CP502
Advanced Fluid Mechanics
Flow of Viscous Fluids
Set 02
Continuity and Navier-Stokes equations
for incompressible flow of Newtonian fluid
υ
R. Shanthini
15 March 2012
ρ
Steady, incompressible flow of Newtonian fluid in an
infinite channel with stationery plates
- fully developed plane Poiseuille flow
Fixed plate
y
z
x
2a
Fluid flow direction
Fixed plate
Step 1: Choose the equation to describe the flow
Navier-Stokes equation will be used since the system
considered is an incompressible flow (Assumption 1) of
a Newtonian fluid (Assumption 2)
Step 2: Choose the coordinate system
Cartesian coordinate system is chosen
R. Shanthini
15 March 2012
Continuity and Navier-Stokes equations
for incompressible flow of Newtonian fluid
in Cartesian coordinates
Continuity:
Navier-Stokes:
x - component:
y - component:
z - component:
R. Shanthini
15 March 2012
Steady, incompressible flow of Newtonian fluid in an
infinite channel with stationery plates
- fully developed plane Poiseuille flow
Fixed plate
y
z
x
2a
Fluid flow direction
Fixed plate
Step 3: Decide upon the functional dependence of the
velocity components
R. Shanthini
15 March 2012
x direction:
y direction:
u = function of (t, x, y, z)
v = function of (t, x, y, z)
z direction:
w = function of (t, x, y, z)
}
(1)
Assumption (3): steady flow and therefore no change in time
Assumption (4): infinite channel and therefore nothing
happens in z-direction
Assumption (5): fully developed flow and therefore no change
in the flow direction (that is, x-direction)
Using the above 3 assumptions, we reduce (1) to the following:
R. Shanthini
15 March 2012
x direction:
y direction:
u = function of (y)
v = function of (y)
z direction:
w=0
}
(2)
Step 4: Use the continuity equation in Cartesian coordinates
v
0
y
u v w


0
x y z
v  constant
v0
or
Flow geometry shows that vv can not
be a constant, and therefore we choose
Fixed plate
v
y
v0
R. Shanthini
15 March 2012
z
x
2a
Fluid flow direction
Fixed plate
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
15 March 2012
}
N-S equation therefore reduces to
p
 2u

  2  g x  0
x
y
p

 g y  0
y
x - component:
y - component:

z - component:
p
 g z  0
z
Ignoring gravitational effects, we get
p
 2u

 2 0
x
y
x - component:
y - component:
z - component:
R. Shanthini
15 March 2012
(4)
p
0
y
p is not a function of y
p

0
z
p is not a function of z

p is a function
of x only

Rewriting (4), we get
 2 u 1 p

2
y
 x
(5)
p is a function of x only and  is a constant and
therefore RHS is a function of x only
u is a function of y only and therefore LHS is a
function of y only
Therefore (5) gives, function of (y) = function of (x) = constant
p
It means p  
= constant
x
That is, pressure gradient in the x-direction is a constant.
R. Shanthini
15 March 2012
LHS = left hand side of the equation
RHS = right hand side of the equation
Rewriting (5), we get
 2 u 1 p  p


2
y
 x

(6)
p
where p  
is the constant pressure gradient in the x-direction
x
Since u is only a function of y, the partial derivative becomes an
ordinary derivative.
Therefore, (5) becomes
d 2u
p

2
dy

R. Shanthini
15 March 2012
(7)
Integrating (6), we get
p y 2
u
 C1 y  C2
 2
(8)
where C1 and C2 are constants to be
determined using the boundary
conditions given below:
u0
u0
at
at
ya
y  a
}
no-slip
boundary
condition
ya
Fixed plate
y
x
2a
Fluid flow direction
y  a
Fixed plate
Substituting the boundary conditions in (8), we get
C1  0
and
p a 2
C2 
 2
Therefore, (8) reduces to
R. Shanthini
15 March 2012
p 2

