Part II

Non-equilibrium Thermodynamics10 Second Law of Thermodynamics and Entropy

Reversibility and the Second Law

Figure 10.1: Transfer of heat from the system to its environment is spontaneous if entropy produc-tion is positive, requiring that the system has a higher temperature.

We start with a definition. Imagine a closed system. The system is not isolated, but rather existsin within a universe, or perhaps to keep things on a more manageable scale, a bigger system. Letssay through a transfer of heat to the system, we change its state from (defined by its state variables)from Si to S f (S just being some arbitrary symbol defining the state). If by taking an equivalentamount of heat away from the system the system returns exactly to its original state Si, then thesystem is said to be reversible. Any process that cannot satisfy these requirements is said to beirreversible.

An example of a reversible process might be an purely cyclical process like the buoyancy os-cillation described by the wave equation (Eq. 9.13). However, the second law of thermodynamicsstates that all natural processes are irreversible. Ultimately, any wave is damped. It loses en-ergy through friction or radiative cooling, and its amplitude must be restored through some otherexternal forcing like orographic forcing or radiative heating.

A corollary of the Second Law of Thermodynamics (which we shall not prove) is

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In a reversible transformation, heat can only be converted to work by moving heatfrom a warmer to a colder body

Another:

In the absence of external work done on a body, heat can only move from warm tocold.

For example, absorbing heat Q2 from a cold reservoir and releasing Q1 to a hot reservoir givesus a refrigerator, which requires an amount of work Q1�Q2 to have been done by some outsideagency.

We can term the change in entropy DS of the system during a transformation.

DS = DQrevT

(10.1)

or per unit mass

Ds = DqrevT

(10.2)

In a complete reversible process, the system is returned to its original state, so the entropy doesnot change and there is no change in the entropy of the surrounding universe (i.e. Ds = 0).

Since all natural processes are irreversible, the total entropy of the universe always increases.

SDS > 0

The zero in this case is simply a recognition that the net change of energy in the universe as awhole is always zero. The reason for the inequality is predicated on conservation of energy andthe condition of an amount of energy DQ moving from a high potential, high temperature systemto a low potential, low temperature environment.

SDS = �DQTsys

+DQTenv

� 0

The inequality holds provided Tsys > Tenv. The inequality is zero under assumed reversible condi-tions of local thermodynamic equilibrium for which Tsys = Tenv.

Note that in the above example that it is possible for the entropy of the system to go down evenwhile the entropy of the universe and the environment to go up. In general, the change in entropy(per unit mass) is

ds = dq/T

Under this assumption, the first law can be written

T ds = dh�ad p = cpdT �ad p (10.3)

ords = cpd lnT �Rd ln p (10.4)

but we can show that from the equation for the potential temperature that

d lnq = d lnT � Rcp

d ln p

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from which we can see thatds = cpd lnq (10.5)

Note, that in the presence of condensation it is better to use qe

ds = cpd lnqe (10.6)

If we are assuming adiabatic motions (which is sometimes a fair assumption in synoptic me-teorology, particularly at high altitudes where phase changes don’t affect entropy as much) thendq/T = ds = 0 and dq = 0. That is why when we say air moves along an isentropic surface (iso-entropic = constant entropy), it is equivalent to saying it is moving along lines of constant q . Fordry air in the absence or balance of diabatic heating (from the sun for example) and cooling (dueto radiation of heat to outer space), this is a decent approximation.

10.1 Examples of changes in entropy1. Increase the sensible heat of an object at constant pressure

T ds = cpdT �ad p✓

∂ s∂ t

◆

p= cp

✓∂ lnT

∂ t

◆

p(10.7)

or, another way this could be expressed, from Eq. 6.7 is✓

∂ s∂ t

◆

p=

1T

✓∂h∂ t

◆

p=

1T

✓∂q∂ t

◆

p(10.8)

Heating at constant pressure increases the entropy of the system.

2. Decrease in the pressure at constant temperature

T ds = cpdT �ad p✓

∂ s∂ t

◆

T=�R

✓∂ ln p

∂ t

◆

p(10.9)

This is very important because it means we can interpret the second law intuitively as arequirement that, absent some external force, the pressure must decrease with time. Remem-ber that maintaining constant temperature requires that heating equals working, i.e. from Eq.6.10 ✓

∂q∂ t

◆

T=✓

∂wdt

◆

T

We showed previously in Eq. 6.11 that work is done when, at constant pressure, the volumegoes up or the pressure goes down

✓∂w∂ t

◆

T=�RT d ln p

dt= RT

d lnadt

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So you can see now that entropy goes up when heating enables work to be done so that thepressure goes down.

