Download AOS 330: Physics of the Atmosphere and Ocean I Class Notes

AOS 330: Physics of the Atmosphere and Ocean I
Class Notes
G.W. Petty
September 4, 2001
Chapter 1
Overview of the Atmosphere
1.1
Composition of the Terrestrial Atmosphere
One may group the constituents of the terrestrial atmosphere into the following four categories:
1. so-called “permanent” gases; principally N2 , O2 , and Ar
2. water (H2 0) in all three of its phases (vapor, liquid, ice)
3. variable gaseous constituents other than water: e.g., CO2 , O3 , SO2 , NO2
4. solid and liquid particles other than water (aerosols)
We will be concerned mainly with (1) and (2). The constituents falling in categories (3) and (4) are often
of great interest chemically, radiatively, or as pollutants, but these have a negligible effect on the bulk
thermodynamic properties of air and will not be considered until later.
Below about 100 km, the permanent gases are present in almost constant proportions, due to efficient mixing
by turbulence. This region of the atmosphere is known as the homosphere. The following tables give the
volume (molar) fractions of the top nine permanent constituents and the top five variable constituents:
The top three permanent constituents — N2 , O2 , and Ar — are seen to account for 99.97% of the permanent
gases in the atmosphere (Table 1.1, p. 5, W&H). The addition of CO2 brings the total up to 99.999% .
Above 100 km, molecular diffusion under the influence of gravity is able to sort gas molecules by weight faster
than they can be remixed by turbulence. As a consequence, constituent proportions are no longer constant
but rather reflect an increase with height in the proportion of lighter gases such as He and H. Furthermore,
intense ultraviolet radiation at high altitudes breaks apart diatomic molecules such as N2 and O2 , so that
these elements are increasingly represented by their monoatomic forms. The region of the atmosphere in
which constituents appear in variable proportions due to diffusive separation is called the heterosphere.
The heterosphere is the subject of a branch of meteorology called aeronomy. We will not concern ourselves
further with the heterosphere in this class.
Water vapor in the homosphere may vary from about 0–7% (by volume) of the air; despite its relatively
small fractional contribution to the total gases of the atmosphere, it is the most important constituent from
a meteorological point of view, owing in part to its very substantial role in the thermodynamics and energy
1
balance of the atmosphere, not to mention the formation of clouds, rain, snow, and other elements of “bad”
weather.
1.2
Thermal Structure of the Atmosphere
Let us now begin to consider the observed vertical structure of the atmosphere. In general, the properties
of the atmosphere, such as pressure and temperature, vary much more rapidly in the vertical direction than
they do in the horizontal direction.
As an example, let’s look at a snapshot in time of the atmosphere at a single location. An actual morning
upper air message transmitted by a certain weather station (72562 = Holdrege, Nebraska) on a certain
summer day reads:
72562
85539
17365
32523
77999
TTAA 76121 72562 99918 25663 17013 00087 ///// /////
27669 17019 70214 12660 19512 50595 06363 35004 40766
28513 30975 33580 28011 25100 451// 32006 20245 541//
15427 611// 32532 10676 671// 23517 88142 631// 32026
51515 10164 00000 10194 ///// 18520=
72562
33669
88390
44128
TTBB 7612/ 72562 00918 25663 11903 32872 22729 15259
10665 44473 08762 55468 09368 66460 10362 77410 15980
19162 99356 23380 11271 40163 22232 489// 33164 601//
637// 55115 617// 66100 671// 31313 110// 2300/=
72562 TTCC 76123 72562 70895 585// 12002 50107 569// 27004
30434 535// 05504 88999 77999=
TTDD 7612/ 72562 11949 691// 22700 585// 33599 597// 44343
519// 55300 535// 66227 469//=
Translated
according
to the coding rules found at http://www.rwic.und.edu/Academics/classdocs/raob.html this message
yields the following information:
2
Pressure (mb)
1000.
918.
903.
850.
729.
700.
669.
500.
473.
468.
460.
410.
400.
390.
356.
300.
271.
250.
232.
200.
164.
150.
128.
115.
100.
94.9
70.
59.9
50.0
34.3
30.0
22.7
Altitude (m)
87
1539
3214
5950
7660
9750
11000
12450
14270
16760
18950
21070
24340
Temperature (◦ C)
--25.6
30.8
27.6
15.2
12.6
10.6
-6.3
-8.7
-9.3
-10.3
-15.9
-17.3
-19.1
-23.3
-33.5
-40.1
-45.1
-48.9
-54.1
-60.1
-61.1
-63.7
-61.7
-67.1
-69.1
-58.5
-59.7
-56.9
-51.9
-53.5
-46.9
Dewpoint(◦ C)
--12.6
10.8
8.6
6.2
2.6
-4.4
-19.3
-20.7
-27.3
-22.3
-24.9
-32.3
-31.1
-53.3
-63.5
-53.1
Very generally, we notice a decrease in temperature as we get to higher altitudes (lower pressures). But
the rate of change of temperature is not quite constant: there are some levels where it changes slowly with
height and others where it changes more rapidly. The rate of change of temperature with height is the
environmental lapse rate and is defined by
Γ=−
∆T
∂T
−
∂z
∆z
(1.1)
So Γ is positive for the usual situation in which temperature decreases with height. It is negative in the less
typical situation in which temperature increases with height.
A layer in which Γ < 0 (i.e., ∂T /∂z > 0) is called an inversion. Inversions thus represent atmospheric layers
in which warm air overlies colder air. If Γ = 0, then the layer is isothermal.
There are several different reasons why inversions form:.
Radiation inversions form as the result of radiational cooling of the ground at night, and consequently
the layer of air directly above it. This effect is most pronounce when the atmosphere above that layer is
relatively transparent to infrared radiation; for example, when the sky is clear and the relative humidity is
low.
3
Subsidence inversions may form when air sinks from a higher altitude, warming by compression as it
goes.
Frontal inversions may appear in a sounding taken on the cold side of a front, since the frontal discontinuity
between the cold and warm air masses will generally slope back over the position where the radiosonde was
released.
Boundary layer inversions frequently delimit the mixed layer of the atmosphere near the surface, which
may be from a few tens of meters thick to a kilometer thick or more. Near the coast, very strong boundary
layer inversions may appear at the top of a shallow layer of cool, moist air flowing in from the ocean .
Finally, an inversion frequently occurs at or above the tropopause, which marks the transition from the troposphere to the stratosphere. In the troposphere, the lapse rate is generally positive (temperature decreasing
with height), whereas in the stratosphere the lapse rate is small or even negative.
Let us clarify the last statement by reviewing the large scale thermal structure of the atmsosphere. Figure
1.8 on p 23 of your textbook by Wallace and Hobbs (hereafter W&H) depicts the typical structure up to
about 100 km (recall also that this altitude roughly corresponds to the transition from the homosphere to
the heterosphere).
The lowest layer is the troposphere, which extends from the surface to around 10 km, give or take a km
or two. Within the troposphere, the general tendency is for the temperature to decrease with height (i.e.,
Γ > 0).
The next layer is the stratosphere, which extends upward to around 40–50 km. Within the stratosphere,
the temperature generally increases with height, so that at the top of the stratosphere, temperatures may
actually be somewhere near the freezing point again!
In the mesosphere, the temperature generally decreases with height again until around 80 km or so. The
temperatures found at the mesopause may be the coldest anywhere in the atmosphere (-93◦ Cin the U.S.
Standard Atmosphere).
The thermosphere (see Figure 1.9, p. 25 of W&H) consists of very thin, hot, ionized gases and has no welldefined upper boundary. The temperature in the thermosphere depends strongly on solar activity; it may
be as “cool” as 600 K when the sun is quiet or may increase in temperature to 2000 K under the influence
of an active sun.
Geometrically, the above layers range in thickness from ∼10 km for the troposphere to 100s of km for the
thermosphere. However, because of the very low air densities found at higher altitudes, most of the mass of
the atmosphere is found in the troposphere. Indeed, the troposphere has about 80% of the total mass, and
the stratosphere has almost all of the remaining 20% . The mesosphere and thermosphere account for only
about 0.1% and 0.001% , respectively.
The troposphere is the most interesting layer for most meteorologists (and for us), not only because it contains
the lion’s share of the mass of the atmosphere, but also because we live in it and it is in the troposphere
that most weather occurs. As we shall see later, the difference between the characteristic lapse rates in the
stratosphere and the troposphere help explain why there is so much more “action” in the troposphere.
Up until now, we have considered only vertical temperature profiles, without regard to geographic and
seasonal variations. A pair of figures on p. 27 of W&H depict the average temperature structure of the
atmosphere both in the horizontal and the vertical for a summer and a winter month, respectively. Several
details are worth noting:
4
• The tropopause is generally lower near the poles than it is near the equator.
• The tropopause is generally lower in wintertime than it is in the summertime.
• The seasonal difference in height is generally greater at middle and high latitudes than in the tropics.
• There is often a discontinuity or “break” in the tropopause near 30–50◦ latitude.
• There is a pronounced inversion near the surface at high latitudes, particularly in the wintertime.
• The pressure at a specified geometric altitude in the upper troposphere and stratosphere is, on average,
significantly greater in the tropics than near the poles.
Of course, these descriptions represent only average conditions. At any given instant in time and at any
given location, an actual profile and/or cross-section of the atmosphere may differ significantly from these
average profiles. One of the most important practical objectives of this class is to give you knowledge you
can use to determine how the day-to-day variations in the thermal structure of the atmosphere can affect
the weather we experience.
5
Chapter 2
Physical Properties of Air
2.1
Some new definitions
Intensive state variables are those which are independent of mass; e.g., temperature, pressure, density.
Extensive variables depend on the total mass of the system; e.g., volume.
Extensive variables may always be converted to intensive variables by dividing by the system’s mass.
2.2
Behavior of Ideal Gases
The state of a homogeneous, gaseous system may be characterized by three variables; for example, density,
pressure, and temperature.
The density ρ is the mass of material in the system divided by the volume of that system (M/V ). Alternatively, one may wish to use specific volume α (volume per unit mass), which is just the inverse of
ρ.
Density and specific volume are expressed in SI units of kg/m3 and m3 /kg, respectively.
The pressure p is the force per unit area exerted by the random motions of the molecules contained within
a system. It has no preferred direction. SI units of pressure are N/m2 or Pascal (Pa).
A fundamental concept in thermodynamics is that of an equation of state. The three variables which
describe a system have been shown experimentally not to be independent of each other. In other words,
they satisfy a relation of the form
f (p, V, T ) = 0
(extensive form)
f (p, α, T ) = 0
(intensive form)
or
6
This means if you know any two of the three variables, the third may be determined.
Boyle’s Law (1660):
At constant temperature, the volume of a given sample of gas varies inversely as the pressure.
p∝
1
V
or
C1
V
p=
or
pV = C1
(2.1)
where C1 is a constant of proportionality [Note: C1 = f (T )].
Charle’s Law (1787):
At constant pressure, the volume of a given sample of gas is proportional to absolute temperature.
V0
VT
=
= C2
T
T0
or
V = C2 T
(2.2)
where C2 is a constant of proportionality [Note: C2 = f (p)].
Charle’s Law and Boyle’s Law are given by the following pair of equations:
pV = C1 (T )
(2.3)
V = C2 (p)T
(2.4)
This can only hold true if there is another constant C such that
pV = CT
(2.5)
C depends on both the size of the gas sample and on the type of gas.
Avogadro found that for fixed pressure and temperature, the number of molecules per unit volume of a gas
is a constant, irrespective of the chemical composition
We therefore introduce an SI unit called the kilomole (kmol) to represent a fixed number of molecules.
A kilomole of substance corresponds to the number of molecules in a sample whose weight in kilograms equals
the standard molecular mass m of the substance. This number is called Avogadro’s constant and has the
value 6.022 × 1026 kmol−1 .
The equation of state of an ideal gas, known as the Ideal Gas Law can therefore be written as
pV = nR∗ T
(2.6)
where n is the number of moles in the sample and R∗ is the Universal Gas Constant.
The value of R∗ was determined experimentally by noting that at 0◦ C and standard atmospheric pressure,
the volume of one mole of an ideal gas is 22.4 liters (1000 liters = 1 m3 ).
Solving for R∗ gives
R∗ = 8314.41 J K−1 kmol−1
Under what conditions does the Ideal Gas Law give an accurate description of the behavior of a gas?
In general, the Ideal Gas Law is valid whenever the density of the gas is low enough (due to a suitable
combination of low pressure and high temperature) so that individual molecules do not experience significant
7
attractive forces, nor does the space occupied by the molecules represent a significant fraction of the total
volume.
Under conditions near the liquefaction point of a gas, the Ideal Gas Law may no longer be sufficiently
accurate. At ordinary atmospheric pressures, air is obeys the Ideal Gas Law quite closely for meteorological
purposes.
In meteorology, it is convenient to deal with a known mass M of a gas rather than a number of kilomoles.
This can be accomplished by replacing nR∗ with M R, i.e.,
pV = M RT
(2.7)
where R is a gas constant which depends on the particular gas.
By this definition,
R ≡ R∗ (n/M ) = R∗ /m
where m is the molecular mass (units kg kmol
−1
(2.8)
= “atomic mass units” or amu)
What do we do when the gas in question is a mixture of molecules of different masses?
No problem: Remember, the form of the Ideal Gas Law using the Universal Gas Constant is always valid.
One can therefore always obtain R for a specific gas simply by using m̄ = (total mass in kg)/(total no. of
kilomoles).
Let’s say a sample contains a mass M1 of one gas and a mass M2 of another. The corresponding molecular
masses are m1 and m2 . The total number of kilomoles, therefore, is given by
n = n1 + n2 = M1 /m1 + M2 /m2
(2.9)
and the total mass M is just M1 + M2 .
For a two-component gas the Ideal Gas Law can thus be written
pV = nR∗ T = (M1 /m1 + M2 /m2 )R∗ T = M RT
(2.10)
where the new gas constant R for the mixture is given by
R ≡ R∗ (n/M ) = R∗
M1 + M2
M1 /m1 + M2 /m2
−1
(2.11)
This approach can be generalized to give us a definition for the mean molecular mass of an arbitrary mixture
of N different gases:
N
Mi
m̄ = N i=1
(2.12)
i=1 Mi /mi
2.2.1
Dalton’s Law
In deriving a form of the Ideal Gas Law which accounts for mixtures of gases of different molecular weights,
we simply assumed that the number of molecules in a gaseous system is what is important for determining
the product pV of the system as a function of its absolute temperature T .
The validity of this assumption is implicit in Dalton’s Law of Partial Pressures:
8
The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures which would be
exerted by each constituent alone if it filled the entire volume at the temperature of the mixture.
That is, for a mixture of k components
p=
k
pi
(2.13)
i=1
where p is the total pressure and the pi are the partial pressures of each gas. These may be assumed to obey
the Ideal Gas Law separately, so that
(2.14)
pi = Mi (R∗ /mi )T /V
where Mi is the total mass (kg) of each constituent and mi is its molecular mass (kg/kmol). Then
p=
k
i=1
pi =
k
Mi (R∗ /mi )T /V = M R∗ T /V
i=1
k
(Mi /mi )/M
(2.15)
i=1
k
where M = i=1 Mi . The above result can be shown to lead to exactly the same definition of a mean
molecular mass m̄ and specific gas constant R as we had already derived earlier without explicitly invoking
Dalton’s Law.
2.2.2
Gas Constant for Dry Air
Armed with the knowledge of the composition of air (see page 5, W&H) , we may compute a mean molecular
weight m̄d and a specific gas constant Rd for dry air. We will not include the contribution of water vapor
at this point because it is so variable in space and time. Later, we will spend a lot of time describing how
water vapor may complicate (and thus make more interesting!) the thermodynamic behavior of air.
The formula we derived earlier for a multi-component gas is useful when the mass fraction of each constituent
is given. In the table above, however, it is the volume (or molar) fraction which is given. You should verify for
yourself that, since volumes are proportional to numbers of molecules, one may calculate the mean molecular
mass m̄ simply as
(2.16)
m̄ =
fv,i mi
where fv,i is the volume fraction of the ith constituent (note that the sum of the fractions used must be
normalized to equal 1).
Taking the volume fractions of only the top four constituents N2 , O2 , Ar, and CO2 we find that
7808(28.013) + 0.2095(31.999) + 0.0093(39.948) + 0.0003(44.010)
(2.17)
m̄d = 28.964 kg/kmol
(2.18)
Rd = 8314/28.96 = 287.06 J/(kgK)
(2.19)
or
Hint: You will use the value of Rd so often that you should commit it to memory as soon as you can!
2.2.3
Example Gas Law Calculations for Dry Air
Having determined Rd , we can proceed to use the Ideal Gas Law to compute state variables for dry air under
a variety of conditions.
9
Example: Standard sea level pressure is 1013.2 hPa and standard surface temperature in the U.S. Standard
Atmosphere is 15◦ C. What is the density of dry air under these conditions? What volume is occupied by
1 kg of dry air?
Solution:
The intensive form of the Ideal Gas Law is
pα = Rd T
(2.20)
p = ρRd T
(2.21)
or, since ρ = 1/α,
2
Substituting p = 1013.2 × 10 Pa and T = 288.15 K and solving for ρ gives us an atmospheric density under
standard conditions of 1.225 kg m−3 or, equivalently, a specific volume α of 0.816m3 kg−1 .
Example: At sea level, the pressure may sometimes be as low as 950 hPa or as high as 1040 hPa. The
temperature, can easily vary between −40 and +50◦ C. Using this information, estimate rough limits on the
range of densities that might be exhibited by air at sea level.
Solution:
The densest air would occur for a combination of p = 1040 hPa and T = −40◦ C. Using the procedure in the
previous example, we find ρ = 1.55 kg m−3 .
The “thinnest” air would occur for p = 950 hPa and T = 50◦ C, for which ρ = 1.03 kg m−3 .
Question: Which of the two variables, pressure or temperature, is more important in causing variations in
the density of air at sea level?
Answer: Density is proportional to pressure and to the inverse of absolute temperature. At sea level, pressure
varies by only about 5% from its standard value while the temperature may vary by more than 20% from
its standard value. Hence, variations in temperature are likely to bring about larger changes in density at
sea level.
Example: At 3 km (about 9,900 ft) altitude, the pressure in the U.S. Standard Atmosphere is 701 hPa and
the temperature is -4.5◦ C(268.7 K). What temperature would a parcel of air at sea level (standard pressure)
have to have in order to match the density of typical air at 3 km altitude?
Solution: The density at 3 km is found to be 0.909 kg m−3 . Substituting this value for ρ into the Ideal Gas
Law along with p = 1013.2 hPa, we find that air would have to be heated to a temperature of 388 K (equals
115◦ Cor 239 ◦ F ) in order to have the same density at standard sea level pressure.
2.2.4
Adding Water Vapor
We previously calculated the Gas Constant Rd for perfectly dry air — i.e., air which contains absolutely
no water vapor, only the permanent gases which are always present in constant proportion. Perfectly
dry air doesn’t exist in the atmosphere; moreover, a dry atmosphere would be extremely uninteresting
meteorologically. Let us therefore begin to examine the behavior of water vapor and, in particular, its
influence on the thermodynamic properties of the air.
First, let’s consider water vapor in isolation; that is, without the added complication of other gases. If we
can treat water vapor as an ideal gas, then the ideal gas law is just as valid here as it was earlier for dry air:
pα = RT
10
(2.22)
For water vapor, however, it is conventional to identify the vapor pressure with the symbol e rather than p.
We will use the symbol ρv to denote the water vapor density, more commonly known to meteorologists as
the absolute humidity.
We can thus write the ideal gas law for water vapor as
e = ρv Rv T
(2.23)
where Rv is the Gas Constant for pure water vapor? What is the value of Rv ? We can calculate this just as
before, using
R∗
(2.24)
Rv =
mv
where mv is the molecular mass of H2 O and equals about 18.016 kg/kmol. Therefore
Rv = 461.5 J kg−1 K−1
(2.25)
Is the ideal gas law as good an approximation for water vapor as it is for dry air? Not really, because water
vapor at typical atmospheric temperatures and pressures is usually much closer to condensation than is the
case for dry air. This implies that attractive forces between the molecules are significant. Nevertheless, the
inaccuracy in assuming that water vapor behaves as an ideal gas is not great enough to worry about in
most meteorological applications. Obviously, water vapor ceases to behave anything like an ideal gas when
it reaches saturation, since an isothermal decrease in volume then no longer implies an increase in pressure.
We will come back to the behavior of water vapor at saturation later.
Now let’s generalize to the case where water vapor and the permanent gases of the atmosphere coexist in
the same volume. Can we still use the relationship e = ρv Rv T ? Of course, provided that we recognize that
the vapor pressure e represents the partial pressure of vapor in the air and that the total pressure is given
by Dalton’s Law as
(2.26)
p = pd + e
Here pd is the partial pressure of the dry air in the mixture and is related to the dry air density ρd and
temperature T by the familiar gas constant Rd = 287.06 J/(kg K):
pd = ρd Rd T
(2.27)
Combining the ideal gas laws for the two components, we have that the total pressure of a moist volume of
air is given by
p = (ρd Rd + ρv Rv )T
(2.28)
Also, the combined density of moist air is obviously just the sum of the densities of the dry air and water
vapor
(2.29)
ρ = ρd + ρv
It is usually inconvenient to use water vapor density ρv or vapor pressure e to express the relative vapor
content of a mass of air, since these quantities are not conserved. That is, the vapor pressure and the density
each increase or decrease when the air containing the vapor is compressed or expanded, respectively. By
contrast, the mixing ratio w of an air mass is conserved, as long as there is no condensation or evaporation
taking place. The mixing ratio is defined by
w≡
Mv
ρv
=
Md
ρd
(2.30)
where Mv is the mass of water vapor mixed into a mass Md of dry air. Because there is usually much more
dry air than vapor in given volume of the atmosphere, it is often convenient to express w in units of grams
vapor per kilograms dry air. For example, in a warm tropical air mass, the mixing ratio may be as high as
20 g/kg. In cooler air masses, w is typically only a few g/kg.
