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 - Eddies 2 km Small plumes Car exhaust Cumulus clouds

Mesoscale 2 - Gravity waves 2,000 km Thunderstorms Tornados Cloud clusters Local winds Urban air pollution

Synoptic scale 500 - High / low pressure systems 10,000 km 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 Dynamical / thermodynamical Gas processes processes Wind speed Gas photochemistry Wind direction Gas-to-particle conversion Air pressure Air density Air temperature Soil temperature Turbulence

Radiative processes Transport processes

Optical depth of Emissions gases / aerosols / Transport cloud drops Gases / aerosols / cloud drops / heat Visibility Dry deposition Infrared radiative transfer Gases / aerosols / Solar radiative transfer cloud drops Sedimentation Aerosols / cloud drops / Aerosol / cloud raindrops processes Nucleation Freezing / melting Coagulation Reversible chemistry Condensation / evaporation Irreversible chemistry Dissolution / evaporation Heterogeneous chemistry Deposition / sublimation Pressure Versus Altitude

Figure 2.1 a.

100

80

60 1 mb (above 99.9%)

10 mb (above 99%) 40 Altitude (km) 100 mb (above 90%)

20 500 mb (above 50%)

0 0 200 400 600 800 1000 Pressure (mb) Density Versus Altitude

Figure 2.1 b.

100

80

60

40 Altitude (km)

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 (N2) 78.08 780,000 Oxygen (O2) 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 (H2O) 0.00001-4.0 0.1-40,000 Carbon dioxide (CO2) 0.0360 360 Methane (CH4) 0.00017 1.7 Ozone (O3) 0.000003-0.001 0.03-10 Fluctuations in Atmospheric CO 2

Figure 2.2.

360

350

340

330 mixing ratio (ppmv) 2 320 CO

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 oC

Thermal conductivity Rate of conduction of energy through a medium

Thermal conductivity of dry air (J m-1 s-1 K-1)

- 5 k d » 0.023807 + 7.1128 ´ 10 (T - 273.15) (2.3)

Table 2.2.

Substance Specific Heat Thermal (J kg-1 K-1) 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

DT H = -k (J m-2 s-1) c d Dz

Example 2.1. -1 -1 -1 Near the surface (T = 298 K, kd = 0.0256 J m s K ) DT = 12 K Dz = 1 mm ----> -2 Hf,c = 307 W m

-1 -1 -1 Free troposphere (T = 273 K, kd = 0.0238 J m s K ) DT = -6.5 K Dz = 1 km ----> -4 -2 Hf,c = 1.5 x 10 W m

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 / Cloud Inversion layer layer

Subcloud layer Altitude Neutral convective Boundary layer mixed layer

Surface layer Daytime temperature Nighttime Boundary Layer

Figure 2.3 b.

Free troposphere

Entrainment zone / Inversion layer

Neutral Altitude residual layer Boundary layer

Stable boundary layer Surface layer Nighttime temperature Temperature Structure of the Lower Atmosphere

Temperature

4 1 k T = M v 2 (2.2) p B 2 a

Figure 2.4

100 0.00032 Thermosphere 90 Mesopause 0.0018 80 0.011

70 Mesosphere 0.052 Pressure (mb) 60 0.22 50 Stratopause 0.8 40 2.9

Altitude (km) Ozone 30 Stratosphere 12 layer 20 55 10 Tropopause Troposphere 265 0 1013 180 200 220 240 260 280 300 Temperature (K) Zonally-/Monthly-Averaged Temperatures

Figure 2.5 a

January

100 160 180 200 80 220

60 240

280 260 40 240 Altitude (km) 220 210 210 20 200 220 0 -80 -60 -40 -20 0 20 40 60 80 Latitude Zonally-/Monthly-Averaged Temperatures

Figure 2.5 b

July

100 180 160 200 140 80 220

60 240 260 260 280 40 240 Altitude (km) 220 20 200 210 220 0 -80 -60 -40 -20 0 20 40 60 80 Latitude Ozone Production / Destruction in the Stratosphere

Natural ozone production

1 O2 + hn O( D) + O l < 175 nm (2.4)

O2 + hn O + O 175 < l < 245 nm (2.5)

1 O( D) + M O + M (2.6)

O + O2 + M O3 + M (2.7)

