INSTITUTE OF AERONAUTICAL ENGINEERING

(Autonomous)

Dundigal, Hyderabad -500 043

CIVIL ENGINEERING

COURSE LECTURE NOTES

Course Name FLUID MECHANICS

Course Code ACEB06

Programme B.Tech

Semester IV

Course

Coordinator Mr. Ch. V. S. S. Sudheer, Assistant Professor

Course Faculty Dr. P. Ram Mohan Rao

Mr. Ch. V. S. S. Sudheer

Lecture Numbers 1-60

Topic Covered All

COURSE OBJECTIVES (COs):

The course should enable the students to:

I Analyse and understand the broad principles of fluid statics, kinematics and dynamics

II Interpret the definitions of the basic terms used in fluid mechanics and their applications

III Summarize classifications of fluid flow and applications of continuity equation

IV Explore the knowledge of continuity, momentum and energy principles

V Obtain the wide applications of dimensional analysis in fluid flow problems

COURSE LEARNING OUTCOMES (CLOs):

Students, who complete the course, will have demonstrated the ability to do the

following:

ACEB06.01 Define fluid and its properties. Describe surface tension and relations in different

conditions.

ACEB06.02 Explain Newton’s law of viscosity. Classify fluids based on Newton’s law of

viscosity and solve problems on Viscosity.

ACEB06.03 Employ capillary principle to calculate capillary rise/fall in a given tube.

ACEB06.04 Understand the concept of surface tension and their applications.

ACEB06.05 Know the concept of capillarity and compressibility.

ACEB06.06 Employ principle of manometry to measure gauge and differential pressure surface.

ACEB06.07 Understand the Archimedes principle to solve numerical examples on fluid

pressure management.

PAGE 1

ACEB06.08 Application and measurement of pressure for different types of pipes such as

inclined, horizontal and vertical.

ACEB06.09 Examine stability of a floating body by determining its metacentric height

ACEB06.10 Understand the various types of flows in closed conduits.

ACEB06.11 Distinguish velocity potential function and stream function and solve for velocity

and acceleration of a fluid at a given location in a fluid flow

ACEB06.12 Establish Euler’s theorem and deduce Bernoulli’s equation for a ideal fluid and

comment on validation assumption made.

ACEB06.13 Examine Bernoulli’s equation for ideal and real fluids and evaluate the direction of

flow

ACEB06.14 Flow and velocity measuring instruments.

ACEB06.15 Employ Darcy-Weischbach and Chezy’s equation to calculate friction losses

ACEB06.16 Describe flow through pipes, and Distinguish between major loss and minor loss in

pipes.

ACEB06.17 Sketch HGL and TEL for a given pipe setting.

ACEB06.18 Distinguish between Drag force and lift force and Examine drag and lift force for a

given set

ACEB06.19 Flow and velocity measuring instruments.

ACEB06.20 Application and analysis of momentum principle in closed pipes to estimate the

force on pipe bends.

ACEB06.21 Understand the Dimensional quantities and analysis for various parameters.

ACEB06.22 Derive the problems based on Rayleigh’s method and Buckingham’s pi theorem

with applications.

ACEB06.23 Explain the concept of similitude with examples and different types of similitude

concepts.

ACEB06.24 Remember the concepts of dimensionless numbers to solve numerical problems

SYLLABUS

Module-I BASIC CONCEPTS AND DEFINITIONS

Distinction between a fluid and a solid; Density, Specific weight, Specific gravity, Kinematic

and dynamic viscosity; variation of viscosity with temperature, Newton law of viscosity; vapour

pressure, boiling point, cavitation; surface tension, capillarity, Bulk modulus of elasticity,

compressibility.

Module-II FLUID STATICS

Fluid Pressure: Pressure at a point, Pascal’s law, pressure variation with temperature, density and

altitude. Piezometer, U-Tube Manometer, Single Column Manometer, U -Tube Differential

Manometer, Micro manometers. Pressure gauges, Hydrostatic pressure and force: horizontal,

vertical and inclined surfaces. Buoyancy and stability of floating bodies.

Module-III FLUID KINEMATICS

Classification of fluid flow: steady and unsteady flow, uniform and non-uniform flow, laminar

and turbulent flow, rotational and irrotational flow, compressible and incompressible flow, ideal

and real fluid flow, one, two- and three-dimensional flows;

Stream line, path line, streak line and stream tube; stream function, velocity potential function.

One-, two and three - dimensional continuity equations in Cartesian coordinates.

Module-IV FLUID DYNAMICS

Surface and body forces; Equations of motion - Euler’s equation; Bernoulli’s equation –

derivation; Energy Principle; Practical applications of Bernoulli’s equation - Venturimeter,

orifice meter and pitot tube; Momentum principle; Forces exerted by fluid flow on pipe bend;

Vortex Flow – Free and Forced;

PAGE 2

Module-V DIMENSIONAL ANALYSIS

Dimensional Analysis and Dynamic Similitude - Definitions of Reynolds Number, Froude

Number, Mach Number, Weber Number and Euler Number; Buckingham’s π-Theorem.

Text Books:

1. C. S. P. Ojha, R. Berndtsson and P. N. Chadramouli, “Fluid Mechanics and Machinery”,

Oxford University Press, 2010.

2. P M Modi and S M Seth, “Hydraulics and Fluid Mechanics”, Standard Book House, 2014.

Reference Books:

1. K. Subramanya, “Theory and Applications of Fluid Mechanics”, Tata McGraw Hill.

2. R.L. Daugherty, J.B. Franzini and E.J. Finnemore, “Fluid Mechanics with Engineering

Applications”, International Student Edition, Tata Mc Graw Hill.

PAGE 3

Fluid Mechanics

Background

• There are three states of matter: solids, liquids and gases.

• Both liquids and gases are classified as fluids.

• Fluids do not resist a change in shape. Therefore fluids assume the shape of the

container they occupy.

• Liquids may be considered to have a fixed volume and therefore can have a

free surface. Liquids are almost incompressible.

• Conversely, gases are easily compressed and will expand to fill a container

they occupy.

• We will usually be interested in liquids, either at rest or in motion.

Liquid showing free surface Gas filling volume

Behaviour of fluids in containers

PAGE

MODULE – I

BASIC CONCEPTS AND DEFINATIONS

PAGE 4

Fluid Mechanics

Definition

The strict definition of a fluid is:

A fluid is a substance which conforms continuously under the action of

shearing forces.

To understand this, remind ourselves of what a shear force is:

Application and effect of shear force on a book

Definition Applied to Static Fluids

According to this definition, if we apply a shear force to a fluid it will deform and

take up a state in which no shear force exists. Therefore, we can say:

If a fluid is at rest there can be no shearing forces acting and therefore all

forces in the fluid must be perpendicular to the planes in which they act.

Note here that we specify that the fluid must be at rest. This is because, it is found

experimentally that fluids in motion can have slight resistance to shear force. This is

the source of viscosity.

PAGE

PAGE 5

Fluid Mechanics

Definition Applied to Fluids in Motion

For example, consider the fluid shown flowing along a fixed surface. At the surface

there will be little movement of the fluid (it will ‘stick’ to the surface), whilst further

away from the surface the fluid flows faster (has greater velocity):

If one layer of is moving faster than another layer of fluid, there must be shear forces

acting between them. For example, if we have fluid in contact with a conveyor belt

that is moving we will get the behaviour shown:

Ideal fluid Real (Viscous) Fluid

When fluid is in motion, any difference in velocity between adjacent layers has the

same effect as the conveyor belt does.

Therefore, to represent real fluids in motion we must consider the action of shear

forces.

PAGE

PAGE 6

Fluid Mechanics

Consider the small element of fluid shown, which is subject to shear force and has a

dimension s into the page. The force F acts over an area A = BC×s. Hence we have a

shear stress applied:

ForceStressAreaFA

τ

=

=

Any stress causes a deformation, or strain, and a shear stress causes a shear strain.

This shear strain is measured by the angle φ .