11
Similar to the mixing ratio w is the specific humidity q which is defined as
q≡
ρv
Mv
=
Md + Mv
ρ
(2.31)
In other words, q gives the mass of water vapor per unit mass of moist air, so that the mass contribution by
the water vapor is included in the denominator.
Note that since the mass of water vapor is typically no more than one or two percent of the total mass, the
numerical values of w and q also differ by no more than one or two percent. For many common applications,
it is fairly unimportant whether one uses w or q in moisture calculations. If an exact conversion is required,
the following relationships may be used
q
w=
(2.32)
1−q
and
w
.
(2.33)
1+w
You should verify that, for realistic values of either w or q, there is little numerical difference between the
two.
q=
Often, it is necessary to convert between mixing ratio w or specific humidity q and vapor pressure e. For w,
this can be accomplished by noting that
w=
εe
εe
ρv
e/Rv T
=
≈
=
ρd
pd /Rd T
p−e
p
(2.34)
mv
Rd
=
= 0.622
Rv
md
(2.35)
ρv
εe
εe
≈
=
ρv + ρd
p − (1 − ε)e
p
(2.36)
εe
≈w
p
(2.37)
where
ε≡
Similarly,
q=
In summary, to a good approximation
q≈
2.2.5
Virtual Temperature
We already saw that the total pressure of moist air is given by
p = pd + e = (ρd Rd + ρv Rv )T
We can factor out the moist air density ρ and the gas constant for dry air Rd by writing
ρd Rd + ρv Rv
ρd
ρv Rv
p = ρRd
+
T = ρRd
T
ρRd
ρ
ρ Rd
Using the definitions of specific humidity q = ρv /ρ and Rd /Rv = ε, it is easy to show that
1
−1 q T
p = ρRd 1 +
ε
(2.38)
(2.39)
(2.40)
This identical to the ideal gas law for dry air, except for the appearance of the factor in brackets on the right
hand side. How does one interpret this factor? To begin with, we know that the ideal gas law for moist air
can be written
p = ρRm T
(2.41)
12
where the gas constant Rm reflects the mean molecular mass of the air, including not only the permanent
gases but also the contribution by water vapor. Clearly, Rm is not a constant but depends on the moisture
content of the air. While we could calculate the mean molecular mass directly using the same formula we
used much earlier in getting Rd , it is easy to see by comparing the last two equations that
1
Rm = Rd 1 +
−1 q
(2.42)
ε
In other words, the term in brackets gives us a convenient means to adjust the gas constant for dry air in
order to account for the presence of water vapor.
However, people don’t usually have an instinctive feel for the physical meaning of a given value of the gas
constant R. As a result, the approach just described for interpreting the water vapor correction term is not
the most common one. Instead, it is conventional to write
1
−1 q T
(2.43)
Tv = 1 +
ε
so that the ideal gas law may be written for moist air as
p = ρRd Tv
(2.44)
The virtual temperature Tv is simply the temperature a dry parcel of air would have to have in order for that
parcel’s density to equal the density of the moist parcel, assuming equal pressures. That is, if a dry parcel of
air has temperature T0 and a moist parcel of air has a virtual temperature Tv which happens to be equal to
T0 , then both parcels have the same density (again, assuming equal pressures).
Note that the quantity 1/ε − 1 is just a constant with the approximate value 0.61; consequently, the most
convenient formula for Tv is just
(2.45)
Tv = (1 + 0.61q) T
By substituting q equal to 0.02 kg/kg (an approximate upper limit for q in the atmosphere), one finds that
in general
<
0 < (Tv − T ) ∼
(2.46)
3.7◦ K
The practical value of Tv is most obvious when the hydrostatic law comes into play (next chapter), which
relates the local density of air to the local rate of change of pressure with height. From this point onward,
anytime you solve problems which depend in some way on the density of air, you should remember that it is
the virtual temperature which determines this density, not the actual temperature (unless of course q = 0).
In many cases, the difference between Tv and T is small enough that we will ignore that difference, but you
should always be at least aware of the difference and know when it is likely to be worth taking into account.
13
Chapter 3
Atmospheric Pressure
3.1
Hydrostatic Balance
Now that we have the Ideal Gas Law at our disposal and know something about how temperature varies with
height in the atmosphere, we are finally equipped to consider in detail how and why atmospheric pressure
varies throughout the atmosphere.
To an excellent approximation under most conditions, the pressure at a given point in the atmosphere is
given simply by the weight of the atmosphere above that point.
For example, if the surface pressure at a given geographic location and time is 1013.2 hPa, then the weight
per square meter of the atmosphere above that point is 101,320 N. Dividing by the acceleration due to gravity
g = 9.81 m s−2 , we find that the mass of the column of air above a square meter of the surface is about
10,328 kg, or a little over 10 tonnes!
How can we be sure that it is this simple? Since typical vertical accelerations within the atmosphere are
observed to be very small compared with the value of g, it follows that any contribution to the pressure by
forces other than gravity must also be relatively small.
To take a relatively extreme case, consider a typical thunderstorm, in which vertical velocities at a middle
level of the troposphere (say 3 km) may be of order 10 m s−1 . A blob of air rising from the surface and
accelerating steadily will require on the order of 10 minutes to reach that speed if it starts out from a
standstill. This implies an acceleration of 0.02 m s−2 , or only about 0.2% of g. Even if the entire vertical
column of the atmosphere underwent this much acceleration, it would give rise to a surface pressure anomaly
of only about 2 millibars. Outside thunderstorms, typical vertical accelerations are far smaller still and may
almost always be safely ignored.
Let us therefore formalize the Hydrostatic Law as follows:
Consider a vertical column of air with unit cross-sectional area (see schematic next page). The mass of the
air between heights z and z + dz in the column is ρ dz, where ρ is the density of the air at height z. The
force acting on this column due to the weight of the air is gρ dz, where g is the acceleration due to gravity
at height z. Now let us consider the net vertical force on the block due to the pressure of the surrounding
air. We will assume that in going from height z to height z + dz the pressure changes by an amount dp
as indicated in the schematic. Since we know that pressure decreases with height, dp must be a negative
quantity, and the upward pressure on the lower face of the shaded block must be slightly greater than the
14
downward pressure on the upper face of the block. Thus the net vertical force on the block due to the vertical
gradient of pressure is upward and given by the positive quantity −dp as indicated. The balance of forces in
the vertical requires that
−dp = ρg dz
(3.1)
or, to give the most common form of the Hydrostatic Equation,
dp
= −ρg
dz
(3.2)
If the pressure at height z is p(z), we have
p(∞)
−
∞
dp =
p(z)
or, since p(∞) = 0
gρ dz
(3.3)
z
∞
gρ dz
p(z) =
(3.4)
z
To summarize, the rate of change of pressure with height is proportional to the density. Furthermore, the
pressure at any given level is approximately proportional to the mass above that level (the proportionality is
not quite exact, because g decreases slightly with altitude).
We can now of course substitute the Ideal Gas Law for ρ and arrive at an expression for the rate of change
of pressure with height as a function of temperature:
pg
dp
= −ρg = −
dz
RT
(3.5)
If we are dealing with a dry atmosphere (i.e., no water vapor), then of course R = Rd =287 J/(kg K) .
Substituting standard sea level values of g = 9.8 m s−2 , p = 1013 hPa and T = 288 K, we find that
dp
dz = 12 Pa/m . In more traditional terms, this translates into a one mb (hPa) change in pressure for every
8.3 m change in elevation. Of course, at higher altitudes you have to change altitude by much more than
this to achieve the same change in pressure, because the air is less dense.
A convenient transformation of the above equation may be obtained simply by dividing both sides by the
pressure p. We then have
g
1 dp
=−
(3.6)
p dz
RT
Since d(ln p) = p1 dp, this can be rewritten as
d ln p
g
=−
dz
RT
(3.7)
In words, the rate of change of the logarithm of pressure with height is inversely proportional to the absolute
temperature, and does not depend on p. As we shall see shortly, this is the same as saying that pressure
generally falls off exponentially with height.
3.1.1
Digression on Gravity
At this point it is worthwhile to backtrack and reconsider the value of g which keeps cropping up in our
equations. As we already know, the value of g is close to 9.81 m s−2 at sea level, and for some purposes,
this value is sufficiently accurate. However, it is important to recognize that g does in fact vary slightly with
15
altitude and latitude. For some purposes, the difference is important, so we will take a closer look at this
quantity and introduce a convention that will then allow us to pretty much forget about it again.
The actual acceleration due to gravity is a function of the distance R from the center of mass of the earth.
Specifically,
GM
(3.8)
g(R) = 2
R
where M is the mass of the earth (M = 5.977 × 1024 kg) and G is the so-called Universal Gravitation
Constant and has a value of 6.6720 × 10−11 N m2 kg−2 . If we wish to consider gravity as a function of the
altitude z above the surface, we can substitute R = R0 + z in the above equation, where R0 is the effective
radius of the earth and is equal to about 6370 km. In this case, we have
g(z) =
GM
(R0 + z)2
By using a Taylor series expansion in (z/R0 ) and discarding higher order terms, we can write
z
g(z) g0 1 − 2
R0
where
g0 =
GM
R02
(3.9)
(3.10)
(3.11)
is the standard acceleration due to gravity at sea level. According to the above formula, for z = 10 km above
sea level (i.e., near the top of the troposphere), g decreases to about 9.78 m s−2 , or about 0.3% less than
its sea level value.
Two things complicate the picture somewhat further: first of all, the earth bulges somewhat at the equator,
so that R0 is not the same for the equator (6378.1 km) as it is for the poles (6356.9 km). For this reason
alone, the sea level value of g is slightly lower (about 0.7%) at the equator than at the poles.
Secondly, we have not considered the minor difference between the true (or “pure”) gravity g and the
apparent gravity g which includes the effects of the rotation of the earth. An adequate approximation
for the latter is given by
(3.12)
g = g − Ω2 R cos2 φ
where Ω is the angular velocity of the earth and equals 2π (radians) per 23.9 hr or 7.29 × 10−5 s−1 , and φ is
the latitude. One can easily calculate that the difference between g and g amounts to only about 0.03 m s−2
at the equator.
Clearly, the range of variability of g due to the oblateness of the earth and due to centrifugal force is no more
than a few tenths of a percent under the conditions of interest to most meteorologists; nevertheless, there
are times when it is important to take into account these differences when performing sensitive calculations.
One way of doing this is to change one’s frame of reference slightly. If you are interested in transformations
of energy in the atmosphere (most meteorologists are, either directly or indirectly), a useful variable is the
geopotential Φ. At any point in the atmosphere, the geopotential is defined as the work that must be done
against the apparent gravitational field in order to raise 1 kg from sea level to that point; i.e.,
z
g (z) dz
(3.13)
Φ=
0
The geopotential is actually the gravitational potential energy per unit mass, that is, the energy
available in a decrease in elevation which may be converted to kinetic energy. As such, it has units of m2 s−2
or J kg−1 . The differential of geopotential is given by
dΦ = g dz
16
(3.14)
Geopotential is often expressed in terms of another quantity, geopotential height Z, defined by
Z=
Φ(z)
g0
(3.15)
It is clear that Z has dimensions of length, usually specified in geopotential meters, but geopotential
meters are slightly larger than real meters at higher altitudes or wherever g < g0 . The advantage is that
by using Z in meteorological calculations instead of the actual height z, one may forget about variability in
the apparent gravity and just use the constant value of g0 everywhere. Indeed, the height values given on
standard constant pressure charts (e.g., the so-called 850 mb chart, 500 mb chart, etc.) are actually in units
of geopotential height. In any case, one should not lose sight of the fact that the geopotential height does
differ slightly from geometric height, and that the former is actually a measure of potential energy and not
of distance.
Throughout the remainder of this course, when we refer to a height or altitude in the atmosphere, it should
automatically be understood (unless otherwise indicated) that we mean geopotential height, which is only
slightly different from the “real” height. This way, we can always utilize a constant effective value of
g ≈ g0 = 9.80665 m/sec2 and not worry about small variations in g from one place to the next.
3.1.2
Hypsometric Equation
Having dispensed with that issue, let us now return to the question of how pressure p varies with (geopotential) height z. We can integrate the hydrostatic equation as follows:
z2
R p1
dz =
T d ln p
(3.16)
g p2
z1
or
∆z = z2 − z1 =
We can simplify this to
RT̄
∆z = z2 − z1 =
g
R
g
p1
p2
p1
T d ln p
(3.17)
p2
p1
RT̄
log
d ln p =
g
p2
(3.18)
if only we define the mean layer temperature T̄ as
p1
T d ln p
p
T̄ ≡ 2p1
d ln p
p2
(3.19)
In words, the thickness ∆z = (z2 − z1 ) of an atmospheric layer between specified pressure levels p2 and
p1 is proportional to the mean layer temperature T̄ (defined as above). The equation for ∆z is known as
the hypsometric equation and is routinely used to derive the heights of pressure levels from atmospheric
temperature and humidity profiles.
Important: Although we used the temperature T in the above derivation, this is strictly valid only for dry
air. If the humidity of the air is significant and/or high accuracy is required, then of course we need to
substitute the virtual temperature profile Tv (z) for the actual temperature profile T (z) in (3.19).
3.1.3
Vertical Structure of Cyclones and Anticyclones
According to the hypsometric equation, the distance between standard pressure levels is smaller in cold air
than in warm air. Pressure features such as lows, highs, troughs, ridges, etc., are thus very closely related to
17
the thermal structure of the atmosphere, and may change shape and intensity, and even disappear altogether
at higher or lower altitudes depending on the horizontal distribution of temperature.
For example, you can easily verify with simple sketches of constant pressure surfaces that:
• A surface warm-core cyclone quickly weakens or disappears with height.
• A surface warm-core anticyclone strengthens with height.
• A warm-core cyclone aloft increases in intensity downward.
• A cold-core cylone at the surface increases in intensity with height.
• A cold-core anticyclone at the surface quickly disappears with height.
• Low pressure centers or troughs are displaced horizontally toward colder air with increasing height.
• At the surface of the earth, high pressure is favored in cold regions and low pressure in warm regions.
3.2
Pressure Profiles Under Idealized Conditions
Up until we have talked in a general way about how temperature, pressure, and altitude are related in
the atmosphere. We showed that if you prescribe any arbitrary temperature profile and surface pressure,
one may use the hypsometric equation to compute the height of any pressure level above the surface, or
conversely, the pressure at any height.
Now we will spend some time talking about some special cases which allow particularly simple mathematical
relationships to be derived and which, hopefully, will also help to illustrate some important concepts.
3.2.1
The Homogeneous Atmosphere (or Ocean)
One of the simplest possible models of an atmosphere is one in which the density ρ is constant everywhere,
irrespective of altitude. Within such an atmosphere, if it existed, pressure would decrease with altitude
according to the hydrostatic law, but the density would remain constant until reaching the top of the
atmosphere, at which point the density would abruptly go to zero. Actually, this model is a far better
representation of an ocean than an atmosphere, since the density of seawater doesn’t change much with
pressure and since it has a sharply defined upper boundary. Nevertheless, let us consider an atmosphere
having these properties and see where it leads us.
If we integrate the hydrostatic equation from sea level, where the pressure is p0 to a height H where the
pressure is zero, we get
H
0
= −ρg
dz
(3.20)
0
p0
or
p0 = ρgH
(3.21)
In other words, the pressure at the bottom of the atmosphere (sea level) is once again just the weight per
unit area of the atmosphere, only now there is no need for any messy integrals, since ρ is constant. Now if
we know both p0 and ρ in addition to g, we can solve for the height H:
H=
18
p0
ρg
(3.22)
If we substitute typical values for p0 and ρ, namely 1013 hPa and 1.25 kg m−3 , respectively, we find that
H 8.3 km. In other words, if our atmosphere had the same mass that it has now but also had a constant
density equal to its sea level density under standard conditions, it would only be 8.3 km deep! (Recall that
most passenger airliners fly at altitudes closer to 10 or 12 km. . . )
It would be easy for an atmosphere to exhibit the above behavior if only air were completely incompressible,
just as water is (well, almost). But this would mean throwing out the ideal gas law and substituting a very
different equation of state (i.e., ρ ≈ constant instead of ρ = p/RT ), which we know is not correct for air.
So let’s consider whether it would, in principle, be possible for an atmosphere to exhibit constant density
throughout its depth and still obey both the ideal gas law and the hydrostatic law.
We can substitute the ideal gas law into the previous equation for H to get
H=
ρRT0
p0
=
ρg
ρg
(3.23)
RT0
g
(3.24)
or
H=
So we can see here that H can be written entirely in terms of the surface temperature T0 and two well-known
constants, R and g. But we still don’t know for sure what’s happening to the temperature above the surface.
However, the pressure is decreasing with height, so it is clear that the temperature must also decrease in
order for the density to stay constant.
Let us write the ideal gas law for dry air, p = ρRT , and differentiate with respect to elevation, holding ρ
constant. We get
dT
dp
= ρR
(3.25)
dz
dz
Substitution into the hydrostatic equation leads to the result:
or
dT
g
= −Γ = −
dz
R
(3.26)
Γ = 34.1◦ K km−1
(3.27)
So in order to have a homogeneous atmosphere, you need a lapse rate which is constant and very large (about
six times as large as what is usually observed in the troposphere). Now let’s pursue this one step further and
find out what the temperature at the top of a homogeneous atmosphere must be. In any atmosphere with a
constant lapse rate Γ, the temperature at an arbitrary height z above the surface may be expressed as
T (z) = T0 − Γz
(3.28)
If we substitute the height of the top of the atmosphere H for z, and use the above formulae for H and Γ in
the homogenous atmosphere, we have
T (z) = T0 −
RT0
g
×
= T0 − T0 = 0
R
g
(3.29)
In other words, in order for an atmosphere obeying the ideal gas law to have constant density all the way to
the top, the temperature at the top must be equal to absolute zero! Clearly, this model of the atmosphere is
quite unrealistic, especially since real air would liquify (and thus cease to obey the ideal gas law) long before
it ever got close to absolute zero. Nevertheless, the concept is useful theoretically. In particular, we shall see
that H = RT /g crops up again even in the mathematical description of more realistic atmospheres, which
we will address now.
19
3.2.2
The Isothermal Atmosphere
Another simple model for the atmosphere is one in which not the density but rather the temperature is
constant with height — that is, Γ = 0. As we learned earlier, such an atmosphere is called isothermal.
Again we start with hydrostatic equation and substitute the ideal gas law:
pg
dp
= −ρg = −
dz
RT
(3.30)
g
1
dp = −
dz
p
RT
(3.31)
This can be rewritten
We can then integrate from sea level (z = 0, p = p0 ) to some arbitrary level z where the pressure is p. Since
T is a constant,
p
z
1
g
dp = −
dz
(3.32)
RT 0
p0 p
or
p
gz
(3.33)
ln
=−
p0
RT
Once again using the definition H = RT /g, we get
z
−
p = p0 e H
(3.34)
where H in this case is interpreted as a scale height; i.e., the vertical distance over which the pressure
decreases by a factor e−1 or to about 37% of its original value.
3.2.3
The Constant Lapse-Rate Atmosphere
Let us assume that the temperature T varies linearly with height; i.e,
T = T0 − Γz
(3.35)
In this case, the hydrostatic equation combined with the ideal gas law becomes
dp
pg
pg
=−
=−
dz
RT
R(T0 − Γz)
or
1
g
dp = −
p
R
dz
T0 − Γz
(3.36)
(3.37)
Once again this can be easily integrated. We shall again integrate between the limits z = 0, where p = p0
and an arbitrary height z, where the pressure is p. The result is
p
g z
1
dz
dp = −
(3.38)
p
R
T
0 − Γz
p0
0
or
ln
p
p0
g
ln
=
RΓ
T0 − Γz
T0
(3.39)
Taking the exponent of both sides, we get
p = p0
T0 − Γz
T0
20
g
RΓ
(3.40)
or
p = p0
T
T0
g
RΓ
(3.41)
In the usual case where T decreases with z (Γ positive), this equation requires that pressure decrease with
elevation, in agreement with the hydrostatic equation. In the less common case that T increases with
elevation (an inversion), the ratio (T /T0 ) is greater than unity above the surface but the exponent of this
ratio is negative, therefore p still decreases with height, as required by the hydrostatic equation. In the
special case of an isothermal layer (Γ = 0), the above formula cannot be used because Γ appears in the
denominator of a fraction and division by zero is undefined.
Note that the exponent in the previous equation is simply the ratio of the constant lapse rate (g/R) in the
homogeneous atmosphere to the actual lapse rate Γ. If the two are equal, exponent becomes unity, and
pressure becomes a linear function of z again, which in turn implies constant density.
An atmosphere with a constant positive lapse rate (decrease of temperature with height) has only a finite
vertical extent. Once the temperature reaches absolute zero, we’ve reached the top of our hypothetical
atmosphere, just as we saw in the case of the homogeneous atmosphere. On the other hand, if the lapse
rate were such that the temperature increased steadily with altitude (i.e., Γ negative), there can be no upper
limit!
3.2.4
U.S. Standard Atmosphere
For operational meteorological or aeronautical calculations involving pressure, temperature, density, etc., at
various altitudes, it is not always possible to predict in advance precisely what conditions will be encountered
in a real-life situation. As we have seen, the vertical temperature structure of the atmosphere, and hence
its density and pressure profiles, may vary markedly from day to day, season to season, and from location
to location. Nevertheless, it is often necessary to assume something about the typical characteristics of
the atmosphere, even if you know that, in any real situation, your assumptions may represent only crude
approximations to reality.
