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Calculating Intensity of Solar Radiation Reaching Earth

Learn how to calculate the intensity of solar radiation reaching Earth and understand the different types of energy transfer and the concept of black-body radiation.

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Calculating Intensity of Solar Radiation Reaching Earth

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  1. Do Now. The sun radiates ~3.9 x 10 26 J /s, Earth av. orbit = 1.5 x 1011 m, calculate intensity of radiation reaching Earth. 3.9 x 10 26 Js-1 4p(1.5 x 1011 m)2, = 1380 W/m2 solar “constant” ~ 1400 W m2.

  2. Solar Intensity E received (any planet) I = power = P I = intensity W/m2, area 4pr2. P = radiated power W. (sphere) r = distance.

  3. Energy Transfer

  4. Name the 3 types of E transfer • Conduction. • Convection • Radiation. • All important in climate study. • Only way for Earth to lose E to space is….

  5. Radiation • All matter can both absorb and emit EM radiation. • Radiation is absorbed by matter at specific frequencies, l. • Solids absorb many more different l than the atmospheric gases.

  6. Black-body radiation • A solid’s color determines the wavelengths it cannot absorb. • A green object reflects (does not absorb) green light. • A black object absorbs and emits all wavelengths. When bathed in white light a solid is the color of the light that it cannot absorb.

  7. EM Emission • Matter above 0 K emits EM radiation. • EM generated by accelerating charges. • Electrons, protons, nuclei, ions. • Emitted radiation is related to the T, and type of surface. As a black body heats up it emits all wavelengths, called black-body radiation.

  8. Demo • What happens as T increases?

  9. Black Body SpectrumTotal intensity goes up.The shorter l more intense as T increases. Sun = 5800 K Vis, IR Earth = 288 K IR only

  10. Peak l is calculated by Wein’s displacement law: b = 2.89 x 10-3 m K

  11. 1. What is the peak wavelength for a lamp that glows at 1800o C? • 1800o C = 2073 K • 2.89 x 10-3 m K 2073 K • 1.39 x 10-6 m.

  12. Stefan-Boltzmann Law - Relates emitted power & to object’s T & area (m2) for black body. s is a constant in data booklet. Emitted P = power Watts. A = surface area m2 (Area sphere = 4 p r2) sun, Earth. T = Kelvin T s = 5.67x 10 – 8 Wm-2 K-4

  13. 2. If the Sun behaves as a perfect black body with T = 6000 K, what is the energy radiated per second? The radius sun is 7 x 108 m. • Area sphere = 4 p r2. P = 4(p)(7 x 108 m)2(5.67x 10 – 8 )(6000 K)4. P = 4.53 x 10 26 W.

  14. 3. A tungsten filament has a length of 0.5 m and a radius of 5.0 x 10-5 m. The power rating is 60 W. Estimate the temperature of the filament if it acts as a black body. (Use surface A = 2prh) . 60 J/s = A s T4. A = (2)(p)(5.0 x 10-5 m)(0.5m) = 1.57 x 10-4m2 T4 = 60 J/s (1.57 x 10-4m2) (5.67x 10 – 8 ) T = 1611 K = 1600 K.

  15. Most objects are not as emissive as a black body.

  16. Emissivity (e) • Is a number from 0 – 1 telling how an object’s emitted radiation compares w/ perfect black body. From Stefan’s law: • e is emissivity = ratio energy emitted/black body energy at a T. • Shiny objects have low e, dark objects have high e.

  17. 4. An object at 500 K with a surface area of 5 m2, emits 5300 W of power. What is its emissivity? • P = eAsT4. • 5300 W = e (5m2) (5.67x 10 – 8) (500 K )4. • e = 0.3

  18. Intensity 3000 4000 5000 1000 2000 Wavelength (nm) 700 600 500 400 Wavelength  / nm Visible Light 5. The sun’s surface has a temperature of 4500K. What is the prevalent wavelength of light? SOLUTION: max= 2.9010-3/ 4500 max= 6.4410-7 m = 644 nm.

