1
Lecture Notes on Fluid Dynamics
(1.63J/2.21J)
by Chiang C. Mei, 2002
7.4
Steady onshore wind in a shallow Sea
Let us model the effect of turbulence by a constant eddy viscosity. Assume that convective
inertia is negligible, the seabed is horizontal and vertical shear is important, the governing
equation in a shallow sea are
∂u ∂v ∂w
+
+
=0
(7.4.1)
∂x ∂y
∂z
∂η
∂2u
∂u
− f v = −g
+ν 2
∂t
∂x
∂z
∂v
∂η
∂2u
+ f u = −g
+ν 2
∂t
∂z
∂z
(7.4.2)
(7.4.3)
The boundary conditons are
u = v = w = 0, z = −h
∂u
∂v
µ
= τxS , µ
= τyS
∂z
∂z
As in a thin boundary layer, the vertical shear dominates.
Integrating over depth and defining the horizontal transport rates,
U=
Z
0
udz,
V =
−h
Z
0
vdz
(7.4.4)
(7.4.5)
(7.4.6)
−h
then the
∂η
∂U
∂V
+
+
=0
∂t
∂x
∂y
∂U
∂η τxS
τB
− f V = −gH
+
− x
∂t
∂x
ρ
ρ
S
τyB
∂η τy
∂V
+ f U = −gH
+
−
.
∂t
∂y
ρ
ρ
where
τxS
τxB
"
∂u
= ρν
∂z
"
∂u
= ρν
∂z
#
#
,
τyS
z=0
,
z=−h
As an order estimate
U∼
τyB
"
#
"
#
∂v
= ρν
∂z
∂v
= ρν
∂z
τS
u2
∼ ∗
ρf
f
(7.4.7)
z=0
(7.4.8)
z=−h
2
where u∗ is the friction velocity. A vertical boundary layer (of Ekman) can exist wherein
Coriolis force ρf U balances the viscous stress
U 1
∂u
∼ ρν
∂z
h δ
τ B ∼ ρν
Therefore, the Ekman layer thickness is
δ=O
s !
ν
f
A typical value of eddy viscosity is ν = 1 cm2 /s and f = 10−4 1/s. Therefore the Ekman
layer thicknss is O(1) m.
Our strategy is to get the horizontal transport, then the details of the boundary layers.
7.4.1
Wind setup due to steady onshore wind
Consider an infinitely long coastline along the x axis. Assume τyS to be a given constant.
Consider the steady state ∂/∂t = 0 and ignore τ B first. Beginning from the equations :
∂V
∂U
+
= 0.
∂x
∂y
∂η
∂x
∂η τyS
+
+f U = −gH
∂y
ρ
−f V
= −gH
with
V = 0, y = 0
(7.4.9)
Because of the infinite coast,
U=
we must have
∂η
= 0,
∂x
∂V
= 0,
∂y
It follows that V = 0 everywhere, and
gH
meaning that there is a sea-level Set-up.
τyS
∂η
=
.
∂y
ρ
(7.4.10)
3
τyS
η =
y + constant
gH
u2
ρu2∗
=
y = ∗ y.
ρgH
gH
If u∗ = 1 cm/sec then τyS = 0.1 Pa. If g = 10 m/sec2 and H = 30 m
2
(10−2 )
u2∗
=
= 3 × 10−7 .
gH
10 · 30
Note that
1
psi.
670
H = 30m the set up is calculated as follows.
1 atm = 105 N/m2 ,
For g = 10m/s2
1N/m2 = 1P a =
∂η
∂y
τyS
0.1 Pa
3 Pa
3 × 10
−7
∆y
100 km
300 km
∆η
3 cm
3 m
Although there is no mean flow (or flux), there is internal flow. We now look at the
detailed distribution in z, by deviding the depth into three parts: the geostrophic interior,
the surface Ekman layer, and the bottom Ekman layer.
7.4.2
Geostrophic core
Outside the boundary layers, we have
−f vg = −g
∂η
= 0,
∂x
∂η
u2
=− ∗
∂y
H
In this geostropic balance, there is a longshore current in the core.
f ug = −g
7.4.3
Surface Ekman layer
Now viscosity is important, so that
∂η
∂2u
+ν 2
∂x
∂z
∂η
∂2v
f u = −g
+ν 2.
∂y
∂z
−f v = −g
(7.4.11)
(7.4.12)
4
For this example
τyS
∂η
ρu2∗
u2
=
=
= ∗
∂y
ρgH
ρgH
gH
∂η
= 0,
∂x
hence
∂2u
(7.4.13)
∂z 2
u2
∂2v
fu = − ∗ + ν 2 .
(7.4.14)
H
∂z
As long as δ/H 1, bounday-layer approximation can be made. Let us make some estimates
based on empirical data, cited from Csanady :
−f v = ν
δ = 0.1
u∗
,
f
ν=
u2∗
,
200f
Re∗ =
u∗ δ
= 20
ν
Pedlosky :
ν = 1 ∼ 103 cm2 /sec
s
1 ∼ 103
cm = 102 cm ∼ 3 × 103 cm.
