CE 1022 Fluid Mechanics Buoyancy - Lecture Note -
BUOYANCY
1. 1 Introduction A fluid exerts a force on any object submerged in it. Such a force due to a fluid in equilibrium is known as the buoyancy or the upthrust. It is often necessary to determine buoyancy in many engineering applications as in the design of ships, boats, buoys etc. The buoyancy has a magnitude equal to the weight of the displaced volume of fluid. It acts upwards through the centre of gravity of the displaced volume of fluid which is known as the centre of buoyancy. This result is often known as the Archimedes principle and can be proved as follows: 2
When an object is submerged in a fluid in equilibrium an equal volume of fluid is displaced. This volume of fluid was in equilibrium under the action of its own weight and the resultant thrust exerted on it by the surrounding fluid which is the same as the buoyancy on the submerged object. Hence the buoyancy should be equal in magnitude to the weight of the displaced volume of fluid and act upwards through its centre of gravity. Alternatively, considering the object ABCD submerged in a fluid in equilibrium as shown in Figure 1, Horizontal thrust on surface ABC = Thrust on left hand side of AC = (say)
surface ADC= Thrust on right hand side of AC = F (say) Total horizontal thrust on ABCD
=0
F
E A
D
B
C Figure 1 3
Downward thrust on surface BAD = Weight of fluid volume BADEF Upward thrust on surface BCD = Weight of fluid volume BCDEF Total upward thrust on ABCD = Weight of fluid volume (BCDEF-BADEF)
= Weight of fluid volume ABCD = Weight of the displaced volume of fluid
F
E A
This result is not restricted D
B
to a fluid of uniform density. C Figure 1
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On a body of volume V, submerged in a fluid of density ρ in equilibrium, the buoyancy U = Vρg, where g = Acceleration due to gravity, and acts through the centre of buoyancy B as shown in Figure 2.
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The position of the center of buoyancy B depends on the shape of the displaced volume of fluid. For a fluid of uniform density, it is at the centroid of the displaced volume of fluid. It should be distinguished from the centre of gravity G of the submerged object, the position of which depends on the way its weight W is distributed as illustrated in Figure 3.
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1.2
Bodies Submerged in Two Immiscible Fluids
A body submerged in two immiscible fluids of densities ρ1 and ρ2 (>ρ1) is shown in Figure 4. The buoyancy U1 on the volume V1
submerged in the upper fluid acts through its centre of buoyancy B1 and the buoyancy U2 on the volume V2 submerged in the lower fluid acts through its centre of buoyancy B2 as shown in Figure 4. Total buoyancy U = U1+U2 = V1ρ1g + V2ρ2g
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U does not pass through the centroid of the volume V1+V2. If the upper fluid is a gas and the lower fluid is a liquid, the buoyancy due to the gas is neglected in many engineering applications. Also, when a body is weighed, the buoyancy due to
atmospheric air is usually neglected.
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1.3 Equilibrium of Submerged Bodies 1.3.1 Fully Submerged Bodies The forces acting on a fully submerged body are the weight W acting through the centre of gravity G and the buoyancy U acting
through the centre of buoyancy B as shown in Figure 5. If U = W, the body will be in equilibrium. If U > W, the body will rise until its average density becomes equal to that of the surrounding fluid or it reaches a floating position on the free surface of the fluid. If U < W, the body will move downwards in the fluid.
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1.3.2 Floating Bodies For a floating body to be in vertical equilibrium, the buoyancy U on the submerged volume should be equal to the weight W of
the body and B and G should be on the same vertical line as shown in Figure 6.
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1.4 Stability of Submerged Bodies Stability or the type of equilibrium is of major importance to a floating body. When a body is given a small displacement and then released, if it returns to its initial equilibrium position, it is said to be in stable equilibrium. If it moves further away from the initial equilibrium position, it is said to be in unstable equilibrium. If it remains at the displaced position, it is said to be in neutral equilibrium. The three types of equilibrium of a body are illustrated in Figure 7.
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1.4.1 Stability of Fully Submerged Bodies The equilibrium position of a fully submerged body in which the centre of buoyancy B is above the centre of gravity G is shown in Figure 8(a). The buoyancy U acts through B and the weight W ( = U ) acts through G.
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As B and G remain at fixed positions relative to the body, an angular displacement θ, as shown in Figure 8(b), produces a restoring moment MR which tends to restore the body at its initial equilibrium position. Hence the body is in stable equilibrium. MR = W(GB)sinθ = W(GB)θ for small values of θ.
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When B is below G, the equilibrium and displaced positions of the body are shown in Figure 9. An overturning moment Mo is produced at the displaced position as shown in Figure 9(b) which tends to move the body further away from the initial equilibrium position. Hence the body is in unstable equilibrium.
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When B G , the equilibrium and displaced positions of the body are shown in Figure 10. As no unbalanced moment is produced when the body is displaced, it remains in equilibrium at the displaced position. Hence the body is in neutral equilibrium.
