(1)p1?3MPa、t?300?C; (2)p1?5MPa、t?155?C; (3)p1?0.3MPa、x?0.92;
试利用水和水蒸气热力性质表查出或计算出各状态的比体积、熵和热力学能。
解:(1)p?3MPat?300℃
v?0.081226m3/kg h?2992.4kJ/kg s?6.5371kJ/(kg?K) u?h?pv?2992.4kJ/kg?3?106Pa?0.081226m3/kg?2748.722kJ/kg
(2)p?5MPa t=155℃
v?0.0010768?0.0010988?0.0010768?(155?140)?0.0010933m3/kg
160?140678.19?592.23h?592.23??(155?140)?656.7kJ/kg
160?1401.9377?1.7345s?1.7345??(155?140)?1.8869kJ/(kg?K)
160?140u?h?pv?656.7?5?103?0.0010933?651.2335kJ/kg
(3) p?0.3MPa x=
v??0.0010732m3/kg v???0.60587m3/kg
h??561.58kJ/kg h???2725.26kJ/kg s??1.6721kJ/(kg?K) s???6.9921kJ/(kg?K)
v?(1?x)v??xv???0.5575m3/kg s?(1?x)s??xs???6.5665kJ/(kg?K)
h?(1?x)h??xh???2552.1656kJ/kg u?h?pv?2384.92kJ/kg
工程热力学作业
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