Question 3 of 7: Composite wall — three-element one-dimensional conduction with convection

Question 3: Composite wall — three-element one-dimensional conduction with convection (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 with a prescribed temperature on the inside face and convection to ambient air on the outside face, discretized with one linear element per layer:

Find. The two interface temperatures $T_2$ and $T_3$, the outer surface temperature $T_4$, and the heat flux through the 8-cm layer.

Approach. Form the two-node conduction matrix for each layer, assemble the global system, add the convection contribution to both the conductance matrix and the load vector at the outer node, impose the prescribed inside temperature, and solve the reduced system; then recover the flux from any element and confirm it against Newton's law of cooling at the outer face.

1. Set up the mesh and element conductances. Four nodes sit at $x = 0$, $0.025$, $0.065$ and $0.145$ m. Each layer is one linear element whose conduction matrix is $$[k^{(e)}] = \frac{K_e A}{L_e}\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix}$$ so the scalar conductances per unit area are $$\frac{K_1}{L_1}=\frac{60}{0.025}=2400,\qquad \frac{K_2}{L_2}=\frac{30}{0.040}=750,\qquad \frac{K_3}{L_3}=\frac{10}{0.080}=125 \ \ \text{W}/(\text{m}^{2}\cdot{}^\circ\text{C})$$
2. Assemble the global conductance matrix. Overlapping the three element matrices at the shared nodes, $$[K] = \begin{bmatrix} 2400 & -2400 & 0 & 0\\ -2400 & 3150 & -750 & 0\\ 0 & -750 & 875 & -125\\ 0 & 0 & -125 & 125\end{bmatrix}$$ where the diagonal entries at the interior nodes are the sums $2400+750$ and $750+125$.
3. Add the convection boundary at node 4. Convection acts on the outer face only, so it contributes a surface term $hA$ to the diagonal of node 4 and a load $hT_\infty A$ to the same equation — there is no lateral surface in this problem and therefore no distributed surface matrix: $$K_{44} \leftarrow 125 + hA = 125+20 = 145,\qquad F_4 = hT_\infty A = (20)(50)(1) = 1000\ \text{W}/\text{m}^{2}$$ with $F_1=F_2=F_3=0$ because there is no internal heat generation.
4. Impose the essential boundary condition and reduce. Node 1 is prescribed at $T_1 = 300\,{}^{\circ}\text{C}$, so its column moves to the right-hand side and its row is set aside for the reaction (the heat input at the inside face). The reduced system in $T_2$, $T_3$, $T_4$ is $$\begin{bmatrix}3150 & -750 & 0\\ -750 & 875 & -125\\ 0 & -125 & 145\end{bmatrix}\begin{Bmatrix}T_2\\ T_3\\ T_4\end{Bmatrix} = \begin{Bmatrix}(2400)(300)\\ 0\\ 1000\end{Bmatrix}=\begin{Bmatrix}720\,000\\ 0\\ 1000\end{Bmatrix}$$
5. Solve the reduced system. Forward elimination and back-substitution give $$\boxed{\;T_2 = 298.257\,{}^{\circ}\text{C},\qquad T_3 = 292.678\,{}^{\circ}\text{C},\qquad T_4 = 259.205\,{}^{\circ}\text{C}\;}$$ The two interface temperatures asked for in part (a) are $T_2$ and $T_3$; $T_4$ is the outer surface temperature, which the convection condition determines as part of the same solution.
6. Cross-check against the series-resistance solution. Because the mesh is one linear element per layer and there is no internal generation, the finite element nodal temperatures are exact. The equivalent thermal resistance per unit area is $$R = \frac{L_1}{K_1}+\frac{L_2}{K_2}+\frac{L_3}{K_3}+\frac{1}{h} = 4.1667\times10^{-4}+1.3333\times10^{-3}+8.0\times10^{-3}+5.0\times10^{-2}=0.059750$$ giving $$q = \frac{T_1-T_\infty}{R} = \frac{300-50}{0.059750} = 4184.10\ \text{W}/\text{m}^{2}$$ and marching that flux through the layers reproduces $T_2 = 300-4184.10/2400 = 298.257$, $T_3 = 298.257-4184.10/750 = 292.678$ and $T_4 = 292.678-4184.10/125=259.205\,{}^{\circ}\text{C}$, identical to the finite element result.
7. Part (b): heat flux through the 8-cm layer. Within element 3 the flux follows from Fourier's law with the linear temperature interpolation, $$q_3 = \frac{K_3}{L_3}\big(T_3-T_4\big) = 125\,(292.678-259.205)$$ $$\boxed{\, q_3 = 4184\ \text{W}/\text{m}^{2}\ \text{(i.e. }4.18\ \text{kW}/\text{m}^{2}\text{, in the}+x\ \text{direction)}\,}$$ The same value is obtained from the convection face, $q = hA(T_4-T_\infty) = 20(259.205-50) = 4184\ \text{W}/\text{m}^{2}$, and from elements 1 and 2 — which is the physical check on the whole solution.

The numbers themselves carry the engineering message of the question. Steady one-dimensional conduction with no generation is a series circuit, so the flux is the same through all three layers and through the air film; asking specifically for the flux “through the 8-cm portion” is a test of whether the candidate recognises that, rather than an invitation to compute something layer-specific. What differs between layers is the temperature drop, which is proportional to each layer's resistance: 1.74 °C across the 2.5-cm high-conductivity layer, 5.58 °C across the 4-cm layer, 33.47 °C across the 8-cm insulating layer and 209.21 °C across the air film.

That last figure is the design insight: the convection film alone accounts for $0.05/0.05975 = 83.7\%$ of the total resistance, against $13.4\%$ for the 8-cm insulating layer and under $3\%$ for the two conductive layers combined. Adding wall thickness would barely change the heat loss, whereas increasing the outside film coefficient — or, conversely, protecting the wall from wind — dominates the result. It also explains why the outer surface sits at $259\,{}^{\circ}\text{C}$: with such a poor film, the wall cannot shed heat fast enough to cool its own outer face, and a touch-safety or material-limit check on that surface would be the next step in a real design.