What did we learn in Ch. 1? Final Exam 2014: 6 Questions. Energy Transfers Link Ch Reversible-Adiabatic-Work. What did we learn in Ch. 3?

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1 Final Exam 0: 6 Questions. First/Second Laws Question. Definitions Question 3. Clausius-Clapeyron with q/h/e) Question. Köhler, Clouds, and Stability Question 5. Climate Model and Sensitivity Question 6. erm Project Answer of Choices) Same as midterm What did we learn in Ch.? What P,, U are for a fluid What an ideal gas is How P,, v relate for an ideal gas and we call this relationship an equation of state) What chemical components constitute the atmosphere for homosphere <0 km) What the hydrostatic balance is How p, vary with z for observed, standard, isopycnic, isothermal, constant lapse-rate atmospheres Key Combined st + nd Law Results st Law: du=dq+dw; u is exact Eq..8 du=dq rev -pdv reversible, expansion only) p. 56 Define Enthalpy: H=U+PV Eq.. dh=du+pdv+vdp also Eq..3: dh=vdp+dη) nd Law: [dq rev /] int.cycle =0 Eq..7 Define Entropy: dη=dq rev / Eq..5a dη=dq rev du=dη-pdv Define Gibbs: G=H-η Eq..33 dg=dh-dη-η=du+pdv+vdp)-dη-η dg=du-dη-pdv)+vdp-η=vdp-η p. 58 δp/δt) g =η/v Eq..0 Energy ransfers Link Ch. - Ch. Atmospheric Composition dry air N, O, Ar, CO ), Water, particles) Ch. First and Second Laws Energy Balance System/Environment) Ch. 3 Stefan-Boltzmann Equation Kirchoff s Law Ch. Water and Phase Changes Latent Heat of Water What did we learn in Ch. 3? Reversible-Adiabatic-Work Radiative transfer definitions Diffuse vs. direct From all directions irradiance F) or one radiance I) Absorption coefficient and optical thickness Blackbody radiation Radiative transfer equations Kirchoff s law gaseous molecules): E λ = A λ Planck s radiation law: F = fcnλ, ) Wien s displacement law: λ ~ 3000/ Stefan-Boltzmann law black body: F bb = σ ) Reversible mass is conserved) Ideal Gas Adiabatic thick walls First Law Internal Energy Δu = Q + W Δu = c v Frictionless Low P, High Reversible, Adiabatic " = P $ ' P R c p dw = pdv p v = R = p v Q = 0 pdv = c v R dv = c $ v ' v dv R v = c v

2 Reversible, Adiabatic Expansion of an Ideal Gas pdv = c v R dv = c $ v ' v dv R v = c v R lnv lnv ) = c v ln ln ) ln v R = ln $ v ' $ ' = v $ ' $ v ' R c v c v R p = $ ' $ ' R $ $ p '' + R c v = p $ ' $ p ' R cv R c v = p R R +cv = p $ ' $ p ' $ p ' = $ ' R c p $ p ' R cv p R cv Potential emperature Condensed Liquid) Water Ch. : Will H O condense? Clausius-Clapeyron describes e sat ) Ch. 5: What will H O condense on? Kohler describes droplet formation Ch. 6: How much H O will condense? Dew point temperature provides metric Ch. 7: How does condensed H O affect stability? Conditional stability is affected by moist adiabats Ch. 8: What happens to condensed H O? Precipitation processes What did we learn in Ch.? Phase equilibrium definitions Criteria of phase equilibria thermal, mechanical, chemical) Degrees of freedom reduced by phases Phase diagram of pure) water Clausius-Clapeyron equations: Will H O condense? Strong dependence of e sat on temperature and L lv ) Doubles every 0C here are two ways to saturate, i.e. H=e/e sat = Increase water vapor in parcel e) Decrease temperature and hence e sat ) What did we learn in Ch. 5? Kohler equation: What will H O condense on? Kelvin effect describes the dependence of surface energy on particle/ droplet size Raoult effect describes the increased energy of dissolved components in mixtures Nucleation a.k.a. activation) Formation of a new phase Role of particles as triggers but not drivers History and recent advances Development of theory Kohler, 936) Measurement of particle nucleation Nenes and Roberts, 005) Implications for albedo womey, 97) What did we learn in Ch. 6? Cloud formation what causes saturation? Cooling to decrease e s ) Increasing water vapor to increase e Mixing of two nearly-saturated parcels Enthalpy balance: How much H O will condense? Latent heat from evaporation causes cooling Latent heat from condensation causes warming Dew point D ) and frost point F ) temperatures Cool isobarically until saturation is reached RH scales with dewpoint depression - D ) L lv affects energy balance Kiehl+renberth, 997)

