Transcript Lect08_similarity

Similarity theory
Turbulent closure problem requires empirical expressions for
determining turbulent eddy diffusion coefficients. The development
of turbulence closure is based on observations not theory.
We need to find an intelligent way of organizing observational data.
Similarity theory is a method to find relationships among variables
based on observations.
1. Buckingham Pi Theorem and examples

We want to find the relationship between
the cruising speed and the weight of airplane.
U velocity (m / s )
a. Define relevant variables and
dimension (m)
their dimensions.
W mass (kg )
U
g acceleration of gravity (m / s 2 )
b. Count number of fundamental
dimensions.
 air density (kg / m3 )
m, s, kg
c. Form n dimensionless groups 1 ,.. n
where n is the number of variables
minus the number of fundamental
dimensions.
Wg
1 
U 2 2
gravitational force
lift force
W
2  3

d. Measure 1 as a function of  2
 1  f ( 2 )
e. Further simplification; assume:
W ~ 3   2  constant  1  constant
i.e.
W ~ U 22 ~ U 2W 2 / 3  W ~ U 6
n 53 2
mass of airplane
mass of
displaced air
Weight (Newtons)
Weight as a function of
cruising speed ( “The
simple science of flight” by
Tennekes, 1997, MIT press)
The great flight diagram
W~U6
Flying objects range
from small insects to
Boeing 747
Speed (m/s)
Procedure of Buckingham Pi Analysis
Step 1, Hypothesize which variables could be important to the flow.
e.g., stress, density, viscosity, velocity, …..
Step 2, Find the dimensions of each of the variables in terms of the
fundamental dimensions. Fundamental dimensions are:
L=length
M=mass
T=time
K=temperature
Dimensions of any other variables can be represented by these
fundamental dimensions.
Example

density
ML-3
U
velocity
LT -1

z0
wind stress
roughness
ML-1T -2
L
H
Boundary Layer height
L

Viscosity Coefficien t
ML-1T -1
Step 3, Count the number of fundamental dimensions in the problem
there are 3 dimensions in this example: L, M, T
Step 4, Pick up a subset of original variables to become “key variables”,
subject to the following restrictions:
•The number of key variables must equal the number of fundamental dimensions.
•All fundamental dimensions must be represented in terms of key variables.
•No dimensionless group is allowed from any combination of key variables.
e.g. Pick up 3 variables:
Invalid set:
 , U, H;
 , U, z 0 ;  ,  , H;
 , H, z 0 ;  ,  , U;
Step 5, Form dimensionless equations of the remaining variables in terms of
the key variables.
e.g.
   a Hb Uc
   dHeUf
z0   g H h Ui
Step 6, Solve for the unknowns a, b, c, d, e, f, g, h, i
e.g.
a b c
  H U
ML-1T -2  (ML-3 ) a (L) b (LT -1 ) c
a  1, b  0, c  2
Step 7, Form dimensionless (PI) groups.
e.g.
z

 1   2 ,  2  UH
,  3  H0 ,
U
Step 8, Form other PI groups if you want as long as the total number is the same.
e.g.

 4   2 ,  5  1 ,
3
3
1 
 ,
U 2

 4  Uz
,  5  zH ,
0
0
Which PI groups are right?
They are all right, but some groups are more commonly used and follow
Conventions.
UH
z
0

,
Reynolds
number;


, relative roughness
3
2

H
1
Next, find relations between
PIs through experiments.
Surface layer similarity (Monin Obukhov similarity)
Surface layer: turbulent fluxes are nearly constant. 20-30 m
Relevant parameters:
z height or eddy size
u *2
g
|  o | /  

(u w ) o2
( m)

1/ 2
2
 ( vw ) o
(m 2s  2 ), frictional velocity
( m 2 s 3 )
( w ' v ') o
v
Say we are interested in wind shear:
u
z
Four variables and two basic units result in two dimensionless numbers, e.g.:
z u
u * z
and
g ( w  v ) 0 z
v
u *3
The standard way of formulating this is by defining:
L
v
u *3
g ( w  v ) 0
Monin-Oubkhov length
PI relation
z u   ( z )   ( ),
m L
m
u * z
  0.35(0.4),
Von - Karman constant
Empirical gradient functions to
describe these observations:
m  (1  16 ) 1 / 4
m  1  5
for   0
for   0
Note that eddy diffusion coefficients
and gradient functions are related:
u w   k m
u
z
Assuming vw   0,
k
m

m
zu *
1
unstable
stable
Now we are interested in the vertical gradient of virtual potential temperature.
 v
z
* 
We can form a new variable
 ( w ') o
u*
Again, four variables and two basic units result in two dimensionless numbers,
z  v
* z
and
z
L
PI relation
z  v   ( z )   ( ),
h L
h
* z
Similarly, we have
z q   ( z )   ( ),
q L
q
q* z
Normally,
h ( )  q ( ),
Surface wind profile
z
L
1. Neutral condition
0
m (0)  1
z u  1
u* z
u
u  * ln(zz )
0
z
z0 
exp(uu )
u* 
*
u
ln(zz )
0
z 0 is the height where winds disappear.
Aerodynamic roughness length
Over land
0.25 N
0.25 N
z0 
 h isi 
 hiwi
S t i 1
L t i 1
Kondo and Yamazawa
(1986)
S t : total area; h i : height of i element; s i : area of i element
L t : total length; w i : width of i element
Over water
z0 
u *2
g
,   0.016
Displacement distance
u
u  * ln(zz-d )
0
z0
If you have observations at three levels,
you may determine displacement as,
u* 
u1
ln(
z1 d
)
z0

