Ghosh - 550 Page 1 4/15/2020
Velocity profiles in laminar boundary layers often are approximated by the equations u
Linear:
U

 y u
Sinusoidal:
U
 sin
 y
2
 u
Parabolic:
U


2
  y

 y
2
Compare the shapes of these velocity profiles by plotting y/

(on the ordinate) versus u/U
(on the abscissa).
1.
Statement of the Problem a) Given

Three approximated velocity profiles in laminar boundary layers, linear, sinusoidal, and parabolic. b) Find

Compare these three approximated velocity profiles by plotting.
2.
System Diagram
It is not necessary for this particular problem.
3.
Assumptions

Laminar boundary layer
4.
Governing Equations
None. Just plot them.
5.
Detailed Solution
There is no detailed discussion for this problem. Just plot and compare them.
Using MatLab the plots look like:

Ghosh - 550 Page 2 4/15/2020
Approximated velocity profiles in laminar boundary layer
1
0.9
0.8
0.7
0.6
Linear
Sinusoidal
Parabolic
0.5
0.4
0.3
0.2
0.1
0
0 0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1 u/U
By common sense, it can be said that the linear profile is not very close approximation to the actual shape of boundary layer velocity profile. The plot for the parabolic profile is the closest approximation to the Blasius solution for velocity profile .
6.
Critical Assessment
Note that the velocity profiles are only useful for 0<y<

. Although the given equations hold outside this range, the curves have no meaning for y>

in boundary layer theory.
Velocity profiles in laminar boundary layers often are approximated by the equations u
Linear:
U

 y u
Sinusoidal:
U
 sin
 y
2
 u
Parabolic:
U

2

  y

 y
2
Calculate
 *
(displacement thickness) and

(momentum thickness) for these velocity profiles and compare the result for each case.
7.
Statement of the Problem

Ghosh - 550 Page 3 4/15/2020 c) Given

Three approximated velocity profiles, linear, sinusoidal, and parabolic, in laminar boundary layers. d) Find




*
(displacement thickness) for three approximated velocity profiles
(momentum thickness) for three approximated velocity profiles

Compare the result for each case
8.
System Diagram
Approximated velocity profiles in laminar boundary layer
1
0.9
0.8
0.7
0.6
Linear
Sinusoidal
Parabolic
0.5
0.4
0.3
0.2
0.1
0
0 0.1
0.2
0.3
0.4
0.5
u/U
0.6
0.7
0.8
0.9
9.
Assumptions

Steady state condition

Laminar boundary layer
10.
Governing Equations

Displacement thickness definition since u

*  
0

1 u
U dy
 
0


1 u
U dy

U at y =

, the integrand is essentially zero for y
 
.

Momentum thickness definition
  
0
 u
U
1 u
U dy

Again, the integrand is essentially zero for y
 
.
 
0
11.
Detailed Solution u
U 
1 u
U dy
1

Ghosh - 550 Page 4
Let
 y
 
, then dy =
  d
 because

=

(x) .

* (Displacement thickness)

Linear velocity profile

* linear
 
0


1
 y
 dy
 
0
1

1
  
   d

  

  1
2
 2


1
0

1
2


Sinusoidal velocity profile

* sin

0
 

 1
 sin

2
 y
  1
0

 1
 sin

2



 dy


   d

  


 
2
 cos

2



1
0

1
2

 

Parabolic velocity profile


* parabolic

0
1

1


2


0





1


 
2
 

2
 
 y
 d


  


   2 
1
3
 3


1
0

1
3


(Momentum thickness)

Linear velocity profile
 y
2 




 dy
4/15/2020

Ghosh - 550 Page 5
 linear
 
0

 y

1
 y
 
0
1  
1
  
   d

  
0
1   
2  d

  


1
2

2 
1
3

3


1
0
 dy

1
6


Sinusoidal velocity profile
 sin
 
0
 sin

2
 y


 1
 sin

2
 y


 dy
  1
0 sin
    1
0 sin

2


2



 1
 sin
 sin
2

2


2



   d

 d

 




 
2
 cos

2






2
 sin 2
4



2

2
 






1
0
   
1
2

2


Parabolic velocity profile


 parabolic

0
1
 
0





2
  y

 y

2
  
2

2
  
2

  
0
1

 
4 
4

3 
5

2 
2





2








1

 d
 d


2
  



1
5

5  
4 
5
3

3  
2


1
0

2
15

Comparison
 y

 y
2 




 dy
4/15/2020

Ghosh - 550 Page 6 4/15/2020
Linear
Sinusoidal
 *
/

1/2 = 50% of B.L.
1-2/

= 36.3% of B.L.

/

1/6 = 16.7% of B.L.
-1/2+2/

= 13.7% of B.L.
Parabolic 1/3 = 33.3% of B.L. 2/15 = 13.3% of B.L.
*** Note: B.L. means "Boundary Layer."
12.
Critical Assessment
Understand the concepts of displacement thickness (

* ) and momentum thickness (

).
This problem illustrates how to calculate them from velocity profiles. The above table shows that

<

*<

for most types of velocity profiles .
Air flows in the entrance region of a square duct, as shown. The velocity is uniform, V
1
= 30
m/s, and the duct is 80 mm square. At a section 0.3 m downstream from the entrance, the displacement thickness,
 *
, on each wall measures 1.0 mm. Determine pressure change between sections

and

.

V
1


*
2
= 1.0 mm
80 mm
80 mm
1.
Statement of the Problem a) Given

Working fluid is air which has

Duct is H = 80 mm square.

Displacement thickness,
 *


Uniform flow at the entrance, V
1 air
= 1.23 kg/m
= 30 m/s .
3 at T = 15

C .
2
= 1.0 mm , on each wall at a section L = 0.3 m downstream from the entrance. b) Find

Pressure change between sections

and

.
2.
System Diagram
V
1
L = 0.3 m
H = 80 mm
 

*
2
= 1.0 mm
H = 80 mm

Ghosh - 550 Page 7 4/15/2020
3.
Assumptions

Steady state condition

Incompressible fluid flow

No frictional effects in freestream

Flow uniform at each section outside

*
2

Flow along a streamline between sections  and 

Negligible elevation changes
4.
Governing Equations

0


 t

CV
 d V
 
CS

V


 d A … Integral version of mass conservation
Incompressible fluid flow problem, the equation above

0
 
V
CS


 d A
1 inlet (  ) and 1 outlet (  )

0
 
V
1
A
1

V
2
A
2

Bernoulli's Equation: p


V
2
2
 gz
 const .
Restrictions:
(1) Steady flow
(2) Incompressible flow
(3) Frictionless flow
(4) Flow along a streamline
5.
Detailed Solution
Use the displacement-thickness concept to find the effective flow area for the freestream flow outside the thin wall boundary layers. Replace the actual boundary-layer velocity profiles with uniform velocity profiles as sketched in the following figures.
V V
H - 2
 *

(a) Actual velocity profile
*
(b) Hypothetical velocity profile
H - 2
 *
(c) Cross section of duct
Apply the continuity and Bernoulli equations to freestream flow outside the boundary-layer displacement thickness, where viscous effects are negligible.
From Bernoulli's equation, we obtain p
1


V
1
2
 gz
1
 p
2


V
2
2
 gz
2

Ghosh - 550 Page 8 4/15/2020 p
1
 p
2

1
2


V
2
2 
V
1
2

From the continuity equation, we have
0
 
V
1
A
1

V
2
A
2

V

2
A
1
A
2
V
1
Substituting this expression into Bernoulli's equation, p
1
 p
2

1
2






A
A
1
2
V
1


2

V
1
2




Areas, A
1
and A
2
, are
A
1

H
2
A
2


H

2

*

2
1
2

V
1
2





A
1
A
2


2

1



… (only effective flow area)
Thus, p
1
 p
2

1
2

V
1
2









H
H

2

*

2
2

1



After plugging in values, p
1
 p
2

58 .
99 Pa
6.
Critical Assessment
To solve this problem, it is critical to understand the meaning and physical interpretation of displacement thickness concept.
The velocity profile in a laminar boundary-layer flow at zero pressure gradient is approximated by the linear expression, this profile to obtain expressions for u
U

 y

/x and C
. Use the momentum integral equation with f
.
1.
Statement of the Problem a) Given

Laminar boundary-layer flow

Zero pressure gradient

Velocity profile is approximated by the linear expression, u
U b) Find
Using the momentum integral equation, obtain expression for

 y
 

Ghosh - 550 Page 9 4/15/2020
 

C
/x f
2.
System Diagram y x, u
3.
Assumptions

Steady state condition

Incompressible fluid flow
4.
Governing Equations

Momentum integral equation

 w
 d dx
 
*
U dU dx where

*  
0



1 u
U
U u
U

 y
 dy … displacement thickness
  
0

 u
U
1 u
U
 dy … momentum thickness
5.
Detailed Solution
For the special case of flow over a flat plate, U = constant. From Bernoulli's equation, we see that for this case, p = constant, and thus dp/dx = 0.
The momentum integral equation then reduces to
 w
 
U
2 d
 dx
 
U
2 d dx

0
 u
U

1 u
U
 dy u
Define
U

 y
   y
   dy
   d
 

     

Ghosh - 550 Page 10 4/15/2020
  w
 
U
2 d dx

0
 u
U

1 u
U
 dy
 
U
2
 
U
2 d dx d
 dx

0
1  
1
  
 
 1
0

   2


 d
 d

 
U
2 d
 dx
 
U
2 d
 dx




1
2

2
1
2

1
3

3


1
0
1
3


0

0



  w

1
6

U
2 d
 dx
On the other hand, the shear stress can be calculated by
 w
 
 u
 y y

0
. u
And
U
 w




 y
 y
 u
U
 y

U
 y

0 y , thus
 
U


 y y

0
 
U
 y

0
  w
 
U

Comparing (equating) this shear stress equation with the previous shear stress expression,

W
 
U


1
6

U
2 d
 dx

Ghosh - 550 Page 11
 d

 dx

 d


 d

6


U


6


U

1

6


U
 dx
 dx
1
2
 2 
6


U
 x
 const
When x = 0 ,

= 0
 const = 0 .

2 
12


U
 x

12


U
 x
2 x
 
12


Ux
 x

  
0


 x

12

Ux


12
Re x

3 .
46
Re x
Skin friction coefficient is defined as C f

 w
1
2

U
2
.
Using the result obtained above,
1

U
2 d

C f
 6
1

U dx
2
2

1
3 d
 dx

1
3 d dx



12


U
 x



4/15/2020

Ghosh - 550 Page 12 4/15/2020

1
3

1
6
12


U

1
2

12


Ux
1 x

1
6
12

Ux


C f

12

6
1
Re x

0 .
577
Re x
6.
Critical Assessment
This problem dealt with linear velocity profile as an approximate solution. The results obtained are rough. However the exercise illustrates the use of the momentum integral method. Practice this method with other types of approximated velocity profile, such as parabolic, sinusoidal, … etc.
Water at 15

C flows over a flat plate at a speed of 1 m/s. The plate is 0.4 m long and 1 m wide. The boundary layer on each surface of the plate is laminar. Assume that the velocity profile may be approximated as linear. Determine the drag force on the plate.
1.
Statement of the Problem a) Given

Working fluid is water at T = 15

U = 1 m/s

L = 0.4 m

C
 
= 999 kg/m
3
&


W = 1 m

The boundary layer on each surface of the plate is laminar

Velocity profile is linear (assuming approximately)
= 1.14

10
-3
N
 s/m
2 b) Find

Drag force on the plate
2.
System Diagram
U = 1 m/s
L = 0.4 m

Ghosh - 550 Page 13
3.
Assumptions

Steady state condition

Incompressible fluid flow

Laminar boundary layer
4.
Governing Equations

Skin friction coefficient definition: C f

 w
1
2

U
2

Reynolds number definition for a flat plate: Re x


Ux

5.
Detailed Solution u
We know that for a linear velocity profile
U
C f


 y

0 .
577

,
Re x
Equating this result and the definition of skin friction coefficient,
C f

0 .
577
Re x

1

 w
U
2
  w

1
2

U
2 
0 .
577
Re x
2
  w

1
2

U
2 
0 .
577

Ux

Drag force on one side of the plate is given by F
D
   w

A p dA .
Since dA = w

dx and 0

x

L , F
D
 
A p
 w
 dA
 
0
L  w
 w
 dx .
F
D
 
0
L
1
2

U
2 
0 .
577

Ux

 w
 dx

0 .
577

U 2 w
2

0 .
577

U
2 w
2
1
 x


0
L
1 x
 dx
1
 x





2 x
1
2



L
0
4/15/2020

Ghosh - 550 Page 14 4/15/2020

F
D

0 .
577

U
2 w
2
1
 x


2 L
Plug in values into this expression obtained above for F
D
,

F
D
For both sides of the plate

F
D , Total

2 F
D

0 .
779 N.

0 .
3894 N.
6.
Critical Assessment
Problem says, "the boundary layer on each surface of the plate is laminar." Let us double check that this is true.
Re
L


UL


592 .
05 << 500,000 = Re cr

Obviously it is a laminar flow.
(Note: This problem could be solved by first obtaining the Overall Skin Friction
Coefficient, C . In that case, the calculation will proceed by f obtaining C f

1
L

C f
( x ) dx , where the integration limits will be set at x = 0 and x = L.
Then C f
  w

1

2
U .
C f
2 area on each face of the plate.)

F
D

2
 w
.
A w
, where, A w
(=W.L) indicates the wet
A flat-bottomed barge, 25 m long and 10 m wide, submerged to a depth of 1.5 m, is to be pushed up a river at the rate of 8 km/hr. Estimate the power required to overcome skin friction if the water temperature is 15

C.
1.
Statement of the Problem a) Given

L = 25 m

W = 10 m

D = 1.5 m

V = 8 km/hr = 2.222 m/s

Working fluid is water at T = 15

C
 
= 999 kg/m
3
&

= 1.14

10
-3
N
 s/m
2 b) Find

Power required to overcome skin friction
2.
System Diagram

Ghosh - 550 Page 15 4/15/2020
Water line
W
V
3.
Assumptions

Model a flat-bottomed barge as a flat plate

Steady state condition

Incompressible fluid flow

Neglect separation
4.
Governing Equations
L
D

Drag Coefficient Definition: C
D

F
D
1
2

V
2
A
5.
Detailed Solution
First of all, calculate Reynolds number:
Re
L


VL



999 kg

/
1 .
14 m

3
 
10
2 .
222 m

3
N
 s /
/ s m
2

25 m

 
4 .
87

10
7
Transition of laminar to turbulent flow occurs at
Re cr

5

10
5 

Vx cr

 x cr


5

10

V
5
 
5


10
5

1 .
14
999 kg / m
3

 
10

3
N

2 .
222 m s
/
/ s m

2


0 .
25678 m << 25 m
This x cr shows that the effect of laminar flow is negligible . It can be said that the flow is turbulent from the leading edge.
For Re
L
< 10 9 , the empirical equation given by Schlichting

Ghosh - 550 Page 16 4/15/2020
C
D

 log
0 .
455
Re
L

2 .
58 fits experimental data very well.
Friction force is (from the definition of drag coefficient)
F
D

1
2

V
2
A

C
D

1
2

V
2
A

 log
0 .
455
Re
L

2 .
58
  
F
D

V

1
2

V
2
A

 log
0 .
455
Re
L

2 .
58

V , where A is the wetted area, A = L

W+2(L

D) .
Finally,
  
1
2

V
2

L

W

2

L

D
 

 log
0 .
455
Re
L

2 .
58

V
Plug in values into this expression
 
= 4200.8 W = 4.20 kW
6.
Critical Assessment
Drag coefficient must be chosen depending upon the value of Reynolds number for a particular flow condition. Some of C
D
expressions are derived by analytical calculation, and others are empirical formulas .
Two hypothetical boundary-layer velocity profiles are shown. Obtain an expression for the momentum flux of each profile. If the two profiles were subjected to the same pressure gradient conditions, which would be more likely to separate first? Why?
U U
 u
 u
U

2

  y

 y
2
(a)
1.
Statement of the Problem a) Given
(b)

Ghosh - 550 Page 17 4/15/2020

Two hypothetical boundary-layer velocity profiles b) Find

Expression for the momentum flux of each profile

Which would be more likely to separate first if the two profiles were subjected to the same pressure gradient conditions? And why?
2.
System Diagram
U u
U

 y
 u
U
 u
U

2

  y

 y
2
(a)
3.
Assumptions

Steady state condition

Incompressible fluid flow
4.
Governing Equations

Definition of Momentum Flux ( mf )
(b) d

V

 
V


 d A

Ghosh - 550 Page 18 4/15/2020
5.
Detailed Solution
Since the flow is 1 - D (positive x direction) and dA = w

dy , the momentum equation can be written as d
 u
   u
 dA
 u
   u
 w
 dy
 mf
  
0 u
   u
 w
 dy
 

0 u
   u
 w
 dy
 The integrand is essentially zero for y
 
.
Linear Velocity Profile u
U

 y
 u

U
 y
 mf linear
  
0
U
 y
  
U
 y
 w
 dy
 
U

2
2 w

0
 y 2  dy
 
U

2
2 w


1
3 y 3



0
Finally, mf linear


U
2 w
3
Parabolic Velocity Profile

. u
U

 y
2
Let
 
 y
. Then u
U
Now, mf parabolic


2
   2
and d

 
0 u
   u
 w
 dy


2
  y

1
 dy 
  
.
  1
0 u
   u
 w
   d

   w
 
   w
  
  1
0 u
 1
0

U
2  d


2
  
2

2
 d

   w
  
U
2   1
0
4

2 
4

3  
4 d


Ghosh - 550 Page 19 4/15/2020
   w
  
U
2 


4
3

3  
4 
1
5

5


1
0
Finally, mf parabolic

8

U
2 w

.
15
Which separates first?
Separation occurs when the momentum of fluid layers near the surface is reduced to zero by the combined action of pressure and viscous forces.
As shown in this figure below, the momentum of the fluid near the surface is greater for the parabolic velocity profile.
Velocity Profiles Momentun-Flux Profiles
1 1
0.8
0.8
Linear
0.6
Linear
0.6
0.4
0.4
Parabolic
Parabolic
0.2
0.2
0
0 0.5
u/U
1
0
0 0.5
(u/U)
2
1
Our previous calculation also shows mf linear


U
2 w

< mf parabolic

8

U
2 w

.
3 15
Consequently, the parabolic velocity profile is better able to resist separation in the same pressure gradient condition.

Linear velocity profile would separate first.
6.
Critical Assessment
Review and understand how the flow separation occurs. Flow separation occurs only when there exists an adverse pressure gradient.
A small sphere (D = 6 mm) is observed to fall through caster oil at a terminal speed of 60
mm/s. The temperature is 20

C. Compute the drag coefficient for the sphere. Determine the density of the sphere. If dropped in water, would the sphere fall slower or faster? Why?

Ghosh - 550 Page 20
1.
Statement of the Problem a) Given

D = 6 mm = 0.006 m

Working fluid is caster oil at T = 20

V t
= 60 mm/s = 0.06 m/s

C

S.G.
oil
= 0.969
&
 oil
= 0.9 N
 s/m
2 b) Find

Drag coefficient for the sphere

Density of the sphere

If dropped in water, would the sphere fall slower or faster? Why?
2.
System Diagram
Caster oil at T = 20

C
D
V t
3.
Assumptions

Steady state condition

Incompressible fluid flow
4.
Governing Equations

Drag Coefficient Definition: C
D

1
F
D

V
2
A
2

Newton's Second Law:
 d P dt
When the mass is constant,



F
 m a

Reynolds Number for Sphere:



F
Re
D
, where



P is momentum.

