Problem Solution To Mechanical Engineering PDF

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This page intentionally left blank Copyright © 2007, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated ...

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Copyright © 2007, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to [email protected]

ISBN (13) : 978-81-224-2551-2

PUBLISHING FOR ONE WORLD

NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at www.newagepublishers.com

Preface Mechanical Engineering being core subject of engineering and Technology, is taught to almost all branches of engineering, throughout the world. The subject covers various topics as evident from the course content, needs a compact and lucid book covering all the topics in one volume. Keeping this in view the authors have written this book, basically covering the cent percent syllabi of Mechanical Engineering (TME102/TME-202) of U.P. Technical University, Lucknow (U.P.), India. From 2004–05 Session UPTU introduced the New Syllabus of Mechanical Engineering which covers Thermodynamics, Engineering Mechanics and Strength of Material. Weightage of thermodynamics is 40%, Engineering Mechanics 40% and Strength of Material 20%. Many topics of Thermodynamics and Strength of Material are deleted from the subject which were included in old syllabus but books available in the market give these useless topics, which may confuse the students. Other books cover 100% syllabus of this subject but not covers many important topics which are important from examination point of view. Keeping in mind this view this book covers 100% syllabus as well as 100% topics of respective chapters. The examination contains both theoretical and numerical problems. So in this book the reader gets matter in the form of questions and answers with concept of the chapter as well as concept for numerical solution in stepwise so they don’t refer any book for Concept and Theory. This book is written in an objective and lucid manner, focusing to the prescribed syllabi. This book will definitely help the students and practicising engineers to have the thorough understanding of the subject. In the present book most of the problems cover the Tutorial Question bank as well as Examination Questions of U.P. Technical University, AMIE, and other Universities have been included. Therefore, it is believed that, it will serve nicely, our nervous students with end semester examination. Critical suggestions and modifications by the students and professors will be appreciated and accorded Dr. U.K. Singh Manish Dwivedi Feature of book 1. Cover 100% syllabus of TME 101/201. 2. Cover all the examination theory problems as well as numerical problems of thermodynamics, mechanics and strength of materials. 3. Theory in the form of questions – Answers. 4. Included problems from Question bank provided by UPTU. 5. Provided chapter-wise Tutorials sheets. 6. Included Mechanical Engineering Lab manual. 7. No need of any other book for concept point of view.

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IMPORTANT CONVERSION/FORMULA 1. Sine Rule

1 80 – β R Q

O

γ

1 80 – α

1 80 – γ P

P Q R = = sin (180 − α ) sin (180 − β) sin (180 − γ )

2. Important Conversion 1N

= = g = 1 H.P. = 1 Pascal(Pa) = 1KPa = 1MPa = 1GPa = 1 bar =

1 kg X 1 m/sec2 1000 gm X 100 cm/sec2 9.81 m/sec2 735.5 KW 1N/m2 103 N/m2 106 N/m2 109 N/m2 105 N/m2

3. Important Trigonometrical Formulas 1. sin (A + B) = sin A.cos B + cos A.sin B 2. sin (A – B) = sin A.cos B - cos A.sin B 3. cos (A + B) = cos A.cos B – sin A.sin B 4. cos (A – B) = cos A.cos B + sin A.sin B 5. tan (A + B) = (tan A + tan B)/(1 – tan A. tan B) 6. tan (A – B) = (tan A – tan B)/(1 + tan A. tan B) 7. sin2 A = 2sin A.cos A 8. sin2A + cos2A = 1 9. 1 + tan2A = sec2A 10. 1 + cot2A = cosec2A 11. 1 + cosA = 2cos2A/2

α

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CONTENTS Preface Syllabus Important Conversion/Formula

v

Part– A: Thermodynamics (40 Marks) 1. 2. 3. 4. 5.

