Basic Mechanical Engineering 9789389583915

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Basic Mechanical Engineering

Basic Mechanical Engineering

The book starts with the law of forces, free-body diagrams, basic information on materials strength including stresses and strains. It further discusses principles of transmission of power and elementary designs of gears, spring, etc. This part concludes with mechanical vibrations — their importance, types, isolation and critical speed.

Part three, Fluid Mechanics and Hydraulics, includes properties of fluids, measurement of pressure, Bernoulli’s equation, hydraulic turbine, pumps and various other hydraulic devices. Part four, Manufacturing Technology, mainly deals with various manufacturing processes, such as metal forming, casting, cutting, joining, welding, surface finishing and powder metallurgy. It further deals with conventional and non-conventional machining techniques, fluid power control and automation including hydraulic and pneumatic systems and automation of mechanical systems. Part five, Automobile Engineering deals with various aspects of IC and SI engines and their classification, etc. Four- and two-stroke engines also find place in this section. Next, systems in automobiles including suspension and power transmission systems, starting, ignition, charging and fuel injection systems. The last section deals with power plant engineering and energy. It includes power plant layout, surface condensers, steam generators, boilers and gas turbine plants. It concludes with renewable, nonrenewable, conventional and non-conventional sources of energy, and energy conversion devices.

Key Features Treatment of topics at depth with a large number of solved problems, diagrams in every chapter Inclusion of section-end multiple choice questions and book-end model questions for ample practice Glossary containing all the important terms with their definitions at the end of the book Kaushik Kumar, B.Tech, [Mechanical Engineering, REC (Now NIT), Warangal], MBA (Marketing, IGNOU) and Ph.D (Engineering, Jadavpur University), is Associate Professor in the Department of Mechanical Engineering, Birla Institute of Technology, Mesra, Ranchi, India. He has 15 years of Teaching & Research and over 11 years of industrial experience. His areas of teaching and research interest are CAD/CAM, Quality Management Systems, Optimisation, Non-conventional machining, Rapid Prototyping and Composites.

Sanghamitra Debta, B.Tech. (Mechanical Engineering, ITER SOA University, Bhubaneswar), is pursuing M.E. (Design of Mechanical Equipment, BIT Mesra). Her areas of interests are Product and Process Design, Strength of Materials, Material Engineering, and Automation.

978-93-89583-91-5

Kumar Roy Debta

Apurba Kumar Roy, B.E. [Mechanical Engineering, REC (Now NIT), Jaipur], M.E. (Mechanical Engineering, Jadavpur University, Kolkata) and PhD (Engineering, IIT Kharagpur), is Associate Professor at the Department of Mechanical Engineering, Birla Institute of Technology, Mesra, Ranchi, India. He has over 27 years of teaching, research and industrial experience. His areas of interest are Fluid Dynamics, Turbo Machines, Multiphase Flow, CFD, Optimisation and Non-conventional Energy, Direct Energy Conversion.

Basic Mechanical Engineering

The second part, Thermal Engineering, deals with basics and laws of thermodynamics; pure substances and their properties. It further includes laws of heat transfer, insulation, and heat exchanges. This part concludes with a detailed discussion on refrigeration and air conditioning.

Kaushik Kumar Apurba Kumar Roy Sanghamitra Debta Distributed by:

9 789389 583915 TM

TM

Basic Mechanical Engineering

©Copyright 2019 I.K. International Pvt. Ltd., New Delhi-110002. This book may not be duplicated in any way without the express written consent of the publisher, except in the form of brief excerpts or quotations for the purposes of review. The information contained herein is for the personal use of the reader and may not be incorporated in any commercial programs, other books, databases, or any kind of software without written consent of the publisher. Making copies of this book or any portion for any purpose other than your own is a violation of copyright laws. Limits of Liability/disclaimer of Warranty: The author and publisher have used their best efforts in preparing this book. The author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness of any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. The accuracy and completeness of the information provided herein and the opinions stated herein are not guaranteed or warranted to produce any particulars results, and the advice and strategies contained herein may not be suitable for every individual. Neither Dreamtech Press nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Dreamtech Press is not associated with any product or vendor mentioned in this book. ISBN: 978-93-89583-91-5 EISBN: 978-93-90457-37-3

Basic Mechanical Engineering

Kaushik Kumar B. Tech (Mech. Engg.), MBA, Ph.D. Birla Institute of Technology-Mesra, Ranchi

Apurba Kumar Roy B.E. (Mech. Engg.), M.E. (Mech Engg.), Ph.D. Birla Institute of Technology-Mesra, Ranchi

Sanghamitra Debta B. Tech (Mech Engg.), M.E. (D.M.E.) Birla Institute of Technology-Mesra, Ranchi

Dedicated to our Parents

Preface The purpose of this book is to present the reader with a text that includes theory and examples. It covers a broad spectrum of critical mechanical engineering topics and helps the reader understand the fundamentals. In this book while explaining the concept, basic concepts and basic principles of mechanical engineering, importance is given to the engineering applications of the subject. Section I comprises four chapters, Chapter 1 gives an introduction to law of forces, free-body diagram. Chapter 2 includes the basic knowledge of material strength including stresses and strains. Chapter 3 deals with principles of transmission of power and elementary design of components like gears, springs, etc. Chapter 4 is on vibrations. Chapter 5 deals with the thermodynamics terms and properties. Chapter 6 deals with the process of conduction, convection and radiation along with heat exchangers. Chapter 7 gives basic definitions and discussions about various refrigerants and their properties. In Chapter 8 fluid mechanics and Chapter 9 hydraulic machines are discussed respectively. Section IV deals with one of the most important part of mechanical engineering which itself is considered a stream for undergraduate studies in many engineering colleges which is manufacturing technologies wherein chapters like Chapter 10 introduction to various manufacturing processes, Chapter 11 Conventional and Chapter 12 Nonconventional machining techniques and Chapter 13 Fluid power control and automation are discussed. Section V is an introduction to automobile engineering where topics like Chapter 14 IC engines and Chapter 15 various automobile systems are the subject of discussion. Finally, in section VI Chapter 16 Power plant engineering focuses on the layout of different power plants and classification along with components explained in detail. Chapter 17 is on the renewable and nonrenewable sources of energy. Glossary contains brief explanation/definitions of important terms used in the book. Kaushik Kumar Apurba Kumar Roy Sanghamitra Debta

Preface

Acknowledgements First and foremost we thank the God. In the process of putting this book together we realized how true this gift of writing is for us. You have given us the power to believe in our passion and pursue our dreams. We could never have done this without the faith we have in you, the Almighty. We are thankful for this kindness. We thank our grandparents, parents and relatives for allowing us to follow our ambitions. Our families showed patience and tolerated us for taking yet another challenge which decreases the amount of time we could spend with them. They were our inspiration and motivation. We would like to thank our colleagues, friends and students of different institutes and organizations and specially Birla Institute of Technology, Mesra. They always supported us throughout our career and also while writing this book. This book was not only inspired by them; it was also improved by their active involvement in its development. Late Prof. Sanat Kumar Mukherjee of Birla Institute of Technology, Mesra, had shown us the importance of learning while working. We appreciate that he believed in us to provide leadership and knowledge to make this small effort a reality. Thanks for his wishes for today’s achievement. Throughout the process of writing this book, many individuals, from different walks of life, have taken time out to help us out. Last, but definitely not least, we would like to thank them all, our well-wishers, for providing us encouragement. Kaushik Kumar Apurba Kumar Roy Sanghamitra Debta

Contents Preface

vii

Acknowledgements

ix Section I Mechanical Engineering

1. Engineering Mechanics 1.1 Introduction 1.2 Classification of Engineering Mechanics 1.3 Laws of Forces 1.4 Lami’s Theorem 1.5 Free-Body Diagrams 1.6 Summary

3 3 4 9 11 12 13

2. Strength of Materials 2.1 Introduction 2.2 Simple Stresses and Strains 2.3 Hooke’s Law of Elasticity 2.4 Relation Between Elastic Constants 2.5 Thermal Stresses 2.6 Creep and Fatigue 2.7 Summary

21 21 24 27 30 30 31 33

3. Machine Design 3.1 Introduction 3.2 Transmission of Power 3.3 Gears 3.4 Springs 3.5 Power Screws 3.6 Thin Cylinders 3.7 Summary

44 44 46 48 53 57 61 61

Contents

xii

Contents

4. Mechanical Vibrations 4.1 Introduction 4.2 Importance of Vibration 4.3 Types of Vibration 4.4 Vibration Isolation and Critical Speed 4.5 Summary

68 68 71 72 79 82

Section II Thermal Engineering 5. Thermodynamics 5.1 Introduction 5.2 Basics of Thermodynamics 5.3 Laws of Thermodynamics 5.4 Properties of Pure Substances 5.5 Summary

97 97 98 103 106 109

6. Heat Transfer 6.1 Introduction 6.2 Basic Definitions 6.3 Basic Laws of Heat Transfer 6.4 Insulations 6.5 Heat Exchangers 6.6 Summary

117 117 119 122 123 124 129

7. Refrigeration and Air Conditioning 7.1 Introduction 7.2 Basic Definitions 7.3 Various Refrigerants and Their Properties 7.4 Summary

135 135 136 140 147

Section III Fluid Mechanics and Hydraulics 8. Fluid Mechanics 8.1 Introduction 8.2 Properties of Fluids 8.3 Measurement of Pressure 8.4 Bernoulli’s Equation 8.5 Summary

161 161 167 168 173 175

Contents

9. Hydraulic Machines 9.1 Introduction 9.2 Hydraulic Turbines 9.3 Hydraulic Pumps 9.4 Hydraulic Devices 9.5 Summary

xiii

193 193 194 198 201 206 Section IV

Manufacturing Technology 10. Manufacturing Processes 10.1 Introduction 10.2 Classification 10.3 Basic Manufacturing Processes 10.4 Metal Forming Processes 10.5 Metal Casting Operations 10.6 Metal Cutting Operations 10.7 Metal Joining Process 10.8 Welding Operations 10.9 Surface Finishing Processes 10.10 Powder Metallurgy 10.11 Summary

215 215 217 217 220 221 224 231 233 236 236 240

11. Conventional Machining Techniques 11.1 Introduction 11.2 Conventional Machining Techniques 11.3 NC and CNC Assisted Conventional Machining Techniques 11.4 Summary

246 246 247 254 256

12. Nonconventional Machining Techniques 12.1 Introduction 12.2 Classification of Nonconventional Machining 12.3 Need of Nonconventional Machining Techniques 12.4 Some Nonconventional Machining Techniques 12.5 Comparison of Conventional and Nonconventional Machining Techniques 12.6 Summary

259 259 260 261 261

13. Fluid Power Control and Automation 13.1 Introduction 13.2 Hydraulic and Pneumatic Systems

277 277 278

271 272

xiv

Contents

13.3 Automation of Mechanical Systems 13.4 Summary

282 282

Section V Automobile Engineering 14. IC Engines 14.1 Introduction 14.2 Classification of IC Engines 14.3 IC and SI Engines 14.4 Four-Stroke and Two-Stroke Engines 14.5 Summary

293 293 296 301 305 308

15. Systems in Automobiles 15.1 Introduction 15.2 Suspension and Power Transmission Systems in Automobiles 15.3 Starting System, Ignition System and Charging Systems in Automobiles 15.4 Fuel Injection System in Automobiles 15.5 Summary

315 315 316 319 322 325

Section VI Power Plant Engineering 16. Power Plant Engineering 16.1 Introduction 16.2 Power Plant Layout 16.3 Surface Condenser and Steam Generators/Boilers 16.4 Gas Turbine Power Plant 16.5 Summary

333 333 334 347 350 353

17. Energy 17.1 Introduction 17.2 Renewable and Nonrenewable Sources of Energy 17.3 Conventional and Nonconventional Sources of Energy 17.4 Energy Conversion Devices 17.5 Summary

354 354 355 362 364 365

Glossary

369

Index

375

Section I Mechanical Engineering

1 Engineering Mechanics

Learning Objectives   

‡ 8QGHUVWDQGLQJWKHWHUPVLQHQJLQHHULQJPHFKDQLFV ‡ 5HYLHZLQJWKHODZVRIIRUFHV ‡ *HWWLQJDQLGHDRQKRZWRGUDZDQGVROYHSUREOHPVRQIUHHERG\GLDJUDPV

1.1

INTRODUCTION

Mechanics is defined as that branch of science, which describes and predicts the conditions of rest or motion of bodies under the action of forces. Engineering mechanics applies the principle of mechanics to design, taking into account the effects of forces. The state of rest and the state of motion of the bodies under the action of different forces has engaged the attention of philosophers, mathematicians and scientists since centuries. The branch of physical science that deals with the state of rest or the state of motion is known as mechanics. Starting from the analysis of rigid bodies under gravitational force and simple applied forces the mechanics has grown to the analysis of robotics, aircraft, spacecraft under dynamic forces, atmospheric forces, temperature forces, etc. Archemedes (287-212 BC), Galileo (1564-1642), Sir Issac Newton (1642-1727) and Einstein (1878-1955) have contributed a lot to the development of mechanics. Contributions by Varignon, Euler, D, Alembert are also substantial. The mechanics developed by these researchers may be grouped as (i) Classical mechanics/Newtonian mechanics, (ii) Relativistic mechanics, and (iii) Quantum mechanics/Wave mechanics. Sir Issac Newton, the principal architect of mechanics, consolidated the philosophy and experimental findings developed around the state of rest and state of motion of the bodies and put forth them in the form of three laws of motion as well as the law of gravitation. The mechanics Basic Mechanical Engineering

4

Basic Mechanical Engineering

based on these laws is called Classical Mechanics or Newtonian Mechanics. Albert Einstein proved that Newtonian Mechanics fails to explain the behaviour of high speed (speed of light) bodies. He put forth the theory of Relativistic Mechanics. Schrodinger (1887-1961) and Broqlie (1892-1965) showed that Newtonian mechanics fails to explain the behaviour of particles when atomic distances are concerned. They put forth the theory of Quantum Mechanics. Engineers are keen to use the laws of mechanics to actual field problems. Application of laws of mechanics to field problems is termed engineering mechanics. For all the problems between atomic distances to high speed distances classical/Newtonian mechanics has stood the test of time and hence that is the mechanics used by engineers. Hence, in this text classical mechanics is used for the analysis of engineering problems.

1.2

CLASSIFICATION OF ENGINEERING MECHANICS

Depending on the body to which the mechanics is applied, Engineering Mechanics is classified as: (a) mechanics of solids, and (b) mechanics of fluids. Solid Mechanics is further classified as mechanics of rigid bodies and mechanics of deformable bodies. The bodies which will not deform or in which deformation can be neglected in the analysis, are called rigid bodies. The mechanics of the rigid bodies dealing with the bodies at rest is termed statics and that dealing with bodies in motion is called ‘dynamics’. The dynamics dealing with the problems without referring to the forces causing the motion of the body is termed kinematics and if it deals with the forces causing motion also is called kinetics. If the internal stresses developed in a body are to be studied, the deformation of the body should be considered. This field of mechanics is called mechanics of deformable bodies/strength of materials/solid Engineering Mechanics

Mechanics of Fluids

Mechanics of Solids

Mechanics of Rigid Bodies

Statics

Mechanics of Deformable Bodies

Dynamics

Kinematics

Theory of Elasticity

1. Ideal Fluid 2. Viscous Fluid 3. Incompressible Fluid

Theory of Plasticity

Kinetics

Figure 1.1

Engineering Mechanics 5

mechanics. This field may be further divided into ‘Theory of Elasticity’ and ‘Theory of Plasticity’. Liquid and gases deform continuously with application of very small shear forces. Such materials are called fluids. The mechanics dealing with the behaviour of such materials is called fluid mechanics. Mechanics of ideal fluids, mechanics of viscous fluids and mechanics of incompressible fluids are further classification in this area. The classification of mechnics is summarised in Fig. 1.1.

1.2.1 Terms Used in Mechanics Mass: The quantity of the matter possessed by a body is called mass. The mass of a body will not change unless the body is damaged and part of it is physically separated. When a body is taken out in a spacecraft, the mass will not change but its weight may change due to change in gravitational force. Even the body may become weightless when gravitational force vanishes but the mass remain the same. Time: Time is the measure of succession of events. The successive event selected is the rotation of the earth about its own axis and this is called a day. To have convenient units for various activities, a day is divided into 24 hours, an hour into 60 minutes and a minute into 60 seconds. Clocks are the instruments developed to measure time. To overcome difficulties due to irregularities in the earth’s rotation, the unit of time is taken as second which is defined as the duration of 9192631770 period of radiation of the caesium-133 atom. Space: The geometric region in which the study of body is involved is called space. At point in the space may be referred with respect to a predetermined point by a set of linear and angular measurements. The reference point is called the origin and set of measurements as ‘coordinates’. If coordinates involve only in mutually perpendicular directions they are known Cartesian coordinates. If the coordinates involve angle and distances, it is termed polar coordinate system. Length: It is a concept for measuring linear distances. The diameter of a cylinder may be 300 mm, the height of a building may be 15 m. Actually metre is the unit of length. However, depending on the sizes involved micro, milli or kilometre units are used. A metre is defined as length of the standard bar of platinum-iradium kept at the International Bureau of Weights and Measures. To overcome difficulties of accessibility and reproduction, now meter is defined as 1690763.73 wavelength of krypton-86 atom. Displacement: Displacement is defined as the shortest distance moved by a body/ particle in the specified direction. Velocity:

The rate of change of displacement with respect to time is defined as velocity.

6

Basic Mechanical Engineering

Acceleration: Acceleration is the rate of change of velocity with respect to time. Thus, a = dv/dt where v is velocity. Momentum: The product of mass and velocity is called momentum. Thus, momentum = mass × velocity Continuum: A body consists of several matters. It is a well known fact that each particle can be subdivided into molecules, atoms and electrons. It is not possible to solve any engineering problem by treating a body as a conglomeration of such discrete particles. The body is assumed to consist of a continuous distribution of matter. In other words, the body is treated as continuum. Statics: The branch of mechanics which is associated with the equilibrium of bodies under action of forces (bodies may be either at rest or move with a constant velocity) is termed statics. Dynamics: The branch of mechanics which is associated with the motion of the body on application of force is termed dynamics. Rigid Body: A rigid body is defined as a definite quantity of matter, the parts of which are fixed in position relative to each other. Physical bodies are never absolutely but deform slightly under the action of loads. If the deformation is negligible as compared to its size, the body is termed rigid. Force: Force might be characterized as any activity that tends to change the condition of rest or movement of a body to which it is connected. It is a vector quantity. Coplanar Forces: Coplanar forces are defined as the force or a set of forces acting on a body are in the same geometric plane. Concurrent Forces: The forces acting or meeting in a single point of action is said to be concurrent force. Non-collinear Forces: Non-collinear force is defined as the force which doesn’t act on the same line of action of that force. Resolution of Forces: Resolution of forces is said to be resolving of forces into its smaller units, or it also can be defined in simple words “splitting of the forces without changing their influence on the body”. Resolution of forces is generally carried out on two mutually perpendicular components, i.e., vertical component along Y-axis and horizontal component along X-axis. Scalars: In scalar quantity only the magnitude of particle changes and the direction remains the same, e.g., time, volume, density, speed, energy, mass. Vectors: The vector quantities possess magnitude as well as direction, and must obey the parallelogram law of addition and the triangle law.

Engineering Mechanics 7

Length, time, and mass are completely independent of each other while force is a derieved idea and is dependent on other concepts. Force acting on a body is associated with the mass of the body and the variation of its velocity with time.

1.2.2 Newton’s Three Laws of Motion Newton’s First Law: It states that a body continues in its state of rest or of uniform motion in a straight line unless it is compelled by an external agency acting on it. This leads to the definition of force as the external agency which changes or tends to change the state of rest or uniform linear motion of the body. Newton’s Second Law: It states that the rate of change of momentum of a body is directly proportional to the impressed force and it takes place in the direction of the force acting on it. Thus, according to this law. Force a rate of change of momentum. But momentum = mass × velocity, as mass does not change, Force a mass × rate of change of velocity i.e., Force a mass × acceleration F a m × a Newton’s Third Law: It states that for every action there is an equal and opposite reaction. Consider the two bodies in contact with each other. Let one body apply a force F on another. According to this law the second body develops a reactive force R which is equal in magnitude to force F and acts in the line same as F but in the opposite direction. Newton’s Law of Gravitation: Everybody attracts the other body. The force of attraction between any two bodies is directly proportional to their masses and inversely proportional to the square of the distance between them. According to this law the force of attraction between the bodies of mass m1 and mass m2 at distance ‘d’ as shown F=G

mm r2

where G is the constant of proportionality and is known as constant of gravitation.

F

F

1

2 m2

m1

d

Figure 1.2

8

Basic Mechanical Engineering

1.2.3 Units Length (L), mass (M) and time (S) are the fundamental units in mechanics. The units of all other quantities may be expressed in terms of these basic units. The three commonly used systems in engineering are: ‡ Metre—Kilogramme—Second (MKS system) ‡ Centimetre—Gramme—Second (CGS system) ‡ Foot—Pound—Second (FPS system) The units of length, mass and time used in the system are used to name the systems. 8VLQJWKHVHEDVLFXQLWVWKHXQLWVIRURWKHUTXDQWLWLHVFDQEHIRXQG)RUH[DPSOHLQ0.6 the units for the various quantities are as shown below: Quality

Unit

Notation

Area

Square metre

m2

Volume

Cubic metre

m3

Velocity

Metre per second

m/sec

Acceleration

Metre per second per second

m/sec2

Unit of forces: Presently, the whole world is in the process of switching over to SI system of units. SI stands for System Internationale d’ units or International System of units. As in MKS system, in SI system also the fundamental units are metre for length, kilogramme for mass and second for time. The difference between MKS and SI system arises mainly in selecting the unit of force. We have force ‫ ן‬mass × Acceleration = K × Mass × Acceleration In SI system unit of force is defined as that force which causes 1 kg mass to move with an acceleration of 1 m/sec2 and is termed 1 newton. Hence, the constant of SURSRUWLRQDOLW\NEHFRPHVXQLW\8QLWRIIRUFHFDQEHGHULYHGDV 8QLW RI IRUFH  NJ ð PVHF2 = kg – m/sec2 In MKS, the unit of force is defined as that force which makes a mass of 1 kg to move with gravitational acceleration ‘g’ m/sec2. This unit of force is called kilogramme weight or kg-w. Gravitational acceleration is 9.81 m/sec2 near the earth surface. In all the problems encountered in engineering mechanics the variation in gravitational acceleration is negligible and may be taken as 9.81 m/sec2. Hence, the constant of proportionality is 9.81, which means 1 kg-wt = 9.81 newtons It may be noted that in daily usage, kg-wt force is called kg only.

Engineering Mechanics 9

1.3 1.3.1

LAWS OF FORCES Law of Transmissibility of Force

According to this law the state of rest or motion of a rigid body is unaltered if a force acting on the body is replaced by another force of the same magnitude and direction but acting anywhere on the body along the line of action of the replaced force. Let F be the force acting on a rigid body at point A as shown in Fig. 1.3. According to the law of transmissibility of force, this force has the same effect on the state of body as the force F applied at point B. In using law of transmissibility of forces it should be carefully noted that it is applicable only if the body can be treated as rigid.

F A B

F

Figure 1.3

1.3.2 Triangle Law If two forces acting on a body are represented one after another by the sides of a triangle, their resultant is represented by the closing side of the triangle taken from the first point to the last point.

P

P

c

c

o

Figure 1.4 Triangle law.

1.3.3

Parallelogram Law of Forces

The parallelogram law of forces enables us to determine the single force called resultant which can replace the two forces acting at a point with the same effect as that of the two forces. This law was formulated based on experimental results. Though Stevinces

10

Basic Mechanical Engineering

employed it in 1586, the credit of presenting it as a law goes to Varignon and Newton (1687). This law states that if two forces acting simultaneously on a body at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through the point of intersection of the two sides representing the forces. In Fig. 1.5 the force F1 = 4 units and force F2 = 3 units are acting on a body at point A. Then to get resultant of these forces, parallelogram ABCD is constructed such that AB is equal to 4 units to linear scale and AC equal to 3 units. Then according to this law, diagonal AD represents the resultant in the direction and magnitude. Thus, the resultant of the forces F1 and F2 on the body is equal to units corresponding to AB in WKHGLUHFWLRQǂWRF1. Force AD is called the resultant of AB and AC and the forces are called its components. F2 = 3 units

C R 3

q A

q

F1 = 4 units

a B

A 4 (b)

(a)

d

a

A

n R spo orre c s D t Uni to A

(c)

Figure 1.5

D= Now applying triangle law

A2 + B2 + 2 AB cos a

Special cases Case I:

If a Ý D=

A2 + B2 + 2 AB cos 0∞

D=

A2 + B2 + 2 AB

D = (A + B)

D

Engineering Mechanics 11

Case II:

If a Ý D=

A2 + B2 + 2 AB cos 180∞

=

A2 + B2 - 2 AB

=

( A - B)2

= (A – B) Case III:

If a Ý D= =

A2 + B2 + 2 AB cos 90∞ A 2 + B2

-1 a = tan

B A

Characteristic of Forces From Newton’s first law, we defined the force as an agency which tries to change the state of stress or state of uniform motion of the body. From Newton’s second law of motion we arrived at practical definition of unit force as the force required to produce unit acceleration in a body of unit mass. Thus, 1 newton is the force required to produce an acceleration of 1 m/sec2 in a body of 1 kg mass. It may be noted that a force is completely specified only when the following four characteristics are specified: ‡ Magnitude ‡ Point of application ‡ Line of action ‡ Direction

1.4

LAMI’S THEOREM

This theorem was given by a great mathematician “Bernard Lamy”, and hence, coined Lami’s theorem. According to this theorem, when three coplanar, concurrent and non-collinear forces act on a body which is in equilibrium then the magnitude of each force is proportional to the sine of angle between other two forces. This theorem can be proved by the sine law. A B C = = sin a sin b sin g

12

Basic Mechanical Engineering A

g b

B a

C

Figure 1.6 Lami’s theorem.

Equllibrium law: Two forces can be in equilibrium only if they are equal in magnitude, opposite in direction and collinear in action.

1.5

FREE-BODY DIAGRAMS

Free-body diagram is necessary to investigate the condition of equilibrium of a body or system. While drawing the free-body diagram all the supports of the body are removed and replaced with the reaction forces acting on it. A physical demonstration of the physical states of a problem is termed a free-body diagram (FBD). At the point, when taking care of any problem, it is fundamental to consider all forces following up on the body and ignoring the forces that are not directly connected to the body. A free-body diagram of a body is a graphical representation or a portrayal of a body in which the body is totally isolated from all the neighbouring bodies, counting any supports, by a nonexistent cut, and the action of each body evacuated on the body being taken into consideration is shown as a constraint on the body when tracing the layout of the diagram. To draw a free-body diagram: 1. pick the free body to be utilized, separate it from other body and draw an outline of it; 2. find all the external forces acting on the free body and write their magnitude and its direction along with the weight of the free body, which is applied at the centre of gravity of the body; 3. find and write all the external forces and responses in the free-body outline; and 4. incorporate all measurements that indicate the location and direction of the applied forces. The free-body diagram of any rigid body can be lessened to that of a particle. The free-body of a particle is utilized to serve as a point and all forces acting on it.

Engineering Mechanics 13

List of Forces and How to Find Them Equation

Name

Direction

Notes

Fg = mg

Gravity; weight

Downwards (towards centre of the earth.

g = 9.8 m/s2 on the earth

FN

Normal force

Perpendicular (i.e., normal) to the surface.

Equal to whatever force is needed to prevent the object from falling through the floor. 8VHWKH)%'UHFLSHWRILQGFN.

FT

Tension

In direction of the rope or string (away from the object).

Is either given already, or needs to be found via the FBD recipe.

Fs = kx

Spring force (Hooke's law)

Opposite to the direction of x

x indicates how much the spring is compressed or stretched, and x = 0 if it's not stretched or compressed at all. k is called the spring constant.

fk = mk FN

Kinetic friction

Opposite to the direction of the velocity

mk is the coefficient of kinetic friction and FN the normal force.

fs

Static friction

Parallel to the surface, but the exact direction depends on what is needed to prevent object from moving

Equal to whatever force is needed to prevent object from moving.

FG = G

m1m2 r2

Fg = rf Vs g

However, the max value of fs is fs, max = ms FN with ms the coefficient of static friction and FN the normal force.

Gravitational force From centre of one mass towards the centre of the other

8VHWKLVHTXDWLRQLQVWHDGRIFg = mg

Buoyance force

rf is the density of the fluid, and Vs the part of the volume of the object that is submerged.

8SZDUGV

when the r is approximately the radius of the earth or larger (i.e., for planets).

When an object of mass m is floating, Fn = rf Vs g = mg.

1.6

SUMMARY

We have seen footballs bouncing and wheels rolling. We may have studied how every one of these movements happens. All these movements are cooperation of various bodies and impact of forces following up on them. The branch of science managing the impact of forces on bodies is called Mechanics.

14

Basic Mechanical Engineering

The standards of Mechanics are precisely relevant to machines, it might be a basic machine, for example, a liver or bike or a machine as perplexing as an air ship. At the point when Mechanics is connected in engineering, plan and examination of instruments and machine, it is called engineering mechanics. At the point when bodies collaborate and forces act between them there are two potential outcomes, they may move or they may stay static. The branch of engineering mechanics, managing the movement of bodies, is called dynamics and the other branch is called statics, in which we concentrate on balance and harmony of bodies. All through the investigation of engineering mechanics the standards of three Newton’s laws of motion are utilized perpetually. So total comprehension of these laws is an unquestionable requirement. Alongside this, as the physical amounts experienced amid building examinations are for the most part vectors, satisfactory information of vector polynomial math is required.

SOLVED EXAMPLES

Example 1.1 Find the tension on each of the ropes AB and AC B

C A

Space diagram

TAB TAC

A

980 N Free-body diagram for point A

Figure 1.7

Engineering Mechanics 15

Solution:

TAB 980 N

TAC

Example 1.2 A rigid body is pivoted to a vertical support and held at 50° to the level by method for a link when a weight of 250 N is suspended as appears in the figure. Find the tension in the link and compression in the bar, neglecting the weight acting on the bar. Solution: A

75°

250 N 50° Space diagram

65° 40°

Free-body diagram for point A

Tension 180 N 75° 65°

40° 250 N

Compression 265 N

16

Basic Mechanical Engineering

The forces may also be calculated using the law of sines Compression in rod Tension in cable 250 N = = sin 75° sin 40° sin 65°

Example 1.3 Find the value of T in figure given below. R

T

W

q

Figure 1.8

Solution: Consider the above figure. Let us give a virtual displacement dS (downward along inclined path). The work done by R is zero, since the motion is perpendicular to it. The virtual work done by T = – TdS The virtual work done by W = W (dS sin q) By virtual work principle, – TdS + W(dS sin q) = 0 T = W sin q

Example 1.4 Spot, the German Shepherd, is being walked by his best pal, Dot, the Siamese cat. Actually, Dot is pulling Spot across the ice rink, and disturbing the neighbourhood hockey game. What are the forces acting on Spot? Solution: Since this is taking place on the earth, we can assume a gravitational force Fmg on Spot. Also, because Spot is on the surface of the ice, the ice is exerting a normal force N on him. We assume that ice is frictionless, and so neither static friction nor kinetic friction acts on Spot (which he probably appreciates). The push/pull force may be misleading. While it is true that Spot is being pulled along by Dot, this is not the force acting on Spot. This is because Dot is actually pulling on the rope. The rope, in turn, through its tension T, exerts a force on Spot.

Engineering Mechanics 17

The FBD is Fpull

N

Fmg

Example 1.5 Given the situation as shown in the figure, determine the acceleration with which the blocks move. Assume there is no friction.

M2 M1 q

Figure 1.9

Solution: The problem asks for the acceleration, and therefore “forces” is the best way to approach this problem (because if we find the net force, we’ve found the acceleration – via SF = ma.). 1. Draw one free-body diagram for each object. We can ignore the pulley (it’s not mentioned that we should take it into account, and we know nothing about it, so we can ignore it). Note that the tension FT is the same for both objects, so we use the same symbol (and different symbols for the weights m1g and m2 g). Also v and a are the same for both objects, and their directions are as indicated. y

FN

v,a

FT

FT

x

y v

x m1 g

q

a m2 g

2. Break the forces up into components. Because of the way we’ve chosen our coordinate system in step 1, we only need to break up the gravitational force of m2 into components.

18

Basic Mechanical Engineering m2 g sin q m2 g cos q

q m2 g



 )RU HDFK REMHFW DQG HDFK GLUHFWLRQ ZULWH GRZQ ƴF = (sum of forces) = ma. We have two objects, but m1 only has forces in one direction. We should therefore write down three equations: Mass m1: SFy = FT – m1 g = m1a

Mass m2: Sfx = m2 g sin q – FT = m2a

Sfy = FN – mg cos q = 0

  

4. Solve this set of equations. If there are N unknowns then you need N equations. We have 3 equations and 3 unknowns (FT, FN, and a), so this is solvable. Often this is the hardest part, because it requires a lot of math and practice. ‡ )LUVW´VROYHµIRUFT using the equation for m1 as: FT = m1a + m1g ‡ 1H[WSOXJWKLVLQWRWKHXSSHUHTXDWLRQIRUm2: m2 g sin q – (m1 a + m1 g) = m2 a ‡ :HQRZKDYHDQHTXDWLRQZLWKRQO\RQHXQNQRZQZKLFKLVHDVLHUWR solve: m2 g sin q - m1 g m2 g sin q – m1 g = m1 a + m2 a = (m1 + m2) a fi a = m1 + m2

Example 1.6 Determine the magnitude and direction of the resultant of the forces shown. 150 N

30° 200 N

Figure 1.10

Solution: 1 Construct a parallelogram by drawing two lines. Each line starts at the tip of one vector and is parallel to the other vector.

Tip

150 N Parallel 30°

30° 200 N

Tip

Engineering Mechanics 19 2

Since opposite sides of a parallelogram are equal in length, the length of each line represents the magnitude of the vector opposite.

200 N 150 N

150 N 30°

30° 200 N

200 N 150 N

150 N

R

30° Tails

200 N 3 The resultant R is drawn from the tails of the vectors to the opposite vertex of the parallelogram.

Adding the forces, we get

20

Basic Mechanical Engineering

EXERCISE 1. Explain the following terms as used in engineering mechanics: (i) continuum (ii) rigid body, and (iii) particle. 2. State and explain Newton’s three laws of motion. 3. State and explain Newton’s law of gravitation. 4. State and explain law of transmissibility of forces. 5. State and explain parallelogram law of forces. From this derive triangle and polygonal laws of forces. 6. Define Lami’s theorem? 7. Into how many component a force can be resolved? 8. Two forces of magnitude 50 N and 100 N are acting on the same geometric plane at angles of 30° and 60° respectively. Find the magnitude of the third force using triangle law of forces. 9. A man of weight W = 712 N holds one end of a rope that passes over a pulley vertically above his head and to the other end of which is attached a weight Q = 534 N. Find the force with which the man’s feet press against the floor. 10. A boat is moved uniformly along a canal by two horses pulling with forces P = 890 N and Q = 1068 N acting under an angle a = 60°. Determine the magnitude of the resultant pull on the boat and the angles b and ǎ. 11. Explain the term force and list its characteristics. 12. Explain the terms—concurrent and non-concurrent force system; and planar and non-planar system of force. 13. The sports car has a mass of 1.5 mg and mass centre at G. If the front two springs each has a stiffness of kA = 58 kN/m and the rear two springs each has a stiffness of kB = 65 kN/m, determine their compression when the car is parked on the 30° incline. Also, what frictional force FB must be applied to each of the rear wheels to hold the car in equilibrium?

Figure 1.11

2 Strength of Materials

Learning Objectives ‡ ‡ ‡ ‡

2.1

8QGHUVWDQGLQJWKHWHUPVLQVWUHQJWKRIPDWHULDOV Basics of stress and strains. Relation between the elastic constants. To understand the thermal stresses in mechanical members.

INTRODUCTION

Strength of materials deals with the relations between the external forces applied to elastic bodies and the resulting deformations and stresses. In the design of structures and machines, the application of the principles of strength of materials is necessary if satisfactory materials are to be utilized and adequate proportions obtained to resist functional forces. Forces are produced by the action of gravity, by accelerations and impacts of moving parts, by gases and fluids under pressure, by the transmission of mechanical power, etc. In order to analyze the stresses and deflections of a body, the magnitudes, directions and points of application of forces acting on the body must be known. Information given in the mechanics section provides the basis for evaluating force systems. The time element in the application of a force on a body is an important consideration. Thus, a force may be static or change so slowly that its maximum value can be treated as if it were static; it may be suddenly applied, as with an impact; or it may have a repetitive or cyclic behaviour. The environment in which forces act on a machine or part is also important. Such factors as high and low temperatures; the presence of corrosive gases, vapours and liquids; radiation, etc., may have a marked effect on how well parts are able to resist stresses. Basic Mechanical Engineering

22

Basic Mechanical Engineering

Nowadays, building of structures, machines and other engineering structures is impossible without projects previously drawn. The project consists of the drawings and explanation notes presenting the dimensions of the construction elements, the materials necessary for their building and the technology involved. The dimensions of the elements and details depend on the characteristics of the used materials and the external forces acting on the structures and they have to be determined carefully during the design procedure. The structure must be reliable as well as economical during the exploitation process. The reliability is guaranteed when the definite strength, stiffness, stability and durability are taken in mind in the structure. The economy of the construction depends on the material’s expenditure, on the new technology and on the cheaper materials. It is obvious that the reliability and the economy are opposite to requirements. Because of that, the strength of materials relies on the experience as well as the theory and is a science in development.

2.1.1 Basic Definitions Strength: The ability of the material to resist deformation is known as strength. Stress: The amount of internal force exerted per unit area is known as stress. s=

F A

where s is the average stress. It is also known as engineering stress or nominal stress. F is the force acting and A is the cross-sectional area. Strain: It is defined as the change in dimension to the original dimension. e=

dl lo

where e is the strain in measured direction, lo is the original length of the material and l is the current length of the material. Total Stress: The internal forces which cause the change in the size or state of a body on application of external force is known as total stress. Unit Stress: Stress acting per unit area is known as unit stress. Normal Stress: The stress acting in a direction perpendicular to section considered is known as normal stress. Ultimate Stress: The ratio of the maximum load which a specimen can sustain to its original area of the cross section is known as ultimate stress.

Strength of Materials

23

Elastic Limit: The limit up to which a material can be stressed and on removal of the stress it is able to return to its original position is known as elastic limit. Yield Point: The point in the stress-strain diagram at which the deformation first increases noticeably without any increase in the applied load is known as yield point. It is always above the proportional limit. Ultimate Strength: The highest amount of stress a material can sustain without failure is known as ultimate strength. Breaking Strength: The stress at which the material starts cracking is known as breaking strength. Modulus of Elasticity (E): The constant that is the ratio of stress to strain for all the values of unit stress within the proportionality limit. It is also called Young’s modulus. Factor of Safety:

The ratio of ultimate strength of the material to the allowable stress.

Elasticity: The ability of a material to deform and return to its original shape once the load is removed is known as elasticity. The amount of deformation is called strain. Ductility: The ability of a material to undergo large permanent deformations in tension, i.e., property which enables a material to be beaten or rolled into thin sheets. Malleability: The ability of a material to undergo large permanent deformation in compression or property which enables a material to be beaten or rolled into thin sheets. This property is important in metalworking. Gold is the most malleable metal followed by aluminium. Toughness: The ability of a material to withstand high unit stress along with great unit deformation without fracture. Stiffness: The ability of a material to resist deformation or deflection under stress. Hardness: The ability of a material to resist very small indentation, abrasion and plastic deformation. In other words, high resistance of a material to various kinds of shape change when force is applied. Strain Hardening: If the material is reloaded from point C, it will follow the previous unloading path and line CB becomes its new elastic region with elastic limit defined by point B. Though the new elastic region CB resembles that of the initial elastic region OA, the internal structure of the material in the new state has changed. The change in the microstructure of the material is clear from the fact that the ductility of the material has come down due to strain hardening. When the material is reloaded, it follows the same path as that of a virgin material and fails on reaching the ultimate strength which remains unaltered due to the intermediate loading and unloading process.

24

Basic Mechanical Engineering s

O

B

C

e

Figure 2.1

2.2 2.2.1

SIMPLE STRESSES AND STRAINS Definition of Stress and Strain

Stress: The force transmitted across any section, divided by the area of that section, is called intensity of stress or stress. X P =

P A

P sA

sA X

Figure 2.2 Stress over section.

where s = stress P = Load A = Area Direct Stress (Tensile / Compressive): Stresses which are normal to the plane on which WKH\DFWDUHFDOOHGGLUHFWVWUHVVHVDQGHLWKHUWHQVLOHRUFRPSUHVVLYH8QLW1P2. Strain: Strain is a measure of the measure of the deformation produced in the member by the load. If a rod of length L is in tension and the elongation produced is L, then the direct strain Elongation DL = e= L Original length Tensile strain will be positive whereas compressive strain will be negative.

Strength of Materials

2.2.2

25

Shear Stress and Strain

Shear Stress: Consider a block or portion of material as shown in Fig. 2.3 (a) subjected to a set of equal and opposite forces Q. (Such a system could be realised in a bicycle brake block when contacted with the wheel.) There is then a tendency for one layer of the material to slide over another to produce the form of failure shown in Fig. 2.3(b). If this failure is restricted, then a shear stress t is set up, defined as follows: Shear stress (t) =

shear load Q = area resisting shear A

This shear stress will always be tangential to the area on which it acts; direct stresses, however, are always normal to the area on which they act. Q

t t

Q

(a)

(b)

Figure 2.3 Shear force and resulting shear stress system showing typical form of failure by relative sliding of planes.

Shear Strain: If one again considers the block of Fig. 2.3 (a) to be a bicycle brake block it is clear that the rectangular shape of the block will not be retained as the brake is applied and the shear forces introduced. The block will in fact change shape or “strain” into the form shown in Fig. 2.4. The angle of deformation g is then termed the shear strain. Shear strain is measured in radians and hence is non-dimensional, i.e., it has no units. t g t

Figure 2.4 Deformation (shear strain) produced by shear stresses.

2.2.3 Strain Energy Impact Loading Consider a rod BC of length L and uniform cross-sectional area A, which is attached at B to a fixed support, and subjected at C to a slowly increasing axial load P. By plotting magnitude P of the load against the deformation x of the rod, we obtain a certain loaddeformation diagram which is characteristics of the rod BC. Let us now consider the

26

Basic Mechanical Engineering

work dU done by the load P as the rod elongates by a small amount dx. This elementary work is equal to the product of the magnitude P of the load and of the small elongation dx. Write dU = Pdx, that is equal to the element of area of width dx located under the load-deformation diagram. The work Ui done by the load as the rod undergoes a deformation Xi is x=x 1

Ui =

Ú

Pdxi

x=0

U is equal to the area under the load-deformation diagram between and x = 0 and x = x1. The work done by the load P as it is slowly applied to the rod must result in the increase of some energy associated with the deformation of the rod. This energy is referred to as the strain energy of the rod. We have, by definition

Ú

Strain energy = Ui =

x

0

pdx

P

U = Area

P

O

x1

x

x

dx

Figure 2.5

Work and energy should be expressed in units obtained by multiplying units of length by units of force. Thus, if SI metric units are used, work and energy are expressed in N.m; this unit is called, a joule (J). Impact Loading: So far we have considered all loadings to be applied to a body in a gradual manner, such that when they reach a maximum value, they remain constant or static. Some loadings, however, are dynamic; that is, they vary with time. A typical example would be caused by the collision of objects. This is called an impact loading. Specifically, impact occurs when one object strikes another, such that large forces are developed between the objects during a short period of time. If we assume no energy is lost during the impact, we can study the mechanics of impact using the conservation of energy. Dmax = 2 D static Impact loading = 2 static loading

Strength of Materials

27

v k

Dmax

Figure 2.6 Impact loading.

2.3

HOOKE’S LAW OF ELASTICITY

The state of a body will change on application of any external force. The elastic bodies change their shape when there is pressure or strain, and come back to their original, or balanced, position when the force is removed (unless the distorting force surpasses the elastic limit of the material). Hooke’s law expresses that if there is a very small change in the dimension of elastic body, the force having a tendency to re-establish the body to its original shape is proportional to the displacement of the body from the equilibrium position. In other words within the elastic limit the stress applied is directly proportional to the strain. On the off chance that a body, which complies with Hooke’s law, is displaced from equilibrium and discharged, the body will experience “simple harmonic motion”. Numerous frameworks, for example, water waves, sound waves, air conditioning circuits and particles in an atom, show this kind of movement. Strain μ Stress s μ e or s = Ee

Figure 2.7 Stress vs strain curve.

2.3.1 Tensile Test In order to compare the strengths of various materials it is necessary to carry out some standard form of test to establish their relative properties. One such test is the standard

28

Basic Mechanical Engineering

Stress s

tensile test in which a circular bar of uniform cross section is subjected to a gradually increasing tensile load until failure occurs. Measurements of the change in length of a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers and a graph of load against extension or stress against strain is produced as shown in Fig. 2.7. This shows a typical result for a test on a mild (low carbon) steel bar; other materials will exhibit different graphs but of a similar general form. For the first part of the test it will be observed that Hooke’s law is obeyed, i.e., the material behaves elastically and stress is proportional to strain, giving the straight-line graph indicated. Some point A is eventually reached, however, when the linear nature of the graph ceases and this point is termed the limit of proportionality. For a short period beyond this point the material may still be elastic in the sense that deformations are completely recovered when load is removed (i.e., strain returns to zero) but Hooke’s law does not apply. The limiting point B for this condition is termed the elastic limit. For most practical purposes it can often be assumed that points A and B are coincident. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will thus be some permanent deformation or permanent set when load is removed. After the points C, termed the upper yield point, and D, the lower yield point, relatively rapid increases in strain occur without correspondingly high increases in load or stress. The graph thus becomes much shallower and covers a much greater portion of the strain axis than does the elastic range of the material. The capacity of a material to allow these large plastic deformations is a measure of the so-called ductility of the material, and this will be discussed in greater detail below. For certain materials, for example, high carbon steels and nonferrous metals, it is not possible to detect any difference between the upper and lower yield points and in some cases no yield point exists at all. In such cases a proof stress is used to indicate the onset of plastic strain or as a comparison of the relative properties with another similar material. This involves a measure of the (b)

sP

Proof stress

O

P

N

Strain e

0.1%

Figure 2.8 Determination of 0.1% proof stress.

Strength of Materials

29

permanent deformation produced by a loading cycle; the 0.1 % proof stress, for example, is that stress which, when removed, produces a permanent strain or set of 0.1% of the original gauge length (see Figure 2.8).

2.3.2

Poisson’s Ratio

&RQVLGHUWKHUHFWDQJXODUEDURI)LJXUHVXEMHFWHGWRDWHQVLOHORDG8QGHUWKHDFWLRQ of this load the bar will increase in length by an amount dL giving a longitudinal strain in the bar of dL e= L P

b dd 2

dd 2

d

P dd 2

dd 2

Figure 2.9

The bar will also exhibit, however, a reduction in dimensions laterally, i.e., its breadth and depth will both reduce. The associated lateral strains will both be equal, will be of opposite sense to the longitudinal strain, and will be given by elateral = -

db dd =b d

Provided the load on the material is retained within the elastic range the ratio of the lateral and longitudinal strains will always be constant. This ratio is termed Poisson’s ratio. dd lateral strain =- d Poisson’s ratio (m) = dL longitudinal strain L

The negative sign of the lateral strain is normally ignored to leave Poisson’s ratio simply as a ratio of strain magnitudes. It must be remembered, however, that the longitudinal strain induces a lateral strain of opposite sign, e.g., tensile longitudinal strain induces compressive lateral strain. Since for most engineering materials the value of m lies between 0.25 and 0.33.

30

Basic Mechanical Engineering

2.4

RELATION BETWEEN ELASTIC CONSTANTS

Young’s Modulus or Modulus of Elasticity: It is the ratio between tensile stress and tensile strain or compressive stress and compressive strain. It is denoted by E. Its units are GN/m2. E = stress/stain = s/e = st/et = sc/ec Modulus of Rigidity or Shear Modulus of Elasticity: It is the ratio of shear stress (t) to shear strain (g). It is represented by C, N or G. Its units are GN/m2. C, N or G = t/g Bulk Modulus or Volume Modulus of Elasticity: It is defined as the ratio of applied pressure (on each face of solid cube) to volumetric strain. It is represented by K. Its units are GN/m2. K = p/ev Poisson’s Ratio: The ratio of lateral strain to linear strain is called Poisson’s ratio. It is denoted by m or v or 1/m. m = lateral strain/linear strain = 1/m The value of m varies from 1/3 to 1/4 depending on the material. Relation between the elastic constants Relation between E and C

:

E = 2 C (1 + m)

Relation between E and K

:

E = 3 K (1 – 2m)

Relation between E, C and K :

2.5

E = 9 KC/(3K + C)

THERMAL STRESSES

7HPSHUDWXUHFKDQJHVFDXVHWKHERG\WRH[SDQGRUFRQWUDFW7KHDPRXQWDžT, is given by dT = aL(TfïTi) = aL DT dT = aL(TfïTi) = aL DT where a is the coefficient of thermal expansion in m/m°C, L is the length in metre, Ti and Tf are the initial and final temperatures, separately in °C. For steel, a = 11.25 × 10–6 m/m°C. If their exansion or contraction in the structure due to change in temperature and is allowed to happen freely, no stress will be generated. Now and again if there is some restriction in temperature change, an internal stress is generated. So under free expansion stress generated is zero. The internal stress thus generated is called thermal

Strength of Materials

31

stresses. For a homogeneous bar mounted between steady backings as appeared, the thermal stress is computed as

L d

Figure 2.10 Expansion of rod due to thermal stresses.

deformation due to temperature changes; dT = aL DT dT = aL DT aL DT =

sL E

s = Ea DTs = Ea DT where s is the thermal stress in MPa, E is the modulus of elasticity of the rod in MPa.

2.6

CREEP AND FATIGUE

Creep is the gradual increase of plastic strain in a material with time at constant load. Particularly at elevated temperatures some materials are susceptible to this phenomenon and even under the constant load mentioned strains can increase continually until fracture. This form of fracture is particularly relevant to turbine blades, nuclear reactors, furnaces, rocket motors, etc. The general form of the strain versus time graph or creep curve is shown in Fig. 2.11 for two typical operating conditions. Fracture x Fracture x

Strain

High stress or temp

Initial strain

Low stress or temp

Primary creep

Secondary creep

Tertiary creep

Time

Figure 2.11 Typical creep curve.

32

Basic Mechanical Engineering

In each case the curve can be considered to exhibit four principal features: (a) An initial strain, due to the initial application of load. (b) A primary creep region, during which the creep rate (slope of the graph) diminishes. (c) A secondary creep region, when the creep rate is sensibly constant. (d) A tertiary creep region, during which the creep rate accelerates to final fracture. It is clearly imperative that a material which is susceptible to creep effects should only be subjected to stresses which keep it in the secondary (straight line) region throughout its service life. This enables the amount of creep extension to be estimated and allowed for in design. Fatigue is the failure of a material under fluctuating stresses each of which is believed to produce minute amounts of plastic strain. Fatigue is particularly important in components subjected to repeated and often rapid load fluctuations, e.g., aircraft components, turbine blades, vehicle suspensions, etc. Fatigue behaviour of materials is usually described by a fatigue life or S-N curve in which the number of stress cycles N to produce failure with a stress peak of S is plotted against S. A typical S-N curve for mild steel is shown in Fig. 2.12. The particularly relevant feature of this curve is the limiting stress S, since it is assumed that stresses below this value will not produce fatigue failure, however, many cycles are applied, i.e., there is infinite life. In the simplest design cases, therefore, there is an aim to keep all stresses below this limiting level. However, this often implies an over-design in terms of physical size and material usage, particularly in cases where the stress may only occasionally exceed the limiting value noted above. This is, of course, particularly important in applications such as aerospace structures where component weight is a premium. Additionally, the situation is complicated by many materials which do not show a defined limit, and modern design

2

Stress MN/m S

s

s or

t t Fatigue loading - typical variations of load or applied stress with time

60 50 40 30 Sn

False zero

Fatigue limit

6

5 10 10 10 Cycles to failure (N)

7

10

8

Figure 2.12 Typical S-N fatigue curve for mild steel.

Strength of Materials

33

procedures therefore rationalise the situation by aiming at a prescribed, long, but finite life, and accept that service stresses will occasionally exceed the value S. It is clear that the number of occasions on which the stress exceeds S, and by how much, will have an important bearing on the prescribed life and considerable specimen, and often full-scale, testing is required before sufficient statistics are available to allow realistic life assessment.

2.7

SUMMARY

Strength of materials (otherwise called mechanics of materials) is the study of the inner impact of external forces on structural members. Stress, strain, distortion avoidance, torsion, flexure, shear force diagram, and bending moment diagram are a portion of the points secured by this subject. The learning of this subject is an absolute necessity in civil engineering, mechanical engineering, materials engineering, electrical engineering, and so on. Select a theme beneath for tackled issues in mechanics and strength of materials. Motivation behind concentrating on strength of materials is to guarantee that the structure utilized will be protected against the most extreme internal impacts that might be delivered by any mix of outside forces. You utilize strength of materials in designing to measure your parts as indicated by the heaps that it will encounter. Basically, you represent the materials quality properties and the strengths following up on it to decide the geometry of the materials to meet the plan necessities. For instance, you have a strong steel posts mounted on a level plane to a divider. Expect that the post is encased in the solid toward the end so it isn’t going any place and you need to hand a 500 lb weight towards the end of it. With strength of materials you can compute the width that you require that post to be to bolster that helps utilizing proper kind of steel.

SOLVED EXAMPLES

Example 2.1 A steel bolt, 2.50 cm in diameter, carries a tensile load of 40 kN. Estimate the average tensile stress at the section a and at the screwed section b, where the diameter at the root of the thread is 2.10 cm. b

a

40 kN 40 kN

2.10 cm diameter 2.50 cm diameter

Figure 2.13

34

Basic Mechanical Engineering

Solution:

The cross-sectional area of the bolt at the section a is Ao =

p (0.025)2 = 0.491 × 10–3 m2 4

The average tensile stress at A is then 40 ¥ 10 3 P = = 81.4 MN/m 2 so = Aa 0.491 ¥ 10 - 3

The cross-sectional area at the root of the thread, section b, is Ab =

p (0.021)2 = 0.346 ¥ 10 - 3 m 2 4

The average tensile stress over this section is sb =

40 ¥ 10 3 P = = 115.6 MN/m 2 -3 Ab 0.346 ¥ 10

Example 2.2 A cylindrical block is 30 cm long and has a circular cross section 10 cm in diameter. It carries a total compressive load of 70 kN, and under this load it contracts by 0.02 cm. Estimate the average compressive stress over a normal cross section and the compressive strain. 70 kN 10 cm diameter

30 cm long

70 kN

Figure 2.14

Solution: The average compressive stress over this cross section is then s=

70 ¥ 10 3 P = = 8.92 MN/m 2 -3 A 7.85 ¥ 10

The average compressive strain over the length of the cylinder is e=

0.02 ¥ 10 - 2 = 0.67 ¥ 10 - 3 30 ¥ 10 - 2

Strength of Materials

35

Example 2.3

A bar is made out of an aluminium section rigidly joined amongst steel and bronze segments, as shown below. Axial load burdens are applied at the positions demonstrated. On the off chance that P = 3000 lb and the cross-sectional area of the bar is 0.5 in2, find the stress in each segment.

Bronze

Steel 4P Aluminium

2 ft

3 ft

P

2.5 ft

Figure 2.15

Solution: For steel: sst Ast = Pst sst Ast = Pst sst (0.5) = 12 sst (0.5) = 12 sst = 24 ksi For aluminium: sal Aal = Pal sal Aal = Pal sal (0.5) = 12 sal (0.5) = 12 sal = 24 ksi For bronze: sbr Abr = Pbr sbr Abr = Pbr sbr (0.5) = 9 sbr (0.5) = 9 sbr = 18 ksi

Example 2.4

Determine the stress in each section of the bar shown in Fig. 2.16 when subjected to an axial tensile load of 20 kN. The central section is 30 mm square cross section; the other portions are of circular section, their diameters being indicated. What will be the total extension of the bar? For the bar material E = 210 GN/m2.

36

Basic Mechanical Engineering 100

250 20 kN

1 20

400

2

3

20 kN 15

30

Not to scale all dimensions mm

Figure 2.16

Solution: Stress =

Force P = Area A

20 ¥ 10 3 80 ¥ 10 3 = = 63.66 MN/m 2 -3 2 -6 Stress in section (1) = p( 20 ¥ 10 ) p ¥ 400 ¥ 10 4

Stress in section (2) =

Stress in section (3) =

20 ¥ 10 3 = 22.2 MN/m 2 30 ¥ 30 ¥ 10 - 6 20 ¥ 10 - 3 80 ¥ 10 3 = = 113.2 MN/m 2 p(15 ¥ 10 - 3 )2 p ¥ 225 ¥ 10 - 6 4

Now the extension of a bar can always be written in terms of the stress in the bar since

i.e.,

E=

Stress s = Strain d /L

d=

sL E

6 \ Extension of section (1) = 63.66 ¥ 10 ¥

6 Extension of section (2) = 22.2 ¥ 10 ¥

250 ¥ 10 - 3 = 75.8 ¥ 10 - 6 m 210 ¥ 109

100 ¥ 10 - 3 = 10.6 ¥ 10 - 6 m 210 ¥ 109

6 Extension of section (3) = 113.2 ¥ 10 ¥

400 ¥ 10 - 3 = 215.6 ¥ 10 - 6 m 210 ¥ 109

Total extension = (75.8 + 10.6 + 215.6) 10–6 = 302 × 10–6 m = 0.302 mm

Strength of Materials

37

Example 2.5 The coupling shown in Fig. 2.17 is constructed from steel of rectangular cross section and is designed to transmit a tensile force of 50 kN. If the bolt is of 15 mm diameter calculate: (a) the shear stress in the bolt; (b) the direct stress in the plate; and (c) the direct stress in the forked end of the coupling. Bolt

50 kN

50 mm

25 kN 6 mm

50 kN 50 kN

6 mm

6 mm 6 mm

25 kN Shear system on bolt

Figure 2.17

Solution: (a) The bolt is subjected to double shear, tending to shear it as shown. There is thus twice the area of the bolt resisting the shear. 50 ¥ 10 3 ¥ 4 P = 2 A 2 ¥ p(15 ¥ 10 - 3 )2

Shear stress in bolt = =

100 ¥ 10 3 = 141.5 MN / m 2 p(15 ¥ 10 – 3 )2

(b) The plate will be subjected to a direct tensile stress given by s=

50 ¥ 10 3 P = = 166.7 MN / m 2 -6 A 50 ¥ 6 ¥ 10

(c) The force in the coupling is shared by the forked end pieces, each being subjected to a direct stress s=

25 ¥ 10 3 P = = 83.3 MN / m 2 A 50 ¥ 6 ¥ 10 - 6

Example 2.6

A circular, metal rod of diameter 1 cm is loaded in tension. When the tensile load is 5 kN, the extension of a 25 cm length is measured accurately and found to be 0.0227 cm. Estimate the value of Young’s modulus, E, of the metal.

Solution: The cross-sectional area is A=

p (0.01)2 = 0.078 ¥ 10 - 3 m 2 4

38

Basic Mechanical Engineering

The tensile stress is then s=

5 ¥ 10 3 P = = 63.7 MN / m2 A 0.0785 ¥ 10 - 3

The measured tensile strain is e=

-2 e 0.0227 ¥ 10 = = 0.910 × 10 – 3 L 25 ¥ 10 - 2

Then Young's modulus is defined by s 63.7 ¥ 106 = 70 GN / m2 E= = e 0.91 ¥ 10 - 3

Example 2.7

What force is required to punch a 20 mm diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2.

Solution: The resisting area is the shaded area along the perimeter and the shear force P is equal to the punching force P. P Puncher 20 mm f 25 mm thick

Punched out 25 mm

20 mm

Figure 2.18

P = tA P = 350[p(20)(25)]; P = 350[p(20)(25)] P = 549778.7 NP = 549778.7 N P = 549.8 kNP = 549.8 kN

Strength of Materials

39

Example 2.8 A pipe conveying steam at 3.5 MPa has an outside distance across of 450 mm and a wall thickness of 10 mm. A gasket is embedded between the rib towards one side of the pipe and a level plate used to top the end. What number of 40 mm measurement bolts must be used to hold the top on if the allowable stress in the bolts is 80 MPa, of which 55 MPa is the initial stress? What circumferential stress is generated in the pipe? Why is it important to fix the bolts at first? Solution: F = sA F = 3.5[14p(4302)] F = 3.5[14p(4302)] F = 508270.42 N Cap

Gasket 10

P

F 430 10 450 mm f

Flange

Figure 2.19

P =F (sbolt tA) n = 508270.42 N (sbolt A) n = 508270.42 N (80 – 55) [14p(40)2] n = 508270.42 or, n = 16.19; say 17 bolts

Example 2.9 (a) A 25 mm diameter bar is subjected to an axial tensile load of 100 kN. 8QGHUWKHDFWLRQRIWKLVORDGDPPJDXJHOHQJWKLVIRXQGWRH[WHQGð– 3 mm. Determine the modulus of elasticity for the bar material. (b) If, in order to reduce weight while keeping the external diameter constant, the bar is bored axially to produce a cylinder of uniform thickness, what is the maximum diameter of bore possible given that the maximum allowable stress is 240 MN/m2? The load can be assumed to remain constant at 100 kN. (c) What will be the change in the outside diameter of the bar under the limiting stress quoted in (b)? (E = 210 GN/m2 and m = 0.3).

