I promised people a little more “Gedanken” about water condensing. I have to admit to still finding various individual discussions in comments at The Air Vent a bit confusing. This may be because the exact geometries and assumptions to each bit is not provided. (That’s a difficulty for discussing any sort of gedanken in comments.) It may be because everyone agrees on some specific geometry and set of assumptions, but I just don’t have the patience to figure out precisely which assumptions or items they are arguing about.
But, I did come across something sufficiently concrete to discuss. This is Weight of Water and Wind, Hurricane Pro’s Weigh in: Konrad’s experiment in comment 212.
This is what Konrad describes:
I have also conducted a simple experiment. While I know that steam will contract dramatically when condensing, I wanted to see if saturated air maintained its volume when condensation occurred. The experiment is simple, all that is required is a clear 2 litre PETG plastic drink bottle, a barbeque match and some hot water.
1. Pour 100ml of hot water into the plastic bottle, put the cap on and shake vigorously for 2 minutes. This will saturate and warm the air in the bottle and also warm the surfaces of the bottle.
2. Release the cap to equalise the pressure and carefully pour out all the water, trying not to disturb the warm moist air in the bottle. keep the bottle at a downward angle.
3. Light then extinguish the barbeque match and place the still smoking end inside the inverted bottle for a short time. The smoke will provide nucleation sites for condensation.
4. Slightly squeeze the bottle before replacing the cap tightly. Release the sides of the bottle once the cap is in place to cause slight expansion of the saturated air in the bottle.
5. Your cloud in a bottle should quickly form. After a few seconds the bottle will start to crumple inward. It will do so even when the sides of the bottle are still warm.
I’ve rigged up cartoons of containers on scales to kinda-sort of mimic the physics of Konrads plastic bottles, with some results below (except the actual weights are realistic. Assume we have tared for the weight of the plastic bottle, it would probably make more sense for me to use 0.01N or something similar.) In my discussion, I’ll assume the container initially hold water vapor with no air. (I’m doing this to simplify discussion of the physics.) When showing illustrations of steam diagrams, I’m going to assume Konrad does this on some planet where the atmospheric pressure Po is greater 100 kPA (1 bar.) These differences will make a quantitative difference in results we would expect, but, oddly enough, very little qualitative difference. My intention is to discuss what happens during an adiabatic expansion (the second part of step 4 in Konrads discussion.)
In the image above left, I’m showing an idealized capped plastic bottle that is in a state equivalent to that after sentence 1 in step 4:
Slightly squeeze the bottle before replacing the cap tightly.
The pressure inside the bottle equals that outside the bottle. The light blue indicates water vapor. The black arrow pointing in are the force of Konrad’s fingers which are pushing in the sides of the bottle. (I’ve made the entire side push in a set amount rather than “pinch” in. ) The red “spring” in the center represents the internal forces in the plastic bottle that make it want to spring back when Konrad unsqueezes. (You can practice squeezing empty open bottles and watch them spring back to the right shape when you release. )
In Konrad’s scenarios, the pressure inside the bottle is equal to that outside the bottle because the bottle was open to the atmosphere, squeezed and then capped while it was still squeezed.
Now, to discuss “what happens” we need to assume we know the various pressure. Let’s assume the water vapor in the container is at “state 0”, with pressure known to be P0=60 bar (that is 6 atmospheres), and temperature of about T0300C. (It’s hot on our planet! Or this is for some reason an industrial process inside a pressure vessel.) I’m going to mark that with a (0) on the Temperature-Entropy (i.e. T-S) diagram below:
You can read the pressure by following the blue curve up to the top of the chart, where I’ve circled “60 bar”. You can read the temperature on the left hand axis. You can read the entropy on the horizontal axis, it’s roughly S=6 KJ/Kg K. While we’re looking at the chart, we could also notice the green curves which indicate the specific volume (this is the inverse of the density.) We could see the point (0) has a specific volume between density 0.02 m3/kg< v< 0.05 m3/kg
(If I needed precision, I’d look it up in the steam tables or a better T-S diagram. This one is good for illustrating ideas.)
At this point, we put the bottle on the scale (even though this has nothing to do with Konrad’s discussion or adiabatic expansion.)
Next, we continue with Konrad’s step 4:
Release the sides of the bottle once the cap is in place to cause slight expansion of the saturated air in the bottle.
Releasing the sides of the bottle is indicated by taking off the black arrows in the image above center. In Konrads case of a real bottle, the depressed region snaps back due to internal forces in the plastic. I replaced this with the red “spring” which pushes until the sum of the force due to the pressure difference on both sides of the bottle walls counter balances the force of the spring.
This whole process happens very quickly, so we could probably correctly assume that the process involves no heat transfer to or from the bottle.
Nevertheless, the volume of the bottle contents increases when the force is released.
This is called an adiabatic expansion. If it happens reversibly, we expect the final state of the contents fall on a line with constant entropy (i.e. constant S). That’s indicated by the black line connecting (0) and (1) on my graph. Let’s assume for the purpose of discussion that the process is reversible– so the state really does end up somewhere on that black line.
Thinking about the force balance, it’s clear that the volume of the bottle and its contents must now be greater than previously. The spring in my image is longer– so it applies a smaller force than previously. But, in my gedanken, the spring still applies some force.
If the spring exerts some force, the pressure inside the bottle will be lower than that outside. We will assume we know this new pressure, and call it P1. Suppose we know the spring constant, the length of the spring and the area of the bottle walls, and can therefor figure out the pressure inside the bottle is now P1=1 bar (or 1 atmosphere), which I’ve indicated at point [1].
Notice the temperature is lower that before– it’s now about 100C. (That’s the boiling point of water at 1 atmosphere.) If you check the green curves, you’ll see the specific volume is larger: near 1 m3 — so the graph agrees the volume of the contents is larger. You’ll can also read something indicated by “X”. This is the vapor fraction. For my gedanken, this is X=80%, which means 20% of the mass inside the bottle condensed into water. I show this as little dots.
So, everything works out the way Konrad reports. All of the following thing happened simultaneously:
- The contents of the bottle expanded,
- water condensed
- the pressure is dropped.
So, in this process, pressure dropped but the contents took up more room, pushing surrounding air away from the point of expansion.
So, what happens next? In Konrad’s real experiment, the bottle began to crumple. But he also reports that if he warms the bottle, he can prevent this from happening.
