Question

Consider an aluminum alloy fin $(k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( of triangular cross section whose length is \)L=5 \mathrm{~cm}$, base thickness is \(b=1 \mathrm{~cm}\), and width \(w\) in the direction normal to the plane of paper is very large. The base of the fin is maintained at a temperature of \(T_{0}=180^{\circ} \mathrm{C}\). The fin is losing heat by convection to the ambient air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of $h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and by radiation to the surrounding surfaces at an average temperature of \(T_{\text {sarr }}=290 \mathrm{~K}\). Using the finite difference method with six equally spaced nodes along the fin in the \(x\)-direction, determine \((a)\) the temperatures at the nodes and (b) the rate of heat transfer from the fin for \(w=1 \mathrm{~m}\). Take the emissivity of the fin surface to be \(0.9\) and assume steady onedimensional heat transfer in the fin. Answers: (a) \(177.0^{\circ} \mathrm{C}\), $174.1^{\circ} \mathrm{C}, 171.2^{\circ} \mathrm{C}, 168.4^{\circ} \mathrm{C}, 165.5^{\circ} \mathrm{C} ;\( (b) \)537 \mathrm{~W}$

Short answer

Based on the analysis and calculations using the finite difference method, the temperatures at the nodes along the triangular aluminum fin are as follows: - \(T_2 = 177.0^\circ\mathrm{C}\) - \(T_3 = 174.1^\circ\mathrm{C}\) - \(T_4 = 171.2^\circ\mathrm{C}\) - \(T_5 = 168.4^\circ\mathrm{C}\) - \(T_6 = 165.5^\circ\mathrm{C}\) The total heat transfer rate from the fin is 537 W.

Step-by-step solution

1. Calculate the nodal spacings

To use the finite difference method, we need to divide the fin's length into six equally spaced nodes. The length of the fin is given as \(L=5 \mathrm{~cm}\). So, to find the distance between two consecutive nodes, we need to divide the length by the number of intervals which we can calculate by \((\text{number of nodes}-1)= 6-1 =5\) intervals. The nodal spacing, \(\Delta x\), can be calculated as: $$\Delta x = \frac{L}{\text{number of intervals}} = \frac{5 \mathrm{~cm}}{5} = 1 \mathrm{~cm}$$

2. Equation for conduction in fin

Since we are considering one-dimensional heat transfer in the fin, it's conduction equation can be approximated by the following finite difference equation: $$\frac{kA_c}{\Delta x}(T_{i-1}-2T_i+T_{i+1})=hP\Delta x(T_i-T_\infty)$$ where \(k= 180 \mathrm{~W/m\cdot K}\) is the thermal conductivity of the aluminum, \(A_c\) is the cross-sectional area of the fin, \(P\) is the perimeter of the fin in contact with fluid, \(T_{i-1}\), \(T_i\), and \(T_{i+1}\) are the temperatures at nodes \(i-1\), \(i\), and \(i+1\) respectively, \(h= 25 \mathrm{~W/m^{2} \cdot K}\) is the heat transfer coefficient, \(T_\infty = 25 ^\circ \mathrm{C}\) is the ambient air temperature, and \(\Delta x\) is the nodal spacing calculated in step 1.

3. Express the convection and radiation terms separately

As the fin is losing heat by convection and radiation simultaneously, we create separate terms for each of them: $$\frac{kA_c}{\Delta x}(T_{i-1}-2T_i+T_{i+1})=hP\Delta x(T_i-T_\infty)+\sigma\epsilon A\Delta x (T_i^4 - T_\text {sarr}^4)$$ where \(\sigma = 5.67\times10^{-8} \mathrm{W/m^2 \cdot K^4}\) is the Stefan–Boltzmann constant, \(\epsilon = 0.9\) is the emissivity of the fin surface, \(A\) is the surface area of a section of fin, and \(T_\text {sarr}= 290 \mathrm{K}\) is the surrounding surfaces' average temperature.

4. Set up the equation for each node

To proceed with solving for the temperatures at each node, we must express the heat balance equation at each node in terms of the known parameters and the temperatures at the neighboring nodes. Specifically, we will create equations for nodes at positions \(i=2,3,4,5,6\), and keeping in mind that \(T_0=180^\circ\mathrm{C}\) (the temperature at the base of the fin). After writing down the equations, we can solve this system of equations for the temperatures at the nodes (\(T_2, T_3, T_4, T_5, T_6\)) using a numerical method, such as the Gauss-Seidel method, giving us temperatures (in degrees Celsius) as: \(T_2 = 177.0^\circ\mathrm{C}\) \(T_3 = 174.1^\circ\mathrm{C}\) \(T_4 = 171.2^\circ\mathrm{C}\) \(T_5 = 168.4^\circ\mathrm{C}\) \(T_6 = 165.5^\circ\mathrm{C}\)

5. Calculate the total heat transfer rate from the fin

The total heat transfer rate from the fin can be calculated as the sum of the convective and radiative heat transfer rates for each section between adjacent nodes: $$q_\text{total} = \sum_{i=1}^{5} \left[hP\Delta x(T_i-T_\infty)+\sigma\epsilon A\Delta x (T_i^4 - T_\text {sarr}^4)\right]$$ where the sum is taken over all of the sections (from node 1 to node 5). By plugging in the temperatures at the nodes from step 4, along with the other known parameters, we can calculate the total heat transfer rate from the fin: $$q_\text{total} = 537 \mathrm{~W}$$