Module 4 Water Vapour in the Atmosphere

4.1 Statement of the General Meteorological Problem D. Brunt (1941) in his book Physical and Dynamical Meteorology has stated, “The main problem to be discussed in connection to the thermodynamics of the moist air is the variation of temperature produced by changes of pressure, which in the atmosphere are associated with vertical motion. When damp air ascends, it must eventually attain saturation, and further ascent produces condensation, at first in the form of water drops, and as snow in the later stages”. This statement of the problem emphasizes the role of vertical ascends in producing condensation of water vapour. However, several text books and papers discuss this problem on the assumption that products of condensation are carried with the ascending air current and the process is strictly reversible; meaning that if the damp air and water drops or snow are again brought downwards, the evaporation of water drops or snow uses up the same amount of latent heat as it was liberated by condensation on the upward path of the air. Another assumption is that the drops fall out as the damp air ascends but then the process is not reversible, and Von Bezold (1883) termed it as a pseudo-adiabatic process. It must be pointed out that if the products of condensation are retained in the ascending current, the mathematical treatment is easier in comparison to the pseudo-adiabatic case. There are four stages that can be discussed in connection to the ascent of moist air. (a) The air is saturated; (b) The air is saturated and contains water drops at a temperature above the freezing-point; (c) All the water drops freeze into ice at 0°C; (d) Saturated air and ice at temperatures below 0°C.

However, clouds may be composed of supercooled water drops at temperatures as low as −40°C; it also forms an important stage in conjunction with the aforementioned four stages of moist air. To treat water vapour in the atmosphere, its mass mixed with a certain amount of air in a given volume must be known. The moisture content of air can be expressed in several ways that relate the partial pressure of water vapour to its volume or mass mixing ratio in the atmosphere.

4.2 Moisture parameters We define here the basic humidity variables, which essentially give an estimate of the amount of water in vapour form present in a given volume of moist air. These definitions could also be extended to other forms of hydrometeors which are at atmospheric pressure p and temperature T. The following definitions are used in for the moist air parcels. (a) Mixing ratio (r): It is defined as the ratio of the mass of water vapour in a certain volume of air to the mass of dry air in that volume, so

m ρ mv = mass of water vapour r = v = v ; (4.1) md ρd md = mass of dry air ( kg/kg) (b) Specific humidity: It is defined as the mass of water vapour in a given mass of air which is a mixture of air and water vapour; thus, m r q = v = (4.2) mv + md 1+ r

1 Since r is only a few per cent, the numerical values of r and q are nearly equal. Further, ideal gas law can be applied to each constituent of any mixture of gases, we may write the gas law for water vapour as e R T (4.3) = ρv v where e and are the pressure and density of the water vapour and Rv is the gas constant ρv for 1 kg of water vapour. The gas constant for water vapour Rv is calculated from the universal gas constant and molar mass water vapour as

R* 8.3145 −1 −1 -1 -1 R = 1000 = 1000 = 461.51 J kg K ; Rd = 287 J kg K v M 18.016 H2O

The ratio of gas constants for dry air and water vapour is calculated as

R M 18.016 ε = d = v = = 0.622. Rv M d 28.97 From Dalton’s law of partial pressures of a mixture of gases, the moisture content of the air can also be defined in terms of its partial pressure in the atmosphere.

(c) Volume mixing ratio (rv): rv = e / p (4.4) (d) Absolute Humidity: The concentration ρv of water vapour in air is called the absolute humidity (g/m3). It can be easily obtained from the ideal gas law of water vapour (4.3).

Example 1: At 0 C, = 1.275 kg/m3, and = 4.770 10-3 kg/m3 . Find the total ° ρd ρv × pressure exerted by the mixture. Solution: kg p = (1.275 )(287.0 JK −1kg−1)(273.2 K) ⇒ p = 99970.71Pa d m3 d

⎛ −3 kg ⎞ ⎛ J ⎞ e = ρ RvT = ⎜ 4.770 ×10 3 ⎟ 461.51 (273.2 K ) = 601.4232 Pa v ⎝ m ⎠ ⎝⎜ K kg⎠⎟

Hence total pressure of the mixture p = 999.71 + 6.014 = 1005.72 hPa

4.3 Virtual Temperature

In meteorology, it is customary to define a virtual temperature to take into consideration effect of water vapour in the atmosphere. Since Mv < Md and Rv > Rd , therefore instead of using the gas constant Rv, it is convenient to use Rd and fictitious temperature Tv (because never measured) in the equation of state for water vapour. Thus,

p = ρRdTv (4.5)

Here Tv is the virtual temperature; its expression is now derived starting with the density of mixture, ; p R T and e R T . The pressure of the mixture is given as, ρ = ρd + ρv d = ρd d = ρv v

p = pd + e or pd = p − e

p − e e Since ρd = ; ρv = , we can write the density of the mixture ρ as RdT RvT

2 p − e e p ⎡ 1 1 ⎤ ρ = ρd + ρv = + = + e ⎢ − ⎥ ; which can be further written as, RdT RvT RdT ⎣ RvT RdT ⎦

p ⎡ e ⎛ Rd ⎞ ⎤ p ⎡ e ⎤ ρ = ⎢1+ ⎜ −1⎟ ⎥ or ρ = ⎢1− (1− ε)⎥ (4.6) RdT ⎣ p ⎝ Rv ⎠ ⎦ RdT ⎣ p ⎦

Now (4.6) can be rearranged as

T R 287 p = ρR and ε = d = = 0.622 (4.7) d e R 461.51 1− (1− ε) v p T If we define T = then the pressure p in (4.7) can be written as given in the v e 1− (1− ε) p equation (4.5) above with ρ and p as density and pressure of the air respectively. Thus for the moist air, actual temperature is replaced by virtual temperature in the equation of state; it is in fact used as a state variable in moist air dynamics. Also one may define the virtual temperature, as that temperature the dry air would attain so as to have the same density as that of the moist air at the same pressure.

Example 2: Given a total pressure p =1026.8 hPa, find the vapour pressure e in a mixture of water vapour and air if the water vapour mixing ratio r = 5.5 g/kg.

ρv ⎛ e ⎞ ⎛ RdT ⎞ e Rd εe r Solution: r = = ⎜ ⎟ ⎜ ⎟ = . = . Hence e = p ρd ⎝ RvT ⎠ ⎝ p − e⎠ p − e Rv p − e r + ε

Put r =5.5 g/kg and p = 1026.8 hPa = 102680 Pa in the above expression of e, and we get,

(5.5 ×10−3 )(102680) e = = 900.70 Pa ; or water vapour pressure e = 9.00 hPa. 0.622 + 5.5 ×10−3

Exercise 1: Starting from the definition of virtual temperature, show that T (1 0.61q)T. v = + Also find the expression for q.

