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

[Jacob White]

Homework Statement

https://www.ioc.ee/~kree/students/iphoTable/files/ipho/1987_EGermany_p1.pdf

Relevant Equations

https://www.ioc.ee/~kree/students/iphoTable/files/ipho/1987_EGermany_p1Sol.pdf

I'm struggling with explanation of part 3. I don't know why they are using adiabatic equation while the gas is constantly heated by condensating vapour. While we are deriving adiabatic equation we use the fact, that there is no additional heat put into the system. Thank you in advance.

 

[archaic]

Moist air is streaming adiabatically across a mountain range as indicated in the figure.

 

[Jacob White]

But from point M1 it is constantly heated by condensating vapour so how it could remain streaming adiabatically?

 

[Chestermiller]

It is adiabatic for the combination of air and water. No external heat enters or leaves any constant-mass parcel of this combination. That is what adiabatic means. When water condenses, its internal energy decreases and is partly balanced by an increase in internal energy of the bone dry air fraction of the combination. So the change in internal energy of the combination is equal to minus the amount of expansion work done by the parcel on surrounding parcels.

 

[Jacob White]

Ok, but how could they use this adiabatic law? Below I have inserted derivation from wikipedia and they are using the fact that energy is expresed in simple way (nCvΔT) but it's not the case in our example.

 

[Chestermiller]

Ok, but how could they use this adiabatic law? Below I have inserted derivation from wikipedia and they are using the fact that energy is expresed in simple way (nCvΔT) but it's not the case in our example.
View attachment 265956

They told you to use the ideal gas law, even though some of the water vapor is condensing. This was done as a first approximation to the temperature at point 2. Then, as an approximate correction to account for the condensing water vapor, they added the additional temperature rise of 6 degrees.l. It isn't exact but, oh well.

 

[Jacob White]

However is it reasonable approximation? Assuming adiabatic flow we are predicting about 14 degrees change and correction is about 6 degrees, almost as half as first aproximation. Shouldn't such a big heat flow diametrically changed the process?

 

[Chestermiller]

Are you asking how I would do it if I wanted to do it more accurately?

 

[Jacob White]

Yes.

 

[Chestermiller]

Yes.

OK. But, to enhance your learning experience, you're going to have to help.

We are going to be tracking what happens to a parcel of moist air containing 1 mole of bone dry air and whatever amount of water vapor it contains as it moves from M1 to M2. So the "basis" of our calculation is 1 mole of bone dry air.

Focusing on M1, this is the location where water vapor just begins to condense (the dew point of the air). That must mean that the partial pressure of the water vapor at this location is equal to the equilibrium vapor pressure of water at the temperature of the parcel, 279 K = 6 C. What is the equilibrium vapor pressure of water at 6 C? If the total pressure is 84.5 kPa, what is the partial pressure of the bone dry air. From Dalton's law, what is the mole fraction of water vapor and of bone dry air? What is the mass fraction of water vapor in this mixture? What is the mass of water vapor in 1 kg of moist air at M1?

We'll continue after you address these questions.

 

[Jacob White]

Is it possible to calculate equilibrium vapor pressure of water? Or can I check it on the internet?

 

[Chestermiller]

Is it possible to calculate equilibrium vapor pressure of water? Or can I check it on the internet?

Check it on the internet

 

[Jacob White]

So at 6 C equilibrium vapor pressure of water is about 0,96kPa hence dry air has partial pressure qual:
P air = 84,5 - 0,96 = 83,54
Due to Dalton's law partial pressure is the same as these gases seperately were occupying the same volume:
P air * V = N air * RT
P w * V = N w *RT
So f = N w/ N air ≈0,01
If water vapour molar mass is: μ w and dry air: μ air then vapour is:
k = (f*μ w)/(f*μ w + μ air)
fraction of total parcel mass.
Using μ air=29 and μ w=18:
k≈6,2*10^-3
So in 1 kg there is 6,2 grams of vapour

 

[Chestermiller]

So at 6 C equilibrium vapor pressure of water is about 0,96kPa hence dry air has partial pressure qual:
P air = 84,5 - 0,96 = 83,54
Due to Dalton's law partial pressure is the same as these gases seperately were occupying the same volume:
P air * V = N air * RT
P w * V = N w *RT
So f = N w/ N air ≈0,01
If water vapour molar mass is: μ w and dry air: μ air then vapour is:
k = (f*μ w)/(f*μ w + μ air)
fraction of total parcel mass.
Using μ air=29 and μ w=18:
k≈6,2*10^-3
So in 1 kg there is 6,2 grams of vapour

Pretty close. I get 0.011 mole fraction of water vapor and 0.989 mole fraction of dry air. So mass fraction water vapor = $\frac{(0.011)(18)}{(0.11)(18)+0.989)(29)}=0.0069$, or 6.9 grams per kg moist air. This compares with the 2.45 grams that eventually rain out.

