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Transcript
```LECTURE NOTES ON
FUNDAMENTALS OF COMBUSTION
Joseph M. Powers
Department of Aerospace and Mechanical Engineering
University of Notre Dame
Notre Dame, Indiana 46556-5637
USA
updated
30 March 2014, 2:12pm
2
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Contents
Preface
11
1 Introduction to kinetics
1.1 Isothermal, isochoric kinetics . . . . . . . . . . . . . . . . . . . . .
1.1.1 O − O2 dissociation . . . . . . . . . . . . . . . . . . . . . .
1.1.1.1 Pair of irreversible reactions . . . . . . . . . . . .
1.1.1.1.1 Mathematical model . . . . . . . . . . .
1.1.1.1.2 Example calculation . . . . . . . . . . .
1.1.1.1.2.1
Species concentration versus time
1.1.1.1.2.2
Pressure versus time . . . . . . .
1.1.1.1.2.3
Dynamical system form . . . . . .
1.1.1.1.3 Effect of temperature . . . . . . . . . . .
1.1.1.2 Single reversible reaction . . . . . . . . . . . . . .
1.1.1.2.1 Mathematical model . . . . . . . . . . .
1.1.1.2.1.1
Kinetics . . . . . . . . . . . . . .
1.1.1.2.1.2
Thermodynamics . . . . . . . . .
1.1.1.2.2 Example calculation . . . . . . . . . . .
1.1.2 Zel’dovich mechanism of NO production . . . . . . . . . .
1.1.2.1 Mathematical model . . . . . . . . . . . . . . . .
1.1.2.1.1 Standard model form . . . . . . . . . . .
1.1.2.1.2 Reduced form . . . . . . . . . . . . . . .
1.1.2.1.3 Example calculation . . . . . . . . . . .
1.1.2.2 Stiffness, time scales, and numerics . . . . . . . .
1.1.2.2.1 Effect of temperature . . . . . . . . . . .
1.1.2.2.2 Effect of initial pressure . . . . . . . . .
1.1.2.2.3 Stiffness and numerics . . . . . . . . . .
1.2 Adiabatic, isochoric kinetics . . . . . . . . . . . . . . . . . . . . .
1.2.1 Thermal explosion theory . . . . . . . . . . . . . . . . . .
1.2.1.1 One-step reversible kinetics . . . . . . . . . . . .
1.2.1.2 First law of thermodynamics . . . . . . . . . . .
1.2.1.3 Dimensionless form . . . . . . . . . . . . . . . . .
1.2.1.4 Example calculation . . . . . . . . . . . . . . . .
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4
CONTENTS
1.2.2
2 Gas
2.1
2.2
2.3
1.2.1.5 High activation energy asymptotics . . . . . . . . . . . . . .
Detailed H2 − O2 − N2 kinetics . . . . . . . . . . . . . . . . . . . . .
mixtures
Some general issues . . . . . . . . . . . . . . . . . . .
Ideal and non-ideal mixtures . . . . . . . . . . . . . .
Ideal mixtures of ideal gases . . . . . . . . . . . . . .
2.3.1 Dalton model . . . . . . . . . . . . . . . . . .
2.3.1.1 Binary mixtures . . . . . . . . . . .
2.3.1.2 Entropy of mixing . . . . . . . . . .
2.3.1.3 Mixtures of constant mass fraction .
2.3.2 Summary of properties for the Dalton mixture
2.3.2.1 Mass basis . . . . . . . . . . . . . . .
2.3.2.2 Molar basis . . . . . . . . . . . . . .
2.3.3 Amagat model . . . . . . . . . . . . . . . . .
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3 Mathematical foundations of thermodynamics
3.1 Exact differentials and state functions . . . . . . . . . . . . . .
3.2 Two independent variables . . . . . . . . . . . . . . . . . . . .
3.3 Legendre transformations . . . . . . . . . . . . . . . . . . . . .
3.4 Heat capacity . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Van der Waals gas . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Redlich-Kwong gas . . . . . . . . . . . . . . . . . . . . . . . .
3.7 Mixtures with variable composition . . . . . . . . . . . . . . .
3.8 Partial molar properties . . . . . . . . . . . . . . . . . . . . .
3.8.1 Homogeneous functions . . . . . . . . . . . . . . . . . .
3.8.2 Gibbs free energy . . . . . . . . . . . . . . . . . . . . .
3.8.3 Other properties . . . . . . . . . . . . . . . . . . . . .
3.8.4 Relation between mixture and partial molar properties
3.9 Frozen sound speed . . . . . . . . . . . . . . . . . . . . . . . .
3.10 Irreversibility in a closed multicomponent system . . . . . . .
3.11 Equilibrium in a two-component system . . . . . . . . . . . .
3.11.1 Phase equilibrium . . . . . . . . . . . . . . . . . . . . .
3.11.2 Chemical equilibrium: introduction . . . . . . . . . . .
3.11.2.1 Isothermal, isochoric system . . . . . . . . . .
3.11.2.2 Isothermal, isobaric system . . . . . . . . . .
3.11.3 Equilibrium condition . . . . . . . . . . . . . . . . . .
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4 Thermochemistry of a single reaction
4.1 Molecular mass . . . . . . . . . . . .
4.2 Stoichiometry . . . . . . . . . . . . .
4.2.1 General development . . . . .
4.2.2 Fuel-air mixtures . . . . . . .
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CC BY-NC-ND. 30 March 2014, J. M. Powers.
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5
CONTENTS
4.3
4.4
4.5
4.6
4.7
First law analysis of reacting systems . . . . . . . . . . . . . .
4.3.1 Enthalpy of formation . . . . . . . . . . . . . . . . . .
4.3.2 Enthalpy and internal energy of combustion . . . . . .
4.3.3 Adiabatic flame temperature in isochoric stoichiometric
4.3.3.1 Undiluted, cold mixture . . . . . . . . . . . .
4.3.3.2 Dilute, cold mixture . . . . . . . . . . . . . .
4.3.3.3 Dilute, preheated mixture . . . . . . . . . . .
4.3.3.4 Dilute, preheated mixture with minor species
Chemical equilibrium . . . . . . . . . . . . . . . . . . . . . . .
Chemical kinetics of a single isothermal reaction . . . . . . . .
4.5.1 Isochoric systems . . . . . . . . . . . . . . . . . . . . .
4.5.2 Isobaric systems . . . . . . . . . . . . . . . . . . . . . .
Some conservation and evolution equations . . . . . . . . . . .
4.6.1 Total mass conservation: isochoric reaction . . . . . . .
4.6.2 Element mass conservation: isochoric reaction . . . . .
4.6.3 Energy conservation: adiabatic, isochoric reaction . . .
4.6.4 Energy conservation: adiabatic, isobaric reaction . . . .
4.6.5 Non-adiabatic isochoric combustion . . . . . . . . . . .
4.6.6 Entropy evolution: Clausius-Duhem relation . . . . . .
Simple one-step kinetics . . . . . . . . . . . . . . . . . . . . .
5 Thermochemistry of multiple reactions
5.1 Summary of multiple reaction extensions . . . . . .
5.2 Equilibrium conditions . . . . . . . . . . . . . . . .
5.2.1 Minimization of G via Lagrange multipliers
5.2.2 Equilibration of all reactions . . . . . . . . .
5.2.3 Zel’dovich’s uniqueness proof . . . . . . . .
5.2.3.1 Isothermal, isochoric case . . . . .
5.2.3.2 Isothermal, isobaric case . . . . . .
5.2.3.3 Adiabatic, isochoric case . . . . . .
5.2.3.4 Adiabatic, isobaric case . . . . . .
5.3 Concise reaction rate law formulations . . . . . . .
5.3.1 Reaction dominant: J ≥ (N − L) . . . . . .
5.3.2 Species dominant: J < (N − L) . . . . . . .
5.4 Irreversibility production . . . . . . . . . . . . . . .
5.4.1 Onsager reciprocity . . . . . . . . . . . . . .
5.4.2 General analysis . . . . . . . . . . . . . . . .
5.4.3 Zel’dovich mechanism example . . . . . . . .
5.4.4 Extended Zel’dovich mechanism example . .
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6
6 Reactive Navier-Stokes equations
6.1 Evolution axioms . . . . . . . . . .
6.1.1 Conservative form . . . . . .
6.1.2 Non-conservative form . . .
6.1.2.1 Mass . . . . . . . .
6.1.2.2 Linear momentum
6.1.2.3 Energy . . . . . .
6.1.2.4 Second law . . . .
6.1.2.5 Species . . . . . .
6.1.2.6 Elements . . . . .
6.2 Mixture rules . . . . . . . . . . . .
6.3 Constitutive models . . . . . . . . .
6.4 Temperature evolution . . . . . . .
6.5 Shvab-Zel’dovich formulation . . . .
CONTENTS
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8 Simple solid combustion: reaction-diffusion
8.1 Simple planar model . . . . . . . . . . . . . . . . . . . . . . .
8.1.1 Model equations . . . . . . . . . . . . . . . . . . . . .
8.1.2 Simple planar derivation . . . . . . . . . . . . . . . . .
8.1.3 Ad hoc approximation . . . . . . . . . . . . . . . . . .
8.1.3.1 Planar formulation . . . . . . . . . . . . . . .
8.1.3.2 More general coordinate systems . . . . . . .
8.2 Non-dimensionalization . . . . . . . . . . . . . . . . . . . . . .
8.2.1 Diffusion time discussion . . . . . . . . . . . . . . . . .
8.2.2 Final form . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.3 Integral form . . . . . . . . . . . . . . . . . . . . . . .
8.2.4 Infinite Damköhler limit . . . . . . . . . . . . . . . . .
8.3 Steady solutions . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.1 High activation energy asymptotics . . . . . . . . . . .
8.3.2 Method of weighted residuals . . . . . . . . . . . . . .
8.3.2.1 One-term collocation solution . . . . . . . . .
8.3.2.2 Two-term collocation solution . . . . . . . . .
8.3.3 Steady solution with depletion . . . . . . . . . . . . . .
8.4 Unsteady solutions . . . . . . . . . . . . . . . . . . . . . . . .
8.4.1 Linear stability . . . . . . . . . . . . . . . . . . . . . .
8.4.1.1 Formulation . . . . . . . . . . . . . . . . . . .
8.4.1.2 Separation of variables . . . . . . . . . . . . .
8.4.1.3 Numerical eigenvalue solution . . . . . . . . .
8.4.1.3.1 Low temperature transients . . . . .
8.4.1.3.2 Intermediate temperature transients
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7
CONTENTS
8.4.2
8.4.1.3.3 High temperature
Full transient solution . . . . . . .
8.4.2.1 Low temperature solution
8.4.2.2 High temperature solution
transients
. . . . . .
. . . . . .
. . . . . .
9.1 Governing Equations . . . . . . . . . . . . . . .
9.1.1 Evolution equations . . . . . . . . . . . .
9.1.1.1 Conservative form . . . . . . .
9.1.1.2 Non-conservative form . . . . .
9.1.1.3 Formulation using enthalpy . .
9.1.1.4 Low Mach number limit . . . .
9.1.2 Constitutive models . . . . . . . . . . . .
9.1.3 Alternate forms . . . . . . . . . . . . . .
9.1.3.1 Species equation . . . . . . . .
9.1.3.2 Energy equation . . . . . . . .
9.1.3.3 Shvab-Zel’dovich form . . . . .
9.1.4 Equilibrium conditions . . . . . . . . . .
9.2 Steady burner-stabilized flames . . . . . . . . .
9.2.1 Formulation . . . . . . . . . . . . . . . .
9.2.2 Solution procedure . . . . . . . . . . . .
9.2.2.1 Model linear system . . . . . .
9.2.2.2 System of first order equations
9.2.2.3 Equilibrium . . . . . . . . . . .
9.2.2.4 Linear analysis near equilibrium
9.2.2.5 Laminar flame structure . . . .
9.2.2.5.1 TIG = 0.2. . . . . . . .
9.2.2.5.2 TIG = 0.076. . . . . . .
9.2.3 Detailed H2 -O2-N2 kinetics . . . . . . . .
10.1 Reactive Euler equations . . . . . . . .
10.1.1 One-step irreversible kinetics . .
10.1.2 Sound speed and thermicity . .
10.1.3 Parameters for H2 -air . . . . . .
10.1.4 Conservative form . . . . . . . .
10.1.5 Non-conservative form . . . . .
10.1.5.1 Mass . . . . . . . . . .
10.1.5.2 Linear momenta . . .
10.1.5.3 Energy . . . . . . . .
10.1.5.4 Reaction . . . . . . . .
10.1.5.5 Summary . . . . . . .
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CC BY-NC-ND. 30 March 2014, J. M. Powers.
8
CONTENTS
10.1.6 One-dimensional form . . . . . . . . . . . . . . . . .
10.1.6.1 Conservative form . . . . . . . . . . . . . .
10.1.6.2 Non-conservative form . . . . . . . . . . . .
10.1.6.3 Reduction of energy equation . . . . . . . .
10.1.7 Characteristic form . . . . . . . . . . . . . . . . . . .
10.1.8 Rankine-Hugoniot jump conditions . . . . . . . . . .
10.1.9 Galilean transformation . . . . . . . . . . . . . . . .
10.1.9.1 Direct approach . . . . . . . . . . . . . . . .
10.1.9.2 Differential geometry approach . . . . . . .
10.1.9.3 Transformed equations . . . . . . . . . . . .
10.2 One-dimensional, steady solutions . . . . . . . . . . . . . . .
10.2.1 Steady shock jumps . . . . . . . . . . . . . . . . . . .
10.2.2 Ordinary differential equations of motion . . . . . . .
10.2.2.1 Conservative form . . . . . . . . . . . . . .
10.2.2.2 Unreduced non-conservative form . . . . . .
10.2.2.3 Reduced non-conservative form . . . . . . .
10.2.3 Rankine-Hugoniot analysis . . . . . . . . . . . . . . .
10.2.3.1 Rayleigh line . . . . . . . . . . . . . . . . .
10.2.3.2 Hugoniot curve . . . . . . . . . . . . . . . .
10.2.4 Shock solutions . . . . . . . . . . . . . . . . . . . . .
10.2.5 Equilibrium solutions . . . . . . . . . . . . . . . . . .
10.2.5.1 Chapman-Jouguet solutions . . . . . . . . .
10.2.5.2 Weak and strong solutions . . . . . . . . . .
10.2.5.3 Summary of solution properties . . . . . . .
10.2.6 ZND solutions: One-step irreversible kinetics . . . . .
10.2.6.1 CJ ZND structures . . . . . . . . . . . . . .
10.2.6.2 Strong ZND structures . . . . . . . . . . . .
10.2.6.3 Weak ZND structures . . . . . . . . . . . .
10.2.6.4 Piston problem . . . . . . . . . . . . . . . .
10.2.7 Detonation structure: Two-step irreversible kinetics .
10.2.7.1 Strong structures . . . . . . . . . . . . . . .
10.2.7.1.1 D > D̃ . . . . . . . . . . . . . . . .
10.2.7.1.2 D = D̃ . . . . . . . . . . . . . . . .
10.2.7.2 Weak, eigenvalue structures . . . . . . . . .
10.2.7.3 Piston problem . . . . . . . . . . . . . . . .
10.2.8 Detonation structure: Detailed H2 − O2 − N2 kinetics
11 Blast waves
11.1 Governing equations . . . . .
11.2 Similarity transformation . . .
11.2.1 Independent variables
11.2.2 Dependent variables .
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CC BY-NC-ND. 30 March 2014, J. M. Powers.
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9
CONTENTS
11.3
11.4
11.5
11.6
11.7
11.2.3 Derivative transformations . . . . .
Transformed equations . . . . . . . . . . .
11.3.1 Mass . . . . . . . . . . . . . . . . .
11.3.2 Linear momentum . . . . . . . . .
11.3.3 Energy . . . . . . . . . . . . . . . .
Dimensionless equations . . . . . . . . . .
11.4.1 Mass . . . . . . . . . . . . . . . . .
11.4.2 Linear momentum . . . . . . . . .
11.4.3 Energy . . . . . . . . . . . . . . . .
Reduction to non-autonomous form . . . .
Numerical solution . . . . . . . . . . . . .
11.6.1 Calculation of total energy . . . . .
11.6.2 Comparison with experimental data
Contrast with acoustic limit . . . . . . . .
Bibliography
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CC BY-NC-ND. 30 March 2014, J. M. Powers.
10
CC BY-NC-ND. 30 March 2014, J. M. Powers.
CONTENTS
Preface
These are lecture notes for AME 60636, Fundamentals of Combustion, a course taught since
1994 in the Department of Aerospace and Mechanical Engineering of the University of Notre
Dame. Most of the students in this course are graduate students; the course is also suitable
for interested undergraduates. The objective of the course is to provide background in
theoretical combustion science. Most of the material in the notes is covered in one semester;
some extra material is also included.
The goal of the notes is to provide a solid mathematical foundation in the physical chemistry, thermodynamics, and fluid mechanics of combustion. The notes attempt to fill gaps in
the existing literature in providing enhanced discussion of detailed kinetics models which are
in common use for real physical systems. In addition, some model problems for paradigm
systems are addressed for pedagogical purposes. Many of the unique aspects of these notes,
which focus on some fundamental issues involving the thermodynamics of reactive gases with
detailed finite rate kinetics, have arisen due to the author’s involvement in a research project
supported by the US National Science Foundation, under Grant No. CBET-0650843. Any
opinions, findings, and conclusions or recommendations expressed in this material are those
of the author and do not necessarily reflect the views of the National Science Foundation.
The author is grateful for this public support. Thanks are also extended to my former
student Dr. Ashraf al-Khateeb, who generated the steady laminar flame plots for detailed
hydrogen-air combustion. General thanks are due to students in the course over the years
whose interest has motivated me to find ways to improve the material.
The notes, along with information on the course itself, can be found on the world wide
web at http://www.nd.edu/∼powers/ame.60636. At this stage, anyone is free to make
copies for their own use. I would be happy to hear from you about errors or suggestions for
improvement.
Joseph M. Powers
[email protected]
http://www.nd.edu/∼powers
\
Notre Dame, Indiana; USA
=
\$
CC
BY:
30 March 2014
11
12
CC BY-NC-ND. 30 March 2014, J. M. Powers.
CONTENTS
Chapter 1
Introduction to kinetics
Poca favilla gran fiamma seconda.1 (A great flame follows a little spark.).
Dante Alighieri (1265-1321), Paradiso, Canto I, l. 34.
————————Let us consider the reaction of N molecular chemical species composed of L elements via
J chemical reactions. Let us assume the gas is an ideal mixture of ideal gases that satisfies
Dalton’s2 law of partial pressures. The reaction will be considered to be driven by molecular
collisions. We will not model individual collisions, but instead attempt to capture their
collective effect.
An example of a model of such a reaction is listed in Table 1.1. There we find a N = 9
species, J = 37 step irreversible reaction mechanism for an L = 3 hydrogen-oxygen-argon
mixture from Maas and Warnatz,3 with corrected fH2 from Maas and Pope.4 The model has
also been utilized by Fedkiw, et al.5 The symbol M represents an arbitrary third body and
is an inert participant in the reaction. We need not worry yet about fH2 , which is known as
a collision efficiency factor. The one-sided arrows indicate that each individual reaction is
considered to be irreversible. Note that for nearly each reaction, a separate reverse reaction is
listed; thus, pairs of irreversible reactions can in some sense be considered to model reversible
reactions. In this model a set of elementary reactions are hypothesized. For the j th reaction
we have the collision frequency factor aj , the temperature-dependency exponent βj , and the
activation energy E j . These will be explained in short order.
1
This verse from La Divina Commedia was selected in a spirit of optimism that the poet’s better known
entreaty from that opus’s more topical Inferno: Lasciate ogne speranza, voi ch’entrate. (All hope abandon,
ye who enter in.), Canto III, l. 1, will prove to be irrelevant.
