## Sunday, January 22, 2012

### ## Lapse rates and entropy.

The dry adiabatic lapse rate (DALR) is in the news again. Willis Eschenbach has led a discussion primarily directed at a theoretical musing on whether gas under gravity but without heat input would have a lapse rate or be isothermal.

I've explained my views on the role of the DALR here and here, and in comments on several other blogs. Just a brief recap of theory; the DALR is the vertical temperature gradient given by:
dT/dz = -g/cp = -9.8 °C/km for air,
where z is altitude, g gravity acceleration, and cp is the (constant pressure) specific heat of air. Any gas cools when it expands adiabatically (ie without exchanging heat with the environment). The rate at which dry air cools as it rises (and the pressure falls) is also the DALR.

The air generally does have a lapse rate which is not quite as large as the DALR. This observed rate is called the environmental lapse rate. The difference is often attributed to the effect of water condensation, and there is reference to a moist adiabatic lapse rate. I think that is only part of the story, as I'll show.

But my view of the role of the DALR, and the answer to Willis' query, is fairly simple. Air motions create a heat pump which move temperatures toward the DALR. If you could prevent all motion, conduction would move the temperature gradient to isothermal. But you can't; in practice the air needs heating to keep it from liquefying, and that will always create motion.

In this post, I'll review the heat pump notions, and add an entropy viewpoint which will quantify some of the issues.

#### The heat pump of moving air.

I've talked about that in the previous posts referenced above. For here, I'll just repeat part of a comment I made at WUWT
When the lapse rate is below the dry adiabat (toward isothermal) it is referred to as convectively stable. Above the adiabat, it is unstable. At the adiabat, it is neutrally stable.

At the adiabat, rising air cools by expansion, at exactly the rate at which the nearby air is becoming cooler (by lapse rate). And falling air warms. There is no buoyancy issue created. Moving air retains the same density as the environment. And no heat is transferred.

But in the stable regime, falling air warms faster than the change in ambient. It becomes less dense, so there is a force opposing its rise. That is why the air is stable. This motion both takes kinetic energy from the air and moves heat downwards (contra your statement). It is a heat pump which works to maintain the lapse rate.

Rising air does the same. It cools faster, and so rises against a buoyancy force. It takes KE from the air and moves “coolness upwards”, ie heat down. It pumps heat just as falling air does.

That is why air in motion tends to the adiabat lapse rate. A heat pump requires energy. Where from?

The atmosphere is famously a heat engine, driven by temperature differences, most notably from equator to pole, but also innumerable local differences, eg land/sea. This provides the kinetic energy that maintains the lapse rate, and it is hard to imagine any planetary atmosphere where the energy would not be available.

The effectiveness of the heat pump tapers as the adiabat lapse rate is approached. Beyond, in the unstable region, everything is reversed. The pump becomes an engine, with heat moving upward creating KE. This of course quickly diminishes the temperature gradient.

#### Entropy

As said above, a heat pump is needed to maintain a lapse rate. The reason is that there is the irreversible process of conduction down the temperature gradient:

Q = -k ∇T

where k is the thermal conductivity (W/m/°K), and Q is the flux. This is a general Fourier law, and k need not be simply molecular conductivity. For present purposes it is augmented by turbulent transfer and also by radiative transfer if there are GHGs.

I'm using "irreversible" in the thermo sense; heat flow down the gradient can be (and is) reversed by a heat pump. But that takes energy.

Since the Fourier law flow is irreversible, it creates entropy:

dEA/dx = Q d(1/T)/dx, where EA is a time rate of increase of entropy per unit area, or more conveniently (see Eq 3):
EV = (k/T2) ∇T • ∇T, where EV is the rate of entropy increase per unit volume.

As you see, EV depends only on k and the temperature and its gradient. It happens inexorably, with or without gravity. You can only reach a steady state if you either
• have k=0 or
• use energy to counter the entropy increase
The adiabatic heat pump does counter the entropy increase. But of course, globally entropy grows. The energy for the heat pump comes from air motion created by the global atmospheric heat engine, which necessarily creates at least as much entropy as the heat pump negates. And that entropy is ultimately exported from the Earth.

#### Environmental lapse rate

The observed lapse rate is usually less than the DALR, and this is often attributed to moisture. There isn't actually a "moist adiabatic lapse rate" that you can cite a value for. But one way of seeing how moisture could matter is if you think of the role of specific heat. This is just the amount of heat needed to change the temperature, and during condensation, you can say that the specific heat experiences a spike - effectively a delta function with area LH. When the SH rises, g/cp diminishes. But the moist adiabat does require that condensation is actually happening, and of course in the air that happens only in some places at some times.

