INTERNATIONAL STUDIES in SCIENCE and ENGINEERING
Roman Weber
COMBUSTION FUNDAMENTALS with Elements of Chemical Thermodynamics
Prof. Dr.–Ing. Roman Weber Technische Universität Clausthal Institut für Energieverfahrenstechnik und Brennstofftechnik (IEVB) Agricolastrasse 4, 38 678 ClausthalZellerfeld, Germany
[email protected] Weber, Roman: Combustion Fundamentals with Elements of Chemical Thermodynamics ClausthalZellerfeld: Papierflieger 2008 ISBN 978–3–89720–921–3 Bibliografische Information der Deutschen Bibliothek Die Deutsche Bibliothek verzeichnet diese Publikation in der Deutschen Nationalbibliografie; detaillierte bibliografische Daten sind im Internet über http://dnb.ddb.de abrufbar.
INTERNATIONAL STUDIES in SCIENCE and ENGINEERING Editor 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 (Germany) 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, ClausthalZellerfeld 2008, Telemannstr. 1, 38678 ClausthalZellerfeld, Tel.: 05323/96746, http://www.papierfliegerverlag.de No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage 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. Marc Muster of TU Clausthal.
To Professor Stanisław Jerzy Gdula who introduced me to the wonderful world 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 text books which accompany a lecture series for students and other books which demonstrate how to apply the knowledge acquired in lecture theatres to industrial practise.
With publishing ”Combustion Fundamentals with Elements of Chemical Thermodynamics” the sixth book in this series was released. At least three further books are 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, PapierfliegerVerlag, ClausthalZellerfeld 2008, ISBN 3–89720–702–8. 2. Jeschar, R.; Kostowski, E.; Alt, R.: Wärmestrahlung in Industrieöfen, PapierfliegerVerlag, ClausthalZellerfeld 2004, ISBN 3–89720–686–2 and Wydawnictwo Politechniki Slaskiej, Gliwice 2004, ISBN 83–7335–232–5. 3. Weber, R.; Alt, R.; Muster, M.: Vorlesungen zur Wärmeübertragung, Teil I, 2. Auflage, PapierfliegerVerlag, ClausthalZellerfeld 2008, ISBN 3–89720– 798–2. 4. Dierkes, P. v.; Bruch, G.: Abfall und Chemie, Teil I, PapierfliegerVerlag, ClausthalZellerfeld 2007, ISBN 3–89720–879–2. 5. Dierkes, P. v.; Bruch, G.: Abfall und Chemie, Teil II, PapierfliegerVerlag, ClausthalZellerfeld 2007, ISBN 3–89720–880–6.
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To the Student How to get the most from these lecture notes These lecture notes in ”Combustion Fundamentals with Elements of Chemical Thermodynamics” have been prepared for undergraduate students attending a fifteen–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 Technology. The notes have been prepared not only for Clausthal but also for the Royal Institute of Technology (Stockholm, Sweden), University of Science and Technology (Beijing, China) and Central South University (Changsha, China). If you are like most students just opening this textbook, you are enrolled on one of the few courses in science and engineering at Clausthal that are delivered in English. Probably English is not your native language but do not worry. This textbook has been written with you in mind in very simple and plain English. Below I provide 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 course if you read each chapter before hearing it. In this way, many of the topics will already 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 chapter. In addition to Examples, Problems are formulated for each chapter and they can be found on the web site (see below). Some Problems will be discussed and solved in the classroom but the principal idea behind Problems is that they are your homework. I suggest you solve them, one by one, and in case of difficulties consult your class instructor. Study together with your fellow students. Many students find it useful to form study groups. You can discuss the challenging topics with one another and have 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 developed a web site to assist you in this course. You can find it by browsing through the IEVB–Institute website of TU Clausthal: http://www.ievb.tuclausthal.de/ (follow ”Science” and ”Lectures”).
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Acknowledgments ”Combustion Fundamentals with Elements of Chemical Thermodynamics” has been written within the scope of a European Community project (CN/ASIALINK/016 (103 187)) of the AsiaLink Programme. The author acknowledges with thanks the European Community financial support.
I would like to thank my colleagues who carefully scrutinized the manuscript making 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 edited this textbook. The Vocabulary attached to these lecture notes has been prepared by Dr. Rüdiger Alt. It has been used by students at TU Clausthal attending lectures on Heat Transfer, Combustion Technology and High Temperature Processes. Although I have made a concerted effort to make this first edition error free, some mistakes may have crept in unbidden. I would appreciate hearing from anyone who finds an error or wishes to comment on the text. You may email or write to me. Roman Weber IEVB – TU Clausthal D38678 ClausthalZellerfeld Agricolastrasse 4 Germany
[email protected]
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Contents 1 Stoichiometry 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Chemical Reactions, Atoms and Molecules in Combustion . 1.2.2 Amount of Substances, Mole and Mass Fractions . . . . . . 1.2.3 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 equivalence 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 Substoichiometric combustion to carbon dioxide and water vapour 1.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 11 11 16 17 17 24 24 29 31 32 34
2 Mass and Energy Balance 2.1 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 SteadyState Conditions 2.1.4 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 . . . . . . . . .
35 35 36 37 38 38 43 46 49
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1 1 4 4 5 6 7 8 8
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Contents
2.3
2.4 2.5 2.6
2.2.3 Energy Balance of Thermal Systems (Machines) . . . . . . . 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 Enthalpies and Formation Enthalpies . . . . . . . . . . . . . . . . . . . 2.3.5 Dependence of LCV on Temperature . . . . . . . . . . . . . 2.3.6 Example of an Energy Balance of a Furnace . . . . . . . . . Temperature of Adiabatic Combustion . . . . . . . . . . . . . . . . Furnace Exit Temperature . . . . . . . . . . . . . . . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Equilibrium Thermodynamics 3.1 Irreversible and Reversible Processes . . . . . . . . . . . . . . . . . 3.2 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Entropy of Liquids and Solids . . . . . . . . . . . . . . . . . 3.2.2 Entropy of Ideal Gases . . . . . . . . . . . . . . . . . . . . . 3.2.3 Entropy of Phase Transition at the Transition Temperature 3.2.4 The Third Law of Thermodynamics . . . . . . . . . . . . . 3.2.5 Absolute Entropy of Pure Substances . . . . . . . . . . . . . 3.3 The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . 3.3.1 The Increase in Entropy Principle . . . . . . . . . . . . . . 3.3.2 Entropy Change for a Continuous Process at SteadyState . 3.3.3 Irreversibility of Processes . . . . . . . . . . . . . . . . . . . 3.4 General Conditions for Thermodynamic Equilibrium . . . . . . . . 3.4.1 Isolated System . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 NonAdiabatic System . . . . . . . . . . . . . . . . . . . . . 3.5 Equilibrium Between Phases . . . . . . . . . . . . . . . . . . . . . . 3.5.1 SingleComponent System Consisting of Two Phases . . . . 3.5.2 Phase Transformations of a Pure Substance . . . . . . . . . 3.5.3 Dependence of Gibbs Free Enthalpy on Temperature and Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.4 Equilibrium in MultiComponent SinglePhase Systems . . 3.5.5 Chemical Potential of Pure Substances . . . . . . . . . . . . 3.5.6 Significance of Chemical Potential . . . . . . . . . . . . . . 3.6 MultiComponent, MultiPhase Systems . . . . . . . . . . . . . . . 3.6.1 The Phase Rule . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Thermodynamics of Mixing . . . . . . . . . . . . . . . . . . . . . .
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56 60 60 62 65 67 69 72 79 81 86 89 90 96 102 102 103 104 104 107 107 110 111 112 112 113 116 116 119 122 126 130 131 134 138 138
Contents 3.8
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
4 Chemical Equilibrium 149 4.1 Definition of Chemical Equilibrium . . . . . . . . . . . . . . . . . . 150 4.2 Single Chemical Reaction . . . . . . . . . . . . . . . . . . . . . . . 150 4.2.1 Extent of a Single Reaction . . . . . . . . . . . . . . . . . . 150 4.2.2 Change of Gibbs Enthalpy as a Chemical Reaction Advances153 4.2.3 Gibbs Enthalpy of Selected Reactions . . . . . . . . . . . . 156 4.2.4 Thermodynamic Equilibrium Constant for a Gaseous Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 4.2.5 Other Equilibrium Constants . . . . . . . . . . . . . . . . . 165 4.2.6 Effect of Pressure and Temperature on Thermodynamic Equilibrium Constant . . . . . . . . . . . . . . . . . . . . . 167 4.2.7 Chemical Equilibrium in Presence of a Solid Phase . . . . . 171 4.2.8 Le Châtelier´s Principle . . . . . . . . . . . . . . . . . . . . 176 4.3 Multiplicity of Chemical Reactions . . . . . . . . . . . . . . . . . . 183 4.3.1 MultiComponent, MultiPhase Systems with Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 4.3.2 Choice of Chemical Reactions . . . . . . . . . . . . . . . . . 188 4.3.3 Exact Number of Chemical Reactions Needed for Equilibrium Determination . . . . . . . . . . . . . . . . . . . . . . 189 4.3.4 Linear Dependence of a Reaction Set . . . . . . . . . . . . . 193 4.3.5 The Phase Rule for a System with Chemical Reactions . . . 196 4.4 Equilibrium Composition . . . . . . . . . . . . . . . . . . . . . . . 200 4.4.1 Systems with a onedimensional reaction basis . . . . . . . . 200 4.4.2 Systems with a twodimensional reaction basis . . . . . . . 209 4.4.3 Systems with a multidimensional reaction basis . . . . . . . 216 4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 5 Elements of Chemical Kinetics 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 5.2 Rate Laws and Reaction Orders . . . . . . . . . . . 5.3 Forward and Reverse Reactions . . . . . . . . . . . 5.4 Elementary Reactions and Reaction Molecularity . 5.5 Rate of Reactions . . . . . . . . . . . . . . . . . . . 5.5.1 Temperature dependence of rate coefficients 5.5.2 Pressure dependence of rate coefficients . . 5.6 Summary . . . . . . . . . . . . . . . . . . . . . . . 6 Mechanisms of Basic Combustion Reactions
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221 . 221 . 222 . 226 . 231 . 237 . 238 . 240 . 241 243
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Contents 6.1 6.2 6.3 6.4 6.5
6.6
Chain Reactions . . . . . . . . . . . . . . Combustion of Carbon Monoxide (CO) . . Combustion of Hydrogen (H2 ) . . . . . . . 6.3.1 Simplified ignition mechanism . . . Combustion of Methane (CH4 ) . . . . . . Methods of Solving Chemical Kinetic Rate 6.5.1 Analytical solutions . . . . . . . . 6.5.2 Numerical Solutions . . . . . . . . Summary . . . . . . . . . . . . . . . . . .
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243 245 246 248 251 253 254 264 279
Gaussian Elimination
283
Vocabulary
287
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List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8
1.9 1.10 1.11 1.12 1.13 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10
Primary Energy Sources in Germany per type of fuel . . . . . . . . Wall Fired Boiler . . . . . . . . . . . . . . . . . . . . . . . . . . . . Modern Reheating Furnace – NKK, Fukuyama Works, Japan . . . Left – pulverised coal flame; Right – MILD combustion of natural gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The subsets of hydrocarbons . . . . . . . . . . . . . . . . . . . . . . Oxygen concentration in dry combustion products as a function of excess air ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Carbon dioxide concentration as a function of excess air ratio . . . Saturation pressure of water vapour as a function of temperature. The plot is obtained using Eq. (1.35) which in the plotted range provides pressure values that are within 2 % accuracy with the values listed in steam tables [7]. . . . . . . . . . . . . . . . . . . . . . Cooling of combustion products (or moist air) at a constant temperature. The T s diagram shows the dewpoint temperature. . . . Effect of sulphur and excess air on acid due point for a crude oil (adapted from [8]). . . . . . . . . . . . . . . . . . . . . . . . . . . . Ash and combustibles in a furnace . . . . . . . . . . . . . . . . . . Burnout of combustibles . . . . . . . . . . . . . . . . . . . . . . . . Composition of dry combustion products for substoichiometric combustion of methane . . . . . . . . . . . . . . . . . . . . . . . . . Control volume for mass and energy balance . . . . . . . . . . . . Mass balance for the boiler . . . . . . . . . . . . . . . . . . . . . Illustration of open, closed and isolated systems . . . . . . . . . . Sankey’s diagram for energy balance for an open or closed system System in translational and rotational motion . . . . . . . . . . . Definition of the sign of work . . . . . . . . . . . . . . . . . . . . Work done to the system . . . . . . . . . . . . . . . . . . . . . . . Specific heats at constant pressure for various molecules . . . . . Physical enthalpies of various gases as a function of temperature Illustration of enthalpy of formation of a compound . . . . . . . .
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2 3 3 4 5 23 23
26 30 30 31 32 34 36 39 44 45 46 51 52 57 58 62
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List of Figures 2.11 Illustration of LCV as an amount of heat extracted from a combustion chamber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12 Illustration of an energy balance . . . . . . . . . . . . . . . . . . 2.13 Example of an energy balance . . . . . . . . . . . . . . . . . . . . 2.14 Energy balance using enthalpy of formation . . . . . . . . . . . . 2.15 Energy balance using LCV . . . . . . . . . . . . . . . . . . . . . . 2.16 Sankey diagram demonstrating the concept of the available heat . 2.17 Fraction of the available heat as a function of the furnace exit temperature and excess air ratio for combustion of pure methane in air. Constant cp values. . . . . . . . . . . . . . . . . . . . . . . 2.18 Fraction of the available heat as a function of the furnace exit temperature and excess air ratio for combustion of pure methane in air. JANAF polynomials for cp have been used. . . . . . . . . . 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19
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66 73 75 77 78 82
. 85
. 86
Examples of irreversible processes in closed and open systems . . . Irreversible process of mixing . . . . . . . . . . . . . . . . . . . . . Typical thermodynamic processes shown using workdiagram . . . Reversible and irreversible gas expansion process . . . . . . . . . . Integration paths from i → f . . . . . . . . . . . . . . . . . . . . . Determination of absolute specific entropy of a pure substance . . . An open system interacting with other bodies by exchanging mass, heat and work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Heat source providing heat to the system . . . . . . . . . . . . . . Mass source providing mass to the system . . . . . . . . . . . . . . A single component system consisting of two phases maintained at constant temperature and pressure . . . . . . . . . . . . . . . . . . Pressuretemperature plot showing the phaseequilibrium curve that defines stability regions for phases 1 and 2 . . . . . . . . . . . . . . The experimentally determined phase diagram for water . . . . . . The variation of the Gibbs enthalpy with temperature for ice, water and water vapour (at 1 bar . . . . . . . . . . . . . . . . . . . . . . . The variation of the Gibbs enthalpy with temperature for ice, water and water vapour at a pressure of 611 Pa . . . . . . . . . . . . . . . Mixing of two ideal gases . . . . . . . . . . . . . . . . . . . . . . . Mixture of two ideal gases A and B . . . . . . . . . . . . . . . . . . Pure speciesA at temperature T and pressure p; SpeciesA in a mixture with species B at Temperature T and pressure p . . . . . . Ideal solution of two liquids in equilibrium with its vapour . . . . . Ideal solution of two liquids A and B; different vapour pressures . .
90 91 94 95 97 106 108 109 110 116 118 120 125 125 139 140 142 143 145
List of Figures 4.1 4.2
Minimisation of Gibbs enthalpy as the reaction advances . . . . . Thermodynamic equilibrium constant K as a function of temperature for reactions listed in Table 4.2 . . . . . . . . . . . . . . . . 4.3 Equilibrium partial pressure of carbon dioxide in calcination reaction CaCO3 → CaO + CO2 . . . . . . . . . . . . . . . . . . . . . 4.4 Accurate and estimated values of the thermodynamic equilibrium constant for Boudouard reaction . . . . . . . . . . . . . . . . . . 4.5 Variation of the partial pressures of CO2 and CO with temperature for total pressure of 1 bar and 10 bar for Boudouard reaction . . 4.6 Equilibrium composition of Boudouard reaction at several total pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Illustration of reaction (4.128) . . . . . . . . . . . . . . . . . . . . 4.8 Gibbs enthalpy as a function of extent of the reaction . . . . . . . 4.9 Gibbs enthalpy as a function of the amount of carbon dioxide . . 4.10 Gibbs enthalpy of the considered system as a function of mCH4 and mH2 O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 5.2 5.3 5.4 5.5 5.6 6.1 6.2 6.3 6.4
6.5
6.6
. 155 . 169 . 175 . 180 . 182 . . . .
183 203 207 209
. 215
Concentration change with time for a firstorder reaction (t0 = 0) . Concentration change with time for a secondorder reaction . . . . The approach of concentrations to their equilibrium values for the −− ⇀ reversible reaction A ↽ − − B that is first order in each direction . . Activation energy of a chemical reaction . . . . . . . . . . . . . . . Arrhenius plot for elementary reactions of halogens with molecular hydrogen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Falloff curves for the reaction C2 H6 −−→ CH3 + CH3 . . . . . . . .
223 225 229 239 240 241
Top – rate constant of CO + OH −−→ CO2 + H; Bottom – rate constant of H + CO2 −−→ CO + OH [19]. . . . . . . . 246 The time behaviour of the chain carriers . . . . . . . . . . . . . . . 251 A simplified scheme of methane oxidation [19] . . . . . . . . . . . . 253 Temporal behaviour of the species concentrations in reactions A1 −→ A2 −→ A3 for k12 = 1 s−1 and k23 = 10 s−1 (reactive intermediate A2 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 Temporal behaviour of the species concentrations in reactions A1 −→ A2 −→ A3 for k12 = 10 s−1 and k23 = 1 s−1 (low reactive intermediate A2 ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 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
6.8 6.9 6.10 6.11 6.12 6.13 6.14
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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) . . . . . . . . . . . . . 260 Euler’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 Numerical solutions using the explicit scheme with different time steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 Numerical solutions using the implicit scheme with different time steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 assuming quasi steady state for A2 species . . . . . . . . 271 Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 , k12 = 1 s−1 , k23 = 10 s−1 , Euler´s Explicit Method . . . 275 Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 , k12 = 1 s−1 , k23 = 100 s−1 , Euler´s Explicit Method . . 276 Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 , Euler´s Implicit Method . . . . . . . . . . . . . . . . . 277
List of Tables 1.1 1.2
Names of aliphatic hydrocarbons . . . . . . . . . . . . . . . . . . . 5 Composition of combustion products . . . . . . . . . . . . . . . . . 12
2.1 2.2 2.3 2.4 2.5 2.6
The incoming and outcoming streams in kmol/h . . . . . . . . . Composition of the combustion products . . . . . . . . . . . . . . The incoming and outcoming flow rates in kg/h . . . . . . . . . . The coefficients for polynomials for T in the range 300–1000 K . . The coefficients for polynomials for T in the range 1000–5000 K . Standard enthalpies of formation and standard entropies of some compounds (JANAF Thermodynamic Tables) . . . . . . . . . . . . LCV of some selected gaseous fuels . . . . . . . . . . . . . . . . . Adiabatic flame temperature Tad for stoichiometric combustion in air. Combustion products contain CO2 and H2 O only. . . . . . . Calculated adiabatic temperature for stoichiometric combustion of pure CH4 with air. . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 2.8 2.9
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41 41 43 57 58
. 64 . 68 . 80 . 85
3.1
Transition temperatures, standard enthalpies and standard entropies of selected substances . . . . . . . . . . . . . . . . . . . . . . . . . 104
4.1
4.4 4.5
Standard enthalpies of formation, standard entropies and Gibbs formation enthalpies of some selected compounds . . . . . . . . . Thermodynamic equilibrium constant K for some reactions important in combustion . . . . . . . . . . . . . . . . . . . . . . . . . . Thermodynamic equilibrium constant K for some reactions important in combustion . . . . . . . . . . . . . . . . . . . . . . . . . . Standard Gibbs enthalpies at T = 1300 K . . . . . . . . . . . . . Standard Gibbs enthalpies at T = 1000 K . . . . . . . . . . . . .
5.1
Elementary reactions in H2 −CO−C1 −O2 system . . . . . . . . . . 233
6.1 6.2
Rate coefficients of the chain initiation reactions . . . . . . . . . . 247 Rate coefficients of the chain branching reactions. The coefficients Ea b are presented in the form k = A T exp − R T . . . . . . . . . . . . 248
4.2 4.3
. 158 . 162 . 163 . 204 . 213
xv
List of Tables 6.3
6.6 6.7 6.8
Comparison of the rate coefficients k of the chain initiation and chain branching reactions. . . . . . . . . . . . . . . . . . . . . . . Simplified mechanism of ignition of the hydrogenoxygen system . Comparison of the rate coefficients k of the chain propagation reactions for methane oxidation. . . . . . . . . . . . . . . . . . . . . Rate coefficients of the Zeldovich mechanism reactions . . . . . . Numerical solution using the explicit scheme with h = 0.3 s. . . . Numerical solution using the implicit scheme with h = 0.3 s. . . .
6.9
Technical vocabulary . . . . . . . . . . . . . . . . . . . . . . . . . . 287
6.4 6.5
xvi
. 249 . 249 . . . .
252 261 266 269
1 Stoichiometry Contents 1.1 1.2
1.3
1.4
1.5
1.6 1.7 1.8
Introduction Definitions 1.2.1 Chemical Reactions, Atoms and Molecules in Combustion 1.2.2 Amount of Substances, Mole and Mass Fractions 1.2.3 Density and Concentration (Molar density) 1.2.4 Equation of State for Gases and Gas Mixtures Combustion Stoichiometry for Gaseous Fuels 1.3.1 Stoichiometric combustion 1.3.2 Excess air ratio (air equivalence ratio) and fuel equivalence ratio 1.3.3 Minimum air requirement for a mixture of gaseous fuels 1.3.4 Composition of combustion products Combustion stoichiometry for liquid and solid fuels 1.4.1 Minimum oxygen and air requirements and excess air ratio 1.4.2 Combustion products Humid Combustion Air 1.5.1 Absolute and relative humidity 1.5.2 Dew Point Temperature of Combustion Products Combustibles burnout for solid fuels Substoichiometric combustion to carbon dioxide and water vapour Summary
1.1 Introduction Nowadays combustion provides more than 90 % of the energy sources. Despite the continuing search for alternative energy sources combustion will remain important
1
1 Stoichiometry for many decades to come, as shown in Fig. 1.1. For most of high temperature processes like power generation, glass manufacturing, cementmaking and steelmaking, combustion of fossil fuels provides the process energy. Any material that can be burned to release thermal energy is called a fuel. Most fuels consist primarily of hydrogen and carbon and they are called hydrocarbons. Fig. 1.2 shows an interior of a combustion chamber of a power plant producing electricity from a hard coal. A sketch of a modern reheating furnace of Fukuyama steelwork in Japan [2] is shown in Fig. 1.3. The furnace is fired with a mixture of natural gas and steelwork gases. Petrol and diesel oil are burned in engines used for transportation. Propellerdriven airplanes are built with propellers powered by engines identical to automobile engines. Civilian and military aircrafts are powered 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 = 1018 J [1]
Industrial flames are essential elements of the processes and flame properties often affect not only the process efficiency but also product quality. Fig. 1.4 (left) shows a typical pulverisedcoal flame for boiler application while Fig. 1.4 (right) shows MILD combustion technology (known also as flameless oxidation) applied to a reheating furnace. Designing efficient combustion processes in car engines and gas turbines has been an ongoing challenge for combustion engineers.
2
1.1 Introduction
Fig. 1.2: Wall Fired Boiler
Fig. 1.3: Modern Reheating Furnace – NKK, Fukuyama Works, Japan [2]
3
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 colliding molecules, for example: H2 + 21 O2 −−→ H2 O The atoms are conserved (they are not created or destroyed) while molecules are not conserved. In the above reaction H, O atoms are conserved while molecules H2 , O2 and H2 O are not. Reactant molecules (H2 and O2 ) are rearranged to become product (H2 O) molecules. Heat is released in this process. Atoms relevant in combustion are: C, H, O, N, S, Cl. Compounds of carbon and hydrogen are called hydrocarbons. Hydrocarbons are classified into: aliphatic hydrocarbons, alicyclic hydrocarbons and aromatic hydrocarbons. The first ten members of the unbranchedchain alkane series are: CH4 C2 H6 C3 H8 C4 H10 C5 H12
4
methane ethane propane butane pentane
C6 H14 C7 H16 C8 H18 C9 H20 C10 H22
hexane heptane octane nonane decane
1.2 Definitions
Fig. 1.5: The subsets of hydrocarbons
Table 1.1: Names of aliphatic hydrocarbons No. of C atoms
alkane
1 2 3 4 5 n
CH4 methane C2 H6 ethane C3 H8 propane C4 H10 butane C5 H12 pentane Cn H2n+2
alkene
C2 H4 C3 H6 C4 H8 C5 H10 Cn H2n
ethene propene butene pentene
alkyne
C2 H2 C3 H4 C4 H6 C5 H8 Cn H2n−2
alkyl group
ethyne propyne butyne pentyne
CH3 C2 H5 C3 H7 C4 H9 C5 H11 Cn H2n+1
methyl ethyl propyl butyl pentyl
Other molecules relevant in combustion are: Haloalkanes Alcohols Aldehydes Amines Ketones Carbocyclic Acids
R−X R−OH R−CHO R−NH2 R−CO−R R−COOH
example: example: example: example: example: example:
CH3 Cl C2 H5 OH CH3 CHO CH3 NH2 CH3 COCH3 CH3 COOH
(chloromethane) (ethanol) (ethanal) (methylamin) (acetone) (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 constant is NA = 6.023 · 1023 atoms or molecules per mole).
5
1 Stoichiometry For a mixture of species: X n= ni
n = total number of moles
(1.1)
i
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 mole of the species (for example: MC = 12 g/mol, MCO2 = 44 g/mol). The mean molar mass (molecular weight) of a mixture is: X Mmean = xi Mi (1.3) i
Frequently mole fractions xi are converted into mass fractions (wi ) and the following relationships hold: wi =
xi Mi n i Mi number of kg of species "i" =P =P n k Mk xk Mk total number of kg in the system k
k
ni number of moles of species "i" in 1 kg of mixture = xi = n total number of moles in 1 kg of mixture wi wi wi Mi Mi P = Mmean = = wk 1 Mi Mk M mean
(1.4)
(1.5)
k
where the summation extends over all the species and wi stands for mass fraction of species i.
1.2.3 Density and Concentration (Molar density) Variables that do not depend on the size (extent) of the system are called intensive variables (for example density, molar density). These are defined as the ratio of the corresponding extensive properties and the system volume V ; density molar density (concentration)
6
ρ= c=
m V n V
(in kg/m3 ) (in kmol/m3 )
1.2 Definitions Thus,
m ρ = = Mmean (1.6) c n Note that the molar density c defined above is in chemistry usually denoted in square brackets [ ] for example cH2 = [H2 ] and chemists prefer to express concentrations 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 and volume: F (p, T, V ) = 0 (1.7) There are several equations of state for gases and gas mixtures [3, 4, 5]. The perfect gas equation of state, called also ClausiusClapeyron equation, is perhaps one of the simplest ones and it reads (1.8)
pV = nRT or c= and ρ=
p RT
p Mmean p P = RT RT i
(1.9)
wi Mi
(1.10)
where p is the pressure (in Pa = N/m2 ), V the volume (in m3 ), n the molar number (in kmol), T the absolute temperature (in K), and R is the universal gas constant R = 8314.3 J/(kmol · K). An ideal gas is an imaginary substance that obeys relation (1.8). It has been experimentally observed that relation (1.8) approximates closely behaviour of real gases at low densities which means at low pressures and at high temperatures. What are a low pressure and a high temperature? The pressure or temperature of a substance is high or low relative to its critical pressure (pcr ) or critical temperature (Tcr ). The useful rules of thumbs are: (a) at very low pressures ( ppcr ≪ 1), gases behave as an ideal gas regardless of temperature; (b) at high temperatures ( TTcr > 2) gases behave also as an ideal gas; (c) deviation from an ideal gas behavior increases upon approaching the critical point.
7
1 Stoichiometry From Eq. (1.8) follows that one kmol of any (ideal) gas under constant temperature and pressure occupies the same volume. One can easily verify that under so called normal conditions: a pressure of p0 = 1 bar (105 N/m2 )and a temperature of T0 = 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 m3n 1 mol of gas = 24.79 dm3n Throughput this lecture course the standard (normal) conditions of p0 = 1 bar (105 N/m2 ) and T0 = 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 other completely forming only carbon dioxide (CO2 ) and water (H2 O). If there is an excess of fuel, the system is fuelrich, and if there is an excess of oxygen, it is called fuellean. Examples are: stoichiometric
CH4 + 2 O2 −−→ 2 H2 O + CO2 CH4 + 3 O2 −−→ 2 H2 O + CO2 + O2 CH4 + O2 −−→ H2 O +
1 2 CO2
+
1 2 CH4
lean (excess of oxygen) rich (excess of fuel)
If the reaction describing complete combustion (products are CO2 and H2 O) is written in such a way that it describes the combustion of 1 kmol of fuel: 1 kmol fuel + ν O2 −−→ products (CO2 + H2 O) one may easily calculate the mole fraction of fuel in a stoichiometric mixture (with 1
8
In older books p0 = 760 Tr (1 Tr = 1 mmHg = 133.322 N/m2 ) and T0 = 0 ◦ C (273.15 K) are used so that 1 kmol of ideal gas occupies a volume of 22.41 m3 .
1.3 Combustion Stoichiometry for Gaseous Fuels oxygen) as follows: xf uel,stoich in oxygen =
1 number of moles of fuel = total number of moles (fuel + oxygen) 1+ν
(1.11)
For example: 2 1 = 1.5 3 1 xCH4 ,stoich in oxygen = 3 2 1 = xCO,stoich in oxygen = 1.5 3
H2 + 21 O2 −−→ H2 O
xH2 ,stoich in oxygen =
CH4 + 2 O2 −−→ CO2 + 2 H2 O CO + 12 O2 −−→ CO2
If dry air is used as an oxidiser it contains only 21 % O2 , 78 % N2 , and 1 % of noble gases. Thus, xf uel,stoich in air =
1 1 ν = 1 + 0.21 1 + 4.762 ν
(1.12)
An example: Combustion of propane in oxygen: C3 H8 + 5 O2 −−→ 3 CO2 + 4 H2 O
and
xC3 H8 ,stoich in oxygen =
1 6
Combustion of propane in air: C3 H8 + 5 ( O2 + 3.762 N2 ) −−→ 3 CO2 + 4 H2 O + 5 · 3.762 N2 and
xC3 H8 ,stoich in air =
1 = 0.0403 1 + 5 · 4.762
Note: The higher the hydrocarbon the lower is the fuel mole fraction at stoichiometric conditions.
1.3.2 Excess air ratio (air equivalence ratio) and fuel equivalence ratio The excess air ratio is defined as: λ=
(wair /wf uel ) (xair /xf uel ) = (xair /xf uel )stoich (wair /wf uel )stoich
(1.13)
9
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: 1 ; 1 + 4.762 · ν · λ xair ; = 0.21 xair = 4.762
xf uel =
xair = 1 − xf uel ;
(1.14)
xO2
xN2 = 3.762 xO2
(1.15)
The combustion process can be divided into: Rich combustion
Φ>1
λ 1): h moisture s n c + 1· + + 1· + + 12 2 18 32 28 1 · 0.79 · lair,min + (λ − 1) · lO2 ,min + (λ − 1) · 0.79 · lair,min c h moisture s n = + + + + + (λ − 1) · lO2 ,min + λ · 0.79 · lair,min 12 2 18 32 28
Vwet = 1 ·
Vwet
H2 O N2 CO2 N2 moisture SO2 excess from s from n oxygen in air The from c from h in fuel above expressions for calculating Vwet can be rearranged into a more elegant form as follows: c h moisture s n + 1· + + 1· + + 0.79 · lair,min + (λ − 1) · lO2 ,min + 12 2 18 32 28 0.79 · (λ − 1) · lair,min c h moisture s n = + + + + + 0.79 · lair,min + 0.21 · (λ − 1) · lair,min + 12 2 18 32 28 0.79 · (λ − 1) · lair,min h moisture s n c + + + + + 0.79 · lair,min + (λ − 1) · lair,min = 12 2 18 32 28 = Vwet,min + (λ − 1) · lair,min
Vwet = 1 ·
Thus, Vwet = Vwet,min + (λ − 1) · lair,min and
c h moisture s n + + + + + 0.79 · lair,min 12 2 18 32 28 is the minimum amount of products (for λ = 1).
Vwet,min = where Vwet,min
(1.28a) (1.28b)
The amount of dry combustion products can be easily calculated by dropping the
18
1.4 Combustion stoichiometry for liquid and solid fuels terms
h 2
moisture 18
and
Vdry =
from the equations above. Thus,
s n c + + + 0.79 · lair,min + (λ − 1) · lO2 ,min + (λ − 1) · 0.79 · lair,min 12 32 28 kmol dry products/kg of fuel (1.29a)
or Vdry = Vdry,min + (λ − 1) · lair,min
kmol dry products/kg of fuel (1.29b)
and Vdry,min =
c s n + + + 0.79 · lair,min 12 32 28 kmol dry products/kg of fuel (1.30)
Composition of the combustion products The composition of the combustion products can be easily calculated realisingthat wet products contain H2 O, CO2 , SO2 , N2 and excess O2 . Thus: xH2 O =
h 2
+
Vwet
xCO2 ,wet =
c 12
Vwet
xSO2 ,wet = xO2 ,wet xN2 ,wet
moisture 18
s 32
Vwet (λ − 1) · lO2 ,min = Vwet n + 0.79 · λ · lair,min = 28 Vwet
kmol H2 O/kmol wet products kmol CO2 /kmol wet products kmol SO2 /kmol wet products kmol O2 /kmol wet products in kmol N2 /kmol wet products
Similarly one can derive simple relationships for calculating composition of dry combustion products (CO2 , SO2 , N2 and excess O2 ): xCO2 ,dry = xO2 ,dry
c 12
Vdry (λ − 1) · lO2 ,min = Vdry
xSO2 ,dry = xN2 ,dry
s 32
Vdry n + 0.79 · λ · lair,min = 28 Vdry
where Vdry is the amount of dry combustion products in kmol/kg of fuel.
19
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 function of excess air ratio. Coal Fettnuss mvb3 (origin – Germany) – coal analysis obtained from a chemical laboratory Basis
Dry ash free Dry As fired
H2 O %wt
Ash %wt c 89.39
Ultimate Analysis %wt h n o 4.68 1.53 3.66
LCV MJ/kg s 0.74
4.4
32.87
3.5
Assumptions: the fuel is combusted to carbon dioxide and water. One begins with calculating the coal composition “as fired” (with moisture and ash). Simple conversions result in the following: Basis
Dry ash free Dry As fired
H2 O %wt
3.5
Ash %wt
4.4 4.246
c 89.39 85.4568 82.4658
Ultimate Analysis %wt h n o 4.68 1.53 3.66 4.47 1.4627 3.499 4.3136 1.4115 3.3765
LCV MJ/kg s 0.74 0.7074 0.6826
34.38 32.87 31.72
The minimum oxygen requirement (λ = 1) is: lO2 ,min = 1 ·
0.006 826 0.033 765 0.824 658 1 0.043 136 + · + 1· − = 12 2 2 32 32 kmol O2 0.0787 kg fuel “as fired”
that corresponds to 0.0787 · 32 = 2.5184 kg 3
mvb  medium volatile bituminous
20
kg O2 of fuel “as fired” .
1.4 Combustion stoichiometry for liquid and solid fuels The minimum air requirement (λ = 1) is: 0.0787 kmol dry air lair,min = = 0.3743 0.21 kg of fuel “as fired” 2.5184 kg of dry air m3n dry air , or = 10.9496 0.3743 · 22.79 = 9.279 kg of fuel 0.23 kg of fuel The amount of combustion products is: Vwet =
0.824 658 0.043 136 0.035 0.006 826 0.014 115 + + + + 12 2 18 32 28 + λ · 0.79 · 0.3743 + (λ − 1) · 0.0787
Vwet = 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) in kmol wet products/kg of fuel “as fired” and Vdry =
0.824 658 0.006 826 0.014 115 + + + λ · 0.79 · 0.3743 + (λ − 1) · 0.0787 12 32 28 = 0.0695 + 0.2957 · λ + (λ − 1) · 0.0787 in kmol dry products/kg of fuel “as fired”
Composition of wet combustion products:
xH2 O
=
xCO2 ,wet =
xN2 ,wet
+
Vwet 0.824 658 12 Vwet 0.006 83 32
0.035 18
=
0.0235 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1)
=
0.0687 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1)
0.0002 Vwet 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) (λ − 1) · lO2 ,min (λ − 1) · 0.0787 = = Vwet 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) 0.79 · λ · lair,min 0.79 · λ · 0.3743 = = Vwet 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1)
xSO2 ,wet = xO2 ,wet
0.043 136 2
=
All the above mole (volume) fractions are in kmol/kg of wet products.
21
1 Stoichiometry Composition of dry combustion products: xCO2 ,dry = xSO2 ,dry = xO2 ,dry xN2 ,dry
0.824 658 12
Vdry 0.006 83 32
=
0.0687 0.0695 + 0.2957 · λ + 0.0787 · (λ − 1)
=
0.0002 0.0695 + 0.2957 · λ + 0.0787 · (λ − 1)
Vdry (λ − 1) · lO2 ,min (λ − 1) · 0.0787 = = Vdry 0.0695 + 0.2957 · λ + 0.0787 · (λ − 1) 0.79 · λ · lair,min 0.79 · λ · 0.3743 = = Vdry 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: xH2 O = 0.0552
xCO2 ,wet = 0.1613
xO2 ,wet = 0.0185
xN2 ,wet = 0.7635
xSO2 ,wet = 0.0005
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 excess air 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 and Fig. 1.7 show the dependence.
22
1.4 Combustion stoichiometry for liquid and solid fuels
Fig. 1.6: Oxygen concentration in dry combustion products as a function of excess air ratio
Fig. 1.7: Carbon dioxide concentration as a function of excess air ratio
23
1 Stoichiometry Comments: (a) For the fuels considered in this example (CH4 , C2 H6 , Groningen natural gas, coal Fettnuss), the relationship between oxygen content in combustion products and excess air ratio (see Fig. 1.6) is almost linear for λ < 1.5. This relationship holds for most of fuels and is used by furnace operators in every day practice. For example a 2 % (dry) oxygen content in the flue gas indicates that the furnace is operated at 1.1 excess air ratio. (b) Carbon dioxide concentration in combustion products reaches maximum at λ = 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 characteristics of the fuel only. End of Example 1.3
1.5 Humid Combustion Air In Section 1.3.3 we have derived Eq. (1.18) for calculating the minimum dry air requirement for combustion of one kmol of a gaseousfuel. A similar expression, Eq. (1.24), has been derived to calculate the minimum dry air requirement for complete combustion of one kilogram of a solid or a liquid fuel. We stress here again 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 contains moisture (water vapour). Accurate calculations of combustion stoichiometry should account for the presence of water vapour. In this paragraph we show how to do it. We recall that air that contains no water vapour is called dry air. Atmospheric or combustion air containing water vapour is named here as combustion air. The temperature of air in combustion applications ranges from about −20 to about 1300 ◦ C. In most of the furnace applications combustion air is supplied at pressures that are slightly higher than ambient air pressure of 1 bar. However, in gas turbines and engines the combustion air is typically compressed to 20–30 bar pressure.
1.5.1 Absolute and relative humidity It is convenient to treat combustion air as a mixture of water vapour and dry air since the composition of dry air remains constant but the amount of water vapour changes. It is certainly convenient to treat the water vapour in combustion air as an ideal gas and for ambient air pressures such an assumption is perfectly valid.
24
1.5 Humid Combustion Air 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 in the range 0.2–6 %. Then the combustion air is treated as an idealgas mixture whose pressure is the sum of the partial pressure of the dry air (pdry air ) and that of the water vapour (pv ) p = pdry air + pv (1.31) The partial pressure of water vapour (pv ) is typically referred to as the vapour pressure. The temperature, however, is uniform throughout the dryair/watervapour mixture so that T = Tdry air = Tv (1.32) Usually the total pressure (p) is known whereas the partial pressure of water vapour (pv ) depends on how much moisture is present in the mixture. For an idealgas 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 various ways. However, the simplest way is to specify the mass of water vapour present in a unit of dry air. This is called absolute or specific humidity and in this lecture series is denoted as ϕ and is expressed in kg of water vapour per kg of dry air since ϕ=
mv Mv · n v = mdry air Mdry air · ndry
= air
18.016 pv pv · = 0.622 · 28.97 p − pv p − pv
(1.34)
where Mv and Mdry air are the molar masses of water and dry air, respectively whilst 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. Now let 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 air can hold no more moisture. At this point the air is said to be saturated with moisture vapour and it is called saturated air. The amount of water vapour in saturated air at a given temperature and total pressure can be determined using Eq. (1.34) by replacing pv by the saturation pressure psat of water at the given temperature. The saturation pressure of water vapour is plotted in Fig. 1.8 as a function of temperature using a relationship4 4
In Chapter 3, Example 3.3, we will derive Eq. (1.35) using ClausiusClapeyron equation.
25
1 Stoichiometry
psat
1 1 = 611 · exp −5304.3 · − T 273.16
in Pa
(1.35)
Water Vapour
psat in kPa
10
5
0 280
290
300
310
320
Temperature in K Fig. 1.8: Saturation pressure of water vapour as a function of temperature. The plot is obtained using Eq. (1.35) which in the plotted range provides pressure values that 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 is called the relative humidity γ pv γ= (1.36) pv,sat where pv,sat stands for the saturation water vapour pressure at the specified temperature. The relative humidity (γ) ranges from 0 for dry air to 1 for saturated air. The amount of moisture that combustion air can hold depends on its temperature. Thus, the relative humidity of air (γ) changes with temperature even when its specific humidity (ϕ) remains constant. Using the relative humidity and the saturation pressure of water vapour, the specific humidity (ϕ) can then be calculated as follows: γ · psat ϕ = 0.662 · (1.37) p − γ · psat
26
1.5 Humid Combustion Air 1.5.1.1 Wet air requirement In the previous paragraphs we have developed simple formulae for calculating the dry air requirements (ldry air ); Eq. (1.17) for gaseous fuels while Eq. (1.23) for liquid and solid fuels. The above considerations on the combustion air humidity allows 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 easily done since hwet air = hdry air + ϕ · hvapour (1.39) where h stands for specific enthalpy in J/g (or kJ/kg). The enthalpy of water vapour at ambient air pressure in the temperature range −10 to 50 ◦ C can be determined 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 be used. Example 1.4 In 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. The objective of this example is to include the moisture and by doing so to examine its effect on the results of the calculations. We assume here that the combustion air is supplied at 1 bar pressure and at a 20 ◦ C temperature. Its relative humidity is 75 %. Assumptions: the fuel is combusted to carbon dioxide and water. One begins with calculating the saturation pressure of water vapour at 20 ◦ C using Eq. (1.35) 1 1 psat = 611 · exp −5304.3 · − = 2298.14 Pa (1.41) 297.16 273.16 The absolute humidity is then ϕ = 0.662 ·
0.75 · 2298.14 γ · psat kg water vapour = 0.662 · 5 = 0.0116 p − γ · psat kg dry air 10 − 0.75 · 2298.14 (1.42)
27
1 Stoichiometry so the minimum amount of wet combustion air is = 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) combustion products so (see Example 1.3)
lwet
air,min
= (1+ϕ) · ldry
air,min
Vwet = 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) +
λ · ϕ · ldry 18
air,min
(1.44)
and kmol wet products kg of fuel “as fired” (1.45) Composition of wet combustion products is therefore as follows: Vwet = 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06
xH2 O =
0.043 136 2
+
0.035 18
+ λ · 0.007 06
Vwet 0.0235 + λ · 0.007 06 = 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06
xCO2 ,wet = = xSO2 ,wet = = xO2 ,wet = = xN2 ,wet = =
0.824 658 12
Vwet 0.0687 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06 0.006 882 6 2
Vwet 0.0002 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06 (λ − 1) · lO2 ,min Vwet (λ − 1) · 0.0787 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06 0.79 · λ · ldry air,min Vwet 0.79 · λ · 0.3743 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06
All the above mole (volume) fractions are in kmol/kmol of wet products.
28
1.5 Humid Combustion Air At 10 % excess air ratio (λ = 1.1) the above formulae provide: xH2 O = 0.0721
xCO2 ,wet = 0.1584
xO2 ,wet = 0.018 14
xN2 ,wet = 0.7498
xSO2 ,wet = 0.000 46
Comments: (a) Taking into account the combustion air moisture has resulted in a 1.2 % correction to the minimum air requirement. (b) However, the mole fraction of water vapour in (wet) combustion products has increased by around 31 %. (c) Obviously neither the amount of the dry combustion products nor its composition 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 natural gas is combusted the water vapour content in the combustion products may be as high as 16 % for λ = 1.2 as shown in Example 1.2. When coal Fettnuss is combusted the water content of around 7 % (see Example 1.4) is expected. While designing combustion systems it is required that combustion products are cooled down to low temperatures before they are released to the atmosphere. Thus, by cooling down the combustion products we may expect that below a certain temperature the water vapour begins to condensate. The dewpoint temperature (Tdp ) is defined as the temperature at which condensation begins when the combustion products (or generally moist air) is cooled at a constant pressure. In other words, Tdp is the saturation temperature of water corresponding to the vapour pressure, as shown in Fig. 1.9. As a matter of fact Eq. (1.35) can be used to determine the duepoint temperature if the pressure is given. Let us assume that there is 7 vol% water vapour content in combustion products of coal Fettnuss combustion. If the combustion products are at 1 bar pressure, the partial pressure of vapour is 7000 N/m2 and using Eq. (1.35) we can estimate that the dewpoint temperature is around 312.4 K (39.3 ◦ C). Thus, in order to avoid condensation it is desired to keep the combustion products at temperatures typically 40–60 ◦ C higher than the due point temperature. Dew points vary with the amounts of O2 , CO2 , SO2 , NOx ,HCl in combustion products. In particular sulphur oxides have a pronounced effect on the dew point temperature as shown in Fig. 1.9 for several excess air ratios.
29
1 Stoichiometry
pv = con s
t.
T
T1
1
Tdp
2
s Fig. 1.9: Cooling of combustion products (or moist air) at a constant temperature. The T s diagram shows the dewpoint temperature.
The excess air curves are not equally spaced since the extra oxygen tends to produce more SO3 which exerts a catalytic effect in raising the dew point. For example the above estimated dewpoint temperature of 39.3 ◦ C would be almost doubled if 2 ppm of SO2 was present. This is the reason that operators of coalfired power station boilers maintain the flue gases at temperatures above 180–200 ◦ C.
Acid dew point, °C
145
20%
140
r ss ai exce 15% 10%
135 130
5% 120 115
1.0
2.0
3.0
4.0
5.0
Weight % sulfur in fuel oil Fig. 1.10: Effect of sulphur and excess air on acid due point for a crude oil (adapted from [8]).
30
1.6 Combustibles burnout for solid fuels
1.6 Combustibles burnout for solid fuels It is not possible to burn 100 % of solid fuels. When firing liquid and solid fuels the major contribution to the “unburns” comes usually from the residual “oilcoke” 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 ash content as shown in Fig. 1.11. Total Combustibles = 1 − ash
Fig. 1.11: Ash and combustibles in a furnace
Composition of the fuel: C0 (Total Combustibles) + ash0 = 1 m0 – mass flow rate of fuel
Composition of the unburned solids: C1 + ash1 = 1 m1 – mass flow rate of unburned solids
Mass balance of combustibles can be written as: m0 · C0 = m1 · C1 + b · m0 · C0 where m0 C0 (in kg/s) stands for the amount of combustibles entering the furnace, m1 C1 (in kg/s) represents the amount of combustibles in the unburned solids leaving the furnace while bm0 C0 (in kg/s) stands for the amount of the combustibles burned whilst b is the fraction of the original combustibles burned. Assuming that ash is an inert and does not enter into any chemical reactions, its mass balance can 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: 0 1 − ash C1 · ash0 (1 − ash1 ) · ash0 ash1 b=1− =1− = C0 · ash1 (1 − ash0 ) · ash1 1 − ash0
(1.46)
31
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 relationship between the combustibles burnout (b) and carbon in ash for several ash contents of the solid fuel.
Fig. 1.12: Burnout of combustibles
1.7 Substoichiometric combustion to carbon dioxide and water vapour Substoichiometric combustion occurs when excess air ratio (λ) is smaller than 1. Obviously, in substoichiometric combustion not all the fuel is combusted since there is not enough oxidiserpresent and some of the fuel remains unburned. If the fuel is combusted to carbon dioxide and water vapour, the combustion calculations leading to the determination of the composition of the combustion products are rather straight forward as shown in Example 1.5. However if some species like, for example, carbon monoxide and hydrogen are to be considered the composition of the combustion products has to be determined using chemical equilibrium calculations underlined in Chapter 4.
32
1.7 Substoichiometric combustion to carbon dioxide and water vapour Example 1.5 Produce a curve showing the composition of combustion products obtained in substoichiometric 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 H2 O The excess air ratio is then λ=
nair 2 0.21
so that
nair = 9.5238 · λ
where nair is the number of moles of air per 1 mole of methane. Thus, when λ < 1 only λ moles of CH4 are combusted to CO2 and H2 O while the remaining 1 − λ moles of CH4 appear directly in combustion products as unburned fuel. Therefore, in the combustion products appear (1 − λ) λ 2λ 0.79 · 9.5238 · λ
moles moles moles moles
of of of of
(unburned) CH4 CO2 H2 O N2
The composition of combustion products is then 1−λ 1 + 9.5238 · λ λ = 1 + 9.5238 · λ 7.5238 · λ = 1 + 9.5238 · λ 2·λ = 1 + 9.5238 · λ
1−λ 1 + 7.5238 · λ λ = 1 + 7.5238 · λ 7.5238 · λ = 1 + 7.5238 · λ
xCH4 ,wet =
xCH4 ,dry =
xCO2 ,wet
xCO2 ,dry
xN2 ,wet xH2 O,wet
xN2 ,dry
Comments: (a) The maximum value of CO2 volume fraction is obtained for λ = 1 as shown in Fig. 1.13. Compare with Fig. 1.7.
33
1 Stoichiometry
Fig. 1.13: Composition of dry combustion products for substoichiometric combustion of methane
End of Example 1.5
1.8 Summary Nowadays more than 90 % of the energy used by human beings is generated by combustion of fossil fuels. Despite the continuing search for alternative energy sources combustion will remain important for many decades to come. In the first lecture the basic concepts of chemical reactions, atoms and molecules have been recalled. The lecture introduces stochiometric, lean and rich combustion as well as the notions of excess air ratio and equivalence ratio. In this lecture we have assumed that fuels are combusted to carbon dioxide and water vapour. Students should know how to calculate the amount of air (oxidiser) needed for complete combustion of a given fuel. Furthermore you should be able to determine composition of wet and dry combustion products for any fuel combusted to carbon dioxide and water vapour at a given excess air. Numerous equations on combustion stochiometry have been derived in this chapter. However, the reader should realise that there is no point in memorising them since they can be easily recreated.
34
2 Mass and Energy Balance Contents 2.1
2.2
2.3
2.4 2.5 2.6
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 SteadyState Conditions 2.1.4 Example of a Mass Balance of a Furnace 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) 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 Enthalpies and Formation Enthalpies 2.3.5 Dependence of LCV on Temperature 2.3.6 Example of an Energy Balance of a Furnace Temperature of Adiabatic Combustion Furnace Exit Temperature Summary
2.1 General Formulation of Mass and Energy Balance Conservation of mass and energy is the basis for any considerations in combustion engineering. The formulation presented in this lecture is based on the work of
35
2 Mass and Energy Balance [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 a control 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 instant and 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 instant of time t, one option involves formulating the law on a rate basis. That is, at any instant there must be a balance between all mass rates expressed in kg/s and simultaneously, there must be a balance between all energy rates expressed in J/s (=W). The mass balance at an instant reads, Fig. 2.1: m ˙ in + m ˙ g −m ˙ out = m ˙ stored ≡
d(m) dt
(2.1)
where m ˙ in , m ˙ g, m ˙ out , m ˙ stored are the rates with which the mass enters the control volume, 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 matter within the volume, respectively. All these terms are expressed in kg/s. Symbol m 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 the mass generation term (m ˙ g ) that is nonzero only if nuclear reactions are considered
36
2.1 General Formulation of Mass and Energy Balance during which matter can be converted into energy. In any other systems with nonnuclear reactions present, the mass generation term is zero and therefore during this combustion course we will use the following equation for the mass balance at an instance: dm (2.2) m ˙ in − m ˙ out = dt The energy balance at an instant reads: dE E˙ in + E˙ g − E˙ out = E˙ stored ≡ dt
(2.3)
where E˙ in , E˙ g , E˙ out , E˙ stored stand for the rates with which energy enters the control 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 control volume, respectively. All these terms are expressed in J/s (W). The symbol E 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 mass balance is formulated in amount of mass (kilograms) whilst the energy balance is formulated in amount of energy (joules). The mass balance over a time interval ∆t = t2 − t1 reads: Zt2
m ˙ in · dt −
t1
Zt2
m ˙ out · dt =
t1
Zt2
d(m) · dt = ∆m dt
(2.4)
t1
or (2.5)
min − mout = ∆m
The terms on the left hand side of Eqs. (2.4) and (2.5) denote the amount of mass entering the control volume over the time period ∆t, and the amount of mass that leaves 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) within the system in this time period. The energy balance over a time interval reads: Zt2
t1
E˙ in · dt +
Zt2
t1
E˙ g · dt −
Zt2
t1
E˙ out · dt =
Zt2
d(E) · dt = ∆E dt
(2.6)
t1
37
2 Mass and Energy Balance or Ein + Eg − Eout = ∆E
(2.7)
where the terms on the left hand side of Eqs. (2.6) and (2.7) stand for the amount of energy entering the control volume, the amount of energy generated in the volume, 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 SteadyState Conditions Under steadystate conditions when there is neither mass nor energy accumulation (storage) in the system we obtain: from Eq. (2.2) for mass balance: m ˙ in = m ˙ out
(2.8)
E˙ in + E˙ g = E˙ out
(2.9)
from Eq. (2.3) for energy balance:
2.1.4 Example of a Mass Balance of a Furnace The purpose of making a mass balance under steady state conditions is to calculate the outcoming amount of mass and to calculate its make up (composition). A correct mass balance is a prerequisite to a subsequent energy balance. There are two basic ways of formulating a mass balance; by expressing the incoming 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 composition in mass fractions. The first way is popular among chemical engineers since it deals with kmol and the rates of chemical reactions are typically expressed using concentrations (kmol/m3 /s). The latter way is common in combustion engineering. Both methods have advantages and disadvantages and they are equivalent. They both lead to the same results. The combustion stoichiometry calculations (see Chapter 1) are required to calculate composition of combustion products leaving the system.
38
2.1 General Formulation of Mass and Energy Balance Example 2.1 A boiler fired with methane is used to generate hot water. The burner of the boiler is fed with 200 kg/h of methane and 4000 kg/h of dry air and it is operated under steady state conditions. Make the mass balance for the boiler assuming that the combustion is complete. Carry out the calculations using the molar flow rates (kmol/h). Repeat the calculations using mass flow rates (kg/h).
Combustion products
200 kg/h CH4
Boiler
4000 kg/h air control volume boundary Fig. 2.2: Mass balance for the boiler
Assumptions: (a) the methane combustion proceeds to completion resulting in CO2 and H2 O, (b) we place the control volume over the boiler (see Fig. 2.2), (c) the boiler operates under steady state and therefore we formulate the mass balance at an instance with the mass accumulation term equal to zero. Mass balance in kmol/h The starting point to any mass balance is an estimation of the incoming (input) streams and writing down proper chemical reactions. Thus, the incoming streams are: m ˙ CH4 = 12.5 kmolCH4 /h and since the molar mass of air is Mair = 0.21 · 32 + 0.79 · 28 = 28.84 kg/kmol we obtain for the incoming stream of 4000kg/h air m ˙ air = 28.84kg/kmol = 138.696(kmol air )/h. The oxidation reaction is: CH4 + 2 O2 −−→ CO2 + 2 H2 O and therefore the minimum air requirement is excess air ratio is λ=
2 0.21
kmol = 9.5238 kmol
138.696 kmol of air/h 12.5 kmol of CH4 /h
9.5238 kmol of air/kmol of CH4
of air of fuel .
