Today s Lecture: Atmosphere finish primitive equations, mostly thermodynamics

Size: px

Start display at page:

Download "Today s Lecture: Atmosphere finish primitive equations, mostly thermodynamics"

Dora Gilbert
5 years ago
Views:

1 Today s Lecture: Atmosphere finish primitive equations, mostly thermodynamics Reference Peixoto and Oort, Sec. 3.1, 3.2, 3.4, 3.5 (but skip the discussion of oceans until next week); Ch. 10

2 Thermodynamic energy equation Combined first and second laws of thermodynamics for a reversible process (all variables are state functions): Since U is a state function, so is the sum (or difference) of U and other state functions: du = T ds p dv (2.24) H = U + pv enthalpy (2.25) dh = T ds + V dp (2.26) G = H TS Gibbs function (2.27) dg = S dt + V dp (2.28) Which differential is most convenient depends on the process under consideration: isentropic S = const T ds = 0 isobaric p = const V dp = 0 isothermal T = const S dt = 0 For our (atmospheric) purposes, the extensive variables are replaced by their intensive counterparts. For example, (2.24) becomes du = T ds p dα (2.29)

3 Heat capacities For an ideal gas, U is a function only of T. The same is true of H = U + pv, since pv can be related to T by the equation of state. At constant volume, any heat added to the system leads directly to an increase in internal energy, since the second term in (2.24) vanishes; we therefore define the heat capacity at constant volume C V = du dt or c V = du V dt (2.30) V At constant pressure, any heat added to the system leads directly to an increase in enthalpy, since the second term in (2.26) vanishes; we therefore define the heat capacity at constant pressure C p = dh dt or c p = dh p dt (2.31) p (Note that the notation d dt and d V dt is not necessary since the differentials are total differentials.) p Differentiating the definition of (specific) enthalpy with respect to temperature yields c p = dh dt = d dt (u + pα) = d dt (u + R dt) = c V + R d (2.32) Furthermore, for dry air at atmospheric temperatures, u 5 2 R dt (see Section 2.6), so c V = 5 2 R d and c V = 7 2 R d (2.33)

4 Thermodynamic energy equation for atmospheric dynamics The most convenient expression of the thermodynamic energy balance for atmospheric dynamics is derived as follows. Recall the differentials form of the first and second laws of thermodynamics in enthalpy form: Taking the time derivative and substituting from (2.31), we find c p dt dt dh = T ds + α dp (2.34) = T ds + α dp = Q + αω, (2.35) dt dt where Q is the diabatic heating rate by clouds and radiation, and αω is the heating due to adiabatic compression. Potential temperature For an adiabatic process, divide (2.35) by T and substitute for α/t from the equation of state: c p d ln T dt which implies conservation of the potential temperature θ, = R d d ln p dt (2.36) θ = T ( p0 p ) κ, κ = R d c p 2 7 (2.37) From (2.35) and (2.37) it can be shown that θ is related to entropy (up to additive constants) by s = c p ln θ (2.38)

5 2.4 Thermodynamics of the moist atmosphere Why is the presence of water in the atmosphere important? Phase changes of water Latent heat release effect on dynamics Cloud formation effect on radiation Precipitation and evaporation coupling to land and ocean

6 Chemical equilibrium In an open system with multiple species (air, water,... ), the internal energy for the system is U = U(S, V, n 1,..., n c), where n k is the number of moles of species k, and the combined first and second law is c du = T ds p dv + µ k dn k, where µ k = U n is called the chemical potential. (2.39) k=1 k S,V,nj k Now consider multiple phases (ϕ) in equilibrium. The extensive variables U, S, V, n k are additive, so U = ϕ U (ϕ), S = ϕ S (ϕ), V = ϕ V (ϕ), n k = ϕ n (ϕ) k (2.40) Equilibrium is reached when the total internal energy of the system is minimized, i.e., any small change in the independent variables δs (ϕ), δv (ϕ), δn (ϕ) leads to k δu = ( T (ϕ) δs (ϕ) p (ϕ) δv (ϕ) + ) µ (ϕ) δn (ϕ) 0 subject to δs (ϕ) = δv (ϕ) = δn (ϕ) = 0 k k k ϕ k ϕ ϕ ϕ (2.41) In the two-phase case (denoted by and ), this requires (T T ) δs (p p ) δv + k (µ k µ k ) δµ k 0 (2.42) which can only be satisfied for arbitrary independent δs, δv, δµ k if the phases satisfy T = T (thermal), p = p (mechanical), µ k = µ k (chemical equilibrium) (2.43)

