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## Weight of Water and Wind, Hurricane Pro’s Weigh in.

Posted by Jeff Id on October 26, 2010

An email from Dr. Kerry Emanuel reproduced with permission from him and George Bryan.  It took a little while to get permission, but this is what I was receiving behind the scenes during the recent wind scuffle.  It sent me down another multi-hour adventure of reading.

—-

Kerry Emanuel:

All:  The neglect of the mass sink owing to condensation has a long history. There are a great many approximations made in models, some of which are less justifiable than others, and after they were first introduced (often in the early days of modeling), people tend to forget about them. (The neglect of the internal heat of condensed water is another, and one of my own pet peeves is the almost ubiquitous neglect of dissipative heating, which is really important to such phenomena as hurricanes.) There are two papers that I know about in the meteorological literature that examine this particular approximation:

Qiu C.-J., J.-W. Bao, and Q. Xu, 1993: Is the mass sink due to
precipitation negligible?Mon. Wea. Rev, 121, 853857

and

Lackmann, G. M., and R. M. Yablonsky, 2004: The importance of the
precipitation mass sink in  tropical cyclones and other heavily precipitating systems.
J. Atmos. Sci., 61, 1674-1692

George Bryan at NCAR did some hurricane simulations for me in which he examined the effect of the condensation mass sink. On the attached plot, compare the blue curve, which uses the full equation set, with the green, which neglects the mass sink owing to condensation. (The other two curves are the same comparison but for a model in which approximations are also made to the thermodynamic equation.) All quantities are graphed as a function of the specified terminal fall speed of precipitation (realistic values of which are in the 5-10 m/s range). Also bear in mind that there are small amplitude chaotic fluctuations in the solutions, so the differences between the curves may in part reflect these.

The bottom line is that while the effect should be included in any model that claims to conserve mass, it is not quantitatively large.

1. ### anna vsaid

Models, models everywhere,
and not ……….

From the plot it looks as a 5% difference. On the other hand CO2 is counted in ppms and look at the havoc coming out of the models. Who knows what this 5% could feed back on 😉

2. ### Tony Hansensaid

All credit to Kerry for replying.
But at which level did he really think about what he wrote?

‘….There are a great many approximations made in models, some of which are less justifiable than others, and after they were first introduced (often in the early days of modeling), people tend to forget about them…..’

I really have some problems accepting ‘approximations …. which are less justifiable than others..’.

3. ### Brian Hsaid

I earlier pointed out also that rain at terminal velocity is transferring potential energy to atmospheric and droplet heat by friction, and to the surface by impact. Is this accounted for?

4. ### Brian Hsaid

P.S. Also before reaching terminal velocity; i.e., its total gravitational p.e. gets transformed.

5. ### Nick Stokessaid

The mass correction looks really tiny – it’s the energy (presumably dissipative) which seems to make the 5% difference.

6. ### Nick Stokessaid

#3 Brian,
I think this is a subset of the dissipative heating that Prof Kerry is talking about. It’s small – the latent heat of 1 Kg water is about 2257 kJ, while the potential energy (which gets converted to heat) is 10 kJ per km.

7. ### Brian Hsaid

Yes, I sort of thought so. But every little bit helps! 🙂

Or not. ;(

8. ### Jeff Idsaid

Brian,

I didn’t reply before about the rainfall but did consider it. I agree with Nick and you about the magnitude.

——–

Reading the above email and considering the effects on models leaves me with some questions. If the latent heat of condensation is so large yet it predicts reasonable sized hurricanes in models with condensation being a weak effect and if the volume loss from condensation which is several magnitudes smaller can also be used to predict hurricane and tornado maximum velocities, doesn’t that mean that most of the latent heat is not being used as work even in the models?

That argument is central to M10’s point.

If the models are inefficiently converting heat by condensation through some factor elsewhere in the equations, they would accidentally treat the condensation volume loss the same way giving the plot above.

I think what is needed is a better physical explanation from M10 of why condensation heat is so much less effective than volume loss.

9. ### Douglas in Ugandasaid

Fascinating topic. Much physical insight on exactly this subject can be gained by looking at the work of my colleagues Anastassia Makarieva, Victor Gorshkov and their co authors. They address the role of condensation in lucid physical detail without the assumptions of simulation models.

They have published these theories in quality physical journals. Visit their site for an overview of some publications: http://www.bioticregulation.ru/pump/pump7.php

I know that several conclusions reached by Dr. Emanuel differ from those of Anastassia et al. but I would welcome a clear discussion of the physical reasoning underlying these differences. If we can agree how to choose among these theoretical positions we can also move the subject forward for clarification, resolution and advance — that is science as it should be …

10. ### Anastassia Makarievasaid

It is really good to see the discussion to have climbed to such a high level. My plan is as follows. (1) I will discuss the meaning of the difference between the red and black lines; (2) I will provide my interpretation what sort of energy conservation is meant by the blue and green lines; (3) I will show that the energetics of mass removal (potential energy release) is (conventionally) neglected in full in the presented consideration, such that the curves shown in the graph do not provide any information about its magnitude. I may not have time to write all the three posts today (although I’ll try to), and tomorrow I’ll have to be offline for the whole day, but on Friday I promise to be ready.

So, let us first understand where the magnitude of the difference between the red and black lines comes from. Let me show that this difference is barycentric velocity discussed in Section 4.3 of M10.

The idea behind barycentric velocity is as follows. If we have a sink in the atmosphere removing gas from it, the velocity it will produce will be right the velocity that will compensate for this sink. That is, as much gas as has disappeared per unit time, as much gas will arrive to the sink location per the same time unit. (I would very much encourage people interested in the topic to contemplate this very idea of accounting for the mass sink. It apparently dispenses with consideration of any pressure gradients, about which M10 got so concerned.)

The sink rate is described by precipitation rate P, which has the dimension of mm/hour or, equivalently, kg H2O/m^2/hour. Considering a hurricane of radius L and height h, the total amount of mass removed from the area per unit time is given by $\pi L^2 P$. This flux must be compensated by the air inflow via circumference $2 \pi L$ of height $h$, with air having average density $\rho$ and horizontal velocity $u$. We thus have

$\rho u 2 \pi L h = \pi L^2 P$ or $u = PL/(2 h \rho)$

Taking characteristic values P = 20 kg/hour/m^2 (in fact, a rather high value), L = 400 km, h = 4 km, $\rho$ = 1 kg/m^3, we obtain u = 0.3 m/s (in fact, this is the upper range). This is right the tiny magnitude that distinguishes the red and black lines; it is so small that it may even be overrun by numerical fluctuations as is carefully noted in the post. Still at very high terminal velocities one can see that the red line is slightly above the black line.

Why barycentric velocity obtained in such a way is not relevant to the mass sink effect, has been discussed by M10 in Section 4.3.

In the meantime, I would very much welcome that somebody who knows corrects me in my interpretations, because the specific graphs made by Dr. Bryan for Dr. Emanuel have never been discussed or shared with me or my colleagues, so my reading comes from my general understanding of hurricane physics and the internal structure of hurricane models.

11. ### Carricksaid

Nick Stokes; <blockquote.The mass correction looks really tiny – it’s the energy (presumably dissipative) which seems to make the 5% difference.

By itself it isn’t a big effect, but it does add substantially to the combined effect of energy+mass. Probably it’s more important for hurricane spin up than for full-fledged storms (there seems to be no effect at high V values).

12. ### Gavinsaid

The Bryan work referred to above is published:

Bryan, George H., Richard Rotunno, 2009: The Maximum Intensity of Tropical Cyclones in Axisymmetric Numerical Model Simulations. Mon. Wea. Rev., 137, 1770–1789.
http://journals.ametsoc.org/doi/abs/10.1175/2008MWR2709.1

(c.f. fig 10).

While I am finding the discussion of M10 fascinating, I am beginning to feel that a language barrier may be causing some confusion. There is mixed talk of condensation, precipitation and mass removal. I am wondering if the word condensation is being substituted for precipitation in some posts. Precipitation can remove mass from an air parcel, condensation can not.

My reading of the M10 paper indicated that what was being proposed was that as water vapour condensed within a moist air parcel, there was a small loss of volume. Previous work indicated that the latent heat released during condensation was transferred to the surrounding air molecules in the air parcel, causing it to maintain or slightly increase its volume and buoyancy. It may be that due to a language barrier my interpretation is wrong. However if it is proposed that an air parcel decreases slightly in volume due to water vapour condensing, this would be a significant change to current thinking. Such a proposal would demand empirical validation. Chalk on blackboards will not suffice.

14. ### Steve Kochsaid

Jeff,

Slightly off topic but if you are looking for interesting stuff to investigate in the GCMs, anything that has been parametrized rather than calculated from first principles is probably interesting.

15. ### Steve Kochsaid

Anastassia Makarieva,

Have you looked at the impact that forests have on the production of nuclei for cloud formation? It seems like an area that had efficient production of cloud formation nuclei would speed up the hydrology cycle compared to an area that had the same humidity but was less efficient at producing cloud formation nuclei.

16. ### Jeff Idsaid

One of the details that is missing from the discussion is the timeframe of models. If each timestep is so long that the condensed rain falls from the volume, it’s simply removed to the modeler. I agree that condensed is being used too liberally in some contexts but am not learned enough to offer much more than that. There is a lot of detail in climate models, and a lot which we should not be impressed with. Condensed yet suspended is not even in the numbers that I can find, of course it can be adjusted for in observed parameters. Just from a few days, I could write several posts on the over-simplified math and I’ve only begun to read.

Blogs don’t cover climate models often, especially skeptic ones. They deserve a bit of spotlight.

17. ### Leonard Weinsteinsaid

Jeff,
Condensation volume loss works because it was forced to work. The many other model “knobs” were adjusted to fit back history, and do a fair job in limited situations. However the latent heat is about an order of magnitude larger effect, so they could do even better with good physics. This reminds me of the crude turbulence models for boundary layer flow that give reasonable results for skin friction and heat transfer for some cases, but use no real physics. Of course they fail for more complex flows, and when we examine higher Reynolds numbers.

18. ### kdk33said

I didn’t read the paper, and I probably never will. But off the top of my head this early morning…

If the rain flux is 10 and the area is 10, then the mass loss due to precipitation is 100 (pick your units, it’s just an example) which is balanced by… air inflow?!

If we do a component balance around water. The water ‘out’ is still 100, but the water in is 100 times the mass% water in air (its humidity), which will be much less than 100%, so water will not balance – the hurricane dries up and blows away. If we add a water generation term – say evaporation – we can balance the water, but now the mass won’t balance.

just a thought

19. ### tallblokesaid

Steve Koch said
October 27, 2010 at 6:16 pm

Have you looked at the impact that forests have on the production of nuclei for cloud formation? It seems like an area that had efficient production of cloud formation nuclei would speed up the hydrology cycle compared to an area that had the same humidity but was less efficient at producing cloud formation nuclei.

Pollen grains are pretty small and very numerous. Could they serve a CCN’s?

20. ### Anastassia Makarievasaid

Let me now turn to the blue and green curves. Thanks to Dr. Schmidt for referring to the paper of Bryan and Rotunno (2009). I note that Fig. 10 in this paper does not show the most interesting red curve that we can see in the post. Also, Fig. 10 shows only one curve for the “conservative equation set”, so I will treat the blue and green lines together as well.

We can express precipitation rate as $P = \rho_l V_t$, where $\rho_l$ is volume density of liquid water in the atmosphere, $V_t$ is terminal velocity of droplets. However, this linear relationship is not universal, because the maximum realistic value of $\rho_l$ is limited by the adiabatic liquid water content (when all condensed vapor remains in the air parcel and travels together with it), $\rho_{lmax} \sim \rho_v$, where $\rho_v$ is density of water vapor.

All the curves in the graph decline with decreasing $V_t$, because the water load $\rho_l$ grows with decreasing $V_t$ and causes lower vertical velocities which impacts all the dynamics. However, the hurricane model in question is peculiar in recycling the dissipated heat (!). Accounting for this recycling should be responsible for the difference of the blue/green curves from the black one.

Let me show this for the limiting case $V_t = 0$. In this case we have $\rho_l \sim \rho_v$, such that the dissipative power caused by droplet friction is equal to $D^+ = \rho_l g h w \sim \rho_l g h P/\rho_v \sim g h P \sim 110$ kg/sec3. Here I took $P \sim w \rho_v = 10$ kg/m2/hour, $h = 4$ km.

The dynamic dissipative power of the hurricane in the model is assumed to be proportional to $D = \rho C_D V^3$, where $C_D \sim 10^{-3}$. At $V_t = 0$ for the traditional equation set (the black curve in the graph) we have $V = 55$ m/s and $D = 166$ kg/sec3. Therefore, the new velocity, enhanced at the expense of heat recycling (!), is expected by Dr. Bryan and colleagues to be $V \sim [(D^++D)/C_D]^{1/3} = 65$ m/sec. As one can see, this just the difference between the blue/green verus black curves at $V_t = 0$.

I have given my reading of these curves here for two reasons — (1) to illustrate that there is never nothing in a model that a physicist would not be able to explain “on the back of the envelope” or, as we Russians say, “with use of two fingers”. It is not that models produce new physics. It is physics (either correct or incorrect) that governs models. If a modeler cannot explain what his model output means in simple physical terms, such a model cannot be trusted. (On the other hand, it is not my model, so I welcome corrections to my interpretations and I would not insist on the utmost accuracy of the numbers I got.)

For the second, I wanted to illustrate that the physical basis of accounting for condensation represented in the graphs above, recycling of dissipated heat from friction (!), has nothing to do with the potential energy release from condensation that forms the basis of our approach.

In response to Jeff

I think what is needed is a better physical explanation from M10 of why condensation heat is so much less effective than volume loss.

There are constraints imposed (at least) by the Carnot efficiency. When you have latent heat release in the adiabatically ascending air in one place, you need a temperature difference with another place (the descending column) to make use of this heat flux. In the case of the hurricane there is no temperature difference altogether. If anything, the descending column is warmer than the ascending one. The appeal to tropospheric temperatures T ~ 200 K is not justified, because heat is not released there (e.g., to space), but is advected horizontally away from the hurricane. If the mass removal at 1/5th of latent heat power is enough to produce the hurricane, you need a temperature difference of at least a few dozen degrees to have a Carnot efficiency of 0.2 (the actual efficiency would be of course much lower). The vertical temperature profiles of the hurricane and its envirionment in the upper atmosphere are practically uniform (which is not surprising given such violent horizontal mixing). The conversion of latent heat to mechanical energy could be occuring at an efficiency not exceeding 0.01 or something like that.

In my next post I will show how the energetics of mass removal should have been correctly accounted for.

21. ### RuhRohsaid

In my field, a significant simulator design challenge is about what to do in the case of singular matrices, for which no inverse matrix exists to do the calculation. In the last decade of the prior millenium, this was probably the dominant issue, a.k.a. ‘convergence failures’.

The other key ingredient is dynamic timestep control. If an equal sized step (as the last one) causes too big of a change in the output, the simulator will try again with smaller and smaller timesteps to try to get through a ‘discontinuity’.
If the same size timestep keeps working (=sufficiently small errors) without ‘hiccups’, it will be increased gradually, up to a user-defined limit.

Some components do indeed have complicated behaviour which is parameterized in their individual model, for computational efficiency. But there is commercial pressure to improve the models for trustworthy simulator accuracy, so the overly-simplistic models do not have enduring ‘legacy’ status.

So, do these “GCM” multi-physix simulators have dynamic timestep behaviours?
Does the ‘singular matrix inverse’ problem bedevil the users, or have the math crunchers somehow banished the inverse problem (or did it never pop up)?
Are lunar and solar ‘atmospheric tidal forces’ embedded in typical GCM embodiments?

Semi humorous AGW thread title; Earth, wind, and fire; (i.e., the water is missing)
Ok, I’ll try to get more sleep and better jokes…

RR

22. ### Nick Stokessaid

RR,
I don’t know very much about GCM’s, but I have experience with Navier-Stokes equations. The problem you describe shows up as an ill-conditioned matrix at the semi-implicit stage. You’re solving a Poisson equation, so the problems are determined by geometry, which doesn’t change over time, so varying the timestep won’t help. The penalty is an increase in the number of steps in the iterative solver that you use. I have a view (expressed here) that the extra work in the solver is equivalent to the work of taking extra timesteps.

GCM’s have a similar semi-explicit stage, but include heat transfer.

23. ### Steve Kochsaid

TallBloke:

There is an interesting paper that came out in Sept, 2010:

“Aerosols control rainfall in the rainforest”

It says that “Researchers show that precipitation-controlling aerosols over the Amazon rainforest mainly originate directly from the forest ecosystem as biological particles”.

It occurred to me that it would not be surprising that trees that are efficient producers of condensation nuclei would have a survival advantage because there would be more clouds, thus more rain. Obviously, the more rain there is, the faster the hydrology cycle cycles (releasing heat on every cycle), which cools the atmosphere down.

24. ### Anastassia Makarievasaid

My apologies, it is my post above, but the images are missing. Here it is again with links for images. The above post can be deleted.

Let us now see what has been overlooked and how the effect of mass removal should have been taken into account. This is a bird’s-eye view of a hurricane:

Precipitation $P$ removes vapor from the hurricane area. The ambient environment delivers air with molar density $N$ at barycentric velocity $u_b$ to the hurricane area, such that the molar density of air inside the hurricane does not change. In our hurricane we set the release of latent heat equal to zero and do not recycle dissipated heat from falling drops. According to the difference between the red and black curves in the graph, in this case all we would have is a “hurricane” with air velocity not exceeding 0.3 m/s.

Let us now have a look at this hurricane in vertical cross-section:

Does it look familiar? It reminds us of a piston that compresses gas. Indeed, while precipitation decreases pressure within the hurricane, the incoming air from the ambient environment compresses the hurricane air to maintain the steady state.

Let us estimate the work the ambient atmosphere performs in this way on the hurricane. This work per unit time is given by $W = p \times 2\pi Lhu_b$, where $p$ is ambient atmospheric pressure. The term $2\pi Lhu_b$ represents the rate at which the volume $V_a$ occupied by the external atmosphere increases as the latter expands into the hurricane area, $dV_a/dt = 2\pi Lhu_b$. Recalling the definition of barycentric velocity at #10 and expressing it in terms of molar density, $u_b = PL/(2hN)$, we conclude that total work performed on the hurricane by its ambient environment per unit time is equal to $W = p\times2\pi Lh PL/(2hN)$. Taken per unit hurricane area $\pi L^2$, this power becomes

$Hurricane \, power \, (H_p) = Pp/N = PRT.$

Here $P$ has the dimension of mol/m2/sec, $R = 8.3$ J/mol/K is the universal gas constant, $T \approx 300$ K is ambient temperature, $p = NRT$ is the ideal gas law. (Remember this formula for $H_p$! It will be taught at schools.)

Now it is time to estimate the power. We take a typical hurricane precipitation rate of 10 mm/hour (e.g., Trenberth et al. 2007). This is equal to $P$ = 10 kg/m2/hour = 0.15 mol/m2/sec, such that we have $H_p$ = 380 W/m2. For a circle of radius 60 km, this gives a total power of $4.3\times 10^{12}$ Watts.

We now go to the page Dr. Chris Landsea has prepared for those interested in hurricane power. We see that an average hurricane, over an area of 60 km radius, has a mechanical power output of around $1.5\times10^{12}$ Watts, i.e., twice less than what we have just obtained. What does it mean? It means that the work caused by vapor mass removal is more than enough to drive the hurricane. In our hurricane paper we outline the details how a more specific (rather than average) spatial power distribution can be obtained.

We conclude that the existing hurricane models (and GCMs) violate the energy conservation law by neglecting the mechanical work associated with the contraction of the mass sink area by the ambient environment. This contraction occurs with a tiny (barycentric) velocity, but in reality represents the most powerful dynamic process that produces huge wind speeds in hurricanes and normal wind speeds elsewhere.

This is not a minor oversight, but one pertaining fundamental physical principles. A very important circumstance consists in the fact that potential energy does not enter the first law of thermodynamics and cannot be retrieved from it. The first law of thermodynamics is not equivalent to energy conservation law. The first law is valid for a specific range of processes (it does not describe a mechanical pendulum or a rotating satellite). It does not describe the process of mass removal from ideal gas either.

Ideal gas is a unique physical phenomenon. While labeled a “fluid” together with liquids, it possesses distinct properties dictated by its equation of state, $pV = RT$, namely that at constant temperature $pdV = -Vdp$. This is a dynamic, not a thermodynamic effect, akin to conversion of kinetic to elastic energy in the oscillating spring etc. This property of ideal gas has been ignored.

The pressure gradient force resulting from the mass sink, Eq. 37 in M10, should have been explicitly included into the list of the volume-specific forces that enter the right-hand side of the Euler (NS) equations. This is the only way to correctly account for the dynamics of gas mass removal without violating the energy conservation law. This is an ultra complex, highly non-linear problem, because of the positive feedback between the pressure gradient force and vertical velocity. In our hurricane paper we showed how this problem could be treated under some reasonable approximations and that the results obtained conform reasonably well to observations.

25. ### Anastassia Makarievasaid

I apologize for some mess. (1) 2s3c is me. (2) The post contained two images, but they disappeared. When I re-submitted the post with links instead of the images, it went to moderation, because there are two many links there. The links to images are:

26. ### Brian Hsaid

Leonard;
Right; if you parameterize and “freeze” the big stuff, all you have left is the fringe effects to play with. And there, anything goes! 😉

27. ### Brian Hsaid

Anastassia Makarieva October 29, 2010 at 12:25 am

Anna;
Homonym error, very common: “two many links” = 2 many links. Should be “too many links”. 😉
_____
Don’t have the link, but someone commented that actual observation of hurricane eyes shows warm air sinking in the center and very fast-rising cold air in the eyewall. If true, there are VERY complex pressure dynamics going on.

28. ### Anastassia Makarievasaid

Thank you, Brian.

Here’s link where this VERY complex pressure dynamics was mentioned.

29. ### Sigue la búsqueda del viento y la discusión con Makarieva « PlazaMoyua.orgsaid

[…] Weight of Water and Wind, Hurricane Pro’s Weigh in. […]

30. ### kimsaid

Models are bitin’.
Ikon or Klasm?
==========

31. ### Anastassia Makarievasaid

I would also like to add that recycling of dissipated heat as hurricane intensifier was criticized by us here. There was a recent comment on that work ascribing our critique to a confusion. Our reply is to the comment is available here.

The physics of pressure gradient associated with the mass sink was also discussed here at Prof. Judith Curry’s blog Climate Etc.

32. ### Richard Yablonskysaid

For those who may wish to do an in-depth literature review of the precipitation mass sink effect, you may be interested in the 14 studies highlighted in section 1.4 of my 2004 master’s thesis on the topic, available here:

http://www.gso.uri.edu/~richard/etd.pdf

Also, the original post mentions our 2004 JAS paper, which includes may of the results discussed in my thesis. There is also a JAS paper written in 2006 on the topic by Wacker, Frisius, and Herbert, available here:

http://journals.ametsoc.org/doi/pdf/10.1175/JAS3754.1

Dr. Makarieva is probably more familiar with the relevant literature written since 2006 than I am because my current tropical cyclone research involves topics other than the precipitation mass sink (although perhaps I will jump back into the fray at some point in the future)…

33. ### Thomassaid

@Anastassia Makarieva

didn’t you suggest in a (rejected) paper, that the mass sink owing to condensation should be considered as the main mechanism that drives hurricanes ( … and tornados …)? The results presented here show clearly, that this is not the case (inclusion results in only a small alteration of v_max)?

34. ### Anastassia Makarievasaid

#33 Thomas,

Thank you for your interest. I suggest that you should read my comments #10, #20 and #24 to see why the account of the mass sink shown in the graph is not correct.

You are right that we suggest that the mass sink from condensation is the main driving force of the hurricane. One our paper was indeed rejected, another one did make it through, and there is another one in between right now. You are welcome to comment on the latter and impact the process.

35. ### Brian Hsaid

Polish study:

“Maxwell assumed that evaporation took place at constant temperature. It is so, if we look at the initial state, that is a liquid, and the final state, that is a vapour. It is true that their temperatures are equal. But during the evaporation process itself, the nature acts in a completely different way,” explains Ph.D. Marek Litniewski from IPC PAS.

The existing description assumed that the heat transfer in the system was stable and the rate of evaporation was limited by the efficiency of the process during which the particles break away from the surface of drops, i.e. diffusion. However, the simulation carried out in the IPC PAS showed that during the evaporation into vacuum or the liquid’s own vapour the system gained mechanical equilibrium very quickly. Particles break away from the surface of a liquid and their mechanical recoil allows the equalisation of the pressure inside the drop. If the rate of evaporation on the surface achieved the maximum value and the system was still unable to equalise the pressures, spaces with new surfaces would open inside the drop and it would start to boil. However, it was observed that the mechanical equilibration of pressure can be insufficient and the temperature on the surface of the liquid decreases: the drop aims at maintaining the pressure equilibrium at the cost of its internal energy. This observation suggests that the factor that is crucial during evaporation is not the diffusion of particles into the environment but the heat transfer and the equality of pressures.

Cites on site.

36. ### douglasinugandasaid

Reading all this through I am so far convinced that Anastassia and co. are on the right track. It is much easier to follow the reasoning through direct physical relationships than to assume that Dr KE’s models with their assumptions and uncertain approximations are somehow superior (why would anyone assume that?). But perhaps I am missing something.
It’d be great if someone would directly address Anastassia’s comments and show us where the flaws in her analyses are (if any) … Can anyone do that?
Best wishes
Douglas

37. ### Nick Stokessaid

36 Douglas,
The standard criticism is that on condensation, air does not contract. The loss of water vapor volume is more than made up by expansion due to latent heat release.

I have seen counter-arguments that LH cannot actually increase the temperature, bedcause if it did, it would stop condensation. But that ignores the dynamics.

Condensation occurs because moist air is rising and cooling by adiabatic expansion. Release of latent heat means that it still cools and condenses as it rises, but at a slower rate than would air without condensation. Consequently air after condensation is warmer and less dense than other air on its level. It has expanded.

38. ### Anastassia Makarievasaid

#37 Nick,

You are missing the crucial point about mass balance.

The gas converges into the hurricane in the low atmosphere at a rate C, diverges from the hurricane in the upper atmosphere at a rate D and disappears within the hurricane at a rate P. This means that the steady state can only be maintained if the net balance is in favor of convergence, 0 = C – D – P.

$C - D = P > 0$.

Therefore, a steady-state mass sink is inseparable from contraction. Heat release/uptake is irrelevant to these mass balance considerations. The difference between C and D is characterized by barycentric velocity, as I described above.

The standard criticism you referred to is not consistent with the laws of thermodynamics, as shown in Section 2 of M10, and misses the same point as well.

39. ### thomassaid

if you really found a new mechanism that is the most important process that drives the atmospheric circulation you should be able to easily build a model that produces much better forecasts than all the state of the art models (ecmwf, gfs, wrf) since they are based on the wrong physical principles (which is difficult to imagine. now the ecmwf model showed skill 10 days in advanced. it is very unlikely that it does this because of the wrong reason). large errors like this should amplify rapidly making a good forecast impossible (this is known since the work of e. lorenz).

40. ### douglasinugandasaid

*37 Nick
Thanks for the reply. This is very valuable — I agree that this has been a standard criticism. But my question now is whether it is true(?). The claim that the a “loss of water vapor volume is more than made up by expansion due to latent heat release” can be evaluated against basic physical laws and appears to fail that test.
That is why M10 starts by establishing (as far as I can see) that this claim is incorrect. It cannot be reconciled with fundamental principles of thermodynamics.
Do you (or anyone) see an error in Section 2 of M10? That’d be really helpful. Thanks again for the feedback.
Best wishes
Douglas

41. ### douglasinugandasaid

*39 Thomas

Great idea. I hope someone tries that.

But for some of us the basic physics can and should be resolved first. It is simpler than building a complex mega-model. The question is if there is a flaw in the physics where is it?

To me the physics is clear and sharply defined. Climate models in contrast are so complex that asessing fit and what to include, approximate, etc. will remain open to interpretation.

It is striking to me coming from outside the climate science community how scientific tests and credibility has become synonymous with models that are too complex to fully grasp … I know it is hard to experiment with the planet (so you do need those models), but surely basic physical reasoning and empirical data trump models in a case like this? (But yes someone should make the models !).

Best wishes

Douglas

42. ### Anastassia Makarievasaid

#39 Thomas,

Thank you for this comment. The existing weather forecast models, as well as hurricane development models, are built relying extensively on empirical parameterizations. That is, the typical characteristics of a number of weather patterns are replicated and predictions are made on the basis of previous statistical data. For this reason, whenever something unusual happens in the atmosphere, the models skill drop rapidly to zero (indeed, they cannot predict what have not been observed before, like hurricanes near the Brazilian coast). For example, this summer during the abnormal heat wave in Russia the 10 days forecast consistently predicted a drop of temperature on the 10th day. This continued for two months during which the temperature never dropped.

It is not accidental either that the skill of the existing models is sharply limited by a few days. This is the time scale of water vapor turnover in the atmosphere. If the basic principles that we are advancing are studied with attention (this is just what we are calling for), there is no doubt that the weather forecast quality will rise sharply. Your compliment about our ability to easily build such a model is well taken, however, building a model does demand some funding and manpower. As I said before, if it were possible for a couple of scientists to solve all atmospheric problems, it would be unclear whether the huge funding currently allocated to weather and climate model development is at all appropriate.

43. ### thomassaid

Douglas,

the thing is, that the operational forecasting models are verified against observations on a daily (!) basis, which really is different to the climate discussion.

my point is, that as you can see from the discussion basic physical reasoning seems sometimes not to be that easy to agree on. but if you are talking about a major effect you should not neet such a complex model to beat the others. and this is really an objective verification of the theorie.

44. ### thomassaid

@Anastassia Makarieva

the models have some skills to even produce hurricanes. which they really shouldn’t if your theorie was true.

“It is not accidental either that the skill of the existing models is sharply limited by a few days.” – no, you shoud read the papers of EDWARD N. LORENZ Deterministic Nonperiodic Flow and others.

I think you really need to look more deeply into real life data or data from operational models.

45. ### douglasinugandasaid

Thomas
Thanks – I accept the point that this is a good way forward IF we accept that the physical reasoning will remain unresolved in the longer term (I dont want to concede that just yet, as I remain optimistic). It would certainly be a valuable exercise if we had the resources to do it properly.
Anyone out there interested to try this out? Please do!
Thanks
Douglas

46. ### Nick Stokessaid

Anastassia,
All you have described in #38 is a mass balance. Whether it results in contraction depends on density, which obviously changes a lot. But the reason why so many people think you are wrong is that the density is affected by the release of latent heat, and in fact reverses the sign. Instead of your
Hp = PRT, the right formula for this effect is
Hp = PRT(1-L/(cp T))
L = LH vap/cond for water, cp = SH for air

By mass, L=2400000, cp=1000, at T=300 approx, so L/(cpT) is about 8.

So Hp is negative. Of course the big heat engine term is the transfer of heat from surface to TOA.

47. ### Steve Fitzpatricksaid

#38, Anastassia,

You say above “Indeed, while precipitation decreases pressure within the hurricane, the incoming air from the ambient environment compresses the hurricane air to maintain the steady state.”

It is hard to directly argue with this, since clearly precipitation reduces the pressure within the hurricane, and surrounding air rushes into the hurricane due to this reduced pressure.

What I am not so sure about is the connection of this drop in pressure purely to mass loss from precipitation. It seems to me that the pressure within the hurricane falls for two reasons (both operating at the same time): 1) mass loss of water that falls as precipitation, and 2) expansion of air entering the hurricane as it rises and expands due to the latent heat released from precipitation in the rising column. The expansion in the rising (warm) air column appears to me to reduce surface pressure much more than the loss of mass due to precipitation. That rising air continuously exits the hurricane from above, looses heat to space, and returns (as a descending dry counter-flow) to the surrounding regions that are supplying moist air to the hurricane, as it must to maintain overall mass balance.

It seems to me the high wind velocities within the hurricane are mostly due to the angular momentum of the air entering the storm, as it is forced into a tight spiral while rising through the hurricane; and the tightest spiral is of course at the eye-wall. I remain puzzled by your focus on the mass loss due to precipitation as the only energy source of the hurricane. It seems to me that the hurricane can be considered a large “air pump” which is powered mainly by the latent heat of condensation, not mass loss from precipitation.

I am no expert in Hurricanes… although I have had several pass directly over my house, which is in Florida. The transition from the violent eye-wall to the calm and sunny eye is simply astounding! 🙂

48. ### Anastassia Makarievasaid

#46 Nick,

All you have described in #38 is a mass balance. Whether it results in contraction depends on density, which obviously changes a lot.

This is incorrect. Remember that we are talking about a steady state, where neither temperature nor density nor pressure change with time. You are once again missing the critical point that neither release nor intake of heat may change the balance of mass.

If you remove gas over some area, but want the pressure to remain steady, you have nothing else to do than to add gas to that area via the vertical cross-section, as shown in the second figure in #24. To do so, you must perform work against the gas that is already there. Formula $H_p = PRT$ follows unambiguously from the energy conservation law that describes this work.

Your formula does not have any justification, it predicts a negative hurricane power and a negative barycentric velocity; both results are unphysical.

49. ### Steve Fitzpatricksaid

Sorry, that should have been not “looses heat to space”, but rather “loses heat to space”. Such errors from a native English speaker reflect, I hope, the early hour here.

50. ### Nick Stokessaid

Anastassia #48,
My formula yields basically the same result as the calculation of Dr Meester on S8981 and Dr Rosenfeld on S12437, discussing your previous paper. You are dismissing a lot of scientific reasoning there. Both emphasised that proper allowance for latent heat reversed the sign. That means that the real power source of a hurricane, which is from the transfer of heat from surface to TOA, has to drive both wind velocities and condensation.

It is late here – I will detail the derivation of my version tomorrow.

51. ### Anastassia Makarievasaid

#47 Steve,

That rising air continuously exits the hurricane from above, loses heat to space, and returns (as a descending dry counter-flow) to the surrounding regions that are supplying moist air to the hurricane, as it must to maintain overall mass balance.

Let me clarify this common misunderstanding. The hurricane air practically does not lose any heat to space (namely this fact makes the low tropospheric temperature irrelevant for consideration of the hurricane as a heat engine). It is easy to calculate that in order to radiate the flux of latent heat in the hurricane to space, the atmosphere should have been hundreds of degrees warmer than it is (see page 17433 in our ACPD08 paper).

In reality what happens is that the latent heat that is released in the region of ascent travels together with the air towards the descending part of the eddy. There, descending by a dry adiabat, it warms the surface. For this reason the ambient environment of the hurricane in the lower atmosphere is warmer than the hurricane itself. Hurricane works as a heat pump to warm the external environment, nothing more. This pump would not have worked if the mass sink were absent and would not drive the air motion.

52. ### Anastassia Makarievasaid

#50 Nick,

You are dismissing a lot of scientific reasoning there. Both emphasised that proper allowance for latent heat reversed the sign. That means that the real power source of a hurricane, which is from the transfer of heat from surface to TOA, has to drive both wind velocities and condensation.

Thank you for pointing us out to the reasoning of Dr. Meesters and Dr. Rosenfeld. This reasoning is incorrect; we explain in our paper why. I notice that you are persistently ignoring the arguments concerning the mass sink and perpetuate the confusion of previous critics. I emphasize the key point: neither release nor intake of heat in the region of a steady-state mass sink does not obviate the need to compress the sink area externally to maintain the steady pressure. I encourage you to take this issue into account in your derivations.

53. ### Steve Fitzpatricksaid

Anastassia #51,

The descending air would of course be warmed along the dry adiabat.

But the air exiting the top of the hurricane must be substantially warmer than the typical air temperature at that altitude, due to substantial latent heat release while rising. If that air does not cool before descent, then the application of the dry adiabat would lead (I believe) to extremely high local temperatures (certainly >40C) in the area surrounding the hurricane. My personal observation is that exceptionally high temperatures like >40C do not happen as the hurricane approaches. Do you not think a significant fraction of the heat has to be lost from that warm air in the upper troposphere prior to descent?

Certainly normal thunder-storm activity carries heat to the upper troposphere; it is hard for me to imagine a hurricane (which contains many intense thunderstorms) does not do the same.

54. ### Steve Fitzpatricksaid

Anastassia,

The analysis of heat flow in your paper does not make a lot of sense to me, and I am reasonably certain that it is not correct.

The total rainfall (averaged over the entire hurricane area) is on the order of 1.5 cm per day (Gray 1981). Much more rain than average falls near the eyewall, but much less in the outer rainbands. This is ~0.0625 cm per hour (on average) for the entire hurricane. If the water vapor content of the rising air is 50 g/M^3, then this is equal to ~0.005 cm of rain from that cubic meter falling over 1 M^2. A total rainfall of ~1.5 cm per day, corresponds to an average rate of rise of ~1.5/0.005 = 300 meters per day, or 0.35 cm per second, while your paper’s calculation suggests an average value of 10 cm per second (and says this is a conservative value!). Clearly this assumed average rate of rise (10 cm per sec) is inconsistent with the measured average rate of rainfall, by a factor of ~29.

Applying your heat loss calculation to the more realistic average vertical flux of 0.35 cm per second, the required loss to space in the upper troposphere would be in fact only a little higher than the normal solar flux in the tropics, and the required blackbody temperature would be modest (~280K); not 600K!. But even this is an overestimate, because the air flow out of the hurricane is by no means limited in area to that of the hurricane itself. The outflow area (and so area available for radiative loss prior to descending return flow) can be considerably more than the area of the hurricane itself, further reducing the required blackbody temperature.

I remain convinced that latent heat is the primary source of energy for hurricanes.

55. ### Anastassia Makarievasaid

#53 Steve,

Hurricane features a finite size, it is not an atmosphere-wide phenomenon. Within the hurricane latent heat is released at a high rate in the rapidly ascending air. As I said, this heat cannot be radiated to space within the hurricane area.

How can a steady temperature within the hurricane can then be maintained? It is maintained due to the fact that the hurricane moves along and leaves the warmer air outside. This air descends much more slowly over a much larger territory. During this process the mechanical power dissipates to heat outside the hurricane and all heat, including latent heat, is released to space at a normal rate. For this reason the ambient surface temperatures do not reach 40 C. The largest rise in temperature is observed near the hurricane outer border where the velocity of air descent should be the highest.

In considerations of hurricane as a heat engine it is presumed that the hurricane leaves heat to the cold upper atmosphere. In reality the hurricane simply gets rid of the warmer air that diverges away from the hurricane area; then the hurricane moves along. It is this mass turnover within the hurricane that keeps the temperature steady. There is no cold body which takes the heat away from the hurricane. Hurricane is a field, not an individual air volume: this pressure field comes, condenses vapor to sustain itself and goes further, if there is enough vapor around. In this sense it can be compared to a particle moving in Wilson chamber and leaving condensed vapor behind.

56. ### Steve Fitzpatricksaid

Anastassia,

I beg you to look at my comment #54 and compare it to the calculation in your paper.