The
= 1.165
Since λ > 1 the combustion products contain CO2 , H2 O, N2 and O2 . The amount of the wet combustion products is 12.5 · (3 + 0.79 · 9.5238 + (1.165 − 1) · 9.5238) = 151.20kmol/h while for the dry combustion products the figure of
39
2 Mass and Energy Balance 12.5 · (1 + 0.79 · 9.5238 + (1.165 − 1) · 9.5238) = 126.20kmol/h is applicable. The composition of the combustion products is as follows: xCO2 ,wet = xH2 O = xN2 ,wet = xO2 ,wet =
12.5 151.20 2 · 12.5 151.20 0.79 · 9.5238 + 0.79 · (1.165 − 1) · 9.5238 109.5654 12.5 · = 151.20 151.20 (1.165 − 1) · 9.5238 · 0.21 4.1250 12.5 · = 151.20 151.20
= 0.0827 = 0.1653 = 0.7246 = 0.0273
and the molar mass of the wet combustion products is: kg Mwet,products = 0.0827 · 44 + 0.1653 · 18 + 0.7246 · 28 + 0.0273 · 32 = 27.7766 kmol
The composition of the dry combustion products is as follows: 12.5 = 0.099 126.20 109.5654 = = 0.8682 126.20 4.1250 = 0.0327 = 126.20
xCO2 ,dry = xN2 ,dry xO2 ,dry
and the molar mass of the dry products is: Mdry,products = 0.099 · 44 + 0.8682 · 28 + 0.0327 · 32 = 29.712kg/kmol Knowing the molar fractions of the species, one may easily calculate the mass fractions since: 0.0827 · 44 = 0.1310 27.7766 0.1653 · 18 = 0.1071 w H2 O = 27.7766 0.7246 · 28 wN2 ,wet = = 0.7304 27.7766
wCO2 ,wet =
40
wCO2,dry =
wN 2,dry =
0.099 · 44 = 0.1466 29.712
0.8682 · 28 = 0.8182 29.712
2.1 General Formulation of Mass and Energy Balance and wO2 ,wet =
0.0273 · 32 = 0.0315 27.7766
wO2,dry =
0.0327 · 32 = 0.0352 29.712
The tables below summarise the mass balance: Table 2.1: The incoming and outcoming streams in kmol/h
IN Species
kmol/h
Methane Nitrogen Oxygen
12.50 109.57 29.13
P
151.20
OUT WET kmol/h Carbon dioxide 12.50 Water vapour 25.00 Nitrogen 109.57 Oxygen 4.13 P 151.20 Species
DRY kmol/h 12.50 − 109.57 4.13 126.20
Table 2.2: Composition of the combustion products
Species Carbon dioxide Water vapour Nitrogen Oxygen P
Molar fraction Wet Dry 0.0827 0.0990 0.1653 − 0.7246 0.8682 0.0273 0.0327 0.9999 0.9999
Mass fraction Wet Dry 0.1310 0.1466 0.1071 − 0.7304 0.8182 0.0315 0.0352 1.0000 1.0000
Mass balance in kg/h
The incoming streams are: m ˙ CH4 = 200 kg/h and m ˙ air = 4000 kg/h. The outcoming flow rate (in kg/h) is easily obtained since: m ˙ products = 200 kg/h + 4000 kg/h = 4200 kg/h. While making the mass balance in kilograms it is convenient to write the oxidation reaction CH4 + 2 O2 −−→ CO2 + 2 H2 O
41
2 Mass and Energy Balance using kilogram rather than kmol: 16 kg CH4 + 64 kg O2 = 44 kg CO2 + 36 kg H2 O The minimum air requirement is 64/16/0.233 = 17.1674 kg of air/kg of methane and the excess air ratio is: λ=
4000 kg of air/h 200 kg of CH4 /h
17.1674 kg of air/kg of CH4
= 1.165
Since λ > 1 the combustion products will contain CO2 , H2 O, N2 and O2 . The combustion products stream contains: Methane Carbon dioxide Water vapour Nitrogen Oxygen TOTAL (wet)
none 550 kg/h 450 kg/h 3068 kg/h 132 kg/h 4200 kg/h
200 · 44 16 = 200 · 36 16 = 0.767 · 4000 = 0.233 · (4000 − 200 · 17.1674) =
and the amount of combustion products dry is then 3750 kg/h. The species mass fractions can be easily calculated: 550 = 0.131 4200 450 w H2 O = = 0.1071 4200 3068 = 0.7305 wN2 ,wet = 4200 132 wO2 ,wet = = 0.0314 4200 wCO2 ,wet =
wCO2 ,dry = 0.1467
wN2 ,dry = 0.8181 wO2 ,dry = 0.0352
The conversion of the mass fractions into the molar fractions (if required) is also rather straightforward: 1 Mproducts,wet
=
0.131 0.1071 0.7305 0.0314 + + + = 0.0360 44 18 28 32 and Mproducts,wet = 27.7795 kg/kmol
42
2.2 The First Law of Thermodynamics 1 0.1467 0.8181 0.0352 + + = 0.033 65 = Mproducts,dry 44 28 32 and Mproducts,dry = 29.716 kg/kmol and xCO2 ,wet = 0.131/44/0.0360 = 0.0827 xCO2 ,dry = 0.1467/44/0.033 65 = 0.0991 xH2 O = 0.1071/18/0.0360 = 0.1653 xN2 ,wet = 0.7305/28/0.0360 = 0.7247 xN2 ,dry = 0.8181/28/0.033 65 = 0.8683 xO2 ,wet = 0.0314/32/0.0360 = 0.0273 xO2 ,dry = 0.0352/32/0.033 65 = 0.0327 The table below summarises the mass balance: Table 2.3: The incoming and outcoming flow rates in kg/h
IN Species kg/h
Comments:
Methane Nitrogen Oxygen
200 3068 932
IN
4200
P
OUT WET kg/h Carbon dioxide 550 Water vapour 450 Nitrogen 3068 Oxygen 132 P OUT 4200 Species
DRY kg/h 550 − 3068 132 3750
(a) The incoming and outcoming molar flow rates are equal as shown in Table 2.1. In the first lecture it has been stressed that in general one cannot balance the molar flows. The question is what makes the considered example so specific that the incoming and outcoming molar flows are equal? (b) Observe that both methods provide the same composition of the wet and dry combustion products, as one would expect. End of Example 2.1
2.2 The First Law of Thermodynamics The first law of thermodynamics is the overall energy balance that is extended into all possible forms of energy. Historically the first law of thermodynamics was formulated after James Prescott Joule observed in 1840 in his
43
2 Mass and Energy Balance experiments that heat and mechanical work were equivalent forms of energy. Before Joule’s experiments, heat was associated with some "fluidic motions" induced by a temperature difference. In thermodynamics we consider exchanges of all possible energy forms between the control volume and the surroundings. The space that is encompassed by the control volume, in which we have a special interest, is called the system. The type of system depends on the characteristics of the boundary (surface) of the control volume, Fig. 2.3. If matter (mass) can be transferred through the boundary the system is classified as open. If matter cannot pass through the boundary the system is classified as closed. Both open and closed systems can exchange energy with the surroundings. An isolated system can exchange neither matter nor energy with its surroundings.
Fig. 2.3: Left – an open system can exchange matter and energy with its surroundings, Middle – a closed system can exchange energy with its surroundings, Right – an isolated system can exchange neither energy nor matter with its surroundings
For an open system that exchanges mass and energy within its surroundings and with no energy generation within the system, the overall energy balance (Eq. (2.7)) over a time period reads: Ein = Eout + ∆E (2.10) and at an instant
d(E) E˙ in − E˙ out = dt
(2.11)
E˙ in · dt − E˙ out · dt = dE
(2.12)
dEin = dEout + dE
(2.13)
Eq. (2.11) can be rearranged:
to obtain
44
2.2 The First Law of Thermodynamics which is a differential form of the first law of thermodynamics. Fig. 2.4 shows a Sankey’s diagram for energy balance expressed by Eqs. (2.10) and (2.13) applicable to both open and closed systems.
Control volume Ein
DE Eout
Fig. 2.4: Sankey’s diagram for energy balance for an open or closed system
It is instructive to consider the following specific cases of Eqs. (2.10) and (2.13): (a) For an isolated system we have, Ein = 0 and Eout = 0, and therefore ∆E = 0, so the energy of an isolated system remains constant. In some (old) thermodynamics text books the reader will find the following formulation of the first law of thermodynamics: "The sum of all energies is constant in an isolated system." (b) If energy of a system remains constant (∆E = 0) then the energy entering the system must equal the energy leaving the system, Ein = Eout . (c) For a system operating in a steadystate (a furnace, a boiler, a turbine) there is no energy accumulation (dE/dt = 0) and therefore from Eq. (2.11) we obtain E˙ in = E˙ out (2.14) The above relationship shows that it is not possible to construct a machine operated continuously that would provide mechanical work E˙ out = L˙ 1 without receiving energy (as mechanical energy, electrical, chemical, nuclear, thermal or in any other form). A device that would operate without a supply of energy is called a perpetuum mobile of the first kind. Therefore, a popular formulation of the first law of thermodynamics is: "It is not possible to construct a perpetuum mobile of the first kind." 1
For the definition of the work L see Section 2.2.2.2.
45
2 Mass and Energy Balance
2.2.1 System Energy Eqs. (2.10) and (2.13) express the first law of thermodynamics formulated over a time interval. In order to use the law we need to learn how to calculate the three terms Ein (dEin ), Eout (dEout ), E (dE) appearing in the equations. We begin with the third term that is with the system energy E (dE). If a system is in motion its energy consists of: the internal energy of the system, the kinetic energy of the motion, and the potential energy: E = U + m·
w2 ω2 +I· + m·g·z 2 2
(2.15)
where U stands for the internal energy, m is the mass of the system, I is the moment of inertia, w and ω are the translational and rotational velocities, respectively, g is the gravity while z stands for the height of the center of gravity of the system. The latter is measured from a reference level (height), see Fig. 2.5. Internal energy U is a sum of kinetic energy, potential energy and electronic energy of all the molecules and atoms of the system. It includes also energy of chemical bonds of the molecules. Internal energy is a state variable and therefore its change from an initial state 1 to a state 2 is: ∆E = U2 − U1 +
I m · (w22 − w12 ) + · (ω22 − ω12 ) + m · g · (z2 − z1 ) 2 2
Fig. 2.5: System in translational and rotational motion
46
(2.16)
2.2 The First Law of Thermodynamics In combustion engineering the kinetic and potential energies of the system can be neglected since they are much smaller than internal energy. Typically, kinetic energies are considered when the velocity of the system is larger than 50 m/s whereas potential energies are taken into account if z2 − z1 is larger than 100 m. Thus, neglecting the kinetic and potential energies we obtain: (2.17)
∆E = U2 − U1
Internal energy is a function of any two of the three state variables (temperature, pressure, volume) and is an extensive property (see textbooks on thermodynamics, for example [3, 5]): U = m · u(T, p) = m · u(T, v) = m · u(p, v)
(2.18)
and its exact differential is: dU = d(m · u) = m · du + u · dm
(2.19)
In the above equations U (in joules) stands for internal energy of the whole system whilst u is specific internal energy expressed in J/kg. Since specific internal energy is a state variable and its differential is exact, so: ∂u ∂u · dT + · dv (2.20) du = ∂T v ∂v T or [3, 5]
∂p du = cv · dT + T · − p · dv ∂T v
(2.21)
where v is specific volume (in m3 /kg) and cv (in J/(kg · K)) is the specific heat at constant volume. It is not possible to calculate an absolute value of specific internal energy. We can calculate its difference between two states by integrating Eq. (2.21):
u(T, v) = u0 +
ZT,v
du
(2.22)
T0 ,v0
where u0 remains unknown. We can prescribe any value to u0 provided that the reference state T0 , v0 is properly selected. While balancing physical phenomena without any phase changes we are completely free in selecting the reference state
47
2 Mass and Energy Balance T0 , v0 . When phase changes occur, the reference state T0 , v0 should be selected so as to allow for going through all the phase changes considered while evaluating the integral in Eq. (2.22). When chemical reactions take place the reference state T0 , v0 should correspond to the state for which reference substances are specified. The specific internal energy is interrelated with the specific enthalpy (h) since h = u + p·v
(2.23)
In thermodynamics a reference value for specific enthalpy (h0 ) is usually specified at a reference state T0 , p0 and therefore, the reference value for specific internal energy u0 must be calculated using u0 = h0 − p0 · v0
(2.24)
where the specific volume v0 is calculated using an equation of state: v0 = v(T0 , p0 )
(2.25)
Specific Internal Energy of Ideal Gases For an ideal gas ClausiusClapeyron equation of state (Eq. (1.8)) is applicable: v=
R·T M ·p
(2.26)
where R = 8, 314.3 J/(kmol · K) is the universal gas constant and M stands for the molecular mass of the gas in question. The reference specific enthalpy (h0 ) is usually prescribed zero value at T0 = 298.15 K and p0 = 1 bar. Following Eq. (2.24) the reference specific internal energy is u 0 = 0 − p0 ·
R · T0 R = − · T0 M · p0 M
For an ideal gas, from Eq. (2.26) we obtain R ∂p = ∂T v v·M
(2.27)
(2.28)
and after inserting into Eq. (2.21), a simple expression for calculating specific internal energy emerges:
48
2.2 The First Law of Thermodynamics
u(T ) = u0 +
ZT
R cv (T ) · dT = − · T0 + M
T0
ZT
cv (T ) · dT
(2.29)
T0
The above expression underlines the fact that internal energy of an ideal gas is a function of temperature only. For real gases however, internal energy is a function of temperature and pressure (or volume) [3, 5].
2.2.2 Energy Entering and Leaving the System In this paragraph we consider the other two terms Ein (dEin ), Eout (dEout ) that appear in the first law of thermodynamics, Eqs. (2.10) and (2.13). There are several means of supplying and removing energy from the system. These are: (a) energy supplied (removed) with a stream of fluid, (b) energy supplied (removed) as mechanical work, (c) energy supplied (removed) as heat. In subsequent paragraphs we deal with all these means of energy supply. 2.2.2.1 Enthalpy as the Energy of a Stream of Fluid In combustion engineering matter and energy are provided into or removed from a system (a furnace, a turbine, a boiler, a chemical reactor) using pipe lines. Streams of fluids carry energy into and from the system. The energy of a fluid stream consists of the internal energy and the product of static pressure and specific volume (see Eq. (2.23) for specific enthalpy). This overall energy of the stream is called enthalpy. Enthalpy is an extensive property H = m · h(T, p) = m · h(T, v) = m · h(p, v)
(2.30)
and its exact differential is: dH = m · dh + h · dm
(2.31)
In the above expression H (in joules) stands for enthalpy of the system whilst h (in J/kg) is specific enthalpy. Specific enthalpy is a state variable and its exact differential reads: ∂h ∂h dh = · dT + · dp (2.32) ∂T p ∂p T 49
2 Mass and Energy Balance or [3, 5] ∂h · dT + dh = ∂T p
! ∂v v−T · · dp ∂T p
(2.33)
Specific Enthalpy of Ideal Gases ∂v p using the ClausiusClapeyron equation (Eq. (2.26)) Calculating the derivative ∂T and after inserting it into Eq. (2.33) we obtain:
h(T ) = h0 +
ZT
cp (T ) · dT
(2.34)
T0
After recalling that for the reference state (T0 = 298.15 K, p0 = 1 bar) we have already chosen h0 = 0 one obtains the final relationship for calculating specific enthalpy of an ideal gas: ZT h(T ) = cp (T ) · dT (2.35) T0
where cp (T ) in J/(kg · K) stands for the specific heat at constant pressure. The above relationship underlines the fact that specific enthalpy of an ideal gas is a function of temperature only. For a real gas, for which the equation of state is different to Clapeyron equation, specific enthalpy is a function of temperature and pressure (or specific volume) [3, 5]. 2.2.2.2 Energy Supplied to the System through Mechanical Work Work is the fundamental property in thermodynamics. Work is done when an object is moved against an opposing force. An example of doing work is an expansion of a gas that pushes out a piston and raises weight. When work is done to a system, for example by moving a piston and the gas is compressed, the capacity of the system to do work is increased. Conversely when the system does work, its energy is reduced because the system can do less work. It is important to realise that work is not energy. Work is a means of transmitting energy. Notion of work makes sense only when the work is actually being done. When we push a piston to compress the gas we are doing work. However, when the gas is already compressed the work does not exist any more whereas the energy of the system is increased and it exists.
50
2.2 The First Law of Thermodynamics It is important to agree upon the sign of work. In this lecture notes work done to the system is positive whilst work obtained from the system is negative, as shown in Fig. 2.6. According to this convention work done to compress a gas is positive. When afterwards the gas expands to a lower pressure, work is obtained from the system and this work is negative.
Fig. 2.6: Work done to the system is positive; Work done by the system (obtained from the system) is negative
In thermodynamics [3, 5] you have learned that the infinitesimal work done to the system is: dL = −p · dV (2.36) and the amount of work (done to the system) which changes parameters of the system from a state 1 to a state 2 is: L=−
Z2
p · dV
(2.37)
1
In the above equation L stands for work done to the system expressed in joules, p is the static pressure (in N/m2 ) while V is the system volume (in m3 ). The corresponding expressions for the specific work done to the system are: dl = −p · dv
(2.38)
Z2
(2.39)
and l=−
p · dv
1
where the specific work is in J/kg and v is the specific volume in m3 /kg. It is easy to see that work is not a state variable. Consider its differential,
51
2 Mass and Energy Balance Eq. (2.38), as dl = −p · dv − 0 · dp. Thus, ∂2l
∂l ∂v p
= −p and
∂2l
∂l ∂p v
= 0. Calculating 2
2
l l 6= ∂p∂ ∂v and the second derivatives ∂v ∂p = −1 and ∂p ∂v = 0 we see that ∂v∂ ∂p therefore the differential of work is not exact (see Example 2.2). Consequently the integrals for evaluating the amount of work L (Eq. (2.37)), and the specific work l (Eq. (2.39)) depend on the integration path. To perform these integrations the dependence of pressure as a function of volume (or specific volume) must be known. In other words to evaluate these integrals the thermodynamic process must be known.
Fig. 2.7: Work done to the system
Example 2.2 In thermodynamics we use frequently functions of two variables. An equation of state, written in a general form as p = p(T, v) is an example of such a function. Other frequently used functions are to calculate internal energy u = u(T, p) , enthalpy h = h(T, p) and entropy s = s(T, p) . These functions are called state functions and the variables appearing in them are called state variables. Work and heat belong to another group of functions that are not state functions. This example is to recall the criteria for determining whether a function is a state function or not. A mathematician expresses it by saying: the differential of a state function is exact. Below you will find what it means. Consider the purely mathematical problem where F (x, y) is a function of two independent variables x and y. If one goes to a neighbouring point x + dx and y + dy, the function F changes by an amount: dF = F (x + dy, y + dy) − F (x, y)
(A1)
dF = A(x, y) · dx + B(x, y) · dy
(A2)
or
52
2.2 The First Law of Thermodynamics where A(x, y) =
∂F (x, y) ∂F (x, y) and B(x, y) = ∂x ∂y
(A3)
In Eq. (A1) the infinitesimal quantity dF is here an ordinary differential; it is also called an exact differential. If one goes from initial point "i" corresponding to (xi , yi ) to a final point "f" corresponding to (xf , yf ) the corresponding change in F is:
∆F =
Zf
dF = Ff − Fi
(A4)
i
The change ∆F can be simply evaluated as F (xf , yf ) − F (xi , yi ) and it is dependent only on the value of the function F at (xf , yf ) and (xi , yi ). The change ∆F may be also written as:
∆F =
Zf
[A(x, y) · dx + B(x, y) · dy]
(A5)
i
Since ∆F does depend only on F (xf , yf ) and F (xi , yi ), the integral (A5) does not depend on the path along it is evaluated, going from the initial point "i" to the final point "f". It can be demonstrated that the necessary and sufficient conditions for the differential (A2) to be an exact differential are:
since
∂A(x, y) ∂B(x, y) = ∂y ∂y
(A6)
∂ 2 F (x, y) ∂ 2 F (x, y) = ∂y · ∂x ∂x · ∂y
(A7)
From the above considerations it follows: (a) if the infinitesimal quantity dF is an exact differential, the function F is a state variable, (b) if the function F is a state variable, its ordinary differential is exact. End of Example 2.2
53
2 Mass and Energy Balance 2.2.2.3 Energy Supplied to the System as Heat Heat is energy in transit due to a temperature difference. The convention is that heat supplied to the system is positive while heat removed from the system is negative. Heat is not a state variable and its differential is inexact. Heat differs from work since work can be fully converted into heat whereas only part of available heat can be converted into work [3, 5] following the second law of thermodynamics. Total heat absorbed by the system consists of heat supplied from the surroundings and heat generated within the system due to friction, so: Qt = Q + Qf
(2.40)
where Qt is the total amount of heat absorbed by the system, Q is the amount of heat supplied from the surroundings, and Qf is the amount of heat generated within the system due to friction. All these terms are expressed in joules. The above relationship can be also expressed in the amount of heat per 1 kg of the matter of the system: qt = q + qf (2.41) Specific Heat Capacities Temperature of a system changes when heat is transferred into or from the system. The specific heat capacity describes the temperature change of 1 kg of a system that results from the heat transfer dqt ; c=
dqt dT
(2.42)
where the specific heat (c) is expressed in J/(kg · K). Equally well one can define specific heat per 1 kmol of a system, and this property is going to be denoted in this lecture series using capital C. Thus, c − denotes specific heat in J/(kg · K) C − denotes specific heat in J/(kmol · K) The heat capacity depends on the conditions during the heat addition or removal (see textbooks on Thermodynamics, for example [3, 5]) and:
54
for v = const process
C = Cv
for p = const process
C = Cp
for T = const process
C=∞
for dqt = 0 (adiabatic) process
C=0
2.2 The First Law of Thermodynamics When the system is at a constant pressure, heat addition increases simultaneously the temperature and the system energy p V through expansion of the system boundary. Thus, the heat capacity at constant pressure (cp ) is larger than the heat capacity at constant volume (cv ). For ideal gases the following relationship applies [3, 5]: Cp − Cv = R (2.43) where R is the universal gas constant R = 8, 314.3 J/(kmol · K). Obviously the following is applicable: R (2.44) cp − cv = M where M is the molecular mass of the matter of the system. Specific heats of ideal gases can be derived using kinetic theory of gases [3, 5]. The general relationship is as follows: Cv =
number of degrees of freedom ·R 2
(2.45)
For a monatomic gas (H,O,N) there are only three degrees of freedom associated with translational motion. Even at high temperatures a monatomic gas can move in three directions only, without experiencing any rotational components. Thus, specific heats of monatomic gases are Cv = 3 R/2 = 12.47 kJ/(kmol · K) and Cp = 5 R/2 = 20.79 kJ/(kmol · K). For a diatomic gas (H2 ,O2 ,N2 ) there are more degrees of freedom. At low temperatures a diatomic gas experiences only translational motion. However, when the temperature increases the diatomic system starts rotating around two symmetry axes and Cv approaches 5 R/2 = 20.78 kJ/(kmol · K) (Cp approaches 7 R/2 = 29.09 kJ/(kmol · K)). At even higher temperatures the diatomic system begins to vibrate and Cv approaches 7 R/2 = 29.09 kJ/(kmol · K) (Cp approaches 9 R/2 = 37.41 kJ/(kmol · K)). Values of specific heat capacities are calculated using statistical thermodynamics however for stable molecules the heat capacities are measured. NIST  JANAF2 tables [12] provide the most comprehensive set of physical enthalpy data from which Cp values can be readlily calculated. The Cp values are expressed as polynomials of the fourth order in T , Cp = Cp,1 +Cp,2 · T +Cp,3 · T 2 +Cp,4 · T 3 +Cp,5 · T 4 R 2
(Cp in kJ/(kmol · K)) (2.46)
NIST  National Institute of Standards and Technology (USA) JANAF  Joint Army Navy Air Force (USA)
55
2 Mass and Energy Balance In order to improve accuracy, usually two different polynomials are used for low (T < 1000 K) and high (T > 1000 K) temperatures. Knowing the coefficients of the polynomials we calculate the specific enthalpy as follows: h(T ) = h(298.15K)+
ZT
′
′′
Cp (T ) · dT = Cp,6 · R+
298.15K
ZT
Cp (T ′ ) · dT ′ (in
kJ ) kmol
298.15K
(2.47) A new symbol h is introduced to denote specific enthalpies expressed per one kmol as opposed to h that stands for the specific enthalpy per one kilogram (in JANAF 0 Tables H T is used for h). In order to compute enthalpies an additional constant h(298.15K) = Cp,6 · R (in kJ/kmol) is needed to make sure that at the reference state conditions (T = 298.15 K, p = 1 bar) the enthalpy is zero. In numerous text books mean specific heats are given. These are quoted for the temperature range from zero to T T 1 C 0 = · T
ZT
C(T ′ ) · dT ′
(2.48)
0
or for specified temperatures T1 and T2 , so T C T21 =
1 · T 2 − T1
ZT2
C(T ′ ) · dT ′
(2.49)
T1
The relationship between Eq. (2.48) and Eq. (2.49) is as follows: T T T2 C 0 2 · T2 − C 0 1 · T1 C T1 = T2 − T1
(2.50)
2.2.3 Energy Balance of Thermal Systems (Machines) In the above paragraphs we have learned how to calculate the individual terms in the general energy balance over a time period, Eq. (2.10). We have also learned how to calculate the differentials in the differential form of the energy balance, Eq. (2.13). If we rearrange Eq. (2.13) so that: dEin − dEout = dE
56
(2.51)
2.2 The First Law of Thermodynamics Table 2.4: The coefficients for polynomials (2.46) for T in the range 300–1000 K
H O N H2 O2 N2 CO H2 O CO2 CH4 H O N H2 O2 N2 CO H2 O CO2 CH4
Cp,1 2.500 000 00 · 10+00 3.168 267 10 · 10+00 0.250 000 00 · 10+01 2.344 331 12 · 10+00 3.782 456 36 · 10+00 0.032 986 77 · 10+02 3.579 533 47 · 10+00 4.198 640 56 · 10+00 2.356 773 52 · 10+00 5.149 876 13 · 10+00 Cp,5 −9.277 323 32 · 10−22 2.112 659 71 · 10−12 0.000 000 00 · 10+00 −7.376 117 61 · 10−12 3.243 728 37 · 10−12 −0.024 448 54 · 10−10 −9.044 244 99 · 10−13 1.771 978 17 · 10−12 −1.436 995 48 · 10−13 1.666 939 56 · 10−11
Cp,2 7.053 328 19 · 10−13 −3.279 318 84 · 10−03 0.000 000 00 · 10+00 7.980 520 75 · 10−03 −2.996 734 16 · 10−03 0.140 824 04 · 10−02 −6.103 536 80 · 10−04 −2.036 434 10 · 10−03 8.984 596 77 · 10−03 −1.367 097 88 · 10−02 Cp,6 2.547 365 99 · 10+04 2.912 225 92 · 10+04 0.561 046 37 · 10+05 −9.179 351 73 · 10+02 −1.063 943 56 · 10+03 −0.102 089 99 · 10+04 −1.434 408 60 · 10+04 −3.029 372 67 · 10+04 −4.837 196 97 · 10+04 −1.024 664 76 · 10+04
Cp,3 −1.995 919 64 · 10−15 6.643 063 96 · 10−06 0.000 000 00 · 10+00 −1.947 815 10 · 10−05 9.847 302 01 · 10−06 −0.039 632 22 · 10−04 1.016 814 33 · 10−06 6.520 402 11 · 10−06 −7.123 562 69 · 10−06 4.918 005 99 · 10−05 Cp,7 −4.466 828 53 · 10−01 2.051 933 46 · 10+00 0.419 390 87 · 10+01 6.830 102 38 · 10−01 3.657 675 73 · 10+00 0.039 503 72 · 10+02 3.508 409 28 · 10+00 −8.490 322 08 · 10−01 9.901 052 22 · 10+00 −4.641 303 76 · 10+00
Cp,4 2.300 816 32 · 10−18 −6.128 066 24 · 10−09 0.000 000 00 · 10+00 2.015 720 94 · 10−08 −9.681 295 09 · 10−09 0.056 415 15 · 10−07 9.070 058 84 · 10−10 −5.487 970 62 · 10−09 2.459 190 22 · 10−09 −4.847 430 26 · 10−08
Fig. 2.8: Specific heats at constant pressure for various molecules (values obtained using JANAF tables). Black diamonds show values predicted by kinetic theory of ideal gases.
57
2 Mass and Energy Balance Table 2.5: The coefficients for polynomials (2.46) for T in the range 1000–5000 K
H O N H2 O2 N2 CO H2 O CO2 CH4 H O N H2 O2 N2 CO H2 O CO2 CH4
Cp,1 2.500 000 01 · 10+00 2.569 420 78 · 10+00 0.241 594 29 · 10+01 3.337 279 20 · 10+00 3.282 537 84 · 10+00 0.029 266 40 · 10+02 2.715 185 61 · 10+00 3.033 992 49 · 10+00 3.857 460 29 · 10+00 7.485 149 50 · 10−02 Cp,5 4.981 973 57 · 10−22 1.228 336 91 · 10−15 −0.203 609 82 · 10−14 2.002 553 76 · 10−14 −2.167 177 94 · 10−14 −0.067 533 51 · 10−13 −2.036 477 16 · 10−14 1.682 009 92 · 10−14 −4.720 841 64 · 10−14 −1.018 152 30 · 10−13
Cp,2 −2.308 429 73 · 10−11 −8.597 411 37 · 10−05 0.174 890 65 · 10−03 −4.940 247 31 · 10−05 1.483 087 54 · 10−03 0.148 797 68 · 10−02 2.062 527 43 · 10−03 2.176 918 04 · 10−03 4.414 370 26 · 10−03 1.339 094 67 · 10−02 Cp,6 2.547 365 99 · 10+04 2.921 757 91 · 10+04 0.561 337 73 · 10+05 −9.501 589 22 · 10+02 −1.088 457 72 · 10+03 −0.092 279 77 · 10+04 −1.415 187 24 · 10+04 −3.000 429 71 · 10+04 −4.875 916 60 · 10+04 −9.468 344 59 · 10+03
Cp,3 1.615 619 48 · 10−14 4.194 845 89 · 10−08 −0.119 023 69 · 10−06 4.994 567 78 · 10−07 −7.579 666 69 · 10−07 −0.056 847 60 · 10−05 −9.988 257 71 · 10−07 −1.640 725 18 · 10−07 −2.214 814 04 · 10−06 −5.732 858 09 · 10−06 Cp,7 −4.466 829 14 · 10−01 4.784 338 64 · 10+00 0.464 960 96 · 10+01 −3.205 023 31 · 10+00 5.453 231 29 · 10+00 0.059 805 28 · 10+02 7.818 687 72 · 10+00 4.966 770 10 · 10+00 2.271 638 06 · 10+00 1.843 731 80 · 10+01
Cp,4 −4.735 152 35 · 10−18 −1.001 777 99 · 10−11 0.302 262 45 · 10−10 −1.795 663 94 · 10−10 2.094 705 55 · 10−10 0.100 970 38 · 10−09 2.300 530 08 · 10−10 −9.704 198 70 · 10−11 5.234 901 88 · 10−10 1.222 925 35 · 10−09
Fig. 2.9: Physical enthalpies of various gases as a function of temperature
58
2.2 The First Law of Thermodynamics we realise that the net energy supply (dEin − dEout ) into the system increases the energy of the system. For a closed system, heat and work are the only means of supplying energy into the system. If we neglect kinetic and potential energies, internal energy is the only system energy, so dQt + dL = dU
(2.52)
dQt − p · dV = dU
(2.53)
or Introducing the specific properties into Eq. (2.53) we obtain m · dqt − p · d(m · v) = d(m · u)
(2.54)
and after some algebra (note that for a closed system dm = 0): m · dqt − m · p · dv = m · du
(2.55)
dqt − p · dv = du
(2.56)
and finally The reader is requested to compare Eqs. (2.53) and (2.56). Both equations are of the same form showing that for a closed system the change in the internal energy equals the sum of the supplied heat and the work done to the system. Now consider an open system that can exchange matter with the surroundings. The change of the internal energy of the system is induced through the heat supplied, the work done to the system and the enthalpy of the stream entering the system, so dQt − p · dV + h · dm = dU (2.57) and further m · dqt − p · d(m · v) + (u + p · v) · dm = d(m · u)
(2.58)
After some algebra we obtain dqt − p · dv = du
(2.59)
The above relationship valid for an open system is identical to Eq. (2.56) that is valid for a closed system. Thus, for any system (opened or closed) the first law of thermodynamics states that the increase in the specific internal energy of the system equals the sum of the heat introduced to the system and the specific work done to the system (−p · dv).
59
2 Mass and Energy Balance We can introduce into Eq. (2.59) the definition of the specific enthalpy h = u+p · v and by doing so we obtain: dqt − p · dv = dh − v · dp − p · dv
(2.60)
dqt + v · dp = dh
(2.61)
and The above equation leads to another formulation of the first law of thermodynamics. For any system (opened or closed) the increase in the system specific enthalpy equals the amount of heat provided into the system per unit of mass plus the product (v · dp). There are two important implications of Eqs. (2.59) and (2.61). In engineering practices furnaces, boilers and chemical reactors are often operated at a constant pressure, so for p = const.
dh = dqt
(2.62)
For autoclaves and chemical reactors operating at constant volumes (batch processes) for v = const.
du = dqt
(2.63)
2.3 Energy Released in Chemical Reactions 2.3.1 Reaction Enthalpy A chemical reaction can be written as follows: ν1 A1 + ν2 A2 + . . . + νn An = 0 or shorter as
n X
νi Ai = 0
i=1
where Ai stands for the chemical symbols and νi are the stoichiometric coefficients; νi < 0 for reactants (substrates) while νi > 0 for products. For example the reaction CH4 + 2 O2 −−→ CO2 + 2 H2 O
60
2.3 Energy Released in Chemical Reactions can be rewritten as: −CH4 − 2 O2 + CO2 + 2 H2 O = 0 and then: A1 = CH4 ,
A2 = O2 ,
A3 = CO2 ,
A4 = H2 O;
ν1 = −1,
ν2 = −2,
ν3 = 1,
ν4 = 2.
The change of the enthalpy in a chemical reaction is called reaction enthalpy and it is calculated as the sum of the enthalpies of the compounds participating in the reaction times the corresponding stoichiometric coefficients: X pressure (2.64) ∆R Htemp = νi hi i
where the overbar indicates that the enthalpy hi 3 is expressed per 1 kmol (or 1 mol) and the reaction enthalpy ∆R H is in joules (or kJ). When the reaction enthalpy is given for a particular reaction, it applies for the stoichiometric coefficients as written. If each of the stoichiometric coefficients is doubled, the reaction enthalpy is doubled. For example, the reaction of ammonia synthesis may be written as 1 2 N2
+ 23 H2 −−→ NH3
0 ∆R H298 = −46, 110J
or N2 + 3H2 → 2NH3
0 ∆R H298 = −92, 220J
In order to calculate the enthalpies hi (in kJ/kmol), there is a need to define the reference state of zero enthalpy. The pure elements in their most stable state at T = 298.15 K and p = 1 bar (the standard state) are prescribed zero enthalpy (by convention) as shown in Table 2.6. Gases like O2 , H2 , N2 are prescribed zero enthalpies and so are carbon as graphite and other pure elements. There is one exception to this general definition of reference state. For phosphorous the reference state is taken to be white phosphorous despite this allotrope not being the most stable form but simple the most reproducible form of the element. 3
In textbooks on chemical thermodynamics capital letter H i is typically used to denote specific (molar) enthalpy. In this lecture we use small letter “h” so that hi and hi stand for the specific enthalpies of species “i” expressed per kg and kmol, respectively. Reaction enthalpy ∆R H is then expressed in kJ (or joules) and is denoted using a capital letter.
61
2 Mass and Energy Balance Using this convention of the reference state absolute enthalpies for every chemical compound can be defined using the concept of standard enthalpy of formation.
2.3.2 Standard Enthalpies of Formation The standard enthalpy of formation of a substance is the reaction enthalpy ∆R HT00 of its formation reaction from the pure elements in their most stable state at the temperature T = 298.15 K and the pressure p = 1 bar (indicated by "0"). Again, 0 the overbar in hf,298 denotes molar values given per one kmol (or mol) of the compound formed whilst subscript f has been added to stress that it concerns the formation enthalpy. For example for the reaction of formation of H (hydrogen atoms) written as: 1 −→ H(g) 2 H2 (g) − 0
0
0 ∆R H298.15 = 1 · (hf,298.15 )H − 21 · (hf,298.15 )H2 =
218 kJ/mol − 12 · 0 kJ/mol = 218 kJ/mol
Fig. 2.10: Illustration of enthalpy of formation of a compound 0
Thus, the enthalpy of formation of H at p = 1 bar and T = 298.15 K is (hf,298.15 ) = 218 kJ/mol of H. If one writes the above reaction as H2 (g) −−→ 2 H(g) the reaction enthalpy is 0
0
0 ∆R H298.15 = 2 · (hf,298.15 )H − 0.5 · (hf,298.15 )H2 =
2 · 218kJ/mol − 0.5 · 0 · kJ/mol = 436kJ
62
2.3 Energy Released in Chemical Reactions however, the formation enthalpy of H atoms remains 218 kJ/mol of H. Note that if the enthalpy of formation of a compound is positive heat has to be supplied to form the compound and the formation reaction is endothermic. If the formation enthalpy is negative heat is generated (released) in the formation reaction that itself is exothermic. Table 2.6 lists standard enthalpies of formation of some compounds. Direct formation of a compound from its elements is often difficult to realise in practice. For example the enthalpy of formation of neither the hydroxyl radical 1 1 −→ OH 2 H2 + 2 O2 − nor methane C + 2 H2 −−→ CH4 have been measured directly. However due to the work of Germain Henri Hess published in 1840 one can determine indirectly the enthalpy of formation. Since enthalpy is a state function it can be determined using reaction enthalpy of oxidation reactions that are relatively easy to measure. Below we will calculate the enthalpy of formation of pure methane knowing the reaction enthalpies of other oxidation reactions. Consider the following reactions: C(s,graphite) + O2 −−→ CO2 H2 (g) + 12 O2 −−→ H2 O CO2 + 2 H2 O(g) −−→ CH4 + 2 O2
∆R1 HT00 = −393.5 kJ/mol
(i)
∆R2 HT00 ∆R3 HT00
= −241.81 kJ/mol
(ii)
= +802.25 kJ/mol
(iii)
By manipulating the above reactions ((i) + 2 (ii) + (iii)) one obtains the reaction of formation of methane since: C + O2 + 2 H2 + O2 + CO2 + 2 H2 O −−→ CO2 + 2 H2 O + CH4 + 2 O2 C + 2 H2 −−→ CH4 Therefore the reaction enthalpy of the last reaction is: ∆CH4 HT00 = ∆R1 HT00 + 2 · ∆R2 HT00 + ∆R3 HT00 = −393.5 + 2 · (−241.81) + 802.25 = −74.87 kJ/mol of CH4 Since only one mole of CH4 is formed in the above reaction, the formation enthalpy 0 of methane is (hf,298 )CH4 = −74.87kJ/mol of CH4 . The reader should check (Table 2.6) that the calculated value is indeed close to the formation enthalpy of methane.
63
2 Mass and Energy Balance
Table 2.6: Standard enthalpies of formation and standard entropies of some compounds (JANAF Thermodynamic Tables) ∆h
0
f,T0 Compound Aggregation state kJ/mol Oxygen O2 (g) 0.0 Hydrogen H2 (g) 0.0 Nitrogen N2 (g) 0.0 Chlorine Cl(g) 0.0 Carbon C(s,graphite) 0.0 Aluminium Al(s) 0.0 Calcium Ca(s) 0.0 Silicon Si(s) 0.0 Sulphur S(s,rhombic) 0.0 Phosphorus P(s,white) 0.0 Oxygen atoms O(g) 249.2 Ozone O3 (g) 142.4 Hydrogen atoms H(g) 218.0 Water vapour H2 O(g) −241.81 Water H2 O(l) −285.83 Hydroxyl radicals OH(g) 39.3 Nitrogen atoms N(g) 472.68 Nitrogen monoxide NO(g) 90.29 Nitrogen dioxide NO2 (g) 33.1 Diamond C(s,diamond) 1.9 Carbon C(g) 716.6 Carbon monoxide CO(g) −110.53 Carbon dioxide CO2 (g) −393.5 Methane CH4 (g) −74.85 Ethane C2 H6 (g) −84.68 Propane C3 H8 (g) −103.85 Ethylene C2 H4 (g) 52.10 Benzene C6 H6 (g) 82.93 Ethanol C2 H5 OH(g) −235.31 g: gaseous, l: liquid, s: solid
64
s0T 0 J/(mol · K)
205.04 130.57 191.50 223.07 5.74 28.33 41.42 18.83 31.80 41.09 160.95 238.8 114.6 188.72 69.95 183.6 153.19 210.66 239.91 2.38 157.99 197.6 213.7 186.10 229.49 269.91 219.45 269.20 282.00
2.3 Energy Released in Chemical Reactions
2.3.3 Lower Calorific Value (LCV) and Gross Calorific Value (GCV) In combustion engineering it is common to use calorific values to describe the chemical enthalpy of the fuel. Both the lower (LCV) and higher (GCV) calorific values are used. The higher calorific value represents the enthalpy of reaction of stoichiometric combustion (oxidation) of fuel in oxygen with CO2 and H2 O being the combustion products and under the condition that water has been condensed. The lower calorific value is the reaction enthalpy with water remaining in gaseous state (vapour). The examples below clarify the concepts. H2 + 12 O2 −−→ H2 O(l)
+ 285 830 kJ/kmol of H2 142 915 kJ/kg of H2 11 530 kJ/mn 3 of H2 GCV
H2 + 21 O2 −−→ H2 O(g)
(oberer Heizwert oder Brennwert)
+ 241 900 kJ/kmol of H2 119 990 kJ/kg of H2 9758 kJ/mn 3 of H2 LCV
(unterer Heizwert oder Heizwert)
Typically both LCV and GCV are given at temperature T = 298.15 K and pressure p = 1 bar. In some older text books LCV and GCV are given either at T = 273.15 K or at T = 293.15 K; the differences are negligible. For the above reactions: GCVH2 − LCVH2 = enthalpy of condensation of water produced in the oxidation reaction. Thus, GCVH2 − LCVH2 = r = 285 830 − 241 900 = 43 930 kJ/kmol of H2 O And in general: GCV (in kJ/kmol of fuel) = LCV (in kJ/kmol of fuel)+ r · x (kmoles of H2 O produced per 1 kmol of fuel) where r = 43 930 kJ/kmol H2 O or
65
2 Mass and Energy Balance
GCV (in kJ/kg of fuel) = LCV (in kJ/kg of fuel)+ r · x (kg of H2 O produced per 1 kg of fuel) where r = 2440.56 kJ/kg H2 O Definitions: Lower Calorific Value (LCV) is the amount of heat obtained from a complete combustion of a unit of fuel (one kg or one kmol) after cooling down the products of combustion to the inlet temperature of the reagents and maintaining the water of the combustion products in vapour state. LCV depends on temperature of the reagents and it is usually reported either at 0 ◦ C, 20 ◦ C or 25 ◦ C.
Fig. 2.11: Illustration of LCV as an amount of heat extracted from a combustion chamber
Gross Calorific Value (GCV) is the amount of heat obtained from a complete combustion of a unit of fuel (one kg or one kmol) after condensing the water of the combustion products and cooling down the products of combustion to the inlet temperature of the reagents. GCV depends on temperature of the reagents and it is usually reported either at 0 ◦ C, 20 ◦ C or 25 ◦ C.
66
2.3 Energy Released in Chemical Reactions
2.3.4 Relationships between Calorific Values, Reaction Enthalpies and Formation Enthalpies The standard formation enthalpies are specified at standard conditions, e.g. pressure of one bar and temperature of 298.15 K. Consider oxidation reaction of a fuel at standard conditions. For example taking hydrogen as the fuel: H2 + 21 O2 −−→ H2 O(g)
(p = 1 bar, T = 298.15 K)
the reaction enthalpy of the above reaction can be easily calculated as: 0 ∆R H298.15 =
X
νi h i = i 0 (hf,298.15 )H2 O
0
0
− (hf,298.15 )H2 − 12 (hf,298.15 )O2 = −241.81 kJ/mol
Thus, 241kJ of heat is liberated per each mole of hydrogen combusted. Recalling the definition of the LCV one observes that the calculated absolute value of the reaction enthalpy equals the LCV for hydrogen (water vapour remains in the gas phase). The values of both LCV and GCV are always given as positive, so: X 0 νi hi LCV (p = 1 bar; T = 298.15 K) = ∆R H298.15 = (2.65) i
for oxidation reaction4 at standard conditions with water in the gas state,
and
X 0 νi hi GCV (p = 1 bar; T = 298.15 K) = ∆R H298.15 =
(2.66)
i
for oxidation reaction at standard conditions with water in the liquid state. Thus, LCVs listed in Table 2.7 are absolute values of enthalpies of oxidation reaction of the fuel at standard conditions with water remaining in the gas phase.
4
The oxidation reaction must be written for 1 unit (1 kmol of 1 kg) of fuel considered
67
2 Mass and Energy Balance Table 2.7: LCV of some selected gaseous fuels (at p = 1 bar, T = 298.15 K)
Fuel
Formula
Hydrogen Carbon oxide Methane Ethane Propane Butane Pentane Ethylene Propylene Butylene Acetylene
H2 CO CH4 C2 H6 C3 H8 C4 H10 C5 H12 C2 H4 C3 H6 C4 H8 C2 H2
Molar mass g/mol 2.016 28.01 16.03 30.05 44.06 58.08 72.15 28.03 42.05 56.06 26.06
LCV
kJ/mol 241.9 283.0 802.5 1428.0 2044.1 2658.5 3272.9 1323.1 1926.0 2542.8 1255.9
kJ/mn 3 9758 11416 32372 57604 82457 107241 132025 53372 77693 102574 50662
kJ/kg 119990 10104 50062 47521 46394 45773 45362 47203 45803 45359 48193
Example 2.3 Calculate LCV of methane using values of formation enthalpies given in Table 2.6. Assumptions: LCV is the absolute value of the enthalpy of the reaction of complete methane oxidation under the conditions that water remains in the gas phase. The oxidation reaction for methane reads: CH4 + 2 O2 − > CO2 + 2 H2 O(g) or −CH4 − 2 O2 + CO2 + 2 H2 O(g) = 0 Formula (2.65) is then used to calculate the reaction enthalpy: X 0 ∆R H298.15 = νi hi = (−1) · hCH4 + (−2) · hO2 + 1 · hCO2 + 2 · hH2 O i
= (−1) · (−74.85) + (−2) · 0 + 1 · (−393.5) + 2 · (−241.81) = −802.27 kJ
Since the values of the formation enthalpies are at p = 1 bar and T = 298.15 K, so the reaction enthalpy ∆R H corresponds also to p = 1 bar and T = 298.15 K.
68
2.3 Energy Released in Chemical Reactions Note that the negative sign of the reaction enthalpy indicates that the heat is generated in the oxidation reaction of methane. Thus, the LCV of methane under p = 1 bar and T = 298.15 K is: LCVT00 = −802.27 = 802.27 kJ/mol of CH4 Comments: Compare the obtained results with the value listed in Table 2.7. End of Example 2.3
2.3.5 Dependence of LCV on Temperature In the previous sections, both Lower and Gross Calorific values are discussed for standard conditions corresponding to a reference temperature of 298.15 K and a reference pressure of 1 bar. The question is what is the LCV of a fuel at a temperature that is different to 298.15 K? To answer the question we will consider the oxidation reaction of the fuel, however our conceptual reactor is operated this time at a constant temperature that is different to the standard temperature. The reactor is fed with one mol of the fuel. Water and carbon dioxide are the only products of the reaction. The water in the products remains in the gaseous phase. If we calculate the (oxidation) reaction enthalpy, the LCV would be equal to the absolute value of the reaction enthalpy. The oxidation reaction may be written as follows: −1 f uel − ν2 O2 + ν3 CO2 + ν4 H2 O = 0 Then, the reaction enthalpy is ∆R H0T
=
4 X
νi h i
i=1
and elaborating it further:
0
∆R H0T = − hf uel,298.15 +
ZT
298.15
Cp,f uel dT −
0
ν2 hO2 ,298.15 +
ZT
298.15
Cp,O2 dT +
69
2 Mass and Energy Balance
0
ν3 hCO2 ,298.15 +
ZT
Cp,CO2 dT +
298.15
0
ν4 hH2 O,298.15 +
ZT
298.15
After rearranging
Cp,H2 O dT = 0
0 0 0 0 ∆R H0T = −hf uel,298.15 − ν2 · hO2 ,298.15 + ν3 · hCO2 ,298.15 + ν4 · hH2 O,298.15 + ZT
−
Cp,f uel dT − ν2 ·
298.15
ZT
298.15 ZT
ν3 ·
Cp,O2 dT +
Cp,CO2 dT + ν4
298.15
ZT
Cp,H2 OdT
298.15
and further ∆R H0T
=
∆R H0298.15
+ −
ZT
ZT
Cp,f uel dT − ν2 ·
298.15
Cp,O2 dT +
298.15
ν3 ·
ZT
Cp,CO2 dT + ν4
298.15
ZT
298.15
and since LCVT0 = −∆R H0T one obtains: all_molecules
LCVT0
=
0 LCV298.15
−
X i
νi
ZT
Cp,H2 O dT
Cp,i dT
298.15
or when splitting all the molecules taking place in the reaction into subtrates and products to make it more transparent:
70
2.3 Energy Released in Chemical Reactions
LCVT0
=
0 LCV298.15
+
substrates X
νi 
i
ZT
Cp,i dT −
products X
νk 
k
298.15
ZT
Cp,k dT (2.67)
298.15
Relationship (2.67) is known as the integral form of the Kirchhoff ’s law (note that Eq. (2.67) is valid for ν1 = −1 – see the combustion reaction). The differential form of the Kirchhoff law can be obtained by differentiating the above equations with respect to temperature: d LCV =− dT
all_molecules
X
νi · Cp,i
(2.68)
i
or again making it more explicit: d LCV = dT
substrates X i
νi  · Cp,i −
products X
νk  · Cp,k
(2.69)
k
Examining the equations describing Kirchhoff law one may observe that the dependence of the LCV with temperature stems from either the changes in the specific heat values with temperature or with the change of the number of mol in the (oxidation) reaction. Example 2.4 Calculate LCV of methane at 1298.15 K (1025 ◦ C) temperature and 1 bar pressure. Assume that the molecules taking place in the oxidation reaction can be treated as ideal gases. Assumptions: (a) LCV at T = 298.15 K at p = 1 bar is 802.5 kJ/mol of CH4 (see Table 2.7) (b) Gases are ideal with constant specific heats as follows: for CH4 , CO2 and H2 O Cp = 33.3 kJ/(kmol · K) for O2 Cp = 29.1 kJ/(kmol · K) The oxidation reaction of methane is as follows: − ⇀ CH4 + 2 O2 − ↽ − − CO2 + 2 H2 O or −CH4 − O2 + CO2 + 2 H2 O = 0
71
2 Mass and Energy Balance The LCV at T = 1298.15 K can be calculated using Kirchhoff law: all_molecules 0 LCV1298.15
=
0 LCV298.15
X
−
νi
i
0 LCV1298.15 K
= 802.5 − (−1) ·
1298.15 Z K
Cp,i · dT
298.15
1298.15 Z K
33.3 · 10−3 · dT −
298.15
(−2) ·
(2) ·
1298.15 Z K
29.1 · 10
298.15 1298.15 Z K
−3
· dT − (1) ·
1298.15 Z K
33.3 · 10−3 · dT −
298.15
33.3 · 10−3 · dT = 802.5 + 2 · (29.1 − 33.3) =
298.15
802.5 − 8.4 = 794.1 kJ/mol of CH4 Comments: (a) Observe that the temperature correction to the LCV is rather small (8.4 kJ/mol) if compared to the LCV value itself (802.5 kJ/mol). Therefore, in many engineering applications such a correction is unnecessary. (b) More accurate calculations should incorporate the specific heat dependence with temperature. End of Example 2.4
2.3.6 Example of an Energy Balance of a Furnace A prerequisite to a correct energy balance is an accurate mass balance (see Example 2.1). There are two different approaches to formulating an energy balance. In the first approach, popular among chemical engineers, enthalpies of each incoming and outcoming streams are calculated using enthalpies of formation. Typically molar flow rates (kmol/h) are used in this case. In the second approach, popular among combustion engineers, calorific values and mass flow rates (kg/h) are used. Again both approaches are equivalent and lead to the same results. For a furnace operated under a constant pressure and at a steady state the energy balance reads (see Fig. 2.12) H˙ in = H˙ out + Q˙ + L˙ (2.70)
72
2.3 Energy Released in Chemical Reactions ˙ done by the combustion products is zero; H˙ in , H˙ out stand for where the work (L) the enthalpy rate (in W) of the incoming and outcoming streams, respectively, whilst Q˙ is the rate of heat removal (in W) from the furnace and this includes the heat transferred to the process (heat sink) as well as the heat losses. The sketch below clarifies further the meaning of the symbols used in Eq. (2.70).
control volume . Hin
Furnace
. Hout . L
Heat Sink (process)
. Q Fig. 2.12: Illustration of an energy balance
In order to calculate H˙ in and H˙ out both the amount and composition of the incoming and outcoming streams have to be known and these are obtainable from a correctly made mass balance. Then the energy balance reads: in_species
X i
ZTin n˙ i · hf,i + Cp,i · dT =
T0
out_species
X k
n˙ k · hf,k +
T Zout
T0
Cp,k · dT + Q˙ (2.71)
where n˙ i stands for the molar flow rate (kmol/s) of each incoming species (type of molecule), hf,i is the formation enthalpy of the species and Cp,i is the molar specific heat at constant pressure. For the outcoming streams the summation extends over all the combustion products including water. Note that the enthalpy of each species consist of two parts since: hi = hf,i +
ZT
Cp,i · dT
(2.72)
T0
where hf,i represents so called "chemical enthalpy" while the integral represents
73
2 Mass and Energy Balance "physical enthalpy". Both are calculated using the standard state conditions as the reference level of zero enthalpy. The formulation (2.72) is consistent with a statement that any (molecule) species has a chemical energy (positive or negative) that is either released or taken during chemical reactions. Only the most stable species (see Table 2.7) under the standard conditions have no (zero) chemical enthalpy as a consequence of the definition of the reference state. For the incoming streams the integral is insignificant at ambient air conditions but it can be important when either the fuel or oxidiser or both are preheated. For the outcoming streams the integral represents the physical enthalpy of combustion products. Combustion engineers calculate the incoming H˙ in enthalpy using Lower Calorific Value rather than enthalpies of formation. The concept here is that the chemical energy associated with the combustion reaction is bound with the fuel only. In other words neither CO2 nor H2 O have chemical enthalpies. After splitting H˙ in into the oxidiser (air) stream and the fuel stream the energy balance reads: ZTin ZTin m ˙ f uel · LCV + cp · dT + m ˙ oxidiser cp,oxidiser · dT =
T0
T0
m ˙ products
T Zout
cp,products · dT + Q˙ (2.73)
T0
and (2.74)
m ˙ f uel + m ˙ oxidiser = m ˙ products
where m ˙ f uel , m ˙ oxidiser and m ˙ products stand for the mass flow rates (kg/h) of the fuel, oxidiser and combustion products, respectively. Note that the reference state (p = 1 bar, T = 298.15 K) for making the energy balance is the same for Eq. (2.71) and (2.73). In Eq. (2.73) the specific heats are expressed per kilogram. Relationship (2.73) can be rewritten as: ZTin ZTin ˙ oxidiser cp,oxidiser · dT = cp · dT + m m ˙ f uel · LCV +
T0
T0
all_products
X k
m ˙k
T Zout
cp,k · dT + Q˙ (2.75)
T0
where m ˙ k stands for the mass flow rate of species k present in the products.
