Overhead Slides for Chapter 2 of Fundamentals of Atmospheric Modeling by Mark Z. Jacobson Department of Civil & Environmental Engineering Stanford University Stanford, CA 94305-4020 January 30, 2002
Scales of Motion Table 1.1. Scale Name Scale Dimension Examples Molecular scale «2 mm Molecular diffusion Molecular viscosity Microscale 2 mm - 2 km Mesoscale 2-2,000 km Synoptic scale 500-10,000 km Eddies Small plumes Car exhaust Cumulus clouds Gravity waves Thunderstorms Tornados Cloud clusters Local winds Urban air pollution High / low pressure systems Weather fronts Tropical storms Hurricanes Antarctic ozone hole Planetary scale > 10,000 km Global wind systems Rossby (planetary) waves Stratospheric ozone loss Global warming
Processes in an Atmospheric Model Figure 1.1 Gas processes Gas photochemistry Gas-to-particle conversion Dynamical / thermodynamical processes Wind speed Wind direction Air pressure Air density Air temperature Soil temperature Turbulence Radiative processes Optical depth of gases / aerosols / cloud drops Visibility Infrared radiative transfer Solar radiative transfer Aerosol / cloud processes Nucleation Coagulation Condensation / evaporation Dissolution / evaporation Deposition / sublimation Transport processes Emissions Transport Gases / aerosols / cloud drops / heat Dry deposition Gases / aerosols / cloud drops Sedimentation Aerosols / cloud drops / raindrops Freezing / melting Reversible chemistry Irreversible chemistry Heterogeneous chemistry
Pressure Versus Altitude Figure 2.1 a. 100 80 Altitude (km) 60 40 20 1 mb (above 99.9%) 10 mb (above 99%) 100 mb (above 90%) 500 mb (above 50%) 0 0 200 400 600 800 1000 Pressure (mb)
Density Versus Altitude Figure 2.1 b. 100 80 Altitude (km) 60 40 20 0 0 0.4 0.8 1.2 Density (kg m -3 )
Composition of the Lower Atmosphere Table 2.1. Volume Mixing Ratio Gas (percent) (ppmv) Fixed Gases Nitrogen (N 2 ) 78.08 780,000 Oxygen (O 2 ) 20.95 209,500 Argon (Ar) 0.93 9,300 Neon (Ne) 0.0015 15 Helium (He) 0.0005 5 Krypton (Kr) 0.0001 1 Xenon (Xe) 0.000005 0.05 Variable Gases Water vapor (H 2 O) 0.00001-4.0 0.1-40,000 Carbon dioxide (CO 2 ) 0.0360 360 Methane (CH 4 ) 0.00017 1.7 Ozone (O 3 ) 0.000003-0.001 0.03-10
Fluctuations in Atmospheric CO 2 360 Figure 2.2. CO 2 mixing ratio (ppmv) 350 340 330 320 310 55 60 65 70 75 80 85 90 95 Year
Specific Heat and Thermal Conductivity Specific heat Energy required to increase the temperature of 1 g of a substance 1 o C Thermal conductivity Rate of conduction of energy through a medium Thermal conductivity of dry air (J m -1 s -1 K -1 ) κ d 0.023807 + 7.1128 10 5 ( T 273.15) (2.3) Table 2.2. Substance Specific Heat (J kg -1 K -1 ) Thermal Conductivity at 298 K. (J m -1 s -1 K -1 ) Dry air at constant pressure 1004.67 0.0256 Liquid water 4185.5 0.6 Clay 1360 0.920 Dry sand 827 0.298
Conductive Heat Flux Equation H c = κ d T z (J m -2 s -1 ) Example 2.1. Near the surface (T = 298 K, κ d = 0.0256 J m -1 s -1 K -1 ) T = 12 K z = 1 mm ----> H f,c = 307 W m -2 Free troposphere (T = 273 K, κ d = 0.0238 J m -1 s -1 K -1 ) T = -6.5 K z = 1 km ----> H f,c = 1.5 x 10-4 W m -2 Consequently, air conductivity is an effective energy transfer process only at the immediate ground surface.
