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z y
x o
Wind stress
Mixed Layer and Wind-induced Ekman layer
Interior geostrophic layer
Bottom Ekman layer
Bottom log-boundary layer
MAR650: Lecture 2: Layer Oceans and Mixed Layer Dynamics
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1) Wind-induced surface Ekman layer
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− fvE = Km∂2u∂z2
fuE = Km∂2v∂z2
Coriolis force = Vertical diffusion
a) Current profile: The Ekman velocity decreases and rotates clockwise with depth Ekman spiral.
hE
€
UE (Ekman volume transport)
€
v E
€
τ
€
τ
€
v E
45o
45o
90o
EU
€
τ
b) The Ekman layer thickness (depth):
fKh m
E2
=Directly proportional to turbulent viscosity coefficient and inversely proportional to the Coriolis parameter.
c) The direction of the surface Ekman current: o
E
E
uv 45,1tan === αα
The angle between the wind stress and surface Ekman current is 45o. On the northern hemisphere, the surface Ekman current is 45o on the right of the wind stress.
d) The total volume transport: The volume transport is always 90o to the direction of the wind stress. On the northern hemisphere, it is on the right of the wind stress.
Ekman Pumping:
xfyV
xUwdz
zw
yv
xu y
hzh
EE
∂
∂=
∂
∂+
∂
∂=⇒=
∂
∂+
∂
∂+
∂
∂−=
−∫
τ
ρ10)(
0
If we extend it to a general case with x and y components of the wind, we get
€
w z= −hE=∂U∂x
+∂V∂y
=1ρf
(∂τ y∂x
−∂τ x∂y
) =1ρo f
k ⋅ curlτ
The vertical velocity at the bottom of the surface Ekman layer is proportional to the vorticity of the surface wind stress and it is independent of the detailed structure of the Ekman flow. the vertical velocity is positive for the positive vorticity of the wind stress, and negative for the negative vorticity of the wind stress.
€
curlτ > 0
hE Ekman pumping
2) Interior geostrophic layer
xP
fv
yP
fu gg ∂
∂−=
∂
∂=
ρρ1,1
Turbulent mixing is weak, the motion is approximately geostrophic
Coriolis force = Pressure gradient force
€
FP
€
Fc
P0
P1
P2
t0 t1 t2
€
Fc
€
Fc
€
Fc€
Vg €
Vg
€
FP(The pressure gradient force)
(The Coriolis force)
Geostrophic adjustment
Where does the energy come here? Vorticity of the wind stress (Ekman pumping)
3) The Bottom Ekman boundary layer
∂
∂+
∂
∂−=
∂
∂+
∂
∂−=−
2
2
2
2
1
1
zvK
yPfu
zuK
xPfv
m
m
ρ
ρ Coriolis force=Pressure gradient force+Vertical diffusion
Surface wind stress drives the surface Ekman layer and also transport the energy the interior layer. The energy driving the bottom boundary layer could be come from the interior geostrophic motion.
a) Total velocity and Ekman velocity:
Total velocity (a sum of geostrophic and Ekman velocities) rotates counterclockwise with depth. It reaches the geostrophic current at the top of the Ekman layer and vanishes at the bottom. Ekman velocity decreases and rotates clockwise with the height (from the bottom. It is equal in magnitude and opposite in direction to the geostrophic velocity at the bottom and vanish at the top of the bottom Ekman layer.
u
v
ug uE -ug 0 0
(bottom) (top)
with depth with height
Total velocity Ekman velocity
d) Total volume transport:
The volume transport is always 90o on the right of the bottom frictional force (on the northern hemisphere). Unlike the volume transport in the surface Ekman layer, the transport in the bottom boundary is related to the Ekman layer thickness and magnitude of the geostrophic current in the interior.
EggE
Eg
E
mbyE huu
hhu
fhK
fU
21
2
2
−=−=−=−=τ
Note: τby is the bottom stress exerting from the fluid on the bottom. The bottom frictional force is equal in magnitude and opposite to the bottom stress.
τby
-τby (bottom frictional force)
(transport) UE
4) The bottom log boundary layer
In a very thin layer close to the bottom, the turbulent viscosity-induced stress is much larger than the Coriolis and pressure gradient forces. In this layer, the stress does not change with depth:
€
∂τ∂z
= 0 This is also called the constant stress boundary layer
velocityfrictional the:)(1*
2* uKu
Kzu
zuK
mmm ==
∂
∂⇒
∂
∂=
ρτ
ρτ
According to the definition of the stress, we have
)(ln*** Czuuzu
lu
zu
+=⇒==∂
∂
κκ
Assume that mixing length , where κ is von Karman constant , then we have zl κ=
Define zo is the roughness height, at which turbulence-induced stress equals to zero, then
ozzuu ln*
κ= mm 60~ mm 0.01:oz
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Surface Mixed Layer
General features of vertical structure of the water temperature:
Mixed layer
Thermoclines
Deep weak stratified layer
0=∂
∂
zT
large zT∂
∂
T
Depth
small zT∂
∂
Solar radiation, wind mixing/cooling
Question 1: Why do the thermoclines exist in the ocean?