But another reason for the lapse rate falling short of the DALR is simply that the heat pump is inadequate. It's effectiveness tapers to zero as the lapse rate approaches the DALR. But in fact the rate of entropy creation that it has to counter is quite high when GHG are present, because of radiative transfer (Rosseland conduction). And as above, to the extent that it is inadequate, the temp gradient tends toward isothermal.

1. Hi Nick – I’ll try to engage here in the WUWT thought experiment set up by Willis’ – an adiabatic control volume containing GHG-free air in a tall column in the presence of gravity field. The question to be answered is when the gas system reaches energy equilibrium is temperature stratified with h above cv bottom ref. h = 0 or isothermal. The Perpetuum Mobile thread caught my interest and I learned the answer is a tough one for even thermo grand masters in the past.

Here, I won’t repeat the thread learning just deal with your writing.

1) My view the Fourier conduction law is written for energy flow and the tall container stratified is in total energy balance at each h, no energy flows, thus the Q = -k ∇T is zero. There is only one system so energy can’t flow when it is in balance at each h & therefore stratified due to ideal gas law. There is no 2nd system T differential to run a heat engine. All 1 system in equilibrium.

2) This means entropy is not declining, good. Entropy is not increasing because no irreversible stuff happens at equilibrium in this 1 system energy field. The molecules move around very fast and with constant entropy.

3) You state “with or without” gravity but show no gravity terms to make that obvious in the math. In particular where is the ∇P term that arises from the gravity field? Seems it would be of opposite sign to the ∇T term. There is a +P gradient existing with the -T gradient.

Trick

2. 1. I don't think there are any energy flow caveats to Fourier's Law. It just says Q = -k ∇T

2&3. I believe that in the absence of heat pumping, entropy would increase according to
E_V = (k/T^2) ∇T &bullet; ∇T
Yes, there's no gravity term. It's just a response to the temperature gradient and the conductivity. It's not a molecular calculation, just based on entropy change = heat flux/T. In fact gravity can't play a role, because it is not part of an irreversible process.

1. 3. The dry adiabatic lapse rate is derived several ways, each of which assumes an isentropic atmosphere.

Example 1. A packet of dry air rises. Let's first assert that it is adiabatic (hopefully not controversial), as it rises it expands and does work on it's environemnt. But that work is reversible. Thus the process is adiabatic and reversible, hence isentropic. By definition.

Example 2. We know from calculus that dx/dy(w)*dy/dw(x)*dw/dx(y) = -1. Let us apply this identity to dT/dp(s), the temperature pressure relationship at constant entropy. We will find that dT/dp=-dS/dP * dT/ds. the second RHS term is Cp/T by the definitoin of entropy. The first RHS term is, by a Maxwell relationship, -dV/dT, which for an ideal gas is R/P (you can google Maxwell relationship). Making the substitutions: dT/dp = RT/PCp = V/Cp for an ideal gas. Lastly we recognize that the gravity imposed pressure gradient is dP/dz = rho*g, but rho is 1/V, so when we multiply: dT/dp * dp/dz = V/Cp * g/V = g/Cp.

Example 3. The state of an ideal gas is fully specified when two variables are specified. We shall choose T and p. The total entropy differential is: dS = dS/dT * dT + dS/dp * dp. at constant entropy, dS=0, then dT/dp = dS/dp * dT/dS and we proceed as above.

So, it seems quite clear to me that the near surface atmosphere is isentropic.

The moist adiabatic lapse rate MALR differs from the DALR because it must account for the latent heat release as water condenses. This extra heat slows the rate of cooling of a rising air packet, so the MALR is less than the DALR - but it is still isentropic as it satisfies all the condition of my example 1, but the math is kinda nasty.

I don't know if the isentropic condition is an equilibrium or steady state result. I suspect it is steady state and arises because heat transfer in the near surface atmosphere is dominated by convection - conduction is not important.

Let's set moisture aside for the moment to simplify an example. Incoming radiation heats the planet surface. Air at the surface is heated and begins to rise, thus convection begins. Convection is attempting to return the atmospheric temperature gradient to zero, but as it rises it experiences an isentropic expansion, which causes it to cool (the reverse also occurs). If convection is the dominate mode of heat transfer (conduction negligible), it cannot drive the temperature gradient to zero, but only to the DALR.

