1. Lecture 1: Global Energy Balance Instructor: Prof. Johnny Luo

3. Climate & Climate Change

4. Big picture: 1.Earth is powered by solar energy (through radiation in the visible).

5. Big picture: 1.Earth is powered by solar energy (through radiation in the visible). 2.Earth can’t just receive w/o giving away: the giving-away part is accomplished through IR radiation.

6. Big picture: 1.Earth is powered by solar energy (through radiation in the visible). 2.Earth can’t just receive w/o giving away: the giving-away part is accomplished through IR radiation. 3.Considering the whole Earth, solar and IR radiation are in balance.

7. Big picture: 1.Earth is powered by solar energy (through radiation in the visible). 2.Earth can’t just receive w/o giving away: the giving-away part is accomplished through IR radiation. 3.Considering the whole Earth, solar and IR radiation are in balance. 4.The balance is not uniformly distributed: tropics receives more than it gives away (surplus); the opposite for the polar region (deficit).

8. Big picture: 1.Earth is powered by solar energy (through radiation in the visible). 2.Earth can’t just receive w/o giving away: the giving-away part is accomplished through IR radiation. 3.Considering the whole Earth, solar and IR radiation are in balance. 4.The balance is not uniformly distributed: tropics receives more than it gives away (surplus); the opposite for the polar region (deficit). 5.There are important consequences for this surplus/deficit distribution of energy - fluid will move. This lecture is going to look at various aspects of the global energy balance & regional imbalance quantitatively

9. Outlines 1.The Sun-Earth system 2.Quantification of radiation: energy flux density 3.Distribution of solar insolation 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 from TOA 8.Consequence of regional energy imbalance: poleward energy flux

10. If you look into the Sun… View from solar optical telescope Facts about the Sun 1.Temperature (appears to us): ~ 5800 K; 2.Source of energy: nuclear fusion - lighter elements made into heavier ones, releasing energy in the process. 3.Lifetime: ~ 11 billion yrs, now about half way 4.Radius: 6.96  10 8 m or 696 million km (Earth radius ~ 6,400 km) 5.Mean distance from Earth: 1.496  10 11 m (~ 200 times of its radius, which is why it looks like a tiny dish) 6.Energy it radiates: 3.9  10 26 Watts 7.Solar energy arriving at the Earth: 4.36  10 16 Watts (~1 out of 10 billion)

11. Energy flux (or luminosity): the Sun radiates out 3.9  10 26 Watts Radiative flux density near the surface of the Sun: 3.9  10 26 Watts / area of the Sun = 3.9  10 26 Watts / [4   (6.96  10 8 ) 2 m 2 ] = 6.4  10 7 Watts m -2 When it strikes the Earth: Energy flux is conserved, but flux density is diluted, so 3.9  10 26 Watts/ area of the larger sphere = 3.9  10 26 Watts/ [4   (1.5  10 11 ) 2 m 2 ] = 1367 Watts m -2 Quantifying Radiative energy d Think-Pair-Share: 1367 Watts m -2 : comparing it to some energy consumption rates of your household electronic devices, is this a small number or a big number? Solar constant

12. Energy flux (or luminosity): the Sun radiates out 3.9  10 26 Watts Radiative flux density near the surface of the Sun: 3.9  10 26 Watts / area of the Sun = 3.9  10 26 Watts / [4   (6.96  10 8 ) 2 m 2 ] = 6.4  10 7 Watts m -2 Quantifying Radiative energy d The world’s energy consumption: 15.5 TW = 1.55  10 13 Watts Evenly spread out over the Earth’s surface: 0.03 W m -2 10% of the surface is occupied by humans, so our energy consumption is 0.3 W m -2. NYC: 100 W m -2. When it strikes the Earth: Energy flux is conserved, but flux density is diluted, so 3.9  10 26 Watts/ area of the larger sphere = 3.9  10 26 Watts/ [4   (1.5  10 11 ) 2 m 2 ] = 1367 Watts m -2 Solar constant

13. Max radiation an object emits (Blackbody radiation) is related to temperature: Stefan-Boltzmann Law: Flux =  T 4, where  = 5.67  10 -8 W m -2 K -4 For the Sun, radiative flux =6.4  10 7 Watts m -2  T 4 = 6.4  10 7 Watts m -2  T = 5796 K 1367 W m -2  6.4  10 7 W m -2  5796 K (the way we infer the T of the Sun) Exercise: calculate the radiation flux density our own body produces (assuming T = 37 0 C) Human body: 524.7 W m -2 Sun: ~ 64,000,000 W m -2

