AOS 330: Physics of the Atmosphere and Ocean I Class Notes. G.W. Petty

Transcription

1 AOS 330: Physics of the Atmosphere and Ocean I Class Notes G.W. Petty September 4, 2001

2 Chapter 1 Overview of the Atmosphere 1.1 Composition of the Terrestrial Atmosphere One may group the constituents of the terrestrial atmosphere into the following four categories: 1. so-called permanent gases; principally N 2,O 2,andAr 2. water (H 2 0) in all three of its phases (vapor, liquid, ice) 3. variable gaseous constituents other than water: e.g., CO 2,O 3,SO 2,NO 2 4. solid and liquid particles other than water (aerosols) We will be concerned mainly with (1) and (2). The constituents falling in categories (3) and (4) are often of great interest chemically, radiatively, or as pollutants, but these have a negligible effect on the bulk thermodynamic properties of air and will not be considered until later. Below about 100 km, the permanent gases are present in almost constant proportions, due to efficient mixing by turbulence. This region of the atmosphere is known as the homosphere. The following tables give the volume (molar) fractions of the top nine permanent constituents and the top five variable constituents: The top three permanent constituents N 2,O 2, and Ar are seen to account for 99.97% of the permanent gases in the atmosphere (Table 1.1, p. 5, W&H). The addition of CO 2 brings the total up to %. Above 100 km, molecular diffusion under the influence of gravity is able to sort gas molecules by weight faster than they can be remixed by turbulence. As a consequence, constituent proportions are no longer constant but rather reflect an increase with height in the proportion of lighter gases such as He and H. Furthermore, intense ultraviolet radiation at high altitudes breaks apart diatomic molecules such as N 2 and O 2, so that these elements are increasingly represented by their monoatomic forms. The region of the atmosphere in which constituents appear in variable proportions due to diffusive separation is called the heterosphere. The heterosphere is the subject of a branch of meteorology called aeronomy. We will not concern ourselves further with the heterosphere in this class. Water vapor in the homosphere may vary from about 0 7% (by volume) of the air; despite its relatively small fractional contribution to the total gases of the atmosphere, it is the most important constituent from a meteorological point of view, owing in part to its very substantial role in the thermodynamics and energy 1

3 balance of the atmosphere, not to mention the formation of clouds, rain, snow, and other elements of bad weather. 1.2 Thermal Structure of the Atmosphere Let us now begin to consider the observed vertical structure of the atmosphere. In general, the properties of the atmosphere, such as pressure and temperature, vary much more rapidly in the vertical direction than they do in the horizontal direction. As an example, let s look at a snapshot in time of the atmosphere at a single location. An actual morning upper air message transmitted by a certain weather station (72562 = Holdrege, Nebraska) on a certain summer day reads: TTAA ///// ///// // // // // // ///// 18520= TTBB 7612/ // // // // // // 2300/= TTCC // // // = TTDD 7612/ // // // // // //= Translated according to the coding rules found at this message yields the following information: 2

4 Pressure (mb) Altitude (m) Temperature ( C) Dewpoint( C) Very generally, we notice a decrease in temperature as we get to higher altitudes (lower pressures). But the rate of change of temperature is not quite constant: there are some levels where it changes slowly with height and others where it changes more rapidly. The rate of change of temperature with height is the environmental lapse rate and is defined by Γ= T z T (1.1) z So Γ is positive for the usual situation in which temperature decreases with height. It is negative in the less typical situation in which temperature increases with height. A layer in which Γ < 0 (i.e., T/ z > 0) is called an inversion. Inversions thus represent atmospheric layers in which warm air overlies colder air. If Γ = 0, then the layer is isothermal. There are several different reasons why inversions form:. Radiation inversions form as the result of radiational cooling of the ground at night, and consequently the layer of air directly above it. This effect is most pronounce when the atmosphere above that layer is relatively transparent to infrared radiation; for example, when the sky is clear and the relative humidity is low. 3

5 Subsidence inversions may form when air sinks from a higher altitude, warming by compression as it goes. Frontal inversions may appear in a sounding taken on the cold side of a front, since the frontal discontinuity between the cold and warm air masses will generally slope back over the position where the radiosonde was released. Boundary layer inversions frequently delimit the mixed layer of the atmosphere near the surface, which may be from a few tens of meters thick to a kilometer thick or more. Near the coast, very strong boundary layer inversions may appear at the top of a shallow layer of cool, moist air flowing in from the ocean. Finally, an inversion frequently occurs at or above the tropopause, which marks the transition from the troposphere to the stratosphere. In the troposphere, the lapse rate is generally positive (temperature decreasing with height), whereas in the stratosphere the lapse rate is small or even negative. Let us clarify the last statement by reviewing the large scale thermal structure of the atmsosphere. Figure 1.8 on p 23 of your textbook by Wallace and Hobbs (hereafter W&H) depicts the typical structure up to about 100 km (recall also that this altitude roughly corresponds to the transition from the homosphere to the heterosphere). The lowest layer is the troposphere, which extends from the surface to around 10 km, give or take a km or two. Within the troposphere, the general tendency is for the temperature to decrease with height (i.e., Γ > 0). The next layer is the stratosphere, which extends upward to around km. Within the stratosphere, the temperature generally increases with height, so that at the top of the stratosphere, temperatures may actually be somewhere near the freezing point again! In the mesosphere, the temperature generally decreases with height again until around 80 km or so. The temperatures found at the mesopause may be the coldest anywhere in the atmosphere (-93 Cin the U.S. Standard Atmosphere). The thermosphere (see Figure 1.9, p. 25 of W&H) consists of very thin, hot, ionized gases and has no welldefined upper boundary. The temperature in the thermosphere depends strongly on solar activity; it may be as cool as 600 K when the sun is quiet or may increase in temperature to 2000 K under the influence of an active sun. Geometrically, the above layers range in thickness from 10 km for the troposphere to 100s of km for the thermosphere. However, because of the very low air densities found at higher altitudes, most of the mass of the atmosphere is found in the troposphere. Indeed, the troposphere has about 80% of the total mass, and the stratosphere has almost all of the remaining 20%. The mesosphere and thermosphere account for only about 0.1% and 0.001%, respectively. The troposphere is the most interesting layer for most meteorologists (and for us), not only because it contains the lion s share of the mass of the atmosphere, but also because we live in it and it is in the troposphere that most weather occurs. As we shall see later, the difference between the characteristic lapse rates in the stratosphere and the troposphere help explain why there is so much more action in the troposphere. Up until now, we have considered only vertical temperature profiles, without regard to geographic and seasonal variations. A pair of figures on p. 27 of W&H depict the average temperature structure of the atmosphere both in the horizontal and the vertical for a summer and a winter month, respectively. Several details are worth noting: 4

