df dz = dp dt Essentially, this is just a statement of the first law in one of the forms we derived earlier (expressed here in W m 3 ) dq p dt dp

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Brett Summers
6 years ago
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1 A problem with using entropy as a variable is that it is not a particularly intuitive concept. The mechanics of using entropy for evaluating system evolution is well developed, but it sometimes feels a bit like magic. It is not obvious how to point at the entropy of an object. Nor are entropy changes easily described for such irreversible processes as scattering. What is more intuitive, and equally valid thermodynamically, is to express irreversibility, or the direction of time, in terms of free or available energy. Free energy represents the potential energy that is available to do work, where work is simply away of expressing the change in energy of something else. Internal contrasts in energy density permit a flow of free energy G from high to low energy density g through the transfer of material with density here in units number per volume) n through constant potential surfaces µ units energy). The Gibbs free energy density g is given by g = nµ Thus, the second law can be stated that for the universe as a whole d g = d g i < 0 i The universe runs downhill to lower average potential surfaces. Or, equivalently, contrasts tend to smooth out on average with time. Note the similarity to the expression for entropy we discussed earlier. Entropy increases with time. Free energy decreases. From this one expression, we can understand things that happen in the atmosphere, like evaporation, droplet nucleation, or atmospheric convection. It also leads very naturally to some familiar statements about local thermodynamic equilibrium in the atmosphere like the hydrostatic equation. Imagine that µ is a potential surface, where the potential can be anything, and nµ)is the density of those potentials. A simple example would be a pyramid n being the density of blocks in some volume enclosing the pyramid and µ being mgh where m is the mass per block. The second law demands that over time, the pyramid falls down such that d g = d d nµ) = n µ µ + µ n n) < 0 Now this is quite clear and intuitive because it states that there are two ways that time moves forward. One is that the total density n of material goes down things spread out as the pyramid collapses and dn/ < 0. The other is that the potential energy goes down things fall and dµ/ < 0. Without thinking very hard, it is usually easy to see how things that you can think of happening have one or both of these things occuring. Now, let s incorporate this with the 1st law of thermodynamics. Sure things fall down, and overall contrasts diminish, but this does not mean the internal energy density at some given potential can t increase or decrease). In units watts per meter cubed du + d g = dq which is essentially stating that d g/ is what we previously called the rate of doing work dw/. This is interesting, actually. Naturally g wants to go down, but we known that contrasts exist in our universe. The atmosphere isn t a homogenous goo, it has clouds, hurricanes, snowflakes and meridional and zonal pressure gradients. What maintains these contrasts? Heating! It is only be 1

2 cause there is a continuous flow of energy into the earth system dq/ that contrasts g and the internal energy u are maintained. Turn off the sun, and indeed everything would homogenize. It might help here to think of the universe as an infinite staircase. Energy is constantly passed from one level of the staircase to a lower level, but there is no bottom to the staircase, no T = 0 K this is sometimes called the third law of thermodynamics), and therefore no fixed point to which we can reference a potential level. This may seem strange, but a potential level µ knows nothing about other potential levels with different values of µ. The level does know however about the local gradient in g = nµ, and it passes material down the hill at the same time it is receiving material from higher up. There is a continuous flow of solar energy into the Earth system at the same time that pressure and temperature drops and energy leaves the system through radiation. What we are talking about, essentially, is the divergence and convergence of energy. In the atmosphere it is this constant concentration and dilution that creates the contrasts that drive the flows that create weather, climate and every other process in the atmosphere. A cascade of energy from high to low pressure or available energy g = nµ occurs at rate a units energy per area per time) and with diffusivity D units area per time). Convergence of energy gets partitioned among increasing the internal energy density u and increasing local contrasts or available energy density gradients g, the sum of which is the enthalpy density h. A continuum solution for evolution of a and g requires solving a = D g = D nµ 1) dh/ = du/ + dg/ = a 2) In this framework, gradieboth a and g continuously evolve at all values of potential µ. One way to think of enthalpy is that it is the sum of kinetic energy the motions, vibrations and rotations in u ) and potential energy the contrasts in g ). These two equations are pretty general and can take us a long, long way. For any situation, all that is now required is an understanding of what form of energy is creating gradients in µ, initial conditions for n and some understanding of the forces determining the diffusivity D. Let s take the example of radiative heating of air in the atmosphere due to the absorption of solar or thermal radiation. What we are aiming for here is the very important equation in Salby) for how radiative flux convergence drives atmospheric heating and lofting. df dz = dp ρc dt p Essentially, this is just a statement of the first law in one of the forms we derived earlier expressed here in W m 3 ) dq = ρc dt dp p ρv = ρc dt p dp Here, we are going to derive it in the context of what we have described above. So, adding radiative energy to a parcel can go both into increase the internal energy of the parcel and increase its contrast with its environment. We re dealing here with oxygen and nitrogen at atmospheric temperatures and pressure, governed by the ideal gas law. For an ideal gas, contrasts in available energy density are due to pressure gradients: g = nµ) = n µ + µ n 2

