One-dimensional heat conduction in a rod with an internal source: 22-Mec-B10 Finite Element Analysis, December 2014, Q2

Question 2: One-dimensional heat conduction in a rod with an internal source (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 circular rod of 4 m length and 0.15 m diameter is modelled with two two-node conduction elements of 2 m each; the left end is held at a fixed temperature, the right end convects to the surroundings, every element generates heat internally, and heat also leaves uniformly through the curved side surface.

Given data
Quantity	Symbol	Value
Rod diameter / length	φ / L	0.15 m / 4 m (2 × 2 m elements)
Conductivity, element 1 / element 2	K_xx	300 / 600 W/(m·°C)
Internal source, element 1 / element 2	Q	400 / 500 W/m³
Outgoing lateral flux	q_out	10 W/m²
Prescribed temperature at node 1	T₁	500 °C
End convection coefficient	h	30 W/(m²·°C)
Ambient temperature	T_∞	100 °C

Find. The nodal temperatures T₂ and T₃, the conduction heat flux carried by element 1, and the heat flow rate crossing node 1 together with its direction.

Question 2 — two-element discretisation of the rod: prescribed temperature at node 1, uniform outgoing flux over the cylindrical surface, and end convection at node 3.

Approach. Build the two conduction matrices, add the end-convection stiffness at node 3, form the nodal load vector from the internal source and the lateral loss, assemble, impose the prescribed nodal temperature and solve; then recover the element flux from the temperature gradient and the nodal heat rate from the first assembled row.

Compute the section properties. Both the cross-sectional area (which scales conduction and the internal source) and the perimeter (which scales the lateral loss) are needed:$$A=\frac{\pi\phi^{2}}{4}=\frac{\pi(0.15)^{2}}{4}=0.017671\ \text{m}^2,\qquad P=\pi\phi=\pi(0.15)=0.471239\ \text{m}$$
Form the element conduction matrices. For a two-node one-dimensional conduction element, $[k_c]=\dfrac{AK_{xx}}{L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$, so$$\frac{AK^{(1)}_{xx}}{L_1}=\frac{0.017671(300)}{2}=2.6507\ \text{W/}^\circ\text{C},\qquad\frac{AK^{(2)}_{xx}}{L_2}=\frac{0.017671(600)}{2}=5.3014\ \text{W/}^\circ\text{C}$$The second element is twice as conductive, so it will sustain half the temperature drop of the first for the same heat flow.
Add the end convection. Convection acts only on the free end face at node 3, an area $A$ rather than a perimeter strip, so it contributes a single diagonal term and a single load term:$$hA=30(0.017671)=0.53014\ \text{W/}^\circ\text{C},\qquad hT_\infty A=30(100)(0.017671)=53.014\ \text{W}$$Because the question states that convection arises only at the right-hand end, no $\dfrac{hPL}{6}\begin{bmatrix}2&1\\1&2\end{bmatrix}$ surface term appears anywhere.
Build the nodal load vector. A uniform internal source is shared equally by the two nodes of its element, $\{f_Q\}=\dfrac{QAL}{2}\begin{Bmatrix}1\\1\end{Bmatrix}$, and the uniform outgoing lateral flux removes heat in the same lumped way, $\{f_q\}=-\dfrac{q_{out}PL}{2}\begin{Bmatrix}1\\1\end{Bmatrix}$. Per node,$$f^{(1)}=\frac{400(0.017671)(2)}{2}-\frac{10(0.471239)(2)}{2}=7.0686-4.7124=2.3562\ \text{W}$$$$f^{(2)}=\frac{500(0.017671)(2)}{2}-\frac{10(0.471239)(2)}{2}=8.8357-4.7124=4.1233\ \text{W}$$Both elements are net heat sources, but only modestly so — the lateral loss cancels roughly two thirds of the generation.
Assemble the global system. Superimposing the two element matrices, the end-convection term and the nodal loads gives$$\begin{bmatrix}2.6507&-2.6507&0\\-2.6507&7.9522&-5.3014\\0&-5.3014&5.8316\end{bmatrix}\begin{Bmatrix}T_1\\T_2\\T_3\end{Bmatrix}=\begin{Bmatrix}2.3562+R_1\\6.4795\\57.1377\end{Bmatrix}$$where $R_1$ is the unknown heat supplied at the prescribed-temperature node and the third load entry is $4.1233+53.014$.
(a) Impose $T_1=500\ ^\circ$C and solve for the free temperatures. Rows 2 and 3 give$$7.9522\,T_2-5.3014\,T_3=6.4795+2.6507(500)=1331.84$$$$-5.3014\,T_2+5.8316\,T_3=57.138$$Solving the pair,$$\boxed{T_2=441.73\ ^\circ\text{C},\qquad T_3=411.37\ ^\circ\text{C}}$$The temperature falls monotonically towards the convecting end, and the drop across the first element (58.3 °C) is roughly double that across the second (30.4 °C), exactly as the conductivity ratio predicts.
(b) Heat flux over element 1. Within a two-node element the temperature is linear, so the gradient is constant and Fourier's law gives a constant flux:$$q_x^{(1)}=-K^{(1)}_{xx}\frac{dT}{dx}=-K^{(1)}_{xx}\frac{T_2-T_1}{L_1}=-300\,\frac{441.73-500}{2}$$$$\boxed{q_x^{(1)}=+8741\ \text{W/m}^2\ \text{(in the}+x\ \text{direction)}}$$The corresponding heat rate carried by the element is $q_x^{(1)}A=8741(0.017671)=154.5$ W.
(c) Heat flow rate and direction at node 1. The first assembled row returns the external heat that must be supplied to hold node 1 at 500 °C:$$R_1=\big[2.6507(500)-2.6507(441.73)\big]-2.3562=154.47-2.36$$$$\boxed{R_1=152.1\ \text{W entering the rod at node 1, flowing in the }+x\ \text{direction}}$$A global energy audit confirms it: generation $31.81$ W plus the node-1 input $152.11$ W equals the lateral loss $18.85$ W plus the end convection $hA(T_3-T_\infty)=0.53014(411.37-100)=165.07$ W. The rod is a net importer of heat because the convecting end runs far above ambient.

