Thermodynamics solutions chapter 1

Technische Thermodynamica(191141010)

Bachelor cursus voor

Werktuigbouwkundestudenten

Answers

Universiteit Twente

Genie Stoffels

2011 - 2012

Thermodynamics: Answers Preface

Preface

This document contains answers to the exercises for the Bachelor course Engineering Thermo-

dynamics for students Mechanical Engineering at the University of Twente. The exercisesare provided in a separate document. The chapter numbers and titles refer to the chapters of the bookused in this course. This book is: Thermodynamics: An Integrated Learning System, by PhilipS. Schmidt, Ofodike A. Ezekoye, John R. Howell en Derek K. Baker. The exercises of this course arebased on exercises given in this book. Beside the book Thermodynamics, An Engineering Approach,by Y.A. Cengel and M.A. Boles is used for inspiration.

Part of the exercises will be treated during the working classes. See the table on the next pagefor the division of the chapters and exercises over the classes. The other exercises can serve as extrapractice for the students. At the beginning of each chapter there are some conceptual questions abouttheoretical aspects. However, most of the questions are about application of the material. The answersto the exercises, ordered in this document, are provided during the course. After the working class(es)of a chapter is (are) finished, the answers of that chapter are made available for the students.

The exercises and answers in the chapters 9, 10 and 11 are representative for the exercises of theexam. However, the exercises of the first eight chapters provide a basis to be able to understand andmake the exercises of the last chapters.

Although answers are present for the greater part of the exercises, some of the exercises (lessthan 5%) do not have answers jet. However, there are always silimar exercises that can be used as anexample, to solve the exersises without answer.

For the exercises taken from the book used in this course the answers are, mostly, taken fromthe solution manual of the book. The answers of the rest of the exercises are made by myself,G.G.M. Stoffels. A lot of them are typed in English, but also a lot of the answers are hand written inDutch and scanned in order to make them available to the students. This is specially the case for themore complicated and elaborate answers in the last chapters (9, 10, 11 and 14). The quality of thescans is not always optimal. It is planned to improve the quality of the hand written answers in thefuture.

Furthermore it has to be noted that in some answers tables from a different book (Cengel andBoles) are mentioned. This mostly concerns the tables with the data for water, the tables 4, 5, 6,and 7 refer to the tables 10, 11, 12 and 13 of the book used in this course. Also this can happen forrefrigerant 134A, the tables 11, 12, 13 and 14 refer to the tables 14, 15, 16 and 17 of the book used inthis course.

Finally I have to mention that there can be mistakes in the answers, I am not perfect, if yousee mistakes please mention them to me. I hope the students can benefit from the answers in thisdocument in the preparation of the exam.

Genie StoffelsMarch 2009

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Thermodynamics: Answers Course content

Content of the course

The course Engineering Thermodynamics is covered in eighteen lecture classes and eighteen workingclasses. The table below contains an overview of the division of the different chapters over the classesas well as an overview of the exercises treated during the working classes. This table follows the orderof the subjects in the book. However from year to year the order of the classes in which the chapters9, 10 and 11 are treated can changed in order to be consistent with project C. Chapter 14, exergy, ismosly treated after chapter 8. At the end of this document the answers of a test exam are included.

