1. Read the section on stability in Wallace and Hobbs. W&H 3.53

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1 Assignment 2 Due Set 5. Questions marked? are otential candidates for resentation 1. Read the section on stability in Wallace and Hobbs. W&H ? Within the context of the Figure, and the 1st law of thermodynamics, exressed as U = Z t 0 dq 0 mr Z t 0 d ln V 0 0 (1) describe what leads to thermodynamic work being done within the atmoshere. Energy that is available to do work in the form of meridional motions is created by a meridional gradient in the zonally averaged imbalance between incoming solar radiation and out-going longwave radiation. he imbalance can be ositive (as in the troics) or negative (as in the oles), all that matters is that there is a gradient. Energy will not be created or destroyed, but it will be transorted from the equators to the oles because a high energy density is maintained by the imbalance at the equator and a low energy density is maintained at the oles. Net heating at the equator and net cooling at the oles drives an exansive flow from the equator to the oles, doing work to move air from high to low energy density or ressure. 1

2 3. Describe the First Law of hermodynamics insofar as it alies to a convective cloud. Do so by aealing to short time scales and long timescales, which you will define. Over short timescales we can resolve the stes of a cycle: solar heating raises the internal energy of surface air so that work can be done to lift the air through the static stability of the atmoshere. Latent heat release continually rovides an extra source of energy along the way. At the to of the cloud there is cooling to sace which lowers the internal energy. Meanwhile, the air mass subsides to the surface. he air would change from cloudy to clear, over the eriod of the cycle. Averaged over longer time scales than this cycle (<1 Hr) we would see none of this cycle. he internal energy of the sky would be fixed. We would see a distribution of eriods of cloudy and clear sky with each one have a fixed ercentage. In other words, we might say artly cloudy. 4.?Heating at constant volume can be exressed = where q and h have units of Joules er kilogram, and are therefore are intensive quantities. (a) But heat q is not a state variable like u, and is instead related to a flow from somewhere else. Exlain in your own words why there is a logical (or hysical) inconsistency in exressing an external flow like heating in terms of an intensive quantity, but that there is no logical inconsistency to state that the above formula in terms of extensive = Heating is a flow from outside of the system to inside the system. he mass of relevance isn t resent outside the system, so it is a bit meaningless to ratio with resect to mass there. Rather there is only dq/. (b) Obviously though, we would like to kee the intensive asect of the exression, since there is nothing wrong with exressing internal energy, at least, in a intensive way. And, in the atmosheric sciences we almost always refer to exress thermodynamics in terms of quantities er unit mass. Can you come u with a way to exress heating at constant ressure that is more grammatically correct, in that it more faithfully describes what is really haening to the temerature when a flow of energy into a system raises its internal energy er unit mass? For guidance think erhas of the flow of hotons into a arcel of air as it is radiatively heated (say by the sun), thereby increasing the energy of the air. For nomenclature you can (and many do) exress flow using the symbol µ ext for the external energy er unit energy carrier in the flow, and j ext is the rate of material flow with units (#/time), and the total energy er unit matter of the system is µ tot = µ int + µ ext =(f +2)µ where µ is the energy er degree of freedom f and there is a fixed number of units of matter in the system N. What is the rate of change of µ tot when there is a flow of energy into 2

3 the system and the system maintains a constant ressure? How can you exress this change in µ tot in terms of temerature? What do you conclude about the relationshi between the rise in temerature and the number of degrees of freedom f? Exressing heating as a flow dq/ = j ext µ ext he resonse at constant ressure is hus, since dh/ = dq/ or dh = dq = j extµ ext N dµ tot = j ext µ ext dµ tot = j extµ ext N where µ tot =7k/2 reresents the total energy er molecule so d = 2 j ext µ tot 7 kn he rise in temerature is inversely related to the number of degrees of freedom. 5. We showed = c or =1/c What is the value Describe in words the exression and describe how it relates to the core of a hurricane From the first law If temerature is constant so or dq = c v d + d dq = R d ln = R d ln dq = R d ln d ln = 1/ (R ) dq If the temerature isn t changing, heating a mass causes a dro in its ressure. An examle might be a hurricane. 3

4 6.?We derived the imortant relationshi At constant ressure, heating increases the enthaly. But this is really rather confusing, isn t it? After all h = u + Pressure and enthaly are linked, so how does enthaly increase without ressure increasing?. How can we increase enthaly along a constant ressure surface? he answer, of course, is that it is density or secific volume that is changing. his is so very imortant, because ultimately it is density (or buoyancy) erturbations, as you know, that lead to atmosheric flows of all kinds. So the question is then, what fraction of the heating goes into creating the density differences that drive the winds (Your answer should be 2/7)? Where does the remainder go? Since dh = du + d + d then at constant ressure we would see an exansion of secific volume, or a dro in density. he dro in density would be due to an increase in temerature since d = Rd. Since c v d goes into the internal energy, it follows that only a fraction R/ (c v + R) =R/c goes into changing the density. A fraction c v /c goes towards increasing the internal energy. 7. How much energy would it take to raise the average temerature of a unit (1 m 2 ) column of the earth s atmoshere by 1 deg? In terms of thermal caacity, the atmoshere is equivalent to a layer of water how many meters dee? What does this tell you about where to look for forced climate change? You may use the hydrostatic equation d = gdz he mass of the atmoshere can be obtained from the hydrostatic equation d g = dz which has units kg/m2, which is what we want. m = d/g =100000/10 = kg/m 2 herefore the amount of energy required to raise the atmosheric column 1 degree is E = mc air = =10 7 J/m2 What deth of water would this corresond to? E/c water = m water =10 7 / (4218 1) = 2371kg/m2 but there is 1000 kg/m3. herefore it equals 2371/1000 = 2.4 m. What this means is that the oceans are caable of transorting far far more heat than the atmoshere. 4

5 8. Show that a dry air mass of constant entroy with ds =0satisfies the exression d ln d ln = c R 9. In atmosheric sciences, integrating the above leads to an exression called the otential temerature = ( 0 /) R/c where reresents the temerature a arcel would have if it were brought dry adiabatically (i.e. dq/ =0) to the surface where = 0 =1000 mbars. and h d are often used interchangeably, even though they are not the same thing, because for adiabatic rocesses where dq/ =0, if d / =0, then dh d / =0. Show that dh d = c d ln Why are constant otential surfaces called isentroes or dry adiabats? We know that So aking the time derivative = ln =ln d ln = d R/c 0 R c ln R d c Multilying by c and noting that d = gdz (hydrostatic equation) and R/ =1/ (ideal gas equation c dln = c d + gdz he RHS is just dh d. herefore dh d = c d ln imlying that if not changing, then the same is true of h d, which would be the case in the absence of net heating, or an adiabat. 10.?We also discussed a situation where dh d / =0because, at constant temerature, working = heating, dq = dw d = = R d ln Show that diabatic heating leads to cross-isentroic flows with vertical seed w = dz = (dq/) c d /dz where d /dz is termed the static stability of the atmoshere. High values inhibit vertical motions in resonse to heating. Hint: start by taking the logarithm of the standard exression for otential temerature. Argue how this exression alies to the motions of the Hadley Cell troical circulation 5

6 As with question 1, we know that So = ln =ln R/c 0 aking the time derivative at constant temerature R c ln d ln = R c d ln his is fine enough for an answer, but taking it further, from the statement of the question d ln 1 dq = R So d ln d = w = dz = d /d dz = = 1 dq c dq c c dq An examle is the Hadley cell circulation where net heating at the equator leads to the ascending branch of the circulation. / d dz 6