p which is the dry adiabatic lapse rate. Above the LCL we can obtain further information about the moist lapse rate. For example, the formula for the slope for the moist adiabat (6.75) can be derived from it.7

Very often a layer of finite thickness is caused to be well mixed by turbulent processes (stirring). For example, in the first kilometer or two above the ground the air is turbulent due to mechanically driven eddies that are induced by the larger scale air flows interacting with the surface features. If the atmospheric profile is stable, buildings, trees and other protuberances above a flat boundary will cause irregularities of the air flow. Moreover, if the air is unstable, convective irregularities will add to the mechanical turbulence. This kind of turbulent, well-mixed layer may

7 See Bohren and Albrecht (1998).

persist up to a few kilometers where it gently changes to the more orderly larger scale flow. The layers above this boundary layer are called the free atmosphere.

In a mixed layer as a whole we do not have a strict thermal equilibrium. That is to say the layer will not reach a uniform temperature as a function of height. The mechanical stirring overrides the tendency for the layer to come to a uniform temperature due to thermal conduction (due to molecular or eddie transport processes). The reason for this is that as parcels rise their temperatures are lowered because of adiabatic expansion. Observations show that for such well-mixed layers, especially near the ground and on gusty days, the temperature profile approaches the dry adiabatic lapse rate. An example is the layer between 850 hPa and 1000 hPa shown in Figure 7.13.

A heuristic proof of the adiabatic lapse rate in a well-mixed layer can be constructed by assuming that the atmosphere is subdivided into horizontal layers, each labeled by an index, i. Now suppose a piece of one of the lower layers is carried to a higher layer and in turn the same amount of mass from the upper layer is carried below by the mechanical stirring mechanism. As the parcel from below is lifted adiabatically and then brought into contact with the layer above, it will be in thermal contact with other parcels in that horizontal layer. The lifted parcel and the others in the upper level will reach a temperature (isobaric mixing) between the original environmental temperature of the layer and that of the adiabatically lifted parcel. The adjustment to equilibrium in this upper horizontal layer represents an increase in entropy of that layer which can be treated as a thermodynamic system. The collection of all the layers can be thought of as a collection of thermodynamic systems which we allow to interact in this peculiar way. Each time a parcel is lifted or lowered and brought into contact with a layer at a different pressure level, the entropy of that system increases and furthermore the entropy of the entire collection increases. As the mixing proceeds in this way, each step preserving the mass at an individual level and preserving the total enthalpy of the system, the system will come to a profile in which further mixing will no longer increase the entropy of the collection. This final state must be the one in which the entropy is homogeneous throughout. This is the state with constant potential temperature (recall S = Mcp ln Q) and this is the adiabatic profile.

Mathematical derivation We can make our argument more compelling by using an analytical approach. First, take the entropy of the whole system of layers to be

We wish to preserve the total enthalpy of the composite system:

We would like to find the extremum of S subject to the constraint that H be held constant. A convenient way to do it is through the use of a Lagrange multiplier, X (almost all calculus books nowadays discuss this technique). We proceed by writing

W (61,..., 6n) = J2 MiCp ln 6i - Xj^ MiCp(6i - Td z) (6.90) ii and set the partial derivatives to zero:

d6j dX

This procedure finds the set of 6i (i = 1,..., n) that will make S extreme. We find

In other words 6i does not depend on i or z; it is a constant.

Of course, it must be kept in mind that the mathematical proof does not ease the assumptions we made about adiabatic lifting and lowering and the assumptions about horizontal (constant pressure) exchanges of heat between the parcel being moved and its environment at the same level (pressure). On the other hand, the fact that such a simple argument reproduces the profile seen in nature so regularly suggests that our assumptions are reasonable.

We have remarked in earlier chapters that the mixing ratio, w, is a conserved quantity under vertical motions below the LCL. We can go through the same argument as above to show that the mixing ratio of water (or that of any other inert chemical species) should become uniform in the layer. Basically, when we bring a parcel into a layer in which the background is different from the mixing ratio in the parcel, the two will mix in such a way that the new mixing ratio will lie between that of the parcel and that of the whole layer in proportion to the masses. This mixing in an individual layer will increase the entropy of that layer. Each exchange of parcels will cause an increase in the entropy of the entire composite system until further exchanges do not increase the entropy. This final configuration will occur when the entire layer or composite of layers is at a uniform mixing ratio, w.

