Page 162 - Advanced thermodynamics for engineers
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148    CHAPTER 7 GENERAL THERMODYNAMIC RELATIONSHIPS




             and hence
                                             vv           R

                                                  ¼                                       (7.33)
                                             vT
                                                p       a     2b
                                                    p    2 1
                                                       v      v
                This can be differentiated again to give
                                               v v     v   vv
                                                2
                                              vT 2  ¼  vT vT
                                                   p          p
             which results in
                                                             2
                                    2
                                  v v        a  2ab      vv    2a  6ab  ¼ 0
                                  vT 2   p    v 2  þ  v 3  þ  vT  v 3     v 4
                                       p                     p
             giving
                                                       2
                                                      vv   2a  6ab
                                           v v           p
                                            2         vT   v 3    v 4
                                                 ¼
                                           vT 2
                                               p          a  2ab
                                                      p    2 þ
                                                          v   v 3
                                    vv

             which, on substituting for  becomes
                                    vT
                                        p
                                                               4
                                                          v
                                                              v
                                          v v                                             (7.34)
                                           2          R 2  2a 3    6ab
                                         vT 2  ¼                   3
                                              p          a     2b
                                                     p    2 1
                                                        v      v
                Hence, for a van der Waals gas, c p ¼ f(p).
                This means that allowance would have to be taken of this when evaluating the change of specific
             heat capacity at constant pressure for a process in which the pressure is changed. This can be evaluated
             from
                                                  p
                                                 Z 2     2
                                                        v v
                                            c p ¼   T     2  dp;                          (7.35)
                                                        vT
                                                            p
                                                 p 1
                      2
                    v v
             where        is calculated from Eqn (7.34).
                    vT 2
                         p
             7.3 Tds RELATIONSHIPS
             Two approaches have been used previously in performing cycle calculations. When evaluating the
             performance of steam plant and refrigeration equipment, much emphasis was placed on the use of
             tables and if the work done between two states was required, enthalpy values at these states were
             evaluated and suitable subtractions were performed. When doing cycle calculations for gas turbines
             and internal combustion engines, the air was assumed to be an ideal gas and the specific heat capacity
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