Evaporative heat flux (Qe)51% of the heat input into the ocean is used for evaporation. Evaporation starts when the air over the ocean is unsaturated with moisture. Warm air can retain much more moisture than cold air.
The rate of heat loss: tee LFQ ⋅=
Fe is the rate of evaporation of water in kg/(m2 s).
Lt is latent heat of evaporation in kJ.
For pure water, kgkJtLt )2.22494( −= . t~ water temperature (oC).
t=10oC, Lt=2472 kJ/kg.t=100oC, Lt=2274 kJ/kg.
In general, Fe is parameterized with bulk formulae:dz
deKF ee ⋅−=
Ke is diffusion coefficient for water vapor due to turbulent eddy transfer in the
atmosphere. It is dependent on wind speed, size of ripples, and waves at sea surface, etc. de/dz is the gradient of water vapor concentration in the air above the sea surface.
In practice: )()(4.1 2daymkgeeVF ase −=
V wind speed (m/s) at 10 m height above sea.
( )( ) 23102.224944.1 mWteeVLFQ sastee−⋅−−==
es is the saturated vapor pressure over the sea-water (unit: kilopascals, 10mb)
The saturated vapor pressure over the sea water (es) is smaller than that over distilled water (ed). For S=35, es=0.98ed(ts). V is wind speed (m/s). Ts is sea surface temperature (oC)
ea is the actual vapor pressure in the air at a height of 10 m above sea level. If
the atmospheric variable is relative humidity (RH), ea=RH x ed(ta).
Example:Ta=15oC, ed = 1.71 kPa = 12.8 mm Hg,
RH=85%, then ea= 1.71 x 0.85 kPa= 1.45 kPa.
m
ee
z
ee
dt
de asas
10
−=
Δ−
−≈ , and VK e 14≈ (very crude parameterization).
This empirical formula is an approximation of eddy diffusion formula because:
1 and 1/2 layer flow
Simplest case of baroclinic flow:
Two layer flow of density 1 and 2.
The sea surface height is =(x,y) (In steady state, =0). The depth of the upper layer is at z=d(x,y)<0. The lower layer is at rest.
312 1 mkg≈−ρρ
)(1
zgp −= ρ ∇×−= kfgV
rr1For z > d,
( ) gzgdgzdgdgp 212121 )()( ρρρηρρηρ −−+=−+−=
If we assume d∇−−=∇1
12ρρρ
The slope of the interface between the two layers (isopycnal) =
100012
1 ≈−ρρρ times the slope of the surface (isobar).
The isopycnal slope is opposite in sign to the isobaric slope.
For z ≤ d,
02 =Vr
A
B
C
D
E
Wind-driven circulation II
●Wind pattern and oceanic gyres
●Sverdrup Relation
●Vorticity Equation
Surface current measurement from ship drift
Current measurements are harder to make than T&SThe data are much sparse.
Surface current observations
Surface current observations
Drifting Buoy Data Assembly Center, Miami, Florida Atlantic Oceanographic and Meteorological Laboratory, NOAA
Annual Mean Surface CurrentPacific Ocean, 1995-2003
Drifting Buoy Data Assembly Center, Miami, Florida Atlantic Oceanographic and Meteorological Laboratory, NOAA
Argo is a global array of 3,000 free-drifting profiling floats that measures the temperature and salinity of the upper 2000 m of the ocean. This allows, for the first time, continuous monitoring of the temperature, salinity, and velocity of the upper ocean, with all data being relayed and made publicly available within hours after collection.
Schematic picture of the major surface currents of the world oceans
Note the anticyclonic circulation in the subtropics (the subtropical gyres)
Relation between surface winds and subtropical gyres
Surface winds and oceanic gyres: A more realistic view
Note that the North Equatorial Counter Current (NECC) is against the direction of prevailing wind.
