Thermodynamics HW Solutions 437

# Thermodynamics HW Solutions 437 - 150 180 200 180 150 150...

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Chapter 5 Numerical Methods in Heat Conduction 5-47 A long solid body is subjected to steady two-dimensional heat transfer. The unknown nodal temperatures are to be determined. Assumptions 1 Heat transfer through the body is given to be steady and two-dimensional. 2 There is no heat generation in the body. Properties The thermal conductivity is given to be k = 45 W/m °C. Analysis The nodal spacing is given to be Δ x = Δ x = l =0.01 m, and the general finite difference form of an interior node for steady two-dimensional heat conduction for the case of no heat generation is expressed as 4 / ) ( 0 4 bottom right top left node 2 node node bottom right top left T T T T T k l g T T T T T + + + = = + + + + There is symmetry about the horizontal, vertical, and diagonal lines passing through the midpoint, and thus we need to consider only 1/8 th of the region. Then, 180 200 180
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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 mirror-image 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. 5-40...
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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.

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