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Heat and Radiation

Heat and Radiation. Naoki Mizukami. Outline. Heat Definition of energy and heat Conduction Advection Latent heat/Sensible heat Surface heat budget 1st law of thermodynamics Definition Thermodynamics equation Potential Temperature Lapse rate Radiation Electromagnetic wave

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Heat and Radiation

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  1. Heat and Radiation Naoki Mizukami CVEEN7920 Hydrometeorology

  2. Outline • Heat • Definition of energy and heat • Conduction • Advection • Latent heat/Sensible heat • Surface heat budget • 1st law of thermodynamics • Definition • Thermodynamics equation • Potential Temperature • Lapse rate • Radiation • Electromagnetic wave • Blackbody radiation • Interaction of EM energy with materials CVEEN7920 Hydrometeorology

  3. 1. Heat – energy & heat Energy: the ability to work on object measured in an unit of work [Joule = kg·m2·s-2] 1 [J] = work done by 1 [N] of force to move 1 kg of object by 1 m distance 1 [cal] = 4.185 [J] rate of work [w =J/s] • Types of energy: • Potential energy: energy associated with position • e.g. gravitational (mgh), magnetic, electric field • Kinetic energy: energy due to motion of object • motion of body (macroscopic) -> ke = 1/2*mv2 • motion of molecule (microscopic) -> Thermal energy Heat: energy transferred between two objects (or system) associated with temperature -> unit [J] Temperature: directly proportional to molecular or atomic motion, measure of thermal energy CVEEN7920 Hydrometeorology

  4. 1. Heat – energy & heat Heat transfer modes • Conduction: transfer within a substance via molecular bouncing • Advection (convection): transfer by movement of medium substance (fluid) • Radiation: transfer by electromagnetic (EM) waves radiation convection conduction CVEEN7920 Hydrometeorology

  5. 1. Heat– energy & heat Conduction (W·m-2) qc >0 W C Where qc: heat flux [W·m-2] k: thermal conductivity [W·m-1·K-1] For air, kair = 0.03 [W·m-1·K-1]) T: Temperature [K] x Flux: quantity transferred into/out of unit area per unit time e.g. energy flux (J·m-2·s-1 = W·m-2) Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere? Ans. 300 [W·m-2] = -0.03 [W·m-1·K-1] x ΔT[K] / 0.001[m] ΔT[K] = -10 [K] This temperature gradient within air mostly happens just above the ground on sunny day CVEEN7920 Hydrometeorology

  6. 1. Heat– energy & heat Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere? z C 1mm W CVEEN7920 Hydrometeorology

  7. 1. Heat– energy & heat Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere? z C Ans. 300 [W·m-2] = -0.03 [W·m-1·K-1] x ΔT[K] / 0.001[m] ΔT[K] = -10 [K] W This temperature gradient within air mostly happens just above the ground on sunny day CVEEN7920 Hydrometeorology

  8. 1. Heat– energy & heat Advection (convection) [K· s-1] Advection: horizontal transfer (x and y direction) Convection: vertical transfer (z direction) e.g. warm advection where u, v and w: wind speed [m·s-1] T: Temperature [K] w c Warm advection -> positive Cold advection -> negative u >0 x This is temperature change rate. Can convert heat transfer rate using specific heat CVEEN7920 Hydrometeorology

  9. 1. Heat– energy & heat Q. Air temperature at a point 50 km north of a station is 3 ° C cooler than station. Wind is blowing 60 ° from north at 20 m s-1. What is temperature change due to advection? c u = ?? v = ?? Wind = 20ms-1 w y x CVEEN7920 Hydrometeorology

  10. 1. Heat– energy & heat Q. Air temperature at a point 50 km north of a station is 5 ° C cooler than station. Wind is blowing 60 ° from north at 20 m s-1 for 1 hr. What is temperature change due to advection? Ans. For x direction; u = 20*sin30° [ms-1], dT/dx = 0 so ΔTx = 0 [° C/sec] c u = ?? v = ?? Wind = 20ms-1 For y direction; v = 20*cos30 ° = 10 [ms-1], dT/dy = -3/50e+3[° C/m] So ΔTy = -10* (-5/50e+3) = 0.001[° C /sec] So 0.001*60*60 = 3.6 [° C] increase w y x For total, 0+ 3.6 [° C] = 3.6 [° C] increase CVEEN7920 Hydrometeorology

  11. 1. Heat– Sensible heat vs. Latent heat Sensible heat: Heat released or absorbed by a substance cause temperature change of a substance, not changing phase. -> Can calculate how much heat is added or released by knowing heat capacity of substance and temperature change (next slide) Latent heat : Heat released or absorbed by a substance causes phase change of substance, not changing temperature. Bowen ratio : Ratio of energy available for sensible heat to energy available for latent heat CVEEN7920 Hydrometeorology

