Chapter 2 Heat Conduction EquationApplying the other boundary condition at rr=1, B. C. at : =1TgkrCCTgkrII=−+→=+&&44122221Substituting this Crelation into Eq. (b) and rearranging give 2TrTgkIwirewire()&(=+−41)(c) Plastic layer The mathematical formulation of heat transfer problem in the plastic can be expressed as 0=⎟⎠⎞⎜⎝⎛drdTrdrdwith Trand TI1=−=−∞kdT rdrhTrT[( )]22The solution of the differential equation is determined by integration to be rdTdrC=1→dTdrCr=1→Cr Cln=+12where Cand C12are arbitrary constants. Applying the boundary conditions give r = r1: Cr11 2211lnln+=→=−r = r2: +−∞kCrhC r CT1212 2[(ln)] →CTTrrkhrI12=−+∞lnSubstituting C1and C2into the general solution, the variation of temperature in plastic is determined to be CrTCrTrrkhrrrIplasticplasticlnln−=+−+∞121We have already utilized the first interface condition by setting the wire and ceramic layer temperatures equal to Tat 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=1TI=1must be the same: −
This is the end of the preview. Sign up
access the rest of the document.
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.