Problem 2.34
[Difficulty: 3]
Given:
Velocity field
Find:
Equation for streamline through point (2.5); coordinates of particle at t = 2 s that was at (0,4) at t = 0; coordinates of
particle at t = 3 s that was at (1,4.25) at t = 1 s;
compare pathline, streamline, streakline
Solution:
Governing equations:
For streamlines
v
u
dy
dx
=
For pathlines
u
p
dx
dt
=
v
p
dy
dt
=
Assumption:
2D flow
Given data
a
2
=
m
s
b
1
=
1
s
x
0
2
=
y
0
5
=
x
1
=
x
x
=
For streamlines
v
u
dy
dx
=
b x
⋅
a
=
So, separating variables
a
b
dy
⋅
x dx
⋅
=
Integrating
a
b
y
y
0
−
(
)
⋅
1
2
x
2
x
0
2
−
⎛
⎝
⎞
⎠
⋅
=
The solution is then
y
y
0
b
2 a
⋅
x
2
x
0
2
−
⎛
⎝
⎞
⎠
⋅
+
x
2
4
4
+
=
=
Hence for pathlines
u
p
dx
dt
=
a
=
v
p
dy
dt
=
b x
⋅
=
Hence
dx
a dt
⋅
=
dy
b x
⋅
dt
⋅
=
Integrating
x
x
0
−
a
t
t
0
−
(
)
⋅
=
dy
b
x
0
a
t
t
0
−
(
)
⋅
+
⎡
⎣
⎤
⎦
⋅
dt
⋅
=
y
y
0
−
b
x
0
t
t
0
−
(
)
⋅
a
2
t
2
t
0
2
−
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⋅
+
a t
0
⋅
t
t
0
−
(
)
⋅
−
⎡
⎢
⎣
⎤
⎥
⎦
⋅
=
The pathlines are
x
x
0
a
t
t
0
−
(
)
⋅
+
=
y
y
0
b
x
0
t
t
0
−
(
)
⋅
a
2
t
2
t
0
2
−
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⋅
+
a t
0
⋅
t
t
0
−
(
)
⋅
−
⎡
⎢
⎣
⎤
⎥
⎦
⋅
+
=
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
For a particle that was at
x
0
0
=
m,
y
0
4
=
m at
t
0
0
=
s, at time
t
2
=
s we find the position is
x
x
0
a
t
t
0
−
(
)
⋅
+
4
=
=
m
y
y
0
b
x
0
t
t
0
−
(
)
⋅
a
2
t
2
t
0
2
−
⎛
⎝
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Lear
 Fluid Mechanics

Click to edit the document details