BOUNDARY-LAYER METEOROLOGY

Han van Dop, September 2008

D 08-01

Contents

1 INTRODUCTION 31.1 Atmospheric thermodynamics . . . . . . . . . . . . . . . . . . . . 31.2 Statistical aspects of fluid mechanics . . . . . . . . . . . . . . . . 6

2 CONSERVATION LAWS 102.1 Governing equations . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 THE REYNOLDS EQUATIONS 143.1 The average flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.2 The mean-flow energy equation . . . . . . . . . . . . . . . . . . . 193.3 The turbulent kinetic energy equation . . . . . . . . . . . . . . . 203.4 Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.5 Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 THE ATMOSPHERIC BOUNDARY LAYER 314.1 Phenomenology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.2 The surface layer; Monin-Obukhov theory . . . . . . . . . . . . . 354.3 The neutral boundary layer . . . . . . . . . . . . . . . . . . . . . 384.4 The convective boundary layer . . . . . . . . . . . . . . . . . . . 444.5 The stable boundary layer . . . . . . . . . . . . . . . . . . . . . . 484.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5 EXCHANGE OF HEAT AND WATER VAPOUR 545.1 The surface-energy budget . . . . . . . . . . . . . . . . . . . . . . 545.2 The profile method . . . . . . . . . . . . . . . . . . . . . . . . . . 565.3 The energy-balance method . . . . . . . . . . . . . . . . . . . . . 585.4 Estimating the evaporation . . . . . . . . . . . . . . . . . . . . . 595.5 Air-Sea interaction . . . . . . . . . . . . . . . . . . . . . . . . . . 61

6 HETEROGENEOUS BOUNDARY LAYERS 706.1 Thermal transitions . . . . . . . . . . . . . . . . . . . . . . . . . 706.2 Roughness transitions . . . . . . . . . . . . . . . . . . . . . . . . 72

7 EXERCISES BOUNDARY LAYERS AND TURBULENCE 77

2

Chapter 1

INTRODUCTION

This reader acts as a follow-up of the lecture notes of the Bachelor Hydro-dynamics and Turbulence course. Some of that material directly relating toboundary-layer meteorology, will be shortly discussed here.

1.1 Atmospheric thermodynamics

First Law of Thermodynamics

The change of energy dQ per unit mass of a closed system equals the sum ofthe change of the internal energy dU , and the amount of work, pdV :

dQ = dU + p dV. (1.1)

The specific heat at constant volume, or constant pressure, cv and cp are:

(dQ)v = (dU)v ≡ cv dT(dQ)p ≡ cp dT

where (. . . )v denotes that volume is held constant, and (. . . )p indicates a con-stant pressure (cv,p in J kg−1K−1).

Equation of stateAgain for a unit mass we have

pV = RT (1.2)

or, using ρ = 1/V :

p = ρRT. (1.3)

3

An alternative for the First Law can now be written as:

dQ = cp dT −1ρdp. (1.4)

where cv +R = cp, and R is the specific gas constant for dry air.

R = R?/ma, with R?, the universal gas constant (8.314 J K−1 mole−1)and ma the molecular weight of (dry) air (0.0288 kg). Thus R equals 288J kg−1 K−1.

Adiabatic lapse rate

The heat exchange between the atmosphere and its surroundings, for exampleby molecular conduction or by radiation, is often negligible, implying dQ = 0.From the first law of thermodynamics and the hydrostatic equation, it followsthat

cp dT + g dz = 0

−dTdz

=g

cp≡ Γd ≈ 10−2Km−1 (1.5)

This vertical temperature gradient is commonly referred to as the (dry) adia-batic lapse rate.

Entropy

The entropy S is defined as

dS ≡ dQ

T= cp

dT

T−Rdp

p

= cp d lnT −R d ln p.

Upon integration, this yields:

S = cp lnT −R ln p. (1.6)

Potential temperature

In adiabatic processes, dS = 0. In a process (1→ 2), this means that :

4

cp lnT1 −R ln p1 = cp lnT2 −R ln p2. (1.7)

If we assign a constant pressure ps (e.g. 1000 mb) and temperature θ (potentialtemperature) to the reference level 2, then (1.7) can be written as:

cp lnT −R ln p = cp ln θ −R ln ps, (1.8)

where the subscript 1 is omitted. This equation can be rewritten to obtain

θ = T (psp

)κ, (1.9)

where κ = R/cp. This expression allows us to determine the potential tempera-ture of any air mass as long as its temperature and pressure are known. In thisway, a vertical profile of potential temperature can be constructed. Using (1.6),the entropy now becomes

S = cp ln θ −R ln ps, (1.10)

leading to the conclusion that, in case of an adiabatic process, there is a directconnection between the potential temperature, θ, and the entropy S. Further-more, Eq. (1.8) implies

dθ

dz≈ dT

dz+ Γd, (1.11)

where the dry adiabatic lapse rate is given by

Γd =θ

T

g

cp≈ g

cp. (1.12)

The above-mentioned equation poses a simple relation between the vertical tem-perature gradient dT/dz and the vertical gradient of the potential temperaturedθ/dz.

Standard atmosphere

Using (1.5), an atmospheric temperature profile can be described:

T

T0= 1− z

H, (1.13)

5

T0 being the surface temperature, and H = cpT0/g. We can now invoke theequation of state (p = ρRT ) and the hydrostatic equation to derive equationsfor vertical atmospheric profiles of pressure and air density:

p

p0= (1− z

H)

cpR (1.14)

ρ

ρ0= (1− z

H)

cpR −1. (1.15)

The atmospheric scale height is denoted by H. This approach implicates thatthe atmosphere has a finite thickness (H ≈ 30 km) which is obviously not correctbut nevertheless yields a reasonable estimate of the thickness of the atmosphere.

1.2 Statistical aspects of fluid mechanics

The length scales relevant in a turbulent flow cover a wide range. The macroscaleand the smallest turbulent scale, known as the Kolmogorov microscale, arerelated through:

l/lK = Re34 (1.16)

In the atmospheric boundary layer, typical values of l = 1000 m and lK = 0.001m, yield a Reynolds number of O(108). Solving all length scales up to 0.001 mon a domain of 10 km x 10 km x 1 km using a numerical program would require1020 grid points, whereas computation on a grid with 1010 points is currentlyfeasible. Thus, turbulence equations cannot be solved numerically in the entiredomain of length scales. As an alternative, turbulence equations could be solvedfor averaged quantities only. In this way, the range of length scales, and thusthe required amount of grid points, can be reduced dramatically.

Stochastic variables, moments, probability density functions

In a turbulent flow, kinematic processes occur over a wide range of scales. Asa first step in a statistical treatment of turbulence, the average of a variable(the first moment) can be determined. More detailed information of the flowcan be obtained by calculating higher moments of variables, or combinations ofvariables (correlations). When analysing a turbulent flow in a statistical way,correlations between flow variables at different locations or at different timesplay a very important role. Examples of such correlations are the variance

u2 ≡ u(xxx, t)u(xxx, t) (1.17)

and the covariance

6

uw ≡ u(xxx, t)w(xxx, t), (1.18)

being statistical properties of flow velocity at location xxx and time t. The overbaris used for the so-called ensemble-average of a quantity, i.e. the average obtainedby taking the average of a large amount of realisations under the same boundaryconditions.If we treat velocity, pressure and other quantities in a turbulent flow as stochasticvariables, it is possible to define the flow by its moments. Let u be any variable,and p(u) its probability density function. The moments are then given by

um ≡∫ +∞

−∞p(u)umdu (1.19)

A stationary stochastic process is defined as a process where the density functionp is invariant under a time translation. If a stochastic process is stationary,p(u1, u2; t1, t2) = p(u1, u2; t1 − t2) applies. The autocorrelation reads

u(xxx, t1) u(xxx, t2) = ρ(xxx, t1 − t2).

Analogously, the spatial autocorrelation is defined as

u(xxx1, t)u(xxx2, t).

In a homogeneous flow this function depends only on the separation:

ρs(xxx1 − xxx2, t) = u(xxx1, t)u(xxx2, t).

where ρs is the (spatial) velocity autocorrelation function.

Reynolds decomposition and averaging processes

Reynolds decomposition

We can split the motion of a turbulent flow into a large scale and a small scalecomponent (here denoted with a prime). In this way, the average of a variableand its fluctuation are discerned. Let uuu be the velocity. It can be split as follows

uuu = uuu+ uuu′ (1.20)

’Averaging’, denoted by the overbar, means ensemble-averaging in this context.We assume that the averaging process is a ’Reynolds operator’ which meansthat

7

u = u (and thus u′ = 0), and that cu = cu (c is a constant)

Averaging

Features of ensemble-averaging are:

• Ensemble-averaging removes all turbulence from the description of theflow: the variables are no longer chaotic. Any information about thestructure of the turbulence vanishes.

• Transport by eddies has to be parameterised, i.e. only the statistical (av-erage) properties of the turbulence are featured in the resulting equations.Such a parameterisation requires a good knowledge of the flow.

• Averages vary much less in time and space than the turbulence itself. Gridpoint distances can therefore be chosen relatively large.

Time-averaging

The average value of a velocity variable over a period T is defined as

UUUT (xxx, t) =1T

∫ t+ 12T

t− 12T

uuu(xxx, t′)dt′, (1.21)

If T incorporates all turbulence time scales, the time-average would be a goodapproximation of the ensemble-average (U will then be independent of T ).

Volume-averaging

Analogous to time-averaging, a variable can also be averaged over a certainvolume. The local values are split into a spatial average and a deviation fromthis average

a = a+ a′ (1.22)

where the volume-average is defined as

a ≡ 1∆V

∫∆V

a dV. (1.23)

In more general terms, the volume-average can also be expressed as

a(x) ≡∫ +∞

−∞G(x′ − x)a(x′)dx′ (1.24)

8

where the filter function G is given by, for example,

G(x− x′) = 1 if |x′ − x| < L

= 0 otherwise.

Important properties of (1.23), that differ from the properties of ensemble-averages, are

ab 6= aba 6= a

(1.25)

In practical applications these inequalities are often ignored, since errors madeby assuming that these quantities are equal appear to be small. Characteristicsof volume-averaging are:

• Volume-averaging only removes the scales smaller than V13 ; it conserves

the larger scales.

• A so-called closure is needed for the scales smaller than V13 . The advantage

of volume-averaging, compared to ensemble-averaging, is that the closurehypothesis is less critical, since the small scales do not carry much energyand have relatively well-known properties.

9

Chapter 2

CONSERVATION LAWS

We summarize the basic equations as follows:

The continuity equation

dρ

dt+ ρ

∂ui∂xi

= 0. (2.1)

The momentum equation

ρ∂ui∂t

+ ρuj∂ui∂xj

= − ∂p

∂xi− ρgδi3 + µ

∂2ui∂xj∂xj

. (2.2)

The temperature equation

∂θ

∂t+ uj

∂

∂xjθ ≡ dθ

dt= κ

∂2θ

∂xj∂xj. (2.3)

2.1 Governing equations

The continuity equation

As a good approximation, the continuity equation can be expressed as

∂ui∂xi

= 0,

also implying that

10

1ρ

dρ

dt≈ 0.

The density along the fluid-particle trajectories is approximately constant. Theflow can therefore be considered incompressible.

In an adiabatic process,

dρ

dt=

1

c2dp

dt

applies, where c (the velocity of sound) stands for (cp

cv

pρ

)12 . When combined

with the Bernoulli equations, this yields

∂ui

∂xi=

1

c2

(d 1

2ui

2

dt+ gu3

).

Choosing U and L as scale sizes of ui and xi respectively, then

L

U

∂ui

∂xi= O(

u2

c2) +O(

g L

c2).

Using the Boussinesq approximations, the right-hand side can be neglected.

Boussinesq approximations

The above approximations, u/c << 1 and L << c2/g, which imply that at-mospheric flow can be considered incompressible, form part of the so-calledBoussinesq approximations. This however does not mean that density differ-ences are dynamically unimportant. They certainly are in combination withgravity. The resulting equations are referred to as the Boussinesq equations.They originate from a number of subtle arguments which will be given below.

We will suppose a reference state of the atmosphere (denoted by the subscript0). The atmosphere is at rest (ui = 0), and horizontally homogeneous regardingtemperature, density and pressure. We further assume that thermodynamicaldeviations from the reference state in a realistic atmosphere are small. Wefurther suppose that the actual atmospheric state does not deviate much fromthe reference state, viz.

p −→ p0 + p

ρ −→ ρ0 + ρ

θ −→ θ0 + θ

ui −→ 0 + ui,

where p, ρ, θ and ui denote small deviations from the reference state. When wesubstitute this in the equation of state, p = ρRθ, (we assume that in the ABL

11

θ ≈ T , see Eq. (1.9)). We obtain in zero order:

p0 = ρ0Rθ0, (2.4)

and to first order

p = R(ρ0θ + ρθ0).

We assume, confirmed by observations and given the approximations alreadymade, that

p << θ, ρ,

so that

ρ

ρ0∼ − θ

θ0. (2.5)

Now we make the same substitutions in (2.2):

(ρ0 + ρ)duidt

= −∂p0

∂xi− ∂p

∂xi− (ρ0 + ρ)gδi3 + µ

∂2ui∂x2

j

.

In zero order this yields

∂p0

∂xi= −ρ0 g δi3, (2.6)

the hydrostatic equation, and to first order:

ρ0duidt

= − ∂p

∂xi− ρg δi3 + µ

∂2ui∂x2

j

,

which can be rewritten, using (2.5), as

duidt

= − 1ρ0

∂p

∂xi+

θ

θ0g δi3 + ν

∂2ui∂x2

j

,

where ν = µ/ρ0, the kinematic viscosity. Back substitution of the originalvariable θ yields

duidt

= − 1ρ0

∂p

∂xi+ (

θ − θ0

θ0)gδi3 + ν

∂2ui∂x2

j

. (2.7)

This is a familiar expression of the Boussineq equations for a shallow boundarylayer. Essential is that density (temperature) deviations are only important incombination with gravity. Otherwise density can be considered constant. Sothe Boussinesq-approximations for the (shallow) ABL include:

12

u

c<< 1

L <<c2

g

θ ≈ Tp

p<< 1

and we summarize the equations in Boussinesq form:

The continuity equation

∂ui∂xi

= 0. (2.8)

The momentum equation

∂ui∂t

+ uj∂ui∂xj

= − 1ρ0

∂p

∂xi+

(θ − θ0)θ0

gδi3 − 2εijkΩjuk + ν∂2ui∂xj∂xj

. (2.9)

For the sake of completeness, the Coriolis term has been added, making theequations apply for a rotating coordinate system, with an angular velocity ΩΩΩ(in rad/s).

