Page 99 - Handbook of Energy Engineering Calculations
P. 99

and 1000°F (537.8°C) to find throttle-steam properties. Applying the symbols
               of the T-S diagram in Fig. 21, we get H  = 1474.5 Btu/lb (3429.7 kJ/kg); S  =
                                                                                                         2
                                                               2
               1.5603 Btu/(lb · °F) [6.5 kJ/(kg · °C)].
























                          FIGURE 21 Irreversible expansion in reheat cycle.



               3. Find the reheat-steam enthalpy
               Assume a constant-entropy expansion of the steam from 2000 to 400 lb/in                      2

               (13,790 to 2758 kPa). Trace this expansion on a Mollier (H-S) chart, where a
               constant-entropy  process  is  a  vertical  line  between  the  initial  [2000  lb/in        2
                                                                     2
               (abs)  or  13,790  kPa]  and  reheat  [400  lb/in   (abs)  or  2758  kPa]  pressures.
               Read on the Mollier chart H  = 1276.8 Btu/lb (2969.8 kJ/kg) at 400 lb/in                     2
                                                   3
               (abs) (2758 kPa).


               4. Compute the actual reheat properties
               The ideal enthalpy drop, throttle to reheat = H  − H  = 1474.5 − 1276.8 =
                                                                         2
                                                                                 3
               197.7 Btu/lb (459.9 kJ/kg). The actual enthalpy drop = (ideal drop) (turbine
               efficiency)  =  H   –  H   =  197.5(0.8)  =  158.2  Btu/lb  (368.0  kJ/kg)  =  W   =
                                          3′
                                                                                                        e1
                                   2
               work output in the high-pressure section of the turbine.
                  Once W  is known, H  can be computed from H  = H  – W  = 1474.5 −
                                              3′
                                                                               3′
                                                                                            e1
                                                                                      2
                            e1
               158.2 = 1316.3 Btu/lb (3061.7 kJ/kg).
                  The steam now returns to the boiler and leaves at condition 4, where P  =
                                                                                                         4
                          2
               400 lb/in  (abs) (2758 kPa); T  = 1000°F (537.8°C); S  = 1.7623 Btu/(lb · °F)
                                                                                 4
                                                   4
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