Sharp vertical gradient of the water temperature
Weak vertical gradient of the water temperature
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h
zTT o γ+=
2hTT o γ−= Mixed layer
Thermoclines
A transition zone between the mixed layer to the deep ocean
An example:
zTT o γ+=
Assume that solar radiation leads to a vertically linear distribution of the water temperate:
Wind mixing and cooling cause a vertical mixing, resulting in a mixed layer of h and a temperature as
2)(1 0 hTdzzT
hT o
ho γγ −=+= ∫
−
An averaged temperature from 0 to -h
For density: pycnoclines; for salinity: thermohalines
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Question 2: Is the oceanic current is a turbulent motion?
Reynold number:
γVLRe =
V: velocity of the motion L: typical scale of the motion γ: fluid’s viscosity
Laboratory experiment: Re > 2000 or 4000 (depending on the property of the fluid), the motion become turbulent!
In the ocean,
Re > 1011 (for example, in the Gulf Stream, V~ 1 m/s, L ~100 km, and γ ~10-6 m2/s)
The oceanic current is a turbulent motion!
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Question 3: What causes turbulence and how is it dissipated?
Shear Instability
Large-scale eddies
Small-scale eddies
Dissipation
Assume that
u2: the typical scale of the kinetic energy of the large-scale eddies;
l: the typical size scale of the large-scale eddies;
The time scale required for the energy transfer of large-scale eddies to small-scale eddies is
€
lu
The energy transfer rate from large-scale eddies to small-scale eddies equals:
€
turbulent kinetic energy time scale
~ u2
l / u=u 3
l
If these small-scale eddy’s energy is dissipated at a rate of ε, then
€
ε ~ u3
lTurbulent dissipation rate
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ρ1 ^ ρ2 ^ ρ3 ^ ρ4
Question 4: How does shear instability cause turbulence?
Moving a blob of water in a stable density gradient leads to vertical oscillations.
The buoyancy force defined as its “weight” equals to:
)( sbg gF ρρ −−=
ρb is the density of the blob; ρs is the density of the surrounding water
The frequency of the blob’s oscillation is related to how fast it changes direction, It is controlled by the buoyancy force. In fact, the frequency of the oscillation is equal to
zgNo ∂
∂−=
ρρ
Brunt-Väisälä frequency
Internal wave frequency:
Nf ≤≤ ω
How could turbulence mixing occur in a stable stratified fluid?
13
One of the most important forces to cause vertical mixing is the vertical shear of the horizontal velocity, i.e,:
zv
zu∂
∂
∂
∂ ,
Example:
Why do the shear of the velocity is so important for mixing?
cigarette
no wind
w > 0
w = 0 w = 0
Smoke plume
In the ocean:
Heavy
Light
Does any shear of the velocity generate turbulence?
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The occurrence of the turbulence mixing is dampened by strong density gradient, but strengthened by strong shear of the velocity. Whether or not turbulence mixing could occur depends on a ratio of vertical stratification to the vertical shear of the velocity.
Richardson number
22 /
∂
∂=
zuNRi
In general, turbulence develops as Ri < 0.25, but recent microstructure measurement shows that critical Richardson number is bigger for the interior ocean mixing or over the slope.
Richardson number is defined as a ratio of the square of the buoyancy frequency to the square of the vertical shear of the horizontal velocity:
Question 5: How is a mixed layer developed? What are the key balance of turbulence energy during the development of a mixed layer?
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A mixed layer is developed with 4 stages:
t
t1/3
t2
t1/3
h
Monin-Obukov Depth
t (time)
h1
h2
h3
h4
Stage 2: the mixed layer continues to deepen, but cooler water below the mixed layer is entrained, the deepening speed of the mixed layer decreases: h ~ t1/3
Stage 1: t << 1/f (time scale << the inertial period), wind stress is limited in a thin layer close to the surface, h ~ t
Stage 3: a strong vertical shear of the velocity is established, shear instability speed up mixing. In this stage, h ~ t2
Stage 4: Energy generated by the vertical shear of the velocity is balanced by the heat-induced buoyancy force, the mixed layer stops deepening. This is a slowly process during which h ~ t1/3.
