CVE312 Fluid Mechanics II (2 Units) Fluid statics: Floatation and stability. Dynamics of fluid flow-conservation. Equation of mass an momentum. Euler and Bernoulli’s equations. Reynolds number. Dimensional analysis, similitude, Buckinggham Pitheorems. Application of hydraulics models. Flow measurements. Flow meters and errors in measurement.

Week One:

Floatation and stability

This topic will focus on the forces that are present in fluids at rest. Force variations (Pressure

variations)- in a static fluid is important to the Civil Engineer. Some examples are water retained by a

dam bounded by a levee, gasoline in a tank truck, and accelerating fluid containers. Furthermore, fluid

statics deals with the stability of floating bodies and submerged bodies and is applied in the design of

ship hull and in load distributions determination for flat bottomed barges. Thus, fluid statics concerns

the forces that are present in fluids at rest, with applications to various practical problems.

The objectives of this chapter are to discuss pressure and pressure measurements , to develop

equations for calculating forces on submerged surfaces, and to examine problems involving the stability

of partially or wholly submerged bodies.

PRESSURE AND PRESSURE MEASUREMENT

Consider a wedge-shaped particle exposed on all sides to a fluid as illustrated in Fig. 1a. Fig.1b is a free-

body diagram of the particle cross section. The dimensions of zyx ,, are small and tend t zero as

the particle shrinks to a point. The only forces considered to be acting on the particle are due to

pressure and gravity. On either of the three surfaces, the pressure force is F=pA. By applying Newton's

second law in the x- and z-directions, we get, respectively,

0

22cos zszz azyx

zyxgyspyxpF

where: px, pz, and ps are average pressures acting on the three corresponding faces ax and az are

accelerations

is the particle density

The net force equals zero in a static fluid. After simplification, with ax=αz=0, these two equations become

px z -pssinθ=0

0)2

(sin zyxyspyzpF xxx

and Pz z -ps s cosθ- 02

zxg

The third term on the left-hand side of the second equation can be neglected because it is a higher-

order term containing zx , which is very small in comparison to the other terms. From the geometry

of the wedge, we find that

z s sinθ

and

cossx

which θ being arbitrarily chosen. Substituting into the pressure equation yields

sz

sx

pp

pp px=pz=ps

which illustrates that pressure at a point is the same in all directions. This concept was shown for a two-

dimensional model, but the proof is easily extended to three dimensions.

From the preceding paragraphs, we have seen that forces acting on a fluid at rest are due to pressure

and gravity. It is therefore important to learn how these forces vary in a static fluid.

Consider an element of a fluid at rest, as illustrated in Figure 1.2 a. The element chosen has a volume dx

dy dz and is sketched in a coordinate system where the positive z-direction is downward, coincident

with the direction of the gravity force. Figure 1.2 b is a view of the element looking in the positive y-

direction; the force acting on the right face is pdydz and that on the left face is ,)/( dydzdxxpp

both normal to their respective surfaces. Summing forces in the x-direction, we have the following for a

static fluid:

dydzdx

x

pppdydzFx )(0 (1)

Simplifying, we get

0

x

p (2)

which means that pressure does not vary with respect to x. A similar argument can be made for the

forces in the y-direction, which would yield

0

y

p (3)

Thus, Equation 1 and 2 show that there is no variation of pressure in any lateral direction.

Fig. 1.2c gives a free-body diagram for the z-direction. Summing forces, we obtain

dxdydz

z

ppgdxdydzpdxdyFx 0

or

dp= gdz (4)

Therefore, pressure does not vary in a static fluid in the z-direction-it increases with depth, as shown by

Equation (4). Integrating both sides yields

2

1

2

1

z

z

p

p

gdzdp (5)

where point 1 is a reference point such as the free surface of a liquid and point 2 is a point of interest.

For incompressible fluids, the density is a constant, and Equation (5) can be easily evaluated to give

zgzzgpp )12(12

where z is the depth below the liquid surface. This relationship is the basic equation of hydrostatics

and is often written as

gzp

Example 1.1

A cylindrical open-topped tank that serves as a reservoir for octane before it is piped to another location

is 140 ft in diameter. Determine the pressure difference between the top and the bottom of the walls

due to the octane when the tank is filled to a depth of 30ft.

Solution

We use the hydrostatic equation

)12(12 zzgpp

Section 1 refers to the free surface of the octane and section 2 is at the bottom . For octane ,

0.701(1.94slug/ft3). By substitution,

p2-p1= p 0.701(1.94slug/ft3)(32.2 ft/s2)(30-0)

or

21310

ft

lbfp

The result is independent of diameter.

Example 2.2

A cup of coffee is 7cm in diameter and filled to a depth of 8cm with coffee (assume properties are the

same as for water ). Calculate the pressure difference between the surface of the coffee and the bottom

of the cup.

SOLUTION

The hydrostatic equation applies with p=1000kg/m3 for water:

2

23

785

)08.0)(81.9)(1000(

mNp

ms

mm

kggzp

The result is independent of the cup diameter

Continuation

Equation (5) was integrated for the case of incompressible fluids (constant density), which is reasonable

for liquids. The hydrostatic equation resulted . Gases on the other hand, are compressible fluids with

properties that are related by the ideal gas law, under simplifying conditions:

RTp

RT

P

Substitute this into Equation (4), to get,

dp= gdzRT

p

or

dzRT

g

p

dp Integrate the above equation from point 1 to point 2, for constant temperature effect

2

1

2

1

z

z

p

p

dzRT

g

p

dp

)(ln 12

1

2 zzRT

g

p

p

Rearranging, we get

1

212 ln

p

p

g

RTzz (8)

In most common example of a compressible fluid, our atmosphere, temperature is not a constant

throughout but varies with height in the troposphere according to;

T=T0-αh (9)

where T =temperature at any point from sea level(where T=T0) to an altitude h of approximately 36,000

ft, or 11km.

