Weber Thermodynamics

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INTERNATIONAL STUDIESin SCIENCE and ENGINEERING

Roman Weber

COMBUSTION FUNDAMENTALS

with

Elements of Chemical Thermodynamics


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Prof. Dr.–Ing. Roman WeberTechnische Universität ClausthalInstitut für Energieverfahrenstechnik und Brennstofftechnik (IEVB)

Agricolastrasse 4, 38 678 Clausthal-Zellerfeld, [email protected]

Weber, Roman:Combustion Fundamentals with Elements of Chemical ThermodynamicsClausthal-Zellerfeld: Papierflieger 2008ISBN 978–3–89720–921–3

Bibliografische Information der Deutschen BibliothekDie Deutsche Bibliothek verzeichnet diese Publikation in der Deutschen National-bibliografie; detaillierte bibliografische Daten sind im Internet über http://dnb.ddb.deabrufbar.

INTERNATIONAL STUDIES in SCIENCE and ENGINEERINGEditor in Chief:Prof. Dr.-Ing. Roman Weber, Clausthal University of Technology (Germany)Editorial Board:Dr.-Ing. Rüdiger Alt, Clausthal University of Technology (Germany)Prof. Dr.-Ing. Ryszard Bialecki, Silesian University of Technology (Poland)Prof. Xu Delong, Xi’an University of Architecture and Technology (China)Prof. Dr. Peter v. Dierkes, former President of Berliner Stadtreinigungsbetriebe (Ger-many)Dipl.-Math. Marc Muster, Clausthal University of Technology (Germany)Prof. Dr.-Ing. Andrzej Nowak, Silesian University of Technology (Poland)

Prof. Dr.-Ing. Reinhard Scholz, Clausthal University of Technology (Germany)

First Edition 2008

Copyright by PAPIERFLIEGER, Clausthal-Zellerfeld 2008, Telemannstr. 1, 38678 Claus-thal-Zellerfeld, Tel.: 05323/96746, http://www.papierflieger-verlag.de

No part of this publication may be reproduced or transmitted in any form or by anymeans, electronic or mechanical, including photocopying, recording, or any informationstorage and retrieval system, without the prior permission in writing from the publisher.

ISBN 978–3–89720–921–3

The cover of this textbook has been designed using photographs provided by Mr. MarcMuster of TU Clausthal.


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ToProfessor Stanisław Jerzy Gdula

who introduced me to the wonderfulworld of thermodynamics

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International Studies in Science

and Engineering

The Editorial Board encourages its colleagues all over the world to publish in the”INTERNATIONAL STUDIES in SCIENCE and ENGINEERING” both textbooks which accompany a lecture series for students and other books whichdemonstrate how to apply the knowledge acquired in lecture theatres to industrialpractise.

With publishing ”Combustion Fundamentals with Elements of Chemical Thermo-dynamics” the sixth book in this series was released. At least three further booksare expected to be published in 2008: one concerns ”Advanced Heat Transfer” ,two more on ”Abfall and Chemie” .

Already published:(the latest editions are listed only)

1. Weber, R.: Lecture Notes in Heat Transfer, 3rd Edition, Papierflieger-

Verlag, Clausthal-Zellerfeld 2008, ISBN 3–89720–702–8.2. Jeschar, R.; Kostowski, E.; Alt, R.: Wärmestrahlung in Industrieöfen,

Papierflieger-Verlag, Clausthal-Zellerfeld 2004, ISBN 3–89720–686–2 andWydawnictwo Politechniki Slaskiej, Gliwice 2004, ISBN 83–7335–232–5.

3. Weber, R.; Alt, R.; Muster, M.: Vorlesungen zur Wärmeübertragung, Teil I,2. Auflage, Papierflieger-Verlag, Clausthal-Zellerfeld 2008, ISBN 3–89720–798–2.

4. Dierkes, P. v.; Bruch, G.: Abfall und Chemie, Teil I, Papierflieger-Verlag,Clausthal-Zellerfeld 2007, ISBN 3–89720–879–2.

5. Dierkes, P. v.; Bruch, G.: Abfall und Chemie, Teil II, Papierflieger-Verlag,

Clausthal-Zellerfeld 2007, ISBN 3–89720–880–6.

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To the StudentHow to get the most from these lecture notesThese lecture notes in ”Combustion Fundamentals with Elements of Chemical Thermodynamics ” have been prepared for undergraduate students attending afifteen–week course (one semester) with 1.5 hour of lecture and 45 minutes of classes per week. This is a typical course at the Clausthal University of Technol-ogy. The notes have been prepared not only for Clausthal but also for the RoyalInstitute of Technology (Stockholm, Sweden), University of Science and Technol-ogy (Beijing, China) and Central South University (Changsha, China). If youare like most students just opening this textbook, you are enrolled on one of thefew courses in science and engineering at Clausthal that are delivered in English.Probably English is not your native language but do not worry. This textbookhas been written with you in mind in very simple and plain English. Below Iprovide you with few suggestions how ” Combustion Fundamentals with Elements of Chemical Thermodynamics ” can help you to succeed in this course.

Read before a lecture. You will get the most out of your combustion courseif you read each chapter before hearing it. In this way, many of the topics willalready be clear in your mind and you will understand the lecture better.

Study Examples and Problems. Each chapter contains several Examples.Study them carefully since they illustrate the subject of each particular chap-

ter. In addition to Examples, Problems are formulated for each chapter and theycan be found on the web site (see below). Some Problems will be discussed andsolved in the classroom but the principal idea behind Problems is that they areyour homework. I suggest you solve them, one by one, and in case of difficultiesconsult your class instructor.

Study together with your fellow students. Many students find it useful toform study groups. You can discuss the challenging topics with one another andhave a good time while doing it. Make sure that you do Problems by yourself since during the exam you will have to prove your skills in problem solving.

Take advantage of the web site. In addition to these lecture notes we have de-veloped a web site to assist you in this course. You can find it by browsing throughthe IEVB–Institute website of TU Clausthal: http://www.ievb.tu-clausthal.de/(follow ”Science” and ”Lectures”).

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Acknowledgments”Combustion Fundamentals with Elements of Chemical Thermodynamics ” hasbeen written within the scope of a European Community project (CN/ASIA-LINK/016 (103 187)) of the Asia-Link Programme. The author acknowledgeswith thanks the European Community financial support.

I would like to thank my colleagues who carefully scrutinized the manuscriptmaking it a better textbook:

Dipl.–Ing. Stefan Brinker, Dipl.–Ing. Sven Gose, Dipl.–Ing. Patrick Schwöppe,Dr.–Ing. Marco Mancini, Dipl.–Math. Marc Muster (all Clausthal University of Technology, Germany), Dr.–Ing. Gabriel Wecel (Silesian University of Technology,Poland). My special thanks go to Stefan Brinker and Marc Muster who editedthis textbook.

The Vocabulary attached to these lecture notes has been prepared by Dr. RüdigerAlt. It has been used by students at TU Clausthal attending lectures on HeatTransfer, Combustion Technology and High Temperature Processes.

Although I have made a concerted effort to make this first edition error free, somemistakes may have crept in unbidden. I would appreciate hearing from anyonewho finds an error or wishes to comment on the text. You may e-mail or write tome.

Roman WeberIEVB – TU ClausthalD-38678 Clausthal-ZellerfeldAgricolastrasse 4Germany

[email protected]

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Contents

1 Stoichiometry 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Chemical Reactions, Atoms and Molecules in Combustion . 41.2.2 Amount of Substances, Mole and Mass Fractions . . . . . . 51.2.3 Density and Concentration (Molar density) . . . . . . . . . 61.2.4 Equation of State for Gases and Gas Mixtures . . . . . . . . 7

1.3 Combustion Stoichiometry for Gaseous Fuels . . . . . . . . . . . . 81.3.1 Stoichiometric combustion . . . . . . . . . . . . . . . . . . . 81.3.2 Excess air ratio (air equivalence ratio) and fuel equivalence

ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3.3 Minimum air requirement for a mixture of gaseous fuels . . 111.3.4 Composition of combustion products . . . . . . . . . . . . . 11

1.4 Combustion stoichiometry for liquid and solid fuels . . . . . . . . . 161.4.1 Minimum oxygen and air requirements and excess air ratio 17

1.4.2 Combustion products . . . . . . . . . . . . . . . . . . . . . 171.5 Humid Combustion Air . . . . . . . . . . . . . . . . . . . . . . . . 241.5.1 Absolute and relative humidity . . . . . . . . . . . . . . . . 241.5.2 Dew Point Temperature of Combustion Products . . . . . . 29

1.6 Combustibles burnout for solid fuels . . . . . . . . . . . . . . . . . 311.7 Sub-stoichiometric combustion to carbon dioxide and water vapour 321.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2 Mass and Energy Balance 352.1 General Formulation of Mass and Energy Balance . . . . . . . . . . 35

2.1.1 Mass and Energy Balance at an Instant . . . . . . . . . . . 362.1.2 Mass and Energy Balance over a Time Interval . . . . . . . 372.1.3 Mass and Energy Balance under Steady-State Conditions . 382.1.4 Example of a Mass Balance of a Furnace . . . . . . . . . . . 38

2.2 The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . 432.2.1 System Energy . . . . . . . . . . . . . . . . . . . . . . . . . 462.2.2 Energy Entering and Leaving the System . . . . . . . . . . 49

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Contents

2.2.3 Energy Balance of Thermal Systems (Machines) . . . . . . . 56

2.3 Energy Released in Chemical Reactions . . . . . . . . . . . . . . . 602.3.1 Reaction Enthalpy . . . . . . . . . . . . . . . . . . . . . . . 602.3.2 Standard Enthalpies of Formation . . . . . . . . . . . . . . 622.3.3 Lower Calorific Value (LCV) and Gross Calorific Value (GCV) 652.3.4 Relationships between Calorific Values, Reaction Enthalpies

and Formation Enthalpies . . . . . . . . . . . . . . . . . . . 672.3.5 Dependence of LCV on Temperature . . . . . . . . . . . . . 692.3.6 Example of an Energy Balance of a Furnace . . . . . . . . . 72

