对于简单林德双压空气液化系统来说,气流的流量比为qmi/qm=0.8,液化器操作压力为0.101MPa和20.2MPa,两台压缩机的进口温度保持在293K,中压为3.03MPa,由空气的压焓图查得:h1(0.101MPa,293K)=295kj/kg,h2(3.03MPa,293K)=286kj/kg,h3(20.2MPa,293K)=259kj/kg,hf(0.101MPa,饱和液体)=126kj/kg,则液化率y为( )。

2023-09-12

A.0.0552
B.0.0684
C.0.0772
D.0.0765

参考答案:B

y=(h1-h3)/(h1-hf)-i[(h1-h2)/(h1-hf)]=(295-259)/(295+126)-0.8×[(295-286)/(295+126)]=0.0684

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