Fundamental concepts, definitions and zeroth law First law of thermodynamics Second law Introduction of I.C. engines Properties of steam and thermodynamics cycle

1 30 50 65 81

Part – B: Engineering Mechanics (40 Marks) 6. Force : Concurrent Force system 7. Force : Non Concurrent force system 8. Force : Support Reaction 9. Friction 10. Application of Friction: Belt Friction 11. Law of Motion 12. Beam 13. Trusses

104 141 166 190 216 242 265 302

Part – C: Strength of Materials (20 Marks) 14. Simple stress and strain 15. Compound stress and strains 16. Pure bending of beams 17. Torsion

331 393 409 432

Appendix 1. Appendix Tutorials Sheets 2. Lab Manual 3. Previous year question papers (New syllabus)

448 474 503

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Fundamental Concepts, Definitions and Zeroth Law /

CHAPTER

1

1

FUNDAMENTAL CONCEPTS, DEFINITIONS AND ZEROTH LAW Q. 1: Define thermodynamics. Justify that it is the science to compute energy, exergy and entropy. (Dec–01, March, 2002, Jan–03) Sol : Thermodynamics is the science that deals with the conversion of heat into mechanical energy. It is based upon observations of common experience, which have been formulated into thermodynamic laws. These laws govern the principles of energy conversion. The applications of the thermodynamic laws and principles are found in all fields of energy technology, notably in steam and nuclear power plants, internal combustion engines, gas turbines, air conditioning, refrigeration, gas dynamics, jet propulsion, compressors, chemical process plants, and direct energy conversion devices. Generally thermodynamics contains four laws; 1. Zeroth law: deals with thermal equilibrium and establishes a concept of temperature. 2. The First law: throws light on concept of internal energy. 3. The Second law: indicates the limit of converting heat into work and introduces the principle of increase of entropy. 4. Third law: defines the absolute zero of entropy. These laws are based on experimental observations and have no mathematical proof. Like all physical laws, these laws are based on logical reasoning. Thermodynamics is the study of energy, energy and entropy. The whole of heat energy cannot be converted into mechanical energy by a machine. Some portion of heat at low temperature has to be rejected to the environment. The portion of heat energy, which is not available for conversion into work, is measured by entropy. The part of heat, which is available for conversion into work, is called energy. Thus, thermodynamics is the science, which computes energy, energy and entropy. Q. 2: State the scope of thermodynamics in thermal engineering. Sol: Thermal engineering is a very important associate branch of mechanical, chemical, metallurgical, aerospace, marine, automobile, environmental, textile engineering, energy technology, process engineering of pharmaceutical, refinery, fertilizer, organic and inorganic chemical plants. Wherever there is combustion, heating or cooling, exchange of heat for carrying out chemical reactions, conversion of heat into work for producing mechanical or electrical power; propulsion of rockets, railway engines, ships, etc., application of thermal engineering is required. Thermodynamics is the basic science of thermal engineering. Q. 3: Discuss the applications of thermodynamics in the field of energy technology.

2 / Problems and Solutions in Mechanical Engineering with Concept Sol: Thermodynamics has very wide applications as basis of thermal engineering. Almost all process and engineering industries, agriculture, transport, commercial and domestic activities use thermal engineering. But energy technology and power sector are fully dependent on the laws of thermodynamics. For example: (i) Central thermal power plants, captive power plants based on coal. (ii) Nuclear power plants. (iii) Gas turbine power plants. (iv) Engines for automobiles, ships, airways, spacecrafts. (v) Direct energy conversion devices: Fuel cells, thermoionic, thermoelectric engines. (vi) Air conditioning, heating, cooling, ventilation plants. (vii) Domestic, commercial and industrial lighting. (viii) Agricultural, transport and industrial machines. All the above engines and power consuming plants are designed using laws of thermodynamics. Q. 4: Explain thermodynamic system, surrounding and universe. Differentiate among open system, closed system and an isolated system. Give two suitable examples of each system. (Dec. 03) Or Define and explain a thermodynamic system. Differentiate between various types of thermodynamic systems and give examples of each of them. (Feb. 2001) Or Define Thermodynamics system, surrounding and universe. (May–03) Or Define closed, open and isolated system, give one example of each. (Dec–04) Sol: In thermodynamics the system is defined as the quantity of matter or region in space upon which the attention is concentrated for the sake of analysis. These systems are also referred to as thermodynamics system. It is bounded by an arbitrary surface called boundary. The boundary may be real or imaginary, may be at rest or in motion and may change its size or shape. Everything out side the arbitrary selected boundaries of the system is called surrounding or environment. Surrounding s