The temperature equation

The temperature equation, neglecting molecular conduction, is changed to

dθ

dt= 0. (2.10)

Together, eqs.(2.8, 2.9 and 2.10) constitute a set of 5 equations (with unknownvariables ui, θ and p) that serves as a starting point for the study of the atmo-spheric boundary layer.

13

Chapter 3

THE REYNOLDSEQUATIONS

We will suppose that a flow consists of a laminar, average flow, and, superim-posed, turbulent fluctuations, the so-called Reynolds decomposition (see sec-tion 1.2).

3.1 The average flow

We will adopt the Navier-Stokes equation, Eq. 2.9,

∂ui∂t

+ uj∂ui∂xj

= − 1ρ0

∂p

∂xi+

(θ − θ0)θ0

gδi3 − 2εijkΩjuk + ν∂2ui∂xj∂xj

. (3.1)

and the continuity equation

∂ui∂xi

= 0. (3.2)

We decompose the velocity ui, the pressure p and the temperature θ in a meanand fluctuating (turbulent) component. After substitution of ui = Ui + ui,p = Π + π and θ = Θ + θ (with this substitution we redefine from hereon θ as atemperature fluctuation) and averaging, we obtain

∂Ui∂t

+Uj∂Ui∂xj

= − 1ρ0

∂Π∂xi

+(Θ− θ0)

θ0gδi3−2εijkΩjUk+ν

∂2Ui∂xj∂xj

− ∂uiuj∂xj

, (3.3)

and

∂Ui∂xi

= 0. (3.4)

14

Figure 3.1: Frictional flow along a flat surface

The extra terms, uiuj , which represent turbulent momentum fluxes, are calledReynolds stresses and are expressed in known variables. This procedure is calledclosure (or ’K-theory’) and will be extensively treated in section 3.5:

uiuj = −Km

(∂Ui∂xj

+∂Uj∂xi

)(3.5)

The proportionality constant Km takes the dimension of viscosity (m2s−1), butis entirely determined by the turbulence itself.Let l and u be the length and velocity scales of the turbulence, then it seemsreasonable to suppose thatKm ∝ lu. This can be nicely illustrated by examininga flow along a flat surface.

Neutral boundary-layer flow

The average flow is stationary and homogeneous in the x-direction (∂U∂t ,∂U∂x = 0).

The equations of motion (meglecting Coriolis forces) are (see 3.3)

∂

∂z

(ν∂U

∂z− uw

)=

1ρ0

∂Π∂x

(3.6)

1ρ0

∂Π∂z− ∂w2

∂z+ g = 0. (3.7)

By differentiating (3.7) to x, we find that 1ρ0∂Π∂x cannot be a function of z. Now,

integrating the first equation yields:(ν∂U

∂z− uw

)∝ z. (3.8)

15

Suppose that h is the ”thickness” of the boundary layer (the height at which∂U/∂z ≈ 0)). Consider two cases:

a. Laminar flow close to the wall (uw = 0).

Eq. (3.8) gives

ν∂U

∂z= az + b, (3.9)

where the constants a and b follow from the boundary conditions at z = 0 andz = h. At z = h we assume ∂U/∂z = 0. b represents the friction force at thesurface (with dimension velocity squared). This defines the ’friction velocity’ u?as

limz→0

ν∂U

∂z≡ u2

?, (3.10)

so that we finally get

ν∂U

∂z∝ u2

?(1−z

h), (3.11)

implying that the velocity profile is linear for z << h. In this (thin) layermolecular friction dominates and thus Re = O(1). If the thickness of that layeris δ we have thus

u?δ

ν≈ 1,

which yields δ ≈ ν/u? for the thickness of the ’laminar sub-layer’.

b. Turbulent flow: the logarithmic wind profile

Above the laminar sub-layer (ν/u? . z . h) turbulent friction dominates:

|uw| >> ν ∂U∂z .

Eq. 3.8 now reads

−uw ∝ z

which, applying the boundary conditions −uw = 0 at z = h and −uw = u2? at

z = ν/u? , yields

−uw ∼ u2?(1−

z

h). (3.12)

(note that ν/u? << h). If z/h << 1, the vertical transport of momentum(−uw ≈ u2

?) is approximately constant. This defines another sub-layer, the’constant flux layer’, ν/u? . z << z/h.

16

We can now use the closure relation (3.5) to express the Reynolds stress in termsof the average gradient in this layer:

−uw = Km∂U

∂z= u2

?. (3.13)

According to (3.13), the velocity profile would be linear (using a constant valuefor Km). This has not been confirmed experimentally: instead, logarithmic pro-files are found. This only follows from (3.13) if the ’eddy’ diffusion coefficientK is proportional to z. If we define

Km = κzu?,

where κ is the Von Karman constant), the solution of (3.13) will be

U

u?=

1κ

lnz

z0, (3.14)

This is a logarithmic profile1, where z0 is an integration constant. The pro-portionality constant κ has a value of ≈ 0.4. The quantity z0 depends onthe roughness of the wall, and is called the roughness length. For an aerody-namically smooth wall, z0 is defined as the height at which the correspondingReynolds number has the value of 1:

(Re)z0 ≡u?z0

ν, (3.15)

leading to z0 = ν/u?. If an aerodynamically rough wall is characterised byirregularities of height h, the Reynolds number is given by

(Re)h ≡u?h

ν. (3.16)

A surface is called smooth when Reh < 1 and rough when Reh > 1. A coupleof typical values for z0 are listed in table 3.1. The roughness of a rough watersurface is treated later in these notes.Under neutral conditions and over a large area of varying surface types, the windprofile is approximately logarithmic from a height z0 to a couple of hundreds ofmetres.If the area is covered with higher objects (trees), it is possible to define a newsurface where the wind speed is 0. The wind profile is then given by

U

u?=

1κ

lnz − dz0

, (3.17)

1There is as yet no exact derivation for the logarithmic wind profile. In section 4.3, it ismade plausible that logarithmic profiles occur in boundary layers

17

Table 3.1: Typical values for the roughness length z0

Surface type Roughness length (m)Smooth water/ice 10−4

Short grass 10−2

Low vegetation 0.05Countryside 0.20Low built-up area 0.6Forests/cities 1− 5

where d is the so-called displacement height. The displacement height equalsroughly 80% of the object height (see figure 3.2). Equation (3.17) is very suitableto calculate the wind speed at a given height, if the wind speed is known on anyother height:

U2

U1=

ln(z2 − d)/z0

ln(z1 − d)/z0.

Drag coefficientIf we square (3.14) and subsequently multiply it by the density, we will find thesurface shear stress:

τ ≡ ρ u2? = ρ

κ2

ln2(z/z0)U2.

The drag coefficient is thus defined as

Cd =κ2

ln2(z/z0), (3.18)

implying that

τ = ρ Cd U2, (3.19)

This is a simple relationship to estimate the surface shear stress from the averagewind speed under neutral circumstances.

Power law

In practice, an algebraic formula for the wind profile is often applied:

18

Figure 3.2: Sketch of the wind speed profile over a homogeneous forest.

U2

U1=(z2

z1

)p. (3.20)

In neutral conditions, the following rule approximately holds:

p ≈ ln−1

(√z1z2

z0

). (3.21)

This relation can be derived by the requirement that at the geometric meanvalue of z1 and z2,

√(z1z2), the wind velocity and the first derivative are equal

in both formulations.A frequently used, but not necessarily correct value of p is 1/7 (based on z1 ∼z2 ∼ 10m and a roughness length of 8 cm).

3.2 The mean-flow energy equation

The starting point is (3.3). We shall neglect the Coriolis force since it does notplay a role in energy budget considerations:

∂Ui∂t

+ Uj∂Ui∂xj

= − 1ρ0

∂Π∂xi

+(Θ− θ0)

θ0gδi3 + ν

∂2Ui∂xj∂xj

− ∂uiuj∂xj

,

Multiplying by Ui yields:

19

∂E

∂t+ Uj

∂E

∂xj

= − 1ρ0Ui∂Π∂xi

+Ui(Θ− θ0)

θ0gδi3 +

∂

∂xjνUi

∂Ui∂xj− Uiuiuj

+ uiuj∂Ui∂xj− ν(

∂Ui∂xj

)2, (3.22)

where E ≡ 12U

2i . If the Reynolds number is high, the viscous terms can be

omitted.When integrated over the volume of the flow, the result is

dEtotdt

=∫V

uiuj∂Ui∂xj

dV +∫V

Ui(Θ− θ0)θ0

gδi3 dV. (3.23)

The integrand of the first integral is negative in a shear flow, and is called thedeformation work. This term provides the translation of the average flow energyto its fluctuations (i.e. the coupling between average flow and turbulence). Thesecond integral represents the conversion of potential into kinetic energy by themean motion in a stratified flow.

3.3 The turbulent kinetic energy equation

The energy transfer from the average flow to turbulence is provided by thedeformation work. We will now derive an expression for the average kineticenergy of the fluctuations, q ≡ 1

2ui2. We shall start from (2.9):

∂ui∂t

+ uj∂ui∂xj

= − 1ρ0

∂p

∂xi+

(θ − θ0)θ0

gδi3 + ν∂2ui∂xj∂xj

,

and for the temperature

∂θ

∂t+ uj

∂θ

∂xj= κ

∂2θ

∂xj∂xj,

As in the derivation of the equation for the mean energy (3.3), we make againthe substitutions ui = Ui + ui, etc. In order to arrive at an equation for thefluctuations, we substract from this equation the equation for the mean flow(3.3). We multiply the resulting equation with ui. Taking the average of thisequation, we finally obtain the equation for the energy of the fluctuation q ≡12ui

2:

∂q

∂t+Uj

∂q

∂xj= −uiuj

∂Ui∂xj

+g

θ0wθ− ∂

∂xj 1

2uiuiuj+1ρ0πuj−ν

∂ 12u

2i

∂xj−ε. (3.24)

20

The first term on the right-hand side of (3.24) is the mechanical production term,supplied by the average flow (normally positive). The second term representsthe thermal production of turbulent fluctuations due to differences in densityof the flow. In a gravity field, this term can transform potential energy intokinetic and vice versa. The ratio between mechanical and thermal productionof turbulence is defined as the flux-Richardson number:

Rif =gθ0wθ

uiuj∂Ui

∂xj

(3.25)

We discern three cases:

wθ > 0 unstable flow Rif < 0

wθ = 0 neutral flow Rif = 0

wθ < 0 stable flow Rif > 0

The use of the flux-Richardson number can be illustrated by the atmosphericboundary layer that is heated by the Earth surface during the day, so thatwθ > 0. At daytime, the boundary layer is thus generally unstable. Duringthe night, the Earth surface will cool down by radiating out its energy. Thus,wθ < 0, meaning that the boundary layer will be stable with little turbulence.We also distinguish the gradient Richardson number, Ri, by applying K-theoryto the fluxes in Eq. 3.25. Similar to Eq. 3.5 we may write for wθ:

wθ = −Kh∂Θ∂z

, (3.26)

so that Eq. 3.25 becomes

Rif =Kh

KmRi, (3.27)

where Ri is given by

Ri =gθ0∂Θ/∂z

(∂Ui/∂xj + ∂Uj/∂xi)∂Ui

∂xj

. (3.28)

Apart from the flux-Richardson number, we also use the bulk-Richardson num-ber, which follows from (3.28), by approximating gradients by finite differencesand assuming a uniform windfield, U(z),

Rib =g

θ0

∆Θ(∆U)2

∆z. (3.29)

21

Returning to Eq. 3.24, we observe that the redistribution term between bracescontains 3 energy fluxes: by turbulent fluctuations, by pressure fluctuations, andby molecular fluctuations (the latter has already been omitted in the equation).The last term represents the energy dissipation. It is the only negative term,and therefore necessarily of the same order of magnitude as the other terms. Ifnot, then q could grow unlimitedly.In a neutral, semi-stationary situation, (3.24) implies

P ≡ −uiuj∂Ui∂xj≈ ν(

∂ui∂xj

)2 ≡ ε,

so P ≈ ε. The production and dissipation of turbulent energy are approximatelybalanced. Also note that P = O(u

3

l ), which means that

ε = O(u3

l). (3.30)

3.4 Spectra

Turbulence can be viewed upon as a set of eddies of different sizes between lKand L. In a turbulent flow, energy of a particular scale is transferred towardslarger and smaller scales. The average flow provides the energy for the turbu-lence at the macroscale. The larger eddies are unstable and disintegrate intosmaller eddies of various sizes. This process is repeated several times. Thesmallest eddies will ultimately lose its energy by molecular viscosity. The neteffect is that eddies tranfer energy from larger to smaller scales, known as theenergy cascade.A turbulent velocity field consists of a superposition of fluctuations with a widerange of temporal and spatial scales. To analyse the contribution of each scale,a Fourier analysis can be exploited. Let the autocorrelation of a continuousvelocity function u(t) be defined as

ρ(τ) = u(t) u(t+ τ) = limT→∞

1T

∫ + T2

−T2

u(t) u(t+ τ) dt. (3.31)

We have assumed stationary conditions so that ρ depends on the time differenceτ only. The energy spectrum S(ω) is, by definition, the Fourier transform ofthe correlation function:

S(ω) ≡ 12π

∫ +∞

−∞e−iωτ ρ(τ) dτ, (3.32)

with the associated inverse operation

ρ(τ) ≡∫ +∞

−∞eiωτS(ω) dω, (3.33)

22

From (3.33), it immediately follows that for τ = 0

u2 ≡∫ +∞−∞ S(ω) dω.

The lefthand side represents the turbulent kinetic energy and this equationshows why S is actually called the energy spectrum, since S(ω) equals thecontribution to u2 of S between the frequencies ω and ω + dω.

23

In practice the energy or power spectrum of the function u(t) is determinedin a more direct way: we rewrite Eq. 3.31 as

ρ(τ) = limT→∞

1

T

Z +∞

−∞v(t) v(t+ τ) dt, (3.34)

where v(t) is defined as

v(t) = u(t) for|t| < T/2

v(t) = 0 otherwise.

The Fourier tranform of v(t) is

v(ω) =1

2π

Z +∞

−∞e−iωt v(t) dt, (3.35)

with the inverse transformation

v(t) =

Z +∞

−∞eiωt v(ω) dω.

We use the last expression to rewrite Eq. 3.34 as

ρ(τ) = limT→∞

1

T

Z +∞

−∞v(t)

Z +∞

−∞v(ω) eiω(t+τ) dω dt.

Rearranging terms we get

ρ(τ) = limT→∞

1

T

Z +∞

−∞v(ω) eiωτ

Z +∞

−∞v(t)eiωt dt dω,

or

ρ(τ) = limT→∞

2π

T

Z +∞

−∞v(ω) v(−ω) eiωτ dω.

