Page 156 - Basic Structured Grid Generation
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Differential models for grid generation  145

                                        !                             "
                                         (Delet ∗ ϕ) j,k+1 − (Delet ∗ ϕ) j,k−1
                                   +0.5
                                               2[(g tt ) j,k + (g oo ) j,k ]
                                                                         
                                           (g ot ∗ ϕ) j+1,k+1 + (g ot ∗ ϕ) j−1,k−1
                                                                         
                                           −(g ot ∗ ϕ) j+1,k−1 − (g ot ∗ ϕ) j−1,k+1
                                                                         
                                   +0.25                                    .             (5.116)
                                                  2[(g tt ) j,k + (g oo ) j,k ]
                                                                         
                                                                         
                        An iterative scheme employing SOR is now given by
                                                 n+1    n       t     n
                                                ϕ   = ϕ   + ω(ϕ   − ϕ   ),                (5.117)
                                                 j,k    j,k     j,k  j,k
                        where ω is an over-relaxation factor which we have taken in the program to be 1.5.
                          The program uses this method to solve Laplace’s equation in the region between con-
                        centric circular arcs shown in Fig. 4.2 of Chapter 4, subject to the Dirichlet boundary
                        conditions:
                                                               1
                                       ϕ = 0  when θ = 0,  ϕ =   sin α  when ϕ = α,
                                                               r
                                     1                         1
                                 ϕ =   sin θ  when r = r 1 ,  ϕ =  sin θ  when r = r 2 .  (5.118)
                                     r 1                       r 2
                                                                       1
                          For these boundary conditions the exact solution ϕ =  sin θ exists, and this enables
                                                                       r
                        us to assess the accuracy of the numerical procedure.
                        5.11.2 More general steady-state equation


                        Suppose that the basic partial differential equation to be solved in physical space for
                        the field variable ϕ has the form

                                               ∇· (vϕ) +∇ · (ν∇ϕ) + S = 0,                (5.119)

                        where S is a source term, ν could be a diffusion coefficient, and v is a vector field
                        (fluid velocity). In cartesian co-ordinates this takes the form

                                               ∂         ∂     ∂ϕ
                                                 (v i ϕ) +   ν     + S = 0,
                                              ∂y i      ∂y i  ∂y i
                        which transforms to the consistent generalized tensor form
                                                  i        ij
                                                (v ϕ) ,i + (νg ϕ ,j ) ,i + S = 0.         (5.120)
                        Using eqns (1.128), (1.122), (1.111), and (1.134), this may be expressed as
                                 i     i     ij
                                v ϕ, i +v ϕ + g (νϕ, j ) ,i + S
                                       ,i
                                       ∂ϕ            ij     ∂     ∂ϕ     k     ∂ϕ
                                      i
                                   = v    + ϕdivv + g        ν     −   ij  ν     + S
                                       ∂x i             ∂x i  ∂x j         ∂x k
                                       ∂ϕ      i  ∂v    ij  ∂     ∂ϕ     2 k   ∂ϕ
                                      i
                                   = v    + ϕg ·     + g      ν      + (∇ x )ν    + S = 0,
                                       ∂x i      ∂x i    ∂x i   ∂x j          ∂x k
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