Presentation Slides
for
Chapter 2
of
Fundamentals of Atmospheric Modeling
2nd Edition
Mark Z. Jacobson
Department of Civil & Environmental Engineering
Stanford University
Stanford, CA 94305-4020
jacobson@stanford.edu
March 10, 2005
Hydrostatic Air Pressure
Weight per unit area of air overhead a given altitude
pa z 
pa
a
g
z

(2.1)
z azgzdz
= air pressure (1 Pa=1 kg m-1 s-2 = 0.01 hPa=0.01 mb)
= air density (kg/m3)
= gravity (m/s2)
= altitude (m)
Typical sea-level pressures:
101,325 Pa =1013.25 hPa = 1013.25 mb = 1.01325 bar
760 mm Hg = 760 torr
29.9 in. Hg
14.7 lb in-2
10,300 kg m-2
100
level
sealevel
Altitude
Altitudeabove
above sea
(km)(km)
100
1 hP a (above 99.9%)
60
10 hPa (above 99%)
40
100 hP a (above 90%)
20
500 hP a (above 50%)
0
Altitude above sea level (km)
80
100
Altitude above sea level (km)
level
sea
Altitude
Altitudeabove
above sea
level
(km)(km)
Pressure, Density, Gravity vs. Altitude
80
60
40
20
0
0
200 400 600 800 1000
Air press ure (hPa)
0
0.4
0.8
1.2
Air dens ity (kg m -3 )
80
60
40
20
0
9.5
9.6 9.7 9.8 9.9
Gravity (m s -2 )
Figs. 2.1a-c
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 Atmosphere
Fixed Gases
Nitrogen (N 2)
Oxygen (O 2)
Argon (Ar)
Variable Gases
Water Vapor (H 2O)
Carbon Dioxide (CO 2)
Methane (CH 4)
Ozone (O 3)
Mixing
Ratio
(%)
Mixing
Ratio
(ppmv)
78.08
20.95
0.93
780,000
209,500
9,300
0.00001-4
0.0375
0.00018
0.0000030.001
0.1-40,000
375
1.8
0.03-10
Table 2.1
mixing ratio (ppmv)
Carbon dioxide
2
CO
(g) mixing ratio (ppmv)
Carbon Dioxide Mixing Ratio
380
370
360
350
340
330
320
310
1960
1970
1980
Year
1990
2000
Fig. 2.2
Definitions of Temperature
From average thermal speed of an air molecule (m/s)
4
1
2
kB T  Mv a

2
va 
8k BT
M
kB
T
M
= Boltzmann’s constant (kg m2 s-2 K-1)
= Absolute temperature (K)
= mass of a single air molecule (kg molec.-1)
From root-mean-square speed
3kB T
3
1
2
v rms 
k BT  Mvrms
M
2
2
From most probable speed
vp 
2k B T
M
1
2
k B T  Mv p
2
(2.3,2.4)
Temperature-Example
Find thermal speeds at T=300 K, T=200K
Average speed
T=300 K --> 468.3 m/s (1685 km/hr)
T=200 K --> 382.4 m/s (1376 km/hr)
Root-mean-square speed
T=300 K --> 508.3 m/s
T=200 K --> 415.0 m/s
Most probable speed
T=300 K --> 415.0 m/s
T=200 K --> 338.9 m/s
Example 2.1
Altitude (km)
Mesopause
0.00032
Thermosphere
0.0018
Mesosphere
Stratopause
Ozone
layer
30
20
10
0
Stratosphere
0.011
0.052
0.22
0.8
2.9
12
55
Pressure (mb)
100
90
80
70
60
50
40
Pressure (mb)
Altitude (km)
Temperature Versus Altitude
Tropopause
180
200
Troposphere 265
1013
220
240
260
280
300
Temperature (K)
Fig. 2.4
The Ozone Layer
Arctic
Ozone Dent
(April - May)
Antarctic
Ozone Hole
(Oct - Nov.)
Stratospheric
Ozone Layer
Stratospheric Ozone Chemistry
• Natural Ozone Production
O2 + hn --> O(1D) + O
O2 + hn --> O + O
O(1D)+M-->O+M
O + O2 + M --> O3 + M
• Natural Ozone Destruction
O3 + hn --> O2 + O(1D)
O3 + hn --> O2 + O
O3 + O --> O2 + O2
l < 175 nm
175 nm < l < 245 nm
l < 310 nm
l > 310 nm
(2.9)-(2.15)
Temperature Structure
Zonally-Averaged Temperatures
July
Altitude (km)
Altitude (km)
January
Fig. 2.5a, b
Boundary Layer
Nighttime
Daytime
Fig. 2.3a, b
Processes Affecting Temperature
Specific 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
Conduction
Important 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
Dry air
 d  0.023807  7.1128 10
(2.5)
5
T  273.15
Water vapor
 v  0.015606  8.3680  10
(2.6)
5
T  273.15 
Moist air
 
