1
Chapter 4 Thermodynamics of Moist Air
4.1 Properties of Water Substance
Water is by far the most important of the variable components of the Earth’s atmosphere, for
a number of reasons:
1
It can comprise up to several per cent of the volume (more than A
rgon).
2
It can exist in the atmosphere in all three phases (hence the generic term “water
substance”).
3
It has large latent and specific heats.
4
It is a major greenhouse gas.
5
The hydrological cycle is vital for life.
At temperatures below 0
C, water can exis
t in the atmosphere as both ice crystals and as
supercooled liquid droplets, as well as a vapour.
It takes energy in the form of latent heat to convert a substance from one phase to another (at
constant
T
). Latent heat is usually slightly temperature depe
ndent. In the case of water, at
0
C, we have
L
il
= 334 kJ/kg
L
iv
= 2834 kJ/kg
L
lv
= 2500 kJ/kg
Note that
L
iv
= L
il
+ L
lv
(of course).
Equation of State for Water Vapour
Because of its importance, we give water vapour pressure its o
wn symbol,
e
. Water vapour
obeys the ideal gas equation, but with its own gas constant (due to its molecular weight).
Thus we have
e =
v
R
v
T
or
e
v
=
R
v
T
Now, for water,
M
v
= 18.016
so that
R
v
=
R*/M
v
= 8314.3/18.016
= 461.5
J/kg/K
(4.1)
Because of this difference in molecular weight, the following symbol is useful:
=
M
v
/
M
d
=
R
d
/
R
v
= 0.622
(4.2)
2
Virtual Temperature
The presence of water vapour in the atmosphere should lead to a change in the
ideal gas
equation, with a modified gas constant which depends on
e
. Instead, we define the virtual
temperature of a moist air sample to be the temperature at which dry air at the same pressure
would have the same density as the moist air. Write the equat
ion of state for moist air as
p
m
=
R
m
T
(2.3)
where
R
m
=
R*/M
m
and
M
m
is the molecular weight of the moist air sample. For a mixture of gases
where
r
=
m
v
/
m
d
(in kg/kg)
is known as the mixing ratio (see section 4.3).
Since
R
v
=
R
d
/0.622 = 1.61
R
d
(1 + 1.61
r
)(1
–
r
+
r
2
)
R
d
= (1 + 0.61
r
–
0.61
r
2
)
R
d
(1 + 0.61
r
)
R
d
Hence
p
m
= (1 + 0.61
r
)
R
d
T
We now choose to rewrite this in the form
p
m
R
d
T
v
where
T
v
= (1 + 0.61
r
)
T
(2.4)
is the virtual temperature. It differs by (at most) a degree or two from the actual temperature,
but it allows us to retain the dry
air gas constant in all equations, and computer programs.
4.2 Saturation
At any temperature below the boiling point, water vapour can attain an equilibrium with
respect to the liquid or solid phase. Technically, this equilibrium is defined with respect
to a
flat
surface of
pure
liquid (or solid). The maximum vapour pressure which can be maintained
with respect to such a surface is the saturation vapour pressure,
e
s
, which varies strongly with
respect to temperature. In general, of course, the actual vapo
ur pressure at a temperature,
T
,
will be less than the saturation vapour pressure,
3
i.e.
e
(
T
)
e
s
(
T
).
Exactly how
e
s
varies with
T
is given be a general result known as the Clausius

Clapeyron
equation, which says
(2.5)
He
re, phase 2 is the vapour, while phase 1 is liquid or solid. In both these cases
2
1
Thus
and hence
so that
ln
e
s
=
L
12
/
R
v
T
+ constant
This constant may now be found by taking the value
of
e
s
at a ‘standard’ temperature, usually
0
C (this is equivalent to integrating from
T
= 0
C to an arbitrary value of
T
). From tables we
find that this value is 6.11 hPa, so finally we obtain
(2.6)
At temperatures belo
w 0
C, water can exist as both solid crystals and liquid droplets, as well
as vapour. The major reason that the droplets do not automatically freeze is that ice has a
definite crystal structure, while a liquid does not. Formation of ice crystals requires f
reezing
nuclei with the correct sized crystal structure.
The saturation vapour pressure over ice is given by the Clausius

Clapeyron equation in
exactly the same form as for the liquid case, except that we must use the appropriate latent
heat value. Since
L
iv
>
L
lv
, we may easily show that for temperatures below 0
C, the
saturation vapour pressure over ice is lower than that over liquid water. The consequences of
this in clouds will be explored in Chapter 6.
Temperatures in the atmosphere range from around 3
0
C in the tropics near the ground, to
around

