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Problem 3.14
[Difficulty: 3]
Air
H
D
Air
H – y
y
y
Given:
Cylindrical cup lowered slowly beneath pool surface
Find:
Expression for y in terms of h and H.
Plot y/H vs. h/H.
Solution:
Governing Equations:
dp
dh
ρ
g
⋅
=
(Hydrostatic Pressure  h is positive downwards)
pV
⋅
MR
⋅
T
⋅
=
(Ideal Gas Equation)
Assumptions:
(1) Constant temperature compression of air inside cup
(2) Static liquid
(3) Incompressible liquid
First we apply the ideal gas equation (at constant temperature) for the pressure of the air in the cup:
⋅
constant
=
Therefore:
⋅
p
a
π
4
⋅
D
2
⋅
H
⋅
=
p
π
4
⋅
D
2
⋅
Hy
−
()
⋅
=
and upon simplification:
p
a
H
⋅
pH y
−
⋅
=
Now we look at the hydrostatic pressure equation for the pressure exerted by the water.
Since
ρ
is constant, we integrate:
pp
a
−
ρ
g
⋅
hy
−
⋅
=
at the waterair interface in the cup.
Since the cup is submerged to a depth of h, these pressures must be equal:
p
a
H
⋅
p
a
ρ
g
⋅
−
⋅
+
⎣
⎦
−
⋅
=
p
a
H
⋅
p
a
y
⋅
−
ρ
g
⋅
−
⋅
−
⋅
+
=
Explanding out the right hand side of this expression:
0p
a
−
y
⋅
ρ
g
⋅
−
⋅
−
⋅
+
=
ρ
g
⋅
h
⋅
H
⋅
ρ
g
⋅
h
⋅
y
⋅
−
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 Fall '07
 Lear
 Fluid Mechanics

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