My claim is this:
If we prevent all heat transfer from the bottle, the bottle will neither crumple nor expand– or more correctly, we will barely notice any change. There here might be a very smallnearly impossible to observe effect as drops fall to the bottom of the bottle, but the rest of the experiment would be totally boring.
So, why do the sides crumble in Konrads experiment?
In Konrad’s specific experiment, the contents cool after expanding because he filled the bottle with warm water. So, the contents of the bottle started out warmer than the air outside the bottle. Konrad reports the sides of the bottle still felt warm as the bottle contracted. As long as the sides of the bottle feel warm, it will be losing heat to the air outside the bottle. More water will condense. Entropy inside the bottle will decline. Even though condensation accompanied expansion during an adiabatic process, condensation will accompany contraction if the contents lose heat.
Because everyone is wondering what happens to pressure inside the bottle as the water condenses: In my gedanken and Konrad’s bottle, the pressure declines as the gas condense. But this is not due to some magical property of condensation. In my gedanken, this due to the presence of the spring. In Konrad’s case, it’s due to ability of Konrad’s plastic bottle to resist deformation. Otherwise, if the force exerted by the spring magically stayed constant as the spring got pushed to in, the volume of the contents would decline at constant pressure. So, if the spring force stayed constant, condensation wouldn’t cause pressure to increase or decrease!
If we’d done a different experiment where the bottle was inside a steam pressure vessel filled with super heated steam at 300C, then closed, the contents of the bottle be at 100C after this expansion in my gedanken. But the steam outside the bottle bottle would till be at 300C, so the contents of the bottle would gain heat from the surroundings, water would evaporate and the bottle would expand while water evaporated.
For those who are aware that this discussion is partly motivated over arguments about whether condensation causes air to expand, contract, gain pressure, lose pressure or remain at constant pressure: The answer is, “That’s an incomplete question.”
To know whether a volume of gas will expand or contract while water is condensing, we need to know why it’s condensing and we also need to know what is happening in the vicinity of the gas with water condensing out. Is this volume of gas losing energy because it’s under going adiabatic expansion? Or is it losing energy owing to heat transfer? In the former case, the gas will expand during condensation, in the latter it will contract during condensation.
I hope those who wanted another gedanken think this one is a little interesting. Let me know if you see another example that needs explanation, have questions about this experiment (which is a big confusing) or want to know about mixtures of gas and air.
“But this is not due to some magical property of condensation. In my gedanken, this due to the presence of the spring. In Konrad’s case, it’s due to ability of Konrad’s plastic bottle to resist deformation.”
How about dry air as control? Heat the bottle in a heat bath, open cap, squeeze, close . Mechanical and cooling forces should be the same.
p.s. I am using CA assistant at the italics no longer work.
AnnaV– Air is qualitatively similar. If you expand air adiabatically, the pressure drops, specific volume increases. If you cool air at constant pressure, volume drops. If you warm air at constant pressure, volume increases. The effect of solids to resist deformation still happens.
The two main differences with air is that
a) you never hit a spot where cooling at constant pressure is also cooling at constant temperature. That happens in steam condensing at constant pressure.
b) specific volume changes can be very dramatic when liquid condenses. So, plastic bottles can explode or crumple very dramatically.
“Within myself still
I can think as I will,
But I laugh, do not cry:
Die Gedanken sind frei!”
Lucia / Anna v-
I still remember a episode of Mr. Wizard that tackled something very similar.
Empty tin can with a screw top, a match, and a pitcher of cold water. The challenge was to crumple the tin can using only the items on the table.
(IIRC) Light the match and drop it into the tin can, wait a few seconds, close the lid, wait a little bit longer, pour cold water over the tin can. The result was that the can crumpled as the water was poured over it.
Explanation was that the match warmed the air inside the can. When the top was put on and the sides of the can were cooled, the drop in temp also dropped the pressure inside the can, the normal outside pressure is now higher than the inside pressure so it pushed in on the now lower pressure air inside the can.
Having seen that show, as I read Konrad’s expirement I could easily guess that the outcome would be for the bottle to crumple.
Lucia,
An interesting analysis. In particular the point that the resistance of the bottle to deformation preventing the pressure from fully equalising. It is possible that the inertia of air surrounding a contracting air parcel could have a slight similar effect in the real atmosphere.
I have carried out further tests including room temperature water and bubble wrap + foil insulation. In all cases the heat lost through radiation and conduction exceeded the latent heat released during condensation. The conditions in the bottle experiment are clearly not adiabatic, even with pre heated insulation. However the test was only to see what would happen to the volume of a warm saturated air parcel as condensation occurred.
As a thought experiment I have considered how much of a 1km diameter saturated air parcel undergoing condensation would have to lose heat diabaticaly to counter the volume increase of the interior in which latent heat is not lost. A crude estimate indicates that if a 25m outer skin of the 1km sphere underwent turbulent mixing with cold air and contracted by 30%, this would be sufficient to counter the buoyancy increase of the interior 950m diameter sphere increasing in volume by 5% due to the latent heat of condensation.
Lucia, you say:
“My claim is this:
If we prevent all heat transfer from the bottle, the bottle will neither crumple nor expand– or more correctly, we will barely notice any change. There here might be a very small nearly impossible to observe effect as drops fall to the bottom of the bottle, but the rest of the experiment would be totally boring. ”
Are you stating that in the adiabatic condition there will be no condensation? Super saturated air will remain supersaturated whether there are nuclei present or not?
Or will the presence of nuclei destroy the adiabatic conditions by the disorder produced by the phase change of condensation? In that case wont there be a decrease in pressure due to the phase change and the removal of a part of the gas from the gas volume? and a concurrent increase due to latent heat release, so the question becomes which has the upper hand?
Example: cloud chambers which remain clear until ionizing radiation provides the nuclei along the path.
p.s.
Once there is condensation, the number of gas molecules changes . Mass is removed from the ideal gas law. Even if the droplets in suspension are treated as molecules it is not the mass that plays a role but the number, no?, and there are zillions of H2O molecules in a droplet.
Lukewarming avoidance
http://www.surveymonkey.com/sr.aspx?sm=ONSUsVTBSpkC_2f2cTnptR6w_2fehN0orSbxLH1gIA03DqU_3d
“If we prevent all heat transfer from the bottle, the bottle will neither crumple nor expand– or more correctly, we will barely notice any change.”
Even if this were true, in “real life for cloud formation” not all of the energy released by condensation goes to heating the surrounding air and at least some of it will be lost to space and/or the ground, surely?
