ATMOSPHERIC PROCESSES—ISOBARIC COOLING
• Dew point, frost point
• Fog formation
Dew Point, Frost Point
Let us consider the isobaric cooling of a parcel of moist air. We will assume that the parcel does
not exchange any water vapour with its environment; hence, both the pressure and the mixing
ratio of the parcel are constant. In consequence, the vapour pressure in the parcel will also
remain constant since e  rP . On the other hand, the saturation vapour pressure, es(T), will

decrease
along
with
the
diminishing
temperature with the result that the relative humidity,
e
u
will increase and may become unity. The temperature at which saturation occurs
e (T )
(i.e., u=1) is called the dewpoint temperature. Since u=1 at the dewpoint temperature, then,
ed(Td)=e (denoting dewpoint by the subscript d) . We take this to be the defining equation of the
dewpoint temperature.
s
If the parcel vapour pressure lies below the triple point vapour pressure, isobaric cooling will lead
to ice saturation before it leads to liquid water saturation. Ice saturation occurs at the
frostpoint temperature, Tf, and it is defined by the equation ei(Tf)=e. The dewpoint and
frostpoint temperatures are illustrated on the Andrews-Amagat (T,e) diagram as follows:
We can derive a relationship between the dewpoint depression, T-Td, and the relative humidity
by integrating the Clausius-Clapeyron equation between points 1 and 2 on the diagram above.
Assuming that over a small interval the latent heat of vapourization can be taken as a constant,
and using the defining equation for the dewpoint temperature, the integration leads to:
es (T ) lv  T  Td
 ln u  ln
 
e
Rv  TTd



(13.1)
Equation 13.1 may be written approximately as:
T  Td  35( log10 u)
(13.2)
where in Eq. 13.2, u is expressed as a fraction, not as a percent. As an example, if u=0.5 then
T-Td=10.5 K. The fact that the dewpoint depression is a function of the relative humidity only,
and not of the initial temperature, can be plainly seen be considering a few cases on the
tephigram. Although thermodynamics can say that the air will be saturated at the dewpoint
and frost points (with respect to liquid water and ice, respectively), it cannot say whether or not
condensation or deposition will occur at these respective temperatures. Condensation and
deposition at saturation will depend on the existence of suitable condensation or deposition
nuclei (aerosol particles). In their absence, the air parcel will simply become supersaturated
with water vapour if further cooling occurs.
We can obtain a relation between the frostpoint and dewpoint temperatures by applying
Eq. 13.1 between the triple point and point 3 on the diagram, and then again between the triple
point and point 4 on the diagram, then subtracting the two equations. This leads to:
et lv  Tt  Td
ln  
e Rv  TtTd
 ls  Tt  T f
  

 Rv  TtT f




(13.3)
With temperatures written in oC, the last equality leads to:
Td ls 9
 
T f lv 8
(13.4)
It should be noted the Tf and Td are invariants under isobaric temperature changes of a closed
air parcel.
FOG FORMATION (CONDENSATION BY ISOBARIC COOLING)
When isobaric cooling of an air parcel results from loss of heat by radiation, a radiation fog
forms at the dewpoint temperature. Further cooling causes the fog to “thicken” (I.e., the liquid
water content in the fog increases).
If the loss of heat occurs through heat transfer to a cold, underlying surface the resulting fog
is called an advection fog.
Both types of fog can be described with the same thermodynamics. As we will see, once a fog
has formed, the release of latent heat substantially reduces the cooling rate with the result that
the overnight minimum temperature with a radiation fog is often not much lower than the
evening dewpoint temperature.
We can calculate the “effective heat capacity” of foggy air as follows. For isobaric processes,
q=dh, so for cloudy air:
q  dh  c p dT  lv drs
(13.5)
We are seeking a relation between q and dT. Consequently, we need to establish a relation
between drs and dT. This can be done using the Clausius-Clapeyron equation with p constant:
rs 
es
p
 drs 

p
des
(13.6)
d ln es
lv
lv es

 des 
dT
2
2
dT
RvT
RvT
(13.7)
Substituting from Eq. 13.7 into Eq. 13.6 and thence into Eq. 13.5, we obtain the desired result:

