Page 147 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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STEADY STATE HEAT CONDUCTION IN MULTI-DIMENSIONS
                                                        2
                                                            a
                                               l h = 1.2 w/cm °C, T = 30°C  k                 139
                                                                        100°C
                                     q = 2 w/cm 2
                                                                                 5 cm





                                               i                     j
                                                        5 cm


                        Figure 5.12 Heat conduction in a square plate. Approximated using a rectangular (square)
                        element


                        again, on simplifying we obtain
                                                               
                                                            5.7
                                                               
                                                              8.3
                                                               
                                                       {f}=                                 (5.45)
                                                           97.7
                                                               
                                                             93.3
                                                               
                           Therefore, the final form of the set of simultaneous equations can be written as
                                                                          
                                             8.0 −2.0 −4.0 −2.0
                                                                  T 1    5.7
                                                                            
                                            −2.0  8.0 −2.0 −4.0           8.3 
                                        1                        T 2
                                                                     =                    (5.46)
                                        6   −4.0 −2.0  20.0  4.0    T 3  97.7
                                                                            
                                            −2.0 −4.0   4.0  20.0        93.3 
                                                                   T 4
                           The temperatures at points 2 and 3 are known. Substitution into the above system results
                        in the following simultaneous equations,
                                                      8T 1 − 2T 4 = 634.2
                                                   −2T 1 + 20T 4 = 559.8                    (5.47)
                                                                               ◦
                                                                                               ◦
                           The solution of the above simultaneous equation gives T 4 = 36.85 C and T 1 = 88.48 C.


                        5.4 Plate with Variable Thickness


                        The conduction heat transfer in a plate with variable thickness is essentially a three-
                        dimensional problem. However, if the thickness variation is small, it is possible to express
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