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Concepts: Buoyancy and Vorticity (IPO-6, 12) Buoyancy Archimedes Principle Weight density of the block W gV ! ! = = ! ! Buoyancy Force density of the water B F gV ! ! = = If W> block sinks If W< block rises B B F F ! ! ! ! " > " < ! !
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Concepts: Buoyancy and Vorticity (IPO-6, 12)

Jan 28, 2022

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Page 1: Concepts: Buoyancy and Vorticity (IPO-6, 12)

Concepts: Buoyancy and Vorticity (IPO-6, 12) Buoyancy

Archimedes Principle

Weight

density of the block

W gV!

!

=

=

!

!

Buoyancy Force

density of the water

BF gV!

!

=

=

If W> block sinks

If W< block rises

B

B

F

F

! !

! !

" >

" <

!

!

Page 2: Concepts: Buoyancy and Vorticity (IPO-6, 12)

A Canonical Ocean Profile of Temperature (T), Density (ρ)

20m-100m

1km

4km

Mixed Layer

PycnoclineThermocline

T ρ

But , , )T S p! != (

Page 3: Concepts: Buoyancy and Vorticity (IPO-6, 12)

Equation of State

0P

!"#

"Note that which shows that seawater is compressible!

The compressibility of seawater allows sound to be propagated at a speedp

c!

"=

"

In this equation p is taken relative to atmospheric pressure

0 0

03

-3

3 3 3

( ) ( )

1027 10 S=35psu

kg kg kg =.15 =.78 = 4.5 10

(m )( ) (m )( ) (m )( )

o

o

a T T b S S kp

where

kgT C

m

a b kC psu decibar

! !

!

0

0

" = " " + " +

= =

#

1500sec

mc !

pc

!

"=

"

Homework ProblemHow much does pressure contribute to seawater density at the bottom of an ocean of depth 4 km?

Page 4: Concepts: Buoyancy and Vorticity (IPO-6, 12)

0 0

03

-3

3 3 3

0 0

-3

4

( ) ( ) p in dbars

1027 10 S=35psu

kg kg kg =.15 =.78 = 4.5 10

(m )( ) (m )( ) (m )( )

( ) ( ) z in meters

kg= 4.5 10

m

o

o

a T T b S S kp

where

kgT C

m

a b kC psu decibar

a T T b S S kz

k

! !

!

! !

!

0

0

0

" = " " + " +

= =

#

" = " " + " +

#

0 0 2

2 2

3

3

N( ) ( ) p in

m

Using

10 101500

4.5 10 sec

pa T T b S S k

g

p kc c

p g

g mc

k

!!

!

! !

!

0

0

"

0

0

"

" = " " + " +

$ $= % = %$ $

#= = =

#

Addendum

Page 5: Concepts: Buoyancy and Vorticity (IPO-6, 12)

22

2

2 2

2

2

0 0

N ( )

Proof

Given ( )

in the equation of state

( ) ( )

g g g g T Sa b

z z c z z

g gT Sa b k

z z c z z c

gkUse c k

p g c

a T T b S S kz

!

!

" "

" " "

" " ""

""

"

" "

0 0 0

0 0

# 0

0

0

$ $ $ $= # = # # = #

$ $ $ $

$ $ $ $= # = # + + #

$ $ $ $

$= = % =$

# = # # + # +

Addendum

Page 6: Concepts: Buoyancy and Vorticity (IPO-6, 12)

Potential Temperature , θ, and Density,

σ Notation

where potential temperature

( , , )

( , ,0)

( , , ) 1000

( , ,0) 1000

t

T S P

T S

T S P

S

! !

!

" #

" #

" #

# !

=

= $1000

= $1000

= $

= $

“potential”.

Adiabatic change (noheat exchanged)

!"

T

θ

!"

Page 7: Concepts: Buoyancy and Vorticity (IPO-6, 12)

ρz

z+δz

)W z gV!"= (

)BF z z gV!" #= ( +

22 2

0 2

0 0 0

25 2

2

{ ) )}

{ ) )}but

where ( ) { } ( )

4.4 10 sec

net B

net

F F W z z z gV

z z z

z z

g g g g T SF zgV N zV N a b

z z z c z z

g

c

! !