 (lv ) 2 es 
dT
q   c p 
2 
pRvT 

(13.8)
The quantity in parentheses can be thought of as the isobaric “effective heat capacity” of foggy
air. Clearly, it is higher than the isobaric specific heat capacity of dry air. Just how much higher
can be demonstrated with an example. If saturated air at 25oC is cooled isobarically, its initial
effective heat capacity is:
cp=1.005x103 Jkg-1 K-1
=0.622
lv=2.4418x106 Jkg-1 (at 25oC)
es=3.167x103 Pa (at 25oC)
p=105 kPa
T=298.15 K
Rv=4.6151x102 Jkg-1 K-1
and so:
cp 
 (lv ) 2 es
pRvT 2
 3.87103
Jkg1K 1
This value is almost four times the isobaric specific heat capacity of dry air (almost as large,
in fact, as the specific heat capacity of liquid water). Consequently, in this case, if the heat flux
remains the same after fog formation, the rate of cooling of the air parcel will be reduced by a
factor of almost four.
-------------------------------------------------Another question we might ask in connection with fog formation is, “How much liquid water
condenses for a given temperature drop?”
If c is the mass of liquid water per unit volume of the air parcel, and v is the vapour density,
then conservation of mass requires that:
dc  dv
(13.9)
Differentiating the ideal gas law for water vapour at saturation, we have:
dv 
e
1
des  s 2 dT
RvT
RvT
(13.10)
where the second term is much smaller that the first and may be neglected (the temperature
squared term in the denominator dominates). Substituting for des in Eq. 13.10 from the
Clausius-Clapeyron equation, and thence into Eq. 13.9, we have:
dc  
lv es
dT
2 3
Rv T
(13.11)
As an example of the use of Eq. 13.11, we can ask what temperature change is required to
produce a liquid water content of 1 gm-3 at 5oC and at 30oC. The results are shown in the table
below:
30oC
Rv=4.6151x102 Jkg-1 K-1
T=303.15 K
lv=2.43x106 Jkg-1
es=4.243x103 Pa
dc=1.0x10-3 kgm-3
dT=0.6oC
5oC
Rv=4.6151x102 Jkg-1 K-1
T=278.15 K
lv=2.4891x106 Jkg-1
es=8.718x102 Pa
dc=1.0x10-3 kgm-3
dT=2.1oC
This example illustrates that a much bigger temperature change is required at low temperatures in order to
obtain a fog of comparable liquid water content. This is why “dense” fogs tend to occur less frequently at low
temperatures. However, one should keep in mind that the visibility in a fog of a particular liquid water content
will depend on the drop size distribution. Reducing the volume median droplet diameter will also tend to
reduce the visibility, even if the liquid water content remains the same. As an exercise, consider the above
examples using the tephigram and check whether you get the same result. Take the pressure to be 100 kPa
and the air density to be 1 kgm-3 , so that the liquid water content and the liquid water mixing ratio are the
same.
ATMOSPHERIC PROCESSES—ADIABATIC/ISOBARIC
• Isobaric equivalent temperature
• Isobaric wet-bulb temperature
• Psychrometric equation
• Wind chill
• Mixing fog
Adiabatic, isobaric processes are isenthalpic since dh=q+vdp=0. For dry air, the result is
uninteresting since dh=cpdT and isenthalpic processes are simply isothermal. Moist air, on the
other hand, will cool if more water vapour is evaporated into it, or it will warm if water vapour
condenses in it (keeping in mind that we have assumed adiabatic conditions, so that the
latent heat must come from or be given to the air itself; i.e., the dry air/water vapour mix is
adiabatic when taken together).
With the possibility of phase changes in mind, we can substitute the latent heat release for q
and write dh=cpdT+lvdr=0, with the result that cpT+lvr=const for such processes. Or, dividing
by cp, then:
T
for adiabatic, isobaric processes.
lv r
 const
cp
(14.1)
ISOBARIC EQUIVALENT TEMPERATURE
Suppose that it were possible to condense all the water vapour in an air parcel adiabatically and
isobarically (actually, this is an impossible process but think of it as an hypothetical possibility).
Then applying Eq. 14.1 to the initial and final states, we have:
Tie  T 
lv r
cp
(14.2)
where Tie is the isobaric equivalent temperature; that is, the temperature in the final, dry state
following condensation of all the water vapour in the parcel (rie=0). Substituting values for the
latent and specific heats, Eq. 14.2 becomes approximately Tie=T+2.5r, where r (in g/kg) is the
actual mixing ratio of the air parcel. The isobaric equivalent temperature cannot be determined
from the tephigram, although a related (and approximately equal) temperature can. This is the
adiabatic equivalent temperature, which we shall encounter in two lectures (16).
ISOBARIC WET-BULB TEMPERATURE
Suppose water is evaporated into an air parcel, adiabatically and isobarically, until it is saturated.
This is a physically realistic process. Examples include rain evaporating into the air below
cloud base, and the wet-bulb thermometer. Once again, writing Eq. 14.1 for the initial and final
states:
T
lv r
l r (T )
 Tiw  v s iw
cp
cp
(14.3)
where Tiw is the isobaric wet-bulb temperature, that is, the temperature of the air parcel after
it has become saturated following the adiabatic, isobaric evaporation of water into it. The
isobaric wet-bulb temperature is usually called simply the wet-bulb temperature, Tw.
NOTES: 1) It should be noted that the isobaric wet-bulb temperature and the dew point temperature are achieved following
very different physical processes. Hence they are not the same unless an air parcel is already saturated, in which case they
are equal to each other as well as to the temperature of the air parcel.
2) Note that Eq. 14.3 is a non-linear equation in Tiw given the temperature dependence of the saturation mixing ratio.
Hence, it must be solved numerically. An iterative process converges, in which one first estimates Tiw, and uses the estimate
to evaluate rs(Tiw). Then Eq. 14.3 can be solved for a new estimate of Tiw, and the process repeated until the result changes
little from iteration to iteration.
PSYCHROMETRIC EQUATION
Since the dew point temperature, Td, and the wet-bulb temperature, Tw, are both measures of the
amount of water vapour in the air, they should be related to each other. This relation is called the
psychrometric equation. It can be derived from Eq. 14.3, which can be re-arranged to give:
r  rs (Tw )  
cp
lv
(T  Tw )
(14.4)
e
r

