4版讲稿例题a466277英

1、A466167英Consider a regenerative cycle using steam as the workingfluid. Steam leaves the boiler and enters the turbine at 4 MPaand 400 . After expansion to 400kPa, some of the steam isextracted from the turbine for purpose of heating the feedwater in an open feedwater heater.The pressure in the feedw

2、ater heater is 400kPa and the water leaving it is saturated liquid at 400kPa. The steam not extracted expands to 10kPa.Determine the cycle efficiency.1Analysis As in example A466167, the process is steady state and kinetic and potential energy changesare negligible.And from the example we have the f

3、ollowing properties:h5 = 3213.6kJ/kg,h7 =2144.1kJ/kg,h1 = 191.8kJ/kgAlso, at p6=400kPa, from the second law,s6 = s5= 6.7960kJ /(kg K)= s6 + x6 (s6- s6 )= 1.7769kJ /(kg K) + x4 (6.8961-1.7769)kJ /(kg K)x4 = 0.975h6 =h6+ x4 (h6- h6 )= 604.87kJ/kg + 0.975(2738.49 - 604.7)kJ/kg= 2685.6kJ/kg2Solution Fir

4、st consider the low-pressure pump Control volume: Low-pressure pumpInlet state:p1known, saturated liquid; state fixed Exit state: p2known.The first law iswPl= h2 - h12Thereforeh2 - h1= 1vdp= v( p2 -p1 )wPl=0.00101m3/kg (400 -10)kPa= 0.4kJh2 =h1 +wPl= 191.8kJ/kg+ 0.4kJ/kg= 192.2kJ/kg Next consider th

5、e feedwater heaterControl volume: Feedwater heater.Inlet states:States2 and 6 both known(as given).Exit state: p3known, saturated liquid; state fixed3The first law gives us Substituting, we obtaina1h6+ (1-a1 )h2= h32685.6kJ/kga1+ (1- a1 )192.2kJ/kg= 604.7kJ/kga1 = 0.1654 For the turbine we haveContr

6、ol volume: turbineInlet state:p5,T5known; state fixed Exit state: p6known; p7known.The first and the second laws arewt=(h5- h6 ) + (1-a1 )(h6- h7 )s5= s6= s7We can calculate the turbine workwt =(h5- h6 ) + (1- a1 )(h6- h7 )= (3213.6 - 2685.6)kJ/kg + (1- 0.1654)(2685.6 - 2144.2)kJ/kg4= 979.9kJ/kg Fin

7、ally for the boilerControl volume: BoilerInlet state:p4,h4known(as given); state fixed.Exit state:State 5 fixed(as given) .The first law isqH = h5- h4Substituting givesqH= h5- h4= 3213.6kJ/kg - 608.6kJ/kg= 2605.0kJ/kgTherefore,ht =wnet qH=975.7kJ/kg 2605.0kJ/kg= 37.5%5Note the increase in efficiency

8、 over the efficiency of the Rankine cycle of example A4 Let us turn now to the high-pressure pumpControl volume: High-pressure pump.Inlet state:state 3 known(as given). Exit state: p4known.The first and the second laws arewPh= h4- h3s3 = s4Substitution leads towPh= h4- h3= v ( p4 -p3 )= 0.001084m3/kg (4000 - 400)kPa = 3.9kJ/kgh4 = h3+ wPh=