2.3 Exercise 1: The Surface Ekman Layer
15
2.3 Exercise 1: The Surface Ekman Layer
2.3.1 Task Description
We consider a water column, 500 m in depth, represented by an equidistant vertical
grid spacing of 1 m. The Coriolis parameter is chosen as f = 1 × 10
−4 s
−1 corresponding to mid-latitudes in the Northern Hemisphere. The water column is initially
at rest. The model is forced via prescription of a southerly wind of a wind stress of
τ y = 0.5 Pa in magnitude.
The total simulation time is 5 days. To avoid the appearance of strong inertial
oscillations, the wind stress is linearly adjusted from zero to its final value over the
first 2 days of simulation. The time step is set to Δt = 5 s.
The surface wind stress enters the finite-difference equations implicitly via the
boundary values u
n
0 and v
n
0 . Using Eq. (2.5), these values are calculated from:
u
n
0 = u
n
1 +
τ
wind
x
ρ o A
+
z
Δz
(2.17)
v
n
0 = v
n
1 +
τ
wind
y
ρ o A
+
z
Δz
(2.18)
where A
+
z = 0.5(A z,0 + A z,1 ) with A z,0 representing vertical eddy viscosity near the
sea surface. The following three different eddy-viscosity scenarios are considered
in this exercise:
1. Eddy viscosity is uniform with a constant value of A z = 5 × 10
−2 m
2 s
−1 ;
2. Same as before, but with a local minimum of A z = 4 × 10
−3 m
2 s
−1 around a
depth of 20 m mimicking a reduction of turbulence levels by an assumed strong
local density stratification;
3. Eddy viscosity is calculated from Prandtl’s mixing-length approach (Prandtl, 1925)
according to:
A z = L
2
(∂u/∂z) 2 + (∂v/∂z) 2
where, for simplicity, the mixing length is set to a constant value of L = 2 m.
Only results of the first scenario are presented here. The other scenarios are
included as options in the FORTRAN 95 code and remain for the reader to be tested.
The resultant steady-state flow pattern is visualised via displacements of neutrally
buoyant floats. To this end, a prediction scheme for neutrally buoyant floats is added
to the code. Initially, floats form a vertical line and lateral displacements are predicted with:
X
n+1
= X
n
+ Δt u
Y
n+1
= Y
n
+ Δt v
15
2.3 Exercise 1: The Surface Ekman Layer
2.3.1 Task Description
We consider a water column, 500 m in depth, represented by an equidistant vertical
grid spacing of 1 m. The Coriolis parameter is chosen as f = 1 × 10
−4 s
−1 corresponding to mid-latitudes in the Northern Hemisphere. The water column is initially
at rest. The model is forced via prescription of a southerly wind of a wind stress of
τ y = 0.5 Pa in magnitude.
The total simulation time is 5 days. To avoid the appearance of strong inertial
oscillations, the wind stress is linearly adjusted from zero to its final value over the
first 2 days of simulation. The time step is set to Δt = 5 s.
The surface wind stress enters the finite-difference equations implicitly via the
boundary values u
n
0 and v
n
0 . Using Eq. (2.5), these values are calculated from:
u
n
0 = u
n
1 +
τ
wind
x
ρ o A
+
z
Δz
(2.17)
v
n
0 = v
n
1 +
τ
wind
y
ρ o A
+
z
Δz
(2.18)
where A
+
z = 0.5(A z,0 + A z,1 ) with A z,0 representing vertical eddy viscosity near the
sea surface. The following three different eddy-viscosity scenarios are considered
in this exercise:
1. Eddy viscosity is uniform with a constant value of A z = 5 × 10
−2 m
2 s
−1 ;
2. Same as before, but with a local minimum of A z = 4 × 10
−3 m
2 s
−1 around a
depth of 20 m mimicking a reduction of turbulence levels by an assumed strong
local density stratification;
3. Eddy viscosity is calculated from Prandtl’s mixing-length approach (Prandtl, 1925)
according to:
A z = L
2
(∂u/∂z) 2 + (∂v/∂z) 2
where, for simplicity, the mixing length is set to a constant value of L = 2 m.
Only results of the first scenario are presented here. The other scenarios are
included as options in the FORTRAN 95 code and remain for the reader to be tested.
The resultant steady-state flow pattern is visualised via displacements of neutrally
buoyant floats. To this end, a prediction scheme for neutrally buoyant floats is added
to the code. Initially, floats form a vertical line and lateral displacements are predicted with:
X
n+1
= X
n
+ Δt u
Y
n+1
= Y
n
+ Δt v