  12. 1. Heat– Heat capacity and Specific heat Heat capacity [J·K-1]: The amount of change in thermal energy due to the corresponding temperature change Specific heat [J·kg-1·K-1]: Heat capacity per unit mass If you know temperature change in a system, can know how much thermal energy increases Q. Which substance undergoes the least temperature changes given the same heat CVEEN7920 Hydrometeorology

  13. qin sublimation qin melting qin evaporation ice vapor qout fusion qout condensation qout deposition water 1. Heat– Latent heat Latent heat [J·kg-1] : the amount of heat released or absorbed by a substance per unit mass due to its phase change CVEEN7920 Hydrometeorology

  14. 1. Heat– Latent heat Computing air temperature increase due to latent heat release Assume that in dry air (mass is mair [kg] ), mlquid [kg] of liquid water is produced due to condensation during Δt [sec]. • The amount of heat released: [J] 2. This heat raises temperature of dry air. Use dry air heat capacity, cp [J·kg-1·K-1] to convert heat [J] to temperature change [K]: 3. Divided by Δt to find rate of temperature change CVEEN7920 Hydrometeorology

  15. 1. Heat– Surface heat balance • Heat flux (W·m-2) on the surface • Radiation (FR) • Heat transfer into ground via conduction (FG) • Latent heat transfer (FL) • Sensible heat transfer into air via conduction and convection (FS) • Sum of all the fluxes is zero (+ : flux upward, - flux downward) FR<0 FR<0 FR>0 FS>0 FL<0 FL>0 FS<0 FS>0 FG<0 FG<0 FG>0 Daytime, dry surface Daytime, moist surface Nighttime, moist surface CVEEN7920 Hydrometeorology

  16. 2. 1st law of thermodynamics - description Conservation of energy of system : Change in thermal energy of the system (e.g. air parcel) is equal to heat added to the system and work done by external forces. heat air parcel • If heat is added to air parcel, the energy is used following way. • Air parcel expand (change volume), meaning air parcel does work against pressure force (external force) • Air parcel gains thermal energy inside the parcel CVEEN7920 Hydrometeorology

  17. 2. 1st law of thermodynamics - equation Thermodynamic energy equation Heat input into the system Rate of change in thermal energy Rate of work done by air parcel against pressure force where cv: specific heat at constant volume = 717 [J·kg-1·K-1] p: pressure [Pa] α: specific volume of dry air [m3·kg-1] CVEEN7920 Hydrometeorology

  18. 2. 1st law of thermodynamics - equation Entropy form of thermodynamic energy equation Equation of state (Ideal gas law) Substitute into thermodynamic equation where cp: specific heat at constant pressure = 1004 [J·kg-1·K-1] cp = cv + R R: dry air gas constant = 287 [J·kg-1·K-1] Divided by T and again use equation of state (verify it) CVEEN7920 Hydrometeorology

  19. 2. 1st law of thermodynamics – potential temperature Definition: Temperature, Θ [K], that a dry air parcel at pressure, p [Pa], and temperature, T [K], would have if it were moved adiabatically to a standard pressure, ps [Pa], typically sea level pressure (~ 1000 [hPa]) From entropy form of thermodynamics equation Adiabatic -> J = 0 Rewrite in differential form Integrate from a state (some height) where pressure is p [Pa], and temperature is T [K], to a state of surface where pressure is ps [Pa], and temperature is Θ [K]. (verify it) CVEEN7920 Hydrometeorology

  20. 2. 1st law of thermodynamics – potential temperature Take logarithm of potential temperature equation and then differentiate with time (verify it) • From this equation, Adiabatic process (no heat exchange, J = 0) -> no potential temperature change • This rule is used to find dry adiabatic lapse rate (next few slides) CVEEN7920 Hydrometeorology

  21. 2. 1st law of thermodynamics – Lapse rate Lapse rate: the rate of air temperature decrease with height • Environmental lapse rate: actual lapse rate measured by sounding etc. • Process lapse rate: lapse rate determined by physical processes -> Dry adiabatic lapse rate: lapse rate determined by moving dry air parcel without any heat exchange from the environment -> Moist adiabatic lapse rate: lapse rate determined by moving saturated air parcel without any heat exchange from the environment, but latent heat of condensation released CVEEN7920 Hydrometeorology

  22. 2. 1st law of thermodynamics – Lapse rate Take logarithm of potential temperature equation and differentiate with height, z And then use hydrostatic equation (dp = - ρgdz) and Ideal gas law (verify) -> Environmental lapse rate For adiabatic process, the left side of the equation is zero -> dry adiabatic lapse rate (~9.8 K/1000m) CVEEN7920 Hydrometeorology

