2001, W. E. Haisler
45
Chapter 3: Conservation of Linear Momentum
Plane Couette Flow between two large parallel plates (one
stationary and one moving)
uo (wall velocity)
d
flow
x
z
L
Couette flow is the case with two parallel plates separated by
a distance d. The bottom plate is stationary and the top plate
move horizontally at a velocity uo. There is no pressure
differential along the length of the plate. The driving force
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
46
is the movement of the top plate. Assume the flow is steady
and incompressible.
a) Assume the following boundary conditions:
steady state    0
t
incompressible    constant
no flow in x or y direction  vx  vy  0
 flow in x direction is prevented by plates
 if no pressure gradient in y direction and plates are long
in y direction, then flow in middle (y direction) can be
considered to have only a y component, i.e., ignore
edge effects of plates in y direction).
body force (gravity) is zero  gx  g y  gz  0
2001, W. E. Haisler
47
Chapter 3: Conservation of Linear Momentum
b) Conservation of mass is given by



















 v






y   v z
    v x 
t
x
y
z

































and reduces to
 (vz)  0 or  vz  0

vz  f (x, y) C
1
z
z
 for plane motion, i.e., we consider only a slice
of the fluid in the middle of the plate (ignore
vz  vz ( y)
edge effects), then 
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
48
Thus vz  vz (x) only!
c) Conservation of linear momentum equations are:
 Sxx  S yx  Szx  P
 vx
 vx
 vx
 vx
(
 vx
 vy
 vz
)   gx 



x
y
z x
t
x
y
z
 Sxy  S yy  Szy
 vy
 vy
 vy
 vy
(
 vx
 vy
 vz
) gy 


 P
x
y
z y
t
x
y
z
 Sxz  S yz  Szz  P

vz

vz

vz

vz
(
 vx
 vy
 vz
)   gz 



x
y
z z
t
x
y
z
Noting that vx  vy  0 , vz  vz (x) , ignoring gravity effects so
that gx  g y  gz  0 , and noting that no pressure gradients
exist ( P / x  0 , etc.), then the COLM equations reduce to:
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
 S xx  S yx  S zx


0
x
y
z
 S xy  S yy  S zy


0
x
y
z
 S xz  S yz  S zz


0
x
y
z
49
x component of linear momentum
y component of momentum
z component of momentum
d) Assume the flow produces no normal stresses. Since
flow is in x-z plane, there is no shear in the x-y or y-z plane
(only shear in x-z plane)
Sxx  S yy  Szz  Sxy  S yx  S yz  Szy  0
 vz
Sxz  Szx  
(fluid property assumption)
x
2001, W. E. Haisler
50
Chapter 3: Conservation of Linear Momentum
The last assumption is observed from experiment. The
traction (shear stress) on the wall is proportional to velocity
 vz
gradient normal to wall:
x
S xz
uo
slope    viscosity
coefficient
vz
x
S xz
S zx
x
flow
z
For a 3-D viscous flow field, we can show that the complete
set of constitutive equations relating stresses and velocity
gradients will be given by:
2001, W. E. Haisler
Sxx
Chapter 3: Conservation of Linear Momentum
vx
 2
x
S yy  2 
v y
y
vz
Szz  2 
z
51
 vx v y 
S xy  S yx   



y

x


 v y vz 
S yz  S zy   



z

y


vz vx 

S zx  S xz   



x

z


From the above, it is clear that if vx  v y  0 and vz  vz ( x) ,
then S xx  S yy  S zz  0 (all normal deviatoric stresses are
zero) and S xy  S yx  S yz  S zy  0 (only non-zero shear
stresses are in the x-z plane).
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
52
e) The 3 linear momentum equations reduce to
00
00
 2v

z  0 (assuming  is a constant)
 x2
Note that since vz  vz (x) , the third momentum equation is
actually an ordinary differential equation:
d 2vz

