Chapter 7
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Heat Conduction Through a Composite Flat Wall

Consider two plane walls in contact (called a composite wall) as shown below. The individual walls are labeled 1 and 2 as are each the thermal conductivity $k$ and thickness $L$. Assume the wall boundaries convect heat to the environment on both sides. Each side may have different convection coefficients $h$ and environmental temperature $T_{\inf}$.

Figure 7.11: Composite Wall With Convection Boundary Conditions

Each layer must satisfy the heat conduction equation $\nabla^2T=\frac{\partial^2T}{\partial x^2}=0$ whose solution is a linear function in $x$. Consequently, we have the following solution for layers 1 and 2:

$\displaystyle \nabla^2T_1=\frac{d^2T_1}{dx^2}=0$ $\textstyle \implies$ $\displaystyle T_1(x)=a_1+b_1x$ (7.22)
$\displaystyle \nabla^2T_2=\frac{d^2T_2}{dx^2}=0$ $\textstyle \implies$ $\displaystyle T_2(x)=a_2+b_2x$  

To evaluate the constants $a_1$, $a_2$, $b_1$, and $b_2$, the boundary conditions at points $A$ and $C$ and the interface conditions at $B$ must be satisfied.

Convective boundary condition at $x=0$: (note direction of $\mathbf{n}$, $\mathbf{n}=-\mathbf{i}$)
    $\displaystyle \left.\mathbf{q}\cdot\mathbf{n}\right\vert _{x=0}=\left. h\left(T_S-T_\infty\right)\right\vert _{x=0}$  
    $\displaystyle \implies\left(-k_1\frac{dT_1}{dx}\mathbf{i}\right)\cdot\left(-\ma...
...eft(T_1-T_{\infty,1}\right)\right\vert _{x=0}
    $\displaystyle \implies k_1b_1=h_1\left(a_1-T_{\infty,1}\right)$ (7.23)

Interface boundary condition between wall 1 and 2 at $x=x_1$:
T_1(x_1)=T_2(x_1)\implies a_1+b_1x_1=a_2+b_2x_1
\end{displaymath} (7.24)

\left(\mathbf{q}\cdot\mathbf{n}\right)_{\text{slab 1}}=\left...
...2}{dx}\mathbf{i}\right)\cdot\mathbf{i}\implies b_1 k_1=b_2 k_2
\end{displaymath} (7.25)

Convective boundary condition at $x=x_2$: (note change in $\mathbf{n}$ direction, $\mathbf{n}=+\mathbf{i}$)

    $\displaystyle \left.\mathbf{q}\cdot\mathbf{n}\right\vert _{x=x_2}=\left.h\left(T_S-T_\infty\right)\right\vert _{x=x_2}$  
    $\displaystyle \implies\left(-k_2\frac{dT_2}{dx}\mathbf{i}\right)\cdot\left(+\ma...
..._{\infty,2}\right)\right\vert _{x=x_2}
    $\displaystyle \implies -k_2b_2=h_2\left(a_2+b_2x_2-T_{\infty,2}\right)$ (7.26)

Consequently we have four equations and four unknowns ($a_1$, $a_2$, $b_1$, and $b_2$) as follows:

$\displaystyle k_1b_1$ $\textstyle =$ $\displaystyle h_1\left(a_1-T_{\infty,1}\right)$  
$\displaystyle a_1+b_1x_1$ $\textstyle =$ $\displaystyle a_2+b_2x_1$ (7.27)
$\displaystyle b_1k_1$ $\textstyle =$ $\displaystyle b_2k_2$  
$\displaystyle k_2b_2$ $\textstyle =$ $\displaystyle -h_2\left(a_2+b_2x_2-T_{\infty,2}\right)$  

The above system of four equations can be solved for $a_1$, $a_2$, $b_1$, and $b_2$ and the result substituted back into equation (7.22) to obtain $T_1(x)$ and $T_2(x)$. The heat flux in the $x$ direction is given by

\end{displaymath} (7.28)

For 1-D slab heat flow, heat can flow only in one direction (in this case, the $x$ direction). Consequently, in the absence of heat sinks/sources in a layer, the heat flux must remain a constant as it passes through the convective air layer on the left, through each slab and finally through the convective air layer on the right.

