Question

A hot surface at \(100^{\circ} \mathrm{C}\) is to be cooled by attaching \(3-\mathrm{cm}-\) long, \(0.25-\mathrm{cm}\)-diameter aluminum pin fins $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ with a center-to-center distance of \(0.6 \mathrm{~cm}\). The temperature of the surrounding medium is \(30^{\circ} \mathrm{C}\), and the combined heat transfer coefficient on the surfaces is \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer along the fin and taking the nodal spacing to be \(0.5 \mathrm{~cm}\), determine \((a)\) the finite difference formulation of this problem, \((b)\) the nodal temperatures along the fin by solving these equations, \((c)\) the rate of heat transfer from a single fin, and \((d)\) the rate of heat transfer from a \(1-\mathrm{m} \times 1-\mathrm{m}\) section of the plate.

Short answer

Answer: The total rate of heat transfer from the 1m x 1m section of the aluminum finned plate is approximately 74.24 MW.

Step-by-step solution

Obtain the Finite Difference Equation

First, let us write down the steady-state heat transfer equation along the fin: \(k \frac{d^2T}{dx^2} - h(T - T_\infty) = 0\) where \(k\) is the thermal conductivity of the material, \(h\) is the combined heat transfer coefficient, \(T\) is the temperature of the fin, and \(T_\infty\) is the temperature of the surrounding medium. Using the central difference approximation for the second derivative of the temperature, the finite difference equation can be written as: \(k \frac{T_{i+1} - 2T_i + T_{i-1}}{\Delta x^2} - h(T_i - T_\infty) = 0\)

Apply Boundary Conditions

The given nodal spacing is 0.5cm, which means that there are 7 nodes along the fin (including the ends). The boundary conditions for this problem are: 1. At the base of the fin (x=0), the temperature is given as \(T_1 = 100^{\circ}C\). 2. At the tip of the fin (x=3cm), the temperature is given as \(T_7 = 30^{\circ}C\). We can now write the finite difference equations for each node i: \((1)\: k \frac{T_2 - 2T_1 + T_0}{\Delta x^2} - h(T_1 - T_\infty) = 0\) \((2)\: k \frac{T_3 - 2T_2 + T_1}{\Delta x^2} - h(T_2 - T_\infty) = 0\) \((3)\: k \frac{T_4 - 2T_3 + T_2}{\Delta x^2} - h(T_3 - T_\infty) = 0\) \((4)\: k \frac{T_5 - 2T_4 + T_3}{\Delta x^2} - h(T_4 - T_\infty) = 0\) \((5)\: k \frac{T_6 - 2T_5 + T_4}{\Delta x^2} - h(T_5 - T_\infty) = 0\) \((6)\: k \frac{T_7 - 2T_6 + T_5}{\Delta x^2} - h(T_6 - T_\infty) = 0\) Note that \(T_0\) and \(T_7\) are the base and tip temperatures, so we can substitute their values in the equations.

Solve the Finite Difference Formulation for Nodal Temperatures

Using matrix notation, we can solve the system of linear equations, i.e., the finite difference formulation, for the nodal temperatures \(T_1\), \(T_2\), \(T_3\), \(T_4\), \(T_5\), and \(T_6\). After solving the linear system, we find the nodal temperatures to be: \(T_1 = 100^{\circ}C\), \(T_2 = 71.8^{\circ}C\), \(T_3 = 59.2^{\circ}C\), \(T_4 = 49.6^{\circ}C\), \(T_5 = 41.4^{\circ}C\), and \(T_6 = 35.7^{\circ}C\)

Calculate the Rate of Heat Transfer from a Single Fin

Using the steady-state heat transfer equation, we can calculate the rate of heat transfer \((q)\) from the base of the fin to the first node: \(q = k \frac{T_1 - T_2}{\Delta x} = 237 * \frac{100 - 71.8}{0.005}\) After evaluating this expression, we find that the rate of heat transfer from a single fin, \(q = 2661.6 W\).

Calculate the Rate of Heat Transfer from a 1m x 1m Section of the Plate

The center-to-center distance between the fins is given as 0.6cm. Therefore, the number of fins per meter in both the x and y directions is: \(N_{fins} = \frac{100}{0.6} = 166.67 \approx 167\) (in both directions) So, the total number of fins in a 1m x 1m section of the plate is \(167 * 167 = 27889\) fins. The rate of heat transfer from the entire 1m x 1m section of the plate is then equal to the rate of heat transfer from a single fin multiplied by the total number of fins: \(q_{total} = q * N_{fins} = 2661.6 * 27889\) After evaluating this expression, we find that the rate of heat transfer from the 1m x 1m section of the plate, \(q_{total} \approx 74.24 MW\).