It is for this reason that the U.S. Standard Atmosphere was created — to provide a standard basis for
computing the typical or average values of operationally significant atmospheric variables. This standard
atmosphere was computed by the U.S. Weather Bureau at the request of the National Advisory Committee
for Aeronautics (NACA). It is meant to represent normal conditions over the United States ast 40◦ N.
The following are the basic specifications of the U.S. Standard Atmosphere up to an altitude of 32 km:
(1) The surface temperature is 15.0◦ Cand the surface pressure is 1013.25 mb (or hPa).
(2) The air is assumed to be dry and to obey the ideal gas law.
(3) The acceleration of gravity is assumed to be constant and equal to 9.80665 m s−2 .
(4) From sea level to 11.0 km the temperature decreases at a constant lapse rate of 6.5◦ Cper km. This region
is the troposphere.
(5) From 11.0 km to 32 km the temperature is constant at −56.5◦ C. This region is the stratosphere.
Note that the U.S. Standard Atmosphere has constant lapse rate within each of the “spheres” (troposphere,
stratosphere, etc.). Thus, the relationships (3.40) and (3.41) applies within those layers, provided only that
21
one remembers (in the case of the stratosphere and higher layers) to substitute the pressure and temperature
at the bottom of the layer in question, and substitute z − zbottom for z.
The following table gives the geopotential heights of a few standard pressure levels in the U.S. Standard
Atmosphere:
Pressure [mb]
1013.25
1000
850
700
500
300
200
3.2.5
Geopotential Height [m]
0
111
1457
3012
5574
9164
11784
Calculation of Standard Pressure Levels for Actual Profiles
The hypsometric equation which allows us to calculate the thickness of the atmospheric layer between
two pressure levels, given an arbitrary (virtual) temperature profile T (p) between the two levels. The
applications of this are obvious: if you know the surface pressure at a certain station and you are able
to obtain temperature profile from a balloon sounding, then you can easily calculate the thickness of each
conveniently sized layer from the surface to the top of the sounding and then add them all up to get the
height (usually in geopotential meters) of each of the so-called mandatory pressure levels — 1000, 925, 850,
700, 500, 400, 300, 250, 200, 150, 100 mb, and so on. Compiled from all the upper air stations around
the country (or the world), these heights are used to produce the widely used constant pressure maps (also
known as upper air charts) , on which the contours represent lines of equal altitude for the specified pressure
level. The 500 mb map in particular is one of the most important analysis products utilized by forecasts,
though other pressure levels also have important roles.
The height of these pressure levels is only meaningful if referenced to a common standard. Therefore, one
must add the station elevation to the heights calculated from the hypsometric equation in order to get the
height of a given pressure level above sea level.
For more info about upper air constant pressure charts,
//http://earthlab.meteor.wisc.edu/el-upair.html.
consult the following web page:
In addition to the constant pressure charts, one of the most important operational meteorological products
used in forecasting is the surface pressure map. This is the kind of map you most often see in the newspaper
with the cold and warm fronts drawn in, as well as the position of lows and highs. The lows and highs
refer to relative minima and maxima in the surface atmospheric pressure. Since altitude has a strong effect
on the pressure actually measured at a meteorological station, the pressures on a surface map must again
be referenced to a common altitude (sea level) in order to be meteorologically meaningful. Otherwise, you
would always find a deep (but meaningless) pressure low over the rocky mountains and other regions of high
terrain.
To calculate the sea level pressure p0 from the observed station pressure, one need only set ∆z in the
hypsometric equation equal to the elevation h of the station above sea level and solve for the pressure at the
bottom of the hypothetical layer of the atmosphere whose top is at the station pressure ps :
p0
RT̄
h=
log
(3.42)
g
ps
22
or
p0 = ps exp
g h
RT̄
(3.43)
But to do this calculations requires an assumption about the mean layer temperature T̄ . This temperature
does not exist, of course, since the layer in question is underground!
The exact assumption used varies from country to country and, sometimes, from station to station. It is
important that whatever method is used, it should not lead to serious differences in the sea level pressure
computed for a given higher altitude station as compared with that of a nearby low level station, since this
would introduce spurious features into a surface pressure map.
One method is to simply assume the station temperature for the top of the hypothetical layer, and then
assume some standard lapse rate for the temperature profile down to sea level. The lapse rate assumed
might reasonably be 6.5◦ C/km, which is the value used by the U.S. Standard Atmosphere; sometimes it is
assumed to be 5.0◦ C/km, which is roughly one-half the dry adiabatic lapse rate and is also fairly typical of
the lapse rate in the free atmosphere. Finally, one might assume a representative lapse rate which has been
determined individually for each station.
One disadvantage of using the current surface temperature alone is that this temperature at a high altitude
station is often much warmer during the day (due to solar heating) and much colder at night (due to
radiational cooling) than the atmosphere at the same altitude above sea level away from the high terrain.
The result can be a serious underestimate of the “true” sea level pressure during the daytime and an
overestimate during the night. One sometimes tries to get the two effects to cancel by instead using the
average of the current surface temperature and that measured 12 hours earlier.
A similar type of correction is needed for mandatory levels other than the surface, if the station pressure
is lower than the pressure for a mandatory level. For example, since an 8-meter change in elevation leads
to about a 1 millibar change in pressure, one need only be around 100 m above sea level in order for the
average pressure to be about 13 mb lower than the average sea level pressure of 1013 mb. In other words, at
very many inland stations, the terrain elevation is such that the station pressure is often (or even always)
less than 1000 mb. The height of the 1000 mb level (and occasionally the 850 mb level) may therefore be
well below the surface of the earth. In such cases, the determination of the height of the constant pressure
surface in geopotential meters is completely analogous to the determination of sea level pressure.
Side note for pilots: The altimeter setting used by pilots is simply the sea level pressure for a given station
expressed in inches of mercury. For example, the standard sea level pressure of 1013 mb is equivalent to an
altimeter setting of 29.92” Hg. Since an altimeter is essentially a barometer calibrated in units of altitude
instead of pressure, an incorrect altimeter setting will lead to an inaccurate estimate of one’s flight altitude.
Question for thought: What happens when a pilot sets his altimeter to reflect a sea level pressure of 29.92”
Hg and subsequently flies at an indicated altitude of 1000’ above the terrain height into a region where
the true sea level pressure (or altimeter setting) is 28.00” Hg ? Also, say the altimeter is correctly set for
an entire flight, but the atmospheric temperature decreases significantly from origin to destination: what
happens to the pilot’s true altitude if he maintains a constant indicated altitude of, say, 10,000’ ?
23
Chapter 4
First Law of Thermodynamics
4.1
Pressure-Volume Work
Having discussed the basic physical properties of the atmosphere — e.g., the balance between gravity and
pressure, the ideal gas law, the observed temperature structure of the atmosphere, etc. — we will now finally
take the plunge into honest-to-goodness atmospheric thermodynamics.
Let us return first of all to the concept of mechanical work: work has dimensions of energy and, in fact, an
amount ∆W of mechanical work done on a system implies the addition of that much energy to the system,
even if we don’t necessarily know in what form that added energy will appear. How can air do work or be the
recipient of energy via mechanical work? Recall that mechanical work is defined as force times displacement;
i.e.,
δW = f) · d)x
(4.1)
or
x1
∆W =
f) · d)x
(4.2)
x0
Since pressure is just force per unit area, it is apparent that air does mechanical work whenever it expands
against an external pressure; it is the recipient of work if it is compressed despite its own internal pressure.
For example, consider a cylindrical piston of cross-sectional area A containing a volume V of air (or any
substance, for that matter) with pressure p. The total force F exerted on the face of the piston is equal to
the product of A and p. If the piston is slowly moved outward by a small amount ds, then the air inside the
cylinder will have done an increment of work equal to
δW = F ds = pA ds
(4.3)
The product A ds in turn represents an incremental change in volume, which we can write as dV , so that
δW = p dV
(4.4)
It turns out that this expression for mechanical work performed by a fluid is valid irrespective of the geometry
of the volume of air considered and the manner in which it expands. It is also valid for any fluid, irrespective
of whether or not it is an ideal gas.
The equation as given above is valid only for an infinitesimal change of volume dV leading to an infinitesimal
amount of work δW . This is because if you make a larger change of volume in the cylinder by moving the
24
piston a finite distance, the pressure within the cylinder is very likely to change, so that a simple product
of p and the change in volume ∆V can no longer give the amount of work done. In such a case, we need to
sum up the products p dV over the entire path taken by the piston; in other words, we need to perform the
integral
V1
∆W =
p dV
(4.5)
V0
In order to perform the above integral, however, we still need to know how the pressure changes as we go
from volume V0 to volume V1 . Based on the information given so far, we can say nothing definitive about the
change of pressure with volume, even for a gas obeying the ideal gas law! This is because the relationship
between p and V in the case of an ideal gas (and most other substances) depends on a third variable, the
temperature T . Without knowing what the temperature is doing during the time that the air expands from
volume V0 to volume V1 , it is impossible to know what the pressure is doing and, consequently, it is impossible
to calculate ∆W .
In short, the total work done by (or on) a system during a transition from one state to the next depends
on how it gets there. For this reason, work cannot be regarded as a state variable: one cannot look at a
system in a particular state (i.e., having known pressure, volume, and temperature) and infer how much total
accumulated mechanical work that state represents, just as one cannot tell what the current odometer reading
of a car is based on the car’s present distance (as the crow flies) from the place where it was manufactured,
In atmospheric science, it is usually more convenient to use intensive units — i.e., work or energy per unit
mass, rather than the total work or energy exchanged between a system and its environment. By dividing
the above equations by the mass M of the system, we have
dV
δW
= δw = p
= p dα
M
M
or
∆w =
(4.6)
α1
δW =
p dα
(4.7)
α0
where α is, as usual, the specific volume. This is the form we will tend to use most often, though you should
be equipped to deal with extensive units when they arise.
It is often helpful to consider the above expression for pressure-volume work in graphical form. One may
plot the path a system takes from one state to another on a graph whose x-axis is specific volume α and
whose y-axis is pressure p. For example, let’s say that the system starts out in the state A = {α0 , p0 } and
winds up in the state B = {α1 , p1 }. The solid curve on the graph below indicates one of many possible paths
the system could have taken to get from the first state to the second state. The corresponding amount of
work ∆W done is given graphically by the area under that solid curve.
Now let’s suppose that the system is returned to its initial state, but this time via the dashed curve. Its final
state is now the same as its initial state; that is, it has the same temperature, pressure and specific volume
that it started out with. However, the pressure-volume work performed during the second path (from B
to A) is now given by the area under the dashed curve, which is different than the area under the solid
curve. In other words, different amounts of work were performed in getting from A to B than from B to A.
Moreover, because the limits of integration have been reversed, the work performed going from A to B has
the opposite sign as that going from B to A. The total work involved in going from A to B and back to A
again is the difference between the amount of work done in each direction separately, and corresponds to the
area enclosed by the two curves.
α
Now let’s clarify the convention we will be using: positive work will be taken to mean α01 p dα > 0, in
other words, work performed by the system on the environment; negative work consequently means work
performed on the system by the environment. This convention is by no means universal – some textbooks
use the reverse. It is only important that one choose a convention, say what it is, and then stick to it.
25
Given the way we have depicted the two paths in the graph, it is clear that this example represents net
positive work; i.e., work done by the system in going from A to B and back to A again. If the system had
followed the reverse path (i.e., counterclockwise), the magnitude of the work done would have stayed the
same but it would now represent negative work — i.e., work done on the system by the environment.
A graph with pressure and specific volume as the two independent variables (as in the above example)
represents one example of a thermodynamic diagram. Loosely defined, a thermodynamic diagram is any
diagram for which the total work involved in an arbitrary cyclic process (one for which the starting and
ending points are the same) is proportional to the area enclosed by the curve representing that process. For
the particular thermodynamic diagram we just used, the work done by the system is positive for a clockwise
path, negative for a counterclockwise path.
4.2
First Law of Thermodynamics
In the previous section, we saw that for a process in which a quantity of air changes volume at finite pressure,
mechanical work is either being done by the air on the environment or is having work done on it by the
environment. We also know by now that mechanical work implies a transfer of energy between the system
and the environment. But energy must be conserved, so it is fair to ask: Where does the energy come from
when a parcel does work, and what becomes of the energy that a parcel receives when work is done on it?
In very general terms, there are two possible sources for the energy an air parcel expends when it does
mechanical work: either the energy is received from the environment and converted to mechanical work or
else some form of stored energy is expended, or a combination of both. Likewise, when work is done on a
parcel, the resulting energy must either somehow be stored in the parcel or else it must be released back into
the environment in another form.
This leads us to define the energy stored by the parcel as its internal energy. In the case of an ideal gas, the
internal energy is simply the total of all the kinetic energy of the molecules within the system in question.
With this definition, it is apparent that internal energy is closely related to the temperature of the parcel.
A very simple statement of the Law of Conservation of Energy could be written as follows:
• work done by system = reduction in internal energy + energy supplied by the environment; or
• work done on system = increase in internal energy + energy transferred to the environment
Mathematically, both statements can be written (in intensive form) as:
δw = −du + δq
(4.8)
where δw is, as usual, the increment of work (per unit mass) done by the system, du is the change in the
internal energy (per unit mass) of the system, and δq represents an increment of heat energy (per unit mass)
transferred from the environment to the system.
Substituting δw = p dα, we can write
δq = du + p dα
(4.9)
This equality is called the First Law of Thermodynamics, which is really just a special case of the Law
of Conservation of Energy.
We already noted that the internal energy per unit mass u is actually the total kinetic energy per unit mass
of the molecules in a system consisting of an ideal gas. It is therefore clear that the absolute temperature
26
of an ideal gas is a direct measure of u. This fact was demonstrated by Joule in 1848, when he showed that
if a gas expands without doing external work (for example, by expanding into a chamber which has been
evacuated) and without taking in or giving out heat (i.e., the chamber is insulated), then the temperature of
the gas does not change. This statement is known as Joule’s law, and is strictly true only for an ideal gas.
Suppose a small quantity of heat δq is given to a unit mass of material and, as a consequence, its temperature
increases from T to T + dT without a phase change occurring. The ratio dq/dT is called the specific heat
(or heat capacity) of the material. However, the specific heat defined in this way can have any number of
values, depending on whether the material does work as it receives the heat. If the volume of the material
is kept constant, a specific heat at constant volume, cv , is defined which is given by
dq
cv ≡
(4.10)
dT α=const
But if the specific volume is constant, then δq = du from the First Law of Thermodynamics; therefore,
du
cv =
(4.11)
dT α=const
But for an ideal gas Joule’s law applies and therefore u depends upon temperature alone and we may write
cv =
du
dT
(4.12)
Therefore, the First Law of Thermodynamics for an ideal gas can be written in the form
δq = cv dT + p dα
We may also define a specific heat at constant pressure, cp , as
dq
cp ≡
dT p=const
(4.13)
(4.14)
where the material is allowed to expand as the heat is added and its temperature rises but its pressure is
kept constant, In this case, a certain amount of the heat added will have to be expended to do work as the
material expands against the constant pressure of the environment. Therefore, a larger quantity of heat must
be added to the material to raise its temperature by a given amount than if the volume is kept constant.
Since p dα = d(pα) − α dp, let us substitute this into the First Law of Thermodynamics:
δq = cv dT + d(pα) − α dp
(4.15)
From the ideal gas law we also know that d(pα) = d(RT ) = R dT . Therefore,
δq = cv dT + R dT − α dp = (cv + R)dT − α dp
(4.16)
At constant pressure, dp = 0, so that
cp = cv + R
(4.17)
The specific heats at constant volume and at constant pressure for dry air are 717 and 1004 J/(K kg),
respectively.
By combining the above equations, we obtain a useful alternate form of the First Law of Thermodynamics:
δq = cp dT − α dp
(4.18)
The two forms of the First Law given by (4.9) and (4.18) will be used over and over again. You would do
well to commit them to memory.
27
Note: In this class, we will be using a very narrow definition of “the system” in this class: the system
is strictly the gaseous portion of the volume of the atmosphere being considered. By this definition, cloud
droplets, raindrops, etc., within a volume of air are considered to be part of the environment, not the system,
so that any energy absorbed or released by liquid water or other non-gaseous constituents is regarded has
having been lost to or received from the environment. Likewise, chemical reactions and/or phase changes
which involve conversion of energy count as external sources or sinks of energy from the perspective of the
parcel of air.
4.2.1
Digression on Exact Differentials
Given an expression A dx + B dy which may be identified as dv, the differential dv is an exact differential if
and only if it is equal to the total differential of some function f (x, y). That is,dv = A dx + B dy is exact if
∂f (x, y) ∂f (x, y) A=
and
B
=
(4.19)
∂x y
∂y x
where the subscripts imply that the indicated variable is held constant during the partial differentiation. If
these equalities do not hold, then dv is inexact.
A useful theorem in dealing with exact and inexact differentials
is Euler’s Theorem which states that an
∂A ∂B expression dv = A dx + B dy is exact if and only if ∂y = ∂x y
x
Proof:
a) If dv is exact
dv =
∂f (x, y) ∂f (x, y) dx
+
dy
∂x y
∂y x
(4.20)
Since
∂2f
∂2f
=
∂x∂y
∂y∂x
∂f and A is identified as ∂f
∂x while B is identified as ∂y then
y
x
b) If
this implies
A=
Since
then therefore
(4.21)
∂A ∂B =
∂y x
∂x y
(4.22)
∂A ∂B =
∂y x
∂x y
(4.23)
∂f ∂x y
and
B=
∂f ∂y x
(4.24)
∂2f
∂2f
=
∂x∂y
∂y∂x
(4.25)
∂f ∂f dv =
dx +
dy
∂x y
∂y x
(4.26)
so that dv must be exact.
28
An important property of an exact differential is that its line integral is a function of only the limits of
integration and not the path taken; i.e.,
dv = vB − vA
(4.27)
C
where the C indicates any curved path in (x, y) and vA and vB denote the values of v at the endpoints of
the curve C. It follows that the line integral around a closed path must equal zero; i.e.,
dv = 0
(4.28)
C
The above is never true for an inexact integral.
So, what was the point of all that? Well, it turns out that the concept of an exact differential gives us a
rigorous basis for determining whether or not a variable in a thermodynamic system is a state function.
From the ideal gas law we know that the state of a given homogeneous parcel of air can be completely
specified with any two of the three variables p, T , and α. In other words, any one of these three can be
regarded as a state function of the other two. Where the definition of exact differentials comes in handy is
when we want to determine whether another variable which is not a member of these original three is also
a state function.
If you choose any two state variables as coordinates (e.g., p and α) and then allow an arbitrary process to
move the system from one state to another [e.g., from (p1 , α1 ) to (p2 , α2 )], any variable which is a function
only of the endpoints and is independent of the path taken is a state variable.
We already saw that an increment of work δw is not an exact differential, and hence is not a state variable,
because ∆w is not zero for a closed path on a graph of p versus α but, rather, it is equal to the area enclosed
by the curve. This is the reason why I have used the notation δw rather than dw to indicated an increment
of work; the δ serves to remind you that w is not a state variable.
What about q, the heat added to a system? The notation I have used in the previous pages already gives it
away, but let’s show it formally: If we integrate the First Law of Thermodynamics around some closed path
in an α,p diagram, we get
δq =
cv dT +
p dα
(4.29)
But for an ideal gas cv dT is zero, since T is a state variable and cv is a constant (it immediately follows
another exact differential, so that the internal energy u is a state
from this that du = cv dT represents
function). And we know that p dα is non-zero, since it is equal to the pressure-volume work ∆w done
along the closed path. So we are left with
δq = p dα = ∆w = 0
(4.30)
confirming that q (heat) is not a state function.
There are a number of very important state variables other than the handful we have already dealt with;
the important thing to remember is that the acid test for a state variable is whether the variable in question
has a unique relationship to p and α and, by extrapolation, to any other pair of legitimate state variables.
4.3
Heat and Heat Capacity
We have already introduced “heat” as the variable q, but do we really understand what is meant? Some
authors, such as Bohren and Albrecht, argue that there is no such thing as heat! In a way, they are right,
but I think the term is still useful if one is careful.
29
Here are a couple of properties of heat that might help clarify things:
• If a temperature difference exists between two systems in thermal contact, thermal energy, measured in
joules, is transferred from the warmer system to the colder system. It is this energy being transferred
from one system to the other that we call “heat.”
• Heat can only appear in the form of thermal energy moving from one part of the universe to another,
during a change of phase, or when one form of energy (e.g., kinetic) is transformed into another (e.g.,
thermal). Thus, a system can have energy (including thermal energy), but it cannot have heat (it is
not a state variable). Of course, this statement might appear to stand in conflict with concepts like
“heat capacity” or “specific heat” — just recall that these really are descriptions of a body’s change
of temperature in response to a unit of added heat.
Because q is not a state variable, you will usually see it expressed as an increment of heat added or subtracted
from a system; i.e., as ∆q or δq.
Common atmospheric processes which can lead to heat transfer:
• Radiation
• Molecular thermal conduction
• Evaporation/condensation (latent heat transfer)
• Frictional dissipation of kinetic energy
Let’s get a little more practice with the relationship between heat, temperature, and heat capacity. Let’s
further simplify things a bit by considering a system which is not allowed to do any pressure-volume work.
