The atmospheric boundary layer: Where the atmosphere meets the surface Utrecht Summer School on Physics of the Climate System

C. H. Tijm-Reijmer

Institute for Marine and Atmospheric Research, Utrecht University, the Netherlands The atmospheric boundary layer Figure 2: Potential temperature pro- file of the clear weather ABL over land during night and day.

Introduction

The atmospheric boundary layer (ABL) is defined as that part of the atmosphere that ex- changes information with the surface on short time scales (less than 1 hour). It is the part of the atmosphere in which we live. ture, wind speed and depth following absorption of solar radiation at the surface. Strong A very important feature of the ABL is that close to the surface of the Earth, the atmo- heat and moisture exchange occurs at the surface. The depth of this ABL varies from 200 spheric flow becomes turbulent and breaks up in eddies. These eddies effectively transport m at night ("stable ABL") to 1500 m at day ("mixed layer"). The neutral ABL over land momentum, heat and moisture in the vertical. The main reasons for turbulence to develop or "Ekman layer" is characterised by the fact that clouds and strong winds prevent strong in the ABL is vertical wind shear and negative buoyancy (d/dz < 0). Through friction at daily cycle. Turbulent heat and moisture exchange at the surface is small. The Ekman ABL the surface, the wind becomes zero at the surface, which makes the flow hydrodynamically acts merely as a friction layer, exchange of momentum. The depth is typically 1000 m. unstable. This type of turbulence is denoted by mechanical produced turbulence. Negative The marine ABL is characterised by a small daily cycle, which is due to mixing in the upper buoyancy (d/ dz < 0) developes through heating of the Earth surface by absorption of ocean layers transporting absorbed solar energy away from the surface. In this ABL the solar radiation (Bouyancy turbulence). roughness is variable (waves). Potentially strong advection effects may occur because of Turbulence is destructed by a stable stratification (positive bouyancy, d/dz > 0). This the long response time of the water surface temperature and because of the strong land- occurs in case of strong surface cooling, i.e. night-time and clear sky. These processes ocean temperature contrast. Otherwise this ABL is comparable to the Ekman ABL. We will are referred to as mechanical production (shear production) of turbulence and buoyancy not specifically treat this ABL in this course. More information on turbulence and different production/destruction of turbulence, respectively. Turbulence is dissipated at the smallest types of boundary layers can be found in Garratt (1992). length scales through viscosity of the air. There are several different types of boundary layers, characterised mainly by stability. We may distinguish: The clear weather ABL over land 1) the clear weather ABL over land, Figure 1 illustrates the development of the clear weather ABL over land. From sunrise the 2) neutral ABL over land (Ekman layer), and Sun starts to heat the surface, and vigorous vertical mixing occurs through strong surface 3) the marine ABL. heating resulting in a daytime mixed ABL. The turbulence is bouyancy and mechanically The clear weather ABL over land is characterised by an outspoken daily cycle in tempera- produced. The maximum depth is 1500 m, topped by a capping surface inversion. After sunset the surface heating disapears⇠ and the nighttime stable ABL developes. The surface cools through longwave cooling resulting in a stable stratification. Turbulence is now only generated by wind shear and desctructed by bouyancy. The ABL only slowly grows in its depth is poorly defined.

Figure 3: The surface energy balance.

Figure 1: The daily cycle in the clear weather ABL over land.

1 2 Unstable stratification (daytime)

Use the following values in Eqs. 1 and 2: 3 = 1.2 kg m 1 V =5ms = 295 K, sr = 300 K 1 1 q = 13 g kg , qsr = 21 g kg (saturated) Ch = 0.001. Figure 4: The diurnal cycle in the surface energy balance 2 2 SHF becomes -25 W m and LHF becomes -120 W m . Note that qsr is the saturation value at 300 K. for a sunny summer day over a moist, dark land surface at middle latitudes. Stable stratification (nighttime)

Use the following values in Eqs. 1 and 2. Compared to the daytime case wind speed is lower and the surface is cooling: 3 = 1.2 kg m 1 V =3ms = 295 K, sr = 290 K 1 1 Figure 2 illustrates the daily cycle in the potential temperature profile. During daytime q = 13 g kg , qsr = 11 g kg (saturated) Ch = 0.0005. the ABL is mixed. The potential temperature profile is constant except for the near surface 2 2 SHF becomes +9 W m and LHF becomes +9 W m . Note that qsr is the saturation value at 290 K. layer, where it is absolutely unstable. From the surface layer air parcels may rise unim- peded to the ABL top where they meet potentially warmer air. During night the ABL is Frame 1: Turbulent fluxes in unstable (daytime) and stable (nighttime) stratification. stably stratified. The surface layer is also stably stratified. Buoyancy destructs turbulence and wind shear must keep the mixing going. As a result the ABL is only slowly growing. of the air and the roughness of the surface. A typical value is 0.001. Frame 1 present two examples for typical values of the heat fluxes in stable and unstable conditions. Exchange of heat and moisture: the surface energy balance The examples in frame 1 and Figure 4 show nighttime fluxes smaller and opposite of sign compared to daytime fluxes. This is often the case. In general the turbulent heat From the above it is clear that the structure of the ABL very much depends on what happens fluxes are smaller than the radiation fluxes. However, annually averaged, they represent at the surface, heating by shortwave radiation, cooling by longwave radiation, heating or a net energy loss for the surface and are therefore important (see Figure 5). In areas cooling by the turbulent heat fluxes. In principle, all energy fluxes through the surface where rainfall exceeds evaporation, surface air is saturated and heat loss through LHF should balance (E = 0), unless energy is used for phase changes of the surface (like melt). (evaporation) dominates the turbulent heat exchange. This energy balance can be written as: The global distribution of SHF and LHF in summer and winter is illustrated in Figure 6. LHF is mostly negative in summer as well as in winter, i.e. evaporation occurs extract- E = Sh (1 )+L +L +SHF + LHF + G, # # " ing heat from the surface. For evaporation to occur moisture must be available. LHF is and is illustrated in Figure 3. Here, the fluxes are defined positive when directed towards therefore largest over the warm ocean in the tropics. Moisture is also available in the polar the surface. The sensible heat flux (SHF) is the result of temperature difference between atmosphere and surface, while the latent heat flux (LHF) describes the heat flux though exchange of moisture between surface and atmosphere, i.e. condensation, evaporation. G is the sub surface energy flux describing transport of heat from or too deeper soil layers. The sensible (SHF) and latent (LHF) heat fluxes at the surface can be expressed as (in 2 Wm ):

SHF = cpChV( sr ) (1) LHF = LChV(q qsr ) , (2) where is the near-surface air density, cp is the specific heat of air at constant pressure 1 1 6 1 (=1005 J kg K ) and L is the latent heat of vaporization (2.5 10 J kg ). Furthermore, V is the absolute wind speed, which is a measure for wind shear⇥ near the surface, i.e. the mechanical turbulence production. In this expression the heat fluxes are proportional to the difference between atmospheric (subscript ) potential temperature () or specific humidity (q), and surface (subscript sr) or q. The atmospheric value is usually a near surface value typically measured at 2 m in e.g. a Stevenson screen. Finally Ch is the bulk Figure 5: The heat balance of the Earth-atmosphere system, globally and annually averaged. heat exchange coefficient for (sensible and latent) heat. It still depends on the bouyancy

3 4 a. b.