It can be concluded that the stability of a fully submerged body depends on the relative positions of B and G as summarized below: Stable equilibrium:
B above G
Unstable equilibrium: B below G Neutral equilibrium
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1.4.2 Stability of Floating Bodies
The equilibrium position of a floating body of weight W acting through the centre of gravity G is shown in Figure 11(a). The buoyancy U (=W) acts through the centre of buoyancy B. The displaced position of the body, when it undergoes a small angular displacement θ is shown in Figure 11(b). The submerged volume remains unchanged but its shape changes during displacement and as a result the centre of buoyancy moves from B to B’. The centre of gravity G usually remains fixed relative to the body.
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The line of action of U through B’ intersects the axis BG at point M which is defined as the metacentre. It has been found that, for small displacements M is fixed in position relative to the body. The distance GM, measured upwards from G, is known as the metacentric height. For the body shown in Figure 11(b), M is above G (or GM > 0) and a restoring moment MR is produced at the displaced position. Hence the body is in stable equilibrium. MR = W (GM) Sinθ = W (GM)θ for small θ and hence GM is a measure of the stability of a floating body.
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For the floating body shown in Figure 11(c), M is below G or GM < 0 and an overturning moment Mo is produced at the displaced position. Hence the body is in unstable equilibrium. If M coincides with G or GM = 0, in a floating body, no unbalanced moment is produced when it is given an angular displacement as shown in Figure 14 and hence the body is in neutral equilibrium.
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It can be concluded that the stability of a floating body depends on the relative positions of the metacentre M and the centre of gravity G or the metacentric height GM. Stable equilibrium Unstable equilibrium Neutral equilibrium
M above G M below G M G
GM > 0 GM < 0 GM = 0
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1.5 Determination of Metacentric Height 1.5.1 Experimental Value The metacentric height of a floating body can be determined experimentally by shifting a known weight by a known distance and measuring the angle of tilt.
The initial equilibrium position of a floating body of total weight W acting through the centre of gravity G is shown in Figure 12(a). It contains a movable weight P at G1. The buoyancy U (=W) acts through the centre of buoyancy B.
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When P is shifted by a distance x, the tilted equilibrium position of the body is shown in Figure 12(b). The new centre of gravity G’ and the centre of buoyancy B’ are shown in Figure 12(b).
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1.5.2 Theoretical value
It is often necessary to determine the metacentric height of a floating body before it is constructed. If the shape of the submerged volume is known, the metacentric height can theoretically be determined. A floating body is shown in Figure 13. The buoyancy U acts through the centre of buoyancy B as shown in Figure 13(a). B has shifted to B’ at the displaced position shown in Figure 13(b). The metacentre M is also shown in Figure 13(b).
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This result is restricted to small angular displacements-usually up to about 80 and the restriction is particularly important when the sides of the floating body are not vertical. As the position of B can be determined for known shapes of submerged volume, the above expression for BM can be used to determine the metacentric height GM. For known positions of G, GM is given by, GM = BM – BG
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The tilting of a floating body about longitudinal and transverse axes are known as rolling and pitching respectively. For a typical cross section of a floating body, as shown in Figure 14, the second moment of area for rolling is smaller than that for pitching. Thus the metacentric height for rolling is less than that for pitching and it is important to check the stability considering the rolling of the floating body.
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1.6 Time Period of Oscillation When a floating body in stable equilibrium is given an angular displacement, it tends to oscillate about its equilibrium position. A floating body in stable equilibrium is shown in Figure 15(a). The buoyancy U acts through the centre of buoyancy B and the weight W acts through the centre of gravity G. As it is in stable equilibrium, the metacentre is above M. When it is oscillating, a displaced position is shown in Figure 15(b) at which the centre of buoyancy has shifted to B’.
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As there is no resultant horizontal force acting on the body, G does not move horizontally during oscillations. The buoyancy and hence submerged volume remains constant during oscillations and therefore O does not move vertically. Thus the body oscillates about A (instantaneous centre of rotation). For small oscillations, A is close to G and it can be assumed that the body oscillates about G.
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As the oscillations describe a simple harmonic motion, the time period of oscillation T can be expressed as, T 2 /
W (GM) IG
W = Mg and IG is usually expressed as IG=Mk2 where M =Total mass and k = Radius of gyration T
2k g (GM )
It has been found that there is better agreement between the theoretical and experimental values of T for rolling than for pitching of floating vessels. In practice, the viscosity of the liquid introduces damping action which suppresses oscillations unless further disturbances cause new angular displacements. 32
Increasing the metacentric height gives greater stability but reduces the time period of oscillation. Thus a floating vessel is less comfortable for passengers and is subjected to high stresses which may damage its structure. Typical values of metacentric height of a floating vessel may be in the range 0.3 m to 1.2 m.
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1.7. Effect of Liquid Cargo The stability of a floating vessel carrying liquid with a free surface is adversely affected by the movement of the centre of gravity of the vessel when it undergoes an angular displacement.