If it is an ideal gas, this can be accomplished simply by sealing the gas in a container so that it cannot
expand. Or we can take the case of a common substance like water whose expansion and contraction is
small enough to be neglected when only ordinary atmospheric pressures are involved. Either way, we can
consider a simplified form of the First Law in which both changes in pressure and changes in volume can be
neglected. In this case
(4.31)
δq = du = cv dT = c dT
If we want to consider a system with some mass M which is not a unit mass, we can (if we want) change
everything to extensive units by defining an extensive heat capacity C = cM , in which case we can write
δQ = dU = mcv dT = C dT
(4.32)
Furthermore, if we assume that C is a constant (this is usually a pretty good approximation) then we can
integrate to get the total amount of heat ∆Q transferred during a finite change in temperature ∆T :
T2
C dT = C(T2 − T1 )
(4.33)
∆Q = δq =
T1
Now let’s consider what happens if two such systems, one with heat capacity C1 and temperature T1 is
brought into thermal contact with another system with heat capacity C2 and temperature T2 . The colder
of the two will warm up and the warmer one will cool down. Eventually, they will both have the same final
temperature (let’s call it Tf ). Energy must be conserved, so the amount ∆Q1 of heat used to warm the
cooler body must equal the amount −∆Q2 which left the warmer one. In other words
∆Q1 + ∆Q2 = 0
(4.34)
C1 (Tf − T1 ) + C2 (Tf − T2 ) = 0
(4.35)
30
We can easily solve for both Tf and ∆Q to find
Tf =
C1 T1 + C2 T2
C1 + C2
∆Q1 = −∆Q2 =
4.4
T2 − T1
1/C1 + 1/C2
(4.36)
(4.37)
Special Cases of the First Law
Let’s repeat the two general forms of the First Law of Thermodynamics that we will be using most frequently:
δq = cv dT + p dα
(4.38)
δq = cp dT − α dp
(4.39)
Certain special cases may now be defined using these equations:
(a) An isobaric process is one in which the pressure does not change. In other words, dp = 0, so
cp
cp
δq = cp dT =
cv dT =
du
cv
cv
(4.40)
(b) An isothermal process is one in which the temperature does not change (dT = 0) :
δq = −α dp = p dα = δw
(4.41)
(c) An isochoric process is one in which the volume does not change, or (dα = 0) :
δq = cv dT = du
(4.42)
(d) An adiabatic process is one in which there is no exchange of heat between the system and its environment;
that is, δq = 0 :
(4.43)
cv dT = −p dα
cp dT = α dp
(4.44)
The adiabatic process is of special significance in meteorology because many of the changes that affect a
moving volume of air in the atmosphere can be approximated as adiabatic. Why is this so? Three of the four
mechanisms of heat exchange mentioned a short while ago are relatively slow acting:
Radiation is most effective at exchanging energy between two objects which (a) have large temperature
differences and (b) are good emitters/absorbers. Sunlight represents a very high temperature radiation
source but air is not a good absorber at solar wavelengths (indeed, it’s almost perfectly transparent); air
absorbs/emits infrared radiation much more effectively, but then the temperature differences between the
source and the recipient (e.g., two levels in the atmosphere) are much smaller. As a result, it takes about a
full day or longer for radiative transfer to directly alter the temperature of the air by more than a degree or
two.
Air is a good thermal insulator, so thermal conduction of heat is efficient only when there are strong differences
in temperature over a very short distance. Large temperature gradients within the free atmosphere are rare.
Typically, therefore, conduction is only an important heat exchange mechanism for air within the few feet
31
nearest to the ground, where solar heating during the day and radiative cooling of the ground at night can
lead to large temperature differences between the surface and the air over short distances.
Frictional dissipation of kinetic energy is an almost negligible source of heat, as can be seen by calculating
the temperature change when all of the kinetic energy of a parcel of air moving 10 m/s (a typical atmospheric
velocity) is converted entirely to heat energy in an isobaric process — the increase in temperature is only
0.05 K.
Finally, the one mechanism of heat exchange which can be very important on short time scales (i.e.,
hours or less) is latent heat release or absorption during the phase changes of water: for example, evaporation/condensation or freezing/melting. Of course, these processes are only found in clouds, so outside
of clouds, and away from the ground, processes in the atmosphere may often be regarded as approximately
adiabatic over time scales of up to a couple of days.
If a process is not adiabatic, then δq in the full form of the First Law is non-zero. This term then represents
what is commonly referred to as the diabatic heating term. Any of the four mechanisms listed above are
potentially sources of diabatic heating, though frictional dissipation is rarely regarded as a very important
term in atmospheric thermodynamics (this does not mean that frictional dissipation is not important in
atmospheric dynamics!).
4.5
Poisson’s Equation and Potential Temperature
From the First Law for an adiabatic process and the ideal gas law, we can write
cp dT = α dp =
RT
dp
p
(4.45)
By dividing through by cp and T , we get
R1
1
dT =
dp
T
cp p
(4.46)
Integrating from some initial temperature T0 and pressure p0 to an arbitrary temperature and pressure T
and p
T
1
R p1
dT =
dp
(4.47)
cp p0 p
T0 T
we get
T
T0
=
p
p0
k
(4.48)
where k = R/cp = 0.286. This equation is called Poisson’s equation for adiabatic processes. It is possible to
derive expressions equivalent to the above relating any two of the thermodynamic state variables pressure,
temperature, and specific volume.
This relationship is extremely important. If you know the initial temperature and pressure of a parcel of air,
and then you change its pressure adiabatically to some new known value, then Poisson’s equation tells you
what the new temperature will be.
Poisson’s equation may be used to define a new thermodynamics state variable, the potential temperature θ.
The potential temperature is defined simply by setting T0 equal to θ and p0 equal to 100 kPa (or 1000 mb):
k
T
p
(4.49)
=
θ
100 kPa
32
or
θ=T
100 kPa
p
k
(4.50)
The potential temperature may be interpreted in a couple of ways: (1) if a parcel of air has a known potential
temperature θ, then the above equation gives its pressure as function of temperature, or vice versa; (2) the
potential temperature is equal to the final temperature of a parcel with known pressure and temperature p
and T if it is compressed or expanded adiabatically to a final pressure of 1000 mb.
In any adiabatic process, θ is constant — that is, no matter how you alter the pressure p, the temperature
T will change in such a way that θ stays the same. We say that potential temperature is a conservative
property in adiabatic processes.
Because θ is approximately conserved in the atmosphere when diabatic heating terms are small, it is sometimes used as a tracer of air masses. For example, if you plot contours of θ on two successive vertical cross
sections of the atmosphere (oriented in the direction of motion of the air), the position of a contour having
a certain value of θ will follow the parcels of air to which it was originally attached. If the contours of θ are
found not to move much over, say, a 24 hour period, yet it is known that there is significant air flow, then
one can often conclude that the motion of the air is parallel to (or along) the θ contours.
On a thermodynamic diagram — i.e., a graph for which the vertical and horizontal axis represent state
variables or simple functions thereof (usually chosen so that the work done in a cyclic process is proportional
to area on the graph) — any point on the diagram is associated with a unique potential temperature θ (this
is because θ is itself a state variable). As a result, one may plot lines of cksonstant θ on a thermodynamic
diagram based on the equation for θ as a function of p and T . These lines are called dry adiabats, since they
represent the permissible paths on the thermodynamic diagram of a dry parcel of air undergoing adiabatic
compression or expansion. Obviously, the value of θ on a given adiabat is equal to the temperature T at the
point where the adiabat crosses the 1000 mb level.
For example, in this class we will sometimes use a chart called a Stüve diagram. Strictly speaking, it is not
a true thermodynamic diagram, because area is not quite proportional to energy; nevertheless, it has often
been used in introductory courses because it is a bit easier for novices to read. On a Stüve diagram, the
vertical coordinate is pressure (actually, pk ) and the horizontal coordinate is temperature T . Dry adiabats
(lines of constant θ) are then straight lines which slope from the lower right to the upper left of the chart,
ultimately converging to a single point at a temperature of absolute zero, which is usually way off the chart.
The Stüve diagram is also sometimes know simply as the pseudoadiabatic chart (e.g., in the textbook by
Wallace and Hobbs).
The so-called Skew-T/Log-P diagram is similar to the Stüve diagram, but the vertical coordinate is proportional to the logarithm of pressure, and the lines of constant temperate (isotherms) are skewed to the left
instead of being vertical. The Skew-T diagram is a true thermodynamic diagram with area proportional to
energy.
33
4.6
Dry Adiabatic Lapse Rate
Poisson’s equation gave us the means to determine how temperature changes with pressure for an adiabatic
process. Now let’s look at how the temperature of a moving parcel of air changes with height.
From the adiabatic version of the First Law, we have the following equation for dry air undergoing an
adiabatic change of pressure:
RT
dp
(4.51)
cp dT =
p
Thus for dry air ascending and expanding we have (from the chain rule)
dT
dT dp
R T dp
=
=
dz
dp dz
cp p dz
(4.52)
The pressure p in an unconfined sample (parcel) of air will immediately adjust to the ambient pressure p ,
so
∂p
dp
=
= −ρ g
(4.53)
dz
∂z
where ρ is the density of the ambient air:
p
ρ =
(4.54)
RT with T the ambient temperature.
Combining these equations yields
RT p
dT
g T
=
g=−
dz
cp p RT
cp T (4.55)
Ordinarily, the temperature T of the parcel is different from ambient temperature T by no more than a
couple of degrees. Therefore T /T 1 and the above result simplifies to
dT
g
= − ≡ −Γd
dz
cp
(4.56)
where Γd = g/cp is the dry adiabatic lapse rate. With cp = 1004 J/(kg K) and g = 9.81 m/s2 , we have
Γd = 0.00977 K/m 9.8 K/km
(4.57)
In summary, Γd gives the rate at which temperature falls off with height for a dry parcel which is lifted
adiabatically. Strictly speaking, the value of Γd changes if the air contains some water vapor or if the
temperature of the parcel is much different from that of the environment, but these effects are usually small
enough to be ignored for most purposes.
4.7
Enthalpy
Another state variable that is used extensively in thermodynamics is enthalpy. Its definition (in intensive
units — i.e., specific enthalpy) is
h ≡ u + pα
(4.58)
where u is the internal energy of a system, and p and α have the usual meanings. Differentiating the above
term by term
dh = du + d(pα) = du + α dp + p dα
(4.59)
34
We may substitute the First Law of Thermodynamics written as
δq = du + p dα
(4.60)
dh = δq + α dp
(4.61)
δq = dh − α dp
(4.62)
so that
Rearranging terms gives us
Comparing this equation with the following form of the First Law
δq = cp dT − α dp
(4.63)
dh ≡ cp dT
(4.64)
it dawns on us that, necessarily,
Now if we assume that the environmental temperature is not too different from the temperature of a parcel
(much as we did for the dry adiabatic lapse rate), then we can use the hydrostatic relation
dp = −ρg dz
(4.65)
δq = dh − α dp = dh − α(−ρg dz)
(4.66)
δq = dh + dΦ = d(h + Φ)
(4.67)
to write
or
where Φ is just the geopotential (gravitational potential energy per unit mass). In other words, for an air
parcel moving about adiabatically in a hydrostatic atmosphere, the quantity (h + Φ) is conserved.
Note: Another commonly used name for enthalpy in meteorology is sensible heat.
35
Chapter 5
Moist Processes
5.1
Water Vapor Saturation
The moisture variables we have discussed so far do not require any assumptions concerning the range of water
vapor content which is permissible in the atmosphere — all we know so far is that if you have a mixture
consisting of this much water vapor and that much dry air, then the corresponding mixing ratio w and the
specific humidity q are defined by the relationships given above. Moreover, expressions for the water vapor
density ρv and the vapor pressure e follow directly from the ideal gas law.
Of course, there is in fact an upper limit to how much water vapor can normally occupy a given volume at
a given temperature, and this limit is exceedingly important in meteorology. It turns out that most of the
remaining important moisture variables are, directly or indirectly, defined in terms of this limit, which we
call the saturation point.
Imagine a sealed container from which all gases have been evacuated — the pressure p in this container is
therefore zero. Now add some water to the container without letting any air in and watch what happens. As
is the case with all materials, molecules will begin to escape from the surface of the water into the vacuum.
Since there are (by definition) no molecules in the vacuum which can go the other way, the flow of water
molecules is initially one way. There is therefore a net loss of liquid water and a net increase in the number
of water molecules occupying what used to be the vacuum. In short, the water begins to evaporate.
As the number of molecules in the vapor phase increases, however, they inevitably impinge on the surface of
the liquid water again and a fraction of these are recaptured. Eventually, the concentration of water vapor
molecules is great enough that the rate at which they are recaptured equals the rate at which new molecules
escape from the surface of the water. Once this occurs (and assuming both vapor and liquid have the same
temperature), the system is in equilibrium, and the water vapor is said to be saturated with respect to the
liquid water.
If you were to increase the concentration of water vapor beyond the saturation point so that the vapor is
supersaturated, then the net flow of molecules would be from vapor to liquid — i.e., more vapor molecules
would be captured by the liquid than would escape. In this way, the system would gradually return to
equilibrium and the vapor would again be exactly saturated.
It turns out that it is the vapor pressure e which most directly determines whether water vapor is saturated
or not: if e is less than the saturation vapor pressure es for the given set of conditions, then the vapor is
unsaturated; if e = es , then the vapor is exactly saturated. Furthermore, the value of es depends only on
36
temperature and is not in any way influenced by the presence of other gases like air.
In summary, if you know or can calculate the actual vapor pressure e in a parcel of air with temperature T ,
then you can compare this value with the saturation vapor pressure es (T ) to determine whether or not the
parcel is saturated. If e < es (T ), then any liquid water present can evaporate until e = es (T ). Conversely,
if e > es (T ) then water will tend to condense out as liquid water until e has again been reduced to the
saturation vapor pressure es .
These observations allow us to define the relative humidity RH as the actual vapor pressure e expressed as
a percentage of the saturation vapor pressure es . That is,
e
× 100%
(5.1)
RH =
es (T )
It is because the relative humidity is a fairly direct measure of the rate at which evaporation can occur (or
whether it can occur at all) that RH plays such an important role in our perception of temperature. If the
RH is high on a warm day, then the perspiration which is supposed to cool our bodies by evaporation simply
accumulates and turns our shirts soggy instead.
Note that the relative humidity is not a very useful measure of the absolute moisture content of the air,
because as the temperature changes, so does the RH, even if the actual vapor pressure (or mixing ratio or
vapor density) remains constant.
One thing you have probably heard many times before is the assertion that “warm air can hold more water
vapor than cold air.” This statement is not really correct because, as noted above, the air present has no
bearing on the amount of water vapor than can occupy a given volume of space, so the air is not “holding” the
water vapor. What is really meant is that the saturation vapor pressure es is larger for warmer temperatures
than it is for colder temperatures; that is es increases with increasing temperature T .
It follows that if e < es (T ) (or equivalently, RH < 100%), then we can always reduce es to equal e —
and thus achieve saturation — simply by reducing the temperature T while keeping the vapor pressure e
constant. In view of this fact, it is logical to define another new moisture variable, the dewpoint Td , by the
following relationship:
(5.2)
es (Td ) = e
In words, the dewpoint Td is the temperature you would have to plug into the function es (T ) in order to get
a saturation vapor pressure equal to the actual vapor pressure.
5.2
Latent Heat
In order to further examine the behavior of water at saturation and/or during phase changes, it is necessary
to first review a couple of fundamental physical concepts. The first of these is relatively straightforward to
understand.
When liquid water evaporates to form water vapor, it is undergoing a phase change from liquid to gas. What
is actually happening is that the more energetic molecules near the surface of the liquid are overcoming the
attractive forces between the molecules which hold the liquid together. A certain minimum amount of energy
is required by each molecule in order to break free (that is, kinetic energy is converted to chemical potential
energy), even though the final temperature of the vapor may be the same as that of the liquid. Conversely,
molecules which leave the gas phase and return to the liquid phase experience an increase in their kinetic
energy as their chemical potential energy relative to the water surface is released again.
The heat energy required to convert a unit mass of a substance from one phase to another while keeping the
pressure and temperature constant is called the latent heat L. In the case of a phase change from liquid to
37
vapor, L is called the latent heat of vaporization. For the reverse process (i.e., gas to liquid) it may be called
the latent heat of condensation, but the magnitude of L is the same in each direction!
There is also latent heat associated with the phase change from solid to liquid (e.g., from ice to water). This
latent is called the latent heat of fusion or latent heat of melting (both terms may be used interchangeably).
If we designate the latent heat of fusion as Lf and the latent heat of vaporization Lv , then the sum of the
two gives the latent heat of sublimation Ls which is associated with a phase change directly from ice to vapor
or vice versa. That is
(5.3)
Ls = Lf + Lv
For the purposes of this course, we will mainly be concerned with the latent heat of vaporization of water
Lv . For this reason, we will drop the subscript and use just L from here on.
It turns out that at a temperature of 0◦ C,
L = 2.50 × 106 J/kg
(5.4)
It is actually not quite constant — L is as high as 2.60 × 106 J/kg at -40◦ Cand as low as 2.40 × 106 J/kg
at +40◦ C. However, this minor variation of L with temperature is not enough for us to worry about in this
class, so we will normally use the value given above for 0◦ C.
5.3
Entropy – a new state variable
In order to derive the functional form of the saturation vapor pressure es (T ), it is helpful to introduce a new
state variable called entropy. For a reversible process — that is, one which represents at most infinitesimal
departures from equilibrium — entropy is defined by the relationship
dφ =
δq
T
(5.5)
where dφ is the increase in entropy (per unit mass) accompanying the addition of heat δq at temperature
T . How can this be a state variable, since we know already that q is not a state variable? The fastest way
to see the truth of this is to invoke Euler’s Theorem, which was explained back on page 37. To refresh your
memory, a differential dv = A dx + B dy is exact (and therefore represents a state variable) if and only if
∂A
∂B
∂y = ∂x . From the First Law, we have
(5.6)
δq = cv dT + p dα
In order for q to be a state variable, we would need to find that
∂cv
=0
∂α
=
∂cv
∂α
R
∂p
=
α
∂T
=
∂p
∂T .
But
(5.7)
So we have just proven once again that q is not exact. But what happens when we divide the First Law by
T ? We then have
p
dT
δq
= cv
+ dα
(5.8)
dφ =
T
T
T
which (with the help of the ideal gas law) reduces to
dφ = cv d ln T + R d ln α
(5.9)
It is immediately obvious that this is an exact differential, because
∂cv
∂R
=0=
∂ ln α
∂ ln T
38
(5.10)
To summarize, the change in specific entropy φ of a gas does not depend on the path taken but only on the
endpoints of a process, unlike the case for q.
We will have occasion later to talk more about the physical meaning of entropy. For now, let us return to
the question of obtaining es (T ) by combining what we now know about latent heat of vaporization L and
entropy φ.
5.4
The Clausius-Clapeyron Equation
Recalling that the latent heat of vaporization is the heat required to convert a unit mass of liquid to vapor,
with pressure and temperature held constant, we may write (using the First Law)
q2
u2
α2
L = ∆q =
δq =
du +
pdα
(5.11)
q1
u1
α1
The pressure in the final integral is just the saturation vapor pressure es , which stays constant, so we can
solve all the integrals and write
(5.12)
L = (u2 − u1 ) + es (α2 − α1 )
Because the temperature is also constant, we can also write
q2
L=T
q1
δq
=T
T
φ2
dφ = T (φ2 − φ1 )
(5.13)
φ1
So we now have two independent expressions for L which we may equate and rearrange so that terms
involving the initial state and the final state are on opposite sides of the equal sign:
u1 + es α1 − T φ1 = u2 + es α2 − T φ2
(5.14)
In other words, the combination G = u + es α − T φ remains constant during an isothermal, isobaric change
of phase. This function is called the Gibbs function of the system, and the above equation can be rewritten
G1 = G2
(5.15)
The Gibbs function is not constant with respect to pressure and temperature, so we may differentiate G as
follows:
(5.16)
dG = du + es dα + α des − T dφ − φ dT
But du + es dα = δq = T dφ, so the last equation reduces to
dG = α des − φ dT
(5.17)
Since G is the same for both the liquid and vapor phases when the two are at equilibrium (i.e., same pressure
and temperature), dG1 = dG2 , or
α1 des − φ1 dT = α2 des − φ2 dT
des
des
− φ1 = α2
− φ2
dT
dT
φ2 − φ1
des
L
=
=
dT
α2 − α1
T (α2 − α1 )
α1
(5.18)
(5.19)
(5.20)
We thus have an exact expression for the rate of change of saturation vapor pressure with temperature. This
equation is called the Clausius-Clapeyron equation.
39
Under typical atmospheric conditions, the specific volume of water vapor α2 is vastly greater than that of
liquid water α1 . Furthermore, we can assume that water vapor behaves as an ideal gas. With both these
facts in mind, we can write
L
des
Les
≈
=
(5.21)
dT
T α2
Rv T 2
This form of the Clausius-Clapeyron equation may be integrated by assuming that L is a constant (as we
saw earlier, this isn’t quite correct but it’s close enough).
es
es0
des
L
=
es
Rv
or
es (T ) = es0 exp
T
T −2 dT
(5.22)
T0
L
Rv
1
1
−
T0
T
(5.23)
where es0 is the value of the saturation vapor pressure at temperature T0 , a constant of integration which
must be determined experimentally. It turns out that if we choose T0 = 273 K, then es0 = 6.11 mb or 611 Pa.
Substituting these values into the expression for es , along with L = 2.50×106 J/kg and Rv = 461.5 J/(kg K),
we get
(5.24)
es (T ) = Ae−B/T
where
A = 2.53 × 108 kPa
and B = 5.42 × 103 K
(5.25)
This expression is simple and reasonably accurate, and you may use it in solving problems in this class unless
otherwise indicated. If in your future meteorological work you require a more accurate formula for es , you
should be aware of the following empirical formula which is accurate to within 0.1% over the temperature
range −30◦ C≤ T ≤ 35◦ C:
17.67 Tc
es (T ) = 6.112 exp
(5.26)
Tc + 243.5
where es is in mb and Tc is the temperature in degrees C (not K!).
