- #36
Jacob White
- 55
- 12
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.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.
Of course.Jacob White said:I think I could do it in C++ or Python. Use the Clausius-Clapeyron equation for ps for each temperature?
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)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:
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.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?
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.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.
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.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.
Yes. In a reversible process, the system passes through a continuous sequence of thermodynamic equilibrium states.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?
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.What about condensation and freezing are these also reversible?
Sure. Then otherwise there would be a finite rate of heat transfer between the moist air and the ice and liquid water.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?
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: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)
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.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?
Very nice Jacob. Using the vapor pressure of ice at the final temperature, what do you now predict for the amount rained per kg?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.
It doesn't seem possible. What vapor pressure did you estimate at the final temperature?Jacob White said:Much more - about 3.737g.
That sounds right, but it implies 2.32 grams/kgJacob White said:0.5084 kPa