Remember that a fluid continuously deforms when under the action of shear. This is

different to a solid: a solid has a single value of φ for each value of τ . So the longer

a shear stress is applied to a fluid, the more shear strain occurs. However, what is

known from experiments is that the rate of shear strain (shear strain per unit time) is

related to the shear stress:

Shear stress Rate of shear strainShear stress Constant Rate of shear strain

∝= ×

PAGE

PAGE 7

Fluid Mechanics

We need to know the rate of shear strain. From the diagram, the shear strain is:

xy

φ =

If we suppose that the particle of fluid at E moves a distance x in time t, then, using

S Rθ= for small angles, the rate of shear strain is:

1xx tyt t y

uy

φ ⎛ ⎞∆= = ⋅⎜ ⎟∆ ⎝ ⎠

=

Where u is the velocity of the fluid. This term is also the change in velocity with

height. When we consider infinitesimally small changes in height we can write this in

differential form, du dy . Therefore we have:

constant dudy

τ = ×

This constant is a property of the fluid called its dynamic viscosity (dynamic because

the fluid is in motion, and viscosity because it is resisting shear stress). It is denoted

µ which then gives us:

Newton’s Law of Viscosity:

dudy

τ µ=

PAGE

PAGE 8

Fluid Mechanics

Generalized Laws of Viscosity

We have derived a law for the behaviour of fluids – that of Newtonian fluids.

However, experiments show that there are non-Newtonian fluids that follow a

generalized law of viscosity:

n

duA Bdy

τ⎛ ⎞

= + ⎜ ⎟⎝ ⎠

Where A, B and n are constants found experimentally. When plotted these fluids

show much different behaviour to a Newtonian fluid:

Behaviour of Fluids and Solids

PAGE

PAGE 9

Fluid Mechanics

In this graph the Newtonian fluid is represent by a straight line, the slope of which is

µ . Some of the other fluids are:

• Plastic: Shear stress must reach a certain minimum before flow commences.

• Pseudo-plastic: No minimum shear stress necessary and the viscosity

decreases with rate of shear, e.g. substances like clay, milk and cement.

• Dilatant substances; Viscosity increases with rate of shear, e.g. quicksand.

• Viscoelastic materials: Similar to Newtonian but if there is a sudden large

change in shear they behave like plastic.

• Solids: Real solids do have a slight change of shear strain with time, whereas

ideal solids (those we idealise for our theories) do not.

Lastly, we also consider the ideal fluid. This is a fluid which is assumed to have no

viscosity and is very useful for developing theoretical solutions. It helps achieve

some practically useful solutions.

PAGE

PAGE 10

Fluid Mechanics

Units

Fluid mechanics deals with the measurement of many variables of many different

types of units. Hence we need to be very careful to be consistent.

Dimensions and Base Units

The dimension of a measure is independent of any particular system of units. For

example, velocity may be in metres per second or miles per hour, but dimensionally,

it is always length per time, or 1L T LT−= . The dimensions of the relevant base units

of the Système International (SI) system are:

Unit-Free SI Units Dimension Symbol Unit Symbol

Mass M kilogram kg Length L metre m Time T second s

Temperature θ kelvin K

Derived Units

From these we have some relevant derived units (shown on the next page).

Checking the dimensions or units of an equation is very useful to minimize errors.

For example, if when calculating a force and you find a pressure then you know

you’ve made a mistake.

PAGE

PAGE 11

Fluid Mechanics

SI Unit Quantity Dimension

Derived Base

Velocity 1LT− m/s 1m s−

Acceleration 2LT− m/s2 2m s−

Force 2MLT− Newton, N 2kg m s−

Pressure

Stress -1 2ML T

Pascal, Pa

N/m2 -1 2kg m s−

Density -3ML kg/m3 -3kg m

Specific weight -2 2ML T− N/m3 -2 2kg m s−

Relative density Ratio Ratio Ratio

Viscosity -1 1ML T− Ns/m2 -1 1kg m s−

Energy (work) 2 2ML T− Joule, J

Nm 2 2kg m s−

Power 2 3ML T−

Watt, W

Nm/s

2 3kg m s−

Note: The acceleration due to gravity will always be taken as 9.81 m/s2.

PAGE

PAGE 12

Fluid Mechanics

SI Prefixes

SI units use prefixes to reduce the number of digits required to display a quantity.

The prefixes and multiples are:

Prefix Name Prefix Unit Multiple

Tera

Giga

Mega

Kilo

Hecto

Deka

Deci

Centi

Milli

Micro

Nano

Pico

T

G

M

k

h

da

d

c

m µ

n

p

1012

109

106

103

102

101

10-1

10-2

10-3

10-6

10-9

10-12

Be very particular about units and prefixes. For example:

• kN means kilo-Newton, 1000 Newtons;

• Kn is the symbol for knots – an imperial measure of speed;

• KN has no meaning;

• kn means kilo-nano – essentially meaningless.

PAGE

PAGE 13

• Water: 1000 kg/m3; • Mercury: 13546 kg/m3; • Air: 1.23 kg/m3; • Paraffin: 800 kg/m3.

times the acceleration due to gravity: gγ ρ=

Relative Density (Specific Gravity)

A dimensionless measure of the density of a substance with reference to the density

of some standard substance, usually water at 4°C:

density of substancerelative densitydensity of water

specific weight of substancespecific weight of water

s s

w w

ρ γρ γ

=

=

= =

PAGE

Mass Density

The mass per unit volume of a substance, usually denoted as ρ . Typical values are:

Specific Weight

The weight of a unit volume a substance, usually denoted as γ . Essentially density

PAGE 14

Fluid Mechanics

Bulk Modulus

In analogy with solids, the bulk modulus is the modulus of elasticity for a fluid. It is

the ratio of the change in unit pressure to the corresponding volume change per unit

volume, expressed as:

Change in Volume Chnage in pressureOriginal Volume Bulk Modulus

dV dpV K

=

−=

Hence:

dpK VdV

= −

In which the negative sign indicates that the volume reduces as the pressure

increases. The bulk modulus changes with the pressure and density of the fluid, but

for liquids can be considered constant for normal usage. Typical values are:

• Water: 2.05 GN/m3; • Oil: 1.62 GN/m3.

The units are the same as those of stress or pressure.

Viscosity

The viscosity of a fluid determines the amount of resistance to shear force.

Viscosities of liquids decrease as temperature increases and are usually not affected

by pressure changes. From Newton’s Law of Viscosity:

shear stressrate of shear straindu dy

τµ = =

Hence the units of viscosity are Pa s⋅ or 2N s m⋅ . This measure of viscosity is

known as dynamic viscosity and some typical values are given:

PAGE

PAGE 15

Fluid Mechanics

Problems - Properties

a) If 6 m3 of oil weighs 47 kN, find its specific weight, density, and relative density.

(Ans. 7.833 kN/m3, 798 kg/m3, 0.800)

b) At a certain depth in the ocean, the pressure is 80 MPa. Assume that the specific

weight at the surface is 10 kN/m3 and the average bulk modulus is 2.340 GPa.

Find:

a) the change in specific volume between the surface and the large depth;

b) the specific volume at the depth, and;

c) the specific weight at the depth.

(Ans. -0.335×10-4 m3/kg, 9.475×10-4 m3/kg, 10.35 kN/m3)

c) A 100 mm deep stream of water is flowing over a boundary. It is considered to

have zero velocity at the boundary and 1.5 m/s at the free surface. Assuming a

linear velocity profile, what is the shear stress in the water?

(Ans. 0.0195 N/m2)

d) The viscosity of a fluid is to be measured using a viscometer constructed of two

750 mm long concentric cylinders. The outer diameter of the inner cylinder is 150

mm and the gap between the two cylinders is 1.2 mm. The inner cylinder is

rotated at 200 rpm and the torque is measured to be 10 Nm.

a) Derive a generals expression for

the viscosity of a fluid using this

type of viscometer, and;

b) Determine the viscosity of the

fluid for the experiment above.

(Ans. 6 × 10-4 Ns/m2)

PAGE

PAGE 16

Fluid Mechanics

Pressure

In fluids we use the term pressure to mean:

The perpendicular force exerted by a fluid per unit area.

This is equivalent to stress in solids, but we shall keep the term pressure.

Mathematically, because pressure may vary from place to place, we have:

0

lim FpA∆→

∆=

∆

As we saw, force per unit area is measured in N/m2 which is the same as a pascal

(Pa). The units used in practice vary:

• 1 kPa = 1000 Pa = 1000 N/m2

• 1 MPa = 1000 kPa = 1 × 106 N/m2

• 1 bar = 105 Pa = 100 kPa = 0.1 MPa

• 1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bars = 1013.25 millibars

For reference to pressures encountered on the street which are often imperial:

• 1 atm = 14.696 psi (i.e. pounds per square inch)

• 1 psi = 6894.7 Pa ≈ 6.89 kPa ≈ 0.007 MPa

MODULE – II

FLUID STATICS

PAGE 17

Fluid Mechanics

Pressure Reference Levels

The pressure that exists anywhere in the universe is called the absolute pressure, absP .

