Atmospheric Thermodynamics

January 7, 2015

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Contents1 Molecular Thermodynamics 3

1.1 Atmospheric application: water vapor and dry air . . . . . . . . . . . . . . . . . . 6

2 Intensive and extensive variables 9

3 Heat, Work and the First Law of Thermodynamics 113.1 Atmospheric Application: the general circulation . . . . . . . . . . . . . . . . . . 13

4 Internal energy in an ideal gas 15

5 Total energy or the Enthalpy 195.1 Atmospheric Application: the dry and moist static energy . . . . . . . . . . . . . . 21

6 Equilibrium solutions 226.1 Case 1: Equilibrium solution for a closed system or parcel . . . . . . . . . . . . . 22

6.1.1 Atmospheric application: the geopotential . . . . . . . . . . . . . . . . . . 226.2 Case 2: Equilibrium solution for a materially closed system at constant pressure . . 23

6.2.1 Atmospheric application: the dry adiabatic lapse rate . . . . . . . . . . . . 246.3 Case 3: Equilibrium in a materially closed system at constant temperature . . . . . 25

6.3.1 Atmospheric application: the hydrostatic equation . . . . . . . . . . . . . 266.3.2 Atmospheric application: atmospheric thickness . . . . . . . . . . . . . . 27

7 Quasi-equilibrium transitions 297.1 Case 1: An ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297.2 Case 2: Phase changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317.3 Atmospheric application: potential temperature . . . . . . . . . . . . . . . . . . . 337.4 Atmospheric application: saturation vapor pressure of water . . . . . . . . . . . . 34

8 Conservation of matter and energy 368.1 Atmospheric application: Adiabatic liquid water content of a cloud . . . . . . . . . 368.2 Atmospheric application: Saturated adiabatic lapse rate and the equivalent poten-

tial temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388.3 Atmospheric application: Stability . . . . . . . . . . . . . . . . . . . . . . . . . . 39

9 Second Law of Thermodynamics and Entropy 459.1 Examples of changes in entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . 479.2 Atmospheric Application: Entropy production in the atmosphere . . . . . . . . . . 48

10 Heat engines 4810.1 Atmospheric Application: A hurricane . . . . . . . . . . . . . . . . . . . . . . . . 5110.2 Atmospheric Application: The general circulation . . . . . . . . . . . . . . . . . . 51

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1 Molecular ThermodynamicsPerhaps the most basic equation in atmospheric thermodynamics is the ideal gas law

p = rRT

where p is pressure, r is the air density, T is temperature, and R is the gas constant for dry air.While the derivation of this equation takes some effort, it is very much worth it because it gives usa deep understanding of what pressure and temperature really are.

Any work the atmosphere can do to move around air is derived from the internal energy ofits component gases. How much of this energy does a unit volume of gas have? Consider thefollowing diagram:

Figure 1.1: A piston

Say you are pushing on a piston filled with a gas. The pressure is simply

p = F/A (1.1)

and the amount of work you would have to do to compress the gas is

dW = F (�dx) (1.2)

butF = pA (1.3)

so from (1.2) and (1.3)dW =�pAdx =�pdV (1.4)

where dV = Adx since you are pushing an area through a differential length to get a differentialvolume.

Alternatively, we could switch the sign to get not the amount of work done on the gas but theamount of work done by the gas

dW = pdV (1.5)

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Where does the pressure come from? The pressure arises from the motion of the molecules ofgas inside the piston banging against the piston wall. Each time they bang against the piston wall,the piston acts like a perfect reflector. If the piston does not move due to this constant banging,then there must be a force balancing the momentum p associated with the banging of the molecules(note bold font to distinguish from pressure – momentum is a vector), that is imparted to the pistonwall with every collision. This force is

F = d p/dt (1.6)

We are assuming here that every molecule that hits the piston leaves with the same energy, i.e.the same average speed v and mass m, and therefore the same momentum. The piston is a perfectreflector. But momentum is a vector! Therefore, the momentum imparted to the piston by a singlemolecule is the change in momentum associated with coming in and bouncing out:

Dpx = mvx� (�mvx) = 2mvx (1.7)

What is Dt though? Suppose we have N molecules in volume V , or n = N/V as the concentra-tion

1. Only molecules within distance vxDt hit the piston

2. The volume occupied by these molecules is DV = vxADt

3. The number of molecules that hit the piston is DN = nDV = nvxADt, where n = DN/DV isthe molecular number concentration

4. The number of molecules that hit per unit time is DN/Dt = nDV/Dt = nvxA

Therefore, for all molecules, we combine number 4 and (1.7) to get

Fx = DN⇥DpxDt

=DNDt

Dpx

SoFx = nvxA ·2mvx

Now, taking (1.1), this leads to the expression for pressure exerted by all molecules in the x-direction

p = Fx/A = 2nmv2xBut if we only want the one half of the x-direction that is in the direction of the piston. Since weare only interested in the average velocity squared, we need to divide by two to get the pressure onthe piston alone

p = Fx/A = nmv2x (1.8)

We could also multiply and divide by two to write

p = 2nmv2x/2 (1.9)

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The term in brackets is obviously merely the average kinetic translational energy of each gasmolecule in the x-direction

u̇trans,x = mv2x/2 (1.10)

(the dot refers to “per molecule”). Since it will come up later, we will also note that, at equilibrium,the kinetic energy equals the potential energy of the molecule. We now have

p = 2nu̇trans,x (1.11)

We also note the gaseous motion is obviously three-dimensional, and there is kinetic energy ineach of these directions. Assuming (through the equipartition principle) that the energy is sharedequally among each degree of freedom:

mv2x2

= mv2y2

= mv2z2

or, since all degrees should share the energy equally, each gets a third of the total

v2x = v2/3

so, the total internal energy is given by

u̇trans = f u̇trans,x = 3u̇trans,x = mv2

2so

p =23

nu̇trans (1.12)

This equation should start to be looking vaguely familiar. We should really be calling u̇trans thetemperature of the gas molecule. However, hindsight is 20/20, and due to some historical misstepswe have defined a temperature T that is related to u̇trans by a constant factor

u̇trans =f2

kT =32

kT (1.13)

where, k = 1.38⇥10�23 J/K/molecule is Boltzmann’s constant and 3kT/2 = mv2/2. The averagekinetic (or potential) energy per degree of freedom f is kT/2.

Therefore, substituting Eq. 1.13 into Eq. 1.12

p = nkT (1.14)

It is revealing here to consider that pressure p has units of J/m3, and therefore it is an expression ofenergy density – the number density of molecules n times kT . What does kT represent? It is not theinternal energy density per molecule, which for translational energy alone is u̇trans = 32kT > kT ,and there is vibrational and rotational energy too, which makes the internal energy even bigger.Perhaps the best way to think about kT is that it is the total energy (kinetic plus potential) carriedalong the direction normal to a surface. Note k.e. = 12kT and p.e =

12kT . Thus kT is the total

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average energy per molecule that is available to do pdV work. p = nkT is then the density of thisavailable energy.

There are several ways to express the ideal gas law, which relates pressure temperature anddensity in the atmosphere (among other places). The most common formulations are

pV = NR⇤T

where N is the number of kilomoles.In the atmospheric sciences, we typically make some simplifications. We replace R⇤ by R =

R⇤/M, where M is the molecular weight of the gas in kg/kmol. In this case

p = rRT (1.15)

Alternatively, we can replace r by the specific volume a = 1/r:

pa = RT (1.16)

Main points

• Pressure is a force over an area where the force is due to the change in momentum associatedwith molecules bouncing off their surroundings

• Pressure has units of energy density

• Temperature has units of energy. The amount of energy kT/2 is the amount of translational(or potential) energy per degree of freedom per molecule.

• There are three degrees of translational freedom so internal translational energy is 3kT/2.

• Pressure has in effect two degrees of freedom due the molecular bounce

• The amount of energy kT is the energy per molecule available to do pdV work.

• Pressure is the density of energy available to do pdV work on the surroundings

Question

What if the volume of gas is an air parcel in a surrounding atmosphere? Do the equations aboveneed to change in any way if the box is removed and there is no piston?

1.1 Atmospheric application: water vapor and dry airThe pressure exerted by a mixture of chemically inert gases is equal to the sum of partial pressuresof the gases. The partial pressure is proportional to the molecular density. As we showed in Eqs.1.14 and 1.15, in total

p = nkT = rRT

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But what if there is a mixture of gases, as in the atmosphere? Dalton’s Law states that we can treatthe gases as if they are independent from one another, making the good assumption that all themolecules have the same temperature (or translational kinetic energy). In other words, we can justadd up the molecular densities ni, or energy densities pi, so that

p = kT Sni = T SriRi = Si pi (1.17)

So it is convenient to separate gases into their major constituents, which in the case of the atmo-sphere, is mostly dry air and moist air.

For dry air (which is mostly nitrogen and oxygen)

pd = rdRdT (1.18)

For water vapor we use the symbol e for pressure and

e = rvRvT (1.19)

The mean molar weight for dry air is Md =28.97 kg/kmol so Rd = R⇤/Md = 287Jdeg�1 kg�1, andfor moist air Mv = 18.0 kg/kmol so Rv = R⇤/Mv = 461Jdeg�1 kg�1.

