Question 1:
Given: Two-dimensional flow in the xy plane
Find: (a) Possible y component for incompressible flow.
Governing equation:
⃗ +
∇. 𝜌𝑉
∂ρ
=0
𝜕𝑡
For incompressible flow this simplifies to: (In rectangular coordinates)
∂u ∂v ∂w
+
+
=0
𝜕𝑥 𝜕𝑦 𝜕𝑧
For two-dimensional flow in the xy plane:
∂u ∂v
+
=0
𝜕𝑥 𝜕𝑦
∂v
∂u
=−
= −2
𝜕𝑦
𝜕𝑥
which gives an expression for the rate of change of v holding x constant.
This equation can be integrated to obtain an expression for v. The result is
∂v
𝑣 = ∫ 𝑑𝑦 + 𝑓(𝑥) = −2𝑦 + 𝑓(𝑥)
𝜕𝑦
Since v=0 along the X-axis:
𝑓(𝑥) = 0 𝑎𝑛𝑑 𝑣 = −2𝑦
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 5.38
[Difficulty: 2]
The velocity field provided above
Given:
Find:
(a) the number of dimensions of the flow
(b) if this describes a possible incompressible flow
(c) the acceleration of a fluid particle at point (1,2,3)
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)
(2) Two dimensional flow (velocity is not a function of z)
(3) Steady flow (velocity is not a function of t)
Assumptions:
The flow is two dimensional.
Based on assumption (2), we may state that:
Based on assumptions (1) and (3), the continuity equation reduces to:
u v

0
x y
This is the criterion against which we will check the flow field.
u  x y
1
2
v   y
3
u v

 y2  y2  0
x y
3
Based on assumptions (2) and (3), the acceleration reduces to:
This could be an incompressible flow field.



V and the partial derivatives of velocity are:
V
v
ap  u
y
x


V

V
2ˆ
ˆ
 y i  yk and
 2 xyiˆ  y 2 ˆj  xkˆ Therefore the acceleration vector is equal to:
x
y

1
1
1
2
a p  xy 2 y 2 iˆ  ykˆ  y 3 2 xyiˆ  y 2 ˆj  xkˆ  xy 4 iˆ  y 5 ˆj  xy 3 kˆ At point (1,2,3), the acceleration is:
3
3
3
3





32 ˆ 16 ˆ
16
1
 1
 2

a p    1  2 4 iˆ    2 5  ˆj    1  2 3 kˆ  iˆ 
j k
3
3
3
3
 3
 3


16
32 ˆ 16 ˆ
a p  iˆ 
j k
3
3
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 