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## Independence

Posted by Jeff Id on October 25, 2010

M10 linked again here, starts with the following equation.

dQ = CvdT + PdV  + L dλ   — Eq 1, from first law of Thermo.

From this equation many on and offline comments were sent to me which claimed heat release was a far greater concern, gavin placed one here although his latent heat number is not correct, in fact it’s kind of all over the place. We’ve learned that several climate models assume exactly this fact though so I tried to improve on Gavin’s calculation below.

One Kg of saturated water at 30 C at constant pressure.   What happens to the temperature when all the water is magically condensed in the atmosphere at constant pressure (isobaric).

Total vapor amount from this table = 0.027Kg which corresponds to 4 percent of 1 atmosphere vapor pressure.

Heat of vaporization/condensation for H20 at 30C is 2,429,000 J/kg.  Converting 0.027 Kg of condensation releases 65,000 Joules of energy.

Q =k*P*(V2 – V1)/(k – 1)

Where k is the ratio Cp/Cv  for air = 1000/713  = 1.4

V2-V1 = Q(k-1)/(kP) = 65000J *(1.4-1)/(1.4*101000) = 0.184 m^3

Saturated air from the table linked above has a density of 1.11 Kg/m^3 or 0.896 m^3/kg.

So 1.1kg/(.896+.184)  = 1.01kg/m^3 density.    The net volume change is 20% by this equation directly but in order to do it we lost 4 percent of the vapor pressure.  So in the end we got about 16 percent expansion.  With a 5 x differential between the effect of mass loss from condensation to the energy released.

This means that the energy content of the water L dλ is the biggest factor in Eq 1.

dQ = CvdT + PdV  + L dλ

Basically this shows that condensation does cause gas expansion, it also causes an additional volume loss which is not usually incorporated in models but is a smaller factor.

Switching topics, in Eq 37 of M10, the formula was derived completely without any latent heat, and when observed values were plugged in, the equation predicted reasonable values for some of the biggest storms on earth – again while apparently completely ignoring latent heat of condensation.  Considering that the latent heat is so much larger, this should be a surprising result.

So far with respect to models we’ve received comments from several experts who agree that the condensation volume loss is a minor factor, so how is it that M10 can predict such good values from basic equations.

Now my question is, is this equation really independent from latent heat when observed values for h gamma and w/u are plugged in.   The fact that γ itself is inserted at 0.03 doesn’t convince me that we still aren’t just seeing mixed effects that happen to give reasonable values for dp/dx which is the reason the post is titled independence.  Would observed hv and w/u already contain the effects of latent heat of condensation?

The relevent paragraph form section 4.1 is here:

The argument Anastassia made by email, was that the thermal heat release was so inefficient that it can be neglected.  Climate and weather models exist that use only the latent heat effect. Maybe someone can provide more insight but somebody is wrong, models or M10.

1. ### Anastassia Makarievasaid

the formula was derived completely without any latent heat, and when observed values were plugged in, the equation predicted reasonable values for some of the biggest storms on earth

Indeed, the biggest storms, Hadley cell, and do not forget, Jeff, the overall efficiency of the general circulation (Section 4.4 in the paper). All due to the simple fact that when the pressure change is due to mass removal, we have pdV = -Vdp, such that all pressure change goes with 100% efficiency into work, i.e., generation of kinetic energy.

While when the pressure change is due to the release of heat, things are no longer as bright as before. Heat is not work, and work is not heat, they are both measured in joules, but these joules have drastically different meanings.

2. ### Jeff Idsaid

It is an interesting concept which we kind of glossed over getting into what the models say. I’ve got to say that the quality of the CAM model’s handling of moisture is diasppointing. Equal spaced grids of highly controlled heat pumps that allow very little possibility for natural feedback mechanisms to reveal themselves isn’t very useful IMO.

If you have other examples which support your results it would be appreciated.

3. ### Paul Linsaysaid

“Basically this shows that condensation does cause gas expansion, it also causes an additional volume loss which is not usually incorporated in models but is a smaller factor.”

A 20% change in volume is significant.

My question from the previous thread still stands: what mechanism transfers the latent heat to the N2/O2 molecules in an open system?

4. ### Jeff Idsaid

Paul, conduction primarily and also radiation from the condensed liquid is my guess but I dont’ know.

Anastassia, the global efficiency calc is fairly convincing IMO as is eq 37. The magnitudes are reasonable. But models are based almost entirely on latent heat, it’s hard to imagine but it may be right that we can ignore latent heat in global circulation models. It sounds a bit crazy is all.

5. ### Anastassia Makarievasaid

#4 Jeff,

Indeed, a little unusual, I agree. It is fun to follow how things develop. True science is a great thrill, you know. It is cruel, but those who survive enjoy it a lot.