This then is the amount of pressure greater than a pure vacuum. The atmosphere on

earth exerts atmospheric pressure, atmP , on everything in it. Often when measuring

pressures we will calibrate the instrument to read zero in the open air. Any measured

pressure, measP , is then a positive or negative deviation from atmospheric pressure.

We call such deviations a gauge pressure, gaugeP . Sometimes when a gauge pressure

is negative it is termed a vacuum pressure, vacP .

The above diagram shows:

(a) the case when the measured pressure is below atmospheric pressure and so is a

negative gauge pressure or a vacuum pressure;

(b) the more usual case when the measured pressure is greater than atmospheric

pressure by the gauge pressure.

PAGE 18

Fluid Mechanics

3.2 Pressure in a Fluid

Statics of Definition

We applied the definition of a fluid to the static case previously and determined that

there must be no shear forces acting and thus only forces normal to a surface act in a

fluid.

For a flat surface at arbitrary angle we have:

A curved surface can be examined in sections:

PAGE 19

Fluid Mechanics

And we are not restricted to actual solid-fluid interfaces. We can consider imaginary

planes through a fluid:

Pascal’s Law

This law states:

The pressure at a point in a fluid at rest is the same in all directions.

To show this, we will consider a very small wedge of fluid surrounding the point.

This wedge is unit thickness into the page:

PAGE 20

Fluid Mechanics

As with all static objects the forces in the x and y directions should balance. Hence:

0xF =∑ : sin 0y sp y p s θ⋅ ∆ − ⋅ ∆ ⋅ =

But ysins

θ ∆=

∆, therefore:

0y s

y s

y s

yp y p ss

p y p y

p p

∆⋅ ∆ − ⋅ ∆ ⋅ =

∆⋅ ∆ = ⋅ ∆

=

0yF =∑ : cos 0x sp x p s θ⋅ ∆ − ⋅ ∆ ⋅ =

But xcoss

θ ∆=

∆, therefore:

0x s

x s

x s

xp x p ss

p x p xp p

∆⋅ ∆ − ⋅ ∆ ⋅ =

∆⋅ ∆ = ⋅ ∆

=

Hence for any angle:

y x sp p p= =

And so the pressure at a point is the same in any direction. Note that we neglected the

weight of the small wedge of fluid because it is infinitesimally small. This is why

Pascal’s Law is restricted to the pressure at a point.

PAGE 21

Fluid Mechanics

Pressure Variation with Depth

Pressure in a static fluid does not change in the horizontal direction as the horizontal

forces balance each other out. However, pressure in a static fluid does change with

depth, due to the extra weight of fluid on top of a layer as we move downwards.

Consider a column of fluid of arbitrary cross section of area, A:

Column of Fluid Pressure Diagram

Considering the weight of the column of water, we have:

0yF =∑ : ( )1 2 1 2 0p A A h h p Aγ+ − − =

PAGE 22

Fluid Mechanics

Obviously the area of the column cancels out: we can just consider pressures. If we

say the height of the column is 2 1h h h= − and substitute in for the specific weight, we

see the difference in pressure from the bottom to the top of the column is:

2 1p p ghρ− =

This difference in pressure varies linearly in h, as shown by the Area 3 of the pressure

diagram. If we let 1 0h = and consider a gauge pressure, then 1 0p = and we have:

2p ghρ=

Where h remains the height of the column. For the fluid on top of the column, this is

the source of 1p and is shown as Area 1 of the pressure diagram. Area 2 of the

pressure diagram is this same pressure carried downwards, to which is added more

pressure due to the extra fluid.

To summarize:

The gauge pressure at any depth from the surface of a fluid is:

p ghρ=

PAGE 23

Fluid Mechanics

Summary

1. Pressure acts normal to any surface in a static fluid;

2. Pressure is the same at a point in a fluid and acts in all directions;

3. Pressure varies linearly with depth in a fluid.

By applying these rules to a simple swimming pool, the pressure distribution around

the edges is as shown:

Note:

1. Along the bottom the pressure is constant due to a constant depth;

2. Along the vertical wall the pressure varies linearly with depth and acts in the

horizontal direction;

3. Along the sloped wall the pressure again varies linearly with depth but also

acts normal to the surface;

4. At the junctions of the walls and the bottom the pressure is the same.

PAGE 24

Fluid Mechanics

Problems - Pressure

1. Sketch the pressure distribution applied to the container by the fluid:

2. For the dam shown, sketch the pressure distribution on line AB and on the

surface of the dam, BC. Sketch the resultant force on the dam.

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Fluid Mechanics

3. For the canal gate shown, sketch the pressure distributions applied to it. Sketch

the resultant force on the gate? If 1 6.0 mh = and 2 4.0 mh = , sketch the

pressure distribution to the gate. Also, what is the value of the resultant force

on the gate and at what height above the bottom of the gate is it applied?

PAGE 26

Fluid Mechanics

3.3 Pressure Measurement

Pressure Head

Pressure in fluids may arise from many sources, for example pumps, gravity,

momentum etc. Since p ghρ= , a height of liquid column can be associated with the

pressure p arising from such sources. This height, h, is known as the pressure head.

Example:

The gauge pressure in a water mains is 50 kN/m2, what is the pressure head?

The pressure head equivalent to the pressure in the pipe is just:

350 10

1000 9.815.1 m

p ghphg

ρ

ρ

=

=

×=

×≈

So the pressure at the bottom of a 5.1 m deep swimming pool is the same as the

pressure in this pipe.

Manometers

A manometer (or liquid gauge) is a pressure measurement device which uses the

relationship between pressure and head to give readings.

In the following, we wish to measure the pressure of a fluid in a pipe.

PAGE 27

Fluid Mechanics

Piezometer

This is the simplest gauge. A small vertical tube is connected to the pipe and its top is

left open to the atmosphere, as shown.

The pressure at A is equal to the pressure due to the column of liquid of height 1h :

1Ap ghρ=

Similarly,

2Bp ghρ=

PAGE 28

Fluid Mechanics

The problem with this type of gauge is that for usual civil engineering applications

the pressure is large (e.g. 100 kN/m2) and so the height of the column is impractical

(e.g.10 m).

Also, obviously, such a gauge is useless for measuring gas pressures.

U-tube Manometer

To overcome the problems with the piezometer, the U-tube manometer seals the fluid

by using a measuring (manometric) liquid:

Choosing the line BC as the interface between the measuring liquid and the fluid, we

know:

Pressure at B, Bp = Pressure at C, Cp

For the left-hand side of the U-tube:

PAGE 29

Fluid Mechanics

1B Ap p ghρ= +

For the right hand side:

2C manp ghρ=

Where we have ignored atmospheric pressure and are thus dealing with gauge

pressures. Thus:

1 2

B C

A man

p pp gh ghρ ρ

=

+ =

And so:

2 1A manp gh ghρ ρ= −

Notice that we have used the fact that in any continuous fluid, the pressure is the

same at any horizontal level.

PAGE 30

Fluid Mechanics

Differential Manometer

To measure the pressure difference between two points we use a u-tube as shown:

Using the same approach as before:

Pressure at C, Cp = Pressure at D, Dp

( )A B manp ga p g b h ghρ ρ ρ+ = + − +

Hence the pressure difference is:

( ) ( )A B manp p g b a hgρ ρ ρ− = − + −

PAGE 31

Fluid Mechanics

Problems – Pressure Measurement

1. What is the pressure head, in metres of water, exerted by the atmosphere?

(Ans. 10.3 m)

2. What is the maximum gauge pressure of water that can be measured using a

piezometer 2.5 m high?

(Ans. 24.5 kN/m2)

3. A U-tube manometer is used to measure the pressure of a fluid of density 800

kg/m3. If the density of the manometric liquid is 13.6 × 103 kg/m3, what is the

gauge pressure in the pipe if

(a) 1 0.5 mh = and D is 0.9 m above BC;

(b) 1 0.1 mh = and D is 0.2 m below BC?

(Ans. 116.15 kN/m2, -27.45 kN/m2)

4. A differential manometer is used to measure the pressure difference between

two points in a pipe carrying water. The manometric liquid is mercury and the

points have a 0.3 m height difference. Calculate the pressure difference when

0.7 mh = .

(Ans. 89.47 kN/m2)

5. For the configuration shown, calculate the weight of the piston if the gauge

pressure reading is 70 kPa.