40

Basic Mechanical Engineering

Solution: (a) We know that PL Ad L

Young’s modulus E =

100 ¥ 10 3 ¥ 200 ¥ 10 - 3 p( 25 ¥ 10 - 3 )2 ) ¥ 0.19 ¥ 10 - 3 4

=

= 214 GN/m2 (b) Let the required bore diameter be d mm; the cross-sectional area of the bar will then be reduced to È p ¥ 252 pd 2 ˘ - 6 p -6 2 2 2 A= Í ˙ 10 = ( 25 - d ) 10 m 4 4 4 Î ˚

4 ¥ 100 ¥ 10 3 P = Stress in bar = A p( 252 - d 2 ) 10 - 6

\

But this stress is restricted to a maximum allowable value of 240 MN/m2. 4 ¥ 100 ¥ 10 3 240 × 10 = p( 252 - d 2 ) 10 - 6 6

\ \

4 ¥ 100 ¥ 10 3 25 × d = 240 ¥ 106 ¥ p ¥ 106

\

d2 = 94.48 and

2

2

d = 9.72 mm

The maximum bore possible is thus 9.72 mm. (c) The chance in the outside diameter of the bar will be obtained from the lateral strain, i.e.,

Lateral strain = But

and \

Poisson's ratio n =

dd d Lateral strain Longitudinal strain

s 240 ¥ 106 = Longitudinaal strain = E 210 ¥ 109 0.3 ¥ 240 ¥ 106 s dd = -v =E 210 ¥ 109 d

Strength of Materials

\

Change in outside diameter = -

41

0.3 ¥ 240 ¥ 106 ¥ 25 ¥ 10 - 3 210 ¥ 109

= – 8.57 × 10–6 m (a reduction)

Example 2.10 Derive an expression for the total extension of the tapered bar of circular cross section shown in Fig. 2.20 when it is subjected to an axial tensile load W. L

LO

D A

d

W

W

r x

dx

Figure 2.20

Solution: d /2 (D - d ) / 2 = L0 L

L0 =

d L (D - d )

Consider an element of thickness dx and radius r, distance x from the point of Stress on the element =

W pr 2

But

d r = 2L o x

\

x( D - d ) Ê D - dˆ x= r = dÁ Ë 2dL ˜¯ 2L

\ Stress on the element =

4 WL2 p(D - d )2 x 2

\ Strain on the element =

s E

and extension of the element = =

sdx E 4 WL2 dx p( D - d ) 2 x 2 E

42

Basic Mechanical Engineering Lo + L

\ Total extension of bar =

Ú

L0

4 WL2 dx p(D - d )2 E x 2 L +L

4 WL2 = p(D - d )2 E

È ÍÎ

4 WL2 = p(D - d )2 E

È Ê 1 ˆ˘ 1 - Á- ˜˙ ÍÎÍ (Lo + L ) Ë L0 ¯ ˙˚

1˘ o x ˙˚ Lo

d L0 = ( D - d ) L

But

Lo + L =

\

(d + d - d) d DL L+L= L= D-d (D - d ) (D - d )

\ Total extension = =

4 WL2 4 WL È ( - d + D) ˘ È (D - d ) (D - d ) ˘ + = 2 Í ˙ p(D - d ) E Î DL dL ˚ p(D - d ) E ÍÎ Dd ˙˚ WL DdE

EXERCISE 1. A certain man’s biceps muscle has a maximum cross-sectional area of 12 cm2 = 1.2 × 10-3 m2 . What is the stress in the muscle if it exerts a force of 300 N? 2. A 25 mm square cross section bar of length 300 mm carries an axial compressive load of 50 kN. Determine the stress set up in the bar and its change of length when the load is applied. For the bar material E = 200 GN/m2. [80 MN/m2; 0.12 mm] 3. A steel tube, 25 mm outside diameter and 12 mm inside diameter, carries an axial tensile load of 40 kN. What will be the stress in the bar? What further increase in load is possible if the stress in the bar is limited to 225 MN/m2? 4. A wire 1.5 m long has a cross-sectional area of 2.4 mm2 . It is hung vertically and stretches 0.32 mm when a 10 kg block is attached to it. Find (a) the stress, (b) the strain, and (c) Young’s modulus for the wire. 5. Define the terms shear stress and shear strain, illustrating your answer by means of a simple sketch. Two circular bars, one of brass and the other of steel, are to be loaded by a shear load of 30 kN. Determine the necessary diameter of the bars (a) in single shear, (b) in double shear, if the shear stress in the two materials must not exceed 50 MN/m2 and 100 MN/m2 respectively.

Strength of Materials

43

6. The rigid bar ABC in Fig. 2.21 is pinned at B and appended to the two vertical bars. At first, the bar is horizontal and the vertical poles are stress free. Determine the stress in the aluminium bar if the temperature of the steel bar is reduced by 40°C. Ignore the weight of bar ABC.

Aluminium L = 1.2 m A = 1200 mm2 9 2 E = 70 × 10 N/m a = 23 mm/(m°C)

Steel L = 0.9 m A = 300 mm2 9 2 E = 200 × 10 N/m a = 11.7 mm/(m°C) A

B

C

0.6 m

1.2 m Figure 2.21

7. A steel rod with a cross-sectional area of 0.25 in2 is stretched between two fixed points. The tensile load at 70°F is 1200 lb. What will be the stress at 0°F? At what temperature will the stress be zero? Assume a = 6.5 × 10–6 in/(in °F) and E = 29 × 106 psi. 8. A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2. 9. State and derive the relation between the elastic constants. 10. A bar ABCD consists of three sections: AB is 25 mm square and 50 mm long, BC is of 20 mm diameter and 40 mm long and CD is of 12 mm diameter and 50 mm long. Determine the stress set up in each section of the bar when it is subjected to an axial tensile load of 20 kN. What will be the total extension of the bar under this load? For the bar material, E = 210 GN/m2.

3 Machine Design

Learning Objectives        

‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡

3.1

*HWWLQJIDPLOLDUZLWKWKHWHUPVLQPDFKLQHGHVLJQ %DVLFFRQFHSWRIGHVLJQLQJHQHUDO &RQFHSWRIPDFKLQHGHVLJQDQGWKHLUW\SHV )DFWRUVWREHFRQVLGHUHGLQPDFKLQHGHVLJQ ,QWURGXFWLRQWRGLIIHUHQWGULYHVXVHGIRUSRZHUWUDQVPLVVLRQLQPDFKLQHV %ULHILGHDDERXWJHDUDQGJHDUWUDLQV ,QWURGXFWLRQWRPHFKDQLFDOVSULQJV ,GHDRQSRZHUVFUHZV

INTRODUCTION

Machine design or mechanical design can be defined as the process by which resources or energy is converted into useful mechanical forms, or the mechanisms so as to obtain useful output from the machines in the desired form as per the needs of the human beings. Machine design can lead to the formation of the entirely new machine or it can lead to upgradation or improvement of the existing machine. For instance, if the existing gearbox is too heavy or cannot sustain the actual loads, entirely new gearbox can be designed. But if the same gearbox has the potential to lift more loads, it can be upgraded by making certain changes in its design. Design is essentially a decision-making process. If we have a problem, we need to find a solution. In other words, to design is to formulate a plan to satisfy a particular need and to create something with a physical reality. Consider for a example, design of a chair. A number of factors need to be considered first: Basic Mechanical Engineering

Machine Design 45

(a) The purpose for which the chair is to be designed such as whether it is to be used as an easy chair, an office chair or to accompany a dining table. (b) Whether the chair is to be designed for a grown up person or a child. (c) Material for the chair, its strength and cost need to be determined. (d) Finally, the aesthetics of the designed chair. Almost everyone is involved in design, in one way or the other, in our daily lives because problems posed need to be solved. Machine design is innovation of new and effective machines and improving the existing ones. A new or effective machine is one which is more economical in the overall cost of production and operation. The design is to formulate a plan for the satisfaction of human needs. In designing a machine component, it is necessary to have a good knowledge of many subjects such as mathematics, engineering mechanics, strength of materials, theory of machines, workshop processes and engineering drawing. Consider two cars of various makes. They may both be reasonable cars and serve the same purpose but the designs are different. The designers consider various factors and come to certain conclusions leading to an optimum design. Market survey gives an indication of what people want. Existing norms play an important role. Once a critical decision is made, the rest of the design features follow. For example, once we decide the engine capacity, the shape and size, then the subsequent course of the design would follow. A bad decision leads to a bad design hence, bad product. Design may be for different products and with the present specialization and knowledge bank, we have a long list of design disciplines, e.g., ship design, building design, process design, bridge design, clothing or fashion design, and so on. We now define a machine as a combination of resisting bodies with successfully constrained relative motions which is used to transform other forms of energy into mechanical energy or transmit and modify available energy to do some useful work. If it converts heat into mechanical energy, we call it a heat engine.

3.1.1 Classification of Machine Design Adaptive Design: In this type of design the designer is concerned with adaptation of existing designs. The designer only makes minor alterations or modifications in the existing designs of the product. Development Design: This type of design requires scientific training and design ability in order to modify the existing designs into a new idea by adopting a new material or different method of manufacture. New Design: This type of design needs lot of research, technical ability and creativity. Only those designers who have personal qualities of a sufficiently high order can take up the work of a new design.

46

Basic Mechanical Engineering

Rational Design: of mechanics.

This type of design depends on mathematical formulae of principle

Empirical Design: This type of design depends on empirical formulae based on the practice and experience. Industrial Design: This type of design depends on the production aspects to manufacture any machine component in the industry. Optimum Design: It is the best design for the given objective function under the specified constraints. It may be achieved by minimizing the undesirable effects. System Design: It is the design of any complex mechanical system like a motor car. Element Design: It is the design of any element of the mechanical system like piston, crankshaft, connecting rod, etc. Computer-Aided Design: This type of design depends on the use of computer systems to assist in the creation, modification, analysis and optimization of a design.

3.1.2 Factors to be Considered in Machine Design Many factors are considered while resolving a design problem. In many cases a common sense approach is required for solving a problem. Some of these factors are as follows: (a) Which device or mechanism should be used? This decides the relative arrangement of the constituent elements. (b) Material (c) Forces on the elements (d) Size, shape and space requirements. The final weight of the product is also a major concern. (e) The method of manufacturing the components and their assembly. (f) How will it operate? (g) Reliability and safety aspects (h) Inspectability (i) Maintenance, cost and aesthetics of the designed product.

3.2

TRANSMISSION OF POWER

3.2.1 Transmission Systems Power is transmitted from one shaft to another shaft by means of belts, ropes, chains and gears. For large distance between the shaft belts, ropes and chains are used. For small distances gears are used.

Machine Design 47

The amount of power transmitted depends on the following factors: 1. The velocity of the belt. 2. The tension under which the belt is placed on the pulleys. 3. The arc of contact between the belt and the smaller pulley. 4. The conditions under which the belt is used. Chain Drives: These are a means of transmitting power like gears, shafts and belt drives. Their major advantage over V-belt drives is that there is no-slip. A V-belt can slip over the pulleys and hence a constant drive speed is not assured. The positive action of the tooth and sprocket in a chain drive means that chain drive assemblies will be more compact than belt drive assemblies.

Figure 3.1 Chain drive.

Characteristics of chain drives are as follows: 1. High axial stiffness 2. Low bending stiffness 3. High efficiency 4. Relatively cheap Belt Drive: Belt drives are called flexible machine elements. Flexible machine elements are used for a large number of industrial applications, some of them are as follows:   8VHGLQFRQYH\LQJV\VWHPVWUDQVSRUWDWLRQRIFRDOPLQHUDORUHVHWFRYHUDORQJ distance.   8VHG IRU WUDQVPLVVLRQ RI SRZHU 0DLQO\ XVHG IRU UXQQLQJ RI YDULRXV LQGXVWULDO appliances using prime movers like electric motors, I.C. engine, etc. 3. Replacement of rigid type power transmission system.

48

Basic Mechanical Engineering Belt motion T2

Friction on belt

T2

T1 >T2 T1

Driven pulley

T1

Friction on pulley Driving pulley

Figure 3.2 Belt drive.

A gear drive may be replaced by a belt transmission system. Flexible machine elements has got an inherent advantage that, it can absorb a good amount of shock and vibration. It can take care of some degree of misalignment between the driven and the driver machines and long distance power transmission, in comparison to other transmission systems, is possible. For all the above reasons flexible machine elements are widely used in industrial applications. The belt drives primarily operate on the friction principle. That is, the friction between the belt and the pulley is responsible for transmitting power from one pulley to the other. In other words, the driving pulley will give motion to the belt and the motion of the belt will be transmitted to the driven pulley. Due to the presence of friction between the pulley and the belt surfaces, tensions on both the sides of the belt are not equal. Velocity Ratio of Belt Drive:

Velocity ratio of belt drive is defined as, dS + t NL (1 - s ) = dL + t NS

where, NL and NS are the rotational speeds of the large and the small pulleys respectively, s is the belt slip and t is the belt thickness. Power Transmission of a Belt Drive: Power transmission of a belt drive is expressed as P = (T1 – T2) v where, P is the power transmission in watt and v is the belt velocity in m/s.

3.3

GEARS

Gears are the toothed wheels which are used to transmit motion between two shafts, when the centre distance between them is small. They help in transmitting power/

Machine Design 49

motion between two shafts by meshing without any slip. This is why gear drives are also known as positive drives. In any pair of gears, the smaller one is called pinion and the larger one is called gear immaterial of which is driving and which is driven. When the pinion is the driver, it results in step-down drive in which the output speed decreases and the torque increases. On the other hand, when the gear is the driver, it results in step-up drive in which the output speed increases and the torque decreases.

3.3.1 Types of Gear Gears are classified according to the shape of the tooth pair and disposition into spur, helical, double helical, straight bevel, spiral bevel and hypoid bevel, worm and spiral gears. Spur Gear: Spur gears have their teeth parallel to the hub and are utilized for transmitting power between two parallel shafts. They are basic in development, simple to produce and are low in cost. They have most elevated proficiency and fabulous accuracy rating. They are utilized as a part of fast and high load application in a wide range of trains and an extensive variety of speed ratios. Consequently, they find wide applications ideal from clocks, household devices, engine cycles, vehicles, and railroads to flying machines. Helical Gear: Helical gears are utilized for parallel shaft drives. They have teeth slanted to the axis of the shaft. Henceforth, for a similar width, their teeth are longer than spur gear arrangement and have higher load conveying limit. Their contact proportion is higher than the spur gear and they are smoother and calmer than spur gears. Their accuracy rating is great. They are prescribed for high speeds and loads. Accordingly, these gears find wide applications in car gearboxes. Worm Gear: Worm and worm gear mesh comprises a worm, which is fundamentally the same as a screw and a worm apparatus, which is a helical gearing arrangement. They are utilized as a part of right-point skew shafts. In these gearings, the engagement happens without any shock. The sliding/slipping is predominant in the framework while bringing about smoother operation which produces impressive frictional heat. High speed reduction ratios in the range 8 to 400 are conceivable. Their efficiency is quite low and lies in the range of 90% to 40%. Higher speed reduction gears are non-reversible. Their exactness rating is reasonable for good. They require high lubrication for heat dissipation and for enhancing the productivity. The drives are extremely smaller. Worm gear finds wide applications in material handling and transportation, machine devices, autos, and so on.

50

Basic Mechanical Engineering

Figure 3.3 Worm and worm gear.

Rack and Pinion Rack: It is a section of a gear of infinite diameter. The tooth will be spur or coiled or spiral. This sort of gearing arrangement is employed for changing rotary motion into translatory motion or vice versa.

Figure 3.4 Rack and pinion.

Spiral Gear: Spiral gears are also known as crossed helical gears. They have high helix angle and transmit power between two non-intersecting non-parallel shafts. They have initially point contact under the conditions of considerable sliding velocities and finally have line contact. Hence, they are used for light load and low speed application such as instruments, sewing machine, etc. Their precision rating is poor.

Machine Design 51

Figure 3.5 Spiral gear.

Law of Gearing: The fundamental law of gearing states that the angular velocity ratio between the gears of a gear set must remain constant throughout the mesh. This amounts to the following relationship w1 n d Z = 1 = 2 = 2 n2 d1 Z1 w2

where, w1 and w2 are speeds in rad/sec n1 and n2 are speeds in rpm d1 and d2 are diameters in mm Z1 and Z2 are number of teeth In order to maintain constant angular velocity ratio between two meshing gears, the common normal of the tooth profiles, at all contact points within mesh, must always pass through a fixed point on the line of centres, called pitch point.

3.3.2

Gear Trains

A gear train is two or extra equipment working collectively by using meshing their tooth and turning each other in a process to generate energy and speed. It reduces speed and increases torque. To create significant gear ratio, gears are linked together to gear trains. They regularly include a couple of gears within the instruct. Probably the most normal of the gear train is the gear pair connecting parallel shafts. The tooth of this style will also be spur, helical or herringbone. The angular speed is effectively the reverse of the tooth ratio. Any combination of gear wheels employed to transmit motion from

52

Basic Mechanical Engineering

one shaft to the other is known as a gear train. The meshing of two gears is also idealized as two smooth discs with their edges touching and there is no slip between them. This diameter is known as the Pitch Circle Diameter (PCD) of the gear. Simple Gear Trains: The typical spur gears as shown in diagram. The direction of rotation is reversed from one gear to another. It has no effect on the gear ratio. All the teeth on the gears must be of same size so if gear A advances one tooth, so does B and C. v v

wA

wB

Gear ‘A’

Gear ‘B’ (Idler gear)

wC

Gear ‘C’

Figure 3.6 Gear trains.

t = Number of teeth on the gear D = Pitch circle diameter N = Speed in rpm m = module = not mesh

D and module must be the same for all gears otherwise they would t

D = rw 2 The velocity v of any point on the circle must be the same for all the gears, otherwise they would slip.

v = linear velocity on the circle, v = w

v

wA

D DA D = w B B = wC C 2 2 2

in terms of rev/min NA tA = NB tB = NC tC

Machine Design 53

Torque and Efficiency: The power transmitted by a torque T N-m applied to a shaft rotating at N rev/min is given by P =

2pNT 60

In an ideal gearbox, the input and output powers are the same so, N1 T1 = N2 T2 N1 T = 2 N2 T1

The efficiency is defined as h =

Power out 2p N 2 T2 ¥ 60 N 2 T2 = = Power in 2p N 1 T1 ¥ 60 N 1 T1

Because the torque in and out is different, a gearbox has to be clamped in order to stop the case or body rotating. A holding torque T3 must be applied to the body through the clamps. The total torque must add up to zero.

3.4

SPRINGS

Spring is a mechanical element which acts as a flexible joint in between two parts or bodies. Following are the objectives of a spring when used as a machine member. 1. Cushioning, absorbing, or controlling of energy due to shock and vibration. Car springs or railway buffers to control energy, springs-supports and vibration dampers. 2. Control of motion maintaining contact between two elements (cam and its follower). In a cam and a follower arrangement, widely used in numerous applications, a spring maintains contact between the two elements. It primarily controls the motion. 3. Measuring forces spring balances, gauges. 4. Storing of energy in clocks or starters the clock has spiral type of spring which is wound to coil and then the stored energy helps gradual recoil of the spring when in operation. Nowadays, we do not find much use of the winding clocks.

3.4.1 Types of Spring Two types of springs which are mainly used are, helical springs and leaf springs. We shall consider in this course the design aspects of two types of springs.

54

Basic Mechanical Engineering

Helical Spring: The figures below show the schematic representation of a helical spring acted upon by a tensile load F and compressive load F. The circles denote the cross section of the spring wire. The cut section, i.e., from the entire coil somewhere we make a cut, is indicated as a circle with shade. F

F

Wire diameter (d)

Shear force (F)

Torsion (T) Coil diameter (D)

Figure 3.7 Helical spring.

If we keenly observe the free body diagram of the shaded region only then we shall see that at the cut section, vertical equilibrium of forces will give us force F as indicated in the figure. This is the shear force. The torque T, at the cut section and its direction is also marked in the figure. There is no horizontal force coming into the picture as there is no horizontal force present. So from the free body diagram one can see if any section of the spring is experiencing a torque and a force. Shear force will always be associated with a bending moment. However, in an ideal situation, when force is acting at the centre of the circular spring and the coils of spring are almost parallel to each other, no bending moment would result at any section of the spring (no moment arm), except torsion and shear force. The major stresses in a helical spring are of two types, shear stress due to torsion and direct shear due to applied load. Leaf Spring

Figure 3.8 Leaf spring.

Machine Design 55

Characteristics of a leaf spring are as follows: 1. Sometimes it is also called a semi-elliptical spring; as it takes the form of a slender arc-shaped length of spring steel of rectangular cross section. 2. The centre of the arc provides the location for the axle, while the tie holes are provided at either end for attaching to the vehicle body. 3. Supports the chassis weight 4. Controls chassis roll more efficiently-high rear moment centre and wide spring base 5. Controls rear end wrap-up 6. Controls axle damping 7. Controls braking forces 8. Regulates wheelbase lengths (rear steer) under acceleration and braking One of the applications of leaf spring of simply supported beam type is seen in automobiles, where, the central location of the spring is fixed to the wheel axle. Therefore, the wheel exerts the force F on the spring and support reactions at the two ends of the spring come from the carriage.

3.4.2 Spring in Series and Parallel Series Combination: When springs are in series, they experience the same force but undergo different deflections. For the two systems to be equivalent, the total static deflection of the original and the equivalent system must be the same.

k1

k2

m

F = mg

Figure 3.9 Spring in series.

56

Basic Mechanical Engineering

Consider two springs with force constants k1 and k2 connected in series supporting a load F = mg Let the force constant of the combination be represented by k For the combination, supporting the load F = mg F = kx = (where x = the total stretch) F x = k For each spring - the bottom supports mg = F and stretches by x1 F = k1x1 mg mg mg = + Therefore, deflection of spring D = keq k1 k2 1 1 1 + = keq k1 k2 Therefore, if the springs are in series combination, the equivalent stiffness is equal to the reciprocal of sum of the reciprocal stiffnesses of individual springs. Springs in Parallel: For the springs in parallel combination, the equivalent spring stiffness can be found out as: each of the individual spring supports part of the load attached to it but both the springs undergo same deflection.

k1

k2

F2

F2

m

F = mg

Figure 3.10 Spring in parallel.

Consider two springs with force constants k1 and k2 connected in parallel supporting a load F = mg Let the force constant of the combination be represented by k For the combination supporting the load F = mg, F = kx (where x = the total stretch)

Machine Design 57

The two individual springs both stretch by x but share the load (F = F1 + F2) and F1 = k1x while F2 = k2 x Thus, the total force is F = F1 + F2 or and

3.5

kx = k1 x + k2 x k = k1 + k2

POWER SCREWS

A power screw is a drive used in machinery to convert a rotary motion into a linear motion for power transmission. It produces uniform motion and the design of the power screw may be such that (a) Either the screw or the nut is held at rest and the other member rotates as it moves axially. A typical example of this is a screw clamp. (b) Either the screw or the nut rotates but does not move axially. A typical example for this is a press. Other applications of power screws are jack screws, lead screws of a lathe, screws for vices, presses, etc. Power screw normally uses square threads but ACME or buttress threads may also be used. Power screws should be designed for smooth and noiseless transmission of power with an ability to carry heavy loads with high efficiency square threads. The thread form is shown in figure below These threads have high efficiency but they are difficult to manufacture and are expensive. The proportions in terms of pitch are: h1 = 0.5 p; h2 = 0.5 p – b; H = 0.5 p + a; e = 0.5 p; a and b are different for different series of threads. p e

a H

h1

h2

b

Figure 3.11

e

58

Basic Mechanical Engineering

Acme or Trapezoidal Threads: The acme thread form is shown in Figure 3.12. These threads may be used in applications such as lead screw of a lathe where loss of motion cannot be tolerated. The included angle 2j = 29° and other proportions are a =

P 2.7

and h = 0.25 p + 0.25 mm.

b

a

h

2f

Figure 3.12

A metric trapezoidal thread form is shown in Fig. 3.12 and different proportions of the thread form in terms of the pitch are as follows: Included angle = 30°; H1 = 0.5 p; z = 0.25 p + H1/2 ; H3 = h3 = H1+ ac = 0.5 p + ac ac is different for different pitch, for example ac = 0.15 mm for p = 1.5 mm; ac = 0.25 mm for p = 2 to 5 mm; ac = 0.5 mm for p = 6 to 12 mm; ac = 1 mm for p = 14 to 44 mm. Buttress Thread: This thread form can also be used for power screws but they can transmit power only in one direction. Typical applications are screw jack, vices, etc. A buttress thread form is shown in Fig. 3.13 and the proportions are shown in the figure in terms of the pitch. On the whole the square threads have the highest efficiency as compared to other thread forms but they are less sturdy than the trapezoidal thread forms and the adjustment for wear is difficult for square threads. In order to analyze the mechanics of the power screw we need to consider two cases: (a) Raising the load (b) Lowering the load. Raising the Load: This requires an axial force P as shown in Fig. 3.14. Here, N is the QRUPDOUHDFWLRQDQGǍN is the frictional force.

Machine Design 59 p/8

p

45° 3 p 4

Figure 3.13

For equilibrium P – m N cos a – N sin a = 0 F + m N sin a – N cos a = 0 This gives N =

F (cos a - m sin a )

P =

F(m cos a + sin a ) (cos a - m sin a ) F mN

P L a

N pdm

Figure 3.14

Lowering the Load: The force system at the thread during lowering the load is shown in Fig. 3.15. For equilibrium P – m N cos a + N sin a = 0 F – N cos a – m N sin a = 0

60

Basic Mechanical Engineering

This gives N =

F (cos a + m sin a )

P =

F(m cos a - sin a ) (cos a + m sin a ) F

mN

P a

L

N pdm

Figure 3.15

Condition for Self-locking: The load would lower itself without any external force if ‡ mpdm < L and some external force is required to lower the load if ‡ mpdm•L This is therefore the condition for self-locking. Efficiency of the Power Screw: Efficiency of the power screw is given by h =

Work output Work input

h =

F tan a p

Here work output = F ◊ L Work input = p ◊ pdm This gives

The above analysis is for square thread and for trapezoidal thread some modification is required. Because of the thread angle the force normal to the thread surface is increased. The torque is therefore given by T = F

dm (mp dm sec j + L ) ( pdm - mL sec j )

This considers the increased friction due to the wedging action. The trapezoidal threads are not preferred because of high friction but often used due to their ease of machining.

Machine Design 61

3.6

THIN CYLINDERS

If the wall thickness is less than about 7% of the inner diameter then the cylinder may be treated as a thin one. Thin walled cylinders are used as boiler shells, pressure tanks, pipes and in other low pressure processing equipment. In general, three types of stresses are developed in pressure cylinders, viz., circumferential or hoop stress, longitudinal stress in closed end cylinders and radial stresses. These stresses are demonstrated in Fig. 3.16.

p t

2r

t

C

(a)

(b)

(c)

Figure 3.16 (a) Circumferential stress (b) Longitudinal stress and (c) Radial stress developed in thin cylinders.

In a thin walled cylinder the circumferential stresses may be assumed to be constant over the wall thickness and stress in the radial direction may be neglected for the analysis. The stress developed in the cylinder P Circumferential stress or hoop stress sC = d 2t Pd Longitudinal stress s1 = 4t

3.7

SUMMARY

The outline procedure is an exciting undertaking that calls upon numerous orders with the end goal for it to create the desired result. The designer must have an assortment of innovative, critical thinking and systematic abilities with the end goal for them to accomplish the ideal outline. The architect must have a consciousness of materials, mechanical components, fabricating forms, stress examination strategies/techniques, physical science included, ergonomics, necessities, costs included and the capacity to make these thoughts on paper or utilizing design instruments. Machine design is the use of: science, kinematics, statics, flow, mechanics of materials, building materials, mechanical innovation of metals and building drawing. It additionally includes utilization of different subjects like thermodynamics, electrical hypothesis, power through pressure,

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Basic Mechanical Engineering

motors, turbines, pumps, and so on. Machine drawing is the essential part of the machine plan, since every one of the segments or the machines that have been planned ought to be attracted to make them according to the details. The chapter essentially discusses the basic concepts of design in general leading to the concept of machine design which involves primarily designing the elements. Different types of design and the factors to be considered have been discussed in detail.

SOLVED PROBLEMS

Example 3.1 Name some of the common flat belt materials. Solution:

Leather, rubber, plastics and fabrics are some of the common flat belt materials.

Example 3.2 What is the correction factors used to modify belt maximum stress? Solution: Correction factor for speed and angle of wrap are used to modify the maximum stress of the belt. This correction is required because stress value is given for a specified drive speed and angle of wrap of 1800. Therefore, when a drive has different speed than specified and angle of wrap is also different from 1800, then the above-mentioned corrections are required.

Example 3.3 Design a flat belt drive for the following data: Drive: AC motor, operating speed is 1440 rpm and operates for over 10 hours. The equipment driven is a compressor, which runs at 900 rpm and the required power transmission is 20 kW. Solution: Let us consider the belt speed to be 20 m/s, which is within the recommended range. The given speed ratio = 1440/900 = 1.6. Let the belt material be leather, which is quite common. Now, 20 =

p ¥ ds ¥ 1440 60 ¥ 1000

\

dS = 265.3 mm

\

dL = 1.6 × 265.3 = 424.5 mm From the standard sizes available, dS = 280 mm and dL = 450 mm. Recalculated speed ratio dL 450 = 1.607 ª 1.61 = dS 280

Therefore, the choice of both the pulley diameters is acceptable. Centre distance, \

C > 2(dL + dL) C > 1460 mm

Hence, let C ª 1500 mm (it is assumed that space is available)

Machine Design 63

Considering an open belt drive, the belt length, Ln = =

1 p ( dL + dS ) + 2C + ( dL - dS )2 2 4C 1 p ( 450 + 280 ) + 3000 + ( 450 - 280 )2 ª 4151 mm 2 6000

As a guideline, to take into consideration the initial tension, the belt length is shortened by 1%. Hence, the required belt length: Lo = 4110 mm. Determination of angle of wrap - 1 Ê dL - dS ˆ = 3.25∞ b = sin Á Ë 2C ˜¯

aL = 180 + 2b = 186.5° = 3.26 rad aS = 180 – 2b = 173.5° = 3.03 rad For the leather belt, the coefficient of friction, m may be taken as 0.4.

Example 3.4 A single square thread power screw is to raise a load of 50 kN. A screw thread of major diameter of 34 mm and a pitch of 6 mm is used. The coefficients of friction at the thread and collar are 0.15 and 0.1 respectively. If the collar frictional diameter is 100 mm and the screw turns at a speed of 1 rev/s, find (a) the power input to the screw. (b) the combined efficiency of the screw and collar. Solution: (a) Mean diameter, dm = dmajor – p/2 = 34 – 3 = 31 mm. Torque T = F

dm (mpdm + L ) d + mc F c 2 ( pdm - mL ) 2

Here F = 5 × 103 N, dm = 31 mm, m = 0.15, mc = 0.1, L = p = 6 mm and dc = 100 mm Therefore, 3 T = 50 ¥ 10 ¥

= 416 Nm

0.031 Ê 0.15p ¥ 0.031 + 0.006 ˆ 0.1 + 0.1 ¥ 50 ¥ 10 3 ¥ Á ˜ 2 Ë p ¥ 0.031 - 0.15 ¥ 0.006 ¯ 2

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Basic Mechanical Engineering

Power input = Tw = 416 × 2p × 1 = 2613.8 watts. (b) The torque to raise the load only (T0) may be obtained by substituting m = mc = 0 in the torque equation. This gives T0 = F

dm Ê L ˆ FL 50 ¥ 10 3 ¥ 0.006 = = = 47.75 2 ÁË pdm ˜¯ 2p 2p

Therefore, h = F

dm (mpdm + L ) d + mc F = C 2 ( pdm - mL )

Example 3.5

The C-clamp shown in Fig. 3.17 uses a 10 mm screw with a pitch of 2 mm. The frictional coefficient is 0.15 for both the threads and the collar. The collar has a frictional diameter of 16 mm. The handle is made of steel with allowable bending stress of 165 MPa. The capacity of the clamp is 700 N. (a) Find the torque required to tighten the clamp to full capacity. (b) Specify the length and diameter of the handle such that it will not bend unless WKHUDWHGFDSDFLW\RIWKHFODPSLVH[FHHGHG8VH1DVWKHKDQGOHIRUFH

Figure 3.17

Solution:

Nominal diameter of the screw,

d = 10 mm. Pitch of the screw, p = 2 mm. Choosing a square screw thread we have the following dimensions: Root diameter, d3 = d nominal – 2h3 = 7.5 mm (since ac = 0.25 mm and h3 = 0.5p + ac); Pitch diameter, d2 = d nominal – 2z = 8 mm. (since z = 0.5 p); Mean diameter, dm = (7.5 + 8)/2 = 7.75 mm.

Machine Design 65

Torque, dm (mpdm sec j + L ) T = F ( pd - mL sec j ) m

Here F = 700 N, m = mc = 0.15, L = p = 2 mm (assuming a single start screw thread) and dc = 16 mm. This gives T = 1.48 Nm. Equating the torque required and the torque applied by the handle of length L we have 1.48 = 15 L since the assumed handle force is 15 N. This gives L = 0.0986 m. Let the handle length be 100 mm. The maximum bending stress that may be developed in the handle is s =

My I

=

32 M pd 3

where d is the diameter of the handle. Taking the allowable bending stress as 165 MPa, we have d = 4.5 mm

Example 3.6

A single square thread power screw is to raise a load of 50 kN. A screw thread of major diameter of 34 mm and a pitch of 6 mm is used. The coefficient of friction at the thread and collar are 0.15 and 0.1 respectively. If the collar frictional diameter is 100 mm and the screw turns at a speed of 1 rev/s, find: (a) the power input to the screw. (b) the combined efficiency of the screw and collar.

Solution: (a) Mean diameter, dm = dmajor – p/2 = 34 – 3 = 31 mm Torque T = F

dm (mpdm + L ) d + mc F = C 2 ( pdm - mL )

Here, F = 5 × 103 N, dm = 31 mm, m = 0.15, mc = 0.1, L = p = 6 mm and dc = 100 mm 3 Therefore, T = 50 ¥ 10 ¥

0.031 Ê 0.15p ¥ 0.031 + 0.006 ˆ 0.1 + 0.1 ¥ 50 ¥ 10 3 ¥ Á ˜ 2 Ë p ¥ 0.031 - 0.15 ¥ 0.006 ¯ 2

= 416 Nm Power input = Tw = 416 × 2p × 1 = 2613.8 watts (b) The torque to raise the load only (T0) may be obtained by substituting m = mc = 0 in the torque equation. This gives

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Basic Mechanical Engineering

T0 = F

dm Ê L ˆ FL 50 ¥ 10 3 ¥ 0.006 = = = 47.75 2 ÁË pdm ˜¯ 2p 2p

Therefore, h =

FL /2p 47.75 = = 0.1147 , i.e., 1147% T 416

Example 3.7 Two spur gears have a diametral pitch of 6. Gear 2 has 24 teeth, and gear 3 has 48. The working pressure angle is 20°, and both the gears are standard involutes. Determine the length of the contact line and the contact ratio. Solution: The addendum for the both gears is given by a=

1 1 = = 0.167 in = a2 = a3 Pd 6

Similarly, the circular pitch for both the gears is given by p p = = 0.524 in Pc = Pd 6 and from Eq. (10.13), the base pitch is related to the circular pitch by Pb = Pc cos = 0.524 cos 20° = 0.492 From Eq. (10.7), the two pitch radii are given by rp2 =

N2 24 = =2 2 Pd 2(6 )

rp3 =

N3 48 = =4 2 Pd 2(6 )

and

The length of the line of contact is given by Eq. (10.17) as 2 2 2 2 l = - rp 2 sin f + a2 + 2 a2 rp2 + rp2 sin 2f - rp3 sin f + a3 + 2 a3 rp3 + rp3 sin 2f

= - 2 sin 20∞ + 0.167 2 + 2(0.167 ) ( 2) + 22 sin 2 20∞ - 4 sin 20∞ + 0.167 2 + 2(0.167 ) ( 4 ) + 4 2 sin 2 20∞ = 0.825 in

We know that the contact ratio is mc =

l 0.825 = = 1.678 Pb 0.492

Machine Design 67

EXERCISE 1. 2. 3. 4. 5. 6. 7. 8.

Define machine design. Suggest briefly the steps to be followed by a designer. What are the advantages of a belt drive? Which one should be the governing pulley to calculate tension ratio?. Distinguish between impulse and reaction turbines. What is the function of a gear drive? What are the major stresses in a helical spring? Two springs, each having stiffness kN/m, are in kept series and parallel arrangement one by one. Find: (i) The equivalent stiffness of spring in series is Ke-series, and (ii) the equivalent stiffness of spring in parallel is Ke-parallel. 9. A force of 5 kg compresses the springs in series 10 cm. What will be the total distance that the springs in parallel are compressed? 10. What is the largest gear that will mesh with a 20° standard full-depth gear of 22 teeth with no interference?

4 Mechanical Vibrations

Learning Objectives:   

‡ ,QWURGXFWLRQWRYLEUDWLRQV ‡ &LWHLPSRUWDQFHRIYLEUDWLRQVDQGW\SHVRIYLEUDWLRQV ‡ 9LEUDWLRQLVRODWLRQDQGFULWLFDOVSHHG

4.1

INTRODUCTION

Vibrations are oscillations in mechanical dynamic systems. Although any system can oscillate when it is forced to do so externally. The term “vibration” in mechanical engineering is often reserved for systems that can oscillate freely without applied forces. Sometimes, these vibrations cause minor or serious performance or safety problems in engineering systems. For instance, when an aircraft wing vibrates excessively, passengers in the aircraft become uncomfortable especially when the frequencies of vibration correspond to natural frequencies of the human body and organs. In fact, it is well known that the resonant frequency of the human intestinal tract (approx. 4-8 Hz) should be avoided at all costs when designing high performance aircraft and reusable launch vehicles because sustained exposure can cause serious internal trauma (Leatherwood and Dempsey, 1976 NASA TN D-8188). If an aircraft wing vibrates at large amplitudes for an extended period of time, the wing will eventually experience a fatigue failure of some kind, which would potentially cause the aircraft to crash resulting in injuries and/ or fatalities. Wing vibrations of this type are usually associated with a wide variety of flutter phenomena brought on by fluid-structure interactions. The most famous engineering disaster of all times was the Tacoma Narrows Bridge disaster in 1940. It failed due to the same type of self-excited vibration behaviour that occurs in aircraft wings. Basic Mechanical Engineering

Mechanical Vibrations

69

In reading books and technical papers on vibration including the previous paragraph, engineering students are usually left with the impression that all vibrations are detrimental because most publicized work discusses vibration reduction in one form or another. But vibrations can also be beneficial. For instance, many different types of mining operations rely on sifting vibrations through which different sized particles are sorted. In nature, vibrations are also used by all kinds of different species in their daily lives. Orb web spiders, for example, use vibrations in their webs to detect the presence of flies and other insects as they struggle after being captured in the web for food. The reason that mechanical systems vibrate freely is because energy is exchanged between the system’s inertial (masses) elements and elastic (springs) elements. Free vibrations usually cease after a certain length of time because damping elements in systems dissipate energy as it is converted back-and-forth between kinetic energy and potential energy. The role of mechanical vibration analysis should be to use mathematical tools for modeling and predicting potential vibration problems and solutions, which are usually not obvious in preliminary engineering designs. If problems can be predicted, then designs can be modified to mitigate vibration problems before systems are manufactured. Vibrations can also be intentionally introduced into designs to take advantage of benefits of relative mechanical motion and to resonate systems (e.g., scanning microscopy). 8QIRUWXQDWHO\ NQRZOHGJH RI YLEUDWLRQV LQ SUHOLPLQDU\ PHFKDQLFDO GHVLJQV LV UDUHO\ considered essential, so many vibration studies are carried out only after systems are manufactured. In these cases, vibration problems must be addressed using passive or active design modifications. Sometimes a design modification may be as simple as a thickness change in a vibrating panel. Added thickness tends to push the resonant frequencies of a panel higher leading to less vibration in the operating frequency range. Design modifications can also be as complicated as inserting magneto-rheological (MR) fluid dampers into mechanical systems to take energy away from vibrating systems at specific times during their motion. The point here is that design changes prior to manufacture are less expensive and more effective than design modifications done later on. The fundamentals of sound and vibrations are part of the broader area of mechanics, with strong connections to classical mechanics, strong mechanics and fluid dynamics. Dynamics is the branch of physics concerned with the movement of bodies under the motion of forces. Vibrations or oscillations are also regarded as a subset of dynamics wherein a method subjected to restoring forces swings from side to side about an equilibrium position. The restoring forces are because of elasticity, or as a result of gravity. The area of sound and vibrations encompasses the generation of sound and vibrations, the distribution and damping of vibrations, how sound propagates in a free subject, and the way it interacts with a closed space, as well as its effect on man and dimension

70

Basic Mechanical Engineering

equipment. Technical purposes span wider disciplines, from applied mathematics and mechanics, to electrical instrumentation and analog and digital signal processing thought, to machinery and constructing design. Most human activities contain vibrations, for example, we hear given that our eardrums vibrate and see due to the fact gentle waves undergo vibration. Respiration is associated with the vibration of lungs and strolling involves (periodic) oscillatory movement of legs and palms. We speak because of the oscillatory motion of larynges (tongue). In many of the engineering applications, vibration is signifies side to side motion, which is undesirable. Galileo learnt the relationship between the length of a pendulum and its frequency and found the resonance of two bodies that have been related via some energy transfer medium and tuned to the equal traditional frequency. Vibrations may be due to the failure of machines or their accessories. The influence of vibration is determined by the magnitude in terms of displacement, pace or acceleration, interesting frequency and the whole duration of the vibration. On this chapter, the vibration of a single-degree-of-freedom (SDOF), two degree of freedom approach with and without damping and introductory multi-degree of freedom approach can be mentioned in this part. Modeling Issues: Modeling is usually 95% of the effort in real-world mechanical vibration problems; however, this course will focus primarily on the derivation of equations of motion, free response and forced response analysis, and approximate solution methods for vibrating systems.

4.1.1 Basic Terminology Oscillating Motions: The study of vibrations is concerned with the oscillating motion of elastic bodies and the force associated with them. All bodies possessing mass and elasticity are capable of vibrations. Most engineering machines and structures experience vibrations to some degree and their design generally requires consideration of their oscillatory motions. Oscillatory systems can be broadly characterized as linear or nonlinear. Linear Systems: The principle of superposition holds and the mathematical techniques available for their analysis are well developed. Nonlinear Systems: The principle of superposition doesn’t hold and the techniques for the analysis of the nonlinear systems are under development (or less well known) and difficult to apply. All systems tend to become nonlinear with increasing amplitudes of oscillations. There are two general classes of vibrations – free and forced. Free Vibration: Free vibration takes place when a system oscillates under the action of forces inherent in the system itself due to initial disturbance, and when the externally applied forces are absent. The system under free vibration will vibrate at one or more

Mechanical Vibrations

71

of its natural frequencies, which are properties of the dynamical system, established by its mass and stiffness distribution. Forced Vibration: The vibration that takes place under the excitation of external forces is called forced vibration. If the excitation is harmonic, the system is forced to vibrate at excitation frequency. If the frequency of excitation coincides with one of the natural frequencies of the system, a condition of resonance is encountered and dangerously large oscillations may result, which results in failure of major structures, i.e., bridges, buildings, or airplane wings, etc. Thus, calculation of natural frequencies is of major importance in the study of vibrations. Because of friction and other resistances vibrating systems are subjected to damping to some degree due to dissipation of energy. Damping has very little effect on natural frequency of the system, and hence the calculations for natural frequencies are generally made on the basis of no damping. Damping is of great importance in limiting the amplitude of oscillation at resonance. Degrees of Freedom (DoF): The number of independent coordinates required to describe the motion of a system is termed degrees of freedom. For example, Particle - 3 DoF (positions), Rigid body-6 DoF (3-positions and 3-orientations). Continuous elastic body - infinite DoF (three positions to each particle of the body). If part of such continuous elastic bodies may be assumed to be rigid (or lumped) and the system may be considered to be dynamically equivalent to one having finite DoF (or lumped mass systems). Large number of vibration problems can be analyzed with sufficient accuracy by reducing the system to one having a few DoF.

4.2

IMPORTANCE OF VIBRATION

Vibration tends to have a lot of benefits and a lot more disadvantages. “Vibration”, a time period that tends to set off the identical in once’s intellect when any individual hears it. Normally, the whole thing vibrates on the planet, some vibrations are just right and valuable, some are relatively negligible, some are tolerable, some annoying and the rest fall beneath the hazardous class. A vibrating guitar tends to produce soothing track for our ears; a vibrating sieve separates particles of first-rate sand and coarse sand; the vibrating mode in our cell cellphone to alert us without disturbing others. Some excellent examples of what priceless things vibration does to us. Some good examples of unwanted matters vibration does to us; an old going for walks engine vibrating so much developing quite a few noise unbearable to us; essentially the most dreaded vibration of the entire “earthquake” a random vibration of the earth. These small critical examples have a tendency to give a glance of the significance that wishes to take delivery of this discipline. Many of the results coming up because of vibration result in failures, damage of apparatus and in worst instances loss of life. Realizing extra about vibration helps us to control injury and take preventive measures from damages to occur. Every computer or appliance that tends to do work by means

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Basic Mechanical Engineering

of reciprocating or rotating has vibration caused into itself as a result of its own characteristics. One of the crucial principal units for the size of vibration are displacement (mm or inches), pace (mm/sec or m/sec) and acceleration (mm/s2).

4.3

TYPES OF VIBRATION

Free Vibration: Free vibration happens when a mechanical mechanism is ready in movement with an initial input and allowed to vibrate freely. Examples of this sort of vibration are pulling a baby back on a swing and letting go, or hitting a tuning fork and letting it ring. The mechanical process vibrates at one or more of its common frequencies and damps down to motionlessness. Free vibration takes situation when a process oscillates below the motion of forces inherent within the system itself as a result of initial disturbance, and when the externally applied forces are absent. The process beneath free vibration will vibrate at a number of its common frequencies, which might be properties of the dynamical approach, headquartered by way of its mass and stiffness distribution. Forced Vibration: Forced vibration is when a time-varying disturbance (load, displacement or velocity) is applied to a mechanical system. The disturbance can be a periodic and steady-state input, a transient input, or a random input. The periodic input can be a harmonic or a non-harmonic disturbance. Examples of these types of vibration include a washing machine shaking due to an imbalance, transportation vibration caused by an engine or uneven road, or the vibration of a building during an earthquake. For linear systems, the frequency of the steady-state vibration response resulting from the application of a periodic, harmonic input is equal to the frequency of the applied force or motion, with the response magnitude being dependent on the actual mechanical system. ‡ The vibration that takes place below the excitation of external forces is referred to as forced vibration. ‡ If excitation is harmonic, the process is compelled to vibrate at excitation frequency. If the frequency of excitation coincides with one of the natural frequencies of the system, a situation of resonance is encountered and dangerously big oscillations may result, resulting in failure of major structures, i.e., bridges, buildings, or aircraft wings and many others. ‡ For that reason calculation of natural frequencies is the most importance part of vibrations. ‡ Considering the fact that of friction and different resistance vibrating techniques are subject to damping as a result of dissipation of energy. ‡ Damping has little or no influence on natural frequency of the system, and hence the calculations for typical frequencies are most likely made on the basis of no damping. ‡ Damping is of great value in limiting the amplitude of oscillation at resonance.

Mechanical Vibrations

73

4.3.1 Single DoF Free Undamped Vibration Free Vibration: Objective of the present section is to write the equation of motion of a system and evaluate its natural frequency, which is mainly a function of mass, stiffness, and damping of the system from its general solution. In many practical situations, damping has little influence on the natural frequency and may be neglected in its calculation. In the absence of damping, the system can be considered conservative and principle of conservation of energy offers another approach to the calculation of the natural frequency. The effect of damping is mainly evident in diminishing the vibration amplitude at or near the resonance.

k

c

Stiffness

Damping m Mass

Figure 4.1 Spring-damper vibration model.

The inertia force model is Fi = mx where m is the mass in kg, x is the acceleration in m/sec2 and Fi is the inertia force in N. The linear stiffness force model is Fs = kx where k is the stiffness (N/m), x is the displacement and Fs is the spring force. The damping force model for the viscous damping is Fd = cx where c is the damping coefficient in N/m/sec, x is the velocity in m/sec and Fd is the damping force. Undamped Free Vibration: A spring mass system as shown below is considered. For simplicity at present the damping is not considered. The direction of x in the downward direction is positive. Also velocity, x , acceleration, x , and force, F, are positive in the downward direction as shown in Fig. 4.2.

74

Basic Mechanical Engineering

Unstretched position

m

D

kD m

k (D + x) x

Static equilibrium position

mg

. .. x x m

m

mg (a)

(b)

(c)

(d)

Figure 4.2

From Figure 4.2 (d) on application of Newton’s second law, we have mx = S F or mx = mg – k(D + x) From Fig. 4.2 (b), we have kD = mg (i.e., spring force due to static deflection is equal to the weight of the suspended mass), so the above equation becomes mx + kx = 0

The choice of the static equilibrium position as reference for x axis datum has eliminated the force due to the gravity. The above-mentioned equation can be written as x + w n2 x = 0

or with

x = - w n2 x w n2 = k m

where wn is the natural frequency (in rad/sec). This equation satisfies the simple harmonic motion condition. Solution of EoM: From the above equation it can be observed that: ‡ The motion is harmonic (i.e., the acceleration is proportional to the displacement). ‡ It is homogeneous, second order, linear differential equation (in the solution two arbitrary constants should be there). ‡ The general solution of equation can be written as x = A sin wnt + B cos wnt

(1)

Mechanical Vibrations

75

where A and B are two arbitrary constants, which depend on initial conditions, i.e., x(0) and x (0 ) . x can be differentiated to give x = Awn cos wnt – Bwn sin wnt

(2)

On application of initial conditions in equation (1) and (2), we get x(0) = B and

x = Awn

x (0 ) wn

or

A=

and

B = x(0)

(3)

On substituting equation (2) into equation (3), we get x=

x (0 ) sin w n t + x(0 ) cos w n t = X cos(w n t - f ) wn

where,

X=

Ï x (0 ) ¸ { x( 0 ) + Ì ˝ Ó wn ˛ 2

(4) 2

tan f =

x (0 ) w n x( 0 )

where X is the amplitude, wn is the circular frequency and f is the phase. The undamped free vibration executes the simple harmonic motion. Since sine and cosine functions repeat after 2p radians (i.e., Frequency × Time period = 2p), we have wnT = 2p (5) The time period (in seconds) can be written as T = 2p

m k

(6)

The natural frequency (in rads/sec or hertz) can be written as f=

1 1 = T 2p

k m

From Fig. 4.2 (b), we have kD = mg fi

k g = m D

(7)

76

Basic Mechanical Engineering

On substituting equation 7 into equation 6, we get

f=

1 2p

g D

Here T, f, wn depend on the mass and stiffness of the system, which are properties of the system. The above analysis is valid for all kinds of SDoF system including beam or torsional members. For torsional vibrations the mass may be replaced by the mass moment of inertia and stiffness by stiffness of torsional spring. For stepped shaft an equivalent stiffness can be taken or for distributed mass an equivalent lumped mass can be taken. The undamped free response can also be written as - iw t iw t x = ae n + be n

= a[cos wnt + i sin wnt] + b[cos wnt – i sin wnt] = (a + b) cos wnt + i(a – b) sin wnt = A cos wnt + B sin wnt where A and B are constants to be determined from initial conditions, which is same as the equation described above.

4.3.2 Damped System Vibration systems may encounter damping of the following types: 1. Internal molecular friction. 2. Sliding friction 3. Fluid resistance Generally, mathematical model of such damping is quite complicated and not suitable for vibration analysis. Simplified mathematical model (such as viscous damping or dash-pot) have been developed which leads to simplified formulation. A mathematical model of damping in which the force is proportional to displacement, i.e., Fd = cx is not possible because with cyclic motion this model will encounter an area of magnitude equal to zero as shown in Fig. 4.3 (a). So dissipation of energy is not possible with this model. The damping force (non-linearly related with displacement) versus displacement curve will enclose an area, referred as the hysteresis loop Fig. 4.3(b), that is proportional to the energy lost per cycle.

Mechanical Vibrations

77

Fd Fd

x O

x

(a) Linear relation

(b) Nonlinear relation

Figure 4.3

Viscously Damped Free Vibration: Viscous damping force is expressed as, Fd = cx

(1)

c is the constant of proportionality and it is called damping coefficient. Figure 4.4 shows spring-damper-mass system with free body diagram. From free-body diagram, we have

S F = mx

(2)

- kx - cx = mx mxð + cx + kx = 0

k

(3)

c

Stiffness

Damping m Mass

Static equilibrium position

. cx

kx x m .. mx

Figure 4.4

. .. x, x, x

78

Basic Mechanical Engineering

Let us assume a solution of equation (3) of the following form x = e st

(4)

where s is a constant (can be a complex number) and t is time. So that x = se st and x = s 2 est , on substituting in equation (3), we get (ms2 + cs + k) est = 0 From the condition that equation (4) is a solution for all values of t, the above equation gives a characteristic equation (Frequency equation) as s2 +

c k s+ =0 m m

(5)

4.3.3 Energy Method In a conservative system (i.e., with no damping) the total energy is constant, and differential equation of motion can also be established by the principle of conservation of energy. ‡ For the free vibration of undamped system: Energy = (partly kinetic energy + partly potential energy). ‡ Kinetic energy T is stored in mass by virtue of its velocity. Potential energy U is stored in the form of strain energy in elastic deformation or work done in a force field such as gravity, magnetic field, etc. (6) ‡ T + U = constant Hence, d (T + U ) = 0 dt

Our interest is to find natural frequency of the system, writing equation for two positions i.e., T1 + U1 = T2 + U2 = constant

(7)

where, 1 and 2 represent two instants of time. Let 1 represent a static equilibrium position (choosing this as the reference point of potential energy, here U1 = 0) and 2 represent the position corresponding to maximum displacement of mass and at this position velocity of mass will be zero and hence T2 = 0. Equation (7) reduces to T1 + 0 = 0 + U2

(8)

Mechanical Vibrations

If mass is undergoing harmonic motion then T1 and U2 are maximum values. Tmax = Umax

4.3.4

79

(9)

Equivalent Stiffness of Series and Parallel Springs

For this system stiffness equation for series and parallel springs is still valid as shown in Figs. 4.5(a) and 4.5(b). keq =

1 1/k1 + 1/k2 k2

k1

Figure 4.5 (a)

keq = k1 + k2 k1

k2

Figure 4.5 (b)

4.4

VIBRATION ISOLATION AND CRITICAL SPEED

4.4.1 Vibration Isolation Vibration isolation schemes are to (1) reduce the propagation of base vibration to the isolated object (machinery) and (2) abate the transmission of vibration energy of machinery to the base. Moreover, in vehicular/marine, industrial machines (such as mechanical presses), as well as seismic applications, isolation systems are also expected to lower the impact of shock from base to the isolated object and/or vice versa. By way of a number of examples, we have demonstrated that the high insertion losses predicted for high frequencies by the crudest models are not actually realized. That is because, as we have already pointed out, the assumptions that underlie the simple models are not valid at high frequencies. Machines and foundations are not rigid, and isolators do not remain compliant. High vibration levels can cause machine failure, as well as objectionable noise levels. A common source of objectionable noise in buildings is the vibration of machines that are mounted on floors or walls. Obviously, the best place to mount a vibrating machine is the ground floor. Vibration isolation prevents the vibration of two interconnected objects to transmit to one another. Passive vibration isolation solutions are commonly realized by placing a set of resilient elements such as elastomeric (rubber), metal, or air springs between the isolated objects (machinery/equipment, buildings, bridges, etc.) and their foundations (bases).

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8QOXFNLO\KRZHYHUIRUWXQDWHO\IRUQRLVHFRQWUROFRQVXOWDQWV WKLVLVQ·WSHUSHWXDOO\ viable. An ordinary obstacle is a rotating desktop (such as a pump, ac compressor, blower, engine, and so on) set up on a roof, or on a floor above the bottom ground. The challenge is mostly most apparent within the immediate neighbourhood of the vibration supply. Nevertheless, mechanical vibrations can transmit for lengthy distances, and by very circuitous routes by means of the constitution of a constructing, sometimes resurfacing 1000s of ft from the source. An associated concern is the isolation of vibrationsensitive machines from the most often occurring disturbances in a constructing (vehicle or bus site visitors, slamming doors, foot site visitors, elevators, etc.). “Passive vibration isolation” refers to vibration isolation or mitigation of vibrations by way of passive systems equivalent to rubber pads or mechanical springs, versus “active vibration isolation” or “electronic drive cancellation” employing electrical vigour, sensors, actuators, and manage methods. Passive vibration isolation is a big discipline, considering there are various varieties of passive vibration isolators used for many special functions. A number of of these applications are for industrial gear equivalent to pumps, motors, HVAC techniques, or washing machines; isolation of civil engineering constructions from earthquakes (base isolation), touchy laboratory gear, useful statuary, and high-finish audio. Figure 4.6 shows a normal example of a vibration source, an enormous reciprocating air conditioning compressor weighing 20,000 kg, set up on a roof. Worrying noise phases at multiples of the compressor rotational frequency, predominantly 60 and 120 Hz, had been measured within the rooms straight below the compressor. Additionally, this variety of compressor (reciprocating) is infamous for prime vibration phases. Centrifugal or scroll kind compressors are a lot quieter.

Figure 4.6 A reciprocating air conditioning compressor and chiller mounted on a flexible roof, note the straight conduit on the left which bypasses the isolators and directly transmits vibration into the roof.

Mechanical Vibrations

81

Design of vibration isolators: There a number of rules are followed when designing vibration isolators. If these are adhered to, the results should be acceptable: (i) The isolator’s (static) stiffness must be chosen so low that the highest mounting resonance falls far below the lowest interesting excitation frequency. (ii) The mounting positions on the foundation should be as stiff as possible. (iii) The points at which the machine is coupled to the isolators should also be as stiff as possible. Rules (ii) and (iii) are normally not difficult to follow at low frequencies; at high frequencies, however, internal resonances make them problematic. (iv) The isolator should, if possible, be designed so that its first internal anti-resonance falls well above the highest excitation frequency of interest. That rule is, in practice, difficult to follow. If it cannot be followed, then one should ascertain by measurements or computations that at least the following alternative rules are fulfilled. (v) The isolator must be designed so that its internal resonances do not coincide with strong components of the excitation spectrum. (vi) The isolator must, therefore, be designed so that its anti-resonance frequencies do not coincide with the resonance frequencies of the foundation. In addition to these rules, there are normally also a number of constraints related to geometric, strength, and stability concerns.

4.4.2

Critical Speed

All rotating shafts even in the absence of outside load, deflect throughout rotation. The combined weight of a shaft and wheel can provide deflection with the intention to create resonant vibration at exact speeds, often called critical velocity. The magnitude of deflection depends on the following: (a) stiffness of the shaft and its support; (b) total mass of shaft and attached instruments; (c) unbalance of the mass with reference to the axis of rotation; and (d) the amount of damping in the approach for that reason, the calculation of critical velocity for fan shaft is imperative. There are two methods to calculate critical velocity, i.e., Rayleigh-Ritz and Dunkerley equations. These equations are an approximation to the first common frequency of vibration, which is assumed to be practically equal to the critical speed of rotation. More commonly, the Rayleigh-Ritz equation overestimates and the Dunkerley equation underestimates the natural frequency. The equation illustrated below is the RayleighRitz equation which suggests that the highest operation speed must not exceed 75% of the critical velocity. Critical velocity is determined by the magnitude or vicinity of the burden or load carried with the help of the shaft, the length of the shaft, its diameter and bearing.

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Basic Mechanical Engineering

The place, g = gravity acceleration (9.81 m/s2) dst = complete highest static deflection In stable mechanics, within the area of rotor-dynamics, the central pace is the theoretical angular velocity that excites the normal frequency of a rotating object, corresponding to a shaft, propeller, lead screw, or equipment. Because the speed of rotation approaches the article’s average frequency, the thing starts to resonate, which dramatically raises simple vibration. The resulting resonance happens in spite of orientation. As a result we will say that after rotational velocity is equal to the numerical value of the natural vibration then that velocity is known as critical velocity.

4.5

SUMMARY

Vibration is a vast topic in field of mechanical engineering. It is associated with all the machines, equipment, structures, etc. We have dealt with vibrations in mechanical systems stating the basic types of vibrations as free and forced vibrations. Isolation of vibrations to full extent is next to impossible. Critical speed also plays a vital role in the vibrations of mechanical systems. In many ways people can expect to obtain an ideal machine viewed from the angle of vibration, which is a machine that produces no vibration at all. Such an ideal machine will greatly save energy because of all the energy given to the whole machine will be used to perform the work alone, whether pumping a fluid, compressing the air, crushing paper, etc., without generating byproducts in the form of vibration. Apparently, it is unlikely because in terms of machining is not possible to get a very homogeneous material, fabrication machines without residual imbalance (residual unbalance) and rotating machinery moving and moving back and forth that does not cause friction with the other part. What is then seen as a result of the onset of engine vibrations, is nothing but the result of abnormal circumstances such as bolts slackness, engine parts, wear faster, is aligned shafts, rotor unbalance, etc. Conditions mentioned above will raise the dissipation energy because of vibration, causing resonance, and dynamic loads on the bearings. Cause and effect that occur will damage (break down) the machine immediately which causes the engine to be shut down or automatically shut itself down because of the protection on the electrical system or its instrumentation. Manufacturing an excellent machine is to minimize imprecision such that the level of vibration is small (fine). A consequence is the price of the machine will be more than machine level imprecision “mediocre”. Even such machines for sometimes cannot be exported out of state as a result an advantage because it has a strategic impact on; politics and national security.

Mechanical Vibrations

83

SOLVED EXAMPLES

Example 4.1 Derive equation of motion of a simple pendulum for large and small oscillations.