Thus the total pressure isp = pd + e = (rdRd +rvRv)T

Note again, that we are saying that the water vapor molecules have the same temperature (kineticenergy) as the dry air molecules. We’ll introduce an important parameter here

e = RdRv

=MvMd

= 0.622 (1.20)

We define the mixing ratio (i.e. the ratio water vapor mass to dry air) in the atmosphere as

w =mvmd

(1.21)

Substitute the ideal gas law (1.18) and (1.19), we get

w =rvrd

=e/RvT

(p� e)/RdT

= e ep� e

w' e ep

(1.22)

An important adjustment that can be made to the ideal gas equation is to adjust for the addedbuoyancy that is associated with water vapor being lighter than dry air so that within a givenvolume, the density of air is given by

r = md +mvV

= rd +rv =p� eRdT

+e

RvT

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or, with some algebra

r = pRdT

1� e

p(1� e)

�=

pRdT

1� (1� e)

ew

�

Rearranging

p =rRdTh

1� (1�e)e wi = rRdTv

where Tv, defined this way, is the “virtual temperature”. It works out that to a good approximation

Tv ' T✓

1+1� e

ew

◆= T (1+0.61w) (1.23)

Figure 1.2: Moist versus dry air

This is a very useful quantity to know since the density of moist air is given by:

r = pRdTv

=p

RdT (1+0.61w)

At constant pressure, air that is either moist or hot will be less dense than air that isn’t, and be moreinclined to rise (Fig. 1.2).

Question

Dalton’s law states that partial densities and pressures are additive. Can you provide a compellingreason why temperatures are not additive as well? Why is there not a partial temperature?

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2 Intensive and extensive variables

Figure 2.1: Intensive (lower case) versus extensive (upper case) variables

When considering physical systems, it is often very useful to stay aware of whether the propertybeing considered is intensive or extensive. An intensive variable is one that does not depend on thevolume of the system, and an extensive variable is one that does. Intensive variables are sometimescalled the bulk properties of the system. For example, consider the following expression of theideal gas law

p = rRT (2.1)

Say we had a box with each of the properties p, r and T , and we subdivided the box, it would notchange the values of these properties, so each of these properties is intensive. However,

r = m/V (2.2)

Subdividing the box, into small boxes with smaller V would lead to a corresponding reduction inthe mass in each subdivided box m. Thus, m and V are extensive variables.

The distinction is important, because often we want to know how much we have of a particularthing. The nature of the thing is defined by one or two of the intensive variables p, r , and T (notethat from 2.1, we only need two to define the third). The amount we have is defined by the massm.

In atmospheric sciences, when we consider the ideal gas equation

p = rRT

where r is an intensive variable and mass m = rV is the extensive variable, where m normallyrefers to either the mass of dry air or the mass of water vapor. The convention used in Wallace andHobbs, one that is commonly employed is to use the lower case for intensive quantities and theupper case for extensive quantities. Thus, for example, the internal energy per unit mass is giventhe symbol u and the internal energy is given the symbol U . In atmospheric sciences, we very often

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thing about intensive quantities, because we reference them to a parcel of air in which the mass ofthe air (but not the volume) does not change with time. Thus

u =Um

In the notes here, I will express intensive variables that are referenced with respect to volume inbold lower case, i.e. the internal energy per unit volume is u. Where the intensive variable is permolecule it is expressed, e.g., as u̇.

Main points

• Intensive variables are independent of how much matter there is.

• Extensive variables depend on how much matter there is.

Question

This distinction between extensive variabiles and intensive variables is particularly important when-ever we talk about flows, of air, radiation, or whatever. Why?

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3 Heat, Work and the First Law of ThermodynamicsWe can express Eq. 1.12 for the ideal gas equation in terms of per unit mass rather than permolecule:

p =23

rutrans (3.1)

whereutrans =

32

RT (3.2)

is the translational energy per kilogram due to molecules whizzing around in three dimensions.There are other types of internal energy we can consider (and will later), but suffice to say we cantalk about a total internal energy per kilogram u. Note that, like T , u is an intensive variable, so theextensive property is

U = mu

where m is the mass of air. We call U, the total internal energy.Our question, is how do we change the total internal energy?

• Some outside agency adds energy to the system (e. g. condensation, absorption of sunlightor terrestrial radiation), through flows of some foreign material that does not define U . Wecall this energy “heat”

• The system changes its energy through flows of whatever material does define U (e. g.through expansion pdV or lifting mgdz). We call this “work”.

Whatever happens, there must be conservation of energy. We express this as the First Law ofThermodynamics

DU = DQ�DW (3.3)

where, DQ is the heat added to the system, and DW is the work done by the system. Both heat andwork can be either positive or negative, so what is the difference between the two? What is mostimportant to recognize is that “heat” and “work” are just names for energy. Our goal is foremostto keep an accurate account of the amount of energy that flows into and out of a system.

That said, there are different types of energy. The best way to distinguish heat from work isthat heat is associated with with flows of a different type of matter than is associated with internalenergy U . So if U = mu, where m is the mass of the material we are interested in (say the airin a parcel of air), there is also an external energy associated with a different type of materialthat we call heat (say photons associated with warming sulight). For example, radiative heatingis associated with photons. Radiative heating can heat a parcel of air composed of molecules.Heating does not directly change the extensive variable of the air parcel m.

In the atmosphere, we think about gases. For a gas, an easy way of visualizing work is toexamine a piston

Imagine there being a chemical explosion in the piston that releases external energy DQ to thegas. The external material flow here is due to a redistribution of electrons in a chemical reaction.This flow of external energy or “heating” adds to the internal energy of the gas. An increase in the

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dx

p A

V1 V2

p1

p2

p

dV

dW = pdV

Figure 3.1: A piston at work

internal energy of the gas is an increase in the amount of energy it has that is available to do work.As a consequence, the differential rate of doing work by the gas on the piston is

dWdt

= pAdxdt

= pdVdt

but pV = mRT sodWdt

=mRT

VdVdt

and noting that x�1dx/dt = d lnx/dt

dWdt

= pdVdt

= mRTd lnV

dtThe amount of work that is done by the gas in the piston is the amount of area under the curve.

DW =Z V2

V1pdV

or, at constant temperature

DW = mRTZ t

0

d lnVdt 0

dt 0

A similar derivative can be applied to the heating

DQ =Z t

0

dQdt

dt 0

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Thus, the change in internal energy is

DU =Z t

0

dQdt

dt 0 �mRTZ t

0

d lnVdt 0

dt 0 (3.4)

Since, in the atmospheric science we do things per unit mass, i.e

a = V/m

dw = pdaThus,

dudt

=dqdt� dw

dt=

dqdt� pda

dt(3.5)

and

Du = Dq�Dw =Z t

0

dqdt

dt 0 �RTZ t

0

d lnadt 0

dt (3.6)

Heating adds to internal energy so that it can do work through expansion, which allows the internalenergy to relax. An important point to be aware of is that the total amount of internal energy wehave DU (or Du) is a consequence of a time integral of heating and doing work. The present islinked to the changes in the past.

Main points

• Total energy is conserved, meaning it can only be shifted from one form to another

• When energy flows into a system, two things can happen. One is that the internal energy cango up. The other is that the system can do work on something else.

• Work, heating, and changes to internal energy happen over time. Current states evolve frompast states.

Question

• When a system does work on something else, where does that energy go? Is one system’swork another systems heat?

3.1 Atmospheric Application: the general circulationWithin the context of Figure 3.2 and Eq. 3.4, we can think about why the atmosphere is ableto do work. Due to an imbalance between incoming shortwave radiation and outgoing longwaveradiation, there is net heating at the equator and there is a net cooling at the poles. Presumablythese two things balance in total, so that the internal energy of the atmosphere as a whole does notchange, so Du = 0 and from Eq. 3.6

Dq = RTZ t

0

d lnadt 0

dt 0 (3.7)

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Figure 3.2: Balance between incoming shortwave radiation and outgoing longwave radiation at thetop of the atmosphere

This equation pretty much sums up climate! Radiative heating Dq permits work to be done throughthe expansive processes of convection and winds

R t0

d lnVdt 0 dt

0. Cooling at the poles leads to workdone on the poles through a shrinking. Meanwhile, the globally-averaged atmospheric temperatureT stays constant. Energy that is available to do work takes the form of meridional motions that wecall the wind. Energy will not be created or destroyed, but it will be transported from the equatorsto the poles because a high energy density is maintained by the imbalance at the equator and a lowenergy density is maintained at the poles. The one thing that is missing in this discussion is thatair does not leave the poles once it gets there, so there must be a return circulation.

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4 Internal energy in an ideal gasWhat if the internal energy is not constant when a gas is heated? In terms of mass, Eq. 1.13 can beexpressed as

utrans =32

RT

where the number three represents the number of degrees of freedom associated with the kineticenergy in the x, y, and z directions. By extension, the total internal energy per kilogram is

u =f2

RT (4.1)

where f is the number of degrees of freedom for a molecule. The total internal energy is not justthe energy derived from the translational motions of the individual gas molecules, but also frominternal motions by the molecules.