57. ### Steve Fitzpatricksaid

Anastassia,

I have much work to do, so I must sing off for now….

58. ### Steve Fitzpatricksaid

Sign off, not sing off.

59. ### DeWitt Paynesaid

As I said, this heat cannot be radiated to space within the hurricane area.

Surely some of it is radiated to space. At the top of of a hurricane, the atmosphere is nearly transparent to LW IR and the cloud surface has an emissivity close to 1 at those wavelengths. IR imaging of tropical storms shows the cloud top temperature near the center of the storm to be close to 220 K.

60. ### Anastassia Makarievasaid

#54 Steve,

The precipitation estimate you take, 1.5 cm/sec, is very low compared to what one actually observes in hurricanes. Have a look at this figure, which shows the rainfall distribution for hurricane Katrina. It drops from 30 mm/HOUR at the windwall to around 4 mm/HOUR at 400 km. Your estimate of 1.5 cm/sec corresponds to 5 m/year, which is very close to the annual mean equatorial precipitation (around 3 m/year). Apparently such an estimate would not single out hurricane as a weather event with intense rainfall.

It is not my goal to immediately persuade all people here. My goal is to present my arguments to a wider readership that would make some people think.

#59 DeWitt,

Of course, the atmosphere does not stop radiating over the hurricane. The point is that this radiative flux is significantly lower than the latent heat flux that is being released within the hurricane. For example, precipitation rate of 4 mm/hour = 4 kg/m^2/hour = 0.06 mol H2O/m^2/sec corresponds, at L = 45 kJ/mol, to a heat flux of 2800 W/m^2, which is an order of magnitude higher than the solar power. This flux can in no way be radiated by an atmosphere at 220 K.

It is a very big problem, to get rid of such a large flux.

61. ### Gavinsaid

Anastasia, if I might offer some advice – do not oversell your ideas. It is beyond question that standard hurricane models (as used in routine forecasts by GFDL, NHC etc.) do produce hurricanes in the right circumstances. While not perfect (and therefore potentially improvable by consideration of the precip mass loss terms you are proposing – and has been suggested by Bryan et al), these models are an undeniable testament that your mechanism is *not* essential to the existence of hurricanes, and that one can get a very good approximation just from using the latent heat terms. Claims to the contrary (i.e #51) will tend to lead people to question the logic of your whole approach.

If you want to make some progress here, I suggest that you point out clearly where in the full equation set of Bryan and Rotunno you would suggest modifications. These equations are valid throughout the domain and not just for vertical ascent along a moist adiabat. This would go a long way to clarifying whether your idea is or is not already included in their ‘conservative’ equation set.

62. ### Brian Hsaid

Re: Anastassia Makarieva (Oct 30 08:25),
Next English lesson:
“neither release nor intake of heat in the region of a steady-state mass sink does not obviate the need”
Double negative; “neither/nor … does not” would logically mean that both do. 😉

63. ### Brian Hsaid

Re: Steve Fitzpatrick (Oct 30 10:05),
“A total rainfall of ~1.5 cm per day, corresponds to an average rate of rise of ~1.5/0.005 = 300 meters per day, or 0.35 cm per second,”

Observe the danger of averages. Who says the rain is falling/generated during the entire rise of the originating air volume? Maybe it happens fast, in a particular vertical band. It would continue to rise, carried by the overall airflow, up to the point where that flow reversed.

64. ### Brian Hsaid

Re: Anastassia Makarieva (Oct 30 10:14),
Next English lesson:
“How can a steady temperature within the hurricane can then be maintained?”
The second “can” is unnecessary and incorrect.
______
How steady is the temperature in reality? That it is needed to make the equations work is dubious justification for assuming it.

65. ### Kenneth Fritschsaid

While I am eminently unqualified to make detailed judgments about the condensation driven calculations presented here, and further from previous exchanges with Anastassia Makarieva that we evidently have major political differences, I have to state that I have been thoroughly impressed with the comments and explanations presented here on a number of physics-applied-to-climate-science topics by Anastassia. She has patiently and thoroughly answered almost everyone’s queries here and from first principles made reasonable estimates of experimental results.

We have had only a very few scientists who were willing to extend this effort at blogs and do it by avoiding the all too frequent snarking and personality issues that arise when controversial issues are discussed. It would appear that those very few scientists are often attempting to explain a newer theory that the consensus, for whatever reasons, has not seen fit to give a platform for a thorough discussion.

These types of discussions get back to a point I have made on a number of occasions about one learning from the replies to posts, but often learning more from a lack of replies to a post (and here I mean a reply of substance and not hand waving gestures). It helps define what the “settled science” really has considered and assumed in its settling of the science.

It is instructive to consider the civil tone, for the most part, that these threads/posts take on when scientists appear motivated to teach and explain and compare them with the all too many threads that get way laid into a food fight. It is these civil threads that, in my mind, show the true potential of the internet, and blogging in particular.

66. ### Steve Fitzpatricksaid

Anastassia #60,

This paper: ams.confex.com/ams/pdfpapers/35695.pdf shows the average rain gauge measurements of all USA landfall hurricanes 1947 to 2000. The average hurricane shows ~8 cm per 24 hours, averaged over the whole of the hurricane field. I guess the measured rainfall rates vary quite a lot. The diagram you linked to says “hours 45-54 of the simulation”. Was this a computer simulation or measured rainfall (either by gauge or radar)?

If the average rainfall rate is 8 cm per 24 hours, then the rate of rise (assuming 50 g/M^3 moisture content) would have to be ~1.8 cm per second; still a factor of >5 lower than the 10 cm per second you describe. The latent heat load would then be ~800 watts per square meter, compared to a normal daily average infrared heat flux in the tropics of about 280 watts per square meter. Once again, there is no reason the outflow field above the hurricane can’t be much larger than the hurricane itself, which reduces the flux rate per square meter. The size of the outflow field is actually indicated by one of the “warning signs” of an approaching hurricane: a rapid increase in high cirrus clouds, well before the arrival of the hurricane itself… yet another indication that the outflow from a hurricane extends far beyond the visible hurricane.

Warm dry air above the hurricane is lower in density than air normally is at that height, so it will not likely sink until it has cooled and become more dense. To force that air downward without cooling would require overcoming significant net buoyancy.. and a source of energy to do it. I just don’t see how the return flow can be as you describe, with little cooling having taken place aloft.

I do not doubt that there is less than 100% loss of the latent heat from the air prior to the descending return, and so, as you say, there is likely some net warming around the hurricane due to the descending dry air. But I think it is perfectly reasonable to expect a significant portion of the total latent heat to be lost to space, thus proving a substantial source of energy for powering the hurricane.

67. ### Anastassia Makarievasaid

Gavin, I very much and sincerely appreciate your continuous input. However, your advice puts me in a difficult position: I am convinced, on scientific grounds, that the existing hurricane models are based on incorrect physics and that latent heat does not drive the hurricanes. That the models reproduce the hurricanes satisfactorily is not a proof that they are physically correct — they could have been artificially fitted to behave satisfactorily. In the particular case of the MPI model I do know how this was done. Perhaps you did not study this model in detail, but I did.

You point to the work of Bryan and Rotunno (2009) as having “equations that are valid throughout the domain”. Let me explain that these equations do not say anything about rainfall altogether. When one goes from a non-condensable gas with zero in the right-hand part of the continuity equation to a condensable gas with some mass sink S, one must develop a theory for S first — before building models. Such a theory has been missing. There is not a single paper in the meteorological literature where it would be considered (!!!). In the work of Bryan and Rotunno, see Eq. (13), S is simply re-written in terms of another unknown variable, terminal velocity $V_t$. This unknown variable is set manually as a parameter. This is NOT a valid account of a mass sink. The main physical feature, the positive feedback between vertical velocity and precipitation, is missing.

I fully respect the opinions of other people, but I am not going to change mine in order to make some progress, whatever you mean by this word. I am sure that if you were in my place you would do the same — stick to science, even if it were difficult in personal terms and against a majority. I am presenting clear, quantitative physical arguments showing that the mass sink IS the main driving force. Hurricane power is given by PRT. No one has been able to refute them. If I meet with a valid physical argument showing that our results are incorrect, I will accept it and admit that we are wrong. But I do not accept reference to models as such an argument. Models are not a supreme authority to me — fundamental physical laws are. I am sending my message to those who share this view.

68. ### Kenneth Fritschsaid

Are the differences here coming down to: on one hand the climate modelers have a model that by parameterization can give what that they judge is a reasonable estimate of the real climate and do it by ignoring some of the physics involved (or not knowing whether it is ignored) and on the other hand an approach from first principles that can reasonably estimate the real climate with some of the physics ignored by the modelers?

It is beyond question that standard hurricane models (as used in routine forecasts by GFDL, NHC etc.) do produce hurricanes in the right circumstances. While not perfect (and therefore potentially improvable by consideration of the precip mass loss terms you are proposing – and has been suggested by Bryan et al), these models are an undeniable testament that your mechanism is *not* essential to the existence of hurricanes, and that one can get a very good approximation just from using the latent heat terms.

It would be most instructive here to provide a few details on measures of just how good these models are in routine forecasting and to what degree they depend on parameterization and details of how those parameters are estimated. Are we talking here about the hurricane formation and/or what happens in a hurricane once initiated? I thought that some hurricane models required hindcasting data.

69. ### Steve Fitzpatricksaid

Anastassia #67,
“Models are not a supreme authority to me — fundamental physical laws are. I am sending my message to those who share this view.”

I most certainly share this view; models are seldom any better than the understanding that they are based on. When that understanding is flawed, so is the model.

But your suggestion that a significant quantity of latent heat from a hurricane is not lost in the outflow field just strikes me as unphysical. Would a measurement of heat loss in the outflow field be convincing?

70. ### Nick Stokessaid

67 Anastassia
“Such a theory has been missing. There is not a single paper in the meteorological literature where it would be considered (!!!).”

I just don’t believe that. When I look again at the CAM3 updraft model I see 4.17 – a budget eqn for wv with the condensation C_u term right there. Or eq 4.15 – heat budget, with the layent heat of condensation. Or even 4.10 – an equation for liquid water, again with condensation. And following 4.29 is the process for solving for C_u, with vapor pressure etc.

71. ### Gavinsaid

#67 Anastasia, in nobodys’ mind are ‘models the supreme authority’. Models are simply the quantification of the assumptions and framework that is written down in the equations and model description (assuming they are correctly coded of course). The existence of plentiful model solutions showing realistic hurricane initiation and evolution is a statement that the underlying equation set and assumptions they use allow for such a solution. If those assumptions do not include your mass balance term, then it is clear that the inclusion of the term is not *necessary* for hurricane development (though it says nothing about whether a new piece of physics might or might not be important). Waving this piece of evidence away with the claim that ‘they might have fixed it’ is simply not a satisfactory answer absent some evidence that this has occurred (and I know of no such evidence).

You comment on Bryan and Rottuno is puzzling. Their dot’qcond’ term is exactly your ‘S’ and is similarly specified. It is clearly included in equation 4 (the pressure term) which I thought you were arguing for. They don’t have a microphysical model for tracking the condensate particles or raindrop formation etc., but then neither does your paper. Thus, it needs to be explained why there equation 4 (with their conservative equation set) is not what you are talking about, because frankly I’m now a little confused.

72. ### Nick Stokessaid

Anastassia #48
“This is incorrect. Remember that we are talking about a steady state, where neither temperature nor density nor pressure change with time.”

You have a confusion with steady state and application of the gas law.

You could apply the ideal gas law to a volume of gas moving with the fluid. It would remain the same mass of gas, and you would correctly budget for the condensation. But that gas is not in steady state. Pressure, temperature etc change as it rises.

Or you could apply the gas law to a volume fixed in space. Now you have steady state, and again condensation. But to your mass budget you must add to the condensation correction also discrepancies of mass flux across the various boundaries. The flow is not divergence-free.

ps in #70 the liquid water eqn was 4.19, not 4.10

73. ### Terrysaid

Re Nick #72 and Anastasia #48

Nick is right, it is not steady state, and the ideal gas law is not applicable. Steve #53 has it correct in that the parcel moves along the adiabat (SALR).

74. ### skysaid

Makarieva’s insistence on understanding the physics from first principles, rather than relying upon the deus ex machina of ad hoc parametrizations found in climate maodels, is highly commendable. I have no problem in accepting the implosion of water wapor into droplets upon condensation aloft. The sink for the released latent heat, however, remains uncertain. And as a gliding enthusiast who seeks out thermals, I am not convinced that differential heating can be neglected in understanding the general thermalization of the troposhere. The unraveling of cyclogenesis is best left to the specialists who proceed dilgently from first principles, instead of being satisfied by occasional handwaving agreement between models and observations.

@24#
Anastassia Makarieva said

October 29, 2010 at 12:20 am

P=NRT is the ideal gas law. (Remember this formula for ! It will be taught at schools.)

Well I don’t think it should be, because when I went to school the ideal gas law was pV=nRT and according to at least some sources it still is. And thinking about the physicality of it it is just as obvious as F=MA. How could it be any other way?

76. ### Nick Stokessaid

75 No, that’s OK. N is molar density = n/V.

76

Ah OK thanks, that would explain the upper case(I wondered about that but not sufficiently obviously) then. I guess the devil really is in the details.

78. ### Steve Fitzpatricksaid

A tough day for Anastassia I think. Which is too bad… she is obviously very smart and wants to advance understanding.

Inclusion of mass loss from precipitation is both correct and needed to improve accuracy. It just does not dominate latent heat.

79. ### Antonio Nobresaid

Dear All

To close the evening, I think the tough undertones of this discussion could be summarized by a quote:

“The biggest problem with models is the fact that they are made by humans who tend to shape or use their models in ways that mirror their own notion of what a desirable outcome would be. (John Firor [1998], Senior Research Associate and former Director of NCAR, Boulder, CO, USA).”

Anastassia has answered every one of the issues raised, with great candor and physical seriousness.

But there is more to come.

Regards

80. ### Steve Fitzpatricksaid

Antonio Nobre,

Anastassia e’ claramente boa demais (e paciente… e esperta). O problema e’ que ela esta’ discontando um facto muito importante… o calor de condensacao. Uma compreensão total via incluir tudos os fatores.

Abracos,

Steve

81. ### Kenneth Fritschsaid

Since I did not receive any replies on my questions about the hurricane models, I linked to these pages with the link below that summarizes the GFDL model that Schmidt mentioned in his post. Apparently that model is used for forecasting hurricane tracks and intensities. It also apparently, like all models of this type, does a better practical job of forecasting the track than the intensity.

I was under the assumption that these models are updated on a frequent basis based on the actual track and intensity of the hurricane. I am having a problem relating the performance of a hurricane model and its workings under these conditions to the discussion in this thread.

http://www.gfdl.noaa.gov/operational-hurricane-forecasting

82. ### Kenneth Fritschsaid

How could it be any other way?

Now you know.

83. ### Antonio Nobresaid

#80 Steve,

I greatly appreciate your kindness and sensitivity in replying to me in my mother tongue. I wish Brian (#64) would follow your example and send a response to Anastassia in her mother tongue (Russian). I am sure she would help him out with his difficulties in her language as he is kindly helping her with her few English mistakes.

I have completely appreciated this discussion, learned a lot, even though I think most recognized experts showing up here missed a great opportunity to discuss these very issues when Anastassia and Viktor published their specific critique on Hurricanes at ACPD (http://www.atmos-chem-phys-discuss.net/8/17423/2008/acpd-8-17423-2008-discussion.html).

The discussion on latent heat, from my point of view, is far from being over as it might have sounded today. I have a convincing history on how latent heat release on condensation cannot explain circulation over the Amazon basin. It will come in a later post(it is really late here, and I travel tomorrow early, back in one week).

At any rate, I commend you specially, but also Gavin and others, for taking the time and showing such a good sport throughout this discussion. Truly amazing what a forum thread can accomplish in one week when people get down to objective scrutiny of ideas.

Brazilian Regards

84. ### Brian Hsaid

I wish Brian (#64) would follow your example and send a response to Anastassia in her mother tongue (Russian). I am sure she would help him out with his difficulties in her language as he is kindly helping her with her few English mistakes.

Elsewhere she specifically thanked me for a small pointer, saying she was eager to get better at English (very understandable since precision of expression in English is critical to making her case, I’d think.) I’m keeping the little edits very short. Don’t take them personally.

85. ### Jim Dsaid

It is wrong at a fundamental level to say condensation leads to contraction. It leads to first a pressure rise, then expansion. As Nick Stokes and others here say, latent heat overcomes vapor loss leading to higher pressure. Contraction would imply an increase in density and negative buoyancy, which everyone knows is not an effect of condensation. How would you explain observed increased buoyancy in condensing/cloudy air? It is because of reduced density, which is because of expansion, which is because of increased pressure, which is because of latent heating dominating over vapor loss. We can phrase this as several statements with which you agree or not.
1. Condensation leads to higher pressure
2. Higher pressure leads to expansion
3. Expansion leads to lower density
4. Lower density leads to positive buoyancy
5. Positive buoyancy leads to upward vertical acceleration
Please indicate the ones that you don’t agree with.

86. ### Brian Hsaid

Re: Anastassia Makarieva (Oct 30 12:40),
English note: “Let me explain that these equations do not say anything about rainfall altogether.”
“altogether” is wrong here; it has the sense of “completely”. Try “whatsoever”, or “at all”.
______
Is this part of missing or “parameterized” treatment of work in the models? Since no locale is individually “in balance” or “conserved thermally”, work moving air and water is likely a big part of the picture, it seems to me.

87. ### Anastassia Makarievasaid

My early-bird Sunday greetings from Russia to all! Thanks for all the comments.

#72 Nick,

You have a confusion with steady state and application of the gas law.

You could apply the ideal gas law to a volume of gas moving with the fluid. It would remain the same mass of gas, and you would correctly budget for the condensation. But that gas is not in steady state. Pressure, temperature etc change as it rises.

Or you could apply the gas law to a volume fixed in space. Now you have steady state, and again condensation. But to your mass budget you must add to the condensation correction also discrepancies of mass flux across the various boundaries. The flow is not divergence-free.

It is you who have a confusion. I do apply the gas law to a volume fixed in space, see my images at #24. The flow is NOT divergence-free, true. But it is not CONVERGENCE-free either. As I told at #38, the air convergence to our fixed volume at a rate C in the lower atmosphere, and diverges from it in the upper atmosphere at a rate D. On top of it, some part of gas disappears within this fixed volume at a rate P. For this reason there must be a NET CONVERGENCE of the gas to our fixed volume, C – D = P, which means contraction.

As you see, heat considerations are not relevant altogether. You cannot compensate loss of mass by excess of heat or vice versa. They do not sum, subtract or “reverse” each other’s sign.

88. ### Nick Stokessaid

#87 Anastassia,

You say that when water condenses it leads to contraction, basically because fewer gas molecules remain. And that the latent heat released by condensation is irrelevant.

Consider a mix of hydrogen and oxygen igmited. There are also fewer gas molecules afterwards, and again heat is released. Would you expect the mixture to contract?

89. ### Anastassia Makarievasaid

#88 Nick,

Consider a mix of hydrogen and oxygen igmited. There are also fewer gas molecules afterwards, and again heat is released. Would you expect the mixture to contract?

Of course I would not, Nick! The mixture will expand — but this is NOT a steady state process. If you have a steady state process with hydrogen and oxygen igniting and forming water (e.g., in an oven), some air will diverge, some will converge, but again there will be net convergence, i.e., positive work of the environment on the fixed volume where the reaction takes place.

90. ### Anastassia Makarievasaid

# 85 Jim,

We can phrase this as several statements with which you agree or not.
1. Condensation leads to higher pressure
2. Higher pressure leads to expansion
3. Expansion leads to lower density
4. Lower density leads to positive buoyancy
5. Positive buoyancy leads to upward vertical acceleration
Please indicate the ones that you don’t agree with.

I disagree with the statement that condensation leads to higher pressure. This is only true for oversaturated air. Under atmospheric conditions condensation routinely occurs due to a temperature drop, not by itself. Why statement 1 is incorrect, is in detail discussed in Section 2 of M10.

Regarding contraction due to condensation, please, see my reply to Nick above.

91. ### Anastassia Makarievasaid

#71 Gavin, let me clarify once again where the problem is with the formulations of Bryan and Rotunno (2009) (and all others). Suppose we have a closed set of equations for a non-condensable gas (dry air). We have as many equations as there are independent variables, such that we can solve the problem and find the dependencies of velocities, pressure, temperature on time and coordinates. One of the equations we have is the continuity (mass conservation) equation. This equation has zero in the right-hand part. Zero is a constant.

Now we go over to a condensable gas. Its mass is not conserved. It decreases/increases at a certain rate S (or .qcond). However, it is not enough just to put S (or .qcond) instead of zero into the right-hand part of the continuity equation or enter S (or .qcond) into other equations as well. Indeed, after we do so, instead of a constant (zero) we now have a new variable S (.qcond). The problem is no longer closed. We are in need of a new equation for S (or .qcond) to close it. Remember that so far we have operated with fundamental equations — Euler equations for motions, mass conservation, energy conservation… Now we are in need of a new equation that would be the same fundamental, an equation that would relate S (or .qcond) to the variables and functions we already have: pressure, velocity, temperature. We are facing an open problem! The field is badly in need of theoretical work — discussions, suggestions, breakthroughs, conferences, workshops! This work has never been performed.

Instead, as I said above, the magnitude of S (.qcond) was simply and silently re-written in terms of new independent variables, like density of liquid rho_l and terminal velocity V_t. You can verify that this is so from Eq. 13, comparing it with our set of eqs. 32-34 at drho/dt = 0. Then the mass sink effect was attempted to be studied by forcing these new variables to take one value or another, without giving any consideration to the possibility that these values could be correlated and there could be fundamental physical relationships between S (.qcond) and other variables present in the equations of hydrodynamics. This is a standard modeling approach. In this particular case it does not provide any insight into the physics of gas mass removal.

92. ### Nick Stokessaid

#91 Anastassia
“This work has never been performed.”

I don’t think that’s true. Again looking at the CAM3 updraft model, they present a closed system of equations which they proceed to show how to solve for C_u (your S) immediately following Eq 4.29.

On the Hydrogen/Oxygen situation, you can create a steady flame. There is still expansion and upward buoyancy.

93. ### Anastassia Makarievasaid

#71 Gavin,

Waving this piece of evidence away with the claim that ‘they might have fixed it’ is simply not a satisfactory answer absent some evidence that this has occurred (and I know of no such evidence).

While I generally agree with your statement, I do not feel that in my case this reprimand is deserved. We did spend a very considerable time discussing the deficiencies of the mainstream theoretical approaches to hurricane description. Our critique of hurricane models in ACPD apparently failed to raise a constructive discussion in the climate community. We published another one in Proceedings A where we also addressed some major problems, see also my comment #31 above. I would also suggest an interesting paper of Smith et al. 2008, where some other problems with the existing hurricane theory were considered.

So I do believe that from our side we did a fair piece of job here showing why we think the existing models do not include the correct physics. But the situation is clearly not reciprocal. It is our approach that is being waived, either by references to models or by statements that condensation increases air pressure. With all due respect, I cannot count this as a serious or responsible consideration. I hope very much that the situation will change and we see a constructive interest from climate physicists to our approach. We are open to cooperation, ready to communicate and otherwise participate in all scientifically meaningful exchanges.

94. ### thomassaid

why do you relate hurricane research to climate research? your’re talking about hurricane dynamics.

especially the fact, that you relate tornados hurricanes and extratropical cyclones to the same mechanisms concerns me. it shows that you are unaware about the literature.

furthermore you claim that you take empirical data into account (data from radiosonds, radar data etc.). i cannot see this anywhere. and of course just talking about a wind profile is not enough.

95. ### douglasinugandasaid

Good morning from Uganda,

A suggestion for progress:

Please take a moment again to see who agrees and who disagrees with the first point on Anastassia’s list (90). Does condensation lead to a reduction in pressure under normal atmospheric conditions?

This simple question seems to underly much of the disagreement and seems to be resolvable based on the physics. That’s section 2 in the new ACP manuscript.

We also know that these ideas (both for and against) have been published and accepted in quality journals. But it seems to me that if we take this step by step we can highlight where in the physics we can agree and disagree and stand a chance to resolve that clearly and without ambiguity. If section 2 in the new ACP manuscript is correct we can move on … if it is false please clarify how (physics not models). I think many of you will agree with the logic if you read it carefully, but then again you may see something I haven’t and be able to show me why I’m wrong. I’d welcome you to try please! No models and assumptions … just basic physics and their implications.

Thanks for taking your time on this. I think we can resolve this and reach a conclusion .. that would be real progress.

Douglas

96. ### Brian Hsaid

Watching the current ‘canes, like Thomas, make right-turns and head north before making continental landfall reminded me of the critique of “Iconoclast” on SA:

14. Iconoclast 05:06 PM 10/23/10

The proposition that the average temperature of the earth’s surface is warming because of increased emissions of human-produced greenhouse gases cannot be tested by any known scientific procedure.

It is impossible to position temperature sensors randomly over the earth’s surface (including the 71% of ocean, and all the deserts, forests, and icecaps) and maintain it in constant condition long enough to tell if any average is increasing. Even if this were done the difference between the temperature during day and night is so great that no rational aveage can be derived.

Measurements at weather stations are quite unsuitable since they are not positioned representatively and they only measure maximum and minimum once a day, from which no average can be derived. They also constantly change in number, location and surroundings. Recent studies show that most of the current stations are unable to measure temperature to better than a degree or two.

The assumptions of climate models are absurd. They assume the earth is flat, that the sun shines with equal intensity day and night, and the earth is in equilibrium, with the energy received equal to that emitted.

(Half of the time there is no sun, where the temperature regime is quite different from the day.

No part of the earth ever is in energy equilibrium, neither is there any evidence of an overall “balance”.)

It is unsurprising that such models are incapable of predicting sny future climate behaviour, even if this could be measured satisfactorily.

There are no representative measurements of the concentration of atmospheric carbon dioxide over any land surface, where “greenhouse warming” is supposed to happen.

After twenty years of study, and as expert reviewer to the IPCC from the very beginning , I can only conclude that the whole affair is a gigantic fraud.

(My bold.) Which means that such minor(!) phenomena as the Trade Winds and Coriolis Effect (global and local, causing rotation) must be put in by hand, if at all. Shear is crucial to the development of hurricanes. How is it modeled?

97. ### Nick Stokessaid

Brian #90,
“The assumptions of climate models are absurd. They assume the earth is flat, that the sun shines with equal intensity day and night, and the earth is in equilibrium, with the energy received equal to that emitted.”
You know nothing about models. Here again is CAM 3:

Flat?
Terrain following spherical coordinates, with of course full accounting for rotational effect.

Day/night?
Here’s their Diurnal Cycle. has the sun in the right place always, each place, each hour, each day. Does perihelion/aphelion and the sunspot cycle.

Equilibrium assumed?
The longwave radiation treatment. A computed solution, with exceedingly detailed gas and cloud calculations. No sign of an equilibrium assumption, though I’m sure they check it.

98. ### Geoff Sherringtonsaid

There would seem to be a fundamental difference between cyclones (as they are known in Australia) that form over water, as the vast majority do, and those that continue over hot, usually dry land, often for several days and thousands of km. For a start, the uptake of water to maintain the downfall of moisture – which happens – is less in the latter case.

At first blush, this would indicate to me that water is more of an incidental factor; and that the driving forces involve heat distribution. We are used to dust storms ‘willy willys’ that arise, twist around and then die away (often from the base upwards). These can happen often over very dry land. They are small, but perhaps they share common explanatory mechanisms. Like initiation by eddy formation, a function of wind geometry.

Would Antonio Nobre at 83 agree generally with this?

99. ### Anastassia Makarievasaid

Let us discuss once again why we are convinced latent heat is not relevant.

Nobody here questions the existence of a thermally driven convection. Those having fireplaces at home know that the air is sucked into the fireplace and that the warm rises from chimneys over the roofs.

The question is quantitative: what sort of vertical velocities would a release of heat produce in the atmosphere? Will it be 0.001 mm/sec or 10 cm/sec, as in hurricanes? How to calculate this? How to prove that the observed velocities have something to do with the latent heat release and disprove the statement that the latent heat release is just a side process that accompanies the mass-sink driven air motion?

We note that earlier researchers were very much concerned about such issues. Riel (1950), pdf available wrote:

Here we shall mention only one objection: a mechanism for organized mass removal aloft is not specified, so that the area of heavy showers and its immediate surroundings will act almost as a closed system. After ascending the air will descend dry-adiabatically in the environment (following the horizontal lines of Fig. 2) and the environment will soon be warmer than the cloudy area. Any incipient circulation must dissipate.

The logic of this concern is clear: in chimneys the air is rising because it is several hundred degrees warmer than the surrounding air. Adiabatic descent will make the surrounding air warmer, such that the ascending air motion in the storm must discontinue.

What do we observe in reality? We do observe that the ascending air of the hurricane is, if anything, COLDER than the surrounding environment. Once again, I refer the readers to Fig. 4C of Montgomery et al. 2006. How can it be? How can our latent heat chimney operate if the air inside is colder than outside? This clearly shows that thermal effects ARE NOT the driving force of the circulation. The ascent of colder air can be explained by the dynamic action of the mass sink only.

The existing hurricane models do not prove that hurricanes are driven by latent heat. In the MPI model the cold sink temperature T_0, that determines the efficiency of the hurricane heat engine, is defined as the outflow temperature, i.e. the temperature at which the air ceases to ascend adiabatically. However, if the air does not move at all, the outflow temperature is simply equal to surface temperature, such that the engine efficiency is zero. A heat engine consideration cannot prove that the air must move. The power of a classical Carnot engine is zero.

In our recent critique we emphasized this deficiency of the thermodynamic approach. This is not something we invented just to criticize somebody, these are well-studied constraints that must be taken into account when one talks of finite-time heat engines. In order to have an engine operating at a finite power, one must specify a dynamic system that brings it into operation. In the case of the atmosphere, this means that one must specify what makes the air move at a given velocity. Replying to our critique, Bister et al. 2010 admitted that they did not consider the constraints stemming from finite time thermodynamics in their work. I can only repeat that there is an apparent need for new approaches in hurricane research and that we hope that this discussion will be considered by the attending scientists as aimed to provoke critical and constructive thinking.

100. ### Gavinsaid

#97 Ditto. Models contain all of those features.

#91 Anastasia, It is obvious that one needs a equation for the condensation rate in any model- and indeed every model contains such equations. In cases of vertical ascent driven by either large scale flow or moist convection (buoyancy effects), dot.qcond is calculated assuming that the parcel stays saturated. In other circumstances, it is driven by large scale condensation/evaporation effects which these days depend on a lot of microphysical detail (cloud condensation nuclei, ice phase/liquid phase dynamics etc.). Some of the considerations are discussed in papers like Del Genio et al 1996. Note too that these papers make a clear distinction between condensation from vapour and precipitation out of the system.

I was under the impression we were discussing the impacts of water vapour condensation on the dynamics, not the reasons why water condenses in the first place. I doubt very much you are claiming that this has not been considered by others!

101. ### anna vsaid

Seems to me that expansion or contraction of air due to condensation is a question that could/should have been answered in a laboratory experiment.

102. ### douglasinugandasaid

Hi All,

Please see my query at 95. Can no-one explain how and why we still disagree on the basic physics like this? If they cannot then we can conclude that Anastassia and co. are correct in their analysis of this point: i.e. condensation always leads to a reduction in atmospheric pressure under natural conditions. Agreed?

Thanks if so as that is an important advance in the discussions. (And much easier to resolve than hurricane and climate models in any broad and general sense). I think we need to debate the arguments brick-by-brick otherwise we are talking past each other most of the time.

Thanks
Douglas

103. ### douglasinugandasaid

Thanks Anna V (our posts were near simultaneous)

I’m not sure about lab tests but we do have some basic physical laws that need to be obeyed.

To clarify, I am talking about section 2 in this text:
http://www.atmos-chem-phys-discuss.net/10/24015/2010/acpd-10-24015-2010.html

Please take a look and see if there is an error in the reasoning. If there are indeed useful tests with real data that would certainly be helpful too.

Best wishes
Douglas

104. ### Nick Stokessaid

#103 Douglas
The problem with Sec 2 is that it treats an unrelated issue. Yes, air held adiabatically in a box cannot undergo condensation, unless it is supersaturated. You don’t need fancy calculations to establish that. But that fact shows why the problem is irrelevant. In the atmosphere, you do get condensation. Something else is happening.

Parcels of air are rising, and cooling through expansion. As they cool, saturation is exceeded, water condenses, and latent heat is released.

Now unsaturated air rises and falls with turbulence. The lapse rate settles at a level which allows this to happen without nett work being done for or against buoyant forces. But saturated air is cooling more slowly than unsaturated air would, and more slowly than the lapse rate. So it becomes warmer than unsaturated air on its level. It expands relative to that air, and displaces some of it upwards. This requires work.

105. ### douglasinugandasaid

#104 Nick
Thanks Nick. It is good of you to address my requests clealy nd directly. I do wonder if that is a general view … that these systems are not adiabatic?
Unfortunately, I have to cross the country and catch a plane now so will be offline a while … but I do consider this discussion and the insights it provides very valuable. I hope the discussion can continue.
Best wishes
Douglas

106. ### anna vsaid

OK, this is an experiment:

What I get from this is that the latent heat of water to ice is absorbed by the ice as it is being created, and it is warmer than the surrounding temperature with which it was in equilibrium before,; it is not absorbed by the ambient background.

In the case of water condensates in the atmosphere it means that the latent heat is carried by the condensate/drops, being in a higher temperature than the background, which falls as warm rain, the air being heated as a black body from the droplets not very efficiently. So the heat is not available for long to heat the air if there is rain. In a cloud I guess there would be an equillibrium, but that is not what is under discussion, I think.

107. ### Anastassia Makarievasaid

#92 Nick,

I don’t think that’s true. Again looking at the CAM3 updraft model, they present a closed system of equations which they proceed to show how to solve for C_u (your S) immediately following Eq 4.29.

I think that you do not think so because you have not analyzed the logic of this model, but just believe that condensation is somewhere there. If you look carefully at the basic equations (4.2) and (4.3) this can give you a hint why this parameterization is not a solution I was talking about at #91.

#100 Gavin,

It is obvious that one needs a equation for the condensation rate in any model- and indeed every model contains such equations.

At #61 you asked me about the model of Bryan and Rotunno (2009) implying that this model also has all the needed equations. At #67 and #91 I explained why although the equations are formally there (and many of them), the problem with condensation rate remains open. If equations of the same type are present in other models as well, they suffer from the same problem as the model of Bryan and Rotunno (2009).

Let me also point out that just some equations will not do for such a fundamental variable as condensation rate. If we google Penman-Montheith equation, we will see how much theoretical work has been put into the task of formalizing the process of evaporation and relate it to basic atmospheric parameters, including wind speeds. Two persons, Penman and Montheith, claim the authority of this equation. What about condensation rate? Where is such an equation? How is it derived? Where can one look at it? What is its physics? Why, if there is such an equation, did not Bryan and Rotunno (2009) use it?

Also, there is a very substantial difference in accounting for latent heat release and for the mass sink. One does not need an extra equation to account for latent heat release, but does need that to account for a mass sink. I bet few people realize how odd this situation is.

#104 Nick,

The problem with Sec 2 is that it treats an unrelated issue. Yes, air held adiabatically in a box cannot undergo condensation, unless it is supersaturated. You don’t need fancy calculations to establish that. But that fact shows why the problem is irrelevant. In the atmosphere, you do get condensation. Something else is happening.

I agree. The goal of Section 2 was to show that calculations of Dr. Rosenfeld made for adiabatic condensation at constant volume are totally irrelevant for atmospheric processes where, indeed, something else is happening. Had it not been for those calculations, the section could have been omitted.

But saturated air is cooling more slowly than unsaturated air would, and more slowly than the lapse rate. So it becomes warmer than unsaturated air on its level.

This is only so if the surface temperature is the same in both the ascending and descending columns. In reality in many cases, including the hurricanes, as I said above at #99, the surface temperature in the descending column is higher, such that there cannot be any temperature-driven expansion of moist rising air. The same happened this summer in Russia, when for two months we had descending air over thousand kilometers that was 10-15 degrees warmer at the surface than the surrounding ascending air.

#66 Steve,
I cannot see how you obtain 800 W/m^2 from 80 mm/day (but I can be mistaken). Let’s check: 80 mm/day = 80 kg/day = 0.05 mol/m^2/sec. With L = 45 kJ/mol, we have 2250 W/m^2, not 800 W/m^2 as you suggested. In any case, both estimates greatly exceed the radiative power of the top of the atmosphere. Our velocity 10 cm/s pertains to the inner hurricane radius rather than average value. I addressed the issue why latent heat is not relevant at #99 above.

#73 Terry,

#62 Brian,
Thanks! I am trying to learn, but what I notice — when I feel strongly about a statement I am delivering, somehow the language always fails, here or there. Even in Russian I often cannot properly end my sentences when excited…

108. ### Steve Fitzpatricksaid

Nick #104,
“It expands relative to that air, and displaces some of it upwards.”

I think rising saturated air displaces surrounding air as it expands mainly in the horizontal direction. This horizontal flow is visible as streaks of cirrus clouds in this satellite image (infrared) http://www.ssd.noaa.gov/goes/flt/t2/ir4-l.jpg.

109. ### Steve Fitzpatricksaid

Anastassia #107,

In your paper, you show a net heat flux of ~4000 watts/M^2 from an assumed average rate of rise of 10 cm per second, or 400 watts/M2 for a rate of rise of 1 cm per second. If the average 24 hour rainfall over the hurricane field (as in the paper I referenced on all USA hurricanes from 1947 to 2000) is ~8 cm, and the water content of near-surface air is 50 g/M^3, then all the water in 1 cubic meter will generate rainfall of 50/10000 = 0.005cm. The total column needed to generate 8 cm is then 8/0.005 = 1600 meters, or 1.85 cm per second rate of rise over 24 hours. Based on the calculated flux in your paper (400 watts/M^2 at 1 cm/sec rate of rise), I get a flux of 1.85 * 400 = 740 watts/M^2.

110. ### Jim Dsaid

To quantify the effect of latent heat on pressure versus vapor loss we can use the gas law (I will use r for density, subscript d for dry air and v for vapor)
p = pd + pv (where pv is commonly known as e, the partial pressure of vapor)

This can be expanded out
p = rd Rd T + rv Rv T (Rd and Rv are the various gas constants)
or
p = r R T (where r is the total density and R is the effective gas constant for moist air, close to the value of Rd in normal cases)

So we have, taking derivatives for small changes,
dp = r R dT + dr R T

dp = r R dT + rd dq Rv T (where now we have used rv = q rd where q is the mixing ratio of vapor)

But for latent heating we know
L dq = – cp dT (where L is the constant for latent heat of vaporization, and cp is the heat capacity of air)

Substituting
dp = r R dT – rd cp dT Rv T / L

The ratio of the two terms is
L R r / rd cp Rv T

If this ratio is larger than 1 the first term dominates and pressure increases.
L=2.5e6 R=287 cp=1005 Rv=461 take T=273 r/rd~1

This gives about 5.7, which is the factor by which heating dominates over vapor loss.