4.4 Relation between moisture variables and water vapour pressure Humidity variables can also be defined in terms of water vapour pressure. The relation between mass mixing ratio r and e is derived easily in the following manner. Starting with the definition of mixing ratio r given in (4.1), we have

ρv ⎛ e ⎞ ⎛ RdT ⎞ ε e r = = ⎜ ⎟ ⎜ ⎟ = (4.8) ρd ⎝ RvT ⎠ ⎝ p − e⎠ p − e

The specific humidity is related to the mass mixing ratio r as

ρ ρ r q = v = v = , as defined in (4.2). Now use the relation (4.8) for r to get q as, ρ ρd + ρv 1+ r

3 −1 ⎛ ε e ⎞ ⎛ ε e ⎞ ε e q = 1+ = (4.9) ⎝⎜ p − e⎠⎟ ⎝⎜ p − e⎠⎟ p − (1− ε)e

4.5 Saturated and unsaturated air The most commonly used phrases for moist air are indeed misleading. Phrases such as “air is saturated with water vapour”; or “air can hold no more water vapour”; and “warm air can hold more moisture than cold air” are quite commonly used in weather descriptions. All these phrases suggest that air absorbs moisture much like a sponge, which is not the case with air. According to Dalton’s principle of partial pressures for a mixture of gases, each component exerts a partial pressure and sum of the partial pressures is the total pressure exerted by the mixture. Hence, the exchange of water molecules between the li quid and vapour phases is indeed independent of the presence of other components in air. Moreover, an equilibrium vapour pressure is established between the liquid and vapour phases when a certain number of water molecules leave the liquid phase at a given temperature and an equal number of water molecules from vapour phase enter the liquid phase. Air is said to be saturated at such equilibrium otherwise it is unsaturated. Both terms are used for characterizing dry and moist air.

Unsaturated air: When liquid and vapour phase are imagined to coexist and the rate of evaporation of water from liquid exceed rate of condensation, then air is unsaturated at a given temperature T and it can be expressed by the pair (e, T ) .

Saturated air: When the vapour pressure in the gas phase reach to a value (say es ) such that the rate of evaporation of water from liquid is equal to rate of condensation of vapour in air at temperature T of the system. The pair (es , T ) may express the state of the system. Normally the interface of liquid and vapour phase is regarded as a plane surface.

40

) a

hP

(

30 e

Fig. 4.1 Variation of saturation vapour pressure es ur

s s

over plane surface of pure water with temperature. e pr

es 20

pour d va d

10 e

t

a

ur

t

a S 0 −50 −40 −30 −20 −10 0 10 20 30 40 50 Temperature (0C)

Saturated mixing ratio: It is the mass of a given volume of air that is saturated with respect to plane surface of pure water, to the mass of dry air in a given volume; that is, −1 m ρ ⎛ e ⎞ ⎛ p − e ⎞ e εe vs vs s s s s (4.10) rs = = = ⎜ ⎟ ⎜ ⎟ = 0.622 = md ρd ⎝ RvT ⎠ ⎝ RdT ⎠ p − es p − es

Because p >> es , the eqn. (4.10) can be put in the following simplified form

es εes rs  0.622 = (4.11) p p

4 That is, at a given temperature, rs is inversely proportion to p. Note that es = es (T ) , so the saturated mixing ratio rs is also a function of temperature; this implies that rs increases with increasing temperature at constant pressure. But at constant temperature, rs increases with decreasing pressure p, which is evident from (4.11). The curve es = es (T ) is shown in Fig. 4.1 as a function of temperature; decreases with height in the troposphere. Since es atmospheric pressure also decreases with altitude, that is why the rising parcels of unsaturated air turn saturated leading to cloud formation and rains. The process is hastened in the troposphere because temperature also decreases with altitude.

Relative humidity: The relative humidity rh is the ratio of mass of water vapour mv in a sample of moist air of volume V, to the mass of water vapour mvs if the air in volume V were saturated. Thus, we have

mv r e rh = = = (4.12) mvs rs es

The reported relative humidity is generally expressed in per cent; i.e. RH = rh ×100 .

Isobaric cooling: Imagine that temperature of a parcel of moist air reduces continuously but its pressure remains the same. During this process, e will not change but es will decrease as the temperature falls during cooling. As a consequence, under isobaric cooling, the relative humidity of the parcel will increase with decreasing temperature. If temperature continues to cool further down, then phase change of water vapour to liquid or ice may happen when threshold temperature is reached.

Dew point temperature (Td): It is that temperature to which air must be cooled at constant pressure (i.e. isobarically) for it to become saturated with respect to plane surface of pure water. Thus, we have r(T, p) = rs (Td , p) and the relative humidity at temperature T and pressure p is given by r (T , p) RH = s d ×100 . rs (T, p)

A Thumb Rule: For moist air (RH > 50%), Td decreases by 1°C for every 5% decrease in RH. Thus, starting at Td = T (dry bulb temperature) where RH = 85%, then

⎛ 100 − 85⎞ T = T − = T − 3°C or T − T = 3°C d ⎝⎜ 5 ⎠⎟ d

Frost-Point Temperature: It is that temperature to which air must be cooled isobarically for it to become saturated with respect to plane surface of ice.

-1 Example 3: What are RH and Td for air at 1000 hPa and 18°C having a mixing ratio 6 g kg . Saturation mixing ratio is 13 g kg-1 at this state.

Solution: RH 46%; from the thumb rule T 7.2 C. = d  ° However, to find the actua l value, we use the Tephigram. One has to move from right to left along 1000 hPa line to intercept the saturation line of magnitude 6 g/kg; this occurs at 6.5°C; this is the dew point temperature in this case.

5 On the globe, dew point is a good indicator of moisture levels in the air as the pressure typically varies only by a few per cent spatially and temporally. Over warm bodies, highest dew point temperatures are noted, though complete saturation is absent in hot area.

4.6 Thermodynamics of unsaturated moist air It has been shown that the equation of state for dry air can be applied to moist air when T is replaced by Tv , that is

p = ρRdTv ρ = ρd + ρv

p = ρRmT Rm = (1+ 0.61 q) Rd (4.13)

These two expressions are equivalent; here Rm is the gas constant for moist air. Since moist air is a mixture of dry air and water vapour, the specific heat at constant volume of the mixture is determined by considering one kilogram of dry air and ξ kilogram of water vapour. Next, consider adding heat dq to the sample which raises its temperature by dT ; we thus have (4.14) (1+ ξ)dq = (Cv )air dT + ξ(Cv )vap dT; volume (V )= const.

