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

Jacob White
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.

JD_PM and etotheipi

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?

Mentor
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.

JD_PM and PhDeezNutz
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.

Mentor
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?

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

PhDeezNutz
Jacob White
Yes.

Mentor
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?

Mentor
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

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Mentor
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?

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

Mentor
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.

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

Mentor
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##

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

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

Mentor
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.

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

Mentor
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

Mentor
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

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

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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

Mentor
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

Mentor
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?

Mentor
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

Mentor
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