111. ### anna vsaid

You are assuming the latent heat of condensation goes to the air, no? From the experiment linked above the heat goes into the condensate primarily. The air is second level.

112. ### Jim Dsaid

Anna v, to be clear, the latent heat of condensation is released in the process. Nearly all of it goes into the air because the small water content has a minuscule heat capacity relative to its surrounding air. It can be quantified, and at most 1% goes into the water droplets. Typical water contents are up to 1g/kg for cloud droplets, and even if they have four times more heat capacity per mass than air, you can’t get a significant amount of heat into them before they get warmer than the air.

113. ### Nic Lsaid

Re Jim D #112

“Nearly all of it goes into the air because the small water content has a minuscule heat capacity relative to its surrounding air. It can be quantified, and at most 1% goes into the water droplets. Typical water contents are up to 1g/kg for cloud droplets, and even if they have four times more heat capacity per mass than air, you can’t get a significant amount of heat into them before they get warmer than the air.”

Understood, but is it possible that the droplets warmed by the latent heat of condensation might become sufficiently large to start falling before they reach temperature equilibrium with the surrounding air? If so, much less than 99% of the latent heat might be taken up by the air at the level where the condensation was occurring.

114. ### anna vsaid

So what happens if they get warmer than the air? They will evaporate so take away the heat they gave? Once they become rain they are large enough not to evaporate much if warmer than the air.

115. ### DeWitt Paynesaid

Once the droplets get warmer than the surrounding air, growth by condensation stops and the droplets don’t get any bigger except by agglomeration. So the temperature difference between the droplets and the air can’t be very large and therefore greater than 90% of the latent heat must go into the air.

116. ### DeWitt Paynesaid

The thermometer in your linked video is measuring the temperature of the water, not the ice. Supercooled water will indeed warm as crystallization is initiated so the latent heat does go into the water, not the ice, see this cooling curve for pure water.

117. ### Gavinsaid

#107 Anastasia, You have now gone from claiming that the impact of condensation is not properly considered in the dynamical equations, to arguing that the calculation of condensation itself is the issue. Forgive me if I don’t seem to be able to keep up, but progress (in transmitting your ideas to others) is greatly helped by disaggregating separate issues. We have now identified at least three factors that may be mis-specified: 1) the local impact of condensation per se (i.e. the impact of cloud formation without mass removal via precipitation), 2) the impact of mass removal via precipitation that reaches the surface (and conversely mass addition via surface evaporation), and now 3) the physics behind condensation itself.

The physics of 1) involves pressure wave effects and do not exist within the framework hydrostatic approximation (corresponding to most climate models) (more correctly, the adjustment is considered to be instant). In non-hydrostatic models (like B&R (2009), and the earlier Bryan and Fritsch (2002)), all the terms are included. However, I think we agreed a while ago that this was not the issue you were raising.

The physics of 2) is included in the ‘conservative equation set’ of Bryan and Rotunno and has a noticeable, but minor effect. I was under the impression that this is what you were discussing. You have claimed their implementation must be wrong, but I do not understand your reasoning here.

The physics of 3) is very complicated and depends on many separate issues (look at the myriad types of clouds for instance), all of which in some sense rely on local saturation, but the specifics are very different if the condensation is being driven by gravity waves, boundary layer turbulence, moist convection, or large scale ascent, and acts in ways that depend on the phase of any existing condensate, the amount of cloud condensation nuclei, radiative effects, auto-conversion rates etc. There is not going to be any master equation that describes this process in general for all circumstances. (Actually, even in calculating surface evaporation, no such equation exists – read Brusaert (1982) for lots of examples). Up until now, these processes were not obviously the point you were trying to make (since your equation for S is simply condensation during vertical ascent and is but one (important) example of atmospheric condensation).

So, I would like to refocus, not on points 1 or 3, but on point 2, and ask again, why you do not think that the treatment of .qcond *in the dynamics* by B&R is satisfactory to you. I am not asking why qcond is changing, but on the effect of a given change in qcond has on the dynamics. If their equation set is incomplete (and looking at the derivation in B&F 2002, it does not appear to be), we should be able to say so clearly. We are not talking about a brand new piece of physics, but rather the consistent application of very well known general principles.

If you disagree with my assessment of the ‘state of the discussion’, please either point out where I have misspoken, or perhaps what I have missed. If you think we are actually in agreement over point 2, then you should definitely point that out.

118. ### anna vsaid

True it is measuring the water, but would it not be in equilibrium in temperature with the created ice?

119. ### DeWitt Paynesaid

As long as the water is supercooled, equilibrium doesn’t exist. The ice layer on the thermometer bulb is indeed at the same temperature as the water.

There are several problems when comparing water freezing and evaporation. The heat of vaporization of water is 6.8 times larger than the heat of fusion. The heat capacity/kg of ice is half the heat capacity of liquid water while the heat capacity of liquid water/kg is four times greater than the heat capacity of moist air. The vapor pressure of water at atmospheric temperatures is a small fraction of the total pressure. The change of volume with pressure and temperature is much greater for air than for water. The density of air is about 3 orders of magnitude less than the density of water. If a kg of liquid water is cooled to -10 C and then seeded and stirred the volume fraction of ice formed will be larger by many orders of magnitude than the volume fraction of liquid water in a kg of air saturated with water vapor at 30 C that is quickly cooled to 20 C before condensation starts.

120. ### Jim Dsaid

The physics of what happens to the less than 1% of latent heat release going into the drops is not of much importance. It will either be transferred to the air down the temperature gradient, or fall out with the rainfall while transferring the heat out to air below. Everything stays close to thermal equilibrium (though I will say that cloud-radiative processes are also considered, as they should be, in the climate/weather models, because longwave and shortwave radiative effects near cloud top can be important).

121. ### Nick Stokessaid

#108
Steve, I agree. But it’s the upward component that requires work.

122. ### Steve Fitzpatricksaid

Nick #121,

I’m not so sure. If we have a volume of supersaturated air in a cylinder with a piston at ambient pressure, and condensation suddenly takes place, the pressure inside the cylinder will rise, and apply force to the piston. Movement of the piston will do work, independent of the orientation.

123. ### DeWitt Paynesaid

I think Steve F is correct on this one. Consider a spherical bag. Any increase in pressure inside the bag will do work against the surrounding pressure in all directions. The calculation of work done is easier, though, if you consider movement of a fixed area plane in only one direction.

124. ### Nick Stokessaid

SteveF #122
I’m thinking Archimedes, not pistons. If updraft air expands, other air has to make way. There will be horizontal movement, but ultimately, the only place to go is up. The work of expansion raises the PE of the adjacent air.

125. ### curioussaid

Apologies if this is way off/too simple but based on my limited understanding, I also think Steve is correct. At Lucia’s, DeWitt, you gave a couple of figures for hurricane eye pressures. Let’s use 900mbar (Wiki has Katrina as 902mbar minimum pressure) and take 1atm as 1000mbar for the sake of argument. This gives a pressure difference across the radius of the hurricane of 100mbar. This will perform work on each parcel of air as it is pushed towards the centre of the hurricane. As an estimate of the work done I’d say take a 1atm, 1kg spherical parcel of air and mulitply it’s projected profile area (pi*rad.parcel^2) by 100mbar and multiply this by the radius of the hurricane (rad.h). In addition I think it will also play a role in generating the high wind speeds as the angular momentum of each parcel of air rotating about the vertical axis of the hurricane will have to be conserved as it travels towards the centre of the hurricane. This will mean that as rad.h tends to zero, v will tend to infinity. As before, apologies if this known and understood by all as the “abc” of hurricanes – or completely wrong!

126. ### Nick Stokessaid

#110 JimD
I think your factor is much the same as I had in #46. I left out the molecumar weight ratio – it should have been:
Hp = PRT(1-rL/(cp T)), r=MW(water)/MW(air)=18/29
Then the arithmetic matches.

127. ### Steve Fitzpatricksaid

Nick Stokes #124,

“The work of expansion raises the PE of the adjacent air.”

On this we agree. In any case, warming/expanding air from condensation does work on the surrounding air.

128. ### Jim Dsaid

#126 Nick, yes, it looked like the discussion needed a derivation because that was just dismissed as ‘incorrect’ in #48, as I am sure mine will be despite only using the gas law for its derivation.

129. ### Geoff Sherringtonsaid

127 Steve Fitzpatrick.

What halts the warming/expanding process eventually?

130. ### Steve Fitzpatricksaid

#129,

When the water vapor is effectively all condensed. (It runs out of steam!) Once the water vapor is gone, the parcel is high in the troposphere and loses heat to space.

But just to be clear: A rising moist parcel of air NEVER warms in an absolute sense, it warms relative to the surrounding air at the altitude of the parcel; its density is lower than that of the surrounding air, and so it rises. The condensation that takes place is due to the continuous cooling from adiabatic expansion.

131. ### DeWitt Paynesaid

The adiabatic assumption also breaks down eventually. Fog formation increases radiative transfer. The packet radiates in all directions at a higher rate than incoming radiation and there will be mixing on the edges from turbulence generated by the vertical velocity difference. Then there’s the tropopause where temperature stops decreasing with altitude.

132. ### anna vsaid

Can you or anybody please give a link to an experiment where supersaturated air expands with condensation?
This piston is a simple experiment to be done and should have been done long ago, but I do not seem to find it or anything similar by googling.

Has anybody heard of verification by experiment?

133. ### Terrysaid

Re Steve #130

There is also a very significant cooling from entrainment of surrounding air, both turbulent and diffusive (which one dominates is primarily dependent on the upwards velocity, in the absence of horizontal effects). This cooling is likewise offset itself by the addition of more water vapor and thus more warming from condensation. Ultimately it asymptotically approaches its final height.

134. ### Geoff Sherringtonsaid

130, 131 Steve & DeWitt, Thank you for the prompt responses. So storm systems do lose energy to space. How do they pick it up at the beginning of the process?

I’m far from expert in these matters, but I’m trying to make logic about cyclones that persist for days over hot dry land. They seem to take a rather long time to run out of steam. One would think the running out process would commence more or less at landfall and be complete in the time it takes for a column of moist air to reach the top of the system, because there is very little more moist air feeding it from below. You both seem to have the figures at fingertips – how long would it take to empty such an unfed storm system dumping say 100 mm a day on the land, not uncommon figures for these cyclones?

135. ### DeWitt Paynesaid

I seriously doubt that there is 100 mm/day over the whole area of the storm. According to Steve F’s post above the number for a tropical storm over the ocean is 8 mm/day over the full area of the storm.

136. ### DeWitt Paynesaid

Dry land could also have a much higher surface temperature than the sea surface. That could be a significant energy source, particularly if the storm center is moving rapidly. Any precipitation could be rapidly recycled back into the storm.

137. ### Geoff Sherringtonsaid

De Witt Payne,

You are probably correct. They tend to report peak rainfalls rather than those averaged over an area defined by convention with a border. But that ground can be mighty thirsty and although I do not have measurements, I don’t feel that much would recycle in the right place at the right time to sustain. And of course the longer ones over land have to cope with both day and night temperatures. Can easily drop to lower than 20 degrees C in air overnight.

Please forgive me, I have never been able to sort the logic and when that happens I wonder if some aspect is being missed, just as you probably do too.

The other peculiar observation is that many cyclones track parallel to the coast some 200 km to 600 km out to sea, then suddenly do a 90 degree turn with no indicator flashing and cross the coast at right angles to its regional direction. It’s fair amazing how often this pattern happens. Almost like the land grabs and drags one side of an eddy toward it. It’s harder to see with the curved geometry of the Gulf of Mexico and nearby.

I try to think if images like this are relevant to explanations (if this is genuine, as I suspect it is). http://i260.photobucket.com/albums/ii14/sherro_2008/Jumbo.jpg

138. ### 2s3csaid

#117 Gavin,

From my point of view, in this thread I have delivered 100% information for a specialist to make all the needed conclusions; additionally, I pointed to the particular equations of Bryan and Rotunno (2009) where the discrepancy with our approach is located. But I do realize that there can be communication barriers of various sorts, so I am very grateful to you for your further interest and questions that allow me to clarify things even further. Here I will present a more detailed analysis of the model of Bryan and Rotunno (2009) [hereafter BR09] with an emphasis on two issues: (1) why condensation rate was accounted for incorrectly and (2) how the model was fitted to nevertheless agree with observations. We are considering posting these materials as a short comment in the ACPD discussion as a response to your questions, so I would very much welcome your further comments. I will deliver my considerations in a series of comments (perhaps 3 or 4).

Part I. Comparing the equations

As I mentioned in #67, a key equation is Eq. 13 of BR09, which in the stationary case takes the form

$\frac{1}{r} \frac{\partial (r\rho_t u)} {\partial r} + \frac{\partial (\rho_t w)}{\partial z} = \frac{\partial (\rho_l V_t )}{\partial z} + D_{q_t}$     [1]

Here $\rho_t = \rho_d + \rho_v + \rho_l$ is the total density the atmosphere, which includes dry air (d), water vapor (v) and liquid water (l), r is radius, u is radial velocity, w is vertical velocity, $V_t$ is terminal velocity of droplets and $D_{q_t}$ is the term describing turbulent diffusion.

In my earlier comments #10 and #20 I explained in great detail the following. The existence of the mass sink in the BR model is taken into account by recycling the dissipated heat from falling droplets. This pertains the term $\partial (\rho_l V_t )/\partial z$ in [1] above. What does it mean? It means that the existence of a non-zero water load, $\rho_l > 0$, is associated with some friction losses. These friction losses grow with decreasing $V_t$ and reach a maximum at $V_t = 0$. In this case the droplets exert maximum dissipative friction on the rising air. Since in the BR09 model the dissipated heat is recycled (!), the greater the dissipation, the greater the hurricane power. This explains the fact that the curves for the conservative equation set in Fig. 10 of BR09 go above the curves for the traditional equation set, with the difference being the largest at $V_t = 0$.

To summarize this step of our analysis: the difference between the curves in Fig. 10 of BR09 does not account for the mass sink per se, but only for the fact that the droplets remain for some time in the atmosphere.

In our work, in line with many conventional approaches, we ignore this effect and consider precipitation to be immediate, such that $\rho_l = 0$ and $\partial (\rho_l V_t )/\partial z = 0$ in [1]. I mentioned several times that the amount of liquid water that remains in the atmosphere is negligible compared to the adiabatic liquid water content, so accounting for the water load will be a minor correction to the overall effect of mass removal.

Let us now put $\rho_l = 0$ and compare [1] to the system of equations 32-34 in M10. We write the latter in the cylindrical coordinates, see [here, Eq. 23] for a (trivial) discussion. Neglecting the (here irrelevant) difference between molar and total density formulations (N and $\gamma$ versus $\rho$ and $q$ in M10 versus BR09, respectively), comparison of the two approaches yields:

$\frac{1}{r} \frac{\partial (r\rho u)} {\partial r} + \frac{\partial (\rho w)}{\partial z} = D_{q_t}$     [2, obtained from BR09 Eq. 13]

$\frac{1}{r} \frac{\partial (r\rho u)} {\partial r} + \frac{\partial (\rho w)}{\partial z} = w\rho\frac{\partial q}{\partial z}$     [3, obtained from M10 Eqs.32-34]

Here $\rho = \rho_d + \rho_v$ is total air density (dry air plus vapor). This comparison carries important information: condensation rate S in M10, Eq. 34, derived from an explicit consideration of the adiabatic ascent, is replaced by $S = D_{q_t}$ in BR09, where $D_{q_t}$ represents the turbulent diffusion flux. This is THE equation I was talking about. As one can see, it does exist and has quite a specific form.

At this point, we emphasize the following: in M10, the expression for S contains a positive feedback between vertical velocity and precipitation. The magnitude of this feedback (i.e., coefficient at w) is quantified in terms of basic atmospheric parameters without any fitting to observational data regarding hurricane velocities. M10 (and earlier Makarieva and Gorshkov) insist that accounting for this positive feedback totally determines all hurricane dynamics. In contrast, in BR09 the dependence of S on atmospheric parameters is delegated to the parameterization of turbulence. How is this parameterization done?

I will continue in my next comment (coming as soon as possible).

139. ### Anastassia Makarievasaid

#117 Gavin,

From my point of view, in this thread I have delivered 100% information for a specialist to make all the needed conclusions; additionally, I pointed to the particular equations of Bryan and Rotunno (2009) where the discrepancy with our approach is located. But I do realize that there can be communication barriers of various sorts, so I am very grateful to you for your further interest and questions that allow me to clarify things even further. Here I will present a more detailed analysis of the model of Bryan and Rotunno (2009) [hereafter BR09] with an emphasis on two issues: (1) why condensation rate was accounted for incorrectly and (2) how the model was fitted to nevertheless agree with observations. We are considering posting these materials as a short comment in the ACPD discussion as a response to your questions, so I would very much welcome your further comments. I will deliver my considerations in a series of comments (perhaps 3 or 4).

Part I. Comparing the equations

As I mentioned in #67, a key equation is Eq. 13 of BR09, which in the stationary case takes the form

$\frac{1}{r} \frac{\partial (r\rho_t u)} {\partial r} + \frac{\partial (\rho_t w)}{\partial z} = \frac{\partial (\rho_l V_t )}{\partial z} + D_{q_t}$     [1]

Here $\rho_t = \rho_d + \rho_v + \rho_l$ is the total density the atmosphere, which includes dry air (d), water vapor (v) and liquid water (l), r is radius, u is radial velocity, w is vertical velocity, $V_t$ is terminal velocity of droplets and $D_{q_t}$ is the term describing turbulent diffusion.

In my earlier comments #10 and #20 I explained in great detail the following. The existence of the mass sink in the BR model is taken into account by recycling the dissipated heat from falling droplets. This pertains the term $\partial (\rho_l V_t )/\partial z$ in [1] above. What does it mean? It means that the existence of a non-zero water load, $\rho_l > 0$, is associated with some friction losses. These friction losses grow with decreasing $V_t$ and reach a maximum at $V_t = 0$. In this case the droplets exert maximum dissipative friction on the rising air. Since in the BR09 model the dissipated heat is recycled (!), the greater the dissipation, the greater the hurricane power. This explains the fact that the curves for the conservative equation set in Fig. 10 of BR09 go above the curves for the traditional equation set, with the difference being the largest at $V_t = 0$.

To summarize this step of our analysis: the difference between the curves in Fig. 10 of BR09 does not account for the mass sink per se, but only for the fact that the droplets remain for some time in the atmosphere.

In our work, in line with many conventional approaches, we ignore this effect and consider precipitation to be immediate, such that $\rho_l = 0$ and $\partial (\rho_l V_t )/\partial z = 0$ in [1]. I mentioned several times that the amount of liquid water that remains in the atmosphere is negligible compared to the adiabatic liquid water content, so accounting for the water load will be a minor correction to the overall effect of mass removal.

Let us now put $\rho_l = 0$ and compare [1] to the system of equations 32-34 in M10. We write the latter in the cylindrical coordinates, see [here, Eq. 23] for a (trivial) discussion. Neglecting the (here irrelevant) difference between molar and total density formulations (N and $\gamma$ versus $\rho$ and $q$ in M10 versus BR09, respectively), comparison of the two approaches yields:

$\frac{1}{r} \frac{\partial (r\rho u)} {\partial r} + \frac{\partial (\rho w)}{\partial z} = D_{q_t}$     [2, obtained from BR09 Eq. 13]

$\frac{1}{r} \frac{\partial (r\rho u)} {\partial r} + \frac{\partial (\rho w)}{\partial z} = w\rho\frac{\partial q}{\partial z}$     [3, obtained from M10 Eqs.32-34]

Here $\rho = \rho_d + \rho_v$ is total air density (dry air plus vapor). This comparison carries important information: condensation rate S in M10, Eq. 34, derived from an explicit consideration of the adiabatic ascent, is replaced by $S = D_{q_t}$ in BR09, where $D_{q_t}$ represents the turbulent diffusion flux. This is THE equation I was talking about. As one can see, it does exist and has quite a specific form.

At this point, we emphasize the following: in M10, the expression for S contains a positive feedback between vertical velocity and precipitation. The magnitude of this feedback (i.e., coefficient at w) is quantified in terms of basic atmospheric parameters without any fitting to observational data regarding hurricane velocities. M10 (and earlier Makarieva and Gorshkov) insist that accounting for this positive feedback totally determines all hurricane dynamics. In contrast, in BR09 the dependence of S on atmospheric parameters is delegated to the parameterization of turbulence. How is this parameterization done?

I will continue in my next comment (coming as soon as possible).

140. ### Nick Stokessaid

#139 Anastassia,
I think there is a misunderstanding of BR09 there. I don’t have the paper, but I don’t think S is delegated to D. They are different – your formulation ignores diffusive flux (which is not unreasonable). BR09’s equation includes it in the left side divergence, which is of all water species. So condensation doesn’t change the water content, which is why he wouldn’t need a precipitation term in [1], even if there were no diffusion.

In your comparison, 2 vs 3, ρ means different things. For 3 it is air + wv, but for 2 it is air + wv + liquid.

In Eq [1], the term with V represents the motion of liquid relative to the flow. In going to 2, you have set V to 0, not ρl.

141. ### Anastassia Makarievasaid

#140 Nick,

I appreciate your immediate corrections, but I would suggest that you should read the paper first.

142. ### Nick Stokessaid

#140
BR09′s equation includes it in the left side divergence, which is of all water species.
That wasn’t very clear.

The density on the LHS of BRO9’s [1] is of air+wv+liquid. So it isn’t changed by condensation, except insofar as liquid leaves. You have assumed that it does so immediately. BR09 adds the term with V on the RHS to allow for liquid efflux. If you omit this term, it sets V to zero, but then the liquid is just carried along with the flow. It doesn’t leave.

143. ### Nick Stokessaid

#141 Anastassia,
The paper is behind a paywall. But your exposition of it is admirably clear, and I believe the issue I point out is worth your investigation.

144. ### curioussaid

137 Geoff – re: hurricane paths, I found this interesting and reasonable:

http://www.sjsu.edu/faculty/watkins/hurricanepaths.htm

145. ### Anastassia Makarievasaid

#143 Nick

just sent you the paper at the email address you left at the acpd thread.

146. ### Gavinsaid

#139 Anastasia, thanks, but I think you have a misunderstanding here. The equation you cite
[1]

is simply the conservation of mass, and has nothing to do with frictional heat losses at all. The term $$\frac{\partial (\rho_l V_t )}{\partial z}$$ is simply the difference between condensate coming in at the top and leaving at the bottom. This is valid whatever the velocity $$V_t$$ is. This has nothing to do with dissipative heat terms (which are controlled by the $$\Theta_3$$). The diffusion terms $$D_*$$ can be zeroed out without any confusion for your argument.

If you want to assume instant removal of condensate, then you need to allow $$V_t$$ to go to infinity (not zero), and so $$q_l$$ will tend to zero. The term $$\frac{\partial (\rho_l V_t )}{\partial z}$$ will not then be zero, but will be exactly the condensation rate == mass removal term $$S$$.

147. ### Gavinsaid

TeX is messed up in previous comment (Jeff, can you fix?). Meantime a quick check of syntax: $$q_l$$ or q_l?

148. ### Gavinsaid

(redo)

#139 Anastasia, thanks, but I think you have a misunderstanding here. Your equation [1] above is simply the conservation of mass, and has nothing to do with frictional heat losses at all. The term $latex>\frac{\partial (\rho_l V_t )}{\partial z}$ is simply the difference between condensate coming in at the top and leaving at the bottom. This is valid whatever the velocity $V_t$ is. This has nothing to do with dissipative heat terms (which are controlled by the $\Theta_3$. The diffusion terms $D_*$ can be zeroed out without any confusion for your argument.

If you want to assume instant removal of condensate, then you need to allow $V_t$ to go to infinity (not zero), and so $q_l$ will tend to zero. The term $\frac{\partial (\rho_l V_t )}{\partial z}$ will not then be zero, but will be exactly the condensation rate == mass removal term $S$. (BR09, eq. 7).

149. ### Gavinsaid

Arrgghh:

The term $\frac{\partial (\rho_l V_t )}{\partial z}$

150. ### Anastassia Makarievasaid

Gavin, thanks, I appreciate your input regarding dissipation, however, my point about the difference between the curves in Fig. 10 being due to frictional dissipation pertains to the physical meaning of the final result, rather than to how it was obtained. True, dissipation rate is formally controlled by $\Theta_3$ in Eq. (5) and $\Pi_3$ in Eq. (4). However, water load does make contribution to total dissipative heating (and the smaller V_t, the greater so), such that the value of $\rho_l > 0$ must make an implicit impact on these dissipative terms.

In the meantime, I have another concern. Actually the terms in the right-hand part of Eq. (13) seem to be of different dimensions (which I did not notice before). $\partial (\rho_l V_t)/\partial z$ has the dimension of kg/sec/m^3, while $D_{q_t}$ apparently has the dimension of 1/sec (cf. Eq. (7), where $\partial q_l/\partial t$ and $\partial D_{q_l}$) have the same units, with q_l being dimensionless). While I know approximately how to remedy this, if you know the correct answer I would appreciate if you share, to avoid ambiguity in my subsequent considerations.

151. ### Steve Fitzpatricksaid

DeWitt #135,

The total average measured rainfall over 24 hours was 8 cm, not 8 mm.

Geoff #134,

Hurricanes/tropical storms do diminish rapidly as they pass from ocean to land. There are effects from both loss of sufficient source of moisture/latent heat (warm ocean) and increased frictional losses due to the rough surface of land compared to ocean. The time to diminish is comparable to the time it takes for the storm to lose inflow of air from the ocean; if the storm system is 800 Km diameter, and moving at 30 Km per hour, it will take ~26 hours to “lose contact” with the ocean. In places like South Florida, the land area is not great enough to fully cut off access to the ocean, and the topography is low, so frictional increases are not so large. Hurricanes will often pass right over the peninsular with only modest loss of intensity.

152. ### Thomassaid

Anastassia Makarieva:

you claim: accounting for this positive feedback totally determines all hurricane dynamics

it all comes down to this:

if the processes you describe are already in the NWPs (numerical weather prediciton models) than you can check how large the effects are by simple switching them off / on (this is what Kerry Emanuel describes in the original email – the result is, that the process is not essential for hurricane dynamics).

if they are not in the models already than they cannot be substantial because the NWPs are able to produce realistic hurricanes.

if you’re theorie is right you should be able to make predictions the established theory can’t. you have to show that you’re theory fits the data better.

153. ### Steve Fitzpatricksaid

Anna V #132,

Filling a cylinder with supersaturated air is an easy thought experiment, but maybe not so easy to do in practice.

I was only trying to illustrate the principle, not suggest a real experiment. Measuring temperature profiles in thunderstorms (versus the surrounding air column) is probably a better confirmation than the cylinder experiment.

154. ### anna vsaid

Thunderstorms?

I have found by googling engineering studies of air conditioners which work with extra heating by utilizing the latent heat of condensation and an extra pressure drop that they say is due to the change in the geometry of the machine outlets due to condensation.

I am sure a simple experiment could be designed to answer this question once and for all, if it has not already been done.

By Zeus, Thunderstorms!

155. ### Dsaid

#52 Thomas

You say “if they are not in the models already than they cannot be substantial because the NWPs are able to produce realistic hurricanes.”

The logic is wrong. Realistic in what sense? It is possible to be very realistic without being right. Look at computer animation. Don’t be fooled by “realistic” simulations — with enough tweaking you can simulate most things in multiple different ways … Realism is not the point.

But you are right about the need for prediction and empirical data. That is where this should be headed. But a new theory still needs to grow its wings to compete with models that have been tweaked and fitted for decades already. BTW This new theory can already say more (be as “realistic”) with less “fitting” is my reading … so Occam’s Razor suggests ….

156. ### DeWitt Paynesaid

Sorry about the typo replacing c with m.

I remember visiting CERN a few years ago and seeing some of the obsolete equipment out on the lawn. One of those pieces was a cloud chamber that was basically a very large cylinder with a piston. My guess is that somewhere in the literature there are papers describing the details of that device which might actually have the information being sought. I would be very surprised, though, if the results were all that different from the fairly simple calculation results.

157. ### Thomassaid

# 155 D

I think you have a wrong concept about tweaking. the atmosphere is a nonlinear system. the global models use one convection parametrization for the whole globe. it is applied for extratropical cyclones and hurricanes so it has to work for everything. there is no room for tweaking it specifically for tropical cyclones. the models are able to forecast hurricane tracks some day in advanced (also the forecast is far from being perfect). this is possible without taking the process into account that determines all hurricane dynamics?!?

Anastassia Makarieva has not presented any comparison with real data so far (besides: “this corresponds well with typical wind speeds” etc.).

158. ### anna vsaid

The CLOUD experiment at CERN which is testing the physics of Svensmark’s cosmic ray hypothesis is using a cloud chamber.
http://cloud.web.cern.ch/cloud/

I will read up the literature and see whether the pressure changes with condensation could be measured there. ( a Wilson cloud chamber went through two cycles of precipitation before looking for tracks in supersaturated air).

I still think that the information must be in the libraries in hard copy, just not online.

159. ### Dsaid

#157 Thomas

Well their papers give profiles of ideal storms derived from physical principles (without the need for a simulation model and all that that entails).

I’m impressed enough to consider this a plauible viable approach … and there ARE lenty of predictions there. See http://www.bioticregulation.ru/common/pdf/pla09b-en.pdf

Some aspects are already realistic: e.g. “The radially symmetric atmospheric circulation patterns
(cyclones) cannot spread over a distance 2L exceeding 3×10^3 km. This conclusion agrees well with observations.” While some need more detailed data.

It’d be great if these ideas could be effectively tested against real data … but I am not sure it is quite fair to require theorists sitting in Russia to have all these killer data quickly to hand.

If someone has contradictory data let them bring it forward for consideration …

160. ### skysaid

In equatorial regions, a near-ground layer of cloud extending tens of meters above the canopy commonly develops on otherwise cloudless nights and persists well after sunrise. It is unaccompanied by any perceptible air movement. Condensation takes place very gradually, and the accompanying microscopic implosions that permeate the air mass seemingly even out very quickly through tiny air flows. The barometric pressure remains steady throughout the night.

By mid-morning, that cloud layer has evaporated under insolation and pop-corn clouds pepper the sky, signifying the stirring of thermally-driven convection. By early afternoon, cumulonimbus develops, eventually bringing isolated thunderstorms. Where there is a discontinuity in air masses aloft, as in Africa during the monsoon, easterly gravity waves may produce more windy line squalls on a different schedule, but their observed intensity nevertheless is tied to local moist convection.

Rather than considering Makarieva’s basic thesis in the highly complicated and less-accessible laboratory of hurricanes, I would suggest testing that thesis in the much simpler setting of barytropic equatorial dynamics. [BTW, Anastassia, hurricanes do not occur in the Amazon Basin and other equatorial regions because the Coriolis effect is too weak there to sustain cyclonic motion.]

I think there is a known resolution to the question of latent heat sinks. Inasmuch as thermal energy cannot exist apart from matter, latent heat that is “released” upon condensation aloft cannot flow instantly into another substance, as some have supposed here. It manifests itself first as sensible heat in the water droplets that form clouds. Only subsequently can the energy be transferred. Unlike gaseous matter, clouds do radiate like a gray body in a continuous spectrum. And they flow out of tops of thunderstorms and hurricanes over a very much wider area than that of intense winds. From those altitudes, virtually no backradiation can reach the surface, given the optical depth of the atmosphere. The ultimate radiative sink is space. But we knew that already, didn’t we?

161. ### skysaid

Please read “barotropic” for “barytropic” in my #160. That’s what happens when typing with heavier thoughts on your mind.

162. ### Nick Stokessaid

#141 Anastassia Makarieva (Nov 1 07:52)
Thanks for the paper, which I am now reading.

163. ### Nick Stokessaid

#158 Anna
On experiments, here is a current evaporation engine I have seen in operation. water is sprayed into a cylinder with warm air. Evaporation cools the air, and despite the addition of extra gas molecules, the air contracts, driving the piston. The same in reverse.

164. ### DeWitt Paynesaid

So what happens when cold rain drops fall into the warmer air below? Isn’t that going to overwhelm any dissipative heat generated by friction, or whatever the proper term, between the falling drops and the surrounding air?

165. ### anna vsaid

Thanks.

It is interesting, though the engine takes the energy from the air, not the latent heat of condensation.

in the links there, “technical report 2009”, a heat pump is described on the same principles:
[3] “Thermodynamic Model for a Two-Stroke Evaporation Engine”, Proc ANZSES Conf, Canberra (2006).

If the inlet air is cool and moist, the Bernoulli effect leads to rapid condensation of microscopic water droplets in the high-speed section and release of latent heat. This causes decreased pressure and density and increased temperature and flow speed. If the droplets are collected and thereby prevented from re-evaporation as the flow is isentropically slowed without flow separation, the outlet temperature will be greater than the inlet temperature.
Bold mine.

[4] “An Evaporation Heat Engine and Condensation Heat Pump”, ANZIAM J, Vol 49 (2008), 503-524.
Also analysed is a related heat pump based on condensation of water vapour in moist air at reduced pressure. These devices operate as two-stroke reciprocating engines, which are their simplest possible embodiments.

Of course this is utilizing the Bernouli effect,(The Bernoulli effect shows there is a drop in pressure, temperature and density when gas accelerates isentropically in a narrowing section of a duct.) but it is interesting that there is a decrease in pressure with condensation.

166. ### Nick Stokessaid

Anastassia,
Thanks, again, I have read the paper. On your dimensional concern, yes, there is an error. The dissipation in (13) should be $D_{\rho_l}$ (ie ρ, not q. The D is a general symbol for the current equation scalar, as defined in the last equation of (14).

But I stay with my earlier argument on the role of S in (13). BR09 spell it out:
“It can easily be shown that total mass inside the domain is conserved in the absence of surface precipitation and surface water vapor flux, and that the so-called precipitation mass sink effect [e.g., studied by Qiu et al. (1993) and Lackmann and Yablonksy (2004)] is included in this model when using the conservative equation set.”
As I said in #141 and #143, if you want to adapt the equation (13), leaving out the term with Vl just sets Vl to 0 – condensate not settling at all. As Gavin says, to move to your approx, you really want Vl=∞. Setting ρl=0 in the logic of (13) specifies no condensation at all, unless you do add an S term.

I think a section which may interest you is the sensitivity test 3d on p 1782, where they vary Vl. Going from Vl=0 to ∞ makes about a 60% difference to hurricane intensity. Vl=∞ corresponds to your case with a precipitation term with instant removal. They attribute the difference to weight of liquid – ie lower buoyancy of air which is carrying the liquid along.

167. ### curioussaid

163 Nick – I’m not sure you have found an equivalent example but I may have been left behind some time ago!

I thought the discussion was regarding an adiabatic transformation of a “closed” parcel of air and moisture. To my mind that is equivalent to Gavin’s thought experiment of a closed container of supersaturated air undergoing spontaneous condensation. My initial view was that this would mean a drop in pressure inside the closed vessel. However following the comments above it seems that there is reasonable agreement that in fact the pressure would rise as the partial pressure contribution of the water vapour would be outweighed by the heating effect on the air from the release of the latent heat of vapourisation. The experiment you link to seems more akin to the previously given example by Konrad (other thread I think)where water cooling is provided as an exernal input to a system.

Apologies if I have misunderstood but I’d be grateful if you could clarify – I noted with interest Terry’s comment at 133. Thanks.

168. ### Nick Stokessaid

Re: DeWitt Payne (Nov 1 16:27),
I think frictional heat is very small. Dominantly, the LH cools the air, which contracts, even though evap makes more gas. That’s just runnung the condensation in reverse.

169. ### Nick Stokessaid

Re: anna v (Nov 1 16:46),
I agree that sentence is odd, and may be wrong – we’d need to see the paper. The analogous sentence in the previous para says that evaporation increases the pressure. But when water is sprayed into the cylinder it clearly decreases the pressure – I have seen the engine working.

170. ### Nick Stokessaid

Re: curious (Nov 1 17:25),
I’m describing the reverse process, where water is sprayed in fine droplets into an enclosed space. Evaporation makes more gas, but this is outweighed by the cooling effect of LH absorption. The gas contracts, driving the engine.

171. ### DeWitt Paynesaid

That’s what I thought. Next question: Is this effect included in the hurricane models?

172. ### curioussaid

170 – thanks Nick, that clarifies. I can see how that would work and it does support the expectation that condensation would result in a net expansion for a given parcel of moist air. However, I’d like to comment that I have reservations about this being analogous to a closed adiabatic process. Also I’d like to comment that IMO the work is done by the external atmosphereric pressure driving the engine.

173. ### Nick Stokessaid

Re: DeWitt Payne (Nov 1 18:22),
Well, it’s in the GCM CAM 3. And yes, LH is there in Bryan and Rotunno, eq (5), and $\dot{q}_{cond}$ in the mass eqns 6 and 7.

174. ### Jim Dsaid

If it can be shown that a pressure drop leads to condensation, that is sufficient to show that the opposite can’t happen. Imagine how a cloud forms, as is well known, during sudden decompression (e.g. in an aircraft). Now imagine that this condensation produces a further pressure drop, which causes more condensation in a positive feedback loop. Eventually you will have zero vapor, which just doesn’t happen spontaneously.
I showed it from the gas law that condensation produces a pressure rise. It is also well known from decompression and cloud chambers that a pressure drop can produce condensation.
In the quoted article, the Bernoulli effect of accelerating air causes the pressure drop and condensation results from that.

175. ### Anastassia Makarievasaid

I continue from #139. My thanks to Gavin who corrected me in my interpretation of Eq. (13). It was incorrect to conclude that a theoretical equation for condensation rate, [2] in #139, does exist in this model (which seemed something new to me). I still cannot believe that there is not any theory for condensation rate, so I was willingly confused to presume that something new was proposed. But I was mistaken.

Indeed, as Gavin said at #148, if we assume that liquid water is immediately removed from the atmosphere, $\rho_l = 0$, and zero the diffusion terms, we obtain that condensation rate S in BR09 versus M10 is:

$S \equiv \rho_d\dot{q}_{cond}$ [BR09]

$S = w\rho\frac{\partial q}{\partial z}$ [M10]

That is, condensation rate is present in the model as an independent variable. Despite BR09 list seven equations and count seven independent variables, they do not count condensation rate and thus hide the fact that the problem is open. But how is it then possible to calculate S in BR09? BR09 write (p. 1774):

“The method to determine $\dot{q}_{cond}$ was described by Bryan and Fritsch (2002, p. 2920). As shown in their study, this numerical model does not precisely conserve total mass and energy, partly because of numerical reasons. However, for simulations using the conservative equation set, we find that artificial loss of mass and energy to be several orders of magnitude lower than for simulations using the traditional equation set. At the end of the simulations (typically at t = 12 days), the artificial loss of mass and energy is less than 0.05% of the initial values.”

