Weber Thermodynamics

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INTERNATIONAL STUDIES in SCIENCE and ENGINEERING

Roman Weber

COMBUSTION FUNDAMENTALS

with

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Prof. Dr.–Ing. Roman Weber Technische Universität Clausthal

Institut für Energieverfahrenstechnik und Brennstofftechnik (IEVB) Agricolastrasse 4, 38 678 Clausthal-Zellerfeld, Germany

[email protected] Weber, Roman:

Combustion Fundamentals with Elements of Chemical Thermodynamics Clausthal-Zellerfeld: Papierflieger 2008

ISBN 978–3–89720–921–3

Bibliografische Information der Deutschen Bibliothek

Die Deutsche Bibliothek verzeichnet diese Publikation in der Deutschen National-bibliografie; 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 (Ger-many)

Dipl.-Math. Marc Muster, Clausthal University of Technology (Germany) Prof. Dr.-Ing. Andrzej Nowak, Silesian University of Technology (Poland) Prof. Dr.-Ing. Reinhard Scholz, Clausthal University of Technology (Germany) First Edition 2008

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

No part of this publication may be reproduced or transmitted in any form or by 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.

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To

Professor Stanisław Jerzy Gdula who introduced me to the wonderful

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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 Thermo-dynamics” 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,

Papierflieger-Verlag, Clausthal-Zellerfeld 2008, ISBN 3–89720–702–8.

2. Jeschar, R.; Kostowski, E.; Alt, R.: Wärmestrahlung in Industrieöfen, Papierflieger-Verlag, Clausthal-Zellerfeld 2004, ISBN 3–89720–686–2 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, Papierflieger-Verlag, Clausthal-Zellerfeld 2008, ISBN 3–89720– 798–2.

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

5. Dierkes, P. v.; Bruch, G.: Abfall und Chemie, Teil II, Papierflieger-Verlag, Clausthal-Zellerfeld 2007, ISBN 3–89720–880–6.

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To the 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 Technol-ogy. The notes have been prepared not only for Clausthal but also for the Royal Institute of Technology (Stockholm, Sweden), University of Science and Technol-ogy (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 chap-ter. 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 de-veloped 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.tu-clausthal.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/ASIA-LINK/016 (103 187)) of the Asia-Link 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 e-mail or write to me. Roman Weber IEVB – TU Clausthal D-38678 Clausthal-Zellerfeld Agricolastrasse 4 Germany [email protected]

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List of Figures

1.1 Primary Energy Sources in Germany per type of fuel . . . 2

1.2 Wall Fired Boiler . . . 3

1.3 Modern Reheating Furnace – NKK, Fukuyama Works, Japan . . . 3

1.4 Left – pulverised coal flame; Right – MILD combustion of natural gas . . . 4

1.5 The subsets of hydrocarbons . . . 5

1.6 Oxygen concentration in dry combustion products as a function of excess air ratio . . . 23

1.7 Carbon dioxide concentration as a function of excess air ratio . . . 23

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 val-ues listed in steam tables [7]. . . 26

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

1.10 Effect of sulphur and excess air on acid due point for a crude oil (adapted from [8]). . . 30

1.11 Ash and combustibles in a furnace . . . 31

1.12 Burnout of combustibles . . . 32

1.13 Composition of dry combustion products for sub-stoichiometric combustion of methane . . . 34

2.1 Control volume for mass and energy balance . . . 36

2.2 Mass balance for the boiler . . . 39

2.3 Illustration of open, closed and isolated systems . . . 44

2.4 Sankey’s diagram for energy balance for an open or closed system . 45 2.5 System in translational and rotational motion . . . 46

2.6 Definition of the sign of work . . . 51

2.7 Work done to the system . . . 52

2.8 Specific heats at constant pressure for various molecules . . . 57

2.9 Physical enthalpies of various gases as a function of temperature . 58 2.10 Illustration of enthalpy of formation of a compound . . . 62

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List of Figures

2.11 Illustration ofLCVas an amount of heat extracted from a

combus-tion chamber . . . 66

2.12 Illustration of an energy balance . . . 73

2.13 Example of an energy balance . . . 75

2.14 Energy balance using enthalpy of formation . . . 77

2.15 Energy balance using LCV . . . 78

2.16 Sankey diagram demonstrating the concept of the available heat . . 82

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. . . 85

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. JANAFpolynomials for cp have been used. . . 86

3.1 Examples of irreversible processes in closed and open systems . . . 90

3.2 Irreversible process of mixing . . . 91

3.3 Typical thermodynamic processes shown using work-diagram . . . 94

3.4 Reversible and irreversible gas expansion process . . . 95

3.5 Integration paths from i → f . . . 97

3.6 Determination of absolute specific entropy of a pure substance . . . 106

3.7 An open system interacting with other bodies by exchanging mass, heat and work . . . 108

3.8 Heat source providing heat to the system . . . 109

3.9 Mass source providing mass to the system . . . 110

3.10 A single component system consisting of two phases maintained at constant temperature and pressure . . . 116

3.11 Pressure-temperature plot showing the phase-equilibrium curve that defines stability regions for phases 1 and 2 . . . 118

