Presentation Slides for

Chapter 9of

Fundamentals of Atmospheric Modeling 2nd Edition

Mark Z. JacobsonDepartment of Civil & Environmental Engineering

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

March 21, 2005

Earth-Atmosphere Energy Balance

Fig. 9.1

Sun

Water Evaporation(latent heat transfer)

4 6 100

16

4

7

50

20 34 6

30

50

30

66

60

23

Atmosphere

Thermal / mechanicalturbulence (sensible

heat transfer)

Infraredradiation

Solarradiation

Energy Transfer From Equator to Poles

Fig. 9.2

-90 -60 -30 0 30 60 90

Radiant energy per year

Latitude

Surplus

Deficit Deficit

Absorbed incoming solar

Emitted

outgoing infrared

Transfer

Rad

iant

ene

rgy

per y

ear -

-->

Radiation in the form of an electromagnetic wave

Electromagnetic Spectrum

Energy per unit photon (J photon-1) (9.3)

Wavelength (9.1)

Radiation in the form of a photon of energy

λ =cν =1

˜ ν

Ep =hν =hcλ

Electromagnetic Spectrum = 0.5 m

--> Ep = 3.97 x 10-19 J photon-1

--> = 5.996 x 1014 s-1

-->= 2 m-1Example 9.1:

˜ ν

˜ ν

= 10 m--> Ep = 1.98 x 10-20 J photon-1

--> = 2.998 x 1013 s-1

--> = 0.1 m-1

RadianceIntensity of emission per incremental solid angle

Planck’s Law

Planck radiance (W m-2 m-1 sr-1) (9.4)

Radiance actually emitted by a substance (9.5)

Bλ,T = 2hc2

λ5 exp hcλkBT

⎛ ⎝ ⎜

⎞ ⎠ ⎟ −1

⎡ ⎣ ⎢

⎤ ⎦ ⎥

eλ =ελBλ,T

Kirchoff’s lawIn thermodynamic equilibrium, absorptivity (a) = emissivity () --> the efficiency at which a substance absorbs equals that at which it emits. --> a perfect emitter is a perfect absorber

Emissivities

Surface Type Emissivity(fraction)

Surface Type Emissivity(fraction)

Liquid water 1.0 Soil 0.9 - 0.98Fresh snow 0.99 Grass 0.9 - 0.95Old snow 0.82 Desert 0.84 - 0.91Liq. water clouds 0.25 - 1.0 Forest 0.95 - 0.97Cirrus clouds 0.1 - 0.9 Concrete 0.71-0.9Ice 0.96 Urban 0.85 - 0.87

Table 9.1

Infrared emissivities of different surfaces

Solid Angle

Fig. 9.3

Radiance emitted from point (O) passes through incremental area dAs at distance rs from the point.

dAs

rssinθdφ

rs

rsdθ

θ

φ

z

x

yO

Incremental surface area (9.7)

dAs = rsdθ( ) rssinθdφ( ) =rs2sinθdθdφ

Incremental solid angle (sr) (9.6)Steradians analogous to radians

dΩa =dAsrs2

=sinθdθdφ

Solid angle around a sphere (9.9)

Ωa = dΩaΩa∫ = sinθdθdφ0

π∫02π∫ =4π

Spectral Actinic Flux Integral of spectral radiance over all solid angles of a sphere

Used to calculate photolysis rate coefficients

Incremental spectral actinic flux (9.10)

Spectral actinic flux (9.11)

Isotropic spectral actinic flux (9.12)

dEλ =IλdΩa

Eλ = dEλΩa∫ = IλdΩaΩa

∫ = Iλ sinθdθdφ0π∫0

2π∫

Eλ =Iλ sinθdθdφ0π∫0

2π∫ =4πIλ

Spectral IrradianceFlux of radiant energy propagating across a flat surface

Used to calculate heating rates

Incremental spectral irradiance (9.13)

Integral of dF over the hemisphere above the x-y plane (9.14)

Isotropic spectral irradiance (9.15)

Spectral irradiance at the surface of a blackbody (9.16)

dFλ =IλcosθdΩa

Fλ = dFλΩa∫ = Iλ cosθdΩaΩa

∫ = Iλ cosθsinθdθdφ0π 2∫0

2π∫

Fλ =Iλ cosθsinθdθdφ0π 2∫0

2π∫ =πIλ

Fλ =πIλ =πBλ,T

Spectral Irradiance v. Temperature

Fig. 9.4

10

-32

10

-24

10

-16

10

-8

10

0

10

8

10

16

10

24

10

32

10

40

10

-6

10

-4

10

-2

10

0

10

2

10

4

10

6

10

8

10

10

10

12

10

-6

10

-4

10

-2

10

0

10

2

10

4

10

6

10

8

10

10

10

12

Radiation intensity (W m

-2

-1

)

Wavgth( )

6000K

15iioK

Gaa

XUV

Visib Ifrard

ShortradioAMradio

Logradio

Tvisio&FMradio

300K

1KIrra

dian

ce (W

m-2

m-1)

Emission Spectra of the Sun and Earth

Fig. 9.5

Irradiance emission versus wavelength for the Sun and Earth when both are considered blackbodies

10

-4

10

-2

10

0

10

2

10

4

0.01 0.1 1 10 100

Radiation intensity (W m

-2

-1

)

Wavgth( )

Su

Earth

Utraviot

Visib

Ifrard

Irra

dian

ce (W

m-2

m-1)

Ultraviolet and Visible Solar Spectrum

Fig. 9.6

Ultraviolet and visible portions of the solar spectrum.