3 or p = nk T + n Available energy is not the internal energy for an ideal gas. Only two of the translational modes are available for pressure driven fluxes through a 2D surface. The vibrational and rotational modes are effectively invisible. Now, for the sake of argument, think of g = g g 0 being the difference between two levels, and that initially the system is in local thermodynamic equilibrium, i.e g = p = 0. One way this could hold is that T = n = 0, but let s consider a more general situation in which there is a positive temperature contrast T = T T 0 > 0, compensated by a negative density contrast n = n n 0 < 0. Now if energy is added to the system through radiative heating, it can do two things. The first is that it can add to the internal energy of the layer by increasing the kinetic energy in the translational, rotational and vibrational modes. The second is that it can stretch the system to higher potential energies. We already know the first du = 5 dt nk 2 = ρc dt v What about the second? Say that we are interested in T and energy goes into increasing contrast by increasing T at constant n. We then get d p n = nk d T = nk d T T 0) = nk dt = ρrdt If we increase contrast by decreasing n at constant T then we can write d p T = d n = dn = dp Adding up all components we now get our desired solution for enthalpy density changes dh = ρ c v + R) dt dp = ρc dt p dp What is interesting here is that contrasts can show up as temperature changes or pressure changes, but that if energy goes into a temperature change and pressure is fixed, the fraction c v /c p must be shared with increasing the internal energy, leaving only the fraction R/c p to increase contrasts themselves. One way of writing this is dw dq = d g dq = dp E dq We re not quite finished though. We still have to address the a. Note that if divergence is positive then there is a net loss of energy from the layer. If the net radiative flux is F, then a negative gradient in F implies that radiative energy is being deposited in the layer. Thus or, alternatively, noting that dp/dz = ρg = R c p df dz = ρc dt p dp df dp = c p dt g 1 dp ρg 3

4 or, df dz = ρc dt dz p + ρg Consider for example a cloud layer of thickness z with radiative temperature T c radiating isotropically both up and down with fluxes σtc 4. Such divergence F = 2σTc 4 should be expected to drive either cooling or subsidence of the layer. In fact the cooling is part of what enables the cloud to sustain itself. But, there is also an upwelling flux from the surface σts 4 and a downwelling flux from the atmosphere above σta 4. These represent a convergence of energy. Thus we have d 2σT 4 c σt 4 a σt 4 s ) dz = ρc p dt + ρg dz If there is net convergence, the cloud will either loft or it will heat and evaporate. It turns out that the balance is very fine, and growing stratiform clouds are just slightly every so slightly in a regime of net divergence. A couple of points to note here. First these radiative temperatures sometimes called blackbody or effective temperatures) are related to the atmospheric temperature T through the emissivity of the layer, e.g. σt 4 a = εσt atmosphere) 4 Also, sky cover would also play a role. An object with a very high radiative temperature like the sun) is thermodynamically equivalent to an object with a low radiative temperature if it only covers some small portion of the solid angle of the sky again like the sun). Temperature contrasts alone do not drive flows, nor do density contrasts alone. Rather it is available energy density contrasts, which must take into account both temperature and density contrasts. Equilibrium scenarios Consider that equilibrium implies no gradients in energy density, i.e. that a µ) = D g µ) = D nµ = 0 3) This is essentially a statement that there are no variations in nµ, something called equipartition. leading to g = nµ) = n µ µ + µ n n = 0 dn n µ = µ µ n assuming and this is important) that µ/µ is small. For the example of an isothermal atmosphere in a gravitational field, there is available energy in two forms: the temperature of molecules and their gravitational potential g = nµ = n + mgz) 4

5 where n is the number density of molecules, m their mass, g gravity and z the vertical distance. Here we are interested in isothermal changes in potential associated with a small isothermal vertical perturbation z around some reference height z = 0. Do large perturbations at your own risk! Other important things may be changing too! implying that µ µ n = mg z d ln n = mg z whose solution leads to the familiar hypsometeric equation ) mg z n = n 0 exp = n 0 exp g z RT A key point here, is that an atmosphere that obeys this structure is in equilibrium, with dg/ = 0, so there are no vertical motions. It is departures from an equilibrium structure that allow for convective circulations. In order to have available potential energy than can allow for dg/ < 0 the free energy has to be higher than the equilibrium value, either by having too high a temperature T or two high a geopotential gz. The more general statement here is ) µ n = n 0 exp which is the familiar Boltzmann distribution. Here, µ can be any variety of things where there is a potential difference. Gravitational potential is one, buoyancy which is related) is another, On a quantum level we get µ = Nhν, giving us the familiar Boltzmann distribution. Our next example, though, is one of the most important relationships in atmospheric sciences: the vapor pressure over water. Take µ = l v, where l v is the latent heating associated with evaporation at a given potential surface. Then the differentials across the liquid-vapor interface is given by dn n = µ µ = l v Solving, the equilibrium molecular number concentration associated with the vapor phase varies as ) n 2 T 1 ) = n 0 exp µ 1 ) = n 0 exp ) l v 1 where n 0 is some reference ground state. Now let s say the temperature is shifted from T 1 to T 2, while maintaining equilibrium, then n 2 T 2 ) = n 0 exp l ) v 2 Dividing one by the other gives us n 2 lv 1 = exp 1 ) ) n 1 k T1 T2 5