Now this is central to atmospheric sciences because it allows us to express how the temperatureof gas is linked to how much we heat it (through, for example, radiative absorption), i.e. thespecific heat. Remember, heating doesn’t change extensive quantities associated with how muchof a particular thing we have (e.g mass or number of molecules). What it does change is theintensive property of the amount of internal energy per unit stuff, or equivalently, the temperature.But, by how much? From the First Law of thermodynamics (Eq. 3.5),

dudt

=dqdt� dw

dtRearranging

dqdt

=dudt

+dwdt

=dudt

+RTd lna

dtIf a substance that is defined by the intensive property u is heated at constant specific volume a(e.g,. we add heat to a box filled with air, then d lna/dt = 0, and the internal energy change isequal to ✓

dudt

◆=

✓dqdt

◆

aWe would expect the temperature to go up in the box at some rate dT/dt. Dividing both sides bydT/dt, we get ✓

dudT

◆=

✓dqdT

◆

a= cv

where cv is what we call the “specific heat at constant volume” (or, equivalently, constant a). Notethat it is much more intuitive to think of it in inverse

✓dTdq

◆=

1cv

Framed this way, cv tells us how much the temperature will rise for a given amount of heating.Now, we can substitute for u, our expression from Eq. 4.1, leading to

cv =✓

∂q∂T

◆

a=

✓dudT

◆=

f2

R (4.2)

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Figure 4.1: Heating at constant volume leads to a temperature and internal energy rise

Just by inspection, we can see that, if we add energy to a volume through heating, and thevolume doesn’t change, then the temperature will rise. But it will not rise by as much as mightbe expected because there are other degrees of freedom in the substance that get a share of theenergy. f is at least three due to the translational modes of the molecules. If f is more thentemperature must rise even less. If you were a French engineer in the 19th century, this would berather important to know because increasing temperature is what is going to drive the piston, andyou would want to know what the increase in temperature is for a given release of energy fromburning coal. Since the engineers needed to figure out how much temperature (or internal energyrises), this led them to the concept of the “specific heat at constant volume” cv.

From an atmospheric science standpoint, radiative heating drives temperature perturbations,although it is not quite the same because the perturbations are not normally considered at constantvolume (unless we are considering the planet as a whole), but rather at constant pressure (see thesection on enthalpy). Still there is a fundamental problem, as will be shown below, which is thatthere is basic disagreement between what classical physics would predict the value of f to be andthe value of cv that is actually measured experimentally. In fact, this was a key point that led19th century physicists to be concerned that the wonderful successes of the kinetic theory of gasesweren’t all that. In commenting on this problem in 1869 Maxwell said in a lecture “I have now putbefore you what I consider to be the greatest difficulty yet encountered by the molecular theory,” aconumdrum that paved the way for the quantum theory of matter.

1. Monotomic Molecule (example Ar)

3 translational degrees of freedom f = 3 , Utot = 32R⇤T . Value of f for Ar implied by obser-

vations at atmospheric temperatures f = 3.

2. Diatomic molecule (examples: N2 and O2)

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3 translational degrees of freedom2 rotational degrees of freedom (whole body)2 vibrational degrees of freedom (P.E. plus K.E. (kT/2 each) associated with change in bondlength)7 degrees of freedom f = 7 , Utot = 72R

⇤T . Value of f for N2 and O2 implied by observationsat atmospheric temperatures f = 5.

There is a big discrepancy here! The classical theory works great for Argon but terrible for nitrogenand oxygen. Air is mostly a diatomic gas (N2 , O2), so for per kilogram of gas

u =52

RT = cvT (4.3)

where we would expect u = 7RT/2 based on the kinetic theory of gases. A curious thing happensthough which is that if the temperature of the gas goes up a lot, say to several thousand degrees,then f does indeed approach a value of 7 as initially predicted. How can f be a function oftemperature? Nitrogen is nitrogen is nitrogen, no matter the temperature. Right? Off hand, itmakes no sense.

One of the first successes of the quantum theory was to resolve this problem. The value kT/2(like RT/2) is a fine representation of the kinetic energy per independent mode provided that kT isnot much less than the smallest possible energy, the quantum energy hn , where n is the oscillationfrequency for that mode! Here, h is the Planck constant. Vibrational modes are higher energy thanrotational or translational modes. What happens is that if the natural frequency of a vibrationalmode n is sufficiently high that kT ⌧ hn then these modes simply will not respond to an additionof energy to the gas. Effectively these vibrational modes are “frozen”because the bonds are toostiff. If the temperature gets sufficiently high, then these modes can be activated again. Otherwise,these degrees of freedom are unavailable at normal atmospheric temperatures. Our atmosphereis composed primarily of nitrogen and oxygen, both of which have stiff bonds with high naturalvibration frequencies. These vibrational modes are simply too high potential energy to interactwith the translational modes we sense as temperature. They are way up in the hills while thetranslational modes live in the valley.

In any case, from Eq. 4.3, we can now rewrite our expression for the first law to be

dqdt

= cv✓

∂T∂ t

◆

a+ p

✓∂a∂ t

◆

T(4.4)

Heating leads to some combination of a temperature rise at constant volume and an expansion atconstant temperature

Main points

• Heating doesn’t all go into expansion work. Some goes into raising the temperature

• Energy is shared among all available degrees of freedom. The more degrees of freedom theless temperature rises.

• Air has 5 degrees of freedom, three translational and two rotational.

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Question

Eq. 4.4 is not actually used very often in the atmospheric sciences to examine temperature changesin an air parcel. What is a limitation for applying the equation to the atmosphere that wouldn’thave been a concern of most 19th century engineers and physicists?

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5 Total energy or the EnthalpySo now we can think of there being two types of energy associated with the volume of a gas. Thereis the internal energy per unit mass u that is due to the rotational and translational motions of themolecules. There is also energy per unit mass associated with the pressure of the molecules ontheir surroundings a p. The total energy is normally called the enthalpy

h = u+a p (5.1)

For an ideal gas, since pa = RTh = cvT +RT

or

h = (cv +R)T =f +2

2RT (5.2)

where f is the number of degrees of freedom for the molecules, or

dhdT

= cv +R (5.3)

For air f = 5, sodhdt

=72

RdTdt

(5.4)

To see why we have introduced the enthalpy, let’s go back to the first law again (Eq. 3.6)

dq = du+dw = du+RT d lna (5.5)

Now, previously we evaluate (∂q/∂T )a to obtain the specific heat at constant volume cv (Eq. 4.2).What about the specific heat at constant pressure? From Eq. 5.5, taking the derivative with respectto temperature: ✓

∂q∂T

◆

p=

✓∂u∂T

◆

p+RT

✓∂ lna

∂T

◆

p

but, since a = RT/p then (d lna)p = (d lnT )p, so✓

∂q∂T

◆

p=

✓∂u∂T

◆

p+RT

✓∂ lnT

∂T

◆

p

Recognizing that u = cvT and RT d lnT = RdT , we get✓

∂q∂T

◆

p= cv +R (5.6)

This is the same expression dervied in Eq. 5.3! Thus

dhdt

=✓

∂q∂ t

◆

p(5.7)

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The reason we have introduced enthalpy as well as heating, even if they change at the same rate, isbecause enthalpy is a property of the system (or a state variable) where as heat is not. You cannotpoint to the heat of a substance but you can point to its enthalpy. Enthalpy is one of the mostimportant variables in atmospheric sciences, because we cannot control for the volume of an airparcel, but we can control for its pressure. When there is is radiative heating of an air parcel atconstant pressure, the temperature rises.

But by how much? The relationship between the enthalpy rise and the temperature rise is givenby the specific heat at constant pressure:

cp =dhdT

=✓

∂q∂T

◆

p(5.8)

whereh = (cv +R)T = cpT (5.9)

Therefore, it is easy to see that, since cv = 5R/2, then cp = 7R/2. For dry air, R = 287J/deg/kg,cv = 717J/deg/kg, so cp = 1004J/deg/kg. For water vapor R = 461J/deg/kg, cv = 1463J/deg/kg,so cp = 1952J/deg/kg. For air, the internal energy has five degrees of freedom and pressure two(due to the bounce). Thus cp : cv : R ' 7 : 5 : 2. This very important, because what it means thatif we add energy to air, five parts of seven go into increasing the internal energy density, and twoparts in seven go into increasing the pressure.

Combining Eq. 5.9 with Eq. 3.5, and noting that d (pa) = pda + ad p gives an alternativeexpression of the First Law

dqdt

= cp✓

∂T∂ t

◆

p�a

✓∂ p∂ t

◆

T(5.10)

If there is heating the temperature rises and the pressure drops.

Main points

• The enthalpy is the internal energy plus the energy the system exerts on its surroundings

• The energy exerted on the surroundings is a pressure volume energy

• The specific heat is related to the number of degrees of freedom associated with the internalenergy plus the number of degrees of freedom associated with the pressure

• The total number of degrees of freedom is 3 + 2 + 2 = 7. Therefore cp : cv : R = 7:5:2 for air.

Questions

Why are we generally more interested in the enthalpy of an air parcel than its internal energy?

20

5.1 Atmospheric Application: the dry and moist static energyIn the atmosphere, we need to consider that there are other forms of internal energy that a parcelof air can have, other than just those associated with translational, vibrational and translationalmotions of a gas. A second form of internal energy is gravitational energy. In our atmosphere, aircan move up and down. Thus, add to the total internal energy the gravitational potential per unitmass

u = cvT +gz (5.11)

Adding pa = RT to this we get a revised enthalpy called the “dry static energy”

hd = gz+ cvT +RT = gz+ cpT (5.12)

so heating at constant pressure yields✓

∂q∂ t

◆

p=

dhddt

= cpdTdt

+gdzdt

(5.13)

A second form of internal energy is that associated with the thermodynamic phase of a substance.The latent heat per unit mass associated with a change in phase from liquid to vapor is Lv =2.5⇥106 J/kg at 0 degrees C. Thus the energy associated with some fraction of dry air being in thevapor phase is Lvw where w = mv/md (Eq. 1.21). Thus, adding the latent heat to the total internalenergy, we get

u = cvT +gz+Lvw (5.14)

which leads us to the moist static energy (MSE in Wallace and Hobbs)

hm = cpT +gz+Lvw (5.15)

so ✓∂q∂ t

◆

p=

dhmdt

= cpdTdt

+gdzdt

+Lvdwdt

(5.16)

Of course, we could keep going and specify any number of things depending on the question wewere interested.