2
John Dalton, 1766-1844, English chemist.
3
Maas, U., and Warnatz, J., 1988, “Ignition Processes in Hydrogen-Oxygen Mixtures,” Combustion and
Flame, 74(1): 53-69.
4
Maas, U., and Pope, S. B., 1992, “Simplifying Chemical Kinetics: Intrinsic Low-Dimensional Manifolds
in Composition Space,” Combustion and Flame, 88(3-4): 239-264.
5
Fedkiw, R. P., Merriman, B., and Osher, S., 1997, “High Accuracy Numerical Methods for Thermally
Perfect Gas Flows with Chemistry,” Journal of Computational Physics, 132(2): 175-190.
13
14
CHAPTER 1. INTRODUCTION TO KINETICS
j
1
2
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37
Reaction
aj
O2 + H → OH + O
OH + O → O2 + H
H2 + O → OH + H
OH + H → H2 + O
H2 + OH → H2 O + H
H2 O + H → H2 + OH
OH + OH → H2 O + O
H2 O + O → OH + OH
H + H + M → H2 + M
H2 + M → H + H + M
H + OH + M → H2 O + M
H2 O + M → H + OH + M
O + O + M → O2 + M
O2 + M → O + O + M
H + O2 + M → HO2 + M
HO2 + M → H + O2 + M
HO2 + H → OH + OH
OH + OH → HO2 + H
HO2 + H → H2 + O2
H2 + O2 → HO2 + H
HO2 + H → H2 O + O
H2 O + O → HO2 + H
HO2 + O → OH + O2
OH + O2 → HO2 + O
HO2 + OH → H2 O + O2
H2 O + O2 → HO2 + OH
HO2 + HO2 → H2 O2 + O2
OH + OH + M → H2 O2 + M
H2 O2 + M → OH + OH + M
H2 O2 + H → H2 + HO2
H2 + HO2 → H2 O2 + H
H2 O2 + H → H2 O + OH
H2 O + OH → H2 O2 + H
H2 O2 + O → OH + HO2
OH + HO2 → H2 O2 + O
H2 O2 + OH → H2 O + HO2
H2 O + HO2 → H2 O2 + OH
(
mole/cm3
1−ν ′
−
)( M,j
s·K
βj
2.00 × 1014
1.46 × 1013
5.06 × 104
2.24 × 104
1.00 × 108
4.45 × 108
1.50 × 109
1.51 × 1010
1.80 × 1018
6.99 × 1018
2.20 × 1022
3.80 × 1023
2.90 × 1017
6.81 × 1018
2.30 × 1018
3.26 × 1018
1.50 × 1014
1.33 × 1013
2.50 × 1013
6.84 × 1013
3.00 × 1013
2.67 × 1013
1.80 × 1013
2.18 × 1013
6.00 × 1013
7.31 × 1014
2.50 × 1011
3.25 × 1022
2.10 × 1024
1.70 × 1012
1.15 × 1012
1.00 × 1013
2.67 × 1012
2.80 × 1013
8.40 × 1012
5.40 × 1012
1.63 × 1013
PN
′
i=1 νij
)
!
βj
0.00
0.00
2.67
2.67
1.60
1.60
1.14
1.14
−1.00
−1.00
−2.00
−2.00
−1.00
−1.00
−0.80
−0.80
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
−2.00
−2.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
Ej
kJ
mole
70.30
2.08
26.30
18.40
13.80
77.13
0.42
71.64
0.00
436.08
0.00
499.41
0.00
496.41
0.00
195.88
4.20
168.30
2.90
243.10
7.20
242.52
−1.70
230.61
0.00
303.53
−5.20
0.00
206.80
15.70
80.88
15.00
307.51
26.80
84.09
4.20
132.71
Table 1.1: Third body collision efficiencies with M are fH2 = 1.00, fO2 = 0.35, and
fH2 O = 6.5.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
15
1.1. ISOTHERMAL, ISOCHORIC KINETICS
j
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Reaction
H2 + O2 ⇋ OH + OH
OH + H2 ⇋ H2 O + H
H + O2 ⇋ OH + O
O + H2 ⇋ OH + H
H + O2 + M ⇋ HO2 + M
H + O2 + O2 ⇋ HO2 + O2
H + O2 + N2 ⇋ HO2 + N2
OH + HO2 ⇋ H2 O + O2
H + HO2 ⇋ OH + OH
O + HO2 ⇋ O2 + OH
OH + OH ⇋ O + H2 O
H2 + M ⇋ H + H + M
O2 + M ⇋ O + O + M
H + OH + M ⇋ H2 O + M
H + HO2 ⇋ H2 + O2
HO2 + HO2 ⇋ H2 O2 + O2
H2 O2 + M ⇋ OH + OH + M
H2 O2 + H ⇋ HO2 + H2
H2 O2 + OH ⇋ H2 O + HO2
aj
(
mole/cm3
1−ν ′
−
)( M,j
s·K
βj
1.70 × 1013
1.17 × 109
5.13 × 1016
1.80 × 1010
2.10 × 1018
6.70 × 1019
6.70 × 1019
5.00 × 1013
2.50 × 1014
4.80 × 1013
6.00 × 108
2.23 × 1012
1.85 × 1011
7.50 × 1023
2.50 × 1013
2.00 × 1012
1.30 × 1017
1.60 × 1012
1.00 × 1013
PN
′
i=1 νij
)
!
βj
0.00
1.30
−0.82
1.00
−1.00
−1.42
−1.42
0.00
0.00
0.00
1.30
0.50
0.50
−2.60
0.00
0.00
0.00
0.00
0.00
Ej
cal
mole
47780
3626
16507
8826
0
0
0
1000
1900
1000
0
92600
95560
0
700
0
45500
3800
1800
Table 1.2: Nine species, nineteen step reversible reaction mechanism for a hydrogen/oxygen/nitrogen mixture. Third body collision efficiencies with M are f5 (H2 O) = 21,
f5 (H2 ) = 3.3, f12 (H2 O) = 6, f12 (H) = 2, f12 (H2 ) = 3, f14 (H2 O) = 20.
Other common forms exist. Often reactions systems are described as being composed
of reversible reactions. Such reactions are usually notated by two sided arrows. One such
system is reported by Powers and Paolucci6 and is listed here in Table 1.2. Both overall
models are complicated.
1.1
Isothermal, isochoric kinetics
For simplicity, we will first focus attention on cases in which the temperature T and volume
V are both constant. Such assumptions are known as isothermal and isochoric, respectively.
A good fundamental treatment of elementary reactions of this type is given by Vincenti and
Kruger in their detailed monograph.7
6
Powers, J. M., and Paolucci, S., 2005, “Accurate Spatial Resolution Estimates for Reactive Supersonic
Flow with Detailed Chemistry,” AIAA Journal, 43(5): 1088-1099.
7
Vincenti, W. G., and Kruger, C. H., 1965, Introduction to Physical Gas Dynamics, Wiley, New York.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
16
1.1.1
CHAPTER 1. INTRODUCTION TO KINETICS
O − O2 dissociation
One of the simplest physical examples is provided by the dissociation of O2 into its atomic
component O.
1.1.1.1
Pair of irreversible reactions
To get started, let us focus for now only on reactions j = 13 and j = 14 from Table 1.1 in
the limiting case in which temperature T and volume V are constant.
1.1.1.1.1 Mathematical model The reactions describe oxygen dissociation and recombination in a pair of irreversible reactions:
13 : O + O + M → O2 + M,
14 : O2 + M → O + O + M,
(1.1)
(1.2)
with
a14
−2
kJ
mole
K
a13 = 2.90 × 10
, β13 = −1.00, E 13 = 0
,
3
cm
s
mole
−1
kJ
K
mole
18
, β14 = −1.00, E 14 = 496.41
.
= 6.81 × 10
3
cm
s
mole
17
(1.3)
(1.4)
The irreversibility is indicated by the one-sided arrow. Though they participate in the overall
hydrogen oxidation problem, these two reactions are in fact self-contained as well. So, let
us just consider that we have only oxygen in our box with N = 2 species, O and O2 , J = 2
reactions (those being 13 and 14), and L = 1 element, that being O. We will take i = 1 to
correspond to O and i = 2 to correspond to O2 .
The units of aj are unusual. For reaction j = 13, we have the forward stoichiometric
′
′
coefficient for species O, i = 1, as νij′ = ν1,13
= 2, and for species O2 , i = 2, as νij′ = ν2,13
= 0.
′
And since an inert third body participates in reaction 13, we have νM,13 = 1. So, for reaction
j = 13, we find the exponent for the (mole/cm3 ) portion of the units for a14 to be
′
1 − νM,j
−
N
X
i=1
′
′
′
νij′ = 1 − νM,13
− ν1,13
+ ν2,13
= 1 − 1 − (2 + 0) = −2.
(1.5)
Similarly for reaction j = 14, we have the forward stoichiometric coefficient for species O,
′
′
i = 1, as νij′ = ν1,14
= 0, and for species O2 , i = 2, as νij′ = ν2,14
= 1. And since an inert
′
third body participates in reaction 13, we have νM,14 = 1. So, for reaction j = 14, we find
the exponent for the (mole/cm3 ) portion of the units for a14 to be
1−
′
νM,j
−
N
X
i=1
′
′
′
νij′ = 1 − νM,14
− ν1,14
+ ν2,14
= 1 − 1 − (0 + 1) = −1.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(1.6)
17
1.1. ISOTHERMAL, ISOCHORIC KINETICS
Recall that in the cgs system, common in thermochemistry, that 1 erg = 1 dyne cm =
10 J = 10−10 kJ. Recall also that the cgs unit of force is the dyne and that 1 dyne =
1 g cm/s2 = 10−5 N. So, for cgs we have
10
10 erg
erg
erg
kJ
= 4.96 × 1012
,
.
(1.7)
E 13 = 0
E 14 = 496.41
mole
mole
kJ
mole
−7
The standard model for chemical reaction, which will be generalized and discussed in
more detail in Chapters 4 and 5, induces the following two ordinary differential equations
for the evolution of O and O2 molar concentrations:
dρO
−E 13
−E 14
β13
β14
= −2 a13 T exp
ρO ρO ρM +2 a14 T exp
ρO2 ρM ,
(1.8)
dt
RT
RT
|
|
{z
}
{z
}
dρO2
dt
|
= a13 T
|
|
=k13 (T )
β13
exp
{z
=k13 (T )
{z
}
=r13
−E 13
RT
{z
=r13
ρO ρO ρM − a14 T
}
|
}
|
β14
|
=k14 (T )
exp
{z
{z
=r14
−E 14
RT
=k14 (T )
{z
=r14
}
ρO2 ρM .
}
(1.9)
}
Here, we use the notation ρi as the molar concentration of species i. Also, a common usage
for molar concentration is given by square brackets, e.g. ρO2 = [O2 ]. The symbol R is the
universal gas constant, for which
7
J
erg
10 erg
R = 8.31441
= 8.31441 × 107
.
(1.10)
mole K
J
mole K
We also use the common notation of a temperature-dependent portion of the reaction rate
for reaction j, kj (T ), where
Ej
βj
.
(1.11)
kj (T ) = aj T exp
RT
Note that j = 1, . . . , J. The reaction rates for reactions 13 and 14 are defined as
r13 = k13 ρO ρO ρM ,
r14 = k14 ρO2 ρM .
(1.12)
(1.13)
We will give details of how to generalize this form later in Chapters 4 and 5. The system
Eq. (1.8-1.9) can be written simply as
dρO
= −2r13 + 2r14 ,
dt
dρO2
= r13 − r14 .
dt
(1.14)
(1.15)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
18
CHAPTER 1. INTRODUCTION TO KINETICS
Even more simply, in vector form, Eqs. (1.14-1.15) can be written as
dρ
= ν · r.
dt
(1.16)
Here, we have taken
ρ =
ρO
ρO2
,
−2 2
ν =
1 −1
r13
r =
.
r14
(1.17)
,
(1.18)
(1.19)
In general, we will have ρ be a column vector of dimension N × 1, ν will be a rectangular
matrix of dimension N × J of rank R, and r will be a column vector of length J × 1. So,
Eqs. (1.14-1.15) take the form
d
−2 2
r13
ρO
=
.
(1.20)
1 −1
r14
dt ρO2
Note here that the rank R of ν is R = L = 1, since det ν = 0, and ν has at least one
non-zero element. Let us also define a stoichiometric matrix φ of dimension L × N. The
component of φ, φli represents the number of element l in species i. Generally φ will be full
rank, which will vary, since we can have L < N, L = N, or L > N. Here, we have L < N
and φ is of dimension 1 × 2:
φ = (1 2).
(1.21)
Element conservation is guaranteed by insisting that ν be constructed such that
φ · ν = 0.
(1.22)
So, we can say that each of the column vectors of ν lies in the right null space of φ. For our
example, we see that Eq. (1.22) holds:
−2 2
φ · ν = (1 2) ·
= (0 0).
(1.23)
1 −1
Let us take as initial conditions
ρO (t = 0) = b
ρO ,
ρO2 (t = 0) = b
ρO2 .
(1.24)
Now, M represents an arbitrary third body, so here
ρM = ρO2 + ρO .
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(1.25)
19
1.1. ISOTHERMAL, ISOCHORIC KINETICS
Thus, the ordinary differential equations of the reaction dynamics, Eqs. (1.8, 1.9), reduce to
−E 13
dρO
β13
= −2a13 T exp
ρO ρO ρO2 + ρO
dt
RT
−E 14
β14
(1.26)
ρO2 ρO2 + ρO ,
+2a14 T exp
RT
dρO2
−E 13
β13
= a13 T exp
ρO ρO ρO2 + ρO
dt
RT
−E 14
β14
(1.27)
ρO2 ρO2 + ρO .
−a14 T exp
RT
Equations (1.26-1.27) with Eqs. (1.24) represent two non-linear ordinary differential equations with initial conditions in two unknowns ρO and ρO2 . We seek the behavior of these two
species concentrations as a function of time.
Systems of non-linear equations are generally difficult to integrate analytically and generally require numerical solution. Before embarking on a numerical solution, we simplify as
much as we can. Note that
dρ
dρO
+ 2 O2 = 0,
(1.28)
dt
dt
d
(1.29)
ρO + 2ρO2 = 0.
dt
We can integrate and apply the initial conditions (1.24) to get
ρO + 2ρO2 = b
ρO + 2b
ρO2 = constant.
(1.30)
The fact that this algebraic constraint exists for all time is a consequence of the conservation
of mass of each O element. It can also be thought of as the conservation of number of
O atoms. Such notions always hold for chemical reactions. They do not hold for nuclear
reactions.
Standard linear algebra provides a robust way to find the constraint of Eq. (1.30). We
can use elementary row operations to cast Eq. (1.20) into a row-echelon form. Here, our
goal is to get a linear combination which on the right side has an upper triangular form. To
achieve this add twice the second equation with the first to form a new equation to replace
the second equation. This gives
d
ρO
−2 2
r13
=
.
(1.31)
0 0
r14
dt ρO + 2ρO2
Obviously the second equation is one we obtained earlier, d/dt(ρO + 2ρO2 ) = 0, and this
induces our algebraic constraint. We also note Eq. (1.31) can be recast as
−2 2
r13
ρO
1 0 d
=
.
(1.32)
0 0
r14
1 2 dt ρO2
| {z }
| {z }
L−1
U
CC BY-NC-ND. 30 March 2014, J. M. Powers.
20
CHAPTER 1. INTRODUCTION TO KINETICS
This is of the matrix form
L−1 · P ·
dρ
= U · r.
dt
(1.33)
Here, L and L−1 are N × N lower triangular matrices of full rank N, and thus invertible.
The matrix U is upper triangular of dimension N × J and with the same rank as ν, R ≥ L.
The matrix P is a permutation matrix of dimension N × N. It is never singular and thus
always invertable. It is used to effect possible row exchanges to achieve the desired form;
often row exchanges are not necessary, in which case P = I, the N × N identity matrix.
Equation (1.33) can be manipulated to form the original equation via
dρ
−1
=P
· L · U} ·r.
| {z
dt
=ν
(1.34)
What we have done is the standard linear algebra decomposition of ν = P−1 · L · U.
We can also decompose the algebraic constraint, Eq. (1.30), in a non-obvious way that
is more readily useful for larger systems. We can write
ρO2 = b
ρO2 −
1
ρO − b
ρO .
2
ρO , we can say
Defining now ξ O = ρO − b
b
ρO
ρO
1
( ξO ) .
= b
+
ρO2
ρ O2
− 12 | {z
}
| {z } | {z } | {z }
=ξ
=D
b
=ρ
=ρ
(1.35)
(1.36)
This gives the dependent variables in terms of a smaller number of transformed dependent
variables in a way which satisfies the linear constraints. In vector form, Eq. (1.36) becomes
b + D · ξ.
ρ=ρ
(1.37)
φ · D = 0.
(1.38)
Here, D is a full rank matrix which spans the same column space as does ν. Note that ν
may or may not be full rank. Since D spans the same column space as does ν, we must also
have in general
We see here this is true:
1
= (0).
(1 2) ·
− 12
| {z }
| {z }
φ
D
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(1.39)
21
1.1. ISOTHERMAL, ISOCHORIC KINETICS
exp(- E j /(R T))
1
T
Ej/ R
Figure 1.1: Plot of exp(−E j /R/T ) versus T ; transition occurs at T ∼ E j /R.
We also note that the term exp(−E j /RT ) is a modulating factor to the dynamics. Let
us see how this behaves for high and low temperatures. First, for low temperature, we have
−E j
lim exp
= 0.
(1.40)
T →0
RT
At high temperature, we have
lim exp
T →∞
−E j
RT
And lastly, at intermediate temperature, we have
−E j
exp
∼ O(1),
when
RT
= 1.
T =O
(1.41)
Ej
R
.
(1.42)
A sketch of this modulating factor is given in Figure 1.1. Note
• for small T , the modulation is extreme, and the reaction rate is very small,
• for T ∼ E j /R, the reaction rate is extremely sensitive to temperature, and
• for T → ∞, the modulation is unity, and the reaction rate is limited only by molecular
collision frequency.
Now, ρO and ρO2 represent molar concentrations which have standard units of mole/cm3 .
So, the reaction rates
dρO2
dρO
and
dt
dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
22
CHAPTER 1. INTRODUCTION TO KINETICS
have units of mole/cm3 /s.
Note that the argument of the exponential to be dimensionless. That is
erg mole K 1
Ej
=
⇒
dimensionless.
mole erg K
RT
(1.43)
Here, the brackets denote the units of a quantity, and not molar concentration. Let us use
a more informal method than that presented on p. 16 to arrive at units for the collision
frequency factor of reaction 13, a13 . We know the units of the rate (mole/cm3 /s). Reaction
13 involves three molar species. Since β13 = −1, it also has an extra temperature dependency.
The exponential of a unitless number is unitless, so we need not worry about that. For units
to match, we must have
mole
mole
mole
mole
= [a13 ]
K −1 .
(1.44)
cm3 s
cm3
cm3
cm3
So, the units of a13 are
[a13 ] =
mole
cm3
−2
K
.
s
For a14 we find a different set of units! Following the same procedure, we get
mole
mole
mole
= [a14 ]
K −1 .
3
3
3
cm s
cm
cm
(1.45)
(1.46)
So, the units of a14 are
[a14 ] =
mole
cm3
−1
K
.
s
(1.47)
This discrepancy in the units of aj the molecular collision frequency factor is a burden of traditional chemical kinetics, and causes many difficulties when classical non-dimensionalization
is performed. With much effort, a cleaner theory could be formulated; however, this would
require significant work to re-cast the now-standard aj values for literally thousands of reactions which are well established in the literature.