Thus, I think the DALR is a steady state condition that arises because convection is the dominate heat transfer mode and in a compressible fluid with a gravity imposed pressure gradient, and convection is constrained by the DALR.

Lastly, if I assume the entire atmosphere is isentropic - all the way to the tippy top - then T1/T2 = P1/P2^0.4. If I solve for T2 letting T1 and P1 be the conditions at the tippy top of the atmosphere, I calculate an enormous value for T2, the surface temperature. Clearly the atmosphere cannot be isentropic all the way up. At some point it becomes non-isentropic.

I think the isentropic condition breaks down when convection ceases to be the dominant mode of heat transfer. As you rise, the atmosphere becomes less dense, convection less effective, until eventually heat transfer is dominated by radiative heat transfer. Radiative heat transfer is not constrained by the DALR and can drive the temperature gradient to zero. Hence the planet surface temperature is not enormous.

Now, I could very well be wrong, or have made mistakes along the way (I do that sometimes - maybe even more than sometimes). But please don't wave your hands at me and tell me to "check the meaning of entropy" - I find that frustrating. I think I've shown sufficient work... Show me my mistake. Anybody. I won't be offended.

T = temperature
p = pressure
V = olume
R = ideal gas constant
Cp = ideal gas heat capacity = 5/2R
g = gravitational constant
rho = density = 1/V
z = vertical spatical coordinate.

4. Above lenghthy comment from kdk33. I can't figure out how to have that name when commenting here.

5. Kdk,
As I said at WUWT, I think your use of isentropic to describe parts of the atmosphere is unorthodox and unhelpful. I assume you mean that it is in a state where you can move gas around isentropically without creating temperature differences. In dry air, that means it is at the DALR. For practical purposes, I think isentropic here means adiabatic.

I agree with your examples. But I don't see how you deduce "the near surface atmosphere is isentropic". To say that, you'd need data about the state of the atmosphere.

I agree that condensing moisture changes things, though you can't put a single number on the MALR. In fact, I think the math can be much simplified by taking a literal view of cp as the amount of heat needed for a 1°C rise, with or without condensation - ie just add LH as a kind of delta function.

I agree about convection driving toward the DALR - that's my argument in this post. That's actually separate from it's role in transporting heat. For that, I think IR is generally dominant.

My view on the reason for breakdown of the ALR is that it reflects the relative size of the wind engine that maintains it, the conductive leakage, and the effect of source/sink - basically absorbed sunlight and radiation to space. Toward the tropopause turbulence drops and conductive process become relatively greater. In the stratosphere, UV absorption becomes a significant source, and reverses the gradient.

BTW, on being anon, check the menu that comes up under "Comment as". The second last option is for you to enter your name. I've been using a Google ID for a long time now - I do find it convenient, and sometimes alloows you to do more things (eg paste in text). Other ID's work too.

6. Nick,

Fair enough.

But if the near surface atmosphere were not isentropic, the DALR would not be g/Cp. You can't have it both ways. Yes, it is adiabatic. Isentropic means adiabatic and reversible. This is exactly the process a rising air packet experiences. That's why the DALR is g/Cp which requires an isentropic atmosphere. And we have now gone full circle.

I'm unsure how being accurate can be unhelpful.

But to each his own.

Good Luck!!

7. KDK, the unhelpful refers to your use of terminology, which I think obscures what you are trying to say. For example "the DALR would not be g/Cp"; The DALR is -g/Cp; that's pretty much its definition. What you mean, I think, is that the lapse rate would not be equal to the DALR.

And yes, isentropic means adiabatic and reversible. Those are words that apply to processes. It doesn't make sense to say that the near surface atmosphere is reversible.

I think its unhelpful because there are good concepts there. In dry air, your notion of isentropic air just comes down to lapse rate=DALR, which is easier to understand. But in moist air, there are also isentropic processes, and what kind of atmosphere would be invariant under those is interesting.

8. Nick,

You seem to agree with me that the near surface atmosphere is isentropic, yet you think it unhelpful to call it so. How can it be helpful to call it something it isn't and not call it what it is?

I accept it would be more clear to say: if the near suface atmosphere were not isentropic, it would not follow the DALR.

...or the MALR for that matter, which is also isentropic. The MALR only differs due to latent heat release - conceptually it is pretty the same thing.

I further accept that the DALR and MALR are idealized models and that the real atmosphere, at least on short time scales, deviates from these.

Isentropic means constant entropy. It has broader application than to just "processes". If the near surface atmosphere were at constant temperature, it would be isothermal. If it is at constant entropy, then it is isentropic. Isothermal also applies to processes.

