Problem 6.6
[3]
Given:
Velocity field
Find:
Expressions for local, convective and total acceleration; evaluate at several points; evaluate pressure gradient
Solution:
The given data is
A
2
1
s
⋅
=
ω
1
1
s
⋅
=
ρ
2
kg
m
3
⋅
=
u
A x
⋅
sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
=
v
A
−
y
⋅
sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
=
Check for incompressible flow
x
u
∂
∂
y
v
∂
∂
+
0
=
Hence
x
u
∂
∂
y
v
∂
∂
+
A sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
A sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
−
=
0
=
Incompressible flow
The governing equation for acceleration is
The local acceleration is then
x
 component
t
u
∂
∂
2
π
⋅
A
⋅
ω
⋅
x
⋅
cos 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
=
y
 component
t
v
∂
∂
2
−
π
⋅
A
⋅
ω
⋅
y
⋅
cos 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
=
For the present steady, 2D flow, the convective acceleration is
x
 component
u
x
u
∂
∂
⋅
v
y
u
∂
∂
⋅
+
A x
⋅
sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
A sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
(
)
⋅
A
−
y
⋅
sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
(
) 0
⋅
+
=
A
2
x
⋅
sin 2
π
⋅
ω
⋅
t
⋅
(
)
2
⋅
=
y
 component
u
x
v
∂
∂
⋅
v
y
v
∂
∂
⋅
+
A x
⋅
sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
0
⋅
A
−
y
⋅
sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
(
)
A
−
sin 2
π
⋅
ω
⋅
t
⋅
(
)
⋅
(
)
⋅
+
=
A
2
y
⋅
sin 2
π
⋅
ω
⋅
t
⋅
(
)
2
⋅
=
The total acceleration is then
x
 component
t
u
∂
∂
u
x
u
∂
∂
⋅
+
v
y
u
∂
∂
⋅
+
2
π
⋅
A
⋅
ω
⋅
x
⋅
cos 2
π
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 Fall '08
 ZOHAR
 Fluid Dynamics, Incompressible Flow, π⋅, A⋅ x⋅ sin

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