1 / 22

# atmospheric processes isobaric cooling dew point, frost point fog formation - PowerPoint PPT Presentation

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'atmospheric processes isobaric cooling dew point, frost point fog formation' - salena

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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 . On the other hand, the saturation vapour pressure, es(T), will

decrease 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 the

dewpoint temperature.

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:

(13.1)

Equation 13.1 may be written approximately as:

(13.2)

where in Eq. 13.2, depression, 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:

(13.3)

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

(13.4)

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

air parcel.

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

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:

(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:

(13.6)

(13.7) depression,

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

(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:

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:

(13.9)

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

(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:

(13.11) capacity of dry air (almost as large,

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:

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.

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

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:

(14.1)

ISOBARIC EQUIVALENT TEMPERATURE capacity of dry air (almost as large,

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:

(14.2)

where Tieis 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:

(14.3)

where capacity of dry air (almost as large,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:

(14.4)

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

(14.5)

Eq. 14.5 is the psychrometric equation. If we can measure capacity of dry air (almost as large,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 capacity of dry air (almost as large,

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:

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:

(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.

Then Eq. 14.6 may be re-arranged to give: should indicate the isobaric, adiabatic

(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 should indicate the isobaric, adiabatic

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):

(14.8)

where H is in Wm-2 , V is the wind speed in m/s, and T is the air temperature inoC. 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,

http://www.msc-smc.ec.gc.ca/education/windchill/index_e.cfm

MIXING FOG should indicate the isobaric, adiabatic

If we mix two air parcels isobarically and adiabatically, their final temperature and vapour

pressure will be give approximately by the mass-weighted averages:

(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

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. should indicate the isobaric, adiabatic

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

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:

(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:

(15.2) should indicate the isobaric, adiabatic

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:

(15.3)

which may be re-arranged to give:

(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 should indicate the isobaric, adiabatic

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):

(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 should indicate the isobaric, adiabatic

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:

(15.7)

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

(15.8)

Hence, the lapse rate of the dewpoint temperature along a dry adiabat is approximately

1.7oC/km.

CLOUD BASE ESTIMATION should indicate the isobaric, adiabatic

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:

(15.9)

The dewpoint temperature lapse rate over the same interval is:

(15.10)

Combining Eqs. 15.9 and 15.10 leads to:

(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.