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Problem 2.33
[Difficulty: 3]
Given:
Velocity field
Find:
Equation for streamline through point (1.1); coordinates of particle at t = 5 s and t = 10 s that was at (1,1) at t = 0;
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
1
5
=
1
s
b1
=
m
s
x
0
1
=
y
0
1
=
t
0
0
=
For streamlines
v
u
dy
dx
=
b
ax
⋅
=
So, separating variables
a
b
dy
⋅
dx
x
=
Integrating
a
b
yy
0
−
()
⋅
ln
x
x
0
⎛
⎜
⎝
⎞
⎠
=
The solution is then
0
b
a
ln
x
x
0
⎛
⎜
⎝
⎞
⎠
⋅
+
5lnx
⋅
1
+
=
=
Hence for pathlines
u
p
dx
dt
=
⋅
=
v
p
dy
dt
=
b
=
Hence
dx
x
ad
t
⋅
=
dy
b dt
⋅
=
Integrating
ln
x
x
0
⎛
⎜
⎝
⎞
⎠
at t
0
−
⋅
=
0
−
bt t
0
−
⋅
=
The pathlines are
xx
0
e
att
0
−
⋅
⋅
=
0
0
−
⋅
+
=
or
0
b
a
ln
x
x
0
⎛
⎜
⎝
⎞
⎠
⋅
+
=
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View Full DocumentFor a particle that was at
x
0
1
=
m,
y
0
1
=
m
at
t
0
0
=
s, at time
t1
=
s we find the position is
xx
0
e
att
0
−
()
⋅
⋅
e
1
5
=
=
m
yy
0
bt t
0
−
⋅
+
2
=
=
m
For a particle that was at
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 Fall '07
 Lear
 Fluid Mechanics

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