1.1.1.1.2 Example calculation Let us consider an example problem. Let us take T =
5000 K, and initial conditions b
ρO = 0.001 mole/cm3 and b
ρO2 = 0.001 mole/cm3 . The
initial temperature is very hot, and is near the temperature of the surface of the sun. This
is also realizable in laboratory conditions, but uncommon in most combustion engineering
environments.
We can solve these in a variety of ways. I chose here to solve both Eqs. (1.26-1.27) without
the reduction provided by Eq. (1.30). However, we can check after numerical solution to see
CC BY-NC-ND. 30 March 2014, J. M. Powers.
23
1.1. ISOTHERMAL, ISOCHORIC KINETICS
if Eq. (1.30) is actually satisfied. Substituting numerical values for all the constants to get
−2 !
mole
K
−E 13
7
β13
= −2 2.9 × 10
(5000 K)−1 exp(0),
−2a13 T exp
3
cm
s
RT
−2
1
mole
,
(1.48)
= −1.16 × 1014
3
cm
s
−1 !
mole
−E
K
14
= 2 6.81 × 1018
2a14 T β14 exp
(5000 K)−1
3
cm
s
RT
erg
−4.96 × 1012 mole
,
× exp
erg
(5000 K)
8.31441 × 107 mole
K
−1
mole
1
10
= 1.77548 × 10
,
(1.49)
3
cm
s
−2
mole
1
−E 13
13
β13
= 5.80 × 10
,
(1.50)
a13 T exp
3
cm
s
RT
−1
1
−E 14
mole
β14
9
.
(1.51)
−a14 T exp
= −8.8774 × 10
3
cm
s
RT
Then, the differential equation system, Eqs. (1.8,1.9), becomes
dρO
= −(1.16 × 1014 )ρ2O (ρO + ρO2 ) + (1.77548 × 1010 )ρO2 (ρO + ρO2 ),
(1.52)
dt
dρO2
= (5.80 × 1013 )ρ2O (ρO + ρO2 ) − (8.8774 × 109 )ρO2 (ρO + ρO2 ),
(1.53)
dt
mole
ρO (0) = 0.001
,
(1.54)
cm3
mole
.
(1.55)
ρO2 (0) = 0.001
cm3
These non-linear ordinary differential equations are in a standard form for a wide variety
of numerical software tools. Numerical solution techniques of such equations are not the
1.1.1.1.2.1 Species concentration versus time A solution was obtained numerically, and a plot of ρO (t) and ρO2 (t) is given in Figure 1.2. Note that significant reaction does
not commence until t ∼ 10−10 s. This can be shown to be very close to the time between
molecular collisions. For 10−9 s < t < 10−8 s, there is a vigorous reaction. For t > 10−7 s,
the reaction appears to be equilibrated. The calculation gives the equilibrium values ρeq
O and
ρeq
,
as
O2
mole
,
cm3
mole
= 0.00127
.
cm3
lim ρO = ρeq
O = 0.0004424
t→∞
lim ρO2 = ρeq
O2
t→∞
(1.56)
(1.57)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
24
CHAPTER 1. INTRODUCTION TO KINETICS
ρO, ρO2 (mole/cm3)
0.00150
O2
0.00100
0.00070
0.00050
O
t HsL
10-11
10-10
10-9
10-8
10-7
10-6
Figure 1.2: Molar concentrations versus time for oxygen dissociation problem.
Here, the superscript eq denotes an equilibrium value. Note that at this high temperature,
O2 is preferred over O, but there are definitely O molecules present at equilibrium.
We can check how well the numerical solution satisfied the algebraic constraint of element
conservation by plotting the dimensionless residual error r
ρ + 2ρ − b
ρO − 2b
ρO2 O
O2
(1.58)
r=
,
b
ρO + 2b
ρ O2
as a function of time. If the constraint is exactly satisfied, we will have r = 0. Any non-zero
r will be related to the numerical method we have chosen. It may contain roundoff error
and have a sporadic nature. A plot of r(t) is given in Figure 1.3. Clearly, the error is small
and has the character of a roundoff error in that it is small and discontinuous. In fact it is
possible to drive r to be smaller by controlling the error tolerance in the numerical method.
1.1.1.1.2.2 Pressure versus time We can use the ideal gas law to calculate the
pressure. Recall that the ideal gas law for molecular species i is
Pi V = ni RT.
(1.59)
Here, Pi is the partial pressure of molecular species i, and ni is the number of moles of
molecular species i. Note that we also have
ni
Pi = RT.
(1.60)
V
Note that by our definition of molecular species concentration that
ni
ρi = .
V
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(1.61)
25
1.1. ISOTHERMAL, ISOCHORIC KINETICS
r
1.5 ´ 10-16
1.0 ´ 10-16
7.0 ´ 10-17
5.0 ´ 10-17
3.0 ´ 10-17
t HsL
10-10
10-9
10-8
10-7
10-6
Figure 1.3: Dimensionless residual numerical error r in satisfying the element conservation
constraint in the oxygen dissociation example.
So, we also have the ideal gas law as
Pi = ρi RT.
(1.62)
Now, in the Dalton mixture model, all species share the same T and V . So, the mixture
temperature and volume are the same for each species Vi = V , Ti = T . But the mixture
pressure is taken to be the sum of the partial pressures:
P =
N
X
Pi .
(1.63)
i=1
Substituting from Eq. (1.62) into Eq. (1.63), we get
P =
N
X
ρi RT = RT
i=1
N
X
ρi .
(1.64)
i=1
For our example, we only have two species, so
P = RT (ρO + ρO2 ).
(1.65)
The pressure at the initial state t = 0 is
ρO + b
ρO2 ),
P (t = 0) = RT (b
= 8.31441 × 107
(1.66)
mole
mole
erg
(5000 K) 0.001
+ 0.001
, (1.67)
3
mole K
cm
cm3
dyne
= 8.31441 × 108
,
(1.68)
cm2
= 8.31441 × 102 bar.
(1.69)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
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CHAPTER 1. INTRODUCTION TO KINETICS
P (dyne/cm2)
8.4 ´ 108
8.2 ´ 108
8. ´ 108
7.8 ´ 108
7.6 ´ 108
7.4 ´ 108
7.2 ´ 108
t HsL
10-11
10-10
10-9
10-8
10-7
10-6
Figure 1.4: Pressure versus time for oxygen dissociation example.
This pressure is over 800 atmospheres. It is actually a little too high for good experimental
correlation with the underlying data, but we will neglect that for this exercise.
At the equilibrium state we have more O2 and less O. And we have a different number
of molecules, so we expect the pressure to be different. At equilibrium, the pressure is
P (t → ∞) = lim RT (ρO + ρO2 ),
(1.70)
t→∞
erg mole
mole
7
= 8.31441 × 10
(5000 K) 0.0004424
+ 0.00127
,
mole K
cm3
cm3
(1.71)
dyne
,
(1.72)
= 7.15 × 108
cm2
= 7.15 × 102 bar.
(1.73)
The pressure has dropped because much of the O has recombined to form O2 . Thus, there
are fewer molecules at equilibrium. The temperature and volume have remained the same.
A plot of P (t) is given in Figure 1.4.
1.1.1.1.2.3 Dynamical system form Now, Eqs. (1.52-1.53) are of the standard
form for an autonomous dynamical system:
dy
= f(y).
dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(1.74)
27
1.1. ISOTHERMAL, ISOCHORIC KINETICS
Here, y is the vector of state variables (ρO , ρO2 )T . And f is an algebraic function of the state
variables. For the isothermal system, the algebraic function is in fact a polynomial.
Equilibrium
The dynamical system is in equilibrium when
f(y) = 0.
(1.75)
This non-linear set of algebraic equations can be difficult to solve for large systems. We will
later see in Sec. 5.2.3 that for common chemical kinetics systems, such as the one we are
dealing with, there is a guarantee of a unique equilibrium for which all state variables are
physical. There are certainly other equilibria for which at least one of the state variables is
non-physical. Such equilibria can be mathematically complicated.
Solving Eq. (1.75) for our oxygen dissociation problem gives us from Eq. (1.8-1.9)
−E 13
−E 14
eq eq eq β13
eq β14
−2a13 exp
ρO ρO ρM T + 2a14 exp
ρeq
= 0,
(1.76)
O2 ρM T
RT
RT
−E 13
−E 14
eq eq eq
eq
β13
β14
a13 T exp
ρO ρO ρM − a14 T exp
ρeq
(1.77)
O2 ρM = 0.
RT
RT
We notice that ρeq
M cancels. This so-called third body will in fact never affect the equilibrium state. It will however influence the dynamics. Removing ρeq
M and slightly rearranging
Eqs. (1.76-1.77) gives
−E 13
−E 14
eq eq
β14
ρO ρO = a14 T exp
ρeq
a13 T exp
O2 ,
RT
RT
−E 13
−E 14
eq eq
β13
β14
a13 T exp
ρO ρO = a14 T exp
ρeq
O2 .
RT
RT
β13
(1.78)
(1.79)
These are the same equations! So, we really have two unknowns for the equilibrium state ρeq
O
and ρeq
O2 but seemingly only one equation. Note that rearranging either Eq. (1.78) or (1.79)
gives the result
−E 14
β14
eq eq
a
T
exp
14
ρO ρO
RT
= K(T ).
=
(1.80)
eq
−E
ρO2
13
β
13
a T exp
13
RT
That is, for the net reaction (excluding the inert third body), O2 → O+O, at equilibrium the
product of the concentrations of the products divided by the product of the concentrations
of the reactants is a function of temperature T . And for constant T , this is the so-called
equilibrium constant. This is a famous result from basic chemistry. It is actually not complete
yet, as we have not taken advantage of a connection with thermodynamics. But for now, it
will suffice. A more complete discussion will be given in Sec. 4.4.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
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CHAPTER 1. INTRODUCTION TO KINETICS
We still have a problem: Eq. (1.78), or equivalently, Eq. (1.80), is still one equation for
two unknowns. We resolve this be recalling we have not yet taken advantage of our algebraic
constraint of element conservation, Eq. (1.30). Let us use this to eliminate ρeq
O2 in favor of
eq
ρO :
1 b
eq
eq
ρO2 =
ρ − ρO + b
ρO2 .
(1.81)
2 O
So, with Eq. (1.81), Eq. (1.78) reduces to
1 b
−E 13
−E 14
eq eq
eq
β13
β14
b
a13 T exp
ρO ρO = a14 T exp
ρ − ρO + ρO2 . (1.82)
2 O
RT
RT
{z
}
|
=ρeq
O
2
Equation (1.82) is one algebraic equation in one unknown. Its solution gives the equilibrium
eq
value ρeq
O . It is a quadratic equation for ρO . Of its two roots, one will be physical. We
note that the equilibrium state will be a function of the initial conditions. Mathematically
this is because our system is really best posed as a system of differential-algebraic equations.
Systems which are purely differential equations will have equilibria which are independent
of their initial conditions. Most of the literature of mathematical physics focuses on such
systems of those. One of the foundational complications of chemical dynamics is the equilibria is a function of the initial conditions, and this renders many common mathematical
notions from traditional dynamic system theory to be invalid. Fortunately, after one accounts for the linear constraints of element conservation, one can return to classical notions
Consider the dynamics of Eq. (1.26) for the evolution of ρO . Equilibrating the right
hand side of this equation gives Eq. (1.76). Eliminating ρM and then ρO2 in Eq. (1.76) then
substituting in numerical parameters gives the cubic algebraic equation
33948.3 − (1.78439 × 1011 )(ρO )2 − (5.8 × 1013 )(ρO )3 = f (ρO ) = 0.
(1.83)
This equation is cubic because we did not remove the effect of ρM . This will not affect
the equilibrium, but will affect the dynamics. We can get an idea of where the roots are
by plotting f (ρO ) as seen in Figure 1.5. Zero crossings of f (ρO ) in Figure 1.5 represent
eq
equilibria of the system, ρeq
O , f (ρO ) = 0. The cubic equation has three roots
ρeq
= −0.003
O
mole
,
cm3
mole
,
cm3
mole
,
= 0.000442414
cm3
ρeq
= −0.000518944
O
ρeq
O
non-physical,
non-physical,
physical.
(1.84)
(1.85)
(1.86)
Note the physical root found by our algebraic analysis is identical to that which was identified
by our numerical integration of the ordinary differential equations of reaction kinetics.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
29
1.1. ISOTHERMAL, ISOCHORIC KINETICS
f(ρO) (mole/cm3/s)
300 000
200 000
100 000
-0.004
-0.003
-0.002
3
0.001 ρO (mole/cm )
-0.001
-100 000
-200 000
Figure 1.5: Equilibria for oxygen dissociation example.
Stability of equilibria
We can get a simple estimate of the stability of the equilibria by considering the slope of
f near f = 0. Our dynamic system is of the form
dρO
= f (ρO ).
(1.87)
dt
3
• Near the first non-physical root at ρeq
O = −0.003 mole/cm , a positive perturbation
from equilibrium induces f < 0, which induces dρO /dt < 0, so ρO returns to its
equilibrium. Similarly, a negative perturbation from equilibrium induces dρO /dt > 0,
so the system returns to equilibrium. This non-physical equilibrium point is stable.
Note stability does not imply physicality!
3
• Perform the same exercise for the non-physical root at ρeq
O = −0.000518944 mole/cm .
We find this root is unstable.
3
• Perform the same exercise for the physical root at ρeq
O = 0.000442414 mole/cm . We
find this root is stable.
In general if f crosses zero with a positive slope, the equilibrium is unstable. Otherwise, it
is stable.
Consider a formal Taylor8 series expansion of Eq. (1.87) in the neighborhood of an equilibrium point ρeq
O:
df d
eq
eq
(ρ − ρO ) = f (ρO ) +
(ρ − ρeq
(1.88)
O) + ...
| {z } dρO ρ =ρeq O
dt O
O
O
=0
We find df /dρO by differentiating Eq. (1.83) to get
df
= −(3.56877 × 1011 )ρO − (1.74 × 1014 )ρ2O .
dρO
8
(1.89)
Brook Taylor, 1685-1731, English mathematician.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
30
CHAPTER 1. INTRODUCTION TO KINETICS
We evaluate df /dρO near the physical equilibrium point at ρO = 0.0004442414 mole/cm3 to
get
df
= −(3.56877 × 1011 )(0.0004442414) − (1.74 × 1014 )(0.0004442414)2,
dρO
1
(1.90)
= −1.91945 × 108 .
s
Thus, the Taylor series expansion of Eq. (1.26) in the neighborhood of the physical
equilibrium gives the local kinetics to be driven by
d
(ρO − 0.000442414) = −(1.91945 × 108 ) (ρO − 0.0004442414) + . . . .
dt
So, in the neighborhood of the physical equilibrium we have
ρO = 0.0004442414 + A exp −1.91945 × 108 t .
(1.91)
(1.92)
Here, A is an arbitrary constant of integration. The local time constant which governs the
times scales of local evolution is τ where
1
= 5.20983 × 10−9 s.
(1.93)
τ=
1.91945 × 108
This nano-second time scale is very fast. It can be shown to be correlated with the mean
time between collisions of molecules.
1.1.1.1.3 Effect of temperature Let us perform four case studies to see the effect of
T on the system’s equilibria and its dynamics near equilibrium.
• T = 3000 K. Here, we have significantly reduced the temperature, but it is still higher
than typically found in ordinary combustion engineering environments. Here, we find
mole
,
cm3
τ = 1.92059 × 10−7 s.
ρeq
= 8.9371 × 10−6
O
(1.94)
(1.95)
The equilibrium concentration of O dropped by two orders of magnitude relative to
T = 5000 K, and the time scale of the dynamics near equilibrium slowed by two orders
of magnitude.
• T = 1000 K. Here, we reduce the temperature more. This temperature is common in
combustion engineering environments. We find
mole
,
cm3
τ = 2.82331 × 101 s.
ρeq
= 2.0356 × 10−14
O
(1.96)
(1.97)
The O concentration at equilibrium is greatly diminished to the point of being difficult
to detect by standard measurement techniques. And the time scale of combustion has
significantly slowed.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
31
1.1. ISOTHERMAL, ISOCHORIC KINETICS
• T = 300 K. This is obviously near room temperature. We find
mole
,
cm3
s.
ρeq
= 1.14199 × 10−44
O
τ = 1.50977 × 1031
(1.98)
(1.99)
The O concentration is effectively zero at room temperature, and the relaxation time
is effectively infinite. As the oldest star in our galaxy has an age of 4.4 × 1017 s
(∼ 13.75 × 109 years9), we see that at this temperature, our mathematical model
cannot be experimentally validated, so it loses its meaning. At such a low temperature,
the theory becomes qualitatively correct, but not quantitatively predictive.
• T = 10000 K. Such high temperature could be achieved in an atmospheric re-entry
environment.
mole
,
cm3
τ = 1.69119 × 10−10 s.
−3
ρeq
O = 2.74807 × 10
(1.100)
(1.101)
At this high temperature, O become preferred over O2 , and the time scales of reaction
become extremely small, under a nanosecond.
1.1.1.2
Single reversible reaction
The two irreversible reactions studied in the previous section are of a class that is common
in combustion modelling. However, the model suffers a defect in that its link to classical
equilibrium thermodynamics is missing. A better way to model essentially the same physics
and guarantee consistency with classical equilibrium thermodynamics is to model the process
as a single reversible reaction, with a suitably modified reaction rate term.
1.1.1.2.1
Mathematical model
1.1.1.2.1.1 Kinetics For the reversible O −O2 reaction, let us only consider reaction
13 from Table 1.2 for which
13 : O2 + M ⇌ O + O + M.
(1.102)
For this system, we have N = 2 molecular species in L = 1 element reacting in J = 1
reaction. Here
−1
mole
cal
11
a13 = 1.85 × 10
. (1.103)
(K)−0.5 ,
β13 = 0.5,
E 13 = 95560
3
cm
mole
9
NASA, http://map.gsfc.nasa.gov/universe/uni age.html, retrieved 2 January 2012.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
32
CHAPTER 1. INTRODUCTION TO KINETICS
Units of cal are common in chemistry, but we need to convert to erg, which is achieved via
7
erg
4.186 J
10 erg
cal
= 4.00014 × 1012
.
(1.104)
E 13 = 95560
mole
cal
J
mole
For this reversible reaction, we slightly modify the kinetics equations to
dρO
1
−E 13
β13
ρO2 ρM −
ρ ρ ρ
,
= 2 a13 T exp
dt
Kc,13 O O M
RT
{z
}
|
dρO2
dt
|
=k13 (T )
= − a13 T
|
β13
|
exp
{z
{z
=r13
−E 13
RT
=k13 (T )
}
ρO2 ρM
1
−
ρ ρ ρ
Kc,13 O O M
{z
=r13
}
(1.105)
.
(1.106)
}
Here, we have used equivalent definitions for k13 (T ) and r13 , so that Eqs. (1.105-1.106) can
be written compactly as
dρO
= 2r13 ,
dt
dρO2
= −r13 .
dt
In matrix form, we can simplify to
d
2
ρO
=
(r13 ).
−1
dt ρO2
| {z }
ν
Here, the N × J=2 × 1 matrix ν is
ν=
2
−1
.
(1.107)
(1.108)
(1.109)
(1.110)
Performing row operations, Eq. (1.109) reduces to
d
2
ρO
=
(r13 ),
0
dt ρO + 2ρO2
(1.111)
or
2
1 0 d
ρO
=
(r13 ).
0
1 2 dt ρO2
| {z }
| {z }
L−1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
U
(1.112)
33
1.1. ISOTHERMAL, ISOCHORIC KINETICS
So, here the N × N = 2 × 2 matrix L−1 is
−1
L
=
1 0
1 2
.
(1.113)
The N × N = 2 × 2 permutation matrix P is the identity matrix. And the N × J = 2 × 1
upper triangular matrix U is
2
U=
.
