Radiative heat transfer - ESPCI ParisRadiative heat transfer Black body radiation P BB(T )= Z 1 0 P e(,T)d = 2⇡5k4 B 15 h3c2 T 4 = T 4 (null)

Radiative heat transfer


Black body radiation



Spec

tral i

nten

sity

(W.m

-2 µ

m-1

)

Pe(�, T ) =2⇡hc2

�5

1

exp(hc/�kBT )� 1<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

Power emitted per unit surface :



Spec

tral i

nten

sity

(W.m

-2 µ

m-1

)

Pe(�, T ) =2⇡hc2

�5

1



�M (m) =2.88⇥ 10�3

T<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

Wien’s displacement law



PBB(T ) =

Z 1

0Pe(�, T )d� =

2⇡5k4B15 h3c2

T 4 = �T 4

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Stefan’s law

Spec

tral i

nten

sity

(W.m

-2 µ

m-1

)

Pe(�, T ) =2⇡hc2

�5

1



�M (m) =2.88⇥ 10�3





PBB(T ) =

Z 1

0Pe(�, T )d� =

2⇡5k4B15 h3c2

T 4 = �T 4


Stefan’s law

Spec

tral i

nten

sity

(W.m

-2 µ

m-1

)

Pe(�, T ) =2⇡hc2

�5

1



Non black bodies

Emissivity ε = absorptivity a

P(λ,T) = Pe(λ,T) ε(λ)

Kirchoff ’s law

�M (m) =2.88⇥ 10�3





Grey bodies

Emissivity independent of λ

ε(λ) = a(λ) = C

PBB(T ) =

Z 1

0Pe(�, T )d� =

2⇡5k4B15 h3c2

T 4 = �T 4


Stefan’s law

Spec

tral i

nten

sity

(W.m

-2 µ

m-1

)

Pe(�, T ) =2⇡hc2

�5

1



Non black bodies

Emissivity ε = absorptivity a

P(λ,T) = Pe(λ,T) ε(λ)

Kirchoff ’s law

�M (m) =2.88⇥ 10�3



Diameter of the Sun: 1,4 106 kmSurface temperature of the Sun : 5800 KDistance to the Sun: 1,5 108 kmDiameter of the Earth: 12700 km

Effective albedo of the Earth: 0.31

The Donald Trump problem at zeroth order : show that with a completely transparent atmosphere, it would be difficult to grow lawn for a golf course on Earth.

Earth and Sun are assumed to be black bodies

-19°C

Diameter of the Sun: 1,4 106 kmSurface temperature of the Sun : 5800 KDistance to the Sun: 1,5 108 kmDiameter of the Earth: 12700 km

Effective albedo of the Earth: 0.31

The Donald Trump problem at zeroth order : show that with a completely transparent atmosphere, it would be difficult to grow lawn for a golf course on Earth.

Earth and Sun are assumed to be black bodies

Radiative flux from the sun, averaged over entire Earth surface : 342 W/m2

Flux adsorbed : 235 W/m2

Corresponding equilibrium temperature : 254°K

The Donald Trump problem at order one : show that with a one layer atmosphere adsorbing in the infrared, it would be easier to grow lawn for a golf course on Earth.

Incoming solar radiation

Reflected solar radiation

Atmosphereemissivity εTa

Te Earth surfaceemissivity 1

Write the energy flux balance for the earth + atmosphere system

Write the energy flux balance for the atmosphere at temperature Ta

Determine Ta and Te as a function of ε

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

emissivity

210

220

230

240

250

260

270

280

290

300

310

Temperature

TearthTatmos

Te =

(1� a)Js�(1� ✏/2)

�1/4

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The Donald Trump problem at order 2 : Evaluate the temperature distribution in a stable « grey » non scattering atmosphere

Atmosphere clear to solar radiation ; all absorption takes place at the ground.Gray atmosphere with no clouds in the IR (no scattering of light)Layers of atmosphere are grey thermal radiatorsPlane parallel or stratified atmosphereEddington approximation for the IR emission : I(z,θ ) = I0(z) + I1(z) cos θ

z

z+dz θ

s

ds

I(s)

I(s+ds)

s

ds

I(s)

I(s+ds)Extinction (absorption, scattering)

dIext = �ext I ds<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

s

ds

I(s)

I(s+ds)Extinction (absorption, scattering)


Augmentation (emission, scattering)

dIaug = aug J ds<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

s

ds

I(s)

I(s+ds)

If no scattering, emission by thermal radiation

dIaug = ext�T 4

⇡= extB(T )


Extinction (absorption, scattering)




s

ds

I(s)

I(s+ds)

If no scattering, emission by thermal radiation

dIaug = ext�T 4

⇡= extB(T )


Extinction (absorption, scattering)




dI

ds= ext(�I +B)


dI

ds= ext(�I +B)


dI

ds= ext(�I +B)


d⌧ = �ext ds<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

Optical path length (dimensionless) τ

dI

ds= ext(�I +B)


d⌧ = �ext ds<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

Optical path length (dimensionless) τ

dI

d⌧= I �B


Js (1-a)

Js (1-a)

Fdown(z)<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

Js (1-a)


Fup(z)<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

Js (1-a)



⌧ =

Z 1

zext dz


Optical path from atmosphere top

Atmosphere top τ = 0

Js (1-a)



⌧ =

Z 1

zext dz




Js (1-a) Fdown(0) = 0<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>



⌧ =

Z 1

zext dz



Ground τ = τ*





⌧ =

Z 1

zext dz



Ground τ = τ*





⌧⇤ =

Z 1

0ext dz


Optical depth of atmosphere

⌧ =

Z 1

zext dz



Ground τ = τ*





Fup = Fup(⌧⇤) = ⇡B(Tg)


Radiation going up :

⌧⇤ =

Z 1

0ext dz



⌧ =

Z 1

zext dz



Ground τ = τ*





Fup = Fup(⌧⇤) = ⇡B(Tg)


Radiation going up :

Jsa+ Fdown(⌧⇤)


Radiation going down :

⌧⇤ =

Z 1

0ext dz



⌧ =

Z 1

zext dz



ext = 0 exp(�z/Hw)<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

Hw = 1.6 km (water vapor)



Ground Temp.

�T 4g = Js(1� a)

✓1 +

3⌧⇤

4

◆




Ground Temp.

�T 4g = Js(1� a)

✓1 +

3⌧⇤

4

◆


⌧⇤ =

Z 1

0ext dz





Ground Temp.

�T 4g = Js(1� a)

✓1 +

3⌧⇤

4

◆


⌧⇤ =

Z 1

0ext dz





Steep temperature gradient leads to convective instability

390 W/m2

Documents

Radiative heat transfer - ESPCI ParisRadiative heat transfer Black body radiation P BB(T )= Z 1 0 P e(,T)d = 2⇡5k4 B 15 h3c2 T 4 = T 4 (null)