1 / 21

Energy Balance Energy in = Energy out + Δ Storage

Energy Balance Energy in = Energy out + Δ Storage. Bio 164/264 January 11, 2007 C. Field. Radiation: Reminders from last time. Energy of a photon depends on 1/wavelength E = hc/ l h is Planck’s constant (6.63*10 -34 Js), c is the speed of light (3*10 8 m s -1 ), and l is wavelength (m).

danno
Download Presentation

Energy Balance Energy in = Energy out + Δ Storage

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Energy BalanceEnergy in = Energy out +Δ Storage Bio 164/264 January 11, 2007 C. Field

  2. Radiation: Reminders from last time • Energy of a photon depends on 1/wavelength • E = hc/l • h is Planck’s constant (6.63*10-34 Js), c is the speed of light (3*108m s-1), and l is wavelength (m). • Thermal radiation depends on T4: Stefan-Boltzmann law • s = 5.67 * 10-8 W m-2 K-4 • Wavelength of maximum energy depends on 1/temperature (Wien Law) • Solar “constant” ~ 1360 W m-2, over sphere = 342 W m-2

  3. Energy balance • Conservation of energy • Energy in = Energy out + ΔStorage • Energy transport • Radiation • Conduction • Convection = Sensible heat • Evaporation = Latent heat • ΔStorage • Change in temperature • Change in the energy stored in chemical bonds • Change in potential energy

  4. Radiation balance SS = 600 W m-2, q = 20 • Thermal • In = IR down + IR up • Out = IR down + IR up • =461 + 346 – 397 – 397 = 63 • SW • In = direct*cosq*a + • diffuse down*a +diffuse up *a= 282 + 120 + 50 W m-2 • Out = reflected up + • reflected down+ • transmitted down+ transmitted up = already included in in Sd = 100 W m-2 T = 10, e = 1.0* ST = 426 W m-2 ST = 365 W m-2 T = 25, e = .95, a = 0.5 ST = 426 W m-2 ST = 486 W m-2 a = 0.6 T = 35, e = .95

  5. Conduction • Not very important in this class.

  6. Convection • Rate of transport = driving force * proportionality factor • Fick’s law – diffusion F’j = -Dj (drj/dz) • D = molecular diffusivity • Fourier’s law – heat transport H = -k (dT/dz) • k = thermal conductivity (m2 s-1) • Darcy’s law – water flow in a porous medium • Jw = -K(y) (dy/dz) • K(y) = hydraulic conductivity

  7. Keeping units straight - Moles • Most of the mass fluxes in this class will be in moles, where 1 mole = m.w. in g • N2 1 mole = 28.01 g • O2 1 mole = 32.00 g • CO2 1 mole = 44.01 g • H2O 1 mole = • Molar density (mol m-3) ® = rj/Mj is the same for all gases • Ideal gas law pjV = njRT • = 44.6 mol m-3 @ 0C and 101.3 kPa (STP) • ® = rj/Mj

  8. First – get mass flux in molar units • Convert Fick’s law to molar units • diffusion F’j = -Dj (drj/dz) • Fj = F’j/Mj= - ®Dj (dCj/dz) • D = molecular diffusivity • Cj = mole fraction of substance j

  9. Convection – moving heat in air • Start with Fourier’s law • Heat transport H = -k (dT/dz) • k = thermal conductivity • cp = molar specific heat of air 29.3 J mol-1 C-1 • k/cp = DH = thermal diffusivity • Heat transport H = - ®cpDH(dT/dz) • In discrete form • Mass Fj = gj (Cjs – Cja) = (Cjs – Cja)/rj • Heat H = gHcp(Ts-Ta) = cp(Ts-Ta)/rH

  10. Conductances and resistances? • Ohm’s law • V = IR • I = V/R • Conductances – mol m-2 s-1 • Resistances -- m2 s mol-1 series parallel

  11. Physics of the conductance gH • Dimensionless groups • Re = ratio of inertial to viscous forces • Pr = ratio of kinematic viscosity to thermal diffusivity • Gr = ratio of bouyant*inertial to viscous2 • Forced convection • gH = (.664®DHRe1/2Pr1/3)/d • gHa = 0.135 √(u/d) (mol m-2 s-1) • Free convection • gH = (.54®DH(GrPr)1/4/d • gHa = .05((Ts-Ta)/d)1/4 (mol m-2 s-1)