3 Parcel now is colder than environment. Condensation starts releasing latent heat so parcel warms. Energy is required to lift parcel. Free Convection CAPE Energy is required to lift parcel. Parcel runs out of water to condense so cools relative to environment). Parcel is now warmer than the environment so lifts due to buoyancy. Hydrostatic Balance Applicable to most atmospheric situations except fast accelerations in thunderstorms) g = p ρ z p = pg R d z Curry and Webster, Ch. Special Cases of Hydrostatic Equilibrium. rho=constant homogeneous) H=8 km =R/g=scale height eq..39. constant lapse rate e.g. if hydrostatic, homogeneous, and ideal gas) -/dz=constant=-g/r=-3/deg/km 3. isothermal =constant and ideal gas) p=p_0*exp-z/h) Homogeneous Atmosphere Density is constant Surface pressure is finite Scale height H gives where pressure=0 g = p ρ z dp = ρgdz 0 H dp = ρgdz p p 0 = ρgh 0) p 0 = ρgh H = p ρg = R d 0 g Curry and Webster, Ch. Hydrostatic + Ideal Gas + Homogeneous Evaluate lapse rate by differentiating ideal gas law Hydrostatic Equilibrium Example Constant Lapse Rate) Ideal gas Density constant Hydrostatic p = ρr d p z = ρr d z p ' * = R d ' * ρ z ) z ) g = p ρ z Γ = z = g = 3. o C/km R d Curry and Webster, Ch. 3

4 Water Saturation Pressures es doubles with every 0C! this is one consequence of Clausius-Clapeyron s equation) Water Vapor Metrics " M v $ Md ' = " R " R $ d ' = 0.6 v $ ' Rv Rd = Mixing ratio Specific humidity Relative humidity C) e S hpa) Water vapor by mass w v = m v = ρ v m d ρ d m v q v = = w v m d + m v + w v H w v w s Water vapor by partial pressure e w v = 0.6 $ p e ' e q v = 0.6 $ p 0.6)e' H = e e s Water saturation Virtual temperature e w s = 0.6 s $ p e s ' e q v = 0.6 s $ p 0.6)e s ' v = q v ) H = Virtual potential temperature θ v = q v ) p 0 $ p ' R d c pd Virtual emperature erminology Review Isotropic Same in all directions, such that F=πI Reflection Change in direction but not energy or wavelength Isentropic Adiabatic+reversible For adiabatic, ideal: p determines and vice versa Potential temperature temperature that air would have if raised/lowered to a reference pressure. What you need to know in Ch. On emperature Axis: Simplified 0 and-layer Earth radiation balance model Earth s actual energy imbalance ocean sink) Detailed energy streams Kiehl and renberth) Atmospheric window Latent heat Major heat-driven features of Earth s circulation Hydrological cycle evaporation-precipitation) Latitudinal differences in heating Meridional heat transfer equator to pole) Zonal heat transfer Walker, monsoonal) z dθ v unstable dθ e stable