6 The tropopause is generally lower near the poles than it is near the equator. The tropopause is generally lower in wintertime than it is in the summertime. The seasonal difference in height is generally greater at middle and high latitudes than in the tropics. There is often a discontinuity or break in the tropopause near latitude. There is a pronounced inversion near the surface at high latitudes, particularly in the wintertime. The pressure at a specified geometric altitude in the upper troposphere and stratosphere is, on average, significantly greater in the tropics than near the poles. Of course, these descriptions represent only average conditions. At any given instant in time and at any given location, an actual profile and/or cross-section of the atmosphere may differ significantly from these average profiles. One of the most important practical objectives of this class is to give you knowledge you can use to determine how the day-to-day variations in the thermal structure of the atmosphere can affect the weather we experience. 5

7 Chapter 2 Physical Properties of Air 2.1 Some new definitions Intensive state variables are those which are independent of mass; e.g., temperature, pressure, density. Extensive variables depend on the total mass of the system; e.g., volume. Extensive variables may always be converted to intensive variables by dividing by the system s mass. 2.2 Behavior of Ideal Gases The state of a homogeneous, gaseous system may be characterized by three variables; for example, density, pressure, and temperature. The density ρ is the mass of material in the system divided by the volume of that system (M/V ). Alternatively, one may wish to use specific volume α (volume per unit mass), which is just the inverse of ρ. Density and specific volume are expressed in SI units of kg/m 3 and m 3 /kg, respectively. The pressure p is the force per unit area exerted by the random motions of the molecules contained within a system. It has no preferred direction. SI units of pressure are N/m 2 or Pascal (Pa). A fundamental concept in thermodynamics is that of an equation of state. The three variables which describe a system have been shown experimentally not to be independent of each other. In other words, they satisfy a relation of the form f(p, V, T ) = 0 (extensive form) or f(p, α, T ) = 0 (intensive form) 6

8 This means if you know any two of the three variables, the third may be determined. Boyle s Law (1660): At constant temperature, the volume of a given sample of gas varies inversely as the pressure. p 1 V or p = C 1 V or pv = C 1 (2.1) where C 1 is a constant of proportionality [Note: C 1 = f(t )]. Charle s Law (1787): At constant pressure, the volume of a given sample of gas is proportional to absolute temperature. V T T = V 0 = C 2 or V = C 2 T (2.2) T 0 where C 2 is a constant of proportionality [Note: C 2 = f(p)]. Charle s Law and Boyle s Law are given by the following pair of equations: pv = C 1 (T ) (2.3) V = C 2 (p)t (2.4) This can only hold true if there is another constant C such that C depends on both the size of the gas sample and on the type of gas. pv = CT (2.5) Avogadro found that for fixed pressure and temperature, the number of molecules per unit volume of a gas is a constant, irrespective of the chemical composition We therefore introduce an SI unit called the kilomole (kmol) to represent a fixed number of molecules. A kilomole of substance corresponds to the number of molecules in a sample whose weight in kilograms equals the standard molecular mass m of the substance. This number is called Avogadro s constant and has the value kmol 1. The equation of state of an ideal gas, known as the Ideal Gas Law can therefore be written as pv = nr T (2.6) where n is the number of moles in the sample and R is the Universal Gas Constant. The value of R was determined experimentally by noting that at 0 C and standard atmospheric pressure, the volume of one mole of an ideal gas is 22.4 liters (1000 liters = 1 m 3 ). Solving for R gives R = J K 1 kmol 1 Under what conditions does the Ideal Gas Law give an accurate description of the behavior of a gas? In general, the Ideal Gas Law is valid whenever the density of the gas is low enough (due to a suitable combination of low pressure and high temperature) so that individual molecules do not experience significant 7

