CP502
Advanced Fluid Mechanics
Flow of Viscous Fluids
and Boundary Layer Flow
Lectures 5 and 6
Continuity and Navier-Stokes equations
for incompressible flow of Newtonian fluid
υ
R. Shanthini
06 Oct 2010
ρ
Steady, fully developed, laminar, incompressible flow of a
Newtonian fluid down an inclined plane under gravity
Exercise 1:
Show that, for steady, fully developed laminar
flow down the slope (shown in the figure), the
Navier-Stokes equations reduces to
z
d 2u
g

sin 
2
dy

θ
where u is the velocity in the x-direction, ρ is the density, μ is
the dynamic viscosity, g is acceleration due to gravity, and θ is
the angle of the plane to the horizontal.
Solve the above equation to obtain the velocity profile u and obtain the
expression for the volumetric flow rate for a flowing film of thickness h.
Exercise 2:
If there is another solid boundary instead of the free-surface at y = h and the
flow occurs with no pressure gradient, what will be the volumetric flow rate?
R. Shanthini
06 Oct 2010
Step 1: Choose the equation to describe the flow
Navier-Stokes equation is already chosen since the system considered
is incompressible flow of a Newtonian fluid.
Step 2: Choose the coordinate system
Cartesian coordinate system is already chosen.
z
Step 3: Decide upon the functional dependence of
the velocity components
x  direction:
y  direction:
z  direction:
u  function(t, x, y, z)
v  function(t, x, y, z)
w  function(t, x, y, z)
}
θ
(1)
Steady, fully developed flow and therefore no change in time and in the flow
direction. Channel is not bounded in the z-direction and therefore nothing
happens in the z-direction.
R. Shanthini
06 Oct 2010
x  direction:
u  function( y )
y  direction:
z  direction:
v  function( y )
w0
}
(2)
Step 4: Use the continuity equation in Cartesian coordinates
v
0
y
u v w


0
x y z
v  constant
or
Flow geometry shows that vv can not
be a constant, and therefore we choose
v0
v
z
v0
R. Shanthini
06 Oct 2010
θ
The functional dependence of the velocity components
therefore reduces to
x direction:
u = function of (y)
(3)
y direction:
v=0
z direction:
w=0
Step 5: Using the N-S equation, we get
x - component:
y - component:
z - component:
R. Shanthini
06 Oct 2010
}
N-S equation therefore reduces to
x - component:
y - component:
z - component:
p
 2u

  2  g x  0
x
y
p

 g y  0
y
p

 g z  0
z
z
θ
No applied pressure gradient to drive the flow. Flow is
driven by gravity alone. Therefore, we get
x - component:
y - component:
z - component:
R. Shanthini
06 Oct 2010
 2u

  gx
2
y

d 2u
g

sin 
2
dy

(4)
p
 g y   g cos
y
p

0
z
p is not a function of z
What was
asked to be
derived in
Exercise 1
Steady, fully developed, laminar, incompressible flow of a
Newtonian fluid down an inclined plane under gravity
Exercise 1:
Show that, for steady, fully developed laminar
flow down the slope (shown in the figure), the
Navier-Stokes equation reduces to
d 2u
g

sin 
2
dy

√done
z
θ
where u is the velocity in the x-direction, ρ is the density, μ is
the dynamic viscosity, g is acceleration due to gravity, and θ is
the angle of the plane to the horizontal.
Solve the above equation to obtain the velocity profile u and obtain the
expression for the volumetric flow rate for a flowing film of thickness h.
Exercise 2:
If there is another solid boundary instead of the free-surface at y = h and the
flow occurs with no pressure gradient, what will be the volumetric flow rate?
R. Shanthini
06 Oct 2010
d 2u
g


sin 
2
dy

(4)
z
Equation (4) is a second order equation in u with
respect to y. Therefore, we require two boundary
conditions (BC) of u with respect to y.
BC 1:
At y = 0, u = 0
BC 2:
At y = h, du  0
h
(no-slip boundary condition)
θ
(free-surface boundary condition)
dy
Integrating equation (4), we get
Applying BC 2, we get

du  g
  
sin   y  A
dy  

 g

A   sin  h
 

Combining equations (5) and (6), we get
R. Shanthini
06 Oct 2010
(5)
(6)