2.4 Temperature of Adiabatic Combustion . . . . . . . . . . . . . . . . 792.5 Furnace Exit Temperature . . . . . . . . . . . . . . . . . . . . . . . 812.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

3 Equilibrium Thermodynamics 89

3.1 Irreversible and Reversible Processes . . . . . . . . . . . . . . . . . 903.2 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

3.2.1 Entropy of Liquids and Solids . . . . . . . . . . . . . . . . . 1023.2.2 Entropy of Ideal Gases . . . . . . . . . . . . . . . . . . . . . 1023.2.3 Entropy of Phase Transition at the Transition Temperature 1033.2.4 The Third Law of Thermodynamics . . . . . . . . . . . . . 1043.2.5 Absolute Entropy of Pure Substances . . . . . . . . . . . . . 104

3.3 The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . 1073.3.1 The Increase in Entropy Principle . . . . . . . . . . . . . . 107

3.3.2 Entropy Change for a Continuous Process at Steady-State . 1103.3.3 Irreversibility of Processes . . . . . . . . . . . . . . . . . . . 111

3.4 General Conditions for Thermodynamic Equilibrium . . . . . . . . 1123.4.1 Isolated System . . . . . . . . . . . . . . . . . . . . . . . . . 1123.4.2 Non-Adiabatic System . . . . . . . . . . . . . . . . . . . . . 113

3.5 Equilibrium Between Phases . . . . . . . . . . . . . . . . . . . . . . 1163.5.1 Single-Component System Consisting of Two Phases . . . . 1163.5.2 Phase Transformations of a Pure Substance . . . . . . . . . 1193.5.3 Dependence of Gibbs Free Enthalpy on Temperature and

Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1223.5.4 Equilibrium in Multi-Component Single-Phase Systems . . 126

3.5.5 Chemical Potential of Pure Substances . . . . . . . . . . . . 1303.5.6 Significance of Chemical Potential . . . . . . . . . . . . . . 131

3.6 Multi-Component, Multi-Phase Systems . . . . . . . . . . . . . . . 1343.6.1 The Phase Rule . . . . . . . . . . . . . . . . . . . . . . . . . 138

3.7 Thermodynamics of Mixing . . . . . . . . . . . . . . . . . . . . . . 138

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Contents

3.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

4 Chemical Equilibrium 1494.1 Definition of Chemical Equilibrium . . . . . . . . . . . . . . . . . . 1504.2 Single Chemical Reaction . . . . . . . . . . . . . . . . . . . . . . . 150

4.2.1 Extent of a Single Reaction . . . . . . . . . . . . . . . . . . 1504.2.2 Change of Gibbs Enthalpy as a Chemical Reaction Advances1534.2.3 Gibbs Enthalpy of Selected Reactions . . . . . . . . . . . . 1564.2.4 Thermodynamic Equilibrium Constant for a Gaseous Reac-

tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1594.2.5 Other Equilibrium Constants . . . . . . . . . . . . . . . . . 1654.2.6 Effect of Pressure and Temperature on Thermodynamic

Equilibrium Constant . . . . . . . . . . . . . . . . . . . . . 1674.2.7 Chemical Equilibrium in Presence of a Solid Phase . . . . . 1714.2.8 Le Châtelier´s Principle . . . . . . . . . . . . . . . . . . . . 176

4.3 Multiplicity of Chemical Reactions . . . . . . . . . . . . . . . . . . 1834.3.1 Multi-Component, Multi-Phase Systems with Chemical Re-

actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1844.3.2 Choice of Chemical Reactions . . . . . . . . . . . . . . . . . 1884.3.3 Exact Number of Chemical Reactions Needed for Equilib-

rium Determination . . . . . . . . . . . . . . . . . . . . . . 1894.3.4 Linear Dependence of a Reaction Set . . . . . . . . . . . . . 1934.3.5 The Phase Rule for a System with Chemical Reactions . . . 196

4.4 Equilibrium Composition . . . . . . . . . . . . . . . . . . . . . . . 2004.4.1 Systems with a one-dimensional reaction basis . . . . . . . . 2004.4.2 Systems with a two-dimensional reaction basis . . . . . . . 2094.4.3 Systems with a multi-dimensional reaction basis . . . . . . . 216

4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

5 Elements of Chemical Kinetics 2215.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2215.2 Rate Laws and Reaction Orders . . . . . . . . . . . . . . . . . . . . 2225.3 Forward and Reverse Reactions . . . . . . . . . . . . . . . . . . . . 2265.4 Elementary Reactions and Reaction Molecularity . . . . . . . . . . 2315.5 Rate of Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

5.5.1 Temperature dependence of rate coefficients . . . . . . . . . 2385.5.2 Pressure dependence of rate coefficients . . . . . . . . . . . 240

5.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

6 Mechanisms of Basic Combustion Reactions 243

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Contents

6.1 Chain Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

6.2 Combustion of Carbon Monoxide (CO) . . . . . . . . . . . . . . . . 2456.3 Combustion of Hydrogen (H2) . . . . . . . . . . . . . . . . . . . . . 2466.3.1 Simplified ignition mechanism . . . . . . . . . . . . . . . . . 248

6.4 Combustion of Methane (CH4) . . . . . . . . . . . . . . . . . . . . 2516.5 Methods of Solving Chemical Kinetic Rate Equations . . . . . . . . 253

6.5.1 Analytical solutions . . . . . . . . . . . . . . . . . . . . . . 2546.5.2 Numerical Solutions . . . . . . . . . . . . . . . . . . . . . . 264

6.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

Gaussian Elimination 283

Vocabulary 287

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List of Figures

1.1 Primary Energy Sources in Germany per type of fuel . . . . . . . . 21.2 Wall Fired Boiler . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Modern Reheating Furnace – NKK, Fukuyama Works, Japan . . . 31.4 Left – pulverised coal flame; Right – MILD combustion of natural

gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.5 The subsets of hydrocarbons . . . . . . . . . . . . . . . . . . . . . . 51.6 Oxygen concentration in dry combustion products as a function of

excess air ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.7 Carbon dioxide concentration as a function of excess air ratio . . . 231.8 Saturation pressure of water vapour as a function of temperature.

The plot is obtained using Eq. (1.35) which in the plotted rangeprovides pressure values that are within 2 % accuracy with the val-ues listed in steam tables [7]. . . . . . . . . . . . . . . . . . . . . . 26

1.9 Cooling of combustion products (or moist air) at a constant tem-perature. The T -s diagram shows the dew-point temperature. . . . 30

1.10 Effect of sulphur and excess air on acid due point for a crude oil(adapted from [8]). . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.11 Ash and combustibles in a furnace . . . . . . . . . . . . . . . . . . 311.12 Burnout of combustibles . . . . . . . . . . . . . . . . . . . . . . . . 321.13 Composition of dry combustion products for sub-stoichiometric

combustion of methane . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.1 Control volume for mass and energy balance . . . . . . . . . . . . . 362.2 Mass balance for the boiler . . . . . . . . . . . . . . . . . . . . . . 392.3 Illustration of open, closed and isolated systems . . . . . . . . . . . 442.4 Sankey’s diagram for energy balance for an open or closed system . 452.5 System in translational and rotational motion . . . . . . . . . . . . 462.6 Definition of the sign of work . . . . . . . . . . . . . . . . . . . . . 512.7 Work done to the system . . . . . . . . . . . . . . . . . . . . . . . . 522.8 Specific heats at constant pressure for various molecules . . . . . . 572.9 Physical enthalpies of various gases as a function of temperature . 582.10 Illustration of enthalpy of formation of a compound . . . . . . . . . 62

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List of Figures

2.11 Illustration of LCV as an amount of heat extracted from a combus-

tion chamber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.12 Illustration of an energy balance . . . . . . . . . . . . . . . . . . . 732.13 Example of an energy balance . . . . . . . . . . . . . . . . . . . . . 752.14 Energy balance using enthalpy of formation . . . . . . . . . . . . . 772.15 Energy balance using LCV . . . . . . . . . . . . . . . . . . . . . . . 782.16 Sankey diagram demonstrating the concept of the available heat . . 822.17 Fraction of the available heat as a function of the furnace exit

temperature and excess air ratio for combustion of pure methanein air. Constant c p values. . . . . . . . . . . . . . . . . . . . . . . . 85

2.18 Fraction of the available heat as a function of the furnace exittemperature and excess air ratio for combustion of pure methane

in air. JANAF polynomials for c p have been used. . . . . . . . . . . 86

3.1 Examples of irreversible processes in closed and open systems . . . 903.2 Irreversible process of mixing . . . . . . . . . . . . . . . . . . . . . 913.3 Typical thermodynamic processes shown using work-diagram . . . 943.4 Reversible and irreversible gas expansion process . . . . . . . . . . 953.5 Integration paths from i → f . . . . . . . . . . . . . . . . . . . . . 973.6 Determination of absolute specific entropy of a pure substance . . . 1063.7 An open system interacting with other bodies by exchanging mass,

heat and work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1083.8 Heat source providing heat to the system . . . . . . . . . . . . . . 109

