Lecture 1b: Global Energy Balance. Instructor: Prof. Johnny Luo
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1 Lecture 1b: Global Energy Balance Instructor: Prof. Johnny Luo
2 Daily average insola>on A few points: 1. Solar constant ~ 1361 W m -2. But averaged over a whole day, we get much less. 2. At NYC in Jan, we get ~ 200 W m -2 daily. However, if you set up a merely 1 m 2 solar panel, it is enough to power a few light bulbs at your home (not yet considering the effects of atmospheric attenuation). 3. At polar region, daily average insolation is high during local summer (>500 W m -2 ) because of the length of the day.
3 Annual mean insolation
4 Average solar zenith angle (sunrise to sunset) Solar zenith angle (θ s ) not only affects the insolation, but also albedo. For polar region, although insolation may be large at local summer, albedo is also large (because of slant angle), which means most of the sunlight will be reflected away.
5 Outlines 1. The Sun-Earth system 2. Quan>fica>on of radia>on: energy flux density 3. Distribu>on of solar insola>on (seasonal & la>tudinal) 4. Energy balance at the top the the atmosphere (TOA) 5. Greenhouse effect 6. Energy balance of the TOA, atmosphere and surface 7. Energy balance in 2D and 3D views 8. Consequence of local energy imbalance: poleward energy flux
6 Absorp>on of solar radia>on occurs only upon the projected area of πr 2, but emission of IR radia>on occurs on the whole surface of the Earth albedo Absorbed solar energy: S 0 (1-α)πr 2 Emi[ed IR energy: σt e 4 4πr 2 T e = Emission Temperature 4 (S 0 /4)(1 α) σ Stefan-Boltzmann law
7 Use the following data to calculate the emission temperature for all nine planets. Does emission temperature monotonically decrease as the distance from the Sun increases? Why? Distance from the Sun (10 6 km) Albedo Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Plato Emission Temperature (K) Solar constant = 1361 W m -2 T e = 4 (S 0 /4)(1 α) σ
8 Quan>fying Radia>ve energy d Energy flux (or luminosity): the Sun radiates out Wa[s Radia?ve flux density near the surface of the Sun: Wa[s / area of the Sun = Wa[s / [4π ( ) 2 m 2 ] = Wa[s m -2 When it strikes the Earth: Energy flux is conserved, but flux density is diluted, so Wa[s/ area of the larger sphere = Wa[s/ [4π ( ) 2 m 2 ] = 1361 Wa[s m -2 Solar constant Solar constant of a planet is scaled by the ratio of the two radii squred: S 0 (d E /d P ) 2, where d E is the Sun-Earth distance, and d P is the Sunplanet distance
9 Use the following data to calculate the emission temperature for all nine planets. Does emission temperature monotonically decrease as the distance from the Sun increases? Why? Distance from the Sun (10 6 km) Albedo Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Plato Emission Temperature (K) Solar constant of a planet = S 0 (d E /d P ) 2, where d E is the Sun-Earth distance, and d P is the Sun-planet distance Solar constant (S 0 ) = 1361 W m -2 T e = 4 (S 0 /4)(1 α) σ
10 Outlines 1. The Sun-Earth system 2. Quan>fica>on of radia>on: energy flux density 3. Distribu>on of solar insola>on (seasonal & la>tudinal) 4. Energy balance at the top the the atmosphere (TOA) 5. Greenhouse effect 6. Energy balance of the TOA, atmosphere and surface 7. Energy balance in 2D and 3D views 8. Consequence of local energy imbalance: poleward energy flux
11 Emission temperature Vs measured surface temperature Distance from the Sun (10 6 km) Albedo Mercury Venus Earth Mars Jupiter Emission Temperature (K) Observed surface T (K) 440 K 735 K 288 K 215 K Saturn Uranus Neptune Plato Solar constant = 1361 W m -2 T e = (S /4)(1 α) 4 0 σ
12
13 Greenhouse Effect: Earth s atmosphere absorbs IR radiation efficiently but is almost transparent to solar radiation. The net effect is that the surface is warmer than if there were no atmosphere. Solar IR Solar/ visible/ shortwave (SW) Terrestrial/ infrared (IR)/ longwave(lw) Earth s Surface
14 A simple model to understand the greenhouse effect Assump>ons: 1. Atmosphere is blackbody for IR radia>on (absorbs all and emits based on Stefan-Boltzmann); 2. Atmosphere is transparent to solar radia>on; Energy balance at the top of the atmosphere (TOA) S 0 (1-α p )/4 = σt A 4 (1) Energy balance near the surface (SRF) S 0 (1-α p )/4 + σt A 4 = σt S 4 (2) From (1), we know T A = 255 K. From (1) & (2), we have 2σT A 4 = σt S4, so T S = (2) 1/4 T A =1.19 T s = 303 K
15 Outlines 1. The Sun-Earth system 2. Quan>fica>on of radia>on: energy flux density 3. Distribu>on of solar insola>on 4. Energy balance at the top the the atmosphere (TOA) 5. Greenhouse effect 6. Energy balance of the TOA, atmosphere and surface 7. Energy balance in 2D and 3D views 8. Consequence of local energy imbalance: poleward energy flux
16
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