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I Heating water and air in a constant volume

  1. Nov 23, 2018 #1
    I found a simulation tool on wolfram alpha where the pressure of a constant volume is calculated. The pressure is affected by the heating of a mixture of water and air inside the container. There is an explanation to how the pressure is calculated which seems pretty straight forward. But there is one thing... in the beginning of the Detail section the final volume is calculated. Is this the Specific volume of water, or something else..?

    The simulation tool with explanation: http://demonstrations.wolfram.com/HeatingWaterAndAirInASealedContainer/
     
  2. jcsd
  3. Nov 23, 2018 #2

    mjc123

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    No, it's the actual volume of water in the container.
     
  4. Nov 23, 2018 #3
    Do you know how to calculate it? I can't find the equation, or more precise the constants they use in the equation in the explanation.
     
  5. Nov 23, 2018 #4

    mjc123

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    I assume they're embedded in the software. As the aim is to demonstrate the software, not enable you to do it yourself, they don't tell you.
     
  6. Nov 23, 2018 #5
    Yeah maybe..?

    Anyhow, I have an approximation i set yp myself that seems to be pretty close to empirical values i found on the e-net. Not fancy pancy as the wolfram alpha stuff though.
     
  7. Nov 23, 2018 #6
    What are your thoughts on how to calculate this? What physical property data do you think is needed?
     
  8. Nov 23, 2018 #7

    BvU

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    If you pick up the player and download the source ("author code") you can open the .nb file and see the expressions, e.g. :

    vol2 = (-1*^-8*T^3 + 6*^-6*T^2 - 2*^-5*T + 0.99695)*vol1;​

    Much more legible than what's in the .nb file itself:
    Code (Text):

          RowBox[{"vol2", "=",
           RowBox[{
            RowBox[{"(",
             RowBox[{
              RowBox[{
               RowBox[{"-", "1*^-8"}], "*",
               SuperscriptBox["T", "3"]}], "+",
              RowBox[{"6*^-6", "*",
               SuperscriptBox["T", "2"]}], "-",
              RowBox[{"2*^-5", "*", "T"}], "+", "0.99695"}], ")"}], "*",
            "vol1"}]}], ";", "\[IndentingNewLine]", "\[IndentingNewLine]",
     
    btw, I was surprised the stuff expands that much. Then again, 250 C is pretty hot. Nice eye-opener.
     
  9. Nov 23, 2018 #8
    How do you parse that?
    For example, what is A? "-1*^-8"??

    I presume it's a polynomial fit to some experimental data set. I looked at the NIST chemistry webbook and it puts the volume increase from 25 to 250°C at just under 25%, which would very nearly fill the container in the simulation.
     
  10. Nov 23, 2018 #9
    I can describe how I would approach this problem, first (a) without dissolution of N2 and O2 in the liquid water and then (b) with dissolution. But I would first like to hear from the OP about his articulation of the physical mechanisms involved.
     
  11. Nov 23, 2018 #10

    BvU

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    Come on ! $$v_2 = (-1*10^{-8}*T^3 + 6*10^{-6}*T^2 - 2*10^{-5}*T + 0.99695)*v_1$$As Clooney would say: What else ?
     
  12. Nov 23, 2018 #11
    The missing "10". It seems so obvious now. Thanks!
     
  13. Nov 24, 2018 #12

    BvU

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    Wholeheartedly subscribe to this. Expressions and equations are useless if not applied properly. Having a tool at your disposal is not equivalent to knowing what goes on !
     
  14. Nov 27, 2018 #13
    My first approximation was ## P_{tot} = P_{air} + P_{water} ##

    I use the ideal gas law for the air pressure and Antoine's equation to calculate the water pressure. Given the simulation tool on wolfram alpha I use some approximation they don't use. In addition to this, I always get a tad nervous when I'm getting closer to the world of chemistry.

    Edit, thanks for all help. I'll try to set up a model based on the polynomial fit used in the wolfram alpha tool.
     
  15. Nov 27, 2018 #14
    My approach to this would be similar to yours, but a little more fleshed out. Let:

    ##\rho_L(T)## represent the density of liquid water at temperature T
    ##P(T)## represent the equilibrium vapor pressure of water at temperature T
    ##V_0## represent the volume of liquid water in the container at 25 C
    ##p_0## represent the pressure in the container at 25 C (in this case 1 atm)

    Then the volume of gas in the container at 25 C is (V-V_0), where V is 1 liter. The mass of air in the container at 25 C (as well as all other temperatures) is then $$m_a=\frac{[p_0-P(25)]M_a}{298R}(V-V_0)$$where ##M_a## is the molecular weight of air. The mass of water in the container at 25 C (as well as all other temperatures) is: $$m_w=V_0\rho_L(25)+\frac{P(25)M_w}{298R}(V-V_0)$$where ##M_w## is the molecular weight of water. The first term on the right hand side of this equation is the mass of liquid water and the second term is the mass of water vapor.

    Now for temperature T. Let f represent the fraction of the water that is liquid at temperature T. Then the mass of water in the gas phase must be ##m_a(1-f)##. And the total volume of liquid and gas must match the volume of the container V. This means that:
    $$\frac{m_wf}{\rho_L(T)}+m_w(1-f)\frac{RT}{P(T)M_w}=V$$The first term on the left hand side of the equation is the volume of liquid water, and the second term is the volume of gas. This equation allows us to solve for f. Once f is known, we can solve for the partial pressure of air at temperature T as: $$p_{air}=\frac{m_aRT}{M_a\left[V-\frac{m_wf}{\rho_L(T)}\right]}$$Total pressure in the container at temperature T is then $$p=p_{air}+P(T)$$
     
    Last edited: Nov 27, 2018
  16. Nov 27, 2018 #15
    I think you meant ##m_w## instead of ##m_a## in a few places.

    Your approach will result in a higher total pressure than in the simulation. On the other hand, whereas you ignore the air going in and out of solution the simulation assumes that it reaches equilibrium effectively instantaneously. If the animation is supposed to be real-time I think that dissolution rates would have to be considered. Am I wrong? Would it actually happen quickly enough as the pressure/temperature rises?
     
  17. Nov 27, 2018 #16
    Really? Like where?
    Well, certainly, the dissolution of air is neglected in my analysis. But this can be included rather readily. As far as the rate effect is concerned, I'm not convinced that the animation describes the variations in real time. Is time included as a parameter in their equations?
    This would have to be quantified with a transient analysis.
     
  18. Nov 27, 2018 #17
    Here and in the denominator of the equation that followed this one:

    $$\frac{m_af}{\rho_L(T)}+m_a(1-f)\frac{RT}{P(T)M_w}=V$$


    I don't remember the simulation mentioning time. It wasn't in the accompanying equations.
     
  19. Nov 27, 2018 #18
    Ah. Thanks. I'm going back and changing that.
     
  20. Nov 27, 2018 #19
    ... and in the sentence preceding the equation. :-)
     
  21. Nov 28, 2018 #20
    Hi, I saw this now. I'm sorry I missed it and that I didn't notice your contribution in the thread. I'll walk through your ideas for solving it, when I have the time. Well now there's tad to much... life. :-)
     
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