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Jacob White

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- #36

Jacob White

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- #37

Chestermiller

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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

Jacob White

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Ok, that was dumb question, sorry for that. Thank you for all the help!

- #39

Chestermiller

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- #40

Jacob White

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- #41

Chestermiller

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Of course.

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Jacob White

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- #43

Chestermiller

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- #44

Chestermiller

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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)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:

- #45

Jacob White

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Thank you again for the help, that was very instructive!

- #46

Chestermiller

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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

Jacob White

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- #48

Chestermiller

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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

Jacob White

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- #50

Chestermiller

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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.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.

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

Jacob White

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- #52

Chestermiller

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Yes. In a reversible process, the system passes through a continuous sequence of thermodynamic equilibrium states.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?

- #53

Jacob White

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- #54

Chestermiller

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Sure. Then otherwise there would be a finite rate of heat transfer between the moist air and the ice and liquid water.

- #55

Jacob White

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Thank you for explanation, so I think now we can go further.

- #56

Chestermiller

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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?

$$d\omega_S+d\omega_L+d\omega_V=0\tag{2}$$

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

Jacob White

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$$\omega_L = \frac {-L_s p_s} {L_f P}, + \frac {273R lnP} {L_f} + c$$

Putting numbers c is approximately -4.029.

Then I solve for 0 moles in wolfram but I got some nonsense value(85kPa)

- #58

Chestermiller

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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:

$$\omega_L = \frac {-L_s p_s} {L_f P}, + \frac {273R lnP} {L_f} + c$$

Putting numbers c is approximately -4.029.

Then I solve for 0 moles in wolfram but I got some nonsense value(85kPa)

$$-0.0028L_F=-L_S(0.606)\left(\frac{1}{P}-\frac{1}{74.1}\right)+273R\ln{(P/74.1)}$$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

Jacob White

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- #60

Chestermiller

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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

Jacob White

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- #62

Chestermiller

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Very nice Jacob. Using the vapor pressure of ice at the final temperature, what do you now predict for the amount rained per kg?

- #63

Jacob White

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Much more - about 3.737g.

- #64

Chestermiller

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It doesn't seem possible. What vapor pressure did you estimate at the final temperature?Much more - about 3.737g.

- #65

Jacob White

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0.5084 kPa

- #66

Chestermiller

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That sounds right, but it implies 2.32 grams/kg0.5084 kPa

- #67

Jacob White

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Sorry again, I forgot about molar masses. 3.737*18/29=2.32

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