To simplify the solution for a composite wall (with no internal heat source), we seek to develop a simplified relation between the overall heat flux through the composite wall and the given temperature gradient from one side of the composite wall to the other:

$\displaystyle q_x=-U\Delta T$ $\textstyle \text{or}$ $\displaystyle q_x=-\left(\frac{1}{R}\right)\Delta T$ (7.29)
$\displaystyle \Delta T$ $\textstyle =$ $\displaystyle T_{\infty,2}-T_{\infty,1}$  

where $U\equiv$ effective heat transfer coefficient of the composite wall, $R=\frac{1}{U}\equiv$ effective thermal resistance of the composite wall and, for the case of convection boundary conditions on each side of the composite wall, the known temperature gradient from left to right is given by $\Delta T=T_{\infty,2}-T_{\infty,1}$.

The solution of the ODE for heat transfer through a single layer with no heat source requires that the temperature variation in the layer is a linear function of $x$: $T(x)=a+bx$ where $a$ and $b$ are constants of integration dependent on boundary conditions. If the temperature on either side of a wall of thickness $L$ is $T_A$ and $T_B$, then $T(x)=T_A+\left[\frac{(T_B-T_A)}{L}\right]x$. For the composite wall shown below, we introduce the following notation. Define the temperature at the left boundary to be $T_A$, at the interface $T_B$, and at the right boundary $T_C$ as shown in the figure below. At this point, $T_A$, $T_B$, and $T_C$ are unknown.

Figure 7.12: Composite Wall With Convective Boundary Conditions

In the absence of a heat source within the body, the temperature in each layer will be a linear function of $x$ so that we may write the following equations for the temperature in each layer:

$\displaystyle T_1(x)$ $\textstyle =$ $\displaystyle T_A+\left(\frac{T_B-T_A}{L_1}\right)x$ (7.30)
$\displaystyle T_2(x)$ $\textstyle =$ $\displaystyle T_B+\left(\frac{T_C-T_B}{L_2}\right)(x-x_1)$  

At $x=L_1$ (the interface), these equations already satisfy the interface condition that $T_1(x)=T_2(x)=T_B$. Therefore, only the heat flux boundary condition needs to be satisfied at the interface and the convective boundary condition at the left and right boundaries of the composite wall. From conservation of energy, the heat energy $Q_x$ through a given area $A$ ($Q_x=q_xA$) must be constant as it enters on the left and leaves on the right boundary (since we assumed there is no internal heat generation, $\Phi$). Since heat flow is normal to wall, each layer has same normal area (so area cancels out). Thus, the heat flux $q_x$ must remain a constant as it passes through the convective air layer on the left, through each slab and finally through the convective air layer on the right and we can write

\end{displaymath} (7.31)

Equation (7.31) may be separated into 4 equations by considering each heat flux term individually to obtain:

$\displaystyle T_{\infty,1}-T_A$ $\textstyle =$ $\displaystyle \frac{1}{h_1}q_x\qquad (1)$  
$\displaystyle T_B-T_C$ $\textstyle =$ $\displaystyle \frac{L_2}{k_2}q_x\qquad (2)$ (7.32)
$\displaystyle T_A-T_B$ $\textstyle =$ $\displaystyle \frac{L_1}{k_1}q_x\qquad (3)$  
$\displaystyle T_C-T_{\infty,2}$ $\textstyle =$ $\displaystyle \frac{1}{h_2}q_x\qquad (4)$  

Add these four equations, (1) through (4) to obtain
\end{displaymath} (7.33)

...rmal resistance of}\\ \text{the composite wall}\\ \end{array}}
\end{displaymath} (7.34)

The fractional term in (7.34) may be defined as the effective heat transfer coefficient $U$:

U=\text{effective heat transfer coefficient}=\frac{1}{\frac{1}{h_1}+\sum\limits^n_{i=1}\frac{L_i}{k_i}+\frac{1}{h_2}}
\end{displaymath} (7.35)

where n is the number of layers in the composite wall. We may also define the effective thermal resistance $R$ by the reciprocal of $U$:
R=\text{effective thermal resistance}=\frac{1}{U}=\frac{1}{h_1}+\sum^n_{i=1}\frac{L_i}{k_i}+\frac{1}{h_2}
\end{displaymath} (7.36)

Consequently, the heat flux $q_x$ through the composite wall with convection boundary conditions on both sides of the wall is given by
$\displaystyle q_x$ $\textstyle =$ $\displaystyle -U\Delta T=-\frac{\Delta T}{R}$ (7.37)
$\displaystyle \text{where}\quad\Delta T$ $\textstyle =$ $\displaystyle T_{\infty,2}-T_{\infty,1}$  

Note that thermal resistance terms (like $\frac{L}{k}$ or $\frac{1}{h}$) are additive similar to resistors in electrical theory.