Using
p and the fact that e=es(Td), Eq. 14.4 becomes:
es (Td )  es (Tw )  
cp p
lv
(T  Tw )
(14.5)
Eq. 14.5 is the psychrometric equation. If we can measure T and Tw with a wet-bulb
thermometer then we can use Eq. 14.5 to infer the dew-point temperature, Td.
For some practical information regarding how to measure the wet-bulb temperature, see
http://www.usatoday.com/weather/wsling.htm
http://www.materialstestingequip.com/psychro.htm
Example
Q: On the night of October 16, 1991, there was a wet snowfall in Edmonton. The pressure was
95 kPa. The following morning the temperature was 0oC and it was quite foggy. What were the
initial temperature and mixing ratio of this air mass?
A: Since it was foggy, we will assume a relative humidity of 100%. Hence, we infer that the
wet-bulb temperature on the morning of October 17 was 0oC. Making use of our tephigram
to determine rs, Eq. 14.3 then becomes:
lv rs (Tw )
4 103
6
Tw 
 0  (2.5 10 ) 
cp
1.005 103
 10  T  2.5r
There is no unique answer for T and r. Several possibilities are listed below. Possibly the air
mass passed through all of these states before becoming saturated.
10oC 0 g/kg 0% R.H.
5oC 2 g/kg 35% R.H.
2.5oC 3 g/kg 52% R.H.
It isn’t entirely obvious that a wet-bulb thermometer should indicate the isobaric, adiabatic
wet-bulb temperature (the similarity of the names notwithstanding). We will demonstrate
this as follows.
Let the temperature measured by the wet-bulb thermometer be called T*. We will assume that
it is in equilibrium with the airstream that is flowing past it, so that the evaporation from the wet
gauze that surrounds the bulb is driven by the heat transfer from the airstream to the thermometer.
The heat balance equation for this process (which we will derive later in conjunction with a
discussion of hailstone thermodynamics) may be written:
 Pr 
h(T  T *)  h 
 Sc 
0.63
lv
pc p
(es (T *)  ues (T ))
(14.6)
where h is the heat transfer coefficient, Pr is the Prandtl number, Sc is the Schmidt number,
u is the relative humidity, and T is the air temperature. (The Prandtl number is the ratio of
momentum diffusivity to thermal diffusivity. The Schmidt number is the ratio of kinematic
viscosity to molecular diffusivity.)
Now the term involving the Prandtl and Schmidt numbers is approximately unity, and we can use
the definitions of relative humidity and dewpoint temperature, viz.
u
e (T )
e
 s d
es (T ) es (T )
Then Eq. 14.6 may be re-arranged to give:
es (Td )  es (T *)  
pcp
lv
(T  T *)
(14.7)
If T*=Tw, then Eq. 14.7 is identical to the psychrometric equation (Eq. 14.5). Hence, we infer
that the wet-bulb thermometer does indeed measure the isobaric wet-bulb temperature.
The psychrometric equation suggests that adiabatic, isobaric processes can be represented as
straight lines on an Andrews-Amagat diagram, as illustrated below:
Note the position of the isobaric equivalent temperature, Tie (at zero vapour pressure) and
its relation to T and es(Tw).
WIND CHILL INDEX
The wind chill index is a measure of the heat flux from bare skin. The Edmonton weather office
uses the following formula (courtesy of Brian Paruk, Jan. 1993):
H  1.1625(10.0V 0.5  10.45  V )(33 T )
(14.8)
where H is in Wm-2 , V is the wind speed in m/s, and T is the air temperature in oC. It is
