space science : atmosphere part-5 planck radiation law local thermodynamic equilibrium: let...
TRANSCRIPT
Space Science : Atmosphere Part-5
Planck Radiation LawLocal Thermodynamic Equilibrium: LETRadiative TransportApproximate Solution in Grey AtmosphereSkin TemperatureGreenhouse EffectRadiative BalanceRadiative Time Constant
Reading Ionosphere for Previous partRadiation TransportGreenhouse Effect
Windows and Absorptions in the Solar Spectrum
0.2 0.6 1.0 1.4 1.8 2.2 2.6 3.0m
Radiation: Solar and Earth Surface
€
Bλ (T) Planck Black Body Emission
Atmosphere is mostly transparent in visible but opaque in UV and IR;
IR window 8-13um
B(T)
Fraction absorbed
Before Discussing Radiation Define Solid Angle
r sin d
r sin d r d
x
z
y
d
dr sin
s)(steradian ddr r
dAdr dV
2 0
d dcos r ddsin r d r
) d sinr ( )dr ( dA
2
222
Ω=
=
≤≤≤≤
→=Ω=
=
πϕπ
ϕϕϕ
€
EMISSION of RADIATION
Ideal Emitter at Temperature T
Photon energy = hν ; ν = c/λ
Planck's Law (written in ν or λ ; Bν dν = Bλdλ ]
Energy flux per unit solid angle between ν and ν + dν from a surface at T
Bν (T) = 2 hν
(c /ν )2
1
exph ν
k T
⎛
⎝ ⎜
⎞
⎠ ⎟−1
;
Peak λmax ≈ 3000μm/T(K) ; ν max ≈ 0.57c/λmax
Brightness peak in ν is at longer wavelengths than peak in λ
Energy peak (hν max ) → λ ≈ 5100μm/T(K)
Bv (T) ν →∞ ⏐ → ⏐ ⏐ 2 c k T
λ4
Bλ (T) λ→∞ ⏐ → ⏐ ⏐ 2 ν 3 h
c2exp[−hν/kT] (like a Boltzmann distribution)
Bλ (T) dλ = Bν (T) dν = 0
∞
∫ 1
π0
∞
∫ σ T4 Total energy flux per unit solid angle
* Net flux across a flat surface :
Integrate over solid angle : dΩ = sinθdθdϕ → dcosθ dϕ
∫ [1
π σ T4 ] cosθ dΩ =
1
π0
1
∫ σ T4 2π cosθ dcosθ
= σ T4
Planck’s Law for Thermal Emission of Photons
GREY ATMOSPHERE Chap 3 G+W , H p10-17
Gray vs. Black vs. Transparent Also, absorption independent of frequencyover the range of relevant frequencies
Processes Surface heated by visible Warm Surface emits IR ~ 3 – 100 m peak ~ 15 m
IR absorbed by CO2, O3, H2O, etc.
Remember why not O2 and N2 ?
Vibrational Bands CO2 (IR active?)
Symmetric Stretch O C O 7.46 m (N)Asymmetric Stretch
O C O 4.26 m (Y) Bending
O C O 15.0 m (Y)
H2O Symmetric Stretch O 2.73 m (Y) H H
Asymmetric Stretch O 2.66 m (Y) H H
Bending O 6.27 m (Y) H H
You can have combination bands
or 2 vib. levels
IR Emission and Absorption
Ground Emits
Primarily Triatomc Molecules Absorb and Re-emit: vibrational and rotational states
To determine T we assume excited molecules heat locally by collisions.
CO2(v=1) + M --> CO2(v=0) + M + K.E.
Slab of Atmosphere Absorption
I + dI
dz
I
€
I = Intensity of the radiation
≡ energy flux per unit solid angle (through the atmosphere)
= power / area / solid angle;
I dΩ∫ = F ; Ω is solid angle
Only Absorbing (notation in C +H)
dI = - k ( I ρ dz )
= - σ abs nabs I dz
Absorption coef.
k = fabs σ abs / m ;
m = average mass of molecules in atmosphere
fabs = fraction of species absorbing
mean free path of a photon for absorption =
= (σ abs nabs )-1 = (k ρ)-1
Solution: Absorption Only
(did earlier; new notation)
€
I(z) = I0 exp[− k ρ dz]∫
k ρ dz∫ = σ abs ∫ nabs dz ≈ σ abs Nabs [ ]
Nabs = column density of
absorbing species
Optical Depth for IR
kρ dzz
∞
∫ ≡ τ IR
Note : k ρ dzz
∞
∫ ≈ k
gdp
z
∞
∫ ≈ k
gp
Therefore, z τ p
What about EMISSIONSlab of atmosphere has a T
emits IRAssume LTE LTE Local Thermodynamic Equilibrium molecular motion and the population
of the vibrational and rotational states are all described by Boltzmann distribution and photons by Planck’s law ---using the same T
Kirchhoff’s LawIn LTE the emissivity of a body (or surface) equals its absorptivity.