# Subject Chapter Exercises

1 Introduction to thermodynamics 1 1.1, 1.5, 1.6, 1.7, 1.8,1.13, 1.14

2 Energy, work, heat transfer, mass transfer, enthalpy 2 2.2, 2.6, 2.9, 2.11,2.12, 2.8

3 State principle, phase transitions, liquids, saturatedmixture, superheated steam, tables en diagrams

3.1 -3.7 3.6, 3.9, 3.10, 3.11,3.12, 3.14

4 Gases, ideal gas law, specific heat, internal energyand enthalpy of gases

3.8 - 3.16 3.17, 3.20, 3,21, 3.22,3.18

5 The first law (conservation of energy) for open andclosed systems

4 4.3, 4.5, 4.7, 4.9,4.11, 4.12, 4.13, 4.17

6 Reversible and irreversible processes, friction andtemperature, entropy

5, 6.1 5.1, 5.2, 5.3, 5.4, 5.5,5.6, 6.1, 6.2, 6.3

7 Entropy change open and closed systems, the secondlaw (entropy always increases), isentropic processes

6 6.4, 6.8, 6.15, 6.16,6.19

8 Applications of the second law, isentropic (ideal) andnon-isentropic systems, Bernoulli

7 7.4, 7.6, 7.12, 7.7, 7.9

9 Thermodynamic cycles, Carnot cycle, efficienciesand COPs, perpetual mobiles

8 8.5, 8.6, 8.7, 8.8,8.10, 8.13

10 Gas cycles, Stirling, Otto en Diesel cycles 9.1 - 9.5 9.2, 9.7, 9.11, 9.5, 9.9

11 Simple Brayton cycle (gas turbine) 9.6 9.16, 9.15, 9.13

12 Brayton cycle with improvements (inter cooling, re-heating, regeneration), jet engines

9.6 - 9.7 9.18, 9.17, 9.19, 9.14

13 Simple Rankine (vapor) cycle (steam turbine), com-pare to Carnot, design parameters

10.1 - 10.3 10.2, 10.4, 10.6, 10.8,10.19

14 Rankine cycle with improvements (reheating, feedwater heating)

10.4 -10.5 10.11, 10.9, 10.12,10.16, 10.17

15 Co-generation, binary cycles, combined cycles, or-ganic Rankine cycle

10.6 10.13, 10.15, 10.18

16 Cooling and heat pumps, reversed Rankine cycle,coefficient of performance (COP)

11 11.2, 11.4, 11.5, 11.7,11.9

17 Exergy (second law analysis), quality of energy,Sankey and Grassmann diagrams

14.1 - 14.2 14.2, 14.4, 14.7, 14.8,14.9, 14.10

18 Exam preparation, questions 1 - 11, 14.2 testexam

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Thermodynamics: Answers Contents

Contents

1 Thermodynamic Concepts and Terminology: Answers 1

1.1 Open, closed and isolated systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 State and thermodynamic variables or properties . . . . . . . . . . . . . . . . . . . . . 11.3 Intensive and extensive properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.6 Mass and volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.7 Ideal gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.8 Galileo thermometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.9 Acceleration and force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.10 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.11 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.12 Pressure on a diver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.13 Pressure and height . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.14 Pressure in a tank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.15 Pressure and manometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.16 Pressure in a cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Energy, Work and Heat Transfer: Answers 9

2.1 Basics of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Energy transfer mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Energy transfer by work and power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Work of a piston-cylinder device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.5 Work done by compression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.6 Work done by lifting a mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.7 Heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.8 Work and heat transfer of a piston-cylinder device . . . . . . . . . . . . . . . . . . . . 122.9 Flow work and enthalpy 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.10 Flow work and enthalpy 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.11 Enthalpy of steam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.12 Energy and power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.13 Energy and Environment 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.14 Energy and Environment 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Thermodynamic Properties of Pure Substances: Answers 15

3.1 Phases of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 T − v diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.3 Incompressible liquid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.4 Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.5 Quality of a mixture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.6 Use of tables: water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.7 Use of tables, refrigerant 134a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.8 Boiling of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.9 Vessel with 2 kg refrigerant 134a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.10 Vessel with refrigerant 134a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.11 Water mixture of liquid and vapor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.12 Vaporization of water at constant pressure . . . . . . . . . . . . . . . . . . . . . . . . . 203.13 Work done by heating 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.14 Work done by heating 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.15 Perfect gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

