Perturbation: Background


Algebraic
Differential Equations
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Perturbation
Original Equation

X 2  25  0
Y  X 2   X  25
Y vs X

15
10
Perturbed equation
X 2   X  25  0
0   1
5
0
-8
-6
-4
-2
0
2
4
6
8
-5
Y
Epsilon=0.8
-10
Epsilon=0.5
Epsilon=0.0
-15
-20
-25
-30
X
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Perturbation


Perturbed equation
X 2   X  25  0
Change in result (absolute values) vs Change in equation
Perturbation in the result
Perturbation in the result (root)
Root -1
-2
-0.15
-0.15
0.06
0.05
=-0.1
=-0.1

Simple (Regular)
Perturbation

Answer can be
in the form
=0.1
=0.1
0.04
0.03
0.02
-0.1
-0.1
0.01
0.01
=-0.01
=-0.01
0
0
-0.05
0
-0.05
0
=0.01
=0.01
0.05
0.05
Epsilon (perturbation)
(perturbation)
Epsilon
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0.1
0.1
0.15
0.15
X  X 0    ( X 0 )   2 ...
Perturbation
Y  X  X  5
2
Original Equation

X 5  0
Y vs X

40
35
Perturbed equation
 X 2  X 5  0
30
Y
25
20
Epsilon=0
15
Epsilon=0.8
10
Two roots instead
of one

Epsilon=1
5

0
-6
-4
-2
-5 0
-10
X
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
4
6
Roots are not close
to the original root
Perturbation
Y  X 2  X  5
Change in result (absolute values) vs Change in equation

Root-1
Root-2
Other root varies
from the original
root dramatically,
as epsilon
approaches zero!

PerturbationininResult
Result
Perturbation
3.5
1200
3
1000
2.5
800
2
600
1.5
400
1

0.5
200
00
00
0.2
0.2
0.4
0.4
0.6
0.6
Epsilon
Epsilon
0.8
0.8
11
1.2
1.2

Singular
perturbation
Answer may NOT be
in the form
X  X 0    ( X 0 )   2 ...
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Differential Equations
dy
x0
dx



y  1, at x  0
2
x
Solution y  1 
2
y  1, at x  0
Perturbation-1 dy
x 
dx
Solution
x2
y  1   x
2
x2
  0, y  1 
2
IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Regular Perturbation
Differential Equations
dy
a
dx



Solution
a 1
yax
Perturbation-1
Solution
y  0, at x  0
y  0, at x  0
dy
 a  y
dx
2 2


a

y  e  1   1   x  x  ...  1
 

  
2
 
2 2

a
 x
   x 
 ...

2

  0, y  ax
a
x
 x

 ax1   ... 
2


IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Regular Perturbation
Differential Equations



Another Regular Perturbation
Perturbation-1a
Solution
dy
 a  x
dx
y  ax 

Solution
 x2
2
  0, y  ax
dy
 a 
dx
Perturbation-1b

y  0, at x  0
y  ax  x
  0, y  ax
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
y  0, at x  0
Differential Equations


Perturbation-2
d 2 y dy
 2  a
dx
dx
a 1
y  0, at x  0
y  1, at x  1
Exact Solution

(eg using Integrating factors method)
x

1 e 
y  ax  1  a  
1
 1 e 







  0, y  ax
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Singular Perturbation
d 2 y dy
 2  a
dx
dx
Differential Equations

Singular Perturbation
a  0.5 (for example)
Y vs X
1.2

Can the solution be of the following
form? (to satisfy the extra boundary
condition?)