Daytime Boundary Layer Figure 2.3 a. Free troposphere Entrainment zone / Inversion layer Cloud layer Altitude Neutral convective mixed layer Subcloud layer Boundary layer Surface layer Daytime temperature
Nighttime Boundary Layer Figure 2.3 b. Free troposphere Entrainment zone / Inversion layer Altitude Neutral residual layer Boundary layer Stable boundary layer Surface layer Nighttime temperature
Temperature Structure of the Lower Atmosphere Temperature 4 π k BT = 1 2 M v a 2 (2.2) Figure 2.4 Altitude (km) 100 90 80 70 60 50 40 Mesopause Stratopause Thermosphere Mesosphere 0.00032 0.0018 0.011 0.052 0.22 0.8 2.9 30 Ozone Stratosphere 12 layer 20 55 10 Tropopause Troposphere 265 0 1013 180 200 220 240 260 280 300 Temperature (K) Pressure (mb)
Zonally-/Monthly-Averaged Temperatures Figure 2.5 a January 100 80 160 180 200 220 Altitude (km) 60 40 20 0 280 260 210 240 240 220 210 200 220-80 -60-40 -20 0 20 40 60 80 Latitude
Zonally-/Monthly-Averaged Temperatures Figure 2.5 b July 100 180 160 140 80 220 200 Altitude (km) 60 40 20 0 240 260 260 280 240 220 200 210 220-80 -60-40 -20 0 20 40 60 80 Latitude
Ozone Production / Destruction in the Stratosphere Natural ozone production O 2 + hν λ < 175 nm O( 1 D) + O (2.4) O 2 + hν O + O 175 < λ < 245 nm (2.5) O( 1 D) + M O + M (2.6) O + O 2 + M O 3 + M (2.7) Natural ozone destruction O 3 + hν O 2 + O( 1 D) λ < 310 nm (2.8) O 3 + hν O 2 + O λ > 310 nm (2.9) O + O 3 2O 2 (2.10)
Equation of State Boyle's Law p 1 V at constant temperature (2.12) Charles' Law V T at constant pressure (2.13) Avogadro's Law V n at constant pressure and temperature (2.14) Ideal gas law (simplified equation of state) p = nr* T V = na V R * T = Nk A B T (2.15)
Equation of State p = nr* T V = na V R * T = Nk A B T (2.15) Example 2.2. Surface p = 1013 mb T = 288 K k B = 1.3807 x 10-19 cm 3 mb K -1 ----> N = 2.55 x 10 19 molec. cm -3 At 48 km altitude p = 1 mb T = 270 K ----> N = 2.68 x 10 16 molec. cm -3
Dalton's Law of Partial Pressure Total atmospheric pressure equals the sum of the partial pressures of all the individual gases in the atmosphere. Total atmospheric pressure (mb) p a = p q = k B T N q = N a k B T (2.17) q q Partial pressures of individual gas (mb) p q = N q k B T (2.16) Total air pressure (mb) p a = p d + p v Dry and Moist Air Number concentration air molecules (molec. cm -3 ) N a = N d + N v
Equation of State for Dry Air p d = n d R* T V = n d m d V R * T = ρ m d R T = n d A R * T = N d V A d k B T (2.18) Dry air mass density (g cm -3 ) ρ d = n d m d V (2.19) Dry air number concentration (molec. cm -3 ) N d = n d A V (2.19) Dry air gas constant (Appendix A) R = R* m d (2.19)
Equation of State Examples Examples 2.3 and 2.4 Dry air, at sea level p d = n d R* T V p d = n d m d R * T = ρ V m d R T (2.18) d = 1013 mb T = 288 K R' = 2.8704 m 3 mb kg -1 K -1 ----> ρ d = 1.23 kg m -3 Water vapor, at sea level p v = n v R* T V p v = n v m v R * T = ρ V m v R v T (2.20) v = 10 mb T = 298 K R v = 4.6189 m 3 mb kg -1 K -1 ----> ρ v = 7.25 x 10-3 kg m -3
Equation of State for Water Vapor p v = n v R* T V = n v m v V (2.20) R * T = ρ m v R v T = n v A v V R * A T = N v k B T Water-vapor mass density (kg m -3 ) ρ v = n v m v V (2.21) Water-vapor number concentration (molec. cm -3 ) N v = n v A V (2.21) Gas constant for water vapor R v = R* m v (2.21)
Volume and Mass Mixing Ratios Volume mixing ratio of gas j (molec. gas per molec. dry air) χ q = N q N d = p q p d = n q n d (2.24) Mass mixing ratio of gas q (g of gas per g of dry air) ω q = ρ q ρ d = m q N q m d N d = m q p q m d p d = m q n q m d n d = m q m d χ q (2.25) Example 2.5. Ozone χ = 0.10 ppmv m q = 48.0 g mole -1 ----> ω = 0.17 ppmm T = 288 K p d = 1013 mb ----> N d = 2.55 x 10 19 molec. cm -3 ----> N q = 2.55 x 10 12 molec. cm -3 ----> p q = 0.000101 mb
Mass Mixing Ratio of Water Vapor Equation of state for water vapor R p v = ρ v R v T = ρ v v R T = ρ v R T R ε (2.22) ε = R = R* R v m d m v R * = m v = 0.622 (2.23) m d Mass mixing ratio of water vapor (kg-vapor kg -1 -dry air) ω v = ρ v ρ d = m v p v m d p d = ε p v p d = εp v p a p v = εχ v (2.26) Example 2.6. p v = 10 mb p a = 1010 mb ----> ω v = 0.00622 kg kg -1 = 0.622%.
