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.

B

F

A

E

C

D

Figure 1

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)

F

surface ADC= Thrust on right hand side of AC

=

Total horizontal thrust on ABCD = 0

(say)

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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

This result is not restricted

to a fluid of uniform density.B

F

A

E

C

D

Figure 1 4

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. 9

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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GB When , 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.

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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 GB

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 M above G GM > 0

Unstable equilibrium M below G GM < 0

Neutral equilibrium M G 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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GI

WT

)GM(/2

)GM(

2

g

kT

As the oscillations describe a simple harmonic motion, the time

period of oscillation T can be expressed as,

W = Mg and IG is usually expressed as IG=Mk2 where M =Total

mass and k = Radius of gyration

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.

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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

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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

is called equipotential surface. From follows

that for equipotential surface:

or

p=const.

01

dp;0 ; dUconstU

dpdU

1

•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

1

2

3 4

5

6A A

B

C

B

C

Fig. 1 Example of absolute equilibrium

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.

g

X

Z

po

u

ax

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:

.xg

az x

• It is equation of inclined plane. Free surface of liquid

represent one of equipotential surfaces and has shape of

plane, inclined by angle (see Fig. 3).

The main law of hydrostatics (3.9) for this case receives

shape

or

const.1

-x pzgxa

,1

gdz-dxa x dp

g

axarctang

Constant of integration may be received from

condition:

p=po when x=z=0.

From there follows and

Solution the equation with respect to p gives result:

Received formula suits for computation of pressure of

liquid acted simultaneously by gravity and inertial forces.

.11

0pp-zgxa x

0

1-const p

.pp o

x

g

azg x

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:

const.2

1

2

1 2222 gzyx

X

Y

Z

zo

po

X

Fig. 3 Relative

equilibrium of fluid

rotated around

vertical axis

p = 0 gives free surface

Rotation of a fluid element in

a rotating tank of fluid

(solid body rotation)

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

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.

ogzgzyx

2

)( 222

.

2

222

0g

yxzz

• Applying indicated acceleration components to the

main law of hydrostatics leads to:

dpgdzydyxdx

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or

where for x=y=0, z=zo and p=po .0

pgzconst

const,2

1

2

1 2222

p

gzyx

• Using this constant in (3.15) and solving it with

respect to p leads to:

(3.16)

where and 2 r2 = u2. Here r

is radius or distance from revolution axis, u is linear

velocity of revolution.

222 ryx

,

2

222

00

g

yxzzgpp

Finally :

Received formula is used for computation of pressure in

rotating vessel and design of vane type hydraulic

machines.

g

uzzgpp oo

2

2

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Then...Additional reading....Additional examples/exercises ....Self learning/life-long learning ....

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