North, Gerald R., Erukhimova Tatiana L. Atmospheric Thermodynamics: Elementary Physics and Chemistry

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5.9 Wet-bulb temperature, LCL 117

Pressure

Temperature

Figure 5.9 Relative positions of the temperature T , dew point T

, and wet-bulb

temperature T

of an air parcel.

5.9 Wet-bulb temperature, LCL

The thermodynamic state of an air parcel is determined by its temperature, pressure

and mixing ratio. However, in some applications it is not convenient to use the

mixing ratio. For example, the mixing ratio is not easily measured directly. There

is another indicator of atmospheric moisture called the wet-bulb temperature (T

)

which can be measured more directly. The wet-bulb temperature lies between the

dew point temperature and the air temperature (see Figure 5.9).

Before introducing the wet-bulb temperature, let us turn again to the dew point

temperature which is also a measure of moisture in the air. Recall that it is the

temperature to which air must be cooled at constant pressure in order to become

saturated with respect to a plane surface of water. As we perform the cooling we

must keep the mixing ratio of the air ﬁxed. At this temperature the actual mixing

ratio becomes equal to the saturation mixing ratio:

w(T

) = 0.622e

)/p. (5.42)

At temperatures higher than the dew point the air contains some moisture, but

less than the saturation value. As the temperature is decreased to the dew point

condensation occurs. Since at the dew point the mixing ratio is equal to the saturation

mixing ratio, it is evident that the dew point temperature is always lower than or

equal to the air temperature. If the two are close together, the relative humidity is

high. If the dew point is far below the air temperature, the relative humidity is low.

The dew point is a good indicator of human discomfort. When temperatures are high

it is more comfortable with a low T

rather than with a high dew point temperature

because of the higher relative humidity. When the relative humidity is high the

rate of evaporation from a moist surface is low (it is actually proportional to 100

minus the relative humidity in percent). Meteorologists often refer to the difference

between temperature and dew point, which is called the dew point depression.

The frost point is deﬁned similarly to the dew point. The frost point is the

temperature to which air must be cooled at constant pressure in order to become

118 Air and water

saturated with respect to a plane surface of ice. Note that the mixing ratio is constant

during a constant pressure cooling of a parcel.

The dew point temperature is difﬁcult to measure directly. It is easy to measure

the wet-bulb temperature. This is the temperature of a wet surface (nominally a wet

cloth wrapped around the bulb of a thermometer) that is immersed in the ambient

air. The wet surface evaporates moisture into the surrounding (typically less than

saturated) air and in so doing the temperature of the wet surface is lowered (as

perspiration into dry air cools the skin). The wet surface will come to an equilibrium

temperature after a short time if the air near the wet surface is continually ventilated

with the dryer ambient air. This equilibrium temperature of the wet cloth is called

the wet-bulb temperature. If the surrounding air is fully saturated, there will be no

net cooling since the rate of evaporation will just equal the rate of condensation onto

the wet surface, leaving the wet-bulb temperature to be the same as the dew point

temperature. For unsaturated air the wet-bulb temperature always falls between the

dew point temperature and the dry-bulb temperature (the difference between the

dry-bulb and wet-bulb temperatures is called the wet-bulb depression). Note that an

evaporating cloud droplet or raindrop has a temperature at the wet-bulb temperature.

The wet-bulb temperature can be measured with a sling psychrometer, which

consists of a thermometer with wet gauze covering its bulb. This hand-held device is

swung around from a short chain to maintain the proximity of fresh ambient air at the

wet surface. Without the swinging, stagnant saturated air would accumulate around

the wet bulb and raise its temperature to an erroneous level. Water molecules leaving

the wet surface diffuse away from the bulb through the thin boundary layer of air

surrounding it. At the same time heat is being conducted from the warmer ambient

air towards the cooler wet surface through the same thin boundary layer. Equilibrium

is established between the enthalpy ﬂux due to evaporation carried by out-ﬂowing

molecules and the in-ﬂowing enthalpy ﬂux. A formula can be derived for the relative

humidity given the wet-bulb and dry-bulb temperatures (see Wet-bulb derivation in

the box below). For practical use the relationship is commonly expressed in tables.

