The Stanar Atmosphere Dr Anrew French 1
The International Stanar Atmosphere (ISA) is an iealize moel of the variation of average air pressure an temperature with altitue. Assumptions: The atmosphere consists of a number of fixe layers, each with a constant temperature graient or lapse rate. Distinct layers which make up the Stanar Atmosphere moel The atmosphere consists of a single ieal gas, whose molecular mass takes into account the actual composition of ifferent gases (e.g. Nitrogen, Oxygen, Carbon Dioxie, water vapour etc). Gas composition of ry air 2
Note in the thermosphere the temperature (i.e. mean kinetic energy of air molecules) can rise to several thousan egrees Celsius. However, the air ensity is extremely low, which means it woul still feel very col inee, as very little heat coul be extracte from a given volume of air. 3
Determining air pressure Consier a 1m 2 horizontal cross section parcel of air of ensity ρ at an altitue z, with vertical with z. The atmospheric pressure change P between altitues z an z+z resulting from the removal of the air parcel from the total weight of air above is: P ρ gz P + P 2 1m z z P Air weight per unit area is 2 1m z g ρ ean sea level 4
Determining air pressure in the ISA Let us assume that the air column is comprise of ry air with molar mass.2896kgmol 1 Let us assume that the air column is an ieal gas. If n moles of gas occupies volume V at pressure P (Pascals) an (absolute) temperature T (Kelvin): PV nrt Ieal gas equation P n RT V The ensity of the air is the mass of n moles ivie by the volume ρ n V Hence: ρ P RT 5
Change in weight of air column Ieal gas P ρ gz ρ P RT P gz P R T P P 1 z g P P R z T z ln P z g z P R z T 6
To account for changes in gravitational fiel strength g with altitue, let us efine a geopotential altitue h. h zgr E ( + ) g z R E R E 6378.137 km 1 1 { 1 cos( 2λ ) cos 2 ( 2λ )} g g a + b g g a b 9.8665ms 9.8616ms.26373.59 2 2 7
So g is now a function of latitue only, an our pressure integral is now in terms of geopotential height h ln P g 1 P R h T h h Now let us efine the temperature T (in Kelvin) to be a linear function of geopotential altitue. The (negative) graient is efine to be the lapse rate L. Note this means a negative lapse rate implies a rise in temperature with height. T T L ( h h ) T h, T ( ) T ( h, T ) h h 8
Case 1: Isothermal layer i.e. L T T P ln - P g RT h h h ( h P ), P P e g - h h RT ( ) P P e - g h h RT ( ) h h 9
Case 2: Constant but non-zero, lapse rate L T T L( h h ) P g ln - h 1 h ( ) h P R T L h h P ln P g h L ln - h P ( L ) R h T + Lh Lh g ( T Lh Lh) h ln + h P LR P g T + Lh Lh ln ln P LR T P T Lh Lh ln ln + P T g LR P h P 1 ( h P ), L h P P 1 T ( h ) L h T ( h ) g LR g LR h 1
http://en.wikipeia.org/wiki/international_stanar_atmosphere 11
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To generate the Pressure an Temperature vs altitue graphs on the previous slie: 1. State a esire range of altitue /km 2. Convert to geopotential altitue using h zgr E ( + E ) 1 cos( 2λ ) cos 2 ( 2λ ) g z R 1 { } g g a + b g 9.8665ms, g 9.8616ms 2 2 1 a.26373, b.59 Note temperature T in egrees Celsius is: 273 + T Kelvin 3. Determine ISA layer base altitue an lapse rate 4. Determine Temperature an Pressure at the base of the layer (i.e. T an P ) 5. Compute pressure an temperature T T T T L( h h ) L P L P e g - h h RT ( ) P P 1 ( h ) L h T g LR 13
What about humiity? i.e. the average pressure, temperature structure of the air column Any sensible escription of non-ari climatology will nee to take into account the impact of variable amounts of water vapour containe within the air column. In the Troposphere at least, the presence of water vapour can have a ramatic influence upon thermoynamic variables such as temperature an pressure, an is obviously a funamental component of weather phenomena such as clou an fog. To moel the effect of humiity upon temperature an pressure, let us moify our original single ieal gas assumption to consier a composite of ry air an water vapour. The molar masses of ry air an water vapour are, respectively: v.2896kgmol.182kgmol 1 1 The respective ieal gas equations are, for n moles of ry air an n v moles of water vapour: P P + P P V V n RT V P V i.e. the overall pressure P is the sum of the partial pressures of the component gases n RT V 14
Let us efine relative humiity U as the ratio of water vapour pressure to that at saturation E s P V UE s An empirical moel for saturation pressure as a pure function of temperature is given by the Aren Buck equation http://en.wikipeia.org/wiki/aren_buck_equation Es where the saturation vapour pressure E s is given in mbar an temperature in egrees Celsius T T 18.678 234.5 T + 257.14 6.1121 e T T E s /mbar There are in fact many empirical formulae for vapour pressure. A goo selection are compare at http://cires.colorao.eu/~voemel/vp.html Note: 1 atm is 113.25 mbar Note the vapour pressure is very small compare to the pressure of ry air when temperatures are low. Therefore to a very goo approximation we can ignore it in the pressure calculation. However, the lapse rate is significantly affecte by humiity... T / o C
The overall ensity of the atmosphere is: n + n ρ n + n V V V V V V From the ieal gas equations: V V nv PV UES RT RT V V V n P P P P UE RT RT RT ( ) ( ) V S Hence: V ρ P U 1 Es ( T ) RT 16
