Problem *3.118
[Difficulty: 2]
Given:
Spinning Utube sealed at one end
Find:
Pressure at A; water loss due to leak
Assumption:
Water is incompressible; centripetal acceleration is constant
Solution:
Basic equation
From the analysis of Example Problem 3.10, solving the basic equation, the pressure
p
at any point (
r
,
z
) in a continuous
rotating fluid is given by
p
p
0
ρ ω
2
⋅
2
r
2
r
0
2
−
⎛
⎝
⎞
⎠
⋅
+
ρ
g
⋅
z
z
0
−
(
)
⋅
−
=
(1)
where
p
0
is a reference pressure at point (
r
0
,
z
0
)
In this case
p
p
A
=
p
0
p
D
=
z
z
A
=
z
D
=
z
0
=
H
=
r
0
=
r
0
r
D
=
L
=
The speed of rotation is
ω
300 rpm
⋅
=
ω
31.4
rad
s
⋅
=
The pressure at
D
is
p
D
0 kPa
⋅
=
(gage)
Hence
p
A
ρ ω
2
⋅
2
L
2
−
(
)
⋅
ρ
g
⋅
0
( )
⋅
−
=
ρ ω
2
⋅
L
2
⋅
2
−
=
1
2
−
1.94
×
slug
ft
3
⋅
31.4
rad
s
⋅
⎛
⎜
⎝
⎞
⎠
2
×
3 in
⋅
(
)
2
×
1 ft
⋅
12 in
⋅
⎛
⎜
⎝
⎞
⎠
4
×
lbf s
2
⋅
slug ft
⋅
×
=
p
A
0.42
−
psi
⋅
=
(gage)
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 Fall '07
 Lear
 Fluid Mechanics

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