3.12 The experimentally determined phase diagram for water . . . 120

3.13 The variation of the Gibbs enthalpy with temperature for ice, water and water vapour (at 1 bar . . . 125

3.14 The variation of the Gibbs enthalpy with temperature for ice, water and water vapour at a pressure of 611 Pa . . . 125

3.15 Mixing of two ideal gases . . . 139

3.16 Mixture of two ideal gases A and B . . . 140

3.17 Pure species-A at temperature T and pressure p; Species-A in a mixture with species B at Temperature T and pressure p . . . 142

3.18 Ideal solution of two liquids in equilibrium with its vapour . . . 143

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List of Figures

4.1 Minimisation of Gibbs enthalpy as the reaction advances . . . 155 4.2 Thermodynamic equilibrium constant K as a function of

tempera-ture for reactions listed in Table 4.2 . . . 169 4.3 Equilibrium partial pressure of carbon dioxide in calcination

reac-tion CaCO₃→ CaO + CO₂ . . . 175 4.4 Accurate and estimated values of the thermodynamic equilibrium

constant for Boudouard reaction . . . 180 4.5 Variation of the partial pressures of CO₂and CO with temperature

for total pressure of 1 bar and 10 bar for Boudouard reaction . . . 182 4.6 Equilibrium composition of Boudouard reaction at several total

pressures . . . 183 4.7 Illustration of reaction (4.128) . . . 203 4.8 Gibbs enthalpy as a function of extent of the reaction . . . 207 4.9 Gibbs enthalpy as a function of the amount of carbon dioxide . . . 209 4.10 Gibbs enthalpy of the considered system as a function of mCH₄ and

mH₂O . . . 215

5.1 Concentration change with time for a first-order reaction (t0 = 0) . 223

5.2 Concentration change with time for a second-order reaction . . . . 225 5.3 The approach of concentrations to their equilibrium values for the

reversible reaction A −−⇀↽−− B that is first order in each direction . . 229 5.4 Activation energy of a chemical reaction . . . 239 5.5 Arrhenius plot for elementary reactions of halogens with molecular

hydrogen . . . 240 5.6 Fall-off curves for the reaction C₂H₆ −−→ CH₃+ CH₃ . . . 241 6.1 Top – rate constant of CO + OH −−→ CO₂+ H;

Bottom – rate constant of H + CO₂ −−→ CO + OH [19]. . . 246 6.2 The time behaviour of the chain carriers . . . 251 6.3 A simplified scheme of methane oxidation [19] . . . 253 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) . . . 257

6.5 Temporal behaviour of the species concentrations in reactions A1−→

A2 −→ A3 for k12= 10 s−1 and k23= 1 s−1 (low reactive

interme-diate A2). . . 257

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

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List of Figures

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

6.8 Euler’s method . . . 265 6.9 Numerical solutions using the explicit scheme with different time

steps . . . 267 6.10 Numerical solutions using the implicit scheme with different time

steps . . . 269 6.11 Temporal behaviour of the species concentrations for reactions A1−→

A2 −→ A3 assuming quasi steady state for A2-species . . . 271

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

A2 −→ A3, k12= 1 s−1, k23= 100 s−1, Euler´s Explicit Method . . 276

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List of Tables

1.1 Names of aliphatic hydrocarbons . . . 5

1.2 Composition of combustion products . . . 12

2.1 The incoming and out-coming streams in kmol/h . . . 41

2.2 Composition of the combustion products . . . 41

2.3 The incoming and outcoming flow rates in kg/h . . . 43

2.4 The coefficients for polynomials for T in the range 300–1000 K . . . 57

2.5 The coefficients for polynomials for T in the range 1000–5000 K . . 58

2.6 Standard enthalpies of formation and standard entropies of some compounds (JANAFThermodynamic Tables) . . . 64

2.7 LCVof some selected gaseous fuels . . . 68

2.8 Adiabatic flame temperature Tad for stoichiometric combustion in air. Combustion products contain CO₂ and H₂O only. . . 80

2.9 Calculated adiabatic temperature for stoichiometric combustion of pure CH₄ with air. . . 85

3.1 Transition temperatures, standard enthalpies and standard entropies of selected substances . . . 104

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

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

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

4.4 Standard Gibbs enthalpies at T = 1300 K . . . 204

4.5 Standard Gibbs enthalpies at T = 1000 K . . . 213

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

6.1 Rate coefficients of the chain initiation reactions . . . 247

6.2 Rate coefficients of the chain branching reactions. The coefficients are presented in the form k = A Tb _{exp −}Ea R T . . . 248

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List of Tables

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

chain branching reactions. . . 249

6.4 Simplified mechanism of ignition of the hydrogen-oxygen system . . 249

6.5 Comparison of the rate coefficients k of the chain propagation re-actions for methane oxidation. . . 252

6.6 Rate coefficients of the Zeldovich mechanism reactions . . . 261

6.7 Numerical solution using the explicit scheme with h = 0.3 s. . . 266

6.8 Numerical solution using the implicit scheme with h = 0.3 s. . . 269

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1 Stoichiometry

Contents

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 equiva-lence ratio

1.3.3 Minimum air requirement for a mixture of gaseous fuels 1.3.4 Composition of combustion products

1.4 Combustion stoichiometry for liquid and solid fuels 1.4.1 Minimum oxygen and air requirements and excess air ratio 1.4.2 Combustion products