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3.16 A balloon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.17 An automobile tire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.18 Two tanks with air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.19 Enthalpy change of nitrogen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.20 Heat transfer and work at constant enthalpy . . . . . . . . . . . . . . . . . . . . . . . . 253.21 Air in a turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.22 Oxygen in a rigid tank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.23 Two closed rigid system in thermal equilibrium . . . . . . . . . . . . . . . . . . . . . . 263.24 Two closed movable systems in equilibrium . . . . . . . . . . . . . . . . . . . . . . . . 27

4 The First Law of Thermodynamics: Answers 29

4.1 First law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Conservation of mass principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.3 Conservation of mass in a nozzle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.4 Conservation of mass in a hair dryer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.5 Energy transfer by mass in a compressor . . . . . . . . . . . . . . . . . . . . . . . . . . 304.6 Cooling a closed system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.7 An insulated rigid tank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.8 Nozzle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.9 Steam turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.10 Refrigerant compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.11 Throttling valve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.12 Feedwater heater . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.13 Steam/water heat exchanger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.14 Steam turbine 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.15 Turbine with Argon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.16 Helium compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.17 Mixing chamber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.18 Refrigerant-134a heat exchanger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5 Reversible and Irreversible Processes: Answers 37

5.1 Reversible and Irreversible Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.2 Coffee . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.3 Insulated piston-cylinder device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.4 Frictionless piston-cylinder device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.5 Rigid adiabatic systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.6 Steady flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

6 Entropy and The Second Law: Answers 41

6.1 Basics of entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.2 Basics of the second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 416.3 Isothermal expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.4 Entropy change of an ideal gas in a rigid tank . . . . . . . . . . . . . . . . . . . . . . . 426.5 Entropy change of compressed air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.6 Radiator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.7 Non adiabatic compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.8 Entropy change of a copper block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466.9 Entropy change of an iron block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.10 Entropy change of an aluminum block . . . . . . . . . . . . . . . . . . . . . . . . . . . 486.11 Entropy change of an ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496.12 Entropy change of ideal gases in an isothermal process . . . . . . . . . . . . . . . . . . 496.13 Entropy change of ideal gases in isentropic process . . . . . . . . . . . . . . . . . . . . 506.14 Entropy change of oxygen in piston-cylinder device . . . . . . . . . . . . . . . . . . . . 50

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6.15 Entropy change of nitrogen in piston-cylinder device . . . . . . . . . . . . . . . . . . . 506.16 Reversible condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506.17 Reversible isothermal expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516.18 Reversible processes for helium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526.19 Melting ice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.20 Adiabatic steady flow device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546.21 Throttling valve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

7 Second Law Applications: Answers 55

7.1 Neon in an insulated container . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.2 Air in the cylinder of an automobile engine . . . . . . . . . . . . . . . . . . . . . . . . 567.3 Isentropic efficiencies of steady devices . . . . . . . . . . . . . . . . . . . . . . . . . . . 587.4 Adiabatic steam turbine 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597.5 Adiabatic steam turbine 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607.6 Refrigerant compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.7 Steam turbine coupled to an air compressor . . . . . . . . . . . . . . . . . . . . . . . . 627.8 Two stage compressor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647.9 Nozzle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667.10 Adiabatic device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 687.11 Steady-state steady-flow adiabatic device . . . . . . . . . . . . . . . . . . . . . . . . . 707.12 Venturi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

8 Analysis of Thermodynamic Cycles: Answers 73

8.1 Thermodynamic cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 738.2 Thermal efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 738.3 Coefficient of performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 738.4 Carnot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 748.5 Geothermal energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 748.6 A heat pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758.7 Carnot engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758.8 Air conditioner . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758.9 Refrigerator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768.10 Heat engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768.11 Nuclear power plant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768.12 Steam cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 778.13 Perpetual - motion machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 798.14 Combined engine and heat pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

