2. Illustration of Atmospheric Greenhouse Effect with Simple Models

Transcription

1 2. Illustration of Atmospheric Greenhouse Effect with Simple Models In the first lecture, I introduced the concept of global energy balance and talked about the greenhouse effect. Today we will address the following questions: What is the magnitude of the greenhouse effect on surface temperature? Why does the surface have to radiate 390 W.m -2 to get rid of 20 W.m -2? Note: All of the numbers today are for global average conditions. We need to introduce 3 concepts: The spectral and temperature dependence of the Planck function Emissivity Radiative Equilibrium

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3 2.1 A Zero-Dimensional Climate Model Recall that: Solar radiation strikes the planet; the planet reflects back to space a fraction A. The absorbed energy heats the planet which then emits IR radiation. This process goes on until a balance is achieved between the emitted energy and the absorbed energy. We now wish to find the temperature of the planet at this equilibrium state. S = Solar Irradiance or Solar Constant A = Albedo = fraction of Irradiance Reflected back to space R = Radius of Planet F = Flux of Longwave Energy emitted by the planet

4 i.solar energy incident upon the planet= πr 2 S ii.solar energy absorbed by the planet = πr 2 S (1 A) iii.energy emitted by the planet = πr 2 F πr 2 is the cross sectional area of the planet intercepted by the solar bean; πr 2 is the total surface area of the planet. For global energy balance: The absorbed solar energy has to balance the emitted energy: πr 2 S( 1 A) = πr 2 F (2.1) Black Body Planet This is simply a statement of global energy balance. To convert it into a climate model, we need to invoke some physical law or assumption. Assumption: Planet emits like a black body with an effective temperature, T e: : (2.2) F = σt e The global energy balance equation (2.1) is rewritten as πr 2 S(1 A) πr 2 σt e = 0.0 σ = Stefan - Bolzmann constant = 5.67 *10-8 W.m 2 K

5 This equation can be solved for T e : T e = S(1 A) σ 1/ 2.3) From measurements: S = 1370 ± 5W.m 2 A = 0.3 T e = 255K Observed Surface Temperature: T s 288 K Deduction: T s is larger because of the atmospheric greenhouse effect. In other words, without the radiatively active atmospheric gases, the surface temperature would be equal to the effective radiating temperture of the planet, T e. 2.2 A One-Dimensional Model: Radiative Equiilibrium Here, a more direct proof (as opposed to the earlier deduction) for the atmospheric greenhouse effect is given. We need one more piece of radiation physics and another dimension in the vertical. What is Radiative Equilibrium (RE)? Radiation is the only mode of energy transfer. Now let us see how the GH effect works. We will start with RE:

6 (S/)(1-A) σt σt T = Atmospheric Temperature Ts= Surface Temperature

7 Energy balance at top: Solution Radiative Equilibrium Energy Balance at the surface (with atmosphere): S (1 A ) T = 255K = T = 0.0 σ (1) Te S σ Ts = (1 A) + σt Replacing σt from Eq. (1) (2) (3) σ T T s s = S = 2 (1 2 1 / A) ; S (1 σ 1 / 1 / A) = 2 Te = 303 K At Surface: (without atmosphere) T s = T e = 255K Referring back to equations above and Fig. 3, we note the following: Without an atmosphere (left-hand panel in Fig. 3), the surface emission has to be exactly equal to the TOA emission, and as a result, Ts = Te=255K. The presence of the atmosphere changes the surface energy balance in an important way (see Eq. 3 above); it adds an extra term, i.e., the downward emission from the atmosphere, or the so-called, back radiation,. This heats the surface in addition to the solar heating!

8 Energy Balance S/ (1-A) 20 S/ (1-A) T=Te=255K Ts=255K σ Ts σ Ts Ts= 303 K No Atmosphere With a Black Atmosphere in the LW Only Why is this a perfect greenhouse example? To proceed further, we need to understand more about atmospheric properties. We assumed a black absorber and emitter and we assumed that all of the solar energy hits the ground. Neither of these assumptions are valid for Earth. Nevertheless, the inclusion of these important details do not invalidate the qualitative conclusions we have arrived at so far.

