meto 621 lesson 8. thermal emission from a surface let be the emitted energy from a flat surface of...

18
METO 621 LESSON 8

Post on 19-Dec-2015

212 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

METO 621

LESSON 8

Page 2: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Thermal emission from a surface• Let

Iνc+ ( ˆ Ω )cosθ dω be the emitted

energy from a flat surface of temperature Ts , within the solid angle d in the direction A blackbody would emit B(Ts)cosd The spectral directional emittance is defined as

)(

)ˆ(

cos)(

cos)ˆ(),ˆ,(

S

e

S

eS TB

I

dTB

dIT

ε

=

=++

Page 3: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Thermal emission from a surface

• In general ε depends on the direction of emission, the surface temperature, and the frequency of the radiation. A surface for which ε is unity for all directions and frequencies is a blackbody. A hypothetical surface for which ε = constant<1 for all frequencies is a graybody.

Page 4: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Flux emittance

• The energy emitted into 2 steradians relative to a blackbody is defined as the flux or bulk emittance

),ˆ,(cos1

)(

)(),ˆ,(cos

)(cos

)ˆ(cos),2,(

S

S

SS

S

e

S

Td

TB

TBTd

TBd

IdT

=

=

∫∫∫

+

+

+

+

+

ε

ε

ε

Page 5: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Absorption by a surface

• Let a surface be illuminated by a downward intensity I. Then a certain amount of this energy will be absorbed by the surface. We define the spectral directional absorptance as:

• The minus sign in - emphasizes the downward direction of the incident radiation

)'ˆ(

)'ˆ(

''cos)'ˆ(

''cos)'ˆ(),'ˆ,(

=

=−−

α

II

dIdI

T aaS

Page 6: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Absorption by a surface• Similar to emission, we can define a flux

absorptance

• Kirchoff showed that for an opaque surface

α(ν ,− ˆ Ω ,TS ) = ε(ν , ˆ Ω ,TS )

• That is, a good absorber is also a good emitter, and vice-versa

−−

−∫ −=−

αα

F

ITd

TS

S1

)'ˆ(),'ˆ,('cos'

),2,(

Page 7: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Surface reflection : the BRDF

Consider a downward beam with intensity Iν− ( ˆ Ω ).

The energy incident on a flat surface is Iν−( ˆ Ω )cosθ dω' .

Let the intensity of the reflected light around the

direction ˆ Ω within a solid angle dω be dIνr− then

ρ (ν ,− ˆ Ω ', ˆ Ω )=dIνr

− ( ˆ Ω )

Iν ( ˆ Ω ')cosθ 'dω'

where ρ (ν ,− ˆ Ω ', ˆ Ω ) is the bidirectional relfectance

distribution function, or BRDF.

Page 8: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

BRDF

FIdI

IdIdI

LLr

L

rr

)()'ˆ('cos')(

and ,)()ˆ,'ˆ,( case In this

surface.Lambert a called isit then ,directions

n observatio and incidence both the oft independen

is which BRDF a has surface reflecting a If

)'ˆ()ˆ,'ˆ,('cos')ˆ()ˆ(

is beams all from

,ˆdirection in theintensity reflected totalThe

ρρ

ρρ

ρ

==

=−

−==

+

−+

+

∫∫

Page 9: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Surface reflectance - BRDF

Page 10: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Collimated incidence

Page 11: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Collimated Incidence - Lambert Surface

• If the incident light is direct sunlight then

I−() Ω ) = F Sδ(

) Ω −

) Ω 0) = F S (cosθ − cosθ0)δ(φ −ϕ 0)

The incident flux is given by F− = F S cosθ0

Hence Ir+ = ρ L cosθ0F

S

For a collimated beam the intensity reflected from

a Lambert surface is proportional to the cosine of

the angle of incidence.

Page 12: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Collimated Incidence - Specular reflectance

• Here the reflected intensity is directed along the angle of reflection only.

• Hence ’ and ’+• Spectral reflection function ρS(

Ir+( ˆ Ω ) = ρ S (θ)F Sδ(cosθ0 − cosθ)δ(φ − φ0 + π[ ])

• and the reflected flux:

Fr+ = ρ S (θ0)F S cosθ0

Page 13: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Absorption and Scattering in Planetary Media

• Kirchoff’s Law for volume absorption and Emission

ε (ν ,T) =α (ν ,T)

k(ν )

The volume emittance is proportional to the

absorption coefficient

Page 14: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Differential equation of Radiative Transfer

• Consider conservative scattering - no change in frequency.

• Assume the incident radiation is collimated• We now need to look more closely at the secondary

‘emission’ that results from scattering. Remember that from the definition of the intensity that

dddtdAIEd )'ˆ(4 Ω=+

Page 15: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Differential Equation of Radiative Transfer

σ ds d4 E '

• The radiative energy scattered in all directions is

• We are interested in that fraction of the scattered energy that is directed into the solid angle d centered about the direction

• This fraction is proportional to

p( ˆ Ω ', ˆ Ω ) dω /4π

Page 16: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Differential Equation of Radiative Transfer

• If we multiply the scattered energy by this fraction and then integrate over all incoming angles, we get the total scattered energy emerging from the volume element in the direction

d4 E = σ (ν ) dV dt dν dω dω'p( ˆ Ω ', ˆ Ω )

4π4π

∫ Iν' ( ˆ Ω ')

• The emission coefficient for scattering is

jνSC ≡

d4 E

dV dt dν dω= σ (ν )

dω'

4π4 π

∫ p( ˆ Ω ', ˆ Ω )Iν ( ˆ Ω ')

Page 17: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Differential Equation of Radiative Transfer

• The source function for scattering is thus

SνSC ( ˆ r , ˆ Ω ) =

jνSC

k(ν )=

σ (ν )

k(ν )

dω'

4π4 π

∫ p( ˆ Ω ', ˆ Ω ) Iν ( ˆ Ω ')

• The quantity σk( is called the single-scattering albedo and given the symbol a(

• If thermal emission is involved, (1-a) is the volume emittance ε

Page 18: METO 621 LESSON 8. Thermal emission from a surface Let be the emitted energy from a flat surface of temperature T s, within the solid angle d  in the

Differential Equation of Radiative Transfer

• The complete time-independent radiative transfer equation which includes both multiple scattering and absorption is

[ ] ∫ +−+−=

τ 4

)ˆ,'ˆ('4

)()()(1 Ipd

aTBaI

d

dI

s