  19. 6. The planet Mercury has a radius of 2.50106 m. Its sunny side has a T 400°C and its shady side -200°C. Treat it as a black-body, find its average power. SOLUTION: Asphere = 4(2.50106)2 = 7.851013 m2. For T use TAVG = (673 + 73) / 2 = 373 K P = AT 4 = (5.6710-8)(7.851013)3734 = 8.621016W. FYI Since no body is at absolute zero (K = 0) it follows from the Stefan-Boltzmann law that all bodies radiate.

  20. Energy from the Sun - Insolation Sun radiates 43% visible, 49 % IR, 8 % UV. Earth receives very small fraction of total solar power ~ 1400 W/m2 - most does not reach surface.

  21. Solar Constant at top of Atmosphere average ~ 1390 W/m2 • 11 year sunspot cycle ~ 0.1 % • Elliptical orbit ~ 7 % • Longer term cycles (Milankovitch)

  22. Absorbed by Earth’s surface? • Some radiation reflected or scattered before absorption.

  23. Albedo - Ability of planet to reflect or scatter radiation. It’s a ratio. Albedo a = total reflected/scattered I total incident Ialways 0-1 1 = high reflectionSee tables.

  24. 7. Given the following values, find the albedo:  • Incident power = 340 W/m2 • Reflected power = 100 W/m2. • Re-radiated power = 2 W/m2. • 0.29 • 29%

  25. 8. Match the surface materials to their correct albedo. • Snow • Ground • Ice • charcoal • 95 • 15 • 1 • 90

  26. Albedo % M ean Earth Albedo = 30%

  27. Kerboo sheet 8.2 solar constant albedo. • End here Wed

  28. Earth’s day/night cycle, tilt, & varying orbital distance affect the insolation hitting surface. Accounting for day/night & seasons could average to ~ 170 W/m2 or less.

  29. To find the exact E reaching the surface or object , we need to know how much is absorbed & reflected by atmosphere & surface.

  30. Natural Greenhouse EffectNatural warming effect due to atmosphere. • The moons av T is -18oC. • Earth is +16 oC. • Same distfr sun but no atmosphere • Atmospheric greenhouse gasses absorb outgoing IR radiation from Earth, re-radiate some back to Earth.

  31. What happens to the radiation? • Sun emits Visible, some IR, &UV. • Visible light gets through atmosphere to Earth. • (UV & IR mostly absorbed in atmosphere) • Earth surface either reflects, or absorbs & later emits E as IR radiation. • Greenhouse gasses absorb, re-radiate IR in all directions, some back to Earth.

  32. Interaction of solar E with matter on Earth

  33. Individual Atoms (gasses): • Can model photon absorption with Bohr • Excite e- to different orbits by difl of photons • Ionization of atom

  34. Molecules More Complexlike to vibrate at specific resonant f. Photons w/ E at the resonant f, are absorbed. KE increases.T increase.Usu absorb IR f.

  35. E Interaction with polyatomic Solidsdif than single atoms or gas molecules. • Solids absorb large range of f • over broader spectrum. • Molc’s vibrate, • Emit low f E IR.

  36. 5 Greenhouse Gases • Natural Resonant f of greenhouse gasses is in IR region-the emission f of solids. Visible light f too high. • When molc absorbs proper IR l / photon resonance occurs. • Molecular KE increase so T increases. • CO2 • H20 • CH4 • N2O • O3.

  37. CO2 absorbs specific IR f’s l, molecular resonance occurs. wavelength

  38. IR Spectrum Methanemore vibrational modes, more l absorbed. wavelength

  39. Gasses absorb & emit specific f, • Have absorption and emission spectrum.

  40. Solids Absorb & emit wide range f ‘s Black bodies – absorb & emit all l.

  41. Film Clip • How do greenhouse gasses work? 3.1 min • https://www.youtube.com/watch?v=sTvqIijqvTg • hi

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