Page 84 - Process Modelling and Simulation With Finite Element Methods
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Partial Differential Equations and the Finite Element Method   71

                                           ac     a2c
                                              =
                              Mass diffusion: - D -
                                           at     ax2
                                           dT     k  a2T
                          Thermal conduction: - --                     (2.9)
                                               =
                                            at   pc,  ax2
                                              do     d"o
                          2-D Vorticity transport: - -                (2.10)
                                                  =
                                                   V
                                              at      ax2
          where  c,T,  and  ~i) are  concentration,  temperature,  and  the  z-component  of
          vorticity in a 2-D flow, respectively, and their corresponding diffusivities are D,
          a,  and V.  This equation is thoroughly  studied in the undergraduate curriculum.
         It has solutions by Fourier and Laplace transforms,  and similarity  solutions for
                                                                 X
          initial  and boundary  conditions that collapse on the variable  77 = -

         doesn't  leave much room for finite element methods - just another technique for
          a tired old problem, right?  Wrong.  FEMLAB can still give this problem a boost
         which is not commonly considered.  FEMLAB solutions are well suited to non-
         constant coefficients, i.e. transport properties that  depend on the field variable.
         For instance, for suitably low pressures and high temperatures, a gas must satisfy
         the ideal gas law:
                                       nM
                                   p=-=-     PM                       (2.11)
                                        V     RT
         where R is the gas constant and M is the relative molecular mass of the species.
         Under these conditions, it is rare to find a gas that has a constant heat capacity.
         For instance, over a range of temperatures, the heat capacity of COz gas is well
         approximated by a quadratic in temperature, ( in MJkg-mol"C), with T in "C:
                      c,  = 36.11 + 0.04233 T  - 2.887~10-~T*         (2.12)

         It follows that

                                k        kR
                               --   -           f  (T)
                               pc,    36.11PM
                                                                      (2.13)
                                         (T + 273)
                       (T)=  l.+ 1.172~10" T  - 7.995~10-~T'
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