5.5
Saturation Mixing Ratio
There is another variable related to saturation that is very important: the saturation mixing ratio, for which
we use the symbol ws . Not surprisingly, ws is completely analogous to the saturation vapor pressure es in
that it represents the mixing ratio w for which air at the specified temperature and pressure is saturated.
Also not surprisingly, one may easily compute ws from es and vice versa using the same relationship we
found earlier between w and e. That is
ws (T, p) =
εes (T )
εes (T )
≈
p − es (T )
p
(5.27)
You should note that whereas es is a function of T only, the expression above shows that ws is a function
of both T and p. Also, since there is a unique value of ws for any specified T and p, the saturation vapor
pressure is a state variable, and lines of constant ws may be drawn on a thermodynamic diagram such as
a Stüve diagram. The lines of constant saturation mixing ratio — usually labeled in units of g/kg — are
extremely useful on such a diagram when processes involving moist air are considered, as we shall see now.
40
5.6
Unsaturated, Adiabatic Processes on a Thermodynamic Diagram
On most typical thermodynamic diagrams, such as the Stüve diagram, there are two principle coordinates. As
we saw much earlier, the horizontal coordinate is usually temperature, and the vertical coordinate is usually
closely related to pressure, so that (for example) changes in altitude correspond to vertical displacements on
the thermodynamic diagram.
If you know the temperature and pressure of a parcel of air, you can plot a point on the thermodynamic
diagram which uniquely identifies the values of those two state variables. The potential temperature θ of
a parcel of air is also a state variable, and lines of constant θ — known as dry adiabats — are plotted as
lines sloping to colder temperatures with decreasing pressure (the position and shape of these lines is of
course governed by Poisson’s Equation). Since potential temperature is conserved in an adiabatic process,
any parcel of air which undergoes only dry adiabatic lifting or sinking must remain on a line of constant θ,
even while its temperature and pressure change.
Now you can also plot a point on the thermodynamic diagram corresponding to a parcel’s pressure and
dewpoint. What good is this? Well, the line of constant saturation mixing ratio ws passing through this
point tells you the mixing ratio w of the parcel, because if you were to cool the parcel isobarically to
the dewpoint Td , then w would equal ws . In the absence of condensation or evaporation, mixing ratio is
conserved. As a result, the dewpoint of the parcel will tend to follow a saturation mixing ratio line, despite
any changes in pressure and/or temperature.
These facts have several useful consequences. First of all, by plotting the temperature and dewpoint of a
parcel having a given pressure, one can immediately read off the values of w and ws . The ratio w/ws is
very close to the ratio e/es ≡ RH (you should verify this for yourself), so you can use the thermodynamic
diagram to quickly and easily obtain the relative humidity of a volume of air, given T , Td , and p.
5.6.1
The Lifting Condensation Level (LCL)
Even more importantly, you can easily investigate how the relative humidity or the degree of saturation
changes with adiabatic changes in pressure. If a parcel of air is lifted adiabatically in the atmosphere, the
temperature will follow a dry adiabat and the dewpoint will follow a line of constant saturation mixing
ratio. At the pressure where the two lines intersect, the temperature equals the dewpoint and the air is
saturated. Any further lifting will cause the air to become supersaturated, in which case water vapor will
begin to condense out as liquid water droplets, thus giving rise to a cloud. The pressure at which saturation
is reached by way of adiabatic lifting is called the lifting condensation level or LCL.
The LCL gives an indication of where you would expect a cloud base to form when a layer of the atmosphere
(often assumed to be the surface layer) is mechanically lifted. An example of this is when air flows over a
mountain range or is lifted by an advancing cold front.
5.7
Wet-bulb Temperature
If you take an ordinary dry thermometer and expose it to a volume of air, it will reach thermal equilibrium
when it has the same temperature as the air. Therefore the temperature you read from the thermometer
will simply be that of the air. Now imagine wrapping the bulb of the thermometer in a cotton wick or
some other porous material and then moistening the wick with water. As long as the wick is kept moist and
41
well-ventilated, and assuming the air is unsaturated, the water will evaporate at a steady rate. As it does so,
heat must be continually supplied in an amount equal to the latent heat of vaporization of the water. This
heat is taken from the air passing over the wick of the thermometer, resulting in a drop in temperature. The
temperature measured in this way is called the wet-bulb temperature and is indicated by the symbol Tw .
Formally, the wet-bulb temperature is the temperature a volume of air attains when water is evaporated into
the air exactly to the point of saturation and assuming that all of the latent heat of evaporation is supplied
by the air.
Obviously, if the air is saturated, no evaporation takes place, in which case the wet-bulb temperature is the
same as the ordinary (or dry-bulb) temperature T If the air is very dry, on the other hand, the wet-bulb
depression, which is the dry-bulb temperature minus the wet-bulb temperature, may be quite substantial.
It is tempting to think that the wet-bulb temperature should be the same as the dewpoint temperature —
after all, evaporation leading to the absorption of latent heat should continue until the temperature of the
wet bulb equals the dewpoint, right? This is not correct, however. The difference between the dewpoint
and the wet-bulb temperature is best understood as follows: in the case of the dewpoint, we are speaking
of the temperature we obtain if we simply cool a volume of air until saturation is reached while keeping the
moisture content constant.. By contrast, the wet-bulb temperature is the temperature we obtain if we cool
the air to saturation by evaporating water into it. The latter provision implies that the moisture content has
increased, and therefore saturation is reached at a higher temperature than the dewpoint. As a result, we
have the following rule:
(5.28)
Td ≤ Tw ≤ T
The equality between Td , Tw , and T holds if and only if the air is saturated.
Not surprisingly, there is a well-defined physical (and mathematical) relationship between the above three
temperatures. As a result, one of the most accurate ways to measure atmospheric humidity is to use two
thermometers — one with a ventilated moistened wick — to simultaneously measure T and Tw . Such
an apparatus combining a wet-bulb and dry-bulb thermometer is called a psychrometer. Based on your
measurement of T and Tw , you may compute the dewpoint Td . From T and Td , one may of course compute
all other relevant moisture variables, such as specific humidity, vapor pressure, relative humidity, mixing
ratio, virtual temperature, LCL, etc.
The exact mathematical formula for dewpoint as a function of the wet-bulb and dry-bulb temperatures is little
complicated. It is generally more convenient to use non-mathematical means to compute the dewpoint from
the temperatures measured by a psychrometer. One way is using a circular slide rule called a psychrometric
computer, sometimes referred to affectionately as a “pizza wheel” or a “psychedelic confuser.” Interestingly,
there is also a simple and very useful graphical relationship between Td , Tw , and T on a thermodynamic
chart like a Stüve diagram. We will describe this relationship shortly, when we have defined the final set of
curves appearing on standard thermodynamic charts.
Question: What happens when you use a psychrometer to measure humidity on a very cold day? Two
things can happen. If the wet-bulb temperature is below freezing, the water on the wick may well freeze.
On the other hand, it might not — it could simply become supercooled. In the latter case, there is no
serious complication; the form of the Clausius-Clapeyron equation describing the saturation vapor pressure
with respect to liquid water (the one we have been using) remains unchanged; moreover, the latent heat
involved in evaporating water from the wick is still the latent heat of vaporization. Consequently, the same
mathematical or graphical relationship (or scale on the psychrometric computer) may be used to compute
dewpoint and other humidity variables.
On the other hand, if the wick freezes, the evaporation taking place is actually sublimation — the direct
conversion of ice to vapor. For this process, the latent heat of sublimation is needed, which is somewhat
larger than the latent heat of vaporization. This fact alters both the definition of water vapor saturation
and the latent heat per unit mass of ice required to bring the air to saturation. Consequently, a different
42
(but completely analogous) set of relationships must be used to calculate humidity when the wet-bulb on a
psychrometer is frozen.
In both of the above cases, we are assuming that the temperature is at or below freezing. Of course, this
implies that the dewpoint will also be found to be below freezing, which leads us to raise another distinction:
the dewpoint, which represents the temperature at which air is saturated with respect to liquid water is, in
general, different than the temperature at which the air is saturated with respect to ice. The latter temperature
is known as the frost point. For temperatures below freezing, the frostpoint is higher than the dewpoint,
which implies that water vapor may be at saturation (or even supersaturated) with respect to an ice surface
even if it is unsaturated relative to liquid water. This fact has important consequences for precipitation
formation and will be discussed later in this course.
5.8
The Pseudoadiabatic Lapse Rate
Let’s return to the Lifting Condensation Level (LCL). This is the pressure level in the atmosphere at which
a parcel of air will reach saturation when lifted dry adiabatically. We already saw that if the air is lifted
above the LCL, it will cool further, and excess water vapor will condense out so that the air remains almost
exactly at saturation (later, we will see how exactly).
But if water is condensing out of the air parcel, then it must be releasing latent heat into the air. This implies
that above the LCL, a rising parcel of air will be warmer than would be predicted by the dry adiabatic lapse
rate.
If we consider the water droplets that form within the rising parcel to be part of the thermodynamic system,
and if we assume that the droplets do not fall out of the parcel, then the latent heat released simply represents
a conversion of energy from one form to another and does not involve the transfer of heat between the parcel
and its surroundings. Seen from this point of view, the process may still be regarded as adiabatic, though not
in the same sense as the dry adiabatic process we considered earlier. Instead, we say that it is a saturatedadiabatic process. A saturated-adiabatic process is reversible — if you begin to lower a parcel which was
previously ascending with condensation, the droplets which formed will simply evaporate again, taking latent
heat with them, and the parcel will experience the same temperature evolution in reverse.
On the other hand, if we assume that all liquid water falls out of the parcel immediately after it is condensed,
then we can no longer speak of a reversible, adiabatic process — in this case we call it pseudoadiabatic. For
calculating the change of temperature of a rising parcel of air with altitude, the difference between a true
saturated-adiabatic process and a pseudoadiabatic process is very slight. A pseudoadiabatic process, however,
is somewhat less complicated to deal with mathematically, because you don’t have to separately account for
the heat energy which is carried by the liquid water droplets.
We can now proceed to look at pseudadiabatic (or moist adiabatic or saturated adiabatic processes) in a more
quantitative way. Earlier, when we talked about enthalpy, we showed that by substituting the hydrostatic
law into the First Law of Thermodynamics we could write
δq = dh + dΦ = cp dT + g dz
(5.29)
where δq is an increment of added heat energy. Now when rising air cools by expansion, the saturation mixing
ratio ws decreases. If we assume that the air does not become supersaturated, then w must remain equal
to ws by allowing excess water vapor to condense out. The amount of latent heat released by condensation
with each incremental decrease dws of saturation mixing ratio may be written
δq = −L dws
43
(5.30)
Substituting this into the form of the First Law given just above, we have
−L dws = cp dT + g dz
(5.31)
Since water vapor is never more than one or two percent of the total mass of a volume of air, we may
safely assume that the heat capacity cp of saturated air is reasonably well-approximated by that of dry air.
Dividing both sides of the above equation by cp dz and rearranging terms, we have the following steps in our
derivation:
L dws
g
dT
=−
−
(5.32)
dz
cp dz
cp
dT
L dws dT
=−
− Γd
dz
cp dT dz
dT
L dws
1+
= −Γd
dz
cp dT
Γs ≡ −
Γd
dT
=
dz
s
1 + cLp dw
dT
(5.33)
(5.34)
(5.35)
where Γs is the saturated adiabatic (or moist adiabatic or pseudoadiabatic) lapse rate, and Γd , as before, is
the dry adiabatic lapse rate given by g/cp .
The derivative
dws
dT
is of course given by
d
dws
=
dT
dT
εes (T )
p − es (T )
(5.36)
s
where p is the environmental pressure. We see that since dw
dT is a function of both temperature and pressure,
s
Γs (unlike Γd ) is also a function of temperature and pressure. Furthermore, since dw
dT is always greater than
zero, the denominator in the expression for Γs is always greater than one, and
Γ s < Γd
(5.37)
as you would expect.
Actual values of Γs typically range from around 4 K/km near the ground in warm, humid air masses to
around 6–7 K/km in the middle troposphere. At very high, cold altitudes (say, near the tropopause), Γs is
only very slightly smaller than Γd , since there is very little moisture left in the parcel to condense out.
Lines on a thermodynamic diagram which show the change in temperature with height of a parcel of air
which is rising moist adiabatically are called pseudoadiabats or moist adiabats.
Key Result: The temperature of a parcel of air which starts out as unsaturated at some pressure p and then
rises adiabatically will follow a dry adiabat on a thermodynamic diagram until it reaches its LCL; thereafter,
it will follow a pseudoadiabat.
5.9
Equivalent Potential Temperature
Earlier we introduced the potential temperature θ, which is defined to be the temperature a parcel would
have if it were compressed (or expanded) dry-adiabatically to a pressure of exactly 1000 hPa. The potential
temperature is conserved in a dry-adiabatic process; that is, if no condensation or evaporation occurs and
no heat is added to or removed from a parcel of air, θ does not change as the parcel moves up and down in
the atmosphere.
44
If condensation does take place, say, because the parcel is lifted dry adiabatically up to the LCL and
pseudoadiabatically thereafter, the potential temperature no longer remains constant but increases because
of the latent heat released by condensation. In situations where saturated adiabatic processes are involved,
it is convenient to define a variable which is analogous to the potential temperature θ but which is conserved
despite the release of latent heat.
One such variable is the equivalent potential temperature, which we designate as θe . The equivalent potential
temperature is defined to be the potential temperature a parcel of air would have if all of the moisture it
contained were condensed out and the resulting latent heat used to warm the air. Clearly, with the above
definition, θe is conserved for both dry and saturated adiabatic processes, since in a dry adiabatic process,
both θ and Lw (the total available latent heat in the parcel) remain unchanged, and in a saturated adiabatic
process, any increase in potential temperature is exactly balanced by a corresponding decrease in Lw.
There are actually two different methods of computing θe , and the results each gives are slightly different.
First there is the isobaric equivalent potential temperature, which is the potential temperature the parcel
would have if all water vapor were condensed out at constant pressure. It is easy to show that this is given
by
Lw
θe = θ 1 +
(5.38)
cp T
since Lw/cp expresses the change of temperature resulting from a complete release of all available latent
heat at constant pressure.
Somewhat more common is the so-called adiabatic equivalent potential temperature, which is the temperature
that an air parcel would have after undergoing a dry-adiabatic expansion until saturation is reached, a
saturated adiabatic expansion up to some extremely cold temperature for which w essentially goes to zero,
and then a dry-adiabatic compression back to 1000 hPa. Graphically, this corresponds to taking the parcel
along a dry adiabat up to the LCL, and then following a pseudoadiabat up to a sufficiently low pressure
(e.g., less than 200 mb) and then following a dry adiabat back down to the 1000 mb level. Because of the
slightly different thermodynamic assumptions involved in pseudoadiabatic processes, the numerical value of
the adiabatic equivalent potential temperature is
θe = θ exp
Lw
cp T
(5.39)
This value is always greater than the isobaric equivalent potential temperature, but not by a great deal,
since cLw
1
pT
Because θe is conserved during a saturated adiabatic process, pseudoadiabats on a thermodynamic diagram
are actually lines of constant θe and are usually labeled with the corresponding values.
5.10
Normand’s rule
A couple of pages back, the wet-bulb temperature Tw was introduced, but no means was given for computing
it. While it would be possible to give a mathematical expression of Tw — one which unfortunately can only
be solved iteratively — it is also possible to conveniently arrive at an approximate value of Tw from a
thermodynamic chart.
Simply put, Normand’s Rule states that the wet-bulb temperature may be determined by finding the temperature of a parcel of air at its LCL and then following a pseudoadiabat from that temperature back down
to the parcel’s actual pressure.
This approach actually defines what is called the adiabatic wet bulb temperature, as contrasted with the
45
isobaric wet bulb temperature corresponding to the temperature measured by the moistened thermometer
in a psychrometer. However, the difference between the two never exceeds a few tenths of a degree and is
therefore usually ignored.
5.11
Wet-bulb Potential Temperature
The wet-bulb potential temperature θw is found graphically by first determining the wet-bulb temperature
Tw using Normand’s Rule and then following the same pseudoadiabat all the way to where it intersects the
1000 hPa level on the thermodynamic diagram.
Like the equivalent potential temperature θe , the wet-bulb temperature θw is conserved during both dry
adiabatic processes and saturated adiabatic ascents.
5.12
The Parcel Method
Our discussion of dry and saturated adiabatic lapse rates, conserved and non-conserved moisture variables,
etc., has implicitly been geared toward the thermodynamic behavior of an isolated “parcel” of air; i.e., an
idealized blob of air which we assumed could experience changes in pressure (e.g., by moving around the
in the atmosphere) without mixing with its environment or otherwise losing its identity. This idealization
is a very useful one, and we will now begin to extend the so-called parcel method to relate the behavior of
individual parcels of air to the vertical temperature and humidity structure of the atmosphere.
When using the parcel method, one typically starts with a hypothetical parcel of air taken from an arbitrary
level in the atmosphere and then move it adiabatically up or down relative to the environment in order to
examine the energetics of such vertical motions. According to the parcel method, one not only assumes that
there is no mixing between the parcel and its environment, one also assumes that the vertical motion of the
parcel does not lead to any alteration in the overall characteristics of the environment. The only way this
assumption can hold is if the volume of the parcel is assumed to be very small, so that upward motion of the
parcel (for example) will not lead to perceptible sinking (for example) of the environmental air surrounding
the parcel.
It should be obvious that the simplistic assumptions entailed by the parcel method are rather unrealistic:
parcels which are small enough to be displaced vertically in the atmosphere without inducing perceptible
compensating motions in the environment are generally too small to move around without completely blending with the surrounding air. About the only common system which closely approximates an idealized parcel
is the air in the bag of a hot air balloon, and then only when the propane flame is turned off (otherwise it’s
not adiabatic!).
Nevertheless, the parcel method provides a very convenient starting point for understanding the effects of
a given vertical temperature and moisture profile on the behavior of the atmosphere, provided one keeps
its limitations in mind. For example, the energetics of thunderstorm updrafts are often analyzed using the
parcel method despite the fact that, in reality, considerable mixing with the environment occurs as well as
compensating downward motion.
46
5.13
Buoyancy
When considering the behavior of a parcel of air moving about in the real atmosphere, its buoyancy relative to the environment is of fundamental importance. Buoyancy is simply the vertical force per unit
mass experienced by a parcel of air, owing to differences between its density and that of the surrounding
environment.
Most people learned in early childhood that warm air tends to rise and cold air tends to sink. A more general
way of stating this observation is that a parcel of something which is less dense than its surroundings will
experience an upward force (helium is also a good example), and vice versa. In the absence of some other
opposing force (e.g., the tension of the string on a helium balloon), a less dense parcel will accelerate upward
and a parcel of denser air will accelerate downward. Eventually, the force of frictional drag will balance the
buoyant force, at which time the acceleration will be zero but the velocity will be nonzero.
Let’s quantify this as follows: Consider a parcel of air with volume V and density ρ. This parcel displaces an
equal volume of environmental air having density ρ . The magnitude of the downward force due to gravity
exerted on the parcel is
(5.40)
Fdown = M g = ρV g
The upward force exerted on the parcel by the surrounding atmosphere represents the difference in pressure
at the bottom and top of the parcel. This difference can be calculated directly from the hydrostatic law,
and we find
∂p
= V ρ g
Fup = −V
(5.41)
∂z
The net buoyant force is thus
FB = Fup − Fdown = V ρ g − V ρg = V g (ρ − ρ)
(5.42)
To find an expression for the buoyant force per unit mass fB (= vertical acceleration when no other forces
are present), we just divide the above by the mass M of the parcel, which is ρV :
ρ −ρ
fB = g
(5.43)
ρ
Now we can use the ideal gas law to express the densities in terms of more commonly measured meteorological
variables
p
p
ρ =
,
(5.44)
ρ=
R d Tv
Rd Tv
substitute these into our expression for fB and cancel variables appearing both the numerator and denominator to get
Tv − Tv
fB = g
(5.45)
Tv
Note that we have used the fact that the pressure p of the parcel is the same as that of the environment at
the same level.
As we already anticipated, the buoyant force is upward when Tv > Tv and downward when Tv < Tv . When
the two virtual temperatures are equal, the parcel is said to be neutrally buoyant.
It is interesting to note from the above relationships that a moist parcel of air is somewhat more buoyant
than a dry parcel with the same temperature. This is simply a consequence of the fact that the molecular
mass of water vapor is much lower than that of dry air (18 versus 29 g/mole).
47
5.14
Stability
In a very general sense, a system is stable if in response to a small perturbation it tends to return to its
initial state. A system is unstable if in response to a similar small perturbation it tends to accelerate away
from its initial state.
As a concrete example, consider a marble resting in the bottom of a round bowl. Left to its own devices, it
will sit at the very lowest point in the bowl. If you gently knock it away from its favorite resting point, it
will roll back downhill toward its starting point, possibly oscillating back and forth until friction brings it
back to rest at the lowest spot in the bowl. This system is stable.
Now consider a marble which has been carefully balanced on top of a globe. If you leave it alone, it just may
stay there forever. However, if you disturb it even slightly, it will roll downhill away from its starting point,
picking up speed as it goes. Without intervention, the marble will never voluntarily return to its starting
point. This system is unstable.
Finally, consider a marble resting on a large, flat, level table. If you give it a nudge, it will move away
from its starting point, but it will not pick up speed, nor will it show any tendency to reverse direction and
return to the starting point. If there is no friction, it will continue rolling without changing speed; if there is
friction, it will eventually come to a stop at a new location which is no better or no worse than the original
location, as far as the marble is concerned. This system is an example of neutral stability.