Qgumh oo
α
ρ 3*2~
=Monin-Obukov Depth:
€
u* = το / f : frictional velocity; ρo: reference density
€
το: surface wind stress, Q: the surface heat flux,
mo: turbulent dissipation coefficient
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The most important properties of a mixed layer is
€
∂V∂ z
= 0; ∂T∂ z
= 0, ∂S∂ z
= 0, ∂ρ∂ z
= 0
Question 6: For a given same wind stress, the wind could drive an Ekman flow, but also wind mixing leads to a mixed layer. Based on the Ekman theory, the wind-induced current has a maximum speed at the surface and decreases exponentially with depth. According to the theory of the mixed layer, no vertical gradient of the velocity exists in a mixed layer. Are these two theories controversial each other?
Answer: No! When mixing occurs, the turbulent viscosity coefficient is very large, turbulent mixing tends to mix density, temperature and salinity, but also tends to mix the momentum. Therefore, the momentum of the Ekman current will be “mix” in the vertical.
In the mathematics, it is equivalent to
−++≈=∂
∂
−++−≈−=∂
∂
)]()][(1[2
cos2
)]()][(1[2
sin2
2
2
2
2
22
2
2
2
2
22
EEEEEo
y
E
hz
Eo
yE
EEEEEo
y
E
hz
Eo
yE
hzO
hz
hzO
hz
fhhze
fhzv
hzO
hz
hzO
hz
fhhze
fhzu
E
E
ρ
τ
ρ
τρ
τ
ρ
τ
As hE→∞ 0=∂
∂=
∂
∂
zv
zu EE
17
Question 7: Is the Ekman theory still valid when vertical mixing occurs?
Answer: Yes.
Why? Mixing tends to re-distribute the momentum but does not change the transport!
The wind-induced Ekman transport:
fU
o
yE ρ
τ= τy
UE Ekman transport
The wind-induced Ekman transport depends only on the wind stress and Coriolis parameter rather than the detailed structure of the Ekman current. Therefore, wind mixing tends to cause the Ekman current to be uniform in the vertical in the mixed layer, but would not be able to change the Ekman
transport!
The Ekman theory is still valid in the mixed layer!
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Mixed Layer Models in the Coastal Ocean
Two types of the mixed layer model:
1) PWP (Price et al., 1986: Price, Weller and Pickel): Mixing is determined based on the criterions of the turbulent motion:
a) static instability; b) mixed layer instability, and c) shear-flow instability
2) Mellor and Yamada 2.5 turbulent closure (Mellor and Yamada, 1974, 1982): Mixing is determined by the turbulent kinetics and mixing length equations—a diffusion
process:
a) Source: velocity shear and buoyancy instability b) Re-distribution: turbulent advection and diffusion c) Sink: turbulent dissipation
PWP is 1-D mixing model which is popular in the open ocean ecosystem studies because it is very simple.
MY2.5 is a 3-D mixing model which is widely used in the coastal ocean.
19
Wind stress + Heat flux
PWP Mixed Layer Model:
Static instability
0>∂
∂
zρ
Mixed layer instability
65.0)( 2 <Δ
Δ=
uhgR
oib ρ
ρ
Shear instability
25.0)/( 2
2
<∂∂
=zu
NRig
Re-distributions of T, S, ρ, U
1) Mixed layer depth 2) Vertical profile of the horizontal velocity; 3) Vertical profile of temperature, salinity, and density
Δρ and Δu are the differences of density and velocity at the bottom of the mixed layer between the mixed and stratified layers.
20
An example: 1-D mixing experiment
∂
∂−−=
∂
∂∂
∂−=
∂
∂∂
∂−=
∂
∂
zfu
tv
zfv
tu
zF
ctT
y
o
x
o
po
τ
ρ
τρ
ρ
1
1
1
F(z) is the short-wave isolation. In this experiment, it changes with diurnal cycle: a diurnal heating
τ
F(z)
21
02:00
22:00
06:00
10:00
14:00
18:00 wind stress
22
Surface heat flux
Wind stress
Observed
Modelled
23
24
MY Turbulent Closure Model
1) Level 2.0 closure model Turbulent production = Turbulent energy dissipation
])()[( 22
zv
zuKP ms ∂
∂+
∂
∂= turbulent shear production
)(z
gKP hb ∂
∂=
ρρ
turbulent buoyancy production
lq /06.0 3=ε turbulent dissipation
where ,2/)( 222 vuq ′+′= zzllκ
κ+
=1max
(Blackadar, 1962, Weatherly and Martin, 1978)
hhmm lqSKlqSK == ,
Therefore,
)1/(068.2688.0)1/(404.152.0
)1/(978.1537.0
jj
jjhm
jjh
RRRR
SS
RRS
−−
−−=
−−=
{
])0346.0316.0(186.0[(725.0 2/12 +−−+= iiij RRRR
25
2) Level 2.5 closure model
lqqbs
qqbs
FzlqK
zEWPPlE
zlqw
ylqv
xlqu
tlq
FzqK
zPP
zqw
yqv
xqu
tq
2)()~
(
)()(2
2
11
2222
22222
+∂
∂
∂
∂+−+=
∂
∂+
∂
∂+
∂
∂+
∂
∂
+∂
∂
∂
∂+−+=
∂
∂+
∂
∂+
∂
∂+
∂
∂
ε
ε
and
lqKlqSKlqSK qhhmm 2.0,, ===
zqglG
GS
GGGS
oh
hh
hh
hm
∂
∂=
−=
−−
−=
ρρ22
676.341494.0,
)127.61)(676.341(354.34275.0
where
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A Simple Mixed Layer Ecosystem Model
€
∂v∂t
+ f k × v =1ρo
∂τ∂z
For a linear case,
In the mixed layer,
€
∂2τ∂z2
= 0
(1)
(2)
Integrating (1) from –h to 0
€
h ∂v∂t
+ f k × v h = τ z= 0 − τ z= −h = τ s − τ z= −h
The stress at the bottom of the mixed layer is related to h !
