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Basic MechanicalEngineering

As per New Syllabus ofBiju Patnaik University of Technology [BPUT]

ODISHA

Dr. S.Ramachandran, M.E., Ph.D.,

Dr. A. Anderson, M.E., Ph.D.,

Sathyabama UniversityJeppiaar Nagar, Chennai - 600 119

and

First Edition: 23-10-16

ISBN : 978-93-84893-55-2

Price : Rs. 150/-

First Semester B.Tech Syllabus for Admission Batch 2016-17

BASIC MECHANICAL ENGINEERING - ODISHA - BPUT

MODULE - 1

Thermodynamics: (9 classes)Systems, Properties, Process, State, Cycle, Internal energy, Enthalpy, Zeroth

Law, First law and Second Law of Thermodynamics, Basic Concept of Entropy,

Properties of ideal gas, Properties of pure substances, Steam formation, Types of

Steam, Enthalpy, Specific volume, Internal energy and dryness fraction of steam, use

of Steam tables, Related numerical.

MODULE - 2: (11 classes)

Application of Thermodynamics:Air compressors, Steam Power Plant, Refrigerators and Heat pump, I.C.

Engines (Brief Description of different components of above mentioned systems and

working principles with Schematic diagram only).

Introduction to Fluid Mechanics and Heat transfer:Fluid properties, Pascal’s law, Buoyancy, Bernoulli’s theorem, pipe flow,

hydraulic turbines and pumps. Different modes of heat transfer, heat exchangers

(basics).

MODULE-3 (8 classes)

Production processesTurning, Casting, Welding and forming (Drawing, Forging, Extrusion)

(working principles with Schematic diagram only)

Engineering materials:Classification of Engineering materials. Mechanical properties of Steel,

Aluminium and Plastics.

MODULE-4 (8 classes)

Fasteners and Power transmission devices:Nut, Bolt, Screw, Rivets, Belt, Rope, Gear drives. Coupling, clutch, brakes.

(Basics, applications, advantages and limitations only).

Mechanical Measurements:Temperature, pressure, velocity, flow, strain, force, torque measurements.

(Working principle only).

Syllabus S.1

Contents

Module 1: Thermodynamics

1.1 Introduction ..................................................................... 1.11.2 Role of Thermodynamics in Engineering and Science 1.21.3 Applications of Thermodynamics .................................. 1.21.4 Basic Concepts ................................................................ 1.31.5 Classical and Statistical Thermodynamics.................. 1.71.6 Thermodynamic Systems ............................................... 1.91.7 Closed System................................................................. 1.91.8 Open System ................................................................... 1.101.9 Isolated System............................................................... 1.101.10 Homogeneous and Heterogeneous Systems .............. 1.111.11 Property.......................................................................... 1.111.12 State, Path and Process .............................................. 1.121.13 Cycle ............................................................................... 1.131.14 Thermodynamic Equilibrium....................................... 1.141.15 Quasi-static (or) Quasi–equilibrium Process............. 1.141.16 Path Function and Point Function............................ 1.161.17 Zeroth Law of Thermodynamics................................. 1.171.18 Measurement of Temperature..................................... 1.17

1.18.1 Thermometry ...................................................... 1.181.18.2 Applications of thermometry............................ 1.181.18.3 Triple point of water ........................................ 1.181.18.4 Comparison of Thermometer ........................... 1.18

1.19 Equation of State ......................................................... 1.191.20 Celsius Temperature Scale.......................................... 1.191.21 Work ............................................................................... 1.21

1.21.1 Work Transfer - a Path Function .................. 1.221.22 Heat ................................................................................ 1.23

1.22.1 Heat Transfer – A Path Function .................. 1.24

Contents C.1

1.23 Heat Capacity ............................................................... 1.251.24 Reversible and Irreversible Process........................... 1.261.25 First Law of Thermodynamics ................................... 1.261.26 Joules Experiment ........................................................ 1.271.27 Perpetual Motion of Machine of First Kind ............ 1.291.28 First Law For A Closed System Undergoing Change of State............................................................. 1.29

1.28.1 Internal Energy ................................................. 1.291.28.2 Energy - A property of the system ................. 1.30

1.29 Specific Heat Capacities .............................................. 1.341.29.1 Specific heat....................................................... 1.341.29.2 Latent heat......................................................... 1.341.29.3 Specific heat at constant volume .................... 1.341.29.4 Enthalpy ............................................................. 1.351.29.5 Specific heat at constant pressure .................. 1.36

1.30 Thermodynamic Processes ........................................... 1.371.31 Non-flow Process (or) Closed System ........................ 1.37

1.31.1 Constant Volume Process ................................. 1.371.31.2 Constant Pressure Process (or) isobaric Process 1.401.31.3 Constant Temperature Process (or) isothermal

Process PV constant; T constant .............. 1.431.31.4 Reversible Adiabatic Process (or) isentropic Process ................................................................ 1.47

1.31.5 Polytropic process: PVn constant n n .... 1.511.32 Steady Flow Process (Open System) ......................... 1.57

1.32.1 Nozzle ................................................................ 1.601.32.2 Diffusor ............................................................. 1.601.32.3 Throttling Device ............................................. 1.621.32.4 Turbine ............................................................. 1.621.32.5 Compressor ....................................................... 1.651.32.6 Heat Exchanger ............................................... 1.66

C.2 Basic Mechanical Engineering - www.airwalkpublications.com

1.33 Limitations of First Law of Thermodynamics ......... 1.671.34 The Second Law of Thermodynamics........................ 1.68

1.34.1 Thermal Energy Reservoirs.............................. 1.691.35 Heat Engines................................................................. 1.701.36 The Second Law of Thermodynamics: Kelvin-PlanckStatement ............................................................................... 1.721.37 The Second Law of Thermodynamics: Clausius Statement...................................................................... 1.741.38 Refrigerators and Heat Pumps................................... 1.751.39 Coefficient of Performance - COP.............................. 1.761.40 Equivalence of The Two Statements ......................... 1.77

1.40.1 Perpetual-Motion Machines.............................. 1.781.41 Carnot Cycle.................................................................. 1.79

1.41.1 The Carnot Principles (or) Carnot Theorem . 1.831.41.2 Corollaries of Carnot’s Theorem ..................... 1.86

1.42 Clausius Inequality....................................................... 1.861.43 Entropy........................................................................... 1.901.44 Third Law of Thermodynamics .................................. 1.991.45 Properties of Ideal Gas................................................ 1.1001.46 Properties of Pure Substances.................................... 1.1081.47 Thermodynamic Properties of Steam......................... 1.111

1.47.1 Property Tables.................................................. 1.1111.47.2 Saturated Liquid and Saturated Vapour States .................................................................. 1.1121.47.3 Saturated Liquid-Vapour Mixture .................. 1.1121.47.4 Superheated Vapour.......................................... 1.113

Module 2: Application of Thermodynamics and Introduction to Fluid Mechanics and Heat transfer

2.1 Air Compressors ............................................................. 2.1

Contents C.3

2.1.1 Applications of Air Compresser......................... 2.12.1.2 Reciprocating Air Compressors.......................... 2.12.1.3 Working Principle of Reciprocating Air Compressors 2.32.1.4 Rotary Compressors ............................................ 2.62.1.5 Roots Blower........................................................ 2.62.1.6 Vane Type Blower Compressor.......................... 2.72.1.7 Centrifugal compressor ....................................... 2.82.1.8 Axial Flow Compressor ...................................... 2.9

2.2 Steam Power Plant ........................................................ 2.102.2.1 Layout of Steam Power Plant........................... 2.102.2.2 Working of Steam Power Plant ........................ 2.132.2.3 Major Thermal Power Plants in India............ 2.13

2.3 Refrigerators and heat pump ....................................... 2.142.3.1 Parts of the household refrigerator .................. 2.152.3.2 Internal parts of the refrigerator ...................... 2.162.3.4 External parts of the refrigerator ..................... 2.17

2.4 IC Engines....................................................................... 2.202.4.1 IC Engine Components and Functions ............ 2.202.4.2 Basic Terms Connected With I.C. Engines ..... 2.252.4.3 Working Principle of Four Stroke Spark IgnitionEngine (4 Stroke Petrol Engine)................................. 2.262.4.4 Four Stroke Diesel Engine (4 Stroke CompressionIgnition Engine) ............................................................ 2.292.4.5 Two Stroke Engine ............................................. 2.312.4.6 Two Stroke Petrol Engine (si Engine) ............. 2.322.4.7 Two Stroke (Compression Ignition) Diesel Engine in Diesel Engine .................................... 2.34

2.5 Fluid Properties .............................................................. 2.402.5.1 Density (or) mass Density.................................. 2.402.5.2 Specific weight (or) Weight density .................. 2.41

2.5.3 Specific Volume v ............................................. 2.41


2.5.4 Specific gravity (or) Relative density s.......... 2.412.5.5 Temperature ......................................................... 2.422.5.6 Viscosity................................................................ 2.42

2.5.6.1 Kinematic Viscosity () .................................... 2.442.5.6.2 Variation of Viscosity with temperature ....... 2.44

2.5.7 Compressibility 1K

...........................................

2.44

2.5.7.1 Relationship between Bulk Modulus K and

pressure P of a Gas for Isothermal and

Isentropic Process................................................ 2.452.5.8 Vapour Pressure .................................................. 2.462.5.9 Cavitation ............................................................. 2.462.5.10 Gas and Gas laws ............................................ 2.462.5.11 Surface Tension................................................. 2.472.5.11.1 Surface Tension on Droplet .......................... 2.492.5.11.2 Surface Tension on a Hollow Bubble ......... 2.502.5.12 Capillarity .......................................................... 2.502.5.12.1 Expression for Capillary Rise ...................... 2.512.5.13 Thermodynamic Properties............................... 2.512.5.14 Newton’s law of viscosity ................................. 2.522.6 Types of Fluid ........................................................ 2.52

2.7 Pascal’s Law.................................................................... 2.572.7.1 Atmospheric Pressure.......................................... 2.572.7.2 Absolute zero Pressure (or) Absolute pressure 2.582.7.3 Gauge Pressure.................................................... 2.582.7.4 Vacuum Pressure................................................. 2.58

2.8 Buoyancy.......................................................................... 2.592.9 Principle of Conservation of Energy............................ 2.602.10 Bernoulli’s Theorem ..................................................... 2.60

2.10.1 Assumptions for derivation of Bernoulli’s theorem............................................................... 2.61

Contents C.5

2.10.2 Bernoulli’s theorem for Real Fluid................. 2.622.10.3 Bernoulli’s Theorem: Applications ................. 2.632.10.3.1 Venturi meter.................................................. 2.642.10.3.2 Orifice meter ................................................... 2.64

2.11 Pipe Flow....................................................................... 2.652.12 Reynolds Number ......................................................... 2.662.13 Loss of Head For A Given Length of Pipe (Hagen Poiseuille Formula) ....................................... 2.692.14 Law of Fluid Friction .................................................. 2.692.15 Frictional Loss in Pipe Flow ...................................... 2.712.16 Hydraulic Turbines....................................................... 2.72

2.16.1 Classification of Hydraulic Turbines ............. 2.722.16.2 Pelton Turbine (or) Pelton Wheel ................... 2.722.16.2.1 Working of A Pelton Wheel .......................... 2.732.16.3 Kaplan Turbine ................................................. 2.732.16.3.1 Working of Kaplan Turbine ......................... 2.742.16.4 Francis Turbine................................................. 2.752.16.4.1 Working of A Francis Turbine..................... 2.752.16.5 Difference Between Kaplan and Francis Turbine 2.752.16.6 Difference Between Impulse and Reaction HydraulicTurbines ......................................................................... 2.77

2.17 Pumps............................................................................. 2.772.17.1 Classification of hydraulic pumps .................. 2.772.17.1.1 Positive displacement pump.......................... 2.772.17.1.2 Non-positive displacement pump.................. 2.78

2.18 Centrifugal Pump (or) Non-positive Displacement Pump ............................................................................. 2.782.19 Reciprocating Pumps (or) Positive Displacement Pump ............................................................................. 2.80

2.19.1 Classification of Reciprocating Pumps........... 2.812.19.2 Single Acting Reciprocating Pump ................. 2.81


2.19.2.1 Working of Reciprocating Pump .................. 2.822.19.3 Double Acting Reciprocating Pump................ 2.832.19.3.1 Working of Double Acting Reciprocating Pump ................................................................ 2.832.19.4 Rotary Pumps .................................................... 2.84

2.20 Different Modes of Heat Transfer ............................. 2.872.20.1 Conduction ......................................................... 2.882.20.2 Convection .......................................................... 2.892.20.3 Radiation............................................................ 2.90

2.21 Heat Exchanger ............................................................ 2.912.21.1 Classification of Heat Exchanger.................... 2.912.21.2 Heat Exchanger Analysis ................................. 2.92

2.22 Direct Contact (or) Open Heat Exchangers ............. 2.922.23 Indirect Contact Type .................................................. 2.932.24 Classification of Heat Exchangers according............ 2.94

2.24.1 Parallel flow ...................................................... 2.942.24.2 Counter flow heat exchanger ........................... 2.962.24.3 Cross flow .......................................................... 2.97

Module 3: Production Process and Engineering Materials

3.1 Lathe ................................................................................ 3.13.2 Turning ............................................................................ 3.2

3.2.1 Plain (or) Straight Turning .............................. 3.23.2.2. Shoulder Turning (or) Step turning ............... 3.33.2.3 Taper Turning ..................................................... 3.33.2.4. Chamfering.......................................................... 3.43.2.5 Facing (or) Face turning.................................... 3.4

3.3 Casting ............................................................................. 3.53.3.1 Parts Made by Casting ...................................... 3.53.3.2 Mould in Casting................................................ 3.5

Contents C.7

3.3.3 Sand Casting ....................................................... 3.53.3.4 Sand Moulds ....................................................... 3.73.3.5 Steps/ Procedure for making sand mould for a two piece pattern ...................................... 3.83.3.6 Pattern .................................................................. 3.113.3.7 Special Casting Processes .................................. 3.113.3.8 Investment casting (Lost wax casting) ............. 3.133.3.9 Centrifugal Casting............................................. 3.14

3.4. Welding ........................................................................... 3.163.4.1 Types of Welding Processes ............................... 3.163.4.2 Welding Terminology .......................................... 3.163.4.3 Welding Processes................................................ 3.173.4.3.1 Oxy – Acetylene Welding (Gas Welding)....... 3.173.4.3.2 Electric Arc Welding Proceses ........................ 3.183.4.3.3 Arc Welding (or) Metal Arc Welding ............ 3.193.4.3.4 Gas Tungusten Arc Welding (GTAW) or Tungusten Inert Gas Welding (TIG) .............. 3.203.4.3.5 Gas Metal Arc Welding (GMAW) or Metal Inert Gas Welding (MIG) ..................... 3.213.4.4 Comparision of Gas Welding and Arc Welding 3.223.4.5 Electrodes ............................................................. 3.223.4.6 Resistance Welding.............................................. 3.223.4.6.1 Spot Welding .................................................... 3.22

3.5 Forming............................................................................ 3.233.6 Drawing............................................................................ 3.23

3.6.1 Classification of drawing operations ................ 3.243.6.1.1 Rod drawing ..................................................... 3.243.6.1.2 Wire drawing.................................................... 3.253.6.1.3 Tube drawing ................................................... 3.253.6.1.4 Deep drawing (or) cupping............................. 3.26

3.7 Extrusion.......................................................................... 3.27


3.8 Forging ........................................................................... 3.293.8.1 Classification of Forging .................................... 3.293.8.2 Smith die forging (or) Open die forging (or) Flat-die forging.................................................... 3.303.8.3 Power Hammers .................................................. 3.333.8.4 Power press (or) Forging press ......................... 3.343.8.5 Impression die forging (or) Closed die forging 3.353.8.6 Comparison between press forging and drop forging (Hammer Forging)................................. 3.35

3.9 Rolling of Metals ............................................................ 3.353.10 Engineering Materials.................................................. 3.383.11 Metals............................................................................. 3.39

3.11.1 Ferrous Metals................................................... 3.393.11.2 Non-Ferrous metals........................................... 3.40

3.12 Mechanical Properties of Materials ........................... 3.413.13 Steel................................................................................ 3.443.14 Effect of Alloying Additions on Steel ........................ 3.44

3.14.1 Effect of Manganese (Mn) in Steels ............... 3.443.14.2 Effect of Silicon (Si) in Steels ........................ 3.453.14.3 Effect of Chromium (Cr) in Steels ................. 3.453.14.4 Effect of Molybdenum (Mo) in Steels............. 3.453.14.5 Effect of Vanadium (V) in Steels ................... 3.453.14.6 Effect of Titanium (Ti) in Steels .................... 3.463.14.7 Effect of Tungsten (W) in Steels..................... 3.46

3.15 Stainless Steels ............................................................. 3.463.16 Tool Steels ..................................................................... 3.473.17 High Strength Low Alloy Steels (HSLA Steels)...... 3.473.18 Maraging Steels (Ultra High Strength Steels) ........ 3.483.19 Cast Iron........................................................................ 3.49

3.19.1 Composition of cast iron .................................. 3.493.19.2 Types of Cast Iron............................................ 3.49

Contents C.9

3.19.3. Grey cast iron................................................... 3.493.19.4 White cast iron.................................................. 3.503.19.5 Malleable Cast Iron.......................................... 3.513.19.6 Spheroidal Graphite Cast Iron: (Ductile iron (or) Nodular iron).................. 3.513.19.7 Alloy Cast Iron.................................................. 3.52

3.20 Non Ferrous Metals ..................................................... 3.533.21 Aluminium and Aluminium Alloys ............................ 3.533.22 Non - Metallic Materials............................................. 3.55

3.22.1 Polymers ............................................................. 3.553.22.2 Classification of Polymers................................ 3.553.22.2.1 Thermoplastic polymers................................. 3.563.22.2.2 Thermosetting Polymers ................................ 3.563.22.3 Commodity and Engineering Polymers .......... 3.57

3.23 Plastics ........................................................................... 3.573.24 Properties and Applications of Some Common Thermoplastics ............................................................... 3.593.25 Properties and Applications of Some Common Thermosetting Polymers ............................................. 3.60

Module 4: Fasteners and Power Transmission Devices & Mechanical Measurements

4.1 Fasteners.......................................................................... 4.14.2 Rivets................................................................................ 4.54.3 Belt Drives ...................................................................... 4.84.4 Wire Ropes/rope Drives ................................................. 4.184.5 Gear Drives ..................................................................... 4.214.5.2 Classification of Gears ................................................ 4.224.6 Gear Trains ..................................................................... 4.28

4.6.1 Classification........................................................ 4.284.6.2 Simple Gear Train.............................................. 4.28


4.6.3 Compound Gear Train ....................................... 4.304.6.4 Epicyclic Gear Train .......................................... 4.32

4.7 Couplings ......................................................................... 4.324.7.2 Features of shaft coupling ................................. 4.334.7.3 Box (or) Sleeve (or) Muff Coupling .................. 4.33

4.8 Clutches............................................................................ 4.344.8.2 Material for friction surface .............................. 4.344.8.3 Types of friction clutches ................................... 4.354.8.5 Single plate clutch .............................................. 4.354.8.6 Cone clutch .......................................................... 4.37

4.9 Brakes .............................................................................. 4.394.9.1 Difference between clutch and brake................ 4.394.9.2 Brake friction materials ..................................... 4.394.9.3 Working principle of braking system ............... 4.394.9.4 Purpose of braking system................................. 4.394.9.5 Types of brakes ................................................... 4.394.9.6 Drum brake.......................................................... 4.404.9.7 Disc brakes........................................................... 4.414.9.8 Mechanical brake ................................................ 4.424.9.9 Hydraulic brake................................................... 4.424.9.10 Pneumatic Braking System.............................. 4.444.9.11 Servo brake system ........................................... 4.454.9.11.1 Vacuum Servo Brakes ................................... 4.454.9.12 Electric brakes ................................................... 4.464.9.13 Antilock Braking Systems (ABS) .................... 4.47

4.10 Temperature Measurement ......................................... 4.484.10.1 Temperature Scales........................................... 4.484.10.2 Temperature Measuring Instruments ............. 4.484.10.3 Bimetallic Strip Thermometer ......................... 4.494.10.4 Thermocouple ..................................................... 4.494.10.5 Thermometer ...................................................... 4.51

Contents C.11

4.10.5.1 A Mercury-In-Glass Thermometer ................ 4.514.10.6 Resistance Temperature Detectors (RTD)....... 4.514.10.7 Thermistor .......................................................... 4.534.10.8 Pyrometers .......................................................... 4.54

4.11 Pressure Measurement Methods ................................ 4.604.11.1 Bourdon gauge (C-Type) - Principle............... 4.604.11.2 Bellows - Principle............................................ 4.614.11.3 Strain Gauge Pressure Transducer ................ 4.624.11.4 Piezoelectric Sensors ......................................... 4.634.11.5 Simple manometers........................................... 4.64

4.12 Velocity Measurement.................................................. 4.734.12.1 Pitot Tube .......................................................... 4.734.12.2 Hot Wire Anemometer ...................................... 4.75

4.13 Flow Measurement ....................................................... 4.774.13.1 Orifice Meter ...................................................... 4.774.13.2 Flow Nozzle........................................................ 4.814.13.3 Electromagnetic Flow Meters........................... 4.824.13.4 Venturimeter - Basic Principle........................ 4.834.13.5 Rotameter (Variable-area Meter) ..................... 4.84

4.14 Strain Measurement..................................................... 4.864.15 Force Measurement ...................................................... 4.874.16 Torque Measurement ................................................... 4.95

4.16.1 Cradled Shaft Bearing Type ........................... 4.954.16.2 Dynamometers.................................................... 4.964.16.3 Strain Gauge Type............................................ 4.98


MODULE 1

Thermodynamics

Thermodynamics: Systems, Properties, Process, State, Cycle, Internalenergy, Enthalpy, Zeroth Law, First law and Second Law of Thermodynamics,Basic Concept of Entropy, Properpties of ideal gas., Propereties of puresubstances, Steam, formation, Types of Steam, Enthalpy, Specific volume,Internal energy and dryness fraction of steam, use of Steam tables, Relatednumerical.