7 Phase transition Consider what happens if we boil water by adding heat to the system; S and V increase while T and p are constant. We can therefore calculate the entropy increase as S = H T or intensively s = h T = L T (2.44) L is the latent heat of the phase change: ice liquid latent heat of fusion, L il liquid vapor latent heat of vaporization, L lv ice vapor latent heat of sublimation, L iv Phase equilibria Now consider two phases in equilibrium (vapor over ice or vapor over liquid) while heat is being added to the system. Invoking the intensive form of the Gibbs Duhem relation (2.52, derived on the next slide) and making use of the phase equilibrium conditions (2.43) for the two phases (again denoted by and ), we have s dt α dp + dµ = 0 (2.45) s dt α dp + dµ = 0 (2.46) or s dt α dp = s dt α dp (2.47)

8 Derivation of the Gibbs Duhem relation U, S, V, n k are extensive, T, p, µ k are intensive. Therefore, for an arbitrary constant λ, U(λS, λv, λn 1,..., λn c) = λu(s, V, n 1,..., n c) (2.48) Denoting the extensive variables by x i and the intensive variables by y i = U/ x i, U(x 1, x 2,... ) = d dλ U(λx 1, λx 2,... ) = i U(λx 1, λx 2,... ) x i = λx i i y i x i (2.49) so that c U(S, V, n 1,..., n c) = TS pv + µ k n k + additive constant (2.50) k=1 Formally, c c c du = d(ts) d(pv) + d(µ k n k ) = T ds + S dt p dv V dp + µ k dn k + n k dµ k (2.51) k=1 k=1 k=1 However, (2.51) is only consistent with the combined first and second law (2.39) if c S dt V dp + n k dµ k = 0 (Gibbs Duhem relation), (2.52) k=1 that is, the state variables cannot all be independent of each other.

9 Clausius Clapeyron equation (2.47) can be transformed to the Clapeyron equation dp dt = s s α α = s α = h T α = L T α (2.53) For the case of water vapor over liquid water or ice, α v α l or α i so that α α v. The equilibrium pressure p is the saturation water vapor pressure e s; if e > e s, water vapor condenses to restore equilibrium, if e < e s liquid water evaporates to restore equilibrium. Substituting for α v using the equation of state for water vapor, we find the Clausius Clapeyron equation d ln e s dt = L R vt 2 (2.54) Approximating L const, (2.54) can be integrated: [ ( L 1 e s = e s,0 exp 1 )] ( ) L T T 0 e s,0 exp R v T 0 T R v T 2 0 (2.55)

10 Consequences of the Clausius Clapeyron equation Exponential increase of saturation pressure with temperature ( 2 for every 10 K) combined with High L of water result in efficient latent heat transport in convection L iv > L lv results in lower saturation pressure over ice than supercooled water es (hpa) e s over ice e s over liquid T ( C)

11 Convection Dry adiabatic convection Consider a parcel of rising air. While the water vapor pressure is below saturation water vapor pressure, adiabatic expansion cools the parcel according to (2.36, with c pd indicating the dry air heat capacity) which, upon substitution of hydrostatic equilibrium, yields the dry adiabatic lapse rate c pd d ln T = R d d ln p (2.56) Γ d = T z = g c pd = const (2.57) Saturated adiabatic convection Once the parcel reaches saturation, further expansion results in condensation of water vapor, which releases latent heat: (c pd + q sc pv) dt + L lv dq s = (R d + q T sr v) dp, where qs is the mixing ratio (2.5) corresponding to es (2.58) p The resulting saturated adiabatic lapse rate, neglecting q sc pv c pd and q sr v R d, is Γ s = T z g = f (T) const (2.59) c pd + L lv (dq s/dt) which is always smaller than the dry adiabatic lapse rate and decreases with increasing temperature.

