Problem 1: Finite difference method
The temperature distribution $T(r)$ in an annular fin of inner radius $r_1$ and outer radius $r_2$ is described by the equation +\begin{equation} +r \frac{d^2 T}{dr^2} + \frac{dT}{dr} - rm^2 (T - T_{\infty}) = 0 \;, +\end{equation} +where $r$ is the radial distance from the centerline (the independent +variable) and $m^2$ is a constant that depends on the heat transfer coefficient, thermal conductivity, and thickness of the annulus. Assuming we choose a spatial step size $\Delta r$,
+
+ a.) Write the finite-difference representation of the ODE (that applies at a location $r_i$), using central differences.
+b.) Based on the last part, write the recursion formula.
+c.) The boundary condition at the outer radius $r = r_2$ is described by convection heat transfer: +\begin{equation} +-k \left. \frac{dT}{dr} \right|_{r=r_2} = h \left[ T(r=r_2) - T_{\infty} \right] \;. +\end{equation} +Write the boundary condition at $r = r_2$ in recursion form (i.e., the equation you would implement into your system of equations to solve for temperature).
+Solution
a.) Replace the derivatives in the given ODE with finite differences, and replace any locations with the $i$ location: +\begin{equation} +r_i \frac{T_{i-1} - 2T_i + T_{i+1}}{\Delta r^2} + \frac{T_{i+1} - T_{i-1}}{\Delta r} - r_i m^2 (T_i - T_{\infty}) = 0 +\end{equation} +or +\begin{equation} +r_i (T_{i-1} - 2T_i + T_{i+1}) + \frac{\Delta r}{2} (T_{i+1} - T_{i-1}) - r_i m^2 \Delta r^2 (T_i - T_{\infty}) = 0 +\end{equation}
+b.) Rearrange and combine terms: +\begin{align} +r_i (T_{i-1} - 2T_i + T_{i+1}) + \frac{\Delta r}{2} (T_{i+1} - T_{i-1}) - r_i m^2 \Delta r^2 (T_i - T_{\infty}) &= 0 \\ +r_i (T_{i-1} - 2T_i + T_{i+1}) + \frac{\Delta r}{2} (T_{i+1} - T_{i-1}) - r_i m^2 \Delta r^2 T_i &= -r_i m^2 \Delta r^2 T_{\infty} \\ +\left(r_i - \frac{\Delta r}{2}\right) T_{i-1} + \left( -2 r_i - r_i m^2 \Delta r^2 \right) T_i + \left( r_i + \frac{\Delta r}{2} \right) T_{i+1} &= -r_i m^2 \Delta r^2 T_{\infty} +\end{align}
+c.) We can use a backward difference to approximate the $dT/dr$ term. $T_n$ represents the temperature at node $n$ where $r_n = r_2$: +\begin{align} +-k \frac{T_n - T_{n-1}}{\Delta r} &= h (T_n - T_{\infty}) \\ +-k (T_n - T_{n-1}) &= h \Delta r (T_n - T_{\infty}) \\ +k T_{n-1} - (k + h\Delta r) T_n &= -h \Delta r T_{\infty} +\end{align}
+Problem 2: eigenvalue
Given the equation $y^{\prime\prime} + 9 \lambda^2 y = 0$ with $y(0) = 0$ and $y(2) = 0$,
+a.) Find the expression that gives all eigenvalues ($\lambda$). What is the eigenfunction?
+b.) Calculate the principal eigenvalue.
+Solution
a.) +\begin{gather} +y(x) = A \sin (3 \lambda x) + B \cos (3 \lambda x) \\ +\text{Apply BCs: } y(x=0) = 0 = A \sin(0) + B \cos(0) = B \\ +\therefore B = 0 \\ +y(x) = A \sin (3 \lambda x) \\ +y(x=2) = 0 = A \sin (3 \lambda 2) \\ +A \neq 0 \text{ so } \sin(3 \lambda 2) = \sin(6 \lambda) = 0 \therefore 6 \lambda = n \pi \quad n=1,2,3,\ldots \\ +\lambda = \frac{n \pi}{6} \quad n=1,2,3,\ldots,\infty +\end{gather}
+The eigenfunction is then the solution function associated with an eigenvalue: +\begin{equation} +y_n = A_n \sin \left( \frac{n \pi x}{2} \right) \quad n = 1, 2, 3, \ldots, \infty +\end{equation}
+b.) The principal eigenvalue is just that associated with $n = 1$: +\begin{equation} +\lambda_p = \lambda_1 = \frac{\pi}{6} +\end{equation}
+
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