✓∂ s∂ t

◆

T=

1T

✓∂q∂ t

◆

T=

1T

✓∂w∂ t

◆

T=�R

✓∂ ln p

∂ t

◆

T(10.10)

3. Move mass at constant temperature and pressure. Entropy increases in a materiallyclosed system as things get hotter at constant pressure, or if there is expansion at constanttemperature. The only thing that remains to be changed is the amount of mass at constanttemperature and pressure. If the system is materially open, so that matter can flow in and outof the constant T and p surface, this is just the same as saying

✓∂S∂ t

◆

T,p=

(∂Q/∂ t)T,pT =

(∂H/∂ t)T,pT

=cpT (∂m/∂ t)T,p

T

= cp✓

∂m∂ t

◆

T,P(10.11)

So, entropy production, provided it is along a surface of constant pressure and temperature,is related to an increase in enthalpy due to an increase in mass! When you (a surface ofconstant temperature and pressure) get fatter, your entropy goes up not down! There isso much confusion here, since many (including some quite famous physicists) think of usas ordered structures, whose growth must correspond to a decrease in entropy. Quite theopposite! Looking at us, by ourselves, it is an increase in entropy.An example of how this works for phase changes is to write:

✓∂S∂ t

◆

T,p=

(∂Q/∂ t)T,pT =

(∂H/∂ t)T,pT

(10.12)

=LT

✓∂m∂ t

◆

T,P

The entropy rises when there is latent heat release due to condensation.

10.2 Atmospheric Application: Entropy production in the atmosphereThe three possibilities for entropy production described above can show up in many differentways in the atmosphere. Obvious examples are as follows. Along a constant pressure surface,the atmosphere responds to radiative heating by increasing molecular motions to give a rise intemperature. Along a constant temperature surface, hot air expands so that the pressure decreases.Locally, the entropy can increase due to convergence of winds to increase the air mass along asurface of constant temperature and pressure (or isentrope). Another possibility is that entropy isproduced through cloud production, since latent heat is released through condensation.

11 Heat enginesNineteenth century engineers like Carnot and Watt were very interested in designing engines thatefficiently burned coal in order to do work. These “heat engines” were the basis of the European

54

and later the American Industrial evolution. Carnot imagined an idealized form of the “heat en-gine” called the Carnot Cycle. In the atmospheric sciences, this cycle is particularly useful to helpus imagine how the atmosphere behaves as a heat engine.

Figure 11.1: Carnot cycle: heating does positive expansion work at constant temperature; coolingdoes negative contraction work; in between there is adiabatic expansion and compression.

Fig. 11.1 shows an illustration of the Carnot cycle. There are four steps in the cycle.

1. From point A to B, there is some adiabatic process that raises the temperature by way ofcompression and an increase in pressure, without any increase in entropy. From Eq. 10.3

cpDT = aDp

or, more generallyDh = aDp

2. Then, from point B to C, there is diabatic heating at a constant high temperature. Energyflows into the engine. From Eq. 6.14 we know this leads to positive work being done by wayof expansion and a drop in pressure.

Dq|T = Dw =�aDp|T

The remaining steps deal with the return phases of the cycle. The same expressions apply,but there is a negative sign.

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3. In step C to D, the heating stops, but there is continued expansion through a decrease inpressure so the temperature adjusts by falling to a lower value.

4. Finally, in step D to A, there is diabatic cooling at a constant low temperature. Energy flowsout of the engine. negative work is done by the heat engine.

Main points

There are some key messages to remember here:

• In a complete cycle, the temperature does not change. The internal energy might changein any given leg of the cycle, but averaged over the course of a complete cycle, there is nochange in internal energy

• Work done by the system in step B to C is offset by work done on the system in step D toA. Since the work is driven by heating at constant temperature, and there is no change ininternal energy the total amount of work done by the system is

Dw = ÂDw = ÂDq = Dqin�DqoutNote that conservation applies: Dqin = Dqout + Dw. Only a portion of the energy that flowsinto the system is available to do work. This implies that we can express an engine efficiency

h = DwDqin

=Dqin�Dqout

Dqin(11.1)

The Second Law prescribes that the overall entropy of the universe must increase for anyprocess:

T ds(universe) > du(system)+Dw(system)Therefore, if the state of the system doesn’t change (Du = 0), the second law says you mustalways put more heat Dqin = T Ds into a system than the amount of work Dw you get out ofit - the efficiency of any process must be less than 100%.

h < 1 (11.2)

• A Carnot Cycle ends up exactly where it started, as if nothing had happened. Entropy iscreated in the universe because there is a flow of heat from high to low temperature. Thetotal increase in entropy is

Ds = DqoutT2

� DqinT1

But, within the engine, there is no change in entropy at all. Thus, the Carnot Cycle is anidealization, albeit a very useful one.

• Work is only the type of energy that we are interested in – it is your choice! There is onlyever energy in and energy out, and these flows are always in balance. There is no objectivedefinition of what counts as work. It may seem silly, but the exhaust of a car does work byforcing away surrounding air. From this perspective, any energy that is wasted in propellingthe vehicle forward would be termed waste heat.