2 10

3

4 10

3

6 10

3

8 10

3

1 10

4

1.2 10

4

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

Radiation intensity (W m

-2

-1

)

Wavgth( )

UV-C

UV-B

UV-A

Visib

Far

UV

Nar

UV

Rd

GrBu

Irra

dian

ce (W

m-2

m-1)

Wien’s Displacement Law

Differentiate Planck's law with respect to wavelength at constant temperature and set result to zero

Peak wavelength of emissions from blackbody (9.17)

Example 9.2:

Sun’s photosphere p = 2897/5800 K = 0.5 m

Earth’s surface p = 2897/288 K = 10.1 m

λp μm( ) ≈2897T K( )

Wien’s Displacement Law

Fig. 9.7

Gives line through peak irradiances at different temperatures.

10

-4

10

-2

10

0

10

2

10

4

0.01 0.1 1 10 100

Spectral irradiance (W m

-2

-1

)

Wavgth( )

6000K

4000K

2000K

1000K

300K

0.5

10

Irra

dian

ce (W

m-2

m-1)

Stefan-Boltzmann LawIntegrate Planck irradiance over all wavelengths

Stefan-Boltzmann law (W m-2) (9.18)

Example 9.3:

T = 5800 K ---> FT = 64 million W m-2

T = 288 K ---> FT = 390 W m-2

Stefan-Boltzmann constant

Fb =π Bλ,T dλ0∞∫ =σBT4

σB = 2kB4π4

15h3c2 =5.67×10−8 W m-2 K-4

Reflection and Refraction

Fig. 9.8

ReflectionAngle of reflection equals angle of incidence

RefractionAngle of wave propagation relative to surface normal changes as the wave passes from a medium of one density to that of another

Water

θ1 θ3

θ2

Air

Icidt Rfctd

Rfractd

Reflection

Table 9.2

Albedo = fraction of incident sunlight reflected

Surface Type Albedo (fraction) Surface Type Albedo (fraction)Earth & atmosphere 0.3 Soil 0.05 - 0.2Liquid water 0.05 - 0.2 Grass 0.16 - 0.26Fresh snow 0.75 - 0.95 Desert 0.20 - 0.40Old snow 0.4 - 0.7 Forest 0.10 - 0.25Thick clouds 0.3 - 0.9 Asphalt 0.05 - 0.2Thin clouds 0.2 - 0.7 Concrete 0.1 - 0.35Sea Ice 0.25 - 0.4 Urban 0.1 - 0.27

Albedos in the non-UVB solar spectrum

RefractionSnell’s law (9.19)

Real part of the index of refraction (≥1) (9.20)Ratio of speed of light in a vacuum to that in a given medium

Real part of the index of refraction of air (9.21)

n2n1

= sinθ1sinθ2

n1 = cc1

na,λ −1=10−8 8342.13+2,406,030130−λ−2 + 15,997

38.9−λ−2⎛ ⎝ ⎜ ⎞

⎠ ⎟

Real Refractive Indices v Wavelength

Table 9.2

Wavelength (m) Air Water

0.3 1.000292 1.3490.5 1.000279 1.3351.0 1.000274 1.32710.0 1.000273 1.218

RefractionExample 9.4: = 0.5 m

θ1 = 45o

---> nair = 1.000279---> nwater = 1.335---> θ2 = 32o

---> cair = 2.9971 x 108 m s-1

---> cwater = 2.2456 x 108 m s-1

n2n1

= sinθ1sinθ2

Water

θ1 θ3

θ2

Air

Icidt Rfctd

Rfractd

Total Internal Reflection

Critical angle (9.22)

Example 9.5:

θ2,c =sin−1 n1n2

sin90o⎛ ⎝ ⎜

⎞ ⎠ ⎟

= 0.5 m---> nair = 1.000279---> nwater = 1.335---> θ2,c = 48.53o

Geometry of a Primary Rainbow

Fig. 9.9

Blue Red

BlueRed

42o 40o

Visibleradiation

Diffraction Around A Particle

Fig. 9.10

Primarywavefronts

Source

Particle

Secondarywavefronts

Diffracted rays

A

Huygens' principleEach point of an advancing wavefront may be considered the source of a new series of secondary waves

Radiation Scattering by a Sphere

Fig. 9.11

Ray A is reflectedRay B is refracted twiceRay C is diffractedRay D is refracted, reflected twice, then refractedRay E is refracted, reflected once, and refracted

A Sidescattering

B

C

D SidescatteringE

Backscattering

Forwardscattering

Forward and Backscattering

Fig. 9.12

Cloud dropletsScatter primarily in the forward direction

Gas moleculesScatter evenly in the forward and backward directions.

Backscattering Forward scattering

Cloud drop

Sunlight

Change in Color of Sun During the Day

Fig. 9.13

Blue

Red

Earth Atmosphere Space

White Noon

Yellow

Afternoon

Sunset /Twilight

Red

Green

Blue

BlueGreen

Red

Gas Absorption

Table 9.4

Gas Absorption wavelengths (m)

Visible/Near-UV/Far-UV absorbersOzone < 0.35, 0.45-0.75Nitrate radical < 0.67Nitrogen dioxide < 0.71

Near-UV/Far-UV absorbersFormaldehyde < 0.36Nitric acid < 0.33

Far-UV absorbersMolecular oxygen < 0.245Carbon dioxide < 0.21Water vapor < 0.21Molecular nitrogen < 0.1

Gas Absorption

Fig 9.14

Fraction of transmitted radiation through all important gases

0

0.2

0.4

0.6

0.8

1

1 10 100 1000

Line-by-line

Model

Fraction transmitted

Wavelength ( )

Evgass

Frac

tion

trans

mitt

ed

Gas Absorption

Fig 9.14

Fraction of transmitted radiation through water vapor

0

0.2

0.4

0.6

0.8

1

1 10 100 1000

Line-by-line

Model

Fraction transmitted

Wavelength ( )