A uniform value of w in a layer exhibits a characteristic shape of the dew point temperature in a thermodynamic diagram. It turns out the dew point curve will lie exactly parallel to one of the saturation lines on the chart.

Ocean mixing There are layers in the ocean in which the mixing theory given above for the atmosphere works. The most important analogous conserved quantity in the ocean is the salinity. This quantity, along with potential temperature, is uniform in the deepest layers of the ocean where enough time has elapsed since their isolation to leave these water masses well mixed.

Notes

Many of the subjects in this chapter are covered in dynamics books such as Holton (1992). The thermodynamic details are discussed in more detail by Bohren and Albrecht (1998) and Irebarne and Godson (1981).

Notation and abbreviations for Chapter 6

A horizontal area of a slab (m2)

g acceleration due to gravity (9.81 m s-2)

Td, Tm, Te lapse rate, -dT/dz of dry air ascending adiabatically, of moist adiabat, of the environment (Km-1) h height above a reference level

H scale height

Hw a scale height for water vapor (m)

k = R/cp (dimensionless)

L = AHvap enthalpy of vaporization (latent heat) (J kg-1)

m, f angular frequency (rad-1), frequency (Hz)

p,p(z),p0 pressure, as a function of z, at a reference level (hPa)

^ (z), , geopotential height as a function of height, at two levels (meters, on charts often in decameters, dm) p, p0, pe density, at a reference level, of the environment (kg m-3) T, T (z), T0 temperature, as a function of z, at a reference level (K) Te(z), Ta(z) temperature of the environment, of an adiabat (K) T vertical average temperature in a layer of air (K)

6, 6e, 6s, 6w potential temperature (K), equivalent potential, saturation equivalent potential, wet-bulb potential w, ws mixing ratio, saturation mixing ratio (kg water vapor per kg dry air)

z, Az vertical distance, increment of it (m)

Zp potential energy per unit mass due to gravity (m)

Problems

6.1 Suppose the temperature of the atmosphere has the dependence T = T0e—z/z0. Find an expression for the pressure p(z).

6.2 A 1m3 parcel of moist air (r = 75%, T = 303 K, p = 1000 hPa) is embedded in surrounding dry air. What is the vertical acceleration of this parcel?

6.3 Suppose a parcel has a vertical acceleration of 0.12 m s—2 (see previous problem). If it starts at rest at the surface, what is its vertical velocity after 5 s, 10 s, 30 s? How long does it take to reach the top of the boundary layer (« 2 km)?

6.4 At a certain level of the (dry) atmosphere z, the temperature is 303 K and the local lapse rate is 12Kkm— Is this layer stable? Suppose a 1 m3 parcel is displaced upwards by 0.5 km adiabatically. What is its acceleration due to buoyancy? How will the answer change if the parcel is displaced isothermally?

6.5 Suppose the atmosphere has its temperature equal to 300 K and pressure 1000 hPa at z = 0. The temperature profile falls linearly with a lapse rate of 6 K km—1 up to 10 km. Above 10 km the temperature is constant. What is the pressure as a function of z?

6.6 Use the results of Problem 6.1 to compute the potential temperature as a function of height z.

6.7 Find the dry lapse rate near the surface for Mars. The mean radius of Mars BMars = 3.40x106m is 0.530 x BEarth; mass of Mars = 0.107MEarth; universal gravitational constant G = 6.67 x 10—nNm2kg—2; for CO2, cp « 0.76kJ kg—1 K—11.

6.8 Suppose an atmospheric profile is given by T(p) = a + b lnp/p0,0 < p < p0. Find an expression for the geopotential height Z(p) as a function of pressure, p.

6.9 What is the thickness of the 1000 to 900 hPa layer if the mean temperature is 300 K?

6.10 What is the acceleration of a dry air parcel whose temperature is 300 K embedded in an environment of 285 K?

6.11 Compute the Brunt-Vaisala frequency for dry air in a layer where dQ /dz = 1K km— 1, Q = 300 K. Give the answer in Hz as well. Compute the period of the oscillations.

6.12 Consider the differential equation:

Show that x = A cos rnt + B sin rnt is a solution for constant values of A and B.

6.13 Relating the last problem to buoyant oscillations of a dry air parcel, find the coefficients A and B for two situations, using dQ/dz =1 Kkm—1 and Q « 300 K: (a) x(0) = 10 m, ^(0) = 0ms—1; (b) x(0) = 0m,«(0) = 1ms—1.

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