Question 24.7: Use the finite-difference approach to simulate the temperatu...
Use the finite-difference approach to simulate the temperature of a heated rod subject to both convection and radiation:
0=\frac{d^2}{d x^{\prime}}+h^{\prime}\left(T_{\infty}-T\right)+\sigma^{\prime}\left(T_{\infty}^4-T^4\right)where σ^′ = 2.7 × 10^{−9} K^{−3} m^{−2}, L = 10 m, h^′ = 0.05 m^{−2}, T_∞ = 200 K, T (0) = 300 K, \text{ and } T (10) = 400 K. Use four interior nodes with a segment length of Δ x = 2 m. Recall that we solved the same problem with the shooting method in Example 24.4.
Using Eq. (24.19) we can successively solve for the temperatures of the rod’s interior nodes. As with the standard Gauss-Seidel technique, the initial values of the interior nodes are zero with the boundary nodes set at the fixed conditions of T_0 = 300 and T_5 = 400. The results for the first iteration are
T_i=\frac{h^{\prime} \Delta x^2 T_{\infty}+\sigma^{\prime} \Delta x^2\left(T_{\infty}^4-T_i^4\right)+T_{i-1}+T_{i+1}}{2+h^{\prime} \Delta x^2} (24.19)
\begin{array}{l} T_1=\frac{0.05(2)^2 200+2.7 \times 10^{-9^{\prime}}(2)^2\left(200^4-0^4\right)+300+0}{2+0.05(2)^2}=159.2432 \\T_2=\frac{0.05(2)^2 200+2.7 \times 10^{-9^{\prime}}(2)^2\left(200^4-0^4\right)+159.2432+0}{2+0.05(2)^2}=97.9674 \end{array}
\begin{array}{l}T_3=\frac{0.05(2)^2 200+2.7 \times 10^{-9^{\prime}}(2)^2\left(200^4-0^4\right)+97.9674+0}{2+0.05(2)^2}=70.4461 \\T_4=\frac{0.05(2)^2 200+2.7 \times 10^{-9^{\prime}}(2)^2\left(200^4-0^4\right)+70.4461+400}{2+0.05(2)^2}=226.8704\end{array}
The process can be continued until we converge on the final result:
T_0 = 300
T_1 = 250.4827
T_2 = 236.2962
T_3 = 245.7596
T_4 = 286.4921
T_5 = 400
These results are displayed in Fig. 24.11 along with the result generated in Example 24.4 with the shooting method.