€
Probability of absorption in dz ≈ k ρ dz ∝ Emission
Therefore, using the Planck flux for emission from a thick material
when the material is in LTE, one writes :
emission from the slab of thickness dz = [ k ρ dz ] B(T)
with B(T) = 1
πσ T4 (power/area/solid angle)
Radiative Transport with Emission + Absorption
∫∫
=Ω
Ω=
−=
+=+−
4T d cos B(T) Also
d cos I surface a acrossFlux
angle solidunit per are B and I :Note
)(simple! B(T)Id
dI
substitute and ) dz (k -=d Use
B(T) ) dz (k I dz) k ( - dI
emission absorption =ChangeIntensity
σθ
θ
τ
ρτ
ρρ
Flux (cont)
€
I [r2 dΩ] = Energy per unit time
across surface area dA
Therefore : I(θ, ϕ ) ⇒ Watts
m2 ster
If isotropic I(θ, ϕ ) = I
Flux across a surface
I(contains speed, c)
Flux = I cosθ dΩ = = ∫ I cosθ dcosθ dϕ∫ = π I∫∫ = F
I
dA = r2dΩ
r
Radiative Transport (cont.)include angles
dz = cos
↓
−
↑
=−=Ω
==Ω
Ω
+
−=
=
+=
∫ ∫∫ ∫
F- I d I
F I d I
d dcos = d and cos= :I isotropican For
fluxes downward upward into Divide
BId
dI
dz k - d
B) cos
dzk (I)
cos
dz k ( - dI
2
0
0
1
2
0
1
0
π
π
πμ
πμ
ϕθθμ
τμ
ρτθ
ρθ
ρ
Radiation Transport (cont.) I and B are isotropic
€
Upward I Downward I
I dΩ = 0
1
∫0
2π
∫ I dΩ = −1
0
∫0
2π
∫ 2 π I
same for B
Now : Use these integrals to integrate
the radiative transport Equation.
μ dI
dτ = I - B
0
2π
∫ →0
1
∫ dF↑
dτ= 2 F↑ - 2 π B Upward flux
0
2π
∫ →-1
0
∫ − dF↓
dτ= 2 F↓ - 2 π B Downward flux
Not Quite Isotropic * 5/3 Use * not in transport eq.
dz
↑+F ↓
+F
↓−
↑− F F
e.unit volumper energy in change dt
dT c
EquationHeat
T ,F ,F unknowns 3but Eqs. 2
T B : Remember
(2) B F d
dF
(1) B F d
dF Solve
p
4
*
*
=
=
−=−
−=∴
↓↑
↓↓
↑↑
ρ
σπ
π
π
Need
Third Equation is the Heat Equation
(3) CF F
)F F (dz
d 0
T state)(steady mequilibriu Find
0 = )F F ( dz
d
dt
dT c
EquationHeat
1
p
=−
−−=
−−=
↓↑
↓↑
↓↑ρ
Radiative Transport Solution (cont.)Use Eqs. (1) and (2) with (3).
€
(1) dF↑
dτ * = F↑ − π B ; (2) −dF↓
dτ * = F↓ − π B
ADD (1) and (2) M SUBTRACT (1) and (2)
d(F↑ − F↓ )
dτ * = F↑ + F↓ − 2πB M
d(F↑ + F↓ )
dτ * = F↑ − F↓
Use Eq. (3) (dz ∝ - dτ*) Use solution to (3)
0 = F↑ + F↓ − 2 πB M d(F↑ + F↓ )
dτ * = C1
(1a) F↑ + F↓ = 2 π B M (2a) F↑ + F↓ = C1τ* + C2
Combine (1a) and (2a)
2 π B(T) = C1 τ* + C2
σ T4 = 1
2[C1 τ
* + C2] (4)
Now need C1 and C2 in order to get T vs. τ (z)
Therefore,
Apply boundary conditions at surface
and the top of the atmosphere
Radiative Transport Solution (cont.)Use Eq. (3) and (4)
€
Solutions
F↑ + F↓ = C1τ* + C2 (4) ; F↑ − F↓ = C1 (3)
Add F ↑ = 1
2C1(τ
* +1) +1
2C2 (5)
Sub. F ↓ = 1
2C1(τ
* −1) +1
2C2 (6)
Boundary conditions on IR radiation
Top : τ * = 0 assume F↓ (0) ≈ 0 WHY ??
