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Unformatted text preview: Chapter 4 Transient Heat Conduction 481 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center temperatures of each geometry in 10, 20, and 60 min are to be determined. Assumptions 1 Heat conduction in the cubic block is threedimensional, and thus the temperature varies in all x, y , and z directions. 2 Heat conduction in the cylindrical block is twodimensional, and thus the temperature varies in both axial x and radial r directions. 3 The thermal properties of the granite are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the oneterm approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of the granite are given to be k = 2.5 W/m. ° C and α = 1.15 × 106 m 2 /s. Analysis : Cubic block: This cubic block can physically be formed by the intersection of three infinite plane walls of thickness 2 L = 5 cm. After 10 minutes : The Biot number, the corresponding constants, and the Fourier number are 400 . ) C W/m. 5 . 2 ( ) m 025 . )( C . W/m 40 ( 2 = ° ° = = k hL Bi → = = λ 1 1 05932 10580 . . and A τ α = = × × = t L 2 6 115 10 0 025 1104 0 2 ( . ( . . . m / s)(10 min 60 s/ min) m) 2 2 To determine the center temperature, the product solution can be written as [ ] { } C 323 ° = = = = θ = θ τ λ ∞ ∞ ) , , , ( 369 . ) 0580 . 1 ( 500 20 500 ) , , , ( ) , , , ( ) , ( ) , , , ( 3 ) 104 . 1 ( ) 5932 . ( 3 1 3 wall block 2 2 1 t T e t T e A T T T t T t t i After 20 minutes τ α = = × × = t L 2 6 115 10 0 025 2 208 0 2 ( . ( . . . m / s)(20 min 60 s / min) m) 2 2 { } C 445 ° = → = = ) , , , ( 115 . ) 0580 . 1 ( 500 20 500 ) , , , ( 3 ) 208 . 2 ( ) 5932 . ( 2 t T e t T After 60 minutes τ α = = × × = t L 2 6 115 10 0 025 6 624 0 2 ( . ( . . . m / s)(60 min 60 s / min) m) 2 2 { } C 500 ° = → = = ) , , , ( 00109 . ) 0580 . 1 ( 500 20 500 ) , , , ( 3 ) 624 . 6 ( ) 5932 . ( 2 t T e t T Note that τ > 0.2 in all dimensions and thus the oneterm approximate solution for transient heat conduction is applicable. 470 T i = 20 ° C Hot gases 500 ° C 5 cm × 5 cm × 5 cm T i = 20 ° C 5 cm × 5 cm Chapter 4 Transient Heat Conduction Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius r o = D /2 = 2.5 cm and a plane wall of thickness 2 L = 5 cm. After 10 minutes : The Biot number and the corresponding constants for the long cylinder are 400 . ) C W/m. 5 . 2 ( ) m 025 . )( C . W/m 40 ( 2 = ° ° = = k hr Bi o → = = λ 1 1 08516 10931 . . and A To determine the center temperature, the product solution can be written as [ ] [ ] { }{ } C 331 ° = → = = = = ∞ ∞ ) , , ( 352 . ) 0931 . 1 ( ) 0580 . 1 ( 500 20 500 ) , , ( ) , , ( ) ,...
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 Spring '11
 ENgel
 Thermodynamics, Heat, Heat Transfer, TI, transient heat conduction

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