Fluid Mechanics Notes

1

Fluid Mechanics I

EG-160/EG160c

Professor M.F. Webster

Room 147

Credit 10

Core module for Civil, Mechanical and Aerospace BEng and MEng

Format: 3 Lectures / week

Fluid Mechanics I

Aims

To create an interest in fluid flow

To show that flow phenomena are amenable to analysis

To show the relevance of fluid mechanics to Engineering

To create confidence and ability in problem-solving in fluid mechanics

Contents

Fluids properties

Hydrostatics

Conservation principles

Viscous flow in pipes

Assessment

1 class test (Blackboard - online) 20% of final mark

Week 4 - Week 5

Summer Exam - 2 hours closed book exam 80%

3

What is a fluid? What is the difference between a solid and a fluid?

• A solid is “hard” and not easily deformed, whereas a fluid is “soft” and is easily deformed • A closer look at the molecular structure a solid (steel, concrete, etc.)

» densely spaced molecules

» large intermolecular cohesive forces

• However, for matter that we normally think of as a liquid (water, oil,etc.) » the molecules are spaced farther apart

» the intermolecular forces are smaller than for solids, • Gases (air, oxygen, etc.) have

» greater molecular spacing and freedom of motion » negligible cohesive intermolecular forces

• A more specific distinction is based on how they deform under the action of an external load. Specifically, a fluid is defined as a substance that deforms continuously when acted on by a shearing stress of any magnitude.

• Common fluids such as water, oil, and air satisfy the definition of a fluid

• Some materials, such as slurries, tar, putty, toothpaste, and so on, are not easily

classified since they will behave as a solid if the applied shearing stress is small, but if the stress exceeds some critical value, the substance will flow. The study of such materials is called rheology

Introduction

Fluid Mechanics I

• The study of fluid mechanics involves the same fundamental laws you have encountered in physics and other mechanics courses. These laws include Newton’s laws of motion, conservation of mass, and the first and second laws of thermodynamics.

• The broad subject of fluid mechanics can be generally subdivided into fluid statics, in which the fluid is at rest, and fluid dynamics, in which the fluid is moving.

5

• ARCHIMEDES 287–212 B.C.Established elementary principles of buoyancy and flotation.

• SEXTUS JULIUS FRONTINUS A.D. 40–103. Wrote treatise on Roman methods of water distribution.

• LEONARDO da VINCI 1452–1519. Expressed elementary principle of continuity; observed and sketched many basic flow phenomena; suggested designs for hydraulic machinery.

• GALILEO GALILEI 1564–164. Indirectly stimulated experimental hydraulics; revised Aristotelian concept of vacuum. • EVANGELISTA TORRICELLI 1608–164. Related barometric height to weight of atmosphere, and form of liquid jet to

trajectory of free fall.

• BLAISE PASCAL 1623–1662. Finally clarified principles of barometer, hydraulic press, and pressure transmissibility. • ISAAC NEWTON1642–1727. Explored various aspects of fluid resistance–inertial, viscous, and wave; discovered jet

contraction.

• HENRI dePITOT 1695–1771. Constructed double-tube device to indicate water velocity through differential head. • DANIEL BERNOULLI1700–1782. Experimented and wrote on many phases of fluid motion, coining name

“hydrodynamics”; devised manometry technique and adapted primitive energy principle to explain velocity head indication; proposed jet propulsion.

• LEONHARD EULER1707–1783. First explained role of pressure in fluid flow; formulated basic equations of motion and socalled Bernoulli theorem; introduced concept of cavitation and principle of centrifugal machinery.

• JEAN le ROND d’ALEMBERT 1717–1783. Originated notion of velocity and acceleration components, differential expression of continuity, and paradox of zero resistance to steady non-uniform motion.

• ANTOINE CHEZY 1718–1798. Formulated similarity parameter for predicting flow characteristics of one channel from measurements on another.

• GIOVANNI BATTISTA VENTURI1746–1822. Performed tests on various forms of mouthpieces–in particular, conical contractions and expansions.

• LOUIS MARIE HENRI NAVIER 1785–1836. Extended equations of motion to include “molecular” forces.

• AUGUSTIN LOUIS de CAUCHY 1789–1857. Contributed to the general field of theoretical hydrodynamics and to the study of wave motion.

• GOTTHILF HEINRICH LUDWIG HAGEN 1797–1884. Conducted original studies of resistance in and transition between laminar and turbulent flow.

• JEAN LOUIS POISEUILLE 1799–1869. Performed meticulous tests on resistance of flow through capillary tubes.

Famous Names

Fluid Mechanics I

• HENRI PHILIBERT GASPARD DARCY 1803–1858. Performed extensive tests on filtration and pipe • resistance; initiated open-channel studies carried out by Bazin.

• JULIUS WEISBACH 1806–1871. Incorporated hydraulics in treatise on engineering mechanics, based on original experiments; noteworthy for flow patterns, non-dimensional coefficients, weir, and resistance equations.

• WILLIAM FROUDE 1810–1879. Developed many towing-tank techniques, in particular the conversion of wave and boundary layer resistance from model to prototype scale.

• ROBERT MANNING 1816–1897. Proposed several formulas for open-channel resistance.

• GEORGE GABRIEL STOKES1819–1903. Derived analytically various flow relationships ranging from wave mechanics to viscous resistance—particularly that for the settling of spheres.

• ERNST MACH1838–1916. One of the pioneers in the field of supersonic aerodynamics.

• OSBORNE REYNOLDS1842–1912. Described original experiments in many fields, cavitation, river model similarity, pipe resistance—and devised two parameters for viscous flow; adapted equations of motion of a viscous fluid to mean conditions of turbulent flow.

• JOHN WILLIAM STRUTT,LORD RAYLEIGH1842–1919. Investigated hydrodynamics of bubble collapse, wave motion, jet instability, laminar flow analogies, and dynamic similarity.

• VINCENZ STROUHAL 1850–1922. Investigated the phenomenon of “singing wires.”

• EDGAR BUCKINGHAM 1867–1940. Stimulated interest in the United States in the use of dimensional analysis. • MORITZ WEBER 1871–1951. Emphasized the use of the principles of similitude in fluid flow studies and formulated a

capillarity similarity parameter.

• LUDWIG PRANDTL 1875–1953. Introduced concept of the boundary layer and is generally considered to be the father of present day fluid mechanics.