H

2

O

Frac

tion

trans

mitt

ed

Gas Absorption

Fig 9.14

0

0.2

0.4

0.6

0.8

1

1 10 100 1000

Line-by-line

Model

Fraction transmitted

Wavelength ( )

CO

2

Frac

tion

trans

mitt

ed

Gas Absorption

Fig 9.14

0

0.2

0.4

0.6

0.8

1

1 10 100 1000

Line-by-line

Model

Fraction transmitted

Wavelength ( )

O

3

Frac

tion

trans

mitt

ed

Gas Absorption

Fig 9.14

0.7

0.75

0.8

0.85

0.9

0.95

1

1 10 100 1000

Line-by-line

Model

Fraction transmitted

Wavelength ( )

CH

4

Frac

tion

trans

mitt

ed

Extinction Coefficient

Fig 9.15

Attenuation of incident radiance, Io, due to absorption as it travels through a column of gas.

I0 I

x0 xdx

Extinction coefficient (s) (cm-1, m-1, or km-1)A measure of the loss of radiation per unit distance

Extinction CoefficientReduction in radiance with distance through a gas (9.23)

Integrate (9.24)

Extinction coefficient due gas absorption (9.25)

Transmission (9.29)

dIλdx =−Nqba,g,q,λ,T Iλ =−σa,g,q,λ,T Iλ

Iλ =I0,λe−Nqba,g,q,λ,T x−x0( ) =I0,λe−σa,g,q,λ,T x−x0( )

σa,g,λ = Nqba,g,q,λ,Tq=1

Nag

∑ = σa,g,q,λ,Tq=1

Nag

∑

Ta,g,q,λ = IλI0

=e−uqka,g,q,λ =e−σa,g,q,λ x−x0( )

Absorption CoefficientExtinction coefficient in terms of mass absorption (9.26)

σa,g,λ = ρqka,g,q,λq=1

Nag

∑ =Nqmq

A ka,g,q,λq=1

Nag

∑ =uq

x−x0( )q=1

Nag

∑ ka,g,q,λ

Fig 9.1610

-3

10

-2

10

-1

10

0

10

1

10

2

6.619 6.6205 6.622 6.6235 6.625

322.15 hPa

22.57 hPa

Absorption coefficient (cm

2

/g)

Wavelength ( )

Abs

orpt

ion

coef

ficie

nt (c

m2 /g

)

Absorption Coefficient

Absorption Coefficient (9.27)

Pressure-broadened half-width (9.28)

ka,g,q,λ = Amq

1π

Sq T( )γq pa,T( )γq pa,T( )2 + ˜ ν λ − ˜ ν q −δq pref( )pa( )[ ]

2

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

γq pa,T( )=TrefT

⎛ ⎝ ⎜ ⎞

⎠ ⎟ nq

γair,q pref,Tref( ) pa−pq( )+γself,q pref,Tref( )pq[ ]

Transmission Example

Monochromatic transmission (9.29)

Exact transmission when two absorption lines (9.30)

Transmission overestimated when lines averaged (9.31)

Ta,g,q,λ uq( ) =IλI0

=e−uqka,g,q,λ =e−σa,g,q,λ x−x0( )

T˜ ν ,˜ ν +Δ˜ ν u( ) =0.5 e−kxu+e−kyu( )

T˜ ν ,˜ ν +Δ˜ ν u( ) =e−0.5 kx+ky( )u

Correlated k-Distribution Method

Exact transmission in wavenumber interval (9.32)

Integration of differential probability is unity (9.33)

Reorder absorption coefs. into cumulative frequency distribution (9.34)

T˜ ν ,˜ ν +Δ˜ ν u( ) = 1Δ˜ ν e−k˜ ν ud˜ ν ˜ ν

˜ ν +Δ̃ ν ∫ ≈ e− ′ k uf˜ ν ′ k ( )d ′ k 0∞∫

f˜ ν ′ k ( )d ′ k 0∞∫ =1

g˜ ν k( ) = f˜ ν ′ k ( )d ′ k 0k∫

Effects on Visibility of Gas AbsorptionMeteorological range (Koschmieder equation)

Meteorological ranges due to Rayleigh scattering and NO2 absorption

x= 3.912σext,λ

Table 9.5

<-- NO2 absorption -->Wavelength Rayleigh Scat. 0.01 ppmv 0.25 ppmv(mm) (km) (km) (km)0.42 112 296 11.80.50 227 641 25.60.55 334 1,590 63.60.65 664 13,000 520

Effects on Visibility of Gas AbsorptionExtinction coefficient due to NO2 and O3 absorption.

Fig. 9.17

10

-6

10

-5

10

-4

10

-3

10

-2

10

-1

10

0

10

1

0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7

Extinction coefficient (km

-1

)

Wavelength ( )

NO

2

(g)(0.25ppv)

O

3

(g)(0.25ppv)

NO

2

(g)(0.01ppv)

O

3

(g)(0.01ppv)

Extin

ctio

n co

effic

ient

(km

-1)

Gas Scattering

Extinction coefficient due to Rayleigh scattering (9.35)

Scattering cross section of a typical air molecule (cm2) (9.36)

Anisotropic correction factor (9.37)

Rayleigh scatterer: 2pr/ <<1

σs,g,λ =Nabs,g,λ

bs,g,λ =8π3 na,λ

2 −1( )2

3λ4Na,02 f δ*( ) ≈32π3 na,λ −1( )2

3λ4Na,02 f δ*( )

f δ*( ) =6+3δ*6−7δ*

≈1.05

Rayleigh Scattering ExampleExample 9.6:

