Consider a nuclear fuel element \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that can be modeled as a plane wall with thickness of \(4 \mathrm{~cm}\). The fuel element generates \(3 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) of heat uniformly. Both side surfaces of the fuel element are cooled by liquid with temperature of \(80^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(8000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using a uniform nodal spacing of \(8 \mathrm{~mm},(a)\) obtain the finite difference equations, \((b)\) determine the nodal temperatures by solving those equations, and (c) compare the surface temperatures of both sides of the fuel element with analytical solution.

#Answer: (a) The finite difference equations for the given system are: At node 1: $\frac{k}{\Delta x^2}(T_2 - T_1) + q'' - \frac{h \Delta x}{2}(T_1 - T_\infty) = 0$ At nodes 2 - 5: $\frac{k}{\Delta x^2}(T_{i+1} - 2T_i + T_{i-1}) + q'' = 0$ At node 6: $\frac{k}{\Delta x^2}(T_6 - T_5) + q'' + \frac{h \Delta x}{2}(T_6 - T_\infty) = 0$ (b) The nodal temperatures, $T_1, T_2, T_3, T_4, T_5,$ and $T_6$, can be obtained by solving the system of linear equations detailed in the solution. (c) To compare the surface temperatures with the analytical solution, we need to find the temperatures $T_0$ and $T_L$ from the conduction-convection equation for a plane wall with volumetric heat generation. Then, compare these values with the obtained nodal temperatures, $T_1$ and $T_6$. The difference between these values shows the accuracy of our numerical solution.