Entirely analogous examples of stability, instability, and neutral stability may be found in the atmosphere.
For example, let’s say a certain parcel of air in the atmosphere is initially at rest and has the same temperature
as the environment at that level. Now let’s say it is given a mild push upward, so that it moves away from
its starting location. After traveling a short distance upward, its temperature falls somewhat owing to
adiabatic expansion. There are three distinct possible outcomes of this process: (1) it might turn out that
the parcel is now denser than the environment at its new level (stable case), in which case buoyant forces
eventually overcome its upward momentum and send it back toward its starting point; (2) it might turn
out instead that the parcel finds itself less dense than its new surroundings (unstable case), in which case it
will continue to accelerate upward, away from its starting point; (3) finally, there may be no change in the
relative temperature of the parcel (neutral case), in which case it will continue upward until friction brings
it to rest at a new level.
How can we tell which of these cases holds in a given situation? In the simplest scenario, both the environment
and the parcel are dry, so that changes in parcel temperature due its vertical motion correspond to the dry
adiabatic lapse rate Γd = 9.8 K/km. Under which conditions will a upward-displaced parcel of environmental
air find itself colder than its surroundings and thus return to its starting point? Clearly, this occurs whenever
the environmental lapse rate Γ at that level is less than the dry-adiabatic lapse rate Γd . Likewise, if Γ > Γd ,
then an upward displacement of environmental air will lead to the displaced air being warmer than its
surroundings, in which case it will continue to accelerate away from its starting location. You should be
able to convince yourself that the criterion for stability for upward displacements is the same as that for
downward displacements.
We therefore have the following stability conditions for dry air:
Γ < Γd
Γ = Γd
Γ > Γd
STABLE
NEUTRAL
UNSTABLE
48
Now what if the air parcel in question is saturated? In that case, it will not cool with the dry adiabatic lapse
rate when it is pushed upward but instead with the pseudoadiabatic lapse rate Γs . Therefore, the parcel will
be unstable under the condition that Γ > Γs . Since Γd > Γs , it is apparent that values of the environmental
lapse rate Γ exist for which stability depends on whether or not the air is saturated. Therefore, for the
general case in which the air may either be saturated or dry, we define the following stability regimes:
Γ < Γs
Γ = Γs
Γ s < Γ < Γd
Γ = Γd
Γ > Γd
STABLE
SATURATED NEUTRAL
CONDITIONALLY UNSTABLE
DRY NEUTRAL
UNSTABLE
Thus, if the environmental lapse rate within a layer of the atmosphere fulfills the criterion for instability
(or conditional instability, if the air is saturated), then even slight vertical motions within the layer will
likely amplify into large vertical motions, with individual parcels of air accelerating away from their point of
origin until they encounter stable conditions at another level. In practice, such overturning will not continue
indefinitely, since the lapse rate eventually responds by becoming neutral.
On the other hand, a stably stratified atmospheric layer will tend to resist overturning. Instead, if there is
a sufficiently forceful disturbance, waves may form in the stable layer (analogous to the waves on a water
surface) as parcels of air bob up and down about their equilibrium points.
In summary, we have derived the criteria for moist and dry hydrostatic stability. We saw that the air at
a level in the atmosphere can be stable, dry neutral, conditionally unstable, saturated neutral, or unstable.
If the appropriate criteria for instability are fulfilled (which depend on whether or not the air is saturated),
then air at the level in question is susceptible to overturning and mixing. If the instability is great enough,
the vertical motions resulting from even slight disturbances can become quite strong, leading to turbulence,
convective cloud formations, and so forth.
It is important to keep in mind that the definitions of hydrostatic stability given above depend only on the
temperature lapse rate at the level of interest and not on the temperature or moisture profile above or below
that level. There are other forms of instability which do depend on temperature and moisture at other levels
in the atmosphere and these are briefly discussed in Section 2.7.3 in Wallace and Hobbs – there will be no
separate printed notes for this material, but you are required to read about it in the textbook.
49
Chapter 6
Adiabatic, Isobaric Mixing
Consider what happens when two masses of air are mixed adiabatically at constant pressure. Such a process
is important virtually anytime there is mixing of dissimilar air masses in the atmosphere, as long as the
vertical displacements involved are negligible.
If we start with a mass m1 of air having temperature T1 and specific humidity q1 and a mass m2 of air having
temperature T2 and specific humidity q2 , conservation of mass tells us the the total mass of the system after
mixing is
(6.1)
m = m1 + m 2
Moreover, by the same principle, the specific humidity of the mixture must be
q=
m1 q 1 + m 2 q 2
m
(6.2)
Finally, because the mixing is isobaric and adiabatic, the total enthalpy of the system is conserved. Thus
cp mT = cp m1 T1 + cp m2 T2 = cp (m1 T1 + m2 T2 )
(6.3)
where we have assumed that the specific heat capacity cp may be taken to be approximately that of dry air
(i.e., differences in cp due to water vapor are assumed small). Solving for the final temperature, we have
T ≈
(m1 T1 + m2 T2 )
m
(6.4)
5e
p
(6.5)
Finally, by using the approximation
q≈
and noting that p is the same for both air masses, we can write
e≈
m1 e1 + m2 e2
m
(6.6)
It is clear that depending on the ratio m1 /m2 of the two initial air masses, it is possible to obtain for a final
value of T any temperature which falls between T1 and T2 . The same holds true for the final vapor pressure
e. However, since the final value of e and of T both depend in a fixed way on the proportion contributed by
each initial air mass, the final value of e is uniquely determined by the final temperature T . We can therefore
combine the formula above for T with the formula for e and eliminate m1 and m2 as variables. The result
may be written
T − T1
(e2 − e1 ) + e1
(6.7)
e = e(T ) =
T2 − T1
50
If you plot this equation on a graph with T on the horizontal axis and e on the vertical axis, you find that
this is a straight line connecting (T1 , e1 ) and (T2 , e2 ).
An arbitrary mixture of the two air masses will always give rise to a final (T, e) which falls on this straight
line between the image points for the two initial air masses.
If we plot the saturation vapor pressure es (T ) on the same graph, it is strongly curved (recall the exponential
form of the Clausius-Clapeyron equation we have been using). Lines of constant relative humidity on the
chart may also be easily drawn by simply plotting a curve of e(T ) which is a constant fraction of es (T ). It
is easy to see that a mixture of two air masses with dissimilar initial temperatures and/or vapor pressures
always leads to a final mixture with higher relative humidity than that of either of the two initial air masses.
It is possible for the straight line connecting (T1 , e1 ) and (T2 , e2 ) to intersect the function es (T ). If that occurs,
then any combination of the two air masses which happens to fall above the es curve will be supersaturated,
and condensation will occur. Since in any real encounter between two air masses, there will result air parcels
which consist of all possible proportions of the original air, the intersection of the straight line with es (T )
may be regarded as the criterion for the occurrence of condensation in an adiabatic, isobaric mixing process.
Examples of where adiabatic, isobaric mixing may commonly lead to condensation include the formation of
contrails from high-flying aircraft, the appearance of fog in your breath on a cold day, and the occurrence of
“steam fog” when cold air overruns warmer water.
51
Chapter 7
Atmospheric Aerosol
7.1
Definition
The term aerosol is used to described any suspended particulate matter in the air, either solid or liquid.
The primary criterion is that the particles be small enough to remain suspended for a “significant” length
of time. This means that the downward settling speed of the particle in still air is small — i.e., of order 1
cm per second or even much less. Furthermore, turbulent motions of the air outdoors are usually sufficient
to completely overwhelm this small settling speed, so that the particle can remain suspended for days or
longer, unless another mechanism acts to remove it.
Question: What balance of forces determines the settling speed (a.k.a. “terminal velocity”) of a suspended
particle in still air? Why do small particles fall slower than larger particles — Galileo’s experiment at the
Leaning Tower of Pisa notwithstanding?
Most aerosols are so small as to be invisible. However, visible dust, smoke, and haze are also made up of
aerosols which are both unusually large and particularly numerous. Suspended pollen is generally invisible,
but nevertheless constitutes a rather large form of aerosol, and one which is significant to hayfever sufferers.
Clouds and fog consist of small droplets of water which, technically speaking, also qualify as aerosols.
However, it is customary to treat aerosols and cloud condensate somewhat separately, though the occurrence
and characteristics of clouds turn out to depend quite strongly on the availability of certain types of aerosols,
as we shall see.
The diameters of aerosol particles may range from 10−4 µm to 10µm, where 1 µm ≡ 10−6 m, or 10−3 mm.
However, particle sizes at the extreme ends of this range are less prevalent because they tend to have short
atmospheric lifetimes.
Question: Explain the short lifetime of aerosols with sizes near 10µm. Estimate the atmospheric lifetime
of an aerosol with a diameter of 103 µm that is released at an altitude of 10 m from the surface.
Given the importance of aerosols in a number of atmospheric physical and chemical processes, it is useful to
develop a framework for characterizing aerosol populations.
Examples of important characteristics:
52
1. number concentration (i.e., count per unit volume of air)
2. distribution of sizes
3. chemical composition
4. physical behavior
7.2
Number concentration
The easiest parameter to define and understand is the number concentration of an aerosol population, which is
simply the total number of particles per unit volume of an air sample. The dimensions of such a concentration
are thus L−3 = “per volume”, since a count has no dimensions.
How might one go about determining the total number of aerosol in a sampel of air? Wallace and Hobbs (p.
144) describe the most common method, called an Aitken nucleus counter. It works by rapidly expanding
a volume of air in a chamber, which leads to strong adiabatic cooling and a consequent high degree of
supersaturation. This supersaturation causes water vapor to condense on virtually every aerosol particle
present, rendering them visible to optical counting devices.
Question: Why does water vapor condense only on aerosol particles and not elsewhere as well? This is a
question we will be looking at much more closely in the next few days!
Using Aitken nucleus counters, aerosol concentrations have been studied at a variety of locations on the
earth’s surface and at higher altitudes. Typical concentrations are found to be as follows
• ∼103 cm−3 over ocean
• ∼104 cm−3 over rural land areas
• ∼105 cm−3 at polluted urban sites
Note that the above values may fluctuate by an order of magnitude or more at any given location.
Question: What does the term “order of magnitude” really mean ?
7.3
Aerosol sizes
As noted above, the range of aerosol sizes is enormous, covering up to 5 orders of magnitude. Aerosols of
different sizes behave differently and are created and removed from the atmosphere by different mechanisms.
In any given sample, how can one characterize the size of the aerosols present? Unfortunately, no two
particles are likely to have exactly the same size. It is therefore impractical to specify the size of each particle
in the sample – you would need as many numbers as there are particles!
It is much more convenient to specify the total number of aerosols (per unit volume) falling in a specific
range of sizes.
Definition: For aerosols, it is convenient to use the diameter D as a measure of the size of the particle.
Strictly speaking, only spheres have a uniquely defined diameter, and many aerosols are far from spherical.
53
Unless otherwise specified, the diameter D refers to the volume-equivalent diameter; that is, to the diameter
of a sphere having the same volume.
Question: What types of aerosols are likely to actually be spherical, and why?
7.4
Size distributions
One way to characterize the sizes of aerosols in a air sample is to tabulate the number per unit volume falling
in a particular size interval, repeating the process for all possible intervals. A histogram or bar chart can
then be used to graphically depict the size distribution.
One simple and widely used breakdown by size involves 3 categories, defined as follows:
• “Aitken”:
D < 0.2µm
• “Large”:
0.2µm < D < 2µm
• “Giant”:
D > 2µm
Almost invariably, the smaller the size, the more particles you find. Aitken nuclei are vastly more numerous
than “large” or “giant” nuclei. Indeed, the reason the smallest category of aerosols are referred to as Aitken
nuclei is because these are the most numerous and therefore contribute the vast majority of aerosols detected
by an Aitken nucleus counter.
Despite being far less numerous, the largest particles may nevertheless contribute a substantial fraction of
the total aerosol mass in an air sample. Keep in mind that a single particle with a diameter of 1 µm has
approximately the same mass as a million particles of 0.01 µm diameter!
Mathematical and graphical descriptions of aerosol size distributions are discussed in Section 4.1.2 of Wallace
and Hobbs. We will not cover that section this semester.
7.5
Sources and Sinks
If the concentration of aerosols of a particular size stays more or less constant in a given air mass for a
prolonged period of time, it is NOT because no aerosols are being added to or removed from the air mass.
On the contrary, the typical lifetime of an individual aerosol particle in the atmosphere is only on the order
of minutes to days, depending on size and meteorological conditions. Average concentrations of aerosols
therefore reflect an approximate balance between sources and sinks; that is, processes that add aerosol to
the air and those that remove them.
Here are the major distinct sources and sinks for aerosols of various sizes:
Note that “diffusion/coagulation” is both a sink for Aitken aerosols and a source for Large aerosols, since
it represents the clumping together of several small, fast moving aerosols into a single larger aerosol, but
doesn’t actually remove any aerosol mass from the atmosphere.
Note also scavenging by precipitation is the single most effective sink for almost all aerosols (though least
effective for Large aerosols). It accounts for 80–90% of the removal of aerosol mass from the atmosphere.
54
Without precipitation to regularly cleanse the atmosphere, the air we breathe might be 5–10 times as dirty
as it is now!
Precipitation scavenging works in several ways. In-cloud scavenging occurs when Giant aerosols serve as
condensation nuclei for the formation of cloud droplets. Additionally, very small Aitken nuclei will diffuse to
and collide with the cloud droplets, so that impurities get concentrated in the droplets. If the cloud rains,
then the cloud droplets containing these impurities aggregate together and fall to the ground. Otherwise,
the droplets eventually evaporate again and the aerosol mass is returned to the atmosphere, albeit in the
form of larger particles. Below-cloud scavenging occurs when falling raindrops collide with Giant aerosols in
their path and carry them to the ground. Smaller aerosols are not effectively removed by this mechanism,
because they more easily follow the airflow around the falling drop and thus avoid collision.
Finally, note that while there are major sources for all three size categories, and efficient sinks for Aitken and
Giant nuclei, there are no particularly efficient sinks for Large nuclei. For this reason, Large nuclei tend to
have a considerably longer atmospheric lifetime (several days to a week, on average) than the other categories
(minutes to a day or so). This also implies that average concentrations of Large nuclei will be somewhat
higher than might otherwise be expected. This so-called accumulation mode is evident as a “hump” near
0.3µm in Fig. 4.3 of Wallace and Hobbs.
Note that section 4.1.4 of Wallace and Hobbs will not be covered in class, but you are encouraged to read it
anyway.
55
Table 7.1: Effective sources and sinks for aerosols in the 3 major size classes
Process
Source/Sink
Size categories affected
Combustion
Gas-to-Particle Conversion
Saltwater foam and spray
Wind blown dust, pollen
Diffusion/coagulation
Diffusion/coagulation
Diffusion near surface
Impaction on obstacles
Gravitational sedimentation
Precipitation scavenging
Source
Source
Source
Source
Source
Sink
Sink
Sink
Sink
Sink
Aitken
Aitken
Giant, Large
Giant
Large
Aitken
Aitken
Giant
Giant
Aitken, Giant
56
Chapter 8
Cloud Formation
8.1
Nucleation of Water Vapor Condensation
One of the most important roles of certain aerosols is as cloud condensation nuclei (CCN). We will look now
at (1) why CCN are necessary for the formation of cloud droplets, and (2) what factors determine whether
an aerosol particle can function as an efficient CCN.
We begin by considering the condensation of a pure water droplet in supersaturated vapor, without the
presence of any foreign substances and aerosols. Such a case is called homogeneous nucleation. Here’s how
it would work:
1. Random collisions of water vapor molecules create small clusters of molecules sticking together which
may be regarded as incipient (or embryonic) cloud droplets
2. Under the right conditions, continued condensation onto these clusters takes place, permitting the
droplet to grow into a typical cloud droplet of ∼10µm radius.
What are the “right conditions”? It turns out that simple supersaturation (i.e., actual vapor pressure greater
than the nominal saturation vapor pressure = relative humidy greater than 100%) is not enough, unless the
degree of supersaturation is extremely great. Why?
Up until now, we have always regarded the saturation vapor pressure es to be a function of temperature
only. This is actually correct to a high degree of accuracy as long as one is only concerned about saturation
with respect to water surfaces which are relatively flat.
In the case of cloud droplets, however, the water surface is far from flat — the radius of curvature may
be just a few µm. Under these conditions, the saturation vapor pressure with respect to the water surface
becomes a function of radius as well as temperature, and it can in fact be a great deal larger than for a flat
water surface.
Key concept: The conventional definition of saturation vapor pressure es (T ) used by meteorologists is with
respect to a plane (i.e., flat) water surface. For very small water droplets, a more general expression must
be used which accounts for the radius of curvature of the water surface. Of course, when the water surface
is flat (radius of curvature equals infinity), the more general formulation simplifies back to the conventional
form.
57
Wallace and Hobbs (p. 159) give one derivation of the more general expression. I will give a less rigorous,
but perhaps more intuitive derivation in the following. You should study both.
To review: For a plane (flat) surface of pure water , the saturation vapor pressure is given approximately by
L
1
1
−
es (T ) ≈ es,0 exp
,
(8.1)
R v T0
T
where L is the latent heat of vaporization of water and has dimensions of energy per unit mass. What we
will show in the following is that, in the case of a very small, spherical droplet, some of the energy required
for evaporation is actually supplied in part by a reduction in the surface area of the droplet. Although there
are a number of ways of looking at the energy balance involved in this reduction, the most intuitive way
perhaps is to consider the pressure-volume work done by the droplet as the surface area shrinks.
It turns out that liquids like water experience surface tension at their interface; that is, the surface behaves
something like an elastic rubber sheet that wants to contract and thus minimize its area. This surface
tension is why soap bubbles and small water droplets are spherical — a sphere has the minimum surface
area possible for a given volume.
The strength of the surface tension on a fluid is denoted σ and has dimensions of force per unit length. This
force can be directly measured with relatively simple equipment (an example was given in class). For pure
water at 0◦ C, σ ≈ 0.076 N/m. (Note that the units are equivalent to joules per square meter!)
Surface tension on a spherical droplet exerts a pressure on the water internal to the droplet. As shown in
class, the excess pressure ∆p (relative to the external atmospheric pressure) can be calculated based on a
balance of forces: the surface tension force, equal to the circumference times σ, tries to collapse the droplet
but is exactly balanced by the pressure times the cross-sectional area. Thus
2πrσ = πr2 ∆P
(8.2)
or
2σ
(8.3)
r
What we see is that the internal pressure can be very large when the radius r is very small! For a 1 µm-radius
droplet, ∆P = 150 kPa, which is about 1.5 times normal atmospheric pressure (the total internal pressure
is thus about 2.5 atmospheres!). If r is smaller, the pressure is even greater. [A similar (nearly) inverse
dependence of internal pressure on radius is the reason it is hard to first begin blowing up a small rubber
balloon!]
∆P =
What are the consequences of the internal pressure for evaporation from a small water droplet? When a
small amount of water evaporates from the droplet, the volume of the droplet decreases proportionally and
the surface area of the droplet thus decreases as well. The surface tension is thus doing positive work and is
therefore contributing a small amount of energy toward the amount L required for vaporization!
The surface tension work per unit mass of water evaporated from the droplet can be easily shown to be
σ
2σ
d
(4πr2 ) =
dm
ρw r
(8.4)
where m = 43 πρw r3 and ρw is the density of pure water.
Thus, the effective latent heat of vaporization is
L = L −
58
2σ
ρw r
(8.5)
If this adjusted value is substituted into the original equation for es (T ) and the result simplified, one obtains
a new formula:
2σ
es (T, r) = es (T ) · exp
(8.6)
Rv ρw T r
which expresses the saturation (or equilibrium) vapor pressure for a droplet of radius r by way of a correction
to the conventional formula for a plane surface. Note that when r becomes very large (thus approximating a
flat surface), the exponential factor goes to 1, so that we wind up with the conventional plane-surface value
again.
On the other hand, when r is very small, the exponential term can become large, implying that much larger
vapor pressures are required in order achieve equilibrium with a small droplet than is necessary to achieve
saturation relative to a plane water surface. In other words, a small enough droplet of water will evaporate
even in a saturated or supersaturate environment!
The implications of this for condensation of water vapor in pure air are very significant:
Since any newly condensed droplet of pure water has to start out almost infinitesimally small, it cannot
possibly survive, let alone grow to reach a larger size, unless the air is extremely supersaturated – that is, the
relative humidity is several hundred percent or more!
Since supersaturations in excess of 1–2% are almost never observed in the real atmosphere, it follows that
homogeneous nucleation of water condensation is virtually impossible. Instead, it is necessary to invoke
heterogeneous nucleation, which entails the condensation of water vapor onto pre-existing hygroscopic aerosol
particles composed of wettable or soluble materials.
Note: for future reference, we may define the saturation ratio S to be the ratio of the actual vapor pressure
to the saturation vapor pressure relative to a plane surface of pure water:
S=
e
es (T )
(8.7)
Expressed in percent, this is really the same as relative humidity. When S = 1, the relative humidity is
100% and the air is exactly saturated.
The degree of supersaturation is denoted s and is given by
s = (S − 1) × 100%.
(8.8)
So a supersaturation of 2% is the same as a relative humidity of 102%.
8.2
The solute effect
Raoult’s law (from freshman chemistry) states that the saturation vapor pressure of a liquid like water is
reduced in proportion to the amount of “stuff” dissolved in that water. In fact, the modified saturation vapor
pressure es is given by
n0
es =
es
(8.9)
n0 + n
where n0 is the number of moles of pure solvent (water, in our case) and n is the number of moles of solute
(e.g., the constituents of a soluble aerosol particle).