3.9 Mass source providing mass to the system . . . . . . . . . . . . . . 1103.10 A single component system consisting of two phases maintained at

constant temperature and pressure . . . . . . . . . . . . . . . . . . 1163.11 Pressure-temperature plot showing the phase-equilibrium curve that

defines stability regions for phases 1 and 2 . . . . . . . . . . . . . . 1183.12 The experimentally determined phase diagram for water . . . . . . 1203.13 The variation of the Gibbs enthalpy with temperature for ice, water

and water vapour (at 1 bar . . . . . . . . . . . . . . . . . . . . . . . 1253.14 The variation of the Gibbs enthalpy with temperature for ice, water

and water vapour at a pressure of 611 Pa . . . . . . . . . . . . . . . 1253.15 Mixing of two ideal gases . . . . . . . . . . . . . . . . . . . . . . . 139

3.16 Mixture of two ideal gases A and B . . . . . . . . . . . . . . . . . . 1403.17 Pure species-A at temperature T and pressure p; Species-A in a

mixture with species B at Temperature T and pressure p . . . . . . 1423.18 Ideal solution of two liquids in equilibrium with its vapour . . . . . 1433.19 Ideal solution of two liquids A and B; different vapour pressures . . 145

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List of Figures

4.1 Minimisation of Gibbs enthalpy as the reaction advances . . . . . . 155

4.2 Thermodynamic equilibrium constant K as a function of tempera-ture for reactions listed in Table 4.2 . . . . . . . . . . . . . . . . . 1694.3 Equilibrium partial pressure of carbon dioxide in calcination reac-

tion CaCO3 → CaO + CO2 . . . . . . . . . . . . . . . . . . . . . . 1754.4 Accurate and estimated values of the thermodynamic equilibrium

constant for Boudouard reaction . . . . . . . . . . . . . . . . . . . 1804.5 Variation of the partial pressures of CO2 and CO with temperature

for total pressure of 1 bar and 10 bar for Boudouard reaction . . . 1824.6 Equilibrium composition of Boudouard reaction at several total

pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1834.7 Illustration of reaction (4.128) . . . . . . . . . . . . . . . . . . . . . 203

4.8 Gibbs enthalpy as a function of extent of the reaction . . . . . . . . 2074.9 Gibbs enthalpy as a function of the amount of carbon dioxide . . . 2094.10 Gibbs enthalpy of the considered system as a function of mCH4 and

mH2O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

5.1 Concentration change with time for a first-order reaction (t0 = 0) . 2235.2 Concentration change with time for a second-order reaction . . . . 2255.3 The approach of concentrations to their equilibrium values for the

reversible reaction A −−⇀↽−− B that is first order in each direction . . 2295.4 Activation energy of a chemical reaction . . . . . . . . . . . . . . . 239

5.5 Arrhenius plot for elementary reactions of halogens with molecularhydrogen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2405.6 Fall-off curves for the reaction C2H6 −−→ CH3 + CH3 . . . . . . . . 241

6.1 Top – rate constant of CO + OH −−→ CO2 + H;Bottom – rate constant of H + CO2 −−→ CO + OH [19]. . . . . . . . 246

6.2 The time behaviour of the chain carriers . . . . . . . . . . . . . . . 2516.3 A simplified scheme of methane oxidation [19] . . . . . . . . . . . . 2536.4 Temporal behaviour of the species concentrations in reactions A1 −→

A2 −→ A3 for k12 = 1 s−1 and k23 = 10 s−1 (reactive intermediateA2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

6.5 Temporal behaviour of the species concentrations in reactions A1 −→A2 −→ A3 for k12 = 10 s−1 and k23 = 1 s−1 (low reactive interme-diate A2). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

6.6 Temporal behaviour of the species concentrations for reactions A1 −→A2 −→ A3 with k12 = k23 = k. . . . . . . . . . . . . . . . . . . . . 259

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List of Figures

6.7 Temporal behaviour of the species concentrations for reactions A1 −→

A2 −→ A3 assuming quasi steady state for A2-species; k12 = 1 s

−1

,k23 = 10 s−1. (to be compared with Fig. 6.4) . . . . . . . . . . . . . 2606.8 Euler’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2656.9 Numerical solutions using the explicit scheme with different time

steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2676.10 Numerical solutions using the implicit scheme with different time

steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2696.11 Temporal behaviour of the species concentrations for reactions A1 −→

A2 −→ A3 assuming quasi steady state for A2-species . . . . . . . . 2716.12 Temporal behaviour of the species concentrations for reactions A1 −→

A2 −→ A3, k12 = 1 s−1, k23 = 10 s−1, Euler´s Explicit Method . . . 275

6.13 Temporal behaviour of the species concentrations for reactions A1 −→A2 −→ A3, k12 = 1 s−1, k23 = 100 s−1, Euleŕ s Explicit Method . . 2766.14 Temporal behaviour of the species concentrations for reactions A1 −→

A2 −→ A3, Euler´s Implicit Method . . . . . . . . . . . . . . . . . 277

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List of Tables

1.1 Names of aliphatic hydrocarbons . . . . . . . . . . . . . . . . . . . 51.2 Composition of combustion products . . . . . . . . . . . . . . . . . 12

2.1 The incoming and out-coming streams in kmol/h . . . . . . . . . . 412.2 Composition of the combustion products . . . . . . . . . . . . . . . 412.3 The incoming and outcoming flow rates in kg/h . . . . . . . . . . . 432.4 The coefficients for polynomials for T in the range 300–1000 K . . . 572.5 The coefficients for polynomials for T in the range 1000–5000 K . . 582.6 Standard enthalpies of formation and standard entropies of some

compounds (JANAF Thermodynamic Tables) . . . . . . . . . . . . . 642.7 LCV of some selected gaseous fuels . . . . . . . . . . . . . . . . . . 682.8 Adiabatic flame temperature T ad for stoichiometric combustion in

air. Combustion products contain CO2 and H2O only. . . . . . . . 802.9 Calculated adiabatic temperature for stoichiometric combustion of

pure CH4 with air. . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

3.1 Transition temperatures, standard enthalpies and standard entropiesof selected substances . . . . . . . . . . . . . . . . . . . . . . . . . 104

4.1 Standard enthalpies of formation, standard entropies and Gibbsformation enthalpies of some selected compounds . . . . . . . . . . 158

4.2 Thermodynamic equilibrium constant K for some reactions impor-tant in combustion . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

4.3 Thermodynamic equilibrium constant K for some reactions impor-tant in combustion . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

4.4 Standard Gibbs enthalpies at T = 1300 K . . . . . . . . . . . . . . 2044.5 Standard Gibbs enthalpies at T = 1000 K . . . . . . . . . . . . . . 213

5.1 Elementary reactions in H2−CO−C1−O2 system . . . . . . . . . . 233

6.1 Rate coefficients of the chain initiation reactions . . . . . . . . . . 2476.2 Rate coefficients of the chain branching reactions. The coefficients

are presented in the form k = A T b exp

− E aR T

. . . . . . . . . . . . 248

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List of Tables

6.3 Comparison of the rate coefficients k of the chain initiation and

chain branching reactions. . . . . . . . . . . . . . . . . . . . . . . . 2496.4 Simplified mechanism of ignition of the hydrogen-oxygen system . . 2496.5 Comparison of the rate coefficients k of the chain propagation re-

actions for methane oxidation. . . . . . . . . . . . . . . . . . . . . . 2526.6 Rate coefficients of the Zeldovich mechanism reactions . . . . . . . 2616.7 Numerical solution using the explicit scheme with h = 0.3 s. . . . . 2666.8 Numerical solution using the implicit scheme with h = 0.3 s. . . . . 269

6.9 Technical vocabulary . . . . . . . . . . . . . . . . . . . . . . . . . . 287

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1 Stoichiometry

Contents1.1 Introduction

1.2 Definitions

1.2.1 Chemical Reactions, Atoms and Molecules in Combustion

1.2.2 Amount of Substances, Mole and Mass Fractions1.2.3 Density and Concentration (Molar density)

1.2.4 Equation of State for Gases and Gas Mixtures

1.3 Combustion Stoichiometry for Gaseous Fuels

1.3.1 Stoichiometric combustion

1.3.2 Excess air ratio (air equivalence ratio) and fuel equiva-lence ratio

1.3.3 Minimum air requirement for a mixture of gaseous fuels

1.3.4 Composition of combustion products

1.4 Combustion stoichiometry for liquid and solid fuels

1.4.1 Minimum oxygen and air requirements and excess air ratio

1.4.2 Combustion products

1.5 Humid Combustion Air

1.5.1 Absolute and relative humidity

1.5.2 Dew Point Temperature of Combustion Products

1.6 Combustibles burnout for solid fuels

1.7 Sub-stoichiometric combustion to carbon dioxide andwater vapour

1.8 Summary

1.1 Introduction

Nowadays combustion provides more than 90 % of the energy sources. Despite thecontinuing search for alternative energy sources combustion will remain important

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1 Stoichiometry

for many decades to come, as shown in Fig. 1.1. For most of high temperature

processes like power generation, glass manufacturing, cement-making and steel-making, combustion of fossil fuels provides the process energy. Any material thatcan be burned to release thermal energy is called a fuel. Most fuels consist pri-marily of hydrogen and carbon and they are called hydrocarbons. Fig. 1.2 showsan interior of a combustion chamber of a power plant producing electricity froma hard coal. A sketch of a modern reheating furnace of Fukuyama steel-work inJapan [2] is shown in Fig. 1.3. The furnace is fired with a mixture of naturalgas and steel-work gases. Petrol and diesel oil are burned in engines used fortransportation. Propeller-driven airplanes are built with propellers powered byengines identical to automobile engines. Civilian and military aircrafts are pow-ered on energy generated through combustion processes occurring in gas turbine

combustors. In short, combustion occurs in our every day life.

Fig. 1.1: Primary Energy Sources in Germany per type of fuel in EJ = 1018J [1]

Industrial flames are essential elements of the processes and flame properties oftenaffect not only the process efficiency but also product quality. Fig. 1.4 (left) showsa typical pulverisedcoal flame for boiler application while Fig. 1.4 (right) showsMILD combustion technology (known also as flameless oxidation) applied to areheating furnace. Designing efficient combustion processes in car engines andgas turbines has been an on-going challenge for combustion engineers.

