Page 124 - Introduction to Computational Fluid Dynamics
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P1: IWV
                           CB908/Date
            0521853265c04
                        EXERCISES
                          ADIABATIC     0 521 85326 5                              May 25, 2005  11:7  103
                          WALL
                                 1 m
                                                         T
                                                           8
                                                              Figure 4.12. Burning from a vertical wall.
                                                        g

                                 1 m
                         VOLATILE
                         FUEL


                            (c) Write initial conditions for each variable assuming pure natural convection
                                heat transfer between x = 0 and x = x 0 .
                        11. In the stagnation-flow carbon-burning problem described in the text, the water
                            vapour was treated as inert and its mass fraction was held constant. However,
                            water vapour can react with carbon, resulting in the following two additional
                            surface reactions:

                                                     ∗
                                                    C + H 2 O → CO + H 2
                                                      C + 2H 2 → CH 4 .
                                                       ∗

                            The reaction rate of the first reaction is about twice that of Reaction 3
                            (i.e., 2k 3 ). For the second reaction, k = 0.035exp(−17,900/T w ). Assuming
                            ω H 2 O,∞ = 0.01, write the equations to be solved along with their source terms
                            and boundary conditions. [Hint: You will need to postulate the following ad-
                            ditional gas-phase reactions to approximately account for the presence of H 2 ,
                            H 2 O, and CH 4 :
                                                            1
                                                    CH 4 →   C 2 H 4 + H 2 ,
                                                            2

                                                    24,962                        M 0.5
                                    = 10 20.32  exp −       ρ 1.97  ω 0.5  ω 1.07  ω 0.4  CH 4  ,
                               R CH 4                        m                   1.07  0.4
                                                      T           CH 4  O 2  H 2  M  M
                                                                                 O 2  C 2 H 4
                                                    C 2 H 4 → 2CO + 2H 2 ,


                                                                                     0.1
                                                  25,164                  −0.37    M
                                   = 10 17.7  exp −       ρ 1.71  ω 0.9  ω 1.18  ω   C 2 H 4  ,
                              R C 2 H 4                    m                            −0.37
                                                    T           C 2 H 4  O 2  CH 4  M 1.18  M
                                                                                  O 2   CH 4
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