PAGE 32

Fluid Mechanics

(Ans. 61.6 kN)

6. A hydraulic jack having a ram 150 mm in diameter lifts a weight W = 20 kN

under the action of a 30 mm plunger. What force is required on the plunger to

lift the weight?

(Ans. 800 N)

PAGE 33

Fluid Mechanics

3.4 Fluid Action on Surfaces

Plane Surfaces

We consider a plane surface, PQ, of area A, totally immersed in a liquid of density

ρ and inclined at an angle φ to the free surface:

Side Elevation

Front Elevation

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Fluid Mechanics

If the plane area is symmetrical about the vertical axis OG, then 0d = . We will

assume that this is normally the case.

Find Resultant Force:

The force acting on the small element of area, Aδ , is:

R p A gy Aδ δ ρ δ= ⋅ = ⋅

The total force acting on the surface is the sum of all such small forces. We can

integrate to get the force on the entire area, but remember that y is not constant:

R gy A

g y A

ρ δ

ρ δ

= ⋅

= ⋅

∫∫

But y Aδ⋅∫ is just the first moment of area about the surface. Hence:

R gAyρ=

Where y is the distance to the centroid of the area (point G) from the surface.

Vertical Point Where Resultant Acts:

The resultant force acts perpendicular to the plane and so makes an angle 90 φ° − to

the horizontal. It also acts through point C, the centre of pressure, a distance D below

the free surface. To determine the location of this point we know:

Sum of moments of forces

Moment of about on all elements about

R OO

=

PAGE 35

Fluid Mechanics

Examining a small element first, and since siny s φ= , the moment is:

( )

( )2

Moment of about sin

sin

R O g s A s

g s A

δ ρ φ δ

ρ φ δ

= ⋅⎡ ⎤⎣ ⎦

= ⋅

In which the constants are taken outside the bracket. The total moment is thus:

2Moment of about sinR O g s Aρ φ δ= ⋅ ⋅∫

But 2s Aδ⋅∫ is the second moment of area about point O or just OI . Hence we have:

2

Moment of about sinsin

sinsin

sin

O

O

O

O

R O g IgAy OC g I

DAy I

IDAy

ρ φρ ρ φ

φφ

φ

= ⋅

× = ⋅

× = ⋅

= ⋅

If we introduce the parallel axis theorem:

( )2

2

sin

O G

G

I I A OG

yI Aφ

= + ×

⎛ ⎞= + ⋅ ⎜ ⎟

⎝ ⎠

Hence we have:

2 2

2

sinsin

G

G

I AyDAy

IyAy

φφ

+= ⋅

= +

Hence, the centre of pressure, point C, always lies below the centroid of the area, G.

PAGE 36

Fluid Mechanics

Plane Surface Properties

PAGE 37

Fluid Mechanics

Plane Surfaces – Example

Problem

Calculate the forces on the hinges supporting the canal gates as shown. The hinges

are located 0.6 m from the top and bottom of each gate.

Plan

Elevation

PAGE 38

Fluid Mechanics

Solution

We will consider gate AB, but all arguments will equally apply to gate BC.

The length of the gate is 3.0 sin30 3.464 mL = = . The resultant pressure on the gate

from the high water side is:

( )1 1 1

3 4.510 9.81 3.464 4.52

344 kN

P gA yρ=

= × × × ×

=

Similarly for the low water side:

( )2 2 2

3 3.010 9.81 3.464 3.02

153 kN

P gA yρ=

= × × × ×

=

The net resultant force on the gate is:

1 2 344 153 191 kNP P P= − = − =

To find the height at which this acts, take moments about the bottom of the gate:

1 1 2 2

4.5 3344 153 363 kNm3 3

Ph Ph P h= +

= × − × =

Hence:

363 1.900 m191

h = =

Examining a free-body diagram of the gate, we see that the interaction force between

the gates, BR , is shown along with the total hinge reactions, AR and the net applied

hydrostatic force, P. Relevant angles are also shown. We make one assumption: the

PAGE 39

Fluid Mechanics

interaction force between the gates acts perpendicular on the contact surface between

the gates. Hence BR acts vertically downwards on plan.

From statics we have Moments about 0A =∑ :

( )sin30 02

12 2

B

B

B

LP R L

PR

R P

⋅ + =

⋅ =

=

Hence 191 kNBR = and the component of BR perpendicular to the gate is 95.5 kN.

By the sum of forces perpendicular to the gate, the component of AR perpendicular to

the gate must also equal 95.5 kN. Further, taking the sum of forces along the gate, the

components of both AR and BR must balance and so 191 kNA BR R= = .

The resultant forces AR and BR must act at the same height as P in order to have

static equilibrium. To find the force on each hinge at A, consider the following figure:

PAGE 40

Fluid Mechanics

Taking moments about the bottom hinge:

( ) ( )

( ),

,

0.6 6 0.6 0.6 0

191 1.900 0.651.7 kN

4.8

A A top

A top

R h R

R

− − − − =

−= =

And summing the horizontal forces:

, ,

, 191 51.7 139.3 kNA A top A btm

A btm

R R R

R

= +

= − =

It makes intuitive sense that the lower hinge has a larger force. To design the bolts

connecting the hinge to the lock wall the direct tension and shear forces are required.

Calculate these for the lower hinge.

(Ans. T = 120.6 kN, V = 69.7 kN)

PAGE 41

Fluid Mechanics

Curved Surfaces

For curved surfaces the fluid pressure on the infinitesimal areas are not parallel and

so must be combined vectorially. It is usual to consider the total horizontal and

vertical force components of the resultant.

Surface Containing Liquid

Consider the surface AB which contains liquid as shown below:

• Horizontal Component

Using the imaginary plane ACD we can immediately see that the horizontal

component of force on the surface must balance with the horizontal force ACF .

Hence:

Force on projection of surfaceonto a vertical planexF =

PAGE 42

Fluid Mechanics

xF must also act at the same level as ACF and so it acts through the centre of

pressure of the projected surface.

• Vertical Component

The vertical component of force on the surface must balance the weight of liquid

above the surface. Hence:

Weight of liquid directly above the surfaceyF =

Also, this component must act through the centre of gravity of the area ABED,

shown as G on the diagram.

• Resultant

The resultant force is thus:

2 2x yF F F= +

This force acts through the point O when the surface is uniform into the page, at

an angle of:

1tan y

x

FF

θ −=

to the horizontal. Depending on whether the surface contains or displaces water

the angle is measured clockwise (contains) or anticlockwise (displaces) from the

horizontal.

PAGE 43

Fluid Mechanics

Surface Displacing Liquid

Consider the surface AB which displaces liquid as shown below:

• Horizontal Component

Similarly to the previous case, the horizontal component of force on the surface

must balance with the horizontal force EBF . Hence again:

Force on projection of surfaceonto a vertical planexF =

This force also acts at the same level as EBF as before.

• Vertical Component

In this case we imagine that the area ABDC is filled with the same liquid. In this

case yF would balance the weight of the liquid in area ABDC. Hence:

PAGE 44

Fluid Mechanics

Weight of liquid which would lie above the surfaceyF =

This component acts through the centre of gravity of the imaginary liquid in area

ABDC, shown as G on the diagram.

The resultant force is calculated as before.

Both of these situations can be summed up with the following diagram:

PAGE 45

Fluid Mechanics

Curved Surfaces – Example

Problem

Determine the resultant force and its direction on the gate shown:

Solution

The horizontal force, per metre run of the gate, is that of the surface projected onto a

vertical plane of length CB:

( )3 610 9.81 6 12

176.6 kN

x CB CBF gA yρ=

⎛ ⎞= × × × × ⎜ ⎟⎝ ⎠

=

And this acts at a depth 2 6 4 m3

h = ⋅ = from the surface. The vertical force is the

weight of the imaginary water above AB:

23 610 9.81 1

4277.4 kN

yF π⎛ ⎞= × ×⎜ ⎟

⎝ ⎠=

PAGE 46

Fluid Mechanics

In which 2 4Rπ is the area of the circle quadrant. The vertical force is located at:

4 4 6 2.55 m3 3

Rxπ π

×= = =

to the left of line BC. The resultant force is thus:

2 2

2 2176.6 277.4328.8 kN

x yF F F= +

= +=

And acts at an angle:

1

1

tan

277.4tan176.6

57.5

y

x

FF

θ −

−

=

=

=

measured anticlockwise to the horizontal. The resultant passes through point C. Also,

as the force on each infinitesimal length of the surface passes through C, there should

be no net moment about C. Checking this:

Moments about 0176.6 4 277.4 2.55 0

706.4 707.4 0

C =

× − × =− ≈

∑

The error is due to rounding carried out through the calculation.