Bryan and Fritsch (2002) have equations (20)-(24) structured similarly to Eqs. (1)-(7) of BR09, with Eq. (20) being for momentum balance and eqs. (21)-(24) describing tendencies in pressure and potential temperature, as well as mass conservation. Eqs. (21)-(24) contain condensation rate, Eq. (20) does not. Bryan and Fritsch (2002) explain how the model equations are integrated:

To account for phase changes, the model uses a saturation adjustment technique, similar to that proposed
by Soong and Ogura (1973). In this technique, the equations are advanced forward in two steps: a dynamical step and a microphysical step. In the dynamical step, the model equations are integrated forward with all terms involving phase changes neglected. Then, the microphysics step is applied, in which only the terms involving phase changes are included. This technique is identical to that used by Klemp and Wilhelmson (1978).

Pressure tendencies due to phase changes are included in the microphysics step. Since changes in pressure affect the saturation vapor pressure, an iterative scheme had to be developed for condensation. In this iterative scheme, equations (21)–(24) are advanced forward using a guess for $\dot{q}_{cond}$. The new values of $\theta$ and $\pi$ are then used to calculate a new value of saturation mixing ratio ($r_{vs}$), which is then used to calculate a new guess for $\dot{q}_{cond}$. This cycle is repeated until the newest value of $\theta$ converges (to within machine accuracy) to the previous value. During each iteration, the value of $\dot{q}_{cond}$ is determined by the following equation from Rutledge and Hobbs (1983):
$\displaystyle \dot{q}_{cond} \equiv \frac{r_v - r_{vs}}{\Delta t (1 + L_v^2r_{vs}/(c_pR_vT^2))}$
where $\Delta t$ is the time step. The iterative technique usually converges in 4–6 iterations.

We can see that the account of condensation rate is made numerically without updating the velocity field during each iteration. This does away with the positive feedback between vertical velocity and condensation rate, which is the core of the theoretical formulation of M10. Accordingly, total energy $E_t$ that is conserved in the model of Bryan and Fritsch (2002) ignores the potential energy release from condensation, see my comment #24 above.

Let us further follow the algorithm of calculating condensation rate. Bryan and Fritsch (2002) referred to Rutledge and Hobbs (1983). These authors, in paper titled “The mesoscale and microscale structure and organization of clouds and precipitation in midlatitude cyclones…”, write (p. 1200, section “condensation and evaporation of cloud water”):

Following Yau and Austin (1979), we express the condensation of water vapor to cloud water (PCOND) as
$\displaystyle PCOND = \rho (q_v - q_{sv}) [\Delta t (1 + L_v^2r_{vs}/(c_pR_vT^2))]^{-1}$ (A6)
If $q_v < q_{sv}$, the cloud water evaporates.

They also issue a warning:

The formulation of microphysical rates, such as Eq. (A6), which has $\Delta t$ in the denominator, may result in the model outputs being sensitive to the chosen time step. To investigate this, we used time steps of 2 and 5 s. Over this range of values for $\Delta t$, there were no changes in the various model outputs. However, in applying models of this type to other situations (i.e., strong convective clouds) it would be wise to formulate microphysical conversion rates that are independent of $\Delta t$.

Indeed! Need for some theoretical work to guess what an equation for condensation rate could look like, as I outlined in , was felt in 1983. Let us now follow this chain further and look what Yau and Austin (1979) had to say, to whom Rutledge and Hobbs (1983) refer. In this paper, titled “A model for hydrometeor growth and evolution of raindrop size spectra in cumulus cells”, it is said (p. 657, section “b. Microphysical processes. 1) Condensation, evaporation and sublimation):

Clark (1973) computed the nucleation of cloud droplets from a spectrum of condensation nuclei and their subsequent growth by diffusion in a dynamic model. The cloud dynamics resulting from this detailed treatment of nucleation and condensation do not deviate substantially from that of a bulk physical model where all supersaturated water condenses instantaneously. In view of this finding, condensation and evaporation of cloud droplets is computed from the dynamic effect alone. Modification of supersaturation mixing ratio through the release of latent heat in a manner proposed by Asai (1965) is adopted. The saturation excess $\delta M = q_v - q_s$ is calculated using Teten’s formula [as given in Asai (1965)] to obtain q_s. A part of $\delta M$, $\delta M_1$, is condensed releasing latent heat which causes an increase of temperature; in turn, this permits the residue, $\delta M_2 = \delta M - \delta M_1$, to remain in vapor form.

Now our ultimate authority is Asai (1965) (i could not found this paper, titled “A numerical study of the air-mass transformation of the Japan Sea in winter”, J. Meteorol. Soc. Japan, 43, 1-15. I was curious to read about some Teten’s formula, but it turned out to be the Clausius-Clapeyron law.

We summarize: THERE HAS BEEN NO THEORY FOR CONDENSATION RATE. In the existing models, it is calculated numerically using approaches dating back to 1965 (or perhaps earlier). This calculation is made within the microphysical domain to the numerical accuracy of the model without simultaneous recalculation of the large-scale dynamic equations. The positive feedback between condensation/precipitation and vertical velocity is ignored.

The potential power resulting from mass removal, equal to the work experienced by a contracting parcel from the ambient environment, is ignored. The unphysical idea that the air becomes supersaturated first (for unknown reason), then suddenly condenses releasing latent heat and warming the environment, as expressed by Dr. Rosenfeld in his calculations, dominates the thinking patterns. The fact that the air condenses because it cools by expanding, and that this is the main process behind condensation, is not given any serious consideration. To me, all this looks as a grandiose physical havoc.

In my next comments, I will explain how it is nevertheless possible, by artificially tuning the latent heat impact on dynamics, to obtain realistic wind speeds.

176. ### anna vsaid

Now imagine that this condensation produces a further pressure drop, which causes more condensation in a positive feedback loop. Eventually you will have zero vapor, which just doesn’t happen spontaneously.

People should be very very careful with feedback arguments. They cannot prove theorems nor substitute for data. If we took your argument seriously then successive expansions would also deplete the air from moisture. The experiments of Wilson when making the cloud chamber(look it up on google) showed that after two expansions which rained out most of the supersaturated humidity , the third time tiny droplets remained suspended, no more precipitation ,and that was the time he used to see ionizing radiation. DATA trumps theory and certainly hand waving.

In the intriguing use of the Bernoulli effect that Nick found , it is the Bernoulli effect that lowers the pressure because of the geometric effect of forcing gas through a narrower channel. Even if condensation lowered pressure, it would not show in such an environment.

The cooling cycle of this engine from the links above :
The Bernoulli effect for a compressible gas shows there will be a drop in pressure, temperature and density when the gas accelerates in a narrowing section of a duct. If the inlet air is hot and dry, spray evaporation of water in the high-speed section leads to increased pressure and density and decreased temperature and air speed.
bold mine.

So , evaporation from spraying increases pressure in this machine and the pressure difference is utilized in a turbine. Condensation decreases pressure in this machine.

It is not good for stating whether condensation lowers pressure, since evaporation increases it in this machine,because extra mass is introduced with the spraying.

One needs a specifically designed experiment for this that controls all factors.

Again, I cannot say it enough, data trumps theory every time.

177. ### Jim Dsaid

#175 Anastassia
The supersaturation is provided by the dynamics, usually through vertical motion that causes the parcel pressure to drop, leading to cooling, and supersaturation, followed by condensation adjustment according to Clausius-Clapeyron. The condensational latent heating allows an ascending parcel to maintain buoyancy, closely following the reversible moist adiabat, as it is observed to do. If a cloud model did not follow Clausius-Clapeyron, it would be worthless, lacking a vital conservation property. You could look into the idea of moist adiabatic lapse rates to learn more about how these models (and clouds) work.

178. ### Anastassia Makarievasaid

#174 Jim

Now imagine that this condensation produces a further pressure drop, which causes more condensation in a positive feedback loop.

Exactly! This is what happens when hurricanes develop: pressure drop due to condensation produces a radial pressure gradient. This gradient accelerates the air and enhances vertical velocity. Vertical velocity enhances condensation. Accumulation of vapor in the atmosphere is equivalent to accumulation of potential energy. Hurricane develops as an avalanche turning all this energy into the kinetic energy of winds. As soon as the moisture store is depleted, hurricane ceases to exist. One has to wait for a while until evaporation prepares the needed amounts of potential energy in the atmosphere for the next hurricane to form.

An essential feature of the hurricane is that it must move — after the local store of potential energy is depleted, it must go to another place where it is still abundant. If the original condensation area is surrounded by relatively dry areas with low vapor content, the hurricane cannot intensify. If there is an area with very high moisture content along the preferential direction of hurricane movement, the probability of hurricane development is high. There is enormous potential here for improving hurricane forecasts.

179. ### Anastassia Makarievasaid

#177 Jim

If one carefully reads my comment, it is not said anywhere that Clausius-Clapeyron law is not used. The problem is very different.

By the way, if the CC law is used, as you say, how can you argue that condensation increases air pressure? Apparently CC law prescribes condensation to occur upon a drop of temperature. Already this drop of temperature lowers pressure, the mass removal only adds to it.

In your calculations in #110, you attempted to compare the effects of mass removal and latent heat release on pressure. However, mass removal changes the amount of matter. This process cannot be in any way compensated by release/heat intake. This is a different process associated with potential energy release and work performed. Heat and work are different quantities in physics. They are not comparable.

180. ### Jim Dsaid

Anna V, I was illustrating how condensation _can’t_ produce a pressure reduction, because it is generally accepted that pressure reductions can cause condensation. I can point to many examples of that including clouds, cloud-chambers, sudden decompressions, of condensation _following_ a pressure reduction. As I stated in my above post, this is the normal sequence in ascending cloud parcels.

181. ### Jim Dsaid

#179 Anastassia
My #110 calculation is not about mass removal, it is about vapor condensation that leads to pressure reduction which is more than compensated by a factor of 5.7 by the pressure increase from latent heating. Net effect is pressure increase.
#178, my #174 was supposed to be why a feedback was impossible. Your idea leads to the feedback, not mine. This feedback makes no sense in the real world, because cause and effect are confused.

182. ### Jim Dsaid

also on #179, yes, a temperature drop leads to condensation. This temperature drop is provided by a pressure drop through adiabatic expansion. This is the necessary sequence for a reversible adiabatic process: expansion, pressure reduction and cooling then condensation.

183. ### Anastassia Makarievasaid

#181 Jim,

My #110 calculation is not about mass removal, it is about vapor condensation that leads to pressure reduction which is more than compensated by a factor of 5.7 by the pressure increase from latent heating. Net effect is pressure increase.

This is an incorrect conclusion precisely because you ignore the mass removal. As you certainly know, air pressure at the surface is equal to the weight of air column. If you remove mass from the column, the air pressure drops. If you do not, it does not. Heating/cooling make no impact on surface pressure.

If you know of a dynamic equation for condensation rate that is used in the system of hydrodynamics equations to close it, please, write it out. Bryan and Rotunno (2009), as well as myself, are apparently unaware of such an equation.

#178, my #174 was supposed to be why a feedback was impossible. Your idea leads to the feedback, not mine. This feedback makes no sense in the real world, because cause and effect are confused.

This feedback does make a perfect sense. There is no confusion, just the inherent instability of condensable vapor in the gravity field and a positive feedback between the rate of mass removal and pressure gradients that it generates. This feedback is caused by this instability.

184. ### Anastassia Makarievasaid

#181 Jim,

Take also notice that rejecting such a feedback as unrealistic, you object not only Makarieva et al., but also many other people, including meteorologists. If you read our paper, you will find citations where the possibility of such a feedback was discussed.

For example, Qiu et al. (1993) wrote:

the mass depletion due to precipitation tends to reduce surface pressure, which may in turn enhance the low-level moisture convergence and give a positive feedback to precipitation

The problem is that the effect was not studied theoretically (an equation for condensation rate has been missing!) but instead evaluated numerically using models that by definition did not include it.

185. ### anna vsaid

I was illustrating how condensation _can’t_ produce a pressure reduction, because it is generally accepted that pressure reductions can cause condensation. I can point to many examples of that including clouds, cloud-chambers, sudden decompressions, of condensation _following_ a pressure reduction. As I stated in my above post, this is the normal sequence in ascending cloud parcels.

My bold emphasizes the problem with climate science at present. Physics does not obey democracy and generally accepted.

You are not pointing to data. You are pointing to a narrative description of observations from long distance and no measurements of the crucial underlying brick of this house.

The fact that condensation follows pressure reduction does not logically exclude that condensation also reduces the pressure further. This must be shown in a controlled laboratory experiment..

186. ### Anastassia Makarievasaid

#160 sky,

[BTW, Anastassia, hurricanes do not occur in the Amazon Basin and other equatorial regions because the Coriolis effect is too weak there to sustain cyclonic motion.]

Thank you for your attention to this important issue. The near-equatorial position (it is not exactly on the equator) of the Amazon does not explain that hurricanes do not happen anywhere near the Brazilian coast or anywhere near South-America.

Let us inspect this famous figure for tropical storms tracks. It is clear that tropical cyclones do happen in many regions that are more close to the equator than some parts of the Amazon basin, e.g., see the typhoons patterns in the Pacific ocean. In the meantime, the Amazon forest is surrounded by thousands of kilometers of calm sea.

Our theory perfectly explains this: the Amazon forest draws excessive vapor from the ocean to land, leaving less potential energy for cyclones to develop. On land, the formation of violent winds is prevented by strict biological control of the condensation process, in line what Steve Koch mentioned at #15 above. In simple words, the forest controls the condensation process not allowing for catastrophic, avalanche-like developments (could be compared to controlled fusion power — in this case, controlled condensation power).

187. ### Anastassia Makarievasaid

Before I continue my consideration of BR09 from #176, I would like to once again focus the reader’s attention at the very peculiar procedure of accounting for condensation rate. As the paper of BR09 is behind the paywall, one can read the same in Bryan and Fritsch (2002) [BF02], which is free.

Once again, we have fewer equations than independent variables (seven versus eight in BR09). We cannot solve the problem. But we want to. To do so, we exclude the dynamics equations from consideration (Eq. 20 in BF02) and perform some magic numerical manipulations with the remaining equations, using guesses for condensation rate, convenient time steps etc. The output of these blackbox calculations is that suddendly we know condensation rate. Then we recall that we also have some other equations (dynamics) and update them accordingly. We move along in this way through all the simulations and obtain a solution describing a circulation.

But how is it possible? Just at the logical level, how is it possible, by whatever numerical solutions and manipulations, to solve a system of equations where there are more variables than constraints on them? This is only possible if there is a hidden controversy in the calculations, when one first sets something to zero (this is the needed extra equation), solves the system, and then puts the solution into it which is already not zero. This is what is being done, it is implicitly assumed that condensation rate has no impact on dynamics. All updates are made using this assumption as the core.

This could have lead to major dynamical discrepancies and absurd results, but to remedy this another procedure is applied, which is to substitute the pressure gradient that should have resulted from condensation by inflated pressure gradient from the latent heat release. This will be the topic of my subsequent comments.

188. ### Anastassia Makarievasaid

#66 Steve,

In your calculations you assumed 50 g/M^3 moisture content. This is too high, and due to this fact you obtained a too low velocity. If we have partial pressure of vapor p_v ~ 0.03 p at the surface and air density rho = 1.2 kg/m^3, the density of vapor at the surface is (p_v/p) * M_v/M * rho = 22 g/m^3. Here M = 29 g/mol and M_v = 18 g/mol are molar masses of air and vapor, respectively. I guess you might have replaced (M_v/M) by (M/M_v) in your calculations.

189. ### Geoff Sherringtonsaid

151 Steve Fitzpatrick – thank you for the interesting reference to possible causes for path changes of such storms. It seems logical enough.

Certainly risking an exposition of my ignorance, I note again the image of the vortex below the jumbo engine in

The purpose of referring to this was to ask if the fundamental assertions about “storms” to use the word generically, are properly founded. There seems to be an inherent assumption that these storms most often start above a warmish part of a tropical ocean and develop in intensity through measures such as are discussed above. Inherent in these are adiabatic processes.

The jumbo jet vortex might be entirely dissimilar. But it seems to be initiated by high velocity winds above it, it seems not to involve much adiabatic lapse change and it does not seem to involve replenishment of water from below. The condensate we see appears to be drawn in laterally.

But then, if the Jumbo analogue is relevant to these storms, we have an entirely different set of drivers and mechanisms, although the observations within and about the storm might be similar. I am asking if the well-documented historical observations can be explained by a different mechanism to that being discussed. Is there any observation of storms that start vortices at high altitude but never reach down to disturb stable ground level weather?

190. ### curioussaid

189 Geoff – this might interest you too:

http://en.wikipedia.org/wiki/Emma_(windstorm)

100mph winds over central Europe land mass. Unfortunately I don’t read German but I’d be interested if anyone can provide more info. on its trajectory:

http://www.met.fu-berlin.de/wetterpate/Lebensgeschichten/Tief_EMMA_28_02_08.htm

Wiki has a discussion of the phenomenon and its difference from hurricanes here:

http://en.wikipedia.org/wiki/Extratropical_cyclone

191. ### curioussaid

189 – Geoff, a bit more: This paper might be of interest –

http://www.engineeringletters.com/issues_v16/issue_3/EL_16_3_20.pdf

Brought up with other interesting material by a google search “jet engine power test inlet vortex”

Also re: Emma trajectory – this was the best I could find:

http://www.guycarp.com/portalapp/publicsite/catdocument.pdf?instratreportid=1677

Apologies if this is straying too far from the central theme of the thread.

192. ### Steve Fitzpatricksaid

Anastassia #188,
“Taking into account that Lv=2.4 kJ g−1 and v50 gm−3,
at vertical velocity of uz0.1ms−1 we obtain FLH4×103Wm−2, i.e., at least 20 times
larger a flux than the flux of solar radiation absorbed by the surface in the tropics”

I simply used your value for moisture… 50 grams per cubic meter. So I assume the value in the paper is wrong.

In any case, a flux rate of 2200 watts per square meter is a long way from 4000 watts per square meter. As best I can determine from ERBE, the typical long wave flux rate in the tropics is ~260 watts per square meter. (The incoming solar flux is higher, about 320 watts per square meter, and the difference carried away by the Hadley circulation and warm ocean currents.) If you assume all the heat of condensation must be lost to space within the hurricane field, then indeed the temperature aloft would have to be very high. But satellite images of hurricanes show obvious bands of cirrus clouds streaming away from the hurricane center, and extending well beyond the area of the hurricane itself. Heat loss could never be confined to the area of the hurricane. I do not suggest that all the heat of condensation is lost to space, but I do think is it very likely that a significant portion of the heat is.

193. ### steven Moshersaid

At this point I think a virtual blackboard that people could write equations on would be hella cool.

194. ### Brian Hsaid

#192, Steve;
Your observation reminds me of a recent study using (IIRC) IR lidar to study clouds, and they found that there is an “invisible” shell up up to 10s of km. around them, where moisture and energy are in very active flux. Seems to me that would apply a fortiori to hurricanes (or any TD).

195. ### Brian Hsaid

#193, steve;
Does Google Docs support anything like that? A shared WP doc with math notations enabled?

196. ### Dsaid

Gavin at #117
Said
“If their equation set is incomplete (and looking at the derivation in B&F 2002, it does not appear to be), we should be able to say so clearly.”

I may be missing the point here – please feel free to correct me. But isn’t this is just what has been shown by Anastassia at #187. That’s what it looks like if I have it right.

So if Bryan and Rotunno do not manage to incorporate condensation into their models in a credible (even transparent?) manner. Right? If that is true would this seems to invalidate all the previous statements that the effect of condensation on dynamics has received due consdiration and can (based on this thorough and exhaustive effort) be considered minor?

Thanks

197. ### PaulMsaid

Condensation can produce a pressure drop.

198. ### Anastassia Makarievasaid

#192 Steve,

Indeed! Now I see that this seems to be a confusion stemming from our paper. Actually there is a double numerical error in that estimate, probably a misprint or something. It was later corrected by us during the discussion (see here p. S11271, second para). But anyway the estimate of 2×10^3 W/m^2 calculated directly from precipitation rates rather than from vertical velocities should be more valid.

One cannot dispute the point that all latent heat is ultimately radiated to space from a large territory.

199. ### DeWitt Paynesaid

A constraint that can be used to solve the condensation problem when the change in pressure is known but change in volume and temperature aren’t known is that the latent heat of water condensation must equal the extra work done by the increased expansion of the parcel of moist air. For a fixed mass of dry air, say 1 kg, the volume and temperature as a function of pressure can be calculated from the adiabatic expansion equation and the ideal gas law. The saturated water vapor mixing ratio, g H2O/kg dry air, can be calculated also. I used this site. The expansion work can be calculated by assuming a fixed area of 1 m2 so the change in volume becomes movement against the average pressure. I then adjust the change in temperature of the moist air so that the latent heat from the condensed water calculated by the change in mixing ratio is equal to the difference of the work done by expansion of the moist air less the work done by the expansion of 1 kg dry air for the same pressure change. Using my usual brute force, inelegant methods, I didn’t try for an exact match, just close. So for 1 kg dry air at 300 K and 100 kPa, a change of pressure to 99 kPa results in 617 J (volume change from 0.861 m3 to 0.8672 m3). The saturated water vapor mixing ratio for the same conditions is 22.902 g/kg and the total volume is 0.8927 m3. At 99 kPa, a temperature reduction of 0.23 K gives a mixing ratio of 21.8189 g/kg. So 0.0831g water condenses and the latent heat release is 207.8 J. The volume increases to 0.90092 m3 for 816.9 J or a difference of 199.7 J. That’s close enough for me. The temperature change for dry air only is 0.86 K. Continuing, I get a lapse rate of ~10 K/km for dry air and an initial lapse rate of ~2.7 K/km for moist air at 300K surface temperature. I’m ignoring the volume and heat capacity of the liquid water. At 5 kPa, only 6.4 g water has condensed.

I think I may have left something out, though. I’ve failed to account for the change in enthalpy of the dry air with temperature. I think I can safely ignore the change in enthalpy of the water vapor. That’s probably why the lapse rate for moist air looks to be too low.

200. ### anna vsaid

I did an experiment:

I took two plastic bags. One I kept over the boiling kettle, filling it like a hot air balloon, then sealed the top capturing the hot steam, and left it on the table to come to room temperature. There was condensation and deflation, but of course the temperature came down. I then resealed the top further down, keeping the first seal, sliding the collar until the bag was inflated tensely.
I then took the second bag with room air, and did the same.

I put both bags in the freezer and looked in half an hour. The one with the condensation is definitely more deflated than the room air one.

I say that the pressure fell with condensation.

Maybe somebody could repeat it , (may be hole in bag? etc). Careful of the steam, use kitchen gloves.

201. ### PaulMsaid

Anna V, correct. This is how early steam engines worked. You can read up about Newcomen and Watt.

202. ### Rob Rsaid

So in the models are hurricanes driven by fudge-factors rather like what appears to be the case with global climate models?

I get the feeling that this discussion has some parallels with the concerns of Spencer and Braswell in relation to negative vs positive feedback from clouds and water vapour!

203. ### Nick Stokessaid

#187 Anastassia,

You have brought up another excellent paper in BF02. It’s well worth reading.

There are, indeed 7 equations (1-5, with 3 components for 1) where a time derivative appears on the left side. I count two other variables – pressure p and condensation rate $\dot{r}_{cond}$. p is related via the equation of state (9), and condensation is constrained by the requirement that saturation not be exceeded. This is expressed via Eq (27).

Just reviewing your earlier complaints about omissions – the conservation equations 4 and 5 allow for mass loss (and gain) by condensation, as they should, and the energy equation (3) has a term for latent heat gain, as it should. I can’t see anything missing. Hurricane models are a lot simpler than GCM’s so it’s easy to track through and see that these terms are retained.

So there are enough equations – it seems now that your concern is about how the ones without explicit derivatives are handled. You mention “magic numerical manipulations”, “guesses for condensation rate”, “convenient time steps”.

But the numerical techniques are standard, nothing “magic”, and were not dreamed up by people on P Jones email list. The general process for solving nonlinear equations from an initial “guess” followed by iteration goes back to Newton and beyond. The Runge-Kutta methods are more than 100 years old.

The “convenient time steps” are physically based – as it happens, I blogged on this a couple of weeks ago.

The saving grace of numerical methods for solving equations is that, while they can seem difficult, you can always go back afterwards and check that your solution does satisfy the equations. In a calc like this, that would have been done.

So it seems to me that they have 9 equations in 9 variables, correctly solved, and that condensation rate is fully accounted for.

204. ### Nick Stokessaid

Re: Rob R (Nov 2 17:15),
Which “fudge-factors” are you referring to?

205. ### Nick Stokessaid

Re: steven Mosher (Nov 2 12:07),
You should check the Latex facility here. TeX button above. It’s good.

206. ### Brian Hsaid

Cross-ref to hang glider’s observations at Momentary Lapse of Reason:

#
Ecoeng October 30, 2010 at 8:15 pm

Hang gliding under over-developed cumulas cloud can be delightful as lift is light, smooth and widespread. It is called ‘cloud suck’.

However, from onset of rain there is invariably significant loss of lift, a likely but not invariable onset of cross winds and in extreme cases downdrafts manifested as increased turbulence. The rule of thumb is – get outta there (or land if too low).

Note most hang gliders fly with basic instruments – almost invariably at least a variometer and usually an air speed indicator and altimeter so the above is a lot more than subjective.

Herer is the only YouTube example I could find – very mild example due to orographic effect location (nice music too) but obvious loss of lift even so.

#
Brian H November 2, 2010 at 4:39 pm

Re: Ecoeng (Oct 30 20:15),
So, rising air under a cumulus, which rapidly reverses, with inflow of air in response to a pressure drop — is how I read that.

#
Brian H November 2, 2010 at 4:40 pm

Re: Brian H (Nov 2 16:39),
That is, reverses as the rain begins, etc.

207. ### Steve Fitzpatricksaid

Nick Stokes #204,

I’m not sure which fudge factors he is referring to in hurricane models. (I’m not seeing any.) In GCM’s it’s mainly assumed aerosol effects….. they allow every diagnosed sensitivity to be ‘correct’; aerosols are as pure a kludge as you will likely ever encounter. Comically bad.

208. ### Nick Stokessaid

Re: Brian H (Nov 2 18:09),
Brian, I don’t see the logic there. A contraction or pressure drop in the cloud above would, if anything, produce an updraft.

209. ### DeWitt Paynesaid

Never mind. I left out the change in entropy. It’s easier to equate enthalpy change and gravitational potential energy. One thing I don’t see in moist adiabats is a slow down at zero degrees C when the liquid water freezes and you start having to add in a contribution to the change in enthalpy from the heat of fusion as well as the heat of vaporization.

210. ### DeWitt Paynesaid

While condensation is occurring there is an updraft. When the rate of condensation slows, the updraft slows and the rain falls. How’s that for hand waving?

211. ### Nick Stokessaid

Re: DeWitt Payne (Nov 2 19:45)
Well, I guess hang gliding is a fancy kind of hand waving. But as I said, I don’t see the logic. In your version, it seems to be that the rate of condensation in the cloud diminished with the onset of rain. Does anyone know that that is true?

It’s actually unclear which way the air would flow anyway. In one version, condensation would, by expansion, force air outward in all directions (downdraft below). In another. the warmer air would rise en masse (at it visually seems to in a cumulus cloud), perhaps drawing air in from below (updraft).

I’m not hand-waving here – I just don’t know. And I don’t think anyone here does.

#200 Anna V,

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.

The experiment shows that some mixtures of air and water vapour will contract when condensation occurs. This raises the question of the actual mix of water vapour and air in a air parcel that forms clouds. The mixture in the bottle would have far more water vapour than normal saturated air, however it may be that the conditions at the top of a rising moist air parcel are close to those in the bottle. Water vapour is lighter than air, and may not be evenly mixed in a moist air parcel.

213. ### skysaid

#186 Anastassia:

Although I agree with most of what you’re positing about the unaccounted effects of the hydrological cycle (and admire your patience in the face of objections from Defenders of the Faith), there is a danger of overreaching in your interpretations of real-world weather.

Hurricanes develop over oceans not land. You will note in the “famous figure” of tropical storm tracks that there is a “deadband” of ~8 degrees latitude that entirely circles the globe. That is the consequence of the vanishing Coriolis effect as one approaches the equator. And the mouth of the Amazon lies almost exactly on it.

The seemingly singular lack of hurricanes off South America (although one did briefly develop off Brazil in recent years) becomes understandable in terms of the oceanic circulation. The northeast-facing coast equatorward from Ceara feeds the warm equatorial current largely into the Antilles Current, leaving the Brazil current to recirculate the cooler waters of the Benguela in a great anticylonic gyre. And off the western coast, the great gyreing of cool Humboldt waters inhibits storm development over perhaps two thousand kilometers of the tropical Pacific.

To convince this old field-going scientist that the “biotic pump” of the Amazonian rain forest has an appreciable effect on hurricane formation, you’d have to explain why that pump doesn’t work in the waters off the rainforest of Mozambique, where tropical cylones are commonplace.

214. ### Jim Dsaid

… and is the cloud evaporating or still condensing as the bottle crumples? I would expect it is by now evaporating, proving that evaporation leads to sufficient cooling for a pressure drop. The cloud is formed only as the bottle expands out.

215. ### Jim Dsaid

#183 Anastassia
I am not talking about hydrostatic pressure. The gas law applies even without gravity and is all that is needed to explain, as I did in #110, how pressure increases when condensation happens. Two equations are needed (1) p = r R T and (2) L dq = – cp dT.
Which of these do you dispute? Or do you dispute the constants I used?

#214 Jim D,

Visible condensation continues to increase as the bottle crumples. Evaporation is unlikely to be occurring during the collapse, as the air in the bottle was fully saturated prior to condensation. If left at room temperature for 30 minutes the condensation will form droplets that pool at the bottom of the bottle, the bottle is now clear but still crumpled.

It is interesting to note that the crumpled bottle with visible condensation can be placed under hot water and reheated. It then expands to its original size and all visible water and fog inside disappears, having been re-evaporated. Once allowed to cool off again, the fog reappears and the bottle re-crumples.

#214 Jim D,

For a slightly different version of the experiment follow steps 1-3 then,

4. Without squeezing the bottle, replace the cap tightly.

5. With the cap tightly on, squeeze the bottle hard, then release. The rapid expansion causes a sudden and dramatic cloud formation. Note that the bottle has not returned fully to its original size.

If the latent heat released by the condensing vapour was able to heat the air in the bottle sufficiently to over compensate for the volume lost due to condensation, the bottle should return to its original size and feel tight. It clearly does not. I suspect that some of the latent heat of condensation is being lost via radiation and failing to heat the air in the bottle. The effect appears to be too rapid to be a result of the bottle surface cooling through conduction.

218. ### Jim Dsaid

Your experiment is interesting, and certainly made me think. It may be that despite the condensation the gas is still cooling by loss of heat from the bottle, which would cause a pressure drop, or the heat is going to the bottle instead of the gas, giving a higher effective heat capacity, also causing the pressure to drop.
It is hard to relate this to anything happening in a cloud because of the other surfaces present.

219. ### Jim Dsaid

A related experiment is this one
http://scifun.chem.wisc.edu/HOMEEXPTS/COLLAPSE.html

Here it is pure vapor, and the can collapses because the heat must be lost to the cold water. My equations in #110 assume all the latent heat goes into the air, which works for the atmosphere because there is nowhere else for it to go.

The collapsing can experiment yields dramatic results because the total volume is water vapour. It is possible that the drink bottles I tested are crumpling because the ratio of water vapour to air is too high. I will have to look for a better way to create warm saturated air. I may also need a way of insulating the outside of the bottle to prevent heat loss through conduction that is also transparent to infra red radiation.

221. ### anna vsaid

What do you mean, nowhere to go?
The heat can stay on the cloud water droplets which have higher heat capacity than the surrounding air, which water droplets en mass as cloud will eventually radiate as a gray body, as someone observed.

222. ### anna vsaid

Thank you for the confirming different experiment.

If only people used their hands and simple tools and then their eyes instead of just imagination, we would not be in such a mess with climate hypotheses.

I cannot see, in the two bag in freezer experiment, how another explanation of the difference in deflation between saturated and unsaturated air content can be made, other than that condensation reduces pressure.

Yes, I also think that the latent heat is lost through the transparency of air and plastic cover to radiation.

experiment tramps visualizations.

223. ### Nick Stokessaid

Re: anna v (Nov 3 01:27),
Anna,
Neither of these exparimewnts is anything like adiabatic. The proposition you’re supposed to be testing is whether the LH of condensation expands the air more than gas mass loss will contract it. If you remove the heat (freezer?) then of course the gas loss will dominate.

Jim is right to say that in the atmosphere the LH has nowhere else to go but to warm the air. Droplets cannot possibly absorb their own LH of condensation. At LH=2500 kJ/kg, and with a SH of 4.2 kJ/kg/K, they would have to heat to about 500C.

224. ### anna vsaid

I should be correct and use the bridge expression.

experiment trumps visualization

225. ### anna vsaid

If the air could be heated from the latent heat, it would be.

Lets see.

You visualize a parcel of air rising, expanding because of lower pressure ,reaching a temperature T where it is saturated , starts condensing, heats the air and it goes further up. No? Pressure is P1 before condensing and you are saying it goes to P2 where P2>P1.

I have a bag at pressure P where the air reaches a temperature T of the freezer where it is saturated, starts condensing, and deflates much more than the neighboring bag with room air.

If your parcel could heat the air at -20C my bag should heat the air at -18C (freezer minimum) and stay inflated, as the room air bag does .
The freezer simulates the high atmosphere temperatures. If the freezer takes away heat, so does the cold bath of the atmosphere. It is a delta time slice of what is happening as your parcel rises to supersaturation temperatures.

What does adiabatic have to do with it at the moment of condensation ?

The excess energy must be radiated away in both cases, not managing to heat the air much . I cannot think of a simple experiment to check this at the moment.

226. ### Nick Stokessaid

Re: anna v (Nov 3 02:09)
If you create condensation by removing heat, including the latent heat as it is released, then of course the air will contract. That’s actually a variant of the argument in M10, where they say you can’t have adiabatic condensation at constant volume.

You have caused condensation precisely by removing the LH.

But in the atmosphere it’s different. The air is cooled adiabatically, by expansion on rising. Any latent heat released is not removed; it stays in the air and counters, but does not stop, the adiabatic cooling that is causing condensation.

227. ### anna vsaid

I cannot see the physics difference between the cool bath of the atmosphere and the cool bath of the freezer.

I did not remove heat with a pump. I let the bag come to equilibrium in the temperature of the freezer. The rising parcel of air finds a temperature and goes to equilibrium with it otherwise how could condensation happen?

There is a difference between the adiabatic expansion to the temperature, and my hand opening the freezer and putting the bag there, but how can it affect in the delta(time) of condensation the actual physics of the situation.

If the freezer removes heat so does the cold atmosphere bath.

228. ### anna vsaid

So as far as I can see, problem number one has been reduced to problem number two, as we used to say in beginning science courses in my time.

It seems you accept that the pressure drops with condensation, but now posit that the latent heat compensates this pressure drop by heating the air parcel simultaneously with the condensation and because there is a lot of energy in latent heat the pressure should rise and there should be expansion. Somehow you think that the freezer bag experiment removes the latent heat from the air, not allowing this expansion. Is this true?

229. ### Anastassia Makarievasaid

#213 Sky,

Thank you for your further comments. Indeed, it is difficult to convince people even in simple things, to say nothing about such far reaching implications as the impact of the Amazon forest on tropical cyclones. As you can see, so far we have not presented such a proof, so what I am talking about is my perspective coming from my understanding. You say that there are no hurricanes because of oceanic circulation. One cannot deny the existence of abiotic factors. If we presume that oceanic circulation is unaffected by atmospheric circulation, this could be accepted as a logically possible explanation. However, if we accept that the atmospheric and oceanic circulations are interconnected, then, should there been no Amazon forest there, the atmospheric (and, consequently) oceanic circulation would have been different. Regarding hurricanes near Mozambique forest — this forest is much smaller than the Amazon forest. No pressure, just thought sharing.

#203 Nick,

Thank you for the valuable information about the age of the Runge-Kutta methods. I also very much appreciate that you have ultimately pointed out to the missing equation for condensation rate, the 9th equation: it looks to you as Eq. 27 in BF02. I am in no way imposing my opinion, but just as a hint — doesn’t this equation look like being derived from the 1st law, which has been already used in the previous eight equations? I think the citation from Yau and Austin (1979) that I quoted at #175 above might be helpful here.

#215 Jim,

Thank you for appreciation that in the gravity field things can be different.

I am not talking about hydrostatic pressure. The gas law applies even without gravity and is all that is needed to explain, as I did in #110, how pressure increases when condensation happens. Two equations are needed (1) p = r R T and (2) L dq = – cp dT.
Which of these do you dispute? Or do you dispute the constants I used?

Would you mind to clarify where the second equation comes from?

#226 Nick,

where they say you can’t have adiabatic condensation at constant volume.

Just to put the right emphasis: it was Dr. Rosenfeld who stated that adiabatic condensation at constant volume is possible. We just settle the matter straight. Otherwise you do describe things correctly at #226.

#200 Anna,

Great experiment! But as for me, personally I was impressed by the phenomenon of medical fire cupping. This allows to disentangle heat release from mass removal very clearly in the ideal gas law. In short, if you take a rod with cotton, moist it with ethanol, ignite and put into the cup, then quickly place at something that can be sucked into (e.g., human skin), you will see that the skin is sucked into the cup due to a pressure drop. This enhances blood circulation and used to be practiced to cure from pneumonia.

The funniest thing is that if you do not moist the cotton with ethanol, but just ignite the cotton and do the same, there will be no “sucking effect” at all. This crucial point, when the cotton is moistened, is not shown in the video I referred to above. The puzzle consists in the chemical reactions that occur, with one leading to less gaseous constituents, and another to more. Heat release (which definitely occurs) is entirely irrelevant. BE VERY CAREFUL WITH FIRE!!! if one attempts to do perform the experiment. I will have to have a break in extensive commenting for a day or two now, but then I will be able to describe in detail what happens, if it will be needed.

230. ### Anastassia Makarievasaid

#199 DeWitte,

At 5 kPa, only 6.4 g water has condensed.

Why did you take 5 kPa as a reference? In the atmosphere, the air parcel rises to several kilometers before reversing the horizontal direction of its movement, such that the pressure fall of relevance is in the range of several dozen kPa.

231. ### anna vsaid

Going back to the adiabatic argument, which is the only one that might introduce a difference with atmospheric conditions.
You are saying that as the air parcel rises, its volume grows due to the drop in outside pressure and thus the temperature falls from expansion within the parcel; the difference with the freezer bag is that the bag comes into equilibrium with an outside temperature .

At delta(time) the mass of air has no way of knowing whether it is cold because of radiative equilibrium or it is cold by adiabatic expansion.

The only thing I can see when one claims adiabatic processes that could affect the argument is
from wiki:
A transformation of a thermodynamic system can be considered adiabatic when it is quick enough that no significant heat is transferred between the system and the outside. (bold mine)
i.e.an adiabatic system acts like a closed system as far as heat/energy transfer is concerned for delta(time).

I have great difficulty in thinking that a parcel of rising air in an atmospheric bath is a closed system, and can trap latent heat, considering convection and radiation let alone precipitation at the instant of condensation.