assumed that skin temperature is 33oC. Comparing Eq. 14.8 with the left hand side of
Eq. 14.6, we see that the expression involving the wind speed gives the heat transfer coefficient.
Note that Eq. 14.8 does not explicitly take into account evaporative cooling.
Environment Canada has a website dedicated to the wind chill index, along with charts,
graphs and a downloadable wind chill calculator:
http://www.msc-smc.ec.gc.ca/education/windchill/index_e.cfm
MIXING FOG
If we mix two air parcels isobarically and adiabatically, their final temperature and vapour
pressure will be give approximately by the mass-weighted averages:
T
e
m1T1  m2T2
m1  m2
m1e1  m2 e2
m1  m2
(14.9)
Thus the final point on an Andrews-Amagat diagram lies along a straight line joining the two air
parcels, as in the diagrams below. Because of the downward curvature of the saturation vapour
pressure curve, the mixed air parcel can have a higher relative humidity than either of the two
original air parcels. In fact, it can be supersaturated. If there are condensation nuclei in the air,
adiabatic, isobaric condensation in the supersaturated air parcel will occur along the
psychrometric equation line (as in the diagram below), giving rise to an amount of condensed
water given by c 
e  es
RvT
Hence, the mixing of two initially unsaturated air parcels can cause a mixing fog. A common
occurrence of such a fog is your visible breath when you exhale in wintertime. Another example
is the condensation trail from jet aircraft.
Mixing Fog on an Andrews-Amagat diagram.
ATMOSPHERIC PROCESSES—ADIABATIC EXPANSION
• Moist adiabatic processes
• Saturation temperature
• Cloud base estimation
Moist Adiabatic Processes
Moist adiabatic expansion can be used to describe the thermodynamics of rising thermals, prior
to the onset of condensation, under the assumption that there is no mixing with the environment
(hence adiabatic). We will ignore the thermodynamic effect of the water vapour, so long as
phase changes do not occur. Hence, we will approximate moist adiabatic expansion with dry
adiabatic expansion which is described, very compactly, by   const
Let us examine how the relative humidity changes during moist adiabatic expansion, with a view
to the possibility that a rising thermal will eventually reach saturation. Differentiating,
logarithmically, the definition of relative humidity (Eq. 8.23), we have:
d ln u  d ln e  d ln es
(15.1)
We would like to express the RHS of Eq. 15.1 in terms of dT, since we know that a
consequence of adiabatic expansion is a drop in temperature. Poisson’s equation for the vapour
may be written as:
Te  const
(15.2)
Note: Eq. 15.2 is a consequence of Poisson’s equation Tp-=const and the fact that the mole fraction of the vapour
remains constant since the air parcel is a closed system. That is, Nv=e/p=const. Alternatively, we may use the fact that
the mixing ratio re/p=const, leading to the same result.
We may logarithmically differentiate Eq. 15.2 and substitute into the first term on the RHS of
Eq. 15.1. Then we may logarithmically differentiate the Clausius-Clapeyron equation and
substitute into the second term on the RHS of Eq. 15.1. The result is:
d ln u 
1