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3.7 The Second Law of Thermodynamics and Entropy 97

It is interesting to note that if, in a graph (called atemperature–entropy diagram40), temperature (in kelvin)is taken as the ordinate and entropy as the abscissa, theCarnot cycle assumes a rectangular shape, as shown inFig. 3.22 where the letters A, B, C, and D correspond tothe state points in the previous discussion. Adiabaticprocesses (AB and CD) are represented by verticallines (i.e., lines of constant entropy) and isothermalprocesses (BC and DA) by horizontal lines. From (3.84)it is evident that in a cyclic transformation ABCDA, theheat Q1 taken in reversibly by the working substancefrom the warm reservoir is given by the area XBCY,and the heat Q2 rejected by the working substance tothe cold reservoir is given by the area XADY.Therefore, the work Q1 ! Q2 done in the cycle isgiven by the difference between the two areas, which isequivalent to the shaded area ABCD in Fig. 3.22.Any reversible heat engine can be represented by aclosed loop on a temperature–entropy diagram, and thearea of the loop is proportional to the net work done byor on (depending on whether the loop is traversedclockwise or counterclockwise, respectively) the enginein one cycle.

Thermodynamic charts on which equal areas rep-resent equal net work done by or on the workingsubstance are particularly useful. The skew T ! ln pchart has this property.

3.7.3 The Clausius–Clapeyron Equation

We will now utilize the Carnot cycle to derive animportant relationship, known as the Clausius–Clapeyron42 equation (sometimes referred to byphysicists as the first latent heat equation). TheClausius–Clapeyron equation describes how thesaturated vapor pressure above a liquid changeswith temperature and also how the melting point ofa solid changes with pressure.

Let the working substance in the cylinder of aCarnot ideal heat engine be a liquid in equilibriumwith its saturated vapor and let the initial state of thesubstance be represented by point A in Fig. 3.23 inwhich the saturated vapor pressure is es ! des at tem-perature T ! dT. The adiabatic compression fromstate A to state B, where the saturated vapor pres-sure is es at temperature T, is achieved by placing thecylinder on the nonconducting stand and compress-ing the piston infinitesimally (Fig. 3.24a). Now let thecylinder be placed on the source of heat at tempera-ture T and let the substance expand isothermallyuntil a unit mass of the liquid evaporates (Fig. 3.24b).

40 The temperature–entropy diagram was introduced into meteorology by Shaw.41 Because entropy is sometimes represented by thesymbol " (rather than S), the temperature–entropy diagram is sometimes referred to as a tephigram.

41 Sir (William) Napier Shaw (1854–1945) English meteorologist. Lecturer in Experimental Physics, Cambridge University, 1877–1899.Director of the British Meteorological Office, 1905–1920. Professor of Meteorology, Imperial College, University of London, 1920–1924.Shaw did much to establish the scientific basis of meteorology. His interests ranged from the atmospheric general circulation and forecast-ing to air pollution.

42 Benoit Paul Emile Clapeyron (1799–1864) French engineer and scientist. Carnot’s theory of heat engines was virtually unknownuntil Clapeyron expressed it in analytical terms. This brought Carnot’s ideas to the attention of William Thomson (Lord Kelvin) andClausius, who utilized them in formulating the second law of thermodynamics.

Fig. 3.22 Representation of the Carnot cycle on a tempera-ture (T)–entropy (S) diagram. AB and CD are adiabats, andBC and DA are isotherms.

Fig. 3.23 Representation on (a) a saturated vapor pressureversus volume diagram and on (b) a saturated vapor pressureversus temperature diagram of the states of a mixture of aliquid and its saturated vapor taken through a Carnot cycle.Because the saturated vapor pressure is constant if tempera-ture is constant, the isothermal transformations BC and DAare horizontal lines.

T1

T2A

X Y

D

Adiabat Adiabat

Isotherm

Isotherm

Tem

pera

ture

, T

Entropy, S

B C

eses – des

B C

DA

Sat

urat

ed v

apor

pre

ssur

e

Sat

urat

ed v

apor

pre

ssur

e

es – des

es

A, D

B, C

T – dT T

(b)Temperature

(a)Volume

P732951-Ch03.qxd 9/12/05 7:41 PM Page 97

Figure 11.2: The Carnot Cycle in S-T space, from Wallace and Hobbs.

Figure 11.3: Illustration of the Carnot cycle

Figures 11.2 and 11.3 show the Carnot Cycle graphically, arguably from a cleaner perspective. Thegoal of these figures is to set up two orthogonal co-ordinates, something we should always try todo when setting up a problem. In Figure 11.2, the cycle axes are set up in a space of entropy andtemperature. The legs of the Carnot Cycle carve out a rectangle whose area is the work done. Tosee how, consider that the area of the rectangle is

Area = (T1�T2)(Ds)= T1Ds�T2Ds= Dqin�Dqout= Dw

From this perspective, because a rectangle is carved out, the following axiom is clear

A Carnot Cycle represents the maximum amount of work that can be done in a cyclethat returns the system to its initial state.

