Page 409 - Advanced thermodynamics for engineers
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17.2 SIMPLE GAS TURBINE CYCLE ANALYSIS          399




                  It will be assumed that reheat occurs to give T 5 ¼ T 3 and also that h C ¼ h T ¼ 1.0. These as-
               sumptions simplify the problem.
                  Then energy added

                                                                                           (17.40)
                                                q 203 ¼ c p T 3   T 20

                                                q 45 ¼ c p T 5   T 4                       (17.41)
                  The heat exchanger effectiveness is given by Eqn (17.21):

                                                       T 3   T 2
                                                   ε ¼
                                                       T 5   T 2
               giving Eqn (17.28)

                                               T 20 ¼ T 2 ð1   εÞþ εT 6
                  Let the pressure ratio of the plant be defined by
                                                           ðk 1Þ=k
                                                s r ¼ p 2 p 1                              (17.42)
               and assume there are no pressure losses.
                  Then

                                                        ðk 1Þ=k

                                                    p 2
                                            T 2 ¼ T 1         ¼ T 1 s r                    (17.43)
                                                    p 1
               and
                                                         ðk 1Þ=k
                                                     p 6         T 5
                                             T 6 ¼ T 5        ¼                            (17.44)
                                                     p 5         s R

                                                     εT 5               εg
                                    T 20 ¼ T 1 s r ð1   εÞþ  ¼ T 1 s r ð1   εÞþ            (17.45)
                                                     s R                s R
                  This results in
                                             q 2 3               εg
                                               0
                                                 ¼ g   s r ð1   εÞ                         (17.46)
                                             c p T 1             s R
                  Similarly

                                                 q 45 ¼ c p ðT 5   T 4 Þ
                  Now
                                                     k 1            k 1

                                          T 4      p 4  k      p 4 p 1  k   s R
                                  T 4 ¼ T 3   ¼ T 3       ¼ T 3         ¼ T 3              (17.47)
                                          T 3      p 3         p 1 p 3      s r

                                                             gs R
                                               q 45 ¼ c p T 1 g                            (17.48)
                                                             s r
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