Page 212 - Introduction to Computational Fluid Dynamics
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P1: IWV
                           CB908/Date
                                        0 521 85326 5
            0521853265c06
                        6.3 UNSTRUCTURED MESHES
                                                                            c 3    May 25, 2005  11:10 191
                                                                                            n
                                                                     b
                                                                                       E  2
                                                              c 4
                        Figure 6.13. Construction of a cell-face con-         c
                        trol volume.
                                                                                               c 2

                                                                      P 2
                                                                                        a

                                                                                c 1


                           Thus, the final discretised pressure correction equation is

                                              NK            NK
                                                                                V
                                                                     
      o
                                     AP p =       AE k p  −     C ck − ρ P − ρ     ,      (6.134)
                                          P            Ek                   P
                                              k=1           k=1                 t
                        where AE k is given by Equation 6.133. Equation 6.134 must be solved with
                        ∂p /∂n | B = 0, which can be accomplished simply by setting AE k = 0 for the

                        boundary face. After solving Equation 6.134, the mass-conserving pressure cor-
                                                                       l
                                                                           l



                        rection is recovered as p = p − p    = p − 0.5(p − p ).
                                              m         sm
                        Evaluation of p
                        Recall that p = 0.5(p + p ), where p       is determined from solution of
                                     P        x 1   x 2         x i
                          2
                               2
                        ∂ p/∂x | P = 0. Thus p , for example, is evaluated from
                               i              x 1
                                                                       NK
                                    2
                            1  	  ∂ p           1  	   1  ∂p       1        1  ∂p
                                        dV =          β        =          β        = 0.   (6.135)
                                                       1                   1
                            V     ∂x  2         V   C    ∂x 1 ck   V         ∂x 1 ck


                                     1 P                               k=1
                        Now, the pressure gradient at the cell face is evaluated by applying Gauss’s theorem
                        over the volume c 1 –c 2 –c 3 –c 4 . Then, it can be shown that

                                   ∂p        x 2,E 2  p E 2  +  x 2,b p b +  x 2,P 2  p P 2  +  x 2,a p a
                                         =                                           ,    (6.136)


                                   ∂x 1 ck                     V ck
                        where
                                                                     ),
                                                  x 2,E 2  = (x 2,c 3  − x 2,c 2
                                                                     ),
                                                   x 2,b = (x 2,c 4  − x 2,c 3
                                                                     ),
                                                   x 2,P 2  = (x 2,c 1  − x 2,c 4
                                                                     ).
                                                   x 2,a = (x 2,c 2  − x 2,c 1
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