d ln T 
lv
dT
2
RvT
(15.3)
which may be re-arranged to give:
du u  c pT  lv 

 
dT T  Rd T 
(15.4)
Hence the relative humidity, u, will increase as the temperature of the adiabatically expanding
air drops, provided T<lv/cp1500 K. This condition is always satisfied under meteorological
conditions.
Can you think of some dramatic examples where air temperature might exceed 1500 K and lead
to saturation in air whose temperature is increasing?
SATURATION TEMPERATURE
So we can be confident that if we continue to let our thermal rise adiabatically, the humidity will
rise and it will eventually become saturated (u=1). We can find the saturation temperature, Ts,
By integrating Eq. 15.3 between the initial point (T,u) and the final point (Ts,1):
T lv  1 1 
ln u  ln    
 Ts Rv  T Ts 
1
(15.5)
This is a non-linear equation for Ts and it needs to be solved numerically. However, we can
determine Ts quite easily using the tephigram. The sketch below indicates how this is done.
VARIATION OF DEW POINT UNDER MOIST ADIABATIC EXPANSION
We can infer the height of the saturation point (i.e., the cloud base) by using the fact that the
temperature and dewpoint temperature decline at different rates under moist adiabatic
expansion. At the saturation point, however, they will be equal. This is illustrated in the
sketches on the previous slide. As we will see later (and have already discovered from the
tephigram), the dry adiabatic lapse rate is about 10oC/km. So let us determine the dewpoint
temperature lapse rate.
The Clausius-Clapeyron equation can be interpreted in terms of the dewpoint temperature as
follows:
2
RvTd
dTd 
d ln e
lv
(15.7)
Substituting from Poisson’s equation (Eq. 15.2), logarithmically differentiated:
2
c pTd dT dT
dTd 

l v T
6
(15.8)
Hence, the lapse rate of the dewpoint temperature along a dry adiabat is approximately
1.7oC/km.
CLOUD BASE ESTIMATION
If we know the initial temperature and the dewpoint temperature of a rising air parcel, we can
estimate the height of the saturation point (the LIFTING CONDENSATION LEVEL or
LCL), as follows. The temperature lapse rate between the initial point and the LCL is:
T  Ts
 10 o C / km
z
(15.9)
The dewpoint temperature lapse rate over the same interval is:
Td  Ts  10 
   C / km
z
6
o
(15.10)
Combining Eqs. 15.9 and 15.10 leads to:
z  0.12(T  Td )
km
(15.11)
This formula will give the base of cumulus clouds which are formed by air rising undiluted
from the level at which the temperature and dewpoint temperature are T and Td, respectively.