Problem 3.79
[Difficulty: 4]
Given:
Sphere with different fluids on each side
Find:
Resultant force and direction
Solution:
The horizontal and vertical forces due to each fluid are treated separately.
For each, the horizontal force is equivalent to that
on a vertical flat plate; the vertical force is equivalent to the weight of fluid "above".
For horizontal forces, the computing equation of Section 35 is
F
H
p
c
A
⋅
=
where A is the area of the equivalent vertical
plate.
For vertical forces, the computing equation of Section 35 is
F
V
ρ
g
⋅
V
⋅
=
where V is the volume of fluid above the curved
surface.
The data is
For water
ρ
999
kg
m
3
⋅
=
For the fluids
SG
1
1.6
=
SG
2
0.8
=
For the weir
D3
m
⋅
=
L6
m
⋅
=
(a) Horizontal Forces
For fluid 1 (on the left)
F
H1
p
c
A
⋅
=
ρ
1
g
⋅
D
2
⋅
⎛
⎝
⎞
⎠
D
⋅
L
⋅
=
1
2
SG
1
⋅
ρ
⋅
g
⋅
D
2
⋅
L
⋅
=
F
H1
1
2
1.6
⋅
999
⋅
kg
m
3
⋅
9.81
⋅
m
s
2
⋅
3m
⋅
()
2
⋅
6
⋅
m
⋅
Ns
2
⋅
kg m
⋅
⋅
=
F
H1
423 kN
⋅
=
For fluid 2 (on the right)
F
H2
p
c
A
⋅
=
ρ
2
g
⋅
D
4
⋅
⎛
⎝
⎞
⎠
D
2
⋅
L
⋅
=
1
8
SG
2
⋅
ρ
⋅
g
⋅
D
2
⋅
L
⋅
=
F
H2
1
8
0.8
⋅
999
⋅
kg
m
3
⋅
9.81
⋅
m
s
2
⋅
⋅
2
⋅
6
⋅
m
⋅
2
⋅
kg m
⋅
⋅
=
F
H2
52.9 kN
⋅
=
The resultant horizontal force is
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 Fall '07
 Lear
 Fluid Mechanics

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