fluid mechanics

1. 81
Find the height H in the shown figure for which the hydrostatic force on the
rectangular panel is the same as the force on the same semicircular panel
below.
H
R
C.G
C.G
Frect
Fsemi.
A
h
g
F
H
R
2
A
&
2
/
H
h





For a vertical rectangular body
For a vertical semi-circular body
A
h
g
F
2
/
R
A
&
3
/
R
4
H
h 2







 Free water surface
Solution
The hydrostatic force on the rectangular panel
2
.
rect H
R
g
)
H
R
2
(
)
2
/
H
(
g
A
h
g
F 








Similarly, the hydrostatic force on the semi-circular panel
2
2
.
rect H
R
g
)
2
/
R
(
)
3
/
R
4
H
(
g
A
h
g
F 










If the hydrostatic force on the rectangular panel is the same as the force
on the same semicircular panel
)
2
/
R
(
)
3
/
R
4
H
(
g
H
R
g 2
2







Worked Example
1

2. 82
Canceling g
 from both sides and rearranging gives
0
3
/
2
)
R
/
H
(
2
)
R
/
H
( 2





Solving for H/R gives
2
)
3
/
2
(
4
)
2
/
(
2
/
R
H
2





 





or
  R
1.918
3
/
2
4
/
4
/
R
H 2






 




 
A steel pipeline conveying gas has an internal diameter of 120 cm and an
external diameter if 125 cm. It is laid across the bed of a river, completely
immersed in water and is anchored at intervals of 3 m along its length.
 Calculate the buoyancy force in Newton per meter and the upward
force in Newton on each anchorage. Density of steel = 7900 kg/m3
,
density of water = 1000 kg/m3
.
Solution
Worked Example
2

3. 83
Upward Force
W
FB
D
i
n
=
1
2
0
c
m
t = 2.5 cm
Steel gas pipeline of S.G = 7.9
Waterline
3.0 m CL to CL
 The buoyancy force in Newton per meter, (the pipe is completely
immersed in water)
m
/
kN
033
.
12
0
.
1
4
25
.
1
81
.
9
1000
)
olume
displacedv
(
g
F
2
water
B









 








 Weight of the steel pipe =
m
/
kN
456
.
7
0
.
1
4
)
20
.
1
25
.
1
(
81
.
9
7900
)
Volume
(
g
2
2
steel









 








 The upward force on each anchorage =
m
/
kN
731
.
13
3
)
456
.
7
033
.
12
(
3
)
W
F
( B 





(The pipe is anchored at intervals of 3 m along its length)

4. 84
A vessel lying in a fresh-water dock has a displacement of 10 000 tones
and the area of the water-line plane is 1840 m2
. It is moved to a sea-water
dock and after removal of cargo its displacement is reduced to 8 500
tones. Assuming that the sides of the vessel are vertical near the waterline
and taking the density of the fresh water as 1000 kg/m3
and that of sea
water as 1025 kg/m3,
 Calculate the alternation in draft?
Solution
Case A: when the vessel is lying in a fresh-water dock
For equilibrium,
)
d
Area
(
g
displaced
volume
g
F
W B 







or
)
Area
(
g
F
W
d B




Thus, for a fresh displacement water of 10 000 tones and the area of the
water-line plane is 1840 m2
 The draft ”d” fresh water =
2
3
B
m
435
.
5
1840
81
.
9
1000
81
.
9
10
000
10
Area
g
W
F









Case B: when the vessel is lying in a sea-water dock
Worked Example
3

5. 85
For a displacement of sea-water of 8 500 tones and the same area of the
water-line plane of 1840 m2
 The draft ”d” sea water =
2
3
B
m
507
.
4
1840
81
.
9
1025
81
.
9
10
500
8
Area
g
W
F









 The alternation in draft = ”d” fresh water - ”d” sea water
= 5.435 – 4.507 = 0.928 m
A rectangular pontoon 10.5 m long, 7.2 m board and 2.4 m deep has a
mass of 70 000 kg. It carries on its upper deck a horizontal boiler of 4.8 m
diameter and a mass of 50 000 kg. The center of gravity of the boiler and
the pontoon may be assumed to be at their centers of figure and in the
same vertical line.
 Find the metacentric height (Density of sea water 1025 kg/m3
).
Solution
The effective center of gravity of the pontoon and the boiler can be
calculated by taking the moment about point “O” at the bottom,
Worked Example
4

