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Problem 3.18
[Difficulty: 2]
Given:
Data on partitioned tank
Find:
Gage pressure of trapped air; pressure to make water and mercury levels equal
Solution:
The pressure difference is obtained from repeated application of Eq. 3.7, or in other words, from Eq. 3.8.
Starting
from the right air chamber
p
gage
SG
Hg
ρ
H2O
×
g
×
3m
⋅
2.9 m
⋅
−
()
×
ρ
H2O
g
×
1
×
m
⋅
−
=
p
gage
ρ
H2O
g
×
SG
Hg
0.1
×
m
⋅
1.0 m
⋅
−
×
=
p
gage
999
kg
m
3
⋅
9.81
×
m
s
2
⋅
13.55
0.1
×
m
⋅
1.0 m
⋅
−
×
Ns
2
⋅
kg m
⋅
×
=
p
gage
3.48 kPa
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics

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