(1.114)
0
Note that ν = L · U or equivalently L−1 · ν = U:
2
2
1 0
=
.
·
1 2
0
−1
| {z } | {z } | {z }
=ν
=U
=L−1
(1.115)
Once again the stoichiometric matrix φ is
φ = (1 2).
And we see that φ · ν = 0 is satisfied:
2
(1 2)·
= (0).
| {z } −1
| {z }
φ
ν
(1.116)
(1.117)
As for the irreversible reactions, the reversible reaction rates are constructed to conserve
O atoms. We have
d
ρO + 2ρO2 = 0.
dt
(1.118)
Thus, by integrating Eq. (1.118) and applying the initial conditions, we once again find
As before, we can say
ρO + 2ρO2 = b
ρO + 2b
ρO2 = constant.
b
ρO
ρO
1
= b
( ξO ) .
+
ρO2
ρ O2
− 12 | {z
}
| {z } | {z } | {z }
ξ
D
b
ρ
ρ
(1.119)
(1.120)
This gives the dependent variables in terms of a smaller number of transformed dependent
variables in a way which satisfies the linear constraints. In vector form, Eq. (1.120) becomes
Once again, φ · D = 0.
b + D · ξ.
ρ=ρ
(1.121)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
34
CHAPTER 1. INTRODUCTION TO KINETICS
1.1.1.2.1.2 Thermodynamics Equations (1.105-1.106) are supplemented by an expression for the thermodynamics-based equilibrium constant Kc,13 which is:
Po
−∆Go13
Kc,13 =
exp
.
(1.122)
RT
RT
Here, Po = 1.01326 × 106 dyne/cm2 = 1 atm is the reference pressure. The net change of
Gibbs10 free energy at the reference pressure for reaction 13, ∆Go13 , is defined as
∆Go13 = 2g oO − g oO2 .
(1.123)
This is the special case of the more general vector form
∆GoT = goT · ν.
(1.124)
Here, ∆Go is a J × 1 vector with the net change of Gibbs free energy for each reaction, and
go is a N × 1 vector of the Gibbs free energy of each species. Both are evaluated at the
reference pressure. For reaction 13, we have
2
o
o
o
.
(1.125)
∆G13 = ( g O g O2 ) ·
−1
| {z }
| {z }
goT
ν
We further recall that the Gibbs free energy for species i at the reference pressure is
defined in terms of the enthalpy and entropy as
o
g oi = hi − T soi .
o
(1.126)
It is common to find hi and soi in thermodynamic tables tabulated as functions of T .
We further note that both Eqs. (1.105) and (1.106) are in equilibrium when
eq
ρeq
O2 ρM =
1 eq eq eq
ρ ρ ρ .
Kc,13 O O M
(1.127)
We rearrange Eq. (1.127) to find the familiar
Kc,13
Q
eq
[products]
ρeq
O ρO
.
= eq = Q
ρO2
[reactants]
(1.128)
If Kc,13 > 1, the products are preferred. If Kc,13 < 1, the reactants are preferred.
Now, Kc,13 is a function of T only, so it is known. But Eq. (1.128) once again is one
equation in two unknowns. We can use the element conservation constraint, Eq. (1.119) to
reduce to one equation and one unknown, valid at equilibrium:
Kc,13 =
10
eq
ρeq
O ρO
.
b
ρO2 + 12 (b
ρO − ρeq
O)
(1.129)
Josiah Willard Gibbs, 1839-1903, American mechanical engineer and the pre-eminent American scientist
of the 19th century.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
35
1.1. ISOTHERMAL, ISOCHORIC KINETICS
Using the element constraint, Eq. (1.119), we can recast the dynamics of our system by
modifying Eq. (1.105) into one equation in one unknown:
−E 13
dρO
β13
= 2a13 T exp
dt
RT


b
×
(ρO2 +
|

1 b
1
1
1
(ρO − ρO )) (b
ρO2 + (b
ρO − ρO ) + ρO ) −
ρO ρO (b
ρO2 + (b
ρO − ρO ) + ρO )
.
2 {z
2 {z
2 {z
|
}|
} Kc,13
}
=ρO2
=ρM
=ρM
(1.130)
1.1.1.2.2 Example calculation Let us consider the same example as the previous section with T = 5000 K. We need numbers for all of the parameters of Eq. (1.130). For O,
we find at T = 5000 K that
erg
o
,
(1.131)
hO = 3.48382 × 1012
mole
erg
soO = 2.20458 × 109
.
(1.132)
mole K
So
erg erg − (5000 K) 2.20458 × 109
,
mole
mole K
erg
.
= −7.53908 × 1012
mole
g oO =
3.48382 × 1012
(1.133)
For O2 , we find at T = 5000 K that
o
So
erg
,
mole
erg
.
= 3.05406 × 109
mole K
hO2 = 1.80749 × 1012
(1.134)
soO2
(1.135)
erg erg − (5000 K) 3.05406 × 109
,
mole
mole K
erg
= −1.34628 × 1013
.
mole
g oO2 =

1.80749 × 1012
(1.136)
Thus, by Eq. (1.123), we have
∆Go13 = 2(−7.53908 × 1012 ) − (−1.34628 × 1013 ) = −1.61536 × 1012
erg
.
mole
(1.137)
Thus, by Eq. (1.122) we get for our system
Kc,13 =
1.01326 × 106 dyne
2
cm
erg
7
8.31441 × 10 mole K (5000 K)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
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CHAPTER 1. INTRODUCTION TO KINETICS
ρO, ρO2 (mole/cm3)
0.00150
O2
0.00100
0.00070
0.00050
O
0.00030
t HsL
10-11
10-9
10-7
10-5
0.001
Figure 1.6: Plot of ρO (t) and ρO2 (t) for oxygen dissociation with reversible reaction.
× exp −
= 1.187 × 10−4
erg
−1.61536 × 1012 mole
erg
8.31441 × 107 mole
(5000 K)
K
!!
,
mole
.
cm3
(1.138)
(1.139)
Substitution of all numerical parameters into Eq. (1.130) and expansion yields the following
dρO
= 3899.47 − (2.23342 × 1010 )ρ2O − (7.3003 × 1012 )ρ3O = f (ρO ),
dt
ρO (0) = 0.001.
(1.140)
(1.141)
A plot of the time-dependent behavior of ρO and ρO2 from solution of Eqs. (1.140, 1.141) is
given in Figure 1.6. The behavior is similar to the predictions given by the pair of irreversible
reactions in Fig. 1.1. Here, direct calculation of the equilibrium from time integration reveals
ρeq
O = 0.000393328
mole
.
cm3
(1.142)
mole
.
cm3
(1.143)
Using Eq. (1.119) we find this corresponds to
ρeq
O2 = 0.00130334
CC BY-NC-ND. 30 March 2014, J. M. Powers.
37
1.1. ISOTHERMAL, ISOCHORIC KINETICS
f(ρO) (mole/cm3/s)
40 000
30 000
20 000
10 000
-0.004
-0.003
-0.002
3
0.001 ρO (mole/cm )
-0.001
-10 000
-20 000
Figure 1.7: Plot of f (ρO ) versus ρO for oxygen dissociation with reversible reaction.
We note the system begins to undergo significant reaction for t ∼ 10−9 s and is equilibrated
when t ∼ 10−7 s.
The equilibrium is verified by solving the algebraic equation suggested by Eq. (1.140):
f (ρO ) = 3899.47 − (2.23342 × 1010 )ρ2O − (7.3003 × 1012 )ρ3O = 0.
(1.144)
This yields three roots:
ρeq
= −0.003
O
mole
,
cm3
mole
,
cm3
mole
,
= 0.000393328
cm3
ρeq
= −0.000452678
O
ρeq
O
non-physical,
non-physical,
physical,
(1.145)
(1.146)
(1.147)
(1.148)
consistent with the plot given in Figure 1.7.
Linearizing Eq. (1.141) in the neighborhood of the physical equilibrium yields the equation
d
(ρ − 0.000393328) = −(2.09575 × 107 ) (ρO − 0.000393328) + . . .
(1.149)
dt O
This has solution
ρO = 0.000393328 + A exp −2.09575 × 107 t .
(1.150)
Again, A is an arbitrary constant. Obviously the equilibrium is stable. Moreover, the time
constant of relaxation to equilibrium is
τ=
1
= 4.77156 × 10−8 s.
2.09575 × 107
(1.151)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
38
CHAPTER 1. INTRODUCTION TO KINETICS
This is consistent with the time scale to equilibrium which comes from integrating the full
equation.
1.1.2
Zel’dovich mechanism of N O production
Let us consider next a more complicated reaction system: that of NO production known
as the Zel’dovich11 mechanism. This is an important model for the production of a major
pollutant from combustion processes. It is most important for high temperature applications.
Related calculations and analysis of this system are given by Al-Khateeb, et al.12
1.1.2.1
Mathematical model
The model has several versions. One is
1:
2:
N + NO ⇌ N2 + O,
N + O2 ⇌ NO + O.
(1.152)
(1.153)
Similar to our model for O2 dissociation, N2 and O2 are preferred at low temperature. As
the temperature rises, N and O begin to appear. It is possible when they are mixed for NO
to appear as a product.
1.1.2.1.1
with
Standard model form Here, we have the reaction of N = 5 molecular species


ρN O
 ρN 



ρ=
 ρN2  .
 ρO 
ρO2
(1.154)
Note that “N” here serves two purposes: one to denote the number of species, the other
to denote elemental nitrogen. The context should allow the reader to distinguish the two.
We have L = 2 elements, with N, (l = 1) and O, (l = 2). The stoichiometric matrix φ of
dimension L × N = 2 × 5 is
1 1 2 0 0
φ=
.
(1.155)
1 0 0 1 2
The first row of φ is for the N atom; the second row is for the O atom.
11
Yakov Borisovich Zel’dovich, 1915-1987, prolific Soviet physicist and father of thermonuclear weapons.
Al-Khateeb, A. N., Powers, J. M., Paolucci, S., Sommese, A. J., Diller, J. A., Hauenstein, J. D., and
Mengers, J. D., 2009, “One-Dimensional Slow Invariant Manifolds for Spatially Homogeneous Reactive Systems,” Journal of Chemical Physics, 131(2): 024118.
12
CC BY-NC-ND. 30 March 2014, J. M. Powers.
39
1.1. ISOTHERMAL, ISOCHORIC KINETICS
We have J = 2 reactions. The reaction vector of length J = 2 is


Ta,1 β1
ρN ρN O − K1c,1 ρN2 ρO
a1 T exp −


T

|
{z
}


r1
k
1
,
r =
=


r2
1
 a2 T β2 exp − Ta,2

ρ
ρ
−
ρ
ρ
N O2
Kc,2 N O O 

T
|
{z
}
k2

 k1 ρN ρN O − K1c,1 ρN2 ρO
.
=  1
k2 ρN ρO2 − Kc,2 ρN O ρO
(1.156)
(1.157)
Here, we have
k1
k2
Ta,1
,
= a1 T exp −
T
Ta,2
β2
= a2 T exp −
.
T
β1
In matrix form, the model can be written as


 
ρN O
−1 1
 ρN   −1 −1   r1
 
d 
 ρN  =  1

0
2 
 r2 .

dt 
 ρO   1
1 
0 −1
ρO2
|
{z
}
=ν
(1.158)
(1.159)
(1.160)
Here, the matrix ν has dimension N ×J = 5×2. The model is of our general form, Eq. (1.16):
dρ
= ν · r.
dt
(1.161)
Note that our stoichiometric constraint on element conservation for each reaction, Eq. (1.22),
φ · ν = 0 holds here:


−1 1
 −1 −1  
0 0
1 1 2 0 0 


.
(1.162)
1
0 =
φ·ν =
0 0
1 0 0 1 2 

 1
1
0 −1
We get 4 zeros because there are 2 reactions each with 2 element constraints.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
40
CHAPTER 1. INTRODUCTION TO KINETICS
1.1.2.1.2 Reduced form Here, we describe non-traditional, but useful, reductions by
using standard techniques from linear algebra to bring the model equations into a reduced
form in which all of the linear constraints have been explicitly removed.
Let us perform a series of row operations to find all of the linear dependencies. Our aim
is to convert the matrix ν into an upper triangular form. The lower left corner of ν already
has a zero, so there is no need to worry about it. Let us add the first and fourth equations
to eliminate the 1 in the 4, 1 slot. This gives


 
−1 1
ρN O

  −1 −1  ρN
 r1
 
d 
 ρN
= 1
(1.163)
0 
2

 r2 .


dt 



0
2
ρN O + ρO
0 −1
ρO2
Next, add the first and third equations to get

 

−1 1
ρN O
  −1 −1  
ρN
 r1
 
d 
 ρN O + ρN  =  0
1 
2 
 r2 .


dt 


0
2 
ρN O + ρO
0 −1
ρO2
Now, multiply the first equation by −1 and add it to the second to get

 

ρN O
−1 1
 −ρN O + ρN   0 −2   r1
 
d 
 ρN O + ρN  =  0
1 
2 
 r2 .


dt 


0
2 
ρN O + ρO
0 −1
ρO2
Next multiply the fifth equation by −2 and add it to the second to get


 
−1 1
ρN O

  0 −2  −ρN O + ρN
 r1
 
d 

= 0
1 
ρN O + ρN2

 r2 .


dt 


0
2 
ρN O + ρO
0
0
−ρN O + ρN − 2ρO2
Next add the second and fourth equations to get

 

−1 1
ρN O
  0 −2  
−ρN O + ρN
 r1
 
d 
= 0

1 
ρN O + ρN2
 r2 .



dt 


0
0 
ρN + ρO
0
0
−ρN O + ρN − 2ρO2
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(1.164)
(1.165)
(1.166)
(1.167)
41
1.1. ISOTHERMAL, ISOCHORIC KINETICS
Next multiply the third equation by 2 and add it to the second to get

 

ρN O
−1 1
  0 −2  
−ρN O + ρN
 r1
 
d 

 ρN O + ρN + 2ρN  =  0
0
2 
 r2 .

dt 
  0

0 
ρN + ρO
−ρN O + ρN − 2ρO2
0
0
Rewritten, this becomes

1 0 0
 −1 1 0

 1 1 2

 0 1 0
−1 1 0
|
{z
=L−1
 

 
0 0
ρN O
−1 1

  
0 0 
 d  ρN   0 −2  r1




0 0   ρN2  =  0
0 
 r2 .
dt




1 0
0
0 
ρO
0 −2
0
0
ρ O2
|
}
{z
}
(1.168)
(1.169)
=U
A way to think of this type of row echelon form is that it defines two free variables, those
associated with the non-zero pivots of U: ρN O and ρN . The remaining three variables ρN2 , ρO
and ρO2 , are bound variables which can be expressed in terms of the free variables. Note that
our set of free and bound variables is not unique; had we formed other linear combinations,
we could have arrived at a different set which would be as useful as ours.
The last three of the ordinary differential equations of Eq. (1.169) are homogeneous and
can be integrated to form
ρN O + ρN + 2ρN2 = C1 ,
ρN + ρO = C2 ,
−ρN O + ρN − 2ρO2 = C3 .
(1.170)
(1.171)
(1.172)
The constants C1 , C2 and C3 are determined from the initial conditions on all five state
variables. In matrix form, we can say


ρ
N
O


 
 ρN 
1 1 2 0 0
C1


 0 1 0 1 0   ρ N  =  C2  .
(1.173)
 2
−1 1 0 0 −2  ρO 
C3
ρ O2
Considering the free variables, ρN O and ρN , to be known, we move them to the right side to
get
 



C1 − ρN O − ρN
ρN2
2 0 0
.
 0 1 0   ρO  = 
C2 − ρ N
(1.174)
C3 + ρN O − ρN
ρ O2
0 0 −2
CC BY-NC-ND. 30 March 2014, J. M. Powers.
42
CHAPTER 1. INTRODUCTION TO KINETICS
Solving for the bound variables, we find
  1


ρN2
C − 12 ρN O − 21 ρN
2 1
.
 ρO  = 
C2 − ρN
1
1
1
− 2 C3 − 2 ρN O + 2 ρN
ρO2
We can rewrite this as

  1
  1
C
ρN2
−2
1
2
 ρO  =  C2  +  0
− 12
− 21 C3
ρ O2

− 21 ρ
N
O
−1 
.
ρN
1
(1.175)
(1.176)
2
We can get a more elegant form by defining ξN O = ρN O and ξN = ρN . Thus, we can say our
state variables have the form
 
 


0
1
0
ρN O
 ρN   0   0
1 
 
 


 ρN  =  1 C1  +  − 1 − 1  ξN O .
(1.177)
2 
  2
 2  2
 ρO   C2   0 −1  ξN
1
ρ O2
− 21
− 12 C3
2
By translating via ξN O = ξ N O + b
ρN O and ξN = ξ N + b
ρN and choosing the constants C1 , C2 ,
and C3 appropriately, we can arrive at
 

 

b
ρN O
1
0
ρN O
 ρN   b
 
1 
  ρN   01

 ξNO
1
 ρN  =  b
 

(1.178)
 2   ρ N2  +  − 2 − 2  ξ N .
 ρO   b



0 −1 | {z }
ρO
1
b
ρO2
− 12
ρO2
=ξ
{z 2 }
| {z } | {z } |
D
b
ρ
ρ
This takes the form of Eq. (1.37):
b + D · ξ.
ρ=ρ
(1.179)
Here, the matrix D is of dimension N × R = 5 × 2. It spans the same column space as
does the N × J matrix ν which is of rank R. Here, in fact R = J = 2, so D has the same
dimension as ν. In general it will not. If c1 and c2 are the column vectors of D, we see that
−c1 − c2 forms the first column vector of ν and c1 − c2 forms the second column vector of
ν. Note that φ · D = 0:


1
0
 0
1 

0
0
1 1 2 0 0 
1
1
−
(1.180)
φ·D =
−2 
= 0 0 .
2
1 0 0 1 2 
 0 −1 
1
− 12
2
CC BY-NC-ND. 30 March 2014, J. M. Powers.
43
1.1. ISOTHERMAL, ISOCHORIC KINETICS
Equations (1.170-1.172) can also be linearly combined in a way which has strong physical relevance. We rewrite the system as three equations in which the first is identical to
Eq. (1.170); the second is the difference of Eqs. (1.171) and (1.172); and the third is half of
Eq. (1.170) minus half of Eq. (1.172) plus Eq. (1.171):
ρN O + ρN + 2ρN2 = C1 ,
ρO + ρN O + 2ρO2 = C2 − C3 ,
1
ρN O + ρN + ρN2 + ρO + ρO2 =
(C1 − C3 ) + C2 .
2
(1.181)
(1.182)
(1.183)
Equation (1.181) insists that the number of nitrogen elements be constant; Eq. (1.182)
demands the number of oxygen elements be constant; and Eq. (1.183) requires the number
of moles of molecular species be constant. For general reactions, including the earlier studied
oxygen dissociation problem, the number of moles of molecular species will not be constant.
Here, because each reaction considered has two molecules reacting to form two molecules,
we are guaranteed the number of moles will be constant. Hence, we get an additional
linear constraint beyond the two for element conservation. Note that since our reaction is
isothermal, isochoric and mole-preserving, it will also be isobaric.
1.1.2.1.3
Example calculation Let us consider an isothermal reaction at
T = 6000 K.
(1.184)
The high temperature is useful in generating results which are easily visualized. It insures
that there will be significant concentrations of all molecular species. Let us also take as an
initial condition
mole
b
.
