AOSC 620 Lecture 2 Physics and Chemistry of the Atmosphere, I

AOSC 620 Lecture 2 Physics and Chemistry of the Atmosphere, I
Professor Russell Dickerson Room 2413, Computer & Space Sciences Building Phone(301) web site Salby Chapter 4. Copyright © R. R. Dickerson

Copyright © 2015 R. R. Dickerson & Z.Q. Li
Survey Results Background 13 AOSC, 3 CHEM, 1 ENCH, 1 AERO Goals/interests 8 Chemistry/Climate 5 Forecasting/Weather 5 Air Pollution 2 Biosphere/atmosphere 1 each atmosphere/ocean, aircraft, cryo, microphysics, remote sensing, aerosols. Copyright © R. R. Dickerson & Z.Q. Li 2

Lecture 2. Thermodynamics of Air, continued – water vapor.
Objective: To find some useful relationships among air temperature, volume, and pressure. Review Ideal Gas Law: PV = nRT Pα = R’T First Law of Thermodynamics: đq = du + đw DW = ∫ pdα Copyright © R. R. Dickerson & Z.Q. Li

Review (cont.) Definition of heat capacity: cv = du/dT = Δu/ΔT cp = cv + R = Reformulation of first law for unit mass of an ideal gas: đq = cvdT + pdα đq = cpdT − αdp Copyright © R. R. Dickerson & Z.Q. Li

For an isobaric process: đq = cpdT For an isothermal process:
Review (cont.) For an isobaric process: đq = cpdT For an isothermal process: đq = − αdp = pdα = đw For an isosteric process: đq = cvdT = du For an adiabatic process: cvdT = − pdα and cpdT = αdp Copyright © R. R. Dickerson & Z.Q. Li

For an adiabatic process: cvdT = − pdα and cpdT = αdp du = đw
Review (cont.) For an adiabatic process: cvdT = − pdα and cpdT = αdp du = đw (T/T0) = (p/p0)K Where K = R’/cp = 0.286 (T/θ) = (p/1000)K Define potential temperature: θ = T(1000/p)K Potential temperature, θ, is a conserved quantity in an adiabatic process. Copyright © R. R. Dickerson & Z.Q. Li

Review (cont.) The Second Law of Thermodynamics is the definition of φ as entropy. dφ ≡ đq/T ჶ dφ = 0 Entropy is a state variable (path independent and no bar on d, đ). Δφ = cpln(θ/θ0) In a dry, adiabatic, process potential temperature doesn’t change, thus entropy is conserved. Copyright © R. R. Dickerson & Z.Q. Li

How does pressure change with altitude?
Start with the Ideal Gas Law PV = nRT or P = ρR’T or P = R’T/α Where R’ = R/Mwt Mwt = Molecular weight (mass) of air. ρ = density of air(g/l) α = specific volume = 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 & Z.Q. Li

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:

Where H₀ = R’T/g we can rewrite this as:. Hypsometric Equation
Where H₀ = R’T/g we can rewrite this as: *Hypsometric Equation* Note: Scale Height: H₀ ~ 8 km for T = 273K For each 8 km of altitude the pressure is down by e⁻¹ or one “e-fold.” Problems for the student: Derive an expression for pressure as a function of altitude using base two and base ten instead of base e. Calculate the scale height for the atmospheres of Venus and Mars. Ans. base 2 = H₀*ln(2) = 5.5 km base 10 = H₀*ln(10) = 18 km

Lapse Rate of Temperature See Salby 2.4.2
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.” Calculating the dry adiabatic lapse rate. First Law Thermodynamics*: dU = đQ – đW Or đQ = dU + đW = dU + PdV *watch out for work done by the system or on the system.

U = Energy of system (also written E) Q = Heat across boundaries
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: Adiabatic (dQ = dH = 0.0) All work PdV work (remember α = 1/ρ) dH = Cp dT – α dP CpdT = α dP dT = (α/Cp) dP

Remember the Hydrostatic Equation OR Ideal Gas Law Result: This quantity, -g/Cp, is a constant in a dry atmosphere. It is called the dry adiabatic lapse rate and is given the symbol γ₀, defined as −dT/dZ. For a parcel of air moving adiabatically in the atmosphere:

Where Z₂ is higher than Z₁, 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, g0, 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, G, 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,

Useful idea - a perfect or exact differential:
If z = f(x,y), dz is a perfect differential iff: ∂2f/∂x∂y = ∂2f/∂y∂x ჶ dz = 0 For example, v = f(T,p) dv = (∂v/∂p)T dp + (∂v/∂T)p dT This is true for dU, dH, dG, but not đw or đq. Copyright © R. R. Dickerson & Z.Q. Li

Various Measures of Water Vapor Content
Vapor pressure Vapor density – absolute humidity Mixing ratio, w (g/kg) Specific humidity Relative humidity Virtual temperature (density temp) Dew point temperature Wet bulb temperature Equivalent temperature Isentropic Condensation Temperature Wet-bulb potential temperature Equivalent potential temperature Copyright © R. R. Dickerson & Z.Q. Li

Virtual Temperature: Tv or T*
Temperature dry air would have if it had the same density as a sample of moist air at the same pressure. Question: should the virtual temperature be higher or lower than the actual temperature? Copyright © R. R. Dickerson & Z.Q. Li

Copyright © 2015 R. R. Dickerson
Alternate derivation: Since r proportional to Mwt Where w is the mass mixing ratio and x (molar or volume mixing ratio) then [H2O] = w/0.622 Copyright © R. R. Dickerson

Copyright © 2015 R. R. Dickerson
Where w is the mass mixing ratio and x (molar or volume mixing ratio) = [H2O] = w/0.622 T* = T (29/(29-11[H2O])) e.g., [H2O] = 1% then T* = T(1.004) Test: if [H2O] = 1% then w(18/29) = 0.01*.62 = T* = T (1 + w/e)/(1+w) = T ( )/( ) = T(1.004) Copyright © R. R. Dickerson

Specific Heats for Moist Air
Let mv = mass of water vapor md = mass of dry air To find the heat flow at constant volume: Copyright © R. R. Dickerson & Z.Q. Li

So Poisson’s equation becomes
For constant pressure So Poisson’s equation becomes (1+0.6w)/(1+0.9w) => (1-0.2w) due to rounding error. Copyright © R. R. Dickerson & Z.Q. Li

Tw ≤ Td because w not constant.
Wet Bulb Temperature The temperature to which air may be cooled by evaporation at constant p. Tw ≤ Td because w not constant. Copyright © R. R. Dickerson & Z.Q. Li

AOSC 620 Lecture 2 Physics and Chemistry of the Atmosphere, I

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