1.5 Humid Combustion Air

1.5.1 Absolute and relative humidity

1.5.2 Dew Point Temperature of Combustion Products 1.6 Combustibles burnout for solid fuels

1.7 Sub-stoichiometric combustion to carbon dioxide and water vapour

1.8 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

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1 Stoichiometry

for many decades to come, as shown in Fig. 1.1. For most of high temperature processes like power generation, glass manufacturing, cement-making and steel-making, combustion of fossil fuels provides the process energy. Any material that can be burned to release thermal energy is called a fuel. Most fuels consist pri-marily of hydrogen and carbon and they are called hydrocarbons. Fig. 1.2 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 steel-work in Japan [2] is shown in Fig. 1.3. The furnace is fired with a mixture of natural gas and steel-work gases. Petrol and diesel oil are burned in engines used for transportation. Propeller-driven airplanes are built with propellers powered by engines identical to automobile engines. Civilian and military aircrafts are pow-ered on energy generated through combustion processes occurring in gas turbine combustors. In short, combustion occurs in our every day life.

Fig. 1.1: Primary Energy Sources in Germany per type of fuel in EJ = 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 on-going challenge for combustion engineers.

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1.1 Introduction

Fig. 1.2: Wall Fired Boiler

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1 Stoichiometry

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

1.2 Definitions

1.2.1 Chemical Reactions, Atoms and Molecules in Combustion A chemical reaction is an exchange and/or rearrangement of atoms between col-liding molecules, for example:

H₂+ 1₂O₂ −−→ H₂O

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 H₂O are not. Reactant molecules (H₂ and O₂) are rearranged to become product (H₂O) 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 unbranched-chain alkane series are:

CH₄ methane C₆H₁₄ hexane C₂H₆ ethane C₇H₁₆ heptane C₃H₈ propane C₈H₁₈ octane C₄H₁₀ butane C₉H₂₀ nonane C₅H₁₂ pentane C₁₀H₂₂ decane

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1.2 Definitions

Fig. 1.5: The subsets of hydrocarbons

Table 1.1: Names of aliphatic hydrocarbons

No. of C atoms

alkane alkene alkyne alkyl group

1 CH4 methane CH3 methyl

2 C2H6 ethane C2H4 ethene C2H2 ethyne C2H5 ethyl

3 C3H8 propane C3H6 propene C3H4 propyne C3H7 propyl

4 C4H10 butane C4H8 butene C4H6 butyne C4H9 butyl

5 C5H12 pentane C5H10 pentene C5H8 pentyne C5H11 pentyl

n CnH2n+2 CnH2n CnH_2n−2 CnH2n+1 Other molecules relevant in combustion are:

Haloalkanes R−X example: CH₃Cl (chloromethane) Alcohols R−OH example: C₂H₅OH (ethanol)

Aldehydes R−CHO example: CH₃CHO (ethanal) Amines R−NH₂ example: CH₃NH₂ (methylamin) Ketones R−CO−R example: CH₃COCH₃ (acetone) Carbocyclic Acids R−COOH example: CH₃COOH (ethanoic acid)

1.2.2 Amount of Substances, Mole and Mass Fractions

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

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

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1 Stoichiometry

For a mixture of species: n =X

i

ni n = total number of moles (1.1)

where n stands for total number of moles, ni is the number of moles of species i,

and the summation extends over all the species. Mole fraction – xi – (mole number) of species i is:

xi =

ni

n (1.2)

The molar mass (molecular weight) in g/mol or kg/kmol is the mass of one mole of the species (for example: MC = 12 g/mol, MCO₂ = 44 g/mol). The mean molar

mass (molecular weight) of a mixture is: Mmean=

X

i

xiMi (1.3)

Frequently mole fractions xi are converted into mass fractions (wi) and the

fol-lowing relationships hold: wi=

number of kg of species "i" total number of kg in the system =

niMi P k nkMk = PxiMi k xkMk (1.4) ni n = xi =

number of moles of species "i" in 1 kg of mixture total number of moles in 1 kg of mixture = wi Mi 1 Mmean = wi Mi Mmean= wi Mi P k wk Mk (1.5)

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 ρ = m

V (in kg/m3)

molar density (concentration) c = n

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1.2 Definitions Thus, ρ c = m n = Mmean (1.6)

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

concen-trations in mol/cm3_.

1.2.4 Equation of State for Gases and Gas Mixtures

For gases and gas mixtures an equation of state relates temperature, pressure 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 Clausius-Clapeyron equation, is perhaps one of the simplest ones and it reads

p V = n R T (1.8) or c = p R T (1.9) and ρ = p Mmean R T = p R T P i wi Mi (1.10) where p is the pressure (in Pa = N/m2_{), V the volume (in m}3_{), 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 ( p

pcr ≪ 1), gases behave as an ideal gas regardless of temperature;

(b) at high temperatures ( T

Tcr > 2) gases behave also as an ideal gas;

(c) deviation from an ideal gas behavior increases upon approaching the critical point.

