Problem 1.35
[1]
Given:
"Engineering" equation for a pump
Find:
SI version
Solution:
The dimensions of "1.5" are ft.
The dimensions of "4.5
x
10
5
" are ft/gpm
2
.
Using data from tables (e.g. Table G.2), the SI versions of these coefficients can be obtained
1.5 ft
⋅
1.5 ft
⋅
0.0254 m
⋅
1
12
ft
⋅
×
=
0.457 m
⋅
=
4.5
10
5
−
×
ft
gpm
2
⋅
4.5 10
5
−
⋅
ft
gpm
2
⋅
0.0254 m
⋅
1
12
ft
⋅
×
1 gal
⋅
4 quart
⋅
1quart
0.000946 m
3
⋅
⋅
60 s
⋅
1min
⋅
⎛
⎜
⎝
⎞
⎟
⎠
2
×
=
4.5 10
5
−
⋅
ft
gpm
2
⋅
3450
m
m
3
s
⎛
⎜
⎝
⎞
⎟
⎠
2
⋅
=
The equation is
Hm
( )
0.457
3450 Q
m
3
s
⎛
⎜
⎝
⎞
⎟
⎠
⎛
⎜
⎝
⎞
⎟
⎠
2
⋅
−
=
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View Full DocumentProblem 3.10
[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
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 Spring '09
 cumming
 slug ft

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