Question 24.1: Use calculus to solve Eq. (24.6) for a 10-m rod with h′ = 0....
Use calculus to solve Eq. (24.6) for a 10-m rod with h^′ = 0.05 m^{−2}[h = 1 J/(m^2 · K · s), r = 0.2 m, k = 200 J/(s · m · K)], T_∞ = 200 K, and the boundary conditions:
T (0) =300 K T (10) = 400 K
0 = \frac{d^2T}{dx^2} + h^′ (T_∞ − T ) (24.6)
This ODE can be solved in a number of ways. A straightforward approach is to first express the equation as
\frac{d^2 T}{d x^2}-h^{\prime} T=-h^{\prime} T_{\infty}
Because this is a linear ODE with constant coefficients, the general solution can be readily obtained by setting the right-hand side to zero and assuming a solution of the form T = e^{λx}. Substituting this solution along with its second derivative into the homogeneous form of the ODE yields
which can be solved for λ = ±\sqrt{h^′} . Thus, the general solution is
T=A e^{\lambda x}+B e^{-\lambda x}
where A and B are constants of integration. Using the method of undetermined coefficients we can derive the particular solution T = T_∞. Therefore, the total solution is
T=T_{\infty}+A e^{\lambda x}+B e^{-\lambda x}The constants can be evaluated by applying the boundary conditions
\begin{array}{l} T_a=T_{\infty}+A+B \\ T_b=T_{\infty}+A e^{\lambda L}+B e^{-\lambda L} \end{array}These two equations can be solved simultaneously for
\begin{array}{l} A=\frac{\left(T_a-T_{\infty}\right) e^{-\lambda L}-\left(T_b-T_{\infty}\right)}{e^{-\lambda L}-e^{\lambda L}} \\ B=\frac{\left(T_b-T_{\infty}\right)-\left(T_a-T_{\infty}\right) e^{\lambda L}}{e^{-\lambda L}-e^{\lambda L}} \end{array}Substituting the parameter values from this problem gives A = 20.4671 and B = 79.5329.
Therefore, the final solution is
T=200+20.4671 e^{\sqrt{0.05 x}}+79.5329 e^{-\sqrt{0.05 x}} (24.7)
As can be seen in Fig. 24.3, the solution is a smooth curve connecting the two boundary temperatures. The temperature in the middle is depressed due to the convective heat loss to the cooler surrounding gas.