1.1 INTRODUCTION

Thermodynamics can be defined as the science which deals with thetransformation of energy and its effects on the properties of the substance. Theword ‘Thermodynamics’ originates from two Greek words, ‘therme’ (heat) and‘dynamis’ (motion or power). Thermodynamics can also be called as the scienceof 3E’s i.e., Equilibrium, Energy and Entropy.

Thermodynamics is a branch of Science that deals with the relationshipamong heat, work and properties of system which are in equilibrium withone another.

One of the most fundamental laws of nature is the conservation ofenergy principle. It states that during an interaction, energy can change fromone form to another but the total amount of energy remains constant. i.e.,Energy can neither be created nor be destroyed, but it can be transformedfrom one form to another. For example,

1. A ball falling from a tower picks speed as a result of its potentialenergy being converted to kinetic energy.

2. A person eating more and working less i.e., a person who has agreater energy input (food) than energy output (exercise) will gain weight (storeenergy in the form of fat)

The first law of thermodynamics is based on the conservation ofenergy principle. The second law of thermodynamics reveals that energy hasquality as well as quantity and actual processes occur in the direction ofdecreasing quality of energy. For example, a cup of hot coffee in a roomeventually cools, but a cup of cool coffee never gets hot by itself.

The first and second laws of thermodynamics emerged simultaneously inthe 1850s, by the works of James Prescott Joule, William Rankine, RudolphClausius and Lord Kelvin.

1.2 ROLE OF THERMODYNAMICS IN ENGINEERING AND SCIENCE

Thermodynamics plays a vital role in the field of Engineering andScience. They are explained as follows.

Thermodynamics is very important in the design, analysis anddevelopment of Engineering systems such as the power-producing systemslike engines, turbines etc. and the power-consuming systems like compressor,pumps, etc.

The working principle of many scientific devices like thermometer,thermistor, thermostat etc... involves thermodynamics.

Good understanding and careful applications of the fundamentals ofthermodynamics leads to improved design, increased efficiency, optimumoperating conditions and decreased levels of environmental pollution etc.

1.3 APPLICATIONS OF THERMODYNAMICSThermodynamic principles are applied in various fields of Engineering

and Technology. Some of the fields are

(i) Internal combustion engine (ii) Gas turbines

(iii) Air conditioner & Refrigerator (iv) Compressor

(v) Steam and Nuclear power plants (vi) Rockets

(vii) Jet engines etc.

Applications of thermodynamics are found in various aspects of life

eg: A commonly encountered thermodynamics application is, ‘thehuman heart, which is constantly pumping blood to all parts of the body.The movement of blood and various energy conversions generates heat, withinthe body. (normally called as body heat). The generated heat is constantlyrejected to the environment and the process is called as metabolic heatrejection. However, this heat rejection can be controlled by adjusting ourclothing to the environment conditions.

Other application of thermodynamics involves increasing the heattransfer rate like exhaust fan in a house, cooling fans in the computer CPUand fins provided on the engine cylinders etc.

1.2 Basic Mechanical Engineering - www.airwalkpublications.com

1.4 BASIC CONCEPTS

Some of the basic concepts of thermodynamics are discussed here. They are.

1.4.1 Units and DimensionsNowadays, the SI (System International) system of units has been used.

Any physical quantity can be characterised by dimensions. The arbitrarymagnitudes assigned to the dimensions are called Units. Some basicdimensions such as mass m, length L, time t and temperature T are selectedas primary (or) fundamental dimensions, while others such as velocity V,energy E, and volume V are expressed in terms of primary dimensions andare called secondary dimensions (or) derived dimensions.

A number of unit systems have been developed over the years. Butnowadays, the SI (System International)

system of units are widely used. The SI is simple and logical system. Somefundamental dimensions and their units in SI are given in the following table.

Dimensions Unit Symbol

Length metre m

Mass kilogram kg

Time second s

Temperature Kelvin K

Electric current Ampere A

Amount of light Candela cd

Amount of matter mole mol

1.4.2 Mass mIt is the quantity of matter contained in a body. Its unit is ‘kg’.

1.4.3 ForceIn SI, the unit for force is derived from Newton’s second law.

i.e., Force Mass Acceleration

F m a

In SI, the force unit is the Newton N and it is defined as the force required

to accelerate a mass of 1 kg at a rate of 1 m/s2. i.e., 1 N 1 kg m/s2

Thermodynamics 1.3

The weight of a body W is the force with which the body is attracted

towards the centre of the earth. It is the product of its mass m and the local

gravitational acceleration g, i.e., W mg, where g 9.80665 9.81 m/s2 atsea level. The mass m of a body will remain same, regardless of its locationin the universe. Its weight, however, will change with a change in gravitationalacceleration. A body will weigh less on top of a mountain since g decreaseswith altitude. At sea level, a mass of 1 kg will weigh 9.807 N.

1.4.4 PressurePressure is the normal force exerted

by a system against unit area of thebounding surface.

Pressure, P ForceArea

FA

The unit for pressure in the SI systemis the Pascal Pa, which is the force of 1

N acting on an area of 1 m2.

1 Pascal 1 Pa 1 N/m2

The unit of pascal is very small. Hence, very often, kilo-pascal (kPa)(or) mega-pascal (MPa) is used. Sometimes bar and standard atmosphereare also used which are not within the SI system of units.

1 bar 105 Pa 100 kPa 0.1 M Pa and

Standard atmosphere 1 atm 101.325 kPa 1.01325 bar

Note: The pressure P at a point in a fluid in equilibrium is the same in alldirections.

Most of the instruments indicate pressure relative to the atmosphericpressure where as the pressure of a system is its pressure above zero (or)relative to a perfect vacuum.

The pressure relative to the atmosphere is called gauge pressure.

The pressure relative to a perfect vacuum is called absolute pressure.

Absolutepressure

Gauge pressure Atmospheric pressure

The SI unit prefixes are used in all branches of engineering. Some standardprefixes in SI units for Mechanical Engineering branch are given in the table.

SY STEM

F n

A

Definition of pressure

Fig. 1.1


Multiple Prefix

1012 Tera, T

109 Giga, G

106 Mega, M

103 Kilo, k

10 3 milli, m

10 6 micro,

10 9 nano, n

10 12 pico, p

1.4.5 TemperatureIt is defined as the degree of intensity of heat or measure of hotness

or coldness of a body.

Temperature - Intensive property - C or K (Centigrade or Kelvin).

Eg: 30C 30 273 303 K

Normally, t is for C and T is for K.

1.4.6 Volume, Specific Volume and Density

Volume V is the space occupied by a substance and is measured in m3.

Volume Total volume V in m3

Specific volume: v

The specific volume v of a substance is defined as the volume per unit

mass and is measured in m3/kg.

Specific volume Volume

mass

Vm

v

Its unit is m3/kg

Density : It is the mass of the substance per unit volume. Its unit is

kg/m3.

Density is the mass per unit volume of a substance and is given

in kg/m3. mV

Thermodynamics 1.5

1.4.7 EnergyEnergy is the ability of the physical system to perform work. Energy

exists in several form such as heat energy, kinetic energy, potential energy,light energy, electrical energy, thermal energy and so on. However, theseenergy transformations are constrained by a fundamental principle known asthe “Conservation of Energy”, Principle. One way to state this principle is

“Energy can neither be created nor destroyed, but may transform fromone form into another form”.

Another approach is that,

“the total energy of the isolated system remains constant”.

In S.I units, energy is represented as Joule (J)

1 Joule 1 Nm kg m

s2 m 1 kg m2

s2

Different forms of energySome forms of energy are explained as follows.

1. Potential Energy, 2. Kinetic Energy, 3. Pressure Energy,4. Internal Energy

1. Potential Energy: (P.E)It is the energy possessed by any system (or body) by virtue of its

height (by virtue of its position above the ground level).

P.E mg Z or mgh in J where Z ht above ground level.

P.E gZ in J/kg

2. Kinetic Energy: (K.E)It is the energy possessed by any system (or body) by virtue of its

velocity.

K.E 12

mV2 in J 12

mC2 in J

where V Velocity in m/sec.

K.E V2

2 in J/kg

C2

2 in J/Kg

To differentiate from volume V, Velocity ‘V’ can be denoted by ‘C’also.


3. Pressure EnergyIt is the energy possessed by any system (or body) by virtue of its

pressure.

Pressure energy Pv N

m2 m3

kg

Nmkg

J kg

Sp. pressure energy Pv in J/kg

4. Internal Energy: (U) in JIt is the energy possessed by any system (or body) due to its molecular

arrangement and motion of molecules. ‘u’ is the specific internal energy inJ/kg.

1.4.8 PowerThe rate of energy transfer (or) storage is called power. The unit of

power is Watts (W), kiloWatts kW (or) MegaWatts (MW).

Power Workdonetime taken

Nm

s

1 W 1 J/s 1 Nm/s; 1kW 1000 W

1.4.9 Dimensional HomogeneityIn engineering, all equations must be dimensionally homogeneous. i.e.,

every term in an equation must have the same unit.

Analyse the following equation, E 35 kJ 8 kJ/kg

where E is the total energy in kJ. Since the two terms on the right hand sidedo not have the same units, they can not be added to obtain the total energy.Multiplying the last term 8 kJ/kg m kg by mass will eliminate thekilograms in denominator, and the whole equation will become dimensionallyhomogeneous i.e., every term in the equation will have the same unit.

i.e., E 35 kJ 8 m kJ

1.5 CLASSICAL AND STATISTICAL THERMODYNAMICS

Normally, a substance consists of a large number of particles calledmolecules and the properties of substance naturally depend on the behaviourof these particles. There are two points of view from which the behaviour ofparticles can be studied. They are

Thermodynamics 1.7

1.5.1 Macroscopic viewpoint (Classical thermodynamics)In this point of view, a certain amount of matter is considered,

neglecting the events occurring at the molecular level. It does not require aknowledge on the behaviour of individual particles. It provides a direct andeasy way to the solution of engineering.

In general, macroscopic point of view is just like, “inventing a deviceor using a instrument, to measure the quantities directly, instead of findingit out through calculation, derivation and formulas”.

eg: ‘Pressure gauge’ is used to measure pressure directly withoutcalculating force on a given unit area.

Likewise, temperature can also be measured using thermometer.

However, macroscopic point of view is limited to those effects thatcan be perceived by human senses.

1.5.2 Microscopic viewpoint (Statistical thermodynamics)In this point of view, behaviour of each molecule has to be studied,

to find out the behaviour of the entire substance. Each molecule at a giveninstant has certain position, velocity and energy, and for each molecule theseparameters changes frequently as a result of collisions. And also, these effectscannot be perceived by human senses.

In general, microscopic point of view requires large number of dataand equations to find out the quantities through calculation and derivation.

eg: Impulse, Enthalpy, Entropy, Gibbs function, Helmhotz function etc.

Classical thermodynamics Statistical thermodynamics

1. It is also called as Macroscopicviewpoint.

It is also called as Microscopicviewpoint.

2. Behaviour of each moleculeneed not to be determined.

Behaviour of each molecule need to bedetermined.

Viewpo in ts

M acroscopic Viewpoint (or) C lassical Therm odynam ics

M icroscopic V iewpoint (or) s ta tis tica l Therm odynam ics


Classical thermodynamics Statistical thermodynamics

3. Molecular effects can beperceived by human senses.

Molecular effects cannot be perceivedby human senses.

4. Classical thermodynamics,however can be derived fromstatistical thermodynamics.

Statistical thermodynamics cannot bedetermined using classicalthermodynamics.

5. eg: Pressure, Temperature eg: Enthalpy, Entropy

1.6 THERMODYNAMIC SYSTEMSThermodynamic system is known as a space

or constrained area upon which our attention isconcentrated on. It is the region to be studied. Thesystem has boundary (or) boundary line. Anythingoutside the boundary is called surroundings. In otherwords, the system and surroundings are separated byboundary. (Boundary may be real or imaginary).

1.6.1 Types of systemThe systems are classified into:

1. Closed system, 2. Open system, 3. Isolated system.

System may be considered to be closed (or) open, depending onwhether a fixed mass (or) a fixed volume in space is chosen for study.

1.7 CLOSED SYSTEMA closed system, is also known as a

control mass, in which only the energy istransferred from the surroundings into the systemand from the system to the surroundings. But thereis no mass transfer involved in closed system.

It consists of a fixed amount of mass andno mass can cross its boundary.

i.e., no mass can enter (or) leave a closedsystem as shown in Fig. 1.3 But energy, in theform of heat (or) work, can cross the boundary.And the volume of a closed system does not have to be fixed.

Eg. Piston and cylinder without valves Fig. 1.4.

SurroundingsBoundary

Surroundings

System

Fig. 1.2

C LOS EDSYS TEM

m = constan t

ene rg y Y ES

M ass N O

M ass can not c ross the bound aries of a closed sys tem , b ut ene rg y can.

Fig. 1.3

Thermodynamics 1.9

1.8 OPEN SYSTEMAn open system (control volume) is

a properly selected region in space.

i.e., Both energy and mass aretransferred from the system to surroundingsand from the surroundings to system.

The boundaries of a control volumeare called as control surface as shown inFig. 1.6 and they can be real (or) imaginary.

(e.g) 1. Air compressor (Fig 1.5), 2. I.C.Engines, 3. Turbines, 4. Boilers, 5. Centrifugal pumps, 6. Nozzles.

Note: Most of the engineering devices areopen systems.

1.9 ISOLATED SYSTEMIn isolated system, neither energy nor mass

transfer occurs between the system and thesurroundings (Fig 1.7).

Practically, no system is isolated.

Closed system(or Non flow system )

Systemboundary

Energyin

Energyout

W ork in

N o m ass tra nsfe rS

u rroundings

System

L .P.

a ir in

Open system(or)Steady flow system

System(A ir

com pressor)

H P

a ir out

W ork in

Fig. 1.4 Fig. 1.5

C ontro l su rfa ce

Bo th m ass and energy can cross the boundaries o f a con tro l volum e.

C O N TRO LVOLU M E

m ass YES

energy YES

Fig. 1.6

Surrounding

No m ass transferNo energy trasfe r

System

Fig. 1.7


1.10 HOMOGENEOUS AND HETEROGENEOUS SYSTEMS

A system, which contains only a single phase, is called asHomogeneous system.

Eg. Mixture of air and water vapour.

A system, which consists of more than one phase, is called asHeterogeneous system.

Eg. Mixture of water and steam.

1.10.1 Pure substanceA Substance with homogeneous and invariable chemical composition

throughout its mass, even though change of phase takes place is called aspure substance. It is a one component system and may exist in one or morephases. Eg. Water.

1.10.2 State postulateThe number of independent intensive properties are required to fix the

state of the system is given by state postulate.

For a pure substance, this is called as two property rule. i.e., twointensive properties are required to fix the state of the pure substance.

Number of independent intensive properties can be found analyticallyusing Gibb’s rule

P F C 2

P - No. of phase, C - No. of components

F - No. of independent intensive properties

eg: For single liquid (or) single gas

P 1, C 1

F C P 2 1 1 2 2

1.11 PROPERTY

It is a characteristic of the system. The system is identified by somequantities like temperature, pressure, volume etc.

1. Intensive Property: The properties which do not depend on the massof the system, are called intensive property.

Intensive property - Independent of the mass of the system: Example- Temperature, pressure, density, surface tension, specific gravity, etc.

Thermodynamics 1.11

2. Extensive Property: Those properties which depend on the mass of thesystem are called extensive property. Example: Volume (If mass is more,then volume will be more) Force, eg: electric charge, weight, energy (heatand work) magnetization. (Fig 1.8).

The extensive property when divided bymass will become intensive property.

(e.g)

Volume m3Extensive

/ mass Specific volume m3/kgIntensive

Enthalpy kJExtensive

/mass Specific enthalpy kJ/kgIntensive

Generally, upper case letters are used todenote extensive properties (with mass ‘m’being a major exception), and lowercase lettersare used for intensive properties (with pressureP and temperature T being the obviousexceptions).

1.12 STATE, PATH AND PROCESSConsider a system does not

undergoing any change. At this point, allthe properties can be measured (or)calculated throughout the system, whichgives us a set of properties that completelydescribes the condition, or the state, of thesystem. At a given state, all the propertiesof a system have fixed values. If the valueof even one property changes, the statewill change to a different one. A systemis shown at different states in the Fig. 1.9.

Thermodynamics deals only withequilibrium states. The word equilibriumimplies a state of balance. In an equilibrium state, there are no unbalancedpotentials (or driving forces) within the system.

Exte ns ivep rop erties

In tens ivep rop erties

C riteria to diffe rentiate intensiveand extens ive properties .

Fig. 1.8

m = 3 k gT 1 = 2 0 Co

= 2 .5 m 3V 1

m = 3 k gT 2 = 2 0 Co

= 3 .5 m 3V 2

( ) State 1a ( ) State 2b

A system at two different states

Fig. 1.9


There are many types ofequilibrium, and a system is not inthermodynamic equilibrium unless theconditions of all the relevant types ofequilibrium are satisfied.

For example, a system is inthermal equilibrium if the temperatureis the same throughout the system, asshown in Fig. 1.10.

i.e., the system experiences notemperature difference, which is thedriving force for heat flow.

Any change that a systemundergoes from one equilibrium state toanother is called a process, and the seriesof states through which a system passesduring a process is called the path of theprocess. To describe a process completely,one should specify the initial and finalstates of the process, as well as the pathit follows.

Refer the Fig. 1.11

At state (1), the pressure is P1 and

the volume is V1. Similarly at state (2),

the pressure is P2 and the volume is V2.

In between 1 and 2, there are so many statesA, B, C, , G.

1.13 CYCLEThe series of processes whose end

states are identical, is called as cycle. ReferFig 1.12.

12 The process A21 The process B121 Cycle

( ) Befo rea ( ) A fterbA c losed sys tem reaching therm a l

equ ilibrium

Fig. 1.10

XX

X

X

XX

X

P 2

P 1

V 2 V 1

V

1

2G

FE

D

CB A

S ta te2 S ta te 1

1- S ta teA - S ta teG - S tate2- S ta te

1 - 2 = P roce ss

Fig. 1.11

2

B

A

1

V

P

Fig. 1.12

Thermodynamics 1.13

The process A and the process B together form a cycle (or)thermodynamic cycle, because, the system has returned to its initial state atthe end of the process. i.e., for a cycle, the initial and final states are identical.

1.14 THERMODYNAMIC EQUILIBRIUM

When mechanical equilibrium, chemical equilibrium and thermalequilibrium are satisfied, then the system is said to be in thermodynamicequilibrium.

When any property remains same with respect to time, it is called asequilibrium or steady state.

Eg. Temperature and pressure at all points of a system are same onlyif the system exists in thermodynamic equilibrium.

1. Mechanical EquilibriumIf the system’s pressure remains same with respect to time, then the

system is said to be in mechanical equilibrium.

2. Chemical EquilibriumIf the system’s chemical composition remains same with respect

to time, then the system is said to be in chemical equilibrium.

3. Thermal EquilibriumIf the system’s temperature remains same with respect to time, then

the system is said to be in thermal equilibrium.

1.15 QUASI-STATIC (OR) QUASI–EQUILIBRIUM PROCESS

When a process proceeds in such a manner that the system remainsinfinitesimally close to an equilibrium state at all times, it is called aquasi-static, (or) quasi-equilibrium process. A quasi-equilibrium process canbe viewed as a sufficiently slow process that allows the system to adjustitself internally so that properties in one part of the system do not changeany faster than those at other parts.

A quasi-static process is an idealized process and is not a truerepresentation of an actual process. But many actual processes closelyapproximate it and they can be modeled as quasi-equilibrium with negligibleerror. Engineers are interested in quasi-equilibrium processes for two reasonsas follows.

1. They are easy to analyze.

2. Work-producing devices (like engine) deliver maximum work whenthey operate on quasi-static processes.


Therefore, quasi-static processes serve as standards to which actualprocesses can be compared.

