Question 24.6: Generate the finite-difference solution for a 10-m rod with ...

Equation (24.18) can be used to represent node 0 as
\left(2+h^{\prime} \Delta x^2\right) T_0-2 T_1=h^{\prime} \Delta x^2 T_{\infty}-2 \Delta x \frac{d T}{d x}                  (24.18)
2.2T_0 − 2T_1 = 40
We can write Eq. (24.16) for the interior nodes. For example, for node 1,
-T_{i-1}+\left(2+h^{\prime} \Delta x^2\right) T_i-T_{i+1}=h^{\prime} x^2 T_{\infty}         (24.16)
−T_0 + 2.2T_1 − T_2 = 40
A similar approach can be used for the remaining interior nodes. The final system of equations can be assembled in matrix form as
\left[\begin{array}{rrrrr}2.2 & -2 & & & \\-1 & 2.2 & -1 & & \\& -1 & 2.2 & -1 & \\& & -1 & 2.2 & -1 \\& & & -1 & 2.2\end{array}\right]\left\{\begin{array}{l}T_0 \\T_1 \\T_2 \\T_3 \\T_4\end{array}\right\}=\left\{\begin{array}{c}40 \\40 \\40 \\40 \\440\end{array}\right\}
These equations can be solved for
T_0 = 243.0278
T_1 = 247.3306
T_2 = 261.0994
T_3 = 287.0882
T_4 = 330.4946
As displayed in Fig. 24.10, the solution is flat at x = 0 due to the zero derivative condition and then curves upward to the fixed condition of T = 400 at x = 10.
For the case where the derivative at x = 0 is set to −20, the simultaneous equations are
\left[\begin{array}{ccccc}2.2 & -2 & & & \\-1 & 2.2 & -1 & & \\& -1 & 2.2 & -1 & \\& & -1 & 2.2 & -1 \\& & & -1 & 2.2\end{array}\right]\left\{\begin{array}{l}T_0 \\T_1\\T_2\\T_3\\T_4\end{array}\right\}=\left\{\begin{array}{c}120 \\40 \\40 \\40 \\440\end{array}\right\}
which can be solved for
T_0 = 328.2710
T_1 = 301.0981
T_2 = 294.1448
T_3 = 306.0204
T_4 = 339.1002
As in Fig. 24.10, the solution at x = 0 now curves downward due to the negative derivative we imposed at the boundary.