TimTheToolMan–
I don’t know what happens in “real life for cloud formation”. There are mechanisms for heat transfer, and presumably, if there are temperature differences, some heat will conduct, convect or radiate away. It should depend on the temperature or the air, drops, ground and space and also some physical properties.
AnnaV
Do you mean with no work?
Close to that. I’m saying once the bottle is at thermal equilibrium after the adiabatic expansion, if there is no heat transfer, the vapor fraction X won’t change. There will be some change from potential energy to thermal energy as drops fall, but in Konrads 1 liter bottle that is trivially small.
My bottle contains only water vapor. At the end of the adiabatic expansion, the air is not super saturated. The vapor fraction X<1.
I don’t understand what you are asking. Are you asking whether the presence of nuclei will result in heat transfer at or through the plastic walls of the bottle? Anyway, I assume there are nuclei. The only reason they might vanish is if water condensed on them and rained the out.
Sure. But what does that have to do with the bottle after the drops have already condensed? In this bottle, either water will condense on the walls (because the bottle is small) or it will condense on nucleai in the bottle.
Are you trying to get to some other issue not captured by this bottle problem? If so, could you describe the issue that concerns you directly instead of trying to argue for something by rhetorical question?
Lucia,
Here is a quote from Roy Spencer that illustrates my point to you.
“Actually, this announcement is a good thing. There has been a persistent refusal on the part of the elitist, group-think, left-leaning class of climate scientists to even debate the global warming issue in public. Maybe they have considered it beneath themselves to debate those of us who are clearly wrong on the global warming issue.
A complaint many of us skeptics have had for years is that those who constitute the “scientific consensus†(whatever that means) will not engage in public debates on global warming. Al Gore won’t even answer questions from the press.”
I know I was specifically talking about Lord Monckton, I should have said “in general” but he is definitely one of the persons who consistently gets shut out. What is your opinion on Roy’s comment? Do you agree that the “consensus” refuses to debate? I do.
Re: lucia (Nov 9 07:19),
Lucia, I am out of my depth with this gedanken and am trying to understand.
I remember an Italian lady professor who at the end of lectures would raise her hand with a strong Italian accent and start with “My questiion iiis…” As students we had a good laugh, and now I find myself in her position /tongue in cheek)
The curves of temperature versus entropy are the adiabatic ones, I suppose.
To keep the adiabatic condition it is necessary not to have energy exchanged out of the volume. I do not know if it sufficient for energy not to be exchanged, to state that the system is following the adiabatic curves.
If condensation happens, this includes water globules within the vapor, not only on the walls or falling (cf clouds as also as in Konrad’s experiment), then the number of particles in the volume that participate in the ideal gas equation changes, which means a shift to a different curve in your graph higher S lower P..
It also means a discontinuity. Are we still talking of adiabatic when there is a discontinuity ?
I think my questions boil down to this: In this gedanken experiment, what is the effect of the diminution of the number of molecules that bounce around in the gas, and give the ideal gas law ,due to condensation. This mass removal because of condensation and subsequent large pressure loss is the lynch pin of Makarieva et al ‘s argument for a new proposal for wind creation . Also where is the latent heat released by condensation? It is supposed to heat the gas and increase the pressure according to standard meteorology.
Lucia,
I don’t know exactly how to say this without insulting people, but I will go ahead. The net fluid mechanics and thermodynamics for the issue are well known. The details of the sequence of events (does the water condense continually over a time or suddenly in a very short time, etc.) don’t really matter for the final result.
The water vapor only occupies a small fraction of the gas volume (typically about 1% to 2% or a bit higher near the ground). The partially moist air rises, and cools from adiabatic expansion of rising to lower pressure. At some point the vapor becomes saturated and eventually condenses out. The energy released from phase change is about 540 cal/g. The specific heat of the air is about 0.22 cal/g deg C. For the following example I assume 1% water vapor by weight (close to 2% by volume) That means 1 gram of condensing moisture would heat 100 g of air by over 24 C, and thus increase volume (at constant pressure) by over 8%. The result is a loss of water vapor volume, but an increase in volume from warming, and the net would be an increase of volume (at constant pressure) or an increase in pressure (at constant volume) by over 6%. This would cause outflow from the condensation region.
The bottle experiment was so badly flawed that it gave a totally misleading result.
AnnaV–
No. Let me explain what it is.
The graph is a type of state diagram for steam. It happens to be the state diagram called the (T,s) diagram. Temperature (T) and specific entropy (s) (that is, per unit mass)
For super heated gases (steam, air whatever) you can specify the state by knowing two thermodynamic properties. Pick whatever you want and the others can be looked up. (This is sophomore year thermo. Maybe even freshman year– but I don’t remember it from freshman year. I know it’s in my first semester sophomore year thermo book.)
The upper right hand region of the T,s diagram is a chart that lets you do look up some thermodynamic properties if you know two others, and you know the phase is vapor. (That particular graph only shows T,s on the axes and then shows lines of constant specific volume,v, and constant density, P. Even cluttered graphs T,s graphs that also show enthalpy,h, exist and are very convenient for engineering problems that might be solved more quickly by reading ‘h’, but I didn’t pick this. There are other graphs and tables that are more convenient for different problems. But all are the same information.)
But in my Gedanken, I said I knew temperature (T) and pressure P of the initial state in the bottle. I located that on the diagram and found it. I labeled it (0). Once I found that point, I can know the specific entropy, s, and specific volume,v, by reading the chart. (You can get better precision by reading a table, or by using program that has all this information coded in. But the graphic is good for explaining.)
So far that’s the chart. I’ll get to the “adiabatic” part in the next comment.
Now, we get to the “adiabatic” part!
The second law of thermo for a closed system tells us
Tds≥q (1)
where q is specific heat added. (That is, heat per unit mass.)
If things are done reversibly we set the inequality to an equality.
Tds=q (2)
So… that’s the 2nd law.
Now, the word “adiabatic” means no heat is added. So q=0. So, if something is both adiabatic and reversible,
ds=0 (3)
So, if we envision that a bottle is filled with “stuff” and there is no heat transfer, ever, ever, ever, ever. We know that “s” can only increase. But if we further assume that things are done reversibly then the state of the contents will remain on a line of constant “s”. Since we already know “s” when we fill the bottle, then if
* no heat is added (adiabatic) and
* all processes are reversible.
All future processes will fall on lines of constant “s”.