232. ### Brian Hsaid

Re: Nick Stokes (Nov 2 18:49),
I would suspect a simple density/lift loss. The internal (column) turbulence and wind inflow suggests rapid re-balancing of a number of variables, including humidity, mass, and density, all more or less via pressure changes, including partial pressure changes.

233. ### Nick Stokessaid

Anna,
The notion of adiabatic cooling is fundamental here. It means that the air cools with no (significant) heat moving anywhere (over the timescale). That means that LH, when released, is not removed.

Here are some lecture notes on the concept. They address the issue of condensation thus:
“So, if condensation is taking place, latent heat is being released to the surrounding air. So you have two opposing trends going on at the same time within this parcel of air. It’s rising and cooling, but it’s also condensing and being warmed. Which one will win out? That is, will the air get colder, or will it get warmer?

Well, what happens is that the air will still cool off, but not as fast. If water vapor in the air is condensing, the adiabatic rate is less. The air is only cooling off at a rate of about 5 degrees C/1000 meters (2.7 deg per 1000 feet). This is called the Saturated adiabatic lapse rate (or the wet adiabatic lapse rate, or the moist adiabatic lapse rate, depending on the textbook you are using). The saturated lapse rate varies with the original temperature of the air parcel, but 5 degrees C/1000 meters is a commonly used value.”

In your freezer, you achieve condensation by conduction etc across the boundary – not adiabatic. It is still the case that LH slows the process down, but you don’t see that, since you only look at the final state, when the heat has gone, and the reduced number of gas molescules is the only effect remaining.

234. ### Brian Hsaid

No, that can’t be a problem. Saturated is saturated; unless you have live steam =>100°C in there with the air.

235. ### HLxsaid

Heat of condensation:

When H_2O(g) condenses to H_2O(l), which form does the heat release take?
Is there a factor of convection in there, or is the majority of the heat released by IR radiation?
If heat release is mostly IR(>>), to which degree does these frequencies get absorbed by O_2, CO_2 and so on?

HLx

236. ### Nick Stokessaid

Re: Anastassia Makarieva (Nov 3 03:06),
No, Eq 27 introduces the dependency of sat wv ratio on pressure and temperature. They predict at the end of a time step, P and T as if precipitation had continued unchanged. Then, using standard tables or a formula, they work out the new $r_{vs}$ (sat vp ratio). That let’s them compute how much extra precipitation there must have been to balance. Then they recompute using this new precipitation estimate. And so on, iterating, until there is no discrepancy. They say that takes 4-6 iterations, and converges to machine accuracy.

I’m not sure what point you are making re Yau and Austin. The formula quoted is the same as (27). The reason Rutledge and Hobbs worried about a Δt dependence seems to be that they didn’t iterate. If you iterate to convergence there should be no such dependence. I note also that when they speak of microphysics in their different situation, they have snow and ice as well as condensation.

The reason Eq 20 for velocity is not used in the microphysics update is simple; there is no direct dependence on condensation rate. Not updating doesn’t negate feedback; it simply means that the feedback is delayed by a fraction of a timestep. Since the timestep is short, being dictated by the speed of sound, this would be insignificant.

237. ### Nick Stokessaid

Re: HLx (Nov 3 04:59)
Condensation happens on a microscopic scale, where conduction dominates. The heat is transferred by molecular collisions to the air.

238. ### Nick Stokessaid

Re: Anastassia Makarieva (Nov 3 03:06),
“The puzzle consists in the chemical reactions that occur, with one leading to less gaseous constituents, and another to more.”

I think it’s just rapid cooling of the gas.

239. ### anna vsaid

In your freezer, you achieve condensation by conduction etc across the boundary – not adiabatic. It is still the case that LH slows the process down, but you don’t see that, since you only look at the final state, when the heat has gone, and the reduced number of gas molescules is the only effect remaining.

But is not that, final state the one that is observed? i.e. rain falling air pressure falling generating winds etc as Anastasia says?

Condensation happens on a microscopic scale, where conduction dominates. The heat is transferred by molecular collisions to the air.

The air is a very bad conductor, and in addition up less dense. Somebody has given enormous numbers for latent heat and pointed out that the droplet could not hold it without evavorating, what collisions by which etc. The only logic is in radiation, so the question is what is the infrared spectrum of the energy released when H2O bonds with another H2O, and how much it is within the absorption spectrum of nitrogen and oxygen.

240. ### Nick Stokessaid

But is not that, final state
No, not in the atmosphere. We’re talking about dynamics. Note the definition of adiabatic that you quoted – “quick enough”. In the long run nothing is adiabatic. But by then the clouds have blown away.

The air is a very bad conductor
The thermal diffusivity of air is 2.2e-5 m^2/s. That’s not exceptionally low, but means that a sphere (of air) radius R in still air would have a cooling time constant of about R^2/2.2e-5 sec. With R=1 m, that’s about 50000 secs. But with R=1 μm it’s about 50 nanosecs. Conduction really scales up as things get smaller.

241. ### Anastassia Makarievasaid

#236 Nick,

No, Eq 27 introduces the dependency of sat wv ratio on pressure and temperature.

Does it really? It reads to me that Eq. 27 introduces a new variable, saturated vapor mixing ratio, r_{vs}. To calculate its magnitude, one uses the Clausius-Clapeyron law (standard tables, as you call it). But Eq. 27 itself is based on a very different assumption. The quote from Yau and Austin (1979) at #175 explains very nicely which one. It should look familiar at this point!

242. ### Thierrysaid

Anastassia you were right.

I have made a rather extensive reference research about condensation rate equations.
I checked some 35 papers, some of them you have already quoted.
All references to condensation rate I found constitute a chain where people quote each other but it always finishes at T.Asai 1965.
So it seems that after T.Asai nobody really looked at the problem and certainly not from the theoretical point of view.
As the T.Asai paper cannot be found on the Net, I have found one from 1975 that uses what is presumably the “Asai equation” (http://onlinelibrary.wiley.com/doi/10.1111/j.2153-3490.1976.tb00672.x/pdf).
The abstract is even given in Russian 🙂

I quote :
“M=alpha.(epsilon.Wcb + 1).(D – k²) where :
M is the condensation rate and Wcb is vertical velocity at the cloud base.
The other undefined quantities (alpha, epsilon, D and k) are model parameters.”

243. ### Nick Stokessaid

r_vs is determined by C-C, so OK, if you want to call it a new variable, then it comes with its own equation.

I do not see your point about Yau, but there is no point in hinting about it. If you think BF02’s use of the equation is wrong in these circumstances, you should say why. My point at this stage is to say that they have enough equations to solve their system.

244. ### DeWitt Paynesaid

There are all sorts of papers on expansion cloud chambers in the literature. Unfortunately, most of them are pay walled. However I did find one that wasn’t. It has transient temperature and pressure data. The volume of the chamber is changed over the course of a few milliseconds. See Figure 5 for an expansion with moisture present. As expected, the temperature drops less than for a dry expansion. The droplets form rapidly. The temperature response is a little slow. But if you look at the figure 3 for a dry expansion, it appears that the temperature measuring device has a fairly slow time constant relative to the time of the experiment. The device isn’t perfectly adiabatic. For the fastest rate of expansion, the experimental gamma value, which they call the polytropic coefficient, is 1.33 instead of 1.40. There does not appear to be any evidence of an initial temperature overshoot and recovery. One would need a very clean chamber with no condensation nuclei present to see that.

245. ### Anastassia Makarievasaid

#243 Nick,

I do not see your point about Yau, but there is no point in hinting about it.

I am really doing all I can. Right now I am running out of time and cannot deliver detailed comments. My point about Yau, see also Thierry’s remark above, as all it goes to Asai (1965), is very simple.

BF02 in their system of equations (21)-(24) use a “normal” version of the 1st law: 0 = c_vdT + pdV +Ldq. In their Eq. (27), as an equation for condensation rate is missing — let me stress it once again, IT IS MISSING! — they use the 1st law for adiabatic condensation at constant volume (or constant pressure) instead, an idea proposed by Asai in 1965: 0 = c_vdT + Ldq or 0 = c_pdT + Ldq. This is where Eq. 27 stems from and Yau and Austin explain it very clearly. First pressure change Vdp is zeroed in Eq. 27, then it is resurrected again in Eqs. (21)-(24).

This is what is meant by the phrase of Yau: “A part of $\delta M$ [saturation excess], $\delta M_1$, is condensed releasing latent heat which causes an increase of temperature; in turn, this permits the residue, $\delta M_2 = \delta M - \delta M_1$, to remain in vapor form.

At #187 I wrote:
“But how is it possible? Just at the logical level, how is it possible, by whatever numerical solutions and manipulations, to solve a system of equations where there are more variables than constraints on them? This is only possible if there is a hidden controversy in the calculations, when one first sets something to zero (this is the needed extra equation), solves the system, and then puts the solution into it which is already not zero. This is what is being done, it is implicitly assumed that condensation rate has no impact on dynamics. All updates are made using this assumption as the core.”

Now the controversy has been pointed out very clearly. It has severe consequences for all conclusions that are made based on such models (and it seems that all the existing models are based on the same approach). As I said, it is a grandiose physical chaos.

246. ### Terrysaid

Anastassia #245,

Are you saying that (at the small time step scale) when adiabatic expansion occurs at either saturated or supersaturated conditions, with the attendant condensation and temperature increase, the starting point for the next iteration on the same parcel of air does not take account of the loss of moisture in the first iteration, ie the conditions are the same as for the first iteration ?

247. ### Nick Stokessaid

Re: Anastassia Makarieva (Nov 3 12:42)
I’m sorry, I cannot understand this latest “controversy” at all. But I too will be away for two days.

248. ### anna vsaid

medical fire cupping was a popular method for treating pleurisy and colds, regularly prescribed before the advent of antibiotics. There was also a version of “cut cupping”, where cuts were made where the cup was applied and blood was drawn.

I will have to have a break in extensive commenting for a day or two now, but then I will be able to describe in detail what happens, if it will be needed.

I will read it with interest, if you have the time to take the trouble.

249. ### DeWitt Paynesaid

I don’t understand either. If one calculates an adiabatic curve around a point with known temperature, pressure and water vapor mixing ratio, doesn’t the condensation rate fall out of that for any given vertical velocity? I’m not convinced that the atmosphere anywhere has so few CCN’s that condensation won’t be diffusion limited.

250. ### HLxsaid

#239:

Anna V
“The air is a very bad conductor, and in addition up less dense. Somebody has given enormous numbers for latent heat and pointed out that the droplet could not hold it without evavorating, what collisions by which etc. The only logic is in radiation, so the question is what is the infrared spectrum of the energy released when H2O bonds with another H2O, and how much it is within the absorption spectrum of nitrogen and oxygen.”

Yeah, we’re thinking in the same line. But, I cant find anything on the subject on the interwebz :(..

Im seriously thinking of looking into my old solid state physics and thermodynamics books …

251. ### skysaid

Anastasia #229:

Atmospheric circulation is, of course, coupled to the oceanic. In fact, it is the winds that drive the currents. The source of thermal energy in cyclogenesis, however, is in the surface waters. High-enough SSTs are a necessary condition.

The extent of rainforest within Mozambique is indeed very much smaller than in the Amazon Basin. But vegetation knows no political borders. A very broad swath of tropical forest extends across all of equatorial Africa, from Gabon, across the Congo Basin, to the mouth of the Zambesi and the eastern slopes of Madagascar, only rarely interrupted by savanna in the highlands. Thus we have an analogous biotic situation on a continental scale, but much different weather regimes.

I simply cannot take more time to comment here. Spasiba for an intelligent discussion. And keep the true believers honest!

252. ### DeWitt Paynesaid

What absorption spectrum of oxygen and nitrogen? They’re effectively transparent in the IR. Why invoke radiation? A molecule of gas at surface pressure and temperature collides with another molecule every few nanoseconds. The number of collisions/second is going to increase exponentially with the size of the droplet. Until the droplet gets big enough that internal conductivity starts to be a problem, it will thermalize to the temperature of the surrounding gas molecules on a very short time scale. Water vapor is a small fraction of the total gas so there are many more collisions with oxygen and nitrogen molecules than with water molecules that release energy by bonding with the droplet.

253. ### Anastassia Makarievasaid

#246 Terry

Are you saying that (at the small time step scale) when adiabatic expansion occurs at either saturated or supersaturated conditions, with the attendant condensation and temperature increase, the starting point for the next iteration on the same parcel of air does not take account of the loss of moisture in the first iteration, ie the conditions are the same as for the first iteration ?

In Eq. 27 of BF02 pressure vapor pv (mixing ratio rv), pressure p and temperature T are calculated first by forcing eqs. 21-24 by one step forward and assuming no phase transitions. This is vapor pressure, pressure and temperature that would have been observed if vapor were non-condensable. Than, for this p and T, saturated pressure and mixing ratio r_{vs} is calculated. But how to find how much of it will condense? Condensation WILL influence both the resulting temperature and pressure. We cannot say that simply r_v – r_{vs} will condense. So the idea behind Eq. 27 is to presume that PRESSURE will not change due to condensation (Vdp = 0), but only temperature will. So for the condensation process it is presumed that Vdp = 0 and 0 = c_pdT + Ldq, while for all other processes it is conventionally held that 0 = c_vdT + pdV + Ldq.

It is not surprising that a study (studieS) based on such a (very well hidden, burried in Asai (1965)) premise ultimately concludes that condensation (mass removal) has no impact on dynamics. Indeed, if we put the condensation-induced Vdp equal to zero from the very beginning, it could not. It is far more remarkable that given the condensation dynamics ideas have been around since 2007 (and, as confirmed by documented evidence, very well known to many very well known climate modelers), this did not spur any reaction from the community aimed at a re-analysis of the basic physics behind the existing models. I think that this phenomenon in the coming future is likely to receive a very detailed consideration from different angles.

254. ### Jim Dsaid

cp dT = -L dq

This is the statement that energy is conserved within the system by all the latent heat going into heating air at constant pressure. The reversibility or energy conservation is important to the definition of an adiabatic process. Cloud parcels follow such processes, and the idea can be used to predict cloud tops accurately.

The next question is why the energy can’t go anywhere, e.g. by radiation. Yes, radiation may influence cloud edges a little, but in the updraft cores, it is completely saturated and radiation does not get very far in those circumstances, besides which we are talking about small temperature differences locally, so the net radiation effect is close to zero.

Once again, condensation has to lead to a pressure rise (as shown in my #110), otherwise it would not lead to positive buoyancy (see my #85 for this chain of argument), and clouds just would not exist for long. Disputing the pressure rise is equivalent to disputing the positive buoyancy. If you have flown through convective clouds, you would experience the buoyancy for yourselves as an upward bump.

255. ### Anastassia Makarievasaid

#254 Jim,

cp dT = -L dq

This is the statement that energy is conserved within the system by all the latent heat going into heating air at constant pressure. The reversibility or energy conservation is important to the definition of an adiabatic process. Cloud parcels follow such processes, and the idea can be used to predict cloud tops accurately.

Indeed. We are interested to know what sort of pressure difference is produced by condensation. To answer this question, we set the pressure constant. Cf. my #253 above. This is an answer to Terry #246.

256. ### Jim Dsaid

#253 Anastassia
The condensation is a two-step process in these models. First the air ascends due to the dynamics leading to pressure reduction (following the air), adiabatic expansion, and cooling. This leaves it very slightly supersaturated. Then condensation adjustment returns it to 100% humidity by removing vapor and adding heat, conserving energy. This is the way the model approximates natural cloudy ascent at 100%, and it is very accurate with the typical time-steps used. You can see that the pressure, temperature, and vapor all change within a time-step.

257. ### DeWitt Paynesaid

Since the atmosphere is not confined, shouldn’t you be working at constant pressure rather than constant volume? Then condensation would lead to a volume increase and lower density = increased buoyancy.

I remember Feynman talking about how physicists prefer the thermodynamic equations holding volume constant and let the pressure change because the equations in that form are more elegant while chemists hold pressure constant and let the volume change because most chemical reactions are carried in apparatus open to the atmosphere.

258. ### Anastassia Makarievasaid

#256 Jim,

Then condensation adjustment returns it to 100% humidity by removing vapor and adding heat, conserving energy.

I have not had much interest in models. I thought that it will be enough to tell people about new physics, scientists will get excited and will study things further themselves. This has not happened. We continue to go along our path alone and, as to ourselves, are OK with this. Science rewards by itself.

But, encouraged by Gavin’s and Nick’s questions, I did have a look at particular models. Indeed, Jim, everything is as you say. First a full version of the first law is applied, 0 = c_vdT + pdV + Ldq, then, for condensation rate, an abridged one: 0 = c_pdT + Ldq. The system is solved simultaneously. The impact of condensation on pressure is ignored. Pressure IS NOT changed when the condensation rate is calculated.

259. ### Jim Dsaid

#257 DeWitt
cp is used, not cv, so it is taken to be constant pressure for the second part of the calculation (in my #256).

260. ### Jim Dsaid

#258 Anastassia,
This is where the model types differ. Hydrostatic models do the calculation as I say, then since temperature has increased, the gas law reduces the density, leading to the buoyancy effect in the column. It shortcuts the pressure increase that causes the density to reduce, which is the effect of assuming sound waves are infinitely fast. Nonhydrostatic compressible models have the full equations with sound waves where heating causes a pressure increase that causes expansion that causes the density to decrease (as I outlined before). You only see the pressure increase with heating in this type of model (or in the real atmosphere), but it is a very temporary situation before the density decreases, and the pressure restores to ambient values.

261. ### Anastassia Makarievasaid

#260 Jim,

Nonhydrostatic compressible models have the full equations with sound waves where heating causes a pressure increase that causes expansion that causes the density to decrease (as I outlined before).

Let us be clear: the matter is NOT about hydrostatic equilibrium. It is about how one calculates condensation rate. As you very well clarified, moist adiabatic ascent is conceptualized as two separate processes: (1) dynamic process with no phase transitions and (2) condensation per se (condensation adjustment).

The first process is allowed to change pressure: 0 = cpdT – Vdp + Ldq. Note that here the change of vdP is not specified, it can be either hydrostatic or non-hydrostatic. The second process, condensation rate, is not allowed to change pressure a priori, 0 = c_pdT + Ldq.

This second process is axiomatically reduced to heating, with mass removal ignored completely. Accordingly, the non-hydrostatic models consider sound waves caused by heating. They, as well as hydrostatic one, ignore sound waves (and everything else) caused by the mass removal.

262. ### Brian Hsaid

Anastassia;
You just inspired an image or POV: rain clouds are a trick the atmosphere uses to transport a mass of water from one patch of surface to another. 😉

263. ### Anastassia Makarievasaid

#262 Brian,

what is POV? Speaking of images, I am a fan of cartoons by Josh. I think that our case makes a very good plot for a cartoon. Several authors are trying to advance a piece of physics to the meteorological literature. Practically nobody speaks to them, except for ANONYMOUS REFEREES.

I imagine this as naked, shy, embarassed authors placed on a round podium (a la scaffold) holding their manuscript in hand and trying to shield themselves by it. The scaffold is surrounded by the ANONYMOUS REFEREES (people in black with black glasses), they are holding spears and poke them at the authors. An essential detail is that the authors have bandages on their eyes, such that they cannot see who is attacking them (even if they would not be able to recognize them anyway).

A thrilling experience, believe me.

264. ### Brian Hsaid

Re: Anastassia Makarieva (Nov 4 02:08),
Point Of View. A way of looking at or interpreting something.
You might find this site useful for understanding all the acronyms thrown around on the Web:
http://www.netlingo.com/acronyms.php

For example, instead of saying “grandiose physical havoc”, just use FUM!

;p

265. ### kimsaid

Round and round the gauntlet carousel goes,
Where it stops only the God of Physics knows.
=======================

266. ### anna vsaid

Do not be discouraged, you are chipping away at a POV ( point of view) that has been prevalent in climate science, so that they come out with statements “everybody knows”, “it is generally accepted”, because within their schools that is the way they have been taught, to work with virtual realities. Somehow, because when working with models physics is what the modeler thinks physics is, they think physics obeys models, and not that models should obey physical laws. I have had people coming back at me with comments like “the physics of the model…”!

And actually, this: “It is generally accepted that the parcel of air does this and such” is a statement that needs clear proof from data. I find it very strange that in an environment with convection, coriollis forces, radiation, atmospheric tides, highs and lows etc one can claim that “generally a parcel of air follows an adiabatic curve”. How probable is it that it will not be disturbed and heat transferred in and out of it?

267. ### Dan Hughessaid

This is interesting; General relationships between pressure, weight, and mass of a hydrostatic fluid.

268. ### DeWitt Paynesaid

Aren’t you leaving out the step where the final volume of the packet is calculated? Once you know the final temperature and how much water has condensed, the final volume is the volume of the dry gas at the temperature and pressure specified plus the volume of the remaining water vapor at the specified temperature and pressure plus the volume of liquid water. I see nothing that says the models calculate the final volume using the water vapor amount before condensation. Unless I’m missing something, that would mean that mass removal is not ignored.

269. ### curioussaid

263, 266 Anastassia and Anna v – I too share the concern over the adiabatic assumption and the idea that within a hurricane there are discrete packets of air whose passage through the “system” can be tracked and modelled. I also have reservations about the limits to some of the underlying assumptions – for example my memeory of Bernoulli is that it is essentially conservation of energy along a streamline and that it is only truly valid for inviscid flow. I do not think this is the case in dynamic air and water systems. However I do not get the time to go back and get properly up to speed on things I haven’t looked at for many years, nor have I read the literature on this, so I put a lot of store by the open and cooperative discussions which take place at tAV. There are clearly a bunch of very bright and experienced people who comment and read here and I think this thread is a great example of what open unmoderated blogging can be. As an author of a paper it is admirable for you to put up your work for open review and comment – yes, criticism may be painful but as an observer I do not see anything personal or malicious in the debate. People are swapping and supporting strongly held views and, in the face of evidence and argument, positions are reconsidered. Stick with it – IMO understanding is advancing as a result of your work. Best wishes C

270. ### curioussaid

220 Konrad – nice experiment. I wonder if insulating the bottle with foil based insulation product such as “thermawrap” would be worth a try? I guess you’d need to do it with a small inspection hole (or two). The other thing that crossed my mind is – can you do the experiment in a sauna?! I might have to renew my sports club membership… 🙂

271. ### Thomassaid

#266 Anna V

here people keep talking about data and evil evil models. but Anastassia Makarieva did not show any (!) data that supports her claims.

I tried to explain that models do not live some where out there disconnected from reality. they are used to forecast the weather on a daily basis and verified against data! and even though forecasts are not perfect they show tremendous skill (500 hPa geopotential for example). today there is even skill 10 days ahead. this is VERY strong evidence, that they are doing something right. so Anastassia Makarieva claim that the whole atmospheric circulation is governed by an effect that is completely absent in the models is very unrealistic. also that she presents a theory for hurricanes AND tornados is worrying. these are very different atmospheric phenomenons. for a first overview I recommend the book Cloud Dynamics by Robert Houze.

272. ### Jim Dsaid

#261 Anastassia
There is a so-called “virtual” temperature effect that accounts for mass removal. Virtual temperature (Tv) includes the contribution of vapor to the effective density. When vapor is lost, virtual temperature reduces, and it is like the effect of cooling on density, so indeed this is accounted for in all types of models with moisture.

Regarding #263
What would you expect to happen if someone said something provably wrong in a paper, and not only that, but claimed all previous papers and textbooks were wrong? I think it would look very much like this. If you found a previous textbook in meteorology you agree with, I would very much like to know what it is, but I think your viewpoints don’t agree with the basic textbooks, even chapter 1 in most cases.

273. ### Anastassia Makarievasaid

#268 DeWitt,

Once you know the final temperature and how much water has condensed, the final volume is the volume of the dry gas at the temperature and pressure specified plus the volume of the remaining water vapor at the specified temperature and pressure plus the volume of liquid water.

You need to determine first how much water has condensed. You do not know that. For this reason you assume Vdp = 0. This is a constraint alike what you were trying to involve at #199. You cannot go without a constraint. If you did not put Vdp = 0, the amount of condensed moisture would have been different.

Moist adiabat is calculated from the 1st law and hydrostatic equilibrium. This does not help determining how much vapor condenses in dynamics. Mass removal in the rising air parcel continuously disturbs the equilibrium. In order to satisfy the equilibrium conditions and to be able to use the moist adiabat, you need to invoke a process that nevertheless keeps the column close to equilibrium despite mass removal. This process is contraction caused by mass removal. It is associated with change of pressure and work performed on the column. By setting Vdp = 0 in calculating condensation rate this work is ignored.

#272 Jim,

There is a so-called “virtual” temperature effect that accounts for mass removal. Virtual temperature (Tv) includes the contribution of vapor to the effective density. When vapor is lost, virtual temperature reduces, and it is like the effect of cooling on density, so indeed this is accounted for in all types of models with moisture.

In your previous comments you explained very well and clearly how the effect of condensation is taken into account. Did you forget about the effects of “virtual” temperature then? I do not think so. All these density effects are included into recalculation of mixing ratios. One does not need to refer to specific terms that people here may not understand to explain things. Virtual temperature simply accounts for the difference in molar masses between dry air and water vapor. It says that if you remove vapor, the molar mass of air will be higher. However, molar mass IS NOT PRESSURE and even not density.

Ideal gas law p = NRT, where N is molar density. Density rho = N*M, where M is molar mass, such that rho = pM/RT. All “virtual temperature” says about is that when there is less vapor, M is higher. This has nothing to do with mass removal, which primarily affects total pressure p, not M. The “virtual temperature” effect would have been absent if dry air Md and vapor Mv had equal molar masses. The effect of mass removal is unrelated to this difference.

What would you expect to happen if someone said something provably wrong in a paper, and not only that, but claimed all previous papers and textbooks were wrong?

I would expect a critical re-thinking of all previous textbooks and a serious discussion. Instead of handwaiving statements that we have everything in models. For example, in this thread at #215 you first claimed that condensation raises air temperature because of cpdT = – Ldq. When I pointed out at #258 that this equation has no physical grounds but is used as an axiom by putting Vdp = 0, you switched the argument to hydrostatic versus non-hydrostatic models at #260. When I explained that this is not an issue here, you have now put forward a totally different and equally irrelevant argument about virtual temperature. Earlier you claimed that we are proposing an unrealistic physical feedback, and insisted twice that it is unrealistic, — but after I pointed out at #187 that meteorological people also spoke of this feedback, you immediately withdrew this argument and I never heard of it again.

To summarize, through all this discussion, I am continuously dissecting your arguments in a very detailed manner. From your side, you do not make any attempt to look into the physics that I am proposing. For example, you persistently ignore the point that a temperature change, whatever it is, cannot compensate for a loss of mass.

However, even in this non-reciprocal form, our dialogue here is FAR MORE civilized than the peer-review process, which would have ended at Ldq = -cpdT (as per Dr. Rosenfeld’s calculations). Here many people can see and follow the arguments.

274. ### anna vsaid

There is a so-called “virtual” temperature effect that accounts for mass removal. Virtual temperature (Tv) includes the contribution of vapor to the effective density.

virtual reality?

What would you expect to happen if someone said something provably wrong in a paper

virtual proof?

Having interacted with theorists in my field for many years, I have found some who used “provably” when they meant “I say so”. Fortunately theories are based in solid mathematics, and mathematical tools offer clear proof of failure. Then, once the theory does not have mathematical holes, it confronts experiment and falls or survives on the test.

Meteorology, which is now strongly identified with modeling in complicated programs, cannot be confused with theory because it has some elements of theory in the construction of the programs, and it still has to confront reality. Yes, the forecasts have improved a lot from before computer times when one had to watch the highs and lows moving across maps.

But what are these computer programs? They take a snapshot every delta(t) solving for the variables of approximations to the solutions of highly nonlinear coupled differential equations. There are many parameters, and as I think von Neuman said: “give me four parameters and I can fit an elephant, with one more he can flap his ears”. By this I mean that the skill of the forecasts in no way proves that the underlying equations are correct. It proves that for a certain number of time steps the simulation is good enough to forsee, not always, the natural path. Just that. It is very useful but in no way substitutes for a mathematical proof because of the plethora of parameters fitted to reality.

275. ### Anastassia Makarievasaid

#274 Anna,

virtual reality?

Thank you for raising this. I was about to write that it is not by accident that the (major) effect of mass removal has been overlooked in atmospheric science. The formulations used in many textbooks are often unphysical and disencourage clear physical thinking in students. For example, the fundamental ideal gas law says that pressure does NOT depend on molar mass of gases, p = NRT, with R being the UNIVERSAL gas constant.

This law in meteorology is re-written via virtual temperature (which depends on molar masses) and individual gas constants for each gas (Rd for dry air, Rv for vapor). This wipes the fundamental message out of the law completely.

Regarding temperature, there are virtual temperature, potential temperature, equivalent temperature, and all of their mixes. In physics one is taught to respect only one temperature — the absolute one. It is the absolute temperature that enters all physical laws. When a student’s thinking is shaped in terms of artificial concepts, he can no longer see the real world, neither explore it.

The many definitions of temperature in meteorology, among which the absolute temperature is simply lost or takes a very tiny place, remind me of a very famous quote from a Russian writer Mikhail Bulgakov. Under Soviet times, the (bad) quality of goods in shops was a continous topic of jokes and sarcasms. In the novel I’ll quote, the hero did not like the quality of fish that he was served in a shop. To excuse himself, the man who was selling the fish told him that “the fish was delivered in the state of second freshness“.

`Second freshness – that’s what is nonsense! There is only one freshness – the first – and it is also the last. And if sturgeon is of the second freshness, that means it is simply rotten.’

276. ### Jim Dsaid

#272 Anastassia
I now have a long list of arguments against the idea that condensation causes contraction. I have tried several routes to the understanding to see what you agree with in basic thermodynamics and meteorology, but so far haven’t found anything you would use as a foundation. This makes the argument very difficult, because in science we start arguments with things we can agree on, and go from there, but it seems impossible to even start here.
Let’s try this: Vdp=0 assumes pressure doesn’t change. OK, so we can have L dq = – cp dT in this situation. This is fine so far.
So condensation causes dq to decrease and temperature to increase at constant pressure. This we can all agree on as long as we hold the pressure constant, which is the situation where you get work pdV expanding the heated air.
On the other hand, we could have done the condensation in a box, and had L dq = – cv dT with dV=0
In this situation you get T increasing at constant volume. What happens to the pressure? You have to look at PV=nRT. Now you are going to say that the decrease in n more than compensates the increase in T, which is provably not true, as I showed in #110. I think we agree on PV=nRT at least?

277. ### Brian Hsaid

Heh. Here in Vancouver, BC, Can., we’re currently having quite an amusing Gong Show of forecasts and revisions. Days forecast as overcast with occasional rain turn out sunny, sunny forecast days move half a week downstream and then vanish, temperatures are up to 5°C incorrect, etc.

We’re used to it, though; we happen to be on a strait sandwiched between a huge mountainous Pacific island and a major inland continental range with portions right near the water, pierced by major river valleys. A true playground for flutterbys in Hong Kong!

How they must laugh as they flap their little wings and mess with our modelers’ minds …

278. ### Anastassia Makarievasaid

#272 Jim,

Indeed, let us agree upon something. What do you propose? We take a closed box with moist saturated air and watch condensation inside it at constant pressure? But what will cause this condensation? Let us agree on this first.

279. ### Jim Dsaid

Anna V, your disagreement with these aspects of meteorology amounts to disagreement with thermodynamics. You are entitled to disbelieve thermodynamics. As long as you’re not one of those engineers who has to use it for their job, you’re fine.

280. ### Jim Dsaid

#278 Anastassia
You would need to cool it diabatically somehow, so your box would have to be a refrigerator. The question would be whether dry air or moist condensing air end up with higher pressure after the same heat has been removed from them. It will be the condensing air due to the latent heat release.

281. ### anna vsaid

Anna V, your disagreement with these aspects of meteorology amounts to disagreement with thermodynamics.

I just described to you how thermodynamics, real theory, is many approximation levels removed in the meteorology programs. Many parameters that can be fitted and non linear solutions to non linear coupled differential equations. If you cannot see this, how can one argue?

Any function can be expanded into “constant+a*x+b(*x^2+….. If the solutions of these differential equations are not symmetric, and I see no reason they could be in this chaotic system, the meteorology programs take the linear term wherever they substitute an average in the numerical calculations, and a few terms at most for variable progress. Non linearity means that after a number of time steps, no matter how good the original fit for “a”‘s is, the true solutions diverge from the modeling because the higher order terms kick in.

Thus, you cannot use the model as proof that meteorology is using thermodynamics correctly. You might be able to do that, my theoretical background is not strong enough to say a definite positive, if the models gave accurate within errors predictions for all times and spaces.

btw my specialty is experimental particle physics, somewhat removed from daily thermodynamics problems, but not from modeling.

282. ### anna vsaid

My non adiabatic two plastic bag experiment showed that the dry air retained the pressure while the humid deflated.

Also Konrad’s bottle experiment shows lowering of pressure.

So it is the “adiabatic” that will turn the pressure up, i.e. that the system does not exchange energy with its surroundings.

283. ### curioussaid

Apologies if this is rather basic but I’m wondering if anyone has any references for any “error bands” on the adiabatic assumption – ie size/composition of parcel and the scale of time step in a fast moving system air/moisture system over which it holds true to say 95% of the ideal case:

http://www.theweatherprediction.com/habyhints2/501/

271 – Thomas, thanks reference on clouds noted: another one for the reading list!

284. ### Thomassaid

Anastassia Makarieva, Anna V, Brian H:

I am convinced now that what you say is right! We really don’t need any stupid computer programs and we really need to go back to basic theory!

so uhm just one last question: could you quickly calculate how the weather will be in 3 days in wien? will it be raining?
if it rains we will have a cyclone?

thanks guys!

285. ### DouglasinUgandasaid

Great to see how this discussion is going.
#284 Thomas
Paradigms can and should be questioned from time to time — that is what science is supposed to do when alternatives can be coherently presented. Finding common language between the groups who see the problem only from within versus outside the paradigm is going to be hard.

The ancient astronomers were able to add their epicycles and predict the locations of the planets convincingly enough (at east to themselves) without noting that the Earth went around the sun. In retrospect we know where that went.

It’d be great if Anastassia and co’s physical mechanisms could be put in a simple weather model to see how well that might work. That might help bridge a communication gap. But someone has to be willing and able to do that.

Best wishes
Douglas

286. ### anna vsaid

Weather programs are extremely useful. They are not theory, they are an application with approximations to the theory and if the true errors were known, the degree of confidence to the predictions could be gauged. That is very valuable.

As Douglas’ epicycles example warns, given enough parameters a model can accommodate data.

Yes, it would be good to have Anastasia’s version in a model and see what comes out. Maybe much more accurate predictions?

287. ### Jeff Idsaid

Anastassia,

I like this:

“Regarding temperature, there are virtual temperature, potential temperature, equivalent temperature, and all of their mixes. In physics one is taught to respect only one temperature — the absolute one. It is the absolute temperature that enters all physical laws. When a student’s thinking is shaped in terms of artificial concepts, he can no longer see the real world, neither explore it.”

In school, what got me through was understanding of the physical meaning of equations. Others would struggle with what seemed simple sometimes because of the way equations were taught.

On another topic, I have still not read all the commentary here because of the time involved. I have read quite a few of the comments and often have found myself stuck in one paper or another for an hour. Regarding hurricane models, it is not clear to me whether pressure is being updated. I see Anastassia’s claim but if the mass removal is being done, it should be a one liner in the program to update pressures. mv + md, T etc… Not having access to models, I was rather hoping Gavin or someone would clear it up.

288. ### Anastassia Makarievasaid

So far I have delivered my arguments to convince the reader that there aren’t any physical grounds behind the claim of Drs. Emanuel, Bryan and Rotunno that the mass removal effect of condensation on atmospheric dynamics is minor. 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. The model of Bryan and Rotunno (2009) (and the entire family of related models) involves a mathematical and physical controversy, when the equations of hydrodynamics are solved assuming simultaneously Vdp = 0 and Vdp $\ne$ 0. See my comments #175, #187 and #245 above on this topic.

I will now show, to keep up with my promises that I first voiced at #139, how it is nevertheless possible, despite neglecting the main physical mechanism, to fit the model to reality based on consideration of latent heat release.

When there is differential heating, there is a pressure gradient. Heat is added to gas, its temperature rises and pressure follows this rise (if nothing else happens). However, each gas volume not only receives heat from a local source (e.g., latent heat release from condensation), but also loses/gains heat via thermal conductivity. If thermal conductivity is large, it can erase the effect of differential heating nearly completely.

Let me emphasize this point, as it is key: if thermal conductivity is large, the pressure gradient produced by differential heating is small. Thermal conductivity represents the inevitable losses that are associated with any process where one attempts to obtain work from heat.

So, if we consider a theory that attempts to explain the hurricane power based on consideration of latent heat release, this theory must evaluate the losses to heat conductivity from the first principles. If, after such an account of thermal conductivity, the differential heating effect will remain sufficiently large (i.e., associated with a sufficiently large pressure gradient), we can say that the theory has explained the hurricane. As is well-known, thermal conductivity in the atmosphere is of turbulent nature, i.e., it is a function of wind speeds. Therefore, one should be able to estimate eddy viscosity $\nu$ in the hurricane. Let us look what is being done in this regard in the work of Bryan and Rotunno (2009).

Using the same approach as Rotunno and Emanuel (1987), BR09 write eddy viscosity $\nu_v$ for horizontal turbulence from dimensional considerations as $v_h = l_h^2 S_h$ (Eq. 17 in BR09). Here deformation $S_h^2$ is a function of coordinates and velocity, $S_h^2 = 2(\partial u/\partial r)^2 + 2(u/r)^2 + (\partial v/\partial r - v/r)^2)$ (Eq. 16 in BR09) and $l_h$ is a horizontal length scale. I do not consider vertical turbulence here, because the process under study is differential heating in the horizontal direction.

Thus, if we knew $l_h$ from basic principles and, using the model of BR09 that ignores mass removal, were able to obtain a realistic hurricane velocity, we could assert for sure: hurricanes are driven by latent heat. However, this is not what actually happens. BR09 write (p. 1776):

There is no quantitative theoretical guidance for how to set $l_h$ and $l_v$ in an axisymmetric model. RE87 [Rotunno and Emanuel 1987] used $l_h$ = 3000 m and $l_v$ = 200 m in their simulations, which they determined by trial and error, and by subjective evaluation of model output.

Bryan and Rotunno (2009) investigate the sensitivity of their model to $l_h$ and, not surprisingly, find it to be very high. In their Fig. 3 they plot maximum hurricane azimuthal speed $V_{max}$ versus $l_h$. It is shown that at $l_h$ = 0 maximum speed is over 100 m/s, but at $l_h$ = 6000 m it drops to 30 m/sec. Larger values of $l_h >$ 6000 m are simply not shown.