PAGE 47

Fluid Mechanics

Problems – Fluid Action on Surfaces

1. You are in a car that falls into a lake to a depth as shown below. What is the

moment about the hinges of the car door (1.0 × 1.2 m) due to the hydrostatic

pressure? Can you open the door? What should you do?

(Ans. 50.6 kNm, ?, ?)

2. A sluice gate consist of a quadrant of a circle of radius 1.5 m pivoted at its

centre, O. When the water is level with the gate, calculate the magnitude and

direction of the resultant hydrostatic force on the gate and the moment required

to open the gate. The width of the gate is 3 m and it has a mass of 6 tonnes.

(Ans. 61.6 kN, 57˚, 35.3 kNm)

PAGE 48

Fluid Mechanics

3. The profile of a masonry dam is an arc of a circle, the arc having a radius of 30

m and subtending an angle of 60˚ at the centre of curvature which lies in the

water surface. Determine: (a) the load on the dam in kN/m length; (b) the

position of the line of action to this pressure.

(Ans. 4280 kN/m, 19.0 m)

4. The face of a dam is curved according to the relation 2 2.4y x= where y and x

are in meters, as shown in the diagram. Calculate the resultant force on each

metre run of the dam. Determine the position at which the line of action of the

resultant force passes through the bottom of the dam.

(Ans. 1920 kN, 14.15 m)

PAGE 49

Fluid Mechanics

Introduction

Hydrostatics involves only a few variables: ρ , g, and h, and so the equations

developed are relatively simple and experiment and theory closely agree. The study

of fluids in motion is not as simple and accurate. The main difficulty is viscosity.

By neglecting viscosity (an ideal fluid), we do not account for the shear forces which

oppose flow. Based on this, reasonably accurate and simple theories can be derived..

Using experimental results, these theories can then be calibrated by using

experimental coefficients. They then inherently allow for viscosity.

As we will be dealing with liquids, we will neglect the compressibility of the liquid.

This is incompressible flow. This is not a valid assumption for gases.

Classification of Flow Pattern

There are different patterns of fluid flow, usually characterized by time and distance:

• Time: A flow is steady if the parameters describing it (e.g. flow rate, velocity,

pressure, etc.) do not change with time. Otherwise a flow is unsteady.

• Distance: A flow is uniform if the parameters describing the flow do not

change with distance. In non-uniform flow, the parameters change from point

to point along the flow.

From these definitions almost all flows will be one of:

MODULE – III

FLUID KINEMATICS

PAGE 50

Fluid Mechanics

Steady uniform flow

Discharge (i.e. flow rate, or volume per unit time) is constant with time and the cross

section of the flow is also constant. Constant flow through a long straight prismatic

pipe is an example.

Steady non-uniform flow

The discharge is constant with time, but the cross-section of flow changes. An

example is a river with constant discharge, as the cross section of a river changes

from point to point.

Unsteady uniform flow

The cross-section is constant but the discharge changes with time resulting in

complex flow patterns. A pressure surge in a long straight prismatic pipe is an

example.

Unsteady non-uniform flow

Both discharge and cross section vary. A flood wave in a river valley is an example.

This is the most complex type of flow.

Visualization

To picture the motion of a fluid, we start by examining the motion of a single fluid

‘particle’ over time, or a collection of particles at one instant. This is the flow path of

the particle(s), or a streamline:

PAGE 51

Fluid Mechanics

At each point, each particle has both velocity and acceleration vectors:

A streamline is thus tangential to the velocity vectors of the particles. Hence:

• there can be no flow across a streamline;

• therefore, streamlines cannot cross each other, and;

• once fluid is on a streamline it cannot leave it.

We extend this idea to a collection of paths of fluid particles to create a streamtube:

Streamlines and streamtubes are theoretical notions. In an experiment, a streakline is

formed by injecting dye into a fluid in motion. A streakline therefore approximates a

streamline (but is bigger because it is not an individual particle).

PAGE 52

Fluid Mechanics

Dimension of Flow

Fluid flow is in general three-dimensional in nature. Parameters of the flow can vary

in the x, y and z directions. They can also vary with time. In practice we can reduce

problems to one- or two-dimensional flow to simplify. For example:

One dimensional flow

A two-dimensional streamtube

Flow over an obstruction

PAGE 53

Fluid Mechanics

Fundamental Equations

To develop equations describing fluid flow, we will work from relevant fundamental

physical laws.

The Law of Conservation of Matter

Matter cannot be created nor destroyed (except in a nuclear reaction), but may be

transformed by chemical reaction. In fluids we neglect chemical reactions and so we

deal with the conservation of mass.

The Law of Conservation of Energy

Energy cannot be created nor destroyed, it can only be transformed from one form to

another. For example, the potential energy of water in a dam is transformed to kinetic

energy of water in a pipe. Though we will later talk of ‘energy losses’, this is a

misnomer as none is actually lost but transformed to heat and other forms.

The Law of Conservation of Momentum

A body in motion remains in motion unless some external force acts upon it. This is

Newton’s Second Law:

Rate of change

Force =of momentum

( )d mvF

dtdvmdt

ma

=

=

=

To apply these laws to fluids poses a problem, since fluid is a continuum, unlike rigid

bodies. Hence we use the idea of a ‘control volume’.

PAGE 54

Fluid Mechanics

Control Volume

A control volume is an imaginary region within a body of flowing fluid, usually at

fixed location and of a fixed size:

It can be of any size and shape so we choose shapes amenable to simple calculations.

Inside the region all forces cancel out, and we can concentrate on external forces. It

can be picture as a transparent pipe or tube, for example.

PAGE 55

Fluid Mechanics

4.2 The Continuity Equation

Development

Applying the Law of Conservation of Mass to a control volume, we see:

Rate of mass Rate of mass Rate of mass

= +entering leaving increase

For steady incompressible flow, the rate of mass increase is zero and the density of

the fluid does not change. Hence:

Rate of mass Rate of mass

=entering leaving

The rate of mass change can be expressed as:

Rate of mass Fluid Volume

=change density per second

×

Using Q for flow rate, or volume per second (units: m3/s, dimensions: L3T-1):

in outQ Qρ ρ=

And as before, assuming that the flow is incompressible:

in outQ Q=

PAGE 56

Fluid Mechanics

Consider a small length of streamtube:

The fluid at 1-1 moves a distance of s vt= to 2-2. Therefore in 1 second it moves a

distance of v. The volume moving per second is thus:

Q Av=

Thus, for an arbitrary streamtube, as shown, we have:

1 1 2 2Av A v=

PAGE 57

Fluid Mechanics

A typical application of mass conservation is at pipe junctions:

From mass conservation we have:

1 2 3

1 1 2 2 3 3

Q Q QAv A v A v

= +

= +

If we consider inflow to be positive and outflow negative, we have:

No. of Nodes

1

0i ii

Av=

=∑

PAGE 58

Fluid Mechanics

Mass Conservation – Example

Problem

Water flows from point A to points D and E as shown. Some of the flow parameters

are known, as shown in the table. Determine the unknown parameters.

Section Diameter

(mm)

Flow Rate

(m3/s)

Velocity

(m/s)

AB 300 ? ?

BC 600 ? 1.2

CD ? 3 42Q Q= 1.4

CE 150 4 30.5Q Q= ?

PAGE 59

Fluid Mechanics

Solution

From the law of mass conservation we can see:

1 2Q Q=

And as total inflow must equal total outflow:

1

3 4

3 3

3

0.51.5

outQ QQ QQ Q

Q

=

= += +

=

We must also work out the areas of the pipes, 2

4i

idA π

= . Hence:

A1 = 0.0707 m3 A2 = 0.2827 m3 A4 = 0.0177 m3

Starting with our basic equation, Q Av= , we can only solve for 2Q from the table:

( )( )2

3

0.2827 1.2

0.3393 m /

Q

s

=

=

We know that 1 2Q Q= and so we can now calculate 3Q from previous:

1 3

313

1.50.3393 0.2262 m /s

1.5 1.5

Q QQQ

=

= = =

PAGE 60

Fluid Mechanics

And so,

334

0.2262 0.1131 m /s2 2

QQ = = =

Thus we have all the flows. The unknown velocities are:

11

1

0.3393 4.8 m/s0.0707

QvA

= = =

44

4

0.1131 6.4 m/s0.0177

QvA

= = =

And lastly, the diameter of pipe CD is:

233

3

0.2262 0.1616 m1.4

QAv

= = =

33

4 0.454 mAdπ

= =

And so it is likely to be a 450 mm ∅ pipe.