80
C at the polar tropopause in winter. In the stratosphere, temperatures rise again.
The Clausius

Clapeyron equation shows that, over this range, the saturation vapour pressure
of water varies significantly,
from as high as 40 hPa (or 4%) to less than 0.1 hPa. As a
consequence, the water vapour distribution in the atmosphere is far from uniform, being
strongly concentrated in the lower altitudes and latitudes. This is despite the fact that storms
are quite com
mon at high latitudes. Clearly, the amount of rain in the tropics (e.g. the
monsoons) is substantial! One consequence of global warming is that there will be more
water vapour in the atmosphere, and hence more rain.
4
Boiling Point
The definition of boilin
g point is that temperature at which
e
s
=
P
(i.e. ambient pressure). For
a pressure of 1013.25 hPa (standard sea level pressure), boiling point for water is 100
C. On
a high mountain, where pressure is lower, the boiling point is lower. By assuming an
expo
nential fall in pressure with height (which we know to be a reasonable assumption), we
may obtain the following approximation for the boiling point:
where
H
is the scale height, and
a
is a constant which will be determined in the as
signment.
4.3 Moisture Parameters
A number of parameters are used to characterize, and indeed measure, the amount of water
vapour in the atmosphere. Many of these can be plotted/extracted on the F160 aerological
chart.
1
Mixing ratio,
r
: mass of vapour/m
ass of dry air. Isopleths of constant mixing ratio
(plotted in g/kg) are plotted on the F160, and are nearly parallel to the isotherms.
(This will be examined below.)
2
Specific humidity,
q
: mass of vapour/mass of moist air.
3
Vapour p
ressure,
e
: that part of the atmospheric pressure exerted by water vapour: the
partial pressure of water vapour (remember Dalton). Note that
Note that we measure
r
in g/kg, and both
e
and
p
in hPa.
4
Saturation mixing ratio,
r
s
: mixi
ng ratio in the case of air which is saturated with
water vapour.
5
Saturation specific humidity,
q
s
: specific humidity of saturated air:
q
s
r
s
.
6
Saturation vapour pressure,
e
s
: vapour pressure in the case of air which is saturated
with water vapour. Agai
n we note that
r
s
= 622
e
s
/
p
7
Relative humidity,
RH
: a measure of humidity relative to saturation.
RH
= (
q
/
q
s
) x 100%
(
r
/
r
s
) x 100% = (
e
/
e
s
) x 100%
8
Dew point,
T
d
: the temperature to which a sample of moist air must be cooled at
constant pressure for it
to become saturated. This is a good measure of human
discomfort. Note that
5
T
–
T
d
is known as the dew point depression.
9
Wet bulb temperature,
T
w
: the lowest temperature to which air may be cooled,
adiabatically and isobarically, by
evaporating water into it. This is much easier to
measure than
T
d
. Remember that, unlike for
T
d
, we are adding extra water vapour to
the (local) atmosphere, so that
T
d
T
w
T
Wet Bulb Temperature
When performing a wet bulb measurement, we evaporate wat
er vapour into the air, which has
two effects: the vapour pressure of water in the air increases (of course); while the
temperature of the air decreases, as it must supply the latent heat needed for the evaporation.
If the mass of evaporated vapour is
m
v
,
then
L m
v
=
m
c
p
(
T
–
T
w
)
But
q
(
T
) = original mass of vapour /
m
(
m
m
d
)
and
q
s
(
T
w
) = final mass of vapour /
m
so
q
s
(
T
w
)
–
q
(
T
) =
m
v
/
m
Thus
q
(
T
) =
q
s
(
T
w
)
–
(
c
p
/
L
) (
T
–
T
w
)
A wet

bulb thermometer is part of the standard set of meteorol
ogical instruments included
inside a Stevenson screen.
4.4
Plotting Moisture on the F160
Saturation mixing ratio isopleths
The fourth set of lines on the F160 chart are the saturation mixing ratio isopleths (that is, lines
of constant saturation mixing rati
o). Mixing ratio is a conserved quantity for an air parcel (in
the absence of precipitation), whereas water vapour partial pressure is not. How do we plot
(saturation) mixing ratios? That is to say, how does
r
s
vary as a function of
p
and
T
?
r
s
=
e
s
(
T
)/
p
and we know
e
s
from the Clausius