Specific Humidity = Moist-air mass mixing ratio (kg-vapor kg -1 -moist air) q v = ρ v ρ = v = ρ a ρ d + ρ v p v R v T p d R T + p v R v T = R p R v v R = p d + p R v v εp v p d + εp v (2.27) Example 2.7. p v = 10 mb p a = 1010 mb ----> p d = 1000 mb ----> q v = 0.00618 kg kg -1 = 0.618%.
Equation of State for Moist Air Total air pressure p a = p d + p v = ρ d R T + ρ v R v T = ρ a R T ρ d + ρ v R v R ρ a (2.28) Gather terms, multiply numerator / denominator by density p a = ρ a R T ρ d +ρ v ε = ρ ρ d + ρ a R T 1+ ρ v v Equation of state for moist (or total) air ( ρ d ε ) = ρ 1 + ρ v ρ a R T 1+ ω v ε d 1 + ω v (2.29) p a = ρ a R m T = ρ a R T v (2.30) Gas constant for moist air (2.31) R m = R 1+ ω v ε = R 1+ 1 ε 1 + ω v ε q v = R 1 +0.608q v ( ) Virtual temperature (2.32) Temperature of dry air having the same density as a sample of moist air at the same pressure as the moist air. T v = T R m R = T 1+ ω v ε = T 1 + 1 ε 1 + ω v ε q v = T 1+ 0.608q v ( )
Molecular Weight of Moist Air Gas constant for moist air R m = R 1+ ω v ε = R 1+ 1 ε 1 + ω v ε q v = R 1 +0.608q v ( ) --> Molecular weight of moist air (> than that of dry air) m a = m d 1+ 0.608q v (2.33) Example 2.8. p d = 1013 mb = 10 mb p v T = 298 K Moist Air Example q v = εp v p d + εp v = 0.0061 kg kg -1 m a = m d 1+ 0.608q v = 28.86 g mole -1 R m = R ( 1+ 0.608q v ) = 2.8811 mb m 3 kg -1 K -1 T v = T( 1+ 0.608q v ) = 299.1 K ρ a = p a R m T = 1.19 kg m-3
Hydrostatic Equation Vertical pressure gradient is exactly balanced by the downward force of gravity. Hydrostatic equation p a z = ρ ag (2.34) Pressure at a given altitude p a z p a,1 p a,0 z 1 z 0 = ρ a,0 g (2.35) Example 2.9. Sea level p a,0 = 1013 mb ρ a,0 = 1.23 kg m -3 100 m altitude ----> p a,100m = 1000.2 mb
Pressure Altimeter Combine hydrostatic equation with equation of state p d z = p d R T g (2.36) Assume temperature decrease with altitude is constant T = T a,s Γ s z Free-tropospheric lapse rate (K km -1 ) Γ s = T z assume 6.5 K km -1 Substitute lapse rate, temperature profile into (2.36) and integrate ln p d = p d,s g Γ s R ln T a,s Γ s z (2.37) T a,s Rearrange z = T a,s Γ s 1 p d p d,s Γ s R g (2.38) Example 2.10 p d = 850 mb = pressure altimeter reading ----> z = 1.45 km
Scale Height Height above a reference height at which pressure decreases to 1/e of its value at the reference height. Density of air from equation of state for moist air ρ a = p a = m d p a R T v R * = p a T v T v A R * m d A p a 1 T v k M = p a M B k B T v (2.39) Mass of one air molecule (4.8096 x 10-23 g) M m d A Combine (2.39) with hydrostatic equation dp a p a = M g dz = dz k B T v H (2.40) Scale height of the atmosphere H = k B T v M g (2.41)
Scale Height Equation Integrate (2.40) at constant temperature p a = p a,ref e z z ref H (2.42) Example 2.11. T = 298 K ----> H = 8.72 km p a,ref = 1000 mb z ref = 0 km z = 1 km ----> p a = 891.7 mb
Energy Capacity of a physical system to do work on matter Kinetic energy Energy within a body due to its motion. Potential energy Energy of matter that arises due to its position, rather than its motion. Gravitational potential energy Potential energy obtained when an object is raised vertically. Internal energy Kinetic and/or potential energy of atoms or molecules within an object. Work Energy added to a body by the application of a force that moves the body in the direction of the force. Radiation Energy transferred by electromagnetic waves.