Consider a system of gas contained in a cylinder Fig 1.13(a). Thesystem initially is in equilibrium state P1, V1 and t1. The weight on the piston

is balanced by the pressure of the gas. If the weight is removed, there willbe an unbalanced force between the system and surroundings, and under thegas pressure, the piston will move up till it hits the stops. The system comesto an equilibrium state - P2, V2 and t2. But the inter-mediate states are

non-equilibrium states and we cannot define its properties. So these two states(1 and 2) are joined by dotted line.

Now instead of single weight, we can place many number of smallweights on the piston Fig 1.13(b). If we remove weights one by one very

Stops

C ylinder

Fina l sta te

W eigh t

In itials tate

Systemboundary

P 1

P 2

1

2

P(P ressure )

V 1 V 2

V(Volum e)

Stops

C ylinder

W eigh ts

Systemboundary

P I S T O N

P V T1 1 1

P(Pressure)

Equilibrium sta tes

Q uas i-sta ticprocess

V (Vo lume)

Fig. 1.13 Quasi-Static process (or) Quasi-Equilibrium process

Thermodynamics 1.15

slowly from the piston, then the piston will move upward very slowly. Thechange of state of system from equilibrium state is infinitesimally small. Soevery state traced by the system is an equilibrium state. The process, joinedby all the equilibrium states passed by the system is called quasi-staticprocess. [‘Quasi’ means ‘almost’].

Quasi-static process is a reversible process.

A process which can be reversed in direction and the system retracesthe same equilibrium states is known as reversible process.

Infinite slowness is the characteristic feature of a quasi-static process.

1.16 PATH FUNCTION AND POINT FUNCTIONRefer the Fig 1.14. In this fig, there

are many quasi-static paths such asA, B or C from state 1 to state 2. Sincearea under each curve represents the workfor each process, the amount of workinvolved in path A, B and C are not equal.The work is not a function of the endstates of the process but it depends on thepath the system follows from state 1 tostate 2. Hence, work is called a path

function and d– W is inexact (or)imperfect differential.

Thermodynamic properties like pressure, volume and temperature arepoint functions, because, for a given state, there is a definite value for eachproperty. The change of property of a system is independent of the path, thesystem follows during the change of state and it depends only on the initialand final states of the system. Hence dV (or) dP is an exact (or) perfectdifferentials.

For example, the integration of property is

V1

V2

dV V2 V1

i.e., Change in volume depends only on the end states of the systemirrespective of the path of the system.

P 1

P 2

A BC

2

V 1 V 2V

W = 10 kJA

W = 8 kJB

W = 6 kJC

p

W ork- a path function

Fig. 1.14


But, workdone in a quasi-static process between two given statedepends on the path followed. And the integration of work transfer is

1

2

dW W2 W1

Rather, 1

2

dW W1 2

1.17 ZEROTH LAW OF THERMODYNAMICS

If the systems A and B are inthermal equilibrium separately and A andC are in thermal equilibrium separately,then B and C should be in thermalequilibrium.

This law is the basis for temperaturemeasurement.

1.17.1 Equality of temperatureWhen two bodies at different temperatures are brought into contact,

after some time, they attains a common temperature and are then said to existin thermal equilibrium.

The zeroth law was first formulated by R.H Fowler. This law explainsthe fundamental principles of thermodynamics, but was recognised after theformulation of first and second laws of thermodynamics. So it was named as‘zeroth law’ since it should have preceded the first and second laws ofthermodynamics.

1.18 MEASUREMENT OF TEMPERATURE

The temperature of a body can be measured, by using a device knownas thermometer. A common thermometer consists of a small amount ofmercury in an evacuated glass tube. There are five different kinds ofthermometer.

1. Constant volume gas thermometer2. Constant pressure gas thermometer3. Electrical resistance thermometer4. Thermocouple 5. Mercury-in-glass thermometer

Therm

al e

quilib

rium

Thermal e quilibrium

B C

A

Fig. 1.15

Thermodynamics 1.17

1.18.1 ThermometryThermometry is the science and practice of temperature measurement.

Any measurable change in the thermometric probe (eg: extension of mercuryin the tube of mercury-in-glass thermometer) can be used to mark temperaturelevels, that should later be calibrated as internationally agreed unit (likecelsius, Fahrenheit, Kelvin, etc...)

Thermometry is broadly divided into two subfields.

(i) Contact thermometry (ii) Non-contact thermometry

1.18.2 Applications of thermometryTemperature is one of the most measured physical parameters in

science and technology, but thermometry is not only applied to measuringtemperature, but also an indirect measure of many material properties likethermal capacities, relative humidity (bymeans of wet bulb (or) dew pointtemperatures) and enthalpy changes (bythermal analysis).

1.18.3 Triple point of waterIt is a point at which three phases

(solid, liquid and gas) i.e., ice, liquidwater and water vapour exist inequilibrium with one another.

The temperature at which, thesestates exist in equilibrium are 273.16 Kor 273.16 degrees Kelvin.

1.18.4 Comparison of Thermometer

Thermometer Thermometric property

1. Constant volume gas thermometer Pressure P2. Constant pressure gas thermometer Volume V

3. Electric resistance thermometer Resistance R4. Thermocouple Thermal e.m.f E5. Liquid-in-glass thermometer Length L

Triple P oin t

Wa te r (L iquid)

Wa te r Vapor(G as)

Ice(So lid)

Pre

ssur

e

Abou t 610 Pa

273 .16 K(0 .01 C )o

Tem perature

Fig:1.16.Triple Point of Water


1.19 EQUATION OF STATEThe equation which relates the properties P, v and T is known as an

Equation of state.

ie f P, v, T 0

The simplest form of equation of state for the ideal gas is given below.

Equation of state: Pv R

T [ R

universal gas constant in kJ kg mol.K

and v molar volume in m3 kg mol ]

Also: Pv RT [ R characteristic gas constant in kJ kg K]

v specific volume in m3 kgThe above equation - Equation of state is also called Characteristic gas equation.

1.20 CELSIUS TEMPERATURE SCALECelsius temperature scale is related to the kelvin scale by

t T 273.15

where t - celsius temperature, T - kelvin temperature

eg: At steam point, T 373.15 K i.e., kelvin temperature of steam at 1 atmis 373.15 K

t 373.15 273.15

t 100 C

Solved ProblemsProblem 1.1: The temperature t on a thermometric scale is defined in termsof a property k by the relation t a ln k b where a and b are constants.

The value of k are found to be 1.83 and 6.78 at the point 0C and steam

point 100C. Determine the temperature corresponding to a reading of kequal to 2.42 on thermometer.

Given:

k 1.83 when t 0; k 6.78 when t 100 k 2.42 when t ? t a ln k b

(i) 0 a ln 1.83 b

0.6043 a b 0 ...(1)

(ii) 100 a ln 6.78 b

1.914 a b 100 ...(2)

Thermodynamics 1.19

on solving, a 76.35 b 46.14

(iii) t 76.35 ln 2.42 46.14

76.35 0.8838 46.14 t

t 21.335 C

Problem 1.2: A New scale N of temperature is divided in such a way thatthe freezing point of ice is 100N and the boiling point 400N what is thetemperature reading on the new scale when the temperature in the celsiusscale is 150C. At what temp both the celsius and New temperature scalewould be same.

Given:

New temperature scale Celsius scale

Ice point 100N 0C

Boiling point 400N 100C

temperature = ? when 150C

Let us take celsius scale on the x - axis and new temperature scale onthe y - axis

Slope, m y

x N

C

400 100

100 0

300100

3

Slope-intercept form equation, is

y mx b (y - intercept)when x 0

N mc b (N 100 when C 0)

N 3C 100

Temperature on the new scale, when temp or celsius scale is 150C N 3 150 100 550 N

050 150100

oC

100

200

300

400

N


Also, the temp at which both scale readings are same.

At particular temperature, N C

From the relation N 3C 100

C 3C 100

C 3C 100 2C 100

C 50 C or N 50 N

So, at 50 both the scale would read the same value.

1.21 WORKIn mechanics, work is said to be

done by a force, as it acts upon a body,which moves the body in the direction ofthe forceWork Force distance moved F x

In thermodynamics, work is said tobe done by a system, if sole effect onthing’s external to system can be reducedto the raising of weight.

Refer Fig.1.17

Let A Cross sectional area of the

piston in m2

P = Pressure of gas at any time.

dl Distance moved by the pistonunder gas pressure.

Work done Force distance travelled

PA dl PAdl Pdv

Consider unit mass.

Then W Pdv [where v specific volume]

When the reversible process (Quasi-static process) takes place betweenstates 1 and 2, then

P1

P2

1

2

V 1 V 2dv

Quasi s ta ticp rocess

Fig. 1.17

Thermodynamics 1.21

Work done 1

2

Pdv

Therefore the work done by thegas during any reversible process is givenby the area under the Pv diagram.(Shaded area).

When the work is done by thesystem, it is positive work Fig 1.18(a).

When the work is done on thesystem, it is negative work Fig 1.18(b).

1.21.1 Work Transfer - a Path Function

Let us consider a system shown inFig 1.19.

The system moves from aequilibrium state 1 to a final equilibriumstate 2 through two different paths

1 A 2 and 1 B 2. The processes arequasi-static.

Since area under each curverepresents the work for each process.1 A 2 gives one value of work and

1 B 2 gives another value of work. Sothe work does not depend on the end stateof the process. But it depends on the pathof the system. So the work is called path

function. Hence d W is an inexact orimperfect differential.

P1

P2

1

2

V 1 V 2dv

Quasi s ta ticp rocess

Fig. 1.19

(a)

(b)

Fig. 1.18


W1 2 1

2

d W 1

2

Pdv

Example: Change in volume V2 V1

Change in pressure P2 P1 ; Change in temperature t2 t1

So the properties of the system depend only on end states andindependent of the path the system follows. So the differentiation of pointfunctions are exact or perfect differentials.

Note:1. Thermodynamic properties are point functions, since they depend

only on the end states.

2. For a cyclic process, the initial and final states of the system arethe same, and hence, the change in any property is zero.

O dV 0, O dP 0, O dT 0

where the sign O denotes the cyclic integral for the closed path. Therefore,

cyclic integral of a property is always zero.

1.22 HEATHeat is defined as the form of

energy that is transferred between twosystems (or a system and its surroundings)by virtue of a temperature difference

The transfer of heat between twobodies in contact is called conduction.The transfer of heat between a wall anda fluid system in motion is calledconvection. The transfer of heat betweentwo bodies separated by empty space (or)gases through electromagnetic waves iscalled radiation.

The temperature difference is the driving force for heat transfer. Thelarger the temperature difference, the higher is the rate of heat transfer ReferFig 1.20.

Normally, the heat is transferred from high temperature system to thelow temperature system. The transfer of heat into a system is frequently

No hea ttransfer

8 J /sHeat

1 6 J /sHeat

Room air 25 Co

25 Co

Water

15 Co 5 Co

Water Water

Fig. 1.20

Thermodynamics 1.23

referred to as heat addition and is taken as positive. The transfer of heatout of a system is called heat rejection and is taken as negative.

Refer Fig. 1.21. A process during which there is no heat transfer iscalled an adiabatic process. An adiabatic process should not be confusedwith an isothermal process. Even though there is no heat transfer during anadiabatic process, the energy content and thus the temperature of a system canstill be changed by other means such as work.

As a form of energy, heat has energy units, kJ. The amount of heattransferred during the process between two states (states 1 and 2) is denoted byQ1 2. Heat transfer per unit mass of a system is denoted by q.

ie q Qm

kJ/kg

The rate of heat transfer i.e., the amount of heat transferred per unittime is denoted by Q

, where the overdot stands for the unit derivative (or)

“per unit time”. The heat transfer rate Q has the unit kJ/s, which is equivalent

to kW.

1.22.1 HEAT TRANSFER – A PATH FUNCTIONHeat transfer is also a path function. It depends on the path the system

follows. It does not depend on the end states. So dQ is inexact differential.

ADIABATICSYSTEM

Insulation

� ��

During an adiabatic process, a system exchanges

no hea t w ith its surroundings.

�=3 0 kJm =2 k g

3 0 kJheat

�=6 k Wq = 15 k J /k g

t =5s

The relationships among , and q � ��,

Fig. 1.21 Fig. 1.22


Refer Fig. 1.23. Area under TS diagram gives the heat transfer. Inthe processes A and B, the end states are same, but the path are different. Sothe heat transfer in two processes A and B are different.

Q1 2 1

2

d Q 1

2

Tds

Note

W1 2 W1 W2; W1 2 Work transfer from 1 to 2.

Q1 2 Q1 Q2; Q1 2 Heat transfer from 1 to 2.

s entropy (which will be explained later).

ds change in entropy.

Path functions have Inexact differentials designated by the symbol

(or) d–. Therefore, a differential amount of heat (or) work is represented by

Q or d– Q, W or d– W, respectively, instead of dQ (or) dW.

1.23 HEAT CAPACITY

Heat capacity (also called as the thermal capacity) is the ratio of theheat added to (or) heat removed from an object to the resulting temperaturechange.

It can also be defined as a product of mass and specific heat.

C mc kg

JkgK

JK

Fig. 1.23

Thermodynamics 1.25

It is denoted by capital letter C, Cp, Cv

The S.I unit of heat capacity is Joule/Kelvin.

1.24 REVERSIBLE AND IRREVERSIBLE PROCESS

Reversible processA process in which both system and its surroundings can be

simultaneously returned to their initial states after the process has beencompleted.

Irreversible processA process in which both system and its surroundings cannot be

simultaneously returned to their initial states after the process has beencompleted.

Reversible process Irreversible process

(i) Reversible process is an idealprocess

Irreversible process is a naturalprocess. (i.e.) all the processesoccurring in nature are irreversible.

(ii) Reversible process attainsequilibrium state at all the stages ofthe operation.

Irreversible process is in equilibriumonly at the initial and final stages ofoperation.

(iii) It is a extremely slow process. It occurs at measurable speed.

(iv) It takes infinite time for theprocess to occur.

It takes place in finite time.

(v) Workdone by a reversibleprocess is greater than thecorresponding irreversible process.

Workdone by an irreversible processis lesser than the correspondingreversible process.

(vi) It does not increase the entropyof the universe.

It increases the entropy of theuniverse (system & surroundings).

1.25 FIRST LAW OF THERMODYNAMICS

The first law of thermodynamics, also known as the conservation ofenergy principle, which provides a basis for studying the relationships amongthe various forms of energy and energy interactions. The first law ofthermodynamics states that energy can be neither created nor destroyed; itcan only change forms.


A rock at some elevation possesses some potential energy, and part ofthis potential energy is converted into kinetic energy as the rock falls.Experimental data show that the decrease in potential energy mg Z exactly

equals to the increase in kinetic energy m V2

2 V12

2 when the air resistance

is negligible;

The increase in the energy of a potato in an oven is equal to theamount of heat transferred to it.

According to First Law of Thermodynamics‘Heat can be converted into work and work can be converted into heat.’

For a closed system, Net heat transfer = Net work transfer.

O d– Q O d– W First law of Thermodynamics for closed system.

O is the cyclic integral for closed path. (i.e for cycle).

1.26 JOULES EXPERIMENT

Consider a closed system as shown in Fig 1.25. It consists of a knownmass of water, insulated container having thermometer and a paddle wheel.

Now a certain amount of work is done on the system by means of apaddle wheel. The work can be measured as

W1 2 Weight distance travelled

Fig. 1.24

Thermodynamics 1.27

The system was initially attemperature t1 and after work transfer, the

system temperature rises to t2. The

pressure is one atmospheric pressure. Theprocess 1-2 is shown in Fig 1.26.

Now the insulation is removed. Theheat transfer takes place between systemand surroundings till the system returns toits original temperature t1. The heat can be

measured as Q2 1 mCpt

The system executes a cycle 1 2 1.

Now we can see W1 2 is equal to Q2 1. It has been found that this

work W1 2 is always proportional to the heat Q2 1 and the constant of

proportionality is called the Joule’s equivalent (or) the mechanical equivalentof heat.

In this example, only two energy transfer occurs. If the cycle has somany heat and work transfers, we can get the same result as

Fig. 1.25

Fig. 1.26


Wcycle JQcycle

where J Joule’s equivalent = 1 Nm/J

Since it is unity, we can write this

Wcycle Qcycle

i.e. O d– W O d– Q

1.27 PERPETUAL MOTION OF MACHINE OF FIRST KIND

PMM 1PMM 1 is an imaginary engine which

develops work without any heat input. Therefore, thePMM 1 violates the I law of thermodynamics, sinceI law states that Q W in a cycle.

Hence this machine is impossible.

1.28 FIRST LAW FOR A CLOSED SYSTEM UNDERGOING CHANGE OF STATE

1.28.1 Internal Energy

Wcycle Qcycle applies only to systems undergoing cycles and

the algebraic sum of all energy transfer across system boundaries is zero. Butif a system undergoes a change of state during which both heat transfer andwork transfer are involved, the net energy transfer will be stored (or)accumulated within the system.

If Q heat transfer and

W work transfer, then the net energy transfer Q W will be storedin the system. Energy in storage is neither heat nor work and is given thename Internal Energy (or) simply energy of the system.

Therefore Q W E

where E is the increase in the energy of the system.

(or) Q E W

The change in energy of the system during a process is the sum of the changesin the internal, kinetic and potential energies and can be expressed as

No Hea t Input

Wo

rk o

utp

ut

ENGINE

Fig. 1.27

Thermodynamics 1.29

E U KE PE

where U m u2 u1; K.E 12

m V22 V1

2; PE mg Z2 Z1

When the initial and final states are specified, the values of the specificinternal energies u1 and u2 can be determined directly from the property tables

(or) thermodynamic property relations.

According to the first law of thermodynamics, when the systemundergoes a process, both heat transfer and work transfer takes place. Thenet energy transfer is stored within the system and is known as stored energy(or) total energy of the system.

Q W E First law of thermodynamics for process.

E P.E K.E UWhen P.E 0; K.E 0, then E U

Then Q W U. This is I law for process when P.E = 0 & K.E = 0. In otherwords, Q W U

1.28.2 Energy - A property of the systemConsider a system which changes its state

from state 1 to state 2 by following the path A,and returns from state 2 to state 1 by followingthe path B. (Refer the following Fig 1.28). Sothe system undergoes a cycle. writing the firstlaw for path A

QA EA WA ... (i)

and for the path B

QB EB WB ... (ii)

The processes A and B together constitute a cycle, for which

Wcycle Qcycle

or WA WB QA QB

or QA WA WB QB ... (iii)

From equations (i), (ii) and (iii), it yields

Fig. 1.28


EA EB ...(iv)

Similarly, if the system returns from state 2 to state 1 by followingthe path C instead of B,

then EA EC ... (iv)

From equations (iv) and (v)

EB EC ...(v)

Therefore, it is proved that change in energy between two states of asystem is the same, whatever path the system may follow in undergoing thatchange of state. Therefore, energy has a definite value for every state of thesystem. Hence, it is a point function and a property of the system.

SOLVED PROBLEMS

Problem 1.3: During a cycle, the sum of all heat transfers is 200 kJ. Thesystem completes 2 cycles per sec. Complete the following table, showingmethod for each item and compute the net rate of work output in kW.

Process Q in kW W in kW E in kW

a b 0 36 –

b c 350 0 –

c d 35 – 610d a – – –

Solution

Process a b, Q W E (I law of thermodynamics for process)

i.e. Q E W

0 E 36

So, Eab 36 kW

Process b c, Q E W

350 E 0 Eb c 350 kW

Process c d, Q E W

35 610 W Wc d 575 kW

Thermodynamics 1.31

Process d a, Qcycle 200 kJ

The system completes 2 cycles/sec, so

Q 200 2 kJ

1sec

400 kW

Q 400 kW Qa b Qb c Qc d Qd a

400 0 350 35 Qd a

So, Qd a 715 kW

Also O dE 0, since cyclic integral of any property is zero.

i.e. Ecycle 0

Ea b Eb c Ec d Ed a 0

36 350 610 Ed a 0

Ed a 296 kW

In Process d a, Q E W

715 296 W

Wd a 1011 kW

Process Q W E

a b 0 36 36b c 350 0 350

c d 35 575 610d a 715 1011 296Total Q 400 W 400 E 0

Problem 1.4: When a system is taken from state a to state b along pathacb, 84 kJ of heat flow into the system and the system does 32 kJ of work.

1. How much will the heat that flows into the system along the pathadb be, if the work done is 10.5 kJ?

2. When the system is returned from b to a along the curved path,the work done on the system is 21 kJ. Does system absorb orliberate heat? and how much of the heat is absorbed or liberated?