Figure 11.3 gives the Carnot Cycle a more 3D perspective in which the legs of the Carnot Cycleare not explicitly resolved, but instead are represented by circulations along isentropes. The circu-lations are sustained by energetic flows in and out of the system. Some fraction of the incomingflow goes to work.

11.1 Atmospheric Application: A hurricaneIn the atmosphere, we can easily think of the sun heating the Earth. This lowers the air’s den-sity, and then the heated airmass does work by expanding upwards or outwards. Meanwhile itradiatively cools to space, and then it contracts. Circulations are along isentropic surfaces. Cross-isentropic flows are due to diabatic heating, as derived in an assignment question.

A good example of this is a hurricane as shown in Fig. 11.4. The ocean heats the air at ahigh constant temperature, following which there is ascent along isentropes within the hurricane

57

Figure 11.4: The Carnot Cycle in a Hurricane. Note that point D above corresponds to point A inFigs. 11.1 and 11.2.

eye-wall. Ascent in the eye-wall really is nearly isentropic. There is cooling as the air expands.Then the hurricane cools efficiently to space through thermal radiation. Finally the cooled air masssinks slowly and compresses quasi-isentropically. It is through this cycle of heating and coolingthat the hurricane is able to perform thermodynamic work in the form of horizontal and verticalmotions.

11.2 Atmospheric Application: The general circulationYou eat food to do work, and the rest of the energy gets lost as heat. The Earth gets heated by thesun, and it uses this Dqin energy to do work in the form of atmospheric motions, with the remainderof the energy Dqout getting lost to space through thermal radiation at the top of the atmosphere.The efficiency in this case is only about 1 to 2% if work is defined in terms of a capacity to createatmospheric motions.

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12 Available energyLet’s imagine that there are two potential surface (like a high and low pressure) and that the totalenthalpy in the system is

H1+2 = H1 +H2 = m1cpT +m2cpT

and the total internal energy is

U1+2 = U1 +U2 = m1cvT +m2cvT

By itself, this doesn’t tell us anything about whether mass flows from 1 to 2, or vice versa, orthere’s no flow at all. It does give us an indication of the total kinetic (or potential) energy inthe system. But that’s it. Because there is no reference to pressure differences, the total enthalpydoesn’t allow for flows.

We could express these ideas in the context of the entropy, which tells us that through thealmighty Second Law that stuff flows from low to high pressure, or high to low density at constanttemperature. A problem with using entropy as a variable is that it is not a particularly intuitiveconcept. The mechanics of using entropy for evaluating system evolution is well developed, but itsometimes feels a bit like magic. Even with the link to mass we outlined, it is still not particularlyobvious how to point at the entropy of an object. Nor are entropy changes easily described for suchimportant irreversible processes in atmospheric sciences as the scattering of light (although theycan be).

What is more intuitive, and equally valid thermodynamically, is to express irreversibility, or thedirection of time, in terms of “free” or “available” energy. Why introduce a new variable? Thinkof it this way. In a materially closed system.

• Internal Energy U is the extensive state variable that changes due to heating at constantvolume at rate dQ/dt

• Enthalpy H is the extensive state variable that changes due to heating at constant pressure atrate dQ/dt

• Available energy G is the extensive state variable that represents a potential and it increasesdue to a heating at constant pressure, it decreases due to a density drop at constant tempera-ture, and it increases if there is material convergence at constant temperature and pressure.

12.1 The Gibbs free energyJosiah Willard Gibbs, unpolluted by an engineer’s desire to build better steam engines, figuredout how to represent potential energy changes thermodynamically in the 1870s. These are nowexpressed in terms of a concept called the Gibbs free energy G. Gibbs defined this energy in termsof

G⌘ H�T S (12.1)Gibbs was one of the greatest American physicists of all time, on par with Boltzmann and Maxwellat a time when American physics barely existed. Einstein called him the “greatest mind in Ameri-can history”. What is remarkable about his work is that it is largely his formulations for statisticalmechanics, thermodynamics and vector calculus that have lasted to this day.

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12.2 A hand waving derivation of available energyFrustratingly, the expression for G is always introduced a bit ad hoc, and rarely clearly explainedin standard texts. Let’s try to hand-wave how much available energy there might be by using thenormal enthalpy of a gas as an example. We have already showed that

✓dqdt

◆

T=�RT d ln p

dt

Heating per mass leads to a pressure change by way of work. We can call RT d ln p the “amountenergy per unit mass that is available to do work”. We know that matter flows from high to lowpressure.

But there are other things that can change the total available energy that we should be able tohand-wave based on intuition. Imagine a ski hill. The total available potential energy might easilybe defined as the total number of skiers multiplied by the vertical distance of the ski slope. Achange in the potential energy could be due to an increase in a number of skiers at a constant ele-vation (i.e. mass), or an increase in the mountain size (i.e. pressure difference). A third componentis that we would have to take into account that many of the skiers would have to be raised to the topof the ski lift before their potential energy became “available”. A way to interpret Eq. 12.1 is thatH is the total amount of energy in the system, in both levels, associated with all degrees of freedomplus those associated with pressure (H = (1+ f /2)mRT ). T S is the portion of this enthalpy that isunavailable for work: total skiers minus those at the bottom of the slope or at the lift.

