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Homework Help: Piston Problem for Engineering Thermodynamics

  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
    cv0 = 0.717 kJ kg-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?


    Attached Files:

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  2. jcsd
  3. Jun 22, 2009 #2
    In order to do the PV = mRT, you'll need to know the pressure and the volume (temperature is known). The pressure can be found using a water table (you probably have one). It's a table showing all of the pressures, densities, and other information at certain temperatures.
    You'll find the volume of the air this way as well since you'll probably find the density of the water in the table.
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