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 doesnot exchange any water vapour with its environment; hence, both the pressure and the mixingratio of the parcel are constant. In consequence, the vapour pressure in the parcel will also remain constant since . On the other hand, the saturation vapour pressure, es(T), willdecrease along with the diminishing temperature with the result that the relative humidity, will increase and may become unity. The temperature at which saturation occurs(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 thedewpoint temperature.

If the parcel vapour pressure lies below the triple point vapour pressure, isobaric cooling will leadto 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:

εrP

e ≅

)(Te

eu

s

=

We can derive a relationship between the dewpoint depression, T-Td, and the relative humidityby 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:

d

d

v

vs

TT

TT

R

l

e

Teu

)(lnln (13.1)

Equation 13.1 may be written approximately as:

)log(35 10 uTT d (13.2)

where in Eq. 13.2, u is expressed as a fraction, not as a percent. As an example, if u=0.5 thenT-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 thetephigram. Although thermodynamics can say that the air will be saturated at the dewpointand frost points (with respect to liquid water and ice, respectively), it cannot say whether or notcondensation or deposition will occur at these respective temperatures. Condensation anddeposition at saturation will depend on the existence of suitable condensation or depositionnuclei (aerosol particles). In their absence, the air parcel will simply become supersaturatedwith water vapour if further cooling occurs.

We can obtain a relation between the frostpoint and dewpoint temperatures by applyingEq. 13.1 between the triple point and point 3 on the diagram, and then again between the triplepoint and point 4 on the diagram, then subtracting the two equations. This leads to:

ft

ft

v

s

dt

dt

v

vt

TT

TT

R

l

TT

TT

R

l

e

eln (13.3)

With temperatures written in oC, the last equality leads to:

8

9

v

s

f

d

l

l

T

T(13.4)

It should be noted the Tf and Td are invariants under isobaric temperature changes of a closedair parcel.

FOG FORMATION (CONDENSATION BY ISOBARIC COOLING)

When isobaric cooling of an air parcel results from loss of heat by radiation, a radiation fogforms at the dewpoint temperature. Further cooling causes the fog to “thicken” (I.e., the liquidwater 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 foghas formed, the release of latent heat substantially reduces the cooling rate with the result thatthe overnight minimum temperature with a radiation fog is often not much lower than theevening dewpoint temperature.

We can calculate the “effective heat capacity” of foggy air as follows. For isobaric processes,q=dh, so for cloudy air:

svp drldTcdhq (13.5)

We are seeking a relation between q and dT. Consequently, we need to establish a relationbetween drs and dT. This can be done using the Clausius-Clapeyron equation with p constant:

sss

s dep

drp

er

εε (13.6)

dTTR

elde

TR

l

dT

ed

v

svs

v

vs22

ln (13.7)

Substituting from Eq. 13.7 into Eq. 13.6 and thence into Eq. 13.5, we obtain the desired result:

dTTpR

elcq

v

svp

2

2)(ε (13.8)

The quantity in parentheses can be thought of as the isobaric “effective heat capacity” of foggyair. Clearly, it is higher than the isobaric specific heat capacity of dry air. Just how much highercan be demonstrated with an example. If saturated air at 25oC is cooled isobarically, its initialeffective heat capacity is:

cp=1.005x103 Jkg-1 K-1

ε=0.622lv=2.4418x106 Jkg-1 (at 25oC)es=3.167x103 Pa (at 25oC)p=105 kPaT=298.15 KRv=4.6151x102 Jkg-1 K-1

and so:

1132

2

1087.3)( KJkgTpR

elc

v

svp

ε

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 fluxremains the same after fog formation, the rate of cooling of the air parcel will be reduced by afactor of almost four.