Since v(−ω) = v?(ω) we have

ρ(τ) = limT→∞

2π

T

Z +∞

−∞|v(ω)|2 eiωτ dω.

Comparing this result with Eq. 3.33 yields

S(ω) = limT→∞

2π

T|v(ω)|2,

which can be rewritten as

S(ω) = limT→∞

2π

T

˛˛Z + T

2

−T2

u(t) eiωt dt

˛˛2

. (3.36)

This equation, known as the Wiener-Khintchin theorem, provides a directrelation between the velocity signal u(t) and its power spectrum. The FastFourier Transform (FFT) is an efficient numerical way to determine thespectrum of an arbitrary stationary process based on a discrete represen-tation of Eq. 3.36.

Some of the properties of S are

• S(ω) = S(−ω)

24

• S(ω) is real and ≥ 0.

• It also follows from the definition that S(0) = 1π

∫∞0ρ(τ) dτ = 1

π (u2 Tu),where Tu denotes the time scale of the signal u(t).

Spectral gap

During the 1950s, the ideas about boundary-layer meteorology were dominatedby the assumption that (small-scale) turbulent kinetic energy mainly occurredat scales comparable to the boundary layer height (1-2 km). At larger scales,hardly any energy would be available, whereas the energy would increase atmesoscale and sub-synoptic scales. This implied that the turbulence energyspectrum would contain a minimum, separating boundary-layer turbulence fromturbulence at larger scales. This would also justify the process of Reynoldsdecomposition. Now that much more experimental data are available, this viewhas become challenged. An example of a spectral analysis of the energy, recordedduring an airplane flight, is given in figure 3.4. In this figure, the spectral energyof the u− and v velocity components continue to increase at larger scales. Otherturbulent variables, like temperature, water vapour content, and liquid waterconcentration show comparable behaviour. Only the spectrum of the verticalvelocity w seems to level off towards larger scales.That the spectral energy at larger scales does not vanish, has some inconvenientimplications for the integral properties of the spectrum. For example, the totalvariance of u, being defined as

u2 ≡∫ ∞

0

Eu(k)dk, (3.37)

cannot be determined if Eu(k) continues to grow for k → 0. In these cases, theogive is defined as

Ogu(k0) ≡∫ ∞k0

Eu(k)dk, (3.38)

so that integral properties of the spectrum can be determined anyway. Thechoice of k0 is an arbitrary one.

3.5 Closure

Expressing the stress in terms of the average flow is called closure. More gener-ally, closure means that higher moments are expressed in terms of lower ones.It is called closure because the set of equations of motion can be solved afterclosure. In the previous sections, some simple examples of 1st-order closure havealready been demonstrated. More elaborate ways of closure exist, which will beshortly discussed below.

25

Figure 3.3: Kaimal’s schematic representation of the energy spectrum. (A)production subrange; (B) the inertial subrange, where production and dissi-pation are not important; (C) dissipation subrange. The second Kolmogorovhypothesis conjectures that in the inertial subrange, the dissipation ε and thewavenumber k are the dominating parameters. The spectrum E(k) can bedetermined using dimensional analysis based on ε and k only. The result isE(k) = CKε

23 k−

53 .

First-order closure (K-theory)

First-order closure is often applied since it is a fast method. It implies that the′K-theory’ is used to determine the Reynolds-stress term (see, e.g., Eq. 3.5 ).The diffusion coefficient K is typical for the flow and may thus change in timeand space. At the surface layer, we acceptably showed that Km = κzu? (seechapter 3.3, Eq. 3.13). Above the surface layer, there are many possible ways tofollow. At the top of the boundary layer, the vertical exchange is small, leavingKm approximatly zero. Somewhere within the boundary layer, Km should havea maximum. A possible formulation for Km is:

Km = l2[(∂U∂z

)2 +(∂V∂z

)2] 12

. (3.39)

In this equation, l is the mixing length, that will be proportional to the heightin the surface layer, l = κz, and converge to a constant value, say λ, for greaterheights. This choice for Km will give the desired behaviour in the surface layer,where V v 0 and −uw are constant (u2

?). It now follows from

uw = −Km∂U

∂z(3.40)

26

Figure 3.4: 1-D spectra of the horizontal (u and v) and vertical (w) wind-speedcomponents as a function of the wave number k. Measurements from an airplaneat approx. 150 m above the sea.

27

after substitution in (3.39), that

∂U

∂z=u?κz,

(see Eq. 3.13) which provides the desired logarithmic wind profile. The height-dependent growth of the length scale can be delimited by choosing the followingparameterisation:

l =κz

1 + κzλ

(3.41)

where λ is an empirical heigth scale (typically ∼100 m). The eddie diffusionparameter Km (3.39) can easily be corrected for the stability by adding anempirical function F :

K = l2[(∂U∂z

)2 +(∂V∂z

)2] 12

F (Ri), (3.42)

F being a function of the stability through the Richardson number. This pa-rameterisation from 1979 is still used in present-day global circulation modelslike the ECMWF-model, that calculate weather forecasts.

1.5-order closure

This type of closure makes use of the energy equation (3.24), which is a second-moment equation. Suppose that Km obeys

Km ≈ lq12 , (3.43)

where q is given by the turbulent kinetic-energy equation (3.24). The termspresent in that equation can be approximated in the following way, if the flowis assumed to be horizontally homogeneous:

−uiuj∂Ui∂xj

= −Km

(∂U∂z

)2g

θ0wθ = − g

θ0

(Kh

∂Θ∂z

)∂

∂xj

(12uiuiuj +

πujρ0

)= − ∂

∂zKm

∂q

∂z

ε = q32 l−1.

Substituting the above equations into the stationary-energy equation yields:

28

0 = Km

(∂U∂z

)2 +g

θ0Kh

∂Θ∂z

+∂

∂zKm

∂q

∂z− q 3

2 l−1. (3.44)

An equation for the turbulent kinetic energy q emerges. The set of equations(3.3, 3.5, 3.41, 3.43 and 3.44) can now be solved. The name ’1.5-order closure’is used, since only one second-order equation is used for the closure algorithm.

Second-order closure

In this case, all second moments, like uiuj , uiθ, are incorporated in the closurealgorithm. These equations contain terms of the third order, which are eitherparameterised or neglected. As a result, the number of equations that are tobe solved increases rapidly. Sometimes, this method yields better results. Onthe other hand, the mathematical complexity, the large number of empiricalconstants and the lack of a solid theory make second-order closure techniquesless favourable.

Large Eddy Simulation (LES)

LES makes use of a fundamentally different approach. The variables in theNavier-Stokes equation are averaged over a small volume. Contrary to thevariables in the Reynolds-averaged equations, the LES-variables describe theturbulent behaviour of the flow at scales larger than the scales of the averagingprocess. The starting point is the momentum equation (see 3.1), where wetake for simplicity a neutral flow and neglect the Coriolis-term and molecularviscosity. The volume-average is:

∂ui∂t

+ uj∂ui∂xj

= − 1ρ0

∂p∂xi

+∂Rij∂xj

, (3.45)

where Rij is a tensor that describes the impact of the flow behaviour withinthe volume on that outside the volume. This is called the subgrid scale stress,defined as Rij = uiuj − uiuj. Comparable to Reynolds-averaging, thisterm is approximated by relating it to volume-averaged variables:

Rij = −νt(∂ui∂xj

+∂uj∂xi

)(3.46)

In a mathematical sense, LES is almost identical to the method of Reynolds-averaging. The difference between the two processes can be found in the waythe eddy viscosity is formulated. Reynolds-averaging appoints an order of mag-nitude u.l to K, so that the Reynolds number, based on K, is of the order of 1.The choice for the eddy viscosity in LES, νt, depends on the volume over whichthe averaging process takes place. This volume is usually of the same order of

29

magnitude as the grid point distance of the numerical scheme, ∆. The Reynoldsnumber, ul/νt ≈ l/∆ is now much larger than 1 (approximatly 102 to 103 usingcurrent computer capacity).A well-known choice for the eddy viscosity is formulated by Smagorinsky (1963),which is based on a mixing hypothesis (like the K-theory):

νt = (Cs ∆)2 |S| (3.47)

where Cs is the Smagorinsky constant (Cs = 0.2), |S| = (2SijSij)12 and

Sij = 12

(∂ui∂xj

+ ∂uj∂xi

),

the deformation tensor.The most important applications of LES can be found in the simulation ofconvective turbulence, where transports are dominated by large-scale motionsin the atmosphere.

Direct Numerical Simulation (DNS)

This technique aims to find solutions for the Navier-Stokes equation itself, with-out introducing closure or averaging. At large Reynolds numbers, this approachfaces difficulties, since the range of turbulence scales increases rapidly. In orderto find numerical solutions, the grid points distances have to be approximatelyequal to the smallest scale. For calculating practical situations, a large numberof grid points are required. Flows with Reynolds numbers up to 100-1000 canbe simulated with currently available computer power.

30

Chapter 4

THE ATMOSPHERICBOUNDARY LAYER

4.1 Phenomenology

Boundary-layer meteorology comprises the dynamics and physics of the atmo-spheric layer that is closest to the Earth surface. Exchange processes betweenthe Earth surface and the ’free atmosphere’ occur in the boundary layer. Thestate of the boundary layer is influenced by flow in the free atmosphere on theone hand, and by boundary conditions imposed by the Earth surface on theother hand.Dominating boundary layer processes are the vertical exchange of momentumτ = −ρ0 uw, heat H = ρ0 cpwθ and water vapour E = ρ0 wq.The free atmosphere (see figure 4.3) is the layer above the uppermost boundaryof the (turbulent) boundary layer.The behaviour of the boundary layer has an enormous impact on values of themaximum and minimum temperature, wind speed gradient (wind shear), fogand cloudiness. In aviation and architecture (wind engineering), both windshear, fog and the occurrence of wind gusts are important factors to keep inmind. Turbulence, radiation, surface properties and thermodynamics are thekey ingredients of boundary-layer meteorology. We will first give a qualitativeoverview.

Slightly above the Earth surface, a thin laminar layer exists: the ’viscous sub-layer’. This layer adjusts the balance between the Earth surface and the surfacelayer. The surface layer (or inner layer) is a couple of tens of metres thick. Thesurface layer is topped by the (planetary) boundary layer, sometimes also re-ferred to as the Ekman layer or outer layer. The transition between the surfacelayer and the Ekman layer is smooth. The scales in the surface layer are verydifferent from those in the Ekman layer.The boundary-layer dynamics are not only influenced by the average horizontal

31

Figure 4.1: Schematic overview of the troposphere.

32

flow, but also by turbulent processes. The most important of these processesare:

a. Mechanical turbulence

b. Buoyant or convective turbulence.

The stability of the atmosphere influences the boundary layer, which has atypical height of 100-2000 m. The flux-Richardson number is a measure forthe atmospheric stability. It is the ratio between buoyancy production andmechanical production

Rif =gθ0wθ

uiuj∂Ui

∂xj

(see section 3.3).The surface layer is heated when the surface heat flux wθ0 is positive. Thishappens during daytime as a consequence of solar radiation. Convection mayalso take place above a warm sea surface, over which cooler air is flowing.If the heat flux is negative, the boundary layer will cool down. This happensduring the night. These two situations differ substantially, having quite someimplications for the dynamics (e.g. the growth of the boundary layer height:if wθ > 0, then Rif < 0, so the atmosphere is unstable. This implies thatgravity is generating turbulence, and the boundary layer height increases as aconsequence. On the other hand, if wθ < 0, so Rif > 0, the boundary layer isstable. Gravity will suppress turbulence. The boundary layer height will notchange notably. We conclude that there is a daily cycle of boundary layer heightover land surfaces (see figure 4.2).

• Unstable boundary layer

Incoming shortwave radiationwill heat the Earth surface which will heatand lift the surface-layer air: convection (thermals) will start to emerge.In this manner, the boundary layer becomes unstable. Normally, there isa stable layer on top of the boundary layer, which limits the growth ofconvection (see figure 4.3). When convective cells penetrate into a stablelayer, they will lose their kinetic energy and won’t ascend any further.These convective cells mix the air of the boundary layer with the airabove. In this way, the thickness of the boundary layer increases. Thisprocess is called entrainment. The strong turbulence makes the mixingeffective, resulting in a rather uniform distribution of momentum, heatand water vapour.

A (moist) convective cell that reaches its LCL (Lifting CondensationLevel) will condensate and form a cumulus cloud. As soon as the cellreaches its LFC (Level of Free Convection) the clouds can grow upwarduntil the next stable atmospheric layer. The limit of this entrainment pro-cess is at the level of the tropopause, an extremely stable layer at a heightof about 10-15 kms.

33

Figure 4.2: Typical diurnal progress of the boundary layer over a land surface.(From: Stull, An Introduction to Boundary-Layer Meteorology, 1988).

Figure 4.3: Characteristic profiles of average potential temperature, wind speed,vapour and a random trace gas in an unstable boundary layer (From: Stull, AnIntroduction to Boundary-Layer Meteorology, 1988).

34

Figure 4.4: Typical profiles of potential temperature and wind speed in a stableboundary layer (From: Stull, An Introduction to Boundary-Layer Meteorology,1988).

• Stable boundary layer

A stable boundary layer emerges when the net longwave radiation of theEarth surface is negative, which cools down the air above it. Verticalmotions are often hampered in a stable boundary layer, since the (nega-tive) buoyancy slows them down. The turbulent kinetic energy and thelength scales are much smaller than in an unstable boundary layer. Theturbulence is not well-structured. The height of a stable boundary layer isdetermined by the balance between turbulence production and dissipation.Since turbulence is suppressed, little mixing takes place in the boundarylayer. Gradients of heat, momentum, and water vapour are therefore muchlarger (see figure 4.4).

4.2 The surface layer; Monin-Obukhov theory

The Monin-Obukhov length

In a non-neutral boundary layer, the surface heat flux (wθ)0 is not zero. To-gether with the variables z and u?, the surface heat flux, or rather the buoyancygθ0

(wθ)0 plays a significant role. Since u? and gθ0

(wθ)0 can be more or less con-stant over an hour or longer, they are considered to have a big influence onthe dynamical structure of the surface layer. Based on these variables, we canderive a stability parameter, the Monin-Obukhov (MO) length scale L:

L = − u?3

κ(g/θ0)(wθ)0

.

The turbulent kinetic energy equation (3.24) can be simplified to

35

0 = −uw∂U∂z

+g

θ0wθ − ε,

and it can be made dimensionless through multiplication by κ z/u3?:

0 = −uwu2?

κ z

u?

∂U

∂z− z

L− φε.

We used the definition of L to arrive at the latter equation, and denoted thedimensionless dissipation of energy by φε. Since in the surface layer −uw ≈ u2

?,we find:

0 = φm −z

L− φε,

in which we wrote the dimensionless wind speed profile as

φm =κ z

u?