Solution: T

q

.. q

m

m mg

q

(b) Free-body diagram

(a) Simple pendulum

Figure 4.7

Figure 4.7 shows a simple pendulum and free-body diagram of a bob. Tension in spring is given by T = mg cos q

(a)

From Newton’s second law the moment balance about centre of rotation O is given as

 = SM = (– mg sin q) l Iq 0

where I is the polar mass moment of inertia, of the mass m about centre O, so that ml 2 q = – mgl sin q

or g  q + sin q = 0 l

(b)

which is the equation of motion of the pendulum for large oscillation. For small amplitude of oscillation equation (b) reduces to (i.e., sin q ª q) g  q + sin q = 0 l

From which the natural frequency of the pendulum can be obtained as w n2 = g l

wn =

w n = 2pf = g rad/sec l

2p T

or f =

1 2p

g cycle/sec l

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Basic Mechanical Engineering

The time period is given as T=

Example 4.2 Solution: Fig. 4.7.

1 = 2p f

l sec g

Obtain equation of motion of a pendulum using energy method. The kinetic and potential energies of the pendulum are given as shown in

T=

1  2 Iq 2

and U = mgl (1 – cos q)

From the principle of conservation of energy, we have d (T + U ) = 0 dt

On substituting energy terms, we get  + mgl I qq

d (1 - cos q) = 0 dt

or

 + mgl(q sin q) = 0 Iqq

(c)

For pendulum I = ml2, hence, equation (c) becomes q (ml 2  q + mgl sin q) = 0

Since q cannot be zero, hence g  q + sin q = 0 l

The above equation is non-linear because of the term sin q.

Example 4.3 Obtain the equation of motion of a spring mass system using energy method. Solution:

For a spring mass system and we have T=

1 mx 2 2

U=

1 2 1 kx + ( - mgx + k Dx ) = kx 2 [\ mg = k D ] 3 2

The loss of potential energy of m due to position x is cancelled by the work done by the equilibrium force of the spring, (kD) in position x = 0.

Mechanical Vibrations

85

For total energy of the system, we have T.E. = T + U, On substituting the above terms in the above equation , we get 1 2 1  + k 2 xx = 0 m xx 2 2

or

(mx + kx ) x = 0

Since x π 0 hence (mx + kx ) = 0 For spring mass system, having harmonic motion at the natural frequency, we have x = a cos pt,

x = – wn sin wnt

x = w n2 a cos w nt = - w n2 x , x max = - wn a att = 0

Tmax =

1 1 m ( - w n a)2 , U max = ka2 2 2

On substituting the above terms in equation, we get wn =

k m

Example 4.4

A flywheel of weight 311.36 N is supported on a knife edge at a distance 15.24 cm from its geometric centre. If the measure period of oscillation was 1.22 sec, determine the mass moment of inertia of the flywheel about its geometric centre.

Solution: 1. The flywheel can be considered a compound pendulum. For which the period is given by T = 2p

l+c g

T = Time period = 1.22 sec. l = The distance between centres of rotation and gravity = 0.1524 m g = Acceleration due to 9.81 m/sec2 c = The distance between centres of gravity and percussion =

T2g -l 4p 2

(1.22)2 ¥ 9.81 - 0.1524 = 0.3698 – 0.1524 = 0.2174 m = 4 ¥ p2

Flywheel weight = 311.36 N The mass moment of inertia of the flywheel about its geometric centre is given as IG = mKG2 = mlc =

311.36 ¥ 0.1524 ¥ 0.2174 = 10.315 kg m 2 9.81

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Basic Mechanical Engineering

EXERCISE 1. 2. 3. 4. 5. 6. 7.

State the important characteristics of mechanical vibrations. How is vibration important for mechanical machines and structures. Explain free and forced vibrations. What is vibration isolation? How does it lead to stability of systems. Write a short note on critical speed and its applications. Explain the role of vibration in mechanical devices. Obtain the natural frequency of the following system. (a) A simply supported beam with a central disc as shown in Fig. 4.8. È ÍAns. w n = Î

48EI ˘ ˙ ml 3 ˚

m

1 2

1 2

Figure 4.8

(b) A simply supported beam with offset disc as shown in Fig. 4.9. È ÍAns. w n = Î a

b

m l

Figure 4.9

3EIl ˘ ˙ ma 2 b 2 ˚

Mechanical Vibrations

87

8. Obtain the natural frequency of a spring-mass system in terms of spring wire diameter d, coil radius R, number of turns of coil n and modulus of rigidity G of spring material. 9. A certain system has stiffness 10 Nm–1 and is heavily damped. The mass is observed to be moving to the left at 10 mm/sec when its position is 30 mm to the right of its equilibrium position. Calculate the resisteance. [Ans. 30 kg/sec] 10. A lightly damped system is set into vibration with initial condition x(0) = 0 and x (0 ) = n0 . Show that the subsequent motion is given as x(t) =

n0 Ê 1 ˆ exp Á - ct ˜ sin w d t Ë 2 ¯ wd

where wd is the damped natural frequency and c is the damping factor.

Multiple-Choice Questions 1. Time dependent permanent deformation is called __________. (a) Plastic deformation

(b) Elastic deformation

(c) Creep

(d) Inelastic deformation

2. Figure out the odd point in the following: (a) Proportional limit

(b) Elastic limit

(c) Yield point

(d) Fracture point

3. If a material is subjected to two incremental true strains, namely, e1 and e2 , then the total true strain is (a) e1 * e2

(b) e1 – e2

(c) e1 + e2

(d) e1 /e2

4. Engineering stress-strain curve and true stress-strain curve are equal up to (a) Proportional limit

(b) Elastic limit

(c) Yield point

(d) Tensile strength point

5. Value of Poisson’s ratio for ionic solids in the range of (a) 0.1

(b) 0.2

(c) 0.3

(d) 0.4

6. Hydrostatic stress results in the following: (a) Linear strain

(b) Shear strain

(c) Both linear and shear strains

(d) None

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Basic Mechanical Engineering

7. High elastic modulus in materials arises from (a) High strength of bonds

(b) Weak bonds

(c) Combination of bonds

(d) None

8. Change in elastic modulus for ordinary materials between 0K and melting point is (a) 10-20% increase

(b) 10-20% decrease

(c) 80-90% decrease

(d) 80-90% increase

9. Bauschinger effect (a) Hysteresis loss during loading and unloading (b) Inelastic deformation (c) Dependence of yield stress on path and direction (d) None 10. Shape of true stress-strain curve for a material depends on (a) Strain

(b) Strain rate

(c) Temperature

(d) All

11. Toughness of a material is equal to the area under __________ part of the stressstrain curve. (a) Elastic

(b) Plastic

(c) Both

(d) None

12. True stress-strain curve needs to be corrected after (a) Elastic limit

(b) Yield limit

(c) Tensile strength

(d) No need to correct

13. The following condition represents onset of necking (a) eu = n

(b) eu = 1 – n

(c) eu = 1 + n

(d) eu = ln (1 + n)

14. As compared with conventional stress-strain curve, the true stress-strain curve is (a) Above and right

(b) Below and right

(c) Above and left

(d) Below and left

15. According to distortion-energy criterion, yielding occurs when (a) Distortion energy reaches a critical value (b) Second invariant of the stress deviator exceeds some critical value (c) Octahedral shear stress reaches a critical value (d) All

Mechanical Vibrations

89

16. von Mises and Tresca criteria give different yield stress for 

D  8QLD[LDOVWUHVV

E  %DODQFHGELD[LDOVWUHVV

(c) Pure shear stress

(d) All

17. Plastic deformation results from the following: (a) Slip

(b) Twinning

(c) Both

(d) None

18. Time dependent recoverable deformation under load is called __________ deformation. (a) Elastic

(b) Inelastic

(c) Elastic after-effect

(d) Viscoelastic

19. Failure due to excessive deformation is controlled by __________. (a) Material properties

(b) Design & dimensions

(c) Both

(d) None

20. Failure due to excessive deformation is controlled by __________. (a) Yield strength

(b) Tensile strength

(c) Young’s modulus

(d) All

21. Time dependent yield is known as



(a) Fracture

(b) Fatigue

(c) Buckling

(d) Creep

 8VXDOO\PDWHULDOVZLWKIROORZLQJFU\VWDOVWUXFWXUHIDLOLQGXFWLOHPRGH (a) FCC

(b) BCC

(c) HCP

(d) None

23. Brittle fracture is more dangerous than ductile fracture because __________. (a) No warning sign (b) Crack propagates at very high speeds (c) No need for extra stress during crack propagation (d) All 24. Fracture voids usually form at (a) Inclusions

(b) Second phase particles

(c) Grain boundary triple points

(d) All

90

Basic Mechanical Engineering

25. Fracture stress (sf) is proportional to (a) crack length (c) (crack length)

(b) 1/crack length 1/2

(d) (crack length)– 1/2

26. Fracture toughness is measured in terms of



(a) Strain energy release rate

(b) Stress concentration factor

(c) Both

(d) None

 7KHIROORZLQJHTXDWLRQGHÀQHVS-N curve (a) Paris equation

(b) Basquin equation

(c) Andrede equation

(d) Garofalo equation

28. Creep rate in ternary stage __________. (a) Decreases

(b) Constant

(c) Increases

(d) None

29. Most often machine components fail by (a) Buckling

(b) Creep

(c) Fatigue

(d) All material science/failure

30. The units of moment of inertia of an area are (a) kg-m²

(b) m4

(c) kg/m²

(d) m³

31. The slope on the road surface generally provided on the curves is known as (a) Angle of friction

(b) Angle of repose

(c) Angle of banking

(d) None of these

32. A force acting on a body may (a) Change its motion (b) Balance the other forces acting on it (c) Retard its motion (d) All of the above 33. On a ladder resting on smooth ground and leaning against vertical wall, the force of friction will be (a) Towards the wall at its upper end (b) Away from the wall at its upper end 

F  8SZDUGVDWLWVXSSHUHQG (d) Downwards at its upper end

Mechanical Vibrations

91

34. The velocity of a body on reaching the ground from a height h, is 

D  ¥JK 

E  ¥JK



F  ¥JK 

G  J¥K

35. Lami’s theorem states that (a) Three forces acting on a point will be in equilibrium (b) Three forces acting on a point can be represented by a triangle, each side being proportional to force (c) If three forces acting on a particle are represented in magnitude and direction by the sides of a triangle, taken in order, they will be in equilibrium (d) If three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two 36. Moment of inertia of a triangular section of base (b) and height (h) about an axis through its base, is (a) bh3/4

(b) bh3/8

(c) bh3/12

(d) bh3/36

37. The forces, which meet at one point and their lines of action also lie on the same plane, are known as (a) Coplanar concurrent forces (b) Coplanar non-concurrent forces (c) Non-coplanar concurrent forces (d) Non-coplanar non-concurrent forces 38. Two springs have spring stiffness of 1500 N/m and 2000 N/m respectively. If they are connected in series, what is the spring stiffness if they are replaced by an equivalent system. (a) 3500 N/m

(b) 1166 N/m

(c) 857.63 N/m

(d) None of the above

39. Which type of vibrations are also known as transient vibrations? 

D  8QGDPSHGYLEUDWLRQV

E  'DPSHGYLEUDWLRQV

(c) Torsional vibrations

(d) Transverse vibrations

40. During transverse vibrations, shaft is subjected to which type of stresses? (a) Tensile stresses

(b) Torsional shear stress

(c) Bending stresses

(d) All of the above

92

Basic Mechanical Engineering

41. In which type of vibrations, amplitude of vibration goes on decreasing every cycle? 

D  'DPSHGYLEUDWLRQV

E  8QGDPSHGYLEUDWLRQV

(c) Both a and b

(d) None of the above

42. The ultimate strength of steel in tension in comparison to shear is in the ratio of (a) 1 : l

(b) 2:1

(c) 3 : 2

(d) 2 : 3

43. The property of a material which enables it to resist fracture due to high impact loads is known as (a) Elasticity

(b) Endurance

(c) Strength

(d) Toughness

44. A hot short metal is (a) Brittle when cold

(b) Brittle when hot

(c) Brittle under all conditions

(d) Ductile at high temperature

45. Tensile strength of a mild steel specimen can be roughly predicted from the following hardness test (a) Brinell

(b) Rockwell

(c) Vicker

(d) Shore’s sceleroscope

46. Resilience of a material is important, when it is subjected to



(a) Combined loading

(b) Fatigue

(c) Thermal stresses

(d) Shock loading

 ,QWKHFDVHRIDQHODVWLFEDUÀ[HGDWXSSHUHQGDQGORDGHGE\DIDOOLQJZHLJKWDW lower end, the shock load produced can be decreased by (a) Decreasing the cross-sectional area of bar (b) Increasing the cross-sectional area of bar (c) Remain unaffected with cross-sectional area (d) Would depend on other factors 48. Stress concentration is caused due to (a) variation in properties of material from point to point in a member (b) pitting at points or areas at which loads on a member are applied (c) abrupt change of section (d) all the above

Mechanical Vibrations



93

 7KHHQGXUDQFHOLPLWRIDPDWHULDOZLWKÀQLVKHGVXUIDFHLQFRPSDULVRQWRURXJK surface is (a) More (b) Less (c) Same (d) More or less depending on quantum of load 50. The fatigue life of a part can be improved by (a) Electroplating

(b) Polishing

(c) Coating

(d) Shot peening

Answers 1. (c)

2. (d)

3. (c)

4. (c)

5. (b)

6. (d)

7. (a)

8. (b)

9. (c)

10. (d)

11. (c)

12. (c)

13. (a)

14. (c)

15. (d)

16. (c)

17. (c)

18. (b)

19. (c)

20 (c)

21. (d)

22. (a)

23. (d)

24. (d)

25. (d)

26. (c)

27. (b)

28. (c)

29. (c)

30. (b)

31. (c)

32. (d)

33. (d)

34. (c)

35. (d)

36. (c)

37. (a)

38. (c)

39. (b)

40. (c)

41. (a)

42. (c)

43. (d)

44. (b)

45. (a)

46. (d)

47. (a)

48. (d)

49. (a)

50 (d)

Section II Thermal Engineering

5 Thermodynamics

Learning Objectives    

‡ ‡ ‡ ‡

5.1

8QGHUVWDQGLQJWKHWHUPWKHUPRG\QDPLFV 9DULRXVGHILQLWLRQVDVVRFLDWHGZLWKWKHUPRG\QDPLFV%DVLFVRIWKHUPRG\QDPLFV 2YHUYLHZRIODZVRIWKHUPRG\QDPLFV &LWHWKHFRQFHSWVRISXUHVXEVWDQFHV

INTRODUCTION

Therme is a Greek word which means heat and dynamic means force. The word thermodynamics originates from these two words. This is the classical history of thermodynamics. In primitive days understanding of thermodynamics centred around the concept of getting power from hot bodies or getting power from heat, abilities of hot body to produce work, which is partly the scope of mechanical engineering thermodynamic student but the scope of thermodynamics is much wider and broader. Before going to define thermodynamics in a formal way, it can be said like this, thermodynamics is probably encountered in our everyday life. Thermodynamics is a fundamental subject that describes the basic laws governing the occurrence of physical processes associated with transfer or transformation of energy and also establishes the relationship between different physical properties which are being affected by these processes. This is the domain of thermodynamics. Now the entire subject thermodynamics is based on laws of natures formed by our observation and common experience. That means if we consider the thermodynamics

Basic Mechanical Engineering

98

Basic Mechanical Engineering

as a table then the legs are nothing but the laws of nature which are sometimes observed in nature by day-to-day experience. Now let us go through the laws of thermodynamics.

5.2

BASICS OF THERMODYNAMICS

5.2.1 Basic Definitions Thermodynamic System: It is defined as a definite area or space where some thermodynamic process is taking place and it is a region where our attention is focused for studying a thermodynamic process. Boundary: It is defined as the envelope which contains the system. The boundary may be fixed like that of a tank enclosing a certain mass of compressed gas or movable like boundary of certain volume of liquid in a pipe. Control Volume: A control volume is a fixed region in space chosen for the thermodynamic study of mass and energy balances for flowing system. The boundary of the control volume may be a real or imaginary envelope. Pure Substance: It is a single substance or mixture of substance which has the same consistent composition throughout, or it is a homogeneous substance and its molecular structure does not vary, e.g., steam, water, air and mixture of water and steam. Steady State: It is that circumstance in which there is no accumulation of mass or energy within the control volume, and the properties at any point within the system are independent of time. Phase: It is the homogeneous part of the substance having the same intensive properties it can be solid, liquid or gas (vapour). Thermodynamic Process: It is a path in which the state of method changes and some properties vary from long-established references. Defined by means of achange in the process, a thermodynamic method is a passage of a thermodynamic process from an initial to a ultimate state of thermodynamic equilibrium. The preliminary and final states are the defining states of the process. The specific direction of the system shouldn’t be the main drawback, and probably is left out. Path P

V = const.

V

Figure 5.1 Outline of thermodynamic process.

Thermodynamics 99

Cycle: Cycle is a number of process in sequence which brings the system to where it started. Properties of a thermodynamic process depend on the thermodynamic state and thus stay same over a cycle. Variables such as heat and work depend on the process hence are not zero over a cycle. a P b

V

Figure 5.2 Thermodynamic cycle.

Volume: Volume is the region or space needed by the system and is measured in m3. In thermodynamics, the volume of parameter for describing the thermodynamic state and is an important extensive property. Density (r): Density is defined as the total mass of a substance divided by the total volume occupied by it. It is measured in (kg/m3). r=

M kg/m 3 V

Specific Volume (v): The specific volume of a substance is defined as the ratio of the volume of a substance to its mass. It is the reciprocal of density. Specific volume can also be stated as the number of cubic metres occupied by one kilogram of a particular substance. It is an intrinsic property of matter. The standard unit is the cubic metre per kilogram It is measured in m3/kg. v=

V m3 M kg

Energy: Energy is defined as the ability of a system to do work or produce heat. However, it’s important to keep in mind that the existance of enegy is not only the single criterion for the work to be done. There are different types of energy: Potential Energy (P.E.): The energy possessed by a body by virtue of its position relative to others is termed potential energy. Example is the gravitational potential energy of an object which depends on its mass and distance from the centre of mass of another object. The unit of energy in the (SI) unit is the joule, which has the symbol J. P.E. = mgz ( J)

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where, m = mass (kg), g = acceleration due to gravity (m/sec2) z = height of body (m) Kinetic Energy (K.E.): Kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to the state of rest. K.E. = mv2 ( J) where, m = mass of the object (kg) v = speed of body (m/s) Internal Energy (U): In thermodynamics, the internal energy of a system is the energy contained within the system, including the kinetic and potential energy as a whole. It keeps account of the gains and losses of energy of the system that are due to changes in its internal state. The internal energy of a system can be changed by transfer of matter and by work and heat transfer. When matter transfer is prevented by impermeable containing walls, the system is said to be closed. Enthalpy (H): It is the measure of energy and is equal to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume. It also involves the internal energy which is the energy required to create space for it by displacing its environment and establishing its volume and pressure. Enthalpy is defined as a state function depending only on the existing state of equilibrium and is identified by the variables: internal energy, pressure, and volume. It is an extensive quantity. The (SI) unit for measurement of enthalpy is the joule. It is a property like pressure, temperature and volume but it cannot be measured directly. It is given by the relation, H = u + pV where, u = specific internal energy (J) of the system being studied p = pressure of the system (Pa) and V = specific volume of the system (m3/kg) Heat: Heat is energy in transit. Heat transfer generally occurs at the molecular level because of the temperature difference. Heat is denoted by Q. A net positive heat indicates that heat is added to the system and negative value indicates that heat is rejected from the system.

Thermodynamics 101

Work: Work is defined for mechanical system as the action of a force on an object through a distance. It equals the product of the force times the displacement (dL). Work is a form of energy, but it is energy in transit. Work is not a property of a system, work is a process done by or on a system, but a system contains no work. Work is a path function. W = PAdL since

AdL = dV = change in volume W = PdV

where, P = pressure of the gas A = area of the piston dV = change in volume

5.2.2 Classification of Thermodynamic System Closed System: The system which has a closed boundary and does not allow the transfer of mass between the system and its surroundings is known as closed system, i.e., it has constant mass, only energy is allowed to transfer across the boundary.

Figure 5.3 Closed system.

Open System: It is defined as a region is which the mass is not necessarily constant besides the mass as well as energy transfer cross its boundary.

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Fuel in

Boundary (control surface)

Drive shaft Exhaust gas out

Figure 5.4 Open system.

Isolated System: It is a thermodynamic system which has a fixed mass and neither mass nor energy crosses its boundary or it is a system which is not at all influenced by the surroundings.

5.2.3 Types of Thermodynamic Equilibrium Equilibrium: When a system is in equilibrium with regard to all the possible changes in state, the system is in thermodynamic equilibrium, the temperature will be the same throughout the system if the gas that comprises the system is in thermal equilibrium. (a) Isothermal equilibrium (Constant temperature) (b) Mechanical equilibrium (Constant pressure) (c) Chemical equilibrium (Constant composition)

5.2.4 Classification of Thermodynamic Properties Extensive Property: The properties of a system whose value for the entire system is equal to the sum of their value for the individual parts of the system. These properties are related to the mass of the system. Like volume, mass, all kinds of energy. Intensive Property: The properties of the system whose value for the entire system is not equal to the sum of their values for the individual parts of the system. These properties do not depend on the mass of the system. (pressure, temperature, specific volume and density).

5.2.5 Classification of Thermodynamic Substances Homogeneous Substances and Homogeneous System: A substance existing at a single phase is called a homogeneous substance. A homogeneous thermodynamic system is defined as the one whose chemical composition and physical properties are the same in all parts of the system, or change continuously from one point to another. A homogeneous system

Thermodynamics 103

can be exemplified by imagining a column of atmospheric air, which is a mixture of a number of gases, mainly nitrogen and oxygen. In a system of this kind, acted upon by the force of gravity, both the composition of the system and its physical properties will continuously change from one point to another. Heterogeneous Substances and Heterogeneous System: A substance that consists of two or more phases is called a heterogeneous substance. A heterogeneous system is defined as one consisting of two or more homogeneous bodies. The homogeneous bodies of a heterogeneous system are referred to as phases. Each phase is separated from other phases by interfaces, or boundaries, and in passing over such a boundary the chemical composition of the substance or its physical properties change abruptly. This phase boundary must not be regarded as a mathematical surface but as a thin layer separating the phases, a layer where the properties of one phase pass, or turn, rapidly into the properties of the other phase. An example of a heterogeneous system is water with ice floating in it. This system has two homogeneous bodies, water and ice. The chemical composition of the two phases is the same, but their physical properties differ. Another example of a heterogeneous system is the content of a sealed steel tube containing liquid mercury, liquid ethyl alcohol, and a mixture of saturated vapours of the alcohol and mercury. This heterogeneous system comprises three phases. The first phase is the liquid mercury, the second is the liquid ethyl alcohol, and the third phase is represented by the mixture of saturated vapours. Here the chemical compositions and the physical properties of all the phases are different.

5.3

LAWS OF THERMODYNAMICS

5.3.1 First Law of Thermodynamics It states that when a closed system undergoes a complete cycle the sum of heat interactions is equal to the sum of work interactions. Mathematically, SQ = SW (over the entire cycle) 1 B

Y

A

2

X

Figure 5.5 Thermodynamic cycle (first law of thermodynamics).

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‡ An isolated system does not interact with the surroundings hence Q = 0 and W = 0. Therefore, E remains constant for such a system. ‡ Let us reconsider the cycle 1-2 along path A and 2-1 along path B as shown in Fig. 5.5. Work done during the path A = Area under 1-A-2-3-4. (Fig. 5.6) ‡ Work done during the path B = Area under 1-B-2-3-4. Since these two areas are not equal, the net work interaction is that shown by the shaded area. ‡ The net area is 1A2B1. Therefore, some work is derived by the cycle. ‡ First law compels that this is possible only when there is also heat interaction between the system and the surroundings. ‡ In other words, if you have to get work out, you must give heat to get work out, you must give heat in.

Figure 5.6 Heat and work (second law of thermodynamics).

5.3.2 Second Law of Thermodynamics The second law of thermodynamics states that a process occurs in a certain direction, not just in any direction. The entropy of the universe is always increasing. It is used to determine the maximum efficiency of any process. A comparison can then be made between the maximum possible efficiency and the actual efficiency obtained. One of the most important areas in second law is the study of power-conversion systems. For example, it’s not possible to transform all of the energy got from a nuclear reactor into electrical power. There ought to be losses in the conversion process. The second law can be utilized to derive an expression for the maximum possible energy conversion taking these losses into consideration. For this reason, the second law denies the possibility of completely changing all the energy to work in a cycle, regardless of

Thermodynamics 105

how flawlessly designed the method is. The second law of thermodynamics is required because the first law of thermodynamics does no longer define the energy conversion system fully. DS =

DQ T

where, DS = entropy change, DQ = heat transferred T = temperature

5.3.3 Third Law of Thermodynamics The third law of thermodynamics deals with the absolute zero temperature. It states that the entropy of a system reaches a constant value as the temperature approaches absolute zero. The entropy of a system at absolute zero is typically zero, and in all cases it is determined only by the number of different ground states it has. Specifically, the entropy of a pure crystalline substance at absolute zero temperature is zero.

5.3.4 Zeroth Law of Thermodynamics The Zeroth law of thermodynamics states that if two bodies are each in thermal equilibrium with a third body, then they are also in equilibrium with each other. Thermal equilibrium means that when two bodies are brought into contact with each other and separated by a barrier that is permeable to heat, there will be no transfer of heat from one to the other.

Figure 5.7 Zeroth law of thermodynamics.

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Basic Mechanical Engineering

PROPERTIES OF PURE SUBSTANCES

5.4.1 Pure Substance A substance that has a fixed or constant chemical composition throughout is known as a pure substance, for example, water, air and nitrogen. A pure substance does not need to be of a single component or compound. A combination of two or more phases of a pure substance is still a pure substance as long as the chemical composition of the substance in all the phases is same. Pure substances show very much characterized physical properties, or properties that are not associated with the substance’s capacity to join with various substances. The temperatures where pure solids melt, known as melting points, are especially sharp, which means the softening happens at a fixed temperature. Likewise, the temperatures where pure fluids start to bubble, or breaking points, happen at fixed temperatures when different elements, similar to pneumatic force like air pressure, are controlled.

5.4.2 Phases of a Pure Substance A pure substance may exist in various phases. There are three main phases of pure substances, namely, solid, liquid and gas. A section is said to have a certain molecular arrangement that’s homogenous all through and isolated from others (assuming any) through effortlessly identifiable boundaries. A substance will have a couple of (one or more than one) phases within the foremost section, each with an alternative atomic structure. For illustration, carbon may just exist as graphite or diamond in the stable section, and ice could exist in seven special phases at excessive pressure. Atomic bonds are the most strong in solids and the weakest in gases. Solid Phase: The atoms are organized in a 3-dimensional sample all by means of the stable bond. The atoms can’t move in respect to each other; nonetheless, they regularly oscillate about their imply role. Liquid Phase: The atomic arrangement in liquid phase is quite the same as that of the solid phase (for the most part marginally higher), apart from the particles are no more at a fixed position in respect to each other. Gaseous Phase: The particles are far separated from each other, and a subatomic order does not exist. Atom of gaseous phase moves arbitrarily and collides with each other and with the boundary or wall of the compartment they are in. Atoms in the gas stage are generally at a higher energy level than they are in liquid or solid phases.

5.4.3 Phase-Change Processes of Pure Substances Consider a procedure where a pure substance begins as a solid and is warmed up at consistent pressure until everything changes into gas. Depending on the pressure acting, the matter will go through different phase changes:

Thermodynamics 107

T

ate

tur

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Sa

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Critical point

dv

Fusion

iqui



P0

ap

ed l



r li

ou

line

urat



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(a) Solid (b) Mixed phase of liquid and solid (c) Sub-cooled or compacted liquid (means it is not going to vaporize) Wet vapour or saturated liquid-vapour mixture, the temperature will stop rising until the liquid is totally vaporized. (d) Superheated vapour (a vapour that is not going to condense). The diagram is the T-V and P-V diagram for the heating method of a pure substance. ‡ $WDJLYHQSUHVVXUHWKHWHPSHUDWXUHDWZKLFKDSXUHVXEVWDQFHVWDUWVERLOLQJ is known as the saturation temperature, Tsat. ‡ /LNHZLVH DW D JLYHQ WHPSHUDWXUH WKH SUHVVXUH DW ZKLFK D SXUH VXEVWDQFH begins to boil is referred to as the saturation pressure, Psat. ‡ 7KURXJKRXWDVHFWLRQDOWHUQDWHSURFHGXUHVWUHVVDQGWHPSHUDWXUHDUHGLUHFWO\ proportional, Tsat = f (Psat). ‡ 7KH UHOHYDQW IDFWRU LV WKH SRLQW DW ZKLFK WKH OLTXLG DQG YDSRXU SKDVHV DUH not distinguishable. The “triple point” is the factor at which the liquid, solid, and vapour phases can exist collectively. On P-V or T-V diagrams, these triple-segment form a kind of line known as the triple point line.

G L S

S+L 2

3

L+G

5

4 Triple point line

1 S+G

V

Figure 5.8 T-V diagram for heating process of a pure substance.

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Basic Mechanical Engineering P Critical point Saturated vapour line Compressed Liquid Region

Superheated Vapour Region

T2 = const. > T1

Saturated LiquidVapour Region

Saturated liquid line

T1 = const

V

Figure 5.9 P-V Diagram of a pure substance.

5.4.4 The P-T or Phase Change Diagram This is called phase diagram since all three phases are separated from each other by three lines. Most pure substances exhibit the same behaviour. One exception is water. Water expands upon freezing. P Substances that contract on freezing

Substances that expand on freezing

Condensation

Melting Liquid

ltin g Me

lting

Me

Freezing

Solid

Va

ri po

za

tio

Critical point

n

Vaporization

Triple point Sublimation

Su

blim

ati

on

Vapour

Sublimation

T

Figure 5.10 Phase change diagram of a pure substance.

Thermodynamics 109

There are two ways that a substance can pass from solid phase to vapour phase. (a) It melts first into a liquid and subsequently evaporates. (b) It evaporates directly without melting (sublimation).  ‡ 7KHVXEOLPDWLRQOLQHVHSDUDWHVWKHVROLGDQGWKHYDSRXU  ‡ 7KHYDSRUL]DWLRQOLQHVHSDUDWHVWKHOLTXLGDQGYDSRXUUHJLRQV  ‡ 7KH PHOWLQJ RU IXVLRQ OLQH VHSDUDWHV WKH VROLG DQG OLTXLG 7KHVH WKUHH OLQHV meet at the triple point.

5.5

SUMMARY

Skills of thermodynamics are required to design any device involving the interchange between heat and work, or the conversion of material to produce heat (combustion). Thermodynamics is the gain knowledge of the connection between work, heat, and energy. It offers with the conversion of heat from one type to another. It also deals with the interplay of a process and its environment. Method identifies the subject of the analysis by defining a boundary and environment.

SOLVED EXAMPLES

Example 5.1 One mole of an ideal gas is heated at constant pressure from 0oC to 200oC. Calculate the work done.

Solution: Work done during heating of gas from 0oC to 200oC Here, T1 = 0oC = 273 K T2 = 200oC = 473 K Now, W = – PDV = – P(V2 – V1) = – P [(nRT2/P) – (nRT1/P)] = – nR (T2 – T1) = – 1 × 1.987 × (473 – 273) = – 397.4 cal Example 5.2 Calculate the final volume of one mole of an ideal gas initially at 0oC and 1 atm pressure, if it absorbs 2000 cal of heat during reversible isothermal expansion.

Solution: The gas is in the standard temperature and pressure condition, i.e., at S.T.P. Hence, V1 = 22.4 dm3 and V2 have to be calculated.

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As the given expansion is isothermal and reversible therefore, DU = 0 We know that DU = q + w But DU = 0 Hence, q = – w = 2000 cal = 2000 × 4.184 J = 8368 J As work done is reversible, isothermal expansion is given by: W = – nRT ln (V2/V1) Therefore, nRT ln (V2/V1) = – w = 8368 J (1 mol) × (8.314 J K – mol–) × (273 K) ln (V2/ 22.4 dm3) = 8368 J V2 = 242.50 dm3 Hence, the final volume of one mole of an ideal gas initially at 0oC and 1 atm pressure is equal to 242.50 dm3.

Example 5.3 Several proposals have been made for building a heat engine that makes use of the temperature differential between the surface waters of the ocean and cooler waters that, being more dense, reside at greater depth. If the exhaust temperature is 5°C, what is the maximum amount of work that could be extracted from 1000 L of surface water at 10°C? (The specific heat capacity of water is 4.184 J g–1K–1.) Solution:

The amount of heat (qH) that must be extracted to cool the water by 5 K is

(4.184 J g–1K–1) (106 g) (5 K) = 2.09 × 107 J. The ideal thermodynamic efficiency is given by 1-

278 = 0.018 283

The amount of work that could be done would be (0.018) (2.09 × 107 J) = 3.7 × 106 J

Example 5.4

A mass m of fluid at temperature T1 is mixed with an equal amount of the same fluid at temperature T2. Prove that the resultant change of entropy of the universe is: [2 mc ln (T1 + T2)/2]/[(T1 T2)1/2] and also prove that it is always positive.

Solution: Mean temperature of the mixture = (T1 + T2)/2 Thus, change in entropy is given by:

Thermodynamics 111

DS = S2 – S1) = mc (T1 + T2)/2 Ú T1 (dT/T) – mc T2 Ú T1 + T2)/2 (dT/T) = mc ln (T1 + T2)/2 T1) – mc ln (2 T2)/(T1 + T2) = mc ln (T1 + T2)/2 T1) + mc ln (T1 + T2)/2 T2) = mc ln (T1 + T2)2/4 T1 T2 = mc ln [(T1 + T2)/2(T1 T2)1/2]2 = 2 mc ln [(T1 + T2)/2 (T1 T2)1/2] = 2 mc ln [(T1 + T2)/2 ]/[(T1 T2)1/2] Hence, resultant change of entropy of universe is: 2 mc ln [(T1 + T2)/2]/[(T1 T2)1/2] The arithmetic mean (T1 + T2)/2 is greater than the geometric mean (T1 T2)1/2 Therefore, ln [(T1 + T2)/2]/[(T1 T2)1/2 ] is always positive. Hence, the entropy of the universe increases.

Example 5.5

A Carnot’s engine whose low temperature reservoir is at 27°C has an efficiency of 40%. What should be the temperature of high temperature reservoir?

Solution:

or

We know that h=

T1 - T2 T1

h=

T1 - T2 T1

or

0.4 T1 = T1ï

or

0.6T1 = 300

or

T1 = 500 K

Example 5.6 Apparatus that liquefies helium is in a laboratory at 296 K. Helium in the apparatus is at 4 K. If 150 mJ of heat is transferred from helium, find the minimum amount of heat delivered to the laboratory? Solution: To obtain the minimum amount of heat delivered to the laboratory, substitute 150 mJ for QL, 296 K for TH and 4.0 K for TL in the equation QH = QL (TH/TL),

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QH = QL(TH/TL) = ((150 mJ) (10–3 J/1 mJ)) (296 K/4.0 K) = 11 J From the above observation we conclude that, the minimum amount of heat delivered to the laboratory would be 11 J.

Example 5.7

What is the significance of Zeroth law of thermodynamics?

Solution: The science of measurement of temperature is called thermometry. Here, body C is used as thermometer to measure the temperature of the bodies. A thermometric property is found, which changes with temperature is found out first and this body C (thermometer) is used to find out the temperature.

Example 5.8:

Tea is a mixture of one or more tasty oils dissolved in hot water. Is this a heterogeneous or homogeneous mixture? Are there any conditions under which this tea can be considered to be a pure substance? Explain.

Solution: If the tea is well mixed, perhaps by stirring, then the mixture can be adequately described as homogeneous. As long as the composition remains same and uniform throughout, a homogeneous mixture can be treated as a pure substance.

Example 5.9 (i) (ii) (iii) (iv)

Calculate the following: the kinetic energy of a body which has a mass of 5 kg and a velocity of 10 m/s; the change in potential energy of a mass of 5 kg when it is raised to a height of 3 m; the strain energy stored in a spring compressed by 18 mm from its free length if the spring constant is 1.50 MN/m; the increase in internal energy of a gas in a closed system during a process in which – 100 J of heat transfer and 400 J of work transfer take place.

Solution: (i) Kinetic energy =

1 1 mc 2 = ¥ 5 ¥ 100 = 250 J 2 2

(ii) Change in potential energy = Mgz = 5 × 9.81 × 3 = 147 J (iii) Strain energy stored =

1 2 kx 2

1 ¥ 1.5 ¥ 106 ¥ 0.018 2 = 243 J 2 (iv) Increase in internal energy = U2 – U1 First law: W + Q = U2 – U1 = 400 – 100 = 300 J

=

Thermodynamics 113

Example 5.10 The piston in figure has no mass and is free to move. The point force F acting on the piston is applied gently so that it moves slowly to the left. As the gas in the cylinder is compressed, the product of pressure and system volume, pV, remains constant. The piston area is 0.02 m2 and when x is x1 the system volume is 0.006 m3. Atmospheric pressure is 1 bar. Calculate the distance (x2 – x1) moved by the piston as F increases from 0 to 70 N. Solution:

Equilibrium of forces on piston: p2 A = 70 + paA p2 =

70 + 10 5 = 1.385 ¥ 10 5 Pa 0.02

Given pV = constant; p1 V1 = p2 V2 V2 =

p1V1 10 5 ¥ 0.006 = 1.385 ¥ 10 5 = 0.00433 m 3 p2 1.385 ¥ 10 5 x1

x2 l2 l1

V2

Distance moved by piston: x2 – x1 = l1 – l2 =

V1 - V2 A

= 0.3 – 0.2166 = 0.0834 m or 83.4 mm

Example 5.11 (i) Calculate the efficiency of a reversible heat engine operating between a hot reservoir at 900 K and a cold reservoir at 500 K. (ii) The temperature of one of the heat reservoirs can be changed by 100 degrees kelvin up or down. What is the highest efficiency that can be achieved by making this temperature change?

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Solution: (i) h = 1 -

T2 500 =1= 44.44% T1 900

(ii) The largest reduction in T2/T1 is produced by subtracting the temperature change from the lower temperature. h = 1-

400 = 55.55% 900

Example 5.12 A perfect gas at a pressure of 58 bars and a temperature of 450 K has a density of 50 kg/m3. The ratio of specific heats g is 1.48. ~, specific heat at constant pressure cp and (i) Calculate the values of molar mass m specific heat at constant volume cv. (ii) Calculate the change in specific entropy of the gas if the pressure is raised to 100 bars and the temperature is lowered to 400 K. Solution: (i) Perfect gas,

\

R=  = m

p = RT r

p 58 ¥ 10 5 = = 257.77 J/kg K rT 50 ¥ 450 R 8.314 ¥ 10 3 = = 322.5 kg/kmol R 257.77

cp =

gR 1.48 = ¥ 257.77 = 794.79 J/kg K g - 1 0.48

cv =

R 257.77 = = 537.02 J/kg K g -1 0.48

(ii) Specific entropy change s2 – s1 = cp ln

p T2 - R ln 2 T1 p1

Ê 400 ˆ Ê 100 ˆ - 257.8 ln Á = 794.8 ln Á Ë 450 ˜¯ Ë 58 ˜¯

= 234.0 J/kgK

Thermodynamics 115

Example 5.13 A perfect gas with a g value of 1.5 can be compressed in either a non-flow or a steady flow process between the same initial and final states. Neglecting any change in kinetic and potential energy, what is the ratio of the specific work done is the two cases if the process is (i) polytropic with an index n of 1.4; (ii) isothermal? (iii) isentropic?

Solution: Ê Wflow ˆ Ratio Á is : Ë Wnon-flow ˜¯ (i) 1.4 (ii) 1 (iii) 1.5

EXERCISE 1. What is entropy and what are its characteristics? 2. Derive entropy of a mixture of ideal gases. 3. An iron cube at a temperature of 400°C is dropped into an insulated bath containing 10 kg water at 25°C. The water finally reaches a temperature of 50°C at steady state. Given that the specific heat of water is equal to 4186 J/kgK. Find the entropy changes for the iron cube and the water. Is the process reversible? If so why? 4. Calculate the efficiency of Carnot’s engine working between 100°C and 0°C. 5. Define the third law of thermodynamics. Discuss its importance. 6. What is irreversible process? Give an example of it. 7. Fog is huge number of liquid water droplets suspended in air. Can fog be considered to be a pure substance? Do you think fog is heterogeneous or homogeneous? Explain. 8. If water vapour condenses on the outside of a cold glass of water, the internal energy of the water vapour has decreased, by an amount equal to the heat of vaporization of the water vapour. Heat energy has left the water vapour, causing it to condense, and heat energy has entered the glass of water, and the air, causing them to get slightly warmer. No work is done, but heat is exchanged.

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9. This definition of efficiency is not useful, because with this definition, if the exhaust heat QL is less than the work done W (which is possible), the “efficiency” would exceed unity. Efficiency should be compared with the heat input, not the heat output. 10. A container holds a mixture of three non-reacting gases: n1 moles of the first gas with molar specific heat at constant volume C1, and so on. Find the molar specific heat at constant volume of the mixture, in terms of the molar specific heats and quantities of the three separate gases. 11. A thermometer of mass 0.055 kg and heat capacity 46.1 J/K reads 15.0°C. It is then completely immersed in 0.300 kg of water and it comes to the same final temperature as the water. If the thermometer reads 44.4°C, what was the temperature of the water before insertion of the thermometer, neglecting other heat losses?

6 Heat Transfer

Learning Objectives     

‡ ‡ ‡ ‡ ‡

6.1

8QGHUVWDQGLQJWKHWHUPVKHDWDQGWHPSHUDWXUH 2YHUYLHZRIPRGHVRIKHDWWUDQVIHU &LWHWKHFRQFHSWVRIFRQGXFWLRQFRQYHFWLRQDQGUDGLDWLRQ 6WDWHPHQWVDQGH[SUHVVLRQVRIODZVRIKHDWWUDQVIHU ,QWURGXFWLRQWRKHDWH[FKDQJHUV

INTRODUCTION

Conduction heat transfer phenomena are found throughout virtually all of the physical world and the industrial domain. The analytical description of this heat transfer mode is one of the best understood ones. Some of the bases of understanding of conduction date back to early history. It was recognized that by invoking certain relatively minor simplifications, mathematical solutions resulted directly. Some of these were easily formulated. What transpired over the years was a vigorous development of applications to a broad range of processes. Perhaps no single work better summarizes the wealth of these studies than does the book by Carslaw and Jaeger (1959). They gave solutions to a broad range of problems, from topics related to the cooling of the earth to the currentcarrying capacities of wires. The general analyses given there have been applied to a range of modern-day problems, from laser heating to temperature-control systems. Today, conduction heat transfer is still an active area of research and application. A great deal of interest has developed in recent years in topics like contact resistance, where a temperature difference develops between two solids that do not have perfect

Basic Mechanical Engineering

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contact with each other. Additional issues of interest include non-Fourier conduction. Also, the problems related to transport in miniaturized systems are generating a great deal of interest. Increased interest has also been directed to ways of handling composite materials, where the ability to conduct heat is very directional. Much of the work in conduction analysis is now accomplished by use of sophisticated computer codes. These tools have given the heat transfer analyst the capability of solving problems in nonhomogenous media, with very complicated geometries, and with involved boundary conditions. It is still important to understand analytical ways of determining the performance of conducting systems. At the minimum these can be used as calibrations for numerical codes. The second law of thermodynamics states that heat transfer takes place from the body at higher temperature to body at lower temperature. In heat exchange issues, we regularly reciprocally utilize the terms heat and temperature. Temperature is a measure of the measure of vitality controlled by the particles of a substance. It shows itself as a level of hotness, and can be utilized to foresee the heating of heat exchange. The standard notation for temperature is T. The scales for measuring temperature in SI units are the Celsius and Kelvin temperature scales. Heat, on the other hand, is vitality in travel. Suddenly, heat spills out of a more smoking body to a colder one. The standard notation for heat is Q. In the SI framework, normal units for measuring heat are the joule and calorie. There are three modes of heat transfer: 1. Conduction 2. Convection 3. Radiation

Figure 6.1 Modes of heat transfer.

Heat Transfer 119

6.2 6.2.1

BASIC DEFINITIONS Conduction

It is the mode of transfer of heat from one substance to another when they are in physical contact. It is a microscopic phenomenon which occurs due to intermolecular interaction. Conduction needs matter and does not require any mass movement of matter. Here the carriers of heat are free electrons. In liquid and gas heat is conducted by collisions and diffusion. Conduction is transfer by means of solids or stationery fluids. When you contact a hot object, the heat you believe is transferred by means of your skin through conduction. Two mechanisms provide an explanation for how heat is transferred by way of conduction, i.e., lattice vibration and particle collision. Conduction through solids occurs through a blend of the two mechanisms; heat is conducted through stationery fluids in particular by using molecular collisions. In solids, atoms are bound to each other via a sequence of bonds, analogous to springs. When there is a temperature difference in the solid, the hot side of the spring experiences extra atomic actions. The vibrations are transmitted by means of the springs to the cooler side of the stable. Finally, they attain an equilibrium, where all of atoms are vibrating with the equal energy. Solids, chiefly metals, have free electrons, which are not sure to any unique atom and may freely move in regard to the solid. The electrons within the hot part of the solid move more rapidly than those on the cooler aspect. As the electrons bear a series of collisions, the turbo electrons supply some of their energy to the slower electrons. Eventually, by way of a series of random collisions, an equilibrium is reached, the place the electrons are relocating at the equal velocity. Conduction via electron collision is more effective than by means of lattice vibration; because of this metals probably are better heat conductors than ceramic materials, which wouldn’t have many free electrons, e.g., heat transfer in metal rod.

Figure 6.2 Conduction.

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6.2.2 Convection It is the mode of transfer of heat from one particle to another particle of the body by convection current. It is macroscopic phenomenon generally seen in fluids (liquids and gases) when heat is transferred through a gas or liquid by the hotter material moving into a cooler area, e.g., ice melting. Heat moves to the ice from the air. This causes melting from a solid to liquid. Hot air balloon — A heater inside the balloon warms the air thus the air moves upwards. This causes the balloon to rise in light of the fact that the hot air gets caught inside. At the point when the pilot wants to come down, he discharges a portion of the hot air and cool air assumes its position, causing the balloon to lower. Convection makes use of the movement of fluids to switch heat. In traditional convective heat transfer, a hot floor heats the surrounding fluid, which is then carried away by way of fluid action such as wind. The warmer fluid is changed via cooler fluid, which is able to draw extra heat far from outside. Considering the heated fluid is constantly replaced by cooler fluid, the cost of heat transfer is better. Traditional convection (or free convection) refers to a case when the fluid action is created by means of the warmer fluid itself. The density of fluid diminishes as it is heated; thus, hot fluids are lighter than cool fluids.

Figure 6.3 Melting of ice (convection).

Figure 6.4

Hot air balloon.

6.2.3 Radiation It is the mode of heat transfer in which energy is continuously emitted in form of electromagnetic waves from the surface of all bodies. It is microscopic phenomenon. Radiation does not need matter.

Heat Transfer 121

Radiative heat transfer does not require a medium to move through; this is why it is one of the most effective forms of heat transfer in vacuum. It uses electromagnetic radiation (photons), which travels at the speed of light and is emitted by using any matter with temperature above zero Kelvin (– 273°C). Radiative heat transfer occurs when the emitted radiation strikes a different physique and is absorbed. We all experience radiative heat exchange every day; our skin can absorb solar radiation which is the reason why we feel warmer in sun than in night. The electromagnetic spectrum classifies radiation consistent with wavelengths of the radiation. Fundamental types of radiation DUH IURP VKRUW WR OHQJWK\ ZDYHOHQJWKV  JDPPD UD\V ;UD\V XOWUDYLROHW 89  YLVLEOH spectrum, infrared (IR), microwaves, and radio waves. Radiation with shorter wavelengths are more energetic and includes extra heat. X-rays, having wavelengths ~10–9 m, are vigorous and can be detrimental to people, whilst noticeable light with wavelengths ~10–7 m include less vigour and hence have little effect on lifestyles. A second characteristic is that radiation with longer wavelengths frequently can penetrate thicker solids. Obvious light, as we all know, is blocked by a wall. As we know that radio waves have wavelengths on the order of meters that’s why it can pass effectively through concrete walls, e.g., heat of sun reaches to us through radiation.

Figure 6.5 Radiation.

6.2.4

Black Body Radiation

The total amount of radiative energy emitted from a surface into all directions above it is termed emissive power. We distinguish between spectral (at a given wavelength l, per unit wavelength) and total (encompassing all wavelengths) emissive power. The magnitude of emissive power depends on wavelength l, temperature T, and a surface property, called emissivity e, which relates to the ability of a surface to emit radiative energy to that of an ideal surface, which emits the maximum possible energy (at a given wavelength and temperature). Such an ideal surface is known as a “black body” or “black surface,” since it absorbs all incoming radiation; i.e., it reflects no radiation and

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is, therefore, invisible (“black”) to the human eye. The spectral distribution of the emissive power of a black surface is given by Planck’s law. C1 , C1 = 3.7419 × 10–16 Wm2, C2 = 14,388 mmK Eb l = 5 C2 /lT l [e - 1] where C1 and C2 are sometimes called Planck function constants.

6.3

BASIC LAWS OF HEAT TRANSFER

6.3.1

Fourier’s Law of Heat Conduction

It states that heat flow dQ through an area dA in a plane normal to the heat transfer in time dT and is given by Q = kA Q A dT dx k

= = = = =

dT dt dx

Amount of heat flow through the body per unit time Surface area of heat flow Temperature difference on the two faces of the body Thickness of the body Proportionality constant

6.3.2 Newton’s Law of Cooling It states that the heat transfer from a hot body to cold body is directly proportional to the surface area and the difference of temperatures between the two bodies and is given by Q = hA (ts – tf ) h=  = Q A= ts = tf = h=

 Q W/m 2 C A(ts - t f )

Rate of convective heat transfer Area exposed to heat transfer Surface temperature Fluid temperature Coefficient of convective heat transfer

6.3.3 Stefan-Boltzmann Law Stefan-Boltzmann law depicts gross heat outflow as opposed to heat exchange. The expression for the real radiation heat exchange rate between surfaces having complex orientation can be quite complex. This law states that the emissive power of a black

Heat Transfer 123

body (i.e., the total radiation emitted by a black body per unit area and time is directly proportional to the fourth power of the absolute temperature. Eb = sT 4 Stefan-Boltzmann constant (s) = 5.67 × 10–8 W/m2 K4 T = Absolute temperature The rate of radiation heat transfer between a small surface and an extensive surrounding is given by: 4 4 q = e ¥ s ¥ A ¥ (Ts - Tsur ) where e is emissivity.

6.4

INSULATIONS

Insulations are used to decrease heat flow and surface temperatures. These materials are found in a variety of forms, typically loose fill, batt, and rigid. Even a gas, like air, can be a good insulator if it can be kept from moving when it is heated or cooled. A YDFXXPLVDQH[FHOOHQWLQVXODWRU8VXDOO\WKRXJKWKHHQJLQHHULQJDSSURDFKWRLQVXODWLRQ is the addition of a low-conducting material to the surface. While there are many chemical forms, costs, and maximum operating temperatures of common forms of insulations, it seems that when a higher operating temperature is required, many times the thermal conductivity and cost of insulation will also be higher. Loose-fill insulations include such materials as milled alumina-silica (maximum operating temperature of 1260°C and thermal conductivities in the range of 0.1 to 0.2 W/m2 K) and perlite (maximum operating temperature of 980°C and thermal conductivities in the range of 0.05 to 1.5 W/m2 K). Batt-type insulations include one of the more common types — glass fibre. This type of insulation comes in a variety of densities, which, in turn, have a profound effect on the thermal conductivity. Thermal conductivities for glass fibre insulations can range from 0.03 to 0.06 W/m2K. Rigid insulations show a wide range of forms and performance characteristics. For example, a rigid insulation in foam form, polyurethane, is lightweight, shows a low thermal conductivity (about 0.02 W/m2 K), but has a maximum operating temperature only up to about 120°C. Rigid insulations in refractory form show quite different characteristics. For example, high-alumina brick is quite dense, has a thermal conductivity of about 2 W/m2 K, but can remain operational to temperatures around 1760°C. Many insulations are characterized in the book edited by Guyer (1989). Often, commercial insulation systems designed for high-temperature operation use a layered approach. Temperature tolerance may be critical. Perhaps a refractory is applied in the highest temperature region, an intermediate-temperature foam insulation is used in the middle section, and a high-performance, low temperature insulation is used on the outer side near ambient conditions. Analyses can be performed including the effects of temperature variations of thermal conductivity. However, the most frequent approach is to assume that the thermal conductivity is constant at some temperature between the two extremes experienced by the insulation.

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Basic Mechanical Engineering

HEAT EXCHANGERS

The two major categories of heat exchangers are shell-and-tube exchangers and compact exchangers. Basic constructions of gas-to-gas compact heat exchangers are plate-fin, tube-fin and all prime surface recuperators (including polymer film and laminar flow exchangers), and compact regenerators. Basic constructions of liquid-to-liquid and liquidto-phase-change compact heat exchangers are gasketed and welded plate-and-frame, welded stacked plate (without frames), spiral plate, printed circuit, and dimple plate heat exchangers. Shell-and-tube exchangers are custom designed for virtually any capacity and operating condition, from high vacuums to ultrahigh pressures, from cryogenics to high temperatures, and for any temperature and pressure differences between the fluids, limited only by the materials of construction. They can be designed for special operating conditions: vibration, heavy fouling, highly viscous fluids, erosion, corrosion, toxicity, radioactivity, multicomponent mixtures, etc. They are made from a variety of metal and nonmetal materials, and in surface areas from less than 0.1 to 100,000 m2 (1 to over 1,000,000 ft2). They have generally an order of magnitude less surface area per unit volume than the compact exchangers, and require considerable space, weight, support structure, and footprint. Compact heat exchangers have a large heat transfer surface area per unit volume of the exchanger, resulting in reduced space, weight, support structure and footprint, energy requirement and cost, as well as improved process design, plant layout and processing conditions, together with low fluid inventory compared with shell-and-tube exchangers. From the operating condition and maintenance point of view, compact heat exchangers of different constructions are used for specific applications, such as for high temperature applications (up to about 850°C or 1550°F), high pressure applications (over 200 bars), and moderate fouling applications. However, applications do not involve both high temperature and pressure simultaneously. Plate-fin exchangers are generally brazed, and the largest size currently manufactured is 1.2 × 1.2 × 6 m (4 × 4 × 20 ft). Fouling is one of the major potential problems in many compact exchangers except for the plate heat exchangers. With a large frontal area exchanger, flow maldistribution could be another problem. Because of short transient times, a careful design of controls is required for startup of some compact heat exchangers compared with shell-and-tube exchangers. No industry standards or recognized practice for compact heat exchangers is available. Heat exchangers are gadgets that facilitate the trade of heat between two fluids which can be at one-of-a-kind temperatures whilst preserving them from mixing with every different. Heat exchangers are more commonly used for various purposes, from heating and air conditioning programmes, to chemical processing and energy production.

Heat Transfer 125

Figure 6.6 Heat exchanger.

A heat exchanger is a gadget that is utilized to exchange heat energy (enthalpy) between at least two liquids, between a strong surface and a liquid, or between strong particulates and a liquid, at various temperatures and in heat contact. In heat exchangers, there are typically no outside heat and work collaborations. Average applications include heating or cooling of a liquid stream of concern and dissipation or buildup of single- or multi-part liquid streams. In different applications, the goal might be to recoup or dismiss heat, or sanitize, purify, fractionate, distil, focus, take shape, or control a procedure liquid. In some heat exchangers, the fluids where heat is exchanged is in direct contact. In most heat exchangers, heat transfer between fluids takes place by way of a separating wall or into and out of a wall in a transient method. In many heat exchangers, the fluids are separated through a heat exchange plate, and ideally they don’t mix or leak. Such exchangers are referred to as direct transfer kind, or readily recuperators. Additionally, exchangers may transfer heat between the hot and cold fluids—via thermal power storage and liberate by means of the exchanger plate or matrix— are referred to as oblique switch form, or regenerators. Such exchangers often have fluid leakage from one fluid movement to the opposite, due to stress variations and matrix rotation/valve switching.

6.5.1

Classification of Heat Exchangers

Broad classification of the heat exchangers are as follows based on several factors.

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Figure 6.7(a) Classification based on transfer process, number of fluids and surface compactness.

Figure 6.7(b) Classification based on construction, flow arrangements and heat transfer mechanisms.

Heat Transfer 127

6.5.2 Basic Types of Heat Exchangers Double Pipe Heat Exchanger: Double pipe heat exchanger is probably one of the simplest configurations. It includes two concentric round tubes with one fluid flowing within the internal tube and the opposite fluid flowing inside the annular area between the tubes. Its fundamental uses are in cooling the fluids in the location where small heat transfer areas are required. It could be designed in a quantity of preparations similar to parallel-flow and counter-flow, and mixed in series or parallel preparations with other heat exchangers to form a process.

Figure 6.8 Double pipe heat exchangers.

Compact Heat Exchanger: Compact heat exchangers are often used in gasoline-togasoline and gas-to-liquid (or liquid-to-fuel) heat exchangers to counteract the low heat transfer coefficient related to fuel float with expanded gas glide with multiplied plate subject. The ratio of the heat transfer surface area of a heat exchanger to its volume is known as the field density. A heat exchanger with b = seven-hundred m2/m3 or (or 200 ft2/ft3) is labelled as being compact. Examples of compact heat exchangers are auto radiators (1000 m2/m3), glass ceramic gas turbine warmness exchangers (6000 m2/m3. Cross-flow Heat Exchanger: In compact heat exchangers, the two fluids more often than not transfer perpendicularly to one another and such waft configuration is known as cross-flow heat exanger. There are two types of cross-flow heat exchangers (a) unmixed and (b) mixed flow.

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Figure 6.9 Cross-flow heat exchangers.

Shell and Tube Heat Exchanger: Shell-and-tube heat exchangers contain a large number of tubes (sometimes several hundred) packed in a shell with their axes parallel to that of the shell. Baffles are commonly placed in the shell to force the shell-side fluid to flow across the shell to enhance heat transfer and to enhance heat transfer and to maintain uniform spacing between the tubes.

Figure 6.10 Shell and tube heat exchanger.

Fouling Factor: The deposition of scale on heat transfer bodies reduces the heat switch and develop the pressure drop and pumping power. The fouling reasons to be used within the design of heat exchangers are more commonly distinct by using the procedure to simulate dirt accumulation on the heat transfer surfaces, but when these are usually not restrained to appropriate phases they can utterly negate any benefits generated by means of skilful design. The fouling factor represents the theoretical resistance to heat

Heat Transfer 129

transfer because of a build-up of a layer of dirt or other fouling substance on the tube surfaces of the heat exchanger however in an attempt to minimise the frequency of cleaning. Definitely they may, if badly chosen, lead to extended cleaning frequency.

6.6

SUMMARY

Thermal energy is generally associated with the change in temperature. For a given material and mass the higher thermal energy will be of the material having higher mass. Heat transfer through a body or between bodies takes place when there is a difference in temperature of the two. When two bodies are at different temperatures, thermal energy transfers from higher temperature body to the one with lower temperature. Heat always moves from hot to cold.

SOLVED EXAMPLES

Example 6.1 The block of 304 stainless steel shown below is well insulated on the front and back surfaces, and the temperature in the block varies linearly in both the x- and y-directions, find the heat fluxes in the x- and y-directions. 5 cm 10°C

15°C 5 cm

10 cm 5ºC

0°C y x

Figure 6.11

Solution: Ax = 10 × 5 = 50 cm2 = 0.0050 m2 Ay = 5 × 5 = 25 cm2 = 0.0025 m2 Q = kA

dT dt = – 14.4 × (– 5/0.05) = 1440 W/m2 dX

Q = kA

dT dt = – 14.4 × (10/0.1) = – 1440 W/m2 dX

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Example 6.2 If a star’s surface temperature is 30,000 K, how much power does a square metre of its surface radiate? Solution: This is a simple question if you understand the Stefan-Boltzmann law, because the term sT 4 is just the power radiated per square metre. This star radiates, sT 4 = 5.67 × 10–8 × 300004 = 4.6 × 1010 Example 6.3 Heat is conducted through a material with a temperature gradient of – 9000°C/m. The conductivity of the material is 25 W/mK. If this heat is convected to surroundings at 30°C with a convection coefficient of 345 W/m2K, determine the surface temperature. If the heat is radiated to the surroundings at 30°C determine the surface temperature. Solution: In this case only convection and conduction are involved. kA

Considering the unit area,

dT = hA(T1 – T2) dx

– 25 × 1 × (– 9000) = 345 × 1 (T1 – 30) Therefore, T1 = 682.17°C In this case conduction and radiation are involved. Heat conducted = Heat radiated ÈÊ T1 ˆ 4 Ê T2 ˆ 4 ˘ – 25 × 1 × (– 9000) = 5.67 ÍÁ ˜ - ÁË ˜ ˙ 100 ¯ ˙˚ ÍÎË 100 ¯

Therefore, T1 = 1412.14 K = 1139°C

Example 6.4 Asbestos layer whose thickness is 20 mm is used as an insulator over a boiler wall. Consider an area of 0.9 m2 and calculate the rate of heat flow over this area if the temperatures on either side of the insulation are 400°C and 30°C. Given k = 0.116 W/mK Solution:

We know that q = - kA

dT dx

= – 0.116 × (30 – 400)/0.02 q = 1566 W/m2 Rate of heat flow Q is expressed by, Q = Heat flux × Area = 1566 × 0.9 Q = 1409.4 W

Heat Transfer 131

Example 6.5 Calculate the rate of heat transfer per square metre of surface of a cork board having 5 cm thickness, and a temperature difference of 75°C is applied across the board. The value of thermal conductivity (k) is – 0.4 W/m°C. Solution:

Given parameters are, k = – 0.4 A = 5 cm dt/dx = 75°C

By substituting in the corresponding formula, we get Q = - kA

dT dx

= – (– 0.4) (5) (75) Hence,

Q = 150 W.

Example 6.6

A slab of copper has a length of 10 cm and an area of 90 cm2. The front side is heated to 150°C and the back to 10°C. Find the heat flux q and the heat flow rate Q in the slab once steady state is reached. Assume dT/dx is constant.

