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The Greenhouse Effect

The Greenhouse Effect. Garver GEO 307. Take home points from Chapter 2. EMR carries energy through space If an object can absorb energy, it can also emit energy. Objects emit at a rate equal to T 4. Boundary to Space. VIS. IR. Atmosphere. Chapter 3: Layer Model.

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The Greenhouse Effect

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  1. The Greenhouse Effect Garver GEO 307

  2. Take home points from Chapter 2 • EMR carries energy through space • If an object can absorb energy, it can also emit energy. • Objects emit at a rate equal to T4

  3. Boundary to Space VIS IR Atmosphere Chapter 3: Layer Model • Algebraic calculation of the effect of an IR absorber (a pane of glass) on the equil. T of the Earth. • Not accurate or detailed • Not used for global forecasts

  4. Bare Rock Model • T of Earth is controlled by the ways that energy comes from the Sun and is re-radiated to space as IR. • Sun’s T is high so its energy flux is high • See SB law

  5. Simple model of Earth T • Layer model • Toy system to learn from • Assumption: energy in = energy out • Fin = Fout • Flux is in (W = J/S) • Incoming sunlight = Iin = 1350 W/m2 • solar constant

  6. Albedo = reflected light =  • Not absorbed and re-radiated as IR • Just ‘bounces’ back • Earth = 0.30 - clouds, snow, ice • Venus = 0.70 - sulfuric acid clouds • But carbon dioxide creates gh effect, T 758 deg K • Mars = 0.15 - no clouds • Some carbon dioxide, T 253 K (afternoon)

  7. Temperature Scales • Kelvin • Celsius • Fahrenheit • Temperature Conversions: ºC = 5/9(ºF-32) K = ºC + 273 Absolute zero at 0 K is −273.15 °C (−459.67 °F)

  8. Incoming solar energy not reflected • 1350 W/m2 (1 - ) = 1000 W/m2 • Want flux for the whole planet (no m2) • Fin(W) = Iin(W/m2) x Area(m2)

  9. Solar Constant Day Night What area do we use? • Sun shines on half the Earth • Light is weaker/stronger • latitude, dawn/dusk

  10. Solar Constant Day Night Sunlight hits Earth from same direction, makes a circular shadow,use the area of a circle, not a sphere.Earth receives influx of energy equal to the intensity of sunlight multiplied by thearea of a circle = r2earth Area (m2) = r2earth

  11. Put them together, total incoming flux is: • Fin = r2earth(1 - ) Iin • Remember: Fin = Fout • Iin = 1350 W/m2 • Reduce by albedo to 1000 W/m2 • Multiply by area of circle to get Flux (W)

  12. First layer model has no atmosphere, just a bare rock! • We’re trying to find a single value for Te of Earth to go along with a single value of the heat fluxes Fin and Fout • Rate at which Earth radiates energy is given by SB law: • Fout= Area x esT4earth • e = emissivity, 0 to 1, unitless • e = 1 would be a blackbody

  13. Emissivity • From Wikipedia • The emissivity of a material (written ε or e) is the relative ability of its surface to emit energy by radiation. It is the ratio of energy radiated by a particular material to energy radiated by a black body at the same temperature. A true black body would have an ε = 1 while any real object would have ε < 1. Emissivity is a dimensionless quantity. • The effective emissivityof earth, about 0.612

  14. As we did for solar energy, need to convert intensity, I, to flux, F, by multiplying by area. • What area do we use? • Energy leaves in all directions • So, we need the area of a sphere; • Asphere = 4r2earth

  15. Total Energy Flux from Earth • Fout= 4r2earthesT4earth • SB law x Area

  16. Fin = Fout r2earth(1 - )Iin = 4r2earthesT4earth (1 - )Iin Flux out Flux in esT4earth

  17. Can cancel out some factors 4r2earthesT4earth = r2earth(1 - )Iin esT4earth = (1 - )Iin 4 No atmosphere

  18. We know everything here except T of earth Now, rearrange and put Tearth alone: esT4earth = (1 - ) Iin Tearth = 4 (1 - ) Iin 4 4es

  19. Tearth = 4 (1 - ) Iin 4es • Now we have a model that shows the relationship between crucial climate quantities: • Solar intensity • Albedo • But, if we calculate Tearthwe get 255 K (-15˚C). • This is too cold, why?