One result of the above law is that water vapor can condense onto salt particles or a concentrated salt
solution even when the air is unsaturated by the conventional definition. (This is what makes the salt in a
59
salt shaker clump up on a humid day. Also, the same basic principal applies to freezing points and is why
salt can be used to melt ice!)
In heterogeneous nucleation of condensation, a small aerosol particle composed of a soluble substance can
help water vapor to begin to condense out even when the air is not quite saturated. When this happens, n0
begins to increase while n remains constant.
Thus, only a small amount of water can condense onto the particle in a sub-saturated environment before
the salt solution is diluted to the point that es becomes equal to the actual vapor pressure. Then the particle
achieves a stable equilibrium, neither growing by condensation nor evaporating as long as the ambient water
vapor pressure remains constant.
An aerosol particle in the above state is referred to as a haze droplet. The droplets become larger under
conditions of high humidity, leading to a noticeable reduction in visibility without the actual occurrence of
clouds.
We may quantify the effects of Raoult’s law for a soluble aerosol particle by noting that
n=
im
Ms
(8.10)
where m is the mass of the dry aerosol particle, Ms is its molecular weight (units of kg per kilomole), and i
is a dissociation constant that accounts for the fact that many ionic molecules break up into 2 or more ions
when they are dissolved in water; e.g., NaCl → Na+ + Cl− .
The number of moles of water depends on the radius of the haze droplet and is given approximately by
n0 =
(4/3)πr3 ρw
Mw
(8.11)
where Mw is the molecular mass of water (about 18.0 kg/kmole). Note that we have assumed that the haze
droplet is dilute; i.e., the mass m of the solute can be neglected relative to the mass of the water, and the
density of the solution is the same as that of pure water ρw .
We may then write
1
es
n0
=
=
≈ 1 − n/n0
es
n0 + n
1 + n/n0
(8.12)
where we have again made use of the assumption that n n0 and thus n/n0 1. Thus
es
3iMw m
≈1−
es
4πρw Ms r3
(8.13)
The following table gives atomic weights for some common elements:
H
N
O
S
Cl
Na
1.01
14.0
16.0
32.1
35.5
23.0
One may then easily determine the molecular weights of compounds containing the above elements. For
example, ammonium sulfate has the formula (NH4 )2 SO4 , so the molecular weight is 132.2 kg/kmole; sodium
chloride is NaCl, molecular weight = 58.5 kg/kmole.
60
8.3
The Köhler Curve
Combining the above solute effect with the curvature effect derived earlier, we may write the following
expression for the equilibrium saturation ratio S of a droplet of water containing a small amount of dissolved
solute:
b
(8.14)
S = ea/r 1 − 3
r
where
a=
2σ
Rv ρw T
(8.15)
b=
3iMw m
4πρw Ms
(8.16)
and
By recognizing that a/r 1, one may make the final approximation that ea/r ≈ 1 + a/r, so that
S = (1 + a/r) · (1 − b/r3 ) ≈ (1 + a/r − b/r3 )
(8.17)
The coefficients a and b are defined in terms of a number of variables, some of which may be treated (at
least approximately) as constants for a droplet composed of water. Examples include σ, Rv , ρw , and Mw .
One may therefore substitute numerical values for each of these and combine all of the constants to yield:
0.33
[µm]
T
(8.18)
4.3 × 1015 im
[µm3 ]
Ms
(8.19)
a=
where T is the absolute temperature in K, and
b=
where m is supplied in units of kg.
Now that we have defined the terms, let’s discuss the implications of the formula
S =1+
b
a
− 3.
r
r
(8.20)
This formula gives the equilibrium saturation vapor pressure for a water droplet of radius r, within which
a fixed mass m of a solute having a molecular mass Ms and dissociation constant i is dissolved. That is to
say, it describes the environmental relative humidity (expressed as the saturation ratio) required to keep the
droplet from either growing or evaporation. It also describes the actual saturation ratio in the extremely
thin film of air exactly adjacent to the droplet.
The term a/r accounts for the so-called curvature effect; that is, the tendency of a very small droplet to
want to evaporate on account of the internal pressure created by surface tension. It is therefore a positive
term.
The term b/r3 contains the so-called solute effect; that is, the ability of a fairly concentrated salt solution to
draw moisture out of the air even when the air is not saturated with respect to pure water. It is a negative
term, because it represents a depression of the equilibrium saturation ratio.
The two terms compete, and one or the other may dominate, depending on the value of r. For small values
of r, the solute term dominates because of the dependence on 1/r3 . For larger values of r, the droplet
becomes so dilute (remember that the mass of the solute is assumed fixed) that it resembles pure water and
the curvature term takes over, leading to S > 1. For very large values of r both the solute and curvature
61
terms become negligible, and the equilibrium saturation ratio goes to 1, which is the expected value for a
flat surface of pure water.
A graph of the above formula is referred to as a Köhler curve. Examples are given in Wallace and Hobbs
and in Rogers and Yau. The precise shape of the curve is determined by the values of a and b, which in
turn depend on the environmental temperature and on the composition and mass of the aerosol particle in
question.
The way to use the Köhler curve is quite simple: compare the equilibrium saturation ratio that it predicts
with the actual environmental saturation ratio. If the former is larger, the droplet will begin to evaporate
and r will decrease. If the latter is larger, then the droplet will begin to grow by condensation and r will
increase. Of course, as r changes, so does the equilibrium saturation ratio, so one must follow the Köhler
curve in the correct direction to see how the degree of supersaturation or undersaturation with respect to the
droplet changes as the droplet grows or shrinks.
In a subsaturated environment (i.e., S < 1), there is always a radius for which the droplet is in equilibrium.
Because of the positive slope of the Köhler curve to the left of the peak, the droplet will always grow or
shrink as necessary to reach that radius, and it will then stay there in stable equilibrium as long as the
environmental saturation ratio doesn’t change. An aerosol droplet that is in stable equilibrium with the
environmental humidity is referred to as a haze droplet. The radius r will grow or shrink in response to
changes in the relative humidity, which is why humid summer days often have poor visibility due to haze.
In a saturated or slightly supersaturated environment, a haze droplet can still achieve equilibrium with the
environment, as long as the ambient supersaturation does not exceed the peak of the Kohler curve and as
long as the radius of the droplet is on the small side of the peak.
If that peak is exceeded, however, the droplet cannot establish equilibrium with the environment and will
therefore grow rapidly, quickly becoming a cloud droplet of perhaps 10 µm radius, rather than a haze droplet
of only ∼ 0.1 µm radius. This process is called activation.
The critical radius r∗ at which activation occurs for a given aerosol particle may be calculated by taken
d/dr of S(r) (above), setting the result equal to zero, and solving for r (you should go through this exercise
yourself).
3b
∗
(8.21)
r =
a
Substituting r∗ back into S(r) yields the critical saturation ratio S ∗ , which represents the height of the peak
of the Köhler curve:
4a3
∗
(8.22)
S =1+
27b
The implications of the above analysis can be summarized as follows:
• In general, the smaller the mass of the solute, the smaller b is and the larger the required supersaturation
for activation.
• Since a is always greater than zero and b is never infinite, S ∗ is always greater than 1, implying that
at least a slight degree of supersaturation is always required to activate any CCN. In other words, a
cloud never forms exactly at the Lifting Condensation Level but rather a short distance (perhaps only
a few meters) above the LCL.
• The total number of cloud droplets in a newly formed cloud will equal the number of activated CCN,
which in turn depends on both the ambient population of aerosols and on the maximum supersaturation
achieved just above cloud base.
62
• Once significant condensation begins (i.e., numerous CCN have been activated and converted to cloud
droplets), it is difficult to maintain significant supersaturation (except perhaps in very intense updrafts).
Consequently, CCN typically have only one chance to activate just after an air parcel reaches saturation;
after that, supersaturation decreases again, and any unactivated CCN remain unactivated.
• Although the total number of cloud droplets in a new cloud is determined by the aerosol population
and by the maximum supersaturation achieved (the latter will generally be greater in a more intense
updraft), the total cloud liquid water content (i.e., mass of water per unit volume cloud) is determined
by the difference between the original water vapor mixing ratio of the parcel of air prior to condensation
and the saturation mixing ratio at its current temperature and pressure. This implies that, all other
factors being equal, a cloud formed from an air mass containing fewer CCN will have fewer, but larger,
cloud droplets.
• It has been observed that marine air masses generally contain about an order of magnitude fewer
CCN ( 102 cm−3 ) than continental air masses ( 103 cm−3 ). This fact, combined with the previous
point, implies that marine clouds tend to have relatively low cloud droplet concentrations relative to
continental clouds, but that their droplet radii are larger on average. It turns out that this difference
makes it far easier for a marine cumulus cloud to produce rainfall, all other factors being equal. It also
affects the efficiency with which the clouds reflect sunlight – marine clouds tend to be less reflective.
63
Chapter 9
Precipitation Formation
9.1
Droplet Growth by Condensation
We now have an understanding of how aerosol particles may become activated to become cloud droplets.
All that is needed is for the environmental humidity to become sufficiently supersaturated so as to exceed
the critical saturation ratio of the particle in question. Once this happens, enough condensation takes place
to more or less eliminate the supersaturation condition, leading to the rapid growth of the CCN particle,
originally having a radius of order 0.1µm, to a droplet with radius ∼10µm. We thus have a ∼100-fold
increase in radius and a ∼ 106 -fold increase in volume. Amazingly, this transition can occur extremely
quickly, as demonstrated for example by the near-instantaneous appearance, under humid conditions, of
visible condensation in the low pressure zones of the flow of air over the wing of a jet airliner.
It would therefore be logical to assume that, under the right conditions, condensational growth of cloud
droplets could proceed to the point that the droplets are large enough to fall out of the cloud as rain. As
we shall see, it is not quite that simple.
The growth of a droplet by vapor condensation is a problem in molecular diffusion. Basically, diffusion
refers to the process by which the motion of molecules in a substance eliminates contrasts in temperature
or composition. A drop of dye added to a glass of water will initially settle to the bottom without much
mixing, but if the glass is left to stand long enough, the dye will eventually be redistributed by molecular
motions until it is uniformly mixed throughout the glass. The transfer of heat by conduction works in the
same way.
A general observation about diffusive processes is that the rate of flow of the material or property under
consideration is proportional to the gradient of that property; that is, the change in the property per unit
distance. In a metal rod of length L whose temperature is held constant at a value T1 at one end and T2
at the other end, heat will be conducted from the warm end to the cold end. Moreover, the rate of flow of
heat will be proportional to (T2 − T1 )/L, so if the temperature difference is doubled, the rate of heat flow
will double. If T1 = T2 , no heat will flow. This kind of behavior may be described mathematically with the
following equation:
dT
(9.1)
ΦH = −C
dx
where x is the position along the length of the rod. The heat flux ΦH in the x direction has dimensions of
energy per unit area per unit time. The constant of proportionality C is called the thermal conductivity,
since it determines how much heat will flow for a given temperature gradient. If C is small, the medium is
a good insulator; if C is large, it is a good conductor. The minus sign in the above equation reflects the fact
64
that heat always flows from warm to cold.
Although it may not be obvious without a detailed look at the thermodynamics of molecular motions in
fluids, a completely analogous equation describes the diffusion of any gaseous component of air, including
water vapor:
dρv
(9.2)
Φv = −D
dx
where ρv is the density of water vapor and D is the diffusion coefficient for water vapor in air. The mass
flux of water vapor Φv has dimensions of mass per unit area per unit time.
How do we use this equation to analyze the growth by condensation of a water droplet? To start with, we
have to recognize that the ρv at the exact surface of the water droplet (but still in the air, not the liquid)
is determined by the saturation vapor pressure of the droplet at the temperature in question. The vapor
density at some distance from the droplet, however, is determined by the environment. When the two are
not equal, then there is a gradient in water vapor density, and vapor will be transported by diffusion in the
direction toward lower vapor density. That is, if the environment is subsaturated relative to the droplet,
then the droplet will lose water by evaporation; if the environment is supersaturated, the droplet will gain
water and grow by condensation. If the environment is exactly saturated relative to the droplet, then it will
neither grow nor evaporate, because the gradient will be zero.
Mathematically, we formulate this problem in terms of a spherical droplet with radius r and mass M . We
define the coordinate x to be the distance from the center of the droplet. When x = r we are at the surface
of the droplet (we don’t care about the case x < r, since there is no vapor diffusion inside the liquid interior
of the droplet).
The rate of condensation of water onto the surface of the droplet is described by
or
dM
= [surface area of droplet] × [flux of vapor at surface]
dt
(9.3)
dρv
dM
= [4πr2 ] × [D
(r)]
dt
dx
(9.4)
Note that we have dropped the minus sign, because
negative x.
dM
dt
is positive when vapor is flowing in the direction of
The second assumption we need to invoke is that the environment surrounding the droplet is not changing
with time; that is, it is at steady state. In order for this to hold, the above equation must be valid not just
for x = r but at any distance x outside the droplet. Otherwise, there would be a net accumulation or loss of
water vapor mass with time at various locations in the surrounding air. Thus, we can generalize the above
to
dρv
dM
= 4πx2 D
(9.5)
dt
dx
This is a first-order Ordinary Differential Equation, which can be easily integrated in the usual way by
collecting the x terms on the left side and leaving dρv on the right:
dM 1
dx = 4πDdρv
dt x2
ρv (∞)
dM ∞ 1
dx
=
4πD
dρv
dt r x2
ρv (r)
∞
dM −1
= 4πD[ρv (∞) − ρv (r)]
dt
x r
dM 1
= 4πD[ρv (∞) − ρv (r)]
dt r
65
(9.6)
(9.7)
(9.8)
(9.9)
Since we have two different variables relating to the size of the droplet (M and r), it is desirable to eliminate
one of them by expressing it in terms of the other. The mass M is simply the density of liquid water times
the volume of the droplet:
4
M = πr3 ρl
(9.10)
3
so by the chain rule
d 4
dr
dM
= ( πr3 ρl ) = ρl (4πr2 )
(9.11)
dt
dt 3
dt
Substituting into the earlier expression, we have
ρl (4πr)
or
r
dr
= 4πD[ρv (∞) − ρv (r)]
dt
D
dr
= [ρv (∞) − ρv (r)]
dt
ρl
(9.12)
(9.13)
Now let’s turn our attention briefly to expressing the term in brackets on the right-hand side in more familiar
meteorological terms. Using the ideal gas law, we can express each of the ρs in terms of temperature and
vapor pressure:
e
es (T )
ρv (r) =
ρv (∞) =
(9.14)
Rv T
Rv T where T represents the temperature at the surface of the droplet, T is the temperature of the surrounding
environment, and e is the environmental vapor pressure.
By assuming T ≈ T and the environmental saturation ratio S ≡ e/es ≈ 1, we may write:
e − es
e
[ρv (∞) − ρv (r)] ≈
Rv T
es
e − es
e
[ρv (∞) − ρv (r)] ≈
es
Rv T
[ρv (∞) − ρv (r)] ≈ (S − 1) ρv (∞)
(9.15)
(9.16)
(9.17)
To wrap things up, we subsitute the above into our growth equation and get
or
dr
1 Dρv (∞)
≈
(S − 1)
dt
r
ρl
(9.18)
1
dr
≈ G(S − 1)
dt
r
(9.19)
where
G≡
Dρv (∞)
ρl
(9.20)
Under typical atmospheric conditions, the value of G falls in the range from 30 µm2 /s to 150 µm2 /s.
What can we already infer from our growth equation?
First, we see that growth is proportional to the degree of supersaturation s ≡ S − 1. In fact, if s < 0
(i.e., the RH is less than 100%), then dr/dt is negative, which implies evaporation. So this equation works
for both condensation and evaporation, provided only that S is not too far from 1 (because of our earlier
approximations).
Second, we see that the rate of growth is inversely proportional to r, all other factors being equal. This
implies that a larger drop will not grow as fast as a small drop.
66
We may integrate the above equation in the usual way to derive an expression for the droplet radius r as a
function of time:
t
r
r dr =
Gs dt,
(9.21)
0
r0
where r0 is the initial radius of the droplet at time t = 0. Then
1 2
r − r02 = Gst,
2
r2 = 2Gst + r02 ,
or finally,
r(t) =
2Gst + r02 .
(9.22)
(9.23)
(9.24)
In class, I will show you some sample calculations for G = 100µm2 /s, s = 0.01 (i.e., 1% supersaturation),
and r0 = 10 µm. The results indicate that even after a full hour of growth under these conditions, the
droplet has grown to a radius of only 85 µm. A typical raindrop has a radius of about 1 mm, or 1000 µm.
How long would it take to reach this size? Using the above conditions, we find that it takes 140 hours, which
is much longer than most individual clouds even exist!
Key Point: Condensation alone can’t explain the formation of raindrops!
So what can? In the next section, we will start by considering the simplest possible case: That of a cloud
whose temperature everywhere is warmer than freezing.
9.2
Warm vs Cold Clouds
In order to prepare for a discussion microphysical processes that lead to precipitation, it is useful to distinguish between clouds in which ice can potentially play a role and those in which it cannot. Clearly, any cloud
which is warmer than freezing everywhere cannot contain any ice! Such a cloud is called a warm cloud. Any
cloud which is not a warm cloud is a cold cloud. A cold cloud is therefore any cloud of which any fraction,
however small, is colder than freezing.
9.3
Growth by Collision/Coalescence
We have already seen that condensation alone is not sufficient to explain the formation of raindrops. Yet
clouds do rain, sometimes very efficiently.
In a warm cloud, there is only one mechanism that that explain the formation of rain drops: the physical merging of lots of small cloud droplets into a few big raindrops. This mechanism is called collision/coalescence, because it involves collisions between droplets followed by their possible coalescence into
a single larger droplet. In the rest of this section, we will derive an approximate growth equation for drops
experiencing collision/coalescence and see how it compares with the formula for condensational growth.
Consider a volume of cloud containing an amount of cloud water wl . Assume that most of this cloud water is
present in the form of small cloud droplets with radius r1 . (Question: what is then the number concentration
N of those droplets, if N has dimensions of length−3 ?). Now let’s imagine introducing a slightly larger droplet
with radius r2 into the mix and follow its progress.
To begin with, we can assume that if all the small droplets have the approximately the same radius, then
they will also have about the same terminal fallspeed vs (r1 ), which means that they will not tend to collide
67
with one another, since one droplet would be unable to overtake another. But the larger droplet has a higher
terminal fallspeed vs (r2 ), so it tends to catch up with the smaller droplets in its path. In fact, the relative
fallspeed is just vs (r2 ) − vs (r1 ).
For the purposes of this notes, we are going to simplify things (and thus bypass a lot of details in Wallace
and Hobbs) by immediately assuming that r1 is much smaller than r2 . Among other things, this means that
we can also assume that vs (r2 ) − vs (r1 ) ≈ vs (r2 ).
Now if we consider the situation in a moving frame of reference (i.e., coordinate system) which follows the
small droplets, We find that the larger droplet “sweeps out” a cylindrical volume per unit time which is the
product of the cross-sectional area of the large droplet and the relative fallspeed which, as we said above,
is approximately vs (r2 ). Also, since we are assuming that r1 is negligible compared with r2 , we can ignore
collisions involving small droplets whose centers fall slightly outside the cylinder swept out by the large
droplet.
Since r1 is small and unimportant, we will simplify our notation by simply referring to r2 as r, and by not
even mentioning r1 anymore.
If every small droplet in the path of the large droplet were struck and absorbed by the latter (later we will
modify this assumption), then we could then write a simple growth equation as follows:
dM
= [volume swept out per unit time] × [cloud water content]
dt
(9.25)
dM
= [cross − sectional area × fallspeed] × [cloud water content]
(9.26)
dt
dM
= πr2 vs (r)wl
(9.27)
dt
Just as we did earlier, we can express the mass of the droplet M in terms of its radius r and thus eliminate
M as a variable:
d 4
dM
dr
= ( πr3 ρl ) = ρl (4πr2 )
(9.28)
dt
dt 3
dt
So
dr
= πr2 vs (r)wl
ρl (4πr2 )
(9.29)
dt
or
vs (r)wl
dr
=
(9.30)
dt
4ρl
All that is needed to complete the picture is an expression for vs (r).
Excursus: fall speeds of droplets
As part of an earlier homework assignment, you derived an expression for the terminal fall speed of spherical
aerosol particles. This expression was proportional to the square of the diameter, and the constant of
proportionality was constructed from things like the viscosity, particle density, acceleration due to gravity,
and so on. The proportionality of fall speed to D2 holds for droplets of water as well, up to radii of about 30
µm. At larger sizes, the aerodynamics of the flow around a droplet change – for example, it becomes more
turbulent and/or the droplet begins to flatten, and so other approximate formulas for fall speed must be
developed to account for these factors. Given here are simple formulas that summarize extensive laboratory
investigations into droplet fall speeds:

<
 k1 r2
30 µm;
if r ∼
<
<
vs (r) ≈ k2 r
(9.31)
if 40 µm ∼
r∼
0.6 mm;

1/2
>
if r ∼ 0.6 mm.
k3 r
68
where
k1 ≈ 1.2 × 106 cm−1 s−1 ,
3
k2 ≈ 8 × 10 s
−1
3
k3 ≈ 2.01 × 10 cm
(9.32)
,
(9.33)
1/2 −1
s
.
(9.34)
(Note that the dimensions of the constants of proportionality k are always such that, when multiplied with
the appropriate power of r, they yield dimensions of velocity.)
In the current problem we are considering, which is to derive an expression for the growth of a droplet by
collision coalescence, we will assume for convenience that the droplet’s radius falls somewhere in the broad
middle range between 40µm and 0.6 mm. The lower end of the range corresponds to a largish cloud droplet;
the upper end is modest raindrop.