Solution:

We have q = – kA (Tback – Tfront)/L = - ( 401 W/mK) ¥

(10∞C - 150∞C ) 0.1 m

= 5.6 × 10(5) W/m2. The heat rate is Q = qA = (90 × 10– 4 m2) = 5.1 × 103 watts or J/s.

Example 6.7

In the general formulation of Fourier’s law what are the vector and scalar quantities? Why is there a minus sign on the right-hand side of the equation?

Solution: Fourier’s law has scalar temperature field and heat flux vector. The minus sign is necessary because heat is always transferred in the direction of decreasing temperature.

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Example 6.8 A furnace wall of 1 m2 consists of 1.2 cm thick stainless steel inner layer covered with 5 cm outside insulation layer of insulation board. The inside surface temperature of the steel is 800 K and the outside surface temperature of the insulation board is 350 K. The thermal conductivity of the stainless steel is 19 W/(mK) and the thermal conductivity of the insulation board is 0.7 W/(mK).

Solution:

The conductive heat transport through the layered wall can be calculated as q=

((800 K) - ( 350 K)) (0.012 m) 2 ((19 W/(mK)) (1 m )) + ((0.05 m)/(0.7 W/(mK)) (1 m 2 ))

= 6245(W)

Example 6.9 What are the physical mechanisms associated with heat transfer by conduction, convection, and radiation? Solution: Conduction – no bulk or macroscopic motion Convection – random molecular motion, diffusion, bulk, or macroscopic motion Radiation – emittance of energy

Example 6.10 A star with the same colour as the sun is found to produce a luminosity 81 times larger. What is its radius compared to the sun? Solution: TStar = TSun (since they have the same colour) LStar = 81 LSun R Star compared to RSun Stefan-Boltzmann law: L = 4pR2 ◊ sT 4 2 4 2 4 2 2 LStar Ê RStar ˆ Ê TStar ˆ Ê RStar ˆ ◊sTStar ◊TStar 4 pRStar RStar = = = ◊ = 2 4 2 4 4 ˜ Á 4 ˜ Á LSun 4 pRSun ◊sTSun RSun ◊TSun Ë TSun ¯ Ë TSun ¯ ÁË TSun ˜¯

\

Ê ˆ LStar = RStar Á LSun Ë TSun ˜¯

2

ÊT ˆ ◊ Á Star ˜ ËT ¯ Sun

2

ÊR ˆ 81 = Á Star ˜ ◊ (1)4 Ë RSun ¯

4

2

ÊT ˆ ◊ Á Star ˜ ËT ¯ Sun

4

Heat Transfer 133

Ê RStar ˆ ÁË R ˜¯

\

2

= 81

Sun

RStar =9 RSun

The star has a radius 9 times that of the sun.

Example 6.11 What fraction of total solar emission falls into the visible spectrum (0.4 to 0.7 mm)?

Solution:

With a solar temperature of 5762 K it follows that for l1 = 0.4 mm, l1Tsun = 0.4 × 5762 = 2304 mmK

and l2 = 0.7 mm, l2Tsun = 0.7 × 5762 = 4033 mmK we can estimate f (l1Tsun) = 12% and f (l2Tsun) = 48%. Thus, the visible fraction of sunlight is 48 – 12 i.e. 36%: with a bandwidth of only 0.3 mm the human eye responds to approximately 36% of all emitted sunlight!

EXERCISE 1. How does heat transfer differ from thermodynamics? Is it true to say that heat transfer is essentially thermodynamic with rated equations added? 2. Distinguish between the conduction, convection and radiation modes of heat transfer. 3. State the laws of heat transfer with their expressions. 4. What are the various types of heat exchangers? Explain with neat sketch. 5. Write a short note on classification of heat exchangers. 6. Explain the role of fouling factor in heat exchangers. 7. A metal (k = 45 W/m-deg) steam pipe 5 cm internal diameter and 6.5 cm external diameter is lagged with a 2.75 radial thickness of high temperature insulation having thermal conductivity of 1.2 W/m-deg. The surface heat transfer coefficient for inside and outside are 4650 and 11.5 W/m2-deg. If the steam temperature is 200°C and the ambient temperature is 25°C. Calculate: (i) heat loss per metre length of pipe (ii) temperature at the interfaces (iii) overall coefficient of heat transfer referred to inside and outside surfaces. [544.15 W, 199.25 C, 198.75 C, 50.46 C: 19.80 and 8.25 W/m2-deg]

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8. Air at 30°C and one atm flows horizontally over a long heated horizontal cylinder (4 cm diameter) whose surface is maintained at 70°C. Find the average heat transfer coefficient if the velocity is 0.3 m/s. [10.83 W/m2 K] 9. At midnight, with the temperature inside your house at 70° F and the temperature outside at 20°F, your furnace breaks down. Two hours later, the temperature in your house falls to 50°F. Assume that the outside temperature remains constant at 20°F. At what time will the inside temperature of your house reach 40°F. [The temperature in the house will reach 40° F a little after 3:30 a.m.] 10. Raja and Ankita are having dinner and each orders a cup of coffee. Raja cools his coffee with three tablespoons of cream. They wait ten minutes and then Ankita cools her coffee with three tablespoons of cream. The two then begin to drink. Who drinks the hotter coffee? (Assume that adding three tablespoons of cream to coffee immediately cools the coffee by 10°F.)

7 Refrigeration and Air Conditioning

Learning Objectives    

‡ ‡ ‡ ‡

7.1

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INTRODUCTION

Air conditioning and refrigeration is a joint procedure. It conditions the air, transports it, and acquaints it with the adapted space. It gives warming and cooling from its focal plant or housetop units. It likewise controls and keeps up the temperature, stickiness, air development, air cleanliness, sound level, and weight differential in a space inside predetermined limits. The final temperature scale is the Celsius scale, founded on the melting point of ice at 0°C and the boiling point of water at air weight at one 100°C. (by means of strict definition, the triple point of ice is 0.01°C at a pressure of 6.1 mbar.) On the Celsius scale, absolute zero is – 273.15°C. Within the study of refrigeration, the Kelvin or complete temperature scale is used. This starts at absolute zero and has the identical degree interims as the Celsius scale, so ice melts at + 273.16 k and water at air weight bubbles at + 373.15 k. Refrigeration is the motion of cooling, and this requires removal of heat and discarding it at a higher temperature. Refrigeration is consequently the science of relocating heat from low temperature to high temperature. In addition to cooling and freezing applications, refrigeration technology is applied in air conditioner

Basic Mechanical Engineering

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and heat pumps. The fundamental principles are those of thermodynamics. Heat is without doubt one of the essential parts of power. The heat of a body is its thermal or internal energy, and a change in this energy may just show as a change of temperature or a change between the solid, liquid and gaseous states. Matter may also produce other types of energy, potential or kinetic, relying on strain, function and movement. Vapour out

Liquid in

Hot region temperature = TH

Heat exchanger

Cold region temperature = TL

Heat leakage Insulation

Figure 7.1 Concept of refrigeration.

7.2

BASIC DEFINITIONS

Refrigeration: The process of keeping an item below room temperature by storing the item in a system or substance designed to cool or freeze. The most common form of refrigeration is provided by systems (i.e., refrigerators) that use a refrigerant chemical to remove heat from items stored inside the system. Refrigerant: Chemicals used as a part of a cooling system, for example, an air conditioner or fridge, as the heat bearer which changes from gas to fluid and the back to gas in the refrigeration cycle. Most basic refrigerants are the chlorofluorocarbons (CFCs) which, in light of their high ozone harming potential, are being eliminated. Air Conditioning: It is a type of refrigeration where thermal energy is taken away from the air in a large space such as a room or a vehicle. Air conditioners are fitted into rooms so that they cool the air inside them. Air conditioners also reduce humidity in

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137

rooms, because the water vapour in the room can condense around the colder parts of the air conditioner. The condensed water can be then drained off. Heat Engine: A heat engine may be described as a gadget that works in a thermodynamic cycle and does a particular measure of net positive work through the exchange of heat from a high temperature body to a low temperature body. Heat Pump: A heat pump is like a cooler, be that as it may, here the required yield is the heat rejected to the high temperature body. Coefficient of Performance: The effectiveness of refrigerator and heat pump is called coefficient of performance and is signified by COP. COP of heat engine is indicated as (COP HE ) and of heat pump is meant as ( COP HP). Vapour Compression Refrigeration: As the name recommends, utilizes a pressure method to raise the pressure of a refrigerant vapour spilling out of an evaporator at pressure P1 to P2, as seen in Fig. 7.2. The refrigerant then courses through a heat exchanger called a condenser at the high pressure, P2 = P3, through a throttling gadget, and back to the low pressure, P1, in the evaporator. The weights P2 = P3 and P4 = P1 compare to refrigerant immersion temperatures, T3 and T1 = T4, individually. These temperatures permit common heat exchange with contiguous hot and cold areas from high temperature to low. That is, T1 is not as much as TL; so that heat, QL, will spill out of the cold region into the evaporator to vaporize the working liquid. Thus, the temperature T3 permits heat, QH, to be exchanged from the working liquid in the condenser to the hot region at TH. This is shown by the points of Fig. 7.2. Hot region temperature = TH Condenser 2 | O H| 3 Throttling device 4

1 Work in Evaporator Compressor

OL Cold region temperature = TL

Figure 7.2 Vapour compression cycle.

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Throttling Gadget: The throttling device, as shown in Fig. 7.2, restrains the flow of the refrigerant from the condenser to the evaporator. Its most important work is to furnish the flow resistance to preserve the pressure difference between the 2 heat exchangers. It additionally serves to manipulate the flow of air from condenser to evaporator. The throttling device is also a thermostatic evaluation valve (TEV) controlled by way of evaporator exit temperature or a protracted bore pipe known as a capillary tube. Psychrometrics: Despite the fact that the mass fraction of water vapour in ambient air more often than not is numerically small, water vapour can have a massive effect on an individual’s thermal comfort and on the design of air conditioner apparatus. Most people can keep in mind instances of affliction related to each excessively dry or moist air. Countless industrial techniques additionally require moisture control to preserve excessive specifications productivity. The engineering fundamentals of coping with moist air and the application of those fundamentals are dealt with here. The study of the behaviour of moist air is called psychrometrics. Humidity Ratio: Properties of moist air are in most cases pointed out as the mass of dry air (Mda), apart from the water vapour, contained in the air. For illustration, the humidity ratio is outlined as the ratio of the mass of water vapour to the mass of dry air in the moist air: W = Mwv/Mda The humidity ratio and the ambient temperature and pressure define the thermodynamic state of moist air, and therefore is also used to outline other properties of moist air. Saturated Air: Moist air is alleged to be saturated when its humidity ratio is the highest for the existing temperature and pressure. The humidity ratio for saturated air is also determined utilizing equations of humidity ratio with the saturation pressure of water vapour acquired from saturated steam tables on the recognized temperature.

7.2.1

Refrigeration Cycle

Basic Vapour Compression Cycle: A liquid boils and condenses – the change between the liquid and gaseous states – at a temperature which depends on its pressure, within the limits of its freezing point and critical temperature. In boiling it must obtain the latent heat of evaporation and in condensing the latent heat must be given up again. The basic refrigeration cycle (Fig. 7.3) makes use of the boiling and condensing of a working fluid at different temperatures and, therefore, at different pressures. Heat is put into the fluid at the lower temperature and pressure and provides latent heat to make it boil and change to a vapour. This vapour is then mechanically compressed to a higher pressure and a corresponding saturation temperature at which its latent heat can be rejected so that it changes back to a liquid. The total cooling effect

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139

will be the heat transferred to the working fluid in the boiling or evaporating vessel, i.e., the change in enthalpies between the fluid entering and the vapour leaving the evaporator. For a typical circuit, using the working fluid refrigerant, R-22, evaporating at – 5°C and condensing at 35°C, the pressures and enthalpies are as shown in Fig. 7.3. Gas at 12.54 bars

Dry saturated gas 5°C 3.21 bars 249.9 kJ/kg Compressor

35°C

5°C Heat in

Fluid in 91.4 kJ/kg

Heat out

Liquid out 35°C 91.4 kJ/kg

Figure 7.3 Basic refrigeration cycle.

Enthalpy of fluid entering evaporator = 91.4 kJ/kg, enthalpy of saturated gas leaving evaporator = 249.9 kJ/kg, cooling effect = 249.9 – 91.4 = 158.5 kJ/kg. A working system will require a connection between the condenser and the inlet to the evaporator to complete the circuit. Since these are at different pressures this connection will require a pressure reducing and metering valve. Since the reduction in pressure at this valve must cause a corresponding drop in temperature, some of the fluid will flash off into vapour to remove the energy for this cooling. The volume of the working fluid therefore increases at the valve by this amount of flash gas, and gives rise to its name, the expansion valve (Fig. 7.4.). High-pressure gas

Low-pressure gas

Discharge Suction Compressor

Condenser

Evaporator Expansion valve Low-pressure (liquid and ash gas)

High-pressure liquid

Figure 7.4

140

Basic Mechanical Engineering

Since the vapour compression cycle uses energy to move energy, the ratio of these two quantities can be used directly as a measure of the performance of the system. This ratio, the coefficient of performance, was first expressed by Sadi Carnot in 1824 for an ideal reversible cycle, and based on the two temperatures of the system, assuming that all heat is transferred at constant temperature (Fig. 7.5). Since there are mechanical and thermal losses in a real circuit, the coefficient of performance (CoP) will always be less than the ideal Carnot figure. For practical purposes in working systems, it is the ratio of the cooling effect to the input compressor power.

Figure 7.5 Ideal reversed Carnot cycle.

7.3

VARIOUS REFRIGERANTS AND THEIR PROPERTIES

Refrigerants are specifically selected substances which have exact primary characteristics together with good refrigeration performance, low flammability and toxicity, compatibility with compressor lubricating oils and metals, and excellent warmth transfer characteristics. They’re most often identified with the aid of a quantity that relates to their molecular composition. Two foremost examples are refrigerants R-12 and R-22. R-12, dichlorodifluoromethane, has two fluorine, one carbon, and two chlorine atoms in a methanekind structure. As a consequence the halogens chlorine and fluorine, exchange hydrogen atoms within the CH4 molecular structure. R-22, monochlorodifluoromethane, has a identical constitution to R-12, besides for a single hydrogen atom changing a chlorine atom.

Refrigeration and Air Conditioning

141

The most often used chlorofluorocarbon (CFC) refrigerants are a rationale of fine quandary, due to the fact their accumulation in the higher surroundings creates a gap. In the ozone layer that frequently shields the earth from sunlight ultraviolet radiation. ,Q  JUHDWHU WKDQ  FRXQWULHV LQFOXGLQJ WKH 8QLWHG 6WDWHV VLJQHG WKH 0RQWUHDO Protocol on materials that fritter away the ozone layer. The Montreal Protocol referred to as for a freeze in 1989 and discount rates within the 1990s on the construction phases of R-11, R-12, R-113, R-114, and R-115. The halocarbon refrigerants, some of them are also generally used as aerosol propellants, foams, and solvents, are actually classified as chlorofluorocarbons (CFCs), hydrochlorofluorocarbons (HCFCs), or hydrofluorocarbons (HFCs). The HFCs, lacking chlorine, aren’t any hazard to the ozone layer but should not be used as refrigerants. CFCs, which contain extra chlorine than do HCFCs, are essentially the most serious offenders, are very stable, and don’t break down speedily in the reduced atmosphere. The Clean Air Act of 1990 mandated termination of FRQVWUXFWLRQ LQ WKH 8QLWHG 6WDWHV RI DOO &)&V FRUUHVSRQGLQJ WR 5 E\ WKH \HDU  Govt. information points out that, considering the structural difference between them, R-12 has twenty times the ozone-depletion capabilities within the upper surroundings of R-22. Nevertheless, R-22 and other HCFCs are additionally scheduled by means of the legislation for phase-out production by the year 2030. For that reason, the need for alternative refrigerants to interchange these in use currently has assumed enormous significance. It is a difficult, luxurious, and continuing task to which industry is vigorously making its efforts.

7.3.1

Refrigerants by Class

Refrigerants may be divided into three classes according to their manner of absorption or extraction of heat from the substances to be refrigerated: ‡ Class 1: This class includes refrigerants that cool by phase change (typically boiling), using the refrigerant’s latent heat. ‡ Class 2: These refrigerants cool by temperature change or ‘sensible heat’, the quantity of heat being the specific heat capacity x the temperature change. They are air, calcium chloride brine, sodium chloride brine, alcohol, and similar non-freezing solutions. The purpose of Class 2 refrigerants is to receive a reduction of temperature from Class 1 refrigerants and convey this lower temperature to the area to be air conditioned. ‡ Class 3: This group consists of solutions that contain absorbed vapours of liquefiable agents or refrigerating media. These solutions function by nature of their ability to carry liquefiable vapours, which produce a cooling effect by the absorption of their heat of solution. They can also be classified into many categories.

Chemical Name

Methyl Fluoride

Methane

Hexachloroethane

Pentachlorofluoroethane

Trichlorotrifluoroethane

Trichlorotrifluoroethane

41

50*

110

111

112

112a

Dichlorotetrafluoroethane

Methyl Chloride

40

114

Methylene Fluoride

32

Trifluoroethane

Chlorofluoromethane

31

Trifluoroethane

CCl3CCl3

Methylenechloride

30

113

CH4

Trifluoromethane

23

113a

160 170*

CH3F

Chlorodifluoromethane

22

C-ClF2CClF2

CCl3CF3

CCl2FCClF2

CCl3CClF2

CCl3FCCl2F

CCl3CCl2F

CH3Cl

CH2F2

CH2ClF

CH2Cl2

CHF3

CHClF2

CHCl2F

Oceafluoropropane

Ethane

Ethyl chloride

Dichloroethane

Dichloroethane

Trichloroethane

Chlorotetrafluoroethane

Trichloroethane

Tetrafluoroethane

Chlorotetrafluoroethane

Pentachloroethane

Chlorotetrafluoroethane

Chlorotetrafluoroethane

Dichlorotrifluoroethane

Pentachloroethane

Hexafluofluoroethane

Chloropentrafluorethane

Dibromotetrafluoroethane

Dichlorotetrafluoroethane

Chemical Name

C318

C317

C316

Octafluorocyclobutane

Chloroheptafluorocyclobutane

Dichlorohexafluorocyclobutane

Oyclic Organic Compounds

218

152a

150a

143a

142b

140a

134a

133a

125

124a

124

123

Dichlorofluoromethane

CHCl3

CF4

CBrF3

120

116

21

Bromotrifluoromethane

13B1

CCl2F2 CClF3

Chloroform

Chlorotrifluoromethane

13

116

Carbontetrafluoride

Dichlorodifluoromethane

12

CCl3F

114B2

114a

Refrigerant Number

20

Trichlorofluoromethane

11

CCl4

Chemical Formula

14

Carbonterrachloride

10

Halocarbon Compund

Refrigerant Number

Table 7.1 (a) Commonly used refrigerants with their chemical name and chemical formula

C4F8

C4ClF3

C4Cl2F6

CF3CF2CF3

CH3CH3

CH3CH2Cl

CH3CHF2

CH3CHCl2

CH3CF3

CH3CClF2

CH2CCl2

CF3CH2F

CH2ClCF2

CHF2CF3

CHF2CClF2

CHClFCF3

CHCl2CF2

CHCl2CCl3

CF3CF3

CClF2CF2

CBrF2CBrF2

CCl2FCF3

Chemical Formula

142 Basic Mechanical Engineering

Retigrants 22: 2 (75/25)

Retigrants 22: 115(488/51.2)

Retigrants 23: 3(40.1/59.9)

Retigrants 32: 31(78.5/22.0)

Retigrants 12: 31(78.5/22.0)

Retigrants 31: 114(52.1/44.9)

501

502

503

504

505

506

717

CH2 = CH2

Methyl Amine

Ethyl Amine

630

631

Nitrogen Compound

Sulfur Dioxide

Nitrous Oxide

Carbon Dioxide

Argon

Oxygen

Air

Nitrogen

Neon

Water

Ammonia

Helium

Dichlordifluoromethane

Vinyl fluoride Ethylene Propylne

1141 1150 1270

C2H3NH2

Vinyl chloride

1140

Vinylidene fluoride

Dichlomethylene

Trichloromethane

Tetrafluoromethane

Chlorofluoromethane

CH3NH2

1132a

1130

1120

C2H2OC2H3 HCOOCH3

Ethyl Ether

Methyl Formate

610

611

1113

1112A 1114

CH3CH = CH2

Chemical Formula

CH3CH = CH2

CH2 - CH2

CH2 - CHF

CH2 - CHCl

CH2 = CF3

CHCl - CHCl

CHCl = CHCl

CHCl = CCl2

CClF - CF2

CCl2 - CF2

SO2

N2O

CO2

A

O2

21O2, ..78N2 ...

N2

Ne

H 2O

NH3

He

Hydrogen (Normal and H3 Para)

Unsaturated Organic Compouncs

764

744A

744

740

Oxygen Compound

Ethylene

Propylene

Isobutane (2 methyl propane) CH(CH3)3

1150

CH2CH2CH2

Butane

600

600a

1270

CH2CH2CH2

Propane

290

CH3CH3

Ethane

CH4

Methane

50

732

729

728

CCl2F2/CH3CF CH2ClF2CClF2CClF2

720

CCl2F2/CH2CF

718

704

702

CHClF2/CClF3 CHF3/CClF3

Chemical Name

Inorganic Compound

Refrigerant Number

CHClF2/CCl2F2

CCl2F2/CH3CHF2

Chemical Formula

170

Macelaneous Organic Compounds

Retigrants 12: 52a (73.8/26.2)

Chemical Name

500

Azeotropes

Refrigerant Number

Table 7.1 (b) Commonly used refrigerants with their chemical name and chemical formula

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7.3.2 Required Properties of Ideal Refrigerants 1. The refrigerant should have low boiling point and low freezing point. 2. It should have low specific heat and excessive latent heat. Considering excessive specific heat decreases the refrigerating influence per kg of refrigerant and high latent heat at low temperature increases the refrigerating effect per kg of refrigerant. 3. Pressures required to be maintained within the evaporator and condenser should be low to reduce the material cost and must prevent leakage of air into the system. 4. It must have high pressure and temperature to prevent massive energy requisites. 5. It should have low specified volume to scale down the size of the compressor. 6. It must have high thermal conductivity to cut back the field of heat transfer in evaporator and condenser. 7. It should be non-flammable, non-explosive, non-toxic and non-corrosive. 8. It should not have any bad effects on the saved material or food, when any leak develops in the method. 9. It ought to have excessive miscibility with lubricating oil and it must not react with lubricating oil within the temperature variety of the approach. 10. It must supply excessive CoP within the working temperature range. That is essential to scale down the strolling price of the procedure. 1. Ammonia (NH3) (R-717) Latent heat = 1312.75 kJ/kg Specific volume = 0.509 m3/kg 2. Dichloro–Difluoro methane (Freon–12) (R-12) [CCl2F2] Latent heat = 162 kJ/kg Specific volume = 0.093 m3/kg 3. Difluoro monochloro methane – or Freon-22 (R-22) [CHClF2] Latent heat = 131 kJ/kg Specific volume = 0.15 m3/kg.

7.3.3 Uses of Refrigerants Refrigerants like ammonia (R-717), carbon dioxide and non-halogenated hydrocarbons don’t deplete the ozone layer and have no (ammonia) or only a low (carbon dioxide, hydrocarbons) to effect global warming. They’re utilized in air conditioning techniques

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for structures, in recreation and leisure facilities, in the chemical/pharmaceutical industry, within the automotive enterprise and especially in the food industry (production, storage, marine delivery). In these settings their toxicity is much less than in dwelling gear. Emissions from automobile air conditioners are a growing concern for they affect WKH FOLPDWH )URP  RQZDUGV WKH (XURSHDQ 8QLRQ SKDVHG RXW UHIULJHUDQWV ZLWK D global warming potential (GWP) of greater than a 150 in automotive air conditioning (GWP = 100 yr warming advantage of 1 kilogram of a gas relative to at least one kilogram of CO2). This may increasingly ban potent greenhouse gases such as the refrigerant +)&DVRPHWLPHVFDOOHG5DLQ1RUWKWKH86$ KDVD*:3RI³WRSURPRWH trustworthy and vigour-effective refrigerants. One promising possible choice is CO2 (R-744). Carbon dioxide is non-flammable, non-ozone depleting, has a global warming advantage of 1. R-744 can be used as a working fluid in climate manipulate systems for cars, residential air conditioning, sizzling water pumps, industrial refrigeration, and merchandising machines. R12 is suitable with mineral oil, even as R134a is suitable with artificial oil that contains esters. GM introduced that it might begin using “hydrofluoroolefin”, HFO-1234yf, in all of its brands in future. Dimethyl ether (DME) can be gaining repute as a refrigerant, however like propane, it is usually highly flammable.

7.3.4

Applications of Refrigeration in Chemical and Process Industries

The enterprises like petroleum refineries, petrochemical plants and paper mash and so forth require substantial cooling limits. The necessity of every industry is distinct, consequently, refrigeration framework must suit individual needs. The primary utilization of refrigeration includes the following classification. Separation of Gases: In petrochemical plant, temperatures as low as –150°C with refrigeration limits as high as 10,000 tonnes of refrigeration (TR) are utilized for detachment of gases by partial refining. Some gases gather promptly at lower temperatures from the blends of hydrocarbon. Propane is utilized as a refrigerant. Condensation of Gases: Some gases that are delivered artificially, are consolidated to fluid state by cooling, so that these can be effectively put away and transported in fluid state. For instance, in smelling salts plant, alkali is dense at –10 to 10°C preceding filling in the chambers, stockpiling and shipment. This low temperature requires refrigeration. Dehumidification of Air: Low moistness air is required in numerous pharmaceutical businesses. It is likewise required for air liquefaction plants. This is likewise required to avert friction based electricity and forestalls short circuits in spots where high voltages are utilized. The air is cooled underneath its dew point temperature, with the goal that some water vapour gathers out and the air gets dehumidified.

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Basic Mechanical Engineering

Solidification of Solute: One of the procedures of partition of a substance or contamination or pollution from fluid blend is by its cementing at low temperature. Greasing up oil is dewaxed in petroleum industry by cooling it beneath – 25°C. Wax hardens at about – 25°C. Storage as Fluid at Low Pressure: Fluid possesses less space than gases. The greater part of the refrigerants are put away at high pressure. This pressure is normally their immersion pressure at air temperature. For some gases, immersion pressure at room temperature is high subsequently these are put away at moderately low pressure and low temperature. For instance, normal gas is put away at 0.7 bar gauge pressure and –130°C. Heat pick up by the chamber dividers prompts to bubbling of a few gas, which is packed, cooled and extended back to 0.7 bar gauge. Removal of Heat of Reaction: In numerous concoction responses, proficiency is better if the response happens underneath room temperature. This requires refrigeration. On the off chance that these responses are exothermic in nature, then more refrigeration limits are required. Generation of rayon, cell acetic acid derivatives and engineered elastic are some of the cases.

7.3.5

Special Applications of Refrigeration

In this category, we consider applications other than chemical uses. These are in manufacturing processes, applications in medicine, construction units, etc. Cold Treatment of Metals: The dimensions of precision parts and gauge blocks can be stabilized by soaking the product at temperature around –90°C. The hardness and wear resistance of carburized steel can be increased by this process. Keeping the cutting tool at –100°C for 15 minutes can also increase the life of cutting tool. In deep drawing process the ductility of metal increases at low temperature. Mercury patterns frozen by refrigeration can be used for precision casting. Medical: Blood plasma and antibiotics are manufactured by freeze-drying process where water is made to sublime at low pressure and low temperature. This does not affect the tissues of blood. Centrifuges refrigerated at –10°C are used in the manufacture of drugs. Localized refrigeration by liquid nitrogen can be used as anaesthesia also. Ice Skating Rings: Due to the advent of artificial refrigeration, sports like ice hockey and skating do not have to depend upon freezing weather. These can be played in indoor stadium where water is frozen into ice on the floor. Refrigerant or brine carrying pipes are embedded below the floor, which cools and freezes the water to ice over the floor. Construction: Setting of concrete is an exothermic process. If the heat of setting is not removed the concrete will expand and produce cracks in the structure. Concrete may be cooled by cooling sand, gravel and water before mixing them or by passing chilled water through the pipes embedded in the concrete. Another application is to freeze the wet soil by refrigeration to facilitate its excavation.

Refrigeration and Air Conditioning

7.4

147

SUMMARY

Considering the fact that refrigeration offers cooling of bodies or fluids to temperatures less than those of environment, it involves absorption of heat at a diminished temperature and rejects higher temperature of the environment. In olden days, the most important rationale of refrigeration was to supply ice, which used to be used for cooling beverages, meals, etc. Refrigeration and air conditioning are treated as a single discipline given that that some of the predominant applications of refrigeration are cooling and dehumidification as required for summer time air conditioning. While refrigeration is required for many applications as opposed to air conditioning and it involves techniques other than cooling and dehumidification.

SOLVED EXAMPLES

Example 7.1

The machine room housing the compressor and condenser of a refrigerant 12 system has dimensions 5 × 4 × 3 m. Calculate the mass of the refrigerant which would have to escape into the space to cause a toxic concentration for a 2 hr exposure.

Solution:

Refrigerant 12 exposure for 2 hr has 20% by volume to become toxic.

Room volume = 5 × 4 × 3 m = 60 m3.

Volume of refrigerant 12. = (0.20)(60) = 12 m2. At atmospheric, 101.325 kPa, Steam Table. Vg = 158.1254 L/kg = 0.1581254 m3/kg Mass of refrigerant 12. = (12 m2)/(0.1581254 m3/kg) = 76 kg

Example 7.2

For water, the latent heat of freezing is 334 kJ/kg and the specific heat capacity averages 4.19 kJ/(kg K). What quantity of heat should be removed from 1 kg of water at 30°C in order to turn it into ice at 0°C

Solution: 4.19(30 – 0) + 334 = 459.7 kJ

Example 7.3

If the latent heat of boiling water at 1.013 bar is 2257 kJ/kg, what quantity of heat should be added to 1 kg of water at 30°C in order to boil it?

Solution: 4.19(100 – 30) + 2257 = 2550.3 kJ

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Basic Mechanical Engineering

Example 7.4

The specific enthalpy of water at 80°C, taken from 0°C base, is 334.91 kJ/kg. What is the average specific heat capacity through the range 0–80°C?

Solution: 334.91/(80 – 0) = 4.186 kJ/(kg K)

Example 7.5

An ideal vapour-compression refrigerant cycle operates at steady state with refrigerant 134a as the working fluid. Saturated vapour enters the compressor at – 10°C, and saturated liquid leaves the condenser at 28°C. The mass flow rate of refrigerant is 5 kg/min. Determine (a) The compressor power, in kW (b) The refrigerating capacity, in tonnes. (c) The coefficient of performance.

Solution: Given: Refrigerant 134a is the working fluid in an ideal vapour-compression refrigerant cycle. Operating data are known. T 2

28°C

3

10°C 4

1

S

Figure 7.6

. Qout T3 = 28°C

3

Condenser

2

Sat. Liquid . WC

Expansion valve

Compressor

Evaporator 4

1 T1 = 10°C . Qin

Sat. vapour

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149

Analysis: First, fix each of the principal states. Compressor: 2 . WC

1

Assumptions: ‡ steady state steady flow process (SSSF) ‡ open system ‡ DKE = DPE = 0 ‡ Neglect heat transfer (q = 0) dEe , v dt

Ê ˆ Vi2  -W  + m  = Q h + Âi i ÁË i 2 + gzi ˜¯ e. v. e. v.

Ê ˆ Ve2  m h + Âe e ÁË e 2 + gze ˜¯

 +m  1 h1 - m  2 h2 0 = W e

 m

i

=

 m

e

1 = m 2 =m  m  = m  ( h2 - h1 ) W e dSev = dt

 j

Qj Tj

+

 m s -  m s i

i

i

e

e

+ s ev

e

 1s1 - m  2 s2 0 = m 1 = m 2 =m  m

s 2 = s1 State 1: T1 = – 10°C, saturated vapour fi h1 = 241.35 kJ/kg, s1 = 0.9253 kJ/kg.K State 2: P2 = Psat 28°C = 7.2675 bars s2 = s1 fi Double interpolation gives: h2 = 267.9 Expansion valve:

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Basic Mechanical Engineering

3

4

Assumptions: ‡ steady state steady flow process (SSSF) ‡ open system ‡ DKE = DPE = 0 ‡ Neglect heat transfer (q = 0) ‡ Neglect work transfer (w = 0) dEe , v

 -W  + m Âi  i e. v. e. v. = Q   dt 0 0

Ê ˆ Vi2 h + + gzi ˜ ÁË i 2 ¯

 e

Ê ˆ V2  e Á he + e + gze ˜ m 2 Ë ¯

 3 h3 - m  4 h4 0 = m

 m

i

=

 m

e

3 = m 4 =m  m

h3 = h4 State 3: T3 = 28°C, saturated liquid fi h3 = 88.61 kJ/kg State 4: Throttling process fi h4 = h3 = 88.61 kJ/kg Evaporator:

1

4

.

Qin

Refrigeration and Air Conditioning

Assumptions: ‡ steady state steady flow process (SSSF) ‡ open system ‡ DKE = DPE = 0 ‡ Neglect work (w = 0) dEe , v dt

 -W  + m Âi  i e. v. e. v. = Q   0

0

Ê ˆ Vi2 h + + gzi ˜ ÁË i 2 ¯

 e

Ê ˆ V2  e Á he + e + gze ˜ m 2 Ë ¯

 +m  4 h4 - m  1 h1 0 = Q in

 m

i

=

 m

e

1 = m 4 =m  m  = m  ( h1 - h4 ) Q in

(a) The compressor power is: Ê kg ˆ  =m  ( h2 - h1 ) = Á 5 W e Ë min ¯˜

Ê 1 min ˆ kJ Ê 1 kW ˆ ÁË 60 s ˜¯ ( 267.9 - 241.35) kg ÁË 1 kJ/s ˜¯ = 2.212 kW

(b) The refrigeration capacity is: kJ Ê 1 tonne ˆ Ê kg ˆ  =m  ( h1 - h4 ) = Á 5 Q ( 241.35 - 88.61) = 3.62 tonnes in ˜ Ë min ¯ kg ÁË 211 kJ/min ˜¯

(c) The coefficient of performance is: b=

 h - h4 ( 241.35 - 88.61) kJ/ kg Q in = 1 = = 5.75  We h2 - h1 ( 267.9 - 241.35) kJ/ kg

EXERCISE 1. 2. 3. 4. 5. 6.

Explain briefly the concept of refrigeration. Define the terms: air conditioning, CoP, psychometrics, humidity ratio. List the required properties of an ideal refrigerant. Explain vapour compression cycle with a neat schematic diagram. Write a note on refrigerants by class. What are the applications of refrigerants? Explain in detail.

151

152

Basic Mechanical Engineering

Multiple-Choice Questions 1. What is the name of the following statement: “When two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other”? (a) First law of thermodynamics

(b) Second law of thermodynamics

(c) Mechanical equivalent of heat

(d) Zeroth law of thermodynamics

(e) Thermal expansion of solids 2. An aluminium plate has a circular hole. If the temperature of the plate increases, what happens to the size of the hole? (a) Increases

(b) Decreases

(c) Stays the same

(d) Increases the top half of the hole

(e) More information is required 

 $ ELPHWDO SODWH FRQVLVWV RI WZR PDWHULDOV RI GLIIHUHQW FRHIÀFLHQWV RI WKHUPDO H[SDQVLRQ 7KH FRHIÀFLHQW RI WKHUPDO H[SDQVLRQ RI WKH WRS SDUW RI WKH SODWH LV less than the bottom part. If the temperature of the entire plate increases, what happens to the plate? (a) Expands

(b) Contracts

(c) Stays the same

(d) Bends down

(e) Bends up 4. Which of the following temperature scales doesn’t have negative numbers? (a) Celsius

(b) Kelvin

(c) Reaumur

(d) Fahrenheit

(e) Galileo 5. Which of the two temperature changes are equivalent? (a) 1 K = 1 F

(b) 1 F = 1 C

(c) 1 Re = 1 F

(d) 1 K = 1 C

(e) 1 Re = 1 K 

 $FRQWDLQHUÀOOHGZLWKDVDPSOHRIDQLGHDOJDVDWWKHSUHVVXUHRIDWP7KHJDV is compressed isothermally to one-fourth of its original volume. What is the new pressure of the gas? (a) 2 atm

(b) 3 atm

(c) 4 atm

(d) 5 atm

(e) 6 atm

Refrigeration and Air Conditioning

153

7. The temperature of an ideal gas increases from 20°C to 40°C while the pressure stays the same. What happens to the volume of the gas? (a) It doubles

(b) It quadruples

(c) It is cut to one-half

(d) It is cut to one-fourth

(e) It slightly increases 8. The state of an ideal gas was changed three times in a way that the volume stays the same. The graph represents three isobaric lines. Which of the following is true about the volume of the gas? (a) V1 > V2 >V3

(b) V1 > V2 < V3

(c) V1 < V2 < V3

(d) V1 = V2 > V3

(e) V1 > V2 = V3 

 $ FRQWDLQHU ZLWK ULJLG ZDOOV LV ÀOOHG ZLWK D VDPSOH RI LGHDO JDV 7KH DEVROXWH temperature of the gas is doubled. What happens to the pressure of the gas? (a) Doubles

(b) Quadruples

(c) Triples

(d) Decreased to one-half

(e) Decreased to one-fourth 10. The absolute temperature of an ideal diatomic gas is quadrupled. What happens to the average speed of molecules? (a) Quadruples

(b) Doubles

(c) Triples

(d) Increases by a factor of 1.41

(e) Stays the same 

 7ZRFRQWDLQHUVDUHÀOOHGZLWKGLDWRPLFK\GURJHQJDVDQGGLDWRPLFR[\JHQJDV The gases have the same temperature. Compare the average speed of hydrogen molecules to the average speed of oxygen molecules. (a) 1/16

(b) 1/4

(c) 16/1

(d) 1/2

(e) 4/1 12. The average molecular kinetic energy of a gas depends on: (a) Pressure

(b) Volume

(c) Temperature

(d) Number of moles

(e) None of the above 13. Kinetic theory is based on an ideal gas model. The following statements about the ideal gas are true EXCEPT:

154

Basic Mechanical Engineering

(a) The average molecular kinetic energy is directly proportional to the absolute temperature (b) All molecules move with the same speed (c) All molecules make elastic collisions with each other and the walls of the container (d) The attractive force between the molecules can be ignored (e) All molecules obey laws of classical mechanics 14. Internal energy of an ideal gas depends on: i) the volume of the ideal gas, ii) the pressure of the ideal gas, and iii) the absolute temperature of the ideal gas (a) I

(b) II

(c) III

(d) I and II

(e) I, II and III 

 $VDPSOHRIDQLGHDOJDVKDVDQLQWHUQDOHQHUJ\8DQGLVWKHQFRPSUHVVHGWRRQH half of its original volume while the temperature stays the same. What is the new LQWHUQDOHQHUJ\RIWKHLGHDOJDVLQWHUPVRI8"



D  8

E  8



F  8

G  8



H  8



 $QLGHDOJDVZLWKDQLQWHUQDOHQHUJ\8LQLWLDOO\DWƒ&LVKHDWHGWRƒ&:KDWLV WKHQHZLQWHUQDOHQHUJ\LQWHUPVRI8"



D  8

E  8



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(a) Newton

(b) Kelvin

(c) Joule

(d) Boltzmann

 $ VLOYHU WHDVSRRQ LV SODFHG LQ D FXS ÀOOHG ZLWK KRW WHD $IWHU VRPH WLPH WKH exposed end of the spoon becomes hot even without a direct contact with the liquid. This phenomenon can be explained by: (a) Thermal expansion

(b) Conduction

(c) Convection

(d) Radiation

(e) Emission 19. The process of heat transfer from one object to another because of molecular motion and interaction is called:

Refrigeration and Air Conditioning

(a) Convection

(b) Conduction

(c) Radiation

(d) Induction

155

(e) None of the above 20. When we touch a piece of metal and a piece of wood that are placed in the same room, the piece of metal feels much colder than the piece of wood. This happens because of the difference in: 

D  6SHFLÀFKHDW

E  7HPSHUDWXUH

(c) Density

(d) Thermal conductivity

(e) Latent heat 21. The process of heat transfer by the movement of mass from one place to another is called: (a) Convection

(b) Conduction

(c) Radiation

(d) Induction

(e) None from the above 22. Which mechanism of heat transfer is involved in heating a pot with water on a stove? (a) Convection

(b) Conduction

(c) Radiation

(d) Induction

(e) None of the above 

 :KLFKPHFKDQLVPRIKHDWWUDQVIHULVLQYROYHGLQKHDWÁRZIURPVXQWRHDUWK" (a) Convection

(b) Conduction

(c) Radiation

(d) Induction

(e) None of the above 24. Which of the following is a characteristic of an adiabatic process? (a) DU = 0

(b) W = 0

(c) Q = 0

(d) DV = 0

(e) DP = 0 25. An ideal heat engine operates between two temperatures 600 K and 900 K. What LVWKHHIÀFLHQF\RIWKHHQJLQH" (a) 50%

(b) 80%

(c) 100%

(d) 10%

(e) 33%

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Basic Mechanical Engineering

 )LQGWKHZURQJVWDWHPHQW6SHFLÀFKHDWRIDPDWHULDOBBBBBBBBBB (a) Constant for a material

(b) Heat capacity per unit mass

(c) Extrinsic property

(d) Has units as J/kg-K.

27. Heat capacity has units as (a) J/kg.K

(b) J/mol.K 2

(c) J.ohm/sec.K 

(d) W/m.K

 8QLWVIRUWKHUPDOFRQGXFWLYLW\ (a) J/kg.K

(b) J/mol.K 2

(c) J.ohm/sec.K

(d) W/m.K

29. Lorentz constant has units as (a) J/kg.K

(b) J/mol.K 2

(c) J.ohm/sec.K

(d) W/m.K

30. Thermal expansion of a material has units as (a) J/kg-K

(b) J/mol-K

(c) J.ohm/sec.K2

(d) 1/°C

31. Polymers have thermal conductivities in the range of





(a) < 1

(b) 1-10

(c) 10-100

(d) >100

 3RO\PHUVKDYHWKHUPDOH[SDQVLRQFRHIÀFLHQWVLQWKHUDQJHRIBBBBBBBBBBð–6. (a) 0.5-15

(b) 5-25

(c) 25-50

(d) 50-400

 &RHIÀFLHQWRIWKHUPDOH[SDQVLRQIRUFHUDPLFVLVWKHUDQJHRIBBBBBBBBBBð–6. (a) 0.5-15

(b) 5-25

(c) 25-50

(d) 50-400

34. Metals have thermal conductivities in the range of (a) < 1

(b) 1-5

(c) 5-25

(d) 20-400

35. Heat capacity of most materials is approximately equal to __________. (a) R

(b) 2R

(c) 3R

(d) R/2

Refrigeration and Air Conditioning

36. With increase in temperature, thermal conductivity of a metal __________. (a) Increases

(b) Decreases

(c) Both (a) and (b)

(d) Depends on metal

37. Thermal conductivity in polymers increases with __________. (a) Increase in crystallinity

(b) Decrease in crystallinity

(c) Both (a) and (b)

(d) None

38. Refrigerant used should be such that its normal boiling point is (a) Greater than the temperature required (b) Less than the temperature required (c) Equal to the temperature required (d) None of the above 39. Pressure of refrigerant in the evaporator should be (a) Equal to the atmospheric pressure (b) Less than the atmospheric pressure (c) Greater than the atmospheric pressure (d) None 40. Latent heat is highest for





(a) Refrigerant-22

(b) Ammonia

(c) Water

(d) None

 6SHFLÀFKHDWDWFRQVWDQWSUHVVXUHLVKLJKHVW (a) Refrigerant-22

(b) Ammonia

(c) Water

(d) None

 6SHFLÀFKHDWVDWFRQVWDQWSUHVVXUHDQGDWFRQVWDQWYROXPHDUHHTXDO (a) Vapor refrigerant-22

(b) Vapour ammonia

(c) Water vapours

(d) None

43. Cp of air is (a) > Cv of air

(b) = Cv of air

(c) < Cv of air

(d) None

44. Primary refrigerant is one which is sensibly (a) Heated in the evaporator (b) Cooled in the evaporator

157

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(c) Neither heated in evaporator nor cooled in condenser (d) None 45. Secondary refrigerant is one which is (a) Cooled by the water

(b) Cooled by the air

(c) Cooled by the primary refrigerant

(d) None

46. Which one acts as a primary, secondary as well as a tertiary refrigerant (a) Water

(b) Ammonia

(c) Freon-22

(d) None

47. Which one acts as a primary, secondary as well as a tertiary refrigerant (a) Water

(b) Air

(c) Freon-22

(d) None

48. Which is the primary refrigerant in central air conditioning plant (a) Air

(b) Water

(c) Freon-22

(d) None

49. Which is the secondary refrigerant in central air conditioning plant (a) Air

(b) Water

(c) Freon-22

(d) None

50. Which is the tertiary refrigerant in central air conditioning plant (a) Air

(b) Water

(c) Freon-22

(d) None

Answers 1. (d)

2. (a)

3. (e)

4. (b)

5. (d)

6. (e)

7. (e)

8. (a)

9. (a)

10. (b)

11. (e)

12. (c)

13. (b)

14. (c)

15. (a)

16. (d)

17. (c)

18. (b)

19. (b)

20. (d)

21. (a)

22. (a)

23. (a)

24. (c)

25. (e)

26. (c)

27. (b)

28. (d)

29. (c)

30. (d)

31. (a)

32. (d)

33. (a)

34. (d)

35. (c)

36. (d)

37. (a)

38. (b)

39. (c)

40. (c)

41. (c)

42. (d)

43. (c)

44. (d)

45. (c)

46. (d)

47. (b)

48. (c)

49. (b)

50. (a)

Section III Fluid Mechanics and Hydraulics

8 Fluid Mechanics

Learning Objectives   

‡ 8QGHUVWDQGLQJWKHWHUPIOXLGDQGLWVSURSHUWLHV ‡ *HWWLQJLGHDDERXWSUHVVXUHDQGSUHVVXUHPHDVXUHPHQW ‡ 2YHUYLHZRI%HUQRXOOL·VSULQFLSOHDQGLWVDSSOLFDWLRQLQIOXLGPHFKDQLFV

8.1

INTRODUCTION

There is air around us, and there are rivers and seas near us. ‘The flow of a river never ceases to go past, nevertheless it is not the same water as before. Bubbles floating along on the stagnant water now vanish and then develop but have never remained.’ So stated Chohmei Kamo, the famous 13th century essayist of Japan, in the prologue of Hohjohki, his collection of essays. In this way, the air and the water of rivers and seas are always moving. Such a movement of gas or liquid (collectively called ‘fluid’) is called the ‘flow’, and the study of this is ‘fluid mechanics’. While the flow of the air and the water of rivers and seas are flows of our concern, so also are the flows of water, sewage and gas in pipes, in irrigation canals, and around rockets, aircraft, express trains, automobiles and boats. And so too is the resistance which acts on such flows. Throwing baseballs and hitting golf balls are all acts of flow. Furthermore, the movement of people on the platform of a railway station or at the intersection of streets can be regarded as forms of flow. In a wider sense, the movement of social phenomena, information or history could be regarded as a flow, too. In this way, we are in so close a relationship with flow that the ‘fluid mechanics’ which studies flow is really a familiar thing to us.

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A fluid is a substance that deforms continuously in the face of tangential or shear stress, irrespective of the magnitude of shear stress. This continuous deformation under the application of shear stress constitutes a flow. In this connection a fluid can also be defined as the state of matter that cannot sustain any shear stress. Fluid mechanics deals with comprehending, anticipating, and controlling the conduct of a fluid. For an engineer, fluid mechanics is an essential field of the connected sciences with numerous practical and exciting applications. Fluid mechanics is concerned with the behaviour of materials which deform without limit under the influence of shearing forces. Even a very small shearing force will deform a fluid body, but the velocity of the deformation will be correspondingly small. This property serves as the definition of a fluid: the shearing forces necessary to deform a fluid body go to zero as the velocity of deformation tends to zero. On the contrary, the behaviour of a solid body is such that the deformation itself, not the velocity of deformation, goes to zero when the forces necessary to deform it tend to zero. We normally recognise three states of matter: solid, liquid and gas. However, liquid and gas are both fluids: in contrast to solids they lack the ability to resist deformation. Because a fluid cannot resist the deformation force, it moves, it flows under the action of the force. Its shape will change continuously as long as the force is applied. A solid can resist a deformation force while at rest, this force may cause some displacement but the solid does not continue to move indefinitely. The deformation is caused by shearing forces which act tangentially on a surface. A

F

A¢

C

B

B¢

F

D

Figure 8.1 Shearing force F acting on a fluid element.

There are two aspects of fluid mechanics which make it different to solid mechanics. 1. The nature of a fluid is much different from that of a solid. 2. In fluids we usually deal with continuous streams of fluid without a beginning or end. In solids we only consider individual elements.

8.1.1 Characteristics of a Fluid Fluids are divided into liquids and gases. A liquid is hard to compress and as in the ancient saying ‘Water takes the shape of the vessel containing it’, it changes its shape according to the shape of its container with an upper free surface. Gas on the other hand is easy to compress, and fully expands to fill its container. There is thus no free

Fluid Mechanics 163

surface. Consequently, an important characteristic of a fluid from the viewpoint of fluid mechanics is its compressibility. Another characteristic is its viscosity. Whereas a solid shows its elasticity in tension, compression or shearing stress, a fluid does so only for compression. In other words, a fluid increases its pressure against compression, trying to retain its original volume. This characteristic is called compressibility. Furthermore, a fluid shows resistance whenever two layers slide over each other. This characteristic is called viscosity. In general, liquids are called incompressible fluids and gases compressible fluids. Nevertheless, for liquids, compressibility must be taken into account whenever they are highly pressurised, and for gases compressibility may be disregarded whenever the change in pressure is small. Although a fluid is an aggregate of molecules in constant motion, the mean free path of these molecules is 0.06 pm or so even for air of normal temperature and pressure, so a fluid is treated as a continuous isotropic substance. Meanwhile, a non-existent, assumed fluid without either viscosity or compressibility is called an ideal fluid or a perfect fluid. A fluid with compressibility but without viscosity is occasionally discriminated and called a perfect fluid, too. Furthermore, a gas subject to Boyle’s-Charles’ law is called a perfect or ideal gas.

8.1.2

Units and Dimensions

All physical quantities are given by a few fundamental quantities or their combinations. The units of such fundamental quantities are called base units and dimensions 7 units, combinations of these are called derived units. The system in which length, mass and time are adopted as the basic quantities, and from which the units of other quantities are derived, is called the absolute system of units. MKS System of Units: This is the system of units where metre (m) is used for the unit of length, kilogram (kg) for the unit of mass, and second (s) for the unit of time as the base units. CGS System of Units: In this system of units centimetre (cm) is used for length, gram (g) for mass, and second (s) for time as the base units. International system of units (SI), the abbreviation of La Systkme International G·8QLWHVLVWKHV\VWHPGHYHORSHGIURPWKH0.6V\VWHPRIXQLWV,WLVDFRQVLVWHQWDQG reasonable system of units which makes it a rule to adopt only one unit for each of the various quantities used in such fields as science, education and industry. There are seven fundamental SI units, namely, metre (m) for length, kilogram (kg) for mass, second (s) for time, ampere (A) for electric current, kelvin (K) for thermodynamic temperature, mole (mol) for mass quantity and candela (cd) for intensity of light. Derived units consist of these units.

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Basic Mechanical Engineering

8.1.3 Density, Specific Gravity, Specific Volume Density or Mass Density: Density (mass density) of a fluid is defined as the ratio of mass of a fluid to the volume of the fluid. In other words, it is defined as the mass per unit volume of a fluid. Mass density (density) = mass/volume r = m/V Specific Weight: Specific weight of a fluid is defined as the ratio of the weight of a fluid to the volume of the fluid. Or weight of a fluid per unit volume is called its specific weight. Specific weight = (weight of fluid)/(volume of fluid) w = mg/V = rg Specific Volume: Specific volume of a fluid is defined as the ratio of the volume of a fluid to the mass of the fluid. In other words, it may also be defined as volume per unit mass of a fluid. Specific volume = (volume of fluid)/(mass of fluid) Specific volume = V/m = 1/r Specific Gravity: Specific gravity is defined as the ratio of the density of a fluid to the density of a standard fluid. For liquids, water is taken as the standard fluid where as for gases air is taken as the standard fluid. It is denoted by the symbol S. Specific gravity, S = (density of liquid)/(density of water) = (density of gas )/(density of air) The density of water is taken as 1000 kg/cubic metre

8.1.4 Viscosity As shown in Fig. 8.2, suppose that liquid fills the space between two parallel plates of area A each and gap h, the lower plate is fixed, and force F is needed to move the upper plate in parallel at velocity U. Whenever Uh/v < 1500 (v = p/p: kinematic viscosity), laminar flow is maintained, and a linear velocity distribution, as shown in the figure, is obtained. Such a parallel flow of uniform velocity gradient is called the Couette flow. In this case, the force per unit area necessary for moving the plate, i.e., the shearing stress (t), is proportional to U and inversely proportional to h 8VLQJ D SURSRUWLRQDO constant m, it can be expressed as follows: t=

F U =m A h

Fluid Mechanics 165

The proportional constant m is called the viscosity, the coefficient of viscosity or the dynamic viscosity. F

U A u

y

t

h

t

x O

Figure 8.2 U In the case of U = 0

dy

y y x

du

O

Figure 8.3

Such a flow where the velocity u in the x direction changes in the y direction is called shear flow. Figure 8.2 shows the case where the fluid in the gap is not flowing. However, the velocity distribution in the case where the fluid is flowing is shown in Fig. 8.3. Extending to such a flow, the shear stress z on the section dy, distance y from the solid wall, is given by the following equation: t= m

dy dx

This relation was found by Newton through experiment, and is called Newton’s law of viscosity.

8.1.5 Surface Tension The surface of a liquid is about to shrink, and its free surface is in such a state where each section pulls another as if an elastic film is being stretched. The tensile strength per unit length of an assumed section on the free surface is called the surface tension. Surface tensions of various liquids are given in Table 8.1.

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Basic Mechanical Engineering

Table 8.1 Liquid

Surface liquid

N/m

Water

Air

0.0728

Mercury

Air

0.476

Mercury

Water

0.373

Methyl alcohol

Air

0.023

A dewdrop on a plant leaf is spherical in shape. This is also because of the tendency to shrink due to surface tension. Consequently, its internal pressure is higher than its peripheral pressure. Putting d as the diameter of the liquid drop, T as the surface tension, and p as the increase in internal pressure, the following equation is obtained owing to the balance of forces: pdT =

pd 2 Dp 4

Dp = 4T/d The same applies to the case of small bubbles in a liquid.

8.1.6 Compressibility As shown in Fig. 8.4, assume that fluid of volume V at pressure p decreased its volume by DV due to the further increase in pressure by Dp. In this case, since the cubic dilatation of the fluid is DV/V, the bulk modulus K is expressed by the following equation: K=

Dp dp = -V DV / V dV

Its reciprocal b b = 1/K is called the compressibility, whose value directly indicates how compressible the fluid is. For water of normal temperature/pressure K = 2.06 × 109 Pa, and for air K = 1.4 × l09 Pa assuming adiabatic change. In the case of water, b = 4.85 × l0–10 Pa, and it shrinks only by approximately 0.005% even if the atmospheric pressure is increased by 1 atm. Putting p as the fluid density and M as the mass, since rV = M = constant, assume an increase in density whenever the volume has decreased by DV, hence K= r

Dp dp =r dr Dr

Fluid Mechanics 167 W

p + Dp

DV V

Figure 8.4

8.1.7 Characteristics of an Ideal Gas Let p be the pressure of a gas, u the specific volume, T the absolute temperature and R the gas constant. Then the following equation results from Boyle’s-Charles’ law: pv = RT This equation is called the equation of state of the gas, and v = l/r

8.2

PROPERTIES OF FLUIDS

Any characteristic of a system is called a property. Some familiar properties are pressure P, temperature T, volume V, and mass m. The list can be extended to include less familiar ones such as viscosity, thermal conductivity, modulus of elasticity, thermal expansion coefficient, electric resistivity, and even velocity and elevation. Properties are considered to be either intensive or extensive.

8.2.1

Intensive Properties

These properties are independent of the mass of a system, such as temperature, pressure, and density.

8.2.2 Extensive Properties These are those properties whose values depend on the size—or extent—of the system. Total mass, total volume V, and total momentum are some examples of extensive properties.

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Basic Mechanical Engineering

An easy way to determine whether a property is intensive or extensive is to divide the system into two equal parts with an imaginary partition, as shown in Fig. 8.4. Each part will have the same value of intensive properties as the original system, but half the value of the extensive properties. Generally, upper case letters are used to denote extensive properties (with mass m being a major exception), and lower case letters are used for intensive properties (with pressure P and temperature T being the obvious exceptions).

8.2.3

Specific Properties

Extensive properties per unit mass are called specific properties. Some examples of specific properties are specific volume (v = V/m) and specific total energy (e = E/m).

Figure 8.5 Criteria to differentiate extensive and intensive properties.

The state of a system is described by its properties. But we know from experience that we do not need to specify all the properties in order to fix a state. Once the values of a sufficient number of properties are specified, the rest of the properties assume certain values. That is, specifying a certain number of properties is sufficient to fix a state. The number of properties required to fix the state of a system is given by the state postulate: The state of a simple compressible system is completely specified by two independent, intensive properties. Two properties are independent if one property can be varied while the other one is held constant. Not all properties are independent, and some are defined in terms of others.

8.3

MEASUREMENT OF PRESSURE

Contrary to temperature, pressure is not a basic magnitude in the International System RI 8QLWV EXW LV D GHULYHG XQLW 3UHVVXUH LV IRUFH DFWLQJ SHU XQLW DUHD HDFK IRUFH DQG

Fluid Mechanics 169

space being derived from the essential magnitudes: time, length, and mass. The pressure measurement reduces to force measuring on a well-defined space. In its flip, forces are often measured by the acceleration of a mass or by the deformation of a body. In summary, elementary pressure scientific discipline could be a subfield of mass scientific discipline (usually split in 2 subfields: mass and density, and force and pressure). Pressure variations don’t imply that pressure could be a relative magnitude like position or time, that depends on the organisation or an initial point set by the observer; pressure, like temperature (or amount, length, or mass), is a magnitude, as are going to be seen below (e.g., the length occupied by a famous quantity of a perfect gas within a perfect cylinderpiston device, are often used either as a measuring instrument or as a piezometer). Moreover, there aren’t any ‘deltas’ within the gas equation. pV = mRT, all variables there are unit absolute. By the way, how will one change the pressure of a system? In a very gaseous system, equation pV = mRT already shows many ways that can be used to change the pressure of a system. Those are ‡ by increasing the mass of gas within; ‡ by increasing the temperature of the system; and ‡ by decreasing the volume of the system. 7KRVHDUHVWDWLFZD\VWKDWDVFROXPQXSDGGLWLRQDOÁXLGRUÀWWLQJDGHFHQWZHLJKW could knock off a liquid. However, to increase pressure in a gaseous system, is with a rotodynamic pump, by initially giving it momentum with associate degree vane, so decelerating it in a diffuser. Giant pressure outputs, however, sometimes demand volumetric-type pumps, wherever the momentum transfer will be performed quasi-statically.

8.3.1 Characteristics of Pressure The pressure has the following three characteristics. 1. The pressure of a fluid always acts perpendicular to the wall in contact with the fluid. 2. The values of the pressure acting at any point in a fluid at rest are equal regardless of its direction. Imagine a minute triangular prism of unit width in a fluid at rest as shown in Fig. 8.5. Let the pressure acting on the small surfaces dA, dA1, and dA2 be p, p1 and p2 respectively. The following equations are obtained from the balance of forces in the horizontal and vertical directions: p1dA1 = pdA sin q p2dA2 = pdA cos q +

1 dA1 dA2 rg 2

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Basic Mechanical Engineering C¢

C p

C p1

A¢

dA1

dA2

B¢

A

dA

B

A p2

B

Figure 8.6

8.3.2 Pressure Unit: “The Pascal” Pressure is represented by small p, and its unit is pascal (Pa), that is, 1 N/m2. The pressure put off level on earth is around 105 Pa, which varies with position and time; its average, around 101.3 ± 0.2 kPa, was established precisely as 1.01325 Pa by international agreement, as given the pressure of a 760 millimetre high mercury column at 0ºC and g = 9.80665 m/s2  ZKDW ZDV QDPHG RQH DWP  DWP ŋ  N3D  7\SLFDO ZHDWKHU modification causes some 1 kPa pressure variations (up to 5 kPa in hurricanes). there’s DWUHQGWRXVH3DEHFDXVHWKHFXVWRPDU\YDOXHRISUHVVXUH,83$&PRGLILHGLQ from one atm to 105 Pa); though the substitution of 1 atm by 1 bar is insignificant in most engineering issues, notice that their distinction is that the typical pressure variation EHFDXVH RI ZHDWKHU FKDQJHV DW D VLWH ZKLFK  EDU   PP+J ZKHUHDV  DWP ŋ  mmHg; the previous pressure units of 1 metric linear unit of mercury column, 1 mmHg, RQH WRUU ŋ  PP+J RQH PHWUH RI ZDWHU FROXPQ RQH NLORSRQG SHU VT FP DQG VR RQ DUHZKROO\RXWGDWHG 7KHEDUEDUŋ3D LVDQRQ6,XQLWSUHVHQWO\DFFHSWHGWREH XVHG ZLWK WKH 6\VWHPH ,QWHUQDWLRQDO G·8QLWHV KRZHYHU ZKRVH XVH LV GLVFRXUDJHG DV well as its submultiples, because the pressure unit, 1 mbar = 100 Pa). Extremely high pressures, from 1 GPa to 10 GPa, are employed in geologic applications, novel food process techniques, water-jet cutting (by micro-cracking and erosion), and so on. Many techniques have been developed for the measurement of pressure and vacuum. Instruments used to measure pressure are called pressure gauges or vacuum gauges. A manometer could also refer to a pressure measuring instrument, usually limited to measuring pressures near the atmospheric. The term manometer is often used to refer specifically to liquid column hydrostatic instruments. A vacuum gauge is used to measure the pressure in vacuum—which is further divided into two subcategories: high and low vacuum (and sometimes ultra-high vacuum). The applicable pressure range of many of the techniques used to measure vacuums have an overlap. Hence, by combining several different types of gauge, it is possible to measure system pressure continuously from 10 mbar down to 10– 11 mbar.

