Zdunkowski W., Bott A. Dynamics of the atmosphere: A course in theoretical meteorology
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13.11 Appendix A: Dimensional analysis 391
13.11 Appendix A: Dimensional analysis
By necessity our discussion must be brief. There exist many excellent books on
dimensional analysis. We refer to Dimensionless Analysis and Theory of Models
by Langhaar (1967), where numerous practical examples are given.
13.11.1 The dimensional matrix
Let us consider a fluid problem involving the variables velocity V , length L,force
F , density ρ, dynamic molecular viscosity µ, and acceleration due to gravity g.
These variables can be expressed in terms of the fundamental variables mass M,
length L,andtimeT as expressed by the following array of numbers, which is
called the dimensional matrix:
VLFρµg
M 001110
L 111−3 −11
T −10−20−1 −2
(13.169)
Consider, for example, the force F . The dimension of force is M
1
L
1
T
−2
, explaining
the entries in the force column. The dimensions of the remaining columns can be
written down analogously.
Any product of the variables V,L,F,...,g has the form
= V
k
1
L
k
2
F
k
3
ρ
k
4
µ
k
5
g
k
6
(13.170)
Whatever the values of the k’s may be, the corresponding dimension of is given
by
[] =
L
1
T
−1
k
1
L
1
k
2
M
1
L
1
T
−2
k
3
M
1
L
−3
k
4
M
1
L
−1
T
−1
k
5
L
1
T
−2
k
6
=
M
k
3
+k
4
+k
5
L
k
1
+k
2
+k
3
−3k
4
−k
5
+k
6
T
−k
1
−2k
3
−k
5
−2k
6
(13.171)
If the product is required to be dimensionless we must demand that the exponents
of the various basic variables add up to zero:
k
3
+k
4
+k
5
= 0,k
1
+k
2
+k
3
−3k
4
−k
5
+k
6
= 0, −k
1
−2k
3
−k
5
−2k
6
= 0
(13.172)
Note that the coefficients multiplying the k
i
, including the zeros, in each equation
are a row of numbers in the dimensional matrix (13.169). Therefore, the equations
for the exponents of a dimensionless product can be written down directly from the
dimensional matrix. This set of three equations in six unknowns is underdetermined,
possessing an infinite number of solutions. In this case we may arbitrarily assign
392 The atmospheric boundary layer
values to three of the k’s, say (k
1
,k
2
,k
3
), and then solve the system of equations
for the remaining k
i
. The solution is given by
k
4
=
1
3
(k
1
+2k
2
+3k
3
),k
5
=
1
3
(−k
1
−2k
2
−6k
3
),k
6
=
1
3
(−k
1
+k
2
)(13.173)
We choose (k
1
,k
2
,k
3
) values such that fractions will be avoided. Three choices of
numbers for (k
1
,k
2
,k
3
) together with the resulting values for (k
4
,k
5
,k
6
)are
k
1
= 1,k
2
= 1,k
3
= 0 =⇒ k
4
= 1,k
5
=−1,k
6
= 0
k
1
=−2,k
2
=−2,k
3
= 1 =⇒ k
4
=−1,k
5
= 0,k
6
= 0
k
1
= 2,k
2
=−1,k
3
= 0 =⇒ k
4
= 0,k
5
= 0,k
6
=−1
(13.174)
All six k
i
are then substituted into (13.170). For each choice of (k
1
,k
2
,k
3
) with
the resulting (k
4
,k
5
,k
6
) we obtain a dimensionless universal -number. These
numbers are known as the Reynolds number Re,thepressure number P ,andthe
Froude number Fr:
1
= Re =
VLρ
µ
,
2
= P =
F
V
2
L
2
ρ
=
p
V
2
ρ
,
3
= Fr =
V
2
Lg
(13.175)
The procedure is quite arbitrary; any values might be chosen for (k
1
,k
2
,k
3
). Suppose
that we choose k
1
= 10,k
2
=−5, and k
3
= 8. Then we get k
4
= 8,k
5
=−16,
and k
6
=−5. The resulting dimensionless number is given as
= V
10
L
−5
F
8
ρ
8
µ
−16
g
−5
(13.176)
This product looks very complicated but it does not really give new information
since we can write this expression as the product
= P
8
Re
16
Fr
5
(13.177)
Regardless of the values we assign to (k
1
,k
2
,k
3
), the resulting dimensionless prod-
uct can be expressed as a product of powers of P, Re,andFr. This fact and the
condition that P, Re,andFr are independent of each other characterize them as
a complete set or group of dimensionless products. We may define a complete set
as follows.