Clapeyron equation as a function of
T
. As we move up the
chart from a pressure of 1000 hPa,
p
decreases, which would lead to an increase in
r
s
.
However,
T
also decreases, and so then does
e
s
, but much more rapidly. Th
us these isopleths
are roughly parallel to the isotherms, but slope gently to the left.
Dew point temperature
Instead of quoting moisture content in terms of mixing ratio (for example), a more common
measure is the dew point temperature,
T
d
. For a given
pressure and mixing ratio, there is a
fixed value of dew point (temperature). Assume that the actual mixing ratio (saturated or
unsaturated) is
r
, and the pressure is
p
. Then the vapour pressure follows immediately from
6
e
=
r p
/
From this, we may imm
ediately find the dew point from
e
s
(
T
d
) =
e
(
T
)
Remember what this means physically: we have a vapour pressure; at what temperature
would this vapour produce saturation? Now we see that these green lines on the F160 chart
can be said to perform a dual
function. Firstly they are labeled (in green) in g/kg, for the
saturation mixing ratio of water vapour for a given
T
and
p
(i.e., a given ‘point on the graph’).
Secondly, we may also read the temperature coordinate at that point, and take this as the dew
point. Thus we see that these two concepts are essentially interchangeable.
Footnote: We can plot the dew point lines (which we now know to also be mixing ratio
isopleths) by inverting the Clausius Clapeyron equation, and replacing
e
by
r
as indicated
abo
ve. This gives the equation
where
e
0
= 6.11 hPa and
T
0
= 273 K. This equation allows us to plot dew point as a function
of pressure, for a given (saturation) mixing ratio.
In the appendix to this chapter, we derive an expression for
the “dew point lapse rate”:
which is approximately 1.8 K per km in the lower atmosphere.
4.5 Saturated Adiabatic Lapse Rate
If a parcel of
saturated
air is raised up, it will expand and cool, in essentially the same way as
a dry
(or unsaturated) parcel. However, as soon as its temperature drops, it will become
supersaturated, as now
T
<
T
d
. As a result, some of the water vapour will condense,
releasing
latent heat. Consequently, the rate of decrease of temperature
–
the lapse rat
e
–
will be
less
than for dry air.
Turning again to the first law of thermodynamics, plus the hydrostatic equation, gives
dh
=
c
p
dT
dp
= c
p
dT + g dz
This time
dh
is not zero, as there is latent heat released:
dh =
L dq
s
(Note t
hat the minus sign is there because we will have a reduction in specific humidity; that
is,
dq
s
will be negative.) Thus
L dq
s
= c
p
dT + g dz
7
(4.7)
This is the saturated ad
iabatic lapse rate (
d
is the dry ALR). The dry ALR is constant, while
the saturated ALR varies considerably due to the
dq
s
/
dT
term. Now
q
s
r
s
=
e
s
/
p
so from the Clausius

Clapeyron equation we have
N
ote that the neglected term in this equation is at least 10 times smaller than the other term.
This expression is clearly positive, so that the saturated ALR is clearly smaller than the dry
ALR. However, as we move high in the troposphere,
e
s
becomes very
small. At this point, the
two ALRs become similar. In the moist lower troposphere,
s
may be close to 5
/km (half the
dry value). A value of 6.5
/km is often taken as “typical” for the low to middle troposphere,
and may often be assumed as a first guess, o
r model value.
Plotting saturated adiabats
These are plotted as dashed green lines which are roughly vertical near the bottom of the
diagram, and then bend to the left at higher elevations, joining on to the dry adiabats at a
pressure (altitude) of aroun
d 200 hPa. How should we unambiguously label these lines,
which depend on both
q
s
and
T
? The obvious solution, as with the dry adiabats, is to label
them with the temperature they pass through at 1000 hPa. This temperature is known as the
wet

