Momentum Equations in a Fluid

V

density of water

(PD) Pressure difference

(Co) Coriolis Force

(Fr) Friction

Total Force acting on a body = mass times its acceleration

1( )i i

i i i i i

F dua PD Co Fr W

m dt

(W) Weight

(PD)

F ma

The forces on a cube of water

Note the subscript “i”.

2

0

2

20

13

0

1 b

1

2 sin( 2 sec ; Latitude

i ij ijk j k i i

j i i

i i

i

uu u f u p u

t

u u pu f u b u

t x x x

f

b g

3

22

2

i

i ix x

Navier Stokes Equation with Gravity and Coriolis

Acceleration

Advection

Coriolis

Pressure Grad

Buoyancy

Friction

2 2 2

2 2 20

2 2 2

2 2 2

2 2 2

2 2 2

1v ( )+ ( )

v v v v 1v ( ) ( )

w w 1v ( ) ( )

u u u u p u u uu w fv

t x y z x x y z

p v v vu w fu

dt x y z y x y z

w w p w w wu w b

t x y z z x y z

Navier Stokes Equation with Gravity and CoriolisComponent Form

Concept of Reynolds Number, Re

2

2

1 ( ) b + i i

j i i ij i i

u u pu f u u

t x x x

Acceleration Advection Pressure Gradient Friction I II III IV

Ignore Coriolis and Buoyancy and forcing

2 ?

U U UU

L L

ULIf IV << I, II Re = 1 Turbulence Occurs

characteristic flow field velocity

L characteristic flow field length scale

Ltime scale =

U

U

Example : a toothpick moving at 1mm/s

Flow past a circular cylinder as a function of Reynolds number From Richardson (1961).

Note: All flow at the same Reynolds number have the same streamlines. Flow past a 10cm diameter cylinder at 1cm/s looks the same as 10cm/s flow past a cylinder 1cm in diameter because in both cases Re = 1000.

Example: a finger moving at 2cm/s

Re<1Re =174

Re = 20

Re = 5,000 Re = 14,480

Turbulent Cases

Re = 80000

Laminar Cases

Example: hand out of a car window moving at 60mph.

Re = 1,000,000

Air

•Water

Ocean Turbulence (3D, Microstructure)Wind

Mixed Layer Turbulence

Thermocline Turbulence

Bottom Boundary Layer turbulence

Turbulent Frictional Effects: The Vertical Reynolds Stress

Turbulence

( )ii i i

i

du pf u g Fr

dt x

x

1( ) v+

1( )

1( )

y

z

du pf Fr

dt x

dv pfu Fr

dt y

dw pg Fr

dt z

or in component form

No

Air

Water

'

mean + fluctuating

u u u u

'u

time

u

u 'u

Three Types of Averages•Ensemble•Time•SpaceErgodic Hypothesis: Replace ensemble average by either a space or time average

Notation q <q>

Mean and Fluctuating Quantities

u ', v ', 'u w

How does the turbulence affect the mean flow?

3D turbulenceMean Flow

u’ w’

Concept of Reynolds Stress

' 'u w

' ' < 0u w

'

v = v v'

'

u u u

w w w

Momentum Equations with Molecular Friction

x

1v ( )+

1( )

1( )

where

v

y

z

du pf Fr

dt x

dv pfu Fr

dt y

dw pg Fr

dt z

du w

dt x y z

2, ,

2

(u,v,w)

molecular viscoisty = sec

x y zFr

m

But

1v ( )

1( )

1( )

where

v

du pf

dt x

dv pfu

dt y

dw pg

dt z

du w

dt x y z

Approach for Turbulence

i i iBut u u u'

u '

v v+v'

w +w'

u u

w

1v ( )

du pf

dt x

ExampleUniform unidirectional wind blowing over ocean surface

1v v ( )

using v 0

( ) ( ) 1( ) ) v ( )

2

u u u u pu w f

t x y z x

u wx y z

u uu vu wu pf

t x y z x

Dimensional Analysis Boundary Layer Flow•Gradient in “x” direction smaller than in “z” direction

1( )

1( )

where

v

dv pfu

dt y

dw pg

dt z

du w

dt t x y z

Example:Mean velocity unidirectional , no gradient in “y” direction

( ') ( ')( w')( ')

u u u u wu u

x z

( ) 1v ( )

u u uw pu f

t x z x

i i iBut u u u'

'

v v+v'

w +w'

u u u

w

Now we average the momentum equation

( ' ' ) 1v

u u u w pu f

t x z x

1 1v

where

= ' ' " "component of Reynolds Stress

x

x

u u pu f

t x x z

u w x

Example: Tidal flow over a mound

U H

26

0

Reynolds Number Re ;

u, L characteristic values of the mean flow, 10sec

For unstratfied flow constant

Turbulence occurs when Re Re ~ 3000c

uL

m

0Unstratfied flow constant

Laminar Flow

Turbulent Flow

2 2 2

2 2 20

I II III IV

1{ v } ( )i i i i i i i i

j

u u u u p u u uu w

t x y z x x y z

3 D Turbulence: Navier Stokes Equation (no gravity, no coriolis effect)

Examples: tidal channel flow, pipe flow, river flow, bottom boundary layer)

