One-dimensional finite element model of a composite wall: 22-Mec-B10 Finite Element Analysis, December 2017, Q5

Question 5: One-dimensional finite element model of a composite wall (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 only. All data are collected below; lengths are converted to metres and the area to square metres for consistency.

Given data
Quantity	Symbol	Value
Inside face temperature	$T_1$	$280\ ^\circ\text{C}$ (prescribed)
Ambient air temperature	$T_\infty$	$60\ ^\circ\text{C}$
Convection coefficient (outside face)	$h$	$15\ \text{W/(m}^2\,{}^\circ\text{C)}$
Conductivity, material 1	$K_1$	$60\ \text{W/(m}\,{}^\circ\text{C)}$
Conductivity, material 2	$K_2$	$30\ \text{W/(m}\,{}^\circ\text{C)}$
Conductivity, material 3	$K_3$	$15\ \text{W/(m}\,{}^\circ\text{C)}$
Thickness, material 1	$L_1$	$3\ \text{cm}=0.03\ \text{m}$
Thickness, material 2	$L_2$	$6\ \text{cm}=0.06\ \text{m}$
Thickness, material 3	$L_3$	$9\ \text{cm}=0.09\ \text{m}$
Cross-sectional area	$A$	$1\ \text{cm}^{2}=1\times10^{-4}\ \text{m}^{2}$
Discretisation	—	3 linear elements, 4 nodes

Find. (a) the two interface temperatures $T_2$ and $T_3$ and the outside face temperature $T_4$; (b) the heat flux through the 6 cm layer (material 2).

Q5: the three-layer wall discretised with one two-node linear element per material. Node 1 carries the prescribed inside temperature; nodes 2 and 3 are the material interfaces; node 4 is the outside face, where convection to the ambient air acts.

Approach. Assemble the one-dimensional conduction stiffness (conductance) matrix from three two-node elements, add the outside-face convection to the last diagonal and the matching term to the load vector, impose the prescribed node-1 temperature by partitioning, solve the $3\times3$ reduced system, and confirm the answer against the series thermal-resistance network.

Form the element conductance matrices. For a two-node linear conduction element of conductivity $K_e$, length $L_e$ and area $A$, $$\left[k^{(e)}\right]=\frac{K_eA}{L_e}\begin{bmatrix}\ \ 1 & -1\\ -1 & \ \ 1\end{bmatrix}.$$ Evaluating the scalar $K_eA/L_e$ for each layer, $$\frac{K_1A}{L_1}=\frac{60\times10^{-4}}{0.03}=0.2000,\quad \frac{K_2A}{L_2}=\frac{30\times10^{-4}}{0.06}=0.0500,\quad \frac{K_3A}{L_3}=\frac{15\times10^{-4}}{0.09}=0.016667\ \ \text{W/}^\circ\text{C}.$$
Assemble the global system and add convection. Overlapping the three element matrices on the shared nodes, and adding $hA$ to the node-4 diagonal together with $hT_\infty A$ to the node-4 load (the free-end convection boundary condition, with $hA = 15\times10^{-4}=1.5\times10^{-3}\ \text{W/}^\circ\text{C}$ and $hT_\infty A = 15\times60\times10^{-4}=0.09\ \text{W}$), $$\left[K\right]=\begin{bmatrix} 0.2000 & -0.2000 & 0 & 0\\ -0.2000 & 0.2500 & -0.0500 & 0\\ 0 & -0.0500 & 0.066667 & -0.016667\\ 0 & 0 & -0.016667 & 0.018333\end{bmatrix},\qquad \{F\}=\begin{Bmatrix}F_1\\ 0\\ 0\\ 0.09\end{Bmatrix}.$$ Because convection acts on the outside face and not on the lateral surface, the correction is the single term $hA$ on the end diagonal, not the surface matrix $\tfrac{hPL}{6}\begin{bmatrix}2&1\\1&2\end{bmatrix}$ that a laterally cooled fin would require.
Impose the prescribed temperature and reduce. With $T_1=280\ ^\circ\text{C}$ known, delete the first row and move the first column to the right-hand side: $$\begin{bmatrix} 0.2500 & -0.0500 & 0\\ -0.0500 & 0.066667 & -0.016667\\ 0 & -0.016667 & 0.018333\end{bmatrix} \begin{Bmatrix}T_2\\ T_3\\ T_4\end{Bmatrix} =\begin{Bmatrix}0.2000\times280\\ 0\\ 0.09\end{Bmatrix} =\begin{Bmatrix}56.00\\ 0\\ 0.09\end{Bmatrix}.$$
Solve the reduced system. Forward elimination on the tridiagonal system gives $$\boxed{\;T_2 = 278.54\ ^\circ\text{C},\qquad T_3 = 272.68\ ^\circ\text{C},\qquad T_4 = 255.12\ ^\circ\text{C}\;}$$ (to five figures, $T_2=278.5366$, $T_3=272.6829$, $T_4=255.1220\ ^\circ\text{C}$). The first interface temperature has dropped only $1.46\ ^\circ$ below the inside face, whereas the outside face sits $195.12\ ^\circ$ above ambient — almost the whole temperature difference is spent across the outside air film.
Confirm with the series-resistance network. A source-free plane wall discretised with linear elements gives exact nodal temperatures, so the answer must reproduce the classical network. Per unit area, $$R = \frac{L_1}{K_1}+\frac{L_2}{K_2}+\frac{L_3}{K_3}+\frac{1}{h} = 0.000500+0.002000+0.006000+0.066667 = 0.0751667\ \frac{\text{m}^{2}\,{}^\circ\text{C}}{\text{W}},$$ $$q'' = \frac{T_1-T_\infty}{R}=\frac{280-60}{0.0751667}=2926.83\ \text{W/m}^{2}.$$ Walking the temperature down layer by layer, $T_2 = 280-2926.83(0.000500)=278.54$, $T_3 = 278.54-2926.83(0.002000)=272.68$, $T_4 = 272.68-2926.83(0.006000)=255.12\ ^\circ\text{C}$ — identical to the finite element solution. The global energy balance closes as well: $h\left(T_4-T_\infty\right)=15(255.12-60)=2926.8\ \text{W/m}^{2}$.
Part (b) — flux through the 6 cm portion. At steady state with no internal generation the same heat rate must cross every layer, so the flux through material 2 is obtained from Fourier’s law across that element and must equal the value already found: $$q''_{2}=K_2\,\frac{T_2-T_3}{L_2}=30\times\frac{278.5366-272.6829}{0.06} =30\times\frac{5.8537}{0.06}$$ $$\boxed{\;q''_{2}=2926.83\ \text{W/m}^{2}\;}$$ Multiplying by the stated cross-section, the heat rate through the modelled strip is $Q = q''A = 2926.83\times10^{-4}= 0.2927\ \text{W}$.
Interpret the resistance split. The share of the total resistance carried by each path is $0.67\%$ (material 1), $2.66\%$ (material 2), $7.98\%$ (material 3) and $88.69\%$ for the outside air film. The outside convection dominates by more than an order of magnitude, so thickening the wall buys very little: doubling every layer would raise $R$ by only $11\%$ and cut the flux by about a tenth, whereas doubling $h$ — for instance by exposing the face to forced rather than natural convection — would nearly double the loss. This is the design conclusion the question is really probing, and it also explains why the three nodal temperatures cluster so tightly near the inside value.

