}(q_x^2+q_xq_y+q_xq_z)(\beta_{xx}k_{xx})+(q_yq_x+q_y^2+q_yq_z)(\beta_{xy}k_{xx})+(q_zq_x+q_zq_y+q_z^2)(\beta_{xz}k_{xx})\\(q_x^2+q_xq_y+q_xq_z)(\beta_{yx}k_{yy})+(q_yq_x+q_y^2+q_yq_z)(\beta_{yy}k_{yy})+(q_zq_x+q_zq_y+q_z^2)(\beta_{yz}k_{yy})\\(q_x^2+q_xq_y+q_xq_z)(\beta_{zx}k_{zz})+(q_yq_x+q_y^2+q_yq_z)(\beta_{zy}k_{zz})+(q_zq_x+q_zq_y+q_z^2)(\beta_{zz}k_{zz}) \end{bmatrix}$$

If we assume the pressure gradient $J$ is collinear with the $x$-axis, then the flow $q$ only occurs along the $x$-axis. Thus,

$$k_{xx}J_x=\mu q_x+\rho q_x^2 \beta_{xx}k_{xx}$$

which, when we divide through by $k_{xx}$, we obtain the commonly cited Forchheimer equation (one-dimensional):

$$J_x=\frac{\mu}{k_{xx}}q_x+\beta_{xx}\rho q_x^2$$