6. 86
m
70
.
2
)
000
50
000
70
(
)
4
.
2
4
.
2
(
000
50
2
.
1
000
70
OG 






CL
d / 2 = 0.77
Sea water of
S.G=1.025
d
O
B
G
M
2.4 m
4.80
m
Waterline
M
G
B
O
2.79
2.70
7.2 m
W = 70 000 kg
W’= 50 000 kg
For equilibrium, B
F
W 
Thus,
The displaced volume of water = 2
B
m
07
.
117
81
.
9
1025
000
120
g
W
F





The draft, d = m
55
.
1
2
.
7
5
.
10
07
.
117
pontoon
the
of
Area
V



and
 The height of center of buoyancy above the bottom OB =
m
775
.
0
2
/
55
.
1
2
/
d 

 The distance m
79
.
2
07
.
117
)
12
/
2
.
7
5
.
10
(
V
I
BM
3





7. 87
 The metacentric height, GM
m
865
.
0
)
775
.
0
70
.
2
(
79
.
2
BG
BM
GM 





A rectangular pontoon 10m by 4m in plane weights 280kN is placed
longitudinally on the deck. A steel tube weighting 34 kN is in a central
position, the center of gravity for the combined weight lies on the vertical
axis of symmetry 250mm above the water surface. Find:
 The metacentric height,
 The maximum distance the tube may be rolled laterally across the
deck if the angle of heel is not to exceed 5o
.
Solution
W’ = 34 KN
water of S.G =1.0
d
O
B
G
M
Overturning
Moment = W’. X
0.025 m
X
Worked Example
5

8. 88
Total weight of the pontoon and the steel tube= 280 + 34 = 314 KN
For equilibrium,
B
F
W 
Thus,
The displaced volume of water = 3
3
B
m
32
81
.
9
1000
10
314
g
W
F






The draft, d = m
80
.
0
4
10
32
pontoon
the
of
Area
V



and
 The height of center of buoyancy above the bottom OB =
m
40
.
0
2
/
80
.
0
2
/
d 

 The distance m
67
.
1
32
)
12
/
4
10
(
V
I
BM
3




 The metacentric height, GM
m
02
.
1
)
40
.
0
25
.
0
(
67
.
1
BG
BM
GM 





 When the tube may be rolled laterally across the deck it causes
the pontoon to heel through an angle (the angle of heel is not to
exceed 5o
).For equilibrium in the heeled position, the righting
moment must equal the overturning moment



 MG
W
X
W'
or )
180
/
5
(
02
.
1
314
X
34 




from which X = 0.822 m

9. 89
 The maximum distance the tube may be rolled laterally across
the deck if the angle of heel is not to exceed 5o
 0.822m
A cylindrical buoy 1.35 m in diameter 1.8 m high has a mass of 770 kg.
The metacentric height,
 Show that it will not float with its axis vertical in sea water of
density 1025 kg/m.
 If one end of a vertical chain is fasten to the base, find the pull
required to keep the buoy vertical. The center of gravity of the
buoy is 0.9 m from its base.
Solution
Worked Example
6
1.80 m
1.35 m
d
O
B
G
O
B
G
M
d /2
h/2
water of S.G =1.025

10. 90
For equilibrium,
B
F
W 
Thus,
The displaced volume of water = 2
B
m
751
.
0
81
.
9
1025
81
.
9
770
g
W
F






The draft, d = m
524
.
0
4
/
3
.
1
751
.
0
pontoon
the
of
Area
V
2




and
 The height of center of buoyancy above the bottom OB =
m
262
.
0
2
/
524
.
0
2
/
d 

 The distance
m
217
.
0
524
.
0
16
35
.
1
d
16
D
d
)
4
/
D
(
)
64
/
D
(
V
I
BM
2
2
2
4











 The metacentric height, GM
m
421
.
0
)
262
.
0
90
.
0
(
217
.
0
BG
BM
GM 






Negative metacentric height indicates that the buoy is unstable
 If one end of a vertical chain is fasten to the base; and “T” = tension in
the chain in Newtowns,
 New buoyancy force
W
T
F'
B 


11. 91
 New displacement volume
3
'
B
'
B
m
81
.
9
1025
F
g
F





 New draft
m
393
14
F
)
4
/
35
.
1
(
81
.
9
1025
F
)
Area
(
g
F '
B
2
'
B
'
B










 New height of buoyancy above “O”
m
786
28
F
393
14
F
5
.
0
'
B
'
B




 m
F
5
.
639
1
)
393
14
/
F
(
16
35
.
1
d
16
D
d
)
4
/
D
(
)
64
/
D
(
V
I
BM '
B
'
B
2
2
2
4


