It is interesting that the geometrical conﬁguration of the wet cloth surrounding the

bulb does not matter because those factors cancel. More discussion of the wet-bulb

temperature can be found in Exercises 7.8 and 7.9 in Chapter 7.

The saturation mixing ratio depends on temperature and air pressure, thus it

is a function of height. When a parcel is lifted adiabatically, its temperature and

pressure both decrease. The temperature dependence of the parcel is linear with

height (we will see in the next chapter that it is 10 K km

−1

). Hence, as the air rises

1 km the temperature will fall about 10 K. The saturation vapor pressure becomes

half its surface value because of this decrease (remember the rule of thumb about

the doubling of vapor pressure for every 10

◦

C). The mixing ratio w

stays the same

for this ascent, while the air pressure p(z) and the vapor pressure e(z) in the parcel

5.9 Wet-bulb temperature, LCL 119

Temperature

Pressure

w = const

Θ = const,

w = const

Lifting Condensation

Level

Figure 5.10 Illustration of the lifting condensation level (LCL).

each fall about 12% (using a scale height H of the atmosphere of about 8 km):

e(z) =

p(z)

0.622

, p(z) = p

−z/H

, w = w

= constant, e

−1/8

= 0.882. (5.43)

So the vapor pressure goes down by ≈12% while the saturation vapor pressure goes

down by 50%. This shows that the saturation vapor pressure is decreasing much

faster than the actual vapor pressure in the parcel. As this continues (the same

percentages for each kilometer of ascent) the curves will cross and condensation

will occur. The level at which an initially unsaturated parcel reaches its saturation

level while being lifted adiabatically is called the lifting condensation level (LCL).

If we know the temperature, pressure and mixing ratio of an air parcel, we can ﬁnd

the LCL, where condensation starts to occur along the ascent.

If we know the mixing ratio, we can easily ﬁnd the dew point and wet-bulb

temperature and vice versa. Figure 5.10 shows the LCL for a parcel at initial

temperature T , pressure p, and dew point T

. It is located at the intersection

of the line with constant potential temperature (the unsaturated parcel ascends

dry adiabatically) and the line of constant saturation mixing ratio starting at the

dew point temperature (this is because the mixing ratio is ﬁxed at its initial value

during this noncondensing part of the ascent). The physical signiﬁcance of these

parameters will be more clear when we start to work with thermodynamic diagrams

in Chapter 7.

Wet-bulb derivation In the preceding we indicated that the geometrical factors

cancel in the wet-bulb depression. To show this we can derive the relationship

between the relative humidity and the wet-bulb depression. For simplicity we take the

bulb to be a sphere of radius R

, but in the end it does not matter what the geometrical

120 Air and water

water molecules

thermal energy flu

Figure 5.11 Water vapor molecules diffuse away from the wet surface as heat is

conducted towards the wet surface. When these ﬂuxes match, the temperature at the

wet surface will be the wet-bulb temperature.

shape is. There are two ﬂuxes from the sphere that have to be computed: (1) the ﬂux

of water vapor molecules leaving the sphere due to evaporation toward less moist air

far away; (2) the heat conducted to the sphere because of the warmer air far away

(Figure 5.11).

The water vapor molecules diffuse away with a ﬂux F

(r) where r is the distance

from the center of the sphere, r ≥ R

. The ﬂux at any value of r passing through a

surrounding sphere is

(r) =−D4π r

(5.44)

where D is the diffusion coefﬁcient for this process (known from tables), n

(r) is the

number density of vapor molecules at radius r. This ﬂux must be independent of r

since otherwise there would be a source of ﬂux other than the wet bulb’s surface.

