LECTURE OUTLINE

1. Ways of reaching saturation • Isobaric processes • Isobaric cooling (dew point ) • Isobaric and adiabatic cooling by of water (wet bulb temperature) Temperatura wilgotnego termometru) • Adiabatic and isobaric mixing

1 /26 WAYS OF REACHING SATURATION FORMATION AND DISSIPATION OF

For simplicity, we assume that clouds form in the atmosphere when the reaches the saturation value, and f=100% (in reality this value should be greater than 100%).

r f = v r p,T s ( )

An increase of relative can be accomplished by:

§ increasing the amount of water vapor in the air (i.e. increasing rv); evaporation of water from a surface or via evaporation of rain falling through unsaturated air,

§ cooling of the air (decrease of rs(p,T)); isobaric cooling (e.g. radiative), adiabatic cooling of rising air, § mixing of two unsaturated parcels of air.

2 /26 THE COMBINED FIRST AND SECOND LAWS WILL BE APPLIED TO THE IDEALIZED THERMODYNAMIC REFERENCE PROCESSES ASSOCIATED WITH CHANGE OF WATER:

1. isobaric cooling 2. adiabatic isobaric processes 3. adiabatic and isobaric mixing 4. adiabatic expansion

3 /26 TEXTBOOKS

• Salby, Chapter 5 • C&W, Chapter 4 • R&Y, Chapter 2

4 /26 ISOBARIC PROCESSES LEADING TO SATURATION THE AIR WITH WATER VAPOR

1. Isobaric cooling (p=const, rv=const) § Dew point temperature

§ Isobaric cooling with (p=const, rt=const) 2. Isobaric and adiabatic cooling/heating by evaporation/condensation of water Izobaryczne i adiabatyczne ochładzanie poprzez odparowanie wody (p=const, a source of water vapor / water) § Wet bulb temperature § Equivalent temperature

3. Isobaric and adiabatic mixing (p=const, rt=const)

5 /26 ISOBARIC COOLING

6 /26 If vertical movement in the atmosphere is insignificant and a departure from the initial state is small, processes can be seen as isobaric processes. If there is no condensation, the First Law of is:

δq = dh = cpdT

cp takes a value for the moist air cp=cpd(1+0.87rv), but can also be approximated by its dry air cpd.

If the air is cooled, the relative humidity (f) will increase: e r f = ≅ v es (T ) rvs ( p,T ) rv does not change; temperature is decreasing therefore the saturated water vapor (es) decreases and because the process is isobaric (p=const) the same is true for rs. Cooling leads to the saturation (f=1) and then condensation starts.

7 /26 DEW POINT TEMPERATURE - TD

The dew point Td is defined as that temperature to which the system must be cooled isobarically to achieve saturation The dew point temperature is unchanged during the isobaric process without condensation.

o If saturation occurs below 0 C, that temperature is the frost point Tf .

8 /26 The dew point temperature is defined as: e=es(Td) lub rv=rs(Td), where e is the actual pressure of water vapor in the air at temperature T, rv=const is the actual mixing ratio.

⎛ L ⎛ 1 1 ⎞ ⎞ e e exp lv Td can be calculated from equation: = s,tr ⎜ ⎜ − ⎟ ⎟ ⎝ Rv ⎝ Ttr Td ⎠ ⎠

The frost point temperature: ⎛ ⎛ ⎞ ⎞ Liv 1 1 e = es,tr exp⎜ − ⎟ R ⎜ T T ⎟ Tf : e=esi(Tf) lub rv=rsi(Tf) ⎝ v ⎝ tr f ⎠ ⎠

TODO 1 (lecture 8)

For given T and rv calculate Td and Tf Show that for negative Td

9 /26 The dew point temperature unit is Kelvin.