9 attractive forces, nor does the space occupied by the molecules represent a significant fraction of the total volume. Under conditions near the liquefaction point of a gas, the Ideal Gas Law may no longer be sufficiently accurate. At ordinary atmospheric pressures, air is obeys the Ideal Gas Law quite closely for meteorological purposes. In meteorology, it is convenient to deal with a known mass M of a gas rather than a number of kilomoles. This can be accomplished by replacing nr with MR, i.e., where R is a gas constant which depends on the particular gas. pv = MRT (2.7) By this definition, R R (n/m) =R /m (2.8) where m is the molecular mass (units kg kmol 1 = atomic mass units or amu) What do we do when the gas in question is a mixture of molecules of different masses? No problem: Remember, the form of the Ideal Gas Law using the Universal Gas Constant is always valid. One can therefore always obtain R for a specific gas simply by using m = (total mass in kg)/(total no. of kilomoles). Let s say a sample contains a mass M 1 of one gas and a mass M 2 of another. The corresponding molecular masses are m 1 and m 2. The total number of kilomoles, therefore, is given by and the total mass M is just M 1 + M 2. For a two-component gas the Ideal Gas Law can thus be written n = n 1 + n 2 = M 1 /m 1 + M 2 /m 2 (2.9) pv = nr T =(M 1 /m 1 + M 2 /m 2 )R T = MRT (2.10) where the new gas constant R for the mixture is given by [ ] 1 R R (n/m) =R M 1 + M 2 (2.11) M 1 /m 1 + M 2 /m 2 This approach can be generalized to give us a definition for the mean molecular mass of an arbitrary mixture of N different gases: N i=1 m = M i N i=1 M (2.12) i/m i Dalton s Law In deriving a form of the Ideal Gas Law which accounts for mixtures of gases of different molecular weights, we simply assumed that the number of molecules in a gaseous system is what is important for determining the product pv of the system as a function of its absolute temperature T. The validity of this assumption is implicit in Dalton s Law of Partial Pressures: 8

10 The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures which would be exerted by each constituent alone if it filled the entire volume at the temperature of the mixture. That is, for a mixture of k components p = k p i (2.13) i=1 where p is the total pressure and the p i are the partial pressures of each gas. These may be assumed to obey the Ideal Gas Law separately, so that p i = M i (R /m i )T/V (2.14) where M i is the total mass (kg) of each constituent and m i is its molecular mass (kg/kmol). Then k k p = p i = M i (R /m i )T/V = MR T/V i=1 i=1 k (M i /m i )/M (2.15) i=1 where M = k i=1 M i. The above result can be shown to lead to exactly the same definition of a mean molecular mass m and specific gas constant R as we had already derived earlier without explicitly invoking Dalton s Law Gas Constant for Dry Air Armed with the knowledge of the composition of air (see page 5, W&H), we may compute a mean molecular weight m d and a specific gas constant R d for dry air. We will not include the contribution of water vapor at this point because it is so variable in space and time. Later, we will spend a lot of time describing how water vapor may complicate (and thus make more interesting!) the thermodynamic behavior of air. The formula we derived earlier for a multi-component gas is useful when the mass fraction of each constituent is given. In the table above, however, it is the volume (or molar) fraction which is given. You should verify for yourself that, since volumes are proportional to numbers of molecules, one may calculate the mean molecular mass m simply as m = f v,i m i (2.16) where f v,i is the volume fraction of the ith constituent (note that the sum of the fractions used must be normalized to equal 1). Taking the volume fractions of only the top four constituents N 2,O 2,Ar,andCO 2 we find that 7808(28.013) (31.999) (39.948) (44.010) (2.17) or m d = kg/kmol (2.18) R d = 8314/28.96 = J/(kgK) (2.19) Hint: You will use the value of R d so often that you should commit it to memory as soon as you can! Example Gas Law Calculations for Dry Air Having determined R d, we can proceed to use the Ideal Gas Law to compute state variables for dry air under a variety of conditions. 9

11 Example: Standard sea level pressure is hpa and standard surface temperature in the U.S. Standard Atmosphere is 15 C. What is the density of dry air under these conditions? What volume is occupied by 1 kg of dry air? Solution: The intensive form of the Ideal Gas Law is pα = R d T (2.20) or, since ρ =1/α, p = ρr d T (2.21) Substituting p = Pa and T = K and solving for ρ gives us an atmospheric density under standard conditions of kg m 3 or, equivalently, a specific volume α of 0.816m 3 kg 1. Example: At sea level, the pressure may sometimes be as low as 950 hpa or as high as 1040 hpa. The temperature, can easily vary between 40 and +50 C. Using this information, estimate rough limits on the range of densities that might be exhibited by air at sea level. Solution: The densest air would occur for a combination of p = 1040 hpa and T = 40 C. Using the procedure in the previous example, we find ρ =1.55 kg m 3. The thinnest air would occur for p = 950 hpa and T =50 C, for which ρ =1.03 kg m 3. Question: Which of the two variables, pressure or temperature, is more important in causing variations in the density of air at sea level? Answer: Density is proportional to pressure and to the inverse of absolute temperature. At sea level, pressure varies by only about 5% from its standard value while the temperature may vary by more than 20% from its standard value. Hence, variations in temperature are likely to bring about larger changes in density at sea level. Example: At 3 km (about 9,900 ft) altitude, the pressure in the U.S. Standard Atmosphere is 701 hpa and the temperature is -4.5 C(268.7 K). What temperature would a parcel of air at sea level (standard pressure) have to have in order to match the density of typical air at 3 km altitude? Solution: The density at 3 km is found to be kg m 3. Substituting this value for ρ into the Ideal Gas Law along with p = hpa, we find that air would have to be heated to a temperature of 388 K (equals 115 Cor 239 F ) in order to have the same density at standard sea level pressure Adding Water Vapor We previously calculated the Gas Constant R d for perfectly dry air i.e., air which contains absolutely no water vapor, only the permanent gases which are always present in constant proportion. Perfectly dry air doesn t exist in the atmosphere; moreover, a dry atmosphere would be extremely uninteresting meteorologically. Let us therefore begin to examine the behavior of water vapor and, in particular, its influence on the thermodynamic properties of the air. First, let s consider water vapor in isolation; that is, without the added complication of other gases. If we can treat water vapor as an ideal gas, then the ideal gas law is just as valid here as it was earlier for dry air: pα = RT (2.22) 10