bulb potentia
l temperature,
w
.
Consider the following process. Imagine taking a parcel of saturated air, and raising it up
until it becomes so cold that all of the water vapour condenses,
and
falls out. Now bring the
parcel adiabatically back to 1000 hPa. It will cle
arly be warmer than it would have been,
because of the release of latent heat. This final temperature is called the equivalent potential
temperature,
e
. It is possible to work out, theoretically, this final temperature. However, it is
very much easier to
use the chart.
Although nature rarely lifts air parcels so high, nor does the water all precipitate out when air
is raised, processes similar to this do occur when air rises to move over a mountain. Some,
but not all, of its moisture rains out near the sum
mit, and the descending air on the lee side
can be much warmer. This is known as a foehn, or chinook. Doing problems of this type is
possible using mathematics, but again much easier on the chart.
8
4.6 Elevation of Moist Air
Consider a parcel of moist a
ir (assumed to be unsaturated at this stage), with a certain amount
of water vapour. Initially we may characterize it by its temperature, and its dew point (or
mixing ratio). Now assume the parcel is forced to rise by whatever mechanism. As long as it
rema
ins unsaturated, it will cool adiabatically: that is, ‘its temperature will follow a dry
adiabat’. We may also express this by saying that its potential temperature will be conserved.
A second conserved quantity is its water vapour mixing ratio (but not t
he water vapour partial
pressure). Thus, as the parcel rises, its mixing ratio will simply follow the relevant isopleth.
Another way to view this is in terms of dew point, which is the quantity which was probably
actually measured. Given the connection bet
ween these two quantities just discussed, we may
in fact say that ‘its dew point follows a mixing ratio isopleth’. At some (pressure) altitude,
these two paths will meet: that is, the parcel’s temperature and dew point will be equal. Now
the parcel is satu
rated, and we have reached the “lifting condensation level” or LCL. If the
parcel is raised any further, it must now follow a saturated adiabat.
Height of the LCL:
We know that the temperature decreases at the dry adiabatic lapse rate of
9.8 K per km, and
that the dew point temperature decreases at the dew point lapse rate of 1.8
K per km. Thus, these two temperatures approach each other at the difference between these
two lapse rates
–
8.0 K per km. (This is the lapse rate of the ‘dew point depression’.) T
hus the
height of the LCL is (to a good approximation)
What about the
wet

bulb
temperature; after all, that is what is usually measured, not the
(more directly relevant) dew point? Our original parcel will have had a wet

bulb readin
g
somewhere between its dew point and its temperature. Now the wet

bulb measurement is, by
definition, a saturated process. Thus it may be understood that, as the parcel rises,
T
w
will
follow a saturated adiabat. (This explains the earlier definition of th
e wet

bulb potential
temperature.)
Clearly, when the parcel reaches the LCL, all three temperatures,
T
,
T
w
and
T
d
, must be equal,
a result known as
Normand’s theorem
. In fact, this result provides a quick way to find any of
these three temperatures, once w
e have the other two: follow these two up the appropriate
lines until they meet (at the LCL), then go back down the third line to the initial pressure
level. (In practice, we often want to find dew point from the other two, as it is the hardest to
measure.
)
If a saturated parcel is raised further, it will follow a SALR. This will move its coordinates to
the left of its original mixing ratio isopleth, indicating that the maximum amount of water
vapour it can hold is less than it held to begin with. This is e
xactly what we expect: we cooled
the parcel to saturation, then cooled it some more. The difference between the new saturation
mixing ratio and the original (saturation) mixing ratio will be the mixing ratio of water which
has condensed to the liquid (or s
olid) phase. Some (or all) of this condensate may or may not
precipitate out.
If the parcel now descends (for example, on the other side of a mountain), it will again follow
a SALR, until any unprecipitated condensate has re

evaporated. After that, it wil
l follow a
DALR. Because of the release of the latent heat of any precipitated water (but not any
condensate which finally re

evaporates), the final temperature will be higher than the initial
temperature (at the same pressure level). With a little practic
e, these problems can be done
quickly and easily (and reasonably accurately) on the F160 chart. They can also be done
mathematically, but why bother?
9
Appendix
Dew point lapse rate
The Clausius

Clapeyron equation states
Now the defin
ition of the dew point is that the air would be saturated at that temperature.
Hence, an identical equation to this must apply for the rate of change of
e
with respect to
T
d
:
Further,
e
is related to
p
via the definition
Since the specific humidity is assumed to be constant,
e
is proportional to
p
, so that:
In this equation, we may replace
dp
(on the left hand side) via the hydrostatic equation;
and we can replac
e
p
(on the right hand side) via the ideal gas equation;
Making these substitutions, and tipping the equation upside down, gives
On the Kelvin scale, temperature and dew point are close enough to write this a
s
Dew point temperatures in the lower troposphere of the middle latitudes don’t differ all that
much from, say, 10 C (283 K), giving a lapse rate of 1.8
0.1 K per km.
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