Concept of Reynolds Number, Re

2

2

1 ( ) b +

i i

j i i i

i 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

U LIf IV < < I , I I R e = 1 T u rb u le n c e O c c u rs

c h a ra c te r i s t ic f lo w f ie ld v e lo c i ty

L c h a ra c te r i s t ic f lo w f ie ld le n g th s c a le

Lt im e s c a le =

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

( )i

i i i

i

d u pf u g F r

d t x

x

1( ) v +

1( )

1( )

y

z

d u pf F r

d t x

d v pfu F r

d t y

d w pg F r

d t z

or in component form

No

Air

Water

'

m e a n + f lu c tu a t in g

u u u

u

'u

time

u

u 'u

Three Types of Averages

•Ensemble

•Time

•Space

Ergodic Hypothesis: Replace ensemble

average by either a space or time average

N o ta t io n q < q >

Mean and Fluctuating Quantities

u ', v ', 'u w

How does the turbulence affect the mean flow?

3D turbulenceMean Flow

u’ w’

C o n cep t o f R eyn o ld s S tre ss

' 'u w

' ' < 0u w

'

v = v v '

'

u u u

w w w

Momentum Equations

with Molecular Friction

x

1v ( ) +

1( )

1( )

w h e re

v

y

z

d u pf F r

d t x

d v pfu F r

d t y

d w pg F r

d t z

du w

d t x y z

2

, ,

2

(u ,v ,w )

m o le c u la r v is c o is ty = s e c

x y zF r

m

But

1v ( )

1( )

1( )

w h e re

v

d u pf

d t x

d v pfu

d t y

d w pg

d t z

du w

d t x y z

Approach for Turbulence

i i iB u t u u u '

u '

v v + v '

w + w '

u u

w

1v ( )

d u pf

d t x

Example

Uniform unidirectional wind blowing over ocean surface

1v v ( )

u s in g v 0

( ) ( ) 1( ) ) v ( )

2

u u u u pu w f

t x y z x

u wx y z

u u u v u w u pf

t x y z x

Dimensional Analysis

Boundary Layer Flow

•Gradient in “x” direction smaller than in “z” direction

1( )

1( )

w h e re

v

d v pfu

d t y

d w pg

d t z

du w

d t 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 u w pu f

t x z x

i i iB u t 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

w h e re

= ' ' " " c o m p o n e n t o f R e y n o ld s S t re s s

x

x

u u pu f

t x x z

u w x

Example: Tidal flow over a mound

UH

2

6

0

R e y n o ld s N u m b e r R e ;

u , L c h a ra c te r i s t ic v a lu e s o f th e m e a n f lo w , 1 0s e c

F o r u n s t r a t f ie d f lo w c o n s ta n t

T u rb u le n c e o c c u r s w h e n R e R e ~ 3 0 0 0c

u L

m

0U n stra tfied flo w co n stan t

Laminar Flow

Turbulent Flow

2 2 2

2 2 2

0

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 . Acceleration

II . Advection (non-linear)

III. Dynamic Pressure

IV. Viscous Dissipation

II R e y n o ld s N u m b e r =

IV

Surface Wind Stress (Unstratified Boundary Layer Flow)

Air

Water

a ir

' 'a ir

a ir a iru w

wind

' 'w

u w

What is the relationship between and ?a ir

w

w a ir

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

2

1 0 1 0

3

1 0

3

1 0

w h e re i s th e w in d s p e e d 1 0 m a b o v e t h e w a te r

m1 0 U < 5

s e c

m 2 .5 1 0 U > 5

s e c

D

D

C U U

C

Empirical Formula for Surface Wind Stress Drag Coefficient D

C

Example. If the wind at height of 10m over the ocean surface

is 10 m/sec, calculate the stress at the surface on the air side and

on the water side. Estimate the turbulent velocity on the air side

and the water side.

, u*=?

, u*=?

