presentation slides for chapter 2 of fundamentals of atmospheric modeling 2 nd edition

Presentation Slides for

Chapter 2of

Fundamentals of Atmospheric Modeling 2nd Edition

Mark Z. JacobsonDepartment of Civil & Environmental Engineering

Stanford UniversityStanford, CA 94305-4020jacobson@stanford.edu

March 10, 2005

Weight per unit area of air overhead a given altitude (2.1)

Hydrostatic Air Pressure

pa = air pressure (1 Pa=1 kg m-1 s-2 = 0.01 hPa=0.01 mb)a = air density (kg/m3)g = gravity (m/s2)z = altitude (m)

Typical sea-level pressures:101,325 Pa =1013.25 hPa = 1013.25 mb = 1.01325 bar 760 mm Hg = 760 torr29.9 in. Hg14.7 lb in-2

10,300 kg m-2

pa z( )= ρa z( )z

∞∫ g z( )dz

Pressure, Density, Gravity vs. Altitude

0

20

40

60

80

100

0 200 400 600 800 1000

Altitude above sea level (km)

Air pressure (hPa)

1 hPa (above 99.9%)

10 hPa (above 99%)

100 hPa (above 90%)

500 hPa (above 50%)

0

20

40

60

80

100

0 0.4 0.8 1.2

Altitude above sea level (km)

Air density (kg m-3)

0

20

40

60

80

100

9.5 9.6 9.7 9.8 9.9

Altitude above sea level (km)

Gravity (m s-2)

Figs. 2.1a-c

Alti

tude

abo

ve s

ea le

vel (

km)

Alti

tude

abo

ve s

ea le

vel (

km)

Alti

tude

abo

ve s

ea le

vel (

km)

Toricelli's Experiment With Mercury Barometer

• 1643 Evangelista Torricelli records the first sustained vacuum and demonstrates that air pressure changes daily.

• Height of fluid = Air pressure / (fluid density x gravity)

• 1648 Blaise Pascal and brother-in-law Florin Périer demonstrate that air pressure decreases with increasing altitude at Puy-de-Dôme, France.

Edgar Fahs Smith Collection, University of Pennsylvania Library

Current Composition of the AtmosphereMixingRatio(%)

Mixing Ratio(ppmv)

Fixed Gases

Nitrogen (N2) 78.08 780,000

Oxygen (O2) 20.95 209,500

Argon (Ar) 0.93 9,300

Variable Gases

Water Vapor (H 2O) 0.00001-4 0.1-40,000

Carbon Dioxide (CO2) 0.0375 375

Methane (CH4) 0.00018 1.8

Ozone (O3) 0.000003-

0.001

0.03-10

Table 2.1

310

320

330

340

350

360

370

380

1960 1970 1980 1990 2000

CO

2

(g) mixing ratio (ppmv)

Year

Carbon Dioxide Mixing RatioC

arbo

n di

oxid

e m

ixin

g ra

tio (

ppm

v)

Fig. 2.2

From average thermal speed of an air molecule (m/s) (2.3,2.4)

Definitions of Temperature

kB = Boltzmann’s constant (kg m2 s-2 K-1)T = Absolute temperature (K)M = mass of a single air molecule (kg molec.-1)

v a =8kBTπM

4π

kBT =12

Mv a2

vrms=3kBTM

32

kBT =12

Mvrms2

vp =2kBT

MkBT =

12

Mvp2

From root-mean-square speed

From most probable speed

Find thermal speeds at T=300 K, T=200K

Temperature-Example

Average speedT=300 K --> 468.3 m/s (1685 km/hr)T=200 K --> 382.4 m/s (1376 km/hr)

Root-mean-square speedT=300 K --> 508.3 m/s T=200 K --> 415.0 m/s

Most probable speedT=300 K --> 415.0 m/s T=200 K --> 338.9 m/s

Example 2.1

Temperature Versus Altitude

0

10

20

30

40

50

60

70

80

90

100

180 200 220 240 260 280 300

1013

265

55

12

2.9

0.8

0.22

0.052

0.011

0.0018

0.00032

Altitude (km)

Temperature (K)

Tropopause

Stratopause

Mesopause

Stratosphere

Troposphere

Mesosphere

Thermosphere

Ozone

layer

Pressure (mb)

Alt

itud

e (k

m) P

ressure (mb)

Fig. 2.4

The Ozone Layer

Antarctic

Ozone Hole

(Oct - Nov.)