I . AccelerationII . Advection (non-linear)III. Dynamic PressureIV. Viscous Dissipation

II Reynolds Number =

IV

Surface Wind Stress (Unstratified Boundary Layer Flow)

Air

Water

air

' 'airair airu w

wind

' 'w u w

What is the relationship between and ? airw

w air

Definition: Stress = force per unit area on a parallel surface

Definition

Concept of Friction Velocity u*

2' ' ( *)

*

u w u

u

u* Characteristic velocity of the turbulent eddies

210 10

310

310

where is the wind speed 10m above the water

m10 U <5

secm

2.5 10 U >5sec

D

D

C U U

C

Empirical Formula for Surface Wind Stress Drag Coefficient DC

Example. If the wind at height of 10m over the ocean surfaceis 10 m/sec, calculate the stress at the surface on the air side andon the water side. Estimate the turbulent velocity on the air side and the water side.

u*=?

u*=?

Since

310

2 3 210 3

2

mU >5 2.5 10

secm

( ) (1.0 )(2.5 10 )(10 )sec

N .25

m

D

air D

air water

C

kgC U

m

2*

air

3

2*

w

3

N.25

mu = = .5sec1.0

N.25

mu = = .016sec1000

air

air

w

w

mkgm

mkgm

i 1 2

1 1( )

where

= ' ' for ' ' ', ' ' v '

= 0 for i = 3 (z)

i i i

j i ij i

i x y

u u pu f u g

t x x z

u w u u u u u

Convention: When we deal with typical mean equations we drop the “mean”Notation!

General Case of Vertical Turbulent Friction

1 1( )

i i ij i i

j i

u u pu f u g

t x x z

Note that we sometimes use 1,2,3 in place x, y, z as subscripts

1 1v v ( )+

v v v v 1 1v ( )

w w 1v ( )

x

y

u u u u pu w f

t x y z x z

pu w fu

dt x y z y z

w w pu w g

t x y z z

Component form of Equations of Motion with Turbulent Vertical Friction

1 1(1) v v ( )+

v v v 1 1(2) v ( )

(3)

x

y

u u u pu f

t x y x z

pu fu

dt x y y z

pg

z

Note: in many cases the mean vertical velocity is small and we can assume w = 0 which leads to the hydrostatic approximation and

Example : Steady State Channel flow with a constant surface slope , (No wind)

Role of Bottom Stress

0 0

1 10=

p

x z

z = 0

z = D

Bottom

Surface

0 ( p g D z z

Flow Direction Why?

Bottom Stress g D

Surface Stress Stress 0

x

Note 0x

00

{ ( } but

g D zpg

x x

( )g D z

z = 0

z = D

Bottom

Surface

0 ( p g D z z

Flow Direction Why?

Bottom Stress g D

0

x

Typical Values

5 6

0

( ) 0

| | 1 /(1 10 ) 10 to 10 & for D =10m

Friction velocity on the bottom is

* | |

* (1 3) / sec

g D zx

cm km to km

u g D

u cm

0 u

Turbulence Case: Eddy Viscosity Assumption

eddy viscositye e

u

z

Note. At a fixed boundary because of molecular friction. In general = (z).

Relating Stress to Velocity

Viscous (molecular) stress in boundary layer flow Low Reynolds Number Flow

2

molecular viscoistysec

u m

z

Note: Viscous Stress is proportional to shear.

Mixing Length Theory: Modeling e ul

l a characteristic length , a characteristic velocity of the turbulenceu

( )g D z Back to constant surface slope example where we found that

2

2

s

( )

( )2

( *) (1 )

2 u* = bottom friction velocity

u ( )2

e

e

uk g D z

zg z z

u Dk

u zz

k D

g Du D

k

z = 0

z = D

If we use the eddy viscosity assumption with constant k

26 5

6 2 22 2

s 25

D =10 m, 10 , 10sec

(9.8 10 *10secu ( )

22*10

sec

.5sec

e

e

mk

mmg D

u Dmk

m

Example Values

Log Layer

Note: in the previous example near the bottom, independent of z

constant

Bottom Boundary Layer

20 0

v

v

0v 0

( *)

