Page 92 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
P. 92

THE FINITE ELEMENT METHOD
                        84
                           Substituting Equation 3.177 into Equation 3.224 and integrating, we have
                                                   2


                                                                    2
                                                          4
                                      2
                               E = 4B − 4Bµ 1 − B + µ − 2Bµ       4       + B 2     8     µ 4  (3.225)
                                                   3                3         15
                           The error is minimized by satisfying ∂E/∂B =0,thatis,
                                        ∂E         4µ 4  16Bµ 4     2  16Bµ 2
                                            = 8B −     +       − 4µ +         = 0          (3.226)
                                        ∂B          3      15             3
                        which gives
                                                         µ 2     µ 2
                                                             1 +
                                                         2       3
                                                  B =               2                      (3.227)
                                                               1   µ
                                                      1 + 2µ 2   +
                                                               3   15
                           Therefore, the solution is given by
                                                               µ 2     µ 2
                                                                   1 +
                                            θ(ζ)           2    2       3
                                                = 1 − (1 − ζ )                             (3.228)
                                                                           2
                                             θ b                   2  1  µ
                                                             1 + 2µ    +
                                                                     3   15
                                                        2
                           For the particular problem where µ = 3, then
                                                   θ(ζ)      15      2
                                                       = 1 −   (1 − ζ )                    (3.229)
                                                    θ b      24
                           Figure 3.27 shows the comparison between all the different weighted residual methods.
                        As seen, the Galerkin method is the most accurate method.
                                    1
                                                               Exact
                                                            Collocation
                                   0.8                     Sub-domain
                                                              Galerkin
                                  Non-dimensional temperature  0.6
                                                           Least-squares




                                   0.4


                                   0.2


                                    0
                                     0         0.2       0.4       0.6       0.8        1
                                                         Distance from fin tip
                        Figure 3.27 Comparison between various weighted residual methods and exact solution
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