Page 187 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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CONVECTION HEAT TRANSFER
                                                           12
                                                              +  ....
                                                          ∂t
                                                   t +∆x
                                                        2
                                                   12
                                                          ∂x
                                                            2
                                                                          ∂p                  179
                                        p                           p +∆x 1   +  ....
                                                                          ∂x 1
                                                                            11
                                                           ∆x       t +∆x      +  ....
                                                                           ∂t
                                       t 11                  2      11   1
                                                                           ∂x 1
                                                     ∆x
                                                       1
                                                      t 12
                        Figure 7.5 Infinitesimal control volume in a flow field. Derivation of conservation of
                        momentum in x 1 direction. Viscous and pressure forces
                           Similarly, the total force due to the deviatoric stress (viscosity or friction) acting on the
                        control volume in the x 1 direction is written as (see Figure 7.5)

                                    ∂τ 11                        ∂τ 12
                               τ 11 +    x 1  x 2 − τ 11  x 2 + τ 12 +   x 2  x 1 − τ 12  x 1  (7.21)
                                     ∂x 1                         ∂x 2
                           Simplifying, we obtain the net force due to the deviatoric stress as
                                                 ∂τ 11        ∂τ 12
                                                     x 1  x 2 +    x 2  x 2                 (7.22)
                                                 ∂x 1         ∂x 2
                           The total force acting on the control volume in the x 1 direction is

                                                         ∂p    ∂τ 11  ∂τ 12

                                                x 1  x 2 −  +      +                        (7.23)
                                                         ∂x 1  ∂x 1  ∂x 2
                           As mentioned before, for equilibrium, the net momentum in the x 1 direction should be
                        equal to the total force acting on the control volume in the x 1 direction, that is,

                                      2
                                  ∂(ρu )   ∂(ρu 1 u 2 )  ∂(ρu 1 )         ∂p   ∂τ 11  ∂τ 12
                                      1
                          x 1  x 2      +          +         =  x 1  x 2 −   +     +        (7.24)
                                   ∂x 1      ∂x 2      ∂t                ∂x 1  ∂x 1   ∂x 2
                           Simplifying, we obtain
                                                  2
                                      ∂(ρu 1 )  ∂(ρu )  ∂(ρu 1 u 2 )  ∂p  ∂τ 11  ∂τ 12
                                                  1
                                            +        +         =−      +     +              (7.25)
                                        ∂t      ∂x 1     ∂x 2      ∂x 1  ∂x 1   ∂x 2
                           Note that the external and body forces (buoyancy) are not included in the above force
                        balance. In the above equations, the deviatoric stresses τ ij are expressed in terms of the
                        velocity gradients and dynamic viscosity as

                                                      ∂u i  ∂u j  2 ∂u k
                                              τ ij = µ    +     −      δ ij                 (7.26)
                                                      ∂x j  ∂x i  3 ∂x k
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