If we had neglected the contribution of the water vapor to the total pressure and total mass, we would have obtained a water vapor mole fraction of 0.93/84.5 = 0.011 and a mass fraction of (0.011)(18)/29 =0.0068, or 6.8 grams per kg. This is a very good approximation and will simplify the subsequent analysis. So we wll continue it: $$\omega_w=\frac{p_s(T)}{P}$$where $p_s(T)$ is the equilibrium vapor pressure at temperature T and P is the total pressure. We will also treat this as the moles of water vapor per mole of dry air.

For the reversible adiabatic expansion of the ideal gas mixture we are dealing with, the first law tells us that $$dU=-PdV$$, or, equivalently,
$$dH=VdP$$where V is the volume of the mixture, neglecting any liquid water that condenses out and neglecting the contribution of the water vapor to the total volume of the parcel. So $$V=\frac{(1)RT}{P}$$. In determining the change in enthalpy for the parcel, we will neglect the sensible heat of the water vapor and liquid water, and only include the latent heat of condensation. So, we write $$dH=(1)C_pdT+Ld\omega_w$$where $C_p$ is the molar heat capacity of air (3.5R) and L is the molar heat of vaporization of the water. So we have $$C_pdT+Ld\omega_w=\frac{RT}{P}dP$$

Questions so far?

 

[Jacob White]

No, you have greatly explained and everything is clear so far.

 

[Chestermiller]

No, you have greatly explained and everything is clear so far.

Excellent. So we have $$d\omega_w=\frac{1}{P}\frac{dp_s}{dT}dT-\frac{p_s}{P^2}dP$$
In the region of interest, $p_s$ can be described by the Clausius-Clapeyron equation: $$p_s(T)=p_s(279)\exp{\left[\left(\frac{L}{RT}\right)\frac{(T-279)}{279}\right]}$$From this, it also follows that $$\frac{dp_s}{dT}=p_s(T)\frac{L}{RT^2}$$So our differential equation now becomes $$\left[C_p+\frac{p_s(T)}{P}\frac{L^2}{RT^2}\right]dT=\left[\frac{RT}{P}+L\frac{p_s(T)}{P^2}\right]dP$$or, equivalently,
$$C_p\left[1+\frac{(\gamma-1)}{\gamma}\left(\frac{L}{RT}\right)^2\frac{p_s(T)}{P}\right]dT=\frac{RT}{P}\left[1+\left(\frac{L}{RT}\right)\frac{p_s(T)}{P}\right]dP$$Note the frequent appearance of the grouping $\frac{L}{RT}$ in this equation and in the Clausius-Clapeyron equation.

Please make some estimates of the magnitudes of the two terms in brackets in this equation over the region between M1 and M2.

 

[Jacob White]

It looks like these complicated terms are close to 0 and both brackets are roughly equal to 1.

 

[Chestermiller]

It looks like these complicated terms are close to 0 and both brackets are roughly equal to 1.

That's not what I get. Let's see a sample calculation.

Here is my sample calculation for conditions at M1:

$p_s=0.93\ kPa$ (from my steam tables)
P = 84.5 kPa
T = 279 K
$\frac{L}{RT}=\frac{45000}{(8.314)(279)}=19.40$

$$term =\frac{0.4}{1.4}(19.4)^2\frac{0.93}{84.5}=1.18$$

 

[Jacob White]

Oh sorry for that. So the term on the right is approximately 1+0,21 ?

 

[Chestermiller]

Oh sorry for that. So the term on the right is approximately 1+0,21 ?

Much better. So the ratio of the right bracketed term to the left bracketed term at location M1 is R = 1.21/2.18 = 0.555. Assuming that the solution they obtained for the temperature at location M2 was approximately correct (271 K), what value do you calculate for this ratio R at M2?

