HW2Soln - Problem 16.1 1 2T 2T T = = 0 , For only radial...

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Problem 16.1 a. Steady State: 0 = t T , For only radial heat conduction: 0 1 2 2 2 2 2 = = z T T r θ Fourier’s field equation reduces to 0 1 1 2 2 = = + r T r r r r T r r T b. Integrating part a twice () 2 1 ln K r K T + = Applying the two boundary conditions 2 1 ln K r K T i i + = ( ) 2 1 ln K r K T o o + = Subtracting these two equations: = o r i r o i T T K ln 1 Using this relationship K 2 is determined as: i o r i r o i i r T T T K ln ln 2 = or o o r i r o i o r T T T K ln ln 2 = The general temperature profile solution is: i i o r i r o i i o r i r o i i o r i r o i T r r T T r T T T r T T T + = + = ln ln ln ln ln ln or o o o r i r o i o o r i r o i o o r i r o i T r r T T r T T T r T T T + = +
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This note was uploaded on 09/07/2009 for the course CBME Kinetics & taught by Professor Lobban during the Spring '09 term at The University of Oklahoma.

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HW2Soln - Problem 16.1 1 2T 2T T = = 0 , For only radial...

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