12 For water vapor, however, it is conventional to identify the vapor pressure with the symbol e rather than p. We will use the symbol ρ v to denote the water vapor density, more commonly known to meteorologists as the absolute humidity. We can thus write the ideal gas law for water vapor as e = ρ v R v T (2.23) where R v is the Gas Constant for pure water vapor? What is the value of R v? We can calculate this just as before, using R v = R (2.24) m v where m v is the molecular mass of H 2 O and equals about kg/kmol. Therefore R v = Jkg 1 K 1 (2.25) Is the ideal gas law as good an approximation for water vapor as it is for dry air? Not really, because water vapor at typical atmospheric temperatures and pressures is usually much closer to condensation than is the case for dry air. This implies that attractive forces between the molecules are significant. Nevertheless, the inaccuracy in assuming that water vapor behaves as an ideal gas is not great enough to worry about in most meteorological applications. Obviously, water vapor ceases to behave anything like an ideal gas when it reaches saturation, since an isothermal decrease in volume then no longer implies an increase in pressure. We will come back to the behavior of water vapor at saturation later. Now let s generalize to the case where water vapor and the permanent gases of the atmosphere coexist in the same volume. Can we still use the relationship e = ρ v R v T? Of course, provided that we recognize that the vapor pressure e represents the partial pressure of vapor in the air and that the total pressure is given by Dalton s Law as p = p d + e (2.26) Here p d is the partial pressure of the dry air in the mixture and is related to the dry air density ρ d and temperature T by the familiar gas constant R d = J/(kg K): p d = ρ d R d T (2.27) Combining the ideal gas laws for the two components, we have that the total pressure of a moist volume of air is given by p =(ρ d R d + ρ v R v )T (2.28) Also, the combined density of moist air is obviously just the sum of the densities of the dry air and water vapor ρ = ρ d + ρ v (2.29) It is usually inconvenient to use water vapor density ρ v or vapor pressure e to express the relative vapor content of a mass of air, since these quantities are not conserved. That is, the vapor pressure and the density each increase or decrease when the air containing the vapor is compressed or expanded, respectively. By contrast, the mixing ratio w of an air mass is conserved, as long as there is no condensation or evaporation taking place. The mixing ratio is defined by w M v = ρ v (2.30) M d ρ d where M v is the mass of water vapor mixed into a mass M d of dry air. Because there is usually much more dry air than vapor in given volume of the atmosphere, it is often convenient to express w in units of grams vapor per kilograms dry air. For example, in a warm tropical air mass, the mixing ratio may be as high as 20 g/kg. In cooler air masses, w is typically only a few g/kg. 11

13 Similar to the mixing ratio w is the specific humidity q which is defined as M v q = ρ v (2.31) M d + M v ρ In other words, q gives the mass of water vapor per unit mass of moist air, so that the mass contribution by the water vapor is included in the denominator. Note that since the mass of water vapor is typically no more than one or two percent of the total mass, the numerical values of w and q also differ by no more than one or two percent. For many common applications, it is fairly unimportant whether one uses w or q in moisture calculations. If an exact conversion is required, the following relationships may be used w = q (2.32) 1 q and q = w 1+w. (2.33) You should verify that, for realistic values of either w or q, there is little numerical difference between the two. Often, it is necessary to convert between mixing ratio w or specific humidity q and vapor pressure e. Forw, this can be accomplished by noting that w = ρ v = e/r vt ρ d p d /R d T = εe p e εe p (2.34) where Similarly, In summary, to a good approximation q = ε R d R v = m v m d =0.622 (2.35) ρ v εe = ρ v + ρ d p (1 ε)e εe p (2.36) q εe p w (2.37) Virtual Temperature We already saw that the total pressure of moist air is given by p = p d + e =(ρ d R d + ρ v R v )T (2.38) We can factor out the moist air density ρ and the gas constant for dry air R d by writing ( ) ( ρd R d + ρ v R v ρd p = ρr d T = ρr d ρr d ρ + ρ ) v R v T (2.39) ρ R d Using the definitions of specific humidity q = ρ v /ρ and R d /R v = ε, it is easy to show that ( ) ] 1 p = ρr d [1+ ε 1 q T (2.40) This identical to the ideal gas law for dry air, except for the appearance of the factor in brackets on the right hand side. How does one interpret this factor? To begin with, we know that the ideal gas law for moist air can be written p = ρr m T (2.41) 12