9 Gas Power Cycles: Answers 81

9.1 Basics of the Stirling cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 819.2 Stirling cycle 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 829.3 Stirling cycle 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 839.4 Basics of the ideal Otto cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849.5 Thermal efficiency Otto engine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 859.6 Otto cycle 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 869.7 Otto cycle 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 879.8 Basics of the ideal Diesel cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 889.9 Thermal efficiency Diesel cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 899.10 Diesel cycle 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 909.11 Diesel cycle 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 919.12 Closed Brayton cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 929.13 Carnot versus Stirling versus Brayton cycle . . . . . . . . . . . . . . . . . . . . . . . . 939.14 Jet propulsion cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

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9.15 Simple open Brayton cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

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Thermodynamics: Answers 1 Thermodynamic Concepts and Terminology

1 Thermodynamic Concepts and Terminology: Answers

1.1 Open, closed and isolated systems

a: A system is a part of the universe that has to be analysed. The system boundary separates thesystem from the rest of the universe (it is the boundary between the system and the surround-ings/environment).

b: In a closed system energy can cross the boundary of a system, but the mass is fixed an no masscan go into or out the system. In an open system both mass and energy are allowed to cross thesystem boundary.

c: In an isolated system no energy or mass can cross the boundary.

d: This is an open system as mass (the water) can leave the radiator (it crosses the boundary ofthe system).

e: The can is a closed system as only heat can cross the boundary of the system.

1.2 State and thermodynamic variables or properties

a: A property is any characteristic that defines a system, a thermodynamic property is a char-acteristic that describes the mass and energy of a thermodynamic system. Some examples offrequently used thermodynamic properties are: temperature, pressure, mass, volume, energy,entropy, enthalpy and the amount of mass. Some other more general properties are: viscosity,thermal conductivity, modulus of elasticity, thermal expansion coefficient, electric resistivity,velocity, specific heat.

b: A thermodynamic state is a set of thermodynamic properties that describes completely thecondition of the thermodynamic system. A thermodynamic state is mathematically describedby a function consisting of several thermodynamic properties. At a given state all the propertieshave fixed values.

c: Two properties are independent if one property can vary while the other one is held constant.Dependent properties are properties that are defined in terms of other ones, one is a functionof the other. Which are dependent and independent properties depends on the situation. Forwater for example the volume is independent of the pressure and the temperature. However fora gas the volume depends on the temperature and pressure.

d: The state postulate: Any two independent intensive thermodynamic properties are sufficient todescribe the state of a system containing a single pure substance.

e: The change of a system to an other state is called a process.

f: A system of which the properties are not changing with time is said to be in a steady state. Inthermodynamics this is called thermodynamic equilibrium. Thermodynamic equilibrium impliesa state of balance of the system where there are no driving forces (like temperature differences)within the system.

1.3 Intensive and extensive properties

a: Intensive variables are independent of the size of the system. Examples are pressure, tempera-ture, chemical potential, density.

b: Extensive properties are directly proportional to the mass or size of the system. They can bewritten as per kg [/kg]. Examples are mass, volume, energy, entropy, enthalpy.

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c: Intensive properties.

d: Dividing an extensive property by mass gives a specific property. It is the property per unitmass. e.g. the specific volume (v) is the volume per unit mass, v = V/m in m3/kg or the specificenergy is the energy per unit mass e = E/m in J/kg.

e: The temperature and pressure are intensive properties, they do not change if the volume ismultiplied. The mass and the energy are extensive properties, they scale with the system, ifthe volume is multiplied by a factor λ the mass and energy are also multiplied by a factor λ.The specific energy, which is the energy per unit mass, does not change as this is an intensiveproperty. Compare if you double the size of a room the temperature and pressure remain thesame but the amount of air and so the mass doubles and also the energy contained in the airdoubles. The specific energy which is the energy per unit mass remains the same as the totalenergy doubles but also the mass.

