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x z y dy dz dx ux of mass in (kg/s) = dz dy u Flux of mass out (kg/s) = dz dy u dz dy dx u x Net Flux of mass in ‘x’ = dz dy dx u x Net Flux of mass in ‘y’ = dz dy dx v y Net Flux of mass in ‘z’ = dz dy dx w z dx u x u , u , w , v u ass per volume per time (kg/(m 2 s) Conservation of Mass

Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

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Page 1: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

x

z

y

dy

dz

dx

Flux of mass in (kg/s) = dzdyu Flux of mass out (kg/s) = dzdyu

dzdydxux

Net Flux of mass in ‘x’ = dzdydxux

Net Flux of mass in ‘y’ = dzdydxvy

Net Flux of mass in ‘z’ = dzdydxwz

dxux

u

, u

, w

, v

u

Mass per volume per time(kg/(m2 s)

Conservation of Mass

Page 2: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

The change of mass per unit time going through the volume element is:

And the change of mass per unit time per unit volume is:

dzdydxtt

M

dzdydxwz

vy

ux

dzdydxt

0

wz

vy

uxt

which is the same as:

0

zw

yv

xu

zw

yv

xu

t

01

zw

yv

xu

DtD

or

Page 3: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

This is the Continuity Equation or Equation of Conservation of Mass

How valid is the Boussinesq approximation in the OCEAN?

How would you determine that?

1 sigma-t throughout one day = 1 / (24*3600.) = 1.1510-5

01

zw

yv

xu

DtD

0DtD

Taking

Boussinesq approximation

0

zw

yv

xu

xu

]O[10km 100

m/s 0.1 6-

DtD

810O1

DtD

But

Page 4: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

0

uzw

yv

xu

zyx,, Nabla or Del operator

Special case of variations in the horizontal only:

yxh ,

Page 5: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

econvergenc 0 uh

x

zu u

ww ww

xdivergence 0 uh

z

u u

ww ww

Page 6: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Example:

What is the vertical velocity in an upwelling region 30 m deep, where the flow accelerates southward by 0.10 m/s in 10 km, and westward by 0.20 m/s in 10 km?

xu

yv

54 102

102.0

54 101

101.0

5103

zw

30

0

5103 dzw

smzw / 10930103|103 45300

5

daymw / 8.77)(86400)109( 4

Page 7: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Continuity Equation in Bulk Form: b0 V E V P R

Continuity assuming: steady stateone dimensional motionflow integrated over a closed surface

Page 8: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

x

z

y

dy

dz

dx

Flux of salt into dydz = dzdyxS

KdzdyuS

Net Flux of Salt in ‘x’ =

dzdydxxS

Kx

dzdydxuSx

, u

, w

, v

uS

Advective flux of salt

Conservation of Salt

xS

K

Diffusive flux of salt

dxuSx

uS

dxxS

Kxx

SK

Flux of salt out of dydz = dzdydxuSx

dzdyuS

dzdydxxS

Kx

dzdyxS

K

Page 9: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Net Flux of Salt in ‘y’ =

dzdydxyS

Ky

dzdydxvSy

Net Flux of Salt in ‘z’ = dzdydxzS

Kz

dzdydxwSz z

Net Salt Flux per unit volume =tS

zS

Kzy

SK

yxS

Kx

wSz

vSy

uSxt

Sz

zw

yv

xu

SzS

wyS

vxS

u

0

zS

Kzy

SK

yxS

Kxz

Sw

yS

vxS

utS

z

Salt Conservation

Page 10: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Conservation of Salt:

zS

zKzy

SK

yxS

Kxz

Sw

yS

vxS

utS

ydiffusivitDt

DS

At steady state and assuming constant diffusivities:

2

2

2

2

2

2

zS

zKyS

KxS

KzS

wyS

vxS

u

Advection-Diffusion Equation

Further assuming motion in one direction and integrated over the volume considered,the statement of CONSERVATION OF SALT may be given as:

Vin Sin = Vout Sout

Page 11: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Continuity Equation in Bulk Form: b0 V E V P R

Sb

S0

Salt Conservation Equation in Bulk Form: VbSb = V0S0

Page 12: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Example of Conservation Principles (Knudsen Relations or Basin Equations)

What is the volume inflow and outflow at the Chesapeake Bay entrance if the mean river discharge is 2,200 m3/s, and the outflowing salinity is 24 and the inflowing salinity is 30?