14 where the gas constant R m reflects the mean molecular mass of the air, including not only the permanent gases but also the contribution by water vapor. Clearly, R m is not a constant but depends on the moisture content of the air. While we could calculate the mean molecular mass directly using the same formula we used much earlier in getting R d, it is easy to see by comparing the last two equations that ( ) ] 1 R m = R d [1+ ε 1 q (2.42) In other words, the term in brackets gives us a convenient means to adjust the gas constant for dry air in order to account for the presence of water vapor. However, people don t usually have an instinctive feel for the physical meaning of a given value of the gas constant R. As a result, the approach just described for interpreting the water vapor correction term is not the most common one. Instead, it is conventional to write T v = [ 1+ ( 1 ε 1 so that the ideal gas law may be written for moist air as ) ] q T (2.43) p = ρr d T v (2.44) The virtual temperature T v is simply the temperature a dry parcel of air would have to have in order for that parcel s density to equal the density of the moist parcel, assuming equal pressures. That is, if a dry parcel of air has temperature T 0 and a moist parcel of air has a virtual temperature T v which happens to be equal to T 0, then both parcels have the same density (again, assuming equal pressures). Note that the quantity 1/ε 1 is just a constant with the approximate value 0.61; consequently, the most convenient formula for T v is just T v =(1+0.61q) T (2.45) By substituting q equal to 0.02 kg/kg (an approximate upper limit for q in the atmosphere), one finds that in general 0 < (T v T ) < 3.7 K (2.46) The practical value of T v is most obvious when the hydrostatic law comes into play (next chapter), which relates the local density of air to the local rate of change of pressure with height. From this point onward, anytime you solve problems which depend in some way on the density of air, you should remember that it is the virtual temperature which determines this density, not the actual temperature (unless of course q = 0). In many cases, the difference between T v and T is small enough that we will ignore that difference, but you should always be at least aware of the difference and know when it is likely to be worth taking into account. 13

15 Chapter 3 Atmospheric Pressure 3.1 Hydrostatic Balance Now that we have the Ideal Gas Law at our disposal and know something about how temperature varies with height in the atmosphere, we are finally equipped to consider in detail how and why atmospheric pressure varies throughout the atmosphere. To an excellent approximation under most conditions, the pressure at a given point in the atmosphere is given simply by the weight of the atmosphere above that point. For example, if the surface pressure at a given geographic location and time is hpa, then the weight per square meter of the atmosphere above that point is 101,320 N. Dividing by the acceleration due to gravity g =9.81 m s 2, we find that the mass of the column of air above a square meter of the surface is about 10,328 kg, or a little over 10 tonnes! How can we be sure that it is this simple? Since typical vertical accelerations within the atmosphere are observed to be very small compared with the value of g, it follows that any contribution to the pressure by forces other than gravity must also be relatively small. To take a relatively extreme case, consider a typical thunderstorm, in which vertical velocities at a middle level of the troposphere (say 3 km) may be of order 10 m s 1. A blob of air rising from the surface and accelerating steadily will require on the order of 10 minutes to reach that speed if it starts out from a standstill. This implies an acceleration of 0.02 m s 2, or only about 0.2% of g. Even if the entire vertical column of the atmosphere underwent this much acceleration, it would give rise to a surface pressure anomaly of only about 2 millibars. Outside thunderstorms, typical vertical accelerations are far smaller still and may almost always be safely ignored. Let us therefore formalize the Hydrostatic Law as follows: Consider a vertical column of air with unit cross-sectional area (see schematic next page). The mass of the air between heights z and z + dz in the column is ρdz, where ρ is the density of the air at height z. The force acting on this column due to the weight of the air is gρ dz, where g is the acceleration due to gravity at height z. Now let us consider the net vertical force on the block due to the pressure of the surrounding air. We will assume that in going from height z to height z + dz the pressure changes by an amount dp as indicated in the schematic. Since we know that pressure decreases with height, dp must be a negative quantity, and the upward pressure on the lower face of the shaded block must be slightly greater than the 14

16 downward pressure on the upper face of the block. Thus the net vertical force on the block due to the vertical gradient of pressure is upward and given by the positive quantity dp as indicated. The balance of forces in the vertical requires that dp = ρg dz (3.1) or, to give the most common form of the Hydrostatic Equation, dp = ρg (3.2) dz If the pressure at height z is p(z), we have or, since p( ) = 0 p( ) p(z) p(z) = dp = z z gρ dz (3.3) gρ dz (3.4) To summarize, the rate of change of pressure with height is proportional to the density. Furthermore, the pressure at any given level is approximately proportional to the mass above that level (the proportionality is not quite exact, because g decreases slightly with altitude). We can now of course substitute the Ideal Gas Law for ρ and arrive at an expression for the rate of change of pressure with height as a function of temperature: dp pg = ρg = dz RT (3.5) If we are dealing with a dry atmosphere (i.e., no water vapor), then of course R = R d =287 J/(kg K). Substituting standard sea level values of g =9.8 m s 2, p = 1013 hpa and T = 288 K, we find that dp dz = 12 Pa/m. In more traditional terms, this translates into a one mb (hpa) change in pressure for every 8.3 m change in elevation. Of course, at higher altitudes you have to change altitude by much more than this to achieve the same change in pressure, because the air is less dense. A convenient transformation of the above equation may be obtained simply by dividing both sides by the pressure p. We then have 1 dp p dz = g (3.6) RT Since d(ln p) = 1 pdp, this can be rewritten as d ln p dz = g RT In words, the rate of change of the logarithm of pressure with height is inversely proportional to the absolute temperature, and does not depend on p. As we shall see shortly, this is the same as saying that pressure generally falls off exponentially with height. (3.7) Digression on Gravity At this point it is worthwhile to backtrack and reconsider the value of g which keeps cropping up in our equations. As we already know, the value of g is close to 9.81 m s 2 at sea level, and for some purposes, this value is sufficiently accurate. However, it is important to recognize that g does in fact vary slightly with 15