Fluid Mechanics 171

Pressure measurements are often used to monitor processes (as for weather forecast) or to regulate them (as in a pressure cooker). The fundamental pressure control is pressure switch (equivalent to the thermostat, however, the word presostat is uncommon in English), used to limit the pressure at a point (by shifting off the pressure building source). Two opposite pressure switches may be accustomed to keep the pressure regulated (switching a pump on and off), although some alternative mechanical devices (like gasometers for gases) are generally more popular. Fluid pressure may be controlled by regulating the flow of the fluid, as seen in relief valves and safety valves. Pressure measurement is usually used as an indirect methodology to judge alternative physical magnitudes, like liquid levels (and diving depth), altitude measurement (atmospheric pressure), speed of the fluid (Pitot tube), fluid flow-rates (Venturi meter), sound level (acoustics), fluid forces on an obstacle (e.g., wind masses on buildings), and so on.

Figure 8.7 Atmospheric pressure measurement.

8.3.3

Absolute, Gauge and Differential Pressures

Although no pressure is an absolute quantity, everyday pressure measurements, such as for tire pressure, are usually made relative to ambient air pressure. In other cases measurements are made relative to a vacuum or reference. When distinguishing between these zero references, the following terms are used. Absolute pressure is zero referenced against a perfect vacuum, so it is equal to gauge pressure plus atmospheric pressure. Gauge pressure is zero referenced against ambient air pressure, so it is equal to absolute pressure minus atmospheric pressure. Negative signs are usually omitted. Differential pressure is the difference in pressure between two points.

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8.3.4

Hydrostatic Pressure

In the presence of a consistent field, applicable to gravity on the planet surface, pressure decreases with height inside a fluid, in accordance with the reduction of the load of the mass. In liquids, pressure increases downward linearly (with the proper liquid model), at a rate of 100 kPa each 10 m of water column roughly, whereas in gases pressure variation are so tiny (e.g., around one kPa each 100 m) that it is neglected in all engineering issues.

8.3.5

Archimedes’ Principle (Buoyancy)

Hydrostatic pressure has many vital effects, just like the proven fact that liquid free surfaces are horizontal at rest, or that we can take out liquid from a deep well with a pump on the bottom (The pump is needed to be installed 10 m above the free surface of water, or higher), or the siphon impact. The most vital impact, notwithstanding, is the appearance of buoyant forces on bodies within the fluids or in different-density fluid masses (as in the case of natural convection). Archimedes’ principle states that any object fully or partly submerged in a fluid experiences a buoyant force, as a result of higher pressures on the bottom of the body compared to the top, which is equal to the magnitude of weight of the fluid displaced by the body. Besides submarines, boats and balloons, Archimedes’ principle is made use of within the experimental determination of density with marked floating bodies (hydrometers or densitometers).

8.3.6 Cavitation Cavitation is the fast formation and collapse of gas pockets by local depressurization in a flowing liquid till vaporisation happens. Cavitation typically takes place in high-speed propellers at low depths (starting at the blade tip), at hydrofoil curve, at pump intakes, at valve outlets, etc. Cavitation is generated by an acoustic pressure field, focusing supersonic waves into a section in a liquid mass. Cavitation produces vibrations, noise, loss of efficiency, etc. The noise in cavitation comes from the abrupt collapse of the cavity. Cavitation may be avoided by increasing the pressure or decreasing the temperature (that is why pumps in heating systems are typically set at the cold return line). In some rare applications cavitation is needed, as when planning for super cavitating crafts, during which a large gaseous region is formed behind a pointed cone-like object forced to travel at extremely high speed; it happens that the hydraulic resistance of this water separating body is way less than that of a solid shell with the form of the entire bubble (a torpedo has been designed following that approach).

Fluid Mechanics 173

8.3.7 Water Hammer Most applications of Bernoulli’s equation deal steady flow. Presently we have a tendency to tend to influence the most necessary unsteady phenomena in pipe flow. The abrupt stopping of flow of liquid by closing a valve downstream, produces large pressure fluctuations causing vibration, noise, and eventually pipe-wall failure. This is often referred to as water hammer.

8.4

BERNOULLI’S EQUATION

Bernoulli equation (named after the Swiss scientist Bernoulli, 1700-1782) is the best well known and most used equation in hydraulics. It’s an exact first integral of the inviscid Navier-Stokes equations (Euler equations) on a contour, relating in an easy manner the fluid motion (its kinetic energy), to the thermodynamical parameters, i.e., temperature, pressure, and density. Moreover, if the flow is irrotational, the constant is the same for all streamlines, and Bernoulli’s equation applies to the total field. The most used type of Bernoulli’s equation simply reads p + rgz + rv2/2 = constant along a streamline in steady inviscous flow of an incompressible fluid in a gravity field, i.e., that total pressure (i.e., ‘static pressure’, ‘head pressure’ and ‘dynamic pressure’), pt ŋ p + rgz + rv2/2, is conserved. This is Bernoulli’s equation in pressure type, as employed in aerodynamics. In applied science it’s typically written in energy terms (dividing by r), and in technology in height terms, named heads (dividing by rg). The two basic functions in pipe-flow calculations are finding the total-pressure losses on the circuit, and matching them with suitable pumping. Notice that, in a piping portion without external work input, total pressure will solely decrease, whereas ‘static pressure’ will increase at the expense of the other terms, disconfirming the often-stated rule that fluids move from high to low pressure.

8.4.1 Applications of the Bernoulli's Equation The Bernoulli equation can be applied to a great many situations not just the pipe flow we have been considering up to now. In the following sections we will see some examples of its application to flow measurement from tanks, within pipes as well as in open channels. Pitot Tube: If a stream of uniform velocity flows into a blunt body, the stream lines take a pattern similar to this (Fig. 8.8):

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2

1

Figure 8.8 Streamlines around a blunt body.

Note how a part of fluid moves to the left and other to the right. But one, in the centre, goes to the tip of the blunt body and stops. It stops because at this point the velocity is zero — the fluid does not move at this one point. This point is known as the stagnation point. From the Bernoulli’s equation we can calculate the pressure at this point. Apply Bernoulli along the central streamline from a point upstream where the velocity is u1 and the pressure p1 to the stagnation point of the blunt body where the velocity is zero, u2 = 0. Also z1 = z2. Venturimeter: Venturimeter 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. It then returns to the original dimensions of the pipe by a gently diverging ‘diffuser’ section. By measuring the pressure differences discharge can be calculated. This is a particularly accurate method of flow measurement as energy losses are very small. about 6° about 20°

2 1

z2 z1

h datum

Figure 8.9 Venturimeter.

Fluid Mechanics 175

Flow Through a Small Orifice: We are to consider the flow from a tank through a hole in the side near the base. The general arrangement and a closeup of the hole and streamlines are shown in Fig. 8.10.

1

Aactual

h

2

Vena contractor

Figure 8.10 Tank and streamlines of flow out of the sharp edged orifice.

The shape of the edges of the holes are sharp to minimise frictional losses by minimising the contact between the hole and the liquid — the only contact is the very edge. Looking at the streamlines you can see how they contract after the orifice to a minimum value when they all become parallel, at this point, the velocity and pressure are uniform across the jet. This convergence is called the vena contracta (from Latin ‘contracted vein’). It is necessary to know the amount of contraction to allow us to calculate the flow. We can predict the velocity at the orifice using the Bernoulli equation.

8.5

SUMMARY

Fluid mechanics is associated with the study of fluids in motion and at rest and is an important subject of many of the branches of engineering. Its numerous branches are fluid statics, fluid mechanics and fluid dynamics. A substance that flows is a fluid. All liquids and volatilised substances are thought to be fluids. Water, oil, and others are important in our daily life as they’re used for numerous applications. For example, water is employed for generation of electricity in power plants and thermal power plants, in atomic energy plants, oil is employed for lubrication of vehicles, etc. Hydraulics is the branch of science that studies the behaviour of fluids once they are in a state of motion or rest. Whether or not the fluid is at rest or motion, it’s subjected to completely different forces and different weather conditions and it behaves in these conditions as per its physical properties. Hydraulics is associated with static, kinematic, and dynamic aspects of the fluid. What’s the requirement of finding out fluids as a facet of engineering? Fluids are an integral a part of our daily life. Engineering permits us to explore the potential of fluids for a variety of applications and functions. Some of these include:

176

Basic Mechanical Engineering

1. Some fluids when burnt, generate large amount of heat, which is used for numerous applications, for example, gas and diesel for vehicles. 2. Some fluids like oil exert extreme high pressure or force. These fluids are often used for lifting heavy masses, for example, fluids employed in hydraulic machines and hydraulic lifters. 3. Fluids having flow properties are used for the lubrication of assorted machines.

SOLVED EXAMPLES

Example 8.1

po

Why does a ship float?

Solution: Fluid pressure acts all over the wet surface of a body floating in a fluid, and the resultant pressure acts in a vertical upward direction. This force is called buoyancy. The buoyancy of air is smaller compared with the gravitational force of the immersed body, so it is normally ignored.

h

Body

po + rgh

Example 8.2

Water flows steadily up the vertical 0.1 m diameter pipe and exits from the nozzle, which is 0.05 m in diameter, discharging to atmospheric pressure. The stream velocity at the nozzle exit must be 20 m/s. Calculate the minimum gauge pressure required at section .

V2 2

4m

1

Solution: 1. Statement of the problem (a) Given  ‡ )OXLGLVZDWHUr = 999 kg/m3 assuming at 15°C).  ‡ 7KHIORZJRHVVWHDGLO\XSDYHUWLFDOFKDQQHO fi Steady state condition and gravity must be concerned.

Fluid Mechanics 177

   

‡ D1 = 0.1 m and D2 = 0.05 m ‡ p2 = atmospheric pressure ‡ V2 = 20 m/s (b) Find ‡ 7KHPLQLPXPJDXJHSUHVVXUHUHTXLUHGDWVHFWLRQ 2. System diagram

V2, z2, p2 2

4m

z1

1

Streamline

   



3. Assumptions ‡ 6WHDG\VWDWHFRQGLWLRQ ‡ ,QFRPSUHVVLEOH IOXLG IORZ ZRUNLQJ IOXLG ZDWHU LV XVXDOO\ DVVXPHG WR EH incompressible) ‡ ,QYLVFLGIOXLGIORZ ‡ 'SUREOHP 4. Governing equations ‡ %HUQRXOOL·VHTXDWLRQ

p V2 + + gz = const. 2 r

Restrictions: 1. Steady flow 2. Incompressible flow 3. Frictionless flow 4. Flow along a streamline 

  ∂ rdV + Ú rV ◊ dA … Integral version of mass conservation (to find Ú CS ∂t CV V1 first)

‡ 0 =

178

Basic Mechanical Engineering

Imagine a control volume (one possibility is shown on the diagram by the box sketched) such that control surface intersects with the inlet and the outlet areas. Now, for an incompressible fluid flow problem, the equation   above fi 0 = Ú V ◊ dA CS

1 inlet (M) and 1 outlet (N) fi 0 = – V1A1 + V2A2 5. Detailed solution The Bernoulli’s equation can be applied between any two points on a streamline provided that the other three restrictions are satisfied. The result is p1 V12 p V2 + + gz1 = 2 + 2 + gz2 2 2 r r

where subscripts 1 and 2 represent any two points on a streamline that correspond to M and in this particular problem. Now, the Bernoulli’s equation can be written as: p1 - p2 1 2 2 = [V2 - V1 ] + g( z2 - z1 ) 2 r

Substituting 0 = – V1A1 + V2A2 fi V1 =

A2 V2 into the above equation, A1

2 p1 - p2 ÊA ˆ ˘ 1È = ÍV22 - Á 2 V2 ˜ ˙ + g( z2 - z1 ) r 2Í Ë A1 ¯ ˙ Î ˚

2 p1 - p2 1 2 È Ê A2 ˆ ˘ = V2 Í1 - Á ˜ ˙ + g( z2 - z1 ) r 2 Í Ë A1 ¯ ˙ Î ˚

p 2 2 A2 4 D2 D22 p1 - p2 1 2 È Ê D22 ˆ ˘ = = = V2 Í1 - Á 2 ˜ ˙ + g( z2 - z1 ) ∵ 2 A1 p 2 D12 r ÍÎ Ë D1 ¯ ˙˚ D1 4

(p1 gauge + patm) – (p2 gauge

Ï1 Ô 2 + patm) = r Ì V2 2 ÔÓ

p1 gauge + p2 gauge

Ï1 Ô = r Ì V22 2 ÓÔ

¸ È Ê D2 ˆ 2 ˘ Ô Í1 - Á 22 ˜ ˙ + g( z2 - z1 )˝ Ë D1 ¯ ˙ Ô˛ ÎÍ ˚

¸ È Ê D2 ˆ 2 ˘ Ô Í1 - Á 22 ˜ ˙ + g( z2 - z1 )˝ Ë D1 ¯ ˙ ÎÍ ˚ ˛Ô

Fluid Mechanics 179

Finally, p2 gauge is 0 kPa because p2 = atmospheric pressure. Thus, p1 gauge

Ï Ô1 – (0 kPa) = (999 kg/m ) Ì ( 20 m/s)2 ÔÓ 2 3

¸ È Ê (0.05 m)2 ˆ 2 ˘ Í1 - Á ˙ + ( 9.81 m/s 2 ) [( 4 m) - (0 m)]Ô˝ 2 ˜ Í Ë (0.1 m) ¯ ˙ Ô˛ Î ˚

Therefore, the minimum gauge pressure required at section

, p1 gauge = 226.5 kPa

Example 8.3

Water flows from a very large tank through a 2 inch diameter tube. The dark liquid in the manometer is mercury. Estimate the velocity in the pipe and the rate of discharge from the tank.

Solution: 1. Statement of the problem (a) Given  ‡ $OOLQIRUPDWLRQLVGHVFULEHGLQWKHILJXUHDERYH (b) Find  ‡ 9HORFLW\LQWKHSLSH  ‡ 5DWHRIGLVFKDUJHIURPWKHWDQN 2. System diagram 3. Assumptions  ‡ 6WHDG\VWDWHFRQGLWLRQ  ‡ ,QFRPSUHVVLEOH IOXLG IORZ ZDWHU ZRUNLQJ IOXLG LV XVXDOO\ FRQVLGHUHG incompressible)

180

 



Basic Mechanical Engineering

‡ ,QYLVFLGIOXLGIORZ ‡ 'SUREOHP 4. Governing equation ‡ %HUQRXOOL·VHTXDWLRQ

p V2 + + gz = const. 2 r

Restrictions: 5. Steady flow 6. Incompressible flow 7. Frictionless flow 8. Flow along a streamline 5. Detailed solution Physical quantities: S.G. = 13.55 mercury

(Specific gravity of mercury)

rwater = 1.94 slug/ft3 at 59°F g = 32.174 ft/s2 Pressure variation in any static fluid is described by the basic pressure-height relation: dp = - rg (This equation is from fluid statics.) dz

Thus, dp = - rg fi dz

Ú

p p0

z

dp = - Ú rgdz fi p – p0 = rg(z0 – z) fi p – p0 = rgh z 0

CRQVLGHUWKHSRUWLRQRIILJXUH$%DQG&DVD8WXEHPDQRPHWHU8VLQJWKLVVWDWLFV relationship (Pascal’s law), pA – pB = – rwater gh1 pB – pC = – rmercury gh2 Adding the two equations gives pA – pC = rmercury gh2 – rwater gh1 Since pC = patm pA – patm = S.G.mercury rwater gh2 – rwater gh1 = rwater g(S.G.mercury h2 – h1) pA = patm = (1.94 slug/ft3) (32.174 ft/s2) [(13.55) (0.5 ft) – (2 ft)] = 298.044 lbf/ft2

Fluid Mechanics 181

This pressure pA is same as the pressure p2 because of the absence of pressure drop for inviscid fluid flow. Thus, p2 = 298.044 lbf/ft2 + patm The Bernoulli’s equation can be applied between any two points on a streamline provided that the other three restrictions are satisfied. The result is: p1 V12 p V2 + + gz1 = 2 + 2 + gz2 2 r 2 r

where subscripts 1 and 2 represent any two points on a streamline that correspond to M and in this particular problem. Now, the Bernoulli’s equation can be written as: p - p2 1 2 [V2 - V12 ] = 1 + g( z1 - z2 ) 2 r È p - p2 ˘ V22 = 2 Í 1 + g( z1 - z2 )˙ + V12 Î r ˚

8sing an approximation V1 § 0 ft/s because Areservoir >> Apipe, and p1 = patm, È p - p2 ˘ V22 = 2 Í atm + g( z1 - z2 )˙ + (0 ft/s)2 r Î ˚ È p - ( 298.044 lbf/ft + patm ) ˘ + ( 32.174 ft/s 2 ) [(12 ft) - (0 ft)]˙ = 464.914 ft /s 2 = 2 Í atm (1.94 slug/ft ) Î ˚

Finally, V2 = 21.56 ft/s Rate of discharge from the tank is: Q2 = V2 A2 = V2

p 2 pÊ 2 D2 = (21.56 ft/s) Á 4 4 Ë 12

2

ˆ ft˜ = 0.470 ft 3 /s ¯

3 y - y 2 in 4 which u is the velocity in m/s at a distance y metre above the plate. Determine the shear stress at y = 0.20 m. Take the dynamic viscosity of the fluid as 0.86 Ns/m2.

Example 8.4

Velocity distribution for a flow over a flat plate is given by u =

Solution: u=

3 y - y2 4

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Basic Mechanical Engineering

du 3 = - 2y dy 4

At y = 0.20 m, du 3 = - 2(0.20 ) dy 4

= 0.75 – 0.40 = 0.35 As dynamic viscosity, m = 0.086 Ns/m2, du Shear stress = m = 0.86 × 0.35 dy = 0.301 Pa

Example 8.5 Determine the dynamic viscosity of a liquid having kinematic viscosity 7 × 10– 4 m2/s and specific gravity 1.8. Solution: Specific gravity of liquid =

Density of liquid Density of water

1.8 =

Density of liquid 1000

Therefore, density of liquid = 1.8 × 1000 = 1800 kg/m3 As kinematic viscosity =

Dynamic viscosity Density

7 × 10–4 =

or,

m 1800

m = 7 × 10– 4 × 1800 = 1.26 Ns/m2 = 12.6 poise

Example 8.6

Glycerin has a density of 1260 kg/m3 and kinematic viscosity of 1.83 × 10– 3 m2/s. What shear stress is required to deform this fluid at a strain of 103/s?

Solution: Given:

r = 1260 kg/m3 n = 1.83 × 10– 3 m2/s du = 103/s dy

Fluid Mechanics 183

Shear stress = r n

du = 1260 ¥ 1.83 ¥ 10 - 3 ¥ 10 3 = 2.305 kPa dy

Example 8.7 A liquid has a specific gravity of 1.8 and kinematic viscosity of 6 stokes. What is its dynamic viscosity? Solution:

Specific gravity s = 1.8

Kinematic viscosity n = 6 stokes = 6 × 10– 4 m2/s S=

rliquid rwater

rliquid = s × rwater = 1.8 × 1000 = 1800 kg/m3 Dynamic viscosity, m = 1800 × 6 × 10– 4 = 1.08 Ns/m2

Example 8.8

Calculate the specific weight, density and specific gravity of one litre of liquid weighing 6 N.

Solution: Weight = 6 N (i) Specific weight, w =

Weight = Volume

6 = 6000 N/m2 1 1000

(ii) Density = 6000/9.81 = 611.62 kg/m3 (iii) Specific gravity =

Density of liquid 611.62 = = 0.6116 Density of water 1000

Example 8.9

Calculate the density, specific weight and weight of the kerosene having specific gravity of 0.82.

Solution: Volume = 1 litre = 0.001 m3 Specific gravity of kerosene, s = 0.82 (i) Density, r = s × 1000 = 0.82 × 1000 = 820 kg/m3 (ii) Specific weight = r × g = 820 × 9.81 = 8044.2 N/m3 (iii) Specific weight, w =

Weight W = Volume 0.001

or, W = 8044.2 × 0.001 = 8.0442 N Example 8.10 A flat plate of area 1.3 m2 is pulled with a speed of 0.3 m/s relative to another plate located at a distance of 0.15 × 10– 3 m from it. Find the force and the power required to maintain the speed if the fluid separating them has dynamic viscosity of 1 poise.

184

Basic Mechanical Engineering

Solution: Area of plate = 1.3 m2 Speed of plate relative to another plate, du = 0.3 m/s Distance between the plates, dy = 0.15 × 10-3 m Dynamic viscosity, m = 1 poise = 0.1 Ns/m Shear stress = m

0.3 du = 0.1 ¥ = 200 N dy 0.15 × 10 - 3

Shear force = Shear stress × area = 200 × 1.3 = 260 N Power required = F × u = 260 × 0.3 = 78 Nm/s = 78 W

Example 8.11 Two square flat plates with side 0.5 m are spaced 25 mm apart. If the lower plate is stationary and the upper plate requires a force of 100 N to keep it moving with a velocity of 2.5 m/s. The oil film is between the plates at the surface of contact. Assume a linear velocity distribution; determine kinematic and dynamic viscosity of oil. Take density of the oil as 950 kg/m3.

Solution: Area of plate, A = 0.5 × 0.5 = 0.25 m2 Speed of upper plate, du = 2.5 m/s Force required for the upper plate, F = 100 N Shear force = m

du 2.5 =m¥ = 100 m dy 0.025

or, 100 = 100 m × 0.25 or, m = 4 Ns/m2 = 40 poise Dynamic viscosity = 40 poise 4 m = = 4.2 × 10– 3 m2/s Kinematic viscosity of oil = r 950

Example 8.12

Ten litres of a liquid of specific gravity 1.3 is mixed with 6 litres of liquid of specific gravity 0.8. If the bulk of the liquid shrinks by 1.5% on mixing, calculate the specific gravity, density, volume and weight of the mixture.

Solution: Weight of 10 litres of liquid specific gravity 1.3 = 10 × 10– 3 × 9810 × 1.3 = 127.53 N Weight of 6 litres of liquid specific gravity 0.8 = 6 × 10– 3 × 9810 × 0.8 = 47.1 N Total volume of liquids before mixing = 10 + 6 = 16 litres On moving the bulk shrinks by 1.5% New total volume = 0.985 × 16 = 15.76 litres Weight of equal volume of water = 15.76 × 10–3 × 981= 154.6 N Weight of mixture = 127.53 + 47.10 = 174.63 N

Fluid Mechanics 185

(i) Specific gravity of mixture = (ii) Density of mixture =

Weight of mixture = 174.63/154.6 = 1.128 Weight of equal volume of water

174.63/9.81 Mass = 1128 kg/m3 = Volume 0.01576

(iii) Volume of mixture = 15.76 litres (iv) Weight of mixture = 174.63 N

Example 8.13 Determine the intensity of shear of a oil having dynamic viscosity of 1 poise. The oil is used for lubricating the clearance between a shaft of diameter 20 cm and its journal bearing. The clearance is 1 mm and shaft rotates at 200 rpm. Solution: Dynamic viscosity, m = 1 poise = 0.1 Ns/m2 Diameter of the shaft, D = 20 cm = 0.2 m Oil film thickness, dy = 1 mm = 1 × 10– 3 m Speed of shaft, N = 200 rpm pDN p × 0.2 × 200 = Tangential speed of shaft, u = = 2.093 m/s 60 60 du , where du = change in velocity between shaft and bearing = u – 0 = u t= m dy t=

1 × 2.093 = 104.65 N/m2 10 ¥ 2 ¥ 0.001

Example 8.14

The space between two square flat parallel plate is filled with oil. Each edge length of the plate is 0.6 m. The thickness of the oil film is 12.5 mm. The upper plate which moves at 2.5 m/s requires a force of 98.1 N to maintain the speed. The specific gravity of the oil is 0.95. Assume a linear velocity distribution. Determine the dynamic and the kinematic viscosity of the oil.

Solution: Specific gravity of the oil, s = 0.95 Each edge length of the square plate = 0.6 m Area of each plate = 0.6 × 0.6 = 0.36 m2 Thickness of the oil film, dy = 12.5 mm = 12.5 × 10– 3 m Velocity of the upper plate, u = 2.5 m/s Change in velocity between the plates, du = u – 0 = u = 2.5 m/s Force required on upper plate = 98.1 N Shear stress, t =

Force 98.1 = = N/m2 Area 0.36

186



Basic Mechanical Engineering

L  8VLQJUHODWLRQ t = m or,

du dy

98.1 2.5 = m 0.36 0.125

Dynamic viscosity, m = (ii) As s =

98.1 × 0.0125 = 1.364 Ns/m2 = 13.64 poise 0.36 × 2.5

roil rwater

or, roil = s × 1000 = 0.95 × 1000 = 950 kg/m3 Kinematic viscosity =

m 1.364 = 0.001435 m2/s = 0.001435 × 104 cm2/s r 950

= 14.35 stokes

Example 8.15

A plate weighing 150 N and measuring 0.8 m and 0.8 m sides down an inclined plane over an oil film of 1.2 mm thickness. For an inclination of 30o and velocity of 0.2 m/s, find the dynamic viscosity of the fluid.

Solution: Weight of plate = 150 N Area of plate = 0.8 m × 0.8 m Velocity of plate, u = 0.2 m/s Thickness of film, t = dy = 1.2 mm Component of weight w along plane = 150 sin 30° t=

150 Area

Force 150 sin 30∞ = Area 0.64

t= m

du u 0.2 =m =m dy t 0.0012

m = 0.7 Ns/m2 Example 8.16 A solid cone of radius ro and vertex angle 2q is to rotate at an angular velocity w. Oil of viscosity μ and thickness h fills the gap between the cone and the housings. Determine torque T to rotate the cone.

Solution:

Shear stress on the inclined wall, t = m

du v rw =m =m dy h h

Fluid Mechanics 187

Considering an elemental area = 2pr × Differential torque, dT = rdF = r(t2pr)

dr = dA sin q dr sin q

È wr dr ˘ ¥ 2pr ¥ = r Ím ˙ sin q ˚ Î h

= m Torque, T =

Ú

ro

0

dT =

2 pwm ¥ h sin q

Ú

ro

0

2pw 1 r 3 dr h sin q

r 3 dr

T=

pwm 2h sin q

Example 8.17

If the pressure of a liquid is increased from 75 N/cm2 to 140 N/cm2, the volume of the liquid decreases by 0.147%. Determine the bulk modulus of elasticity of the liquid.

Solution: Initial pressure = 75 N/cm2 Final pressure = 140 N/cm2 Increase in pressure = dp = 140 – 75 = 65 N/cm2 Decrease in volume = 0.147% We know

dv 0.147 = v 100

K=

dp 65 = = 4.42 ¥ 10 4 N/cm 2 dv 0.147 v 100

Example 8.18

A liquid is compressed in a cylinder having volume of 0.113 m3 at 6.87 MN/m2 pressure and a volume of 0.0112 m3 at 13.73 MN/m2 pressure. What is its bulk modulus of elasticity?

Solution: dp = 13.73 – 6.87 = 6.86 MN/m2 dv = 0.0112 – 0.0113 = – 0.0001 m3 V = 0.0113 m3

188

Basic Mechanical Engineering

K=

dp 6.86 ¥ 0.0113 = = 7.75 × 108 N/m2 = 0.775 GN/m2 dv 0.0001 v

Example 8.19

At a depth of 2 km in the ocean the pressure is 840 N/cm2. Assume the specific weight at the surface as 1025 N/m3 and the average bulk modulus of elasticity is 24000 N/cm2 for the pressure range. (i) What will be the change in specific volume, that is, at the surface and at the depth? (ii) What will be the specific volume at the depth? (iii) What will be the specific weight at the depth?

Solution: K=

dp = 24000 N/cm2 dv v

dv dp 840 = = - 0.035 = v k 24000

The negative sign corresponds to decrease in the volume with increase in pressure. The specific volume of the water at the surface of the ocean = 1/1025 m2/N. Change in the specific volume, that is, at the surface and at the depth dv = 0.035/1025 = 3.41 × 10– 4 m3/N 1 0.035 = 9.41 ¥ 10 - 4 m 3 /N The specific volume at the depth V = 1025 1025 The specific weight of water at that depth is =

1 1 = = 1063 N/m3 V 9.41 ¥ 10 - 4

Example 8.20 What will be the pressure within a drop plate of water 0.05 mm in diameter at 9°C. The pressure outside the drop plate is at 1.03 N/cm2. Given sigma = 0.075 N/m for water at 20°C. Solution: P=

2s 12

s = 0.075 N/m = 0.00075 N/cm R = 0.025/10 cm P = 2¥

0.075 10 ¥ = 0.60 N/cm2 100 0.025

Fluid Mechanics 189

The pressure outside the drop plate of water = 1.03 N/cm2 The pressure intensity within the droplet of water = 1.03 + 0.60 = 1.63 N/cm2

Example 8.21

Surface tension of water in contact with air at 20C is 0.0725 N/m. The pressure inside a drop plate of water is to be 0.02 N/cm2 greater than the outside pressure. Calculate the diameter of drop plate of water.

Solution: Surface tension, s = 0.0725 N/m Pressure intensity, P = 0.02 N/m2 d = dia of drop plate P = 4s/d d=

4 ¥ 0.0725 = 1.45 mm 200

Example 8.22 Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m2 above the atmospheric pressure. Diameter of the bubble is 40 mm. Solution: As for soap bubble, p = 2.5 = or,

8s 8s = d 0.040

s=

2.5 ¥ 0.040 = 0.0125 N/m 8

Example 8.23 Find the gauge pressure with a jet of water 4 mm in diameter. Surface tension of water is 0.07 N/m. Solution: P=

2s 2 ¥ 0.07 = = 35 N/m2 = 35 Pa d 0.004

Example 8.24

To form a stream of air bubbles, air is induced through a nozzle into a tank of water at 20°C. If the process requires 2 mm diameter bubbles to be formed, then by how much should the air pressure at the nozzle must exceed that of the surrounding water. Take surface tension of water at 20°C as 0.075 N/m.

Solution: d = 0.002 s = 0.075 N/m P=

4s 4 ¥ 0.0075 = = 147 N/m2 d 0.002

190

Basic Mechanical Engineering

Example 8.25

Calculate the capillary rise in a glass tube of 2.5 mm diameter when immersed vertically in (a) Water (b) Mercury. Take surface tension of water as 0.0725 N/m and for mercury 0.52 N/m. Contact with air the specific gravity for mercury is given as 13.6, angle of contact is 130° with air.

Solution: d = 2.5 mm = 0.025 m Surface tension

swater = 0.0725 N/m smercury = 0.52 N/m

and

Specific gravity of mercury = 13.6 So density of mercury is 13.6 × 1000 = 13600 kg/m3 (a) Capillary rise for water q = 0° h=

4 ¥ 0.0725 4s = = 1.18 cm rgd 1000 ¥ 9.81 ¥ 0.0025

(b) For mercury q = 130° h=

4 ¥ 0.52 ¥ cos 130∞ 4s = = - 0.4 cm rgd 13.6 ¥ 1000 ¥ 981 ¥ 0.0025

Negative sign indicates depression.

Example 8.26

Calculate the capillary effect in mm in a glass tube of 4 mm diameter, when immersed in (a) Water (b) Mercury The temperature of liquid is 20°C and values of surface tension of water and mercury at 20°C in contact with air are 0.0735 N/m and 0.48 N/m respectively. The contact angle of water q = 0° and mercury q = 130°

Solution: The rise or depression (h) of a liquid in capillary tube (h) = (a) Capillary effect in water, surface tension s = 0.0735

N m

and q = 0°

4s cos q r gd

Fluid Mechanics 191

rg = 9800 N/m2 at 20°C h=

4s cos q 4 ¥ 0.0735 = = 7.5 × 10– 3 m. 9800 ¥ 0.004 rgd

(b) Capillary effect in mercury, surface tension, s = 0.48

N m

and q = 130°

rg = (9800 × 13.6) N/m2 at 20°C h=

4s cos q 4 ¥ 0.48 cos 130 = = – 2.31 × 10– 3 m (depression) rgd 9800 ¥ 13.6 ¥ 0.004

Example 8.27 At standard atmospheric condition, determine the increase in pressure necessary to cause: 1. 1% reduction in the volume of water 2. 1% reduction in the volume of air when subjected to isentropic compression 3. 1% reduction in the volume of air subjected to isothermal compression. Take average value of bulk modulus of elasticity of seawater is 1.96 × 109 N/m2. Solution: As bulk modulus of elasticity, Ev = -

dp dv v

1. Reduction of volume by 1% results in = 0.01 Increase in pressure of water to produce 1% reduction in its volume dp = 1.96 × 109 × 0.01 = 1.96 × 107 N/m2 = 19.6 MPa 2. For air subjected to isentropic process, the bulk modulus is given by

P = kP = 1.4 ¥ 101.3 ¥ 10 3 = 142 kPa r Increase in pressure, dp = 142 × 1000 × 0.01 = 1.42 kPa 3. For air under isothermal process given bilk modulus Ev = p = 101.3 kN/m3 Increase in pressure, dp = 101.3 × 1000 × 0.01 = 1.01 kPa Example 8.28 To determine the depth of sea at a place, a detonator was exploded at 100 m below the seawater surface. The first reflected wave was recorded after 2.5 seconds at the surface. Find out the depth of seabed assuming the sea has a flat bottom. Take the average value of bulk modulus of elasticity of seawater as 1.96 × 109 N/m2 and its specific weight as 10 × 1000 N/m3. Ev = rk

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Basic Mechanical Engineering

Solution: Ev = 1.96 × 109 N/m2, d = 10000/9.81 = 1020 kg/m3 Ev = 1385 m/s d Let the depth of the sea below the seawater surface be d m. The distance travelled by the reflected sound wave = (d – 100) + d = 2d – 100 Time taken by the first reflected wave to reach the surface is

Velocity of sound in seawater =

2.5 = or,

2d - 100 1385

d = 1781.25 m.

EXERCISE 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

What do you understand by the term fluid in relevance with fluid mechanics? Explain the properties of a fluid. Derive the SI unit of force from base units. Express the viscosity and the kinematic viscosity in SI units. The density of water at 4°C and 1 atm is 1000 kg/m3. Obtain the specific weight of the liquid. Distinguish between extensive and intensive properties. Define pressure and its unit. State the terms used for pressure in zero reference. Explain Archimedes’ principle and define buoyancy. What is cavitation? How can we avoid it? What do you understand by the term water hammer? Explain Bernoulli’s principle. What are the applications of Bernoulli’s principle?

9 Hydraulic Machines

Learning Objectives  

‡ &LWLQJWKHFRQFHSWRIWXUELQHVDQGSXPSV ‡ ,QWURGXFWLRQWRK\GUDXOLFSUHVVUDPDFFXPXODWRUDQGLQWHQVLILHU

9.1

INTRODUCTION

Controlled movement of parts or a controlled application of a force is a common requirement in the industries. These operations are performed mainly by using electrical machines or diesel, petrol and steam engines as a prime mover. These prime movers can provide various movements to the objects by using some mechanical attachments like screwjack, lever, rack and pinions, etc. However, these are not the only prime movers. The enclosed fluids (liquids and gases) can also be used as prime movers to provide controlled motion and force to the objects or substances. The specially designed enclosed fluid systems can provide both linear as well as rotary motion. The high magnitude controlled force can also be applied by using these systems. This kind of enclosed fluid based systems using pressurized incompressible liquids as transmission media are called hydraulic systems. Hydraulic systems works on the principle of Pascal’s law which says that the pressure in an enclosed fluid is uniform in all the directions. What is a fluid machine? Now, you can define a fluid machine as the system or contrivance where the stored energy in the fluid is converted to mechanical energy or vice versa, that is, the mechanical energy is converted to stored energy in fluid. Now, from this very basic definition the importance of a fluid machine in the engineering field is obvious because of the need of mechanical energy for generating electrical energy Basic Mechanical Engineering

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and for different engineering applications. So, the importance of fluid machine lies in the importance of mechanical energy, as we need today. Now, next question comes about the nature of the stored energy in a fluid and what is the nature of the mechanical energy for the conversion from one to other is taking place through this system fluid machine. The stored energy in a fluid is known as the internal energy, that is, the energy stored in a system. We can divide it into two parts, that is, when the fluid is at rest, and the system is at rest, then the only way the energy is stored are intermolecular energy and potential energy. Intermolecular is the energy due to molecular motions and the potential energy of the molecules, that is, the kinetic energy and potential energy of the molecules. 8VXDOO\ LW LV WKH NLQHWLF HQHUJ\ RI WKH PROHFXOHV IRU LGHDO JDVHV WKLV HQHUJ\ RI WKH molecules is due to the temperature. So, for all system or fluid at rest, the intermolecular energy is there by virtue of its temperature. So, therefore they have the kinetic energy of the molecules and the potential energy of the molecules, which keeps raising to intermolecular energy.

9.2

HYDRAULIC TURBINES

A hydraulic machine is a device in which mechanical energy is transferred from the liquid flowing through the machine to its operating member (runner, piston, etc.) or from the operating member of the machine to the liquid flowing through it. Hydraulic machines in which, the operating member receives energy from the liquid flowing through it and the inlet energy of the liquid is greater than the outlet energy of the liquid are referred as hydraulic turbines. Hydraulic machines in which energy is transmitted from the working member to the flowing liquid and the energy of the liquid at the outlet of the hydraulic machine is less than the outlet energy are referred to as pumps. It is well known from Newton’s law that to change momentum of a fluid, a force is required. Similarly, when momentum of fluid is changed, a force is generated. This principle is made use in hydraulic turbines. In a turbine, blades or buckets are provided on a wheel and directed against water to alter the momentum of water. As the momentum is changed with the water passing through the wheel, the resulting force turns the shaft of the wheel which generates power. A hydraulic turbine uses potential energy and kinetic energy of water and converts them into usable mechanical energy. The mechanical energy made available at the turbine shaft is used to run the electric power generator which is directly coupled to the turbine shaft. The electric power which is obtained from the hydraulic energy is known as hydroelectric energy. Hydraulic turbines belong to the category of rotodynamic machinery. The hydraulic turbines are classified according to the type of energy available at the inlet of the turbine, direction of flow through vanes, head at the inlet of the turbines and specific speed of the turbines.

Hydraulic Machines 195

9.2.1 According to the Type of Energy at Inlet Impulse Turbine: In the impulse turbine, the total head of the incoming fluid is converted into a large velocity head at the exit of the supply nozzle, that is, the entire available energy of the water is converted into kinetic energy. Although there are various types of impulse turbine designs, perhaps the easiest to understand is the Pelton wheel turbine. It is most efficient when operated with a large head and lower flow rate.

Figure 9.1 Impulse turbine.

Reaction Turbine: Reaction turbines, on the other hand, are best suited for higher flow rate and lower head situations. In this type of turbines, the rotation of runner or rotor (rotating part of the turbine) is partly due to impulse action and partly due to change in pressure over the runner blades; therefore, it is called reaction turbine. For, a reaction turbine, the penstock pipe feeds water to a row of fixed blades through casing. These fixed blades convert a part of the pressure energy into kinetic energy before water enters the runner. The water entering the runner of a reaction turbine has both pressure energy and kinetic energy. Water leaving the turbine is still left with some energy (pressure energy and kinetic energy). Since the flow from the inlet to tail race is under pressure, casing is absolutely necessary to enclose the turbine. In general, Reaction turbines are medium to low-head, and high-flow rate devices. The reaction turbines in use are Francis and Kaplan.

Figure 9.2 Francis turbine.

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9.2.2 According to the Direction of Flow Through Runner Tangential Flow Turbine: In this type of turbines, the water strikes the runner in the direction of tangent to the wheel. Example: Pelton wheel turbine. Radial Flow Turbines: In this type of turbines, the water strikes in the radial direction. accordingly, it is further classified as: (a) Inward flow turbine: The flow is inward from the periphery to the centre (centripetal type). Example: old Francis turbine. (b) Outward flow turbine: The flow is outward from the centre to the periphery (centrifugal type). Example: Fourneyron turbine. Axial Flow Turbine: The flow of water is in the direction parallel to the axis of the shaft. Example: Kaplan turbine and propeller turbine. Mixed Flow Turbine: The water enters the runner in the radial direction and leaves in axial direction. Example: Modern Francis turbine.

9.2.3 According to the Head at Inlet of Turbine High Head Turbine: In this type of turbines, the net head varies from 150 m to 2000 m or even more, and these turbines require a small quantity of water. Example: Pelton wheel turbine. Medium Head Turbine: The net head varies from 30 m to 150 m. These turbines require moderate quantity of water. Example: Francis turbine. Low Head Turbine: The net head is less than 30 m. These turbines require large quantity of water. Example: Kaplan turbine.

9.2.4 Radial Flow Turbines Radical flow turbines are those turbines in which the water flows in radial direction. The water may flow radically from outwards to inwards or from inwards to outwards. If the water flows from outwards to inwards through the runner, the turbine is known as inward radial flow turbine. If the water flows from inwards to outwards, the turbine is known as outward radial flow turbine. Reaction turbine means that the water at the inlet of turbine possesses kinetic energy as well as pressure energy. The main parts of a radial flow reaction turbine are: Casing: The water from penstocks enters the casing which is of spiral shape in which area of cross section of casing goes on decreasing gradually. The casing completely surrounds the runner of the turbine.

Hydraulic Machines 197

Guide Mechanism: It consists of a stationary circular wheel all round the runner of the turbine. The stationary guide vanes are fixed on guide mechanism. The guide vanes allow the water to strike the vanes fixed on the runner without shock at inlet. Runner: It is a circular wheel on which a series of radial curved vanes are fixed. The surfaces of the vanes are made very smooth. The radial curves are so shaped that the water enters and leaves without shock. Draft Tube: The pressure at the exit of the runner of reaction turbine is generally less than the atmospheric pressure. The water exit cannot be directly discharged to the tailrace. A tube or pipe of gradually increasing area is used for discharging water from the exit of turbine to the tailrace. This tube of increasing area is called draft tube.

9.2.5 Axial Flow Turbines If the water flows parallel to the axis of the rotation of the shaft, the turbine is known as axial flow turbine. If the head at the inlet of the turbine is the sum of pressure energy and kinetic energy and during the flow of water through the runner, a part of pressure energy is converted into kinetic energy, the turbine is known as reaction turbine. For the axial flow reaction turbines, the shaft of the turbine is vertical. The lower end of the shaft is made larger which is known as hub. The vanes are fixed on the hub and hence the hub acts as runner for axial flow reaction turbine. Table 9.1 shows the comparison between impulse and reaction turbine: Table 9.1 Comparison between impulse and reaction turbines Impulse turbine

Reaction Turbine

The entire available energy of water is converted into kinetic energy.

Only a portion of the fluid energy is converted into kinetic energy before the fluid enters the turbine runner.

The work is done only by the change in the kinetic energy of the jet.

The work is done partly by the change in the velocity head, but almost entirely by the change in pressure head.

Flow regulation is possible without loss.

It is not possible to regulate the flow without loss.

8QLWLVLQVWDOOHGDERYHWKHWDLOUDFH

8QLW LV HQWLUHO\ VXEPHUJHG LQ ZDWHU EHORZ WKH tail race.

Casing has no hydraulic function to perform, because the jet is unconfined and is at atmospheric pressure. Thus, casting serves only to prevent splashing of water.

Casing is absolutely necessary, because the pressure at the inlet to the turbine is much higher WKDQ WKH SUHVVXUH DW WKH RXWOHW 8QLW KDV WR EH sealed from atmospheric pressure.

It is not essential that the wheel should run full and air has free access to the buckets.

Water fills the vane passage completely.

198

9.3

Basic Mechanical Engineering

HYDRAULIC PUMPS

The combined pumping and driving motor unit is known as hydraulic pump. The hydraulic pump takes hydraulic fluid (mostly some oil) from the storage tank and delivers it to the rest of the hydraulic circuit. In general, the speed of the pump is constant and the pump delivers an equal volume of oil in each revolution. The amount and direction of fluid flow is controlled by some external mechanisms. In some cases, the hydraulic pump itself is operated by a servo controlled motor but it makes the system complex. The hydraulic pumps are characterized by their flow rate capacity, power consumption, drive speed, pressure delivered at the outlet and efficiency of the pump. The pumps are not 100% efficient. The efficiency of a pump can be specified in two ways. One is the volumetric efficiency which is the ratio of actual volume of fluid delivered to the maximum theoretical volume possible. Second is power efficiency which is the ratio of the output hydraulic power to the input mechanical/electrical power. The typical efficiency of pumps varies from 90-98%. The hydraulic pumps can be of two types: ‡ Centrifugal pump ‡ Reciprocating pump. Centrifugal pump uses rotational kinetic energy to deliver the fluid. The rotational energy typically comes from an engine or electric motor. The fluid enters the pump impeller along or near the rotating axis, accelerates in the propeller and flungs out to the periphery by centrifugal force as shown in Fig. 9.3. In centrifugal pump the delivery is not constant and varies according to the outlet pressure. These pumps are not suitable for high pressure applications and are generally used for low-pressure and high-volume flow applications. The maximum pressure capacity is limited to 20-30 bars and the specific speed ranges from 500 to 10000. Most of the centrifugal pumps are not selfpriming and the pump casing needs to be filled with liquid before the pump is started.

Outlet pipe

Inlet pipe

Figure 9.3 Centrifugal pump.

Hydraulic Machines 199

Figure 9.4 Reciprocating or positive displacement pump.

The reciprocating pump is a positive plunger pump. It is also known as a positive displacement pump or piston pump. It is often used where relatively small quantity is handled and the delivery pressure is quite large. The construction of these pumps is similar to the four-stroke engine as shown in Fig. 9.4. The crank is driven by some external rotating motor. The piston of pump reciprocates due to crank rotation. The piston moves down in one half of crank rotation, the inlet valve opens and the fluid enters the cylinder . In the second half crank rotation the piston moves up, the outlet valve opens and the fluid moves out from the outlet. At a time, only one valve is opened and another is closed so there is no fluid leakage. Depending on the area of cylinder the pump delivers constant volume of fluid in each cycle independent of the pressure at the output port.

9.3.1 Classification of Hydraulic Pumps These are mainly classified into two categories: Non-positive displacement pumps and positive displacement pumps. Non-positive Displacement Pumps: These pumps are also known as hydrodynamic pumps. In these pumps the fluid is pressurized by the rotation of the propeller and the fluid pressure is proportional to the rotor speed. These pumps cannot withstanding high pressures and generally used for low-pressure and high-volume flow applications. The fluid pressure and flow generated due to inertia effect of the fluid. The fluid motion is generated due to the rotating propeller. These pumps provide a smooth and continuous

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flow but the flow output decreases with increase in system resistance (load). The flow output decreases because some of the fluid slips back at higher resistance. The fluid flow is completely stopped at a large system resistance and thus the volumetric efficiency becomes zero. Therefore, the flow rate not only depends on the rotational speed but also on the resistance provided by the system. The important advantages of non-positive displacement pumps are lower initial cost, less operating maintenance because of less moving parts, simplicity of operation, higher reliability and suitability with wide range of fluid, etc. These pumps are primarily used for transporting fluids and find little use in the hydraulic or fluid power industry. Centrifugal pump is the common example of non-positive displacement pumps. Positive Displacement Pumps: These pumps deliver a constant volume of fluid in a cycle. The discharge quantity per revolution is fixed in these pumps and they produce fluid flow proportional to their displacement and rotor speed. These pumps are used in most of the industrial fluid power applications. The output fluid flow is constant and is independent of the system pressure (load). The important advantage associated with these pumps is that the high-pressure and low-pressure areas (means input and output regions) are separated and hence the fluid cannot leak due to higher pressure at the outlets. These features make the positive displacement pump most suited and universally accepted for hydraulic systems. The important advantages of positive displacement pumps over non-positive displacement pumps include capability to generate high pressures, high volumetric efficiency, high power to weight ratio, change in efficiency throughout the pressure range is small and wider operating range pressure and speed. The fluid flow rate of these pumps ranges from 0.1 and 15,000 gpm, the pressure head ranges between 10 and 100,000 psi and specific speed is less than 500. It is important to note that the positive displacement pumps do not produce pressure but they only produce fluid flow. The resistance to output fluid flow generates the pressure. It means that if the discharge port (output) of a positive displacement pump is opened to the atmosphere, then fluid flow will not generate any output pressure above the atmospheric pressure. But, if the discharge port is partially blocked, then the pressure will rise due to the increase in fluid flow resistance. If the discharge port of the pump is completely blocked, then an infinite resistance will be generated. This will result in the breakage of the weakest component in the circuit. Therefore, the safety valves are provided in the hydraulic circuits along with positive displacement pumps. Important positive displacement pumps are gear pumps, vane pumps and piston pumps. The details of these pumps are discussed in the following sections.

9.3.2 Gear Pumps Gear pump is a robust and simple positive displacement pump. It has two meshed gears revolving about their respective axes. These gears are the only moving parts in the pump. They are compact, relatively inexpensive and have a few moving parts. The rigid

Hydraulic Machines 201

design of the gears and housings allow for very high pressures and the ability to pump highly viscous fluids. They are suitable for a wide range of fluids and offer self-priming performance. Sometimes gear pumps are designed to function as either a motor or a pump. These pumps include helical and herringbone gear sets (instead of spur gears), lobe-shaped rotors similar to root blowers (commonly used as superchargers), and mechanical designs that allow the stacking of pumps. Based upon the design, the gear pumps are classified as: ‡ External gear pumps ‡ Lobe pumps ‡ Internal gear pumps ‡ Gerotor pumps Generally, gear pumps are used to pump ‡ Petrochemicals: Pure or filled bitumen, pitch, diesel oil, crude oil, lube oil, etc. ‡ Chemicals: Sodium silicate, acids, plastics, mixed chemicals, isocyanates, etc. ‡ Paint and ink ‡ Resins and adhesives ‡ Pulp and paper: acid, soap, lye, black liquor, kaolin, lime, latex, sludge, etc. ‡ Food: Chocolate, cacao butter, fillers, sugar, vegetable fats and oils, molasses, animal food, etc.

9.4

HYDRAULIC DEVICES

9.4.1 Hydraulic Press Hydraulic press is one of the oldest of the basic machine tools. In its modern form, it is used to presswork ranging from coining jewellery to forging aircraft parts. Modern hydraulic presses are, in some cases, better suited to applications where the mechanical press has been traditionally more popular. Factors that may favour the use of hydraulic presses over their mechanical counterparts may include the following: (i) Depending on the application, a hydraulic press may cost less than an equivalent mechanical press. (ii) In small lot production where hand feeding and single stroking occurs, production rates equal to mechanical presses are achieved. (iii) Single stroking does not result in additional press wear. (iv) Die shut heights variations do not change the force applied. (v) There is no tonnage curve derating factor. (vi) Forming and drawing speeds can be accurately controlled throughout the stroke.

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(vii) Hydraulic presses with double actions and or hydraulic die cushions are capable of forming and drawing operations that would not be possible in a mechanical press. The press shown in Fig. 9.5 has a frame and bolster that are similar to the construction used for open back stationary (OBS) mechanical presses. The frame is of robust construction to limit both angular and total deflection. The bolster and ram provide a surface to mount tooling. The ram is actuated by a large hydraulic cylinder in the centre of the upper part of the frame. Additional alignment is provided by two round guiderods. The motor drives a rotary pump, which draws oil out of the reservoir housed in the machine frame. The control system has electrically-actuated valves which respond to commands to advance and retract the slide or ram. A pressure regulator is either manually or automatically adjusted to apply the desired amount of force. In most hydraulic presses, full force is available throughout the stroke. Figure 9.5 illustrates why the rated force capacity of a mechanical press is available only near the bottom of the stroke. The full force of a hydraulic press can be delivered at any point in the stroke. This feature is an important characteristic of most hydraulic presses. Deep drawing and forming applications often require large forces very high in the press stroke. Some mechanical presses do not develop enough force high enough in the downward stroke to permit severe drawing and forming applications such as inverted draw dies to be used without danger of press damage.

Figure 9.5 Outline of a hydraulic press.

Hydraulic Machines 203

9.4.2 Hydraulic Ram A hydraulic ram, or hydram, is a cyclic water pump powered by hydropower. It takes in water at one “hydraulic head” (pressure) and flow rate, and outputs water at a higher hydraulic head and lower flow rate. The device uses the water hammer effect to develop pressure that allows a portion of the input water that powers the pump to be lifted to a point higher than where the water originally started. The hydraulic ram is sometimes used in remote areas, where there is both a source of low-head hydropower and a need for pumping water to a destination higher in elevation than the source. In this situation, the ram is often useful, since it requires no outside source of power other than the kinetic energy of flowing water. A simplified hydraulic ram is shown in Fig. 9.6. Initially, the waste valve is open, and the delivery valve is closed. The water in the inlet pipe starts to flow under the force of gravity and picks up speed and kinetic energy until the increasing drag force closes the waste valve. The momentum of the water flow in the inlet pipe against the now closed waste valve causes a water hammer that raises the pressure in the pump, opens the delivery valve, and forces some water to flow into the delivery pipe. Because this water is being forced uphill through the delivery pipe farther than it is falling downhill from the source, the flow slows; when the flow reverses, the delivery check valve closes. Meanwhile, the water hammer from the closing of the waste valve also produces a pressure pulse which propagates back to the source where it converts to a suction pulse that propagates back down the inlet pipe. This suction pulse, with the weight or spring on the valve, pulls the waste valve back open and allows the process to begin again. A pressure vessel containing air cushions the hydraulic pressure shock when the waste valve closes, and it also improves the pumping efficiency by allowing a more constant flow through the delivery pipe. Although the pump could in theory work without it, the efficiency would drop drastically and the pump would be subject

Figure 9.6 Hydraulic ram with 1. Inlet – drive pipe 2. Free flow at waste valve 3. Outlet – delivery pipe 4. Waste valve 5. Delivery check valve 6. Pressure vessel

204 Basic Mechanical Engineering

to extraordinary stresses that could shorten its life considerably. One problem is that the pressurized air will gradually dissolve into the water until none remains. One solution to this problem is to have the air separated from the water by an elastic diaphragm (similar to an expansion tank); however, this solution can be problematic in developing countries where replacements are difficult to procure. Another solution is to have a mechanism such as a snifting valve that automatically inserts a small bubble of air when the suction pulse mentioned above reaches the pump. Another solution is to insert an inner tube of a car or bicycle tire into the pressure vessel with some air in it and the valve closed. This tube is in effect the same as the diaphragm, but it is implemented with more widely available materials. The air in the tube cushions the shock of the water the same as the air in other configurations does.

9.4.3

Hydraulic Accumulator

A hydraulic accumulator is a pressure storage reservoir in which a non-compressible hydraulic fluid is held under pressure that is applied by an external source. The external source can be a spring, a raised weight, or a compressed gas. An accumulator enables a hydraulic system to cope with extremes of demand using a less powerful pump, to respond more quickly to a temporary demand, and to smooth out pulsations. It is a type of energy storage device. Compressed gas accumulators, also called hydropneumatic accumulators, are by far the most common type.

Figure 9.7 Hydraulic accumulator.

Hydraulic Machines 205

In modern times, often mobile, hydraulic systems the preferred item is a gas charged accumulator, but simple systems may be spring loaded. There may be more than one accumulator in a system. The exact type and placement of each may be a compromise due to its effects and the costs of manufacture. An accumulator is placed close to the pump with a non-return valve preventing flow back to the pump. In the case of piston-type pumps this accumulator is placed in the ideal location to absorb pulsations of energy from the multi-piston pump. It also helps protect the system from fluid hammer. This protects system components, particularly pipework, from both potentially destructive forces. An additional benefit is the additional energy that can be stored while the pump is subject to low demand. The designer can use a smaller capacity pump. The large excursions of system components, such as landing gear on a large aircraft, that require a considerable volume of fluid can also benefit from one or more accumulators. These are often placed close to the demand to help overcome restrictions and drag from long pipe work runs. The outflow of energy from a discharging accumulator is much greater, for a short time, than even large pumps could generate. An accumulator can maintain the pressure in a system for periods when there are slight leaks without the pump being cycled on and off constantly. When temperature changes cause pressure excursions the accumulator helps absorb them. Its size helps absorb fluid that might otherwise be locked in a small fixed system with no room for expansion due to valve arrangement. The gas pre-charge in an accumulator is set so that the separating bladder, diaphragm or piston does not reach or strike either end of the operating cylinder. The design precharge normally ensures that the moving parts do not foul the ends or block fluid passages. Poor maintenance of pre-charge can destroy an operating accumulator. A properly designed and maintained accumulator should operate trouble-free for years.

9.4.4 Hydraulic Intensifier A hydraulic intensifier is a hydraulic machine for transforming hydraulic power at low pressure into a reduced volume at higher pressure. Such a machine may be constructed by mechanically connecting two pistons, each working in a separate cylinder of a different diameter. As the pistons are mechanically linked, their force and stroke length are the same. If the diameters are different, the hydraulic pressure in each cylinder will vary in the same ratio as their areas: the smaller piston giving rise to a higher pressure. As the pressure is inversely proportional to the area, it will be inversely proportional to the square of the diameter. The working volume of the intensifier is limited by the stroke of the piston. This, in turn, limits the amount of work that may be done by one stroke of the intensifier. These are not reciprocating machines (i.e., continually running multi-stroke machines) and so their entire work must be carried out by a single stroke. This limits their usefulness somewhat, to

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Figure 9.8 Outline of hydraulic intensifier.

machines that can accomplish their task within a single stroke. They are often used where a powerful hydraulic jack is required, but there is insufficient space to fit the cylinder size that would normally be required, for the lifting force necessary and with WKHDYDLODEOHV\VWHPSUHVVXUH8VLQJDQLQWHQVLILHUPRXQWHGRXWVLGHWKHMDFNDOORZV a higher pressure to be obtained and thus a smaller cylinder used for the same lift force. Intensifiers are also used as part of machines such as hydraulic presses, where a higher pressure is required and a suitable supply is already available. Some small intensifiers have been constructed with a stepped piston. This is a double-ended piston, of two different diameters, each end working in a different cylinder. This construction is simple and compact, requiring an overall length little more than twice the stroke. It is also still necessary to provide two seals, one for each piston, and to vent the area between them. A leak of pressure into the volume between the pistons would transform the machine into an effective single piston with an equal area on each side, thus defeating the intensifier effect.

9.5

SUMMARY

Hydraulic machines are machinery and tools that use liquid fluid power to do simple work. Heavy equipment is a common example. In this type of machine, hydraulic fluid is transmitted throughout the machine to various hydraulic motors and hydraulic cylinders and becomes pressurised according to the resistance present. The fluid is

Hydraulic Machines 207

controlled directly or automatically by control valves and distributed through hoses and tubes. The popularity of hydraulic machinery is due to the very large amount of power that can be transferred through small tubes and flexible hoses, and the high power density and wide array of actuators that can make use of this power. Hydraulic machinery is operated by the use of hydraulics, where a liquid is the powering medium. Here, in this chapter, we have discussed hydraulic turbines, pumps and principles of working of some hydraulic devices.

EXERCISE 1. 2. 3. 4. 5. 6. 7. 8.

What do you understand by hydraulic devices? Explain hydraulic turbines in detail. Distinguish between impulse and reaction turbines. What is the working principle of a hydraulic pump? Explain its types in detail. Summarize the operational principle of hydraulic ram with a neat sketch. What are the advantages of hydraulic presses? Write its applications in short. Explain any two hydraulic devices in detail with a neat sketch. What are the criteria for classification of hydraulic pumps?

