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Planck Equation, continued

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  1. Planck Equation, continued • Planck’s law of radiation can be also rewritten as: I(L,T) = 3.74*10^8/{L^5*[exp(1.44*10^4)/(L*T)]-1]} Where the units are w/m2 per micrometer wavelength interval and L the wavelength is given in micrometer (10^-6 m) • Planck’s equation describes how heat energy is transformed into radiant energy • According to Planck’s law, an object will emit radiation in all wavelengths but not equally

  2. As an example, the radiance for the sun at L = 0.5 micrometers, given that the sun’s surface temperature is T = 5800 K. I(L,T) = 8.41*10^7 W/m^2/micrometers

  3. Wien’s law of radiation L(Imax) = 2897/T With L in micrometers.

  4. Examples using Wien’s Displacement Equation Tsun= 5800K Peak of Sun’s radiation = 2898mmK / 5800K = 0.5 mm Tearth = 288K Peak of Earth’s radiation = 2898mmK / 288K = 10 mm

  5. Stefan-Boltzmann Law of radiation I(T) = s*T^4 Where s is a constant equals to 5.67*10^-8 This law is one of the most important (if not the most important) laws for climate. It is used to solve for the temperature of the planet once the incoming radiation is known

  6. What do they mean?? • Planck equation gives the radiance of an object at a given temperature at any wavelength • Stefan-Boltzmann equation describes the total amount of energy being radiated • Wien’s equation describes the wavelength of maximum radiation

  7. We will see how this is done in the next lectures. However, we mention here that the basic assumption for computing the planets temperature is that the incoming radiation must equal the outgoing radiation on average. • For Earth, direct sunlight I = 1367 but this is different from the average value • The average value turns out to be ¼ of the solar constant because:

  8. Because the sunlight effectively strikes over a circular area (pR^2) of the same radius of the planet but must be spread over its spherical surface area (4 pR^2), 4 times larger than the circular shadow. • Therefore the average value is I/4

  9. If we want to compute the balance temperature for Earth we have: 1367/4 = 5.67*10^-8*T^4  T = 279 K Which is close, but still different, from the actual value of 288 K. We’ll see why …

  10. Albedo • The albedo is the ratio between incoming and reflected radiation • The average albedo for the Earth is ~ 30 % • If we incorporate the albedo in computing the Earth’s temperature we have: 1367*(1-alb)/4 = 1367*0.7*0.25 = 5.67*10^-8*T^4 Which yields to T = 255 K = -18 C

  11. So, improving the theory leads us to poorer results !!! • In reality, we neglected an important effect, which is the Greenhouse Effect which will fix things !

  12. Surfaces with high albedo (bright) are colder than surfaces with low albedo (dark): Try to stand with your feet on a black surface at noon in summertime or do the same on a white surface. Which one would you enjoy less ?

  13. Albedo of typical Earth substances: New snow  0.8-0.9 Old snow  0.6 -0.8 Thawing snow  0.35 – 0.65 White Chalk  0.45 Green Forest  0.16 -0.27 Green Grass  0.08 – 0.27 Pine Forest  0.06 – 0.19

  14. After melting, snow grains tend to bound each other with a consequence decrease in the visible/infra-red albedo

  15. Clean snow Dirty snow

  16. The albedo of Earth, hence, varies with latitude as a consequence of different surface properties and varying cloud conditions. • Therefore, it also varies with season

  17. Satellite measured reflected short-wave (no clouds)

  18. Satellite measured reflected short-wave (with clouds)

  19. Question: What is the temperature of the Earth if albedo = 0.3 and emissivity = 0.94 ? Stefan-Boltzmann  I = (1367/4)*(1-0.3) = 0.94*5.67*10^-8*T^4  solving for T  T = 258.8 K Reduced emissivity (from 1 to 0.94) has the consequence of warming the Earth (from 255 K to 258.8 K). Reduced emissivity  reduced absorptivity

  20. Emissivity of the atmosphere (Infrared) • The atmospere emits and absorbs infrared radiation. One (the best ?) way to treat the atmosphere is to use the gray body approach. • The emissivity of the atmosphere depends on the content of water vapor, carbon dioxide and pressure at each level. • The higher the humidity or the more carbon dioxide , the higher the emissivity and the more the atmosphere retains heat, and, hence, the stronger the Greenhouse Effect

  21. Atmospheric Composition • Mechanical Mixture of GasesGasVolume % (dry air) • Nitrogen 78.1 • Oxygen 20.9 • Argon 0.9 Also important trace gases: • CO2 ~370ppmv (1999) • Water Vapour • Ozone