6. ### RuhRohsaid

Does the cloud ‘color’ say anything?
Is it just my impression that ‘nimbus’ clouds are darker than fleecy white non-rain clouds?
Why are clouds white anyway?
On an counterintuitive tangent here…
RR

7. ### RBsaid

RR: density and local contrast.

8. ### Andrewsaid

6-The white color is a result of scattering essentially the entire visible spectrum:

http://sci.odu.edu/sci/Scire/05Edition/whitecloud.html

9. ### Terrysaid

There is one additional term that I dont see in the equation, but Im sure that the CFD experts here can elaborate on it. When a warm air parcel rises it entrains cooler air from its edges, this occurs repeatedly until the parcel asymptotically rises to an “equilibrium” height and thus temperature. It is most clearly seen for example in cooling tower plumes that are miniature cloud cells. The degree of entriainment in plume models is determined empirically, but not in CFD models (how do they do it ?). Thus heat is lost from both doing work (rising vertically) plus heating the cooler entrained air mass.

10. ### Gavinsaid

You are correct, I picked the latent heat of melting instead of condensation in my example. The corrected numbers in my examples are:

Temperature changes from 288 to 288+0.00313*24290000/1005 = 295.6 (ie. by 7.6 deg C). The new density is then 100000/(286.9*295.6)*1.01687/1.027 = 1.168 kg/m3 i.e. significantly less dense than before. The change in density due to the change in q would have been +0.002 kg/m3, and the change from the change in temperature is -0.028 kg/m3 (14 times as big and in the other direction).

so much more important than I initially calculated. So thanks.

As to why the M10 equation appears to work, is it not simply because it implicitly includes the moist adiabat? Had the dry adiabat been used (which is what you would have if L=0), the vertical gradients in gamma and p_v would be very different (and unrealistic).

11. ### Brian Hsaid

“One Kg of saturated water at 30 C at constant pressure. What happens to the temperature when all the water is magically condensed in the atmosphere at constant pressure (isobaric).

Total vapor amount from this table = 0.027Kg which corresponds to 4 percent of 1 atmosphere vapor pressure.”

S/b “air” above, methinks. Having a hard time imagining saturated water — 😉 :O

12. ### Nick Stokessaid

#9
Terry, CFD models do it through the continuity equation. The horizontal pressure (from density) gradient then produces horizontal acceleration.

That’s my problem with AM’s eq 34, leading on to 37. It seems to be continuity with the horizontal components forced to zero, or very small.

13. ### Terrysaid

Nick re #9, thanks. I may be completely out of depth here, but Eqns 32 and 33 dont seem to make sense to me. From a simple conservation of mass point of view d(Nd)/dx should equal d(Nd)/dz. And unless du/dx = dw/dz then I cant see how eqn 32 holds.

14. ### John Norrissaid

re Gavin #10: https://noconsensus.wordpress.com/2010/10/25/345435/#comment-39392
Perhaps I am missing something here. Typically there is a little more humility from someone who asserted such authority on a topic, and then got handed the correct formula.

15. ### Brian Hsaid

Re: John Norris (Oct 25 21:05), Humility and warmism have never met, much less gotten acquainted. Any warmist, like JC, who expresses humility is subject to immediate fratrophagy.

16. ### Nick Stokessaid

13 Terry,
I think you’re thinking of diffusion there, but this is advection – just accounting for stuff carried along by the fluid. In vector terms, steady state, it is div(v N_v)=0. It’s related to the Gauss divergence theorem. With v as vector velocity, V N is the mass flux (of N_v), and the integral of its divergence is the nett flow into a unit volume.

17. ### Anastassia Makarievasaid

#10 Gavin,

As to why the M10 equation appears to work, is it not simply because it implicitly includes the moist adiabat? Had the dry adiabat been used (which is what you would have if L=0), the vertical gradients in gamma and p_v would be very different (and unrealistic).

I very much welcome this discussion. To set the scene, L enters the coefficient of proportionality between relative changes of saturated vapor pressure and temperature in Clausius-Clapeyron law, Eq. 3 in M10. If L = 0, water would not be a condensable gas, vapor mass would not be removed, latent heat would not be released.

As L is not zero, and as adiabatic lapse rate is large enough, vapor is a condensable gas under terrestrial conditions. Two processes occur as it condenses: mass is removed and heat is released. Both exert some influence on air pressure and lead to formation of some gradients (due to the fact that the condensation process is essentially spatially non-uniform). There has been some confusion first, but now all seem to have agreed, as I can judge, that both latent heat release and mass removal will cause a lower pressure in the ascending column — heat release due to divergence aloft, Fig. 1c, mass removal directly.

In our work, given the real value of L, we calculate the effect of real mass removal on air pressure. So, if anything, we could only underestimate the resulting pressure difference associated with condensation. But as we obtained approximately what is observed, there is little room for the effect of latent heat release left. Although I believe from what I have learnt so far it would make a contribution at the level of 10%. This is, by the way, a much more generous share than the zero allocated to mass removal until recently.