ρN O = b
ρN = b
ρN2 = b
ρO = b
ρO2 = 1 × 10−6
cm3
(1.185)
For this temperature and concentrations, the pressure, which will remain constant through
the reaction, is P = 2.4942 × 106 dyne/cm2 . This is a little greater than atmospheric.
Kinetic data for this reaction are adopted from Baulch, et al.13 The data for reaction 1
is
−1
mole
1
13
a1 = 2.107 × 10
,
β1 = 0,
Ta1 = 0 K.
(1.186)
3
cm
s
For reaction 2, we have
a2 = 5.8394 × 10
9
mole
cm3
−1
1
K 1.01 s
,
β2 = 1.01,
Ta2 = 3120 K.
(1.187)
13
Baulch, et al., 2005, “Evaluated Kinetic Data for Combustion Modeling: Supplement II,” Journal of
Physical and Chemical Reference Data, 34(3): 757-1397.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
44
CHAPTER 1. INTRODUCTION TO KINETICS
Here, the so-called activation temperature Ta,j for reaction j is really the activation energy
scaled by the universal gas constant:
Ta,j =
Ej
.
R
(1.188)
Substituting numbers, we obtain for the reaction rates
k1
k2
−1
mole
1
−0
13
= 2.107 × 10
,
(1.189)
= (2.107 × 10 )(6000) exp
3
6000
cm
s
−1
−3120
mole
1
9
1.01
13
= (5.8394 × 10 )(6000) exp
= 2.27231 × 10
.
3
6000
cm
s
(1.190)
13
0
We will also need thermodynamic data. The data here will be taken from the Chemkin
database.14 Thermodynamic data for common materials is also found in most thermodynamic texts. For our system at 6000 K, we find
g oN O = −1.58757 × 1013
g oN = −7.04286 × 1012
g oN2 = −1.55206 × 1013
g oO = −9.77148 × 1012
g oO2 = −1.65653 × 1013
erg
,
mole
erg
,
mole
erg
,
mole
erg
,
mole
erg
.
mole
(1.191)
(1.192)
(1.193)
(1.194)
(1.195)
We use Eq. (1.124) to find ∆Goj for each reaction:
∆GoT = goT · ν,
( ∆Go1
∆Go2 ) = ( g oN O
Thus, for each reaction, we find ∆Goj :
g oN
g oN2
g oO


−1 1
 −1 −1 


.
1
0
g oO2 ) 


 1
1 
0 −1
(1.196)
(1.197)
(1.198)
∆Go1 = g oN2 + g oO − g oN − g oN O ,
13
12
12
13
= −1.55206 × 10 − 9.77148 × 10 + 7.04286 × 10 + 1.58757 × 10 ,
14
R. J. Kee, et al., 2000, “The Chemkin Thermodynamic Data Base,” part of the Chemkin Collection Release 3.6, Reaction Design, San Diego, CA.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
45
1.1. ISOTHERMAL, ISOCHORIC KINETICS
=
∆Go2 =
=
=
(1.199)
erg
−2.37351 × 1012
,
(1.200)
mole
g oN O + g oO − g oN − g oO2 ,
(1.201)
−1.58757 × 1013 − 9.77148 × 1012 + 7.04286 × 1012 + 1.65653 × 1013 ,
(1.202)
erg
−2.03897 × 1012
.
(1.203)
mole
At 6000 K, we find the equilibrium constants for the J = 2 reactions are
−∆Go1
Kc,1 = exp
,
RT
2.37351 × 1012
= exp
(8.314 × 107 )(6000)
= 116.52,
−∆Go2
,
Kc,2 = exp
RT
2.03897 × 1012
= exp
,
(8.314 × 107 )(6000)
= 59.5861.
(1.204)
(1.205)
(1.206)
(1.207)
(1.208)
(1.209)
Again, omitting details, we find the two differential equations governing the evolution of
the free variables are
dρN O
= 0.72331 + 2.2181 × 107 ρN + 1.1452 × 1013 ρ2N
dt
−9.4353 × 105 ρN O − 3.1960 × 1013 ρN ρN O ,
dρN
= 0.72331 − 2.3266 × 107ρN − 1.1271 × 1013 ρ2N
dt
+5.8187 × 105 ρN O − 9.9994 × 1012 ρN ρN O ,
mole
,
ρN O (0) = 1 × 10−6
cm3
mole
ρN (0) = 1 × 10−6
.
cm3
(1.210)
(1.211)
(1.212)
(1.213)
Solving numerically, we obtain the solution shown in Fig. 1.8. The numerics show a relaxation
to final concentrations of
lim ρN O = 7.336 × 10−7
t→∞
lim ρN = 3.708 × 10−8
t→∞
mole
,
cm3
mole
.
cm3
(1.214)
(1.215)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
46
CHAPTER 1. INTRODUCTION TO KINETICS
ρNO, ρN (mole/cm3)
1 ´ 10-5
5 ´ 10-6
1 ´ 10-6
NO
5 ´ 10-7
1 ´ 10-7
5 ´ 10-8
N
10-10
10-9
10-8
10-7
10-6
t s
10-5
Figure 1.8:
NO and N concentrations versus time for T = 6000 K, P = 2.4942 ×
106 dyne/cm2 Zel’dovich mechanism.
Equations (1.210-1.211) are of the form
dρN O
= fN O (ρN O , ρN ),
dt
dρN
= fN (ρN O , ρN ).
dt
(1.216)
(1.217)
At equilibrium, we must have
fN O (ρN O , ρN ) = 0,
fN (ρN O , ρN ) = 0.
(1.218)
(1.219)
We find three finite roots to this problem:
mole
,
non-physical, (1.220)
cm3
mole
,
non-physical, (1.221)
2 : (ρN O , ρN ) = (−5.173 × 10−8 , −2.048 × 10−6 )
cm3
mole
3 : (ρN O , ρN ) = (7.336 × 10−7 , 3.708 × 10−8 )
,
physical.
(1.222)
cm3
1 : (ρN O , ρN ) = (−1.605 × 10−6 , −3.060 × 10−8 )
Obviously, because of negative concentrations, roots 1 and 2 are non-physical. Root 3
however is physical; moreover, it agrees with the equilibrium we found by direct numerical
integration of the full non-linear equations.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
47
1.1. ISOTHERMAL, ISOCHORIC KINETICS
We can use local linear analysis in the neighborhood of each equilibria to rigorously
ascertain the stability of each root. Taylor series expansion of Eqs. (1.216-1.217) in the
neighborhood of an equilibrium point yields
∂fN O d
∂fN O eq
eq
(ρ − ρN O ) = fN O |e +
(ρ − ρN O ) +
(ρ − ρeq
N) + ...,
| {z } ∂ρN O e N O
dt N O
∂ρN e N
=0
∂fN d
∂fN eq
eq
(ρN − ρN ) = fN |e +
(ρN O − ρN O ) +
(ρN − ρeq
N) + ....
|{z} ∂ρN O e
dt
∂ρN e
(1.223)
(1.224)
=0
Evaluation of Eqs. (1.223-1.224) near the physical root, root 3, yields the system
d ρN O − 7.336 × 10−7
ρN O − 7.336 × 10−7
−2.129 × 106 −4.155 × 105
=
.
2.111 × 105 −3.144 × 107
ρN − 3.708 × 10−8
dt ρN − 3.708 × 10−8
{z
}
|
∂f
=J= ∂ρ
|e
(1.225)
This is of the form
∂f d
eq
· (ρ − ρeq ) = J · (ρ − ρeq ) .
(ρ − ρ ) =
dt
∂ρ e
(1.226)
It is the eigenvalues of the Jacobian15 matrix J that give the time scales of evolution of the
concentrations as well as determine the stability of the local equilibrium point. Recall that
we can usually decompose square matrices via the diagonalization
J = S · Λ · S−1 .
(1.227)
Here, S is the matrix whose columns are composed of the right eigenvectors of J, and Λ is
the diagonal matrix whose diagonal is populated by the eigenvalues of J. For some matrices
(typically not those encountered after our removal of linear dependencies), diagonalization
is not possible, and one must resort to the so-called near-diagonal Jordan form. This will
not be relevant to our discussion, but could be easily handled if necessary. We also recall the
eigenvector matrix and eigenvalue matrix are defined by the standard eigenvalue problem
J · S = S · Λ.
(1.228)
We also recall that the components λ of Λ are found by solving the characteristic polynomial
which arises from the equation
det (J − λI) = 0,
15
(1.229)
after Carl Gustav Jacob Jacobi, 1804-1851, German mathematician.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
48
CHAPTER 1. INTRODUCTION TO KINETICS
where I is the identity matrix. Defining z such that
S · z ≡ ρ − ρeq ,
(1.230)
and using the decomposition Eq. (1.227), Eq. (1.226) can be rewritten to form
d
(S · z) = |S · Λ{z· S−1} · (S · z),
| {z }
dt
J
ρ−ρeq
dz
S·
= S · Λ · z,
dt
dz
= S−1 · S · Λ · z,
S−1 · S ·
dt
dz
= Λ · z.
dt
(1.231)
(1.232)
(1.233)
(1.234)
Eq. (1.234) is in diagonal form.This has solution for each component of z of
z1 = C1 exp(λ1 t),
z2 = C2 exp(λ2 t),
..
.
(1.235)
(1.236)
(1.237)
Here, our matrix J, see Eq. (1.225), has two real, negative eigenvalues in the neighborhood
of the physical root 3:
1
,
s
1
= −2.132 × 106 .
s
λ1 = −3.143 × 107
(1.238)
λ2
(1.239)
Thus, we can conclude that the physical equilibrium is linearly stable. The local time
constants near equilibrium are given by the reciprocal of the magnitude of the eigenvalues.
These are
τ1 = 1/|λ1| = 3.181 × 10−8 s,
τ2 = 1/|λ2| = 4.691 × 10−7 s.
(1.240)
(1.241)
Evolution on these two time scales is predicted in Fig. 1.8. This in fact a multiscale problem.
One of the major difficulties in the numerical simulation of combustion problems comes in
the effort to capture the effects at all relevant scales. The problem is made more difficult as
the breadth of the scales expands. In this problem, the breadth of scales is not particularly
challenging. Near equilibrium the ratio of the slowest to the fastest time scale, the stiffness
ratio κ, is
κ=
4.691 × 10−7 s
τ2
=
= 14.75.
τ1
3.181 × 10−8 s
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(1.242)
49
1.1. ISOTHERMAL, ISOCHORIC KINETICS
Many combustion problems can have stiffness ratios over 106 . This is more prevalent at
lower temperatures.
We can do a similar linearization near the initial state, find the local eigenvalues, and
the local time scales. At the initial state here, we find those local time scales are
τ1 = 2.403 × 10−8 s,
τ2 = 2.123 × 10−8 s.
(1.243)
(1.244)
So, initially the stiffness, κ = (2.403 × 10−8 s)/(2.123 × 10−8 s) = 1.13 is much less, but the
time scale itself is small. It is seen from Fig. 1.8 that this initial time scale of 10−8 s well
predicts where significant evolution of species concentrations commences. For t < 10−8 s, the
model predicts essentially no activity. This can be correlated with the mean time between
molecular collisions–the theory on which estimates of the collision frequency factors aj are
obtained.
We briefly consider the non-physical roots, 1 and 2. A similar eigenvalue analysis of root
1 reveals that the eigenvalues of its local Jacobian matrix are
λ1 = −1.193 × 107
λ2 = 5.434 × 106
1
,
s
1
.
s
(1.245)
(1.246)
Thus, root 1 is a saddle and is unstable.
For root 2, we find
1
,
s
1
= 4.397 × 107 − i7.997 × 106 .
s
λ1 = 4.397 × 107 + i7.997 × 106
(1.247)
λ2
(1.248)
The eigenvalues are complex with a positive real part. This indicates the root is an unstable
spiral source.
A detailed phase portrait is shown in Fig. 1.9. Here, we see all three finite roots. Their
local character of sink, saddle, or spiral source is clearly displayed. We see that trajectories
are attracted to a curve labeled SIM for “Slow Invariant Manifold.” A part of the SIM is
constructed by the trajectory which originates at root 1 and travels to root 3. The other
part is constructed by connecting an equilibrium point at infinity into root 3. Details are
omitted here and are given in a related problem by Al-Khateeb, et al.16
1.1.2.2
Stiffness, time scales, and numerics
One of the key challenges in computational chemistry is accurately predicting species concentration evolution with time. The problem is made difficult because of the common presence
16
Al-Khateeb, A. N., Powers, J. M., Paolucci, S., Sommese, A. J., Diller, J. A., Hauenstein, J. D., and
Mengers, J. D., 2009, “One-Dimensional Slow Invariant Manifolds for Spatially Homogeneous Reactive Systems,” Journal of Chemical Physics, 131(2): 024118.
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50
CHAPTER 1. INTRODUCTION TO KINETICS
x 10
-7
5
0
ρN (mole/cm 3)
3 sink
SIM
1
SIM
-5
-10
-15
spira l
source
-20
2
-4
-3
-2
-1
0
ρNO (mole/cm 3)
1
2
x 10
-6
Figure 1.9: NO and N phase portraits for T = 6000 K, P = 2.4942 × 106 dyne/cm2
Zel’dovich mechanism.
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51
1.1. ISOTHERMAL, ISOCHORIC KINETICS
ρN, ρNO (mole/cm3)
10-6
NO
10-8
10-10
10-12
10-11
N
10-9
10-7
t s
10-5
0.001
0.1
10
Figure 1.10: ρN O and ρN versus time for Zel’dovich mechanism at T = 1500 K, P =
6.23550 × 105 dyne/cm2 .
of physical phenomena which evolve on a widely disparate set of time scales. Systems which
evolve on a wide range of scales are known as stiff, recognizing a motivating example in
mass-spring-damper systems with stiff springs. Here, we will examine the effect of temperature and pressure on time scales and stiffness. We shall also look simplistically how different
numerical approximation methods respond to stiffness.
1.1.2.2.1 Effect of temperature Let us see how the same Zel’dovich mechanism behaves at lower temperature, T = 1500 K; all other parameters, including the initial species
concentrations are the same as the previous high temperature example. The pressure however, lowers, and here is P = 6.23550×105 dyne/cm2 , which is close to atmospheric pressure.
For this case, a plot of species concentrations versus time is given in Figure 1.10.
At T = 1500 K, we notice some dramatic differences relative to the earlier studied
T = 6000 K. First, we see the reaction commences in around the same time, t ∼ 10−8 s. For
t ∼ 10−6 s, there is a temporary cessation of significant reaction. We notice a long plateau
in which species concentrations do not change over several decades of time. This is actually
a pseudo-equilibrium. Significant reaction recommences for t ∼ 0.1 s. Only around t ∼ 1 s
does the system approach final equilibrium. We can perform an eigenvalue analysis both
at the initial state and at the equilibrium state to estimate the time scales of reaction. For
this dynamical system which is two ordinary differential equations in two unknowns, we will
always find two eigenvalues, and thus two time scales. Let us call them τ1 and τ2 . Both
these scales will evolve with t.
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52
CHAPTER 1. INTRODUCTION TO KINETICS
At the initial state, we find
τ1 = 2.37 × 10−8 s,
τ2 = 4.25 × 10−7 s.
(1.249)
(1.250)
The onset of significant reaction is consistent with the prediction given by τ1 at the initial
state. Moreover, initially, the reaction is not very stiff; the stiffness ratio is κ = 17.9.
At equilibrium, we find
mole
,
cm3
mole
,
= 4.2 × 10−14
cm3
lim ρN O = 4.6 × 10−9
t→∞
lim ρN
t→∞
(1.251)
(1.252)
and
τ1 = 7.86 × 10−7 s,
τ2 = 3.02 × 10−1 s.
(1.253)
(1.254)
The slowest time scale near equilibrium is an excellent indicator of how long the system
takes to relax to its final state. Note also that near equilibrium, the stiffness ratio is large,
κ = τ2 /τ1 ∼ 3.8 × 105 . Because it is large, the scales in this problem are widely disparate,
and accurate numerical solution becomes challenging.
In summary, we find the effect of lowering temperature while leaving initial concentrations
constant:
• lowers the pressure somewhat, slightly slowing the collision time, and slightly slowing
the fastest time scales, and
• slows the slowest time scales many orders of magnitude, stiffening the system significantly, since collisions may not induce reaction with their lower collision speed.
1.1.2.2.2 Effect of initial pressure Let us maintain the initial temperature at T =
1500 K, but drop the initial concentration of each species to
mole
b
.
ρN O = b
ρN = b
ρN2 = b
ρO2 = b
ρO = 10−8
cm3
(1.255)
With this decrease in number of moles, the pressure now is
P = 6.23550 × 103
dyne
.
cm2
(1.256)
This pressure is two orders of magnitude lower than atmospheric. We solve for the species
concentration profiles and show the results of numerical prediction in Figure 1.11. Relative to
the high pressure P = 6.2355 × 105 dyne/cm2 , T = 1500 K case, we notice some similarities
CC BY-NC-ND. 30 March 2014, J. M. Powers.
53
1.1. ISOTHERMAL, ISOCHORIC KINETICS
ρN, ρNO (mole/cm3)
10-8
NO
10-10
10-12
10-14
N
10-16
10-8
10-6
10-4
t s
0.01
1
100
Figure 1.11: ρN O and ρN versus time for Zel’dovich mechanism at T = 1500 K, P =
6.2355 × 103 dyne/cm2 .
and dramatic differences. The overall shape of the time-profiles of concentration variation
is similar. But, we see the reaction commences at a much later time, t ∼ 10−6 s. For
t ∼ 10−4 s, there is a temporary cessation of significant reaction. We notice a long plateau
in which species concentrations do not change over several decades of time. This is again
actually a pseudo-equilibrium. Significant reaction recommences for t ∼ 10 s. Only around
t ∼ 100 s does the system approach final equilibrium. We can perform an eigenvalue analysis
both at the initial state and at the equilibrium state to estimate the time scales of reaction.
At the initial state, we find
τ1 = 2.37 × 10−6 s,
τ2 = 4.25 × 10−5 s.
(1.257)
(1.258)
The onset of significant reaction is consistent with the prediction given by τ1 at the initial
state. Moreover, initially, the reaction is not very stiff; the stiffness ratio is κ = 17.9.
Interestingly, by decreasing the initial pressure by a factor of 102 , we increased the initial
time scales by a complementary factor of 102 ; moreover, we did not alter the stiffness.
At equilibrium, we find
lim ρN O = 4.6 × 10−11
t→∞
lim ρN = 4.2 × 10−16
t→∞
mole
,
cm3
mole
,
cm3
(1.259)
(1.260)
(1.261)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
54
CHAPTER 1. INTRODUCTION TO KINETICS
and
τ1 = 7.86 × 10−5 s,
τ2 = 3.02 × 101 s.
(1.262)
(1.263)
By decreasing the initial pressure by a factor of 102 , we decreased the equilibrium concentrations by a factor of 102 and increased the time scales by a factor of 102 , leaving the stiffness
ratio unchanged.
In summary, we find the effect of lowering the initial concentrations significantly while
leaving temperature constant
• lowers the pressure significantly, proportionally slowing the collision time, as well as
the fastest and slowest time scales, and
• does not affect the stiffness of the system.
1.1.2.2.3 Stiffness and numerics The issue of how to simulate stiff systems of ordinary
differential equations, such as presented by our Zel’dovich mechanism, is challenging. Here,
a brief summary of some of the issues will be presented. The interested reader should consult
the numerical literature for a full discussion. See for example the excellent text of Iserles.17
We have seen throughout this section that there are two time scales at work, and they are
often disparate. The species evolution is generally characterized by an initial fast transient,
followed by a long plateau, then a final relaxation to equilibrium. We noted from the phase
plane of Fig. 1.9 that the final relaxation to equilibrium (shown along the green line labeled
“SIM”) is an attracting manifold for a wide variety of initial conditions. The relaxation onto
the SIM is fast, and the motion on the SIM to equilibrium is relatively slow.
