Fluid Mechanics - Buoyancy
Short Description
Fluid mechanics for students...
Description
BUOYANCY; PRINCIPLE OF ARCHIMEDES W
L.S. (γ,
N m3
)
Body of volume, V
F b submerged ged in a liquid liquid of specifi specificc weight weight γ is buoyed buoyed Principle of Archimedes: “ a body submer up by a force force equal to the weight of the displaced liquid,”
F b V γ where where:: V = volum volume e of of the submer submerged ged body body or volum volume e displa displaced ced liquid liquid γ = spec specifi ificc weig weight ht of the liquid liquid Note: F b is called the “ buoyant force” and its direction is vertically upward.
Actual Weight of a Body ( Weight in Air )
W V B γ b V B s B γ w where:
volume of the body V B γ B specific weight of the body s B specific gravity of the body γ w specific weight of water
Apparent Weight of a Body (Weight in Liquid) W’ = W - Fb
Flotation; Flotation; Stability of Floating Bodies W G O
Bo
F b
(a) The body is in in Uprigh Uprightt Positi Position on W F b V γ
G = Center of gravity gravity of the body Bo = Center Center of buoyancy buoyancy ( centroid centroid of the submerg submerged ed portion )
( b) The body is in Tilted M (Metacenter)
Position
W x
B A
G
A’ B’
O
B
B1
o
F b
θ r
B1 = new center of Buoyancy
r = horizontal shifting of Bo x = moment arm of W or F b
θ = angl angle e of of tilt tilt
(b) Tilted Position M (Metacenter) W x G
O
B A
B1
The Righting or Overturning Couple, C
_____ C W x W MG sin θ
A’ B’
Bo
θ r
NOTE: C is a righting moment if M falls above G, an overturning moment if M falls below G. MG is known as the “metacentric height”.
M (Metacenter)
W
x
F b
G O
B A
B1
F b
A’ B’
Bo
F b
θ r S
The shifting of the original upward buoyant force F b in the wedge A’OB’ to F b in the wedge AOB causes a shift in F b from Bo to B1 , a horizontal distance r Hence,
F b r F b S
γ V r γν S
ν
r
V
S
_____
Also,
_____
Then,
Note:
r MBo sin θ
_____
ν
MBo sin θ _____
MBo
V
S
ν S
r MBo sinθ
an d F b ν γ
Where ν is the additional volume AOB. S is the distance between the centroids of AOB and A’OB’.
V sin θ
For small angle θ, _____
MBo
where:
(approximat e y ) V θ
ν = volume of the wedge A’OB’ V = volume of the submerged body S = horizontal distance between the centroid of A’OB’ and AOB θ = angle of tilt
______
Metacentric Height, MG _____
_____
_____
MG MBo GBo
+ if Bo is above G - if Bo is below G
_____
NOTE: GB o is usually a known value _____
If θ is ne li ible MB is iven as _____
MBo
I o V
Where: I o is the moment of inertia of the waterline section relative to a line through O.
_____
Derivation of MB o
I o M (Metacenter) W
V
Consider now a small prism of the wedge AOB, dA at a distance x from O, having a horizontal area dA. For small angles the length of this prism = xθ x (approximately). The buoyant force produced G O B By this immersed prism is γ x θ d A , an d A 2 The moment of this force about O is γ x θ d A . B B1 The sum of all these moments for both wedges o F b Must be e ual to νS or
γθ
x 2 d A γν S γ Vr
F b A’ B’
F b
_____
But for small angles r MBo θ ( approximat ely ) Hence _____
θ
x d A V MBo 2
r S
2 x d A is the moment of inertia, I o , of the water-line section about the longituBut dinal axis through O (approximately constant for small angles of heel). Therefore
_____
MB o
I o V
,
The metacentric height
_____
_____
_____
MG MBo GBo
VESSEL WITH RECTANGULAR SECTION (B/2) (cosθ)
M (Metacenter)
F b
x
B A
(B/2)(secθ)