3. If Ua 0 and Ud 42 kJ, find the heat absorbed in the processesad and db.

Solution: Given Qacb 84 kJ [+ sign heat flows into the system]

Wacb 32 kJ [+ sign work is done by the system]

Qacb Ub Ua Wacb

84 Ub Ua 32

Ub Ua 52 kJ

1. To Find Qadb

Qadb Ub Ua Wadb

52 10.5 62.5 kJ

2. To Find Qba (along curved path)

Qba Ua Ub Wba

Wba 21 kJ [Since work is done on the system]

Qba 52 21 73 kJ[ . . . Ub Ua 52

Ua Ub 52 ]

So the heat is liberated (73 kJ) from the system to surroundings.

3. If Ua 0 and Ud 42 kJ, Then Qad ?; Qdb ?

Wadb 10.5 kJ Wadb Wad Wdb 10.5 kJ

Here Wdb 0 since it is constant volume process.

[ . . . W pdv; During db process, dv 0 ]

So, Wad 10.5 kJ

To Find Qad

Qad Ud Ua Wad

42 0 10.5 Qad 52.5 kJ

Heat absorbed during ad process 52 kJ

Thermodynamics 1.33

To Find Qdb

Qdb Ub Ud Wdb

Qadb 62.5 Qad Qdb 52.5 Qdb

So, Qdb 62.5 52.5 10 kJ

Heat absorbed during db process 10 kJ

1.29 SPECIFIC HEAT CAPACITIES

1.29.1 Specific heatThe specific heat of a substance is defined as the amount of heat

required to raise a unit mass of the substance through a unit rise intemperature. The symbol C will be used for specific heat.

C Q

m t J/kg K

where Q amount of heat transfer (J)

m mass of the substance kg; t rise in temperature K

For gases, if the process is at constant pressure, it is Cp , and if the

process is at constant volume, it is Cv. The product of mass and specific heat

(mC) is called the heat capacity of the substance.

1.29.2 Latent heatThe latent heat is the amount of heat transfer required to cause a

phase change in unit mass of a substance at a constant pressure andtemperature.

The latent heat of fusion is the amount of heat transferred to meltunit mass of solid into liquid.

The latent heat of vapourization is the quantity of heat required tovaporize unit mass of liquid into vapour.

The latent heat of sublimation is the amount of heat transferred toconvert unit mass of solid to vapour.

1.29.3 Specific heat at constant volume:The specific heat of a substance at constant volume Cv is defined as

the rate of change of specific internal energy with respect to temperature

when the volume is held constant, ie Cv u

T v


For a constant volume process uv T1

T2

Cv dT

The first law may be written for a closed stationary system composedof a unit mass of a pure substance

Q u W

or d– Q du dW

For a process in the absence of work other than Pdv work

d– W Pdv

d– Q du Pdv

when the volume is held constant

Qv uv

Qv T1

T2

Cv dT

ie Heat transferred at constant volume increases the internal energy ofthe system.

If the specific heat of a substance is defined in terms of heat transfer, then

Cv Q

T v

Since u, T and v are properties, Cv is a property of the system. The

product mCv is called the heat capacity at constant volume (J/K).

1.29.4 Enthalpy

The enthalpy h of a substance, is defined as h u Pv. It is aproperty of a system and intensive property.

From first law d– Q du Pdv

At constant pressure, Pdv d Pv

d– Qp du d Pv d u Pv dh

where h u Pv is the specific enthalpy, a property of the system.

Thermodynamics 1.35

Using the definition of enthalpy, and the equation of state of an idealgas, we have

h u Pv and Pv RT then h u RT

Since R is constant, and u f T, it follows that the enthalpy of anideal gas is also a function of temperature only.

h f T

1.29.5 Specific heat at constant pressure:The specific heat at constant pressure Cp is defined as the rate of

change of enthalpy with respect to temperature when the pressure is heldconstant.

Cp h

T p

For a constant pressure process,

hp T1

T2

Cp dT

... (i)The first law for a closed stationary system of unit mass

d– Q du Pdv

Again h u Pv

dh du Pdv vdP d– Q vdP

d– Q dh vdP

d– Qp dh

or Qp hp

ie Qp T1

T2

Cp dT

The heat capacity at constant pressure is equal to mCP (J/K)


1.30 THERMODYNAMIC PROCESSESThermodynamic processes are broadly classified into two types i.e.,

1.31 NON-FLOW PROCESS (OR) CLOSED SYSTEM

The following non-flow processes are reversible.

1. Constant volume process (Isochoric process) V constant n

2. Constant pressure process (Isobaric process) P constant n 0

3. Constant temperature process (or) Isothermal process t constant or Pv constant) n 1

4. Isentropic process (or) Reversible adiabatic process

Pv constant n 5. Polytropic process Pvn constant n n

1.31.1 Constant Volume ProcessDuring this heating or cooling process, the volume remains constant.

W1 2 Pdv[Since volume is constant]

dV 0; W1 2 0

Q u W u 0

u mCvT2 T1; So Q mCvT2 T1

Thermodynam ics Processes

Non - FlowProcesses

F lowProcesses

Steady FlowProcesses

Transient (o r)Unsteady Flow

Processes

Thermodynamics 1.37

For any process Pvn constant

Here n infinitive

So, PV C; V C P

V C p1/ 1

i.e. V constant

Change in enthalpy H mCpT2 T1

1

2

dH 1

2

dU 1

2

dPV

H2 H1 U2 U1 P2V2 P1V1

mCvT2 T1 mRT2 T1 Cv R mT2 T1

H2 H1 mCpT2 T1 [ . . . Cp Cv R ]

Change in entropy S mCv ln T2

T1

Fig. 1.29 Reversible constant volume process


or S mCv ln P2

P1

Problem 1.5: Air occupies a space of 0.3 m3 at a pressure of 2 bar and atemperature of 80C. It is heated at constant volume, until the pressure is 8bar. Determine (1) temperature at the end of the process (2) mass of air (3)change in internal energy (4) change in enthalpy (5) change in entropy duringthe process.

Solution

V1 0.3 m3; P1 200 kPa; T1 80 273 353 K; P2 800 kPa

Constant volume processTo Find T2 Temp. at the end of the process

P1V1

T1

P2V2

T2

P1

T1

P2

T2 [... V1 V2]

T2 P2 T1

P1 800

353200

1412 K

To Find m - mass of air

P1V1 mRT1

m P1V1

RT1

200 0.3

0.287 353 0.59224 kg

[ . . . R 0.287 kJ/kg K for air]

To Find U change in internal energy

U mCvT2 T1 0.59224 0.7181412 353 450.32 kJ

[ . . . Cv 0.718 kJ/kg K for air]

To Find H change in enthalpy

H mCpT2 T1 0.59224 1.0051412 353 630.32 kJ

[ . . . Cp 1.005 kJ/kg K for air]

Thermodynamics 1.39

To Find S change in entropy

S mCv ln T2

T1 0.59224 0.718 ln

1412353

0.5895 kJ / kg K

Problem 1.6: A rigid tank contains air at 600 kPa and 150C. As a resultof heat transfer to the surroundings, the temperature and pressure inside thetank are 70C and 400 kPa, respectively. Determine the boundary work doneduring this process.

Solution:

A sketch of the system and the P-Vdiagram of the process are shown in Fig.

The boundary work can bedetermined as

W 1

2

PdV 0

because dV 0 ie a rigid tank has aconstant volume and dV 0 in the aboveW equation. Therefore, there is noboundary work done during this process.That is, the boundary work done during aconstant - volume process is always zero.This is also evident from the PV diagram of the process (ie the area underthe process curve is zero).

1.31.2 Constant Pressure Process (or) isobaric Process [P C; n 0]

During this process, (Heating (or) Cooling process) the pressureremains constant.

W1 2 1

2

PdV P 1

2

dV PV2 V1

[ . . . P constant ]Apply First law

Q U W mCvT2 T1 PV2 V1


mCvT2 T1 mRT2 T1 mCpT2 T1 [ ... mRT PV ]

[ ... Cp Cv R, i.e Cp Cv R ]

Also, Q U W U2 U1 PV2 V1

U2 P2V2 U1 P1V1 H2 H1

So, Q H2 H1 mCpT2 T1

Heat transfer = Change in enthalpy in constant pressure process.

Also change in entropy S mCp ln T2T1

(or) S mCp ln V2

V1

For any process PVn constant

In constant pressure process n 0

So, PV0 C P C

Fig. 1.30 Constant Pressure Process (or) Isobaric Process

Thermodynamics 1.41

Problem 1.7: A frictionless piston - cylinder device contains 5 kg of water

vapour at 0.40 MPa and specific volume 0.5342 m3/kg. Heat is now

transferred to the system until the volume reaches 0.5951 m3/kg at constantpressure. Determine the workdone by the steam during this process.

Solution: A sketch of the system and P-V diagram of the process are shownhere

Assumption: The expansionprocess is quasi-equilibrium

W 1

2

Pdv P v2 v1

W mP v2 v1

5 0.4 103 0.5951 0.5342 121.8 kJ

The positive sign indicates that the work is done by the system. Thatis, steam uses 121.8 kJ of its energy to do this work. The magnitude of thiswork could also be determined by calculating area under the process curve(ie rectangle) on the P-V diagram, which is simply PoV for this case.

Problem 1.8: 1 kg of gas with initial volume 2.5 m3 is heated at constantpressure from 25C to 220C. Estimate the heat added, ideal work done,change in internal energy, change in enthalpy, change in entropy. Alsocalculate final pressure and final volume. Take Cp 0.984 kJ/kg K;

Cv 0.728 kJ/kg K.

Solution: Given: V1 2.5 m3; m 1 kg; T1 25 273 298 K;

T2 220 273 493 K; Cp 0.984; Cv 0.728

Constant pressure process.

To Find R Characteristic Gas Constant

For any gas Cp Cv R ; R 0.984 0.728 0.256 kJ/kg K

To Find W1 2

W1 2 PdV PV2 V1 mRT2 T1 [ ... PV mRT ]

1 0.256493 298

H O2

m =0.5 k gP =0 .4

Hea

t

A re a = w b

P ,M Pa

0 .41 P 0 = 0 .4 2

M p a

Schem atic and d iag ram P -v

1 =0.53 42 =0 .5951

, / m k g3


49.92 kJ

To Find U

U mCvT2 T1 1 0.728493 298

141.96 kJ

To Find QHeat added Q mCpT2 T1 1 0.984493 298

191.88 kJ H

[ . . . Q H for constant pressure process]

To Find Final Pressure P2

Since it is constant pressure process, P1 P2

P1V1 mRT1

P1 1 0.256 298

2.5 30.52 KPa P2

To Find Final Volume V2

P1V1

T1

P2V2

T2 , since P1 P2

V1

T1

V2

T2; V2

V1

T1 T2

V2 2.5298

493 4.136 m3

1.31.3 Constant Temperature Process (or) isothermal Process PV constant ; T constant

When a piston expands from high pressure to low pressure, thetemperature will fall. But in this process, heat is added continuously tomaintain the temperature constant. Similarly, during compression process, thepressure is raised. To avoid the temperature rise, the heat is liberatedcontinuously. So the temperature will remain constant.

When the system temperature is maintained constant during a process,the process is called isothermal process or constant temperature process.

Thermodynamics 1.43

P1V1

T1

P2V2

T2

Since T1 T2,

P1V1 P2V2 i.e. PV constant

(or) P CV

W1 2 1

2

PdV 1

2

CV

dV C 1

2

dVV

C [ ln V ]12

PV ln

V2

V1 mRT

ln

V2

V1

(or) W1 2 PV ln

P1

P2 mRT

ln

P1

P2

. . .

P1

P2

V2

V1

Q U W

Q mCvT2 T1 W

Q 0 W

Fig. 1.31 Reversible isothermal process

Fig. 1.31(c)


So, Q W H mCpT2 T1 0 [ . . . T2 T1 ]

S Change in Entropy

S mR ln V2

V1 kJ/kg

or S mR ln P1

P2 kJ/kg K

Problem 1.9: A piston - cylinder device initially contains 0.5 m3 of air at

100 kPa and 80C. The air is now compressed to 0.1m3 in such a way thatthe temperature inside the cylinder remains constant. Determine the workdoneduring this process.

Solution: A sketch of the system and the PV diagram of the process areshown in Fig.

Assumptions: The compression process is quasi-equilibrium. Air can beconsidered as an ideal gas.

For an ideal gas at constant temp.

PV mRT C [constant]

(or) P CV

W 1

2

PdV 1

2

CV

dV

C 1

2

1V

dV C lnV 1

2 C ln

V2

V1

W1 2 P1V1 ln V2

V1

In the above equation, P1V1 can be replaced

by P2V2 (or) mRT. Also, V2V1

can be replaced by

P1P2

because P1V1 P2V2 ie V2V1

P1

P2

Thermodynamics 1.45

Substituting numerical values in the above equation.

W1 2 P1V1 ln V2

V1 100 0.5

ln

0.10.5

W1 2 80.472 kJ

The negative sign indicates that this work is done on the system (awork input), which is always the case for compression process.

Problem 1.10: A quantity of air has a volume of 0.5 m3 at a pressure of5 bar and 80C. It is expanded in a cylinder at a constant temperature toa pressure of 1 bar. Determine the amount of work done and heat transferduring expansion.

Solution:

Given: V1 0.5 m3; P1 5 bar; P2 1 bar T1 80 273 353 K;

Constant Temperature Process

To Find mP1V1 mRT1

m P1V1

RT1

5 102 0.5

0.287 353

2.468 kg

To Find V2

P1V1 P2V2

V2 P1V1

P2

5 102 0.5

1 102 2.5 m3

To Find W W mRT1ln V2

V1 2.468 0.287 353

ln

2.50.5

402.42

W1 2 402.42 kJ

Q W in Isothermal process

So Q W 402.42 kJ

Note: For any process PVn C

For isothermal process n 1

So, PV constant


Problem 1.11: A mass of 1.5 kg of air is compressed in a quasi staticprocess from 0.1 MPa to 0.7 MPa for which PV Constant. The initial

density of air is 1.16 kg/m3. Find the work done by the piston to compressthe air.

Solution:Given data: m 1.5 kg ; PV C Isothermal process

P1 0.1 MPa 100 kN/m2 ; P2 0.7 MPa 700 kN/m2

Initial density 1 1.16 kg/m3

v1 1

1.16 0.8621 m3/kg

V1 m v1 1.5 0.8621 1.2931 m3

To find Work done W:

For isothermal process P1V1 P2V2

V2 P1V1

P2

100 1.2931700

0.1847 m3

Work done for isothermal process W P1V1ln V2

V1

W 100 1.2931 ln 0.18471.2931

251.6

W 251.6 KJ

ve sign indicates that the work is done on the system)

1.31.4 Reversible Adiabatic Process (or) isentropic Process

PV C; S constant

Reversible - No friction

Adiabatic - No heat transfer Q 0

If there is no friction and no heat transfer, then there is no entropy change.

So reversible adiabatic process is isentropic process.

Q U W

Thermodynamics 1.47

W1 2 1

2

PV

Here PV C

P C

V

W 1

2

C

V dV C

1

2

1

V dV PV

1

2

V dV PV V2 1 V1

1

1

P2V2

V21 P1V1

V11

1

P2V2 P1V1

1

mRT2 T11

W1 2 P2V2 P1V1

1

mRT2 T11

Fig. 1.32 Reversible adiabatic process


Aliter:

Q U W [Q 0 since it is adiabatic]

So, 0 U W

W U

Also W U [ mCvT2 T1 ]

Note: Cv R

1, So W

mR 1

T2 T1

W mR T2 T1

1

Change in Enthalpy H mCpT2 T1Change in Entropy S 0

For any process PVn C

For this process PV C

So, n

Problem 1.12: Air at 1.02 bar, 22C initially occupies a cylinder volume

of 0.015 m3 is compressed isentropically by a piston to a pressure of 6.8 bar.Determine, (i) the final temperature (2) the final volume (3) the work done.

Solution

P1 1.02 bar;

T1 22 273 295 K

V1 0.015 m3; P2 6.8 bar

Isentropic compression

To Find T2

T2T1

P2

P1

1

[We can get this relation from pV C]

Thermodynamics 1.49

T2 T1 P2P1

1

T2 295

6.81.02

0.41.4

507.3 K[ . . . 1.4 for air ]

2. To Find Final Volume V2 P1V1 P2V2

V2

V1

P1

P2

V2

V1 P1

P2

1

V2 V1 P1P2

1 0.015

1.026.8

11.4

3.869 10 3m3

3. Work done

W P2V2 P1V1

1 6.8 102 3.869 10 3 1.02 102 0.015

1 1.4

2.7523 kJ

Note

P1V1 P2V2

; V2

V1

P1P2

V2

V1

P1

P2

V2

V1 P1

P2

1

P1V1

T1

P2V2

T2


T2

T1

P2V2

P1V1

P2

P1 P1

P2

1

. . .

V2

V1 P1P2

1

P2

P1 P2P1

1

P2

p1

1 1 P2

P1

1

So, T2

T1 P2

P1

1

1.31.5 Polytropic process: PVn constant n nGenerally all the processes are polytropic process. When the index of

expansion (or) compression ‘n’ is changed, the process name is changed.

When index n , it is called constant volume processwhen n 0, it is constant pressure process

when n 1, it is isothermal processwhen n , it is isentropic process.

Fig 1.33 (a)

Thermodynamics 1.51

When n n, then it is polytropic process.The n varies from 0 to .

During actual expansion and compression processes of gases, pressure

and volume are often related by PVn C, where n and C are constants. Aprocess of this kind is called a polytropic process. Let us derive a generalexpression for the workdone during a polytropic process.

We know, PVn C for polytropic process. p C

Vn CV n

Substitute the relation in W 1

2

PdV equation

We get W 1

2

PdV 1

2

CV n dv

C V n 1

1 n 1

2

PVn V2

1 n V11 n

1 n

P2V2

nV21 n P1V1

nV11 n

1 n

W P2V2 P1V1

1 n [. . . PVn P2V2

n P1V1n]

Note here that C PVn P1V1n P2V2

n

For an ideal gas, PV mRT,

P

P 1

P 2

V 1 V 2 V

2

1 P V P V1 1 2 2 = n n

P n = const.V

Schem atic and d iagram for a polytropic p rocess.P -V

G AS

= const.P V = C n

Fig. 1.33 (b)


The workdone equation can be written as W mR T2 T1

1 n and n 1

This PVn can be used for any process by substituting appropriate value for n. Polytropic Workdone can be calculated by replacing in the isentropicprocess workdone by n.

1. W P2V2 P1V1

1 n

mRT2 T11 n

(or) W P1V1 P2V2

n 1

mRT1 T2n 1

Q 0

Q U W mCvT2 T1 mRT2 T1

1 n

Note V2

V1 P1

P2

1n

; T2

T1 P2

P1

n 1n

2. U mCvT2 T1 mRT2 T1

1

Q U W . . . Cv

R 1

mRT2 T1

1

mRT1 T2n 1

mRT2 T1

1

mRT2 T1n 1

n 1 1 1 1

1 n 1 [ mRT2 T1 ]

n

1 n 1 [ mRT2 T1 ]

n 1

mRT1 T2

n 1 n 1

W . . . W

mRT1 T2n 1

Insert table from Pg 2.30, 2.31 (Kerala book Thermodynamics)

Thermodynamics 1.53

For

all

pro

cess

es,

P1V

1

T1

P

2V2

T2

an

d P

V

mR

T

ie P

1V1

mR

T1;

P2V

2

mR

T2

....

Pro

cess

pVn

CW

UQ

H S

Rem

arks

Wor

k do

ne

W

p

dv

Ch

ange

in

Inte

rnal

en

ergy

Hea

t tr

ansf

erC

hang

e in

En

thal

pyC

han

ge i

nE

ntr

opy

1.C

onst

ant

volu

me

proc

ess

n

V

C

0 (z

ero)

mC

vT

2

T1

mC

vT

2

T1

mC

pT

2

T1

mC

vln

T2

T1 (

or)

mC

vln

P2

P1

Pd

V

0

sinc

e

dV

0

Q

U

V1

V2

2.C

onst

ant

pres

sure

proc

ess

n

0

P

CW

P V

2

V1

(or)

W

mR T

2

T1

mC

vT

2

T1

mC

pT

2

T1

mC

pT

2

T1

mC

pln

T2

T1

(or)

mC

pln

V2

V1

Q

H

P1

P2

3.C

onst

ant

tem

pera

ture

proc

ess

(or)

Isot

herm

alpr

oces

s

n

1 P

V

CT

C

PV

ln V

2

V1 or

PV

ln P

1

P2

or

mR

T ln

V2

V1

0 (z

ero)

(s

ince

T1

T2)

Q

W

P

V ln

V2

V1

P

V ln

P1

P2

0 (z

ero)

(sin

ce

T1

T2)

.m

R ln

V2

V1 (

or)

mR

ln P

1

P2

Q

W

P1V

1

P2V

2

T1

T2

4.Is

entr

opic

proc

ess

(or)

Rev

ersi

ble

Adi

abat

icpr

oces

s

n

PV

C

T2

T1

P2

P1

1

T2

T1

V1

V2

1

P1V

1

P2V

2

1

(or)

mR T

1

T2

1

mC

V T

2

T1

or

U

W

Zero

(No

Hea

tT

ran

sfer

)

mC

p T

2

T1

0 (z

ero)

si

nce

s 1

s2

P1V

1 P

2V2 ;

T2

T1

P2

P1

1

;

V2

V1

P1

P2 1 ;

T2

T1

V1

V2

1

5.P

olyt

ropi

cpr

oces

sn

n; p

Vn

C

T2

T1

p 2 p 1 n

1

n

V2

V1

p 1 p 2 1 n

P1V

1

P2V

2

n

1

(or)

mR T

1

T2

n

1

mC

vT

2

T1

n

1

W

(or)

mC

nT

2

T1

mC

pT

2

T1

mC

nln T

2

T1

(or)

n

1

mR

ln V

2

V1

Pol

ytro

pic

sp.

heat

Cn

Cn

Cv

n

1

n ;

P1V

1n P

2V2n

T2

T1

V1

V2 n

1

The

rmod

ynam

ics

1.55

1.54

Bas

ic M

echa

nica

l E

ngin

eerin

g -

ww

w.a

irwal

kpul

icat

ions

.com

3. So in polytropic process Q n

1 W (or)

Q mCn T2 T1 where Cn CV n1 n

4. Change in enthalpy H mCpT2 T1

5. Change in entropy S S n 1

mR ln V2

V1 (or) S mCn ln

T2

T1

By studying all the non-flow processes we can draw a tabular column toknow W, Q, U, H, S for ready made use.