As shown in Fig. 3.2, heating can be related to the radiative flux divergence through

1r

✓∂Fnet

∂ z

◆

p=

✓∂q∂ t

◆

p

So the energy that is available to create changes in T , z and w comes from the absorption andemission of radiation. The conversion of radiative flux divergence to a change in hm is the startingpoint for radiative and climate processes.

21

6 Equilibrium solutionsVery often we want to consider situations that are adiabatic, which strictly means that the internalenergy and enthalpy do not change because there is no heating. But this is a bit boring, and in factit is never possible (as will be shown). All objects radiate for example. What we look at far morefrequently is equilibrium conditions where energetic flows are in balance. Most of the textbookequations in atmospheric sciences are derived from this principle.

The starting point is three expressions of the first law (Eq. 3.7), all basically equivalent. Thefirst expression is

dudt

=DqDt� dw

dt(6.1)

or, if the work is done by an ideal gas.

dudt

=DqDt� pda

dt(6.2)

And from Eq. 5.10dhdt

=DqDt

+a d pdt

(6.3)

So again, as in Eq. 5.7, if energy is added to a parcel of air at constant pressure, it’s enthalpy willgo up by the same amount.

6.1 Case 1: Equilibrium solution for a closed system or parcelAs will be shown, constant entropy s is a short hand for no net heating of the system, i.e. is isadiabatic and Dq = 0, but work is being done on or by the system. In this case, from Eq. 6.1, weobtain ✓

∂u∂ t

◆

s=�

✓∂w∂ t

◆

s(6.4)

Main points

• For a materially closed system, the adiabatic or equilibrium condition is that work done on asystem raises its internal or potential energy. Work done by the system lowers it.

• Alternatively, a rise in the potential energy can be balanced by an ability to do work onsomething else. Potential energy is an ability to do work.

6.1.1 Atmospheric application: the geopotential

What is the potential energy of the atmosphere? There are many possible forms of potential energybut here we look at the gravitational potential. Remember from Eq. 1.2 that work is required tomove a system a distance through a force field. Here, the force field is Fg = mg, and the distancemight be d z. Thus, the work done on the system to increase its height, per unit mass is gd z. At

22

constant temperature, from Eq. 5.11, du = gdz. Thus, we can term a potential energy that is theamount of energy required to lift a mass m through a gravitation acceleration g through a distanced z. The intensive variable is the geopotential, which has units of energy per mass

F(z) =Z z

0g(z)dz

Basically, the geopotential is the potential energy per unit mass that comes from lifting through avariable gravitational field at constant temperature (see Table 3.1 in Wallace and Hobbs). For mostpurposes, we can assume gravity is a constant, so that

F(z) = gZ z

0dz

or equivalently, we can take the derivative to get

dF(T ) = gdz (6.5)

6.2 Case 2: Equilibrium solution for a materially closed system at constantpressure

Figure 6.1: An abstract illustration of where heating of a system is balanced by an equivalent rateof dissipation or cooling, so that all primary system state variables stay fixed.

23

Figure 6.2: Heating at constant pressure leads to an enthalpy or total energy rise

Now we can consider a second system in which net heating of the system at constant pressureis zero. Energy may be entering the system through heating, and leaving the system throughdissipation or cooling, but there is no net flux convergence or divergence. In other words, from Eq.6.3:

dhdt

=✓

∂q∂ t

◆

p= 0 (6.6)

Transformations can still occur however, but they must obey the “adiabatic” condition above. Adi-abatic here doesn’t mean no heating, but rather no net heating.

Main points

• For a materially closed system at constant pressure, the adiabatic or equilibrium conditionis that energetic flows in must be balanced by energetic flows out, such that total energy perunit mass h (in whatever form) does not change.

• The different components of total energy can still vary, but there must be a balance betweenso that gains in one are offset by loss of another

6.2.1 Atmospheric application: the dry adiabatic lapse rate

For example, if the air is dry, and we consider the dry static energy, a change in hd at equilibriumis given by (Eq. 5.13)

dhddt

= cpdTdt

+gdzdt

= 0

24

A parcel that rises or falls without exchanging energy with its surroundings at a given pressure willhave dhd/dt = 0 and will change its temperature at rate

cp✓

∂T∂ t

◆

p=�g

✓∂ z∂ t

◆

p

or ✓∂T∂ z

◆

p=� g

cp(6.7)

This is the “dry adiabatic lapse rate” of about Gd = �g/cp = 9.8 K/km. Of course we could dosomething similar for the moist static energy, but the derivation is more complicated (and outlinedin Wallace and Hobbs problem 3.50). Note that cp is not a function of pressure, and g is only avery very weak function of pressure (the atmosphere largely disappears by the time we care), so toa very close approximation:

dTdz

=� gcp

6.3 Case 3: Equilibrium in a materially closed system at constant tempera-ture

Figure 6.3: An abstract illustration of where heating of a system leads to working by a systemwhile the internal energy or enthalpy stay fixed

The final condition is constant temperature. A system doesn’t have to be adiabatic to be inequilibrium. Alternatively, the system might be heated, but it sufficiently fast that the temperature,or energy per degree of freedom, stays constant. In this way heating balances working. From Eq.6.1

25

Figure 6.4: Heating at constant temperature leads to an equivalent amount of working

✓∂q∂ t

◆

T=

✓∂w∂ t

◆

T= p

✓∂a∂ t

◆

T(6.8)

Substituting the ideal gas law

✓∂q∂ t

◆

T=

✓∂w∂ t

◆

T= RT

✓∂ lna

∂ t

◆

T=�RT

✓∂ ln p

∂ t

◆

T= p

✓∂a∂ t

◆

T=�a

✓∂ p∂ t

◆

T(6.9)

Main points

• For a materially closed system at constant temperature, the adiabatic or equilibrium conditionis that energetic flows in must be balanced by expansion work, such that total energy per unitmass h or u (in whatever form) does not change.

• If temperature does not change, then net heating leads to an increase in volume or a drop ofpressure

6.3.1 Atmospheric application: the hydrostatic equation

Since the total internal energy is given by u = cpT + gz (Eq. 5.12), at constant temperature du =gdz. Thus, from Eq. 6.9

gdz|a = ad p|a

26

68 Atmospheric Thermodynamics

vertical force that acts on the slab of air between zand z ! "z due to the pressure of the surroundingair. Let the change in pressure in going from height zto height z ! "z be "p, as indicated in Fig. 3.1.Because we know that pressure decreases withheight, "p must be a negative quantity, and theupward pressure on the lower face of the shadedblock must be slightly greater than the downwardpressure on the upper face of the block. Therefore,the net vertical force on the block due to the verticalgradient of pressure is upward and given by the posi-tive quantity #"p, as indicated in Fig. 3.1. For anatmosphere in hydrostatic balance, the balance offorces in the vertical requires that

or, in the limit as ,

(3.17)$p$z

% #!&

"z : 0

#"p % !&"z

Equation (3.17) is the hydrostatic equation.14 Itshould be noted that the negative sign in (3.17)ensures that the pressure decreases with increasingheight. Because & % 1!' (3.17) can be rearranged togive

(3.18)

If the pressure at height z is p(z), we have, from(3.17), above a fixed point on the Earth

or, because p(() % 0,

(3.19)

That is, the pressure at height z is equal to the weightof the air in the vertical column of unit cross-sectional area lying above that level. If the mass ofthe Earth’s atmosphere were distributed uniformlyover the globe, retaining the Earth’s topographyin its present form, the pressure at sea level wouldbe 1.013 ) 105 Pa, or 1013 hPa, which is referred toas 1 atmosphere (or 1 atm).

3.2.1 Geopotential

The geopotential * at any point in the Earth’satmosphere is defined as the work that must bedone against the Earth’s gravitational field to raisea mass of 1 kg from sea level to that point. In otherwords, * is the gravitational potential per unitmass. The units of geopotential are J kg#1 or m2 s#2.The force (in newtons) acting on 1 kg at height zabove sea level is numerically equal to !. The work(in joules) in raising 1 kg from z to z ! dz is !dz;therefore

or, using (3.18),

(3.20)d* " !dz % #'dp

d* " !dz

p(z) % #(z

!&dz

##p (()p (z)

dp % #(z

!&dz

!dz % #'dp

Column with unit cross-sectional area

Pressure = p + δp

Pressure = p

Ground

z

–δp

gρδz

δz

Fig. 3.1 Balance of vertical forces in an atmosphere inwhich there are no vertical accelerations (i.e., an atmospherein hydrostatic balance). Small blue arrows indicate the down-ward force exerted on the air in the shaded slab due to thepressure of the air above the slab; longer blue arrows indicatethe upward force exerted on the shaded slab due to the pres-sure of the air below the slab. Because the slab has a unitcross-sectional area, these two pressures have the samenumerical values as forces. The net upward force due to thesepressures (#"p) is indicated by the upward-pointing thickblack arrow. Because the incremental pressure change "p is anegative quantity, #"p is positive. The downward-pointingthick black arrow is the force acting on the shaded slab dueto the mass of the air in this slab.

