SIO 217aAtmospheric and Climate Sciences I:Atmospheric ThermodynamicsFall 2008 Midterm Exam (No calculators, notes, books, PDAs.) KEYCurry and Webster, Ch. 1-4 (and Section12.1)Here are some numerical values, some of which may be useful on this exam:Average radius of Earth: 6370 kmMean molecular weight of dry air: 29 g/moleMean reflectivity of the Earth: 0.31Mean molecular weight of water vapor: 18 g/moleGas constant for dry air: 287 J deg-1 kg-1Gas constant for water vapor: 461 J deg-1 kg-1Specific heat at constant pressure: 1004 J deg-1 kg-1Specific heat at constant volume: 717 J deg-1 kg-1Solar luminosity 3.92×1026 WEarth-sun distance 1.50×1011 mStefan-Boltzmann constant 5.67×10-8 W m-2 K-41. Do you think the average mass of water vapor per unit mass of atmosphere in tropical regionsis about the same as in polar regions, or more, or less, and why? State your assumptions andany applicable thermodynamic relationships.The tropical regions contain much more water vapor than the polar regions, as Fig. 1.1 on page 4shows dramatically, and the key reason is the strong constraint provided by the Clausius-Clapeyron equation, which expresses the dramatic monotonically increasing dependence ofsaturation vapor pressure on temperature. Note in Appendix D (page 440) that for a range oftypical surface atmospheric temperatures, say from – 10 deg C to + 30 deg C, the saturationvapor pressure approximately doubles for every 10 deg C increase in temperature. Indetermining the difference between the moisture content of the tropical and polar atmosphere,this strong upper bound on vapor pressure is far more important than variations in relativehumidity or other factors that might affect average water vapor concentrations.2. Define the following terms in 10 words or less; an equation, graph, or sketch may be added ifappropriate:a) equivalent temperatureThe temperature that an air parcel would have if all of the water vapor were to condense in anadiabatic isobaric process; Te=T+( Llv/cpd)*wv.b) specific humidityThe mass fraction of water to air; qv=mv/(mv+md)=wv/(wv+1) [Eqn. 1.20].c) Wien’s lawλmax=2890/T; The maximum wavelength emitted is proportional to the inverse temperature.d) AdiabaticA path in which no heat is lost or gained during the process.e) Carnot cycleReversible cyclic process resulting in the maximum possible efficiency; it consists of twoisothermal and two adiabatic steps.f) WorkThe quantity of energy transferred from one system to another without an accompanyingtransfer of entropy; examples include mechanical work (dw=F*dx) and expansion work(dw=-p*dv).3. Water exists simultaneously in three phases in the Earth’s atmosphere.a. For an atmosphere consisting only of pure water, what does the coexistence of threephases imply about the temperature and pressure of the system? Be as specific as youcan, stating any assumptions and equations that you use.If we assume that all three phases are in thermal, chemical, and mechanical equilibrium, thenthe Gibbs phase rule applies. Degrees of freedom by Gibbs phase rule: f=χ-ϕ+2=1-3+2=0.Thus there is only one temperature and pressure; in addition, the temperature and pressure isthe triple point, T=273.16K; p=6.11 hPa.b. If air is also present in the system, how does that change your answer to part (a)? Be asspecific as you can, stating any assumptions and equations that you use.If we assume that all three phases are in thermal, chemical, and mechanical equilibrium, thenthe Gibbs phase rule applies. Assume we can treat air as one homogeneous mixture, and it istherefore one component. Degrees of freedom by Gibbs phase rule: f=χ-ϕ+2=2-3+2=1.There is now one degree of freedom, so a range of temperature and pressure, over which allthree phases co-exist.c. Do the three phases of water in the Earth’s atmosphere meet the conditions necessary forGibbs Phase Rule to apply? State the assumptions and how you know if they are met ornot.In general, the three phases of water in the Earth’s atmosphere do not meet the conditions ofthermal, chemical, and mechanical equilibrium. There is no thermal equilibrium becausethere is a vertical gradient in temperature. There is no mechanical equilibrium becausepressure varies. Water fraction also varies so there is no chemical equilibrium.4. Consider the effect on the temperature of the Earth if mirrors were positioned in space toreflect 1% of the solar radiation incident on the Earth (this is a current multi-trillion dollargeo-engineering idea). State and simplify the equations needed to determine the equivalentblack-body emission temperature of this planet. State all assumptions and approximations.Solve the equations and identify the values of all constants but you do not need to evaluatethe temperature of this planet. Do you expect that this planet will be hotter than the Earth orcolder? Discuss the reasons.Assume that: (1) the earth behaves as a blackbody, (2) atmosphere is transparent to non-reflectedportion of the solar beam; (3) atmosphere in radiative equilibrium with surface; (4) noatmosphere. Then, at equilibrium, the incoming shortwave flux and outgoing longwave flux areequal (i.e. there is no accumulation) so for the normal solar luminosity we can write:FL= σTsurf4 (assumption 1; Eqn. 3.20)FS = FL(assumption 2-4; Eqn. 3.20)0.25*0.99*S0(1- αp) = 0.25*0.99*S0 = σTsurf4 (Eqn. 3.20, Eqn. 12.)Tsurf = [0.25*0.99*S0/σ]0.25where S0 = L0/(4πd2) = 1.3938x103 W m-2 (Eqn. 12.), αp=0, σ=5.67×10-8 W m-2 K-4Tsurf = 254.2 K (previously 254.8K)The planet is cooler than the Earth with no atmosphere by a very small amount (0.6K) but (If yousolve it with a simplified atmosphere, it is much hotter than all of these, but the relative change isstill one of cooling!)5. The saturation vapor pressure (of water) at a temperature of 30°C is 42.4 hPa. Consider moistair at 30°C, a pressure of 1,000 hPa, and a relative humidity of 25%. Find the values:a. vapor pressuree=H*es = 0.25*42.4 = 10.6 hPa [Eqn. 4.34a].b. mixing ratiowv=mv/md=(Mv/Md)*(e/(p-e))=0.622*(10.6/(1000-10.6))=0.00666 [Eqn. 4.36].c. specific humidityqv=mv/(mv+md)=wv/(wv+1)=0.00662 [Eqn. 1.20].d. virtual temperatureTv=(1+0.608qv)*T=(1+0.608qv)*T=304.2K [Eqn. 1.25].6. Consider air with the same specific humidity as in problem (5), but at a temperature of 50°C.State how you would find the values below, including any laws, equations, and assumptionsused, and simplifying as much as possible:a. saturation vapor
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