No!
y  1, at x  1
1
Y
0.8
0.6
Y=ax

0.4
0.2
0
0
0.2
0.4
0.6
0.8
X
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
1
1.2
Based on the perturbed equation
Differential Equations
Y vs X

At the limit
1.2
  0, y  ax  (1  a), for 0  x  1
1
  0, 0  y  (1  a), for x  0
Y
0.8
0.6
Y=ax
Epsilon 0.01

Method to find solution

Transform variables (x,y,)
Called “Stretching
Transformation”
Zooms in the ‘rapidly
varying domain’
Obtain “inner solution”
0.4
Epsilon 0.1
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2

X

IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Method to find solution

Differential Equations

Let  =0 and simplify eqn
 2nd order equation,
satisfying only one
boundary condition (x=0)
 one constant remains
arbitrary
 Valid only near x=0
Y vs X

1.2
Inner
solutions
1
Y
0.8
Outer Soln
0.6
Y=ax
0.4
0.2

0
0
0.2
0.4
0.6
Inner solution:
0.8
X
1
1.2
Obtain outer solution, for
first order equation,
satisfying one Boundary
Condition (x=1)

IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Valid everywhere, except
near x=0
Differential Equations

Y vs X

1.2
Inner
solution
1
Y
0.8
Outer Soln
0.6
Y=ax
Method to find solution
Match the two solutions in
the segment in between, by
choosing the remaining
constant

0.4
Match the value and the
slopes
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
X

IIT-Madras, Momentum Transfer: July 2005-Dec 2005
x


1 e  
y  ax  1  a  

1




Close to the exact solution
Numerical Solution (to BL)


“Real” solution

Approx solution



IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Grid generation
Structured grid vs
unstructured grid
Uniform vs non-uniform grid
What about placing more grid
everywhere?
More grid points near surface
Similar to “stretching
transformation”
Boundary Layer theory

Situations we have seen so far






V0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Laminar flow in cylinder
Fully developed (entrance
effects are negligible)
Steady State
Unsteady State
Again, entrance effects are
negligible
Movement of infinite plate, in
a semi-infinite medium
Boundary Layer theory

Semi-infinite plate
Inviscid flow (irrotational)
Will NOT satisfy ‘no slip’
condition at the plate


Fluid Velocity V0



Flow over cylinder (inviscid)
Flow over sphere (2D, viscous
flow) (tutorial problem)
Flow over any other shape
(while accounting for no-slip condition
and not assuming fully developed flow)
is treated with “boundary
layer theory”
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Boundary Layer theory

Semi-infinite plate

Fluid Velocity V0



IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Away from plate, inviscid
solution is valid (and will
satisfy the boundary
condition). This is “outer
solution”
Near the plate, different
solution (including viscosity)
will be found using ‘stretching
transformation’. [Inner
Solution]
Inner solution will satisfy the
boundary condition (no slip)
Match both solutions to find
the other constant
Boundary Layer theory

Solid Boundary
INF
INF
Velocity V0
Velocity V0
INVISCID FLOW
ASSUMPTION OK HERE
No Slip
00
FRICTION CANNOT BE
NEGLECTED HERE
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Velocity 0
Boundary Layer theory

Solid Boundary
INF
INF
d
0
x
B L thickness
99% Free Stream Velocity
d
0
What happens to d when you
move in x?
B L thickness
increases
with x
IIT-Madras, Momentum Transfer: July 2005-Dec
2005
Momentum Transfer
Boundary Layer theory

Draw d vs x

Analytical Expression, for velocity vs (x,y), below BL:

Continuity

Navier Stokes Equation
d
d
y
x
B L thickness increases with x
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Boundary Layer theory


 .V   0
t
Steady,incompressible, two dimensional (semi infinite plate)
 


 Vx   Vy   Vz   0
t x
y
z

Hence
Vx Vy

0
x
y
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
1
N-S Eqn

DV
2

 P   V  g
Dt
Consider only X and Y equations (2D assumption)
  2Vx  2Vx  2Vx 
 Vx
Vx
Vx
Vx 
P
  
 
 Vx
 Vy
 Vz
   2  2  2    gx
x
y
z 
x
y
z 
 t
 x
  2Vy  2Vy  2Vy 
Vy
Vy
Vy 
 Vy
P
  
 
 Vx
 Vy
 Vz
  2  2  2    gy
x
y
z 
y
y
z 
 t
 x

Steady flow, gravity can be incorporated in Pressure term (or
assume gravity is in Z direction, for example)