α is called a lapse rate (3.60F/1000ft or 6.50C/km)

The stratosphere, the layer above the troposphere can be described by the same ideal gas equation. A

lapse rate equation is not necessary for the stratosphere because it is approximately isothermal.

Whether the fluid is compressible or incompressible, it is important to note that pressure variations in

static fluids are the result of gravity. The pressure increases with depth in either case.

Example 2.3

Graph the relationship between pressure and elevation in the stratosphere assuming it to isothermal at

-570C. The stratosphere begins at an altitude of approximately 11 000m, where the pressure is 22.5

kPa. Extend the graph to an elevation of 20000 m.

Solution

In this case, Equation (8) applies, but it must first be modified. Equation (8) was derived from

Equation(4), but Equation(4) is based on the assumption that the positive z-direction is coincidental with

the direction of the gravity force. In this example, we are dealing with the atmosphere, and we are

quoting measurements of altitude using the earth's surface as a reference. So the positive z-direction is

upward, which is opposite from the direction of the gravity force. Therefore, if we use Equation(8), we

must account for this discrepancy. We can do so by using a negative g. Thus,

1

212 ln

p

p

g

RTzzz

We select as our reference, z1=11 000m, where p1=22.5 kPa. From standard tables of the physical

properties of gases at room temperature and pressure, R=286.8J/(kg.K) for air, and we were given

T=570C=216K. In applying the preceding equation, let z1=z=any value ranging from 11 to 20km and

p2=p=the corresponding pressure.

Substitution gives

22500ln

81.9

)216(8.28611000

pz

or

z=11000-6.315ln22500

p

Rearranging yields

22500ln

315.6

11000 pz

Solving for pressure, we get

p=22500exp

315.6

11000 z

Plot altitude, z(cm) against pressure, p (kPa).

THE STABILITY OF BODIES IN FLUIDS

a. The stability of submerged bodies

For a body not otherwise restrained, it is important to know not only whether it will rise or fall in the

fluid, but also whether an originally vertical axis in the body will remain vertical. We are not here

concerned with effects of a fluid in motion but with states of equilibrium. We must, however ,

distinguish three types of equilibrium . A body in STABLE EQUILIBRIUM will, if given a small displacement

and then released, return to its original position. If, on the other hand, the equilibrium is unstable, the

body will not return to its original position but will move further from it. In neutral equilibrium, the body,

having been given a small displacement and then released, will simply adopt its new position.

For a body wholly immersed in a single fluid-as, for example the balloon and gondola illustrated in Fig. 3-

the conditions for stability of equilibrium are simple. An angular displacement from the normal position

(a) brings into action the couple Wx, which tends to restore the system to position (a). This, then, is a

stable arrangement. If, however, the centre of gravity G were above the centre of buoyancy B, the

couple arising from a small angular displacement would be such as to cause the assembly to topple

over. So far a completely immersed body the condition for stability is simply that G be below B. If B and

G coincide, neutral equilibrium is obtained.

b. The stability of floating bodies

The condition for angular stability of a body floating in a liquid is a little more complicated. This is

because, when the body undergoes an angular displacement about a horizontal axis, the shape of the

immersed volume in general changes, so the centre of buoyancy moves relative to the body. As a result

stable equilibrium can be achieved even when G is above B.

Figure 4a illustrates a floating body -a boat, for example -in its equilibrium position. The net force is

zero, so the buoyancy is equal in magnitude to the weight W of the body. There must be no moment on

the body, so the weight acting vertically downwards through the centre of gravity G must be in line with

the buoyancy acting vertically downwards through the centre of buoyancy B. Figure 4b shows the

situation after the body has undergone a small angular displacement or angle of heel, θ. It is assumed

that the position of the centre . It is assumed that the position of the centre of gravity G remains

unchanged relative to the body. (This is not always a justifiable assumption for a ship since some of the

cargo may shift during an angular displacement) . The centre of buoyancy B, however, does not remain

fixed relative to the body . During movement , the volume immersed on the right-hand side increases

while that on the left-hand side decreases, so the centre of buoyancy (i.e. the centroid of the immersed

volume ) moves to a new position B'. Suppose that the line of action of the buoyamcy (which is always

vertical) intersects the axis BG at M. For small values of θ, the point M is practically constant and is

known as the metacentre. For the body shown in the Figure, M is above G, and the couple acting on the

body in its displaced position is a restoring couple, that is, it tends to restore the body to its original

position. If M were below G, the couple would be an overturning couple and the original equilibrium

would have been unstable.

The distance of the metacentre above G is known as the metacentric height, and for stability of the body

it must be positive (i.e. M above G). Neutral equilibrium is of course obtained when the metacentric

height is zero and G and M coincide. For floating body, stability is not determined simply by relative

positions of B and G. The magnitude of the restoring couple is W(GM) sinθ and the magnitude of GM

therefore serves as a measure of the stability of a floating body.

Example 4

A uniform, closed cylindrical buoy, 1.5 m high, 1.0m diameter and of mass 80kg is to float with its axis

vertical in sea-water of density 1026 kg.m-3. A body of mass 10 kg is attached to the centre of the top

surface of the body. Show that, if the buoy floats freely, initial instability will occur.