Use of common numerical techniques can often mask or obscure the actual dynamics.
Numerical methods to solve systems of ordinary differential equations can be broadly categorized as explicit or implicit. We give a brief synopsis of each class of method. We cast
each as a method to solve a system of the form
dρ
= f(ρ).
dt
(1.264)
• Explicit: The simplest of these methods, the forward Euler method, discretizes Eq. (1.264)
as follows:
ρn+1 − ρn
= f(ρn ),
(1.265)
∆t
so that
ρn+1 = ρn + ∆t f(ρn ).
(1.266)
Explicit methods are summarized as
17
A. Iserles, 2008, A First Course in the Numerical Analysis of Differential Equations, Cambridge University Press, Cambridge, UK.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
55
1.1. ISOTHERMAL, ISOCHORIC KINETICS
– easy to program, since Eq. (1.266) can be solved explicitly to predict the new
value ρn+1 in terms of the old values at step n.
– need to have ∆t < τf astest in order to remain numerically stable,
– able to capture all physics and all time scales at great computational expense for
stiff problems,
– requiring much computational effort for little payoff in the SIM region of the phase
plane, and thus
– inefficient for some portions of stiff calculations.
• Implicit: The simplest of these methods, the backward Euler method, discretizes
Eq. (1.264) as follows:
ρn+1 − ρn
= f(ρn+1 ),
∆t
(1.267)
ρn+1 = ρn + ∆t f(ρn+1 ).
(1.268)
so that
Implicit methods are summarized as
– more difficult to program since a non-linear set of algebraic equations, Eq. (1.268),
must be solved at every time step with no guarantee of solution,
– requiring potentially significant computational time to advance each time step,
– capable of using very large time steps and remaining numerically stable,
– suspect to missing physics that occur on small time scales τ < ∆t, and
– for many purposes, especially when fine transients are unimportant, better performers than explicit methods,
– potentially dangerous when fine transients are necessary to capture high frequency
physical instabilities.
A wide variety of software tools exist to solve systems of ordinary differential equations.
Most of them use more sophisticated techniques than simple forward and backward Euler
methods. One of the most powerful techniques is the use of error control. Here, the user
specifies how far in time to advance and the error that is able to be tolerated. The algorithm,
which is complicated, selects then internal time steps, for either explicit or implicit methods,
to achieve a solution within the error tolerance at the specified output time. A well known
public domain algorithm with error control is provided by lsode.f, which can be found in
the netlib repository.18
18
Hindmarsh, A. C., 1983,“ODEPACK, a Systematized Collection of ODE Solvers,”
Scientific Computing, edited by R. S. Stepleman, et al., North-Holland, Amsterdam, pp. 55-64.
http://www.netlib.org/alliant/ode/prog/lsode.f.
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56
CHAPTER 1. INTRODUCTION TO KINETICS
∆t (s)
102
101
100
10−1
10−2
10−3
10−4
10−5
10−6
Explicit
Ninternal
106
105
104
103
102
101
100
100
100
Explicit
∆tef f (s)
10−4
10−4
10−4
10−4
10−4
10−4
10−4
10−5
10−6
Implicit
Ninternal
100
100
100
100
100
100
100
100
100
Implicit
∆tef f (s)
102
101
100
10−1
10−2
10−3
10−4
10−5
10−6
Table 1.3: Results from computing Zel’dovich NO production using implicit and explicit
methods with error control in dlsode.f.
Let us exercise the Zel’dovich mechanism under the conditions simulated in Fig. 1.11,
T = 1500 K, P = 6.2355 × 103 dyne/cm2 . Recall in this case the fastest time scale near
equilibrium is τ1 = 7.86 × 10−5 s ∼ 10−4 s at the initial state, and the slowest time scale is
τ = 3.02 × 101 s at the final state. Let us solve for these conditions using dlsode.f, which
uses internal time stepping for error control, in both an explicit and implicit mode. We
specify a variety of values of ∆t and report typical values of number of internal time steps
selected by dlsode.f, and the corresponding effective time step ∆tef f used for the problem,
for both explicit and implicit methods, as reported in Table 1.3.
Obviously if output is requested using ∆t > 10−4 s, the early time dynamics near t ∼
10−4 s will be missed. For physically stable systems, codes such as dlsode.f will still provide
a correct solution at the later times. For physically unstable systems, such as might occur in
turbulent flames, it is not clear that one can use large time steps and expect to have fidelity
to the underlying equations. The reason is the physical instabilities may evolve on the same
time scale as the fine scales which are overlooked by large ∆t.
1.2
It is more practical to allow for temperature variation within a combustor. The best model
for this is adiabatic kinetics. Here, we will restrict our attention to isochoric problems.
1.2.1
Thermal explosion theory
There is a simple description known as thermal explosion theory which provides a good
explanation for how initially slow exothermic reaction induces a sudden temperature rise
accompanied by a final relaxation to equilibrium.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
57
Let us consider a simple isomerization reaction in a closed volume
A ⇌ B.
(1.269)
An isomerization reaction is one in which the molecular structure may rearrange, but the
elemental composition, and thus the molecular mass, of the molecule does not change. Let
us take A and B to both be calorically perfect ideal gases with identical molecular masses
MA = MB = M and identical specific heats, cvA = cvB = cv ; cP A = cP B = cP . We can
consider A and B to be isomers of an identical molecular species. So, we have N = 2 species
reacting in J = 1 reactions. The number of elements L here is irrelevant.
1.2.1.1
One-step reversible kinetics
Let us insist our reaction process be isochoric and adiabatic, and commence with only A
present. The reaction kinetics with β = 0 are
dρA
−E
1
= − a exp
ρA −
ρ ,
(1.270)
dt
Kc B
RT
|
{z
}
=k
|
{z
}
=r
−E
1
dρB
= a exp
ρA −
ρ ,
(1.271)
dt
Kc B
RT
{z
}
|
=k
|
{z
}
=r
ρA (0) = b
ρA ,
ρB (0) = 0.
(1.272)
(1.273)
For our alternate compact linear algebra based form, we note that
−E
1
r = a exp
,
ρA −
ρ
Kc B
RT
and that
d
dt
ρA
ρB
=
−1
1
(r).
(1.274)
(1.275)
Forming the appropriate linear combinations by replacing the second equation with the sum
of the two equations yields
d
ρA
−1
=
(r).
(1.276)
0
dt ρA + ρB
Expanded, this is
1 0
1 1
d
dt
ρA
ρB
=
−1
0
(r).
(1.277)
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CHAPTER 1. INTRODUCTION TO KINETICS
Integrating the second of Eq. (1.276) yields
Thus, Eq. (1.270) reduces to
d
(ρ + ρB ) = 0,
dt A
ρA + ρB = b
ρA ,
b
ρB = ρA − ρA .
(1.279)
(1.280)
(1.281)
dρA
= −a exp
dt
−E
RT
1 b
.
ρA −
ρ − ρA
Kc A
Scaling, Eq. (1.281) can be rewritten as
1
ρA
E
ρA
ρA
1
d
= − exp −
−
.
1−
b
b
d(at) b
RTo T /To
ρA
ρA Kc
ρA
1.2.1.2
(1.278)
(1.282)
First law of thermodynamics
Recall the first law of thermodynamics, and neglecting potential and kinetic energy changes:
dE
= Q̇ − Ẇ .
dt
(1.283)
Here, E is the total internal energy. Because we insist the problem is adiabatic Q̇ = 0.
Because we insist the problem is isochoric, there is no work done, so Ẇ = 0. Thus, we have
dE
= 0.
dt
(1.284)
E = Eo .
(1.285)
Thus, we find
Recall the total internal energy for a mixture of two calorically perfect ideal gases is
(1.286)
E = nA eA + nB eB ,
n
n
A
B
= V
eA +
eB ,
(1.287)
V
V
= V (ρA eA + ρB eB ) ,
(1.288)
PA
PB
+ ρB hB −
,
(1.289)
= V ρA hA −
ρA
ρB
= V ρA hA − RT + ρB hB − RT ,
(1.290)
o
o
= V ρA cP (T − To ) + hTo ,A − RT + ρB cP (T − To ) + hTo ,B − RT ,(1.291)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
59
= V
= V
= V
= V
= V
= V
o
o
(ρA + ρB )(cP (T − To ) − RT ) + ρA hTo ,A + ρB hTo ,B ,
o
o
(ρA + ρB )((cP − R)T − cP To ) + ρA hTo ,A + ρB hTo ,B ,
o
o
(ρA + ρB )((cP − R)T − (cP − R + R)To ) + ρA hTo ,A + ρB hTo ,B ,
o
o
(ρA + ρB )(cv T − (cv + R)To ) + ρA hTo ,A + ρB hTo ,B ,
o
o
(ρA + ρB )cv (T − To ) + ρA (hTo ,A − RTo ) + ρB (hTo ,B − RTo ) ,
(ρA + ρB )cv (T − To ) + ρA eoTo ,A + ρB eoTo ,B .
Now, at the initial state we have T = To , so
o
o
b
b
Eo = V ρA eTo ,A + ρB eTo ,B .
(1.292)
(1.293)
(1.294)
(1.295)
(1.296)
(1.297)
(1.298)
So, we can say our caloric equation of state is
E − Eo = V (ρA + ρB )cv (T − To ) + (ρA − b
ρA )eoTo ,A + (ρB − b
ρB )eoTo ,B , (1.299)
o
o
b
b
b
b
= V (ρA + ρB )cv (T − To ) + (ρA − ρA )eTo ,A + (ρB − ρB )eTo ,B . (1.300)
As an aside, on a molar basis, we scale Eq. (1.300) to get
e − eo = cv (T − To ) + (yA − yAo )eoTo ,A + (yB − yBo )eoTo ,B .
(1.301)
And because we have assumed the molecular masses are the same, MA = MB , the mole
fractions are the mass fractions, and we can write on a mass basis
e − eo = cv (T − To ) + (YA − YAo )eoTo ,A + (YB − YBo )eoTo ,B .
(1.302)
Returning to Eq. (1.300), our energy conservation relation, Eq. (1.285), becomes
0 = V (b
(1.303)
ρA + b
ρB )cv (T − To ) + (ρA − b
ρA )eoTo ,A + (ρB − b
ρB )eoTo ,B .
Now, we solve for T
0 = (b
ρA + b
ρB )cv (T − To ) + (ρA − b
ρA )eoTo ,A + (ρB − b
ρB )eoTo ,B ,
ρ −b
ρA o
ρ −b
ρB o
0 = cv (T − To ) + A
eTo ,A + B
eT ,B ,
b
b
ρA + b
ρB
ρA + b
ρB o
b
ρ − ρA eoTo ,A b
ρ − ρB eoTo ,B
T = To + A
+ B
.
b
b
ρA + b
ρB cv
ρA + b
ρB cv
ρB = 0, giving also ρB = b
ρA − ρA ,
Now, we impose our assumption that b
b
ρA − ρA eoTo ,A ρB eoTo ,B
T = To +
−
,
b
b
cv
ρA
ρA cv
b
ρ − ρA eoTo ,A − eoTo ,B
.
= To + A
b
cv
ρA
(1.304)
(1.305)
(1.306)
(1.307)
(1.308)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
60
CHAPTER 1. INTRODUCTION TO KINETICS
o
ρA :
o
In summary, realizing that hTo ,A − hTo ,B = eoTo ,A − eoTo ,B we can write T as a function of
T = To +
(b
ρA − ρA ) o
o
(hTo ,A − hTo ,B ).
b
ρ A cv
(1.309)
o
o
We see then that if hTo ,A > hTo ,B , that as ρA decreases from its initial value of b
ρA that T
will increase. We can scale Eq. (1.309) to form
!
o
o
hTo ,A − hTo ,B
T
ρA
=1+ 1−
.
(1.310)
b
To
cv To
ρA
We also note that our caloric state equation, Eq. (1.301) can, for yAo = 1, yBo = 0, be
rewritten as
e − eo = cv (T − To ) + (yA − 1)eoTo ,A + yB eoTo ,B ,
= cv (T − To ) + ((1 − yB ) − 1)eoTo ,A + yB eoTo ,B ,
= cv (T − To ) − yB (eoTo ,A − eoTo ,B ).
(1.311)
(1.312)
(1.313)
Similarly, on a mass basis, we can say,
e − eo = cv (T − To ) − YB (eoTo ,A − eoTo ,B ).
For this problem, we also have
Kc = exp
with
∆Go =
=
=
=
−∆Go
RT
,
(1.315)
g oB − g oA ,
o
o
hB − T soB − (hA − T soA ),
o
o
(hB − hA ) − T (soB − soA ),
o
o
(hTo ,B − hTo ,A ) − T (soTo ,B − soTo ,A ).
So
o
Kc = exp
(1.314)
o
hTo ,A − hTo ,B − T (soTo ,A − soTo ,B )
RT
o
o
!
,
hTo ,A − hTo ,B − T (soTo ,A − soTo ,B )
cv To
= exp
cv To
RT
= exp
1 1
γ − 1 TTo
o
o
(1.316)
(1.317)
(1.318)
(1.319)
(1.320)
!!
,
hTo ,A − hTo ,B
T (soTo ,A − soTo ,B )
−
cv To
To
cv
(1.321)
!!
.
(1.322)
Here, we have used the definition of the ratio of specific heats, γ = cP /cv along with R =
cP − cv . So, we can solve Eq. (1.281) by first using Eq. (1.322) to eliminate Kc and then
Eq. (1.309) to eliminate T .
CC BY-NC-ND. 30 March 2014, J. M. Powers.
61
1.2.1.3
Dimensionless form
Let us try writing dimensionless variables so that our system can be written in a compact
dimensionless form. First, let us take dimensionless time τ to be
τ = at.
(1.323)
Let us take dimensionless species concentration to be z with
z=
ρA
.
b
ρA
(1.324)
Let us take dimensionless temperature to be θ with
θ=
T
.
To
(1.325)
Let us take dimensionless heat release to be q with
o
o
hT ,A − hTo ,B
q= o
.
cv To
(1.326)
Let us take dimensionless activation energy to be Θ with
Θ=
E
.
RTo
(1.327)
And let us take the dimensionless entropy change to be σ with
(soTo ,A − soTo ,B )
σ=
.
cv
(1.328)
So, our equations become
1
dz
Θ
z−
= − exp −
(1 − z) ,
dτ
θ
Kc
θ = 1 + (1 − z)q,
1 1
Kc = exp
(q − θσ) .
γ −1θ
(1.329)
(1.330)
(1.331)
It is more common to consider the products. Let us define for general problems
λ=
ρB
ρB
.
=
b
ρA + ρB
ρA + b
ρB
Thus, λ is the mass fraction of product. For our problem, b
ρB = 0 so
λ=
ρB
ρc − ρA
= A
.
b
b
ρA
ρA
(1.332)
(1.333)
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62
CHAPTER 1. INTRODUCTION TO KINETICS
Thus,
λ = 1 − z.
(1.334)
We can think of λ as a reaction progress variable as well. When λ = 0, we have τ = 0, and
the reaction has not begun. Thus, we get
1
Θ
dλ
(1 − λ) −
= exp −
λ ,
(1.335)
dτ
θ
Kc
θ = 1 + qλ,
(1.336)
1 1
Kc = exp
(q − θσ) .
(1.337)
γ −1θ
1.2.1.4
Example calculation
Let us choose some values for the dimensionless parameters:
Θ = 20,
σ = 0,
q = 10,
7
γ= .
5
With these choices, our kinetics equations, Eqs. (1.335-1.337), reduce to
dλ
−25
−20
(1 − λ) − λ exp
,
λ(0) = 0.
= exp
dτ
1 + 10λ
1 + 10λ
(1.338)
(1.339)
The right side of Eq. (1.339) is at equilibrium for values of λ which drive it to zero.
Numerical root finding methods show this to occur at λ ∼ 0.920539. Near this root, Taylor
series expansion shows the dynamics are approximated by
d
(λ − 0.920539) = −0.17993(λ − 0.920539) + . . .
dτ
(1.340)
Thus, the local behavior near equilibrium is given by
λ = 0.920539 + C exp (−0.17993 τ ) .
(1.341)
Here, C is some arbitrary constant. Clearly the equilibrium is stable, with a time constant
of 1/0.17993 = 5.55773.
Numerical solution shows the full behavior of the dimensionless species concentration
λ(τ ); see Figure 1.12. Clearly the product concentration λ is small for some long period of
time. At a critical time near τ = 2.7 × 106 , there is a so-called thermal explosion with a
rapid increase in λ. Note that the estimate of the time constant near equilibrium is orders
of magnitude less than the explosion time, 5.55773 ≪ 2.7 × 106 . Thus, linear analysis
here is a poor tool to estimate an important physical quantity, the ignition time. Once the
ignition period is over, there is a rapid equilibration to the final state. The dimensionless
temperature plot is shown in Figure 1.13. The temperature plot is similar in behavior to
CC BY-NC-ND. 30 March 2014, J. M. Powers.
63
λ
1.0
0.8
0.6
0.4
0.2
0
1. ´ 106
2. ´ 106
3. ´ 106
4. ´ 106
5. ´ 106
τ
Figure 1.12: Dimensionless plot of reaction product concentration λ versus time τ for adiabatic isochoric combustion with simple reversible kinetics.
θ
10
8
6
4
2
1. ´ 106
2. ´ 106
3. ´ 106
4. ´ 106
5. ´ 106
τ
Figure 1.13: Dimensionless plot of temperature θ versus time τ for adiabatic, isochoric
combustion with simple reversible kinetics.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
64
CHAPTER 1. INTRODUCTION TO KINETICS
the species concentration plot. At early time, the temperature is cool. At a critical time,
the thermal explosion time, the temperature rapidly rises. This rapid rise, coupled with the
exponential sensitivity of reaction rate to temperature, accelerates the formation of product.
This process continues until the reverse reaction is activated to the extent it prevents further
creation of product.
1.2.1.5
High activation energy asymptotics
Let us see if we can get an analytic prediction of the thermal explosion time, τ ∼ 2.7 × 106 .
Such a prediction would be valuable to see how long a slowing reacting material might take
to ignite. Our analysis is similar to that given by Buckmaster and Ludford in their Chapter
1.19
For convenience let us restrict ourselves to σ = 0. In this limit, Eqs. (1.335-1.337) reduce
to
dλ
−q
Θ
(1 − λ) − λ exp
,
(1.342)
= exp −
dτ
1 + qλ
(γ − 1)(1 + qλ)
with λ(0) = 0. The key trouble in getting an analytic solution to Eq. (1.342) is the presence
of λ in the denominator of an exponential term. We need to find a way to move it to the
numerator. Asymptotic methods provide one such way.
Now, we recall for early time λ ≪ 1. Let us assume λ takes the form
λ = ǫλ1 + ǫ2 λ2 + ǫ3 λ3 + . . .
(1.343)
Here, we will assume 0 < ǫ ≪ 1 and that λ1 (τ ) ∼ O(1), λ2 (τ ) ∼ O(1), . . ., and will define ǫ
in terms of physical parameters shortly. Now, with this assumption we have
1
1
=
.
2
1 + qλ
1 + ǫqλ1 + ǫ qλ2 + ǫ3 qλ3 + . . .
(1.344)
Long division of the term on the right side yields the approximation
1
= = 1 − ǫqλ1 + ǫ2 (q 2 λ21 − qλ2 ) + . . . ,
1 + qλ
= 1 − ǫqλ1 + O(ǫ2 ).
(1.345)
(1.346)
So,
exp −
Θ
1 + qλ
∼ exp −Θ(1 − ǫqλ1 + O(ǫ2 )) ,
∼ exp(−Θ) exp ǫqΘλ1 + O(ǫ2 ) .
(1.347)
(1.348)
We have moved λ from the denominator to the numerator of the most important exponential
term.