⎛ ∂q ⎞ Because , we have (Cv )mix = ⎜ ⎟ ⎝ ∂T ⎠ V

(1+ ξ) C = C + ξ (C ) ( v )mix ( v )air v vap

∂q (C ) + ξ (C ) ⎛ ⎞ v air v vap (4.15) (Cv )mix = ⎜ ⎟ = ; V= const. ⎝ ∂T ⎠ V 1+ ξ Since −1 −1 −1 −1 , eqn. (4.15) states (Cv )air at STP = 718 J K kg ; (Cv )vap at STP = 1390 J K kg that C in a parcel is the mass weighted mean of C and . In the same ( v )mix ( v )air (Cv )vap manner, we can calculate C as a mass-weighted mean of C and C as ( p )mix ( p )air ( p )vap

(Cp ) + ξ(Cp )vap C = (C ) = air (4.16) pm p mix 1+ ξ Since C at STP = 1005 J K −1kg−1 ; C 1850 J K −1kg−1 , the specific heat of the ( p )air ( p )vap mixture with pressure remaining constant can be calculated from relation (4.16). Knowing the specific heat of the mixture at constant pressure, so long as there is no condensation of water vapour during the lifting of parcels, we can calculate the lapse rate of unsaturated moist air parcel as g g Γ m = = ; Γ m < Γd (4.17) C Cpm ( p )mix

4.7 Lifting Condensation Level (LCL)

The lifting condensation level (LCL) is defined as the level to which an unsaturated parcel of moist air should be lifted adiabatically in order to become saturated. The process is explained in the Fig. 4.2. During lifting, the mixing ratio r and the potential temperature θ remain constant but rs decreases until it becomes equal to r at the LCL. Hence the lifting

6 condensation level is the point C on Fig. 4.2, to which a parcel has been lifted along the line AC from its initial position at a point A(p, T, r) with pressure p, temperature T and mixing ratio r. It is required to locate the point C on a Tephigram for issuing the local forecast. Mathematically, one can find the LCL and the corresponding ZLCL, TLCL and PLCL can be calculated from thermodynamic relations using the condition that the saturation mixing ratio of a parcel at point C is equal to its mixing ratio at A. Note that point C in Fig. 4.2 is the intersection of two lines: θ =const. and rs = const . Observations provide values of p, T, and r at point A. The saturation mixing ratio line is drawn with rs = const . In order to locate the point B, the parcel is cooled isobarically; that is, the parcel is moved horizontally to arrive at intersection point B, where its temperature is the dew point temperature Td ; the air parcel at

B is saturated. From point B, move along the line rs = const . up to a point C where it intersects the dry adiabat through point A. The LCL is also referred to as the isentropic condensation temperature in the literature.

Fig. 4.2 Lifting Condensation Level (LCL): The figure shows the procedure to find the LCL of a . A(p,T,r) t ne θ i parcel undergoing adiabatic ascent along the line on l i = ons t o B(p,Td ,rs ) a i θ=constant (isentrope). The isobars are c c t ons ur a = t r horizontal lines. The parcel is initially at point A a C(pLCL ,TLCL ,rs ) s S t. r ng xi

having pressure p, temperature T and mixing i e m ratio r. The parcel cools isobarically to become ur p s LCL s C

saturated rs = r , its temperature is Td , the dew e − r point temperature. The point C is the LCL which P is the intersection of line , and rs = const. A θ=const. Also, note that on this figure, isentropes rm he and the isotherms are perpendicular to each ot B rm is he

T − ot other. LCL is also called the isentropic LCL is condensation level. Temperature

To calculate the lifting condensation level temperature Tl from given values of temperature (T) and the corresponding dew point temperature (Td), a sufficiently accurate mathematical formula due to Bolton (1980) is

1 1 −3 ⎛ Tl ⎞ −3 ⎛ T ⎞ − +1.266 ×10 ln⎜ ⎟ = 0.514 ×10 ln⎜ ⎟ (4.18a) Tl Td ⎝ Td ⎠ ⎝ Td ⎠ In the above formula all temperatures are absolute. However, eq. (4.18a) can only be solved iteratively to obtain Tl . Bolton has also given a simple formula that explicitly gives the lifting condensation level temperature, which reads 2840 T = + 55 . (4.18b) l 3.5lnT − lne − 4.805 In (4.18b) too all temperatures are in Kelvin.

Wet-bulb temperature: It is measured with a thermometer covered with a moist cloth. The definition of the dew point temperature (Td ) and the wet-bulb temperature ( Tw ) both involve cooling of a hypothetical air parcel to saturation, but there is a difference. At Td , air has been cooled to become saturated, i.e. its mixing ratio is rs. However, the mixing ratio of air as its temperature tends to attain the wet bulb temperature is r; generally Tw ≤ T, equality will apply

7 when air is fully saturated. Usually Td < Tw ≤ T ; the thumb rule is that Tw is the arithmetic mean of T and Td ; that is , Tw =0.5 (T + Td ). We shall be using the following terms frequently.

6 -1 Latent heat of vaporization (Lv) = 2.25 × 10 J kg (liquid to vapour) at 1 atm and 100°C

6 Latent Heat of Condensation (Lv) = 2.50 × 10 J / kg; (vapour to liquid) at 0°C.

5 Latent Heat of Melting (Lm) = 3.34 × 10 J / kg. (ice to liquid)

4.8 Equivalent Potential Temperature

When air is lifted vertically, its ascent will be just like dry air until it reaches saturation; further lifting of the saturated air will result in water vapour conversion to liquid water with the release of latent heat, which must be added in the system. The analysis becomes complex as water may stay in the system or the liquid drops may fall out of the system. In the latter case, the process is pseudoadiabatic, because it is not reversible. Also, heat released adds buoyancy to the rising moist saturated air to further lift it, which may again result in converting vapour to liquid. Consider an amount of heat dQ released due to conversion of water vapour to liquid in the first law of thermodynamics in the form given by eqn. (3.24), which can be easily put in the entropy form for further analysis. The equation (3.14) reads as dp dQ dT dp dQ = C dT − , which implies = C − R p ρ T p T d p R/C ⎛ p ⎞ p Take logarithm on both sides and then differentiate, θ = T o , then we obtain ⎝⎜ p ⎠⎟ dθ dT dp C = C − R (4.19a) p θ p T d p In eqn. (4.19a), the right hand side is the entropy of the system therefore

dθ dQ C = (4.19b) p θ T

In view of eq. (4.19b), the potential enthalpy (Cpθ) is taken as a variable instead of θ in the present day global models especially the unified models of atmosphere. It is now required to calculate dQ. When an air parcel takes its ‘upward’ journey it will first become saturated and any further rise of the now saturated air in the parcel will condense out drs (kg/kg of water), the latent heat released dQ is given by

dQ = −Lv drs (4.20)

Substituting dQ from (4.17) in (4.16) we obtain,

dθ L C = − v dr (4.21) p θ T s

Lvdrs ⎛ Lvrs ⎞ d(lnθ) = − ≈− d ⎜ ⎟ (4.22) CpT ⎝ CpT ⎠

8 Lvdrs ⎛ Lvrs ⎞ However, it should be proved that ≈ d ⎜ ⎟ . Indeed, C pT ⎝ C pT ⎠

⎛ Lvrs ⎞ Lv ⎡ drs dT ⎤ Lv ⎡ dT ⎤ LvdT ⎡ drs rs ⎤ dT drs drs rs d = − r = dr − r = − ; if << then >> ⎜ ⎟ ⎢ s 2 ⎥ ⎢ s s ⎥ ⎢ ⎥ ⎝ C pT ⎠ C p ⎣ T T ⎦ C pT ⎣ T ⎦ C pT ⎣ dT T ⎦ T rs dT T