Multiple-Choice Questions 1. Pascal-second is the unit of



(a) Pressure

(b) Kinematic viscosity

(c) Dynamic viscosity

(d) Surface tension

 $QLGHDOÁXLGLV (a) One which obeys Newton’s law of viscosity (b) Frictionless and incompressible (c) Very viscous (d) Frictionless and compressible 3. The unit of kinematic viscosity is



(a) gm/cm-sec2

(b) dyne-sec/cm2

(c) gm/cm2-sec

(d) cm2/sec

 ,IWKHG\QDPLFYLVFRVLW\RIDÁXLGLVSRLVHDQGVSHFLÀFJUDYLW\LVWKHQWKH NLQHPDWLFYLVFRVLW\RIWKDWÁXLGLQVWRNHVLV (a) 0.25

(b) 0.50

(c) 1.0

(d) none of the above

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Basic Mechanical Engineering

5. The viscosity of a gas (a) Decreases with increase in temperature (b) Increases with increase in temperature (c) Is independent of temperature (d) Is independent of pressure for very high pressure intensities 6. Newton’s law of viscosity relates (a) Intensity of pressure and rate of angular deformation (b) Shear stress and rate of angular deformation (c) Shear stress, viscosity and temperature (d) Viscosity and rate of angular deformation 









 $QRSHQWDQNFRQWDLQVPGHHSZDWHUZLWKFPGHSWKRIRLORIVSHFLÀFJUDYLW\ 0.8 above it. The intensity of pressure at the bottom of tank will be (a) 4 kN/m2

(b) 10 kN/m2

(c) 12 kN/m2

(d) 14 kN/m2

 7KHÁXLGSURSHUW\GXHWRZKLFKPHUFXU\GRHVQRWZHWWKHJODVVLV (a) Surface tension

(b) Viscosity

(c) Cohesion

(d) Adhesion

 /DPLQDUÁRZRID1HZWRQLDQÁXLGFHDVHVWRH[LVWZKHQWKH5H\QROGVQXPEHU exceeds (a) 4000

(b) 2100

(c) 1500

(d) 30003

 :KHQWKHPRPHQWXPRIRQHÁXLGLVXVHGIRUPRYLQJDQRWKHUÁXLGVXFKDGHYLFH is called a/an (a) Jet pump

(b) Blower

(c) Acid egg

(d) None of these

 7KHQRUPDOVWUHVVLVWKHVDPHLQDOOGLUHFWLRQVDWDSRLQWLQDÁXLGZKHQWKHÁXLG is (a) Non-viscous (b) Incompressible (c) Both (a) and (b)



G  +DYLQJQRPRWLRQRIRQHÁXLGOD\HUUHODWLYHWRWKHRWKHU

Hydraulic Machines 209

12. Head developed by a centrifugal pump depends on its (a) speed (b) impeller diameter (c) both (a) and (b) 

G  QHLWKHUD QRUE 7KHKHDGORVVLQWXUEXOHQWÁRZLQDSLSHYDULHV 13. According to Archimedes’ principle, if a body is immersed partially or fully in a ÁXLGWKHQWKHEXR\DQF\IRUFHLVBBBBBBBBBBWKHZHLJKWRIÁXLGGLVSODFHGE\WKH body. (a) equal to

(b) less than

(c) more than

(d) unpredictable

14. When is a liquid said to be not in a boiling or vaporized state? (a) If the pressure on the liquid is equal to its vapour pressure (b) If the pressure on the liquid is less than its vapour pressure (c) If the pressure on the liquid is more than its vapour pressure 

G  8QSUHGLFWDEOH 15. Bulk modulus is the ratio of (a) Shear stress to volumetric strain (b) Volumetric strain to shear stress (c) Compressive stress to volumetric strain (d) Volumetric strain to compressive stress 16. The body whose surface does not coincide with the streamline when placed in a ÁRZLVFDOOHG





(a) Streamline body

(b) Wave body

(c) Bluff body

(d) Induced body

 7KHVXPRIFRPSRQHQWVRIVKHDUIRUFHVLQWKHGLUHFWLRQRIÁRZRIÁXLGLVFDOOHG (a) Shear drag

(b) Friction drag

(c) Skin drag

(d) All of the above

 7KH FRPSRQHQW RI WKH WRWDO IRUFH H[HUWHG E\ D ÁXLG RQ D ERG\ LQ WKH GLUHFWLRQ parallel to the direction of motion is called (a) Lift

(b) Drag

(c) Both (a) and (b)

(d) None of the above

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Basic Mechanical Engineering

19. Boundary layer thickness is the distance from the boundary to the point where YHORFLW\RIWKHÁXLGLV (a) Equal to 10% of free stream velocity (b) Equal to 50% of free stream velocity (c) Equal to 90% of free stream velocity (d) Equal to 99% of free stream velocity 20. Which type of body is an airfoil? (a) Streamline body

(b) Wave body

(c) Bluff body

(d) Induced body

21. Which of the following number is applicable in open hydraulic structure such as spillways, where gravitational force is predominant? (a) Reynolds Number

(b) Euler’s Number

(c) Weber’s Number

(d) Froude’s Number

22. Which of the following is a dimensionless equation? (a) Reynold’s equation

(b) Euler’s equation

(c) Weber’s equation

(d) All of the above

23. Which of the following equations is not dimensionally homogeneous? Consider standard symbols for quantities. (a) (Force) F = m × a (b) (Head loss due to friction) hf = (f L V2)/(2 g d) (c) (Torque) T = F × Distance (d) None of the above 24. Match the following physical quantities in Group 1 with their dimensions in Group 2. 1. Work done (Energy) (W) __________________ A. [ML2 T–3 ] 2. Power (P) ________________________________ B. [ML–1 T–1] 3. Momentum (M) __________________________ C. [ML2 T–2] 4. Modulus of elasticity (E) ___________________ D. [MLT–1] 5. Dynamic viscosity (m) ______________________ E. [ML–1 T–2] (a) 1-(C), 2-(A), 3-(D), 4-(E), 5-(B)

(b) 1-(A), 2-(C), 3-(D), 4-(E), 5-(B)

(c) 1-(C), 2-(A), 3-(E), 4-(B), 5-(D)

(d) 1-(D), 2-(E), 3-(B), 4-(A), 5-(C)

Hydraulic Machines 211

25. What is a syphon? (a) A long bend pipe used to carry water from a reservoir at a higher level to another reservoir at a lower level when two reservoirs are separated by a hill (b) A long bend pipe used to carry water from a reservoir at a lower level to another reservoir at a higher level with some work input when two reservoirs are separated by a hill (c) A long bend pipe used to carry water from one reservoir to another reservoir when two reservoirs are at same elevation 

G  8QSUHGLFWDEOH

Answers 1. (c)

2. (b)

3. (d)

4. (c)

5. (b)

6. (b)

7. (d)

8. (a)

9. (b)

10. (a)

11. (d)

12. (c)

13. (a)

14. (c)

15. (c)

16. (c)

17. (d)

18. (b)

19. (d)

20. (a)

21. (d)

22. (d)

23. (a)

24. (b)

25. (a)

Section IV Manufacturing Technology

10 Manufacturing Processes

Learning Objectives    

‡ ‡ ‡ ‡

10.1

8QGHUVWDQGLQJWKHWHUPVXVHGLQPDQXIDFWXULQJ &ODVVLILFDWLRQDQGWKHPHWDOFDVWLQJSURFHVVHV 0HWDOMRLQLQJDQGPHWDOFXWWLQJSURFHVVHV 3RZGHUPHWDOOXUJ\

INTRODUCTION

Manufacturing is the backbone of any industrialized nation. Manufacturing and technical staff in industry must know the various manufacturing processes, materials being processed, tools and equipment for manufacturing different components or products with optimal process plan using proper precautions and specified safety rules to avoid accidents. Beside above, all kinds of the future engineers must know the basic requirements of workshop activities in terms of man, machine, material, methods, money and other infrastructural facilities needed to be positioned properly for optimal shop layouts or plant layout and other support services effectively adjusted or located in the industry or plant within a well planned manufacturing organization. The complete understanding of basic manufacturing processes and workshop technology is highly difficult for any one to claim expertise over it. The study deals with several aspects of workshop practices also for imparting the basic working knowledge of the different engineering materials, tools, equipment, manufacturing processes, basic concepts of electromechanical controls of machine tools, production criteria, characteristics and uses of various testing instruments and measuring or inspecting devices for checking Basic Mechanical Engineering

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components or products manufactured in various manufacturing shops in an industrial environment. It also describes and demonstrates the use of different hand tools (measuring, marking, holding and supporting tools, cutting, etc.), equipment, machinery and various methods of manufacturing that facilitate shaping or forming the different existing raw materials into suitable usable forms. It deals with the study of industrial environment which involves the practical knowledge in the area of ferrous and nonferrous materials, their properties and uses. It should provide the knowledge of basic workshop processes, namely, benchwork and fitting, sheet metal, carpentry, pattern making, mould making, foundry, smithy, forging, metal working and heat treatment, welding, fastening, machine shop, surface finishing and coatings, assembling inspection and quality control. It emphasizes on basic knowledge regarding composition, properties and uses of different raw materials, various production processes, replacement of or improvement over a large number of old processes, new and compact designs, better accuracy in dimensions, quicker methods of production, better surface finishes, more alternatives to the existing materials and tooling systems, automatic and numerical control systems, higher mechanization and greater output. Manufacturing is derived from the Latin word manufactus, "means made by hand". In modern context it involves making products from raw material by using various processes, by making use of hand tools, machinery or even computers. It is therefore a study of the processes required to make parts and to assemble them in machines. Process engineering, in its application to engineering industries, shows how the different problems related to development of various machines may be solved by a study of physical, chemical and other laws governing the manufacturing process. The study of manufacturing reveals those parameters which can be most efficiently influenced to increase production and raise accuracy. Advance manufacturing engineering involves the following concepts: 1. Process planning. 2. Process sheets. 3. Route sheets. 4. Tooling. 5. Cutting tools, machine tools (traditional, numerical control (NC), and computerized numerical control (CNC). 6. Jigs and fixtures. 7. Dies and moulds. 8. Manufacturing information generation. 9. CNC part programmes. 10. Robot programmers. 11. Flexible manufacturing systems (FMS), group technology (GT) and computer integrated manufacturing (CIM).

Manufacturing Processes 217

Manufacturing is the process of converting raw materials into products; encompasses the design and manufacturing of goods using various production methods and techniques. The advancement and the success of human development are represented and judged chiefly by change and support of way of life through accessibility or generation of sufficient and quality merchandise and ventures for men’s material welfare in all regards covering housing, attire, instruction, transport, and furthermore entertainment. The fruitful formation of men’s material welfare depends for the most part on. ‡ The availability of natural resources in abundance ‡ Execution of ability of human both physically and mentally ‡ Development and utilization of force devices and machines Creation or assembling can be essentially characterized as esteem expansion forms by which crude materials of low utility and esteem because of its deficient material properties and poor or sporadic size, shape and complete are changed over into high utility and esteemed items with unmistakable measurements, structures and wrap up some practical capacity.

10.2

CLASSIFICATION

Manufacturing processes can be classified as processing operations and assembly operations.

10.2.1 In Processing Operation The work material is transformed from one state to other advanced state. Through this operation value is added to the work material by changing the geometry; shape SURSHUWLHVDSSHDUDQFHHWFRIWKHVWDUWLQJZRUNPDWHULDO8VXDOO\SURFHVVLQJRSHUDWLRQV are performed on individual components. But in some cases like aerospace industry, the processing operations are performed on assembled items also.

10.2.2 In Assembly Operation Two or more components are joined to create a new entity. The new entity is called assembly, sub-assembly based on its state in the product. If the entity is an intermediate state of the product, it is called sub-assembly. Some other terms are also referred based on the joining process. The assembly created by welding operation is called weld met.

10.3

BASIC MANUFACTURING PROCESSES

Manufacturing process is that part of the production process which is directly concerned with the change of form or dimensions of the part being produced. It does not include the transportation, handling or storage of parts, as they are not directly concerned with

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the changes into the form or dimensions of the part produced. It is described as the manufacturing process which creates or adds value to a product.

10.3.1 Classification of Manufacturing Processes For producing of products materials are needed. It is therefore important to know the characteristics of the available engineering materials. Raw materials used manufacturing of products, tools, machines and equipment in factories or industries are extracted from ores. The ores are suitably converted into a metal by reducing or refining processes.

10.3.2 Primary Shaping Processes Primary shaping processes are manufacturing of a product from an amorphous material. Some processes produce finished products or articles into its usual form whereas others do not, and require further working to finish component to the desired shape and size. Castings need re-melting of scrap and defective ingots in cupola or in some other melting furnace and then pouring of the molten metal into sand or metallic moulds to obtain the castings. Thus, intricate shapes can be manufactured. Typical examples of the products that are produced by casting process are machine beds, automobile engines, carburettors, flywheels, etc. The parts produced through these processes may or may not require to undergo further operations. Some of the important primary shaping processes is: 1. Casting 2. Powder metallurgy 3. Plastic technology 4. Gas cutting 5. Bending 6. Forging

10.3.3 Secondary or Machining Processes As large number of components require further processing after the primary processes. These components are subjected to one or more number of machining operations in machine shops, to obtain the desired shape and dimensional accuracy on flat and cylindrical jobs. Thus, the jobs undergoing these operations are the roughly finished products received through primary shaping processes. The process of removing the undesired or unwanted material from the workpiece or job or component to produce a required shape using a cutting tool is known as machining. This can be done by a manual process or by using a machine called machine tool (traditional machines, namely, lathe, milling machine, drilling, shaper, planner, slotter). In many cases these operations are performed on rods, bars and flat surfaces in machine shops. These secondary processes

Manufacturing Processes 219

are mainly required for achieving dimensional accuracy and a very high degree of surface finish. The secondary processes require the use of one or more machine tools, various single or multi-point cutting tools (cutters), job holding devices, marking and measuring instruments, testing devices and gauges, etc., for getting desired dimensional control and required degree of surface finish on the workpieces. The example of parts produced by machining processes include hand tools, machine tools, instruments, automobile parts, nuts, bolts and gears, etc. Lot of material is wasted as scrap in the secondary or machining processes. Some of the common secondary or machining processes are: Solidi cation processes Particulate processing

Shaping processes

Deformation processes Processing operation

Material removal Property enhancing

Surface processing

Manufacturing processes

Heat treatment

Cleaning & surface treatment Coating and deposition Welding

Permanent joining

Brazing and soldering

Adhesive bonding Assembly operation Mechanical fastening

Threaded fasteners Permanent fastening

Figure 10.1 Classification of manufacturing processes.

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‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡

10.4

Turning Threading Knurling Milling Drilling Boring Planning Shaping Slotting Sawing Broaching Hobbing Grinding Gear cutting Thread cutting 8QFRQYHQWLRQDOPDFKLQLQJSURFHVVHVQDPHO\PDFKLQLQJZLWK1XPHULFDO&RQWURO (NC) machines tools or Computer Numerical Control (CNC) machines tools using (&0/%0$-0860VHWXSVHWF

METAL FORMING PROCESSES

Forming processes encompass a wide variety of techniques, which make use of suitable force, pressure or stresses, like compression, tension and shear or their combination to cause a permanent deformation of the raw material to impart required shape. These processes are also known as mechanical working processes and are mainly classified into two major categories, i.e., hot working processes and cold working processes. In these processes, no material is removed; however, it is deformed and displaced using suitable stresses like compression, tension, and shear or combined stresses to cause plastic deformation of the materials to produce required shapes. Such processes lead to production of directly usable articles which include kitchen utensils, rods, wires, rails, cold drink bottle caps, collapsible tubes, etc. Some of the important metal forming processes are discussed below:

10.4.1 Hot Working Processes (1) Forging, (2) rolling, (3) hot spinning, (4) extrusion, (5) hot drawing and (6) hot spinning.

Manufacturing Processes 221

10.4.2 Cold Working Processes (1) Cold forging, (2) cold rolling, (3) cold heading, (4) cold drawing, (5) wire drawing, (6) stretch forming, (7) sheet metal working processes such as piercing, punching, lancing,

notching, coining, squeezing, deep drawing, bending, etc. 10.5

METAL CASTING OPERATIONS

Casting is one of the earliest metal shaping techniques. It means pouring molten metal into a refractory mould cavity and allowing it to solidify. The solidified object is taken out from the mould either by breaking or taking the mould apart. The solidified object is called casting and the technique followed is known as casting process. The casting process was discovered probably around 3500 BC in Mesopotamia. In many parts of the world during that period, copper axes (wood cutting tools) and other flat objects were made in open moulds using baked clay. These moulds were essentially made in single piece. The Bronze Age 2000 BC brought forward more refinement into casting process. For the first time, the core for making hollow sockets in the cast objects was invented. The core was made of baked sand. Also the lost wax process was extensively used for making ornaments using the casting process. Casting technology was greatly improved by Chinese from around 1500 BC. For this there is evidence of the casting activity found in China. For making highly intricate jobs, a lot of time in making the perfect mould to the last detail was required on the casting made from the moulds. Indus Valley civilization was also known for their extensive use of casting of copper and bronze for ornaments, weapons, tools and utensils. But there was not much of improvement in the casting technology. From various objects that were excavated from the Indus Valley sites, they appear to have been familiar with all the known casting methods such as open mould and piece mould. This chapter describes the fluidity of molten metal, different casting techniques and various casting defects occurring in casting processes. Generally, in all the processes of casting a metal alloy is melted and then poured or forced into a mould where it takes the shape of the mould and is allowed to solidify. When the melt solidifies the casting is taken out of the mould. Some castings require finishing due to the cast appearance, tolerance, or surface finish requirements. As the mould solidifies the metal starts shrinking (gray cast iron is an exception) so moulds must be made slightly oversize in order to accommodate the shrinkage and still achieve the desired final dimensions. There are two basic types of molds used in castings, namely, expendable moulds (sand casting and investment casting) that are destroyed to remove the part, and permanent molds (die casting). Expendable moulds are created using either a permanent pattern (sand casting) or an expendable pattern (investment casting). Permanent moulds, of course, do not require a pattern. Two major advantages for

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selecting casting as the process of choice for creating a part are the wide selection of alloys available and the ability, as in injection moulding, to create complex shapes. However, not all alloys can be cast by all processes. The most common metal casting processes are sand casting, investment casting, and die casting. Common examples: door handles, locks, the outer casing or housing for motors, pumps, etc., wheels of many cars. Casting is also heavily used in the toy industry to make parts, e.g., toy cars, planes, and so on.

Figure 10.2 Casting process.

The probable causes and suggested remedies of various casting defects is given in table below. Table 10.1 Probable causes and suggested remedies of various casting defects S. No. 1.

Name of Casting Defect Blow holes

Probable Causes 1. Excess moisture content in moulding sand.

Suggested Remedies 1. Control of moisture content.

2. Rust and moisture on chills, 8VH RI UXVWIUHH FKLOOV FKDSOHW chaplets and inserts. and clean inserts. 3. Cores not sufficiently baked.

3. Bake cores properly.

4. Excessive use of organic binders.

4. Ram the moulds less hard.

5. Moulds not adequately vented. 5. Provide adequate venting in mould and cores. 6. Moulds not adequately vented. 7. Moulds rammed very hard.

Manufacturing Processes 223 2.

Shrinkage

1. Faulty gating and risering 1. Ensure proper directional system. solidification by modifying gating, risering and chilling. 2. Improper chilling.

3.

Porosity

1. High pouring temperature.

1. Regulate pouring temperature.

2. Gas dissolved in metal charge. 2. Control metal composition. 3. Less flux used.

3. Increase flux proportions.

4. Molten metal not properly 4. Ensure effective degassing. degassed. 5. Slow solidification of casting. 6. High moisture and permeability in mould. 4.

5.

Misruns

Hot tears

5. Modify gating and risering.

low 6. Reduce moisture and increase permeability of mould.

1. Lack of fluidity in molten 1. Adjust proper pouring metal. temperature. 2. Faulty design.

2. Modify design.

3. Faulty gating.

3. Modify gating system.

1. Lack of collapsibility of core.

1. Improve core collapsibility.

2. Lack of collapsibility of mould. 2. Improve mould collapsibility.

6.

Metal penetration

3. Faulty design.

3. Modify casting design.

4. Hard ramming of mould.

4. Provide softer ramming.

1. Large grain size and used.

8VH VDQG KDYLQJ ILQHU JUDLQ size.

2. Soft ramming of mould.

2. Provide hard ramming.

3. Moulding sand or core has low 3. Suitably adjust strength. temperature.

pouring

4. Moulding sand or core has high permeability 5. Pouring temperature of metal too high. 7.

8.

Cold shuts

Cuts and washes

1. Lack of fluidity in molten 1. Adjust proper metal. temperature.

pouring

2. Faulty design.

2. Modify design.

3. Faulty gating.

3. Modify gating system.

1. Low strength of mould and 1. Improve core. strength.

mould

and

core

2. Lack of binders in facing and 2. Add more binders to facing core stand. and core sand. 3. Faulty gating.

3. Improve gating.

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10.6

METAL CUTTING OPERATIONS

Metal cutting or traditional machining processes are also known as conventional machining processes. These processes are commonly carried out in machine shops or tool rooms for machining a cylindrical or flat jobs to a desired shape, size and finish on a rough block of job material with the help of a wedge-shaped tool. The cutting tool is constrained to move relative to the job in such a way that a layer of metal is removed in the form of a chip. General metal cutting operations are shown in Fig. 10.3. These machining processes are performed on metal cutting machines, more commonly termed machine tools using various types of cutting tools (single or multi-point). A machine tool is a power driven metal cutting machine which assists in managing the needed relative motion between the cutting tool and the job that changes the size and shape of the job material. In metal cutting (machining) process, working motion is imparted to the workpiece and cutting tool by the mechanisms of machine tool so that the work and tool travel relative to each other and machine the workpiece material in the form of shavings (or swarf) known as chips.

Figure 10.3(a) Metal cutting operation.

Orthogonal Cutting:

When chip flows along orthogonal plane, p0, i.e., rc = 0

Oblique Cutting: When chip flow deviates from orthogonal plane, i.e., rc   %XW practically rc may be zero even if l = 0 and rc may not be exactly equal to l even if l   %HFDXVH WKHUH DUH VRPH RWKHU WKDQ l) factors also which may cause chip flow deviation.

Manufacturing Processes 225

Geometry of single point turning tools both material and geometry of the cutting tools play very important roles on their performances in achieving effectiveness, efficiency and overall economy of machining. Cutting tools may be classified according to the number of major cutting edges (points) involved as follows: ‡ 6LQJOH SRLQW HJ WXUQLQJ WRROV VKDSLQJ SODQQLQJ DQG VORWWLQJ WRROV DQG ERULQJ tools ‡'RXEOHWZR SRLQWHJGULOOV ‡ 0XOWLSRLQW PRUH WKDQ WZR  HJ PLOOLQJ FXWWHUV EURDFKLQJ WRROV KREV JHDU shaping cutters, etc., (i) Concept of rake and clearance angles of cutting tools. The word tool geometry is basically referred to some specific angles or slope of the salient faces and edges of the tools at their cutting point. Rake angle and clearance angle are the most significant for all the cutting tools. The concept of rake angle and clearance angle will be clear from some simple operations shown in Fig. 10.3 (b).

Figure 10.3 (b)

Rake Angle (g): Angle of inclination of rake surface from reference plane. Rake angle is provided for ease of chip flow and overall machining. It may be positive, or negative or even zero. Clearance Angle (a): Angle of inclination of clearance or flank surface from the finished surface. Clearance Angle: It is essentially provided to avoid rubbing of the tool (flank) with the machined surface which causes loss of energy and damages of both the tool and the job surface. Hence, clearance angle is a must and must be positive (3° ~ 15° depending on tool-work materials and type of the machining operations like turning, drilling, boring, etc.)

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ASA and ORS Tool Geometry

Figure 10.4 Tool angles in ASA system.

ASA system Rake Angles: gx = side (axial rake: angle of inclination of the rake surface from the reference plane (pR) and measured on machine reference plane, pX. gy = back rake: angle of inclination of the rake surface from the reference plane and measured on machine transverse plane, pY. Clearance Angles: [Fig. 10.4] ax = side clearance: angle of inclination of the principal flank from the machined surface (or VC) and measured on pX plane. ay = back clearance: same as ax but measured on pY plane. Cutting angles: [Fig. 10.4] js = approach angle: angle between the principal cutting edge (its projection on pR) and pY and measured on pR je = end cutting edge angle: angle between the end cutting edge (its projection on pR) from pX and measured on pR Nose radius, r (in inch) r = nose radius Curvature of the tool tip: It provides strengthening of the tool nose and better surface finish.

Orthogonal Rake System (ORS) This system is also known as ISO – old. The planes of reference and the coordinate axes used for expressing the tool angles in ORS are: pR - pC - pO and Xo - Yo - Zo respectively which are taken in respect of the tool configuration

Manufacturing Processes 227

‡ ASA System – gy, gx, ay, ax, je, js, r (inch) ‡ ORS System – l, go, ao, ao¢, j1, j, r (mm) where, SR = Reference plane; plane perpendicular to the velocity vector SX = Machine longitudinal plane; plane perpendicular to SR and taken in the direction of assumed longitudinal feed SY = Machine transverse plane; plane perpendicular to both SR and SX [This plane is taken in the direction of assumed cross feed]  O = inclination angle; angle between SC from the direction of assumed longitudinal feed [SX] and measured on SC ‡ Clearance angles  DO = orthogonal clearance of the principal flank; angle of inclination of the principal flank from SC and measured on SO DO' = auxiliary orthogonal clearance; angle of inclination of the auxiliary flank from auxiliary cutting plane, SC' and measured on auxiliary orthogonal plane, SO' ‡ Cutting angles  I = principal cutting edge angle: angle between SC and the direction of assumed longitudinal feed or SX and measured on SR I1 = auxiliary cutting angle: angle between SC' and SX and measured on SR ‡ Nose radius, r (mm) r = radius of curvature of tool tip Tool life generally indicates the amount of satisfactory performance or service rendered by a fresh tool or a cutting point till it is declared failed. Tool life is defined in two ways: (a) In R&D: Actual machining time (period) by which a fresh cutting tool (or point) satisfactorily works after which it needs replacement or reconditioning. The modern tools hardly fail prematurely or abruptly by mechanical breakage or rapid plastic deformation. Those fail mostly by wearing process which systematically grows slowly with machining time. In that case, tool life means the span of actual machining time by which a fresh tool can work before attaining the specified limit of tool wear. Mostly tool life is decided by the machining time till flank wear, VB reaches 0.3 mm or crater wear, KT reaches 0.15 mm. (b) In industries or shop floor: The length of time of satisfactory service or amount of acceptable output provided by a fresh tool prior to it is required to replace or recondition. Assessment of tool life for R&D purposes, tool life is always assessed or expressed by span of machining time in minutes, whereas, in industries besides machining time in minutes some other means are also used to assess tool life, depending on the situation, such as:

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Basic Mechanical Engineering

‡ 1XPEHURISLHFHVRIZRUNPDFKLQHG ‡ 7RWDOYROXPHRIPDWHULDOUHPRYHG ‡ 7RWDOOHQJWKRIFXW Measurement of tool wear: (i) by loss of tool material in volume or weight, in one lifetime – this method is crude and is generally applicable for critical tools like grinding wheels. (ii) by grooving and indentation method – in this approximate method wear depth is measured indirectly by the difference in length of the groove or the indentation outside and inside the worn area. (iii) using optical microscope fitted with micrometer – very common and effective method. (iv) using scanning electron microscope (SEM) – used generally, for detailed study; both qualitative and quantitative. (v) Talysurf, specially for shallow crater wear.

Taylor’s Tool Life Equation Wear and hence tool life of any tool for any work material is governed mainly by the level of the machining parameters, i.e., cutting velocity, (CV), feed, (so) and depth of cut (t). Cutting velocity affects maximum and depth of cut minimum. Taylor derived the simple equation as VT n = C where, n is called, Taylor’s tool life exponent. The values of both n and c depend mainly on the tool-work materials and the cutting environment (cutting fluid application). The metal components nowadays are made into different shapes and dimensions using various metal working processes. Metal working processes are classified into two major groups. They are: (i) Non-cutting shaping or chips less or metal forming process — forging, rolling, pressing, etc. (ii) Cutting shaping or metal cutting or chip forming process — turning, drilling, milling, etc.

Material Removal Process/Machining Machining is an essential process of finishing by which workpieces are produced to the desired dimensions and surface finish by gradually removing the excess material from the preformed blank in the form of chips with the help of cutting tool(s) moved past the work surface(s).

Machine Tools A machine tool is a non-portable power operated and reasonably valued device or system of devices in which energy is expended to produce jobs of desired size, shape

Manufacturing Processes 229

and surface finish by removing excess material from the preformed blanks in the form of chips with the help of cutting tools moved past the work surface(s). Machine tools basically produce geometrical surfaces like flat, cylindrical or any contour on the preformed blanks by machining work with the help of cutting tools.

10.6.1 Types of Machine Tools Centre lathe, shaping machine, drilling machine, milling machine, etc. Lathe: It is one of the most versatile and widely used machine tools all over the world. It is commonly known as the mother of all other machine tool. The main function of a lathe is to remove metal from a job to give it the required shape and size. The job is secure1y and rigidly held in the chuck or in between centres on the lathe machine and then turn it against a single point cutting tool which will remove meta1 from the job in the form of chips. Figure 10.5 shows the working principle of lathe. An engine lathe is the most basic and simplest form of the lathe. It derives its name from the early lathes, which obtained their power from engines. Besides the simple turning operation as described above, lathe can be used to carry out other operations also, such as drilling, reaming, boring, taper turning, knurling, screwthread cutting, grinding, etc.

Figure 10.5 Lathe

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Figure 10.6 Drilling machine.

Drilling Machine: Drilling is an operation of making a circular hole by removing a volume of metal from the job by cutting tool called drill. A drill is a rotary end-cutting tool with one or more cutting lips and usually one or more flutes for the passage of chips and the admission of cutting fluid. A drilling machine is a machine tool designed for drilling holes in metals. It is one of the most important and versatile machine tools in a workshop. Besides drilling round holes, many other operations can also be performed on the drilling machine such as counter-boring, counter-sinking, honing, reaming, lapping, sanding, etc. Shaping Machine: Shaper is a reciprocating type of machine tool in which the ram moves the cutting tool backwards and forwards in a straight line. The basic components of a shaper are shown in Fig. 10.6. It is intended primarily to produce flat surfaces. These surfaces may be horizontal, vertical, or inclined. In general, the shaper can produce any surface composed of straight-line element. A shaper is used to generate flat (plane) surfaces by means of a single point cutting tool similar to a lathe tool.

Manufacturing Processes 231

Figure 10.7 Shaping machine.

Milling Machine: A milling machine is a machine tool that removes metal as the work is fed against a rotating multi-point cutter. The milling cutter rotates at high speed and it removes metal at a very fast rate with the help of multiple cutting edges. One or more number of cutters can be mounted simultaneously on the arbor of milling machine. This is the reason that a milling machine finds wide applications in production work. Milling machine is used for machining flat surfaces, contoured surfaces, surfaces of revolution, external and internal threads, and helical surfaces of various cross sections. In many applications, due to its higher production rate and accuracy, milling machine has even replaced shapers and slotters.

10.7

METAL JOINING PROCESS

Many products observed in day-to-day life, are commonly made by putting many parts together, may be in subassembly. For example, a ball pen consists of a body, refill, barrel, cap, and refill operating mechanism. All these parts are put together to form the product as a pen. More than 800 parts are put together to make various subassemblies and final assembly of car or aeroplane. A complete machine tool may also require to assemble more than 100 parts in various subassemblies or final assembly. The process of putting the parts together to form the product, which performs the desired function, is called assembly. An assemblage of parts may require some parts to be joined together using various joining processes. But assembly should not be confused with the joining process. Most of the products cannot be manufactured as single unit they are manufactured as different components using one or more of the above manufacturing processes, and these components are assembled to get the desired product. Joining processes are widely used in fabrication and assembly work. In these processes two or

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more pieces of metal parts are joined together to produce desired shape and size of the product. The joining processes are carried out by fusing, pressing, rubbing, riveting, screwing or any other means of assembling. These processes are used for assembling metal parts and in general fabrication work. Such requirements usually occur when several pieces are to be joined together to fabricate a desired structure of products. These processes are used for developing steam or water-tight joints. Temporary, semipermanent or permanent type of fastening to make a good joint is generally created by these processes. Temporary joining of components can be achieved by use of nuts, screws and bolts. Adhesives are also used to make temporary joints. Some of the important and common joining processes are: 1. Welding (plastic or fusion) 2. Brazing 3. Soldering 4. Riveting 5. Screwing 6. Press fitting 7. Sintering 8. Adhesive bonding 9. Shrink fitting 10. Explosive welding 11. Diffusion welding 12. Keys and cotters joints 13. Coupling 14. Nut and bolt joints.

10.7.1 Importance of Joining Process (i) Some (even simple) products are too large to be made by individual processes: 3-D hollow structural member, 5 m. (ii) Easier, more economical to manufacture and join individual components: cooking pot with handle. (iii) Products to be disassembled for maintenance like appliances: engines. (iv) Varying functionality of product: carbide inserts in tool steel (brakeshoes). (v) Transportation + assembly is less costly: shelving units; machinery

Manufacturing Processes 233

Figure 10.8 Various joining operations

10.8

WELDING OPERATIONS

Welding is a process for joining two similar or dissimilar metals by fusion. It joins different metals/alloys, with or without the application of pressure and with or without the use of filler metal. The fusion of metal takes place by means of heat. Heat may be generated either from combustion of gases, electric arc, electric resistance or by chemical reaction. During some type of welding processes, pressure may also be employed, but this is not an essential requirement for all welding processes. Welding provides a permanent joint but it normally affects the metallurgy of the component. It is therefore usually accompanied by post-weld heat treatment for most of the critical components. Welding is widely used as a fabrication and repairing process in industries. Some of the typical applications of welding include the fabrication of ships, pressure vessels, automobile bodies, off-shore platform, bridges, welded pipes, sealing of nuclear fuel and explosives, etc. Most of the metals and alloys can be welded by one type of welding process or the other. However, some are easier to weld than others. To compare this ease in welding term ‘weldability’ is often used. The weldability may be defined as

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property of a metal which indicates the ease with which it can be welded with other similar or dissimilar metals. Weldability of a material depends on various factors like the metallurgical changes that occur due to welding, changes in hardness in and around the weld, gas evolution and absorption, extent of oxidation, and the effect on cracking tendency of the joint. Plain low carbon steel (C-0.12%) has the best weldability amongst metals. Generally, it is seen that the materials with high castability usually have low weldability. Welding is the process of joining of metal where two or more parts are joined together at their contacting surfaces by application of heat or pressure. Sometimes parts are united together by application of pressure only without external heat. In some welding processes a filler material is added to facilitate the joining of the metals. Welding is most commonly associated with metallic parts but it can also be used for plastics.

10.8.1 Types of Welding Processes Solid State Welding Processes: In solid state welding the surfaces to be joined are brought into close proximity by heating the surfaces without getting melted and applying normal pressure or by providing relative motion between the two surfaces and applying light normal pressure or by applying high pressure without heating, e.g., forge welding, friction welding. Liquid State Welding Processes: Arc welding, resistance welding, oxy-fuel gas welding and other processes. There are two inherent problems with fusion welding. ‡ Effect of localized heating and rapid cooling localized heating on the microstructure and properties of the parent metals. ‡ Effect of residual stresses developed in the parent metals due to restrained ex-metals expansion or contraction.

10.8.2 Welding Defects Cracks: This causes significant reduction in the strength of weldment. Welding cracks are caused by embrittlement or low ductility of the weld and/or base metal combined with high restraint during contraction. Cavities:

These include porosity and shrinkage voids.

Solid Inclusions: These are metallic or non-metallic solid material particles entrapped in the weld metal. The most common form is slag inclusion or metallic oxides. Incomplete Fusion: A similar defect is lack of penetration.

Manufacturing Processes 235

Figure 10.9 Welding operation.

10.8.3 Advantages of Welding        

‡ :HOGLQJLVPRUHHFRQRPLFDODQGLVDPXFKIDVWHUSURFHVVDVFRPSDUHGWRRWKHU processes (riveting, bolting, casting, etc.) ‡ :HOGLQJLISURSHUO\FRQWUROOHGUHVXOWVLQSHUPDQHQWMRLQWVKDYLQJVWUHQJWKHTXDO or sometimes more than base metal. ‡ /DUJHQXPEHURIPHWDOVDQGDOOR\VERWKVLPLODUDQGGLVVLPLODUFDQEHMRLQHGE\ welding. ‡ *HQHUDOZHOGLQJHTXLSPHQWLVQRWFRVWO\ ‡ 3RUWDEOHZHOGLQJHTXLSPHQWFDQEHHDVLO\PDGHDYDLODEOH ‡ :HOGLQJSHUPLWVFRQVLGHUDEOHIUHHGRPLQGHVLJQ ‡ :HOGLQJFDQMRLQZHOGLQJMREVWKURXJKVSRWVDVFRQWLQXRXVSUHVVXUHWLJKWVHDPV end-to-end and in a number of other configurations. ‡ :HOGLQJFDQDOVREHPHFKDQL]HG

10.8.4 Disadvantages of Welding     

‡ ,WUHVXOWVLQUHVLGXDOVWUHVVHVDQGGLVWRUWLRQRIWKHZRUNSLHFHV ‡ :HOGHGMRLQWQHHGVVWUHVVUHOLHYLQJDQGKHDWWUHDWPHQW ‡ :HOGLQJJLYHVRXWKDUPIXOUDGLDWLRQVOLJKW IXPHVDQGVSDWWHU ‡ -LJVDQGIL[WXUHVPD\DOVREHQHHGHGWRKROGDQGSRVLWLRQWKHSDUWVWREHZHOGHG ‡ (GJHVSUHSDUDWLRQRIWKHZHOGLQJMREVDUHUHTXLUHGEHIRUHZHOGLQJ

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‡ 6NLOOHGZHOGHULVUHTXLUHGIRUSURGXFWLRQRIJRRGZHOGLQJ ‡ +HDWGXULQJZHOGLQJSURGXFHVPHWDOOXUJLFDOFKDQJHVDVWKHVWUXFWXUHRIWKHZHOGHG joint is not same as that of the parent metal.

10.9

SURFACE FINISHING PROCESSES

Surface finishing processes are utilized for imparting intended surface finish on the surface of a job. By imparting a surface finishing process, dimension of the part is not changed functionally; either a negligible amount of material is removed from certain material is added to the surface of the job. These processes should not be misunderstood as metal removing processes in any case as they are primarily intended to provide a good surface finish or a decorative or protective coating on to the metal surface. Surface cleaning process is also called a surface finishing process. Some of the commonly used surface finishing processes are: 1. Honing 2. Lapping 3. Super finishing 4. Belt grinding 5. Polishing 6. Tumbling 7. Organic finishes 8. Sanding 9. Deburring 10. Electroplating 11. Buffing 12. Metal spraying 13. Painting 14. Inorganic coating 15. Anodizing 16. Sheradising 17. Parkerizing 18. Galvanizing 19. Plastic coating 20. Metallic coating 21. Anodizing 22. Sand blasting.

10.10

POWDER METALLURGY

Powder metallurgy is used for manufacturing products or articles from powdered metals by placing these powders in moulds and compacting the same using a heavy compressive force. Typical examples of such articles or products are grinding wheels, filament wire, magnets, welding rods, tungsten carbide cutting tools, self-lubricating bearings electrical contacts and turbines blades having high temperature strength. The manufacture of parts by powder metallurgy process involves the manufacture of powders, blending, compacting, profiteering, sintering and a number of secondary operations such as sizing, coining, machining, impregnation, infiltration, plating, and heat treatment. The compressed articles are then heated to temperatures much below their melting points to bind the particles together and improve their strength and other properties. Few non-metallic materials can also be added to the metallic powders to provide adequate bond or impart some needed properties. The products made through this process are

Manufacturing Processes 237

costly on account of the high cost of metal powders as well as of the dies used. The powders of almost all metals and a large quantity of alloys, and nonmetals may be used. The application of powder metallurgy process is economically feasible only for mass production. Parts made by powder metallurgy process exhibit properties, which cannot be produced by conventional methods. Simple shaped parts can be made to size with high precision without waste, and completely or almost ready for installation. Powder metallurgy is the process of making powder metals which are then used for making finished and semifinished products from the mixture of metal alloy powders or may be with the addition of nonmetallic constituents to it. Following are the steps in powder metallurgy: powder production, compaction, sintering, and secondary operations.

10.10.1

Powder Production

The raw material used in powder metallurgy is the powder. Powders can be of different kind. It can be pure element, pre-alloyed powder. Methods for making powder: (a) Atomization: Produces powders of both ferrous and nonferrous powders like stainless steel, superalloys, Ti alloy powders. (b) Reduction of compounds: Production of iron, Cu, tungsten, molybdenum (c) Electrolysis: For making Cu, iron, silver powders along with additives are mixed, using mixers. Lubricants are added prior to mixing to facilitate easy ejection of compact and to minimize wear of tools; waxes, metallic stearates, graphite, etc.

10.10.2

Compaction

The process of compaction is done using dies which are machined to close tolerances. Dies are made of cemented carbide, die/tool steel, pressed using hydraulic or mechanical presses. The basic purpose of compaction is to obtain a green compact with sufficient strength to withstand further handling operations. The green compact is then taken for sintering.

10.10.3

Sintering

This process is performed at controlled atmosphere for the metallurgical bonding of the atoms. Bonding occurs at 70% of abs. melting point of materials by the diffusion of atoms. It serves to consolidate mechanically bonded powders into a coherent body having desired on service behaviour. Densification occurs during the process and improvement in physical and mechanical properties are seen.

10.10.4

Secondary Operations

The operations which includes repressing, grinding, plating are done by secondary operation. They are used to ensure close dimensional tolerances, good surface finish, increase density, corrosion resistance, etc.

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10.10.5

Advantages of Powder Metallurgy

1. The processes of powder metallurgy are quite and clean. 2. Articles of any intricate or complicated shape can be manufactured. 3. The dimensional accuracy and surface finish obtainable are much better for many applications and hence machining can be eliminated.   8QOLNHFDVWLQJSUHVVIRUPLQJPDFKLQLQJQRPDWHULDOLVZDVWHGDVVFUDSDQGWKH process utilizes full raw material. 5. Hard to process materials such as diamond can be converted into usable components and tools through this process. 6. High production rates can be easily achieved. 7. The phase diagram constraints, which do not allow an alloy formation between mutually insoluble constituents in liquid state, such as in case of copper and lead are removed in this process and mixtures of such metal powders can be easily processed and shaped through this process. 8. This process facilitates production of many such parts, which cannot be produced through other methods, such as sintered carbides and self-lubricating bearings. 9. The process enables an effective control over several properties such as purity, density, porosity, particle size, etc., in the parts produced through this process. 10. The components produced by this process are highly pure and bear longer life. 11. It enables production of parts from such alloys, which possess poor castability. 12. It is possible to ensure uniformity of composition, since exact proportions of constituent metal powders can be used. 13. The preparation and processing of powdered iron and nonferrous parts made in this way exhibit good properties, which cannot be produced in any other way. 14. Simple shaped parts can be made to size with 100 micron accuracy without waste. 15. Porous parts can be produced that could not be made in any other way. 16. Parts with wide variations in compositions and materials can be produced. 17. Structure and properties can be controlled more closely than in other fabricating processes. 18. Highly qualified or skilled labour is not required in powder metallurgy process. 19. Super-hard cutting tool bits, which are impossible to produce by other manufacturing processes, can be easily manufactured using this process. 20. Shapes of components obtained possess excellent reproducibility. 21. Control of grain size, relatively much uniform structure and defect such voids and blowholes in structure can be eliminated.

Manufacturing Processes 239

10.10.6

Limitations of Powder Metallurgy

1. Powder metallurgy process is not economical for small-scale production. 2. The cost of tool and die of powder metallurgical set-up is relatively high. 3. The size of products as compared to casting is limited because of the requirement of large presses and expensive tools which are required for compacting. 4. Metal powders are expensive and in some cases difficult to store without some deterioration. 5. Intricate or complex shapes produced by casting cannot be made by powder metallurgy because metallic powders lack the ability to flow to the extent of molten metals. 6. Articles made by powder metallurgy in most cases do not have as good physical properties as wrought or cast parts. 7. It may be difficult sometimes to obtain particular alloy powders 8. Parts pressed from the top tend to be less dense at the bottom. 9. A completely deep structure cannot be produced through this process. 10. The process is not found economical for small-scale production. 11. It is not easy to convert brass, bronze and a numbers of steels into powdered form.

10.10.7

Applications of Powder Metallurgy

The powder metallurgy process has provided a practical solution to the problem of producing refractory metals, which have now become the basis of making heat-resistant materials and cutting tools of extreme hardness. Another important and useful item of the products made from powdered metals is porous self-lubricating bearing. Some of the powder metal products are given as under. 1. Porous products such as bearings and filters. 2. Tungsten carbide, gauges, wire drawing dies, wire-guides, stamping and blanking tools, stones, hammers, rock drilling bits, etc. 3. Various machine parts are produced from tungsten powder. Highly heat and wear resistant cutting tools from tungsten carbide powders with titanium carbide, powders are used for die manufacturing. 4. Refractory parts such as components made out of tungsten, tantalum and molybdenum are used in electric bulbs, radio valves, oscillator valves, X-ray tubes in the form of filament, cathode, anode, control grids, electric contact points, etc. 5. Products of complex shapes that require considerable machining when made by other processes, namely, toothed components such as gears. 6. Components used in automotive part assembly such as electrical contacts, crankshaft drive or camshaft sprocket, piston rings and rocker shaft brackets,

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door, mechanisms, connecting rods and brake linings, clutch facings, welding rods, etc. 7. Products where the combined properties of two metals or metals and non-metals are desired such as non-porous bearings, electric motor brushes, etc. 8. Porous metal bearings which are later impregnated with lubricants. Copper and graphite powders are used for manufacturing automobile parts and brushes. 9. The combinations of metals and ceramics, which are bonded by similar process as metal powders, are called cermets. They combine in them useful properties

of high refractoriness of ceramics and toughness of metals. They are produced in two forms, namely, oxides based and carbide based. 10.11

SUMMARY

Manufacturing becomes successful by understanding how the system works, how goods are controlled, and the decision making at the correct level. Engineers must possess a broad fundamental knowledge of design, metallurgy, processing, economics, accounting and human relations. Manufacturing procedures are relevant in all aspects of our lives, so much that we frequently don’t understand or consider it. From the autos we drive, the holders our sustenance comes in, the TVs, PCs and different gadgets we utilize, control instruments, warmers, ventilation systems, the channels that convey our water and the rundown continues endlessly to incorporate pretty much everything characterizing our present-day society. These things are all made or worked from fabricated segments. Producing gear itself should likewise be fabricated. The assembling procedure utilized is dictated by an assortment of variables. The major thought of assembling or generation is to make, (or create), something that has a valuable frame. This frame is doubtlessly foreordained and figured, with a specific physical geometry. Typically, this geometry has certain resistances that it must meet keeping in mind the end goal to be viewed as adequate.

SOLVED EXAMPLES

Example 10.1 A certain mould has a sprue whose length is 20 cm and the cross-sectional area at the base of the sprue is 2.5 cm2. The sprue feeds a horizontal runner leading into a mould cavity whose volume is 1560 cm3. Determine: (a) velocity of the molten metal at the base of the sprue, (b) volume rate of flow, and (c) time to fill the mould. Solution: (a) The velocity of the following metal at the base of the sprue is given by v=

2 (981) ( 20 ) = 198.1 cm/s

Manufacturing Processes 241

(b) The volumetric flow rate is Q = (2.5 cm2) (198.1 cm/s) = 495 cm2/s (c) Time required to fill a mould cavity of 1560 in3 at this flow rate is MFT = 1560/495 = 3.2 s

Example 10.2 A cylindrical riser must be designed for a sand-casting mold. The casting itself is a steel rectangular plate with dimensions 7.5 cm x 12.5 cm x 2.0 cm. Previous observations have indicated that the total solidification time (TST) for this casting = 1.6 min. The cylinder for the riser will have a diameter-to-height ratio = 1.0. Determine the dimensions of the riser so that its TST = 2.0 min. Solution: First determine the V/A ratio for the plate. Its volume V = 7.5 × 12.5 × 2.0 = 187.5 cm3 and its surface area A = 2(7.5 × 12.5 + 7.5 × 2.0 + 12.5 × 2.0) = 267.5 cm2. Given that TST = 1.6 min, we can determine the mold constant cm, using a value of n = 2 in the equation. cm =

TST 1.6 = = 3.26 min/cm 2 (V/A)2 (187.5 / 267.5)2

Next we must design the riser so that its total solidification time is 2.0 min, using the same value of mold constant. Since both the casting and the riser are in the same. The volume of the riser is given by V=

nD2 h 4

2pD2 = 1.5 pD2 and 4 Thus, the V/A ratio = D8VLQJWKLVUDWLRLQ&KYRULQRY VHTXDWLRQZHKDYH 2 A = pD +

Ê Dˆ TST = 2.0 = 3.23 Á ˜ 2 = 0.09056 D2 Ë 6¯

D2 = 2.0/0.09056 = 22.086 cm2 D = 4.7 cm Since H = D, then H = 4.7 cm also. The riser represents waste metal that will be separated from the cast part and remelted to make subsequent castings. It is desirable for the volume of metal in the riser to be a minimum. Since the geometry of the riser is normally selected to maximize the V/A ratio, this tends to reduce the riser volume as much as possible. Note that the volume of the riser in our example problem is V = p(4.7)3/4 = 81.5 cm3 only 44% of the volume of the plate (casting), even though its total solidification time is longer by 25%.

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Example 10.3 What are the limitations of a milling machine? Solution: The major limitations of a milling machine are as follows: 1. Milling machine is mostly used for machining jobs of smaller size. 2. Its speed is slow for machining long jobs. 3. Cutting tool is costlier.

Example 10.4 In a turning operation on stainless steel with hardness = 200 HB, cutting speed = 200 m/min, feed = 0.25 mm/rev, and depth of cut = 7.5 mm. How much power will the lathe draw in performing this operation if its mechanical efficiency = 90%. Solution: 3

3

U = 2.8 N-m/mm = 2.8 J/mm (from Standard Table) RMR = vfd = (200 m/min)(103 mm/m)(0.25 mm)(7.5 mm) = 375,000 mm3/min = 6250 mm3/s Pc = (6250 mm3/s)(2.8 J/mm3) = 17,500 J/s = 17,500 W = 17.5 kW Accounting for mechanical efficiency, Pg = 17.5/0.90 = 19.44 kW

Example 10.5 Name the three most common machining processes. Solution: The three common machining processes are: 1. turning 2. drilling 3. milling

Example 10.6 What are the two basic categories of cutting tools in machining? Give two examples of machining operations that use each of the tooling types.

Solution: The two categories are: 1. single-point tools, used in operations such as turning and boring; and 2. multiple-edge cutting tools, used in operations such as milling and drilling.

Example 10.7 Name and briefly describe the four types of chips that occur in metal cutting.

Solution: The four types are: 1. discontinuous, in which the chip is formed into separate segments; 2. continuous, in which the chip does not segment and is formed from a ductile metal; 3. continuous with built-up edge, which is the same as (2) except that friction at the tool-chip interface causes adhesion of a small portion of work material to the tool rake face, and

Manufacturing Processes 243

4. serrated, which are semicontinuous in the sense that they possess a saw-tooth appearance that is produced by a cyclical chip formation of alternating high shear strain followed by low shear strain.

Example 10.8 How does a boring operation differ from a turning operation? Solution: Boring produces an internal cylindrical shape from an existing hole, while turning produces an external cylindrical shape.

Example 10.9 In an orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness before the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm. Calculate: (a) the shear plane angle, and (b) the shear strain for the operation. Solution: (a) r = to/tc = 0.30/0.65 = 0.4615 j = tan– 1 (0.4615 cos 15/(1 – 0.4615 sin 15)) = tan–1(0.5062) = 26.85° (b) Shear strain, g = cot 26.85 + tan (26.85 – 15) = 1.975 + 0.210 = 2.185

Example 10.10 A cylindrical workpart 200 mm in diameter and 700 mm long is to be turned in an engine lathe. Cutting speed = 2.30 m/s, feed = 0.32 mm/rev, and depth of cut = 1.80 mm. Determine (a) cutting time, and (b) metal removal rate.

Solution: (a) N = v/(pD) = (2.30 m/s)/0.200p = 3.66 rev/s fr = Nf = 6.366(.3) = 1.17 mm/s Tm = L/fr = 700/1.17 = 598 s = 9.96 min Alternative calculation using Eq. (16.5), Tm = 200(700)p/(2,300 × 0.32) = 597.6 sec = 9.96 min (b) RMR = vfd = (2.30 m/s)(103)(0.32 mm)(1.80 mm) = 1320 mm3/s 16.2

Example 10.11 In a production turning operation, the foreman has decreed that a single pass must be completed on the cylindrical workpiece in 5.0 min. The piece is 400 mm long DQGPPLQGLDPHWHU8VLQJDIHHG PPUHYDQGDGHSWKRIFXW PPZKDW cutting speed must be used to meet this machining time requirement?

Solution: Starting with Tm = pDoL/vf. Rearranging to determine cutting speed: v = pDoL/fTm v = p(0.4)(0.15)/(0.30)(10-3 )(5.0) = 0.1257(103 ) m/min = 125.7 m/min

Example 10.12 A drilling operation is to be performed with a 12.7 mm diameter twist drill in a steel workpart. The hole is a blind hole at a depth of 60 mm and the point angle is 118°. The cutting speed is 25 m/min and the feed is 0.30 4 mm/rev. Determine (a) the cutting time to complete the drilling operation, and (b) metal removal rate during the operation, after the drill bit reaches full diameter.

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Solution: (a) N = v/pD = 25(103)/(12.7p) = 626.6 rev/min fr = Nf = 626.6(0.30) = 188 mm/min A = 0.5 D tan (90 – /2) = 0.5(12.7) tan(90 – 118/2) = 3.82 mm Tm = (d + A)/fr = (60 + 3.82)/188 = 0.339 min (b) MRR = 0.25 pD2 fr = 0.25 p(12.7)2 (188) = 23,800 mm3/min

Example 10.13 If in turning of a steel rod by a given cutting tool (material and geometry) at a given machining condition (so and t) under a given environment (cutting fluid application), the tool life decreases from 80 min to 20 min due to increase in cutting velocity, VC from 60 m/min to 120 m/min, then at what cutting velocity the life of that tool under the same condition and environment will be 40 min?

Solution: Assuming Taylor’s tool life equation, VT n = C V1T1 = V2T2 = V3T3 = C Here, V1 = 60 m/min; T1 = 80 min. V2 = 120 m/min; T2 = 20 min V3 = ? (to be determined); T3 = 40 min. Taking, V1T n1 = V2T 2n Ê T1 ˆ ÁË T ˜¯

i.e.,

n

ÊV ˆ = Á 2˜ Ë V1 ¯

n

Ê 120 m/min ˆ = Á Ë 60 m/min ˜¯

2

Ê 80 min ˆ ÁË 20 min ˜¯

or from which, n = 0.5 Again

V3T n3 = V3T n3 i.e.,

or

Ê T1 ˆ Ê V3 ˆ ÁË V ˜¯ = ÁË T ˜¯ 3 1 Ê 80 ˆ V3 = Á ˜ Ë 40 ¯

n

0.5

¥ 60 = 84.84 m/min Ans.

Manufacturing Processes 245

EXERCISE 1. Define manufacturing processes. 2. With the help of flowchart show and explain the classification of manufacturing process. 3. What is metal joining process? Explain with examples. 4. Briefly explain the metal casting process with examples. 5. Name the various defects which occur in sand castings and state their probable causes and remedies? 6. List the defects generally occurring from the following, stating the precautions necessary to prevent them: (i) Improper pouring technique  LL  8VHRIGHIHFWLYHJDWLQJV\VWHP (iii) Poor or defective cores (iv) High moisture content in sand. 7. Discuss briefly the causes and remedies of the following casting defects: (i) Blow holes (ii) Porosity (iii) Hot tears (iv) Shrinkage cavities (v) Scabs (vi) Gas porosity 8. What are various types of welding defects? Explain any two of them in brief. 9. Explain powder metallurgy and briefly explain the steps involved in it. 10. Describe the process of blending, compacting and sintering in detail. 11. What are the effects of sintering on the powder compact produced by pressing? 12. Name the products of powder metallurgy. 13. During turning a metallic rod at a given condition, the tool life was found to increase from 25 min to 50 min when VC was reduced from 100 m/min to 80 m/ min. How much will be the life of that tool if machined at 90 m/min?

11 Conventional Machining Techniques

Learning Objectives   

‡ 8QGHUVWDQGLQJWKHWHUPVLQFRQYHQWLRQDOPDFKLQLQJWHFKQLTXHV ‡ 2YHUYLHZ1&DQG&1&PDFKLQLQJWHFKQLTXHV ‡ &LWHWKHFRQYHQWLRQDOPDFKLQLQJSURFHVVHV

11.1

INTRODUCTION

Machining is a widely used term used to describe the process of material removal from a workpiece and it includes several processes, which are described below. ‡ Cutting process which generally involves single-point or multi-point cutting tools, each with a defined geometry. ‡ Abrasive processes, such as grinding. ‡ Nontraditional machining processes, using electrical, chemical, and different sources of energy. It is vital to see machining, and also all the manufacturing processes as a framework comprising the workpiece, the cutting tool and the machine. The customary machining process incorporates introduction on turning, milling, drilling, and grinding. In addition it incorporates computer applications which are bolstered by the preliminaries. The non-customary machining incorporates introduction on the themes like ECM, EDM, $)08600DFKLQLQJLVWKHPRVWHVVHQWLDORIWKHDVVHPEOLQJIRUPV0DFKLQLQJFDQEH characterized as the way towards expelling material from a workpiece as chips. The term metal cutting is utilized when the material is metallic. Most machining has low Basic Mechanical Engineering

Conventional Machining Techniques 247

set-up cost contrasted with shaping, embellishment, and throwing forms. Be that as it may, machining is substantially more costly for high volumes. Machining is important where tight resistances on measurements and completions are required.

11.2

CONVENTIONAL MACHINING TECHNIQUES

The material removal processes is mainly divided into two groups and they are “conventional machining processes” and “non-conventional manufacturing processes”. Examples of conventional machining processes are turning, boring, milling, shaping, broaching, slotting, grinding, etc. In conventional machining the shape of a workpiece LVFKDQJHGXVLQJDQLPSOHPHQWPDGHRIDKDUGHUPDWHULDO8VLQJFRQYHQWLRQDOPHWKRGV to machine hard metals and alloys means increased demand of time and energy and therefore increases in costs. Conventional machining may not be feasible in all the cases. With constantly expanding interest for products of hard combinations and metals, for example, Inconel 718 or titanium, more intrigue has inclined toward non-routine machining strategies. Customary machining can be characterized as a procedure utilizing mechanical (movement) energy. Regardless of the type of lathe, three key parameters determine productivity and part quality. These parameters are: ‡7KHFXWWLQJVSHHG The cutting speed in a machining operation refers to the peripheral speed of a point on the surface of the machine in contact with the work. It is usually expressed in metres/ min. The cutting speed (Cs) may be calculated as: Cs = ((22/7) × D × N)/1000 where, D = diameter of the in mm, and N = rpm. ‡7KHIHHGUDWH The feed is the distance the machine moves into the job at each revolution of the spindle. It is expressed in millimetre. The feed may also be expressed as feed per minute. The feed per minute may be defined as the axial distance moved by the machine into the work per minute. The feed per minute may be calculated as: F = Fr × N

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where, F = Feed per minute in mm. Fr = Feed per revolution in mm. N = RPM ‡The depth of cut The cutting speed is the speed of the work as it rotates past the cutting tool. The feed rate is the rate at which the tool advances into the work. The depth of cut is the amount of material removed as the work revolves on its axis. Other factors include the machinability of the stock, the type and the geometry of the cutting tool, the angle of the tool to the work, and the overall condition and power of the lathe itself.

11.2.1 Turning Turning is the metal removal process where metal is removed from the outer diameter of a rotating cylindrical workpiece. It is used to reduce the diameter of the workpiece, usually to a given dimension and also to produce a smooth finish on the metal. The workpiece is often turned to have different diameters adjacent sections. Turning is one of the most common of metal cutting operations. In turning, a workpiece is rotated about its axis as single-point cutting tools are fed into it, shearing away unwanted material and creating the desired part. Turning can occur on both external and internal surfaces to produce an axially-symmetrical contoured part. Parts ranging from pocket watch components to large diameter marine propeller shafts can be turned on a lathe. The capacity of a lathe is expressed in two dimensions. The maximum part diameter, or swing, and the maximum part length, or distance between centres.

Tapers and Taper Turning A taper is defined as a uniform increase or decrease in diameter of a piece of work measured along its length. In a lathe machine, taper turning means to produce a conical surface by gradual reduction in diameter from a cylindrical job. Taper in the British system is expressed in taper per foot or taper per inch. Taper per inch = (D – d) /l where, D = diameter of the large end of cylindrical job, d = diameter of the small end of cylindrical job, and l = length of the taper of cylindrical job, all expressed in inches. When the taper is expressed in taper per foot, the length of the taper l is expressed in foot, but the diameters are expressed in inches. A taper is generally turned in a lathe by feeding the tool at an angle to the axis of rotation of the workpiece. The angle formed

Conventional Machining Techniques 249

by the path of the tool with the axis of the workpiece should correspond to the half taper angle. A taper can be turned by any one of the following methods: 1. by swivelling the compound rest, 2. by setting over the tailstock centre, 3. by a broad nose form tool, 4. by a taper turning attachment, 5. by combining longitudinal and cross feed in a special lathe, and 6. by using numerical control lathe.

11.2.2 Drilling Drilling is a hole making process for which drill is used as a cutting tool for producing round holes of different sizes and depths. For producing holes in jobs on lathe, the job is held in a chuck or on a face plate. The drill is held in the position of tailstock and which is brought nearer the job by moving the tailstock along the guide ways, the thus drill is fed against the rotating job as shown in Fig. 11.1.

Figure 11.1 Drilling operation.

Types of Drills: A drill is a multi-point cutting tool used to produce or enlarge a hole in workpiece. It usually consists of two cutting edges set at an angle with the axis. Broadly, there are three types of drills: 1. Flat drill, 2. Straight-fluted drill, and 3. Twist drill.

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Basic Mechanical Engineering

Flat drill is usually made from a piece of round steel which is forged to shape and ground to size, then hardened and tempered. The cutting angle is usually 90° and the relief or clearance at the cutting edge is 3 to 8°. The disadvantage of this type of drill is that each time the drill is ground the diameter is reduced. Twist drill is the most common type of drill in use today.

11.2.3 Boring Boring is the process of enlarging a hole that has already been drilled or cast. The process is used to achieve greater accuracy of the diameter of a hole, and can be used to cut a tapered hole.

11.2.4 Milling Milling is the most common form of machining, a material removal process, which can create a variety of features on a part by cutting away the unwanted material. The milling process requires a milling machine, workpiece, fixture, and cutter. The workpiece is a piece of pre-shaped material that is secured to the fixture, which itself is attached to a platform inside the milling machine. The cutter is a cutting tool with sharp teeth that is also secured in the milling machine and rotates at high speeds. A milling machine is a machine tool that removes metal as the work is fed against a rotating multi-point cutter. The milling cutter rotates at high speed and it removes metal at a very fast rate with the help of multiple cutting edges. One or more number of cutters can be mounted simultaneously on the arbour of milling machine. This is the reason that a milling machine finds wide applications in production work. Milling machine is used for machining flat surfaces, contoured surfaces, surfaces of revolution, external and internal threads, and helical surfaces of various cross sections. Principle of Milling: In milling machine, the metal is cut by means of a rotating cutter having multiple cutting edges. For cutting operation, the workpiece is fed against the rotary cutter. As the workpiece moves against the cutting edges of milling cutter, metal is removed in chip form of trochoid shape. Machined surface is formed in one or more passes of the work. The work to be machined is held in a vice, a rotary table, a three jaw chuck, an index head, between centres, in a special fixture or bolted to machine table. The rotatory speed of the cutting tool and the feed rate of the workpiece depend on the type of material being machined. Milling Methods: There are two distinct methods of milling classified as follows: 1. up-milling or conventional milling, and 2. down milling or climb milling.

