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I Pressure and temperature of expanding humid air

  1. May 11, 2017 #1
    As a follow-on to this thread, which in turn followed this closed thread, I'm starting a new thread on a related real-world problem.

    Say I have a 2-liter plastic soda bottle filled partway with water, and pressurized. It has been sitting inverted (with the opening pointed down, and sealed) for long enough to reach a state of equilibrium in terms of pressure, temperature, and humidity.

    The cap is removed and only the water is allowed to escape before the cap is replaced. What is the final temperature and pressure of the air in the bottle?

    Assumptions:
    • The bottle is rigid and insulating.
    • The water escapes rapidly (not a slow leak, it's gone less than 0.1 sec),
    • Ignore anything else that may have negligible effects (not sure what they would be at the moment).

    Defining some values:
    ##V_F## = final volume of air in the bottle = 2.17 liters = 0.00217 m3
    ##w## = volume fraction of water (1/3 full for this exercise)
    ##V_0 = (1-w)V_F## = initial volume of air in the bottle (then ##V_0=\frac{2}{3}0.00217## = 0.001445 m3)
    ##P_0## = initial pressure = 100 psi over 1 atm (1 atm = 101,325 Pa, therefore P0=790,801 Pa)
    ##T_0## = initial temperature = 20°C (293 K)
    ##P_F## = final pressure in the bottle when the cap is replaced
    ##T_F## = final temperature of the air in the bottle when the cap is replaced
    ##\gamma = 1.4## = gas constant used in formulas, actually more like 1.398 for humid air according to literature I found, but we'll use 1.4.​

    There are various answers to this problem that don't account for humidity.

    Plugging in the values above into the isentropic (adiabatic) expansion formulas, we get:
    $$P_F = P_0 \left ( \frac{V_0}{V_F} \right ) ^ \gamma = 448,270 \text{ Pa}\tag{1}$$
    $$T_F = T_0 \left ( \frac{P_F}{P_0} \right ) ^ \frac{\gamma - 1}{\gamma} = 249\text{ K} = -44\text{°C}\tag{2}$$
    This is saying that even though the bottle was only 1/3 full of water initially, the pressure dropped to almost half of the original pressure instead of 2/3 the original pressure predicted by the isothermal ideal gas law. And the temperature may well be sub-zero for dry air, but I'm skeptical that's what happens in the real world with humid air forcing out the water.

    Example 2, equations [5] and [7], in https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/ gives these results, first solving for ##P_F## in [7]:

    $$P_F = P_0 \left [ \frac{V_0}{V_0 + \gamma (V_F - V_0)}\right] = 465,177\text{ Pa}\tag{3}$$
    $$T_F=T_0 \left [ 1-\frac{(\gamma -1)}{\gamma}\frac{(P_0-P_F)}{P_0} \right ] = 259\text{ K} = -34\text{°C}\tag{4}$$
    (I may be misapplying these formulas to this problem.) The final pressure and temperature lie between the isentropic expansion and isothermal ideal gas values: ##P_F = P_0 \frac{V_0}{V_F} = 535,427\text{ Pa}## and ##T_F=T_0##.

    The above results are for dry air. They aren't realistic because the air isn't dry. Final temperatures so far below 0°C are unlikely. So I'm wondering how to account for the presence of humidity in these pressure and temperature results. Is there a formula for this? I know it can get complicated, I found information on Clausius-Clapeyron equation and Kirchoff's equation, involving latent heat at the interface between liquid and gas phases.

    All I really need is a reasonable approximation that gives me a number that looks agreeable to real-world observations, in which one might observe fog forming in the bottle, but it's unlikely to be ice at sub-zero temperature due to the modulating effect of the water vapor in the air.
     
    Last edited: May 11, 2017
  2. jcsd
  3. May 12, 2017 #2
    I didn't check your calculations. This addresses your question about hot much heat can be supplied by condensing and freezing the water vapor in the air. At room temperature, the equilibrium vapor pressure of water vapor in air is 0.0234 bars. This is 2371 Pa. This compares the initial pressure of the air of 790701 Pa. So, if the air is 100% relative humidity to begin with, the initial mole fraction of water in the air is 0.00300. If we take as a basis 1.0 moles of air/water vapor mixture, we would have 0.00300 moles of water vapor and 0.997 moles of air. From the steam tables, the specific enthalpy of water vapor at 20 C is 2538 J/gm and the enthalpy of liquid water at 0 C is 0 J/gm. The heat of fusion of water is 334 J/gm. So the enthalpy of ice at 0 C is -334 J/gm. The heat capacity of ice is 2.06 J/gm-C. So cooling ice from 0 C to -44 C would require 91 J/gm. So the enthalpy of ice at -44 C is -334-91=-425 J/gm. So the change in enthalpy between water vapor at 20 C and ice at -44 C is 2963 J/gm, or 53334 J/mole. Since we have only 0.003 moles of water vapor in the air, the amount of heat that this can remove from the air is only (53334)(0.003)=160 J. The molar heat capacity of air is 3.5R = 29 J/mole-C of air. So the heating that will occur if 160 J is used to heat 1 mole of air is only 5.5 C. This is if the air is 100 % humidity. If the air is only 70 % humidity, the heating will be only 3.9 C.
     
