Thermodynamics

Chapter 10 | 573State 3:State 4:P 3 1.2 MPaf v 3 v f @ 1.2 MPa 0.001138 m 3 >kgSat. liquid h 3 h f @ 1.2 MPa 798.33 kJ>kgP 4 15 MPas 4 s 3w pump II,in v 3 1P 4 P 3 2 10.001138 m 3 1 kJ>kg23115,000 12002 kPa4a1 kPa # b m3 15.70 kJ>kgh 4 h 3 w pump II,in 1798.33 15.702 kJ>kg 814.03 kJ>kgState 5:State 6:P 5 15 MPaT 5 600°C f h 5 3583.1 kJ>kgs 5 6.6796 kJ>kg # KP 6 1.2 MPaf h 6 2860.2 kJ>kgs 6 s 5 1T 6 218.4°C2State 7:P 7 10 kPas 7 s 5x 7 s 7 s f 6.6796 0.6492 0.8041s fg 7.4996h 7 h f x 7 h fg 191.81 0.8041 12392.12 2115.3 kJ>kgThe energy analysis of open feedwater heaters is identical to the energyanalysis of mixing chambers. The feedwater heaters are generally well insulated(Q . 0), and they do not involve any work interactions (W . 0). Byneglecting the kinetic and potential energies of the streams, the energy balancereduces for a feedwater heater toE # in E # out S ainm # h aoutm # horwhere y is the fraction of steam extracted from the turbine (m . 6 /m. 5 ). Solvingfor y and substituting the enthalpy values, we findThus,andy h 3 h 2 798.33 193.01h 6 h 2 2860.2 193.01 0.2270q in h 5 h 4 13583.1 814.032 kJ>kg 2769.1 kJ>kgq out 11 y2 1h 7 h 1 2 11 0.22702 12115.3 191.812 kJ>kg 1486.9 kJ>kgh th 1 q outq inyh 6 11 y2h 2 1 1h 3 21486.9 kJ>kg 1 0.463 or 46.3%2769.1 kJ>kg