74
2.3 Energy Released in Chemical Reactions Formulations (2.71) and (2.73) are equivalent. They both have advantages and disadvantages. The chemical engineering formulation (2.71) is general in the sense that there is no need to distinguish between the fuel and the oxidiser. They both take part in the reactions and as long as the reactions are written correctly, calculating H˙ in and H˙ out enthalpies is relatively straightforward. The generality of Eq. (2.71) makes it ideal for development of general computer packages for computer simulations of industrial processes. However, in Eq. (2.71) there is no explicit term that would show how much energy enters into the system. Thus, answering the question "what is the thermal input of the system considered?" requires some further calculations. On the other hand, Eq. (2.73) requires a priori determination of the fuels and therefore it is less general. The left hand side of Eq. (2.73) represents explicitly the total (chemical + physical enthalpy) thermal input, the process variable that is essential in thermal engineering. Furthermore, the outcoming flow rate of combustion products is rather straight forward to obtain since the flow rates are in kg/h. Example 2.5 Consider again the boiler of Example 2.1. Make the energy balance of the boiler knowing that 2 MW of heat is extracted by the water tube wall (heat sink). Both the fuel and the air at the burner are at ambient temperatures (Tin = 298.15 K). Calculate the temperature at the boiler exit and the efficiency of the process. Carry out the energy balance using enthalpies of formation and repeat the calculations using LCV. For the sake of simplicity treat the gases as ideal with constant (independent of temperature) specific heats. Assumptions: (a) the boiler operates at a steady state, (b) the control volume encompasses the boiler,
control volume 200 kg/h CH4
Boiler
Combustion products
4000 kg/h air Heat Sink
. Q = 2MW Fig. 2.13: Example of an energy balance
75
2 Mass and Energy Balance (c) the gases are treated as ideal with constant specific heats as follows: for CH4 , CO2 and H2 O Cp = 33.3 kJ/(kmol · K) for O2 , N2 Cp = 29.1 kJ/(kmol · K) Energy balance formulated using enthalpies of formation The molar flow rates are given in Table 2.1 while the standard enthalpies of formation are taken from Table 2.6. For the incoming streams (CH4 , O2 and N2 ) the left hand side of Eq. (2.71) is as follows: RT 12.5 · 103 + T00 33.3dT = −259.9 kJ/s Methane 3600 · −74.85 R T0 109.57 = 0 kJ/s Nitrogen 3600 · T0 29.1dT R T0 29.13 Oxygen = 0 kJ/s 3600 · T0 29.1dT TOTAL INPUTS
−259.9 kJ/s
For the outcoming streams (CO2 , H2 O, N2 , O2 ) (the right hand side of Eq. (2.71)) the following is applicable:
Heat extracted
12.5 −393.5 · 103 + 33.3 · (Tout − T0 ) = 3600 −1366.32 + 0.115 63 · (Tout − T0 ) 25 −241.81 · 103 + 33.3 · (Tout − T0 ) = 3600 −1679.24 + 0.231 25 · (Tout − T0 ) 109.57 (29.1 · (Tout − T0 )) = 0.8856 · (Tout − T0 ) 3600 4.13 (29.1 · (Tout − T0 )) = 0.0334 · (Tout − T0 ) 3600 = 2000.00
TOTAL OUTPUTS
= −1045.56 + 1.2659 · (Tout − T0 )
Carbon dioxide:
Water vapour:
Nitrogen: Oxygen:
Thus, the energy balance reads: −259.9 = −1045.56 + 1.2659 · (Tout − T0 ) (both sides in kJ/s) and Tout can be readily calculated to be Tout = 918.81 K.
76
2.3 Energy Released in Chemical Reactions
259.9 kJ/s Chemical Enthalpy of CH4 Nil Physical Enthalpy Furnace
control volume 1045.56 kJ/s Chemical Enthalpy 785.69 kJ/s Physical Enthalpy of Combustion Products
Tin =298.15K
Tout =918.81K Heat Sink
(process)
. Q = 2MW Fig. 2.14: Energy balance using enthalpy of formation
The thermal input into the furnace consist of the physical enthalpy and the chemical enthalpy. Since the inlet temperature of the substrates (methane and air) equals the reference temperature the physical enthalpy at the inlet is zero. The chemical enthalpy, expressed using heats of formation, is: −259.9 kJ s . The combustion products enthalpy consists of the physical and chemical enthalpies. The chemical enthalpy includes the carbon dioxide component (−1366.32kJ/s) and the water vapour component (−1679.24kJ/s), so its value is −3045.56kJ/s. The physical enthalpy of the combustion products which includes all the components (CO2 , H2 O, N2 and O2 ) equals 1.2659 · (918.81 − 298.15) = 785.69kJ/s. The efficiency of the boiler can then be calculated as Ef f iciency = =
heat extracted by the water tube walls enthalpy of inlet streams − chemical enthalpy of combustion products 2000kJ/s = 0.718 = −259.9kJ/s − (−3045.56)kJ/s
Energy balance formulated using LCV The mass flow rates of the streams are given in Table 2.3 whilst the LCV of CH4 is taken from Table 2.7 to be 50 062 kJ/kg of CH4 . The enthalpy of the incoming streams is as follows (left hand side of Eq. (2.73)): R T0 200 33.3dT /16 = 2781.22 kJ/s · 50 062 + Methane 3600 T0 R T0 39 068 Nitrogen = 0 kJ/s 3600 · T0 29.1dT /28 R T0 932 Oxygen = 0 kJ/s 3600 · T0 29.1dT /32 TOTAL INPUTS
2781.22 kJ/s
77
2 Mass and Energy Balance The figure of 2.78 MW is the total thermal input into the furnace that consists of the chemical enthalpy of the fuel (2.78 MW) and the physical enthalpy of the fuel and of the air stream. For the outcoming streams (CO2 , H2 O, N2 , O2 ) (the right hand side of Eq. (2.73)) the following is applicable: 550 33.3 · (Tout − 298.15) = 0.115 63 · (Tout − 298.15) Carbon dioxide: 3600 44 450 33.3 Water vapour: · (Tout − 298.15) = 0.231 25 · (Tout − 298.15) 3600 18 3068 29.1 Nitrogen: · (Tout − 298.15) = 0.8856 · (Tout − 298.15) 3600 28 132 29.1 · (Tout − 298.15 = 0.0334 · (Tout − 298.15) Oxygen: 3600 32 Heat extracted = 2000.00 TOTAL OUTPUTS
= 2000.00 + 1.2659 · (Tout − 298.15)
Thus, the energy balance reads: 2781.22 = 2000.00 + 1.2659 · (Tout − T0 ) (in kJ/s) and Tout = 915.28 K. The physical enthalpy of the combustion products can now be calculated as 1.2659 · (915.28 − 298.15) = 780.87 kJ/s and the energy balance closes since: Input = 2.78 MJ/s Output = 2 MJ/s (heat extracted) 0.78 MJ/s (physical enthalpy of products at Tout = 915.28 K) control volume
2.786MW Chemical Enthalpy of CH4 Nil Physical Enthalpy
0.786MW Physical Enthalpy of Combustion Products Furnace Tout =918.81K
Tin =298.15K Heat Sink (process)
. Q = 2MW Fig. 2.15: Energy balance using LCV
78
2.4 Temperature of Adiabatic Combustion The efficiency of the boiler is then Ef f iciency =
2M W heat extracted by the water tube walls = = 0.718 total energy input 2.78M W (2.76)
Comments: (a) Observe that when enthalpies of formation are used, the total enthalpy of the incoming streams can be negative; total enthalpy of the outcoming streams can also be negative. (b) When LCV is used enthalpies of the incoming streams and outcoming streams are always positive. (c) When enthalpies of formation are used, the thermal input into the furnace is “hidden” in the energy balance Eq. (2.71). (d) When LCV is used, the thermal input into the furnace is explicitly given since it is just the left hand side of Eq. (2.73). (e) The furnace exit temperature is calculated to be 918.81 K when the energy balance is formulated using the enthalpies of formation. When LCV is used, the furnace exit temperature is 915.28 K. What is the reason for the 3.52 K difference? End of Example 2.5
2.4 Temperature of Adiabatic Combustion In an adiabatic system (Q˙ = 0) the combustion products after complete combustion carry the physical enthalpy equal to the total thermal input, as shown by Eq. (2.73). If combustion is stoichiometric, the temperature of the combustion products is the highest achievable for this specific fuel. This temperature is called the temperature of adiabatic combustion (adiabatic flame temperature) under stoichiometric conditions and is a property of the fuel. Inserting into Eq. (2.73) the relationships Q˙ = 0 (adiabatic combustion) and m ˙ oxidiser = lair,min · m ˙ f uel (stoichiometric combustion) one obtains: ZTin ZTin m ˙ f uel · LCV + cp · dT + lair,min · m ˙ f uel cp,oxidiser · dT =
T0
T0
79
2 Mass and Energy Balance ZTad cp,products · dT (m ˙ f uel + lair,min · m ˙ f uel ) · T0
and after dividing both sides by m ˙ f uel ZTin ZTin cp,oxidiser · dT = LCV + cp · dT + lair,min T0
T0
ZTad cp,products · dT (1 + lair,min )
(2.77)
T0
Eq. (2.77) can be used to calculate the adiabatic combustion temperature Tad under stoichiometric conditions. Note that Tad is uniquely defined for any fuel and therefore is regarded as a fuel property. Table 2.8 lists adiabatic flame temperatures for some fuels which are calculated using Eq. (2.77); the numbers refer to Tin = 298.15 K, p = 1 bar. Table 2.8: Adiabatic flame temperature Tad for stoichiometric combustion in air (Tin = 298.15 K, p = 1 bar). Combustion products contain CO2 and H2 O only.
H2 C2 H2 C3 H8
2473 K 2936 K 2400 K
CH4 C2 H6 CO
2285 K 2357 K 2624 K
Table 2.8 shows the calculated values of Tad providing CO2 and H2 O are the only combustion products. For temperatures in excess of 2273 K dissociation of CO2 and H2 O takes place according to the reactions: −− ⇀ CO2 ↽ − − CO + O
and
−− ⇀ H2 O ↽ − − OH + H
Both reactions are endothermic therefore the real adiabatic flame temperatures are lower than those listed in Table 2.8. For temperatures in excess of 2273 K, the combustion products contain not only CO2 and H2 O but also CO, H2 , O, OH and N radicals. In such high temperatures the combustion products composition should be determined using chemical equilibrium considerations (see Chapter 4).
80
2.5 Furnace Exit Temperature
2.5 Furnace Exit Temperature The energy balance for any nonadiabatic combustion system can be written as: ZTin ZTin m ˙ f uel · LCV + cp,f uel · dT + λ · lair,min · m ˙ f uel cp,air · dT
T0
T0
= (m ˙ f uel + λ · lair,min · m ˙ f uel )
T Zout
cp,products · dT + Q˙ (2.78)
T0
In the above equation we assume that the fuel is oxidised completely to CO2 and H2 O and air is used as oxidiser. However by replacing lair,min with an equivalent variable, one may write such an equation for any oxidiser namely oxygen or enriched air. The left hand side of Eq. (2.78) can be rewritten as:
TTI = m ˙ f uel · LCV +m ˙ f uel
FUEL THERMAL INPUT
ZTin ZTin cp,f uel · dT + λ · lair,min · m ˙ f uel cp,oxidiser · dT
T0
@ I @ @
T0
PHYSICAL ENTHALPY (PREHEAT)
where TTI stands for total thermal input into the system. Dividing both sides of Eq. (2.78) by m ˙ f uel · LCV and after some algebra one obtains: R Tout (1 + λ · lair,min ) 298.15 cp,products · dT Q˙ =1− + m ˙ f uel · LCV LCV R Tin,air R Tin,f uel cp,air · dT λ · lair,min 298.15 298.15 cp,f uel · dT + LCV LCV
(2.79)
˙ m The ratio Q/( ˙ f uel · LCV ) is the fraction of the fuel thermal input extracted by the process (including heat losses). For a given fuel and a given preheat, the ratio is a function of the excess air (λ) and the furnace exit temperature Tout only, so that Q˙ = f (λ, Tout ) (2.80) m ˙ f uel · LCV Combustion engineers use this function to estimate the amount of the energy
81
2 Mass and Energy Balance available for the process. It is worth observing that for Q˙ → 0
Tout → Tad
and if Tin = 298.15 K (no preheat) Eq. (2.79) simplifies further to: RT (1 + λ · lair,min ) T0out cp,products · dT Q˙ =1− X= m ˙ f uel · LCV LCV ˙
(2.81)
The ratio X = m˙ f uelQ· LCV is a fraction of the fuel thermal input that is available for the process and therefore it is often called in short "percentage available heat". A Sankey diagram shown in Fig. 2.16 visualises the energy balance of a furnace operated without air preheat. The figure provides a further explanation for the "available heat" concept.
Fig. 2.16: Sankey diagram demonstrating the concept of the available heat
82
2.5 Furnace Exit Temperature Example 2.6 Generate a graph showing relationship (2.81) for combustion of pure methane in air. Assumptions: (a) LCV of methane at T = 298.15 K and at p = 1 bar is 50 062 kJ/kg of CH4 (see Table 2.7), (b) the combustion products contain CO2 and H2 O and thermal dissociation is ignored, (c) as a first estimation we assume that the gases are ideal with constant specific heat capacities as follows: for CH4 , CO2 and H2 O
Cp = 33.3 kJ/(kmol · K)
for O2 , N2
Cp = 29.1 kJ/(kmol · K)
(d) we repeat the calculations using Cp polynomials to examine the effect of the assumption (c) The oxidation reaction of CH4 is −− ⇀ CH4 + 2 O2 ↽ − − CO2 + 2 H2 O or 16 kg CH4 + 64 kg O2 = 44 kg CO2 + 36 kg H2 O The minimum air requirement is 64/16/0.233 = 17.1674 kg of air/kg of methane. For λ > 1 the combustion products contain CO2 , H2 O, N2 , and O2 . Per 1 kg of CH4 , the combustion products contain: Methane Carbon dioxide Water vapour Nitrogen Oxygen TOTAL (wet) =
44/16 = 36/16 = 0.767 · λ · 17.1674 = 0.233 · (λ − 1) · 17.1674 =
none 2.75 kg 2.25 kg λ · 13.1674 kg (λ − 1) · 4 kg 1 + λ · 17.1675 kg
83
2 Mass and Energy Balance Composition of the wet combustion products is: wCO2 = 2.75/(1 + 17.1675 · λ) wH2 O = 2.25/(1 + 17.1675 · λ) wN2 = 13.1675 · λ/(1 + 17.1675 · λ) wO2 = 4 · (λ − 1)/(1 + 17.1675 · λ) The above expressions are needed for evaluating the specific heats of combustion products since cp,products = wCO2 ·
Cp,CO2 Cp,H2 O Cp,N2 Cp,O2 + w H2 O · + w N2 · + wO 2 · 44 18 28 32
and the physical enthalpy of the wet combustion products is then T Zout
cp,products · dT =
298.15
(wCO2 ·
33.3 29.1 29.1 33.3 + wH2 O · + w N2 · + wCO2 · ) · (Tout − 298.15) (B1) 44 18 28 32
Using Cp polynomials given in Table 2.5, one can evaluate the physical enthalpy of the combustion products more accurately: T Zout
cp,products · dT =
298.15
X
R
products_species
wi 1 2 {(Cpi,1 (Tout − 298.15) + Cpi,2 (Tout − 298.152 )+ Mi 2
1 1 3 4 Cpi,3 (Tout − 298.153 ) + Cpi,4 (Tout − 298.154 )+ 3 4 1 5 Cpi,5 (Tout − 298.155 )} (B2) 5
Now expression (2.81) can be put together as: RT (1 + λ · 17.1674) · T0out cp,products · dT Q˙ X= =1− m ˙ f uel · LCV 50 062
(B3)
where the integral can be evaluated (for any Tout and λ > 1) using either simplified
84
2.5 Furnace Exit Temperature expression (B1) or accurate Eq. (B2). Relationship (B3) for combustion of pure methane is shown in Fig. 2.17 and Fig. 2.18. Obviously for specific heat capacities taken as independent of temperature, the relationship is linear for any excess air ratio (λ). For a given furnace exit temperature the amount (fraction) of the energy extracted is a function of λ and therefore any uncontrolled inleakage of air into the furnace should be avoided. The point of intersection of the line with the Xaxis shows the adiabatic combustion temperature for given λ (Tin = 298 K). Note for example that for stoichiometric combustion of pure methane we have calculated the adiabatic combustion temperature around 2800 K when constant cp values were used (Fig. 2.17). When the specific heat values are evaluated using the JANAF polynomials the relationships become nonlinear as shown in Fig. 2.18. More importantly the results differ substantially from those shown in Fig. 2.17. The adiabatic temperature of stoichiometric combustion decreases by 525 K as summarised in the table below. Table 2.9: Calculated adiabatic temperature for stoichiometric combustion of pure CH4 with air (Tin = 298 K, p = 1 bar). Compare with Table 2.8
Tad
constant cp values 2810 K
cp from JANAF tables 2285 K
˙
Q Fig. 2.17: Fraction of the available heat (X = m˙ f uel cdotLCV ) as a function of the furnace exit temperature and excess air ratio for combustion of pure methane in air. Constant cp values.
85
2 Mass and Energy Balance
˙
Fig. 2.18: Fraction of the available heat (X = m˙ f uelQ· LCV ) as a function of the furnace exit temperature and excess air ratio for combustion of pure methane in air. JANAF polynomials for cp have been used.
Comments: (a) Note the pronounced effect of using Cp polynomials. (b) At temperatures of around 2300 K some (not much) CO2 and H2 O may dissociate so precise calculations should be carried out using a chemical equilibrium procedure (see Chapter 4). End of Example 2.6
2.6 Summary Conservation of mass and energy is the basis of combustion engineering. Identification of the control volume and the specification of an appropriate time basis are essential skills in formulating mass and energy balances. A correct mass balance of a system is a prerequisite to a subsequent energy balance. In this lecture the first law of thermodynamics has been recalled in order to identify various forms of energy supplied to and removed from a system. It has been shown that in combustion engineering, or more general in thermal engineering, the internal energy is typically the largest energy and the kinetic and potential
86
2.6 Summary energies can be neglected. The student should realise that the first law of thermodynamics, written for any systems (open or closed) takes the form dqt − p · dv = du expressing the fact that the internal energy (du) can be altered either by the (total) heat supplied to the system (dqt ) or the work done to the system (−p · dv) or both. Equivalently, the above relationship can be written as dqt + v · dp = dh where h stands for specific enthalpy defined as u + p · v = h. In engineering practises furnaces, boilers and chemical reactors are often operated at a constant pressure and then dqt = dh
when
p = const.
For autoclaves and chemical reactors operating at a constant volume (batch processes) one obtains dqt = du when v = const. It is essential to realise that it is not possible to calculate absolute values of energy. Instead human beings have defined a reference state for calculating energy as a state of pure elements in their most stable state at T0 = 298.15K and p0 = 1bar. By convention this state is prescribed zero enthalpy (see Table 2.6). Thus, enthalpies at any other states are calculated in reference to this reference state as follows ZT,p 0 h(T, p) = hf,298 + dh T0 ,p0
0 hf,298
where is the molar formation enthalpy. In this way the total specific enthalpy of a species consists of the chemical enthalpy and the physical enthalpy. For furnaces and boilers which operate at a steady state and at a constant pressure, Eq. (2.71) formulates the energy balance, or more precisely enthalpy balance, assuming that the gases are ideal. Formulation (2.71) is general in the sense that there is no need to distinguish between the fuel and the oxidiser. However in formulation (2.71) there is no explicit term that would show the thermal input into the system. For a combustion engineer the thermal input into the system is of paramount importance. Furthermore, for a combustion engineer all the chemical enthalpy is associated with the fuel only and carbon dioxide and wa
87
2 Mass and Energy Balance ter vapour molecules which are the products of the fuel combustion contain no chemical enthalpy. Following this concept, the notion of Calorific Values (Lower and Higher/Gross) has been developed leading to the enthalpy balance equation (2.73). In this equation the left hand side shows explicitly the thermal input into the furnace. Both approaches, through Eq.(2.71) or through Eq.(2.73) are equivalent and the student must command them. In this lecture we have also introduced the temperature of adiabatic combustion. If adiabatic combustion is stoichiometric, the temperature of the combustion products is the highest achievable for this specific fuel and therefore this temperature is a property of the fuel. While considering the first law of thermodynamics, we have reiterated that both internal energy and enthalpy are state variables and their differentials are exact. Work and heat are not state variables and their differentials are inexact.
88
3 Equilibrium Thermodynamics Contents 3.1 3.2
3.3
3.4
3.5
3.6
Irreversible and Reversible Processes Entropy 3.2.1
Entropy of Liquids and Solids
3.2.2
Entropy of Ideal Gases
3.2.3
Entropy of Phase Transition at the Transition Temperature
3.2.4
The Third Law of Thermodynamics
3.2.5
Absolute Entropy of Pure Substances
The Second Law of Thermodynamics 3.3.1
The Increase in Entropy Principle
3.3.2
Entropy Change for a Continuous Process at SteadyState
3.3.3
Irreversibility of Processes
General Conditions for Thermodynamic Equilibrium 3.4.1
Isolated System
3.4.2
NonAdiabatic System
Equilibrium Between Phases 3.5.1
SingleComponent System Consisting of Two Phases
3.5.2
Phase Transformations of a Pure Substance
3.5.3
Dependence of Gibbs Free Enthalpy on Temperature and Pressure
3.5.4
Equilibrium in MultiComponent SinglePhase Systems
3.5.5
Chemical Potential of Pure Substances
3.5.6
Significance of Chemical Potential
MultiComponent, MultiPhase Systems 3.6.1
The Phase Rule
3.7
Thermodynamics of Mixing
3.8
Summary
89
3 Equilibrium Thermodynamics
3.1 Irreversible and Reversible Processes In nature some things happen spontaneously (naturally), some do not. For example a hot body cools down to a temperature of the surroundings in a spontaneous process of giving up heat. After a sufficiently long time equilibrium is reached and the body’s temperature reaches surroundings temperature. A compressed gas stored in a cylinder expands spontaneously into surroundings as soon as the interaction with the surroundings is made possible by opening the cylinder valve. Again after a long enough time equilibrium is reached; the gas pressure in the cylinder equals the surroundings pressure and the gas outflow from the cylinder ceases. A cigarettes smoke mixes into the air by a (spontaneous) process of molecular diffusion. All these spontaneous processes proceed naturally, in one direction, and there is no need to do any work to bring the changes about. These spontaneous processes are irreversible. By calling a process irreversible we mean that the initial state, or any other past state, cannot be reached by any spontaneous process that would proceed in the opposite direction.
surroundings
surroundings valve
Tsurr Q hot body
T p T
Tsurr psurr
surroundings cigarette smoke
compressed gas
Fig. 3.1: Examples of irreversible processes in closed and open systems. Left – hot body cooling down to surroundings temperature; Middle – compressed gas expanding to surroundings pressure (throttling); Right – cigarette smoke diffusing into surroundings.
Consider again our three examples: the hot body being cooled down, the compressed gas expanding to surroundings pressure (the process is called throttling) and the cigarette smoke diffusing into the air. The first example is a case of a closed system while the second and the third examples are open systems, see Fig. 3.1. During theses spontaneous processes the system (the hot body, the compressed gas, the cigarette smoke) interacts with surroundings. The irreversibility of open or closed processes means that it is not possible to get back to any previous states of the system and the surroundings. After cooling the hot body to surroundings temperature, it is not possible to heat it up again using surroundings, even though doing so would not violate the first
90
3.1 Irreversible and Reversible Processes law of thermodynamics. Similarly after expanding the gas to surroundings pressure it is not possible to compress the gas back to any previous states using only the surroundings. As soon as the cigarette smoke disperses into the air, there is no way of bringing it back to the cigarette tip using only surroundings. It does not mean that the original situation can never be restored. It can, provided that the system plus the surroundings are not kept constant but are allowed to interact with other systems. Obviously the hot body that has been cooled down to the surroundings temperature can be heated up back to its initial temperature. However to accomplish this we cannot use the heat that the body has given up to surroundings but we must employ another source of energy for example a burner or an electrical heater. Similarly, we can compress the gas to its initial pressure using a compressor however, it is not possible to achieve it using energy of surroundings only.
Fig. 3.2: Irreversible process of mixing. An isolated system consisting of a box divided by a partition. The left hand side is filled with a gas whilst the right hand side is empty.
As an example of an isolated system consider a box divided by a partition into two equal parts, each of volume V , Fig. 3.2. The left hand side is filled with gas while the right hand side is empty. Suppose now that the partition is removed. As a result of collisions with the walls and with each other, the molecules will very quickly redistribute themselves over the entire volume of the box. The final equilibrium state, where the density of molecules is uniform throughout the entire box is attained rather quickly. The question is what determines the direction of this spontaneous mixing process? It is certainly not the total energy of this isolated system since it remains constant through the whole process. However when a change within the box occurs, the energy is parcelled out (distributed) differently. This irreversible mixing process proceeds towards the greater chaotic dispersal of the total energy of the system. The irreversibility of an isolated system means that it is not possible to reach any previous states of the system and the process proceeds in one direction only. The spontaneous,
91
3 Equilibrium Thermodynamics irreversible processes are always associated with a redistribution of energy into a more disordered form. There is another observation to be made in conjunction with irreversible processes taking place in open or closed systems. The more spontaneous and violent the process is, the more difficult it is to bring the system to the initial state since the changes ("damages") to surroundings are larger. When the body is hotter or more precise when the temperature difference between the body and the surroundings is larger, the cooling process proceeds faster. To bring the body back to the initial state, we would have to use an energy source (a burner, an electrical heater) capable of heating the body to a higher temperature. The second law of thermodynamics governs the direction of irreversible processes. Historically there have been many formulations of the law. Probably the simplest formulation is due to Rudolf Clausius who in 1850 stated that "Heat cannot of itself pass from a colder to a hotter body." Other more elaborate formulation due to William Thomson (Lord Kelvin) reads: "A transformation whose only final result is to transform into work heat extracted from a source which is at the same temperature throughout is impossible." Both of these formulations are not applicable in engineering practise. They are just statements indicating in which direction heat can (or rather cannot) be transferred. After introducing the notion of entropy another formulation can be derived (see Section 3.3.1) which allows the second law of thermodynamics to be used in engineering. Irreversible processes reach a different "degree of irreversibility" depending on how the processes proceed. Thus, if we alter the way of carrying out the process we may minimise the changes ("damages") to the surroundings. Perhaps we can even carry out the process in such a way that the degree of irreversibility would be zero and the process would be reversible. A reversible process is able to bring the system to any previous states, and finally to the initial state, without any changes to surroundings. This is possible only if the net heat and net work exchange between the system and the surroundings is zero for the combined (original and reverse) process. Reversible processes do not occur in nature. However we mentally (conceptually) construct reversible processes in order to compare real, irreversible processes against these ideal conceptual ones. Reversible processes can be approximated by actual devices but they can never be achieved. Engineers are interested in
92
3.1 Irreversible and Reversible Processes reversible processes because devices producing work (engines, turbines) deliver maximum work when they realise reversible processes. Devises like pumps and compressors require minimum work when they realise reversible processes. Thus, reversible processes can be regarded as theoretical limits of the corresponding real processes. Consider a closed system (of constant mass) that is at an initial state marked in Fig. 3.3 as state 1. By applying (adding) an infinite number of combinations of heat and work (see Eq. (2.56)) to the system, the matter of the system can reach an infinite number of states. Among these infinite number of thermodynamic processes the following five are typically considered: (a) isothermal process (either expansion or compression); T = const., (b) processes proceeding under constant volume; V = const., (c) processes proceeding under constant pressure; p = const., (d) adiabatic processes; Q = 0, (e) polytropic processes, pV α = const. with α = const. where α is called polytropic exponent. Fig. 3.3 shows these typical processes using a pV (work) diagram. All of them might be carried out as reversible processes if additions of the infinitesimal amounts of heat and work is carried out in a series of quasistatic steps (see below). However there are two thermodynamic processes that are highly irreversible and cannot be reversed. These are: throttling (expansion without any work) and mixing. Fig. 3.1 (middle) shows an example of throttling while Fig. 3.1 (right) and Fig. 3.2 show a highly irreversible process of mixing. Chemical reactions are highly irreversible. Processes shown in Fig. 3.3, can be carried out reversibly if the infinitesimal amounts of heat and work are added (or removed) into the system in a process which is carried out so slowly that the system remains arbitrarily close to equilibrium at all stages of the process. Such a process is said to be "quasistatic", or "quasiequilibrium" for the system. A quasistatic process can be viewed as a sufficiently slow process which allows the system to adjust itself internally so that properties in one part of the system do not change any faster than in another part.
93
3 Equilibrium Thermodynamics
p
V=const.
1 p=const.
2
T=c o 2
V1
. Q
nst.
12 =0
2 2
V2
V
L12 . Q12 Fig. 3.3: Typical thermodynamic processes shown using workdiagram; T = const., V = const., p = const., Q = 0, (pV α = const. not shown).
How slowly one must proceed to make sure that the process is quasistatic depends on the time (called "relaxation time") that the system requires to attain equilibrium if it is suddenly disturbed. To be slow enough to be quasistatic implies that the process proceeds slowly compared to the relaxation time. For example, if the gas in Fig. 3.3 returns to equilibrium within a time of 10−5 seconds after the piston is suddenly moved back from position 2 to position 1, then a process wherein the piston is moved in 0.1 second can be considered quasistatic to a good approximation. Consider a pistoncylinder, Fig. 3.4, that contains a gas at initial pressure p1 . When the piston is moved suddenly to the right the molecules near the face of the piston move to the space made just available, and therefore the pressure near the piston decreases. Because of this low pressure, the system (gas) is not longer in equilibrium since a higher pressure exists at the other (left hand) end of the cylinder. Such a process is called non quasiequilibrium. However, if the piston is moved slowly, the molecules have sufficient time to redistribute and the pressure in the entire volume of the gas remains always uniform. Since equilibrium within the gas prevails at all times, this is a quasiequilibrium or quasisteadystate process. In order to carry out a reversible expansion of the compressed gas a series of infinitesimal reversible steps would have to be realised as shown in Fig. 3.4.
94
3.1 Irreversible and Reversible Processes
Fig. 3.4: Gas expansion process; Left – Irreversible (nonquasiequilibrium) expansion, Right – Reversible (quasiequilibrium) expansion.
The work done to the system (gas) during the process is
Lrev = −
final Z state
p · dV = Lmax
(3.1)
initial state
The maximum work available from such an expansion from initial to final states is obtained when the process is carried out reversibly. The irreversibility of a process is caused by the occurrence of the followings: (a) friction, (b) heat transfer through a finite temperature difference, (c) throttling (gas expansion without any work obtained), (d) nonquasi equilibrium compression or expansion, (e) electrical resistance (dissipation of heat), (f) chemical reactions (destruction of chemical bonds).
95
3 Equilibrium Thermodynamics
3.2 Entropy In Chapter 2 we have introduced the first law of thermodynamics and one of the most useful formulations of this law (see Eq. (2.52)) is: dQt + dL = dU
(3.2)
where U stands for the internal energy and L is the absolute work done to the system. Symbol Qt (t stands for total) represents the total heat absorbed by the system: dQt = dQf + dQ (3.3) where Q is the amount of heat supplied from the surroundings or any other heat source, and Qf is the amount of heat generated within the system due to friction. All these terms are expressed in joules. We have already learned that internal energy (U ) is a state variable and its differential is exact while neither heat (Q) nor work (L) are state variables and their differentials are inexact. However, T1 is an integrating factor for dQt turning it into an exact differential called entropy (S): dS =
dQt T
(3.4)
where dQt is the infinitesimal amount of heat exchanged and T is the temperature of the quasiequilibrium, reversible process (see Example 3.1). Entropy is an extensive property so S = m·s (3.5) where S is the total entropy of the system in J/K, m is the mass of the system in kilograms and s is the specific entropy of the system in J/kg · K. Therefore ds =
dqt T
(3.6)
Example 3.1 The objective of this example is to demonstrate how an inexact differential can be transformed into an exact one. We wish to improve the understanding of relationships (3.4) and (3.6) which transform an inexact differential of heat into an exact differential of entropy. Consider the infinitesimal quantity: dG ¯ = α · dx + β ·
96
x · dy = α · dx + β · x · d(ln y) y
3.2 Entropy Is dG ¯ an exact differential? To check this, we have to calculate the second derivatives: ∂ x β ∂ (α) = 0 and β· = ∂y ∂x y y 2
2
∂ G ∂ G 1 since ∂y ∂x 6= ∂x ∂y the infinitesimal quantity is not an exact differential . Thus, the integral of dG ¯ should be path dependent. Let "i" denote the initial point (1, 1) and "f " the final point (2, 2). If one then calculates the integral of dG ¯ along the path i → a → f passing through the point "a" with coordinates (2, 1) one obtains:
Z
i→a→f
dG ¯ =
Zf
α dx + β
x 2 dy = α · (2 − 1) + 2 β ln = α + 2 β ln 2 y 1
(A1)
i
Fig. 3.5: Integration paths from i → f
Let us calculate the same integral along the path i → b → f : Z Z 2 x dG ¯ = α dx + β dy = α · (2 − 1) + β · 1 · ln = α + β ln 2 (A2) y 1 i→b→f
i→b→f
We can also calculate the integral along a direct path from (1, 1) to (2, 2). To this 1
The crossbar over letter d¯ indicates that the differential dG ¯ is inexact. In some textbooks the inexact differentials of heat and work are marked as dQ ¯ and dL ¯ to distinct them from the exact differentials of internal energy dU and enthalpy dH [3, 5].
97
3 Equilibrium Thermodynamics end, we introduce a parametric variable t so: x=t
1≤t≤2
y=t Z
i→line→f
Z t dG ¯ = α dt + β dt = (α + β) · (2 − 1) = (α + β) t
(A3)
As it can be clearly seen, we have obtained three different answers (A1), (A2) and (A3), depending on the integration path. Thus, the integrals are different, the quantity dG ¯ is not an exact differential, and G is not a state function in the x, y space. Let us make a new infinitesimal quantity dF by multiplying dG ¯ by dF =
1 x
so:
α β dG ¯ = dx + dy. x x y
The new infinitesimal dF is an exact differential since: ∂ β ∂ α =0 and =0 so ∂y x ∂x y
∂2F ∂2F = ∂x∂y ∂y∂x
Integrating dF along any path between "i" (1, 1) and "f " (2, 2) should give us the same answer. To confirm this, let us calculate the integral along several paths. Z Z α β dF = dx + dy = α ln 2 + β ln 2 = (α + β) ln 2 (B1) x y i→a→f i→a→f Z Z α β dF = dx + dy = β ln 2 + α ln 2 = (α + β) ln 2 (B2) x y i→b→f
i→b→f
Z
i→line→f
dF =
Z
i→line→f
β α dx + dy x y
=
Z2 1
= α ln 2 + β ln 2 = (α + β) ln 2
α β dt + dt t t
(B3)
The above integral does not have to be integrated along a straight line path. The integral can be evaluated along any path. Let us calculate the integral along a parabola: y = (x − 1)2 that goes through points "i" (1, 1) and "f " (2, 2). In order
98
3.2 Entropy to perform the integration we introduce a parametric variable t so that: x=t
1≤t≤2 2
y = (t − 1) + 1;
dy = 2 (t − 1) dt
Thus, the integration gives: Z Z dF = i→parabola→f
i→parabola→f
=
Z2
α β dx + dy x y
α β dt + · 2 (t − 1) dt t (t − 1)2 + 1
1
= α ln 2 + β
Z2
2 (t − 1) dt = α ln 2 + β ln 2 = (α + β) ln 2 (t − 1)2 + 1
1
(B4) R The integral i→f dF has been evaluated along four different paths and all the answers are identical ((B1), (B2), (B3), (B4)), as one would expect. The quantity dF is an exact differential and F is a state function in the x, y space. Since we know that dF is an exact differential, we can easily calculate the function F (x, y). Knowing that ∂F (x, y) α = ∂x x
and
∂F (x, y) β = ∂y y
then F (x, y) =
Z
α dx + f (y) = α · ln x + f (y) x
and ∂F (x, y) ∂ ∂f (y) β = [α · ln x + f (y)] = = ∂y ∂y ∂y y so f (y) =
Z
β dy = β · ln y y
99
3 Equilibrium Thermodynamics Thus, the function F (x, y) is of the form: F (x, y) = α · ln x + β · ln y + const. Knowing the function F (x, y) we can check that: Zf
dF = F (f ) − F (i) = F (2, 2) − F (1, 1)
i
It is so since: F (f )−F (i) = α · ln 2+β · ln 2+const.−(α · ln 1+β · ln 1+const.) = (α+β) · ln 2 The reader should note that F (f ) − F (i) = (α + β) · ln 2 is in full agreement with the results of the integration along several paths (B1, B2, B3, B4). There is another important observation to be made in conjunction with this example. The infinitesimal differential dG ¯ is not exact. However, when the differential 1 dG ¯ has been multiplied by x we have obtained a new differential dF that is exact. In mathematics the factor x1 is called the integrating factor. The integrating factor transforms an inexact differential into an exact differential. In the same way an inexact differential of heat may be transformed into an exact differential (entropy) using the integrating factor T1 . End of Example 3.1 Now our task is to derive appropriate mathematical relationships that would allow for calculating specific entropy as a function of temperature and pressure of the system s = s(T, p) and as a function of temperature and specific volume s = s(T, v). We begin with deriving the exact differential for specific entropy as a function of temperature and pressure s = s(T, p). To this end we introduce into the definition of the specific entropy (Eq. (3.6)) the first law of thermodynamics (Eq. (2.61)): dh − v · dp (3.7) ds = T Introducing relationship (2.33) into the above equation we obtain: ! 1 v ∂v 1 ∂h · dT + · v − T · · dp − · dp (3.8) ds = · T ∂T p T ∂T p T 100
3.2 Entropy and after simple algebra: cp ∂v · dp · dT − ds = T ∂T p
(3.9)
Thus, knowing the equation of state v = v(T, p) one can evaluate the derivative ∂v ∂T p and calculate the exact differential of specific entropy as a function of temperature and pressure. The procedure for deriving the exact differential of entropy as a function of temperature and specific volume is similar. Into the definition of specific entropy (Eq. (3.6)) we introduce the first law of thermodynamic (Eq. (2.59)) obtaining: ds =
du + p · dv T
(3.10)
and making use of Eq. (2.21) and after some algebra one obtains: cv ∂p ds = · dv · dT + T ∂T v
(3.11)
Eqs. (3.9) and (3.11) are the basis for calculating the entropy of any substance as a function of temperature and pressure, or temperature and specific volume, respectively. Similarly to enthalpy and internal energy we can only calculate the entropy change with respect to a certain reference state s0 = s(T0 , p0 ) since ZT,p
s(T, p) = s0 +
T0 ,p0
and s(T, v) = s0 +
ZT,v
T0 ,v0
ZT,p
ds = s0 +
T0 ,p0
ds = s0 +
ZT,v
T0 ,v0
! cp ∂v · dp · dT − T ∂T p cv ∂p · dv · dT + T ∂T v
(3.12)
(3.13)
Thus, to calculate entropy we need to know the specific heats and the equation of state of the substance. The integration can be carried out along any path from (T0 , p0 ) to (T, p) or (T, v).
101
3 Equilibrium Thermodynamics
3.2.1 Entropy of Liquids and Solids Liquids and solids are incompressible and therefore ∂v =0 and cp = cv = c ∂T p
(3.14)
Eq. (3.12) can be then simplified to
s = s0 +
ZT
c · dT T
(3.15)
T0
and if the specific heat is independent of temperature we obtain: s = c · ln
T T0
(3.16)
assuming that s0 = 0.
3.2.2 Entropy of Ideal Gases Invoking again the equation of state for ideal gases: p·v =
R·T M
(3.17)
from which we can easily obtain: R ∂v = ∂T p p · M
(3.18)
and inserting Eq. (3.18) into Eq. (3.12) we obtain:
s = s0 +
ZT,p
cp R · dT − · dp T M ·p
T0 ,p0
(3.19)
Since cp of ideal gases is independent of pressure the above integral can be written as: Zp ZT cp R dp · dT − · (3.20) s = s0 + T M p T0
102
p0
3.2 Entropy If then cp is independent of temperature we obtain: s = s0 + cp · ln
R p T − · ln T0 M p0
(3.21)
In the above equations s stands for specific entropy in J/kg · K, T is the temperature in Kelvin, cp is the specific heat at constant pressure in J/kg · K. In many text books on thermodynamics, the reader can find similar expressions for specific entropy (s) expressed in J/kmol · K written as: s = s0 + Cp · ln
T p − R · ln T0 p0
(3.22)
where Cp is the specific heat at constant pressure in J/kmol · K. Using Eq. (3.13) one may derive similar expressions for calculating specific entropy as a function of temperature and specific volume: s = s0 +
ZT
T0
R v cv · dT + · ln T M v0
(3.23)
and if the specific heat at constant volume is independent of temperature the following is applicable: s = s0 + cv · ln
T R v + · ln T0 M v0
(3.24)
3.2.3 Entropy of Phase Transition at the Transition Temperature We consider a system and its surroundings at a temperature at which two phases are in equilibrium. Such a temperature is called transition temperature and is denoted by Ttrs . For example, this temperature is 0 ◦ C for ice in equilibrium with water at 1 bar, and 100 ◦ C for water in equilibrium with its vapour at 1 bar. At the transition temperature any transfer of heat between the system and its surroundings is reversible since the two phases in the system are in equilibrium. Because our system is at a constant pressure, the change of the specific entropy of the system is: ∆trs s =
∆trs h Ttrs
or
∆trs s =
∆trs h Ttrs
(3.25)
103
3 Equilibrium Thermodynamics where ∆trs h and ∆trs h are the enthalpy of the transition in J/kg and J/kmol, respectively. Standard (at p = 1 bar) transition enthalpies and transition temperatures are listed in the literature; Table 3.1 is an extract from reference [13]. If the phase transition is exothermic (condensation or freezing) then the entropy change is negative. This reflects the fact that during condensation or freezing the system is becoming more ordered. For endothermic transitions the entropy change is positive and the system becomes more disordered. Table 3.1: Transition temperatures, standard enthalpies and standard entropies (p=1bar) of selected substances [13].
Substance
H2 O O2 N2 CO2 CH4 C2 H6 Cl2
Temperature K 273.15 54.36 63.15 90.68 89.85 172.1
Melting ∆trs h
∆trs s
kJ/mol 6.008 0.444 0.719
J/mol · K 22.00 8.17 11.39
0.941 2.86 6.41
10.38 31.83 37.25
Vaporization Tempe∆trs h ∆trs s rature K kJ/mol J/mol · K 373.15 40.656 108.95 90.18 6.820 75.63 77.35 5.586 72.22 194.6 25.23 129.65 111.7 8.18 73.23 184.6 14.7 79.63 332.4 29.45 88.60
3.2.4 The Third Law of Thermodynamics At T = 0 K all energy of thermal motion has been ceased and substances are in solid state. At T = 0 pure substances are arranged in regular, crystal forms in a perfect order. The third law of thermodynamics states that entropy of a pure crystalline substance at absolute zero temperature is zero2 : s(T = 0) = 0
(3.26)
3.2.5 Absolute Entropy of Pure Substances The third law of thermodynamics specifies zero entropy at absolute zero temperature for pure substances. By measuring specific heat capacities cp at different 2
For detailed discussions on the third law of thermodynamics the reader is requested to consult textbooks [3, 5, 13].
104
3.2 Entropy temperatures and evaluating the integrals in Eq. (3.20) the specific entropy (called also absolute entropy) of any pure substance can be evaluated:
s(T, p) =
ZT
cp R · dT − · T M
Zp
dp p
(3.27)
p0
0
with the second integral vanishing for liquids and solids. The entropy of phase transition must be added for each phase transition between T = 0 and the temperature of interest T , thus s(T, p0 = 1 bar) =
ZTm 0
cp ∆melt h · dT + + T Tm
ZTb
∆boiling h cp · dT + + T Tb
Tm
ZT
cp · dT (3.28) T
Tb
c
Fig. 3.6 shows a typical dependence of Tp with temperature as well as specific c entropy as a function of temperature. The area under the Tp curve is the specific entropy. Near T = 0 K it is difficult to measure cp and in this region an extrapolation, called Debye extrapolation [3, 5, 13], is used. The reader may recall that for enthalpies we have chosen a reference state at T0 = 298.15 K and p = 1 bar. Furthermore we have agreed that h(T0 , p0 ) = 0 and the formation enthalpies of pure O2 , H2 , N2 , pure graphite and pure sulphur are given zero values (see Table 2.6). Using Eq. (3.28) we can evaluate specific entropies of theses substances, and any others, at the reference state T0 = 298.15 K and p0 = 1 bar. Example 3.2 [13] shows a method of obtaining the specific entropy for nitrogen at the reference state. Table 2.6 lists specific entropies (s0298 )3 at standard conditions for several pure substances. These entropies are called thirdlaw entropies. In Paragraph 2.2.2.3 we have recommended the Cp polynomials (see Eq. (2.46)) obtained using Joint ArmyNavyAir Force (JANAF) tables for enthalpies (see Table 2.4 and Table 2.5) of gaseous species. The specific entropy C can be easily obtained upon integrating Tp , thus s0T
=
s0298
+
ZT
298
′
Cp · dT = Cp,7 · R +
ZT
Cp · dT ′ T′
(3.29)
298
where an additional constant Cp,7 · R (in kJ/kmol · K) is needed to make sure that at the reference temperature of 298.15 K the integration results in a value 3
In textbooks on chemical thermodynamics the thirdlaw entropies (in J/mol · K) are usually 0 denoted as S298 .
105
3 Equilibrium Thermodynamics corresponding to the thirdlaw entropy.
Boiling
(b) S
Dboil s
Solid
Liquid
Debye approximation
Melting
(a)
cp/T
Dmelt s Gas S(0)
Tm
Tb
T
0
Tm
Tb
T
Fig. 3.6: Determination of absolute specific entropy of a pure substance [13]. c Left – the variation of Tp with the temperature; Right – the specific entropy
Example 3.2 Evaluate the standard entropy of nitrogen gas at T0 = 298.15 K and p = 1 bar. The following data is needed: Debye extrapolation Integration from 10 K to 35.61 K Phase transition at 35.61 K Integration from 35.61 K to 63.14 K Melting at 63.14 K Integration from 63.14 K to 77.32 K Boiling at 77.32 K Integration from 77.32 K to 298.15 K Total
s0T0 in J/mol · K 1.92 25.25 6.43 23.38 11.42 11.41 72.13 39.20 191.14
Compare the value obtained with the value listed in Table 2.6. End of Example 3.2
106
3.3 The Second Law of Thermodynamics
3.3 The Second Law of Thermodynamics 3.3.1 The Increase in Entropy Principle While considering reversible and irreversible processes (Section 3.1) we have already quoted two formulations of the second law of thermodynamics. However, none of these formulations have proven general enough for determining direction of the thermodynamic processes. To this end the notion of entropy is needed. Since we have just learned how to calculate entropy we can reformulate the second law of thermodynamics as follows: In a real irreversible process the sum (π) of the changes of the entropy of all bodies participating in the process, including the surroundings, is positive: π>0 (3.30) For reversible processes: (3.31)
π=0
However it is not possible to carry out a process for which π is less than zero. Thus, π>0
for irreversible processes,
π=0
for reversible processes,
π g2 then the minimum value of G is achieved when all the substance transforms into phase 2, so that G = m · g2 . Phase 2 is then the stable one. If T and p are such that g1 = g2 , then Eq. (3.66) is satisfied and any amount m1 of phase 1 can coexist with the remaining amount m2 = m − m1 of phase 2. Thus, the G value remains unchanged when m1 is varied. The locus of points where T and p are such that conditions (3.70) are satisfied is the phaseequilibrium curve, Fig. 3.11. Along such a curve the two phases can coexist in equilibrium. The phaseequilibrium curve divides the p, T plane into two regions: one with g1 < g2 , so that phase 1 is the stable phase, and the other region where g1 > g2 , so that phase 2 is the stable one. What happens if g1 > g2 ? Obviously the two phases are not in equilibrium. Since the process must proceed 6
One should demonstrate that G is actually minimum. For this the reader is refered to textbooks on thermodynamics.
117
3 Equilibrium Thermodynamics so as to satisfy the requirement dG < 0 then (g1 −g2 ) · dm1 ≤ 0 ((g1 −g2 ) · dm1 = 0 at the equilibrium only), so dm1 must be negative, which means that some amount of the substance in phase 1 must undergo transition to phase 2 until g1 = g2 . We have just made an important observation. The difference in the Gibbs specific free enthalpy, namely g1 −g2 , is the driving force for phase change of a single substance, just as the temperature difference is the driving force for heat transfer [15].
Fig. 3.11: Pressuretemperature plot showing the phaseequilibrium curve that defines stability regions for phases 1 and 2
It is possible to derive a differential equation of the phaseequilibrium curve. We begin the derivation with recalling that the conditions along the phaseequilibrium curve are: g1 (T, p) = g2 (T, p) (3.71) or dg1 (T, p) = dg2 (T, p)
(3.72)
Invoking the definition of the Gibbs free enthalpy: g = h − T ·s
(3.73)
dg = dh − T · ds − s · dT
(3.74)
we obtain its differential as
Using the first law of thermodynamics (see Eq. (2.61)) we can replace dh with
118
3.5 Equilibrium Between Phases dqt + v · dp so, dg = dqt + v · dp − T · ds − s · dT
(3.75)
that simplifies further to dg = v · dp − s · dT
(3.76)
Since conditions (3.72) must be fulfilled at equilibrium we obtain: v1 · dp − s1 · dT = v2 · dp − s2 · dT
(3.77)
and after simple algebra (v1 − v2 ) · dp = (s1 − s2 ) · dT
(3.78)
s2 − s1 ∆s dp = = dT v2 − v 1 ∆v
(3.79)
The above equation is called ClausiusClapeyron equation. It relates the slope of the phaseequilibrium curve to the entropy change ∆s and (specific) volume change ∆v of the substance that undergoes a change of phase at a point lying on the phaseequilibrium curve. Eq. (3.79) is written for one kilogram of phase 1 and ∆s phase 2, but since the right hand side ( ∆v ) is a ratio, the equation is also valid for ∆S extensive properties ∆V . Since the entropy change due to the phase transition is ∆s =
∆htrs Ttrs
(3.80)
where ∆htrs is the latent heat of the phase change (per kg of the substance) and Ttrs is the temperature of the phase change. Thus, ∆htrs dp = dT Ttrs · ∆v
(3.81)
3.5.2 Phase Transformations of a Pure Substance Fig. 3.12 shows a phaseequilibrium diagram for water with phasecurves separating solid from liquid, liquid from gas and solid from gas. The gas phase is sometimes called the vapour phase. The three phaseequilibrium curves meet a common point called the triple point. The triple point of water lies at 273.16 K and 611 Pa and at this unique temperature and pressure arbitrary amounts of all three phases can coexist in equilibrium with each other (This property makes the triple point of water suitable as an easy reproducible standard for temperature).
119
3 Equilibrium Thermodynamics
Fig. 3.12: The experimentally determined phase diagram for water. Note the change of scale at 2 atm [13]
120
3.5 Equilibrium Between Phases At the critical point the liquidgas equilibrium curve ends since the volume change ∆v between liquid and gas approaches zero. Beyond the critical point there is no further phase transformation since there exists only one "fluid phase" which can be characterised by the very dense gas that is indistinguishable from the liquid. Note that at high pressures there exist several ice forms, all solid phases, which are named ice II, III, V, and VI. These specific ice phases are formed as a result of modifications between ice molecules as a result of stress. During melting processes (going from solid to liquid), the entropy of the substance almost always increases (degree of disorder increases) and the corresponding latent heat is positive indicating that the phase change is endothermic. In most cases the solid expands upon melting so ∆v > 0. However, water contracts upon melting so that ∆v < 0 and therefore the slope of the melting curve for water is negative.
Example 3.3 In this Example we will reproduce the phaseequilibrium curve for water in the vicinity of the triple point. We will deal with equilibrium between liquid water and its vapour. Using ClausiusClapeyron equation we will derive an approximate expression for the pressure of a vapour in equilibrium with the liquid at a temperature T . We begin with recalling ClausiusClapeyron equation: dp l = dT T · ∆v
(A1)
where l is the latent heat of evaporation (2.45 · 103 kJ/kg) and ∆v = vgas − vliquid is the change in the specific volume (in m3 /kg). Since the gas (vapour) is much less dense than liquid, vgas ≫ vliquid and we may write ∆v ∼ = vgas . We may assume that the gas can be treated as an ideal gas, so that its equation of state is simply R·T p·v = (A2) M Inserting (A2) into (A1) leads to l · M dT dp = · 2 p R T
(A3)
The above equation can be easily integrated: ln p = −
l·M + constant R·T
(A4)
121
3 Equilibrium Thermodynamics The integration constant can be evaluated knowing the parameters of the triple point, so constant = ln ptr +
l·M R · Ttr
(A5)
Thus, the expression for the vapour pressure as a function of temperature is: 1 l·M 1 p =− · − (A6) ln ptr R T Ttr and finally: p = ptr · e
− l ·RM
1 T
− T1
tr
(A7)
For ptr = 611 N/m2 , Ttr = 273.16 K, l = 2.45 · 103 kJ/kg, M = 18 kg/kmol, R = 8.314 kJ/kmol · K we obtain: 1 1 p = 611 · exp −5304.3 · − (A8) T 273.16 The reader can easily verify the correctness of this formula for low and medium temperatures using the phase diagram of water shown in Fig. 3.12. For example for T = 373.16 K the formula yields p = 1.111 · 105 Pa that is close to the expected pressure of 1 atmosphere. The formula breaks down for temperatures approaching the critical point since the ideal gas equation is not longer valid. End of Example 3.3
3.5.3 Dependence of Gibbs Free Enthalpy on Temperature and Pressure Since Gibbs free enthalpy is such an important property it is instructive to examine how it varies with temperature and pressure. Thus, we wish to examine ∂G ∂G the dependence of ∂T p and ∂p T with T and p. We begin with writing down the general form of the differential of G(T, p) as follows ∂G ∂G · dT + · dp dG = ∂T p ∂p T
(3.82)
Differentiating Eq. (3.58), which is the definition of Gibbs free enthalpy, we obtain dG = dH − T · dS − S · dT
122
(3.83)
3.5 Equilibrium Between Phases and introducing the first law of thermodynamics (dH = dQt +V · dp) into Eq. (3.83) we obtain: (3.84)
dG = −S · dT + V · dp
The above expression states that the change in G is proportional to the change in dT and dp with (−S) and (V ) being the slopes of the plots of G against T and p, respectively. When this expression is compared to Eq. (3.82) we observe that: ∂G = −S ∂T p
and
∂G =V ∂p T
(3.85)
Similar considerations on specific Gibbs free enthalpy (g) leads to the following relationships ∂g = −s ∂T p
and
∂g =v ∂p T
(3.86)
Relationships (3.85) and (3.86) are important since they determine the dependence of Gibbs free enthalpy with temperature and pressure. Example 3.4 Make graphs showing the dependence of the specific Gibbs enthalpy as a function of temperature and pressure for ice, water and water vapour. Locate the regions where each of these phases is stable. Assume that the specific heats of ice and water are equal Cp = 75.3 kJ/kmol · K while for water vapour Cp = 33.58 kJ/kmol · K. The table below shows the thermodynamic data. Thermodynamic data of ice, water and water vapour
Phase Ice Water Water vapour
Standard formation enthalpy kJ/mol −291.83 −285.83 −241.81 (see also Table 2.6)
Standard entropy J/mol · K 47.98 69.95 188.72
The molar Gibbs potential (in kJ/kmol) of ice as a function of temperature and
123
3 Equilibrium Thermodynamics pressure can be calculated as follows: g ice (T, p) =
0 (hf,298 )ice
+
ZT
T0
T 0 Cp,ice · dT − T · (s298 )ice + Cp,ice · ln T0 ′
(C1)
0
where (hf,298 )ice = −291.83 · 103 kJ/kmol and Cp,ice = 75.3 kJ/kmol · K and (s0298 )ice = 47.98 kJ/kmol · K. Similarly for the molar Gibbs potential for water we obtain: g water (T, p) =
0 (hf,298 )water +
ZT
T0
T 0 Cp,water · dT −T · (s298 )water + Cp,water · ln T0 ′
0 where (hf,298 )water = −285.83 · 103 kJ/kmol and (s0298 )ice = 69.95 kJ/kmol · K.
and
Cp,water
(C2) = 75.3 kJ/kmol · K
Assuming that water vapour can be treated as an ideal gas one obtains: g vapour (T, p) =
0 (hf,298 )vapour +
ZT
Cp,vapour · dT ′
T0
−T ·
p T − R ln (s0298 )vapour + Cp,vapour · ln T0 p0 (C3)
0
where (hf,298 )vapour = −241.81 · 103 kJ/kmol and Cp,vapour = 33.58 kJ/kmol · K, (s0298 )vapour = 188.72 kJ/kmol · K and R = 8.314 kJ/kmol · K. In Eqs. (C1), (C2) and (C3) the reference state parameters are T0 = 298.15 K and p0 = 1 bar (105 N/m2 ). Fig. 3.13 shows the dependence of the specific Gibbs enthalpies with temperature for a pressure of 1 bar. It is interesting to observe that the curves for ice and water cross at the melting temperature of 273 K while the curves for water and water vapour cross at the boiling temperature of 373 K. The regions where the Gibbs potential for ice is the lowest among the three potentials corresponds to conditions under which ice is the only phase stable. Similarly the region for water being the only stable phase can be easily identified.
124
3.5 Equilibrium Between Phases
Fig. 3.13: The variation of the Gibbs enthalpy with temperature for ice (function (C1)), water (function (C2)) and water vapour (function (C3)) at 1 bar.
Fig. 3.14: The variation of the Gibbs enthalpy with temperature for ice (function (C1)), water (function (C2)) and water vapour (function (C3)) at a pressure of 611 Pa.