Since

3

1 0

2 3 2

1 0 3

2

mU > 5 2 .5 1 0

s e c

m( ) (1 .0 ) ( 2 .5 1 0 ) (1 0 )

s e c

N .2 5

m

D

a ir D

a ir w a te r

C

k gC U

m

2*

a i r

3

2*

w

3

N.2 5

mu = = .5

s e c1 .0

N.2 5

mu = = .0 1 6

s e c1 0 0 0

a ir

a ir

w

w

m

k g

m

m

k g

m

i 1 2

1 1( )

w h e r e

= ' ' f o r ' ' ', ' ' v '

= 0 f o r i = 3 ( z )

i i i

j i i

j 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 i

j 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

d t 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

d t 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 , a. (No wind)

Role of Bottom Stress

0 0

1 10 =

p

x z

z = 0

z = D

z

Bottom

Surface

0( p g D z z z

Flow Direction Why?

Bottom Stress

g D a

Surface Stress Stress0

x

N o te 0x

za

0

0

{ ( } b u t

g D zpg

x x

z a

( )g D z a

a

z = 0

z = D

z

Bottom

Surface

0( p g D z z z

Flow Direction Why?

Bottom Stress

g D a

0

x

Typical Values

5 6

0

( ) 0

| | 1 / (1 1 0 ) 1 0 to 1 0 & fo r D = 1 0 m

F r ic t io n v e lo c i ty o n th e b o t to m is

* | |

* (1 3 ) / s e c

g D zx

c m k m to k m

u g D

u c m

z a a

a

a

0 u

Turbulence Case: Eddy Viscosity Assumption

e d d y v is c o s i tye 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

m o le c u la r v is c o is tys e c

u m

z

Note: Viscous Stress is proportional to shear.

Mixing Length Theory: Modeling

e

u l

l a characteristic length , a characteristic velocity of the turbulenceu

( )g D z a

Back to constant surface slope example where we found that

2

2

s

( )

( )2

( * ) (1 )

2

u * = b o t to m f r ic t io n v e lo c i ty

u ( )2

e

e

uk g D z

z

g z zu D

k

u zz

k D

g Du D

k

a

a

a

z = 0

az = D

If we use the eddy viscosity assumption

with constant k

2

6 5

6 2 2

22

s 2

5

D = 1 0 m , 1 0 , 1 0s e c

( 9 .8 1 0 * 1 0s e c

u ( )2

2 * 1 0s e c

.5s e c

e

e

mk

mm

g Du D

mk

m

a

a

Example Values

Log Layer

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

co n s tan t

Bottom Boundary Layer

2

0 0

v

v

0

v 0

( * )

*

.4 , V o n K a rm a n 's c o n s ta n t

*ln ( ) th e ro u g h n e s s p a ra m e te r

uu

z

u u

z z

u zu z

z

z

v

E d d y s iz e to d is ta n c e f ro m b o tto m

k *u z

Note we have used the fact that

0

ln ( )1

z

z

z z

Typical Ocean Profile of temperature (T), density (

20m-

100m

1km

4km

Mixed Layer

pycnoclinethermocline

T

But , , )T S p (

0

10

D wp g

D t z

Stratified Flow

Vertical Equation:

Hydrostatic condition

No stratification

0

0 0

0

( )10

D wp g

D t z

p p g z

Vertical Equation:

Hydrostatic condition

Stratification

Horizontal Equation

1 1

w h e re ' ' &

h

D uf u p

D t z

Du w u

D t t

Buoyancy

Archimedes Principle

Weight

d e n s i ty o f th e b lo c k

W g

Buoyancy Force

d e n s i ty o f th e w a te r

BF g

If W > b lo c k s in k s

If W < b lo c k r is e s

B

B

F

F

z

z+dz)W z g

(

)B

F z z g

d (

2

2 2

2

2

5 2

2

{ ) ) }

{ ) ) }b u t

w h e r e ( ) { }

4 .4 1 0 s e c

n e t B

n e t

F F W V z z z g

z z z

z z

g g gF V z g V N z N

z z z c

g

c

d

d

d

d d

( (

( (

Concept of Buoyancy frequency N

Gradient Richardson Number

Turbulence in the Pycnocline

Velocity Shear

u

z

g

Nz

2

2( )

g

NR i

u

z

Turbulence occurs when

1

4g

R i

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( )