Note that in a problem such as this the individual calculations do not pose a problem.

It is the strategy used to solve it that is key. In this problem, we started from some

knowns and one equation. Even though we couldn’t see all the way to the end from

Step 1, with each new calculation another possibility opened up. This is the ‘art of

problem solving’ and it can only be learned by practice!

PAGE 61

Fluid Mechanics

The real behaviour of fluids flowing is well described by an experiment carried out

by Reynolds in 1883. He set up the following apparatus:

The discharge is controlled by the valve and the small ‘filament’ of dye (practically a

streamline) indicates the behaviour of the flow. By changing the flow Reynolds

noticed:

• At low flows/velocities the filament remained intact and almost straight. This

type of flow is known as laminar flow, and the experiment looks like this:

MODULE – IV

FLUID DYNAMICS

PAGE 62

Fluid Mechanics

• At higher flows the filament began to oscillate. This is called transitional flow

and the experiment looks like:

• Lastly, for even higher flows again, the filament is found to break up

completely and gets diffused over the full cross-section. This is known as

turbulent flow:

Reynolds experimented with different fluids, pipes and velocities. Eventually he

found that the following expression predicted which type of flow was found:

Re vlρµ

=

In which Re is called the Reynolds Number; ρ is the fluid density; v is the average

velocity; l is the characteristic length of the system (just the diameter for pipes), and;

µ is the fluid viscosity. The Reynolds Number is a ration of forces and hence has no

units.

Flows in pipes normally conform to the following:

• Re 2000< : gives laminar flow;

PAGE 63

Fluid Mechanics

• 2000 Re 4000< < : transitional flow;

• Re 4000> : turbulent flow.

These values are only a rough guide however. Laminar flows have been found at

Reynolds Numbers far beyond even 4000.

For example, if we consider a garden hose of 15 mm diameter then the limiting

average velocity for laminar flow is:

( ) ( )3

3

Re

10 0.0152000

0.55 100.073 m/s

vl

v

v

ρµ

−

=

=×

=

This is a very low flow and hence we can see that in most applications we deal with

turbulent flow.

The velocity below which there is no turbulence is called the critical velocity.

PAGE 64

Fluid Mechanics

Characteristics of Flow Types

For laminar flow:

• Re < 2000;

• ‘low’ velocity;

• Dye does not mix with water;

• Fluid particles move in straight lines;

• Simple mathematical analysis possible;

• Rare in practical water systems.

Transitional flow

• 2000 < Re < 4000

• ‘medium’ velocity

• Filament oscillates and mixes slightly.

Turbulent flow

• Re > 4000;

• ‘high’ velocity;

• Dye mixes rapidly and completely;

• Particle paths completely irregular;

• Average motion is in the direction of the flow;

• Mathematical analysis very difficult - experimental measures are used;

• Most common type of flow.

PAGE 65

Fluid Mechanics

4.3 The Energy Equation

Development

We apply the Law of Conservation of Energy to a control volume. To do so, we must

identify the forms of energy in the control volume. Consider the following system:

The forms of energy in this system are:

• Pressure energy:

The pressure in a fluid also does work by generating force on a cross section

which then moves through a distance. This is energy since work is energy.

• Kinetic energy:

This is due to the motion of the mass of fluid.

• Potential energy:

This is due to the height above an arbitrary datum.

PAGE 66

Fluid Mechanics

Pressure Energy

The combination of flow and pressure gives us work. The pressure results in a force

on the cross section which moves through a distance L in time tδ . Hence the pressure

energy is the work done on a mass of fluid entering the system, which is:

1 1m A Lρ=

And so the pressure energy at the entry is:

1 1PrE pAL p A L= =

Kinetic Energy

From classical physics, the kinetic energy of the mass entering is:

2 21 1 1

1 1KE2 2

mv A Lvρ= =

Potential Energy

The potential energy of the mass entering, due to the height above the datum is:

1 1 1PE mgz A Lgzρ= =

Total Energy

The total energy at the entry to the system is just the sum:

* 21 1 1 1 1 1 1 1 1

12

H p A L A Lv A Lgzρ ρ= + +

Be careful to

distinguish the

density ρ and the

pressure p.

PAGE 67

Fluid Mechanics

Final Form

It is more usual to consider the energy per unit weight, and so we divide through by

1 1mg gA Lρ= :

*1

11 1

21 1 1 1 1 1 1 1

1 1 1 1 1 12

1 11

1

12

2

HHgA L

p A L A Lv A LgzgA L gA L gA L

p v zg g

ρ

ρ ρρ ρ ρ

ρ

=

= + +

= + +

Similarly, the energy per unit weight leaving the system is:

2

2 22 2

2 2p vH z

g gρ= + +

Also, the energy entering must equal the energy leaving as we assume the energy

cannot change. Also, assuming incompressibility, the density does not change:

1 2

2 21 1 2 2

1 22 2

H H

p v p vz zg g g gρ ρ

=

+ + = + +

And so we have Bernoulli’s Equation:

2 2

1 1 2 21 2 constant

2 2p v p vz z Hg g g gρ ρ

+ + = + + = =

PAGE 68

Fluid Mechanics

Comments

From Bernoulli’s Equation we note several important aspects:

1. It is assumed that there is no energy taken from or given to the fluid between

the entry and exit. This implies the fluid is frictionless as friction generates

heat energy which would be an energy loss. It also implies that there is no

energy added, say by a pump for example.

2. Each term of the equation has dimensions of length, L, and units of metres.

Therefore each term is known as a head:

• Pressure head: pgρ

;

• Kinetic or velocity head: 2

2vg

;

• Potential or elevation head: z .

3. The streamtube must have very small dimensions compared to the heights

above the datum. Otherwise the height to the top of a cross-section would be

different to the height to the bottom of a cross-section. Therefore, Bernoulli’s

Equation strictly only applies to streamlines.

We have derived the equation from energy considerations. It can also be derived by

force considerations upon an elemental piece of fluid.

PAGE 69

Fluid Mechanics

Energy Equation – Example

Problem

For the siphon shown, determine the discharge and pressure heads at A and B given

that the pipe diameter is 200 mm and the nozzle diameter is 150 mm. You may

neglect friction in the pipe.

PAGE 70

Fluid Mechanics

Solution

To find the discharge (or flow) apply Bernoulli’s Equation along the streamline

connecting points 1 and 2. To do this note:

• Both 1p and 2p are at atmospheric pressure and are taken to be zero;

• 1v is essentially zero.

2 21 1 2 2

1 20 00

22

1 2

2 2

2

p v p vz zg g g g

vz zg

ρ ρ= ==

⎛ ⎞⎛ ⎞ ⎛ ⎞+ + = + +⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

− =

Hence, from the figure:

22

2

1.22 0.152 9.815.18 m/s

v

v

+ =×

=

And using continuity:

( )2 2 2

2

3

0.155.18

40.092 m /

Q A v

s

π

=

= ⋅

=

For the pressure head at A, apply Bernoulli’s equation from point 1 to A:

2 2

1 11

0 02 2A A

Ap v p vz zg g g gρ ρ= =

⎛ ⎞⎛ ⎞+ + = + +⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

PAGE 71

Fluid Mechanics

Hence:

( )2

1 2A A

Ap vz zg gρ

= − −

Again using continuity between point 2 and A and the diameter of the pipe at A:

( )

2

2

0.092

0.20.092

42.93 m/s

A

A A

A

A

Q QA v

v

v

π

==

⋅ =

=

Hence the kinetic head at A is just 2

0.44 m2

Avg

= , and so:

2.44 0.44

2.88 m

Apgρ

= − −

= −

This is negative gauge pressure indicating suction. However, it is still a positive

absolute pressure.

Similarly to A, at B we have, B Av v= and 1 1.22 mBz z− = and so:

( )2

1 21.22 0.440.78 m

B BB

p vz zg gρ

= − −

= −=

PAGE 72

Fluid Mechanics

Discharge Measurement in Pipelines

We consider two kinds of meters based on constricting the flow: the Venturi meter

and the Orifice meter, as shown.

The constriction in these meters causes a difference in pressure between points 1 and

2, and it is this pressure difference that enables the discharge to be measured.