On that graph, these lines of constant “s” are all vertical lines. So, if processes are reversible, then lines of constant “s” are “adiabatic” lines. (It’s best not to think of them as that because some processes are irreversible. For that reason, there is no such thing as an “adiabatic” line on a state diagram. But as a short cut, a line of constant entropy is the closest thing someone might call an “adiabat” on a state diagram. )
I’ll discuss Konrads bottle in the next comment.
In Konrads bottle, we knew the first point– labeled (0) above. He was squishing the bottle, then released. If we assume the bottle springs back quickly and condensation is reversible (both are assumptions) then we might imagine heat transfer during the spring back is small.
So, we assume q=0, because there isn’t sufficient time for much heat transfer to occur through the sides of the plastic bottle. (As much as you like can happen inside the bottle.)
If we assume that condensation is reversible then dS=0. So, point (1) after the bottle springs back is on that line of constant entropy, s, (or, if you prefer the not quite right mental short cut “an adiabatic line.”)
We know the volume of the bottle is larger after the bottle springs back. If we knew the new volume, v1, we could find the intersection of that line of constant volume with the vertical line for “s”. It would be at point (1).
Reading, we notice this is at a lower pressure– p1. (My discussing this in the blog post involved knowing the pressure–by way of reading the lenght of the spring. But either way of reasoning gets us to the same place.)
So, after an adiabatic expansion (to higher volume) the pressure is lower.
Now.. you will also notice that for my gedanken, I picked point (1) in the saturated region. So, in this case, at equilibrium, the contents of the bottle will consist of liquid water and saturated vapor. The mass fraction of vapor is “X”, and in my example, that x=80%.
My picture shows droplets– but that’s not important to the thermo. The thermo works even if the liquid condenses on the walls. All we need for equilibrium is for x=80%. But if there are sufficient nucleation sites, it’s likely the drops will form.
To avoid complication, my example is water only not air and water. The “number of particles in the volume” is irrelevant to determining the equilibrium state. The water can condense on the walls, into big drops, little drops. Whatever, if
* The bottle expands too quickly for heat transfer to occur between the outside of the bottle and the bottle.
* The inside is at thermodynamic equilibrium after expansion (this is an assumption — and it’s more likely to be true if drops form than if all condensation is at the walls. So, water temperature = air temperature, no gradients exist in the bottle.)
* The condensation is reversible.
Then “s” will be the same before and after the expansion. The curve already accounts for “per unit mass” of water in the bottle. The condensed water did not leave the bottle, so it hasn’t vanished from the problem.
I don’t know what you are asking me. You seem to be jumping ahead to some other more complicated geometry. There is no dimunition of molecules in Konrads bottle. The liquid water is still in the bottle.
At the present time, I do not understand Makarieva’s proposed mechanism for wind creation. I have read the paper a little, and I am trying to understand what, precisely she proposes for the mechanism, but I don’t know what it is.
This may be because she seems to be describing a series of “gedankens”, but I’m not sure I know what they are. (The fault could be me. But I can’t explain what happens when water drops out of “something” if I don’t know what that “something” is, I don’t know it’s boundary conditions and etc. I have not been able to give sufficient time to her paper to quite figure that out. If you can describe the specific geometry she proposes that could help me.)
Leonard–
I’m not saying the bottle experiment tells us much about the other issue of condensation in a cloud. (Assuming that’s the issue over at the Markeiva thread.) I’m pretty sure it doesn’t tell us a whole lot. But the bottle discussion went on a bit over there, and even there, some issues were getting muddled about the bottle experiment as a bottle experiment!
Now to clouds:
Yes.
If we have a finite volume of air in a balloon (with zero surface temperature) and no heat transfer occurs across the balloon surface as it rises, and we look at that only, yes, this must happen. (I don’t know how far it rises, but I did notice the gravity terms are considered over in TAV thread. )
Presumably, in this thought experiment, the pressure of the air in the balloon equals that outside the balloon? Right? That’s why it expands.
Yes. At some point during expansion, the temperature reaches a point where the relative humidity would exceed 100%, and, provided there are nucleation sites, water will condense. If this were a real balloon, the water falls to the bottom of the balloon. Right? (Or in the gedanken, is the balloon is water permeable but holds air? I’m not trying to be a butt here, I’m trying to make sure I understand the assumptions directly instead of inferring from the conclusions in a range of people’s comments over at TAV.)
This all sounds like ideas relative to an air only filled balloon? IF the argument really involves these two balloons, we can presumably just do the two balloon problems, and show the result using the phase diagrams. (That is: The engineering psychometrics way.)
Do you want me to look at two balloons rising… through… something? And talk about what happens to the two balloons? I can do that. It’s worth it if that’s the problem someone cares about. Also, I’ll state the various assumptions as we go along, and then we can discuss if they matter.
But right now, what I’m finding difficult in the papers (and is comments) is figuring out what the assumptions are in all the various “thought” experiments. Everyone has them– but it seems they sometimes need to be teased out from people’s conclusions.
“lucia (Comment#59526) November 8th, 2010 at 3:12 pm
… you never hit a spot where cooling at constant pressure is also cooling at constant temperature.”
Cooling at constant temperature proves how unique climatology is.
Re: lucia (Nov 9 10:30),
From what I have gathered from her explanation in the Air Vent discussion, her point is that once the H2O turns into liquid, either droplets or rain, the number of molecules that comprise the gas is reduced . In the bottle experiment it makes no difference that the mass exists , it is not there as part of the ideal gas:The derivation of the ideal gas law in statistical mechanics does not depend on the type of molecules, just on the number of molecules,
pV=NkT
“Here k is the Boltzmann constant, and N is the actual number of molecules”
Thus if N goes to N1, where N1<N something has to give. In statistical mechanics T is related to the kinetic energy, and the kinetic energy of the rest of the gas molecules does not change at the moment of condensation, (though it will change from latent heat distribution at a second stage) so pV has to drop.
The usual meteorological argument ignores this mass drop from the ideal gas, and uses the latent heat to assert expansion with condensation because of heating.
That is as far as I have gone in understanding the issue, and my interest in the experiments ( I also have a two bag experiment ) is to see whether one can reproduce their statements in a real experiment.
So far I have isolated the argument of mainstream meteorology to:” it is in adiabatic situations that condensation increases pressure due to latent heat release.” I thought Konrad’s experiment gave a handle and am not sure yet it cannot be utilized : insulate two bottles, one for control with dry air, one with supersaturated and smoke for nucleation. See if there is a difference in how much the bottle compresses if at all.