--------------------------------------------------

Another question we might ask in connection with fog formation is, “How much liquid watercondenses 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:

vddc −= (13.9)

Differentiating the ideal gas law for water vapour at saturation, we have:

dTTR

ede

TRd

v

ss

vv 2

1−= (13.10)

where the second term is much smaller that the first and may be neglected (the temperaturesquared term in the denominator dominates). Substituting for des in Eq. 13.10 from theClausius-Clapeyron equation, and thence into Eq. 13.9, we have:

dTTR

eldc

v

sv32−= (13.11)

As an example of the use of Eq. 13.11, we can ask what temperature change is required toproduce a liquid water content of 1 gm-3 at 5oC and at 30oC. The results are shown in the tablebelow:

30oCRv=4.6151x102 Jkg-1 K-1

T=303.15 Klv=2.43x106 Jkg-1

es=4.243x103 Pa

dc=1.0x10-3 kgm-3

dT=0.6oC

5oCRv=4.6151x102 Jkg-1 K-1

T=278.15 Klv=2.4891x106 Jkg-1

es=8.718x102 Pa

dc=1.0x10-3 kgm-3

dT=2.1oCThis 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 lowtemperatures. 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 isuninteresting since dh=cpdT and isenthalpic processes are simply isothermal. Moist air, on theother hand, will cool if more water vapour is evaporated into it, or it will warm if water vapourcondenses in it (keeping in mind that we have assumed adiabatic conditions, so that thelatent heat must come from or be given to the air itself; i.e., the dry air/water vapour mix isadiabatic when taken together).

With the possibility of phase changes in mind, we can substitute the latent heat release for qand write dh=cpdT+lvdr=0, with the result that cpT+lvr=const for such processes. Or, dividingby cp, then:

constc

rlT

p

v

for adiabatic, isobaric processes.

(14.1)

ISOBARIC EQUIVALENT TEMPERATURE

Suppose that it were possible to condense all the water vapour in an air parcel adiabatically andisobarically (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:

p

vie c

rlTT (14.2)

where Tie is the isobaric equivalent temperature; that is, the temperature in the final, dry statefollowing condensation of all the water vapour in the parcel (rie=0). Substituting values for thelatent and specific heats, Eq. 14.2 becomes approximately Tie=T+2.5r, where r (in g/kg) is theactual mixing ratio of the air parcel. The isobaric equivalent temperature cannot be determinedfrom the tephigram, although a related (and approximately equal) temperature can. This is theadiabatic 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 belowcloud base, and the wet-bulb thermometer. Once again, writing Eq. 14.1 for the initial and finalstates:

p

iwsviw

p

v

c

TrlT

c

rlT

)( (14.3)

where Tiw is the isobaric wet-bulb temperature, that is, the temperature of the air parcel afterit 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 theamount of water vapour in the air, they should be related to each other. This relation is called thepsychrometric equation. It can be derived from Eq. 14.3, which can be re-arranged to give:

)()( wv

pws TT

l

cTrr (14.4)

Using and the fact that e=es(Td), Eq. 14.4 becomes: p

er

ε

)()()( wv

pwsds TT

l

pcTeTe

ε (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, seehttp://www.usatoday.com/weather/wsling.htmhttp://www.materialstestingequip.com/psychro.htm

Example

Q: On the night of October 16, 1991, there was a wet snowfall in Edmonton. The pressure was95 kPa. The following morning the temperature was 0oC and it was quite foggy. What were theinitial 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 thewet-bulb temperature on the morning of October 17 was 0oC. Making use of our tephigramto determine rs, Eq. 14.3 then becomes:

rT

c

TrlT

p

wsvw

5.210

10005.1

104)105.2(0

)(3

36

There is no unique answer for T and r. Several possibilities are listed below. Possibly the airmass 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 demonstratethis 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

*)(63.0

TueTepc

l

SchTTh ss

p

v

ε(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 ofmomentum 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.

)(

)(

)( Te

Te

Te

eu

s

ds

s

If T*=Tw, then Eq. 14.7 is identical to the psychrometric equation (Eq. 14.5). Hence, we inferthat the wet-bulb thermometer does indeed measure the isobaric wet-bulb temperature.