= 0.5 mpa = 1 atm (sea level) T = 288 K

---> ss,g,= 1.72 x 10-7 cm-1

---> x = 227 km

= 0.55 m---> ss,g,= 1.17 x 10-7 cm-1

---> x = 334 km

Imaginary Index of Refraction Measure of extent to which a substance absorbs radiation

Attenuation of incident radiance, I0, due to absorption

Fig 9.18

x0 x

I0 I

dx

Equation for attenuation (9.38)

Integrate (9.39)

dIdx =−4πκ

λ I

I =I0e−4πκ x−x0( ) λ

Complex Index of Refraction(9.40)

Real and imaginary refractive indices at = 0.5 and 10 m

mλ =nλ −iκλ

Table 9.6

<-- 0.5 m --> <-- 10 m -->Substance Real Imaginary Real ImaginaryLiquid water 1.34 1x10-9 1.22 0.05Black carbon 1.82 0.74 2.4 1.0Organic matter 1.45 0.001 1.77 0.12Sulfuric acid 1.43 1x10-8 1.89 0.46

Transmission

Light transmission through particles at = 0.5 m

Table 9.6

<-- Transmission (I/I0) -->Diameter (m) Black carbon Water(=0.74) (=1x10-9)

0.1 0.16 0.9999999971.0 8x10-9 0.9999999710 0 0.9999997

Imaginary Refractive Index of Liquid Nitrobenzene

Fig. 9.19

0

0.1

0.2

0.3

0.4

0.5

0.2 0.3 0.4 0.5 0.6 0.7

Imaginary index of refraction

Wavelength ( )

Imag

inar

y in

dex

of re

frac

tion

Particle Extinction CoefficientsParticle absorption/scattering extinction coefficients (9.41)

Particle absorption/scattering cross sections (9.42)

σa,a,λ = niba,a,i,λi=1

NB∑

σs,a,λ = nibs,a,i,λi=1

NB∑

ba,a,i,λ =πri2Qa,i,λ

bs,a,i,λ =πri2Qs,i,λ

Tyndall Absorption / ScatteringRayleigh regime (di<0.03 or ai,<0.1)

Tyndall absorption efficiency (linear with ri/) (9.44)

Size parameter (9.43)αi,λ =2πri λ

Qa,i,λ =−42πriλ Im

mλ2 −1

mλ2 +2

2⎛

⎝ ⎜ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⎟

≈2πriλ

24nλκλnλ

2 +κλ2( )

2+4 nλ

2 −κλ2 +1( )

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

κλ → 0 Qa,i,λ =2πriλ

24nλκλnλ

2+2( )2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

(9.45)

--> --> linear with

Tyndall Absorption / Scattering

Tyndall scattering efficiency [linear with (ri/)4] (9.46)

Example 9.7 (Liquid water):

= 0.5 m ri = 0.01 m

---> = 1.34---> = 1.0 x 10-9

---> Qs,i, = 2.9 x 10-5

---> Qa,i, = 2.8 x 10-10

Qs,i,λ =83

2πriλ

⎛ ⎝ ⎜ ⎞

⎠ ⎟ 4 mλ

2 −1mλ

2 +2

2

Mie Absorption / ScatteringMie regime (0.03< di <32 or 0.1< ai,<100)

Single particle Mie absorption efficiency

Single particle Mie scattering efficiency (9.47)

Single particle total extinction coefficient (9.48)

Qs,i,λ = 2αi,λ

2k+1( ) ak2+bk

2( )

k=1

∞∑

Qa,i,λ =Qe,i,λ −Qs,i,λ

Qe,i,λ = 2αi,λ

2k+1( )Re ak +bk( )k=1

∞∑

Soot Absorption/Scattering Efficiencies

Fig. 9.20

Single Particle Absorption/Scattering Efficiency at = 0.50 m

0

0.5

1

1.5

2

0

0.5

1

1.5

2

0.01 0.1 1 10 100 1000

Particle diameter ( )

Q

a

Q

s

Q

f

Mirgi Gotric

rgi

Rayighrgi

Ray

leig

h re

gim

e

Water Absorption/Scattering Efficiencies

Fig. 9.21

Single Particle Absorption/Scattering Efficiency at = 0.50 m.

0

1

2

3

4

5

10

-9

10

-7

10

-5

10

-3

10

-1

10

1

10

3

0.01 0.1 1 10 100 1000

qsc

Particle diameter ( )

Q

a

Q

s

Q

f

Mirgi Gotric

rgi

Rayighrgi

Ray

leig

h re

gim

e

Geometric Absorption / Scattering

In the limit (ai,-->∞), scattering efficiency is constant (9.49)

Example 9.8:

Geometric regime (di >32 or ai,>100) --> significant diffraction

Also, as ai,-->∞, Qs,i,≈Qs,i, regardless of how weak the imaginary index of refraction is.

limαi ,λ→ ∞

Qs,i,λ =1+ mλ −1mλ +2

= 0.5 m---> = 1.34 for liquid water---> Qs,i, = 1.1 as ai,-->∞ from9.49 ---> Qa,i, = 1.1 as ai,-->∞ from Fig. 9.21

Mixing Rules

Volume average dielectric constant (9.54)

Volume average (9.50)

Complex dielectric constant (9.51)

mλ =υqυ mλ,q

q=1

NV∑ =υqυ nλ −iκλ( )

q=1

NV∑

mλ2 =

υqυ mλ,q

2q=1

NV∑ =υqυ ελ,q

q=1

NV∑ =υqυ εr,λ −iεi,λ( )

q=1

NV∑

ελ =mλ2 = nλ −iκλ( )2 =nλ

2 −κλ2 −i2nλκλ =εr,λ −iεi,λ

Mixing Rules

Real and imaginary refractive indices (9.53)