 '
B
'
B
F
5
.
639
1
786
28
F
90
.
0
OM
OG
G
M
and
m
90
.
0
OG
1.80 m
d
O
B
G
O
B
G
M
d/2
h/2
T
w
M
FB


12. 92
For equilibrium; taking the moment about G gives,






















 '
B
'
B
'
B
'
B
F
5
.
639
1
786
28
F
90
.
0
F
BG
F
90
.
0
T
or























 '
B
'
B
'
B
'
B
'
B
F
5
.
639
1
786
28
F
90
.
0
F
BG
F
)
W
F
(
90
.
0
Solving for '
B
F gives,
N
632
4
81
.
9
770
12186
T
and
N
12186
F'
B 




Consider a homogeneous right circular cylinder of height h, radius R, and
specific gravity SG, floating in water (SG = 1.0).
 Show that the body will be stable with its axis vertical if
)
SG
1
(
SG
2
h
R


Solution
Assume that the cylinder has a length L, radius R, specific gravity S.G and
floats stable in water with a draft d. Therefore, the water line area relative
to tilt axis =  R2
. Meanwhile,
Worked Example
7

13. 93
 The distance
d
4
R
d
R
4
/
R
V
I
BM
2
2
4








 …….. (1)
To find the relation between h (height of the cylinder) and the draft d, for
equilibrium it follows that,
B
F
W 
or
d
R
g
G
.
S
h
R
g
G
.
S 2
2













h
G
.
S
d 

 …….. (2)
and
2
/
d
OB
&
2
/
h
OG 

h
D
=
2R
d
O
B
G
O
B
G& M
d /2
h/2
water of S.G =1.0
Cylinder of of S.G
 The metacentric height MG is positive,
i.e. The distance, )
d
h
(
5
.
0
OG
OB
BG
MG 



 ………….... (3)
S.G = 1.0

14. 94
Substituting Eq. (2) into Eq. (3) gives
)
G
.
S
1
(
2
h
BG 


 …….. (4)
For stability the metacentric height GM = BM – BG must be positive, so
that BM must be greater than GM or,
From Eq. (1) and Eq. (4)
)
SG
1
(
2
h
h
G
.
S
4
R2







or
)
SG
1
(
G
.
S
2
h
R



A hollow wooden cylinder of S.G 0.55 has an outer diameter of 0.60 m, an
inner diameter of 0.30 m and has its ends open. It is required to float in oil
of S.G 0.84
 Calculate the maximum height of the cylinder so that it shall be
stable when floating with its axis vertical and the depth to which it
will sink.
Solution
Worked Example
8

15. 95
 The distance
d
16
)
D
D
(
4
/
)
D
D
(
d
64
/
)
D
D
(
V
I
BM
2
in
2
out
2
in
2
out
4
in
4
out 









d
16
45
.
0
d
16
30
.
0
60
.
0
BM
2
2



 …….. (1)
 The distance, BG = OB – OG = )
d
h
(
5
.
0 
 …….. (2)
To find the relation between h (height of the cylinder) and the draft d,
For equilibrium,
B
F
W 
or
d
4
/
)
D
D
(
g
h
4
/
)
D
D
(
g 2
in
2
out
oil
2
in
2
out
wood 












d
527
.
1
d
h
wood
oil





 …….. (3)
Eq. (2) and (3) gives
d
264
.
0
)
0
.
1
527
.
1
(
d
5
.
0
BG 



 …….. (4)
For stability the metacentric height GM = BM – BG must be positive, so
that BM must be greater than GM or,
From Eq. (1) and Eq. (4)
d
264
.
0
d
16
45
.
0



16. 96
h
d
O
B
M &G
S. G = 0.84
S. G = 0.55
0.60 m
0.15 m
Neutral equilibrium
 The depth to which the cylinder will sink m
327
.
0
d  and,
 The maximum height of the cylinder so that it shall be stable when
floating with its axis vertical
m
50
.
0
327
.
0
527
.
1
d
527
.
1
h 




A pontoon is to be used as a working platform for diving activities
associated with a dockyard scheme. The pontoon is to be rectangular in
both plane and elevation, and is to have the following specification:
 Width = 6.0 m,
 Mass = 300 000 kg,
 Metacentric height  1.50 m,
Worked Example
9