17 altitude and latitude. For some purposes, the difference is important, so we will take a closer look at this quantity and introduce a convention that will then allow us to pretty much forget about it again. The actual acceleration due to gravity is a function of the distance R from the center of mass of the earth. Specifically, g(r) = GM R 2 (3.8) where M is the mass of the earth (M = kg) and G is the so-called Universal Gravitation Constant and has a value of Nm 2 kg 2. If we wish to consider gravity as a function of the altitude z above the surface, we can substitute R = R 0 + z in the above equation, where R 0 is the effective radius of the earth and is equal to about 6370 km. In this case, we have g(z) = GM (R 0 + z) 2 (3.9) By using a Taylor series expansion in (z/r 0 ) and discarding higher order terms, we can write ( )] z g(z) g 0 [1 2 R 0 (3.10) where g 0 = GM R0 2 (3.11) is the standard acceleration due to gravity at sea level. According to the above formula, for z =10kmabove sea level (i.e., near the top of the troposphere), g decreases to about 9.78 m s 2, or about 0.3% less than its sea level value. Two things complicate the picture somewhat further: first of all, the earth bulges somewhat at the equator, so that R 0 is not the same for the equator ( km) as it is for the poles ( km). For this reason alone, the sea level value of g is slightly lower (about 0.7%) at the equator than at the poles. Secondly, we have not considered the minor difference between the true (or pure ) gravity g and the apparent gravity g which includes the effects of the rotation of the earth. An adequate approximation for the latter is given by g = g Ω 2 R cos 2 φ (3.12) where Ω is the angular velocity of the earth and equals 2π (radians) per 23.9 hr or s 1,andφis the latitude. One can easily calculate that the difference between g and g amounts to only about 0.03 m s 2 at the equator. Clearly, the range of variability of g due to the oblateness of the earth and due to centrifugal force is no more than a few tenths of a percent under the conditions of interest to most meteorologists; nevertheless, there are times when it is important to take into account these differences when performing sensitive calculations. One way of doing this is to change one s frame of reference slightly. If you are interested in transformations of energy in the atmosphere (most meteorologists are, either directly or indirectly), a useful variable is the geopotential Φ. At any point in the atmosphere, the geopotential is defined as the work that must be done against the apparent gravitational field in order to raise 1 kg from sea level to that point; i.e., Φ= z 0 g (z) dz (3.13) The geopotential is actually the gravitational potential energy per unit mass, that is, the energy available in a decrease in elevation which may be converted to kinetic energy. As such, it has units of m 2 s 2 or J kg 1. The differential of geopotential is given by dφ =g dz (3.14) 16

18 Geopotential is often expressed in terms of another quantity, geopotential height Z, defined by Z = Φ(z) (3.15) g 0 It is clear that Z has dimensions of length, usually specified in geopotential meters, but geopotential meters are slightly larger than real meters at higher altitudes or wherever g <g 0. The advantage is that by using Z in meteorological calculations instead of the actual height z, one may forget about variability in the apparent gravity and just use the constant value of g 0 everywhere. Indeed, the height values given on standard constant pressure charts (e.g., the so-called 850 mb chart, 500 mb chart, etc.) are actually in units of geopotential height. In any case, one should not lose sight of the fact that the geopotential height does differ slightly from geometric height, and that the former is actually a measure of potential energy and not of distance. Throughout the remainder of this course, when we refer to a height or altitude in the atmosphere, it should automatically be understood (unless otherwise indicated) that we mean geopotential height, which is only slightly different from the real height. This way, we can always utilize a constant effective value of g g 0 = m/sec 2 and not worry about small variations in g from one place to the next Hypsometric Equation Having dispensed with that issue, let us now return to the question of how pressure p varies with (geopotential) height z. We can integrate the hydrostatic equation as follows: or z2 z 1 dz = R g p1 z = z 2 z 1 = R g We can simplify this to z = z 2 z 1 = R T g if only we define the mean layer temperature T as p1 p 2 p 2 Tdln p (3.16) p1 p 2 Tdln p (3.17) d ln p = R T g log [ p1 p 2 ] (3.18) T p1 p 2 p1 p 2 Tdln p d ln p (3.19) In words, the thickness z =(z 2 z 1 ) of an atmospheric layer between specified pressure levels p 2 and p 1 is proportional to the mean layer temperature T (defined as above). The equation for z is known as the hypsometric equation and is routinely used to derive the heights of pressure levels from atmospheric temperature and humidity profiles. Important: Although we used the temperature T in the above derivation, this is strictly valid only for dry air. If the humidity of the air is significant and/or high accuracy is required, then of course we need to substitute the virtual temperature profile T v (z) for the actual temperature profile T (z) in (3.19) Vertical Structure of Cyclones and Anticyclones According to the hypsometric equation, the distance between standard pressure levels is smaller in cold air than in warm air. Pressure features such as lows, highs, troughs, ridges, etc., are thus very closely related to 17

