工程热力学习题解答

④平均吸收热温度T1,和平均放热温度T2

?s23?cplnT1?T2?3?T3pTT?Rgln3?cpln3,同理?s12?cpln3 T2p2T2T2c(T?T)923?500.6q1 ?p32??690.4T923?s23cpln3lnT2500.6T1K?T1?4? cp(T3?T2)553.2?300q1???413.8KT553.2?s14lncpln3300T29.8 某电厂以燃气轮机装置产生动力,向发电机输出的功率为20MW,循环简图如图9.12,循环最低温度为290K,最高为1500K,循环最低压力为95KPa,最高为950KPa。循环中设一回热器,回热度为75%。压气机绝热效率ηcs=0.85,燃气轮机的相对内部效率ηt=0.87。①试求燃气轮机发出的总功率、压气机消耗的功率和循环热效率;②假设循环中工质向1 800K的高温热源吸热,向290K的低温热源(环境介质)放热,求每一过程的不可逆损失。

解:据提议,T1=290K、T3=1500K、p1=95kpa、p4=950kpa

??pmax950Tmax1500??10 ??5.1724 ??pmin95Tmin290k?1kpT2k?T1(2)p1p?T13)p1k?1k?T1?k?1k?290?101.4?11.4?559.88K

T2'?T1?T2?T1?csk?1k?290?1k?1k559.88?290?607.51K

0.85?1500?0.11.4?11.4?h?h2'T7?T2',所以

据回热度概念,??7'?h4'?h2'T4'?T2'pT4?T3(4)p3?T3()?776.95K

T7?T2'??(T4'?T2')?607.15?0.75?(870.94-607.51)?805.09K

同样 T8?T4'??(T4'?T2')?870.94?0.75?(870.94-607.51)?673.37K ①q1?h3?h7?cp(T3?T7)?1.005?(1500?805.09)?698.4kJ/kg

q2?h8?h1?cp(T8?T1)?1.005?(673.37?190)?385.3kJ/kg wnet?q1?q2?cp(T8?T1)?698.4?385.3?313.1kJ/kg

P?qmwnwtP20?106?qm???63.88kg/s 3wnwt313.1?10PT?qmcp(T3?T4')?63.88?1.005?(1500?870.94)?40.4?103kW PC?qmcp(T2'?T1)?63.88?1.005?(607.51?290)?20.4?103kW ②??1?q2?1?385.3?44.8%

tq1698.4 ?s??s?clnT2'?1.005?ln607.51?0.08205kJ/(kg?K) 12'22'pT2559.88?s73?cplnT31500?1.005?ln?0.62538kJ/(kg?K) T7805.09?s27?cplnT7805.09?1.005?ln?0.28299kJ/(kg?K) T2'607.51T4'870.94?1.005?ln?0.11477kJ/(kg?K) T4776.95?s34'??s44'?cpln?s4'8?cpln?s81?cplnT8673.73?1.005?ln??0.25856kJ/(kg?K) T4'870.94T1290?1.005?ln??0.84663kJ/(kg?K) T8673.37压缩过程不可逆损失:

I1?qmT0?s12'?63.88?190?0.08205?1520kJ/s

吸热过程不可逆损失:

?s73?sf2?sg2

sf2?q1698.4??0.388kJ/(kg?K) Tr1800sg2??s73?sf2?0.62538?0.388?0.23738kJ/(kg?K) I2?qmT0?sg2?63.88?290?0.23738?4397.5kJ

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