1.4 Processes

a: An isothermal process is a process in which the temperature is constant. An isobaric processis a process in which the pressure is constant. An isochoric processes is a process in which thevolume is constant.

b: An adiabatic process is a process during which there is no heat transfer to the surroundings.

c: A steady-flow process, is a process that remains constant in time, so time independent.

d: In a cyclic process one returns to the original starting point. After a cycle the properties of thestate are not changed.

e: During a quasi-equilibrium (or quasi-static) process, the system remains practically in equilib-rium at all times. Note that a quasi-equilibrium process is an idealized process and not a truerepresentation of an actual process. But many actual processes closely approximate it and theycan be modeled as quasi-equilibrium with negligible error. Engineers are interested in such pro-cesses as they are easy to analyze. Furthermore work-producing devices deliver the most workwhen they operate on quasi-equilibrium processes.

1.5 Dimensions

a: The seven fundamental dimensions are:

1: Mass, in general denoted by M in kilogram [kg].

2: Length, in general denoted by x in meters [m].

3: Temperature, in general denoted by T in Kelvin [K].

4: Time, in general denoted by t in seconds [s].

5: Amount of matter, in general denoted by n in mole [mol].

6: Electric current, in general denoted by I in ampere [A].

7: Amount of light, in general denoted by I in candela [cd].

b: The first four.

c: Force = Mass × acceleration ⇒ F = M × a

[N ] = [kg]

[

m

s2

]

=

[

kgm

s2

]

(1)

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Pressure = Force per area ⇒ P = F/A

[Pa] =

[

N

m2

]

=

[

kgms2

m2

]

=

[

kg

ms2

]

(2)

Energy = Force × distance ⇒ E = F × x

[J ] = [N ][m] =

[

kgm

s2

]

[m] =

[

kgm2

s2

]

(3)

Or energy = Mass × velocity of light squared ⇒ E = M × c2

[J ] = [kg]

[

(

m

s

)2]

=

[

kgm2

s2

]

(4)

Power = energy per time ⇒ E = E/t

[W ] =[J ]

[s]=

[

kgm2

s2

]

×1

s=

[

kgm2

s3

]

(5)

d: Kinetic energy = half × mass × velocity squared ⇒ Ekin = 12Mv2

[kg]

[

(

m

s

)2]

=

[

kgm2

s2

]

= [J ] (6)

Potential energy = mass × acceleration of gravity × height ⇒ Epot = Mgh

[kg]

[

m

s2

]

[m] =

[

kgm2

s2

]

= [J ] (7)

1.6 Mass and volume

a: The mass of the substance is Msubstance = ρV = ρ(lbh) [kg] and the mass of the combined systemis Mcomb = Mtank +Msubstance = Mtank + ρV = Mtank + ρ(lbh) [kg].The specific volume (v) is defined as the total volume (V ) per mass (M). The specific volumeof the substance is vsubstance = Vsubstance/Msubstance = lbh/(ρlbh) = 1/ρ [m3/kg]. Note that thespecific volume is the reciproce of the density.The specific volume of the combined system is the volume of the combined system divided bythe mass of the combined system vcomb = Vcomb/Mcomb = lbh/(Mtank + ρ(lbh)) [m3/kg].

b: ρwater = 1000 kg/m3 =⇒ Mwater = 1000(1·0.4·0.5) = 200 kg and Mcomb = 3+1000(1·0.4·0.5) =3 + 200 = 203 kg.The specific volume of the water is vwater = 1/ρ = 1/1000 = 10−3 m3/kg. The specific volumeof the combined system is vcomb = Vcomb/Mcomb = (1 · 0.4 · 0.5)/203 = 9.85 · 10−4 m3/kg.

c: ρair = 1.225 kg/m3 =⇒ Mair = 1.225(1 ·0.4 ·0.5) = 0.245 kg and Mcomb = 3+1.225(1 ·0.4 ·0.5) =3 + 0.245 = 3.245 kg. In the case of air the weight of the air can be neglected compared to theweight of the tank. Whereas in the case of water the weight of the tank can be neglectedcompared to the water.The specific volume of air is vair = 1/ρair = 0.816 m3/kg and the specific volume of the combinedsystem is vcomb = Vcomb/Mcomb = (1 · 0.4 · 0.5)/3.245 = 0.0616 m3/kg.