 v  nv 
 a   d 1 1.17  1.02



n

n
 
d  v
d 
n=number of moles
(2.7)
Turbulence
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
Wind
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/Divergence
Convergence: horizontal net inflow of air into a region
L
Divergence: horizontal net outflow of air from a region
H
Convection
Vertical air motion
Free convection: vertical air motion due to
thermal turbulence
Forced convection: vertical air motion
due to mechanical turbulence
L
Equation of State
• Boyle’s Law
p~1/V at const. T
• Charles’ Law
V~T
Lifting of a parcel of air
Low T, Low P, High V
at const. p
• Avogadro’s Law
V~n
at const. p, T
High T, High P, Low V
• Ideal Gas Law
pV=n R*T
(2.17)-(2.20)
Jacques Charles (1746-1823)
Edgar Fahs Smith Collection, U. Penn. Lib.
•
June 4, 1783 Montgolfier brothers launch hotair 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 hydrogenballoon 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
Franklin
Equation of State
p
R*
A
kB
nR* T nA R* 
(2.20)
p

 T  Nk BT
V
V  A 
= air pressure
n = number of moles of gas
= gas const. (cm3 hPa mol-1 K-1) T = absolute temperature (K)
= Avogadro’s num. (molec. mol-1) N = gas conc. (molec. cm-3)
= Boltzmann’s constant
(1.380658x10-19 cm3 hPa K-1)
Example 2.3
-->
p = 1013 hPa
T = 288 K
N = 2.55 x 1019 molec. cm-3
-->
p = 1 hPa
T = 270 K
N = 2.68 x 1016 molec. cm-3
Dalton’s Law of Partial Pressure
Total air pressure equals the sum of the partial
pressures of all the individual gases in the atmosphere
Total air pressure (hPa)
(2.22)
pa   pq  k BT  N q  N a k B T
q
q
Partial pressure of an individual gas
(2.21)
pq  NqkBT
Total air pressure equals partial pressures of dry plus moist air
pa  pd  pv
Number concentration of total air
N a  Nd  Nv
Equation of State for Dry Air
Partial pressure of dry air (hPa)
nd R* T nd m d
pd 

V
V
(2.23)
 R* 
nd A R * 
 T   d RT 
 T  Nd k BT
V  A 
md 
Dry air mass density (g cm-3)
nd md
d 
V
Dry air number concentration (molec. cm-3)
nd A
Nd 
V
Dry air gas constant (Appendix A)
R*
R
md
Equation of State for Water Vapor
Partial pressure of water vapor (hPa)
(2.25)
nv R* T nvm v  R* 
nv A  R* 
pv 

 T   v Rv T 
 T  Nv k B T
V
V mv 
V  A 
Water vapor mass density (g cm-3)
nv mv
v 
V
Water vapor number concentration (molec. cm-3)
nv A
Nv 
V
Gas constant for water vapor
R*
Rv 
mv
Equation of State Examples
Example 2.4
-->
pd  d R T
pd
T
R’
d
= 1013 hPa
= 288 K
= 2.8704 m3 hPa kg-1 K-1
= 1.23 kg m-3
pv  v Rv T
Example 2.5
-->
pv
T
Rv
v
= 10 hPa
= 298 K
= 4.6189 m3 hPa kg-1 K-1
= 0.00725 kg m-3
Volume and Mass Mixing Ratio
Volume mixing ratio (molec. of gas per molecule of dry air)
N q pq nq
(2.29)
q 