Your gedanken says there will not be, but I would like to see an experiment.
Anna v
In my bottle experiment, after water condenses the number of molecules that are gas also is reduced relative to the number of molecule that were gas before.
It happens that if you had detailed chart (of a steam table) you could look up the specific volume of the liquid water and the specific volume of the water vapor as a function of T. You would find the specific molar volume of the water vapor will come close to obeying v/N = pRT where p is the pressure in the bottle and N is the number of molecules of gas. The specific volume of liquid water will be much smaller (by a factor of about 1000.)
I don’t see how this differ qualitatively (or quantitatively) from what Anastasia is saying.
I don’t know if the usual meteorological argument does or does not ignore this mass drop nor do I really know what Anastasia is arguing about the consequences of the obvious fact that if a known fixed Mass initially consisting of Mass=(water0+vapor0) becomes (water1+vapor1+liquid1) water,
then
(water0+vapor0) > (water1+vapor1)
Yes. This is so. But I need to know more about what’s happening to the water, and the fixed mass to figure out more. Did the water condense because the fixed mass did work on the surrounding? Did it condense owing to heat transfer? Did it condense because it was previously not at equilibrium and something caused the water to spontaenously condense?
Any of these can be analyzed. But we’ll get different answers for final states depending on why the water suddenly condensed. So… other than water condenses, what are the assumptions? Presumably, they are made and can be stated? (I know they may have been and I may have missed them. But… what are they?)
Do you mean relative to cases with only air?
Do you want me to do the analysis of Konrads water bottle with air and water? Only the expansion?
I can do the following– provided it addresses your question:
1) Picking a set of initial Po and To in Konrad’s bottle, assume it contains a volume V or air only. Allow it to expand to P1, T1, V1 where we know V1, the final state.
2) Repeat but this time, have the bottle filled with air and water.
The problem is do able–I might have to iterate, but its doable. We can compare the final pressure in both the cases. (Then those who think they know what this relates to can decide what this means in terms of Makarieva. Because I don’t. But maybe it has something to do with it. At least we can figure out the answers to certain gedankens that people seem to think have something to do with it.)
Anna v–
Your problem in comment 200 here seems to involve
1) bagged filled with steam at air pressure, then pressurized. (Adiabatic expansion.) The cooled along a line of Pressure=Patm+ P(R) where P(R) will be the pressure exerted by the bag which will be a function of the bags level of inflation. If you give me a law for the pressure v (R) of the bag, this could be traced on the graph above.
2) Bagged filled with air room temperature, then cooled along a line of Pressure=Patm+ P(R) where P(R). I could trace these along the line too. (I assume both bags are equally elastic. )
To make it easier…. let’s assume you cool to a volume where the bag exerts no pressure– ok? Then we can do the problem.
But if we do this problem, what do you learn? What I can tell you is: How much heat was tranferred from two bags, the final volume as a function of T. Etc. Whatever you want to know, it’s solvable.
Alex
We cool steam leaving turbines at constant temperature and pressure all the time in power plants. This is not unique. It’s thermo common to engineering and climate science.
Annav
I’d like to clarify this:
My gedanken doesn’t say everything in the two problems is quantitively the same. There are qualitative similarities. But if you provide a set of assumption about
a) A steam filled bottle,
b) an dry air filled bottle and
c) an wet air filled bottle.
I can compare things. (I suspect you only want b compared to c? “c” is a bit of a PITA, but it can be done. The thing is, you need to tell me what you assume happens to the bottle. Is it expanded like Konrads? That’s all? Or more?)
Thank you Lucia.
When I did the two bags experiment it was because people were saying that condensation increases pressure, a blanket statement. I convinced myself that in this case condensation decreases pressure, and they accepted this with the caveat that it was not an adiabatic experiment since the freezer cooled both bags. Your conclusion also is that it depends on the circumstances :
To know whether a volume of gas will expand or contract while water is condensing, we need to know why it’s condensing and we also need to know what is happening in the vicinity of the gas with water condensing out. Is this volume of gas losing energy because it’s under going adiabatic expansion? Or is it losing energy owing to heat transfer? In the former case, the gas will expand during condensation, in the latter it will contract during condensation.
I am satisfied with the qualitative difference between dry air and saturated air with the bags.
The meteorological problem, at the heart of the discussion in Air Vent, is reduced to the adiabatic expansion and how condensation increases pressure and creates all the winds, turbulances, hurricanes,… observed and fitted by models.
It is still an open question for me but I do not think gedanken experiments can solve it. It probably will need real balloon experiments.
If I get any insights on the Makarieva et al paper relevant to the thread I will come back, if it is still alive.
Re: lucia (Nov 9 12:34),
I can compare things. (I suspect you only want b compared to c? “c†is a bit of a PITA, but it can be done. The thing is, you need to tell me what you assume happens to the bottle. Is it expanded like Konrads? That’s all? Or more?)
You make me smile 🙂 .
The difference between a theorist and an experimentalist :).
I want to see what happens in the real world. Will the b contract less than the c?
Makarieva et al get a “+”
Will the c inflate more than the b? meteorology status quo gets a “+”
You want to calculate :
Both should be “triggered” by a slight squeeze, a cap, a release, c should have nuclei for condensation.
AnnaV–
At TAV, I think I have a qualitative understanding of the issues discussed in these two papers:
They seem to be primarily concerned with the following issue:
We assume there is a fairly stable structure where wet air) rises in a particular location (say the eyewall of a hurricane). As the air rises, it expands, and water condenses out. When water condensate rains out, that results in a decrease in mass in the column of of air where the rain falls out. The pressure at the bottom of the column is known to be equal to the weight of material above it. Because the mass of air rained out, that pressure is lower than it otherwise would be if the water drops remained magically suspended.
Now, we look to the surface: Because flow is actually 3-d, and we know the pressure is lower in the region where air rises is lower than other regions. So, we expect air to be drawn into the center owing to this pressure difference. (Of course, we know air will flow in anyway, owing to continuity effects. But this pressure difference encourages this even more. )
They then do some sensitivity experiments accounting for the effect of the mass sink on the hydrostatic pressure in one set and ignoring it in others. The difference exists (as it must) but the difference is small.
I guess Marakeiva says these miss the point in some way? Or not? I don’t know. That’s the bit I don’t get.
