Page 213 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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CONVECTION HEAT TRANSFER
                        a solution for the above equation. However, it is possible to introduce an artificial com-
                        pressibility formulation to avoid the implicit treatment of pressure. This will be discussed
                        in a later section.                                                   205
                           It has been stated in the literature that, even though the pressure terms in Equation 7.136
                        need to be treated implicitly, the scheme is really an explicit one. However, in our own
                        publications, the scheme is referred to as being semi-implicit because the implicit solution
                        to the pressure equation is used in the second step.
                        Step 3 Velocity or momentum correction: The velocity correction has already been derived
                        in the previous step (Equation 7.133). This involves the pressure and intermediate velocity
                        field, and is written as

                                  u n+1  −˜u 1  1 ∂p  n    ∂     1 ∂p    n  ∂     1 ∂p    n
                                   1
                                           =−        + u 1            + u 2
                                      t        ρ ∂x 1     ∂x 1  ρ ∂x 1    ∂x 2  ρ ∂x 1
                                  u n+1  −˜u 2  1 ∂p  n    ∂     1 ∂p    n  ∂     1 ∂p    n
                                   2
                                           =−        + u 1            + u 2                (7.137)
                                      t        ρ ∂x 2     ∂x 1  ρ ∂x 1    ∂x 2  ρ ∂x 2
                           The higher-order terms in the above equations may be neglected as these terms have
                        very little influence on the velocity correction.

                        Step 4 Temperature calculation: Applying the CG procedure to the temperature equation,
                        we get

                                                                                   n
                                                                                 !

                                                                        2
                                                                               2
                                     T  n+1  − T  n  ∂T  n    ∂T  n    ∂ T    ∂ T
                                               =−u 1     − u 2    + α      +
                                          t          ∂x 1     ∂x 2      ∂x 2  ∂x 2
                                                                          1     2
                                                       t ∂      ∂T  n   ∂T  n
                                                  + u 1      u 1    + u 2
                                                       2 ∂x 1  ∂x 1     ∂x 2
                                                       t ∂      ∂T  n   ∂T  n
                                                  + u 2      u 1    + u 2                  (7.138)
                                                       2 ∂x 2  ∂x 1     ∂x 2
                           All four preceding steps will now be summarized.
                        Step 1: Intermediate velocity
                        intermediate x 1 momentum equation

                                                                       2
                                                                             2
                                      ˜ u 1 −˜u n  ∂u 1  n  ∂u 1  n     ∂ u 1  ∂ u 1  ! n
                                            1
                                             =−u 1      − u 2    + ν      +
                                         t         ∂x 1     ∂x 2      ∂x 2   ∂x 2
                                                                        1      2
                                                                          n

                                                     t ∂      ∂u 1   ∂u 1

                                                + u 1      u 1   + u 2
                                                     2 ∂x 1   ∂x 1   ∂x 2
                                                                          n
                                                     t ∂      ∂u 1   ∂u 1


                                                + u 2      u 1   + u 2                     (7.139)
                                                     2 ∂x 2   ∂x 1   ∂x 2
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