14 In accordance with Eq. (1.3), the left-hand side of (3.17) is written in partial differential notation, i.e., $p!$z, because the variation ofpressure with height is taken with other independent variables held constant.

P732951-Ch03.qxd 9/12/05 7:41 PM Page 68

Figure 6.5: Fig 3.1 from Wallace and Hobbs. A downward gravitational pressure must be balancedby an upward pressure in order to maintain the air mass in hydrostatic equilibrium

since a = 1/r , we obtain the equation✓

∂ p∂ z

◆

T= rg

which suggests that pressure would increase with height. But, the atmosphere doesn’t fall down.There must be a balancing pressure pushing upwards as shown in Fig. 6.5. No vertical motionsrequires that the total vertical pressure gradient at any location is zero. By changing sign we obtainthe hydrostatic equation ✓

∂ p∂ z

◆

T=�rg (6.10)

Since density is always positive, the air pressure must decrease with height.

6.3.2 Atmospheric application: atmospheric thickness

We can now substitute the ideal gas law. Assuming that the atmospheric layer is isothermal (i.e.constant temperature), then the density is (Eq. 1.23)

r = p(z)RT

=p(z)RdTv

27

Thus, substituting we get ✓∂ p∂ z

◆

T=� p(z)g

RdTv(6.11)

Suppose we want to get the amount of potential energy per unit mass found between two layers,we are looking for a geopotential difference F2�F1. This is useful, because air flows from highto low potential energy, so we can use this difference in two locations to figure out where the airwill go. How do we find out this quantity? Re-arranging

d pp

=�gdzRdTv

Since at constant temperature dF = gdz (Eq. 6.5) and d p/p = d ln p

RdTvd ln p =�gdz (6.12)

Integrating between two levelsZ z2

z1gdz =

Z F2

F1dF =�RdTv

Z p2

p1d ln p = RdTv

Z p1

p2d ln p

Giving us

DF = F2�F1 = RdTv lnp1p2

(6.13)

This is really important because differences in DF gives us the amount of energy per unity massthat is available to do thermodynamic work through atmospheric flows. It is one possible form ofthe potential g we described earlier. Note that if we integrate Eq. 6.9 and change sign to accountfor the hydrostatic equation we can also obtain Eq. 6.13

DF = gDz = RdTv lnp1p2

(6.14)

Hurricanes are driven by a net heating of an atmospheric column, through a flux of energy fromthe ocean and a radiative cooling to space. The heating permits work to be done by way of anexpansion of the column and a pressure drop.

An expression that is related to Eq. 6.13 is to talk about the thickness of an atmospheric layerbetween two pressure surfaces. To get this we just divide Eq. 6.13 by gravity to get the hypsometricequation:

DZ = Z2�Z1 =RdTv

gln

p1p2

(6.15)

It is easy to see from this equation a very intuitive result. A warm air column with higher Tv will bethicker than a cold air column, provided that the column is defined as lying between two pressuresurfaces p1 and p2. Warm air is less dense as illustrated in Fig. 6.6.

28

3.2 The Hydrostatic Equation 71

Solution: From the hypsometric equation (3.29)

where p0 is the sea-level pressure and the rela-tionship has been used.Substituting into this expression, andrecalling that Zsea level ! 0 (Table 3.1), gives

Therefore, with p0 ! 1014 hPa, the geopotential heightZ1000 hPa of the 1000-hPa pressure surface is found tobe 112 m above sea level. !

3.2.3 Thickness and Heights of ConstantPressure Surfaces

Because pressure decreases monotonically withheight, pressure surfaces (i.e., imaginary surfaces onwhich pressure is constant) never intersect. It can beseen from (3.29) that the thickness of the layerbetween any two pressure surfaces p2 and p1 is pro-portional to the mean virtual temperature of thelayer, . We can visualize that as increases, the airbetween the two pressure levels expands and thelayer becomes thicker.

Exercise 3.4 Calculate the thickness of the layerbetween the 1000- and 500-hPa pressure surfaces(a) at a point in the tropics where the mean virtualtemperature of the layer is 15 °C and (b) at a pointin the polar regions where the corresponding meanvirtual temperature is "40 °C.

Solution: From (3.29)

Therefore, for the tropics with , #Z !5846 m. For polar regions with

. In operational practice, thickness is rounded tothe nearest 10 m and is expressed in decameters (dam).Hence, answers for this exercise would normally beexpressed as 585 and 473 dam, respectively. !

4730 m#Z !Tv ! 233 K,

Tv ! 288 K

#Z ! Z500 hPa " Z1000 hPa !RdTv

!0 ln !1000500 " ! 20.3Tv m

TvTv

Z1000 hPa # 8 (p0 " 1000)

H # 8000ln (1 $ x) # x for x %% 1

! H ln !1 $ p0 " 10001000 " # H !p0 " 10001000 "Z1000 hPa " Zsea level ! H ln ! p01000"

Before the advent of remote sensing of the atmos-phere by satellite-borne radiometers, thickness wasevaluated almost exclusively from radiosonde data,which provide measurements of the pressure, tempera-ture, and humidity at various levels in the atmosphere.The virtual temperature Tv at each level was calculatedand mean values for various layers were estimatedusing the graphical method illustrated in Fig. 3.2. Usingsoundings from a network of stations, it was possible toconstruct topographical maps of the distribution ofgeopotential height on selected pressure surfaces.These calculations, which were first performed byobservers working on site, are now incorporated intosophisticated data assimilation protocols, as describedin the Appendix of Chapter 8 on the book Web site.

In moving from a given pressure surface toanother pressure surface located above or below it,the change in the geopotential height is related geo-metrically to the thickness of the intervening layer,which, in turn, is directly proportional to the meanvirtual temperature of the layer. Therefore, if thethree-dimensional distribution of virtual temperatureis known, together with the distribution of geopoten-tial height on one pressure surface, it is possible toinfer the distribution of geopotential height of anyother pressure surface. The same hypsometric rela-tionship between the three-dimensional temperaturefield and the shape of pressure surface can be used ina qualitative way to gain some useful insights into thethree-dimensional structure of atmospheric distur-bances, as illustrated by the following examples.

i. The air near the center of a hurricane is warmerthan its surroundings. Consequently, the intensityof the storm (as measured by the depression ofthe isobaric surfaces) must decrease with height(Fig. 3.3a). The winds in such warm core lows

Fig. 3.3 Cross sections in the longitude–height plane. Thesolid lines indicate various constant pressure surfaces. Thesections are drawn such that the thickness between adjacentpressure surfaces is smaller in the cold (blue) regions andlarger in the warm (red) regions.

(a) (b)

W

A

R

M

WARM

COLD

12

0

Hei

ght (

km)

P732951-Ch03.qxd 9/12/05 7:41 PM Page 71

Figure 6.6: Fig 3.3 from Wallace and Hobbs. Where air is warm, the pressure difference requiredto fill up a vertical thickness is less; or, within a warm column, and relative to the surface, thepressure is lower at a given height.

7 Quasi-equilibrium transitionsWe have defined a two important equilibrium conditions. The first is one is heating at constantpressure and ✓

dqdt

◆

p=

dhdt

(7.1)

Assuming equilibrium, we were able to derive, for example the dry adiabatic lapse rate (∂T/∂ z)p =�g/cp.

The second sense is that there is heating at constant temperature is balanced by working througha change in pressure at constant temperature. This leads to the following key result

✓∂q∂ t

◆

T=

✓∂w∂ t

◆

T= RT

✓∂ lna

∂ t

◆

T=�RT

✓∂ ln p

∂ t

◆

T= p

✓∂a∂ t

◆

T=�a

✓∂ p∂ t

◆

T(7.2)

From which we were able to show that heating increases the atmospheric thickness.Now we ask “what will happen to the relationship between heating and enthalpy if the pressure

is changed a little bit for the constant pressure condition (Eq. 7.2)” or “what will happen to therelationship between heating and working if the temperature is changed a little bit for the constanttemperature condition (Eq. 7.2)”.

Effectively, we are asking what is the sensitivity of these equilibrium conditions to smallchanges. We are still looking for an equilibrium condition. However, we now want to knowhow the equilibrium solution is different at different temperatures. Effectively what we are seek-ing is an expression for “quasi-equilibrium transitions”. Conditions are changing slowly enoughthat something very close to equilibrium is always maintained.

7.1 Case 1: An ideal gasSuppose for example, there is no heating at constant pressure, then, from Eq. 7.1 and Eq. 5.10,

cpdT |p�ad p|T = 0

29

Figure 7.1: An abstract illustration of transitions from one equilibrium state to another

But if pressure is constant then we only get the trivial solution that temperature is constant too.Not very interesting.

But, let’s integrate to see how a very small change in pressure would relate to a very smallchange in temperature at equilibrium:

cp (T �T0) = a (p� p0)

Now, we keep pressure p0 constant as we promised, and T0 too, but now we take the derivative ofa small change with respect to temperature

cpd (T �T0)

dT= a d (p� p0)

dT

We end up with the expression

cp = ad p|TdT |p

(7.3)

that could have obtained easily before, but how we got there matters as will become more clearbelow. Since a = RT/p

d ln p|Td lnT |p

=cpR

(7.4)

30

or more generally from Eq. 5.8

d ln p|Td lnT |p

=

⇣∂q∂T

⌘

pd(pa)

dT

=dh

d (pa)(7.5)

Both the numerator and the denominator on the RHS are related to the number of degrees offreedom in the system. Eq. 7.5 is a bit of a “master equation” that we can take to almost anysituation. It’s really useful.