The pressure integral is now more complex, but can be evaluate using a numerical metho: P V ρ gh ρ P U 1 E ( T ) RT P g V P U 1 Es ( T ) h RT s T T L( h h ) A simple iterative numeric solution scheme might be to use a finite geopotential altitue change h, an start from a known temperature an pressure e.g. 15 o C, 113.25mbar. h h + h T T L h E s 6.1121 e T T 18.678 234.5 T + 257.14 g V P P U 1 Es ( T ) h RT P P + P v.2896kgmol.182kgmol 1 1 17
Now the partial pressure of water vapour is typically very small compare to ry air. Therefore we can ignore it in pressure calculations. However, the lapse rate L is significantly affecte by the presence of water vapour. A moel for L is: T L h L g v c p 2 +.2896kgmol.182kgmol g 9.876ms Lapse rate r H RsT s 2 R T s 1+ ( H ) 2 1 1 http://en.wikipeia.org/wiki/lapse_rate s V r r P T is in Kelvin E s E s Specific latent heat of vaporization of water H s 2,51, J kg -1 s Specific heat of ry air at constant pressure is: c p 13.5 Jkg -1 K -1 Specific gas constant for ry air R s 287 J kg -1 K -1 R s R 8.314 287.2896 Essentially the molecular fraction of water to ry air 18
Hence rule of thumb : ry air lapse rate is about 1 o /km, wet air lapse rate is about 6 o /km So humiity only makes a significant ifference at low pressures i.e. at high altitues. At this point the assumption of a significant relative humiity is likely to be invali anyway! 19
The ew point is the temperature to which a given parcel of air must be coole, at constant barometric pressure, for water vapour to conense into water. The August-Roche agnus approximation efines the ew point to be efine by the following expression in terms of relative humiity U (with values 1) an ambient air temperature T. i.e. the air parcel being coole will be coler than the ambient air! Note in the formula below, T is efine in egrees Celsius. T a b at b lnu + b + T at The ew point is the temperature at a lnu b + T 17.625 243.4 which the water vapour in a sample of air at constant barometric pressure conenses into liqui water at the same rate at which it evaporates http://en.wikipeia.org/wiki/dew_point http://anrew.rsmas.miami.eu/bmcnoly/humiity.html 2
Why clous form Clous contain saturate air, i.e. where liqui water has conense out of its vapour phase. So how is this buoyant? Surely the ensity must excee the air aroun it? aiabatic means no heat transfer The explanation is that the air within a clou is, on average, warmer than the air outsie it. This is because energy is liberate uring the phase transition from vapour to liqui. A clou therefore contains ry air that is less ense than the surrouning air. This warmer air acts as a sort of floatation evice for the conense water within the clou. The clou top is when the overall ensity insie the clou equals to the (ry air) ensity outsie. The clou base is when the local air temperature reaches the ew point an conensation occurs. One can moel temperatures this by consiering a moist lapse rate for the clou an a ry lapse rate for the surrouning air. 21
What about boiling point? To boil water, it must unergo a phase transition from liqui to gas. This requires a certain amount of heat, the latent heat of vaporization, to break the inter-molecular bons inherent in the water. 22
The graient of a phase transition line in the P,T iagram is given by the Clausius-Clapeyron equation: Pressure in Pascals P T Temperature in Kelvin Latent heat /Jmol -1 H T V Volume change of 1 mole of substance uring the phase transition In a liqui to gas transition we can assume the volume change is sufficiently large as to ignore the original flui volume. If one assumes the resulting gas is ieal: RT V P Hence: P H 2 P T RT 23
We can use this relationship to etermine the liqui-to-gas line in the P,T iagram if the latent heat of vaporization is assume to be temperature inepenent. (In reality this is not the case, but is less of an issue at low temperatures see next slie). P H 2 P T RT P 1 H T 1 P P* P R T* T P H 1 1 ln P R T T P T * * Pe * H 1 1 R T T* 2 T 1 R P ln T* H P * 1 Since the temperature correspons to the liqui-to-gas transition line, we can therefore erive an expression for the boiling point of a liqui relative to ambient temperature T an pressure P, as long as one fixe boiling point T * an corresponing pressure P * is known. 24
T boil 1 R P ln T* H P * 1 Boiling point of water at ifferent atmospheric pressures /mbar. Latent heat of vaporization of water is H 4.7 kj mol -1 at 1 o C an 113.25 mbar ambient air pressure. At typical troposphere temperatures, a higher value of 45.7 kj mol -1 is use in some calculations, such as the lapse rate moel quote earlier. http://en.wikipeia.org/wiki/lapse_ rate http://en.wikipeia.org/wiki/enthal py_of_vaporization http://en.citizenium.org/wiki/heat_ of_vaporization 25
In summary, we can moel the variation of pressure, temperature, lapse rate boiling point an ew point with altitue using the following iterative scheme: T 15 C T 1 C P P 113.25mbar h.1km o o * * h h + h T T L h E s 6.1121e T T 18.678 234.5 T + 257.14 Es r P E g V P P U 1 Es ( T ) h RT P P + P 1 at 1 R P b ln U + Tboil ln b + T T* H P Tew * at a lnu b + T s L g c p + r H RsT s 2 R T s 1 + ( H ) 2 s V r Note this works out T in Kelvin a b 17.625 243.4 26
Hence rule of thumb: ry air lapse rate is about 1 o /km, wet air lapse rate is about 6 o /km 27
Boiling point of water vs altitue 28