• LEWIS FERRY MOODY1880–1953. Provided many innovations in the field of hydraulic machinery. Proposed a method of correlating pipe resistance data which is widely used.

• THEODOR VON KÁRMÁN 1881–1963. One of the recognized leaders of twentieth century fluid mechanics. Provided major contributions to our understanding of surface resistance, turbulence, and wake phenomena.

• PAUL RICHARD HEINRICH BLASIUS 1883–1970. One of Prandtl’s students who provided an analytical solution to the boundary layer equations. Also, demonstrated that pipe resistance was related to the Reynolds number.

7

Primary Units

• The four primary units of the SI system are shown in the table below:

• Notice how the term ’Dimension’ of a unit has been introduced in this table. This is not a property of the individual units, rather it tells what the unit represents. For example a metre is a length which has a dimension L but also, an inch, a mile or a kilometre are all lengths so have dimension of L.

• The above notation uses the MLT system of dimensions, there are other ways of writing dimensions

Units

Fluid Mechanics I

Derived Units

• There are many derived units all obtained from combination of the above primary units. Those most used are shown in the table below:

• The above units should be used at all times. Values in other units should NOT be used without first converting them into the appropriate SI unit. • If you do not know what a particular

unit means, find out else your guess will probably be wrong.

• One very useful tip is to write down the units of any equation you are using. If at the end the units do not match you know you have made a mistake. • For example is you have at the end of

a calculation, 30 kg/m s = 30 m you have certainly made a mistake

9

Examples

• During a study of a certain flow system the following equation relating the pressure p₁ and

p₂at two points was developed

In this equation Vis a velocity, Lthe distance between the two points, Da diameter, g the acceleration of gravity, and f a dimensionless coefficient. Is the equation dimensionally consistent?

• If Vis a velocity, L a length, W a weight, and µa fluid property having dimensions of FL-2_T

determine the dimensions of (a) VLW/

µ

, (b)WL

µ

, (c) V

µ

/Land (d) VL2

µ

_/W

Dg

fLV

p

p

₂

=

₁

+

Units Fluid Mechanics I

• Before we can proceed, however, it will be necessary to define and discuss certain

fluid properties

Measures of Fluid Mass and Weight

Density

• The densityof a fluid, designated by the Greek symbol ρρρρ, is defined as its

mass per unit volume.

Density is typically used to characterize the mass of a fluid system. In SI the units are kg/m3

• The value of density can vary widely between different fluids

» For liquids, variations in pressure and temperature generally have only a small effect on the value of density

» For gas, the density is strongly influenced by both pressure and temperature

Specific Volume

• The specific volume, is the volume per unit mass

and is therefore the reciprocal of the density

Fluids Properties

ρ

1

=

11

Specific Weight

• The specific weightof a fluid, designated by the Greek symbol γγγγ, is defined as its

weight per unit volume. Thus, specific weight is related to density through the equation where g is the local acceleration of gravity

• Just as density is used to characterize the mass of a fluid system, the specific weight is used to characterize the weight of the system. In SI the units are N/m3

Specific Gravity

• The specific gravityof a fluid, designated as SG, is defined as the ratio of the density of the fluid to the density of water at some specified temperature. Usually the specified temperature is taken as 4o_{C and at this temperature the density of water is 1000 kg/m}3

• and since it is the ratio of densities, the value of SG does not depend on the system of units used.

It is clear that density, specific weight, and specific gravityare all interrelated, and from a knowledge of any oneof the three the otherscan be calculated.

ρ

γ

=

g

Fluids Properties

1000

/

ρ

=

SG

Fluid Mechanics I

It is clear that the previous properties are not sufficient to uniquely characterize

how fluids behave since two fluids such as water and oil can have approximately

the same value of density but behave quite differently when flowing.

There is apparently some additional property that is needed to describe the

“

fluidity

” of the fluid which we will see later in the course.

Examples

• Find the density of mercury if its specific gravity is 13.55

• A reservoir of glycerine has a mass of 1200 kg and a volume of 0.952 m3_{. Find the} glycerine's weight, mass density, specific weight and specific gravity (Ans: 11.77kN, 1261 kg/m3_{, 12.36 kN/m}3_{, 1.26)}

• The specific gravity of ethyl alcohol is 0.79. Calculate its specific weight and mass density (Ans: 7.73 kN/m3_{, 790 kg/m}3₎

• The specific weight of a substance is 8.2 kN/m3_{, what is its mass density (Ans: 836} kg/m3₎

13

Pressure

• Pressure in a fluid at rest is defined as the normal force per unit area exerted on a plane surface (real or imaginary) immersed in a fluid and is created by the bombardment of the surface with the fluid molecules.

• From the definition, pressure has the dimension FL-2 _{and in SI units is expressed}

as N/m2_{. In SI, 1N/m}2_{is defined as a pascal, abbreviated as Pa, and pressures are}

commonly specified in pascals

• The equations of motion (Newton’s second law) in the

F

=

m

a

y and z directions are

z s z z y s y y

a

z

y

x

z

y

x

s

x

p

y

x

p

F

a

z

y

x

s

x

p

z

x

p

F

2

2

cos

2

sin

δ

δ

δ

ρ

δ

δ

δ

γ

θ

δ

δ

δ

δ

δ

δ

δ

ρ

θ

δ

δ

δ

δ

=

−

−

=

∑

=

−

=

∑

Fluids Properties Fluid Mechanics I • Note that • Hence,

• Since we are interested in the pressure at a point, we take the limit as

δ

x,

δ

y

and

δ

z tend to zero and it follows that

• we can conclude that the pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shearing stresses present. This important result is known as Pascal’s law named in honour of Blaise Pascal

θ

δ

δ

θ

δ

δ

y

=

s

cos

z

=

s

sin

(

)

2

2

z z s z y y s y

a

p

p

a

p

p

δ

γ

ρ

δ

ρ

+

=

−

=

−

s y z

p

p

p

=

=

Fluids Properties

15

Compressibility of Fluids

• This measure how easily can the volume (and thus the density) of a given mass of the fluid be changed when there is a change in pressure . A property that is commonly used to characterize compressibility is the bulk modulus, defined as

• The bulk modulus (or the bulk modulus of elasticity)has dimensions of pressure. • Large values for the bulk modulus indicate that the fluid is relatively incompressible

Examples

• A liquid compressed in a cylinder has a volume of 1000 cm3 _{at 1MN/m}2_{and volume of} 995 cm3 _{at 2MN/m}2_{. What is its bulk modulus of elasticity?}

• If the bulk modulus of elasticity for water is 2.2 GPa, what pressure is required to reduce a volume by 0.6 precent? (Ans: 13.2 Mpa)

ρ

ρ

d

dp

V

dV

dp

E

_v

=

−

=

Fluids Properties Fluid Mechanics I

Basic Equation for Pressure Field

• Consider a cylindrical element of fluid inclined at an angleθto the vertical. The pressure at the end with height z is p and at the end of height z+δzis p+δp

• Resolving the forces in the direction along the central axis gives

• Or in differential form

• If θ=90οthen

s

is in the x or y directions, (i.e. horizontal),so

or in homogenous domain, Pressure in the horizontal direction is constant.