7.2 Case 2: Phase changesBut what about phase changes where changes are more abrupt? What we have discussed thusfar is the internal energy associated with “sensible heat”, i.e. the energy that can be associatedwith m < v2/2 > temperature T . A second form of internal energy is that associated with thethermodynamic phase of a substance. It plays an extremely important role in the atmospherebecause water is found in the liquid, ice and vapor phase, and rapidly changes between all three,releasing or absorbing vast amounts of energy.

From Eq. 7.2, the equilibrium solution is that the change in volume for a given heating is equalto inverse of the current pressure ✓

∂a∂q

◆

T=

1p

When we look at phase changes, we are concerned with the amount of heat that must be appliedto a liquid to cause it go from a dense liquid phase to a gas that occupies much more space permolecules. Because we are looking at a “jump”, since this is a phase change and the differencebetween a gas and a liquid is very large, we are interested in D differences. Thus, the relationshipis no longer a differential, but rather ✓

DaDq

◆

T=

1p

The amount of heat that must be applied to enable a phase change is the latent heat of vaporization

Dq = Lv

At equilibrium, the pressure in the gas and liquid phase are equal. Taking the pressure in the vaporphase, pvav = RvT , so

Da|T =LvavRvT

How does this equilibrium expression for constant temperature change if the temperature changesa small amount? Taking the derivative with respect to temperature and keeping the pressure fixed

dDa|TdT |p

=� LavRvT 2

31

Here, we make a simplification. For specific volume, the difference between vapor and liquid is

Da = av�al=

1rv� 1

rlBut the density of vapor is typically tiny compared to that of liquid, by about a factor of 1000, so

Da ' 1rv

= av

Thus, the expression for the quasi-equilibrium transition to another temperature becomes

dav|TdT |p

=� LvavRvT 2

ord lnav|Td lnT |p

=� LvRvT

(7.6)

Normally, this is expressed a little differently. Recognizing that d lnav|T =�d ln p|T , we get

d ln p|Td lnT |p

=Lv

RvT(7.7)

which again is similar to Eq. 7.5, except

d ln p|Td lnT |p

=Dhpav

(7.8)

where Dh = Lv and pav = RvT .

Main points

• We are often not only interested in the equilibrium solution at constant pressure or tempera-ture, but how the equilibrium solution changes as pressure or temperature change.

• The “quasi-equilibrium” solution for an ideal gas is related to the ratio of the gas constant tothe specific heat at constant pressure

• The quasi-equilibrium solution for a phase change is related to the ratio of the the gas con-stant times the temperature to the latent heat

• Either expression is dimensionless

Question

What are the similarities between Eq. 7.5 and Eq. 7.7? Can you get to Eq. 7.7 directly from Eq.7.5?

32

7.3 Atmospheric application: potential temperatureIntegrating Eq. 7.4 from initial state i to final state f gives us

Tf = Ti✓

p fpi

◆R/cp

We can use the above equation to derive an expression for the potential temperature q , whichrepresents the temperature a parcel would have if it were brought adiabatically from from (p,T ) tostandard pressure p0, which is typically set to 1000 mb. In this case

q = T✓

p0p

◆R/cp

The potential temperature is an important concept as it represents a line of thermodynamic equilib-rium for dry air in which there is no net exchange of energy with the environment through radiativeabsorption or emission, or through turbulent mixing, such that dhd = 0. In quasi-equilibrium tran-sitions an air parcel may rise or sink, or move around horizontally, simultaneously changing bothits temperature and pressure. But if there is no energy exchange, q is constant. Effectively, allparcels lying along a constant theta surface are in local thermodynamic equilibrium (Fig. 7.3).

For example, suppose an air parcel rises from the surface at 1000 mb to 700 mb without mixingwith its surroundings. If it’s surface temperature is 25 C, what is its final temperature?

Tf = Ti✓

p fpi

◆R/cp= 269K

Actually this equation can be related to the dry adiabat on a Skew T log p plot, if we consider fromabove that

lnT = R/cp ln p+ constOn the Skew T log p plot ln p is the ordinate. Since T not lnT is the abscissa, adiabats are notstraight but slightly curved from the lower right to upper left of the diagram Dry adiabats on aSkew-T are the same as lines of constant q .

It is interesting to look at potential climatologically too since air motions tend to follow surfacesof constant potential temperature. This is often used in air quality modeling to see where air isgoing or coming from (look up HYSPLIT for example).

At least, the potential temperature is very useful for being a tracer of air motions in systems thatare not changing rapidly. What do we mean by rapidly? Well, consider the following importanttimescales in the atmosphere.

• Free troposphere with no condensational processes – 1 to 2 weeks is the timescale for radia-tive cooling to have an influence

• Synoptic disturbances: timescale – 1 to 2 weeks – the same as the above, which is not acoincidence since radiative cooling and heating drive atmospheric motions

• Moist convection: timescales – minutes to hoursThe implication is that an assumption of thermodynamic equilibrium, and motion along constant qsurfaces is a useful guide over shorter time scales of perhaps a few days, provide there is no moistconvection. Where would this work? Not the tropics!

33

Figure 7.2: A Skew T diagram for SLC

7.4 Atmospheric application: saturation vapor pressure of waterThe famous Clausius Clapeyron equation gives the sensitivity of the saturation (equilibrium) vaporpressure over a plane surface of pure water eS (T ). From Eq. 7.7, we have

d lneSd lnT

=Lv

RvT(7.9)

One would be tempted to derive a solution for this equation as

eS (T ) = eS0 (T )exp✓

LvRv

✓1T0� 1

T

◆◆(7.10)

At freezing the latent heat is 2.5⇥ 106 J kg�1 and Rv = 461 J/K/kg. This is fine, but only if thedifference between T and T0 is small. The expression d ln p/d lnT is extraordinarily powerful, butit must not be solved without care, because it is a sensitivity and other things that we have assumedto be constant might be changing with temperature also. The latent heat Lv does vary somewhat bya few percent over the normal range of atmospheric temperatures, and this does bad things to theaccuracy of our nice exponential solution for e(T ). A more accurate solution takes into account asolution for Lv (T ).

es(T ) = 611.2exp(17.67T

T +243.5)

34

Figure 7.3: Zonal mean temperature and potential temperatures in the troposphere (FromHoughton, 3rd Ed.).

where, T is in C. The reason for the slightly different form of the equation is that Lv itself is a weakfunction of temperature.

Often you will hear the expression that at warm temperatures the air can hold more watervapor. Certainly from the C-C equation this seems to be correct, but semantically it is misleading.Nowhere in our discussion of the derivation have we needed to mention air at all. Rather, the C-Cequation would hold even in the absence of dry air. That said, in our atmosphere at least, havingair present is what enables the temperature to be as high as it is, since it is molecular collisions thatenable radiative absorption to show up as a measurable temperature.

35

3.5 Water Vapor in Air 81

(a) Unsaturated (b) Saturated

Water

·· · ·

·

· ··

· ·

·

··· · ·

·

·

··

·

····

· · ··

·

·

·T, e T, es

··

·

Water

Fig. 3.8 A box (a) unsaturated and (b) saturated with respectto a plane surface of pure water at temperature T. Dots repre-sent water molecules. Lengths of the arrows represent therelative rates of evaporation and condensation. The saturated(i.e., equilibrium) vapor pressure over a plane surface of purewater at temperature T is es as indicated in (b).

26 For further discussion of this and some other common misconceptions related to meteorology see C. F. Bohren’s Clouds in a Glass ofBeer, Wiley and Sons, New York, 1987.

27 As a rough rule of thumb, it is useful to bear in mind that the saturation vapor pressure roughly doubles for a 10 °C increase intemperature.

to a plane surface of pure water at temperature T, andthe pressure es that is then exerted by the water vaporis called the saturation vapor pressure over a planesurface of pure water at temperature T.

Similarly, if the water in Fig. 3.8 were replaced by aplane surface of pure ice at temperature T and therate of condensation of water vapor were equal tothe rate of evaporation of the ice, the pressure esiexerted by the water vapor would be the saturationvapor pressure over a plane surface of pure ice at T.Because, at any given temperature, the rate of evapo-ration from ice is less than from water, es(T) ! esi(T).

The rate at which water molecules evaporatefrom either water or ice increases with increasingtemperature.27 Consequently, both es and esi increasewith increasing temperature, and their magnitudes

depend only on temperature. The variations withtemperature of es and es " esi are shown in Fig. 3.9,where it can be seen that the magnitude of es " esireaches a peak value at about "12 °C. It followsthat if an ice particle is in water-saturated air it willgrow due to the deposition of water vapor upon it.In Section 6.5.3 it is shown that this phenomenon

It is common to use phrases such as “the air is sat-urated with water vapor,” “the air can hold nomore water vapor,” and “warm air can hold morewater vapor than cold air.” These phrases, whichsuggest that air absorbs water vapor, rather like asponge, are misleading. We have seen that thetotal pressure exerted by a mixture of gases isequal to the sum of the pressures that each gaswould exert if it alone occupied the total volumeof the mixture of gases (Dalton’s law of partialpressures). Hence, the exchange of water mole-

cules between its liquid and vapor phases is(essentially) independent of the presence of air.Strictly speaking, the pressure exerted by watervapor that is in equilibrium with water at a giventemperature is referred more appropriately to asequilibrium vapor pressure rather than saturationvapor pressure at that temperature. However, thelatter term, and the terms “unsaturated air” and“saturated air,” provide a convenient shorthandand are so deeply rooted that they will appear inthis book.