To keep things simple at this stage we’ll do everything in terms of unit mass, i.e. h = u+ pa =cvT + RT = cpT . Suppose again that we have two air masses with total energies h1 and h2, andwith internal energies u1 and u2. We want to know how much energy is available to drive flowsbetween airmass 1 and airmass 2. As a starting point, suppose that u1 = u2 because temperaturesof the two systems are equivalent. Then the available (or Gibbs free) energy Dg = h2�h1 = aDp.Well, that is easy, and makes sense. There is energy available to drive flows if there is a pressuredifference at constant temperature.

But, now suppose that the two air masses not only have different pressures, but different tem-peratures as well. How much energy is available then? Suppose that the airmass 1 is colder thanairmass 2. Then, the amount of available energy is less than aDp because a fraction of this energymust go into first raising the temperature of airmass 1 until it is in equilibrium with airmass 2. Inother words

Dg = aDp� (h2�h1)= aDp�Dh= aDp|T � cpDT |p

Or, in differential form

dg =�dh|p +ad p|h =�cpdT |p +ad p|T (12.2)

Ah, very nice. Notice that Eq. 12.2 is just the negative of Eq. 10.3! In other words

dgdt

=�T dsdt

(12.3)

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The law of the universe is that the entropy increases and the available energy decreases. Note thatif we wanted to be properly general in how we interpret pressure and temperature, we might wishto write Eq. 12.2 considering Eq. 5.3 as

dg =�d Âgint +dgext

where Âgint is the sum of all potentials that, unlike gext , do not drive flows.

12.3 A formal derivation of the available energyIn Gibbs original treatment

H =Z

T dS

Taking as a reference point that S2 = 0 in the upper level, and S = S1 > 0 in the unavailable lowerlevel, then DS = S1�S2 = S1 and

ZT dS = T DS = T S1 = H1

H1 is the portion of the total enthalpy that is associated with high entropy matter with S = S1. Thus,

G = H�T S = H1+2 (S = 0)�H1 (S = S1)

G is the amount of enthalpy (which could always be revised to moist static energy or whatever)that remains as being available to do work through flows from the upper level to the lower level.

Now lets say we have two systems, each with their own lower level S1 and their own temper-ature and pressure. And suppose that we are interested in the amount of energy that is availablefor flows between the two systems. We get this by taking the total derivative of G to obtain DG,assuming that DG is small.

DG = DH�D(T S)

= (DU + pDV |p +V Dp|V )�T DS|T �SDT |S +✓

∂G∂m

◆

T,pDm

= (DQ+V Dp|V )�DQ|T �SDT |S +g(T, p)Dm= �SDT |S +V Dp|V +g(T, p)Dm

where the D refers to the (small) difference between the two levels, and DU = DQ� pDV |p. Notethat we are allowing for difference in mass too through Dm so that

DG(T, p) = (∂G/∂m)T,p Dm = g(T, p)Dm

Gibbs expressed g(T, p) as the chemical potential µ (T, p) = (∂G/∂N)T,p. Indeed this is what isused most commonly in other sciences. However, in the atmospheric sciences, we tend to referencepotential with respect to mass rather than molecular number, hence g(T, p) not µ (T, p).

Notice that the nice thing about this expression is that it makes clear that, in the absence of anycontrasts in mass temperature or pressure, DG is a minimum of 0. Remember too that g(T, p) caninclude many different sources of energy, such as gz or Lvqv.

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Also, as we mentioned before, we can make the transformation DS (T, p) = cpDm, or startingfrom zero:

S = cpm(T, p)

In which case, we can write the more general expressions for the the amount of energy that isavailable to do enable energetic transformations

DG = G2(T2, p2,m2)�G1(T1, p1,m1) =�DH|m,p +V (T,m)Dp|m,T +g(T, p)Dm|T,p (12.4)

The total differential for DG can be expressed as

DG = mDg|m +g(T, p)Dm|g(T,p) (12.5)

Substituting Eq. 12.5 into Eq. 12.4, we obtain

mDg|m =�DH|m,p +V (T,m)Dp|m,T (12.6)

which is known as the Gibbs-Duhem equation. Dividing by mass, the available energy per unitmass is

Dg =�Dh|p +aDp|T (12.7)

which is the same as the expression we got through hand-waving in Eq. 12.2.What this formalizes is how the potential per unit mass of a system can be changed by a

change in pressure or temperature without changing the potential energy DG of the system. In theatmospheric sciences, we might write

dg =�dh|p +1r

d p|T (12.8)

where, as always, we could replace h with hd or hm by adding other potentials. Note that thederivative in Eq. 12.8 could be, for example, with respect to space dx or time dt.