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1.7 Ideal gases

Use the ideal gas law to calculate the specific volume v, v = RT/P . In the formula T should bein Kelvin 42◦C = 315 K. The pressure should be in Pascal, 1.2 Bars = 120 kPa = 120 kN/m2.This results in v = (0.14304 · 315)/120 = 0.375 m3/kg. That the outcome is the specific volume inm3/kg follows by looking to the dimensions, (kJ/kgK)(K)(m2/kN) = m3/kg. The total volume V isV = v · m = 0.375 · 1.5 = 0.563 m3. The density of the gas is the reciproce of the specific volumeρ = 1/v, ρ = 1/0.375 = 2.667 kg/m3.

1.8 Galileo thermometer

a: If the outside temperature is 30 degree Celsius all bubbles are too heavy and they are all on thebottom. If the outside temperature is only 10 degree Celsius all bubbles are too light and allthe bubbles are on the the top.

b: The bubble that indicates the right temperature floats in the water somewhere in the middlebetween the bubbles that have a too low temperature tag on the bottom and the bubbles thathave a too high temperature tag on the top.

c: The bubble with the 18 degree Celsius tag is the heaviest and the bubble with the 24 temperaturetag is the lightest.

d: The 20 degree Celsius bubble is floating in the middle of the thermometer. Therefore, the upwardforce, Fu, that the bubble encounters is equal to the weight of water displaced by the bubble.The weight of the displaced water is the product of the density of water at 20 degree, ρ20, thegravitational acceleration, g and the volume of the bubble, Vb = 4/3πr3

→ Fu = ρgVb = 4/3πρ20gr3.

As the bubble floats the upward force is equal to the gravitational force, Fg, which is equal tothe mass of the bubble, Mb20, times the gravitational acceleration→ Fg = Mb20g.

Equate the two forces together results in: Fu = 4/3πρ20gr3 = Fg = Mb20g

→ Mb20 = 4/3πρ20r3. Filling in r = 1

2d = 0.0125 m and ρ20 = 998.2071 kg/m3 gives→ Mb20 = 8.1666 · 10−3 kg = 8.1666 grams.

e: Lowest temperature tag is the 18 degree bubble, its mass is: Mb18 = 4/3πρ18r3 = 8.1698 grams.

Highest temperature tag is the 24 degree bubble, its mass is: Mb24 = 4/3πρ24r3 = 8.1591 grams.

The mass difference between both bubbles is: Mb18 −Mb24 = 0.0107 grams = 10.7 milligrams.

The mass of the 19 degree bubble is: Mb19 = 4/3πρ19r3 = 8.1682 grams.

The mass of the 23 degree bubble is: Mb23 = 4/3πρ23r3 = 8.1611 grams.

Mass difference between 18 and 19 degree bubble: Mb18−Mb19 = 0.0016 grams = 1.6 milligrams.Mass difference between 23 and 24 degree bubble: Mb23−Mb24 = 0.0020 grams = 2.0 milligrams.

As the relation between the temperature and the density of the water is not linear (you caneasily see this in the graph, the line is not straight but a bit curved) the mass difference betweentwo successive bubbles is not constant. The mass difference between the bubbles increases withtemperature.

The relative mass difference of the 18 and 19 degree bubble is: 0.00168.1591 × 100% = 0.0196%

The relative mass difference of the 23 and 24 degree bubble is: 0.00208.1611 × 100% = 0.0245%

So, also the relative mass difference between the bubbles increases with temperature, and variesaround 0.02% of the mass of the bubbles.