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1.1 Introduction

Fig. 1.2: Wall Fired Boiler

Fig. 1.3: Modern Reheating Furnace – NKK, Fukuyama Works, Japan [2]

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1 Stoichiometry

Fig. 1.4: Left – pulverised coal flame; Right – MILD combustion of natural gas

1.2 Definitions

1.2.1 Chemical Reactions, Atoms and Molecules in Combustion

A chemical reaction is an exchange and/or rearrangement of atoms between col-liding molecules, for example:

H2 + 12 O2 −−→ H2O

The atoms are conserved (they are not created or destroyed) while molecules arenot conserved. In the above reaction H, O atoms are conserved while moleculesH2, O2 and H2O are not. Reactant molecules (H2 and O2) are rearranged tobecome product (H2O) molecules. Heat is released in this process.

Atoms relevant in combustion are: C, H, O, N, S, Cl. Compounds of carbon andhydrogen are called hydrocarbons. Hydrocarbons are classified into: aliphatichydrocarbons, alicyclic hydrocarbons and aromatic hydrocarbons. The first tenmembers of the unbranched-chain alkane series are:

CH4 methane C6H14 hexaneC2H6 ethane C7H16 heptaneC3H8 propane C8H18 octaneC4H10 butane C9H20 nonaneC5H12 pentane C10H22 decane

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1.2 Definitions

Fig. 1.5: The subsets of hydrocarbons

Table 1.1: Names of aliphatic hydrocarbons

No.of C

atoms

alkane alkene alkyne alkyl group

1 CH4 methane CH3 methyl2 C2H6 ethane C2H4 ethene C2H2 ethyne C2H5 ethyl3 C3H8 propane C3H6 propene C3H4 propyne C3H7 propyl4 C4H10 butane C4H8 butene C4H6 butyne C4H9 butyl5 C5H12 pentane C5H10 pentene C5H8 pentyne C5H11 pentyl

n CnH2n+2 CnH2n CnH2n−2 CnH2n+1

Other molecules relevant in combustion are:

Haloalkanes R−X example: CH3Cl (chloromethane)Alcohols R−OH example: C2H5OH (ethanol)Aldehydes R−CHO example: CH3CHO (ethanal)Amines R−NH2 example: CH3NH2 (methylamin)Ketones R−CO−R example: CH3COCH3 (acetone)Carbocyclic Acids R−COOH example: CH3COOH (ethanoic acid)

1.2.2 Amount of Substances, Mole and Mass Fractions

Atoms and molecules are counted in amount of substances or moles. 6.023 · 1023

particles (atoms, molecules) are called one mole of the substance (Avogadro con-stant is NA = 6.023 · 1023 atoms or molecules per mole).

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1 Stoichiometry

For a mixture of species:

n =

i

ni n = total number of moles (1.1)

where n stands for total number of moles, ni is the number of moles of species i,and the summation extends over all the species.Mole fraction – xi – (mole number) of species i is:

xi = ni

n (1.2)

The molar mass (molecular weight) in g/mol or kg/kmol is the mass of one moleof the species (for example: M C = 12 g/mol, M CO2 = 44 g/mol). The mean molar

mass (molecular weight) of a mixture is:

M mean =

i

xi M i (1.3)

Frequently mole fractions xi are converted into mass fractions (wi) and the fol-lowing relationships hold:

wi = number of kg of species "i"

total number of kg in the system =

ni M ik

nk M k=

xi M ik

xk M k(1.4)

nin = x

i = number of moles of species "i" in 1 kg of mixture

total number of moles in 1 kg of mixture

=

wiM i

1M mean

= wiM i

M mean =

wiM i

k

wkM k

(1.5)

where the summation extends over all the species and wi stands for mass fractionof species i.

1.2.3 Density and Concentration (Molar density)

Variables that do not depend on the size (extent) of the system are called intensivevariables (for example density, molar density). These are defined as the ratio of the corresponding extensive properties and the system volume V ;

density ρ = mV

(in kg/m3)molar density (concentration) c = n

V (in kmol/m3)

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1.2 Definitions

Thus,ρ

c =

m

n = M mean (1.6)Note that the molar density c defined above is in chemistry usually denoted insquare brackets [ ] for example cH2 = [H2] and chemists prefer to express concen-trations in mol/cm3.

1.2.4 Equation of State for Gases and Gas Mixtures

For gases and gas mixtures an equation of state relates temperature, pressure andvolume:

F ( p, T, V ) = 0 (1.7)There are several equations of state for gases and gas mixtures [3, 4, 5]. Theperfect gas equation of state, called also Clausius-Clapeyron equation, is perhapsone of the simplest ones and it reads

p V = n R T (1.8)

orc =

p

R T (1.9)

and

ρ =

p M mean

R T =

p

R T

iwiM i (1.10)

where p is the pressure (in Pa = N/m2), V the volume (in m3), n the molarnumber (in kmol), T the absolute temperature (in K), and R is the universalgas constant R = 8314.3 J/(kmol · K). An ideal gas is an imaginary substancethat obeys relation (1.8). It has been experimentally observed that relation (1.8)approximates closely behaviour of real gases at low densities which means atlow pressures and at high temperatures. What are a low pressure and a hightemperature? The pressure or temperature of a substance is high or low relativeto its critical pressure ( pcr) or critical temperature (T cr). The useful rules of thumbs are:

(a) at very low pressures ( p pcr ≪ 1), gases behave as an ideal gas regardless of temperature;

(b) at high temperatures ( T T cr

> 2) gases behave also as an ideal gas;(c) deviation from an ideal gas behavior increases upon approaching the critical

point.

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1 Stoichiometry

From Eq. (1.8) follows that one kmol of any (ideal) gas under constant tempera-

ture and pressure occupies the same volume. One can easily verify that under socalled normal conditions: a pressure of p0 = 1 bar (105 N/m2)and a temperatureof T 0 = 25 ◦C (298.15 K) a 1 kmol of any ideal gas occupies a volume of 24.79 m3.It is important to realisethat normal cubic meter (Nm3 or m3n) is a unit of mass(substance) – not volume. Thus,

1 kmol of gas = 24.79 m3n1 mol of gas = 24.79 dm3n

Throughput this lecture course the standard (normal) conditions of p0 = 1 bar(105 N/m2) and T 0 = 25 ◦C (298.15 K) are used1.

1.3 Combustion Stoichiometry for Gaseous

Fuels

1.3.1 Stoichiometric combustion

Combustion is said to be stoichiometric if fuel and oxidiserconsume each othercompletely forming only carbon dioxide (CO2) and water (H2O). If there is anexcess of fuel, the system is fuel-rich, and if there is an excess of oxygen, it iscalled fuel-lean.

Examples are:

CH4 + 2 O2 −−→ 2 H2O + CO2 stoichiometric

CH4 + 3 O2 −−→ 2 H2O + CO2 + O2 lean (excess of oxygen)

CH4 + O2 −−→ H2O + 12 CO2 +

12 CH4 rich (excess of fuel)

If the reaction describing complete combustion (products are CO2 and H2O) iswritten in such a way that it describes the combustion of 1 kmol of fuel:

1 kmol fuel + ν O2 −−→ products (CO2 + H2O)

one may easily calculate the mole fraction of fuel in a stoichiometric mixture (with

1In older books p0 = 760 Tr (1 Tr = 1 mmHg = 133.322 N/m2) and T 0 = 0

◦C (273.15 K) areused so that 1 kmol of ideal gas occupies a volume of 22.41 m3.

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oxygen) as follows:

xfuel,stoich in oxygen = number of moles of fuel

total number of moles (fuel + oxygen) = 1

1 + ν (1.11)

For example:

H2 + 12 O2 −−→ H2O xH2,stoich in oxygen =

1

1.5 =

2

3

CH4 + 2 O2 −−→ CO2 + 2 H2O xCH4,stoich in oxygen = 1

3

CO + 12 O2 −−→ CO2 xCO,stoich in oxygen = 1

1.5 =

2

3

If dry air is used as an oxidiser it contains only 21 % O2, 78 % N2, and 1 % of noble gases. Thus,

xfuel,stoich in air = 1

1 + ν 0.21=

1

1 + 4.762 ν (1.12)

An example:Combustion of propane in oxygen:

C3H8 + 5 O2 −−→ 3 CO2 + 4 H2O and xC3H8,stoich in oxygen = 1

6

Combustion of propane in air:

C3H8 + 5 ( O2 + 3.762N2) −−→ 3 CO2 + 4 H2O + 5 · 3.762N2

and xC3H8,stoich in air = 1

1 + 5 · 4.762 = 0.0403

Note: The higher the hydrocarbon the lower is the fuel mole fraction at stoichio-metric conditions.

1.3.2 Excess air ratio (air equivalence ratio) and fuel equivalence

ratio

The excess air ratio is defined as:

λ = (xair/xfuel)

(xair/xfuel)stoich=

(wair/wfuel)

(wair/wfuel)stoich(1.13)

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1 Stoichiometry

whilst the fuel equivalence ratio Φ is defined as: Φ = 1/λ. The above equation

can be rewritten to allow the calculation of the mole fuel fraction in a mixture of known Φ or λ:

xfuel = 1

1 + 4.762 · ν · λ; xair = 1 − xfuel; (1.14)

xO2 = 0.21 xair = xair4.762

; xN2 = 3.762 xO2 (1.15)

The combustion process can be divided into:

Rich combustion Φ > 1 λ 1Example 1.1

Five hundred m3n/h of ethane are combusted with five thousand five hundredm3n/h of dry air. Calculate the excess air ratio and the fuel equivalence ratio.Assumptions: the fuel (ethane) is combusted to carbon dioxide and water.