Natural ozone destruction

O + hn 1 l < 310 nm 3 O2 + O( D) (2.8)

O3 + hn O2 + O l > 310 nm (2.9)

O + O3 2O2 (2.10) Equation of State

Boyle's Law

1 p µ at constant temperature (2.12) V

Charles' Law

V µ T at constant pressure (2.13)

Avogadro's Law

V µ n at constant pressure and temperature (2.14)

Ideal gas law (simplified equation of state)

nR*T nA æR * ö p = = ç ÷T = NkBT (2.15) V V è A ø Equation of State

nR*T nA æR * ö p = = ç ÷T = NkBT (2.15) V V è A ø

Example 2.2 .

Surface p = 1013 mb T = 288 K -19 3 -1 kB = 1.3807 x 10 cm mb K ----> N = 2.55 x 1019 molec. cm-3

At 48 km altitude p = 1 mb T = 270 K ----> N = 2.68 x 1016 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)

pa = å pq = kBT å Nq = NakBT (2.17) q q

Partial pressures of individual gas (mb)

pq = NqkBT (2.16)

Dry and Moist Air

Total air pressure (mb)

pa = pd + pv

Number concentration air molecules (molec. cm-3)

Na = Nd + Nv Equation of State for Dry Air

* æ * ö æ * ö nd R T ndmd R nd A R pd = = ç ÷T = r d R¢ T = ç ÷T = Nd kBT V V è md ø V è A ø (2.18)

Dry air mass density (g cm-3)

n m r = d d (2.19) d V

Dry air number concentration (molec. cm-3)

n A N = d (2.19) d V

Dry air gas constant (Appendix A)

R* R¢ = (2.19) md Equation of State Examples

Examples 2.3 and 2.4

Dry air, at sea level

* æ * ö nd R T ndmd R pd = = ç ÷T = r d R¢ T (2.18) V V è md ø pd = 1013 mb T = 288 K R' = 2.8704 m3 mb kg-1 K-1 ----> -3 r d = 1.23 kg m

Water vapor, at sea level

* æ * ö nvR T nvmv R pv = = ç ÷T = r vRvT (2.20) V V è mv ø pv = 10 mb T = 298 K 3 -1 -1 Rv = 4.6189 m mb kg K ----> -3 -3 r v = 7.25 x 10 kg m Equation of State for Water Vapor

* æ * ö æ *ö nvR T nvmv R nvA R pv = = ç ÷T = r vRvT = ç ÷T = NvkBT V V è mv ø V è A ø (2.20)

Water-vapor mass density (kg m-3)

n m r = v v (2.21) v V

Water-vapor number concentration (molec. cm-3)

n A N = v (2.21) v V

Gas constant for water vapor

R* Rv = (2.21) mv Volume and Mass Mixing Ratios

Volume mixing ratio of gas j (molec. gas per molec. dry air)

Nq pq nq c q = = = (2.24) Nd pd nd

Mass mixing ratio of gas q (g of gas per g of dry air)

r q mq Nq mq pq mqnq mq wq = = = = = c q (2.25) r d md Nd md pd md nd md

Example 2.5. Ozone c = 0.10 ppmv -1 mq = 48.0 g mole ----> w = 0.17 ppmm T = 288 K pd = 1013 mb ----> 19 -3 Nd = 2.55 x 10 molec. cm ----> 12 -3 Nq = 2.55 x 10 molec. cm ----> pq = 0.000101 mb Mass Mixing Ratio of Water Vapor

Equation of state for water vapor

æR ö r R ¢T p = r R T = r ç v ÷ R¢ T = v (2.22) v v v v è R ¢ø e

* R¢ R æm v ö mv e = = ç * ÷ = = 0.622 (2.23) Rv md è R ø md

Mass mixing ratio of water vapor (kg-vapor kg-1-dry air)

rv mv pv pv epv wv = = = e = = ecv (2.26) r d md pd pd pa - pv

Example 2.6. pv = 10 mb pa = 1010 mb ----> -1 wv = 0.00622 kg kg = 0.622%. Specific Humidity

= Moist-air mass mixing ratio (kg-vapor kg-1-moist air)

pv R¢ pv r v r v RvT Rv epv qv = = = = = r r + r pd pv R ¢ p + ep a d v + pd + pv d v R¢ T RvT Rv (2.27)