∂U

∂z. (4.1)

In this simplified form, the energy equation can be interpreted as a balancebetween mechanical and buoyancy production on the one hand, and the dissi-pation of energy on the other. The equation is expressed in only a few scales(z, L, and u?). From dimensional analysis, we can also define a dimensionlesstemperature gradient φh:

φh(z/L) =κz

θ?

dΘdz. (4.2)

The temperature scale θ? is defined as θ? ≡ −wθ0/u?.The functions φm and φh have been determined experimentally. In case ofinstability (z/L < 0), the following relations have been found:

φm = (1− 16z/L)−1/4

φh = (1− 16z/L)−1/2

and for the stable case (z/L > 0):

φm = φh = 1 + 5z

L.

An analogous function for the water vapour profile is often set equal to theexpression for φh.

36

Figure 4.5: A summary of dimensionless gradients of wind speed (a) and tem-perature (b) as they have been experimentally determined for the surface layer(From: Yaglom, blm 1977).

Upon integration of (4.1) and (4.2), we obtain

U(z) =u?κln(z/z0)−Ψm(z/L)

Θ(z)−Θ0 =θ?κln(z/z0)−Ψh(z/L)

If z/L < 0, the expressions for Ψm and Ψh become

Ψm = 2 ln(

1 + x

2

)+ ln

(1 + x2

2

)− 2 tan−1 x+ π/2

Ψh = 2 ln(

1 + y

2

),

and otherwise (if z/L > 0)

Ψm = Ψh = −5z/L,

where x = (1− 15z/L)1/4 and y = (1− 9z/L)1/2 (see figure 4.5).If, for whatever practical reason, a power law for the wind speed profile ispreferred, the exponent p is derived from the previous equations using (3.21):

p =φm(z12/L)

ln z12/z0 −Ψ(z12/L),

where z12 =√z1 z2.

Turbulence in the surface layer

37

The following (empirical) relations hold in a stable, flat surface layer:

σu/u? = 2.39± 0.03σv/u? = 1.92± 0.03σw/u? = 1.25± 0.03

In unstable conditions, this relation becomes

σwu?

= 1.25(

1− 3zL

)1/3

.

In the unstable situation, the horizontal-velocity fluctuations are too stronglyinfluenced by variations on a larger scale. The Monin-Obukhov theory starts tolose its validity. For the temperature variance we have

σθ/|θ?| = 0.95(z

−L)−

13 .

Expressions for the unstable situation, that also apply to the surface layer, willbe given in section 4.4. (see also Garratt section 3.5)

Spectra

There exists a plethora of spectra and spectral data in the surface layer, forexample in ?. For our purposes, it suffices to know that

fSu(f)u2?

=102 n

(1 + 33n)5/3

fSv(f)u2?

=17 n

(1 + 9.5n)5/3

fSw(f)u2?

=2.1 n

(1 + 5.3n)5/3,

(4.3)

in which n is the dimensionless frequency (n = fz/U).

4.3 The neutral boundary layer

The Ekman layer

In this section, we will consider a neutral, homogeneous and stationary bound-ary layer. Unlike previously, the Coriolis force will also be taken into account.We will assume that the flow above the boundary layer is geostrophic, i.e. the

38

Figure 4.6: Force balance in the Ekman layer. P is the pressure gradient, Cothe Coriolis force, and Fr is friction.

pressure gradient and the Coriolis force balance each other. From the momen-tum equation, Eq. 3.3, neglecting the buoyancy term, it follows that

0 = f(V − Vg)−∂uw

∂z

0 = −f(U − Ug)−∂vw

∂z(4.4)

(see figure 4.6). Ug and Vg are the components of the geostrophic wind, bydefinition being equal to the wind speed while neglecting the turbulent stress in(4.4). These components are expressed as

Ug = − 1ρ0f

∂Π∂y

Vg =1ρ0f

∂Π∂x

.

The Coriolis parameter, f , equals 2Ω sinφ, where φ represents the latitude, Ωthe angular velocity (rad/s) of the Earth axis. In order to solve eqs. (4.4),K-theory is applied. Suppose that

uw = −Km∂U

∂z

vw = −Km∂V

∂z.

After substitution into (4.4), we obtain

0 = f(V − Vg) +Km∂2U

∂z2

0 = −f(U − Ug) +Km∂2V

∂z2,

39

Figure 4.7: The Ekman spiral, or the Leipzig wind profile (Mildner, 1932).

with the solution

U = Ug(1− e−γz cos γz)V = Ug e

−γz sin γz. (4.5)

(γ =√f/2Km).

These equations are solved by multiplying one of them by −i (=√−1) and

add it to the other. Using the substitution w = U − Ug + i(V − Vg), theresult is

∂2w

∂z2= −i

f

Kmw,

In a coordinate system where GGG is chosen along the x-axis (so that Vg = 0;see figure 4.7), the boundary conditions for w are given by w = −Ug atz = 0, and w = 0 at z = ∞. The solution which satisfies the boundaryconditions is

w = −Ug exp (i− 1)γz,

with γ equal topf/2Km. Equating the real and imaginary parts yields

the solution (4.5).

The Ekman spiral is only sporadically observed, since the conditions of thetheoretical derivation are seldomly met in practice. The assumption that Km

is constant in the entire layer is incorrect, as are the assumptions of first-orderclosure. (For example, Km would rather be proportional to z close to thesurface).Using the parameter γ defined in the box above, we can define the boundarylayer height, for example as the height at which the wind speed is within 5 % ofthe geostrophic wind. In this way, we can choose a boundary layer height like

h ∝ (Km/f)1/2.

If, on the other hand, the eddy diffusivity parameter is supposed to be theproduct of a length scale (h) and a velocity scale (u?), Km = hu?, then theboundary layer height can be expressed as:

40

h = c1u?/f. (4.6)

If we substitute the values c1 = 0.3, f ∼ 10−4s−1, and u? = 0.3m/s, then theneutral boundary layer height amounts to approximately 1000 m.The equations for the Ekman profile can now be rewritten by using (4.6):

U − Ug = −Ug e−√c1/2 z/h cos(

√c1/2 z/h)

V = Ug e−√c1/2 z/h sin(

√c1/2 z/h).

As a result, the wind profile can be written as the product of the geostrophicwind speed and dimensionless functions of z/h. This solution doe not hold forthe surface layer, where a logarithmic wind speed profile is found, both in theoryand in practice. Apparently, other parameters apply close to the surface. Wewill come back to this is the next section.

Resistance laws

In this paragraph, we will discuss the relation betwen the surface layer param-eter u?, and the external pressure gradient G. A stationary, homogeneous andneutral boundary layer is considered. Resistance laws for non-neutral boundarylayers will be shortly touched later on.The starting point is again a stationary, homogeneous and neutral boundarylayer. The negative x-axis is directed towards the shear stress. The equationsof motion are

0 = f(V − Vg)−∂uw

∂z

0 = −f(U − Ug)−∂vw

∂z, (4.7)

The boundary conditions that apply are given in table 4.1. Note that thelower boundary of the layer is denoted by z = z0, the roughness length. Weare looking for a relation between the friction at the Earth surface (u?), and

the geostrophic wind speed G ≡√Ug

2 + Vg2. Dimensional analysis shows that,

apart from G/u?, only one other dimensionless parameter can be found, namelyRo = u?/fz0, the so-called friction Rossby number, implying that

Ugu?

= kx(Ro) (4.8)

Vgu?

= ky(Ro)

where kx,y, are unknown function of ρ0.

41

Table 4.1: Boundary conditions

z = z0 z = hU = 0 U = UgV = 0 V = Vg

−uw = u?2 uw = 0

vw = 0 vw = 0

These relations are called resistance laws. We can construct these by adaptingthe solution for z ≈ h to the solution for z = z0. This procedure is called’matching’. Let us define two dimensionless height coordinates as η = z/h, andξ = z/z0 where h = u?/f , the boundary layer height. We implicitly assumethat z0 << h = u?/f , being equivalent to taking Ro → ∞. When z ≈ h, thedynamical equations (see 4.4) in this coordinate system are:

U − Ugu?

= − dvw/u?2

dzf/u?≡ gx(η) (4.9)

V − Vgu?

=duw/u?

2

dzf/u?≡ gy(η). (4.10)

In case z0 . z << h and τττ(−uw0, 0) (note that the shear stress at the surfaceis chosen along the negative x-axis):

U

u?= fx(ξ) (4.11)

V

u?≈ 0. (4.12)

If we subtract (4.9) from (4.11), we are left with

Ugu?

= fx(ξ)− gx(η), (4.13)

which can be proven to be equal to

Ugu?

=1κ

(lnRo−A0) . (4.14)

42

Proof: using Eq. 4.8, Eq. 4.13 can be written as

kx(Ro) = fx(Ro η)− gx(η). (4.15)

Note that ξ = Ro η. Now, f, g and k must be continuous and logarithmicfunctions. This can be demonstrated for f by first differentiating (4.15) toRo and subsequently to η. We then obtain the differential equation

0 = f′x +Ro η f

′′x ,

that has a logarithmic solution. A similar procedure shows that also g isa logarithmic function. Substituting the logarithmic solutions into (4.9)yields:

1

κln ξ −

Ug

u?=

1

κln η + a constant.

which can be rewritten to

Ug

u?=

1

κ(lnRo−A0) , (4.16)

where A0 is a constant.

For Vg we have

Vgu?

= ky(Ro)

which for Ro → ∞, must converge to a constant. If we assign to this constantthe value −B0/κ, the equation can be written as

Vgu?

= −B0

κ(4.17)

Combining the above-mentioned equations and substituting G =√Ug

2 + Vg2

gives:

u?G

=κ√

(ln Ro−A0)2 +B20

. (4.18)

This is the resistance law under neutral circumstances. A0 ≈ 1.7 and B0 ≈ 4.5.The relation between frictional velocity and the geostrophic wind forcing isonly roughly approximated by resistance laws. If the layer is not neutral, the’constants’ A0 and B0 in (4.18) are simply assumed to relate to h/L, implyingthe following empirical relation (see figure 4.8):

u?G

=κ√

(ln Ro−A(h/L))2 +B2(h/L),

in which h stands for u?/f , Ro = u?/(z0 f) and τ =√uw2 + vw2.

43

The angle between the geostrophic wind and the shear stress at the surface isgiven by α = arctan(Vg/Ug). Substitution leads to

α = arctanB

lnRo−A.

Thus, the angle α is a measure for the wind turning in the boundary layer. Theconstants A and B have been determined experimentally (see figure 4.8).

4.4 The convective boundary layer

The dynamics in the convective boundary layer are determined by wind shearand convection (wθ > 0). As soon as the convection becomes so strong thatthe contribution of the wind shear to the turbulence becomes negligible, theconvection is usually called ’free convection’. The velocity scale u? is then ofminor importance. Relevant parameters in this layer are the boundary layerheight (h) and the surface heat flux. Using these, we can define a convectivevelocity and temperature scale as follows:

w? =(g

θ0(wθ)0h

)1/3

T? = (wθ)0/w?.

Besides these, z and h are the length scales. The time scale h/w? amountsto approximately 1000 s, which is one order of magnitude smaller than 1/f .The influence of the Coriolis force on the dynamics in the convective boundarylayer is thus only important in processes that play at a scale of hours or more.Concerning the wind profile in the convective layer, it can only be remarkedthat the wind profile gradient will be small due to the strong mixing that willoccur. Usually, a uniform (constant) wind is assumed.

Turbulence characteristics

In the convective boundary layer, the following expressions are used. They arederived using dimensional analysis, the constants follow from observations:

σw/u? = (−z/L)1/3,

and in the upper part of the boundary layer:

σw/u? = (−h/L)1/3.

44

Figure 4.8: The functions A and B are dependent on the stability parameter µ ≡h/L, given that L ≥ 0. The curves numbered 1-5 are based on empirical data.The curves labeled 6 are the relations A(µ) = 1.7 + ln(1 + 3

õ

3.4 )− 2.55√µ and

B(µ) = 4.5 + 3√µ

1.7 . The hatched areas represent the spread in the observations.(from: ?).

45

Figure 4.9: Schematic representation of the flow, temperature and eddy struc-ture in the convective boundary layer.

The horizontal variances are expressed empirically as follows:

σuu?≈ σvu?

=(

12− 0.5h

L

)1/3

These equations also apply to the surface layer.

For the temperature variance we have

|σθ|θ?

= 1.34( zh

)−1/3

,

which applies in the lower half of the CBL.

Dynamics

In a convective boundary layer, the air is lifted locally (thermals). When thesepockets of air reach the uppermost boundary layer, they penetrate into a morestable part of the atmosphere. As a consequence, they will start to descendagain. The net vertical transport is approximately zero. This process also causeswarmer air to be pulled into the turbulent layer. This is called entrainment. Inthis way, the boundary between strong and weak turbulence, the inversion layer,can move upward. The mechanical turbulence may also contribute to the decayof the stable layer above, and thus giving way to the growth of the mixing layer.Because the convection is controlled by the heat flux at the surface, which hasa strong diurnal cycle, this process is highly unstationary. In the course of themorning, a convective layer will build up if sufficient solar radiation is available.

46

Figure 4.10: Structure of the unstable boundary layer (From: Holtslag andNieuwstadt, 1986).

Its vertical size will reach a maximum at the end of the afternoon. For thisgrowth and decay, a simple mathematical models exists, that is based on threesimplified dynamical equations: one for the boundary layer temperature (θ), theinversion strength (∆), and the boundary layer height (h). In figure 4.11a twosubsequent temperature profiles, separated by a time difference ∆t, in the CBLare drawn. The surface ∆ δh represents the amount of heat that is captured bythe boundary layer by entrainment, so that the total amount of heat absorbedby the boundary layer in a period ∆t equals

hδθ = (wθ)0δt+ ∆ δh, (4.19)

where the surface heat flux, (wθ)0, is supposed to be constant. In a differentialform:

h∂θ

∂t= (wθ)0 + ∆

∂h

∂t. (4.20)

From geometric considerations, the two subsequent profiles imply:

∂∆∂t

= γ∂h

∂t− ∂θ

∂t, (4.21)

47

where γ is the temperature gradient above the inversion (∂θ∂z ). Concludingly, wewill assume that the ratio between entrainment and surface heat flux, ε, remainsconstant, leaving us with

∆∂h

∂t≡ −(wθ)i = −ε(wθ)0. (4.22)

These equations, containing the unknown variables h, ∆ and θ provide an ex-pression for h:

(h

h0

)2

− 1− 2ε[(h

h0)1/ε − 1

]=

2(1− 2ε)(wθ)0

γh20

t. (4.23)

In this equation, h(t) represents the inversion height, h0 its initial value, γ theinitial temperature gradient (= dθ/dz), and ε is the ratio between the heat fluxat the top and the bottom of the boundary layer (see figure 4.11), (wθ)i/(wθ)0.In the derivation of (4.23), it is assumed that ∆(t = 0) = 0.If the mechanical turbulence cannot be neglected as a fraction of the thermallyproduced turbulence, the entrainment process is parameterised using:

(wθ)i = ε (wθ)0 + εmu3?

hg/θ0.

