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Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.1. The Gulf Stream flows northward along the east coast of the United States with a
surface current of average magnitude 2 m/s. If the flow is assumed to be in geostrophic balance,
find the average slope of the sea surface across the current at a latitude of 45°N. [Answer: 2.1 cm
per km]
Solution 13.1. Given v = 2 m/s, and f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1. For
geostrophic balance:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.2. A plate with water (
ν
= 10−6 m2/s) above it rotates at a rate of 10 revolutions per
minute. Find the depth of the Ekman layer, assuming that the flow is laminar.
Solution 13.2.
Given Ω = 10 rpm = 20
π
rad./min. = 1.047 s–1, the Coriolis frequency is f = 2Ω = 2.09 s–1. From
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.3. Assume that the atmospheric Ekman layer over the earth’s surface at a latitude of
45ºN can be approximated by an eddy viscosity of
ν
V = 10 m2/s. If the geostrophic velocity
above the Ekman layer is 10 m/s, what is the Ekman transport across isobars? [Answer: 2203
m2/s]
Solution 13.3.
Given
ν
V = 10 m2/s, f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, and U = 10 m/s,
the Ekman layer height from (13.28) is
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.4. a) From the set (13.45) – (13.47), develop the following equation for the water
surface elevation
η
(x,y,t):
∂
∂t
∂2
∂t2+f2−gH ∂2
∂x2+∂2
∂y2
#
$
%&
'
(
)
*
+
,
-
.
η
(x,y,t)=0
b) Using
η
(x,y,t)=ˆ
η
exp i(kx +ly −
ω
t)
{ }
show that that the dispersion relationship reduces to
ω
= 0 or (13.82).
c) What type of flows have
ω
= 0?
Solution 13.4. The starting-point equation set is:
∂η
∂
t+H
∂
u
∂
x+
∂
v
∂
y
!
"
#$
%
&=0
,
∂
u
∂
t−fv =−g
∂η
∂
x
,
∂
v
∂
t+fu =−g
∂η
∂
y
. (13.45, 13.46, 13.47)
Apply ∂/∂t to (13.46), multiply (13.47) by f, and add the results to find:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.5. Find the axis ratio of a hodograph plot for a semidiurnal tide in the middle of the
ocean at a latitude of 45°N. Assume that the mid-ocean tides are rotational surface gravity waves
of long wavelength and are unaffected by the proximity of coastal boundaries. If the depth of the
ocean is 4 km, find the wavelength, the phase velocity, and the group velocity. Note, however,
that the wavelength is comparable to the width of the ocean, so that the neglect of coastal
boundaries is not very realistic.
Solution 13.5. Given f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1,
ω
= 2 rev./day =
1.45 x10–4 s–1, and H = 4 km, the axis ratio (as given at the top of page 733) is
axis ratio =
ω
/f = 1.45/1.03 = 1.414.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.6. An internal Kelvin wave on the thermocline of the ocean propagates along the
west coast of Australia. The thermocline has a depth of 50 m and has a nearly discontinuous
density change of 2 kg/m3 across it. The layer below the thermocline is deep. At a latitude of
30°S, find the direction and magnitude of the propagation speed and the decay scale
perpendicular to the coast.
Solution 13.6. The given information is: f = –2Ωsin30° = –0.73x10–4 s–1, H = 50 m, and Δ
ρ
= 2
kg/m3. This is the case of a shallow layer of lighter water, overlying a deep sea. From equation
(8.115), the internal gravity wave speed is:
c=!
g H =gΔ
ρ
ρ
o
H=(9.81ms−1)2kgm−3
103kgm−3(50m)=0.99ms−1
.
These waves propagate southward along the west coast of Australia. The decay scale
perpendicular to the coast is the Rossby radius:
Λ=c
f=0.99ms−1
0.73×10−4s−1=1.36 ×104m=13.6km
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.7. Derive (13.96) for the vertical velocity w from (4.10), (13.48), (13.49), (13.51),
(13.95) by eliminating all other dependent variables.