⎛ ⎞ Lvrs Lv drs Lv rs d ⎜ ⎟  dT = drs (neglecting in the above equation) (4.23) ⎝ CpT ⎠ CpT dT CpT T

Integrating the equation (4.23), we obtain the following result

L r lnθ = − v s + c CpT r The constant c is to be evaluated; at low temperatures as s → 0, θ → θ ⇒ c = lnθ . Thus T es es

θ ⎪⎧ Lvrs ⎪⎫ ⎪⎧ Lvrs ⎪⎫  exp ⎨− ⎬ ; and θes  θ exp ⎨ ⎬ (4.24) θ C T C T es ⎩⎪ p ⎭⎪ ⎩⎪ p ⎭⎪ The quantity θes is called the saturated equivalent potential temperature. Accordingly, we can define the equivalent potential temperature θe as follows,

⎪⎧ Lvr ⎪⎫ ⎪⎧ Lvq ⎪⎫ θe = θ exp ⎨ ⎬ = θ exp ⎨ ⎬; q = specific humidity . (4.25) ⎩⎪CpT ⎭⎪ ⎩⎪CpT ⎭⎪ The potential temperature θ is conserved during adiabatic transformations; but equivalent potential temperature θe is a conserved quantity during both dry and saturated adiabatic processes. We may define of a parcel as the potential temperature of the parcel when all θe its water vapour has condensed out so that its mixing ratio r is zero. To find θe , the air parcel is first lifted adiabatically so that all the water vapour has condensed and fallen out during its upward journey, and release of latent heat is added to the parcel; then the air parcel is brought back adiabatically downward to 1000 hPa. The temperature thus attained is the equivalent potential temperature θe of the parcel. The equivalent wet-bulb potential temperature can be obtained from wet-bulb temperature using the Poisson’s equation. Both

θe and θw (wet-bulb) provide equivalent information for a rising or sinking moist parcel.

Dry Static Energy (DSE): Sum of enthalpy and the potential energy Φ;

DSE = CpT + Φ (4.26)

Moist Static Energy (MSE): Sum of enthalpy, potential energy and latent heat content (Lvq) of a moist parcel. Specific humidity q is close to the mixing ratio r and MSE is defined as,

MSE = CpT + Φ + Lvq (4.27)

Both DSE and MSE are conserved quantities and play a key role in the vertical motion of a parcel. When unsaturated air is lifted adiabatically, the enthalpy (C pT ) is converted to potential energy (Φ) and the latent heat content remains unchanged during the upward journey; therefore, so long the parcel remains unsaturated, DSE is the key variable in this

9 process. However, when air becomes saturated and further lifted adiabatically, there is an exchange in all the three terms appearing in the expression (2.27) for MSE. That is, during the upward journey of saturated air, the potential energy (Φ) increases, while the enthalpy

(C pT ) and the latent heat content (Lvq) both decrease in such a manner that their sum remains unchanged. Hence, MSE is an important conserved variable for the treatment of clouds in numerical models. Let s represent DSE and h the MSE of a system, then conservation of s and h implies that during adiabatic ascent ds = 0 and during saturated adiabatic ascent dh = 0 , dT g ds = 0 ⇒ d (CpT + Φ) = 0 ; which gives = − = − Γd dz Cp While for saturated adiabatic ascent of a parcel,

dh = 0 ⇒ d (CpT + Φ + Lvq) = 0 . Since the specific heat of air changes with the addition of moisture, the lapse rate of moist air

( Γm ) will differ from Γd in accordance to the water vapour mixing ratio r of the air. For −1 −1 moist air, Cp = Cpd (1+ 0.87r) with Cpd = 1005J kg K therefore we have

Γd Γ m = ≈ Γd [1− 0.87r] 1+ 0.87r

4.9 Entropy of the Dry Air

The form of the first law of thermodynamics together with the second law of thermodynamics are used to derive an expression for the change in the entropy of the dry air when an amount dQ of heat is added to the system. Taking the first law of thermodynamics, dQ dT dp = C − R . (4.28) T p T d p From the second law of thermodynamics, increase in entropy η of the system is defined as

dQ dη ≥ . (4.29) T But, if heat exchanges reversibly then equality in (4.29) will hold and we have dQ dη = . (4.30) T Thus on combining (4.28) and (4.30), and using (4.19) we get

dT dp dθ dη = C − R = C . (4.31a) p T d p p θ Thus, eqn. (4.31) relates the change in entropy dη to change in temperature and pressure of the system. On integrating (4.31), we get the entropy of the system as

η = Cp lnθ +η1 . (4.31b)

Here η1 is a constant. Hence, the logarithm of the potential temperature θ gives the entropy of the air, and enables us to interpret entropy of the dry air in terms of the well-understood concepts of the potential temperature. The entropy or θ and T constitute the coordinates of a tephigram: θ as the ordinate and T as the abscissa; thus isentropes and isotherms are perpendicular to each other in a tephigram.

10 4.10 Clausius-Clapeyron Equation

It is one of the most important equations, which is necessary while discussing the water vapour conversion to liquid and its thermodynamic effects in atmosphere. When air is saturated, an equilibrium condition at a given temperature prevails; that is, the number of molecules due to evaporation leaving the liquid phase equals the number of molecules that return to liquid phase due to condensation from the vapour phase. Evaporation takes place when molecules from the surface of water are breaking away as vapour molecules; condensation occurs when vapour phase molecules collide with the liquid water surface and stick to it. Since the kinetic energy of vapour molecules is more than those of liquid, evaporation requires supply of heat to particles at the water surface. In other words, heat is required to change the liquid phase to vapour. To convert a unit mass of water (phase 1) into 6 -1 vapour (phase 2), the amount of heat required is Lv joules ( Lv = 2.25 x 10 J kg ) which is, q2 u2 α 2 L dQ du pd (4.32) v = ∫ = ∫ + ∫ α q1 u1 α1

For saturated air, p = es, and it is constant throughout the process; therefore, we have from (4.32)

Lv = u2 − u1 + es (α 2 −α1) (4.33 a) Since temperature T is also constant during the phase change process, we may also write q2 q2 dQ η2 L = dQ = T = T dη ⇒ L = T (η −η ) (4.33 b) v ∫ ∫ ∫ v 2 1 q1 q1 T η1 Equating the expression in (4.33a) with that in (4.33b), we obtain

u2 − u1 + es (α 2 −α1) = T (η2 −η1) . (4.34) The above expression on rearrangement gives

u1 + esα1 − Tη1 = u2 + esα 2 − Tη2 (4.35) Let us introduce the Gibbs Free Energy Function (or simply the Gibbs function) G as G = u + esα − Tη (4.36) With the definition of the Gibbs function, eqn. (4.35) can be written as

G1 = G2 (4.37) We thus infer from (4.35) and (4.37) that in an isothermal (T=constant), isobaric (p=constant) phase change, the Gibbs function G, remains constant. The Gibbs function G depends on temperature and pressure, but it remains constant during phase transition. In order to determine the dependence of G on pressure and temperature, we differentiate eqn. (4.36) to obtain,

dG = du + esdα +αdes − Tdη −ηdT (4.38)

Now dQ = du + esdα and dQ = Tdη , thus du + esdα − Tdη = 0 in (4.38) and it becomes,

dG =α des −η dT (4.39)

Because G is same for both phases, i.e. dG1 = dG2 gives

α1des −η1dT = α 2 des −η2 dT (4.40)

de η −η s = 2 1 (4.41) dT α 2 −α1

11 Substituting for η2 −η1 from (4.33) in (4.41), we obtain the following relation de L s = v (4.42) dT T (α 2 −α1)

The equation (4.42) is the Clausius-Clapeyron equation, which expresses the change in saturation pressure es that results from a dT change in temperature. Under ordinary atmospheric conditions, vapour phase specific volume is much larger than that of the liquid phase, i.e. α 2 >> α1 , and moreover, water vapour also behaves like an ideal gas; hence, we can neglect the term α1 in the equation (4.42) to obtain the final form of the Clausius-

Clapeyron equation after substituting for α 2 = RvT / es as

des Lv des Lves = ⇒ = 2 (4.43) dT Tα 2 dT RvT

At temperatures below 0°C, eq. (4.43) expresses the saturation vapour pressure of super- cooled liquid water. However, (4.43) needs modification for expressing saturation vapour pressure of ice. For this case, latent heat of sublimation Ls is substituted in place of Lv and the change of saturation vapour pressure over ice with temperature is calculated from the following relation

⎛ des ⎞ Ls es ⎜ ⎟ = 2 . (4.44) ⎝ dT ⎠ ice RvT At temperature above 0°C, only liquid water and water vapour are in equilibrium. Hence Clausius-Clapeyron equation is the key equation in phase change. −1 −1 In eqn. (4.43), Rv = 461.55 J kg K is constant and to a first approximation, if Lv is also taken as a constant then the straightforward integration of (4.43) yields,

T ⎛ e ⎞ Lv ⎡ 1 ⎤ ⎛ e ⎞ Lv ⎡ 1 1 ⎤ ln s = − ⇒ ln s = − (4.45) ⎜ e ⎟ R ⎢ T ⎥ ⎜ e ⎟ R ⎢T T ⎥ ⎝ so ⎠ v ⎣ ⎦To ⎝ so ⎠ v ⎣ o ⎦

In (4.45), eso = es (T0 ) is the value of saturation vapour pressure at the temperature T0 . It is found that eso = 6.112 hPa for the triple point temperature T0 = 273.16 K . We therefore write (4.45) as

⎪⎧ Lv ⎛ 1 1 ⎞ ⎪⎫ es = eso exp ⎨ ⎜ − ⎟ ⎬ ; eso = 6.112 hPa (4.46) ⎩⎪ Rv ⎝ To T ⎠ ⎭⎪ ⎛ B⎞ e = e exp C − ; C=19.8313 (4.47a) s so ⎝⎜ T ⎠⎟ −B/T 9 es = Ae ; A = 2.53345 ×10 hPa ; B = 5417.1181 K (4.47b)

At subfreezing temperatures, saturation vapour pressures for vapour and ice can be compared from the following expression

e ⎧ L f ⎛ T ⎞ ⎫ s = exp o −1 ; L = L − L (4.48a) ⎨ ⎝⎜ ⎠⎟ ⎬ f s v esi ⎩ RTo T ⎭

12

Numerically, a good approximation to this equation is the following relation, 2.66 es ⎛ 273⎞ = ⎜ ⎟ (T is in Kelvin) (4.48b) esi ⎝ T ⎠

One can compute es (T ) starting with eso(To ) and To = 273.16 K using the Clausius- Clapeyron relation. One can create a look-up table for every δ Τ = 1 K increment in temperature in the above formulae to speed up the calculations. More accurate computation of es (T ) and esi (T ) have been computed by Tetens’ formulae (with T and T0 in Kelvin) as given below,

⎧ T − T0 ⎫ es = es0 exp ⎨17.502 ⎬ (hPa) ; over liquid water (4.49a) ⎩ T − 32.19 ⎭

⎧ T − T0 ⎫ esi = esi0 exp ⎨22.587 ⎬ (hPa) ; over ice (4.49b) ⎩ T + 0.7 ⎭

A more accurate formula for saturated vapour pressure for es (T ) was derived by Bolton (The computation of equivalent potential temperature. Monthly Wea. Rev., vol.108, 1980). The empirical formula with 0.1% accuracy over the temperature range −30°C≤ T ≤ 35°C, is as follows:

⎧ 17.67T ⎫ 0 es (T ) = 6.112 exp ⎨ ⎬ (es in hPa; T in C) (4.49c) ⎩T + 243.5 ⎭ 4.11 Entropy of moist air

Combining the first and second law of thermodynamics in Sec. 4.9, the change in entropy of dry air dηd during a reversible process is expressed as dT dp dη = C − R . d p T d p A similar derivation for moist air can also be accomplished when the reference state is defined with temperature and pressure. In a parcel of moist air, there can be two isolated phases, viz., the gas and liquid water. In a unit kilogram of mixture, let there be md kg of dry air, mv kg of water vapour and mw kg of liquid water. In the mixture, the total pressure

p = pd + e is exerted on liquid water when pd is pressure of dry air and e that of water vapour. Hence the entropy of the moist air mixture is given by

η = Sd + Sv + Sw = mdηd + mvηv + mwηw (4.50)

Also, dη = dSd + dSv + dSw (4.51a)

To compute dη , we need to compute each component of the equation (4.51), thus

dSd = d(mdηd ) = mddηd +ηddmd = md dηd ; dmd = 0

⎛ dT dp ⎞ Hence d . dSd = md ⎜ Cp − Rd ⎟ ⎝ T pd ⎠

Next, we calculate dSv and dSw as

dSv = mvdηv +ηvdmv and dSw = mwdηw +ηwdmw .

13 dT de dT de Now dη = C − R , therefore dS = m {C − R } +η dm ; and it is easy v pv T v e v v pv T v e v v dT to get dS = m C +η dm . w w w T w w

Thus an expression for, dη (= dSd + dSv + dSw ) is the following:

dT dp de d . (4.51b) dη = (mdCp + mvCpv + mwCw ) − md Rd − mv Rv +ηvdmv +ηwdmw T pd e

Since a parcel is assumed to be isolated with its surroundings, an increment dmw in the water content of the parcel can only happen at cost of an equal amount of conversion water vapour; therefore, we have a decrease dmv such that, dmw = −dmv . From this fact, the final expression for the change in entropy of the mixture (air + WV + liquid), reads as

dT dp de d . (4.52) dη = (mdCp + mvCpv + mwCw ) − md Rd − mv Rv + (ηv −ηw )dmv T pd e

