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Lecture Notes on Fluid Dynamics (1.63J/2.21J) by Chiang C. Mei, 2006
7.6 Transient longshore wind
[Ref]: Chapter 14, p. 195 ff, Cushman-Roisin Csanady: Circulation in the Coastal Ocean
Figure 7.6.1: Longshore wind
Let the x be the coast, and the sea is in y > 0, as seen in Figure 7.6.1. In view of the last section, we ignore the bottom stress. Assume that the wind is uniform in space but transient in time, so that ∂/∂x = 0, The flux equations are
∂η ∂t
∂V
∂y
∂U
∂t
− f V =
τ (^) xS ρ
∂V
∂t
∂η ∂y
The boundary condition on the coast x = 0 : U = 0.
7.6.1 Sudden long-shore wind
Let the wind stress be
τ (^) xS =
{ 0 , t ≤ 0 , T, t > 0. (7.6.4)
the initial conditions are η, U, V = 0, t = 0, ∀y. (7.6.5)
This initial-boundary value problem can be solved by Laplace transform (Cr´epon, 1967). The solution consists of two parts: one part is oscillatory and decays with time; the other
part increases monotonically with time. To avoid the complex mathematics we only examine the latter which is the dominant part for large time,
U = t U¯ (y), V = V¯ (y), η = tη¯(y) (7.6.6)
The oscillatory part is needed to ensure the initial condition on U. This solution clears fails for very large t, indicating the there is no steady- state solution in the linearized approxi- mation. For a realistic theory nonlinearity must be accounted for, or the wind duration has to be finite. It is easy to see from (7.6.1) to (7.6.3) that
η¯ +
d V¯ dy
f U¯ = −gh
dη¯ dy
U¯ − f V¯ = T /ρ (7.6.9)
These three equations can be combined into one :
d^2 V¯ dy^2
f 2 gh
V¯ = f T ρgh
The solution satisfies no flux on the coast is
V¯ = − T ρf
( 1 − e−x/Ro
) (7.6.11)
where
Ro =
gh f
is called the Rossby radius of deformation. If we take f = 10−^4 1/s in a shallow sea of h = 10 m the Rossby radius is about 10^5 m = 100 km. It is easy to find that η = tη¯ = t
T
ρgh
e−y/Ro^ (7.6.13)
and
U = t U¯ = t
T
ρf
e−y/Ro^ (7.6.14)
Clearly when y/Ro 1, the coast line has no influence. The flux is U = 0, V = −T /ρf , and is inclined to the right of the wind by 90 degrees, as predicted by the Ekman layer theory. The sea surface rises if T > 0 (coast is on the right of wind), and sinks near the coast if T < 0 (coast is on the left of wind).
Low frequency: ω < f
The solution to (7.6.22) bounded at infinity is
η 0 = A e−y/R^0 (7.6.23)
where
R 0 =
√ gh f 2 − ω^2
is the modified Rossby radius. Applying the B.C. on the coast : V 0 = 0, y = 0, we get from (7.6.20),
dη 0 dy
f ρgH
τ 0 ω
(7. 6 =.23) −A
R 0
and,
A = −
τ 0 ω
f ρgh
R 0 (7.6.25)
Hence
η 0 = −
f τ 0 ρωgh
R 0 e−y/Ro
and finally
η = −
f τ 0 ρωgh
R 0 e−y/R^0 e−iωt^ (7.6.26)
Its real part is the solution. Now from Eqn. (7.6.20), we get
U =
τoω ρ(f 2 − ω^2 )
( 1 −
f 2 ω^2
e−y/R^0
) e−iωt. (7.6.27)
From Eqn. (7.6.21)
V 0 = −
iτ 0 f ρ(f 2 − ω^2 )
( 1 − e−y/R^0
) (7.6.28)
(7.6.29)
Hence,
V = −
τ 0 f ρ (f 2 − ω^2 )
( 1 −
f 2 ω^2
e−x/R^0
) sin ωt. (7.6.30)
If τ 0 > 0 (wind blows eastward, along the coast), the sea level near the coast sinks at the start.
High frequency : ω > f
Of the two possible oscillatory solutions to (7.6.22), we must choose the one that represents outgoing waves at infinty (the radiation condition),
η 0 =^ Beikx,^ (7.6.31)
where the wavenumber is the inverse of the modified Rossby radius of deformation,
k =
√ ω^2 − f 2 gh
It is easy to show that from the no-flux confditon on the coast thata
B =
−iτ 0 f ρghωk
hence, in complex form,
η = −
iτ 0 f ρghωk
eikx−iωt^ (7.6.34)
or, in real form,
η = −
τ 0 f ρghωk
sin(kx − ωt), (7.6.35)
The final formulas for U, V are left as an exercise. Homework: For the oscillating-wind problem, find the flow in the inviscid interior and then the flow in the surface boundary layer.