Solution 13.7. The five starting-point equations are:
∂u
∂x+∂v
∂y+∂w
∂z=0
, (4.10)
∂u
∂t−fv =−1
ρ
0
∂#
p
∂x
,
∂v
∂t+fu =−1
ρ
0
∂#
p
∂y
, (13.48, 13.49)
∂"
ρ
∂t−
ρ
0
N2
gw=0
, and (13.51)
∂w
∂t=−1
ρ
0
∂#
p
∂z−g#
ρ
ρ
0
. (13.95)
Apply ∂/∂t to (13.48), multiply (13.49) by f, and add the results to find:
∂2
∂t2+f2
"
#
$%
&
'u=−1
ρ
0
∂2)
p
∂x∂t+f∂)
p
∂y
"
#
$%
&
'
. (a)
Multiply (13.48) by –f, apply ∂/∂t to (13.49), and add the results to find:
∂2
∂t2+f2
"
#
$%
&
'v=−1
ρ
0
∂2)
p
∂y∂t−f∂)
p
∂x
"
#
$%
&
'
. (b)
Apply ∂/∂x to (a) and ∂/∂y to (b), and add the results to reach:
∂2
∂t2+f2
"
#
$%
&
'∂u
∂x+∂v
∂y
"
#
$%
&
'=∂2
∂t2+f2
"
#
$%
&
'−∂w
∂z
"
#
$%
&
'=−1
ρ
0
∂2
∂x2+∂2
∂y2
"
#
$%
&
'∂)
p
∂t
, (c)
where the first equality follows from (4.10). Now apply ∂/∂t to (13.95), and use (13.51) to
substitute for ∂
ρ
´/∂t:
∂2w
∂t2=−1
ρ
0
∂2#
p
∂z∂t−g
ρ
0
∂#
ρ
∂t=−1
ρ
0
∂2#
p
∂z∂t−g
ρ
0
ρ
0
N2
gw
$
%
&'
(
)=−1
ρ
0
∂2#
p
∂z∂t−N2w
. (d)
Use the definitions
∇H
2≡ ∂2∂x2+∂2∂y2
to rewrite the final equality of (c), and rewrite the ends
of the extended equality (d):
∂2
∂t2+f2
"
#
$%
&
'∂w
∂z=1
ρ
0
∇H
2∂)
p
∂t
, and
∂2
∂t2+N2
"
#
$%
&
'w=1
ρ
0
∂2(
p
∂z∂t
(e,f)
Apply ∂/∂z to (e) and
∇H
2
to (f), and add the results to find:
∂2
∂t2+f2
"
#
$%
&
'∂2w
∂z2+∇H
2∂2
∂t2+N2
"
#
$%
&
'w=1
ρ
0
∇H
2∂2)
p
∂z∂t−1
ρ
0
∇H
2∂2)
p
∂z∂t=0
.
Rearrange the terms noting that
∇H
2+∂2∂z2≡ ∇2
:
∇2∂2w
∂t2+f2∂2w
∂z2+N2∇H
2w=0
,
which is (13.96).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.8. Using the dispersion relation m2 = k2(N2 − ω2)/(ω2 − f2) for internal waves, show
that the group velocity vector is given by
cgx,cgz
!
"#
$=(N2−f2)km
(m2+k2)3 2 (m2f2+k2N2)1 2 m,−k
[ ]
.
[Hint: Differentiate the dispersion relation partially with respect to k and m.] Show that cg and c
are perpendicular and have oppositely directed vertical components. Verify that cg is parallel to
u.]
Solution 13.8. Start with m2 = k2(N2 − ω2)/(ω2 − f2), and algebraically rearrange it to find:
ω
2 = f2 + k2(N2 − ω2)/m2. (1)
Here we note that c = (cx, cz) =
ω
( )
, so the components of the group velocity must be
( )
( )
To find cgy, differentiate (1) with respect to the vertical wave number m:
2
ω
∂
ω
∂m=k2
m2−2
ω
∂
ω
∂m
#
$
%&
'
(−2k2N2−
ω
2
( )
1
m3
.
