Page 124 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
P. 124

STEADY STATE HEAT CONDUCTION IN ONE DIMENSION
                        116
                                                                          T w
                                                      r i
                                                         r o


                                                      h, T a

                                 Figure 4.10  Radial heat conduction in an infinitely long cylinder

                        where T w is the inside wall temperature, T o is the outside wall temperature, k is the
                        thermal conductivity, h is the heat transfer coefficient at the outside surface and T a is the
                        atmospheric temperature.
                           Integrating Equation 4.57, we obtain

                                                                                            (4.59)
                                                    kT = C 1 ln r + C 2
                        Subjecting the above equation to the boundary conditions of Equation 4.58 results in
                                        C 1 =−hr o (T o − T a )  and C 2 = kT w − C 1 ln r i  (4.60)

                           Substituting the constants and rearranging Equation 4.59, we obtain the exact solution as
                                                   (T − T w )  hr o  r i
                                                            =     ln                        (4.61)
                                                   (T o − T a )  k  r o
                           With the use of the finite element method and assuming a linear variation of temperature,
                        the resulting stiffness matrix is given by


                                            T                  T
                                  [K] =   [B] [D][B]d
 +   h[N] [N]dA s
                                        
                A s
                                              1
                                              
                                         r o
                                                     1 1                N i
                                            −
                                             l 
                                     =          k −       2πr dr +   h      N i N j dA s
                                              1       l  l              N j
                                           
                                        r i                         A s
                                              l
                                       2πk (r i + r j )     1 −1      00
                                     =                      + 2πr o h                       (4.62)
                                         l    2     −1   1          01
                        per unit length of a cylinder. In the above equation, the variation of r is expressed as
                        r = N i r i + N j r j . The surface area per unit length is A s = 2πr o . The loading vector is

                                                          T               0
                                            {f}=    hT a [N] dA s = hT a 2πr o              (4.63)
                                                                          1
                                                  A s
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