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Unformatted text preview: 150 180 200 180 150 150 180 200 180 150 180 200 180 1 2 3 4 5 6 7 8 9 8 6 4 2 9 7 3 1 T T T T T T T T = = = = = = Therefore, there are there are only 3 unknown nodal temperatures, , and thus we need only 3 equations to determine them uniquely. Also, we can replace the symmetry lines by insulation and utilize the mirrorimage concept when writing the finite difference equations for the interior nodes. 5 3 1 and , , T T T 2 2 5 1 5 2 2 1 4 / 4 : (interior) 3 Node 4 / ) 2 200 ( : (interior) 2 Node 4 / ) 2 180 180 ( : (interior) 1 Node T T T T T T T T = = + + = + + = Solving the equations above simultaneously gives C 190 C 185 = = = = = = = = = 8 6 5 4 2 9 7 3 1 T T T T T T T T T Discussion Note that taking advantage of symmetry simplified the problem greatly. 540...
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This note was uploaded on 01/22/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.
 Fall '10
 Dr.DanielArenas
 Thermodynamics, Mass, Heat

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