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Worked Out Examples
(Internal Fluid Flows)
Example 1. (Poiseuille Flows)
A fluid flows steadily between two parallel plates. The flow is fully developed and laminar.
The distance between the plates is h.
(a) Derive an equation for the shear stress as a function of y. Plot this function.
(b) For  = 2.4  10-5 lbfs/ft2, p/x = -4.0 lbf/ft2/ft, and h = 0.05 in., calculate the maximum
shear stress, in lbf/ft2.
1. Statement of the Problem
a) Given
 Laminar, fully developed flow between two parallel plates
 Gap between the plates is h = 0.05 in. = 4.167  10-3 ft.
  = 2.4  10-5 lbfs/ft2
 p/x = -4.0 lbf/ft2/ft
b) Find
 Expression for the shear stress as a function of y.
 Plot the shear stress.
 Calculate the maximum shear stress, in lbf/ft2.
2. System Diagram
Velocity Profile
yx
y
x
h
xy
xy
yx
3. Assumptions
 Steady state condition
 Constant fluid properties (density and viscosity)
 2-D problem (xy plane)
 Laminar, fully developed flow
4. Governing Equations

2-D Incompressible Continuity Equation:
u v

0
x y
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

 
DV
Navier-Stokes Equation: 
 g  p   2V
Dt
 v u 
Shear Stress (for xy plane):  xy   yx   
 
 x y 
5. Detailed Solution
Shear stress as a function of y
u = u(y) and v  0 because the fluid flows between two parallel plates.
Considering this fact, the continuity equation becomes:
0
u v
u

0
0
x y
x
This tells that the flow is fully developed, which is given in the problem and assumed to be,
because the velocity doesn't change in the x direction. Therefore, the velocity u is function of
only y, u = u(y).
Navier-Stokes equation in y direction is useless on this problem because v  0.
Navier-Stokes equation in x direction:
Continuity equation
gx = 0
  2u  2u 
 u
u
u 
p
   u  v   g x     2  2 
x
y 
x
y 
 t
 x
Steady state
v0
Continuity equation
Therefore,
p
 2u
0    2
x
y
 2 u 1 p
d 2 u 1 p



because u = u(y).
y 2  x
dy 2  x
 du 
1 p
 d  dy     x dy
du 1 p

y  Const
dy  x
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This problem deals with the flow between two parallel plates, so the maximum velocity
occurs at y = 0 (centerline). This implies
du
dy
 0 due to the symmetry. Considering
y 0
this fact, Const on the above expression becomes 0.
du 1 p
du p

y  

y
dy  x
dy x
 v u 
Shear stress is defined as  xy   yx   
  .
 x y 
Now, we have:
From the problem geometry, we are interested in yx. Since v  0, the shear stress becomes:
 yx  
u
du
  yx  
because u = u(y).
y
dy
We already have this expression above; therefore, the shear stress is:
 yx  
du p

y
dy x
Plot the shear stress
Using MatLab, the shear stress distribution looks like:
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2.5
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Shear Stress Distribution
x 10
Upper Wall
2
1.5
1
y (ft)
0.5
yx = (p/x)y
0
-0.5
-1
-1.5
-2
Lower Wall
-2.5
-0.01 -0.008 -0.006 -0.004 -0.002
0
0.002 0.004 0.006 0.008
0.01
yx (lbf/ft )
2
Maximum shear stress
The maximum shear stress occurs at the wall where y =  h/2.
 yx ,max 
p  h 
 
x  2 
 4.167  10 3 ft 

  4.0lbf / ft 2 / ft   
2


2
=  0.008333 lbf/ft


6. Critical Assessment
The shear stress is negative for y > 0 (see the plot). This is correct because the way shear
stress is defined. See the "System Diagram." The upper shear stress is positive toward
right, but on the upper wall, the shear stress works toward left in reality (since the shear
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force must be positive and the upper wall is a negative surface). This is why it is
negative.
[Note: Instead of making the fully developed flow assumption, if only the parallel flow
assumption is made, the analysis will follow the same pattern since parallel flows are
fully developed. Simplifications will be easier since u0, but v=w=0.]
Example 2. (Use of velocity profiles):
An incompressible fluid flows between two infinite stationary parallel plates. The velocity
profile is given by u = umax(Ay2 + By + C), where A, B, and C are constants and y is
measured from the center of the gap. The total gap width is h units. Use appropriate
boundary conditions to express the magnitude and units of the constants in terms of h.
Develop an expression for volume flow rate per unit depth and evaluate the ratio V / u max .
1. Statement of the Problem
a) Given
 Velocity profile: u = umax(Ay2 + By + C)
where A, B, and C are constants and y is measured from the center of the gap.
 Total gap width is h units.
b) Find
 Magnitude and units of the constants in terms of h.
 Volume flow rate per unit depth.
 Ratio V / u max .
2. System Diagram
u(y)
y
x, u
h
umax
3. Assumptions
 Steady state condition
 Incompressible fluid flow
4. Governing Equations
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
Volume Flow Rate Definition: Q  V  dA
A
For a flow between two infinite stationary plates,
y2
Q   u  b  dy where b is the depth in z direction.
y1