Example 2.7. pv = 10 mb pa = 1010 mb ----> pd = 1000 mb ----> -1 qv = 0.00618 kg kg = 0.618%. Equation of State for Moist Air

Total air pressure

r d + r vRv R ¢ pa = pd + pv = rd R ¢T + r vRvT = r aR¢ T r a (2.28)

Gather terms, multiply numerator / denominator by density

r d +r v e 1+ r v (r de) 1+ wv e pa = r aR¢ T = r a R¢ T = ra R¢ T r d + rv 1 + r v r d 1 + wv (2.29) Equation of state for moist (or total) air

pa = r aRmT = r aR ¢T v (2.30)

Gas constant for moist air (2.31)

1+ wv e æ 1- e ö Rm = R ¢ = R¢ ç 1 + qv÷ = R¢ (1 +0.608qv) 1 + wv è e ø

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.

Rm 1+ wv e æ 1 - e ö Tv = T = T = Tç1 + qv÷ =T(1+0.608qv ) R¢ 1 + wv è e ø Molecular Weight of Moist Air

Gas constant for moist air

1+ wv e æ 1- e ö Rm = R ¢ = R¢ ç 1 + qv÷ = R¢ (1 +0.608qv) 1 + wv è e ø

--> Molecular weight of moist air (> than that of dry air)

md ma = (2.33) 1+ 0.608qv Moist Air Example

Example 2.8. pd = 1013 mb pv = 10 mb T = 298 K

epv -1 qv = = 0.0061 kg kg pd + epv

md -1 ma = = 28.86 g mole 1+ 0.608qv

3 -1 -1 Rm = R ¢(1 + 0.608qv ) = 2.8811 mb m kg K

Tv = T(1+ 0.608qv ) = 299.1 K

pa -3 r a = = 1.19 kg m RmT Hydrostatic Equation

Vertical pressure gradient is exactly balanced by the downward force of gravity.

Hydrostatic equation

¶p a = -r g (2.34) ¶z a

Pressure at a given altitude

¶pa pa,1- pa,0 » = -r a,0g (2.35) ¶z z1- z0

Example 2.9.

Sea level pa,0 = 1013 mb -3 r a,0 = 1.23 kg m

100 m altitude ----> pa,100m = 1000.2 mb Pressure Altimeter

Combine hydrostatic equation with equation of state ¶p p d = - d g (2.36) ¶z R¢ T

Assume temperature decrease with altitude is constant

T = Ta,s - Gsz

Free-tropospheric lapse rate (K km-1) ¶T G = - assume » 6.5 K km-1 s ¶z

Substitute lapse rate, temperature profile into (2.36) and integrate

æ p ö g æT a,s - Gsz ö lnç d ÷ = lnç ÷ (2.37) è pd,s ø GsR ¢ è Ta,s ø

Rearrange

é GsR ¢ù Ta,s ê æ p ö g ú z = ê1 - ç d ÷ ú (2.38) Gs ê è pd,s ø ú ë û

Example 2.10 pd = 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

pa md pa pa æ A ö md pa æ 1 ö paM r a = = * = ç *÷ » ç ÷M = R¢ T v R Tv Tv è R ø A Tv è kBø kBTv (2.39)

Mass of one air molecule (4.8096 x 10-23 g)

m M » d A

Combine (2.39) with hydrostatic equation

dp M g dz a = - dz = - (2.40) pa kBTv H

Scale height of the atmosphere

k T H = B v (2.41) M g Scale Height Equation

Integrate (2.40) at constant temperature

- æz - z ö H p = p e è ref ø (2.42) a a,ref

Example 2.11. T = 298 K ----> H = 8.72 km pa,ref = 1000 mb zref = 0 km z = 1 km ----> pa = 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 Water Vapor Freezing Condensation Ice crystal Water Drop Melting Evaporation

Sublimation Latent Heat of Evaporation

Latent heat of evaporation (J kg-1) dL e = c - c (2.43) dT p,V W

Le = Le,0 - (cW - cp,V)(T - T0) (2.47)

6 Le » 2.501´ 10 - 2370Tc (2.48)

Example 2.12. T = 273.15 K ----> 6 -1 -1 Le = 2.5 x 10 J kg (about 600 cal g )