The psychrometric equation suggests that adiabatic, isobaric processes can be represented asstraight lines on an Andrews-Amagat diagram, as illustrated below:

*)(*)()( TTl

pcTeTe

v

psds

ε

Then Eq. 14.6 may be re-arranged to give:

(14.7)

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 officeuses the following formula (courtesy of Brian Paruk, Jan. 1993):

)33)(45.100.10(1625.1 5.0 TVVH (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 vapourpressure will be give approximately by the mass-weighted averages:

21

2211

21

2211

mm

ememe

mm

TmTmT

(14.9)

Thus the final point on an Andrews-Amagat diagram lies along a straight line joining the two airparcels, as in the diagrams below. Because of the downward curvature of the saturation vapourpressure curve, the mixed air parcel can have a higher relative humidity than either of the twooriginal 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

Hence, the mixing of two initially unsaturated air parcels can cause a mixing fog. A commonoccurrence of such a fog is your visible breath when you exhale in wintertime. Another exampleis the condensation trail from jet aircraft.

TR

eec

v

s

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, priorto 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 asphase changes do not occur. Hence, we will approximate moist adiabatic expansion with dryadiabatic expansion which is described, very compactly, by const

Let us examine how the relative humidity changes during moist adiabatic expansion, with a viewto the possibility that a rising thermal will eventually reach saturation. Differentiating,logarithmically, the definition of relative humidity (Eq. 8.23), we have:

sededud lnlnln (15.1)

We would like to express the RHS of Eq. 15.1 in terms of dT, since we know that aconsequence of adiabatic expansion is a drop in temperature. Poisson’s equation for the vapourmay be written as:

constTe (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 rεe/p=const, leading to the same result.

We may logarithmically differentiate Eq. 15.2 and substitute into the first term on the RHS ofEq. 15.1. Then we may logarithmically differentiate the Clausius-Clapeyron equation andsubstitute into the second term on the RHS of Eq. 15.1. The result is:

dTTR

lTdud

v

v2

ln1

ln (15.3)

which may be re-arranged to give:

TR

lTc

T

u

dT

du

d

vp ε(15.4)

Hence the relative humidity, u, will increase as the temperature of the adiabatically expandingair drops, provided T<εlv/cp1500 K. This condition is always satisfied under meteorologicalconditions.

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 willrise 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):

sv

v

s TTR

l

T

Tu

11ln1

ln (15.5)

This is a non-linear equation for Ts and it needs to be solved numerically. However, we candetermine 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 thetemperature and dewpoint temperature decline at different rates under moist adiabatic expansion. At the saturation point, however, they will be equal. This is illustrated in thesketches on the previous slide. As we will see later (and have already discovered from thetephigram), the dry adiabatic lapse rate is about 10oC/km. So let us determine the dewpointtemperature lapse rate.

The Clausius-Clapeyron equation can be interpreted in terms of the dewpoint temperature asfollows:

edl

TRdT

v

dvd ln

2

(15.7)

Substituting from Poisson’s equation (Eq. 15.2), logarithmically differentiated:

6

2dT

T

dT

l

TcdT

v

dpd

ε (15.8)

Hence, the lapse rate of the dewpoint temperature along a dry adiabat is approximately1.7oC/km.

CLOUD BASE ESTIMATION

If we know the initial temperature and the dewpoint temperature of a rising air parcel, we canestimate the height of the saturation point (the LIFTING CONDENSATION LEVEL orLCL), as follows. The temperature lapse rate between the initial point and the LCL is:

kmCz

TT os /10

(15.9)

The dewpoint temperature lapse rate over the same interval is:

kmCz

TTo

sd /6

10

(15.10)

Combining Eqs. 15.9 and 15.10 leads to:

kmTTz d )(12.0 (15.11)

This formula will give the base of cumulus clouds which are formed by air rising undilutedfrom the level at which the temperature and dewpoint temperature are T and Td, respectively.