Ipho 1987, thermodynamics problem: Moist air ascending over a mountain range

In summary, the conversation discusses the use of the adiabatic equation in a system where moist air is constantly heated by condensing water vapor. The ideal gas law is used as a first approximation to calculate the temperature change, with a correction added for the condensing water vapor. However, the accuracy of this approximation is questioned. To make a more accurate calculation, the equilibrium vapor pressure of water at a given temperature is needed. This can be found online and used in conjunction with Dalton's law to find the partial pressure of dry air and water vapor. The fraction of water vapor in the total parcel mass is then calculated, leading to the conclusion that in 1kg of moist air, there is approximately
  • #36
And I have last question about 5). They treat this air as it wouldn't condensate anymore from M2 to M3. But at M2 there is still some vapour in the gas and I don't why it should stop condensating at M2.
 
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  • #37
Jacob White said:
And I have last question about 5). They treat this air as it wouldn't condensate anymore from M2 to M3. But at M2 there is still some vapour in the gas and I don't why it should stop condensating at M2.
It has already equilibrated at M2, and the liquid water has rained out. In going from M2 to M3, the parcel gets recompressed and reheats. The liquid water can no longer evaporate (since it is no longer there) so the parcel heats up more than if the liquid water were still there to evaporate and modulate the reheating.
 
  • #38
Ok, that was dumb question, sorry for that. Thank you for all the help!
 
  • #39
To get an even more accurate numerical solution to the differential equation, I'm thinking of integrating it numerically using a Forward Euler 1st order accurate finite difference integration scheme with 100 pressure steps of 0.145 bars each. I can do this on an Excel spreadsheet, but I have to decide whether this is really worth the effort (or whether we have already beat this one to death).
 
  • #40
I think I could do it in C++ or Python. Use the Clausius-Clapeyron equation for ps for each temperature?
 
  • #41
Jacob White said:
I think I could do it in C++ or Python. Use the Clausius-Clapeyron equation for ps for each temperature?
Of course.
 
  • #42
So I get 270.27K as a final temperature. I have done calculations using dP step equal 0.01 Pa and I have verified it by comparing result to 0.001Pa step. Suprisingly I get relation very close to linear function:
 

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  • #43
It took me 15 min to implement the integration with Excel that I indicated in post #39. For the temperature at M2, I got 270.3 K, compared to the value 270.5 K that we got with the simpler method. The predicted amount of water condensing was 2.42 g/kg vs the 2.45 g/kg in the problem statement. Not bad!
 
  • #44
Jacob White said:
So I get 270.27K as a final temperature. I have done calculations using dP step equal 0.01 Pa and I have verified it by comparing result to 0.001Pa step. Suprisingly I get relation very close to linear function:
I just noticed that we both got the same result for the final temperature (and, thus, for the rainout). Very satisfying. (Actually, for an additional significant figure, I also got 270.27 K)
 
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  • #45
Thank you again for the help, that was very instructive!
 
  • #46
There is another interesting aspect of this physical system that we have, up to now, neglected. You will notice that the calculated final temperature is close to 270 K, which is about 3 degrees C below the freezing point of water. Our numerical solution indicates that this temperature occurs at a pressure of about 74 kPa, compared to the final pressure at M2 of 70 kPa. So the question is: how should we proceeded if we had properly taken this into account?

If the liquid water rained out immediately as it formed, we should have simply changed the differential equation to use the heat of sublimation (and the vapor pressure relationship for ice) at subsequent lower pressures. But what about if the condensed water remained entrained in the parcel of moist air being tracked until the parcel reached M2? I offer this up as a challenge problem. It took me a few hours to figure out how to proceed below 74 kPa with both ice and liquid water being taken into account. Good luck.
 
  • #47
So a little bit more vapour would stop being vapour and the question is what is the fraction of disappearing vapor that is sublimating and how much more would change phase?
 
  • #48
Jacob White said:
So a little bit more vapour would stop being vapour and the question is what is the fraction of disappearing vapor that is sublimating and how much more would change phase?
It issue is, what happens with the liquid water. When the parcel temperature hits 0 C (at about 74 Pa), the parcel consists of a mixture of liquid water fog and moist air. If the liquid water freezes suddenly to ice, what would happen to the heat given off? If it reheated the ice, it would just change back to liquid, and if part of the heat went into the air, the temperature would go above 0 C; plus, the temperature increase would cause the air pressure to increase; but it can't increase because the pressure is controlled externally.