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1 Stoichiometry

From Eq. (1.8) follows that one kmol of any (ideal) gas under constant tempera-ture and pressure occupies the same volume. One can easily verify that under so called normal conditions: a pressure of p0 = 1 bar (105N/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 m}3

n) is a unit of mass

(substance) – not volume. Thus,

1 kmol of gas = 24.79 m3_n 1 mol of gas = 24.79 dm3_n

Throughput this lecture course the standard (normal) conditions of p0 = 1 bar

(105_N/m2_{) and T}

0 = 25◦C (298.15 K) are used1.

1.3 Combustion Stoichiometry for Gaseous

Fuels

1.3.1 Stoichiometric combustion

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

Examples are:

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

CH₄+ 3 O₂ −−→ 2 H₂O + CO₂+ O₂ lean (excess of oxygen) CH₄+ O₂ −−→ H₂O + ₂1CO₂+1₂CH₄ rich (excess of fuel)

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

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

one may easily calculate the mole fraction of fuel in a stoichiometric mixture (with 1_{In older books p}

0 = 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

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1.3 Combustion Stoichiometry for Gaseous Fuels

oxygen) as follows: xf uel,stoich in oxygen=

number of moles of fuel

total number of moles (fuel + oxygen) = 1 1 + ν (1.11) For example: H₂+ 1₂O₂ −−→ H₂O xH₂,stoich in oxygen= 1 1.5 = 2 3 CH₄+ 2 O₂ −−→ CO₂+ 2 H₂O xCH₄,stoich in oxygen= 1 3 CO + 1₂O₂ −−→ CO₂ xCO,stoich in oxygen= 1 1.5 = 2 3

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

xf uel,stoich in air= 1 1 + ν 0.21 = 1 1 + 4.762 ν (1.12) An example:

Combustion of propane in oxygen:

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

1 6 Combustion of propane in air:

C₃H₈+ 5 ( O₂+ 3.762 N₂) −−→ 3 CO₂+ 4 H₂O + 5 · 3.762 N₂ and xC₃H₈,stoich in air=

1

1 + 5 · 4.762 = 0.0403 Note: The higher the hydrocarbon the lower is the fuel mole fraction at

stoichio-metric conditions.

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

The excess air ratio is defined as: λ = (xair/xf uel)

(xair/xf uel)stoich

= (wair/wf uel) (wair/wf uel)stoich

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1 Stoichiometry

whilst the fuel equivalence ratio Φ is defined as: Φ = 1/λ. The above equation can be rewritten to allow the calculation of the mole fuel fraction in a mixture of known Φ or λ: xf uel = 1 1 + 4.762 · ν · λ; xair = 1 − xf uel; (1.14) xO₂ = 0.21 xair= xair 4.762; xN2 = 3.762 xO2 (1.15) The combustion process can be divided into:

Rich combustion Φ > 1 λ < 1 Stoichiometric combustion Φ = 1 λ = 1 Lean combustion Φ < 1 λ > 1

Example 1.1 Five hundred m3

n/h of ethane are combusted with five thousand five hundred

m3

n/h of dry air. Calculate the excess air ratio and the fuel equivalence ratio.

Assumptions: the fuel (ethane) is combusted to carbon dioxide and water. The combustion reaction is then

C₂H₆+7₂O₂ −−→ 2 CO₂+ 3 H₂O Thus, (xair/xf uel)stoich= 3.5 · 4.7619/1 = 16.6667_{kmol fuel}kmol air

500 m3

n/h of C2H6 is equivalent to 500/24.79 = 20.169 kmol of C2H6/h

5500 m3_n/h of dry air is equivalent to 5500/24.79 = 221.864 kmol of air/h Thus, xair

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

(a) Since the combustion is fuel rich, (1 − 0.66) · 500 = 170 m3

n/h of ethane will

not be combusted.

(b) In the above example the conversion from m3

n/h into kmol/h is superficial.

End of Example 1.1

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

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1.3.3 Minimum air requirement for a mixture of gaseous fuels

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

xH₂+ xCO+ xCH₄+ xCnHm+ xO2 + xCO2+ xN2 = 1 (1.16) where xi’s stand for mole fractions.

The following combustion reactions are applicable: H₂+1₂O₂ −−→ H₂O CO +1₂O₂ −−→ CO₂

CH₄+ 2 O₂ −−→ CO₂+ 2 H₂O C₂H₆+7₂O₂ −−→ 2 CO₂+ 3 H₂O C_nH_m+ (n + m₄)O₂ −−→ n CO₂+ m₂H₂O Thus, minimum oxygen requirement for a mixture of gases is:

lO₂,min = 1 2xH2+ 1 2xCO+ 2 xCH4 + X n + m 4 xCnHm− xO2 (1.17) (lO₂,min in kmol O2/kmol dry gaseous fuel)

whilst the minimum dry air requirement is: ldry air,min =

lO₂,min

0.21 = 4.7619 · lO2,min (1.18) Thus, the excess air ratio is

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

ldry air

ldry air,min

(1.19)

