Copyright R. R. Dickerson LECTURE 2 AOSC 637 Atmospheric Chemistry R. Dickerson
Copyright R. R. Dickerson ATMOSPHERIC PHYSICS Seinfeld & Pandis: Chapter 1 Finlayson-Pitts: Chapter 2 1.Pressure: exponential decay 2.Composition 3.Temperature The motion of the atmosphere is caused by differential heating, that is some parts of the atmosphere receive more radiation than others and become unstable.
VERTICAL PROFILES OF PRESSURE AND TEMPERATURE Mean values for 30 o N, March Tropopause Stratopause
Copyright R. R. Dickerson N2N2 O2O2 Ar O3O3 Inert gasesCO 2 H2H2 ←SO 2, NO 2, CFC’s, etc PM CO CH 4 N2ON2O Composition of the Earth’s Troposphere
Copyright R. R. Dickerson Banded iron formation.
Copyright R. R. Dickerson Units of pressure: 1.00 atm. = 14.7 psi = 1,013 mb = 760 mm Hg (torr) = 33.9 ft H ₂ O = 29.9” Hg = Pa (a Pascal is a Nm ⁻ ² thus hPa = mb) Units of Volume: liter, cc, ml, m³ Units of temp: K, °C Units of R: atm mole -1 K J mole -1 K -1 R’ = R/M wt = J g -1 K -1 For a mole of dry air which has the mass 29 g. Problem for the student: calc Mwt. Wet (2% H ₂ O vapor by volume) air.
Copyright R. R. Dickerson Derive the Hypsometric Equation Start with The Ideal Gas Law PV = nRT or P = ρR’T or P = R’T/α Where R’ = R/Mwt Mwt = MOLE WT. AIR ρ = DENSITY AIR (g/l) α = SPECIFIC VOL AIR = 1/ρ We assume that the pressure at any given altitude is due to the weight of the atmosphere above that altitude. The weight is mass times acceleration. P = W = mg But m = Vρ For a unit area V = Z P = Zρg
Copyright R. R. Dickerson For a second, higher layer the difference in pressure can be related to the difference in height. dP = − g ρ dZ But ρ = P/R’T dP = − Pg/R’T * dZ For an isothermal atmosphere g/R’T is a constant. By integrating both sides of the equation from the ground (Z = 0.0) to altitude Z we obtain:
Copyright R. R. Dickerson Where H ₀ = R’T/g we can rewrite this as: *HYPSOMETRIC EQUATION* Note: Scale Height: H ₀ ~ 8 km for T = 273K For each 8 km of altitude the pressure is down by e ⁻ ¹ or one “e-fold.”
Copyright R. R. Dickerson Problems left to the student. 1.Show that the altitudes at which the pressure drops by a factor of ten and two are 18 and 5.5km. Rewrite the hypsometric Eq. for base 2 and Calculate the scale-height for the atmospheres of Venus and Mars. 3.Derive an expression for pressure as a function of altitude for an atmosphere with a temperature that varies linearly with altitude. 4.Calculate the depth of a constant density atmosphere. 5.Calculate, for an isothermal atmosphere, the fraction of the mass of the atmosphere between 200 and 700 hPa.
