Problem *3.93
[Difficulty: 2]
Given:
Geometry of steel cylinder
Find:
Volume of water displaced; number of 1 kg wts to make it sink
Solution:
The data is
For water
ρ
999
kg
m
3
⋅
=
For steel (Table A.1)
SG
7.83
=
For the cylinder
D
100 mm
⋅
=
H1
m
⋅
=
δ
1mm
⋅
=
The volume of the cylinder is
V
steel
δ
π
D
2
⋅
4
π
D
⋅
H
⋅
+
⎛
⎜
⎝
⎞
⎟
⎠
⋅
=
V
steel
3.22
10
4
−
×
m
3
⋅
=
The weight of the cylinder is
WS
G
ρ
⋅
g
⋅
V
steel
⋅
=
W
7.83
999
×
kg
m
3
⋅
9.81
×
m
s
2
⋅
3.22
×
10
4
−
×
m
3
⋅
Ns
2
⋅
kg m
⋅
×
=
W
24.7N
=
At equilibium, the weight of fluid displaced is equal to the weight of the cylinder
W
displaced
ρ
g
⋅
V
displaced
⋅
=
W
=
V
displaced
W
ρ
g
⋅
=
24.7 N
⋅
m
3
999 kg
⋅
×
s
2
9.81 m
⋅
×
kg m
⋅
2
⋅
×
=
V
displaced
2.52L
=
To determine how many 1 kg wts will make it sink, we first need to find the extra volume that will need to be dsiplaced
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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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