Arctic

Ozone Dent

(April - May)

Stratospheric

Ozone Layer

Stratospheric Ozone Chemistry

• Natural Ozone Production

O2 + h --> O(1D) + O < 175 nm

O2 + h --> O + O 175 nm < < 245 nm

O(1D)+M-->O+M

O + O2 + M --> O3 + M

• Natural Ozone Destruction

O3 + h --> O2 + O(1D) < 310 nm

O3 + h --> O2 + O > 310 nm

O3 + O --> O2 + O2 (2.9)-(2.15)

Temperature Structure

.

270

180160

210

290

140200

220

240

220

260

260

240

220

200 210

-80 -60 -40 -20 0 20 40 60 80

100

80

60

40

20

0

Latitude (deg)

Altitude (km)

Zonally-Averaged Temperatures

.

200

280

180160

290

190

240

220210

220200

240

260

220

210

-80 -60 -40 -20 0 20 40 60 80

100

80

60

40

20

0

Latitude (deg)

Altitude (km)

January July

Alti

tude

(km

)

Alti

tude

(km

)

Fig. 2.5a, b

Boundary Layer

Daytime

Fig. 2.3a, b

CloudlayerEntrainment zone /Inversion layerFree troposphere

Neutral convectivemixed layerSurface layer

Subcloud layer

Daytime temperature

Boundary layerAltitude

Nighttime

Entrainment zone /Inversion layerFree troposphere

Surface layerNighttime temperature

Neutralresidual layerStableboundarylayer

Boundary layerAltitude

Processes Affecting TemperatureSpecific heat (J kg-1 K-1)

Energy required to increase the temperature of 1 kg of a substance 1 K

= 1004.67 for dry air = 4185.5 for liquid water= 1360 for clay= 827 for sand

Lower specific heat --> substance heats up faster upon addition of energy--> soil heats up during the day more than does water

Processes Affecting Temperature Energy Transfer Processes

Conduction

Transfer of energy from molecule to molecule

Convection

Transfer of energy by the vertical mass movement of a fluid

Advection

Horizontal propagation of the mean wind

Radiation

Energy transferred in the form of electromagnetic waves

ConductionImportant only near the ground

Conductive Heat Flux (W m-2)

Hc=- T / z (2.8)

= thermal conductivity (W m-1 K -1) = 0.0256 for dry air = 0.6 for liquid water = 0.92 for clay = 0.298 for sand

= change in temperature (oC)z = change in altitude (m)

Near the ground T / z = -12 K / 0.001 m--> Hc =307 W m-2

Free troposphere

T / z = -6.5 K / 1000 m--> Hc=0.00015 W m-2

Thermal Conductivity

κd ≈0.023807+7.1128×10−5 T −273.15( )

κv ≈0.015606+8.3680×10−5 T −273.15( )

κa ≈κd 1− 1.17−1.02κvκd

⎛

⎝ ⎜

⎞

⎠ ⎟

nvnv +nd

⎡

⎣ ⎢

⎤

⎦ ⎥

Dry air (2.5)

Water vapor (2.6)

Moist air (2.7)

n=number of moles

Turbulence

Wind

Buoyancy: lifting of low-density (warm) air in bath of cold air

Wind shear: variation of wind speed with height or distance

Eddy: swirling motion of air due to wind shear

Turbulence: chaotic air motion from eddies of different sizes

Thermal turbulence: turbulence due to buoyancy

Mechanical turbulence: turbulence due to wind shear or convergence/divergence

Convergence/DivergenceConvergence: horizontal net inflow of air into a region

L

H

Divergence: horizontal net outflow of air from a region

ConvectionVertical air motion

Free convection: vertical air motion due to thermal turbulence

L

Forced convection: vertical air motion due to mechanical turbulence

Equation of State• Boyle’s Law

p~1/V at const. T

• Charles’ Law

V~T at const. p

• Avogadro’s Law

V~n at const. p, T

• Ideal Gas Law

pV=n R*T

Low T, Low P, High V

High T, High P, Low V

Lifting of a parcel of air

(2.17)-(2.20)

Jacques Charles (1746-1823)• June 4, 1783 Montgolfier brothers launch hot-

air balloon in Annonay, France

• August 27, 1783 Charles launches silk balloon filled with hydrogen in Paris

• “The country people who saw it fall were frightened and attacked it with stones and knives so that it was much mangled” - Benjamin Franklin

• November 21, 1783 Montgolfiers organize first manned hot-air balloon flight.

• December 1, 1783 Charles in hydrogen-balloon flight.

