UNSTEADY VISCOUS FLOW



Du
2

  g   p   u
Dt
Assume linear,
horizontal motion
u
t
 
1 p
 x
 u
2

y
2
Viscous effects confined to within some finite area near the boundary →
boundary layer
In unsteady viscous flows at low Re the boundary layer thickness δ grows
with time; but in periodic flows, it remains constant
If the pressure gradient is zero, Navier-Stokes equation (in x) reduces to:
u
t
 u
2

y
2
u
t
 u
2

y
2
Heat Equation– parabolic partial differential equation - linear
Requires one initial condition and two boundary conditions
u x , y ,t  0   0
@ y  
u  0
@ y  0
u U
Impulsively
started
plate –
Stokes first
problem
y
U
Total of three conditions
u
t
 u
2

y
2
Heat Equation– parabolic partial differential equation
Can be solved by “Separation of Variables”
Suppose we have a solution: u  y , t   Y  y  T t 
Substituting in the diff eq:
May also be written as:

Y  y  T t   
t

2
y
2
Y  y  T t 
Y  y  T  t    Y   y  T t 
Moving variables to same side:
T  t 
 T t 

The two sides have to be equal for any choice
of y and t ,
The minus sign in front of k is for convenience
Y   y

Y y 

T  t 
 T t 

Y   y


Y y 
k
T  t 
 T t 

Y   y


Y y 
Y   kY  0
 Y
T   k T  0
y
t
t
T
T
u  x , y , t  0   0 u  y , t   Y  y  T t 
2
T
T
This equation contains a pair of ordinary
differential equations:
k
2
 kY  0
@ y  
 k T  0 @ y  0
0  Y  
u  0
U  Y 0 
u U
0  T 0 
 k T  0
  k t
 Y

T
T
  k
2
y
2
 kY  0

ln T   k  t  A
t
Y  B cos


k y  C sin

ky
T  Ae

 k t
Y  B cos


k y  C sin

ky
 n 
Applying B.C.,
kn  

B = 0; C =1;
L



u  y , t   Y  y  T t 
2
2
u  An e
 n 

  t
 L 
 ny 
sin 

L


u  Ue
increasing time

y
2
4 t
Alternative solution to“Separation of Variables” – “Similarity Solution”
u
t
 u
u x , y ,t  0   0
y
@ y  
2

2
u  0
@ y  0
New independent variable:  
y
t


t

y

2
d u
Substituting into 

2
2t d 
4 t d 
heat equation:
 d
2
 
t d

2
y

u
from:
2  t
η is used to transform heat equation:
 du
u U

 d

y d
1
d

d
2
d
2t d 
1
d
2  t d
2
2
2
y
2
4 t d 
d u

u
 2
du
d
 0
2
d u
d
2
 2
du
d
u  0
 0
as
  
@  
@  0
2 BC turn into 1
u  0
u U
du
To transform second
f

order into first order:
d
df
d
 2 f  0
With solution:
df
f
e
  2 d 
f  Ae


2
erf 

2

Integrating to u  A e    d    B

obtain u:
0
2
Or in terms of the error
function:
erf   

2



e
 
2
d 
u  U 1  erf 

0
For η > 2 the error function is nearly 1, so that u → 0
u  U 1  erf 
  For η > 2 the error function is nearly 1, so that u → 0
Then, viscous effects are confined to the region η < 2
This is the boundary
layer δ
 
y
2  t
2 

2  t
   4  t
δ grows as the squared
root of time
increasing time
u  U 1  erf 
 
y
2  t

UNSTEADY VISCOUS FLOW
Oscillating Plate – Stokes’ second problem
u
t
 u
2

y
2
@ y  
u  bounded
@ y  0
u  U cos  t 
y
Ucos(ωt)

Look for a solution of the form: u  Y  y  cos  t   Re Y  y e i  t
e
i t
 cos  t   i sin  t 
Euler’s formula


u  Re Y  y e
Fourier’s transform in the time domain:
u
Substitution into:

t
Ye
i t

i  Ye
i t
t
 u
y
2
y

Y  A exp   1  i 


Y    0

e
 Y
2

U  Y 0 
Y    0
y
2
i t
i

y
 Y
2
 Y
 i Y  
2
Y  0
i  
y
1  i 
2



y  B exp  1  i 


2



 B  0

2
2
2
B.C. in Y
 u
2

i t
Y 0   U

Y  U exp   1  i 


 A  U

y 

2



y 

2


2

Y  U exp   1  i 


Most of the motion is confined to region
 
within:

y 

2


2



u  Re Ye
i t
  Ue
y

y 

cos   t 




@ y  
y
Ue

 Ue
1
 0 . 37 U
@ y  4  /  
y
Ue

 Ue
4
2
 Ue

 0 . 06 U
y
Ucos(ωt)
u  Ue
y

y 

cos   t 

 



UNSTEADY VISCOUS FLOW
Oscillating Plate
u
t
 u
2

y
2
@ y W
u  0
W
y
@ y  0
u  U cos  t 
Ucos(ωt)

Look for a solution of the form: u  Y  y  cos  t   Re Y  y e i  t
e
i t
 cos  t   i sin  t 
Euler’s formula


u  Re Y  y e
Fourier’s transform in the time domain:
u
Substitution into:

t
Ye
i t

i  Ye
i t
t
 u
y
2
y

Y  A exp   1  i 


A  B  U
 
2


e
 Y
2


Y 0   U
Y W
2
y
i t
i

 Y
y
2
Y  0
0
 Y
2
 i Y  
i  
y
2
1  i 
2



y  B exp  1  i 


2




0  A exp   1  i 



2
2
2
B.C. in Y
 u
2

i t

y 

2




W   B exp  1  i 


2




W 

2


A  B  U
0  Ae
W 

   1 i 

 

0  U  B e
 B  
Y U
Ue
 Be
  

2 sinh 

  1  i 
W 

  1 i 

 

 Be
e
 

e


e
 A  U  1 
2 sinh 

sinh (1  i )( W  y ) 
sinh (1  i )W 


W

 2 sinh 
 Y  Ae

 

 y
 Be
 y

@W  
sinh (1  i )W 

(1  i )W

Y U
sinh (1  i )( W  y ) 
sinh (1  i )W 


Y  U
W  y
W
y 

 U 1

W 

y 

u  U 1
 cos  t 
W 

@ W  
e

e

sinh (1  i )W
 2 sinh 


1
( 1 i ) W
2e

 (1  i ) y 
 Y  U exp  




Same solution as
for unbounded
oscillating plate
y 

u  U 1
 cos  t 
W