Phase Changes of Water Figure 2.6. Deposition Ice crystal Freezing Melting Water Drop Condensation Evaporation Water Vapor Sublimation
Latent Heat of Evaporation Latent heat of evaporation (J kg -1 ) dl e dt = c p,v c W (2.43) L e = L e,0 ( c W c p,v )( T T 0 ) (2.47) L e 2.501 10 6 2370T c (2.48) Example 2.12. T = 273.15 K ----> L e = 2.5 x 10 6 J kg -1 (about 600 cal g -1 ) T = 373.15 K ----> L e = 2.264 x 10 6 J kg -1 (about 540 cal g -1 ) Variation of c W with temperature (Figure 2.7) 6000 (J kg -1 K -1 ) c W 5500 5000 4500 4000-40 -30-20 -10 0 10 20 30 40 Temperature ( o C)
Latent Heats of Melting, Sublimation Latent heat of melting (J kg -1 ) dl m dt = c W c I (2.44) L m 3.3358 10 5 + T c ( 2030 10.46T c ) (2.49) Example 2.13. T = 273.15 K ----> L f = 3.34 x 10 5 J gk -1 (about 80 cal g -1 ) T = 263.15 K ----> L f = 3.12 x 10 5 J kg -1 (about 74.6 cal g -1 ) Supercooled water Water that exists as a liquid when T < 273.15 K Latent heat of sublimation (J kg -1 ) dl s dt = c p,v c I (2.44) L s = L e + L m 2.83458 10 6 T c ( 340 +10.46T c ) (2.50)
Clausius-Clapeyron Equation Clausius-Clapeyron equation dp v,s dt = ρ v,s T L e (2.51) Density of water vapor over the particle surface (kg m -3 ) ρ v,s = p v,s R v T Combine Clausius-Clapeyron equation and density dp v,s dt = L ep v,s R v T 2 (2.52) Substitute latent heat of evaporation dp v,s p v,s = 1 R v A h T 2 B h dt (2.53) T Integrate p v,s = p v,s,0 exp A h 1 1 R v T 0 T + B h ln T 0 (2.54) R v T A h = 3.15283 x 10 6 J kg -1 B h = 2390 J kg -1 K -1 p s,0 = 6.112 mb at T 0 = 273.15 K
Saturation Vapor Pressure over Liquid Water Derived saturation vapor pressure 1 p v,s = 6.112exp 6816 273.15 1 +5.1309ln 273.15 T T Example 2.14. T = 253.15 K (253.15 K) ----> p v,s = 1.26 mb (2.55) T = 298.15 K (98.15 K) ----> p v,s = 31.6 mb Alternative parameterization 17.67T p v,s = 6.112exp c T c + 243.5 (2.56) Example 2.15. T c = -20 o C (253.15 K) ----> p v,s = 1.26 mb T c = 25 o C (298.15 K) ----> p v,s = 31.67 mb
Saturation Vapor Pressure Over Liquid Water / Ice Figure 2.8 a and b 120 Vapor pressure (mb) 100 80 60 40 20 Over liquid water 0-20 -10 0 10 20 30 40 50 Temperature ( o C) Vapor pressure (mb) 8 7 6 5 4 3 2 Over liquid water 1 Over ice 0-50 -40-30 -20-10 0 Temperature ( o C)
Saturation Vapor Pressure Over Ice Clausius-Clapeyron equation dp v,i dt = L sp v,i R v T 2 (2.57) Substitute latent heat of sublimation and integrate p v,i = p v,i,0 exp A I 1 1 R v T 0 T + B I ln T 0 + C I T R v T R 0 T v A I = 2.1517 x 10 6 J kg -1 B I = -5353 J kg -1 K -1 C I = 10.46 J kg -1 K -2 p I,0 = 6.112 mb at T 0 = 273.15 K T 273.15 K ( ) (2.58) Example 2.16. T = 253.15 K (-20 o C) ----> p v,i = 1.04 mb ----> p v,s = 1.26 mb
Condensation / Evaporation Figure 2.9 a and b. Water vapor Water droplet Water vapor Water droplet
Bergeron Process Figure 2.10 Water vapor Water droplet Ice crystal
Relative humidity Relative Humidity ( ) f r =100% ω v = 100% p v p a p v,s ω v,s p v,s p a p v ( ) 100% p v p v,s (2.60) Saturation mass mixing ratio of water vapor ω v,s = εp v,s p a p v,s εp v,s p d (2.61) Example 2.17 T = 288 K p v = 12 mb ----> p v,s = 17.04 mb ----> f r = 100% x 12 mb / 17.04 mb = 70.4%
