# Thermal Conductivity of a Layered Composite

I have this question assigned as part of my homework, and I am having trouble figuring out the solution. Below is the question and my attempts at solving it. I need to have the homework ready tomorrow morning, so I would really appreciate any help or direction in finding the solution.

Question:

Suppose that a composite solid consists of alternating materials $A$ and $B$, with layer thickness $L_A$ and $L_B$, respectively. Both materials are isotropic, with thermal conductivities $k_A$ and $k_B$. It is desired to predict the average, steady-state heat flux in a sample that contains many layers. The flux in the composite is written as $$\boldsymbol{ Q=-k\cdot G}$$ where $\boldsymbol{Q}$ and $\boldsymbol{G}$ are the average heat flux and temperature gradient, respectively, and $\boldsymbol{k}$ is the effective thermal conductivity. With coordinate axis of material interfaces aligned with normal along the $y$-axis, the effective conductivity will be of the form $$\boldsymbol{k}=\left[\begin{array}{ccc} k_{xx} & 0 &0\\0&k_{yy}&0\\0&0&k_{zz}\end{array}\right].$$ (a) Evaluate $k_{xx}$, $k_{yy}$, and $k_{zz}$ in terms of $k_A$, $k_B$, and $\phi$, where $\phi=L_A/(L_A+L_B)$ is the volume fraction of material $A$ ($L_A$ is the length of material $A$, and $L_B$ is the length of material $B$). (Hint: The steady heat flux $q$ through a slab of thickness $L$ and thermal conductivity $k$ is $q=k\Delta T/L$, where $\Delta T$ is the temperature difference between surfaces.)

(b) Show that, in general, the vectors $\boldsymbol{Q}$ and $\boldsymbol{-G}$ will not be parallel to one another. That is, the directions of the heat flux and temperature gradient will tend to differ.

Work:

We can consider the orientation of axes such that the positive $x$-axis comes out of the page towards us, the positive $y$-axis goes to the right of the page, and the positive $z$-axis goes to the top of the page. We can also consider a segment of the composite solid with material $A$ on the left and material $B$ on the right along the $y$-axis, i.e., material $A$ is from $0$ to $L_A$ on the $y$-axis, and material $B$ is situated from $L_A$ to $L_B$ along the $y$-axis.

The vector equation $\boldsymbol{Q=-k\cdot G}$ can be broken down into 3 equations (using the definition $\boldsymbol{G}=\nabla T$): $$q_x=-k_{xx}\frac{\partial T}{\partial x}\\q_y=-k_{yy}\frac{\partial T}{\partial y}\\q_z=-k_{zz}\frac{\partial T}{\partial z}$$

Regarding $k_{xx}$ and $k_{zz}$, since the difference in material is only along the $y$-axis, I would think that $k_{xx}$ and $k_{zz}$ would not depend on $\phi$, $L_A$, or $L_B$, rather only on the thermal conductivity of the particular material $A$ or $B$, depending on the location along the $y$-axis. However, if the materials are assumed to be infinite along the $x$-axis and $z$-axis (this may be a logical inference, since the height and width of the composite solid are not given), then the flux along the $x$ and $z$ directions should be $0$. If this assumption is correct, then $\frac{\partial T}{\partial x}=\frac{\partial T}{\partial z}=0$, so this tells me nothing about $k_{xx}$ or $k_{zz}$. Is this assumption correct? Did I miss something regarding the evaluation of $k_{xx}$ and $k_{zz}$?

To figure out $k_{yy}$, we can construct equations for q for each material along the $y$-axis: $$q_A=k_A\Delta T_A /L_A\\q_B=k_B\Delta T_B/L_B$$ Then, using the given hint, we can consider temperature to the left of material $A$ to be $T_0$, the temperature at the interface between materials $A$ and $B$ to be $T_1$, and the temperature to the right of material $B$ to be $T_2$. Accordingly, $\Delta T_A=T_0-T_1$ and $\Delta T_B=T_1-T_2$, so the equations become the following: $$q_A=k_A\left(T_0-T_1\right)/L_A\\q_B=k_B\left(T_1-T_2\right)/L_B$$ The Equations can be rearranged: $$T_0-T_1=q_A L_A/k_A\\T_1-T_2=q_B L_B/k_B$$ Then the two equations can be added together to form $\Delta T=T_0-T_2$: $$\Delta T=\frac{q_A L_A k_B +q_B L_B k_A}{k_A+k_B}$$ From here, however, I am not sure how I would proceed; $\phi$ does not explicitly appear, and $q_A$ and $q_B$ do appear but the question implies they should not appear.

An alternate approach would be to add the two equations together: $$q_A+q_B=\frac{k_A L_B \left(T_0-T_1\right)+k_B L_A \left(T_1-T_2\right)}{L_A+L_B}$$ From the definition of $\phi$, we also know that $1-\phi=L_B/(L_A+L_B)$. Using this fact, the above equation can be rearranged as follows: $$q_A+q_B=k_A(T_0-T_1)+\phi\left[k_A(T_1-T_0)+k_B(T_1-T_2)\right]$$ However, here again I do not know how to proceed; I still have $T_0$, $T_1$, and $T_2$ separate, and this equation is only useful if we assume $q_y=q_A+q_B$. Is this last assumption correct? Is there a simplification of the equation that I missed?