1.7.1. Liquid Cargo in a Single Compartment A vessel floating in a liquid of density ρ is shown in Figure 16(a). It carries a liquid of density ρ1 in a single compartment. The total weight W acts through the centre of gravity G and the weight of liquid cargo acts through G1. When the vessel tilts by an angle θ, as shown in Figure 16(b).
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The (restoring) moment produced at the displaced position MR = W(NM)Sinθ Hence NM can be considered as the effective metacentric height. Reduction of metacentic height = GN NM = GM – GN 35
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CE 1022 Fluid Mechanics Buoyancy Relative Equilibrium - Lecture Note -
Conception of equipotential surface • Surface at any point of which force potential U is constant, i.e. : dU=0 1
is called equipotential surface. From dU dp follows that for equipotential surface:
U const ;
dU 0;
or
1 dp 0
p=const. •Thus equipotential surface simultaneously is surface of constant pressure. Such surface is of great interest for many engineering problems of hydrostatics, when pressure action to surfaces of structures are analyzed.
Absolute equilibrium • Absolute equilibrium is called equilibrium case, when fluid is acted by gravity forces only. If axis x is director upward, components of gravity forces acceleration would have magnitudes: ax=ay=0 and az=-g. Applying them in (3.13) it obtains such shape:
-gdz=0 Solution of received differential equation gives result: -gz =const or z=const.
The last equality means equation of horizontal surface, at any point of which pressure is of the same magnitude (Fig. 1). Free surface of liquid (plane A) is horizontal, therefore it is one of infinite large number of equipotential surfaces (planes B, C). At all points of these surfaces pressure is of the same magnitude, i.e. p1=p6, p2=p5, p3=p4 A
B C
1
6 2
A
5 3
4
Fig. 1 Example of absolute equilibrium
B C
Relative equilibrium of horizontally moving liquid • Liquid in vessel moving in horizontal direction with acceleration (see Fig. 2) is acted by gravity and inertial forces, characterised by accelerations ax and g and also by surface forces. Z po X
ax g
u
Fig. 2 Relative equilibrium of a liquid moving with acceleration in horizontal direction
If axis Z is directed upward and axis X – in vessel motion direction, in expression of (3.12) ax=-ax, ay=0 and az=-g. From it follows such equation of equipotential surface dU=-axdx-gdz=0, or -xax-zg=const. x=0 when y=0, from what follows const=0. Thus equipotential surface equation receives such expression:
ax z x. g
• It is equation of inclined plane. Free surface of liquid represent one of equipotential surfaces and has shape of
ax plane, inclined by angle arctang g
(see Fig. 3).
The main law of hydrostatics (3.9) for this case receives shape
a x dx - gdz or
xa x - zg
1
1
dp,
p const.
Constant of integration may be received from condition: p=po when x=z=0.
const -
From there follows
xa x -zg
1
p
1
1
p0
and
p0 .
Solution the equation with respect to p gives result:
ax p p o g z g
x .
Received formula suits for computation of pressure of liquid acted simultaneously by gravity and inertial forces.
Relative equilibrium of rotating liquid • Liquid in rotating vessel is acted by centrifugal and gravity forces (Fig. 3). Centrifugal force may be characterised by centrifugal acceleration ac=r2, components of which along axis x and y are ax=x2 and ay=y2. The liquid is acted also by gravity force, which is characterised by gravity acceleration along axis z, i.e. az=-g. Equation of equipotential surface (Fig. 3) in this case obtains expression: 2xdx+2ydy—gdz=0 solution of which leads to: 1 2 2 1 2 2 x y gz const. 2 2
Z
po
X zo Y
X
Fig. 3 Relative equilibrium of fluid rotated around vertical axis
Rotation of a fluid element in a rotating tank of fluid (solid body rotation)
p = 0 gives free surface
For point on axis on free surface of the liquid x=y=0 and z=zo (Fig. 3) and const=-gzo. Now equipotential surface obtains shape:
or
2 ( x 2 y 2 ) gz gzo 2
z z0
x y 2
2
2g
2
.
It is equation of parabolic. Free surface of the liquid being one of equipotential surfaces has shape of concave meniscus with the lowest point in the middle and the highest in periphery.
• Applying indicated acceleration components to the main law of hydrostatics leads to:
xdx ydy gdz 2
2
1
dp
or
1 2 2 1 2 2 p x y gz const, 2 2 where for x=y=0, z=zo and p=poconst
gz
p0
.
• Using this constant in (3.15) and solving it with respect to p leads to:
x y , p p 0 g z 0 z 2g 2
where
x
2
y
2
r
2
and
2
2
2 r2 = u2.
(3.16)
Here r
is radius or distance from revolution axis, u is linear velocity of revolution.
Finally :
2 u p po g z o z 2 g Received formula is used for computation of pressure in rotating vessel and design of vane type hydraulic machines.
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Then... Additional reading.... Additional examples/exercises .... Self learning/life-long learning ....
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