Dew Point Dew point ( ) ( ) T D = 4880.357 29.66 ln p v = 4880.357 29.66 ln ω v p d ε 19.48 ln p v 19.48 ln ω v p d ε Ambient mass mixing ratio of water vapor ω v = εp v p d (2.62) Example p v = 12 mb ----> T D = 282.8 K
Morning and Afternoon Dew Point and Temperature Profiles at Riverside Figure 2.11 a and b 700 700 750 Temperature 750 Temperature Pressure (mb) 800 850 900 Pressure (mb) 800 850 900 3:30 p.m. Dew point 950 3:30 a.m. 1000 260 270 280 290 300 310 Temperature (K) 950 Dew point 1000 260 270 280 290 300 310 Temperature (K)
First Law of Thermodynamics dq * = du * + dw * (2.63) dq* du* dw * = change in energy due to energy transfer (J) = change in internal energy of the air (J) = work done by (+) or on (-) the air (J) In terms of energy per unit mass of air (J kg -1 ) dq = du + dw (2.65) where dq = dq* M a du = du* M a dw = dw* M a (2.64) M = mass of a parcel of air (kg).
Work and Energy Transfer Work done by air during adiabatic expansion (dv > 0) dw * = p a dv Work done per unit mass of air dw = dw* M a = p a dv M a = p a dα a. (2.66) Specific volume of air (cm 3 g -1 ) α a = V M a = 1 ρ a (2.67) Diabatic energy sources (dq > 0) condensation deposition freezing solar heating infrared heating Diabatic energy sinks (dq < 0) evaporation sublimation melting infrared cooling
Internal Energy Change in temperature of the gas multiplied by the energy required to change the temperature 1 K, without affecting the work done by or on the gas and without changing its volume. du = Q T dt = c v,m dt (2.68) α a Specific heat of moist air at constant volume (J kg -1 K -1 ) Change in energy required to raise the temperature of 1 g of air 1 K at constant volume. c v,m = Q = M dc v,d + M v c v,v T α M a d + M v = c v,d + c v,v ω v = c 1+ω v,d ( 1+ 0.955q v ) v (2.70) derived from ( M d + M v )dq= ( M d c v,d + M v c v,v )dt (2.69)
Variations of First Law First law of thermodynamics dq = c v,m dt + p a dα a (2.71) Equation of state for moist air p a α a = R m T Differentiate p a dα a +α a dp a = R m dt (2.72) Combine (2.72) with (2.70) dq = c p,m dt α a dp a (2.73) Specific heat of moist air at constant pressure (J kg -1 K -1 ) Energy required to increase the temperature of one gram of air one degree Kelvin without affecting the pressure of air c p,m = dq dt = M d c p,d + M v c p,v p a M d + M v = c p,d + c p,v ω v = c 1+ω p,d ( 1+ 0.856q v ) v (2.74)
First Law in Terms of Virtual Temperature Change in internal energy in terms of virtual temperature du = Q T dt v = c v,d dt v v α a Change in work in terms of virtual temperature dw = p a dα a = R dt v α a dp a Relationship between c p,d and c v,d c p,d = c v,d + R Substitute into dq = du + dw dq c p,d dt v α a dp a (2.76)
Applications of First Law Isobaric process (dp a = 0) dq = c p,m dt = c p,m c v,m du (2.77) Isothermal process (dt = 0) dq = α a dp a = p a dα a = dw (2.78) Isochoric process (dα a = 0) dq = c v,m dt = du (2.79) Adiabatic process (dq = 0) c v,m dt = p a dα a (2.80) c p,m dt =α a dp a (2.81) c p,d dt v = α a dp a (2.82)