9 Goody, 196

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11 RADIATION PHYSICS

12 3. Radiation Physics 3.1 Black Body Radiation Absorbs all energy incident on it Black Body Emits maximum possible energy at all wavelengths for that temperature A λ = fraction of radiation absorbed = absorptivity = 1.0 = 1.0 for black body < 1 for all others The emission of all natural substances is governed by Planck s law, and is given by the Planck Function: W e λ = m. 2 Wavelength( m) = e = C 1 λ C 2 λ 5 λt e 1 L: Planck s Const; K = Boltzmann s Const; C: Speed of light C 1 = 3.7 *10 16 W.m +2 C 2 = 1.39 *10 2 m. k λ = wavelength; usually also denoted by microns (µm) Wavelength unit: m (for meters) or 10-6 m (for micrometers)

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14 For example: λ = 1X10 5 m(middleir);10µm λ = 5.X10 7 (visible);0.5µm Wavelength of maximum radiation: m 2898 λ ( micron) = T eλnormalized Ts=6000K 0.5 µm 10 µm Ts=288K Figure 2

15 Integrated Black Body Radiation - Stefan Boltzmann s Law e = 8 2 e dλ = σt ; σ = 5.67X10 W. m K λ ε Emissivity λ ε ( T ) emission of = λ = eλ ( T ) emission of a the body black body e λ = λ 5 C 1 { C2/ λt } e 1 Note also that, emissivity equals Absorptivity In general: 1 ε λ Α = λ ε λ For a black body ε λ Α λ = 1.0 = 1.0

16 Fig. 3; Goody, 196

17 3.2 Selective Radiative Characteristics of the Atmosphere Important Radiatively Active Constituents: H 2 0, Clouds, CO 2 and Ozone I) Solar Radiation: Absorb and scatter solar radiation in selective wavelengths: Spectral Composition: UV : 0.1 λ 0. µm 9% Rayleigh Scattering by Air; Scattering by Clouds absorption Visible: 0. < λ 0.75 µm 5% Weak Rayleigh Scattering & 0 3 absorption and Scattering by Clouds & aerosols Absorption by soot and other pollutants Near IR:.76 < λ 0.9 µm 12% Cloud Scattering & absorption; H 2 O + O 2 absorption Near IR: 0.9 < λ µm 33% Strong absorption by Clouds, H 2 O, CO 2

18 3.3 Solar Absorption The gobal average estimates for the absorption and reflection of radiation by the atmosphere and surface are shown below Air Scattering Atmospheric Absorption 68 Reflection by Clouds Surface Reflection Clouds 0 to to 172 Figure 5

19 Fractional Absorption by Atmospshere = 68 = 0.2 = as 33 Fractional Reflection by Surface- Atmosphere System (Albedo) A = = 0.3 Fractional Absorption at Surface = = 0.5

20 3. Non-Black Atmosphere Longwave Length Radiation: λ 100 µm λ 8 µm H 2 O + clouds; absorbs nearly 100% 8 λ 9 µm Weak continuum H 2 O 9 λ 10 µm Strong O 3 absorption in stratosphere 10 λ 12.5 µm Weak H 2 O continuum 12.5 λ 18 µm CO 2 15 µm bands; strong 18 λ 200 µm H 2 O; very strong Measure of Absorption: K the absorption coefficient ρ g ( gm / m 3 ) Ζ Figure 6

21 K λ = m z 2 gm 1 gm τ = optical depth : K λ * ρ g m 3 Z (thickness m) τ is unitless Consider absorption by the slab shown in fig. 5. The intensity (Wm -2 sr -1 ) of radiation incident t the bottom of the slab is I λ. The optical depth is τ λ. If there is no scattering, the difference between the intensity at the bottom and at the top is the absorbed energy. I λ e τ λ τ λ τ λ = K λ ρg Ζ I λ Lecture 3- Figure 7

22 Absorption = Absorptivity = I λ I λ e τ λ A λ = 1 e τ λ [ ] ; fractional absorption Thus the optical depth, in a non-scattering atmosphere, is a fundamental measure of absorption. For monochromatic radiation, radiation of one wavelength, the fractional absorption is 1 e τ λ ; for total absorption over many wavelengths, we obtain: ( ) A = λ 2 A λ d λ λ1 Emission: Consider an isothermal layer ε = Emissivity = e λ (T)A λ d λ σt T = const oo o e λ (T)Α λ d λ Z Figure 8 τ λ oo o e λ Α λ d λ

23 Show emissivity plots for H 2 O, CO- 2. Let us assume we know the emissivity for an isothermal layer: εσt σt s (1-ε) T emitted = 2εσT Absorbed = εσt s Z; ε εσt σt s T s Figure 8

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26 Show emissivity plots for H 2 O, CO- 2. Let us assume we know the emissivity for an isothermal layer: Net Longwave Heating T εσt σt s (1-ε) emitted = 2εσT Absorbed = εσt s Z; ε (energy going into the layer): T s = εσ 2εσT = εσt s 1 2 T Ts εσt σt s Figure 8

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