δ δ δ δ θ τ ρ θ τ μ δ δ δ δ δ δ δ

ID:‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Quiz: No. 8 Time: 15 minutes Name: ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Course: 58:160, Fall 2010 The exam is closed book and closed notes. A laminar boundary layer velocity profile is approximated by
u / U  [2  ( y /  )]( y /  ) for y   ,
y   . (a) Show that this profile satisfies the appropriate boundary conditions. (b) Use the
momentum integral equation to determine the boundary layer thickness    ( x) . Then write the
and u  U for
boundary layer thickness in the form of

x

A
. What would be the value of constant A? Assume that
Re x
 ( x  0)  0 .
Note:
i.
 w  U 2

ii.
iii.
u
U
0
 
 

iv.

0
 u
1  dy
 U
du
dy
   1 
*
d
dx
u
dy
U
Solution:
(a)
The velocity profile has to satisfy some conditions:
1. velocity on wall has to be zero
2. velocity at far from the plate has to be U.
3. velocity at far from the plate has to reach to uniform velocity condition i.e. du/dy=0
at y  0 :
at y   :
u / U  [2  ( y /  )]( y /  )  0
u / U  [2  1](1)  1  u  U
at y   :
du / dy  U [2(1/  )  2( y /  2 )]  0
(b)
OK
OK
OK
(1.5 point)
ID:‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Quiz: No. 8 Time: 15 minutes Name: ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ Course: 58:160, Fall 2010 
u
U
0
 
 u
1 
 U

u

dy  
U

0
 u
1 
 U

u

dy  

 U
 u
1  
 U


U
 U
  [2( y /  )  5( y /  ) 2  4( y /  )3  ( y /  ) 4 ]dy  
0

 2
0



y2
y3
y4
y5
dy  5 2 dy  4  3 dy   3 dy  0
2
3
4
5
0
0
0
5
1
2
        
3
5
15
w  
du
dy
 U [2(1/  )  2( y /  2 )] y 0  2U /  (2.5 point)
y 0
d
d 2
 2 U /   U 2 (  )
dx
dx 15

1

dx   d
U
15
(1.5 point)

1

dx    d 
U
15

1

x  2 C
U
30
 ( x  0)  0  C  0 (1 point)
 w  U 2

1
x  2
30
U
 2 1 2

x  
30
Ux


1
x2
  2 or :
Re x 30
(1 point)

x

30

Re x
A  30
 U
1 
 U

 dy
 (2.5 point)