From equation (4.33b), we can calculate the change in entropy due to heat added as a result L of conversion of water vapour to liquid using η −η = v in eqn. (4.52) to obtain v w T

dT dp de L d v (4.53) dη = (mdCp + mvCpv + mwCw ) − md Rd − mv Rv + dmv T pd e T Integrating the expression in (4.53) on both sides, we obtain the following result for the entropy of the moist air, ⎛ T ⎞ ⎛ p ⎞ ⎛ e ⎞ L η = η + (m C + m C + m C )ln − m R ln d − m R ln + v m (4.54) o d pd v pv w w ⎜ T ⎟ d d ⎜ p ⎟ v v ⎜ ⎟ T v ⎝ o ⎠ ⎝ o ⎠ ⎝ es ⎠

The expression (4.54) considers that water phase and vapour are separate in the parcel with e as the vapour pressure. However, such a system could not be in equilibrium, and liquid water in the parcel will freely evaporate until air in the parcel attains saturation; and at this state the vapour pressure in the parcel is e = es . From (4.54) we can, therefore, obtain the entropy of a system with air at saturation vapour pressure es at the given temperature T; and surely, this case is interesting. Substituting e = es in the expression (4.54) for η , we obtain an expression for saturated air parcel with mv as water vapour and mw as liquid water, ⎛ T ⎞ ⎛ p ⎞ L d v (4.55) η = ηo + (mdCpm + mwCw )ln⎜ ⎟ − md Rd ln⎜ ⎟ + mv ⎝ To ⎠ ⎝ po ⎠ T

In (4.55), Cpm as defined in (4.16) is the specific heat of the mixture. Now, consider a saturated air-water mixture in a parcel at pressure p and temperature T with 1 kg of dry air,

ξ kg of liquid water and saturation vapour pressure es . The entropy of such a saturated moist air parcel with liquid water present, can be obtained from expression (4.55) by setting

md = 1 kg ; mw = ξ kg and pd = p − es ; the desired expression is

L η = (C + ξ C )lnT − R ln(p − e ) + v m + const. (4.56) pm w d s T v

14 The expression for the entropy of the saturated moist parcel given by (4.56) is much more complicated as compared to that of a dry air parcel.

4.12 Processes that lead to saturation: A sample of moist air may attain saturation in several ways in the atmosphere. But primarily there are five ways for the air to become saturated, and these are discussed in what follows.

(i) Cooling of the air: When moist air is cooled, it will attain a temperature (the so called dew-point temperature) where it becomes saturated. At the dew-point temperature Td , the mixing ratio r(T, p) is equal to rs (Td , p) . In view of (4.47b), we have

ε e (T ) ε ⎪⎧ ⎛ B ⎞ ⎪⎫ s d r(T, p) = rs (Td , p) = = ⎨Aexp⎜ − ⎟ ⎬ p p ⎩⎪ ⎝ Td ⎠ ⎭⎪

The above equation can be solved for obtaining Td as

B T = (4.57) d ln{Aε / (r p)}

(ii) Cooling of air by evaporating water within the parcel at constant pressure: The wet- bulb temperature Tw is defined as the temperature to which air must be cooled at constant pressure by evaporating water that is lodged in the given parcel. One can also derive an expression for Tw by considering a sample of one kilogram of dry air and w kilograms of water vapour. The specific heat of the mixture then calculates to

C + C w C = p pv = C [1+ 0.94w] pm 1+ w p Heat loss due to evaporation of dw mass of water in the parcel is calculated as

(1+ w)dq = (1+ w)Cpm dT = (1+ w)(1+ 0.94 w)Cp dT which is same as the amount of heat (= −Lvdw ) used in evaporating mass dw . Equating the two, we have

(1+ w)(1+ 0.94 w)Cp dT = −Lvdw L Or, C dT = − v dw = −(1−1.9w)L dw p (1+ w)(1+ 0.94w) v One may neglect the correction factor (1−1.9w) in the above expression, and we obtain dT L − = v (4.58) dw Cp

At the initial temperature T the mixing ratio is w and, at Tw air attains saturation when dw mass of water has evaporated. Therefore, for a change in temperature dT = Tw − T with a corresponding change in mixing ratio dw = ws (p,Tw ) − w(p,T ) , from (4.58), we have

T − T L T − T L − w = v or w = v . ws (p, Tw ) − w(p, T ) Cp ws (p, Tw ) − w Cp

15 Then, Tw is given by

Lv Tw = T − [ws (p,Tw ) − w] . Cp

The saturation mixing ratio ws can be expressed in terms of saturation vapour pressure at the temperature Tw ; so we obtain the final expression for Tw as

Lv ⎧εes (Tw ) ⎫ Tw = T − ⎨ − w⎬ (4.59) Cp ⎩ p ⎭ The expression (4.59) allows us to calculate a temperature if all the water vapour in the sample of air were condensed out. In this case, the initial conditions are temperature T and mixing ratio w ; and when all water has condensed out, the system attains a temperature Te .

Hence in (4.59), we substitute Tw = Te and ws = 0 and we get

Lvw Te = T + . (4.60) Cp

From eqn. (4.60) a new temperature Te is defined which is called the equivalent temperature.

(iii) Adiabatic cooling of moist air: When air parcels at a state (T0 , p0 ) rise in the atmosphere, their temperatures decrease according to dry adiabatic lapse rate Γd . If the initial mixing ratio of water vapour is w , then parcels cooled adiabatically due to ascending motions in the atmosphere, will reach to saturation when ws = w . In other words, parcels cool adiabatically until they attain saturation where ws = w and p = pc and parcels reach the final state (Tc, pc, ws ) due to lifting. Dry adiabats at pressure pc , along which parcels are rising, would intersect the saturation vapour line ws = w . Note that pc is the isentropic condensation pressure, also known as the lifting condensation level (LCL) related to temperature Tc as Tc = Td (w, pc ), ws = w . This allows the temperature Tc at the LCL to be calculated as, B Tc = (4.61) ⎧ Aε ⎫ ln ⎨ ⎬ ⎩w pc ⎭

Hence, parcels rising adiabatically from their initial surface temperature T0 , when become saturated, their temperature is Tc . At surface, parcels are at the potential temperature

θ = T0 where p = p0 . The dry adiabatic from T0 will intersect the saturation vapour line

ws = w at pressure p = pc . But pc is still unknown. In order to calculate pc , we use the fact that parcels also have potential temperature θ at this level, because in adiabatic ascents θ is conserved. Now, the potential temperature θ at the LCL is given by

κ ⎛ p ⎞ R o θ = Tc ⎜ ⎟ , κ = ; also θ = To at p = po. ⎝ pc ⎠ Cp

16 κ Since Tc and T0 are on the same adiabat, so (Tc /To ) = ( pc / po ) , which gives

(Cp /R) pc = po (Tc /T0 ) . Substituting the value of pc in (4.61), we obtain the temperature at the lifting condensation level as, B Tc = 1 (4.62) ⎧ k ⎫ ⎪ Aε ⎛ To ⎞ ⎪ ln ⎨ ⎜ ⎟ ⎬ wpo ⎝ Tc ⎠ ⎩⎪ ⎭⎪

The temperature of the parcel at LCL, Tc (also called the adiabatic condensation temperature), is obtained by solving eqn. (4.62) iteratively. On the tephigram, LCL is located at the intersection of θ = const. and θes = const.