1.3.4 Composition of combustion products

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

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1 Stoichiometry

Table 1.2: Composition of combustion products

Fuel component

Component of combustion products CO₂ H₂O N₂ O₂ H₂ − 1 − − CO 1 − − − CH₄ 1 2 − − C₂H₆ 2 3 − − C_nH_m n m/2 − − O₂ − − − − N₂ − − 1 −

amount (at stoichiometric conditions) of wet combustion products is: Vwet,min= 1 xH₂ + 1 xCO+ 3 xCH₄ + 5 xC₂H₆+ X n + m 2 xCnHm+ 1 xN2+ 0.79 lair,min

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

while for lean combustion:

Vwet = Vwet,min+ (λ − 1) · lair,min

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

Vdry,min = 1 xCO+ 1 xCH₄ + 2 xC₂H₆ +

X

(n · xCnHm) + 1 xN2+ 0.79 lair,min (Vdry,min is in kmol dry products/kmol fuel)

while for lean combustion:

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Composition of combustion products (wet): xCO₂,wet =

1 xCO+ 1 xCH₄+ 2 xC₂H₆ + n xCnHm Vwet

kmol CO₂ kmol wet products xH₂O,wet =

1 xH₂+ 2 xCH₄+ 3 xC₂H₆+1₂m xCnHm Vwet

kmol H₂O kmol wet products xN₂,wet =

1 xN₂+ 0.79 λ lair,min

Vwet

kmol N₂ kmol wet products xO₂,wet =

0.21 (λ − 1) lair,min

Vwet

kmol O₂ kmol wet products

Composition of combustion products (dry):

The amount of dry combustion products per 1 kmol of fuel can be calculated as: Vdry= Vwet 1 − xH₂O,wet = Vwet− 1 xH₂+ 2 xCH₄+ 3 xC₂H₆+ 1₂m xCnHm

in kmol dry products/kmol of fuel Thus, composition of dry combustion products can be calculated as follows:

xCO₂,dry =

1 xCO+ 1 xCH₄ + 2 xC₂H₆ + n xCnHm Vdry

kmol CO₂ kmol dry products xN₂,dry =

1 xN₂ + 0.79 λ lair,min

Vdry

kmol N₂ kmol dry products xO₂,dry =

0.21 (λ − 1) lair,min

Vdry

kmol O₂ kmol dry products Example 1.2

Calculate composition of wet and dry products of combustion of Dutch (Gronin-gen) natural gas at 20 % excess air [6].

Composition of Dutch natural gas:

CH₄ 81 (vol%) C₂H₆ 3 (vol%) C₃H₈ 1 (vol%) N₂ 15 (vol%) Assumptions: the fuel is combusted to carbon dioxide and water.

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1 Stoichiometry

Thus, the relevant combustion reactions are:

CH₄+ 2 O₂ −−→ CO₂+ 2 H₂O || 0.81 C₂H₆+7₂O₂ −−→ 2 CO₂+ 3 H₂O || 0.03 C₃H₈+ 5 O₂ −−→ 3 CO₂+ 4 H₂O || 0.01 Minimum oxygen requirement:

lO₂,min = 0.81 · 2 + 0.03 · 3.5 + 0.01 · 5 = 1.775 kmol O₂ kmol fuel _m₃ n of O2 m3 n of fuel

Minimum air requirement: lair,min =

1.775

0.21 = 8.452

kmol dry air kmol fuel The amount of air for λ = 1.2;

lair= 1.2 · 8.452 = 10.14

kmol dry air kmol fuel Combustion products: Fuel component Component of combustion products molar fraction in dry fuel CO₂ H₂O N₂ O₂ xi CH₄ 1 2 − − 0.81 C₂H₆ 2 3 − − 0.03 C₃H₈ 3 4 − − 0.01 N₂ − − 1 − 0.15

Minimum amount of wet combustion products (for λ = 1): Vwet,min= 3 · 0.81 + 5 · 0.03 + 7 · 0.01 + 1 · 0.15 + 0.79 · 8.452 =

9.477kmol wet products kmol of fuel The amount of combustion products for lean combustion (for λ ≥ 1)

Vwet = 9.477 + (λ − 1) · 8.452 and for λ = 1.2 Vwet= 11.17

kmol wet products kmol of fuel

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Composition of wet combustion products: xCO₂,wet=

