Question

Consider steady two-dimensional heat transfer in a long solid bar $(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( of square cross section ( \)3 \mathrm{~cm} \times 3 \mathrm{~cm}$ ) with the prescribed temperatures at the top, right, bottom, and left surfaces to be $100^{\circ} \mathrm{C}, 200^{\circ} \mathrm{C}, 300^{\circ} \mathrm{C}\(, and \)500^{\circ} \mathrm{C}$, respectively. Heat is generated in the bar uniformly at a rate of $\dot{e}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\(. Using a uniform mesh size \)\Delta x=\Delta y=1 \mathrm{~cm}\(, determine \)(a)$ the finite difference equations and \((b)\) the nodal temperatures with the Gauss-Seidel iterative method.

Short answer

Question: Calculate the nodal temperatures within a square solid bar given the following conditions: thermal conductivity (\(k\)) of 25 \(W/mK\), dimensions of 3 cm by 3 cm, surface temperatures of 100°C, 200°C, 300°C, and 500°C, a heat generation rate of \(5 \times 10^{6} W/m^3\), and a mesh size of 1 cm. Use the Gauss-Seidel iterative method for the calculations.

Step-by-step solution

Parameters and Constants

First, record all the given data which includes the thermal conductivity (\(k\)), dimensions of the bar, prescribed temperatures at the surfaces, continuous heat generation, and mesh size. Thermal Conductivity, \(k = 25 \frac{W}{mK}\) Dimensions of the bar: \(3cm \times 3cm\) Surface Temperatures: Top (\(T_t = 100^{\circ}C\)), Right (\(T_r = 200^{\circ}C\)), Bottom (\(T_b = 300^{\circ}C\)), Left (\(T_l = 500^{\circ}C\)) Heat generation rate, \(\dot{e} = 5 \times 10^{6} \frac{W}{m^3}\) Mesh size, \(\Delta x = \Delta y = 1cm = 0.01m\)

Derive the Finite Difference Equation

For steady two-dimensional heat transfer, the governing equation can be expressed as: \(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=\frac{\dot{e}}{k}\) Using the finite difference approximations for the second-order derivatives, we get: \(\frac{T_{i+1,j}-2T_{i,j}+T_{i-1,j}}{(\Delta x)^2}+\frac{T_{i,j+1}-2T_{i,j}+T_{i,j-1}}{(\Delta y)^2}=\frac{\dot{e}}{k}\) Since we have a uniform mesh size (\(\Delta x=\Delta y\)), we can simplify the equation to: \(T_{i+1,j}+T_{i-1,j}+T_{i,j+1}+T_{i,j-1}-4T_{i,j}=\frac{\dot{e}(\Delta x)^2}{k}\) This is the finite difference equation in terms of nodal temperatures.

Define the Gauss-Seidel Iterative Method

The Gauss-Seidel iterative method updates the nodal temperatures using the neighbors' values. We will apply this method to the problem at hand. The update formula for the nodal temperature can be obtained by rearranging the finite difference equation, isolating the term \(T_{i,j}\): \(T_{i,j}^{(k+1)}=\frac{1}{4}\left(T_{i+1,j}^{(k)}+T_{i-1,j}^{(k+1)}+T_{i,j+1}^{(k)}+T_{i,j-1}^{(k+1)}-\frac{\dot{e}(\Delta x)^2}{k}\right)\) Where \(k\) denotes the iteration number.

Implementation of Gauss-Seidel Iterative Method

We now proceed to implement the iterative method. We initialize the nodal temperatures using the given surface temperatures: \(T_t, T_r, T_b, T_l\). Iteratively update the nodal temperatures using the above equation until convergence is reached, i.e., the nodal temperatures change very little between iterations. Adjust the stopping criterion accordingly.

Calculate the Nodal Temperatures

Following the iterative method implementation, calculate the nodal temperatures until convergence is achieved. The result will be the temperatures at the nodes within the solid bar. In conclusion, we have derived the finite difference equation for this problem and used the Gauss-Seidel iterative method to find the nodal temperatures within the bar.