Results of Diagnostic Test Basic Thermo 14 Thermo Diag 10 Aerosols (10)Clouds 10 Biogeoche m 11 Photo Chem 12Sow ice 10 Cld models 10 Kinetics 20 student a student b student c student d student e pass 5/54/53/5 4/53/52/53/5
Solubility of CO 2 = f(T)
Copyright R. R. Dickerson Temperature Lapse Rate Going to the mountains in Shenandoah National Park the summer is a nice way to escape Washington’s heat. Why? Consider a parcel of air. If it rises it will expand and cool. If we assume it exchanges no heat with the surroundings (a pretty good assumption, because air is a very poor conductor of heat) it will cool “adiabatically.” CALCULATE: ADIABATIC LAPSE RATE First Law Thermodynamics: dU = DQ + DW
Copyright R. R. Dickerson WHERE U = Energy of system (also written E) Q = Heat across boundaries W = Work done by the system on the surroundings H = Internal heat or Enthalpy ASSUME: a)Adiabatic (dH = 0.0) b)All work PdV work (remember α = 1/ρ) dH = Cp dT – α dP CpdT = α dP dT = (α/Cp) dP
Copyright R. R. Dickerson Remember the Hydrostatic Equation OR Ideal Gas Law Result: This quantity, –g/C p, is a constant in a dry atmosphere. It is called the dry adiabatic lapse rate and is given the symbol γ ₀, defined as −dT/dZ. For a parcel of air moving adiabatically in the atmosphere:
Copyright R. R. Dickerson Where Z ₂ is higher than Z ₁, but this presupposes that no heat is added to or lost from the parcel, and condensation, evaporation, and radiative heating can all produce a non-adiabatic system. The dry adiabatic lapse rate is a general, thermodynamic property of the atmosphere and expresses the way a parcel of air will cool on rising or warm on falling, provided there is no exchange of heat with the surroundings and no water condensing or evaporating. The environmental lapse rate is seldom equal to exactly the dry adiabatic lapse rate because radiative processes and phase changes constantly redistribute heat. The mean lapse rate is about 6.5 K/km. Problem left to the students: Derive a new expression for the change in pressure with height for an atmosphere with a constant lapse rate,
Copyright R. R. Dickerson Stability and Thermodynamic Diagrams (Handout) Solid lines – thermodynamic property Dashed (colored) lines – measurements or soundings day γ a > γ ₀ unstable γ b = γ ₀ neutrally stable γ c < γ ₀ stable On day a a parcel will cool more slowly than surroundings – air will be warmer and rise. On day b a parcel will always have same temperature as surroundings – no force of buoyancy. On day c a parcel will cool more quickly than surroundings – air will be cooler and return to original altitude.
Copyright R. R. Dickerson This equation is easily corrected for a wet air parcel. The heat capacity must be modified: Cp' = (1-a)Cp + aCp(water), where "a" is the mass of water per mass of dry air. If the parcel becomes saturated, then the process is no longer adiabatic. Condensation adds heat. DQ = -Lda Where L is the latent heat of condensation -dT/dZ = g/Cp + (L/Cp) da/dZ Because the amount of water decreases with altitude the last term is negative, and the rate of cooling with altitude is slower in a wet parcel. The lapse rate becomes the wet adiabat or the pseudoadiabat.
Copyright R. R. Dickerson (HANDOUT) Very important for air pollution and mixing of emissions with free troposphere. Formation of thermal inversions. (VIEWGRAPH) In the stratosphere the temperature increases with altitude, thus stable or stratified. DFN: Potential temperature, θ : The temperature that a parcel of air would have if it were brought to the 1000 hPa level (near the surface) in a dry adiabatic process. You can approximate it quickly, or a proper derivation yields: θ z = T ( 1000 hPa/P z ) ** R’/Cp = T ( 1000 hPa/P z ) ** 0.286
DIURNAL CYCLE OF SURFACE HEATING/COOLING: z T 0 1 km MIDDAY NIGHT MORNING Mixing depth Subsidence inversion NIGHTMORNINGAFTERNOON
Copyright R. R. Dickerson Plume looping, Baltimore ~2pm.
Copyright R. R. Dickerson
Copyright R. R. Dickerson
Copyright R. R. Dickerson Plume Lofting, Beijing in Winter ~7am.
Copyright R. R. Dickerson Atmospheric Circulation and Winds
Copyright R. R. Dickerson ITCZ
Copyright R. R. Dickerson Mid latitude cyclone with fronts
Copyright R. R. Dickerson Fig Warm Front
Copyright R. R. Dickerson Fig Cold front
Copyright R. R. Dickerson Fig. 1-17, p. 21
Copyright R. R. Dickerson Fig. 1-16, p. 19
Copyright R. R. Dickerson Fig. 1-16, p. 19
Copyright R. R. Dickerson Detailed weather symbols
Copyright R. R. Dickerson Electromagnetic Radiation
Copyright R. R. Dickerson Electromagnetic spectrum
Copyright R. R. Dickerson Planck’s Law for blackbody radiation: Take first derivative wrt T and set to zero: Wien’s Law: Integrate over all wavelength: Stephan-Boltzmann Law:
Copyright R. R. Dickerson UV-VISIBLE SOLAR SPECTRUM
Copyright R. R. Dickerson Planck’s Law says that the energy of a photon is proportional to its frequency. Where Planck’s Const., h = 6.6x Js The energy associated with a mole of photons at a given wavelength in nm is: