Presentation Slides for
Chapter 2of
Fundamentals of Atmospheric Modeling 2nd Edition
Mark Z. JacobsonDepartment of Civil & Environmental Engineering
Stanford UniversityStanford, CA [email protected]
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
