Page 261 - Introduction to Computational Fluid Dynamics
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                                                                       NUMERICAL GRID GENERATION
                               The dot products are now easily evaluated from Equations 8.17 and 8.18. Thus
                                                               2         2
                                                    1
                                           1   1         ∂x 1      ∂x 2          2
                                            a · a =             +          = α/J ,
                                                   J  2  ∂ξ 2       ∂ξ 2
                                                     1
                                           1   2         ∂x 1 ∂x 1  ∂x 2 ∂x 2        2
                                            a · a =−             +           =−β/J ,
                                                     J 2  ∂ξ 1 ∂ξ 2  ∂ξ 1 ∂ξ 2
                                                               2         2
                                                    1
                                           2   2         ∂x 1      ∂x 2          2
                                            a · a =             +          = γ/J .             (8.24)
                                                   J  2  ∂ξ 1       ∂ξ 1
                            Employing these relations and using Equations 8.13 and 8.14, we can show that
                                                 2
                                                            2
                                                                      2
                                          1     ∂          ∂         ∂          ∂       ∂
                                    2
                                  ∇   =       α      − 2β        + γ       + P     + Q     .   (8.25)
                                          J  2  ∂ξ 2      ∂ξ 1 ∂ξ 2  ∂ξ 2       ∂ξ 1    ∂ξ 2
                                                  1                     2
                                                                           2
                                                                                   2
                               We now replace   by x 1 and x 2 and note that ∇ x 1 =∇ x 2 = 0. Then, the
                            equations for x 1 and x 2 will read as
                                         2          2         2
                                        ∂ x 1      ∂ x 1     ∂ x 1     2    ∂x 1     ∂x 1
                                      α   2  − 2β        + γ    2  =−J    P     + Q       ,    (8.26)
                                        ∂ξ        ∂ξ 1 ∂ξ 2  ∂ξ             ∂ξ 1     ∂ξ 2
                                          1                     2
                                         2          2         2
                                        ∂ x 2      ∂ x 2     ∂ x 2     2    ∂x 2     ∂x 2
                                      α   2  − 2β        + γ    2  =−J    P     + Q       ,    (8.27)
                                        ∂ξ        ∂ξ 1 ∂ξ 2  ∂ξ             ∂ξ 1     ∂ξ 2
                                          1                     2
                            where
                                                         ∂x 1 ∂x 2  ∂x 2 ∂x 1
                                                     J =         −         .                   (8.28)
                                                         ∂ξ 1 ∂ξ 2  ∂ξ 1 ∂ξ 2
                               To determine functions (8.1), therefore, Equations 8.26 and 8.27 must be solved
                            simultaneously with the boundary conditions specified at ξ 1 = 0, ξ 1 = ξ 1max , ξ 2 =
                            0, and ξ 2 = ξ 2max . Note that Equations 8.26 and 8.27 are coupled and nonlinear
                            because α, β, and γ are themselves functions of dependent variables x 1 and x 2 .
                            Further, we note that the equations contain both the first and second derivatives and,
                                 2
                                           2
                            if −J P and −J Q are regarded as velocities, the equations have the structure of
                            a general transport equation.
                               It might appear that Equations 8.26 and 8.27 can be easily discretised and solved.
                            However,thereisadifficultyassociatedwiththeapplicationofboundaryconditions.
                            The difficulty can be understood as follows. In fluid flow problems, we would often
                            desire that the grid lines intersect orthogonally with the boundary. Thus, at the north
                            and south boundaries, for example, we would desire that ∂x 1 /∂ξ 2 = 0. However,
                            once this specification is made, we cannot specify x 1 on these boundaries. This is
                            because if Dirichlet and Neumann boundary conditions are specified at the same
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