Solution

Moments of mass about horizontal axis through O:

(10 kg) (1.5m)+(80 kg)

m

2

5.1={(80+10)kg} (OG)

OG=0.8333 m

FIGURE

For vertical equilibrium, buoyancy =weight.

kgggmkghm )1080(.1026)1(4

32

whence h=0.1117m

From Fig. 5, Ak2 of a circle about a centroidal axis =64

4d

h

dhdd

VAkBM

164/

64

224

2

= mm 560.01117.016

12

and

GM=OB+BM-OG= m

8333.0560.0

2

1117.0

=-0.2175m

Since this is negative (i.e. M is below G),the buoy is unstable.

Stability of a body subject to an additional force

When an unconstrained body is in equilibrium in a fluid the only forces relevant to its stability are the

weight of the body and its buoyancy. If, however, an additional force is provided -by, for example , an

anchor chain-stability is determined by the lines of action of the buoyancy and the resultant downward

force.

Example 5

For the buoy considered in Example 4, calculate the least vertical downward force applied at the centre

of the base that would just keep the buoy upright. What would then be the depth of immersion?

Solution

A vertically downward force F applied at O increases the total downward force from W (the total weight

of the buoy) to W+F. To maintain vertical equilibrium , the buoyancy too is increased to W+F, and so the

new depth of immersion h' is given by

FWhdg '4

2

Taking moments of forces about a horizontal axis through O gives the requirement for the restoring

couple to be just zero:

FIGURE

W(OG)=(W+F)(OB+BM)=

'162

''

4

22

h

dhhdg

that is,

(90 kg)g (0.8333m)=(1026kg.m-3)g

16

)1(

2

)'()1(

4

222 mh

m

whence

h'=0.2473m

and

F= 22 14

81.91026('4

Whdg

=1072N

c. Stability of a fluid itself

In the preceding sections, we have considered the stability of separate, identifiable, bodies wholly or

partly immersed in a fluid. We now turn attention to the stability of parts of the fluid itself that, perhaps

because of uneven heating or cooling, have a density slightly different from that of neighboring fluid.

These differences of density are the cause of fluid motion known as convection currents, which are

frequently encountered in both liquids and gases.

If, for example, only the lower layers of a certain bulk of fluid are heated, an unstable condition results.

This is because if some of the warmer fluid is displaced upwards, it finds itself surrounded by cooler,

and therefore denser, fluid. The buoyancy force exerted on the warmer fluid by its surroundings is

equal in magnitude to the weight of an equal volume of the surrounding denser fluid. As this buoyancy

is greater than the weight of the newly arrived fluid, there is a net upward force on the warmer fluid,

which therefore continues to rise. Heavier fluid then flows downwards to take the place of the less

dense fluid that has moved up and thus free convection is started.

If, however, the lower layers of fluid are cooled, the conditions are stable. Fluid displaced downwards

would be surrounded by cooler, denser, fluid; it would therefore experience a buoyancy force greater

than its own weight and would return upwards to its original position.

Such movements occur on a large scale in the atmosphere. The lower part of the atmosphere is

continually being mixed by convection, which is largely due to the unequal heating of the earth's

surface.

EQUILIBRIUM OF MOVING FLUIDS

In certain instances the methods of hydrostatics may be used to study the behavior of fluids in motion.

For example, if all the fluid concerned moves uniformly in a straight line, there is no acceleration and

there are no shear forces. Thus no force acts on the fluid as a result of the motion and, in these

circumstances, the hydrostatic equations apply without change.

If the fluid concerned, is undergoing uniform acceleration in a straight line, no layer moves relative to

another, so there are still no shear forces. There is however, an additional force acting to cause the

acceleration. Nevertheless, provided that due allowance is made for this additional force, the system

may be studied by the methods of hydrostatics. Fluids in such motion are said to be in relative

equilibrium.

The Principles Governing Fluids in Motion

Introduction

In this section, we lay the foundations of the analysis of fluid flow by considering first the description of

motion in terms of displacement, velocity and acceleration but without regard to the forces causing it.

The Principle of Conservation of Mass is introduced; then the inter-relation between different forms of

energy associated with the fluid flow is examined; and finally some simple applications of these results

are considered.

Acceleration of a Fluid Particle

The velocity of a fluid particle is a function both of position and of time. As the particle moves from say,

point A to point B, its velocity changes for two reasons. One is that particles at B have a velocity

different from particles at A, even at the same instant of time; the other reason is that during the time

the given particle moves from A to B, the velocity at B changes. If B is at only a small distance s from A,

the particle's total increase of velocity u is the sum of the increase due to its change of position and the

increase due to the passing of a time interval t :

tt

us

s

uu

and so, in the limit, as ,0t the acceleration as in the direction of flow is given by

t

u

sdt

uds

dt

duas

or, since udt

ds ,

dt

u

s

uu

dt

duas

(1)

The full rate of increase dt

du for a given particle is often termed the substantial acceleration. The term

tu

represents only the local or temporal acceleration, that is, the rate of increase of velocity with

respect to time at a particular point in the flow. The term )( suu is known as the convective

acceleration, that is, the rate of increase of velocity due to the particle's change of position. Although in

steady flow tu is zero, the convective acceleration is not necessarily zero, so that substantial

acceleration is not necessarily zero.