Yes isentropic just comes down to lapse rate. The DALR is -g/Cp because it is isentropic. if it were not isentropic it would be something else. It is not good enough to say it is adiabatic; adiabatic is an incomplete description.

"isentropic" is a helpful concept because it is correct and adiabatic is not. The four canonical (I think that's the right word) thermodynamic variables are P,V,T, and S. If the atmosphere were at constant P, V, or T would you object to my calling isobaric, isochoric, or isothermal? Recognizing that one of thse four is constant is rather kinda important.

9. I haven't thought the following through, but if you'd like to ponder something interesting.

Looking at Wiki (fount of all knowledge) under thermodynamic equilibrium, it says that for a completely isolated system, the equilibrium condition is constant entropy. So, given a really tall cylinder of an ideal gas in a gravity field, it will have a gravity imposed pressure gradient and, to be at equilibrium (constant S), it must have a corresponding temperature gradient. The temperature gradient will be the one that imposes constant entropy; it will be the DALR.

Or not. I tend to think the DALR is a steady state condition brought about because convection dominates.

Perhaps you can guide me here.

10. Kdk,
I disagree. It only makes sense to speak of a continuum as being iso- for an intensive property like temperature or pressure. There's a list here. But volume and entropy are extensive variables. There's no continuum word for iso-volume. Isochoric?. Again, it's for a process, not a continuum.

When Wiki says that "For a completely isolated system, ΔS = 0 at equilibrium." they are referring to the system entropy - an extensive variable.

11. I don't really know what Wiki means, that's why I asked. OTOH, your answer doesn't make sense to me. when you say delta-S, I assume you are comparing two entropies S2 and S1, can you define these for me.

I see why there is confusion. volume can be m3, in which case it is extensive, or it can be M3/mol, in which case it is intensive. Simlarly for entropy.

The four energies are U=internal energy, H=enthalpy, G=Gibbs free energy, A=Hemholtz free energy.

dU = TdS - PdV
dH = TdS + VdP
dG = VdP - SdT
dA = -PdV - SdT

These are all intensive variables.

12. I think the extensive / intensive modifier is specific. So, enthalpy is extensive; specific enthalpy is intensive. Ditto for entropy etc.

13. Kdk,
I think Wiki means that there is no change in total entropy for the system. No irreversible processes. Usually it's easier to define an entropy change than a suitable absolute value.

This is what I was trying to use in this post. If there's a temp gradient, you can write down the associated rate of entropy change. You can't have equilibrium if entropy keeps increasing.

I think your U,H,G and A are all extensive variables. You can define specific versions. But if people refer to a change in enthalpy, say, they are referring to the extensive variable.

14. Nope. U, H, G, A, S, T, V, P are all intensive variables provided you use the "specific" form. For exampole: the units of enthalpy are BTU and this is extensive; the units of specifric enthlpy are BTU/mol and this is intensive. I'm positive. And the link you provided confirms this if you scroll down a bit.

I'm still thinking on the Wiki delta-S bit.

15. As I said, you can define specific versions. But people generally talk about the extensive form, and statements made about them don't necessarily apply to the specific form. For example, energy during a process is conserved, but energy/mol? You have to be careful which mol you are talking about and whether there is internal exchange.

16. Yes Nick, energy per mol is conserved. I have a process, it contains 1 mol of gas...

What people generally speak of probably depends on your background. In my experience, I rarely speak of the extensive property, preferring the intensive instead. Which is why it took me a bit to identify the confusion.

17. Here is an interesting thought experiment:

Imagine an infinitely tall cylinder filled with an ideal gas and placed in a gravity field (a simplified slice of an ideal gas atmosphere).

I define two cases: 1) the cylinder has a gravity imposed pressure gradient and a temperature gradient and follows the DALR. From this I calculate the total volume occupied by my gas. 2) the cylinder has a gravity imposed pressure profile, but is isothermal. I set the temperature by requiring case 1 and 2 to have total constant internal energy (the extensive U).

Which configuration has the greatest energy? Case 2 right? Well, maybe not so fast. With a different temperature profile the gas occupies a different volume. Is that volume more or less. If the volume changes, the entropy changes. If the volume of case 2 is less than case 1, this will partially (or completely) offset the entropy increase from the temperature change.

Any thoughts?

18. "Yes Nick, energy per mol is conserved. I have a process, it contains 1 mol of gas... "
No, that's exactly the difficulty. You've made it an extesive variable again.