The constants ε and εm have the values 0.2 and 2.5, respectively.

4.5 The stable boundary layer

In a stable boundary layer, the turbulence is suppressed by Archimede or buoy-ancy forces. Especially during nighttime above a land surface, the cooling ofthe Earth surface induces a negative heat flux (directed to the surface). Theturbulent energy will decrease dramatically, and vertical motion in the bound-ary layer will be suppressed. This process will ultimately result in a cooledboundary layer and a stable potential-temperature profile (dθ/dz > 0).

Nocturnal boundary-layer height

The vertical exchange being small, the growth of the stable layer is much weakerthan the growth of an unstable layer. Observations indicate that a stable layerrapidly builds up in the beginning of the evening, which remains approximatelythe same during the rest of the night. The thickness of the layer can be estimatedusing dimensional analysis:

h = c3√u∗0L/f.

48

Figure 4.11: Geometry of an expanding boundary layer. In the uppermost panel,the idealized potential temperature profile is drawn at two subsequent moments.In the rightmost picture, the heat flux as a function of height is depicted. Theentrainment process is symbolically depicted by the hatched area. The middleand lower panes give a more realistic picture of the entrainment process (From:?).

49

Figure 4.12: Structure of the stable boundary layer (From: Holtslag and Nieuw-stadt, 1986).

The constant c3 approximately equals 0.4.A more general estimate would be the solution of

(f h

Cnu?

)2

+h

Cs L+N h

Ciu?= 1,

where Cn,s,i are 0.5, 10 and 20, respectively, andN ≡√

gθ0∂Θ∂z , the Brunt-Vaisala

frequency.

The local length scale (Λ)

In a stable boundary layer, the vertical exchange is small. As a consequence,the dynamics are governed by local parameters. We will therefore introduce thelocal Monin-Obukhov length:

Λ ≡ − τ3/2

κ(g/θ0)wθ,

where τ and wθ are now dependent on z.As a result of the high stability, the vertical exchange over large distances isvery limited. Small eddies having a more isotropic structure dominate the tur-bulent processes. The y-component of the shear stress will be significant as well.

50

For that reason, the total shear stress τ (=√uw2 + vw2) is preferred to the

frictional velocity u?.Vertical exchange within the stable boundary layer can be so small that thedistance to the surface becomes a meaningless parameter. In theoretical treat-ments, the Richardson number is assigned its maximal constant value of 0.25.Another occurence is called intermittant turbulence. In that case, flow canbecome laminar at times. Wavelike processes are sometimes observed as well.

Wind maximum in the stable boundary layer

It regularly happens that, especially during nighttime above a land surface,the wind speed relatively close to the surface can be 10-15 m/s, while at thesurface, it is very calm. This is when the so-called low-level jet occurs. It is aconsequence of the transition from a convective turbulent boundary layer to astable boundary layer at nightfall. The boundary layer rapidly becomes stable,reducing the vertical transport of momentum strongly. As starting equations,we will take the time-dependent boundary layer equations in a horizontallyhomogeneous situation:

∂U

∂t= fV

∂V

∂t= −f(U −G).

We chose a coordinate system with the positive x-axis directed towards thegeostrophic wind, implying GGG = (G, 0)).Multiply the latter equation by i, add both, and introduce the new variablew ≡ U + iV . This will result in:

U −G = V0 sin ft+ (U0 −G) cos ft (4.24)V = V0 cos ft− (U0 −G) sin ft.

As starting condition, it is assumed that at t = 0 : (U, V ) = (U0, V0). Squaringand adding these equations will lead to (U − G)2 + V 2 = (U0 − G)2 + V0

2.This equation can be represented by a circle in a (U,V)-coordinate system, asdepicted in figure 4.13.The wind vector will move in a clockwise direction over the circle in a pe-riod 2π/f (approx. 15 hours at middle latitudes). It can thus become super-geostrophic (|UUU | > G).

The radius of the circle (√V0

2 + (U0 −G)2) increases when the difference be-tweenUUU0 and the geostrophic wind is larger (close to the surface). The boundarycondition UUU = 0 at z = 0, will cause a maximum for the wind speed at a certainheight (see figure 4.14).

51

U

V

(U0,V0)

G

Figure 4.13: Graphical representation of the behaviour of the wind vector duringa low-level jet. The initial situation is depicted (subscript 0). The wind vectorwill start to move clockwise over the circle, and will in the course of the processbecome larger than the geostrophic wind GGG.

Figure 4.14: Simple representation of the low-level jet.

52

4.6 Summary

An overview of the most important parameters in the atmospheric boundarylayer is given in the following table.

The wind profile

It is not easy to formulate general expressions describing the wind profile in theboundary layer. It does turn out, however, that the wind in the surface layercan be extrapolated into the layers above satisfactorily. An empirical relationthat performs quite well is that from ?:

U2 = U1

(ln z2/z0 −Ψm(z2/L)

ln(z1/z0)−Ψm(z1/L)

).

This expression can be used up to 200 m, both in neutral and in diabatic situ-ations. The deviations from this equation are usually less than 1 m/s.

Table 4.2: Anatomy of the atmospheric boundary layer.

STABILITY PARAMETERS HEIGHT INTERVALviscous sublayer all u?, ν 0− ν/u?turbulent sublayer all u?, z0 ν/u? − z0

surface layer all z, u?, θ? z0 << z < Lquasi-neutral boundary layer neutral h, z, u?, f 0.1h - hconvective boundary layer unstable h,w? 0.1h - hfree convection layer unstable z, w? 0.01h - 0.1hlocal scaling layer stable z, τ, wθ 0.1h-0.5hz-less scaling layer very stable τ, wθ 0.1h-hintermittency layer very stable N 0-hentrainment layer stable ∆U,∆θ, u?, w? 0.8h - 1.2h

53

Chapter 5

EXCHANGE OF HEATAND WATER VAPOUR

5.1 The surface-energy budget

Radiation is the driving force of heating and evaporation at the Earth sur-face (see figure 5.1). It is composed of longwave radiation originating from theEarth and the atmosphere itself, and of shortwave solar radiation. Furthermore,a distinction is made between incoming (downward) and outgoing (upward) ra-diation. The incoming shortwave radiation (S ↓) equals the sum of the directlyincoming solar radiation and the diffuse radiation due to scattering processes(by clouds and atmospheric molecules). The outgoing shortwave radiation (S ↑)is the upward reflected shortwave solar radiation. The longwave incoming radi-ation at the Earth surface (L ↓) has its origin in the atmospheric water vapour.It is mainly determined by the amount of water vapour and the temperatureof the atmosphere. The outgoing longwave radiation at the Earth surface isdirectly linked to the surface temperature of the Earth.The net radiation (RN ) at the surface is defined positive when the radiation isdirected towards the surface, and given by the sum of all radiation components:

RN = S ↓ −S ↑ +L ↓ −L ↑ .

The ratio of outgoing and incoming shortwave radiation is called the albedo, a,in formula:

a =S ↑S ↓

.

The albedo (reflectivity) of a grass surface is approximately 0.2, it is 0.1-0.5 for awater surface and 0.7-0.95 for fresh snow. The incoming and outgoing radiationexhibit a daily cycle. For a clear day, the typical behaviour of RN is shown infigure 5.3.

54

Figure 5.1: Longwave and shortwave radiation in the atmosphere.

Rn

H

λE

G

Figure 5.2: The energy balance.

55

At climatological time scales, we assume that the energy budget at the Earthsurface is approximately in a state of thermal equilibrium. The components ofthe energy balance are the net radiation (RN ) on the one hand, and on the otherhand the energy that is removed by flow or diffusion: the sensible heat flux (H),evaporation (E), and the surface heat flux (G). The total energy balance at theEarth surface can then be expressed as:

RN = H + λE +G, (5.1)

(see figure 5.3), where λ represents the latent heat of evaporation, and E is thewater vapour flux. The terms on the right-hand side of 5.1 are given by

H = ρ0 cp (wθ)0

E = ρ0 wq0

G = k

(∂Ts∂z

)0

,

where k is the molecular diffusion coefficient of the soil. RN has a daily cycleabove a land surface, see fig. 5.3. As a result of this, λE,H and G also exhibita daily cycle. During the day, RN is positive. A typical value at a clear summerday would be 400 Wm−2. H is also positive during daytime, as well as λE. G ispositive, but very small. It amounts to around 10 % of the net radiation. Duringnighttime, RN is negative, typically ∼ 50 Wm−2 in a clear night. Because thereis only little evaporation during night (rather would there be dew formation),λE is practically zero. At night, H is negative, while G can be relatively large(50 % of the net radiation).Several possibilities exist to determine λE and H from measurements of otherquantities. For example, temperature and vapour profile measurements can beused, or measuring all other components of the energy balance. Other estimateshave been made by Penman and Monteith. A summary will be given below.

5.2 The profile method

When using the profile method, temperature and humidity are measured at twolevels above the surface. Wind speed is measured at one level, and the surfaceroughness is supposed to be known.Using Monin-Obukhov similarity theory, we find that:

U(z2) =u?κln(z2/z0)−Ψm(z2/L)

Θ(z2)−Θ(z1) =θ?κln(z2/z1)−Ψh(z2/L) + Ψh(z1/L) (5.2)

q(z2)− q(z1) =q?κln(z2/z1)−Ψq(z2/L) + Ψq(z1/L).

56

Figure 5.3: Daily cycle of the components of the energy balance on a clear dayover a maize crop. Source: ?.

u2T2

q2

q1

T1

z0

Figure 5.4: The profile method experimental setup. At level 1, temperature andhumidity are measured. At level 2, also the wind speed is recorded.

57

Figure 5.5: The setup for the Bowen-ratio method. Net radiation, surface heatflux, temperature, and humidity (at two levels) are measured.

By measuring wind, temperature and humidity at two levels, the three unknownparameters (u?, θ? and q? ) can be determined. From these equation, the heatand water-vapour fluxes immediately follow:

H = −ρ0cpu?θ?

E = −ρ0u?q?

5.3 The energy-balance method

This method will also exploit the energy balance equation(Eq. 5.1). The ex-perimental setup is as follows: at two levels, temperature and humidity aremeasured. At one level, net radiation is measured. At the surface, the surfaceheat flux G is recorded.We introduce the Bowen ratio, β ≡ H/(λE). From the energy balance, it followsthat:

λ E =1

1 + β(RN −G)

H =β

1 + β(RN −G).

The Bowen ratio is calculated from the temperature and humidity-differencemeasurements. From the definition of the Bowen ration, it can be concludedthat:

β = γ∆θ∆q

,

58

with γ = cp

λ and where (5.2) and (5.3), as well as the assumption that Ψm ≈ Ψq.

5.4 Estimating the evaporation

In this section, we will present some methods to calculate the evaporation. Thiscan applied usefully in, for example, agriculture. In all different formulations,the type and state of the surface play an important role: the Penman formulaecan be applied for a wet surface without vegetation, and the Penman-Monteithformulae can be used for vegetation. Other practical evaporation formulae havebeen developed by Priestly and Taylor. All of these formulations will be dis-cussed below.

The Penman method

It is assumed that RN −G is known, and that the evaporation pressure of a wetsurface equals the saturation value, q0 = qs(T0). The temperature and humidityof the atmosphere and the surface, as well as the wind speed, are supposed tobe known. From (5.2), the following applies in a stratified surface layer:

θ?u?

=∆θU

q?u?

=∆qU.

Note that ∆q = q1 − q0, which combines with the definitions of H and λE togive:

H = −ρ0 cp ∆θu2?

U(5.3)

λE = −ρ0 λ ∆qu2?

U. (5.4)

If we define a drag coefficient as cd ≡ (u?/U)2, we get:

H = −ρ0cp ∆θ cd U (5.5)λE = −ρ0 λ ∆q cd U. (5.6)

Using the following equations, we can obtain a simple estimate of the evapo-ration rate. We will use Taylor expansion (subscripts 0 and 1 refer to the twomeasurements levels):

qs(T1) = qs(T0) + (T1 − T0)(∂qs∂T

)0 + . . . , (5.7)

which we’ll rewrite using the notation s ≡ ∂qs

∂T 0, and the substitution T1− T0 ≈

θ1 − θ0 ≡ ∆θ = −H/(ρ0cpcdU):

qs(T1)− qs(T0) = ∆θ∂qs∂T 0

,

59

yielding

qs(T0) = qs(T1) +sH

ρ0 cp cd U. (5.8)

The difference in relative humidity, ∆q, can now be written as:

∆q = q(T1)− qs(T0) = q(T1)− qs(T1)− sH

ρ0cpcdU. (5.9)

Substituting this equation into(5.4), replacing H by RN−G−λE and reorderingyields:

λE =γ

s+ γρ0λcdU [qs(T1)− q(T1)] +

s

s+ γ(RN −G). (5.10)

The term cp/λ is denoted as γ. The first term on the right-hand side in (5.10) iscalled the aerodynamical evaporation, whereas the second term represents theso-called thermodynamical evaporation.In order to calculate the evaporation, in addition to RN −G, both temperature,humidity and wind speed have to be known at the reference level (1). These arestandard meteorological data that are collected every 1 to 3 hours worldwide.

The Penman-Monteith method

This approach is best suited for a surface with complete coverage of vegetation.Heat exchange and evaporation within the vegetation layer is not taken intoaccount. On the other hand, evaporation is assumed to depend on the vegetationtype. The crops have their own resistance against evaporation through stomata(pores) in the leaves. The vapour level is saturated inside the stomata of thecrop leaves. Depending on the amount of available water, the stomata will bemore or less closed. In this way, there will be more or less resistance againstevaporation. This does not apply to the exchange of heat. If, analogous toOhm’s law, the heat and vapour difference is written as the product of the heat(or vapour) flux and a resistance, we can rewrite (5.6) as

H = −ρ0 cp ∆θrh

(5.11)

λE = −ρ0 λ ∆qrh

,

where rh, the ’resistance’, equals 1/(cdU). As a next step, we have to realizethat transport of moisture within the crops is an extra resistance. This can beeasily accounted for by replacing rh with rh + rs in the second equation, wherers is the so-called crop resistance. Using the procedure of the previous section,we get an expression for the evaporation:

60

λE = − γδ

sδ + γρ0λcd U [q(T1)− qs(T1)] +

sδ

sδ + γ(RN −G), (5.12)

where δ = rh/(rh + rs).If sufficient water is available in the upper soil layer (where the plant roots are),the plant will optimize its evaporation by minimizing the crop resistance. Eachcrop is characterized by its own crop resistance. The maximum evaporation isthe potential evaporation rate of the crop. At this rate, the crops grow fastest.This is an important quantity in agriculture: if there is drought, the crops haveto be irrigated until the potential evaporation rate is reached. Any irrigationsurplus is useless since the plant has already reached the potential evaporationrate.