1 · 0.81 + 2 · 0.03 + 3 · 0.01 9.477 + (λ − 1) · 8.542

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

xH₂O,wet =

2 · 0.81 + 3 · 0.03 + 4 · 0.01 9.477 + (λ − 1) · 8.542

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

xN₂,wet =

1 · 0.15 + λ · 8.452 · 0.79 9.477 + (λ − 1) · 8.542

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

xO₂,wet =

(λ − 1) · 8.452 · 0.21 9.477 + (λ − 1) · 8.542

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

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

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

7.7271 + (λ − 1) · 8.452

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

Composition of dry combustion products: xCO₂,dry =

1 · 0.81 + 2 · 0.03 + 3 · 0.01 7.727 + (λ − 1) · 8.542

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

xN₂,dry =

1 · 0.15 + λ · 8.452 · 0.79 7.727 + (λ − 1) · 8.542

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1 Stoichiometry

xO₂,dry =

(λ − 1) · 8.452 · 0.21

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

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

End of Example 1.2

1.4 Combustion stoichiometry for liquid and

solid fuels

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

The following applies:

c + h + o + n + s + moisture + ash = 1 (1.20) where small letters stand for mass fractions. In the above relationship the mass fractions of the elements (c, h, s, o, n) are “as fired” (sometimes called also “as received”), it means water and ash are present in the fuel. Fuel composition is often expressed on a dry basis (without water, but with ash) and then the following conversion is applicable:

cdry=

cas received

1 − moisture (1.21)

When fuel composition is expressed on a dry ash-free basis the following conversion is applicable:

cdry,ashf ree =

cas received

1 − moisture − ash (1.22) Obviously, similar conversions can be applied to hydrogen, sulfur, oxygen, and nitrogen mass fractions.

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1.4 Combustion stoichiometry for liquid and solid fuels

1.4.1 Minimum oxygen and air requirements and excess air ratio The following reactions determine the minimum oxygen and air requirements:

C + O₂−−→ CO₂ S + O₂−−→ SO₂ H₂+1₂O₂−−→ H₂O

N + O₂−−→ NO₂2

Thus, the minimum oxygen requirements (in kmol O₂ per kg of fuel “as received”) is lO₂,min= 1 · c 12+ 1 2· h 2 + 1 · s 32+ 1 · n 28 − o 32 kmol O₂ kg of fuel as received (1.23) or lO₂,min= 24.79 · 1 · c 12+ 1 2· h 2 + 1 · s 32 + 1 · n 28− o 32 m3 n O2 kg of fuel as received and the minimum air requirement is:

lair,min=

lO₂,min

0.21 = 4.762 · lO2,min (1.24) The excess air ratio is then calculable as:

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

ldry air

ldry air,min

(1.25)

1.4.2 Combustion products

Products of complete combustion of fuels contain CO₂, H₂O, SO₂, N₂ and excess O₂. If the temperature of the combustion products is above the dew point the water vapor does not condense. Combustion products can be considered as dry or wet:

Vwet= Vdry+ VH₂O (1.26)

2_{It has been assumed that sulphur and nitrogen present in the fuel are oxidisedto SO}

2andNO2,

respectively. In reality, other sulphur and nitrogen oxides also occur in combustion products. However, since the nitrogen and sulphur contents in the fuel are low, such an assumption is justifiable for calculating the minimum oxygen and air requirements.

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1 Stoichiometry

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

fuel while Vdryis the amount of dry products and the VH₂Ostands for the amount

of water in the products. The minimum amount (at stoichiometric combustion, λ = 1) of wet combustion products is:

Vwet,min= 1 · c 12+ 1 · h 2 + moisture 18 + 1 · s 32 + n 28 + 1 · 0.79 · lair,min

in kmol wet products/kg of fuel (1.27) and at lean combustion (λ > 1):

Vwet = 1 · c 12 + 1 · h 2 + moisture 18 + 1 · s 32 + n 28 +

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

Vwet = c 12+ h 2 + moisture 18 + s 32+ n 28+ (λ − 1) · lO2,min+ λ · 0.79 · lair,min CO₂ from c H₂O

from h moisturein fuel

SO₂ from s

N₂

from n excessoxygen N₂

in air The above expressions for calculating Vwetcan be rearranged into a more elegant form

as follows: Vwet= 1 · c 12 + 1 · h 2 + moisture 18 + 1 · s 32 + n 28 + 0.79 · lair,min+ (λ − 1) · lO2,min+ 0.79 · (λ − 1) · lair,min = c 12 + h 2 + moisture 18 + s 32 + n 28+ 0.79 · lair,min+ 0.21 · (λ − 1) · lair,min+ 0.79 · (λ − 1) · lair,min = c 12 + h 2 + moisture 18 + s 32 + n 28+ 0.79 · lair,min+ (λ − 1) · lair,min = Vwet,min+ (λ − 1) · lair,min Thus,

Vwet = Vwet,min+ (λ − 1) · lair,min (1.28a)

and Vwet,min= c 12+ h 2 + moisture 18 + s 32 + n 28 + 0.79 · lair,min (1.28b) where Vwet,min is the minimum amount of products (for λ = 1).