du  g
  sin  h  y 
dy  

(7)
z
h
θ
 g

y2 
sin   hy    B
Integrating equation (7), we get u  

2 


Applying BC 1, we get
B=0
Combining equations (8) and (9), we get
R. Shanthini
06 Oct 2010
(8)
(9)
 g

y2 
u   sin   hy  
2 
 

(10)
 g

y2 
u   sin   hy  
2 
 

(10)
z
Volumetric flow rate through one unit width
fluid film along the z-direction is given by
h
h
Q   u dy
θ
0
 g

y2 
Q    sin   hy   dy

2 

0
h
h
 g
 y
  h 3 h 3  gh3
y   g
Q  
sin   h
   
sin      
sin 
6 0  
3
 
 2
 2 6 
2
R. Shanthini
06 Oct 2010
3
(11)
Steady, fully developed, laminar, incompressible flow of a
Newtonian fluid down an inclined plane under gravity
Exercise 1:
Show that, for steady, fully developed laminar
flow down the slope (shown in the figure), the
Navier-Stokes equation reduces to
d 2u
g

sin 
2
dy

√done
z
θ
where u is the velocity in the x-direction, ρ is the density, μ is
the dynamic viscosity, g is acceleration due to gravity, and θ is
the angle of the plane to the horizontal.
Solve the above equation to obtain the velocity profile u and obtain the
expression for the volumetric flow rate for a flowing film of thickness h.
Exercise 2:
√done
If there is another solid boundary instead of the free-surface at y = h and the
flow occurs with no pressure gradient, what will be the volumetric flow rate?
R. Shanthini
06 Oct 2010
d 2u
g


sin 
2
dy

(4)
Equation does not change.
BCs change.
z
BC 1:
At y = 0, u = 0
(no-slip boundary condition)
h
BC 2:
At y = h, du  0
(free-surface boundary condition)
dy
u=0
θ
(no-slip boundary condition)
Integrating equation (4), we get

du  g
  
sin   y  A
dy  

 g
 y2
Integrating equation (12), we get u   
sin  
 Ay  B
 
 2
Applying the BCs in (13), we get B = 0 and
R. Shanthini
06 Oct 2010
 g
h
A   sin  
 
2
(12)
(13)
Therefore, equation (13) becomes
 g
 hy y 2 
u   sin    
2
 
 2
(14)
Volumetric flow rate through one unit width
fluid film along the z-direction is given by
z
h
h
θ
Q   u dy
0
 g
 hy y 2 
Q    sin     dy

2 
 2
0
h
h
 g
  hy
  h 3 h 3  gh3
y   g
Q  
sin   
   
sin      
sin 
6 0  
 
 4
  4 6  12
2
R. Shanthini
06 Oct 2010
3
(15)
Summary of Exercises 1 and 2
Gravity flow through two planes
Free surface gravity flow
z
z
d
d
θ
 g

y2 
u   sin   hy  
2 
 

gh3
Q
sin 
3
θ
(10)
(11)
 g
 hy y 2 
u   sin    
2
 
 2
gh3
Q
sin 
12
(14)
(15)
Why the volumetric flow rate of the free surface gravity flow is
R. Shanthini
4 2010
times larger than the gravity flow through two planes?
06 Oct
Any clarification?
R. Shanthini
06 Oct 2010
Steady, fully developed, laminar, incompressible flow of a
Newtonian fluid down a vertical plane under gravity
Exercise 3:
A viscous film of liquid draining down the side of a wide
vertical wall is shown in the figure. At some distance down
the wall, the film approaches steady conditions with fully
developed flow. The thickness of the film is h. Assuming that
the atmosphere offers no shear resistance to the motion of
the film, obtain an expression for the velocity distribution
across the film and show that
 3Q 
h  

 g 
(1 / 3 )
where ν is the kinematic viscosity of the liquid, Q is the
volumetric flow rate per unit width of the plate and g is
acceleration due to gravity.
R. Shanthini
06 Oct 2010
y
h
Workout Exercise 3 in 5 minutes!
R. Shanthini
06 Oct 2010
d 2 u V du