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Up-Milling or Conventional Milling Procedure: In the up-milling or conventional milling, as shown in Fig. 11.2, the metal is removed in the form of small chips by a cutter rotating against the direction of travel of the workpiece. In this type of milling, the chip thickness is minimum at the start of the cut and maximum at the end of the cut. As a result the cutting force also varies from zero to the maximum value per tooth movement of the milling cutter. The major disadvantages of up-milling process are the tendency of cutting force to lift the work from the fixtures and poor surface finish obtained. But being a safer process, it is commonly used method of milling.

Figure 11.2 Up-milling.

Down-Milling or Climb Milling: Down-milling is shown in Fig. 11.3. It is also known as climb milling. In this method, the metal is removed by a cutter rotating in the same direction of feed of the workpiece. The effect of this is that the teeth cut downward instead of upwards. Chip thickness is maximum at the start of the cut and minimum in the end. In this method, it is claimed that there is less friction involved and consequently less heat is generated on the contact surface of the cutter and workpiece. Climb milling can be used advantageously on many kinds of work to increase the number of pieces per sharpening and to produce a better finish. With climb milling, saws cut long thin slots more satisfactorily than with standard milling. Another advantage is that slightly lower power consumption is obtainable by climb milling, since there is no need to drive the table against the cutter.

Figure 11.3 Down-milling.

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11.2.5 Shaping The main functions of shaping machines are to produce flat in different planes. The cutting motion provided by the linear forward motion of the reciprocating tool and the intermittent feed motion provided by the slow transverse motion of the job along with the bed result in producing a flat surface by gradual removal of excess material layer by layer in the form of chips. Working Principle of Shaper: A single-point cutting tool is held in the tool holder, which is mounted on the ram. The workpiece is rigidly held in a vice or clamped directly on the table. The table may be supported at the outer end. The ram reciprocates and thus cutting tool held in tool holder moves forward and backward over the workpiece. In a standard shaper, cutting of material takes place during the forward stroke of the ram. The backward stroke remains idle and no cutting takes place during this stroke. The feed is given to the workpiece and depth of cut is adjusted by moving the tool downward towards the workpiece. The time taken during the idle stroke is less as compared to forward cutting stroke and this is obtained by quick return mechanism.

11.2.6 Broaching Broaching is one of the most precise and productive processes in the metalworking area and despite the high costs of tooling, its largely applied in the automobiles industry. The broaching is a machining operation which uses a tool called broach moving it over the workpiece to remove material, cutting a predetermined shape. Commonly circular or odd shapes, both internal or external, are obtained by broaching.

11.2.7 Slotting The slotting machine is a reciprocating machine tool in which, the ram holding the tool reciprocates in a vertical axis and the cutting action of the tool is only during the downward stroke. The slotter or slotting machine is also a reciprocating type of machine tool similar to a shaper or a planer. It may be considered a vertical shaper. The chief difference between a shaper and a slotter is the direction of the cutting action. The machine operates in a manner similar to the shaper, however, the tool moves vertically rather than in a horizontal direction. The job is held stationary. The slotter has a vertical ram and a hand or power operated rotary table. Operations Performed on a Slotting Machine: A slotter is an economical machine tool when used for certain classes of work given as under. (i) It is used for machining vertical surfaces. (ii) It is used an angular or inclined surfaces. (iii) It is used to cut slots, splines, keyways for both internal and external jobs such as machining internal and external gears.

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(iv) It is used for works as machining concave, circular, semicircular and convex surfaces. (v) It is used for shaping internal and external forms or profiles. (vi) It is used for machining of shapes which are difficult to produce on shaper. (vii) It is used for internal machining of blind holes. (viii) It is used for machining dies and punches, and since a slotter works slowly. It has less use in mass production work. It can be substituted by the broaching machine.

11.2.8 Grinding Grinding is an abrasive machining material removal process that involves the use of abrasive cutting tools. Grinding is a material removal process in which abrasive particles are kept in a bonded grinding wheel that operates at very high surface speeds. The grinding wheel is usually disc shaped and is precisely balanced for high rotational speeds. Grinding is generally called fine machining or finishing operations of removing materials from surface usually 0.25-0.50 mm in most operations through the use of grinding wheel. Grinding wheel is highly useful in removing extra unwanted metal and sharpening cutting tools such as chisels, drill, taps, and other cutting tools. It may be used to finish almost all surfaces, which have been previously roughly shaped by some other processes or to remove the extra material which is too hard to be removed by other machining processes. The accuracy in fine grinding is in few microns or even less. In grinding, the work is held pressed against the high speed rotating grinding wheel and the metal gets reduced by abrasion. Grinding wheel is generally made from silicon carbide or aluminium oxide. It is generally made up of particles of hard substance called the abrasive and is embedded in a matrix called the bond. These abrasives form the cutting points in a wheel and are termed grains. The abrasives are of generally two types, namely, natural and artificial. Emery and corundum are two natural abrasives, while carborundum and aloxite are artificial abrasives. The hardness or softness of the wheel is dependent on the amount and kind of the bonding material. Generally, hard wheels of aloxite are used for grinding soft materials and soft wheels of carborundum for grinding hard materials using various types of grinding machines known as grinders. In wet grinding, large amount of coolant over the work and on wheel face is provided. Coolant will remove heat generated during grinding and promotes long wheel life and produces very good surface finish. The cutting face of a grinding wheel should be kept in a true, clean and sharp conditioned shape for obtaining efficient cutting. Suitable dressers are also employed periodically for reconditioning and dressing of glazed or blunt wheels. Grinder may be of various types such as cylindrical grinder, surface grinder, pedestal grinder, tool and cutter grinder, centre-less grinder, internal grinder and jig grinder and profile grinder.

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11.3

NC AND CNC ASSISTED CONVENTIONAL MACHINING TECHNIQUES

11.3.1 Definition Computer Numerical Control (CNC) is one in which the capacities and movements of a machine instrument are controlled by method for a readied program containing coded alphanumeric information. CNC can control the movements of the workpiece or apparatus, the information parameters, for example, feed, depth of cut, speed, and other functional capacities, for example, turning shaft on/off, turning coolant on/off.

11.3.2 Applications The uses of CNC incorporate both for machine components and also non-machine apparatus. In the machine instrument class, the operations where CNC is generally utilized are for lathe, drilling, milling, grinding, laser, sheet-metal press working machine, tube twisting machine and so forth. Exceptionally computerized machine instruments, for example, turning focus and machining focus which change the cutting apparatuses consequently under CNC control have been produced. In the non-machine device classification, CNC applications incorporate welding machines (curve and resistance), coordinate measuring machine, electronic assembly, tape laying and filament winding machines for composites and so forth.

11.3.3 Advantages and Limitations The benefits of CNC are:  ‡ +LJKDFFXUDF\LQPDQXIDFWXULQJ  ‡ 6KRUWSURGXFWLRQWLPH  ‡ *UHDWHUPDQXIDFWXULQJIOH[LELOLW\  ‡ 6LPSOHUIL[WXULQJ  ‡ &RQWRXUPDFKLQLQJWRD[LVPDFKLQLQJ  ‡ 5HGXFHG KXPDQ HUURU 7KH GUDZEDFNV LQFOXGH KLJK FRVW PDLQWHQDQFH DQG WKH requirement of skilled part programmer.

11.3.4 Elements of A CNC Machine A CNC system consists of three basic components: (a) Part program  E  0DFKLQHFRQWUROXQLW0&8  (c) Machine tool (lathe, drill press, milling machine, etc.)

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11.3.5 Part Program The part program is a defined set of commands which are to be followed by the machine tool. Each command specifies a position in the Cartesian coordinate system (x, y, z) or motion (workpiece travel or cutting tool travel), machining parameters and on/off function. Part programmers should be well versed with machine tools, machining processes, effects of process variables, and limitations of CNC controls. The part program is written manually or by using computer assisted language such as APT (automated programming tool).

11.3.6 Machine Control Unit It is a typical numerical control system for a milling machine. Machine control unit 0&8 LVDPLFURFRPSXWHUWKDWVWRUHVWKHSURJUDPDQGH[HFXWHVWKHRUGHUVLQWRDFWLYLWLHV E\WKHPDFKLQHLQVWUXPHQW7KH0&8FRPSULVHVWZRSULPDU\XQLWVWKH'DWD3URFHVVLQJ 8QLW'38 DQG&RQWURO/RRSXQLW&/8 7KH'38SURJUDPPLQJLQFRUSRUDWHVFRQWURO framework programming, computation calculations, interpretation programming that FKDQJHV RYHU WKH SDUW SURJUDP LQWR D XVDEOH DUUDQJHPHQW IRU WKH 0&8 LQVHUWLRQ calculation to accomplish smooth movement of the cutter, altering of part program (in WKH HYHQW RI EOXQGHUV DQG FKDQJHV  7KH '38 IRUPV WKH LQIRUPDWLRQ IURP WKH SDUW SURJUDP DQG JLYHV LW WR WKH &/8 ZKLFK ZRUNV WKH GULYHV MRLQHG WR WKH PDFKLQH OHDG screws and gets criticism motions on the genuine position and speed of every one of the tomahawks. A driver (dc engine) and an input gadget are connected to the leadVFUHZ7KH&/8FRPSULVHVWKHFLUFXLWVIRUSRVLWLRQDQGVSHHGFRQWUROFLUFOHVGHFHOHUDWLRQ and backfire take up, capacity controls, for example, shaft on/off.

Figure 11.4 Machine control unit.

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11.3.7 Machine Tool The machine tool could be one of the following: lathe, milling machine, laser, plasma, coordinate measuring machine, etc. Figure 11.5 shows that a right-hand coordinate system is used to describe the motions of a machine tool. There are three linear axes (x, y, z), three rotational axes (i, j, k), and other axes such as tilt are possible. For example, a 5-axis. machine implies any combination of x, y, z, i, j, and k.

Figure 11.5 Machine tool axis.

11.4

SUMMARY

Machining of composites is vital for completing parts to the required resilience and for get ready segments for ensuing assembly. Routine machining procedures, for example, turning, processing, boring, rough cutting, and pounding are utilized to create complex components by expelling material as modest chips. The machining procedure happens as a consequence of drawing in an inflexible cutting instrument into the workpiece and endorsing the important relative movements between the tool-work combine that will bring about material expulsion and producing new surfaces. The size and state of the chip evacuated, material expulsion rate and coming about perfect surface complete are firmly identified with the kinematic connections between the cutting apparatus and the workpiece.

SOLVED EXAMPLES

Example 11.1 Tool life tests on a lathe have resulted in the following data: (1) at a cutting speed of 375 ft/min, the tool life was 5.5 min; (2) at a cutting speed of 275 ft/min, the tool life was 53 min. (a) Determine the parameters n and C in the Taylor tool life equation (b)

Conventional Machining Techniques 257

based on the n and C values, what is the likely tool material used in this operation? (c) 8VLQJ\RXUHTXDWLRQFRPSXWHWKHWRROOLIHWKDWFRUUHVSRQGVWRDFXWWLQJVSHHGRIIW min. (d) Compute the cutting speed that corresponds to a tool life T = 10 min.

Solution: (a) (VT)n = C; Two equations: 1. 375(5.5)n = C and 2. 275(53)n = C 375(5.5)n = 275(53)n 375/275 = (53/5.5)n 1.364 = (9.636)n ln 1.364 = n ln 9.636 0.3102 = 2.2655 n n = 0.137 C = 375(5.5)0.137 = 375(1.2629) C = 474 (b) The likely tool material is high speed steel. (c) At v = 300 ft/min, T = (C/v)1/n = (474/300)1/0.137 = (1.579)7.305 = 28.1 min (d) For T = 10 min, v = C/T n = 474/100.137 = 474/1.371 = 346 ft/min

Example 11.2 In a production turning operation, the workpart is 125 mm in diameter and 300 mm long. A feed of 0.225 mm/rev is used in the operation. If cutting speed = 3.0 m/s, the tool must be changed every 5 workparts; but if cutting speed = 2.0 m/s, the tool can be used to produce 25 pieces between tool changes. Determine the Taylor tool life equation for this job.

Solution: 1. Tm = p(125 mm)(0.3 m)/(3.0 m/s)(0.225 mm) = 174.53 s = 2.909 min T = 5(2.909) = 14.54 min (2) Tm = p(125 mm)(0.3 m)/(2.0 m/s)(0.225 mm) = 261.80 s = 4.363 min T = 25(4.363) = 109.08 min (1) v = 3 m/s = 180 m/min 2. v = 2 m/s = 120 m/min (1) 180(14.54)n = C (2) 120(109.08)n = C 180(14.54)n = 120(109.08)n ln 180 + n ln(14.54) = ln 120 + n ln(109.08) 5.1929 + 2.677 n = 4.7875 + 4.692 n 5.1929 - 4.7875 = (4.692 – 2.677) n 0.4054 = 2.0151 n n = 0.2012 C = 180 (14.54)0.2012 C = 308.43

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EXERCISE 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

What do you understand by the term machining? Define conventional machining techniques. Differentiate between shaping and milling. What are the advantages of using a CNC machine? Explain machine tool in CNC machining operation. State the working principle of a drilling machine. Explain the principal parts of a drilling machine and sketch the mechanism of a drilling machine. State the working principle of a milling machine. How will you classify milling machines? 8VLQJQHDWVNHWFKGHVFULEHWKHSULQFLSDOSDUWVRIWKHPLOOLQJPDFKLQHE\QHDW sketches. Differentiate between up-milling and down-milling. State the working principle of a shaper. What are the operations performed on a slotting machine?

12 Nonconventional Machining Techniques

Learning Objectives    

‡ ‡ ‡ ‡

12.1

8QGHUVWDQGLQJWKHWHUPQRQFRQYHQWLRQDOPDFKLQLQJWHFKQLTXHV &ODVVLILFDWLRQRIQRQFRQYHQWLRQDOPDFKLQLQJSURFHVVHV ,QWURGXFWLRQWRVRPHRIWKHQRQFRQYHQWLRQDOPDFKLQLQJWHFKQLTXHV &RPSDULVRQVEHWZHHQFRQYHQWLRQDODQGQRQFRQYHQWLRQDOWHFKQLTXHV

INTRODUCTION

Manufacturing processes can be broadly divided into the primary and secondary manufacturing processes. In the primary manufacturing processes basic shape and size is given to the material as per the requirement of the designer, for example, casting, forming, powder metallurgy, etc. In the secondary manufacturing processes similarly the final shape and size with tighter control on dimension surface characteristics is given to the material. The material removal processes generally come under the secondary manufacturing processes. These are further divided into two groups, i.e.,“conventional machining processes” and “nontraditional manufacturing processes”. Examples of conventional machining processes are turning, boring, milling, shaping, broaching, VORWWLQJ JULQGLQJ HWF 6LPLODUO\ $EUDVLYH -HW 0DFKLQLQJ $-0  8OWUDVRQLF 0DFKLQLQJ 860 :DWHU-HWDQG$EUDVLYH:DWHU-HW0DFKLQLQJ:-0DQG$:-0 (OHFWURGLVFKDUJH Machining (EDM) are some of the nontraditional machining (NTM) processes.

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12.2

CLASSIFICATION OF NONCONVENTIONAL MACHINING

For the classification of nontraditional machining (NTM) processes, we need to study and analyse the differences and similarities between both the processes. Conventional machining processes deal mainly with the removal of material in the form of chips by the application of forces on the work material with a wedge-shaped cutting tool that is harder than the work material under machining conditions. Because of such forces the material undergoes plastic deformation within the workpiece leading to shear deformation along the shear plane and chip formation. Figure 12.1 shows some of the non-traditional machining processes.

Figure 12.1 NTM operations.

Classification of NTM processes depends on the nature of energy used for material removal. Given below is the broader classification. 

(i) Mechanical Processes Abrasive Jet Machining (AJM)  8OWUDVRQLF0DFKLQLQJ860 Water Jet Machining (WJM) Abrasive Water Jet Machining (AWJM)

Nonconventional Machining Techniques 261

(ii) Electrochemical Processes Electrochemical Machining (ECM) Electrochemical Grinding (ECG) Electro Jet Drilling (EJD) (iii) Electrothermal Processes Electro-discharge Machining (EDM) Laser Jet Machining (LJM) Electron Beam Machining (EBM) (iv) Chemical Processes Chemical Milling (CHM) Photochemical Milling (PCM), etc.

12.3

NEED OF NONCONVENTIONAL MACHINING TECHNIQUES

Following are the conditions where NTM processes are preferred over the conventional machining processes: (a) To provide intricate shape to blind holes, e.g., square hole of 15 mm × 15 mm with a depth of 30 mm can be carved using this process. (b) For machining materials which are generally difficult to machine, e.g., same example as above in Inconel, Ti-alloys or carbides. (c) For grinding low stress materials. Like electrochemical grinding is preferred to conventional grinding for softer materials. (d) To dig a deep hole with small hole diameter, e.g., j 1.5 mm hole with l/d = 20. (e) For the machining of composites which are again sometimes hard to be machined.

12.4

SOME NONCONVENTIONAL MACHINING TECHNIQUES

12.4.1 Abrasive Jet Machining In abrasive jet machining (AJM), abrasive particles are made to impinge on the work material at a high velocity. The jet of abrasive particles is carried by carrier gas or air. The high velocity stream of abrasive is generated by converting the pressure energy of the carrier gas or air to its kinetic energy and hence high velocity jet. The nozzle directs the abrasive jet in a controlled manner onto the work material, so that the distance between the nozzle and the workpiece and the impingement angle can be set desirably. The high velocity abrasive particles remove the material by micro-cutting action as well as brittle fracture of the work material. Figure 12.2 shows the material removal process.

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Figure 12.2 AJM process.

AJM is different from standard shot or sand blasting, as in AJM, finer abrasive grits are used and the parameters can be controlled more effectively providing better control over product quality. In AJM, generally, the abrasive particles of around 50 μm grit size would impinge on the work material at velocity of 200 m/s from a nozzle of I.D. of 0.5 mm with a stand-off distance of around 2 mm. The kinetic energy of the abrasive particles would be sufficient to provide material removal due to brittle fracture of the workpiece or even micro cutting by the abrasives.

Advantages of Abrasive Jet Machining (AJM) ‡ This process is quite suitable for machining brittle, heat resistant and fragile materials like, glass, ceramic, germanium, mica, etc. ‡ It can be utilized for cutting, drilling, polishing, deburring, cleaning, etc. of the materials. ‡ The depth of damage to the surface is negligible. ‡ Holes of intricate shapes could be produced efficiently. ‡ The surface machined can have good finish (by controlling the grain size mainly).

Disadvantages of AJM The materials removal rate is low. For example, for glass, it is 0.0164 cm3/min. ‡ The tapering of hole especially, when the depth of the hole is more, becomes almost inevitable.

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‡ A dust collecting chamber is a basic requirement to prevent atmospheric pollution to cause health hazards. ‡ The abrasive particles may remain embedded in the work surface. ‡ Abrasive particles cannot be reused.

Applications of AJM ‡ Abrasive jet machining is best suited for machining brittle and heat-sensitive materials like glass, quartz, sapphire, ceramics, etc. ‡ It is used for drilling holes, cutting slots, cleaning hard surfaces, deburring, polishing, etc.

12.4.2 Ultrasonic Machining (USM) 8OWUDVRQLF PDFKLQLQJ LV D QRQWUDGLWLRQDO PDFKLQLQJ SURFHVV )LJXUH  EULHIO\ VKRZV WKH860SURFHVV,QXOWUDVRQLFPDFKLQLQJDWRRORIGHVLUHGVKDSHYLEUDWHVDWDQXOWUDVRQLF frequency (19 ~ 25 kHz) with an amplitude of around 15-50 μm over the workpiece. Generally, the tool is pressed downward with a feed force, F. Between the tool and workpiece, the machining zone is flooded with hard abrasive particles generally in the form of a water based slurry. As the tool vibrates over the workpiece, the abrasive particles act as the indenters and indent both the work material and the tool. The abrasive particles, as they indent, the work material, would remove the same, particularly if the work material is brittle, due to crack initiation, propagation and brittle fracture of the PDWHULDO +HQFH 860 LV PDLQO\ XVHG IRU PDFKLQLQJ EULWWOH PDWHULDOV ZKLFK DUH SRRU conductors of electricity and thus cannot be processed by electrochemical and electrodischarge machining.

Figure 12.3 USM process.

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Advantages of USM ‡ ‡ ‡ ‡

Machines all sorts of hard materials Produces fine finish and structured results Produces less heat Various hole cut shapes due to vibratory motion of the tool

Disadvantages of USM Requires a high degree of integrity and skills No certified record of radiography 8QQHFHVVDU\ODUJHJUDLQVL]HVFDXVHGHIHFWV Additional repairs might be required due to spurious signs and misunderstanding of the process 8OWUDVRQLF PDFKLQHV DUH WKH IXWXUH RI PDFKLQLQJ ,W LV XVHG DOO RYHU WKH ZRUOG IRU creating hard and brittle forms of materials for the industrial uses. A lot of operations can be performed with the ultrasonic machines which can benefit the industrialists in a variety of ways. The advanced technology creates solution which helps in opening up the market opportunities and has made things easier. ‡ ‡ ‡ ‡

Applications of USM 8VHG IRU PDFKLQLQJ KDUG DQG EULWWOH PHWDOOLF DOOR\V VHPLFRQGXFWRUV JODVV FHUDPLFV carbides, etc. ‡ 8VHGIRUPDFKLQLQJURXQGVTXDUHLUUHJXODUVKDSHGKROHVDQGVXUIDFHLPSUHVVLRQV ‡ Machining, wire drawing, punching or small blanking dies.

12.4.3 Electrochemical Machining (ECM) In the electrochemical machining (ECM) process which is a nontraditional machining (NTM) process and comes under the category of electrochemical, is opposite to galvanic coating or deposition process. Thus, ECM can be considered a controlled anodic dissolution at atomic level of the workpiece that is electrically conductive by a shaped tool due to flow of high current at relatively low potential difference through an electrolyte which is quite often a water based neutral salt solution. Figure 12.4 shows the basic principle of ECM. ECM technique removes material by atomic level dissolution of the same by electrochemical action. Thus, the material removal rate or machining is not dependent on the mechanical or physical properties of the work material and it only depends on the atomic weight and valency of the work material and the condition that it should be electrically conductive. Thus, ECM can machine any electrically conductive work material

Nonconventional Machining Techniques 265

Figure 12.4 ECM process.

irrespective of their hardness, strength or even thermal properties. Moreover, as ECM leads to atomic level dissolution, the surface finish is excellent with almost stress-free machined surface and without any thermal damage. ECM is used for die sinking, profiling and contouring, trepanning, grinding, drilling and micro-machining.

Applications of Electrochemical Machining ‡ Machining of hard materials. The process parameters and the tool life do not depend on the hardness of the workpiece, therefore, electrochemical machining is often used for machining hard materials. Turbine blades and rifle barrels are fabricated by electrochemical machining. ‡ Producing holes and cavities which cannot be obtained by conventional machining methods. ‡ Die sinking. Electrochemical machining is often used as an alternative to the cavity type electric discharge machining (EDM). ‡ Fabrication of thin walled parts. Electrochemical machining does not produce surface stress in the workpiece, therefore even very brittle and easily deformed materials may be machined in thin walled shapes. ‡ Grinding of a workpiece by a rotating wheel, which performs grinding operation through an electrolyte. The wheel is conductive and cathodically connected. Non-conductive hard particles are set on the wheel surface. The particles provide a constant gap through which an electrolyte is continuously fed. Hard and brittle materials are ground by the method.

Advantages of Electrochemical Machining ‡ ‡ ‡ ‡

The rate of machining does not depend on the hardness of the workpiece material. The tool does not wear. Soft materials (e.g., copper) may be used for tool fabrication. No stresses are produced on the surface of the workpiece. No burrs form in the machining operation.

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‡ High surface quality may be achieved. ‡ High accuracy of the machining operation.

Disadvantages of Electrochemical Machining ‡ ‡ ‡ ‡ ‡

Higher cost. Electrolyte may cause corrosion of the equipment. Large production floor is required. Only electrically conductive materials may be machined. Not environment-friendly process.

12.4.4 Electro-discharge Machining (EDM) Electro-discharge Machining (EDM) is an electrothermal nontraditional machining process, where electrical energy is used to generate electrical spark and material removal mainly occurs due to thermal energy of the spark. EDM is mainly used to machine difficult-to-machine materials and high strength temperature-resistant alloys. EDM can be used to machine difficult geometries in small batches or even on job-shop basis. Work material to be machined by EDM has to be electrically conductive.

Figure 12.5 EDM process.

Advantages of EDM ‡ Tolerances of +/- 0.005 can be achieved ‡ Material hardness does not affect the process – Tungsten Carbide – Stellite – Hastelloy – Nitralloy – Waspaloy – Nimonic – Inconel all can be successfully machined ‡ Cutting complex shapes and thin walled configurations without distortion ‡ EDM is a no-contact and no-force process, making it well suited for delicate or fragile parts that cannot take the stress of traditional machining ‡ The EDM process leaves no burrs

Disadvantages of EDM ‡ Only able to machine conductive materials ‡ More expensive than conventional milling or turning

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12.4.5 Electro Beam and Laser Beam Machining Techniques Electron beam machining (EBM) and laser beam machining (LBM) of NTM machininig techniques are important processes. They consider thermal processes mechanisms for the material removal. However, electrical energy is used to generate high-energy electrons in case of electron beam machining (EBM) and high-energy coherent photons in case of laser beam machining (LBM). Thus, these two processes are often classified as electrooptical-thermal processes. There are different jet or beam processes, namely, abrasive jet, water jet, etc. These two are mechanical jet processes. They are also called thermal jet or beams. Other examples include oxyacetylene flame, welding arc, plasma flame, etc. EBM as well as LBM are such thermal beam processes. Figure 12.6 shows the variation in power density vs. the characteristic dimensions of different thermal beam processes. Characteristic length is defined as the diameter over which the beam or flame is active. In case of oxyacetylene flame or welding arc,

Figure 12.6 EBM process.

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the characteristic length is in mm to tens of mm and the power density is typically low. Electron beam may have a characteristic length of tens of microns to mm depending on the degree of focusing of the beam. In case of defocused electron beam, power density would be as low as 1 watt/mm2. But in case of focused beam the same can be increased to tens of kW/mm2. Similarly, as can be seen in Fig. 12.6, laser beams can be focused over a spot size of 10-100 μm with a power density as high as 1 MW/mm2. Electrical discharge typically provides even higher power density with smaller spot size.

Advantages of Laser Beam Machining      

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Disadvantages of Laser Beam Machining     

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Applications of Laser Beam Marching     

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Characteristics of EBM Process The significant characteristics of an EBM process are as follows:  ‡ (OHFWULFDOO\FRQGXFWLQJDQGQRQFRQGXFWLQJPDWHULDOVFDQEHHDVLO\PDFKLQHG Examples: Aluminium, copper, nickel, ceramics, plastics, leather, etc.

Nonconventional Machining Techniques 269

         

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Advantages of Electron Beam Machining (EBM)       

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Disadvantages of Electron Beam Machining (EBM)        

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Table 12.1: General Characteristics of Advanced Machining Processes S. No.

Process

1.

Chemical machining (CM)

Shallow removal on large flat or curved 0.0025-0.1 mm/min surfaces; blanking of thin sheets; low tooling and equipment cost; suitable for low-production runs.

2.

Electrochemical machining (ECM)

Complex shapes with deep cavities; highest rate of material removal among other nontraditional processes; expensive tooling and equipment; high power consumption; medium-to-high production quantity.

3.

Electrochemical grinding (ECG)

Cutting off and sharpening hard A: 1-3 A/mm2, typically materials, such as tungsten-carbide tools; 25 mm3/s per 1000 A. also used as honing process; higher removal rate than grinding.

4.

Electrical-discharge machining (EDM)

Shaping and cutting complex parts made V: 50-380; A: 0.1-500; of hard materials; some surface damage typically 300 mm3/min may result; also used as a grinding and cutting process; expensive tooling and equipment.

5.

Wire electric discharge machining

Contour cutting of flat or curved surfaces; Varies with material and expensive equipment. thickness.

6.

Laser beam machining (LBM)

Cutting and hole making on thin 0.50-7.5 mm/min materials; very small holes and slots; heat-affected zone, requires a vacuum; expensive equipment.

7.

Electro-beam machining (EBM)

Cutting and hole making on thin 1-2 mm3/mim materials; very small holes and slots; heat-affected zone; requires a vacuum; expensive equipment.

8.

Water jet machining (WJM)

Cutting all types of nonmetallic materials; Varies considerably with suitable for contour cutting of flexible material. materials; no thermal damage; noisy.

9.

Abrasive water jet machining (AWJM)

Single-layer or multi-layer cutting of 8SWRPPLQ metallic and nonmetallic materials.

Abrasive jet machining (AJM)

Cutting, slotting, deburring, etching and Varies considerably with cleaning of metallic and nonmetallic material. materials; tends to round off sharp edges; can be hazardous.

10.

Characteristics

Process Parameters and MRR or cutting Speed

V: 5-25 DC; A: 1.5-8 A/mm2, 2.5-12 mm/min, depending on the current density.

Nonconventional Machining Techniques 271

12.5

COMPARISON OF CONVENTIONAL AND NONCONVENTIONAL MACHINING TECHNIQUES

The process of conventional machining generally involves changing the shape of a ZRUNSLHFH XVLQJ DQ LPSOHPHQW PDGH RI D KDUGHU PDWHULDO 8WLOL]LQJ FRQYHQWLRQDO machining processes to machine hard metals and amalgams implies expanded request of time and energy and along these lines increments in expenses; now and again routine machining may not be attainable. Ordinary machining additionally costs as far as tool wear and in loss of value in the item because of the development of residual stress in the material. With ever expanding interest for products of hard alloys and metals, for example, Inconel 718 or titanium, more intrigue has inclined towards nontraditional machining techniques. Ordinary machining can be characterized as a procedure utilizing mechanical (movement) energy. Nontraditional machining uses different types of energy. The three main forms of energy used in nonconventional machining processes are: ‡ Thermal energy ‡ Chemical energy ‡ Electrical energy One such example of machining which uses the thermal energy is laser. Thermal methods have many advantages over conventional machining, but there are a few of disadvantages, such as: ‡ 8VLQJWKHUPDOPHWKRGVUHTXLUHVKLJKHQHUJ\LQSXWIRUWKHVHPDWHULDOVDV,QFRQHO 718, titanium and other hard metals and alloys have a high melting point. ‡ Concentrating heat onto any material greatly affects its microstructure and normally causes cracking, which is not desirable. ‡ Safety requirements for thermal methods, especially laser, are demanding in terms of time and cost. ‡ Machining large areas or many surfaces at the same time using thermal methods is not normally possible. Electro-discharge machining (EDM) and anodic machining (AM) generally use electrical energy for the removal of material. The EDM process is said to be spark erosion and it uses pulsed voltage to remove material from a workpiece and a non-conductive medium for discarding the remaining dirt. Since the medium is electrically inert no complicated tool design criteria are required and the tool is a direct reverse of the workpiece. One problem in EDM is the shock of spark erosion can affect the microstructure on the surface of the workpiece. This process (EDM) has a lower material removal rate than AM.

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The chemicals used in AM are non-toxic and the energy required is less than other nonconventional machining processes. It has no effect on the microstructure of the workpiece. The electrolyte can even be common seawater, enabling AM to be used in a sub-sea capacity. The hardness and thermal resistivity of the workpiece material do not matter therefore hard metals and alloys can be machined using tools made from softer materials. The only disadvantage is that tool design is a little more complex than that of EDM, but software is being developed to make this easier. The controllability, environmental versatility, speed, safety and absence of change in workpiece microstructure make AM a competitive manufacturing process.

12.6

SUMMARY

Nontraditional machining procedures are characterized as a group of procedures that evacuate overabundance material by different systems including mechanical, thermal, electrical or synthetic energy or mix of these energies. However, it does not utilize a sharp slicing instrument as is utilized for conventional machining forms. To a great extent hard and fragile materials are hard to machine by conventional machining procedures, for example, turning, drilling, moulding and processing. Nonconventional machining forms, likewise propelled producing procedures, are utilized where customary machining procedures are not practical, acceptable or sparing because of different reasons as discussed above.

SOLVED EXAMPLES

Example 12.1

Material removal rate in AJM is 0.5 mm3 /sec. Calculate MRR/impact if the mass flow rate of abrasive is 3 gm/min, density is 3 gm/CC and grit size is 60 microns. Also calculate the indentation radius. Solution Data Given: Material removal rate = 0.5 mm3/sec. Abrasive grain size = 60 microns = 6 × 10–3 mm Mass flow rate of abrasives rg = 3 gm/min = 3 × 10–3/60 kg/sec Mass of grit

⎡⎛ π 3 ⎞ ⎤ = ⎢⎜ d g ρg ⎟ ⎥ ⎠⎦ ⎣⎝ 6

MRR = Volume of the material removed × È 3 Í 1000 ¥ 60 2 Í p( r ) 3 × Í 0.5 × 10–6 = 3 3 3 p Ê 60 ˆ Í Á ˜ –6 ÍÎ 6 Ë 10000 ¯ 1000 ¥ 10

Mass flow rate of abrasives Mass of abrasive grit

˘ ˙ ˙ ˙ = 10 microns ˙ ˙˚

Nonconventional Machining Techniques 273

Ï Volume of the material ¸ ÏNumber of impacts made ¸ MRR = Ì ˝¥Ì ˝ Óremoved per grit per cycle.˛ Ó by abrasives per second ˛ È ˘ 3 ¥ 10 -3 ¥ 6 Í ˙ 60 ˙ = 254648 Number of impact/time = Í -6 3 Í p ¥ ( 50 ¥ 10 ) ¥ 3000 ˙ ÍÎ ˙˚

Example 12.2 During AJM, mixing ratio used is 0.2. calculate mass ratio, if the ratio of density of abrasives and density of carrier gas is equal to 20. Solution Mixing ratio (MR) =

Mass ratio (a) =

Volume flow rate of abrasive particle Volume flow rate of carrier gas Abrasive mass flow rate Combined mass flow rate of abrasive and carrier gas

Va Ma raVa = MR = V also a = M a + g ra Va + r g Vg g

or

1 = ra Va + r g Vg = 1 + È r g ˘ ÈVg ˘ = 1 + 1 ¥ 1 Í ˙Í ˙ ra Va 20 0.2 Î ra ˚ Î Va ˚ a

i.e.

1 = 1.25 or a = 0.80 a

Example 12.3

Turbine section of HAL wishes to produce complex die for making a part for aircraft application. 250 numbers of parts are required to be produced. Part is made up of hard nickel-based alloy. The profile can be generated on a copy milling machine or an ECM SODQW3UHVXPLQJWKDWWZRPDFKLQHVDUHIXOO\XWLOL]HG8VLQJ%($6XJJHVWZKLFKPDQXIDFturing method is economical to produce 250 parts. The following data are provided: Copy milling

ECM

Capital cost

Rs. 1500000

3750000

Amortization period

10 years

5 years

Labour cost / hour

Rs 100

Rs 100

Fixture and tooling cost

Rs 225000

Rs 1125000

Electric power consumption per hour

Rs 8

Rs 80

Production time/Component

40 hours

1 hour

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Basic Mechanical Engineering

Assume 2500 working hours in a year, Solution: Machine cost/hour = Capital cost / (Amortization period × 2500) Machine cost/hour (for ECM) = 3750000/5 × 2500 = Rs 300/hour Machine cost/hour (for Copy milling) = 1500000/10 × 2500 = Rs 60 hour Let x be number of components, Cost of which on the two machines are equal. Solution Various costs per component are: Copy milling

ECM

Machining cost/Component

60 × 40 = Rs 2400

Rs 300

Labour cost

100 × 40 = Rs 4000

Rs 100

Power cost

8 × 40 = Rs 320

Rs 80

Tooling cost/Component

225000/x

1125000/x

2400 + 400 + 320 + 225000/x = 300 + 100 + 80 + 1125000/x Solving we get x = 144. ECM is more economical than copy milling machine. In the above calculations, it is implied that both machines are fully utilized; however, this might not always be true. Break even analysis chart is shown below.

Example 12.4

Assuming no losses, determine water jet velocity, when the water pressure is 4000 bar, being issued from an orifice of diameter 0.3 mm. Solution Vw =

2p = rw

2 ¥ 4000 ¥ 10 5 = 894 m / s 1000

Nonconventional Machining Techniques 275

Example 12.5

Determine the mass flow rate of water for the given problem assuming all related coefficients to be 1. Solution p mw = rw ◊ Qw = rw do2 vw 4 p ¥ (0.3 ¥ 10 -3 )2 ¥ 894 4 = 0.0631 kg / s = 0.0631 ¥ 60 = 3.79 kg / min = 1000 ¥

Example 12.6 If the mass flow rate of abrasive is 1 kg/min, determine the abrasive water jet velocity assuming no loss during mixing process using the above data (data of Questions 12.4 and 12.5) Ê ˆ Ê ˆ Solution Á Ê 1 ˆ 1 ˜ 1 Á ˜ vwj = Á Vawj =Á ˜ vwj = Á ˜ ¥ 894 = 707 m / s ˜ 1 m + 1 R Ë ¯ Á 1 + abr ˜ Á1+ ˜ ÁË Ë 3.79 ¯ mw ˜¯

Example 12.7 In a certain electrochemical dissolution process of iron, a MRR of 2 cm3/min was desired. Determine the amount of account of current required for the process. Solution Assume atomic weight of iron = 56 gm Valancy at which dissolution occur = 2 Density of iron = 7.8 gm/cm. Grams equivalent weigh of iron (ECE) = m/v = 28; MRR =

(ECE)lt Pp

F = 1609 amp/min; (ECE)I MRR = FR 28 I 1609 ¥ 7.8

or,

2=

or,

I = 896 amp

Example 12.8

Diameter of nozzle is 1.0 mm and jet velocity is 200 m/s. Find the volumetric flow rate (cm3/sec) of carrier gas and abrasive mixture. Solution Cross-sectional area of nozzle = (p × 0.52 × 10–2) = p × 25 × 10–4 cm2 Volumetric flow rate of carrier gas and abrasive mixture is = area × velocity = p × 25 × 2 = 50 p cm3/sec

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Basic Mechanical Engineering

Example 12.9 Estimate the MRR in AJM of a brittle material with flow strength of 4 GPA. The abrasive flow rate is 2 gm/min, velocity is 200 m/s, density of abrasive is 3 gm/sec. Solution Flow strength of work material, sw = 4 × 109 N/m2 = 4000 N/mm2 Abrasive grain density = 3 gm/CC = 3 × 10–3 × 10–3 = 3 × 10–6 kg/mm3 Mass flow rate of abrasives, rg = 2 gm/min = 2× 10–3/60 kg/sec Velocity, V = 200 × 1000 mm/sec Since the material is brittle. We need to use the MRR formula corresponding to the brittle material. MRRBrittle

È Ma v Í = Í 1 3 Í (r g ) 4 (s w ) 4 Î

3 ˘ È -3 ˙ Í 2 ¥ 10 ¥ 200000 2 1 3 ˙=Í ˙ ÍÎ ( 3 ¥ 10 -6 ) 4 ( 4 ¥ 109 ) 4 ˚

˘ ˙ ˙ ˙˚

= 48 mm3/minutes

EXERCISE 1. 2. 3. 4. 5.

What do you mean by nonconventional machining? Classify the NTM in brief. Write a short note on AJM technique. Explain the mechanisms involved in EDM and ECM techniques. How do you differentiate between conventional and nonconventional machining techniques? 6. What are the advantages and disadvantages of ECM machining? 7. What are the applications of EDM? 8. Write the applications of abrasive machining.

13 Fluid Power Control and Automation

Learning Objectives   

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13.1

INTRODUCTION

Fluid power is the use of fluids under pressure to generate, control, and transmit power. Fluid power is subdivided into hydraulics using a liquid such as mineral oil or water, and pneumatics using a gas such as air or other gases. Compressed-air and water-pressure systems were once used to transmit power from a central source to industrial users over extended geographic areas. Fluid power systems today are usually within a single building or mobile machine. In the industry we utilize three strategies for transmitting power starting with one point then onto the next. Mechanical transmission is through shafts, gears, chains, belts, and so forth. Electrical transmission is through wires, transformers, and so on. Fluid power is through liquids or gases in a confined space. In this section, we talk about part of structure of hydraulic and pneumatic systems. We will likewise examine the advantages and disadvantages and compare hydraulic, pneumatic, electrical and mechanical systems. Fluid power is the technology that deals with the generation, control and transmission of forces and movement of mechanical element or system with the use of pressurized fluids in a confined system. Both liquids and gases are treated as fluids. Fluid power system includes a hydraulic system (hydra meaning water in Greek) Basic Mechanical Engineering

278

Basic Mechanical Engineering

and a pneumatic system (pneuma meaning air in Greek). Oil hydraulic employs pressurized liquid petroleum oils and synthetic oils, and pneumatic employs compressed air that is released to the atmosphere after performing the task.

13.2

HYDRAULIC AND PNEUMATIC SYSTEMS

13.2.1 Hydraulic Systems A hydrostatic system uses fluid pressure to transmit power. Hydrostatics deals with the mechanics of stationary fluids and uses the theory of equilibrium conditions in fluid. The system creates enormous pressure, and through a transmission line and a control element, this pressure drives an actuator (linear or rotational). The pump used in hydrostatic systems is a positive displacement pump. The relative spatial position of this pump is arbitrary but should not be large due to losses (must be less than 50 m). An example of pure hydrostatics is the transfer of force in hydraulics. Hydrodynamic systems use fluid motion to transmit power. Power is transmitted by the kinetic energy of the fluid. Hydrodynamics deals with the mechanics of moving fluid and uses flow theory. The pump used in hydrodynamic systems is a non-positive displacement pump. The relative spatial position of the prime mover (e.g., turbine) is fixed. An example of pure hydrodynamics is the conversion of flow energy in turbines in hydroelectric power plants. In oil hydraulics, we deal mostly with the fluid working in a confined system, that is, a hydrostatic system. Hydraulic systems are power-transmitting assemblies employing pressurized liquid as a fluid for transmitting energy from an energy-generating source to an energy-using point to accomplish useful work. Figure 13.1 shows a simple circuit of a hydraulic system with basic components. The following are the functions of the parts as shown in the Figure 13.1.

Figure 13.1 Components of a hydraulic system.

Fluid Power Control and Automation 279

1. The hydraulic actuator is a device used to convert the fluid power into mechanical power to do useful work. The actuator may be of the linear type (e.g., hydraulic cylinder) or rotary type (e.g., hydraulic motor) to provide linear or rotary motion, respectively. 2. The hydraulic pump is used to force the fluid from the reservoir to the rest of the hydraulic circuit by converting mechanical energy into hydraulic energy. 3. Valves are used to control the direction, pressure and flow rate of a fluid flowing through the circuit. Motor 1 – Off 2 – Forward 3– Return 3 2 1 Load Direction control valve Pump Oil tank Filter Actuator Pressure regulator. 4. External power supply (motor) is required to drive the pump. 5. Reservoir is used to hold the hydraulic liquid, usually hydraulic oil. 6. Piping system carries the hydraulic oil from one place to another. 7. Filters are used to remove any foreign particles so as keep the fluid system clean and efficient, as well as avoid damage to the actuator and valves. 8. Pressure regulator regulates (i.e., maintains) the required level of pressure in the hydraulic fluid. The piping shown in Figure 13.1 is of closed-loop type with fluid transferred from the storage tank to the one side of the piston and returned from the other side of the piston to the tank. Fluid is drawn from the tank by a pump that produces fluid flow at the required level of pressure. If the fluid pressure exceeds the required level, then the excess fluid returns back to the reservoir and remains there until the pressure acquires the required level. Cylinder movement is controlled by a three-position change over a control valve. (i) When the piston of the valve is changed to upper position, the pipe pressure line is connected to port A and thus the load is raised. (ii) When the position of the valve is changed to lower position, the pipe pressure line is connected to port B and thus, the load is lowered. (iii) When the valve is at centre position, it locks the fluid into the cylinder (thereby holding it in position) and dead-ends the fluid line (causing all the pump output fluid to return to tank via the pressure relief).  8VLQJWKHV\PEROVDOVRZHFDQGHPRQVWUDWHWKHK\GUDXOLFV\VWHPWKDWLVGHSLFWHG in Fig. 13.2. Here all the components have their own symbols to be used so that the understanding of the system becomes easier. The hydraulic system discussed above can be broken down into four main divisions that are analogous to the four main divisions in an electrical system. (a) The power device parallels the electrical generating station. (b) The control valves parallel the switches, resistors, timers, pressure switches, relays, etc.

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Basic Mechanical Engineering

Figure 13.2 Components of hydraulic system using symbols.

(c) The lines in which the fluid power flows parallel to the electrical lines. (d) The fluid power motor (whether it is a rotating or a non-rotating cylinder or a fluid power motor) parallels the solenoids and electrical motors.

13.2.2 Pneumatic Systems A pneumatic system carries power by employing compressed gas, generally air, as a fluid for transmitting energy from an energy-generating source to an energy-using point to accomplish useful work. Figure 13.3 shows a simple circuit of a pneumatic system with basic components. The functions of various components shown in Fig. 13.3 are as follows: 1. The pneumatic actuator converts the fluid power into mechanical power to perform useful work. 2. The compressor is used to compress the fresh air drawn from the atmosphere. 3. The storage reservoir is used to store a given volume of compressed air. 4. The valves are used to control the direction, flow rate and pressure of compressed air.

Fluid Power Control and Automation 281

Figure 13.3 Components of pneumatic systems.

5. External power supply (motor) is used to drive the compressor. 6. The piping system carries the pressurized air from one location to another. Air is drawn from the atmosphere through an air filter and raised to required pressure by an air compressor. As the pressure rises, the temperature also rises; hence, an air cooler is provided to cool the air with some preliminary treatment to remove the moisture. The treated pressurized air then needs to get stored to maintain the pressure. With the storage reservoir, a pressure switch is fitted to start and stop the electric motor when pressure falls and reaches the required level, respectively. The three-position changeover valve delivering air to the cylinder operates in a way similar to its hydraulic circuit. Table 13.1 Comparison between hydraulic and pneumatic systems S. No.

Hydraulic System

Pneumatic System

1.

It employs a pressurized liquid as a fluid

It employs a compressed gas, usually air, as a fluid

2.

An oil hydraulic system operates at pressures up to 700 bars

A pneumatic system usually operates at 5-10 bars

3.

Generally designed as closed system

8VXDOO\GHVLJQHGDVRSHQV\VWHP

4.

The system slows down when leakage occurs

Leakage does not affect the system much

5.

Valve operations are difficult

Valve operations are easy

6.

Heavier in weight

Lighter in weight

7.

Pumps are used to provide pressurized liquids

Compressors are used to provide compressed gases

8.

The system is unsafe to fire hazards

The system is free from fire hazards

9.

Automatic lubrication is provided

Special arrangements for lubrication are needed

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Basic Mechanical Engineering

13.3

AUTOMATION OF MECHANICAL SYSTEMS

There are three basic methods of transmitting power: electrical, mechanical and fluid power. Most applications actually use a combination of the three methods to obtain the most efficient overall system. To properly determine which method to use, it is important to know the salient features of each type. For example, fluid systems can transmit power more economically over greater distances than mechanical types. However, fluid systems are restricted to shorter distances compared to electrical systems. Table 13.2 lists the salient features of each type. Table 13.2 Comparison of different automated power systems Property

Mechanical

Electrical

Pneumatic

Hydraulic

Input energy source

IC engines electric IC engines water/ IC engines motor gas turbines Pressure tank

IC engines Electric motor Air turbine

Energy transfer element

Levers, gears, shafts

Electrical cables and magnetic field

Pipes and hoses

Pipes and hoses

Energy carrier

Rigid and elastic objects

Flow of electrons

Air

Hydraulic liquids

Power-to-weight ratio

Poor

Fair

Best

Best

Torque/inertia

Poor

Fair

Good

Best

Stiffness

Good

Poor

Fair

Best

Response speed

Fair

Best

Fair

Good

Dirt sensitivity

Best

Best

Fair

Fair

Relative cost

Best

Best

Good

Fair

Control

Fair

Best

Good

Good

Motion type

Mainly rotary

Mainly rotary

Linear or rotary

Linear or rotary

13.4

SUMMARY

The automation market in India is estimated to be 1/10th of China. If India has to become one of the leading economies in the world, based on manufacturing, it will have to attain higher technological standards and higher level of automation in manufacturing. In the past 30 years, fluid power technology rose as an important industry. With increasing emphasis on automation, quality control, safety and more efficient and green energy systems. Fluid power technology should continue to expand in India. Fluid power industry is gaining a lot of importance in Indian industry. According to a recent survey, it has shown a growth of 20% over the last 10 years and the size of market is estimated to be close to 5000 crore per annum. This makes it a sizable industry segment in India. The growth rate of this industry in India is typically about twice the growth of economy.

Fluid Power Control and Automation 283

One of the major segments for hydraulic industry in India is mobile hydraulics. Because of massive programmes on road construction, there is a major expansion of construction machinery industry as well. In addition to this, a trend towards the usage of more sophisticated hydraulics in tractors and farm equipment is witnessed. The manufacturing industry in India is working towards higher automation and quality of output. As Indian industry moves toward modernization to meet the productivity and compete in the global market, an excellent potential for the pneumatic industry is expected in India. Another area of interest for fluid power industry would be the opportunities in defence equipment. Defence is a major market segment in Indian fluid power industry and contributes to over 40% of the market demand. There is also a move towards products with miniature pneumatics, process valves, servo drives, hydraulic power steering with new controls and sophisticated PLC, microprocessor controls. However, the key input required for the effective utilization of fluid power is education and training of users. So there is a need for education and training in design application and maintenance of fluid power systems. REXROTH recently opened many competence centres in India to train the manpower and to create awareness about the use of fluid power in Indian industry.

EXERCISES 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Define the term fluid power. Differentiate between fluid transport and fluid power systems. Differentiate between hydraulics and pneumatics. List the six basic components used in a hydraulic system. List the six basic components used in a pneumatic system. List ten applications of fluid power in the automotive industry. Name ten hydraulic applications and ten pneumatic applications. List five advantages and five disadvantages of hydraulics. List five advantages and five disadvantages of pneumatics. List the main components of a fluid power system and their functions. Discuss in detail the future of fluid power industry in India. Compare different power systems used in industries.

Multiple-Choice Questions 1. The capacity of a lathe is expressed as: (a) Horsepower and chuck diameter (b) Swing and distance between centres (c) Bed length and spindle speed (d) Tool post size and travel

284

Basic Mechanical Engineering

2. The carriage of the lathe travels along which axis: (a) The “B” axis (b) The “Z” axis (c) The “Y” axis (d) The “X” axis 3. The basic turning lathe is: (a) The turret lathe (c) The engine lathe



(b) The automatic lathe (d) The Swiss type lathe

4. “Forming” on a lathe is accomplished by: (a) Specially shaped cutting tools (b) Compound rest travel F  5RWDWLQJÀ[WXUHV G  %\FXWWLQJJURRYHVDQGFKDPIHUV 5. Large diameter holes are produced on a lathe by: (a) Step drilling (b) Facing (c) Reaming (d) Boring 6. Vertical lathes are used for work requiring a: (a) Larger swing (b) More powerful engine (c) Greater distance between centres (d) Multiple jaw chuck 7. The Swiss-type automatic lathe is used primarily for: (a) Heavy turnings (b) Slug turnings (c) Long thin turnings (d) Threading 8. Feed rate refers to: (a) Spindle speed (c) Tool advancement into the work



(b) Chuck rotation (d) Depth of cut

9. Work study is concerned with D  ,PSURYLQJSUHVHQWPHWKRGDQGÀQGLQJVWDQGDUGWLPH (b) Motivation of workers (c) Improving production capability (d) Improving production planning and control (e) All of the above 10. Basic tool in work study is (a) Graph paper (c) Planning chart (e) Analytical mind

(b) Process chart (d) Stopwatch

11. What does symbol ‘O’ imply in work study (a) Operation (b) Inspection (c) Transport (d) Delay/temporary storage (e) None of the above

Fluid Power Control and Automation 285

12. What does symbol ‘D’ imply in work study (a) Inspection (b) Transport (c) Delay/temporary storage (d) Permanent storage (e) None of the above 13. What does symbol ‘V’ employ in work study (a) Operation (b) Inspection (c) Delay/temporary storage (d) Permanent storage (e) None of the above 14. Which one of the following welding processes uses nonconsumable electrode? (a) Gas metal arc welding (b) Submerged arc welding (c) Gas tungsten arc welding (d) Flux coated arc welding 15. The method of joining metal surface by introducing a nonferrous alloy with melting point above 400°C is known as (a) Soldering (b) Brazing (c) Welding (d) None of the above 16. Cutting forces at the cutting tool can be measured by (a) A dynamometer (b) A viscosity meter (c) A sine bar (d) A combination set 



17. Extrusion is a process of D  3XVKLQJWKHKHDWHGELOOHWRIDPHWDOWKURXJKDQRULÀFH (b) Producing a hole by a punch (c) Making cup-shaped parts from the sheet metal (d) None of the above 18. The clearance angle is provided on the tools with a view to (a) Strengthen the tool (b) Shear off the metal F  )DFLOLWDWHHDV\ÁRZRIFKLSV (d) Prevent the tool from rubbing on the workpiece 19. By cold working of materials, the fatigue strength (a) Increases (b) Decreases (c) Remains same (d) None of the above



20. Investment casting is used for D  6KDSHVZKLFKDUHPDGHZLWKGLIÀFXOW\XVLQJFRPSOH[SDWWHUQVLQVDQGFDVWLQJ (b) Mass production

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(c) Shapes which are very complex and intricate and can’t be cast by any other method (d) There is nothing like investment casting 21. Blow holes are casting defects (a) Which occur due to some sand shearing from the cope surface (b) Which takes the form of internal voids of surface depression due to excessive gaseous material not able to escape (c) Which occur due to discontinuity in metal casting resulting from hindered contraction (d) Caused by two streams of metals that are too cold to fuse properly

 

22. Laser is produced by (a) Graphite (c) Diamond

(b) Ruby (d) Emerald

23. The major problem in hot extrusion is (a) Design of punch (c) Wear and tear of die

(b) Design of die (d) Wear of punch

24. File used for woodwork is D  6LQJOHFXWÀOH F  5DVSFXWÀOH

E  'RXEOHFXWÀOH G  1RQHRIWKHDERYH

25. An example of fusion welding is (a) Arc welding (c) Thermit welding

(b) Gas welding (d) Forge welding

26. Gases used in tungsten inert gas welding are (a) Hydrogen and oxygen (b) CO2 and H2 (c) Argon and neon (d) Argon and helium 

 2[\DFHW\OHQHÁDPHLVXVHGWRZHOG (a) Steel (c) Stainless steel

(b) Copper alloys (d) Cast iron

28. Orthogonal cutting system is also known as (a) One-dimensional cutting system (b) Two-dimensional cutting system (c) Three-dimensional cutting system (d) None of the above 29. In metal cutting operations discontinuous chips are produced while machining (a) Brittle material (b) Ductile material (c) Hard material (d) Soft material

Fluid Power Control and Automation 287

30. Term CLA (Centre Line Average) is used for 

D  6XUIDFHURXJKQHVVÀQLVK 

E  6XUIDFHKDUGQHVV

(c) Cutting tool hardness

(d) None of the above

31. Knurling is an operation of 

D  &XWWLQJVPRRWKFROODUV

E  8QGHUFXWWLQJ

(c) Roughing the surface for hand grip

(d) None of the above

32. Poor fusion in a weld is due to 

(a) High welding speed

(b) Dirty metal surface

F  ,PSURSHUFXUUHQW

G  /DFNRIÁX[

33. Counterboring is the operation of (a) Enlarging the end of a hole cylindrically (b) Cone-shaped enlargement of the end of a hole (c) Smoothing and squaring the surface around a hole 

G  6L]LQJDQGÀQLVKLQJDKROH 34. A connecting rod is made by (a) Casting

(b) Drawing

(c) Forging

(d) Extrusion

35. Preheating before welding is done to (a) Make the steel softer (b) Burn away oil, grease, etc., from the plate surface (c) Prevent plate distortion (d) Prevent cold cracks 36. In Electro-discharge machining (EDM), the tool is made of (a) Copper

(b) High speed steel

(c) Cast iron

(d) Plain carbon steel

37. Which of the following is a single-point cutting tool? (a) Hacksaw blade

(b) Milling cutter

(c) Grinding wheel

(d) Parting tool

38. In ASA system, if the tool signature is 8-6-5-5-10-15-2 mm, then the side rake angle will be (a) 5°

(b) 6°

(c) 8°

(d) 10°

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Basic Mechanical Engineering

39. Cold working of metal increases (a) Tensile strength (c) Yield strength

(b) Hardness (d) All of the above

40. Seamless tube can be produced by (a) Two high rolling mills (b) Ring rolling combined with stretch forming (c) Piercing (d) Steam hammering forging 41. In electrochemical machining (ECM) the material removal is due to (a) Corrosion (b) Erosion (c) Fusion (d) Ion displacement 42. Crater wear occurs mainly due to (a) Abrasion (c) Oxidation

(b) Diffusion (d) Adhesion

43. Tool life of the cutting tool is most affected by (a) Cutting speed (b) Tool geometry (c) Cutting feed and depth (d) Microstructure of material being cut 44. Metal in machining operation is removed by (a) Tearing chips (b) Shearing the metal across a zone (c) Distortion of metal (d) Cutting the metal across a zone 

 8VXDOFDVWLQJPHWKRGIRUPDNLQJGHQWDOFURZQV (a) Sand casting (b) Die casting (c) Continuous casting (d) Investment casting 46. The mechanism of material removal in EDM process is (a) Melting and evaporation (b) Melting and corrosion (c) Erosion and cavitation (d) Cavitation and evaporation 47. If the metals are ductile and the cutting speed is high, then (a) Continuous chips are formed (b) Discontinuous chips are formed (c) Continuous chips with built-up edges are formed (d) None of these

Fluid Power Control and Automation 289

48. The point at which the cutting tool reaches, beyond which it will not function satisfactorily until it is reground, is called (a) Tool wear

(b) Tool failure

(c) Too diffusion

(d) None of the above

49. Which cutting condition affects the cutting temperature predominantly? (a) The shear zone

(b) The chip-tool interface zone

(c) The tool-work interface zone

(d) None of the above

50. The forces required for metal cutting operation (a) Increase with increase in the feed of the tool and decrease with increase in the depth of cut (b) Decrease with increase in the feed of the tool and increase with increase in the depth of cut (c) Increase with increase in both the feed of the tool and the depth of cut (d) Decrease with increase in both the feed of the tool and the depth of cut

Answers 1. (b)

2. (b)

3. (c)

4. (a)

5. (d)

6. (a)

7. (c)

8. (c)

9. (a)

10. (d)

11. (a)

12. (c)

13. (d)

14. (c)

15. (b)

16. (a)

17. (a)

18. (d)

19. (a)

20. (c)

21. (b)

22. (b)

23. (c)

24. (c)

25. (a)

26. (d)

27. (a)

28. (b)

29. (a)

30. (a)

31. (c)

32. (c)

33. (c)

34. (c)

35. (d)

36. (a)

37. (d)

38. (b)

39. (d)

40. (c)

41. (d)

42. (b)

43. (a)

44. (b)

45. (d)

46. (a)

47. (a)

48. (b)

49. (a)

50. (c)

Section V Automobile Engineering

14 IC Engines

Learning Objectives   

‡ 8QGHUVWDQGLQJWKHFODVVLILFDWLRQRI,&HQJLQHV ‡ ,QWURGXFWLRQWR&,DQG6,HQJLQHVZLWKFRPSDULVRQV ‡ 2YHUYLHZRIWZRVWURNHDQGIRXUVWURNHHQJLQHVZLWKFRPSDULVRQV

14.1

INTRODUCTION

Internal combustion (IC) engines are seen on a daily basis in cars, trucks, and buses. The name internal combustion refers conjointly to gas turbines except that the name is sometimes applied to reciprocating internal combustion engines just like the ones found in everyday cars. There are primarily 2 varieties of IC ignition engines, those which require a sparking plug, and those that consider compression of a fluid. Spark ignition engines take a combination of fuel and air, compress it, and ignite it employing a sparking plug. Figure 14.1 shows a piston and a few of its basic elements. The name ‘reciprocating’ is given owing to the motion that the crank mechanism goes through. The piston cylinder engine is basically a crank-slider mechanism, where the slider is that the piston in this case. The piston is moved up and down by the movement of the 2 arms or links. The shaft rotates which makes the two links rotate. The piston is encapsulated inside a combustion chamber. The bore is the diameter of the chamber. The valves on top represent induction and exhaust valves necessary for the intake of an air-fuel mixture and exhaust of chamber residuals. In a spark ignition engine a sparking plug is needed to transfer an electrical discharge to ignite the mixture. In compression ignition engines the mixture ignites at high temperatures and pressures. The lowest point where the piston reaches Basic Mechanical Engineering

294

Basic Mechanical Engineering

is called bottom dead centre. The highest purpose where the piston reaches is called top dead centre. The ratio of bottom dead centre to top dead centre is called the compression ratio. The compression ratio is extremely vital in several aspects of each compression and spark ignition engines, by definition the efficiency of engines.

Figure 14.1 Piston.

The intake in compression ignition engines is the atmospheric air, which is then compressed to high pressure and temperature, at this time combustion happens. These engines are high in power and fuel economy. Engines also are divided into four-stroke and two-stroke engines. In four-stroke engines the piston accomplishes four different strokes for each of the two revolutions of the crankshaft. In a two-stroke engine there are two different strokes in one revolution. Figure 14.2 shows a P-v diagram for the particular method of a four stroke internal combustion (IC) engine. Once the piston starts at bottom dead centre (BDC) the valve opens. A combination of fuel and water then is compressed to top dead centre (TDC), wherever the plug is used to ignite the mixture. This is often called the compression stroke. After striking TDC the air and fuel mixture have lighted and combustion happens. The growth stroke, or the ability stroke, supplies the force necessary to drive the rotating shaft. after the power stroke the piston

IC Engines 295

moves to BDC wherever the valve opens. The exhaust stroke is where the exhaust residuals leave the combustion chamber. So as for the exhaust residuals to depart the combustion chamber the pressure must be greater than the atmospheric. Then the piston returns to TDC wherever the valve closes. The next stroke is the intake stroke. Throughout the intake stroke the valve remains open which allows the air and fuel mixture to enter the combustion chamber and repeat the same process.