Use Eq. (6) to get
C1 = C2
Therefore,
F ↑ = 1
2C1(τ
* + 2) (5) ; F ↓ = 1
2C1τ
* (6)
Bottom : τ ≡ τ g use the ground temperature, Tg, for emission :
i.e. surface flux in IR is F↑ (τ g*) = π B(Tg ) = π Bg
Therefore, evaluate (5) at τ g * and use C1 = C2
π Bg = 1
2C1[τ g
*+2]
or
C1 /2 = π Bg / [ τ g* + 2 ]
Radiative Transport: Solution use C1 in (5) and (6)
€
F ↑ = π Bg [τ * + 2]
[τ g* + 2]
F ↓ = π Bg
τ *
[τ g* + 2]
Use earlier result
[F↑ + F↓ ]/2 = π B = σ T4 = π Bg [τ * + 1]
[τ g* + 2]
Also use : π Bg = σ Tg4
Therefore, we get solution for T
T4 = Tg4
[τ * + 1]
[τ g* + 2]
Note : T = temp. of air
τ g* = ground ,
T(z = 0) = To = air temp at surface
To4 = Tg
4 τ g
* +1
τ g* + 2
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
In radiative equilibrium no conductive contact
between the air and ground, only radiative; so there is a
discontinuity! Obviously not realistic; need conductive also.
Solution to the radiative transfer equations for a grey atmosphere
ConductiveTransport(Adiabatic Lapse Rate)Radiative
TransportBecomes radiative dominated near tropopause
* Optical Thickness in IR
g*
Finally: we do not know Tg
we know only Te for emission to space!
This is the Green House EffectGround T exceeds T for
emission to space
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+=
=
+=
=
↑
↑
2 1 T T
T B using Therefore,
]2[
2 B (0)F :solutionour From
T of Definition ;T (0) F
*g4
e4
g
4gg
*g
g
e4
e
τ
σπ
τπ
σ
A Real Green House
How do you get IR out equal to Visible light absorbed inside: RAISE T
Note: For a real green house convection may be as important:
i.e. glass a thermal barrier
IR
Visible
OutsideInside
Greenhouse Effect is Complex
PLANETARY ENERGY BALANCE G+W fig 3-5
113 86Convective30
IR Radiation To Space67
GROUND
€
Absorbed
Clouds 21
Atmos. 22
Ground 24
67
Back to space
Reflect Ground 7
Clouds 26
Albedo 33
Incoming solar radiation
mesopause
Radiation Transport (Review)
dz Atmospheric Slab
IR
4
abs
T B
BId
dI *
dzn dz k d
IR in thedepth Optical
Bcos
k I
cos
k
dz
dI
Emission Absorption Intensity of Change
Equations General
Flux downward F
Flux upward F
intersity I Review
σπ
σρ
ρ
ρ
=
−=
−=−=
+−=
+−=
=
=
=
↓
↑
Integrate (*) for upward moving and downward moving IR photons
4g
*g
4e
p
*
*
*
T )(F
T 0)=( F :Top (3)
right). quite(not in coming IR no 0 0)= ( F :Top )2(
ConditionsBoundary
T , F , F
unknowns Three (1)
mEquilibriuin 0 ]F[Fdz
d
dt
dT c .3
EquationHeat
B Fd
dF 2.
B Fd
dF 1.