  20. Need a model with a single pane of glass as its atmosphere • Simple model lacks gh effect

  21. Layer Model with GH Effect • Energy diagram for a planet with a single pane of glass for an atmosphere. • Glass is transparent to incoming VIS but a blackbody to outgoing IR IR VIS IR IR

  22. Incoming VIS passes thru atm, absorbed by surface • Surface radiates IR as esT4ground • Iup,ground, which is entirely absorbed by atm • Atm has a top and bottom, so it radiates energy up and down: • Iup,atm • Idown ,atm esT4ground

  23. 2esT4atm esT4ground • The model assumes that the energy budget is in a steady state: energy in = energy out • This applies to individual pieces of the model as well. • So energy budget for atm: Iup,atm+ Idown,atm = Iup,ground (units are Watts/Area) Or, 2esT4atm = esT4ground

  24. esT4ground • Budget for the ground is differentnow because we have heat flowing down from atm. • We still assume the energy budget is in a steady state: Iin = Iout • So the component fluxes are: Iup,ground= Iin,solar+ Idown,atm (units are Watts/Area) Or, esT4ground = (1-a)Isolar + esT4atm 2esT4atm 4

  25. esT4ground Finally, a budget for the Earth overall: • Draw a boundary above the atm and figure that if energy gets across the line in, it’s flowing out at the same rate. • Iup,atm = Iin,solar • The intensities are comprised of individual fluxes from the Sun and the atmosphere. esT4atm = (1-a)Isolar In = Out 2esT4atm 4 Page 25

  26. esT4ground Budget for the Earth overall: 4 esT4atm = (1-a)Isolar One unknown is Tatm If we solve for Tatmwe get the same answer as solving for Tearth in the bare planet model. This is an important point! In = Out 2esT4atm

  27. It tells us that the place in the Earth system where the T is most directly controlled by the rate of incoming solar energy is the T at the location that radiates to space. This is called the skin temperature of the Earth. (it’s equal to the outer most Tatm) Now that we know this we can plug that into the budget eqn for the atm. Page 25 - 26 In = Out 2esT4atm esT4ground

  28. And see that: (page 26) 2esT4atm = esT4ground Or, Tground = 2Tatm This means that the T of the ground must be warmer than the skin T by a factor of the fourth root of 2, an irrational number that = 1.189. (~19%) Fourth root of 2 = ± 1.189207 Tatm 4 2esT4atm Warmer by about 19% esT4ground

  29. Bottom Line! • In our model where we slide in an atmosphere we have shown that: • The T of the ground must be warmer than the skin T (at top of atm) by roughly ~19%.

  30. Final points for Chapter 3 Result from bare rock model, this T is more similar to skin T at top of atm Recorded T Data, not from a model Result from layer model (w/greenhouse) Too cold Pretty close Too warm 253 x 1.189 = 300

  31. Atm is not an energy source like a giant heat lamp in the sky. So, how does it change the T of the ground?? Analogy: Equilibriumwater level in a steadily filling/draining sink. 1. Water flows in, hangs out awhile, drain out. energy in = energy out 2. Drains faster as the level rises due to inc. pressure (weight). energy flows out faster as T rises 3. Eventually water level gets to a point where flow out equals flow in. This is the equlibrium T Sink Water Level Drain

  32. 4.Now, constrict the drain with a penny. Water flows out more slowly, The water level rises until the higher water level pushes more water down drain to balance the flow from the faucet again. Greenhouse gases are the ‘penny in the drain’. They make it more difficult for heat to escape the Earth, and as a result the equilibrium temperature has to go up until the fluxes balance each other again. Sink Water Level Rises Drain Constrict flow

  33. Take Home Points • The outflow of IR energy from a planet must balance heating from the Sun. • The planet accomplishes this balance by adjusting its temperature. • Absorption of outgoing IR by the atmosphere warms the surface of the planet, as the planet strives to balance its energy budget

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