Hydrostatics

θ

ρ

δ

δ

θ

δ

ρ

δ

θ

δ

ρ

δ

cos

cos

0

cos

)

(

g

s

p

s

g

p

s

gA

A

p

p

pA

−

=

⇒

−

=

=

−

+

−

θ

ρ

g

cos

ds

dp

−

=

0

=

=

=

dy

dp

dx

dp

ds

dp

17 • If θ=0οthen

s

is in the z directions, (i.e. vertical),so

• Integrating the above equation gives • Integrating between z₁ znd z₂ will lead to

• Thus in a fluid under gravity,

pressure decreases with increase in height

Pressure and Head

• In a liquid with a free surface the pressure at any depth z is normally measured from the free surface so that z = -h. This gives:

• At the surface the pressure is the atmospheric pressure, p_atmospheric

g

dz

dp

ds

dp

₌

₌

₋

_ρ

)

(

_{2 1} 1 2

p

g

z

z

p

−

=

−

ρ

−

constant

gz

p

=

−

ρ

+

constant

gh

p

=

ρ

+

c atmospheri c atmospheri

p

h

p

p

gh

p

+

=

+

=

γ

ρ

Hydrostatics Fluid Mechanics I

• The lower limit of any pressure is zero- that is pressure in Vacuum. Pressure measured above this datum is known as absolute pressure

• Since everything is under this pressure, it is convenient to take the atmospheric pressure as datum, hence, pressure quotes in this condition is called the Gauge pressure

• Since gis constant, the gauge pressure can be given by stating the vertical height of any fluid of density

ρ

which is equal to the this pressure, this vertical height is know as head

of fluid.

Example

What is the pressure of 500 KN/m2 _{in terms of the height of water,}

ρ

_=1000kg/m3 _{, and in} terms of Mercury,

ρ

=13600 kg/m3_.

h

gh

p

_guage

=

ρ

=

γ

19

Examples

• Because of a leak in a buried gasoline storage tank, water has seeped in to the depth shown. If the specific gravity of the gasoline is SG= 0.68, determine the pressure at the gasoline-water interface and at the bottom of the tank. • For the open tank, with piezometers attached on the side,

containing two different immiscible liquids. Find (a) the elevation of the liquid surface in piezometer A (b) the elevation of the liquid surface in piezometer B (c) Total pressure at te bottom of the tank

(Ans:2.0m, 0.82m, 18.9kPa)

• The reading of an automobile fuel gauge is proportional to the gauge pressure at the bottom of the tank.

If the tank is 32cm deep and is contaminated with 3cm of water, how many cm of air remains at the top when the gauge indicates full? Use γ_gasoline = 6670 N/m3_and

γair= 11.8 N/m3. (Ans: 1.4cm) 5m 1m Hydrostatics Fluid Mechanics I

Pressure Measurement

• The relation between pressure and head is used to measure pressure with manometer, liquid gauge.

The Piezometer Tube Manometer

• A tube which is attached to the top of a vessel

containing fluid at a pressure higher than atmospheric • The pressure measured is relative to atmospheric,

hence it is gauge pressure

• This method can only be used for liquids. And must not be too small or too large.

1 1 1

gh

h

p

p

_A

=

=

ρ

=

γ

Hydrostatics

21

The “U” Tube Manometer

• The “U” tube measure the pressure of both liquids and gases • The “U” tube is filled with a fluid called the Manometric fluid • The density of the fluid whose pressure to be measured must

be less than that of the manomatric fluid • We know that,

• However,

• Since we are measuring gauge pressure • Hence,

• If the fluid to be measure is gas, the , and the gauge pressure

3 2 1

and

p

p

p

p

_A

=

=

1 2

p

h

p

=

_A

+

γ

2 3

h

p

=

γ

_man 1 2

h

h

p

_A

=

γ

_man

−

γ

ρ

ρ

man

>>

p

A

=

γ

man

h

2 Hydrostatics Fluid Mechanics I

Measurement of pressure difference using the “U” Tube Manometer

• The “U” tube is connected at the two points where the

pressure difference is to be measured • Using the figure

• And

• If the fluid is the same at the two points then • Again if the fluid is gas then

5 1

and

p

p

p

p

_A

=

_B

=

3 3 1 1 2 3 3 2 1 1 3 2

h

h

h

p

p

h

h

p

h

p

p

p

man B A man B A

γ

γ

γ

γ

γ

γ

+

−

=

−

+

+

=

+

=

)

(

_{3 1} 1 2

h

h

h

p

p

_A

−

_B

=

γ

_man

+

γ

−

2

h

p

p

_A

−

_B

=

γ

_man Hydrostatics

23

Inclined Tube Manometer

• If the pressure difference to be measured is small, one leg of the tube is inclined at an angle

θθθθ

• In this case

• If the fluid is the same at the two points then • Again if the fluid is gas then

)

(

sin

_{1 3 1} 2

h

h

l

p

p

_A

−

_B

=

γ

_man

θ

+

γ

−

θ

γ

l

₂

sin

p

p

_A

−

_B

=

_man 3 3 1 1 2

sin

h

h

l

p

p

_A

−

_B

=

γ

_man

θ

−

γ

+

γ

Hydrostatics Fluid Mechanics I

Examples

• A closed tank contains compressed air and oil (SG. 0.9). A U-tube manometer using mercury is connected to the tank. For column heights h₁=90cm, h₂= 15cm and h₃=22.5cm determine the pressure reading of the gage.

