Page 417 - Advanced thermodynamics for engineers
P. 417

17.3 AIRCRAFT GAS TURBINES         407




                                           1
                                               2              3    4
                                                    Combustion
                                                     chambers            5
                                 m                                          m






               FIGURE 17.21
               Single spool turbojet.
                  Solution
                  From the international standard atmosphere table, at 5000 m

                                        p a ¼ 0:5405 bar  and  T a ¼ 255:7K:
                  The stagnation conditions after the inlet may be obtained as follows:
                                          V a 2     270 2
                                             ¼                ¼ 36:3K
                                          2c p  2   1:005   1000
                                                 V 2 a
                                       T 01 ¼ T a þ  ¼ 255:7 þ 36:3 ¼ 292 K
                                                 2c p
                                                     k=ðk   1Þ
                                                T 01
                                        p 01 ¼ p a           ¼ 0:8602 bar:
                                                T a
                  At the outlet from the compressor

                                             p 02
                                      p 02 ¼     p 01 ¼ 8:0   0:8602 ¼ 6:88 bar
                                             p 01
                                              ðk   1Þ=k
                                                          !

                                    T 01  p 02                292    0:286
                          T 02   T 01 ¼                  1  ¼     8:0      1 ¼ 272:7K
                                    h C   p 01                0:87
                                            T 02 ¼ 292 þ 272:5 ¼ 564:7K
                                              w T ¼ w C =h m  and hence

                                               T 02   T 01  1:005   272:7
                                           c p a
                                 T 03   T 04 ¼          ¼             ¼ 241:4K
                                                          1:147   0:99
                                                c p g
                                           T 04 ¼ 1200   241:4 ¼ 958:6K
                                                                    4
                                          T 04
                                           0   k=ðk   1Þ     931:8
                                 p 04 ¼ p 03          ¼ 6:88         ¼ 2:501 bar:
                                          T 03               1200
                  The nozzle pressure ratio is therefore
                                               p 04  2:501
                                                  ¼       ¼ 4:627
                                               p a  0:5405
   412   413   414   415   416   417   418   419   420   421   422