  12. Heat transport by convection • If: • Ta = 20,Tl = 25, u = 2, d = .2 • Then • gHa = .135(3.16) = .427 • H = gHa*2*cp*(Tl-Ta) = .427*2*29.3*5 = 125 W m-2

  13. Latent heat: Energy carried by water • Latent heat of vaporization (l): energy required to convert one mol of liquid water to a mol of water vapor l is a slight function of temp, but is about 44*103 J mol-1 at normal ambient • (this is 585 cal/g!) • Latent heat of fusion: energy required to convert one mol of solid water to a mol of liquid water 6.0*103 J mol-1 • Latent heat plays a dramatic role in temperature control. • Water temperature won’t rise above boiling • Frozen soil or snow won’t rise above zero • Evaporating water requires a large amount of energy. • 1 mm/day = 1kg/m2day, requires 2.45*106 J/m2 • since a day is 86,400 s and a Watt is a J/s, this amounts to 2.45*106/8.64*104 = 28.3 W/m2 • when the atmosphere is dry, evaporation can be 6 mm/day, or even more

  14. Evaporation • Here, we can return directly to Fick’s law • Fj = F’j/Mj= - ®Dj (dCj/dz) • Fj = gj (Cjs – Cja) = (Cjs – Cja)/rj • Where the driving gradient (Cjs – Cja) is the difference between the water vapor inside and outside the leaf (mol mol-1) • And gw is a theme for another lecture

  15. Water vapor concentration • The amount of water vapor the air can hold is a function of temperature = saturation vapor pressure • Relative humidity = ratio of actual vapor pressure to saturation vapor pressure

  16. Saturation vapor pressure where t = 1 - (373.16/T) • T = absolute temperature = T (ºC) + 273.16 • Vsat is in Pascals – 101325 Pascals = 1 atm • Vapor pressure of the air V = Vsat*RH • Vapor pressure deficit = Vsat – V • Mol fraction (wi) = V/P where P = atmospheric pressure

  17. Evaporation and Latent heat • E = gw(wl – wa) • Latent heat = lE • Example • If gw = .5 mol m-2 s-1, wl = 0.03 mol mol-1, wa = 0.01 mol mol-1 • Then E = .5*.02 = .01 mol m-2 s-1 • lE = .01*44*10^3 = 440 W m-2

  18. Energy balance • Net radiation + Convection + Latent heat + D storage = 0 • Or • Rn + H + lE + D storage = 0

  19. Functional role of energy balance • Ehleringer, J., O. Björkman, and H. A. Mooney. 1976. Leaf pubescence: effects on absorptance and photosynthesis in a desert shrub. Science 192:376-377.

  20. Energy balance classics – leaf scale • Parkhurst, D. F., and O. L. Loucks, 1972: Optimal leaf size in relation to environment. Journal of Ecology, 60, 505-537. • Mooney, H. A., J. A. Ehleringer, and O. Björkman, 1977: The energy balance of leaves of the evergreen desert shrub Atriplex hymenelytra. Oecologia, 29, 301-310. • Gates, D. M., W. M. Heisey, H. W. Milner, and M. A. Nobs, 1964: Temperatures of Mimulus leaves in natural environments and in a controlled chamber. Carnegie Inst. Washington Ybk., 63, 418-426.

  21. Energy balance classics – large scale • Charney, J., P. H. Stone, and W. J. Quirk. 1975. Drought in the Sahara: A biogeophysical feedback. Science 187:434-435. • Shukla, J., and Y. Mintz. 1982. Influence of land-surface evapotranspiration on the earth's climate. Science 215:1498-1501. • Bonan, G. B., D. B. Pollard, and S. L. Thompson. 1992. Effects of boreal forest vegetation on global climate. Nature 359:716-718. • Sellers, P. J., L. Bounoua, G. J. Collatz, D. A. Randall, D. A. Dazlich, S. Los, J. A. Berry, I. Fung, C. J. Tucker, C. B. Field, and T. G. Jenson. 1996. A comparison of the radiative and physiological effects of doubled CO2 on the global climate. Science 271:1402-1405.

More Related