Final results — Question 2
Quantity	Value
Temperature at node 2	441.73 °C
Temperature at node 3	411.37 °C
Heat flux in element 1	+8741 W/m² (towards +x)
Conduction rate in element 1	154.5 W
Heat flow rate at node 1	152.1 W, entering the rod (+x)
Heat rejected by end convection	165.1 W

Topic. Finite element formulation of one-dimensional steady conduction.

Key relations. $[k_c]=\dfrac{AK_{xx}}{L}[1,-1;-1,1]$; end convection $hA$ with load $hT_\infty A$; source $\{f_Q\}=\dfrac{QAL}{2}\{1,1\}^T$; specified outgoing surface flux $\{f_q\}=-\dfrac{q^{*}PL}{2}\{1,1\}^T$; Fourier $q_x=-K_{xx}\,dT/dx$.

Why this works. The weak form of the steady conduction equation splits into a conductivity term that is identical in structure to a bar stiffness, boundary terms that add a diagonal stiffness wherever the surface temperature itself drives the heat transfer (convection), and load terms wherever a known quantity of heat is added or removed (volumetric source, specified surface flux). Because the element shape functions are linear, the consistent source and flux vectors reduce to a simple half-and-half split between the two nodes. The distinction that decides the whole assembly here is geometric: the internal source scales with the cross-sectional area, the lateral flux with the perimeter, and the end convection with the end-face area alone.

Common pitfall. Applying the convection surface matrix hPL/6 [2 1; 1 2] along the rod when the paper says convection acts only on the end face, and losing the sign of the lateral flux. A specified outgoing flux is a negative nodal load; entering it positively turns a 19 W loss into a 19 W gain and shifts every temperature.

Source. Logan, A First Course in the Finite Element Method (6th ed.), §13.4–13.5 (one-dimensional conduction with convection); Reddy, An Introduction to the Finite Element Method (4th ed.), Ch. 3.

Question 2 of 7: One-dimensional heat conduction in a rod with an internal source

Question 2: One-dimensional heat conduction in a rod with an internal source (20 marks)