Page 260 - Introduction to Computational Fluid Dynamics
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                        8.3 DIFFERENTIAL GRID GENERATION
                        direction ξ i ) is defined as                               May 10, 2005  16:28 239
                                                   r
                                                 d       ∂x 1    ∂x 2   ∂x 3
                                              a i =  = i     + j    + k     .              (8.17)
                                                 d ξ i   ∂ξ i    ∂ξ i   ∂ξ i
                        Similarly, the contravariant base vector (normal to coordinate surface ξ i = constant)
                        is defined as
                                       i
                                        a =∇ ξ i = i    ∂ξ i  + j    ∂ξ i  + k    ∂ξ i  =  j ×  k /J,  (8.18)
                                                                             a
                                                                         a
                                                   ∂x 1    ∂x 2   ∂x 3
                        where J is the Jacobian. Now, from Green’s theorem [70], for any quantity (vector
                        or scalar)  ,


                                      3                      3               3
                                   1      ∂                      ∂    i          i  ∂
                            ∇   =               a j × a k ·   =       a   =       a        (8.19)
                                   J     ∂ξ i                   ∂ξ i               ∂ξ i
                                     i=1                    i=1             i=1
                                i
                        since ∂  /∂ξ i = 0. Therefore,
                               a

                                                 3             3
                              2                      i  ∂          l  ∂
                             ∇   =∇ · ∇   =           a    .        a
                                                      ∂ξ i          ∂ξ l
                                                i=1           l=1
                                                3   3                      3   3       l
                                                                                     a
                                                              ∂    ∂                ∂  ∂
                                                        i   l                      i
                                                           a
                                            =           a ·             +           a      .
                                                             ∂ξ i  ∂ξ l             ∂ξ i ∂ξ l
                                               i=1 l=1                    i=1 l=1
                                                                                           (8.20)
                        If we now set   = ξ l (a scalar), then
                                                            3       l
                                                                  a
                                                     2          i  ∂
                                                   ∇ ξ l =      a    .                     (8.21)
                                                                 ∂ξ i
                                                           i=1
                        Substituting Equation 8.21 in Equation 8.20 gives,

                                             3   3                       3
                                                           ∂    ∂               ∂
                                      2              i   l                   2
                                                        a
                                     ∇   =            a ·            +     ∇ ξ l   .       (8.22)
                                            i=1 l=1        ∂ξ i  ∂ξ l   l=1     ∂ξ l
                        In two dimensions (∂/∂x 3 = ∂/∂ξ 3 = 0), Equation 8.22 will read as
                                                                   2
                                                   2
                                                                                  2
                                                  ∂               ∂              ∂
                                            1
                                     2
                                                                           2
                                                           1
                                                                          a
                                                                              a
                                    ∇   = a · a 1     + 2   ·  2        +  ·   2
                                                          a
                                                              a
                                                  ∂ξ  2          ∂ξ 1 ∂ξ 2       ∂ξ 2
                                                    1                              2
                                                  ∂          ∂
                                               2          2
                                           +∇ ξ 1     +∇ ξ 2     .                         (8.23)
                                                  ∂ξ 1       ∂ξ 2
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