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M.Sc. in Meteorology Physical Meteorology Part 2 Atmospheric

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M.Sc. in Meteorology Physical Meteorology Part 2 Atmospheric M.Sc. in Meteorology Physical Meteorology Part 2 Prof Peter Lynch Atmospheric Thermodynamics Mathematical Computation Laboratory Dept. of Maths. Physics, UCD, Belfield. 2 Atmospheric Thermodynamics Outline of Material Thermodynamics plays an important role in our quanti- • 1 The Gas Laws tative understanding of atmospheric phenomena, ranging from the smallest cloud microphysical processes to the gen- • 2 The Hydrostatic Equation eral circulation of the atmosphere. • 3 The First Law of Thermodynamics The purpose of this section of the course is to introduce • 4 Adiabatic Processes some fundamental ideas and relationships in thermodynam- • 5 Water Vapour in Air ics and to apply them to a number of simple, but important, atmospheric situations. • 6 Static Stability The course is based closely on the text of Wallace & Hobbs • 7 The Second Law of Thermodynamics 3 4 We resort therefore to a statistical approach, and consider The Kinetic Theory of Gases the average behaviour of the gas. This is the approach called The atmosphere is a gaseous envelope surrounding the Earth. the kinetic theory of gases. The laws governing the bulk The basic source of its motion is incoming solar radiation, behaviour are at the heart of thermodynamics. We will which drives the general circulation. not consider the kinetic theory explicitly, but will take the thermodynamic principles as our starting point. To begin to understand atmospheric dynamics, we must first understand the way in which a gas behaves, especially when heat is added are removed. Thus, we begin by studying thermodynamics and its application in simple atmospheric contexts. Fundamentally, a gas is an agglomeration of molecules. We might consider the dynamics of each molecule, and the inter- actions between the molecules, and deduce the properties of the gas from direct dynamical analysis. However, consider- ing the enormous number of molecules in, say, a kilogram of gas, and the complexity of the inter-molecular interactions, such an analysis is utterly impractical. 5 6 I. The Gas Laws Again, the gas law is: pV = mRT The pressure, volume, and temperature of any material are related by an equation of state, the ideal gas equation. For The value of R depends on the particular gas. most purposes we may assume that atmospheric gases obey For dry air, its value is R = 287 JK−1 kg−1. the ideal gas equation exactly. Exercise: Check the dimensions of R. The ideal gas equation may be written Since the density is ρ = m/V , we may write pV = mRT p = RρT . Where the variables have the following meanings: Defining the specific volume, the volume of a unit mass of p = pressure (Pa) gas, as α = 1/ρ, we can write 3 V = volume (m ) pα = RT . m = mass (kg) T = temperature (K) R = gas constant (J K−1 kg−1) 7 8 Special Cases Avogadro’s Hypothesis Boyle’s Law: We may write One mole (mol) of a gas is the molecular weight in grams. mRT One kilomole (kmol) of a gas is the molecular weight in V = . p kilograms. For example, the molecular weight of nitrogen For a fixed mass of gas at constant temperature, mRT is N2 is 28 (we ignore the effects of isotopic variations). So: constant, so volume is inversely proportional to pressure: One mole of N2 corresponds to 28 g V ∝ 1/p . One kilomole of N2 corresponds to 28 kg According to Avogadro’s Hypothesis, equal volumes of dif- ferent gases at a given temperature and pressure have the Charles Law: We may write same number of molecules; or, put another way, gases with mR V = T. the same number of molecules occupy the same volume at p a given temperature and pressure. For a fixed mass of gas at constant pressure, mR/p is con- stant, so volume is directly proportional to temperature: V ∝ T. 9 10 The number of molecules in a mole of any gas is a universal Then the gas law may be written in the form normally found constant, called Avogadro’s Number, NA. The value of NA in texts on chemistry: 23 is 6.022 × 10 . So: pV = nR∗T. 28 g of nitrogen contains N molecules of N A 2 ∗ −1 −1 3 with n the number of moles of gas and R = 8.3145 JK mol . 