u
a  y 2  Parabolic
velocity profile
2
Steady, incompressible flow of Newtonian fluid in an
infinite channel with one plate moving at uniform velocity
- fully developed plane Couette-Poiseuille flow
p y 2
u
 C1 y  C2
 2
y
where C1 and C2 are constants to be
determined using the boundary
conditions given below:
u U
u0
at
at
ya
y  a
ya
(8)
}
no-slip
boundary
condition
Moving plate
at velocity U
x
2a
Fluid flow direction
y  a
Fixed plate
Substituting the boundary conditions in (8), we get
U
C1 
2a
and
U p a 2
C2  
2
 2
Therefore, (8) reduces to
R. Shanthini
15 March 2012
Parabolic velocity profile
p 2
U
2

a  y  is super imposed on a
u
a  y 
2
2a
linear velocity profile
Steady, incompressible flow of Newtonian fluid in an
infinite channel with one plate moving at uniform velocity
- fully developed plane Couette-Poiseuille flow
Starting from plane Coutte-Poiseuille flow
p 2
U
2

a  y 
u
a  y 
2
2a
ya
Moving plate
at velocity U
y
x
If zero pressure gradient is maintained
in the flow direction, then ∆p = 0.
2a
Fluid flow direction
y  a
Fixed plate
Therefore, we get
u
R. Shanthini
15 March 2012
U
a  y  Linear velocity profile
2a
Point to remember:
- Pressure gradient gives parabolic profile
- Moving wall gives linear profile
Let us get back to the velocity profile of
fully developed plane Couette-Poiseuille flow
u
p 2
U

a  y 
a  y2 
2
2a
Let us non-dimensionalize it as follows:
Rearranging gives
Dividing by U gives
a 2 p 
y2  U 
y
1  2   1  
u
2  a  2  a 
u a 2 p 
y2  1 
y
1  2   1  

U 2 U 
a  2
a
u
y
Introducing non-dimensional variables u 
and y 
a
U
a 2 p
1
2

u
1  y   1  y 
2U
2
R. Shanthini
15 March 2012
, we get
Plot the non-dimensionalized velocity profile of the
fully developed plane Couette-Poiseuille flow
a 2 p
1
2

u
1  y   1  y 
2U
2
p
2 U / a 2
1
y
y
a
1
0.25
0
-0.5
R. Shanthini
15 March 2012
0
-1
0.5
1
1.5
u
u
U
0
-0.25
-1
Plot the non-dimensionalized velocity profile of the
fully developed plane Couette-Poiseuille flow
a 2 p
1
2

u
1  y   1  y 
2U
2
1
y
y
a
p
2 U / a 2
BACK FLOW
0
0
-0.25
0.75
-0.5
R. Shanthini
15 March 2012
u
u
U
-1
Plot the non-dimensionalized velocity profile of the
fully developed plane Couette-Poiseuille flow
a 2 p
1
2

u
1  y   1  y 
2U
2
p
2 U / a 2
1
y
y
a
0
-0.25
0
-1
R. Shanthini
15 March 2012
0
-1
1
u
u
U
-0.5
-1.5
Determine the volumetric flow rate of the
fully developed plane Couette-Poiseuille flow
Let us determine the volumetric flow rate
through one unit width fluid film along zdirection, using the velocity profile
ya
y
Moving plate
at velocity U
x
p 2
U
2