A particle may also have an acceleration in a direction perpendicular to the direction of flow. When a

particle moves in a curved path, it changes direction and so has an acceleration towards the centre of

curvature of the path, whether or not the magnitude o the velocity is changing. If the radius of the path-

line is rp, the particle's acceleration towards the centre of curvature is .2

pru Alternatively, if the

streamline has a radius of curvature rs, the particle's acceleration an towards the centre of curvature of

the streamline has in general a convective part sru 2 and a temporal part ,tun where un represents

the component of velocity of the particle towards the centre of curvature. Although, at that moment, un

is zero, it is, unless the flow is steady, increasing at the rate .tun Thus

t

u

r

ua n

s

n

2

(2)

The Continuity Equation

The principle of the conservation of mass expresses the fact that matter can neither be created nor

destroyed. The continuity equation is a mathematical statement of that principle. Applying the principle

to a fixed region within a fluid (See the Figure 3.1), we can write:

The rate at which mass enters the region=The rate at which mass leaves the region+ The rate of

accumulation of mass in the region

If the flow is steady(i.e. unchanging with time), the rate at which mass is accumulated within the region

is zero. The expression then reduces to

The rate at which mass enters the region=The rate at which mass leaves the region

This relation may now be applied to a stream-tube whose cross-section is small enough for there to be

no significant variation of velocity over it. A length s of the stream-tube is considered between the

cross-sectional planes B and C (Fig.3.2 ), s being so small that any variation in the cross-sectional area

A along that length is negligible. Then, the volume of fluid contained in that small piece of the stream-

tube is ( sA ) . (We recall that the cross-section by definition is perpendicular to the length ). If the

fluid initially between planes B and C passes through the plane C in a short time interval t , then the

rate at which fluid volume passes through C is ,/ tsA or, in the limit, ./ dtdsA But dtds is the

linear velocity there, say u, so the rate of volume flow is .uA As in calculating a volume, a length must

be multiplied by the area of a surface perpendicular to that length, so in calculating the rate of volume

flow (frequently termed the discharge and represented by the symbol Q) the velocity must be multiplied

by the area of a surface perpendicular to it. The rate of mass flow is given by the product of the

discharge and the density.

The rate at which a mass of fluid enters a selected portion of a stream tube-where the cross-sectional

area is ,1A the velocity of the fluid u1 and its density 1 -is therefore .111 uA For steady flow, there

is no accumulation of mass within the stream-tube, so the same mass must pass through all cross-

sections of the tube in unit time. Thus

...221111 uAuA constant (3)

For the entire collection of stream-tubes occupying the cross-section of a passage through which the

fluid flows, eqn (3) may be integrated to give

A

udA constant

where u is everywhere perpendicular to the elemental area A . If and u are constant over the entire

cross-section, the equation becomes

Au constant

For a fluid of constant density the continuity relation reduces to

A

udA constant

which may be written

uA constant= Q

whereu represents the mean velocity and Q is the volumetric flow rate. For the flow of an

incompressible fluid along a stream-tube, eqn 3 indicates that Au constant, so as the cross-sectional

area A decreases, the velocity increases, and vice versa. This fact at once allows a partial

interpretation of the pattern formed by streamlines in steady flow: in regions where the streamlines are

close together, the velocity is high, but where the same streamlines are more widely spaced the velocity

is lower. This conclusion, which applies to incompressible fluids, does not necessarily apply to the flow

of compressible fluids in which large density changes occur.

BERNOULLI'S EQUATION

The velocity of a fluid in general varies from one point to another even in the direction of flow. Since, by

Newton's First Law, a change of velocity must be associated with a force, it is to be expected that the

pressure of the fluid also changes from point to point.

The relation between these changes may be studied by applying Newton's Second Law to a small

element of the fluid over which the changes of velocity and pressure are very small. The element is so

chosen that it occupies part of a stream of a stream -tube of small cross-section (see Fig. 3.3). The ends

of the element are plane and perpendicular to the central streamline, but may be of any geometrical

shape.

The forces under investigation are those due to the pressure of the fluid all round the element, and to

gravity. The other possible forces are assumed to be negligible.

The behavior of an ideal fluid is thus often remarkably similar to that of an ideal, inviscid one. In the

absence of shearing forces, any force acting on a surface is perpendicular to it, whether the surface is

that of a solid boundary or that of an element of fluid. It is also assumed that the flow is steady.

The element is of length ,s where s represents the distance measured along the stream-tube in the

direction of flow. The length ,s is so small that the curvature of the streamlines over this distance is

neglected.

The pressure, velocity and so on will (in general) vary with s, but, as the flow is steady, quantities at a

particular point do not change with time and so, for the stream-tube considered, each variable may be

regarded as a function of s only.

At the upstream end of the element, the pressure is p, and at the downstream end p+ p (where p

may of course be negative). At the sides of the element, the pressure varies along the length, but a

mean value of p+k p may be assumed, where k is a fraction less than unity. The pressure at the

upstream end (where the cross-sectional area is A) results in a force pA on the element in the direction

of flow; the pressure at the downstream end (where the cross-sectional area is (A+ A ) causes a force

(p+ p )(A+ A ) on the element in the opposite direction.

Since the force in any direction is given by the product of the pressure and the projected area

perpendicular to that direction, the net axial force downstream due to the pressure at the sides of the

element is (p+k p ) A , since A is the net area perpendicular to the flow direction.