• the volume rate of flow, Q, through a pipe can be determined by means of a flow nozzle located in the pipe as illustrated. The nozzle creates a pressure drop, along the pipe which is related to the flow through the equation

where K is a constant depending on the pipe and nozzle size. The pressure drop is frequently measured with a differential U-tube manometer. Determine an equation for p_A– p_Bin terms of the specific weight of the flowing

fluid, γ₁the specific weight of the gage fluid, γ₂and the various heights indicated. For γ₁= 9.8 kN/m3_, γ

2= 15.6 kN/m3 ,h1= 1.0m and h2= 0.5m what is the value of the

pressure drop, p_A– p_B ? B A

p

p

K

Q

=

−

Hydrostatics

25

Problems

• A Differential manometer is attached to two tanks. Calculate the pressure difference between chambers

A and B. Take SG_Mercury= 13.6, SG_Oil= 0.89 and

SG_{Tetrachloride}= 1.59. (Ans:-37kN/m2₎

• Calculate the level h of the oil in the right hand tube (Ans: 0.18m)

• The liquid at A and B is water and the manometer liquid is oil with SG = 0.8, h₁ = 300mm, h₂= 200mm and

h₃= 600mm. (a) determine p_A-p_B. (b) Ifp_B= 50 kPa and the barometer reading is 730 mmHg, find the absolute pressure at A in meters of waters. (Ans:-1.37kPa, 14.9m)

Hydrostatics

Fluid Mechanics I

Forces on Submerged Surfaces in Static Fluids

We have seen that:

• Hydrostatic vertical pressure distribution, the pressure varies linearly with depth • Pressures at any equal depth in a continuous fluid are equal

• Pressure at a point acts equally in all directions (Pascal’s law)

• Forces from a fluid on a boundary acts at right angles to that boundary • Pressure is defined as force per unit area

Fluid Pressure on a Surface

• The determination of the forces developed on the surface due to the fluid is important in the design of storage tanks, ships, dams, and other hydraulic structures.

• For a horizontal surface, the magnitude of the resultant force is F_R=p A

• where p = γh is the uniform pressure on the bottom • If atmospheric pressure acts on both sides of the surface,

the resultant force is due to the liquid in the tank. • Since the pressure is constant and uniformly distributed

over the bottom, the resultant acts through the centroid of the area

27 • For a general case, assuming that the fluid surface is open to the atmosphere and using

the x–y coordinate system shown.

• We wish to determine the direction, location, and magnitude of the resultant force acting on one side of this area due to the liquid.

• At any given depth, h, the force acting on dA is

dF = γhdA and is perpendicular to the surface. • Thus, the magnitude of the resultant

• The integral is the first moment of the area

• y_cis the ycoordinate of the centroid, thus

• The resultant force is equal to the pressure at the centroid multiplied by the total area

∫

∫

∫

=

=

=

_{A A A} R

hdA

y

dA

ydA

F

γ

γ

sin

θ

γ

sin

θ

A

y

ydA

_c A

=

∫

c c R

Ay

Ah

F

=

γ

sin

θ

=

γ

Hydrostatics Fluid Mechanics I

• Since all differential forces are perpendicular to the surface , the resultant force must also be perpendicular to the surface

• To find the location of the resultant force we consider the moment around the x axis

• Since

• The numerator is the second moment of the area, I_x. Using the parallel axis theorem

• I_xcis the second moment of the area with respect to an axis passing through the centroid and parallel to the x axis. Thus,

∫

∫

=

=

_{A A} R R

y

ydF

y

dA

F

γ

2

sin

θ

A

y

dA

y

y

c A R

∫

=

2

θ

γ

_c

sin

R

Ay

F

=

2 c xc x

I

Ay

I

=

+

c c xc R

y

A

y

I

y

=

+

Hydrostatics

29 • It is clear that the resultant force does not pass through the centroid but always belowit

• Similarly the x coordinate of the resultant force can be obtained by summing moment about the yaxis

• I_xyis the product of inertia with respect to the xand yaxes. I_xycis the product of inertia with respect to an orthogonal system passing through the centroid

c c xyc c xy c A R

x

A

y

I

A

y

I

A

y

xydA

x

=

∫

=

=

+

Hydrostatics Fluid Mechanics I

Examples

• The 4m-diameter circular gate is located in the inclined wall of a large reservoir containing water γ= 9.80 kN/m3_{The gate is mounted on} a shaft along its horizontal diameter. For a water depth of 10 m above the shaft determine: (a) the magnitude and location of the resultant force exerted on the gate by the water, and (b) the moment that would have to be applied to the shaft to open the gate.

• A pressurized tank contains oil (S.G=0.9) and has a square, 0.6m by 0.6m plate bolted to its side. When the pressure gage on the top of the tank reads 50 kPa, what is the magnitude and location of the resultant force on the attached plate?

31

Hydrostatic Forces on a Curved Surfaces

• Integration method can be used, however, it can be tedious

• The easiest way is to consider the fluid volume enclosed by the curved surface

• The magnitude and location of the forces on the horizontal and vertical surfaces can be determined from the relationship of planer surfaces

• In order for force system to be in equilibrium the horizontal components must be equal and collinear and the vertical components must be equal and collinear

• The location of the resultant force is found by Summing moment about an appropriate axis

2 2 1 2

,

V R

(

H

)

(

V

)

H

F

F

F

W

F

F

F

F

=

=

+

⇒

=

+

Hydrostatics Fluid Mechanics I

Examples

• The 3m long cylinder floats in oil and rest against a wall Determine the horizontal force the cylinder exerts on the wall at the point of contact.

• The 2m-diameter drainage conduit is half full of water at rest.

Determine the magnitude and line of action of the resultant force that the water exerts on a 1m length of the curved section BC of the conduit wall.