We can then set F

(r) = F

= constant. This allows us to integrate each side from

to ∞ after dividing through by 4πr

D. This produces:

∞

− n

=−

4πR

. (5.45)

We can recognize that

= n

sat

) (5.46)

where n

sat

) is the saturation number density for T = T

. If we divide through by

sat

) we have

RH −

sat

)

−F

4πR

sat

)

(5.47)

5.10 Equilibrium vapor pressure over a curved surface 121

where RH = n

∞

sat

) is the relative humidity away from the wet bulb. So far we

do not know the value of F

in the above formula. Another measurement is required.

We must now turn to condition (2), the heat conduction. Heat ﬂows from inﬁnity

toward the wet bulb by heat conduction. The ﬂux of heat crossing a spherical surface

of radius r is

H (r) = 4πr

= H

= constant (5.48)

where κ

is the (available from tables) thermal conductivity coefﬁcient and the same

argument is used as above for the constancy of the ﬂux passing through spheres of

different radii.

As above we can integrate after dividing through by the factor 4π r

. We get:

− T

4πR

. (5.49)

There is one other condition, that the ﬂux of heat has to be related to the ﬂux of

vapor molecules. Each molecule of water vapor leaving to inﬁnity cools the wet bulb

by  =

L/N

where L is the enthalpy of vaporization per mole of water vapor and N

is Avogadro’s number. This says that

= F

. (5.50)

Now we can substitute for H

and divide the two equations above to obtain:

RH =

sat

)

sat

)

−

D

− T

)

sat

)

. (5.51)

Using the psychrometer we measure both T

and T

. We can then calculate n

sat

)

and n

sat

) from the Clausius–Clapeyron relation. The only unknown above is RH,

which can now be calculated. The computation is tedious, hence, the tables. Note that

the geometric factors 4π R

cancelled out. It can be shown that no matter what the

geometrical shape of the web bulb, these geometrical quantities will cancel out and

the wet-bulb temperature is independent of the shape of the bulb. So why swing

the psychrometer around? The reason is to have fresh environmental air within a

millimeter or two of the wet bulb to insure that the moist air at inﬁnity is

representative and not contaminated by the wet bulb’s evaporation.

5.10 Equilibrium vapor pressure over a curved surface

So far our discussion of the saturation vapor pressure has been restricted to that

over a plane surface of water. However, in atmospheric physics we also encounter

situations where the surface is curved. This is the case, for example, in cloud droplet

formation. Cloud droplets have approximately spherical shape, which means that

growth of a droplet implies an increase of the surface area of the drop. Increasing

122 Air and water

surface area of a liquid requires work (consider blowing a soap bubble). Growth of a

droplet then requires consideration of surface tension. Thus to ﬁnd the equilibrium

vapor pressure over a droplet we have to include surface tension in the energy

balance. The energy required for an increase of surface area dA is σ dA, where σ

is the surface tension (surface energy per unit area), in J m

−2

(the value for water

is 0.0761 J m

−2

at 0

◦

C). The energy required to create a spherical drop

of radius

a is σ 4π a

Consider the formation of a cloud droplet from pure water vapor (no aerosols

or other impurities present). Such a process is called homogeneous nucleation as

opposed to heterogeneous nucleation, when small aerosol particles take part in the

droplet formation. Suppose that initially, at t = t

, we have water vapor of mass M

at partial pressure e and temperature T . The Gibbs energy of the system, G

initial

,is

initial

= Mg

(5.52)

where g

is the speciﬁc (per unit mass) Gibbs energy of the water vapor. It depends

on the vapor pressure and the temperature. Suppose that at some later time, t = t

an embryonic droplet starts to form. It grows by occasional sticking collisions by

water vapor molecules and at some moment develops a radius a and mass

The spherical surface area of the droplet is A = 4πa

. The total mass is conserved,

so if

is the mass of water vapor remaining after condensation, then

+ M

= M. (5.53)

The total Gibbs energy of the system at time t

, G

ﬁnal

,is

ﬁnal

= g

+ g

+ σ A. (5.54)

The ﬁrst term on the right-hand side of (5.54) is the Gibbs energy of the water

vapor, the second is the Gibbs energy of the liquid, and the last term is due to

surface tension.

