Thermodynamics HW Solutions 409

Thermodynamics HW Solutions 409 - for the interior nodes 1...

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Chapter 5 Numerical Methods in Heat Conduction Solving the 5 equations above simultaneously for the 5 unknown nodal temperatures gives T 1 =177.0 ° C, T 2 =174.1 ° C, T 3 =171.2 ° C, T 4 =168.4 ° C, and T 5 =165.5 ° C ( b ) The total rate of heat transfer from the fin is simply the sum of the heat transfer from each volume element to the ambient, and for w = 1 m it is determined from ] ) 273 [( ) ( 4 surr 4 5 0 surface, 5 0 surface, 5 0 element, fin T T A T T hA Q Q m m m m m m m m + + = = = = = εσ Noting that the heat transfer surface area is wx Δ /cos θ for the boundary nodes 0 and 5, and twice as large
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Unformatted text preview: for the interior nodes 1, 2, 3, and 4, we have [ ] W 533 = ]} ) 273 [( ] ) 273 [( 2 ] ) 273 [( 2 ] ) 273 [( 2 ] ) 273 [( 2 ] ) 273 {[( cos ) ( ) ( 2 ) ( 2 ) ( 2 ) ( 2 ) ( cos 4 surr 4 5 4 surr 4 4 4 surr 4 3 4 surr 4 2 4 surr 4 1 4 surr 4 5 4 3 2 1 fin T T T T T T T T T T T T x w T T T T T T T T T T T T x w h Q − + + − + + − + + − + + − + + − + Δ + − + − + − + − + − + − Δ = ∞ ∞ ∞ ∞ ∞ ∞ & 5-12...
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