Real/imaginary complex dielectric constant (9.52)

εr,λ =nλ2 −κλ

2 εi,λ =2nλκλ

nλ =εr,λ2 +εi,λ

2 +εr,λ2 κλ =

εr,λ2 +εi,λ

2 −εr,λ2

Mixing Rules

Bruggeman (9.56)

Maxwell Garnett (9.55)

mλ2 =ελ =ελ,M 1+

3υA,q

υελ,A,q −ελ,Mελ,A,q +2ελ,M

⎛ ⎝ ⎜ ⎜

⎞ ⎠ ⎟ ⎟

q=1

NA∑

1− υA,qυ

ελ,A,q −ελ,Mελ,A,q+2ελ,M

⎛ ⎝ ⎜ ⎜

⎞ ⎠ ⎟ ⎟

q=1

NA∑

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

υA,qυ

ελ,A,q −ελελ,A,q +2ελ

⎛ ⎝ ⎜ ⎜

⎞ ⎠ ⎟ ⎟

q=1

NA∑ + 1−υA,q

υq=1

NA∑⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

ελ,M −ελελ,M +2ελ

⎛ ⎝ ⎜

⎞ ⎠ ⎟

Absorption Cross Section Enhancement

Fig. 9.22

Absorption cross section enhancement due to internal mixing

0

1

2

3

4

5

6

0.01 0.1 1 10 100

0.04

0.08

0.12

Total particle (soot core+S(VI) shell) diameter ( )

Absorptiocrosssctiohactfactor

=0.5 d

BC

( )

Abs

orpt

ion

cros

s se

ctio

nEn

hanc

emen

t fac

tor

Modeled Extinction Coefficient Profile

Fig. 9.23a

0 0.1 0.2 0.3 0.4 0.5

300

400

500

600

700

800

900

1000

0 0.1 0.2 0.3 0.4 0.5

(km-1

)

Pressure (hPa)

Claremont

8/27/97

11:30

0.32

Gasscat.

Ar.scat.

Ar.abs.Gasabs.

Pres

sure

(hPa

)

Modeled Extinction Coefficient Profile

Fig. 9.23b

0 0.1 0.2 0.3 0.4 0.5

300

400

500

600

700

800

900

1000

0 0.1 0.2 0.3 0.4 0.5

(km-1

)

Pressure (hPa)

Claremont

8/27/97

11:30

0.61

Gasscat.

Ar.scat.

Ar.abs.

Gasabs.

Pres

sure

(hPa

)

Visibility DefinitionsMeteorological range

Distance from an ideal dark object at which the object has a 0.02 liminal contrast ratio against a white backgroundLiminal contrast ratio

Lowest visually perceptible brightness contrast a person can seeVisual range

Actual distance at which a person can discern an ideal dark object against the horizon skyPrevailing visibility

Greatest visual range a person can see along 50 percent or more of the horizon circle (360o), but not necessarily in continuous sectors around the circle.

Visibility DefinitionsThe intensity of radiation increases from 0 at point x0 to I at point x due to the scattering of background light into the viewer’s path

x0 xdx

Scattering out of path

Scattering into path

I0 I

Fig. 9.24

Meteorological Range

Change in object intensity along path of radiation (9.58)

Contrast ratio (9.57)

Total extinction coefficient (9.59)

Cratio=IB −IIB

dIdx =σt IB −I( )

σt =σa,g +σs,g+σa,p +σs,pIntegrate (9.58) and substitute into (9.57) (9.60)

Cratio=IB −IIB

=e−σtx

Meteorological range (Koschmieder equation) (9.61)

x=3.912σt

Meteorological Range

(Larson et al., 1984)Table 9.8

Meteorological Range (km)Gas

scatteringGas

absorptionParticle

scatteringParticle

absorptionAll

Polluted day

366 130 9.59 49.7 7.42

Less-polluted

day

352 326 151 421 67.1

Optical Depth

Incremental distance versus incremental path length (9.63)

Total extinction coefficient (9.62)

Incremental optical depth (9.66)

Optical depth as a function of altitude (9.63)

σλ =σs,g,λ +σa,g,λ +σs,a,λ +σa,a,λ +σs,h,λ +σa,h,λ

dz =cosθsdSb =μsdSb

dτλ =−σλdz =−σλμsdSb

τλ = σλdz∞z∫ = σλμsdSb∞

Sb∫

Optical Depth

Sun

dzdt dSb

θs

t=t z=0 Sb=0

t=0 z=if. Sb=if.

Fig. 9.22

Relationship between optical depth, altitude, solar zenith angle, and pathlength

Solar Zenith Angle

Solar declination angle (d)Angle between the equator and the north or south latitude of the subsolar point

Subsolar pointPoint at which the sun is directly overhead

Local hour angle (Ha)Angle, measured westward, between longitude of subsolar point and longitude of location of interest.

Cosine of solar zenith angle (9.67)

cosθs =sinϕsinδ+cosϕcosδcosHa

Solar Zenith Angle

Fig. 10.26 b

Geometry for zenith angle calculations.

Surfacenormal

Sun

d

Ha

f

θs

Solar Declination Angle

Obliquity of the ecliptic (9.69)Angle between the plane of the Earth's equator and the plane of the ecliptic, which is the mean plane of the Earth's orbit around the Sun.