G O
B A’ B’
A
B/2 __
B1
F b
Bo
θ r S
x
Recall: _____
B
O
θ
MBo where:
S
2
ν S V sin θ
V BDL 1 B B v tan θ L 2 2 2 1 2 v B L tan θ 8
Considering triangle AOB
Multiplying both sides by 2 we obtain,
(B/2) (cosθ) (B/2)(secθ) B
O
θ B/2
x
S
cos θ secθ
3 B 1 s cos θ 3 cosθ
A __
B
s
s
B cos θ 1 2
3 cosθ
Then from the formula, From geometry, the centroid of the triangle is defined by the coordinates of the vertices: __
x
s
2
_____
MBo
x1 x2 x3
0
3 B
2
cos θ 3
B
2
sec θ
_____
MBo
ν S V sin θ
2 cos θ 1 1 2 B B L tan θ 8 3 cosθ
BDL sin θ
_____
MBo
2 cos θ 1 1 2 B B L tan θ 8 3 cosθ
BDL sin θ
_____
MBo _____
3
B L sin θ _____
MBo 2 4 _____
MBo _____
MBo _____
MBo
B
2 cos θ 1
cosθ cosθ BDL sin θ
1 1 2 2 4 D cos θ 2
2 4 D
MBo
1 sec θ 2
2
2 4 D B
2
2
θ 1
2 tan θ 2
2 4 D B
1 tan
2
12 2 D
2 tan θ 2
2 2 B tan θ 1 MBo 12 D 2 _____
2
B
_____
cos 1
2 4 D cos2 θ B
MBo
B
SAMPLE CALCULATIONS Position of weight on the mast M _____ _____ G (center of gravity) MB GB __
Y h
Y 1
__
Y
O
MB
h
2 h
B
h
_____
2
Sketch showing various distances on the pontoon
DETERMINATION OF THEORETICAL METACENTRIC HEIGHT FROM THE GEOMETRY OF THE PONTOON:
I o
Lb 3
12
400mm 2 00mm3 12
Displaced Volume
V
W
ρ g
2 . 52 kg
1, 000
kg m3
0. 40m0. 2 0m3 2 . 67 x104 m4 12
9 . 81 N kg
9 . 81
m s2
2 . 52 x10 3 m 3
_____
MB
I V
2 . 67 x 10
4
2 . 52 x 10
3
m
4
m
3
0 . 1059 m 106 mm
Depth of displaced water 3 4 2 . 52 x10 m
V
0 . 0315 m h Lb 0. 4m 0. 2 0m
_____
The center of buoyancy force below the water surface and the distance OB will be _____
OB
h 0 . 0315 m
0 . 01575 m 15 . 75 mm _____
The Metacenter is above the water surface and distance MO is _____
_____
____
MO MB OB
106 15 . 75
90 . 2 5 mm
In the case when the height of the mast, Y 1
100__mm an d th e
height of the center of gravity ( by experiment) , Y 69 mm _____
Thus, the theoretical metacentric height MGth _____
_____
_____
Position of weight on the mast M (Metacenter) _____ _____ G (center of gravity) MB GB
MGth MB GB
__ h MGth MB Y 2 _____
_____
.
106 69 2 37. 02 mm _____
__
Y
Y h
__
Y
h
B
h
2 h 2
Because MG th is positive, this shows that the pontoon is stable.
_____
MB
Determination of Metacentric Height by Experiment M x
G w O
G d
w
B 1
B
θ
F b W
_____
The metacentric height is determined experimentally as shown in the figure above. When shifting the jockey weight w to the left side of the pontoon at a distance x, the pontoon tilts to a small angle θ causing the metacentric height to rotate slightly around the longitudinal axis of the pontoon . Likewise, the buoyancy force F b shifted a horizontal distance d from G. Hence, the moment produced by w must be equal to _____ moment of F b , w x MG
wcosθ x d W
_____ MG sin θ W
MG
W tan θ
w W
x
θ
(For small angle of tilt)
METACENTRIC HEIGHT APPARATUS
Mast Vertical scale
Vertical sliding weight oc ey we g Balancing weight
Tilt angle scale Plumb bob
Pontoon
_____
DETERMINATION OF METACENTRIC HEIGHT, MG BY EXPERIMENT Typical Data: In the case of vertical sliding weight on the mast is at the height, Y 1
100 mm .
Distance of jockey weight w from center of pontoon , x = 80 mm Angle of tilt, θ = 6.80˚ Convert angle of tilt into radian 6. 80
π 180
6. 80 0. 11868radian
Then, From equation (2), the experimental metacentric height is,
x
θ
80 mm 0 . 11868 radian
_____
MG
ex p
w W
x
θ
674 . 06 mm 0 . 2 0 kg 2 . 52 kg
674 . 06 mm 53 . 49 mm
_____
MG ex p is positve, this shows that the pontoon at that tilt angle is stable.