The Priestley-Taylor method

The Priestley-Taylor evaporation method makes use of the Penman-Monteithequation (5.12). The crop resistance is supposed to be zero, and the relativehumidity is chosen at 100 %: q(T1) = qs(T1) . From (5.12), it can be deducedthat the equilibrium evaporation is given by:

λE =s

s+ γ(RN −G). (5.13)

5.5 Air-Sea interaction

The surface layer theory also applies over a water surface. There are someaspects that have to be kept in mind, however. When estimating the stabilityof the surface layer, evaporation plays an important role, since it influences theatmospheric density (at a constant temperature, humid air has a lower densitythan dry air). To simulate this density effect, a somewhat higher temperatureis assigned to humid air, related to the water vapour content in the humid air.This temperature is called virtual temperature, Tv. Without deriving it here,the following expression is generally valid for the virtual temperature:

Tv = T (1 + 0.61 q) ≈ T + 0.61θ0 q, (5.14)

where q is the water vapour mixing ration (in kg/kg), and θ0 is a referencetemperature (e.g. 287 K). In the surface layer, the same formulation holds forthe virtual potential temperature. θv:

θv = θ(1 + 0.61 q) ≈ θ + 0.61θ0 q, (5.15)

If we replace θ by θv in all Navier-Stokes equations, we also take dynamicaleffects of humid air into account. This excludes the mechanism of condensa-tion! Should one want to formulate a stability criterion, it is again possible

61

to use (wθ)v > 0 for an unstable atmosphere, etcetera (cf. section 4.1). TheRichardson number is now given by:

Rif =gθ0wθv

uiuj∂Ui

∂xj

. (5.16)

(see Eq. (3.25)). The virtual-temperature flux follows from (5.15):

wθv = wθ + 0.61θ0 w q.

In order to determine the potential temperature, the water-vapour flux wq hasto be known. It can be estimated from the water-surface temperature and thehumidity (see section 5.4). The atmosphere will usually be neutral above a seasurface, since the air flow above it will gradually adapt to the water temperature,which only varies slightly in time and place.The exchange above a sea surface can also be crudely estimated by using dragcoefficient relations, where transport is proportional to temperature and to hu-midity differences. If we take the wind shear stress ρ0 u?

2 and the drag coeffi-cient cd ≡ (u?/U10)2 (using wind speed at 10 m), the momentum exchange canbe described by:

τ ≡ ρ0u?2 = ρ0 cd U10

2.

Measurements show that the drag coefficient is≈ 1.3 10−3 for neutral conditions.In a stable surface layer, we find lower values, and higher values can be foundunder unstable circumstances. A simple stability correction is cd = cdn(1−Ri),where the Richardson number can be estimated from:

Ri = 10gθ0

(T10 − Ts)U10

2

again assuming the reference height at 10 m.Similar to a land surface, the heat and water-vapour transport can be estimatedusing:

H = −ρ0 cp ch U10 (T10 − Ts) (5.17)

and

λE = −ρ0 λ cq U10 (q10 − qs) (5.18)

The coefficients ch and cq both adapt a constant value of 1.4 10−3.

62

Unlike at open sea, stability effects do play a role near the coast. We will comeback to that later. A complicating difference between land and sea surfaces isthe waves at the sea surface. The observed wave pattern usually is a result ofboth wind-generated waves (wind waves) and waves that have been generatedelsewhere (swell). On top of the wave pattern lies a turbulent layer, againcharacterized by a vertical momentum flux ρ0 uw, analogous to the situationabove a land surface. Also at sea, a logarithmic wind profile in the surface layeris a suitable assumption:

U(z)u?

=1κ

lnz

r+ C. (5.19)

(see 3.14). We chose on purpose another parameters , since the length scale rand the constant C will depend on the state of the sea surface and atmosphericturbulence. Because the exact theory is very complicated, we will use scaleanalysis. A spectral analysis of wind waves can give us a wave number kp thatcorresponds to the characteristic wave length of the wind waves, 2π/kp (e.g. thevalue where the spectrum shows a maximum). In this way, the phase velocity Cpis determined: Cp =

√g/kp. Its dimensionless counterpart Cp/u? is sometimes

called wave age.The average wave height of a stationary windy sea, Hav, can be derived usingscale analysis: Hav ∝ u2

?/g.Another important wind wave parameters is the so-called fetch, the length ofthe trajectory that a wave has travelled from its ’birth’ to its full-grown size.There exists a dimensionless relation between the fetch and the phase velocity.The presence of waves enhances the momentum exchange from the atmosphereto the ocean. Short, steep waves play a relatively important role, since pressureforces can easily clamp to the waves (unlike the land surface, where only theshear stress accounts for the momentum exchange).

Momentum exchange above sea

The functions C and r in (5.19) depend on u?, g and Cp, according to theprevious section.An alternative formulation of 5.19 would be:

U(z)u?

=1κ

lngz

u2?

+ C, (5.20)

where r ∝ u2?/g from dimensional considerations. The proportionality constant

has been included in the new constant C, and may still depend on the wave age(Cp/u?). A constant value of 11.3 turns out to apply reasonably well, as will beshown later.If C is assumed constant, it is also possible to include C into the logarithm, andrewrite (3.17) analogous to the land surface case:

63

Figure 5.6: An ensemble of measurements of the wind-shear stress relation.(a) lower wind speeds, and (b) moderate to strong winds. The dashed linerepresents Eq. (5.20). (From ?)

U(z)u?

=1κ

lnz

z0,

where z0 = 0.011 u2?/g represents the roughness of a windy sea. This formulation

(Charnock’s Law) combines the parameters r and C without taking into accounttheir functional dependence on Cp/u?. Under non-neutral circumstances, theseformulations can be adjusted in very much the same way as over a land surface(see section 4.2).Charnock’s formulation is pretty succesful, despite its simplicity (see figure 5.6).We can conclude that Charnock’s Law (including stability correction!) appliesfor a relatively large range of wind speeds (6-25 m/s), giving a satisfying relationbetween U and u?.The influence of wave age has been studied in more detail. We then assumethat z0 = Ac u

2?/g, where Ac depends on Cp/u?. A large number of experiments

support the applicability of the following empirical relation (see figure 5.7):

Ac = 1.89(Cpu?

)−1.59

[1.0 + 47.2 (

Cpu?

)−2.59 + 11.8 (Cpu?

)−4.59

]. (5.21)

A simple model for ocean-atmosphere exchange

We will assume a homogeneous, thin, and well-mixed layer (for example, the

64

Figure 5.7: Several measurements of Cp and z0. The solid line represents therelation in 5.21. (From ?)

65

H LE

Shallow Lake d

Ta

Tsρs

Uqa

Figure 5.8: Schematic representation of ocean-atmosphere exchange.

oceanic mixing layer, or a shallow lake). Heat and water vapour transport arethe most important processes. The water temperature Ts can be described by:

ρs cps d∂Ts∂t

= −(H + λE), (5.22)

where the subscript s indicates sea-related variables (see fig. 5.8).Using the definition from the previous section (5.17, 5.18), we know that:

H + λE = ρ0 cp ch U10

Ts − Ta +

λ

cp(qs(Ts)− qa)

. (5.23)

We will introduce the wet-bulb temperature, Tw ≡ Ta − λcp

(qs(Tw)− qa).

This expression is directly obtained by integrating the First Law of ther-modynamics for an adiabatic process, cp dT + λdq = 0, from the initialcondition T, qa until the state Tw, qs(Tw). This describes the physical pro-cess in an air parcel in which liquid water evaporates until it saturates.

Developing qs into a Taylor series

qs(Ts) = qs(Tw) + (Ts − Tw) (∂qs∂T

)s + . . . , (5.24)

and substituting into (5.23) yields

ρs cpsd∂Ts∂t

= −ρ0 ch U10(Ts − Tw) (cp + sλ). (5.25)

This is a differential equation exhibiting an exponential behaviour if all coeffi-cients are assumed constant. From (5.25), we will find the time constant τ :

τ =ρs cps

d

ρ0 ch U10 (cp + sλ), (5.26)

66

which is proportional to d/U10. The cooling rate is proportional to the layerthickness, and inversely proportional to the wind speed.We can also include the heat loss through radiation. The energy budget thenbecomes:

∂Ts∂t

= Q− Ts − Twτ

(5.27)

where Q represents the net longwave radiation term due to radiation loss of thewater layer (∝ T 4

s ) and the atmospheric back-radiation (∝ T 4a ), implying

∂Ts∂t

= αT 4a − βT 4

s −Ts − Tw

τ. (5.28)

T 4s can be approximated as

T 4s = (Tw + Ts − Tw)4 ≈ T 4

w + 4 T 3w (Ts − Tw),

yielding

∂Ts∂t

= αT 4a − βT 4

w −Ts − Tw

τ ′, (5.29)

with a time constant τ ′ = τ/(1 + 4βT 3w τ). The response time turns out to

decrease due to the longwave radiation. The value of τ ′ on mid-latitudes turnsout to be 1 day for a depth of 1 m, and 10 days for a depth of 10 m. At thesedepths, water temperatures can apparently fluctuate quite rapidly.

Finally, we will give an example of the effect of a periodic forcing of the watertemperature by the atmosphere, q cosωt, where q is the temperature forcing(in Km−1):

∂Ts∂t

= q cosωt− Ts − Teqτ

, (5.30)

where ω may represent the annual cycle of temperature, and Teq the annualaverage temperature. The general solution of this equation is:

Ts − Teq = Ce−t/τ − qτ

1 + (ωτ)2e−t/τ +

qτ√1 + (ωτ)2

cos(ωt+ φ), (5.31)

where C is an integration constant, and φ = arctanωτ . If t >> τ , this relationcan be rewritten as

67

Figure 5.9: Phase difference between the annual atmosperic temperature cycleand the water temperature at two depths (5 and 15 m).

68

T eqs − Teq =qτ√

1 + (ωτ)2cos(ωt+ φ). (5.32)

We can conclude that the water temperature is out of phase with respect tothe radiative atmospheric forcing. There are two extreme cases, ωτ → 0 andωτ →∞. In the first case, the phase difference is 0, and in the latter, it is 90o.In reality, values in between these extremes will be found (see figure 5.9).

69

Chapter 6

HETEROGENEOUSBOUNDARY LAYERS

In the previous sections, the Earth surface has always been assumed to be flatand uniform. However, the largest part of the land surface on Earth is notflat. The geometry of the terrain may be very irregular, and the characteristicsof the surface may change. We will discuss two types of change in terraincharacteristics: water-vapour and heat-exchange characteristics, and roughnesschanges.

6.1 Thermal transitions

A thermal transition occurs when the air mass flows from a warm surface toa cold surface or v.v. As the thermal characteristics of the underlying surfacechange, the atmospheric stability can change dramatically. Turbulence andmomentum exchange are influenced by the underlying surface as well. Thermaltransitions typically happen at a land-sea boundary, if warm and convective airmass flows out over a relatively cold sea (offshore flow or land breeze). A new,more stable boundary layer will emerge within the existing, unstable boundarylayer. On the other hand, relatively cold air may also flow over a warm seasurface. The warmer sea water will start a convection process and create a fast-growing, turbulent internal boundary layer. These processes will of course alsooccur over a land surface, when the wind is onshore (sea breeze) (see fig. 6.1).In both cases, the growth of the internal boundary layer will be proportional to√x, x being the distance to the coast.

Both the diurnal cycle of land temperature and the seasonal cycle will influencethese thermal transition processes. Moreover, the presence of clouds can have amoderating effect on these processes. In general, the meteorological situation ina coastal area (over sea and land) will be highly complex and variable. Althoughchanges in the turbulent structure may be significant, changes in the averageflow field are usually neglected.

70

Figure 6.1: The development of an internal boundary layer (a) onshore wind. Astable layer will develop into a convective boundary layer over land, hb(x). (b)The inverse situation, where a stable boundary layer emerges from a convectivelayer over sea (From: ?).

71

Figure 6.2: Numerical simulation of land-sea breeze circulation. (from: A.B.C.Tijm, PhD dissertation, (IMAU) Universiteit Utrecht, 1999)

A typical feature involving thermal transitions is the land-sea breeze circulation,which can be witnessed when wind speeds are low and when the differencebetween land and sea temperatures is large. The different rates of warmingand cooling of both surfaces lead to a sea breeze during daytime, and to a landbreeze at night. This local circulation may extend tens of kilometres from thecoast (see figure 6.2).

6.2 Roughness transitions

Transitions in the terrain roughness are relevant in wind turbine technology,since large changes in the wind field may arise due to changes in terrain rough-ness. In this case, we will suppose a sudden transition from one uniform surfaceroughness to another. Furthermore, the wind is considered to blow perpen-dicular to the boundary between both surfaces. We will have to discriminatebetween the transition smooth→ rough and rough→ smooth. In the first case,the air flow will slow down, in the latter it will accelerate. Initially, this effectis only felt at the surface, but it will gradually expand to higher altitudes, asa new internal boundary layer is developed (see figure 6.3). At the boundarybetween the surfaces, pressure effects will arise causing vertical accelerations,being upward if the transition is from smooth to rough, and downward whenrough → smooth. At a certain distance from the boundary, pressure effects areneglected.

Dynamics of the internal boundary layer

We will make an approximation that is once again based on dimension analysis.

72

Let us assume a neutral boundary layer (subscript 1) having a logarithmic windprofile:

U(z)u∗1

=1κ

lnz

z01.

After the terrain transition (indicated by subscript 2), a new boundary layer willdevelop. Its growth in the vertical direction is determined by the new velocityscale u∗2. The growth can simply be described by

dh

dx= B

u∗2U(h)

.

where x denotes the horizontal coordinate relative to the surface boundary, andB is a constant (≈ 1.3). If we substitute the logarithmic wind profile of terrain2 into U(h), a simple differential equation emerges, having an implicit solutionfor h(x):

1 +h

z02

[ln

h

z02− 1]

= Bκx

z02

If z < h(x), the wind profile has adapted to the new terrain roughness, whereasthe wind profile remains unchanged for z > h(x). This is schematically depictedin figure 6.4.Empirical approximations are usually exploited to smoothen the sudden tran-sition in the wind profiles (figure 6.4). A fairly satisfying agreement turns outto exist between measured and modelled profiles, as is shown in figure 6.5. Itis nonetheless difficult to determine the new shear stress u∗2 from the new,smoothened profiles. The distance from the surface boundary should be largerin order to do so.

