48
K. Myrberg and A. Lehmann
They can be written in the following form for a sea with finite depth:
A v
d 2 q
dz 2 − if q = 0.
(2.8)
Their general solution is
q = C 1 e
λz
+ C 2 e
−λz , λ=
1 + i
2
f
2A v
.
(2.9)
In the classical Ekman layer theory the length D = π
√
2A v /f is defined as the
Ekman depth. When the depth of the sea H D, the velocity profile draws the famous Ekman spiral. At a depth of z = D the current speed is e −π times smaller than
the surface speed (that is, ≈4 % of the surface speed) and the direction has rotated
clockwise by the angle of π ; consequently, the current at this depth is opposite to
the surface current. At the sea bottom there is also an Ekman layer where the velocity decreases to zero at the bottom with a spiral profile. For a representative eddy
viscosity in the whole Ekman layer, A v ∼ 10 −3 –10 −2 m 2 /s, we have D ∼ 10–40 m.
In the Baltic Sea the eddy viscosity is large and H ∼ D. This means that the speed
decays faster with depth and the rotation is less evident than in the deep ocean, and
the surface and bottom Ekman layers may merge together. The boundary conditions
for Eq. (2.8) are:
q z=0 = 0,
A v =
∂q
∂z
z=H
=
τ a
ρ
,
(2.10)
where z = 0 is the sea bottom, z = H is the sea surface and τ a is the (scalar) wind
stress. The relevant particular solution of Eq. (2.8) is
q = q s
sinh λz
sinh λH
= 0, q s =
τ a
ρA v λ
tanh λH,
(2.11)
where q s is the surface velocity. The resulting ‘truncated’ spiral approaches the
Ekman spiral as λ −1 H (Fig. 2.6). The vertically integrated velocity of the flow
in the Ekman layer (called Ekman transport) is
Q =
H
0
q dz = i
τ a − τ 0
ρf
,
(2.12)
where τ 0 is the bottom drag. If τ 0 = 0, Eq. (2.12) expresses the classical result of
Ekman transport that is perpendicular to the wind stress. In the Baltic Sea this comes
true for the upper layer in deep areas where the halocline exists. In shallow areas
the bottom friction reduces both the transport volume and its veering from the wind
direction. Using the velocity solution (2.11) we have
Q = −i
τ a
ρf
1 −
1
cosh λH
.
(2.13)
K. Myrberg and A. Lehmann
They can be written in the following form for a sea with finite depth:
A v
d 2 q
dz 2 − if q = 0.
(2.8)
Their general solution is
q = C 1 e
λz
+ C 2 e
−λz , λ=
1 + i
2
f
2A v
.
(2.9)
In the classical Ekman layer theory the length D = π
√
2A v /f is defined as the
Ekman depth. When the depth of the sea H D, the velocity profile draws the famous Ekman spiral. At a depth of z = D the current speed is e −π times smaller than
the surface speed (that is, ≈4 % of the surface speed) and the direction has rotated
clockwise by the angle of π ; consequently, the current at this depth is opposite to
the surface current. At the sea bottom there is also an Ekman layer where the velocity decreases to zero at the bottom with a spiral profile. For a representative eddy
viscosity in the whole Ekman layer, A v ∼ 10 −3 –10 −2 m 2 /s, we have D ∼ 10–40 m.
In the Baltic Sea the eddy viscosity is large and H ∼ D. This means that the speed
decays faster with depth and the rotation is less evident than in the deep ocean, and
the surface and bottom Ekman layers may merge together. The boundary conditions
for Eq. (2.8) are:
q z=0 = 0,
A v =
∂q
∂z
z=H
=
τ a
ρ
,
(2.10)
where z = 0 is the sea bottom, z = H is the sea surface and τ a is the (scalar) wind
stress. The relevant particular solution of Eq. (2.8) is
q = q s
sinh λz
sinh λH
= 0, q s =
τ a
ρA v λ
tanh λH,
(2.11)
where q s is the surface velocity. The resulting ‘truncated’ spiral approaches the
Ekman spiral as λ −1 H (Fig. 2.6). The vertically integrated velocity of the flow
in the Ekman layer (called Ekman transport) is
Q =
H
0
q dz = i
τ a − τ 0
ρf
,
(2.12)
where τ 0 is the bottom drag. If τ 0 = 0, Eq. (2.12) expresses the classical result of
Ekman transport that is perpendicular to the wind stress. In the Baltic Sea this comes
true for the upper layer in deep areas where the halocline exists. In shallow areas
the bottom friction reduces both the transport volume and its veering from the wind
direction. Using the velocity solution (2.11) we have
Q = −i
τ a
ρf
1 −
1
cosh λH
.
(2.13)