• “I had a pocket glass, with which I followed it until I lost sight, first of the men, then of the car, and when I last saw the balloon it appeared no bigger than a walnut” - Benjamin FranklinEdgar Fahs Smith Collection, U. Penn. Lib.

Equation of State(2.20)p =

nR*TV

=nAV

R*

A

⎛

⎝ ⎜

⎞

⎠ ⎟ T =NkBT

p = air pressure n = number of moles of gasR* = gas const. (cm3 hPa mol-1 K-1) T = absolute temperature (K) A = Avogadro’s num. (molec. mol-1) N = gas conc. (molec. cm-3)kB = Boltzmann’s constant

(1.380658x10-19 cm3 hPa K-1)

Example 2.3 p = 1013 hPaT = 288 K

--> N = 2.55 x 1019 molec. cm-3

p = 1 hPaT = 270 K

--> N = 2.68 x 1016 molec. cm-3

Dalton’s Law of Partial Pressure

pa = pqq∑ =kBT Nq

q∑ =NakBT

Total air pressure equals the sum of the partial pressures of all the individual gases in the atmosphere

Total air pressure (hPa) (2.22)

Partial pressure of an individual gas (2.21)

Total air pressure equals partial pressures of dry plus moist air

pa =pd +pv

Na =Nd +Nv

Number concentration of total air

pq =NqkBT

Equation of State for Dry Air

Dry air mass density (g cm-3)

Dry air number concentration (molec. cm-3)

Dry air gas constant (Appendix A)

ρd =ndmd

V

Nd =ndAV

′ R =R*

md

pd =ndR*T

V=

ndmdV

R*

md

⎛

⎝ ⎜

⎞

⎠ ⎟ T =ρd ′ R T =

ndAV

R*

A

⎛

⎝ ⎜

⎞

⎠ ⎟ T =NdkBT

(2.23)Partial pressure of dry air (hPa)

Equation of State for Water Vapor

Water vapor mass density (g cm-3)

Water vapor number concentration (molec. cm-3)

Gas constant for water vapor

Partial pressure of water vapor (hPa) (2.25)

pv =nvR*T

V=

nvmvV

R*

mv

⎛

⎝ ⎜

⎞

⎠ ⎟ T =ρvRvT =

nvAV

R*

A

⎛

⎝ ⎜

⎞

⎠ ⎟ T =NvkBT

Rv =R*

mv

Nv =nvAV

ρv =nvmv

V

Equation of State Examples

Example 2.4 pd = 1013 hPaT = 288 KR’ = 2.8704 m3 hPa kg-1 K-1

--> d = 1.23 kg m-3

pd =ρd ′ R T

pv =ρvRvT

Example 2.5 pv = 10 hPaT = 298 KRv = 4.6189 m3 hPa kg-1 K-1

--> v = 0.00725 kg m-3

Volume and Mass Mixing RatioVolume mixing ratio (molec. of gas per molecule of dry air)

Example 2.6 - ozone = 0.1 ppmv mO3 = 48 g mol-1 --> O3 = 0.17 ppmmT = 298 Kpd = 1013 hPa --> Nd = 2.55 x 1019 molec. cm-3

--> NO3 = 2.55 x 1012 molec. cm-3

--> pO3 = 0.000101 hPa

χq =NqNd

=pqpd

=nqnd

(2.29)

Mass mixing ratio (mass of gas per mass of dry air)

(2.30)ωq =ρqρd

=mqNqmdNd

=mqpqmdpd

=mqnqmdnd

=mqmd

χq

Mass Mixing Ratio of Water VaporEquation of state for water vapor

Example 2.7 pv = 10 hPa pd = 1013 hPa --> v = 0.00622 kg kg-1 (0.622%)

Mass mixing ratio of water vapor (mass-vapor per mass dry air)

(2.27)pv =ρvRvT =ρvRv

′ R ⎛ ⎝ ⎜

⎞ ⎠ ⎟ ′ R T =

ρv ′ R Tε

ε =′ R

Rv=

R*

md

mvR*

⎛

⎝ ⎜

⎞

⎠ ⎟ =

mvmd

=0.622

(2.31)ωv =ρvρd

=mvpvmdpd

=εpvpd

=εpv

pa−pv=εχv

Specific Humidity= Moist-air mass mixing ratio (mass of vapor per mass moist air)

Example 2.8 pv = 10 hPa pa = 1010 hPa--> pd = 1000 hPa --> qv = 0.00618 kg kg-1 (0.618%)