14. Think-Pair-Share: are we hotter than a 100 Watts light bulb? Let’s model lightbulb as a sphere (miniature Sun) of 5 cm radius. The energy flux density (100 Watts) is: 100 W / (4π×0.05 2 m 2 ) = 3183 W m -2 Stefan-Boltzmann Law: Flux =  T 4, where  = 5.67  10 -8 W m -2 K -4 So, T ~ 487 K or 214 0 C

15. Outlines 1.The Sun-Earth system 2.Quantification of radiation: energy flux density 3.Distribution of solar insolation 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. The same Sun appears much colder at a winter noon than at a summer noon. So, solar constant is not so constant when we study the distribution of solar energy at a specific location. Solar constant S 0 =1367 W m -2 Insolation = S 0 cos  s Solar radiation incident on the TOA depends on 1) latitude, 2) season, and 3) time of the day. These three parameters collectively determines the solar zenith angle  s.

17. 1.Local zenith direction: Z 2.Solar zenith angle: θ s 3.Subsolar point:ss 4.Declination angle:δ 5.Latitude: Φ 6.Hour angle:h Let’s pick a random point on Earth: Χ

18. 1.Local zenith direction: Z 2.Solar zenith angle: θ s 3.Subsolar point:ss 4.Declination angle:δ 5.Latitude: Φ 6.Hour angle:h Let’s pick a random point on Earth: Χ At a given location on Earth at a given season and time: θ s = θ s (δ, Φ, h): 1)hour of the day; 2)a specific season 3)a specific location

19. 1.Local zenith direction: Z 2.Solar zenith angle: θ s 3.Subsolar point:ss 4.Declination angle:δ 5.Latitude: Φ 6.Hour angle:h Let’s pick a random point on Earth: Χ At a given location on Earth at a given season and time: θ s = θ s (δ, Φ, h): 1)hour of the day; 2)a specific season 3)a specific location

20. Radian: Measure of the angle (radian) = arc/radius Radian of a full circle is 2π because 2πr/r = 2π  Hour angle (h).  Latitude .  Declination angle (  : -23.5 0 to +23.5 0 ) Solar zenith angle  s depends on: Radian to degree conversion: radians = degrees × π / 180 So, 360 0 = 2π, 180 0 = π, 90 0 = π/2

21. Arc P-X = 90 -  Arc P-ss = 90 -  Arc X-ss =  s  h (hour angle): ss point relative to its position at noon Known: , ,  h Unknown:  s This is a unit sphere, i.e., radius =1, so length of an arc equals the angle (in radian).

22. Spherical trigonometry: law of cosines (on unit sphere) a, b, c are arcs on great circles

23. Spherical trigonometry: law of cosines (on unit sphere) cos(  s )=cos(90-  )cos(90-  )+ sin(90-  )sin(90-  )cos(h) cos(  s )=sin(  )sin(  )+ cos(  )cos(  )cos(h) X-ss P-XP-ss a, b, c are arcs on great circles

24. cos(  s )=sin(  )sin(  )+ cos(  )cos(  )cos(h) Implications of the solar zenith angle equation Local sunrise & sunset time: mathematical way of saying “sunrise” or “sunset” is  s = 90 0 or cos  s = 0. Knowing that tan(a)=sin(a)/cos(a), we have cos(h 0 )= - tan(  )tan(  ) You can check the declination angle of the Sun (  ) online: http://www.wsanford.com/~wsanford/exo/sundials/DEC_Sun.html http://www.wsanford.com/~wsanford/exo/sundials/DEC_Sun.html On Feb 6,  = -15 0 52’ or -0.28 in radian, so tan(  )= -0.29. In NYC,  = 40 0 43 ’ or 0.71 in radian so tan(  )=0.86 cos(h 0 )=0.29  0.86 = 0.25 so h 0 = 1.3 or 75 0. Noon (12 pm) to 6 am (6 pm) is 90 0, so 75 0 means 7 am or 5 pm.

25. NYC’s latitude is about 40 0 This is solar constant (1367 W m -2 ) midnight noonmidnight noon midnight

26. Daily average insolation So, daily average insolation is a function of: 1)latitude, 2)declination angle of the Sun or in other words, day of the year, 3)Length of the daytime (but it is determined by the former two factors).

27. Daily average insolation A few points: 1. Solar constant ~ 1367 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 lamps at your home (not considering the effects of air and clouds). 3. At polar region, daily average insolation is high during local summer (>500 W m -2 ) because of the length of the day. T-P-S: where is the maximum and how do you explain it.