It is instructive to plot functions (C1), (C2) and (C3) for pressure of 611 Pa that
125
3 Equilibrium Thermodynamics is the triple point pressure. Such a plot is shown in Fig. 3.14. Not surprisingly the three curves cross at one point corresponding to the presence of the three phases. The regions where ice and water vapour are the only stable phases can be easily identified. Comments: (a) Specific Gibbs potential for ice (solid) and water (liquid) is independent of pressure. (b) Specific Gibbs potential for water vapour (gas) is strongly dependent on pressure. End of Example 3.4
3.5.4 Equilibrium in MultiComponent SinglePhase Systems In this paragraph we consider singlephase systems consisting of minimum two components that do not react. An example of such a system is a process of mixing of two (or more) gases. Another example concerning liquid phase is a process of making solutions like wateralcohol mixtures. For a multicomponent system Gibbs free enthalpy is a function of temperature, pressure and the amounts of different components so G = G(T, p, m1 , m2 , . . . , mk , . . . , mN )
(3.87)
where mk stands for the amount (in kg) of kcomponent while G is the Gibbs free enthalpy (in joules) for the whole system. We can also write Eq. (3.87) as g = g(T, p, w1 , w2 , . . . , wk , . . . , wN )
(3.88)
where g stands for specific Gibbs enthalpy (in J/kg) while wk is the mass fraction of kcomponent (equally well we may use here molar quantities replacing the specific Gibbs enthalpy g with molar Gibbs enthalpy g, and replacing the mass fractions wk with the molar fractions xk , respectively). The exact differential of Gibbs free energy expressed using Eq. (3.87) can be written as:
dG
126
=
∂G · dT ∂T p,mk
+
∂G · dp ∂p T,mk
+
∂G · dm1 + ∂m1 p,T,mk ,k6=1
3.5 Equilibrium Between Phases ∂G ∂G · dm2 + . . . + · dmk + ∂m2 p,T,mk ,k6=2 ∂mk p,T,ml ,l6=k
∂G · dmN ... + ∂mN p,T,mk ,k6=N
and using Eq. (3.85)
N X ∂G dG = −S · dT + V · dp + · dmk ∂mk p,T,mj = −S · dT + V · dp +
k=1 N X
µk · dmk
j 6= k
(3.89)
(3.90)
k=1
Similarly for specific Gibbs energy we obtain: N X ∂g dg = −s · dT + v · dp + · dck ∂ck p,T,mj k=1
= −s · dT + v · dp +
N X
µk · dck
j 6= k
(3.91)
j 6= k
(3.92)
k=1
Eqs. (3.90) and (3.91) introduce a new variable µk ∂G ∂g µk = = ∂mk p,T,mj ∂ck p,T,wj
which is known as the chemical potential. Expressions (3.90) and (3.91) are the fundamental equations of equilibrium and chemical thermodynamics. The chemical potential µk which plays an important role in chemical thermodynamics is an intensive variable that is in general a function of the state of the system, as given by p and T , and the wk ’s. Even though a species is not present in a system, its chemical potential nevertheless needs not to be zero since there is always the possibility of introducing it into the system. In this case G and therefore also g will be altered and the value of the corresponding µk must be different from zero. Gibbs potential G is an extensive property as shown by Eq. (3.87). Consider a system that is αtimes larger than that described by Eq. (3.87), that means that amounts of the species have been increased by a factor of α, G = G(T, p, α · m1 , α · m2 , . . . , α · mN )
(3.93)
127
3 Equilibrium Thermodynamics We expect that G(T, p, α · m1 , . . . , α · mk , . . . , α · mN ) = α · G(T, p, m1 , . . . , mk , . . . , mN ) (3.94) Differentiating the above equation with respect to α (at constant T and p) gives, N X k=1
∂(α · mk ) ∂G · = G(T, p, m1 , . . . , mk , . . . , mN ) ∂(α · mk ) ∂mk
(3.95)
and using Eq. (3.92) we obtain N X
µk (T, p) · mk = G(T, p, m1 , . . . , mk , . . . , mN )
(3.96)
k=1
and after dividing the above equation by the mass of the system: N X
µk (T, p) · wk = g(T, p)
(3.97)
k=1
The above equation allows to calculate specific Gibbs potential (in J/kg) knowing the chemical potentials (in J/kg) of the components of the system and their mass fractions wk . Alternatively, one may calculate molar Gibbs energy g (in J/kmol) using molar chemical potentials µk (in J/kmol) and molar fractions xk : N X
µk (T, p) · xk = g(T, p)
(3.98)
k=1
GibbsDuhem Equation The exact differential of Gibbs potential has been already formulated by Eq. (3.90) and is here copied for your convenience: dG = −S · dT + V · dp +
N X
µk · dmk
(3.99)
k=1
The differential shows how the Gibbs potential changes upon varying the temperature (T ), pressure (p) and amounts of each component (mk ) of the singlephase considered. Thus, relationship (3.99) is indeed general. Relationship (3.96) which is to calculate Gibbs potential of the system knowing the chemical potentials of all the components and their amounts is also general. It can be differentiated
128
3.5 Equilibrium Between Phases giving N N N X X X mk · dµk µk · dmk + µk · mk ) = dG = d( k=1
k=1
k=1
(3.100)
At equilibrium, in both relationships (3.99) and (3.100), dG = 0 and therefore −S · dT + V · dp +
N X
µk · dmk =
N X
µk · dmk +
k=1
k=1
N X
mk · dµk
(3.101)
k=1
then finally −S · dT + V · dp −
N X
mk · dµk = 0
(3.102)
k=1
The above relationship is referred to as GibbsDuhem equation for the system at equilibrium. It states that the intensive properties (T, p, µk ) of the system at equilibrium are not independent but they are interrelated. For a singlephase system consisting of N components, there are N+2 intensive properties and N+1 are independent only. The GibbsDuhem equation can be regarded as the source of the phase rule (see below). GibbsDuhem equation shows also that in a singlephase equilibrium mixture, the chemical potential of a component of a mixture cannot change independently of the chemical potential of the other components since N X
mk · dµk = 0
for
p = const and T = const
(3.103)
k=1
For example, in a binary mixture consisting of species A and B, if chemical potential of species A is increasing, chemical potential of species B must decrease, since mB dµA = − · dµB (3.104) mA Phase Rule for SinglePhase MultiComponent System For a singlephase system consisting of N (nonreacting) components there exist N+2 independent intensive variables which uniquely describe the thermodynamic state of the system. At thermodynamic equilibrium there are however only N+1 independent variables since GibbsDuhem equation (3.102) must be satisfied. Thus, the simplest form of the phase rule is given by N IV = N + 1 for a single phases system of Ncomponents where N IV stands for a Number of Independent Variables.
129
3 Equilibrium Thermodynamics
3.5.5 Chemical Potential of Pure Substances Eq. (3.92) defines the chemical potential µk . The chemical potential of a pure substance shows how the Gibbs energy of a system changes as the substance is added to it since (∂m · g) =g (3.105) µ= ∂m Thus, chemical potential of a pure substance is simple its specific Gibbs enthalpy and it is expressed in kJ/kg or kJ/kmol. In the latter case it is called molar Gibbs energy. We have already learned how to calculate the specific Gibbs enthalpy (see Eq. (3.65)) (3.106)
g = h − T ·s
and therefore chemical potential of pure substances can be easily calculated since
g(T, p) =
ZT,p
dh − T ·
T0 ,p0
ZT,p
(3.107)
ds
T0 ,p0
Chemical potential for an ideal gas can be readily calculated as follows: µ(T, p) = g(T, p) =
h0f,298 +
ZT
′
cp · dT +T
· (s0298 +
T0
ZT
T0
cp R p · dT ′ − · ln ) (3.108) T M p0
which for a constant specific heat simplifies into µ(T, p) = g(T, p) = h0f,298 +cp · (T −T0 )+T · (s0298 +cp · ln
p T R − · ln ) (3.109) T0 M p0
The above relationship is usually expressed as: µ(T, p) = µ0 (T ) +
p R·T · ln M p0
(3.110)
so the term µ0 (T ) is a function of the temperature only. The term µ0 (T ) is called the standard chemical potential at 1 bar expressed in kJ/kg (or alternatively in kJ/kmol). In order to calculate the chemical potential of a pure liquid we consider the liquid to be in equilibrium with its vapour. Under such conditions the chemical potential of the pure liquid must be equal to the chemical potential of its vapour (see
130
3.5 Equilibrium Between Phases Section 3.5.1) so: µpure
liquid (T, p)
= µ0pure
liquid (T )
+
pvapour RT · ln M p0
(3.111)
Chemical potential of a solid state substance is a function of temperature only since µsolid (T ) = µ0solid (T ) (3.112) It is important to realise that Eqs. (3.108) and (3.110) are applicable to calculating chemical potential of ideal gases. For real gases the true pressure p is replaced by an effective pressure called the fugacity. For further considerations the reader is requested to consult textbooks on physical chemistry [13] or chemical thermodynamics [5, 14, 4].
3.5.6 Significance of Chemical Potential While considering nonadiabatic systems that interact with surroundings by exchanging heat (Q), we have already observed (see Section 3.4.2) that it is not possible to derive general conditions for such systems at thermodynamic equilibrium. It is so because for any nonadiabatic system which is in an initial state, there is an infinite number of ways for the system to reach the (final) equilibrium state. Each way (process path) is associated with an amount of heat exchanged with the surroundings and consequently there exist an infinite number of Qvalues each corresponding to a different path. For each path inequality (3.50) must be satisfied. Realising that we are not able to derive general conditions for thermodynamic equilibrium we have subjected our considerations to some constraints (restrictions) accepting that the conclusions from such considerations are valid under these constraints only. We have already considered nonadiabatic systems of a given (constant) mass interacting with the surroundings so that both the pressure and the temperature of the system remain constant throughout the process of approaching equilibrium. Under the constraints of constant temperature and pressure, we have derived the following conditions for the system to reach equilibrium: for T = const. and p = const. G → Gmin ;
dG = 0
;
N X
µk · dmk = 0
(3.113)
k=1
where
∂G ∂g µk = = ∂mk p,T,mj ∂wk p,T,wj
Consider now another type of constraints; equilibrium at constant temperature
131
3 Equilibrium Thermodynamics and volume. Under such constraints the system reaches equilibrium when F → Fmin or dF = 0 as shown by Eq. (3.64). Helmholtz potential can be written as (3.114)
F = U − T ·S = H − p·V − T ·S = G − p·V and its differential is then
(3.115)
dF = dG − p · dV − V · dp Introducing (3.99) into (3.115) we obtain dF = −S · dT − p · dV +
N X
(3.116)
µk · dmk
k=1
and by writing down the exact differential of F = F (T, V, m1 , . . . , mk , . . . , mN ) as N X ∂F ∂F ∂F dF = · dT + · dV + · dmk (3.117) ∂T p,mk ∂V T,mk ∂mk T,V,mj ,j6=k k=1
we can see that at equilibrium obtained at T = const. and V = const. the following is applicable M X
µk · dmk = 0
k=1
∂F ∂f with µk = = ∂mk T,V,mj ∂wk T,v,wk
j 6= k
(3.118)
where f and wk stand for the specific Helmholtz potential (in kJ/kg) and mass fraction of kspecies, respectively. Summarising, under constraints of constant temperature and volume we have obtained the following equilibrium conditions: for T = const. and V = const. F → Fmin ;
dF = 0
;
N X
(3.119)
µk · dmk = 0
k=1
where
∂f ∂F = µk = ∂mk p,T,mj ∂wk p,T,wj
j 6= k
We can also consider another type of constraints approaching equilibrium under constant entropy and volume. By considerations similar to these presented above, it is possible to demonstrate that the equilibrium conditions for S = const. and
132
3.5 Equilibrium Between Phases V = const. are U → Umin ;
dU = 0
;
N X
(3.120)
µk · dmk = 0
k=1
where
∂u ∂U = µk = ∂mk p,T,mj ∂wk p,T,wj
j 6= k
where U and u are the internal energy (in kJ) and specific internal energy of the system (in kJ/kg), respectively. Now it is time to examine the equilibrium conditions obtained. Eqs. (3.113), (3.119) and (3.120) have been derived following three different ways of reaching equilibrium or in other words by considering three7 different constraints. Remarkably, the same relationship N X
µk · dmk = 0
(3.121)
k=1
appears in each of the equilibrium conditions of Eqs. (3.113), (3.119) and (3.120). Thus, we may expect that when the equilibrium is reached, relationship (3.121) must be satisfied, regardless of the path the singlephase system follows while interacting with surroundings. This is a powerful observation; thermodynamic equilibrium does not depend on the path the system follows from the initial to the final (equilibrium) state. It is remarkable that in Eqs. (3.113),(3.119) and (3.120) the same function, the chemical potential µk appears. We have already learned that the chemical potential shows how the Gibbs function changes when the composition changes (under constant temperature and pressure), as shown by Eq. (3.92). By deriving Eqs. (3.119) and (3.120) we have learned that the chemical potential also shows how Helmholtz potential and internal energy of the system change although again under different set of conditions. This is why the chemical potential is central to equilibrium thermodynamics and, as we will demonstrate in the next chapter, it is central to equilibrium chemistry or generally to chemistry.
7
We may consider many other constraints but relationship (3.121) always emerges.
133
3 Equilibrium Thermodynamics
3.6 MultiComponent, MultiPhase Systems without Chemical Reactions In this paragraph we consider multicomponent and multiphase nonreacting systems. Any property of a component will have two indices; one describing the phase and the other describing the component according to the convention: iconcerns phase – i=1,2,...,β
(3.122)
kconcerns component – k=1,2,...,α
We begin with systems at constant temperature T and pressure p, so T (1) = T (2) = T (3) = . . . = T (i) = . . . = T (β) = T p
(1)
=p
(2)
=p
(3)
= ... = p
(i)
= ... = p
(β)
=p
(3.123) (3.124)
The methodology of developing equilibrium conditions is going to be rather straightforward. We will formulate a Gibbs function for multiphase, multicomponent systems and we will find the conditions that minimise this function. To find a minimum of such a function we will need a special mathematical tool that is called the method of Lagrangean multipliers. The method, explained in Example 3.5, is used to find a minimum of a function with constraints. Example 3.5 Find a minimum of a function G(x, y) = x2 + y 2 under a constraint x + y = 1. It is easy to see that the global minimum of G, without any constraint, is x = 0 and y = 0. Under the constraint x + y = 1, the minimum is located somewhere else. First we rewrite the constraint as 1−x−y =0
(D1)
Now, we build a new function Y (x, y) as follows: Y (x, y) = G(x, y) + λ (1 − x − y)
(D2)
Y (x, y) = x2 + y 2 + λ (1 − x − y)
(D3)
where λ is called a Lagrangean multiplier. The necessary conditions for function Y (x, y) to have an extremum (either minimum or maximum) under the specified constraint are (The Method of Lagrangean
134
3.6 MultiComponent, MultiPhase Systems multipliers): λ ∂Y (x, y) =0 2x − λ = 0 x= (D4) ∂x 2 ∂Y (x, y) λ =0 2y − λ = 0 y= (D5) ∂y 2 λ λ 1 1 ∂Y (x, y) = 0 1 − x − y = 0 1 − − = 0; λ = 1 and x = ; y = (D6) ∂λ 2 2 2 2 Thus, the extremum of G(x, y) under constrain x + y = 1 is at x = 21 , y = 12 . One can easily verify that it is a minimum. It is worth noting that function Y (x, y) is identical with G(x, y), because the constraints are zero. However, the partial derivatives of Y and G with respect to x and y are different because function Y incorporates the constraints. End of Example 3.5 The Gibbs function for a phase of the system is: Gi = Gi (T, p, mi1 , . . . , mik , . . . , miα )
(3.125)
and for all phases: G=
β X
Gi (T, p, mi1 , . . . , mik , . . . , miα )
(3.126)
i=1
Thus, we should find a minimum of G with respect to the amounts of each component (mik ). However, the minimum should be found with constraints imposing the conservation of mass of each of the components. The constraints reflect the fact that the system, while approaching equilibrium changes mik , however the total amount of each component must remain conserved and equal to the initial amount. So the constraints are mass balances of each species so that: β X
i=1 β X
mi1 = m01 mi2 = m02
(3.127)
i=1
...
135
3 Equilibrium Thermodynamics β X
miα = m0α
(3.128)
i=1
m01 , m02 , m0k , m0α
where indicate the initial amounts of each component. It is easy to see that the number of constraints equals the number of species (α). Now we are going to use the Gibbs function (Eq. (3.126)), α Lagrangean multipliers together with α constraints to build a new function Y (see Example 3.5):
Y =
β X
Gi (T, p, mi1 , . . . , mik , . . . , miα )+
i=1
λ1 ·
β X
mi1
−
i=1
λk ·
m01
!
β X
+ λ2 ·
β X
mi2
−
m02
i=1
mik
i=1
−
m0k
!
!
+ ...+
+ ...+
λα ·
β X
miα − m0α
i=1
The necessary conditions for function Y to have a minimum are: ∂Y =0 – satisfied, since T = const. ∂T p,mi k ∂Y =0 – satisfied, since p = const. ∂p T,mi k ∂Y ∂G1 + λ1 = 0 µ11 = −λ1 = ∂m11 T,p,mi ,k6=1 ∂m11 k ∂G2 ∂Y = + λ1 = 0 µ21 = −λ1 2 ∂m1 T,p,mi ,k6=1 ∂m21
!
(3.129)
(3.130) (3.131) (3.132) (3.133)
k
.. .
∂Y ∂Gβ = + λ1 = 0 ∂mβ1 T,p,mik ,k6=1 ∂mβ1
µβ1 = −λ1
(3.134)
So that µ11 = µ21 = µ31 = · · · = µβ1 meaning the chemical potential of the first species in all the phases must be the same.
136
3.6 MultiComponent, MultiPhase Systems Differentiating further with the respect to mi2 one obtains ∂G1 ∂Y + λ2 = 0 µ12 = −λ2 = 1 ∂m2 T,p,mi ,k6=2 ∂m12 k ∂G2 ∂Y + λ2 = 0 µ22 = −λ2 = ∂m22 T,p,mi ,k6=2 ∂m22
(3.135) (3.136)
k
.. .
∂Y ∂Gβ = + λ2 = 0 β ∂m2 T,p,mik ,k6=2 ∂mβ2
µβ2 = −λ2
(3.137)
So that µ22 = µ22 = µ32 = · · · = µβ2 meaning the chemical potential of the second species in all the phases must be the same. Finally, differentiation with respect to miα gives ∂Y ∂G1 = + λα = 0 µ1α = −λα (3.138) ∂m1α T,p,mi ,k6=α ∂m1α k ∂G2 ∂Y = + λα = 0 µ2α = −λα (3.139) ∂m2α T,p,mi ,k6=α ∂m2α k
.. .
∂Y ∂Gβ = + λα = 0 ∂mβα T,p,mik ,k6=α ∂mβα
µβα = −λα
(3.140)
Thus µ1α = µ2α = µ3α = · · · = µβα the chemical potential of the αspecies in all the phases must be the same. The above equations formulate the general conditions for a multicomponent, multiphase system to be in equilibrium as: T (1) = T (2) = T (3) = . . . = T (i) = . . . = T (β) = T p(1) = p(2) = p(3) = . . . = p(i) = . . . = p(β) = p µ1k = µ2k = µ3k = . . . = µβk
for
k = 1, 2, . . . , α
(3.141)
The general conditions for a multicomponent multiphase system to be at equilibrium are equality of temperatures and pressures of each phase as well as equality of the chemical potentials in each phase for each component.
137
3 Equilibrium Thermodynamics
3.6.1 The Phase Rule We have already demonstrated in Section 3.5.1 that a singlecomponent twophase system may exist in equilibrium at different temperatures (or pressures). However, once the temperature is fixed, the system will reach equilibrium state and all intensive (specific) properties of each phase (except their relative amounts) will be fixed. Thus, a singlecomponent twophase system has one degree of freedom (independent variable) which may be either temperature or pressure. For a singlephase system consisting of N (nonreacting) components there exist N+1 independent variables according to the GibbsDuhem equation (3.102). Consider now a system consisting of α chemically nonreacting components (species) and β phases. The number of possible degrees of freedom is two, for T and p, and α · β for components so: NIV
= 2 + α·β
(3.142)
Eq. (3.141) specifies (β − 1) · α conditions8 to be fulfilled at equilibrium. Furthermore for each phase GibbsDuhem equation must be satisfied taking up β independent variables. Thus, at thermodynamic equilibrium, the number of independent extensive variables (NIV) is given by: NIV
= 2 + α · β − [α · (β − 1) + β] = α − β + 2
(3.143)
For a singlecomponent (α = 1) twophase (β = 2) system discussed in Section 3.5.1, one independent intensive property needs to be specified (N IV = 1). At the triple point however, β = 3 and therefore N IV = 0 that means that none of the properties of a pure substance at the triple point can be varied.
3.7 Thermodynamics of Mixing Mixing of substances is an essential process in chemical engineering and combustion. In this paragraph we are going to examine what happens when one mixes gases and later on liquids. Since the mixing process is spontaneous at thermodynamic equilibrium Gibbs potential must reach a minimum.
8
Since the overall Gibbs potential of the considered system is kept constant, upon specifying α · (β − 1) chemical potential conditions the chemical potential of each species in the last phase is automatically determined. Thus, here we have α · (β − 1) conditions and not α · β.
138
3.7 Thermodynamics of Mixing Mixing of Ideal Gases Consider two containers containing two ideal gases in amounts nA and nB (n is expressed in kmol). Both containers are at temperature T and pressure p. The containers are separated by a partition as shown in Fig. 3.15.
Fig. 3.15: Mixing of two ideal gases
Before mixing, the chemical potential of the two gases have their pure values and the Gibbs enthalpy of the whole system is p p 0 0 G = nA · µA + nB · µB = nA · µA + R T · ln + nB · µB + R T · ln p0 p0 (3.144) where nA and nB are the amounts (in kmol) of gas A and B, respectively whilst µA and µB are their molar chemical potentials, respectively. After mixing, the partial pressure of gases are pA and pB , respectively, however the total pressure remains the same, since p = pA + pB (3.145) During the mixing process the state variables T and p remain constant. Only partial pressures pA and pB change so as to minimise the value of G. Actually, there is only one independent variable (either pA or pB ) since the relationship (3.145) holds. Thus, the Gibbs enthalpy of the mixture can be expressed as a function of pA (or pB ): pA p − pA 0 0 + nB · µB + R T · ln (3.146) G(T, p, pA ) = nA · µA + R T · ln p0 p0 At minimum of G dG(T, p, pA ) = nA · R T ·
dpA dpA − nB · R T · =0 pA p − pA
(3.147)
139
3 Equilibrium Thermodynamics and therefore
nB nA = pA p − pA
(3.148)
and further pA =
nA · p = xA · p nA + nB
and
pB = xB · p
(3.149)
Relationship (3.149) allows for calculating the partial pressure of Acomponent, pA that minimises G function and therefore corresponds to the equilibrium composition. Eq. (3.149) shows that ideal gases mix in all proportions as shown in Fig. 3.16. The system of two gases, shown in Fig. 3.15, has got two components (α = 2) and one single phase (β = 1). Thus, N IV = α − β + 2 = 2 − 1 + 2 = 3 and therefore the pressure, the temperature and the mole fraction of one of the components can be changed without invoking a phase change.
Fig. 3.16: Mixture of two ideal gases A and B
The relationship pi = xi · p valid for mixtures of ideal gases has been used by students for long. Now it has become apparent that this simple relationship has got the basis in equilibrium thermodynamics. An ideal gas is a model gas comprised of imaginary molecules of zero volume. The molecules do not interact. Each chemical species in a mixture of ideal gases has got its own properties which
140
3.7 Thermodynamics of Mixing are uninfluenced by the presence of other species. Consequently any partial molar property (other than the volume) of a consistent species in a mixture of ideal gases is equal to the corresponding molar property of the species as a pure ideal gas at the mixture temperature but at a pressure equal to its partial pressure in the mixture. Therefore for an ideal gas for which its internal energy and its enthalpy are independent of pressure we have X X umixture = xi · ui and hmixture = xi · hi (3.150)
where umixture and hmixture are specific internal energy and enthalpy of mixtures of ideal gases (in kJ/kmol), respectively.
ui and hi stand for the specific internal energy and enthalpy of pure species (in kJ/kmol), respectively while xi is the molar fraction of ispecies. However chemical potential of an ideal gas mixture is a function of the partial pressures of the components since µmixture =
X i
xi · µ =
X
xi · (µ0i + R · T · ln
i
=
X
pi ) p0
xi · µ0i + R · T ·
X
xi · ln
i
xi · p p0
(3.151)
where p stands for the total pressure. It is rather easy to relate chemical potential of a species in a mixture to its potential as a pure substance. Such a relationship is derived in Example 3.6 and it reads (µi )in_mixture = (µi )as_pure_substance + R · T · ln xi (3.152) In chemical thermodynamics, Eq. (3.152) is often used as the definition of an ideal mixture. Example 3.6 In this example we consider nA moles of an ideal gas (A) filling up a container at a temperature T and pressure p, as shown in Fig. 3.17 (left). Its chemical potential is then (µA )pure_substance = (µ0A )pure_substance + R · T · ln
p p0
(E1)
where we use pA = p since only Acomponent is present. Now, we add into the container nB moles of another ideal gas (B), as shown in Fig. 3.17 (right) (since we wish to keep the total pressure p unaltered, the volume of the container increases
141
3 Equilibrium Thermodynamics accordingly). We formulate two questions (a) how has the chemical potential of species A been altered upon addition of species B?, (b) can we relate chemical potential of species A in the mixture (with species B) to the chemical potential of pure species A?
Fig. 3.17: Pure speciesA at temperature T and pressure p; SpeciesA in a mixture with species B at Temperature T and pressure p
Chemical potential of species A in the mixture with species B, at temperature T and total pressure p, is (µA )in_mixture = (µ0A )pure_substance + R · T · ln
pA p0
= (µ0A )pure_substance + R · T · ln
xa · p p0
(E2)
Calculating (µ0A )pure_substance from (E1) and inserting it into (E2) gives (µA )in_mixture = (µA )pure_substance + R · T · ln xA
(E3)
Comments: (a) Upon addition of speciesB, chemical potential of speciesA has decreased, as shown by Eq. (E3). (b) Relationship (E3) is to calculate chemical potential of a species in a mixture of ideal gases. (c) When xA → 1 then (µA )in_mixture = (µA )pure_substance . End of Example 3.6
142
3.7 Thermodynamics of Mixing Ideal Mixtures of Liquids Mixtures of liquids are called solutions. Consider a solution of two liquids A and (l) (l) B so that xA and xB are molar fractions of species A and B in the mixture (l) (l) (superscript (l) indicates the liquid phase). Obviously xA + xB = 1. By analogy to gas mixtures, the liquid solution is regarded as ideal when (l)
(3.153)
(l)
(3.154)
(µA )in_solution = (µA )pure_substance + R · T · ln xA and also
(µB )in_solution = (µB )pure_substance + R · T · ln xB
Fig. 3.18: Ideal solution of two liquids, A and B, in equilibrium with its vapour
Consider now this ideal solution of liquids A and B in equilibrium with its vapour at a given temperature T , as shown in Fig. 3.18. We have two components (α = 2) and two phases (β = 2) in the system. All together there are six variables describing the system; the temperature, the pressure, the mole fractions of species (1) (1) A and B in the liquid phase, xA and xB , and mole fractions of species A and B in (v) (v) the gas (vapour) phase, xA and xB where superscript (v) indicates the vapour phase. Among these six variables, only two are independent since, following the phase rule, N IV = α − β + 2 = 2 − 2 + 2 = 2. Thus, specifying any two of the six variables determines uniquely the equilibrium. At equilibrium the (l) (v) chemical potentials of Aspecies in both phases must be equal, so that µA = µA . (l) (v) Moreover, for Bspecies µB = µB is applicable. Let us first exploit the condition (l) (l) (v) µA = µA . Using (3.153) we can calculate µA as follows (l)
(1)
(µA )in_solution = (µA )pure_substance + R · T · ln xA = p∗ (l) (µ0A )pure_substance + R · T · ln A + R · T · ln xA (3.155) p0
143
3 Equilibrium Thermodynamics where relationship (3.110) has been used und p∗A stands for the partial pressure of vapour of Aspecies at equilibrium with pure Aliquid. The chemical potential of Aspecies in the vapour phase is simply (v)
(µA )in_vapour_mixture = (µ0A )pure_substance + R · T · ln
pA p0
(3.156)
where pA is the partial pressure of Aspecies in the vapour. Equating (3.155) and (3.156), and after some algebra we obtain (1)
xA · p∗A = pA
(3.157)
By carrying out similar considerations for Bspecies, we also obtain (1)
xB · p∗B = pB
(3.158)
Obviously, the total vapour pressure is (1)
(1)
p = pA + pB = xA · p∗A + xB · p∗B
(3.159)
Relationships (3.157) and (3.158) express Raoult’s law. The French chemist Francois Raoult found experimentally that the ratio of the partial vapour pressure of each component to its vapour pressure as a pure liquid ( ppA ∗ ) is approximately A
(1)
equal to the mole fraction of Acomponent in the liquid mixture (xA ). Fig. 3.19 (Top) shows the vapour pressures of an ideal binary solution. We wish to stress again that Fig. 3.19 is applicable only to an ideal solution for which the Raoult’s law applies. Many real solutions depart from this ideal behaviour.
The mole fraction of Acomponent in the vapour phase is easily calculable since (1)
xA · p∗A
(v)
xA =
(1) xA · p∗A
+ (1 −
(1) xA ) · p∗B
1
=
(1)
1+
1−xA (1)
xA
(3.160)
p∗
· pB∗
A
p∗
The above relationship is plotted in Fig. 3.19 (Bottom) for several pB∗ ratios. The A essence of the above relationship is that the vapour is richer than the liquid in the more volatile component.
144
3.7 Thermodynamics of Mixing
Vapour pr essur e
p*B p = pA+ pB pA*
(1)
*
pA = xA · pA
0.0
0.2
0.4
0.6
Mole fraction of A,
0.8
1.0
(1) xA
Mole fr action of A in vapou r, xA
(V)
1.0
0.05
0.8
0.1
0.5
0.6
1 2
0.4
10 20
0.2
0.0 0.0
0.2
0.4
0.6
0.8
1.0
(1)
Mole fraction of A in liquid, xA
Fig. 3.19: Ideal solution of two liquids A and B; (Top) Vapour pressures are proportional to the mole fractions of A and B in the liquid phase, (Bottom) the mole fraction of A in the vapour as a function of its mole fraction in the p∗ liquid, the small labels represent the pressure ratio pB ∗ ; the vapour is richer A than the liquid in the more volatile component.
145
3 Equilibrium Thermodynamics
3.8 Summary In this chapter several concepts essential to equilibrium thermodynamics have been recalled. Most of them should be already known to students from lectures on chemical thermodynamics and physical chemistry. The key to understanding equilibrium is the second law of thermodynamics and the associated concepts of irreversible and reversible processes. The law, expressed as the entropy increase principle, says that in a real irreversible process the sum of the changes of the entropy of all bodies participating in the process, including the surroundings, is positive. Using this formulation it is relatively easy to demonstrate that for an isolated system thermodynamic equilibrium is reached when the system entropy reaches a maximum value. If we regard our Universe as an isolated system (is our Universe the only one?) then its entropy is continuously increasing from the big bang through the present state until finally, in thermodynamic equilibrium (dead state), a maximum value of entropy will be reached. Engineers seldom deal with isolated systems. To the contrary. Furnaces, turbines, engines interact with the surroundings by exchanging matter and heat. Thus, the above maximum entropy principle applicable to isolated systems is not of a much help in engineering. Fortunately both experiments and theory teach that thermodynamic equilibrium does not depend on the path the system follows from the initial to the final (equilibrium) state. Instead, equilibrium is uniquely determined by the initial state and any two state variables (p and T ; p and V ; V and T ) at the final equilibrium state. At equilibrium, reached under constant temperature and pressure, Gibbs potential of the system defined as G = H − T · S reaches a minimum value. For a singlephase multicomponent system the fundamental equation of equilibrium reads 0 = −S · dT + V · dp +
N X
µk · dmk
k=1
where µk is chemical potential of kcomponent. When the equilibrium is reached under a constant temperature (dT PN= 0) and a constant pressure (dp = 0) the above relationship simplifies to k=1 µk · dmk = 0. When the equilibrium is reached under other conditions e.g. constant P temperature and volume, or constant entropy and volume, the same relationship N k=1 µk · dmk = 0 is obtained. This is why the chemical potential which is the specific Gibbs function, is essential in thermodynamics of equilibrium states. In this chapter we have proven that the equilibrium conditions for a multi
146
3.8 Summary component (k = 1, 2, . . . , k, . . . , α) multiphase (i = 1, 2, . . . , i, . . . , β) system are
T (1) = T (2) = T (3) = . . . = T (i) = . . . = T (β) = T p(1) = p(2) = p(3) = . . . = p(i) = . . . = p(β) = p µ1k = µ2k = µ3k = . . . = µβk
for k = 1, 2, . . . , α
Thus, equality of temperatures and pressures of each phase as well as equality of the chemical potentials in each phase for each component are required at equilibrium. At thermodynamic equilibrium, the number of independent extensive variables is given by: N IV = α − β + 2 where N IV stands for the Number of Independent Variables9 , α is the number of chemically nonreacting components and β is the number of phases present. The above equation is a well known phase rule for a nonreacting system. The principle of Gibbs potential minimisation has been used to analyse mixing of ideal gases and liquids. It has been demonstrated that gases mix in all proportions and the relationship pi = xi · p ( where p and pi stand for the total pressure and partial pressure, respectively while xi is icomponent mole fraction) has got a sound basis in equilibrium thermodynamics. The text also contains a derivation of the Raoult’s law applicable to mixtures of ideal liquids.
9
also called: Degrees of Freedom
147
3 Equilibrium Thermodynamics
148
4 Chemical Equilibrium Contents 4.1
Definition of Chemical Equilibrium
4.2
Single Chemical Reaction
4.3
4.4
4.5
4.2.1
Extent of a Single Reaction
4.2.2
Change of Gibbs Enthalpy as a Chemical Reaction Advances
4.2.3
Gibbs Enthalpy of Selected Reactions
4.2.4
Thermodynamic Equilibrium Constant for a Gaseous Reaction
4.2.5
Other Equilibrium Constants
4.2.6
Effect of Pressure and Temperature on Thermodynamic Equilibrium Constant
4.2.7
Chemical Equilibrium in Presence of a Solid Phase
4.2.8
Le Châtelier´s Principle
Multiplicity of Chemical Reactions 4.3.1
MultiComponent, MultiPhase Systems with Chemical Reactions
4.3.2
Choice of Chemical Reactions
4.3.3
Exact Number of Chemical Reactions Needed for Equilibrium Determination
4.3.4
Linear Dependence of a Reaction Set
4.3.5
The Phase Rule for a System with Chemical Reactions
Equilibrium Composition 4.4.1
Systems with a onedimensional reaction basis
4.4.2
Systems with a twodimensional reaction basis
4.4.3
Systems with a multidimensional reaction basis
Summary
149
4 Chemical Equilibrium
4.1 Definition of Chemical Equilibrium We begin this chapter with recalling the definition of chemical equilibrium which we have developed in Section 3.4. The system is in chemical equilibrium if no changes occur in the chemical composition of the system. The methodology of this chapter is simple; we begin with the simplest case, namely with a singlephase system in which only one reaction occurs. Later on we extend our considerations to a singlephase system of multiplicity of reactions. Finally, we derive equilibrium criteria for a multiphase, multicomponent system with multiplicity of reactions. The goal of this chapter is to provide the knowledge required for calculation of chemical composition at equilibrium. Most of the combustion processes are carried out to equilibrium and therefore this topic is essential to combustion technology. For some, not too many, combustion processes which are not carried out to equilibrium, combustion chamber design is based primarily on reaction rates (see Chapter 6). However, the choice of the operating conditions may still be influenced by equilibrium considerations. Furthermore, chemical equilibrium is essential to understanding pollutants formation and destruction mechanisms. In Chapter 1 we have learned how to carry out stoichiometric combustion calculations. There the assumption was that the fuel is completely combusted to products containing carbon dioxide and water vapour. The methods of this chapter allow for calculating equilibrium composition of combustion products which many contain not only carbon dioxide and water vapour but also unburned fuel, carbon monoxide or even radicals like H, O, OH or N.
4.2 Single Chemical Reaction 4.2.1 Extent of a Single Reaction We consider a closed system in which a single chemical reaction ν1 · A1 + ν2 · A2 + . . . + να−1 · Aα−1 + να · Aα = 0
(4.1)
occurs. The substrates of this reaction are assigned negative stoichiometric coefficients while the products are given positive coefficients (see Section 2.3.1). The above reaction can also be written as:
150
4.2 Single Chemical Reaction
α X
(4.2)
νk · Ak = 0
k=1
where α stands for the number of chemical species present in the system. If a species present in the system does not take part in the reaction, its stoichiometric coefficient is zero. Let n0k be the number of moles of species k in the initial state of the system. When the reaction proceeds, the variations of the number of moles of each species are not independent. Instead they change as determined by the stoichiometric coefficients of reaction (4.1), so dn1 dn2 dnα−1 dnα = = ... = = = dξ ν1 ν2 να−1 να
(4.3)
where nk indicates the number of moles of each species and the variable ξ relates the changes in the amount of the chemical species present. This variable is called the extent of reaction1 and is expressed in moles. By integrating (4.3) we obtain: nk = n0k + νk · ξ (4.4) The above relationship relates the number of moles of each species to the extent of the reaction. For the initial state of the system we take usually ξ = 0. Since all the numbers of moles are always not negative, the extent of the reaction varies from zero to a value that corresponds to the equilibrium state. If the initial number of moles n0k for the substrates are sufficiently large so that ξ = 1 is an allowed value, then this value of ξ corresponds to the conversion of a number of moles equal to the stoichiometric coefficients. Thus, an increase of ξ by 1 mol is equivalent to the conversion of number of moles of reactants to number of moles of products corresponding to the stoichiometric coefficients of the reaction. Summation of (4.4) over all species yields n=
X k
nk =
X k
n0k + ξ ·
X k
νk =
X
(4.5)
k
where n is the overall number of moles in the system, of moles at the beginning of the reaction. The term 1
n0k + ξ · ∆ν
P
k
n0k is the overall number
The extent of reaction ξ has been given other names such as: reaction coordinate, degree of reaction and progress variable.
151
4 Chemical Equilibrium
∆ν =
X
(4.6)
νk
k
represents the algebraic sum of the stoichiometric coefficients and it can be either negative, positive or zero. The mole fractions xi of the species present are related to the extent of the reaction through xk =
n 0 + ξ · νk nk = P k0 n k nk + ξ · ∆ν
(4.7)
Example 4.1 Consider a system in which the following reaction occurs C2 H6 + 3 12 O2 −−→ 2 CO2 + 3 H2 O Initially in the system there are present 2 mole C2 H6 , 1 mole O2 , 0.5 mole CO2 . Derive expressions for the mole fraction of each component as functions of the extent of the reaction. Assumptions: none For the above reaction we obtain for propane
nC2 H6 = n0C2 H6 + νC2 H6 · ξ = 2 − 1 · ξ
for oxygen
nO2 = n0O2 + νO2 · ξ = 1 − 3.5 · ξ
for carbon dioxide
nCO2 = n0CO2 + νCO2 · ξ = 0.5 + 2 · ξ
for water vapour
nH2 O = n0H2 O + νH2 O · ξ = 0 + 3 · ξ
for all the species
n = 3.5 + 0.5 · ξ
Thus, the mole fractions are 2−ξ 3.5 + 0.5 · ξ 0.5 + 2 · ξ = 3.5 + 0.5 · ξ
xC2 H6 = xCO2
1 − 3.5 · ξ 3.5 + 0.5 · ξ 3·ξ = 3.5 + 0.5 · ξ
xO2 = xH2 O
Comments: P (a) Since ∆ν = k νk = 0.5 > 0 , the number of moles on the right hand side of the reaction is larger than the number of moles on the left hand side; the difference is 0.5 mole.
152
4.2 Single Chemical Reaction (b) The mole fractions cannot be negative. Therefore, considering propane, one obtains ξ ≤ 2 while considering oxygen ξ ≤ 0.2857. Consequently, one expects that the extent of the reaction at equilibrium will take a value somewhere in the range 0 < ξ eq ≤ 0.2857. (c) Consider now the same chemical reaction with the following initial number of moles: n0C H = 5 mol, n0O = 10 mol, n0CO = 1 mol and n0H O = 2 mol. 2 6 2 2 2 The number of moles as functions of the extent of the reaction ξ are then: nC2 H6 = 5 − ξ, nO2 = 10 − 3.5 · ξ, nCO2 = 1 + 2 · ξ , nH2 O = 2 + 3 · ξ. In this case an increase of ξ by 1 mol is possible. For ξ = 1 mol one obtains nC2 H6 = 4 mol, nO2 = 6.5 mol, nCO2 = 3 mol, nH2 O = 5 mol and therefore 1 mol of C2 H6 and 3.5 mol of O2 have been converted producing 2 mol of CO2 and 3 mol of H2 O. Note that the four numbers 1, 3.5, 2 and 3 correspond to the stoichiometric coefficients of the chemical reaction in question. This example clarifies the statement ". . . an increase of ξ by 1 mol is equivalent to the conversion of number of moles of reactants to number of moles of products corresponding to the stoichiometric coefficients of the reaction". End of Example 4.1
4.2.2 Change of Gibbs Enthalpy as a Chemical Reaction Advances In Section 3.5.4 we have considered equilibrium in multicomponent singlephase systems without chemical reactions. However the methodology of deriving equation (3.90) that is the fundamental equation of equilibrium and chemical thermodynamics, remains unaltered when chemical reactions are considered. This is bacause after expressing Gibbs free enthalpy as a function of temperature, pressure and amounts of species (see Eq. (3.87)) G = G(T, p, m1 , m2 , . . . , mk , . . . , mα )
(4.8)
it is of no relevance whether the amounts of species mk change due to chemical reactions or mixing. Thus, the general conditions for the chemical equilibrium are the same as for mixing, namely (see Eq. (3.90)) dG = −S · dT + V · dp +
α X
µk · dmk = 0
(4.9)
k=1
153
4 Chemical Equilibrium Therefore, at constant temperature and pressure, the differential of Gibbs enthalpy is: α X µk · dmk (4.10) dG = k=1
For a chemically reacting system however, the changes in the amounts of the reacting species occur according to the stoichiometric reaction scheme (4.1). These changes can be readily expressed as a function of the extent of the reaction since, using Eq. (4.4), we obtain mk = (n0k + νk · ξ) · Mk
(4.11)
dmk = νk · Mk · dξ
(4.12)
and Inserting (4.12) into (4.10) we obtain dG =
α X
Mk · µk · νk · dξ = ∆R GpT · dξ
(4.13)
k=1
where ∆R GpT =
α X k=1
Mk · µk · νk =
α X k=1
µk · ν k =
α X
g k · νk
(4.14)
k=1
is the Gibbs reaction enthalpy named also as free enthalpy of reaction. In the above equation Mk is the molecular mass of species k, µk = gk is the specific Gibbs energy per kilogram of the species. This variable has already been introduced as chemical potential; µk = g k is the chemical potential expressed per mol (or kmol). Gibbs reaction enthalpy is an intensive variable and similarly to chemical potentials it depends on the composition of the system. To obtain a deeper meaning of Gibbs reaction enthalpy, ∆R GpT , we can express relationship (4.8) using the extent of reaction as follows: G = G(T, p, m01 , m02 , . . . , m0k , . . . , m0α , ξ)
(4.15)
where m0k is the initial amount of species k and ξ stands for the extent of reaction. The exact differential of Gibbs enthalpy can then be written as: ∂G dG = −S · dT + V · dp + dξ (4.16) ∂ξ T,p
154
4.2 Single Chemical Reaction Comparison of Equations (4.16), (4.9) and (4.13) shows that α α X X ∂G = Mk · νk · µk = νk · µk = ∆R GpT ∂ξ p,T k=1
(4.17)
k=1
Gibbs enthalpy, G
Thus, Gibbs reaction enthalpy equals to the derivative of Gibbs potential of the whole system with respect to the extent of the reaction. More explicitly, the Gibbs reaction enthalpy is the change in the Gibbs potential of the system when the extent of the reaction (ξ) changes by 1 mol (or kmol). Consequently, Gibbs reaction enthalpy is expressed in joules (or related units) per one mol (or kmol) of the extent of the reaction. Using the above relationship a new important meaning of Gibbs enthalpy of reaction can be formulated as follows (see Fig. 4.1):
DRG < 0
DRG > 0
DRG=0 Extent of reaction, x Fig. 4.1: Minimisation of Gibbs enthalpy as the reaction advances. Note how the slope of the Gibbs enthalpy changes. Equilibirium corresponds to zero slope.
If ∆R GpT < 0, then the reaction takes place spontaneously from left to right since dG < 0 implies that dξ > 0; If ∆R GpT > 0, then the reaction takes place spontaneously from right to left since dG < 0 implies that dξ < 0; If ∆R GpT = 0, then the reaction reached equilibrium.
155
4 Chemical Equilibrium Thus, the necessary conditions for a chemical reaction to be at equilibrium at constant temperature and pressure are: ∆R GpT =
α X
∆R GpT =
or
Mk · νk · µk = 0
α X
ν k · µk = 0
(4.18)
k=1
k=1
4.2.3 Gibbs Enthalpy of Selected Reactions Introducing (into Eq. (4.14)) the definition of the specific Gibbs enthalpy g = h − T · s we obtain ∆R GpT =
α X
νk · (g pT )k =
k=1
α X
p
νk · [(hT )k − T · (spT )k ]
k=1
=
α X
p
νk · (hT )k − T ·
νk · (spT )k (4.19)
k=1
k=1
and
α X
∆R GpT = ∆R HTp − T · ∆R STp
(4.20)
where ∆R HTp =
α X
p
νk · (hT )k
and
∆R STp =
α X
νk · (spT )
(4.21)
k=1
k=1
At this point we reiterate that ∆R HTp is the change of the enthalpy of the system when the extent of reaction (ξ) advances by 1 mol. Thus, ∆R HTp is expressed in joule (or related units) per 1 mol of the extent of the reaction. Similarly ∆R STp is the change of the entropy of the system when ξ advances by 1 mol and is therefore expressed in J/K · mol or related units. Obviously, for standard conditions we have 0 0 ∆R G0298 = ∆R H298 − 298.15 · ∆R S298
(4.22)
and accordingly for a formation reaction at standard conditions one obtains 0 0 ∆f G0298 = ∆f H298 − 298.15 · ∆f S298
(4.23)
In Chapter 2 we have used Table 2.6 which lists both enthalpies of formation and standard (third law) entropies of some selected substances. Now, using re
156
4.2 Single Chemical Reaction lationship (4.23), we can calculate Gibbs formation enthalpies for the formation reactions. These Gibbs enthalpies of the formation reactions ∆f G0298 are listed in the fifth column of Table 4.1 (see also Example 4.2). For the reference state species like for example O2 (g) and H2 (g), the Gibbs enthalpy of the formation reaction is zero. This is rather obvious, since the formation reaction for O2 can be written as O2 (g) −−→ O2 (g) (4.24) and therefore ∆R GpT = 0 for any temperature and pressure. Thus, for all the reference state species the Gibbs reaction enthalpy is always zero. Example 4.2 Calculate the Gibbs formation enthalpy of nitrogen oxide (NO) at standard conditions. Assumptions: none We begin with writing the NO formation reaction 1 2 O2
+ 12 N2 →NO
or − 12 O2 − 12 N2 + NO = 0
(A1)
Thus, 0 0 ∆f G0298 = ∆f H298 − 298.15 · ∆f S298 = h i 0 0 0 −0.5 · (h298 )O2 − 0.5 · (h298 )N2 + (h298 )NO − 298.15 · −0.5 · (s0298 )O2 − 0.5 · (s0298 )N2 + (s0298 )NO =
90.29 kJ/mol − 298.15 · [−0.5 · 205.15 − 0.5 · 191.61 + 210.76] · 10−3 kJ/mol = 86.60kJ/mol
Comments: (a) Check the obtained value against the one listed in Table 4.1. (b) Whenever oxygen and nitrogen molecules react so that the extent of the reaction is 1 mol, the Gibbs enthalpy changes by 86.60 kJ. End of Example 4.2
157
4 Chemical Equilibrium
Table 4.1: Standard enthalpies of formation, standard entropies and Gibbs formation enthalpies of some selected compounds; g: gaseous, l: liquid, s: solid
Compound Oxygen Hydrogen Nitrogen Carbon Calcium Iron
Aggregation State O2 (g) H2 (g) N2 (g) C(s g raphite) Ca (crystal,α) Fe (reference)
0
hf ,298 (kJ/mol) 0 0 0 0 0 0
s0298 (J/molK) 205.15 130.68 191.61 5.74 41.59 27.32
∆f G0298 (kJ/mol) 0 0 0 0 0 0
logK 0 0 0 0 0 0
Oxygen atoms O(g) 249.17 161.06 231.74 40.60 Ozone O3 (g) 142.67 238.93 163.18 28.59 Hydrogen atoms H(g) 218.00 114.72 203.28 35.61 Water vapour H2 O(g) 241.83 188.83 228.58 40.05 Water H2 O(l) 285.83 69.95 237.14 41.55 Hydroxyl radicals OH(g) 39.00 183.71 34.28 6.01 Nitrogen atoms N(g) 472.68 153.30 455.54 79.81 Nitrogen monoxide NO(g) 90.29 210.76 86.60 15.17 Nitrogen dioxide NO2 (g) 33.10 240.03 51.26 8.98 Carbon C(g) 716.67 158.10 671.24 117.60 Calcium oxide CaO(crystal) 635.09 38.21 603.50 105.73 Calcium carbonate+ CaCO3 (calcite) 1206.90 92.90 1128.80 197.77 Iron oxide FeO(crystal) 272.04 60.75 251.43 44.05 Carbon monoxide CO(g) 110.53 197.65 137.16 24.03 Carbon dioxide CO2 (g) 393.52 213.80 394.40 69.10 Methane CH4 (g) 74.87 186.25 50.77 8.90 Ethane* C2 H6 (g) 83.82 229.12 31.92 5.59 Propane* C3 H8 (g) 104.68 270.20 24.39 4.27 Ethylene* C2 H4 (g) 52.51 219.20 68.44 13.09 Benzene* C6 H6 (g) 82.88 269.30 129.60 22.71 Methanol* CH3 OH(g) 200.94 239.88 162.32 28.44 Ethanol* C2 H5 OH(g) 234.95 280.64 167.85 29.41 NISTJANAF, Thermochemical Tables, American Chemical Society and the American Institute of Physics for the National Institute of Standards and Technology, 1998 * Perry’s Chemical Enginering Handbook, 7th Edition, McGrawHill, 1997 + P. Atkins, Physical Chemistry, 6th Edition, W.H. Freeman and Company, 1997
158
4.2 Single Chemical Reaction
4.2.4 Thermodynamic Equilibrium Constant for a Gaseous Reaction Consider reaction (4.1) taking place in a closed system at constant temperature and pressure. All the reaction substrates and products are in gaseous phase. When the system is at equilibrium, the Gibbs potential of the system reaches minimum: G → Gmin . Equivalently Gibbs reaction enthalpy ∆R GpT is zero; it means Eq. (4.18) must be satisfied. Since all our species are in gaseous form we can use Eq. (3.110) to calculate their chemical potential: µk (T, pk ) = µ0k + RT · ln(
pk ) p0
(4.25)
where µk (T, pk ) is the chemical potential of species k expressed in kJ/kmol, µ0k (T, pk ) is the standard chemical potential at 1 bar (in kJ/kmol), R is the gas constant, T is the temperature and p0 is the standard (reference state) pressure of 1 bar. It is important to note that pk is the partial pressure (in bars) of species k in the gaseous mixture (Relationship (4.25) is valid for ideal gases; for real gases the partial pressure pk should be replaced with fugacity [16, 17]). Inserting relationship (4.25) into Eq. (4.18) we obtain: ∆R GpT =
α X
νk · µk =
k=1
α X
νk · (µ0k + RT · ln(
k=1
pk )) = 0 p0
(4.26)
The above relationship can be easily rearranged into: α X
α
νk · RT · ln(
X pk νk · µ0k )=− p0
(4.27)
k=1
k=1
and further RT · ln
α Y pk ν k
k=1
p0
= −∆R G0T
(4.28)
The above relationship is usually written as −∆R G0T R·T where K is known as the standard equilibrium constant defined as: ln K =
(4.29)
159
4 Chemical Equilibrium
K=
α Y pk ν k ( ) p0
(4.30)
k=1
The standard equilibrium constant is a dimensionless quantity. For a given reaction, the value of this constant depends on temperature, on the standard (reference) state pressure (p0 ) and temperature (T0 ). Using relationship (4.29) the standard equilibrium constant can be calculated using the standard Gibbs enthalpies listed in thermodynamic tables. The latter ones can also be calculated knowing the formation enthalpies and standard entropies of species participating in the reaction, as shown by Eqs. (4.19) and (4.20). The standard equilibrium constant is often designated as the thermodynamic equilibrium constant. In chemical engineering literature Eq. (4.30) is known as the law of mass action. The thermodynamic equilibrium constant K (see Eq. (4.29)) has been derived directly from the standard Gibbs reaction enthalpy ∆R G0T and therefore is a function of temperature only. The constant K is related, through the law of mass action (Eq. (4.30)), to the partial pressures of the species at the equilibrium. We have already calculated Gibbs formation enthalpies (in kJ/mol) for several chemical species and these are listed in the fifth column of Table 4.1. Knowing these enthalpies one can readily calculate the thermodynamic equilibrium constant for the formation reactions at standard conditions using Eq. (4.29). Take nitrogen oxide (NO) as an example for which the formation reaction is 1 2 N2
+ 12 O2 = NO
(4.31)
with the Gibbs reaction enthalpy of 86.60 kJ/mol. The thermodynamic equilibrium constant for the above reaction at standard conditions is then ln K = −
86.60 · 103 kJ/kmol = −34.936 8.314kJ/kmol/K · 298.15K
(4.32)
and therefore K = 6.7219 · 10−16 and log K = −15.17 (compare with Table 4.1). Since the thermodynamic equilibrium constant is very small indeed, the equilibrium of reaction (4.31) at standard temperature is very much to the left so that the extent of the reaction is indeed very close to zero. The reader is requested to practice calculations of the thermodynamic equilibrium constant using Table 4.1 and locate the species for which the equilibrium of their formation reactions at standard temperature is far to the right. Table 4.2 lists values of the thermodynamic equilibrium constants (K) at different
160
4.2 Single Chemical Reaction temperatures for some reactions important in combustion. All but the water gas shift reaction CO2 + H2 ←→ CO + H2 O are formation reactions for which values of the Gibbs reaction enthalpy are tabulated [12] for various temperatures. Thus, for these reactions, the thermodynamic equilibrium constants can be readily evaluated using relationship (4.29). However, for the water gas shift reaction which is not a formation reaction, relationship (4.20) is used to evaluate ∆R G0T and then using (4.29) the equilibrium constant is calculated. Below we show such calculations for the reference temperature of 298.15K. Let rewrite the water gas shift reaction as −CO2 −H2 + CO + H2 O = 0
(4.33)
The Gibbs enthalpy for the above reaction is
∆R G0298 = −(∆g 0f,298 )CO2 − (∆g 0f,298 )H2 + (∆g 0f,298 )CO + (∆g 0f,298 )H2 O
(4.34)
and using Table 4.1
∆R G0298 = −(−394.40) − 0 + (−137.16) + (−228.58) = 28.66kJ/mol
(4.35)
Then ln K =
−∆R G0298 −28.66 · 103 kJ/kmol = = −11.562 R · 298.15 8.314kJ/(kmol · K) · 298.15K
(4.36)
and K = 9.522 · 10−6 so that log K = −5.021. Alternatively, one can obtain the thermodynamic equilibrium constant by manipulating the Kvalues of appropriate formation reactions (see Example 4.3). Example 4.3 Calculate the thermodynamic equilibrium constant K of the homogeneous water gas shift reaction −CO2 −H2 + CO + H2 O = 0 (B0) using Kvalues of other reactions listed in Table 4.2.