Nd
pd nd
Mass mixing ratio (mass of gas per mass of dry air)
wq 
q
d

mq Nq
m d Nd
Example 2.6 - ozone

= 0.1 ppmv
mO3 = 48 g mol-1
T
= 298 K
pd
= 1013 hPa

mq pq
md pd

m q nq
md nd

mq
md
q
(2.30)
--> wO3 = 0.17 ppmm
--> Nd = 2.55 x 1019 molec. cm-3
--> NO3 = 2.55 x 1012 molec. cm-3
--> pO3 = 0.000101 hPa
Mass Mixing Ratio of Water Vapor
Equation of state for water vapor
v R T
 Rv 
pv  v Rv T   v  RT 
 R 

(2.27)
R R* mv  mv


 0.622
 *  
Rv md  R  md
Mass mixing ratio of water vapor (mass-vapor per mass dry air)
v
mv pv
pv
pv
wv 



  v (2.31)
 d md pd
pd
pa  pv
Example 2.7
-->
pv
pd
wv
= 10 hPa
= 1013 hPa
= 0.00622 kg kg-1 (0.622%)
Specific Humidity
= Moist-air mass mixing ratio (mass of vapor per mass moist air)
v
qv 
a
(2.32)
R
pv
v
Rv
pv

 p


p
R

 d  v
d  v
pd  pv
pd 
pv
RT RvT
Rv
pv
Rv T
Example 2.8
-->
-->
pv
pa
pd
qv
= 10 hPa
= 1010 hPa
= 1000 hPa
= 0.00618 kg kg-1 (0.618%)
Equation of State for Moist Air
Total air pressure
(2.34)
d   v Rv R 
pa  pd  pv  d R T   v RvT  a RT
a
Gather terms
(2.35)
1  v  d  
d  v 
1 wv 
pa   a RT
  a RT
 a RT
d  v
1  v d
1  wv
Total air pressure
(2.36)
pa  a RmT
Gas constant for moist air
(2.37)
1 wv 
 1   
Rm  R 
 R1 
q v   R1  0.608 q v 


1  wv

Equation of State for Moist Air
Gas constant for moist air
(2.37)
1 wv 
 1   
Rm  R 
 R1 
q v   R1  0.608 q v 


1  wv

Molecular weight of total air less than that of dry air
(2.39)
R*
R*
Rm 
 R1 0.608q v  
1  0.608q v 
ma
md
md
ma 
1 0.608qv
Virtual Temperature
Equation of state for moist air
(2.36)
pa   a Rm T   a R Tv
Virtual temperature
Temperature of dry air having the same density as a sample of
moist air at the same pressure as the moist air.
(2.38)
Rm
1 wv 
 1   
Tv  T
T
 T 1 
q v   T 1 0.608q v 


R
1  wv

Moist Air Example
Example 2.9
-->
pd
pv
T
= 1013 hPa
= 10 hPa
= 298 K
pv
qv 
 0.0061 kg kg -1
pd  pv
md
ma 
 28.86 g mol -1
1 0.608qv
Rm  R 1 0.608q v  2.8811 hPa m3 kg-1 K-1
Tv  T1 0.608q v   299.1 K
pa
a 
 1.19 kg m -3
Rm T
Hydrostatic Equation
Vertical equation of motion in absence of vertical acceleration
Upward pressure gradient balances downward gravity (2.40)
dpa  agdz
Use equation to calculate pressure at a given altitude
(2.41)
pa,k  pa,k 1  a,k1gk1z k  zk1
Example 2.10
At sea level,
pa,k+1
a,k+1
gk+1
pa,100m
= 1013.25 hPa
= 1.225 kg m-3
= 9.8072 m s-2
= 1001.24 hPa
-->
At 100 m,
-->
Air pressure decreases about 1 hPa per 10 m altitude
Pressure Altimeter
Measures pressure at unknown altitude with aneroid barometer
Combine hydrostatic equation with equation of state
pa
pa