u 2
ln(
z 2 d
)
z0
d
u 3

ln(
z 3 d
)
z0
u 2  u1
z d
z d
ln( z3 d )  ln( z2 d )
1
1
u 3  u1
2. Non-neutral condition

z
L
0
m  (1  16 ) 1 / 4
m  1  5
for   0
for   0
q  h  (1  16 ) 1 / 2
for   0
q  h  1  5
for   0
Integral form of wind and temperature profiles in the surface layer
u
u  * [ln(zz )  m ( )]
0
m ( ) 
2
1

x
1

x
2 ln( 2 )  ln( 2 )  2 tan 1
x  2 , x  (1  16 )1 / 4 ,
m ( )  5 ,
for   0
for   0
ln(zz )
0
L  0,   0
L  ,   0
L  0,   0
u/u *
Integral form of wind and temperature profiles in the surface layer
 ( v  v 0 )
z )   ( )

ln(
h
*
zt
1 y
),
2
h ( )  2 ln(
y  (1  16 )1 / 2 ,
h ( )  5 ,
for   0
 v   v0 at z  z t
Normally, z t  z 0
Similarly,
 ( q q 0 )
q*
for   0
 ln(zz )  q ( )
q
q ( )  h ( )
q  q 0 at z  z q
Normally, z t  z 0
Bulk transfer relations
How to estimate surface fluxes using conventional surface observations,
surface winds (10m), surface temperature (2m),…?
u *2  C D u 2 ,
( w  v ) 0  C H u( v   v0 ),
( w q ) 0  CQ u(q q 0 ).
C D , C H , CQ : Drag coefficient of momentum, heat, and moisture.
CD 
u
( u* ) 2

2
[ln( z/z 0 )  m ( )]2
,
C DN 
u
( u* ) 2

2
[ln( z/z 0 )]2
,
2
C H  [ln( z/z )   ( )][ln( z/z )   ( )] ,
0
m
t
h

C HN  [ln( z/z )][ln(
,
z/z t )]
0
2
CQ  [ln( z/z
2
0 )  m ( )][ln( z/z q )  q ( )]
2
CQN  [ln( z/z )][ln(
z/z
0
CD
C DN
,
CH
C HN
z
z0
1.5
1.0
q )]
,
z
z0
 102
z
z0
1.5
z
z0
 105
 102
 105
1.0
z
L
-0.5
0
0.5
z
L
-0.5
0
0.5
Over Land
A new perspective on MOS
Surface layer (constant flux layer) :
1. Steady neutral condition :
Kolmogorov -5/3 power law:
Example of spectrum of energy density from the SCOPE data
η estimated from the SCOPE data
Best nonlinear fitting curve
  c13 / 2c2  [1  c3 exp( c4u )],   0.35, c3  5.0, c4  0.5.
Weak wind
’staircase’-like
Strong wind
’elevator’-like
After Hunt
and Carlotti
(2001)
2. Steady non-neutral condition :
Unstable condition (ς<0):
Stable condition (ς<0):
Flux footprint
General concept of the flux footprint. The darker the red color,
the more contribution that is coming from the surface area certain
distance away for the instrument.
Relative contribution of the land surface area to the flux for two
different measurement heights at near-neutral stability.
Relative contribution of the land surface area to the flux for two different surface
roughnesses at near-neutral stability.
Relative contribution of the land surface area to the flux for two different cases
of thermal stability.
Problem: Assuming we have wind observations but no temperature
observations at two levels, say, 5 m and 10 m, in the surface layer,
can we estimate surface roughness and stability?
u10
u*
z
 ( u10  u 5 )
u*
 ( u10  u 5 )
Neutral :
Stable :
u*
u*
u*
u*
z
z
z
z
 ln( z10 )   ( L10 )   ( L5 ),
5
z
 ln( z10 )  0
z
 ln( z10 )  0,
5
 ( u10  u 5 )
 ( u10  u 5 )
z
5
5

ln(
)


(
),
u*
z0
L
5
 ( u10  u 5 )
Unstable :
u 5
z
 ln( z10 )   ( L10 ),
0
z
 ( u10  u 5 )
u*
z
 ln( z10 )
5
x  (1  16 Lz )1 / 4
z
z
 ln( z10 )   ( L10  L5 )
5
z
 ln( z10 )  0,
5
 {ln[
(1 x10 ) 2
(1 x 5 )
2
1 x10 2
]  ln[
1 x 5
2
]  2(tan 1 x10  tan 1 x 5 )}