Rearrange (2.82) Dry Adiabatic Lapse Rate dt v = α a c dp a. p,d Differentiate with respect to altitude --> Dry adiabatic lapse rate in terms of virtual temperature Γ d = T v z d = α a c p,d p a z = α a c ρ ag = p,d g c p,d = +9.8 K km -1 (2.83) Rearrange (2.81) dt = α a c dp a. p,m Differentiate with respect to altitude --> Dry adiabatic lapse rate in terms of temperature Γ d = T z d = g c p,m = g c p,d 1+ ω v 1+ c p,v ω v c (2.84) p,d
Potential Temperature Substitute α a = R m T p a into (2.81) dt T = Integrate R m c dp a --> (2.85) p,m p a R m p T = T a c p,m 0 = T0 p a,0 Exponential term κ = p a p a,0 ( ) ( ) p T0 a R 1+0.608q v c p,d 1+0.859q v p a,0 κ( 1 0.251q v ) (2.86) R = c p,d c v,d = 0.286 (2.87) c p,d c p,d p a,0 = 1000 mb --> T 0 = potential temperature of moist air (θ p,m ) 1000 mb θ p,m = T p a κ( 1 0.251q v ) (2.88) Potential temperature of dry air κ 1000 mb θ p = T p d (2.89) Example 2.20. p d = 800 mb T = 270 K ----> θ p = 287.8 K
Morning and Afternoon Potential Temperature Soundings at Riverside Figure 2.12. 700 750 Pressure (mb) 800 850 900 3:30 a. m. 950 3:30 p. m. 1000 280 290 300 310 320 330 Potential temperature (K)
Potential Virtual Temperature Potential virtual temperature ( ) θ v = T 1 +0.608q v κ 1000 mb p a κ 1000 mb = T v p a (2.90) Virtual potential temperature 1000 mb θ p,v = θ p,m ( 1+ 0.608q v )= T v p a κ( 1 0.251q v ) (2.91) Exner function c p,d P Potential temperature factor κ p P = a 1000 mb (2.92) Temperature and virtual temperature T = θ p P T v = θ v P (2.93)
Change in entropy Isentropic Surfaces ds = dq T Figure 2.13 Isentropic surfaces (surfaces of constant potential temperature) between the equator and the North Pole Increasing θ v θ v,1 Isentropic surfaces θ v,2 θ v,3 θ v,4 Altitude Decreasing θ v Equator Latitude N. Pole
Stability Criteria for Unsaturated Air Figure 2.14. Altitude (km) 2.2 2 1.8 1.6 1.4 1.2 1 0.8 Unstable Stable 0 5 10 15 20 25 30 Virtual temperature ( o C) > Γ d unsaturated unstable Γ v = Γ d unsaturated neutral < Γ d unsaturated stable (2.94)
Stability Criteria from Potential Virtual Temperature Figure 2.15. Altitude (km) 2.2 2 1.8 1.6 1.4 1.2 Unstable Stable 1 0.8 10 15 20 25 30 35 40 Potential virtual temperature ( o C) θ v z < 0 unsaturated unstable = 0 unsaturated neutral > 0 unsaturated stable (2.95)
Stability Criteria from Potential Virtual Temperature Differentiate (2.90) κ 1000 dθ v = dt v p a + T v κ 1000 κ 1 p a 1000 p a 2 dp a = θ v dt T v κ θ v dp v p a a (2.96) Differentiate (2.96), substitute hydrostatic equation and Γ v θ v z = θ v T v T v z κ θ v p a p a z = θ v R θ Γ T v + v ρ v c p,d p a g (2.97) a Substitute equation of state for air and definition of Γ d θ v z = θ v T v Γ v + θ v g T v c p,d = θ v T v ( Γ d Γ v ) (2.98) Example 2.23. p a = 925 mb T v = 290 K Γ v = +7 K km -1 ----> θ v = 296.5 K θ v z = 3.07 K km -1
Rewrite (2.98) Brunt-Väisälä Frequency lnθ v z = 1 T v ( Γ d Γ v ) (2.99) Multiply by g --> Brunt-Väisälä (buoyancy) frequency 2 lnθ N bv = g v z = g ( Γ T d Γ v ) (2.100) v Period of oscillation τ bv = 2π N bv Stability criteria 2 N bv < 0 unsaturated unstable = 0 unsaturated neutral > 0 unsaturated stable (2.101) Example 2.24. T v = 288 K Γ v = +6.5 K km -1 ----> N bv = 0.0106 s- 1 ----> τ bv = 593 s