Thermodynamics

Chapter 10 | 563(a) This is the steam power plant discussed in Example 10–1, except thatthe condenser pressure is lowered to 10 kPa. The thermal efficiency isdetermined in a similar manner:State 1:State 2: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 >kgP 2 3 MPas 2 s 11 kJw pump,in v 1 1P 2 P 1 2 10.00101 m 3 >kg2313000 102 kPa4a1 kPa # b m3 3.02 kJ>kgh 2 h 1 w pump,in 1191.81 3.022 kJ>kg 194.83 kJ>kgState 3:P 3 3 MPaT 3 350°C fh 3 3116.1 kJ>kgs 3 6.7450 kJ>kg # KState 4:Thus,andTherefore, the thermal efficiency increases from 26.0 to 33.4 percent as aresult of lowering the condenser pressure from 75 to 10 kPa. At the sametime, however, the quality of the steam decreases from 88.6 to 81.3 percent(in other words, the moisture content increases from 11.4 to 18.7 percent).(b) States 1 and 2 remain the same in this case, and the enthalpies atstate 3 (3 MPa and 600°C) and state 4 (10 kPa and s 4 s 3 ) are determinedto beThus,andP 4 10 kPa1sat. mixture2s 4 s 3x 4 s 4 s f 6.7450 0.6492 0.8128s fg 7.4996h 4 h f x 4 h fg 191.81 0.8128 12392.12 2136.1 kJ>kgq in h 3 h 2 13116.1 194.832 kJ>kg 2921.3 kJ>kgq out h 4 h 1 12136.1 191.812 kJ>kg 1944.3 kJ>kgh th 1 q outq inh 3 3682.8 kJ>kgh 4 2380.3 kJ>kg1x 4 0.9152q in h 3 h 2 3682.8 194.83 3488.0 kJ>kgq out h 4 h 1 2380.3 191.81 2188.5 kJ>kgh th 1 q outq in1944.3 kJ>kg 1 0.334 or 33.4%2921.3 kJ>kg2188.5 kJ>kg 1 0.373 or 37.3%3488.0 kJ>kg