161
4 Chemical Equilibrium
162
C(s) + O2 → CO2
C(s) + 12 O2 → CO
H2 + 12 O2 → H2 O(g)
CO2 + H2 → CO + H2 O(g)
298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3200 3400 3600 3800 4000 4500 5000 5500 6000
C(s) + 2H2 → CH4
Temperature in K
Table 4.2: Thermodynamic equilibrium constant K for some reactions important in combustion [12]
logK 8.894 8.813 5.492 3.420 1.993 0.943 0.138 0.500 1.018 1.447 1.807 2.115 2.379 2.609 2.810 2.989 3.147 3.289 3.416 3.531 3.636 3.732 3.819 3.899 3.973 4.042 4.105 4.164 4.219 4.319 4.407 4.485 4.555 4.619 4.753 4.863 4.955 5.034
logK 69.095 68.670 51.539 41.299 34.404 29.505 25.829 22.969 20.679 18.805 17.242 15.919 14.784 13.800 12.939 12.128 11.502 10.896 10.351 9.858 9.406 8.998 8.622 8.275 7.955 7.658 7.383 7.126 6.886 6.450 6.064 5.721 5.413 5.134 4.544 4.068 3.675 3.344
logK 24.030 23.911 19.110 16.236 14.320 12.948 11.916 11.109 10.461 9.928 9.481 9.101 8.774 8.488 8.236 8.013 7.813 7.633 7.470 7.322 7.186 7.062 6.946 6.840 6.741 6.648 6.562 6.481 6.404 6.265 6.140 6.027 5.924 5.830 5.627 5.458 5.313 5.188
logK 40.047 39.785 29.238 22.884 18.631 15.582 13.287 11.496 10.060 8.881 7.897 7.063 6.346 5.724 5.179 4.698 4.269 3.885 3.540 3.227 2.942 2.682 2.443 2.223 2.021 1.833 1.658 1.495 1.344 1.068 0.825 0.608 0.414 0.239 0.131 0.428 0.672 0.877
logK 5.018 4.974 3.191 2.179 1.453 0.975 0.626 0.364 0.158 0.004 0.136 0.245 0.336 0.412 0.476 0.583 0.580 0.622 0.659 0.691 0.722 0.746 0.767 0.788 0.807 0.823 0.837 0.850 0.862 0.883 0.901 0.914 0.925 0.935 0.952 0.962 0.966 0.967
4.2 Single Chemical Reaction
298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000 3200 3400 3600 3800 4000 4500 5000 5500 6000
logK 8.980 8.944 7.517 6.672 6.114 5.717 5.420 5.188 5.003 4.851 4.724 4.615 4.522 4.441 4.370 4.307 4.251 4.201 4.155 4.114 4.077 4.042 4.011 3.982 3.955 3.930 3.908 3.886 3.866 3.831 3.799 3.772 3.747 3.726 3.681 3.648 3.622 3.602
logK 79.810 79.298 58.710 46.341 38.084 32.180 27.746 24.294 21.530 19.266 17.377 15.778 14.407 13.217 12.175 11.256 10.437 9.705 9.046 8.448 7.905 7.409 6.954 6.535 6.148 5.790 5.457 5.147 4.857 4.332 3.867 3.455 3.085 2.751 2.046 1.480 1.015 0.625
logK 40.599 40.330 29.469 22.936 18.570 15.446 13.098 11.269 9.803 8.603 7.601 6.752 6.024 5.392 4.839 4.350 3.915 3.525 3.175 2.857 2.568 2.304 2.062 1.839 1.633 1.442 1.265 1.100 0.946 0.666 0.420 0.200 0.004 0.173 0.547 0.846 1.092 1.296
NO
+ 21 O2 →
logK 6.005 5.963 4.265 3.246 2.569 2.085 1.724 1.444 1.222 1.041 0.981 0.764 0.656 0.563 0.482 0.410 0.347 0.291 0.240 0.195 0.153 0.116 0.082 0.050 0.021 0.005 0.030 0.053 0.074 0.112 0.145 0.174 0.200 0.223 0.270 0.307 0.335 0.357
1 2 N2
+ 21 H2 → 1 2 O2
logK 35.613 35.378 25.876 20.158 16.335 13.597 11.538 9.932 8.644 7.587 6.705 5.956 5.313 4.754 4.264 3.831 3.446 3.100 2.788 2.506 2.249 2.014 1.798 1.599 1.415 1.245 1.087 0.939 0.801 0.551 0.330 0.133 0.044 0.203 0.539 0.809 1.029 1.213
OH
→H 1 2 H2
→O 1 2 O2
→N 1 2N2
O2 → 1 2 N2 + N O2
Temperature in K
Table 4.3: Thermodynamic equilibrium constant K for some reactions important in combustion [12]
logK 15.172 15.074 11.143 8.784 7.210 6.086 5.243 4.587 4.063 3.633 3.275 2.972 2.712 2.487 2.290 2.116 1.926 1.824 1.699 1.587 1.484 1.391 1.306 1.227 1.154 1.087 1.025 0.967 0.913 0.815 0.729 0.653 0.585 0.524 0.396 0.295 0.213 0.146
163
4 Chemical Equilibrium Assumptions: none Using the following three reactions (see Table 4.2) −C−O2 + CO2 = 0
(B1)
−H2 − 21 O2 + H2 O = 0
(B2)
−C− 21 O2 + CO = 0
(B3)
with the thermodynamic equilibrium constants KB1 , KB2 , KB3 , we can compose reaction (B0) since − (B1) + (B2) + (B3) = C + O2 − CO2 − H2 − 1/2O2 + H2 O − C − 1/2O2 + CO = − CO2 − H2 + H2 O + CO = 0 Adding up Gibbs enthalpies of the above three reactions, we obtain − ∆GB1 + ∆GB2 + ∆GB3 = − [−gC − gO2 + gCO2 ] + [−gH2 − 1/2gO2 + gH2 O ] + [−gC − gO2 + gCO ] = − gCO2 − gH2 + gH2 O + gCO = ∆GB0 so that ln KB0 =
−∆GB0 = R·T −(−∆GB1 + ∆GB2 + ∆GB3 ) −∆GB1 −∆GB2 −∆GB3 =− + + R·T R·T R·T R·T
Therefore ln KB0 = − ln KB1 + ln KB2 + ln KB3 and ln KB0 = ln Thus, finally KB0 =
164
KB2 · KB3 KB1
KB2 · KB3 KB1
(B4)
4.2 Single Chemical Reaction and log KB0 = log KB2 + log KB3 − log KB1
(B5)
Comments: (a) For T = 298K, using Table 4.2 we obtain log K = 40.047 + 24.030 − 69.095 = −5.018. (b) The values listed in column six of Table 4.2 have been obtained using the logarithms of thermodynamic constants listed in columns three, four and five, using relationship (B5). (c) Note that manipulating the logarithms of the thermodynamic equilibrium constants is equivalent to adding or subtracting Gibbs enthalpies. (d) Derive relationship (B4) using the law of mass actions, Eq. (4.30). End of Example 4.3
4.2.5 Other Equilibrium Constants Other forms of the law of mass action are also used. It is possible to define an equilibrium constant (Kp ) relative to pressures as follows: Kp =
α Y
pνkk
(4.37)
k=1
where by comparison with (4.30) Kp = K · (p0 )∆v
(4.38)
When partial pressures pk are expressed in bars
Kp = K
(numerically)
(4.39)
since p0 = 1 bar. The equilibrium constant Kp is in general not dimensionless unless ∆ν is zero. Other forms of the law of mass action are encountered when different means of expressing concentrations are used. Using mol (volume) fractions xk and assuming ideal gas behaviour
165
4 Chemical Equilibrium
K=
α Y pk ν k
k=1
p0
α Y xk · p νk
=
k=1
p0
=
p p0
∆ν Y α
(xk )
νk
=
k=1
p p0
∆ν
Kx (4.40)
or
Kx = K ·
p p0
−∆v
(4.41)
where p is the total pressure of the system (in bars) and Kx is the equilibrium constants expressed in molar (volume) fractions
Kx =
α Y
(xk )νk
(4.42)
k=1
We can also use the following derivation
K=
α Y pk ν k
k=1
p0
=
α Y xk · p νk
k=1
p0
α Y nk · p νk
k=1
n · p0
=
=
p n · p0
∆ν Y α
νk
(nk )
k=1
=
p n · p0
∆ν
Kn (4.43)
or
Kn = K ·
p n · p0
−∆v
(4.44)
where nk is the number of kmol of species k, n is the total number of kmol in the system and Kn is defined as follows Kn =
α Y
(nk )νk
(4.45)
k=1
Very often there is a need to use concentrations ([Ak ] in mol/m3 ) of species. In this case the following derivation is applicable assuming the perfect gas law:
166
4.2 Single Chemical Reaction
K=
α Y pk ν k
k=1
p0
α α Y [Ak ] · RT νk RT ∆ν RT ∆ν Y νk Kc ([Ak ]) = = = p0 p0 p0 k=1 k=1 (4.46)
or Kc = K ·
p −∆v 0 R·T
(4.47)
where Kc (c stands for concentration) is defined as
Kc =
α Y
([Ak ])νk
(4.48)
k=1
It should be emphasised that while the thermodynamic equilibrium constant K is dimensionless, the other constants (Kp , Kn , Kc ) are generally not dimensionless. Only when ∆ν = 0 the other constants are also dimensionless. Values of the equilibrium constants lie in the range from zero to infinity and the larger the Kvalue the further to the right is the equilibrium shifted. In other words, the larger the Kvalue the more complete the reaction.
4.2.6 Effect of Pressure and Temperature on Thermodynamic Equilibrium Constant As shown by Eq. (4.29) the thermodynamic equilibrium constant (K) depends on the value of Gibbs reaction enthalpy ∆R G0T which is defined at standard pressure of 1 bar. The value of ∆R G0T and subsequently K are therefore both independent of the pressure at which the chemical equilibrium is established. Mathematically we can express this independence as
∂K ∂p
=0
(4.49)
T
The effect of temperature on the thermodynamic equilibrium constant can be obtained by differentiating Eq. (4.29):
167
4 Chemical Equilibrium
∂ ln K ∂T
p
1 ∂ =− R ∂T
∆R G0T T
p
1 ∂ =− R ∂T
Pα
k=1 νk
=
· µ0k (T )
T
=
p
α ¯0 ∆R HT0 1 X h νk k2 = R T R·T2
(4.50)
k=1
¯ 0 is the enthalpy of formation per kmol of species k while ∆R H 0 is the where h T k reaction enthalpy. The relationship ∆R HT0 d ln K = dT R·T2
(4.51)
is known as the van’t Hoff equation that can be also written in another form: ∆R HT0 d ln K = − R d T1
(4.52)
Eq. (4.52) shows that a graphical representation of ln K as a function of 1/T allows the determination of the standard enthalpy of reaction. If we assume that in a temperature range from T1 to T2 the standard enthalpy of reaction is constant then ∆R HT0 K(T2 ) ln = · K(T1 ) R
1 1 − T1 T2
(4.53)
For significant changes of temperature the variation of the standard enthalpy with temperature has to be taken into account when integrating Eq. (4.51), so K(T2 ) = ln K(T1 )
ZT2
∆R HT0 · dT RT 2
(4.54)
T1
where the dependence of ∆R HT0 with temperature is obtained using Kirchoff’s equation (see Section 2.3.5) T2
∆R HT02
=
∆R HT01
+
α Z X k=1 T
1
168
νk · Cp,k (T ) · dT
(4.55)
4.2 Single Chemical Reaction where Cp,k (in kJ/kmol/K) is the specific molar heat at constant pressure of species k. Relationship (4.52) implies that a plot of ln K versus the reciprocal of absolute temperature is a straight line. Fig. 4.2, a plot of log K versus 1/T for the reactions listed in Table 4.2, illustrates this near linearity. Thus, Eq. (4.53) provides reasonable accurate relation for the extrapolation and interpolation of values of thermodynamic equilibrium constant. If reaction enthalpy ∆R HT0 is negative , i.e. if the reaction is exothermic, the thermodynamic equilibrium constant decreases as the temperature increases. Conversely, K increases with T for an endothermic reaction. The above explained dependence of the thermodynamic equilibrium constant on temperature can be explicitly seen by rearranging Eq. (4.29) as K = exp
−∆R HT0 R·T
· exp
∆R ST0 R
= K∞ · exp
−∆R HT0 R·T
(4.56)
Fig. 4.2: Thermodynamic equilibrium constant K as a function of temperature for reactions listed in Table 4.2. Note: ln K = 2.302585 · log K.
Example 4.4 Calculate thermodynamic equilibrium constant for the reaction −− ⇀ C + O2 ↽ − − CO2
169
4 Chemical Equilibrium at T = 1000K knowing that at 298K, log K = 69.095 (see Table 4.2). Assumptions: none Relationship (4.54) is to be used, so that K(1000) ln = K(298)
1000 Z
∆R HT0 · dT RT 2
(C1)
298
In order to perform the above integration the dependence of the reaction enthalpy ∆R HT0 with temperature should be known and it can be obtained using Kirchoff’s law (see Eq. (4.55)). In order to simplify the calculations we assume that the reaction enthalpy is independent of temperature and is equal to 0 = −393.52 kJ/mol as listed in Table 4.1. Then, after performing the ∆R H298 integration we obtain (see Eq. (4.53))
0 K(1000) ∼ ∆R H298 1 1 ln · − = = K(298) R 298 1000 1 −393.52 · 103 kJ/kmol 1 · − = −111.5007 (C2) 8.314 kJ/kmol/K 298 1000 Then K(1000) = 3.76 · 10−49 K(298)
(C3)
log K(1000) − log K(298) = −48.424
(C4)
and
Finding in Table 4.2, log K(298) = 69.095
(K(298) = 1.02 · 1030 ) we obtain
log K(1000) = −48.424 + 69.095 = 20.671 and K(1000) = 4.688 · 1020
(C5)
Comments: (a) The obtained value of log K(1000) = 20.671 is very close to the value of 20.679 listed in Table 4.2.
170
4.2 Single Chemical Reaction (b) In the temperature range from 298K to 1000K, the dependence of the reaction enthalpy ∆R HT0 with temperature is indeed weak. (c) At both temperatures the equilibrium is shifted far to the right meaning that carbon, producing carbon dioxide, consumes almost all available oxygen . End of Example 4.4
4.2.7 Chemical Equilibrium in Presence of a Solid Phase In Section 4.2.4 we have derived the law of mass action, Eq. (4.30), for a chemical reaction taking place in a gas phase. In this paragraph we consider systems when gas phase and solid phase are present. Examples of such reactions are: C(s) + O2 (g) −−→ CO2 (g) CaCO3 (s) −−→ CaO(s) + CO2 (g) FeO(s) + CO(g) −−→ Fe(s) + CO2 (g) Before proceeding further we may recall that chemical potential of a gaseous species is calculated as given by Eq. (3.110). The chemical potential of solids is calculable as (see Eq. (3.112)): 0
s0298 + Cs · ln µ ¯s (T, p) = hf,298 + Cs · (T − T0 ) + T · (¯
T )=µ ¯0s (T ) T0
(4.57)
0
where Cs is the molar specific heat of the solid, hf,298 is its molar formation enthalpy, s¯0298 is its standard molar entropy and T0 = 298.15 K. Since solids are incompressible, their chemical potential is a function of temperature only, as shown in Eq. (4.57), and is identical to their standard potential. Consider now the first of the reactions given above: C(s) + O2 (g) −−→ CO2 (g) There are two phases, gaseous and solid, present. There is one species present in the solid phase (C(s)) and two species (O2 (g) and CO2 (g)) present in the gaseous phase. Recalling our considerations on equilibrium in multiphase, multicomponent systems (see Section 3.6) we observe that if the chemical potential of a species present in two phases is not the same, then the system is not at equilibrium. Spontaneous transport takes place from the phase where the species
171
4 Chemical Equilibrium has the highest chemical potential towards a phase where its chemical potential is lower, thus altering the composition of the system until equilibrium is reached. At equilibrium the chemical potential of each of the species present in the solid phase is equal to its chemical potential in the gas phase, so µ ¯C(g) = µ ¯C(s) = µ ¯0C(s)
(4.58)
The equilibrium condition in the gaseous phase is dG = 0 or
∆R G =
3 X
νk · µ ¯k = 0
(4.59)
k=1
and therefore 0 = −¯ µC(s) − µ ¯0O2(g) − RT · ln
pCO2 pO2 +µ ¯0CO2(g) + RT · ln p0 p0
(4.60)
After some algebra we obtain
where
RT · ln K = −∆R G0T = − µ ¯0CO2(g) − µ ¯0O2(g) − µ ¯0C(s)
K=
pCO2 p0 pO2 p0
!
=
pCO2 = Kp pO2
(4.61)
(4.62)
It is important to note that in expression (4.61) for calculating the thermodynamic equilibrium constant K, the standard Gibbs enthalpy stays the same as for a gaseous system. However, expression (4.62) that expresses the law of action, includes only the species that are exclusively present in the gas phase. The above observations can be generalised and if all gases behave as ideal ones, we have [17]:
RT · ln K =
−∆R G0T
=−
α X
νk · µ0k where the summation extends over
k=1
all the species participating in the reaction, (4.63)
172
4.2 Single Chemical Reaction
Kp = K · (p0 )∆ν where ∆ν =
X
νk and the summation extends over
k
the species exclusively present in the gas phase, (4.64)
Kp =
Y
pνkk where the multiplication extends over
k
the species exclusively present in the gas phase. (4.65)
Example 4.5 Generate a graph showing the dependence of carbon dioxide partial pressure as a function of temperature for calcination reaction at equilibrium CaCO3 −−→ CaO + CO2
(D1)
Assumptions: calcium carbonate is chemically pure We begin with calculating the thermodynamic equilibrium constant for reaction (D1) using the data listed in Table 4.1. The Gibbs enthalpy for the reaction in question is
¯ CaCO3 + h ¯ CO2 + h ¯ CaO − ∆G0T = ∆HT0 − T · ∆ST0 = −h − T · [−¯ sCaCO3 + s¯CO2 + s¯CaO ] =
[− (−1206.90) + (−393.52) + (−635.09)] kJ/mol − T · [−92.90 + 213.80 + 38.21] · 10−3 kJ/mol = = 178.29 − T · 159.15 · 10−3 in kJ/mol So, the thermodynamic equilibrium constant is
ln K =
−∆G0T −(178.29 − T · 159.15 · 10−3 ) · 103 kJ/kmol 21, 444.55 = = 19.142− R·T 8.314kJ/(kmol · K) · T · K T
At the equilibrium K =
pCO2 p0
so that
pCO2 = p0 · exp(19.142 − 21, 444.55/T )
(D2)
173
4 Chemical Equilibrium Since p0 = 1 bar, the partial pressure of carbon dioxide at equilibrium is pCO2 = exp(19.142 − 21, 444.55/T )
in bar
(D3)
Fig. 4.3 shows the dependence of the equilibrium partial pressure of carbon dioxide with temperature. Comments: (a) We have calculated that the reaction enthalpy is 178.29 kJ/mol at 298K. We have assumed that ∆HT0 does not change with temperature. Using Kirchoff’s law (see Eq. (4.55)) one can easily account for the dependence of ∆HT0 with temperature. However, such a dependence is often weak. For example, for the calcination reaction, ∆HT0 at 1200K is 167 kJ/mol. (b) At ambient temperature (300K), the equilibrium partial pressure of CO2 is 1.95 · 10−23 bar which is a very low value. The partial pressure of CO2 in the atmosphere is much larger than 1.95 · 10−23 bar so CaO when exposed to the atmosphere spontaneously absorbs CO2 . This is often used in analytical chemistry to generate a CO2 free environment. (c) It is instructive to consider calcination reaction CaCO3 −−→ CaO + CO2 in terms of the phase equilibrium conditions which we have derived in Section 3.6 (see Eq. (3.141)). We have here two phases; the solid phase and the gas phase, and three components. The first two conditions (Eq. (3.141) (A) and (B)) are obviously satisfied. The third condition (Eq. (3.141) (C)) stating that the chemical potential of each component in the gas phase must be equal to its chemical potential in the solid phase is rewritten below as (g)
(s)
(D4)
(g)
(s)
(D5)
(g)
(s)
(D6)
µCaCO3 = µCaCO3 µCaO = µCaO µCO2 = µCO2
Relationships (D4) and (D5) are satisfied since the temperature of the gas phase is the same as the temperature of the solid phase. The left hand side of (D6) is the chemical potential of carbon dioxide in the gas phase which is a function of the temperature and the CO2 partial pressure. The right hand side of (D6) stands for the CO2 chemical potential in the solid phase and is a
174
4.2 Single Chemical Reaction
Fig. 4.3: Equilibrium partial pressure of carbon dioxide in calcination reaction CaCO3 → CaO + CO2
175
4 Chemical Equilibrium function of the temperature only. Thus, for a given temperature there exists only one CO2 partial pressure that satisfies relationship (D6). Therefore in Fig. 4.3, on the equilibrium partial pressure line, relationship (D6) is quoted. When, for a given temperature, the CO2 partial pressure is lower than the equilibrium pressure then CO2 undergoes a spontaneous phase change from the solid phase to the gas phase since such a phase change minimises the (g) (s) Gibbs potential of the whole system (µCO2 < µCO2 ). Conversely, when CO2 partial pressure exceeds the equilibrium pressure, CO2 is absorbed in the solid (g) (s) phase since µCO2 > µCO2 . End of Example 4.5
4.2.8 Le Châtelier´s Principle Consider a single chemical reaction (4.1) that has reached equilibrium. We name this equilibrium state as an "old" equilibrium. In this paragraph we examine what happens to the system when, after reaching this old equilibrium, we impose some changes to the system by altering for example the temperature, the pressure or we add some inert components. The question is whether the old equilibrium is retained or the system moves to a "new" equilibrium state. If the latter is true, the question is in which direction will the equilibrium be shifted; towards the products or towards the substrates? In the previous paragraphs we have learned how to calculate the thermodynamic equilibrium constant K (see Eq. (4.29)) and how to relate its value to the partial pressures of the species at equilibrium using the law of mass action (see Eq. (4.30)). Using this knowledge, we will deduct the information about the system response to the imposed changes. Anticipating the results, it will be soon apparent, that the system response can be deducted using Le Châtelier’s principle which states that Any system initially in an equilibrium state when subjected to a change will shift in composition in such a way so as to minimise the change. Below we analyse the effect of the temperature change, the pressure change, the volume change and the effect of adding inert species on the equilibrium state. Effect of temperature As a matter of fact, in Section 4.2.6 we have already examined the effect of temperature on the thermodynamic equilibrium constant K. For convenience we copy here Eq. (4.56)
176
4.2 Single Chemical Reaction
K = exp
−∆R HT0 R·T
· exp
∆R ST0 R
= K∞ · exp
−∆R HT0 R·T
(4.66)
and we reiterate that for endothermic reactions (∆R HT0 > 0) K increases with increasing temperature. Conversely, for exothermic reactions (∆R HT0 < 0) K decreases with increasing temperature. Fig. 4.2 illustrates the point further. The question is how these observations agree with the principle of Le Châtelier. In an exothermic reaction the temperature increases with the extent of the reaction. Any removal of heat from the system which is equivalent to lowering the system temperature will result in shifting the system composition (extent of the reaction) so as to compensate for the heat removal. Thus, the new equilibrium state will be shifted to the right; e.g. the extent of the reaction increases. By the same reasoning we can conclude that an endothermic reaction is favoured by a higher system temperature. Effect of pressure In Section 4.2.6 we have concluded that the thermodynamic equilibrium constant K is independent of pressure (see Eq. (4.49)). This does not mean that the equilibrium state (composition) is independent of the pressure. Consider then a system that is in equilibrium at given a temperature and a pressure. Now, maintaining the temperature, we are increasing the (total) pressure by decreasing the volume. We recall here relationship (4.40)
K=
p p0
∆ν Y α
k=1
(xk )νk =
p p0
∆ν
Kx
(4.67)
and bearing in mind that the left hand side (K) is independent of the pressure we observe if ∆ν = 0 the equilibrium is unaffected, if ∆ν < 0 when p ↑ the equilibrium is shifted to the right if ∆ν > 0 when p ↑ the equilibrium is shifted to the left. It is worthy observing that the algebraic sum of the stoichiometric coefficients (∆ν) plays a decisive role. In other words the stoichiometry of the reaction is important. As an example consider a reaction C(s) + CO2 −−→ 2 CO
177
4 Chemical Equilibrium for which ∆ν = 1 > 0. The total pressure in the system increases as the reaction progresses since two moles of a gaseous component (CO) are produced per one mole of a gaseous substrate (CO2 ). Then, an increase in pressure implies a shift of the equilibrium to the left since the system attempts to reduce the number of moles and opposes the invoked pressure increase in agreement with the principle of Le Châtelier. It is also possible to increase the total pressure by adding a nonreacting (inert) species into the system so that the temperature and the volume remain unaltered. However, by adding an inert component, the partial pressures of the reacting species do not change and therefore, following the law of mass action, Eq. (4.30), the equilibrium composition does not change. Effect of volume Now we examine what happens to the system which is in equilibrium at given a temperature and a pressure and the system volume (V ) changes. The volume change takes place at a constant temperature. An increase in the system volume results in a decrease of the total pressure; conversely a volume decrease results in a corresponding total pressure increase. Since we have just established the effect of the total pressure on the system, the effect of the volume change on the equilibrium is as follows: if ∆ν = 0 the equilibrium is unaffected, if ∆ν < 0 when V ↓ the equilibrium is shifted to the right if ∆ν > 0 when V ↓ the equilibrium is shifted to the left. where V is the total volume. Again the stoichiometry of the system (∆ν) controls the observed shifts in the equilibrium state. If ∆ν > 0 then a decrease in the volume implies an increase in the pressure and therefore the system shifts the equilibrium to the left, opposing the pressure increase, according to the principle of Le Châtelier. Alternatively, one can derive the observed effects of the volume change directly from Eq. (4.43) that is here copied for your convenience K=
p n · p0
∆ν
Kn =
R·T V · p0
· Kn
(4.68)
Effect of addition of an inert gaseous species As we have already observed, an addition of an inert gas, so that the temperature and the volume remain unaltered, does not change the partial pressures of gaseous species and therefore does not alter the equilibrium state. However, if the addition is at a constant pressure and a constant temperature then the system volume
178
4.2 Single Chemical Reaction changes and the equilibrium shifts as described above under the heading "Effect of volume". Example 4.6 Consider the following reaction CO2 + C −−→ 2 CO
(E1)
which in combustion and processengineering is known as Boudouard reaction. Determine the partial pressures of gaseous species at equilibrium. Examine the effect of pressure on the equilibrium. Assumptions: (a) CO2 and CO behave as ideal gases, (b) to examine the effect of pressure on the equilibrium, we intend to calculate the equilibrium CO2 and CO partial pressures at total pressures of 0.01, 0.1, 1, 5 and 10 bar. We begin with calculating the thermodynamic equilibrium constant for reaction (E1) as a function of temperature. To this end, we use Table 4.2 where the equilibrium constants of the following two reactions are listed C + 21 O2 −−→ CO
(E2)
C + O2 −−→ CO2
(E3)
Since 2 · (E2) − (E3) = (E1) we have log KE1 = 2 · log KE2 − log KE3
(E4)
Alternatively, we can use data listed in Table 4.1 to estimate the equilibrium constant KE1 . The Gibbs reaction enthalpy is then
¯ CO2 − h ¯ C + 2 · hCO − T · [−¯ ∆G0T = −h sCO2 − s¯C + 2 · s¯CO ] =
= [−(−393.52) − 0 + 2 · (−10.53)] − T · [−213.80 − 5.74 + 2 · 197.65] · 10−3 = = 172.46 − T · 0.175 76 kJ/mol (E5)
Here again, being a bit lazy, we have ignored the effect of temperature on the reaction enthalpy ∆HT0 (see again Kirchoff’s law, Eq. (4.55)). Then
179
4 Chemical Equilibrium
−∆G0T −172.46 + T · 0.17576 · 103 kJ/kmol ln KE1 ∼ = = R·T 8.314 kJ/(kmol · K) · T · K −20, 743.33 = + 21.14 (E6) T Thus, relationship (E4) is exact while (E6) is an estimate of KE1 constant. Fig. 4.4 displays both relationships showing that the approximated relationship (E6) is indeed very accurate. The law of mass action provides the link between the equilibrium constant and the equilibrium composition since
KE1 =
2
pCO p0 pCO2 p0
(E7)
Fig. 4.4: The accurate values (Eq.(E4)) and the estimated values (Eq.(E6)) of the thermodynamic equilibrium constant for Boudouard reaction CO2 + C → 2CO
180
4.2 Single Chemical Reaction The above relationship is accompanied by pT = pCO2 + pCO
(E8)
where pT stands for the total pressure. Simple algebra leads to the following quadratic equation for pCO2 p2CO2 − pCO2 · [2 · pT + KE1 · p0 ] + p2T = 0
(E9)
The roots of (E9) can be easily found as
(pCO2 )1 =
(2 · pT + KE1 · p0 ) −
q 2 · p2 4 · pT · KE1 · p0 + KE1 0
(E10)
q 2 · p2 4 · pT · KE1 · p0 + KE1 0
(E11)
2
and (pCO2 )2 =
(2 · pT + KE1 · p0 ) +
2
Since the second root, given by (E11), is always larger than the total pressure pT it is rejected and finally at equilibrium we have q 2 · p2 (2 · pT + KE1 · p0 ) − 4 · pT · KE1 · p0 + KE1 0 pCO2 = (E12) 2 and pCO = pT − pCO2
(E13)
Thus, for a given total pressure pT , we can produce a curve showing the partial pressures of carbon dioxide and carbon monoxide as a function of temperature. Fig. 4.5 shows such curves for pressures of 1 and 10 bar. Observe the shift in the equilibrium with pressure. The information presented in Fig. 4.5 can be more elegantly shown using a single curve as depicted in Fig. 4.6. Here a single variable is plotted against temperature for total pressures of 0.01, 0.1, 1, 5 and 10 bar. Fig. 4.6 is an equilibrium curve for CO2 + C −−→ 2 CO reaction which can be found in many textbooks on combustion or chemicalengineering.
181
4 Chemical Equilibrium
Fig. 4.5: Variation of the partial pressures of CO2 and CO with temperature for total pressure of 1 bar (top) and 10 bar (bottom) for Boudouard reaction CO2 + C → 2CO
182
4.3 Multiplicity of Chemical Reactions
Fig. 4.6: Equilibrium composition of Boudouard reaction CO2 + C → 2CO at several total pressures
Comments: (a) Fig. 4.6 shows that at a constant total pressure, higher temperatures favour production of carbon monoxide it means with an increasing temperature the equilibrium shifts towards CO. This is in agreement with the principle of Le Châtelier since reaction CO2 + C −−→ 2 CO is endothermic. (b) For a fixed temperature, the equilibrium shifts to the left with an increasing pressure in accordance with the principle of Le Châtelier. End of Example 4.6
4.3 Multiplicity of Chemical Reactions In Section 4.2 we have considered a single chemical reaction while in this paragraph we are going to be more general. We consider here a multicomponent system existing in a several phases with a multiplicity of chemical reactions occurring. Our task is to derive the conditions for such a system to be at equilibrium under a constant temperature and a constant pressure. To some extent
183
4 Chemical Equilibrium this paragraph is going to be similar to Section 3.6 where we have also considered a multicomponent and multiphase system. However, this time the system may find its equilibrium by both phase changes and chemical reactions occurring simultaneously.
4.3.1 MultiComponent, MultiPhase Systems with Chemical Reactions Since we are considering here a multicomponent and multiphase system, any property of a component will have two indices; one describing the phase and other the component (species) according to the convention introduced in Section 3.6: iconcerns phase – i=1,2,...,β
kconcerns component – k=1,2,...,α
(4.69)
Thus, there are α components (species) (k = 1, 2, . . . , α) and β (i = 1, 2, . . . , β) phases in the system. There are R (r = 1, 2, . . . , R) simultaneous chemical reactions occurring and for each of them a stoichiometric equation can be written as α X νkr · Ak = 0 for r = 1, 2, . . . , R (4.70) k=1
where here we use again the convention that the substrates are assigned negative coefficients while the products are given positive values. Since the system temperature and pressure are specified and kept constant we have T (1) = T (2) = T (3) = . . . = T (i) = . . . = T (β) = T
(4.71)
p(1) = p(2) = p(3) = . . . = p(i) = . . . = p(β) = p
(4.72)
The methodology of developing equilibrium conditions for the system is going to be the same as in Section 3.6. We will formulate a Gibbs function for multiphase, multicomponent systems and we will find the conditions that minimise this function. To find a minimum of such a function we will use the method of Lagrangean multipliers. We merely recall that the method which you have learned in Chapter 3 (see Example 3.5) is to find a minimum of a function with constraints.
184
4.3 Multiplicity of Chemical Reactions The Gibbs function for a singlephase of the system is:
i
G =G
i
(T, p, mi1 , mi2 , ..., miα )
=
α X
µik · mik
(4.73)
k=1
and for all phases:
G=
β X
Gi =
i=1
β X α X
µik · mik
(4.74)
i=1 k=1
Thus, we should find a minimum of G with respect to the amounts of each component (mik ) under constraints. The equations formulating these constraints are going to be based on the following observations; firstly the overall mass of the system remains conserved2 and secondly the amount of each species changes in accordance with the stoichiometric equations (4.70). We formulate as many constraints as there are species in the system. Each constraint equation should describe how the amount of the species in question varies when the chemical reactions proceed. For a single reaction we have already developed relationship (4.11) that says precisely what we want. This relationship has to be reformulated for a system containing β phases and R chemical reactions. Thus,
β X
i=1 β X
mi1
=
m01
+ M1 ·
ν1r · ξr
for the first species
ν2r · ξr
for the second species
(4.75)
r=1
mi2 = m02 + M2 ·
R X r=1
i=1
...
β X
R X
miα = m0α + Mα ·
i=1
R X
ναr · ξr
for the αspecies.
r=1
Now we use Gibbs function (4.74), α Lagrangean multipliers (λ1, λ2 , ...λα ) together 2
Nuclear reactions are excluded.
185
4 Chemical Equilibrium with αconstraints (4.75) to build a new function Y3
Y =
β X α X
µik
· mik
β X
+ λ1 ·
mi1
−
m01
− M1 ·
+ λ2 ·
β X
ν1r · ξr
r=1
i=1
i=1 k=1
R X
mi2 − m02 − M2 ·
R X
ν2r · ξr
r=1
i=1
β X
+ λα ·
!
!
+
+ ...+
miα − m0α − Mα ·
R X
ναr · ξr
r=1
i=1
!
(4.76)
The exact differential of Yfunction at equilibrium reached under a constant temperature and a constant pressure must be zero, so dY = 0 =
β X α X
µik · ∂mik + λ1 ·
β X
∂mi1 − λ1 · M1 ·
+ λ2 ·
β X
∂mi2
ν1r · dξr +
r=1
i=1
i=1 k=1
R X
− λ 2 · M2 ·
R X
ν2r · dξr + ....
r=1
i=1
+ λα ·
β X
∂miα
− λ α · Mα ·
R X
ναr · dξr (4.77)
r=1
i=1
By rearranging the above equation we obtain 0=
β X α X
µik
· ∂mik R X r=1
3
∂mi1
+ λ2 ·
i=1
i=1 k=1
− λ 1 · M1 ·
+ λ1 ·
β X
ν1r · dξr − λ2 · M2 ·
β X i=1
R X r=1
∂mi2
+ ... + λα ·
β X
∂miα +
i=1
ν2r · dξr − ... − λα · Mα ·
R X
ναr · dξr
r=1
If you have forgotten how to build the Yfunction see Section 3.6 and Example 3.5.
186
(4.78)
4.3 Multiplicity of Chemical Reactions and further 0=
β X α X
µik
· ∂mik
i=1 k=1
+
α X
λk ·
k=1
− λ 1 · M1 ·
R X
β X
∂mik +
i=1
ν1r · dξr − λ2 · M2 ·
R X
ν2r · dξr − ... − λα · Mα ·
ναr · dξr
r=1
r=1
r=1
R X
(4.79)
and after some further algebra we obtain
0=
β X α X i=1 k=1
R α X X νkr · dξr λ k · Mk µik + λk · ∂mik − k=1
(4.80)
r=1
Thus, the conditions for the system to be at equilibrium are
µik = −λk
for k = 1, 2, ..., α
(4.81)
and
0=
R α X X
µik
· Mk · νkr · dξr =
k=1 r=1
α R X X
µik
· Mk · νkr · dξr =
R X
∆r GpT · dξr (4.82)
r=1
r=1 k=1
or more explicitly 0 = ∆1 GpT · dξ1 + ∆2 GpT · dξ2 + ... + ∆R GpT
(4.83)
and therefore ∆r GpT = 0
for r = 1, 2, ..., R
(4.84)
To summarise, we rewrite here the general conditions for a multicomponent, multiphase system with a multiplicity of chemical reactions to be in equilibrium
187
4 Chemical Equilibrium as: T (1) = T (2) = T (3) = ... = T (i) = ... = T (β) = T p
(1)
=p
(2)
=p
(3)
= ... = p
µ1k = µ2k = µ3k = ... = µβk
(k)
= ... = p
(α)
=p
for k = 1, 2, . . . α
and α α X X ∂G p µ ¯k · νkr = 0 µk · Mk · νkr = = ∆r G T = ∂ξr k=1
(4.85) (4.86) (4.87) (4.88) (4.89)
k=1
The general conditions for a multicomponent, multiphase system with chemical reactions taking place to be at equilibrium are equality of temperatures and pressures of each phase as well as equality of the chemical potentials in each phase for each component. Furthermore, Gibbs enthalpies of all the chemical reactions are zero. The just derived conditions should be compared with Eq. (3.141) which we have derived for a nonreacting system. The conclusion is obvious. The same conditions are needed for phase equilibrium. Additionally Gibbs enthalpies of all chemical reactions must be zero.
4.3.2 Choice of Chemical Reactions In Section 4.2 we have considered single chemical reactions. More precisely we have considered systems that could be described by a single chemical reaction. While considering an equilibrium in a system containing CO2 , CO and C it has been "obvious" to us that Boudouard reaction CO2 + C −−→ 2 CO which contains all the species in question should be used. However, when a large number of species are present it is not longer obvious what and how many reactions should be considered. In Section 4.3.1 we have considered a multiplicity of chemical reactions (r = 1, 2, . . . , R) proceeding simultaneously. We have just demonstrated that at equilibrium Rrelationships (4.89) hold. We can easily come up with other reactions which can be generated by a linear combination of two or more reactions already considered by the set of the stoichiometric equations (4.70). Since the equilibrium condition (4.89) is satisfied for all R reactions, it is also satisfied for
188
4.3 Multiplicity of Chemical Reactions a linear combination of any of these R reactions. However, such a linear combination does not provide any new relation among the chemical potentials of the system species. Therefore, it is necessary to determine the exact number of chemical reactions needed for the determination of the equilibrium state of the system.
4.3.3 Exact Number of Chemical Reactions Needed for Equilibrium Determination We consider a system of α chemical species (A1 , A2 , . . . , Aα ) which are known to be the only species present. We wish to determine both the number of reactions (R) as well as the reactions themselves. Any chemical reaction which occurs in the system can be expressed by a stoichiometric equation of the following form ν1 A1 + ν2 A2 + ... + νk Ak + ... + vα Aα = 0
(4.90)
In the above equation, the stoichiometric coefficients νk are unknown and there are α such coefficients to be determined. Each species Ak is build up out of (atoms) elements like H, C, O and others. Let us assume that there are γ elements in the system. For each element we can write down a balance equation so that γbalance equations can be formulated. They can be written as a α × γ element balance matrix with αrows and γcolumns. If we denote the rank of this matrix as ρ then the number of chemical reactions to consider is R = α − ρ where α is the number of species and ρ is the rank of the element balance matrix. Sounds complicated? Do not worry; it is simple. The following example clarifies the procedure. The following chemical species H2 , CO, CO2 , CH3 OH, H2 O are the components of methanol production. Thus, we have identified five (α = 5) chemical species and we wish to determine the equilibrium state for a given temperature and pressure. To this end we would like to use the knowledge which we have developed in the previous paragraphs. At this stage we have no clue as to what and how many chemical reactions shall we consider. However, we realise that each of the reactions must be of the following form ν1 H2 + ν2 CO + ν3 CO2 + ν4 CH3 OH + ν5 H2 O = 0
(4.91)
The five species H2 , CO, CO2 , CH3 OH, H2 O are build out of three (γ = 3) elements H,C and O. For each element we can formulate a balance of atoms as
189
4 Chemical Equilibrium follows H
2 · ν1 + 0 · ν2 + 0 · ν3 + 4 · ν 4 + 2 · ν 5 = 0
(4.92a)
O
0 · ν1 + 1 · ν2 + 2 · ν3 + 1 · ν 4 + 1 · ν 5 = 0
(4.92b)
C
0 · ν1 + 1 · ν2 + 1 · ν3 + 1 · ν 4 + 0 · ν 5 = 0
(4.92c)
The above set of equations can be written in a matrix notation as A × ~ν = ~0 or
2 0 0 4 2 0 1 2 1 1 × 0 1 1 1 0
(4.93) ν1 ν2 ν3 ν4 ν5
0 = 0 0
(4.94)
The important number which is now emerging is the rank ρ of matrix A. There are several methods for determination of the rank of a matrix and here you should consult textbooks on linear algebra. Perhaps the most common definition of a rank of a matrix is the following: "the rank counts the number of independent rows in the matrix ". Since the rank of the matrix A is ρ = 3 (see below), the number of chemical reactions is R = α − ρ = 5 − 3 = 2. Thus, for the considered system, consisting of the five species, we have to select two chemical reactions. The next question is obvious: what reactions? To answer the question we have to solve the set of three linear equations (4.94) of five unknowns. To this end we recommend Gauss elimination method (see Appendix A). The method is constantly used to solve large sets of linear equations. Say, we have to solve a set of M equations. The idea of Gauss is simple; multiples of the first equation are subtracted from the other equations, so as to remove the first unknown from those equations. It leaves a smaller system of M −1 equations. The process is repeated until there is only one equation and one unknown which can be solved explicitly. Then, we proceed backwards and find other unknowns (see Appendix A). Now we will apply Gauss elimination to equation set (4.94). We begin with a matrix (see Eq. (4.94)) 2 0 0 4 2 0 0 1 2 1 1 0 0 1 1 1 0 0
190
(4.95)
4.3 Multiplicity of Chemical Reactions In order to remove the second number (1) from the third row, we subtract the second row from the third row to obtain (the first and the second rows remain unaltered) 2 0 0 4 2 0 0 1 2 1 1 0 (4.96) 0 0 −1 0 −1 0 In our case, the Gauss elimination happens to be completed by one operation only. We mention in passing that at the end of the Gauss elimination we have obtained a matrix which have three nonzero rows4 . Now we begin the process of backward substitution. Since we have five unknowns ν1 , ν2 , ν3 , ν4 , ν5 and three equations, two out of the five unknowns can be freely selected while the remaining three must be obtained by solving matrix (4.96). To take a full benefit of matrix (4.96), we select the pair (ν4 , ν5 ) and calculate the remaining three unknowns (ν1 , ν2 , ν3 ). Indeed, we have a full freedom in selecting the pair (ν4 , ν5 ) with one limitation; we should not take ν4 = 0 and ν5 = 0 since it would remove both the fourth species (CH3 OH) and the fifth species (H2 O) from the system. Thus, for any pair (ν4 , ν5 ) we calculate the triple (ν1 , ν2 , ν3 ) and by doing so we determine the chemical reaction (4.90). Since we need to determine two reactions, we select two pairs (ν4 , ν5 ). For a free selection of the pair (ν4 , ν5 ) we chose ν4 = 1 and ν5 = 0. By examining the third row of matrix (4.96) we obtain − ν3 − ν5 = 0
so
ν3 = 0
while examining the second row brings ν2 + 2 · ν 3 + ν4 + ν5 = 0
so
ν2 = −1
so
ν1 = −2
and finally inspecting the first row provides 2 · ν1 + 4 · ν4 + 2 · ν5 = 0
Thus, the first reaction, determined by the five stoichiometric coefficients taking values of ν1 = −2, ν2 = −1, ν3 = 0, ν4 = 1, ν5 = 0, can be written as −2H2 − CO + CH3 OH = 0 4
Another definition of the rank of a matrix is "the rank ρ equals the number of nonzero rows in the final matrix of Gauss elimination process".
191
4 Chemical Equilibrium or 2H2 + CO = CH3 OH If we make another selection of the pair (ν4 , ν5 ) so that ν4 = 0 and ν5 = 1, we obtain the second set of the stoichiometric coefficients ν1 = −1, ν2 = 1, ν3 = −1, ν4 = 0, ν5 = 1. This set determines the second chemical reaction −H2 + CO − CO2 + H2 O = 0
(4.97)
or H2 + CO2 = CO + H2 O We have just determined the first set of two reactions required for considering chemical equilibrium in a system containing the five species H2 , CO, CO2 , CH3 OH, H2 O . These are 2H2 + CO = CH3 OH
(4.98a)
H2 + CO2 = CO + H2 O
(4.98b)
The reader can verify that for (ν4 , ν5 )pairs taking values of (1,0) and (1,1), we obtain another set of the two reactions 2H2 + CO = CH3 OH
(4.99a)
3H2 + CO2 = CH3 OH + H2 O
(4.99b)
Again, for (0,1) and (1,1) we obtain H2 + CO2 = CO + H2 O
(4.100a)
3H2 + CO2 = CH3 OH + H2 O
(4.100b)
Indeed, there exist infinitely many pairs of reactions since there are infinitely many combinations of numbers that can be used for ν4 and ν5 . It is important to realise that any plausible chemical reaction that may occur among the five species H2 , CO, CO2 , CH3 OH, H2 O can be expressed by a linear combination of set (4.98), or set (4.99), or set (4.100). In this lecture course each of these sets is called a reaction basis. In terms of linear algebra, the above sets of two reactions span the space of all plausible reactions in the five species system. Thus, every reaction in this reactions space is a linear combination of a reaction basis. Here we stress again that the two reactions which form a reaction basis are independent. This
192
4.3 Multiplicity of Chemical Reactions is an essential feature of a reaction basis. There exist also some other methods for reaction basis determination however they are more cumbersome. The above described procedure is logical, consistent and simple. After gaining some experience, you will be able to "guess" an appropriate reaction basis without following the above procedure. However, you must make sure that the reactions you have chosen for your reaction basis are independent. The next paragraph deals with the issue of checking the reaction independence.
4.3.4 Linear Dependence of a Reaction Set For the sake of arguments, let us assume that somebody, who considers the five species system H2 , CO, CO2 , CH3 OH, H2 O of methanol production, has come up with an idea that the following four chemical reactions have to be taken into account for the determination of the system equilibrium 2H2 + CO = CH3 OH
R1
(4.101a)
H2 + CO2 = CO + H2 O
R2
(4.101b)
H2 + 2CO + H2 O = CH3 OH + CO2
R3
(4.101c)
3H2 + 3CO + H2 O = 2CH3 OH + CO2
R4
(4.101d)
We wish to find out if the above four reactions are independent. We begin with rewriting the stoichiometric equations using convention (4.70), so that −2H2 −CO +CH3 OH = −H2 +CO −CO2 +H2 O = −H2 −2CO +CO2 +CH3 OH −H2 O = −3H2 −3CO +CO2 +2CH3 OH −H2 O =
0 0 0 0
(4.102)
Examining whether the above four reactions are linearly independent means checking whether there exist four numbers λ1 ,λ2 , λ3 ,λ4 , not all of them equal to zero, so that λ1 · (−2H2 − CO + CH3 OH) + λ2 · (−H2 + CO − CO2 + H2 O)+ + λ3 · (−H2 − 2CO + CO2 + CH3 OH − H2 O)+ + λ4 · (−3H2 − 3CO + CO2 + 2CH3 OH − H2 O) = 0 (4.103)
193
4 Chemical Equilibrium or in a matrix form −2 −1 0 λ1 + 1 0 or more compact
−1 1 −1 0 1
λ2 +
−1 −2 1 1 −1
λ3 +
−3 −3 1 2 −1
λ4 =
B · ~λ = ~0
0 0 0 0 0
(4.104)
(4.105)
where B matrix is ~1 R
B =
~2 R
~3 R
~4 R
−2 −1 −1 −3 −1 1 −2 −3 0 −1 1 1 1 0 1 2 0 1 −1 −1
(4.106)
Note that the number of columns of B matrix is equal to the number of reactions so each reaction is represented by a column vector. The number of rows equals the number of species so that the first row corresponds to H2 , the second one to CO and so on. If ~λ = ~0 is the only solution of equation (4.105) the reactions are independent. Let us look for the solution of the system (4.105) using Gauss elimination. The elimination procedure is described in the following steps:
194
4.3 Multiplicity of Chemical Reactions
−2 −1 −1 −3 −1 1 −2 −3 0 −1 1 1 1 0 1 2 0 1 −1 −1
0 0 0 0 0
−2 −1 −1 −3 0 0 −1 1 1 0 0 1 −1 −1 0 −1 1 −2 −3 0 1 0 1 2 0
−2 −1 −1 −3 0 0 −1 1 1 0 0 1 −1 −1 0 0 1.5 −1.5 −1.5 0 0 −0.5 0.5 0.5 0 −2 −1 −1 −3 0 0 −1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
exchanging rows row(4)0.5row(1) and row(5)+0.5row(1)
row(3)+row(2); row(4)+1.5row(2) and row(5)0.5row(2) (4.107)
As it can be easily seen from (4.107), the rank of matrix B is two. It means that only two of the four reactions considered are independent. If we wish, we can determine the relationships between the four reactions. Matrix (4.107) is equivalent to −2 · λ1 − λ2 − λ3 − 3 · λ4 = 0
(4.108)
−λ2 + λ3 + λ4 = 0
(4.109)
We select λ3 = 1 and λ4 = 0 to obtain λ2 = 1 and λ1 = −1. Thus, −1 1 for λ3 = 1 and λ4 = 0, we have ~λ = 1 so 0 ~2 + 1 · R ~3 = 0 ~1 + 1 · R −1 · R
(4.110)
195
4 Chemical Equilibrium and therefore
~3 = R ~1 − R ~2 R
(4.111)
The correctness of (4.111) can be easily verified recalling (4.102) so that
(−2H2 − CO + CH3 OH) − (−H2 + CO − CO2 + H2 O) =
(4.112)
−H2 − 2CO + CO2 + CH3 OH − H2 O
(4.113)
If we wish to establish the relationship between R4 reaction and R1 and R2 we need to find another solution to Eqs. (4.108) and (4.109). If we select λ3 = 0 and λ4 = 1 we obtain λ2 = 1 and λ1 = −2 so for
−2 1 ~ ~ ~ λ3 = 0 and λ4 = 1 we have ~λ = 0 so − 2 · R1 + 1 · R2 + 1 · R4 = 0 (4.114) 1
and therefore ~4 = 2 · R ~1 − R ~2 R
(4.115)
Again the above relationship is easily verifiable since
2 · (−2H2 − CO + CH3 OH) − (−H2 + CO − CO2 + H2 O) = − 3H2 − 3CO + CO2 + 2CH3 OH − H2 O
4.3.5 The Phase Rule for a System with Chemical Reactions In Chapter 3 we have derived the phase rule for nonreacting systems of β phases and α chemical species as (see Eq. (3.143)) N IV = α − β + 2
(4.116)
The rule must be modified for systems in which chemical reactions occur. As we have discussed in Section 3.6.1 the number of possible degrees of freedom is 2 +
196
4.3 Multiplicity of Chemical Reactions α · β. From this number, α · (β − 1) + β combinations must be subtracted already for a nonreacting system (see Section 3.6.1). However, for each independent chemical reaction, Eqs. (4.89) provide R additional relations that must be satisfied at equilibrium. Thus, the phase rule for reacting systems is N IV = α − β + 2 − R
(4.117)
where R stands for the number of independent reactions or in other words R is the dimension of a reaction basis spanned over the chemical species considered. While deriving Eq. (4.117) we have considered the phaseequilibrium equations and the chemicalreactionequilibrium equations which interrelate the variables appearing in the phase rule. However, in certain circumstances, special additional constraints must be taken into account. If the number of these additional constraints is denoted by S then the most general form of the phase rule is N IV = α − β + 2 − R − S
(4.118)
It is not straight forward to realise when these additional (special) constraints, marked by S, must be applied. This is for example the case when one of the reactions is the decomposition of a pure species resulting in more than one gaseous product. Then, a special relation exists between the partial pressures of the gaseous products of the decomposition and this relation formulates an additional constraint. Example 4.7 clarifies further the issue. ′
Often a new variable α = α − R − S is introduced so that the phase rule remains as ′
N IV = α − β + 2
(4.119)
′
and the number α is known as the number of independent species. Example 4.7 We revisit all Examples of this chapter to demonstrate usefulness of the phase rule. We determine the number of independent variables (N IV ) for the following systems: (a) a system consisting of the gases C2 H6 , O2 , CO2 , H2 O in chemical equilibrium (Example 4.1), (b) a system consisting of the gases O2 , N2 , NO in chemical equilibrium (Example 4.2),
197
4 Chemical Equilibrium (c) a system consisting of the gases CO2 , H2 , CO, H2 O in chemical equilibrium (Example 4.3), (d) a system consisting of C(s), O2 and CO2 in chemical equilibrium (Example 4.4), (e) a system consisting of CaCO3 (s), CaO(s) and CO2 in chemical equilibrium (Example 4.5), (f) a system consisting of C(s), CO2 and CO in chemical equilibrium (Example 4.6). Solution (a). The system contains four (α = 4) C2 H6 , O2 , CO2 , H2 O species, all in a single gas phase (β = 1). The species are built up out of three elements C, H, O so (γ = 3). The rank of A matrix (Eq. (4.93), Paragraph 4.3.4) is three so that ρ = 3 . Thus, the number of independent chemical reactions is R = α − ρ = 4 − 3 = 1 and one of them can be written as C2 H6 + 3 21 O2 −−→ 2 CO2 + 3 H2 O There are no special constraints, so S = 0. Thus, N IV = α − β + 2 − R − S = 4 − 1 + 2 − 1 − 0 = 4 This result means that one is free to specify four phaserule variables for example temperature, pressure and two mass (mole) fractions in an equilibrium mixture of these four chemical species. Solution (b). The system contains three (α = 3) O2 , N2 , NO species, all in a single gas phase (β = 1). There are two (γ = 2) elements (O and N) in the system and A matrix is of the rank two , ρ = 2. Since R = α − ρ = 3 − 2 = 1 there is only one chemical reaction (R = 1) that forms the reaction basis. If we take the reaction 1 2 O2
+ 12 N2 = NO
as the reaction basis, we see that if in its initial state, the system was made up of pure NO, the stoichiometry of the reaction implies that the partial pressure of oxygen equals the partial pressure of nitrogen. This relation between the partial pressures of oxygen and nitrogen provides an additional special constraint (S = 1). Thus, N IV = α − β + 2 − R − S = 3 − 1 + 2 − 1 − 1 = 2
198
4.3 Multiplicity of Chemical Reactions This result means that one is free to specify two phaserule variables. Solution (c). N IV = 4 (see Solution (a)) Solution (d). The system contains three (α = 3) C(s), O2 and CO2 species, in two (β = 2) phases. There are two (γ = 2) elements (C and O) and the rank of Amatrix is two, ρ = 2. Thus, there is only one chemical reaction R = α − ρ = 3 − 2 = 1 that forms the reaction basis. We may take the reaction C + O2 −−→ CO2 as the reaction basis. Thus, N IV = α − β + 2 − R − S = 3 − 2 + 2 − 1 = 2 This result means that one is free to specify two phaserule variables. Solution (e). The system contains three (α = 3) CaCO3 (s), CaO(s) and CO2 species, in two solid phases (CaCO3 and CaO) and one gas phase (CO2 ) so that β = 3. There are three (γ = 2) elements (Ca, O, C) and the rank of matrix A is two, ρ = 25 . Therefore R = α − ρ = 3 − 2 = 1 and only one reaction forms the reaction basis. Typically one takes CaCO3 −−→ CaO + CO2 as the reaction basis. There are no specific constraints, so S = 0. Thus, N IV = α − β + 2 − R − S = 3 − 3 + 2 − 1 = 1 This result means that one is free to specify only one phaserule variable in an equilibrium mixture of these three species. We have just demonstrated why calcium carbonate exerts a fixed decomposition pressure at a given temperature, as shown in Fig. 4.3. Solution (f). The system contains three (α = 3) C(s), CO2 , CO species, in two phases (β = 2). There are two (γ = 2) elements (C, O) and the rank of matrix A is two, ρ = 2. Therefore R = α − ρ = 3 − 2 = 1 and only one reaction forms the reaction basis. We take 5
Although the number of elements is three (γ = 3), the rank of the element balance matrix A is two, ρ = 2.
199
4 Chemical Equilibrium
C + CO2 −−→ 2 CO as the reaction basis. There are no specific constraints, S = 0. Thus, N IV = α − β + 2 − R − S = 3 − 2 + 2 − 1 = 2 This result means that one is free to specify two phaserule variables in an equilibrium mixture of these three species, as shown in Fig. 4.6. End of Example 4.7
4.4 Equilibrium Composition In the previous paragraphs of this chapter, we have acquired comprehensive knowledge on chemical equilibrium. In this paragraph we use this knowledge to calculate equilibrium composition of reacting mixtures. We reiterate here that the equilibrium composition depends on the temperature and pressure at equilibrium as well as on the initial concentration of the chemical species of the reacting mixture. Any analysis of chemical equilibrium begins with identification of the species which are present in the system. What follows is a determination of the initial concentrations (amounts, mass or mole fractions, partial pressures ets.) of these species together with a specification of the temperature and total pressure of the system. At this stage it is useful to identify the elements (atoms) present. From this point onwards, there are several methods of calculating the equilibrium composition. We are going to examine most of these methods using examples. We begin with a system whose reaction basis can be described by a single chemical reaction, it means the reaction basis is onedimensional. What follows is an example of a system whose reaction basis is twodimensional. Finally, we demonstrate how to determine the equilibrium composition without considering chemical reactions at all.