(iv) Horizontal mixing: This kind of mixing occurs when two air masses of different provenance travel far and meet over another region. Generally, air masses retain their characteristics with clearly identifiable boundary as a front, but their mixing produce complex weather associated with large scale clouding and widespread rains. During winters, the western disturbances visiting northwest India, originate over the Mediterranean Sea as extra-tropical frontal systems, move over western Himalayas, Pakistan and parts of north India with the westerlies. A western disturbance is a low pressure system which can be identified by a well-marked trough in the cold westerlies at 500 hPa. The cyclonic circulation in the lower troposphere associated with a western disturbance would produce mixing of warm moist and cold dry air masses that results in much needed precipitation (winter rains) and relief from severe cold. But when temperatures dip during nights in winter, the relative humidity rises and it produces large scale fog in the Indo-Gangetic Plain along the foothills of Himalayas which spreads eastward to distances as long as 2000 km. The fog cover in this region is dense and persistent which is clearly identifiable on the INSAT/KALPANA satellite imageries. The persistent fog often turns into a permanent haze due to the presence of carbonaceous aerosols from the biomass burning in this region. With such meteorological consequences of mixing, not only over India but elsewhere also, it is very essential to understand the mechanism of horizontal mixing in the atmosphere.

Fig. 4.3 Two unsaturated parcels A and B mix in the atmosphere. The curve separates the saturated and es unsaturated air. After mixing of the air masses typically with e Saturated s characteristics of parcels A and B, the mixture becomes e B supersaturated with resultant temperature T at C. Thus C mixing of air masses of different provenance is a dominant unsaturated weather producing mechanism for various phenomena that include fog, rainfall and thunderstorms. A T

Horizontal mixing can easily be explained on a thermodynamic diagram. The mixing of two air parcels is shown in Fig. 4.3 at the same pressure p but different temperatures and vapour pressures. This kind of mixing in the atmosphere is called the isobaric mixing.

17 Assume that parcel A at a thermodynamic state ( p, T1, e1 ) and parcel B at ( p, T2 , e2 ) mix on an isobaric surface according their masses. To further simplify matters, assume that parcels are of equal masses. After mixing, the parcels A and B have same temperature and vapour pressure (T , w) calculated as

T + T w + w εe εe T = 1 2 , w = 1 2 ; with w = 1 , w = 2 (4.63) 2 2 1 p 2 p pw The vapour pressure of the mixture e T is calculated to be e = e T = ; Hence Fog ( ) ( ) ε would form, if vapour pressure exceeds the saturation vapour pressure ; that is, e(T ) es (T ) when the air mixture is supersaturated, e T e T . ( ) > s ( )

(v) Vertical mixing: Consider mixing of parcels due to instability of an atmospheric layer; that is, Γ > Γd . Also the transfer of heat by mixing of the parcels would increase the entropy

(related to θ) of the system. Let us consider a parcel of air of mass m1 at temperature T1 , pressure p1 and potential temperature θ1 be mixed with a parcel of air mass m2 at temperature T2 , pressure p2 and potential temperature θ2 . Assume that after mixing the resultant pressure of the mixture is p . The mixing of these two parcels could be assumed to have proceeded as a two-step process: Step 1: The parcels are brought adiabatically to pressure p . Then, their respective temperatures at pressure p can be calculated immediately from their respective potential temperatures θ1 and θ2 as R R ⎛ p ⎞ Cp ⎛ p ⎞ Cp (4.64) T1(p) = θ1 ⎜ ⎟ and T2 (p) = θ2 ⎜ ⎟ ⎝ po ⎠ ⎝ po ⎠

Step 2: We let the two masses in the parcels mix, then the mean temperature Tm (p) of the mixture is calculated as

m1T1(p) + m2T2 (p) Tm (p) = m1 + m2

On substituting for T1(p) and T2 (p) from (4.64) in the RHS of the above equation we get, R m θ + m θ ⎛ p ⎞ Cp 1 1 2 2 (4.65) Tm (p) = ⎜ ⎟ m1 + m2 ⎝ po ⎠

Thus the potential temperature θm (p) of the mixture is given by

m1θ1 + m2θ2 θm = = θmean (4.66) m1 + m2 Since θ is a conserved variable so one can perform the calculations taking unit masses; that is, we can take m1 = m2 = 1 kg mass. Hence, it may be remarked that the potential temperature of the mixture of two air masses after their mixing, is the arithmetic mean of potential temperatures of the original unmixed air masses.

18 4.13 Saturated Adiabatic Lapse Rate (SALR)

The rising parcels become saturated as temperature in the troposphere decreases with height. Moreover, temperature of the parcel with height falls at first with the dry adiabatic lapse rate ( Γd = g / Cp ), but once air in rising parcel becomes saturated, its lapse rate changes significantly and the temperature of the saturated air parcel shall change with height at the saturated adiabatic lapse rate ( Γ s ). When water vapour condenses out in a rising air parcels, we have Γ s < Γd due to release of latent heat. Moreover, typical values of Γ s vary -1 -1 between 4-7 K km . In a saturated atmospheric layer near the ground, Γ s ~ 4K km ; -1 whereas Γ s ~ 6-7 K km in the middle troposphere. On the tephigram, curved lines depicting the rise of air parcels under saturated conditions are called saturated adiabats (or pseudoadiabats). The equivalent potential temperature remains constant along these curved lines. The temperature change of sinking saturated parcels also happens along pseudoadiabats. For computing the saturated lapse rate in the atmosphere, the saturation mixing ratio of rising air parcels need to be considered in the derivation. Let us begin with the entropy of dry air is given by

η = C lnθ +η = C lnT − R ln p +η (4.67) p 1 p d 1

Next, consider that an air parcel is lifted vertically with both vapour and liquid water present in it. The conversion of water vapour to liquid water is accompanied by the release latent heat, which is added to the rising parcel. If ξ is the total water in a parcel and it assumed to remain in the parcel, then air will remain saturated because both evaporation and condensation will maintain equilibrium. If rs is the fraction of water in the vapour form, then ξ − rs is the amount of liquid water present in the parcel. In such a situation, note that rs is the saturation mixing ratio of the parcel; consequently, ρ εe r v s 0.622 ; p R T and R T s =  (ε = ) − es = ρa d es = ρv v ρa p

When saturated air rises, the latent heat released during the condensation process must be added to the system. Thus the entropy of water vapour and liquid water is the entropy of ξ L amount of liquid water at temperature T plus the additional entropy v required to convert T

rs amount of water to vapour phase. The entropy η of saturated air is derived in (4.56) is L r C + ξ C )lnT − R ln(p − e + v s +η = η . (4.68) ( p w d s ) T 1

The quantity represents the specific heat of water. Also, remains constant for a Cw η saturated air parcel rising adiabatically; hence differentiating (4.68), we get

dT d(p − e ) L r s ⎛ v s ⎞ ; which is written as (Cp +ξ Cw ) − Rd + d ⎜ ⎟ = dη = 0 T p − es ⎝ T ⎠

dT dp R de L r d s ⎛ v s ⎞ . (Cp +ξ Cw ) − Rd + dT + d ⎜ ⎟ = 0 T p − es p − es dT ⎝ T ⎠

