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.


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.


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!


1 Answer 1


The thought process is not entirely correct in the work section. The heat fluxes are not additive, rather they should be equal due to conservation of energy.

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.


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