1m

960N/m3

1m

Hydrostatics

33

Problems

• Isosceles triangular gate AB is hinged at A. Compute the horizontal force P at point B for equilibrium, neglecting the weight of the gate. (Ans: 22.47 kN)

• The triangular channel is hinged at A and held together by cable BC at the top. If cable spacing is 1m into the paper, what is the cable tension. (Ans: 88.5 kN)

• Determine the pivot locaion yof the square gate so that it will rotate open when the liquid surface is shown. (Ans: 0.833m)

• Compute the air pressure required to keep the gate closed. The gate is a circular plate of diameter 0.8m and weight

2.0kN.(Ans: 43 kPa)

Hydrostatics

Fluid Mechanics I

Problems

• Compute the horizontal and vertical components of the hydrostatic force on the quarter-circle face of the tank. (Ans: 308kN, 289kN)

• Compute the hydrostatic force and its line of action on semicylindrical indentation ABCper meter of width into the paper. (Ans: 115.1 kN, φ= 10.6o₎

• The face of the dam retaining water to depth 10m is shaped as shown. Determine the magnitude of the hydrostatic force acting on the curve portion AB per meter width of the dam and the moment of this force about A. (Ans: 658 kN, 2470kNm)

35

Fluid Dynamics

Fluid Dynamics

This section discusses the analysis of fluid in motion - fluid dynamics. The motion of fluids can be predicted in the same way as the motion of solids are predicted using the fundamental laws of physics together with the physical properties of the fluid

Objectives

• Introduce concepts necessary to analyse fluids in motion

• Identify differences between Steady/unsteady uniform/non-uniform compressible/incompressible flow

• Demonstrate streamlines and stream tubes

• Introduce the Continuity principle through conservation of mass and control volumes • Derive the Bernoulli (energy) equation

• Demonstrate practical uses of the Bernoulli and continuity equation in the analysis of flow • Introduce the momentum equation for a fluid

• Demonstrate how the momentum equation and principle of conservation of momentum is used to predict forces induced by flowing fluids.

Fluid Mechanics I Fluid Dynamics

Flow Classification

It is possible - and useful - to classify the type of flow which is being examined into small number of groups. The following terms describe the states which are used to classify fluid flow:

• uniform flow: If the flow velocity is the same magnitude and direction at every point in the fluid it is said to be uniform

• non-uniform: If at a given instant, the velocity is not the same at every point the flow is

non-uniform. (In practice, by this definition, every fluid that flows near a solid boundary will be non-uniform – as the fluid at the boundary must take the speed of the boundary, usually zero. However if the size and shape of the of the cross-section of the stream of fluid is constant the flow is considered uniform)

• steady: A steady flow is one in which the conditions (velocity, pressure and cross-section) may differ from point to point but DO NOT change with time

• unsteady: If at any point in the fluid, the conditions change with time, the flow is described as unsteady. (In practise there is always slight variations in velocity and pressure, but if the average values are constant, the flow is considered steady

37

Fluid Dynamics

Combining the above we can classify any flow in to one of four type

:

• Steady uniform flow.Conditions do not change with position in the stream or with time. An example is the flow of water in a pipe of constant diameter at constant velocity.

• Steady non-uniform flow. Conditions change from point to point in the stream but do not change with time. An example is flow in a tapering pipe with constant velocity at the inlet -velocity will change as you move along the length of the pipe toward the exit.

• Unsteady uniform flow. At a given instant in time the conditions at every point are the same, but will change with time. An example is a pipe of constant diameter connected to a pump pumping at a constant rate which is then switched off

• Unsteady non-uniform flow. Every condition of the flow may change from point to point and with time at every point. For example waves in a channel.

steady flow

is by far the most simple of the four. You will then be pleased to hear

that this course is restricted to only this class of flow

Fluid Mechanics I Fluid Dynamics

Three-dimensional flow

• Although in general all fluids flow three-dimensionally, in many cases the greatest changes only occur in two directions or even only in one.

• Flow is one dimensionalif the flow parameters (such as velocity, pressure, depth etc.) at a given instant in time only vary in the direction of flow and not across the cross-section. An example of one-dimensional flow is the ideal flow in a pipe.

• Note that since flow must be zero at the pipe wall - yet non-zero in the centre – there is a difference of parameters across the cross-section. Which is only necessary if very high accuracy is required.

• Flow is two-dimensionalif it can be assumed that the flow parameters vary in the direction of flow and in one direction at right angles to this direction.

39

Fluid Dynamics

Streamlines and Streamtubes

• The motion of each fluid particle is described in terms of its velocity vector, V

• If it is steady flow,each successive particle that passes through a given point will follow the same path. For such cases the path is a fixed line in the x–z plane.

• For steady flows each particle slides along its path, and its velocity vector is everywhere tangent to the path. The lines that are tangent to the velocity vectors throughout the flow field are called streamlines

• Close to a solid boundary streamlines are parallel to that boundary

• The fluid is moving in the same direction as the streamlines, hence, fluid can not cross it • Streamlines can not cross each other

• Any particle starting on one streamline will stay on the same streamline • Streamlines are two dimension while streamtubes are three dimensions

Fluid Mechanics I Fluid Dynamics

Flow rate

• Mass flow rate: is the mass of fluid flowing per unit time

• Volume flow rate – Discharge: is the volume of fluid flowing per unit time. (It is also commonly, but inaccurately, simply called flow rate). The symbol normally used for discharge is Q. Multiplying this by the density of the fluid gives us the mass flow rate

Mean Velocity:

• This is the discharge divided by the area cross section. • This does not imply that the velocity is constant across

the cross section

Inviscid Flow:

• That is the fluid is assumed to have zero viscosity.

• In practice there are no inviscid fluid. However, for many flow situation the viscous effect is small compared to other forces such as pressure gradient and gravitation.

41

Conservation Principles

Continuity

• Matter cannot be created or destroyed. This principle is know as the conservation of mass.The principle is applied to fixed volumes, known as control volumes

(or surfaces)

Mass entering / unit time = Mass leaving / unit time + Increase of mass in the control volume/unit time • For steady state

Mass entering / unit time = Mass leaving / unit time

• For incompressible flow ρ = constant, hence

constant

V

A

V

A

_{1 m1}

=

_{2 2 m2}

=

1

ρ

ρ

Q

V

A

V

A

_{1 m1}

=

_{2 m2}

=

Fluid Mechanics I

Examples

• For the pipe contraction shown, determine the velocity of water at section 2 if the velocity at section 1 is 2.1 m/s and the surface area of section 1 and section 2 are

0.01 m2_{and 0.003 m}2_respectively • For the pipe expansion shown, determine

the velocity of water at section 1 if the velocity at section 2 is 3.0 m/s and the diameters of section 1 and section 2 are 30 mm and 40 mm respectively

• If the mean velocity in pipe 1 is 2 m/s and its diameter is 50mm and pipe 2 diameter is 40 mm and takes 30% of the total discharge and pipe 3 diameter is 60mm, determine the values of discharge and the mean velocity in each pipe

43

Energy – The Bernoulli’s Equation

• Mass passing in 1 sec = • Weight passing in 1 sec =

• Kinetic Energy passing in 1 sec = • KE per unit weight =

• Potential Energy passing in 1 sec = • PE per unit weight =

• Work done by pressure in 1 sec = force * Distance = • Work done per unit weight =