4.4.1 Systems with a onedimensional reaction basis The mixture of 3 kmol of CO2 and 1 kmol of H2 is heated up to 1300 K at a pressure of 1 bar and we expect that at equilibrium in addition to CO2 and H2 also H2 O and CO are present. Thus, there are four species (α = 4) CO2 , H2 , CO,
200
4.4 Equilibrium Composition H2 O in the system. The task is to determine the composition of the equilibrium mixture. There are three elements (H, C, O) in the system (γ = 3). Following the procedure underlined in Section 4.3.3 we can easily check that the rank of the element balance matrix is three (ρ = 3) and therefore there is only one reaction R = α − ρ = 4 − 3 = 1 that forms the reaction basis. We chose the following reaction CO2 + H2 = CO + H2 O
(4.120)
as a reaction basis, so that νCO2 = −1, νH2 = −1, νCO = 1, νH2 O = 1 and ∆ν = 1 + 1 − 1 − 1 = 0. In this way, we have identified the species present and we have selected an appropriate reaction basis. Furthermore, the initial amounts of the species have been given as n0CO2 = 3 kmol, n0H2 = 1 kmol, n0CO = 0 kmol, n0H2O = 0 kmol while T = 1300 K and p = 1 bar. Thus, the preliminary work has been done and we step into calculating the equilibrium composition. Method 1  Based on the extent of the reaction and the equilibrium constant We express the concentrations of all four species participating in the water gas shift reaction (4.120) as a function of the extent of the reaction: nCO2 (ξ) = n0CO2 − ξ nH2 (ξ) = n0H2 − ξ nCO (ξ) = n0CO + ξ nH2 O (ξ) = n0H2 O +
(4.121)
ξ
and the total number of kmol of the mixture is nT (ξ) = nCO2 (ξ)+nH2 (ξ)+nCO (ξ)+nH2 O (ξ) = n0CO2 +n0H2 +n0CO +n0H2 O (4.122) Since n0H2 = 1 kmol, the extent of the reaction ξ cannot be larger than 1. Thus, we expect that ξeq , expressed in kmol, is in the range 0 < ξeq ≤ 1. From Table 4.2 we find that K = 1.75792361 at T = 1300 K (alternatively we could calculate K using Eq. (4.29)). Since in this example we have expressed the concentrations of reacting species in kmol, it is convenient to relate K to Kn . To this end we use relationships (4.43) to obtain:
201
4 Chemical Equilibrium
K=
νH2 O νCO  nCO (ξeq ) · nH O (ξeq ) 2 νH2 νCO  2 nCO (ξeq ) · nH (ξeq ) 2
·
p nT (ξeq ) · p0
∆ν
(4.123)
2
Substituting (4.121) and (4.122) into (4.123) provides K=
ξeq · ξeq (3 − ξeq ) · (1 − ξeq )
(4.124)
By solving Eq. (4.124) for ξ, with K = 1.75792361, we obtain ξeq = 0.8230 kmol (since Eq. (4.124) is quadratic there exists another solution, ξeq = 8.455 kmol, that is however not applicable to our problem). Thus, the equilibrium composition of the reacting mixture is: neq CO = 3 − 0.8230 = 2.177 kmol 2
(4.125)
neq H2 = 1 − 0.8230 = 0.1770 kmol neq CO = 0 + 0.8230 = 0.8230 kmol neq H2 O = 0 + 0.8230 = 0.8230 kmol If needed, the mole fractions at the equilibrium can be easily calculated
2.177 = 0.54425 4 0.8230 = 0.20575 = 4
xeq CO2 = xeq CO
0.1770 = 0.04425 4 0.8230 = = 0.20575 4
xeq H2 = xeq H2O
(4.126) (4.127)
It is important to note that the equilibrium composition for the water gas shift reaction is independent of the total pressure since ∆v = 0. Method 2  Based on the balance of elements and the equilibrium constant The stoichiometric equation (4.120) is obviously valid. However, we can write the actual reaction that reaches equilibrium as: 3 CO2 + 1 H2 = xCO2 + yH2 + zCO + wH2 O
202
(4.128)
4.4 Equilibrium Composition where xCO2 +yH2 indicates the leftover (unreacted) reactants whilst zCO+wH2 O represents the reaction products, as shown in Fig. 4.7. The task is to determine x, y, z, and w that correspond to equilibrium at T = 1300K and p = 1bar. There are three elements C, H and O in the system and for each of them we can formulate a balance equation as follows:
C balance
3=x+z
or
z =3−x
(4.129a)
O balance
6 = 2x + z + w
or
w =3−x
(4.129b)
H balance
2 = 2y + 2w
or
y =x−2
(4.129c)
Before reaction
After reaction
Initial amounts
At equilibrium x kmol CO2 y kmol H2
3 kmol CO2 1 kmol H2
z kmol CO w kmol H2O
} leftover reactants }leftover products
Fig. 4.7: Illustration of reaction (4.128)
The total number of kmol is then (4.130)
nT = x + y + z + w Using again Eq. (4.123) we obtain ν
K=

ν

CO nCO (ξeq ) · nHH2O (ξeq ) 2O
ν

ν

CO2 nCO2 (ξeq ) · nHH2 (ξeq ) 2
·
p nT (ξeq ) · p0
∆ν
z·w · = x·y
p (x + y + z + w) · p0
0
(4.131)
Eqs. (4.129a), (4.129b), (4.129c) and (4.131) formulate a system of four equations with four unknowns (x, y, z, w). Inserting (4.129a), (4.129b) and (4.129c) into
203
4 Chemical Equilibrium (4.131) we obtain: 1.75792361 =
(3 − x) · (3 − x) x · (x − 2)
(4.132)
There are two solutions to the above equation; x = −5.454568 and x = 2.17699. Obviously only the second solution is applicable to our problem. Thus, at the equilibrium: neq CO2 = x = 2.177 kmol
(4.133)
neq H2 = y = 0.1770 kmol neq CO = z = 0.8230 kmol neq H2 O = w = 0.8230 kmol The above solution is identical to the equilibrium composition given by (4.125). Method 3  Minimization of Gibbs enthalpy of the system using the extent of the reaction Consider, for the third time, the mixture of three kmol of carbon dioxide and one kmol of hydrogen that is heated up to 1300K. Our task is to determine the equilibrium composition of the reacting mixture by minimising Gibbs enthalpy of the reacting mixture consisting of CO2 , H2 , CO and H2 O. We assume that the gases behave as ideal ones. We begin with finding the Gibbs standard enthalpies at T = 1300K for the species considered. Table 4.4 lists the values taken from the JANAF tables [12]6 . Table 4.4: Standard Gibbs enthalpies at T = 1300 K [12]
Species CO2 H2 CO H2 O(g)
∆f G01300 kJ/mol 396.177 0 226.509 175.774
We pause here to check whether the data listed in Table 4.4 are consistent with the previously used value of the equilibrium constant K = 1.75792361. Using Table 4.4 we can calculate the Gibbs energy for reaction (4.120) as follows: 6
In Table 4.1 Gibbs enthalpies at 298K are listed only. To find the values at T = 1300K one needs either the whole JANAF tables or one calculates ∆R G01300 using ∆R G0298 .
204
4.4 Equilibrium Composition
∆R G01300 = −226.509 − 175.774 + (−1) · (−396.177) + (−1) · 0 = −6.106 kJ/mol (4.134) and therefore
K(1300 K) = exp(−
−6.106 · 103 kJ/kmol ∆R G01300 ) = exp(− ) = 1.757926 R·T 8.314kJ/(kmol · K) · 1300K (4.135)
which is close enough to the value of K = 1.75792361 previously used. As in Method 1, we express the number of kmol of the reacting species as a function of the extent of the reaction using relationships (4.121) which we will copy here for the sake of completeness: nCO2 (ξ) = n0CO2 − ξ nH2 (ξ) = n0H2 − ξ nCO (ξ) = n0CO + ξ nH2 O (ξ) = n0H2 O + ξ
(4.136)
and the total number of kmol of the mixture is nT (ξ) = nCO2 (ξ)+nH2 (ξ)+nCO (ξ)+nH2 O (ξ) = n0CO2 +n0H2 +n0CO +n0H2 O (4.137) where n0CO2 = 3 kmol, n0H2 = 1 kmol, n0CO = 0 kmol , n0H2 O = 0 kmol and n0T = 4 kmol. The Gibbs enthalpy of the system can be readily calculated as follows: G(ξ) = GCO2 (ξ) + GH2 (ξ) + GCO (ξ) + GH2 O (ξ)
(4.138)
and the above equation is the heart of this method (compare with Eq. (4.15)). The four terms appearing in the above equations are calculated using Eq. (3.110) applicable to ideal gases pCO2 GCO2 (ξ) = (3 − ξ) · 10 · −396.177 + RT · ln = p0 (3 − ξ) · p 3 (3 − ξ) · 10 · −396.177 + RT · ln (4.139) 4 · p0 3
205
4 Chemical Equilibrium
pH2 GH2 (ξ) = (1 − ξ) · 10 · 0 + RT · ln = p0 (1 − ξ) · p 3 (4.140) (1 − ξ) · 10 · RT · ln 4 · p0 3
pCO GCO (ξ) = ξ · 10 · −226.509 + RT · ln = p0 ξ·p ξ · 103 · −226.509 + RT · ln (4.141) 4 · p0 3
pH2 O GH2 O (ξ) = ξ · 10 · −175.774 + RT · ln = p0 ξ·p 3 ξ · 10 · −175.774 + RT · ln (4.142) 4 · p0 3
In the above equations p0 is the reference pressure (1bar) and p is the total pressure (in bars) at equilibrium while the universal gas constant is R = 8.314 · 10−3 molkJ· K . Fig. 4.8 shows the plot of the Gibbs enthalpy (4.138) as a function of the extent of the reaction for T = 1300K and p = 1bar. The minimum of the Gibbs function is at ξeq = 0.823, as expected. And the equilibrium composition of the mixture is then given by (4.125) or (4.133). Method 4  Without considering chemical reactions The starting point to this method is the expression for the Gibbs enthalpy of the system G = GCO2 + GH2 + GCO + GH2O = mCO2 · µCO2 + mH2 · µH2 + mCO · µCO + mH2O · µH2O (4.143) with GCO2 = mCO2 ·
206
RT −9.004 + · ln MCO2
mCO2 · M p · m · MCO2 p0
· 103
(4.144)
4.4 Equilibrium Composition 1.210
G ib b s en th alp y (G J )
1.215
CO2/H2/CO/H2O mixture 1.220
T = 1300 K
1.225
n CO = 3 kmol
0
2
0
n H = 1 kmol 2
1.230 1.235 1.240 1.245 0.0
0.2
0.4
0.6
0.8
1.0
Extent of the reaction (kmol) Fig. 4.8: Gibbs enthalpy as a function of extent of the reaction for a CO2 /H2 /CO/H2 O mixture at T = 1300K, p = 1bar; initial conditions: n0CO2 = 3 kmol, n0H2 = 1 kmol
RT mH2 · M p = mH2 · 0 + · ln · · 103 MH2 m · MH2 p0
(4.145)
mCO · M p RT · ln · · 103 = mCO · −8.090 + MCO m · MCO p0
(4.146)
GH2
GCO
GH2O = mH2O ·
mH2O · M p RT · ln · · 103 −9.765 + MH2O m · MH2O p0
(4.147)
where mCO2 , mH2 , mCO , mH2O stand for the amounts (in kg) of the species and MCO2 , MH2 , MCO , MH2O are the molecular masses in kg/kmol. While approaching the equilibrium, the system changes the four variables mCO2 , mH2 , mCO , mH2O and at the equilibrium the function (4.143) reaches its minimum. However, the changes in the four variables mCO2 , mH2 , mCO , mH2O occur so that the overall mass (m) remains constant m = m0CO2 + m0H2 + m0CO + m0H2O = 3 · 44 + 1 · 2 = 134 kg
(4.148)
207
4 Chemical Equilibrium and m = mCO2 + mH2 + mCO + mH2O
(4.149)
In Eqs. (4.144)  (4.147), M stands for the molecular mass of the mixture which is related to the four unknowns through the relationship mCO2 mH2 mCO mH2O 1 = + + + M m · MCO2 m · MH2 m · MCO m · MH2O
(4.150)
Mathematically speaking we should find values of the four variables mCO2 , mH2 , mCO , mH2O which minimise function (4.143). However, while looking for the minimum we are not allowed to vary the unknowns freely; we have to make sure that the number of C atoms, the number of O atoms and the number of H atoms is the same as in the initial mixture. Thus, the following equations which formulate the balance of elements must be satisfied
balance of C balance of O balance of H
12 12 · m0CO2 = · mCO2 + 44 44 32 32 · m0CO2 = · mCO2 + 44 44 2 m0H2 = mH2 + · mH2O 18
12 · mCO 28 16 16 · mCO + · mH2O 28 18
(4.151) (4.152) (4.153)
We mention in passing that summing up Eqs. (4.151), (4.152) and (4.153) provides Eq. (4.149), as one would expect. Now, using again the language of mathematics, we are facing a problem of finding a minimum of function (4.143) with the three constraints formulated by Eqs. (4.151), (4.152) and (4.153). The Lagrangean multiplier method is suitable for such problems and it can be applied here, as pointed out earlier. However, our current problem is rather simple and it can also be solved graphically. Using Eqs. (4.151)  (4.153) we can relate mH2 , mCO , mH2O to mCO2 mCO = 84 − 0.63636 · mCO2 mH2O = 54 − 0.4091 · mCO2 mH2 = −4 + 0.04546 · mCO2
(4.154) (4.155) (4.156)
We plot relation (4.143) as a function of a single variable mCO2 and we find that the minimum is at mCO2 = 95.788kg, as shown in Fig. 4.9. It is easy to see
208
4.4 Equilibrium Composition that the minimization of the Gibbs enthalpy of the system with respect to CO2 amount with constraints (4.154), (4.155) and (4.156) has provided the equilibrium mixture composition which is in full agreement with previously obtained results using Methods 1,2 and 3. We complete Method4 with a remark. The reader should notice that for the first time in this textbook, we have obtained an equilibrium composition of a system within which a chemical reaction occurs, without considering the reaction itself. This is an important observation since it invokes immediately a question "Do we need to consider at all chemical reactions for the determination of the equilibrium of a chemically reactive mixture?". We will elaborate on this important issue in the forthcoming paragraphs. 1.210
G ib b s en th alp y (G J )
1.215
CO2/H2/CO/H2O mixture 1.220
T = 1300 K
1.225
n CO = 3 kmol (132 kg)
0
2
0
n H = 1 kmol (2 kg) 2
1.230 1.235 1.240 1.245 80
90
100
110
120
130
140
mCO (kg) 2
Fig. 4.9: Gibbs enthalpy as a function of the amount of carbon dioxide for a CO2 /H2 /CO/H2 O mixture at T = 1300K and p = 1bar (compare with Fig. 4.8)
4.4.2 Systems with a twodimensional reaction basis As an example of a system with a twodimensional reaction basis, we consider a gas phase system containing the species: CH4 , H2 O, CO, CO2 , and H2 . The task is to calculate the equilibrium composition at T = 1000K and p = 1bar; the initial unreacted mixture contains 2kmol of CH4 and 3kmol of H2 O.
209
4 Chemical Equilibrium There are five (α = 5) species and three elements (γ = 3) in the system. All possible reactions can be written as ν1 CH4 + ν2 H2 O + ν3 CO + ν4 CO2 + ν5 H2 = 0
(4.157)
The balance of elements provides a matrix 1 0 1 1 0 0 0 1 1 2 0 0 4 2 0 0 2 0
(4.158)
where the first, second and the third rows show the balance of carbon, oxygen and hydrogen, respectively. Gauss elimination process applied to matrix (4.158) provides 1 0 1 1 0 0 1 0 1 1 0 0 0 1 1 2 0 0 2 0 0 → 0 1 1 0 0 −6 −8 2 0 0 2 −4 −4 2 0
(4.159)
Since the rank of matrix (4.159) is three (ρ = 3) then the reaction basis is two dimensional (R = α − ρ = 5 − 3 = 2) it means two independent reactions form a basis. Selecting ν4 = 1, ν5 = 0 we obtain ν3 = −4/3, ν 2 = −2/3, ν 1 = 1/3, so the first reaction is 1 2 4 3 CH4 − 3 H2 O− 3 CO
+ CO2 = 0
or CH4 + 3 CO2 = 4 CO + 2 H2 O
(4.160)
Selecting ν4 = 0, ν5 = 1 and obtaining from matrix (4.159) ν3 = 1/3, ν 2 = −1/3, ν 1 = −1/3 so that − 13 CH4 − 13 H2 O + 31 CO + H2 = 0 or CH4 + H2 O = CO + 3 H2
210
(4.161)
4.4 Equilibrium Composition Thus, we have selected reactions (4.160) and (4.161) as a reaction basis. Below we present two methods of finding the equilibrium composition. Only in the first methods we use the selected reaction basis. Method 1  Based on the extent of the reactions and the equilibrium constants We begin with calculating the thermodynamic equilibrium constants for the selected reaction basis. To this end we use the first four reactions listed in Table 4.2 which are here copied for completeness −C−2 H2 + CH4 = 0
(R1)
−C−O2 + CO2 = 0
(R2)
−C− 21 O2 + CO = 0
(R3)
−H2 − 12 O2 + H2 O = 0
(R4)
Since reaction (4.160) can be composed using reactions R1R4 as follows R(4.160) = −R1 − 3 · R2 + 4 · R3 + 2 · R4 log K(4.160) = −(−1.018) − 3 · 20.679 + 4 · 10.461 + 2 · 10.060 = 0.945 and K(4.160) = 8.8105. Reaction (4.161) can be created using reactions R1R4 as follows R(4.161) = −R1 + R3 − R4 and therefore log K(4.161) = −(−1.018) + 10.461 − 10.060 = 1.419 and K(4.161) = 26.2422. We pause here to summarise. We have the following two reactions into which we prescribe two extents, ξ1 and ξ2 , so that CH4 + 3CO2 = 4CO + 2H2 O
∆ν = 6 − 4 = 2
K = 8.8105
ξ1
CH4 + H2 O = CO + 3H2
∆ν = 4 − 2 = 2
K = 26.2422
ξ2
211
4 Chemical Equilibrium with the initial number of kmol: n0CH4 = 2kmol and n0H2O = 3kmol. Now we express the number of kmol of each species as a function of ξ1 and ξ2 so that nCH4 = n0CH4 − ξ1 − ξ2 = 2 − ξ1 − ξ2 nH2O = n0H2O + 2 · ξ1 − ξ2 = 3 + 2 · ξ1 − ξ2 nCO2 = −3 · ξ1 nCO = 4 · ξ1 + ξ2 nH2 = 3 · ξ2 nT = n0CH4 + n0H2O + 2 · ξ1 + 2 · ξ2 = 5 + 2 · ξ1 + 2 · ξ2
Using relationship (4.44) to calculate equilibrium constant Kn and after some algebra we obtain for the first reaction (4 · ξ1 + ξ2 )4 · (3 + 2 · ξ1 − ξ2 )2 − 8.8105 · (5 + 2 · ξ1 + 2 · ξ2 )2 = 0 (2 − ξ1 − ξ2 ) · (−3 · ξ1 )3
(4.162)
and for the second reaction (4 · ξ1 + ξ2 ) · (3 · ξ2 )3 − 26.2422 · (5 + 2 · ξ1 + 2 · ξ2 )2 = 0 (2 − ξ1 − ξ2 ) · (3 + 2 · ξ1 − ξ2 )
(4.163)
It is not a trivial matter to find the solution of Eqs. (4.162) and (4.163) since they are highly nonlinear. There exist several pairs ξ1 and ξ2 which satisfy the equations and therefore the more we know about the solution the better. After examining the interrelationships between the number of kmol and the reaction extends ξ1 and ξ2 , we expect ξ1eq < 0 and ξ2eq > 0 at the equilibrium. One of the easiest methods for finding a solution of Eqs. (4.162) and (4.163) is to frame the problem in terms of residuals. If we mark the left hand side of Eq. (4.162) as F1 (ξ1 , ξ2 ) and the left hand side of Eq. (4.163) as F2 (ξ1 , ξ2 ), finding the solution is equivalent to minimising the residuals of a function F12 (ξ1 , ξ2 ) + F22 (ξ1 , ξ2 ). To minimise this function we have used a general gradient method which is easily accessible since it is incorporated into the Solver option of the Excel program. The
212
4.4 Equilibrium Composition Solver needs an initial guess of ξ1 and ξ2 . For the initial values of ξ1 = −0.5 and ξ2 = 2.0 the solution has been found to be ξ1eq = −0.1065364 and ξ2eq = 1.931939. eq Consequently, the equilibrium mixture contains: neq CH4 = 0.17460 kmol, nH2O = eq eq eq 0.85499 kmol, nCO2 = 0.31961 kmol, nCO = 1.50579 kmol, nH2 = 5.79582 kmol eq and neq T = 8.65081kmol. Thus, the mole fractions at the equilibrium are xCH4 = eq eq eq eq 0.02018, xH2O = 0.09883, xCO2 = 0.03695, xCO = 0.17406, xH2 = 0.66997. Method2 Minimization of Gibbs enthalpy of the system The method presented in this paragraph is almost identical to Method4 of the previous paragraph. The main difference is in that the minimization procedure is going to be performed with respect to two variables. The Gibbs enthalpy at T = 1000K for the five species present in the system are given in Table 4.5. Table 4.5: Standard Gibbs enthalpies at T = 1000 K [12]
Species CH4 H2 O(g) CO CO2 H2
∆f G01000 kJ/mol 19.492 192.590 200.275 395.886 0
The starting point is the expression for the Gibbs enthalpy of the whole system G = GCH4 + GH2O + GCO + GCO2 + GH2
(4.164)
RT mCH4 · M p = mCH4 · 1.218 + · ln · · 103 MCH4 m · MCH4 p0
(4.165)
mH2O · M p RT · ln · · 103 = mH2O · −10.699 + MH2O m · MH2O p0
(4.166)
RT mCO · M p = mCO · −7.153 + · ln · · 103 MCO m · MCO p0
(4.167)
with
GCH4
GH2O
GCO
213
4 Chemical Equilibrium
GCO2 = mCO2 ·
GH2
RT −8.997 + · ln MCO2
mCO2 · M p · m · MCO2 p0
· 103
mH2 · M p RT · ln · · 103 = mH2 · 0 + MH2 m · MH2 p0
(4.168)
(4.169)
where mCH4 , mCO2 , mH2 , mCO , mH2O stand for the amounts (in kg) of the species and MCH4 , MCO2 , MH2 , MCO , MH2O are the molecular masses in kg/kmol and M stands for the molecular mass of the mixture which is related to the species amounts through the relationship
1 mCH4 mH2O mCO mCO2 mH2 = + + + + M m · MCH4 m · MH2O m · MCO m · MCO2 m · MH2
(4.170)
The overall mass of the system is m = m0CH4 + m0H2O = 2 · 16 + 3 · 18 = 86 kg
(4.171)
Now the task is to find values of the five variables mCH4 , mH2 O , mCO , mCO2 , mH2 which minimise function (4.164). The following equations formulate the constraints of the minimization problem:
balance of C balance of O balance of H
12 12 · m0CH4 = · mCH4 + 16 16 16 16 · m0H2O = · mH2O + 18 18 4 2 · m0CH4 + · m0H2O = 16 18
12 12 · mCO + · mCO2 (4.172a) 28 44 16 32 · mCO + · mCO2 (4.172b) 28 44 4 2 · mCH4 + · mH2O + mH2 16 18 (4.172c)
Since we have five variables mCH4 , mH2 O , mCO , mCO2 , mH2 and three constraints, we express7 mCO , mCO2 , mH2 as a function of mCH4 and mH2 O so that mCO = 28 − 3.5 · mCH4 + 1.5556 · mH2O 7
(4.173)
You can practice Gauss elimination to obtain relationships (4.173), (4.174) and (4.175).
214
4.4 Equilibrium Composition
mCO2 = 44 + 2.75 · mCH4 − 2.4445 · mH2O
(4.174)
mH2 = 14 − 0.25 · mCH4 − 0.1111 · mH2O
(4.175)
Fig. 4.10: Gibbs enthalpy of the considered system as a function of mCH4 and mH2 O . The mixture contains five species CH4 , H2 O, CO, CO2 , H2 at T = 1000K and p = 1bar. Initial nonreacted mixture of 2 kmol CH4 and 3 kmol H2 O.
Fig. 4.10 shows a plot of function (4.164) in the vicinity of the minimum for the system in question. It is to observe that the minimum of function (4.164) is somewhere around mCH4 ≈ 2.5 kg and mH2O ≈ 15 kg. To find the exact values of mCH4 which minimise function (4.164) we have used a general gradient method which is incorporated into the Solver option of the Excel program. For an initial guess of mCH4 = 2.5 kg and mH2O = 15 kg, the solver finds the exact solution as eq meq CH4 = 2.7901 kg and mH2O = 15.3835 kg. The remaining species are present at the equilibrium in amounts determined by Eqs. (4.173), (4.174) and (4.175) eq eq so meq CO = 42.165 kg, mCO2 = 14.068 kg and mH2 = 11.593 kg. Recalculating
215
4 Chemical Equilibrium these values into mole fractions we obtain the following equilibrium composition: eq eq eq eq xeq CH4 = 0.02016, xH2O = 0.09879, xCO = 0.17407, xCO2 = 0.03696, xH2 = 0.670. Obviously the just calculated values are in close agreement with the solution obtained using Method1.
4.4.3 Systems with a multidimensional reaction basis In Section 4.4.1 we have considered systems with a onedimensional reaction basis (R = α − ρ = 1). We have observed that when a single chemical reaction is sufficient to describe the system (to form reaction basis) we could determine the equilibrium composition by using several methods namely Method 1, Method 2 and Method 3. All these three methods have used reaction (4.120) and they are rather simple. Indeed, it is likely that you have already used Method 1 (or Method 2) in your high school chemistry course. However, Method 4, based on the minimization of Gibbs enthalpy, is perhaps the most interesting since we have succeeded in calculating the equilibrium composition without considering any chemical reactions whatsoever. In Section 4.4.2 we have considered systems with a twodimensional reaction basis (R = α − ρ = 2). Method1, based on extents of the reaction basis and the equilibrium constants, has already become a bit more cumbersome. Firstly, we had to identify the two reactions forming the reaction basis. Secondly we had to calculate the equilibrium constants of the reactions and thirdly the resulting Eqs. (4.162) and (4.163) were highly nonlinear. All together, the procedure underlined in Method1 was doable but tedious and perhaps it was at the limit of our patience. Imagine what would happen when a system of, let say, twenty species would have to be considered. In short, Method1 does not seem to be easy to generalise to tackle systems of many reacting species. To the contrary, Method2 based on Gibbs enthalpy minimization retains its simplicity even if systems of many species are considered. Most importantly, we are not bothered with the chemical reactions so the question of what and how many reactions should we chose is irrelevant. We just have to make sure that both Gibbs system enthalpy and constraints are properly formulated. Having done that, we need a mathematical procedure for finding a minimum of the Gibbs enthalpy with all constraints satisfied. As a matter of fact, we have already used in Section 3.6 and Section 4.3.1 the Lagrangean multipliers method which does just this. Method  Minimization of Gibbs system enthalpy using Lagrangean multipliers Although the theoretical background of the method has already been provided in
216
4.4 Equilibrium Composition Section 3.6 (see Example 3.5) we demonstrate here its applicability for calculating the equilibrium composition of the system containing the five species: CH4 , H2 O, CO, CO2 and H2 . Again we calculate the equilibrium composition at T = 1000K and p = 1bar; the initial unreacted mixture contains 2kmol of CH4 and 3kmol of H2 O. The starting point is again relationship (4.164) which we copy for the sake of completeness G = GCH4 + GH2O + GCO + GCO2 + GH2 (4.176) with the constraints (4.172a), (4.172b), (4.172c) 24 − 0.75000 · mCH4 − 0.42857 · mCO − 0.27273 · mCO2 = 0
(4.177)
48 − 0.88889 · mH2O − 0.57143 · mCO − 0.72727 · mCO2 = 0
(4.178)
14 − 0.25000 · mCH4 − 0.11111 · mH2O − mH2 = 0
(4.179)
The new function Y (see Example 3.5) is formulated as Y = GCH4 + GH2O + GCO + GCO2 + GH2 + + λ1 · [24 − 0.75000 · mCH4 − 0.42857 · mCO − 0.27273 · mCO2 ] + + λ2 · [48 − 0.88889 · mH2O − 0.57143 · mCO − 0.72727 · mCO2 ] + + λ3 · [14 − 0.25000 · mCH4 − 0.11111 · mH2O − mH2 ] = 0 (4.180) The necessary conditions for a minimum of Yfunction are: ∂Y = µCH4 − 0.75000 · λ1 − 0.25000 · λ3 = ∂mCH4 mCH4 · M p RT · ln · − 0.75000 · λ1 − 0.25000 · λ3 = 0 1.218 + MCH4 m · MCH4 p0 (4.181) mH2O · M p RT ∂Y = −10.699 + · ln · ∂mH2O MH2O m · MH2O p0 − 0.88889 · λ2 − 0.11111 · λ3 = 0 (4.182) ∂Y = ∂mCO mCO · M p RT · ln · − 0.42857 · λ1 − 0.57143 · λ2 = 0 −7.153 + MCO m · MCO p0 (4.183)
217
4 Chemical Equilibrium ∂Y = ∂mCO2 RT mCO2 · M p −8.997 + − 0.27273 · λ1 − 0.72727 · λ2 = 0 · ln · MCO2 m · MCO2 p0 (4.184)
∂Y ∂mH2
=
mH2 · M p RT 0+ · ln · − λ3 MH2 m · MH2 p0
=
0 (4.185)
∂Y = 24 − 0.75000 · mCH4 − 0.42857 · mCO − 0.27273 · mCO2 = 0 (4.186) ∂λ1 ∂Y = 48 − 0.88889 · mH2O − 0.57143 · mCO − 0.72727 · mCO2 = 0 (4.187) ∂λ2 ∂Y ∂λ3
=
14 − 0.25000 · mCH4 − 0.11111 · mH2O − mH2
=
0 (4.188)
Inserting into (4.181)  (4.188) T = 1000K, p = 1bar, m = 86kg and appropriate values for the molecular masses we obtain mCH4 · M 2.3438 + ln − 1.4432 · λ1 − 0.4811 · λ3 = 0 (4.189) 1.376 · 103 mH2O · M − 23.1616 + ln − 1.9243 · λ2 − 0.2405 · λ3 = 0 (4.190) 1.548 · 103 mCO · M − 24.0869 + ln − 1.4432 · λ1 − 1.9243 · λ2 = 0 (4.191) 2.408 · 103 mCO2 · M − 1.4432 · λ1 − 3.8486 · λ2 = 0 (4.192) − 47.6128 + ln 3.784 · 103 mH2 · M 0 + ln − 0.2405 · λ3 = 0 (4.193) 172 24 − 0.75000 · mCH4 − 0.42857 · mCO − 0.27273 · mCO2 = 0 (4.194) 48 − 0.88889 · mH2O − 0.57143 · mCO − 0.72727 · mCO2 = 0
(4.195)
14 − 0.25000 · mCH4 − 0.11111 · mH2O − mH2 = 0
(4.196)
The above equations are accompanied by relationship (4.170) which is here written
218
4.5 Summary as mCH4 mH2O mCO mCO2 mH2 1 = + + + + 3 3 3 3 M 1.376 · 10 1.548 · 10 2.408 · 10 3.784 · 10 172
(4.197)
Perhaps it is time to pause and summarise. We have formulated eight equations (4.189)  (4.196); five equations (4.189)  (4.193) stem from the differentiating of the Gibbs enthalpy with respect to the amounts of the five species while the remaining three equations are the constraints. Thus, all together we have eight equations and eight unknowns (five unknown amounts of the species mCH4 , mH2O , mCO , mCO2 , mH2 and three multipliers λ1 , λ2 , λ3 ). Finding a solution of the eight equations is equivalent to minimising the residuals of a function F12 + F22 + ... + F82 which at the solution should take near zero value. To minimise this function we used again the Solver option of the Excel program. For an initial guess of mCH4 = 10 kg and mH2O = 20 kg, the solver finds the exact solution as meq CH4 = eq eq eq 2.7906 kg and mH2O = 15.3855 kg, mCO = 42.166 kg, mCO2 = 14.065 kg and meq H2 = 11.593 kg. Obviously the just calculated values are in close agreement with the solutions obtained previously.
4.5 Summary This chapter is concerned with chemical equilibrium which has been defined as the state of the system if no changes occur to its chemical composition. The ultimate goal of this chapter is to provide the knowledge required for calculation of chemical composition at equilibrium. Extent of a reaction (ξ), expressed in mol or kmol, is a single variable which relates the changes in the amount of species participating in the chemical reaction. It varies from zero to a value that corresponds to the equilibrium state. At constant a temperature and a pressure, when a reaction proceeds towards equilibrium, Gibbs enthalpy of the system approaches a minimum. The derivative of Gibbs enthalpy of the system with respect to the extent of the reaction is the Gibbs reaction enthalpy (∆R GpT ). More explicitly, the Gibbs reaction enthalpy is the change in Gibbs enthalpy of the system when the extent of the reaction changes by 1 mol (or kmol). At the equilibrium the Gibbs reaction enthalpy is zero. On the basis of this statement the thermodynamic equilibrium constant is derived which is a dimensionless quantity. Several other equilibrium constants have been derived from the thermodynamic constant. The general conditions for a multicomponent, multiphase system with chemical
219
4 Chemical Equilibrium reactions to be at equilibrium are: T (1) = T (2) = T (3) = . . . = T (i) = . . . = T (β) = T p(1) = p(2) = p(3) = . . . = p(i) = . . . = p(β) = p µ1k = µ2k = µ3k = . . . = µβk and
∂G ∂ξr
= ∆r GpT =
α X k=1
for k = 1, 2, . . . , α
µk · Mk · νkr =
α X
µk · νkr = 0
k=1
Thus, equality of temperatures and pressures of each phase as well as equality of the chemical potentials in each phase for each component are required. Furthermore, Gibbs enthalpies of all the chemical reactions must be zero at chemical equilibrium. A considerable part of this chapter has been dedicated to methods used for determination of equilibrium composition for given both a pressure and a temperature and when the composition of the initial (unreacted) mixture is given. This is a classical problem in combustion, process and chemicalengineering and must be well understood. In highschools you have learned how to determine the equilibrium composition for a single chemical reaction using both the extent of the reaction and the equilibrium constant. Here you have extended this knowledge into more complex systems supporting a multiplicity of chemical reactions. A procedure for the determination of both the number of reactions and the reactions themselves has been underlined in this chapter. However, if the number of reaction exceeds two, the calculation methods have become tedious and they exceed the upper limit of patience for a reasonably human being. Perhaps the most important message of this chapter is that there is no need to consider chemical reactions at all to calculate the chemical equilibrium. This should not be too surprising since from the thermodynamic point of view the equilibrium state is uniquely determined by the temperature, pressure and the initial (unreacted) mixture. The procedure of Gibbs enthalpy minimization is highly recommended. In this procedure, the question of what chemical reactions are involved never enters directly into any of the equations. However, the choice of a set of species is entirely equivalent to the choice of a reaction basis among the species. In any case, a set of species must always be assumed and different assumptions produce different results. Students should master the method of Lagrangean multipliers used to determine a minimum of functions describing Gibbs enthalpy of the system.
220
5 Elements of Chemical Kinetics Contents 5.1
Introduction
5.2
Rate Laws and Reaction Orders
5.3
Forward and Reverse Reactions
5.4
Elementary Reactions and Reaction Molecularity
5.5
Rate of Reactions
5.6
5.5.1
Temperature dependence of rate coefficients
5.5.2
Pressure dependence of rate coefficients
Summary
5.1 Introduction In Chapter 4 we have learned that equilibrium composition can be found without any considerations as to the mechanisms of chemical reactions. This is, because the equilibrium composition for a given temperature and a pressure is uniquely determined by the initial conditions. The calculations carried out in Chapter 4 do not provide any hint as to how long it takes to reach the equilibrium. A discipline called Chemical Kinetics aims both at providing insights into the reaction mechanisms and at determining the rates of individual reactions. Thus, Chemical Kinetics provides means of estimating the time needed for reactions to reach a certain extent. Essential features of chemical kinetics, which occur frequently in combustion phenomena, are reviewed in this chapter. For a thorough coverage of the subject one should refer to books on chemical kinetics [18, 19].
221
5 Elements of Chemical Kinetics
5.2 Rate Laws and Reaction Orders Let’s write a chemical reaction as: νA · A + νB · B + νC · C + . . . −→ νD · D + νE · E + νF · F + . . .
(5.1)
where A, B, C, . . . and D, E, F, . . . denote chemical species involved in the reaction. A rate law is an empirical formula for the reaction rate; it means for the rate of formation (for example in mol/s) or consumption of a species. In chemical kinetics, the rate of consumption of species A is expressed as: d[A] = −k · [A]a · [B]b · [C]c · . . . dt
(5.2)
In the above expression a, b, c, . . . are reaction orders with respect to species A, B, C, . . . and k is the rate coefficient of the reaction. The sum of the all exponents a, b, c, . . . is the overall reaction order. The convention used in this chapter is that square parenthesis around a chemical symbol denotes the concentration of that species in moles (or grams) per cubic centimeter. If for example concentrations of species B and C remain nearly constant during the course of the reaction (species B and C are in excess), one may obtain a simplified version of expression (5.2) d[A] = −kexper · [A]a dt
(5.3)
where kexper denotes the rate constant for a particular experiment: kexper = k · [B]b · [C]c
(5.4)
Example 5.1 When exponent "a" in Eq. (5.3) equals one (a = 1) the reaction is called a firstorder reaction. Calculate the rate of change of the Aspecies concentration with time for a firstorder reaction. Assumptions: none For a = 1 the rate of the consumption of Aspecies is: d[A] = −kexper · [A] dt
222
5.2 Rate Laws and Reaction Orders A separation of the variables leads to: d[A] = −kexper · dt [A] and a subsequent integration gives: ln
[A](t) = −kexper · (t − t0 ) [A]0
Fig. 5.1: Concentration change with time for a firstorder reaction (t0 = 0)
Comments: (a) For firstorder reactions, logarithm of normalised concentration of the reagent [A] [A]0 is a linear function of time. (b) The slope is determined by the rate constant k. End of Example 5.1 It is informative to compare how quickly the concentration of the reacting species (A) decays with time, depending on the reaction order and the initial concentra
223
5 Elements of Chemical Kinetics tion of [A]0 . The equations derived in Examples 5.1 and 5.2 yield: [A](t) = [A]0 · exp(−kexper · (t − t0 )) 1 [A](t) = 1 [A]0 + kexper · (t − t0 )
for a firstorder reaction for a secondorder reaction
while [A](t) =
s
1 1 [A]20
+ 2 kexper · (t − t0 )
for a thirdorder reaction.
As for any timedependent process one can calculate the time constant for any single reaction. The time constant is the time required for the initial concentration [A]0 to drop to [A]e 0 , where e = 2.718 281 83 is the base of natural logarithm. For example, for a firstorder reaction: [A](t) = [A]0 · exp(−kexper · (t − t0 )) 1 [A](tc ) = = exp(−kexper · (tc − t0 )) [A]0 e
(5.5) (5.6)
and tc = t0 +
1 kexper
(5.7)
It is easy to verify that time constants for the second and thirdorder reactions are given by: (e − 1) kexper · [A]0 e2 − 1 tc = t0 + kexper · [A]20 tc = t0 +
for a secondorder reaction
(5.8)
for a thirdorder reaction
(5.9)
Example 5.2 When exponent "a" in Eq. (5.3) equals two (a = 2) the reaction is called a secondorder reaction while for a = 3 the reaction is thirdorder. Calculate the rate of change of the Aspecies concentration with time for such reactions. Assumptions: none
224
5.2 Rate Laws and Reaction Orders For a = 2 the rate of the consumption of Aspecies is: d[A] = −kexper · [A]2 dt and after separation of variables: d[A] = −kexper · dt [A]2 and integration, one obtains: 1 1 − = kexper · (t − t0 ) [A](t) [A]0
Fig. 5.2: Concentration change with time for a secondorder reaction
Comments: (a) Fig. 5.1 and Fig. 5.2 show a good method of determining the reaction order using measured data. End of Example 5.2
225
5 Elements of Chemical Kinetics
5.3 Forward and Reverse Reactions At chemical equilibrium the rate of consumption equals the rate of formation of chemical species and the forward and backward reactions are of the same rate. While expression (5.2) holds for the forward reaction, the following is applicable for the backward reaction: d[A] = kb · [D]d [E]e [F ]f . . . dt
(5.10)
At chemical equilibrium one has kf · [A]a [B]b [C]c . . . = kb · [D]d [E]e [F ]f . . .
(5.11)
where kf and kb are the rate constants for the forward and backward reactions, respectively. Rearranging Eq. (5.11) one obtains: kf [D]d [E]e [F ]f . . . = a b c kb [A] [B] [C] . . .
(5.12)
The left hand side of the above equation is the equilibrium constant which we have already derived in Chapter 4, thus Kc =
kf kb
(5.13)
Example 5.3 Consider a firstorder reaction −− ⇀ A↽ − −B close to equilibrium. Derive a formula for calculating A and B species concentrations with time. Assume that initially (at t = 0) only Aspecies is present and its initial concentration is [A]0 . Calculate both the equilibrium composition and the time required to reach the equilibrium for kf = 2 s−1 , kb = 2 s−1 and kf = 3 s−1 , kb = 1 s−1 . Assumptions: Both forward and backward reactions are first order. We begin with noting that the concentration of Aspecies is reduced by the forward reaction and it is increased by the backward reaction. The net rate change is therefore: d[A] = −kf [A] + kb [B] dt
226
5.3 Forward and Reverse Reactions Since the initial concentration of A is [A]0 and no Bspecies is present, then at any instant [A] + [B] = [A]0 and d[A] = −kf [A] + kb ([A]0 − [A]) = −(kf + kb ) [A] + kb [A]0 dt or d[A] + (kf + kb ) [A] = kb [A]0 dt The above equation is a firstorder linear differential equation of a general form: y ′ + p(x) · y = r(x) where p(x) is a constant equal (kf + kb ) and r(x) is also a constant equal kb [A]0 . The solution of such an equation can R be found upon multiplying both sides of the equation by an integrating factor e p(x) dx obtaining: e
R
p(x) dx ′
y + p(x) · y · e
R
p(x) dx
= r(x) e
R
p(x) dx
Observing the product rule of differentiation one obtains: ′ R R y · e p(x) dx = r(x) · e p(x) dx
and now integrating with respect to x, we obtain the general form of the solution: Z R R p(x) dx = r(x) · e p(x) dx dx + C y·e
Now we apply this general solution to our chemical reaction rate equation. We recall that: p(x) = (kf + kb ) and r(x) = kb [A]0 while yvariable represents [A](t) and xvariable represents the time t, so e Z
r(x) · e
R
p(x) dx
R
p(x) dt
dx =
Z
=e
R
(kf +kb ) dt
= e(kf +kb ) t
kb [A]0 · e(kf +kb ) t dt =
kb [A]0 · e(kf +kb ) t (kf + kb )
227
5 Elements of Chemical Kinetics and [A](t) =
C kb [A]0 + (k +k ) t (kf + kb ) e f b
We can easily calculate the integration constant C using the initial conditions: [A](0) = [A]0 so kb · [A]0 +C [A]0 = (kf + kb ) and C = [A]0
kf (kf + kb )
Thus, the final form of the solution of the chemical rate equation is: [A](t) =
kb + kf · e−(kf +kb ) t · [A]0 (kf + kb )
for Aspecies (C1)
and [B](t) = [A]0 − [A](t) = [A]0 ·
kb + kf · e−(kf +kb ) t 1− (kf + kb )
!
for Bspecies (C2)
Fig. 5.3 shows the plot of the Aspecies and Bspecies concentrations with time for different values of kf and kb rate constants.
The above equations can be used to calculate the equilibrium composition of the reacting species. This is obtained upon examining what happens to [A](t) and [B](t) when t → ∞. Such an analysis leads to the following: [A](∞) = lim [A](t) = [A]0 · lim · t→∞
t→∞
[B](∞) = lim [B](t) = [A]0 · lim t→∞
t→∞
kb + kf · e−(kf +kb ) · t kb = [A]0 · (kf + kb ) kf + kb ! −(k +k ) · t kf kb + kf · e f b = [A]0 · 1− (kf + kb ) kf + kb
For kf = 2 s−1 and kb = 2 s−1 , the equilibrium composition is [B]eq [A]0 [A]eq [A]0
228
1 2
= =
1 4
[A]eq [A]0
=
1 2
and
while for kf = 3 s−1 and kb = 1 s−1 the equilibrium composition is and
[B]eq [A]0
=
3 4
as shown in Fig. 5.3.
5.3 Forward and Reverse Reactions
Fig. 5.3: The approach of concentrations to their equilibrium values for the reversible − ⇀ reaction A − ↽ − − B that is first order in each direction. The solid line: kf = 2 s−1 and kb = 2 s−1 ; the dash dot line: kf = 3 s−1 and kb = 1 s−1 .
We will now calculate the time required to reach the equilibrium. Strictly speaking an infinitely long time is needed to reach the equilibrium as shown in the above analysis. However, as shown in Fig. 5.3, practically after around 1.3 s the concentrations of the reacting species do not vary any more with time for kf = kb = 2 s−1 . Thus, we will calculate the time for the Aspecies concentration to be within 1 % from the equilibrium concentration. To this end we begin with the relationship describing the dependence of the Aspecies concentration with time: kb + kf · e−(kf +kb ) t [A](t) = · [A]0 (kf + kb ) and 1.01 ·
kb + kf · e−(kf +kb ) t0.01 kb · [A]0 · [A]0 = kf + kb (kf + kb )
and 0.01 kb = kf · e
−(kf +kb ) t0.01
t0.01 =
kb ln 0.01 kf
−(kf + kb )
Thus, for kf = 2 s−1 and kb = 2 s−1 the time required for Aspecies to reach a
229
5 Elements of Chemical Kinetics concentration that is 1 % different from the equilibrium concentration is 1.15 s. For kf = 3 s−1 and kb = 1 s−1 a value of 1.43 s is applicable. To complete our exercise we will carry out simple thermodynamic calculations to obtain the equilibrium −− ⇀ composition. The equilibrium constant for the reaction A ↽ − − B reads: Kc =
kf [B]eq = [A]eq kb
The equilibrium concentrations can be expressed in terms of the extent of the reaction ξ (see Section 4.2.1) by drawing up the following table:
Initial concentration Change to reach equilibrium Concentration at equilibrium
Aspecies [A]0 −ξ [A]0 [A]0 (1 − ξ)
Bspecies 0 ξ [A]0 ξ [A]0
The equilibrium constant is therefore: Kc =
kf ξeq [A]0 ξeq = = kc (1 − ξeq ) [A]0 1 − ξeq
For kf = 2 s−1 and kb = 2 s−1 , the equilibrium constant Kc equals 1 and ξeq equals 12 , so the equilibrium composition is [A]eq = 12 [A]0 and [B]eq = 12 [A]0 while for kf = 3 s−1 and kb = 1 s−1 the equilibrium composition is [A]eq = 41 [A]0 and [B]eq = 43 [A]0 . These values are in agreement with our kinetic calculations shown in Fig. 5.3. Comments: (a) “Mathematically speaking” an infinitely long time is needed to reach equilibrium. (b) Relationship (C1) which has been derived on the basis of chemical kinetics, provides the equilibrium concentration of Aspecies upon invoking the condition t → ∞; similarly relationship (C2) gives the equilibrium concentration of Bspecies upon t → ∞. (c) Generally, when t → ∞ chemical kinetics meets equilibrium thermodynamics. (d) Note that relationship (5.13), namely Kc = only.
kf kb ,
End of Example 5.3
230
is valid at the equilibrium
5.4 Elementary Reactions and Reaction Molecularity
5.4 Elementary Reactions and Reaction Molecularity If a reaction occurs on a molecular level exactly as it is described by the reaction equation (Eq. (5.1)) then it is called an elementary reaction. The reaction of hydroxyl radicals (OH) with molecular hydrogen (H2 ) forming water and hydrogen atoms is an elementary reaction, OH + H2 −−→ H2 O + H
(5.14)
Hydroxyl radical molecules collide with hydrogen molecules and in the case of reactive collisions water molecules and hydrogen molecules are formed. In the case of nonreactive collisions the molecules bounce away. The reaction 2 H2 + O2 −−→ 2 H2 O
(5.15)
is not an elementary reaction. According to this reaction two molecules of hydrogen should collide with one molecule of oxygen to produce two molecules of water. Detailed studies show that many reactive intermediates like H, O and OH are formed. Thus, reaction (5.15) describes an overall effect of numerous elementary reactions; reactions (5.15) are called overall reactions. Often these overall reactions have complicated rate laws with the reactions orders a, b, c, . . . (see Eq. (5.2)) being usually not integers that can be negative. Table 5.1 (adapted from [20]) shows some elementary reactions in H2 −CO−C1 −C2 −O2 system. There are several advantages in using elementary reactions. The reaction order of these reactions is constant (not dependent on time and experimental conditions) and it can be easily determined knowing the reaction equation. In order to be able to use the elementary reactions one has to establish the molecularity of the reaction. The molecularity is the number of species that form the reaction complex that is the transition state between reactants and products. Only three values of the reaction molecularity have been observed experimentally. These are unimolecularity, bimolecularity and trimolecularity. Unimolecular reactions proceed according to the equation: A −→ products
(5.16)
and they describe dissociation of a molecule. The rates of these reactions are described by a firstorder equation (see Example 5.1).