Sverdrup RelationConsider the following balance in an ocean of depth h of flat
bottom
∫ +=+∫=∂∂
− −
000
0
hxyx
hfMvdzfdz
xp ττρ
(1)
∫ +−=+∫−=∂∂
− −
000
0
hyxy
hfMudzfdz
yp ττρ
(2)
∫=−
0
hx udzM ρ
∫=−
0
hy vdzM ρ
zvf
xp x
∂∂+=
∂∂ τρ
zuf
yp y
∂∂+−=
∂∂ τρ
Integrating vertically from –h to 0 for both (1) and (2), we have(neglecting bottom stress and surface height change)
where
(3)
(4)
are total zonal and meridional transport of mass
sum of geostrophic and ageostropic transports
Differentiating , we have
000=
∂∂−
∂∂+−
∂∂+
∂∂−
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
yxdydfM
yM
xMf xy
yyx ττ
€ P=pdz−h0∫Define We have
€ ∂p∂xdz=∂∂xpdz−h0∫ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥=∂P∂x−h0∫
€ ∂P∂x=fMy+τx0€
∂P∂y=−fMx+τy0(3) and (4) can be written as
(5) (6)
€ ∂6()∂x−∂5()∂y
€ ∂2P∂y∂x−∂2P∂x∂y=−f∂Mx∂x+∂τy0∂x−f∂My∂y−Mydfdy−∂τx0∂y=0
000=
∂∂−
∂∂+−
∂∂+
∂∂−
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
yxdydfM
yM
xMf xy
yyx ττ
Using continuity equation 0=∂∂+
∂∂
yM
xM yx
And define
dydf=β
( )ττττβz
xyy curlk
yxM =⋅×∇=
∂∂−
∂∂= ⎟
⎠⎞⎜
⎝⎛
rr00
Vertical component of the wind stress curl
We have Sverdrup equation
€ ∂τy0∂x−∂τx0∂y=0If
€ My=0The line provides a natural boundary that separate the circulation into “gyres”
€ My=Myg+MyEis the total meridional mass transport
€ Myg=ρvgdz=1f∂p∂xdz=1f∂P∂x−h0∫−h0∫ Geostrophic transport
€ MyE=ρvEdz=−τx0f−h0∫ Ekman transport
Order of magnitude example:At 35oN, 1-4 s-1, β2 10-11 m-1 s-1, assume τx10-1 Nm-2 τy=0
€ curlzτ()=−∂τx0∂y≈−10−1Nm−21000km≈−10−7Nm−3
€ MyE=−τx0f≈−103kgm−1s−1
€ My=Myg+MyE=curlzτ()β≈−10−72×10−11=−5×103kgm−1s−1€ Myg=−4×103kgm−1s−1
Alternative derivation of Sverdrup Relation
xp
gfv ∂∂=
ρ1
ypfu
g ∂∂−=
ρ1
Construct vorticity equation from geostrophic balance
(1)
(2)
zw
fv gg ∂
∂=β
Integrating over the whole ocean depth, we have
€ f∂ug∂x+f∂vg∂y+βvg=−1ρ∂2p∂y∂x+1ρ∂2p∂x∂y=0€
∂2()∂x+∂1()∂y€ βvg=−f∂ug∂x+∂vg∂y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟=f∂wg∂z
Assume ρ=constant
€ βVg=βvgdz=fwgz=0()−wgz=−h()[ ]−h0∫
∫ ==−
0
hEgg fwdzvV ββ
kf
wE
rr⋅×∇=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
ρτ
where is the entrainment rate from the surface Ekman layer
⎟⎟⎠
⎞⎜⎜⎝
⎛=+= ρτ
β curlVVV Eg1
The Sverdrup transport is the total of geostrophic and Ekman transport.The indirectly driven Vg may be much larger than VE.
( )6tan ≈===
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
ϕβτ
βτ
LR
LfO
f
curlO
VV
E
gat 45oN
€ βVg=βvgdz=fwgz=0()−wgz=−h()[ ]−h0∫
€ wgz=0()=wE€ wgz=−h()≈0€ Vg=fβρ∂∂xτyf ⎛ ⎝ ⎜ ⎞ ⎠ ⎟−∂∂yτxf ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟=fρβ1f∂τy∂x−1f∂τx∂y+βτxf2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟€ Vg=1ρβ∂τy∂x−∂τx∂y+βτxf ⎛ ⎝ ⎜ ⎞ ⎠ ⎟=1ρβ∂τy∂x−∂τx∂y ⎛ ⎝ ⎜ ⎞ ⎠ ⎟−VE
€ βMy=∂τy0∂x−∂τx0∂y
€ f=2Ωsinφ€ dy=Rdφ( )
⎟⎠⎞⎜
⎝⎛ Ω
=R
zcurlM y
ϕ
τ
cos2then
( ) ( )⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎠⎞⎜
⎝⎛ −∂
∂Ω−=∂
∂−=∂∂ τϕτϕ zz
yx curlcurly
RyM
xM tan
cos21
€ β=dfdy=d2Ωsinφ( )Rdφ=2ΩsinφR
Since , we have
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
∂
∂+
∂∂
Ω−=∂∂
yyR
xM xxx
τϕ
τϕ
tancos21
2
2
set x =0 at the eastern boundary,
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
∫ ∫∂∂+∂
∂Ω−=
0 0
2
2tan
cos21
x x
xxx dx
ydx
yRM τϕτ
ϕ
yRM x
y ∂∂
Ω= τϕcos2
€ τy≈0
Further assume€ τx≈τxy()In the trade wind and equatorial zones, the 2nd derivative term dominates:
€ Mx≈xR2Ωcosφ∂2τx∂y2€ Mx=x2Ωcosφ∂τx∂ytanφ+∂2τx∂y2R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟
Mass Transport
Since 0=∂∂+
∂∂
yM
xM yx
Let y
M x ∂∂−= ψ
,
xM y ∂
∂= ψ,
( )βτψ zcurl
x=
∂∂
( )∫=0
x
z dxcurlβτψ
where ψ is stream function.
Problem: only one boundary condition can be satisfied.
1 Sverdrup (Sv) =106 m3/s