Page 187 - Introduction to Computational Fluid Dynamics
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                                                                2D CONVECTION – COMPLEX DOMAINS
                            Using Equation 5.2, it is now possible to replace Cartesian fluxes q 1 and q 2 . After
                            some algebra, it can be shown that

                                    ∂(ρ )     ∂               
 eff  2  ∂
                                 rJ        +      ρ rU f1   − r   dA 1
                                      ∂t     ∂ξ 1               J     ∂ξ 1
                                              ∂               
 eff   ∂
                                           +      ρ rU f2   − r   dA 2 2
                                             ∂ξ 2               J     ∂ξ 2
                                              ∂     
 eff    ∂       ∂     
 eff   ∂
                                           =      r    dA 12     +       r    dA 12     + rJ S,
                                             ∂ξ 1    J      ∂ξ 2    ∂ξ 2    J      ∂ξ 1
                                                                                               (6.20)
                            where
                                                         2


                                                      dA = β   1 1    2  + β 2 1    2  ,
                                                         1
                                                         2


                                                      dA = β   1    2  + β 2    2  ,
                                                         2     2       2
                                                              1  1    2  2
                                                     dA 12 = β β + β β                         (6.21)
                                                              1  2    1  2
                            and the contravariant flow velocities are given by
                                                    1       2      ∂x 2     ∂x 2
                                             U f1 = β u f1 + β u f2 =  u f1 −   u f2 ,         (6.22)
                                                            1
                                                    1
                                                                   ∂ξ 2     ∂ξ 1
                                                    1       2      ∂x 1     ∂x 1
                                             U f2 = β u f1 + β u f2 =  u f2 −   u f1 ,         (6.23)
                                                    2       2
                                                                   ∂ξ 1     ∂ξ 2
                            where u f1 and u f2 are the Cartesian velocity components.
                            6.2.3 Interpretation of Terms
                            Several new terms appearing in Equation 6.20 can be interpreted using vector
                            mathematics.

                            Elemental Area
                            The elemental area dA i normal to the (ξ j ,ξ k ) plane is given by

                                                                   r
                                                            r
                                                           ∂     ∂

                                                   d A i =     ×      dξ j dξ k ,              (6.24)
                                                           ∂ξ j  ∂ξ k
                            where the position vector   = ix 1 + jx 2 + kx 3 . For our 2D case, if we set i = 1,


                                                   r

                                                   r
                             j = 2, and k = 3 then ∂ /∂ξ 3 = ∂x 3 /∂ξ 3 = 1 because the x 3 and ξ 3 directions
                            coincide and are normal to the (ξ 1 ,ξ 2 ) plane. Thus, taking unit dimension in the x 3
                            direction gives

                                                          ∂x 2
                                              ∂ r                 ∂x 1         1

                                      dA 1 =        dξ 2 = i  − j                     2
                                                                      dξ 2 = i β + j β dξ 2
                                                                               1      1

                                             ∂ξ 2         ∂ξ 2    ∂ξ 2
                                                        %
                                                          
  1   2  
  2   2
                                                      =    β   + β    dξ 2 .                   (6.25)
                                                            1      1
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