The change of the Gibbs energy due to the droplet formation can be found by

subtracting (5.52) from (5.54) and taking into account (5.53):

ﬁnal

− G

initial

= (g

− g

+ σ A. (5.55)

The next step is to ﬁnd the difference g

−g

. We know that at a constant temperature

(in our case the temperature is ﬁxed) the change in the Gibbs energy is dg =

v dp,

The example of a soap bubble is helpful. Soapy water has a higher surface tension coefﬁcient than pure water

so the surface tension is very important. Also one must remember that the bubble has twice the surface area

because there is an inside surface as well as an outside surface, in contrast to the water droplet.

5.10 Equilibrium vapor pressure over a curved surface 123

vapor pressure

vapor

droplet

liquid with flat surface

Figure 5.13 Schematic diagram illustrating the notation for equilibrium vapor

pressure over a droplet, e, and that over a ﬂat surface, e

where v is the speciﬁc volume (v = (density)

−1

). Then, for the vapor we have

= v

de, for the liquid dg

= v

de, and the difference is equal to

d(g

− g

) = (v

− v

)de ≈ v

de (5.56)

since

 v

. Substituting speciﬁc volume v

from the Ideal Gas Law, we obtain

d(g

− g

) = R

= R

T d(ln e) (5.57)

and, after the integration,

− g

= R

T ln e + constant. (5.58)

We can ﬁnd the constant of integration by taking into account that in equilibrium

(see Figure 5.13), along the phase boundary where e = e

(T ), g

= g

(see (5.7)).

Then,

− g

= R

T ln





. (5.59)

Substituting (5.59) into (5.55) gives

ﬁnal

− G

initial

=−R

T ln





+ σ A. (5.60)

The mass of a spherical water droplet with radius a and density ρ

is M

πρ

, and the surface area is A = 4πa

. Then the change in the Gibbs energy

due to the droplet formation, G = G

ﬁnal

− G

initial

, is:

G =−

πa

T ln





+ 4πa

σ . (5.61)

From (5.61) we see that in the subsaturated air, when e < e

,ln(e/e

) is negative,

and G is always positive (see Figure 5.14). From Section 4.8.2 we know that

equilibrium occurs when the minimum of the Gibbs energy is achieved; or, in other

124 Air and water

∆G

e < e

e > e

Figure 5.14 The change of Gibbs energy due to the formation of a water droplet

as a function of the radius of the droplet.

words, the system tends spontaneously toward its equilibrium state by having the

Gibbs energy diminish more and more until it reaches its minimum. We conclude

that there are no favorable conditions for cloud droplet (a > 0) formation in

unsaturated air.

The situation is different when the air is supersaturated, which means that e > e

and ln (e/e

) is positive (see Figure 5.14). In this case G increases as the radius

a increases, then it reaches a maximum at some radius, and then decreases with the

further increase of a. If the cloud droplet has a radius less than the critical radius a

∗

it will disappear by evaporation. If, however, the cloud droplet reaches the critical

radius a

∗

, then it will continue to grow. We can ﬁnd an expression for the critical

radius by equating the derivative ∂G/∂a to zero. The result is

∗

2σ

T ln (e/e

)

(5.62)

which is called Kelvin’s formula. It allows one to ﬁnd the radius a

∗

of a droplet

which is in equilibrium with air with the vapor pressure e. This equilibrium state

is unstable. This becomes evident if we consider a slight growth of the droplet.

This growth might be due to the condensation of water vapor in the vicinity of the

droplet, which means that the relative humidity decreases just above the surface.