The weight of the element, W, equals sgA (the second order of small quantities being neglected) and

its component in the direction of motion is - sgA cosθ, where represents the density of the fluid and

θ the angle shown between the vertical and the direction of motion. Thus, in the absence of other

forces, such as those due to viscosity, the total force acting on the element in the direction of flow is

cos)())(( sgAApkpAApppA

When the second order of small quantities is neglected, this reduces to

cossgApA (6)

Since the mass of the element is constant, this net force must, by Newton's Second Law, equal the mass

multiplied by the acceleration in the direction of the force, that is, ).( dtdusA

We may write coss as z , where z represents height above some convenient horizontal datum and

z the increase in level along the length of the element. Then dividing by sA and taking the limits s

0 , we obtain

01

ds

dzg

dt

du

ds

dp

(7)

From equation (1)

dt

u

s

uu

dt

du

However, for steady flow the local acceleration

0 dtu and so )( dsduudtdu (the full derivative now taking the place of the partial because for

this stream -tube u is a function of s only). We then have

01

ds

dzg

ds

duu

ds

dp

(8)

as the required equation in differential form. This is often referred to as Euler's equation, after the Swiss

mathematician Leonard Euler (1707-83). It cannot be completely integrated with respect t s unless is

either constant or a known function of p. For a fluid of constant density, however, the result of

integration is

gzup

2

2

constant (9)

or, if we divide by g,

zg

u

g

p

2

2

constant (10)

This result (in either form) is usually known as Bernoulli's equation or the Bernoulli equation in honour

of another Swiss mathematician, Daniel Bernoulli (1700-82), who in 1738 published one of the first

books on fluid flow (Equations 9 and 10, however not developed until some years later).

Z is the elevation above some horizontal plane arbitrarily chosen as a base of measurement. The

Bernoulli Equation assumes that the flow is steady and hence not applicable to unsteady flows.

To sum up, the conditions to which Bernoulli's equation applies are: the fluid must be frictionless

(inviscid) and of constant density; the flow must be steady; and the relation holds in general only for a

single streamline.

The significance of the terms in Bernoulli's equation

The derivation of Bernoulli's equation is based on the application of Newton's Second Law o Motion,

which relates the rate of change of momentum of a body to the sum of the applied forces. However, as

noted previously, the analysis incorporates a number of important simplifications.

It is assumed that the fluid is inviscid and incompressible, that the flow is steady, and the relations have

been derived along a single streamline. If, in addition, there is no heat transfer along the streamline and

no shaft work is done (say, by a pump or turbine), then, the equation takes on exactly the same form as

the corresponding energy equation. In these circumstances, the terms of the Bernoulli equation can be

interpreted as contributions in an energy balance

Equation (9) states that the sum of three quantities is constant. Consequently the separate quantities

must be interchangeable and thus of the same kind. The second term 22u , represents the kinetic

energy of a small element of the fluid divided by mass of the element. The third term gz, also represents

energy/ mass and corresponds to the work that would be done on the fluid element in raising it from

datum level to the height z divided by the mass of the fluid element.

Similarly, the term p must also represent an amount of work divided by the mass of the fluid.

Each of the terms in p , 22u and gz represents energy/mass or energy/weight. The quantities in

Eq.(10) are therefore usually referred to respectively as pressure head (or static head), velocity head and

gravity head or elevation, and their sum as the total head.

General Energy Equation for Steady Flow of any Fluid

For a fluid at pressure p1 and with velocity at a section where the (average ) elevation is z1 and leaves

with pressure p2 and velocity u2 where the (average) elevation is z2. As the fluid moves from inlet to

outlet , its properties, in general, change from one point to another. However, we assume they do not

change with time.

From the expression, the Steady-Flow Energy Equation was derived as

weegzup

gzup

q

121

2

1

1

12

2

2

2

2

2

1

2

1

(13)

where q represents the net heat transferred to the fluid divided by mass and w represents the net shaft

work done by the fluid divided by mass; w is the net amount of work done by the fluid.

For the above Eq.(13) to be valid, the following assumptions must be followed strictly;

1. The flow is steady and continuous, that is, the rate at which mass enters the region considered equals

that at which mass leaves the region and neither varies with time.

2. Conditions at any point between the inlet and outlet sections 1 and 2 do not vary with time.

3. Heat and work are transferred to or from the fluid at a constant net rate.

4. Quantities are uniform over the inlet and outlet cross-sections 1 and 2.

5. Energy due to electricity, magnetism, surface tension or nuclear reaction is absent. If energy due to

any of these phenomena is, in fact, involved, appropriate additional terms will appear in the equations.

The Steady-Flow Energy Equation in Practice

The Steady-Flow Energy Equation (SFEE) applies to liquids, gases and vapours, and accounts for viscous

effects. In many applications, it is considered simplified because some of the terms are zero or cancel

with others. If no heat energy is supplied to the fluid from outside the boundaries, and if the

temperature of the fluid and that of its surroundings are practically identical (or if the boundaries are

well insulated) q may be taken as zero.

If an incompressible fluid with zero viscosity flows in a stream-tube across which there is no transfer of

heat or work, the temperature of the fluid remains constant. Therefore the internal energy is also

constant and the equation reduces to

1

2

1

1

12

2

2

2

2

1

2

10 gzu

pgzu

p

This is identical to the Bernoulli's Equation (9).

Real fluids have viscosity, and the work done in overcoming the viscous forces corresponds to the so-

called fluid friction. The energy required to overcome the friction is transformed into thermal energy.

The temperature of the fluid rises above the value for frictionless flow; the internal energy increases

and, in general, the heat transferred from the fluid to its surroundings is increased. The increase of

temperature, and consequently of internal energy, is generally of no worth (the temperature rise is

normally only a very small fraction of a degree) and thus corresponds to a loss of useful energy.

