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Chapter 6: Conservation Equations
                           and the chemical composition matrix for this reaction is represented by
                                      Molecular species → C 2 H 6  O 2  H 2 OCO 2           75
                                                         ⎡                ⎤
                                      carbon              2    0    0    1
                                      hydrogen           ⎣ 6   0    2    0 ⎦
                                      oxygen              0    2    1    2
                           Equation (6.21) can be used to determine stoichiometric coefficients and the number
                           of independent reactions that are needed to describe any given system (Cerro et al.,
                           2004). In addition, it represents the fundamental axiom concerning conservation of
                           mass for reacting systems since it can be used to derive Eq. (6.6).

                           6.1.3  Mechanics
                           To describe multicomponent systems we use four axiomatic statements concerning
                           (1) the linear momentum of species A, (2) the angular momentum of species A, (3) the
                           total diffusive force in a mixture, and (4) the rate of production of linear momentum
                           owing to chemical reactions. These axioms take the form:

                           AXIOM I:
                                       d
                                             ρ A v A dV =  ρ A b A dV +  t A(n) dA       (6.23)
                                       dt
                                         V A (t)      V A (t)      A A (t)
                                                            B=N


                                                      +         P AB dV +   r A v A dV
                                                             B=1
                                                        V A (t)         V A (t)
                           AXIOM II:

                                  d
                                        r × ρ A v A dV =  r × ρ A b A dV +  r × t A(n) dA  (6.24)
                                  dt
                                    V A (t)         V A (t)          A A (t)
                                                           B=N


                                                     +        r × P AB dV +   r × r A v A dV
                                                           B=1
                                                      V A (t)            V A (t)
                           AXIOM III:
                                                      A=N B=N

                                                             P AB = 0                    (6.25)
                                                      A=1 B=1
                           AXIOM IV:
                                                       A=N

                                                           r A v A = 0                   (6.26)
                                                       A=1
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