Page 188 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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                        where δ ij is the Kroneker delta, which is equal to unity if i = j and equal to zero if i  = j.
                        From the previous expression, τ 11 is expressed as
                                                                    CONVECTION HEAT TRANSFER

                                                    ∂u 1  ∂u 1  2 ∂u 1  2 ∂u 2
                                           τ 11 = µ    +     −       −                      (7.27)
                                                    ∂x 1  ∂x 1  3 ∂x 1  3 ∂x 2
                           Note that i = j = 1 in the above equation and k = 1, 2 for a two-dimensional flow.
                        The above equation may be simplified as follows:

                                                          4 ∂u 1  2 ∂u 2
                                                 τ 11 = µ      −                            (7.28)
                                                          3 ∂x 1  3 ∂x 2
                           Similarly, τ 12 is

                                                           ∂u 1  ∂u 2
                                                   τ 12 = µ    +                            (7.29)
                                                           ∂x 2  ∂x 1
                           Substituting Equations 7.28 and 7.29 into Equation 7.25, we obtain the x 1 component
                        of the momentum equation as
                                                       2
                                          ∂(ρu 1 )  ∂(ρu )  ∂(ρu 1 u 2 )
                                                       1
                                                 +       +          =
                                            ∂t      ∂x 1      ∂x 2
                                                   ∂p    ∂       4 ∂u 1  2 ∂u 2
                                                 −     +     µ       −
                                                   ∂x 1  ∂x     3 ∂x 1  3 ∂x 2
                                                    ∂       ∂u 2  ∂u 1
                                                 +      µ      +                            (7.30)
                                                   ∂x 2    ∂x 1  ∂x 2
                           The momentum component in the x 2 direction can be derived by the following steps,
                        which are similar to the derivation of the x 1 component of the momentum equation. The
                        x 2 momentum equation is

                                                                  2
                                          ∂(ρu 2 )  ∂(ρu 1 u 2 )  ∂(ρu )
                                                                  2
                                                 +         +        =
                                            ∂t       ∂x 1      ∂x 2
                                                      ∂p    ∂       ∂u 1  ∂u 2
                                                   −     +      µ      +
                                                     ∂x 2  ∂x 1    ∂x 2  ∂x 1
                                                      ∂       4 ∂u 2  2 ∂u 1
                                                   +      µ       −                         (7.31)
                                                     ∂x 2    3 ∂x 2  3 ∂x 1
                           For a constant density flow (incompressible flow), the momentum equations can be fur-
                        ther reduced by taking the density term out of the differential signs. In addition, substitution
                        of the conservation of mass equation (Equation 7.11) into the momentum equation leads to
                        a further simplification of the momentum equation. After simplification (see Appendix D
                        for the detailed derivation), the momentum equations are

                                                                          2     2
                                       ∂u 1    ∂u 1    ∂u 1     ∂p       ∂ u 1  ∂ u 1
                                    ρ      + u 1   + u 2    =−     + µ       +              (7.32)
                                        ∂t     ∂x 1    ∂x 2     ∂x 1     ∂x 2   ∂x 2
                                                                           1      2
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