Since our differential equation cannot be integrated analytically, our game plan will be to integrate numerically (approximately) between M1 and M2 using the average value $\bar{R}$ over the interval for the integration:
$$\frac{d\ln{T}}{d\ln{P}}=\frac{(\gamma-1)}{\gamma}\bar{R}$$where the right hand side is now treated as approximately constant. This is called a Crank-Nicholson second-order accurate finite difference approximation.

 

[Jacob White]

So I got R≈5,888 at M2 and after averaging:
ln T = ln P * 0,572 * 0,4/1,4

 

[Chestermiller]

So I got R≈5,888 at M2 and after averaging:
ln T = ln P * 0,572 * 0,4/1,4

Yyou mean 0.588? What do you get for ps(T2)?
Constant of integration or integration limits? What do you get for T2 from the integration?

 

[Jacob White]

Yes, 0,588 I used 0,61kPa. ln T = ln P *0,572*0,4/1,4 + c. Plugging values from M1 I got c≈3.77

 

[Chestermiller]

Yes, 0,588 I used 0,61kPa. ln T = ln P *0,572*0,4/1,4 + c. Plugging values from M1 I got c≈3.77

So what do you get for T2 from the integration?

 

[Jacob White]

268,6 K so indeed it is close to the first approximation

 

[Chestermiller]

268,6 K so indeed it is close to the first approximation

I get 270.5. Please show your calculation.

 

[Jacob White]

At M1:
c≈ ln(279)-ln(84500)*0,572*0,4/1,4≈3,777
So at M2:
s≈ln(70000)*0,572*0,4/1,4 + 3,777≈5,593
T = e^s≈268,6

 

[Chestermiller]

At M1:
c≈ ln(279)-ln(84500)*0,572*0,4/1,4≈3,777
So at M2:
s≈ln(70000)*0,572*0,4/1,4 + 3,777≈5,593
T = e^s≈268,6

I get 5.60 for s

 

[Chestermiller]

I did it this way: $$\frac{T_2}{T_1}=\left(\frac{P_2}{P_1}\right)^{\frac{(\gamma-1)}{\gamma}\bar{R}}$$
$$T_2=279\left(\frac{70}{84.5}\right)^{0.1634}=270.5$$

 

[Jacob White]

So we get almost the same result as their's simple approximation. How we could predicted before that this simpler method would also work?

 

[Chestermiller]

So we get almost the same result as their's simple approximation. How we could predicted before that this simpler method would also work?

I don't think we could have. In my judgment, it was just fortuitous. I would never have done it their way.

Incidentally, from the Clausius-Clapeyron equation, what is the final amount of water vapor per kg of moist air at M2, and how does the loss of rainfall predicted from that compare with their estimate of the amount of rainfall?

 

[Jacob White]

I get ps(T)≈0,522 and further 4,628g of vapour in 1 kg. So 6,9g-4,628g=2,72g have condensated so about 10% more than their's 2,45g

 

[Chestermiller]

I get ps(T)≈0,522 and further 4,628g of vapour in 1 kg. So 6,9g-4,628= 2.72g have condensated so about 10% more than their's 2,45g

check your subtraction.
They needed the 2.45 to end up with a final temperature of 271 using their method. Our value of 2.27 is self-consistent with this temperature. Thoughts?

 

[Jacob White]

Oh sorry again for that mistake. Now indeed it is consistent. These addtional 0,18g exactly contributes to missing 0,5 degrees.

 

[Chestermiller]

I think they knew the final temperature and specified whatever rainout amount necessary to make their crazy method seem to work.

 

[Jacob White]

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.

 

[Chestermiller]

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]

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

 

[Chestermiller]

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).

 

[Jacob White]

I think I could do it in C++ or Python. Use the Clausius-Clapeyron equation for ps for each temperature?

 

[Chestermiller]

I think I could do it in C++ or Python. Use the Clausius-Clapeyron equation for ps for each temperature?

Of course.

 

[Jacob White]

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:

 

[Chestermiller]

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!

 

[Chestermiller]

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)

 

[Jacob White]

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

 

[Chestermiller]

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.

 

[Jacob White]

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?

 

[Chestermiller]

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?

 

[Jacob White]

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.

 

[Chestermiller]

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.