! ! !

! !

" # "

" " # "

#

" " "# " #

" " "

$ $

= $ = ( + $ (

% ( + $ (&

%

% % % % %= = $ = $ = $ + = $

% % % % %

& '

Concept of Buoyancy Frequency N

Page 8: Concepts: Buoyancy and Vorticity (IPO-6, 12)

Homework ProblemCalculate the and draw the density profile , ρ(z) and buoyancy profile N(z) for the temperature profile shown. Assume a constant salinity of S = 35.5 psu. Repeat your calculation for S = 36 psu. Comment on how the N profile changes when the salinity changed.

100m

1km

4km

20o

T C=

8o

T C=

2o

T C=

z =0m

Page 9: Concepts: Buoyancy and Vorticity (IPO-6, 12)

Vorticity

Concept of Conservation of angular momentum

Definition of Vorticityω= 2 x angular velocityω units of 1/sec

Page 10: Concepts: Buoyancy and Vorticity (IPO-6, 12)

10

Vorticity velocity angular velocity =

2 radius

v

2 r

!"

=

= =

Component form

v

v

x

y

z

w

y z

u w

z x

u

x y

!

!

!

" "= #" "

" "= #" "

" "= #" "

u! ="#! !

But vorticity is a vector! (why?)

r

Most important for large scale flow >~10km)

Counter clockwise ω > 0Clockwise ω < 0

Page 11: Concepts: Buoyancy and Vorticity (IPO-6, 12)

2

4

( )

but ( )

( )

I Amr

m V r H

I A Hr

! !

" " #

! #"

2

=

= =

$ =

1!

1 1 1 1 1 2

1 2

1 2

Using &

I I V V

H H

! !

! !

= =

" =

Conservation of Angular MomentumFollowing a Fluid Parcel (Case of f =0)

2

2

vorticity

Moment of Inertia

V Volume H r

I Amr

!

"= =

=

1H

1r

1 1 2 2I I! !=

2H

2!

2r

Page 12: Concepts: Buoyancy and Vorticity (IPO-6, 12)

Conservation of Potential Vorticity 0f !

1 1 2 2

1 2

2

{ } 0

{( ) } 0

f f

H H

d f

dt H

df N

dt

! !

!

!

+ +=

+=

+ =

= relative vortivity

2 sin( planetary vortivity

total vortivity

potential vorticity

f

f

f

H

!

"

!

!

= # ) =

+ =

+$ = =

Terminlogy

ω2>ω1

ω1

Page 13: Concepts: Buoyancy and Vorticity (IPO-6, 12)

Homework: von Arx (1962) has suggested that to understandconservation of potential vorticity we consider what happens to abarrel of water as it moves around the earth. Take the shaded circles asthe initial rest position ( no relative vorticity , ω = 0 ) of a barrel ofwater. Assume that the shape and volume of the barrel is constant, i.e.H, r constant. Using the concept of conservation of potential vorticityexplain what happens to the relative vorticity and its direction ofrotation (clockwise, counterclockwise) as it moves along paths #1,2,3,4,5,6,7.

#1#2

#3

#4

#5

#7#6 H

!

r

Page 14: Concepts: Buoyancy and Vorticity (IPO-6, 12)

relative vorticity of the flow

relative vorticity from input from wind stress

relative vorticity of WBC b

!

"

"

"

=

=

=

Why are there Western boundary Currents (WBC)?

Answer: Conservation of Potential Vorticity (PV)

ω

ω

ωb

ωτ

ωω

ωτ

No WBC Unbalanced PV in western side

WBC Balanced PV both east and west

ωτ

Page 15: Concepts: Buoyancy and Vorticity (IPO-6, 12)

Homework: Why does Barotropic flow over asub-sea ridge turned equatorward to conservepotential vorticity.

From Dietrich, et al. (1980).

Example of PV Conservation: Barotropic Flow over a Sub-Sea Ridge