Bryan and Rotunno (2009) elaborate:

One might wonder whether we can determine values for $l_h$ and $l_v$ that yield reasonably realistic hurricanes as compared to observations. Based on the estimated observed maximum intensity of 70 m/s, as well as comparisons of maximum radial inflow to observations, it seems that $l_h$ = 1500 m and $l_v$ = 100 m are appropriate. Additionally, observations of azimuthally average properties can be useful in this regards. In an analysis of a category-5 tropical cyclone, Montgomery et al. (2006) found the radial gradient of moist entropy in the eyewall to be $-1.7 \times 10^{-3}$ m/sec^2/K; the same value occurs in our simulations if $l_h$ = 1500 m. Despite this encouraging comparison between model output and observations, we cannot say with confidence that these values of lh and l will be appropriate for all cases.

In summary, the key parameter that determines hurricane intensity is fitted by Bryan and Rotunno (2009) (as earlier by Rotunno and Emanuel (1987)) to observations. It is taken to be precisely such that the latent heat release (more correctly, its spatial gradient related to vertical velocities) produces a realistic wind speed. No considerations exist whatsoever why it should be 1500 m and not, say, 8000 m (when maximum wind speeds will fall to a dozen meters per second). The work of Bryan and Rotunno (2009) does not constitute a hurricane theory (and, to be fair, is not claimed to be one). It is a model based on an (incorrect, as we show) premise that latent heat release is the driving force of the hurricane and artificially fitted to observations via parameterization of turbulence. I think that similar parameterization schemes are employed in GCMs.

Once again: we take $l_h$ = 1500 m and get 70 m/sec. We take $l_h$ = 8000 m and get 10 m/sec. We do not know which value to take, but as far as we do not know any other mechanism except latent heat release, we take 1500 m, to satisfy the observations.

To summarize, there are no scientific grounds to state (see #61 Gavin):

While not perfect (and therefore potentially improvable by consideration of the precip mass loss terms you are proposing – and has been suggested by Bryan et al), these models are an undeniable testament that your mechanism is *not* essential to the existence of hurricanes, and that one can get a very good approximation just from using the latent heat terms.

These models are not such a testament.

In our work we derive hurricane velocities without fitting the key parameters of the theory to observations. (It can well be a long talk, but the steady state homogeneous turbulence employed by BR09 is not applicable to hurricanes. Turbulent eddies with mass removal (condensation) have a source of potential energy neglected in the conventional turbulent kinetic energy (TKE) budget. The TKE budget considered by BR09 only includes the potential energy associated with buyoancy effects (the Brunt-Vaisala frequency).)

289. ### DeWitt Paynesaid

I still don’t understand why assuming pressure is constant for the purpose of calculation during a time step is wrong. As far as I can tell, it’s no different than breaking the atmosphere into layers for calculating radiative transfer. Those equations can’t be solved in a continuous form either. If you had a gas in a cylinder with a piston and reduced the pressure on the piston at a constant rate, what’s wrong with calculating the temperature, pressure and volume of the gas at each time step assuming the pressure at that time step is constant? I’ve done it in a spreadsheet and it seems to give the correct answer.

290. ### Anastassia Makarievasaid

#289 DeWitt,

I think it would be more productive if this argument were more specific. There is no constant pressure at each time step in the non-stationary model of Bryan and Rotunno (2009). During each time step pressure is updated. The task is to calculate the magnitude of this update — how much will the pressure change? Pressure change can be due to both dynamics (e.g., divergence of air away from the considered volume) as well as local processes like condensation. In order to include the impact of condensation rate on pressure, we need to know how much vapor has condensed during this time step. As we do not have any independent considerations re how to calculate this, we do this calculation assuming that condensation per se does not change the pressure but only impacts temperature, 0 = cpdT – Ldq. Having calculated the condensation rate in this manner, we use the obtained value of dq to calculate the total pressure change during this time step.

Please, pay attention to the fact that the matter is not only about a correct calculation the number of molecules that has condensed. In our paper, see Section 4.2, we show that changing S for Sd in the expression for condensation rate zeroes the horizontal pressure gradient. One can err a little bit in calculating S and at the same time be 100% mistaken in calculating the associated potential energy release.

#280 Jim,

You would need to cool it diabatically somehow, so your box would have to be a refrigerator. The question would be whether dry air or moist condensing air end up with higher pressure after the same heat has been removed from them. It will be the condensing air due to the latent heat release.

You first say at #276 that “I now have a long list of arguments against the idea that condensation causes contraction”. You suggest that we should agree on something basic and offer Ldq = -cpdT as such a basic point. When I ask what causes condensation Ldq you want to talk about, you now suggest to cool it diabatically somehow.

However, “somehow” will not do in a basic physic argument. Moreover, diabatic cooling is inconsistent with Ldq = -cpdT, which presumes dQ = 0. So if you do have a list of arguments, please, lay them out at the background of a specific description of a particular process you have in mind using the well-established laws. If you want to bring up the point that latent heat release makes the moist adiabat shallower than the dry adiabat, I am aware of this fact. Moreover, in our paper this effect of latent heat is considered in great detail, see Figure 1C.

291. ### Anastassia Makarievasaid

I have made another promise in this thread: to explain what happens during fire cupping, see #229.

Fire cupping picturesquely disentangles mass removal from heat release in the ideal gas law.

Let us take a rod, wrap some cotton around it, moisten it with ethanol and ignite. We then put the flame rapidly into a glass cup, rapidly remove it and place the open end of the cup on the human body. We will see that the skin is “sucked” into the cup. Obviously, pressure has dropped there.

Let us see what happens when ethanol is burning:
$\rm CH_3CH_2OH + 3O_2 = 2CO_2 + 3H_2O$

The amount of gaseous vapor is limited from above by the saturated concentration. Therefore, most water formed during the reaction is liquid and does not make a contribution to total gas pressure in the cup. We thus start with four gas molecules and end with two (CO2). This mass removal effect causes pressure to drop immediately overcoming the simultaneous release of heat. Note that this is a great illustration of the ideal gas law, which says that pressure depends on the number of molecules per unit volume, not on their molecular masses.

Let us now see what happens if we do not moist the cotton with ethanol. In this case it is cellulose (the cotton itself) that is burning:
$\rm (CH_2O)n + nO_2 = nCO_2 + nH_2O$

Cotton is not a gas, vapor undergoes condensation as in the previous case. So, from n oxygen molecules we get n molecules of carbon dioxide (plus heat release). The number of gas molecules in the cup does not change. Pressure does not drop. The cup will not “suck” anyone’s skin in. You can test this yourself (BUT PLEASE BE CAREFUL)!

Note that Nick immediately ascribed the pressure drop effect to temperature effects. That is, air inside the cup rapidly cools, pressure drops and the skin is sucked in. This confident explanation is wholly in line with the traditional meteorological paradigm that pressure is all about temperature. It is also wrong.

As we see, temperature cannot explain the fact that the skin is sucked in immediately nor the fact that the pressure drop is missing without ethanol moistening. There are stories of unexperienced medical stuff who thought along the same lines: they tried to warm the cups as much as they could (but forgetting to moisten the cotton with ethanol). In the result, the cups did not work, but the patient received burns from overly hot cups. It is incorrect (and sometimes dangerous) to concentrate all explanations of gas pressure on heat release and ignore the physics of mass removal.

292. ### Anastassia Makarievasaid

Let me say that so far this has been a great discussion. Many thanks to Jeff for hosting this and for all the participants who took the time to look into our stuff, to criticize it or approve of it. I’ve been around here from October 15th, it’s been exciting but also very demanding. I apologize in advance if in the coming days I will not be able to comment extensively, although I’ll try to react if any new arguments are put forward. Otherwise we have an invitation from Dr. Curry to speak of this physics on Climate Etc., so — hopefully — in some not very distant future there will be an opportunity to resume this discussion (if everything goes as planned and nothing interferes). On another Jeff’s thread I made a promise in response to a comment of Lucia to discuss the speed of sound as related to our work — this will be done in that post.

293. ### Nick Stokessaid

Re: Anastassia Makarieva (Nov 5 12:50)
Out of curiosity, what is your explanation for fire-cupping? The gas formed in combustion is well over 1000C. Water is then a gas. You have more gas before combustion than after.

If the gas cools enough to allow condensation, it contracts by a factor of at least four. That’s what happens first. Then, and only then, you might get extra contraction from condensation. That would be by a factor of at most 1.5.

So while I don’t really know what your explanation is, I don’t think the immediacy of the contraction supports you there.

294. ### Nick Stokessaid

Re: Nick Stokes (Nov 5 15:24)
I’d forgotten nitrogen. In fact, as it cools to about 80C, the air contracts by a factor of at least four thermally, while still having more gas by volume than before. Only when the temp has dropped below 80 is there some deficiency in gas molecules, but this amounts to at most 50% of the original oxygen. At most a 10% contraction of gas due to condensation.

295. ### Phil Rsaid

anna v said
November 5, 2010 at 2:18 am

Re: Jim D (Nov 5 01:50),

My non adiabatic two plastic bag experiment showed that the dry air retained the pressure while the humid deflated.

Also Konrad’s bottle experiment shows lowering of pressure.

So it is the “adiabatic” that will turn the pressure up, i.e. that the system does not exchange energy with its surroundings.

I’ve been following this fascinating conversation for several days now. I don’t have the expertise to comment on the physics or the models, but I became curious about the “bottle” experiments and conducted one of my own during lunch yesterday. I can’t say whether it was “adiabatic” or not, or even whether it means anything, but the experiments discussed earlier included heating and boiling and inversion and submergence, etc.

I took a 1-liter Pepsi bottle and filled it with hot tap water out of the faucet. I then emptied most of the water out, leaving maybe 1/2 inch (sorry, not metric) of water in the bottom. I then put the top on and shook it up real hard (don’t know if this did anything, but thought it might add moisture to the air). I then opened the top briefly to equilabrate the pressure (and I’m sure it let some ambient air in too), then screwed the top on tightly.

If I understand correctly, I had a bottle full of saturated air in equilibrium with local pressure and very close to ambient temperature (I could be wrong on this). I then turned the tap to cold water and washed over the bottle. Whatever happened, this was enough to reduce the pressure inside the bottle and cause it to collapse (although maybe not as spectacularly as a heated soda can).

Don’t know if this means anything, but I’m happy to say that this thread made me do an experiment.

296. ### Jim Dsaid

#290 Anastassia
OK, I was giving an example of how to cool a gas at constant pressure in answer to your question which was how to get condensation at constant pressure, but that was a sidetrack.

I will return to #110 and add the V dp term that you object to my leaving out so it is L dq = – cp dT – V dp. Now instead of
dp = r R dT – rd cp dT Rv T / L

we have after rearranging to collect dp terms,
dp ( 1 + rd Rv T / r L ) = r R dT – rd cp dT Rv T / L

This modified the left side with an extra factor ( 1 + rd Rv T / r L ), which is actually 1.05 when you put in the numbers. So you can see I was justified in neglecting the effect of V dp as it has no influence on the sign of dp. The rest follows as before.
Please check my derivation, as I did it without a textbook.

297. ### anna vsaid

The volume is constant, it is fast, so there is no external heat exchange ( walls of bottle still warm).

Your experiment is more like my two bags, non adiabatic because you cooled the bottle and gas contracts with cooling too.For a control of dry air you could heat a bottle from outside in a pan of hot water as hot as the water you put in the first bottle, open the top to equalize pressure, and cool as before.

One of my two bags shows that dry air does not contract as much with cooling as the bag with moisture, but is not adiabatic because heat is exchanged
with the surroundings.

So I say again what about Konrad’s simple experiment? Is it not adiabatic? If it is adiabatic that does the trick, why does the bottle contract?

While thinking of other ways to gather empirical data to check the claims of the M10 paper, the thought occurred that a weather balloon with an internal air bladder and small air pump could control its altitude to stay within a rising moist air parcel. A quick check of the web showed that similar devices called smart balloons have already been constructed.

All that would need to be added is a sensitive anemometer such as this one –

mounted within a venturi tube with a vertical orientation. The venturi duct would allow the instrument sensitivity to be significantly increased. Two such setups may be needed if the instrument cannot detect negative (reversed) wind speeds This would allow the smart balloon to maintain near zero vertical airspeed with regard to the air parcel it was travelling within. In addition to carrying normal a radiosonde air data transmitter, an optical sensor aimed at a black card could detect the moment of condensation. IR sensors (with sun shields) to detect radiation at the frequencies of air and water molecules could also be added.

It would appear that relatively inexpensive equipment could be used to empirically examine the claims of M10 in the real world environment.

299. ### Nick Stokessaid

Re: anna v (Nov 6 01:30)
You can tell it is not adiabatic because there is no adiabatic expansion which can produce cooling by doing PV work. Instead the condensation is achieved by removing heat. There is nothing else in the experiment that would cause it.

If condensation is not achieved by heat loss to the environment, then why does the system need to start hot?

#299 Nick Stokes

I have tried the experiment at lower temperatures and the bottle still crumples but no dramatic cloud formation occurs. I would agree that heat has been lost for pressure reduction to occur. This may be through radiation or conduction through the sides of the bottle. On paper the latent heat of condensation should be able to heat the air molecules sufficiently to over compensate for the pressure loss of water vapour condensing. With the bottle experiment I am trying to check what happens to an air parcel in the real world. Just as with the bottle, an moist air parcel is surrounded by colder air at condensation altitude. The air parcel can lose heat through conduction and radiation just like the bottle.

I would encourage you to try the experiment yourself. I have achieved better results using a smouldering ( light then extinguish) post-it note ( light then extinguish) rolled into a 5mm tube. The smoke from the paper is finer that that of the timber match. Version 2 of the experiment gives the most dramatic cloud.

Over 300! Is this the Air Vent or WUWT?

302. ### anna vsaid

You can tell it is not adiabatic because there is no adiabatic expansion which can produce cooling by doing PV work. Instead the condensation is achieved by removing heat. There is nothing else in the experiment that would cause it.

I think you are wrong here. In supersaturated air condensation can happen when condensing nuclei are provided and then the system comes down to the humidity appropriate to the temperature. These are introduced, fast, by the smoke. It is a delta(time) interval in the adiabatic curve, that is collapsed by condensation. No need for cooling to be invoked.

If condensation is not achieved by heat loss to the environment, then why does the system need to start hot?

Hot is a way to create supersaturated air easily and have a strong effect. More H2O in the air as temperatures go up the more mass is removed and the pressure drop is greater.

303. ### anna vsaid

I would agree that heat has been lost for pressure reduction to occur.

Think again? Of course some heat is lost in radiation and conduction, which will be similar to what happens to the air parcel in real life, once condensation starts the adiabatic assumption crumbles.

Heat, latent, is released in the cloud, but still pressure is reduced because there are less gas molecules in the air, the globules of condensation do not count as gas.

304. ### Nick Stokessaid

I have calculated here the behaviour expected for condensation in a cylinder with a piston at 300°K, 1 bar. It shows the rate of temperature rise, condensation and volume expansion as pressure reduces. Latent heat does dominate.

305. ### anna vsaid

I guess I am saying in 302 and 303 that the phase transition of condensation is a discontinuity. In the PV=NkT curve, because N is suddenly reduced by x% since H2O molecules cluster in globules. Since N is reduced, the volume is constant and the temperature is constant (the kinetic energy of the rest of the molecules does not change at delta (time) because some coalesce into globules) and k is a constant P has to reduce appropriately.

306. ### anna vsaid

I cannot check your calculations, lets hope Anastasia can have a look at them.

I will repeat, experiment trumps calculations, and Konrad’s experiment is adiabatic to first approximation. No heat loss to first approximation.

He supersaturates by squeezing and expanding the bottle after capping:
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 have to find a large enough plastic bottle to repeat the experiment to see how fast condensation and crumbling happens.Maybe wrap it with aluminum foil and then insulator to reduce heat transfers to surroundings.

307. ### Jeff Idsaid

304 Nick,

Nobody disputes that the heat release is not the dominant energy factor in condensation. The claim in a large hadley cell though is that the heat drives the air up and then as it cools the air falls back down. Anastassia has shown that in a large region such as a hurricane, the heat release based density change isn’t creating enough gradient to drive the process. It does create a mass of warm air which all circulates together as driven by condensation volume based pressure loss.

She has also demonstrated that the effect is big enough to calculate basic speeds for hurricanes and tornadoes without considering the latent heat release as well as how it get’s accidentally parametrized out of hurricane models.

I’ve been reviewing her foundation papers which started down this path. There is one (linked by her somewhere in this thread) where she describes a cylinder being heated and cooled that I can’t make any sense out of at all. I’ll look for the link later but it was unreadable for me and heating a gas in an enclosed cylinder should make perfect sense.

308. ### DeWitt Paynesaid

Konrad’s experiment is not adiabatic. He squeezes the bottle. At the higher pressure, the water vapor mixing ratio is higher plus the air in the bottle warms so water evaporates. But since the bottle is now warmer than the surroundings, some heat will be lost through the walls. Depending on how long the pressure is maintained, the temperature in the bottle could return to near ambient. He releases the bottle, the pressure drops, the gas in the bottle cools and water condenses. The temperature in the bottle is now less than ambient so the pressure in the bottle is also less than ambient and the walls flex inward slightly. We know this is the case because after the water vapor condenses, the total volume of liquid water is exactly the same as the starting conditions. If the bottle is allowed to stand for a while, the pressure should return to ambient unless the cap leaked while the bottle was being squeezed, a distinct possibility btw. Screw on caps on plastic bottles never seal as well the second time. Before doing the experiment, try squeezing the sealed bottle with the cap under water to see if there’s a leak.

309. ### DeWitt Paynesaid

A bottle with flexible walls is approximately a constant pressure system so it should be: when the added pressure is released, the temperature drops and the volume returns to somewhat less than the original volume at somewhat less than the original temperature. If the experiment were adiabatic, the final conditions should be exactly the same as the initial conditions by definition. Since they aren’t, it isn’t.

310. ### anna vsaid

Konrad says he squeezes slightly, and the collapse is not described as slight.The air in the bottle is supersaturated and nucleation molecules have been provided with the smoke. The slight increase in pressure triggers nucleation, is my explanation. A phase transition.

Seems to me you do not get the mass point.
The N, number of molecules, in the ideal gas does not care what type of molecule is in the air. Once N becomes N-w_c because globules have appeared, initially the pressure has to drop since the volume cannot, except by collapse, which is what happens and pressures are then equalized. To go back into H2O gas the water needs to be heated to evaporate again and the pressure to increase to blow out the bottle.

Nick has calculations that show the opposite, but I cannot check them. I can try to repeat the experiment to my satisfaction.

This is hand waving on my part until I repeat the experiment.

311. ### anna vsaid

1) there is no energy exchange
2) The pressure is P1 temperature T1
3) slight pressure dP1, slight volume compression , in PV=NkT T same
4)pressure released back to P1 , T1, condition the same except nucleation triggered.

312. ### Jim Dsaid

A simple way to tell if a system is adiabatic, is if it is reversible. The best example is a gas with a piston. If it is expanded slowly it cools, and when compressed slowly goes back to the original temperature, as long as energy has not leaked out through the walls.
Clouds are adiabatic, because expansion cools them and causes condensation, but if the same air was compressed again, it would warm and evaporate returning to the original state. Note that if you remove the drops (e.g. by precipitation), it may become non-adiabatic, or irreversible, because you may not have enough water to evaporate back to the original state. Another condition is that it all has to happen in vapor/water equilibrium (i.e. 100% humidity). Condensing supersaturated water, or evaporating water in subsaturated conditions are irreversible processes, as you might be able to imagine, and so are not adiabatic. [Actually this means they are associated with an entropy increase, that can’t be reversed owing to the second law of thermodynamics.]

313. ### anna vsaid

Yes, after nucleation the process is no longer adiabatic, it is a phase transition after all,but that should hold on the sky as well as in the bottle.

314. ### DeWitt Paynesaid

The N, number of molecules, in the ideal gas does not care what type of molecule is in the air. Once N becomes N-w_c because globules have appeared, initially the pressure has to drop since the volume cannot, except by collapse, which is what happens and pressures are then equalized. To go back into H2O gas the water needs to be heated to evaporate again and the pressure to increase to blow out the bottle.

You’re neglecting an important point. Where did the number of water molecules that condensed come from? If the bottle was at equilibrium and not super-saturated, then they could only come from evaporation of liquid water in the bottle during the compression phase. If the air in the bottle was super-saturated when the bottle was sealed, then of course the final volume will be lower than the initial volume and the volume of liquid water in the bottle will be higher than at the start.

As I said above and Jim D seconded, an adiabatic process is reversible by definition. If Konrad’s experiment was adiabatic, the final volume and temperature would be identical to the initial volume and temperature. It isn’t, which means the experiment is not adiabatic. QED

315. ### curioussaid

289 – DeWitt, please could you give more detail on your spreadsheet model? What are you modelling, with what time step size and what are you validating it against? Thanks

312 – Jim D, yes your description of an adiabatic process tallies with my memory. However when you say “…but if the same air was compressed again….” I find it hard to imagine this is really possible in the atmosphere. My expectation is that the atmosphere is not well suited to respecting the boundaries of theroetical closed parcels of air and water which are the basis of the adiabatic assumption. This view is supported by the web reference I gave in comment 283.

316. ### curioussaid

309 – Dewitt, sorry I’d also exercise some caution with the constant pressure assumption for a PET bottle. I think this may hold true over a small range of deflection but I think structural and material considerations of the bottle will work to (attempt to) restore it’s undeformed shape for significant deflections. Apologies if you already have this in mind with your qualification “approximately”.

317. ### Brian Hsaid

#293, Nick;
It seems to me you continue to ignore the “immediately” part of the pressure drop with the ethanol cup, vs. no or very slow drop with just cotton combustion products. Your post also seems to postulate a fast loss of nitrogen temp. Exactly what would induce that?

318. ### Steve Fitzpatricksaid

Jeff #307,

I do not think the energy that manifests itself as high wind speeds has to come from loss of water vapor mass at all. It is the coriolis force combined with huge air flow into the base of the hurricane that is generating the high wind velocities. Hurricanes only form far enough north or south of the equator for there to be significant coriolis induced spin, but close enough to the equator for warm ocean water to provide the latent heat that produces relatively slow (compared to the horizontal wind velocity!) upward convective flow near the center of the hurricane. You need only look at satellite images of hurricanes to see the rapid anti-cyclonic outflow of air above the hurricane in striated cirrus cloud patterns. There is clearly a huge volume of air being pumped.

319. ### Brian Hsaid

#314, DeWitt;
the initial press and release is very fast: a few seconds, or less. No opportunity for heat loss. And the fog forms instantly on release; no time for transport from the wet inner surfaces. Observe it yourself a few times; it is very hard to escape the conclusion that the vapor is spontaneously condensing throughout the volume on sudden pressure drop.

320. ### Jim Dsaid

#315 Curious
Yes, how can you compress a cloud in nature to retrieve its initial state? Good question, because often a cloud gains buoyancy by latent heating making it unlikely to descend again and compress. This can probably only be done in the lab with a condensing gas and a piston. In nature, you might imagine a situation, where even despite the latent heat release it remains negatively buoyant due to a very stable environmental profile, so it could descend again and evaporate. An example would be stationary wave clouds over mountains where the flow through them goes up and down while condensing and evaporating.
Regarding your #283 link, I would agree with what it says. Adiabatic theory is a useful approximation to what happens deep in clouds, and obviously won’t work when mixing occurs at cloud edges. Parcel theory, as applied to vertical profiles, based on this adiabatic idea, has proved useful in meteorology, because this idealized approximation holds very well according to observations of clouds, so it is a good predictor of cloud base and top heights.

321. ### curioussaid

318 – Steve F, My feeling is the Coriolis effect, which I understand as angular momentum conservation about the Earth’s rotational axis, combined with conservation of angular momentum about the hurricane’s vertical axis are the significant factors in inducing very high wind speeds. As I commented in 125 above IMO the potential energy in the atmospheric pressure field is significant. Anastassia’s comment at 24 (and elsewhere) implies she is aware of this – Anastassia, if you are still reading, clarification would be welcome.

320 – JimD, thanks, extra comments appreciated.

322. ### curioussaid

308 – DeWitt, I was reminded of this product:

http://www.amazon.co.uk/Airzound-Rechargeable-Horn-worlds-loudest/dp/B000PTIJB8

Check the comments for info. on the pressure capabilities of PET bottles – I’m not sure if the caps for this horn have special seals but I just stood on a standard bottle with a standard cap and it easily supported an 80kg load.

323. ### DeWitt Paynesaid

I used to work for a major player in the PET bottle business. The restoring force for a very small deflection, i.e. no buckling of the wall would be significant but the pressure change for a small deflection isn’t much. Soft drink bottles are pressurized to 60 psi gauge pressure when filled because the rate of diffusion of CO2 through the walls is significant. The shelf life is on the order of a few months. Oxygen diffuses fairly rapidly too, which is why you don’t see too many PET beer bottles. The ones you do see have an oxygen scavenger layer between two PET layers or two scavenger layers between 3 PET layers. Anyway, an 80 kg or ~180 lb load isn’t many square inches at even 30 psi compared to the ~ 100 sq. in. surface area of a 2 liter bottle.

The spread sheet is for calculating equilibrium temperature and density vs pressure for 1 kg dry air. The density is then used to calculate the altitude for each pressure level in a 1 g field. Time isn’t involved. The pressure steps are 1,000 Pa from 100,000 to 20,000 Pa. For the calculation the initial temperature is 300 K. It’s still a work in progress because I don’t calculate the water vapor mixing ratio but use an online app and manually iterate. For 0% RH initially, the lapse rate is 9.8 K/km so I must be doing something right. For 100% initial humidity, I equate the change in enthalpy to the change in gravitational potential energy which then determines the temperature. I’ve recently included the change in heat of vaporization with temperature, but haven’t recalculated. The change isn’t much. I also include the heat of fusion starting at 273.15 K. That gives a flat spot on the curve assuming all water remains with the packet. I haven’t included the heat capacity of the liquid water yet either.

324. ### Brian Hsaid

#320, Jim;
Not sure if compression is involved, but it has recently been observed that ‘popcorn’ clouds over the ocean vanish and reappear, alternating with the gaps. Explanations are in process.

325. ### DeWitt Paynesaid

Re: Brian H (Nov 6 13:55), From Konrad above:

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.

[my emphasis]

The air is then super-saturated compared to ambient. What happens when you squeeze and release is that the temperature/pressure pulse is enough to create nuclei similar to adding smoke. The pressure then drops as the water vapor condenses and heat is released to the environment through the walls. If heat weren’t being lost, the water wouldn’t condense. Period.

For a more nearly adiabatic experiment, put room temperature water in the bottle, cap, shake, pour out the water and then with the cap slightly loose, allow the system to equilibrate to room temperature again for an hour or two. Shaking by itself will add a small amount of heat to the system. Now tighten the cap and see if squeezing and releasing causes a cloud to form. I don’t have any 2 liter bottles around or I would try it myself.

326. ### DeWitt Paynesaid

Or try an insulated container sufficiently large to submerge a 2 liter bottle. Fill it with hot water ( I’m assuming that hot water means from a home water heater at about 55 C). Submerge and fill the bottle. Empty it and, with the bottle mostly submerged, cap it and submerge it the rest of the way. Now see if you can make a cloud by squeezing and releasing with the bottle submerged. Now take the bottle out of the hot water and see how fast it collapses. That’s how fast the heat content of the bottle is lost to the environment.

327. ### DeWitt Paynesaid

A quick and dirty calculation says a bottle filled with air at 55 C saturated with water will lose about 20% of its volume when brought to 25 C, about half due to condensation and half due to reduction in the volume of the dry air.

328. ### Nick Stokessaid

“as well as how it get’s accidentally parametrized out of hurricane models. “

Jeff, this seems to be an article of faith around here – they’ve left something out. What it is, and where it was omitted, is hard to pin down.

Anastassia pointed to Bryan and Fritsch (2002) [BF02], which is a good clear reference. I went through the equations here. I believe everything is included in the right place. No-one seems to have checked.

“Nobody disputes that the heat release is not the dominant energy factor in condensation.”
Well, AM didn’t exactly say energy factor but:
As you see, heat considerations are not relevant altogether. You cannot compensate loss of mass by excess of heat or vice versa.

As I mentioned earlier, I’ve done a cylinder calculation, but adiabatic, with expansion.

329. ### anna vsaid

I am puzzled by the insistence that when condensation happens heat must be lost.

Take a cloud chamber. It is brought to a supersaturate condition and rests there with nothing happening until a cosmic muon ( or nuclear decay from the gas) starts ionizing the gas on its track. Around these ionized nuclei droplets form. For sure heat is released by ionization, so it is not heat loss that starts nucleation.

One would expect it is chemical bonds that induce nucleation, together with vapor saturation at a given temperature.

330. ### curioussaid

323 DeWitt – thanks for the extra info. You describe PET bottles as I understand them but I think the induced pressure would be 80kg/area of contact of my foot with the bottle. I agree this is still well below the pressure capability of the bottle/cap system – the comments on Amazon report 80psi as the max pressure recommended for the air horn – ie over 5 atm.

Comments on the spread sheet appreciated too – please could you update us when you have the moist condition calculated?

As far as the experiment goes, please can I check that we are trying to test the proposition that Nick put forwards in comment 37? Your 326 and 327 seem reasonable.

328 Nick, thanks for your cylinder calc. reference. I missed that first time round.

331. ### Steve Fitzpatricksaid

Curious #321,

“My feeling is the Coriolis effect, which I understand as angular momentum conservation about the Earth’s rotational axis, combined with conservation of angular momentum about the hurricane’s vertical axis are the significant factors in inducing very high wind speeds.”

Yes, as I had noted in #47 above.

332. ### Jim Dsaid

Anna V,
Supersaturated vapor does not require cooling to initiate condensation. It just requires a particle to condense on. Being supersaturated, it is already cool enough. Generally to bring it to supersaturation, cooling would have been involved.

333. ### curioussaid

331 Steve F – Thanks, noted, I wasn’t claiming prior art! 🙂 Have you seen/done any rough order of magnitude calcs?

334. ### DeWitt Paynesaid

There are various types of cloud chambers. All involve cooling to achieve super-saturation. In the diffusion cloud chamber you put dry ice on the bottom and a volatile solvent soaked fabric on the top. After a while, the nuclei rain out and a super-saturated layer forms near the bottom. The tracks from ionizing radiation happens in the super-saturated layer as the ionized air or solvent molecules produced by the radiation act as condensation nuclei. A charged grid can be used to sweep droplets from the super-saturated zone.

In the expansion type chamber, gas is saturated with water vapor in a cylindrical chamber. Then the volume of the chamber is increased by rapidly moving a piston. This creates super-saturation at the new low pressure. Run a few cycles to wash out the condensation nuclei and carefully control wall temperature and super-saturation can be maintained for long enough to get useful information.

Cloud chambers are obsolete now. Bubble chambers based on superheated liquid hydrogen supplanted them, but they in turn have been replace with first spark chambers and now wire chambers.

335. ### Noel Bartonsaid

To step back a bit, let me observe that condensation during expansion can be exploited in a heat pump. On the other hand, evaporative cooling at reduced pressure can be the basis for a heat engine. Nick Stokes mentioned my heat engine work in post 163.

These processes are analysed in my paper: N.G. Barton “An Evaporation Heat Engine and Condensation Heat Pump”, ANZIAM J, Vol 49 (2008), 503-524. My analysis uses the set of variables: V, T, P_air, P_vap, rho_air, rho_vap. The analysis is not limited to one temperature, and the latent heat is taken as a tabulated function of T. The ideal gas law is assumed to hold. Some details are mentioned at http://www.sunoba.com.au.

For condensation during expansion, there are competing effects of latent heat (that acts to warm the air and slow down the pressure decrease) and removal of water vapour (acting to accelerate the pressure decrease). Latent heat is dominant.

336. ### anna vsaid

You say:

For condensation during expansion, there are competing effects of latent heat (that acts to warm the air and slow down the pressure decrease) and removal of water vapour (acting to accelerate the pressure decrease). Latent heat is dominant.

How do you explain Konrad’s experiment then?

Seems to me that for the bottle to collapse after the letting go of the squeeze on the side of the bottle (pressure decrease) shows that lapse energy released is not dominant. Do you have an experiment that shows latent heat is dominant?

337. ### anna vsaid

Sorry that should read “latent energy releases is not dominant”.

338. ### anna vsaid

Sorry, just woke up and it shows :), that should read “latent energy released is not dominant”.

339. ### Anastassia Makarievasaid

#296 Jim,
If you want to make a convincing analysis, let us do it accurately. Please, write it out how you go from 0 = cpdT – vdp + Ldq to dp ( 1 + rd Rv T / r L ) = r R dT – rd cp dT Rv T / L. I am especially interested to see how you treat the dq term. It would be also good to explain what you mean by each variable.

I tell you in advance that it is possible to calculate the surface-specific flux of latent heat and the surface-specific flux of work caused by contraction only if one knows condensation rate. (This is where you had to involve the observed data). This rate cannot be derived from the 1st law, which it enters as an independent variable. As I showed in calculations at #60 and #24, for precipitation of 10 mm/hour, these fluxes are about 7,000 and 400 W/m^2, respectively. There is nothing sensational in this comparison, because, as I several times emphasized, WORK is not equivalent to HEAT. Mechanical power of the hurricane has to do with WORK. It is not by accident that the observed mechanical power output of the hurricane is closely around 400 W/m^2.

#307 Jeff,
The paper about cylinder makes the following (main) point: one cannot increase an engine’s power by recirculating heat.

#125, #321 Curious,
The whole point under discussion is how vapor condensation produces the observed pressure gradients (which represents a store of potential energy). When you have a radial pressure gradient, there is no problem to have a hurricane. What sustains this pressure gradient despite air convergence? Why its magnitude is such as it is? This is what we explain by consideration of mass removal. Coriolis force does not make the air move (remember, it is proportional to velocity). It only guides that for a given pressure gradient most kinetic energy will be associated with tangential (rather than radial) velocity. Apparently, where there is no angular momentum (at the equator) the condensation-induced air movement will take different forms.

A couple of words concerning contraction/expansion. To avoid confusion, one should keep the relevant processes clearly in mind. p = NRT, pV = RT, from this it follows that if you just remove a few molecules from a given volume (as Anna pointed out above), volume V (which is molar volume), will increase. Molar volume has increased, but no real expansion/contraction has occurred, no one can observe them. But pressure p has dropped. To keep the steady state and regain the steady-state molar density N (and the original number of molecules per unit volume), the remaining gas must undergo contraction. Contraction comes into play when one considers a steady state.

340. ### anna vsaid

could you do a parallel control bottle experiment? ie. as dry as possible but brought to the same temperature ( a bath?) as the original, and do the squeeze. Is there contraction comparable to the one with saturated air? This will answer whether the pressure drop is due to energy loss or also something else ( condensation).

I will try it also once I locate large squeezable bottles.

341. ### DeWitt Paynesaid

Konrad’s experiment is in principle the same as the collapsing soda can but at a lower temperature smaller volume change. The pressure in the can and the bottle go down because heat is lost the the surroundings. Put Konrad’s bottle in an oven at the same temperature as the water and it won’t collapse nor will a cloud form.

342. ### Brian Hsaid

#341;
DeWitt, have you tried this? As I pointed out above, the brief squeeze-release is very fast, no time for significant thermal cooling. The fog forms instantly upon release, but the bottle surface does not heat as one would expect if a surge of heat flux from interior water/vapor/gas to the surrounding air was occurring.

IOW, there is no plausible mechanism or time for your internal cooling, nor any evidence of it.

343. ### Brian Hsaid

#339, Anastassia;
I’m having a units problem with “p = NRT, pV = RT”.
If you divide the 1st equation by the second, you get 1/V = N.
Density = the simple inverse of volume? Volume is measured in such units as m^3, while density is in such units as gm/m^3. They are not equivalent.

344. ### DeWitt Paynesaid

Let’s look at numbers. If we assume that hot water is at 55C or 328 K, the saturated vapor pressure is 15650 Pa. If the atmospheric pressure is 101325 Pa, then the 2 liter bottle contains 0.309 L of water vapor and 1.691 l of air at 328 K. At 25 C or 298 K an 101325 Pa, the vapor pressure of water has been reduced to 3142 Pa which has a volume of 0.051 L and the air has a volume of 1.536 L for a total volume of 1.587 L or about 80% of the original volume.

You only get a loss in volume like that if heat is lost from the system. But if I’m reading AM correctly, she neglects heat loss. No heat loss, no shrinkage. If you remove condensed water from the air packet, you can’t return the air to the surface without doing a lot of work because the air will be much hotter and more buoyant than the surrounding air as it descends.

345. ### DeWitt Paynesaid

but the bottle surface does not heat as one would expect if a surge of heat flux from interior water/vapor/gas to the surrounding air was occurring.

Of course the surface doesn’t heat. It can never exceed the initial temperature. Second Law and all that. The bottle walls are thin so they conduct heat fairly readily. The wall will stay warm because there is water condensing on the wall as well as in the volume of the container. To see how fast the bottle loses heat, take a dry empty bottle, seal it and put it in the refrigerator. Or instead of doing Konrad’s experiment with hot water, use room temperature water and put the bottle in the refrigerator. It won’t be as dramatic because there’s a lot less water vapor in the bottle, but it will still condense most of the water vapor in the bottle and cool the air. I’m betting you will be surprised how fast that happens.

346. ### Anastassia Makarievasaid

#342 Brian,
“I’m having a units problem with p = NRT, pV = RT.”

V is molar volume (volume occupied by one mole of gas). Ideal gas law is not about density, it is about molar density. N = 1/V is molar density, number of gas moles per unit volume. Therefore, if you reduce N, molar volume V grows, but there is no actual expansion of gas.

In the 1st law as well, V stands not for actual volume, but for volume occupied by a unit amount of matter (e.g., 1 mol or 1 kg). Term Ldq, where dq is dimensionless and L is either in kJ/mol or in kJ/kg, specifies this unit. This seems to be causing some confusion in this thread. One should not confuse molar volume with actual volume. Molar volume can rise without any expansion.

347. ### DeWitt Paynesaid

Therefore, if you reduce N, molar volume V grows, but there is no actual expansion of gas.

But the pressure will have to go down or the temperature go up or some combination of both to avoid expansion: N = p/RT

348. ### Anastassia Makarievasaid

#347 DeWitt,

Yes, pressure drops, this is what mass removal is about. The original citation at #339 is

A couple of words concerning contraction/expansion. To avoid confusion, one should keep the relevant processes clearly in mind. p = NRT, pV = RT, from this it follows that if you just remove a few molecules from a given volume (as Anna pointed out above), volume V (which is molar volume), will increase. Molar volume has increased, but no real expansion/contraction has occurred, no one can observe them. But pressure p has dropped. To keep the steady state and regain the steady-state molar density N (and the original number of molecules per unit volume), the remaining gas must undergo contraction. Contraction comes into play when one considers a steady state.

I should add that if no contraction occurs, the potential energy associated with the pressure gradient that has formed, remains in the atmosphere in the form of this gradient.