12.4 Components of the available energyThe energy barrier

We’ve got three expressions on the right of Eq. 12.4. The first �DH (m, p) represents the potentialenergy barrier or activation energy that must first be overcome before flows can happen. Likelifting skiers up the lift before they can ski, we must break bonds by raising the temperature beforethe bonds become chemically accessible. If you think of pressure and temperature being twoorthogonal axes, we have to get the temperature difference DT back to the origin (aka T ) before wehave diffusional equilibrium and the pressure difference Dp becomes available to enable materialflows. In a Carnot Cycle, it might represent steps A to B or steps C to D. Perhaps in the atmosphere,you could interpret this as the energy that needs to be added through convection before air can“diffuse” from high to low pressure from the equator towards the poles in the Hadley cell. In acar, there is an enthalpy barrier that must be overcome through external ignition before chemicalenergy becomes available to create a pressure change. (The fact that one has to happen first is

62

a reason to object to differential expressions like d p/dT which imply the changes happen at thesame time). The amount of energy per unit matter required to over come the enthalpy barrier is:

Dg(p) = cpDT |p = Dh|p (12.9)

The amount of energy available to do work through material flows

The second term V Dp(m,T ) for DG states that, like a bigger mountain, we get greater contrasts andavailable potential energy if there are large pressure differences. Suppose that we are in a materiallyclosed system without any (or very small) temperature contrasts, then the available energy per unitmatter is

Dg(T,m) =✓

∂G∂m

◆

T=✓

∂V∂m

◆

TDp|T,m = aDp =

1r

Dp = RT D ln p =�RT D lnr (12.10)

We have seen this expression previously. If we are at constant temperature (i.e. no heating),then the gibbs energy per unit material goes down when work is done by way of expansion

and a drop of pressure. Effectively dg is the potential energy available per unit material that isavailable to do pda work. Since global entropy Ds = RD ln p always increases with time, thismeans that, absent an external inflow of energy, the total atmosphere spontaneously loses potentialenergy as the pressure drops and the specific volume expands. Stuff flows downhill from high tolow pressure and spreads out. A low pressure system has mass flowing to it from the high pressuresystem (pressure has units energy per volume). One expression we’ll come back to a lot whendiscussing evaporation is the molecular form of Eq. 12.10

Dµ(T,N) = kT Dlnp (12.11)

The amount of material available to dissipate energy through material flows

The third term yields g(T, p)Dm, the increase in potential associated with an increase in mass. Onthe face of it, this is pretty obvious. The more we have the more potential we have.

12.5 The available energy at equilibriumAt equilibrium, where DG is not changing, Eq. 12.5 must be equal to zero. Thus we get a veryimportant balance equation for equilibrium conditions,

mDg|m =�gDmg(T,p) (12.12)

Increasing the potential Dg must be balanced by a decrease �Dm in the amount of mass at thatpotential. Expressed in terms of a number and potential, this would be

nDµ|n =�µDnµ(T,p) (12.13)

which is the same as the expression we previously derived in Eq. 8.7 that yielded the power lawexpression d lnn/d ln µ .

The important point here is that we may not be able to resolve internal gradients along isen-tropes defined by surfaces of a constant relationship between T and p. However, we can assume acertain statistical distribution for the likelihood that something has a certain amount of energy.

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13 Non-equilibrium flows

13.1 The Universal StaircaseIf both pressure and temperature (or enthalpy) are fixed then from Eq. 12.8

✓∂g∂ t

◆

T,p= 0 (13.1)

Unlike the extensive variable G, which depends on mass or number, the intensive variables g cannotchange at constant T and p. So constant g is the same as saying we’re along an isentrope, whereprocesses are in the idealized state of being reversible. Nothing exciting happens along isentropes.Temperature and pressure are in a fixed relation. In fact, being reversible, time isn’t even resolved.But this is super useful because if we define this surface, then all we have to (and can) look atis irreversible, time dependent flows of matter in and out of this surface. So Eq. 13.1 may seemrather boring, but it really lets us hone in on problems in a way that was more difficult before.Another phrase you will see for describing such surfaces is that they are at Local ThermodynamicEquilibrium.

I find that a helpful way to think of the universe is as a staircase in a space of g(T, p), wherethe size of each step m is constantly changing due to flows in to the step from higher values of gand out of the step to lower values of g or chemical potential µ .

Spontaneous processes occur, by the second law, when total potential energy drops (or entropyincreases), so that for the system as a whole

dGdt

=d Âi migi (T, p)

dt< 0 (13.2)

Let’s suppose that the system as a whole is materially closed. The only way the second law can besatisfied is for matter to get distributed to lower specific potential surfaces gi (T, p) = (∂G/∂m)T,p.There can always be an increase in the total potential G along any given specific potential g(T, p).As things flow downhill, there could be net convergence of matter m at any given level.