By referring the tabular column in the nearby pages, we can solve anytype of problem in non-flow process.

Problem 1.13: Air at 300C and 10 bar expands to 3 bar reversibly

following the law PV1.35 C. Determine the work done per kg of air, heattransfer and change in entropy.

Solution

Given: T1 300 273 573 K; P1 10 bar 1000 KPa

P2 300 KPa; PV1.35 C; mass 1 kg

Polytropic Process

To Find W

W mRT1 T2

n 1

T2T1

P2

P1

n 1n

; T2 T1 P2

P1

n 1n

T2 573

3001000

0.351.35

419.4 K

W 1 0.287573 419.4

0.35 125.98 kJ

Q n

1 W

1.4 1.351.4 1

125.98 15.747 kJ/kg


[Aliter: Cn Cv n

1 n 0.718

1.4 1.35

1 1.35 0.10257 kJ/kgK

Q mCnT2 T1 1 0.10257419 573 15.755 KJ

S mCnln T2

T1 1 0.10257 ln

419.4573

S 0.032 kJ/kg K

1.32 STEADY FLOW PROCESS (OPEN SYSTEM)

When there is a mass transfer across the boundary, the system is calledas open system. Most of the engineering devices are open system, involvingthe flow of mass through them.

We can mark an imaginary box (broken line around the open system)and this is called control surface. The volume inside the control surface isknown as control volume. The mass of fluid is flowing through this controlvolume.

Steady flow means that the rate of flow of mass and energy areconstant. The properties will have fixed value at particular location and willnot alter with time.

m

1 m

2 m mass rate of flow in kg/sec

m 1 A1 C1 2 A2 C2

m

A1 C1

v1

A2 C2

v2

where density of fluid, C Velocity of fluid, v specific volume

A Area of cross section through which mass flows.

According to Ist Law, Q E W Q E2 E1 W

E1 Q E2 W

E1 energy at position 1 for one Kg of mass

E1 PE1 KE1 Flow work u1

P.E. = Potential energy

K.E. = Kientic energy; Flow work P1 v1

u Internal energy

Thermodynamics 1.57

E1 Z1 g C1

2

2 P1 v1 u1; E2 Z2 g

C22

2 P2 v2 u2

The steady flow energy equation on mass basis is given below.

Z1 g C1

2

2 P1 v1 u1 Q Z2 g

C22

2 P2 v2 u2 W

Here all energy values are in J/kg.

The steady flow energy equation for m

kg/sec of fluid is given asfollows. (Time basis)

m [Z1 g

C12

2 h1] Q

m

[Z2 g

C22

2 h2] W

where Q in J/s i.e., Watts [ h1 u1 P1 v1]

W

in J/s i.e., Watts h2 u2 P2 v2]

Z1 ht above datum at 1 in m

C1 Velocity of fluid at 1 in m/sec

h1 enthalpy of fluid at 1 in J/kg

A1 Area at inlet in m2

Similarly Z2, C2, h2 and A2

The steady flow energy equation

h1 C1

2

2 Z1 g Q h2

C22

2 Z2 g W in J/kg (or)

m h1

C12

2 Z1 g Q

m

h2

C22

2 Z2 g W

in Watts which can be

applied for

1. Nozzles and Diffuser 2. Throttling Device

3. Turbine and Compressor

4. Heat exchangers (Boiler, Condensor and evaporator)


Formulae to Remember1. The Steady Flow Energy Equation (SFEE) is given by per unit mass

h1 C1

2

2 Z1 g Q h2

C22

2 Z2 g W in J/kg (or)

m h1

C12

2 Z1 g Q

m

h2

C22

2 Z2 g W

in Watts

2. For boiler SFEE is Q h2 h1

3. For condenser Q h2 h1

4. For nozzle SFEE is C2

2 C12

2 h1 h2 [where h1, h2 are in J/kg ]

Final velocity C2 2 h1 h2 C12 m/s

If initial velocity is neglected

C2 2 h1 h2

5. For turbine SFEE is W h1 h2 J/kg

6. For rotary Compressor W h2 h1 J/kg

7. For Reciprocating Compressor W Q h2 h1 J/kg

Problem 1.14 A mass of 0.8333 kg/sec enters the control volume of a steadyflow system at 2 bar and 100C and an elevation of 100 m above the datum.The same mass leaves the control volume at 150 m elevation with a pressureof 10 bar and temperature of 300C. The entrance velocity is 40 m/sec andthe exit velocity is 20 m/sec. During the process 13.8889 kW of heat aretransferred to the control volume and the rise in enthalpy is 8 kJ/kg. Calculatethe power developed.

Solution

Given: m 0.8333 kg/sec; P1 2 Bar; T1 100 273 372 K ; Z1 100 m;

Z2 150 m; C1 40 m/sec; C2 20 m/sec; P2 10 bar;

T1 300 273 573 K ; Q 13.8889 kW 13.8889 103 W;

h2 h1 8000 J/kg; h1 h2 8000 J/kg

Steady flow Energy equation gives

Thermodynamics 1.59

m h1

C12

2 Z1 g Q

m

h2

C22

2 Z2g

W

Q is in Watts and W

is in Watts

W

Power m h1 h2

C1

2 C22

2 Z1 g Z2 g

Q

0.8333 8000

402 202

2 [100 150 9.81]

13.8889 103

7313.75 Watts

1.32.1 Nozzle: Nozzle is a device which is used to increase the velocity of fluid at

the expense of its pressure drop.

1.32.2 Diffusor: Diffusor is a device which is used to increase the pressure of fluid at

the expense of its kinetic energy (by velocity drop)

Problem 1.15: The velocity and enthalpy of fluid at the inlet of a certainnozzle are 50 m/sec and 2800 kJ/kg respectively. The enthalpy at the exit ofnozzle is 2600 kJ/kg. The nozzle is horizontal and insulated so that no heattransfer takes place from it. Find 1. Velocity of fluid at exit of the nozzle,

2. mass flow rate, if the area at inlet of nozzle is 0.09 m2 and specific

volume is 0.185 m3/kg 3. Exit area of the nozzle, if the specific volume at

the exit of the nozzle is 0.495 m3/kg


Solution

Applying steady flow energy equation for the nozzle

h1 C1

2

2 Z1 g Q h2

C22

2 Z2 g W

[mass basis]

Here Z1 g Z2 g, Q 0 (Insulated; so no heat transfer)

W 0 (no work transfer during flow through nozzle)

So the equation becomes h1 C1

2

2 h2

C22

2 C2

2 2 h1 h2 C12

C2 2 h1 h2 C12

2 2800 103 2600 103 502 634.43 m/sec

To find Exit Area A2 and m

m mass flow rate

A1 C1

v1

A2 C2

v2

0.09 500.185

A2 634.43

0.495 A2 0.018979 m2

m

0.09 500.185

24.324 kg/sec

Thermodynamics 1.61

1.32.3 Throttling Device: It is the device used to reduce the pressure when its enthalpy remains constant.

When a fluid is passing through restricted passage as shown in fig.then it is throttled.

S.F.E.E (Steady Flow Energy Equation)

h1 C1

2

2 Z1 g Q h2

C22

2 Z2 g W

During flow through throttling device,

Q 0; W 0; No change in velocity ... C1 C2

No change in height (So Z1 Z2

So the equation becomes h1 h2

So the enthalpy during throttling process remains constant.

1.32.4 Turbine: The turbine is used to produce power. Expansion of working fluid

takes place in the turbine and the work is done.

Fig 1.34 Throttling Process


Problem 1.16: A turbine operates under steady flow condition receivingsteam with the following data. Pressure P1 1.2 MPa; Temperature

t1 188C; enthalpy h1 2785 kJkg

; Velocity C1 33.3 m/sec and elevation

Z1 3 m. The steam leaves the turbine at 100 m/sec velocity, zero elevation,

pressure = 20 kPa, 2512 kJ/kg of enthalpy. Heat is lost to the surroundingat the rate of 0.27 kJ/sec. Determine the power output of the turbine, if themass flow rate of steam is 0.42 kg/sec.

Solution

P1 1.2 MPa 1.2 106 Pa P2 20 kPa 20 103 Pa

T1 188 273 461 K T2 ?

h1 2785 kJ/kg

2785 103 J/kg

h2 2512 103 J/kg

C1 33.3 m/sec C2 100 m/sec

Z1 3 m Z2 0

Q 0.27 103 J/sec 270 J/sec m

0.42 kg/sec

Applying S.F.E.E

m h1

C12

2 Z1 g

Q

m

h2

C22

2 Z2 g

W

Thermodynamics 1.63

W

m h1 h2

C12 C2

22

Z1 g Z2 g Q

0.42 2785 103 2512 103

33.32 1002

2 3 9.81 0

270

112535 J/sec 112.535 kW

Problem 1.17: In a Gas turbine installation, the gases enter the turbine atthe rate of 5 kg/sec with a velocity of 50 m/sec and enthalpy of 900 kJ/kgand leave the turbine with 150 m/sec. and enthalpy of 400 kJ/kg. The lossof heat from the gases to the surroundings is 25 kJ/kg. AssumeR 0.285 kJ/kgK Cp 1.004 kJ/kg K and inlet conditions to be at 100 kPa

and 27C. Determine the diameter of the inlet pipe.

Given m 5 kg/s, C1 50 m/s, h1 900 kJ/kg

C2 150 m/s, h2 400 kJ /kg

R 0.285 kJ /kg K, Cp 1.004 kJ /kg K, Q 25 kJ /kg

P1 100 kPa; t1 27 CThe steady flow equation is given by

h1

C12

2 Z1g

Q

h2

C22

2 Z2g

W

assume Z1 Z2 h1

C12

2 Q

h2

C22

2 W

900

502

2 1000 25

400

1502

2 1000 W

876.25 411.25 W

W 465 kJ/kg

W

m W 5 465 2325 kW

To find d1, mass flow rate m

A1C1

V1

we know, P1V1 mRT1

V1 RT1

P1

0.285 27 273100

0.855 m3


m 1 A1 C1

A1 C1V1

A1 mV1

C1

5 0.85550

A1 0.0855 m2

(ie) 4

d12 0.0855 d1

2 0.10886 m2

d1 0.3299 m

1.32.5 Compressor: Compressor is a device used to increase the pressure of the fluid by

taking work input.

Problem 1.18: Air enters a compressorwith a velocity of 60 m/sec, pressure100kPa, temp 40C and leaves thecompressor with a velocity of 90 m/sec,500kPa and 120C. Consider the system isadiabatic. Find the power of motor for themass flow rate of 40kg/min. Write theassumptions made.

Solution

Given

C1 60 m/sec C2 90 m/sec

P1 100 103 Pa P2 500 103 Pa

T1 40 273 313 K T2 120 273 393 K

Q 0 adiabatic

m 40 kg/min

4060

0.6667 kg/sec

S.F.E.E

m h1

C12

2 Z1 g Q

Air Compressor

(1)

W

(2)

Q =0(a

diab

atic

)

Thermodynamics 1.65

m h2

C22

2 Z2 g W

Assume Z1 g Z2 g

(Since nothing is given about P.E.)

W

m h1 h2

C12 C2

2

2

h1 h2 Cp T1 T2 1.005 313 393

80.4 kJ/kg 80.4 103 J/kg

W

0.6667 80.4 103

602 902

2

55102.76 J/sec 55.103 KW

[– sign indicates work is done on the system]

1.32.6 Heat Exchanger: It is a device used to heat the fluid (or) cool the fluid. No work

transfer, change in kinetic energy is negligible. Eg. Boiler cooling plant.

(i) Boiler:

It is a device in which water (or) fluid is heated.

Problem 1.19: In a cooling plant water enters with enthalpy of 240 kJ/kgand leaves with enthalpy of 192 kJ/kg. Pump work is negligible.The exit pipe is 20 m above the inlet piple. Change in K.E. is negligible.Find the heat transfer from the water.

Solution S.F.E.E. on mass basis

h1 C1

2

2 Z1 g Q h2

C22

2 Z2 g W

W 0

C12

2

C22

2 ; Z2 Z1 20 m

The S.F.E.E. equation becomes h1 Z1 g Q h2 Z2 g

Q h2 h1 Z2 g Z1 g 192 103 240 103 20 9.81


Q 47804.8 J/kg 47.803 kJ / kg

[– sign indicates heat is flowing from the water to surroundings]

Problem 1.20: The boiler generates 800 kg/hr of steam. The enthalpy of feedwater is 170 kJ/kg and the enthalpy of steam is 2800 kJ/kg. Neglectingpotential and kinetic energies, find the rate of heat transfer.

Solution

m

8003600

0.222 kg sec

S.F.E.E on time basis

m h1

C12

2 Z1 g Q

m

h2

C22

2 Z2 g W

Change in K.E. = 0; Change in P.E. = 0

W

0

The S.F.E.E equation becomes m h1 Q

m

h2

Q m

h2 h1 0.222 2800 103 170 103

584444.4 J/sec 584.44 kW

58.444 kW of heat is added to boiler water.

1.33 LIMITATIONS OF FIRST LAW OF THERMODYNAMICS

Limitations Description Examples

(i) No restriction in thedirection of heat flow.

First law does notspecify the direction ofthe process. It onlyestablishes a relationshipbetween heat and work,but failed to mention,whether the heat flowsfrom cold body to hotbody or not.

Consider heating of aroom by means of anelectric room heater.Room gets heated bypassing current through aresistor. But, transferringthe heat from room willnot generate electricalenergy.

Thermodynamics 1.67

Limitations Description Examples

(ii) Feasibility of theprocess is not specified.

It does not give anyinformation, whether theprocess is possible or not.

Consider two bodies Aand B. A is at highertemperature than B.When brought together,either the body A losesheat to body B or thebody A gains heat frombody B. First law failsto explain the processthat is possible.

(iii) Amount of energyconversion is notspecified.

Work can be convertedinto 100% equivalentheat but heat cannot beconverted into 100%equivalent work.

An electric heater canconvert 1 kJ of work to1 kJ heat but an ICengine cannot convert 1kJ heat to 1kJ work.

So heat and work arenot completelyinterchangeable forms ofenergy.

This is because someamount of heat getsdissipated duringcombustion process.

1.34 THE SECOND LAW OF THERMODYNAMICSA process must satisfy the first law to occur.

However, satisfying the first law alone does notensure that the process will actually take place. Acup of hot coffee left in a cooler room eventuallycools off (Fig. 1.35). This process satisfies the firstlaw of thermodynamics since the amount of energylost by the coffee is equal to the amount gained bythe surrounding air. Now let us consider the reverseprocess–the hot coffee getting even hotter in a coolerroom as a result of heat transfer from the room air.We all know that this process never takes place. Yet,doing so would not violate the first law as long as the amount of energy lostby the air is equal to the amount gained by the coffee.

Consider a paddle-wheel mechanism that is operated by the fall of amass (Fig. 1.36). The paddle wheel rotates as the mass falls and stirs a fluid

Fig. 1.35 A cup of hotcoffee does not get

hotter in a cooler room


within an insulated container. As a result, thepotential energy of the mass decreases, and theinternal energy of the fluid increases inaccordance with the conservation of energyprinciple. However, the reverse process, raisingthe mass by transferring heat from the fluid tothe paddle wheel, does not occur in nature,although doing so would not violate the firstlaw of thermodynamics.

The first law places no restriction on thedirection of a process, but satisfying the firstlaw does not ensure that the process will actually occur. This inadequacy ofthe first law to identify whether a process can take place is remedied byintroducing another general principle, the second law of thermodynamics. Aprocess will not occur unless it satisfies both the first and the second lawsof thermodynamics. Before studying second law of thermodynamics, weshould know the following terms.

1.34.1 Thermal Energy ReservoirsA hypothetical body with a relatively

large thermal energy capacity (mass specific heat) that can supply or absorb finiteamounts of heat without undergoing anychange in temperature is called a thermalenergy reservoir, or just a reservoir. Inpractice, large bodies of water such as oceans,lakes, and rivers as well as the atmosphericair can be considered as thermal energyreservoirs because of their large thermalenergy storage capabilities or thermal masses(Fig. 1.37). The atmosphere, for example,does not warm up as a result of heat losses from residential buildings inwinter.

A reservoir that supplies energy in the form of heat is called a source,eg: Furnace. A reservoir that absorbs energy in the form of heat is called asink eg: atmosphere (Fig. 1.38). Thermal energy reservoirs are often referredto as heat reservoirs since they supply or absorb energy in the form of heat.

Fig. 1.37 Bodies with relativelylarge thermal masses can beconsidered as thermal energy

Fig. 1.36 Transferring heat toa paddle wheel will not

cause it to rotate

Thermodynamics 1.69

1.35 HEAT ENGINESWork can easily be converted to other forms

of energy, but converting other forms of energy towork is not that easy. The mechanical work done bythe shaft shown in Fig. 1.38 is first converted to theinternal energy of the water. This energy may thenleave the water as heat. Any attempt to reverse thisprocess will fail. That is, transferring heat to the waterwill not cause the shaft to rotate. From thisobservations, we conclude that work can be convertedto heat directly and completely, but converting heatto work requires the use of some special devices.These devices are called heat engines.

1. Heat engines receive heat from ahigh-temperature source (solar energy, oilfurnace, nuclear reactor etc.).

2. They convert part of this heat to work (usually in the form of arotating shaft).

Fig. 1.38 A sourcesupplies energy in theform of heat and sink

absorbs it.

Fig. 1.39 Work can always be convertedto heat directly and completely, but the

reverse is not true.

Fig. 1.40 Part of the heat received by a heat engine is

converted to work, while the restis rejected to a sink.


3. They reject the remaining waste heat to a low-temperature sink(the atmosphere, rivers, etc.).

4. They operate on a cycle.

1.35.1 Heat Engine CycleA heat engine cycle is a thermodynamic cycle in which there is a net

heat transfer to the system and a net work transfer from the system. Herethe heat engine is considered as a system.

The work-producing device that best fits into the definition of a heatengine is the steam power plant, which is an external-combustion engine.That is, combustion takes place outside the engine, and the thermal energyreleased during this process, is transferred to the steam as heat. The schematicof a basic steam power plant is shown in Fig 1.41

Q1 Qin Amount of heat supplied to water in boiler from a

high-temperature source (furnace)

Q2 Qout Amount of heat rejected from steam in condenser to a

low-temperature sink (the atmosphere, a river, etc.)

Wout Amount of work delivered by steam as it expands in turbine

Win Amount of work required to compress water to boiler pressure

Fig. 1.41 Schematic of a steam power plant

Thermodynamics 1.71

The net work output of this power plant is simply the differencebetween the total work output of the plant and the total work input.

Wnet Wout Win (kJ) WT WP

The efficiency of a heat engine or a heat engine cycle is defined as

Network output of the cycleTotal heat input to the cycle

Wnet

Q1

WT WP

Q1

By the first law of thermodynamics Q W

For a heat engine, Q1 Q2 WT WP

Q1 Q2

Q1

1 Q2

Q1

This is also known as the thermal efficiency of a heat engine.

1.36 THE SECOND LAW OF THERMODYNAMICS: KELVIN-PLANCK STATEMENT

Even under ideal conditions, a heat engine must reject some heat to alow-temperature reservoir in order to complete the cycle. That is, no heatengine can convert all the heat it receives to useful work. This limitation onthe thermal efficiency of heat engines forms the basis for the Kelvin-Planckstatement of the second law of thermodynamics, which is expressed asfollows:

It is impossible for any device that operates on a cycle to receive heatfrom a single reservoir and produce a net amount of work.

Efficiency Wnet

Q1

is always less than unity because heat transfer to the system cannot

be completely converted into work transfer. So, 100%. There is alwaysheat rejection.