(2.32)

qv =ρvρa

=ρv

ρd +ρv=

pvRvT

pd′ R T

+pv

RvT

=

′ R Rv

pv

pd +′ R

Rvpv

=εpv

pd +εpv

Equation of State for Moist AirTotal air pressure (2.34)

Gather terms (2.35)

pa =pd +pv =ρd ′ R T +ρvRvT =ρa ′ R Tρd +ρvRv ′ R

ρa

pa =ρa ′ R Tρd +ρv ερd +ρv

=ρa ′ R T1+ρv ρdε( )1+ρv ρd

=ρa ′ R T1+ωv ε1+ωv

Total air pressure (2.36)

Rm= ′ R 1+ωv ε1+ωv

= ′ R 1+1−ε

εqv

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = ′ R 1+0.608qv( )

Gas constant for moist air (2.37)

pa =ρaRmT

Equation of State for Moist Air

Molecular weight of total air less than that of dry air (2.39)

ma =md

1+0.608qv

Rm=R*

ma= ′ R 1+0.608qv( ) =

R*

md1+0.608qv( )

Rm= ′ R 1+ωv ε1+ωv

= ′ R 1+1−ε

εqv

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = ′ R 1+0.608qv( )

Gas constant for moist air (2.37)

Virtual Temperature

Virtual temperatureTemperature of dry air having the same density as a sample of moist air at the same pressure as the moist air. (2.38)

Tv =TRm

′ R =T

1+ωv ε1+ωv

=T 1+1−ε

εqv

⎛ ⎝ ⎜ ⎞

⎠ ⎟ =T 1+0.608qv( )

pa =ρaRmT =ρa ′ R Tv

Equation of state for moist air (2.36)

Moist Air ExampleExample 2.9 pd = 1013 hPa

pv = 10 hPaT = 298 K

-->

Rm= ′ R 1+0.608qv( )=2.8811 hPa m3 kg-1 K-1

qv =εpv

pd +εpv=0.0061 kg kg-1

ma =md

1+0.608qv=28.86 g mol-1

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

ρa =pa

RmT=1.19 kg m-3

Hydrostatic Equation

Use equation to calculate pressure at a given altitude (2.41)

Upward pressure gradient balances downward gravity (2.40)

Vertical equation of motion in absence of vertical acceleration

dpa =−ρagdz

pa,k ≈pa,k+1−ρa,k+1gk+1 zk −zk+1( )

Example 2.10 At sea level, pa,k+1 = 1013.25 hPa

a,k+1 = 1.225 kg m-3 gk+1 = 9.8072 m s-2

--> At 100 m, pa,100m= 1001.24 hPa

--> Air pressure decreases about 1 hPa per 10 m altitude

Pressure Altimeter

Assume free-tropospheric lapse rate

Combine hydrostatic equation with equation of state (2.42)

Measures pressure at unknown altitude with aneroid barometer

∂pa∂z

=−pa

RmTg

T =Ta,s −Γsz

Assume constant temperature decrease with altitude

Γs =−∂T∂z

=6.5 K km-1

Assume standard atmosphere surface pressure, temperature

pa,s = 1013.25 hPa Ts = 288 K

Pressure AltimeterIntegrate from pa,s to pa (2.43)

Example 2.11 Pressure-altitmeter reading pa = 850 hPa --> z = 1.45 km

Rearrange for altitude as function of pressure (2.44)

lnpapa,s

⎛

⎝ ⎜

⎞

⎠ ⎟ =

gΓsRm

lnTa,s −Γsz

Ta,s

⎛

⎝ ⎜

⎞

⎠ ⎟

z=Ta,sΓs

1−papa,s

⎛

⎝ ⎜

⎞

⎠ ⎟

ΓsRmg

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

Scale Height

Combine (2.45) with hydrostatic equation (2.46)

Density of air from equation of state (2.45)

Height above a references height at which pressure decreases to 1/e its value at the reference height

Mass of one air molecule

Scale height (2.47)

ρa =pa′ R Tv

=mdR*

paTv

=paTv

A

R*⎛

⎝ ⎜

⎞

⎠ ⎟

mdA

≈paTv

1kB

⎛

⎝ ⎜

⎞

⎠ ⎟ M =

paM kBTv

M ≈mdA

dpapa

=−M gkBTv

dz=−dzH

H =kBTvM g

Scale Height EquationIntegrate (2.46) at constant pressure (2.46)

pa =pa,refe− z−zref( ) H

Example 2.12 T = 298 K

--> H = 8.72 kmpa,ref = 1013.25 hPazref = 0 kmz = 1 km

--> pa = 903.5 hPa

EnergyKinetic energy

Energy within a body due to its motion

Potential energyEnergy that arises due to an object’s position rather than motion