Solar declination angle (9.68)

δ =sin−1 sinεobsinλec( )

εob =23o.439−0o.0000004NJD

Solar Declination Angle

Mean longitude of the sun (9.72)

Ecliptic longitude of the sun (9.71)

Mean anomaly of the sun (9.72)

λec =LM +1o.915singM +0o.020sin2gM

LM =280o.460+0o.9856474NJD

gM =357o.528+0o.9856003NJD

Solar Zenith Angle

Example:At noon, when sun is directly overhead, Ha = 0 --->

Local hour angle (9.73)

When the sun is over the equator, d = 0 --->

Ha = 2πts86,400

cosθs =sinϕsinδ+cosϕcosδ

cosθs =cosϕcosHa

Solar Zenith Angle

Example 9.9:1:00 p.m., PST, Feb. 27, 1994, φ = 35 oN

---> DJ = 58---> NJD = -2134.5

---> gm = -1746.23o

---> Lm = -1823.40o

---> ce = -1821.87o

---> ob = 23.4399o

---> d = -8.52o

---> Ha = 15.0o

--->

---> θs = 45.8o

cosθs =sin35o( )sin−8.52o( ) +cos35o( )cos−8.52o( )cos15.0o( )

Solstices and Equinoxes

Fig. 9.27

Solar declinations during solstices and equinoxes. The Earth-Sun distance is greatest at the summer solstice.

.

N.H. winter, S.H. summersolstice, Dec. 22

23.5o

23.5o

Vernal equinox, Mar. 20

Autumnal equinox, Sept. 23

Tropic ofCancer

Topic ofCapricorn

Earth

Sun

Sun

Sun

Equator

Axis ofrotation

N.H. summer, S.H. wintersolstice, June 22

Radiative Transfer Equation

Scattering of radiation out of the beam (9.75)

Change in radiance / irradiance along a beam of interest

Absorption of radiation along the beam (9.76)

Change in radiance along incremental path length (9.74)

dIλ =−dIso,λ −dIao,λ +dIsi,λ +dISi,λ +dIei,λ

dIso,λ =Iλσs,λdSb

dIao,λ =Iλσa,λdSb

Radiative Transfer Equation

Single scattering of direct solar radiation into beam (9.78)

Emission of infrared Planckian radiation into beam (9.79)

Multiple scattering of diffuse radiation into the beam (9.77)

dIsi,λ = σs,k,λ4π Iλ, ′ μ , ′ φ Ps,k,λ,μ, ′ μ ,φ, ′ φ d ′ μ −1

1∫ d ′ φ 02π∫⎛

⎝ ⎜ ⎞ ⎠ ⎟ k∑⎡

⎣ ⎢ ⎤ ⎦ ⎥ dSb

dISi,λ = σs,k,λ4π Ps,k,λμ,−μs,φ,φs

⎛ ⎝ ⎜ ⎞

⎠ ⎟ k∑⎡ ⎣ ⎢

⎤ ⎦ ⎥ Fs,λe−τλ μsdSb

dIei,λ =σa,λBλ,TdSb

Extinction Coefficients

Extinction due to total scattering plus absorption (9.81)

Extinction due to total scattering (9.80)

Extinction due to total absorption (9.80)

σs,λ =σs,g,λ +σs,a,λ +σs,h,λ

σa,λ =σa,g,λ +σa,a,λ +σa,h,λ

σλ =σs,λ +σa,λ

Scattering Phase Function

redirects diffuse radiation from ’, f’ to , f

Gives angular distribution of scattered energy vs. direction

Scattering phase function for diffuse radiation

Scattering phase function for direct radiation

redirects direct solar radiation from -, f to , f

Ps,k,λ,μ, ′ μ ,φ, ′ φ

Ps,k,λ,μ,−μs,φ,φs

Scattering Phase FunctionSingle scattering of direct solar radiation and multiple scattering of diffuse radiation.

Fig. 9.28

Sun

(,φ)

(',φ')

(-s,φs)Fs

Qs

Q

Mutipscattrig

Diffus

Sigscattrig

Dirct

Ba

Scattering Phase Function

Substitute --> (9.83)

Scattering phase function defined such that (9.82)

Q = angle between directions ’, f’ and , f

14π Ps,k,λ Θ( )4π∫ dΩa =1

dΩa =sinΘdΘdφ

14π Ps,k,λ Θ( )sinΘdΘdφ0

π∫02π∫ =1

Scattering Phase Function

Phase function for Rayleigh scattering (9.85)

Phase function for isotropic scattering (9.84)

Henyey-Greenstein function (9.86)

Ps,k,λ Θ( )=1

Ps,k,λ Θ( )=34 1+cos2Θ( )

Ps,k,λ Θ( )=1−ga,k,λ

2

1+ga,k,λ2 −2ga,k,λ cosΘ( )

32

Scattering Phase FunctionScattering phase functions for (a) isotropic and (b) Rayleigh scattering

Fig. 9.29

-1.5

-1

-0.5

0

0.5

1

1.5

-2-1.5-1-0.500.511.52

Q

( Q )Ps

-1.5

-1

-0.5

0

0.5

1

1.5

-2-1.5-1-0.500.511.52

Q

( Q )Ps

(a) (b)

Asymmetry Factor

Expand with ---> (9.89)

First moment of phase function -- relative direction of scattering (9.88)

Isotropic Scattering ---> ---> (9.90)

(9.87)

ga,k,λ>0 forward(Mie) scattering=0 isotropicor Rayleighscattering<0 backwardscattering

⎧ ⎨ ⎪

⎩ ⎪

ga,k,λ = 14π Ps,k,λ Θ( )cosΘdΩa4π∫

dΩa =sinΘdΘdφ

ga,k,λ = 14π Ps,k,λ Θ( )cosΘsinΘdΘdφ0

π∫02π∫

Ps,k,λ Θ( )=1

ga,k,λ = 14π cosΘsinΘdΘdφ0

π∫02π∫ =0

Asymmetry Factor

Mie scattering ---> (9.92)

Rayleigh scattering ---> (9.91)

