工程热力学课后题答案

第 章第三章3-3由热力学第一定律QUWWmgh450kg9.8m/s230m132300J因为绝热Q=0U132300JmCVT4.186103(T220)5kgT226.32OC3-4根据热力学第一定律，循环0QW过程Q（kJ)W(kJ)△E1-21100011002-30100-1003-4-9500-9504-1050-503-13热力学能的变化①采用定值比热容U CV T, h CpTMCV7R078.31429.099kJ/(kmolgK)22MCP(29.0998.314)kJ/(kmolgK)37.413kJ/(kmolgK)UMCV(600900)198.34kJ44.014MCp(600900)255.007kJh44.014②采用平均比热容6009001.104kJ/(kggK),CPm01.040kJ/(kggK),CPm0h1.1049001.040600369.6kJU(1.040.0.1889)600(1.1040.1889)900312.9kJ3-17忽略流体动能、位能变化m1'h1 m2'h2 (m1' m2')h3''120(500273)210(500273)mT11m2T2T3m2'120210m1'582.09K309.09oC第四章4-21）由可逆绝热定熵方程得S 01WmRT11P20.287(273150)1(0.1)1P11.410.52）

1.411.4 111.87kJWUmCV(T2T1)0.7174(300423)88.25kJSCPlnT2Rlnp2(0.2870.7174)ln3000.287ln0.10.117kJ/(kggK)T1p14230.5(3)WRTlnp20.287423ln0.5195.3kJp10.1SRlnp10.462kJ/(kgK)p2(4)n1RT11n(0.1)WP20.2874231n1P210.51SCPlnT2Rlnp20.462kJ/(kggK)T1p1

2167kJ4-61）T2T130oCWmRTlnP160.287303.15ln0.3573.5kJP20.1qW573.5kJ2）11.41T2T1P2303.15(0.1)1.4221.48oCP10.3Wm1R(T1T2)6110.287(303.15221.48)351.5kJ11.4q0kJ3）n11.21nT2T1P2303.15(0.1)1.2252.4oCP0.31Wm1R(T1T2)610.287(303.15252.4)436.7kJn11.21qnk(T2T1)1.21.460.7174(252.4303.15)218.45kJnmCV1.214-71nln(P2/P1)ln(0.12/0.6)1.6094ln(v1/v2)ln(0.236/0.815)1.31.2393W1(Pv1Pv2)1(0.61030.2360.121030.815)146kJn1121.31UCV(T2T1)0.7174(0.60.2361030.121030.815)109.5kJ/kg0.2870.287QPv11T1,P2v2T2RRhCP(T2T1)153.3kJ/kgSCPlnv2CvlnP20.1144ln0.8150.7174ln0.120.089kJ/(kggK)vP0.2360.6114-11①PvRTv1RT10.287573.150.274m3/kg111P0.6mPa,T573.185K1②v23v10.822m3/kgv1111.41T2T1v2573.15(3)369.3KP2RT20.287369.3128.95kPav20.822③v3v10.274m3/kgT2T2369.3KP3RT30.287369.3386.8kPav30.274W1R(T1T2)RT2lnP21P3110.287(573.15369.3)1.287369.3ln128.9529.8kJ/kg1.4386.84-16nln(P2/P1)ln(P2/P1)ln(v1/v2)ln(T1gP2)T2P1n1WnPv[1(P2)n]s,nn111P1QmPv110.1106400RT1287289.15

ln(0.5/0.1)1.13ln(0.5g289.15)0.1248.15400[11.1311.130.1106(0.5)1.13]1178.7kW1.131600.1482kgHmCP(T2T1)4821.0044(348.15289.15)28563kJQHWS285631178.76042159kJ/min第五章5-31-2过程：定压吸热Q12mCP(T2T1)1.011000300707kJW12RT2T10.2871000300200.9kJ2-3过程：可逆绝热膨胀Q230,T3T1300KW231RT2T311000300502.25kJ10.2871.413-1过程：定温放热1T3p3p31.478kPaT2p2W31Q31RTlnp30.287300ln1.478362.817kJp1100净功W0Q12Q31344.18kJW0344.1848.68%