• Energy per unit weight =

• Along a streamtube If there is no energy dissipation

AV

ρ

gAV

ρ

(

)

2 2

V

AV

2

1

2

mV

_ρ

=

2g V2

(

AV

)

gh

mgh

=

ρ

h

( )

pA

V

g

p

ρ

2g

V

h

g

p

₊

₊

2

ρ

Datum h A V

Constant

2g

V

h

g

p

₊

₊

2

₌

ρ

Conservation Principles

Fluid Mechanics I Conservation Principles

Energy – The Bernoulli’s Equation

• In the direction of the streamtube we have the following forces » Pressure force at upstream end =

» Pressure force at downstream end =

» Pressure forces around the circumference = zero » Weight force =

• Using Newton 2nd_Law » deviding by

» Noting that gives

» This is known as the Euler’s Equation, for incompressible fluid this can be integrated to yield

pA

A

p

p

)

(

+

δ

−

(

)

dL

V

d

dL

dV

V

dt

dL

dL

dV

dt

dV

₌

₌

₌

2

/

2

)

/

(

)

(

p

p

A

A

hg

A

L

dV

dt

pA

m

⇒

−

+

δ

−

ρ

δ

=

ρ

δ

=

a

F

hg

A

Lg

A

mg

θ

=

−

ρ

δ

θ

=

−

ρ

δ

−

cos

cos

0

1

1

₊

₊

₌

⇒

dL

dh

dt

dV

g

dL

dp

g

L

gA

ρ

δ

ρ

0

2

1

1

₊

2

₊

₌

dL

dh

dL

dV

g

dL

dp

g

ρ

Constant

h

2g

V

g

p

₊

2

₊

₌

ρ

45

Energy – The Bernoulli’s Equation

• Note that all the individual terms in the Bernoulli’s equation have units of length • The term is know as potential head

• The term is know as velocity head • The term is know as pressure head

• The term is know as total head

Example

• A fluid of constant density of 960 kg/m3_{is flowing} steadily through the tube shown. The diameter at section 1 is 100 mm and at section 2 is 80 mm. The pressure gauge at section 1 indicated a pressure of 200 kN/m2_{and the velocity was 5 m/s.}

Determine the pressure at section 2.

2g

V

2

h

g

p

ρ

2g V h g p H 2 + + = ρ Conservation Principles Fluid Mechanics I

Application of the Energy Equation

Pitot Tube

• If a stream of uniform velocity flows into a blunt body, some move to the left and some to the right. But one, in the centre, goes to the tip of the blunt body and stops. This point is known as the stagnation point

• Applying the Bernoulli’s equation between point 1 and 2 gives

• The term is called dynamic pressure

• Knowledge of the static and stagnation pressure will enable the calculation of the velocity of the fluid

• This is the principle on which the Pitot-static tube is based

• Two concentric tubes are attached to two pressure gauges 2 1 1 2

2

1

V

p

p

=

+

ρ

2 1 2 1 V

ρ

Conservation Principles

47

Example

• For the Pitot tube shown, show that the relation between the fluid velocity and the manometer reading h

Venturi Meter

• The Venturi meter is a device for measuring discharge in a pipe.

• It consists of a rapidly converging section which increases the velocity of flow and hence reduces the pressure.

• By measuring the pressure differences the discharge can be calculated.

(

)

ρ

ρ

ρ

−

=

gh

man

V

₁

2

Conservation Principles h h Fluid Mechanics I

• Using the Bernoulli’s equation between 1 and 2

• Using the continuity equation

• However,

• Hence,

• The theoretical discharge is

• To get the actual discharge we account for losses due to friction, we include a coefficient of discharge 2 2 2 2 1 2 1 1

2

2

g

z

V

g

p

z

g

V

g

p

+

+

=

+

+

ρ

ρ

gh

h

z

g

p

gz

p

₁

+

ρ

₁

=

₂

+

ρ

(

₂

−

)

+

ρ

_man 2 1 1 2 2 2 1 1

A

A

V

V

A

V

A

V

Q

=

=

⇒

=

1

2

2 2 1 1

−

























₋

=

A

A

gh

V

man

ρ

ρ

ρ

1 1

A

V

Q

_ideal

=

ideal d actual

C

Q

Q

=

Conservation Principles A _B h

49

Flow Through a Small orifice

• At the surface velocity is negligible and the pressure atmospheric.

• At the orifice the jet is open to the air so again the pressure is atmospheric

• If we take the datum line through the orifice then

z1 = h and z2 =0.

• Hence,

• This is the theoretical value of velocity.

• To incorporate friction we use the coefficient of velocity

• Each orifice has its own coefficient of velocity, they usually lie in the range (0.97 - 0.99) • The actual area of the jet is the area of the vena contractanot the area of the orifice. We

obtain this area by using a coefficient of contraction for the orifice

gh

V

g

V

h

2

2

2 2 2

⇒

=

=

l theoretica v actual

C

V

V

=

l theoretica c actual

C

A

A

=

Conservation Principles Fluid Mechanics I

• So the discharge through the orifice is given by

gh

A

C

Q

V

A

C

C

V

A

Q

orifice d actual l theoretica orifice v c actual actual actual

=

=

=

Conservation Principles

51

Flow Over Notches and Weirs

• A notch is an opening in the side of a tank or reservoir which extends above the surface of the liquid.

• It is usually a device for measuring discharge.

• A weir is a notch on a larger scale - usually found in rivers. • Weir can be sharp crested but also may have a substantial

width in the direction of flow - it is used as both a flow measuring device and a device to raise water levels. • We will assume that the velocity of the fluid approaching

the weir is small so that kinetic energy can be neglected. • We will also assume that the velocity through any elemental

strip depends only on the depth below the free surface. • These are acceptable assumptions for tanks with notches or

reservoirs with weirs, but for flows where the velocity approaching the weir is substantial the kinetic energy must be taken into account

Conservation Principles

Fluid Mechanics I

• To determine the theoretical discharge over a weir we will consider a strip of thickness δh, width land at depth h

• Velocity through the strip • Discharge through the strip

• Integrating from the free surface to the weir crest

Rectangular Weir

• The width is constant, b, Hence

• To obtain the actual discharge we introduce the coefficient of discharge

Triangular Weir

• The width b at depth his , hence

gh

V

=

2

gh

h

l

AV

Q

δ

2

δ

=

=

∫

=

g

H

lh

dh

Q

0 2 / 1

2

2 / 3 0 2 / 1

2

3

2

2

g

h

dh

b

g

H

b

Q

H

=

=

_∫

2 / 3

2

3

2

H

g

b

C

Q

_actual

=

_d

(

/

2

)

tan

)

(

2

H

h

θ

b

=

−

(

)

5/2

2

2

/

tan

15

8

H

g

C

Q

_actual

=

_d

θ

Conservation Principles

53

Examples

• Water flows along a circular duct from A to B where conditions are those shown. Assuming no losses, estimate the pressure at B.