349. ### DeWitt Paynesaid

The walls of a polyester soft drink bottle are about 0.5 mm thick and the thermal conductivity is 0.14 W m-1 K-1. That makes the wall conductivity 280 W/m2 K. The wall area of 100 sq in is 0.0645 m2 so the bottle conductivity is 18 W/K. For a delta T of 30 K that means the initial conductivity is 540 W. We’re only talking about 2 g of air and less than 0.2g of water so the total heat needed to be lost is on the order of 500 J or so. That really won’t take much time even allowing for diffusion from the center of the bottle.

350. ### DeWitt Paynesaid

But you can’t remove mass by condensation at constant temperature without removing heat as well. If there is condensation but no heat removal, the temperature and the pressure of the gas will go up not down. In adiabatic expansion of a moist gas, the increase in volume from latent heat of condensation exceeds the loss of volume from loss of water vapor mass by about a factor of 5.

351. ### Anastassia Makarievasaid

#350 DeWitt

But you can’t remove mass by condensation at constant temperature without removing heat as well.

Heat is of course removed. Let us look at the atmospheric column in the steady state, where some mass removal (precipitation) and latent heat release occurs. We speak of a particular atmospheric volume, a place in the atmosphere, not about a traveling air parcel. What happens? All latent heat that is released in the column goes away from the considered volume together with the diverging warm air in the upper atmosphere. Only this can keep the temperature of the column constant. If there is no heat outflow, temperature would be rising due to continuous heat release. In reality, as much heat as is released, as much is transported away. This is what concerns temperature balance of the column — it is achieved by heat export.

Let us now turn to the mass balance — again, FOR A PARTICULAR ATMOSPHERIC VOLUME. Some air arrives to this volume at a rate C (in the lower part of the column), some air leaves this volume at a rate D (in the upper part of the column), plus there is local mass removal due to precipitation at a rate P. So, in the steady state we have C – D – P = 0. I repeat the point made at #87: the net effect of this divergence/convergence/precipitation balance is contraction, $C - D > 0$, of this volume by the external environment. This contraction is associated with work performed on this volume. This work goes to the kinetic energy of air.

352. ### Thomassaid

Anastassia Makarieva:

btw. it does not rain inside a tornado 😉

353. ### Nick Stokessaid

Re: Anastassia Makarieva (Nov 7 06:02)
“So, in the steady state we have C – D – P = 0. etc”
No, this is a nonsensical argument. It contains no information about the flows across the boundaries – and air can flow across any of the boundaries. It implies, for example, that if there were no precipitation, you could not have a contracting or expanding flow at all.

And you can. The flame emerging from a Bunsen burner, for example, is a steady expanding flow, and it isn’t because of the mass balance of the reaction. It is because combustion, like condensation, releases heat.

354. ### Anastassia Makarievasaid

#353 Nick,

It contains no information about the flows across the boundaries – and air can flow across any of the boundaries.

It contains information precisely about the flows across the boundaries. Take a volume and surround it with a boundary. Everything that comes into the volume through this boundary is C (convergence). Everything that goes away from the volume through the boundary is D (divergence). Precipitation P is what happens within the boundary. Whatever you do, to keep a steady state convergence must exceed divergence. And to add an amount of air into a given non-empty volume does demand work.

No, this is a nonsensical argument. … It implies, for example, that if there were no precipitation, you could not have a contracting or expanding flow at all.

This is what looks to me a truly nonsensical argument, Nick, with the logic of causes and effects dramatically confused. C – D + P = 0 does not imply that at all. It only implies that if $C - D > 0$, you will have a circulation. And it allows one to calculate the power of this circulation by consideration of work associated with contraction. We do that and find that what we get is precisely the hurricane power.

There can be other types of circulations, who disputes that. But to calculate the power of a circulation caused by differential heating is by far not that straightforward, see my comment #288.

355. ### anna vsaid

That is my two plastic bags experiment.The dry air one did not deflate as far as the eye could judge. The saturated one did.

356. ### Jim Dsaid

#339 Anastassia,

I will re-post my #110 with by #296 modification, and a sign correction that you hinted was needed. So this should be complete now. Thanks for checking.

To quantify the effect of latent heat on pressure versus vapor loss we can use the gas law (I will use r for density, subscript d for dry air and v for vapor)
p = pd + pv (where pv is commonly known as e, the partial pressure of vapor)

This can be expanded out
p = rd Rd T + rv Rv T (Rd and Rv are the various gas constants)
or
p = r R T (where r is the total density and R is the effective gas constant for moist air, close to the value of Rd in normal cases)

So we have, taking derivatives for small changes,
dp = r R dT + dr R T

dp = r R dT + rd dq Rv T (where now we have used rv = q rd where q is the mixing ratio of vapor)

But for latent heating we know
L dq = – cp dT + V dp (where L is the constant for latent heat of vaporization, and cp is the heat capacity of air, and V dp is corrected from #296)

Substituting for dq
dp = r R dT – (rd cp dT Rv T / L) + (rd Rv T dp / L r)

Putting dp terms on left
dp ( 1 – rd Rv T / r L ) = r R dT – rd cp dT Rv T / L

The factor on dp is ~.95, so it is positive and modified dp little.
The ratio of the two terms on the right is
L R r / rd cp Rv T

If this ratio is larger than 1 the first term dominates and pressure increases.
L=2.5e6 R=287 cp=1005 Rv=461 take T=273 r/rd~1

This gives about 5.7, which is the factor by which heating dominates over vapor loss.

357. ### Anastassia Makarievasaid

#356 Jim,

It is good when one can follow the derivation from beginning to the end. So, you conclude that under terrestrial conditions condensation is always accompanied by a pressure rise (take T = 273 K or T = 303 K, your conclusion regarding the ratio of the two terms will stand). This is remarkable. I thought that if one takes moist saturated air and cools it by 10 degrees, its pressure will fall, not rise. Your calculations show the opposite. But I cannot see how

dp = r R dT + rd dq Rv T

follows from

dp = r R dT + dr R T

This implies that dr R = rd dq Rv. However, r = rd + rv, such that dr = drd + drv. Even if we assume that drv = rd dq, what happened to the drd term?

358. ### DeWitt Paynesaid

Re: anna v (Nov 7 10:58), From your post above:

sliding the collar until the bag was inflated tensely. I then took the second bag with room air, and did the same.

How do you know you actually did the same to each bag and that the pressure inside the two bags were equal before you put them in the refrigerator? You don’t. Also, it wouldn’t take much of a pressure excess in the dry bag such that even in the freezer, the pressure will be above ambient and the bag won’t contract visibly. Removing heat lowers pressure. Removing heat from moist air lowers the pressure even more. Nobody disputes that.

359. ### anna vsaid

It was plastic bags without much elasticity. Once they were stretched and shiny I stopped sliding the ring for both.

Removing heat from moist air lowers the pressure even more. Nobody disputes that.

And this is not because of condensation?

360. ### DeWitt Paynesaid

If this ratio is larger than 1 the first term dominates and pressure increases.

Ignoring the second term on the right, dp = rRdT

If dT is negative, the pressure will decrease, not increase. It will decrease less though, not more, because of condensation.

I think AM may have a point about leaving out the drd term unless you can show it’s much smaller then the drv term.

361. ### DeWitt Paynesaid

Of course it’s because of condensation, but your diabatic experiment is not relevant to the discussion of adiabatic experiments. You state here:

As has been stated by several people including myself, it isn’t. In principle it’s no different than your bag experiment. Btw, the volume of a room temperature (298 K, 77F) bag put in a freezer at 266.5 K (20F) should decrease by ~10%, depending on the relative humidity of the room, if it’s at ambient pressure initially. That should be enough to see for a 2 liter plastic bottle. Any pressure above ambient in the bag will reduce the loss in volume. High relative humidity will increase the loss.

362. ### Jim Dsaid

#357 Anastassia and #360 DeWitt

This calculation holds volume constant, like in a box, so the dry air mass stays the same. It shows that pressure rises in such a situation, and that would finally lead to expansion in the real atmosphere, since it is now pushing the walls of the box more than before condensation.

You could do a similar calculation at constant pressure, which would show that the volume increases. Maybe that would be better to do to approximate the atmosphere, and I will look at it, but I chose this way because the paper talked about pressure change, not volume change.

363. ### Jim Dsaid

#357 Anastassia
I can confirm that at constant pressure (dp = 0) and dry air mass (rd V = constant), the dry density (rd) will decrease (so volume increases) with the same condition that
L R r is greater than rd cp Rv T

Regarding cooling moist saturated air by 10 degrees, you can only adiabatically cool it by reducing the pressure to allow the temperature to drop. You interpreted this pressure drop as a result of the cooling rather than the cause. Obviously dry air would cool much more for the same pressure drop.

364. ### DeWitt Paynesaid

Regarding cooling moist saturated air by 10 degrees, you can only adiabatically cool it by reducing the pressure to allow the temperature to drop.

Which is why you want to work at constant pressure rather than constant volume. At constant volume and constant mass, you have no way to cause condensation or any other change in the system, assuming initial equilibrium, other than by adding or removing heat from the system.

365. ### Jim Dsaid

#364 DeWitt
Condensation could not occur in adiabatic conditions unless you reduce the pressure first.
It needs the cooling, and this can only be achieved by pressure reduction.

366. ### DeWitt Paynesaid

If you’re going to work at constant volume and change the mixing ratio, you should substitute for dp, not dq.

dp = (L dq + Cp dT)/V

V drops out if we specify dq and dT as being per unit volume

in the rearranged equation

dq = (dp – r R dT)/(rd Rv T)

dq (rd Rv T – L) = Cp dT – r Rd T

dq ( 1- L/( rd Rv T)) = dT/T ( Cp/(rd Rv) – 1)

At 273 K, Rv = 461, L = 2.5E6, Cp = 1005, r = rd = 1.275

L/(rd Rv T) is large compared to 1 and Cp/ (rd Rv ) is greater than 1

dq = -dT/L (Cp -rd Rv )

Then for dq negative, dT is positive.

dT = -L dq /(Cp – rd Rv)

dp = L dq – L dq Cp /(Cp-rd Rv)

dp = L dq (1- Cp/(Cp-rd Rv))

and dp is positive if dq is negative unless rd is large enough that rd Rv is greater than Cp.

I suspect I’ve made some errors here, especially leaving out drd. That probably removes the singularity at rd Rv = Cp.

367. ### Nick Stokessaid

#366 DeWitt, Jim,
Here’s my version of the algebra, based on the post here.

If you have a volume V of saturated air, which you expand adiabatically, then the state variables are:
1 T Temp
2 w number of moles of water vapor
3 V volume
4 P pressure
There are 3 equations (energy, gas, C-C) relating dT, dw, dV and dP, so you can get a ratio expression.
If L= LH vap (per mole)
R = gas const
b=L/(R*T)
c=(L*w)/(P*V)
then
dT/T = dw/w/b = -dV/V/(2.5+b*c) = dP/P/(3.5+c+b*c)

You can regard any of the state variables as the one you control.

368. ### Noel Bartonsaid

# 336: Anna V

I’m new to the blog, so will have missed many things. But some comments based on my previous work might be useful:

In early studies for my evaporation heat engine, I ran some experiments in which a steel container (about 100 litres, constant volume) was filled with dry nitrogen at 20-25 deg C. I sprayed fine water droplets into the container and measured the pressure drop as a function of time. Typically the pressure drop was around 2.5-3.0 kPa after about 1-2 sec when the droplets had mixed and evaporated to saturation. This was what I was looking for with my evaporation heat engine concept. However, subsequently the pressure would rise back towards the initial value. I investigated this and am confident it was caused by heat transfer back into the chamber from the surrounding steel walls (of course weighing much more than the gas inside).

The point I learned was that transient heat transfer from container walls is very important. An adiabatic assumption has to be made with great caution.

Back to condensation during expansion … I’m happy to provide my 2008 paper on request (containing analysis of both evaporation heat engine and condensation heat pump). If it would help, I can also run simulations of an expansion process with condensation. For example, if you provide initial values for V, T, P_air, P_vap (but easiest to make it saturated) and also give the expansion ratio r, I can give V, T, P_air, P_vap, rho_air, rho_vap throughout the expansion. Note however that I make the adiabatic assumption – there is no heat transfer across the boundaries of the container. Other features of the analysis are constant specific heat capacities for air and vapour, ideal gas laws for air and vapour, and latent heat is given as a tabulated function of temperature.

369. ### Jim Dsaid

#366 DeWitt
Your Rv should have been R towards the end there. Then you get (if I also keep V)
V dp = L dq ( 1 – [cp / cp – R])
We can note that cp – R = cv
where cv is the heat capacity at constant volume.
cp / cv is a constant which is about 7/5 for air.
V dp = L dq ( 1 – 1.4) = -0.4 L dq
So the conclusion is correct that dp has the opposite sign to dq, and thus it agrees that condensation which has dq negative leads to dp positive.
Nick, I won’t try to derive what you have.

#340 Anna V,
Yes I am still reading, but have been away from the computer for a day. I do intend to conduct further experiments, however local weather conditions mean that I will have to wait for access to dry air. One experiment I wish to try is evaporating water into dry air of the same temperature and observing pressure changes.

371. ### DeWitt Paynesaid

I had R at first and thought it was wrong and changed it to Rv. Fixed mass of dry air in a fixed volume so drd = 0 only drv can change. I knew that Cv had to be involved too, but hadn’t worked it out yet. Thanks.

I investigated this and am confident it was caused by heat transfer back into the chamber from the surrounding steel walls (of course weighing much more than the gas inside).

The expansion cloud chamber papers that I found by searching on line talked about wall temperature control that had to be precise and rapid so I’m sure that’s the correct answer.

There has been much discussion on this thread of adiabatic vs. diabatic systems. With the bottle experiments I am not attempting to replicate an adiabatic system. I am simply trying to model a warm moist air mass undergoing expansion and condensation. Version 1 of the experiment is perhaps the most valid.

The results linked to by Dewitt Payne #244 and Nick Stokes calcs #304 relate to adiabatic systems. However we know that the conditions in a rising moist air parcel are not wholly adiabatic as heat can leave the air parcel. Firstly the air parcel is largely transparent to IR radiation. Secondly the air parcel will experience turbulent mixing with colder air at its outer edges as it rises. Thirdly the vertical velocity of an air parcel is not solely a function of its buoyancy at a given altitude, but also dependant on momentum acquired as it rises.

I feel that there is something in Noel Barton’s comment (#368) “An adiabatic assumption has to be made with great caution.”

The question should not be “Are the equations for condensation in an adiabatic system correct?” but rather “What volume changes occur in a rising moist air parcel in the real atmosphere?”

The experimental apparatus I described at #300 would be one way of finding out.

373. ### DeWitt Paynesaid

Meteorology has been around for quite some time. They assume adiabatic expansion as a first approximation because it works.

374. ### DeWitt Paynesaid

For anyone with sufficient math and physics background interested in current thinking in Physical Meteorology and don’t want to spend money on a textbook, there’s Rodrigo Caballero’s Lecture Notes on Physical Meteorology, which looks like a textbook in the making ( 28 MB pdf file). Be warned, meteorologists have numerous thermodynamic conventions that are very different than those taught in physics or chemistry courses.

375. ### Jim Dsaid

While clear air is quite transparent to IR, cloudy air is almost completely opaque to it. This is why only cloud edges are subject to radiative effects (and mixing). As determined from cloud tops, the core cloudy air makes it up to levels consistent with adiabatic theory.

376. ### anna vsaid

I have been thinking of utilizing a hair drier, sure dries the hair 🙂 .

What about running the dehumidifier mode on the air conditioning unit?

377. ### anna vsaid

My monkey is not working at the moment 😦

#373 DeWitt Payne,
This assumption seems to be made on the basis of the observation that the height of cloud tops in strongly developed cumulus is consistent with adiabatic theory as Jim D points out at #373. One thing we should have all learned from climate blogs is that correlation does not prove causation and science is rarely settled. While I would not be excepting the claims of M10 without empirical evidence, I am seeing a lack of empirical evidence in contradiction.

A strong thunderstorm is developing outside my house atm. Through breaks in the low cloud the towering cumulus are visible. These have wide bases, a thinner tower and a mushroom top. This would seem to be consistent with the core of the rising air mass behaving as an adiabatic system and showing increased buoyancy/volume after condensation.

However these forms could also be consistent with a more diabatic process. Towering cumulus are formed from air that acquires a strong vertical momentum before condensation occurs. The wide base may also be consistent with diabatic cooling. The thin tower could be consistent with the initial momentum of the air mass enhanced by inward moving air. The mushroom top may be caused by moist air carried rapidly by momentum into lower pressure at altitude.

After cooling at altitude a warm moist air parcel should occupy less volume than a warm dry air parcel raised to the same altitude and cooled. There may be some basis to the M10 paper, but what is lacking is empirical data. Less money for GCMs. More money for smart balloons.

379. ### anna vsaid

OK, monkey working after restarting Firefox :).

To get dry air, put your bottle tied with a plastic bag upside down in a refrigerator. Water should drain in the bag. Tie off the bag and keep bottle capped, (maybe have the cap in the bag from the beginning).

If it is a sufficient condition that adiabatic conditions prevail when there is no heat transfer, you could try insulating the bottle with aluminum foil for radiation and plastic foam. Those bags one gets at the supermarket to bring home frozen products should do.

380. ### Jim Dsaid

Obviously the observations of cloud tops is not the only evidence of adiabatic ascent. In-cloud aircraft also show unmixed air with properties of the boundary layer in the updrafts. I could also say models show this, but I know that would be no proof to this audience, so I won’t say that.

#379 Anna V,
I am prepared to wait for the weather for the dry air experiments. In the meantime I have found a rigid clear 2L container. I intend to place a sealed 1/4 full freezer bag of saturated air in the rigid container, then depressurize the container and see what results.

382. ### anna vsaid

What about nucleation? Trust in preexisting nuclei?

383. ### Nick Stokessaid

369 Jim,
I’ve added to the end of my post more details of the algebra, including a verification by substitution. It is set out so you can get all kinds of rate of change.

#382 Anna V,
Sadly I didn’t get that far. The container was slightly weaker than I has supposed. The results were quite loud, although there was not too much fragmentation. The cat however is unimpressed. I expect that he should be back around dinner…

385. ### kimsaid

Three blind physicists.
Three blind physicists.
See how they speculate.
See how they speculate.
They all suppose about the air.
The cat ran out for the butcher knife.
See how they speculate.
==============

386. ### Terrysaid

The easiest way to get dry air is to suck it out of the freezer. The air inside the freezer is “for all practical purposes, dry”. Just make sure to flush the dry “vessel” a couple of times with the dry air to ensure there is no moisture left.

387. ### DouglasinUgandasaid

Great to see all this careful and detailed attention but I now worry that we are neglecting the point that the laws of thermodynamics prohibit adiabatic condensation at constant volume in M10 Section 2. (Take a look at ACPD:
http://www.atmos-chem-phys-discuss.net/10/24015/2010/acpd-10-24015-2010.html).
Perhaps the Clausius-Clapeyron equation which governs the partial pressure of saturated water needs attention (equation 3 & 4 on in M10).

The entropy factor in section 2.3 is also worth a look. There we show that for any process where entropy increases, dS=dQ/T>0, water vapor condensation (d <0) is accompanied by drop of air pressure (i.e., dp<0).

The conclusions diverge with many of those presented, but rather than debating derivation details we should agree first which basic physical laws operate here.

Anyway I am a bit confused how Jim D managed to derive all his derivations without any reference to the Clausius-Clapeyron equation/law. I admit I may have missed something as its been a busy thread!
Thanks
Douglas

388. ### DouglasinUgandasaid

#387 was aimed at Jim D # 362 and subsequent
Douglas

389. ### DeWitt Paynesaid

I now worry that we are neglecting the point that the laws of thermodynamics prohibit adiabatic condensation at constant volume

And we are aware of that fact.

See here and here for example.

There we show that for any process where entropy increases, dS=dQ/T>0, water vapor condensation (d <0) is accompanied by drop of air pressure (i.e., dp<0).

Adiabatic expansion and compression are isentropic, dQ=0 so dS = 0. No one has said that there isn’t a pressure drop when heat is removed from the system.

I am a bit confused how Jim D managed to derive all his derivations without any reference to the Clausius-Clapeyron equation/law.

For the purpose of the derivation it’s not important how q changes as a function of temperature and pressure (Clausius-Clapeyron) just that it does change. Nick Stokes has derived the full set of equations using C-C, the ideal gas law and conservation of energy. The conclusion is the same: the volume of moist air increases faster with a change of pressure than for dry air because the conversion of latent to sensible heat by condensation causes a lot more expansion than the loss in volume of the condensed water vapor. C-C, like the ideal gas law, only gets you close, btw. The actual change of water vapor pressure with temperature in air has to be determined empirically and fitted to an empirical model because no real substance has ideal behavior. See here for the Hardy model based on the ITS-90 temperature scale. Part of the reason for that is that we don’t actually know the temperature. We measure some property of matter like the electrical resistance of a coil of platinum wire and use a model that relates that property to temperature. But the models aren’t perfectly accurate. There are defined fixed points and a defined method of interpolation between the fixed points. The current standard is ITS-90.

It’s still possible that the models don’t calculate adiabatic expansion of moist air correctly. There is a loss of volume from condensation which, if not included, would lead to a too large volume change with pressure. But no one has shown me actual code that proves that it is done incorrectly. Or even better, run the code and looked at the numbers. On the contrary, Nick Stokes has looked at the code from the CAM3 model and says it looks correct and all the important factors are included.

390. ### Jim Dsaid

#387 Douglas
Yes, it is surprising C-C doesn’t come into it. C-C gives an additional constraint on how dp relates to dT in a moist adiabatic process, so between these two equations, dp and dT are completely defined for a moist adiabat. C-C only relates to the equilibrium between vapor and water, so it has nothing to say about the interaction with dry air.
I have also said here that the pressure drop causes condensation, so it is not wrong to say that condensation is associated with one, but it is not the cause of the pressure drop.

391. ### The Blackboard » More Gedanken:said

[…] 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. […]

392. ### Nick Stokessaid

Re: DouglasinUganda (Nov 8 06:34)
Douglas,
2.1. Does an analysis similar to the one I posted, with slightly different results. That doesn’t matter – at least part of the small difference is due to something I left out (a pressure derivative in the C-C equation). It’s the broad conclusion that is quite wrong. It concludes that pressure decreases as condensation proceeds, and so someone is refuted.

But of course it does – that’s how condensation was induced. Rising air. You would also have found that volume increased – an expanding flow!

The point is whether the air expanded faster than it would have without condensation (yes). Sec 2.1 says nothing about this.

2.2 Just shows that adiabatic condensation cannot occur at constant volume (or pressure). That is obvious from the structure of the equations.

2.3 has the same fallacy as 2.1. It’s just the non-isentropic version.

393. ### Noel Bartonsaid

Re #389 DeWitt Payne and Nick Stokes (various)

“Nick Stokes has derived the full set of equations using C-C, the ideal gas law and conservation of energy.”

“It’s still possible that the models don’t calculate adiabatic expansion of moist air correctly. There is a loss of volume from condensation which, if not included, would lead to a too large volume change with pressure. But no one has shown me actual code that proves that it is done incorrectly. Or even better, run the code and looked at the numbers.”

I think Nick’s model does not consider the effect of water vapour in the equation of state for the air and also assumes that the latent heat is a constant. Both effects can be included, and I did so in my analysis (referenced in #335).

I provide below output from two cases. Case (1) is V = 1 m^3, T = 25 degC = 298.15 K, Pair = 98131 Pa, Pvap = 3169 Pa (exactly saturated during expansion); Case (2) is V = 1 m^3, T = 25 degC = 298.15 K, Pair = 100,800 Pa, Pvap = 500 Pa (will remain unsaturated even after expansion). Maybe these might help the discussion? The expansions are continued until the temperature gets down to freezing, at which stage the condensation model needs to be changed.

Just to repeat the effects included: adiabatic expansion, ideal gas laws for dry air and water vapour, constant specific heat capacities for air and vapour, saturation vapour pressure and latent heat given as tabulated function of temperature, loss of volume of vapour during condensation.

Sorry for providing the data in tabular form; perhaps someone can advise how to include graphs in this post?

Given the earlier discussions, I don’t think you’ll see any surprises in these results.

Case (1) (saturated, moist adiabat, delmv is the amount of water condensed (indicated by – sign); units are respectively m^3, degree C, Pa, Pa, Pa, kg/m^3, kg/m^3, kg per kg dry air)

V T Pair Pvap Ptot rhoair rhovap delmv
1 25.00 98,131 3,169 101,300 1.1468 0.0230 0.0000
1.05 22.90 92,800 2,821 95,621 1.0922 0.0206 -0.0012
1.1 21.04 88,026 2,512 90,538 1.0425 0.0185 -0.0023
1.15 19.24 83,683 2,243 85,925 0.9972 0.0166 -0.0034
1.2 17.40 79,692 2,010 81,701 0.9557 0.0150 -0.0044
1.25 15.75 76,069 1,800 77,869 0.9174 0.0135 -0.0054
1.3 14.07 72,719 1,617 74,335 0.8822 0.0122 -0.0063
1.35 12.39 69,614 1,455 71,070 0.8495 0.0110 -0.0071
1.4 10.85 66,767 1,309 68,076 0.8191 0.0100 -0.0079
1.45 9.32 64,117 1,179 65,296 0.7909 0.0090 -0.0086
1.5 7.72 61,629 1,065 62,694 0.7645 0.0082 -0.0093
1.55 6.25 59,328 961 60,289 0.7399 0.0075 -0.0100
1.6 4.87 57,191 865 58,056 0.7168 0.0067 -0.0107
1.65 3.31 55,147 784 55,931 0.6950 0.0061 -0.0112
1.7 1.87 53,246 709 53,955 0.6746 0.0056 -0.0118
1.75 0.53 51,473 639 52,112 0.6553 0.0051 -0.0124

Case (2) (unsaturated, dry adiabat but with water vapour included in the equation of state, no water condensed; units are respectively m^3, degree C, Pa, Pa, Pa, kg/m^3, kg/m^3)

V T Pair Pvap Ptot rhoair rhovap
1 25.00 100,800 500 101,300 1.1780 0.00363
1.025 22.07 97,377 483 97,860 1.1493 0.00355
1.05 19.25 94,148 467 94,615 1.1219 0.00346
1.075 16.51 91,098 452 91,550 1.0958 0.00338
1.1 13.87 88,215 438 88,652 1.0709 0.00330
1.125 11.30 85,484 424 85,908 1.0471 0.00323
1.15 8.82 82,895 411 83,306 1.0243 0.00316
1.175 6.41 80,437 399 80,836 1.0025 0.00309
1.2 4.07 78,102 387 78,490 0.9817 0.00303
1.225 1.79 75,881 376 76,257 0.9616 0.00297

394. ### Nick Stokessaid

Noel,
I was interested to compare your table results (sat) with the revised numbers I recently posted. My numbers were
DT/DP=0.106, Dnw/DP=.916, DV/DP= -0.861

Your corresponding (2nd order) difference results, at 25°C (mine were 300°K) are:
DT/DP=0.126, Dnw/DP=0.971, DV/DP= -0.847
Pretty close.

It’s true I didn’t have T-varying L; I was calculating slopes at a single state (T=300°K). But I believe I did allow for condensation in the equation of state. I subtracted the change in nw from the number of moles n.

You can’t show pictures in coomments here – I usually store on somewhere like tinypic (you could use your own blog) and link. You can use the HTML < table >.

395. ### Anastassia Makarievasaid

#390 Jim,

Yes, it is surprising C-C doesn’t come into it.

At #356 you claimed to present a complete derivation showing the effect of condensation on pressure. You concluded that pressure always rises upon condensation (when the second term is larger, and it always is). Then at #362 you admitted that your calculation is for constant volume. However, as DeWitt at #389 points out in response to Douglas and Nick at #392 confirms, people here are aware of the fact that adiabatic condensation at constant volume is not possible. Therefore, your derivation is not correct for any relevant process. If you know of another derivation to support your conclusions, let us write it out accurately and discuss it.

#392 Nick,

2.1. Does an analysis similar to the one I posted, with slightly different results.

I am sure you agree that it would be more appropriate to say that after reading our paper you were able to derive a similar result in your post. Unless it is already done, I am sure that you will refer in your post to the original result, which is Eq. 9 in M10, that is valid for all cases, including non-adiabatic.

In my opinion, words like nonsensical, fallacy, low standard etc. could have been used much more sparingly in your comments on our work without compromising your scientific message in the slightest way. In Section 2 of M10, apart from deriving Eq. 9 that is used in subsequent sections, it is shown that condensation cannot be accompanied by a pressure rise. The question of cause/effect is not discussed. It is not relevant, because the point is to refute the claim of Dr. Rosenfeld (which is identical to Jim’s) that condensation leads to pressure rise. As we show that condensation is never accompanied by a pressure rise, it clearly suffices to refute the claim that it never leads to it. Jim’s comments show vividly that while rather obvious, such an analysis is nevertheless useful as setting the matter straight given the predominant thinking patterns.

Turning to back to science, regarding the adiabatic approximation and comparison of moist vs dry air. As I noted above at #51 and Steve Fitzpatrick emphasized at #53, if the air were descending dry adiabatically after having risen moist adiabatically, the surface temperature in the region of descent would have been dozens degrees higher than in the region of ascent. This clearly does not happen. This makes the comparison of the moist adiabat with the dry adiabat, to which the effects of condensation have been conventionally reduced, even more irrelevant for discussion of the mass removal effect of condensation on air pressure than it actually is.

396. ### Noel Bartonsaid

# 394 (Nick Stokes)

Thanks Nick. I’ll run simulations tomorrow with finer resolution near T = 300 K so as to make a better comparison for the derivatives.

In the meantime, I put some plots for Case (1) and Case (2) on my website: http://www.sunoba.com.au (follow link to “blog for 2010”.

#396 Noel Barton
Thanks for putting the graphs up on your site. If I am interpreting the second graph correctly dry air cooling from 25 degrees to 0 will reduce in pressure by 25% while the same volume of saturated air will decrease in pressure by almost 50%. Am I reading this correctly?

#395 Anastassia
“The question of cause/effect is not discussed. It is not relevant..[ ]”
Anastassia,
I feel many reading this thread would appreciate some description of the physical processes of the creation of horizontal pressure gradients through condensation that you propose. While I can accept that present modeling of condensation is based on adiabatic assumptions, these do seem match observations.

398. ### Nick Stokessaid

Noel,
I ran my own progran with constants which I thought were closer to yours. Some of mine were rough, as I was only looking for whether there was an expansion. This time I used:
T=298.15
N=1000
Nw=31.28 #0
P=98131
R=8.314
V=N*R*T/P
Cv=R*2.5
L=2300000*0.018
I’m not sure if L=2300000 is what you used, but it is a figure you mentioned. Then I got
DT/DP= 0.1218, Dnw/DP=1.0356, DV/DP= -0,8457
vs your
DT/DP= 0.1262, Dnw/DP=0.9718, DV/DP= -0,8474
Closer, but not quite there.

399. ### Nick Stokessaid

Re: Anastassia Makarieva (Nov 9 02:32)
Anastassia,
I was responding to #387, where your co-author suggested that the important results of Sec 2 should be looked at. I responded that those results are obvious, and to suggest that falling pressure associated with condensation was important was a fallacy. Now you say that the section was included to continue an argument with a referee of a previously rejected paper.

This was the basis of my earlier complaint of “low standard” citation. Referee comments are not published in the normal sense, and are not themselves refereed. So they would not normally be cited in a serious paper, let alone form the basis for a whole section.

It’s not a worthy topic for a research paper, but a blog comment can be permitted to note that you are misrepresenting Dr Rosenfeld. He prefaced his remarks with:
“The authors assert that condensational removal of water vapor that is held in an air parcel at a constant volume causes a decrease of its pressure when not considering the released latent heat that must occur with condensation. When the restriction of constant volume is removed, the decreased pressure causes air from the outside to move in, and so cause the convergence that can energize a storm.

For the sake of the argument, let’s adopt this position of the authors, while considering along the same physical reasoning the consequence of the latent heat that must occur with the adiabatic condensational removal of the vapor.”

So he’s not asserting that constant volume condensation is possible – he is attributing that to you (perhaps wrongly) and explicitly using it as a framework for showing that latent heat dominates mass removal, which is his (and Jim’s) real point.

It’s true that he made some odd remarks about isobaric (not constant volume) condensation. But it’s not something to write research papers about.

As to citing your eq 9, this is undergraduate maths, and neither you nor I can claim to be the inventor. But I did say “on re-reading sec 2…”, and indeed that is when I noticed the similarity. And differences – so far from citing Eq 9, my equation is not the same, and I don’t believe yours is correct. I do not see how you can use a molar volume form of the ideal gas law when mass is being removed.

I note too that Noel Barton had earlier done a more complex analysis (which I’m checking my numbers against) for a full engine cycle – not just gradients at a single point. And even he might not claim to have originated the method of analysis from the combined equations, although his engine application is indeed original.

400. ### DouglasinUgandasaid

Thanks to DeWitt, Jim and Nick for taking the time to reply to my comments and requests for clarification at #387. The details help a lot.

Nick at 399 — you may be right but I feel you are being a little harsh on section 2. It is reassuring that you find it obvious, but I am not convinced that everyone else finds this either trivial or irrelevent — those that do can jump to section 3. Anastassia and co. have had their theories quesioned/dismissed due to sometimes “odd” judgments (the citation is an “e.g.” only). Apparently these basic principles can and do cause confusions — even among some experts. These confusions recur so there is a legitimate value in establishing some basic principles. This thread illustrates that point too. So unless (nearly) everyone agrees with you I think that some brief basic clarifications are justified.

Our paper is not “about” section 2 per-se, but like many theory based presentations starts by establishing a number of basic results that help establish more valuable results at a later point in the paper. Indeed we made a conscious choice to go back to first principles to ensure people could follow the reasoning. I’m sure it can be improved and I am grateful for your help with that (good sharp criticism is valuable). But we did make a conscious choice to go through the arguments step-by-step.

Konrad at 397 asks the question which is our real point (and I had been encouraging us to postpone as we worked out way through the individual bricks of the underlying arguments)re-the development of horizontal pressure gradients.

Konrad take a look at the new paper at ACPD: http://www.atmos-chem-phys-discuss.net/10/24015/2010/acpd-10-24015-2010.html

There are also hurricane papers at Victor and Anastassia’s web site. See http://www.bioticregulation.ru/pubs/pubs2.php.

Best wishes
Douglas

401. ### Nick Stokessaid

400 Douglas
My main suggestion about Sec 2 is that you could take it further. The derivatives do tell you whether condensation actually increases or decreases volume expansion relative to gas without condensation. Just basing the conclusion on the sign of d(wv)/dP is a let-down.

402. ### Anastassia Makarievasaid

#399 Nick,

Referee comments are not published in the normal sense, and are not themselves refereed. So they would not normally be cited in a serious paper, let alone form the basis for a whole section.

As I already pointed out, the comment of Dr. Rosenfeld was carefully scrutinized and approved, including what you characterize as “odd remarks about isobaric condensation”, by the entire ACP Executive Committee and the Editor-in-Chief of the Atmospheric Chemistry and Physics journal of the European Geosciences Union. As such, this comment accurately reflects the dominant thinking patterns among leading meteorologists. Neither Dr. Rosenfeld, nor the ACP Executive committee, nor Bryan and Fritsch (2002), nor Bryan and Rotunno (2009), nor Jim consider either adiabatic condensation at constant volume or adiabatic condensation at constant pressure “odd”. For example, if Dr. Rosenfeld realized that adiabatic condensation at constant volume is impossible and thought, as he apparently did, that we were considering such a process, it would be possible for him just to refute our work on the basis that we involved a process prohibited by physical laws. Instead, Dr. Rosenfeld based his conclusions on a numerical consideration of this prohibited process.

As I said several times, all this shows that the (basic) consideration that we give in Section 2 is very useful, to set the matter straight. This is undergraduate math, you are right, but quite fundamental physics where, apparently, many people get confused. My claim over priority concerns this particular discussion. Relating your post to our paper which you claimed you had read, you did not realize that Eq. 9 is similar. Based on your analysis, DeWitt Payne at #389 informed Douglas that “Nick Stokes has derived the full set of equations”. I inform the reader that there is nothing conceptually new compared to M10 that you have done in your post and nothing that has not been considered in our paper; numerical results are shown in Figure 1C. I believe that now when you have realized this, a reference to our Eq. 9 in your post (whether you fully agree with it or not) would be in order; but this is what I would have done in your place, you are of course free to do on your blog everything you think is appropriate.

Importantly, no “real point”, as you say, can be made based on a consideration of a process that does not exist. The claim that “latent heat dominates mass removal” has been put forward many times in this thread, with various numerical factors, however, its origin and relevance to the real atmospheric processes and formation of pressure gradients has not been revealed. My attempts to retrieve such an explanation from Jim has so far ended with a consideration of adiabatic condensation at constant volume.

Concerning your point about “not just gradients at a single point” that Noel Barton calculated. Full engine cycle concerns temporal changes in the working body, not atmospheric circulation or spatial gradients. Unlike M10, you have not presented an expression for a horizontal pressure gradient associated with condensation. I tell you in advance that you will be unable to do so for reasons I outlined at #288. Until someone comes forward with a basic derivation of a horizontal pressure gradient based on heat release and shows that it is larger than the gradient caused by mass removal as quantified by M10, the general claim that “latent heat dominates” lacks a proof.

403. ### Nick Stokessaid

Re: Anastassia Makarieva (Nov 9 07:49)
My development in the blog post is a complete set of all possible derivatives – your Eq 9 is just DT/DP. And Eq 9 nothing new – I have included an update on my post which notes that DT/DP is given by Eq 3.74 of Rodrigo Caballero’s Phys Met lecture notes (with hydrostatic dp = g dz). RC says that it is approximate (his Eq 3.72), and in fact the approximation is equivalent to the omission of the last term in my IG eqn (Eq 2 in my post). But I believe in general that term should be included.

And of course by blog post is just a blog post. I’m not claiming it as a novel research result.

404. ### HLxsaid

Just tried to calculate an abstract problem.

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?

405. ### Anastassia Makarievasaid

I feel many reading this thread would appreciate some description of the physical processes of the creation of horizontal pressure gradients through condensation that you propose. While I can accept that present modeling of condensation is based on adiabatic assumptions, these do seem match observations.

Thank you for this question. You mention adiabatic assumptions. I repeat from my comment at #402 that it is not possible to derive a horizontal pressure gradient based on adiabatic assumptions. Vertical temperature lapse rates are very often neither moist adiabatic in the region of ascent, nor dry adiabatic in the region of descent. The mean tropospheric lapse rate is 6.5 K/km, which is neither of the two but in between. There is strong mixing in the horizontal dimension, which makes the adiabatic assumptions implausible for considerations of horizontal pressure gradients.

If someone is really keen to understand what we are talking about (rather than to put us in the corner to defend ourselves against the alleged “latent heat dominance” [quite a militant claim, I would say]), I would advise him or her to forget about heat altogether, to avoid confusions. Open your mind to consider a totally different process. Consider our atmosphere, which is very thin as compared to wide. A tiny several kilometers high and a thousand kilometers wide, this is what a typical circulation is about.

