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Thermodynamic properties - Isochronic process - First law

  1. Feb 11, 2016 #1
    1. The problem statement, all variables and given/known data
    An amount of water substance having a mass of 1 kg is contained in a vessel at a pressure of 1 MPa. It is heated for some time after which the pressure is 3 MPa. The volume of the vessel is 0.1 m3. Assuming no work is done by or to the substance, calculate the final temperature of, and the heating done to, the contents of the vessel.

    2. Relevant equations
    I am not sure about them. I believe, they would be:

    ## m=m_g + m_f ## , ##V=v_f m_f +v_g m_g## and ##PV=nRT## , ##ΔQ+ΔW=ΔU##
    Thermodynamic and transport properties of fluids.

    3. The attempt at a solution

    I have stated that ##m_f+m_g=m=1kg## and ##v_g m_g + v_f m_f = 0.1 m^3##. It is also known that ##P_1=1MPa=10 bar## and ##P_2=30 bar##.

    I have attempted to read the value of a temperature from 'Saturated Water and Steam section' for ##P=10bar## which would give me ##179.9°C##. Assuming that volume is constant throughout the whole process as the mixture is in a vessel it gave me ##{\frac{P_2 T_1}{P_1}}=T_2## but it gave me the wrong answer.

    I have then tried to obtain the value of ##v_g## for 'Saturated Water and Steam' and it the value would be 0.1944 ##[kg/m^3]##. Solving for ##m_f## and ##m_g## by assuming that since ##m_f+m_g=1kg## and so ##dryness fraction=x=mg/1=mg## did actually either got me nowhere or I just do not know what to do with these values.

    I need help to know what kind of assumptions should I use and have I chosen appropriate equations for solving this problem. Thanks
     
  2. jcsd
  3. Feb 11, 2016 #2
    You got off to a very good start, but then you hit a snag. Let's confine attention to the initial state. Your determination of the dryness fraction was not correct. You are aware that there is both liquid and vapor in the container to begin with, correct? So you need to take into account both the specific volume of the liquid and the specific volume of the vapor. Let x (kg) represent the mass of vapor in the tank and 1-x represent the mass of liquid in the tank. In terms of x, what is the volume of vapor and what is the volume of liquid. These have to add up to 0.1 m^3. So, you can use this to solve for x. Once you know x, you can determine the initial internal energy of the water in the tank using the specific internal energy of the vapor and the specific internal energy of the liquid. Try this, and let's see what you get.

    Chet
     
  4. Feb 12, 2016 #3
    Hi Chet,

    Thanks for replying.

    So I attempted again and I think I again have hitten a wall:

    Reading from the tables for Saturated Water and Steam for ##P=10 bar## : ##v_g=0.1944## and ##v_f=0.001128## and assuming that ##V=v_f (1-x) + v_g x##
    ##0.1= (1-x) 0.001128 + x 0.1994##
    ##x=0.511##

    Then using equation ##e=(1-x)e_f+xe_fg## where ##e_g## and ##e_f## are values for ##P=10 bar## read from Saturated Water and Steam table I have got:
    ##e=1693##

    Now I do not know with what do I compare that? I noted that final pressure is above the critical pressure point for water. Should it guide me somewhere?

    Thanks
     
  5. Feb 12, 2016 #4
    Sure. Look in the Superheated tables. But, first you need to calculate the average specific volume of the tank contents (if you have not already done so).
     
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