0
dx 2
We assume a “no slip” Boundary Condition for the fluid at
each wall. Since the lower wall is stationary, then
2001, W. E. Haisler
53
Chapter 3: Conservation of Linear Momentum
vz (x  0)  0 . The upper wall moves at a velocity uo so that
vz (x  d )  uo . Integrating the third momentum equation and
assuming  is a constant, we obtain vz  C x C .
1
2
Substituting boundary conditions for velocity yields
at x=0: vz (x  0)  0  C (0)  C
 C 0
1
2
2
at x=d: vz (x  d )  uo  C (d )  C  uo / d
1
1
or
vz (x)  uo  x 
d 
Thus, the velocity profile is a straight line for the plane
Couette flow problem.
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
54
The shear stress is obtained by substituting the velocity into
the constitutive equation:
u 
vz
S xz  
   o 
 d 
x




which is constant from bottom to top. Thus, as uo increases,
the shear stress increases. As d increases, the shear stress
decreases. At the wall, the shear stress acting on the wall
must be equal and opposite to the shear stress acting on the
fluid!
BE CAREFUL! THE RESULTS ABOVE APPLY ONLY
FOR COUETTE FLOW!
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
55
Some real world examples of Couette flow:
a) Wing moving through calm air at speed uo. At some
distance far away from the wing (normal to direction
wing is moving), the air is motionless – think of this
point as a fixed boundary where the fluid velocity is zero.
At the surface of the wing, the fluid velocity is uo if we
assume a no-slip condition – think of this as the moving
boundary. So, looks just like Couette flow.
b) Piston moving up and down in the cylinder of an engine.
Between the piston and cylinder wall is lubrication oil
with a thickness of d. The cylinder wall is the fixed
boundary and the piston wall is the moving boundary.
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
56
Some viscosity coefficient values:
Air at standard sea level conditions: =1.79 x 10-5 kg/(m s)
Water: = 1.005 x 10-3 kg/(m s) at 20C
Motor oil: = 1.07 kg/(m s)
at 20C
Note: 1 centipoise = 10-3 kg/(m s) = 6.72 x 10-4 lbm/(ft s)
See: http://www.lmnoeng.com/fluids.htm
Some exercises
1. A flat-bottomed boat with a wetted surface area of 25 sq.ft.
moves through the water. Assume the boat does not “push”
any water in front of it. How much force in lbs. is required to
propel the boat at 15 mph in water that is 3 ft. deep?
2. A piston with a diameter of 3 in. moves in a cylinder of
diameter 3.01 in. Oil with viscosity of 1,000 centipoise fills
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
57
the gap. How much force is required to move the cylinder if
its velocity is 4 ft/sec?
2001, W. E. Haisler
58
Chapter 3: Conservation of Linear Momentum
Poiseulle Flow between two parallel plates (both stationary)
x
P =higher
L
pressure
flow
z
L
d P =lower
R
pressure
P PR  PL

dz
L
Poiseulle flow is the case of fluid flow between two fixed
parallel plates separated by a distance d and a pressure
gradient in the z direction. The driving force is the pressure
P
differential from left to right ( ). Assume the flow is
z
steady and incompressible.
2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
59
a) Assume the following boundary conditions:
steady state    0
t
incompressible    0
t
no flow in x or y direction  vx  vy  0
 flow in x direction is prevented by plates
 if no pressure gradient in y direction and plates are long
in y direction, then flow in middle (y direction) can be
considered to have only a y component, i.e., ignore
edge effects of plates in y direction).
 body force (gravity) is zero in y and z directions so that
 g y  gz  0
2001, W. E. Haisler
60
Chapter 3: Conservation of Linear Momentum
b) Conservation of mass is given by



















 v






y   v z
    v x 
t
x
y
z

































and reduces to
 (vz)  0 or  vz  0
z
z
vz  f (x, y) C
1
 for plane motion, i.e., we consider only a slice
of the fluid in the middle of the plate (ignore
vz  vz ( y)
edge effects), then 
Thus
vz  vz (x)
only!