The combustion reaction is then

C2H6 + 72 O2 −−→ 2 CO2 + 3 H2O

Thus, (xair/xfuel)stoich = 3.5 · 4.7619/1 = 16.6667 kmol airkmol fuel

500 m3n/h of C2H6 is equivalent to 500/24.79 = 20.169 kmol of C2H6/h5500 m3n/h of dry air is equivalent to 5500/24.79 = 221.864 kmol of air/h

Thus, xairxfuel

= 221.864/20.169 = 11 and λ = 11/16.6667 = 0.66 (Φ = 1.5152).Comments:

(a) Since the combustion is fuel rich, (1 − 0.66) · 500 = 170 m3n/h of ethane willnot be combusted.

(b) In the above example the conversion from m3n/h into kmol/h is superficial.

End of Example 1.1

Formula (1.11) is very useful and simple when a single fuel is combusted. How-ever, if the gaseous fuel is a mixture of hydrogen, CO, hydrocarbons and othermore complex compounds and the air contains moisture, the calculations of excessair ratio is more troublesome. Therefore in combustion engineering the excess airratio for mixture of technical gases is calculated using the minimum air require-ment.

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1.3.3 Minimum air requirement for a mixture of gaseous fuels

Composition of gaseous fuels is usually expressed in mol (volume) fractions. Achemical analysis of a dry gaseous fuel includes the mol fractions of hydrogen,carbon monoxide, carbon dioxide, methane, ethane, propane, higher hydrocarbonsas well as oxygen. The following applies:

xH2 + xCO + xCH4 + xCnHm + xO2 + xCO2 + xN2 = 1 (1.16)

where xi’s stand for mole fractions.The following combustion reactions are applicable:

H2 +

1

2 O2 −−→ H2OCO + 12 O2 −−→ CO2

CH4 + 2 O2 −−→ CO2 + 2 H2O

C2H6 + 72 O2 −−→ 2 CO2 + 3 H2O

CnHm + (n + m

4 )O2 −−→ n CO2 + m

2 H2O

Thus, minimum oxygen requirement for a mixture of gases is:

lO2,min = 1

2 xH2 +

1

2 xCO + 2 xCH4 +

n +

m

4

xCnHm − xO2 (1.17)

(lO2,min in kmol O2/kmol dry gaseous fuel)

whilst the minimum dry air requirement is:

ldry air,min =lO2,min

0.21 = 4.7619 · lO2,min (1.18)

Thus, the excess air ratio is

λ = amount of dry air supplied per kmol of fuelminimum dry air requirement per kmol of fuel

= ldry airldry air,min

(1.19)

1.3.4 Composition of combustion products

The table below shows the number of moles of combustion products produced incomplete combustion of gaseous fuels: Two types of combustion products are con-sidered in combustion engineering namely wet and dry products. The minimum

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1 Stoichiometry

Table 1.2: Composition of combustion products

Fuelcomponent

Component of combustion productsCO2 H2O N2 O2

H2 − 1 − −CO 1 − − −CH4 1 2 − −C2H6 2 3 − −CnHm n m/2 − −

O2 − − − −N2 − − 1 −

amount (at stoichiometric conditions) of wet combustion products is:

V wet,min = 1 xH2 + 1 xCO + 3 xCH4 + 5 xC2H6+n +

m

2

xCnHm + 1 xN2 + 0.79 lair,min

(V wet,min is in kmol wet products/kmol fuel)

while for lean combustion:

V wet = V wet,min + (λ − 1) · lair,min

The minimum (at stoichiometric conditions) amount of dry combustion productsis then

V dry,min = 1 xCO + 1 xCH4 + 2 xC2H6 +

(n · xCnHm) + 1 xN2 + 0.79 lair,min

(V dry,min is in kmol dry products/kmol fuel)

while for lean combustion:

V dry = V dry,min + (λ − 1) · lair,min

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Composition of combustion products (wet):

xCO2,wet =1 xCO + 1 xCH4 + 2 xC2H6 + n xCnHm

V wet

kmol CO2kmol wet products

xH2O,wet =1 xH2 + 2 xCH4 + 3 xC2H6 +

12 m xCnHm

V wet

kmol H2Okmol wet products

xN2,wet =1 xN2 + 0.79 λ lair,min

V wet

kmol N2kmol wet products

xO2,wet = 0.21 (λ − 1) lair,min

V wet

kmol O2kmol wet products

Composition of combustion products (dry):

The amount of dry combustion products per 1 kmol of fuel can be calculated as:

V dry = V wet

1 − xH2O,wet

= V wet −

1 xH2 + 2 xCH4 + 3 xC2H6 + 12 m xCnHm

in kmol dry products/kmol of fuel

Thus, composition of dry combustion products can be calculated as follows:

xCO2,dry =1 xCO + 1 xCH4 + 2 xC2H6 + n xCnHm

V dry

kmol CO2kmol dry products

xN2,dry =1 xN2 + 0.79 λ lair,min

V drykmol N2

kmol dry products

xO2,dry = 0.21 (λ − 1) lair,min

V dry

kmol O2kmol dry products

Example 1.2Calculate composition of wet and dry products of combustion of Dutch (Gronin-

gen) natural gas at 20 % excess air [6].

Composition of Dutch natural gas:

CH4 81 (vol%) C2H6 3 (vol%)C3H8 1 (vol%) N2 15 (vol%)

Assumptions: the fuel is combusted to carbon dioxide and water.

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1 Stoichiometry

Thus, the relevant combustion reactions are:

CH4 + 2 O2 −−→ CO2 + 2 H2O || 0.81

C2H6 + 72 O2 −−→ 2 CO2 + 3 H2O || 0.03

C3H8 + 5 O2 −−→ 3 CO2 + 4 H2O || 0.01

Minimum oxygen requirement:

lO2,min = 0.81 · 2 + 0.03 · 3.5 + 0.01 · 5 = 1.775 kmol O2kmol fuel

m3n of O2m3n of fuel

Minimum air requirement:

lair,min = 1.7750.21

= 8.452 kmol dry airkmol fuel

The amount of air for λ = 1.2;

lair = 1.2 · 8.452 = 10.14 kmol dry air

kmol fuel

Combustion products:

Fuelcomponent

Component of combustion products

molar fractionin dry fuel

CO2 H2O N2 O2 xiCH4 1 2 − − 0.81C2H6 2 3 − − 0.03C3H8 3 4 − − 0.01

N2 − − 1 − 0.15

Minimum amount of wet combustion products (for λ = 1):

V wet,min = 3 · 0.81 + 5 · 0.03 + 7 · 0.01 + 1 · 0.15 + 0.79 · 8.452 =

9.477

kmol wet products

kmol of fuel

The amount of combustion products for lean combustion (for λ ≥ 1)

V wet = 9.477 + (λ − 1) · 8.452 and for λ = 1.2 V wet = 11.17 kmol wet products

kmol of fuel

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Composition of wet combustion products:

xCO2,wet = 1 · 0.81 + 2 · 0.03 + 3 · 0.01

9.477 + (λ − 1) · 8.542

and for λ = 1.2 one obtains xCO2,wet = 0.081

xH2O,wet = 2 · 0.81 + 3 · 0.03 + 4 · 0.01

9.477 + (λ − 1) · 8.542

and for λ = 1.2 one obtains xH2O,wet = 0.156

xN2,wet = 1 · 0.15 + λ · 8.452 · 0.79

9.477 + (λ − 1) · 8.542

and for λ = 1.2 one obtains xN2,wet = 0.7309

xO2,wet = (λ − 1) · 8.452 · 0.21

9.477 + (λ − 1) · 8.542

and for λ = 1.2 one obtains xO2,wet = 0.0318

The amount of dry combustion products (for λ ≥ 1):

V dry = 1 · 0.81 + 2 · 0.03 + 3 · 0.01 + 1 · 0.15 + 0.79 · 8.452 + (λ − 1) · 8.452 =7.7271 + (λ − 1) · 8.452

and for λ = 1.2 one obtains V dry = 9.4175 kmol products/kmol fuel

Composition of dry combustion products:

xCO2,dry = 1 · 0.81 + 2 · 0.03 + 3 · 0.01

7.727 + (λ − 1) · 8.542

and for λ = 1.2 one obtains xCO2,dry = 0.0956

xN2,dry = 1 · 0.15 + λ · 8.452 · 0.79

7.727 + (λ − 1) · 8.542

and for λ = 1.2 one obtains xN2,dry = 0.8667

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1 Stoichiometry

xO2,dry = (λ − 1) · 8.452 · 0.21

7.727 + (λ − 1) · 8.542 and for λ = 1.2 one obtains xO2,dry = 0.0377

Comments:O2, CO2 and N2 concentrations (dry and wet) are functions of excess air ratio.Make a graph showing this dependence (see Fig. 1.6 and Fig. 1.7).

End of Example 1.2

1.4 Combustion stoichiometry for liquid andsolid fuels

For solid and liquid fuels the elemental composition (ultimate analysis) is usuallyexpressed in mass fraction (percentage). A complete fuel analysis includes thenmass fractions of carbon, hydrogen, sulphur, oxygen and nitrogen as elements,and water (moisture) and ash as compounds.

The following applies:

c + h + o + n + s + moisture + ash = 1 (1.20)

where small letters stand for mass fractions. In the above relationship the massfractions of the elements (c, h, s, o, n) are “as fired” (sometimes called also “asreceived”), it means water and ash are present in the fuel. Fuel compositionis often expressed on a dry basis (without water, but with ash) and then thefollowing conversion is applicable:

cdry = cas received1 − moisture

(1.21)

When fuel composition is expressed on a dry ash-free basis the following conversionis applicable:cdry,ashfree =

cas received1 − moisture − ash

(1.22)

Obviously, similar conversions can be applied to hydrogen, sulfur, oxygen, andnitrogen mass fractions.