Now consider the fact that gas is continuously removed from a certain volume in this atmosphere. We do not know how this happens and why. We just know that in this region the gas disappears (perhaps to a black hole). Apparently, this region will be characterized by a lower pressure than the surroundings. How to calculate the pressure difference between this region and the environment where no mass removal takes place?

Apparently, the resulting pressure gradient will be proportional to the rate of mass removal: the higher this rate, the greater the pressure difference (other things being equal). It is clear that if the process of mass removal were infinitely fast, then the pressure in the region where it acts would be simply zero (vacuum).

However, it is also clear that the air reacts to any pressure disturbance. So, as soon as a pressure drop has formed, the air will start moving towards the region of low pressure. Apparently, the faster the air arrives to the area of mass removal, the smaller value of the steady-state pressure gradient that can be maintained (other things being equal). Therefore, we conclude that the resulting steady-state pressure gradient should be inversely proportional to the horizontal velocity u.

That’s the effect of mass removal in a nutshell: dp/dx = SRT/u, where S is the volume-specific rate of mass removal (mol/m^3/sec), R is the universal gas constant (J/mol/K), T is absolute temperature (K). This is what our final equation 37 in M10 essentially says.

Please, consider this as food for thought. There are several caveats here, e.g., it is not specified on which scale this equation is valid. As I said above, I hope to be able to speak more of this in not so distant future. At the moment please refer to M10, Section 4.

406. ### Anastassia Makarievasaid

#403 Nick,

And of course by blog post is just a blog post. I’m not claiming it as a novel research result.

Why do you presume that a blog post cannot carry a novel research result? I, for one, claim that the formula H_p = PRT for hurricane power at #24 is a novel research result first published at Jeff’s. We have never put this in this form before (although essentially the idea was there in our previous work on hurricanes).

I believe that true science has no borders. It can be absent from a “serious paper” and present in a blog comment and vice versa. See how much time people, including yourself, have already spent discussing our non-reviewed paper and our propositions. People are apparently interested and provoked. Finally, let us not forget that the first Millenium prize in mathematics was awarded for research that had never made it to a peer-reviewed journal but was published in the form of non-refereed electronic preprints.

Anyway, my colleagues are signaling me that I should consider shutting up here and doing some “serious research”, too, besides blog discussions.

407. ### Thomassaid

@Anastassia Makarieva

the air will move because of differential heating. this is what the atmospheric circulation is all about…

and by the way Grigori Perelman’s research was not rejected but he just did not bother to publish it…

I really recommend reading a book about theoretical meteorology before continuing the discussion (for example An Introduction to Dynamic Meteorology by holton).

408. ### Dan Hughessaid

Two-fluid formulation of the cloud-top mixing layer for direct numerical simulation

Phase transition at the cloud boundaries often compounds the difficulty in understanding turbulent entrainment [15,21]. There are many different aspects of the problem, which can be considered. One of them, the role of buoyancy reversal due to the evaporative cooling that is promoted by the evaporation of the droplets under certain mixing conditions, has been long debated using theory, field and laboratory measurements, and numerical simulations [7, 14, 25, 28, 29, 36, 46, 51, 54, 61].

409. ### curioussaid

407 Thomas – I am not sure if this is your direct assertion:

“the air will move because of differential heating. this is what the atmospheric circulation is all about…”

so forgive me if I’ve misunderstood, but there have been several comments in this thread on the role of fluid momentum wrt the atmospheric system. My view is this cannot easily be ignored – the earth, is after all, a spinning body with biological life on it exerting it’s own gravitational field in space with an equatorial tangential velocity of approx 1000mph. Entrained around this is a gaseous air and water system which is subject to a daily and seasonally varying radiative heat input, gravitational effects from other bodies, land and water mass geometry effects and no doubt others that I’m not aware of. IMO there are several factors at work here and to concentrate simply on differential heating IMO does not
seem reasonable. All theory needs to accurately reflect observation, and the limits of assumptions need to be acknowledged at each stage in order to keep perspective on the certainty associated with any claims made. Apologies if this is acknowledged in the literature and I am simply showing my ignorance but, as an engineer, I find it hard to envisage a dynamic moving system where momentum and continuity effects are not significant. (I note you have provided another reading reference, the title of which refers to “Dynamic Meteorology” – thank you)

410. ### DouglasinUgandasaid

Thomas at 407

“this is what the atmospheric circulation is all about…”
Anyone following the thread will recognise that that is not agreed. Why are you certain? My feeling is that scientists have to keep an open mind and avoid dogma. Once a theory becomes unquestionable it is no longer science. So can we question the differential heating model and advance an alternative? It’d be interesting to hear under what circumstances you would allow that.

By the way no one said that Perelman’s research was rejected … seems an odd comment. If it is meant as a reference to Anastassia and co’s publication efforts please note that they have had publications on these topics in high class physics journals — the challange is in persuading climate scientists in climate journals. My guess is that some is about clear communication and some may be about challanging doctrines. Perhaps you have some insight on that?

Why should Anastassia read a theoretical meteorology text before continuing the discussion? I am not sure whether to read that as a well intended but poorly specified suggestion or an impatient jibe. If it is the former please clarify. I know that we worked through Curry and Webster’s book and Lorentz’s reviews to make sure we were building from accepted foundations — the goal was to reach and communicate with climate and meteorological scientists. We remain keen to learn.

Thanks
Best wishes
Douglas

411. ### Howardsaid

I am really struggling with this topic, but it is interesting none-the-less.

Is the M10 point that differential heating causes moist, warm air to vertically rise. Then, once that moisture condenses, horizontal winds develop to fill the void?

412. ### DouglasinUgandasaid

Howard at 411 – not quite right. M10 is the ACPD ref I gave at 400. Hope you can follow it.
For some less technical “why it matters” overviews please see the blog and links at
http://blog.mongabay.com/2010/10/20/which-came-first-the-forest-or-the-rain/
Hope that helps — good luck.
Best wishes
Douglas

413. ### Anastassia Makarievasaid

#411 Howard,

Thank you for the question.

To understand M10, one should temporarily forget about differential heating to avoid confusions. (By the way, warm air rises quite infrequently despite the nearly ubiquitous belief in the opposite. In hurricanes, it is the colder air that is rising, see my comment #99 above, as well as here.)

The air rises because the vertical distribution of saturated vapor is non-equilibrium, see here for an animated graph. Therefore, any occasional upward displacement of a volume of moist saturated air leads to condensation. This lowers local air pressure and disturbs hydrostatic equilibrium. This pressure disturbance is rapidly relaxed in the vertical direction (which is very small compared to horizontal dimension), such that the hydrostatic equilibrium is recovered. This relaxation consists in the fact that the shortage of vapor caused by condensation in the ascending air is compensated by excess of dry air coming from below.

In the result of this hydrostatic adjustment, air pressure at the surface drops. This initiates horizontal convergence. Convergence enhances the initial vertical motion and sustains condensation.

Therefore, you are right that the air streams towards the condensation area “to fill the void”. But as long as it brings vapor that continues to condense, it does not appear possible to actually fill the void (new “void” is continuously being created by condensation within the incoming air). But if the incoming air lacks water vapor, the void is rapidly filled and the circulation stalls.

414. ### Nick Stokessaid

Re: Dan Hughes (Nov 9 12:03)
That’s a good paper. Also good is a review article on moist convection, written by one of the co-authors, Bjorn Stevens.

415. ### Howardsaid

Douglas(412): Thanks, interesting site. Uganda looks beautiful and overwhelming.

Do forests contain a very large surface area of condensation nuclei over hundreds of feet above ground and thereby create the suck for winds to develop?

Anastassia(413): Thanks for hosting such a fine discussion.

I read the links and also looked at the hurricane Isabel paper. Is the theory that tropical forests create a condensation vacuum that is the primary power-source for the Hadley cell? Are the mid-latitude and polar cells similarly generated from Taiga (Boreal forests)?

416. ### Noel Bartonsaid

My simulations are for adiabatic processes, i.e. no heat transfer between parcel of air and surrounding environment. So, with reference to your interpretation:

“If I am interpreting the second graph correctly dry air cooling from 25 degrees to 0 will reduce in pressure by 25% while the same volume of saturated air will decrease in pressure by almost 50%. Am I reading this correctly?”

I’d put temperature changes in terms of pressure drops – for dry air, reducing the pressure by about 25% sees the temperature of the parcel of air drop from 25 deg C to freezing; for saturated air, a pressure drop of almost 50% is required for the same temperature drop.

417. ### DouglasinUgandasaid

415 Howard
Nuclei may help (and there is some evidence of these) but the key point is the high evaporation which causes a positive feedback, driven by condensation, that generates a pressure gradient which draws in air. This is not immediately intuitive so I suggest you work through one or more of the simple articles and the diagrams. The Bioscience article is not free but there is a text version at http://lomaprieta.sierraclub.org/forestprotection/FeatureArticles/HowForestsAttractRain.pdf and I could send you a full PDF if you send me your email.

Nick and Noel – thanks for the texts — gives us some homework.
Best wishes
Douglas

418. ### Noel Bartonsaid

# 394, 398 (Nick Stokes)

Nick

So as to enable a better comparison between results, I ran two more simulations for Case (1) with small expansion ratios. See below (same layout and units for the table as previously).

V T Pair Pvap Ptot rhoair rhovap delmv
1 25.00 98,131 3,169 101,300 1.1468 0.0230 0.0000
1.01 24.56 97,016 3,096 100,112 1.1355 0.0225 -0.0002
1.02 24.13 95,926 3,024 98,950 1.1243 0.0220 -0.0005

If you want finer resolution, just let me know.

For the value of L at 25 C, the tabulated value I used was 2304900.

To explain better my comment (#393) “I think Nick’s model does not consider the effect of water vapour in the equation of state for the air” … in your equation (1)

n Cv dT + L dnw + P dV = 0

you are considering 1000 mols of air PLUS water, but I think your Cv is that for dry air. In the corresponding part of my analysis the Cv is taken as that appropriate for the given mixture of dry air plus vapour. I don’t expect this to cause a big difference.

As you said at an earlier post (I forget the number), as the expansion continues one needs to numerically track along the moist adiabat, not just use the starting values. That is done in my analysis/code.

419. ### Howardsaid

Douglas (417)

Thanks, I read the text. The forests are sources of storage and evaporation. Evaporation is enhanced during the heat of the day to produce afternoon showers. Anyone who has spent time in South Florida can set their watch by the summer afternoon thundershowers.

This forest evaporation process seems to be driven by temperature and the buoyancy of of warm humid air. In aviation, we learn and experience that performance of piston airplanes are diminished on hot and humid days in an effect called density altitude. The rule of thumb is that the pressure altitude at sea-level changes by 1,000-feet for every 10’C change in temperature.

I think I keep getting stuck back here with the rest of the skeptics (like Nick, lol) where differential heating causes warm moist air to rise followed by condensation at higher altitude due to the lapse rate.

Cheers

420. ### DeWitt Paynesaid

What table are you using? That value for L looks to be too low. The value at 100 C is 2257000 and from this table it’s 2305420 at 81.35 C.

421. ### Jim Dsaid

#395 Anastassia
We know condensation cannot occur adiabatically if either pressure or volume are held constant. On the other hand, it makes no sense to look for a pressure effect of condensation if you don’t hold something constant, such as volume. So I held volume constant and looked at the effect of condensation which removes vapor and adds heat, and this showed a pressure rise.
This can be achieved by having supersaturated vapor in a box that then condenses, so it is a physically realizable system, but it is not adiabatic or reversible, and therefore C-C would not hold.

If you want to restrict the discussion to reversible processes, the condensation can only be caused by a pressure drop as there is no other way to reversibly cool the air. For dry air, a pressure drop accompanies cooling, but we don’t say the cooling causes the pressure to drop. It is the same thing with saturated air, but it cools less for a given pressure drop than dry air due to the latent heating.

It comes down to cause and effect. How do you get reversible condensation? By a temperature drop. How do you get the temperature to drop reversibly? By a pressure drop.

Your statement at the end of section 2.1 is not wrong.
“This proves that water vapor condensation in any adiabatic process is necessarily accompanied by reduced air pressure”,
but it is key to understand the difference between “accompanied by” and “causes”.
Similarly the abstract is correct to say “condensation is associated with a decline in air pressure”. No one would argue with these statements for reversible processes again because “associated with” does not say anything about which causes which.

What you do in section 3.3 is not related to the reversible processes at all. The pressure drop in the moist column is completely due to removal of condensed water, and the paper correctly says this is the cause. This is not the same thing as a reversible process discussed in the earlier sections, because removing the vapor via condensate is not reversible. This pressure drop is just a statement of mass conservation, not adiabatic thermodynamics. This condensation was actually done by diabatically cooling the column at constant volume, but the pressure drop is just from mass removal, and certainly is nothing to do with C-C in this diabatic process.

Section 4.1 begins with
“We have shown that condensation of water vapor produces a drop of air pressure in the lower atmosphere up to an altitude of a few kilometers”.
This condensation was achieved by diabatically cooling the air many degrees at constant volume, which is not a natural process, and cannot be compared with the C-C equation.

In the real atmosphere, the moist adiabatic profile comes from lifting surface air through the troposphere, which needs vertical motion, and the pressure reduces as you follow air parcels in a lagrangian sense. This pressure reduces whether or not you remove the water from the parcel as it ascends, so you have two separate pressure reduction processes confused. One is hydrostatic reduction due to mass loss, one is adiabatic reduction that causes condensation.

I’ll stop there.

422. ### Noel Bartonsaid

# 420 DeWitt Payne

The values you cite are for change in enthalpy, whereas my value is for the change in internal energy. The difference is made up of the delta (Pv) contribution, which Nick’s analysis accounts for explicitly (as does mine).

423. ### Nick Stokessaid

Re: Noel Barton (Nov 9 19:06),
Thanks, Noel
It is close, but I can’t get much closer. DT/DP discrepancy could be just the precision of T in your table.

DT/DP   Dnw/DP   DV/DP
NB table    0.1258775    0.9562824    -0.8434638
NS theory    0.1215239   1.0339561    -0.8461305

Still, pretty good. I tried all the things that might be wrong, eg the extra term I had in the Gas Eqn, or the last correction I made. But the formulation currently on the blog gives by far the best result.

424. ### Noel Bartonsaid

# 421 (Jim D)

Let me add some comments from the standpoint of my analysis and simulation results (# 393, 418).

In the simplest form, my simulations are for a piston withdrawn from a cylinder. There is no heat transfer through the boundaries. The process is adiabatic, and should be reversible if the piston is withdrawn slowly enough. So the condensation that is observed should be reversible, with the entropy at the start equal to the entropy at the finish.

This was a key check for my code. For the condensation case, the entropy of the air component of the air-vapour-droplet mixture increases as the piston is withdrawn from the cylinder. (To explain: there is an increase in entropy due to the increase in air volume and a decrease in entropy due to the drop in air temperature. The former effect is bigger than the latter.) On the other hand, there is a decrease in entropy of the vapour-droplet component of the air-vapour-droplet mixture. (To explain, the dominant effect is the loss in entropy as vapour condenses.)

I evaluated these entropy changes carefully for a test case and was mightily pleased to see they cancelled out. (Details are given in my cited reference, see # 335). Entropy was constant. Condensation can be reversed by pushing the piston back into the cylinder.

If anyone is interested, I can provide a 2.5 MB video clip I took showing alternating condensation/evaporation processes within a piston-in-cylinder experiment. This shows the process reversing about 10 times in 2 secs.

425. ### Noel Bartonsaid

# 423 (Nick)

Thanks Nick. In the table below, I provide the T values to greater precision. Can I encourage you to try the checks again?

V T Pair Pvap Ptot rhoair rhovap delmv
1 25.00000 98,131 3,169 101,300 1.1468 0.0230 0.0000
1.01 24.55919 97,016 3,096 100,112 1.1355 0.0225 -0.0002
1.02 24.12925 95,926 3,024 98,950 1.1243 0.0220 -0.0005

426. ### Nick Stokessaid

Re: Noel Barton (Nov 10 01:54)
Thanks Noel,
I tried, but it didn’t help. I also tried using lookup values for the coef in the Clapeyron eqn rather than theoretical, but no joy there.

427. ### Thomassaid

@DouglasinUganda

evaporation over oceans is much larger than over forests….

428. ### Douglasin Ugandasaid

Howard at 419
See what Anastassia was suggesting at 413 — you “should temporarily forget about differential heating”. Then if you feel able try M10 or one if the papers at Anastassia and Victors site.
The evaporation process is not temperature driven.

Jim D at 421
I dont think we disagree on anything here other than the suggestion that there is a confusion between adiabatic and non-adiabatic processes. I am not sure who is confused on that. We flag that difference though we can be more specific if you think the reader needs it. Thanks for raising the concern though.

Thomas at 427
per unit area? No, that is wrong. See 410 please — there are questions for you.

Best wishes
Douglas

429. ### anna vsaid

For sure it will be an interesting video if you can put it up on your site.

430. ### Dan Hughessaid

Here are two that investigate the speed of sound in misty atmospheres:

J. E. Cole III and R. A. Robbins, “Propagation of Sound through Atmospheric Fog”, Journal of the Atmospheric Sciences, Vol. 27, pp. 426-434, 1970.

G. A. Davidson, “Sound Propagation in Fogs”, Journal of the Atmospheric Sciences, Vol. 32, pp. 2201-2205, 1975.

The Google gets many more hits in the acoustics area.

431. ### DeWitt Paynesaid

How are you calculating water vapor pressure? The Hardy model using the ITS-90 temperature scale gives slightly different values (see here).

temp Pvap Pvap’
25.00 3169.92 3182.94
24.55919 3087.56 3100.10
24.12925 3009.04 3021.13

Pvap’ is with atmospheric enhancement factor:

In the presence of a carrier gas, the equilibrium vapor pressure for pure water is slightly increased by an enhancement factor, which depends upon the type of the carrier gas, the overall absolute gas pressure and the dew-point temperature.

432. ### Noel Bartonsaid

# 431 DeWitt Payne

I used Table A-4 of Cengel & Boles, Thermodynamics, An Engineering Approach. Their source was van Wylen & Sonntag, Fundamentals of Classical Thermodynamics.

In the temperature range that we are discussing (around 25 deg C), the saturation vapour pressures are given to 1 Pa accuracy at 5 deg C intervals. I used linear interpolation between the given values.

The values at 20 deg C and 25 deg C are 2339 Pa and 3169 Pa respectively. Thinking about the accuracy of linear interpolation, I wouldn’t be surprised to see errors of 10-20 Pa in my interpolated values. There is no mention of the atmospheric enhancement factor.

The latent heat was interpolated in the same way.

433. ### Noel Bartonsaid

# 429 (Anna V)

Thanks for your interest. The video clip should now be available at http://www.sunoba.com.au (follow the link to “Blog for 2010”). I also installed another short clip showing my evaporation engine in action.

434. ### DeWitt Paynesaid

I think Nick is using C-C rather than linear interpolation so that may explain part of the difference. For heat of vaporization, I found this non-pay walled article on calculating enthalpy of vaporization for temperatures between the critical point and the triple point of water. It probably doesn’t make much difference in the final calculation.

lucia has an new post on deriving equation 34 in M10. She can derive a similar equation, but it’s not the same.

435. ### Noel Bartonsaid

# 426 (Nick Stokes), # 434 (DeWitt Payne)

Thanks to Nick for working on the comparison and to DeWitt for comments. I think we’ll have to be satisfied with good but not perfect.

Exact equality between results cannot be expected. Our analyses have some common features (ideal gas laws, constant specific heat), some differences in physics (especially I think consideration of sensible heat in the vapour component of the air-vapour mixture, which will have an effect on isentropy), different numerical approaches (prediction of properties at one point, T = 25 C, as opposed to tracking along the expansion path), and different implementations even when the physical processes are the same (C-C versus tabulation/interpolation).

436. ### anna vsaid

Thanks.

Bear with me since I am a simple physicist whose thermodynamics semester course was taken back in 1960. The piston shows sequential condensation and evaporation according to pressure, which means that heat is not leaving the system (approximately of course, since some heat leaves by radiation and conduction with the walls)because evaporation will take back the latent heat from the previous cycle.

So my question is, when one is trying to model the air motions in the atmosphere, is an adiabatically rising air parcel the best simulation? Latent heat should be retained in the parcel to presume adiabaticity and then how can it be available as energy for wind creation?

Once the system is open the argument of latent heat increasing pressure is not self evident.

Back to your piston experiment: Would you have a way of measuring an increase in pressure ( over the decrease you supply) due to latent heat release?

437. ### DeWitt Paynesaid

Adiabatic expansion doesn’t mean work isn’t done. Expansion moves the piston against a force. Think of a vertical piston and cylinder in a vacuum with a weight holding down the piston. Put a smaller weight on the piston and it goes up, lifting the weight. That movement against constant pressure is work. Similarly, you have to do work to compress adiabatically. If the work is done reversibly, the net result of a full cycle is no change. But that’s for a piston. In the atmosphere the work is done by converting kinetic energy into gravitational potential energy. If you lift 1 kg 1 km against the force of gravity, you have done 98,000 J work. The heat capacity of dry air is about 1005 J/kg K. So if you lift 1 kg air adiabatically by 1 km, the temperature drops by 9.8 K, otherwise known as the dry adiabatic lapse rate.

438. ### anna vsaid

And then?

Does the argument about condensation boil down to

a) differential heating :the latent heat released within the parcel will increase the lift and so more air will come underneath the parcel to fill the void
or
b) condensation pressure drop: defeats the lift and the parcel falls and air comes from above to fill the void.

?

?

439. ### Makarieva sigue, y sigue bien. « PlazaMoyua.orgsaid

[…] several caveats here, e.g., it is not specified on which scale this equation is valid. As I said above, I hope to be able to speak more of this in not so distant future. At the moment please refer […]

440. ### plazaemesaid

I woul like to thank very much Anastassia Makarieva, Jeff Id, and *all* the participants for such an instructive thread(s). I guess I can do it also in the name of the (I hope many) readers.

441. ### Anastassia Makarievasaid

I am sure Jeff will allow me to mention that right now the discussion continues at the Blackboard. In particular, the meaning of Equation 34 in M10 is under investigation.

442. ### Anastassia Makarievasaid

My apologies, I’ve overlooked that DeWitt Payne has already mentioned that at #434.

443. ### curioussaid

433 Noel – just watched your videos, thanks for posting.

444. ### Anastassia Makarievasaid

I would like to draw readers’ attention to the following anonymous comment Dr. Curry posted at Climate Etc. in the thread discussing our work. This very helpful comment sounds like made made by a high-level climate modeller. He/she says, in particular, that no model that had a dynamic core first described by Lin (2004) was able to correctly account for a mass sink effect:

But I would like to point out that the “vertically Lagrangian control-volume discretization” used the the finite-volume dynamical core (see4 Lin 2004, Mon. Wea. Rev.) allows us to have a local mass sink/source due to moisture changes. All other climate models that I know can’t accomplish this because the vertical coordinate is tied directly to surface pressure.

This suggests that Qiu et al. 1993 (the first reference Dr. Emanuel provided in his email) could not correctly account for a mass sink even from the point of view of a climate modeller.

The researcher continues:

I did some sensitivity experiments with and without this “moist” mass-sink effect, and we did not find a strong positive feedback mechanism we suspected. This is perhaps due to our experiments design in which we prescribed the SST. Nevertheless, it is important to have this effect for correct mass conservation and also for accuracy reason.

This is a claim very similar to what was made in this thread concerning the study of Bryan and Rotunno (2009). However, in this particular case (concerning GCM) the researcher does not give any references where the results of his/her sensitivity experiments were published. Neither is Dr. Emanuel apparently aware of such references. I commented at Climate Etc. that if such a reference is provided (which should be a post 2004 paper), it will be able to scrutinize the physical propositions that underlay these sensitivity analyses and see whether they were correct.

445. ### Anastassia Makarievasaid

At #444, it should have been, of course, “no model that did not have the dynamic core…”

446. ### Jeff Idsaid

Anastassia,

In reply #444, that was my impression of the CAM model. They never update the pressure for the effect of mass removal. While the mass isn’t there, the model runs right on as though nothing changed.

447. ### Nick Stokessaid

Re: Jeff Id (Nov 12 09:49),
Jeff,
I think that is true for the equations that they apply to the main grid, and use for normal weather. But their
updraft model, which they presumably use for sub-grid convection (including hurricanes etc) does have a normal treatment of mass removal (eg 4.19).

448. ### Anastassia Makarievasaid

(I am posting this here to summarize the discussions of Eq. 34 on the Blackboard. Perhaps due to great length, I cannot post it there. But I will do here, and give a link there.)

Since the publication of our paper on October 15th, I’ve spent about a month in intense blog discussions. It is exciting to see that the novel concept of condensation-induced dynamics leaves very few people indifferent. I’ve been unfortunately unable to closely follow the discussion the last several days and I will unlikely have significantly more time the coming one-two months. There is one month left for the public discussion of M10 in ACPD. So I encourage the Blackboard readers feeling that there are unanswered questions and having time to formulate a coherent scientific argument to follow Nick Stokes’ example and submit their comments to that discussion.

Here I will give some considerations pertinent to discussions of Eq. 34 on the Blackboard. The objections against Eq. 34 can be classified in two categories. The first one is “I do not understand where Eq. 34 comes from”. Obviously, this can be either a problem of the authors or a problem of the readers. Everyone a bit acquainted with theoretical problems of hydrodynamics does know that to propose a novel theoretical framework coherently explaining observations is — well, something. That not all people will understand everything at once is expectable, and the authors should be prepared to be patient (and to some extent relaxed) about that (while ready to provide clarifications to informed inquiries). (In my opinion, the best way to understood M10 lies through the attempt to appreciate that disappearance of mass creates a store of potential energy.)

Another category is the one the authors must attend to. These are statements of the kind: “Eq. 34 is incorrect because of that and that”, with specific arguments. Let us consider the logic of such arguments in the light of what has been said so far. We start with writing out 2D Euler equations for a hydrostatic atmosphere:
$\displaystyle w \frac{\partial w}{\partial z} + u \frac{\partial w}{\partial x} = 0$ (1)

$\displaystyle \rho w \frac{\partial u}{\partial z} + \rho u \frac{\partial u}{\partial x} = -\frac{\partial p}{\partial x}$ (2)

We also have the continuity equation:
$\displaystyle \frac{\partial \rho u}{\partial x} + \frac{\partial \rho w}{\partial z} = S$ (3)

M10 considered the following problem. Suppose there are no other drivers of circulation in the atmosphere except mass removal S. In this case when S = 0, velocities u, w, and the horizontal pressure gradient are zero. Nothing happens. The existence of a mass sink (or source) must disturb the system and bring about some flow. It cannot be that there is a mass source at some point, but no pressure gradient.

As one can see from Eqs. (1)-(3), the information about pressure gradient $P_S \equiv \partial p/\partial x$ can be retrieved from Eq. (3) only. (Moreover, it is clear that as far as $P_S (S) = 0$, at small S we will observe that $P_S$ is directly proportional to S, which is what M10 have demonstrated). But the main point is that it is by setting S that we retrieve $P_S$. The magnitude of the mass sink or source, together with boundary conditions, will determine the resulting horizontal pressure gradient. The system receives a unique non-trivial solution when the value of S is set.

From this consideration it is clear that if we take an engineered flow, with pressure gradient $P_{E}$ determined not by condensation, but by something else, and put this $P_{E}$ into the system (1)-(3) with a given S, we will observe a controversy. This controversy is not a proof that Eq. (34) is incorrect. This controversy is an indication that the engineered flow characterized by $P_{E}$ is NOT caused by condensation described by Eq. (34). I repeat this point: not every pressure gradient is compatible with a particular expression for S.

Similarly, an argument has been put forward that a different expression for S can be obtained from M10 Eqs. (32)-(33) by setting the horizontal flow equal to zero, $\partial Nu/\partial x = 0$. However, this condition is incompatible with the existence of a steady-state mass sink. In a closed circulation this flux at the boundary (which can be theoretically placed at an infinite distance from the sink) is necessarily zero (which, by the way, means that there must be a source somewhere as well or that the circulation pattern must move as a whole). But near the sink the flux reaches a non-zero value (otherwise stationary conditions cannot be observed). Therefore, as I several times pointed out to Nick Stokes, neither the condition $\partial Nu/\partial x = 0$ nor any observations derived from it say anything about the real processes caused by condensation. These are conditions incompatible with the existence of a steady-state mass sink in the atmosphere and hence cannot be used to analyze its magnitude.

Thus, while clearly helpful as thought experiments, considerations of engineered flow performed so far do not falsify Eq. (34) for the simple reason that they are not relevant to it. I feel that Lucia has come to the same conclusion, hence we have not seen the demonstration of an engineered flow that had been promised. The existence of a mass sink imposes strict limitations on the possible velocity and pressure fields.

Note that the system (1)-(3) is purely dynamical. There are no explicit heat sources. In atmospheric hydrodynamics one conventionally puts S = 0 but adds the 1st law of thermodynamics instead. If a correct pressure gradient (actually produced by condensation) is formally deduced from the differential heating (e.g., by fitting the unknown proportionality coefficients to observations), then resurrecting S in Eq. (3) will not produce ANY effect on the system (because the right gradient $P_S$ is already there!). This helps explain why the previous attempts to investigate the impact of S on dynamics did not find anything significant.

I will conclude my consideration by noting that an expression for S cannot be derived from the continuity equations. When we have a two-component atmosphere, we have a new equation to be added to system (1)-(3) (for water vapor):
$\displaystyle \frac{\partial \rho_v u}{\partial x} + \frac{\partial \rho_v w}{\partial z} = S$ (4)
However, this equation comes with a new variable $\rho_v$.We can add one more equation with yet another variable (for dry air).
$\displaystyle \frac{\partial \rho_d u}{\partial x} + \frac{\partial \rho_d w}{\partial z} = 0$ (5)
Still it is clear to us that there is not enough equations to determine S. Therefore, Lucia’s expression for S (Eq. 4 in the post) says nothing about S in the same way as Eqs. (3)-(5) above say nothing about it. Lucia’s Eq. 4, being a simple algebraic consequence of Eqs. (3)-(5), does not specify the magnitude of sink or source in any way. To check this, just put S = 0 and perform all the derivations to see that the left-hand part of Eq. 4 is zero. As mentioned by several readers, the value of sink S must be determined externally. It has nothing to do with the left-hand part of the continuity equation (however similar it might seem to it to readers unwilling to look into the physics). As I emphasized several times, atmospheric hydrodynamics lacks any theory on S.

A good point that has been put forward is that Eqs. (32)-(34) are valid in the region where vapor is saturated. In fact, this necessitates the existence of turbulent eddies and local evaporation, which keep the column on average saturated despite condensation occurs. Local evaporation will increase the upward flux of vapor. However, as discussed in Section 4, moisture evaporated in the region of ascent precipitates locally as well and does not contribute to the large-scale horizontal pressure gradient. Therefore, assuming that Nv is saturated takes into account the existence of such eddies in Eq. 34. (At the same time it can be shown that the column cannot be saturated everywhere. But the observations show it is pretty close to it.)

I would also like to dwell on the semi-empiricism of M10. This is true in that sense that M10 have not proved that a condensation-induced circulation is stable. That is, we have not proved that if we take an Earth with a liquid ocean, a circulation similar to what we observe will self-organize. However, M10 estimated the release of potential energy that is caused by condensation that does occur according to observations. One can no longer ignore this energy as it has been ignored before. It does exist and must be taken into account.

Once again, I would like to thank all the discussion participants for their interest. My special thanks are due to Lucia for dedicating several threads to our work. I hope that the readers have enjoyed the discussion as much as I did. If someone feels strongly about an unanswered comment which I might have overlooked, I look forward to reading your comments in the ACPD thread. (It is free, but a registration is needed.)

PS As I am not sure that Latex formulae will parse here, I will post the same comment at tAV in the Hurricane pro’s thread.

449. ### DeWitt Paynesaid

Revisiting heat dissipation in a hurricane:

If you calculate the heat dissipation from a hurricane as an open system with all the water vapor coming in as moist air at the surface and all the heat of condensation leaving as hot air at altitude, then there is far too much heat released to possibly be dissipated to space within the area of the hurricane with a cloud top temperature of 200K. 8 cm/day rainfall = 2316 W/m2. But does all that energy have to be dissipated to space and is a hurricane really a completely open system? Well, no. Hydrostatic equilibrium over a large area requires that just as much air comes down as goes up. It may be violated locally, but not over large areas. We know this is true because downdrafts as as well as updrafts are observed in hurricanes and thunderstorms. If it weren’t true, then any equation derived using the hydrostatic equilibrium assumption would be wrong.

A first approximation back of the envelope calculation: A 1 kg of dry air packet saturated with water vapor at 300 K (22.8 g water/kg air) is lifted to 10 km at a lapse rate of 5 K/km. The air temperature is 250 K. All the condensed water is allowed to fall to the ground. We’ll assume that’s all 22.8 g. Now the 1 kg dry air is moved back down to the surface at the dry lapse rate of 10 K/km (that takes work, but this is a first approximation, so I’ll ignore it). The air temperature is now 350 K. At 1005 J/kg K heat capacity, that’s 50,250 J more than dry air at 300 K. So what now? The excess energy is used to evaporate water from the surface. At 300 K, L =2437 J/g so 20.6 g water will be evaporated and the air will be nearly saturated again and ready for another cycle. If we assume that the well mixed surface layer is 50 m thick (from this link), the surface temperature only drops by 0.0002 K.

In fact, then, one only needs to dissipate enough heat to maintain the circulation and account for the energy lost from the falling rain drops. For the raindrops, that’s a maximum of 22.8 g lifted 10 km against gravity or 2234 J. Even at a vertical velocity of 10 m/sec, that’s only 2.2 W/m2. The full 22.8 g is not lifted the full 10 km either. The energy dissipation from wind velocity over a 60 km radius of a typical hurricane from this link is 1.5 E12 W or 133 W/m2. At 220 K, the cloud tops radiate 130 W/m2 and the clouds cover a lot bigger area than a radius of 60 km, more like 600 km.

A hurricane is then a water pump driven by heat from the surface water dissipated to space. Storm tracks do leave cooler water in their wake. The rate of cooling for a 50 m layer would be greater than 3600*24*133/(4137*1000*10) = 0.085 K/day. Greater than because the efficiency of the heat engine will be much less than 100%. For the temperature drops caused by Katrina and Rita in 2005, see here. The local temperature drop over four days was as large as 5 K.

450. ### Steve Fitzpatricksaid

DeWitt,

“A hurricane is then a water pump driven by heat from the surface water dissipated to space.”

What?!?!
Do you not understand that heat transport is purely incidental to the energy flow in a hurricane… and in Earth’s atmosphere in general? Don’t you not know that thermal energy flux has nothing to do with the flow of air and water in the atmosphere? Clearly you have missed the entire point of AM(10)… that is, all historical analysis of atmospheric energy flow over the last 75 years is 100% mistaken, and wholly and completely new physical ‘laws’…. like the famous Eq.34 (offered without justification, which is obviously not needed, because the profound insight of Eq 34 is on a par with basic thermodynamics). Really DeWitt, I am disappointed in you.

451. ### DeWitt Paynesaid

I’m disappointed with me too. The hair shirt wearing and self-flagellation will commence immediately. Mea culpa mea maxima culpa.

Science of Doom says he has Matlab. Does anyone know if it’s possible to obtain Matlab legally for a reasonable price? \$2,000 for the basic program is way too much for me.

452. ### phinniethewoosaid

payne

octave is the open source alternative and from what I read it is compatible in some ways.
Haven’t used it but the bigger open source sw developments are very good and actually easier to learn as they have less of the nice-too features.

453. ### anna vsaid

In your back of the envelope, what happens to volume? 1 kg at surface occupies much less space than 1kg at 5 km. Just curious.

As I have said, this is a model. AM is proposing another model under test. If it cannot be rejected theoretically, and both models fit the observations, one needs hands on data, those pesky measurements, to vote.

454. ### DeWitt Paynesaid

Of course the volume changes with altitude, that’s why I specified a mass rather than a volume. Adiabatic expansion holds $PV^\gamma$ constant where γ for air is 1.4. The temperature of the dry packet won’t follow a dry adiabat when it’s descending and be as high as 350 because the surrounding temperature is too low. The volume would not decrease as rapidly as for a dry adiabat as the pressure has to equal the surface pressure at the surface. But the energy content would still be high enough to evaporate the ~20g of water.

Now to see if I got the TeX syntax right.

455. ### anna vsaid

To continue on this vane, your model has a number of free variables too, which in computer models are put in by hand.

AM says they predict hurricane velocities without computer models which use hand inputs, and the only ad hoc is equation 34 if I understand correctly. That is one set of data that could have rejected an alternate model.

I think it is necessary to float instrumented balloons in hurricanes to measure all thermodynamic variables, not only wind velocities.
Then one could solidly discuss what the real world is doing and if it is modeled correctly or what should change in the modeling.

The interesting outputs if AM is proven to be correct is that a lot of new stuff comes out.

456. ### anna vsaid

In vain I try to tap the true vein of english language, instead I find a weather vane :).

My mother had a better one:
“may the tough cough plough me through when I laugh”

457. ### DeWitt Paynesaid

There’s a difference between free variables and adjustable parameters. I have no adjustable parameters in my calculations. I use known parameters like gas constants, heat of vaporization and fusion of water and the acceleration of gravity and known physical relationships like the ideal gas law, adiabatic expansion and conservation of energy. The only adjustment is to the free variables, temperature, pressure and volume, so that the known physical relationships are satisfied.

On further thinking, there is loss of volume in the system, it’s just not where AM says it is. The loss of volume happens when the hot dry air returns to the surface and evaporates water, not when the water condenses.

458. ### anna vsaid

In my experimentalist’s book, adjustable parameters are variables until they are measured. At a second level functional assumptions for the specific problem are also under scrutiny until they fit measurements. example of this last :ideal gas” how ideal is the gas under consideration, etc.

459. ### mixomatosesaid

there is this multilingual (kind of delusional) bloke in an Umberto Eco movie I forgot the title of now (something with a library) who might just be the alter ego of Dewitt. His cousin. One is in languages the other in scientific narratives. Or Doppelgaengers.

loss of volume,omg. Time is eaten away by the fallen angels. it was all in the davinci code, look here : a differential and 2 integral signs . The demons. Thee is many scientific reports that warned about this onset, for example:Michael Lesouris 1996, Arnold Canardo 2004

460. ### Pierre Kuntersaid

Following are some basic rules that can be helpful – you need 450 cfm per ton of cooling. Also 1 ton of cooling = 12000 BTU/h = 0.00029*12000 kW = 3.48 kW. Hope this helps…good luck

461. ### DouglasinUgandasaid

Dear All

You may be interested that my M10 co-authors and myself just posted two comments at the M10 review website. These address some common concerns.

(a) show why previous model based studies of the precipitation mass sink have incorrectly gauged the magnitude and importance of condensation driven pressure gradients;
(b) clarify what eq. 34 (in M10) represents by examining the vertical distribution vapor and dry air in a column of moist air at equilibrium. (There will be more on this soon as we are also drafting our response to Nick Stokes.)

You can find the comments here

Best wishes

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