In this context, if the lapse rate were larger than the moist adiabatic lapse rate (and kept at this value by some process), this would cause greater cooling of the ascending air and would lead to unrealistically high values of mass removal rate. This can be seen from comparing formulae (24) and (25), which show that with increasing lapse rate $\Gamma$ the height scale of water vapor $h_v$ (24) decreases, and the height scale of condensation rate $h_\gamma$ (25) at small $h_v < h$ decreases as well. This produces a greater pressure gradient in Eq. (37). It would be unrealistic, but it would be unrealistically high, not low.

To summarize, the value of L is used in M10 to derive a realistic theoretical estimate of condensation rate. Once it is done, the value of L can be put to zero, and nothing changes in our derivations and result Eq. (37). For example, if the same rate of mass removal were caused by a chemical reaction turning atmospheric gases to solid products accompanied by intake of heat, not release, we would obtain the same pressure gradient (37).

18. ### Anastassia Makarievasaid

Let me also point out the following. Latent heat is released because vapor condenses. So the rate of latent heat release is always equal to $LS$ kJ/m^3/sec, where S is volume-specific condensation rate. This means that if one reduces S (and, hence, the pressure gradient (37) caused by mass removal), rate of latent heat will be reduced as well. By the way, as latent heat release is proportional to L, with decreasing L it will decrease more rapidly than S.

So the key point is that latent heat release is just heat, while mass removal is work. There is a huge heat flux from the heating tubes in our homes, still the work it produces is negligible. It is unphysical to compare fluxes of heat to fluxes of work. Nothing can be gained from such a comparison. The fact that the argument has been put forward by many scientists shows once again how little attention has been given to the physics of mass removal in the atmospheric theory.

19. ### kuhnkatsaid

” It is cruel, but those who survive enjoy it a lot”

One could say the same for Gladiators.

20. ### skysaid

My (incomplete) understanding of condensation aloft is that the release of latent heat does NOT change the temperature of the surrounding air parcel. Since enthalpy is conserved, I wonder if that is simply the proportionate consequence of the reduction in pressure described in M10.

21. ### Brian Hsaid

#20;
Impossible. Condensation does not occur spontaneously. At the dew point, the vapour next to a “seed” (dust, large molecule, charged atom from a cosmic ray strike, etc.) joins with it, imparting its latent heat energy to the nearest atmospheric molecules or radiating it. Ask an operator of a bootleg still what happens to the cooling tubes as the alcohol (which has a latent heat just over 1/3 that of water) condenses in it. Heat equivalent to the latent heat must be withdrawn from the molecule or it will not attach to another or to a “seed”.

A pressure reduction can occur only with an expansion of volume or reduction of the total kinetic energy of the molecules. A pressure reduction proportionate to the latent heat of the water molecule would require that it had “occupied” a volume 2287X the change in volume caused by a 1°C change in temperature.

IMO.

22. ### Nic Lsaid

#21. Why is it impossible that the release of latent heat on condensation doesn’t change the temperature of the surrounding air parcel, at least when the condensation is onto an existing water droplet?

My understanding is that condensation and evaporation occurring at a liquid/ gas interface does indeed NOT change the temperature of the gas. Rather, it is the temperature of the liquid that changes, increasing upon condensation occurring and reducing upon evaporation occurring.

In the case of condensation, that is because in order for the gas molecule hitting the surface of the liquid to be captured, it must give up the kinetic energy it has gained upon approaching other molecules in the liquid phase (it loses potential energy doing so, thus must gain kinetic energy). The molecule can only lose the extra kinetic energy by imparting it to other molecules in the liquid via collisions, which raises the temperature of the liquid.

The temperature of the gas is, I think, unaffected by condensation (or evaporation). Whilst I do not claim to be an expert on this subject, my understanding of what happens seems to be supported by Feynman – see his Lectures on Physics, volume 1, at 42-1.

23. ### DeWitt Paynesaid

My understanding is that condensation and evaporation occurring at a liquid/ gas interface does indeed NOT change the temperature of the gas. Rather, it is the temperature of the liquid that changes, increasing upon condensation occurring and reducing upon evaporation occurring.

Cause and effect are reversed here. At equilibrium, evaporation and condensation are balanced. If we lift a parcel of air containing dispersed water droplets by a small distance delta x, expansion does work and reduces the kinetic energy of the gas molecules. Those gas molecules then collide with the water droplets transferring energy from the droplets to the gas molecules. Evaporation and condensation are now out of balance because the droplet temperature has been reduced. Condensation increases raising the temperature of the droplet until the temperature of the droplets and the gas molecules are equal and condensation and evaporation are balanced again. The net result is larger (or more) water droplets a lower temperature and a larger volume for the parcel of air.

24. ### An appeal to the climate science blogosphere | Watts Up With That?said

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