Backscatter ratio (9.93)

ga,k,λ = 14π

34 1+cos2Θ( )cosΘsinΘdΘdφ=00

π∫02π∫

ga,k,λ =Q f,i,λ Qs,i,λ

ba,k,λ =Ps,k,λ Θ( )sinΘdΘdφπ 2

π∫02π∫

Ps,k,λ Θ( )sinΘdΘdφ0π∫0

2π∫

Energy Emitted by Sun to Earth

Lp = emissions from Sun's photosphere = 3.9 x 1026 WRp = radius from Sun center to photosphere = 6.96 x 108 mTp = temperature of photosphere = 5796 K

Summed irradiance (W m-2) emitted at Sun's photosphere (9.94)

Example 9.10:

Tp = 5796 K--->Fp ≈ 6.4 x 107 W m-2

Fp =Lp

4πRp2 =σBTp

4

Solar Constant

Calculated ≈ 1379 W m-2

Mean summed irradiance at top of Earth's atmosphere (9.95)

Measured ≈ 1365 W m-2

Varies by +/- 1 W m-2 over each 11 year sunspot cycle

F s =RpR es

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 2

Fp =RpR es

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 2σBTp

4

F sF s

Solar ConstantDaily solar flux depends on Earth-Sun distance (9.96)

Empirical formula (9.97)

Fs = ResR es

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 2

F s

ResR es

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 2

≈1.00011+0.034221cosθJ +0.00128sinθJ

+0.000719cos2θ J +0.000077sin2θJ

θJ =2πDJ DY

Incident Solar FluxExample 9.11:December 22---> Fs = 1365 x 1.034 = 1411 W m-2

June 22---> Fs = 1365 x 0.967 = 1321 W m-2

--> Irradiance varies by 90 W m-2 (6.6%) between Dec. and June

Cumulative solar irradiance as sum of spectral irradiances (9.98)

Fs = Fs,λΔλ( )λ∑ = Res

R es

⎛ ⎝ ⎜

⎞ ⎠ ⎟

2F s,λΔλ( )

λ∑ = Res

R es

⎛ ⎝ ⎜

⎞ ⎠ ⎟ 2F s

SeasonsRelationship between the Sun and Earth during the solstices and equinoxes

Fig. 9.30

Obliquity23.5o

Sun

Autumnal equinoxSept. 23

Vernal equinoxMar. 20

147 million km 152 million km

N.H. winterS.H. summer

solsticeDec. 22

N.H. summerS.H. winter

solsticeJune 22

Equilibrium Earth Temperature

Energy (W) absorbed by the Earth-atmosphere system (9.99)

Energy emitted by the Earth's surface (9.100)

Equate --> temperature without greenhouse effect (9.101)

Ein =F s 1−Ae,0( ) πRe2( )

Eout=εe,0σBTe4 4πRe

2( )

Te=F s 1−Ae,0( )4εe,0σB

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

14

Equilibrium Earth Temperature

Example 9.12: = 1365 W m-2

Ae,0 = 0.3

---> Te = 254.8 K

Actual average surface temperature on earth ≈ 288 K --> difference due to absorption by greenhouse gases

F s

Radiative Transfer Equation(9.102)

Single scattering albedo (9.103)

Total extinction coefficient

Optical depth

dIλ,μ,φdSb

=−Iλ,μ,φ σs,λ +σa,λ( )+ σs,k,λ4π Iλ, ′ μ , ′ φ Ps,k,λ,μ, ′ μ ,φ, ′ φ d ′ μ −1

1∫ d ′ φ 02π∫⎛

⎝ ⎜ ⎞ ⎠ ⎟

k∑

+Fs,λe−τλ μs σs,k,λ4π Ps,k,λ,μ,−μs,φ,φs

⎛ ⎝ ⎜ ⎞

⎠ ⎟ k∑ +σa,λBλ,T

ωs,λ =σs,λσλ

=σs,g,λ +σs,a,λ +σs,h,λ

σs,g,λ +σa,g,λ +σs,a,λ +σa,a,λ +σs,h,λ +σa,h,λ

σλ =σs,λ +σa,λ

dτλ =−σλμsdSb

Radiative Transfer Equation

where (9.105)

Rewrite radiative transfer equation (9.104)

(9.106)

(9.107)

μdIλ,μ,φdτλ

=Iλ,μ,φ−Jλ,μ,φdiffuse−Jλ,μ,φ

direct−J λ,μ,φemis

J λ,μ,φdiffuse= 1

4πσs,k,λ

σλIλ, ′ μ , ′ φ Ps,k,λ,μ, ′ μ ,φ, ′ φ d ′ μ −1

1∫ d ′ φ 02π∫⎛

⎝ ⎜ ⎞ ⎠ ⎟

k∑

J λ,μ,φdirect= 1

4π Fs,λe−τλ μs σs,k,λσλ

Ps,k,λ,μ,−μs,φ,φs⎛ ⎝ ⎜

⎞ ⎠ ⎟

k∑

J λ,μ,φemis = 1−ωs,λ( )Bλ,T

Beer’s Law

Solution (9.109)

Consider only absorption in downward direction (9.108)

−μdIλ,−μ,φdτa,λ

=Iλ,−μ,φ

Iλ,−μ,φ τa,λ( ) =Iλ,−μ,φ τa,λ,t( )e− τa,λ −τa,λ,t( ) μ

Schwartzchild’s Equation

Solution (9.112)

Consider absorption and infrared emissions (9.111)