231
5 Elements of Chemical Kinetics Bimolecular reactions proceed according to the equations: A + A −→ products
(5.17)
A + B −→ products
(5.18)
or
and they have a secondorder rate law. Trimolecular reactions are proceeding according to the equations: A + A + A −→ products A + A + B −→ products
(5.19)
A + B + C −→ products and these are usually recombination reactions. It is important to distinguish molecularity from order: reaction order is an empirical quantity obtained from the experimental rate law, while the molecularity refers to an elementary reaction proposed as an individual step in a mechanism. In general the molecularity equals the order for elementary reactions. Consider a system containing N species (n = 1, 2, . . . , N ) taking part in R elementary reactions (r = 1, 2, . . . , R). Each of these elementary reactions r can be written as: N X
n=1
k
r (s) νr,n An −→
N X
(p) νr,n An
for r = 1, 2, . . . , R
n=1
(5.20)
where the superscripts (s) and (p) represent substrates and products, respectively. The rate of formation of the nth species in reaction r is given by: N Y (e) ∂[An ] νr,n (p) (e) [A ] = ν − ν · k · n r r,n r,n ∂t rreaction
for n = 1, 2, . . . , R
n=1
232
(5.21)
5.4 Elementary Reactions and Reaction Molecularity The summation for all the elementary reactions gives the overall rate of formation of nth species as: R X ∂[An ] ∂t r=1
= rreaction
R X r=1
N Y (e) (p) (e) [An ]νr,n νr,n − νr,n · kr ·
for n = 1, 2, . . . , N
n=1
(5.22)
Table 5.1: (adapted from [20]). Elementary reactions in H2 −CO−C1 −O2 system at p = 1 bar for high temperature (T > 1200 K); rate coefficients are presented in the form k = A T b exp − RET , [M∗ ] = [H2 ] + 13 [ H2 O] + 52 [ O2 ] + 25 [ N2 ] + 34 [ CO] + 32 [ CO2 ] + 3 [ CH4 ]; 2 only forward reaction is considered; reverse reaction to be calculated using (5.13). Reaction — 01.–04. H2 −CO Oxidation — 01. H2 −O2 Reactions (HO2 , H2 O2 not included) O2 +H =OH +O H2 +O =OH +H H2 +OH =H2 O +H OH +OH =H2 O +O H +H +M∗ =H2 +M∗ ∗ O +O +M =O2 +M∗ ∗ H +OH +M =H2 O +M∗ — 02. HO2 Formation/Consumption H +O2 +M∗ =OH +O +M∗ HO2 +H =OH +OH HO2 +H =H2 +O2 HO2 +H =H2 O +O HO2 +O =OH +O2 HO2 +OH =H2 O +O2 — 03. H2 O2 Formation/Consumption HO2 +HO2 =H2 O2 +O2 OH +OH +M∗ =H2 O2 +M∗ H2 O2 +H =H2 +HO2 H2 O2 +H =H2 O +OH H2 O2 +O =OH +H2 O H2 O2 +OH =H2 O +HO2 — 04. CO Reactions CO +OH =CO2 +H CO +HO2 =CO2 +OH CO +O +M∗ =CO2 +M∗ CO +O2 =CO2 +O — 10.–19. C1 Hydrocarbons Oxidation
A [cm,mol,s]
b
E [kJ/mol]
2.00 · 1014 5.06 · 104 1.00 · 108 1.50 · 109 1.80 · 1018 2.90 · 1017 2.20 · 1022
0.00 2.67 1.60 1.14 −1.00 −1.00 −2.00
70.30 26.30 13.80 0.42 0.00 0.00 0.00
2.30 · 1018 1.50 · 1014 2.50 · 1013 3.00 · 1013 1.80 · 1013 6.00 · 1013
−0.80 0.00 0.00 0.00 0.00 0.00
0.00 4.20 2.90 7.20 −1.70 0.00
2.50 · 1011 3.25 · 1022 1.70 · 1012 1.00 · 1013 2.80 · 1013 5.40 · 1012
0.00 −2.00 0.00 0.00 0.00 0.00
−5.20 0.00 15.70 15.00 26.80 4.20
6.00 · 106 1.50 · 1014 7.10 · 1013 2.50 · 1012
1.50 0.00 0.00 0.00
−3.10 98.70 −19.00 200.00
233
5 Elements of Chemical Kinetics Table 5.1: (continued) Reaction — 10. CH CH CH CH CH CH — 11. CHO CHO CHO CHO CHO CHO CHO — 12. CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 — 13. CH2 O CH2 O CH2 O CH2 O CH2 O CH2 O CH2 O — 14. CH3 CH3 CH3 CH3 CH3 CH3 O CH3 CH3 CH3
234
CH Reactions +O +O2 +M∗ +CO2 +H2 O +H2 O +M∗ +OH CHO Reactions +M∗ +H +O +O +OH +O2 +CHO CH2 Reactions +H +O +CH2 +CH2 +CH3 +O2 +O2 +M∗ +O2 +H2 + CH3 CH2 O Reactions +M∗ +H +O +OH +HO2 +CH3 +O2 CH3 Reactions +M∗ +M∗ +O +H +OH +H +O2 +HO2 +HO2
A [cm,mol,s]
b
E [kJ/mol]
=CO =CHO =CHO =CH2 O =CHO =CHO
+H +O +CO +H +O +CO
4.00 · 1013 6.00 · 1013 3.40 · 1012 3.80 · 1012 6.00 · 1012 3.40 · 1013
0.00 0.00 0.00 0.00 0.00 0.00
0.00 −0.00 2.90 −3.20 −3.20 0.00
=CO =CO =CO =CO2 =CO =CO =CH2 O
+H +M∗ +O +CO +H +O +CO +CO
7.10 · 1012 9.00 · 1013 3.00 · 1013 3.00 · 1013 1.00 · 1014 3.00 · 1013 3.00 · 1013
0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 −0.00 2.90 −3.20 −3.20 0.00 0.00
=CH =CO =C2 H2 =C2 H2 =C2 H4 =CO =CO2 =3 CH2 =CO =CH3 =C2 H4
+H2 +H +H +H2 +H +H +H +OH +H +H2 +M∗ +OH +H +H +H
6.00 · 1012 8.40 · 1012 1.20 · 1013 1.10 · 1014 4.20 · 1013 1.30 · 1013 1.20 · 1013 1.20 · 1013 3.10 · 1013 7.20 · 1013 1.60 · 1013
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
−7.50 0.00 3.40 3.40 0.00 6.20 6.20 0.00 0.00 0.00 −2.38
=CHO =CHO =CHO =CHO =CHO =CHO =CHO
+H +M∗ +H2 +OH +H2 O +H2 O2 +CH4 +HO2
5.00 · 1016 1.60 · 1013 1.60 · 1013 1.60 · 1013 1.60 · 1013 1.60 · 1013 1.60 · 1013
0.00 1.05 0.57 1.20 0.00 0.00 0.00
320.00 13.70 11.60 −1.90 54.70 25.50 171.00
=CH2 =CH2 =CH2 O =CH4 =CH3 O =CH3 =CH2 O =CH3 O =CH4
+H +M∗ +H +M∗ +H
6.90 · 1014 1.00 · 1016 8.43 · 1013 1.93 · 1036 2.26 · 1014 4.75 · 1016 3.30 · 1011 1.80 · 1013 3.60 · 1012
0.00 0.00 0.00 −7.00 0.00 −0.13 0.00 0.00 0.00
345.00 379.00 0.00 38.00 64.80 88.00 37.40 0.00 0.00
+H +OH +OH +OH +O2
5.4 Elementary Reactions and Reaction Molecularity Table 5.1: (continued) Reaction CH3 +CH3 CH3 +CH3 — 15a. CH3 O Reactions CH3 O +M∗ CH3 O +H CH3 O +O2 CH2 O +CH3 O CH3 OH +CHO CH3 O +O CH3 O +O — 15b. CH2 OH Reactions CH2 OH +M∗ CH2 OH +H CH2 OH +O2 — 16. CH3 O2 Reactions CH3 O2 +M∗ CH3 +O2 +M∗ CH3 O2 +CH2 O CH3 O2 H +CHO CH3 O2 +CH3 O CH3 O +CH3 O CH3 O2 +HO2 CH3 O2 H +O2 CH3 O2 +CH3 O2 CH3 O2 +CH3 O2 — 17. CH4 Reactions CH4 +H CH4 +O CH4 +OH CH4 +HO2 CH4 +CH CH4 +CH2 — 18. CH3 OH Reactions CH3 OH CH3 OH +H CH3 OH +O CH3 OH +OH CH3 OH +HO2 CH2 OH +H2 O2 CH3 OH +CH3 CH3 O +CH3 OH CH2 OH +CH3 OH CH3 OH +CH2 O CH3 O +CH3 O
A b [cm,mol,s] 1.00 · 1016 0.00 1.69 · 1053 −12.00
E [kJ/mol] 134.00 81.20
=C2 H4 =C2 H6
+H2
=CH2 O =C2 H6 =C2 H6 =CHO =CH2 O =O2 =OH
+H +M∗ +H2 OH +HO2 +CH3 OH +CH3 O +CH3 +CH2 O
5.00 · 1013 1.80 · 1013 4.00 · 1010 6.00 · 1011 6.50 · 1009 1.10 · 1013 1.40 · 1012
0.00 0.00 0.00 0.00 0.00 0.00 0.00
105.00 0.00 8.90 13.80 57.20 0.00 0.00
=CH2 O =CH2 O =CH2 O
+H +M∗ +H2 +HO2
5.00 · 1013 3.00 · 1013 1.00 · 1013
0.00 0.00 0.00
105.00 0.00 30.00
=CH3 =CH3 O2 =CHO =CH3 O2 =CH3 O =CH3 O2 =CH3 O2 H =CH3 O2 =CH2 O =CH3 O
+O2 +M∗ +M∗ +CH3 O2 H +CH2 O +CH3 O +CH3 +O2 +HO2 +O2 +CH3 OH +CH3 O +O2
7.24 · 1016 1.41 · 1016 1.30 · 1011 2.50 · 1010 3.80 · 1012 2.00 · 1010 4.60 · 1010 3.00 · 1012 1.80 · 1012 3.70 · 1012
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
111.00 30.00 37.70 42.30 −5.00 0.00 −10.90 163.00 0.00 9.20
=H2 =OH =H2 O =H2 O2 =C2 H4 =CH3
+CH3 +CH3 +CH3 +CH3 +H +CH3
1.30 · 1004 6.92 · 1008 1.60 · 1007 1.10 · 1013 3.00 · 1013 1.30 · 1013
3.00 1.56 1.83 0.00 0.00 0.00
33.60 35.50 11.60 103.00 −1.70 39.90
=CH3 =CH2 OH =CH2 OH =CH2 OH =CH2 OH =HO2 =CH4 =CH2 OH =CH3 O =CH3 O =CH3 OH
+OH +H2 +OH +H2 O +H2 O2 +CH3 OH +CH2 OH +CH3 OH +CH3 OH +CH3 O +CH2 O
9.51 · 1029 4.00 · 1013 1.00 · 1013 1.00 · 1013 6.20 · 1012 1.00 · 1007 9.00 · 1012 2.00 · 1011 2.20 · 1004 1.53 · 1012 3.00 · 1013
−4.30 0.00 0.00 0.00 0.00 1.70 0.00 0.00 1.70 0.00 0.00
404.00 25.50 19.60 7.10 81.10 47.90 41.10 29.30 45.40 333.00 0.00
235
5 Elements of Chemical Kinetics Table 5.1: (continued) Reaction — 19. CH3 O2 H Reaction CH3 O2 H =CH3 O +OH OH +CH3 O2 H =H2 O +CH3 O2 — 20. – 29. C2 Hydrocarbons Oxidation — 20. C2 −H Reactions C2 H +O =CO +CH C2 H +O2 =HCCO +O — 21. HCCO Reactions HCCO +H =C2 +CO HCCO +O =CO +H +CO HCCO +CH2 =C2 H3 +CO — 22. C2 H2 Reactions C2 H2 +M∗ =C2 H +H +M∗ C2 H2 +O2 =HCCO +OH C2 H2 +H =C2 H +H2 C2 H2 +O =CH2 +CO C2 H2 +O =HCCO +H C2 H2 +OH =H2 O +C2 H C2 H2 +C2 H =C4 H2 +H — 23. CH2 CO Reaction CH2 CO +M∗ =CH2 +CO +M∗ CH2 CO +H =CH3 +CO CH2 CO +O =CHO +CHO CH2 CO +OH =CH2 O +CHO — 24. C2 H3 Reactions C2 H3 =C2 H2 +H C2 H3 +OH =C2 H2 +H2 O C2 H3 +H =C2 H2 +H2 C2 H3 +O =C2 H2 +OH C2 H3 +O =CH3 +CO C2 H3 +O =CHO +CH2 C2 H3 +O2 =CHO +CH2 O — 25a. CH3 CO Reactions CH3 CO =CH3 +CO CH3 CO +H =CH2 CO +H2 — 25b. CH3 CHO Reactions CH2 CHO +H =CH2 CO +H2 — 26. C2 H4 Reactions C2 H4 +M∗ =C2 H2 +H2 +M∗ ∗ C2 H4 +M =C2 H3 +H +M∗ C2 H4 +H =C2 H3 +H2 C2 H4 +O =H +CH2 CHO C2 H4 +O =CHO +CH3 C2 H4 +OH =C2 H3 +H2 O
236
A [cm,mol,s]
b
E [kJ/mol]
4.00 · 1015 2.60 · 1012
0.00 0.00
180.00 0.00
1.00 · 1013 3.00 · 1012
0.00 0.00
0.00 0.00
1.50 · 1014 9.60 · 1013 3.00 · 1013
0.00 0.00 0.00
0.00 0.00 0.00
3.60 · 1016 2.00 · 1008 6.02 · 1013 1.72 · 1004 1.72 · 1004 6.00 · 1013 3.00 · 1013
0.00 1.50 0.00 2.80 2.80 0.00 0.00
446.00 126.00 116.00 2.10 2.10 54.20 0.00
1.00 · 1016 3.60 · 1013 2.30 · 1012 1.00 · 1013
0.00 0.00 0.00 0.00
248.00 14.10 5.70 0.00
4.73 · 1040 5.00 · 1013 1.20 · 1013 1.00 · 1013 1.00 · 1013 1.00 · 1013 5.40 · 1012
−8.80 0.00 0.00 0.00 0.00 0.00 0.00
194.00 0.00 0.00 0.00 0.00 0.00 0.00
2.32 · 1026 2.00 · 1013
−5.00 0.00
75.10 0.00
2.00 · 1013
0.00
0.00
7.50 · 1017 8.50 · 1017 5.67 · 1014 1.40 · 1006 2.42 · 1006 2.11 · 1013
0.00 0.00 0.00 2.08 2.08 0.00
320.00 404.00 62.90 0.00 0.00 24.90
5.5 Rate of Reactions Table 5.1: (continued) Reaction — 27. CH3 CHO Reactions CH3 CHO +M∗ CH3 CHO +H CH3 CHO +H CH3 CHO +O CH3 CHO +O CH3 CHO +O2 CH3 CHO +OH CH3 CHO +HO2 CH3 CHO +CH2 CH3 CHO +CH3 — 28. C2 H5 Reactions C2 H5 C2 H5 H C2 H5 O C2 H5 O C2 H5 O2 C2 H5 CH3 C2 H5 C2 H5 — 29. C2 H6 Reactions C2 H6 H C2 H6 O C2 H6 OH C2 H6 HO2 C2 H6 O2 C2 H6 CH2 C2 H6 CH3
A [cm,mol,s]
b
E [kJ/mol]
=CH3 =CH3 CO =H2 =CH3 CO =OH =CH3 CO =CH3 CO =CH3 CO =CH3 CO =CH3 CO
+M∗ +H2 +CH2 CHO +OH +CH2 CHO +HO2 +H2 O +H2 O2 +CH3 +CH4
7.00 · 1015 2.10 · 1009 2.00 · 1009 5.00 · 1012 8.00 · 1011 4.00 · 1013 2.30 · 1010 3.00 · 1012 2.50 · 1012 2.00 · 10−06
0.00 1.16 1.16 0.00 0.00 0.00 0.73 0.00 0.00 5.64
343.00 10.10 10.10 7.60 7.60 164.00 −4.70 50.00 15.90 10.30
=C2 H4 =CH3 =H =CH2 O =C2 H4 =C2 H4 =C2 H4
+H +CH3 +CH3 CHO +CH3 +HO2 +CH4 +C2 H6
1.02 · 1043 3.00 · 1013 5.00 · 1013 1.00 · 1013 1.10 · 1010 1.14 · 1012 1.40 · 1012
−9.10 0.00 0.00 0.00 0.00 0.00 0.00
224.00 0.00 0.00 0.00 −6.30 0.00 0.00
=C2 H5 =C2 H5 =C2 H5 =C2 H5 =C2 H5 =C2 H5 =C2 H5
+H2 +OH +H2 O +H2 O2 +HO2 +CH3 +CH4
1.40 · 1009 1.00 · 1009 7.20 · 1006 1.70 · 1013 6.00 · 1013 2.20 · 1013 1.50 · 10−07
1.50 1.50 2.00 0.00 0.00 0.00 6.00
31.10 24.40 3.60 85.90 217.00 36.30 25.40
5.5 Rate of Reactions Consider a reaction A + 2 B −→ 3 C + D
(5.23)
for which, at any instant, the rate of consumption or formation of the reacting species fulfil the following relationship: d[D] 1 d[C] d[A] 1 d[B] = =− =− dt 3 dt dt 2 dt
(5.24)
so there are several rates connected with the reaction. In order to avoid the problem of having several possible different rates to describe the same reaction
237
5 Elements of Chemical Kinetics the unique rate of reaction is defined as: r=
1 d[J] · νj dt
(5.25)
where νj is the stoichiometric number of jspecies with νj negative for reactants and positive for products.
5.5.1 Temperature dependence of rate coefficients Rate coefficients (k) depend strongly on temperature. This temperature dependence is described by a simple law, named Arrhenius law, Ea (5.26) k = A · exp − RT According to some more recent measurements the dependence of k on temperature is described with a better accuracy using the following function: Ea b (5.27) k = A · T exp − RT Table 5.1 lists values of A, b and Ea for many elementary reactions. The coefficient A · T b is called the preexponential factor. The activation energy Ea corresponds to an energy barrier which has to be overcome during the reaction, see Fig. 5.4. The maximum of the activation energy corresponds to the bond energies in the molecule. The activation energy can be very small, or even zero if new bonds are formed simultaneously with the breaking of the old bonds. Fig. 5.5 shows the temperature dependence of some elementary reactions of halogen atoms with molecular hydrogen. This is a typical Arrhenius plot with the logarithm of the rate coefficient plotted against the reciprocal of temperature. A linear dependence is obtained since log k = log A T b − RETa and a weak dependence of the term A T b is often obscured by the experimental error. As shown by Eq. (5.27) for activation energies very low or for very high temperatures, the exponent term goes to one, and the rate constant k goes to A T b . In chemical kinetics linear extrapolations of Arrhenius plots are common methods used to estimate rate coefficients at temperatures where they have not been measured directly. However, Arrhenius plots may also be curved and then both preexponential factor and activation energy increase with temperature.
238
(f)
Ea
5.5 Rate of Reactions
(b)
Ea
Potential energy
Ureactants
Uproducts
Extent of reaction Fig. 5.4: Activation energy of a chemical reaction
In trimolecular reactions listed in Table 5.1, M ∗ represents an inert collision partner called sometimes a third body factor. Although M ∗ appears on both sides of the chemical equation it cannot be left out. Its appearance in the chemical equation is associated with the role of other (third body) molecules as collisional energisers and deenergisers. Consider for example the 6th reaction in Table 5.1 O + O + M ∗ = O2 + M ∗
(5.28)
If two oxygen atoms collide and form an oxygen molecule it will immediately dissociate because of the energy excess of the newly formed O2 . The third collision partner [M ∗ ] takes away the energy excess stabilising the O2 molecule. The concentration of this collision partner is calculated as
[M ∗ ] = [H2 ] + 6.5[H2 O] + 0.4[O2 ] + 0.4[N2 ] + 0.75[CO] + 1.5[CO2 ] + 3[CH4 ] (5.29) showing that each of the stable molecules play a role in the energy removing process but with a different efficiency (weighting factor).
239
5 Elements of Chemical Kinetics
Fig. 5.5: Arrhenius plot for elementary reactions of halogens with molecular hydrogen [20]
5.5.2 Pressure dependence of rate coefficients For a unimolecular reaction, A −→ products, the rate of formation of products can be written as: d[products] = k · [A] (5.30) dt and the rate coefficient k depends on pressure and temperature. The theory of kinetics of unimolecular reactions yields falloff curves. An example of such a curve is given in Fig. 5.6. For bimolecular and trimolecular reactions the dependence of the rate constant k on temperature is rather complex and the reader is referred to textbooks on chemical kinetics [18, 16, 19].
240
5.6 Summary
Fig. 5.6: Falloff curves for the reaction C2 H6 −−→ CH3 + CH3 [20]
5.6 Summary In this chapter a discipline of Chemical Kinetics has been introduced. Principally, Chemical Kinetics is concerned with two issues: with the rate of chemical reactions and with mechanisms of chemical reactions. Chapter 5 underlines the basis for calculating the rates of chemical reactions while in the forthcoming Chapter 6 we will deal with combustion mechanisms. One should become familiar with the equation d[A] = −k · [A]a · [B]b · [C]c . . . dt describing the rate of consumption of Aspecies in a chemical reaction. The rate constant k which is strongly temperature dependent, is calculable using Arrhenius equation k = A · T b · exp(−
Ea ) RT
241
5 Elements of Chemical Kinetics Reaction order and reaction molecularity are inevitable associated with chemical kinetics and so is the concept of an elementary reaction. Firstorder, secondorder and thirdorder reactions have been considered in this chapter. The concept of the forward and reversible reactions have been introduced. It has been emphasised that a correctly formulated system of kinetic equations describing the rates of both forward and reversible reactions must lead to the chemical composition at equilibrium when t → ∞. In short, when t → ∞ chemical kinetics meets equilibrium thermodynamics and then and only then Kc =
242
kf kb
6 Mechanisms of Basic Combustion Reactions Contents 6.1 6.2 6.3
Chain Reactions Combustion of Carbon Monoxide (CO) Combustion of Hydrogen (H2 ) 6.3.1 Simplified ignition mechanism 6.4 Combustion of Methane (CH4 ) 6.5 Methods of Solving Chemical Kinetic Rate Equations 6.5.1 Analytical solutions 6.5.2 Numerical Solutions 6.6 Summary
6.1 Chain Reactions In our calculations of combustion stoichiometry (see Chapter 1) we used an overall reaction for hydrogen burning 2 H2 + O2 −−→ 2 H2 O
(6.1)
Even such a simple overall reaction is a chain process, in which O, H and OH radicals are active centres. Table 5.1 lists a number of elementary reactions taking place in combustion of hydrogen (H2 ) carbon monoxide (CO), methane (CH4 ) and C2 hydrocarbons (ethane – C2 H6 ; ethene – C2 H4 ; ethyne – C2 H2 ). In most instances, two reacting molecules do not react directly but one of them dissociates first to form radicals. These radicals, that are very reactive, initiate a chain of steps (chain of elementary reactions). Consider reactions taking place in H2 /O2 system (reactions listed in Table 5.1 in groups 01 to 03). Reaction (6.1) appears nowhere in the table since this is
243
6 Mechanisms of Basic Combustion Reactions an overall reaction and not an elementary reaction. Two hydrogen molecules do not react directly with an oxygen molecule to form two water molecules. Instead either hydrogen molecule or oxygen molecule dissociate forming radicals, so O2 + M∗ −−→ O + O + M∗
(6.2)
H2 + M∗ −−→ H + H + M∗
(6.3)
or Reactions (6.2) and (6.3) are chain initiation steps (reactions). In chain initiation reactions radicals (reactive species) are formed from stable species. The newly formed O and H radicals attack the oxygen molecule and the hydrogen molecule according to the reactions: O + H2 −−→ OH + H
(6.4)
H + O2 −−→ OH + O
(6.5)
and Both above reactions are chain branching steps (reactions). In chain branching reactions a reactive species (radical) react with a stable species (molecule) forming two reactive species e.g. two radicals. Table 5.1 shows that radicals H, O and OH take part in so called chain propagation reactions, for example: OH + H2 −−→ H2 O + H
(6.6)
CO + OH −−→ CO2 + H
(6.7)
H2 O2 + H −−→ OH + H2 O
(6.8)
or or In chain propagation reaction a reactive species (a radical) reacts with a stable species forming another reactive species as exemplified in reactions (6.6), (6.7) and (6.8). In chain termination reactions, reactive species react to stable species and the chain is terminated: H + O2 + M∗ −−→ HO2 + M∗
(6.9)
HO2 + OH −−→ H2 O + O2
(6.10)
or
244
6.2 Combustion of Carbon Monoxide (CO) or O + O + M∗ −−→ O2 + M∗
(6.11)
6.2 Combustion of Carbon Monoxide (CO) Consider oxidation of carbon monoxide that can be described using the "overall" equation 2 CO + O2 −−→ 2 CO2 (6.12) In a reacting system containing only CO and O2 , without presence of any water molecules, the only elementary reactions leading to CO2 are (see Table 5.1): (6.13)
CO + O2 −−→ CO2 + O ∗
∗
CO + O + M −−→ CO2 + M
(6.14)
where reaction (6.13) is chain initiation step and reaction (6.14) is chain termination. Using the above mechanism (reactions (6.13) and (6.14)), one may write appropriate rate equations and calculate the concentrations of all the species taking part in the reaction as a function of time, until equilibrium is reached. However, the rate of CO oxidation is substantially affected by the presence of OH radicals. Thus, in the presence of hydrogen containing molecules another reaction CO + OH −−→ CO2 + H
(6.15)
opens up a so called "wet" route of CO oxidation. As it is easy to see, any elementary reactions leading to formation and destruction of hydroxyl radicals (OH) may be important. In the presence of water vapour the reaction H2 O + M∗ −−→ H + OH + M∗
(6.16)
is important since, by producing H and OH radicals, it opens up new possibilities of OH radicals formation and destruction (see the first three groups of reactions in Table 5.1). In combustion of hydrocarbons almost all CO is oxidised by OH radicals. Fig. 6.1 shows the rate constants of reaction (6.15) and its backward counterpart.
245
6 Mechanisms of Basic Combustion Reactions
Fig. 6.1: Top – rate constant of CO + OH −−→ CO2 + H; Bottom – rate constant of H + CO2 −−→ CO + OH [19].
6.3 Combustion of Hydrogen (H2) A complete mechanism of combustion of hydrogen with oxygen is listed in Table 5.1 in groups 01–03. The mechanism is relatively simple [22]. There are only eight chemical species (O2 , H2 , H, OH, O, H2 O, HO2 , H2 O2 ) and nineteen elementary reactions involved. It is possible to write down rate equations (5.22)
246
6.3 Combustion of Hydrogen (H2 ) for each species. Using special numerical procedures it is possible to solve the equations and by doing so obtaining information about the change of the concentration of each species in time. In this paragraph we intend to analyse only a few steps of the mechanism indicating the most important reactions. The chain initiation reactions are: O2 + M∗ −−→ O + O + M∗
(6.17)
with its backwards counterpart: O + O + M∗ −−→ O2 + M∗
(6.18)
H2 + M∗ −−→ H + H + M∗
(6.19)
H + H + M∗ −−→ H2 + M∗
(6.20)
and with The rate constants and the activation energies for these chain initialization reactions are given in Table 6.1. The rate constants and the activation energies of O2 and H2 dissociation reactions are similar and so are the rate constants for the Oradicals and Hradicals recombination reactions. Thus, one may anticipate that for thesame initial concentrations of H2 and O2 both reactions will produce H and Oradicals in proximately equal amounts and with equal rates. Table 6.1: Coefficients of the rate constant equation k = A T b exp − REaT for the chain initiation reactions. [M∗ ] = [H2 ] + 13 [ H2 O] + 52 [ O2 ] + 25 [ N2 ] + 43 [ CO] + 32 [ CO2 ] + 3 [ CH4 ] 2
Reaction O2 + M∗ −−→ O + O + M∗ O + O + M∗ −−→ O2 + M∗ H2 + M∗ −−→ H + H + M∗ H + H + M∗ −−→ H2 + M∗
A [cm,mol,s] 3.48 · 1014 2.9 · 1017 6.19 · 1014 1.8 · 1018
b 0 1 0 1
Ea [kJ/mol] 451 0.0 402 0.0
After H and Oradicals are formed through the chain initialization reactions they attack oxygen molecules and hydrogen molecules, respectively through the chain branching reactions: O2 + H −−→ OH + O (6.21) and H2 + O −−→ OH + H
(6.22)
247
6 Mechanisms of Basic Combustion Reactions Table 6.2 lists the rate coefficients for the chain branching reactions and their backward counterparts. It is important to notice that the activation energies for the chain branching reactions are substantially lower than these for the chain initialization reactions of dissociation of hydrogen and oxygen molecules. There are no direct measurements of the fourth reaction listed in Table 6.2. However, it is possible to estimate the rate of this reaction knowing the equilibrium constant for the third reaction in Table 6.2. Table 6.2: Rate coefficients of the chain branchingreactions. The coefficients are presented in the form k = A T b exp − REaT .
Reaction O2 + H −−→ OH + O OH + O −−→ H + O2 H2 + O −−→ OH + H OH + H −−→ O + H2
A [cm,mol,s] 2.0 · 1014 1.8 · 1013 5.06 · 1014 ?
b 0 0 2.67 ?
Ea [kJ/mol] 70.3 0.0 26.3 ?
It is instructive to compare the rate coefficients of the chain initialization reactions (Table 6.1) with the rates of the chain branching reactions(Table 6.2). Table 6.3 shows such a comparison for several temperatures. It is easy to see that the chain branching reactions are overwhelmingly faster than the dissociation reactions. Furthermore Hradical consumed in reaction (6.21) is produced in reaction (6.22) and vice versa Oradical consumed in reaction (6.22) is regenerated in reaction (6.21). Consequently, the most important radical in the initial stages of combustion of hydrogen molecules is the hydroxyl (OH) radical which is produced in reactions (6.21) and (6.22). Therefore instead using reactions (6.17) and (6.19) as the chain initiation reactions, the sum of reactions (6.21) and (6.22) which can be written as: O2 + H2 −−→ 2 OH (6.23) is often used, in the literature on gas reactions kinetics, as the chain initiation reaction. Reaction (6.23) cannot be found in Table 5.1 since it is an overall reaction and not an elementary reaction. Reaction (6.23) allows to simplify the mechanism and ignore unimportant hydrogen and oxygen dissociation reactions (they might be of some importance at very high temperatures).
6.3.1 Simplified ignition mechanism Following the considerations of the previous paragraph and using Table 5.1, we may derive a simplified chain mechanism for ignition of the hydrogenoxygen system. The most important reactions are shown in Table 6.4 [22, 23, 28].
248
6.3 Combustion of Hydrogen (H2 ) Table 6.3: Comparison of the rate coefficients k of the chain initiation and chain 3 branching reactions. The k values, given in cm /mol · s, are calculated from k = A T b exp − REaT for specified temperatures.
Reaction O2 + M∗ −−→ O + O + M∗ H2 + M∗ −−→ H + H + M∗ O2 + H −−→ OH + O H2 + O −−→ OH + H
500 K 2.7 · 10−33 6.2 · 10−28 9.0 · 106 1.5 · 109
1000 K 9.6 · 10−10 6.2 · 10−7 4.3 · 1010 2.2 · 1011
2000 K 5.8 · 102 2.0 · 104 2.9 · 1012 6.8 · 1012
3000 K 4.9 · 106 6.2 · 107 1.2 · 1013 3.4 · 1013
Table 6.4: Simplified mechanism of ignition of the hydrogenoxygen system
(0)
H 2 + O2
=
2 OH
(1) (2) (3) (4) (1+2+3)
OH + H2 H + O2 O + H2 H + O2 + M∗ 2 H 2 + O2
= = = = =
H2 O + H OH + O OH + H HO2 + M∗ H2 O + OH + H
chain initiation (reaction rate R) chain propagation chain branching chain branching chain termination
If one sums up the chain propagation and chain branching reactions (1+2+3), one can see that H and OH radicals are produced from the reactants. The chain reactions go on until no radicals are produced any more. The radicals are completely consumed in chain termination reactions. We can write down the rate laws for the three radicals: d[H] = k1 [OH] · [H2 ] − k2 [H] · [O2 ] + k3 [O] · [H2 ] − k4 [H] · [O2 ] · [M∗ ] dt d[OH] = 2 R − k1 [OH] · [H2 ] + k2 [O2 ] · [H] + k3 [O] · [H2 ] dt d[O] = k2 [H] · [O2 ] − k3 [O] · [H2 ] dt
(6.24) (6.25) (6.26)
These three radicals, H, OH and O are chain carriers and the term 2 R is called the rate of the chain initiation. By summing up the three equations we obtain the rate of formation of the free valences (H, OH, and O). Oxygen atoms however should be counted twice since there are two free valences per oxygen radical, d([H] + [OH] + 2 [O]) = 2 R + (2 k2 [O2 ] − k4 [O2 ] · [M∗ ]) · [H] dt
(6.27)
249
6 Mechanisms of Basic Combustion Reactions To simplify the notation, we mark the concentration of the free valences as [n], so (6.28)
[n] ≡ [H] + [OH] + 2 [O]
As a crude approximation we may replace the concentration of the hydrogen radicals as one fourth of the concentration of the free valences, so 1 [H] ≈ [n] 4
(6.29)
With the above introduced simplifications, Eq. (6.27) can be written as 1 d[n] = 2 R + · (2 k2 [O2 ] − k4 [O2 ] · [M∗ ]) · [n] dt 4
(6.30)
and the dependence of [n] with time can be easily obtained as [n] = 2 R · t
for g = f
(6.31)
2R 1 [n] = 1 · (f − g) · t 1 − exp 4 4 · (g − f )
for g 6= f
(6.32)
where f = 2 k2 [O2 ] and g = k4 [O2 ] · [M∗ ]. The solution (Eqs. (6.31) and (6.32)) is shown schematically in Fig. 6.2. For g > f the exponential term in Eq. (6.32) tends to zero with increasing time. For time being large enough, a timeindependent solution is obtained for the chain carriers 8R (6.33) [n] = (g − f ) When g = f a linear increase of the chain carriers concentration is obtained. For g < f , after a short time, the exponential term in Eq. (6.32) is larger than one and the result is an exponential growth of the carriers concentration 1 8R · exp · (f − g) (6.34) [n] = (g − f ) 4 thus an explosion is observed. It should be realised that the ignition scheme shown in Table 6.4 is a simplification of the complete mechanism listed in Table 5.1. The scheme has been used to illustrate the importance of H, O, OH radicals in the initiation of the hydrogen combustion. Having at the disposal the complete kinetic scheme (Table 5.1) and a general numerical procedure for solving ordinary (stiff) differential equations,
250
6.4 Combustion of Methane (CH4 ) it is possible to examine whether the combustion is extinct, propagates through the mixture or an explosion takes place.
Fig. 6.2: The time behaviour of the chain carriers (Eqs. (6.31) and (6.32))
6.4 Combustion of Methane (CH4) Combustion of hydrocarbons is a complicated chemical process. Methane is the most common hydrocarbon and it is the basic component of natural gas. Energies of the C−H bonds in methane exceed those of other hydrocarbons and therefore methane is the least reactive of all hydrocarbons. Table 5.1, groups 10–19, lists the most important elementary reactions for methane oxidation. Since a CH4 molecule contains only one atom of carbon, no aldehyde groups (R−CHO) are formed in contrast to elementary reactions for oxidation of C2 hydrocarbons (see Table 5.1 groups 20–29). Methane molecules are first attacked by H, O and OH radicals [19, 23] CH4 + H −−→ CH3 + H2
(6.35)
CH4 + O −−→ CH3 + OH
(6.36)
CH4 + OH −−→ CH3 + H2 O
(6.37)
Table 6.5 lists the rate constants of these reactions for several temperatures. All these reactions proceed with similar rates and therefore they are of equal importance. The methyl radical (CH3 ) is formed as a result of these chain propagation
251
6 Mechanisms of Basic Combustion Reactions reactions. The methyl radical is almost exclusively removed by a reaction with oxygen atoms CH3 + O −−→ CH2 O + H (6.38)
Table 6.5: Comparison of the rate coefficients k of the chain propagation reactions for methane oxidation. The k values, given in cm3 /mol · s, are calculated from k = A T b exp − REaT for specified temperatures.
Reaction CH4 + H −−→ CH3 + H2 CH4 + O −−→ CH3 + OH CH4 + OH −−→ CH3 + H2 O
500 K 5.0 · 108 3.6 · 109 8.5 · 1010
1000 K 2.28 · 1011 5.9 · 1011 1.2 · 1012
2000 K 1.4 · 1013 1.2 · 1013 8.8 · 1012
3000 K 9.1 · 1013 4.4 · 1013 2.3 · 1013
The formaldehyde (CH2 O) is destroyed again by reactions with the radicals: CH2 O + OH −−→ CHO + H2 O
(6.39)
CH2 O + H −−→ CHO + H2
(6.40)
CH2 O + O −−→ CHO + OH
(6.41)
The CHO radical is converted to CO in the following reactions: (6.42)
CHO + O2 −−→ CO + HO2 ∗
∗
CHO + M −−→ CO + H + M CHO + H −−→ CO + H2
(6.43) (6.44)
The conversion of methane into CH3 via reaction (6.37) competes with the basic reaction of CO oxidation (reaction (6.15)) as soon as carbon monoxide is produced. Reaction (6.37) is around one order faster than reaction (6.15) and combustion of methane, as with combustion of other hydrocarbons, proceeds in two distinct steps. Firstly a rapid conversion of CH4 into CO, H2 , H2 O and free radicals take place. Secondly, a relatively slow afterburning of CO takes place. For this second step, the mechanisms of CO and H2 oxidations discussed previously are valid. Fig. 6.3 shows a simplified scheme of methane oxidation. In Chapter 1 on combustion stoichiometry calculations we used an overall reaction for combustion of methane CH4 + 2 O2 −−→ CO2 + 2 H2 O (6.45) In light of the above considerations a better overall reaction scheme for methane
252
6.5 Methods of Solving Chemical Kinetic Rate Equations oxidation would be CH4 + O2 −−→ CO + H2 + H2 O CO + H2 +
1 2 O2 1 2 O2
(6.46)
−−→ CO2
(6.47)
−−→ H2 O
(6.48)
or perhaps reaction (6.46) could be combined with a radical mechanism of CO−H2 oxidation.
Fig. 6.3: A simplified scheme of methane oxidation [19]
6.5 Methods of Solving Chemical Kinetic Rate Equations Complete mechanisms for oxidation of hydrocarbons may consist of several hundred elementary reactions with as many as a hundred species. On the other
253
6 Mechanisms of Basic Combustion Reactions hand, a simple mechanism of CO oxidation, in waterfree environment, consists of two reactions with only four species involved. In order to determine how the concentration of each species changes with time we formulate a set of ordinary differential equations of the type d[Ai ] = fi ([A1 ], [A2 ], . . . , [AN ]; k1 , k2 , . . . , kR ) dt
for i = 1, 2, . . . , N
(6.49)
with initial conditions [Ai ](t = t0 ) = [Ai ]0
(6.50)
We write down the rate law (Eq. (6.49)) for each species of the mechanism considered, thus N stands for a number of species while R represents a number of elementary reactions. In above equations the time is the independent variable, the species concentrations are the dependent variables while ki are parameters of the system and [Ai ]0 denote the initial conditions at time t0 . Eqs. (6.24), (6.25) and (6.26) are such rate laws describing the concentrations of the three radicals for the simplified ignition mechanism of hydrogen. We assume that the temperature, needed to calculate the rate coefficients ki is given, thus Eqs. (6.49) and (6.50) are valid for an isothermal system (we can easily include an energy balance equation to calculate the temperature as a function of time, if desired). We wish to possess a general mathematical procedure to solve the system (6.49) with the initial conditions (6.50) for any temperature and pressure. In some specific cases, when a number of species is perhaps not larger than three or four, we may attempt to solve the system analytically. However, we expect that the general solver is numerical.
6.5.1 Analytical solutions In Chapter 5 (Examples 5.1 and 5.2) we have derived solutions to the rate law equations for a firstorder, a secondorder or generally a nth order (ninteger) irreversible reaction (the solution for a zeroorder reaction may be easily found). Consider a simple reaction chain consisting of two elementary steps; a chain initiation and chain termination k
k
A1 −−12 → A2 −−23 → A3 The rate laws for the three species A1 , A2 and A3 are
254
(6.51)
6.5 Methods of Solving Chemical Kinetic Rate Equations
d[A1 ] = −k12 [A1 ] dt d[A2 ] = k12 [A1 ] − k23 [A2 ] dt d[A3 ] = k23 [A2 ] dt
(6.52) (6.53) (6.54)
At time zero the only component is A1 so the initial conditions are: [A1 ](0) = [A1 ]0 ; [A2 ](0) = 0 and [A3 ](0) = 0. The solution of Eq. (6.52) is easy to find (6.55)
[A1 ](t) = [A1 ]0 exp(−k12 t) Inserting (6.55) into (6.53) leads to the following equation for [A2 ] d[A2 ] + k23 [A2 ] = k12 [A1 ]0 exp(−k12 t) dt
(6.56)
Eq. (6.56) is an ordinary differential equation of a type y ′ + k23 y = r(t)
(6.57)
which we have considered in ChapterR 5. The solution is found upon multiplying Eq. (6.56) by the integrating factor e k23 dt = ek23 t so d[A2 ] k23 t ·e + k23 · ek23 t · [A2 ] = k12 · [A1 ]0 · exp(−k12 t) · ek23 t dt d [A2 ] · ek23 t = k12 · [A1 ]0 · exp(−k12 t) · ek23 t dt
Integration of the above equation leads to Z k23 t [A2 ] · e = k12 · [A1 ]0 exp(−k23 t) · ek23 t dt + C1
(6.58) (6.59)
(6.60)
and after some algebra [A2 ](t) =
k12 · [A1 ]0 ·
and [A2 ](t) =
R
e(−k12 +k23 )t dt
ek23 t
+
C1 ek23 t
k12 · [A1 ]0 · e(k23 −k12 )t C1 + k t (k23 − k12 ) · e 23 ek23 t
(6.61)
(6.62)
255
6 Mechanisms of Basic Combustion Reactions Using the initial condition [A2 ](0) = 0 we can calculate the integration constant C1 as k12 · [A1 ]0 (6.63) C1 = − (k23 − k12 ) so finally, the solution for the species A2 reads: [A2 ](t) =
i k12 · [A1 ]0 h −k23 t e − e−k12 t k12 − k23
(6.64)
Knowing the dependence of A2 species concentration with time, Eq. (6.64), we may easily calculate the dependence of A3 species with time. To this end we insert Eq. (6.64) into (6.54), and we perform the integration obtaining Z i k12 k23 · [A1 ]0 h −k23 t [A3 ](t) = · e − e−k12 t dt + C2 (6.65) (k12 − k23 ) −k12 t e e−k23 t k12 k23 · [A1 ]0 · − [A3 ](t) = + C2 (6.66) (k12 − k23 ) k12 k23 We obtain C2 integration constant using the initial condition [A3 ](0) = 0, so (6.67)
C2 = [A1 ]0 and after some algebra [A3 ](t) = [A1 ]0 · 1 +
k12 k23 e−k12 t − e−k23 t k12 − k23 k12 − k23
(6.68)
Thus, the solution to the problem formulated by Eqs. (6.51)–(6.54) is given by formula (6.55) for A1 species, formula (6.64) for A2 species, and formula (6.68) for A3 species. Fig. 6.4 shows the temporal behaviour of the concentrations of species A1 , A2 and A3 in reaction (6.51) for k12 = 1 s−1 and k23 = 10 s−1 . In this case, when the k23 ≫ k12 the concentration of the intermediate A2 species remains low. This situation corresponds to A2 species being very reactive and therefore its rate of consumption is approximately equal to its rate of formation. Fig. 6.5 shows the temporal behaviour of the concentration of the three species for k12 = 10 s−1 and k23 = 1 s−1 . This situation corresponds to a low reactive intermediate A2 species which, as shown in Fig. 6.5, is present in rather large quantities and its concentration varies with time.
256
6.5 Methods of Solving Chemical Kinetic Rate Equations
Fig. 6.4: Temporal behaviour of the species concentrations in reactions A1 −→ A2 −→ A3 for k12 = 1 s−1 and k23 = 10 s−1 (reactive intermediate A2 )
Fig. 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 intermediate A2 ).
257
6 Mechanisms of Basic Combustion Reactions Example 6.1 Consider again a simple reaction system given be Eqs. (6.51)–(6.54). Derive appropriate formulae for calculating the concentrations of the A1 , A2 and A3 species as a function of time for k12 = k23 = k. The initial conditions are again: [A1 ](0) = [A1 ]0 ; [A2 ](0) = 0; [A3 ](0) = 0. We have already derived formulae (6.55), (6.64) and (6.68) for calculation the species concentration as a function of time. Relationship (6.55) can be used putting k12 = k, so [A1 ](t) = [A1 ]0 · e−kt
(E1)
However in formulae (6.64) and (6.68) singularity appears for k12 = k23 = k. Thus, we have to step back to Eq. (6.61) that for k12 = k23 = k gives: [A2 ](t) =
k · [A1 ]0 · t C1 + kt ekt e
Since [A2 ](0) = 0, C1 constant is also equal zero and the formula for the concentration of A2 species is [A2 ](t) = k · [A1 ]0 · t · e−kt (E2) Inserting the above relationship into Eq. (6.54) gives: d[A3 ] = k 2 · [A1 ]0 · t · e−kt dt and 2
[A3 ](t) = k · [A1 ]0
Z
te−kt dt + C2
While performing the integration we use the integration by parts method Z Z u(x)v ′ (x) dx = uv − u′ (x)v(x) dx to obtain Z −kt e te−kt + dt + C2 = [A3 ](t) = k · [A1 ]0 · − k k 2
k 2 · [A1 ]0 ·
(tk + 1) · e−kt + C2 − k2
Using the initial condition [A3 ](0) = 0 we calculate the integration constant to be C2 = [A1 ]0 and therefore the formula for calculating the concentration of
258
6.5 Methods of Solving Chemical Kinetic Rate Equations A3 species reads:
[A3 ](t) = [A1 ]0 1 − (kt + 1) · e
−kt
(E3)
Fig. 6.6 shows the temporal behaviour of the concentrations of species A1 , A2 and A3 as predicted using Eqs. (E1), (E2) and (E3) valid for k12 = k23 = k.
Fig. 6.6: Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 with k12 = k23 = k.
End of Example 6.1 6.5.1.1 QuasiSteady State Assumption In engineering practise we are often interested in the change of the substrates and products with time. The intermediate species are of lesser importance and if possible we would like to omit them in our considerations. Consider again the reaction sequence (6.51) where A1 represents substrates, A3 products and A2 stands for intermediates. Fig. 6.4 and Fig. 6.5 show the temporal behaviour of the three components in the reaction sequence (6.51) for k23 ≫ k12 and k12 ≫ k23 , respectively. In the later case (Fig. 6.5), the concentration of the A2 intermediate reaches relatively high values and what is even more important, it varies with time as rapidly as the concentration of substrates A1 or products A3 . Mathematically, 2] the derivative d[A dt is far from zero throughout the course of the reaction and it is
259
6 Mechanisms of Basic Combustion Reactions not possible to omit this species. In the former case (Fig. 6.4) the concentration of A2 intermediates is low (since it is a reactive intermediate) and it does not 2] vary too much with time. Mathematically, we can say that the derivative d[A dt is almost zero, thus d[A2 ] = k12 · [A1 ] − k23 [A2 ] ≈ 0 (6.69) dt With this simplification the rate equations describing our system read: [A1 ](t) = [A1 ]0 · exp(−k12 t)
(6.70)
k23 · [A2 ] = k12 · [A1 ]
(6.71)
[A3 ](t) = k23 · [A2 ] = k12 · [A1 ] = k12 · [A1 ]0 · exp(−k12 t)
(6.72)
or after integration of Eq. (6.72) [A3 ](t) = k12 · [A1 ]0 · (1 − exp(−k12 t))
(6.73)
Fig. 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)
Fig. 6.7 shows the temporal behaviour (Eqs. (6.70), (6.71) and (6.73)) obtained using the quasisteady state assumption for A2 intermediate for k12 = 1 s−1 and k23 = 10 s−1 . It is easy to see that the simplified solution (Eqs. (6.70), (6.71) and
260
6.5 Methods of Solving Chemical Kinetic Rate Equations (6.73)) is only marginally different from the exact solution (Eqs. (6.55), (6.64), (6.68)) shown in Fig. 6.4. This remains to be so as long as k23 ≫ k12 . We consider formation of nitric oxide (NO) in order to provide an example of a steadystate approximation. In 1946 Zeldovich postulated the mechanism of NO formation by three elementary reactions [22, 23, 20]: O + N2 −−→ NO + N
(6.74)
N + O2 −−→ NO + O
(6.75)
N + OH −−→ NO + H
(6.76)
The first two reactions (6.74) and (6.75) are called Zeldovich mechanism and here the role of hydroxyl radicals is neglected. All the three reactions are named extended Zeldovich mechanism. Table 6.6 lists the rate constants and the rates of the reactions at several temperatures. Reaction (6.74) is substantially slower than reactions (6.75) and (6.76). Its activation energy of 316 kJ/mol is very high since there is a strong triple bond in the N2 molecule to break. Thus, this reaction may be of some importance at high temperatures only and this is why the mechanism is named as thermal mechanism of NO formation. The rate of NO formation for the extended Zeldovich mechanism is d[NO] = k1 · [O] · [N2 ] + k2 · [N] · [O2 ] + k3 · [N] · [OH] dt
(6.77)
where k1 , k2 and k3 are the rate constants of reactions (6.74), (6.75), (6.76), respectively. These are given in Table 6.6. Table 6.6: Rate coefficients of the Zeldovich mechanism reactions [25]. The rate coefficients are in the form k = A T b exp − REaT .
Reaction
O + N2 −−→ NO + N N + O2 −−→ NO + O N + OH −−→ NO + H
A cm3 /mol/s 7.6 · 1013 6.4 · 109 1.0 · 1014
b 0 1 0
E kJ/mol 316 26.2 0
500 K
1000 K
2000 K
7.4 · 10−20 2.4 · 10−03 4.2 · 1005 5.9 · 1009 2.7 · 1011 2.6 · 1012 1.0 · 1014 1.0 · 1014 1.0 · 1014
3000 K
2.4 · 108 6.7 · 1012 1.0 · 1014
Table 6.6 shows that reaction (6.74) is the rate controlling step of the thermal NOformation mechanism. Consider the nitrogen radicals. They are formed in reaction (6.74) with a relatively low rate. However, they are rapidly removed in reactions (6.75) and (6.76). Here we come across a situation analogue to the reaction sequence A1 −→ A2 −→ A3 with k23 ≫ k12 . We can assume the steady state approximation for the Nradicals. The rate formation of Nradicals for the
261
6 Mechanisms of Basic Combustion Reactions extended Zeldovich mechanism reads: d[N] = k1 · [O] · [N2 ] − k2 · [N] · [O2 ] − k3 · [N] · [OH] ≈ 0 dt
(6.78)
] and the derivative d[N dt is almost zero. Inserting Eq. (6.78) into (6.77) results in a simple expression for the thermal NOformation rate:
d[NO] = 2 · k1 · [O] · [N2 ] dt
(6.79)
The equation shows that the important factors affecting the NOformation rate are: the temperature (through k1 ), the Oradicals concentration and N2 concentration. 6.5.1.2 Partial Equilibrium Relationship (6.79) can be used to calculate the NO concentration as a function of time. However, in order to proceed with the integration we must know (in addition to given initial conditions) the temperature (to calculate k1 ), the Oradical concentration and the N2 concentration at any instant. Alternatively we may write down a reaction mechanism for Oradicals formation and destruction and formulate appropriate rate equations. However, it is clear that numerical integration would be needed to solve such a system. Partial equilibrium assumption offers a mathematically simpler method. We will apply the partial equilibrium concept to solve Eq. (6.79). In order to estimate the Oradicals concentration we assume that they are in equilibrium with O2 molecules ∗ − ⇀ O2 + M∗ − (6.80) ↽ − −O+O+M so Kc = and therefore
[O]2 [O2 ]
(6.81)
q d[NO] (6.82) = 2 · k1 · Kc · [O2 ] · [N2 ] dt Using the rate coefficients of the O2 dissociation reactions (Table 6.1) and the rate coefficient of reaction (6.75) for NOformations (Table 6.6) it is easy to verify that at high temperatures the equilibrium between the molecular oxygen and oxygen atoms is reached quickly if compared to the progress of the ratecontrolling reaction (6.74). Thus, the assumption of the partial equilibrium in reaction (6.80)
262
6.5 Methods of Solving Chemical Kinetic Rate Equations seems to be justified. However, it has been observed that the equilibrium, particularly at low pressures, underpredicts [O] by a factor of up to five. This is particularly true at flame fronts where oxygen atoms are generated by many other routes and are not at all at equilibrium. A better approximation for [O] radicals can be made assuming the equilibrium of the following elementary reactions: −− ⇀ H + O2 ↽ − − OH + O −− ⇀ O + H2 ↽ − − OH + H
(A)
(6.83)
(B)
(6.84)
−− ⇀ OH + H2 ↽ − − H2 O + H
(C)
(6.85)
Thus, at equilibrium the following is applicable: [OH] · [O] [H] · [O2 ] [OH] · [H] = [O] · [H2 ] [H O] · [H] = 2 [OH] · [H2 ]
Kc,A =
(6.86)
Kc,B
(6.87)
Kc,C
(6.88)
where Kc,A , Kc,B and Kc,C are the equilibrium constants of reactions (6.83), (6.84) and (6.85), respectively. Using these equations one can express the concentrations of the intermediates (O, H, OH) using the concentrations of the stable species: [O] = Kc,A · Kc,C ·
[O2 ] · [H2 ] [H2 O]
(6.89) 1
3
[O ] 2 · [H2 ] 2 [H] = Kc,A · Kc,B · Kc,C · 2 [H2 O] 1 2
1 2
1
1
1 2
1
1
2 2 · Kc,B · [O2 ] 2 · [H2 ] 2 [OH] = Kc,A
(6.90) (6.91)
If we now insert Eq. (6.89) into (6.79) we obtain [O ] · [H2 ] · [N2 ] d[NO] = 2 · k1 · Kc,A · Kc,C · 2 dt [H2 O]
(6.92)
One should note that k1 rate constant corresponds to reaction (6.74) while the equilibrium constants Kc,A and Kc,C refer to reactions (6.83) and (6.85). It is important to stress that the partial equilibrium assumption provides satisfactory results only at high temperatures (typically > 1800 K).
263
6 Mechanisms of Basic Combustion Reactions Most combustion systems entail oxidation mechanisms containing numerous individual reaction steps. Under certain circumstances a group of reactions may proceed rapidly and reach a quasiequilibrium whilst other reactions may proceed slowly. If the rate constant of this slow reaction is to be determined, the concentrations of these reactive intermediates (radicals) can be estimated using the partial equilibrium assumption. Thus, the partial equilibrium assumption looks like the steadystate approximation we considered earlier. There is however one essential difference between these approaches; in the steady state approximation one is concerned with a particular species. In the partial equilibrium approach one is concerned with particular reactions [23].
6.5.2 Numerical Solutions We are concerned with solving N coupled ordinary differential Eqs. (6.49) with initial conditions (6.50). For a number of specific cases we have succeeded in finding analytical solutions when only a few chemical species are present in the system. However, we realise that a general solver for the system of Eqs. (6.49) has to be a numerical one. 6.5.2.1 One Ordinary Differential Equation Prior to considering a set of coupled ordinary differential equations we will analyse just one equation: dy = f (t, y) (6.93) dt with the initial condition y(0) = y0 (6.94) where the function on the righthand side of Eq. (6.93) is known. Mathematicians call this problem an initial value problem since yvalue is given at a starting point t0 , and it is desired to find the y’s at some final point tf inal or at some discrete list of points. The underlying idea of any numerical procedure for solving the initial value problem is always the same. Write the dy and dt in Eq. (6.93) as finite steps ∆y and ∆t, and multiply Eq. (6.93) by ∆t. In this way we obtain an algebraic formula for the change in the function when the independent variable t is increased by one timestep ∆t. In the limit of making the timestep very small, a good approximation to the underlying differential equation is achieved. Literal implementation of this procedure results in Euler’s method (see below). However we
264
6.5 Methods of Solving Chemical Kinetic Rate Equations do not recommend this method for any practical use. The method is conceptually important since all practical methods originate from this idea. Euler’s explicit and implicit methods The formula for the Euler method applied to Eq. (6.93) is yn+1 = yn + h · f (yn , tn )
(6.95)
which advances a solution from tn to tn+1 = tn + h, see Fig. 6.8. This method is a stepbystep method that is, we start from the given y0 and proceed stepwise computing approximate values of the solution y(t) at the discrete points tn+h . Formula (6.95) is unsymmetrical since it advances the solution through the timestep h using derivative information (f (yn , tn )) only at the beginning of that time interval. The method is called explicit (or forward) because the new value yn+1 is given explicitly in terms of the old value yn . One may wish to evaluate the derivative at the new value yn+1 obtaining a formula yn+1 = yn + h · f (yn+1 , tn+1 )
(6.96)
for the implicit (or backward) Euler’s method. This method requires solving Eq. (6.96) for yn+1 .
Fig. 6.8: Euler’s method
The main reason that Euler’s methods are not recommended for practical use is that there are more accurate and faster methods available that run at the
265
6 Mechanisms of Basic Combustion Reactions equivalent timestep size. Furthermore, the explicit Euler’s scheme may become unstable if the size of the time step is not small enough (see Example 6.2). Example 6.2 Consider the rate law for a first order irreversible chemical reaction (see Example 5.1) d[A1 ] = −k12 · [A1 ] (F1) dt and the initial condition [A1 ](0) = [A1 ]0 . Solve the equation using the explicit Euler’s scheme. Vary the size of the timestep h. For the rate constant use a value k12 = 1 s−1 . The explicit Euler scheme for integrating Eq. (F1) with timestep h is yn+1 = yn + h · f (tn , yn ) = yn − k · h · yn = (1 − h · k) · yn
(F2)
We expect that with n → ∞ the solution yn+1 = (1 − h · k)n+1 · y0 → 0, so ˙ < 1 and therefore the size of the time step should be chosen so −1 < (1 − hk) that 2 h< (F3) k The solution of Eq. (F1) using the explicit scheme (F2) for h = 0.3 s is shown in Table 6.7. Fig. 6.9 shows both the solutions for several values of h and the exact solution. The numerical solution approaches the exact one when the timestep h decreases. The numerical solution for h = 1.5 s oscillates around the correct solution and for t → ∞ it converges to the correct answer. For h = 2.14 s the numerical procedure is unstable as predicted by requirement (F3). Table 6.7: Numerical solution of Eq. (F1) using the explicit scheme (F2) with h = 0.3 s.
n 0 1 2 3 4 5 6 7 8 9
266
tn 0.0 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4
yn 1 0.4 0.16 0.064 0.0256 0.01024 0.004096 0.0016384 0.00065536 0.00026214
6.5 Methods of Solving Chemical Kinetic Rate Equations
Fig. 6.9: Numerical solutions of Eq. (F1) using the explicit scheme (F2) with different time steps. An instability has been encountered in integrating using a too large timestep.
Comments: (a) The method provides an accurate solution only when the time step is sufficiently small. (b) For a too large time step, the method is unstable (see Eq. (F3)). (c) The faster the reaction, the smaller the time step required for obtaining an accurate numerical solution. If this method were applied to reactions listed 2 14 ) sec. in Table 6.5, the time step would have to be smaller than ( 10 End of Example 6.2 Stability and accuracy are important features of numerical schemes. In Examples 6.2 and 6.3 we consider a simple linear equation for which the Euler’s explicit scheme becomes unstable when a too large timestep is used while the implicit scheme is unconditionally stable. This nice feature of implicit methods holds only for linear systems but even in the general case implicit methods give better stability. Now, we will consider accuracy of the Euler’s scheme (6.95). We may expand our unknown function y(t) around a point tn as suggested by the Taylor series: y(tn + h) = y(tn ) + h · y ′ (tn ) + h2 · y ′′ (tn ) + . . .
(6.97)
267
6 Mechanisms of Basic Combustion Reactions For a small value of h, the higher powers of h2 , h3 , . . . are very small so a crude approximation is (6.98) y(tn + h) ∼ = y(tn ) + h · y ′ (tn ) which is exactly the same expression as the Euler’s explicit formula (6.95). The Euler’s method is a firstorder method since in Eq. (6.98) we take only the constant terms and the term containing the first power of h. The omission of the further terms causes an error which is called the truncation error of the method. For small h, the third and higher powers of h are small compared with h2 in the first neglected term in Eq. (6.97). Therefore we say that the truncation error per step (or local truncation error) is of the order h2 . Any arithmetic operation among floating numbers introduces an additional error called roundoff error that is computer (hardware) dependent (More precisely, the error depends on how many bits are in the mantissa for the floating point representation of numbers). Roundoff errors accumulate with increasing amounts of calculations. As a general rule there is not much that a programmer can do about roundoff errors, other than to choose algorithms that do not magnify them unnecessarily. Example 6.3 Consider again the rate law for a first order irreversible chemical reaction (see Example 6.2) d[A1 ] = −k12 · [A1 ] (G1) dt and the initial condition [A1 ](0) = [A1 ]0 . Solve the equation using the implicit Euler’s scheme. Vary the size of the timestep h. For the rate constant use a value k12 = k = 1 s−1 . Assumptions: None In the implicit Euler’s scheme we evaluate the right hand side of Eq. (G1) at the new yn+1 corresponding to the new time tn+1 so the scheme reads (see Eq. (6.96)) yn+1 = yn + h · f (yn+1 , tn+1 ) = yn − k · h · yn+1 and yn+1 =
yn (1 + h · k)
(G2) (G3)
It is easy to see that the method is absolutely stable since even at n → ∞, yn+1 = (1+h y·0k)n+1 → 0 which is in fact the correct solution of Eq. (G1). Thus, for late times the implicit method converges to the true equilibrium solution even for large timesteps. The solution of Eq. (G1) using the implicit scheme (G2) for
268
6.5 Methods of Solving Chemical Kinetic Rate Equations h = 0.3 s is shown in Table 6.8. Fig. 6.10 shows both the solutions for several values of h and the exact solution. No instability has been encountered in the numerical integration and the numerical solution converges nicely towards the exact solution. Of course if the time step is large in following the evaluation towards equilibrium we give up accuracy but we maintain stability. Table 6.8: Numerical solution of Eq. (G1) using the implicit scheme (G2) with h = 0.3 s.
n 0 1 2 3 4 5
tn 0.0 0.6 1.2 1.8 2.4 3.0
yn 1 0.625 0.390625 0.24414063 0.15258789 0.09536743
n 6 7 8 9 10
tn 3.6 4.2 4.8 5.4 6.0 0
yn 0.05960464 0.0372529 0.02328306 0.01455192 0.00909495
Fig. 6.10: Numerical solutions of Eq. (G1) using the implicit scheme (G2) with different time steps. No instability has been encountered.
Comments: (a) The method is unconditionally stable. (b) We control the accuracy of the method by varying the time step. End of Example 6.3
269
6 Mechanisms of Basic Combustion Reactions RungeKutta Method (of Fourth Order) In the Euler’s schemes (see Eq. (6.95) and (6.96)), the derivative of the unknown function y(t) is evaluated either at the beginning (explicit scheme) of the time interval or at the end (implicit scheme). There are many other ways of evaluating the derivative and therefore there exist a number of methods for integration of the ordinary differential equations. For details the reader should consult textbooks on numerical methods [26, 27]. By far the most popular and arguable even most useful is the fourthorder RungeKutta method. The algorithm of the method is as follows: l1 = h · f (tn , yn ) l1 h l2 = h · f tn + , yn + 2 2 h l2 l3 = h · f tn + , yn + 2 2 l4 = h · f (tn + h, yn + l3 ) l1 l2 l3 l4 yn+1 = yn + + + + + O(h5 ) 6 3 3 6
(6.99)
The RungeKutta method treats every timestep in a sequence of five substeps described in the algorithm (6.99). For serious computing an adaptive time stepsize control is essential [26]. The purpose of this adaptive stepsize control is to achieve some predetermined accuracy in the solution with minimum computational effort. 6.5.2.2 Two consecutive elementary reactions We have already considered (see Section 6.5.1) a simple chain consisting of two elementary steps; a chain initiation and chain termination k
k
A1 −−12 → A2 −−23 → A3
(6.100)
The rate laws for A1 , A2 and A3 species have been given by Eqs. (6.52), (6.53) and (6.54). The analytical solutions showing the temporal behaviour of species A1 , A2 and A3 have been obtained in Section 6.5.1 (see Eq. (6.55), (6.64) and (6.68)) and they are quoted here for compactness: [A1 ](t) = [A1 ]0 exp(−k12 t) i k12 · [A1 ]0 h −k23 t −k12 t e −e [A2 ](t) = k12 − k23 270
(6.101) (6.102)
6.5 Methods of Solving Chemical Kinetic Rate Equations [A3 ](t) = [A1 ]0 · 1 +
k23 k12 e−k12 t − e−k23 t k12 − k23 k12 − k23
(6.103)
Fig. 6.11 shows the temporal behaviour of the concentrations of species A1 , A2 and A3 for k12 = 1 s−1 and k23 = 10 s−1 . Fig. 6.11 shows the behaviour for k12 = 1 s−1 and k23 = 100 s−1 . Both figures correspond to A2 species being very reactive if compared to the reactivity of A1 species. The concentration of A2 intermediate decreases rapidly with increasing k23 and in Eqs. (6.102) and (6.103) the terms e−k23 t can be neglected.
