Generate the finite-difference solution for a 10-m rod with Δ x = 2 m, h′ = 0.05 m^−2, T∞ = 200 K, and the boundary conditions: Ta′ = 0 and Tb = 400 K. Note that the first condition means that the slope of the solution should approach zero at the rod’s left end. Aside from this case, also generate

Question

Generate the finite-difference solution for a 10-m rod with [latex]Δx = 2 m, h^′ = 0.05 m^{−2}, T_∞ = 200 K,[/latex] and the boundary conditions: [latex]T _a^′ = 0 ext{ and } T_b = 400 K.[/latex] Note that the first condition means that the slope of the solution should approach zero at the rod’s left end. Aside from this case, also generate the solution for dT/dx = −20 at x = 0.

Accepted Answer

Equation (24.18) can be used to represent node 0 as
[latex]\left(2+h^{\prime} \Delta x^2 ight) T_0-2 T_1=h^{\prime} \Delta x^2 T_{\infty}-2 \Delta x \frac{d T}{d x}[/latex] (24.18)
[latex]2.2T_0 − 2T_1 = 40[/latex]
We can write Eq. (24.16) for the interior nodes. For example, for node 1,
[latex]-T_{i-1}+\left(2+h^{\prime} \Delta x^2 ight) T_i-T_{i+1}=h^{\prime} x^2 T_{\infty}[/latex] (24.16)
[latex]−T_0 + 2.2T_1 − T_2 = 40[/latex]
A similar approach can be used for the remaining interior nodes. The final system of equations can be assembled in matrix form as
[latex]\left[\begin{array}{rrrrr}2.2 & -2 & & & \-1 & 2.2 & -1 & & \& -1 & 2.2 & -1 & \& & -1 & 2.2 & -1 \& & & -1 & 2.2\end{array} ight]\left\{\begin{array}{l}T_0 \T_1 \T_2 \T_3 \T_4\end{array} ight\}=\left\{\begin{array}{c}40 \40 \40 \40 \440\end{array} ight\}[/latex]
These equations can be solved for
[latex]T_0[/latex] = 243.0278
[latex]T_1[/latex] = 247.3306
[latex]T_2[/latex] = 261.0994
[latex]T_3[/latex] = 287.0882
[latex]T_4[/latex] = 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
[latex]\left[\begin{array}{ccccc}2.2 & -2 & & & \-1 & 2.2 & -1 & & \& -1 & 2.2 & -1 & \& & -1 & 2.2 & -1 \& & & -1 & 2.2\end{array} ight]\left\{\begin{array}{l}T_0 \T_1\T_2\T_3\T_4\end{array} ight\}=\left\{\begin{array}{c}120 \40 \40 \40 \440\end{array} ight\}[/latex]
which can be solved for
[latex]T_0[/latex] = 328.2710
[latex]T_1[/latex] = 301.0981
[latex]T_2[/latex] = 294.1448
[latex]T_3[/latex] = 306.0204
[latex]T_4[/latex] = 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.

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

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