TEST PROCEDURES: Data recording: - Pontoon weight,
W = 2.50 kg
- Jockey weight,
w = 0.20 kg
- Adjustable vertical weight
= 0.40 kg
- Pontoon width,
D = 200 mm
- Pontoon length,
L = 400 mm
Determining the Center of Gravity of the Pontoon
Center of gravity ( CG)
Scale
Adjustable vertical weight Mast
Procedures: 1. Tilt the pontoon as shown in figure. 2. Attach the plum bob on the angle scale. 3. Move or adjust the vertical weight to a required distance and record that distance from the scale on the mast. 4. Place knife edge support under the mast and move it to a position of equilibrium and record the height ( center of gravity) where the knife edge is position on the scale.
Taking Readings with the Pontoon in a Water Tank 1. Initial Set Up When placing the pontoon in the water ensure that the position of the jockey weight horizontal adjustments is in the middle of the pontoon and the pontoon is sitting level in the water. The pontoon should be in a vertical position and have no angle of tilt ( zero degrees in the tilt angle scale). If not, adjust the balancing weight until the angle of tilt is “0”.
2. The jockey weight can change the position of the pontoon in the water and in order to take some experimental readings we move the jockey weight in steps from its central position horizontally and record the tilt angle of the pontoon from the scale on the pontoon in degrees.
3. Each time we move the jockey weight from its central position we must record on the data sheets supplied the distance measured from its central position and the angle of tilt. 4. We also change the adjustable vertical weight height on the mast and record its measurement along with the jockey weight distance from its central position, the angle of tilt at different values and record all the data on the sheets provided. 5. Step (3) and (4) can be repeated many times to obtain a satisfactory conclusion.
SAMPLE DATA SHEET METACENTRIC HEIGHT APPARATUS Position of jockey weight in a horizontal position (cm.) 2
4
6
8
10
12
14
16
18
Distance x of the jockey weight measured from the center of the pontoon (mm) 80
Height of weight on the Tilt Angle mast, θ ____mm. (degrees)
x/θ
Height of (mm/rad.) center of Metacentr gravity, ic Height (mm) ____mm.
60
40
20
0
20
40
60
80
Example 1. An iceberg weighing 8.95 kN/m3 floats in sea water, γ = 10.045 kN/m3, with a volume of 595 m 3 above the surface. What is The total volume of the iceberg? Solution:
W
kN
kN V 8 . 95 3 m 3 m
V s 10 . 045
bu t W.S.
3
.
1 . 095 V
Let Vs = volume submerged Fb V = total volume (a) ∑Fv = 0, Fb = W Vsx γw = V x γi where: γw = specific weight of sea water γi = specific weight of iceberg
kN m3
V s V 595 m kN m 3
.
3
kN m3
5 , 976 . 775 kN
3 V 5 , 458 . 2 42 m
Example 2. A sphere 0.90 m in diameter floats half submerged in a tank of oil ( s=0.80). (a) What is the total vertical pressure on the sphere? (b) What is the minimum weight of an anchor weighing 24 kN/m3 that will be required to submerge the sphere completely? Solution: O.S.
W
W 0.45 m
0.45 m
O.S.
Wa
Figure (a)
Figure (b)
Fb F ba
(a) Consider Figure (a)
F v 0 , F v W 0 F v W V γ
kN 2 3 π 0 . 45 m 0 . 80 x 9 . 81 3 m 3
W
W 1 . 498 kN
O.S.
W V a 0 . 093 m 3
0.45 m
therefore Fb Wa
W a V a γ a
Figure (b)
F ba
W a 0 . 093 m 2 4 3
kN 3
m
W a 2 . 2 32 kN
(b) Consider Figure (b),
F v 0, F b F b a W W a 0,
V γ V a γ 1. 498 kN V a γ a 0 kN kN 4 kN 3 π 0 . 45 m 0 . 80 x 9 . 81 3 V a 0 . 80 x 9 . 81 3 1 . 498 kN V a 2 4 3 0 m m m 3
16 . 152
kN m
3
V a 1 . 498 kN
Example 3. A cylinder weighing 500 N and having a diameter of 0.90 m floats in salt water ( s=1.03) with its axis vertical as shown in the figure. The anchor consists of 0.30 m 3 of concrete weighing 24 kN/m3. What rise in tide r, will be required to lift the anchor off the bottom? W
Solution: new W.S.