The dew point temperature is not a measure of temperature, but a es(T) measure of humidity in the air (relative humidity, f) . We will integrate the Clousius-Clapeyron equation between temperatures T and Td ( at temperature T the saturated water vapor e=es(Td) pressure is es(T), at temperature Td the saturated wapor pressure is equal to the actual pressure of water vapor in the air e=es(Td)): Td T

dp L p d ln p L ⎛ d ln p L ⎞ TODO 2 (lecture 8) lv lv lv = 2 ⇒ = 2 ⎜ = 2 ⎟ dT RvT dT RvT ⎝ dTd RvTd ⎠ Plot f(T-Td) for different T e d L temperatures. dln p = lv dT ∫ ∫ R T 2 Fig. 6.2 C&W es T v L ⎛ ⎞ ⎛ e ⎞ lv 1 1 ln = ln f = − We assume Llv=const . If Llv takes a mean value for ⎜ e ⎟ ⎜ ⎟ ⎝ s ⎠ Rv ⎝ T Td ⎠ temperatures between T and Td, the assumption is justified! ⎛ ⎞ Llv ⎛ T − Td ⎞ f = exp⎜ − ⎜ ⎟ ⎟ ⎝ Rv ⎝ T ⋅Td ⎠ ⎠ T-Td is the dew point deficit or the dew point spread.

10 /26 11 /26 ISOBARIC COOLING WITH CONDENSATION

If the air is cooled isobarically below the dew point temperature (Td), the condensation starts.

First Law of thermodynamics : dh = dq + vdp

Enthalpy for two-component system: dh = cpdT + Llvdrv For isobabaric process dp=0: dh = dq

dq = cpdT + Llvdrv

Let’s assume that condensation occurs always at saturation state (f=1) , therefore

§ the mixing ratio is equal to its value at saturation state (rv=rs),

§ rt=rs+rl.

In a closed system rt does not change, i.e. drt=0 , and drl=-drs .

Let’s calculate the amount of water drl , that condenses during isobaric cooling: es rs = ε ε ε L e ε 2 L e p dr = −dr = − de = − lv s dT = − lv s dT l s s 2 2 de L e p p RvT pRdT s lv s = 2 dT RvT

Rd Rv = ε 12 /26 In the First Law equation for an isobaric process dq = cpdT + Llvdrv

2 ε Llves § we express drv as a function of dT (see previous slide): drv = −drl = 2 dT pRdT to find how the temperature changes if dq is added:

! 2 2 $ ε Llves δq = #cp + 2 &dT " pRdT %

Before condensation occurs dq =cpdT. After condensation starts the temperature decreases slower in response to isobaric cooling. It is due to the release of of condensation.

δq dT = 2 2 ε Llves cp + 2 pRdT 13 /26 In the First Law of thermodynamics for isobaric process dq = cpdT + Llvdrv

§ we express dT as a function of drv (see previous slide): 2 pRdT dT = − 2 drl ε Llves to find how much heat should be taken out of the system to condense a given amount of water drl.

2 pRdT δq = −cp 2 drl − Llvdrl ε Llves " ε 2L e % dr lv s q l = −$ 2 2 2 'δ # cp pRdT +ε Llves &

The relation must be integrated numerically, because es depends on temperature.

14 /26 Once condensation begins, the dew-point temperature descreases, since the water vapor mixing ratio is decreasing as the water is condensed. Relative humidity remains constant, at f=1.

Isobaric cooling is a primary formation mechanism for certain types of and stratus clouds.

The equations derived are equally applicable for isobaric heatnig. In this instance, an existing or fog can be dissipated by evaporation that ensues from isobaric heating (e.g. solar radiation).

15 /26 ADIABATIC AND ISOBARIC COOLING AND MOISTENING BY WATER EVAPORATION

16 /26 WET-BULB TEMPERATURE, TW Consider a system composed of unsaturated moist air plus rain falling through the air. Because the air is undersaturated, the rain will evaporate. If there is no external heat sources (Dq=0), and the evaporation occurs isobarically (dp=0), the is conserved (dh=0):

the water vapor content (dr ) chenges by dh = 0 = c dT + L dr = c dT - L dr v p lv v p lv l evaporation of liquid water (-drl)

cp=cpd(1+0.87rv) or can be approximated as the dry-air value cpd

If we allow just enough liquid water from the rain to evaporate so that the air becomes (rv=rs), we can integrate the :

§ from the state where there is rl of liquid water at temperature T

§ to the state where all liquid water evaporates (rl=0) and the temperature decreased to Tw:

Tw 0 c dT = L dr p ò ò lv l T rl rl is the amount of water that must be evaporated to bring the air to saturation at temperature Tw.

17 /26 During the evaporation process, latent heat is drown from the atmosphere, and the final temperature, referred to as the wet-bulb temperature, Tw , is cooler than the original temperature. The wet bulb temperature – temperature to which air may be cooled by evaporating water into it at constant pressure, until saturation is reached.

