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Homework Help: Thermodynamics rigid vessel

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A rigid vessel of volume 2.5 m3 contains steam at a pressure of 6 bar and a dryness fraction of 0.75. If the contents of the vessel are heated, determine:
    i) the pressure at which the steam becomes dry saturated,

    ii) the temperature of the steam when the pressure reaches 10.0 bar,

    iii) the heat transfer to the steam in attaining state b).

    2. Relevant equations
    Just notation
    Vg= specific volume of saturated vapour

    3. The attempt at a solution
    So i'm just really confused about this question
    Firstly its rigid so i know that the volume is a constant.
    So surely when the steam becomes dry saturated M*Vg = 2.5m^3
    To find mass i did 0.75 x Vg at 6 bar, i found this in steam tables to be 0.3156
    Vg = 0.2367 so mass of wet vapour = 2.5/0.2367 = 10.56 Kg, but mass of the whole contents is 10.56/0.75 = 14.08 kg
    When the steam becomes dry saturated M*vg = 2.5 so 2.5/14.08 = Vg
    Vg= 0.1775, vg@11 bar = 0.1774 so i made the conclusion that it must be at 11 bar.

    I then took a glance at my teachers solution he put up today, he has done something a lot different.
    He stated that Vg=0.75x0.3156 as did i.
    ∴Vg = 0.2367
    He then stated that this Vg lies between 8 and 9 bar on the steam tables
    So he did this:
    @9 bar vg = 0.2149
    @8 bar vg= 0.2403
    P=0.2403-0.2367/0.2403-0.2149 = 0.14
    So P = 8.14 Bar.
    I really don't follow what he's done here, i don't see any equation i've been given in this form.
  2. jcsd
  3. May 11, 2013 #2


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    Homework Helper

    Your teacher is doing a linear interpolation to find the pressure at which vg = 0.2367.
    He is assuming that the saturation line between P = 8 bar and P = 9 bar is a straight line.
  4. May 11, 2013 #3
    So with linear interpolation :


    d is the variable i want, ie pressure.
    g is the variable i've been given vg

    D1=8 bar
    D2=9 bar

    8+(0.2367-0.2403/0.2149-0.2403)*(9-8) = 8.142

    I don't see why the specific volume wouldn't change, surely when it becomes a dry vapour its volume will change?
  5. May 17, 2013 #4
    I'm not sure, but I don't think you did it quite right. Let M be the total mass of liquid water plus vapor in the vessel. Then, initially, the mass of liquid water is 0.25M, and the mass of vapor is 0.75M. Let vil represent the initial saturated specific volume of the liquid water and viv represent the initial saturated specific volume of the vapor. Then,


    This allows you to solve for the total mass of water in the vessel.

    The final specific volume is M/2.5
  6. May 17, 2013 #5
    Honestly my first instincts was to do this. Whats confused me in my teacher solution is that he's said the specific volume of the initial condition(where there is a dryness fraction) will be constant through the whole process as it goes from 8 bar and to 10 bar. But surely the specific volume can still change even if it is rigid, as the temperature increases and decreases the dryness fraction is going to change, at which one point it will be 1 and hence the specific volume will be its maximum value?
  7. May 17, 2013 #6
    What he is saying is that the specific volume of the combination of liquid and vapor (i.e., averaged over the liquid and vapor) does not change.
  8. May 18, 2013 #7
    Oops. In this post, I meant to write that the final specific volume is 2.5/M, not M/2.5.
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