They could build these thermometers in Galileo’s time, as the mass differences between theindividual bubbles are in the order of milligrams. These differences can be determined using asimple balance scale which was already know in Galileo’s time.

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1.9 Acceleration and force

a: Force = Mass × acceleration, F = M × a with a = 6g =⇒ Faircraft = 6Mg [N].

b: Fmoon = 0.17Mg [N].

c: g = 9.8 m/s2 =⇒ Fmanaircraft = 5292 N, Fwoman

aircraft = 3234 N, Fmanmoon = 150 N, Fwoman

aircraft = 91.6 N.

1.10 Temperature

a: The absolute zero temperature is the lowest theoretically attainable possible temperature, thisis the temperature at which all movements of molecules are frozen. Absolute temperatures usea reference state of zero at absolute zero. They can never be negative. The unit is Kelvin.

b: A relative temperature is a temperature measured relative to a non-absolute zero temperature.It can be negative.

c: The most common temperature scales in the SI system are are the Celsius scale and the Kelvinscale. The are related by a factor of 273.15 → T (K) = T (C) + 273.15.

d: The boiling and freezing point of water.

1.11 Pressure

a: Pressure is defined as the normal force per unit acting on a surface, P = force [N]area [m2]

= FA . The

unit of pressure is N/m2=Pa.

b: Vacuum pressure is zero pressure and atmospheric pressure is the pressure of the atmosphereoutside. The absolute pressure is the pressure measured related to the vacuum pressure andthe gauge pressure is the pressure measured related to the atmospheric pressure. They can berelated Pabsolute = Patmosphere + Pgauge.

1.12 Pressure on a diver

a: The relative pressure on the diver is the pressure exerted by the water Prel = Pwater = ρgh.The absolute pressure is the pressure of the water and the atmospheric pressure Pabs = Patm +Pwater = Patm + ρgh.

b: At h = 15 m → Prel = 1000 · 9.81 · 15 = 147150 Pa = 147.2 kPa and Pabs = 248.2 kPa.At h = 30 m → Prel = 1000 · 9.81 · 30 = 294300 Pa = 294.3 kPa and Pabs = 395.2 kPa.

c: The relative pressure is doubled if the diver goes two times deeper below the water level. Theabsolute pressure is not doubled.

d: The density of seawater is 1030 kg/m3.At h = 15 m → Prel = 1030 · 9.81 · 15 = 151565 Pa = 151.6 kPa and Pabs = 252.6 kPa.At h = 30 m → Prel = 1030 · 9.81 · 30 = 303129 Pa = 303.1 kPa and Pabs = 403.1 kPa.

e: Patm = Pread − ρseawatergh → Patm = 398800− 1030 · 9.81 · 30 = 95671 Pa = 95.7 kPa.

1.13 Pressure and height

a: Pbarometric = gρairh

b: ∆P = gρair∆h → ∆h = ∆Pgρair

= (930−780)·102

·9.7·1.2 = 1287 m.

d: Assume the tower is 40 m (13 floors and each floor about 3 meter) → ∆Pbarometric = gρair∆h =9.81 · 1.2 · 40 = 471 Pa = 4.71 mBar.

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1.14 Pressure in a tank

a: It is assumed that the pressure in the tank is uniform (i.e. its variation with elevation is negligibledue to its low density), and thus we can determine the pressure at the water-air interface.Starting with the pressure at the water-air interface and moving along the tube by adding orsubtracting the ρgh terms until we reach the outside mercury-air interface and setting the resultequal to Patm since the tube is open to the atmosphere gives

Pair−tank + ρwatergh1 + ρoilgh2 − ρmercurygh3 = Patm. (8)

We are interested in the gauge pressure, which is the pressure in the tank relative to the atmo-spheric pressure and thus we write:

Pgauge = Pair−tank − Patm = ρmercurygh3 − ρwatergh1 − ρoilgh2. (9)