Average Velocity Definition
V 
Q
A
5. Detailed Solution
It is obvious to tell the units of constants (A, B, and C) by observing the given velocity profile
function: u = umax(Ay2 + By + C). Note that each term inside the parenthesis has to have no
units. Therefore the units on A, B and C are as follows.
Units
m-2
A
m-1
B
1
C
Note: This will be verified after the constants (A, B, and C) are evaluated in terms of h.
Evaluation of Constants
Realizing the fact that the maximum velocity occurs at y = 0 in this problem condition and
geometry, the constant C can be evaluated as follows:
u = umax(Ay2 + By + C)
umax = umax[A(0)2 + B(0) + C]
C=1
Because the maximum velocity occurs at y = 0, this relation must be satisfied:
du
dy
du
dy
 0 since shear stress is zero on the centerline (or, line of symmetry).
y 0

y 0


d
u max  Ay 2  By  C
dy

y 0
 u max  2 Ay  B  y 0  B  u max  0
 u max  0, B  0
Velocity must be 0 (non-slip condition) at the wall where y = h/2. Thus,
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0  u max
2/6/2016
  h 2

h
  A   B   C 
2
  2 

1 2 1
Ah  Bh  C
4
2
0
Substituting B = 0 and C = 1 into this expression, 0 
1
4
Ah 2  1  A   2 .
4
h
Finally, the velocity profile becomes:
 4

u  u max  2 y 2  1
 h

or
u
u max
 y
 1  4 
h
2
Volume Flow Rate per Unit Depth
h
Q   2h u  b  dy

2
 4

  2h u max   2 y 2  1  b  dy

 h

2
h
h
 u max
 4 1
2
 b   2  y 3  y 
 h 3
 h
2
2
 u max  b  h
3

Q 2
 hu max
b 3
Ratio V / u max
V 

Q Q Q1 2
1 2


 hu max  u max
A bh b h 3
h 3
V
2

u max 3
6. Critical Assessment
The problem illustrates how to determine velocity profiles from boundary conditions
and calculate volumetric flow rate and average velocity. An important point on this
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problem is to realize where the maximum velocity occurs and how the mathematical
expression can be written for it.
Example 3. (Couette Flows)
A sealed journal bearing is formed from concentric cylinders. The inner and outer radii are
25 and 26 mm, the journal length is 100 mm, and it turns at 2800 rpm. The gap is filled with
oil in laminar motion. The velocity profile is linear across the gap. The torque needed to
turn the journal is 0.2 Nm. Calculate the viscosity of the oil. Will the torque increase or
decrease with time? Why?
1. Statement of the Problem
a) Given
 r1 = 25 mm, r2 = 26 mm, a = r2 - r1 = 1 mm
 L = 100 mm
  = 2800 rpm
 Oil in laminar motion
 Linear velocity profile
 T = 0.2 Nm
b) Find
 Viscosity of the oil
 Will the torque increase or decrease with time? Why?
2. System Diagram
a
U
r1
r2
a
u
y

x
3. Assumptions
 Steady state condition
 Incompressible fluid flow
 Laminar flow
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




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Fully developed flow
p/x = 0 (since the flow is symmetric in the actual bearing at no load)
Torque on the journal is due to viscous shear in the oil film. The gap width is small, so
the flow may be modeled as flow between infinite parallel plates.
Infinite width (L/a = 100 mm / 1 mm = 100, so this is a reasonable assumption)
2-D problem (L is in z direction, but the main analysis can be done in xy plane.)
4. Governing Equations



u v

0
x y


 
DV
Navier-Stokes Equation: 
 g  p   2V
Dt
 v u 
 
Shear Stress (for xy plane):  xy   yx   

x
y 

2-D Incompressible Continuity Equation:
5. Detailed Solution
Viscosity of the Oil
The problem gives the velocity profile to be linear, so we can simply say that the velocity
profile is (See "Critical Assessment" for a detailed discussion of this velocity profile):
u  u( y) 
 v
U
y
a
u 
The shear stress is:  xy   yx   
 
 x y 
On this problem, we are interested in yx and v  0   yx  
 yx  
u
 U
 
y
y  a
r
U
 1 …
a
a
Torque is given by T  F  R   yx  A  R   yx  2    r1  L   r1 … 
2    r13      L
Equating  and  gives: T 
…
a
Therefore,

y .

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

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Ta
2    r13    L
2/6/2016
0.2 N  m 1  10 3 m 
3 
rev  1 min  2rad 
3
2    25  10 3 m  2800


 100  10 m
min  60 s  1rev 





= 0.0695 Ns/m2
Will the torque increase or decrease with time? Why?
Torque will decrease with time.
Reason: This bearing is going to be heated up by the shear stress effect (temperature
increases). Viscosity decreases as the temperature increases (check a plot or graph of
viscosity as a function of temperature). If the viscosity decreases, then the torque decreases
by the equation  above.
6. Critical Assessment
The problem gives the velocity profile as linear, so we simply said that the velocity
profile is
u  u( y) 
U
y
a
But here is the detailed demonstration of how this linear velocity profile was obtained.
Linear Velocity Profile
u = u(y) and v  0 because the fluid flows between two parallel plates.
Considering this fact, the continuity equation becomes:
0
u v
u

0
0
x y
x
This shows that the flow is fully developed, which is assumed to be, because the velocity
doesn't change in the x direction. Therefore, the velocity u is function of y only, u = u(y).
Navier-Stokes equation in y direction is useless on this problem because v  0.
Navier-Stokes equation in x direction:
Continuity equation
gx = 0
  2u  2u 
 u
u
u 
p
 u  v   g x     2  2 
x
y 
x
y 
 t
 x
 
Steady state
v0
Continuity equation
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Therefore,
0
p
 2u
 2
x
y
 2 u 1 p
d 2 u 1 p

because u = u(y).


y 2  x
dy 2  x
Integrating this equation twice gives: u 
Boundary Conditions
u = 0 at y = 0  C2 = 0
u = U at y = a  C1 
1 p 2
y  C1 y  C 2
2 x
U
1 p

a
a 2 x
Finally,
2
U
a 2  p   y   y 
u  u( y)  y 
      
a
2  x   a   a 
u  u( y) 
U
p
y because
 0 (See assumptions listed above).
a
x
(This result matches with the given velocity profile, linear.)
Example 4. (Special boundary conditions in channel flow):
A continuous belt, passing upward through a chemical bath at speed U0, picks up a liquid
Udensity
film of thickness h,
, and viscosity . Gravity tends to make the liquid
0
drain down, but the
movement of the belt keeps the liquid from
Assume that the flow is fully
running off completely.
g
developed and laminar
with zero pressure gradient, and
h
that the atmosphere
produces no shear stress at the
outer surface of the film.
State clearly the boundary conditions to be
satisfied by the velocity
at y = 0 and y = h. Obtain an expression for
dx
the velocity profile.
p = patm
x
dy
y
Bath
Belt
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1. Statement of the Problem
a) Given
 Belt speed, U0
 Liquid properties: h, , and 
 Gravity exists
 Flow is fully developed, laminar, with zero pressure gradient (p/x = 0)
 Atmosphere produces no shear stress at the outer surface of the film
b) Find
 State the boundary conditions to be satisfied by the velocity at y = 0 and y = h.
 Obtain an expression for the velocity profile.
2. System Diagram
U0

g
xy
h
p = patm
yx
yx
x
y
xy
, 
3. Assumptions
 Steady state condition
 Constant property fluid
 v0
 2-D problem (xy plane). That is, the z-depth is infinite  u
0
z
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4. Governing Equations
 Navier-Stokes Equation

2/6/2016


 
DV

 g  p   2V
Dt
Shear Stress (for xy plane)
 v
u 
 xy   yx     
 x y 
5. Detailed Solution
The problem gives the flow is fully developed. Thus,
u
 0 . Also, . u
 u(y) only.
0
x
z
(This relationship could be derived from the continuity equation.)
Navier-Stokes equation (2-D) in y direction is useless because v  0 and gy = 0.
Navier-Stokes equation (2-D) in x direction:
v0
Steady state
Fully developed
  2u  2u 
 u
u
u 
p
 u  v   g x     2  2 
x
y 
x
y 
 t
 x
 
Fully developed
Given
Therefore,
0  g  
x
d u since u(y) only.
dy
2
2
Shear Stress
We are interested in yx and v  0 in this problem. Thus,
 v
u 
u
 xy   yx        yx  
y
 x y 
Boundary Conditions
@ y = 0, u = U0
@ y = h,  yx  
u
u
0 
 0 because the atmosphere produces no shear stress at
y
y
the outer surface of the film.
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Going back to the equation obtained from Navier-Stokes equation:
0  g  
x
d u
d u
0   g  
dy
dy
2
2
2
2
because gx = -g (check the directions of the gravity and the x coordinate).
Thus,
d u g

dy

2
2
Integrating this equation once gives du g
.

yC
dy 
1
Using the second boundary condition, C1 can be evaluated as C1  
g
h.

