Consider a straight rectangular fin with length L = 3 cm, and thickness r= 1 cm.

Question

Consider a straight rectangular fin with length L = 3 cm, and thickness r= 1 cm. (The width of the fin may be considered as one unit length.) The fin is made of a material with a thermal conductivity of k-67 Wm. K. It is exposed to ambient air at To = 12°C with a convective heat transfer coefficient of h = 120 W/m2, K. The base surface temperature of the fin is constant at T'b = 85°C, while the fin tip is fully-insulated. To solve for the temperature distribution of the fin using the finite difference method, the fin is meshed with uniform grid of Ar Ay 5 mm. Adiabatic 13 14 15 16 17 18 (a) Use Table 4.2 of the textbook, and determine the case that is applicable for each node (e.g., Case 1 for node 1, Case 2 for node 2, etc.). b) Write the finite-difference equation for each node (c) Solve the finite-difference equations numerically, using MS Excel or other software, to (d) Use the appropriate analytical equation from Chap. 3 (Table 3.4) to calculate the determine the temperature for each node. temperature distribution T(x) of the fin. Compare the analytical results with the numerical results from part (c): 1.) Present the results in a table, similar as shown below; 2.) Plot the T(x) vs. x by using a solid line (for the analytical results, and symbol )for the numerical data points; 3.) Briefly discuss the comparison between the analytical and the numerical results. What are the differences in percentage? Analytical Ta) Node Numerical T) 10 12 Note: Make sure appropriate units are presented in all the results.)

Explanation / Answer

We consider the following assumptions for the analysis of heat flow through the fin:

Answer for Part a:

The following table show the nodes mapped with the respective cases

This node experiences convection, (i,e) Heat Exchange happens between the Atmosphere and Fin

Answer for Part b

Referring the Table 4.2 and the Solution Table for Part a, The Finite difference equations for each node can be obtained by substituting Hear Flux or Heat transfer co-effiecient to zero

At Node 7: Tb + T1 + T8 + T13 - 4T7 = 0 The same procedure repeats for nodes 8, 9, 10, 11.

At Node 1: 2T7 + TB + T2 - 4T1 = 0 The same procedure repeats for nodes 2,3,4,5

At Node 13: 2T7 + TB + T14 - 4T13 = 0 The same procedure repeats for nodes 14, 15, 16, 17

At Node 12: 2T11 + T6 + T18 - 2T12 = 0

Case Nodes Explanation Case 1- Interior Node 7, 8, 9, 10, 11 Nodes which are surrounded by other Nodes, present in the middle(in this problem's Mesh model) Case 5 - Node at plane surface with uniform Heat Flux 1,2,3,4,5,13,14,15,16, 17 We have to consider the assumption that, The Heat flow is unidirectional so there is no Convection laterally. Case 3 - Node at Plane surface with Convection 12

This node experiences convection, (i,e) Heat Exchange happens between the Atmosphere and Fin

Case 4 - Node at External Corner with Convection 6, 18 These nodes are present at the corner of the external tip of the fin which is insulated.

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