Page 184 - Fundamentals of The Finite Element Method for Heat and Fluid Flow
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                                    Flow direction
                                                         Plate      CONVECTION HEAT TRANSFER



                                   Leading edge                              Trailing edge
                                                           (a)


                                              Heater


                                                        Air circulation





                                                           (b)

                            Figure 7.3  Forced flow over a flat plate and natural convection inside a room

                        in the x 1 direction and
                                                                        2 2
                                                         x 2 ∂(ρu 2 )   x 2 ∂ (ρu 2 )
                                                      +           +              + ··· .     (7.3)
                                   (ρu 2 ) x 2 + x 2  = (ρu 2 ) x 2
                                                         1!  ∂x 2     2!   ∂x 2
                                                                             2
                        in the x 2 direction. From an inspection of the control volume shown in Figure 7.2, we can
                        write the difference between the total mass entering and exiting the control volume as
                                                                            2 2
                                                             x 1 ∂(ρu 1 )   x 1 ∂ (ρu 1 )

                           x 2 (ρu 1 ) x 1  − (ρu 1 ) x 1 + x 1  =− x 2  +       2  + ···    (7.4)
                                                             1!  ∂x 1    2!    ∂x
                                                                                 1
                           Similarly, in the x 2 direction
                                                                            2 2
                                                             x 2 ∂(ρu 2 )   x 2 ∂ (ρu 2 )

                                                   =− x 1             +             + ···    (7.5)
                           x 1 (ρu 2 ) x 2  − (ρu 2 ) x 2 + x 2                  2
                                                             1!  ∂x 2    2!    ∂x
                                                                                 2
                           Note that the total mass is calculated as being the mass flux rate times the perpendicular
                        area to the following regime. For instance, the total mass entering the control volume in
                        the x 1 direction is  x 2 × 1 × ρu 1 . A unit thickness is assumed in the x 3 direction.
                           Adding Equations 7.4 and 7.5 gives the total mass stored inside the control volume.
                        Neglecting the second- and higher-order terms, the total mass stored inside the control
                        volume is

                                                         ∂(ρu 1 )  ∂(ρu 2 )
                                               − x 1  x 2       +                            (7.6)
                                                           ∂x 1     ∂x 2
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