dy2  dy
Steady, fully developed, laminar, incompressible flow of a
Newtonian fluid over a porous plate sucking the fluid
Exercise 4:
An incompressible, viscous fluid (of kinematic viscosity ν) flows between
two straight walls at a distance h apart. One wall is moving at a constant
velocity U in x-direction while the other is at rest as shown in the figure.
The flow is caused by the movement of the wall. The walls are porous and
a steady uniform flow is imposed across the walls to create a constant
velocity V through the walls. Assuming fully developed flow, show that the
velocity profile is given by
u
1  exp(Vy /  )
U
1  exp(Vh /  )
Also, show that
(i) u approaches Uy/h for small V, and
(ii) u approaches U exp V h  y  /   for very
large Vh/ν.
R. Shanthini
06 Oct 2010
V
U
z
h
U
Step 1: Choose the equation to describe the flow done
Step 2: Choose the coordinate system done
Step 3: Decide upon the functional dependence of the velocity components
Steady, fully developed flow and therefore no change in time and in the flow
direction. Channel is not bounded in the z-direction and therefore nothing
happens in the z-direction.
x  direction:
u  function( y )
y  direction:
z  direction:
v  function( y )
w0
}
(1)
Step 4: Use the continuity equation in Cartesian coordinates
u v w


0
x y z
v
0
y
V
v  constant or v  0
R. Shanthini
06 Oct 2010
v V
U
z
h
U
The functional dependence of the velocity components
therefore reduces to
x direction:
u = function of (y)
(2)
y direction:
v=V
z direction:
w=0
Step 5: Using the N-S equation, we get
x - component:
y - component:
z - component:
R. Shanthini
06 Oct 2010
}
N-S equation therefore reduces to
x - component:
y - component:
z - component:
u
p
 2u
V

 2
y
x
y
p
 g
y
p
0
z
No applied pressure gradient to drive the flow. Flow is caused by the
movement of the wall. Therefore, we get
x - component:
d 2 u V du

2
dy
 dy
(3)
V
U
R. Shanthini
06 Oct 2010
z
h
U
d 2 u V du
du
V



where


dy2  dy
dy

(3)
Equation (3) is a second order equation in u with respect to y. Therefore, we
require two boundary conditions (BC) of u with respect to y.
BC 1:
At y = 0, u = 0
(no-slip boundary condition)
BC 2:
At y = h, u = U
(no-slip boundary condition)
Integrating equation (3), we get
du
 exp(y  A)
dy
Integrating equation (4), we get
u
1

(4)
exp( y  A)  B
(5)
Applying the BCs in equation (5), we get
0
R. Shanthini
06 Oct 2010
U
1

1

exp( A)  B
exp( h  A)  B
(6)
(7)
From equations (6) and (7), we get
exp(A) 
U
exp(h)  1
B
1

exp(A)  
U
exp(h)  1
Substituting the above in equation (5), we get
u
U
U
1  exp(y )
exp(y ) 

U
exp(h)  1
exp(h)  1 1  exp(h)
u
1  exp(Vy /  )
U
1  exp(Vh /  )
(8)
V
U
R. Shanthini
06 Oct 2010
z
h
U
u
1  exp(Vy /  )
U
1  exp(Vh /  )
(8)
(i) For small V, expand exp(Vy/ν) and exp(Vh/ν) using Taylor series as follows:


(Vy /  ) 2 (Vy /  ) 2
1  1  (Vy /  ) 

 
2!
3!

U
u


(Vh /  ) 2 (Vh /  ) 2
1  1  (Vh /  ) 

 
2!
3!


For small V, we can ignore the terms with power. We then get
u
Vy / 
y
U U
Vh / 
h
V
U
Could you recognize the above profile?
R. Shanthini
06 Oct 2010
z
h
U
u
1  exp(Vy /  )
U
1  exp(Vh /  )
(8)
For very large Vh/ν, exp(Vh/ν) goes to infinity. Therefore. Divide equation (8) by
exp(Vh/ν). We then get
u
exp(Vh /  )  exp(Vy /  ) exp(Vh /  )
exp(Vh /  )  1
For very large Vh/ν, exp(-Vh/ν) goes to zero. Therefore, we get
 exp(Vy /  ) exp(Vh /  )
u
U
( 1)
V
u  U exp V h  y  /  
R. Shanthini
06 Oct 2010
U
z
h
U