So, from the general expression for G, along a constant potential surface, the potential energyincreases with mass. ✓

∂G∂ t

◆

T,p= g(T, p)

✓∂m∂ t

◆

T,p(13.3)

Remember, at constant T and p, g (or µ) is constant. What we have now is the result that theextensive variable G is increased due to net convergence of material into a surface at constanttemperature and pressure. This is super-important because it will call for us to apply the divergencetheorem later.

13.2 Hand-wavingWe established previously that, at any given level of temperature (where temperature could beexpanded more broadly to include things like gravitational potential) and pressure g, the potentialenergy G is determined by g(T, p)m.

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The amount of mass at any given potential level m is determined by flows. Flows throughspace are driven by gradients in the density of potential energy. It is the divergence of flows arounda point of constant potential (pressure and temperature) that determines whether the amount ofmaterial there is increasing or decreasing. We are now getting to the question of why and howthings happen. What matters?

Figure 13.1: Illustration of flows through potential energy levels leading to convergence along

As in Fig. 13.1, think of a cascading waterfall with a series of three pools, each with its owng(T, p), and g points upwards. Intuitively we might say the following

• The balance of flows of water into and out of the pool is determined by✓

∂m∂ t

◆

g(T,p)= ~jin�~jout

where ~j is a current (units mass/time) defined so that it is positive pointing downhill. Moregenerally, in a continuum of space, we could say that the change is determined by the diver-gence of flows ✓

∂m∂ t

◆

g(T,p)=�g(T, p)— ·~j =�g(T, p) d

~jdg

We see expressions like this all the time. For example g could be a height co-ordinate in agravitational potential field gz. Notice that if flows ~j are decreasing as we move down the

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spatial co-ordinate defined by —· then more must be coming into intermediate levels thanflowing out. Hence the negative sign, since this leads to dm/dt being positive.

• The Second Law demands that the average matter-weighted potential of all pools hGi =Âi migi/m = ÂGi/m must go down with time. We showed that

�Dh+RT D ln p = Dg|T,p

so at constant temperature or internal enthalpy (i.e., there is no energy barrier that mustfirst be overcome to enable flows), the potential difference is related to a pressure differencebetween pools through Dg = RT D ln p = Dp/r . Since the pressure difference is what enableshGi to drop, we’d expect the current to be proportional to the pressure difference betweenthe pools

~j µ�Dp

where the negative is because the current is in the opposite direction of the pressure gradient.Note that we have discretized the potential difference. The potential difference could also betreated as a continuum in space, in which case

~j µ�—p

Putting the above two points together, and noting that at constant T , d p = RT dr✓

∂m∂ t

◆

g(T,p)µ�— ·~j µ�— · (�—p)⌘ —2 p µ —2r

So there is a LaPlacian involved here. A gradient is required to drive flows. But unless thegradient itself changes with distance, then there is no convergence or divergence of flows atany given level. This is what we call the “steady-state” condition, for which —2r|T = 0

• We might expect some cross-sectional interface between the two pools to determine the totalmagnitude of flows

• There would be a speed of flow that should matter too

13.3 Formal solutionThe Gibbs-Duhem relationship gives us the relationship between pressure and temperature gradi-ents and potential gradients

�SDTS,p +V DpT,V = mDg

If there is no energy barrier to initiating flows, we can constrain further for constant temperature

mDg = V Dp

or

Dg = 1r

Dp (13.4)

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or as, a continuum expression

—g = 1r

—p (13.5)

So higher pools have higher pressure.How is this linked to down gradient flows? This is where we must add some additional physics.

What we would like is an expression relating the flow of mass into level g(T, p) due to a pressuregradient. To get an idea of what this should look like, previously we showed that in a steady-statesystem at constant temperature where total potential isn’t changing:

m✓

∂g∂ t

◆

m=�g

✓∂m∂ t

◆

T,p(13.6)

which means thatd lnm

dt=�d lng

dtwhich, since Dg µ Dp, dm can be rewritten as

dm|T,p =�mp

d p|T,m (13.7)

Also, since m = rV , and d ln p|T = d lnr|T

dm|T,p = �rVp

d p|T,m

= �V dr|T,m

So, the rate of increase in mass along a given potential level should be of form✓

∂m∂ t

◆

T,p=�rV

p

✓∂ p∂ t

◆

T,m(13.8)

or alternatively ✓∂m∂ t

◆

T,p=�V

✓∂r∂ t

◆

T,m(13.9)

These two equations are saying that as things fall down through pressure or density surfaces inthe system as a whole (meaning at constant temperature and mass), the mass in the lower surfaceincreases.

We now examine three ways of expressing Equations 13.8 and 13.9, all essentially equivalent.

