Question 1 of 7: Composite wall — three conduction elements with a convective face

Question 1: Composite wall — three conduction elements with a convective face (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A three-layer plane wall is discretised with one linear conduction element per material, the inner face held at a known temperature and the outer face losing heat by convection to still air.

Find. The two interface temperatures $T_2$ and $T_3$ (and the outer surface temperature $T_4$ that the convective boundary condition delivers along with them), then the heat flux passing through the third layer.

Approach. Build one two-node conduction element per layer with conductance $K A/L$, assemble the four-node global system, add the convective terms to the node-4 row only, impose the prescribed inner temperature and solve; then confirm the result against the series-resistance network, which must agree exactly because the wall has no internal heat generation.

1. Form the element conductances. A two-node linear conduction element of length $L_e$, area $A$ and conductivity $K_e$ has $$[k^{(e)}] = \frac{K_e A}{L_e}\begin{bmatrix}\ \ 1 & -1\\ -1 & \ \ 1\end{bmatrix}$$ so that, with $A = 1\times10^{-4}\ \text{m}^2$, $$k_1=\frac{(60)(10^{-4})}{0.02}=0.3000,\qquad k_2=\frac{(30)(10^{-4})}{0.04}=0.0750,\qquad k_3=\frac{(10)(10^{-4})}{0.06}=0.016667\ \text{W}\,{}^\circ\text{C}^{-1}$$ The three conductances fall by roughly a factor of four at each step, which already signals that almost the whole temperature drop will occur across layer 3 and across the outside film.
2. Assemble the global conductance matrix. Overlapping the three element matrices on nodes $1\!-\!2$, $2\!-\!3$ and $3\!-\!4$ gives the tridiagonal array $$[K] = \begin{bmatrix} k_1 & -k_1 & 0 & 0\\ -k_1 & k_1+k_2 & -k_2 & 0\\ 0 & -k_2 & k_2+k_3 & -k_3\\ 0 & 0 & -k_3 & k_3\end{bmatrix}$$ with a zero load vector, because the wall contains no distributed source.
3. Apply the convective boundary condition at node 4. Convection acts on the end face of area $A$ only — there is no lateral surface in this one-dimensional model — so it contributes a single stiffness term and a single load term: $$K_{44} \leftarrow K_{44} + hA = 0.016667 + (20)(10^{-4}) = 0.018667\ \text{W}\,{}^\circ\text{C}^{-1},\qquad F_4 = hT_{\infty}A = (20)(65)(10^{-4}) = 0.1300\ \text{W}$$ This is the step most often mishandled: the surface-convection matrix $\dfrac{hPL}{6}\begin{bmatrix}2&1\\1&2\end{bmatrix}$ belongs to a fin losing heat around its perimeter $P$, not to a wall convecting from one end face.
4. Impose the prescribed inner temperature and reduce. With $T_1 = 320\ ^\circ\text{C}$ known, the first column moves to the right-hand side and the first row is set aside, leaving $$\begin{bmatrix} k_1+k_2 & -k_2 & 0\\ -k_2 & k_2+k_3 & -k_3\\ 0 & -k_3 & k_3+hA\end{bmatrix}\begin{Bmatrix}T_2\\ T_3\\ T_4\end{Bmatrix} = \begin{Bmatrix} k_1T_1\\ 0\\ hT_{\infty}A\end{Bmatrix} = \begin{Bmatrix} 96.0\\ 0\\ 0.13\end{Bmatrix}$$
5. Solve the reduced system. Forward elimination and back substitution give $$\boxed{\,T_2 = 318.53\ ^\circ\text{C},\qquad T_3 = 312.63\ ^\circ\text{C},\qquad T_4 = 286.10\ ^\circ\text{C}\,}$$ The two interface temperatures the question asks for are $T_2$ (between materials 1 and 2) and $T_3$ (between materials 2 and 3); $T_4$ is the outer surface temperature, which the same solve returns and which part (b) needs.
6. Confirm against the series-resistance network. Because the wall is source-free and the elements are linear, the finite element temperatures are exact, so the classical thermal-resistance chain must reproduce them. Per unit area, $$R'' = \frac{L_1}{K_1}+\frac{L_2}{K_2}+\frac{L_3}{K_3}+\frac{1}{h} = 0.000333+0.001333+0.006000+0.050000 = 0.0576667\ \text{m}^2\,{}^\circ\text{C}\,\text{W}^{-1}$$ $$q'' = \frac{T_1-T_{\infty}}{R''} = \frac{320-65}{0.0576667} = 4421.97\ \text{W}\,\text{m}^{-2}$$ and marching the drops $q''L_i/K_i$ inwards from $320\ ^\circ\text{C}$ returns $318.53$, $312.63$ and $286.10\ ^\circ\text{C}$ to every digit carried, which closes the check.

Part (b) follows immediately, and its point is a physical one. Since no heat is generated or stored anywhere in the wall, the same flux crosses every layer, so the flux through the third portion can be read from that layer's own Fourier law:

The convective face gives the same number independently, $q'' = h(T_4-T_{\infty}) = 20(286.10-65) = 4421.97\ \text{W m}^{-2}$, and over the stated $1\ \text{cm}^2$ section this is a heat rate of only $Q = q''A = 0.4422\ \text{W}$. Setting the four resistance shares side by side shows where the design leverage lies: layer 1 carries $0.58\%$ of the total resistance, layer 2 carries $2.31\%$, layer 3 carries $10.40\%$, and the outside air film alone carries $86.71\%$. Almost nine-tenths of the $255\ ^\circ\text{C}$ available drop is spent getting the heat off the outer surface, which is why the outer face sits at $286\ ^\circ\text{C}$ rather than anywhere near ambient, and why increasing $h$ — forced convection, fins, a swept face — would change this wall's performance far more than thickening any of the three materials.