Real boundary

Boundary Su rro u

Cylinder

ndings

Fig. 1.1 The system

Convenient imaginary boundary

System Piston

Piston

Fig. 1.2 The real and imaginary boundaries

The union of the system and surrounding is termed as universe. Universe = System + Surrounding

System

Fundamental Concepts, Definitions and Zeroth Law /

3

Types of system The analysis of thermodynamic processes includes the study of the transfer of mass and energy across the boundaries of the system. On the basis the system may be classified mainly into three parts. (1) Open system (2) Closed System (3) Isolated system (1) Open system The system which can exchange both the mass and energy (Heat and work) with its surrounding. The mass within the system may not be constant. The nature of the processes occurring in such system is flow type. For example 1. Water Pump: Water enters at low level and pumped to a higher level, pump being driven by an electric motor. The mass (water) and energy (electricity) cross the boundary of the system (pump and motor). Heat Transfer Mass in

Mass may change Boundary free to move

Mass Out

Fig. 1.3

Work Transfer

2.Scooter engine: Air arid petrol enter and burnt gases leave the engine. The engine delivers mechanical energy to the wheels. 3. Boilers, turbines, heat exchangers. Fluid flow through them and heat or work is taken out or supplied to them. Most of the engineering machines and equipment are open systems. (2) Closed System The system, which can exchange energy with their surrounding but not the mass. The quantity of matter thus remains fixed. And the system is described as control mass system. The physical nature and chemical composition of the mass of the system may change. Water may evaporate into steam or steam may condense into water. A chemical reaction may occur between two or more components of the closed system. For example 1. Car battery, Electric supply takes place from and to the battery but there is no material transfer. 2. Tea kettle, Heat is supplied to the kettle but mass of water remains constant. Heat Transfer Mass may change Boundary free to move

Work Transfer Fig 1.4

3. Water in a tank 4. Piston – cylinder assembly. (3) Isolated System In an Isolated system, neither energy nor masses are allowed to cross the boundary. The system has fixed mass and energy. No such system physically exists. Universe is the only example, which is perfectly isolated system.

4 / Problems and Solutions in Mechanical Engineering with Concept

Other Special System 1. Adiabatic System: A system with adiabatic walls can only exchange work and not heat with the surrounding. All adiabatic systems are thermally insulated from their surroundings. Example is Thermos flask containing a liquid. 2. Homogeneous System: A system, which consists of a single phase, is termed as homogeneous system. For example, Mi×ture of air and water vapour, water plus nitric acid and octane plus heptanes. 3. Hetrogeneous System: A system, which consists of two or more phase, is termed as heterogeneous system. For example, Water plus steam, Ice plus water and water plus oil. Q. 5: Classified each of the following systems into an open or closed systems. (1) Kitchen refrigerator, (2) Ceiling fan (3) Thermometer in the mouth (4) Air compressor (5) Pressure Cooker (6) Carburetor (7) Radiator of an automobile. (1) Kitchen refrigerator: Closed system. No mass flow. Electricity is supplied to compressor motor and heat is lost to atmosphere. (2) Ceiling fan: Open system. Air flows through the fan. Electricity is supplied to the fan. (3) Thermometer in the mouth: Closed system. No mass flow. Heat is supplied from mouth to thermometer bulb. (4) Air compressor: Open system. Low pressure air enters and high pressure air leaves the compressor, electrical energy is supplied to drive the compressor motor. (5) Pressure Cooker: Closed system. There is no mass exchange (neglecting small steam leakage). Heat is supplied to the cooker. (6) Carburetor: Open system. Petrol and air enter and mi×ture of petrol and air leaves the carburetor. There is no change of energy. (7) Radiator of an automobile: Open system. Hot water enters and cooled water leaves the radiator. Heat energy is extracted by air flowing over the outer surface of radiator tubes. Q. 6: Define Phase. Sol: A phase is a quantity of matter, which is homogeneous throughout in chemical composition and physical structure. If the matter is all gas, all liquid or all solid, it has physical uniformity. Similarly, if chemical composition does not vary from one part of the system to another, it has chemical uniformity. Examples of one phase system are a single gas, a single liquid, a mi×ture of gases or a solution of liquid contained in a vessel. A system consisting of liquid and gas is a two–phase system. Water at triple point exists as water, ice and steam simultaneously forms a three–phase system. Q. 7: Differentiate between macroscopic and microscopic approaches. Which approach is used in the study of engineering thermodynamics. (Sept. 01; Dec., 03, 04) Or Explain the macroscopic and microscopic point of view.Dec–2002 Sol: Thermodynamic studies are undertaken by the following two different approaches. l. Macroscopic approach–(Macro mean big or total) 2. Microscopic approach–(Micro means small) The state or condition of the system can be completely described by measured values of pressure, temperature and volume which are called macroscopic or time–averaged variables. In the classical