13.4 Flows between discrete surfacesOften we treat things as if our steps are discretized (like a high and low pressure zone), such thatwe have fixed intervals in p that are crossed with characteristic time t

✓∂m∂ t

◆

T,p=�rV

pDpt

=�V Drt

(13.10)

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or ✓∂r∂ t

◆

T,p=�Dr|T,m

t(13.11)

For a volume encompassing everything, local mass or density can change due to flows down po-tential gradients. The above two equations will prove essential to understanding why whateverevaporates in sub-saturated environments, where there is a jump in vapor pressure or density. Notethat we can’t say anything in the above about what determines the time scale t . Figuring out thistime scale requires we know more about how the atmosphere works.

13.5 Continuum flows expressed in terms of a velocityWhat if we approach the problem as a continuum in space? More accurately we might consider aflux due to a speed versus forced by a pressure (or density) gradient so that

✓∂m∂ t

◆

T,p=�rV

p

✓∂ p∂ t

◆

T,m=�rV

pd pd~x

· d~xdt

=�rVp

~u · —p =�V~u · —r (13.12)

So if the local density or pressure gradient is positive in the direction of~x, then flows must have anegative velocity ~u for there to be a local convergence of matter. In terms of density

✓∂r∂ t

◆

T,p=�~u · —r (13.13)

which is the advection equation. If the local gradient Notice that if we bring the RHS to the LHS,then we get the expression for the total derivative used in dynamics. It expresses how fluid flowsat speed ~u down a density gradient lead to a local convergence of fluid.

13.6 Continuum flows to particles expressed in terms of a diffusivityNormally we think of flows as being through a cross-sectional area density A/V in the total volumeV , in which case we replace the velocity with a diffusivity D (units area/time), so that (A/V )D =AD ⌘ V~u. The diffusivity is like a velocity, except that it is a scalar rather than a vector quantity,and is always towards to system of interest, so that there is a reversal of sign.

✓∂m∂ t

◆

T,p= AD—r (13.14)

In discrete form we get ✓∂m∂ t

◆

T,p= D

ADrDx

(13.15)

where Dx is some length scale separating high and low density r . This is the equation we will useto explore growth of a droplet.

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13.7 Non-equilibrium flows and the diffusion (or heat) equationTo be more complete, we need to recognize that we may have gradient that itself is changing thusthe current ~j = AD—r may be spatially variable along some differential distance d~x. Imaginethat flows are across a cross-section A in and out of an elemental unit of volume DV = ADx thatcontains a differential of mass at constant temperature and pressure Dm = rDV (e.g. a sphere withcross-section 4pr2 has DV = 4pr2Dr). A is normal to the flow and Dx is in the direction of theflows. Further, consider that we can reference the the flows in terms of the cross-section so that wecan define a flux ~F by

~F = ~j/A

We’ll eventually treat radiation this way. From Gauss’s divergence theorem✓

∂m∂ t

◆

T,p=Z

V

⇣— ·~F

⌘dV

Thus, since

~F =~jA

=AD—r

A= D—r

and(— · —r) =�

�—2r

�

then ✓∂m∂ t

◆

T,p=Z

V

⇣— ·~F

⌘dV = DVD—2r

Thus, ✓∂r∂ t

◆

T,p=✓

∂ (m/DV )∂ t

◆

T,p= D—2r (13.16)

This is the diffusion equation used throughout physics. There can only be a local convergenceof matter when the LaPlacian of density —2r is positive, meaning that the density gradient isincreasing away from the object. If the gradient itself isn’t changing with distance, then flowsinto an isentrope will be present, but they will be equivalent to flows out. This is the Steady Statecondition described by: ✓

∂r∂ t

◆

T,p= D—2r = 0 (13.17)

The time scale t that is required to reach steady-state can be estimated using dimensional analysis:Drt⇠ D

Dx2

t ⇠ Dx2

Dor, since AD ⌘V~u

t ⇠ ADx2

Vu⇠ Dx

uThese expressions are presented because very often it makes things much easier if we can assumesteady state. What is required is that t be short compared to the timescales of interest.

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Main points

These are the key equations, all of which say the same thing in different ways.

• In a closed system, the density drops as mass collects at lower levels:

✓∂m∂ t

◆

T,p=�V

✓∂r∂ t

◆

T,m(13.18)

• Equilibrating density differences has a characteristic time t✓

∂r∂ t

◆

T,p=�Dr|T,m

t

• Equilibrating density gradients is done with speed ~u

✓∂r∂ t

◆

T,p=�~u · —r

• Particle mass with area A at a lower potential grows due to a density difference at rate

✓∂m∂ t

◆

T,p= AD—r

• Where there are flows in and out of a system, it is the gradient of gradients, or LaPlacian thatdetermines whether things grow or shrink:

✓∂r∂ t

◆

T,p= D—2r

• The above expressions all came from the same place – the Gibbs-Duhem equation. All could(and perhaps should) be modified to include an energy barrier in enthalpy or temperature,but assuming this barrier has been crossed, the above equations are seen everywhere. Thekey thing to recognize is that they all express the same idea given by the balance equation

m✓

∂g∂ t

◆

m=�g

✓∂m∂ t

◆

T,p(13.19)

Mass flows are from high to low potential energy per unit mass.

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