Page 94 - Handbook of Energy Engineering Calculations
P. 94

FIGURE 19 Mollier chart plot of the cycle in Fig. 18.

                  Read on the Mollier chart H  = 1346.7 Btu/lb (3132.4 kJ/kg). Since the
                                                      3
               engine  or  turbine  efficiency  e   =  H   −  H /(H   −  H )  =  0.8  =  1474.5  −
                                                                     3
                                                                          2
                                                             2
                                                     e
                                                                                  3
               H /(1474.5 − 1346.7); H  = actual enthalpy of the steam at the inlet to heater
                                             3
                  3
               1 = 1474.5 − 0.8(1474.5 − 1346.7) = 1372.2 Btu/lb (3191.7 kJ/kg). Plot this
                                                        2
               enthalpy  point  on  the  750-lb/in   (abs)  (5171.3-kPa)  pressure  line  of  the
               Mollier chart, Fig. 19. Read the entropy at the heater inlet from the Mollier
                                                                                                    2
               chart  as  S   =  1.5819  Btu/(lb  ·  °F)  [6.6  kJ/(Kg  ·  °C)]  at  750  lb/in   (abs)
                            3′
               (5171.3 kPa) and 1372.2 Btu/lb (3191.7 kJ/kg).
                                                                                          2
                  Assume constant-S expansion from H  to H  at 200 lb/in   (abs)  (1379.0
                                                                         4
                                                                 3′
               kPa), the inlet pressure for feedwater heater 2. Draw the vertical straight line
               3′−4 on the Mollier chart, Fig. 19. By using a procedure similar to that for
               heater 1 , H  = H  − e (H  − H ) = 1372.2 − 0.8(1372.2 − 1230.0) = 1258.4
                                               3′
                                           e
                                                      4
                                     3′
                             4′
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