热工基础复习题2013计算题答案

.word格式,1B精品文档放心下载CA上感谢阅读C1B2ABC、1和2和，感谢阅读kk谢谢阅读谢谢阅读将和k精品文档放心下载图1260kg精品文档放心下载；；）谢谢阅读图210Pa51b1V=A*L=100*10*10m31×10Pa感谢阅读52b2T=T21Vm感谢阅读32感谢阅读m=p11V/(RTm=p1g122V/(RT)2g2,专业..word格式,V=pV/p=2.00910*1010=1.4145m=m21553312mu-mu2211m=mT=TΔ感谢阅读12K21p*A*10*100×J52感谢阅读3感谢阅读300K的谢谢阅读3感谢阅读图3精品文档放心下载ABT=T感谢阅读2UUU(TT)200B1AB即：mT)mc(TcAVA1BV

2T=T=T=300KT=T=300K谢谢阅读112p精品文档放心下载2(mm)RTRT

ppVpVABgV22()2g2A1A1B1B1RTRTV

Vg1gA1B1

1T(pp)2VVT0.15MPa1A1B1B1(V)VA1B1RppVp）2A11.918J/pSAmAgKA11ATpA1

2ppVp

SR2B11.352JpBmg/KB1B1B1BTpB12SSS0.566J/KAB,专业..word格式,P=15MPat谢谢阅读11P=1.4MPat=tP=0.005MPa感谢阅读AR12q)(hh)(hh114感谢阅读RA(153.9)(3694.82892.8)4231.37kJ/kgqhh2401.1122263.22kJ/kg231q12263.22q46.5%t4231.3721感谢阅读qqtPm1Pqq4231.37mt1/h76.24kg/s274t4C=1.4kJ/(kg.K)C=1.14kJ/(kg.K)谢谢阅读pvQ1；Q2；。,专业..word格式,图4Q1mCp(t2t1)4116KJQ2t1QQ21TT21Q2T0Q1TT0mCdTpT1TQTT22Tp2389.2kJ20TTT10p

1WQQ21726.810.6MPa精品文档放心下载？精品文档放心下载t=20感谢阅读0100m/h精品文档放心下载3谢谢阅读,专业..word格式,sgCpTT21Rgp0.62ln0(kg)p0.11。）T2sT1(ppk12)k488.9K1C(T)(2s)C(T)Cp21Tlnp2730.6sCR221.004ln0.287lnTglng0.0678kJ1pp202730.11iT2930.067819.866kJ/skg0g(kg)）T1PqCm=p1p=0.1p2(T2T=MPa)C1pq(Tp2T1v1RTg10.1103100)1(25020)7.6KW0.287293谢谢阅读按附表1取空气的定值比热容为：kJ/(kg·K)，kJ/(kg·K)谢谢阅读气体常数kJ/(kg·K)（）定温过程对等温过程：，则：,专业..word格式,（）定熵过程对定熵过程：＝，＝Δ；＝Δh由等熵过程的过程方程知kJ/kgkJ/kg（）多变过程kJ/kgkJ/kgkJ/kgp=0.1t=50℃，压缩比6谢谢阅读11q=750kJ/kg精品文档放心下载1,专业..word格式,(a)1.4(b)0t,v1101111.4151.16%W0q1tvkg7500.5116387.7kJ/T3T2，因120150)6TV2T11V2qCT)q1v032q750TT1T3max

20.718661.711706.28KCv0PP和3223PT即P3而P233PT22PmaxPV022.578611.4P则P311V222.578612.286031.6807barT=300Kp=0.15MPa感谢阅读11感谢阅读感谢阅读,专业..word格式,对谢谢阅读对V'V

WqRTlnRTlnTT2

1VV11287.1V220.91872m3/VP10.152287.1WT10.5742m3/kgP0.1510.2871300ln/kguT0PPVVs'TRlnRln1Rln1

Rln2s0s01'1'PPV'V111110.2871)/(kgK)对2Wv=0qvuu2u1344.70914.07130.63kJ/kg图a图bqm8kg/s4精品文档放心下载：01.4，T3T29541180K,专业..word格式,W010011.410.28711180111.41RT011.41123602.73kJ/WCS01.40.28712951.40111012111.411.41

0psmwTswcs8602.73306.472370.11kWtp111011121.450.83%1.41051MPa精品文档放心下载0.12106kJ/h谢谢阅读8谢谢阅读图5,专业..word格式,[解]：参考图、8-11及8-12，由P'10.2MPa，查h图得h'1560kJ/kg由P'10.2MPa沿等熵线向上与P'2MPa线交于2'点查得h2s'1800kJ/kg因为：CSW理Wh2s'h'2h'1h'1则h2'h3h15701800hh15601'2s'1'0.80cs440kJ/kglgP-hh图）1860kJ/kg(1)氨的：QQ0.1210/36006m0.02980kg/s22qhh156044021'4'(2)压气机出口温及耗功：h1860kJ/kgT'2P'线交点，查lgP-h图可得：T'2386KP/m(hCh)/s0.29801560)/0.808.94kW2MPa等压(3)制系数论：q2Wcs实际：'csh1560h4404'180015601'hh2's1'4.66170.803.73364.6617(4)却水却水带走的热应等于氨气放出的热，即由热平衡方程得：Q11m(hh)m'C3pt,专业..word格式,m'm(hh)0.02980(186044.0)/s34.18784.1878可见所需却水是相当大的。'mm谢谢阅读砖，各层