Problem 3.22
[Difficulty: 2]
Given:
Two fluid manometer contains water and kerosene.
With both tubes
open to atmosphere, the difference in free surface elevations is known
H
o
20 mm
⋅
=
SG
k
0.82
=
(From Table A.1, App. A)
Find:
The elevation difference, H, between the free surfaces of the fluids
when a gage pressure of 98.0 Pa is applied to the right tube.
Solution:
We will apply the hydrostatics equation.
Governing Equations:
dp
dh
ρ
g
⋅
=
(Hydrostatic Pressure  h is positive downwards)
ρ
SG
ρ
water
⋅
=
(Definition of Specific Gravity)
Assumptions:
(1) Static liquid
(2) Incompressible liquid
When the gage pressure
Δ
p is applied to the right tube, the water in the
right tube is displaced downward by a distance, l.
The kerosene in the
left tube is displaced upward by the same distance, l.
Under the applied gage pressure
Δ
p, the elevation difference, H, is:
h
H
A
B
l
l
H
0
H
1
A
B
Δ
p
HH
o
2l
⋅
+
=
Since points A and B are at the same elevation in the same fluid, their
pressures are the same.
Initially:
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 Fall '07
 Lear
 Fluid Mechanics

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