−μdIλ,−μ,φdτa,λ

=Iλ,−μ,φ−Bλ,T

Iλ,−μ,φ τa,λ( ) =Iλ,−μ,φ τa,λ,t( )e− τa,λ −τa,λ,t( ) μ

+1μ Bλ,T ′ τ a,λ( )e

− τa,λ − ′ τ a,λ( ) μ⎡ ⎣ ⎢

⎤ ⎦ ⎥ τa,λ,t

τa,λ∫ d ′ τ a,λ

Two-Stream Method

Substitute (9.114) into (9.105) (9.115)

Divide phase function into upward (+) and downward component (9.114)

14π Iλ, ′ μ , ′ φ Ps,k,λ,μ, ′ μ ,φ, ′ φ d ′ μ −1

1∫ d ′ φ 02π∫ ≈

1+ga,k,λ( )2 I++

1−ga,k,λ( )2 I− upward

1+ga,k,λ( )2 I−+

1−ga,k,λ( )2 I+ downward

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

14π

σs,k,λσλ

Iλ, ′ μ , ′ φ Ps,k,λ,μ, ′ μ ,φ, ′ φ d ′ μ −11∫ d ′ φ 0

2π∫⎛ ⎝ ⎜

⎞ ⎠ ⎟

k∑ ≈

ωs,λ 1−bλ( )I++ωs,λbλI−

ωs,λ 1−bλ( )I−+ωs,λbλI+

⎧ ⎨ ⎪

⎩ ⎪

Two-Stream Method

Integrated fraction of backscattered energy (9.116)

Integrated fraction of forward scattered energy (9.116)

Effective asymmetry parameter (9.117)

1−bλ =1+ga,λ2

bλ =1−ga,λ2

ga,λ =σs,a,λga,a,λ +σs,c,λga,h,λσs,g,λ +σs,a,λ +σs,h,λ

Modeled Asymmetry Parameter

0 0.2 0.4 0.6 0.8 1

300

400

500

600

700

800

900

1000

0 0.2 0.4 0.6 0.8 1

Pressure (hPa)

Claremont

8/27/97

11:30

0.32

Witharosos

Noarosos

g

a

ω

s

Pres

sure

(hPa

)

Fig. 9.28

0 0.2 0.4 0.6 0.8 1

300

400

500

600

700

800

900

1000

0 0.2 0.4 0.6 0.8 1

Pressure (hPa)

Claremont

8/27/97

11:30

0.61

Witharosos

Noarosos

g

a

ωs

Pres

sure

(hPa

)

Two-Stream Approximation

Downward radiance equation (9.120)

Upward radiance equation (9.119)

Irradiances in terms of radiance for two-stream approximation

μ1dI ↑dτ =I ↑−ωs 1−b( )I ↑−ωsbI ↓−ωs

4π 1−3gaμ1μs( )Fse−τ μs

−μ1dI ↓dτ =I ↓−ωs 1−b( )I ↓−ωsbI ↑−ωs

4π 1+3gaμ1μs( )Fse−τ μs

F ↑=2πμ1I ↑ F ↑=2πμ1I ↑

Two-Stream Approximation

Solar irradiance (9.121)

Substitute irradiances and generalize for different phase function approximations

Surface boundary condition (9.123)

dF ↑dτ =γ1F ↑−γ2F ↓−γ3ωsFse−τ μs

dF ↓dτ =−γ1F ↓+γ2F ↑+1−γ3( )ωsFse−τ μs

F ↑NL +12=AeF ↓NL +12 + AeμsFse−τNL +12 μs solarεπBT infrared

⎧ ⎨ ⎩

Two-Stream ApproximationCoefficients for two stream approximations using two techniques

Infrared irradiance (9.122)

Table 9.10

Approximation

Quadrature

Eddington

γ1 γ2 γ31−ωs 1+ga( ) 2

μ1ωs 1−ga( )

2μ11−3gaμ1μs

27−ωs 4+3ga( )

4 −1−ωs 4−3ga( )

42−3gaμs

4

dF ↑dτ =γ1F ↑−γ2F ↓−2π 1−ωs( )BT

dF ↓dτ =−γ1F ↓+γ2F ↑+2π 1−ωs( )BT

Delta Functions

--> adjust terms with delta functions (9.124)

Quadrature and Eddington solutions underpredict forward scattering because expansion of phase function is too simple to obtain the strong peak in scattering efficiency.

′ g a = ga1+ga

′ ω s =1−ga

2( )ωs

1−ωsga2

′ τ = 1−ωsga2( )τ

Heating RatesNet flux divergence equation (9.125)

Net downward minus upward radiative flux

Partial derivative term (9.126)

Temperature change (9.127)

∂T∂t

⎛ ⎝ ⎜ ⎞

⎠ ⎟ r

= 1cp,m

dQsolardt +dQir

dt⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 1cp,mρa

∂Fn∂z

Fn = F ↓λ −F ↑λ( )dλ0∞∫

∂Fn,k∂z ≈

F ↓λ,k−12 −F ↑λ,k−12( )− F ↓λ,k+12 −F ↑λ,k+12( )[ ]λ∑zk−12 −zk+12

ΔTk ≈ 1cp,mρa

∂Fn∂z h

Photolysis CoefficientsPhotolysis rate (s-1) at bottom of layer k (9.128)

Radiance at bottom of layer k (photons cm-2 m sr-1 s-1) (9.129)

Example 9.14: I = 12 W m-2 in band 0.495 m < < 0.505 m

---> mean = 0.5 m---> Ip, = 3.02 x 1015 photons cm-2 s-1

J q,p,k+12 = 4πIp,λ,k+120∞∫ ba,g,q,λ,TYq,p,λ,Tdλ

Ip,λ,k+12 = I ↓λ,k+12 −I ↑λ,k+12( ) 10−10 m3

cm2μm⎛ ⎝ ⎜

⎞ ⎠ ⎟ λhc