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Thermodynamics reversible isothermal process

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data
    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)


    2. Relevant equations

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

    3. 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
     
  2. jcsd
  3. Nov 25, 2012 #2
    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?
     
  4. Nov 26, 2012 #3

    rude man

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    Gold Member

    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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