  22. Solar Emittance Curve Radiation leaving the surface of the sun Solar radiation at sea level

  23. The formula for the emissivity of water vapor is: • Compute the depth of precipitable water in each layer • Multiply this value by [p(avg)/10^5]^0.5 to obtain u • Use the following formula for the emissivity: u<10^-4  .182*log[1+12.6*u] 10^-4<u<10^-3  .167*log[u]+0.440 10^-3<u<10^-2  .195*log[u]+0.491 10^-2<u<10^-1  .235*log[u]+0.527 10^-1<u<10^0  .259*log[u]+0.542 10^0<u  .219*log[u]+0.542

  24. For carbon dioxide: eC02 = .185*[1-exp(-.0356*dp^0.4)] Where dp is the pressure thickness of the layer

  25. The total emissivity is expressed as: • = eCO2 + eH20- eH20* eCO2 This because both CO2 and H2O absorb and emit at some of the same wavelength and the same radiation cannot be counted twice

  26. For example: What are the emissivities for the layer from 800 to 1000 mb with a mixing ratio of .003 ? • The precipitable water is .003*2m = 0.6 cm • u = 0.5*(900/1000)^.5 • eH20 = .259*log[.569]+0.542 = .396 • eCO2=.185*[1-exp(-.0356*dp^0.4)]=.156 • e= eCO2 + eH20- eH20* eCO2 = .490

  27. The Greenhouse effect It is the property of the atmosphere to allow the sun’s radiation with great ease but to make it rough for the upgoing radiation to get out

  28. ea*Ta^4 Io Alb*Io (1-ea)*Tg^4 euv*Io ea*Tg^4 ea*Ta^4 Tg^4 (1-alb-euv)*Io The simplest model of Greenhouse effect is a simple two layers balance temperature model. Each layer is assumed to have its own balance temperature. Atmosphere layer  ea*s*Tg^4 = 2* ea*s*Ta^4 Ground layer  ea*s*Ta^4 + I(S) = s*Tg^4 Where I(S) is the mean incoming solar radiation after the Earth’s albedo has been included.

  29. Example: What is the balance temperature for the Earth and atmosphere when the emissivity of the atmosphere is 0.8 ? Atm) .8*5.67*10^-8*Tg^4 = 2*.8*5.67*10^-8*Ta^4 Ground) .8*5.67*10^-8*Ta^4+239 = 5.67*10^-8*Tg^4 Which yields to Ta = 243.4 K and Tg = 289.5 K

  30. A simple problem: The current global average absorbed radiance is 1367*(1-0.3)/4 = 239.225 W/m2 with a temperature valance of 265.86 K. If I increases by 1.9 to 241.125 W/m2 then the new balance temperature becomes T = 255/37 K. Thus, if no other changes take place the average temperature should increase 0.51 C. This is the increase observed since 1900.

  31. Nevertheless, the problem of global temperature is more complicated. One complication arises from the depth of the ocean. • The oceans, indeed, absorb the heat for a long time before they are heated to the depths. • It is important to account for the physics of cooling and warming bodies

  32. Cooling and warming bodies Cooling rates can be computed by combining the Stefan-Boltzmann with the first law of thermodynamics: I = e*s*T^4 = m*c*dT/(A*dt)=r*V*c*dT/(A*dt) When a radiating object is a column of atmosphere with a certain pressure this becomes: dT/dt = e*g*s*T^4/[c*dr]

  33. Example: Hw far will the temperature of the atmosphere fall in 10 days if it receives no heat from the sun or the earth and its original temperature is 255 K and its emissivity is 0.8 ? dT = [0.8*9.8*5.67*10^-8*255^4/[1004*10^5]]*8.64*10^5 = -16.16 And the final temperature is 255 – 16.1 = 238.9 K

  34. However there is one thing wrong with this example ! If we pick up dt = 1000 days the final temperature would be -1616 K. Impossible ! • The mistake is that the temperature T is not a constant but it is a variable ! • Thus, we need to integrate !

  35. dT/T^4 = e*g*s*T^4/[c*dr] Integration yields to T(f)^-3 = T(i)^-3+3* e*g*s*t/[c*dr] Therefore in the previous example we have: T(f)^-3 = 255^-3+3*.8*5.67*10^-8*9.8/[10^5*1004]*8.64*10^5  T(f) = 240.61 K

  36. When an object is a sphere with radius r, if we assume that the sphere has the same temperature and cools at the same rate, then we can re-write the cooling law as: dT/dt = 3* e*g*s*T^4/[r*c*r]

  37. Readings: Handouts 2ab