Moreover, as we have defined q as the heat transferred to the fluid divided by mass of the fluid, a loss

of heat from the system is represented by -q and so the total loss (divided by the mass of the fluid) is e2-

e1-q. For a fluid of constant density it is usual to express this loss of useful energy, resulting from friction,

as head loss due to friction, hf. Therefore

hf=(e2-e1-q)/g

Then for a constant-density fluid with no other heat transfer and no shaft work performed the Steady-

Flow Energy Equation reduces to

2

2

221

2

11

22z

g

u

g

phz

g

u

g

pf

(16)

Here u1 and u2 represent mean velocities over the cross-sectional (1) and (2) respectively . If we assume

further that the flow occurs in a horizontal pipe of uniform cross-section, then u1=u2 and z1=z2 and so

.21 / fhgpp That is, the displacement work done on the fluid in the pipe is entirely in

overcoming friction.

Example

A pump delivers water through a pipe 150mm in diameter. At the pump inlet A, which is 225 mm

diameter, the mean velocity is 1.35 ms-1 and the pressure 150 mmHg vacuum. The pump outlet B is 600

mm above A and is 150 mm diameter. At a section C of the pipe, 5 m above B, the gauge pressure is 35

kPa. If friction in the pipe BC dissipates energy at the rate of 2.5 kW and the power required to drive the

pump is 1.27 kW, calculate the overall efficiency of the pump. (Relative density of mercury=13.56).

Solution

Mean velocity at A=uA=1.35ms-1

Therefore, by continuity,

CB

ACBArea

AAreauuu

,)(

)(

11

2

038.3150

22535.1

msms

Steady-Flow Energy Equation:

CCC

AAA gzu

p

timeMass

timefrictiontolossEnergy

timeMass

timepumpbyaddedEnergygzu

p 22

2

1

/

/

/

/

2

1

Time

pumpbyaddedEnergy

Time

frictiontolossEnergyzzguupp

Time

MassACACAC

)(

2

1 22

=

kWkW

msNmsm

6.85.2

.6.581.91000)35.1038.3(10002

1)150.0(81.9135603500035.1)225.0(

4

12212

Therefore, efficiency of pump=8.6/12.7=67.7%

Notice that pA, is a vacuum pressure, is negative.

Energy transformation in a constant-density fluid

The concept of head, that is, energy divided by weight of a constant-density fluid, is of great value in

allowing a geometrical representation of energy change. From the system shown Fig 7, piezometer

tubes are connected at certain points to a pipe conveying liquid from a large reservoir. At a point where

the (gauge) pressure in the pipe is p, the liquid will rise in the piezometer tube to a height

gp .

At points in the reservoir far from the outlet, the velocity of the liquid is so small as to be negligible. At

such a point1 at a depth h1 below the free surface, the pressure is therefore given by the hydrostatic

relation 11 ghp , so the sum of the three terms in Bernoulli's expression is

Hzhzgggh 111

2

1 20

Thus H is the total head for the streamline on which the point 1 lies. If no energy is dissipated by friction,

the total H is constant along that streamline and may therefore be represented by a line parallel to the

datum plane.

At point 2 in the pipe, the pressure is indicated by the rise gp 2 of the liquid in the piezometer tube.

The amount by which the sum of gp 2 and z2 falls short of the total head corresponds to the velocity

head gu 22

2 for the streamline considered. There is a similar state of affairs at point 3, although here

the cross-section of the pipe is smaller and so the mean velocity is greater than at 2 by virtue of the

continuity equation uA =constant.

In practice, friction leads to a loss of mechanical energy, so the total head line drops. The height of any

point on this line above the datum plane always represents the total head ( gp )+( gu 22

2 )+z of the at

the point in question. Another line that may be drawn is that representing the sum of the pressure head

and elevation only: ( gp )+z. This line, which would pass through the surface levels in the piezometer

tubes of Fig 7, is known as the pressure line or hydraulic grade line. The geometrical representation that

these lines afford is frequently useful, and it is therefore important to distinguish clearly between them.

The one-dimensional continuity relation shows that, for a fluid of constant density, a reduction in the

cross-sectional area causes an increase in the mean velocity. Equation (16) shows that in the absence of

additional energy input to the fluid, the increase in velocity will be accompanied by a decrease of

pressure (provided that the change of elevation z is small). Conversely, an increase of cross-sectional

area of the flow gives rise to a decrease of velocity and an increase of pressure.

The energy equation (16) also shows , however, that for a given elevation, the velocity cannot be

increased indefinitely by reducing the cross-sectional area. Apart from exceptional circumstances, not

encountered in normal engineering practice, the absolute pressure can never become less than zero.

Any further reduction of the cross-sectional area would not bring about an increase of velocity, and

therefore the discharge (i.e. area x mean velocity) would be reduced. There would then be a consequent

decrease in the velocity at other sections. This is known as choking

With liquids, however, difficulties arise before the pressure becomes zero. At low pressures, liquids

vaporize and pockets of vapour may thus be formed where the pressure is sufficiently low. These

pockets may suddenly collapse-either because they are carried along by the liquid until they arrive at a

region of higher pressure, or because the pressure increases again at the point in question. The forces

then exerted by the liquid rushing into the cavities cause very high localized pressures, which can lead to

serious erosion of the boundary surfaces. This action is known as cavitation. Furthermore, the flow may

be considerably disturbed when cavitation.

In ordinary circumstances, liquids contain some dissolved air. The air is released as the pressure is

reduced, and it too may form pockets in the liquid which are often known as air locks. To avoid all these

undesirable effects, the absolute pressure head in water, for example, should not be allowed to fall

below about 2m (equivalent to about 20 kPa).