17. 97
 The pontoon center of gravity = 0.30 m above geometrical center,
 Freeboard (height from water level to deck)  750 mm.
 Estimate: The overall length, L, and the overall height, h for the
pontoon if it is floating in fresh water (density = 1000 kg/m3
)
Solution
 The given specifications:
 The center of buoyancy, B, is at the center of gravity of the
displaced water = 2
/
d
OB 
 The pontoon center of gravity = 0.30 m above geometrical center.
i.e.
m
30
.
0
2
/
h
OG 
 and )
m
(
75
.
0
d
h 
 at least
m
675
.
0
30
.
0
)
d
75
.
0
d
(
5
.
0
30
.
0
)
d
h
(
5
.
0
BG 










 m
175
.
2
50
.
1
675
.
0
GM
BG
BM 



 …………………….… (1)
 The volume of water displaced =
Mass /  = 300 000 / 1000 = 300 m2
and
L
18
12
/
6
L
12
/
B
L
I 3
3





 The distance L
06
.
0
300
L
18
V
/
I
BM 


 …………… (2)

18. 98
6.0 m
d
Freeboard
 0.75 m
B
G
O
L
=
?
?
h
=
?
?
M
Metacentric height = 1. 50 m
Mass =
300 000 kg
S.G. = 1.0
CL
CL
From Eq. (1) and (2),
The overall length, L = 2.175 / 0.06 = 36.25 m
The draft of the pontoon = )
Length
width
(
/
ed
terdisplac
volumeofwa 
m
38
.
1
)
25
.
36
6
(
/
300
)
L
B
(
/
V 



The overall height, h for the pontoon
)
m
(
15
.
2
75
.
0
38
.
1
75
.
0
d
h 





19. 99
A ship has displacement of 5000 metric tons. The second moment of the
waterline section about a force and aft axis 12 000 m4
and the center of
buoyancy is 2 m in below the center of gravity. The radius of gyration is
3.7 m.
 Calculate the periodic time of oscillations. (Sea water has a
density of 1025 kg/m3
).
Solution
Given:
Displaced volume = 5000 metric tons, I = 12 000 m4, the distance
m
0
.
2
BG  and the radius of gyration is 3.7 m.
 The displaced volume of water = 2
B
m
4878
81
.
9
1025
81
.
9
5000
g
W
F






 The distance m
46
.
2
4878
000
12
V
I
BM 


 The metacentric height, GM
m
46
.
0
0
.
2
46
.
2
BG
BM
GM 





 the periodic time of oscillations ,t
.
sec
94
.
10
81
.
9
46
.
0
70
.
3
2
g
GM
k
2
2
2







Worked Example
10

20. 100
A solid cylinder 1.0 m in diameter and 0.8 m high is of uniform relative
density 0.85.
 Calculate the periodic time of small oscillations when the cylinder
floats with its axis vertical in still water.
(Test your understanding)
Solution
Worked Example 10

21. 101
Pressu
re
Absolute
zero
Vacuum or Negative gauge pressure (2)
Absolute
Pressure (1)
Absolute
Pressure (2)
Gauge Pressure (1)
"1"
"2"
Atmosphe
ric
Pressure

22. 102

23. 103
B/2 B/2
X
y
a/2
a/2
a
B
A 

12
B
a
I
,
12
a
B
I
3
yc
3
xc




X
y
R
2
R
A 

4
R
I
I
4
yc
xc



Circle Rectangular
c c
X
y
d
a
(B +d)/ 3
b
a/3
2
b
a
A 
36
a
b
I
3
xc 
c
Triangular
X
y
y
R
4
R
A
2


4
yc
xc R
0549
.
0
I
I 

R

3
R
4

3
R
4
X
2
R
A
2


R
3927
.
0
I
,
R
1098
.
0
I 4
yc
4
xc 


3
R
4
Semicircle Quarter circle
c
c
Fluid “1”
1.8 m
1.5 m
2.1 m
0.9 m
Fluid “2”
Fluid “3”
Worked Example

24. 104
D
B
C
Fluid “1”
1.8 m
1.5 m
2.1 m
0.9 m
Fluid “2”
Fluid “3”
A
Worked Example
Fluid “1”
1.8 m
1.5 m
2.1 m
0.9 m
Fluid “2”
Fluid “3”
Worked Example
D
B
C
Fluid “1”
1.8 m
1.5 m
2.1 m
0.9 m
Fluid “2”
Fluid “3”
A
Worked Example

25. 105