Solution

Problem 5.5
[Difficulty: 2]
Given:
x component of velocity
Find:
y component for incompressible flow; Valid for unsteady? How many y components?
Solution:
Basic
Equation:

x

x

y
( ρ v ) 

z
( ρ w) 

t
ρ0
Incompressible flow (ρ is constant)
Flow is only in the x-y plane
Assumptions:
Hence
( ρ u ) 
u 

y
v 0
or



2
3


v   u   3  x  y  y  6  x  y
y
x
x

2
v ( x y )   6  x  y dy  3  x  y  f ( x )

This basic equation is valid for steady and unsteady flow (t is not explicit)
Integrating
There are an infinite number of solutions, since f(x) can be any function of x. The simplest is f(x) = 0
v ( x y )  3  x  y
2
Problem 5.10
Given:
Approximate profile for a laminar boundary layer:
U y
u
δ  c x (c is constant)
δ
Find:
(a) Show that the simplest form of v is
v
[Difficulty: 2]
u y

4 x
(b) Evaluate maximum value of v/u where δ = 5 mm and x = 0.5 m
Solution:
We will check this flow field using the continuity equation
Governing
Equations:

u    v    w    0 (Continuity equation)
x
y
z
t
Assumptions:
(1) Incompressible flow (ρ is constant)
(2) Two dimensional flow (velocity is not a function of z)
u v

0
x y
Based on the two assumptions listed above, the continuity equation reduces to:
u u d
Uy 1 
Uy
v
Uy
u

  2  cx 2  


3 Therefore from continuity:
3
x  dx
2

y
x
2cx 2
2cx 2
1
The partial of u with respect to x is:
Integrating this expression will yield the y-component of velocity:

v




U y
3
2  c x
2
U y
2
Now due to the no-slip condition at the wall (y = 0) we get f(x) = 0. Thus: v 
v
δ
The maximum value of v/U is where y = δ: v ratmax 

u
4 x
v ratmax 
dy  f ( x ) 
3
4  c x
U y
3
4  c x
U y


y
1 4 x
2
5  10
2
c x

 f ( x)
2
u y
4 x
(Q.E.D.)
v
u y

4 x
2
3
m
4  0.5 m
v ratmax  0.0025
Problem 5.42
Given:
Find:
[Difficulty: 2]
The velocity field provided above
(a) if this describes a possible incompressible flow
(b) the acceleration of a fluid particle at point (x,y) = (0.5 m, 5 mm)
(c) the slope of the streamline through that point
We will check this flow field against the continuity equation, and then apply the definition of acceleration
Solution:
Governing

u    v    w    0 (Continuity equation)
Equations:
x
y
z
t






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
Dt
x
(1) Incompressible flow (ρ is constant)
Assumptions:
(2) Two-dimensional flow (velocity is not a function of z)
(3) Steady flow (velocity is not a function of t)
Based on the assumptions above, the continuity equation reduces to:
u
A U y
1
x
2
v
A U y
2
3
4 x
2
u v
1 AUy
AUy


2 3 0
3
x y
2 2
x
4x 2
Based on assumptions (2) and (3), the acceleration reduces to:

AUy
3 AUy 2 ˆ
V
  3 2 iˆ 
j
x
2x
8x 5 2
u v

0
x y
This is the criterion against which we
will check the flow field.
This represents a possible incompressible flow field.



V and the partial derivatives of velocity are:
V
v
ap  u
y
x

V
AU
AUy

and
 1 2 iˆ  3 2 ˆj Therefore the acceleration vector is equal to:
y
2x
x

AUy  AUy ˆ 3 AUy 2 ˆ  AUy 2  AU ˆ AUy ˆ 
A 2U 2 y 2 ˆ A 2U 2 y 3 ˆ
At (5 m, 5 mm):


ap  1 2  3 2 i 
j 
i  3 2 j  
i
j
52
32  12
2
3
x  2x
8x
2x
4x
4x

 4x  x
2
2
2
2
2
3
 1  141  

m   0.005   ˆ  1  141  
m   0.005   ˆ
a p      1 2    0.240   
 i    
   0.240   
 j
s   0.5    4  m1 2  
s   0.5  
 4  m  

The slope of the streamline is given by: slope 
v
u

A U y
3
4 x
2

x
2
A U y



m
a p  2.86 10  2 iˆ  10  4 ˆj 2
s
1
y
4 x
2
Therefore, slope 
0.005
4  0.5
slope  2.50  10
3
Problem 5.46
Given:
[Difficulty: 2]
Duct flow with incompressible, inviscid liquid
U  5
m
s
L  0.3 m
u ( x )  U  1 

x


2 L 
Find:
Expression for acceleration along the centerline of the duct
Solution:
We will apply the definition of acceleration to the velocity.
Governing
Equation:
Assumptions:






V V (Particle acceleration)
V
DV
V

w
v
ap 
u
t
z
y
Dt
x
(1) Incompressible flow (ρ is constant)
(2) One-dimensional flow along centerline (u = u(x) only)
(3) Steady flow (velocity is not a function of t)
Based on assumptions (2) and (3), the acceleration reduces to:
apx  u 

x
u  U  1 
 
2
    U    U   1  x 
 



2 L 
2 L 
2 L   2 L 
x
2
apx  
U
2 L
  1 

x


2 L 
Problem 5.63
[Difficulty: 3]
Flow between parallel disks through porous surface
Given:
Find:
(a) show that V r = vor/2h
(b) expression for the z-component of velocity (v o<<V)
(c) expression for acceleration of fluid particle in the gap
Solution:
We will apply the continuity equation to the control volume shown:
 


d
V


V

  dA
t CV
CS



 V

DV
ap 
 V  V 
Dt
t
Governing
Equations:
0

(Continuity)

(Particle Accleration)
(1) Steady flow
(2) Incompressible flow
(3) Uniform flow at every section
(4) Velocity in θ-direction is zero
Assumptions:
Based on the above assumptions the continuity equation reduces to:
1 

 r Vr  Vz  0
r r
z
 
We apply the differential form of continuity to find Vz :


Vz  


r
2
0  ρ v 0  π r  ρ Vr 2  π r h Solving for Vr: Vr  v 0 
2 h
v0
z
dz  f ( r)  v 0   f ( r) Now at z = 0: Vz  v 0
h
h
v0
1 

 r Vr 
  Vz
r r
h
z
 
Therefore we can solve for f(r):
0
v 0  v 0   f ( r) f ( r)  v 0
h
So we find that the z-component of velocity is:
Based on the above assumptions the particle acceleration reduces to:
v0
Vr 
2 h
r


z
Vr  0

r
Vz  0
v0
Vz  
h
z



V
V
a p  Vr
 Vz
r
z
Therefore:
z
Vz  v 0   1  
h


So the accelerations are:
v0
apr  Vr Vr  Vz Vr  v 0 

 v 0   1 
2

h
2

h

r
z


r
apz  Vr Vz  Vz Vz  v 0 
 0  v 0   1 
2 h

r
z


r
2
z
 0
apr 
h
z
v0
 
h h
apz 
v0
h
2
 
z
h
v0  r
4 h
2
 1


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