Page 43 - Introduction to Computational Fluid Dynamics
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                               Now to evaluate the time derivative, we write   1D HEAT CONDUCTION

                                                   n
                                                            o
                                              (CT ) = (CT ) +  t   ∂(CT )      +· · · ,        (2.13)
                                                   P
                                                            P
                                                                      ∂t    P
                            or
                                                                   n        o
                                                              (CT ) − (CT )
                                                   ∂(CT )          P        P
                                                            =                .                 (2.14)

                                                     ∂t              t
                                                          P
                               In Equation 2.13, derivatives of order higher than 1 are ignored; therefore,
                            Equation 2.14 is only a first-order-accurate representation of the time derivative. 2
                               Inserting Equations 2.11 and 2.12 in Equation 2.7 and Equation 2.14 in Equation
                            2.8 and employing Equation 2.6, we can show that
                                          n
                                   ρ VC                          n           n        n
                                             + ψ (AE + AW)     T = ψ AE T + AW T        + S, (2.15)
                                                                P           E         W
                                       t
                                            P
                            with
                                          2             x w d (kA)         x
                                  AE =        (kA) P +                           ,             (2.16)
                                         x e            2    dx      ( x e +  x w )
                                                                  P
                                          2             x e d (kA)         x
                                  AW =         (kA) P −                          ,             (2.17)
                                         x w            2    dx      ( x e +  x w )
                                                                  P
                                                ,n           ,o                    o        o
                                    S = ψ q     + (1 − ψ)q      V + (1 − ψ) AE T + AW T
                                            P             P                       E         W
                                                   o
                                            ρ VC                               o
                                        +              − (1 − ψ)(AE + AW)    T ,               (2.18)
                                                                               P
                                                t     P
                            where  V = A  x. Note that if the cell faces were midway between adjacent nodes,
                            2 x =  x e +  x w . Before leaving the discussion of the TSE method, we make
                            the following observations:
                            1. Calcuation of coefficients AE and AW requires evaluation of the derivative
                               d (kA)/dx | P . This derivative can be evaluated using expressions such as (2.11)
                               in which T is replaced by kA.
                            2. For certain variations of (kA) and choices of  x e and  x w , AE and/or AW can
                               become negative.
                                                                        o
                            3. For certain choices of  t, the multiplier of T in Equation 2.18 can become
                                                                        P
                               negative.
                                                                   o
                            4. In steady-state problems,  t =∞ and T has no meaning. Therefore, in such
                               problems, ψ always equals 1.

                            2  Clearly, it is possible to represent the time derivative to a higher-order accuracy. However, the
                                                                   0
                                                                      00
                                                                n
                              resulting expression will involve reference to T , T , T , and so on.
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