Figure 7: Vertical profiles of potential temperature illustrating the entrainment process (left) and the encroachment process (right).

boundary layer. The heat transported into the layer by SHF is completely used to increase the temperature over the complete layer (conservation of energy): c. d. dm SHF = . (3) Figure 6: Global distribution of LHF (a,b) and SHF (c,d) for June, July, August (a,c) and December, dt cphm January, February (b,d) (in W m 2). Here, m is the vertically integrated mixed layer potential temperature. This change of temperature decreases the temperature inversion by dm/dt. Entrainment will add en- regions as snow and ice. Evaporation is however small due to the low temperatures. Desert ergy to the boundary layer which is not used to heat or cool but to increase its height hm. regions are also clearly visible as regions with low amount of evaporation. Temperature is When hm increases, the temperature inversion will increase as well. The increase depends high enough for significant evaporation, there is however no moisture available. Over the on the lapse rate above the boundary layer. This results in an expression for the change in large land masses a clear annual cycle is visible with higher vales in LHF in summer and temperature inversion over time: lower values in winter. SHF is positive in winter and negative in summer. Negative values indicate strong heat- d dm d dhm dm dhm = + = + ing of the surface by solar radiation resulting in unstable conditions in which the surface dt dt dz dt dt dt heats the atmosphere. There is a strong annual cycle over land in areas with seasonal Combining these two results and rewritting the expression to obtain an expression for the snow cover. In winter, when the surface is snow covered large part of the solar radiation change in boundary layer height due to entrainment results in: is reflected by the surface and not used to heat the surface. The atmosphere is relatively dhm 1 d SHF warm and sensible heat transport is directed towards the surface. In summer, the snow has = + (4) dt dt c h dissapeared and the surface warms up, heating the atmosphere from below.  p m Note that in these expression SHF is defined positive upwards. Growth of the mixed layer Encroachment During daytime the clear weather ABL over land is well mixed resulting in the potential temperature profile in Figure 2b. After sunrise, the Sun starts to heat the surface and SHF Encroachment is the growth of the boundary layer arising purely from surface heating alone. will become negative, i.e. directed upwards, heating the atmosphere from below. The It takes place in the absence of a capping inversion (Figure 7). As a result the increase in temperature increases and the mixed layer starts to grow. The growth of the mixed layer height of the ABL Eq. 4 reduces to: occurs at the top of the boundary layer and is described by entrainment and encroachment. dhm SHF = dt cphm Entrainment Encroachment results in a slower increase of the depth of the ABL compared to entrainment. Entrainment is the exchange of air over a capping inversion from above to within the bound- ary layer. This is usually the means by which an atmospheric boundary layer topped by a The neutral ABL over land (Ekman or friction layer) capping inversion increases its height. With the exchange of air over the capped inver- sion, heat is transported into the boundary layer. First assume no increase in height of the In the neutral ABL (d/dz = 0) no exchange of heat takes place with the surface, only exchange of momentum, hence the term friction layer. The exchange of momentum at the

5 6 Reynolds decomposition is a mathematical technique to separate the average and fluctuating parts of a quantity.

For example describe two quantities X and Y as an average (X) plus high frequency fluctuations (0):

X = X + 0 ,Y= Y + y0 The overbar denotes the time average, the prime denotes the fluctuations. Some calculus rules that apply on this description are:

0 = 0 , the time average of the fluctuations is zero. X = X , the time average of the mean equals the mean. cX = cX , the time average of a constant times the mean equals the constant times the mean. Furthermore: And: (XY)=(X + 0)(Y + y0) (XY)=XY = XY + Xy0 + 0Y + 0y0 Xy Xy 0 Figure 8: The effect of friction on a geostrophic flow. ( 0)= 0 = = (XY)+(Xy0)+(0Y)+(0y0) = XY + 0 + 0 + ( y ) surface, friction, results in a decrease in the flow speed. In addition, friction results in a 0 0 = XY + ( y ) turning of the wind vector, e.g. a geostrophic flow along isobars is deflected in the direction 0 0 Frame 2: Reynolds decomposition. of the low pressure. This follows from the balance of forces in a stationary flow. In the case without friction the pressure gradient and coriolis force balance for a geostrophic flow. In the previous the transport of momentum / impuls is described by a linear relation Adding friction distroys this balance (Figure 8a). To restore the balance the flow is deflected between friction and flow speed. A more sophisticated way of quantifying the transport of towards the low pressure (Figure 8b). momentum (impuls) is by looking at the contributions to the wind variation of signals with Using the momentum equations: various frequencies. This can be done by describing the wind (and other variables) as a d 1 p mean plus high frequency variations due to turbulence. This technique is called Reynolds = + ƒ + friction dt decomposition and described in Frame 2. Reynolds decomposition in boundary layer me- d 1 p teorology embodies substituting the mean plus a high frequency variation term in the con- = ƒ + friction (5) dt y servation equations (Eqs 5) and subsequently taking the time average of the resulting an expression for both wind components can be derived. Assume that friction is linearly expressions. The resulting Reynolds decomposed momentum equations are: dependent on wind speed (k,k), assume the flow is stationary (d/dt = 0) and assume a d 1 p (00) = + ƒ pressure gradient only in the -direction: dt z 1 p d 1 p (00) 0 = + ƒ k = ƒ (6) dt y z 0 = ƒ k The last term on the right hand side in both equations is the friction term, or the vertical The minus sign results from the fact that friction allways acts parallel and opposite to the divergence of turbulent impuls. Note that the left hand side of both equations is the total flow. Combining these two equations result in an expression for the and -component of derivative and therefore still includes advection of the mean flow: the wind speed: d + + + . k p ƒ p dt ⌘ t y z ƒ + = and = = k2 + ƒ 2 k k2 + ƒ 2 In case of Figure 8 p/ > 0, which results in < 0 and > 0. Thus, wind velocity obtains a component in the direction of the low pressure, filling up the low pressure system and draining the high pressure system (Figure 9). Note also that in case friction is 0, (k = 0), = 0.

Figure 10: Left, mean wind profile. Right, the corresponding profile of the turbulent impuls flux.

Figure 9: Frictional convergence and divergence.

7 8 The momentum equations without friction read:

1 p + + + = + ƒ t y z 1 p + + + = ƒ t y z y

Substitute: p = p + p0, = + 0, = + 0, = + 0, assume to be constant and average over time. Make use of the calculus rules from Frame 2 and the different terms become:

( + 0) = = Figure 11: The solution for the Ekman layer on the North- t t t ern Hemisphere. ( + 0) = ( + ) 0

