An improved approximation for hydraulic conductivity for pipes of triangular cross-section by asymptotic means

Abstract

In this paper, we explore single-phase flow in pores with triangular cross-sections at the pore-scale level. We use analytic and asymptotic methods to calculate the hydraulic conductivity in triangular pores, a typical geometry used in network models of porous media flow. We present an analytical formula for hydraulic conductivity based on Poiseuille flow that can be used in network models contrasting the typical geometric approach leading to many different estimations of the hydraulic conductivity. We consider perturbations to an equilateral triangle by changing the length of one of the triangle sides. We look at both small and large triangles in order to capture triangles that are near and far from equilateral. In each case, the calculations are compared with numerical solutions and the corresponding network approximations. We show that the analytical solution reduces to a quantitatively justifiable formula and agrees well with the numerical solutions in both the near and far from equilateral cases.

Appendix A: Separation of variables in triangular coordinates

Consider the eigenvalue problem

$$\begin{aligned} \nabla ^2 \phi = -\lambda ^2 \phi \quad \text {with} \; \phi =0 \; \text {on} \; W=0, \end{aligned}$$

(60)

where W is the equilateral triangle domain given as in Fig. 18.

Fig. 18

Equilateral triangle domain

The problem with separation of variables in this domain is that the boundaries of the triangle are not constant in x or y. Transforming to a triangular coordinate system is useful as each point in space is measured relative to the bisectors of the angles through the midpoint of the opposite sides, with the origin being the centre of the inscribed circle of the triangle, with radius $r_{\text {inc}} = {\sqrt{3}}/{6}$.

Fig. 19

Triangular coordinate system

Considering Fig. 19, each point P in (x, y, z) is projected onto (u, v, w) in this new coordinate system, giving the following relations:

$$\begin{aligned} u&= r_{\text {inc}} - y, \end{aligned}$$

(61a)

$$\begin{aligned} v&= \frac{\sqrt{3}}{2}x +\frac{1}{2}(y-r_{\text {inc}}), \end{aligned}$$

(61b)

$$\begin{aligned} w&= -\frac{\sqrt{3}}{2}x +\frac{1}{2}(y-r_{\text {inc}}) \end{aligned}$$

(61c)

with positive orientation directed towards the boundary. Adding these three equations it can be seen that there is a linear dependence in the variables,

$$\begin{aligned} u+v+w=0, \end{aligned}$$

(62)

which is consistent with ${\mathbb {R}}^2$ being uniquely spanned by two vectors. Now, in this new coordinate system, the boundaries of the triangle are now $u=r_{\text {inc}}$, $v=r_{\text {inc}}$ and $w=r_{\text {inc}}$, which are constant values and so allow us to proceed with separation of variables.

Rearranging the equations in (61) we see that

$$\begin{aligned} x&= \frac{v-w}{\sqrt{3}}, \end{aligned}$$

(63a)

$$\begin{aligned} y&= r_{\text {inc}}-u \end{aligned}$$

(63b)

and so defining $\xi = u$ and $\eta =v-w$ provides a good orthogonal system. In this new coordinate system, the Laplace operator can be written as

$$\begin{aligned} \nabla ^2 = \frac{\partial ^{2}}{\partial x^{2}} + \frac{\partial ^{2}}{\partial y^{2}} = \frac{\partial ^{2}}{\partial \xi ^{2}} +3 \frac{\partial ^{2}}{\partial \eta ^{2}} = {\hat{\nabla }}^2, \end{aligned}$$

(64)

and the system (60) becomes

$$\begin{aligned} {\hat{\nabla }}^2 \phi = -\lambda ^2 \phi , \end{aligned}$$

(65)

subject to $\phi = 0$ on $\xi = r_{\text {inc}}$, $\eta =u +2 r_{\text {inc}}$, and $\eta = -u-2r_{\text {inc}}$. Seeking a separable solution of the form $\phi (\xi ,\eta ) = f(\xi )g(\eta )$ and substituting this into (65) yields

$$\begin{aligned} f_{\xi \xi }+ \alpha ^2 f&= 0, \end{aligned}$$

(66a)

$$\begin{aligned} g_{\eta \eta }+ \beta ^2 g&= 0, \end{aligned}$$

(66b)

$$\begin{aligned} \alpha ^2 + 3\beta ^2&= \lambda ^2. \end{aligned}$$

(66c)

