Ch.  4  Radia*on    

Text:  Wallace  and  Hobbs,  Ch.  4,  Radia*ve  Transfer  p113-­‐152  

•  Radiation can be viewed as a ensemble of electromagnetic waves propagating at the speed of light, c*=2.998X108 m/s.

•  The wave length (λ), wave frequency( ) and wave number υ obey the following relationships:

4.1  The  spectrum  of  radia*on:  

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˜ v

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υ =1/λ˜ v = c*υ = c* /λ

4.2 Radiance, irradiance, and flux

•  Radiance: monochromatic or spectral intensity, I: the radiative energy transferred in a specific direction through a unit area (normal to the direction considered) per unit time at a specific wavelength (or wave number).

•  Q: Does Iλ vary with incident •  angle and wavelength?

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Iλ = Iλdλλ1

λ2∫radiance also can be expressed as a function of frequencyIυ = λ2Iλ

Flux density-irradiance •  Monochromatic flux density or

irradiance: rate of radiative energy transfer per unit area with a given wave length through a plane surface with a specified orientation in 3D space (integrated incoming radiance for all directions).

•  Q: Is Fλ zenith and azimuth angle dependent?

€

Fλ = Iλ2π∫ cosθdω

where dw is an elemental arc of soild angle, θ is the anglebetween the incident radiaton and the direction normal to dA, the surface.cosθ : intensity dilution due to slanted orientationrelative to the surface €

If Iλ is isotropic, then

Fλ = Iλ2π∫ cosθdω = Iλ cosθ sinθdθdϕθ =0

π / 2∫

φ =0

2π∫

= 2πIλ cosθ sinθdθθ =0

π / 2∫ = −2πIλ

12

cosθ( )2 |0π / 2

= πIλ

Zenith  angle:  θ

dA  

azimuth  angle:  φ

What  is  the  difference  between  radiance  and  irradiance?  

•  Radiance:  wavelength  and  angle  dependent  •  Irradiance:  wavelength  dependent,  integrated  over  all  direc*ons.  

Estimate the solar irradiance at the top of the earth’s atmosphere:

•  We know that solar radiance is 2.00X107 W/m2/sr, the distance between the sun and the earth is 1.50X1011 m. What would be the solar irradiance at the top of the earth’s atmosphere at the zenith?

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At the top of the earth's atmosphere

Fso = Isoδω∫ cosθdω because δω is small

Fso = Iso cosθδω ⇒ Fso = Isoδω at zenithThe fraction of the sky (the solid angle) is

occupied by the sun : δω2π

=πRs

2

2πd2

Fso = Isoδω = IsoπRs

2

d2

= 2.00 ×107Wm−2sr−1 × 3.14 7.00 ×108m1.50 ×1011m⎛

⎝ ⎜

⎞

⎠ ⎟

2

sr

=1368Wm−2

earth  

d  

2πd2  

The  solar  irradiance  at  the  top  of  the  earth’s  atmosphere  at  the  zenith  is  1368  W/m2.    

4.3  Blackbody  radia*on  

•  What is a blackbody? –  A surface that completely absorbs all

incident radiation. A blackbody also emits 100% of the absorbed radiation and it’s radiation is isotropic.

–  Radiation emitted by a blackbody is determined by it’s surface temperature as described by three laws:

–  The Plank’s law –  The Wien’s law –  The Stefan-Boltzmann’s law

Nothing  reflects  back-­‐black  

The Planck’s function:

•  What is the planck’s function? –  The spectrum and wavelength

of maximum radiative energy emitted by a blackbody is determined solely by its temperature.

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Bλ (T) =c1λ

−5

π (ec2 /λT −1)

where C1 = 3.74 ×1016 Wm-2,C2 =1.45 ×10-2 mKλ : wavelength, T : temperature, Bλ : radiance of emitted by the blackbody

•  The  wavelength  of  peak  emission  for  a  blackbody  is  inversely  related  to  the  surface  temperature  of  the  blackbody,  λm=2897/T  –  Where  λm  is  wavelength  in  unit  of  µm,  T  is  temperature  in  unit  of  K.  

•  Q:  If  you  were  an  alien  travel  through  the  solar  system.      •  a)  You  first  spot  the  earth  and  measured  that  the  wavelength  of  maximum  emission  at  11.4  µm.    

What  would  be  the  effec*ve  temperature  of  earth’s  atmosphere?    (effec*ve  temperature  is  the  temperature  corresponding  to  the  maximum  emission);    

•  b)  As  you  con*nued  to  travel,  you  measured  the  wavelength  of  peak  emission  of  the  Mars  at  13.4  µm.    What  would  be  the  effec*ve  temperature  of  Mar’s  atmosphere?    