Gravitational potential energyPotential energy obtained when an object is raised vertically

Internal energyKinetic and/or potential energy of atoms or molecules within an

object

WorkEnergy added to a body by the application of a force that moves

the body in the direction of the force

RadiationEnergy transferred by electromagnetic waves

Latent HeatEnergy required to change a substance from one state to another

Condensation, freezing, depositionrelease energy --> warm the air

Evaporation, melting, sublimationabsorb energy --> cool the air

Fig. 2.6

Water DropFreezingMelting

CondensationEvaporation

Sublimation

DepositionWater VaporIce crystal

Latent Heat of EvaporationChange of latent heat of evaporation with temperature (2.49)

Example 2.13 T = 273.15 K

--> Le = 2.5x106 J kg-1

T = 373.15 K--> Le = 2.3x106 J kg-1

dLedT

=cp,V −cW

Le =Le,0 − cW −cp,V( ) T −T0( )

Substitute constants (J kg-1) (2.54)

Le ≈2.501×106 −2370Tc

Integrate (2.53)

Specific Heat of Liquid Water

Fig. 2.7

4000

4500

5000

5500

6000

-40 -30 -20 -10 0 10 20 30 40

(J kg

-1 K

-1)

Temperature (oC)

cWCW (

J kg

-1 K

-1)

Latent Heat of MeltingChange of latent heat of melting with temperature (2.50)

Example 2.14 T = 273.15 K

--> Lm = 3.34x105 J kg-1

T = 263.15 K--> Lm = 3.12x105 J kg-1

Integrate and substitute constants (J kg-1) (2.55)

dLmdT

=cW −cI

Lm≈3.3358×105+Tc 2030−10.46Tc( )

Latent Heat of SublimationChange of latent heat of sublimation with temperature (2.50)

Integrate and substitute constants (J kg-1) (2.56)

dLsdT

=cp,V −cI

Ls =Le+Lm≈2.83458×106 −Tc 340+10.46Tc( )

Clausius-Clapeyron EquationChange of saturation vapor pressure with temperature (2.57)

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

dpv,sdT

=ρv,sT

Le

ρv,s =pv,sRvT

Combine density with Clausius-Clapeyron equation (2.58)

dpv,sdT

=Lepv,s

RvT2

Substitute latent heat of evaporation (2.59)dpv,spv,s

=1Rv

AhT2 −

BhT

⎛

⎝ ⎜

⎞

⎠ ⎟ dT

Clausius-Clapeyron EquationIntegrate (2.60)

Substitute constants (2.61)

pv,s =pv,s,0expAhRv

1T0

−1T

⎛

⎝ ⎜

⎞

⎠ ⎟ +

BhRv

lnT0T

⎛ ⎝ ⎜

⎞ ⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

pv,s =6.112exp68161

273.15−

1T

⎛ ⎝ ⎜ ⎞

⎠ ⎟ +5.1309ln

273.15T

⎛ ⎝ ⎜ ⎞

⎠ ⎟

⎡

⎣ ⎢ ⎤

⎦ ⎥

Example 2.15 T = 253.15 K

--> pv,s = 1.26 hPa

T = 298.15 K--> pv,s = 31.6 hPa

Saturation Vapor PressureEmpirical parameterization (2.62)

Example 2.16 Tc = -20 oC (253.15 K)

--> pv,s = 1.26 hPa

Tc = 25 oC (298.15 K)--> pv,s = 31.67 hPa

pv,s =6.112exp17.67Tc

Tc +243.5

⎛

⎝ ⎜

⎞

⎠ ⎟

0

20

40

60

80

100

120

-20 -10 0 10 20 30 40 50

Vapor pressure (hPa)

Temperature (oC)

Over liquidwater

Vap

or p

ress

ure

(hP

a)

Saturation Vapor Pressure Over IceChange of saturation vapor pressure with temperature (2.63)

Substitute latent heat of sublimation and integrate (2.64)

dpv,IdT

=Lspv,I

RvT2

pv,I =6.112exp

46481

273.15−

1T

⎛ ⎝ ⎜ ⎞

⎠ ⎟

−11.64ln273.15

T⎛ ⎝ ⎜ ⎞

⎠ ⎟ +0.02265273.15−T( )

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

Saturation Vapor Pressure Over IceEmpirical parameterization (2.65)

Example 2.17 Tc = -20 oC (253.15 K)

--> pv,s = 1.26 hPa--> pv,I = 1.04 hPa

pv,I =6.1064exp21.88T −273.15( )