T = 373.15 K ----> 6 -1 -1 Le = 2.264 x 10 J kg (about 540 cal g )

Variation of cW with temperature (Figure 2.7)

6000 ) -1 5500 K -1 5000 (J kg W c 4500

4000 -40 -30 -20 -10 0 10 20 30 40 Temperature (oC)

Latent Heats of Melting, Sublimation

Latent heat of melting (J kg-1)

dL m = c - c (2.44) dT W I

5 Lm » 3.3358´ 10 + Tc(2030 - 10.46Tc ) (2.49)

Example 2.13. T = 273.15 K ----> 5 -1 -1 Lf = 3.34 x 10 J gk (about 80 cal g )

T = 263.15 K ----> 5 -1 -1 Lf = 3.12 x 10 J kg (about 74.6 cal g )

Supercooled water Water that exists as a liquid when T < 273.15 K

Latent heat of sublimation (J kg-1)

dL s = c - c (2.44) dT p,V I

6 Ls = Le + Lm » 2.83458 ´ 10 - Tc (340 +10.46Tc) (2.50) Clausius-Clapeyron Equation

Clausius-Clapeyron equation

dpv,s r v,s = L (2.51) dT T e

Density of water vapor over the particle surface (kg m-3)

pv,s r v,s = RvT

Combine Clausius-Clapeyron equation and density

dpv,s Lepv,s = 2 (2.52) dT RvT

Substitute latent heat of evaporation

dpv,s 1 æA h Bh ö = ç 2 - ÷d T (2.53) pv,s Rv èT T ø

Integrate

é ù Ah æ 1 1 ö Bh æT 0 ö pv,s = pv,s,0 expê ç - ÷ + lnç ÷ú (2.54) ë Rv è T0 T ø Rv è T ø û

6 -1 Ah = 3.15283 x 10 J kg -1 -1 Bh = 2390 J kg K ps,0 = 6.112 mb at T0 = 273.15 K Saturation Vapor Pressure over Liquid Water

Derived saturation vapor pressure

é æ 1 1 ö æ273.15 öù pv,s = 6.112exp 6816ç - ÷ +5.1309lnç ÷ êë è 273.15 T ø è T øúû (2.55) Example 2.14. T = 253.15 K (253.15 K) ----> pv,s = 1.26 mb

T = 298.15 K (98.15 K) ----> pv,s = 31.6 mb

Alternative parameterization

æ 17.67Tc ö pv,s = 6.112expç ÷ (2.56) èT c + 243.5 ø

Example 2.15. o Tc = -20 C (253.15 K) ----> pv,s = 1.26 mb

o Tc = 25 C (298.15 K) ----> pv,s = 31.67 mb Saturation Vapor Pressure Over Liquid Water / Ice

Figure 2.8 a and b

120

100

80

60 Over liquid 40 water

Vapor pressure (mb) 20

0 -20 -10 0 10 20 30 40 50 Temperature (oC)

8 7 6 5 4 3 Over liquid water 2 Vapor pressure (mb) 1 Over ice 0 -50 -40 -30 -20 -10 0 Temperature (oC) Saturation Vapor Pressure Over Ice

Clausius-Clapeyron equation

dpv,I Lspv,I = 2 (2.57) dT RvT

Substitute latent heat of sublimation and integrate

é ù AI æ 1 1 ö BI æT 0 ö CI pv,I = pv,I,0 expê ç - ÷ + lnç ÷ + (T0 - T )ú ë Rv è T0 T ø Rv è T ø Rv û (2.58) 6 -1 AI = 2.1517 x 10 J kg -1 -1 BI = -5353 J kg K -1 -2 CI = 10.46 J kg K pI,0 = 6.112 mb at T0 = 273.15 K T £ 273.15 K

Example 2.16. T = 253.15 K (-20 oC) ----> pv,I = 1.04 mb ----> pv,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

p p - p w v v( a v,s ) pv fr =100% ´ = 100%´ » 100%´ wv,s pv,s( pa - pv) pv,s

(2.60)

Saturation mass mixing ratio of water vapor

epv,s epv,s wv,s = » (2.61) pa - pv,s pd

Example 2.17 T = 288 K pv = 12 mb ----> pv,s = 17.04 mb ----> fr = 100% x 12 mb / 17.04 mb = 70.4% Dew Point