As an intensive variable, conserved would have to mean conserved in every sub-volume. And that would generally fail if there is any internal energy transport.

19. Yes nick, energy is conserved in every sub-volume.

I can define the intensive variable as BTU/lb-mol, BTU/kg-mol, BTU/g-mol, BTU/1000 molecules. And energy conservation always applies.

Furthermore, the math works. The first law for enthalpy is:

dH = TdS + VdP

These are all intensive variables.

You don't believe me, you don't believe Wiki, you don't believe math. At this point, I must conclude you are being willfully obtuse. It's not an important enough point to continue...

On more interesting matters. Any thoughts on my proposed experiment above.

20. Kdk,
I'm not sure of your setup, but it seems to me the energies are the same. In the energy expression the hydrostatic requirement takes out both the PdV and ρgz - they balance. That leaves just c_v T, which it seems to me you'r requiring to have equal integrals in both cases.

21. Yes, that is correct. I am requiring the systems to have the same energy. But...

They will have different temperatujre profils - the DALR or a uniform temperature. As a results, they will also have different pressure profiles - the governing equation, dp/dz = rho*g, will be the same, but the density profile will differ due to the different temperature profiles.

I don't know if the total volume will be the same.

I'd like to figure out which configuration, DALR or uniform temperature - has the most entropy. I think the higher entropy state is the most stable one.

22. Hi again Nick – Your site locked me out completely after posting and I just came back & it let me in to see your response/thread. If you are still interested in the topic, I am also back after reading through the 1998 Bohren&Albrecht “Atmospheric Thermodynamics” text to which the Verkley et. al. 2004 paper refers.

You write in the top post about “general” Fourier Law. The B&A text points out it is not a “law” – it is a heat conduction formula generally applicable to solids and generally not applicable to ideal gases with molecules having outside forces acting on their mass such as gravity (or electrodynamic, et. al.).

The B&A text chapter 4.4 shows the general integration from 1st principles of non-constant temperature w/height in an adiabatic control volume of ideal dry gas column in a gravity field from maximizing entropy techniques. It concludes under those conditions temperature is indeed not constant with height (i.e. non-isothermal, isentropic). The integration is definitely non-trivial. Turn off the gravity field and it then shows the ideal gas does revert to isothermal.

I may be locked out again if the site treats me the same but I see my 1st post appeared despite the software lock-up.

Trick

23. Hello Trick,
Sorry about the problems - your comment went to the spam filter for some reason. I'll try to get access to the B&A text.

The Fourier Law is the general law of conduction. For fluids, the matter is complicated by advection. Turbulence leads to behaviour similar to Fourier's law,and a turbulent heat conductivity is often measured.

Radiation also provides an alternative heat path. But all these mechanisms, whether exact Fourier Law or not, transfer heat from hotter places to colder, and won't cease until there are no temperature differences. It takes a heat pump mechanism to maintain a temperature differential in the presence of conductive pathways, and that takes energy. Gravity only provides energy while there is a nett downward movement of mass.

24. I understand there is a long running paradox & this can be debated; however the details in this particular idealization using the rule of entropy maximization have, to me, much stronger physical merit than the intuitive (from conduction in solids) ceasing of heat flow when temperature differences equalize as you write.

Here the heat flow ceases when there are no entropy differences by random strong mixing of the molecules w/constant enthalpy. B&A & Verkley part b authors explain why this is so and why it is counterintuitive. The math supports constant entropy not constant temperature at equilibrium. Very interesting result. Any debate supporting the constant temperature solution has to deal with constant KE of the ascending molecule mass in gravity field and why that situation doesn’t break energy conservation laws.

Verkley et. al. part a shows the condition that is necessary for the isothermal solution to hold: work is done on the air column immediately above and below control volume (cv). They also show experimental evidence strongly supports the max. entropy non-isothermal, isentropic solution is way closer to real atmosphere than the isothermal solution allowing work to cross cv.

I predict you will enjoy the B&A text; it is a fun read for such a “dry” topic, ha.

Trick

25. I have lotta doubts about Lapse Rate Model, although it is accepted generally by the science.

Where are implied the properties of the air in the Lapse Rate, excluding Cp.

1) Lets say about Molar Mass, where it is dissapearing? So we find that molar mass of the air does not have influence in Lapse Rate? hmmmm

2) Lapse Rate Model is too much esotheric because we dont see in it any sample of a mass. We describe only a entropy per mass that is nothing ! Entropy in it definition have to contain the MASS