If the liquid water froze gradually, on the other hand, we know that ice and water can exist simultaneously only at 0 C, so the mixture temperature would have to stay constant at 0 C while the pressure continued to decrease and the mixture expanded; but how can the mixture expansion be adiabatic and isothermal at the same time? (This is just a rhetorical question, since I worked this all out before I posted the challenge).

So the question is, mechanistically, what actually would happen and how can we quantify this?
 
  • #49
Previously I was thinking that this liquid water quickly leaves the parcel and go out as rain. But if it stays in the parcel, I don't see a contradiction. Mixture of ice and liquid water could stay at 0 C while the rest of the parcel continues cooling. Then from condensating vapor, new portion of water(at lower temperature) would be constantly appearing and heating this portion with heat flow to the rest of the parcel could take the energy from freezing. This would complicate our equations however I don't see a contradiction.
 
  • #50
Jacob White said:
Previously I was thinking that this liquid water quickly leaves the parcel and go out as rain. But if it stays in the parcel, I don't see a contradiction. Mixture of ice and liquid water could stay at 0 C while the rest of the parcel continues cooling.
Since the parcel is experiencing an adiabatic reversible expansion, at any time during the change, the parcel will be essentially in a state of thermodynamic equilibrium. For this to happen, the air would also have to stay at 0 C during this part of the change.
Then from condensating vapor, new portion of water(at lower temperature) would be constantly appearing and heating this portion with heat flow to the rest of the parcel could take the energy from freezing. This would complicate our equations however I don't see a contradiction.
Actually, this is not what would be happening. Until all the water has frozen to ice, the temperature of the ice, liquid water, and moist air will all be stationary at 0 C. The continued expansion cooling will exactly balance the latent heat effect of liquid changing to ice. So, even though the parcel expansion continues as the pressure continues to decrease, the temperature of the entire parcel will not be changing. The only thing that will be happening energy-wise will be that the proportions of liquid water, ice, and water vapor in the parcel will be changing.

So as the pressure continues to drop and the liquid water continues to freeze to ice, the parcel will stay at 0 C until a pressure is reached where all the liquid water has frozen. After that point, the parcel temperature can resume decreasing again with further decreases of pressure.

Are you comfortable with this description? If so, I'll continue with how to quantitatively analyze this situation.
 
  • #51
I'm still not confident about it's temperature. So when the temperature of all components is equal then the process is reversible, so only the assumption of reversibility has given us this argument of equal temperatures? What about condensation and freezing are these also reversible?
 
  • #52
Jacob White said:
I'm still not confident about it's temperature. So when the temperature of all components is equal then the process is reversible, so only the assumption of reversibility has given us this argument of equal temperatures?
Yes. In a reversible process, the system passes through a continuous sequence of thermodynamic equilibrium states.
What about condensation and freezing are these also reversible?
Sure. Have you ever heard of an ice bath, with ice and liquid water, both at 0 C? All the ice doesn't suddenly change to liquid water and all the liquid water doesn't suddenly change to ice. And you can obviously have steam and water both in contact at 100 C and 1 atm pressure, and all the steam doesn't suddenly change to water and all the water suddenly doesn't all change to steam.
 
  • #53
But it's strictly the assumption of reversibility that ensure us that moist air will stay at the temperature of ice and liquid water, because otherwise it would be irreversible?
 
  • #54
Jacob White said:
But it's strictly the assumption of reversibility that ensure us that moist air will stay at the temperature of ice and liquid water, because otherwise it would be irreversible?
Sure. Then otherwise there would be a finite rate of heat transfer between the moist air and the ice and liquid water.
 
  • #55
Thank you for explanation, so I think now we can go further.
 
  • #56
The variation of the parcel enthalpy with pressure during the interval of declining pressures over which the temperature remains constant at 0 C is still given by: $$dH=VdP\tag{1}$$Since the temperature of the bone dry air does not change, its contribution to dH is zero. Only the relative changes in the amounts of water vapor, liquid water, and ice determine dH. Let ##\omega_S##, ##\omega_L##, and ##\omega_V## represent the number of moles of ice, liquid water, and water vapor present in our parcel at any pressure P in the interval. Then, from a mass balance on the water, we have: $$\omega_S+\omega_L+\omega_V=\omega_{total}$$where ##\omega_{total}=\frac{p_s(279)}{84.5}## is the total number of moles of water in the parcel initially, a constant. From this it follows that the changes in the relative molar amounts are given by:
$$d\omega_S+d\omega_L+d\omega_V=0\tag{2}$$