Question 5 — final results
Quantity	Value
Element conductances $K_eA/L_e$	$0.2000$, $0.0500$, $0.016667\ \text{W/}^\circ\text{C}$
End convection terms	$hA = 1.5\times10^{-3}\ \text{W/}^\circ\text{C}$; $hT_\infty A = 0.09\ \text{W}$
Inside face $T_1$ (prescribed)	$280.00\ ^\circ\text{C}$
(a) Interface 1–2, $T_2$	$278.54\ ^\circ\text{C}$
(a) Interface 2–3, $T_3$	$272.68\ ^\circ\text{C}$
(a) Outside face, $T_4$	$255.12\ ^\circ\text{C}$
Total resistance $R$ (per unit area)	$0.075167\ \text{m}^{2}\,{}^\circ\text{C/W}$
(b) Flux through the 6 cm layer	$2926.83\ \text{W/m}^{2}$
Heat rate through $A=1\ \text{cm}^2$	$0.2927\ \text{W}$
Outside film share of $R$	$88.69\%$

Topic. One-dimensional steady heat conduction by the finite element method: element conductance matrices, convective end boundary conditions, and the equivalence with the series thermal-resistance network.

Key relations. $\left[k^{(e)}\right]=\dfrac{K_eA}{L_e}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$; end convection adds $hA$ to the terminal diagonal of $[K]$ and $hT_\infty A$ to that load row; Fourier’s law $q''=-K\,dT/dx$; series network $R=\sum L_i/K_i + 1/h$ with $q''=\Delta T/R$.

Why this works. A two-node linear element assumes a straight-line temperature profile across the layer, which for steady conduction with no internal generation is the exact solution, so one element per material is not a compromise — the nodal temperatures are exact and refining the mesh would change nothing. The finite element machinery earns its keep by handling the convective boundary in the same framework: the outside face contributes a surface integral $\int h N_i N_j\,dS$ that, for a one-dimensional end face, degenerates to the single scalar $hA$ on the last diagonal, and a load contribution $\int h T_\infty N_i\,dS = hT_\infty A$. Cross-checking against the resistance network is not optional bookkeeping; it is the cheapest way to catch an assembly slip, because both must agree exactly for this class of problem.

Common pitfall. Applying the lateral surface-convection matrix $\tfrac{hPL}{6}\begin{bmatrix}2&1\\1&2\end{bmatrix}$ when the problem states convection at the outside face only — that would introduce a perimeter $P$ the wall does not have. The second pitfall is dropping $A$ because it “cancels”: it does cancel from the temperatures, but only if it is carried consistently in both the conductances and the convective terms. Finally, part (b) is not asking for a different number from part (a); recognising that the flux is the same through every layer is the point of the question.

Source. D. L. Logan, A First Course in the Finite Element Method, 6th ed., Ch. 13 (heat transfer and mass transport, one-dimensional conduction with convection); J. N. Reddy, An Introduction to the Finite Element Method, 4th ed., Ch. 3 (second-order one-dimensional problems).

Question 5 of 7: One-dimensional finite element model of a composite wall

Question 5: One-dimensional finite element model of a composite wall (20 marks)