  23. Dew points Actual temperature Dry adiabatic lapse rate Moist adiabatic lapse rate CVEEN7920 Hydrometeorology

  24. 3. Radiation– Electromagnetic energy Electromagnetic wave c λ c: speed of light 3e+6 [m·s-1] λ: Wavelength [m] ν: frequency [s-1] A A: amplitude - energy is proportional to A2 T: period [s] –time required to travel by full wavelength, ν = 1/T CVEEN7920 Hydrometeorology

  25. 3. Radiation– Electromagnetic energy Any objects above zero absolute temperature ( 0K = -273 °C) emit EM energy Blackbody: a body that emit the maximum energy it can emit or absorb given by body temperature [K] Plank’s law: EM energy flux, Eλ(T) [W·m-2·μm-1], emitted from blackbody whose temperature is T [K] as a function of λ[μm] -> where h: plank’s constant 6.624e-34 [J·s] c: speed of light 3e+6 [m·s-1] k: Boltzmann constant 1.381e-23 [J·K-1] CVEEN7920 Hydrometeorology

  26. 3. Radiation– Electromagnetic energy Blackbody radiation curve CVEEN7920 Hydrometeorology

  27. 3. Radiation– Electromagnetic energy Electromagnetic spectral Peak of solar radiation Peak of earth radiation CVEEN7920 Hydrometeorology

  28. 3. Radiation– Electromagnetic wave Stefan-Boltzmann law: Total energy flux, E(T) [W·m-2], emitted from blackbody whose absolute temperature is T [K] σ: Stefan-Boltzmann constant (5.67e-8 [W·m-2·K-4]) CVEEN7920 Hydrometeorology

  29. 3. Radiation– Electromagnetic wave Q: How does total energy flux change if temperature increases by 1 K for earth (275 [K]) and sun (6000 [K])? CVEEN7920 Hydrometeorology

  30. 3. Radiation– Electromagnetic wave Q: How does total energy flux change if temperature increases by 1 K for earth (275 [K]) and sun (6000 [K])? Ans: Write Stefan-Boltzmann equation in derivative form For earth, dE = 3 x 5.67e-8 x 275^3 x 1 = 2.22 [W·m-2] For sun, dE = 3 x 5.67e-8 x 6000^3 x 1 = 2.31e4 [W·m-2] CVEEN7920 Hydrometeorology

  31. 3. Radiation– Electromagnetic wave Blackbody emits theoretically maximum EM energy. But Actual emission, I, is less than blackbody emission. Emissivityε: a fraction blackbody radiation that is actual emitted depends on wavelength and substances (0 ~1) Emissivity at a particular wavelength Emissivity over all range of wavelength CVEEN7920 Hydrometeorology

  32. 3. Radiation– interaction of radiation with materials In vacuum, characteristics of EM wave (energy) is intact. if EM wave encounter the matter (air, water, other terrestrial materials), it is partitioned into three processes Reflection (reflectivity): Absorption (absorptivity): Transmission (transmissivity): All the coefficients depends on wavelength Kirchhoffs law: ε λ = aλ Objects that absorb EM energy of a particular wavelength emit EM energy of the same wavelength in the same efficiency CVEEN7920 Hydrometeorology

  33. 3. Radiation– interaction of radiation with materials Absorption in atmosphere CVEEN7920 Hydrometeorology

  34. Reference Texts A first course in atmospheric Radiation, G.W. Petty Meteorology today, C.D. Ahrens Introduction to Dynamic meteorology, J.R. Holton CVEEN7920 Hydrometeorology

  35. Q1 Latent heat Consider an air column of which height is h [m]. The average of condensation heating rate [K/s] over the air column is estimated by measuring rainfall rate on the ground. • Drive the equation for this using rainfall rate, P [ms-1], liquid water density, ρliquid [kg·m-3], dry air density, ρair [kg·m-3] and h [m]. • Using this equation, find average latent heat heating for 5 [mm·h-1]. Use the following parameters • h = 11 km, ρair= 0.689 [kg·m-3] , ρliquid = 1025 [kg·m-3], Lv = 2.5e+6 [J·kg-1], and cp = 1004 [J·kg-1·K-1] Hint; See slide 9 CVEEN7920 Hydrometeorology

  36. Q.2. (Lapse rate) Wind blowing from west to east is pushing air up the mountain. On the windward side of the mountain, air temperature decrease with height at 5 °C per 1000 m. On the leeward side of the mountain, air temperature increase with decreasing height at dry adiabatic lapse rate. Surface air temperature was measured as 15 °C at A. What is the surface temperature at B ?? Top: 3000 m wind B, ?? °C A, 15 °C 2000 m 1400 m CVEEN7920 Hydrometeorology

  37. Q.3. (Radiation) Suppose air polluted air reflects 30% and transmits 50% of the incoming solar radiation (Assume 1368 [W·m-2] ). A) How much is absorbed by this air? B) How much is emitted by this air, assuming Tair = 270 [K] (hint: use Kirchhoffs law to find emissivity of this air) CVEEN7920 Hydrometeorology

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