Since the air in the vicinity of the droplet becomes drier, there is a diffusive ﬂux of

moist air toward the droplet, and the condensation process continues, leading to the

further growth of the droplet. On the other hand, if the droplet evaporates slightly,

the relative humidity just above the droplet surface increases, the water vapor starts

to diffuse from the droplet, and the droplet will continue to evaporate to maintain

the relative humidity corresponding to radius a

∗

5.10 Equilibrium vapor pressure over a curved surface 125

We can rewrite (5.62) in order to determine the equilibrium vapor pressure e over

a droplet with radius a,

e = e

exp



2σ



= e

b/a

(5.63)

where the parameter b is deﬁned by

b =

2σ

. (5.64)

If we substitute σ = 0.076 J m

−2

, ρ

= 1000 kg m

−3

, and R

= 461.5 J kg

−1

we obtain

b =

3.30 × 10

−7

(5.65)

(in meters). At 273 K, b = 1.21 nm (typically b  a). Formula (5.63) shows that

the equilibrium vapor pressure over a spherical droplet is not equal to the saturation

vapor pressure as determined over a plane surface of water. This happens because

of the surface tension. If the radius of the droplet goes to inﬁnity, a →∞, which

corresponds to a plane surface, we obtain the result for a ﬂat surface e = e

(see

Figure 5.13).

Example 5.8 Water droplets are in equilibrium with surrounding vapor at a

temperature of 2

◦

C. Calculate the vapor pressure and relative humidity over the

droplet with radius 0.008 µm.

Answer: The saturation vapor pressure at 2

◦

C is: e

= 2.497 ×10

exp(−5417/

275.2) (hPa) = 7.06 hPa. From (5.63) we obtain e = 7.06 × exp(3.3 ×10

−7

(275.2 ×0.008 ×10

−6

)) = 8.2 hPa. RH = 116%. A supersaturation of 16% is

required for the creation of a cloud droplet with radius 0.008 µm by homogeneous

nucleation at 2

◦

C. 

To obtain the relative humidity, divide both sides of (5.63)bye

= e

b/a

. (5.66)

This relationship is illustrated in Figure 5.15 for a temperature of 5

◦

C. As shown

in the ﬁgure, the formation of a droplet with radius 0.01 µm requires supersaturation

of 112%.At the same time, in real clouds the relative humidity rarely exceeds 101%.

As we see from Figure 5.15, 1.0% supersaturation is required to form a droplet with

radius larger than 0.1 µm. Large droplets in a cloud simply cannot form by random

collisions of molecules on their surfaces. From the above discussion it follows

that the process of homogeneous nucleation is unlikely. In nature cloud droplets

form by heterogeneous nucleation on atmospheric aerosols. Consider then how the

126 Air and water

0.10 1.00 10.000.01

100

102

104

106

108

110

112

Droplet radius, µm

Relative humidity,%

Figure 5.15 The relative humidity as a function of droplet radius. The curve is

plotted for a temperature of 5

◦

C. This graph shows the relative humidity necessary

to form a droplet of the size indicated on the abscissa.

equilibrium vapor pressure over the droplet changes if the droplet contains dissolved

electrolytes (usually from aerosols, e.g. sea salt). We will consider hygroscopic

particles (those that are soluble in water). The most common of these are sodium

chloride (NaCl) and ammonium sulfate ((NH

)

) which dissolve when water

condenses onto them. In this case a water droplet can be treated as a solution with

the water as a solvent and the salt as a solute.

Chemistry refresher: Raoult’s Law The ratio of the equilibrium vapor pressure

over a solution, e



, to the equilibrium vapor pressure over pure solvent, e, is equal to

the mole fraction of the solvent in the solution, f , e



/e = f . The presence of salt in

solution always lowers the vapor pressure.

Consider a cloud droplet that contains n

salt

molecules of salt and n

molecules

of water per unit volume. If e



is the equilibrium vapor pressure over the solution

(see Figure 5.16) then, according to Raoult’s Law,



+ n

salt

1 + n

salt

(5.67)