Thermodynamics HW Solutions 153

Thermodynamics HW Solutions 153 - Chapter 2 Heat Conduction...

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Chapter 2 Heat Conduction Equation Applying the other boundary condition at rr = 1 , B. C. at : = 1 T g k rC CT g k r II =− + = + && 44 1 2 22 2 1 Substituting this C relation into Eq. ( b ) and rearranging give 2 TrT g k I wire wire () & ( =+ 4 1 ) ( c ) Plastic layer The mathematical formulation of heat transfer problem in the plastic can be expressed as 0 = dr dT r dr d with Tr and T I 1 = −= k dT r dr hTr T [( ) ] 2 2 The solution of the differential equation is determined by integration to be r dT dr C = 1 dT dr C r = 1 C r C ln = + 12 where C and C 1 2 are arbitrary constants. Applying the boundary conditions give r = r 1 : Cr 11 2 2 11 ln ln += = r = r 2 : + k C r hC r C T 1 2 12 2 [( ln ) ] C TT r r k hr I 1 2 = + ln Substituting C 1 and C 2 into the general solution, the variation of temperature in plastic is determined to be C r T C r T r r k hr r r I plastic plastic ln ln = + + 1 2 1 We have already utilized the first interface condition by setting the wire and ceramic layer temperatures equal to T at the interface . The interface temperature is determined from the second interface condition that the heat flux in the wire and the plastic layer at I = 1 T I = 1 must be the same:
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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.

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