*

.4, Von Karman's constant

*ln( ) the roughness parameter

uu

zu u

z z

u zu z

z

z

v

Eddy size to distance from bottom

k *u z

0

ln( )1

zz

z z

Typical Ocean Profile of temperature (T), density

20m-100m

1km

4km

Mixed Layer

pycnoclinethermocline

T

But , , )T S p

0

10

Dwp g

Dt z

Stratified Flow

Vertical Equation: Hydrostatic condition No stratification

0

0 0

0

( )10

Dwp g

Dt z

p p gz

Vertical Equation: Hydrostatic condition Stratification

Horizontal Equation

1 1

where ' ' &

h

Duf u p

Dt z

Du w u

Dt t

Buoyancy

Archimedes Principle

Weight

density of the block

W g

Buoyancy Force

density of the waterBF g

If W> block sinks

If W< block risesB

B

F

F

z

z+z)W z g

)BF z z g

22 2

2

25 2

2

{ ) )}

{ ) )}but

where ( ) { }

4.4 10 sec

net B

net

F F W V z z z g

z z z

z z

g g gF V zg V N z N

z z z c

g

c

Concept of Buoyancy frequency N

Gradient Richardson Number

Turbulence in the Pycnocline

Velocity Shear

u

z

gN

z

2

2( )g

NRi

uz

Turbulence occurs when

1

4gRi

Billow clouds showing a Kelvin-Helmholtz instability at the top of a stable atmospheric boundary layer. Photography copyright Brooks Martner, NOAA Environmental Technology Laboratory.

1( )

4gRi

Depth(m)

Distance (m)

Turbulence Observed in an internal solitary wave resulting inGoodman and Wang (JMS, 2008)

1( )

4gRi

Temperature (Heat)Equationwith Molecular Diffusion

2

27

where

v

molecular diffusivity of T

= 1.4 1sec

T

T

dTT

dt

du w

dt x y z

m

Approach for Turbulence

0

But ' & '

v 0

v ' ' 0

dT

dtT T T w w w

T T T Tu w

t x y z

T T T Tu w w T

t x y z z

' ' 0T T

w w Tt z z

' '

Tw T k

zT

H k cz

Eddy Diffusivity Model

Case of Vertical Advection and Turbulent Flux

Note: Heat Flux is given by

Advection Diffusion Equation drop bar notation

2

2

2

2

00

0

0

Steady State Case 0

0

:

( ) exp[ ( )]

( ) exp( ) where

(T(z)=T ( ) {1 exp(

T

T

sT

Ts

s s

T T Tw

t z zT

t

T Tw w upwelling velocity

z zSolution

T T wD z

z z

T D zz

z z w

Tz

z

0

00

)}

T(z)=T {1 exp( )} where ( )s s

D z

z

D z TT T z

z z

T(z)

w

3s

Heat Transfered

H = ( ) where c specific heat of water 4.2 10T S

T Jc

z oC kg

sT

Z=0

Z=D Surface (s)

u u

Example: Suppose the heat input is in water of depth 50m . The turbulent diffusivity is (a) For the case of no upwelling what is the heat transferred, H, at the surface, mid depth, and the bottom? What is the water temperature at the surface mid depth and the bottom? (b) Suppose there was an upwelling velocity of .1 mm/sec how would the results in part (a) change?

s 2H = 500 , 20 o

s

WattsT C

m

2310

sec

mk

sT

Z=0

Z=D Surface (s)

s 2H =500 , 20 o

s

WattsT C

m

2

2

23 3 3

3

0 ( ) exp[ ( )] = ( )

500 at all depths!

500( ) ( )

10 4.2*10 *10sec

.12 at all depths!

s sT

s

ss

To

o

T T w Tw D z

z z z

WattsH H

mWatts

HT T mkg J mz z cm kg C

C

m

0

0

0 000

s

T(z)=T {1 exp( )}

As T(z)= lim[T ( ) {1 exp( )}]

T(z)=T ( ) ( )

at: z =50, T=T 20

z =25, T=20 .12 (25 ) 17

z =0, T=20 .12 (50 ) 14

s

s sz

s s

o

oo o

oo o

D zT

z

k T D zz z

w z z

TD z

z

C

CC m C

m

CC m C

m

(a) No Upwelling w=0

sT

Z=0

Z=D Surface (s)s 2

H =50 , 20os

WattsT C

m

0

0

T(z)=T {1 exp( )}

( ) 10 *.12 1.2

at: z =50, T= 20

25z =25, T=20 1.2 [1 exp( )] 18.9

1050

z =0, T=20 1.2 [1 exp( )] 18.810

10at: z =10,T=20 1.2 [1 exp( )] 19.2

10

s

o

s

o

o o o

o o o

o o o

D zT

z

T CT z m

z m

C

C C C

C C C

C C C

(a)Upwelling w= .1 mm/sec

23

04

0

s0

s 2

2 2

2 2

10sec 10

10sec

(z)= ( ) ( ) exp( )

(z)=H exp( )

50; H=H 500

2525 ; H=500 exp( ) 41

1050

0 ; H=500 exp( ) 3.310

T

T T s

m

z mmw

T T D zH c c

z z z

D zH

z

Wattsz

mWatts Watts

z mm m

Watts Wattsz m

m m

wu u