g
z
Rm T
(2.42)
Assume constant temperature decrease with altitude
T  Ta,s  s z
Assume free-tropospheric lapse rate
T
-1
s  
 6.5 K km
z
Assume standard atmosphere surface pressure, temperature
pa,s = 1013.25 hPa
Ts = 288 K
Pressure Altimeter
Integrate from pa,s to pa
(2.43)
 pa 
Ta,s  s z 
g
ln 
ln 
 

p

R
T
 a,s 
s m 
a,s

Rearrange for altitude as function of pressure
(2.44)
s R m 

Ta,s   pa  g 
z
1 


s   pa,s 



Example 2.11
Pressure-altitmeter reading
pa
= 850 hPa
-->
z
= 1.45 km
Scale Height
Height above a references height at which pressure decreases to
1/e its value at the reference height
Density of air from equation of state
(2.45)
pa
m d pa pa  A  md
pa  1 
pa M
a 
 *


 * 
 M 
RTv
Tv R  A
Tv k B 
k B Tv
R Tv
Mass of one air molecule
md
M
A
Combine (2.45) with hydrostatic equation
(2.46)
dpa
Mg
dz

dz  
pa
k BTv
H
Scale height
(2.47)
k B Tv
H
Mg
Scale Height Equation
Integrate (2.46) at constant pressure
pa  pa,ref e

(2.46)
 z z ref
H
Example 2.12
-->
-->
T
H
pa,ref
zref
z
pa
= 298 K
= 8.72 km
= 1013.25 hPa
= 0 km
= 1 km
= 903.5 hPa
Energy
Kinetic energy
Energy within a body due to its motion
Potential energy
Energy that arises due to an object’s position rather than 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
Latent Heat
Energy required to change a substance from one state to another
Fig. 2.6
Condensation, freezing, deposition
release energy --> warm the air
Evaporation, melting, sublimation
absorb energy --> cool the air
Latent Heat of Evaporation
Change of latent heat of evaporation with temperature (2.49)
dLe
 c p,V  cW
dT
Integrate


(2.53)
Le  Le,0  c W  c p,V T  T0 
Substitute constants (J kg-1)
(2.54)
Le  2.501  106  2370Tc
Example 2.13
-->
-->
T
Le
T
Le
= 273.15 K
= 2.5x106 J kg-1
= 373.15 K
= 2.3x106 J kg-1
Specific Heat of Liquid Water
cCWW (J
(J kg
kg-1-1KK-1-1))
6000
5500
5000
4500
4000
-40 -30 -20 -10 0 10 20 30 40
o
Temperature C)
(
Fig. 2.7
Latent Heat of Melting
Change of latent heat of melting with temperature
(2.50)
dL m
 cW  c I
dT
Integrate and substitute constants (J kg-1)
(2.55)
Lm  3.3358 10 5  Tc 2030 10.46Tc 
Example 2.14
-->
T
Lm
= 273.15 K
= 3.34x105 J kg-1
-->
T
Lm
= 263.15 K
= 3.12x105 J kg-1
Latent Heat of Sublimation
Change of latent heat of sublimation with temperature (2.50)
dL s
 c p,V  c I
dT
Integrate and substitute constants (J kg-1)
(2.56)
Ls  Le  Lm  2.83458 106  Tc 340 10.46Tc 
Clausius-Clapeyron Equation
Change of saturation vapor pressure with temperature (2.57)
dpv,s  v,s

Le
dT
T
Density of water vapor over particle surface (kg m-3)
pv,s
 v,s 
RvT
Combine density with Clausius-Clapeyron equation
(2.58)
dpv,s Le pv,s

2
dT
Rv T
Substitute latent heat of evaporation
dpv,s
1 Ah Bh 

 2 
dT
pv,s
Rv T
T 
(2.59)
Clausius-Clapeyron Equation
Integrate
Ah
pv,s  pv,s,0 exp 
Rv
 1 1 
   
T0 T 
Substitute constants
Bh
Rv
(2.60)
T0 
ln  
 T 
(2.61)