Note that jumping horizontally from one tube to the next and realizing that pressure remainsthe same in the same fluid simplifies the analysis considerably.

b: Filling in the numbers gives: Pgauge = 13600 · 9.81 · 0.46 − 1000 · 9.81 · 0.2 − 850 · 9.81 · 0.3 =56907.81 Pa = 56.9 kPa.

c: If the oil in the tube is replaced by mercury and you assume that the values of the heightstay the same the pressure inside the tank is lower Pgauge = ρmercuryg(h3 − h2) − ρwatergh1 =13600 · 9.81 · (0.46− 0.3)− 1000 · 9.81 · 0.2 = 19384.56 Pa = 19.4 kPa.If you assume that the pressure inside the tank does not change if you replace the oil by mercurythe level of h1 and h3 changes (h2 has no meaning anymore). How the levels change can not becalculated.

d: If both oil and mercury are taken out and the level h1 stays the same the pressure inside thetank drops to Pgauge = −ρwatergh1 = −1000 ·9.81 ·0.1 = −981 Pa = - 0.98 kPa. This pressure isnegative, that means that the pressure inside the tank is larger than the atmospheric pressure.That is easily seen by the fact that the water in the tube is at a lower level than the water inthe tank.If is assumed that the pressure inside the tank does not change the water level h1 becomes:h1 = −

Pgauge

ρwaterg= − 57907.81

1000·9.81 = −5.9 m. This is 5.9 meters above the water surface!

e: Oil has a lower density than water while mercury has a higher density than water. This impliesthat in a manometer tube the oil reaches a height hinger than water while the mercury reachesa height lower than water. If the pressure is very high it is preferred to use mercury as the tubecan be shorter than in the case of water. If the pressure is very low oil has the preference as theaccuracy that can be reached is higher (the oil will be higher than water of mercury and thus iseasier to read).

1.15 Pressure and manometer

a: The pressure in the duct is above the atmospheric pressure as the pressure in the duct is thefluid in the right leg of the manometer is higher. Pressure in the duct atmospheric pressure plusthe pressure exerted by the fluid.

b: Prel = ρgh and Pabs = ρgh+ Patm.

c: h = 15 mm → Prel = 147.15 Pa and Pabs = 100147.15 Pa.h = 30 mm → Prel = 294.30 Pa and Pabs = 100294.30 Pa.The relative pressure doubles if the height doubles while the absolute pressure increases only alittle bit, only about 0.15%. In the absolute pressure the atmospheric pressure dominates.

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d: h = 15 mm → Prel = 2001.24.15 Pa and Pabs = 102001.24 Pa.h = 30 mm → Prel = 4002.48 Pa and Pabs = 104002.48 Pa.The relative pressure in the duct is much higher than in the case the fluid is water. This isbecause the density of mercury is about 14 times higher than the density of water. The relativepressure is about 14 times higher. The difference with the absolute pressure in c is less thanwith the relative pressure as the atmospheric pressure has a large contribution to the absolutepressure compared to the relative pressure.

e: If the absolute pressure in the duct is two times the atmospheric pressure the relative pressurein the duct is equal to the atmospheric pressure. Prel = Patm = ρgh → h = Patm

ρg .

For water → h = 1000001000·9.81 = 10.1 m.

For mercury → h = 10000013600·9.81 = 0.75 m.

For mercury you need a much shorter tube.

1.16 Pressure in a cylinder

a: Pcylinder = Patm + Ppiston + Pspring = Patm + MgA + F

A = Patm + Mg+FA .

b: Pcylinder = 95000 + 4·9.81+6035·10−4 = 123354 Pa. The pressure inside the cylinder is 123.4 kPa.

c: Pcylinder = Patm + Mg+FA = 2Patm → Patm = Mg+F

A → F = PatmA−Mg.F = 95000 · 35 · 10−4 − 4 · 9.81 = 293.26 N.

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Thermodynamics solutions chapter 1

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