0 0 = + + + 0 0

= + 0 0

ƒ = ƒ ( + 0)=ƒ + ƒ 0 = ƒ + 0 = ƒ

The other terms are derived similarly. The term in the box is new and appears for each advection term. The momentum equations for turbulent flow now become: The Ekman layer

d 0 0 0 1 p + + + = + ƒ An interesting application of Eq. 6 is the Ekman layer. The Ekman layer describes a neutral dt 0 0 y 0 z boundary layer in which to a first approximation the Ekman solution for wind speed applies. d 0 0 0 1 p + 0 + 0 + 0 = ƒ To describe the Ekman solution we assume stationary flow (d/dt = 0). We use the notation dt y z y for geostrophic flow V~g =(g, g): The new turbulent terms can be rewritten by noting that: 1 p 1 p 0 0 0 0 , , (7) = + , which reslts in : g = g =+ y 0 y 0 y ƒ y ƒ and choose the axis of our reference frame such that the geostrophic flow above the bound- 0 0 0 00 00 00 0 0 0 + + = + + + + ary layer is along the -axis ( 0). In Frame 4 it is shown how the following expression 0 0 y 0 z y z 0 y z g =  for the two wind components (z) and (z) can be derived: 00 = z (z)=g 1 exp z ƒ/2K cos z ƒ/2K In the last step two assumptions are made. Firstly we assume incompressibility for the wind speed fluctuations, î Ä ä Ä äó and secondly, we assume horizontal homogeneity in the ABL: (z)=g exp z ƒ/∆2K sin z ƒ/2∆K (8) î Ä ∆ ä Ä ∆ äó 0 0 0 00 00 with K the eddy diffusivity, assumed constant with height. These equations show that in + + = 0 , nd = = 0respectively. y z y case K = 0, then = 0 and = g, i.e. pure geostrophic flow. Furthermore, (z) = 0 at The end result is Eq. 6. z / ƒ/2K 1 km in case K =5m2 s 1. And finally at z = 0 the wind vector points 45 = ⇠= degrees west of the geostrophic wind. Figure 11 illustrates this solution. Frame 3: Reynolds decomposition of the momentum equations (Eq. 5). p In reality the Ekman solution is very rare. This is partly because the turbulent momentum The derivation of Eq. 6 is given in Frame 3. fluxes are only to a first approximation proportional to the gradient in the mean momentum How to interpret this friction term in the momentum equations (Eq. 6). Figure 10 shows (fist order closure), and partly because the eddy viscosity K is not constant with height. on the left a mean wind speed profile. A parcel with mean wind speed is brought upwards, Especially near the surface K must vary rapidly. thus 0 > 0. At the higher level the horizontal wind speed from this parcel is smaller than The nocturnal low level jet the mean at that level, thus 0 < 0. Combined, (00) < 0. Similarly a parcel with mean wind speed is brought downwards, thus 0 < 0. At the lower level the horizontal wind An interesting feature related to the rotation of the wind close to the surface due to friction speed from this parcel is larger than the mean at that level, thus 0 > 0. And again com- effects as described in the Ekman layer, is the nocturnal low level jet (LLJ). The LLJ developes bined, (00) < 0. With increasing altitude the vertical difference in wind speed becomes due to the development of an inertial oscillation in the remains of a convective boundary smaller, thus (00) less negative resulting in (00)/z>0. layer after sunset. After sunset dry-convective turbulent mixing almost ceases and a stable boundary layer starts to develope from the surface. As a result, in the remnants of the

9 10 Assume stationary flow (d/dt = 0), use the notation for geostrophic flow (Eq. 7) and choose the axis of the reference frame such that the geostrophic flow above the boundary layer is along the -axis (g = 0). The momentum equations (E. 6) then reduce to:

(00) (00) 0 =+ƒ , 0 = ƒ ( g) z z where we have left out the average bars over the independent variables. We now have a set of two equations with four unknown variables , , (00) and (00). To close this system of equations we must find an expression for (00) and (00) in terms of and . A first order closure assumption is to assume that the (00) and (00) are linearly related to the vertical gradient in and : K , K ( 0 0)= ( 0 0)= Figure 12: Illustration of the rotation of the wind vector resulting in a low level jet. z z with K the eddy diffusivity, assumed constant with height. Substitute in the momentum equations:

2 period of the oscillation is P equals 2/ƒ, which is about 15 hours for The Netherlands. 0 =+ƒ + K =+ƒ + K z z z2 Depending on the initial angle between (0, 0) and the geostrophic wind (g, g), the 2 maximum wind speed is reached about 6 to 8 hours after sunset (for The Netherlands). 0 = ƒ ( g)+ z K z = ƒ ( g)+K z2 1/2 Combine these two equations by introducing the complex velocities Vˆ + and Vˆg g + g with =( 1) , ⌘ ⌘ Assume no friction, use the notation for geostrophic flow (Eq. 7) and choose the axis of the reference frame such then multiply the second equation by and add the first equation. Note that 1/ = . The result is: that the geostrophic flow is along the -axis (g = 0). The momentum equations (E. 6) then reduce to: 2Vˆ ƒ 0 Vˆ Vˆ = 2 ( g) d z K =+ƒ dt ˆ ƒ ƒ A general solution of this equation is: V(z)=A exp K z + B exp K z + C d = ƒ ( g) År ã Å r ã dt The boundary conditions require that at the surface (z = 0) = 0 and = 0 thus Vˆ(0)=0, while far from the We now have a set of two equations that describe the evolution in time of the wind components , . Combine ground (z ) g, g 0 and Vˆ Vˆg g. Substitution of the general solution in the equation results = = = ( )= = 1/2 ! these two equations by introducing the complex velocities Vˆ and Vˆg g g g with 1 , then in C g, and the boundary conditions provide (check), A 0 and B g. The solution now becomes: + + = =( ) = = = ⌘ ⌘ 2 multiply the second equation by and add the first equation. Note that 1/ = and = 1. The result is: ƒ ˆ dVˆ V(z)= g exp z + g ˆ ˆ 0 v K 1 = ƒ (V Vg) u dt t @ A ˆ The last step is to separate the Imaginary and Real part of the exponent in order to find a solution for (z) separate A general solution of this equation is: V(t)=A exp ( ƒt) + B from (z). Make use of the fact that exp( )=cos sin and p =( + 1)/p2. The solution now becomes The boundary conditions require that at t = 0 = 0 and = 0, thus Vˆ(0)=Vˆ0, and for large t = g and = g (check): thus Vˆ(t)=Vˆg = g. Substitution of the general solution in the equation results in B = g. The boundary conditions ƒ ƒ ƒ provide (check), A = 0 g + 0 The solution now becomes: Vˆ(z)=g 1 exp z cos z sin z 2 0 v 2K 1 0 v 2K v 2K 13 Vˆ t exp ƒt u u u ( )= 0 g + 0 ( ) + g t t t 4 @ A @ A5 The last step is to separate the Imaginary and Real part of the exponent in order to find a solution for (t) separate of which (z)=Re(Vˆ), the Real part and (z)=m(Vˆ), the Imaginary part, given in Eq. 8. from (t). Make use of the fact that exp( )=cos sin . The solution now becomes (check): Frame 4: The Ekman layer. Vˆ(t)= 0 g + 0 (cos ƒt sin ƒt) + g of which (t)=Re(Vˆ), the Real part and (t)=m(Vˆ), the Imaginary part, given in Eq. 9. convective boundary layer, friction is no longer a significant force. The wind in this remnant layer starts to rotate to adjust itself to the geostrophic flow. In the absence of other forces, Frame 5: The Low Level Jet. an inertial oscillation developes. To describe the oscillation and the LLJ we use the momentum equations 5 assuming friction to be 0 and the notation for geostrophic flow (Eq. 7). As in the decription of the Ekman layer we choose the axis of our reference frame such that the geostrophic flow is along the -axis (g = 0). In Frame 5 it is shown how the following expression for the two wind components (z) and (z) can be derived:

(z)=g + V0 sin ƒt+(0 g) cos ƒt (z)=0 cos ƒt (0 g) sin ƒt (9) with (0, 0) the wind vector at t = 0, the moment friction force becomes neglegible. The

11 12 Avogadro’s number The mass of a gas is often expressed in terms of the molar (or molecular) mass, 23 which is the number of grams in one mole of gas. One mole is defined by Avogadro’s number: NA 6.02 10 molecules per mole gas. This number of molecules in one mole of gas is defined such that one mole⇡ of gas is the amount of molecules needed to obtain a mass in grams equal to the molecular mass of a certain gas. For Appendix A example, the molecular mass of Carbon (C) is 12, and therefore, 1 mole = 6.02 1023 molecules of 12C contains 12 g of mass.