Vz=0, Vx and Vyare not functions of z
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Eqn




Obtain “order of magnitude” idea
Can be used to ignore small terms (simplify eqn by removing
‘regular’ perturbations)
Can be used to non-dimensionalize equations
example:
Vy
x
y
'
Vx
'
'
'
Vy 
x 
y 
Vx 
U2
U1
L1
L2
P' 
P
U12
U 1t
'
t 
L1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Steady State
N-S Eqn

Write the NS-eqn in “usual” form, for steady state
  2Vx  2Vx 
 Vx
Vx 
P
  
 Vx
 Vy
  2  2 
y 
x
y 
 x
 x
  2V y  2V y 
V y 
 V y
P
  
 Vx
 Vy
  2  2 
x
y 
y
y 
 x

IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Eqn

What are the relevant scales for the lengths (eg what are L1,
L2 in this particular case?
x
x 
L
y 
'
'
y
d
L2  d
L1  L
Note : d is a function of x
d
y
x
L
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
N-S Eqn
  2Vx  2Vx 
 Vx
Vx 
P
  
 Vx
 Vy
  2  2 
y 
x
y 
 x
 x

What are the relevant scales for the velocity?

Vx varies from 0 to Vo (or we can call it VINF)
Vx ~ V


~ means “Order of ”
Note: Some books show it as Vx ~ OV 
y
x

IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Similarly
Vy ~ U 2
  2Vx  2Vx 
 Vx
Vx 
P
  
 Vx
 Vy
  2  2 
y 
x
y 
 x
 x
N-S Eqn

What are the relevant scales for the derivatives?
Vx V  0
~
x
L0
Vx V
~
x
L
V y
y
Note: The sign is not important here

Continuity

Vy
V
~ 
y
L
~?
x0
V y
Vx

0
x
y
Vy ~
1
V
d
L
xL
Vx  V
Vx  0
y
x
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
  2Vx  2Vx 
 Vx
Vx 
P
  
 Vx
 Vy
  2  2 
y 
x
y 
 x
 x
N-S Eqn

What are the relevant scales for the derivatives?
Vx V
~
y
d
 2Vx V
~ 2
2
y
d
 2Vx V
~ 2
2
x
L
Thin Boundary Layer assumption d  L
 2Vx
 2Vx
 2
2
x
y
d
y
x
L
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
  2Vx 
 Vx
Vx 
P
  
 Vx
 Vy
  2 
y 
x
 x
 y 
N-S Eqn
Can we approximate pressure drop?
Assume that pressure drop is similar to inviscid flow


P
V
  V 
x
x
Vx V2
Vx
~
x
L
From Bernoulli’s eqn
Vx  V d  V

Vy
~ 
y  L  d
P
V
V
  V   ~ 
x
x
L
2

 V2
 ~
 L
 V 
 2Vx
 2 ~   2 
y
d 
d
Claim: as  -->0,
y
x
L
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
V 
 2Vx
 2 ~   2   non  zero
y
d 
  2Vx  V2
 Vx
Vx 
V
  V
 Vx
 Vy
  2  ~
y 
x
 x
 y  L
N-S Eqn

For the Y component of N-S equation
  2V y  2V y 
V y 
 V y
P
  
 Vx
 Vy
  2  2 
x
y 
y
y 
 x

 V
 d
Vy
Vx
~ V  L
x
 L


 V
Vy  V  L d
Vy
~  d 
y  L  d



 V2
~  d
 L



 V2
 ~ 2 d 2
 L


Each term is small compared to the
equivalent in X-eqn P
0

==>
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
y

V
d V
 Vy
L ~ d
~
x 2
L2
L3
2
V
d V
 Vy
L ~ 
~
y 2
d2
Ld
2

 2Vy
y 2
~
V
Ld
Prandtl BL eqn (steady state)
  2Vx 
 Vx
Vx 
V
  V
 Vx
 Vy
  2 
y 
x
 x
 y 
V y
Vx