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terms h

2 and moisture18 from the equations above. Thus,

Vdry= c 12 + s 32 + n

28 + 0.79 · lair,min+ (λ − 1) · lO2,min+ (λ − 1) · 0.79 · lair,min kmol dry products/kg of fuel (1.29a) or

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

and Vdry,min = c 12 + s 32 + n 28 + 0.79 · lair,min

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 H₂O, CO₂, SO₂, N₂ and excess O₂. Thus:

xH₂O= h 2 + moisture 18 Vwet

kmol H₂O/kmol wet products xCO₂,wet =

c 12

Vwet

kmol CO₂/kmol wet products xSO₂,wet=

s 32

Vwet

kmol SO₂/kmol wet products xO₂,wet=

(λ − 1) · lO₂,min

Vwet

kmol O₂/kmol wet products xN₂,wet =

n

28+ 0.79 · λ · lair,min

Vwet

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 = c 12 Vdry xSO₂,dry = s 32 Vdry xO₂,dry = (λ − 1) · lO₂,min Vdry xN₂,dry = n 28+ 0.79 · λ · lair,min Vdry

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1 Stoichiometry

Example 1.3

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

Coal Fettnuss mvb3 _{(origin – Germany) – coal analysis obtained from a chemical}

laboratory

Basis H2O Ash Ultimate Analysis LCV

%-wt %-wt %-wt MJ/kg

c h n o s

Dry ash free 89.39 4.68 1.53 3.66 0.74

Dry 4.4 32.87

As fired 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 H2O Ash Ultimate Analysis LCV

%-wt %-wt %-wt MJ/kg

c h n o s

Dry ash free 89.39 4.68 1.53 3.66 0.74 34.38

Dry 4.4 85.4568 4.47 1.4627 3.499 0.7074 32.87

As fired 3.5 4.246 82.4658 4.3136 1.4115 3.3765 0.6826 31.72

The minimum oxygen requirement (λ = 1) is: lO₂,min = 1 · 0.824 658 12 + 1 2· 0.043 136 2 + 1 · 0.006 826 32 − 0.033 765 32 = 0.0787 kmol O2

kg fuel “as fired” that corresponds to 0.0787 · 32 = 2.5184 kg O2

kg of fuel “as fired”.

3

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The minimum air requirement (λ = 1) is: lair,min=

0.0787

0.21 = 0.3743

kmol dry air kg of fuel “as fired” 0.3743 · 22.79 = 9.279m 3 n dry air kg of fuel , or 2.5184 0.23 = 10.9496 kg of dry air kg of fuel

The amount of combustion products is: Vwet= 0.824 658 12 + 0.043 136 2 + 0.035 18 + 0.006 826 32 + 0.014 115 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 12 + 0.006 826 32 + 0.014 115 28 + λ · 0.79 · 0.3743 + (λ − 1) · 0.0787 = 0.0695 + 0.2957 · λ + (λ − 1) · 0.0787

in kmol dry products/kg of fuel “as fired” Composition of wet combustion products:

xH₂O = 0.043 136 2 +0.03518 Vwet = 0.0235 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) xCO₂,wet= 0.824 658 12 Vwet = 0.0687 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) xSO₂,wet= 0.006 83 32 Vwet = 0.0002 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) xO₂,wet = (λ − 1) · lO₂,min Vwet = (λ − 1) · 0.0787 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) xN₂,wet = 0.79 · λ · lair,min Vwet = 0.79 · λ · 0.3743 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) All the above mole (volume) fractions are in kmol/kg of wet products.

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1 Stoichiometry

Composition of dry combustion products: xCO₂,dry= 0.824 658 12 Vdry = 0.0687 0.0695 + 0.2957 · λ + 0.0787 · (λ − 1) xSO₂,dry= 0.006 83 32 Vdry = 0.0002 0.0695 + 0.2957 · λ + 0.0787 · (λ − 1) xO₂,dry = (λ − 1) · lO₂,min Vdry = (λ − 1) · 0.0787 0.0695 + 0.2957 · λ + 0.0787 · (λ − 1) xN₂,dry = 0.79 · λ · lair,min Vdry = 0.79 · λ · 0.3743 0.0695 + 0.2957 · λ + 0.0787 · (λ − 1) All the above mole (volume) fractions are in kmol/kg of dry products. Composition of combustion products for 10 % excess air ratio (λ = 1.1) Using the above relationships one may easily obtain:

Composition of wet products for λ = 1.1 is:

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

xO₂,wet = 0.0185 xN₂,wet = 0.7635

Composition of dry products for λ = 1.1 is:

xCO₂,dry = 0.1706 xSO₂,dry = 0.0005

xO₂,dry = 0.0195 xN₂,dry = 0.8079

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

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

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1 Stoichiometry

Comments:

(a) For the fuels considered in this example (CH₄, C₂H₆, Groningen natural gas, coal Fettnuss), the relationship between oxygen content in combustion prod-ucts and excess air ratio (see Fig. 1.6) is almost linear for λ < 1.5. 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 con-tains 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. Atmo-spheric 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.

