Problem 2.23
[Difficulty: 3]
Given:
Velocity field
Find:
Plot of pathline of particle for t = 0 to 1.5 s that was at point (1,1) at t = 0; compare to streamlines through same
point at the instants t = 0, 1 and 1.5 s
Solution:
Governing equations:
For pathlines
u
p
dx
dt
=
v
p
dy
dt
=
For streamlines
v
u
dy
dx
=
Assumption:
2D flow
Hence for pathlines
u
p
dx
dt
=
a x
⋅
=
a
1
5
=
1
s
v
p
dy
dt
=
b y
⋅
t
⋅
=
b
1
25
=
1
s
2
So, separating variables
dx
x
a dt
⋅
=
dy
b y
⋅
t
⋅
dt
⋅
=
dy
y
b t
⋅
dt
⋅
=
Integrating
ln
x
x
0
⎛
⎜
⎝
⎞
⎠
a t
⋅
=
x
0
1
=
m
ln
y
y
0
⎛
⎜
⎝
⎞
⎠
b
1
2
⋅
t
2
⋅
=
y
0
1
=
m
Hence
x t
( )
x
0
e
a t
⋅
⋅
=
y t
( )
y
0
e
1
2
b
⋅
t
2
⋅
⋅
=
Using given data
x t
( )
e
t
5
=
y t
( )
e
t
2
50
=
For streamlines
v
u
dy
dx
=
b y
⋅
t
⋅
a x
⋅
=
So, separating variables
dy
y
b t
⋅
a x
⋅
dx
⋅
=
which we can integrate for any given t (t is treated as a constant)

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