Thermodynamics reversible isothermal process

AI Thread Summary
A reversible isothermal process involving 10 kg of water starts as saturated vapor at 300°C and ends at a pressure of 1 MPa. The specific volumes were calculated using thermodynamic tables, yielding values of v1 = 0.02167 and v2 = 0.25794. The heat transfer (Q) during the process was calculated to be 2362.7 kJ, with the assumption that the process is adiabatic. The discussion also raises questions about the equilibrium vapor pressure of water at 300°C and its comparison to 1 MPa, as well as the energy required to convert 10 kg of saturated vapor to saturated liquid at that temperature. Understanding these thermodynamic principles is crucial for solving similar problems in thermal systems.
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Homework Statement


A system containing 10kg of water undergoes a reversible isothermal process. The initial state can be characterized as saturated vapor at 300°C. The pressure in the final state is 1MPa. The heat transfer during the process in (inKJ)


Homework Equations



Q = m(U2-U1)+W
W = PdV
v1 = V/m
v1 = specific volume at initial state
V = volume

The Attempt at a Solution


The specific volume values were found using the thermodynamic tables for water. I figured that since it was an isothermal process and would be adiabatic. I wanted somebody to see if my answer is correct.

v1=.02167
v2=.25794
V1= 2.5795
V2=2.5794
mass=1.91
T1 =T2 =300°C
Q = mCvo(T2-T1)+W
Q = .717(300-300)+PdV

Q = 0 +1000(2.5794-.2167)
Q = 2362.7
 
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What is the equilibrium vapor pressure of water at 300C? How does that compare with a pressure of 1 MPa? What is the volume of the water vapor at the equilibrium vapor pressure?
 
I would go as follows:

1. Assume liquid water is incompressible.
2. What is the vapor pressure of saturated vapor/water at 300C? Is it above or below 1 MPa?
3. What's it take in joules to change 10 kg of saturated vapor to saturated liquid (at 300C)?
 
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