Aula Teórica 14
Escoamento de Couette, i.e.,
escoamento entre duas placas
planas.
Couette Flow
•
•
We will assume stationary, incompressible and that the
plates have infinite length and depth along the direction
normal to the paper. This is equivalent to say that we are
far away from the entrance and the exit and from the walls
perpendicular to x3. In this case the velocity as a single
component, along x1.
The BC are u=0 at y=0 and u=U0 at y=h
• The continuity equation becomes:
ui
0
xi
u1 u2 u3


0
x1 x2 x3
u1
0
x1
• The NS equation becomes:
 ui
dui
ui 
p




uj


 t
dt
x j 
xi x j

 ui

 x
j

p   u 
0        g x
x y  y 

  g i


y
p   u 
0        g x
x y  y 
gx
z z
 tg 

gz
x x
z
gx
g
𝜶
𝜶
gy
0
p   u 
z
     g
x y  y 
x
0
  p  gz    u 
   
x
y  y 
0
 P    u 
   
x y  y 
x
Integrating along yy
0
  p  gz    u 
   
x
y  y 
 u 
 P 
    
y  C1

y

x


1  P  2
u
y  C1 y  C2
2  x
•
The BC are u=0 at y=0 and u=U0 at y=h
2
1   p  gz  2  y  y  
y
u
h      U0
2
x
h
 h  h  
C2  0
1  P  2
h  C1h
2  x
U
1  P 
C1  0 
h
h 2  x
1  P  2 U 0
1  P 
u
y 
y
hy
2  x
h
2  x
U0  
Interpretation
2
1  p  gz  2  y  y  
y
u
h      U0
2
x
h
 h  h  
Defining:
1  p  gz  h 2
 
2
x
U0
  0 is the straight line
This flow could exist between:
• The piston of a car and the cylinder,
• A tire and the road,
• A foot and the soil,
• …..
Flow over a inclined plane
h
• We will assume stationary,
incompressible that the plates have
infinite length along the direction
normal to the paper and that we are
far from the entrance and the exit.
In this case the velocity as a single
component, along x1.
• The BC are u=0 at y=0 and τ=0 at y=h
• The continuity equation becomes:
ui
0
xi
u1 u2 u3


0
x1 x2 x3
u1
0
x1
• The NS equation becomes:
 ui
dui
ui 
p




uj


 t
dt
x j 
xi x j

0
  u 
    g x
y  y 
 ui

 x
j


  g i


Integrating along yy
0
 gz    u 
   
x
y  y 
 u 
 gz 
    
y  C1

y

x


1  gz  2
u
y  C1 y  C2
2  x
•
The BC are u=0 at y=0 and τ=0 at y=h
1  gz  2  1 y 2 y 
u
h 
 
2
 x
2
h
h

•
The velocity profile is a parabola
and shear stress is a straight line
C2  0
 gz 
0
h  C1
x
 gz 
C1 
h
x
1  gz  2  gz 
u
y 
hy
2  x
x
Average velocity, Maximum shear
 1  gz  2  1 y 2 y  
Uh   udy    
h 
  dy 
2
 x
h 
2 h
0
0
h
h
1  gz  2  1 y 1 y 
h  2

 x
6
h
2
h

0
1  gz  2
U
h
3 x
Uh  
3
2
h
 u 
 gz 
      
y  C1

y

x


 gz   y 
 gz 
 
h 1  
 max  
h
x
h

x


•
Does it make sense maximum shear to be
independent of the viscosity?