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morganhenderson6706
07.04.2020 •
Physics
Five kg of water is contained in a piston–cylinder assembly, initially at 5 bar and 360°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 2660 kJ and kinetic and potential energy effects are negligible. Determine the final volume, in m3, and the work for the process, in kJ.
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Ответ:
Explanation:
Given that,
Mass = 5kg of water
Initial pressure P•i = 5bar = 500 kPa
Pressure is constant P = 5bar
Initial temperature T•i = 360°C
Heat transferred Q = 2660 KJ
Kinetic and potential energy are negligible
a. Final volume
The initial enthalpy and specific volume can be determine from the steam table.
h•1 = 3188.83 KJ/kg
v•1 = 0.5796 m³/kg.
So, we can calculate the final enthalpy by using
Q = m(h2—h•1)
2660 = 5 (h2—3188.83)
2660/5 = h2—3188.83
h2 — 3188.83 = 532
h2 = 520 + 3188.83
h2 = 3720.83 KJ/kg.
Using steam table, we can determine the final temperature and final specific volume
T•f = 608°C
v•f = 0.8115 m³/kg
Then, specific volume = volume /mass
v•f = v2/m
v2 = vf × m
v2 = 0.8115 × 5
v2 = 4.0575m³
b. Workdone?
W = m•P•(vf —vi)
W = 5 × 500(0.8115 — 0.5796)
W = 579.75 KJ
Ответ: