Ghosh - 550
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Worked Out Examples
(Kinematics of Fluid Flows)
Example 1. (Use of Continuity equation and Acceleration):


Consider the velocity field V  Axy  i 
1 2 
Ay  j in the xy plane, where A  0.25m 1  s 1 ,
2
and the coordinates are measured in meters. Is this a possible incompressible flow field?
Calculate the acceleration of a fluid particle at point ( x, y )  (2,1) .
1. Statement of the Problem
a) Given



Velocity field: V  Axy  i 
1 2 
Ay  j , where A  0.25m 1  s 1 .
2
b) Find
 Whether the velocity field is a possible incompressible flow field or not.
 Acceleration of a fluid particle at point ( x, y )  (2,1) .
2. System Diagram
It is not necessary for this problem.
3. Assumptions
 Steady state condition
 2 - D flow field problem
4. Governing Equations


 u v w



0
t
x
y
z
u v
Incompressible 2 - D continuity equation:

0
x y





 DV V
V
V
V
Acceleration: a 

u
v
w
Dt
t
x
y
z
Continuity Equation:
2 - D problem:
Du u
u
u
u

u
v
w
Dt t
x
y
z
Dv v
v
v
v
ay 

u v w
Dt t
x
y
z
u

u
(
x
,
y
)
v

v
(
x
,
y
)
Since
&
,
u
u
ax  u
v
x
y
v
v
ay  u  v
x
y
ax 
5. Detailed Solution
Ghosh - 550
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

Velocity field V  Axy  i 
u  u ( x, y )  Axy
1
v  v( x, y )   Ay 2
2
2/6/2016
1 2 
Ay  j shows the components to be:
2
The continuity equation, incompressible version of continuity equation in this case, must be
satisfied to have a valid incompressible flow field.
So, check the incompressible continuity equation:
u v 
  1


 Axy   Ay 2   Ay   Ay   0  Continuity equation is satisfied.
x y x
y  2

Therefore, it can be said that there exists a possible incompressible flow field.
Acceleration is:
u
u

 1
 
 1

v
 Axy  Axy   Ay 2   Axy  Axy  Ay    Ay 2   Ax
x
y
x
 2
 y
 2

 1
 1
 a x  A 2 xy 2   A 2 xy 2   A 2 xy 2
 2
 2
ax  u
v
v
  1
  1
   1

 1

v
 Axy   Ay 2    Ay 2    Ay 2   0   Ay 2    Ay 
x
y
x  2
  2
 y  2

 2

1 2 3
 ay  A y
2
ay  u
Therefore, at ( x, y )  (2,1) , the acceleration is:


2
1 2 2 1
2
A xy  0.25m 1  s 1  2m   1m   0.0625m / s 2
2
2
2
1 2 3 1
3
a y  A y  0.25m 1  s 1  1m   0.03125m / s 2
2
2
ax 







 a  a x  i  a y  j  (0.0625m / s 2 )  i  (0.03125m / s 2 )  j
6. Critical Assessment
Incompressible 2 - D continuity equation has been satisfied; therefore, the fluid flow is
incompressible in this flow field.
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Example 2. (Use of Integral Continuity + Acceleration)
Consider the incompressible flow of a fluid through a nozzle as shown. The area of the
nozzle is given by A = A0(1 - bx) and the inlet velocity varies according to U = C(1 + at),
where A0 = 1 ft2, L = 4 ft, b = 0.1 ft-1, a = 2 s-1, and C = 10 ft/s. Find the acceleration of a fluid
particle on the centerline at x = L/2 for t = 0 and 0.5 s. Plot ax on the centerline as a function
of x/L for t = 0 and 0.5 s.
U
Ao
L
x
1. Statement of the Problem
a) Given
 Incompressible flow of a fluid through a nozzle ( = constant).
 Area of the nozzle, A = A0(1 – bx)
 Inlet velocity, U = C(1 + at)
 Constants
A0 = 1 ft2
L = 4 ft
b = 0.1 ft-1
a = 2 s-1
C = 10 ft/s
b) Find
 Acceleration of a fluid particle on the centerline at x = L/2 for t = 0 and 0.5 s.
 Plot of ax on the centerline as a function of x/L for t = 0 and 0.5 s.
2. System Diagram
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Control Volume
U

Ao

: at the inlet
: at a general x
L
x
3. Assumptions
 Incompressible but unsteady fluid flow problem
 1 – D problem along x direction because the interest is an acceleration, ax, on the
centerline; therefore, u = u(x,t) only.
4. Governing Equations







 DV V
V
V
V
a

u
v
w
Dt
t
x
y
z
Du u
u
u
u
1 - D problem  a x 

u
v
w
Dt t
x
y
z
u
u
u
Because u = u(x,t)  a x 
t
x

t



dV   V  dA … Integral version of mass conservation
CV
CS
 
Incompressible fluid flow problem  0   V  dA
0
CS
1 inlet () and 1 outlet () on 1 - D problem  0  u0 A0  u1 A1
5. Detailed Solution
Obtain u = u(x,t) from 0  u0 A0  u1 A1 .
0  U  A0  u ( x, t )  A
u( x, t ) 
A0
U
A
[ is at general x, so u1  u( x, t ) .]
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A0
 C (1  at )
A0 (1  bx)
1  at
C
 u  u ( x, t ) 
1  bx
u( x, t ) 
Acceleration ax can be obtain …
u
u
u
t
x
C
 1  at

ax 
 (0  a )  
 C   (1  at )  C  (1)  (1  bx)  2  (0  b)
1  bx
 a  bx 
ax 

 ax 
aC
(1  at ) 2
… [Check units! It must have ft/s2.]
 bC 2 
1  bx
(1  bx) 3
Finally the acceleration is …


(2 s 1 )(10 ft / s )
a x ( x  L / 2, t  0 s ) 
 (0.1 ft 1 )(10 ft / s ) 2
 (4 ft ) 
1  (0.1 ft 1 ) 
 2 



 a ( x  L / 2, t  0s )  a x  i  (44.53 ft / s 2 )  i
1  (2s
Plot of ax vs. x/L
ax 
aC
(1  at ) 2
 bC 2 
1  bx
(1  bx) 3
aC
(1  at ) 2
 bC 2 
3
x


x


1  bL 
1  bL L 
L
 

At t = 0 s:
1  a(0s)
aC
ax 
 bC 2 
3
x

 x 
1  bL 
1  bL L 
L
 

2
ax 
aC
1
 bC 2 
3
x

 x 
1  bL 
1  bL L 
L
 

1
)(0 s )

2

1  ( 4 ft ) 
1  (0.1 ft ) 2 



(2 s 1 )(10 ft / s )
a x ( x  L / 2, t  0.5s ) 
 (0.1 ft 1 )(10 ft / s ) 2
(
4
ft
)


1  (0.1 ft 1 ) 

2




 a ( x  L / 2, t  0.5s )  a x  i  (103.1 ft / s 2 )  i
ax 

1  (2s
1
)(0.5s )
3

2

1  ( 4 ft ) 
1  (0.1 ft ) 2 



3
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At t = 0.5 s:
1  a(0.5s)
aC
 bC 2 
3
x

 x 
1  bL 
1  bL L 
L
 

2
ax 
Using MatLab, the plot looks like …
ax vs. x/L at t = 0 and 0.5 s
250
t=0s
t = 0.5 s
200
a
x
150
100
50
0
0
0.1
0.2
0.3
0.4
0.5
x/L
0.6
0.7
0.8
0.9
1
6. Critical Assessment
Note how the problem used the integral control volume analysis to obtain u(x,t). The
selection of the control surface  must be at a general x-location. Once u(x,t) is known,
finding ax is a matter of applying the formula correctly.
Example 3. (Test for incompressibility and irrotationality):




A flow is represented by the velocity field V  10 x  i  10 y  j  30  k . Determine if the
field is (a) a possible incompressible flow and (b) irrotational.
1. Statement of the Problem
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a) Given


2/6/2016


 Velocity field: V  10 x  i  10 y  j  30  k
b) Find
If the field is
 a possible incompressible flow
 irrotational
2. System Diagram
It is not necessary for this problem.
3. Assumptions
 Steady state condition
4. Governing Equations


 u v w



0
t
x
y
z
u v w
Incompressible continuity equation:


0
x y z

    w v   u w   v u 
   j  
Vorticity:   2    V  i 
  k   
 z x 
 y z 
 x y 
Continuity Equation:
5. Detailed Solution




The velocity field is V  10 x  i  10 y  j  30  k .
This shows the components to be:
u  10x
v  10 y
w  30
The continuity equation, incompressible version in this case, must be satisfied to have a
possible incompressible flow field. Checking the incompressible continuity equation:
u v w 




 10 x   10 y   30  10   10  0  0
x y z x
y
z
It can be said that the flow field is valid (possible) since the incompressible continuity
equation is satisfied.
The next is to check the vorticity to see if the flow is rotational or irrotational.

 w
v 
 u
w 
 v
u 
  i 
   j  
  k   
 z x 
 y z 
 x y 
x component:
w v 


 30   10 y   0
y z y
z
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u w 


 10 x   30  0
z x z
x
v u 

z component:

  10 y   10 x   0
x y x
y
y component:





Therefore,   i 0  j 0  k 0  0  This shows there is no rotation in the flow.
The flow is irrotational.
6. Critical Assessment
Incompressible continuity equation has been satisfied; therefore, the fluid flow is
incompressible in this flow field. Also the fluid has no rotation, as it satisfies the zero
vorticity requirement.
Example 4 (Rotation & Stream Function):
Consider a velocity field for motion parallel to the x axis with constant shear. The shear rate is

du/dy = A, where A = 0.1 s-1. Obtain an expression for the velocity field, V . Calculate the rate of
rotation. Evaluate the stream function for this flow field.
1. Statement of the Problem
a) Given
 Velocity field parallel to the x axis with constant shear
 Shear rate du/dy = A, where A = 0.1 s-1
b) Find




Expression for the velocity field, V
Calculate the rate of rotation
Evaluate the stream function for this flow field
2. System Diagram
y
yx
xy
xy
u = u(y)
yx
x
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3. Assumptions
 Steady state condition
 Incompressible fluid flow
 2-D xy plane problem
4. Governing Equations

Fluid rotation about z axis on the xy plane
1  v

Shear stress on the xy plane
 v

u 
 z    
2  x y 
u 
 xy   yx     
 x y 
Stream function (incompressible fluid flow version) definition
u


& v
x
y
5. Detailed Solution

Expression for the velocity field, V
The shear rate is given as du/dy = A. This can be integrated as follows:
 du   Ady  A dy because A is constant.
Therefore,
u = Ay + Const.

Const. = 0 because u = 0 at y = 0.
Finally, the velocity field is



V   Ay   i  0.1s 1  y  i


Calculate the rate of rotation
A
0




   z  k   A  k    0.1s 1   k  0.05s 1   k

1
2
1
2
Evaluate the stream function for this flow field
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Using the definition of stream function,

 Ay … 
y

v
0 …
x
u
 becomes:   Ay  y 
A
    Ay  y    2 y
2
 f ( x)
where f(x) is any function of x including constants.
Using this , take


df ( x)
 0
. That will be
.…
x
dx
x
Comparing  with ,
Finally,  
df ( x)
 0 .  f(x) = constant.
dx


A 2
0.1s 1 2
y  const .   
y  const.    0.05s 1 y 2  const.
2
2


6. Critical Assessment
The stream function, , exists; therefore, the velocity field must satisfy the continuity
equation,
u v

 0 . Let us double-check that on this problem.
x y
u v 


 Ay  0  0  Continuity is satisfied.
x y x
y