Question: How to determine the stress at the bottom of the mixed layer?
27
z t=0 t1=t+Δt
Δh Δh t=0
t1=t+Δt
€
τ z= −h = −ρv' ′ w Stress at –h:
When the mixed layer deepens from h to h+ Δh over a time interval Δt, water below the mixed layer intrudes into the mixed layer to participate in mixing, the change rate of the total momentum equals to the difference of the turbulent momentum flux at the h and h+ Δh surfaces, i.e.,
€
v' ′ w = 0 (no motion)
€
v' ′ w ≠ 0h
h+Δh
A
€
d(ρvΔ h A)dt
= ρ A(−v' ′ w z= −h
+ v' ′ w z= −h−Δh
)0
28
As Δh→0, we have
€
τ z= −h = −ρv' ′ w z= −h
= ρv ∂h∂ t
When the mixed layer becomes shallower, no deep water’s intrusion, so
€
d(ρvΔ hA)dt
= 0; τ z= −h= −ρv' ′ w z= −h
= 0
deeper
shallower
Therefore,
€
h ∂v∂ t
+ f k × v h =τ sρ− v ∂h
∂ tH(∂h
∂ t)
0 as 0 0 as1)(
≤=
>>
ξ
ξξH
For example, τx=0, and τy=τo (constant)
o
ofuhtvhfvh
tuh
ρ
τ−=+
∂
∂=−
∂
∂ ,0 (3)
29
The solution is
−=
−=
ftf
vh
ftf
uh
o
o
sin
)1(cos
ρτ
ρτ
Ekman transport Inertial oscillations
Properties:
1) Velocity in the mixed layer decreases as the mixed layer depth increases, but the total transport is determined by the Ekman transport which is independent of the velocity in the mixed layer,
2) The motion consists of two parts: 1) time-dependent inertial oscillations and 2) wind-induced steady Ekman transport. The period of the oscillation is 2π/f.
hvhu ,
0 π/2f π/f 3π/2f 2π/f
t (time)
foo
ρ
τ
30
Consider the nutrient transport in the mixed layer: we have
No QthH
thNN
yNvh
xNuh
tNh =
∂
∂
∂
∂−+
∂
∂+
∂
∂+
∂
∂ )()(
where QN is the nutrient flux at the surface, No is the nutrient concentration in the stratified layer below the mixed layer
1) Case 1: 0,0,0 >∂
∂=
∂
∂=
∂
∂=
th
yN
xNQN ,we have
0)( =∂
∂−+
∂
∂
thNN
tNh o
then,N If o N>
0)2()(
2
>∂
∂−=
∂
∂ ht
NNtN
o
When the mixed layer deepens, the deeper nutrients will entrain the mixed layer to increase the nutrient concentration in the mixed layer.
31
o
o
NhhNhhN
hhNN
N
)(
then,at Let
constant)N-h(N
have weconstant, a is When
111
11
o
−+=
==
=
2) Case 2: 0and0,0 =∂∂== th/ NQ oN
0=∂
∂+
∂
∂+
∂
∂
yNv
xNu
tN
Assume that
)()( yxetNN +−= λ
)1sin(cosln0)( −−=∂
∂⇒=+−
∂
∂ ftftfht
NNvutN o
ρλτ
λ
Yield:
€
dNdt
= 0
32
Solution:
)()cos(sin2 yxftft
fhtfh eeCeN
oo
+−+
= λρ
λτ
ρ
λτ
At a local position, the high nutrient center will decrease exponentially with time in the mixed layer. The time scale required to mix all the nutrients is
)/(~
fhTo
N ρτλ
Since in this case the nutrient is conservative when it is advected, the decrease of nutrients at a local position is resulted from the advective process due to the Ekman transport.
Note: one probably could find that N varies locally at inertial frequency.