6 Noting that e = n we re using e for vapor pressure out of convention), and assuming that T is small, we have the saturation vapor pressure over water e s as a function of temperature which can be expressed equivalently as e s T 2 ) = e s T 1 ) exp lv k 1 T1 1 T2 ) ) Lv 1 e s T 2 ) = e s T 1 ) exp 1 )) R v T1 T2 At freezing the latent heat is J kg 1 and R v = 461 J/K/kg. Or expressed in differential form d ln e s d ln T = l v = L v R v T A more precise form of the exponential form of the equation is e s T ) = exp 17.67T T ) where, T is in C. The reason for the slightly different form of the equation is that L v itself is a weak function of temperature. Often you will hear the expression that at warm temperatures the air can hold more water vapor. Certainly from the C-C equation this seems to be correct, but semantically it is misleading. Nowhere in our dicussion of the derivation have we needed to mention air at all. Rather, the C-C equation would hold even in the absence of dry air. That said, in our atmosphere at least, having air present is what enables the temperature to be as high as it is, since it is molecular collisions that enable radiative absorption to show up as a measurable temperature. We could actually arrive at the Clausius-Clapeyron equation a little more directly and rigorously albeit roughly equivalently). Consider lifting a spring. There are two degrees of freedom. The first is the gravitational potential of the spring µ 1, and the second is the potential energy in the spring itself µ 2. The potentials are zero for an unstretched spring on the the ground. Now suppose you have a bunch of springs, all linked together, so that the energy density in spring vibrations can flow to the energy density in the height of the springs. For any system in general with two degrees of freedom 2 g = n i µ i i=1 If there is local thermodynamic equilibrium between the energy density in the two modes, then dg = µ 1 dn 1 + µ 2 dn 2 + n 1 dµ 1 + n 2 dµ 2 = 0 We could make a variety of useful shortcuts here depending on the situation we are interested in. Suppose for example we are interested in the familiar problem of dry adiabatic motions. Here there 6

7 e s mb) T C) Figure 1: Clausius-Clapeyron Equation for water vapor over liquid water 7

8 are actually three degrees of freedom: changing the internal energy, creating a pressure change and changes in height. But as shown above, the first two are linked, so effectively there are just two degrees of freedom, from changes in temperature µ 1 = 7 2 ), and changes in height µ 2 = gz). Molecules that change temperature are the same ones that change height so n 1 = n 2 n. Also, if the number of molecules in the parcel doesn t change, then dn = 0. This leads us to dg = n gdz nkdt So that at equilibrium ndµ 1 = ndµ 2 dµ 1 dµ 2 = n n = 1 7kdT 2 gdz = 1 Expressing things in more familiar units of per mass c p dt gdz = 1 dt dz = g c p which, of course is the adiabatic lapse rate, which for our atmosphere is -9.8 K/km. Let s choose now another problem with two degrees of freedom: evaporation for which the degrees of freedom are latent heat release µ 1 ) and changes in the vapor pressure µ 2 ). Let s say we want to know how the saturation equilibrium) vapor pressure varies with temperature, we would do the following. First, we need to establish the equilibrium condition at a particular temperature, then test the sensitivity to changing temperature. So, taking µ 2 =, we are assuming dµ 2 = 0. Second, in this case is a bit special in that there is no continuum in latent heat release in the same way that there is a continuum in geopotenial. A molecule is either evaporated or it isn t. There are only two options n 1 = 0 and µ 1 = 0 or n 1 = n v and µ 1 = l v. Thus d n 1 µ 1 ) = n v l v. And, of course, as before n 1 = n 2 = n sat v. The same molecules that are evaporating are creating the vapor pressure. Thus dg = n µ 1 + µ 2 dn = 0 d ln n = µ 1 µ 2 which conveniently leads to the Boltzmann distribution n = n 0 exp µ 1 µ 2 which we have already applied to molecules and the density distribution of the atmosphere. But, sticking to the case of evaporation ) ln n sat v T ) /n sat v,0 T ) = l v 8

9 Not quite sure what n sat v,0 means, but its some baseline. It doesn t matter, because if we now take the first derivative with respect to temperature, for the general case: ln n µ 2 ) /n 0 µ 2 )) = µ 1 µ 2 then d lnn µ 2 ) /n 0 µ 2 )) = d ln n µ 2) = µ 1 dµ 2 dµ 2 µ 2 2 Leading to the general form of the Clausius-Clapeyron equation d ln n µ 2 ) d ln µ 2 = µ 1 µ 2 Taking in this case µ 1 = l v and µ 2 =, and recognizing that for a particular temperature de = dn d ln e s d ln T = l v = L v RT By this point, I hope you can appreciate that there is great power in being able to establish some time and spatial scale where things are at equilibrium, i.e. not changing. A wide variety of powerful physical relationships can follow in a rather straight-forward fashion. 9

Quasi-equilibrium transitions

Quasi-equilibrium transitions We have defined a two important equilibrium conditions. he first is one in which there is no heating, or the system is adiabatic, and dh/ =0, where h is the total enthalpy