Definition: A set of dimensionless products of given variables is complete, if each
product in the set is independent of the others and every other dimensionless product
of the variables is the product of powers of dimensionless products in the set.
Application of linear algebra to the theory of dimensional analysis resulted in
the following theorem.
13.11 Appendix A: Dimensional analysis 393
Theorem:
The number of dimensionless products in a complete set is equal to the
total number n of variables minus the rank r of the dimensional matrix.
This theorem does not tell us the exact form of the dimensional products but it
does tell us the number of universal products we should be looking for. The theorem
is of great help. For the present case we have n = 6 variables. The rank of a matrix is
defined as the order of the largest nonzero determinant that can be obtained from the
elements of the matrix. In case of the dimensional matrix (13.169) the rank is r = 3,
so the complete set consists of three independent dimensionless numbers. These
are the products Re, P ,andFr which are considered to be universal numbers.
13.11.2 The Buckingham -theorem
Much of the theory of dimensional analysis is contained in this celebrated the-
orem which applies to dimensionally homogeneous equations. In simple words,
an equation is dimensionally homogeneous if each term in the equation has the
same dimension. Empirical equations are not necessarily dimensionally homo-
geneous.
Theorem: If an equation is dimensionally homogeneous, it can be reduced to a
relationship among members of a complete set of dimensionless products.
If n variables are connected by an unknown dimensionally homogeneous equa-
tion Buckingham’s -theorem allows us to conclude that the equation can be
expressed in the form of a relationship among n −r dimensionless products, where
n −r is the number of products in the complete set. It turns out that, in many cases,
r is the number of fundamental dimensions in a problem. This was also the case
above.
13.11.3 Examples from boundary-layer theory
Example 1: Observational data show that the wind profile in the neutral Prandtl
layer is determined by z, u
∗
,anddu/dz. Our task is to find the functional relation
among these three quantities. The dimensional matrix
zu
∗
du/dz
L 11 0
T 0 −1 −1
(13.178)
allows us to write down two linear equations from which the k
i
may be determined:
k
1
+ k
2
= 0, −k
2
− k
3
= 0(13.179)
394 The atmospheric boundary layer
The number of variables is n = 3andtherankisr = 2. The rank and the number
of fundamental variables coincides. With n − r = 1 we expect one dimensionless
universal -number. Choosing k
1
= 1, we obtain k
2
=−1andk
3
= 1. The
universal number is taken as 1/k,wherek is the Von Karman constant. The result
is the logarithmic wind profile
=
z
u
∗
du
dz
=
1
k
(13.180)
Example 2: For the non-neutral Prandtl layer an additonal variable is needed,
which is the MO stability length L
∗
. Now the dimensional matrix must be found
from the four variables z, u
∗
,du/dz,andL
∗
:
zu
∗
du/dz L
∗
L 11 0 1
T 0 −1 −10
(13.181)
Since the rank is r = 2, we expect two universal -numbers. We have two equations
in four unknowns:
k
1
+ k
2
+ k
4
= 0, −k
2
− k
3
= 0(13.182)
We specify (k
1
,k
2
) and obtain (k
3
,k
4
):
k
1
= 1,k
2
= 0 =⇒ k
3
= 0,k
4
=−1,
1
=
z
L
∗
k
1
= 1,k
2
=−1 =⇒ k
3
= 1,k
4
= 0,
2
=
kz
u
∗
du
dz
(13.183)
For convenience we have included the constant k in
2
. Formal application of the
Buckingham -theorem gives
f
1
(
1
,
2
) = 0or
2
= S(
1
) =⇒
kz
u
∗
du
dz
= S
z
L
∗
(13.184)
From boundary-layer theory we know that, for the present problem, the MO func-
tion S(z/L
∗
) is a universal number.
13.12 Appendix B: The mixing length
The mixing length was introduced by equation (13.51) on purely dimensional
grounds. Prandtl (1925) introduced this concept in analogy to the mean free path
of the kinetic gas theory. For a given density of molecules (number of molecules
13.12 Appendix B: The mixing length 395
per unit volume) there exists an average distance that a molecule may traverse
before it collides with another molecule. This is the mean free path, which is about
10
−7
m for standard atmospheric conditions. During each collision momentum will
be exchanged. Moreover, molecular collisions cause the molecular viscosity or
internal friction.