Fig. 6.11: Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 assuming quasi steady state for A2 species; k12 = 1 s−1 , k23 = 100 s−1 (compare with Fig. 6.4).
Euler’s Explicit Method In engineering practise, we are often interested in temporal behaviour of substrates and products while intermediates are of lesser importance. Consequently we would be happy with an integration procedure that would allow for calculating accurately only the substrates and products while we might accept some inaccuracy on the intermediates. Fig. 6.12 shows the numerical integration of Eqs. (6.52), (6.53) and (6.54) using Euler’s explicit method (see Example 6.2) for k12 = 1 s−1 , k23 = 10 s−1 . Since we are mainly interested in concentrations of A1  and A3 species, we perform the integration using a 0.2 s timestep that on the basis of Example 6.2 is small enough for achieving both a good accuracy and stability. Fig. 6.12(top) shows however that while the A1 species is calculated with good enough an accuracy, the A3 product concentration shows an unstable behaviour. The A2 intermediate concentration exhibits also oscillations. When the timestep of the integration is decreased to 0.1 s, accurate and stable numerical solutions for all the
271
6 Mechanisms of Basic Combustion Reactions species participating in the reactions are obtained as shown in Fig. 6.12(bottom). Consider now a case when k12 = 1 s−1 and k23 = 100 s−1 , so the reactivity of the intermediate species (A2 ) is increased by a factor of ten. As shown in Fig. 6.11, the A2 species concentration is indeed very small and it can be certainly neglected while considering A1  and A3 species. Fig. 6.13, top and bottom, show the results of the numerical integration when the Euler method is used with an integration timestep of 0.02 s and 0.01 s, respectively. Also in this case we had to reduce the integration step to a value of 0.01 s to obtain accurate and stable results. Integration using a timestep larger than 0.02 s would lead to instability. So far we have considered two situations; in the first case k12 = 1 s−1 and k23 = 10 s−1 (Fig. 6.12) while in the second case k12 = 1 s−1 and k23 = 100 s−1 (Fig. 6.13). In both cases the size of the integration timestep, needed for obtaining a stable numerical solution, has been determined by the second rate constant, namely by k23 . In order to obtain a stable numerical solution using the Euler’s explicit scheme we used a timestep equal to k123 , as predicted by Eq. (F3) of Example 6.2. When the integration step is larger than k223 the Euler’s explicit method "explodes". We observe that the size of the timestep is determined by the fastest reaction even though the contribution of the intermediate component A2 to the concentration of the A1  and A2 species is negligible. This is a frustrating observation. In Table 5.1 there are reactions that differ in their rate constants, and so in their rates, by as much as 1015 . Thus, if we were to develop a general numerical procedure for solving a set of Eqs. (6.49) and the solver would be based on the Euler explicit scheme, the size of the integration timestep would be based on the fastest reaction even though they might be of little importance in formation of the (final) stable products. The question is do we really have to resolve all these small scales even if the intermediates are of no interest to us? Example 6.4 Develop a numerical scheme for integration of Eqs. (6.52), (6.53) and (6.54) using Euler’s explicit method. The initial conditions are [A1 ](0) = [A1 ]0 , [A2 ](0) = 0 and [A3 ](0) = 0. Assumptions: None In order to simplify the notation we introduce y1 , y2 and y3 in place of [A1 ], [A2 ] and [A3 ] so Eqs. (6.52), (6.53) and (6.54) read: dy1 = −k12 · y1 dt dy2 = k12 · y1 − k23 · y2 dt
272
(H1) (H2)
6.5 Methods of Solving Chemical Kinetic Rate Equations dy3 = k23 · y2 dt
(H3)
and the initial conditions are y1 (0) = y10 , y2 (0) = 0 and y3 (0) = 0. Here we recall that Euler’s explicit method is based on (see Eq. (6.95)) the following relationship yn+1 = yn + h · f (yn , tn ) (H4) where yn+1 and yn are values of the independent variable at timesteps tn+1 and tn , respectively. Therefore, for y1 we obtain: y1,n+1 = y1,n + h · (−k12 · y1,n ) = y1,n · (1 − k12 · h)
(H5)
for y2 y2,n+1 = y2,n + h · (k12 · y1,n − k23 · y2,n ) = y2,n · (1 − k23 · h) + h · k12 · y1,n (H6) and finally for y3 y3,n+1 = y3,n + h · k23 · y2,n
(H7)
The formulae (H5), (H6) and (H7) constitute the Euler’s explicit method for the linear ordinary differential Eqs. (H1), (H2) and (H3). This scheme has been used to produce numerical solutions shown in Fig. 6.12 and Fig. 6.13. End of Example 6.4 Euler’s Implicit Method So far, the numerical solutions describing the temporal behaviour of the two consecutive reactions (6.100) have been obtained using Euler’s explicit scheme. Fig. 6.14top shows the solution obtained using Euler’s implicit scheme (see Example 6.5) for k12 = 1 s−1 and k23 = 10 s−1 . Note, that even for such a large timestep as 1 s, a stable solution has been obtained. Fig. 6.14bottom shows that when k23 is increased to 100 s−1 , Euler’s implicit scheme is stable for all the used time steps. Already in Example 6.3 we have predicted that Euler’s implicit scheme is unconditionally stable when one ordinary, linear differential equation is considered. Fig. 6.14 demonstrates the same point for the reaction system (6.100) described by three Eqs. ((6.52)–(6.54)). In this way we have demonstrated superiority of the implicit scheme over the explicit one. It is generally true that for any system of linear ordinary differential equations the implicit schemes are unconditionally stable and therefore they are superior over explicit schemes.
273
6 Mechanisms of Basic Combustion Reactions Example 6.5 Develop a numerical scheme for integration of Eqs. (6.52), (6.53) and (6.54) using Euler’s implicit method. The initial conditions are [A1 ](0) = [A1 ]0 , [A2 ](0) = 0 and [A3 ](0) = 0. Assumptions: None In order to simplify the notation we introduce y1 , y2 and y3 in place of [A1 ], [A2 ] and [A3 ] so Eqs. (6.52), (6.53) and (6.54) read: dy1 = −k12 · y1 dt dy2 = k12 · y1 − k23 · y2 dt dy3 = k23 · y2 dt
(I1) (I2) (I3)
and the initial conditions are y1 (0) = y10 , y2 (0) = 0 and y3 (0) = 0. Here we recall that Euler’s implicit method is based on (see Eq. (6.96)) the following relationship yn+1 = yn + h · f (yn+1 , tn+1 ) (I4) where yn+1 and yn are values of the independent variable at timesteps tn+1 and tn , respectively. Therefore, for y1 we obtain y1,n (I5) (1 + h · k12 ) y2,n + h · k12 · y1,n+1 = (1 + h · k12 ) (I6)
y1,n+1 = y1,n + h · (−k12 · y1,n+1 )
so y1,n+1 =
y2,n+1 = y2,n + h · (k12 · y1,n+1 − k23 · y2,n+1 )
so y2,n+1
y3,n+1 = y3,n + h · k23 · y2,n+1
(I7)
The formulae (I5), (I6) and (I7) constitute the Euler’s implicit method for the linear ordinary differential Eqs. (I1), (I2) and (I3). This scheme has been used to produce numerical solutions shown in Fig. 6.14. End of Example 6.5
274
6.5 Methods of Solving Chemical Kinetic Rate Equations
Fig. 6.12: Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 , k12 = 1 s−1 , k23 = 10 s−1 . Numerical solutions obtained using Euler’s Explicit Method. Top – h = 0.2 s integration time step, Bottom – h = 0.1 s timestep. (see Example 6.4).
275
6 Mechanisms of Basic Combustion Reactions
Fig. 6.13: Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 , k12 = 1 s−1 , k23 = 100 s−1 . Numerical solutions obtained using Euler’s Explicit Method. Top – h = 0.02 s; Bottom – h = 0.01 s. (see Example 6.4).
276
6.5 Methods of Solving Chemical Kinetic Rate Equations
Fig. 6.14: Temporal behaviour of the species concentrations for reactions A1 −→ A2 −→ A3 ; Top curves – k12 = 1 s−1 , k23 = 10 s−1 ; Bottom curves – k12 = 1 s−1 , k23 = 100 s−1 Numerical solutions obtained using Euler’s Implicit Method. (see Example 6.5).
277
6 Mechanisms of Basic Combustion Reactions 6.5.2.3 Set of NonLinear Equations In previous paragraphs we have considered solutions of either one (Paragraph 6.5.2.1) or three (Paragraph 6.5.2.2) ordinary differential equations. We have observed the superiority of the implicit numerical scheme over an explicit scheme. In order to simplify the notation we rewrite Eqs. (6.49) using a matrix notation: y′ = F(y)
(6.104)
for which the initial conditionsvector y(0) is given. The application of Euler’s implicit scheme to (6.104) gives: yn+1 = yn + h · F(yn+1 )
(6.105)
In general this is some nasty set of nonlinear equations that has to be solved iteratively at each timestep. Usually we can get away with linearization the equation: " # ∂F yn+1 = yn + h · F(yn ) + (yn+1 − yn ) (6.106) ∂y yn Here ∂F ∂y is the matrix of the partial derivatives of the righthand side, and so at each time step we have to invert the matrix 1 − h · ∂F ∂y to find yn+1 . As we have previously noted, elementary reactions have greatly different reaction rates and reaction time constants (scales). The ratio between the largest and the smallest time constants of the reactions may be used to determine a degree of stiffness of the reaction system. The ∂F ∂y matrix is called Jacobian matrix of the system under considerations. Eigenvalues and eigenvectors of the Jacobian matrix reveal information about the time scales of the chemical reactions. These widely different time scales present classical methods (such as RungeKutta method) with the following difficulty; to ensure stability of the numerical solution, these methods are restricted to using very short time steps that are determined by the fastest reactions (the smallest time constant). However, the time for all chemical species to reach nearequilibrium values is determined by the slowest reactions (the largest time constant). Stiffness is an efficiency issue. If we were not concerned with how much time a computation takes, we wouldn’t be concerned about stiffness. Nonstiff methods can solve stiff problems. They just take a long time to do it. Thus, for a fast and robust solver, special numerical procedures suitable for solving stiff equations are required [27].
278
6.6 Summary
6.6 Summary In this chapter an insight into oxidation mechanisms of carbon monoxide, hydrogen and methane has been given. One should become familiar with chain initiation, chain branching and chain termination reactions. Using hydrogen as an example, the role of O, H and OH radicals not only in initiating but also in propagating combustion process has been underlined. Indeed, oxidation of carbon monoxide proceeds mainly though the "wet" route due to the reaction with OHradical. Oxidation of methane which is the most stable hydrocarbon, is initiated by the attack of O, H and OH radicals which produces reactive methyl radical (CH3 ). Then, through several intermediate reactions, carbon monoxide is produced which is further oxidised by OH to carbon dioxide. It has been stressed that carbon monoxide is an important intermediate in methane oxidation and the overall methane oxidation reaction CH4 + 2 O2 −−→ CO2 + 2 H2 O is an oversimplification. The following overall scheme CH4 + O2 −−→ CO + H2 + H2 O CO + 12 O2 −−→ CO2 H2 + 21 O2 −−→ H2 O is better since it accounts for COintermediate. Studying mechanisms of combustion reactions involves both formulation and solving chemical kinetic rate equations which form a set of ordinary nonlinear differential equations of the type d[Ai ] = fi ([A1 ], [A2 ], . . . , [AN ]; k1 , k2 , . . . , kR ) dt
for i = 1, 2, . . . , N
with initial conditions [Ai ](t = t0 ) = [Ai ]0 As a matter of fact in Chapter 5 we have already found analytical solutions to simple ordinary differential equations that describe first, second and third order reactions (see Section 5.2). In this chapter we have developed analytical solutions to a system k k A1 −−12 → A2 −−23 → A3
279
6 Mechanisms of Basic Combustion Reactions which contains not only substrates (A1) and products (A3) but also intermediates (A2). For firstorder reactions we have developed analytical formulae for calculating the variation of the concentrations of species A1,A2 and A3 with time. Then, we have analysed how the species concentrations vary with time for three cases: k12 ≫ k23 , k12 ≪ k23 and k12 = k23 . We have observed that when k12 ≪ k23 we could have simplified the solution (without compromising the accuracy too much) by invoking a quasisteady state assumption for A2intermediates. The quasisteady state approximation has been applied to Zeldovich mechanism of nitric oxide (NO) formation. The assumption of partial equilibrium for some reactions has been invoked to estimate the concentration of Oradicals. Students should understand both concepts important in kinetics; quasisteady state assumption and partial equilibrium. A comprehensive mechanism of methane combustion may involve up to a hundred species with perhaps as many as two hundred radical reactions. Thus, a general solver for solving chemical kinetic rate equations must be numerical. Euler’s explicit and implicit methods are presented in the lecture however we do not recommend them for any practical use. The methods are conceptually important since all practical methods originate from this idea. Using these methods we have introduced the notions of stability and accuracy which are two important features of numerical schemes. It has been underlined that the major problem associated with the simultaneous numerical integration of large sets of chemical kinetics rate equations is that of stiffness. For an efficient and robust solver special numerical schemes applicable to stiff equations are needed.
280
Bibliography [1] E. Thöne, U. Fahl. Energiewirtschaftliche Gesamtsituation. BWK: das Energie  Fachmagazin,Bd. 59 (2007), 4, S.3450. [2] H.Tsuji, A.K. Gupta, T. Hasegawa, M. Katsuki, K. Kishimoto and M. Morita. High Temperature Air Combustion, CRS Press, 2003. [3] Y.A. Cengel and M.A. Boles. Thermodynamics  An Engineering Approach, 4th Edition, McGrawHill, 2002. [4] Ch. Lüdecke and D. Lüdecke. Thermodynamik, Springer, 2000. [5] H.D. Baehr. Thermodynamik, 9th Edition, Springer, 1996. [6] W. Leuckel. Technische Verbrennung, Vorlesungsskript, TU Karlsruhe, 1992. [7] G.J. Van Wylen and R.E. Sonntag. Fundamentals of Classical Thermodynamics, EnglishSI version, 3rd Edition, John Wiley & Sons, New York, 1965. [8] North American Combustion Handbook, North American Mfg. Co., Cleveland, Ohio, USA, Volume II, 3rd Edition, 1995. [9] F.P. Incropera and D.P. DeWit, Fundamentals of Heat and Mass Transfer, 4th Edition, John Wiley & Sons, New York, 1996. [10] R. Weber, Lecture Notes in Heat Transfer, 2nd Edition, Papierflieger, ClausthalZellerfeld, 2004. [11] R. Weber, R. Alt and M. Muster Vorlesungen zur Wärmeübertragung, Teil I: Grundlagen, 3rd Edition, Papierflieger, ClausthalZellerfeld, 2007. [12] NISTJANAF Thermochemical Tables. 4th Edition. Journal of Physical and Chemical Reference Data. Monograph No. 9. Published by the American Chemical Society and the American Institute of Physics for the National Institute of Standards and Technology, 1998. [13] P. Atkins. Physical Chemistry, 6th Edition, W.H. Freeman and Company, New York, 1998. [14] J.M. Smith, H.C. Van Ness, M.M. Abbott. Introduction to Chemical Engineering Thermodynamics, 6th Edition, McGrawHill, 1996.
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Bibliography [15] F. Reif. Fundamentals of Statistical and Thermal Physics. McGrawHill, 1965. [16] J. M. Smith, H.C. Van Ness, M.M. Abbott, Introduction to Chemical Engineering Thermodynamics , 6th Edition, McGrawHill, 1996. [17] M. Graetzel, P. Infelta The Bases of Chemical Thermodynamics , Universal Publishers, Parkland, Florida, 2002. [18] S.W. Benson. The Foundations of Chemical Kinetics, McGrawHill, New ork, 1960. [19] W.C. Gardiner. Combustion Chemistry, SpringerVerlag, 1984. [20] J. Warnatz, U. Maas, R.W. Dibble. Combustion. 2nd Edition, SpringerVerlag, 1999. [21] D.L. Baulch, D.D. Drysdale, J. Duxbury and S.J. Grant, Evaluated Kinetic Data for High Temperature Reactions , Vol.1: Homogeneous Gas Phase Reactions of the H2O2 System , Butterworths, London, 1976. [22] J. Chomiak. Combustion; A Study in Theory, Fact and Application , Abacus Press, 1990. [23] I. Glassman. Combustion, 2nd Edition, Academic Press, 1987. [24] R. K. Wilk, LowEmission Combustion , Wydawnictwo Politechniki Slaskiej, Gliwice, 2002. [25] Miller I.A. and T. Bowmann. Mechanism and Modeling of Nitrogen Chemistry in Combustion. Progress in Energy and Combustion Science, 15, 287338, 1989. [26] W.H. Press, B.P. Flannery, S.P. Teukolsky, W.T. Vetterling. Numerical Recipes, Cambridge University Press, 1988. [27] E. Hairer and G. Wanner, Solving Ordinary Differential Equations II. Stiff and DifferentialAlgebraic Problems, 2nd Revised Edition, SpringerVerlag, 2002. [28] C.K. Law, Combustion Physics, Cambridge University Press, 2006.
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Gaussian Elimination There are two wellestablished ways to solve linear equations. The first one, which is a rather sophisticated way, introduces the idea of determinants. There is an exact formula called Cramer’s rule, which gives the correct values of the unknowns as a ratio of two M by M determinants. However for a reasonably human being M = 3 is about the upper limit of patience. The second method is called Gaussian elimination algorithm, or in short Gaussian elimination, which we highly recommend. The algorithm is constantly used to solve large systems of linear equations. The idea of Gauss is extremely simple: multiples of the first equation are subtracted from the other equations, so as to remove the first unknown from those equations. It leaves a smaller system of M − 1 equations in M − 1 unknowns. The process is repeated until there is only one equation and one unknown, which can be solved explicitly. Then, we proceed backwards and find other unknowns in reverse order. The example below will clarify the algorithm. We begin with a system of M = 3 equations:
2x + 3y + z = 5 x − y = −1 −4x + 2y + z = −3
(Ap1)
which can be written in matrix notation 5 x 2 3 1 1 −1 0 × y = −1 −3 z −4 2 1
(Ap2)
The task is to find the unknowns x, y, z using Gauss elimination. We subtract multiples of the first equation from the others, so as to eliminate x from the last two equations. To this end, we (a) subtract 0.5 times the first equation from the second equation,
283
Gaussian Elimination (b) subtract (2) times the first equation from the third equation. The result is the following system of equations:
2x + 3y + z = 5 1 7 5 − y− z=− 2 2 2 8y + 3z = 7 or
5 x 2 3 1 0 −2.5 −0.5 × y = −3.5 7 z 0 8 3
(Ap3)
(Ap4)
The coefficient 2, which appears in front of the first unknown x, in the fist equation, is called the first pivot. Elimination is constantly dividing the pivot into the numbers underneath it, to find out the right multipliers. At the second stage of elimination; we ignore the first equation. The other two equations contain only y and z unknowns and the same elimination process can be applied to them. The pivot for this elimination stage is (−5/2). A multiplicity of the second equation will be subtracted from the remaining equations (in this case only the third equation remains). We multiply the second equation by 8 · 25 = 16 5 and add it to the third one. In other words, we (c) subtract (−16/5) times the second equation from the third one:
2x + 3y + z = 5 1 7 5 − y− z=− 2 2 2 7 21 z=− 5 5
(Ap5)
or 5 x 2 3 1 0 −2.5 −0.5 × y = −3.5 −4.2 z 0 0 1.4
284
(Ap6)
The elimination process is now complete. Observe that the final matrix of the elimination process contains zeros in all places located below the diagonal. The order in which to solve the above system (Ap6) is obvious. The last equation gives z = −3. Substituting into the second equation, we find y = 2. Then the first equation provides x = 1. The last process is called backsubstitution. Note that under completion of the elimination process we obtain a triangular system which is easy to solve by backsubstitution. The elimination process, which in our case includes three steps, has produced three pivots; 2 – the first stage pivots, (−5/2)  the second stage pivot and (7/5) – the third stage pivot. The reader will notice that pivots cannot be zero. We need to divide by them. An important number which we use is the rank ρ of a matrix; say matrix A of Eq. (Ap(2)). There are several definitions of the rank and they are equivalent. Three most important definitions are quoted here for completeness: the rank counts the number of independent rows in the matrix, the rank is the number of pivots in the Gauss elimination process, the rank is the number of nonzero rows in the final matrix of the Gauss elimination process.
285
Gaussian Elimination
286
Vocabulary Table 6.9: Technical vocabulary. The vocabulary has been prepared for students of TU Clausthal attending lectures on Combustion Technology, Heat Transfer, Advanced Heat Transfer and High Temperature Processes.
English Ability absorb absorptance absorption capacity absorptivity acceleration accordingly accumulation accuracy accurate achieve acid acid rain adiabatic adjacent advanced advantage affect ageing agreement aggregation state aim (at) air blower air equivalence ratio airflow system air stream
Deutsch Fähigkeit absorbieren Absorptionsgrad Absorptionsvermögen Absorptionsgrad, vermögen Beschleunigung demgemäß, folglich Ansammlung, Anhäufung, Speicherung Genauigkeit genau vollenden, erhalten, erreichen Säure; sauer saurer Regen adiabat angrenzend, benachbart erweitert Nutzen, Vorteil beeinflussen, sich auswirken Alterung Abkommen, Zustimmung, Einigung Aggregatzustand zielen (auf), anstreben, beabsichtigen Luftgebläse, Ventilator Luft(verhältnis)zahl (z.B. 1,1 ... 1,15) Belüftungssystem Luftstrom
287
Vocabulary Table 6.9: Technical vocabulary (continued)
English aligned alloy alloy steel aluminum ambient ambiguity ammonia amount amount of heat "Q" (in J) anodized annular annulus aperture application appreciate approach appropriate approximation a priori determination arbitrary area ash assess assessment associated assumption atomization attenuation audit augmented available average averaged avoid
Deutsch ausgerichtet Legierung legierter Stahl Aluminium umgebend, UmgebungsZweideutigkeit, Mehrdeutigkeit Ammoniak Menge Wärmemenge (in J) eloxiert ringförmig Ringspalt Öffnung Anwendung, Einsatz abschätzen Annäherung passend, geeignet Näherung Vorherbestimmung willkürlich, beliebig, allgemein Fläche Asche abschätzen, bewerten, ansetzen, festlegen Wertung, Abschätzung zugeordnet Annahme Zerstäubung Abschwächung, Verminderung Prüfung vermehrt verfügbar Durchschnitt gemittelt vermeiden
Baffle
Umlenkung, Blende, Trennwand
288
Table 6.9: Technical vocabulary (continued)
English balance bank of tubes bar basis basement beam bell mouth benchmark benefit bent blackbody blast furnace blast furnace gas blower body force boiler boiling bond booster bottom bottom line boundary boundary condition boundary layer bounding curve branching brass brick bright bubble buffer bulb bulk bulk flow bulk motion
Deutsch Bilanz Rohrbündel Balken Grundfläche, Basis Keller Strahl Schalltrichter Bewertung, Maßstab, Richtwert Nutzen; nützen gebogen Schwarzer Körper Hochofen Hochofengas, Gichtgas Lüfter, Gebläse Volumenkraft Kessel Sieden Bindung (chemisch) Zusatzantrieb Boden das Entscheidende Grenze, Rand, Begrenzung Randbedingung Grenzschicht Grenzkurve Verzweigung Messing Ziegelstein blank, glänzend Blase Zwischenspeicher, Dämpfer Glühbirne Volumen, Menge, Masse, Schüttung, Klumpen Massenströmung, Kollektivströmung (von Molekülen) Massenbewegung
289
Vocabulary Table 6.9: Technical vocabulary (continued)
English bulk temperature bumpy bundle buoyancy burner burnout, degree of combustion, loss on ignition (LOI)
Calculation
calibration curve capacitance capacity capsule carbon carbon dioxide carbonic acid carbonization cartoon representation cavity cement kiln center of gravity ceramic fiber chain chain reaction channel char chemical engineering chemistry chimney chlorine chromium circuit circumference clarify classes
290
Deutsch mittlere Temperatur, kalorische Temperatur uneben Bündel Auftrieb Brenner Ausbrand
Berechnung Eichkurve Kapazität, Belastbarkeit Vermögen, Fähigkeit, Leistung Kapsel Kohlenstoff Kohlendioxid Kohlensäure Kohlenstoffanreicherung, Verkokung Prinzipdarstellung Holhlraum Zementofen Schwerpunkt Keramikfaser Ablauf, Kette Kettenreaktion Kanal Holzkohle Chemieingenieurwesen Chemie Schornstein Chlor Chrom Kreis, Anlage, Schaltung, Schaltkreis, Schaltbild Umfang klären Übungsstunden
Table 6.9: Technical vocabulary (continued)
English clay clay brick closed system cluster coal coarse coat coating coflow, cocurrent flow, (parallel flow) coke cokeoven gas cold blast collapse collide collision collision factor colloidal column combustible combustion combustion chamber combustion engineering combustion product compact comparison compass composite composition compound comprehensive compression compression work comprise concentration concern
Deutsch Lehm, Ton Lehmziegel geschlossenes System Anhäufung, Ballen, der Cluster Kohle grob beschichten Beschichtung, Belag Gleichstrom Koks Koksofengas, Kokereigas kaltgeblasen Zusammenbruch, Einsturz; zusammenfallen kollidieren, zusammenstoßen Zusammenstoß Stoßfaktor fein verteilt Säule, Spalte Brennbares, Brennmaterial Verbrennung Brennkammer, Brennraum Verbrennungstechnik Verbrennungsprodukt dicht Vergleich, Abgleich, Gegenüberstellung Zirkel zusammengesetzt; Verbundwerkstoff Zusammensetzung Verbindung, Mischung umfassend, ausgedehnt Kompression, Verdichtung Druckänderungsarbeit zusammenfassen Konzentration sich befassen mit
291
Vocabulary Table 6.9: Technical vocabulary (continued)
English concrete condensation condense condenser condition conduction conductivity conductor (electric) cone confidence intervall configuration factor conjecture conservation constrain constrained constraint consumption contact surface contaminated soils continuity equation contribute contribution control volume convection (convective) heat transfer coefficient "h" (in W/m2 K) convenient conversion cool cooling cooling rate coordinate copper correlation corrugated
292
Deutsch Beton Kondensation, Verdichtung kondensieren, niederschlagen Kondenser, Verflüssiger Bedingung Leitung Leitfähigkeit Leiter (elektrischer) Kegel Vertrauensbereich Einstrahlzahl, Winkelverhältnis, Formfaktor Vermutung Erhaltung einschränken zwangsläufig Nebenbedingung, Einschränkung Verbrauch Berührungsfläche Altlasten Kontinuitätsgleichung beitragen, beisteuern, liefern Beitrag Kontrollvolumen Konvektion, "Mitführung" von Energie durch kleinste Strömungsteilchen (konvektiver)Wärmeübergangskoeffizient "α" (in W/m2 K) passend, geeignet, bequem Umformung, Umwandlung kühlen, abkühlen Kühlung Kühlrate Koordinate Kupfer Wechselbeziehung gewellt
Table 6.9: Technical vocabulary (continued)
English corrugated iron corrugation counter balance counterflow, countercurrent flow coupled differential equations courtesy of cross flow cross section crucial cumulative cure current curvature cyclic integral
Dashed line
data gathering decay decline decrease deduce deep freezer deficient deficiency defrost degree of freedom delivery denominator dense density departure dependence depletion derivation desalination
Deutsch Wellblech Welle, Riffelung Gegengewicht; ausgleichen Gegenstrom gekoppelte Differentialgleichungen freundlicherweise zur Verfügung gestellt Kreuzstrom Querschnitt ausschlaggebend anwachsend trocknen, aushärten (elektrischer) Strom; laufend, Krümmung Umlaufintegral gestrichelte Linie Datenerfassung Zerfall, Abnahme abnehmen, neigen abnehmen, vermindern herleiten, folgern Tiefkühltruhe unzureichend Mangel auftauen Freiheitsgrad Zuführung, Lieferung Nenner dicht Dichte Abweichung Abhängigkeit Abreicherung, Substanzverringerung Herleitung, Ableitung Entsalzung
293
Vocabulary Table 6.9: Technical vocabulary (continued)
English design determination determine development deviate device devolatilization dew point diameter diatomic diffraction diffuse dihedral dilute dimensionless direction disadvantage discovery dispatch displacement disposal dissipate dissociation distinct distinguish distribution disturb dot ("e dot") double pipe heat exchanger downstream draft, draught drag coefficient drag force
294
Deutsch Plan, Konstruktion, Entwurf, Auslegung konstruieren, auslegen Bestimmung bestimmen Entwicklung Abweichen Apparat, Einrichtung, Vorrichtung, Gerät Entgasung, Entfernung flüchtiger Bestandteile Taupunkt Durchmesser zweiatomig Beugung, Ablenkung diffus zweiflächig abschwächen, verdünnen dimensionslos Richtung Nachteil Entdeckung Beseitigung, Bericht Verschiebung Beseitigung, Deponie, Entsorgung, Verwendung, Verfügung (Wärme) abführen, zerstreuen Dissoziation verschieden unterscheiden Verteilung stören Punkt ("e˙ ") Doppelrohrwärmeübertrager stromabwärts, nachgeschaltet Zug, Schornstein, Kaminzug Widerstandsbeiwert (hydrodynamische) Widerstandskraft
Table 6.9: Technical vocabulary (continued)
English drawback driving force droplet dry duct due to dummy variable durability duration dust
Deutsch Nachteil, Hindernis Antriebskraft, treibende Kraft Tröpfchen trocken Kanal, Röhre, Schacht beruhend auf Integrationsvariable Haltbarkeit Zeitdauer Staub
Edge
Ecke, Rand Strahlungstemperatur Wirkungsgrad, Wirtschaftlichkeit Abfluß, Ausfluß Anstrengung, Leistung z. B. Glühbirne elektromagnetische Wellen Emission, Aussendung Energiestromdichte, spezifische Ausstrahlung Emissionsgrad, vermögen aussenden betonen anwenden, einsetzen einkapseln Hohlraum, Umhüllung, Einschluß, abgeschlossener Raum einschließen, enthalten endotherm, wärmeaufnehmend, verbrauchend Energie Energiebilanz Maschine Technik, Technologie erhöhen
effective temperature efficiency efflux effort e.g. (=exempli gratia, for instance) electrical bulb electromagnetic waves emission emissive power emissivity emit emphasise employ encapsulate enclosure encompass endothermic energy energy balance engine engineering enhance
295
Vocabulary Table 6.9: Technical vocabulary (continued)
English enriched ensure enthalpy enthalpy of formation entrain entrainment entropy environment equalise equally spaced equation equation of state equilibrium equipment equivalence estimate estimation ethene ethyne evacuate evaluate evaporation exceed excess excess air ratio, excess air factor excess temperature exchange exhaust exhaust gas exothermic expand expansion experimenter explanation expose
296
Deutsch angereichert sicherstellen, ergeben Enthalpie Bildungsenthalpie mitführen Mitführen, Mitreißen, Eintrag Entropie Umgebung ausgleichen äquidistant Gleichung Zustandsgleichung Gleichgewicht Ausrüstung, Einrichtung Gleichwertigkeit schätzen, berechnen, bestimmen Bewertung Äthen Äthin, Azetylen evakuieren berechnen Verdampfung übersteigen Überschuß Luftzahl, überschuß (z.B. 0,1 ... 0,15) Übertemperatur Austausch absaugen Abgas exotherm, wärmeabgebend ausdehnen Expansion, Ausdehnung, Entspannung Experimentator Erklärung aussetzen
Table 6.9: Technical vocabulary (continued)
English exposure extended extent external flow
Deutsch Bestrahlung erweitert Ausdehnung überströmt
Factory
Betrieb, Fabrik Ausfall vertraut Föhn, Ventilator Gebläseleistung Fehler, Defekt Durchführbarkeit, Wahrscheinlichkeit Eigenschaft, Merkmal Rohstoff, Ausgangsmaterial Speisewasservorwärmer Faser Abbildung, Zeichen, Ziffer Glühfaden, Wendel Rippe, Kühlrippe gerippt Schamottestein einfügen, befestigen, anpassen Ausgleichskurve flockenartig eben, flach Mehl Bewegung, Durchfluß, Strömung; strömen Durchsatz, Volumenstrom Schwankung Strömungsgleichungen Wirbelschicht, Fließbett Fluß, Stromdichte konzentrieren auf Folie
failure familiar fan fan capacity fault feasibility feature feedstock feedwater fiber figure filament fin finned fire clay brick fit fitting curve flaketype flat flour flow flow rate fluctuation fluid flow equations fluidised bed flux focus on foil
297
Vocabulary Table 6.9: Technical vocabulary (continued)
English forced convection forced flow forecast fouling fraction fractional emissive power free convection free jet free stream frequency friction frustum, frustrum fuel furnace
Gage, gauge
gaseous gas phase, gaseous phase gathering general generate generation globe goal goggles govern governing equation gradient graph graphite gravitational force gravity gray (grey) body grid grip
298
Deutsch erzwungene Konvektion, Zwangskonvektion erzwungene Strömung Vorhersage; vorausberechnen Beschlagen, Verschmutzung Bruchteil Bruchteilfunktion freie Konvektion Freistrahl freie, unbehinderte Strömung Frequenz, Häufigkeit Reibung Stumpf, Kegelstumpf Brennstoff Ofen Meßinstrument gasförmig Gasphase Sammeln, Erfassung allgemein erzeugen Erzeugung Erdkugel Zielpunkt, setzung Brille beschreiben, beeinflussen, bestimmen, herrschen prozeßbestimmende Gleichung Gradient, Neigung, Steigung Graphik Graphit Gravitationskraft, Schwerkraft Schwerkraft Grauer Körper Gitter, Raster Druckwalze
Table 6.9: Technical vocabulary (continued)
English gross calorific value "GCV" gypsum plaster
Deutsch Brennwert, oberer Heizwert "ho " Gipskartonplatte
Halflife time
Halbwertzeit Halbwertsbreite unbedenklich, unschädlich schraffiert Wärmekapazität Wärmeleitfähigkeit "λ" (in W/m · K) Wärmeleitungsgleichung
half width harmless hatched heat capacity heat conductivity "k" (in W/m · K) heat conduction (or: diffusion) equation heat diffusion heater heat exchanger ˙ (in J/s= W) heat flow "Q" heat flux "q" ˙ (in W/m2 ) heat generation heating rate heating time heating zone heat loss heat recovery heat removal heat sink heat source heat transfer heat transfer coefficient "h" (in W/m2 · K) ˙ (in J/s= W) heat (transfer) rate "Q" hemispherical heterogeneous high alloy steel hollow cylinder homogeneous honeycomb
Wärmeausbreitung Heizkörper Wärmeübertrager ˙ (in J/s= W) Wärmestrom "Q" Wärmestromdichte "q" ˙ (in W/m2 ), Wärmefluß Wärmequelle Aufheizgeschwindigkeit Erwärmungszeit Heizzone, Wärmzone Wärmeverlust Wärmerückgewinnung Wärmeabfuhr Wärmesenke Wärmequelle Wärmeübertragung Wärmeübergangskoeffizient "α " (in W/m2 · K) ˙ (in J/s= W) Wärmestrom "Q" hemisphärisch, Halbkugelheterogen, ungleichartig hochlegierter Stahl Hohlzylinder homogen Wabe
299
Vocabulary Table 6.9: Technical vocabulary (continued)
English hot blast hot blast stove hotwire anemometer humidity hydrocarbon hydrochlorid acid hydrodynamic hydrodynamic boundary layer hydrodynamic entrylength hydrogen
Deutsch heißgeblasen Winderhitzer Hitzdrahtanemometer Feuchtigkeit, Luftfeuchte Kohlenwasserstoff Salzsäure hydrodynamisch Strömungsgrenzschicht hydrodynamische Anlaufstrecke Wasserstoff
Ideal gas
Ideales Gas Entzünden, Zündung eintauchen Aufprall, Auswirkung Prall, Zusammenstoß wichtig Verbesserung, Fortschritt Verunreinigung weißglühend (Strahlungs, Licht) Einfall, Einfluß einfallend, auftreffend Veraschung, Veraschen eintretend berücksichtigen wachsen, zunehmen (temperatur)unabhängig hervorrufen, veranlassen Saugzuggebläse Ungleichung Trägheitskraft infinitesimal, unendlich klein Ablenkung Wendepunkt, Knickpunkt Einfließen Block Anfangs
ignition immerse impact impingement important improvement impurity incandescent incidence incident incineration incoming incorporate increase independent (of temperature) induce induced draft fan inequality inertial force infinitesimal inflection inflection point inflow ingot initial
300
Table 6.9: Technical vocabulary (continued)
English initial condition inleakage inlet temperature insignificant instant (of time) insulating plate insulation integer intensity intercept
inviscid irradiation, irradiance irreversible isolated system isolation isotherm isotropic issue
Deutsch Anfangsbedingung Leckluft Eintrittstemperatur unwichtig, bedeutungslos Zeitpunkt, Augenblick Isolierplatte Isolation ganze Zahl Intensität, Strahldichte Achsenabschnitt; auffangen, empfangen Schnittstelle, Grenzschicht dazwischenliegend Innere Energie durchströmt untereinander zusammenhängen Wechselbeziehung Schnittpunkt, Schnittmenge, Durchschnitt reibungsfrei Bestrahlung nicht umkehrbar isoliertes ("abgeschlossenes") System Isolation Isotherme; isotherm isotrop, richtungsunabhängig Ausgabe, Ergebnis, Thema, Problem
Jet
Strahl
Kernel of integral equation
Kern der Integralgleichung Trockenofen, Brennofen kinetische Energie
Laminar flow
laminare Strömung Lampe
interface intermediary internal energy internal flow interrelate interrelation intersection
kiln kinetic energy
lamp
301
Vocabulary Table 6.9: Technical vocabulary (continued)
English landfill lattice layer lead leading edge leakage lean combustion least square method leave lignin limestone linearization line graph liquid localised log mean temperature difference "LMTD" low alloy steel lower calorific value "LCV" lubricating oil lumped lumped capacitance method lumped thermal capacity
Magnification maintain make up malfunction manufacturing manure margin masonry wall mass balance mass flow mass flow rate mass fraction
302
Deutsch Ablagerung, Deponie Gitter Schicht Blei Stirnseite, Anströmkante Leck, Undichtigkeit, Falschluft überstöchiometrische Verbrennung Methode der kleinsten (Fehler) Quadrate verlassen Lignin, Holzstoff Kalkstein Linearisierung Liniendiagramm Flüssigkeit; flüssig punktförmig mittlere logarithmische Temperaturdifferenz niedriglegierter Stahl (unterer) Heizwert "hu " Schmieröl punktförmig verteilt, konzentriert "Newtonsches Abkühlungsgesetz" thermische Blockkapazität Vergrößerung (aufrecht)erhalten Zusammensetzung Fehlfunktion, Störung Erzeugung, Produktion Dünger Rand gemauerte Wand Massenbilanz Massenstrom Massendurchsatz Massenbruch, Massenanteil
Table 6.9: Technical vocabulary (continued)
English mat matching material property matter mean free path measure measurement mechanism merger merit function methane mill mineral fibre minimum air requirement moisture molar mass molar weight mole fraction, molar fraction molecularity molten moment of inertia momentum momentum balance equation monatomic
Deutsch matt Abgleich, Anpassung Materialeigenschaft Materie mittlere freie Weglänge Maß; messen Messung Mechanismus Verschmelzung Gütefunktional, MeritFunktion Methan Walze Mineralfaser minimaler Luftbedarf Feuchte, Feuchtigkeit molare Masse (in kg/kmol) Molekulargewicht Molenbruch, Molanteil Molekularität geschmolzen Trägheitsmoment Impuls Impulsbilanzgleichung einatomig
Natural convection
natürliche (freie) Konvektion Erdgas notwendig vernachlässigen vernachlässigbar netto Stickstoff Edelgas senkrecht (zu) Norm(al)bedingung, Standardbedingung Haftbedingung
natural gas necessary neglect negligible net nitrogen noble gas normal (to) normal condition noslip condition
303
Vocabulary Table 6.9: Technical vocabulary (continued)
English nozzle nucleate boiling nucleation site numerator numerically
Deutsch Düse Blasensieden Keimstelle Zähler numerisch, in Zahlen
Objective
Ziel beobachtbar erhalten offensichtlich vorkommen, stattfinden, sich ereignen, auftreten Knick, räumliche Verschiebung Anfang, Umschlag zu Undurchsichtigkeit opak, undurchsichtig offenes System Betriebsbedingung optische Dicke Größenordnung Erz ausgerichtet entstehen, ausgehen von Oszillator, Schwingungserzeuger Ergebnis austretend Abfluss Ausgangsspannung Ofen geteilt durch: z geteilt durch n ( nz ) Gesamtwärmeverlust Wärmedurchgangskoeffizient "k" (in W/m2 · K) Überdruck Oxidierung Oxidationsmittel (Sauerstoff, Luft) Verbrennung mit reinem Sauerstoff
observable obtain obvious occur
offset onset opacity opaque open system operating condition optical thickness order of magnitude ore oriented originate oscillator outcome outcoming outflow output voltage oven over: z over n overall heat loss overall heat transfer coefficient "U" (in W/m2 · K) overpressure oxidation oxidiser oxy fuel combustion
304
Table 6.9: Technical vocabulary (continued)
English oxygen
Deutsch Sauerstoff
Parallelepiped
Quader Ballen, Bündel Partialdruck Muster Pellet Pendel vorletzte prozentualer Anteil Ideales Gas IdealesGasGesetz, Zustandsgleichung für Ideale Gase leisten, erfüllen, durchführen Verhalten, Durchführung, Leistung Umfang Permeabilität, Durchlässigkeit Normale, Senkrechte; senkrecht (zu) gehören zu sachdienlich, passend Phasenwechsel Roheisen Bolzen Rohr Rohrleitung Kolben Neigung, Steigung, Abstand eben ebene Wand Anlage Platte Plattenwärmeübertrager Rauchfahne, Abwasserfahne betonen, hervorheben, auf etwas hinweisen Schadstoff; verschmutzt Verunreinigung
parcel partial pressure pattern pellet pendulum penultimate percentage perfect gas perfect gas equation of state perform performance perimeter permeability perpendicular pertain to pertinent phase change pig iron pin pipe pipeline piston pitch planar, plane plane wall plant plate plate heat exchanger plume point out pollutant pollution
305
Vocabulary Table 6.9: Technical vocabulary (continued)
English polynomial polystyrene pool boiling porous potential energy power power engineering power generation power plant power station precisely preferential preheat preheating zone preparation prerequisite pressure pressure drop prevail previous primitive printed board product profile propagation property proposal propulsion protective atmosphere provide provide for pulverised coal pump purchasing purify purpose
306
Deutsch Polynom Polystyrol Behältersieden porös potentielle Energie Leistung Energietechnik Stromerzeugung Kraftwerk, Triebwerk Kraftwerk, Elektrizitätswerk genau bevorzugt vorwärmen Vorwärmzone Bereitstellung, Aufbereitung Vorbedingung, Voraussetzung Druck Druckverlust, abfall, absenkung vorherrschen vorhergehend Stammfunktion, (unbestimmtes) Integral Leiterplatine Produkt, Erzeugnis Profil Wachstum Eigenschaft Ansatz, Vorschlag Antrieb Schutzatmosphäre, Schutzgas beschaffen, zuführen, liefern gewährleisten pulverisierte Kohle, Kohlenstaub Pumpe Beschaffung, Erwerb reinigen Zweck
Table 6.9: Technical vocabulary (continued)
English
Deutsch
Quantum number
Quantenzahl Wirkungsquantum abschrecken ruhend angeben, zitieren
Radiance
Strahlstärke Strahlung Strahler, Heizkörper Helligkeit regellose, zufällige Bewegung Bereich schnell Strom, Menge je Zeiteinheit, pro Zeit Geschwindigkeitskoeffizient Geschwindigkeitskonstante Transportansatz Geschwindigkeitsgesetz ˙ (in J/s=W) Wärmestrom "Q" Umsetzungsgeschwindigkeit Rohmaterial Reaktant, Reaktionspartner, Edukt Reaktion Reaktionsenthalpie Reagens Heck umstellen, umordnen vernünftig Argumentation, Schlußfolgerung empfohlen, vorgeschlagen in Einklang bringen beziehen auf Raffinierung, Verbesserung reflektieren, spiegeln Reflexionsgrad, vermögen
quantum of action quench quiescent quote
radiation radiator radiosity random motion range rapid rate rate coefficient rate constant rate equation rate law ˙ (in J/s=W) rate of heat transfer "Q" rate of consumption raw material reactant reaction reaction enthalpy reagent rear rearrange reasonable reasoning recommended reconcile refer to refinement reflect reflectivity
307
Vocabulary Table 6.9: Technical vocabulary (continued)
English refractory refractory brick refractive index refrigerant refuse derived fuel (RDF) regenerative burner regenerator region reheat reheating furnace relation relationship release reliability removal replacement require rescale resistance responsible resource result retardation reversible rich combustion right circular cone rocket booster rod rolling mill roughly roughness roundoff error, rounding error rubbish
Sample
saturated
308
Deutsch feuerfestes Material; feuerfest feuerfester Stein, Schamottestein Brechungskoeffizient, Brechungsindex, Brechzahl Kältemittel, Kühlflüssigkeit Sekundärbrennstoff aus Abfall Regenerativbrenner Regenerator Bereich wiedererwärmen Nachwärmofen Beziehung Verhältnis, Beziehung freisetzen, abgeben Zuverlässigkeit, Sicherheit Abtransport Ersetzung fordern, anfordern neuen Maßstab festsetzen Widerstand verantwortlich Ressource, Quelle, Hilfsmittel Resultat, Ergebnis Verzögerung umkehrbar unterstöchiometrische Verbrennung gerader Kreiszylinder Zusatztriebwerk Stab Walzwerk grob gesagt Rauhigkeit Rundungsfehler Abfall, Schutt Probe, Exemplar, Muster gesättigt
Table 6.9: Technical vocabulary (continued)
English saturation temperature saving scale scatter scope screen sealing separation series series expansion sewage sludge shape shape factor shear stress shell shellandtubeheat exchanger shield shrouding side wall silicium carbide silver simultaneous sink sinter site sketch skilfully slab slag slide slope sludge slurry smelt smooth
Deutsch Sättigungstemperatur Einsparung Maßstab streuen Bereich Bildschirm, Blende, Sieb; durchsieben, überprüfen Dichtung Ablösung Reihe, Folge Reihenentwicklung Klärschlamm Form, Gestalt Formfaktor, Einstrahlzahl, Winkelverhältnis Schubspannung, Scherspannung Hülle, Mantelfläche, Ofenwanne Rohrbündelwärmeübertrager Schirm Abdeckung Seitenwand Siliziumkarbid Silber gleichzeitig Senke Sinter Ort, Lage, Gebiet Skizze geschickt Platte Schlacke Dia, Folie Steigung, Gefälle Klärschlamm Schlamm, wässrige Masse schmelzen flach, gleichmäßig; glätten
309
Vocabulary Table 6.9: Technical vocabulary (continued)
English soak soaking zone soften soils (contaminated soils) solid solid angle solidification solid material solid state solution soot sought source space shuttle sparse specific heat capacity (in J/kg · K) species sphere sponge spring square (r square = r2 ) stability stable stack staged air staggered stagnant air stagnation point stainless steel standard enthalpy of formation state equation state variable static pressure stationary steady state
310
Deutsch ausgleichen Ausgleichszone weich machen Böden (Altlasten) fest Raumwinkel Erstarrung Feststoff, Schleudergut Festkörper Lösung Ruß gefragt, gesucht Quelle Raumfähre spärlich, wenig spezifische Wärmekapazität (in J/kg · K) Spezies, Art, Substanz, Stoff Kugel Schwamm Feder Quadrat Stabilität stabil, beständig Speicher, Stapel gestufte Luft versetzt, gestaffelt ruhende Luft Staupunkt rostfreier Stahl Standardbildungsenthalpie Zustandsgleichung Zustandsvariable, größe statischer Druck stationär, ortsfest stationärer Zustand; stationär, zeitunabhängig
Table 6.9: Technical vocabulary (continued)
English steam steel steelmaking steelwork steep stoichiometry storage store straw stream stressstrain relation subscript substance substantial substrate subtend sufficient suggest suitable sulphur, sulfur superficial supply support surface surrounding suspension swirl symmetry axis symmetric plane
Deutsch Wasserdampf Stahl Stahlerzeugung Stahlwerk steil Stöchiometrie Speicherung speichern Stroh Strom SpannungsDehnungsBeziehung Index Stoff wesentlich Substrat gegenüberliegen hinreichend vorschlagen brauchbar, geeignet Schwefel oberflächlich Versorgung; zuführen Halterung, Fassung, Lager Oberfläche Umgebung Aufschwemmung Wirbel Symmetrieachse Symmetrieebene
Tackle
bewältigen, in Angriff nehmen Teerpappe Temperatur Temperaturdifferenz Temperaturverteilung Temperaturabfall, abnahme Tetraeder
tar paper temperature temperature difference temperature distribution temperature drop tetrahedron
311
Vocabulary Table 6.9: Technical vocabulary (continued)
English thermal thermal boundary layer thermal capacity thermal conductivity "k" (in W/m · K) thermal diffusivity "α "(in m2 /s) thermal expansion coefficient thermal radiation thickness thin walled throttling time constant tin torque total thermal input "TTI" total thermal resistance "TTR" (in W/K) to the power: T to the power 4 trailing edge transfer transient transition transmission transmissivity transmit transparency transparent transverse transverse pitch trap treat as treatment trickle
312
Deutsch thermisch, WärmeTemperaturgrenzschicht Wärmekapazität Wärmeleitfähigkeit, koeffizient "λ" (in W/m · K) Temperaturleitfähigkeit, koeffizient "α" (in m2 /s) Wärmeausdehnungskoeffizient Temperatur, Wärmestrahlung Dicke dünnwandig Drosselung Zeitkonstante Zinn Moment, Drehmoment insgesamt zugeführte thermische Energie Gesamtwärmewiderstand hoch: T hoch 4 (T4 ) Hinterkante übertragen zeitlich veränderlich Übergang (Wärme)Durchgang (Durchlaß), Transmissionsgrad, vermögen transmittieren, durchlassen Transparentbild, Diapositiv, Folie durchsichtig quer Querteilung auffangen, einschließen behandeln als Aufbereitung, Bearbeitung, Verfahren tröpfeln, rieseln
Table 6.9: Technical vocabulary (continued)
English trigger tripping trip wire tube tube bank tubular heat exchanger tundish tungsten turbulent flow
Deutsch auslösen Auslösung Reißleine Rohr Rohrbündel RohrbündelWärmeübertrager Gießwanne Wolfram turbulente Strömung
Unaltered
unverändert nichtbrennbare Anteile Bildunterschrift; unterstreichen, betonen gleichverteilt Vereinigung (von Mengen) einheitlich, eindeutig, einzig Einheit Einheitsvektor sperrig, unhandlich stromaufwärts, vorgeschaltet
unburnts underline
uniform union (of sets) uniquely unit unit vector unwildy upstream
Valid
validate valve vapor, vapour vaporization velocity vibration vicinity view factor viscous viscous force voidage
gültig bestätigen, für gültig erklären Ventil Dampf Verdampfung Geschwindigkeit Schwingung Umgebung Einstrahlzahl, Winkelverhältnis, Formfaktor viskos, zäh, dickflüssig Zähigkeitskraft Leerraum, Porosität, relatives Porenvolumen
313
Vocabulary Table 6.9: Technical vocabulary (continued)
English volatile voltage volumetric flow rate volumetric heat capacity: "cp " voluntary vortex
Wake walking beam waste waste water water wavelength wave number wedge wet width wire wooden lagging work
Deutsch Flüchtige; flüchtig (elektrische) Spannung Volumenstrom volumenbezogene Wärmekapazität: "cp " willkürlich Wirbel Nachlauf, Kielwasser, Spur, Wirbelschleppe Hubbalken Abfall Abwasser Wasser Wellenlänge Wellenzahl Keil benetzen; feucht Breite Draht Holzverschalung Arbeit
Xaxis Xrays
xAchse, Abszisse Röntgenstrahlen
Yaxis
yAchse, Ordinate
Zinc
Zink
314
Acronyms NIST
National Institute of Standards and Technology (USA)
JANAF
Joint ArmyNavyAir Force
GCV
Gross Calorific Value (oberer Heizwert oder Brennwert)
LCV
Lower Calorific Value (unterer Heizwert oder Heizwert)
TTI
Total Thermal Input
NIV
Number of Independent extensive Variables
315
Acronyms
316
Nomenclature Latin Letters C
specific heat capacity
c
concentration
c
specific heat capacity
J kmol · K mol m3 J kg · K J mol · K J kg · K J mol · K J kg · K
Cp
specific heat at constant pressure
cp
specific heat at constant pressure
Cv
specific heat at constant volume
cv
specific heat at constant volume
d¯ E˙
inexact differential energy rate
W
E
energy amount
J
Ea
activation energy
F
Helmholtz free energy or free internal en
J mol
J
ergy g
specific free enthalpy
G
Gibbs thermodynamic potential or free
J mol
J
enthalpy J kg
g
Gibbs specific free enthalpy
g H˙
earth’s gravity
9.806 65 m/s2
enthalpy rate
W
h
specific enthalpy
J mol
H
enthalpy
J
h
specific enthalpy
J kg
I
Irreversibility
W
317
Nomenclature moment of inertia
K
standard equilibrium constant rate constant for a
k k, kf , kb L˙
ldry
kg · m2
I
1st
order reaction
rate constants for a reaction
— 1 s
depends on reaction order
mechanical power
W
L
work
J
l
specific work
air,min
minimum amount of air required per unit
J kg kg air kmol fuel
of fuel ldry
air
m ˙
mass rate
m
mass
Mi
molar mass of species i
Mmean
average molar weight of a mixture
kg air kmol fuel kg s
kg g mol g mol kmol s 6.023 · 1023 mol−1
n˙ i
molar flow rate
NA
Avogadro constant
N
Loss of power
W
n
amount of substance
mol
ni
amount of substance i
mol
p
pressure
Pa
pi
partial pressure of component i
Pa
saturation pressure
Pa
rate of heat removal
W
Q
heat
J
q
specific heat
J kg
R
universal gas constant
r
enthalpy of condensation of water
s
specific entropy
S
entropy
s
specific entropy
T
temperature
K
t
time
s
psat Q˙
318
amount of air supplied per unit of fuel
8314.3 J/(kmol · K) 43 940 kmol kJ of H
2O
J mol · K J K J kg · K
Nomenclature umixture
specific internal energy
U
internal energy
u
specific internal energy
V
volume
v
specific volume
Vdry,min
minimum dry amount of combustion
kJ kmol
J J kg m3 m3 kg kmol dry products kmol fuel
products Vdry Vwet,min
dry amount of combustion products minimum wet amount of combustion
kmol dry products kmol fuel kmol wet products kmol fuel
products Vwet
wet amount of combustion products
kmol wet products kmol fuel m s
w
translational velocity
wi
mass fraction of species i
1
xi
mole fraction of species i
1
z
height of center of gravity
m
Greek Letters ∆R GpT
Gibbs reaction enthalpy
kJ mol
γ
relative humidity
λ
excess air ratio
µ
chemical potential
µ
chemical potential
π˙
rate of entropy change
Φ
fuel equivalence ratio
π
sum of the entropy change in process
ϕ
specific humidity
ρ
density
ν
stoichiometric factor
ω
rotational velocity
s−1
ξ
extent of reaction
mol
1 J kg J mol J K·s
1 J K kg water vapour kg dry air kg m3
319
Nomenclature
Superscripts 0
denotes standard pressure of 1 bar
Subscripts 0
starting point, initial condition
cr
critical
c
constant
eq
equilibrium
f
formation
f
friction
g
generated
i
species
k
species
n
normal or standard condition
t
total
trs
320
transition
Nomenclature
321