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1.4.1 Minimum oxygen and air requirements and excess air ratio

The following reactions determine the minimum oxygen and air requirements:

C + O2 −−→ CO2S + O2 −−→ SO2

H2 + 12 O2 −−→ H2O

N + O2 −−→ NO22

Thus, the minimum oxygen requirements (in kmol O2 per kg of fuel “as received”)is

lO2,min = 1 · c

12 + 1

2 ·h

2 + 1 · s

32 + 1 · n

28 − o

32

kmol O2kg of fuel as received (1.23)

or

lO2,min = 24.79 ·

1 ·

c

12 +

1

2·

h

2 + 1 ·

s

32 + 1 ·

n

28 −

o

32

m3n O2

kg of fuel as received

and the minimum air requirement is:

lair,min =lO2,min

0.21 = 4.762 · lO2,min (1.24)

The excess air ratio is then calculable as:

λ = amount of dry air supplied per kg of fuelminimum dry air requirement per kg of fuel

= ldry airldry air,min

(1.25)

1.4.2 Combustion products

Products of complete combustion of fuels contain CO2, H2O, SO2, N2 and excessO2. If the temperature of the combustion products is above the dew point thewater vapor does not condense. Combustion products can be considered as dryor wet:

V wet = V dry + V H2O (1.26)

2It has been assumed that sulphur and nitrogen present in the fuel are oxidisedto SO 2 andNO2,respectively. In reality, other sulphur and nitrogen oxides also occur in combustion products.However, since the nitrogen and sulphur contents in the fuel are low, such an assumption is

justifiable for calculating the minimum oxygen and air requirements.

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1 Stoichiometry

where V wet is the amount (in kmols or m3n) of wet combustion products per kg of

fuel while V dry is the amount of dry products and the V H2O stands for the amountof water in the products. The minimum amount (at stoichiometric combustion,λ = 1) of wet combustion products is:

V wet,min = 1 · c

12 + 1 ·

h

2 +

moisture

18 + 1 ·

s

32 +

n

28 + 1 · 0.79 · lair,min

in kmol wet products/kg of fuel (1.27)

and at lean combustion (λ > 1):

V wet = 1 · c

12 + 1 ·

h

2 +

moisture

18 + 1 ·

s

32 +

n

28 +

1 · 0.79 · lair,min + (λ − 1) · lO2,min + (λ − 1) · 0.79 · lair,min

V wet = c

12 +

h

2 +

moisture

18 +

s

32 +

n

28 + (λ − 1) · lO2,min + λ · 0.79 · lair,min

CO2from c

H2Ofrom h

moisturein fuel

SO2from s

N2from n

excessoxygen

N2in air The

above expressions for calculating V wet can be rearranged into a more elegant formas follows:

V wet = 1 · c

12 + 1 ·

h

2 +

moisture

18 + 1 ·

s

32 +

n

28 + 0.79 · lair,min + (λ − 1) · lO2,min+

0.79 · (λ − 1) · lair,min

= c

12 +

h

2 +

moisture

18 +

s

32 +

n

28 + 0.79 · lair,min + 0.21 · (λ − 1) · lair,min+

0.79 · (λ − 1) · lair,min

= c

12 +

h

2 +

moisture

18 +

s

32 +

n

28 + 0.79 · lair,min + (λ − 1) · lair,min

= V wet,min + (λ − 1) · lair,min

Thus,V wet = V wet,min + (λ − 1) · lair,min (1.28a)

andV wet,min =

c

12 +

h

2 +

moisture

18 +

s

32 +

n

28 + 0.79 · lair,min (1.28b)

where V wet,min is the minimum amount of products (for λ = 1).

The amount of dry combustion products can be easily calculated by dropping the

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terms h2 and moisture

18 from the equations above. Thus,

V dry = c

12 +

s

32 +

n

28 + 0.79 · lair,min + (λ − 1) · lO2,min + (λ − 1) · 0.79 · lair,min

kmol dry products/kg of fuel (1.29a)

or

V dry = V dry,min + (λ − 1) · lair,min kmol dry products/kg of fuel (1.29b)

and

V dry,min = c

12

+ s

32

+ n

28

+ 0.79 · lair,min

kmol dry products/kg of fuel (1.30)

Composition of the combustion productsThe composition of the combustion products can be easily calculated realisingthatwet products contain H2O, CO2, SO2, N2 and excess O2. Thus:

xH2O =h2 +

moisture18

V wetkmol H2O/kmol wet products

xCO2,wet =c

12

V wetkmol CO2/kmol wet products

xSO2,wet =

s

32V wet kmol SO2/kmol wet products

xO2,wet =(λ − 1) · lO2,min

V wetkmol O2/kmol wet products

xN2,wet =n28 + 0.79 · λ · lair,min

V wetin kmol N2/kmol wet products

Similarly one can derive simple relationships for calculating composition of drycombustion products (CO2, SO2, N2 and excess O2):

xCO2,dry =c

12

V dryxSO2,dry =

s32

V dry

xO2,dry =(λ − 1) · lO2,min

V dryxN2,dry =

n28 + 0.79 · λ · lair,min

V dry

where V dry is the amount of dry combustion products in kmol/kg of fuel.

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1 Stoichiometry

Example 1.3

A coal of the composition given in the table below is combusted with air at 10 %excess air. Calculate the composition of the combustion products (wet and dry)and produce a curve showing the CO2 and O2 mole fractions (dry) as a functionof excess air ratio.

Coal Fettnuss mvb3 (origin – Germany) – coal analysis obtained from a chemicallaboratory

Basis H2O Ash Ultimate Analysis LCV %-wt %-wt %-wt MJ/kg

c h n o s

Dry ash free 89.39 4.68 1.53 3.66 0.74Dry 4.4 32.87As fired 3.5


One begins with calculating the coal composition “as fired” (with moisture andash). Simple conversions result in the following:

Basis H2O Ash Ultimate Analysis LCV %-wt %-wt %-wt MJ/kg

c h n o sDry ash free 89.39 4.68 1.53 3.66 0.74 34.38Dry 4.4 85.4568 4.47 1.4627 3.499 0.7074 32.87As fired 3.5 4.246 82.4658 4.3136 1.4115 3.3765 0.6826 31.72

The minimum oxygen requirement (λ = 1) is:

lO2,min = 1 ·0.824 658

12 +

1

2·

0.043136

2 + 1 ·

0.006826

32 −

0.033 765

32 =

0.0787 kmol O2

kg fuel “as fired”that corresponds to 0.0787 · 32 = 2.5184 kg O2kg of fuel “as fired” .

3mvb - medium volatile bituminous

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1 Stoichiometry

Composition of dry combustion products:

xCO2,dry =0.824658

12

V dry=

0.0687

0.0695 + 0.2957 · λ + 0.0787 · (λ − 1)

xSO2,dry =0.006 83

32

V dry=

0.0002

0.0695 + 0.2957 · λ + 0.0787 · (λ − 1)

xO2,dry =(λ − 1) · lO2,min

V dry=

(λ − 1) · 0.0787

0.0695 + 0.2957 · λ + 0.0787 · (λ − 1)

xN2,dry =0.79 · λ · lair,min

V dry=

0.79 · λ · 0.3743

0.0695 + 0.2957 · λ + 0.0787 · (λ − 1)

All the above mole (volume) fractions are in kmol/kg of dry products.

Composition of combustion products for 10 % excess air ratio (λ = 1.1)Using the above relationships one may easily obtain:Composition of wet products for λ = 1.1 is:

xH2O = 0.0552 xCO2,wet = 0.1613 xSO2,wet = 0.0005

xO2,wet = 0.0185 xN2,wet = 0.7635

Composition of dry products for λ = 1.1 is:

xCO2,dry = 0.1706 xSO2,dry = 0.0005

xO2,dry = 0.0195 xN2,dry = 0.8079

CO2 and O2 concentrations in dry combustion products as a function of excessair ratio

The dependence of carbon dioxide and oxygen concentrations as a function of excess air ratio can easily be derived from the above relationships. Fig. 1.6 andFig. 1.7 show the dependence.

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Fig. 1.6: Oxygen concentration in dry combustion products as a function of excess airratio

Fig. 1.7: Carbon dioxide concentration as a function of excess air ratio

23


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1 Stoichiometry

Comments:

(a) For the fuels considered in this example (CH4, C2H6, Groningen natural gas,coal Fettnuss), the relationship between oxygen content in combustion prod-ucts and excess air ratio (see Fig. 1.6) is almost linear for λ < 1.5. Thisrelationship holds for most of fuels and is used by furnace operators in everyday practice. For example a 2 % (dry) oxygen content in the flue gas indicatesthat the furnace is operated at 1.1 excess air ratio.

(b) Carbon dioxide concentration in combustion products reaches maximum atλ = 1.0 and decreases almost linearly with excess air ratio as shown in Fig. 1.7.This maximum carbon dioxide concentration at λ = 1.0 is a characteristicsof the fuel only.

End of Example 1.3


In Section 1.3.3 we have derived Eq. (1.18) for calculating the minimum dry airrequirement for combustion of one kmol of a gaseousfuel. A similar expression,Eq. (1.24), has been derived to calculate the minimum dry air requirement forcomplete combustion of one kilogram of a solid or a liquid fuel. We stress hereagain that Eqs. (1.18) and (1.24) are for calculating the dry air requirements.However, combustion or atmospheric air is seldom completely dry and it con-

tains moisture (water vapour). Accurate calculations of combustion stoichiometryshould account for the presence of water vapour. In this paragraph we show howto do it. We recall that air that contains no water vapour is called dry air. Atmo-spheric or combustion air containing water vapour is named here as combustionair. The temperature of air in combustion applications ranges from about −20 toabout 1300 ◦C. In most of the furnace applications combustion air is supplied atpressures that are slightly higher than ambient air pressure of 1 bar. However, ingas turbines and engines the combustion air is typically compressed to 20–30 barpressure.