28 kg contains 10 × NA molecules. For a gas of molecular weight M, with mass m (in kilograms) The gas constant for a single molecule of a gas is also a the number n of kilomoles is universal constant, called Boltzmann’s constant, k. Since ∗ m the gas constant R is for NA molecules (the number in a n = . M kilomole), we get R∗ So, we use m = nM in the gas law to write it k = NA pV = n(MR)T Now, for a gas containing n0 molecules per unit volume, the By Avogadro’s hypothesis, equal volumes of different gases equation of state is at a given temperature and pressure have the same number p = n0kT . of molecules. Therefore, the value of MR is the same for any gas. It is called the universal gas constant, denoted: R∗ = MR = 8.3145 JK−1 mol−1 = 8314.5 JK−1 kmol−1 . 11 12 The numerical values of Rd and Rv are as follows: Virtual Temperature ∗ ∗ R −1 −1 R −1 −1 The mean molecular weight M of dry air is about 29 Rd = = 287 JK kg ,Rv = = 461 JK kg . d Md Mv (average of four parts N2 (28) and one part O2 (32)). We define the ratio of these as: The molecular weight Mv of water vapour (H2O) is about Rd Mv 18 (16 for O and 2 for H2). ε ≡ = ≈ 0.622 . Rv Md Thus, the mean molecular weight, Mm, of moist air, which is a a mixture of dry air and water vapour, is less than that, For moist air, which is a mixure of dry air and water vapour, M , of dry air and more than that of water vapour: d the mean molecular weight Mm, and therefore also the gas Mv < Mm < Md ‘constant’ Rm, depends on the amount of moisture in the air. The gas constant for water vapour is larger than that for It is inconvenient to use a gas ‘constant’ which varies in dry air: R∗ R∗ this way. It is simpler to retain the constant R = Rd for dry R = , and R = air, and to use a modified temperature, T , in the ideal gas d M v M v d v equation. We call this the virtual temperature. so that Mv < Md =⇒ Rv > Rd . 13 14 Mixing Ratio Mixing Ratio and Vapour Pressure Let’s consider a fixed volume V of moist air at temperature By the ideal gas law, the partial pressure pressure exerted T and pressure p which contains a mass m of dry air and by a constituent of a mixture of gases is proportional to d the number of kilomoles of the constituent in the mixture. a mass mv of water vapour. The total mass is m = md + mv. Thus: The mixing ratio is defined by ∗ pd = ndR T dry air ∗ mv e = nvR T water vapour w = . ∗ md p = nR T moist air where pd is the pressure due to dry air, e the pressure due The mixing ratio is a dimensionless number. It is usually to water vapour and p the total pressure. given as grams of water vapour per kilogram of air. Therefore, In middle latitudes, w is typically a few grams per kilogram. e nv nv mv/Mv −1 = = = In the tropics it can be greater than 20 g kg . If there p n nv + nd mv/Mv + md/Md is no evapouration or condensation, the mixing ratio is a Dividing by Mv/md, this gives conserved quantity. e w = p w + ε 15 16 Problem: If the mixing ratio is 5.5 g kg−1, and the total Virtual Temperature pressure is p = 1026.8 hPa, calculate the vapour pressure. Solution: We have The density of the mixture of air and water vapour is w w md + mv e = p ≈ p ρ = = ρd + ρv w + ε ε V where ρd is the value the density would have if only the where ε = 0.622. Substituting w = 5.5 g kg−1 = 0.0055 g g−1, we mass md of dry air were present and ρv is the value the find that e = 9 hPa. density would have if only the mass mv of water vapour were present. We apply the ideal gas law to each component: pd = RdρdT e = RvρvT where pd and e are the partial pressures exerted by the dry air and water vapour respectively. 17 18 By Dalton’s law of partial pressure, Again, p = RdρTv p = pd + e . where the virtual temperature Tv is defined by Combining the above results, T pd e T = . ρ = ρd + ρv = + v RdT RvT 1 − (e/p)(1 − ε) p − e Rd e = + The great advantage of introducing virtual temperature is RdT Rv RdT p e e that the total pressure and total density of the mixture are = − + ε RdT RdT RdT related by the ideal gas equation with the gas constant the p e same as that for dry air, R . = 1 − (1 − ε) . d RdT p The virtual temperature is the temperature that dry air must have in order to to have the same density as the moist We may write this equation as air at the same pressure. Note that the virtual temperature p = RdρTv is always greater than the actual tempeature: where the virtual temperature Tv is defined by Tv ≥ T.
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