Conservation of Mass: Vout = Vin + R

Conservation of Salt: Vout Sout = Vin Sin

ininoutin SVSRV

ininoutoutin SVRSSV

outoutinin RSSSV

outin

outin SS

SRV

sVin /m 88006

242200 3

sVout /m 11000 3

Page 13: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Residence Time

Time it takes to flush entire volume of system (this is one definition).

May be determined from volume of basin and volume of water that enters the basin per unit time. From the above example:

days 53/m 8800

m 1043

310

sRT

Page 14: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Conservation of Heat:

zT

zzyT

yxT

xzT

wyT

vxT

utT

Earth doesn’t gain or lose heat, but the ocean exchanges it with the atmosphere

pa

iBls

pa CQQQQ

CQ

zT

z

Q heat flux through air-water interface (Watts/m2)

rhoa density of air (1.2 kg/m3)

Cp specific heat of moist air [~1000 joules/(kg oC)]

Qs sensible heat or direct thermal transfer

Ql latent heat flux or evaporation

QB long wave radiation from the ocean

Qi short-wave radiation or incident solar radiation

Page 15: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Heating and Cooling Processes

100

30 reflected from clouds, by water and land, and backscattered by air(shortwave radiation)

19

51

51 absorbedin ocean

To Space

21 long-waveradiation

6 to space

23 Evaporation 7 Sensible

Absorbed byAtmosphere

15

64 emitted byclouds, water vapor and CO2

(long-wave radiation)

QiQl

QsQB

1370 W/m2

)7.01)(1(

)6.01(

)(

)(

42

10

10

csioi

scB

sl

ass

QQ

TQ

WqqQ

WTTQ

Ts sea surface temperatureTa air temperatureW10 wind speed at a height of 10 mqs saturation specific humidityq specific humidity of aireta c cloud cover (%)Qio net downward flux of solar radiationαs surface albedo

Page 16: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

From Kallberg et al 2005ERA-40 atlas. ERA-40 Project Report Series No. 19. © 2005 Robert H. Stewart

Short Wave Radiation )7.01)(1( csioi QQ

Page 17: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

From Kallberg et al 2005ERA-40 atlas. ERA-40 Project Report Series No. 19. © 2005 Robert H. Stewart

Long Wave Radiation42 )6.01( scB TQ

Page 18: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

From Kallberg et al 2005ERA-40 atlas. ERA-40 Project Report Series No. 19. © 2005 Robert H. Stewart

Latent Heat Flux 10)( WqqQ sl

Page 19: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

From Kallberg et al 2005ERA-40 atlas. ERA-40 Project Report Series No. 19. © 2005 Robert H. Stewart

Sensible Heat Flux 10)( WTTQ ass

Page 20: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

Annual mean net heat flux

From Kallberg et al 2005ERA-40 atlas. ERA-40 Project Report Series No. 19. © 2005 Robert H. Stewart

Page 21: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

zC

zzyC

yxC

xzC

wyC

vxC

utC

Advection-Diffusion for a Non-conservative Tracer

C could be chlorophyll_a

dissolved oxygen

nutrient species

pollutant

planktonic species

sediment concentration

Page 22: Flux of mass in (kg/s) = Flux of mass out (kg/s) = Net Flux of mass in ‘x’ = Net Flux of mass in ‘y’ = Net Flux of mass in ‘z’ =, u, w, v Mass per volume

zC

zzyC

yxC

xzC

wyC

vxC

utC

At oceanic station S1 the [DIN] is 200 µg/l and at station S2, 5 km to the E, it is 300 µg/l. If biochemical processes add 100 µg/l to the entire area each day and mixing/diffusion can be neglected, how much would you expect [DIN] to change in time if the region is swept by a 0.5 m/s eastward current?

slg

xC

utC

86400/100

slg

mlg

smtC

864/ 1

5000/100

/ 5.0

slg

tC

/

009.0