Example Questions on Buoyancy and Metacentric Height

Example 1

A wooden block of width 1.25m, depth 0.75m, and length 3.0m is floating in water. Specific weight of

the wood is 6.4kN/m3. Find

(i) Volume of water displaced, and

(ii) Position of centre of buoyancy

Solution

Width of the wooden block=1.25m

Depth of the wooden block=1.25m

Depth of the wooden block=3.0m

Volume of the block=1.25x0.75x3=2.812m3

Specific weight of wood, w=6.4kN/m3

Weight of the block=6.4x2.812=18kN

(I) Volume of water displaced:

For equilibrium, the weight of water displaced=weight of wooden block=18N

Volume of water displaced= 3835.181.9

18m

waterofdensityweight

displacedwaterofweight

(where density of water =9.81 kN/m3)

(ii) Position of centre of buoyancy:

We know that,

Volume of wooden block in water=volume of water displaced

or 1.25xhx3.0=1.835

(where h=depth of wooden block in water)

Therefore, h= m489.00.325.1

835.1

Hence centre of buoyancy = 244.02

489.0 from the base.

Example 2

A wooden block of specific gravity 0.7 and having a size of 2m x 0.5 m x0.25 m is floating in water.

Determine the volume of concrete of specific weight 25 kN/m3, that may be placed which will immerse

the (I) block completely in water and (ii) block and concrete completely in water.

Solution

Size of the block=2 m x0.5 m x0.25 m

Therefore, volume of the block =0.25m3

Specific gravity of the block=0.7 x9.81 =6.867 kN/m3

Weight of the block =6.867 x 0.25=1.716kN

(Therefore, specific weight of water =9.81 kN/m3)

Let Wc =weight of concrete required to be placed on the block, and

Vc=Volume of concrete required to be placed on the block

Total weight of the block=Wc+1.716kN (i)

(i) Immersion of the block only:

When the block is completely immersed, the volume of water displaced =0.25m3

Therefore, upward thrust at the time of complete immersion

=0.25 x9.81=2.45kN (ii)

Now, equating (i) and (ii), we get

Wc+1.716=2.45

or Wc=0.734kN

Volume of concrete, Vc=30294.0

25

734.0

.m

weightsp

weight

(ii) Immersion of block and concrete:

Total weight of the block and concrete=25Vc+1.716...................................(i)

and upward thrust=(Vc+0.25)x9.81.............................................................(ii)

Equating (i) and (ii), we get

25Vc+1.716=(Vc+0.25)x9.81

or 25Vc+1.716=9.81Vc+2.45 or 15.19Vc=0.734

or Vc=0.0483m3.

Example 3

A cylinder of mass 10 kg and area of cross-section 0.1m2 is tied down with string in a vessel containing

two liquids as shown in Figure a below. Calculate gauge pressure on the cylinder bottom and the tension

in the string. Density of water=1000kg/m3. Specific gravity of A=0.8. Specific gravity of B(water)=1.0.

Solution

Given: Mass of cylinder, m=10kg

Area of cross-section =0.1m2

Density of water (liquid B)=1000kg/m3

Density of liquid A=0.8x1000=800kg/m3

Tension in String, T:

Volume of liquid A displaced =0.1x0.1=0.01m3

Therefore mass of liquid A displaced, mA=0.01x800=8kg

Volume of liquid B displaced=0.1x0.125=0.0125m3

Mass of liquid B displaced mB=0.0125x1000=12.5kg

Total mass of liquid displaced=mA +mB=8+12.5=20.5kg

Upward thrust =20.5x9.81=201.1N

Weight of cylinder=mg=10x9.81=98.1N

Net upward thrust=201.1-98.1=98.1N

Therefore, tension in the string T=103N

Pressure (gauge)on the cylinder bottom, p:

p=2/1030

1.0

103

secmN

tioncrossofArea

thrustupwardNet

Example 4

A wooden block of specific gravity 0.75 floats in water. If the size of the block is 1mx0.5mx0.4m, find its

metacentric height.

Solution

Size (or dimensions) of the block=1mx0.5mx0.4m.

Specific gravity of wood=0.75

Specific weight w=0.75x9.81=7.36kN/m3

Weight of wooden block=specific weight x volume

=7.36x1x0.5x0.4=1.472kN

Let depth of immersion=h meters

Weight of water displaced=specific weight of water x volume of the wood submerged in water

=9.81 x1x 0.5x h=4.9h kN

Now for equilibrium,

Weight of wooden block=weight of water displaced; 1.472=4.9h

h= m3.09.4

472.1 ,

Therefore, distance of centre of buoyancy from bottom, i.e

OB= mh

15.02

3.0

2

and OG= m2.02

4.0

BG=OG-OB=0.2-0.15=0.05m

Also BM=V

I

where I=Moment of inertia of a rectangular section and

=3

3

014.012

5.01m

and V=volume of water displaced (or volume of wood in water)

=1x0.5xh=1x0.5x0.3=0.15m3

BM= mV

I069.0

15.0

0104.0

We know that the metacentric height,

GM=BM-BG (Therefore, G is above B)

=0.069-0.05=0.019m

Example 5

A solid cylinder 2 m in diameter and 2m high is floating in water with its axis vertical. If the specific

gravity of the material of cylinder is 0.65, find its metacentric height. State also whether the equilibrium

is stable or unstable.

