Thermodynamics HW Solutions 485

Thermodynamics HW Solutions 485 - node (node 5) is...

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Chapter 5 Numerical Methods in Heat Conduction Therefore, any time step less than 179 s can be used to solve this problem. For convenience, we choose the time step to be Δ t = 60 s. Then the mesh Fourier number becomes 072 . 0 ) m 1 . 0 ( s) /s)(60 m 10 12 ( 2 2 6 2 = × = Δ = l t α τ (for Δ t = 60 s) Using the specified initial condition as the solution at time t = 0 (for i = 0), sweeping through the 3 equations above will give the solution at intervals of 1 min. Using a computer, the solution at the center
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Unformatted text preview: node (node 5) is determined to be 217.2C , 302.8C, 379.3C, 447.7C, 508.9C, 612.4C, 695.1C, and 761.2C at 10, 15, 20, 25, 30, 40, 50, and 60 min, respectively. Continuing in this manner, it is observed that steady conditions are reached in the medium after about 6 hours for which the temperature at the center node is 1023C . 5-88...
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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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