Am I misunderstanding the papers cited at Jeff’s? Do you understand what the “key” bit that Marakeiva is saying is wrong is? Or am I way off here?
anna v–
My ph.d. was experiments and many of my projects were experiments. 🙂
But some of this is undergraduate thermo, and you still need to know what theory you are testing to devise the appropriate experiment and figure out what we want to control and what we want to measure. Right?
Do you want me to do the problem? I’ll do contract both to atmospheric pressure– that is, the bags don’t exert force after contraction. Do you want me to fill them both with the same mass and same pressure? Same volume and pressure? Or do you want me to just pick, and then after do a different choice if you think that matters. (Either way is fine with me. Then we’ll see the answer. )
To make this easier though…. I’d like to change the problem so I don’t have a spring and we assume we know the pressure at state (0) and state (1). (I don’t think this matter for what you want to figure out. In fact, I think no spring will be clearer. It’s just the experiment you would do wouldn’t be done with a PET bottle we squeeze.)
Re: lucia (Nov 9 13:41),
I think the basic difference between the model you describe and Makarieva at al’s model is that they believe they are demonstrating that there is a large pressure drop because of removal of mass from the gas, at the time of condensation, before it rains down and generates pressure differences due to mass/weight changes. This mass loss affects pressure not because of its collective weight, but because it introduces changes in the ideal gas law, the number of molecules available to play ball and keep the law going. It is there as tiny droplets with zillions of molecules but it is missing in the numbers balance for pV=Nkt and they prove that this lowers the pressure.
The derivation of the formulae in their paper is over my head, considering that my thermodynamics course was only a semester back in 1960.
Re: lucia (Nov 9 13:51),
Whatever you choose is fine as long as one can see if under adiabatic conditions condensation increases pressure or decreases pressure or nothing happens.
Makarieva et al say at 2.2. that” adiabatic condensation cannot occur at constant volume ” , so the volume should be allowed to vary. ( in Konrad’s bottle it would be allowed to collapse).
Anna V–
The recent submitted paper I read has some extensive discussions of how pressure varies if condensation happens at constant this, or constant that, and what limits happen if entropy must increase and what not. I can perfectly well follow them but I don’t understand what point she is trying to make.
Yes. If we require condensation to happen with “S” decreasing only, we find some restrictions. Yes, if condensation happens at constant volume something else is required. And so on.
I get the impression she says someone adiabiatic condensation happens at constant volume and she shows (maybe) this can’t happen. (I’d have to look to check if she only shows it can’t happen at constant “s” and constant “v” at the same time– that would be trivially true.)
What I don’t know is whether any one, model or anything ever assumes adiabatic condensation happens at constant “v”.
So, it’s not really the math or physics I think I don’t get (though maybe I don’t get that.) My impression is, depending on where in the paper I am reading, I don’t understand either
a) How a particular proof or analysis fits into some rhetorical argument or
b) What the geometry and assumptions connected to a particular sub analysis are or
c) How this fits into what someone else, somewhere thinks.
(The “c” bit could very well be me because I’m not a climatologist. The rest could be me too. But it’s not the thermo or fluids I don’t follow! At least, I don’t htink it is.)
Just tried to calculate an imaginary case.
Initial conditions:
Mix of 1g water vapor and N2(gas) is kept in a 1 m^3 chamber, with totally insulating walls.
The mix holds a temp of 3oo Kelvin, the pressure is 1oo,ooo N/m^2.
Suddenly all the water vapor condenses:
1g water vapor -> removes 1/18.o15 mol gas from the chamber -> until now H2O has contributed to a partial pressure of ~138.5 N/m^2 -> N2 (gas) must, until now, have contributed with ~99,861.5 N/m^2 pressure
Condensation of water releases (wikipedia[Latent heat, formula takes Celsius]:Lwater(T) = − 0.0000614342T3 + 0.00158927T2 − 2.36418T + 2500.79 =) 2,437.27 kJ/kg (Joule/g) -> Condensation of 1g water releases 2,437.27 Joule
Using pV=nRT and partial pressure of N2 gives 4o.o352 mol N2 (Calculated under mixed conditions)
Heat capacity N2 [wiki]: 29.124 J·mol−1·K−1 (at 25*C; close enough).
When H2O initially is removed, temperature drops to (99,861.5/1oo,ooo=99.8615%) 299,5846 Kelvin.
Using heat capacity 29.124 J·mol−1·K−1, the received heat from condensation of 2,437.27 Joule and the amount of 4o.o352 mol N2 ->
Using pV=nRT again, new temp becomes 3o1,67 Kelvin with a pressure of 1oo,419.o8 N/m^2
Is this totally wrong reasoning, or are there instances where pressure actually increase?
Are we talking of the same paper?
section 5 seems a fair summary of what they are presenting as an alternative to the existing model of how the atmosphere behaves. They claim they do not have to fit parameters ( in contrast to the currently accepted differential heating model), but wind velocities come out of the calculations with correct magnitudes.
As I see it their aim is to bring forth a new model of atmospheric circulation based on strong pressure differentials appearing because of condensation and subsequent drop in pressure rather than rise. The claim is that it will give better predictive and descriptive power.
So whether adiabatic condensation gives rise to higher or lower pressure seems to be important to distinguish between differential heating on the one hand and condensation induced circulation on the other.
p.s. way past my bedtime, it is close to midnight here
a demain
anna
Ok Anna. I’ll switch this around to compare adiabatic conditions with condensation. (I’ll need an adiabatic expansion for step 2.)
Lucia : “There are mechanisms for heat transfer, and presumably, if there are temperature differences, some heat will conduct, convect or radiate away.”
Then I reckon its vital to understand precisely what form the energy takes upon condensation. If its mostly radiated (initially) and a significant portion of it is in the non-atmosphere-absorbing wavelengths, then the atmosphere isn’t going to recapture that heat.
Simplistic assumptions like the heat released by water expands the air are possibly entirely invalid.
Hopefully someone here knows the answer 🙂
Anna v–
Same paper. Section 2 seems to be discussing adiabatic condensation of various sorts. That’s the section I was thinking of when I wrote: “paper I read has some extensive discussions of how pressure varies if condensation happens at constant this, or constant that, and what limits happen if entropy must increase and what not. I can perfectly well follow them but I don’t understand what point she is trying to make.”
Yes. We can do lots of thermo examples… So? But presumably this is there for some reason which is revealed eventually.
So, you think “section 5 seems a fair summary of what they are presenting as an alternative to the existing model of how the atmosphere behaves. ”
I see a qualitative discussion on page 24044 and so on. But… what are the geometries in section 4?