The change in enthalpy of the overall parcel in the pressure interval over which the temperature remains constant at 0 C is then given by: $$dH=h_Sd\omega_S+h_Ld\omega_L+h_Vd\omega_V\tag{3}$$
where ##h_S##, ##h_L##, and ##h_V## are the molar enthalpies of saturated ice, saturated liquid water, and water vapor relative to an arbitrary reference state. If we eliminate ##d\omega_S## between Eqns. 2 and 3, we obtain: $$dH=(h_L-h_S)d\omega_L+(h_V-h_S)d\omega_V$$where ##(h_l-h_S)=L_F=6200\ J/mole## is the heat of fusion and ##(h_V-h_S)=L_S=(L+L_F)=45000 + 6200 = 51200\ J/mole## is the heat of sublimation. So we have: $$dH=L_Fd\omega_L+L_Sd\omega_V\tag{4}$$During the pressure interval where the temperature is constant, the mole fraction of water vapor in the parcel is given by $$\omega_V=\frac{p_s(273)}{P}$$So $$d\omega_V=-\frac{p_s(273)}{P^2}dP$$and thus our equation for the enthalpy change becomes:
$$L_Fd\omega_L=L_S\frac{p_s(273)}{P^2}dP+\frac{273R}{P}dP\tag{5}$$Because this equation is not a function of temperature, it can be integrated analytically to get the change in the moles of liquid water as a function of the decreasing pressure. It should be integrated from the pressure at which the temperature first hits 0C to the pressure at which the moles of liquid water has dropped to zero.

From your numerical solution, what is the pressure at which the temperature first hits 273 K? How many moles of liquid water are present in the parcel at this pressure?
 
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  • #57
So I get the pressure 74.108kPa and 0.008 moles of vapour. I have integrated equation(5) and get:
$$\omega_L = \frac {-L_s p_s} {L_f P}, + \frac {273R lnP} {L_f} + c$$
Putting numbers c is approximately -4.029.
Then I solve for 0 moles in wolfram but I got some nonsense value(85kPa)
 
  • #58
Jacob White said:
So I get the pressure 74.108kPa and 0.008 moles of vapour. I have integrated equation(5) and get:
$$\omega_L = \frac {-L_s p_s} {L_f P}, + \frac {273R lnP} {L_f} + c$$
Putting numbers c is approximately -4.029.
Then I solve for 0 moles in wolfram but I got some nonsense value(85kPa)
At 74.1, I get 0.0082 moles water in the vapor, and in the initial moist air I get 0.0110 moles water in the vapor. So the amount of liquid water at 74.1 kPa is 0.0028 moles. So, integrating, I get:
$$-0.0028L_F=-L_S(0.606)\left(\frac{1}{P}-\frac{1}{74.1}\right)+273R\ln{(P/74.1)}$$or, equivalently,
$$L_S(0.606)\left(\frac{1}{P}-\frac{1}{74.1}\right)+273R\ln{(74.1/P)}=0.0028L_F$$The solution to this for P must be less than 74.1
 
  • #59
So at 73.535kPa there would be no water molecules left. So from this pressure we could just go back to our first numerical solution and simply change the condensation heat to sublimation heat?
 
  • #60
Jacob White said:
So at 73.535kPa there would be no water molecules left. So from this pressure we could just go back to our first numerical solution and simply change the condensation heat to sublimation heat?
Yes, and also use this heat of sublimation in calculation of subsequent values for the vapor pressure (of ice), starting at 0.606 kPa at 273 K.
 
  • #61
So now after these corrections the final temperature I get is 270.89K which is about 0.6K higher than without considering ice. And that makes sense since we get some additional energy from freezing water.
 
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  • #62
Jacob White said:
So now after these corrections the final temperature I get is 270.89K which is about 0.6K higher than without considering ice. And that makes sense since we get some additional energy from freezing water.
Very nice Jacob. Using the vapor pressure of ice at the final temperature, what do you now predict for the amount rained per kg?
 
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  • #63
Much more - about 3.737g.
 
  • #64
Jacob White said:
Much more - about 3.737g.
It doesn't seem possible. What vapor pressure did you estimate at the final temperature?
 
  • #65
0.5084 kPa
 
  • #66
Jacob White said:
0.5084 kPa
That sounds right, but it implies 2.32 grams/kg
 
  • #67
Sorry again, I forgot about molar masses. 3.737*18/29=2.32
 
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