Kelvin-Planck Statement in simple wordsIt is impossible to construct a

heat engine which will convert all heatenergy into equal amount of work in acyclic process. i.e., All heat cannot beconverted into work. Some heat will berejected to the surroundings (Sink).

That is, a heat engine mustexchange heat with a low-temperaturesink as well as a high-temperaturesource to keep operating. TheKelvin-Planck statement can also beexpressed as no heat engine can havea thermal efficiency of 100 percent(Fig. 1.42), or as for a power plant tooperate, the working fluid must exchange heat with the environment aswell as the furnace.

Perpetual motion of machine kind II - PMM2A heat engine which will produce work in a complete cycle by

exchanging heat with only one reservoir, is called PMM2. This engine is notpossible since it violates second law - Kelvin Planck Statement.

Fig. 1.42 A heat engine that violatesthe Kelvin-Planck statement of the

second law

Fig. 1.43 PPM2 and Possible Engine

Thermodynamics 1.73

1.37 THE SECOND LAW OF THERMODYNAMICS: CLAUSIUS STATEMENT

It is impossible to construct a device that operates in a cycle andproduces no effect other than the transfer of heat from a lower-temperaturebody to a higher-temperature body.

In otherwords, it is impossible toconstruct a device which will transfer theheat from cold body to hot body withoutexternal work.

[i.e., Heat cannot flow from cold body to hotbody without any external work]

It is a common knowledge that heatdoes not flow from a cold medium to awarmer one. The Clausius statement simplystates that a refrigerator, will not operateunless its compressor is driven by a externalpower source, such as an electric motor (Fig.1.44). This way, the net effect on thesurroundings involves the consumption ofsome energy in the form of work, in addition

Fig. 1.44 (a)&(b)

Fig. 1.44(c) A refrigerator thatviolates the Clausius statement

of the second law.


to the transfer of heat from a colder body to a warmer one. Therefore, ahousehold refrigerator is in complete compliance with the Clausius statementof the second law.

Both the Kelvin- Planck and the Clausius statements of the second laware negative statements. Until now, no experiment has been conducted thatcontradicts the second law, and this should be taken as sufficient evidenceof its validity.

1.38 REFRIGERATORS AND HEAT PUMPS

1.38.1 RefrigeratorWe all know that heat

flows in the direction ofdecreasing temperature, that is,from high-temperature medium tolow-temperature one. This heattransfer process occurs in naturewithout requiring any devices. Thereverse process, however, cannotoccur by itself. The transfer ofheat from a low-temperaturemedium to a high-temperature onerequires special devices calledrefrigerators.

Refrigeration is the process of maintainingthe temperature of a body below that of itssurroundings.

The working fluid used for this purpose iscalled Refrigerant, the equipment used is calledRefrigerator.

The refrigerant is compressed to higher pressureand

high temperature in the compressor and it getscooled in the condenser at constant pressure.

The high pressure refrigerant vapour isexpanded in the expander and the required quantityof refrigerant evaporates in the evaporator byabsorbing heat from the space to be cooled and thecycle is repeated.

Condenser

Evaporator

Expander

Com pressor

Fig 1.45 (a)

Fig. 1.45 (b)

Thermodynamics 1.75

Refrigerators, like heat engines, are cyclicdevices. The working fluid used in the refrigerationcycle is called a refrigerant.

A refrigerator is shown schematically in Fig.1.45 (a). In Fig. 1.45 (b), Q2 is the magnitude of the

heat removed from the refrigerated space attemperature T2. Q1 is the magnitude of the heat

rejected to the warm environment at temperature T1

and Win is the net work input to the refrigerator.

1.38.2 Heat PumpHeat pump is a device which maintains

temperature of a body greater than that of itssurroundings.

Another device that transfers heat from alow-temperature medium to a high-temperature oneis the heat pump, shown schematically in Fig. 1.46Refrigerators and heat pumps operate on the samecycle but differ in their objectives. The objective of a refrigerator is tomaintain the refrigerated space at low temperature by removing heat from it.

The objective of a heat pump, is to maintain a heated space at hightemperature. This is accomplished by absorbing heat from a low- temperaturesource, such as well water or cold outside air in water, and supplying thisheat to the high-temperature medium such as a house.

1.39 COEFFICIENT OF PERFORMANCE - COPThe efficiency of a refrigerator is expressed in terms of the Coefficient

of Performance (COP). It is a performance parameter. The objective of arefrigerator is to remove heat Q2 from the refrigerated space. To accomplish

this objective, it requires a work input of Win. Then the COP of a refrigerator

can be expressed as

COPR Desired outputRequired input

Q2

Win

This relation can also be expressed in rate form by replacingQ2 by Q

2 and Win by W

in.

Fig.1.46 The objectiveof a heat pump is to

supply heat Q1 into thewarmer space


The conservation of energy principle for a cyclic device requires that,Win Q1 Q2

Then the COP relation can also be expressed as

COPR Q2

Q1 Q2

1Q1/Q2 1 ... (i)

Notice that the value of COPR can be greater than unity. That is the

amount of heat removed from the refrigerated space can be greater than thatof the amount of work input.

The measure of performance of a heat pump is also expressed in termsof the Coefficient Of Performance COPHP, defined as

COPHP Desired outputRequired input

Q1

Win

which can also be expressed as

COPHP Q1

Q1 Q2

11 Q2/Q1 ...(ii)

A comparison of Equations (i) and (ii) reveals that

COPHP COPR 1

for fixed values of Q2 and Q1

1.40 EQUIVALENCE OF THE TWO STATEMENTS

The Kelvin-Planck and the Clausius statements are equivalent in theirconsequences, and either statement can be used as the expression of thesecond law of thermodynamics. Any device that violates the Kelvin-Planckstatement also violates the Clausius statement, and vice versa. This can bedemonstrated as follows.

Consider a heat enginerefrigerator combination shown in Fig. 1.47 (a)operating between the same two reservoirs. The heat engine is assumed toviolate the Kelvin-Planck statement and have a thermal efficiency of 100percent, and therefore it converts all the heat Q1 to work W. This work is

now supplied to a refrigerator that removes heat in the amount of Q2 from

the low-temperature reservoir and rejects heat in the amount of Q2 Q1 to

the high-temperature reservoir. During this process, the high-temperaturereservoir receives a net amount of heat Q2 (the difference between

Thermodynamics 1.77

Q2 Q1 and Q1). Thus, the combination of these two devices can be viewed

as a refrigerator, as shown in Fig. 1.47 (b), that transfers heat in an amountof Q2 from a cooler body to a warmer one without requiring any input from

outside. This is clearly a violation of the Clausius statement. Therefore, aviolation of the Kelvin-Planck statement results in the violation of the Clausiusstatement.

It can also be shown in a similar manner that a violation of theClausius statement leads to the violation of theKelvin-Planck statement. Therefore, theClausius and the Kelvin-Planck statements aretwo equivalent expressions of the second lawof thermodynamics.

1.40.1 Perpetual-Motion MachinesA process cannot take place unless it

satisfies both the first and second laws ofthermodynamics. Any device that violateseither law is called a Perpetual-MotionMachine.

A device that violates the first law ofthermodynamics (by creating energy) is calleda Perpetual-Motion Machine of the first

Fig. 1.47 Proof that the violation of the Kelvin-Planck statement leads to theviolation of the Clausius statement

PUM PTURBINE

Sys tem bou ndary

W ne

t

.

Q��

.

A perpetual-motion machinethat v io lates the second law of therm odynamics (PM M2).

Fig. 1.48


kind (PMM1), and a device that violates the second law of thermodynamicsis called a Perpetual- Motion machine of the second kind (PMM2).

1.41 CARNOT CYCLEThe best known reversible cycle is the Carnot cycle, first proposed in

1824 by French engineer Sadi Carnot. The theoretical heat engine that operateson the Carnot cycle is called the Carnot heat engine. The Carnot cycle iscomposed of four reversible processes-two isothermal and two adiabatic–and itcan be executed either in a closed or a steady-flow system.

However, Carnot cycle is an imaginary cycle. It gives maximumefficiency. We are studying carnot cycle to compare the efficiencies of othercycles (Air Standard cycles).

The efficiency of Carnot Cycle can be calculated as follows:The Carnot Cycle consists of following processes.

Process 1-2: Reversible Isothermal expansion process

PV C ; P1V1 P2V2

Process 2-3: Isentropic expansion process. P2V2 P3V3

; PV C

Carnot Cycle

Carnot Cycle

1

2

3

4

Q 1

Q 2

P-V diagram

Isotherm al Exp ansion

Isen trop icExp ansion

Isen trop iccom pressio n

Isotherm alcom pressio n

T 2

T 1

(P)

W c

W E

S ource T1

Q 1

Q 2

S ink T 2

S ystem A diaba tic

W E

W C

T 2

T 1

T

1 2

34

T-S diagram

s

Therm al C onductor(D iatherm ic C over)

Therm al Insulator(Adiabatic Cover)

V

T = const1

T = const2

Fig. 1.49 (a)

Thermodynamics 1.79

Process 3-4: Reversible Isothermal compression process.

P3V3 P4V4 ; PV C

Process 4-1: Isentropic Compression process. P4V4 P1V1

The area within the PV diagram represents the networkWnet WE WC or Q1 Q2 of this cycle. The net effects of the cycle are

(i) Heat is added to gas from the high temperature reservoir (source)at T1.

(ii) Heat is rejected from the gas to the low temperature reservoir(sink) at T2.

(iii) Work is produced.

First law states that

Wnet Heat added Heat rejected, Wnet Q1 Q2

Heat added to cylinder at T1, Q1 mRT1 ln V2

V1 (Isothermal process)

Heat rejected from the cylinder at T2, Q2 mRT2 ln V2

V1 (Isothermal process)

Efficiency of carnot cycle Work done

Heat supplied

Q1 Q2

Q1

mRT1 ln V2

V1 mRT2 ln

V2

V1

mRT1 ln V2

V1

T1 T2

T1

So Carnot cycle efficiency carnot Q1 Q2

Q1

T1 T2

T1

WnetQ1

Note: Carnot cycle requires the piston to travel very slowly duringisothermal expansion process to maintain the temperature constant at T1.

During remaining process (adiabatic expansion) of this cycle, it requires thepiston to travel quickly (very fast). This is not practically possible. So carnotcycle is an ideal cycle which gives maximum efficiency.


Problem 1.21: A Carnot engine working between 400C and 40C produces130 kJ of work. Calculate (i) the cycle efficiency; (ii) the heat added; (iii) theentropy changes during heat rejection process.

Solution:

Source temperature T1 400 273 673 K T2

Sink temperature T3 T4 40 273 313 K

Work produced Wnet 130 kJ

To Find Carnot Cycle Efficiency

carnot T1 T2

T1

673 313673

0.535 53.5%

To Find Heat Added

carnot Wnet

Q1

(where Q1 heat added,

Q2 heat rejected and

Wnet Q1 Q2

Q1 Wnet

carnot

1300.535

243 kJ

To Find Entropy Change during Heat Rejection Process s3 s4

Q T ds

Heat rejected Q2 T3 s3 s4

s3 s4 Q2

T3

But Wnet Q1 Q2

Q2 Q1 Wnet 243 130 113 kJ

Now s3 s4 Q2

T3

113313

0.361 kJ K

Carnot Cycle

1

2

3

4

Q 1

Q 2

P-V diagram

Isothe rm al Expansion

Isen trop icExpansion

Isen trop iccom press ion

Isothe rm alcom press ion

T 2

T1

(P)

W c

W E

T2

T1

T

1 2

34

T-S diagrams

V

Carnot Cycle

Thermodynamics 1.81

Problem 1.22: In a carnot cycle, the maximum temperature and pressureare given as 410C and 18 bar respectively. The ratio of isentropiccompression is 6 and isothermal expansion is 1.5. Assuming the volume of

air at the beginning of isothermal expansion as 0.18 m3, determine (i) thetemperature and pressure at all cordinal points (ii) change in entropy duringisentropic expansion.

Solution: Maximum pressure P1 18 bar ;

Max. temperature T1 T2 410 273 683 K;

Isentropic compression ratio V4V1

6

Isothermal expansion ratio V2V1

1.5

Volume of air at the beginning of isothermal expansion V1 0.18 m3

To Find T and P at all Cordinal PointsProcess 4-1 Isentropic Compression Process

T1

T4 V4V1

1

60.4

T4 T1

60.4 683

60.4 333.2 K

T3 T4 333.2 K

Also T1

T4 P1

P4

1

P1

P4 T1

T4

1

P4 P1

T1

T4

1

18

683333.2

1.40.4

1.46 bar

1

2

3

4

Q 1

Q 2

P-V diagram

Iso the rm a l E xpa nsion

Isen trop icE xpa nsion

Isen trop icco m pression

Isothe rm a lco m pression

T 2

T 1

P

W c

W E

T 2

T 1

T1 2

34

T-S diagram

s

V


Process 1-2 Isothermal Process P1V1 P2V2 T1 T2 683 K

P2 P1V1

V2

181.5

12 bar . . . Given

V2

V1 1.5

Process 2-3 Isentropic Process

P2V2 P3V3

P3 P2 V2

V3 P2

V2

V3

P2 V1V4

12 16

1.4

0.97 bar . . .

V4

V1

V3

V2

ResultP1 18 bar ; T1 T2 683 K; 12 bar ; P3 0.97 bar

P4 1.46 bar ; T3 T4 333.2 K

To Find Change in Entropy (during Isothermal Expansion)

Process 1-2

T1 s2 s1 heat added Q1 mRT1 ln V2

V1

s2 s1 P1V1

T1 ln

V2

V1 [ . . . P1V1 mRT1 ]

18 105 0.18

683 ln 1.5 192 J K 0.192 kJ K

1.41.1 The Carnot Principles (or) Carnot TheoremTwo conclusions pertain to the thermal efficiency of reversible and

irreversible (i.e. actual) heat engines, and they are known as the Carnotprinciples (or) Carnot theorem (Fig. 1.49 (b)) expressed as follows:

1. The efficiency of an irreversible heat engine is always less thanthe efficiency of a reversible one operating between the same tworeservoirs.

2. The efficiencies of all reversible heat engines operating betweenthe same two reservoirs are the same.

Thermodynamics 1.83

1. Carnot’s theorem (I)All heat engines operating between

a given constant temperature, source anda given constant temperature sink, nonehas a higher efficiency than reversibleengine.

Carnot engine is a reversible enginehaving higher thermal efficiency than anyother engines.

Carnot engines

Carnot engine Q1 Q2

Q1

Wnet

Q1

T1 T2T2

actual engine Q1 Q2

Q

Wnet

Q1

T1 T2

T1

carnot

actual irreversible heat engine

actual reversible heat engine

actual impossible heat engine

If carnot actual, then it is actual engine possible engine irreversible engine

if carnot actual, then it is reversible engine

if carnot actual, then it is not possible

2. Carnot Theorem IIAll reversible engines working between same temperature limits will

have same efficiency.

3. Carnot Theorem IIIEfficiency of ideal engine (Carnot engine) working between two fixed

temperature is independent of working substances, but dependent only ontemperature.

High-tem perature reservoirat T 1

1Irrev.HE

2Rev.HE

3Rev.HE

Low-tem perature reservoirat T 2

th,1 th ,2< th,2 th ,3=

Fig. 1.49(b) The Carnot principles.


Problem 1.23: A carnot (reversible) engine works between 300C and

30C. The heat supplied to the engine is 20 kJ. Determine 1. efficiency;2. Work output; 3. Heat rejection

Solution

T1 300 273 573 K

Q1 20kJ

T2 30 273 303 K

carnot T1 T2

T1

573 303573

0.4712 47.12%

carnot 0.4712 WQ1

Work output carnot Q1 0.4712 20 9.4241 kJ

Heat rejection Q2,W Q1 Q2

Q2 Q1 W 20 9.4241 10.5792 kJ

Problem 1.24: A refrigerator plant operates on reversed carnot heat enginecycle. It is maintained at a temperature of 5C and the heat is rejected at

the rate of 5 kW. The atmospheric temperature is 25C. Calculate the powerrequired to drive the plant.

Solution T2 5 273 268 K

T1 25 273 298 K;

Q2 5kW W Q1 Q2

COP T2

T1 T2

Q2

Q1 Q2

268

298 268 8.9333

Power Work in kW

Carnot COP Q2

W

H.E

T 2

Q 2

Q 1

T1 = 573 K

= 20 kJ

= 303 K

W

Q 2Q 1Ref

T2

Q 2

Q 1

T1 = 298

= 268 K

W = -

=5kW

Thermodynamics 1.85

Power

W Q2

Carnot COP

58.9333

0.5597 kW

1.41.2 Corollaries of Carnot’s Theorem

Corollary 1All reversible heat engines operating between the same thermal

reservoirs must have the same efficiency.

i.e., the efficiency of reversible heat engine is independent of the typeof engine, amount and nature of the working fluid used and also other factors.

Corollary 2The efficiency of a reversible heat engine is a function of their

respective temperature of the two thermal reservoirs. It can be evaluated byreplacing the ratio of heat transfers Q2 and Q1 by the ratio of temperature

T2 and T1 of their respective heat reservoirs.

Therefore, the efficiency of reversible heat engine

rev WQ1

Q1 Q2

Q1 1

Q2Q1

1 T2

T1

Q1

Q2 Rev

T1

T2

1.42 CLAUSIUS INEQUALITY

An important inequality that has major consequences inthermodynamics is the Clausius inequality. It was first stated by the Germanphysicist R. J.E. Clausius (1822-1888), one of the founders ofthermodynamics, and is expressed as

O QT

0

Here the cyclic integral of Q/T is always less than or equal to zero.This inequality is valid for all cycles, reversible or irreversible. The symbol

O (integral symbol with a circle in the middle) is used to indicate that the

integration is to be performed over the entire cycle.

The inequality of Clausius is used to find whether the cycle isreversible (or) irreversible.


If O dQT

0, the cycle is reversible

O dQT

0, the cycle is irreversible and possible.

O dQT

0, the cycle is impossible, since it violates the second law.

Problem 1.25: A heat engine is supplied with 19 kW of heat from 565 Kreservoir and rejects heat to 282.5 K reservoir. Which of the following engineis irreversible engine, reversible engine and impossible engine:Case (a): If 14.0833 kW of heat is rejected.Case (b): If 4.75 kW of heat is rejected.Case (c): If 9.5 kW of heat is rejected.Use Carnot method and Clausius inequality method.

Solution

By Carnot Method

carnot T1 T2

T1

565 282.5

565

0.49558 0.5

Case (a): Q2 14.0833 kW

actual Q1 Q2

Q1

19 14.083319

0.258774

Since actual 0.25 carnot 0.5, it is irreversible engine.

Case (b): If Q2 4.75 kW

actual 19 4.75

19 0.75

Since actual0.75 carnot0.5, it is impossible engine.

Q 2Q 1H E

T 2

Q 2

Q 1

T 1 =5 65 K

= 282 .5 K

W = -

=19kW

Thermodynamics 1.87

Case (c): If Q2 9.5 kW

actual 19 9.5

19 0.5

Since actual carnot, it is reversible engine.

By Clausius Inequality Method

If QT

‘’, then irreversible engine.

If QT

0, then reversible engine.

If QT

‘’, then impossible engine.

Case (a): Q2 14.0833 kW

QT

Q1

T1

Q2

T2

19565

14.0833

282.5

0.0162, so it is irreversible engine.

Case (b): Q2 4.75 kW

QT

19

565

4.75282.5

0.0168 so it is impossible engine

Case (c): Q2 9.5 kW

QT

19565

9.5

282.5

0, so it is reversible engine.

Problem 1.26: An inventor claims to have developed an engine which givesan efficiency of 55% when working under a source of 1000 K and sink of500 K. Check whether his claim is correct or not.


Solution

carnot T1 T2

T1

1000 500

1000

0.5

Since actual0.55 carnot, it is not possible.

The inventor is telling lie.

Problem 1.27: A refrigerator removes heat from a refrigerated space at2C at a rate of 300 kJ/min and rejects heat to kitchen air at 26C at arate of 345 kJ/min. Verify whether it violates II law by (i) Clausius inequality(ii) Carnot theorem. (FAQ)

Solution

T2 2 273 275 K; Q2 30060

5 kW

T1 26 273 299 K; Q1 34560

5.75 kW

Clausius Inequality

QT

Q1

T1

Q2

T2

5.75

299

5275

[Q1 is‘’ because heat is removed

from the system]

1.048 10 3

Since it is ‘’ ve, it is irreversible refrigerator.

Carnot Principle

COPcarnot T2

T1 T2

275299 275

11.4583

COPactual Q2

Q1 Q2

55.75 5

6.6667

HE

T2

Q 2

Q 1

T1 =1000 K

= 500 K

W

R ef

T 2

Q 2

Q 1

T 1 = 299 K

= 2 75

W

= 5 .7 5kW

= 5kW

Thermodynamics 1.89

Since C OPactual6 . 6 6 6 7

C OPcarnot11.4583

, it is possible irreversible refrigerator.