19 the thermal structure of the atmosphere, and may change shape and intensity, and even disappear altogether at higher or lower altitudes depending on the horizontal distribution of temperature. For example, you can easily verify with simple sketches of constant pressure surfaces that: A surface warm-core cyclone quickly weakens or disappears with height. A surface warm-core anticyclone strengthens with height. A warm-core cyclone aloft increases in intensity downward. A cold-core cylone at the surface increases in intensity with height. A cold-core anticyclone at the surface quickly disappears with height. Low pressure centers or troughs are displaced horizontally toward colder air with increasing height. At the surface of the earth, high pressure is favored in cold regions and low pressure in warm regions. 3.2 Pressure Profiles Under Idealized Conditions Up until we have talked in a general way about how temperature, pressure, and altitude are related in the atmosphere. We showed that if you prescribe any arbitrary temperature profile and surface pressure, one may use the hypsometric equation to compute the height of any pressure level above the surface, or conversely, the pressure at any height. Now we will spend some time talking about some special cases which allow particularly simple mathematical relationships to be derived and which, hopefully, will also help to illustrate some important concepts The Homogeneous Atmosphere (or Ocean) One of the simplest possible models of an atmosphere is one in which the density ρ is constant everywhere, irrespective of altitude. Within such an atmosphere, if it existed, pressure would decrease with altitude according to the hydrostatic law, but the density would remain constant until reaching the top of the atmosphere, at which point the density would abruptly go to zero. Actually, this model is a far better representation of an ocean than an atmosphere, since the density of seawater doesn t change much with pressure and since it has a sharply defined upper boundary. Nevertheless, let us consider an atmosphere having these properties and see where it leads us. If we integrate the hydrostatic equation from sea level, where the pressure is p 0 to a height H where the pressure is zero, we get 0 H = ρg dz (3.20) p 0 0 or p 0 = ρgh (3.21) In other words, the pressure at the bottom of the atmosphere (sea level) is once again just the weight per unit area of the atmosphere, only now there is no need for any messy integrals, since ρ is constant. Now if we know both p 0 and ρ in addition to g, we can solve for the height H: H = p 0 ρg (3.22) 18

20 If we substitute typical values for p 0 and ρ, namely 1013 hpa and 1.25 kg m 3, respectively, we find that H 8.3 km. In other words, if our atmosphere had the same mass that it has now but also had a constant density equal to its sea level density under standard conditions, it would only be 8.3 km deep! (Recall that most passenger airliners fly at altitudes closer to 10 or 12 km...) It would be easy for an atmosphere to exhibit the above behavior if only air were completely incompressible, just as water is (well, almost). But this would mean throwing out the ideal gas law and substituting a very different equation of state (i.e., ρ constant instead of ρ = p/rt ), which we know is not correct for air. So let s consider whether it would, in principle, be possible for an atmosphere to exhibit constant density throughout its depth and still obey both the ideal gas law and the hydrostatic law. We can substitute the ideal gas law into the previous equation for H to get H = p 0 ρg = ρrt 0 (3.23) ρg or H = RT 0 (3.24) g So we can see here that H can be written entirely in terms of the surface temperature T 0 and two well-known constants, R and g. But we still don t know for sure what s happening to the temperature above the surface. However, the pressure is decreasing with height, so it is clear that the temperature must also decrease in order for the density to stay constant. Let us write the ideal gas law for dry air, p = ρrt, and differentiate with respect to elevation, holding ρ constant. We get dp dt = ρr (3.25) dz dz Substitution into the hydrostatic equation leads to the result: dt dz = Γ = g (3.26) R or Γ=34.1 Kkm 1 (3.27) So in order to have a homogeneous atmosphere, you need a lapse rate which is constant and very large (about six times as large as what is usually observed in the troposphere). Now let s pursue this one step further and find out what the temperature at the top of a homogeneous atmosphere must be. In any atmosphere with a constant lapse rate Γ, the temperature at an arbitrary height z above the surface may be expressed as T (z) =T 0 Γz (3.28) If we substitute the height of the top of the atmosphere H for z, and use the above formulae for H and Γ in the homogenous atmosphere, we have T (z) =T 0 g R RT 0 g = T 0 T 0 = 0 (3.29) In other words, in order for an atmosphere obeying the ideal gas law to have constant density all the way to the top, the temperature at the top must be equal to absolute zero! Clearly, this model of the atmosphere is quite unrealistic, especially since real air would liquify (and thus cease to obey the ideal gas law) long before it ever got close to absolute zero. Nevertheless, the concept is useful theoretically. In particular, we shall see that H = RT/g crops up again even in the mathematical description of more realistic atmospheres, which we will address now. 19

21 3.2.2 The Isothermal Atmosphere Another simple model for the atmosphere is one in which not the density but rather the temperature is constant with height that is, Γ = 0. As we learned earlier, such an atmosphere is called isothermal. Again we start with hydrostatic equation and substitute the ideal gas law: dp pg = ρg = dz RT (3.30) This can be rewritten 1 p dp = g dz (3.31) RT We can then integrate from sea level (z =0,p= p 0 ) to some arbitrary level z where the pressure is p. Since T is a constant, p z 1 p 0 p dp = g RT 0 or ( ) p ln = gz p 0 RT Once again using the definition H = RT/g, weget dz (3.32) (3.33) p = p 0 e z H (3.34) where H in this case is interpreted as a scale height; i.e., the vertical distance over which the pressure decreases by a factor e 1 or to about 37% of its original value The Constant Lapse-Rate Atmosphere Let us assume that the temperature T varies linearly with height; i.e, T = T 0 Γz (3.35) In this case, the hydrostatic equation combined with the ideal gas law becomes or dp dz = pg RT = pg R(T 0 Γz) 1 p dp = g ( ) dz R T 0 Γz (3.36) (3.37) Once again this can be easily integrated. We shall again integrate between the limits z = 0, where p = p 0 and an arbitrary height z, where the pressure is p. The result is p 1 p 0 p dp = g R or ( ) p ln = g p 0 RΓ ln Taking the exponent of both sides, we get z 0 dz T 0 Γz ( ) T0 Γz T 0 ( ) g T0 RΓ Γz p = p 0 T 0 20 (3.38) (3.39) (3.40)