3.3 Can Air Be Saturated with Water Vapor?26

Temperature (°C)

es–esi

es

e s –

esi (

hPa)

0.28

0.24

0.20

0.16

0.12

0.08

0.04

0

60

50

40

30

20

10

0

Sat

urat

ion

vapo

r pr

essu

re e

s ov

er p

ure

wat

er (

hPa)

–50 403020100–10–20–30–40

Fig. 3.9 Variations with temperature of the saturation (i.e.,equilibrium) vapor pressure es over a plane surface of purewater (red line, scale at left) and the difference between esand the saturation vapor pressure over a plane surface of iceesi (blue line, scale at right).

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Figure 7.4: Fig 3.9 from Wallace and Hobbs. Saturation vapor pressure over liquid es and thedifference of saturation vapor pressures between liquid and ice es� esi.

8 Conservation of matter and energyFrom Eq. 6.6, we stated that the enthalpy (or total energy) of an adiabatic parcel does not changewith time. We also alluded to the fact that the system is “materially closed”, requiring that theamount of matter in the system is also constant.

8.1 Atmospheric application: Adiabatic liquid water content of a cloudThe basic physics we need to consider for understanding the formation of cloud in a parcel thatrises and falls adiabatically is that heat and energy in the parcel are conserved

• Q = c +w = const. (Conservation of mass)

• hm = cpT +gz+Lvw = const. (Conservation of energy)

36

where w is the water vapor mixing ratio and c is the liquid water mixing ratio. Q is the totalwater content, the quantity that is conserved if the parcel does not gain or lose mass. In a pseudo-adiabatic process, as the air expands, conservation of mass requires that the saturated mixing ratiodecreases by an amount dws as vapor is converted to condensed liquid. If the air remains saturated

Q = c +ws (8.1)

dQ = 0 = dc +dws (8.2)

dc =�dws (8.3)The liquid water content has units of g/m3, so

LWC = rc (8.4)

where r is the air density. Typical values range from 0.05 g/m3in thin stratus to in excess of 1g/m3in cumulus towers.

In a saturated adiabatic process

dhm = 0 = cpdT +gdz+Lvdws (8.5)

Therefore�Lvdws = cpdT +gdz

dwsdz

=�cpLv

dTdz

+gcp

�

dwsdz

=�cpLv

(Gd�Gs)' const. (8.6)

or, from Eq. 8.3 and 8.4dLWC

dz=

rcpLv

(Gd�Gs)' const. (8.7)

So we have just shown that LWC increases approximately linearly with height. Lets try sometypical values for stratocumulus

Gs ' 6⇥10�3

Gd = 9.8⇥10�3

r ' 1cp = 1004

Lv = 2.5⇥106

which gives usdLWC

dz' 1.5⇥10�3 g/m3/km

which is about right. A 200 m thick stratus cloud typically has a cloud top value of LWC of 0.3g/kg.

37

8.2 Atmospheric application: Saturated adiabatic lapse rate and the equiv-alent potential temperature

But what is the moist adiabatic lapse rate Gs?Again, in a moist adiabatic process where mass is also conserved

dhm = cpdT +gdz+Lvdws = 0

We can use this to derive the saturated adiabatic lapse rate,

Gs ⌘�dTdz

=gcp

+Lvcp

dwsdz

but ws is a function of temperature and pressure since ws ' ees (T )/p, where es (T ) is given by theClausius Clapeyron equation (Eq. 7.9)

desdT

=LvesRvT 2

(8.8)

Thusdws =

✓∂ws∂ p

◆

Td p+

✓∂ws∂T

◆

pdT

dwsdz

=✓

∂ws∂ p

◆

T

d pdz

+✓

∂ws∂T

◆

p

dTdz

But the hydrostatic equation tells us that d p/dz =�rg. Also Gs =�dT/dz

dwsdz

=�✓

∂ws∂ p

◆

Trg�

✓∂ws∂T

◆

pGs (8.9)

Equating this with Eq. 8.6, and solving for Gs, we get

Gs =gcp

1�rL⇣

∂ws∂ p

⌘

T

1+ Lcp⇣

∂ws∂T

⌘

p

And recognizing ws ' ees (T )/p, and the C-C equation for es (T ) we get

Gs =gcp

1+ LwsRdT1+ eL

2wscpRdT 2

(8.10)

It is often useful to refer to a thermodynamic chart for a first guess. The point of learning themis to see how the essential ingredients of cloud physics can be combined: conservation of energyand mass, the hydrostatic equation, and the Clausius-Clapeyron equation for the saturation vaporpressure over a liquid surface.

Another expression, that is useful, one that is derived in Wallace and Hobbs, is the EquivalentPotential Temperature, which is conserved along moist adiabats. As before when we said that it isequivalent to say

38

dhddt

= 0,dqdt

= 0

Here

qe = T✓

p0p

◆R/cpexp

wsLvcpT

�

qe = q exp

wsLvcpT

�

and

dhmdt

= 0,dqedt

= 0

On the Skew-T plot shown in Fig. 7.2, moist adiabats are the purple lines. Note that they donot cool with height as rapidly as the orange dry adiabats. Note too that in the region around 600mb where the dewpoint and the temperature are almost the same, the temperature nearly parallelsthe nearest moist adiabat line. This suggests a well-mixed cloud layer, perhaps altocumulus.

8.3 Atmospheric application: StabilityWe are going to try to address the problem of how a parcel of air rises within its environmentwhen there is no exchange of energy or matter. The basic conceptual idea is that a parcel startssomewhere in the atmosphere with some updraft velocity, perhaps due to orographic lifting orlocalized convergence of some sort. As the parcel rises its further acceleration or deceleration isdue to its temperature relative to its environment. If the parcel is warmer than its environment, abuoyancy force will accelerate the parcel upwards towards regions of similar low environmentaldensity aloft. However, as the parcel rises, it cools. In an idealized model in which the parceldoes not mix with its surroundings, it cools with height at the dry adiabatic lapse rate. Since theenvironment is cooling at slower rate with height, eventually the parcel is colder and less buoyantthan its surroundings, and it begins to sink.

What we will show is that an unstable dry atmosphere is characterized by �dT/dz = G > Gdor dq/dz < 0. A stable dry atmosphere is characterized by G < Gd or dq/dz > 0

The physics is as follows:Within both the parcel and the environmental air, from the hydrostatic equation (Eq. 6.10)

d pdz

=�rg

39

88 Atmospheric Thermodynamics

3.6 Static Stability3.6.1 Unsaturated Air

Consider a layer of the atmosphere in which theactual temperature lapse rate ! (as measured, forexample, by a radiosonde) is less than the dry adia-batic lapse rate !d (Fig. 3.12a). If a parcel of unsat-urated air originally located at level O is raised tothe height defined by points A and B, its tempera-ture will fall to TA, which is lower than the ambienttemperature TB at this level. Because the parcelimmediately adjusts to the pressure of the ambientair, it is clear from the ideal gas equation that thecolder parcel of air must be denser than thewarmer ambient air. Therefore, if left to itself,the parcel will tend to return to its original level. Ifthe parcel is displaced downward from O itbecomes warmer than the ambient air and, if left toitself, the parcel will tend to rise back to its originallevel. In both cases, the parcel of air encounters arestoring force after being displaced, whichinhibits vertical mixing. Thus, the condition ! " !dcorresponds to a stable stratification (or positivestatic stability) for unsaturated air parcels. In gen-eral, the larger the difference !d # !, the greaterthe restoring force for a given displacement and thegreater the static stability.34

Exercise 3.11 An unsaturated parcel of air has den-sity $% and temperature T%, and the density and tem-perature of the ambient air are $ and T. Derive anexpression for the upward acceleration of the air par-cel in terms of T, T%, and !.

Solution: The situation is depicted in Fig. 3.13. Ifwe consider a unit volume of the air parcel, its massis $%. Therefore, the downward force acting on unitvolume of the parcel is $%!. From the Archimedes35principle we know that the upward force acting onthe parcel is equal in magnitude to the gravitationalforce that acts on the ambient air that is displaced bythe air parcel. Because a unit volume of ambient airof density $ is displaced by the air parcel, the magni-tude of the upward force acting on the air parcel is$!. Therefore, the net upward force (F) acting on aunit volume of the parcel is

F & ($ # $%) !

34 A more general method for determing static stability is given in Section 9.3.4.35 Archimedes (287–212 B.C.) The greatest of Greek scientists. He invented engines of war and the water screw and he derived the

principle of buoyancy named after him. When Syracuse was sacked by Rome, a soldier came upon the aged Archimedes absorbed in study-ing figures he had traced in the sand: “Do not disturb my circles” said Archimedes, but was killed instantly by the soldier. Unfortunately,right does not always conquer over might.

TemperatureTATB

Hei

ght

Γd

Γ

(b)

AB

(a)Temperature

TA TB

ΓdΓ

O

BA

Hei

ght

O

Fig. 3.12 Conditions for (a) positive static stability (! " !d)and (b) negative static instability (! ' !d) for the displace-ment of unsaturated air parcels.