1 
 1
273.15 

pv,s  6.112 exp 6816 
   5.1309 ln 
273.15 T 
 T 

Example 2.15
-->
T
pv,s
= 253.15 K
= 1.26 hPa
-->
T
pv,s
= 298.15 K
= 31.6 hPa
Saturation Vapor Pressure
Empirical parameterization
(2.62)
 17.67 Tc 
pv,s  6.112 exp 

T

243
.5
 c

Example 2.16
Tc = -20 oC (253.15 K)
--> pv,s = 1.26 hPa
oC
Tc = 25
(298.15 K)
--> pv,s = 31.67 hPa
(hPa)
Vapor
Vapopressure
r pressure (hPa)
120
100
80
60
40
Over liqu id
water
20
0
-20 -10
0 10 20 30 40 50
Temperature (o C)
Saturation Vapor Pressure Over Ice
Change of saturation vapor pressure with temperature (2.63)
dpv,I Ls pv,I

2
dT
RvT
Substitute latent heat of sublimation and integrate
(2.64)

1
1 


 
4648 

273.15 T 

pv,I  6.112 exp 
273.15 
11.64 ln 

 0.02265 273.15 T 


 T 


Saturation Vapor Pressure Over Ice
Empirical parameterization
(2.65)
21.88T  273.15
pv,I  6.1064 exp 

T  7.65


Example 2.17
Tc = -20 oC (253.15 K)
--> pv,s = 1.26 hPa
--> pv,I = 1.04 hPa
(hPa)
Vapor
Vapopressure
r pressure (hPa)
8
7
6
5
4
3
Over liqu id
water
2
1
0
-50
Over ice
-40
-30 -20 -10
Temperature (o C)
0
Condensation/Evaporation
Condensation when pv > pv,s
Evaporation when pv < pv,s
Fig. 2.9a,b
Formation of Rain in Cold Clouds
Vaporpressure
pressure (hPa)
(hPa)
Vapor
Ice Crystal (Bergeron) Process
8
7
Over liquid water
Over ice
water
droplet
6
gas
molecules
5
4
3
2
1
0
-50 -40 -30 -20 -10 0
Temperature ( oC)
10
• 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
ice
crystal
Relative Humidity
Relative humidity (percent)
wv
fr  100% 
 100% 
w v,s

pv pa  pv,s
  100% 
pv,s pa  pv 
Saturation mass mixing ratio
pv,s
pv,s
w v,s 

pa  pv,s
pd
To increase relative humidity,
 increase partial pressure of water
 decrease temperature, which decreases pv,s
To decrease relative humidity,
 lower partial pressure of water
 increase temperature
(2.66)
pv
pv,s
(2.67)
Relative Humidity Example
Vapor pressure (hPa)
If T = 35°C and pv=20 hPa
--> find pv,s and fr
pv,s=56.2 hPa
pv=20 hPa
fr = 100% x 20 hPa / 56.2 hPa
= 35.6%
Temperature (oC)
T=35°C
If T = 24°C and fr =80%
--> at what temperature
does fog appear upon
cooling the air
Vapor pressure (hPa)
Relative Humidity Example
pv = 80% x 29.6 hPa / 100%
= 23.7 hPa
T=20°C
pv,s=29.6 hPa
pv=23.7 hPa
T=24°C
Temperature (oC)
Dew Point
Temperature to which air must be cooled at constant water vapor
partial pressure to be saturated over a liquid water surface
Dew point (K)
(2.68)
4880 .357  29 .66 ln pv 4880 .357  29.66 ln wv pd  
TD 