Fundamental Concepts, Definitions and Zeroth Law /

5

thermodynamics, macroscopic approach is followed. The results obtained are of sufficient accuracy and validity. Statistical thermodynamics adopts microscopic approach. It is based on kinetic theory. The matter consists of a large number of molecules, which move, randomly in chaotic fashion. At a particular moment, each molecule has a definite position, velocity and energy. The characteristics change very frequently due to collision between molecules. The overall behaviour of the matter is predicted by statistically averaging the behaviour of individual molecules. Microscopic view helps to gain deeper understanding of the laws of thermodynamics. However, it is rather complex, cumbersome and time consuming. Engineering thermodynamic analysis is macroscopic and most of the analysis is made by it. These approaches are discussed (in a comparative way) below: Macroscopic approach 1. In this approach a certain quantity of matter is considered without taking into account the events occurring at molecular level. In other words this approach to thermodynamics is concerned with gross or overall behaviour. This is known as classical thermodynamics.

2. The analysis of macroscopic system requires simple mathematical formulae.

3. The values of the properties of the system are their average values. For example, consider a sample of a gas in a closed container. The pressure of the gas is the average value of the pressure exerted by millions of individual molecules. Similarly the temperature of this gas is the average value of transnational kinetic energies of millions of individual molecules. these properties like pressure and temperature can be measured very easily. The changes in properties can be felt by our senses. 4. In order to describe a system only a few properties are needed.

Microscopic approach 1. The approach considers that the system is made up of a very large number of discrete particles known as molecules. These molecules have different velocities and energies. The values of these energies are constantly changing with time. This approach to thermodynamics, which is concerned directly with the structure of the matter, is known as statistical thermodynamics. 2. The behaviour of the system is found by using statistical methods, as the number of molecules is very large. so advanced statistical and mathematical methods are needed to explain the changes in the system. 3. The properties like velocity, momentum, impulse, kinetic energy, and instruments cannot easily measure force of impact etc. that describe the molecule. Our senses cannot feel them.

4. Large numbers of variables are needed to describe a system. So the approach is complicated.

6 / Problems and Solutions in Mechanical Engineering with Concept Q. 8: Explain the concept of continuum and its relevance in thermodynamics. Define density and pressure using this concept. (June, 01, March– 02, Jan–03) Or Discuss the concept of continuum and its relevance. (Dec–01) Or Discuss the concept of continuum and its relevance in engineering thermodynamics. (May–02) Or What is the importance of the concept of continuum in engineering thermodynamics. (May–03) Sol: Even the simplification of matter into molecules, atoms, electrons, and so on, is too complex a picture for many problems of thermodynamics. Thermodynamics makes no hypotheses about the structure of the matter of the system. The volumes of the system considered are very large compared to molecular dimensions. The system is regarded as a continuum. The system is assumed to contain continuous distribution of matter. There are no voids and cavities. The pressure, temperature, density and other properties are the average values of action of many molecules and atoms. Such idealization is a must for solving most problems. The laws and concepts of thermodynamics are independent of structure of matter. According to this concept there is minimum limit of volume upto which the property remain continuum. Below this volume, there is sudden change in the value of the property. Such a region is called region of discrete particles and the region for which the property are maintain is called region of continuum. The limiting volume up to which continuum properties are maintained is called continuum limit. For Example: If we measure the density of a substance for a large volume (υ1), the value of density is (ρ1). If we go on reducing the volume by δv’, below which the ratio äm/äv deviates from its actual value and the value of äm/äv is either large or small. Thus according to this concept the design could be defined as ρ = lim δv– δv’ [δm / δv...