Figure 14.2 Actual cycle.

14.1.1 Basic Connecting-Rod Crank Mechanism

Figure 14.3

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Bore = cylinder diameter stroke = piston movement (BDC to TDC) = 2 × crank radius Bore = ‘squareness’ Stroke

The force on the piston is affected by the quantity of air induced. It is therefore affected by the gas dynamics (breathing) of the engine, which is mainly determined by the engine speed. The mechanical losses (friction) depend on the piston forces and RPM. Friction power loss is proportional to RPM³ (approx.) Thus, the fundamental output of a reciprocating IC engine can be considered to be torque. It is theoretically produced even when the engine speed (RPM) is zero. The torque fluctuations are smoothed out by: (i) multi-cylinder arrangement (ii) the use of a flywheel so that the torque output from the engine’s drive shaft is ‘smooth’ at any given condition of speed and load. Thus, the fundamental output of a reciprocating IC engine can be considered to be torque. It is theoretically produced even when the engine speed (RPM) is zero. Coefficient of speed fluctuation = This coefficient is decreased by increasing the moment of inertia (I) of the flywheel. (wmax – wmin)/2 Wm.

14.2

CLASSIFICATION OF IC ENGINES

IC engines can be classified on the following basis: 1. According to thermodynamic cycle  ‡ 2WWRF\FOHHQJLQHRUFRQVWDQWYROXPHKHDWDGGLWLRQF\FOH  ‡ 'LHVHOF\FOHHQJLQHRUFRQVWDQWSUHVVXUHKHDWDGGLWLRQF\FOH  ‡ 'XDOFRPEXVWLRQF\FOHHQJLQH

IC Engines 297

                    

2. According to the fuel used ‡ 3HWUROHQJLQH ‡ 'LHVHOHQJLQH ‡ *DVHQJLQH 3. According to the cycle of operation ‡ 7ZRVWURNHHQJLQH ‡ )RXUVWURNHHQJLQH 4. According to the method of ignition ‡ 6SDUNLJQLWLRQHQJLQHV6, ‡ &RPSUHVVLRQLJQLWLRQHQJLQHV&, 5. According to number of cylinders ‡ 6LQJOHF\OLQGHUHQJLQHV ‡ 0XOWLF\OLQGHUHQJLQHV 6. According to arrangement of cylinders ‡ +RUL]RQWDOHQJLQHV ‡ 9HUWLFDOHQJLQHV ‡ 9HQJLQHV ‡ ,QOLQHHQJLQHV ‡ 5DGLDOHQJLQHV 7. According to the method of cooling ‡ $LUFRROHG ‡ :DWHUFRROHG 8. According to their applications ‡ 6WDWLRQDU\HQJLQHV ‡ $XWRPRELOHHQJLQHV ‡ $HURHQJLQHV ‡ /RFRPRWLYHHQJLQHV ‡ 0DULQHHQJLQHV

14.2.1 Ideal Engine Cycles Otto Cycle: The Otto cycle is a model of the real cycle that assumes heat addition at the top dead centre. The Otto cycle consists of four internally reversible cycles, that describe the process of an engine. Figure 14.4, shows the P-v and T-s diagram for the Otto cycle.

298

Basic Mechanical Engineering

Figure 14.4 Otto cycle.

Process 1-2 is an isentropic compression of air and fuel, which occcurs when the piston moves from bottom dead centre (BDC) to TDC. In this process air and fuel are compressed for the second process. Process 2-3 is a constant volume heat addition process where the air to fuel mixture is ignited. Process 3-4 is an isentropic expansion, where work is done on the piston, but no heat is added. This process is referred to as the power stroke. The next process, 4-1, is a constant volume heat removal that ends at BDC. Work and heat are important aspects of engines, that are represented by Fig. 2.1. On the T-s diagram the area 1-4-a-b-1 corresponds to the heat rejected per unit of mass. Area 2-3-a-b-2 corresponds to the heat added per unit of mass. The enclosed area shown represents the net heat added during the process. The area 1-2-a-b-1 in the P-v diagram corresponds to the work input per unit mass and area 3-4-b-a-3 corresponds to work output per unit mass. The net work done is interpreted by the enclosed region in Figure above in the T-s diagram. In the Otto cycle there are therefore two processes that involve work but no heat transfer and two different processes that involve heat transfer but no work. The energy transfer can be expressed in the following form: W12 = u2 – u 1 m W34 = u3 – u 4 m Q23 = u3 – u 2 m Q41 = u4 – u 1 m

IC Engines 299

The efficiency for the engine can be expressed as the net work done over the heat added. The net work per unit mass is expressed as: Wcycle m

=

W34 W12 + m m

Therefore, the efficiency for the engine is: h = 1-

m 4 - m1 u3 - u2

Diesel Cycle: The Diesel cycle is similar to the Otto cycle, except that heat addition and rejection occur at different conditions. The Diesel cycle is also an ideal cycle meaning that it does not give an exact representation of the actual process. The Diesel cycle consists of four internally reversible processes. Process 1-2 is an isentropic compression. Process 2-3 is a constant pressure heat addition. This process makes the rest part of the power stroke. Process 3-4 is an isentropic expansion, which makes up the rest of the power stroke. Process 4-1 finishes the cycle with a constant volume heat rejection with the piston at BDC. The P-v and T-s diagram for the Diesel cycle is shown below.

Figure 14.5 Diesel cycle.

Since the Diesel cycle consists of four internally reversible processes, areas on the P-v and T-s diagram represent work and heat. On the T-s diagram the area 2-3-a-b-2 represents the heat added to the system. Area on the T-s curve, 1-4-a-b-1 represents the heat rejected. The enclosed area represents the net heat added during the process. On the P-v diagram the area 1-2-a-b-1 represents the work input and area 2-3-4-b-a-2 represents the work done as the piston moves to the BDC. The net work done is interpreted by the enclosed region in the P-v diagram. The efficiency for the engine is expressed as the net work done over the heat added. The efficiency is therefore:

300

Basic Mechanical Engineering

h=

Wcycle /m Q23 /m

=1-

u4 - u1 h3 - h2

The compression ratio of a diesel engine plays a greater significance than in a spark ignition engine. The thermal efficiency of a compression ignition (CI) engine increases as the compression ratio increases. The cut-off ratio, rc , is defined as: rc =

V3 V2

Dual Cycle: The dual cycle is a better description of the actual pressure variation in the engine. There are several differences in the Otto and diesel cycles. In the dual cycle there are five processes. Process 1-2 is an isentropic compression where there is no heat transfer but there is work done. Process 2-3 is a constant volume heat addition process where no work is done. Process 3-4 is another heat addition process but with constant pressure. This process is also known as the power stroke. Process 4-5 is an isentropic expansion that finishes with the remainder of the power stroke. Finally, process 5-1, is a constant volume heat rejection process. Figure below shows a P-v and T-s diagram of the dual cycle. Since the dual cycle is composed of the same processes in the Otto and diesel cycles, the efficiency is equal to the net work done divided by the heat input. The efficiency, therefore can be expressed as: h = 1-

Q51 Q23 + Q34

Figure 14.6 Dual cycle.

IC Engines 301

14.3

IC AND SI ENGINES

14.3.1 Terminology in IC Engines Cylinder cover

Valve

Clearance volume

Extreme position of piston at top (TDC) Piston

Stroke

Cylinder Bone Extreme position of piston at bottom (BDC)

Figure 14.7 IC engine terminology.

(i) Bore: The internal diameter of the cylinder is termed the bore. (ii) Stroke: The linear distance between the two limiting points of the piston on the axis of the cylinder is termed stroke. (iii) Top dead centre (TDC): The highest position of the piston towards cover end aspect of the cylinder” is termed top dead centre. Just in case of horizontal engine, it’s known as as inner dead centre. (iv) Bottom dead centre (BDC): The bottom position of the piston towards the crank end aspect of the cylinder is termed bottom dead centre. As in case of horizontal engine, it’s known as outer dead centre (ODC). (v) Clearance volume: The volume contained within the cylinder on top of the piston, once the piston is at the top dead centre is termed clearance volume. (vi) Compression ratio: It is the ratio of total cylinder volume to clearance volume.

14.3.2 Spark Ignition Engines (SI Engines) Internal combustion engines are divided into spark ignition engines and compression ignition engines. The majority of vehicles nowadays use spark ignition engines while trailers and a few massive trucks use compression ignition engines. The main distinction between the two is that the method in which the air to fuel mixture is ignited, and also the style of the chamber that leads to certain power and efficiency characteristics. Spark

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ignition engines use an air to fuel mixture that’s compressed at high pressures. At this high pressure the mixture should be close to stoichiometric to be chemically inert and ready to ignite. Stoichiometric implies that there’s a one to at least one ratio between the air and fuel mixture. Therefore, the mixture so as to ignite should not be either with too much of fuel or an excessive amount of air but rather have an overall even amount. There are many elements to a spark ignition engine. Chamber style, mixture and also the injection system are the most vital aspects of an spark ignition engine.

14.3.3 Compression Ignition Engines (CI Engines) Compression ignition engines differ from spark ignition engines in a variety of ways but the most obvious one being the way in which the air and fuel mixture is ignited. As stated above, a spark plug is used to create a spark in the combustion chamber which ignites the mixture. In a compression ignition engine there is no spark to create the flame but rather high temperatures and pressures in the combustion chamber cause a flame to initiate at different sites of the combustion chamber. Combustion increases with increasing pressure and temperature. Compression ignition engines are divided into direct and indirect ignition engines. Diesel engines require fuel injection systems to inject fuel into the combustion chamber. Fuel injection systems are either linear or rotary. Rotary fuel injectors are used in indirect ignition engines because of low pressures. Direct injection engines use pressures of up to 1000 bars to inject fuel into the combustion chamber. High pressure is needed because the heat addition process takes place at a compressed state, so in order for the fuel to inject well the pressure has to be greater than the one that has been accumulated through compression. There are several engineered direct injection combustion chambers. This shows that the actual design of compression ignition engines is not as critical as the design considered for spark ignition engines. Swirl is the most important air motion in the diesel engine. The importance of swirl is that it mixes the air and fuel so that combustion can increase. The direction of swirl is at a downward angle so that proper mixing can take place. The compression ratio for direct ignition engines is usually between 12 : 1 and 16 : 1. Indirect ignition engines have a pre-combustion chamber where the air-to-fuel mixture is stored. The purpose of the separate chamber is to speed up the combustion process in order to increase the engine output by increasing the engine speed. The two basic combustion systems are the swirl and pre-combustion chambers. Pre-combustion chambers depend on turbulence to increase the combustion speed and swirl chambers depend on the fluid motion to raise combustion speed. In divided chambers the pressure required is not as high as the pressure required for direct ignition engines. The pressure required for both types of divided chambers is only about 300 bars. With all diesel engines there is some type of aid to help combustion. Electrical components aid in the initiation of the combustion process by using an electrical source, such as a car battery, to heat themselves and transfer the energy to the mixture for combustion. Cold starting a diesel engine is

IC Engines 303

difficult without the use of these tabs that conduct an electric current. When electrical elements heat up and the air-to-fuel mixture comes in close contact with the tab then combustion occurs. The diesel engine has high thermal efficiencies, and therefore, low fuel consumption. The disadvantage of diesel engines is their low power output, relative to their weight, as compared with spark ignition engines. Compression ignition engines are different from the spark ignition engines in a variety of ways that however, the most obvious one being the method in which the air and fuel mixture is ignited. As expressed above, a sparking plug is employed to form a spark within the combustion chamber that ignites the mixture. In a compression ignition engine there is no spark to create the flame, however, rather high temperatures and pressures within the combustion chamber cause a flame to initiate at completely different sites of the combustion chamber. Combustion increases with increasing pressure and temperature. Compression ignition engines are divided into direct and indirect ignition engines. Diesel engines need fuel injection system to inject fuel into the combustion chamber. Fuel injection systems are either linear or rotary. Rotary fuel injectors are utilized in indirect ignition engines owing to low pressures. Direct injection engines use pressures nearly about 1000 bars to inject fuel into the combustion chamber. High pressure is required as a result of the heat addition process takes place at a compressed state, thus in order for the fuel to inject well the pressure has got to be larger than the one that has been accumulated through compression. There are many designed direct injection combustion chambers. This goes to point out that the particular style of compression ignition engines isn’t as crucial because the design thought-about for spark ignition engines. Swirl is the most vital air motion within the ICE. The importance of swirl is that it mixes the air and fuel so combustion will increase. The direction of swirl is at a downward angle in order that proper mixing can occur. The compression ratio for direct ignition engines is typically between 12:1 and 16:1. Indirect ignition engines have a pre-combustion chamber where the air-to-fuel mixture is kept. The requirement of the separate chamber is to fasten up the combustion method so as to extend the engine output by increasing the engine speed. The two basic combustion systems are the swirl and pre-combustion chambers. Pre-combustion chambers rely on turbulence to extend the combustion speed and swirl chambers rely on the fluid motion to lift combustion speed. In divided chambers the pressure needed isn’t as high as the pressure needed for direct ignition engines. The pressure needed for each kind of divided chambers is just about 300 bars. With all diesel engines there’s some kind of aid to assist combustion. Electrical components aid within the initiation of the combustion method by exploitation of an electrical supply, like a car battery, to heat themselves and transfer the energy to the mixture for combustion. Cold starting an ICE is extremely tough without the employment of those tabs that conduct an electrical current. Once electrical parts heat up and the air-to-fuel mixture comes in close contact with the tab then combustion happens. The

304

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ICE has high thermal efficiencies, and so low fuel consumption. The disadvantage of diesel engines is their low power output, relative to their weight, as compared with spark ignition engines. Table 14.1 Comparison between SI and CI Engines SI Engine ‡,WZRUNVRQ2WWRF\FOH

CI Engine It works on diesel or dual combustion cycle.

‡$IXHOKDYLQJKLJKHUVHOILJQLWLRQWHPSHUDWXUH A fuel having lower self-ignition temperature is is desirable, such as petrol (gasoline). desirable such as diesel. ‡$LUDQGIXHOPL[WXUHLQJDVHRXVIRUPLV inducted through the carburettor in the cylinder during the suction stroke.

Only air is introduced into the cylinder during the suction stroke and therefore the carburettor is not required. Fuel is injected at high pressure through fuel injectors direct into the combustion chamber.

‡7KHWKURWWOHYDOYHRIWKHFDUEXUHWWRUFRQWUROV the quantity of the charge. The quality of the charge remains almost fixed during normal running conditions at variable speed and load. So, it is a quantity governed engine.

The amount of air inducted is fixed but the amount of fuel injected is varied by regulating the quantity of fuel in the pump. The air-fuel ratio is varied at varying loads. So, it is a quality governed engine.

‡6SDUNLVUHTXLUHGWREXUQWKHIXHO)RUWKLVDQ ignition system with spark plugs is required. Because of this it is called a spark-ignition (SI) engine.

Combustion of fuel takes place on its own without any external ignition system. Fuel burns in the presence of highly compressed air inside the engine cylinder.

‡$FRPSUHVVLRQUDWLRRIWRLVHPSOR\HG The upper limit is fixed by the anti-knock quality of fuel. The engine tends to knock at higher compression ratios.

A compression ratio of 14 to 22 is employed. The upper limit of compression ratio is limited by the rapidly increasing weight of the engine. Engine tends to knock at lower compression ratios.

‡3DUWORDGHIILFLHQF\LVSRRUVLQFHHYHQDWSDUW load the air/fuel ratio is not much varied. In order to improve the part load efficiency of the SI engine, the MPFI technique of fuel supply is used in modern engines.

Part load efficiency is good. As the load decreases, the fuel supply to the engine can also be reduced and lean mixture to the engine is then supplied.

‡7KHFRVWRISHWUROLVKLJKHUWKDQWKDWRIGLHVHO

The cost of diesel oil is less than that of petrol. Moreover, as fuel is sold on volume basis and diesel has higher specific gravity, more weight is obtained in one litre.

‡1RLVHDQGYLEUDWLRQDUHOHVVEHFDXVHRIOHVV engine weight.

Noise and vibrations are more because of heavier engine components due to higher compression ratio.

‡7KHPDLQSROOXWDQWVDUHFDUERQPRQR[LGH (CO), oxides of nitrogen (NO2) and hydrocarbons (HC).

A part from CO, NO2, and HC, soot or smoke particles are also emitted to the atmosphere.

IC Engines 305

14.4

FOUR-STROKE AND TWO-STROKE ENGINES

14.4.1 Four-stroke Engines The four-stroke cycle petrol engines operate on Otto (constant volume) cycle as shown in Fig. 14.4. Since ignition in these engines is due to a spark, they are also called spark ignition engines. The four different strokes are: (i) Suction stroke (ii) Compression stroke (iii) Power or working or expansion stroke (iv) Exhaust stroke

Pressure

D

C

Isen

trop

ic

A O

E B

Volume

Figure 14.8 Theoretical Otto cycle.

Figure 14.9 Construction and working of a four-stroke petrol engine.

306

Basic Mechanical Engineering

Suction Stroke: During suction stroke, the piston is enraptured from the top dead centre to the bottom dead centre by the crankshaft. The crankshaft is rotated either by the momentum of the regulator or by the electrical starting motor. The water valve remains open and is closed during this stroke. The proportionate air-petrol mixture is sucked into the cylinder as a result of the downward movement of the piston. This operation is described by the path AB on the P-v diagram. Compression Stroke: During compression stroke, the piston moves from bottom dead centre to the top dead centre, so pressing air gasoline mixture. Because of compression, the pressure and temperature are accumulated and is shown by the path BC on the P-v diagram. Just before the top of this stroke the spark-plug initiates a spark, that ignites the mixture and combustion takes place at constant volume as shown by the path CD. Both the water and exhaust valves stay closed throughout this stroke. Operating Stroke: The expansion of hot gases exerts a pressure on the piston. Because of this pressure, the piston moves from top dead centre to bottom dead centre and so the work is obtained during this stroke. Both the water and exhaust valves stay closed throughout this stroke. The expansion of the gas is shown by the curve DE. Exhaust Stroke: During this stroke, the water valve remains closed and the exhaust valve opens. The larger part of the burnt gases escapes due to their own expansion. The drop in pressure at constant volume is described by the path EB. The piston moves from bottom dead centre to top dead centre and pushes the remaining gases to the atmosphere. Once the piston reaches the top dead centre the valve closes and cycle is completed. This stroke is described by the path BA on the P-v diagram. The operations are continual over and over again in running the engine. So a four-stroke engine completes one operating cycle, during this the crank rotate by two revolutions.

14.4.2 Two-stroke Engines In two-stroke cycle engines, the suction and exhaust strokes are eliminated. There are only two remaining strokes, i.e., the compression stroke and power stroke and these are sometimes known as upward stroke and downward stroke severally. Also, rather than valves, there is a body of water and exhaust ports in two-stroke cycle engines. The burnt exhaust gases are forced out through the exhaust port by a recent charge that enters the cylinder nearly at the tip of the operating stroke through the body of water port. The method of removing burnt exhaust gases from the engine cylinder is known as scavenging. The principle of two-stroke cycle fuel engine is shown in Fig. 14.10. Its two-strokes are represented as follows:

IC Engines 307

Figure 14.10 Working of two-stroke petrol engine.

Upward Stroke: During the upward stroke, the piston moves from bottom dead centre to top dead centre, compressing the air-petrol mixture within the cylinder. The cylinder is connected to a closed crank chamber. As a result of upward movement of the piston, a partial vacuum is formed within the housing, and a new charge is drawn into the crankcase through the uncovered inlet port. The exhaust port and transfer port are covered when the piston is at the top spatial relation position as shown in Fig. 14.6 (b). The compressed charge is ignited in the combustion chamber by a spark provided by the spark plug. Downward Stroke: As soon as the charge is ignited, the hot gases force the piston to move downwards, rotating the crankshaft, thus doing the useful work. During this stroke the inlet port is covered by the piston and the new charge is compressed in the crankcase as shown in Fig. 14.6(c). Further, downward movement of the piston uncovers first the exhaust port and then the transfer port as shown in Fig. 14.6(d). The burnt gases escape through the exhaust port. As soon as the transfer port opens, the compressed charge from the crankcase flows into the cylinder. The charge is deflected upwards by the hump provided on the head of the piston and pushes most of the exhaust gases. It may be noted that the incoming air-petrol mixture helps the removal of burnt gases from the engine cylinder. If in case these exhaust gases do not leave the cylinder, the fresh charge gets diluted and efficiency of the engine will decrease. The cycle of events is then repeated.

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Table 14.2 Comparison between four-stroke and two-stroke engines Four-stroke Engine

Two-stroke Engine

‡2QHSRZHUVWURNHLVREWDLQHGLQHYHU\WZR revolutions of the crankshaft as the cycle is completed in four strokes of the piston or in the revolutions of the crankshaft.

One power stroke is obtained in each revolution of the crankshaft as the cycle is completed in two strokes or in one revolution of the crankshaft.

‡2QHSRZHUVWURNHLQWZRUHYROXWLRQVRIWKH crankshafts makes the turning movement of the shaft non-uniform and hence a heavier flywheel is needed to rotate the shaft uniformly.

The turning movement of the shaft is more uniform and hence a lighter flywheel is needed to rotate the shaft uniformly.

‡3RZHUSURGXFHGIRUWKHVDPHVL]HRIWKH engine is less and for the same power output, the engine is larger in size, because only one power stroke is obtained in two revolutions.

Power produced for the same size of the engine is more and for the same power output, the engine is smaller in size, because one power stroke is obtained in every revolution.

‡,WFRQWDLQVYDOYHVDQGYDOYHPHFKDQLVP

It has ports. Some engines are fitted with exhaust valve or reed valve.

‡%HFDXVHRIKHDY\ZHLJKWDQGFRPSOLFDWHG valve mechanism, the initial cost is high.

It is light in weight and has no valve mechanism. Its initial cost is therefore low.

‡'XHWRSRVLWLYHVFDYHQJLQJDQGJUHDWHUWLPH of induction, its thermal efficiency and volumetric efficiency are higher.

It has lower thermal efficiency and volumetric effiency. Some of the fresh charge escapes unburnt during scavenging in petrol engines.

‡8VHGZKHUHKLJKHIILFLHQF\LVLPSRUWDQWDVLQ automobiles, power generation and aeroplanes.

8VHGZKHUHORZFRVWORZZHLJKWDQG compactness are important as in scooters, mopeds, lawnmowers, motorcycles, etc. Two-stroke diesel engines are used in very large sizes for ship propulsion because of low weight and compactness.

‡1RUPDOO\ZDWHUFRROHGWKHZHDUDQGWHDULV therefore less. It consumes less amount of lubricant. The lubricant is placed in the crankcase. It is not mixed with the fuel.

Normally, air-cooled, the wear and tear is therefore more. It requires more amount of OXEULFDQW8VXDOO\PRELORLOLVPL[HGZLWKIXHO

14.5

SUMMARY

An internal combustion engine (ICE) is a heat engine where the combustion of a fuel occurs with an oxidizer (usually air) in a combustion chamber that is an integral part of the working fluid flow circuit. In an internal combustion engine the expansion of the high-temperature and high-pressure gases produced by combustion apply direct force to some component of the engine. The force is applied typically to pistons, turbine blades, rotor or a nozzle. This force moves the component over a distance, transforming chemical energy into useful mechanical energy. An IC engine is one in which the heat transfer to the working fluid occurs within the engine itself, usually by the combustion of fuel with the oxygen of air. In external

IC Engines 309

combustion engines heat is transferred to the working fluid from the combustion gases via a heat exchanger, e.g., steam engines, Stirling engines. IC engines include spark ignition (SI) engines using petrol as a fuel, and compression ignition (CI) engines (usually referred to as diesel engines) using fuel oil, DERV, etc., as a fuel. In these engines there is a sequence of processes: compression, combustion, expansion, exhaust/induction There are two basic mechanical designs to achieve these four processes in either: The basic difference between the petrol engine and the diesel engine is in the method of ignition and the combustion process, both of which will be discussed later. Four strokes of the piston—hence the 4-stroke engine, or two strokes of the piston—hence 2-stroke engines. Here, in this chapter the basics of IC engines is covered citing the importance of working phenomenon of four-stroke and two-stroke engines in detail and listing the basic fundamental differences between SI & CI engines and four-stroke and two-stroke engines.

SOLVED EXAMPLES

Example 14.1 A four-cylinder car engine has bore x stroke = 79 mm × 77 mm. What is the capacity of the engine in cc? Solution: Capacity in cc = N.(p/4).B2 . S Here, N = number of cylinders; B = bore diameter in cm; S = stroke length in cm. Therefore, engine capacity in cc = 4 × (p ð ð  §FF

Example 14.2

Calculate the air standard cycle efficiencies of an Otto cycle engine and a Diesel cycle with a compression ratio of 10:1. Take r = 1.5 for the diesel engine.

Solution:

Otto cycle efficiency, 1 r k -1 1 1 = 1 - 1.4 - 1 = 1 - 0.4 = 0.6019 = 60.19% Ans. 10 10

hotto = 1 -

Diesel cycle efficency,

hdiesel = 1 -

1 È rk - 1 ˘ Í ˙ r k - 1 Î k(r - 1) ˚

È 1.51.4 - 1 ˘ Í ˙ = .0.5654 = 56.54% Ans. 101.4 - 1 Î 1.4(1.5 - 1) ˚ Example 14.3 Calculate the air standard Diesel cycle efficiency of the engine with a compression ratio of 11:1; if the fuel supply is cut-off at 6% of the stroke (or swept or displacement volume).

= 1-

1

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Solution: ncut = 0.06 nstroke

Given,

Now from Fig. 14.5, we can write, n3 - n2 = 0.06 n 4 - n2

or,

n3 - n2 = 0.06 n1 - n2

or,

n2 ( n3 / n2 - 1) = 0.06 n2 ( n1 / n2 - 1)

or,

r-1 = 0.06 r -1

or,

r-1 = 0.06 11 - 1

Diesel cycle efficiency, hdiesel = 1 -

hdiesel = 1 -

1 È rk - 1 ˘ Í ˙ r k -1 Î k(r - 1) ˚ 1 111.4 - 1

È 1.61.4 - 1 ˘ Í ˙ = 0.5753 Î 1.4(1.6 - 1) ˚

h = 57.53% Ans.

Example 14.4

A four-stroke engine producing 160 N.m from 2 litres of displacement. What will be it brake mean effective pressure (BMEP) ?

Solution:

BMEP is given by, pmep =

Tnc 2p Vd

Here, T = 160 N-m Vd = 2 litres = 2 × 10–3 m3 nc = 2 for a 4-stroke engine

IC Engines 311

So, BMEP = (160 N.m)(4p)/(0.002 m3) = 1,005,000 N/m2 = 1005 kPa (10.05 bars)

Example 14.5

If the same engine (i.e., four-stroke, 2 litres) as above produces 76 kW at 5400 rpm (90 Hz), find its BMEP.

Solution: Here, P = 76 × 103 W

We have,

w = 2pN/60 = 2p(5400)/60 = 565.5 rad/s 3

T = P/w = (76 × 10 )/565.5 = 134.4 N.m Now,

where Vd = 2 litres = 2 × 10–3 m3 nc = 2 for a 4-stroke engine pmep =

Tnc 2p Vd

So, BMEP = (134.4 N.m)(4p)/(0.02 m3) = 844460 N/m2 = 844.5 kPa (8.34 bars) Example 14.6 The following observations are recorded during a test on a four-stroke petrol engine; FC = 3000 of fuel in 12 sec; speed of the engine is 2500 rpm; BP = 20 kW; air intake orifice diameter = 35 mm; pressure across the orifice = 140 mm of water; coefficient of discharge of orifice = 0.6; piston diameter = 150 mm; stroke length = 100 mm; Density of the fuel = 0.85 gm/cc; r = 6.5; Cv of fuel = 42000 kJ/kg; Barometric pressure = 760 mm of Hg; Room temperature = 24°C Determine: (i) Volumetric efficiency on the air basis alone (ii) Air-fuel ratio (iii) The brake mean effective pressure (iv) The relative efficiency on the brake thermal efficiency

Solution: Given data: Fuel consumption = 30 cc in 12 sec Speed (N) = 2500/60 rps Brake power = 20 kW

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Basic Mechanical Engineering

Orifice diameter (do) = 0.035 m Pressure across the orifice (Po) = 140 mm of water Coefficient of discharge (Cd) = 0.6 Piston diameter (d) = 150 mm = 0.15 m; stroke length (l) = 0.1 m Density of fuel (r) = 0.85 gm/cc; compression ratio (r) = 6.5; room temperature (Ta) = 297 K Barometric pressure = 760 mm of Hg = 101.325 KK/m2 = 10.34 m of water Find: (i) Volumetric efficiency on the air basis alone (ii) Air-fuel ratio (iii) The brake mean effective pressure (iv) The relative efficiency on the brake thermal efficiency 10.34 m of water = 101.325 kN/m2 Pressure head =

ro r¥ g

Po = 0.14 m of water =

101.325 ¥ 0.14 10.34

Po = 1372 N/m2 Density of gas, r = P/RT =

101.325 0.287 ¥ 297

= 1.1887 kg/m3 13.72 1.1887 ¥ 9.81

Pressure head, h =

= 117.6557 m Qair = C d ¥ a ¥ 2 gh = 0.6 ¥

p (0.035)2 2 ¥ 9.81 ¥ 117.6557 4

= 0.02774 m3/sec

IC Engines 313

Number of suction strokes per second = Air consumption per stroke =

N 2500 = = 20.8333 2 60 ¥ 2

0.02774 20.8333

= 0.001332 m3 Stroke volume (Vs) =

p ¥ (0.15)2 ¥ 0.1 = 0.001767 m 3 4

Volumetric efficiency, hvol =

0.001332 ¥ 100% 0.001767

hvol = 75.382% Volume of air consumed, Vair = Qair = 0.02774 m3/sec = 0.02774 × 3600 m3/hr Mass of air consumed, ma = Va × ra = 99.864 × 1.1887 = 118.71 kg/hr Fuel consumption = 9000 cc/hr Mass of the fuel consumed, mf = 9000 × 0.85 = 7.65 kg/hr Air-fuel ratio =

mg mf

=

118.71 = 15.518 : 1 7.65

Brake power, BP = 20 kW = Pmb × 1 × a × n × k 20 0.001767 ¥ 20.833 ¥ 1

Pmb =

= 543.294 kN/m2 Air standard efficiency, hair = 1 -

1 ( r )g - 1

= 1-

1 (6.5)1.4 - 1

= 52.703% Brake thermal efficiency, hBT =

BP ¥ 3600 = 22.4% F.C ¥ C .V

Relative efficiency on brake thermal efficiency basis, hrel = hBT/hair

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EXERCISE 1. 2. 3. 4. 5. 6. 7.

8.

9.

10.

What is an IC engine? Write a short note on IC engines. How can IC engines be broadly classified? With a neat sketch define various terminologies associated with IC engines. Write a short note on CI and SI engines. Draw the P-v diagram and schematic diagram of construction of four-stroke engine, explain its working in detail. Write down the working of two-stroke engine with relevant diagrams. Differentiate between the following: (a) SI and CI engines (b) Four-stroke and two-stroke engines. In an automobile device working on Otto cycle, energy generated per cycle is twice as much rejected through the exhaust. Calculate the thermal efficiency and compression ratio. Consider the working fluid as an ideal gas. An engine, bore 7.5 cm and stroke 10 cm, has a compression ratio of 5:1. To increase the compression ratio,1.25 mm is machined off the cylinder head face. Calculate the new compression ratio. An air standard Otto cycle is designed to operate with the following data: Max cycle pressure and temperature: 5 MPa and 2250 K Min cycle pressure and temperature: 0.1 MPa and 300 K Determine the net work output per unit mass of working fluid and the thermal efficiency.

15 Systems in Automobiles

Learning Objectives    

‡ ‡ ‡ ‡

15.1

8QGHUVWDQGLQJWKHWHUPVLQDXWRPRELOHV 2YHUYLHZVXVSHQVLRQDQGSRZHUWUDQVPLVVLRQLQDXWRPRELOHV &LWHWKHFRQFHSWVVWDUWLQJLJQLWLRQDQGFKDUJLQJV\VWHPVLQDXWRPRELOHV ,QWURGXFWLRQWRYDULRXVIXHOLQMHFWLRQV\VWHPVLQDXWRPRELOHV

INTRODUCTION

Every automobile is the after-effect of joint work of various frameworks. Every framework, however, principally independent, is affected by the impact of different frameworks communicating with it. Before talking about the collaboration of different frameworks, let us first count the different frameworks that are available in a car.

15.1.1 Power Plant A vehicle which goes under the class automobile must deliver its own particular force sufficiently adequate to start and keep up the drive. The force is created from inside the car, normally from a reduced motor put either in the front or back. In most of the cases the engine is an internal combustion type that converts chemical energy of fuel into mechanical energy. This conversion is done inside a piston cylinder arrangement where controlled explosion of fuel-air mixture is done which produces a very high pressure inside. This high pressure drives the piston out from the cylinder. The linear displacement of the piston is converted to rotary motion with the help of a reciprocating motion mechanism. The output from the engine is available through a shaft. Basic Mechanical Engineering

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Basic Mechanical Engineering

15.1.2 Drive Train Drive train helps to transfer power produced from the engine to the wheels with the help of intermediate linkages. The set of linkages in between the engine and the wheels constitutes the drive train. It includes the clutch, the gearbox, the universal joints and the drive shaft and the differential arrangement.

15.1.3 Slowing Mechanism To slow down or to completely stop a vehicle one needs a braking system. Brakes absorb the kinetic energy and dissipate or store it in some other form (usually heat or electricity).

15.1.4 Suspension It provides the vehicle a smooth ride even when the wheels traverse uneven terrain. This is achieved by damping of the vibrations that get transmitted to the chassis through the wheels. Thus, a suspension system is employed between the wheel (axle) and the chassis. All types of suspension systems absorb the energy when a jerk/impact tries to set the suspension in motion. The absorbed energy may be dissipated or converted to other form. By providing a smoother working condition, the suspension enhances the life of all components which are mounted on the chassis, viz., the drive train, the engine and all bolting. A suspension system also helps improve the fuel efficiency by maintaining a continuous contact of the wheels with the road and thereby preventing rolling slip.

15.1.5 Electrical The electrical system on a modern automobile is tightly integrated with the engine, steering and suspension system. It draws power from an alternator (AC) or dynamo (DC) which is coupled with the engine. The power thus produced is either directly utilized for various activities or is stored in a suitable battery pack for future use. In spark ignition engines, this system also takes care of maintaining the spark and its timing. A high voltage sub-circuit is used for operating the spark plug. This system also powers and maintains the electronic stability programme (ESP) which combines the suspension, braking and engine control.

15.2

SUSPENSION AND POWER TRANSMISSION SYSTEMS IN AUTOMOBILES

15.2.1 Suspension System Suspension refers to the use of front and rear springs to suspend a vehicle’s frame, body, engine and power train above the wheels. The vehicle suspension system is responsible

Systems in Automobiles 317

for driving comfort and safety as the suspension carries the vehicle body and transmits all forces between the body and the road. In order to positively influence these properties, semi-active and/or active components are introduced. These enable the suspension system to adapt to various driving conditions. By adding a variable damper and/or spring, driving comfort and safety are considerably improved compared to suspension setups with fixed properties. This strategy requires that the control behaviour of these components is known and that laws on how to adapt free parameters depending on the driving excitations are known. This also requires the identification and fault detection of the involved components resulting in a mechatronic design. The vehicle suspension system consists of wishbones, the spring, and the shock absorber to transmit and also filter all forces between the body and road. The spring carries the body mass and isolates the body from road disturbances and thus contributes to drive comfort. The damper contributes to both driving safety and comfort. Its task is to damp the body and wheel oscillations, where avoidance of wheel oscillations directly refers to drive safety, as a non-bouncing wheel is the condition for transferring road-contact forces. The following are some of the functions of a suspension system. (a) Provide vertical compliance so the wheels can follow the uneven road, isolating the chassis from roughness in the road. (b) Maintain the wheels in the proper steer and camber attitudes to the road surface. (c) React to the control forces produced by the tyres – longitudinal (acceleration and braking) forces, lateral (cornering) forces, and braking and driving torques. (d) Resist roll of the chassis.

Figure 15.1 Vehicle suspension systems in automobiles.

318

Basic Mechanical Engineering

15.2.2 Power Transmission System Transmission is a speed reducing mechanism, equipped with several gears (Fig. 15.2). It may be called a sequence of gears and shafts, through which the engine power is transmitted to the tractor wheels. The system consists of various devices that cause forward and backward movement of tractor to suit different field conditions. The complete path of power from the engine to the wheels is called power train. The following are some of the functions of the power transmission system. (a) Function of power transmission system to transmit power from the engine to the rear wheels of the tractor. (b) To make reduced speed available, to rear wheels of the tractor. (c) To alter the ratio of wheel speed and engine speed in order to suit the field conditions. (d) To transmit power through right angle drive, because the crankshaft and rear axle are normally at right angles to each other.

Figure 15.2 Power transmission systems in automobiles.

Constituents of Power Transmission System: The power transmission system consists of the following. Combination of all these components is responsible for transmission of power.

Systems in Automobiles 319

Clutch: Clutch is a device, used to connect and disconnect the tractor engine from the transmission gears and drive wheels. Clutch transmits power by means of friction between driving members and driven members. Transmission Gears: Often the term transmission refers simply to the gearbox that uses gear and gear trains to provide speed and torque conversions from a rotating power source to another device. If hp of the engine is constant, it is obvious that for higher torque at wheels, low speed is required and vice versa. So the gearbox is fitted between the engine and the rear wheel for variable torque and speed. This is done by suitable design of gear and shafts. Speed varies according to the field requirements and so a number of gear ratios are provided to suit the varying conditions. Gears are usually made of alloy steel. Differential: Differential unit is a special arrangement of gears to permit one of the rear wheels of the vehicle to rotate slower or faster than the other. While turning the vehicle on a curved path, the inner wheel has to travel less than the other at the turning point to the tractor to move faster. Final Drive: It is the last part of the transmission system in a motor vehicle that delivers the necessary torque to turn the wheels of a vehicle. It is connected to the differential axle which is further connected to the wheels. Rear Axle: Rear axle is the last member of a power train. In most of automobiles, rear axle is the driving axle. It lies between the driving wheels and the differential gear and transmit power from the differential to the driving wheels. It consists of two half shaft connected to the differential gear—one for one wheel.

15.3

STARTING SYSTEM, IGNITION SYSTEM AND CHARGING SYSTEMS IN AUTOMOBILES

15.3.1 Starting System The starting system uses battery power and an electric dc motor to turn engine’s crankshaft for starting the engine. It changes electrical energy to mechanical energy and provides gear reduction/torque multiplication. When the ignition key is turned on, the current flows through the solenoid coil. This closes the contacts, connecting battery to the starter motor. The starting system in automobiles consists of the following: (a) Battery provides the current to turn the starter motor. (b) Fuse protects the circuit. (c) Ignition switch closes the circuit. (d) Relay uses small amount of current to control large amount. (e) Neutral safety switch opens the circuit until the vehicle is in neutral (manual transmission), or parked (automatic).

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Basic Mechanical Engineering

(f) Solenoid does the same thing as relay, but performs mechanical operation. It is an electromagnetic switch. (g) Starter motor engages pinion gear to ring gear (mounted on flywheel, or torque converter).

Figure 15.3 Starting systems in automobiles.

15.3.2 Ignition System The ignition system of an internal combustion engine is an important part of the overall engine system. All conventional petrol (gasoline) engines require an ignition system. By contrast, not all engine types need an ignition system, for example, a diesel engine relies on compression-ignition, that is, the rise in temperature that accompanies the rise in pressure within the cylinder is sufficient to ignite the fuel spontaneously. It provides for the timely burning of the fuel mixture within the engine. The ignition system is usually switched on/off through a lock switch, operated with a key or code patch. The following are the types of ignition system. Glow Plug Ignition: Glow plug ignition is used on some kinds of simple engines, such as those commonly used for model aircraft. A glow plug is a coil of wire made from nichrome that glows red hot when an electric current is passed through it. This ignites the fuel on contact, once the temperature of the fuel is already raised due to compression. The coil is electrically activated for starting the engine, but once running, the coil retains sufficient residual heat on each stroke due to the heat generated on the previous stroke. Glow plugs are also used to aid starting of diesel engines. Magneto System: The simplest form of spark ignition is that using a magneto. The engine spins a magnet inside a coil, and also operates a contact breaker, interrupting the current and causing the voltage to be increased sufficiently to jump a small gap. The spark plugs are connected directly from the magneto output. Magnetos are not used in modern cars, but because they generate their own electricity they are often found on small engines such as mopeds, lawnmowers, snow blowers, chainsaws, etc., where there

Systems in Automobiles 321

is no battery, and also in aircraft piston engines, where their simplicity and self-contained nature confers a generally greater reliability as well as lighter weight in the absence of a battery and generator or alternator. Battery Operated Ignition: With the universal adaptation of electrical starting for automobiles, and the concomitant availability of a large battery to provide a constant source of electricity, magneto systems were abandoned for systems which interrupted current at battery voltage, used an ignition coil (a type of autotransformer) to step the voltage up to the needs of the ignition, and a distributor to route the ensuing pulse to the correct spark plug at the correct time. Mechanical Ignition: Most four-stroke engines use a mechanically timed electrical ignition system. The heart of the system is the distributor which contains a rotating cam running off the engine’s drive, a set of breaker points, a condenser, a rotor and a distributor cap. External to the distributor is the ignition coil, the spark plugs, and wires linking the spark plugs and ignition coil to the distributor. Electronic Ignition: The disadvantage of the mechanical system is the use of breaker points to interrupt the low voltage high current through the primary winding of the coil; the points are subject to mechanical wear where they ride the cam to open and shut, as well as oxidation and burning at the contact surfaces from the constant sparking. They require regular adjustment to compensate for wear, and the opening of the contact breakers, which is responsible for spark timing, is subject to mechanical variations. In addition, the spark voltage also depends on contact effectiveness, and poor sparking can lead to lower engine efficiency. Electronic ignition (EI) solves these problems.

Figure 15.4 Battery ignition systems in automobiles.

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Basic Mechanical Engineering

15.3.3 Charging System The purpose of the charging system is to provide the electrical energy to charge the battery and to power all the electrical components and systems on the automobile. When the engine is not running, the battery provides this electrical energy. When the engine is running, the charging system takes over. The alternator is the heart of the charging system. It is an alternating-current generator mounted on the engine, which is driven by a belt from the crankshaft. The alternator develops alternating current, which is changed to direct current. Alternating current changes from positive (+) to negative (–) at a regular cycle. Direct current does not change from positive (+) to negative (–). Only direct current can be used to charge a battery. A voltage regulator, either inside or outside the alternator, senses the electrical needs of the vehicle and adjusts the output of the alternator accordingly. An indicator light on the instrument panel allows the driver to observe whether the system is operating properly. The battery is connected electrically to the alternator, so that either one may supply the electrical needs, and so that the alternator can charge the battery. Two basic types of charging systems were used. The first was a dc generator, which was discontinued in the 1960s. Since that time the ac alternator has been the predominant charging device. The dc generator and the ac alternator both use similar operating principles. As the battery drain continues, and engine speed increases, the charging system is able to produce more voltage than the battery can deliver. When this occurs, the electrons from the charging device are able to flow in a reverse direction through the battery’s positive terminal. The charging device now supplies the electrical system’s load requirements; the reserve electrons build up and recharge the battery. The entire charging system consists of the following components: Battery, ac or dc generator, drive belt, voltage regulator, charge indicator (lamp or gauge), ignition switch, cables and wiring harness, starter relay (some systems), fusible link (some systems). All charging systems use the principle of electromagnetic induction to generate electrical power. Electromagnetic principle states that a voltage will be produced if motion between a conductor and a magnetic field occurs.

15.4

FUEL INJECTION SYSTEM IN AUTOMOBILES

For the engine to run smoothly and efficiently, it needs the right quantity of fuel/air mixture according to its demand. Petrol engine cars use indirect fuel injection. A fuel pump sends the petrol to the engine bay, and it is then injected into the inlet manifold by an injector. There is either a separate injector for each cylinder or one or two injectors into the inlet manifold. The fuel injection system is a huge development in automotive engineering since it has to be able to deliver the benefits of making your vehicle less polluting and more fuel efficient. The fuel injectors have an important role because they are responsible for delivering gasoline to the cars. A fuel injector is a valve with electronic

Systems in Automobiles 323

controls that receives signals from the engine’s computer system. The pump in the gas tank delivers the pressurized gasoline to the nozzle in the fuel injector. Once the fuel reaches the fuel injector it sprays a fine mist of gasoline into the motors intake manifold. It passes through the manifold and is delivered to the engine where it is mixed with the oxygen creating combustion. By having the gasoline go into the engine in a mist format makes for a more efficient way of burning the gas versus having it go in by a droplet form. The more gasoline the motor requires, the longer the valves in the fuel injector will stay open. Not only does the fuel injection system make your car run more efficiently, it also adds more power to the motor and eliminates vapour lock. The following are types of fuel injection systems.

15.4.1 Single-point Fuel Injection System In single-point fuel injection or throttle body injection (TBI) system the fuel is injected into the throttle body. The throttle body fuel injection system, works using a single or pair of injectors. The throttle looks like a carburettor without the fuel bowl, the metering jets or the float. This type of fuel injection system consists of only two major castings — the fuel body and the throttle body. The fuel body supplies the fuel while the throttle body has a valve that controls the flow of air. On the throttle, there are ports that gather signals to relay to the manifold absolute pressure sensor and to the emission control system.

Advantages of TBI Fuel Injection   

 



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Section 4

Manufacturing Technology

Cutting Tool: Tool materials for cutting tool: 1. High carbon steel 2. High speed steel (W + Cr + V) 3. Carbide (W Carbide + Ti Carbide + Co Carbide) Fixture: It is a device which holds and locates the workpiece. It is used in miling, grinding, planning and turning. Heat Treatment: Operation involving heating and cooling of a metal in solid state for obtaining desirable condition without changing the chemical composition. Its objective is to increase hardness of metal, increase quality of metal (heat, corrosion, wear resistance quality) improve machinability. Jig: It is a device which holds and locates the workpiece and also guide and control cutting tool. It is used in drilling, reaming and tapping. Machine Tool: It is the power driven tool. It cuts and forms all kinds of metal parts. Examples - Lathe, Drill Press, Shaper, Planer, Grinding, Miling, Broaching, Boring Welding: It is the process of joining two similar or dissimilar metals by fusion.

Section 5

Automobile Engineering

Rating of Fuel S.I. Engine: Octane number. Octane number indicates ability of fuel to resist knock. C.I. Engine: Cetane number. Cetane number indicates ability of ignition of diesel fuel. That means, how much fast ignites diesel fuel. Scavenging: It is the process of removing burnt gas from combustion chamber of engine cylinder. Stoichiometric Ratio: It is the chemically correct air-fuel ratio by volume. By which theoretically sufficient oxygen will be gotten to burn all combustible elements in fuel completely.

Glossary 373

Supercharging: Actually, power output of an engine depends on what amount of air enters into the engine through intake manifold. When amount of entry air is increased, the engine speed will increased. Amount of air will be increased by increasing inlet air density. The process of increasing inlet air density is supercharging. The device which is used for supercharging is called supercharger. Supercharger is driven by a belt from engine crakshaft. It is installed in the intake system. Turbocharging: Turbocharging is similar to the supercharging. But in that case tubocharger is installed in the exhaust system whereas supercharger is installed in intake system. Turbocharger is driven by the force of exhaust gas. Generally, turbocharger is used for 2-stroke engines by utilizing exhaust energy of the engine, it recovers energy otherwise which would go waste.

Section 6

Power Plant Engineering

Air-Preheater: It is a part of boiler. Its function is to preheats the air to be supplied to furnace and it recovers heat from exhaust gas. Generator: It is a clossed vessel which is made of steel. Its function is to transfer heat to water to generate steam. Calorific Value of Fuel: It is the total amount of heat obtained from burning 1 kg solid or liquid fuel Economiser: It is a part of a boiler. Its function is to heat feed water which is supplied to boiler. Superheater: It is a part of boiler. Its function is to increase temperature of steam into a boiler.

Index A Abrasive jet machining 260, 261 advantages of 262 applications of 263 disadvantages of 262 Absolute pressure 171 Acme 58 Advanced machining processes characteristics of 270 Archimedes’ principle 172 ASA system 226 Atomic reactor technology 342 Automobiles, fuel injection system in 322 suspension system 317 systems in 315, 316 $[LDOÁRZWXUELQH

B %DVLFFRQQHFWLQJURGFUDQNPHFKDQLVP Basic manufacturing processes 217 Basic vapour compression cycle 138 Battery operated ignition 321 Belt drive 47 Bernoulli’s equation 173 Biofuels 360 Biomass energy 358 Black body radiation 121 Boring 250 Broaching 252 Buttress thread 58

C Cavitation 172 CGS system of units 163 Chain drives 47 Charging system 322 CI engine 304

Climb milling 251 CNC assisted conventional machining techniques 254 CNC machine, elements of 254 &ROGZRUNLQJSURFHVVHV &RPELQHGSRZHUF\FOHV Compact heat exchanger 127 Compaction 237 Compressibility 166 Compression ignition engines 302 &RQGXFWLRQ Convection 120 Conventional machining techniques 246, 247, 271 Conventional milling procedure 251 Conventional sources of energy 362 Creep 31 Critical speed 81 &URVVÁRZKHDWH[FKDQJHU Curve, principal features of 32

D Damped system 76 'HÁHFWLRQPDJQLWXGHRI Density 164 'LHVHOF\FOH 'LHVHOSRZHUSODQW advantages of 344 disadvantages of 344 Differential pressure 171 Direct stress 24 Double pipe heat exchanger 127 'RZQPLOOLQJ 'ULOOLQJ Drilling machine 230 'ULOOVW\SHVRI

376

Index

Drive train 316 Dual cycle 300 Dynamics 4

*ORZSOXJLJQLWLRQ Grinding 253

E

Heat 100 Heat exchangers 124  FODVVLÀFDWLRQRI types of 127 +HDWWUDQVIHUEDVLFODZVRI +HOLFDOJHDU Helical spring 54 Heterogeneous substances 103 Heterogeneous system 103 +LJKKHDGWXUELQH High pressure boilers 346, 348 +RRNH·VODZRIHODVWLFLW\ +RWZRUNLQJSURFHVVHV +\GHOSRZHUSODQWOD\RXWRI Hydraulic accumulator 204 Hydraulic devices 201 +\GUDXOLFLQWHQVLÀHU +\GUDXOLFPDFKLQHV Hydraulic press 201 +\GUDXOLFSXPSV  FODVVLÀFDWLRQRI Hydraulic ram 203 Hydraulic systems 278, 281 +\GUDXOLFWXUELQHV Hydroelectric energy 362 Hydrostatic pressure 172

EBM process, characteristics of 268 advantages of 266 disadvantages of 266 (IÀFLHQF\ (ODVWLFFRQVWDQWVUHODWLRQEHWZHHQ Electrical system 316 Electro beam 267 Electrochemical machining 264 advantages of 265 applications of 265 disadvantages of 266 Electro-discharge machining 266 Electron beam machining  DGYDQWDJHVRI  GLVDGYDQWDJHVRI Electronic ignition 321 Energy conversion devices 364 Energy from biomass 364 Energy method 78 (QJLQHHULQJPHFKDQLFVFODVVLÀFDWLRQRI Enthalpy 100 (TXOOLEULXPODZ

F Fatigue 31 )LUVWODZRIWKHUPRG\QDPLFV Fluid, characteristics of 162 properties of 167 Fluidized bed boilers 347, 350 )OXLGPDFKLQH )OXLGPHFKDQLFVVSHFLÀFSURSHUWLHVRI Forced vibration 72 Forces 13 characteristic of 11 unit of 8 Fouling factor 128 )RXULHU·VODZRIKHDWFRQGXFWLRQ Four-stroke engine 305, 308 )UDQFLVWXUELQH Free-body diagrams 12 Free vibration 73

G *DVWXUELQHSRZHUSODQW Gas turbines, types of 352 Gauge pressure 171 Gear pumps 200 Gear trains 51 *HDUW\SHVRI *HRWKHUPDOHQHUJ\

H

I IC engines  FODVVLÀFDWLRQRI terminology in 301 ,GHDOHQJLQHF\FOHV Ideal gas, characteristics of 167 Ideal refrigerants, properties of 144 Ignition system 320 Impact loading 26 ,PSXOVHWXUELQH Insulations 123 ,QWHUQDOFRPEXVWLRQ,& HQJLQHV Internal energy 100 ,QZDUGÁRZWXUELQH

J Joining process, importance of 232

K Kinematics 4 Kinetic energy 100 Kinetics 4

L Lami’s theorem 11

Index 377 Laser beam machining, advantages of 268 applications of 268 disadvantages of 268 techniques 267 /DZRIJHDULQJ /DZRIWUDQVPLVVLELOLW\RIIRUFH /DZVRIIRUFHV Leaf spring 54 characteristics of 55 Linear systems 70 /LTXLGVWDWHZHOGLQJSURFHVVHV /RZKHDGWXUELQH

M Machine design 44  FODVVLÀFDWLRQRI Machine design factors 46 Machine tools 228  W\SHVRI Machining 228 Machining processes 218 Magneto system 320 Manufacturing processes 215, 217  FODVVLÀFDWLRQRI Mass density 164 Material removal process 228 Mechanical design 44 Mechanical ignition 321 Mechanical systems, automation of 282 Mechanical vibrations 68 basic technology 70 Mechanics 3 terms used in 5 units in 8 0HFKDQLFVRIÁXLGV Mechanics of solids 4 0HGLXPKHDGWXUELQH Metal casting operations 221, 224 Metal forming processes 220 Metal joining process 231 Milling 250 principle of 250 Milling machine 231 Milling methods 250 0L[HGÁRZWXUELQH MKS system of units 163 Multi-point fuel injection system 324 advantages of 324 disadvantages of 325

N NC 254 Nonconventional machining techniques 261 1HZWRQ·VODZRIFRROLQJ 1HZWRQ·VODZRIJUDYLWDWLRQ 1HZWRQ·VWKUHHODZVRIPRWLRQ

1RQFRQYHQWLRQDOPDFKLQLQJFODVVLÀFDWLRQRI Nonconventional machining techniques 261, 271 Nonconventional sources of energy 363 1RQSRVLWLYHGLVSODFHPHQWSXPSV 1RQUHQHZDEOHVRXUFHVRIHQHUJ\ Nuclear energy 363 1XFOHDUSRZHUSODQWOD\RXWRI Nuclear reactor technology 345

O Orthogonal rake system 226 Oscillating motions 70 Oscillations 68 2WWRF\FOH 2XWZDUGÁRZWXUELQH

P 3DUDOOHORJUDPODZRIIRUFHV 3DUDOOHOVSULQJVHTXLYDOHQWVWLIIQHVVRI Pascal 170 Phase change diagram 108 3LVWRQ Pneumatic systems 278, 280, 281 3RLVVRQ·VUDWLR Positive displacement pumps 200 3RZGHUPHWDOOXUJ\ advantages of 238  DSSOLFDWLRQVRI  OLPLWDWLRQVRI 3RZGHUSURGXFWLRQ 3RZHUSODQW 3RZHUSODQWOD\RXW 3RZHUVFUHZV 3RZHUWUDQVPLVVLRQRI 3RZHUWUDQVPLVVLRQV\VWHP constituents of 318 functions of 318 3UHVVXUHFKDUDFWHULVWLFVRI measurement of 168 Primary shaping processes 218 Pure substance, phases of 106 phase-change processes of 106 properties of 106

R Rack and pinion 50 5DGLDOÁRZWXUELQHV Radiation 120 Raising the load 58 5HDFWLRQWXUELQH Refrigerants, properties of 140 and air conditioning 135 applications of 145  EDVLFGHÀQLWLRQV by class 141 concept of 136

378

Index

special applications of 146 uses of 144 Refrigeration cycle 138 5HQHZDEOHVRXUFHVRIHQHUJ\ Rigid bodies 4

S Secondary operations 237 6HFRQGODZRIWKHUPRG\QDPLFV 6HULHVVSULQJVHTXLYDOHQWVWLIIQHVVRI Shaping machine 230, 252 Shear strain 25 Shear stress 25 Shell and tube heat exchanger 128 SI engine 304 Simple gear trains 52 Simple stresses 24 Single DoF free undamped vibration 73 Single-point fuel injection system 323 Sintering 237 Slotting 252 Solar energy 356, 364 Solid mechanics 4 6ROLGVWDWHZHOGLQJSURFHVVHV Spark ignition engines 301 6SHFLÀFJUDYLW\ 6SHFLÀFYROXPH 6SHFLÀFZHLJKW Spiral gear 50 Spring in parallel 56 Spring in series 55 Springs 53 6SXUJHDU 6WDUWLQJV\VWHP Steam generators/boilers 348 6WHDPSRZHUSODQWOD\RXWRI 6WHIDQ%ROW]PDQQODZ Strain 24 6WUHVVGHÀQLWLRQRI Strain energy impact loading 25 Strength of materials 21  EDVLFGHÀQLWLRQV Stress vs strain curve 27 6XSHUFULWLFDOERLOHU Surface condenser 347 6XUIDFHÀQLVKLQJSURFHVVHV Surface tension 165

T 7DQJHQWLDOÁRZWXUELQH Tapers 248 Taper turning 248 Taylor’s tool life equation 228 TBI fuel injection, advantages of 323 disadvantages of 323

Tensile test 27 Thermal electric energy 363 Thermal stresses 30 Thermodynamic equilibrium, types of 102 Thermodynamic properties  FODVVLÀFDWLRQRI Thermodynamics  EDVLFVRI  ODZVRI Thermodynamic substances FODVVLÀFDWLRQRI 7KHUPRG\QDPLFV\VWHPFODVVLÀFDWLRQRI Thin cylinders 61 7KLUGODZRIWKHUPRG\QDPLFV Tidal energy 363 Torque 53 Transmission systems 46 Trapezoidal threads 58 7ULDQJOHODZ Turning 248 7ZRVWURNHHQJLQH

U Ultrasonic machining 263 advantages of 264 applications of 264 disadvantages of 264 Undamped free vibration 73 Up-milling 251

V Vapour compression refrigeration 137 Venturimeter 174 Vibration, importance of 71 9LEUDWLRQLVRODWLRQ Vibrations 68 types of 72 Viscosity 164 Viscously damped free vibration 77

W Water hammer 173 Welding advantages of 235 disadvantages of 235 Welding defects 234 Welding operations 233 Welding processes, types of 234 Wind energy 357, 363 Work 101 :RUPJHDU

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Basic Mechanical Engineering

Basic Mechanical Engineering

The book starts with the law of forces, free-body diagrams, basic information on materials strength including stresses and strains. It further discusses principles of transmission of power and elementary designs of gears, spring, etc. This part concludes with mechanical vibrations — their importance, types, isolation and critical speed.

Part three, Fluid Mechanics and Hydraulics, includes properties of fluids, measurement of pressure, Bernoulli’s equation, hydraulic turbine, pumps and various other hydraulic devices. Part four, Manufacturing Technology, mainly deals with various manufacturing processes, such as metal forming, casting, cutting, joining, welding, surface finishing and powder metallurgy. It further deals with conventional and non-conventional machining techniques, fluid power control and automation including hydraulic and pneumatic systems and automation of mechanical systems. Part five, Automobile Engineering deals with various aspects of IC and SI engines and their classification, etc. Four- and two-stroke engines also find place in this section. Next, systems in automobiles including suspension and power transmission systems, starting, ignition, charging and fuel injection systems. The last section deals with power plant engineering and energy. It includes power plant layout, surface condensers, steam generators, boilers and gas turbine plants. It concludes with renewable, nonrenewable, conventional and non-conventional sources of energy, and energy conversion devices.

Key Features Treatment of topics at depth with a large number of solved problems, diagrams in every chapter Inclusion of section-end multiple choice questions and book-end model questions for ample practice Glossary containing all the important terms with their definitions at the end of the book Kaushik Kumar, B.Tech, [Mechanical Engineering, REC (Now NIT), Warangal], MBA (Marketing, IGNOU) and Ph.D (Engineering, Jadavpur University), is Associate Professor in the Department of Mechanical Engineering, Birla Institute of Technology, Mesra, Ranchi, India. He has 15 years of Teaching & Research and over 11 years of industrial experience. His areas of teaching and research interest are CAD/CAM, Quality Management Systems, Optimisation, Non-conventional machining, Rapid Prototyping and Composites.

Sanghamitra Debta, B.Tech. (Mechanical Engineering, ITER SOA University, Bhubaneswar), is pursuing M.E. (Design of Mechanical Equipment, BIT Mesra). Her areas of interests are Product and Process Design, Strength of Materials, Material Engineering, and Automation.

978-93-89583-91-5

Kumar Roy Debta

Apurba Kumar Roy, B.E. [Mechanical Engineering, REC (Now NIT), Jaipur], M.E. (Mechanical Engineering, Jadavpur University, Kolkata) and PhD (Engineering, IIT Kharagpur), is Associate Professor at the Department of Mechanical Engineering, Birla Institute of Technology, Mesra, Ranchi, India. He has over 27 years of teaching, research and industrial experience. His areas of interest are Fluid Dynamics, Turbo Machines, Multiphase Flow, CFD, Optimisation and Non-conventional Energy, Direct Energy Conversion.

Basic Mechanical Engineering

The second part, Thermal Engineering, deals with basics and laws of thermodynamics; pure substances and their properties. It further includes laws of heat transfer, insulation, and heat exchanges. This part concludes with a detailed discussion on refrigeration and air conditioning.

Kaushik Kumar Apurba Kumar Roy Sanghamitra Debta Distributed by:

9 789389 583915 TM

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