12 Thermodynamic variables for saturated atmospheres Equivalent potential temperature From the adiabatic energy balance equation (2.58), this time retaining q sc pv c pd and q sr v R d, we see that the equivalent potential temperature ( ) L θ e = θ lv q s d exp (2.60) c pd T is conserved in saturated convection. Static stability Depending on the environmental lapse rate Γ, a parcel of air of a certain water vapor mixing ratio will be stable or unstable against vertical perturbation: Γ > Γ d Γ d > Γ > Γ s Γ < Γ s Static energy unconditionally unstable: the lifted parcel will always be more buoyant than the ambient air conditionally unstable depending on lifting condensation level and entrainment unconditionally stable: the lifted parcel will always be less buoyant than the ambient air Static energy is the sum of potential and thermal energy, including the latent heat released in the condensation of water vapor ( static because it does not include the macroscopic kinetic energy): h = c pt + gz + L lv q l (2.61)

13 2.5 Exchange processes of the atmosphere with land and ocean 107 Reflected Solar Radiation 107 W m Incoming Solar Radiation 342 W m Outgoing Longwave Radiation 235 W m -2 Reflected by Clouds Aerosol and Atmosphere 77 Reflected by Surface Absorbed by Surface Absorbed by Atmosphere Thermals Latent Heat Evapotranspiration Emitted by Atmosphere Surface Radiation Atmospheric Window Back Radiation 324 Absorbed by Surface Greenhouse Gases Radiative fluxes from surface into atmosphere Momentum dissipation (= momentum source for ocean) Fluxes of sensible heat, latent heat ( moisture flux) Orography also influences general circulation Figure: Kiehl and Trenberth, 1996

14 Boundary layer the dynamic view Viscous layer Friction by molecular viscosity: relative motion at the surface is zero, and viscous media resist shearing Mixed layer Friction by turbulent mixing: momentum transport against the gradient Ekman layer Both Coriolis force and turbulent friction are important; mean wind speed increases with height, mean wind direction becomes increasingly geostrophic Free troposphere Geostrophic balance in (mostly) non-turbulent flow Figure: Peixoto and Oort

15 Viscous transport z Velocity Wall Viscous friction is due to molecular effects that resist shear in flow. At the surface, the wind velocity must be zero, while in the free stream, the wind velocity will have some arbitrary magnitude. Since random motion of air molecules results in exchange of mass between the low-momentum layers and the high-momentum layers, resulting in the velocity profile shown on the left. The macroscopic friction is the result of the molecular-scale transport of momentum from the free stream into the boundary layer. The shear stress in the fluid in the viscous layer is τ zx = µ u z (2.62) (and similarly for τ zy and v), where the subscripts denote the stress is in the x direction in the z = const plane. The friction in the x direction is F x = 1 ( τxx ρ x + τyx y 1 ( µ u ρ z z ) + τzx z (2.63) ) = ν 2 u z 2 (2.64) ν = O ( 10 5 m 2 s 1), so the depth of the viscous layer is 1 m. Figures: Stewart 2008, Holton 2004