The  Wien’s  displacement  Law  

•  The  wavelength  of  peak  emission  for  a  blackbody  is  inversely  related  to  the  surface    temperature  of  the  blackbody,  λm=2897/T  

•  Q:  If  you  were  an  alien  first  spot  the  earth  and  you  measured  that  the  wavelength  of  maximum  emission  at  11.4  µm.    What  is  the  effec*ve  temperature  of  earth’s  atmosphere?    (effec*ve  temperature  is  the  temperature  corresponding  to  the  maximum  emission)  –  Based  on  Wien’s  law,  T=2897/λm=2897/11.4=254K  for  the  earth  –  T=2897/13.4=216K  for  Mars’  atmosphere.  

The  Wien’s  displacement  Law  

The  Stefan-­‐Boltzman  Law:  •  The  blackbody  flux  density  (irradiance)  integrated  over  all  wavelengths  using  the  Planck  func*on,  F,  is  propor*onal  to  T4.  F  =σ  T4                      

where  σ=5.67X10-­‐8  Wm-­‐2K-­‐4,  T:  temperature  in  K.  

The  Stefan-­‐Boltzman  law  is  commonly  used  in  es*mate  radia*ve  energy  balance  because  its  simple  rela*on  with  T.  

Exercise:  

•  The  effec*ve  temperature  of  the  Venus  atmosphere  is  225K.      – A)  What  is  the  radia*ve  energy  emihed  by  the  Venus  atmosphere?  

– B)  The  solar  irradiance  at  the  top  of  the  Venus  atmosphere  is  2639  W/m2.  How  much  solar  radia*on  has  to  be  reflected  in  order  to  balance  the  radia*ve  energy  emihed  by  the  Venus  atmosphere?  

Exercise:  •  The  effec*ve  temperature  of  the  Venus  atmosphere  is  225K.      

–  A)  What  is  the  radia*ve  energy  emihed  by  the  Venus  atmosphere?  

–  B)  The  solar  irradiance  at  the  top  of  the  Venus  atmosphere  is  2639  W/m2.  How  much  solar  radia*on  has  to  be  reflected  in  order  to  balance  the  radia*ve  energy  emihed  by  the  Venus  atmosphere?  

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a) Based on the Stefan - Boltzman law :Femit =σT4 = 5.67X10−8Wm−2K −4 × (2.25 ×102K)4

=145 Wm−2

Venus atmosphere emits 145 Wm-2 longwave radiative energy

b) The solar radiation absorbed by the Venus' atmosphere hasto be balanced by the longwave emission. Thus

Fs(1− a)⋅ πRv2 = 4πRv

2Femit ⇒ α =1- 4FemitFs

=1− 4X145Wm2

2639Wm2 = 0.78

78% of the solar radiation has to be reflected back to space to balancelongwave emission by the Venus' atmosphere.

Non-­‐blackbody:  

•  Earth’s  atmosphere  and  surface  is  not  exactly  a  blackbody.  Earth  emits  on  average  about  61%  of  the  absorbed  radia*on.  

•  Emissivity,  ελ=Iλ/Bλ(T) –  The ratio of emitted radiation intensity vs. that corresponding

to the blackbody radiation. •  Reflectivity, Rλ=Iλ(reflected)/Iλ(incident) –  The ratio of reflected radiation intensity vs. that of incident

radiation. •  Transmissivity, Tλ=Iλ(transmitted)/Iλ(incident)  

•  The  Kirchhoff’s  law:  the  emissivity  must  equal  to  the  absorp*vity  to  maintain  radia*ve  equilibrium  at  each  and  all  wavelengths.    

Ελ=αλ      

•  Kirchhoff’s  law  is  applicable  to  gases  below  al*tude  of  60  km,  where  frequency  of  molecular  collision  >>  frequency  of  molecular  absorb  and  emit  radia*on.  

The Greenhouse effect:

•  How  does  greenhouse  effect  increases  the  surface  temperature?  

•  Conceptually,  we  can  approximately  treat  the  earth’s  atmosphere  as  mul*ple  isothermal  layers  that  let  sun  light  penetrate  through  but  trap  all  infrared  radia*on.      

•  The  more  layers  we  have,  the  stronger  the  greenhouse  effect  is  on  the  surface  temperature.  