Applying the energy equation between the inlet and the constriction:

2 2

1 1 2 2

2 2p v p vg g g gρ ρ

+ = +

Thus the difference in height in the piezometer is:

2 2

1 2 1 2

2p p v vh

g gρ− −

= =

And from continuity 1 1 2 2Q Av A v= = , and using 1 2k A A= we get:

PAGE 73

Fluid Mechanics

( )2 2 2 2 2 22 1 2 1 1v A A v k v= =

And also:

2 21 22v gh v= +

Which after substituting for 22v and rearranging gives:

1 2

21

ghvk

=−

Hence the discharge is:

1 1

1 2

21

Q Av

ghAk

=

=−

This equation neglects all losses. The actual discharge requires the introduction of the

coefficient of discharge, dC :

1 2

21actual d

ghQ C Ak

=−

For properly designed Venturi meters, dC is about 0.97 to 0.99 but for the Orifice

meter it is much lower at about 0.65.

PAGE 74

Fluid Mechanics

4.4 The Momentum Equation

Development

We consider again a general streamtube:

In a given time interval, tδ , we have:

1 1

2 2

momentum enteringmomentum leaving

Q t vQ t v

ρ δρ δ

==

From continuity we know 1 2Q Q Q= = . Thus the force required giving the change in

momentum between the entry and exit is, from Newton’s Second Law:

( )d mvF

dt=

( )

( )

2 1

2 1

Q t v vF

tQ v v

ρ δδ

ρ

−=

= −

This is the force acting on a fluid element in the direction of motion. The fluid exerts

an equal but opposite reaction to its surroundings.

PAGE 75

Fluid Mechanics

Application – Fluid Striking a Flat Surface

Consider the jet of fluid striking the surface as shown:

The velocity of the fluid normal to the surface is:

cosnormalv v θ=

This must be zero since there is no relative motion at the surface. This then is also the

change in velocity that occurs normal to the surface. Also, the mass flow entering the

control volume is:

Q Avρ ρ=

Hence:

PAGE 76

Fluid Mechanics

( )

( )( )2

cos

cos

d mvF

dtAv v

Av

ρ θ

ρ θ

=

=

=

And if the plate is perpendicular to the flow then:

2F Avρ=

Notice that the force exerted by the fluid on the surface is proportional to the velocity

squared. This is important for wind loading on buildings. For example, the old wind

loading code CP3: Chapter V gives as the pressure exerted by wind as:

20.613 sq v= (N/m2)

In which sv is the design wind speed read from maps and modified to take account of

relevant factors such as location and surroundings.

PAGE 77

Fluid Mechanics

Application – Flow around a bend in a pipe

Consider the flow around the bend shown below. We neglect changes in elevation

and consider the control volume as the fluid between the two pipe joins.

The net external force on the control volume fluid in the x-direction is:

1 1 2 2 cos xp A p A Fθ− +

In which xF is the force on the fluid by the pipe bend (making it ‘go around the

corner’). The above net force must be equal to the change in momentum, which is:

( )2 1cosQ v vρ θ −

Hence:

( )( )

( ) ( )

1 1 2 2 2 1

2 1 1 1 2 2

2 2 2 1 1 1

cos cos

cos cos

cos

x

x

p A p A F Q v v

F Q v v p A p A

Qv p A Qv p A

θ ρ θ

ρ θ θ

ρ θ ρ

− + = −

= − − +

= + − +

PAGE 78

Fluid Mechanics

Similarly, for the y-direction we have:

( )( )

( )

2 2 2

2 2 2

2 2 2

sin sin 0

sin 0 sin

sin

y

y

p A F Q v

F Q v p A

Qv p A

θ ρ θ

ρ θ θ

ρ θ

− + = −

= − +

= +

The resultant is:

2 2x yF F F= +

And which acts at an angle of:

1tan y

x

FF

θ −=

This is the force and direction of the bend on the fluid. The bend itself must then be

supported for this force. In practice a manhole is built at a bend, or else a thrust block

is used to support the pipe bend.

PAGE 79

Fluid Mechanics

Application – Force exerted by a firehose

Problem

A firehose discharges 5 l/s. The nozzle inlet and outlet diameters are 75 and 25 mm

respectively. Calculate the force required to hold the hose in place.

Solution

The control volume is taken as shown:

There are three forces in the x-direction:

• The reaction force RF provided by the fireman;

• Pressure forces PF : 1 1p A at the left side and 0 0p A at the right hand side;

• The momentum force MF caused by the change in velocity.

So we have:

M P RF F F= +

The momentum force is:

PAGE 80

Fluid Mechanics

( )2 1MF Q v vρ= −

Therefore, we need to establish the velocities from continuity:

( )

3

1 21

5 100.075 4

1.13 m/s

QvA π

−×= =

=

And

( )

3

2 25 100.025 4

10.19 m/s

vπ

−×=

=

Hence:

( )( )( )

2 1

3 310 5 10 10.19 1.13

45 N

MF Q v vρ−

= −

= × −

=

The pressure force is:

1 1 0 0PF p A p A= −

If we consider gauge pressure only, the 0 0p = and we must only find 1p . Using

Bernoulli’s Equation between the left and right side of the control volume:

PAGE 81

Fluid Mechanics

2 2

1 1 0 0

02 2p v p vg g g gρ ρ =

⎛ ⎞+ = +⎜ ⎟

⎝ ⎠

Thus:

( )

( )

2 21 1 0

32 2

2

2

10 10.19 1.132

51.28 kN/m

p v vρ⎛ ⎞= −⎜ ⎟⎝ ⎠⎛ ⎞

= −⎜ ⎟⎝ ⎠

=

Hence

( ) ( )1 1 0 0

23 0.075

51.28 10 04

226 N

PF p A p A

π

= −

⎛ ⎞= × −⎜ ⎟⎜ ⎟

⎝ ⎠=

Hence the reaction force is:

45 226181 N

R M PF F F= −= −= −

This is about a fifth of an average body weight – not inconsequential.

PAGE 82

INTRODUCTION:

Many practical real flow problems in fluid mechanics can be solved by using equations and analytical procedures. However, solutions of some real flow problems depend heavily on experimental data.

Sometimes, the experimental work in the laboratory is not only time-consuming, but also expensive.

So, the main goal is to extract maximum information from fewest experiments. In this regard,

dimensional analysis is an important tool that helps in correlating analytical results with experimental data and to predict the prototype behavior from the measurements on the model.

Dimensions and Units:

Dimensional analysis is a mathematical technique which makes use of the study of dimensions as an

aid to the solution of several engineering problems. It deals with the dimensions of the physical

quantities are measured by comparison, which is made with respect to an arbitrarily fixed value.

• Length L, mass M and Time T are three fixed dimensions which are of importance in fluid mechanics. If in any problem of fluid mechanics, heat is involved then temperature is also

taken as fixed dimension.

• These fixed dimensions are called fundamental dimensions or fundamental quantity. The

following are the Fundamental Dimensions (MLT)

• Mass - kg - M

• Length - m - L

• Time - s - T

• Secondary dimensions are those quantities which possess more than one fundamental

dimension. 1. Geometric dimensions

• Area - m2 - L2

• Volume - m3 - L3

2. Kinematic dimensions

• Velocity - m/s - L

T or L.T-1

• Acceleration - m/s2 - L

T2 or L.T-2

3. Dynamic dimensions

• Force -N - ML

T or M.L.T-1

• Density -kg/m3 - M/L3 or M.L-3

Use of Dimensional Analysis:

1. Conversion from one dimensional unit to another

2. Checking units of equations (Dimensional Homogeneity) 3. Defining dimensionless relationship using

• Rayleigh’s Method

• Buckingham’s π-Theorem

4. Model Analysis

MODULE – V

DIMENSIONAL ANALYSIS

PAGE 83

Dimensional Homogeneity:

Dimensional Homogeneity means the dimensions in each equation on both sides equal.

Consider the following equation – 𝑉 = √2 𝑋 𝑔 𝑋 ℎ

Dimensions of L. H. S = 𝑉 = 𝐿

𝑇= 𝐿 𝑋 𝑇−1

Dimensions of R. H. S = √2 𝑋 𝑔 𝑋 ℎ = √𝐿

𝑇−2 𝑋 𝐿 = 𝐿 𝑋 𝑇−1

⸫ Dimensions of L. H. S = Dimensions of R. H. S

Hence the above equation is said to be dimensionally homogenous as Dimensions of L. H. S and R. H. S are equal.

Methods of dimensional analysis:

Rayleigh’s Method:

To define relationship among variables. This method is used for determining the expression for a variable which depends upon maximum three or four variables only.

