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Unformatted text preview: k r g dr dT r dr d & and then integrating once with respect to r , 1 2 2 C k r g dr dT r + − = & Rearranging the differential equation again r C k r g dr dT 1 2 + − = & and finally integrating again with respect to r , we obtain 2 1 2 ln 4 ) ( C r C k r g r T + + − = & where C and C 1 2 are arbitrary constants. Applying the boundary conditions give r = r 1 : 2 1 1 2 1 1 ln 4 ) ( C r C k r g r T + + − = & r = r 2 : 2 2 1 2 2 2 ln 4 ) ( C r C k r g r T + + − = & Substituting the given values, these equations can be written as 2 1 2 ) 15 . ln( ) 20 ( 4 ) 15 . )( 894 , 37 ( 60 C C + + − = 2 1 2 ) 20 . ln( ) 20 ( 4 ) 20 . )( 894 , 37 ( 80 C C + + − = Solving for C simultaneously gives C 2 C 2 1 and 2 . 257 34 . 98 2 1 = = C C Substituting C into the general solution, the variation of temperature is determined to be 1 and 279...
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This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.
 Fall '10
 Dr.DanielArenas
 Thermodynamics, Mass, Heat

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