Now, the equation is du g
g .

y
h
dy 

Integrating this equation gives u 
g 2 gh
y 
y  C2 .
2

Using the first boundary condition, C2 can be evaluated as C2 = U0.
Therefore, the velocity profile is
u( y) 
1 g 2 gh
y 
y U0
2 

or
g 2  1  y   y 
h        U 0

 2  h   h 
2
u( y) 
6. Critical Assessment
Here, the velocity profile has been obtained by a direct application of Navier-Stokes
equation. The same result can be obtained by applying the momentum equation for
integral control volume:



  

FS  FB   VdV   VV  dA
CS
t CV
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Try to obtain the same result using this approach.
Also try simplifying the problem by assuming parallel flow instead of fully-developed
flow.
Example 5. (Electrical Analogy):
Resistance to fluid flow can be defined by analogy to Ohm's law for electric current. Thus
resistance to flow is given by the ratio of pressure drop (driving force) to volume flow rate
(current). Show that resistance to laminar flow is given by
Resistance =
128L
D 4
which is independent of flow rate. Find the maximum pressure drop for which this relation
is valid for a tube 50 mm long with 0.25 mm inside diameter for both kerosine and caster oil
at 40 C.
1. Statement of the Problem
a) Given
 Resistance analogy between Ohm's law and fluid flow
Ohm's Law
Fluid Flow
Driving Force
V (voltage)
p (pressure drop)
Current
I (current)
Q (volume flow rate)



Resistance
R (resistance)
Rf (flow resistance)
L = 50 mm = 0.05 m
D = 0.25 mm = 2.5  10-4 m
Kerosine (at 40 C)
  = 1.1  10-3 Ns/m2
  = 1.35  10-6 m2/s
 Caster Oil (at 40 C)
 S.G. = 969
  = 2.4  10-1 Ns/m2
  = 2.48  10-4 m2/s
b) Find
 Expression of flow resistance for laminar flow using the analogy to Ohm's law
 Maximum pressure drop in the pipe for both kerosine and castor oil as working fluids
2. System Diagram
Ohm's Law
Fluid Flow
I
Q
R
V
p
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3. Assumptions
 Steady state condition
 Constant fluid properties
 Laminar, fully developed flow
4. Governing Equations

Velocity Profile for a Pipe Flow
u

1  p  2
2
 r  R 
4  x 
Volume Flow Rate Definition
 
Q   V  dA
A
5. Detailed Solution
Flow Resistance
The problem gives the resistance to flow is R f 
Volume flow rate Q can be calculated as follows:
p
.
Q
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 
Q   V  dA
A
R
  u  2r  dr
0
1 p 2
r  R 2  2r  dr
0 4  x
 p R 3

r  R 2 r dr
2  x 0


R



R
 p  1 4 R 2 2 

r 
 r 
2  x  4
2
0

 p  R 4 


2  x  4 
 p  D 

 
8 x  2 
4
D 4 p
Q  
128 x
Because
p
p p
p
p p  p
 lim



x
x
L
L
L
2
1
1
2
drop
, the volume flow rate Q can
x 0
be written as:
D 4  p  D 4 p
Q


128  L  128L
Finally, the flow resistance is
Rf 
p
p
128L


4
Q D p
D 4
128L
This expression, flow resistance, is independent of the flow rate because there is no Q or V
included. It is consisted of only fluid property and the geometry of a pipe.
Maximum Pressure Drop
Rf 
p
 128L 
 128L    2  32L
 V  A  
 V  D 
V
 p  R f  Q  
4 
4  
4
Q
D2
 D 
 D  

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Examining this pressure drop expression, one can say that the faster the average velocity is,
the bigger the pressure drop is; therefore, to obtain the maximum pressure drop, we want the
fastest flow velocity in laminar flow constraint.
Consider the Reynolds number Re 
VD

 Re cr . This gives the allowable fastest flow
velocity in laminar flow constraint. Recr = 2300 for a pipe flow.
Re 
VD

 Re cr  V 
Re cr 
D
Substituting this fastest average flow velocity into the pressure drop expression, we get:
p 
32L Re cr  32L Re cr


D
D2
D3
After plugging in values into this pressure drop expression, we obtain:
p for kerosine is 350 kPa.
p for castor oil is 14.0 GPa.
6. Critical Assessment
Notice the big difference of pressure drops between kerosine and castor oil. This
difference is directly from the viscosity and density of fluids, keeping other parameters
constant.
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