Well… yes. But that’s also why it’s important to state assumptions and admit you made them.
Well, the difficulty is that if the only thing that happens is the latent heat of condensation goes into the air, then, yes, the air will expand relative to the situation where all other condtions match but there is no condensation.
But, to conclude whether the air explands or contracts, I need to know more. Was the entire blob of “water vapor+liquid water+air” being cooled by heat transfer? Or was that blob of mass doing work?
Nick–
That’s inline with the “bottle” results. Except what is said to “cause” something and what is said to be an “effect” is switched around.
In the bottle proble: We release our hold on the bottle, causing an expansion– and as a result of this expansion, the contents of the bottle “do” ‘P dV’ type work on the bottle. (Or the bottle “does” -‘P dV’ work on the contents. )
This happens expansion and work doesn’t happen “because” the gas condenses, but because Work on the contents is done at the boundaries of the container.
As a result of the work (of whatever sign convention you like) is done by the contents on the universe, the internal energy of the contents drops. If there was no heat transfer, and things are reversible, the water condenses.
So, they way I see this, condensation doesn’t “cause” the expansion. The expansion “causes” the condensation!
But still, I think we both agree: If a fixed volume of gas+water vapor expands with no time for heat transfer, then water will condense. What happens afterwards depends on heat transfer (or whether the blob does more work etc.). We need to know more about that.
HLx (Comment#59685)–
Why does the water suddenly condense in the chamber? Was heat removed? Was work done?
HLx–
I should explain why I ask.
We can engineer a system where we make a box of dry air at some temperature and spray in water at that same temperature. We can put a little fan in to keep it all well mixed. The water would evaporate until the partial pressure of water reaches the vapor pressure at that temperature. After that, water will no longer evaporate. We could insulate the box or do something to make the thing adiabatic.
If we did this, we could compute the change in entropy for the process– it’s going to be positive. But it’s really difficult to make water vapor in equilibrium with the air suddenly condense without either cooling the box or forcing it to expand. So, why does the condensation occur?
“So, they way I see this, condensation doesn’t “cause†the expansion. The expansion “causes†the condensation! “
Yes, indeed. But I think the problem is that there are two “expansions” being discussed. One is the expansion that any gas experiences on being uplifted in the atmosphere. The other is what you specified as
“the air will expand relative to the situation where all other condtions match but there is no condensation.”
That’s what counts here. Globally, when saturated air rises, other air (probably unsaturated) descends. If there’s to be energy to drive something, you have to get the nett effect.
(ps did you mean to respond on the other thread?)
Nick…
Oh.. maybe…. But to a large extent, people are wording it as “condensation causes X” as if condensation just magically occurs and then “causes” something to happen. In real life, a “X” (whatever it might be) imposed by something other than condensation results in a change of state, which happens to be accompanied by condensation.
Re: lucia (Nov 9 17:29),
Well, to clarify the two expansions – expansion 1 (uplift) causes condensation, which then causes expansion 2 (relative to dry air in the same conditions).
Nick–
Are you discussing the problem at your blog? I’ll go there. Right now, reading quickly:
At the end of the uplift, the pressure of a blob of mass M=(water vapor + air + liquid water) is presumably already matches local conditions, right? It’s temperature is whatever one you get after this mass M undergoes an adiabatic expansion, right?
Then, that blob somehow stays at that location, and, if the temperature of the blob differs from that of its new surroundings, heat transfer occurs.
So, if the temperature of the new surrounding air is higher than the temperature of the blob that just rose, the blob will gain heat from the surrounding conditions and… some water will evaporate and the blob will expand even more, while… evaporation occurs. (Or at least, that’s what happens with steam only. And air expands if it’s heated too.)
So…. I’ll go look at your analysis. Or, we are somehow comparing “A” to “B”, but I don’t know what “A” and “B” are. (I mean… the physics of a blob of moist air doesn’t about conditions of dry air that is no where near the most blob. So, I don’t know why it would expand to whatever conditions of dry air, nor do I know which are “the same”. Same P and T? )
Lucia,
I’m about to go out for a while. I’ve only done the adiabatic cylinder style analysis. But the whole atmosphere idea is that when one lot of air rises, another lot descends. And while you can’t actually identify what that is, it’s reasonable to say that if sat air ascending arrives warmer and lighter than other air on it’s level (after starting at ambient), then a larger mass of air (same volume, but colder)) will go downward. That will then occupy a larger volume when the descent is complete. Nett result, a larger volume overall. The atmosphere, in total, has gained PE, not lost, as in the AM scenario,
Continuity requires this. Just like any number of non-atmosphere problems. (Pots of water in which I boil spaghetti etc. )
Well…same volume, denser.
Doesn’t it adiabatically compress? (Flip side of adiabatic expansion?)
Well… something is wrong here. The atmosphere doesn’t perpetually increase in volume…
Well… I’ll have to go look at your page tomorrow. But, presumably, over the long term, the atmosphere’s PE, volume etc all stay pretty constant. (There must be dissipation or something.)
Re: lucia (Nov 9 20:58),
Lucia,
Yes, of course any volume changes due to a hurricane are tempeorary, Everything eventually settles back.
Lucia : ” the air will expand relative to the situation where all other condtions match but there is no condensation.”
Quite right and I’d badly worded my statement which ought to have read
“Simplistic assumptions like ALL the heat released by water expands the air are possibly entirely invalid.”
So are you aware of any literature on the wavelengths radiated by condensing water? Is it like a black body with radiation across the spectrum? Or is it radiated concentrated around its absorbtion wavelengths?
Sorry if I have my terminology all wrong and my questions are obvious and dumb.
Lucia, thank you for your interest in our work. What you say at Comment#59679 is a good description of the problem [I do not know how to make electronic links to comments here, my apologies].
Qiu et al., Lackmann and Yablonsky, Bryan and Rotunno (2009) all attempted to investigate the precipitation mass sink. However, our claim is that they did so incorrectly, as I explained at Jeff’s here, here and here. See also here for our explanation why the model results nevertheless conformed to observations.
But otherwise the problem is like you described: mass removal creates a pressure shortage in the region of ascent. The resulting pressure difference drives horizontal winds.
In another comment you try to visualize the x-z circulation we are talking about. Here I placed a figure PDF to help this (note that there is also legend there). Our website is quite modest and not counted to cope with a heavy traffic, so please feel free to place it elsewhere if you found this appropriate.