Dew point

4880.357 - 29.66 ln pv 4880.357 - 29.66 ln(wvpd e) TD = = 19.48 - ln pv 19.48 - ln(wv pd e) (2.62)

Ambient mass mixing ratio of water vapor

epv wv = pd

Example pv = 12 mb ----> TD = 282.8 K Morning and Afternoon Dew Point and Temperature Profiles at Riverside

Figure 2.11 a and b

700 700

Temperature 750 750 Temperature

800 800

850 850 3:30 p.m. Pressure (mb) Pressure (mb) 900 900

Dew point 950 950 Dew point 3:30 a.m. 1000 1000 260 270 280 290 300 310 260 270 280 290 300 310 Temperature (K) Temperature (K) First Law of Thermodynamics

dQ* = dU* + dW* (2.63) dQ* = change in energy due to energy transfer (J) dU* = change in internal energy of the air (J) dW * = 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* dU* dW* dQ = dU = dW = (2.64) Ma Ma Ma

M = mass of a parcel of air (kg). Work and Energy Transfer

Work done by air during adiabatic expansion (dV > 0)

* dW = pa dV

Work done per unit mass of air

* dW pa dV dW = = = pa da a . (2.66) Ma Ma

Specific volume of air (cm3 g-1)

V 1 a a = = (2.67) Ma r a

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.

æ¶ Qö dU = ç ÷ dT = c dT (2.68) è ¶T ø v,m a 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.

æ¶ Q ö Mdcv,d + Mvcv,V cv,d + cv,V wv c = ç ÷ = = = c (1+ 0.955q ) v,m è ¶T ø M + M 1+w v,d v a a d v v (2.70) derived from

(Md + Mv )dQ= (Mdcv,d + Mvcv,V )dT (2.69) Variations of First Law

First law of thermodynamics

dQ = cv,mdT + pada a (2.71)

Equation of state for moist air

paa a = RmT

Differentiate

pada a +a adpa = RmdT (2.72)

Combine (2.72) with (2.70)

dQ = cp,mdT - a adpa (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

ædQ ö Mdcp,d +Mvcp,V cp,d +cp,Vwv c = ç ÷ = = = c (1+ 0.856q ) p,m è dT ø M +M 1+w p,d v pa d v v (2.74) First Law in Terms of Virtual Temperature

Change in internal energy in terms of virtual temperature

æ¶ Q ö dU = ç ÷ dTv = cv,ddTv è ¶Tv ø a a

Change in work in terms of virtual temperature

dW = pada a = R ¢d Tv - a adpa

Relationship between cp,d and cv,d

cp,d = cv,d + R¢

Substitute into dQ = dU + dW

dQ » cp,d dTv - a adpa (2.76) Applications of First Law

Isobaric process (dpa = 0)

cp,m dQ = cp,mdT = dU (2.77) cv,m

Isothermal process (dT = 0)

dQ = -a adpa = pada a = dW (2.78)

Isochoric process (da a = 0)

dQ = cv,mdT = dU (2.79)

Adiabatic process (dQ = 0)

cv,mdT = - pada a (2.80)

cp,mdT =a adpa (2.81)

cp,ddTv = a adpa (2.82) Dry Adiabatic Lapse Rate

Rearrange (2.82)

æ ö a a dTv = ç ÷d pa. èc p,d ø

Differentiate with respect to altitude --> Dry adiabatic lapse rate in terms of virtual temperature

æ ö æ ö æ¶ Tv ö a a ¶pa a a g -1 Gd = - ç ÷ = - ç ÷ = ç ÷r ag = = +9.8 K km è ¶z ød è cp,d ø ¶z è cp,dø cp,d (2.83)

Rearrange (2.81)

æ ö a a dT = ç ÷d pa. è cp,m ø

Differentiate with respect to altitude --> Dry adiabatic lapse rate in terms of temperature

æ ö æ¶ T ö g g 1+ wv Gd = - ç ÷ = = ç ÷ (2.84) è ¶z ød cp,m cp,d è1 +cp,Vwv cp,d ø Potential Temperature

Substitute a a = RmT pa into (2.81)