r
Fb Wa
F ba
0.30 m
F v 0, F b F ba W W a 0
V γ V aγ 500 N V aγ a 0
W
new W.S.
r
Fb
0.30 m
Wa
F ba V γ V aγ 500 N V aγ a 0
π 4
0. 90 m 2 0. 30 m r 1. 03 x9810
N
N 3 0 . 3 1 . 03 9810 m x 3 m 3 m
N 500 N 0 . 3m 3 2 4 , 000 3 0 m r 0 . 42 6 m
Example 4. Timber AC hinged at A having a length of 10 ft., cross sectional area of 3 in.2 and weighing 3 lbs. Block attached to the end C having a volume of 1 ft.3, and weighing 67 lbs. Required: Angle θ for equilibrium. (2) Buoyant force on the block, csc θ A
F b B V γ w 1 62 . 4 62 . 4lb
10 – csc θ
1’
10’ WT
θ
5cosθ 1 10 10 csc θ 2
C B=
bT cos θ
10 csc θ cos θ 2
F b B
10cosθ
Solution: (1) Buoyant force on the timber,
F bT V γ w
3
10 cscθ 62 . 4
144 F bT 1. 310 cscθ
s
csc θ
csc2 θ 6. 15
A 10 – csc θ
1’
cscθ 2 . 48 WT
5cosθ
θ
F bT
1 10 10 csc θ cos θ 2 10 csc θ cos 2
sin θ 0. 40
C
θ 2 3. 8
WB=67 lbs
b B
10cosθ
(3) ∑MA = 0,
10 cscθ cosθ F b 10 cosθ 0 2 10 cscθ 35 6710 1. 310 cscθ 62 . 410 0 2 15 670 0. 65100 csc2 θ 62 4 0
W T 5 cosθ W B 10cosθ F bT
B
Example 5. A vessel going from salt into fresh water sinks two inches, then after burning 112,500 lb of coal, rises one inch. What is the original weight of the vessel? W
W
W – 112,500 lb
d + 2/12
d + 1/12
d
Fb (a) Salt water ( γ = 64 lb/ft3)
Fb
Fb
(b) Fresh water (γ = 62.4 lb/ft3)
(c ) Fresh water after losin 112 500 lb
Solution: 1. In figure (a), submerged volume is, Va = Axd ft3 where: A = cross-sectional area of the vessel ( ft2 ) 2. In figure (b), submerged volume is Vb = Va + (2/12)(A) 3. In figure (c), submerged volume is Vc = Va + (1/12)(A)
W
W
W – 112,500 lb
d + 2/12
d + 1/12
d
Fb (a) Salt water ( γ = 64 lb/ft3)
Fb
Fb
(b) Fresh water (γ = 62.4 lb/ft3)
(c ) Fresh water after losing 112,500 lb
4. In salt water, W = F W = Va ( 64 )
( 1)
5. In fresh water, W = Fb W = [Va + (2/12)(A)](62.4) 6. In fresh water after losing 112,500 lb, W – 112,500 = [Va +(1/12)(A)](62.4)
(2)
(3)
W 1 A 62 . 4 0. 975W 10 . 4 A 64 6 (4) 0. 02 5W 10 . 4 A
7. Substitute eq. 1 to eq. 2 and eq. 3, W
W 1 A 62 . 4 0. 975W 5 . 2 A 64 12
W 112 , 500
0 . 02 5 W 5 . 2 A 112 , 500
8. Solve eqs. (4) and (5) simultaneously, we obtain W 9 x10 lb 6
(5)
Example 6. A ship of 4,000 tons displacement floats in sea water with its axis of symmetry vertical when a weight of 50 tons is midship. Moving the weight 10 feet towards one side of the deck causes a plumb bob, suspended at the end of a string 12 feet long, to move 9 inches. Find the metacentric height. Solution:
9 1. Solve the angle of tilt, θ A rc tan 12 12
3. 58
12’
2. Righting Moment = W (MG x sinθ) θ
50 x10 4050 MGxsin 3. 58 9”
MG 1 . 977 ft
Example 7. A ship with a horizontal sectional area at the waterline of 76,000 ft2 has a draft of 40.5 ft. in sea water (γs =64 lb/ft3). In fresh water it drops 41.4 ft. Find the weight of the ship. With an available depth of 41 ft. in a river above the sills of a lock, how many long tons of the cargo must the ship be relieved off so that it will pass the sills with a clearance of 0.60 ft.? Solution: W
