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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
a2
=
m
s
b1
=
1
s
x
0
2
=
y
0
5
=
x1
=
xx
=
For streamlines
v
u
dy
dx
=
bx
⋅
a
=
So, separating variables
a
b
dy
⋅
xdx
⋅
=
Integrating
a
b
yy
0
−
()
⋅
1
2
x
2
x
0
2
−
⎛
⎝
⎞
⎠
⋅
=
The solution is then
0
b
2a
⋅
x
2
x
0
2
−
⎛
⎝
⎞
⎠
⋅
+
x
2
4
4
+
=
=
Hence for pathlines
u
p
dx
dt
=
a
=
v
p
dy
dt
=
⋅
=
Hence
dx
a dt
⋅
=
dy
b x
⋅
dt
⋅
=
Integrating
0
−
at t
0
−
⋅
=
dy
b x
0
0
−
⋅
+
⎣
⎦
⋅
dt
⋅
=
0
−
0
tt
0
−
⋅
a
2
t
2
t
0
2
−
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⋅
+
at
0
⋅
0
−
⋅
−
⎡
⎣
⎤
⎦
⋅
=
The pathlines are
0
0
−
⋅
+
=
0
0
0
−
⋅
a
2
t
2
t
0
2
−
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
⋅
+
0
⋅
0
−
⋅
−
⎡
⎣
⎤
⎦
⋅
+
=
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View Full Document For a particle that was at
x
0
0
=
m,
y
0
4
=
m at
t
0
0
=
s, at time
t2
=
s we find the position is
xx
0
at t
0
−
()
⋅
+
4
=
=
m
yy
0
bx
0
tt
0
−
⋅
a
2
t
2
t
0
2
−
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics

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