5 On Skew- Axis: Simplified Ch. 8: Main Cloud ypes z unstable dθ v dθ e stable. Cirrus Ci). Cirrocumulus Cc) 3. Cirrostratus Cs). Altocumulus Ac) 5. Altostratus As) 6. Nimbostratus Ns) 7. Stratocumulus Sc) 8. Stratus St) All high clouds Middle clouds Grayish, block the sun, sometimes patchy Low clouds 9. Cumulus Cu) 0. Cumulonimbus Cb) Sharp outlines, rising, bright white GFDL AMp5 vs NCAR CAM Ch. 8: Cloud ypes and Drop Sizes Frequency distributions of the mean cloud droplet size for various cloud types B. Soden Ch. Simplified Climate Model Atmosphere described as one layer Albedo α p ~0.3: reflectance by surface, clouds, aerosols, gases Shortwave flux absorbed at surface F S =0.5*S 0 - α p ) Earth behaves as a black body emperature e : equivalent black-body temperature of earth Longwave flux emitted from surface F L =σ e F S Incident on projected disc πr Simplified Climate Model Incoming shortwave = Outgoing longwave Energy absorbed = Energy emitted F S = 0.5*S 0 - α p ) F L = σ e F L Emitted from sphere surface πr F S = F L Curry and Webster, Ch. pp ; also Liou, 99 5

6 Simplified Climate Model At thermal equilibrium why?) F S = F L 0.5*S 0 - α p ) = σ e e = [0.5*S 0 - α p )/σ] 0.5 e ~ 55K Observed surface temperature = 88K What s missing? Add an Atmosphere! Atmosphere is transparent to non-reflected portion of the solar beam Atmosphere in radiative equilibrium with surface Atmosphere absorbs all the IR emission F S F surf F atm F atm OA: F S = F atm 0.5*S 0 - α p ) = σ atm atm = 55K Atmos: Fsurf = F atm σ surf = σ atm surf = 303K Section 3.3 Water Vapor Feedback Key points: his is a strongly positive feedback, nearly doubling the effect of carbon dioxide alone. Relative humidity seems to be approximately constant under climate change p. 359). Relatively dry regions, such as the upper troposphere and polar regions, are especially sensitive p. 36).. Section 3. Cloud-radiation Feedback Key points: Clouds affect both shortwave low clouds) and longwave high clouds). Present climate has cloud cooling dominating cloud warming pp ). Many different mechanisms, including those involving aerosol-cloud interactions, may be important, but the sign and magnitude of cloud feedbacks is still largely unknown p. 37). Clouds have big effects in models Section 3.5 Snow/Icealbedo Feedback Key points: his feedback is large and positive in high northern latitudes. Observations show that this effect is occurring now. Melting ice on land has another large effect, unrelated to albedo: it causes sea level to rise. Solar Constant Luminosity of the sun Irradiance at earth S 0 = L 0 /πd ) =.x0 3 W/m L 0 ~ 3.9x06 W p. 33) Area = πd d =.5 0 m p.37) 6

