Problem 3.11
[Difficulty: 2]
Given:
Properties of a cube suspended by a wire in a fluid
Find:
The fluid specific gravity; the gage pressures on the upper and lower surfaces
Solution:
From a free body analysis of the cube:
Σ
F0
=
Tp
L
p
U
−
()
d
2
⋅
+
Mg
⋅
−
=
where
M
and
d
are the cube mass and size and
p
L
and
p
U
are the pressures on the lower and upper surfaces
For each pressure we can use Eq. 3.7
pp
0
ρ
g
⋅
h
⋅
+
=
Hence
p
L
p
U
−
p
0
ρ
g
⋅
Hd
+
⋅
+
⎣
⎦
p
0
ρ
g
⋅
H
⋅
+
−
=
ρ
g
⋅
d
⋅
=
SG
ρ
H2O
⋅
d
⋅
=
where
H
is the depth of the upper surface
Hence the force balance gives
SG
⋅
T
−
ρ
H2O
g
⋅
d
3
⋅
=
SG
2 slug
⋅
32.2
×
ft
s
2
⋅
lbf s
2
⋅
slug ft
⋅
×
50.7 lbf
⋅
−
1.94
slug
ft
3
⋅
32.2
×
ft
s
2
⋅
lbf s
2
⋅
slug ft
⋅
×
0.5 ft
⋅
3
×
=
SG
1.75
=
From Table A.1, the fluid is Meriam blue.
The individual pressures are computed from Eq 3.7
0
ρ
g
⋅
h
⋅
+
=
or
p
g
ρ
g
⋅
h
⋅
=
SG
ρ
H2O
⋅
h
⋅
=
For the upper surface
p
g
1.754
1.94
×
slug
ft
3
⋅
32.2
×
ft
s
2
⋅
2
3
×
ft
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 Fall '07
 Lear
 Fluid Mechanics

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