T −7.65⎡

⎣ ⎢ ⎤

⎦ ⎥

0

1

2

3

4

5

6

7

8

-50 -40 -30 -20 -10 0

Vapor pressure (hPa)

Temperature (oC)

Over liquidwater

Over iceVap

or p

ress

ure

(hP

a)

Condensation/EvaporationCondensation when pv > pv,s

Condensation

Particle

Gas

Fig. 2.9a,b

Evaporation when pv < pv,s

Evaporation

Particle

Gas

Formation of Rain in Cold CloudsIce Crystal (Bergeron) Process

• pv,s over ice is less than that over liquid water• Water droplets evaporate, vapor flows to ice crystals• Ideal precipitation if 100,000 droplets per ice crystal

0

1

2

3

4

5

6

7

8

-50 -40 -30 -20 -10 0 10

Over liquid waterOver ice

Vapor pressure (hPa)

Temperature (oC)

Vap

or p

ress

ure

(hP

a)

gas

molecules

water

droplet

ice

crystal

Relative Humidity

To increase relative humidity, increase partial pressure of waterdecrease temperature, which decreases pv,s

To decrease relative humidity, lower partial pressure of water increase temperature

fr =100%×ωvωv,s

=100%×pv pa−pv,s( )

pv,s pa −pv( )≈100%×

pvpv,s

Relative humidity (percent) (2.66)

Saturation mass mixing ratio (2.67)

ωv,s =εpv,s

pa−pv,s≈

εpv,spd

Vap

or p

ress

ure

(hP

a)

Temperature (oC)

Relative Humidity Example

If T = 35°C and pv=20 hPa

--> find pv,s and fr

fr = 100% x 20 hPa / 56.2 hPa = 35.6%

pv,s=56.2 hPa

pv=20 hPa T=35°C

Vap

or p

ress

ure

(hP

a)

Temperature (oC)

Relative Humidity Example

If T = 24°C and fr =80% --> at what temperature does fog appear upon cooling the air

pv = 80% x 29.6 hPa / 100% = 23.7 hPa

pv,s=29.6 hPa

pv=23.7 hPaT=24°C

T=20°C

Dew Point

Dew point (K) (2.68)

Ambient water vapor mixing ratio

Temperature to which air must be cooled at constant water vapor partial pressure to be saturated over a liquid water surface

TD =4880.357−29.66ln pv

19.48−ln pv=

4880.357−29.66ln ωvpd ε( )19.48−ln ωvpd ε( )

ωv =εpvpd

Example 2.17 pv = 12 hPa

--> TD = 282.8 K

Vap

or p

ress

ure

(hP

a)

Temperature (oC)

Dew Point Example

pv=20 hPa

TTD

T = 30°C and

pv=20 hPa --> find TD

Vap

or p

ress

ure

(hP

a)

Temperature (oC)

Dew Point Example

TD = 16°C and fr =83%

--> find pv and pv,s

pv,s = pv x100% / fr

= 21.3hPa

pv,s=21.3 hPa

pv=17.7 hPa

TD

Morning/Afternoon Temperature/Dew Point at Riverside, Calif.

260 270 280 290 300 310

700

750

800

850

900

950

1000

Temperature (K)

Pressure (hPa)

Temperature

Dew point

3:30 p.m.

Air

pre

ssur

e (h

Pa)

260 270 280 290 300 310

700

750

800

850

900

950

1000

Temperature (K)

Pressure (hPa)

3:30 a.m.

Temperature

Dew pointAir

pre

ssur

e (h

Pa)

Figs. 2.11a,b

First Law of ThermodynamicsFirst law for atmosphere (2.69)

dQ* =dU* +dW*

dQ* = change in energy (J) due to energy transferdU* = change in internal energy (J) of the airdW* = work done by (+) or on (-) the air (J)

First law in terms of energy per unit mass (2.71)

dQ=dU+dW

dQ=dQ*

MadU=

dU*

MadW =

dW*

Ma

Ma = mass of air (kg)

Change in WorkWork done by air during expansion (dV>0) or contraction of air

Work done per unit mass of air (2.72)

dW* =padV

dW =dW*

Ma=

padVMa

=padαa

Specific volume of air (2.73)

αa =VMa

=1

ρa

Change in Internal Energy

Change in internal energy (2.74)

-->Specific heat moist air at const. volume (J kg-1 K-1) (2.76)Change in energy required to change temperature of 1 kg

air 1K at constant volume

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.

