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Unformatted text preview: 2 = L g V & Integrating Eq. (2) twice gives 4 3 2 3 2 ) ( C y C y k g y T C y k g dy dT + + = + = & & Applying the two boundary conditions give B.C. 1: y =0 3 = = = C dy dT k y B.C. 2: y = L 2 4 2 ) ( L k g T C T L T & + = = Substituting, the temperature distribution becomes + = 2 2 2 1 2 ) ( L y k L g T y T & Maximum temperature occurs at y = 0, and it value is k L g T T T 2 ) ( 2 max & + = = which is equivalent to the result k T T T 2 ) ( 2 max V + = = obtained in Prob. 643. 620...
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 Fall '10
 Dr.DanielArenas
 Thermodynamics, Convection, Mass, Heat

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