10−4
Let the total velocity in the surface boundary layer be
δ=
u = ug + uE = −
u∗ 2
+ uE , v = v g + v E = v E .
fH
(7.4.15)
so that (uE , vE ) are the boundary layer corections. Then for z < 0, we have
∂ 2 uE
−f vE = ν
∂z 2
f uE = ν
(7.4.16)
∂ 2 vE
∂z 2
(7.4.17)
The boundary conditions are
ν
∂uE
= 0,
∂z
ν
∂vE
= u2∗
∂z
uE , v E → 0
on z = 0
z → −∞.
This is the Ekman boundary-layer problem. The solution is best obtained by introducing
the complex velocity,
qE = uE + ivE
then
qE = ν
∂ 2 qE
∂z 2
5
if
d 2 qE
−
qE = 0
2
dz
ν
or
Let the solution be of the form,
qE ∝ eDz
then
D2 −
Since
if
=0
ν
1+i
(i)1/2 = ±eiπ/4 = ± √ .
2
We get


1+i z 
z/δ
qE = A exp  √ q
= A e(1+i) .
2
ν/f
Let
(7.4.18)
s
2ν
(7.4.19)
f
denote the boundary layer thickness. Apply the boundary condition on the sea surface,
δ=
∂qE 2
= i u∗
ν
∂z 0
hence
A=
i u2∗ δ
.
(1 + i)ν
The solution is
2
iδ u2∗ (1+i)z/δ
δ u∗
qE = u E + i v E =
e
=
(1 + i) ez/δ
(1 + i)ν
2ν
z
z
cos + i sin
δ
δ
(7.4.20)
Separating real and imaginary parts, we get the velocity components,
δ u2∗ z/δ
z
z
uE =
e
cos − sin
2 ν
δ
δ
(7.4.21)
δ u2∗ z/δ
z
z
vE =
e
cos + sin
.
2 ν
δ
δ
The hodograph is shown in Figure 7.4.1,
(7.4.22)
1. Physical Remark #1:
Maximum velocity occurs on z = 0:
!
uE (0)
vE (0)
u∗
U
u∗
=
=
=
.
u∗
u∗
fδ
u∗ h
fh
and is 45 degrees to the right of wind.
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Figure 7.4.1: Ekman Boundary layer.
2. Physical Remark #2 :
The total flux in Ekman layer is
UxE + iVyE =
Z
0
dz (uE + i vE ) =
−∞
δ u2∗
Z
0
Z
0
dz qE
−∞
e(1+i)z/δ dz
2ν
−∞
δ u2∗
δ
δ 2 u2∗
u2
z/δ 0
=
(1 + i) ·
· e(1+i) =
= ∗.
−∞
2ν
1+i
2ν
f
=
where
(1 + i)
2ν
.
f
Therefore the total mass flux in Ekman layer is 90 degrees inclined with respect to
wind.
δ2 =
3. Physical remark # 3:
Note that the flux in the surface Ekman layer is counter balanced by the geostrophic
return flow beneath. This implies the assumption that the bottom Ekman layer is very
weak and contributes little to the flux. Let us check.
7.4.4
Bottom Ekman layer
The total flow is governed by
−f v = ν
∂2u
∂z 2
f u = f ug + ν
∂2v
.
∂z 2
7
Let
uE = u − u g
so that
vE = v
(7.4.23)
∂ 2 vE
∂ 2 uE
,
f
u
=
ν
(7.4.24)
E
∂z 2
∂z 2
Let us shift to new coordinates with the origin on the sea bed so that the boundary conditions
are
z → ∞, uE , vE → 0
(7.4.25)
−f vE = ν
and
z = 0, uE = −ug , vE = 0
Let
qE = uE + ivE
qE = A e−(1+i)z/δ
Since there is no slip at z = 0
qE = −ug .
We conclude,
A = −ug .
and
qE = (uE + ivE ) = −ug e−(1+i)z/δ
z
z
.
= −uG e−z/δ cos − i sin
δ
δ
Therefore,
z
uE = −ug e−z/δ cos
δ
z
−z/δ
vE = ug e
sin .
δ
The bottom shear stress is
ν
at z = 0
∂qE
1+i
=
ug e−(1+i) z/δ
∂z
δ
∂qE 1+i
1 + i u2∗
ν
=
u
.
g = −
∂z 0
δ
δ fH
The total flux in bottom Ekman layer is
UE + iVE =
Z
∞
0
= −ug
(uE + ivE ) dz
Z
∞
0
e−(1+i/δ)z dz
∞
1
e−(1+i/δ)z 0
−(1 + i)/δ
2
δ
u
δ
δ
u2
= −ug
= ∗
= (1 − i) ∗
1+i
fH 1 + i
2
fH
= −ug
(7.4.26)
(7.4.27)
8
It is of order δ and is directed at 135 degrees to the right of wind.
7.4.5
Summary
:
• Total Ekman layer flux on top is u2∗ /f
• Total geostropic flux is −u2∗ /f
• Total bottom Ekman layer flux is very small O
h
u2∗
f
i
δ
H
.