Avogadro’s law states that at given pressure and temperature, the same number of molecules of different Physical properties of the atmosphere gases occupy the same volume, independent of their mass. Under standard atmospheric conditions: Pressure p0 = 101300 Pa = 1013 hPa = 1 atm

Temperature T0 =0C = 273.16 K 3 this volume (V0) is found to be 22.4 m for 1 kmole of molecules.

Introduction Newton’s first law of motion pertains to the behavior of objects for which the forces acting on it are balanced. It states that an object at rest tends to stay at rest and an object in motion tends to stay in motion with A first step in understanding the Atmospheric boundary layer is understanding the physical the same speed and in the same direction unless acted upon by an unbalanced force. properties of the Earth’s atmosphere, i.e. its composition and structure. We will first discuss The first law of thermodynamics is the application of the conservation of energy principle to heat the dry atmosphere, then extend the description to include moisture. and thermodynamic processes. The change in internal energy of a system (dE) is equal to the heat added to the system (dQ) minus the work done by the system (dW).

dE = dQ dW (A.1) General structure of the atmosphere The lowest 150 km of the atmosphere is divided in the thermosphere (>90 km), the meso- Dalton’s law states that the total pressure exerted by a mixture of gasses is equal to the sum of the sphere (50 to 90 km), the stratosphere (15 to 50 km), and the troposphere (to 15 km) (Fig- pressures that would be exerted by the gasses if they alone were present and occupied the total volume: p = p. ure A.1). The daily variations in weather mainly occur in the troposphere. The different P layers are separated by the mesopause, the stratopause and the tropopause, respectively. Frame 6: Fundamentals. These are the regions where the sign of the temperature gradient changes. In the troposphere the temperature generally decreases with altitude. Solar radiation reaches the Earth surface. Above the stratosphere the air becomes extremely thin and in is absorbed at the surface and heats the atmosphere from below. In the stratosphere the thermosphere the air temperature may vary between 200 and 2000 K, depending on the temperature increases again, which is due to the absorption of solar radiation, mainly the solar activity. Above the stratopause the vertical exchange is suppressed. in the ozone layer at about 20 km altitude. This ozone layer is also very important for life on Earth because the layer strongly reduces the amount of ultraviolet radiation that Atmospheric density and temperature

To describe the state of the atmosphere in terms of pressure and temperature we make use of equations of state. In general, for a given substance or mixture of substances, equations Table A.1: Composition of the atmo- of state attempt to describe the relationship between temperature, pressure, and volume sphere (lowest 100 km) of that substance. One of the most simple equations of state is the ideal gas law. The ideal

Gas Molar Molar Molar gas law is reasonably accurate for gasses at low pressures and high temperatures, but it mass (volume) mass becomes increasingly inaccurate at higher pressures and lower temperatures. Gasses that (M) fraction fraction behave according to the ideal gas law are called ideal gasses. The atmosphere is a mixture (M/Mtot) of gasses and behaves to a good approximation as an ideal gas. For a single component H2O 18 <0.030 gas the ideal gas law is given by: N2 28 0.7808 0.7552 O2 32 0.2095 0.2315 Ar 40 0.0093 0.0128 p = RT , CO2 44 0.0003 0.0005 3 where p is pressure in Pa, density in kg m , R is the specific gas constant for the single 1 1 component gas (in J kg K ) and T is temperature in K. However, the atmosphere is a mixture of gasses. Table A.1 presents the most important permanent gasses present in the atmosphere with some of their properties, such as the mo- Figure A.1: Schematic view of the atmospheric tempera- lar mass M, molar (volume) fraction and molar mass fraction. Each gas in the atmosphere ture structure. behaves to a good approximation as an ideal gas. Using the Law of Dalton (p = p), the P 13 14 Consider a column of air between two vertical levels with pressures p at z, and A single component gas: The ideal gas law describes p of a single component gas in terms of T and : p dp at z dz (note dp < 0) (see Figure). In absence of a vertical acceleration + + nR mR Newton’s first law requires that the forces acting on the column balance, i.e. the p = T = T = RT . V MV total vertical pressure force balances the downward gravitational force (downward p pressure Pa force is defined negative, and note that pressure is force per unit surface, Pa = m Nm 2): n = M number of moles kmol m mass kg 1 pdA (p + dp)dA mg = 0 M molecular weight kg kmol 1 1 (p p dp)dA = mg R universal gas constant J K kmol 3 V volume m dpdA = dVg T temperature K dpdA = dzdAg 3 = m/V density kg m 1 1 dp = dzg R = R /M specific gas constant J kg K dp R follows from the standard conditions (see Avogadro’s law in Frame 6): p0 = 101300 Pa, T0 = 273.16 K, V0 = = g dz 22.4 m3, n = 1 kmol, substituted in a rearanged ideal gas law: p V pdA is the pressure force acting on the column from the air below, dA is the surface area of the column, p 0 0 3 1 1 ( + R = = 8.314 10 JK kmol . dp dA is the pressure force acting on the column from above, mg is the gravitational force, g is the gravitational nT0 ⇥ ) acceleration, m is mass, is the air density, dV the volume of the column, dz de height of the column defined positive in the upward direction. A mixture of gasses: Dalton’s law (p = p) is used to derive the equation of state. Assume that each The result is the hydrostatic balance, Eq. A.3. gas (i) is characterized by mass m ( m /V), pressure p and gas constant R , then the ideal gas law for a = P mixture of gasses (excluding water vapor) becomes: Frame 8: The hydrostatic balance.

p = p

XmR = T The hydrostatic balance P V m = T R The air is not evenly distributed over the vertical (and horizontal) extent of the atmosphere, mtot X but the amount of air decreases with increasing altitude. Air pressure is defined as the = RdT, force acted on a unit surface by the air and it decreases upwards in the atmosphere. If we mtot where mtot m, = V , and Rd is the gas constant of dry air, which can be written as: ⌘ assume a homogenic isothermal atmosphere without motion, the upward pressure gradient m m P R M force is balanced by gravity. The resulting balance is known as the hydrostatic balance and Rd R = R = . ⌘ mtot mtot Md basically states that air pressure is the weight of the column of air above you. Hydrostatic X X with Md the weighted average molecular mass of dry air. balance requires an isothermal (constant temperature) motionless fluid. This means that the atmosphere never quite satisfies hydrostatic equilibrium. But for most atmospheric Frame 7: The ideal gas law. motions it is a very good approximation. It only gives problems in areas where there are ideal gas law for a single gas can be rewritten for a mixture of gasses. The equation of state strong vertical accelerations, like in thunderstorms. then becomes: The hydrostatic balance is derived from Newton’s first law (Frame 6) with pressure and gravity acting as the balancing forces (Frame 8). The hydrostatic balance is given by: p R T, (A.2) = d dp g. (A.3) = where Rd is the gas constant of dry air, which is the universal gas constant (R ) divided by dz the molecular weight of dry air (see Frame 7). This end result shows that the ideal gas law According to the equation of state, pressure depends on the density and temperature of for the dry atmosphere is similar to that for a single gas if R is replaced by Rd as defined in the gas. Therefore, the combination of the equation of state for a dry atmosphere (Eq. A.2) Frame 7. and the hydrostatic balance (Eq. A.3) provides an expression which can be used to derive The weighted average molecular mass of dry air (Md) based on the four major con- the vertical variation in pressure (Frame 9): stituents of the dry atmosphere, Nitrogen (N2), Oxygen (O2), Argon (Ar) and Carbon dyox- g z dz ide (CO2), can be calculated using the numbers for the molecular mass and mass fraction p(z)=p0 exp R T z given in Table A.1, and is 28.9. Given this value the gas constant of dry air (Rd = R /Md) is  d Zz=0 ( ) 1 1 287.05 J kg K . Furthermore, we can also calculate the density of dry air () if we know This relation shows that pressure decreases exponentially with height. Assuming an isother- the air pressure and air temperature. The next step is to look at the vertical distribution of mal atmosphere (constant T) this equation can be solved: temperature and pressure. zg p(z)=p0 exp , R T ✓ d ◆

