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