a  y 
u
a  y 
2
2a
2a
Fluid flow direction
y  a
Fixed plate
a
 p  2
 p 2

U
y3  U 
y 2 
2
a  y dy    a y     ay  
Q   udy    a  y  
2
2a
3  2a 
2  a

a
a 
 2 
a
a
 p  3 a 3  U  2 a 2   p  3 a 3  U  2 a 2 
 a       a   
  a  
Q    a   
3  2a 
2   2  
3  2a 
2 
 2 
R. Shanthini
15 March 2012
2a 3 p
Q
 Ua
3 
Plot the non-dimensionalized volumetric flow rate of the
fully developed plane Couette-Poiseuille flow
Non-dimensionalizing Q, we get
Q
2 a 2 p
 1
Ua
3 U
4
Q
Ua
Why the flow rate becomes zero?
3
2
1
0
-4
-2
0
-1
R. Shanthini
15 March 2012
2
a 2 p
U
4
Steady, incompressible flow of Newtonian fluid in a pipe
- fully developed pipe Poisuille flow
φ
Fixed pipe
r
z
Fluid flow direction
2a
Workout on your own
R. Shanthini
15 March 2012
2a
Steady, incompressible flow of Newtonian fluid between
a stationary outer cylinder and a rotating inner cylinder
- fully developed pipe Couette flow
Step 1: Continuity and Navier-Stokes equations are used
Step 2: Cylindrical polar coordinate is chosen
Step 3: functional dependence of the velocity components are
determined
r direction: ur = function of (t, r,  , z)
 direction: u = function of (t, r,  , z)
z direction: uz = function of (t, r,  , z)
a
b
R. Shanthini
15 March 2012
r
z
aΩ
Assumptions:
- Steady flow
no time dependence
- fully developed
no change in the axial (z)-direction
- axi-symmetric flow
no change in the tangential ( )-direction
Flow condition: no flow in the axial (z-) direction is assumed.
r direction: ur = function of (r)
 direction: u = function of (r)
z direction: uz = 0
R. Shanthini
15 March 2012
a
b
r
z
aΩ
Step 4: Use the continuity equation
ur = constant/r or ur = 0
Flow geometry shows that velocity component in the r-direction can not
be a constant divided by r in order to satisfy the “no flow through the
walls” boundary conditions, and therefore we choose ur = 0
r direction: ur = 0
 direction: u = function of (r)
z direction: uz = 0
R. Shanthini
15 March 2012
a
b
r
z
aΩ
r direction: ur = 0
 direction: u = function of (r)
z direction: uz = 0
Step 5: Using the N-S equation, ignoring the gravity, we get
Radial component:
Tangential component:
Axial component:
R. Shanthini
15 March 2012
Radial component of the N-S equation reduces to
p


r
r
u2
which gives
u2
p

r
r
(9)
Can you identify the centrifugal force???
Axial component of the N-S equation reduces to
p
0
z
R. Shanthini
15 March 2012
p is not a function of z
Tangential component of the N-S equation reduces to
 1   u
1 p
 r
0
 
r 
 r r  r
 u 
  2 
 r 
(10)
which gives
 1 d  du
p
 r
 r 

 r dr  dr
Starting from
which gives
 u 
  2  = function of r only = f(r)
 r 
(say)
p
 f (r ) , we get p   f (r )

p  0 at   0 and p  2  f (r) at   2
Pressure takes different values at   0 and at   2 , which is the
same point in the tangential direction. Since pressure is continuous,
it is impossible.
p
0
Therefore,
we conclude f(r) = 0, which gives
R. Shanthini
15 March 2012

(11)
Combining (10) and (11), we get
1 d  du
 r
r dr  dr
 u
  2
 r
(12)
Radial component was
u2
p

r
r
(9)
There are two unknowns and two equations.
Solve (12) to get the tangential velocity component u
Then solve (9) to get the pressure p
R. Shanthini
15 March 2012
C2
Integrating (12) gives u  C1r 
r
(13)
where C1 and C2 are constants to be determined using the
no-slip boundary conditions given below:
u  a at r  a
u  0 at r  b
Substituting the boundary conditions in (13) gives
a 2
C1   2
b  a2
a 2b2
and C2  2
b  a2
Therefore (13) could be written as
a 2
u  2
2
b

a
R. Shanthini
15 March 2012
 b2

  r 
 r

(14)
a
b
r
z
aΩ
a 2
u  2
b  a2
 b2

  r 
 r

(14)
Let us introduce the following non-dimensional variables:
u
r
u 
and r 
a
a
Using the non-dimensional, (14) is reduced to the following:
 (b / a ) 2

1

u 
 r 
2

(b / a )  1  r

R. Shanthini
15 March 2012
(15)
This simple circular flow exists only at low rotational speed
of the inner cylinder.
As the rotational speed increases, this simple steady flow is
replaced by another steady flow in which the space between
cylinders is filled with (Taylor) vortices.