W’
W original fresh W.S
sea W.S
fresh W.S
1 ft
ΔV
ΔV1 0.90 ft 40.5 ft
41.4 ft
41 ft
41.4 ft
sills
F bs (a) SEA WATER
F b f (b) FRESH WATER
F b ' f
0.60 ft
(c) SHIP IN THE LOCK
W – original weight of the ship (including cargo) W’ – new weight of the ship when part of the cargo has been disposed F bs - buoyant force in sea water, F b f - buoyant force in fresh water
W
W’
W original fresh W.S
sea W.S
fresh W.S
ΔV2
1 ft
ΔV1 0.90 ft 41.4 ft
40.5 ft
41 ft
41.4 ft
sills
F bs F b ' f W '
(a) SEA WATER
F b f
F b ' f
0.60 ft
(b) FRESH WATER
(c) SHIP IN THE LOCK
ΔV1 = additional submerged volume = 0.90(76,000) = 68,400 f t3 ΔV2 = volume of the ship at the waterline which rose up when it was relieved off the cargo = 1 (76,000) = 76,000 ft 3 V = original volume submerged ( in sea water ) (1) Using position (a); (2) Using position (b)
F bs W
F b f W
V γ s W
V V 1 γ f W
V 64 W
(1)
V 68, 400 62 . 4 W
(2)
(3) Solve equations (1) and (2) simultaneously, 64V 62 . 4V 68, 400 V 2 , 667, 600 ft
3
and
The ship’s displacement in sea water
V V 1 2 , 736, 000 ft
3
Therefore,
W 64V
The ship’s displacement in fresh water W’
1. 7072 6 x10 lb 8
original fresh W.S
or
W 62 . 4V V 1 1 . 7072 6 x10
8
ΔV2
1 ft
lb
41.4 ft
(4) Using position (c);
41 ft
W ' F b ' f V V 1 V 2 62 . 4
2 , 736, 000 76, 000 62 . 4
sills
1 . 65984 x10 lb 8
0.60 ft
(5) Weight of disposed cargo = W – W’
(c) SHIP IN THE LOCK
W W ' 0 . 04742 x10 8 lb
4 , 742 , 000 lb 4 , 742 , 000 lbx
1 LT 2 , 2 00 lb
F b ' f
2 ,155 . 5 LONG TONS
Example 8.
W.S
W G
A
4’
A
Bo
9’
8’
4’ 15’
Fb
W.S
A’
Required : Righting or overturning moment when one side, as shown, is at the point of submergence
15’
M
A
G O Bo
Given: Rectangular scow 50’ x 30’ x 12’ as shown with the given draft and center of gravity.
C W MG sin θ
where: W V γ
A’ θ A
4’
W 50 30 8 62 . 4 W 748 , 800 lb
4 14 . 93 15 MG MBo GBo θ A rc tan
MG MBo 5
MBo
vS V sinθ
2 15 450 30 2 3 MBo 50308 sin14. 93
S
W.S
G O
A’
Bo
Then,
C 748 , 800 ( 4 . 7 sin 14 . 93 ) C 906 , 600 ft lb
30'
Θ = 14.93°
A’ A
4’
8’
MBo 9. 7 ft
MG 9. 7 ft 5 ft 4. 7 ft
3
A
1
Therefore,
2
30’
Example 9. A rectangular scow 10 m wide, 16 m long, and 4.0 m high has a draft in sea water (s = 1.03) of 2.5 m. Its center of gravity is 2.80 m above the bottom of the scow. Determine the following: (a) The initial metacentric height, (b) The righting or overturning moment when the scow tilts until one side is just at the point of submergence. Solution:
MBo 3. 333m G
S.W.
GBo 2 . 80m 1. 2 5m .
Bo
2.5 m
2.8 m 1.25 m
GBo 1. 55 m The initial metacentric height , MG
10 m
MG MBo GBo
(a) Initial Metacentric Height 2 2 B tan θ 1 MBo 12 D 2
where θ = 0
tan2 0 1 MBo 12 2 . 5 2 102
MG 3. 333 1. 55
MG 1. 783m
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