Atmospheric Thermodynamics

Transcription

1 Atmospheric Thermodynamics R. Wordsworth February 12, Objectives Derive hydrostatic equation Derive dry and moist adiabats Understand how the theory relates to observed properties of real atmospheres 2 Airless energy balance equations uminosity of the Sun is in W. Hence solar flux for a given semi-major axis d is F = /4πd 2 in W/m 2. Total flux intersected by a planet of radius r is πr 2 F. Total flux absorbed is (1 A)πr 2 F for a planetary albedo A. ongwave emitting area is 4πr 2 so flux absorbed per unit area is 1 4 (1 A)F. Stefan-Boltzmann law for longwave emission then gives σts 4 = 1 (1 A)F (1) 4 on average, in thermal equilibrium. T s can generally be written as the equilibrium temperature T e. Equilibrium temperature of Earth is around 255 K. Can do similar exercises for Mars, Venus. For exoplanets one of the first things you want to work out is equilibrium temperature. In similar way substellar, dayside average temperatures are sometimes useful. 3 Energy balance of a toy atmosphere Completely isothermal atmosphere. Surface energy balance: σt 4 s = 1 4 (1 A)F + e aσt 4 a. (2) a a is atmospheric absorptivity. Atmospheric energy balance: remember radiation travels up and down! e a is atmospheric emissivity. By Kirchoff s aw e a = a a so e a σt 4 a + e a σt 4 a = a a σt 4 s. (3) 2T 4 a = T 4 s (4) 1

2 If atm is very thin (e a << 1), T s is just the equilibrium temperature. Then T a = T skin 2 1/4 T e = 0.84T e (5) where T skin is the skin temperature. Skin layer is isothermal because each level is only in contact with space and the surface (interlayer exchange negligible because e a a a = e 2 a is so small). T skin 214 K for Earth. Caveats: real atmospheres are a) non-gray b) absorb in UV up high. More on this later. What is surface temperature if there is a greenhouse effect? σt 4 s = 1 4 (1 A)F e aσt 4 s (6) and when e a is small Ts 4 = 1 F (1 A) 4σ e a (7) T s T e ( e a) 1/4. (8) e a cannot be greater than 1, by energy conservation. Hence there is a maximum achievable surface temperature given an isothermal atmosphere. This is not true in general, as real atmospheres have temperatures that vary with altitude. 4 Dry thermodynamics ocal thermodynamic equilibrium local temperature. Holds very well for most of the atmosphere. ater, we ll consider NTE. Ideal gas law: p = k B nt (9) p = (R /m)ρt (10) m is molecular weight [kg/mol in SI units]. Ideal gas law no longer holds when energy in interactions between molecules (v.s. KE) no longer negligible. Theory gets more difficult rapidly, but one example is Van der Waals equation ( ) p + n2 a V 2 (V nb) = nrt (11) a measures attraction between particles, b measures volume excluded by one mole of particles, V is total volume. Further nonlinear equations possible. But it s rarely a concern: only deep Venus and gas giant atmospheres go non-ideal, and they re also opaque down there at most wavelengths. Related concept: collision-induced absorption in radiation (later in the course). 2

3 Some definitions Molar mixing ratio f 1,2 = p 1 p 2 (12) Molar concentration f 1 = p 1 i p i Not the same thing! But interchangeable for trace gases. Mass mixing ratio Mass concentration (sometimes called specific concentration) Mass columns (13) α 1,2 = ρ 1 ρ 2 = m 1 m 2 p 1 p 2 (14) q 1 = ρ 1 i ρ. (15) i For a pure atmosphere M = p g (16) in kg/m 2. Just remember F = ma and p = F/A. For a component of a mixed atmosphere M i = m i p i m g (17) with m the mean mass of the mixture. Remember it well! 5 Hydrostatic equation Consider a column of air with density ρ(z) under gravity g. Pressure at base is p(z), pressure at top is p(z + δz). Then by Taylor s theorem Pressure is force per unit area so net upwards force is simply p(z + δz) p(z) + dp δz (18) δf p = dp δzδa (19) This must balance the gravitation force, i.e. δf g = gδm = gρδzδa. Hence dp δzδa = gρδzδa (20) dp = gρ. (21) 3