Set cgz = ∂
ω
/∂m, and rearrange the last equation to find:
cgz =−k2N2−
ω
2
( )
m
ω
m2+k2
( )
. (3)
Here again,
ω
can be eliminated from (3) using (1) to find:
cgz =−k2m N 2−f2
( )
m2+k2
( )
3 2
m2f2+k2N2
( )
1 2
.
Thus we can write: cg =
cgx,cgz
( )
=(N2−f2)km
(m2+k2)3 2 (m2f2+k2N2)1 2 m,−k
( )
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.9. Suppose the atmosphere at a latitude of 45°N is idealized by a uniformly stratified
layer of height 10 km, across which the potential temperature increases by 50°C.
a) What is the value of the buoyancy frequency N?
b) Find the speed of a long gravity wave corresponding to the n = 1 baroclinic mode.
c) For the n = 1 mode, find the westward speed of nondispersive (i.e., very large wavelength)
Rossby waves. [Answer: N = 0.01279 s−1; c1 = 40.71 m/s; cx = −3.12 m/s]
Solution 13.9. The given info. is f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, H = 10
km, and ΔT = 50 °C. At absolute temperature T, the coefficient of thermal expansion for a
perfect gas is 1/T.
a) Therefore the buoyancy frequency is given by:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.10. Consider a steady flow rotating between plane parallel boundaries a distance L
apart. The angular velocity is Ω and a small rectilinear velocity U is superposed. There is a
protuberance of height h << L in the flow. The Ekman and Rossby numbers are both small: Ro
<< 1, E << 1. Obtain an integral of the relevant equations of motion that relates the modified
pressure and the stream function for the motion, and show that the modified pressure is constant
on streamlines.
Solution 13.10. For Rossby and Ekman numbers both much less than unity (Ro << 1 and E <<
1), the Taylor-Proudman theorem gives
2Ω×u=−∇p
ρ
for the momentum equation, where p includes the hydrostatic pressure. And, for incompressible
flow, we must have:
∇ ⋅ u=0
. The boundary conditions w = 0 on z = 0 and L require w = 0
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 13.11. Consider an atmosphere of height H that initially contains quiescent air and N
different cyclonic disks of height H and radius Ri inside which the air rotates at rate Ωi. After
some time, the various cyclonic disks merge into one because of the reverse energy cascade of
geostrophic turbulence. Show that the radius Rf and rotation rate Ωf of the single final disk is
<Begin Equation>
Rf
2=Ωi
2Ri
4
i=1
N
∑Ωi
2Ri
2
i=1
N
∑
and,
Ωf
2=Ωi
2Ri
2
i=1
N
∑
( )
2
Ωi
2Ri
4
i=1
N
∑
.
</End Equation>
by conserving energy and enstrophy. How are these answers different if all of the energy but
only a fraction
ε
(0 <
ε
< 1) of the enstrophy is retained after the merging process? Assume the
relevant horizontal area is the same at the start and end of the disk-merging process.
Solution 13.11. This exercise may solved similarly to Example 13.14. For an atmosphere of
height H, the kinetic energy KE of rotating disks of air with radii Ri undergoing solid body
rotation with rates Ωi is:
KE =
π
4
ρ
oH
i
∑Ωi
2Ri
4=
π
4
ρ
oHΩf
2Rf
44
.
where
ρ
o
is the altitude-averaged density and the second equality applies to the final fully-
Simultaneous solution of these two equations leads to:
Rf
2=Ωi
2Ri
4
i=1
N
∑Ωi
2Ri
2
i=1
N
∑
is ,
Ωf
2=Ωi
2Ri
2
i=1
N
∑
( )
2
Ωi
2Ri
4
i=1
N
∑
.
When only a fraction
ε
of the enstropy is retained, then two equations for simultaneous
