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Piston Problem

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Problem shown in attach picture

    A cylinder of total internal volume 0.10 m3 has a
    frictionless internal piston (of negligible mass and
    thickness) which separates 0.50 kg of water below
    the piston from air above it. Initially the water exists
    as saturated liquid at 120°C and the air, also at
    120°C, exerts a pressure such that it exactly
    balances the upward force exerted by the water.
    (refer to the figure right). heat is then transferred into
    the entire cylinder, such that the two substances are
    at the same temperature at any instant, until a final
    State 2 having a uniform temperature of 180°C is
    reached. The air may be assumed to behave ideally,
    having R = 0.287 kJ kg-1
    K-1; cp0 = 1.004 kJ kg-1
    K-1;
    cv0 = 0.717 kJ kg-1
    K-1.
    (a) What is the mass of air in the cylinder?
    (b) Determine both the volume below the piston, and the dryness fraction of
    the water occupying that volume when State 2 is reached.
    (c) What is the change in the total internal energy of:
    (i) the air; and
    (ii) the water
    during the entire heat addition process from State 1 to State 2.
    (d) How much heat has been transferred into the complete system comprising
    both water and air?


    2. Relevant equations

    I really have no idea, I was trying with pV=mRT

    3. The attempt at a solution

    Im totally stuned by this question, I tried doing pV=mRT but that was wrong the a) is 0.175kg. COuld someone please help me how I could start going through this problem?
     
  2. jcsd
  3. Jun 22, 2009 #2

    Astronuc

    User Avatar

    Staff: Mentor

    What is the saturation pressure of water at 120°C? Think about how to use the mass of water, 0.50 kg, and the specific volume of the sat liquid at 120°C. I'm assuming that the water is completely liquid in state 1.

    When heat is added the water will change phase. The steam (vapor) occupies more volume than the liquid. As liquid transforms to vapor, the piston pushes on the air, which is compressed.


    The question in State 2 - does the water at 180°C remain saturated, or is it superheated or is it a compressed liquid?
     
  4. Jun 22, 2009 #3
    Thanks for the reply so, for saturated water at T=120 degress the Psat=0.1985MPa and v=0.001060m^3 given the mass of the saturated liquid is 0.50kg and the total volume of the piston is 0.10m3. So the air is taking up 0.09894m3 of volume. How can I link it to how much mass it is?Do I have to convert specific volume back to volume first? I could just do mv=V right?

    Thanks for the help
     
  5. Jun 24, 2009 #4

    Astronuc

    User Avatar

    Staff: Mentor

    Remember the specific properties are give per unit mass, so v=0.001060 m3/kg. The liquid will occupy a volume of V = m v = kg * m3/kg. Be careful in writing units.

    The volume of liquid and volume of air must = 0.10 m3, so as the volume of liquid/vapor increases the volume of air must decrease.
     
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