19.48  ln pv
19.48  lnwv pd  
Ambient water vapor mixing ratio
pv
wv 
pd
Example 2.17
pv = 12 hPa
--> TD = 282.8 K
T = 30°C and
pv=20 hPa --> find TD
Vapor pressure (hPa)
Dew Point Example
pv=20 hPa
TD
Temperature (oC)
T
TD = 16°C and fr =83%
--> find pv and pv,s
pv,s = pv x100% / fr
= 21.3hPa
Vapor pressure (hPa)
Dew Point Example
TD
pv,s=21.3 hPa
pv=17.7 hPa
Temperature (oC)
Morning/Afternoon Temperature/Dew
Point at Riverside, Calif.
700
700
Temperature
(hPa)
Air pressure
Pressure (hPa)
(hPa)
Air pressure
Pressure (hPa)
750
Temperature
800
850
900
950
Dew point
750
800
850
3:30 p.m.
900
950
Dew point
3:30 a.m.
1000
260
270
280 290 300
Temperature (K)
310
1000
260
270
280 290 300
Temperature (K)
310
Figs. 2.11a,b
First Law of Thermodynamics
First law for atmosphere
(2.69)
dQ *  dU*  dW*
dQ* = change in energy (J) due to energy transfer
dU* = change in internal energy (J) of the air
dW* = work done by (+) or on (-) the air (J)
First law in terms of energy per unit mass
(2.71)
dQ  dU dW
dQ *
dQ 
Ma
Ma = mass of air (kg)
dU *
dU 
Ma
dW*
dW 
Ma
Change in Work
Work done by air during expansion (dV>0) or contraction of air
dW *  pa dV
Work done per unit mass of air
(2.72)
dW* pa dV
dW 

 pa d a
Ma
Ma
Specific volume of air
V
1
a 

Ma  a
(2.73)
Change in 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.
Change in internal energy
(2.74)
Q 
dU    dT  c v,m dT
 T 
a
Conservation of energy
(2.75)
Md  Mv dQ = Mdcv,d  Mvcv,V dT
-->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
M d c v,d  Mv cv,V c v,d  c v,V w v
Q 
c v,m    

 cv,d 1 0.955q v 
T 
Md  Mv
1 w v
a
Different Forms of First Law
First law of thermodynamics
(2.77)
dQ  c v,m dT  pa d a
Equation of state for moist air
pa a  RmT
Differentiate equation of state
(2.78)
pad a   adpa  RmdT
Combine (2.77) and (2.78) and cp,m= cv,m +Rm
dQ  cp,mdT   adpa
(2.79)
Different Forms of First Law
Specific 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
Md c p,d  M vc p,V c p,d  c p,V wv
dQ 

c p,m    

dT p
M d  Mv
1 w v
a
 c p,d 1  0.856q v 
Virtual temperature
(2.38)
Tv  T1 0.608q v 
First law in terms of virtual temperature
(2.82)
dQ  cp,mdT   adpa
1  0.856q v
=
c p,d dTv   a dpa  c p,d dTv   a dpa
1 0.608q v
Different Forms of First Law
Isobaric process (dpa=0)
dQ  c p,mdT 
Isothermal process (dT=0)
(2.83)
c p,m
c v,m
dU
(2.84)
dQ   adpa  pad a  dW
Isochoric process (da=0)
(2.85)
dQ  c v,m dT  dU
Adiabatic process (dQ=0)
c v,m dT  pa d a
cp,mdT   adpa
(2.86)
cp,ddTv  adpa
Adiabatic/Diabatic Processes
Adiabatic process (dQ=0)
Process by which no energy is
exchanged between a system
(parcel of air) and its
surroundings
(atmosphere)
Parcel
of air
Diabatic processes (dQ>0 or <0)
Condensation/evaporation
Deposition/sublimation
Freezing melting
Radiative heating/cooling
Atmosphere
Dry adiabatic Expansion in Rising Air
278.2 K
1. Rising air expands
2. Expanding air cools
Rising air cools
1 km
Unsaturated air cools +9.8 K
per 1 km rise in altitude
288 K
 dry adiabatic lapse rate
d = +9.8 K/km
Stability in Terms of Temperature
Unstable
Stable
Compare parcel
temperature with
environmental
temperature to
determine stability
Fig. 2.14
Dry Adiabatic Lapse Rate
Rewrite first law for adiabatic process


c p,d dTv   a dpa  dTv   a c p,d dpa
Differentiate with respect to altitude
(2.89)
--> Dry adiabatic lapse rate in terms of virtual temperature
   p
  