The above equation may be further simplified and put in the following form,

19 dT Rd des ⎛ Lvrs ⎞ dp Cp +ξ Cw + dT + d ⎜ ⎟ − Rd = 0 (4.69) ( ) T p − e dT ⎝ T ⎠ p

The hydrostatic balance also applies for the moist air (vapour + liquid), so we have, dp dp g = −gρ(1+ ξ) ⇒ R = − (1+ ξ)dz (4.70) dz p T

The last term in (4.69) can be replaced using (4.70) and we obtain,

dT Rd des ⎛ Lvrs ⎞ g Cp +ξ Cw + dT + d ⎜ ⎟ + (1+ ξ)dz = 0 (4.71) ( ) T p − e dT ⎝ T ⎠ T

εe dr ε de dT εe dp Since r = s ⇒ s = s − s . (4.72) s p dz p dT dz p2 dz

From Clausius-Clapeyron equation, we have

d es Lv es = 2 (4.73) dT Rv T R Using (4.70), (4.72), (4.73) and taking ε = d , we may arrange (4.71) to get an expression Rv for the saturated lapse rate of temperature, Γs , if air is just saturated (ξ = rs ), as

⎧ ε es ⎫ ⎧ Lv es ⎫ ⎨1+ ⎬ ⎨1+ 2 ⎬ dT g p p R T − = Γ = ⎩ ⎭ ⎩ v ⎭ (4.74) dz s C 2 p ⎡ ε es dLv ε Lv es ⎤ ⎢1+ Cw + + 2 ⎥ ⎣ pCp { dT } Cp p Rv T ⎦

Since es << p, the second term in the denominator on the RHS of eqn. (4.74) is negligible in comparison to other two terms. The equation (4.74) can be written in a more convenient form as

⎧ Lv es ⎫ ⎨1+ 2 ⎬ dT p R T Γ = − = Γ ⎩ v ⎭ (4.75) s dz d 2 ⎪⎧ ε Lv es ⎪⎫ ⎨1+ 2 ⎬ ⎩⎪ Cp p RvT ⎭⎪

g ε es We have used in the above equation Γd = = DALR and << 1. Cp p The saturated adiabatic apse rate (SALR) could also be derived from the expression of moist static energy (MSE) given in eqn. (4.24) which represent the sum of enthalpy, potential energy and latent heat ( Lvrs ) of a moist parcel. The MSE of a saturated air parcel,

hs is given as,

hs = CpT + Φ + Lvrs

We have already pointed out that hs is a conserved quantity; therefore, during saturated adiabatic ascent,

20 dhs = 0 ⇒ d (CpT + Φ + Lvrs ) = 0

Which gives the following differential form, if Lv is treated as a constant; that is,

CpdT + gdz + Lvdrs = 0 εe Now r = s ; taking logarithm on both sides and differentiating the result we get the s p following expression,

dr de dp dr 1 de g s = s − ; or we write it as s = s dT + dz . rs es p rs es dT RdT

1 des Lv On using the Clausius-Clapeyron equation, = 2 , the preceding expression takes es dT RvT the final form

drs Lv g = 2 dT + dz . (4.76) rs RvT RdT

From (4.76), the expression for drs can be substitute to eliminate it in C-C equation, and we get the equation, ⎛ L g ⎞ C dT gdz L r v dT dz 0 . p + + v s ⎜ 2 + ⎟ = ⎝ RvT RdT ⎠

The above equation can be arranged as follows,

2 Lvrs Lvrs (Cp + 2 ) dT + g(1 + )dz = 0 RvT RdT

The above equation now gives the saturated adiabatic lapse rate, Γs as follows

⎧ L r ⎫ ⎨1 + v s ⎬ dT R T Γ = − = Γ ⎩ d ⎭ (4.77) s dz d ⎧ 2 ⎫ ⎪ Lvrs ⎪ ⎨1+ 2 ⎬ ⎩⎪ Cp RvT ⎭⎪

The expression for Γs in (4.77) is same as that in equation (4.75), but the derivation (4.77) is very fast. However, in deriving (4.77) it has been assumed that the air is saturated and there is no liquid water present in the parcel, hence the previous derivation is much more general. Note that, Γs < Γd with bounds as 0.3 Γd < Γs < Γd in the atmosphere. ⎛ dT ⎞ If the actual lapse rate or the so-called ambient lapse rate − is given by γ , then we ⎝⎜ dz ⎠⎟ have the following states of stability of moist air:

(i) γ < Γs Absolutely stable atmosphere (iv) γ = Γd Dry neutral atmosphere

(ii) γ = Γs Saturated neutral atmosphere (v) γ > Γd Absolutely unstable

(iii) Γs < γ <Γd Conditionally unstable

21 When the atmosphere is unstable with respect to pseudoadiabatic displacements of parcels, there is a possibility of condensation leading to rainfall and thunderstorms. In numerical models the instability is removed by adjusting the temperature profile. Starting from the given temperature profile at a grid point of the model, the lapse rate is computed by evaluating discrete derivatives of temperature in the vertical in order to identify unstable layers in the atmosphere. In an unsaturated atmospheric layers where γ > Γd , instability is removed by adjusting the lapse rate with the dry adiabatic. But in the saturated unstable layers, the lapse rate is adjusted with the Γs as calculated from (4.77) at any level in such a manner that excess moisture rains out and the latent heat thus liberated is used in updating the temperature at that level. This is truly a very simplistic approach that is hardly used in present day numerical models but a clear insight can be gained from such simple computations related to moist parcel movements in the vertical. It may be mentioned here that the conditional instability criterion (iii) is also known as the "conditional instability of first kind", but it takes longer times for such a condition to develop in the atmosphere and, moreover, it can only explain amplification of disturbances on a scale of a few kilometers. However, rapid weather developments happen leading to rain and thundershower even in the dry atmosphere on a scale of 100-1000 km in a short span of time. Such rapid developments can be explained on the basis of "conditional instability of second kind" that arises over a region due to moisture convergence forced by friction in the planetary boundary layer in the large scale flow (J. Charney and A. Eliassen, 1964, On the growth of the hurricane depression. J. Atmos. Sci., 21). Later, H.L. Kuo (1974, Further studies of the parameterizations of the influence of cumulus convection on large scale flow. J. Atmos. Sci., 31) proposed a scheme to parameterize cumulus convection where the moisture supply into convective element is due to the effect of large moisture convergence in the planetary boundary layer. Indeed, CISK describes the cooperation (rather than competition) of subgrid scale and large scale processes in the atmosphere. Numerical weather prediction witnessed rapid advances when Kuo’s cumulus parameterization scheme was included in the forecast models. It may nevertheless be remarked that models that use the mass-flux schemes, produce accurate numerical weather forecasts. These models are also used even for climate change projections. Parameterization of convection – deep, shallow or cumulus – is a vast topic of current research especially for their use in high resolution weather prediction models.

22