7 Radiation Balance Example It has been estimated from satellite observations that variations in solar radiance during the last 0 years amounted to 0. W m -, or less than 0. of the incoming shortwave radiation. Calculate the approximate change in the temperature at the Earth s surface for a 0. decrease in solar luminosity for a simplified climate model. State all assumptions, simplifications, and equations used. Values of constants that you may need are Earth s albedo 0.3, solar luminosity 3.9x0 6 W, Earth-sun distance.50x0 m, Stefan-Boltzmann constant 5.67x0-8 W m - K -. Assume that: ) the earth behaves as a blackbody, ) atmosphere is transparent to non-reflected portion of the solar beam; 3) atmosphere in radiative equilibrium with surface; ) atmosphere absorbs all the infrared emission. hen, at equilibrium, the incoming shortwave flux and outgoing longwave flux are equal i.e. there is no accumulation) so for the normal solar luminosity we can write: F L= σ surf assumption ; Eqn. 3.0) F S = F Lassumption -) 0.5*S 0- α p) = σ surf Eqn..b) surf = [0.5*S 0- α p)/σ] 0.5 where S0 = L 0/πd ) =.3938x0 3 W m - p. 33), αp=0.3, σ=5.67x0-8 W m - K - surf = 5.8K Reducing L to 99.9 of its value, we get S0 = 0.999*.3938x0 3 =.39x0 3 W m - surf = 5.78K he resulting cooling of 0.06K associated with a 0. reduction in solar radiation is negligible for this simplified climate model. Hydrostatic Equilibrium Example Consider a planet with an atmosphere in hydrostatic equilibrium. Assume that the atmosphere is an ideal gas. Also assume that the temperature is a maximum at the surface of the planet, and, as height increases, the temperature in the atmosphere decreases linearly in other words, temperature decreases with height at a constant rate). Derive a formula for atmospheric density as a function of height in this atmosphere. From the hydrostatic equation for an ideal gas Eqn..) p = pg R z and a constant lapse rate Γ = dz dp = pg R dz ' dz * ' Γ * = pg ΓR ) d dp p = g ' ΓR d * ) dp p = g ' ΓR d ln p p 0 = g ' ΓR d * ) * ln ) 0 we get g ' * Rd Γ) p = p 0 ' * 0 ) which is Eqn.8. hen dividing both sides by R and noting that for an ideal gas ρ = p, we get R g g p R = ρ = p 0 RdΓ p ' * = 0 0 Γz RdΓ ' * R 0 ) R 0 Γz) 0 ) Degrees of Freedom Example Clausius Clapeyron Example he saturation vapor pressure at a temperature of 30 C is. hpa. he gas constant for dry air is 87 J K - kg -. he gas constant for water vapor is 6 J K - kg -. In addition to the constants given above, here is one more: the saturation vapor pressure at a temperature of 0 C is 73.8 hpa. Assuming that the latent heat of vaporization is constant, use this information to calculate the numerical value for this latent heat. he Clausius Clapeyron equation can be integrated if L is assumed constant, and the result is Eqn..3. Using 30 C=303K and 0 C=33K, and knowing saturation vapor pressure values for each, the only unknown is L. Solving Eqn..3, ) e = e exp L lv, +. * R v $ '- L lv = R v ln e = 6 ln 73.8 =. 0 6 $ ' $ e ' $ ' $.' L=.x0 6 J/kg. Definition Example Define the following terms, briefly and clearly, in light of their use in the kinetic theory of gases and the first and second laws of thermodynamics: a) an ideal gas: vapor whose molecules have collisions with perfect elasticity, typical at low pressures for air atm) and high temperatures typically 300K); vapor that satisfies pv=r and has the properties that dh=c p, du=c v, c p-c v=r p. ). b) temperature: the intensive property describing the internal energy of a gas, which for an ideal gas depends only on the average speed of the molecules. c) entropy: a state property whose differential describes the amount of energy that is not available for doing work for a reversible process in which maximum work is done) and satisfies the criteria of an exact differential; it can be evaluated from Eqn..5a and.6b: dη = dq = c pdln) Rdln p). $ ' rev d) exact differential: a function ξ for which dξ has the properties ) for any closed path dξ = 0, and ) for ξx,y) where x and y are independent, then dξ = ξ dx + ξ dy Mdx + Ndy M = N. $ x ' y $ y ' y x $ ' $ x ' x y e) enthalpy: a state property whose differential describes the change in heat for a constant-pressure process and satisfies the criteria of an exact differential; it is defined as H=U+pV Eqn..). Ch. : Problem 5 Consider moist air at a temperature of 30 C, a pressure of,000 hpa, and a relative humidity of 50. Find the values of the following quantities: a) vapor pressure b) mixing ratio c) specific humidity d) specific heat at constant pressure e) virtual temperature he saturation vapor pressure at a temperature of 30 C is. hpa. he gas constant for dry air is 87 J K - kg -. he gas constant for water vapor is 6 J K - kg -. a) vapor pressure: e=h*e s = 0.50*. =. hpa [Eqn..3a] b) mixing ratio: w v=m v/m d=m v/m d)*e/p-e))=0.035 [Eqn..36] c) specific humidity: q v=m v/m v+m d)=w v/w v+)=0.033 [Eqn..0] d) specific heat at constant pressure: c p~7r/)+0.87q v)=06 J/K/kg [Eqn..65] e) virtual temperature: v=+0.608q v)*=305.6 K [Eqn..5] 7