Change of frictional velocity

The change of the frictional velocity is determined by wind tunnel measure-ments, complemented by numerical simulations. A compilation of measure-ments and calculations is shown in figure 6.7. The fast adaptation to the newsituation is striking, as is the overshoot that occurs. A simple estimation thatturns out to perform fairly well, is obtained by demanding u1(h) = u2(h) atz = h, yielding:

u∗2u∗1

= 1− ln(z01/z02)lnh/z02)

.

The ratio will only depend on h(x).

73

Figure 6.3: Simple representation of a transition from a smooth to a roughsurface.

100

10

1

0.1

0.01

z

U(z)z01

z02

100

10

1

0.1

0.01

z

U(z)z01

100

10

1

0.1

0.01

z

U(z)z01

z02

h(x)

x

Figure 6.4: Schematic representation of a smooth → rough transition. The ver-tical scale is logarithmic, so the logarithmic profiles are represented by straightlines.

74

Figure 6.5: Wind profiles as measured in a wind tunnel at several distances fromthe surface boundary. (a) smooth → rough, z01 = 0.02 mm and z02 = 2.5 mm;(b) rough → smooth, z01 = 2.5 mm and z02 = 0.02 mm (From: ?)

Turbulence changesWe will have a look at momentum transport and the variance as a function ofheight after a transition from a smooth to a rough surface has occurred. We canrely on wind tunnel measurements. In figure 6.6, the variance σu ≡ u2, σw ≡ w2

and the momentum transport uw are shown. We can see that the value of uwincreases strongly at the surface, followed by a decrease to an equilibrium value(compare with figure 6.7). A maximum can be found at a certain height, thatseems to move away from the surface. This behaviour cannot be seen whenlooking at the variances. The development of h is very similar in both graphsand resembles the development of h(x) that followed from the wind profile.We can formulate the equation for the average flow using (3.3):

U∂U

∂x= −∂u

2

∂x− ∂uw

∂z.

We have assumed that the flow is stationary, and that the pressure gradientoccurring at the roughness transition can be neglected. Figure 6.6 shows thatthe right-hand side of the equation is smaller than zero, causing the averageflow to slow down.

75

Figure 6.6: Development of the shear stress (a) and the variance (b) in a smoothto rough transition, as a function of the distance from the surface boundary. Thevariables have been normalized with respect to the wind tunnel velocity. (From:?)

Figure 6.7: Some calculations (lines) and wind tunnel data (points) of the changein surface shear stress (u∗2/u∗1)2 (From: ?)

76

Chapter 7

EXERCISES BOUNDARYLAYERS ANDTURBULENCE

CHAPTER 1CHAPTER 1CHAPTER 1

Exercise 1Exercise 1Exercise 1The first law of Thermodynamics states:

dQ = (dU)v + pdV

a) Why is in this form of the first law of Thermodynamics dU = (dU)v used?b) From this derive:

dQ = (cv +R)dT − 1ρdp

c) Derive cv +R = cp and equation 1.4.

Exercise 2Exercise 2Exercise 2Derive the pressure and density profile for an isothermal atmosphere (T (z) =T0).

77

Exercise 3Exercise 3Exercise 3a) Show from equation 1.4 that for an adiabatic atmosphere we obtain equation1.5.b) Use this to obtain the adiabatic temperature profile (equation 1.13).c) Use again the adiabatic form of equation 1.4 and the equation of state toobtain (eqs. 1.14, 1.15).

Exercise 4Exercise 4Exercise 4a) Derive from equation 1.8:

dθ

dz=

θ

T(dT

dz+

g

cp)

b) Show now that eqs. 1.11, 1.12 are valid for the boundary layer.

Exercise 5Exercise 5Exercise 5Consider a stochastic 1-D velocity field u(x). We define the velocity correlationbetween two arbitrary points located at x1 and x2 as

ρ0(x1, x2) ≡ u(x1)u(x2)

The field is called homogeneous if its statistical properties are the same in allregions of the field. Thus

ρ0(x1 + L, x2 + L) = ρ0(x1, x2),

where L is an arbitrary displacement. Hence the correlation is invariant underthe translation L.

a Show that in a homogeneous field ρ0 is only a function of the separationr = x1 − x2.

b Show for the homogeneous case that the velocity variance σ defined asσ = ρ0(x, x) = u2(x) must be constant.

c Make a sketch of the velocity correlation in a turbulent flow as functionof r. Explain the shape.

d Give a similar consideration for stationarity of a stochastic process u(t).

78

CHAPTER 2CHAPTER 2CHAPTER 2

Exercise 6Exercise 6Exercise 6Starting with dQ = cpdT − 1

ρdp derive for an adiabatic process:

dρ

dt=

1c2dp

dt

with c2 = cp

cv

pρ

CHAPTER 3CHAPTER 3CHAPTER 3

Exercise 7Exercise 7Exercise 7The Boussinesq momentum equation (eq. 2.9) (without the Coriolis term) isgiven by:The momentum equation

∂ui∂t

+ uj∂ui∂xj

= − 1ρ0

∂p

∂xi+

(θ − θ0)θ0

gδi3 + ν∂2ui∂xj∂xj

. (7.1)

a Use this equation to substitute the Reynolds decomposition (we call thisequation A), and take the average of this whole equation to obtain

∂Ui∂t

+ Uj∂Ui∂xj

= − 1ρ0

∂Π∂xi

+(Θ− θ0)

θ0gδi3 + ν

∂2Ui∂xj∂xj

− ∂uiuj∂xj

,

which is the first equation in section 3.2

b Subtract this equation from equation A to obtain the fluctuation equation:

∂ui∂t

+uj∂Ui∂xj

+Uj∂ui∂xj

+uj∂ui∂xj− ∂uiuj

∂xj= − 1

ρ0

∂π

∂xi+θ

θ0gδi3 +ν

∂2ui∂xj∂xj

c Check that the average of this fluctuation equation is zero.

d Multiply this whole fluctuation equation by ui, and take the average ofthis result to obtain equation (3.24). Explain all the terms.

79

Exercise 8Exercise 8Exercise 8In general our chosen frame of reference forms a curvilinear co-ordinate systemwhich is fixed to the Earths surface. As a result two apparent forces are intro-duced, the curvature terms and the Coriolis forces. If we take the co-ordinatesystem such that x is positive to the East, y positive to the North and z positiveupwards, we can write the momentum equations:

du

dt= − 1

ρ0

∂p

∂x− uw

R+uv tanϕ

R+ fv − lw + ν

∂2u

∂x2j

dv

dt= − 1

ρ0

∂p

∂y− vw

R− u2 tanϕ

R− fu+ ν

∂2v

∂x2j

dw

dt= − 1

ρ0

∂p

∂z− g +

u2 + v2

R+ lu+ ν

∂2w

∂x2j

where R is the radius of the Earth, ρ0 the constant density, f = 2ω sinϕ andl = 2ω cosϕ (see e.g. page 35 in J. R. Holton, An introduction to dynamicmeteorology).a) Derive an equation for the kinetic energy (u2 + v2 +w2)/2 for the above setof momentum equations.b) Discuss why the curvature terms and Coriolis forces do not appear in thisequation.

Exercise 9Exercise 9Exercise 9The flow in a vegetation canopy with height h0 is considered with the assumptionthat the leaves are diffusely distributed, so that they act as a continuum-likesink for momentum. The horizontal equation of motion in the absence of apressure gradient is then

−∂τ∂z−Dc = 0

where τ(z) is the horizontal shear stress in the air and Dc is the momentum sinkterm, or the drag experienced by the foliage per unit volume of air. Further itis assumed that the shear stress is proportional to the velocity gradient

τ = −ρ0K∂U

∂z

and that Dc is proportional to U2 as

Dc = ρ0CdAc12U2

with the the drag coefficient, Cd ≈ 0.25 and Ac is the surface area of leaves perunit volume of air.

80

a) With the assumption of a constant mixing length l in the canopy so thatK = l2|∂U/∂z|, and Cd, Ac is constant, solve the velocity profile in the canopy(hint: substitute an exponential function). Use U = U(h0) at z = h0.b) Solve the corresponding expression for the exchange coefficient.c) Assuming that the velocity profile and its derivative match at h0 with thelogarithmic wind profile kU/u? = ln((z − d)/z0). Derive an expression for z0 interms of the displacement height (d), h0, etc. What is a typical value for z0/h0?Take d/h0 = 2/3 and h0(CdAc/(4l2))1/3 ∼ 3.

Exercise 10Exercise 10Exercise 10In the RANS approximation the buoyancy production, B, in the turbulent ki-netic energy equation is given by:

B =g

θ0wθ

The buoyancy production is equal to the release of potential energy of the sys-tem.a) Give an expression for the vertically integrated potential energy (P) of theatmosphere.b) Of course we are only interested in the change of potential energy in time.Therefore, we only consider the change in potential energy compared to somereference state (P0). Show that P−P0 per unit area is approximately given by:

P − P0 = −ρ0

∫ z

0

g

Θ0(θ − θ0) z′ dz′

c) Using the Reynolds averaged equation for the potential temperature (horizon-tally homogeneous boundary layer, see Section 3.3), derive the expression for thechange of potential energy of the system in terms of the buoyancy production(B):

P − P0 = −ρ0

∫ t

0

∫ z

0

z′∂B

∂z′dz′ dt

d) Apply this result to the convective boundary layer (without shear) with agiven surface heat flux. Discuss the vertically integrated kinetic energy equationand its relation with the change in potential energy of the boundary layer.

81

Exercise 11Exercise 11Exercise 11The goal is to ’derive’ the mixing length expression for the exchange coefficients(Km and Kh). Neglecting the material derivative (dq/dt) and neglecting thetransport term in the Turbulent Kinetic Energy (TKE) equation we obtain:

0 = −uiuj∂Ui∂xj

+g

θ0wθ − ε

We parametrize the momentum flux, potential temperature flux and viscousdissipation as:

−ui uj = Km

(∂Ui∂xj

+∂Uj∂xi

)≡ KmSij

−wθ = Kh∂Θ∂z

ε = cεq

32

l

where l is a length scale, q the turbulent kinetic energie and cε a constant.a) If we assume that the exchange coefficients Km,h can be written as theproduct of a length and a velocity scale: Km,h = cm,h l q

12 , then solve the

kinetic energy q in terms of l, cε, cm,h, Rig and the deformation tensor Sij .Where Rig is given by:

Rig =gθ0∂Θ∂z

12S

2ij

b) Give the corresponding expression for Km,h.c) Which value of the critical gradient Richardson number is obtained if ch/cm ≈3?

Exercise 12Exercise 12Exercise 12In a 3-dimensional Large Eddy Simulation (LES), the large scale turbulent mo-tions (anisotropic) are explicitly resolved whereas the smaller scale motions(isotropic) have to be parametrized. In an LES model with grid resolution∆, the subgrid fluxes for momentum and heat are parametrized as:

−uiuj = Km

(∂Ui∂xj

+∂Uj∂xi

)−wθ = Kh

∂Θ∂z

where Km,h are the turbulence exchange coefficients. Uppercase variables de-note averages over the grid volume. The resolution (∆) of the LES has to bein the inertial subrange (see Fig. 3.3). The exchange coefficients are written

as the product of a length and a velocity scale: Km,h = cm,h∆q12m, where qm

is the sub-grid kinetic energy and cm,h are constants. For an LES model this

82

expression can exactly be derived from the inertial subrange theory, for whichthe turbulence is isotropic.The spectrum of the turbulence kinetic energy, E(k) and potential temperaturevariance Eh(k) in the inertial subrange is given by:

Em(k) = αε23m k−

53

Eh(k) = βεhε− 1

3m k−

53

where α ≈ 1.5 and β ≈ 1.0 are constants, k is the wavenumber, εm is the dissi-pation rate of turbulence kinetic energy and εh the dissipation rate of potentialtemperature variance. The subgrid turbulence kinetic energy (qm) and potentialtemperature variance (qh) are given by:

qm =∫ ∞kc

Em(k) dk (2a)

qh =∫ ∞kc

Eh(k) dk (2b)

where kc(= 2π/λc) and λc ≈ 2∆ is the cut-off wavenumber. For isotropicturbulence dissipation rates are given by:

εm = 2Km

∫ kc

0

k2 Em(k) dk (3a)

εh = Kh

∫ kc

0

k2 Eh(k) dk (3b)

a) Derive from (2a,b) an expression for qm and qh in terms of kc, εm and εh.b) Use (3a,b) to derive an expression for Km and Kh in terms of kc and εm,and give the ratio Km/Kh.c) Note that the expressions for the turbulence exchange coefficient can be

written as: Km,h = cm,h∆q12m. Derive the values of cm,h in terms of α and β.

The results obtained above can be substituted in the following simplified subgridturbulence kinetic energy equation:

0 = −ui uj∂Ui∂xj

+g

θ0wθ − ε

d) Which critical gradient Richardson number is then obtained?

83

Exercise 13Exercise 13Exercise 13Consider the TKE (3.24).a) Give the physical interpretation of all terms and derive the equation for astationary, homogeneous surface-layer. Neglect the transport term. Note thatin the surface layer we have

−uw = u2? = constant

−θw = θ?u? = constant.

b) The flux Richardson number is defined as

Rif =gθ0wθ

uw dUdz

.

Interpret this number in view of the results obtained under a. Show that Rif ≤ 1and express Rif as a function of z/L. Make a sketch of this relationship usingthe empirical functions

φm ≡κz

u?

dU

dz= (1− 16

z

L)−

14

φm = 1 + 5z

L,

for z/L < 0 and z/L > 0, respectively.c) Motivate the following closure hypotheses

u2? = Km

dU

dz

ε =q

32

l,

where Km = l q12 and l a length scale.

d) Express Km in the parameters u?, l and Rif .e) Assume l = κz. Show that in a neutral case U(z) ∝ ln z. Derive an asymp-totic expression for Km in the limit u? → −∞.

CHAPTER 4CHAPTER 4CHAPTER 4

Exercise 14Exercise 14Exercise 14Make a sketch and discuss in an unstable and stable boundary-layer the verticalprofiles ofa) mean wind and potential temperatureb) variances of velocity and temperature fluctuationsc) momentum and heat flux

84

Focus on differences between stable and unstable conditions. List a number ofscaling parameters in the various regions of the ABL.