g
K
Tv 
a
a
a
d       

 

ag 
 9.8




 z d
c p,d
km
c p,d  z
c p,d 
Rewrite first law for adiabatic process


c p,m dT   a dpa  dT   a c p,m dpa
Differentiate with respect to altitude
(2.89)
--> Dry adiabatic lapse rate in terms of actual temperature

g
g 
1 wv
T 
d,m     




 z d c p,m c p,d 1 c p,V wv c p,d 

Potential Temperature
Substitute
(2.91)
Rm T
dT  Rm  dpa
a 
into c p,m dT   a dpa 
 


pa
T c p,m 
 pa
Integrate
(2.92)
R 10.608q v 
Rm
 1 0.251q v 
 p c p,m
 p c p,d 1 0.856q v 
 p  
T  T0  a 
 T0  a 
 T0  a 
pa,0 
 pa,0 
 pa,0 
Exponential term
(2.93)
c p,d  cv,d


 0.286
c p,d
c p,d
R
Potential Temperature
Potential temperature of moist air (p,m)
Assume pa,0=1000 hPa --> T0=p,m
1 0.251q v 
 p 
T  T0  a 
 pa,0 
10.251q v 
1000 hPa 
  p,m  T

p


a
Potential temperature of dry air (p)

1000 hPa 
 p  T 

p


d
Example 2.21
-->
(2.94)
pd
T
p
= 800 hPa
= 270 K
= 287.8 K
(2.95)
700
700
750
750
(hPa)
Air pressure
Pressure (hPa)
(hPa)
Air pressure
Pressure (hPa)
Temperature vs. Potential
Temperature
Temperature
800
850
900
950
Dew point
800
850
900
950
3:30 p. m.
3:30 a.m.
1000
260
270
280 290 300
Temperature (K)
3:30 a. m.
310
1000
280
290 300 310 320 330
Potential temperature (K)
Figs. 2.11a, 2.12
Potential Virtual Temperature
Potential 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


1000 hPa 
1000 hPa 
 v  T 1  0.608q v 
  Tv 

p
p




a
a
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
1 0.251q v 
1000 hPa 
 p,v   p,m 1 0.608qv  Tv 

p


a
Stability in Terms of v
Altitude
Altitude (km)
(km)
2.2
2
1.8
1.6
1.4
Unstable
1.2
1
0.8
10
 0
 v 
 0
z 
 0
Stable
15
20
25
30
35
o
Potential v irtual temp erature ( C)
uns aturated unstable
uns aturated neutral
uns aturated s table
40
Fig. 2.15
Stability Criteria For v
Potential virtual temperature

1000 hPa 
 v  Tv 

p


a
(2.96)
Differentiate
(2.101)

1 
1000 
1000 
d v  dTv 
 Tv  


p
p
 a 
 a 
1000 
v
v
 2 dpa 
dTv  
dpa
Tv
pa
 pa 
Take partial derivative with respect to height
and substitute ∂pa /∂z=-a g and v=- ∂Tv /∂z
 v  v Tv


z
Tv z
(2.102)
v pa
v
R  v

v 
a g
pa z
Tv
c p,d pa
Stability Criteria For v
Substitute R’/pa=1/aTv and d=g/cp,d
(2.103)
 v
v
R   v
v
 vg
v

v 
ag  
v 


d  v 
z
Tv
c p,d pa
Tv
Tvc p,d Tv
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
Brunt-Väisälä Frequency
Rewrite (2.103)
(2.105)
 v  v
ln v
1

d  v  

d  v 
z
Tv
z
Tv
Brunt-Väisälä frequency
(2.106)
ln  v
g
2
Nbv  g

d  v 
z
Tv
Stability criteria
(2.107)
 0
unsaturated unstable
2 
Nbv  0
unsaturated neutral

unsaturated stable
 0
Example 2.25
-->
--> period
Tv
v
bv
bv
= 288 K
= +6.5 K km-1
= 0.0106 s-1 --> stable
= 2π/ bv = 593 s
Isentropic Surfaces
Change in entropy
dQ
dS 
T
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