2001, W. E. Haisler
Chapter 3: Conservation of Linear Momentum
61
c) Conservation of linear momentum equations are:
 Sxx  S yx  Szx  P
 vx
 vx
 vx
 vx
(
 vx
 vy
 vz
)   gx 



x
y
z x
t
x
y
z
 Sxy  S yy  Szy
 vy
 vy
 vy
 vy
(
 vx
 vy
 vz
) gy 


 P
x
y
z y
t
x
y
z
 Sxz  S yz  Szz  P

vz

vz

vz

vz
(
 vx
 vy
 vz
)   gz 




x

y

z
z
t
x
y
z
Noting that vx  vy  0 , vz  vz (x) , and considering gravity
effects only in the x direction so that g y  g z  0, the COLM
equations reduce to:
62
 2001, W. E. Haisler
 S yx  S

S
 P   g  xx 
zx

x x
x
y
z
 Sxy  S yy  S zy
 P


y x
y
z
 S yz  S

S
 P  xz 
 zz
z x
y
z
x component of linear momentum
y component of momentum
z component of momentum
d) Assume the flow produces no normal stresses. Since
flow is in y-z plane, their is no shear in the x-y or y-z plane
(only shear in x-z plane)
Sxx  S yy  Szz  Sxy  S yx  S yz  Szy  0
63
 2001, W. E. Haisler
 vz
Sxz  Szx  
x
The last assumption is observed from a Couette-type
experiment. The traction (shear stress) on the wall is
 vz
proportional to the velocity gradient normal to wall,
x
(see Couette flow problem)
S xz
uo
slope    viscosity
coefficient
vz
x
S xz
S zx
x
z
flow
64
 2001, W. E. Haisler
For a 3-D viscous flow field, we can show that the complete
set of constitutive equations relating stresses and velocity
gradients will be given by:
Sxx
vx
 2
x
S yy  2 
v y
y
vz
Szz  2 
z
 vx v y 
S xy  S yx   



y

x


 v y vz 
S yz  S zy   



z

y


vz vx 

S zx  S xz   


 x z 
From the above, it is clear that if vx  v y  0 and vz  vz ( x) ,
then S xx  S yy  S zz  0 (all normal deviatoric stresses are
65
 2001, W. E. Haisler
zero) and S xy  S yx  S yz  S zy  0 (only non-zero shear
stresses are in the x-z plane).
e) With all the assumptions and the constitutive equation,
the 3 linear momentum equations reduce to
P
  gx
x
P
0
y
Integrate  P( x)   g x x  f ( y, z )   gh
Integrate  P  f ( x, z )
 2v z
P

z
 x2
We assume a “no slip” Boundary Condition for the fluid at
each wall. Since both walls are stationary, then
 2001, W. E. Haisler
66
vz (x  0)  0 and vz (x  d )  0 .
We further take the pressure gradient dP to be a given value
dz
(Boundary Condition). Integrating the third (z) momentum
equation and assuming  is a constant, we obtain
2
v z  ( 1 dP ) x  C1x  C2 . Substituting boundary conditions
2 dz
for velocity yields:
vz ( x  0)  0  ( 1 dP )(0)2  C1(0)  C2  C2  0
2 dz
d dP
2
dP
1
vz ( x  d )  0  (
)(d )  C1(d )  C2  C1  
2  dz
2  dz
67
 2001, W. E. Haisler
Substituting C1 and C2 into the v z equation above gives the
following result:
vz ( x)
2
v z ( x )  1 dP ( x  dx )
d
x
2 dz
z
Thus, the velocity profile is a quadratic in x for Poiseulle
Flow between two stationary parallel plates.
The shear stress is given by substituting the velocity into the
constitutive equation to obtain:
dvz dP
S xz ( x)  

( x  d / 2)
dx dz
 2001, W. E. Haisler
68
The shear stress is a maximum at either wall, and zero at the
center. Interestingly, the shear stress does NOT depend on
the viscosity coefficient, .
A photograph of velocity profiles of fluid starting from
rest and flowing from left to right is shown below.