4g

R i

Depth(m)

Distance (m)

Turbulence Observed in an internal solitary wave resulting in

Goodman and Wang (JMS, 2008)

1( )

4g

R i

Temperature (Heat)Equation

with Molecular Diffusion

2

2

7

w h e r e

v

m o le c u la r d i f f u s iv i ty o f T

= 1 .4 1s e c

T

T

d TT

d t

du w

d t x y z

m

Approach for Turbulence

0

B u t ' & '

v 0

v ' ' 0

d T

d t

T 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

z

TH k c

z

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

0

0

0

0

S te a d y S ta te C a s e 0

0

:

( ) e x p [ ( ) ]

( ) e x p ( ) w h e r e

(T ( z ) = T ( ) {1 e x p (

T

T

s

T

T

s

s s

T T Tw

t z z

T

t

T Tw w u p w e l l in g v e lo c i ty

z z

S o lu t io n

T T wD z

z z

T D zz

z z w

Tz

z

0

0

0

) }

T ( z ) = T {1 e x p ( ) } w h e r e ( )s s

D z

z

D z TT T z

z z

T(z)

w

3

s

H e a t T r a n s f e r e d

H = ( ) w h e r e c s p e c if ic h e a t o f w a te r 4 .2 1 0T S

T Jc

z o C k g

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? Note change in numbers from notes!!

s 2H = 5 0 0 , 2 0

o

s

W a ttsT C

m

2

31 0

s e c

mk

sT

Z=0

Z=D Surface (s)

s 2H = 5 0 0 , 2 0

o

s

W a ttsT C

m

2

2

2

3 3 3

3

0 ( ) e x p [ ( ) ] = ( )

5 0 0 a t a ll d e p th s !

5 0 0

( ) ( )

1 0 4 .2 * 1 0 * 1 0s e c

.1 2 a t a ll d e p th s !

s s

T

s

s

s

T

o

o

T T w Tw D z

z z z

W a t t sH H

m

W a t ts

HT T m

k g J mz z c

m k g C

C

m

0

0

0 00

0

s

T ( z ) = T {1 e x p ( ) }

A s T ( z ) = l im [ T ( ) {1 e x p ( ) } ]

T ( z ) = T ( ) ( )

a t : z = 5 0 , T = T 2 0

z = 2 5 , T = 2 0 .1 2 ( 2 5 ) 1 7

z = 0 , T = 2 0 .1 2 ( 5 0 ) 1 4

s

s sz

s s

o

o

o o

o

o 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 = 5 0 , 2 0o

s

W a ttsT C

m

0

0

T ( z ) = T {1 e x p ( ) }

( ) 1 0 * .1 2 1 .2

a t : z = 5 0 , T = 2 0

2 5z = 2 5 , T = 2 0 1 .2 [1 e x p ( ) ] 1 8 .9

1 0

5 0z = 0 , T = 2 0 1 .2 [1 e x p ( ) ] 1 8 .8

1 0

1 0a t : z = 1 0 ,T = 2 0 1 .2 [1 e x p ( ) ] 1 9 .2

1 0

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

2

3

0

4

0

s

0

s 2

2 2

2 2

1 0s e c

1 0

1 0s e c

(z )= ( ) ( ) e x p ( )

( z )= H e x p ( )

5 0 ; H = H 5 0 0

2 52 5 ; H = 5 0 0 e x p ( ) 4 1

1 0

5 00 ; H = 5 0 0 e x p ( ) 3 .3

1 0

T

s

m

z mmw

T T D zH c k c k

z z z

D zH

z

W a t tsz

m

W a t ts W a t tsz m

m m

W a t ts W a t tsz m

m m

wu u