15 16 p Combine the eqtion of stte : p = RdT = Use of the first law of thermodynamics (Eq. A.1, Frame 6) and describe each term in the equation. ! RdT dp dW The work done by a parcel to expand is defined by the force used by the parcel to expand times the distance nd the hydrosttic blnce : = g dp = gdz . dz ! the parcel expanded, or pressure p times volume change dV. Assume that V is the volume of a unit mass of 1 kg, dp gdz which is the specific volume , and is per definition equal to the inverse of the density = 1. to obtin : = p RdT 1 dW = pd = pd Integrate this expression between the surface z=0 (p = p0 at z = 0) and a given level z (p = p(z) at z) to obtain an expression for the vertical variations in pressure as a function of the vertical variations in temperature: 1 In the atmosphere d or d is difficult to determine, therefore we express dW in terms of parameters we can easily measure: temperature change dT and pressure change dp. Use the equation of state (Eq. A.3) for a parcel p(z) dp z g dz = with unit mass (p = RdT). Partially differentiate: p R T z Zp0 Zz=0 d ( ) dp pd R dT g z dz + = d ln(p(z)) ln(p0)= R T z and substitution in the above equation for dW results in: d Zz=0 ( ) p z g z dz ( ) dW pd R dT dp ln = = = d p R T z ✓ 0 ◆ d Zz=0 ( ) g z dz dIE is written in terms of the heat capacity (c) of the parcel, which is the amount of energy necessary to p(z)=p0 exp R T z  d Zz=0 ( ) change the temperature of the parcel, or the change in internal energy with a change in temperature at constant volume: b d b Note tht: = ln nd ln ln b = ln dE b c Z dT ⌘  dV=0 where c is the heat capacity at constant volume. Thus, the change in internal energy d E c dT. Substitute Frame 9: The vertical pressure distribution. ( )= this, and dW in Eq. A.1 results in: which is called the barometric equation. An example of the barometric relation is given in dQ = c dT + RddT dp = cpdT dp , (A.4) where c c R , the heat capacity of the air parcel at constant pressure (for air: c = 717 J kg 1K 1, c = Figure A.2 for an isothermal atmosphere of -18C. The relation can also be solved for a linear p = + d p 1004 J kg 1K 1). decreasing temperature (see exercises). Furthermore, the term RdT/g in the barometric equation provides a scaling height for pressure. A scaling height is the height at which a dQ Assume the process is adiabatic, i.e. there is no heat exchange between parcel and surroundings, dQ = 0. parameter has decreased by factor 1/e. For an atmosphere with an average temperature dT 1 of 15C, the scaling height for pressure is about 8.4 km. 0 = cpdT dp = = ! dp cp cp

The dry-adiabatic lapse rate (d) of air is now given by: The dry adiabatic lapse rate dT dT dp g g d = = = = . dz dp dz c c  dQ=0  dQ=0 p p In the previous section the assumption is made that the atmosphere is isothermal. This is in which we make use of the hydrostatic balance, Eq. A.3. in reality not the case, temperature varies with elevation. For a dry atmosphere, the dry- Frame 10: The dry adiabatic lapse rate.

adiabatic lapse rate describes the rate of temperature change with elevation. Using the first law of thermodynamics and the assumption that there is no heat exchange between parcel and surroundings results in the dry-adiabatic lapse rate d (Frame 10): g 1 d = = 9.8Kkm , cp with g the gravity acceleration and c the heat capacity of air at constant pressure (1004 J kg 1K 1). Figure A.2: Vertical pressure distribution as- p suming a constant atmospheric temperature of There are two general restrictions in the use of the dry-adiabatic lapse rate: 1) no phase changes are allowed that lead to internal heat production or loss, and 2) heat exchange -18C (red curve) and assuming a surface tem- with the surroundings must be small compared to heat loss or gain due to expansion or perature of 15C and a constant lapse rate of 1 -0.006 K m (blue) (see exercises). compression. Therefore, d is only applicable for processes involving fast vertical displace- ments. Using the dry-adiabatic lapse rate we can study the static stability of the unsaturated T (dry) atmosphere. To do so, we compare the observed atmospheric lapse rate = z with 1 the dry-adiabatic lapse rate of d = -9.81 K km of a fictitious air parcel that we move

17 18 This equation calculates the temperature an air parcel would have when brought adiabati- cally from a location p with temperature T to a reference pressure p0. p0 is usually taken to be 1000 hPa or the surface pressure. Furthermore, is a conserved quantity for adiabatic processes, as is shown by taking the ratio of 1 when brought from p1 to a reference level, to 2 when brought from p2 to the same reference level:

R /c p0 d p Rd/cp Rd/cp Rd/cp 1 T1 p T1 p2 p1 p2 1 1 . = R /c = = = 2 Ä p0 ä d p T2 p1 p2 p1 T2 p2 ✓ ◆ ✓ ◆ ✓ ◆ Ä ä Because potential temperature is a conserved quantity for adiabatic processes it can be a b used as an indicator of air masses, because it is insensitive to vertical movements as long Figure A.3: a) Temperature profile, and b) profile of the temperature lapse rate of the profile in a). as no condensation or evaporation takes place. The potential temperature can also be used to study the static stability of the atmo- dry-adiabatically up or downwards. We can distinguish three different situations: sphere. The potential temperature lapse rate is defined as: = z . For dry-adiabatic < d unstable stratification movements the potential temperature lapse rate is 0. The static stability using potential = d neutral stratification temperature is then defined similarly to the temperature lapse rate with 3 different cases: > stable stratification d < 0 unstable stratification In the stable case a parcel dry-adiabatically forced upwards wants to move back to its = 0 neutral stratification original position. In the unstable case a parcel forced upwards will accelerate even further > 0 stable stratification upwards. In this case an air parcel also is able to spontaneously move upwards resulting The next step is adding moisture to our atmosphere. in convection. In reality this happens mainly when the atmosphere is strongly heated from below so that becomes larger than the dry-adiabatic lapse rate. Finally, in the neutral Atmospheric moisture case, the initial| movement| will not change. The parcel will move further without being ac- celerated or decelerated. Figure A.3 presents an example of a temperature profile and its In the previous section the atmosphere was assumed to be dry. However, the atmosphere corresponding lapse rate profile. In the example the observed lapse rate is over almost the is mixture of dry air and water, in solid, liquid and vapour state. The concentration of complete profile larger than the dry-adiabatic lapse rate indicating stable conditions. water in the atmosphere is very variable in space and time, but seldom exceeds several %. Furthermore, compared to the oceans, the atmosphere is only a small water reservoir; Potential temperature only 0.001% of the total amount of water on Earth on average resides in the atmosphere However, moisture in the atmosphere has a big influence on weather and climate through: We will now introduce a variable that is conserved under adiabatic motion, the potential - Its ability to absorb shortwave and longwave radiation. temperature. The potential temperature can also be derived using the first law of ther- - Its ability to deplete heat through evaporation. modynamics (see Frame 11) and is defined as: - Its ability to supply heat through condensation (on average 25% of S0/4). - Its ability to form clouds with their strong radiative properties. R /cp p0 d - The transport of moisture and heat from one place to another. = T . (A.5) p ✓ ◆ Measures of humidity Use the first law of thermodynamics rewritten as Eq. A.4, assuming adiabatic conditions (dQ = 0), and use the equation of state, then the specific volume can be eliminated from Eq. A.4 resulting in an expression in which There are several parameters that express the amount of moisture in the atmosphere, temperature and pressure are the only unknown quantities: e.g. water vapor pressure, absolute humidity, mixing ratio, specific humidity and relative cp dT dp = humidity. It depends on the application which ones are used. Rd T p The water vapor pressure (e, in Pa) is the partial atmospheric pressure which is ex- Integration of this expression between T1 at p1 and T2 at p2 gives: erted by water vapor. It results from the general fact that from a substance in liquid form, R /cp T2 p2 d = some of the molecules will escape into the vapor phase; evaporation occurs. The partial T1 p1 ✓ ◆ pressure due to these molecules per cubic metre is known as the vapor pressure, or in case which can be used to define the potential temperature , Eq. A.5. of water, the water vapor pressure. From this follows that water vapor pressure is propor- Frame 11: The potential temperature. tional to the number of water molecules in the atmosphere, and therefore, water vapor pressure is a possible measure of humidity. Another possible measure of humidity is the

19 20 3 absolute humidity (, in kg m ), which is the mass of water vapor per cubic meter, or using the definition of specific humidity q in terms of mass m: q = m/(m + md). the density of the water vapor. mdRd + mR As the other gasses in the atmosphere, water vapor behaves to a good approximation Rm = md + m as an ideal gas, and using the definitions for water vapor pressure and absolute humidity, 1 the equation of state for water vapor is defined as: = Rd 1 q 1 ✓ ✓ ◆◆ e R T, (A.6) Rd(1 + 0.61q) = ⇡ Substitution of this result into equation A.7 gives: where R is the specific gas constant for water vapor. The value of R follows from the specific gas constant for a single gas, which is defined by the universal gas constant divided p = RmT = Rd(1 + 0.61q)T = RdT , 1 1 by the molecular weight (M) of the gas, thus R = R /MH2O = 462 J kg K where MH2O ⇠ where T =(1 + 0.61q)T, the virtual temperature. The virtual temperature is a fictitious = 18. Note that since the water vapor is lighter than air (mass number 18 vs. 29), an temperature at which a dry parcel, in the same conditions, would have the same density equal-volume mixture of moist air is lighter than dry air. as the moist parcel. Furthermore, this also shows that the ideal gas law for the moist The mixing ratio and the specific humidity are very similar measures of humidity. The atmosphere is similar to that for the dry atmosphere if T is replaced by T as defined mixing ratio (r, in g kg 1) is the ratio of the mass of water vapor to the mass of dry air. It above. can be written as: Using the above result, a new expression for the specific humidity can be derived. Apply e m R T Rd e e the ideal gas law to the wet and dry air fraction separately and use the definition of q in r = = = = = pd terms of density, q = / = /( + d): md d R T R pd p e d e = RT = qRT in which m is the mass of the water vapor and md is the mass of the dry air. Since the volume is the same for both masses, r also equals the ratio of the densities. Substituting pd = dRdT =(1 q)RdT the equations of state for water vapor (Eq. A.6) and dry air (Eq. A.2) then results in an In the next step is eliminated and q isolated: expression in terms of the vapor pressure and total pressure of the volume, where is q eRd e e the ratio of the specific gas constant of dry air to the specific gas constant of water vapor q (A.8) = = ⇠= ( R /R 0.62). The specific humidity (q, in g kg 1) is defined similar to the mixing 1 q pdR pd ! p = d ratio, as the⇡ ratio of the mass of water vapor to the mass of air. It can similarly be written In the last step of this equation two assumptions are made: firstly 1 q 1 and secondly ⇡ as: pd p. The error this introduces to the estimated q does not exceed a few %. The prob- ⇡ lem with this expression is that partial water vapor pressure can not be readily measured. m m r q d r, Therefore, moisture is often measured relative to saturation by determining e.g. the dew = = = = = = m md mtot d 1 1 r + + + d + ⇡ point temperature, the wet-bulb temperature or the relative humidity. To explain what saturation actually is, consider an air parcel in which vapor molecules in which mtot is the total mass of the air package and the density of the moist air. As collide with each other. Some may stick to form tiny water droplets that are also constantly a result the specific humidity can be written in terms of the mixing ratio and to a good breaking apart. If enough vapor molecules are present, there may be enough collisions to approximation they are the same. However, we will use the specific humidity because the form a stable population of liquid water droplets: this is called saturation and the vapor conservation equations are expressed in it. Note that the specific humidity can also be used pressure at this point is the saturation vapor pressure (es). Note that the saying "warm air in the equation of state for water vapor pressure: e = RT = qRT. can hold more water vapor than cold air" is strictly speaking not correct. It is not the air Since water vapor behaves to a good approximation as an ideal gas the equation of state that is holding the water vapor. It therefore does not matter what and if other gases are for a moist atmosphere can now be derived. The moist atmosphere is a mixture of dry air present, the saturation water vapor pressure would not change. and water vapor. Following Dalton’s law the total pressure is the pressure for dry air plus The above mentioned three parameters are defined as follows. The dew point temper- the pressure for water vapor. ature (Td, in K) is the temperature at which after isobaric cooling (at constant pressure) m R m R d d condensation occurs; e reaches its saturation value es and e(T)=es(Td). Note that the p = pd + p = T + T V V specific humidity remains constant during cooling. The dew point temperature is in fact mdRd + mR the minimum temperature that can be reached due to longwave radiative cooling. It can = T m m ✓ d + ◆ be determined by using a cooled mirror. Another temperature based on saturation is the = RmT (A.7) wet-bulb temperature (T, in K), which is the temperature an air parcel can reach when moisture is added till saturation occurs. In other words, the wet-bulb temperature is the Note that =(md + m)/V. From this a specific gas constant for moist air can be derived equilibrium temperature of a surface kept wet in a well ventilated environment. The spe- cific humidity now increases to saturation point qs. A graphic representation of Td and T

21 22 Combine the first law of thermodynamics (Eq. A.1) rewritten as Eq. A.4, with the hydrostatic equation (Eq. A.3) in the form dp = gdz. Add heat from condensation to the parcel, dQ = Lcdqs where dqs is the change in saturation specific humidity qs when heat is added to the parcel.