Fig. 3.13 The box represents an air parcel of unit volumewith its center of mass at height z above the Earth’s surface.The density and temperature of the air parcel are $% and T%,respectively, and the density and temperature of the ambientair are $ and T. The vertical forces acting on the air parcel areindicated by the thicker arrows.

ρg

Surface

ρ′g

ρ, T ρ′, T ′

z

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Figure 8.1: Fig 3.13 from Wallace and Hobbs. Vertical forces acting on an air parcel due to adensity and temperature difference between an air parcel (primes) and its environment.

But the density r is different between the parcel and the environmental air. If r 0 and T 0, and r

40

and T are the density and temperature of the parcel and the environment respectively 8.1, then theupward positive force experienced by the parcel is

��r 0 �r

�g

per unit volume, or

�(r0 �r)r

g

per unit mass. In other words, a parcel denser than its environment will accelerate downwards.Substituting the ideal gas equation, and assuming that the environmental temperature is close tothe parcel temperature, we get

ab =(T 0 �T )

Tg

for the upward buoyancy force per mass (or acceleration).

Figure 8.2: The period of the oscillations resulting from an initial perturbation is 2p/N

Actually, we should really be taking into account the fact that humidity also contributes to thebuoyancy of the parcel, and use the virtual temperature Tv instead of the temperature T (Eq. 1.23)

Tv ' T (1+0.61w)

where w is the mixing ratio. So the buoyancy acceleration is then

ab =(T 0v �Tv)

Tvg

This leads us to Eq. 3.73 in Wallace and Hobbs, since acceleration is the second time derivative ofposition

d2zdt2

=(T 0v (z)�Tv (z))

Tv (z)g (8.11)

41

We can improve on this expression by noting that as the parcel goes up or down, its going to coolor warm at the dry adiabatic lapse rate Gd . Thus

T 0v = Tv0�Gdz0

where z0 = z� z0 is the displacement from the equilibrium position. Similarly, the environmentaltemperature at the same level is given by

Tv = Tv0�Gz0

This leads to(T 0v �Tv)

Tv=

(Tv0�Gdz0)� (Tv0�Gz0)Tv0�Gz0

=�(Gd�G)z0

Tv0�Gz0

Provided that the vertical temperature perturbation is small compared to the actual temperature weget

(T 0v �Tv)Tv

=�(Gd�G)z0

Tv0�Gz0' �(Gd�G)z

0

Tv0Substituting now into Eq. 8.11, we get

d2z0

dt2=�(Gd�G)z

0

Tv0g (8.12)

88 Atmospheric Thermodynamics

3.6 Static Stability3.6.1 Unsaturated Air

Consider a layer of the atmosphere in which theactual temperature lapse rate ! (as measured, forexample, by a radiosonde) is less than the dry adia-batic lapse rate !d (Fig. 3.12a). If a parcel of unsat-urated air originally located at level O is raised tothe height defined by points A and B, its tempera-ture will fall to TA, which is lower than the ambienttemperature TB at this level. Because the parcelimmediately adjusts to the pressure of the ambientair, it is clear from the ideal gas equation that thecolder parcel of air must be denser than thewarmer ambient air. Therefore, if left to itself,the parcel will tend to return to its original level. Ifthe parcel is displaced downward from O itbecomes warmer than the ambient air and, if left toitself, the parcel will tend to rise back to its originallevel. In both cases, the parcel of air encounters arestoring force after being displaced, whichinhibits vertical mixing. Thus, the condition ! " !dcorresponds to a stable stratification (or positivestatic stability) for unsaturated air parcels. In gen-eral, the larger the difference !d # !, the greaterthe restoring force for a given displacement and thegreater the static stability.34

Exercise 3.11 An unsaturated parcel of air has den-sity $% and temperature T%, and the density and tem-perature of the ambient air are $ and T. Derive anexpression for the upward acceleration of the air par-cel in terms of T, T%, and !.

Solution: The situation is depicted in Fig. 3.13. Ifwe consider a unit volume of the air parcel, its massis $%. Therefore, the downward force acting on unitvolume of the parcel is $%!. From the Archimedes35principle we know that the upward force acting onthe parcel is equal in magnitude to the gravitationalforce that acts on the ambient air that is displaced bythe air parcel. Because a unit volume of ambient airof density $ is displaced by the air parcel, the magni-tude of the upward force acting on the air parcel is$!. Therefore, the net upward force (F) acting on aunit volume of the parcel is

F & ($ # $%) !

34 A more general method for determing static stability is given in Section 9.3.4.35 Archimedes (287–212 B.C.) The greatest of Greek scientists. He invented engines of war and the water screw and he derived the

principle of buoyancy named after him. When Syracuse was sacked by Rome, a soldier came upon the aged Archimedes absorbed in study-ing figures he had traced in the sand: “Do not disturb my circles” said Archimedes, but was killed instantly by the soldier. Unfortunately,right does not always conquer over might.

TemperatureTATB

Hei

ght

Γd

Γ

(b)

AB

(a)Temperature

TA TB

ΓdΓ

O

BA

Hei

ght

O

Fig. 3.12 Conditions for (a) positive static stability (! " !d)and (b) negative static instability (! ' !d) for the displace-ment of unsaturated air parcels.

Fig. 3.13 The box represents an air parcel of unit volumewith its center of mass at height z above the Earth’s surface.The density and temperature of the air parcel are $% and T%,respectively, and the density and temperature of the ambientair are $ and T. The vertical forces acting on the air parcel areindicated by the thicker arrows.

ρg

Surface

ρ′g

ρ, T ρ′, T ′

z

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Figure 8.3: Fig 3.12 from Wallace and Hobbs. Conditions for (a) positive static stability and (b)negative static stability

So notice here that if the displacement z0 is positive and the parcel is above its equilibriumpoint, and G < Gd (which is the normal case), then the RHS is negative because the air parcel iscolder than its surroundings (Fig. 8.3). The parcel has inertia, of course, but eventually the negativeacceleration or buoyancy pushes the parcel back towards its equilibrium point. Again because ithas inertia, it will overshoot the equilibrium, and the displacement will be negative, making the

42

parcel positively buoyant, which accelerates the parcel upwards. And so on. If this sounds like anoscillation, this is true. We can rewrite Eq. 8.12 in the form of the wave equation

d2z0

dt2+N2z0 = 0 (8.13)

whereN2 =

gTv

(Gd�G) (8.14)

is the square of the Brunt-Vaisala frequency. This is a hugely important number in atmosphericsciences, as it is like the second hand in the atmospheric clock. Rule of thumb is that N =0.01radianss�1.

N =r

9.8255

(9.8�6)⇥10�3 = 0.012

So if a full cycle is 2p radians then an oscillation period is

t = 2pN' 600s

or about 10 minutes. Clouds are the easiest place to see this. They rise and decay roughly over thistime scale, as they go through a full buoyancy oscillation.

The solution to Eq. 8.13 is, of course, sinusoidal. If the amplititude of the wave is z00, and theparcel is at peak positive displacement at t = 0, then in complex notation

z0 (t) = z00e�iNt = z00 (cosNt� isinNt)

Looking just at the real componentz0 (t) = z00 cosNt

The advantage of using the complex solution, is that something becomes very apparent for thespecial case that the atmosphere is absolutely unstable, in which case G > Gd . From Eq. 8.14

N =r

gTv

(Gd�G) =r

(�1) gTv

(G�Gd) = ir

gTv

(G�Gd)

where i =p�1 is the imaginary number. Since i2 =�1, this then implies that

z0 (t) = z00e�iNt = z00 exp

�i2

rgTv

(G�Gd)t�

= z00 expr

gTv

(G�Gd)t�

Since the expression in the brackets is positive, any initial displacement upwards results in theparcel continuing to rise exponentially. There is no restoring force to bring the parcel back towardsequilibrium.

43

In terms of potential temperatureEarlier, in an assignment, we derived the expression relating the dry static energy to the potentialtemperature

dhddt

= cpTd lnq

dtThis can also be expressed in terms of a gradient with respect to height

dhddz

= cpTd lnq

dz

Since hd = cpT +gz, it follows that

cpdTdz

+g = cpTd lnq

dz

Dividing by cpT and substituting expressions for G and Gd

d lnqdz

=1T

(dTdz

+gcp

) =1T

(Gd�G)

Sodqdz

=qT

(Gd�G)

This is very useful since we know that if G < Gd then the atmosphere is stable and meaning thatdq/dz > 0. So we can look at a sounding on a Skew-T plot and compare the temperature profile tothe lines of constant potential temperature. If the temperature sounding is going to higher valuesof potential temperature q as it goes up with height, then the atmosphere is stable. Of course, ifdq/dz < 0 then the atmosphere is unstable, and a storm is sure to happen because air must rise torestore stability, taking high q air up with it, and restoring the stable situation of dq/dz.

Because the atmosphere is moist, the more general expression for moist static stability is

dqedz

=qeT

(Gs�G) > 0

It’s actually very rare to see situations where dq/dz < 0 which are called absolutely unstable, sincethey cannot last long at all. However, we can see situations quite frequently when dqe/dz < 0.These are easily identified on an atmospheric sounding on a Skew-T plot by comparing the tem-perature sounding to the moist-adiabatic. We call such situations convectively unstable, becausethey lead to cloud formation.

44

9 Second Law of Thermodynamics and Entropy

Reversibility and the Second Law

Figure 9.1: 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. 8.13). 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

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

Another:

45

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

(9.1)

or per unit mass

Ds = DqrevT

(9.2)

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