Regarding (b), I presume that once I find $\boldsymbol{Q}$ and $\boldsymbol{-G}$, I can show $\boldsymbol{Q\cdot -G}\neq1$, but I do not yet have $\boldsymbol{Q}$ or $\boldsymbol{-G}$ (or even $\nabla T$, from which I can find $\boldsymbol{-G}$) to figure this out.

Any help or direction on any part of this problem would be much appreciated! As I mentioned, I need to have the homework ready tomorrow morning, so I am kind of in a rush. Thank you in advance!

Thus, in the $y$ direction, $q_y=q_{yA}=q_{yB}$. Since $q=k\frac{\Delta T}{L}$, \begin{align} q_{yA}&=k_A\frac{\Delta T_A}{L_A}\\ q_{yB}&=k_B\frac{\Delta T_B}{L_B}. \end{align} Therefore, \begin{align} q_{y}&=\frac{k_A}{L_A}(T_0-T_1)\\ q_{y}&=\frac{k_B}{L_B}(T_1-T_2). \end{align} Rearranging the equations and replacing in for $T_1$, $$T_0-\frac{L_A}{k_A}q_y=T_2+\frac{L_B}{k_B}q_y,$$ from which $$T_0-T_2=q_y\left(\frac{L_A}{k_A}+\frac{L_B}{k_B}\right).$$ Defining $\Delta T=T_0-T_2$, $$q_y=\frac{1}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}\Delta T.$$ Since $G_y=-\frac{\Delta T}{L_A+L_B}$, $$q_y=\frac{L_A+L_B}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}\frac{\Delta T}{L_A+L_B}=-\frac{L_A+L_B}{\frac{L_A}{k_A}+\frac{L_B}{k_B}}G_y.$$ Rearranging and using the definition $\phi=\frac{L_A}{L_A+L_B}$, $$q_y=-\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}G_y.$$ Since $q_y=-k_{yy}G_y$, $$k_{yy}=\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}.$$ For $k_{xx}$ and $k_{zz}$, the paths through $A$ and $B$ are parallel, so $\frac{\Delta T}{L}$ is the same for $A$ and $B$, and $G_x=G_z=-\frac{\Delta T}{L}$ is the same for $A$ and $B$. Since the paths are parallel, \begin{align} q_x&=\frac{L_A}{L_A+L_B}q_{xA}+\frac{L_B}{L_A+L_B}q_{xB}\\ q_z&=\frac{L_A}{L_A+L_B}q_{zA}+\frac{L_B}{L_A+L_B}q_{zB}. \end{align} Now, \begin{align} q_{xA}&=q_{zA}=k_A\left(\frac{\Delta T}{L}\right)_A\\ q_{xB}&=q_{zB}=k_B\left(\frac{\Delta T}{L}\right)_B, \end{align} but $$\left(\frac{\Delta T}{L}\right)_A=\left(\frac{\Delta T}{L}\right)_B=-G_x=-G_z,$$ so \begin{align} q_x&=\phi k_A(-G_x)+(1-\phi)k_B(-G_x)=-\left[\phi k_A+(1-\phi)k_B\right]G_x\\ q_z&=\phi k_A(-G_z)+(1-\phi)k_B(-G_z)=-\left[\phi k_A+(1-\phi)k_B\right]G_z. \end{align} Since \begin{align} q_x&=-k_{xx}G_x\\ q_z&=-k_{zz}G_z, \end{align} $$k_{xx}=k_{zz}=\phi k_A+(1-\phi)k_B.$$ Therefore, $$\mathbf{k}=\left( \begin{array}{ccc} k_{zz}=\phi k_A+(1-\phi)k_B&0&0\\ 0&\frac{k_Ak_B}{(1-\phi)k_A+\phi k_B}&0\\ 0&0&k_{zz}=\phi k_A+(1-\phi)k_B \end{array} \right).$$
For part b, in order for two vectors $\mathbf{a}$ and $\mathbf{b}$ to be parallel, the ratio $a_i/b_i$ must be constant for all $i$. Here, \begin{align} \mathbf{Q}&=-k_{xx}G_x\hat{x}-k_{yy}G_y\hat{y}-k_{zz}G_z\hat{z}\\ -\mathbf{G}&=-G_x\hat{x}-G_y\hat{y}-G_z\hat{z}. \end{align} \begin{align} \frac{-k_{xx}G_x}{-G_x}&=k_{xx}=k_{zz}=\frac{-k_{zz}G_z}{-G_z}\\ \frac{-k_{yy}G_y}{-G_y}&=k_{yy}, \end{align} but unless $k_A=k_B$, $k_{yy}\neq k_{xx}$, so the ratio does not hold and the vectors are not parallel.