Substituting k2 r for vs (r) in our growth equation, we get
dr
k2 wl
=
r
dt
4ρl
(9.35)
Now we need to return to our original assumptions concerning the collision/coalescence process. Recall that
in deriving the above, we assumed two things: (1) every small droplet in the path of the larger droplet will
be involved in a collision; (2) every time two droplets collide, they will coalesce (merge). Neither assumption
is necessarily valid.
To begin with, it is possible for small droplets in the path of the larger drop to “dodge” a collision, if the air
flow around the larger drop sweeps the smaller drop around it. This is more likely for droplets already near
the edge of the cylindrical volume swept out by the falling big drop than it is for droplets near the center
of the volume. For this reason, it is convenient to multiply the right hand side of the growth equation by a
factor which we will call the collision efficiency E, which is a function of both r1 and r2 :
Key Point: The Collision Efficiency E describes the actual likelihood of a collision between between two
droplets, relative to what one would expect from geometric considerations alone (i.e., ignoring aerodynamic
effects).
Normally,
<
1
(9.36)
0<E∼
However, when two drops are nearly the same size, it is actually possible for E to be slightly greater than 1!
This is because drops not directly in the path of a falling drop, but nearby and falling at nearly the same
speed, can actually get “sucked” into the wake of the larger drop, catch up, and collide!
On the other hand, when r1 is much smaller than r2 , E may also be small because the small drops get easily
pushed out of the way of the larger drops.
OK, so we now have a way to correct for aerodynamic effects on the likelihood of collisions – all we have
to do is multiply the right hand side of our growth equation by E and then supply an appropriate value.
There is now one other wrinkle we must consider: not all droplet that collide actually coalescence. That is,
it is possible for two droplets to collide but then bounce off of each other. This possibility is accounted for
by adding a coalescence efficiency E .
Key Point: The Coalescence Efficiency E describes the probability that two droplets will coalesce (or
merge), given that they have actually collided, where
0 < E < 1.
(9.37)
Unlike the case for E, it is impossible for E to be greater than 1, because droplets cannot merge unless they
first collide.
69
Now we can combine the two separate efficiency factors to define a single collection efficiency Ec as follows:
Ec ≡ E · E (9.38)
Key Point: The collection efficiency describes the fraction of cloud droplets in the path of a larger droplet
that are actually collected by that droplet.
It turns out that E is usually close enough to 1 that we can ignore it for the purposes of this course and
write:
(9.39)
Ec ≡ E · E ≈ E
With the above definitions and approximations, we can complete our approximate growth equation for
collision/coalescence:
dr
Ek2 wl
=
r
(9.40)
dt
4ρl
Integrating gives
r
r0
1
dr =
r
t
0
Ek2 wl
dt
4ρl
Ek2 wl
t
4ρl
Ek2 wl
r(t) = r0 exp
t
4ρl
log(r/r0 ) =
(9.41)
(9.42)
(9.43)
Key Point: The above equation shows that in the case of growth by collision/coalescence (unlike that case
for growth by condensation) droplet growth is exponential. This indicates that the larger the droplets are,
the more rapidly they grow. This makes collision/coalescence a good prospect for producing rainfall fairly
quickly and efficiently from a warm cloud with the right characteristics.
What are those characteristics? From the expression for r(t), we see that growth by collision/coalescence
is most rapid when the cloud water density wl is large, and when the collection efficiency E is close to its
maximum value of approximately 1. Of course, the constants k2 and ρl don’t vary significantly from cloud
to cloud.
Let’s take typical values for the various parameters in the growth equation and see how large a droplet can
grow in, say, 1/2 hour: We’ll assume E ≈ 1, wl = 1g m−3 , ρl = 1000 kg m−3 , and k2 = 8 × 103 s−1 .
Substituting, we find that
(9.44)
r(t) = r0 exp 2 × 10−3 (sec−1 )t
When t = 3600 sec (i.e., 1/2 hour), we get
r = 36.6r0 .
(9.45)
So if the droplet started out with a radius r0 = 100µm, then after 1/2 hour, it has grown to 3.7 mm. This
is a pretty substantial raindrop.
Note that our assumption of wl = 1 g m−3 corresponds to a fairly “wet” cloud, which in turn is most likely
in a cloud with a fairly strong updraft, since it is adiabatic cooling that tends to create liquid condensate.
A strong updraft also tends to keep a growing raindrop within the cloud where it can continue to grow;
otherwise it might fall out of the cloud base while it is still only drizzle-size or smaller. For the same reason,
a deep cloud favors precipitation growth because a droplet starting near cloud top has a longer period of
time to grow before it falls out of the cloud base.
<
1/2 hour in a warm cloud,
To summarize: Collision/coalescence can lead to production of rainfall in ∼
provided only that
70
1. wl is reasonably large. This is most likely when a strong updraft is present.
2. The cloud is sufficiently deep and/or a strong updraft is present, so that the growing drop does not
exit the cloud too soon.
3. A mix of larger and smaller droplets is present in the cloud, so that the collision/coalescence process
can get started.
The last of these conditions ties in with the drop size spectrum of a cloud, which in turn depends on
the environment within which the cloud formed. We already saw that marine clouds tend to have a smaller
number of drops, owing to the lower concentrations of CCN, but that these drops compensate by being larger
on average. This explains why marine warm clouds usually precipitate much more easily than continental
warm clouds, all other factors being equal.
9.4
Precipitation Processes in Cold Clouds
9.5
Formation of ice
Recall that a cold cloud is defined as any cloud which is at least partially colder than 0◦ C. The occurrence of
sub-freezing temperatures in a cloud admits the possibility of processes involving ice particles. Just because
a cloud is colder than freezing however does not necessarily mean that cloud condensate is all in the form of
ice. On the contrary, the simple act of cooling cloud water to below freezing does not normally cause it to
freeze; rather, the usual response is for the liquid water to become supercooled – i.e., to stay in liquid form
despite being colder than freezing.
The reason for the common failure of cloud droplets to freeze when they fall below the freezing point is similar
to the reason why droplets of water do not normally condense out of even a supersaturated atmosphere unless
CCN are present. That is, it is usually necessary for a foreign particle, called an ice nucleus (IN) to trigger
ice formation. Large volumes of water almost always freeze right at the freezing point, because it only takes
one IN somewhere in the volume to trigger the freezing of the entire volume. Cloud droplets are so small,
however, that the odds of an IN being present within a given droplet (or coming into contact with that
droplet) are very small.
The only exception to the above rule occurs at around −40◦ C: At this temperature, all liquid water freezes
spontaneously even without the help of IN. In light of this observation, we can identify two mechanisms for
the initiation of freezing:
• Homogeneous nucleation: The spontaneous formation of ice in supercooled water without the help
of foreign materials. Generally occurs at temperatures near −40◦ C.
• Heterogeneous nucleation: The initiation of ice formation by a foreign substance, usually an insoluble substance whose crystal structure at the molecular level mimics that of ice.
Different substances have a greater or lesser ability to nucleate ice formation. A measure of this ability is the
activation (or threshold) temperature. Above this temperature, the substance will not nucleate ice formation;
at colder temperatures it will. A very good ice nucleating material will have a threshold temperature just a
few degrees below freezing; a poor ice nucleating material will require much colder temperatures. As noted
above, however, water will always freeze spontaneously when it reaches about −40◦ C, after which point the
presence of IN is irrelevant.
71
Table 9.1 in Rogers and Yau gives examples of ice nucleating substances and their threshold temperatures.
Since we have seen that freezing of condensate is not automatic in a cold cloud, we can proceed to divide
cold clouds into three categories:
• Supercooled: A cold cloud in which no ice is present – all condensate which is colder than freezing
exists in the form of supercooled liquid water.
• Glaciated: A cloud which contains only ice, no supercooled liquid water.
• Mixed: A cloud in which both ice and supercooled liquid water coexist.
Clouds which are colder than −40◦ Care always glaciated. These include cirrus clouds, which form very high
in the troposphere. It is the fact that they are glaciated that gives them their distinctive fuzzy or fibrous
appearance. Also, growing cumulus towers become known as cumulonimbus when their tops glaciate at
altitudes above the −40◦ Cisotherm.
At temperatures warmer than −40◦ C, clouds may be supercooled, mixed, or glaciated. The warmer the
cloud, the more likely it is to be supercooled, since there are fewer IN whose threshold temperatures will
have been reached. Also, the older a cold cloud is, the more likely that ice will have begun to appear. As we
shall see, once a cloud reaches the “mixed” stage, it will quickly evolve toward an ever more glaciated state,
as the presence of ice particles (which can be viewed as IN with a threshold temperature of 0◦ C) causes the
rapid conversion of supercooled liquid water to ice.
Figure 5.11 in Wallace and Hobbs shows how the fraction of clouds found by instrument aircraft to contain
at least some ice varies with cloud top temperature. In this figure about half the clouds with tops colder
than −10◦ Cwere found to be entirely supercooled; by −20◦ C, the vast majority were found to contain at
least some ice.
9.6
Relative predominance of IN
We have already noted the importance of IN in the creating of ice particles in many cold clouds. It is very
important to also note that IN are far rarer than CCN. Typical concentrations of CCN are on the order of
105 per liter, whereas typical concentrations of active IN are often closer to 1 per liter (depending of course
on the temperature). This implies that for every 105 cloud droplets in a newly formed cold cloud, there
may be only one ice particle. This disparity between ice particle and supercooled droplet concentrations has
profound implications for the formation of precipitation, owing to the large difference between the saturation
vapor pressure with respect to liquid water and that with respect to ice at the same temperature.
9.7
The Bergeron/Wegener/Findeisen Process
Recall the integrated form for the Clausius-Clapeyron equation giving the saturation vapor pressure with
respect to pure water:
L
1
1
es (T ) ≈ es,0 exp
−
(9.46)
R v T0
T
where L is the latent heat of vaporization (≈ 2.5 × 106 joules/kg) for liquid water, and if we choose T0 = 273
then es,0 = 611 Pa.
72
What if we want to know the equilibrium (saturation) vapor pressure with respect to ice? Since the ClausiusClapeyron equation is just as valid for solid to vapor phase changes as for liquid to vapor changes, all we
have to do is substitute the latent heat of sublimation Ls for L in the above equation, and (if necessary)
supply a different laboratory-measured value for es,0 .
Energy conservation requires that the latent heat of sublimation be given by
Ls = Lf + L
(9.47)
where Lf is the latent heat of freezing (melting) and equals 3.3 × 105 joules/kg. Thus, Ls > L and is about
equal to 2.83 × 106 joules/kg.
Making this replacement, we have an expression for the saturation vapor pressure with respect to ice:
Ls
1
1
−
ei (T ) ≈ ei,0 exp
.
(9.48)
R v T0
T
Moroever, it turns out that when T0 is chosen to be 273 K, then ei,0 = es,0 = 611 Pa.
Following the approach used on an earlier homework assignment, we can write the following convenient
formulas for ei (T ) and es (T ):
ei (T ) = Ai e−Bi /T
A = 3.41 × 1012 Pa, B = 6130 K
(9.49)
es (T ) = Ae−B/T
A = 2.53 × 1011 Pa, B = 5420 K
(9.50)
It is easy to verify that for temperatures T below freezing, ei (T ) < es (T ) (at T = 273, the two become
equal). This leads to the following very important result:
At subfreezing temperatures, an environment which is saturated with respect to liquid water will be supersaturated with respect to ice. Conversely, an environment which is saturated with respect to ice will be
subsaturated with respect to supercooled liquid water at the same temperature.
This implies that a mixed cloud (one containing both supercooled liquid and ice) will not be microphysically
stable. Instead, the ice will add water vapor from its surroundings to its own mass (i.e., it will grow by vapor
deposition) while the liquid tries to replenish it by evaporating. The net result will be a steady transfer
of water mass from liquid to solid. If allowed to continue to completion, all supercooled liquid water will
evaporate and be redeposited on the ice particles which are present.
Now it becomes apparent why the imbalance between the number of IN and CCN is so important: if there
are N cloud droplets for every one ice particle, then the above process will result in the conversion of lots
of small supercooled cloud droplets to a single ice particle which is N times as massive. As noted earlier, a
typical value for N is on the order of 105 !
The rate at which an ice particle grows by vapor deposition is governed by the same physical reasoning that
we used earlier to investigate the growth by condensation of a liquid water drop. The main difference is
that we can’t count on an ice particle to have a spherical shape. Therefore we write the following modified
equation:
C
dM
= D [ρv (∞) − ρv,c ]
(9.51)
dt
50
where ρv (∞) is the vapor density of the environment and ρv,c is the saturate vapor density with respect to
the particle in question. The factor C/50 may be regarded as a “size and shape factor” that reflects the
geometry of the growing particle. In fact, C turns out to be the same as the electrical capacitance of a
conducting having the same size and shape as our hydrometeor particle. In the case of a spherical raindrop,
C = 4πr50 , which is consistent with our earlier derivation.
73
We can simplify the previous equation to
C
dM
= Gi si
dt
50
(9.52)
Gi = Dρv (∞),
(9.53)
by defining
where si is the environmental supersaturation with respect to ice. An example of how Gi si varies with
temperature in a supercooled cloud (assumed saturated with respect to liquid water) is given in Fig. 4.31
of W&H. We see that it is maximum near −15◦ C, which implies that snow formation is favored in mixed
clouds having this temperature.
In the case of a growing liquid water droplet, we earlier concluded that condensation is too slow to produce
raindrops. This was for supersaturations on the order of 2% or so. However, for ice particles suspended in a
supercooled liquid water cloud, the effective supersaturations can be much higher, owing to the potentially
large difference between ei and es .
In summary, the appearance of a few ice particles in an otherwise supercooled cloud often leads to the
efficient formation of precipitation in the form of snow. In fact, much of the precipitation that reaches the
surface of the earth in the middle latitudes originates in this way. The same cloud may not be capable of
producing precipitation by way of collision/coalescence alone, though the latter process may help intensify
the precipitation as it falls through lower layers of the cloud.
The production of precipitation due to the above described imbalance between the saturation vapor pressures
of ice and water is known as the Bergeron process (or, in Wallace and Hobbs, the Bergeron/Wegener/Findeisen
process). The direct result of the deposition (or sublimation) of water vapor onto ice particles is snow crystals,
which have a characteristic six-sided structure. Snow crystals may take many forms, of which the broadest
categories include hexagonal plates, columns, needles, and dendrites. The last of these refers to the fairly
complex six-sided fern-like structures that you see depicted on Christmas cards, etc. Which of these shapes
(or habits) develops in any given case depends on the environmental conditions of temperature and humidity.
9.8
Riming/accretion - formation of graupel
Once precipitation-size ice particles (snow crystals) have started being produced in a mixed cloud by the
Bergeron process, additional possibilities for hydrometeor growth come into play. The most important of
these occurs when fairly high levels of supercooled liquid water are maintained in the cloud, despite the
presence of ice. This is most likely when there is an updraft creating a continual supply of new liquid water
condensate to replace what is converted to ice.
An ice particle falling through a “wet” cold cloud will collide with supercooled cloud droplets, exactly as
occurs in a warm clouds during collision/coalescence. The difference is that the liquid water now freezes on
contact with the falling ice particle, so that the the result is a growing ice particle rather than a rain drop.
Because the growth occurs by riming, the result of this growth process is not a snow crystal, but rather a
graupel particle which usually has the appearance of an irregular, opaque white pellet of rime ice. Common
synonyms for graupel include soft hail, snow pellets, and tapioca snow.
The equation governing the growth of a graupel particle is exactly the same as that for collision/coalescence,
except for the difference density ρg of graupel as compared with pure liquid water, and except for possible
differences in the geometry of the growing particle. The latter differences are easily accommodated by always
starting with the mass-growth form of the equation, which you should recall as:
dM
= [collection efficiency] × [cross-sectional area] × [relative fallspeed] × [cloud water content]
dt
74
(9.54)
This form is valid for all types of growth by collision/coalescence and riming/accretion, provided only that
you substitute the right expressions for each term. And by starting with an appropriate relationship between
M and D or r, one can always eliminate M as a variable and instead write the final growth equation in
terms of one of the latter two dimensions, if desired.
9.9
Aggregation of snow crystals - formation of snowflakes
If a cloud produces significant concentrations of snow crystals via the Bergeron process, there is the possible
that some of these will clump,or aggregate together to form snowflakes. The tendency for this to occur
depends strongly on the shapes of the individual snow crystals and on the temperature. Complicated
shapes like dendrites are more likely to get entangled with one another rather than bouncing off each other;
temperatures close to freezing make ice particles stickier so that they are also more likely to aggregate. The
growth equation for aggregation of snow crystals is identical to that given in the previous section, except that
the density of snow crystals wi (mass of ice per unit volume of air) should be substituted for the supercooled
cloud liquid water content.
9.10
Summary of precipitation processes in cold clouds
We have now identified several mechanisms for the production of precipitation in cold clouds:
vapor deposition → snow crystals
(9.55)
accretion/riming → graupel
(9.56)
aggregation of snow crystals → snow flakes
(9.57)
All of the products of cold cloud precipitation processes are initially frozen; however, if the surface temperature is much warmer than freezing, then the frozen particles melt before reaching the surface. Additionally,
precipitation produced in cold clouds which subsequently falls into warmer clouds at lower altitudes may
continue to grow by collision/coalescence.
Key point: The majority of rainfall reaching the surface in the United States originates as frozen precipitation produced at high altitudes in cold clouds.
Note that different conditions in cold clouds favor different kinds of growth. In order for identifiable snow
crystals to emerge from a cloud, there can’t be too much supercooled cloud liquid water present, otherwise
these particles would quickly become rimed and emerge as graupel instead. In order for aggregated snow
flakes to emerge, there must be a comparatively high concentration of snow crystals, and conditions must
favor their clumping together.
9.11
Relevance to cloud seeding
Sometimes a supercooled cloud may form in which there are not enough IN present to trigger the formation
of any ice particles. This is most likely when the cloud top is only a few degrees below freezing, since fewer IN
are active at warmer temperatures. Under such conditions, the introduction of artificial IN with fairly warm
threshold temperatures might be enough to cause the cloud to precipitate via the Bergeron process. Silver
iodide (AgI) is commonly used as the seeding material cause of its warm threshold temperature (−4◦ C).
75
Note that seeding a cloud that already contains a significant concentration of ice particles is unlikely to
enhance precipitation. On the contrary, a cloud which becomes too glaciated will not produce ice particles
which are large enough to precipitate, because the available moisture is being shared by a larger number of
particles.
Another form of seeding relies on the fact that the conversion of supercooled liquid water to ice releases
latent heat of fusion. If enough water is induced to freeze by seeding, then the warmth released may actually
enhance the vertical development of a convective cloud, thus improving the conditions for the formation of
precipitation.
Note: Despite the simple theory behind it, and despite the potential value of a technique that could increase
rainfall in certain areas, it has been very difficult to prove to everyone’s satisfaction that cloud seeding (in
the form described above) really works in the real world! Weather modification through cloud seeding and
other means remains a very controversial subject.
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Chapter 10
Precipitation at the Surface
Regardless of the process by which it is produced, most precipitation in the world reaches the surface in the
form of rain, except of course when the surface temperature is near or below 0◦ C.
10.1
Precipitation intensity
Most commonly, the characteristic of precipitation that is directly measured is its accumulation over a
specified period of time, usually expressed as
depth of water =
mass of water collected
volume of liquid water collected
=
collection area
ρl × collection area
(10.1)
Even in the case of snowfall, the equivalent water depth– i.e., the depth of the liquid water obtained by melting
the snowfall – is usually of greatest interest for both hydrology (e.g., runoff estimates) and atmospheric
energetics (latent heat release).
Unfortunately, there is no one-to-one relationship between snow accumulation and equivalent water depth –
the ratio can vary from about 1:5 to 1:20, depending on how loose or fluffy the snow is. The ratio also tends
to decrease as a snow layer ages and becomes more densely packed with time.
Although it is usually accumulation over some period of time that is directly measured by rain gauges, the
intensity or rate of precipitation is also of considerable interest for many purposes. Formally, the average
rainfall intensity R̄ during a period of time ∆t is given by
R̄ =
accumulation during time interval ∆t
∆t
(10.2)
The instantaneous precipitation intensity R is estimated by allowing the measurement interval ∆t to approach
zero, or at least become small enough so that the accumulation rate doesn’t vary during the interval.
Conventional units for R or R̄ are either inches/hour or mm/hour. Note that a millimeter depth of precipitation is equivalent to one liter of water per square meter.
Note that the above definition of “heavy” rain does not do justice to the range of rainfall intensities that
are observed! It is not at all uncommon to observed rain rates here in Lafayette in excess of 25 mm/hour
77
(1 inch per hour). In fact, as much as 150 mm/hour (6 inches per hour) is commonly observed in intense
thunderstorms and tropical downpours.
Of course, the maximum average rainfall rate observed during a period of time depends in part on the length
of that period. The following table gives a sampling of world record accumulations over various observing
periods:
78
Table 10.1: Standard definitions of rain intensity used in meteorological reports in the U.S.
SA Symbol
Name
Intensity Range
R−
R
R+
Light rain
Moderate rain
Heavy rain
R < 0.1 inch/hour = 2.5 mm/hour
0.1 < R < 0.3 inch/hour
R > 0.3 inch/hour = 7.5 mm/hour
Table 10.2: Some world record rainfall accumulations
Record Rainfall
Observing period
R̄
Location
20.5 cm (8.1”)
135 cm (53”)
188 cm (74”)
930 cm (366”)
2647 cm (1042”)
20 min
12 hr
24 hr
1 month
12 months
615 mm/hr
112 mm/hr
74 mm/hr
13 mm/hr
3.0 mm/hr
Romania
Reunion Island
Reunion Island
Cherrapunji, India
Cherrapunji, India
79