  4. May 12, 2017 #3
    Thanks for that. It took me a bit of studying but I am following it. My physics education never involved steam tables so this is a first-time look for me.
    Just clarifying: Are you using 18 g/mole for air? I thought it was more like 29 g/mole, in which case I'd get 85927 J/mole for the total change in enthalpy, and 258 J of heat that can be removed.
    Molar heat capacity of air is a constant? I thought there was a pressure dependency on that. Likely I am confused about something I read in the past.

    Anyway, if I'm correct about the molar density of air being 29 g/mole, the heating would be more like 10°C.

    Either way, the effect of the moisture isn't as big as I thought. That's surprising. I wouldn't expect the temperatures to go so low in real life. If only fog forms in the bottle and not sub-zero ice, I can think of two possibilities that would contribute to a more moderate reduction in temperature:
    • The bottle wall (something like 250 microns of PET plastic) may actually be a poor insulator, even for the 60 milliseconds or so duration of this process.
    • The thermal mass of water in the bottle is huge compared to the air, and this can prevent the air temperature from going too low.
    Couple of questions:
    1. How would this change of temperature affect the final pressure? Would I take this final adjusted temperature and back-calculate pressure from the adiabatic formula, as ##P_F = P_0 \left ( \frac{T_f}{T_0} \right ) ^ \frac{\gamma}{\gamma-1}## ?
    2. Is it correct for to assume this is an irreversible process, and that your example 2 I quoted earlier would be more appropriate?
     
  5. May 12, 2017 #4
    I used 18 g/mole for water.
    No. In this region of low pressures, the heat capacity of air is not significantly affected by pressure. It is a very weak function of temperature only, and, over this temperature range, the temperature-dependence can be neglected.

    Sorry, but you're not correct.
    It's bigger than I expected.
    1. It's not worth the trouble.
    If the water leaves very fast, then example 2 is closer.
     
  6. May 12, 2017 #5
    Ah, I get it. This is the heat contribution of the WATER VAPOR in the air. Not the air that has water vapor in it. Now it all makes sense.

    Thank you for the clarification.

    "It's not worth the trouble" to calculate the resulting pressure...? In this example, an inverted pressurized bottle that ejects its water mass in 60 milliseconds will take off like a rocket. If the purpose here is to predict the altitude reached by such a rocket a few seconds later, a small difference in pressure in the thrust phase makes a noticeable difference in final altitude. Running my simulation using the isothermal ideal gas law for pressure, and again using the adiabatic expansion formula for pressure, makes a difference of about 12% in final altitude (84 versus 94 meters), which is significant.

    In my simulation I'm accounting for a number of things, among them air drag, thrust from the final burst of air, and additional ballast mass -- with a bottle weighing only 50 g with a large frontal area, and having a drag coefficient of about 0.3, the air resistance almost slams it to a stop when the water is exhausted, so it needs some ballast to coast up higher. There's also an initial optimum mass of water -- too little and the thrust is too short in duration, too much and the rocket is too heavy to get very high. Slightly less than 1/3 full seems to be a sweet spot when starting with 100 psi of pressure.
     
    Last edited: May 12, 2017
  7. May 12, 2017 #6
    If I wanted to take into account the water vapor contribution in a simple way, I would fake out the math by modifying the heat capacity of the air a little. A little higher heat capacity for the air would result in a little lower value for the specific heat ratio gamma. So, I would try some modified values until the final temperature came out a few degrees higher. Probably, something like 1.38 would do the trick.
     
  8. May 12, 2017 #7
    Fantastic! I had thought of that a few days ago but didn't try it because (a) I had no idea how to calculate the temperature difference due to water vapor, and (b) I'm hesitant to make an arbitrary change to a known physical constant without knowing if that's a reasonable approach. You solved both of those problems today.

    I found the fulltext of an article tabulating values of gamma at different temperatures and relative humidity values at 1 atm - http://asa.scitation.org/doi/10.1121/1.391597 - which showed a value γ=1.398 for 100% RH at 20°C, and I noticed that it raised the final temperature slightly versus the value of 1.4, as well as raised the final pressure.

    For the problem initially described, raising the final temperature 5.5° requires setting γ=1.346 for the adiabatic expansion formula, and 1.329 for the irreversible process formula from your example 2.

    And, it turns out also that in my simulation, now that I've tested this, the difference in maximum altitude of slightly less than 1 meter, which isn't significant after all. There is still a significant difference between these gas expansion calculations and my initial model that assumed P×V=constant, however.
     
  9. May 13, 2017 #8
    Yeah. The way to figure out what value of gamma to use is something like this. If the 160 J resulted from cooling by 60 C, that would be equivalent to a 3.7 J/mole C average increase in the heat capacities of the air at const. volume and const pressure. So, Cv would increase to about 20.8+3.7=24.5, and Cp would increase to about 29.1+3.7=32.8. So, the equivalent gamma over our temperature range would be about 32.8/24.5=1.34. That agrees pretty well with your estimate.

    Assuming the PV is constant is the same as saying that the gas temperature is constant during the rapid expansion. Do you think that's a better approximation?
     
  10. May 13, 2017 #9
    That's good to know. I got it by iteration.
    No, that's worse because it doesn't model reality. The air in the bottle does get pretty cold, cold enough for water vapor to condense inside, and cold enough to affect the pressure pushing the water out. Assuming constant temperature results in significantly higher thrust, causing the simulation to over-estimate the final altitude. The current world record altitude for an unreinforced 2-liter bottle rocket is 86 meters. If that's near the theoretical maximum, it agrees almost perfectly with my simulation using the adiabatic expansion formula and the modified gamma.
     
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