22 or p = p 0 ( T T 0 ) g RΓ (3.41) In the usual case where T decreases with z (Γ positive), this equation requires that pressure decrease with elevation, in agreement with the hydrostatic equation. In the less common case that T increases with elevation (an inversion), the ratio (T/T 0 ) is greater than unity above the surface but the exponent of this ratio is negative, therefore p still decreases with height, as required by the hydrostatic equation. In the special case of an isothermal layer (Γ = 0), the above formula cannot be used because Γ appears in the denominator of a fraction and division by zero is undefined. Note that the exponent in the previous equation is simply the ratio of the constant lapse rate (g/r) inthe homogeneous atmosphere to the actual lapse rate Γ. If the two are equal, exponent becomes unity, and pressure becomes a linear function of z again, which in turn implies constant density. An atmosphere with a constant positive lapse rate (decrease of temperature with height) has only a finite vertical extent. Once the temperature reaches absolute zero, we ve reached the top of our hypothetical atmosphere, just as we saw in the case of the homogeneous atmosphere. On the other hand, if the lapse rate were such that the temperature increased steadily with altitude (i.e., Γ negative), there can be no upper limit! U.S. Standard Atmosphere For operational meteorological or aeronautical calculations involving pressure, temperature, density, etc., at various altitudes, it is not always possible to predict in advance precisely what conditions will be encountered in a real-life situation. As we have seen, the vertical temperature structure of the atmosphere, and hence its density and pressure profiles, may vary markedly from day to day, season to season, and from location to location. Nevertheless, it is often necessary to assume something about the typical characteristics of the atmosphere, even if you know that, in any real situation, your assumptions may represent only crude approximations to reality. It is for this reason that the U.S. Standard Atmosphere was created to provide a standard basis for computing the typical or average values of operationally significant atmospheric variables. This standard atmosphere was computed by the U.S. Weather Bureau at the request of the National Advisory Committee for Aeronautics (NACA). It is meant to represent normal conditions over the United States ast 40 N. The following are the basic specifications of the U.S. Standard Atmosphere up to an altitude of 32 km: (1) The surface temperature is 15.0 Cand the surface pressure is mb (or hpa). (2) The air is assumed to be dry and to obey the ideal gas law. (3) The acceleration of gravity is assumed to be constant and equal to m s 2. (4) From sea level to 11.0 km the temperature decreases at a constant lapse rate of 6.5 Cper km. This region is the troposphere. (5) From 11.0 km to 32 km the temperature is constant at 56.5 C. This region is the stratosphere. Note that the U.S. Standard Atmosphere has constant lapse rate within each of the spheres (troposphere, stratosphere, etc.). Thus, the relationships (3.40) and (3.41) applies within those layers, provided only that 21

23 one remembers (in the case of the stratosphere and higher layers) to substitute the pressure and temperature at the bottom of the layer in question, and substitute z z bottom for z. The following table gives the geopotential heights of a few standard pressure levels in the U.S. Standard Atmosphere: Pressure [mb] Geopotential Height [m] Calculation of Standard Pressure Levels for Actual Profiles The hypsometric equation which allows us to calculate the thickness of the atmospheric layer between two pressure levels, given an arbitrary (virtual) temperature profile T (p) between the two levels. The applications of this are obvious: if you know the surface pressure at a certain station and you are able to obtain temperature profile from a balloon sounding, then you can easily calculate the thickness of each conveniently sized layer from the surface to the top of the sounding and then add them all up to get the height (usually in geopotential meters) of each of the so-called mandatory pressure levels 1000, 925, 850, 700, 500, 400, 300, 250, 200, 150, 100 mb, and so on. Compiled from all the upper air stations around the country (or the world), these heights are used to produce the widely used constant pressure maps (also known as upper air charts), on which the contours represent lines of equal altitude for the specified pressure level. The 500 mb map in particular is one of the most important analysis products utilized by forecasts, though other pressure levels also have important roles. The height of these pressure levels is only meaningful if referenced to a common standard. Therefore, one must add the station elevation to the heights calculated from the hypsometric equation in order to get the height of a given pressure level above sea level. For more info about upper air constant pressure charts, consult the following web page: // In addition to the constant pressure charts, one of the most important operational meteorological products used in forecasting is the surface pressure map. This is the kind of map you most often see in the newspaper with the cold and warm fronts drawn in, as well as the position of lows and highs. The lows and highs refer to relative minima and maxima in the surface atmospheric pressure. Since altitude has a strong effect on the pressure actually measured at a meteorological station, the pressures on a surface map must again be referenced to a common altitude (sea level) in order to be meteorologically meaningful. Otherwise, you would always find a deep (but meaningless) pressure low over the rocky mountains and other regions of high terrain. To calculate the sea level pressure p 0 from the observed station pressure, one need only set z in the hypsometric equation equal to the elevation h of the station above sea level and solve for the pressure at the bottom of the hypothetical layer of the atmosphere whose top is at the station pressure p s : h = R T g log [ p0 p s ] (3.42) 22