Prandtl introduced a mixing length l for turbulent motion, analogous to the
free path on the molecular scale, by assuming that a portion of fluid or an eddy
originally at a certain level in the fluid suddenly breaks away and then travels a
certain distance. While the eddy is traveling it conserves most of its momentum
until it mixes with the mean flow at some other level. Let us assume that we have
an average horizontal velocity at level z. An eddy originating at this level carries
the horizontal momentum of this level. After traversing the mixing length l
at the
level z + l
it will cause the turbulent fluctuation
v
h
=
v
h
(z) −
v
h
(z + l
) =−l
∂
v
h
∂z
(13.185)
if the Taylor expansion is discontinued after the linear term. According to (13.9)
and (11.36) the stress vector of the eddy is defined by
T = i
3
· R =−ρw
v
=−ρw
v
h
(13.186)
We have ignored the molecular contribution and the density fluctuations and have
retained only the horizontal part of the fluctuation vector. On substituting the
fluctuation v
h
into (13.186) we find that the stress vector of the eddy may be
expressed in terms of the vertical gradient of the mean velocity:
T =
ρl
w
∂
v
h
∂z
(13.187)
For continuity of mass we require
v
h
=
w
(13.188)
For the upward and downward eddies we must have
w
> 0,l
> 0orw
< 0,l
< 0(13.189)
so that the product of the velocity fluctuation w
and the mixing length l
is always
a positive quantity and the signs of the corresponding w
and l
are always identical.
From (13.185) it follows that the fluctuation may be expressed as
w
= l
∂
v
h
∂z
(13.190)
396 The atmospheric boundary layer
Thus the stress vector of the eddy may be written as
T =
ρl
l
∂
v
h
∂z
∂
v
h
∂z
=
ρl
2
∂
v
h
∂z
∂
v
h
∂z
with l
2
= l
l
(13.191)
where l is the mean mixing length. Taking the scalar product i
1
· T and assuming
that the mean flow is along the x-axis only, we obtain
i
1
· T = τ = ρl
2
∂u
∂z
2
= A
v
∂u
∂z
with A
v
= pl
2
∂u
∂z
=
ρK
v
(13.192)
This expression is the definition (13.52). A
v
is known as the Austauschkoeffizient.
The similarity to the molecular situation becomes even more apparent on com-
parison with (1.12) defining the molecular stress tensor. On choosing the velocity
vector v
a
=u(z)i
1
we find for the molecular stress the expression
τ
mol
= i
1
· (i
3
· J) = µ
∂u
∂z
(13.193)
which is identical in form with (13.192).
Our faith in Prandtl’s formulation of l should not be unlimited since the stress
vector depends only on the local vertical gradient, which is not always the case, as
follows from the transilient-mixing formulation mentioned briefly in Chapter 11.
This appendix follows Pichler (1997) to some extent. Similar treatments may be
found in many other textbooks.
13.13 Problems
13.1: Apply the Prandtl layer conditions to equation (11.45) for the mean motion.
Show that this results in the condition τ = constant.
13.2: In case of thermal turbulence for windless conditions (local free convection)
the dimensional matrix may be constructed from the variables ∂
θ/∂z,H/(
ρc
p,0
),
g/
T ,andz.Findthe-number.
13.3: Let us return to the previous chapter. In the inertial subrange can be
expressed as (k) = f
1
(k,
M
).
(a) Use Buckingham’s -theorem to find the -number and then set = κ
K
to
obain .
(b) Now include the dissipation range so that (k) = f
2
(k,
M
,ν). Find the -
numbers.
13.13 Problems 397
13.4:
(a) Show that (13.88) satisfies the differential equation (13.87).
(b) Suppose that ξ → 0. Is S
l
(0) defined and what is its value? Use
equation (13.88) to prove your result.
13.5: Find S
l
from (13.84) and (13.88) for L
∗
= 8 m representing a stable at-
mosphere. Choose z = 10 m. Discuss your results by comparing them with the
empirical value S
l
= 0.872.
13.6: It has been postulated that, within the Prandtl layer,
u
w
=−k
2
∂u
∂z
4
∂
2
u
∂z
2
2
Assume that −ρ
u
w
= τ . Integrate this expression to find the logarithmic wind
profile for neutral conditions.
13.7: Solve the Ekman-spiral problem with the conditions stated in the text. Instead
of u(z = 0) = 0, v(z = 0) = 0 use the lower boundary condition
z → 0: u + iv = C
∂
∂z
(u + iv)
where C is a real constant. The cross-isobar angle α
0
at the ground may be specified.