0
x
y
Unsteady State
  2Vx 
 Vx
Vx
Vx 
V
  V
 
 Vx
 Vy
  2 
x
y 
x
 t
 y 
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Prandtl BL eqn : flow over Flat plate
  2Vx 
 Vx
Vx 
V
  V
 Vx
 Vy
  2 

x

y

x


 y 
d
No pressure Drop
Steady State
2D-flow (Stream Function)

Vx 
y
V y
Vx

0
x
y
y
x
L
 
Vy 
x
Stretching Transformation (near the boundary)
d  x
a
d 
b
Non-dimensionalize y
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
 1
d 
V
 




c
 V
y
 ~
  ~ y
 x
d 

 x
d 
V
 




1
2
y

2
 V

 x





1
2




1
2
Prandtl BL eqn : flow over Flat plate
  2Vx  V2
 Vx
Vx 
    2  ~
 Vx
 Vy
y 
 x
 y  L
Another perspective for the choice of d
 V V2
~
2
d
L
d2 ~
1
 L
V
d
y
x
L
 x
d ~

If we write the BL eqn in stream function
V 

  
  2   2
 3

 3
Boundary Conditions

2
y xy x y
y

2
y  0, Vx 
y  0, V y 
y
0
 
0
y
y  , Vx 

 V
y
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
x  0, Vx 

 V
y
Prandtl BL eqn : flow over Flat plate
V y
Vx

0
x
y
  2Vx 
 Vx
Vx 
V
  V
 Vx
 Vy
  2 

x

y

x


 y 
f ( ) 
d
 ( x, y )
1
 x V  2
   x V 
y
x
L
1
2
f


  x V 
Vx 

y
y
y
 
2
 V

 x





1
2
1
2
f


1  V
 
y
2
 x
IIT-Madras, Momentum Transfer: July 2005-Dec 2005

  x V 






1
1
2
f
 
y
2

1 y  V


x
2x 2 
 x




1
2

1

2x
Prandtl BL eqn : flow over Flat plate
  2Vx 
 Vx
Vx 
V
  V
 Vx
 Vy
  2 
y 
x
 x
 y 
V
Vx 
f
2
 
Vy 
x
d
1

1  V  2 

Vy 
  f   f
2  x  


Vx
 V 

f 
x
4x
Vx
V  V
  
y
4 
 x
2
V
8x




1
2
f 
f   f f   0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005

y
x
L
 2Vx
V

f 
2
y
8x
2
Some books may have -ve sign, or
a factor of 2, in the equation,
depending on the definition of
Stream function and
transformations used
Prandtl BL eqn : flow over Flat plate
f   f f   0

Boundary Conditions:
  0, f  f   0






  , f   2
No solution in ‘usual’ form
Blasius Solution: Series solution, valid for small 
f 


0
An n

n!
For large , asymptotic series that matches with the boundary
condition
1
Numerical values tabulated (f,f’,f’’...)
Plot of Vx/VINF vs 
Note: definition of  may be slightly
different in various books (usually by
a factor of 2)
Vx
V
0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0

3
Prandtl BL eqn : flow over Flat plate

Blasius Solution

Valid for high Reynolds Number
Re

Local:( dV/)
 More useful (convenient): (X V / ) (sometimes, this is referred to
as “local” Reynolds number)
 105 or more



Not valid very near x= 0 (at the point x=0,y=0)
Another way1to express boundary1 layer thickness
 x
d ~
V 

  

2
 
d

~
 xV
L






2
 1 
~

Re


1
2
Reynolds number high ==> Boundary layer is thin
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Prandtl BL eqn : flow over Flat plate







Boundary layer thickness
Drag estimate
Other definitions (for thickness)
Similarity
Effect of pressure variation (Loss of similarity and separation)
Thermal vs momentum Boundary Layer
von Karman method
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
References:






Introduction to Mathematical Fluid Dynamics by Richard E
Meyer
Perturbation methods in fluid dynamics, by Van Dyke
BSL
3W&R
Fluid flow analysis by Sharpe
Introduction to Fluid Mechanics by Fox & McDonald
IIT-Madras, Momentum Transfer: July 2005-Dec 2005