Page 99 - Introduction to Computational Fluid Dynamics
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P1: IWV
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                            to the equation                                    2D BOUNDARY LAYERS
                                                     ∂             ∂
                                                        r ˙ m   − r 
   = 0.                   (4.32)
                                                    ∂y              ∂y
                            Then, it follows that

                                                                           /2) − 1
                                                    d
                                                          d

                                                                   exp (P c n
                                              n =   +   −       d                  ,           (4.33)
                                                    P     N     P
                                                                     exp (P c n ) − 1

                                                                          /2) − 1
                                                          d
                                                    d

                                                                   exp (P c s
                                               s =   +   −      d S                ,           (4.34)
                                                    S
                                                          P
                                                                    exp (P c s ) − 1
                            where, the cell Peclet numbers are evaluated using the harmonic mean (see Equa-
                            tion 2.58):

                                                   ˙ m n  y n    y n − y P  y N − y n
                                                =                       +          ,           (4.35)
                                            P c n         = ˙ m n
                                                    
 n            
 P       
 N

                                                   ˙ m s  y s    y s − y S  y P − y s
                                                =                       +          .           (4.36)
                                             P c s        = ˙ m n
                                                     
 s           
 S      
 P
                               Thus, substituting Equations 4.33–4.36 in Equation 4.30 and combining the
                            latter with Equations 4.27 and 4.29, we can show that the discretised version of
                            Equation 4.26 takes the following form:
                                                d
                                                          d
                                                                            u
                                                                   d
                                           AP   = AN   + AS   + AU   + S  V,                   (4.37)
                                                                            P
                                                          N
                                                P
                                                                   S
                            where
                                                    r n ˙ m n  x
                                             AN =             ,                                (4.38)
                                                          − 1
                                                   exp P c n
                                                   r s ˙ m s  x exp P c s
                                             AS =                 ,                            (4.39)
                                                            − 1
                                                     exp P c s
                                                     u
                                             AU = ψ  ω,       AP = AU + AN + AS.               (4.40)
                                                    EI
                            In deriving the AP coefficient, use is made of the mass conservation equation. Thus,
                                              	  n  ∂              	  n  ∂
                                                     (ρ ru)dy =−          (ρ r v)dy
                                                  ∂x                   ∂y
                                               s                    s
                                                              =−(r n ˙ m n − r s ˙ m s )       (4.41)
                                                                 ∂  	  n  ∂ψ
                                                              =            dy
                                                                 ∂x  s  ∂y
                                                                  ω 
  d     u
                                                              =      ψ − ψ   EI  .             (4.42)
                                                                       EI
                                                                  x
                               Finally, the node-indexed version of Equation 4.37 can be written as
                                                                               u
                                        AP j   j = AN j   j+1 + AS j   j−1 + AU j   + S j  V j  (4.43)
                                                                               j
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