Page 155 - Advanced Thermodynamics for Engineers, Second Edition
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142    CHAPTER 7 GENERAL THERMODYNAMIC RELATIONSHIPS




                The chemical potential terms will be omitted in the following analysis, although similar equations
             to those below can be derived by taking them into account.
                It can be seen from Eqns (7.2) and (7.2a) that the specific internal energy can be represented by a
             three-dimensional surface based on the independent variables of entropy and specific volume. If this
             surface is continuous then the following relationships can be derived based on the mathematical
             properties of the surface. The restriction of a continuous surface means that it is ‘smooth’. It can be
             seen from Fig. 2.5 that the p–v–T surface for water is continuous over most of the surface, but there are
             discontinuities at the saturated liquid and saturated vapour lines shown in Fig. 2.6. Hence, the
             following relationships apply over the major regions of the surface, but not across the boundaries. For
             a continuous surface,

                                    z ¼ zðx; yÞ where z is a continuous function:
                Then,
                                                 vz        vz

                                           dz ¼      dx þ      dy:                         (7.3)
                                                 vx  y     vy  x
                Let

                                               vz             vz
                                         M ¼        and N ¼       :                        (7.4)
                                               vx             vy
                                                   y             x
                Then,
                                               dz ¼ Mdx þ Ndy:                             (7.5)
                                                   2        2
                For continuous functions the derivatives  v z  and  v z  are equal, and hence  vM  ¼  vN  .
                                                 vxvy     vyvx                    vy  x   vx  y
             Consider also the expressions obtained when z ¼ z(x,y) and x and y are themselves related to additional
             variables u and v, such that x ¼ x(u,v) and y ¼ y(u,v). Then,

                                       vz      vz    vx      vz   vy
                                            ¼            þ                                 (7.6)
                                       vu      vx    vu      vy   vu
                                          v        y    v       x     v
                Let z ¼ v and u ¼ x, then x ¼ x(z) and

                                           vz             vx
                                                ¼ 0; and       ¼ 1:                        (7.7)
                                           vu             vu
                                              v               v
                Hence

                                                vz      vz   vy
                                           0 ¼      þ                                      (7.8)
                                                vx      vy   vx
                                                   y       x     z
             giving
                                            vz    vy   vx

                                                            ¼ 1:                           (7.9)
                                            vx    vz   vy
                                                y    x     z
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