Solutions to (66a) and (66b) are simple harmonic oscillators. We first consider the solution to (66a). The boundary condition to be satisfied is $\phi = 0$ on $\xi =r_{\text {inc}}$, but we need a second boundary condition. Considering that we would like the top corner of the triangle to be consistent with the boundary conditions on the triangle sides, we impose that $\phi = 0$ on $\xi = -2r_{\text {inc}}$.

Thus, we require $f(r_{\text {inc}})=f(-2r_{\text {inc}})=0$. Let’s consider a shift in $f(\xi )$,

$$\begin{aligned} f(\xi ) = \cos {\alpha (\xi -\gamma )}+ \sin {\alpha (\xi -\gamma )}, \end{aligned}$$

(67)

for some constant $\gamma$. Taking $\gamma =-2r_{\text {inc}}$ we see that

$$\begin{aligned} f(\xi )= \sin {\bigg (\frac{l \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )}, \ \ \ \ \ l \in {\mathbb {N}}, \ l \ne 0; \ \ \ \ \ \alpha = \frac{l \pi }{3 r_{\text {inc}}}. \end{aligned}$$

(68)

To solve the original problem, we will consider separately the even and odd functions of $\eta$, denoted $\phi _e$ and $\phi _o$ respectively,

$$\begin{aligned} \phi _e&= \sin {\bigg (\frac{l \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )} \cos {\left( \beta \eta \right) }, \end{aligned}$$

(69a)

$$\begin{aligned} \phi _o&= \sin {\bigg (\frac{l \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )} \sin {\left( \beta \eta \right) }, \end{aligned}$$

(69b)

noting that if both $\phi _e$ and $\phi _o$ satisfy (60) then their sum will automatically satisfy (60) also. Now we must satisfy the boundary conditions at $v=r_{\text {inc}}$ and $w=r_{\text {inc}}$, which are equivalent to $\eta = u + 2r_{\text {inc}}$ and $\eta = -u-2r_{\text {inc}}$. By symmetry, if $\phi = 0$ is satisfied at one of these two boundaries, then it is satisfied at the other also.

Considering $\phi _e$, we need to satisfy $\phi _e = 0$ at $\eta = u+2r_{\text {inc}}$. Recall that $\xi = u$,

$$\begin{aligned} \phi _e = \sin {\bigg (\frac{l \pi }{3 r_{\text {inc}}} (u + 2r_{\text {inc}})\bigg )} \cos {\big (\beta (u+2r_{\text {inc}}) \big )} =0. \end{aligned}$$

(70)

Now (70) cannot be zero for all u. But we know that $\phi _e$ is a solution for all non-zero integers. This then motivates consideration of additive solutions at $\eta = u+2r_{\text {inc}}$ of the form

$$\begin{aligned} \phi _e = \sin {\bigg (\frac{l \pi }{3 r_{\text {inc}}} (u + 2r_{\text {inc}})\bigg )} \cos {\big (\beta _l (u+2r_{\text {inc}}) \big )} + \sin {\bigg (\frac{m \pi }{3 r_{\text {inc}}} (u + 2r_{\text {inc}})\bigg )} \cos {\big (\beta _m (u+2r_{\text {inc}}) \big )}=0. \end{aligned}$$

(71)

Trigonometric identities can be used to write (71) as a sum of sines, which then indicates that for $\phi _e=0$ to be satisfied we have $\phi _e=0$, so the two mode solution is also inadequate. We consider adding more integers,

$$\begin{aligned} \begin{aligned} \phi _e&= \sin {\bigg (\frac{l \pi }{3 r_{\text {inc}}} (u + 2r_{\text {inc}})\bigg )} \cos {\big (\beta _l (u+2r_{\text {inc}}) \big )} + \sin {\bigg (\frac{m \pi }{3 r_{\text {inc}}} (u + 2r_{\text {inc}})\bigg )} \cos {\big (\beta _m (u+2r_{\text {inc}}) \big )} \\&\quad + \sin {\bigg (\frac{n \pi }{3 r_{\text {inc}}} (u + 2r_{\text {inc}})\bigg )} \cos {\big (\beta _n (u+2r_{\text {inc}}) \big )} =0. \end{aligned} \end{aligned}$$