The last result may be expanded to include various boundary conditions on the left and right side of the composite wall. For example, for a composite wall with 3 layers we obtain the following summary of results:

Summary of Conduction Through Composite Walls

Figure 7.13:

    $\displaystyle R=\frac{1}{h_1}+\sum^N_{i=1}\frac{L_i}{k_i}+\frac{1}{h_2}\qquad
R=\sum^N_{i=1}\frac{L_i}{k_i}$ (7.38)
    $\displaystyle q_x=-\left(T_{\infty,2}-T_{\infty,1}\right)\frac{1}{R}\qquad

where the heat flux in each layer is given by:
$\displaystyle q_x$ $\textstyle =$ $\displaystyle -h_1\left(T_A-T_{\infty,1}\right)$  
$\displaystyle q_x$ $\textstyle =$ $\displaystyle -\frac{k_1}{L_2}\left(T_B-T_A\right)$  
$\displaystyle q_x$ $\textstyle =$ $\displaystyle -\frac{k_2}{L_2}\left(T_C-T_B\right)$ (7.39)
$\displaystyle q_x$ $\textstyle =$ $\displaystyle -\frac{k_3}{L_3}\left(T_D-T_C\right)$  
$\displaystyle q_x$ $\textstyle =$ $\displaystyle -h_2\left(T_{\infty,2}-T_D\right)$  

Note: the fist and last terms below represent heat flux through the fluid layers where convection occurs (terms with $h$) and the 2 ${}^{\text{nd}}$ through 4 ${}^{\text{th}}$ terms represent heat flux through the solid layers where conduction occurs (terms with $k$).

Considering the definition of $R$ (7.38) for the various cases of different boundary conditions we note that when there is convection on the left and right, the terms $h_1$ and $h_2$ appear in $R$. When there is convection only on the right, only $h_2$ appears, etc. For three solid layers, we have $\frac{L}{k}$ for each of the three layers. This suggests the following simplified definition of $R$:

R=\text{effective thermal resistance}=\frac{1}{U}=\left\lang...
\end{displaymath} (7.40)

where $\left\langle \right\rangle$ means to include the $h$ term only if there is convection on left ($h_1$) or right ($h_2$)

The general solution procedure then consists of three steps:

Evaluate effective thermal resistance $R$ using (7.40)
Evaluate the heat flux $q_x$ for the composite wall using $q_x=-\left(\frac{1}{R}\right)\Delta T$ where
$\Delta T=(\text{right most temperature})-(\text{left most temperature})$.
Evaluate the temperatures for each layer using (7.39) working from left to right through the layers. $T_A$ can be obtained from the first equation in (7.39), $T_B$ from second equation, etc.

Example 7-5

Consider a two-layer composite wall with 1-D heat transfer through the layers and free convection of air on either side with $h=5\;\frac{\text{BTU}}{(\text{hr ft}^2\;{}^\circ\text{F})}$. Assume the thickness of each layer is $L_1=L_2=10$ cm. The temperature difference from left to right is $\left(T_{\infty,2}-T_{\infty,1}\right)=50\;{}^\circ$C. Find $q_x$ for the following situations:

Material 1-glass; Material 2-glass
Material 1-copper; Material 2-glass
Material 1-copper; Material 2-teflon

$h$ is first converted to metric:

...rac{\text{J}}{\text{m}^2\text{ s}\cdot{}^\circ\text{C}}\right] \end{displaymath}

...text{free convection}\\ \text{glass/glass}\\


...text{free convection}\\ \text{copper/glass}\\ \end{array}\right]


...ext{free convection}\\ \text{copper/teflon}\\ \end{array}\right]

Note that in case c), the introduction of teflon, which is a good insulator with a relatively low coefficient of thermal conductivity $k$, yields a higher effective resistance $R$ and correspondingly lower heat flux, $q_x$.

Example 7-6

Consider a two layer composite wall of copper and teflon as shown below. The copper has a thickness of 10 cm but the thickness of the teflon is to be determined. The temperature on the left boundary is equal to $200\;{}^\circ$C and on the right boundary $25\;{}^\circ$C. Determine the thickness of the teflon layer so that the heat flux is equal to $200\;\frac{\text{W}}{\text{m}^2}$.