dQ = cpdT dp Lcdqs = cpdT + gdz ! Figure A.4: Saturated vapor pressure over Now consider dqs by twice using Eq. A.8 and partially integrating: water, ice and their difference as a function es es des dp of temperature. Included is a graphic repre- dqs = d = des dp = qs p p p2 e p ✓ ◆ ✓ s ◆ sentation of the dew point temperature Td, Substitution of the hydrostatic equation in which density is eliminated using the ideal gas law results in: the wet-bulb temperature T and the abso- lute temperature of a parcel. dp g des g = dz Lcdqs = Lcqs + dz p RT e RT ! ✓ s ◆ Substitution of this result in the rewritten form of the first law of thermodynamics and solving it for dT/dz results

in the moist adiabatic lapse rate s:

Lcdqs = cpdT + gdz is given in Figure A.4. Finally, the relative humidity (RH, in %) is the ratio of actual vapor des g Lcqs + dz = cpdT + gdz e RT pressure to saturation vapor pressure (e/es 100%). Using the ideal gas law it can also be ✓ s ◆ ⇥ L q de dT L q g dT written as the ratio of vapor density to saturation vapor density (/s 100%) or as the ra- c s s c s = cp + g ⇥ es dT dz RT dz tio of specific humidity to saturation specific humidity (q/qs 100%). RH can be measured 1 Lcqs by using changing characteristics of materials. ⇥ dT g + RT = dz cp 1Ä Lcqs deäs To determine the actual moisture content from the above parameters knowledge of the + cpes dT saturation vapor pressure is necessary. The Clausius Clapeyron equation relates the sat- ⇣ ⌘ If we neglect the pressure dependence of dqs we can write: urated vapor pressure (es) to the absolute temperature (T). The relation follows from the dT g 1 1 first law of thermodynamics and the ideal gas law (see Frame 12) and reads: . s = = L dq = d L dq dz cp 0 1 c s 1 0 1 c s 1 + cp dT + cp dT L 1 1 @ A @ A es = e0 exp , (A.9) R T T  ✓ 0 ◆ Frame 13: The moist adiabatic lapse rate. where e0 is a known saturation vapor pressure es at given T0. The value for es at melting 5 1 point of ice is often taken for e0 at T0, e0 = 610.78 Pa at T0 = 273.16 K. L is the phase Lm ice water 3.34 10 J kg melt ! ⇥ 6 1 change from state 1 to 2. Values of L for the phase change from ice to water to vapor are: Ls ice vapor 2.84 10 J kg sublimation ! ⇥ 6 1 L water vapor 2.51 10 J kg evaporation ! ⇥ Combine the first law of thermodynamics (Eq. A.1) written as Eq. A.4, and the ideal gas law for moist air (Eq. A.6): The Clausius Clapeyron equation shows that the saturation water vapor pressure es (and hence saturation specific humidity qs) increases sharply with temperature (Figure A.4). The dQ c dT pd = + Figure also shows that there is a significant difference between the saturation vapor pres- e R T, = sure over water and over ice, which has implications for the formation of precipitation. Assume there is no change in internal energy of the parcel, c dT = 0. Eq. A.4 reduces to dQ = pd, where in this case p is the vapor pressure e. Substituting the ideal gas law to eliminate vapor pressure and using the derivative of pressure to temperature from the ideal gas law (de/dT = R) to eliminate density, results in: Stability of the moist atmosphere dQ de = T d dT The presence of moisture in the atmosphere has its effect on the stability of the atmo- The last step is to consider phase changes from state 1 to 2. dQ is then the heat necessary for the change and sphere. It does so in two different ways: equals the latent heat for a phase change from 1 to 2, L12. d is the difference in specific volumes or densities 1. Moisture decreases the mean molecular weight of air compared to dry air. between the two phases, 2 1. Furthermore, since the phase change occurs at saturated conditions e can be 2. Upon condensation, moisture releases heat into the air (latent heat). replaced by es, the saturated vapor pressure: Both effects increase the bouyancy of moist air compared to dry air but because the la- des 1 L12 L12es . = = 2 tent heat of condensation Lc (= L) is considerable, the second effect is generally more dT T 2 1 ⇠ R T important. In the last step, the approximation is made that 1 2 and the ideal gas law is used again the remove density or specific volume from the equation. Integration of⌧ this expression between a known saturation vapor pressure The release of heat during condensation changes the temperature lapse rate in the es at given T, written as e0 at T0 results in the Clausius Clapeyron equation, Eq. A.9. atmosphere. Consider an unsaturated moist air parcel that rises. It will cool according to 1 the dry adiabatic lapse rate d = -9.8 K km . At a certain level the temperature reaches the Frame 12: The Clausius Clapeyron relation. dew point temperature (T = Td) where saturation (e = es) and thus condensation occurs.

23 24 a b c a b c Figure A.5: Illustration of stability in a moist atmosphere in terms of the dry, moist and actual lapse Figure A.6: a) Measured profile of temperature and corresponding calculated saturation specific rate. a) absolutely stable, b) conditionally unstable, and c) absolutely unstable. humidity. b) Potential en equivalent potential temperature lapse rate. c) Temperature lapse rate.

The level where this occurs is the Lifting Condensation Level (LCL). When the parcel is now in which Te is the corresponding equivalent temperature. Note that in this expression T is lifted further the release of heat associated with condensation slows down the cooling of the temperature at saturation (at LCL). A good approximation of this equation is given by: the parcel so that in general the moist (or pseudo) adiabatic lapse rate s > d or since both lapse rates are negative s < d . The term pseudo is often used because a small part Lc | | | | e = + q, of the heat released during condensation is removed by rain. However, we will assume, all cp heat is added to the parcel. where a Taylor expansion around 0 is taken from the exponent in equation A.11, assuming The moist adiabatic lapse rate can be derived similarly to the dry adiabatic lapse rate the argument in the exponent is small. The equivalent potential temperature is a conserved by using the first law of thermodynamics and adding heat from condensation (Frame 13). quantity for moist adiabatic processes. Note that the equivalent potential temperature is When neglecting the pressure dependence of the change in saturation specific heat of the always larger than or equal to the potential temperature, the presence of vapor in the parcel (dqs) we can write: atmosphere increases the potential temperature. 1 The atmospheric static stability can also be expressed in terms of the equivalent poten- s = d . L dqs tial temperature. Again there are three cases: 0 1 c 1 + cp dT e,s < 0 conditionally unstable stratification @ A This relation shows that s depends on temperature through dqs/dT. This also results in e,s = 0 saturated neutral stratification Lc dqs s not being constant with height as is d. Furthermore, since is larger than zero, cp dT e,s > 0 conditionally stable stratification s > d (or s < d ). Finally, s asymptotically approaches d with height because In combination with the potential temperature lapse rate () the stratification is said to | | | | of the decrease of temperature with height in combination with the decrease in pressure be absolutely unstable when and e are negative and absolute stable when both are becoming important as well. positive. The term conditional refers to the fact that the stability or instability is conditional In terms of the stability of the vertical structure of the atmosphere we can now distin- to the saturation of the air parcel. guish three cases: An example of a vertical profile of temperature and moisture with corresponding lapse < d absolutely unstable rates is plotted in Figure A.6. d < < s conditionally unstable s < absolutely stable The addition of the word conditionally refers to that for unstability to occur enough moisture References must be present. Figure A.5 presents a schematic view of these three cases. Garratt, J.. 1992. The Atmospheric Boundary Layer. Cambridge University Press, Cam- When condensation or evaporation occurs in an air parcel and heat is added to or ex- bridge, UK. 316 pp. tracted from the parcel, the potential temperature is no longer conserved. The potential temperature is therefore also no longer a sufficient parameter to determine atmospheric static stability. Again we can use the first law of thermodynamics to derive a parameter that is conserved under moist adiabatic motion (Frame 14). The result is the equivalent potential temperature (e, in K) defined as the temperature of an air parcel when first lifted to the level were saturation occurs, then all water is condensed out, and the parcel is adiabatically lowered to a reference pressure level:

