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1、直径为1m，高为2m的桶内充满p=102.4kPa，t=32，75%的湿空气经复杂过程后达终态若终态时=75%的湿空气，经复杂过程后达终态，若终态时t=18，=90%，求：1）终态压力p2；2）凝结水量；）凝水量；3）过程中空气放热量。解：1）2231m2m1.57m44Vd L=查表32 C4 759kPatp=o查表32 C4.759kPastp=1110.75 4.759kPa3.56925kPavspp=18 C2.0626kPastp=o0 9 2 0626kP1 85634kP2220.9 2.0626kPa1.85634kPavspp=a11v1102.4kPa3.56925kP
2、a98.83075kPappp=()3a1a1g 198830.75Pa 1.57m1.7726kg287J/(kg K)27332 Kp VmR T=+a2g2a1g2m R Tm R T忽略凝水体积()a2g2a1g2a221 7726kg287kJ/(kg K)273 18 KpVV=+()31.7726kg287kJ/(kg K)273 18 K94294.4Pa1.57m+=94 2944kPa1 85634kPa96 15kPappp+2）3v13569.25Pa 1.57m0 0398kgp Vm=2a2v294.2944kPa1.85634kPa96.15kPappp=+=+=
3、2）v1g,v 13v20.0398kg461.5J/(kg K)305K1856.34Pa 1.57m0 0217kmR Tp V=v2v2g,v20.0217kg461.5J/(kg K)209KpmR T=v2v10.0217kg0.0398kg0.0181kglmmmm=即析水0.0181kg或者v113.56925kPa0.6220.6220.02246kg/kg98 83075kPpd=干空气1a1g g98.83075kPap21 85634kPap空气()ddv22a21.85634kPa0.6220.6220.01201kg/kg96.15kPapdp=干空气()()a211
4、.7726kg0.01201 0.02246 kg/kg0.01852kgmmdd=干空气3）取闭口系0QUWW=+=Q()()jiQmumu=aa2v2v2aa1v1v1llQm um umum um u=+()aa2a1v2v2v1v1=+llmuum umum uuu()aa2a1v2v2v1v1ll()=uuc TT218 C132 Cvvuuuuoopvhu=a2a121()=vuuc TT18 Cluu=opvhu()1.7726kg 0.7176kJ/(kg K)1832C0.0217kg2400.2kJ/kg0.0181kg 75.58kJ/kg0.0398kg2419.3kJ/kg60.6kJQ=+=ogggg