æ ö dT Rm dpa = ç ÷ --> (2.85) T è cp,m ø pa

Integrate

R R ¢( 1+0.608q ) m v k 1- 0.251q æ ö æ ö æ ö ( v ) pa cp,m pa cp,d (1+0.859q v ) pa T = T0 ç ÷ = T0ç ÷ » T0ç ÷ è pa,0 ø è pa,0 ø è pa,0 ø (2.86) Exponential term

R¢ cp,d - cv,d k = = = 0.286 (2.87) cp,d cp,d pa,0 = 1000 mb --> T0 = potential temperature of moist air (qp,m)

k 1- 0.251q æ1000 mb ö ( v ) q p,m = Tç ÷ (2.88) è pa ø

Potential temperature of dry air

k æ1000 mbö q p = Tç ÷ (2.89) è pd ø

Example 2.20. pd = 800 mb T = 270 K ----> qp = 287.8 K Morning and Afternoon Potential Temperature Soundings at Riverside

Figure 2.12.

700

750

800

850

Pressure (mb) 3:30 a. m. 900

950 3:30 p. m. 1000 280 290 300 310 320 330 Potential temperature (K) Potential Virtual Temperature

Potential virtual temperature

k k æ1000 mbö æ1000 mbö q v = T (1 +0.608qv)ç ÷ = Tvç ÷ (2.90) è pa ø è pa ø

Virtual potential temperature

k 1- 0.251q æ1000 mb ö ( v ) q p,v = q p,m(1+ 0.608qv )= Tvç ÷ (2.91) è pa ø

Exner function

cp,dP

Potential temperature factor

k æ p ö P = ç a ÷ (2.92) è1000 mb ø

Temperature and virtual temperature

T = q pP Tv = qvP (2.93) Isentropic Surfaces

Change in entropy

dQ dS = T

Figure 2.13 Isentropic surfaces (surfaces of constant potential temperature) between the equator and the North Pole

Increasing q v Isentropic surfaces qv,1 qv,2 qv,3 qv,4

Altitude

Decreasing qv

Equator Latitude N. Pole Stability Criteria for Unsaturated Air

Figure 2.14.

2.2 2 1.8 1.6 Unstable Stable 1.4

Altitude (km) 1.2 1 0.8 0 5 10 15 20 25 30 Virtual temperature (oC)

ïì > Gd unsaturated unstable Gví = Gd unsaturated neutral (2.94) ï î < Gd unsaturated stable Stability Criteria from Potential Virtual Temperature

Figure 2.15.

2.2 2 1.8 1.6 1.4 1.2 Unstable Stable Altitude (km) 1 0.8 10 15 20 25 30 35 40 Potential virtual temperature (oC)

< 0 unsaturated unstable ¶q ïì v í = 0 unsaturated neutral (2.95) ¶z ïî > 0 unsaturated stable Stability Criteria from Potential Virtual Temperature

Differentiate (2.90)

k k- 1æ ö æ1000 ö æ1000 ö 1000 qv qv dq v = dTv ç ÷ + Tvkç ÷ ç - 2 ÷d pa = dTv - k dpa è pa ø è pa ø è pa ø Tv pa (2.96)

Differentiate (2.96), substitute hydrostatic equation and Gv

¶q v qv ¶Tv qv ¶pa q v R¢ q v = - k = - Gv + r ag (2.97) ¶z Tv ¶z pa ¶z Tv cp,d pa

Substitute equation of state for air and definition of Gd

¶q v qv q vg qv = - Gv + = (Gd - Gv ) (2.98) ¶z Tv Tvcp,d Tv

Example 2.23. pa = 925 mb Tv = 290 K -1 Gv = +7 K km ----> -1 qv = 296.5 K ¶q v ¶z = 3.07 K km Brunt-Väisälä Frequency

Rewrite (2.98)

¶ lnq v 1 = (Gd - Gv ) (2.99) ¶z Tv

Multiply by g --> Brunt-Väisälä (buoyancy) frequency

2 ¶ lnq v g Nbv = g = (Gd - Gv ) (2.100) ¶z Tv

Period of oscillation

2p t bv = Nbv

Stability criteria

ì < 0 unsaturated unstable 2 ï Nbv í = 0 unsaturated neutral (2.101) ïî > 0 unsaturated stable

Example 2.24. Tv = 288 K -1 Gv = +6.5 K km ----> 1 Nbv = 0.0106 s- ----> t bv = 593 s