16 Turbulent transport A more efficient method of momentum transport is turbulence. Here the transport is accomplished by small-scale disturbances (u, v, w, θ ) in the mean flow (ū, v, w, θ ). The turbulent motions span greater length scales than molecular motion and can support deeper boundary layers. For a vertically sheared layer, the turbulent transport terms take the form u w, v w for the vertical transport of horizontal momentum, θ w for the vertical transport of potential temperature (and thus buoyancy). The transport is against the gradient. Gradient-flux ansatz Unlike in the case of viscosity, there is no theoretically well founded way to derive u, v, w, θ. The best we can do is make a reasonable guess and see how well it works. The English expression for this is ansatz. In the gradient-flux ansatz, the turbulent transport is assumed to be proportional to the gradient: ( ) 1 τzx = v ρ τ w v = K m zy z and θ w = K h θ z with constant turbulent diffusion coefficients for mass and heat K m and K h. This formulation is by analogy to viscous friction, but with much larger viscosity : K m = O ( 10 1 m 2 s 1). However, a constant K m and K h do not represent the atmospheric boundary layer well because the turbulent length scale grows with distance from the surface. Mixing-length approach The next-better refinement is the mixing-length approach, where K m is itself a function of the gradient and a characteristic length scale for mixing. The surface atmosphere interaction can then be formulated in terms of coefficients of transfer for momentum, sensible heat and water vapor (latent heat) (2.65) τ = ρc D v(z) v(z), F SH = ρc pc H v(z) {θ(z) θ(0)}, E = ρc W v(z) {q(z) q(0)} (2.66)

.. Free troposphere by an entrainment layer Processes Sensible and latent heat flux at the surface and.

17 Boundary layer the cloud-process view Vertical structure Boundary layer is well mixed and capped by a... Cloud layer which maintains a temperature inversion by cloud-top cooling and is weakly coupled to the... Free troposphere by an entrainment layer Processes Sensible and latent heat flux at the surface and... Radiative cooling at cloud top destabilize the airmass; this results in... Convection which mixes the layer vertically and horizontally Figure: Wood 2012

18 2.6 Appendix: Why does dry air have c v = 5 2 RT? Recall the partition function (Zustandssumme) of the canonical ensemble from statistical mechanics, which is the sum over all microstates weighted by the (Boltzmann) probability of occupying each microstate: Z = exp( βh(p i, q i )) dp i dq i, β = 1 (2.67) kt i where q i are the canonical positions and p i the canonical momenta, and H is the hamiltonian of the system. The expectation value of the internal energy U is U = H = 1 H(p i, q i ) exp( βh(p i, q i )) dp i dq i, (2.68) Z i which can be calculated from the partition sum: U = 1 β ln Z β. (2.69) For degrees of freedom that enter the hamiltonian as quadratic terms H = i (A iqi 2 + B i pi 2 ) (for example H = 1 2m (p2 x + p2 y + p2 z )), Z = i ( ) exp βa i qi 2 dq i exp( βb i p 2 i ) dp i = i π βa i π βb i (2.70)

19 Since 1 β π = 1 β βa i 2 1 β = 1 2 kt A i, B i (2.71) the total internal energy is 1 NkT for each degree of freedom that enters quadratically into the hamiltonian. This is the 2 equipartition theorem. How many degrees of freedom does dry air have? At the 99% level, air is a diatomic gas (N 2, O 2 ). A diatomic molecule has three translational H trans = 1 2m (p2 x + p2 y + p2 z ) two rotational and one vibrational H rot = 1 2 (I 1ω I 2ω 2 2 ) H vib = 1 2 k x2 degrees of freedom. Although the energy levels are quantized, the translational and rotational energy levels are closely enough spaced that the classical continuum limit is valid. The vibrational energy levels are much higher than atmospheric temperatures [O(10 3 ) K]; the vibrational degree of freedom contributes negligibly to the internal energy. Therefore, U = 5 ( ) u 2 NkT and cv = = 5 T V 2 R d (2.72)

Thermodynamics Review [?] Entropy & thermodynamic potentials Hydrostatic equilibrium & buoyancy Stability [dry & moist adiabatic]

Thermodynamics Review [?] Entropy & thermodynamic potentials Hydrostatic equilibrium & buoyancy Stability [dry & moist adiabatic] Entropy 1. (Thermodynamics) a thermodynamic quantity that changes in a