• If the velocity of the water jet at point A is 20 m/s,

what is the pressure at point B? Neglect all losses due to viscous effects and assume that the nozzle outlet is at the same height as point B.

• A horizontal venturi tube, for measuring the flow of water, tapers from 300 mm diameter at the inlet to 100 mm diameter at the throat and has a discharge coefficient of 0.98. If the differential U-tube manometer, containing water over mercury (specific gravity 13.6), connecting the inlet and the throat, shows a difference in mercury levels of 55 mm, determine the volume flow.

Conservation Principles

Fluid Mechanics I

Problems

• A pitot-static tube used to measure the air speed in a wind tunnel is coupled to a water manometer. If the dynamic pressure is h mm of water, obtain an expression for the air speed in m/s. (Ans: 3.99h0.5₎

• Water collects in the bottom of a rectangular oil tank as shown. How long will it take for the water to drain from the tank through a 0.02- m-diameter drain hole in the bottom of the tank? Assume quasisteady flow. (Ans: 2.45 hr)

• A triangular orifice is cut in the vertical side of a tank containing a liquid. The base of the orifice is horizontal and of breadth b. The apex of the orifice is at a height d above the base and is located at a depth d below the liquid surface level. If the coefficient of • discharge is unity, derive an expression for the volume flow rate. (Ans: 0.91b(gd3₎0.5₎

55

Problems

• Water flows through the pipe contraction shown. For the given 0.2-m difference in manometer level, determine the flowrate as a function of the diameter of the small pipe, D. (Ans: 0.0156 m3/_s)

• Water flows steadily through the pipe shown such that the pressures at sections (1) and (2) are 300 kPa and 100 kPa, respectively. Determine the diameter of the pipe at section (2), D₂, if the velocity at section 1 is 20 m/s and viscous effects

are negligible. (Ans: 0.0688 m)

Conservation Principles

Fluid Mechanics I Conservation Principles

The Momentum Equation

• Moving fluids exerting forces on whatever it hits

• In fluid mechanics the analysis of motion is performed in the same way as in solid mechanics by the use of Newton’ s Second Law of motion

• The momentum equation is a statement of Newton’s Second Law and relates the sum of the forces acting on an element of fluid to its acceleration or rate of change of momentum. • Newton’s 2nd_{Law can be written:}

The Rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force

• We start by assuming that we have steady flow

• In time δt

» momentum of fluid entering stream tube » momentum of fluid leaving stream tube » Applying Newton’s Second Law

dt

m

d

( V

)

F

=

1 V t V A ₁δ ρ1 1 2 V t V A ₂δ ρ2 2 t t V A t V A 2 1 δ δ ρ δ ρ V2 V1 F= 2 2 − 1 1 Note 1 represent inflow 2 represent outflow

57

Conservation Principles

» Assuming fluid with a constant density and using the continuity equation

» Since the velocity have components in the x, y,and zdirection, it is more convenient to consider each direction separately

• The force F is made up of following components:

» F_R= Force exerted on the fluid by any solid body touching the control volume » F_B= Force exerted on the fluid body (e.g. gravity)

» F_P= Force exerted on the fluid by fluid pressure outside the control volume

• When using the momentum equation, the following steps need to be considered: » Draw a control volume

» Decide on co-ordinate axis system » Calculate the total force

» Calculate the pressure force » Calculate the body force » Calculate the resultant force

2 1 Q AV V A Q1= 1 2= 2 ) ( 2V2 1V1 F=ρ Q −Q ) ( ) ( ) ( _{2 x2 1 x1 y 2 y2 1 y1 z 2 z2 1 z1} x QV QV F QV QV F QV QV F =ρ − =ρ − =ρ −

Fluid Mechanics I Conservation Principles

Application of the Momentum Equation

The Force Due to the Flow Around a Pipe Bend

• Consider a pipe bend with a constant cross section lying in the horizontal plane and turning through an angle ofθ

• Control Volume : The control volume include the faces at the inlet and outlet of the bend and the pipe walls • Co-ordinate system

It is convenient to choose the co-ordinate axis so

that one is pointing in the direction of the inlet velocity. • Calculate the total force:

• Calculate the pressure force:

• Calculate the body force:The only body force is that exerted by gravity and have not component in the x and y directions

)

0

sin

(

)

(

)

cos

(

)

(

2 1 2 1 2 1 2

−

=

−

=

−

=

−

=

θ

ρ

ρ

θ

ρ

ρ

V

Q

V

V

Q

F

V

V

Q

V

V

Q

F

y y Ty x x Tx

θ

θ

sin

0

cos

2 2 2 2 1 1

A

p

F

A

p

A

p

F

Py Px

−

=

−

=

59

Conservation Principles

• Calculate the resultant force:

» Hence

» the force on the bend is the same magnitude but in the opposite direction

θ

θ

ρ

θ

θ

ρ

sin

sin

cos

)

cos

(

2 2 2 2 2 1 1 1 2

A

p

QV

F

A

p

A

p

V

V

Q

F

F

F

F

F

F

F

F

F

Ry Rx By Py Ry Ty Bx Px Rx Tx

+

=

+

−

−

=

+

+

=

+

+

=

















=

+

=

− Rx Ry Ry Rx R

F

F

F

F

F

1 2 2

tan

φ

F_Rx FRx F_R Φ V5_5.mov

Fluid Mechanics I Conservation Principles

Force on a Pipe Nozzle

• Because the fluid is contracted at the nozzle forces are induced in the nozzle

• Control Volume and Co-ordinate system are shown

• Calculate the total force:

» Using the continuity equation

• Calculate the pressure force:

» We use the Bernoulli equation and noting that the pressure outside is atmospheric

• Calculate the body force:gravity have no component in the x direction

• Calculate the resultant force:

)