(72)

Writing (72) as a sum of sines, it can then be seen that $\phi _e=0$ if

$$\begin{aligned} \frac{l \pi }{3 r_{\text {inc}}}- \beta _l&= -\frac{m \pi }{3 r_{\text {inc}}}-\beta _m, \end{aligned}$$

(73a)

$$\begin{aligned} \frac{l \pi }{3 r_{\text {inc}}}+ \beta _l&= -\frac{n \pi }{3 r_{\text {inc}}}+\beta _n, \end{aligned}$$

(73b)

$$\begin{aligned} \frac{m \pi }{3 r_{\text {inc}}}- \beta _m&= -\frac{n \pi }{3 r_{\text {inc}}}-\beta _n, \end{aligned}$$

(73c)

where it should be noted that these choices are not unique. We notice that adding the above expressions

$$\begin{aligned} l + m + n =0, \end{aligned}$$

(74)

which echoes the linear dependence $u+v+w=0$. Adding (73a) and (73c), substituting m based on (74) we see

$$\begin{aligned} \beta _n - \beta _l = \frac{(l+n)\pi }{3 r_{\text {inc}}}. \end{aligned}$$

(75)

Then, due to the degeneracy in (74), we recall the condition (66c), which simplifies as

$$\begin{aligned} \beta _l^2-\beta _n^2 = \frac{(n^2-l^2) \pi ^2}{27 r_{\text {inc}}^2}. \end{aligned}$$

(76)

Combining (75) and (76) we can solve for $\beta _n$, and then we can use (73) to determine $\beta _l$ and $\beta _m$,

$$\begin{aligned} \beta _n&=\frac{(l-m)\pi }{9 r_{\text {inc}}}, \end{aligned}$$

(77a)

$$\begin{aligned} \beta _l&=\frac{(m-n)\pi }{9 r_{\text {inc}}}, \end{aligned}$$

(77b)

$$\begin{aligned} \beta _m&=\frac{(n-l)\pi }{9 r_{\text {inc}}}, \end{aligned}$$

(77c)

with corresponding eigenvalue,

$$\begin{aligned} \lambda _{mn}^2=\frac{4 \pi ^2}{27 r_{\text {inc}}^2} (m^2+mn+n^2). \end{aligned}$$

(78)

The odd solution $\phi _o$ can be determined in a similar manner, where, in fact, due to the choice of conditions in (73), we find that the odd solution is of the same form. Thus the solution to (65) subject to the homogeneous Dirichlet boundary conditions is given by

$$\begin{aligned} \phi _e^{mn}&= \sin {\bigg (\frac{l \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )} \cos {\bigg (\frac{(m-n)\pi }{9r_{\text {inc}}} \eta \bigg )} + \sin {\bigg (\frac{m \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )} \cos {\bigg (\frac{(n-l)\pi }{9r_{\text {inc}}} \eta \bigg )} \nonumber \\&\quad + \sin {\bigg (\frac{n \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )} \cos {\bigg (\frac{(l-m)\pi }{9r_{\text {inc}}} \eta \bigg )} =0, \end{aligned}$$

(79a)

$$\begin{aligned} \phi _o^{mn}&= \sin {\bigg (\frac{l \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )} \sin {\bigg (\frac{(m-n)\pi }{9r_{\text {inc}}} \eta \bigg )} + \sin {\bigg (\frac{m \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )} \sin {\bigg (\frac{(n-l)\pi }{9r_{\text {inc}}} \eta \bigg )} \nonumber \\&\quad + \sin {\bigg (\frac{n \pi }{3 r_{\text {inc}}} (\xi + 2r_{\text {inc}})\bigg )} \sin {\bigg (\frac{(l-m)\pi }{9r_{\text {inc}}} \eta \bigg )} =0. \end{aligned}$$

(79b)

More detail can be found in McCartin [65].