19
J. D. Buckmaster and G. S. S. Ludford, 1983, Lectures on Mathematical Combustion, SIAM, Philadelphia.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
65
Now, let us take the limit of high activation energy by defining ǫ to be
ǫ≡
1
.
Θ
(1.349)
Let us let the assume the remaining parameters, q and γ are both O(1) constants. When Θ
is large, ǫ will be small. With this definition, Eq. (1.348) becomes
Θ
1
exp −
∼ exp −
exp qλ1 + O(ǫ2 ) .
(1.350)
1 + qλ
ǫ
With these assumptions and approximations, Eq. (1.342) can be written as
d
1
exp qλ1 + O(ǫ2 )
(ǫλ1 + . . .) = exp −
dτ
ǫ
−q
.
× (1 − ǫλ1 − . . .) − (ǫλ1 + . . .) exp
(γ − 1)(1 + qǫλ1 + . . .)
(1.351)
Now, let us rescale time via
1
τ∗ = exp
ǫ
−1
ǫ
τ.
(1.352)
With this transformation, the chain rule shows how derivatives transform:
d
1
dτ∗ d
=
=
dτ
dτ dτ∗
ǫ exp
1
ǫ
d
.
dτ∗
(1.353)
With this transformation, Eq. (1.351) becomes
1
1
d
2
(ǫλ
+
.
.
.)
=
exp
−
exp
qλ
+
O(ǫ
)
1
1
ǫ
ǫ exp 1ǫ dτ∗
× (1 − ǫλ1 − . . .) − (ǫλ1 + . . .) exp
−q
.
(γ − 1)(1 + qǫλ1 + . . .)
(1.354)
This simplifies to
d
(λ1 + . . .) = exp qλ1 + O(ǫ2 )
dτ∗
× (1 − ǫλ1 − . . .) − (ǫλ1 + . . .) exp
−q
(γ − 1)(1 + qǫλ1 + . . .)
.
(1.355)
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66
CHAPTER 1. INTRODUCTION TO KINETICS
Retaining only O(1) terms in Eq. (1.355), we get
dλ1
= exp (qλ1 ) .
dτ∗
(1.356)
This is supplemented by the initial condition λ1 (0) = 0. Separating variables and solving,
we get
exp(−qλ1 )dλ1 = dτ∗ ,
1
− exp(−qλ1 ) = τ∗ + C.
q
(1.357)
(1.358)
Applying the initial condition gives
1
− exp(−q(0)) = C,
q
1
= C.
−
q
(1.359)
(1.360)
So
1
1
− exp(−qλ1 ) = τ∗ − ,
q
q
exp(−qλ1 ) = −qτ∗ + 1,
1
,
exp(−qλ1 ) = −q τ∗ −
q
1
−qλ1 = ln −q τ∗ −
,
q
1
1
λ1 = − ln −q τ∗ −
.
q
q
(1.361)
(1.362)
(1.363)
(1.364)
(1.365)
For q = 10, a plot of λ1 (τ∗ ) is shown in Fig. 1.14. We note at a finite τ∗ that λ1 begins to
exhibit unbounded growth. In fact, it is obvious from Eq. (1.356) that as
1
τ∗ → ,
q
that
λ1 → ∞.
That is, there exists a finite time for which λ1 violates the assumptions of our asymptotic
theory which assumes λ1 = O(1). We associate this time with the ignition time, τ∗i :
1
τ∗i = .
q
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(1.366)
67
λ1
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.00
0.02
0.04
0.06
0.08
0.10
0.12
τ*
Figure 1.14: λ1 versus τ∗ for ignition problem.
Let us return this to more primitive variables:
1
1
−1
τi =
exp
,
ǫ
ǫ
q
ǫ exp 1ǫ
τi =
,
q
exp Θ
.
τi =
Θq
(1.367)
(1.368)
(1.369)
For our system with Θ = 20 and q = 10, we estimate the dimensionless ignition time as
τi =
exp 20
= 2.42583 × 106 .
(20)(10)
(1.370)
This is a surprisingly good estimate, given the complexity of the problem. Recall the numerical solution showed ignition for τ ∼ 2.7 × 106 .
In terms of dimensional time, ignition time prediction becomes
exp Θ
,
aΘq
1 RTo
=
a
E
ti =
(1.371)
cv To
o
o
hTo ,A − hTo ,B
!
exp
E
RTo
.
(1.372)
Note the ignition is suppressed if the ignition time is lengthened, which happens when
CC BY-NC-ND. 30 March 2014, J. M. Powers.
68
CHAPTER 1. INTRODUCTION TO KINETICS
• the activation energy E is increased, since the exponential sensitivity is stronger than
the algebraic sensitivity,
o
o
• the energy of combustion (hTo ,A − hTo ,B ) is decreased because it takes longer to react
to drive the temperature to a critical value to induce ignition,
• the collision frequency factor a is decreased, which suppresses reaction.
1.2.2
Detailed H2 − O2 − N2 kinetics
Here is an example which uses multiple reactions for an adiabatic isothermal system is given.
Consider the full time-dependency of a problem similar to the thermal explosion problem
just considered. We choose a non-intuitive set of parameters for the problem. Our choices
will enable a direct comparison to a detonation of the same mixture via the same reaction
mechanism in a later chapter; see Sec. 10.2.8.
A closed, fixed, adiabatic volume, V = 0.3061251 cm3 , contains at t = 0 s a stoichiometric
hydrogen-air mixture of 2 × 10−5 mole of H2 , 1 × 10−5 mole of O2 , and 3.76 × 10−5 mole of
N2 at Po = 2.83230 × 106 P a and To = 1542.7 K.20 Thus, the initial molar concentrations
are
ρH2 = 6.533 × 10−5 mole/cm3 ,
ρO2 = 3.267 × 10−5 mole/cm3 ,
ρH2 = 1.228 × 10−4 mole/cm3 .
The initial mass fractions are calculated via Yi = Mi ρi /ρ. They are
YH2 = 0.0285,
YO2 = 0.226,
YN2 = 0.745.
To avoid issues associated with numerical roundoff errors at very early time for species
with very small compositions, the minor species were initialized at a small non-zero value
near machine precision; each was assigned a value of 10−15 mole. The minor species all have
ρi = 1.803 × 10−16 mole/cm3 . They have correspondingly small initial mass fractions.
We seek the reaction dynamics as the system proceeds from its initial state to its final
state. We use the reversible detailed kinetics mechanism of Table 1.2. This problem requires
a detailed numerical solution. Such a solution was performed by solving the appropriate
equations for a mixture of nine interacting species: H2 , H, O, O2 , OH, H2 O, HO2 , H2 O2 ,
and N2 . The dynamics of the reaction process are reflected in Figs. 1.15-1.17.
20
This temperature and pressure correspond to that of the same ambient mixture of H2 , O2 and N2 which
was shocked from 1.01325 × 105 P a, 298 K, to a value associated with a freely propagating detonation.
Relevant comparisons of reaction dynamics will be made later in Sec. 10.2.8.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
69
0
H
O
H2
O2
OH
H2 O
H2 O 2
HO2
N2
10
−5
Y
i
10
−10
10
−15
10
−10
−5
10
10
0
10
t (s)
Figure 1.15: Plot of YH2 (t), YH (t), YO (t), YO2 (t), YOH (t), YH2 O (t), YHO2 (t), YH2 O2 (t), YN2 (t),
for adiabatic, isochoric combustion of a mixture of 2H2 + O2 + 3.76N2 initially at To =
1542.7 K, Po = 2.8323 × 106 P a.
4000
3500
3000
T (K)
2500
2000
1500
1000
500
0
−10
−5
10
10
t (s)
Figure 1.16: Plot of T (t), for adiabatic, isochoric combustion of a mixture of 2H2 +O2 +3.76N2
initially at To = 1542.7 K, Po = 2.8323 × 106 P a.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
70
CHAPTER 1. INTRODUCTION TO KINETICS
7
P (Pa)
10
6
10
−10
−5
10
10
t (s)
Figure 1.17: Plot of P (t), for adiabatic, isochoric combustion of a mixture of 2H2 + O2 +
3.76N2 initially at To = 1542.7 K, Po = 2.8323 × 106 P a.
At early time, t < 10−7 s, the pressure, temperature, and major reactant species concentrations (H2 , O2 , N2 ) are nearly constant. However, the minor species, e.g. OH, HO2 ,
and the major product, H2 O, are undergoing very rapid growth, albeit with math fractions
whose value remains small. In this period, the material is in what is known as the induction
period.
After a certain critical mass of minor species has accumulated, exothermic recombination
of these minor species to form the major product H2 O induces the temperature to rise, which
accelerates further the reaction rates. This is manifested in a thermal explosion. A common
definition of the end of the induction period is the induction time, t = tind , the time when
dT /dt goes through a maximum. Here, one finds
tind = 6.6 × 10−7 s.
(1.373)
A close-up view of the species concentration profiles is given in Fig. 1.18.
At the end of the induction zone, there is a final relaxation to equilibrium. The equilibrium mass fractions of each species are
YO2
YH
YOH
YO
Y H2
Y H2 O
CC BY-NC-ND. 30 March 2014, J. M. Powers.
=
=
=
=
=
=
1.85 × 10−2 ,
5.41 × 10−4 ,
2.45 × 10−2 ,
3.88 × 10−3 ,
3.75 × 10−3 ,
2.04 × 10−1 ,
(1.374)
(1.375)
(1.376)
(1.377)
(1.378)
(1.379)
71
0
10
Y
i
−5
10
H
O
H2
O2
OH
H2 O
H2 O 2
−10
10
HO2
N2
2
4
6
t (s)
8
10
−7
x 10
Figure 1.18: Plot near thermal explosion time of YH2 (t), YH (t), YO (t), YO2 (t), YOH (t),
YH2 O (t), YHO2 (t), YH2 O2 (t), YN2 (t), for adiabatic, isochoric combustion of a mixture of
2H2 + O2 + 3.76N2 initially at at To = 1542.7 K, Po = 2.8323 × 106 P a.
YHO2 = 6.84 × 10−5 ,
YH2 O2 = 1.04 × 10−5 ,
YN2 = 7.45 × 10−1 .
(1.380)
(1.381)
(1.382)
We note that because our model takes N2 to be inert that its value remains unchanged.
Other than N2 , the final products are dominated by H2 O. The equilibrium temperature is
3382.3 K and 5.53 × 106 P a.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
72
CHAPTER 1. INTRODUCTION TO KINETICS
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 2
Gas mixtures
One is often faced with mixtures of simple compressible substances, and it the thermodynamics of such mixtures upon which attention is now fixed. Here, a discussion of some of
the fundamentals of mixture theory will be given. In general, thermodynamics of mixtures
can be a challenging topic about which much remains to be learned. In particular, these
notes will focus on ideal mixtures of ideal gases, for which results are often consistent with
intuition. The chemical engineering literature contains a full discussion of the many nuances
associated with non-ideal mixtures of non-ideal materials. Relevant background for this
chapter is found in standard undergraduate texts.1234 Some of these notes on mixtures are
adaptations of material found in these texts, especially Borgnakke and Sonntag.
2.1
Some general issues
Consider a mixture of N components, each a pure substance, so that the total mass and
total number are
m = m1 + m2 + m3 + . . . + mN =
N
X
mi ,
mass (g),
(2.1)
number (mole).
(2.2)
i=1
n = n1 + n2 + n3 + . . . + nN =
N
X
ni ,
i=1
1
Borgnakke, C., and Sonntag, R. E., 2009, Fundamentals of Thermodynamics, Seventh Edition, John
Wiley, New York.
2
Sandler, S. I., 1998, Chemical and Engineering Thermodynamics, Third Edition, John Wiley, New York.
3
Smith, J. M., Van Ness, H. C., and Abbott, M., 2004, Introduction to Chemical Engineering Thermodynamics, Seventh Edition, McGraw-Hill, New York.
4
Tester, J. W. and Modell, M., 1997, Thermodynamics and Its Applications, Third Edition, Prentice Hall,
73
74
CHAPTER 2. GAS MIXTURES
Recall 1 mole = 6.02214129 × 1023 molecules. The mass fraction of component i is defined
as Yi :
mi
Yi ≡
,
mass fraction, dimensionless.
(2.3)
m
The mole fraction of component i is defined as yi :
ni
mole fraction, dimensionless.
(2.4)
yi ≡ ,
n
Now, the molecular mass of species i is the mass of a mole of species i. It units are typically g/mole. Molecular mass is sometimes called “molecular weight,” but this is formally
incorrect, as it is a mass measure, not a force measure. Mathematically, the definition of Mi
corresponds to
g mi
Mi ≡
,
molecular mass
.
(2.5)
ni
mole
One gets mass fraction in terms of mole fraction as
mi
,
(2.6)
Yi =
m
ni Mi
=
,
(2.7)
m
ni Mi
,
(2.8)
= PN
m
j
j=1
ni Mi
,
(2.9)
= PN
j=1 nj Mj
=
=
ni Mi
,
PN n
1
j=1 nj Mj
n
ni Mi
PN n nj Mj ,
j=1 n
yi Mi
= PN
.
j=1 yj Mj
Similarly, one finds mole fraction in terms of mass fraction by the following:
ni
yi =
,
n
(2.10)
(2.11)
(2.12)
(2.13)
mi
= PNMi
(2.15)
=
(2.16)
=
CC BY-NC-ND. 30 March 2014, J. M. Powers.
mj ,
j=1 Mj
mi
Mi m
PN mj ,
j=1 Mj m
Yi
Mi
P N Yj .
j=1 Mj
(2.14)
75
2.1. SOME GENERAL ISSUES
The mixture itself has a mean molecular mass:
m
M ≡
,
n
PN
i=1 mi
,
=
n
N
X
ni Mi
=
,
n
i=1
=
N
X
yi Mi .
(2.17)
(2.18)
(2.19)
(2.20)
i=1
Example 2.1
Air is often modelled as a mixture in the following molar ratios:
O2 + 3.76N2.
(2.21)
Find the mole fractions, the mass fractions, and the mean molecular mass of the mixture.
Take O2 to be species 1 and N2 to be species 2. Consider the number of moles of O2 to be
n1 = 1 mole,
and N2 to be
n2 = 3.76 mole.
The molecular mass of O2 is M1 = 32 g/mole. The molecular mass of N2 is M2 = 28 g/mole. The
total number of moles is
n = 1 mole + 3.76 mole = 4.76 mole.
So, the mole fractions are
1 mole
= 0.2101.
4.76 mole
3.76 mole
y2 =
= 0.7899.
4.76 mole
y1 =
Note that
N
X
yi = 1.
(2.22)
i=1
That is to say, y1 + y2 = 0.2101 + 0.7899 = 1. Now, for the masses, one has
g = 32 g,
m1 = n1 M1 = (1 mole) 32
mole
g m2 = n2 M2 = (3.76 mole) 28
= 105.28 g,
mole
So, one has
m = m1 + m2 = 32 g + 105.28 g = 137.28 g.
The mass fractions then are
Y1 =
32 g
m1
=
= 0.2331,
m
137.28 g
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CHAPTER 2. GAS MIXTURES
Y2 =
Note that
m2
105.28 g
=
= 0.7669.
m
137.28 g
N
X
Yi = 1.
(2.23)
i=1
That is, Y1 + Y2 = 0.2331 + 0.7669 = 1. Now, for the mixture molecular mass, one has
M=
m
137.28 g
g
=
= 28.84
.
n
4.76 mole
mole
Check against another formula.
M=
N
X
i=1
yi Mi = y1 M1 + y2 M2 = (0.2101) 32
g g g
+ (0.7899) 28
= 28.84
.
mole
mole
mole
Now, postulates for mixtures are not as well established as those for pure substances.
The literature has much controversial discussion of the subject. A strong advocate of the
axiomatic approach, C. A. Truesdell, proposed the following “metaphysical principles” for
mixtures, which are worth considering.5
1. All properties of the mixture must be mathematical consequences of properties of the
constituents.
2. So as to describe the motion of a constituent, we may in imagination isolate it from the
rest of the mixture, provided we allow properly for the actions of the other constituents
upon it.
3. The motion of the mixture is governed by the same equations as is a single body.
Most important for the present discussion is the first principle. When coupled with fluid
mechanics, the second two take on additional importance. The approach of mixture theory
is to divide and conquer. One typically treats each of the constituents as a single material
and then devises appropriate average or mixture properties from those of the constituents.
The best example of this is air, which is not a single material, but is often treated as such.
2.2
Ideal and non-ideal mixtures
A general extensive property, such as energy E, for an N-species mixture will be such that
E = E(T, P, n1 , n2 , . . . , nN ).
5
C. A. Truesdell, 1984, Rational Thermodynamics, Springer-Verlag, New York.
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(2.24)
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2.3. IDEAL MIXTURES OF IDEAL GASES
A partial molar property is a generalization of an intensive property, and is defined such that
it is the partial derivative of an extensive property with respect to number of moles, with T
and P held constant. For internal energy, the partial molar internal energy is
∂E ei ≡
.
(2.25)
∂ni T,P,nj ,i6=j
Pressure and temperature are held constant because those are convenient variables to control
in an experiment. One also has the partial molar volume
∂V vi =
.
(2.26)
∂ni T,P,nj ,i6=j
It shall be soon seen that there are other natural ways to think of the volume per mole.
Now, in general one would expect to find
ei = ei (T, P, n1 , n2 , . . . , nN ),
vi = vi (T, P, n1 , n2 , . . . , nN ).
(2.27)
(2.28)
This is the case for what is known as a non-ideal mixture. An ideal mixture is defined as a
mixture for which the partial molar properties ei and vi are not functions of the composition;
that is to say
ei = ei (T, P ),
vi = vi (T, P ),
if ideal mixture,
if ideal mixture.
(2.29)
(2.30)
An ideal mixture also has the property that hi = hi (T, P ), while for a non-ideal mixture
hi = hi (T, P, n1 , . . . , nN ). Though not obvious, it will turn out that some properties of an
ideal mixture will depend on composition. For example, the entropy of a constituent of an
ideal mixture will be such that
si = si (T, P, n1, n2 , . . . , nN ).
2.3
(2.31)
Ideal mixtures of ideal gases
The most straightforward mixture to consider is an ideal mixture of ideal gases. Even here,
there are assumptions necessary that remain difficult to verify absolutely.
2.3.1
Dalton model
The most common model for a mixture of ideal gases is the Dalton6 model. Key assumptions
define this model:
6
John Dalton, 1766-1844, English physicist.
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CHAPTER 2. GAS MIXTURES
• Each constituent shares a common temperature.
• Each constituent occupies the entire volume.
• Each constituent possesses a partial pressure which sums to form the total pressure of
the mixture.
These features characterize a Dalton model for any gas, ideal or non-ideal. One also takes
for convenience
• Each constituent behaves as an ideal gas.
• The mixture behaves as a single ideal gas.
It is more convenient to deal on a molar basis for such a theory. For the Dalton model,
additional useful quantities, the species mass concentration ρi , the mixture mass concentration ρ, the species molar concentration ρi , and the mixture molar concentration ρ, can be
defined. As will be seen, these definitions for concentrations are useful; however, they are not
in common usage. Following Borgnakke and Sonntag, the bar notation, ·, will be reserved for
properties which are mole-based rather than mass-based. As mentioned earlier, the notion
of a partial molal property is discussed extensively in the chemical engineering literature
and has implications beyond those considered here. For the Dalton model, in which each
component occupies the same volume, one has
Vi = V.
The mixture mass concentration, also called the density is simply
g m
.
ρ= ,
V
cm3
The mixture molar concentration is
mole
n
.
ρ= ,
V
cm3
For species i, the equivalents are
mi
,
V
ni
=
,
V
ρi =
ρi
g ,
3
cm mole
.
cm3
(2.32)
(2.33)
(2.34)
(2.35)
(2.36)
One can find a convenient relation between species molar concentration and species mole
fraction by the following operations, beginning with Eq. (2.36):
ni n
,
(2.37)
ρi =
V n
ni n
,
(2.38)
=
nV
= yi ρ.