R /cp Lcqs p0 d e = exp = Te , (A.10) c T p ✓ p ◆ ✓ ◆

25 26 Exercises

Take the first law of thermodynamics (Eq. A.1) rewritten as Eq. A.4, substitute the gas law (Eq. A.2) and add heat Growth of a convective boundary layer: from condensation dQ = Lcdqs. the formation of severe thunderstorms dQ = cpdT dp RdT Lcdqs = cpdT dp p Lcdqs dT Rd dp = cpT T cp p Lcdqs Rd = d ln T d ln p cpT cp

Substitute the total derivative of the potential temperature in the above result:

Rd /cp At sunrise, the vertical tem- p0 Rd potential = T dln = dlnT dlnp, p c perature profile in the atmosphere ✓ ◆ ! p results in an expression that gives the potential temperature change in a saturated displacement: schematically looks like this:

dQ Lcdqs dln = = . (A.11) cpT cpT Now condensate all water vapor out by lifting the parcel to the level were saturation occurs, and then lower the parcel adiabatically to a reference pressure level:

e 0 Lc d ln = c T Z p Zqs a. Discuss the vertical stability of the different layers. Focussing on the possible develop- Lc ment of thunderstorms. ln e ln = (0 qs) cpT e Lcqs The heat conservation equation is given by: ln = cpT d 1 R e Lcqs net ( 0 0) = exp = LE + cpT dt c z z ✓ ◆ p ✓ ◆ Lcqs e = exp where is the potential temperature, t time, air density, cp heat capacity of air at constant c T ✓ p ◆ Rd /cp pressure, LE latent heat, Rnet the net radiation, z elevation, and ( ) is the turbulent Lcqs p0 0 0 = T exp c T p sensible heat flux. ✓ p ◆✓ ◆ Rd /cp p0 b. Derive an expression for the average potential temperature change of the Convective = Te p ✓ ◆ Boundary layer (CBL) in terms of the surface sensible heat flux (SHF = cp(00)s) i.e. where e is the equivalent potential temperature and Te is the equivalent temperature. derive equation 3 from the heat conservation equation. Assume that advection, phase changes and radiation effects can be neglected. Frame 14: The equivalent potential temperature. Hint 1: Assume a constant CBL potential temperature and integrate the reduced heat conservation equation from the surface to the CBL top at height hm. Hint 2: Use Leibniz’ rule for variable limits: d b b b ƒ (z,t)dz = ƒ (z,t)dz + ƒ (b, t) ƒ (,t) dt t t t Z Z Hint 3: Note that when assuming a constant CBL potential temperature, the turbulent heat flux must be linear function of height.

27 28 c. Derive an expression for the growth of the Convective Boundary layer in terms of the Answer surface SHF assuming encroachment. a. 0 - 2 km: d/dz>0, absolute stable (dry air!) As soon as the surface energy balance becomes positive at t 0, the growth of the CBL will = 2 km - tropopause: d/dz 0, conditionally unstable (saturated air) start. The turbulent sensible heat flux SHF is estimated by a sinus-function with a period P above tropopause:, d/dz>⇡ 0, absolute stable. of 24 hours: A stable nocturnal ABL (SBL) containing dry air is capped by an unstable, humid air layer. Convection up to the tropopause is inhibited only by the stable ABL, which must be re- SHF = Amp sin (t) moved before thunderstorms can develop. In other words: clouds won’t appear unless 2 2 a forcing is present. If the daytime atmospheric boundary layer reaches 2000 m, The amplitude is SHFm = 500 W m and SHF =0Wm at t = 0. Use that (cp) equals 1 3 becomes small and the boundary layer can grow quickly up to the height of the tropo- 1300 J K m . sphere. Since the air above 2000 m is saturated, it is likely that in that case showers will d. Can we expect thunderstorms to develope on this day? occur. This can be achieved by heating the SBL from below, forming a CBL through encroach- ment (no mechanical mixing). The encroachment model assumes the potential temper- ature profile drawn in figure 7b.

b. If we neglect advection, radiation processes and phase changes, the heat conservation equation:

d 1 Rnet (00) = + + + = LE + dt t y z c z z p ✓ ◆ can be written as:

(00) = t z

Perform a vertical integration from the surface (neglecting the contribution made by the thin surface layer) to the top of the CBL.

h m ( 0 0)hm z = (00) (A.12) 0 t Z Z( 0 0)s

First the left hand side. Use Leibniz’ rule for limits that are variable with ƒ (z,t)=, = 0 and b = hm:

hm hm d hm 0 z = dz (hm) + (0) t dt t t Z0 Z0 d dh hm m = [mz]0 (hm) dt dt d dhm = (mhm m0) (hm) dt dt m hm = hm ((hm) m) t t m = hm t

Where we used

1 hm m = (z)dz = constnt h m Z0

29 30 and Integrate between hm(t = 0)=0 and hm(t) results in: d h h m m m m hm t = hm + m SHFm sin(t) dt t t hmhm = t c Z0 Z0 p Now the right hand side of equation A.12. Assuming a constant mixed-layer potential 1 hm SHF cos t t 2 m ( ) temperature must be related to a linear turbulent heat flux profile, since the change in hm = 2 0 cp time must be uniform with height. So we can write:   0 2SHFm(cos(t) cos(0)) h2 ( )h 0 m = 0 0 m 0 cp ( )= = ( ) = ( )s 0 0 0 0 ( )s 0 0 2SHF 1 cos t 0 0 2 m( ( )) Z( 0 0)s Z( 0 0)s h = î ó m c Combine with the result for the left hand side of equation A.12: p

hm For t P/2 12 hr., h reaches its maximum, which is 2057 m (substitute given values) m = = m z = hm (h reaches a maximum when dh /dt 0). This value assumes that above 2000 m t t m m = Z0 is equal to below 2000 m, which is clearly incorrect. Nevertheless, the ABL reaches results in: the conditional unstable layer, so showers are indeed likely to occur.

m (00)s SHF = = t hm cphm c. Changes in depth and temperature of the CBL are linked one-to-one through the en- croachment relation:

m d dhm = + t t dz dt m dhm = + = 0 t dt

Where is the potential temperature lapse rate. Thus:

m dhm = t dt Combined with the expression of the result b.:

m (00)s SHF = = t hm cphm yields an expression for the growth rate of the CBL in terms of the turbulent surface sensible heat flux:

hm SHF = t cphm d. In other words the question is: Is SHF large enough to let the CBL grow larger than 2000 m? Thus: calculate the max height of the CBL. The Sensible heat flux expressed as:

SHF = Amp sin(t) Substitution in the (rearranged) equation for boundary layer height (result c.)(hm with = 2/24hr. and Amp = SHFm:

hm SHFm sin(t) hm = , t cp

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