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1 Module 4 Water Vapour in the Atmosphere 4.1 Statement of The 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 ) .
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