Problem 1.28: An inventor claims that his engine develops 3 kW of workfor a heat addition of 240 kJ/min. The highest and lowest temperature of thecycle are 1527C and 327C. Would you agree his claim. [Use clausiusinequality method]. (FAQ)

Solution

W 3 kW Q1 24060

4 kW

T1 1527 273 1800 K;

T2 327 273 600 K

Clausius Inequality Method

QT

Q1

T1

Q2

T2

4

1800

1600

5.55 10 4

It violates the II law.

So it is impossible.

Extra: Carnot Principle

carnot T1 T2

T1

1800 6001800

0.6667

actual Q1 Q2

Q1

4 14

0.75

Since actual carnot, it is impossible. It violates the Carnot theorem

1.43 ENTROPY

Entropy can be viewed as a measure of molecular disorder, ormolecular randomness. As a system becomes more and more disordered, theposition of the molecules become less predictable and the entropy of thesystem increases. Thus, it is not surprising that the entropy of a substance islowest in the solid phase and highest in the gas phase (Fig. 1.50). In thesolid phase, the molecules of a substance continually oscillate about their

H E

T 2

Q 2

Q 1

T 1 = 1800 K

= 600

W

= 4 kW

= 3 kW

= Q -W = 1 kW1


equilibrium positions, but they cannotmove relative to each other, and theirposition at any instant can be predictedwith good certainty. In the gas phase,however, the molecules move about atrandom, collide with each other, andchange direction, making it extremelydifficult to predict accurately themicroscopic state of a system at anyinstant. Associated with this molecularchaos is a high value of entropy.

Therefore, from a microscopic pointof view, the entropy of a system increaseswhenever the molecular randomness oruncertainty (i.e., molecular probability) of asystem increases. Thus, entropy is ameasure of molecular disorder, and themolecular disorder of an isolated systemincreases anytime if it undergoes a process.

The gas molecules will not rotate apaddle wheel inserted into the container andproduce work. This is because the gasmolecules, and the energy they possess,are disorganized. Probably the number ofmolecules trying to rotate the wheel in onedirection at any instant is equal to thenumber of molecules that are trying torotate it in the opposite direction, causingthe wheel to remain motionless. Therefore,we cannot extract any useful work directlyfrom disorganized energy (Fig. 1.51)

Now consider a rotating shaft shownin Fig. 1.52. This time the energy of themolecules is completely organized sincethe molecules of the shaft are rotating inthe same direction together. This organizedenergy can readily used to perform usefultasks such as raising a weight or generatingelectricity. Being an organized form of

SO LID

L IQ U ID

G A S

T h e le v e l o f m o le c u la r d iso rd er(e n tro p y) o f a su bs tan ce in c re a s e s

a s i t m elts o r e vap o ra tes .

E n trop y,kJ /kg K.

Fig 1.50

LOA D

D isorganized energy does not create m uch useful effect, no m atter how large it is.

Fig. 1.51 Disorganized energydoes not create much useful

effect, no matter how large it is

W E IGH T

W sh

Fig. 1.52 In the absence offriction, raising a weight by arotating shaft does not create

any disorder (entropy) and thusenergy is not degraded during

this process

Thermodynamics 1.91

energy, work is free of disorder or randomness and thus free of entropy. Thereis no entropy transfer associated with energy transfer as work. Therefore, inthe absence of any friction, the process of raising a weight by a rotating shaft(or a fly wheel) will not produce any entropy. Any process that does notproduce a net entropy is reversible, and thus the process described above canbe reversed by lowering the weight. Therefore, energy is not degraded duringthis process, and no potential to do work is lost.

1.43.1 Concept of EntropyEntropy is an extensive property of a system and sometimes it is also

referred as total entropy. Entropy per unit mass, designated as s, is anintensive property and has the unit kJ/kg K. The term entropy is generallyused refer to both total entropy and entropy per unit mass since the contextusually clarifies which one is meant.

The entropy change of a system during a process can be determinedby integrating Eq. (i) between the initial and the final states:

S S2 S1 1

2

QT

int rev

kJ/K

Note that the entropy is a property, andlike all other properties, it has fixed values atfixed states. Therefore, the entropy changeS between two specified state is the sameno matter what path, (reversible orirreversible) is followed during a process(Fig. 1.53).

There is a close link between entropyand molecular disorder. When work isdissipated into internal energy, the moleculardisorder increases. So entropy increases.When two gases are mixed together, there isa molecular disorder. So entropy increases. Soentropy can be defined as the measure of thedegree of molecular disorder existing in thesystem.

When heat is added to the system,molecular disorder increases so entropy increases.

The entropy of the isolated system either increases or remains constant

1

2

0.3 0.7

Revers ibleprocess

S ,kJ /K

T

Irrevers ib leprocess

S S S= - =0.4 kJ/k2 1

Fig. 1.53 The entropy changebetween two specified state isthe same whether the process

is reversible or irreversible


dSisolated 0.

The combination of the system and the surroundings is known as the universe.

So dSuniverse dSsystem dSsurroundings 0

1.43.2 Characteristics of entropy1. Entropy of the system increases when heat is added to system.

2. Entropy of the system decreases, when heat is removed from thesystem.

3. Entropy is constant in isentropic process (Reversible adiabaticprocess).

Change in entropy S2 S1 1

R

2

dQT

where R represents reversible path.

1 Initial State, 2 Final State.Let a system is changed from initial state1 to final state 2 through a reversible path,R1. Then the system is again changed

back from 2 to 1 by another reversiblepath R2. Refer the Fig. 1.54. Now the two

reversible paths R1 and R2 form a cycle.

By using Clausius theorem

OR1 R2

dQT

0

The above cyclic integral can be replaced by the summation of twointegrals as follows.

1

R1

2

dQT

2

R2

1

dQT

0

P

R 1

R 2

2

V

1

Two revers ib le paths R and Rbetween tw o equ ilibrium

states 1 and 2

1 2

Fig. 1.54

Thermodynamics 1.93

(or) 1

R1

2

dQT

2

R2

1

dQT

Since R2 is a reversible path, we can write as follows.

1

R1

2

dQT

1

R2

2

dQT

Since R1 and R2 are any two reversible paths, the value 1

R

2

dQT

is

independent of the reversible path connecting 1 and 2. Therefore 1

R

2

dQT

represents Change in one property of the system and that property is knownas entropy.

S1 entropy at initial state;

S2 entropy at final state

1

R

2

dQT

change in entropy S2 S1

If two end states are very near, then we can write,

dQR

T dS

The unit of entropy S is J/K

Specific entropy is s and the unit is J/kgK

. . . s

Sm

J/kgK

Refer Fig. 1.55. When the system is brought from an initial state 1 toa final state 2 by an irreversible path, the irreversible path can be replacedby reversible path to integrate for evaluation of change in entropy in the


irreversible process. Because, the entropyis a point function. It is a property. It isindependent of the path followed. So wecan replace irreversible path by reversiblepath and calculate the entropy.

Heat which is supplied to a hightemperature substance is converted moreinto work than the heat supplied to a lowtemperature substance.

Entropy shows the possibility ofconversion of heat into work. The changein entropy is more when heat is added ata lower temperature and the change in entropy is less when heat is added ata higher temperature. i.e., for a maximum entropy change, there is aminimum availability for conversion into work and for a minimum entropychange, there is a maximum availability for conversion into work.

Solved problemsProblem 1.29: When water is flowing through turbine, its temperature israised from 35C to 40C due to friction. No heat transfer occurs duringthis process. Assume it is a constant volume process. Find the change inentropy of water while it is passing through the turbines.

Solution

Since friction causes the temperatureraise, this process is taken as irreversibleprocess. Refer the T S diagram. Theirreversible process is shown by dotted line.We can replace the irreversible process byreversible process, since the entropy changedepends on only the two end states. It isindependent of the path the system follows.

T2 40 273 313 K T1 35 273 308 K

We know dQrev TdS

dS dQrev

T

T

T

R

2

1

ds

ss1 s 2

Rev.pathw hich replaces

the irreversible path

Integration can be done onlyon a reversible path

Fig. 1.55

T

irreve rs ible path 2

dss

dQ rev

R evers ib le pa thconnecting the in itia l a nd fina l equ ilibr iumstates

1

Thermodynamics 1.95

dS m cv dT

T

Integrating on both sides, we get

S2 S1 mcv ln T2

T1

S2 S1 1 4.187 ln T2

T1

... mass is taken as 1 kg . Cv for water 4.187 kJ/KgK

1 4.187 ln

313308

0.0674

Change in entropy of water S2 S1 0.0674 kJ / K

Problem 1.30: In a closed system air is at a pressure of 1 bar, temperature

of 300 K and volume of 0.025 m3. The system executes the following processesduring the completion of thermodynamic cycle : 1-2; constant volume heataddition till pressure reaches 3.8 bar, 2-3; constant pressure cooling of air,3-1; isothermal heating to initial state. Determine the change in entropy ineach process. Take Cv 0.718 kJ/kg K, R 287 J/kg K. (FAQ)

Solution:

T3 T1 300 K, P1 1 bar 1 105 N/m2

V1 V2 0.025 m3

P2 3.8 bar 3.8 105 N/m2

R 287 J/kgK ; CV 0.718 kJ/kgK

Change in entropy in each process

(i) Process 1 2 constant volume process

S2 S1 mCv ln T2

T1

we know, m P1V1

RT1

1 105 0.025

287 300 0.029 kg

for a perfect gas at constant volume


P1

T1

P2

T2

T2 P2

P1 T1

3.8 105

1 105 300 1140 K

S2 S1 0.029 0.718 ln 1140300

0.02779 kJ/K

(ii) Process 2 3 constant pressure process

S3 S2 mCp ln T3

T2

We know CP Cv R CP R Cv 0.287 0.718

CP 1.005 kJ/kg K.

S3 S2 0.029 1.005 ln

3001140

0.0389 kJ/K

Negative sign indicates there is a decrease in entropy.

Thermodynamics 1.97

(iii) Process 3 1 constant temperature process

S1 S3 S2 S3 S2 S1 0.0389 0.02779

0.0111 kJ/K

Problem 1.31: An heat exchanger circulates 5000 kg/hr of water to cooloil from 150C to 50C. The rate of flow of oil is 2500 kg/hr. The averagespecific heat of oil is 2.5 kJ/kgK. The water enters the heat exchanger at21C. Determine the net change in the entropy due to heat exchange process,and the amount of work obtained if cooling of oil is done by using the heatto run a Carnot engine with sink temperature of 21C. (FAQ)

SolutionThe Heat transfer by oil Qoil moil Cpoil Toil

25003600

2.5 150 50 173.611 kJ

Heat exchange by water mwCpw Hw

Heat lost by oil Heat gained by water

173.611 50003600

4.184 t2 21

t2 50.87 C

Final temperature of water 50.87 C

entropy change of oil soil mCpln T2

T1

25003600

2.5 ln 323423

0.46826 kJ/K

entropy change of water swater mCpln t2t1

50003600

4.184 323.87

294

6.4015 kJ/K

net change in entropy soil swater 5.93324 kJ/K


Calculation of amount of work

we know that, T1 T2

T1

423 294

423

0.30496 30.496%

we also know, WQ1

W Q1

0.30496 173.611 52.945 kJ

1.44 THIRD LAW OF THERMODYNAMICS

As we discussed earlier that, for a carnot engine to have a maximumefficiency, the sink temperature T2 must be lowered. To achieve progressively

lower temperatures, the process becomes more and more difficult. This meansto attain absolute zero, infinite number of operations have to be performedwithout violating the other laws of thermodynamics. This observation iscontained in the third law of thermodynamics.

“It is impossible to attain the absolute zero by a finite number ofoperations”.

Just like second law, the third law has an equivalent number ofstatements.

(i) FowlerGuggenheim statement of the third law:

It states that “It is impossible by any procedure, no matter howidealized, to reduce any system to the absolute zero of temperature in a finitenumber of operations”.

(ii) Nernst-Simon statement of the third law: It states that “The entropies of all systems and the entropy changes in

all reversible isothermal process tend to zero as the temperature approacheszero”.

Mathematically,

lim T 0

S 0; lim T 0

S 0

Source

150 Co

S ink

21 Co

RevH.E

W

Q =173.611kJ1

Thermodynamics 1.99

This was experimentally discovered by Nernst. He observed that thenet change in entropy of a system is very small, when there is a change fromone lower temperature equilibrium state to another.

For example,

Take separate samples of sodium and chlorine at 4C and convert itinto sodium chloride at the same temperature.

This means that, near the absolute zero, all the systems are highlyordered and the entropy of all states is the same and the values of entropyare extremely small or zero.

In simple words, the principles of first law and second law can beused to calculate the entropy difference only, but to find the absolute valueof entropy, the third law of thermodynamics is required.

Note: Efficient people lead low-entropy (highly organized) lives. They havea place for everything (minimum uncertainty), and it takes minimum energyfor them to locate something. Inefficient people, on the other hand, aredis-organized and lead high-entropy lives. It takes them long time (if nothours) to find something they need, and they are likely to create a biggerdisorder as they are searching since they will probably conduct the searchin a disorganized manner. People leading high-entropy lifestyles are alwayson the run, and never seem to catch up.

1.45 PROPERTIES OF IDEAL GAS

A hypothetical gas which obeys the law Pv R

T (equation of state) iscalled an ideal gas.

An ideal gas is a theoritical gas consists of four gas variables i.e.,Pressure P, Volume V, Temperature T and number of mole of a gas

n. Lastly, the constant in the equation is R

, which is the universal gasconstant R

8.314 kJ/kg. mol. k

PV

R

T ...(1)

V

molar specific volume m3

kg. mole

Divide the above equation by molecular weight , then the equationbecomes,

Pv RT

where Pvm

RT; R R


molecular weight kgkg

mol

R characteristic gas constant (kJ/kg k)

Also, specific volume, v V

Also, m n

n number of mole of a gas (kg moles)

m mass of the gas (kg)

Therefore,

PV n R

T ...(2)

PV n R T

PV m RT ...(3)

Equation 1,2,3 is called the ideal gas equation of state.

At very low pressure or density, all gases and vapour approaches idealgas behaviour.

Practically, no ideal gas (perfect gas) exists in nature. However,hydrogen, oxygen, nitrogen and air behave as an ideal gas under normalcondition.

Any equation that relates the pressure, temperature and specific volumeof a substance is called an equation of state which is given here as

Pv RT

where the constant of proportionality R is called the gas constant.A gas that obeys this relation is called an ideal gas.

The ideal gas equation is very simple and thus very convenient to use.But, gases deviate from ideal-gas behaviour significantly. This deviation fromideal gas behaviour at a given temperature and pressure can accurately beaccounted by the introduction of a correction factor called the compressibilityfactor Z defined as

Z PVRT

or PV ZRT

Thermodynamics 1.101

It can also be expressed as

Z Vactual

Videal where Videal

RTP

Obviously, Z 1 for ideal gases.

For real gases, Z can be greater than (or) less than 1 when the gasdeviates from ideal - gas behaviour

But real gases do not conform to this equation of state with completeaccuracy. Ideal gas do not have intermolecular attractive forces. These idealgases obeys the equation of state at all ranges of pressure and temperature.

Practically, no ideal gas (perfect gas) exists in nature. However, hydrogen,oxygen, nitrogen and air behave as an ideal gas under normal condition.

1.45.1 Equation of stateThe equation which relates the properties P, v and T is known as an

Equation of state.

ie f P, v, T 0

The simplest form of equation of state for the ideal gas is given below.

Equation of state: Pv R

T [ R

universal gas constant in kJ kg mol.K

and v molar volume in m3 kg mol ]

Also: Pv RT [ R characteristic gas constant in kJ kg K ]

v specific volume in m3 kgThe above equation - Equation of state is also called Characteristic gas equation.

1.45.2 Relationship Among PropertiesA mole of a substance has a mass numerically equal to the molecular

weight of the substance.

1 gm mol of oxygen has a mass of 32 gm, 1 kg mol of oxygen hasa mass of 32 kg and so on.

For a certain gas, if m is its mass in kg, and is its molecular weight,then the number of kg moles of the gas, n, is given by,

n m

kg.moles

The functional relationship among the properties, pressure P, molar orspecific volume v and temperature, T, known as the equation of state and itis expressed as


f P, v, T 0

If two of these properties of a gas are known, the third can beevaluated from the equation of state.

1.45.3 Mole Fraction

Mole fraction x of a gas is defined as the ratio between the numberof k mol of the gas and the total number of k mol of the gas mixture.

If ni No. of k mol of ith gas component

k No. of gases in the mixture

Therefore, xi ni

1

k

ni

xi ni

n

1

k

ni n1 n2 n3 nk n

1.45.4 Mass Fraction

Mass fraction y of a gas in a gas mixture is defined as the ratiobetween the mass of the gas and the total mass of the gas mixture.

If, mi Mass of ith gas component

k No. of gases in the mixture

yi mi

1

k

mi

yi mi

m

1

k

mi m1 m2 m3 mk m

1.45.5 Relationship Between Mass and Mole Fractions

The mass fraction yi of the component i is given by

yi mi

m, Where mi Mass of the component i

Similarly, the mole fraction xi of component i is given by


xi ni

n, Where ni Number of moles of component i

All the mass fractions yi and mole fraction xi are not independent and

there exists a relationship between them.

yi 1 and xi 1

The mass mi of a component i is related to the no. of moles ni and

the molar mass Mi of the component i.

mi niMi

Where the molar mass of a mixture M, can be defined as the ratioof the mass m of the mixture to the total no. of the moles n of the mixture.

M mn

imi

ini

iniMi

ini

i 1

xiMi

eg.If dry air consists of nitrogen and oxygen mixtures only, then

yN2 0.80 y02 0.25, then the molar mass,

M 0.8 28 0.25 32 30.4 kg/kmol

Problem 1.32: Calculate the molecular volume of all gases at

250 kN/m2 and 30 C.

Solution: P 250 kN/m2 ; T1 30 273 303 K.

We know universal gas constant R

8.3143 kJ/kg mol K

Equation of state: P v R

T

molar volume v

R

TP

8.3143 303

250 10.077 m3/kg mole

Problem 1.33: Calculate the molar volume of all gases at N.T.P. condition

Solution: N.T.P. Normal Temperature and Pressure

0 C and 760 mm of Hg (1.01325 bar i.e., 101.325 kN/m2)

We know universal gas constant R

8.3143 kJ/kg mol K

molar volume v

R

TP

8.3143 273

101.325 22.4 m3/kg mole


Problem 1.34: A cylinder of 120 lit capacity contains CO2 at 100 bar and

20 C. Determine (1) mass of the gas (2) molar volume (3) density of thegas (4) specific volume.

Solution:

R

8.3143 kJ/kg mole K ; V 120 10 3 m3

P 100 102 kN/m2 ; T 20 273 293 K

Molecular weight. for CO2 44;

Characteristic gas constant R

R for CO2 R

CO2

8.3143

44 0.18896 kJ/kgK

Mass of the gas m

PV mRT

m PVRT

100 102 120 10 3

0.18896 293 21.674 kg

Molar volume v Pv R

T v

R

TP

8.3143 293

100 102 0.2436 m3/kg mole

Density of the gas

Density of the gas mv

21.674

120 10 3 180.61 kg/m3

Specific volume v 1

1

180.61 5.54 10 3 m3/kg

1.45.6 Avogadro’s LawAvogadro’s law states that the volume of a g.mol of all gases is the

same at N.T.P. and is equal to 22.4 litres. [N.T.P. means Normal Temperature0 C and Pressure (760 mm of Hg is 1.01325 bar )].

1 g mol of gas has a volume of 22.4 litres.

1 kg mol of gas has a volume of 22.4 m3.

A mole of a substance has a mass equal to molecular weight ofthe substance.


So 1 kg mol of Oxygen has a mass of 32 kg [ . . . O2 32 ]

Similarly 1 kg mol of Nitrogen has a mass of 28 kg[ . . . N2

28 ]

For most of the gases, the number of kg moles of the gas n is given by

n mkg

kg

kg mol

m

kg moles

The molar volume v

Vn

m3/kg mol.

According to Avogadro’s law, the molar volume is same for all gasesat any given pressure and temperature and v

v

The molar volume of any gas at N.T.P. [ N.T.P. 0 C and 1.01325 bar ] isgiven by

v 22.4 m3/kg mole v

where molecular weight ; v specific volume of the gas

1.45.7 Universal Gas Constant R

and Characteristic Gas Constant R

From Avogadro’s law, when P 1.01325 105 N/m2, (760 mm of

Hg), T 273 K and v 22.4 m3/kg mol, the equation of state: P v

R

T

Universal gas constant R

P v

T

1.01325 105 22.4273

8314.3 Nm/kg mol K

8.3143 kJ/kg mol K

When molar volume v v,

then equation of state can be written as

P v R

T Pv R

T P

Vm

R

T

PV m R

T PV mRT

where R characteristic gas constant R


For any gas, we can use the equation R R

To find RO2, RO2

R

O2

8.3143

32 0.2598 kJ/kg K

and To find R for air, Rair R

air

8.314328.96

0.287 kJ/kg K

Therefore equation of state of an ideal gas can be written as

PV mRT Pv RT

Pv R

T [ v molar volume v ]

PV n R

T [ n number of kg moles ]

PV N KT [ where K R

A and

A Avogadro number 6.023 1026 molecules/kg mol

N Total number of molecules nA ]

1.45.8 Comparison between Ideal and Real Gases

Ideal gas Real gas

(i) An Ideal gas is an hypothetical gas As the name implies, these gasesare real and exist in nature.