1.5.1 Absolute and relative humidity

It is convenient to treat combustion air as a mixture of water vapour and dry airsince the composition of dry air remains constant but the amount of water vapourchanges. It is certainly convenient to treat the water vapour in combustion air asan ideal gas and for ambient air pressures such an assumption is perfectly valid.

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Even for pressures up to 30 bar and temperatures up to 1300 ◦C the ideal gas

assumption is justifiable and the maximum departure from reality is typically inthe range 0.2–6 %. Then the combustion air is treated as an ideal-gas mixturewhose pressure is the sum of the partial pressure of the dry air ( pdry air) and thatof the water vapour ( pv)

p = pdry air + pv (1.31)

The partial pressure of water vapour ( pv) is typically referred to as the vapourpressure. The temperature, however, is uniform throughout the dry-air/water-vapour mixture so that

T = T dry air = T v (1.32)

Usually the total pressure ( p) is known whereas the partial pressure of watervapour ( p

v) depends on how much moisture is present in the mixture. For an

ideal-gas mixture, the mole fraction of water vapour is then

xv = pv p

(1.33)

The amount of water vapour in the combustion air can be specified in variousways. However, the simplest way is to specify the mass of water vapour presentin a unit of dry air. This is called absolute or specific humidity and in this lectureseries is denoted as ϕ and is expressed in kg of water vapour per kg of dry airsince

ϕ =

mv

mdry air =

M v · nv

M dry air · ndry air =

18.016

28.97 ·

pv

p − pv = 0.622 ·

pv

p − pv (1.34)

where M v and M dry air are the molar masses of water and dry air, respectivelywhilst nv and ndry air stand for the number of moles of water vapour and dry air,respectively.

Dry air contains no water vapour and therefore its specific humidity is zero. Nowlet us add some water vapour to this dry air so the specific humidity (ϕ) increases.If we add more water vapour the specific humidity keeps increasing until the aircan hold no more moisture. At this point the air is said to be saturated withmoisture vapour and it is called saturated air. The amount of water vapour insaturated air at a given temperature and total pressure can be determined usingEq. (1.34) by replacing pv by the saturation pressure psat of water at the giventemperature. The saturation pressure of water vapour is plotted in Fig. 1.8 as afunction of temperature using a relationship4

4In Chapter 3, Example 3.3, we will derive Eq. (1.35) using Clausius-Clapeyron equation.

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1 Stoichiometry

psat = 611 · exp

−5304.3 · 1

T −

1

273.16

in Pa (1.35)

Water Vapour

p

i n

k P

a

s a

t

Temperature in K

10

5

0

280 290 300 310 320

Fig. 1.8: Saturation pressure of water vapour as a function of temperature. The plot isobtained using Eq. (1.35) which in the plotted range provides pressure valuesthat are within 2 % accuracy with the values listed in steam tables [7].

The ratio of the amount of moisture the air holds to the maximum amount of moisture the air can hold (at the saturation state) at the same temperature iscalled the relative humidity γ

γ = pv

pv,sat(1.36)

where pv,sat stands for the saturation water vapour pressure at the specified tem-perature. The relative humidity (γ ) ranges from 0 for dry air to 1 for saturatedair. The amount of moisture that combustion air can hold depends on its tem-

perature. Thus, the relative humidity of air (γ ) changes with temperature evenwhen its specific humidity (ϕ) remains constant. Using the relative humidity andthe saturation pressure of water vapour, the specific humidity (ϕ) can then becalculated as follows:

ϕ = 0.662 · γ · psat

p − γ · psat(1.37)

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1.5.1.1 Wet air requirement

In the previous paragraphs we have developed simple formulae for calculating thedry air requirements (ldry air); Eq. (1.17) for gaseous fuels while Eq. (1.23) forliquid and solid fuels. The above considerations on the combustion air humidityallows for inclusion of water vapour since

lwet air = ldry air + ϕ · ldry air = (1 + ϕ) · ldry air (1.38)

If there is a need to calculate the enthalpy of wet combustion air it can be easilydone since

hwet air = hdry air + ϕ · hvapour (1.39)

where h stands for specific enthalpy in J/g (or kJ/kg). The enthalpy of watervapour at ambient air pressure in the temperature range −10 to 50 ◦C can bedetermined approximately using

hvapour(T ) ∼= 2501.3 + 1.82 · T in kJ/kg where T in ◦C (1.40)

For higher temperatures and higher pressures values from steam tables should beused.

Example 1.4In Example 1.3 we have calculated the air requirement and the composition of dry

and wet combustion products as a function of the excess air ratio for coal Fettnuss.

In Example 1.3 we have ignored the moisture content of the combustion air. Theobjective of this example is to include the moisture and by doing so to examineits effect on the results of the calculations. We assume here that the combustionair is supplied at 1 bar pressure and at a 20 ◦C temperature. Its relative humidityis 75 %.


One begins with calculating the saturation pressure of water vapour at 20 ◦C usingEq. (1.35)

psat = 611 · exp −5304.3 · 1

297.16 −

1

273.16 = 2298.14 Pa (1.41)The absolute humidity is then

ϕ = 0.662 · γ · psat

p − γ · psat= 0.662 ·

0.75 · 2298.14

105 − 0.75 · 2298.14 = 0.0116

kg water vapourkg dry air

(1.42)

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1 Stoichiometry

so the minimum amount of wet combustion air is

lwet air,min = (1+ϕ) · ldry air,min = 1.0116 · 10.9496 = 11.0766 kg wet air/kg of fuel(1.43)

The moisture supplied with the combustion air stream occurs in the (wet) com-bustion products so (see Example 1.3)

V wet = 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · ϕ · ldry air,min

18 (1.44)

and

V wet = 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.00706 kmol wet productskg of fuel “as fired”

(1.45)Composition of wet combustion products is therefore as follows:

xH2O =0.043136

2 + 0.035

18 + λ · 0.00706

V wet

= 0.0235 + λ · 0.00706

0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.00706

xCO2,wet =0.824658

12

V wet

= 0.0687

0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.00706

xSO2,wet =0.0068826

2

V wet

= 0.0002

0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.00706

xO2,wet =(λ − 1) · lO2,min

V wet

= (λ − 1) · 0.0787

0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.00706

xN2,wet = 0.79 · λ · ldry air,min

V wet

= 0.79 · λ · 0.3743

0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.00706

All the above mole (volume) fractions are in kmol/kmol of wet products.

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At 10 % excess air ratio (λ = 1.1) the above formulae provide:

xH2O = 0.0721 xCO2,wet = 0.1584 xSO2,wet = 0.00046

xO2,wet = 0.01814 xN2,wet = 0.7498

Comments:

(a) Taking into account the combustion air moisture has resulted in a 1.2 % cor-rection to the minimum air requirement.

(b) However, the mole fraction of water vapour in (wet) combustion products hasincreased by around 31 %.

(c) Obviously neither the amount of the dry combustion products nor its com-position is affected by combustion air moisture content.

End of Example 1.4

1.5.2 Dew Point Temperature of Combustion Products

Typically combustion products contain water vapour. When for example a nat-ural gas is combusted the water vapour content in the combustion products maybe as high as 16 % for λ = 1.2 as shown in Example 1.2. When coal Fettnuss iscombusted the water content of around 7 % (see Example 1.4) is expected. Whiledesigning combustion systems it is required that combustion products are cooleddown to low temperatures before they are released to the atmosphere. Thus,

by cooling down the combustion products we may expect that below a certaintemperature the water vapour begins to condensate. The dew-point temperature(T dp) is defined as the temperature at which condensation begins when the com-bustion products (or generally moist air) is cooled at a constant pressure. In otherwords, T dp is the saturation temperature of water corresponding to the vapourpressure, as shown in Fig. 1.9.

As a matter of fact Eq. (1.35) can be used to determine the due-point temperatureif the pressure is given. Let us assume that there is 7 vol% water vapour content incombustion products of coal Fettnuss combustion. If the combustion products areat 1 bar pressure, the partial pressure of vapour is 7000 N/m2 and using Eq. (1.35)we can estimate that the dew-point temperature is around 312.4 K (39.3 ◦C).

Thus, in order to avoid condensation it is desired to keep the combustion productsat temperatures typically 40–60 ◦C higher than the due point temperature.

Dew points vary with the amounts of O2, CO2, SO2, NOx,HCl in combustionproducts. In particular sulphur oxides have a pronounced effect on the dew pointtemperature as shown in Fig. 1.9 for several excess air ratios.

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1 Stoichiometry

T

T1

Tdp

s

p =

c o n s t

.

v

1

2

Fig. 1.9: Cooling of combustion products (or moist air) at a constant temperature.

The T

-s

diagram shows the dew-point temperature.

The excess air curves are not equally spaced since the extra oxygen tends toproduce more SO3 which exerts a catalytic effect in raising the dew point. Forexample the above estimated dew-point temperature of 39.3 ◦C would be almostdoubled if 2 ppm of SO2 was present. This is the reason that operators of coal-firedpower station boilers maintain the flue gases at temperatures above 180–200 ◦C.

1.0 2.0 3.0 4.0 5.0

Weight % sulfur in fuel oil

Aciddew

point,°C

135

140

130

120

115

145

2 0 % e x c e

s s a i r

1 5 %

1 0 %

5 %

Fig. 1.10: Effect of sulphur and excess air on acid due point for a crude oil (adaptedfrom [8]).

