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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