) 5/3 ( Equations 2 :Result
σ
σ
ρ
π
π
=
=
=
=−−=
−=−
−=
≈
↑
↑
↓
↓↑
↓↑
↓↓
↑↑
(Review continued)
fluxsolar the
with balance thefrom comes T and
z offunction a is *remember
3. ]1[2
T T
T of instead T of in terms rature)(air tempe T Write
2. ][2
T F
1. ]2[2
T F
T of in terms T and Fluxes Can write
T B Remember
e
*4
e4
ge
*4
e
*4
e
e
4
σπ
+=
=
+=
=
↓
↑
Ground T (review)
€
In terms of Te (determined from solar flux and albedo)
Tg4 = Te
4 [ 1 + τ g *
2]
τ g * = optical depth of atmosphere + IR
This is the Green house effect
Tg > Te
T in terms of Tg
(rearrange the equations )
T4 = Tg4
[1+ τ*]
[2 + τ g*]
This is air temperature relative to ground temperature
T(z = 0) ≠ Tg for radiative only solution
What have we ignored :
absorption is in bands which have a width
molecule motion causes doppler shifts affecting the absorption efficiencies
collisions also affect the absorption efficiences
population of the levels affects absorption efficiency
G + W (simple version; 4 layers)
0
1
2
3
4
Ground
SpaceF1
F1
F2
F2
F3
F3
F4
F4
Fg
agreement closely Fortuitous
layers ofnumber =
] (5/6) 1 [ T= ] /2* 1 [ T T
assolution wour
F ] layers ofnumber 1 [ F can write :Note
F 5 F F 2 F F F F 2 4
F 4 F F 2 F F F F 2 3
F 3 F F 2 F F F F 2 2
F 2 F 2 F F F 2 1
T F F 0
1) = ( s thicknesoptical one layer Each
layers into atmosphere Divide
g
g4
eg4
eg
outg
out34gg34
out234423
out123312
out1221
4eout1
σ
++=
+=
=−=→+==−=→+==−=→+=
==→===
=
FVIS=Fout
Earth g 2
€
Te = 250K
Implies
Tg = 330 K ! No Way
Again - - - it is clear that at the surface
convection dominates
Before Finishing
If τ g ≈ 2 ≈ σ abs fabs N
Using N = 2 × 1025 mol/cm2 (see early lecture)
fabs = 1% (H2O, CO2, O3 ....)
σ abs ≈ 10-23 cm2
IR absorbers have small cross sections relative to UV
VENUS (Problem for set 2)
Te = 230 Tg = 750
Therefore: g* = ?
Therefore: Use cross section from previous slide, pure CO2
N = ?
TopWhy isn’t Te = T() at the top?
=emissivety
Te4(1-) T4
T4
€
Te4 = 2 ε T4
T = 1
21/4 Te = asymptotic value of T (skin T)
€
T4 (z) = Te
4
2(1+ τ*)
τ * → 0 at top of atmosphere
Earlier we calculated that Te ≈ 250 K at Earth
T(∞) = 1
21/4Te ≈ 210 K
This, of course, ignores the direct thermosphere heating
Direct means of obtaining T(∞)
Goody + Walker (Skin T)
How Thin Layer at Top is Heated
Thermal Structure Tropopause to Mesopause
small iscontent heat but the
hot'' is reThermosphe :also Note
radiativepurely not isit Suggests
2) (K 330 (calc.)T :T Surface
)T(bracket they heating ozone theIgnoring
km 12 215 se)T(Tropopau
km 80 200 e)T(Mesopaus
K 210 )T( Find
K 255 T Start with
g
e
≈≈
∞≈≈
≈∞≈
TIME CONSTANT FOR RADIATIVE EQUILIBRIUM
€
Initially : Equilibrium
ρ cp ∂T
∂t = − OUT + IN = 0
Imbalance : suddenly add heat can write as
σ T4 → σ [T + ΔT]4 in a layer of atmosphere Δz
ρ cp dT
dt = -
1
Δz[2σT4 − 2σ (T + ΔT)4 ]
dT
dt ≈
8 σ T3
Δz ρ cp
ΔT
time ≈ T/[dT/dt] ≈ Δz ρ cp
8 σ T3
cp ≈ 1000 J/kg K ρ = 1.3 kg/m3 σ = 5.7 × 10-8 J
m2
1
K4 s Troposphere Δz → H
t rad ≈ 9 days
Therefore, thermal changes are slow in the radiative region
vs what we are familiar with in the troposphere
Carbon concentrationvs. time
Carbon Concentration Long Term
Later we will look at the carbon cycle
GREEN HOUSE EFFECT
.
Constants Time andcarbon for Reservoirs :discuss to
evolution catmospheri discuss weback when come willWe
OCEAN ATMOSPHERE !K 2 - 1
ONLY ATMOSPHERE !K 5 - 2.5 T
1800) since increase 30% ~ ( ?CO Double
! T decrease25K z
T increase30K z O Increase :Note
K 288 K 33 K 255
T Te
K 33
K 3 ),CH O,Other(N
K 2 O
K 7 CO
K 21 OH
2
3
42
3
2
2
+==Δ
<>
⇒
=+Δ+
++++
L
However,
#4 Summary Things you should know
Planck Radiation LawLocal Thermodynamic Equilibrium:
LETRadiative TransportGreenhouse Effect Surface temperatureSkin TemperatureRadiative Time Constant