Tw 0 c dT = L dr p ∫ ∫ lv l T rl c T T L r , r r (T ) r p ( w − ) = − lv l l = s w − v L T T lv r (T ) r w = − ⎣⎡ s w − v ⎦⎤ cp

Temperature dependence of L(T) has been neglected. Given rv and T, this expression is implicit for Tw and must be solved numerically. However, if T and T are given, then r is easily determined. w v TODO 3 (lecture 8)

Calculate Tw for given T and rv or f (relative humidity). 18 /26 The wet bulb temperature Tw is measured with the wet bulb thermometer. It consists of two thermometers one of which is surrounded by a wet cloth and is ventilated.

Assman’s psychrometer 19 /26 The wet-bulb temperature in the atmosphere is conservative with respect to evaporation of falling rain.

Calculations for given values of T and rv show that Td

If ice is the evaporating phase, the ice-bulb temperature Ti can be T analogousely determined: L T T iv r (T ) r i = − ⎣⎡ s i − v ⎦⎤ cp

It is easily shown that Ti > Tw

20 /26 EQUIVALENT TEMPERATURE TE

Equivalent temperature (Te) is defined as the temperature a sample of moist air would attain if all the moisture were condensed out at constant pressure without any heat transfer from/to the environment. We use the same equation as before:

0 = dh = cpdT − Llv drl = cpdT + Llv drs

The equation is integrated

§ from a state in which there is rs of water vapor (saturated) at temperature e T

§ to a state in which all water vapor condensed (rl=0) and the temperature increased to Te:

Te 0 c dT = − L dr p ∫ ∫ lv s T rs

c T T L r p ( e − ) = lv s

Llvrs Te = T + cp 21 /26 ADIABATIC AND ISOBARIC MIXING

22 /26 In some conditions isobaric mixing of different undersaturated air masses can lead to fog formation (saturated mass).

Let’s consider the isobaric mixing of two undersaturated air masses (Y1 and Y2) having different temperatures and water vapor content. Assume that there is no condensation.

To describe the adiabatic and isobaric mixing we will use the First Law of thermodynamic (in enthalpy form). e Y1 dH = δq +Vdp — 0 = dH ≈ m c dT + m c dT Y 1 pd 1 2 pd 2 — — Y’ dT1 and dT2 correspond to the temperature change that occurs in — Y2 response of the mixing process. T We neglect the input of water vapor to the .

23 /26 We will integrate the equation form the initial state defined by temperatures T1 and T2 to the final state having the temperature T. — Y1

Y — — m c T T m c T T 0 Y’ 1 pd ( − 1 ) + 2 pd ( − 2 ) ≈ m m — Y T ≈ 1 T + 2 T 2 m + m 1 m + m 2 1 2 1 2 T2 T T1 T

The total mass m=m1+m2 of the system remains unchanged during the mixing process. The specific humidity of the mixture, qv, is a waited average of initial specific qv1 and qv2.

m1 m2 qv = qv1 + qv2 m1 + m2 m1 + m2

Because qv»rv, we can assume that mixing ratios are also mixed linearly.

Partical presssures of water vapor mixe also linearly because rv »ee/p, and the process is isobaric.

24 /26 e — Y1 The values (e,T) at point Y’ can be find by solving simultaneously the former equation and the Clausiusa Clapeyron equation. Y — The amount of water condensed in this process is: — Y’

ε — Y2 Δr = ⎡e(Y) − e(Y ')⎤ l p ⎣ ⎦ TODO 4 (lecture 8) T Calculate e and T at point Y’

An example of cloud formation resulting from adiabatic and isobaric mixing: § contrails § mist/haze formed in exhaled air

A plane flying at 200 mb ejects water vapor into the atmosphere at the temperature and water represented by point A (600K, 4mb). For atmospheric temperatures less than 47ºC (226 K), the water vapor will condense forming condensation trails. Rys 6.4 C&W 25 /26 TODO LIST

TODO 1 (lecture 8)

For given T and rv calculate Td and Tf Show that for negative temperatures Td

TODO 2 (lecture 8) Plot f(T-Td) for different temperatures. Fig. 6.2 C&W

TODO 3 (lecture 8)

Calculate Tw for given T and rv or f (relative humidity).

TODO 4 (lecture 8) Calculate e and T at point Y’

26 /26