Methodology:

1. Let X is a function of X1, X2, and X3 and mathematically it can be written as

X = f (X1, X2, X3)

2. This can be also written as X = K (X1a , X2b , X3c ) where K is constant and a, b and c are arbitrarily powers.

3. The values of a, b and c are obtained by comparing the powers of the fundamental dimension

on both sides.

Buckingham’s π-Theorem:

This method of analysis is used when number of variables are more.

Theorem:

If there are n variables in a physical phenomenon and those n variables contain m dimensions, then

variables can be arranged into (n – m) dimensionless groups called Φ terms.

Explanation:

1. If f (X1, X2, X3, ……… Xn) = 0 and variables can be expressed using m dimensions then, f (π1, π2, π3, ……… πn - m) = 0, where, π1, π2, π3, … are dimensionless groups.

2. Each π term contains (m + 1) variables out of which mare of repeating type and one is of non

– repeating type. 3. Each π term being dimensionless, the dimensional homogeneity can be used to get each π

term. π denotes a non-dimensional parameter.

Selecting Repeating Variables:

1. Avoid taking the quantity required as the repeating variable.

2. Repeating variables put together should not form dimensionless group.

3. No two repeating variables should have same dimensions. 4. Repeating variables can be selected from each of the following properties.

• Geometric property → Length, height, width, area

• Flow property → Velocity, Acceleration, Discharge

• Fluid property → Mass density, Viscosity, Surface tension

PAGE 84

Model Analysis:

For predicting the performance of the hydraulic structures (such as dams, spillways etc.) or hydraulic

machines (such as turbines, pumps etc.) before actually constructing or manufacturing, models of the

structures or machines are made and tests are conducted on them to obtain the desired information. Model is a small replica of the actual structure or machine. The actual structure or machine is called

as Prototype. Models can be smaller or larger than the Prototype. Model Analysis is actually an

experimental method of finding solutions of complex flow problems.

Similitude or Similarities:

Similitude is defined as the similarity between the model and prototype in every aspect, which means that the model and prototype have similar properties.

Types of Similarities:

• Geometric Similarity →Length, Breadth, Depth, Diameter, Area, Volume etc.,

• Kinematic Similarity → Velocity, Acceleration etc.,

• Dynamic Similarity → Time, Discharge, Force, Pressure Intensity, Torque, Power

Geometric Similarity:

The geometric similarity is said to be exist between the model and prototype if the ratio of all

corresponding linear dimensions in the model and prototype are equal. Mathematically,

Kinematic Similarity

The kinematic similarity is said exist between model and prototype if the ratios of velocity and

acceleration at corresponding points in the model and at the corresponding points in the prototype are

the same. Also the directions of the velocities in the model and prototype should be same.

Dynamic Similarity

The dynamic similarity is said exist between model and prototype if the ratios of corresponding forces

acting at the corresponding points are equal.

It means for dynamic similarity between the model and prototype, the dimensionless numbers should be same for model and prototype.

P P

m

V a and V ar rV am

where is Velocity RatioVr

where is Acceleration Ratioa r

= =

P

m

r

FFr

F

where is Force RatioF

=

P P

m

P P P

r

m m m m

2 3VB AL D and and L L Lr rVL B D A

where is Scale RatioLr

= = = = =

PAGE 85

Forces acting on moving fluid:

Following are the different types of forces to be considered when the liquid is in motion.

• Inertia force, Fi

• Viscous force, Fv

• Gravity force, Fg

• Pressure force, Fp

• Surface tension force, Fs

• Elastic force, Fe

1. Inertia Force, Fi :

• It is the product of mass and acceleration of the flowing fluid and acts in the direction opposite to the direction of acceleration.

• It always exists in the fluid flow problems.

2. Viscous Force, Fv:

• It is equal to the product of shear stress due to viscosity and surface area of the flow.

• It is important in fluid flow problems where viscosity is having an important role to play.

3. Gravity Force, Fg:

• It is equal to the product of mass and acceleration due to gravity of the flowing fluid.

• It is present in case of open surface flow.

4. Pressure force, Fp:

• It is equal to the product of pressure intensity and cross-sectional area of flowing fluid.

• It is present in case of pipe-flow.

5. Surface Tension Force, Fs:

• It is equal to the product of surface tension and length of surface of the flowing fluid.

6. Elastic Force, Fe:

• It is equal to the product of elastic stress and area of the flowing fluid

Dimensionless Numbers:

Dimensionless numbers are obtained by dividing the inertia force by viscous force or gravity force or

pressure force or surface tension force or elastic force. Following are some of the dimensionless numbers described below. The detailed applications of these dimensionless numbers shall be

discussed detailed in further sections.

e

FR

oeyn

cold’

is num

ober, R

Inertia r e VL VDor

V sc us Force

=

=

u

iE

ule

or’s

cnumbe

rr, E  

Inert a F r e V

P ess r 

ure Fo ce p /==

e

Inertia Force V

Surface Tension Force / LWeber’s number, W ==

e

Inertia Force V

Gravity Force LgFroude’s number, F ==

PAGE 86

Model Laws:

The laws on which the models are designed for dynamic similarity are called model laws or laws of

similarity.

• Reynold’s Model Law

• Froude Model Law.

• Euler Model Law.

• Weber Model Law

• Mach Model Law

1. Reynold’s Model Law:

If the viscous forces are predominant, the models are designed for dynamic similarity based on

Reynold’s number. Models based on Reynolds’s Number includes:

• Pipe Flow

• Resistance experienced by Sub-marines, airplanes, fully immersed bodies etc.

• Mathematically, it can be expressed as:

•

2. Froude Model Law:

If the gravity force is predominant, the models are designed for dynamic similarity based on Froude number. Froude Model Law is applied in the following fluid flow problems:

• Free Surface Flows such as Flow over spillways, Weirs, Sluices, Channels etc.,

• Flow of jet from an orifice or nozzle

• Where waves are likely to formed on surface

• Where fluids of different densities flow over one another

3. Euler Model Law:

Euler Model Law is applied in the following cases:

• Closed pipe in which case turbulence is fully developed so that viscous forces are

negligible and gravity force and surface tension is absent where phenomenon of

cavitation’s takes place.

4. Weber Model Law:

Weber Model Law is applied in the following cases:

• Capillary rise in narrow passages

• Capillary movement of water in soil

• Capillary waves in channels

• Flow over weirs for small heads

Inertia Force V

Elastic Force CMach’s number, M = =

m Pm Pm P

m p

R Re e m p

ρ ρV VL L

=

=

e rm p

pm

m pm p

FF

VV

X L X Lg g

=

=

PAGE 87

5. Mach Model Law:

Mach Model Law is applied in the following cases:

• Flow of aero plane and projectile through air at supersonic speed ie., velocity more than

velocity of sound.

• Aero dynamic testing

• Underwater testing of torpedoes, and

• Water-hammer problems

Problems:

1. Find Dimensions for the following:

• Stress / Pressure

• Work

• Power

• Kinetic Energy

• Dynamic Viscosity

• Kinematic Viscosity

• Surface Tension

• Angular Velocity

• Momentum

• Torque

2. Check Dimensional Homogeneity of the following:

• Q = AV

• EK = v2/2g 3. Water flowing through a pipe of diameter 30 cm at a velocity of 4 m/s. Find the velocity of oil

flowing in another pipe of diameter 10cm, if the conditions of dynamic similarity is satisfied

between two pipes. The viscosity of water and oil is given as 0.01 poise and 0.025 poise. The

specific gravity of oil is 0.8. 4. In 1 in 40 model of a spillway, the velocity and discharge are 2 m/s and 2.5 m3/s. Find

corresponding velocity and discharge in the prototype

5. In a 1 in 20 model of stilling basin, the height of the jump in the model is observed to be 0.20m. What is height of hydraulic jump in the prototype? If energy dissipated in the model is 0.1kW,

what is the corresponding value in prototype?

6. A 7.2 m height and 15 m long spillway discharges 94 m3/s discharge under a head of 2m. If a 1:9 scale model of this spillway is to be constructed, determine the model dimensions, head over

spillway model and the model discharge. If model is experiencing a force of 7500 N, determine

force on the prototype.

7. A Dam of 15 m long is to discharge water at the rate of 120 cumecs under a head of 3 m. Design a model, if supply available in the laboratory is 50 lps

8. A 1:50 spillway model has a discharge of 1.5 cumecs. What is the corresponding discharge in

prototype?. If a flood phenomenon takes 6 hour to occur in the prototype, how long it should take in the model.

PAGE 88