Chapter 2 Heat Conduction EquationAssumptions 1Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2Thermal conductivity is constant. 3 There is no heat generation in the pipe. PropertiesThe thermal conductivity is given to be k= 7.2 Btu/h⋅ft⋅°F. Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as 0=⎟⎠⎞⎜⎝⎛drdTrdrdand −=−∞kdT rdrhTTr()[(11)]T22160== °F(b) Integrating the differential equation once with respect to r gives Steam 250°F h=1.25L= 15 ftT =160°F C=1rdTdrDividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, dTdrCr=1C r Cln=+12where Cand C12are arbitrary constants. Applying the boundary conditions give r = r1: −+∞kCrCr C1111 2ln)]2C2r = r2: Tln2122=Solving for Csimultaneously gives 1
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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.