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Problem 3.38
[Difficulty :2]
Fluid 1
Fluid 2
σπ
D
cos
θ
ρ
1
g
Δ
h
π
D
2
/4
Given:
Two fluids inside and outside a tube
Find:
(a) An expression for height
Δ
h
(b) Height difference when
D
=0.040 in for water/mercury
Assumptions:
(1) Static, incompressible fluids
(2) Neglect meniscus curvature for column height and
volume calculations
Solution:
A freebody vertical force analysis for the section of fluid 1 height
Δ
h
in the tube below
the "free surface" of fluid 2 leads to
F
∑
0
=
Δ
p
π
D
2
⋅
4
⋅
ρ
1
g
⋅
Δ
h
⋅
π
D
2
⋅
4
⋅
−
π
D
⋅
σ
⋅
cos
θ
()
⋅
+
=
where
Δ
p
is the pressure difference generated by fluid 2 over height
Δ
h
,
Δ
p
ρ
2
g
⋅
Δ
h
⋅
=
Hence
Δ
p
π
D
2
⋅
4
⋅
ρ
1
g
⋅
Δ
h
⋅
π
D
2
⋅
4
⋅
−
ρ
2
g
⋅
Δ
h
⋅
π
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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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