Conservation of energy (2.75)

dU=∂Q∂T

⎛ ⎝ ⎜

⎞ ⎠ ⎟ αa

dT =cv,mdT

cv,m =∂Q∂T

⎛ ⎝ ⎜

⎞ ⎠ ⎟ αa

=Mdcv,d +Mvcv,V

Md +Mv=

cv,d +cv,Vωv1+ωv

=cv,d 1+0.955qv( )

Md +Mv( )dQ= Mdcv,d +Mvcv,V( )dT

Different Forms of First LawFirst law of thermodynamics (2.77)

Equation of state for moist air

Differentiate equation of state (2.78)

dQ=cv,mdT +padαa

paαa =RmT

padαa +αadpa =RmdT

Combine (2.77) and (2.78) and cp,m= cv,m +Rm (2.79)

dQ=cp,mdT −αadpa

Different Forms of First LawSpecific heat of moist air at const. pressure (J kg -1 K-1) (2.80)

Energy required to increase the temperature of 1 kg of air 1K without affecting air pressure

Virtual temperature (2.38)

Tv =T 1+0.608qv( )

dQ=cp,mdT −αadpa

cp,m=dQdT

⎛ ⎝ ⎜ ⎞

⎠ ⎟

pa

=Mdcp,d +Mvcp,V

Md +Mv=

cp,d +cp,Vωv1+ωv

=cp,d 1+0.856qv( )

First law in terms of virtual temperature (2.82)

=1+0.856qv1+0.608qv

cp,ddTv −αadpa ≈cp,ddTv −αadpa

Different Forms of First LawIsobaric process (dpa=0) (2.83)

Isothermal process (dT=0) (2.84)

Isochoric process (da=0) (2.85)

Adiabatic process (dQ=0) (2.86)

dQ=cp,mdT =cp,mcv,m

dU

dQ=−αadpa =padαa =dW

dQ=cv,mdT =dU

cv,mdT =−padαa cp,mdT =αadpa cp,ddTv ≈αadpa

Adiabatic/Diabatic Processes

Parcelof air

Atmosphere

Adiabatic process (dQ=0)Process by which no energy is exchanged between a system (parcel of air) and its surroundings (atmosphere)

Diabatic processes (dQ>0 or <0)Condensation/evaporationDeposition/sublimationFreezing meltingRadiative heating/cooling

1. Rising air expands2. Expanding air cools

Rising air cools

Unsaturated air cools +9.8 K per 1 km rise in altitude

dry adiabatic lapse rated = +9.8 K/km

Dry adiabatic Expansion in Rising Air

288 K

278.2 K

1 km

Stability in Terms of Temperature

Compare parcel temperature with environmental temperature to determine stability

Unstable Stable

Fig. 2.14

Dry Adiabatic Lapse RateRewrite first law for adiabatic process

Differentiate with respect to altitude (2.89)--> Dry adiabatic lapse rate in terms of virtual temperature

Differentiate with respect to altitude (2.89)--> Dry adiabatic lapse rate in terms of actual temperature

Γd,m=−∂T∂z

⎛ ⎝ ⎜

⎞ ⎠ ⎟ d

=g

cp,m=

gcp,d

1+ωv1+cp,Vωv cp,d

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Rewrite first law for adiabatic process

cp,mdT =αadpa → dT = αa cp,m( )dpa

cp,ddTv ≈αadpa → dTv = αa cp,d( )dpa

Γd =−∂Tv∂z

⎛ ⎝ ⎜

⎞ ⎠ ⎟ d

≈−αacp,d

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

∂pa∂z

=αacp,d

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ρag =

gcp,d

=+9.8Kkm

Potential TemperatureSubstitute (2.91)

T =T0pa

pa,0

⎛

⎝ ⎜

⎞

⎠ ⎟

Rmcp,m

=T0pa

pa,0

⎛

⎝ ⎜

⎞

⎠ ⎟

′ R 1+0.608qv( )cp,d 1+0.856qv( )

≈T0pa

pa,0

⎛

⎝ ⎜

⎞

⎠ ⎟

κ 1−0.251qv( )

κ =′ R

cp,d=

cp,d −cv,dcp,d

=0.286

αa =RmTpa

into cp,mdT =αadpa →dTT

=Rm

cp,m

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

dpapa

Integrate (2.92)

Exponential term (2.93)

Potential TemperaturePotential temperature of moist air (p,m) (2.94)

Assume pa,0=1000 hPa --> T0=p,m

T ≈T0pa

pa,0

⎛

⎝ ⎜

⎞

⎠ ⎟

κ 1−0.251qv( )

→ θp,m =T1000 hPa

pa

⎛

⎝ ⎜

⎞

⎠ ⎟

κ 1−0.251qv( )

θp =T1000 hPa

pd

⎛

⎝ ⎜

⎞

⎠ ⎟

κPotential temperature of dry air (p) (2.95)

Example 2.21 pd = 800 hPa T = 270 K --> p = 287.8 K

Temperature vs. Potential Temperature

260 270 280 290 300 310

700

750

800

850

900

950

1000

Temperature (K)

Pressure (hPa)

3:30 a.m.