Figure 7.14:

Find: $L_2$


$\displaystyle q_x$ $\textstyle =$ $\displaystyle -U\Delta T=-\frac{1}{R}\Delta T$  
$\displaystyle R$ $\textstyle =$ $\displaystyle -\frac{\Delta T}{q_x}=-\frac{25-200}{200}=0.875\;\frac{{}^\circ\text{C - }\text{m}^2}{\text{W}}$ (7.41)
$\displaystyle R$ $\textstyle =$ $\displaystyle \frac{L_1}{k_1}+\frac{L_2}{k_2}=\frac{0.1\;\text{m}}{398\;\frac{\...
....875\;\frac{{}^\circ\text{C - }\text{m}^2}{\text{W}}\implies L_2=0.22\;\text{m}$  

Example 7-7

Consider steady-state heat conduction through a cylindrical wall with convection on both sides of the cylindrical wall. Find the temperature of the wall.

Figure 7.15:

The heat transfer equation in cylindrical coordinates is given by

\left.\begin{array}{l} \nabla^2T=0\\ T(r_1)\\ \end{array}\ri...
...{d^2T}{dr^2}+\frac{1}{r}\frac{dT}{dr}=0\implies T=C_1\ln r+C_2
\end{displaymath} (7.42)

In the absence of internal a heat source in the solid, the solution provided above will always hold.
at A

\end{displaymath} (7.43)

at B
\end{displaymath} (7.44)

Substituting (7.42) into the previous two boundary condition equations yields:
$\displaystyle kC_1\frac{1}{r_A}=h_1\left(C_1\ln r_A+C_2-T_{\infty,1}\right)$     (7.45)
$\displaystyle kC_1\frac{1}{r_B}=-h_2\left(C_1\ln r_B+C_2-T_{\infty,2}\right)$      

Equations (7.45) may be solved for $C_1$ and $C_2$ and substituted into (7.42) to obtain the solution for the temperature distribution $T(r)$.

Example 7-8

Consider steady-state heat conduction through a cylindrical wall with specified temperature on the boundaries of the cylindrical wall. Find the temperature of the wall.

Figure 7.16:

The solution of the heat flow equation in cylindrical coordinates is given by

\begin{displaymath}\left.\begin{array}{l} \nabla^2T=0\\ T(r_1)\\ \end{array}\rig...
...2T}{dr^2}+\frac{1}{r}\frac{dT}{dr}=0\implies T(r)=C_1\ln r+C_2 \end{displaymath}

Applying the boundary conditions at the inner and outer radius gives

\begin{eqnarray*}T(r_i)=T_i&=&C_1\ln r_i+C_2\,\,\,.......\,\,\,(1)\\
T(r_o)=T_o&=&C_1\ln r_o+C_2\,\,\,.......\,\,\,(2)

Subtracting equation (2) from equation (1) gives:

\begin{displaymath}T_i-T_o=C_1\ln\left(\frac{r_i}{r_o}\right) \end{displaymath}


\begin{displaymath}C_1=\frac{T_i-T_o}{\ln\left(\frac{r_i}{r_o}\right)} \end{displaymath}

Substituting $C_1$ into equation (1) above gives

\begin{displaymath}C_2=T_i-\frac{T_i-T_o}{\ln\left(\frac{r_i}{r_o}\right)}\ln(r_o) \end{displaymath}

Substituting $C_1$ and $C_2$ into $T(r)$ yields

\begin{displaymath}T=\frac{T_i-T_o}{\ln\left(\frac{r_i}{r_o}\right)}\ln r
...ln\left(\frac{r}{r_i}\right)}{\ln\left(\frac{r_i}{r_o}\right)} \end{displaymath}


\begin{displaymath}T(r)=T_i-(T_i-T_o)\frac{\ln\left(\frac{r}{r_i}\right)}{\ln\left(\frac{r_o}{r_i}\right)} \end{displaymath}

The heat flux in the radial direction is given by:

\begin{displaymath}q_r(r)=-k\frac{dT}{dr}=+k\left(\left(T_i-T_o\right)\frac{1}{r}\frac{1}{\ln\left(\frac{r_o}{r_i}\right)}\right) \end{displaymath}


\begin{displaymath}q_r(r)=k\frac{1}{r}\frac{T_i-T_o}{\ln\left(\frac{r_o}{r_i}\right)} \end{displaymath}

Note that the heat flux $q_r$ is a function of radial position $r$. This is necessary because the area through which the heat flows increases as $r$ increases. The radial flow $Q_r$ for time $\Delta t$ is given by $Q_r=q_r A\Delta t=q_r(2\pi r)\Delta t=2\pi\Delta t k\frac{T_i-T_o}{\ln\left(\frac{r_0}{r_i}\right)}$. Note that $Q_r$is independent of $r$ (as it should be) since there is no internal heat source and thus the heat flow must be the same at all radial positions, $r$.


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