30


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It is not possible to burn 100 % of solid fuels. When firing liquid and solid fuels themajor contribution to the “unburns” comes usually from the residual “oil-coke” or

“coal char”, although the incompletely combusted gases (CO and hydrocarbons)and soot particles can also make a contribution. In the considerations that follow,the mass fraction of total combustibles is simply calculated as 1 minus the ashcontent as shown in Fig. 1.11.

Total Combustibles = 1 − ash

Fig. 1.11: Ash and combustibles in a furnace

Composition of the fuel: Composition of the unburned solids:C 0 (Total Combustibles) + ash0 = 1 C 1 + ash1 = 1m0 – mass flow rate of fuel m1 – mass flow rate of unburned solids

Mass balance of combustibles can be written as:

m0 · C 0 = m1 · C 1 + b · m0 · C 0

where m0C 0(in kg/s) stands for the amount of combustibles entering the furnace,m1C 1 (in kg/s) represents the amount of combustibles in the unburned solids leav-ing the furnace while bm0C 0 (in kg/s) stands for the amount of the combustiblesburned whilst b is the fraction of the original combustibles burned. Assuming thatash is an inert and does not enter into any chemical reactions, its mass balancecan be written as

m0 · ash0 = m1 · ash1

where the left hand side is the ash input (in kg/s) while the right hand side stands

for ash flow leaving the furnace.

From the above relationships one may obtain:

b = 1 − C 1 · ash0C 0 · ash1

= 1 − (1 − ash1) · ash0(1 − ash0) · ash1

=1 − ash0

ash1

1 − ash0(1.46)

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1 Stoichiometry

where b shows the degree of burnout and (1 − ash1) is often called “carbon in

ash” (in the USA it is called “loss on ignition”). Fig. 1.12 shows the relationshipbetween the combustibles burnout (b) and carbon in ash for several ash contentsof the solid fuel.

Fig. 1.12: Burnout of combustibles

1.7 Sub-stoichiometric combustion to carbon

dioxide and water vapour

Sub-stoichiometric combustion occurs when excess air ratio (λ) is smaller than1. Obviously, in sub-stoichiometric combustion not all the fuel is combustedsince there is not enough oxidiserpresent and some of the fuel remains unburned.If the fuel is combusted to carbon dioxide and water vapour, the combustioncalculations leading to the determination of the composition of the combustionproducts are rather straight forward as shown in Example 1.5. However if somespecies like, for example, carbon monoxide and hydrogen are to be consideredthe composition of the combustion products has to be determined using chemicalequilibrium calculations underlined in Chapter 4.

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1.7 Sub-stoichiometric combustion to carbon dioxide and water vapour

Example 1.5

Produce a curve showing the composition of combustion products obtained insub-stoichiometric combustion of methane in air as a function of excess air ratio.

Assumptions: methane is combusted to carbon dioxide and water vapour.

We begin with writing down the oxidation reaction for methane

CH4 + 2 O2 −−→ CO2 + 2 H2O

The excess air ratio is then

λ = nair

20.21

so that nair = 9.5238 · λ

where nair is the number of moles of air per 1 mole of methane. Thus, when λ


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1 Stoichiometry

Fig. 1.13: Composition of dry combustion products for sub-stoichiometric combustionof methane

End of Example 1.5

1.8 Summary

Nowadays more than 90 % of the energy used by human beings is generated bycombustion of fossil fuels. Despite the continuing search for alternative energysources combustion will remain important for many decades to come.

In the first lecture the basic concepts of chemical reactions, atoms and moleculeshave been recalled. The lecture introduces stochiometric, lean and rich combus-tion as well as the notions of excess air ratio and equivalence ratio. In this lecturewe have assumed that fuels are combusted to carbon dioxide and water vapour.Students should know how to calculate the amount of air (oxidiser) needed forcomplete combustion of a given fuel. Furthermore you should be able to deter-

mine composition of wet and dry combustion products for any fuel combusted tocarbon dioxide and water vapour at a given excess air.

Numerous equations on combustion stochiometry have been derived in this chap-ter. However, the reader should realise that there is no point in memorising themsince they can be easily recreated.

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2 Mass and Energy Balance

Contents2.1 General Formulation of Mass and Energy Balance

2.1.1 Mass and Energy Balance at an Instant

2.1.2 Mass and Energy Balance over a Time Interval

2.1.3 Mass and Energy Balance under Steady-State Conditions2.1.4 Example of a Mass Balance of a Furnace

2.2 The First Law of Thermodynamics

2.2.1 System Energy

2.2.2 Energy Entering and Leaving the System

2.2.3 Energy Balance of Thermal Systems (Machines)

2.3 Energy Released in Chemical Reactions

2.3.1 Reaction Enthalpy

2.3.2 Standard Enthalpies of Formation

2.3.3 Lower Calorific Value (LCV) and Gross Calorific Value (GCV)

2.3.4 Relationships between Calorific Values, Reaction En-thalpies and Formation Enthalpies

2.3.5 Dependence of LCV on Temperature

2.3.6 Example of an Energy Balance of a Furnace

2.4 Temperature of Adiabatic Combustion

2.5 Furnace Exit Temperature

2.6 Summary

2.1 General Formulation of Mass and EnergyBalance

Conservation of mass and energy is the basis for any considerations in combustionengineering. The formulation presented in this lecture is based on the work of

35


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[9, 10, 11]. In formulating mass and energy balances we need first of all to identify

the control volume. Control volume, Fig. 2.1, is a region of space bounded by acontrol surface (boundary) through which energy and matter may be transferred.

Fig. 2.1: Control volume for mass and energy balance

Once the control volume is identified an appropriate time basis must be specified.There are two options for formulating the balances – a formulation at an instantand a formulation over a time interval.

2.1.1 Mass and Energy Balance at an Instant

Since the mass and energy balances must be satisfied at each and every instantof time t, one option involves formulating the law on a rate basis. That is, atany instant there must be a balance between all mass rates expressed in kg/sand simultaneously, there must be a balance between all energy rates expressedin J/s (=W).

The mass balance at an instant reads, Fig. 2.1:

ṁin + ṁg − ṁout = ṁstored ≡ d(m)

dt (2.1)

where ṁin, ṁg, ṁout, ṁstored are the rates with which the mass enters the controlvolume, the rate of mass generation inside the control volume, the rate with which

matter leaves the control volume and the rate of accumulation (storage) of matterwithin the volume, respectively. All these terms are expressed in kg/s. Symbolm stands for the mass of the system expressed in kg.

Eq. (2.1) is the most general form of the mass balance at an instant. It contains themass generation term ( ṁg) that is non-zero only if nuclear reactions are considered

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2.1 General Formulation of Mass and Energy Balance

during which matter can be converted into energy. In any other systems with non-

nuclear reactions present, the mass generation term is zero and therefore duringthis combustion course we will use the following equation for the mass balance atan instance:

ṁin − ṁout = dm

dt (2.2)

The energy balance at an instant reads:

Ė in + Ė g − Ė out = Ė stored ≡ dE

dt (2.3)

where Ė in, Ė g, Ė out, Ė stored stand for the rates with which energy enters the con-trol volume, the rate of energy generation in the volume, the rate with which the

energy leaves the volume and the rate of accumulation (storage) within the con-trol volume, respectively. All these terms are expressed in J/s (W). The symbolE represents the system energy expressed in joules.

2.1.2 Mass and Energy Balance over a Time Interval

The second option is to formulate balances over a time interval, thus the massbalance is formulated in amount of mass (kilograms) whilst the energy balanceis formulated in amount of energy (joules). The mass balance over a time interval∆t = t2 − t1 reads:

t2 t1

ṁin · dt −t2

t1

ṁout · dt =t2

t1

d(m)

dt · dt = ∆m (2.4)

ormin − mout = ∆m (2.5)

The terms on the left hand side of Eqs. (2.4) and (2.5) denote the amount of massentering the control volume over the time period ∆t, and the amount of mass thatleaves the control volume over this interval, respectively. The right hand side of Eqs. (2.4) and (2.5) represents the amount of mass stored (accumulated) withinthe system in this time period.

The energy balance over a time interval reads:

t2 t1

Ė in · dt +

t2 t1

Ė g · dt −

t2 t1

Ė out · dt =

t2 t1

d(E )

dt · dt = ∆E (2.6)

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or

E in + E g − E out = ∆E (2.7)where the terms on the left hand side of Eqs. (2.6) and (2.7) stand for the amountof energy entering the control volume, the amount of energy generated in thevolume, and the amount of energy leaving the volume in the time period ∆t =t2 − t1. The right hand side of Eqs. (2.6) and (2.7) represents the energy stored(accumulated) in the control volume over the time interval ∆t = t2 − t1.

2.1.3 Mass and Energy Balance under Steady-State Conditions

Under steady-state conditions when there is neither mass nor energy accumulation

(storage) in the system we obtain:

from Eq. (2.2) for mass balance:

ṁin = ṁout (2.8)

from Eq. (2.3) for energy balance:

Ė in + Ė g = Ė out (2.9)

2.1.4 Example of a Mass Balance of a Furnace

The purpose of making a mass balance under steady state conditions isto calculate the out-coming amount of mass and to calculate its makeup (composition). A correct mass balance is a prerequisite to a subsequentenergy balance.

There are two basic ways of formulating a mass balance; by expressing the in-coming and outcoming flows in kmol/h (kmol/s) and their composition in mol(volume) fractions or by expressing the streams in kg/h (kg/s) and their compo-sition in mass fractions. The first way is popular among chemical engineers sinceit deals with kmol and the rates of chemical reactions are typically expressed usingconcentrations (kmol/m3/s). The latter way is common in combustion engineer-ing. Both methods have advantages and disadvantages and they are equivalent.They both lead to the same results. The c

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Weber Thermodynamics