Solution

Given: Diameter of cylinder d=2m; height of cylinder, h=2m; specific gravity=0.65

Depth of cylinder in water =sp.gravityxh

=0.65x2.0=1.3m

Distance of centre of buoyancy (B) from O, i.e.,

OB= m65.02

3.1

Distance of centre of gravity (G) from O, i.e.,

OG= m0.12

0.2

BG=OG-OB=1.0-0.65=0.35m

Also, BM=V

I

where, I= Moment of inertia of the plan of the body about Y-Y

= 344 785.026464

md

and, V =volume of cylinder of water

= 322 084.43.1244

mwaterincylinderofdepthd

BM= mV

I192.0

084.4

785.0

We know that the metacentric height,

GM=BM-BG=0.192-0.35

=-0.158M

-ve sign means that the metacentric (M) is below the centre of gravity (G). Thus the cylinder is in

unstable equilibrium.

Summary of Fluid Dynamics

1. The science which deals with the geometry of motion of fluids without reference to the forces causing

the motion is known as "hydrokinematics'' (or simply kinematics).

2. The science which deals with the action of the forces in producing or changing motion of fluids is

known as " hydrokinetics" (or simply kinetics). Hence, the study of fluids motion involves the

consideration of both the kinematics and kinetics.

3. In fluid mechanics, the basic equations are :(1) Continuity equation, (2) Energy equation, and (3)

Impulse-momentum equation

4. The different types of head (or Energies) of a Liquid in Motion are:

(i) Potential head or potential energy

This is due to configuration or position above some suitable datum line. It is denoted by z.

(ii) Velocity head or kinetic energy

This is due to velocity of flowing liquid and is measured as g

V

2

2

where V is the velocity of flow and 'g'is

the acceleration due to gravity (g=9.81)

(iii) Pressure head or pressure energy

This is due to the pressure of liquid and reckoned as w

p where p is the pressure and w is the weight

density of the liquid.

Total head/energy

Total head of a liquid particle in motion is the sum of its potential head, kinetic head and pressure head.

Mathematically,

Total head,

H= liquidofmw

p

g

Vz )(

2

2

Total Energy=z+ liquidofkgNmw

p

g

V)/(

2

2

5. Bernoulli's Equation

It states that " In an ideal incompressible fluid, when the flow is steady and continuous , the sum of

pressure energy, kinetic energy and potential (or datum) energy is constant along a stream line.''

Mathematically zg

V

w

p

2

2

=constant

where,

energypressurew

p

andenergyKineticg

V,

2

2

z=Datum(or elevation) energy

Examples Questions under Bernoulli's equation

1. In a pipe of 90mm diameter water is flowing with a mean velocity of 2m/s and at a gauge pressure of

350 kN/m2. Determine the total head, if the pipe is 8 meters above the datum line.

Neglect friction

Solution

Diameter of the pipe =90 mm

Pressure, p=350kN/m2

Velocity of water, V=2m/s

Datum head, z=8m

Specific weight of water, w=9.81kN/m3

Total head of water, H:

H=w

p

g

Vz

2

2

=8+ m88.4381.9

350

81.92

22

H=43.88m

2. A pipeline (Fig c) is 15cm in diameter and it is at an elevation of 100m at section A. At section B it is an

elevation of 107 m and has diameter of 30cm. When a discharge of 50 litre/sec of water is passed

through this pipeline, pressure at A is 35kPa. The energy loss in pipe is 2m of water. Calculate pressure

at B, if flow is from A and B.

Solution

Given DA=15cm=0.15m; DB=30cm=0.3m;

pA=35kPa; Q=50litres/sec=0.05m3/s;

hf=2m of water; Direction of flow; from A to B

Pressure at B, pB

sm

D

QV

A

A /829.2

15.04

05.0

4

22

sm

D

QV

B

B /707.0

)3.0(4

05.0

4

22

Applying Bernoulli's equation between section A and B, we get

fBBB

AAA hz

g

V

w

Pz

g

V

w

P

22

22

or fBABAAB hzz

g

VV

w

P

w

P

)(

2

22

or

fBA

BAAB hzz

g

VVwpp )(

2

22

= 2)107100(81.92

707.0829.2

1000

)81.91000(35

22

35+9.81(0.3824-7-2)=-49.54kPa.

i.e., pB= -49.54kPa. This shows that the given pressure at A, 35kPa is gauge pressure and hence there is a

vacuum at point B.

3. Water flows in a circular pipe. At one section, the diameter is 0.3m, the static pressure is 260kPa

gauge, the velocity is 3 m/s and the elevation is 10 m above ground level. The elevation at a section

downstream is 0m, and the pipe diameter is 0.15m. Find out the gauge pressure at the downstream

section.

Frictional effects may be neglected. Assume density of water to be 999 kg/m3.

Solution

Refer to Fig.d. D1=0.3m; D2=0.15m; z1=0; z2=10m; p1=260kPa, V1=3m/s; 3/999 mkg

From continuity equation, A1V1=A2V2,

12

2

2

1

1

2

13

4

4 V

D

D

VA

AV

= smVD

D/123

15.0

3.02

1

2

2

1

Weight density of water,

w= g=999x9.81=9800.19N/m3

From Bernoulli's equation between section 1 and 2 (neglecting friction effects as given), we have

2

2

221

2

11

22z

g

V

w

pz

g

V

w

p

1081.92

)3(

19.9800

1000260 2

081.92

)12(

19.9800

2

2

p

26.53+0.459+10= 34.719.9800

2 p

or p2= 290566 N/m2=290.56kPa