(

_{x2 x1} Tx T

F

Q

V

V

F

=

=

ρ

−

2 2 1 1

A

p

A

p

F

_Px

=

−













−

=

⇒

=

=

1 2 2 2 2 1 1

1

1

A

A

Q

F

V

A

V

A

Q

_T

ρ













−

=

₂ 1 2 2 2 1

1

1

2

A

A

Q

p

ρ

1 2 1 2 2 2 1 2 2

)

1

1

(

2

)

1

1

(

A

A

A

Q

A

A

Q

F

F

F

F

F

Rx Bx Px Rx Tx

−

−

−

=

+

+

=

ρ

ρ

61

Conservation Principles

Force Due to an Inclined Jet Hitting a Plate

• Control Volume and Co-ordinate system are shown

• Calculate the total force:

» Using the energy equation and noting that z₁=z₂=z₃and the pressure is all atmospheric we can prove that v₁= v₂= v₃ = v

• Calculate the pressure force: all zero as the pressure is everywhere atmospheric

• Calculate the body force:gravity have no component in the x and y directions

• Calculate the resultant force:

)

sin

(

cos

1 2 3 1

θ

ρ

θ

ρ

Q

Q

Q

V

F

F

F

F

F

V

Q

F

F

F

F

F

Ry By Py Ry Ty Rx Bx Px Rx Tx

+

−

−

=

⇒

+

+

=

−

=

⇒

+

+

=

)

sin

)

((

)

)

((

cos

)

)

((

1 1 3 3 2 2 1 1 3 3 2 2 1 1 1 1 3 3 2 2

θ

ρ

ρ

θ

ρ

ρ

V

Q

V

Q

V

Q

V

Q

V

Q

V

Q

F

V

Q

V

Q

V

Q

V

Q

F

y y y Ty x x x Tx

−

−

=

−

+

=

−

=

−

+

=

V2 V1 V3 Fluid Mechanics I

Force on a Pelton Wheel Blade

• The above analysis of impact of jets be extended and applied to analysis of turbine blades • One clear demonstration of this is with the blade of a turbine called the Pelton wheel

• A narrow jet is fired at blades which stick out around the periphery of a large metal disk • The jet is deflected by the blade and the change of its momentum transfer a force to the

blade and hence a torque to the drive shaft • Calculate the total force:

• Calculate the pressure force: all zero as the pressure is everywhere atmospheric

• Calculate the body force:gravity have no component in the x and y directions

• Calculate the resultant force:

Conservation Principles

)

cos

(

V

2

V

1

Q

F

F

F

F

F

Tx

=

Rx

+

Px

+

Bx

⇒

Rx

=

ρ

β

+

)

cos

(

)

)

2

2

((

1 2 1 2 2

V

V

Q

V

Q

V

Q

V

Q

F

_{Tx x x x}

+

=

−

+

=

β

ρ

ρ

63

Examples

• Air flows from a 600 mm diameter pipe, through a nozzle which is bolted on to the end of the pipe, and discharges into the atmosphere. The outlet diameter of the nozzle is 300 mm. A U-tube manometer, connected to the pipe, shows a pressure difference of 250mm of water. Assuming that there are no losses, estimate the speed of the

air at the outlet of the nozzle and the force in the bolts required to hold the nozzle in position. (Ans: 65.2 m/s; 415 N)

• A horizontal pipeline has a bend which changes the direction of the water flowing through it by 45° and at the same time changes in diameter from 0.5 m upstream to 0.25 m downstream. The gauge pressure upstream is 2x1O5_N/m2 _{and the volume flow is O.4m}3_/s. Neglecting losses, determine the force required to hold the bend in position. (Ans:R=33kN)

Conservation Principles

Fluid Mechanics I

Problems

• A nozzle providing a horizontal jet of water 25 rnm diameter at a speed of 10 m/s is supplied by a pipe from an open reservoir whose free surface is 7 m above the nozzle. The jet powers a simple turbine made up of flat plate blades which the jet strikes at 90°. The blades are connected to a shaft so that the point of impact between the jet and the blades is 300 mm from the centre of the shaft

Mate an estimate of: (i) the loss of head in the pipe. (ii) the force exerted by the jet if the blades are stationary.

If the shaft is allowed to rotate at 200 rev/min, calculate: (iii) the force now exerted on the blades;(iv) the power available at the shaft; and (v) the overall efficiency, taking the free surface of the reservoir as the input

(Ans: (i) 1.9 m; (ii) 49.1 N (iii) 18.3 N; (iv) 115 W; (v) 34%)

• A horizontal streamlined nozzle issues a jet of fluid of density ρ, cross sectional area a and velocity U. Ignoring any viscous losses, derive an expression for the force required to hold the nozzle in position at the end of a pipeline of cross sectional area A.

65

Problems

• Air of density 1.22 kg/m3_{flows in a duct of internal diameter} 600 mm and is discharged to the atmosphere. At the outlet end of the duct, and co-axial with it, is a cone with a base diameter greater than 600 mm and a vertex angle of 90°. The flow through the duct is controlled by moving the cone into the duct, the air then escaping along the sloping

sides of the cone. The mean velocity in the duct upstream of the cone is 15 m/s and the air leaves the cone with a mean velocity of 60 m/s parallel to the sides. Neglecting viscous effects, calculate the force exerted by the air on the cone. (Ans: 441N)

• In the system shown, air is drawn from the atmosphere into a 250 mm diameter duct by a fan and flows out past a 200 mm diameter obstacle with a speed of 30 m/s. If the air is stationary in the atmosphere and there are no losses in the duct, calculate: (i) the speed of the air in the duct

(ii) the pressure in the duct upstream of the fan

(iii) the force F required to hold the obstacle in position (iv) the power delivered to the air by the fan (Ans: (i)10.8 m/s; (ii) -71.7 N/m2 _{gauge; (iii) 11.1 N; (iv) 293 W)}

Conservation Principles

Fluid Mechanics I

Viscosity

• It is clear that the previous properties are not sufficient to uniquely characterize how fluids behave since two fluids such as water and oil can have approximately the same value of density but behave quite differently when flowing.

• There is apparently some additional property that is needed to describe the “fluidity” of the fluid.

• From the definition of fluid, deforms continuously when subjected to shear forces, if a fluid is at rest there are no shearing forces.

• Shear stresses develop if the particles of moving fluid move relative to one another. • At all solid boundaries the flow particles have

zero relative velocity to the boundaries and it will increase as we move toward the centre • Since we are concerned with flow past solid

boundaries; cars, aeroplanes, pipes and channels, shear forces will be present