Effect of void shape and highly conducting boundary on 2D conductivity of porous materials

Abstract

In this paper, the heat conductivity of two-dimensional (2D) media made of an arbitrarily thermal anisotropic material and containing pores with arbitrary shape and superconductive boundary is considered. In addition to the bulk behavior, the line conduction model is used for the boundary behavior. Such idealized mathematical model can be seen as the limit case of very thin material layer with very high conductivity. The fundamental heterogeneity problem in the micromechanics of a single void embedded in an infinite matrix with both boundary and bulk behavior is then investigated and solved with the complex variable and the Conformal Mapping (CM) techniques. The heterogeneity problem results are then used to obtain the effective heat conductivity of the porous material with different homogenization schemes.

Appendices

Appendix A: Expressions in transformed coordinates

For heat transfer problem, the temperature T must satisfy the following energy conservation equation of the matrix phase with thermal conductivity tensor $\mathbf {K}^0 = \begin{bmatrix} K^0_{11} &{} 0 \\ 0 &{} K^0_{22} \end{bmatrix}$

$$\begin{aligned} K^{0}_{11}\frac{\partial ^2 T}{\partial x^2} + K^{0}_{22}\frac{\partial ^2 T}{\partial y^2} = 0. \end{aligned}$$

(42)

In the conventional complex plane where $z=x+i\sqrt{K^{0}_{11}/K^{0}_{22}}y$, the temperature T can be expressed in the following form

$$\begin{aligned} T=\mathfrak {R}\{ \varphi (z)\}=\frac{1}{2} (\varphi (z)+\overline{\varphi (z)}) \end{aligned}$$

(43)

where $\varphi (z)$ is an analytical function of z. The heat flux q can be expressed by $q = q_x - i\sqrt{K^{0}_{11}/K^{0}_{22}}q_y$ with

$$\begin{aligned}&q_x= -\frac{K^{0}_{11}}{2}[\varphi '(z)+ \overline{\varphi '(z)}],\quad q_y=-i\frac{\sqrt{K^{0}_{11}K^{0}_{22}}}{2}[\varphi '(z)- \overline{\varphi '(z)}]. \end{aligned}$$

(44)

Next, by introducing the new variable $\zeta$ and by using the transformation $z=\omega (\zeta )$, the function $\varphi (z)$ becomes

$$\begin{aligned} \varphi (z)=\varphi (\omega (\zeta ))=\varphi _1(\zeta ). \end{aligned}$$

(45)

In general, the mapping function $z=\omega (\zeta )$ is chosen to map a void plate with boundary $\varGamma$ in z plane into the unit disk in $\zeta$ plane ($|\zeta |\le 1$) and solve the problem with $\zeta$ as variable.

On the unit circle mapped from $\varGamma$, we pose $\zeta =e^{-i\theta }$ with $\theta$ running from 0 to $2\pi$. This parameterization also guarantees the arc coordinate s on $\varGamma$ goes in the positive (counter-clockwise) direction. Differential calculus yields the following result

$$\begin{aligned}&\text {d}\zeta =-i e^{-i\theta } d\theta =-i\zeta \text {d}\theta ,\quad {\text {d}}\bar{\zeta } =i e^{i\theta } d\theta =i\bar{\zeta } {\text {d}}\theta \nonumber \\&{\text {d}}z = \omega '(\zeta ) {\text {d}}\zeta = -i\omega '(\zeta ) \zeta \text {d}\theta ,\quad {\text {d}}s=|\text {d}z|=|\omega '(\zeta )|d\theta \end{aligned}$$

(46)

We can immediately obtain

$$\begin{aligned}&\frac{\text {d}\zeta }{{\text {d}}s} =-\frac{i\zeta }{|\omega '|},\quad \frac{\text {d}\bar{\zeta }}{{\text {d}}s} =\frac{i\bar{\zeta }}{|\omega '|},\quad \frac{\text {d}z}{{\text {d}}s}=\frac{i\omega '}{|\omega '|} \end{aligned}$$

(47)

and hence the three identities of (22, 23).

Appendix B: Detailed dimensions of voids

The details dimensions of all the voids are summarized in Fig. 8 where L is the basic dimension and for arbitrary shape, $L_1=4.6L$ and $L_2=2.5L$. All the results are to the value of L.

Fig. 8

Void dimensions (a Circle, b square, c ellipse, d octagon, f arbitrary)