Thermodynamics

Chapter 10 | 567(b) To determine the thermal efficiency, we need to know the enthalpies atall other states:State 1:P 1 10 kPaf h 1 h f @ 10 kPa 191.81 kJ>kgSat. liquid v 1 v f @ 10 kPa 0.00101 m 3 >kgState 2:P 2 15 MPas 2 s 1w pump,in v 1 1P 2 P 1 2 10.00101 m 3 >kg21 kJ 3115,000 102kPa4 a1 kPa # b m3 15.14 kJ>kgh 2 h 1 w pump,in 1191.81 15.142 kJ>kg 206.95 kJ>kgState 3:State 4:ThusP 3 15 MPaT 3 600°C fh 3 3583.1 kJ>kgs 3 6.6796 kJ>kg # KP 4 4 MPaf h 4 3155.0 kJ>kgs 4 s 3 1T 4 375.5°C2q in 1h 3 h 2 2 1h 5 h 4 2 13583.1 206.952 kJ>kg 13674.9 3155.02 kJ>kg 3896.1 kJ>kgq out h 6 h 1 12335.1 191.812 kJ>kg 2143.3 kJ>kgT, °C15 MPa360015 MPa3Reheating5Boiler4High-PturbineLow-Pturbine15 MPa4ReheaterP 4 = P 5 = P reheat26 10 kPa10 kPa515 MPaCondenser16Pump10 kPa2 s1FIGURE 10–13Schematic and T-s diagram for Example 10–4.