F   F  

F

T

Te  

T(z)  

z

F  

Surface  receives  heat  of  2F  

F

T

Te2  

T(z)  

z  F  

Surface  receives  heat  of  3F  

F  

F  

2F  

Te1  

2F  3F   3F  

F  

F   F  

F

T

Te  

T(z)  

z

F  

Surface  receives  heat  of  2F  

F

T

Te2  

T(z)  

z  

F  

Surface  receives  heat  of  3F  

F  

F  

2F  

Te1  

2F   3F  

Exercise:      Calculate  the  surface  effec*ve  temperature  assuming      

a)  the  atmosphere  is  one  isothermal  layer;    

b)  the  atmosphere  is  approximately  two  isothermal  layers,  respec*vely.          Assuming  that  the  incident  solar  radia*on  is  342  W/m2  at  the  top  of  the  atmosphere.  30%  of  it  is  reflected  back  to  space  by  the  earth.      Assuming  the  isothermal  layers  and  the  surface  act  like  blackbodies.      

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a) If the atmosphere were one isothermal layer :Fsfc,a↓ = 2Fs(1− 0.3) =σTsfc,a

4

Tsfc,a =2Fs⋅ 0.7

σ

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 4

=2X0.7X342W /m2

5.67 ×10−8W /m2 /K 4

⎡

⎣ ⎢

⎤

⎦ ⎥

1/ 4

= 303K

b) If we approximately treat the atmosphere as two isothermal layers : Fsfc,b↓ = 3Fs(1− 0.3) =σTsfc,b

4

Tsfc,b =3Fs⋅ 0.7⋅ 0.9

σ

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 4

=3X0.7X342W /m2

5.67 ×10−8W /m2 /K 4

⎡

⎣ ⎢

⎤

⎦ ⎥

1/ 4

= 335Kmulti − isothermal layers with greenhouse gases are very effective in trapping longwave radiation and leadingto warmer surface temperature.

Discussion:  

•  Tsfc,a=303K  for  one  isothermal  layer  •  Tsfc,b=335K  for  two  isothermal  layers  

Why  are  these  es*mated  Tsfc  values  much  higher  than  observed  global  mean  Tsfc=288K?  

Discussion:  •  Tsfc,a=303K  for  one  isothermal  layer  •  Tsfc,b=335K  for  two  isothermal  layers  

Why  are  these  es*mated  Tsfc  values  much  higher  than  observed  global  mean  Tsfc=288K?  

•  Energy  leaked  through  the  atmospheric  window  is  not  accounted  for.  •  Atmospheric  absorp*on  change  with  height.  •  In  reality,  the  surface  solar  radia*on  is  balanced  by  latent  and  sensible  heat  flux,  in  

addi*on  to  the  longwave  radia*on.  •  When  surface  T  become  sufficiently  high,  lapse  rate  become  unstable  and  

convec*ve  adjustment  will  reduce  surface  temperature  to  neutral  lapse  rate.  

Discussion:  

•  Climate skeptics claims that a further increase of greenhouse gases would not increase surface temperature because CO2 absorption is saturated in the troposphere. However, observations have shown a strong increase of CO2 in the upper troposphere, where CO2 absorption is not saturated. Do you think that a further increase of CO2 would increase the surface temperature? Why or why not?

Source:  Charles  Jackson’s  Tech  talk,  F09  

Summary-­‐1:  •  What  are  differences  between  radiance,  irradiance,  and  radia*ve  flux?  – Radiance  is  the  radia*ve  energy  transmihed  in  a  specific  direc*on  at  a  specific  wavelength  per  unit  area  and  *me  interval.    The  unit  is  W/m2/Sr.  It  is  wavelength,  zenith  and  azimuth  angle  dependent.  

–  Irradiance:  radiance  integrated  over  all  incident  angles  (direc*ons).    It  is  wavelength  dependent.  

– Radia*ve  flux:  irradiance  integrated  for  all  wavelengths.    

Summary-­‐2:  •  What  is  blackbody?  What  determines  the  emission  of  radia*ve  energy  for  a  blackbody?  – A  blackbody  is  an  object  absorbs  all  incident  radia*on  and  emit  all  the  absorbed  radia*on.  

– The  wavelength  of  maximum  emission  and  total  radia*ve  flux  emihed  by  a  blackbody  is  a  func*on  of  its  surface  temperature,  following  the  Planck’s  func*on,  Wien’s  Law  and  Stefan-­‐Boltzman’s  law.  

Summary-­‐3:  •  What  is  the  greenhouse  effect?  –  The  greenhouse  effect  is  due  to  absorp*on  and  re-­‐emission  of  the  longwave  radia*ve  energy  by  greenhouse  gases.    These  trace  gases  allow  solar  radia*on  pass  through  the  atmosphere,  but  opaque  for  longwave  radia*on  emihed  by  the  earth’s  surface.  

– How  does  increase  of  greenhouse  gases  increase  the  surface  temperature?  

   It  increases  temperature  in  the  upper  troposphere,  which  in  turn,  increases  downwelling  longwave  radia*on  (F~Te4),  and  warm  the  surface  temperature.