- #1

liquidFuzz

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*Specific volume*of water, or something else..?

The simulation tool with explanation: http://demonstrations.wolfram.com/HeatingWaterAndAirInASealedContainer/

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- Thread starter liquidFuzz
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- #1

liquidFuzz

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The simulation tool with explanation: http://demonstrations.wolfram.com/HeatingWaterAndAirInASealedContainer/

- #2

mjc123

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No, it's the actual volume of water in the container.

- #3

liquidFuzz

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- #4

mjc123

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- #5

liquidFuzz

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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.

- #6

Chestermiller

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What are your thoughts on how to calculate this? What physical property data do you think is needed?

- #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. :I can't find the equation, or more precise the constants they use in the equation in the explanation

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:

```
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.

- #8

JT Smith

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

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.

- #9

Chestermiller

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- #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 ?How do you parse that?

- #11

JT Smith

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The missing "10". It seems so obvious now. Thanks!

- #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 !But I would first like to hear from the OP about his articulation of the physical mechanisms involved

- #13

liquidFuzz

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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.

- #14

Chestermiller

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My approach to this would be similar to yours, but a little more fleshed out. Let:

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.

##\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:

- #15

JT Smith

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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?

- #16

Chestermiller

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Really? Like where?I think you meant ##m_w## instead of ##m_a## in a few places.

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?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?

This would have to be quantified with a transient analysis.Would it actually happen quickly enough as the pressure/temperature rises?

- #17

JT Smith

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Really? Like where?

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.

- #18

Chestermiller

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Ah. Thanks. I'm going back and changing that.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

JT Smith

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... and in the sentence preceding the equation. :-)

- #20

liquidFuzz

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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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