Temperature

Dew pointAir

pre

ssur

e (h

Pa)

280 290 300 310 320 330

700

750

800

850

900

950

1000

Potential temperature (K)

Pressure (hPa)

3:30 a. m.

3:30 p. m.A

ir p

ress

ure

(hP

a)

Figs. 2.11a, 2.12

Potential Virtual TemperaturePotential virtual temperature (v) (2.96)

Found by converting all moisture in a parcel to dry air, then bringing the parcel to 1000 hPa and determining its temperature

Virtual potential temperature (p,v) (2.97)Found by bringing a parcel to 1000 hPa, then converting all

moisture to dry air and determining the parcel's temperature

θv =T 1+0.608qv( )1000 hPa

pa

⎛

⎝ ⎜

⎞

⎠ ⎟

κ

=Tv1000 hPa

pa

⎛

⎝ ⎜

⎞

⎠ ⎟

κ

θp,v =θp,m1+0.608qv( )=Tv1000 hPa

pa

⎛

⎝ ⎜

⎞

⎠ ⎟

κ 1−0.251qv( )

Stability in Terms of v

0.8

1

1.2

1.4

1.6

1.8

2

2.2

10 15 20 25 30 35 40

Altitude (km)

Potential virtual temperature (oC)

StableUnstable

Alt

itud

e (k

m)

Fig. 2.15

∂θv∂z

<0 unsaturated unstable

=0 unsaturated neutral

>0 unsaturated stable

⎧

⎨ ⎪

⎩ ⎪

Stability Criteria For vPotential virtual temperature (2.96)

Differentiate (2.101)

θv =Tv1000 hPa

pa

⎛

⎝ ⎜

⎞

⎠ ⎟

κ

dθv =dTv1000pa

⎛

⎝ ⎜

⎞

⎠ ⎟

κ

+Tvκ1000pa

⎛

⎝ ⎜

⎞

⎠ ⎟

κ−1

−1000

pa2

⎛

⎝ ⎜

⎞

⎠ ⎟ dpa =

θvTv

dTv −κθvpa

dpa

Take partial derivative with respect to height (2.102)and substitute ∂pa /∂z=-a g and v=- ∂Tv /∂z

∂θv∂z

=θvTv

∂Tv∂z

−κθvpa

∂pa∂z

=−θvTv

Γv +′ R

cp,d

θvpa

ρag

Stability Criteria For v

Example 2.21 pa = 925 hPa Tv = 290 K

v = +7 K km-1 --> v = 296.5 K--> ∂v / ∂z = 3.07 K km-1 --> stable

Substitute R’/pa=1/aTv and d=g/cp,d (2.103)

∂θv∂z

=−θvTv

Γv +′ R

cp,d

θvpa

ρag=−θvTv

Γv +θvg

Tvcp,d=

θvTv

Γd −Γv( )

Brunt-Väisälä Frequency

Example 2.25 Tv = 288 Kv = +6.5 K km-1

--> bv = 0.0106 s-1 --> stable--> period bv = 2π/ bv = 593 s

Rewrite (2.103) (2.105)∂θv∂z

=θvTv

Γd −Γv( ) →∂lnθv

∂z=

1Tv

Γd −Γv( )

Nbv2 =g

∂lnθv∂z

=gTv

Γd −Γv( )

Brunt-Väisälä frequency (2.106)

Stability criteria (2.107)

Nbv2 <0 unsaturated unstable

=0 unsaturated neutral>0 unsaturated stable

⎧ ⎨ ⎪

⎩ ⎪

Isentropic SurfacesChange in entropy

LatitudeN. PoleEquator

v,4v,3v,2v,1 Isentropic surfaces

DecreasingvAltitude

Increasingv

dS=dQT

Isentropic surfaces occur when dS=0, which occurs when dQ=0 (adiabatic process), which occurs when v is constant with distance or height. Isentropic surfaces slant northward in the Northern Hemisphere.

Fig. 2.13