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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