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

assumption is justifiable and the maximum departure from reality is typically in the range 0.2–6 %. Then the combustion air is treated as an ideal-gas 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 dry-air/water-vapour 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

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

pv

p (1.33)

The amount of water vapour in the combustion air can be specified in 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 mdry air

= Mv· nv Mdry air· ndry air

= 18.016 28.97 · pv p − pv = 0.622 · 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

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1 Stoichiometry psat= 611 · exp −5304.3 · 1 T − 1 273.16 in Pa (1.35) Water Vapour p in kP a sat Temperature in K 10 5 0 280 290 300 310 320

Fig. 1.8: Saturation pressure of water vapour as a function of temperature. The plot 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

γ = pv pv,sat

(1.36) where pv,sat stands for the saturation water vapour pressure at the specified

tem-perature. The relative humidity (γ) ranges from 0 for dry air to 1 for saturated air. The amount of moisture that combustion air can hold depends on its tem-perature. 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:

ϕ = 0.662 · γ · psat p − γ · psat

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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) psat= 611 · exp −5304.3 · 1 297.16 − 1 273.16 = 2298.14 Pa (1.41) The absolute humidity is then

ϕ = 0.662 · γ · psat p − γ · psat = 0.662 · 0.75 · 2298.14 105_{− 0.75 · 2298.14} = 0.0116 kg water vapour kg dry air (1.42)

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1 Stoichiometry

so the minimum amount of wet combustion air is

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

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

Vwet= 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) +

λ · ϕ · ldry air,min

18 (1.44)

and

Vwet = 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06

kmol wet products kg of fuel “as fired”

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

xH₂O= 0.043 136 2 +0.03518 + 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 = 0.824 658 12 Vwet = 0.0687 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06 xSO₂,wet = 0.006 882 6 2 Vwet = 0.0002 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06 xO₂,wet = (λ − 1) · lO₂,min Vwet = (λ − 1) · 0.0787 0.0929 + 0.2957 · λ + 0.0787 · (λ − 1) + λ · 0.007 06 xN₂,wet = 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.

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

xH₂O = 0.0721 xCO₂,wet = 0.1584 xSO₂,wet = 0.000 46

xO₂,wet = 0.018 14 xN₂,wet = 0.7498

Comments:

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

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

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

End of Example 1.4

1.5.2 Dew Point Temperature of Combustion Products

Typically combustion products contain water vapour. When for example a nat-ural gas is combusted the water vapour content in the combustion products 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 dew-point temperature (Tdp) is defined as the temperature at which condensation begins when the

com-bustion products (or generally moist air) is cooled at a constant pressure. In 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 due-point 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 dew-point 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.

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1 Stoichiometry T T₁ T_dp s p=c ons t. v 1 2

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

The excess air curves are not equally spaced since the extra oxygen tends to produce more SO₃ which exerts a catalytic effect in raising the dew point. For example the above estimated dew-point temperature of 39.3◦_{C would be almost}

doubled if 2 ppm of SO₂was present. This is the reason that operators of coal-fired power station boilers maintain the flue gases at temperatures above 180–200◦_C.

1.0 2.0 3.0 4.0 5.0

Weight % sulfur in fuel oil

Aciddewpoint,°C 135 140 130 120 115 145 20%excess air 15% 10% 5%

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

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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 “oil-coke” or “coal char”, although the incompletely combusted gases (CO and hydrocarbons) and soot particles can also make a contribution. In the considerations that follow, the mass fraction of total combustibles is simply calculated as 1 minus the 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: Composition of the unburned solids: C0 (Total Combustibles) + ash0= 1 C1+ ash1 = 1

m0 – mass flow rate of fuel m1 – mass flow rate of unburned solids

Mass balance of combustibles can be written as: m0· C0 = m1· C1+ b · m0· C0

where m0C0(in kg/s) stands for the amount of combustibles entering the furnace,

m1C1(in kg/s) represents the amount of combustibles in the unburned solids

leav-ing the furnace while bm0C0 (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: b = 1 −C1· ash0 C0· ash1 = 1 − (1 − ash1) · ash0 (1 − ash0) · ash1 = 1 − ash0 ash1 1 − ash0 (1.46)

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1 Stoichiometry

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

ash” (in the USA it is called “loss on ignition”). Fig. 1.12 shows the 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 Sub-stoichiometric combustion to carbon

dioxide and water vapour

Sub-stoichiometric combustion occurs when excess air ratio (λ) is smaller than 1. Obviously, in sub-stoichiometric 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.

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1.7 Sub-stoichiometric combustion to carbon dioxide and water vapour

Example 1.5

Produce a curve showing the composition of combustion products obtained in sub-stoichiometric combustion of methane in air as a function of excess air ratio. Assumptions: methane is combusted to carbon dioxide and water vapour. We begin with writing down the oxidation reaction for methane

CH₄+ 2 O₂ −−→ CO₂+ 2 H₂O The excess air ratio is then

λ = nair₂

0.21

so that nair = 9.5238 · λ

where nairis 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 H₂O while the remaining 1 − λ moles of CH₄ appear directly in combustion products as unburned fuel.

Therefore, in the combustion products appear

(1 − λ) moles of (unburned) CH₄ λ moles of CO₂

2λ moles of H₂O 0.79 · 9.5238 · λ moles of N₂ The composition of combustion products is then

xCH₄,wet = 1 − λ 1 + 9.5238 · λ xCH4,dry = 1 − λ 1 + 7.5238 · λ xCO₂,wet = λ 1 + 9.5238 · λ xCO2,dry = λ 1 + 7.5238 · λ xN₂,wet = 7.5238 · λ 1 + 9.5238 · λ xN2,dry = 7.5238 · λ 1 + 7.5238 · λ xH₂O,wet = 2 · λ 1 + 9.5238 · λ 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.