(ii) It has no definite volume Real gas has definite volume

(iii) It has no mass It has mass

(iv) Ideal gas follows the equationPV

R

TReal gas follows the equation P

a

v2 v b RT

(v) Ideal gas follows gas laws at alltemperatures pressures

Real gas do not follow gas lawsat all temperatures & pressures

(vi) No force of attraction or repulsionbetween molecules

Force of attraction or repulsionbetween molecules is significant

(vii) Ideal gases don’t exist in natureand they don’t behave like real gasat any temperatures & pressures

Real gases behave like an idealgas at high temperatures and lowpressures


1.46 PROPERTIES OF PURE SUBSTANCESA substance with a fixed chemical composition throughout is called

pure substance. For example water, nitrogen, helium and carbondioxide areall pure substance.

A pure substance does not need to be of a singlechemical element or compound. However, a mixture ofvarious chemical elements or compounds also qualifies asa pure substance as long as the mixture is homogeneous.Since air is a mixture of several gases, it is consideredas a pure substance because it has a uniform chemicalcomposition (Fig. 1.56). However a mixture of oil andwater is not a pure substance. Since oil is not soluble inwater, it will come up on top of the water, forming twochemically dissimilar regions.

A mixture of two or more phases of a puresubstance is still a pure substance as long as the chemicalcomposition of all phases is the same (Fig. 1.57) Amixture of ice and liquid water, for example,is a pure substance because both phaseshave the same chemical composition. Amixture of liquid air and gaseous air,however, is not a pure substance since thecomposition of liquid air is different fromthe composition of gaseous air, and thus themixture is no longer chemicallyhomogeneous. This is due to differentcomponents in air condensing at differenttemperatures at a specified pressure.

1.46.1 THERMODYNAMIC PROPERTIES OF PURE SUBSTANCES IN SOLID, LIQUID AND VAPOUR PHASES

Consider a piston-cylinder device containing liquid water at 20C and1 atm pressure (state 1. Fig. 1.58). Under these conditions, water exists inthe liquid phase, and it is called a compressed liquid, or a subcooled liquid,meaning that is not about to vaporize. Heat is now transferred to the wateruntil its temperature rises to, say, 40C. As the temperature rises, the liquidwater expands slightly, and so its specific volume increases. To accommodatethis expansion, the piston will move up slightly. The pressure in the cylinderremains constant at 1 atm during this process since it depends on the outside

N 2

A IR

Fig 1.56 Nitrogenand gaseous air

are pure

VAP O R

LIQUID

(a) H O2

VAP O R

LIQUID

(b) A IR

Fig 1.57 A mixture of liquid andgaseous water is a pure

substance, but a mixture ofliquid and gaseous air is not


barometric pressure and the weight of thepiston, both of which are constant. Wateris still a compressed liquid at this statesince it has not started to vaporize.

As more heat is transferred, thetemperature will keep rising until itreaches 100C (state 2, Fig. 1.59). At thispoint water is still a liquid, but any heataddition will cause some of the liquid tovaporize. That is, a phase-change processfrom liquid to vapour is about to takeplace. A liquid that is about to vaporizeis called a saturated liquid. Therefore,state 2 is saturated liquid state.

1.46.2 Saturated Vapour and Superheated VapourOnce boiling starts, the temperature will stop rising until the liquid

is completely vaporized. That is, the temperature will remain constant duringthe entire phase-change process if the pressure is held constant. During aboiling process, the only change we will observe is a large increase in thevolume and a steady decline in the liquid levelas result of more liquid turning to vapour.

Midway about the vaporization line (state3, Fig. 1.60), the cylinder contains equalamounts of liquid and vapour. As we continuetransferring heat, the vaporization process willcontinue until the last drop of liquid is vaporized(state 4, Fig. 1.61). At this point, the entirecylinder is filled with vapour. Any heat lossfrom this vapour will cause some of the vapourto condense (phase change from vapour toliquid). A vapour that is about to condense iscalled a saturated vapour. Therefore, state 4 isa saturated vapour state. A substance at statesbetween 2 and 4 is often referred to as asaturated liquid-vapour mixture since the

STATE 1

P = 1a tmT = 20 CO

Heat

Fig 1.58 At 1atm and 20C,water exists in

the liquidphase

(compressed

STATE 2

P = 1atmT = 100 C

O

Heat

Fig 1.59 At 1atm pressure

and 100C, waterexists as a

liquid that isready to vaporize(saturated liquid)

Heat

Satu ratedvapor

Sa tu ratedliqu id

STATE 3

Fig 1.60 As more heat istransferred, part of the

saturated liquid vaporizes(saturated liquid vapour

mixture


liquid and vapour phases coexist in equilibrium at thesestates.

Further transfer of heat will result in an increase inboth the temperature and the specific volume (Fig. 1.62) Atstate 5, the temperature of the vapour is, let us say,300C; and if we transfer some heat from the vapour, thetemperature may drop somewhat but no condensation willtake place as long as the temperature remains above100C (for P 1 atm. A vapour that is not about tocondense (i.e., not a saturated vapour) is called asuperheated vapour. Therefore, water at state 5 is asuperheated vapour. This constant-pressure phase-changeprocess is illustrated on a T v diagram in Fig 1.63.

If the entire process described here is reversed bycooling the water while maintaining the pressure at the samevalue, the water will go back to state 1, retracing the samepath, and in doing so, the amount of heat released willexactly match the amount of heat added during the heatingprocess.

STATE 4

P = 1atmVapo ur

T = 100 CO

Heat

Fir 1.61 At 1atm pressure,

the temperatureremains

constant at100C until the

last drop ofliquid is

vaporized(satured vapour)

P = 1atm

Super heatedVapour

T = 300 CO

H eat

Fig 1.62A s m ore hea t is trans fe rred, thetem perature o f the vapor s ta rts

to rise ( ).supe rheated vapo r

STATE 5

TOC

300

100

20

Com

pre s

s ed

liqui

d

1

2 S a tu ra te d m ix tu re

3 Supe

rhea

ted

vapo

rP

= 1

a tm

5

4

Fig 1.63T-v diagram for the heating process of

water at constant pressure.

v


1.46.3 Saturation Temperature and Saturation Pressure

Water started to boil at 100C. Strictly speaking, the statement “waterboils at 100C” is incorrect. The correct statement is “water boils at 100C”

and 1 atm pressure 1.01325 bar. If the pressure inside the cylinder wereraised to 500 kPa by adding weights on top of piston, the water would startboiling at 151.9C. That is, the temperature at which water starts boilingdepends on the pressure; therefore, if the pressure is fixed, so is the boilingtemperature.

At a given pressure, the temperature at which a pure substance changesphase is called the saturation temperature Tsat. Likewise, at a given

temperature, the pressure at which a pure substance changes phase is calledthe saturation pressure Psat. At a pressure of 101.325 kPa, Tsat is 100C.

Conversely, at a temperature of 100C, Psat is 101.325 kPa.

Saturation tables that list the saturation pressure against the temperature(or the saturation temperature against the pressure) are available for practicallyall substances. For water, steam table is available.

The amount of energy absorbed or released during a phase-changeprocess is called the latent heat. More specifically, the amount of energyabsorbed during melting is called the latent heat of fusion and is equivalentto the amount of energy released during freezing. Similarly, the amount ofenergy absorbed during vaporization is called the latent heat of vaporizationand is equivalent to the energy released during condensation. The magnitudesof the latent heats depend on the temperature or pressure at which the phasechange occurs. At 1 atm pressure, the latent heat of fusion of water is 333.7kJ/kg and the latent heat of vaporization is 2257.1 kJ/kg.

1.47 THERMODYNAMIC PROPERTIES OF STEAM

1.47.1 Property TablesFor most substances, the relationship among thermodynamic properties

are too complex to be expressed by simple equations. Therefore, propertiesare frequently presented in the form of tables. Some thermodynamic propertiescan be measured easily, but others cannot and are calculated by using therelations between them and measurable properties. The results of thesemeasurements and calculations are presented in tables in a convenient format.In the following discussion, the steam tables will be used to demonstrate theuse of thermodynamic property tables.


1.47.2 Saturated Liquid and Saturated Vapour StatesThe subscript f is used to denote properties of a saturated liquid, and

the subscript g to denote the properties of saturated vapour. Another subscriptcommonly used is fg, which denotes the difference between the saturatedvapour and saturated liquid values of the same property. For example,

vf specific volume of saturated liquid

vg specific volume of saturated vapour

vfg difference between vg and vf (that is, vfg vg vf)

The quantity hfg is called the enthalpy of vaporization (or latent heat

of vaporization). It represents the amount of energy needed to vaporize a unitmass of saturated liquid at a given temperature or pressure. It decreases asthe temperature or pressure increases, and becomes zero at the critical point.

1.47.3 Saturated Liquid-Vapour MixtureDuring a vaporization process, a substance exists as part liquid and

part vapour. That is, it is a mixture of saturated liquid and saturated vapour.To analyze this mixture properly, we need to know the proportions of theliquid and vapour phases in the mixture. This is done by defining a newproperty called the quality x as the ratio of the mass of vapour to the totalmass of the mixture:

T

S

Sat

l iqu

id li

n e

W et steam

Critica l point

Dry satu rated steam

linefg

h

Critica l point

W et steamSaturated

liqu idline

Superhea tedregion

Dry sa turated

steam

S

Superhea tedregion

Fig 1.64


x mvapour

mtotal

Where mtotal mliquid mvapour

Quality has significance for saturated mixtures only. It has no meaningin the compressed liquid or superheated vapour regions. Its value is between0 and 1. The quality of a system that consists of saturated liquid is 0 (or 0percent), and the quality of a system consisting of saturated vapour is 1 (or100 percent)

Let us consider a mixture of saturatedliquid water and water vapour in equilibriumat pressure P and temperataure T. Thecomposition of the mixture by mass will begiven by its quality x, and its state will bewithin the vapour dome (Fig. 1.65) Theproperties of mixture are as follows.

v vf x vfg; u uf x ufg

h hf x hfg; s sf x sfg

where vf, vfg, hf, hfg, sf and sfg are the

saturation properties at the given pressureand temperature.

1.47.4 Superheated VapourThe region to the right of the saturated vapour line and at temperature

above the critical point temperature, a substance exist as superheated vapour.Since the superheated region is a single-phase region (vapour phase only),temperature and pressure are no longer dependent properties and they canconveniently be used as the two independent properties in the tables.

Superheated vapour is characterized by

Lower pressures (P Psat at a given T)

Higher temperatures (T Tsat at a given P)

Higher specific volumes (v vg at a given P or T)

Higher internal energies (u ug at a given P or T)

Higher enthalpies (h hg at a given P or T)

T

sP=C

constant pressure line

P=C ,T=C

P=C

x

sf s sg

Fig 1.65 Property in two phaseregion


Higher entropies (s sg at a given P or T )

1.47.5 Thermodynamic Properties of SteamImportant Notes:

At critical point, the water suddenly flashes into vapour.

Critical pressure 221.2 bar

Critical temperature 374.15C; Critical volume 0.00317 m3/kg

Critical point is a point in which saturation liquid and saturation vapourcoexists.

Dryness Fraction: xIt is defined as the mass of steam present in 1 kg of mixture

(mixture = water + steam)

x mass of steam

mass of water mass of steam

Wetness Fraction: 1 xIt is defined as the mass of water present in 1 kg of mixture.

1 x mass of water

mass of mixture

If x 0.1 means 10% of steam present in the 100% mixture.

x 0.9 means 90% of steam present in the 100% mixture.

f stands for liquid (or) water.

hf enthalpy of liquid in kJ/kg.

g stands for gas, vapour (or) steam.

hg enthalpy of dry steam (simply steam) in kJ/kg.

Similarly, vf, vg, sf, sg, uf, ug, etc.

All the values vf, vg, hf, hg, sf and sg for different saturated pressure

and saturated temperature are given in steam table.

To Find u (Internal Energy)

h u Pv can be used. ie., u h Pv


Wet steamDry saturated

steamSuper heated steam

x 1, t tsat x 1, t tsat x 1, t tsat

v vf xvfg v vg v for P and t in super heated steamtable.

h hf xhfg h hg h for P and t in super heated steamtable.

s sf xsfg s sg s for P and t in super heated steamtable.Degree of superheat t tsat

hfg means hg hf = Latent heat of vapourization.

vfg means vg vf; sfg means sg sf

NoteIf v vg, it is dry saturated steam.

If v vg, it is wet steam. If v vg, it is superheated steam.

Similarly, If s sg, it is dry saturated steam.

If s sg, it is wet steam; If s sg, it is superheated steam.

Similarly, If h hg, it is dry saturated steam.

If h hg, it is wet steam.

If h hg, it is superheated steam.

Similarly,

If t tsat, it is dry saturated steam (or) Wet stream.

If t tsat, it is superheated steam.


Properties Wet steam Dry saturated steam Superheated steamProperties can betaken from steamtable (pressuretable (or)temperature table)once pressure (or)temperature isgiven

Properties can betaken from steamtable (pressuretable (or)temperature table)once pressure (or)temperature isgiven

Once pressure andtemperature is given, wecan take all propertiesfrom superheated steamtable.

SpecificEnthalpy inkJ/kg

h hf x hfg h hgh can be taken for givenP and t from superheatedsteam table.(or)

h hg Cp steam t tsat

where Cp steam 2.1

to 2.3 kJ/kg KSpecificVolume in

m3/kg

v vf x vfg

or v x vg

[ . .

. vf ~ 0 ]

v vgv can be taken fromsuper heated steam table

(or) v vg

TTsat

Specificentropy in

kJkg K

s sf x sfg s sg(From super heated

steam table) (or) s

sg Cp steam ln

TTsat

Problem 1.35: Find the saturation temperature, changes in specific volumeand entropy during evaporation and the latent heat of vaporization of steamat 10 bar.

Solution

From steam table - pressure table,

for 10 bar, tsat 179.9C

P

bartsC

vf

m3/kg

vg

m3/kg

hf

kJ/kg

hfg in

kJ/kg

hg in

kJ/kg

sf in

kJ/kgK

sfg in

kJ/kgK

sg in

kJ/kgK

10

bar179.9 0.001127 0.1943 762.6 2013.6 2776.2 2.138 4.445 6.583

vf 0.001127 m3/kg; vg 0.1943 m3/kg


Change in specific volume vg vf 0.1943 0.001127

0.193173 m3/kg

Change in entropy during evaporation, sfg 4.445 kJ/kg K

Latent heat of vaporization hfg 2013.6 kJ/kg

Problem 1.36: Saturated steam has an entropy of 6.66 kJ/kg K. What are itspressure, temperature, specific volume and enthalpy?

From steam table, for sg 6.66 kJ/kg K; P 8 bar;

t2 170.4C; vg 0.24026 m3/kg ; hg 2767.4 kJ/kg.

Problem 1.37: Find the specific volume of steam at 12 bar absolute, whenthe condition of steam is (a) wet steam with dryness fraction 0.9, (b) drysaturated steam and (c) super heated to the temperature of 300C.

Solution: P 12 bar

(a) Wet steam with x 0.9

For 12 bar, tsat 188C

From steam table, pressure table, vf 0.001139 m3/kg, vg 0.16321 m3/kg

Specific volume v vf x vfg 0.001139 0.90.16321 0.001139

0.147003 m3/kg

(b) Dry saturated steam

Specific volume vg 0.16321 m3/kg

(c) Super heated steam with t 300CFrom super heated steam table

For 12 bar and 300C

Specific volume v 0.2140 m3/kg

Degree of super heat t tsat 300 188 112C

Problem 1.38: Find the volume, enthalpy and internal energy of steam (i)500 kPa and 0.75 dry (ii) 1 MPa and 425 C


Solution: Given data:

(i) P1 500 kPa 5 bar (ii) P1 1 MPa 10 bar

x 0.75 t1 425 CCase (i) From steam table at 5 bar

vf 0.001093 m3/kg ; vg 0.37466 m3/kg

hf 640.1 kJ/kg ; hfg 2107.4 kJ/kg

sf 1.86 kJ/kg K ; sfg 4.959 kJ/kg K

Volume, v vf x vg vf 0.001093 0.75 0.37466 0.001093

v 0.28126825 m3/kg

Enthalpy, h hf xhfg 640.1 0.75 2107.4 2220.65

h 2220.65 kJ/kg

Internal Energy, u h pv 2220.65 500 0.28126825

u 2080 kJ/kg

Case (ii) From steam table at 10 bar and 425 C

tsat for 10 bar 179.9 C

Since t 425 C tsat 179.9 C,

it is super heated steam.

So take value from superheated steam table.

Interpolation

v400 0.3065 m3/kg h400 3264.4 kJ/kg

v500 0.3540 m3/kg h500 3478.3 kJ/kg

v425 v400 v500 v400

100 25 h425 h400

h500 h400

100 25

0.3065 0.3540 0.3065

100 25 3264.4

3478.3 3264.4

100 25

0.318375 m3/kg hsup 3317.88 kJ/kg

By interpolation,

v425 0.318375 m3/kg; h425 3317.85 kJ/kg


Internal energy u

u h Pv 3317.85 10 100 0.318375 2999.475 kJ/kg

u 2999.475 kJ/kg [ . . . 100 for making bar into kPa ]

Problem 1.39: Find the enthalpy and entropy of steam when the pressure

is 20 bar and the specific volume is 0.09 m3/kg.

Solution

P 20 bar; v 0.09 m3/kg

For 20 bar from steam table,

vf 0.001177 and vg 0.09955 m3/kg

Since v0.09

vg0.09955

, it is wet steam.

So v vf xvg vf at 20 bar

0.09 0.001177 x0.09955 0.001177

x 0.90293

Dryness fraction ‘x’ 0.90293

To Find Enthalpy h

h hf x hfg at 20 bar

From steam table for 20 bar,

hg 908.5 kJ/kg; hfg 1888.7 kJ/kg

So, h 908.5 0.90293 1888.7 2613.86 kJ/kg

To Find Entropy s

s sf x sfg

From steam table for 20 bar,

sf 2.447 kJ/kg K; sfg 3.890 kJ/kg K

s 2.447 0.902933.89 5.9794 kJ/kg K

Problem 1.40: Find the state of steam (i) t 90 C ; s 7.276 kJ /kg K(ii) P 15 bar ; u 2300 kJ /kg (iii) P 6 bar ; h 2727 kJ/kg


Solution:Case (i) From steam table at 90 C[ s 7.276 kJ/kg K ]

sf 1.193 kJ/kg ; sfg 6.287 kJ/kg ; sg 7.48 kJ/kg

Since s 7.276

sg 7.48

, the steam is wet steam.

To find x

s sf x sfg 7.276 1.193 x 6.287

x 7.276 1.193

6.287 0.97

x 0.97

The steam is wet and dryness fraction is 0.97.

Case (ii) From steam table at 15 bar

hf 844.6 kJ/kg ; hfg 1945.2 kJ/kg ; hg 2789.9 kJ/kg ;

vf 0.001154 m3/kg ; vg 0.13167 m3/kg

We know, u h Pv

uf hf Pvf 844.6 1500 0.001154

uf 842.87 kJ/kg

ug hg P vg 2789.9 1500 0.13167 2592.395

ug 2592.395 kJ/kg

[ Given data u 2300 kJ/kg]

Since u 2300

ug 2592.4

, it is wet steam.

To find x

u uf x ufg 2300 842.87 x 2592.395 842.87

x 2300 842.87

2592.395 842.87

1457.131749.525

x 0.83

The steam is wet with dryness fraction of 0.83.


Case (iii) From steam table at 6 bar

hf 670.4 kJkg

h 2717

kJkg

hfg 2085.1 kJkg

hg 2755.5 kJkg

Since h2717

hg2755.5

; it is wet steam

h hf x hfg 2717 670.4 x 2085.1

x 2717 670.4

2085.1 0.981

The steam is wet with dryness fraction of 0.981.

Problem 1.41: 2 kg of water at 200C are contained in a 20 m3 vessel.Determine the pressure, enthalpy, mass and volume of vapour within thevessel. (FAQ)

Solution

Given: mf 2 kg, tsat 200 C, V 20 m3

From steam tables corresponding to tsat 200 C,

The saturation pressure Psat 15.549 bar

and vg 0.12716 m3/kg

But we know that, mg Vvg

20

0.12716 157.28 kg

The dryness fraction of given steam,

x mg

mf mg

157.282 157.28

0.987

The enthalpy of the vapour is given by h hf xhfg

From steam tables, corresponding to temperature 200C

hf 852.4 kJ/kg; hfg 1938.5 kJ/kg

h 852.4 0.987 1938.5 2765.69 kJ/kg


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