Anastassia
A strictly layperson view of this discussion of condensation effects on hurricanes/wind can be summarized by AM’s criticism of the existing view of hurricane power as noted in her comments at TAV below. The “bottle” and mind experiments discussed here and at TAV and the mathematical manipulations the gas and thermodynamics laws can give insights as to what can dominant under given conditions but the important consideration has to be what simulates the atmospheric conditions that hold in explaining what powers hurricanes.
I surely would like to hear some legitimate counter evidence to what AM says below and further her assertion that: “The conclusions of Bryan and Rotunno (2009) are based on a controversy pertaining the description of condensation rate that is made assuming constant pressure. This description, dating back to Asai (1965) and never re-considered ever since, does not allow one to study the effect of mass removal in theory.” What she claims is rather damning for the current state of affairs for hurricane theory and/or modeling.
Re: Kenneth Fritsch (Nov 10 09:50),
Her critique of BR09 may well be valid. However, as far as I can tell, she and her co-authors proceed to throw out the baby with the bath water by ignoring latent heat release entirely or at least hand waving it away. Condensation of moist air in the real world happens on a very short time scale, on the order of milliseconds. You can see it happen if the conditions are right during a Formula One race. If the air is humid enough, the flow over the rear wing will generate sufficient change in pressure to cause short vapor trails at the ends of the wing. If the car is going 180 mph and the vapor trail is 1 foot long, the time scale for condensation and re-evaporation is less than 0.004 seconds. A rising parcel of air expands as the pressure drops with altitude. A rising moist parcel of air expands more than a rising dry parcel of air. In other words: what contraction?
DeWitt Payne (Comment#59725),
Yes. That is the issue. There is never contraction during the condensation process, only expansion. The total mass of air moved by convection is many times greater than the mass of water vapor lost to condensation. It is the overall motion (circulation) of air that is important.
Anastasia–
Thank you. I’m going to have more questions about the assumption in section 4:
Tom-
Note: Equation 34 it is the discussion in the section find troubling. The problem isn’t the math it’s the assumptions. I’m trying to keep track of the all– since they don’t seem to b stated explicitly.
One way to achieve this might be:
1. Rig an aquarium pump up to two bottles in series.
2. In the first bottle, add water and a dip tube that will bubble air through the water.
3. Run that air through the second bottle.
4. In the second bottle will be a container of anhydrous calcium sulfate. The container is sealed but can be opened from the outside of the bottle using a magnet.
5. After the air has run for a while, seal the second bottle.
6. Open the container of calcium sulfate which will then absorb the water vapor.
7. Allow the temperature to return to room temperature, as the CaSO4 might heat as it absorbs the water.
8. Observe the bottle for compression or better yet, just rig it up with a thermometer and pressure gage from the get-go.
DeWitt–
I’m just trying to look at section 4 at a much more elementary level. There seems to be some progressive revelation of assumption, and I’m not entirely certain they aren’t self contradictory. The reason I’m not certain is I’m not sure what they all are.
DeWitt Payne, when you say the following, I have a difficult time relating this statement to AM’s argument: “A rising moist parcel of air expands more than a rising dry parcel of air. In other words: what contraction?”
And that is because I thought the disagreement was what effect(s) of condensation dominates in powering a hurricane: the expansion due latent heat or the contraction due to gas molecules becoming a liquid. I thought all agreed that condensation is what powers hurricanes, yet my simple minded view sees no condensation referenced in your comment.
I suspect that all agree that the energy of latent heat is larger than the dpV (or is it Pdv) work of condensation. I suppose that the work aspect cannot be dissipated but the latent heat can be. Again, in my simple minded view, the question comes down to the dissipation of latent heat.
Re: Kenneth Fritsch (Nov 10 09:50)
Kenneth,
<i?"I surely would like to hear some legitimate counter evidence to what AM says below"
Well, I’d like to see the original evidence. I asked here what is actually wrong with the equations as laid out in BF02 (same as BR09, but not behind a paywall). It seems to me that their the equations have mass removal treated in all the right places. There’s nothing missing.
AM then found some fault with the equation that simply relates precipitation to wv remaining, but I just can’t see that at all. In any case, it’s a big change from the original claim that there are untreated variables.
Kenneth Fritsch,
“I suspect that all agree that the energy of latent heat is larger than the dpV (or is it Pdv) work of condensation.”
.
I think you suspect correctly, save for perhaps the authors of AM10. The latent heat and associated thermal expansion along the moist adiabat swamp ‘volume contraction’ due to condensation of water vapor. I think this whole discussion is starting to resemble ‘Through the Looking Glass’; really very odd. There is no possible way that latent heat does not contribute substantially to cyclonic storm energy.
SteveF when you say “I think you suspect correctly, save for perhaps the authors of AM10” could you point me to where AM has made this statement here or at TAV about latent heat versus work from pressure-volume changes. She has talked about not mixing and making equivalent heat and work energy, but I must have missed the implication you give in your comment above.
On this specific issue, I cannot believe how many bright and talented people are missing it.
Let’s go back to 6th grade science… when heat increases, the molecules in a substance move faster. These molecules slam against and bounce away from other molecules harder and faster which is what causes a gas to expand as it heats (change in pressure).
In the case of the bottle experiment, the pressure of the air is decreasing as the air on the inside of the bottle cools. Since the temperature of the air on the outside of the bottle is not changing, neither is pressure on the outside of the bottle. Therefore, the pressure on the inside of the bottle decreases causing the pressure on the outside of the bottle to push the bottle inward, thus crimpling the bottle.
The same begins the condensation process. As the water molecules cool, their speed slows to a point to where molecular forces take over, but it is the surface of the smoke particle which cools enough, and is an attractant substance for the water molecule. Once the water molecules begin to form a bond with the smoke particle, other water molecules are attracted to those water molecules until there is a droplet large enough so that gravity will pull it downward.
So the bottle experiment fails because there is no change in volume due to condensation, but due to change in pressure.
Re: Nick Stokes Comment 59736
Nick, my answer to your question was given here. There it is also shown that it is consistent with my statement about a missing equation for condensation rate and controversies that it must entail. To find condensation rate, BF02 use the 1st law of thermodynamics twice during each update: first with Vdp non-zero (dynamics), then with Vdp = 0 (adiabatic condensation at constant pressure). There is a difference between writing out an extra equation for condensation rate and using one and the same equation (1st law) twice in a non-compatible way.
To be accurate, I told you about this (although in a brief form) even earlier, namely here.