Thermodynamics reversible isothermal process

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SUMMARY

The discussion focuses on a reversible isothermal process involving 10 kg of water, initially in a saturated vapor state at 300°C, transitioning to a final pressure of 1 MPa. The heat transfer (Q) during this process is calculated using the equation Q = m(U2-U1) + W, where specific volumes v1 and v2 were determined from thermodynamic tables. The calculated heat transfer value is 2362.7 kJ, and the equilibrium vapor pressure of water at 300°C is compared to the final pressure of 1 MPa to assess the state of the water.

PREREQUISITES
  • Understanding of thermodynamic processes, specifically isothermal and reversible processes.
  • Familiarity with thermodynamic tables for water properties.
  • Knowledge of the first law of thermodynamics and related equations.
  • Basic concepts of specific volume and heat transfer calculations.
NEXT STEPS
  • Research the properties of water at various temperatures using thermodynamic tables.
  • Learn about the implications of adiabatic processes in thermodynamics.
  • Study the concept of equilibrium vapor pressure and its significance in phase changes.
  • Explore advanced thermodynamic equations and their applications in real-world scenarios.
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Students and professionals in thermodynamics, mechanical engineers, and anyone involved in heat transfer and phase change processes in fluids.

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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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