4 This equation holds to a very good approximation in almost all atmospheres. Only exception is high up in escaping atmospheres, where some new terms come in. We ll cover this briefly at the end of the course. Now, if we use the ideal gas law we can write which we can write as dp = pg RT dln[p] = g RT ln[p] ln[p 0 ] = p(z) = p 0 e z 0 z 0 (22) (23) g RT (24) g RT (25) p(z) = p 0 e z/h. (26) with scale height H RT/g in simplest limit of a thin isothermal atmosphere. This is height over which pressure of atmosphere decays by e 1. 6 Dry adiabats Fluid blob / gas parcel moving upwards in an atmosphere. First law of thermodynamics: δq = c v dt + pdv (27) v is specific volume (1/ρ). c v is specific heat at constant volume. First term is increase in internal energy of the system (δu = c v dt ), second term is work done on the system, δq is heat supplied to system. Apply chain rule δq = c v dt + pd(1/ρ) (28) then using ideal gas law we can write δq = c v dt + d(p/ρ) dp ρ (29) δq = c v dt + d(rt ) dp ρ (30) δq = c p dt dp ρ. (31) Finally divide all by T and replace δq with ds, where s is entropy, a state variable. ds = c p dlnt Rdlnp (32) Along a dry adiabat, the change in entropy of the gassy parcel is zero. No longer true if there s radiative cooling or diffusion, for example. Solving this we get dlnt = dlnp R c p (33) 4

5 R/c p 0.28 for N 2. I leave derivation of the atmospheric lapse rate T = T 0 ( p p 0 ) R/cp (34) Γ = dt = g c p (35) as an exercise. T vs. p is generally more useful than T vs. z because p determines the atmospheric physics. 7 Moist adiabats (ideal case) Same fun as before but with new terms. Condensate is often H 2 O but can really be anything. First we have saturation vapour pressure, often given by Clausius-Clayperon equation (which I m not going to derive here). p sat (T ) = p sat (T 0 )e Rv (T 1 T 1 0 ) (36) p 0, T 0 can conveniently be defined as the triple point (e.g. 611 Pa, K for H 2 O). These quantities and the latent heat vary widely with compound, leading to wildly different regimes. C.f. Earth with active H 2 O cycle and T surf in the 260 to 310 K range, and Titan with T surf of order 90 K but an active methane cycle. For the moist adiabat, we start by writing the previous dry equation with subscripts and multiply by M n M n δq = M n c p,ndt M n dp n ρ n (37) n for non-condensable. M n is mass of non-condensing part of fluid blob (e.g. N 2 ). Now we add a condensing component denoted by subscript v. On its own, such a gas would obey M v δq = M v c p,v dt M v dp v ρ v + dm v. (38) But the 1st law of thermodynamics must be satisfied for the whole system, so we write (M v + M n )δq = M n c p,n dt M n dp n ρ n + M v c p,v dt M v dp v ρ v + dm v (39) where dm v is the critical latent heat term. Divide whole thing by M n T and define mass mixing ratio α v M v /M n. (40) Then (1 + α v ) δq T = c p,ndlnt dp n T ρ n + α v [c p,v dlnt dp v T ρ v ] + T dα v (41) Then assume ideal gas law holds for both components (p v = ρ v R v T, p n = ρ n R n T ) (1 + α v ) δq T = c p,ndlnt R n dlnp n + α v [c p,v dlnt R v dlnp v ] + T dα v. (42) 5

6 Note if there is no condensation, dα v = 0 and this is just the dry adiabat for 2 species again. Now key assumption of saturation: p v = p sat. Also adiabat so set δq/t = ds = 0 as before. 0 = c p,n dlnt R n dlnp n + α sat [c p,v dlnt R v dlnp sat (T )] + T dα sat (43) Our aim is to get T (p), where p = p n + p v is the total pressure. First collect terms: Note that Hence Now note that 0 = (c p,n + α sat c p,v )dlnt R n dlnp n α sat R v dlnp sat (T ) + T dα sat (44) dα v = d(ρ v /ρ n ) = ɛd(p v /p n ) = ɛ p v p n (dlnp v dlnp n ) = α v (dlnp v dlnp n ) (45) 0 = (c p,n + α sat c p,v )dlnt (R n + α sat T )dlnp n + ( T R v)α sat dlnp sat. (46) dlnp sat = from the Clausius Clayperon equation. Hence or Remember Rn c p,n 0 = [ (c p,n + α sat c p,v ) + ( T R v)α sat R v T dlnt = R n 1 + [ dlnp n c p,n cp,v 1 + c p,n + ( T R v 1) dlnt (47) R v T ] dlnt (R n + α sat T )dlnp n (48) R nt α sat c p,nt 0.28 for N 2. Take α sat 0 and we get the dry adiabat again. ] α sat. (49) 6