Find the height of the geostrophic wind level. Hint: Observe that the integration
constants in equation (13.120) in general are complex numbers.
14
Wave motion in the atmosphere
14.1 The representation of waves
It is well known that the nonlinear system of the atmospheric equations includes
a great number of very complex wave motions occurring on various spatial and
time scales. In this chapter we wish to treat some simple types of wave motion that
can be isolated from the linearized system of atmospheric equations. This system
is obtained with the help of perturbation theory. Before introducing this theory
we will briefly discuss the wave concept. For simplicity we are going to employ
rectangular coordinates.
A periodic process occurring in time or in space is called an oscillation. If both
time and space are involved we speak of wave motion. Let us consider the scalar
wave equation
∇
2
U =
1
c
2
∂
2
U
∂t
2
(14.1a)
where U describes some atmospheric field or a component of a vector field and
c the speed of propagation of the wave. For simplicity let us consider only the
z-direction of propagation so that (14.1a) reduces to
∂
2
U
∂z
2
=
1
c
2
∂
2
U
∂t
2
(14.1b)
A solution to this equation is given by
U(z, t) = U
0
cos(kz − ωt) with ω/k = c (14.2)
provided that the ratio of the constants ω and k is equal to the constant c.The
particular solution (14.2) is known as a plane harmonic wave whose amplitude
is U
0
.
A graph of the function U(z, t) is shown in Figure 14.1. For a given value of
the spatial coordinate z thewavefunctionU (z, t) varies harmonically in time. The
398
14.1 The representation of waves 399
Fig. 14.1 A graph of U versus z at times t and t + t.
angular frequency of the variation with time is the constant ω. At a given instant of
time the wave varies sinusoidally with z. Since the argument of the trigonometric
function must be dimensionless, the constant k, known as the wavenumber,must
have the dimension of an inverse length. Thus, k is the number of complete wave
cycles in a distance of 2π units. From a study of the graph of U (z, t) we see that,
at a certain instant in time, the curve is a certain cosine function whereas at t +t
the entire curve is displaced by the distance z = ct in the z-direction. z is
the distance between any two points of equal phase, as shown in Figure 14.1. For
this reason c is called the phase speed.
Let us now return to the three-dimensional scalar wave equation (14.1) which is
satisfied by the three-dimensional plane harmonic wave function
U(x,y,z, t) = U
0
cos(k · r − ωt)withk = i
1
k
x
+ i
2
k
y
+ i
3
k
z
(14.3)
Here r is the position vector and k the propagation vector or wave vector.Often
the solution is written in the following equivalent form
U = U
0
cos ω
k · r
kc
− t
= U
0
cos ω
n · r
c
− t
(14.4)
where n = k/k is the unit normal.
In order to interpret equation (14.3) let us consider constant values of the argu-
ment of the cosine function
k · r − ωt = k
x
x + k
y
y + k
z
z − ωt = constant (14.5)
which describes a set of planes called surfaces of constant phase. Consider the
plane shown in Figure 14.2, where the vector r points to a general point on
the plane while k is a vector that is perpendicular to the plane. The vector u in the
plane is perpendicular to k so that their scalar product vanishes:
k · u = k · (r − k) = 0(14.6a)
400 Wave motion in the atmosphere
Fig. 14.2 The equation of the plane k · r = k
2
.
Since k · k = k
2
x
+ k
2
y
+ k
2
z
= k
2
we obtain
k · r = k
x
x + k
y
y + k
z
z = k
2
(14.6b)
The normal form of the equation of the plane is given by
k
x
k
x +
k
y
k
y +
k
z
k
z = k (14.6c)
Hence equation (14.5), indeed, is the equation representing plane surfaces of con-
stant phase.
Let us return to the argument of the cosine function of equation (14.2), which is
called the phase of the harmonic wave. There is no reason why the magnitude of
the wave could not be anything one would like it to be at time t = 0andatz = 0.
This can be achieved by shifting the cosine function by introducing an initial phase
so that the phase is given by
ϕ = kz − ωt + (14.7)
Without loss of generality we will set = 0. When we envision a harmonic wave
sweeping by, we determine its speed by observing the motion of a point at which
the magnitude of the disturbance remains constant. Thus, the speed of the wave is
the speed at which the condition of constant phase travels, or
dz
dt
ϕ=constant
=−
∂ϕ
∂t
z
∂ϕ
∂z
t
=
ω
k
=
2πν
k
=
L
τ
= c with L =
2π
k
,τ=
1
ν
(14.8)
Here L is the wavelength and τ the period of the wave.