Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=20 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(95^{\circ} \mathrm{C}\), while the right side loses heat by convection to the surrounding air at \(T_{\infty}=15^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(10 \mathrm{~cm},(a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the nodal temperatures by solving those equations, and (c) evaluate the rate of heat transfer through the wall.

To determine the finite difference formulation for all nodes, we'll first calculate the number of nodes, based on the given nodal spacing. We have the thickness L = 0.4 m and the nodal spacing as 10 cm (0.1 m). Thus, the total number of nodes is \(n = \frac{L}{d} + 1 = \frac{0.4}{0.1} + 1 = 5\). The heat conduction equation for 1D steady-state condition is given by: \(\frac{d^2T}{dx^2} = 0\), where T is the temperature, and x is the spatial coordinate. Using finite difference approximation, we can rewrite the equation for the interior nodes (i=2, 3, 4) as: \(\frac{T_{i-1} - 2T_i + T_{i+1}}{d^2} = 0\) For the left boundary (i=1), we have the Dirichlet boundary condition of a constant temperature T1 = 95°C. For the right boundary (i=5), we have a Neumann boundary condition that accounts for convection at the wall: \(-\frac{T_5 - T_4}{d} = -h (T_5 - T_{\infty})\)

We'll form a system of linear equations for the nodal temperatures, based on our finite difference equations for each node: 1. \(T_1 = 95^{\circ}C\) 2. \(\frac{T_1 - 2T_2 + T_3}{d^2} = 0\) 3. \(\frac{T_2 - 2T_3 + T_4}{d^2} = 0\) 4. \(\frac{T_3 - 2T_4 + T_5}{d^2} = 0\) 5. \(-\frac{T_5 - T_4}{d} = -h (T_5 - T_{\infty})\) Substituting the given values of T1, d, h, and T_infinity and solving the system of equations, we obtain the nodal temperatures: \(T_2 = 75.6^{\circ}C, T_3 = 56.2^{\circ}C, T_4 = 36.8^{\circ}C, T_5 = 17.4^{\circ}C\)

With the nodal temperatures calculated, we can now determine the rate of heat transfer through the wall (Q). We'll calculate the heat flux at the left boundary (using the temperature difference between nodes 1 and 2) and multiply it by the surface area (A) to obtain the heat transfer rate: \(q = -k \frac{T_1 - T_2}{d}\) \(Q = A * q = A * (-k \frac{T_1 - T_2}{d})\) Substituting the given values and calculated nodal temperatures: \(Q = 20 \mathrm{~m}^{2} * (-2.3 \frac{95 - 75.6}{0.1}) = -886 \mathrm{~W}\) The negative sign indicates that the heat transfer is from left to right, as expected, with constant temperature higher on the left side. The rate of heat transfer through the wall is 886 W.