Exercise 15Exercise 15Exercise 15Consider the scaling of local turbulence parameters in the convective boundarylayer (h/− L >> 1).a) Which scaling parameters apply in the Monin-Obukhov theory? In whichpart of the CBL does this theory apply? Apply this scaling to the vertical

velocity variance, σw ≡ w212 .

b) Which parameters are important in convective scaling and in which part ofthe CBL does this apply. Get again an expression for σw in this region.c) How would you model σw in the intermediate region?

Exercise 16Exercise 16Exercise 16a) What is Monin-Obukhov scaling? Apply it to the following quantities:b) the wind and temperature gradient, dU/dz and dΘ/dz, respectively,Show that with the results obtained in a, the momentum and heat flux at thesurface can be obtained from measurements of mean temperature and windspeed.

Exercise 17Exercise 17Exercise 17The momentum equations for a homogeneous boundary layer in a rotating frameare

∂U

∂t= fV − 1

ρ0

∂Π∂x− ∂uw

∂z

∂V

∂t= −fU − 1

ρ0

∂Π∂y− ∂vw

∂z

0 = − 1ρ0

∂Π∂z− g

a) Introduce the definition of the geostrophic wind,−→Vg. Show that in a barotropic

ABL the geostrophic wind does not vary with z.b) Show that in a flow with

−→Vg = 0 the mean kinetic energy vanishes. (Formulate

the energy equation, integrate it over the interval z = 0−h and use U(0), V (0) =0, uw(h), vw(h) = 0 and the closure: uw = −KdU/dz, vw = −KdV/dz).

Exercise 18Exercise 18Exercise 18For steady state conditions and neglecting the turbulent transport and advection

85

terms, the TKE equation reduces in horizontally homogeneous conditions to:

0 = −uw∂U∂z− vw∂V

∂z+

g

θ0wθ − ε (4)

which states that the turbulence kinetic energy budget is completely determinedby local parameters. We express the fluxes in terms of their local gradients usingthe turbulent exchange coefficient Km :

uw = −Km∂U

∂z

vw = −Km∂V

∂z

Using the definition

τ = (uw2 + vw2)12

and multiplying (4) with κz/τ32 , we obtain (κ is von Karman’s constant ∼ 0.4):

0 = φm −z

Λ− φε

where

φm =κz

τ12

[(∂U

∂z)2 + (

∂V

∂z)2]

12

Λ = − τ32

κ gθ0wθ

φε =κz

τ32ε

This equation states that the local dimensionless wind shear (φm) and dissi-pation (φε) only depend upon the stability parameter z/Λ. This is the mostgeneral form of local scaling and is used in the neutral and stable boundarylayer and the surface layer.Typically stability is discussed in terms of the stability parameter z/Λ or thegradient Richardson number (Ri).

a) Derive that the gradient Richardson number is related to z/Λ throughRig = z

Λφh

φ2m

.b) In the very stable limit (z/Λ >> 1 ), it is typically assumed that both φmand φh increase linearly with z/Λ, φm,h ∼ βm,h z/Λ (with βm,h ∼ 5). Whatdoes this mean for the gradient and flux Richardson number in the very stablelimit?c) Sometimes it is argued that the flux Richardson number (Rif ) has a criticalbehaviour and the gradient Richardson number has not . What could be thephysical argument for this?(d) We (empirically) assume a critical Rif number and that Rig is related tothe turbulent Prandtl number (Prt) through Prt = 1 + 4Rig (Kim and Mahrt,1992).

86

What does this imply for φm, φh and the critical gradient Richardson number?Reference: Kim J. and L. Mahrt, 1992; Simple formulation of turbulent mixingin the stable free atmosphere and nocturnal boundary layer. Tellus 44A, 381-394.)

Exercise19Exercise19Exercise19Consider a stationary and horizontally homogeneous boundary layer over a flatsurface with geostrophic wind components (Ug, 0). Given the momentum fluxprofile with components

−uw =u2?(1− b zh ) for z < h

0 for z > h

−vw = 0

with b a constant > 1.Sketch the mean wind profile (U, V ) between z = 0 and z = 2h.

Exercise 20Exercise 20Exercise 20In case of a geostrophic wind balance, the wind blows parallel to the isobars,this is typically observed in the free troposphere. Due to surface friction thewind in the boundary layer (for the northern hemisphere) is backing relativeto the geostrophic wind. For a high and low pressure system the wind in theboundary layer is schematically shown below. For the low pressure situationwe get a convergence of the horizontal wind in the boundary layer, and as aresult vertical advection. In order to study the relative importance of verticaladvection and turbulent mixing we consider a stationary and horizontally homo-geneous boundary layer over a flat surface with geostrophic wind (Ug, Vg). Wetake the x-axis parallel to the wind at the surface such that the y-component ofthe surface stress is zero. The angle between the surface and geostrophic windis denoted by α.a) Show from the momentum equations

0 = f(V − Vg)−∂uw

∂z

0 = −f(U − Ug)−∂vw

∂z

that the net transport of air mass (integrated over the whole boundary layer(height h of about 500 m) and relative to the geostrophic wind) has a componentin the y-direction only. Give the net transport also in the ’geostrophic wind’co-ordinate system.b) Assume a circle symmetric flow and a geostrophic wind in a solid-body ro-tation (Vg = ωr with ω = 10−5s−1). Derive an equation for the horizontal

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Figure 7.1: Wind field of the boundary layer (northern hemisphere) with a lowand high pressure system.

mean-vertical velocity at the top of the boundary layer from the continuityequation which reads in cylindrical coordinates:

∂w

∂z+

1r

∂ru

∂r= 0

c) Assume that the integral of U with height is proportional to ωrh. Estimatethe upwelling rate (w) at the top of the boundary layer due to ’Ekman pumping’.d) What are the relative time scales of Ekman pumping and turbulence in theseboundary layers. Or, in other words: how long does it take to bring materialfrom the surface to the top of the boundary layer?

Exercise 21Exercise 21Exercise 21In this exercise we will study the mixed-layer dynamics of the horizontally ho-mogeneous dry boundary layer driven by a constant surface heat flux (wθ)0

and subsidence. Assuming a mean vertical velocity (W (z) ), the mixed layerequations can be summarised as:

∂h

∂t= we +Wh

∂Θm

∂t=

(wθ)0 − (wθ)hh

∂∆∂t

= γθ we −∂θm∂t

where the entrainment velocity is defined as we ≡ −(wθ)h/∆, θm is the constantpotential temperature in the mixed layer, ∆ is the potential temperature jumpat the inversion, h the boundary layer depth, γθ the potential temperature lapserate above the inversion, Wh is the mean vertical velocity at h and β a constant(about 0.2). In order to close the set of equations we assume

(wθ)h = −β(wθ)0

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a) Assume that the large scale subsidence is given by: W (z) = −D z, where D(equal to the divergence of the horizontal wind, typically about 10−5 s−1) is aconstant larger than zero. If we assume h constant in time what is the solutionfor h and ∆?b) Give a physical interpretation of the obtained solution.c) In order to investigate the effect of the surface friction on the wind jumps weconsider the corresponding mixed-layer equations for the wind (positive x-axisin the mean mixed-layer wind direction):

∂Um∂t

= f(Vm − Vg) +(uw)0 − uwh

h(5a)

∂Vm∂t

= −f(Um − Ug) +(vw)0 − vwh

h(5b)

where we assume the geostrophic wind components (Ug, Vg) constant with heightand time. Find the stationary solution for the wind jumps (∆U,∆V ) at theboundary layer height from (5a,b) in terms of D,h, f and (uw)0 . Give a physicalinterpretation of the obtained solution.

Exercise 22Exercise 22Exercise 22Here we address the mixed-layer dynamics of the horizontally homogeneous dryboundary layer driven by a given surface sensible heat flux wθ0. The surfacesensible heat fluxes is positive. The mixed-layer equations for potential temper-ature θm can be summarised as:

∂h

∂t= we (6a)

h∂θm∂t

= (wθ)0 − (wθ)h (6b)

∂∆∂t

= γθ we −∂θm∂t

(6c)

where the entrainment velocity is defined as we ≡ −(wθ)h/∆, θm is the potentialtemperature in the mixed layer, ∆(t) is the potential temperature jump at theinversion, h(t) the boundary layer depth, γθ the potential temperature lapserate above the inversion and β a constant (about 0.2). In order to close the setof equations we assume

(wθ)h = −β(wθ)0

a) We assume that the initially that θm(0) = 302 K, ∆(0) = 7K, h(0) = 800mand γθ = 4K/km. Discuss the time evolution of the potential temperatureprofile and the relevant physical processes.b) Show that the differential equations given by (6ab,c) can be combined togive:

∂∆∂t

+1 + β

β

∆h

∂h

∂t= γθ

∂h

∂t(7)

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c) Note that equation (7) can be written as

∂∆∂h

+1 + β

β

∆h

= γθ (8)

Solve equation (8) and obtain ∆(t) as a function of h(t).

Exercise 23Exercise 23Exercise 23Consider the mixed-layer growth of the horizontally homogeneous boundarylayer driven by a constant heat flux wθ0 . Without subsidence the mixed layerequations can be summarised as:

∂h

∂t= we

h∂θm∂t

= wθ0 − wθh∂∆∂t

= γθ we −∂θm∂t

where the entrainment velocity is defined as we ≡ −(wθ)h/∆, θm is the potentialtemperature in the mixed layer, ∆(t) is the potential temperature jump at theinversion, h(t) the boundary layer depth, γθ the potential temperature lapserate above the inversion and β a constant (about 0.2). In order to close the setof equations we assume

wθh = −βwθ0

a) Give a physical interpretation of the terms in the equations.b) Derive a differential equation in terms of the unknowns h and ∆ (eq. 6.16 ofGarratt), and solve this equation (there exists a more general solution than eq.6.17 of Garratt).c) The differential equation has a homogeneous and a particular solution Thehomogeneous solution has an unknown constant (for instance denoted by A).Determine A from the initial value of ∆, γθ and h. Under which conditions isA = 0? Show these results graphically.d) Determine for A = 0 the evolution of h and ∆ as a function of time.

Exercise 24Exercise 24Exercise 24Describe the structure of the CBL. Identify the scaling parameters and theregion in the CBL where they apply to. Express σu, σw and σθ in these scalesand find the explicit relationship in the intermediate region.

Exercise 25Exercise 25Exercise 25

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Derive the equations of motion for a pure convective boundary layer, where thebuoyancy flux is the only source of turbulence production (no mean flow).

Exercise 26Exercise 26Exercise 26We consider an atmospheric boundary layer.a) Which parameter tells us whether the boundary layer is convective?b) Give an expression for σθ in the surface layer.c) Do the same in the for the mixed layer.c) And again for the intermediate layer, just above the surface layer.

Exercise 27Exercise 27Exercise 27Consider a dry horizontally homogeneous stable boundary layer (SBL) drivenby a constant (in time and space) geostrophic wind (Ug, Vg). If we neglect theradiative flux divergence and molecular diffusion terms in the SBL the Reynoldsaveraged thermodynamic equation becomes:

∂Θ∂t

= −∂wθ∂z

In steady state conditions and in case we can neglect the turbulent transportterm the TKE equation reduces to:

0 = −uw∂U∂z− vw∂V

∂z+

g

θ0wθ − ε

a) How does the heat flux change with height if we assume a quasi-stationaryboundary layer (∂/∂t(∂Θ/∂z) = 0 )? Denote the surface heat flux as wθ (< 0)and take the heat flux at the boundary layer depth (h) equal to zero.b) How is the flux Richardson number defined?c) Write down an equation for the net work against gravity (conversion of kineticenergy into potential energy) and shear production averaged over the wholeboundary layer. Use the stationary horizontal momentum equations to rewritethe shear production term.d) If we assume the flux Richardson number constant throughout the SBL andequal to its critical value (about 0.25), then derive a relationship between h, thefriction velocity at the surface (u?0), (wθ)0 and the geostrophic wind from theresults obtained above. Discuss the relationship between h and the surface heatflux for a given u?0 and geostrophic wind.

Exercise 28Exercise 28Exercise 28a) What could be the cause for the occurrence of the so-called ’low-level jet’during a cloudless night?b) Formulate the equations of motion in this case and solve these.

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c) Make a sketch of wind speed as a function of time for a 24 h period (cloudlessconditions and over land) at a height of 200 and 10 m.

CHAPTER 5CHAPTER 5CHAPTER 5

Exercise 29Exercise 29Exercise 29a) Discuss the various terms in the surface energy balance.b) Make a sketch of the diurnal cycle of these terms on a cloudless day overland.c) Define the Bowen ratiod) Assume that we know the net radiation and the soil heat flux. Which addi-tional measurements do we need to determine the Bowen ratio?

Exercise 30Exercise 30Exercise 30On 30 November 1996 most components of the surface energy balance weremeasured near the 200 m high meteorological tower at Cabauw. The surface inCabauw is relatively flat and at that time of the year covered with short grass.Conservation of energy at the surface requires that:

Rn = G+H + LE

where Rn is the net radiation (positive downwards), G is the heat flux intothe soil, H is the sensible heat flux and LE is the latent heat flux. The heatflux (measured with a sonic anemometer) and net radiation are measured at3m height. The latent heat flux is estimated by the Bowen ratio method. Thesoil heat flux is calculated from the volumetric heat capacity of the soil andtemperature measurements at different depths in the soil. The surface energybalance on 30 November is shown in the figure.a) Discuss the diurnal variation of the different components of the energy balanceshown in the left figure.b) Discuss the relationship between the heat flux as a function of height and thetemporal evolution of the temperature profile after sunset (shown in the figureon the right).c) Which other physical process(es), besides turbulence, can cool the air in theupper part of the boundary layer?d) What is the Bowen ratio method?e) Describe three other (empirical) methods from which the latent heat flux canbe estimated.

Exercise 31Exercise 31Exercise 31We design an experiment in horizontally homogeneous terrain in order to de-termine the turbulent fluxes of heat and momentum in the surface layer. The

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Figure 7.2: The hourly averaged components of the (left) surface energy bal-ance and (right) temperature profiles after 1500 UTC on 30 November 1996 asmeasured in Cabauw. Local time is one hour plus UTC. The arrows indicatetime of sunrise and sunset, respectively.

aerodynamic roughness of the terrain z0 is known. We measure mean wind andtemperature only. Give a minimum configuration of the measurement. Make asketch and discuss it.

Exercise 32Exercise 32Exercise 32Consider the top-layer of the ocean with thickness d and temperature Ts. Thetemperature respons due to solar heating is predicted by

dTsdt

= −Ts − Teτ

+Q(t)ρ0cpd

.

The equilibrium temperature is Te, the response time τ (∼ 105 s) and Q is aperiodic forcing with mean zero. we investigate the response by Fourier seriesexpansions of Q and Ts. Analyze the response in amplitude and phase shift asa function of the frequency of the forcing. Take d= 20 m. discuss both thediurnal and yearly cycle.

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