1 - D in y direction


VD ma y
 
F y
4/15/2020

Ghosh - 550 Page 21 4/15/2020
5.
Detailed Solution
Drag Coefficient
First of all, calculate Reynolds number:
Re
D

 oil
V t

D


S .
G .
oil
 
 water


V t
D


0 .
969
 
1000 kg /
0 .
9 m
N
3


 s
0 .
06
/ m
2 m

/ s

0 .
006 m


Re
D
= 0.3876 < 1

There is no flow separation from a sphere. The wake is laminar and the drag is predominantly friction drag.
Stokes has shown analytically, for very low Reynolds number flows where inertia forces may be neglected, that drag force on a sphere of diameter D, moving speed V, through a fluid of viscosity

, is given by
F
D

3

VD
The drag coefficient, C
D
, is then
C
D

1
2

F
D
V
2
A

1
2

3

VD
V
2



4
D
2




24
VD


24
Re
D
(Note: For sphere, the area, A, is just a cross-sectional area, which is

4
Thus,
D
2
.)
C
D

24
Re
D

24
0 .
3876

61 .
92
Density of the Sphere
Free Body Diagram
F
D
F
B
W
 mg
  s
V s g
F
D

3
 oil
V t
D y
F
B

W oil displaced
  oil
V s g
W
The sphere reached the terminal speed
 a y
= 0 ma y
 
F y
0
0


F
D

3

F
B
 oil
V t
D

W
  oil
V s g
  s
V s g
 s

3
 oil
V t
D

V s g
 oil
V s g

Ghosh - 550 Page 22 4/15/2020
 s

3
 oil
V t
D


S .
G .
oil



4
3


2
 water





4
3

D
3



 g
D
2
3



 g
After plug in values into this expression,
 s
= 3721 kg/m
3
.
If the working fluid is water at T = 20

C
  w
= 998 kg/m 3 &
 w
= 1

10 -3 N
 s/m 2
Because
 w
= 1

10
-3
N
 s/m
2
<<
 oil
= 0.9 N
 s/m
2
, the author guesses the sphere drops faster in water than in caster oil.
If it's faster and
 w
<<
 oil
, Re
D water

Re oil
D

0 .
3876 .

F
D
= 3

VD cannot be used because the equation works only for very low Reynolds number which we don't know whether this is appropriate or not any more for this case.
Now, guess a value of C
D
from Figure 9.11 (Drag coefficient of a smooth sphere as a function of Reynolds number) and calculate V tW
. Then calculate Re
D
and verify the chosen C
D was appropriate or not.
Guess C
D
= 0.4

F
D

1
2
Free Body Diagram (again)

2 w
V tW
A

C
D
W
 mg
  s
V s g
F
D
F
B y
F
D
F
B

1
 w
V tW
2
A

C
D
2

W water displaced
  w
V s g
 w
2
V tW
A

C
D
The sphere reaches a new terminal speed, V tW
 a y
= 0 ma y
 
F y

0

F
D

F
B

W
0

1
2
W
  w
V s g
  s
V s g

Ghosh - 550 Page 23 4/15/2020
V tW

  s
  w

V s g
1
2
 w
AC
D

  s

1
2

 w w





4



4
3

D
2
D
2



C
D
3



 g
After plugging values into this expression, V tW
= 0.732 m/s .
With this new terminal speed, Reynolds number is Re
D

 w
V tW

D

4383 .
Figure 9.11 says when C
D
= 0.4
, Re
D w

4

10 3 , which is about right for this case. This shows the new terminal speed is a valid number.
V tW
= 0.732 m/s > V t
= 0.06 m/s

The sphere drops faster in water than in caster oil.
6.
Critical Assessment
Drag coefficient depends upon the value of Reynolds number. Be careful with choosing a right C
D
depending on a particular flow condition.