(2.39)
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2.3. IDEAL MIXTURES OF IDEAL GASES
A similar relation exists between species molar concentration and species mass fraction via
ni m Mi
,
V m Mi
=m
z }|i{
m ni Mi
=
,
V mMi
ρi =
(2.40)
(2.41)
=Y
i
z}|{
mi 1
= ρ
,
m Mi
Yi
= ρ .
Mi
(2.42)
(2.43)
The specific volumes, mass and molar, are similar. One takes
V
,
n
V
vi = .
ni
V
,
m
V
=
,
mi
v=
v =
vi
(2.44)
(2.45)
Note that this definition of molar specific volume is not the partial molar volume defined in
the chemical engineering literature, which takes the form vi = ∂V /∂ni |T,P,nj ,i6=j .
For the partial pressure of species i, one can say for the Dalton model
P =
N
X
Pi .
(2.46)
i=1
For species i, one has
Pi V
Pi
N
X
Pi
|i=1{z }
=P
= ni RT,
ni RT
=
,
V
N
X
ni RT
=
,
V
i=1
P =
N
RT X
ni .
V i=1
| {z }
(2.47)
(2.48)
(2.49)
(2.50)
=n
So, for the mixture, one has
P V = nRT.
One could also say
P =
n
RT = ρRT.
V
(2.51)
(2.52)
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CHAPTER 2. GAS MIXTURES
Here, n is the total number of moles in the system. Additionally R is the universal gas
constant with value
J
.
(2.53)
R = 8.3144621
mole K
Sometimes this is expressed in terms of kb the Boltzmann7 constant and NA , Avogadro’s8
number:
R = kb NA ,
(2.54)
NA = 6.02214129 × 1023
kb = 1.3806488 × 10−23
molecule
,
mole
J
.
K molecule
(2.55)
(2.56)
With the Boltzmann constant and Avogadro’s number, the ideal gas law, Eq. (2.51), can be
rewritten as
P V = nNA kb T,
(2.57)
where nNA is the number of molecules.
Example 2.2
Compare the molar specific volume defined here with the partial molar volume from the chemical
engineering literature.
The partial molar volume vi , is given by
∂V .
vi =
∂ni T,P,nj ,i6=j
For the ideal gas, one has
PV
V
∂V ∂ni T,P,nj ,i6=j
= RT
=
=
=
=
N
X
nk ,
k=1
P
RT N
k=1 nk
RT
P
PN
(2.59)
,
∂nk
k=1 ∂ni
P
PN
(2.60)
,
(2.61)
k=1 δki
,
P

 =0
=0
=1
=0
z}|{
z}|{
z}|{ z}|{
RT  δ1i + δ2i + . . . + δii + . . . + δN i 
RT
(2.58)
7
P
(2.62)
,
(2.63)
Ludwig Boltzmann, 1844-1906, Austrian physicist.
Lorenzo Romano Amedeo Carlo Bernadette Avogadro di Quaregna e di Cerreto, 1776-1856, Italian scientist.
8
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81
2.3. IDEAL MIXTURES OF IDEAL GASES
vi
=
=
RT
,
P
V
PN
k=1
=
(2.64)
nk
,
(2.65)
V
.
n
(2.66)
Here, the so-called Kronecker9 delta function has been employed, which is much the same as the identity
matrix:
δki
δki
=
=
k=
6 i,
k = i.
0,
1,
(2.67)
(2.68)
Contrast this with the earlier adopted definition of molar specific volume
vi =
V
.
ni
(2.69)
So, why is there a difference? The molar specific volume is a simple definition. One takes the
instantaneous volume V , which is shared by all species in the Dalton model, and scales it by the
instantaneous number of moles of species i, and acquires a natural definition of molar specific volume
consistent with the notion of a mass specific volume. On the other hand, the partial molar volume
specifies how the volume changes if the number of moles of species i changes, while holding T and P
and all other species mole numbers constant. One can imagine adding a mole of species i, which would
necessitate a change in V in order to guarantee the P remain fixed.
2.3.1.1
Binary mixtures
Consider now a binary mixture of two components A and B. This is easily extended to a
general mixture of N components. First, the total number of moles is the sum of the parts:
n = nA + nB .
(2.70)
Now, write the ideal gas law for each component:
PA VA = nA RTA ,
PB VB = nB RTB .
(2.71)
(2.72)
But by the assumptions of the Dalton model, VA = VB = V , and TA = TB = T , so
PA V
PB V
= nA RT,
= nB RT.
(2.73)
(2.74)
One also has
P V = nRT.
9
(2.75)
Leopold Kronecker, 1823-1891, German mathematician.
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82
CHAPTER 2. GAS MIXTURES
Solving for n, nA and nB , one finds
PV
,
RT
PA V
=
,
RT
PB V
=
.
RT
n =
nA
nB
(2.76)
(2.77)
(2.78)
Now, n = nA + nB , so one has
PA V
PB V
PV
=
+
.
RT
RT
RT
P = PA + PB .
(2.79)
(2.80)
That is the total pressure is the sum of the partial pressures. This is a mixture rule for
pressure
One can also scale each constituent ideal gas law by the mixture ideal gas law to get
PA V
nA RT
,
=
PV
nRT
nA
PA
=
,
P
n
= yA ,
PA = yA P.
(2.83)
(2.84)
PB = yB P.
(2.85)
(2.81)
(2.82)
Likewise
Now, one also desires rational mixture rules for energy, enthalpy, and entropy. Invoke Truesdell’s principles on a mass basis for internal energy. Then, the total internal energy E (with
units J) for the binary mixture must be
E = me = mA eA + mB eB ,
m
mB A
= m
eA +
eB ,
m
m
= m (YA eA + YB eB ) ,
e = YA eA + YB eB .
(2.86)
(2.87)
(2.88)
(2.89)
For the enthalpy, one has
H = mh = mA hA + mB hB ,
m
mB A
= m
hA +
hB ,
m
m
= m (YA hA + YB hB ) ,
h = YA hA + YB hB .
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(2.90)
(2.91)
(2.92)
(2.93)
83
2.3. IDEAL MIXTURES OF IDEAL GASES
It is easy to extend this to a mole fraction basis rather than a mass fraction basis. One can
also obtain a gas constant for the mixture on a mass basis. For the mixture, one has
PV
= nRT ≡ mRT,
(2.94)
PV
≡ mR = nR,
T
= (nA + nB )R,
mA
mB
=
+
R,
MA MB
R
R
,
+ mB
=
mA
MA
MB
= (mA RA + mB RB ) ,
m
mB
A
R =
RA +
RB ,
m
m
R = (YA RA + YB RB ) .
(2.95)
(2.96)
(2.97)
(2.98)
(2.99)
(2.100)
(2.101)
For the entropy, one has
S = ms = = mA sA + mB sB ,
m
mB A
= m
sA +
sB ,
m
m
= m (YA sA + YB sB ) ,
s = Y A sA + Y B sB .
(2.102)
(2.103)
(2.104)
(2.105)
Note that sA is evaluated at T and PA , while sB is evaluated at T and PB . For a calorically
perfect ideal gas, one has
T
PA
o
,
(2.106)
sA = sTo ,A + cP A ln
−RA ln
To
Po
|
{z
}
≡soT,A
=
soTo ,A
sB =
soTo ,B
+ cP A ln
T
To
− RA ln
yA P
Po
.
(2.107)
−RB ln
yB P
Po
.
(2.108)
Likewise
|
+ cP B ln
{z
≡soT,B
T
To
}
Here, the “o” denotes some reference state. As a superscript, it typically means that the
property is evaluated at a reference pressure. For example, soT,A denotes the portion of the
entropy of component A that is evaluated at the reference pressure Po and is allowed to vary
with temperature T . Note also that sA = sA (T, P, yA) and sB = sB (T, P, yB ), so the entropy
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CHAPTER 2. GAS MIXTURES
of a single constituent depends on the composition of the mixture and not just on T and P .
This contrasts with energy and enthalpy for which eA = eA (T ), eB = eB (T ), hA = hA (T ),
hB = hB (T ) if the mixture is composed of ideal gases. Occasionally, one finds hoA and hoB
used as a notation. This denotes that the enthalpy is evaluated at the reference pressure.
However, if the gas is ideal, the enthalpy is not a function of pressure and hA = hoA , hB = hoB .
If one is employing a calorically imperfect ideal gas model, then one finds for species i
that
yi P
o
si = sT,i − Ri ln
,
i = A, B.
(2.109)
Po
2.3.1.2
Entropy of mixing
Example 2.3
Initially calorically perfect ideal gases A and B are segregated within the same large volume by
a thin frictionless, thermally conducting diaphragm. Thus, both are at the same initial pressure and
temperature, P1 and T1 . The total volume is thermally insulated and fixed, so there are no global heat
or work exchanges with the environment. The diaphragm is removed, and A and B are allowed to mix.
Assume A has mass mA and B has mass mB . The gases are allowed to have distinct molecular masses,
MA and MB . Find the final temperature T2 , pressure P2 , and the change in entropy.
The ideal gas law holds that at the initial state
VA1 =
mA RA T1
,
P1
VB1 =
mB RB T1
.
P1
(2.110)
At the final state one has
V2 = VA2 = VB2 = VA1 + VB1 = (mA RA + mB RB )
T1
.
P1
(2.111)
Mass conservation gives
m2 = m1 = mA + mB .
(2.112)
One also has the first law
E2 − E1
=
E2
m2 e 2
=
=
E2 − E1
=
(mA + mB )e2 =
0 =
0 =
T2
=
=
Q − W,
(2.113)
0,
(2.114)
E1 ,
mA eA1 + mB eB1 ,
(2.115)
(2.116)
mA eA1 + mB eB1 ,
mA (eA1 − e2 ) + mB (eB1 − e2 ),
(2.117)
(2.118)
mA cvA (T1 − T2 ) + mB cvB (T1 − T2 ),
mA cvA T1 + mB cvB T1
,
mA cvA + mB cvB
T1 .
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(2.119)
(2.120)
(2.121)
85
2.3. IDEAL MIXTURES OF IDEAL GASES
The final pressure by Dalton’s law then is
P2
=
=
=
=
=
=
PA2 + PB2 ,
mA RA T2
mB RB T2
+
,
V2
V2
mA RA T1
mB RB T1
+
,
V2
V2
(mA RA + mB RB ) T1
,
V2
(mA RA + mB RB ) T1
,
(mA RA + mB RB ) PT11
(2.122)
(2.123)
(2.124)
(2.125)
(2.126)
P1 .
(2.127)
So, the initial and final temperatures and pressures are identical.
Now, the entropy change of gas A is
TA2
PA2
sA2 − sA1 = cP A ln
− RA ln
,
TA1
PA1
T2
yA2 P2
= cP A ln
− RA ln
,
T1
yA1 P1
yA2 P1
T1
,
−RA ln
= cP A ln
T1
yA1 P1
| {z }
=0
yA2 P1
= −RA ln
,
(1)P1
= −RA ln yA2 .
(2.128)
(2.129)
(2.130)
(2.131)
(2.132)
Likewise
sB2 − sB1
= −RB ln yB2 .
(2.133)
So, the change in entropy of the mixture is
∆S
=
=
=
mA (sA2 − sA1 ) + mB (sB2 − sB1 )
−mA RA ln yA2 − mB RB ln yB2 ,
R
R
ln yA2 − (nB MB )
ln yB2 ,
− (nA MA )
| {z } MB
| {z } MA
| {z }
| {z }
=mA
=mB
=RA
=
=
≥



nB
nA


+nB ln
−R nA ln
,

nA + nB
nA + nB 
{z
}
{z
}
|
|
0.
(2.136)
=RB
−R(nA ln yA2 + nB ln yB2 ),

≤0
(2.134)
(2.135)
(2.137)
(2.138)
≤0
(2.139)
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CHAPTER 2. GAS MIXTURES
For an N-component mixture, mixed in the same fashion such that P and T are constant,
this extends to
∆S = −R
= −R
N
X
k=1
N
X
(2.140)
nk
n ln yk ,
n
(2.141)
nk
ln yk ,
n
(2.142)
k=1
N
X
= −Rn
= −R
nk ln yk ,
k=1
N
X
m
M
= −Rm
yk ln yk ,
k=1
N
X
ln ykyk ,
(2.143)
(2.144)
k=1
yN
= −Rm (ln y1y1 + ln y2y2 + . . . + ln yN
),
y1 y2
yN
= −Rm ln (y1 y2 . . . yN ) ,
!
N
Y
= −Rm ln
ykyk ,
(2.145)
(2.146)
(2.147)
k=1
∆s
∆s
= − ln
=
R
R
N
Y
k=1
ykyk
!
.
(2.148)
Note that there is a fundamental dependency of the mixing entropy on the mole fractions.
Since 0 ≤ yk ≤ 1, the product is guaranteed to be between 0 and 1. The natural logarithm
of such a number is negative, and thus the entropy change for the mixture is guaranteed
positive semi-definite. Note also that for the entropy of mixing, Truesdell’s third principle
is not enforced.
Now, if one mole of pure N2 is mixed with one mole of pure O2 , one certainly expects
the resulting homogeneous mixture to have a higher entropy than the two pure components.
But what if one mole of pure N2 is mixed with another mole of pure N2 . Then, we would
expect no increase in entropy. However, if we had the unusual ability to distinguish N2
molecules whose origin was from each respective original chamber, then indeed there would
be an entropy of mixing. Increases in entropy thus do correspond to increases in disorder.
2.3.1.3
Mixtures of constant mass fraction
If the mass fractions, and thus the mole fractions, remain constant during a process, the
equations simplify. This is often the case for common non-reacting mixtures. Air at moderate
values of temperature and pressure behaves this way. In this case, all of Truesdell’s principles
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87
2.3. IDEAL MIXTURES OF IDEAL GASES
can be enforced. For a calorically perfect ideal gas, one would have
e2 − e1 = YA cvA (T2 − T1 ) + YB cvB (T2 − T1 ),
= cv (T2 − T1 ).
(2.149)
(2.150)
cv ≡ YA cvA + YB cvB .
(2.151)
h2 − h1 = YA cP A (T2 − T1 ) + YB cP B (T2 − T1 ),
= cP (T2 − T1 ).
(2.152)
(2.153)
cP ≡ Y A cP A + Y B cP B .
(2.154)
where
Similarly for enthalpy
where
For the entropy
s2 − s1 = YA (sA2 − sA1 ) + YB (sB2 − sB1 ),
(2.155)
y A P2
T2
y B P2
T2
− RA ln
+ YB cP B ln
− RB ln
,
= YA cP A ln
T1
y A P1
T1
y B P1
P2
T2
P2
T2
− RA ln
+ YB cP B ln
− RB ln
(, 2.156)
= YA cP A ln
T1
P1
T1
P1
T2
P2
= cP ln
− R ln
.
(2.157)
T1
P1
The mixture behaves as a pure substance when the appropriate mixture properties are defined. One can also take
cP
γ= .
(2.158)
cv
Note that some intuitive definitions do not hold: with γA = cP A /cvA , γB = cP B /cvB , γ 6=
YA γA + YB γB .
2.3.2
Summary of properties for the Dalton mixture model
2.3.2.1
Mass basis
Listed here is a summary of mixture properties for an N-component mixture of ideal gases
on a mass basis:
M =
ρ =
N
X
i=1
N
X
yi Mi ,
(2.159)
ρi ,
(2.160)
i=1
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CHAPTER 2. GAS MIXTURES
1
v = PN
1
i=1 vi
e =
h =
N
X
i=1
N
X
1
= ,
ρ
(2.161)
Yi ei ,
(2.162)
Yi hi ,
(2.163)
i=1
=R
R
R =
=
M
N
X
Yi Ri =
i=1
N
X
i=1
z }| {
yi Mi Ri
N
X
yj Mj
j=1
cv =
N
X
| {z } =M
Yi cvi ,
N
R X
yi ,
=
M i=1
| {z }
(2.164)
=1
(2.165)
i=1
cv = cP − R,
N
X
cP =
Y i cP i ,
if ideal gas
(2.166)
(2.167)
i=1
PN
Y i cP i
cP
= Pi=1
γ =
,
N
cv
i=1 Yi cvi
s =
N
X
Y i si ,
(2.168)
(2.169)
i=1
Yi =
Pi =
ρi =
vi =
V =
T =
hi =
hi =
hi =
yi Mi
,
M
yi P,
Yi ρ,
v
1
= ,
Yi
ρi
Vi ,
Ti ,
hoi ,
if ideal gas,
Pi
ei +
= ei + Pi vi = ei + Ri T ,
| {z }
ρi
if ideal gas
Z T
cP i (T̂ ) dT̂ ,
hoTo ,i +
To
{z
}
|
if ideal gas
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(2.170)
(2.171)
(2.172)
(2.173)
(2.174)
(2.175)
(2.176)
(2.177)
(2.178)
89
2.3. IDEAL MIXTURES OF IDEAL GASES
si =
soTo ,i
|
si
+
Z
T
cP i (T̂ )
T̂
To
{z
=soT,i
dT̂ −Ri ln
}
Pi
,
Po
(2.179)
{z
}
yi P
Pi
o
o
= sT,i − Ri ln
= sT,i − Ri ln
,
Po
Po
|
{z
} |
{z
}
|
if ideal gas
if ideal gas
(2.180)
if ideal gas
Ri T
Pi = ρi Ri T = ρRi T Yi =
,
vi
|
{z
}
(2.181)
if ideal gas
N
X
RT
Yi
=
,
P = ρRT = ρRT
M
v
i
i=1
|
{z
}
(2.182)
if ideal gas
h =
N
X
Yi hoTo ,i
+
i=1
|
h = e+
Z
T
cP (T̂ ) dT̂ ,
(2.183)
}
(2.184)
To
{z
if ideal gas
P
= e + P v = e| +{zRT} ,
ρ
if ideal gas
s =
N
X
Yi soTo ,i +
|i=1
Z
T
cP (T̂ )
T̂
To
dT̂ − R ln
{z
P
Po
if ideal gas
− R ln
N
Y
i=1
yiyi
!
}
(2.185)
These relations are not obvious. A few are derived in examples here.
Example 2.4
Derive the expression h = e + P/ρ.
Start from the equation for the constituent hi , multiply by mass fractions, sum over all species,
and use properties of mixtures:
N
X
i=1
Pi
,
ρi
hi
=
ei +
Yi hi
=
Yi ei + Yi
Yi hi
=
(2.186)
Pi
,
ρi
N
N
X
X
Pi
Yi ,
Yi ei +
ρi
i=1
i=1
(2.187)
(2.188)
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90
CHAPTER 2. GAS MIXTURES
N
X
Yi hi
N
X
=
Yi ei +
i=1
i=1
| {z }
N
X
ρi Ri T
,
ρi
Yi
i=1
| {z }
(2.189)
=e
=h
h
=
e+T
N
X
Yi Ri ,
(2.190)
i=1
| {z }
=R
=
e + RT,
P
e+ .
ρ
=
(2.191)
(2.192)
Example 2.5
Find the expression for mixture entropy of the ideal gas.
si
Yi si
s=
N
X
Yi si
=
=
=
soTo ,i +
Yi soTo ,i
N
X
Z
T
cP i (T̂ )
To
+ Yi
Z
+
=
N
X
Yi soTo ,i
i=1
=
T
cP i (T̂ )
N
X
i=1
i=1
i=1
T̂
To
Yi soTo ,i
soTo +
Z
T
To
+
dT̂ − Ri ln
Z
T̂
Z
T
Yi
cP i (T̂ )
T̂
Yi cP i (T̂ )
T̂
To i=1
cP (T̂ )
T̂
dT̂ −
Pi
Po
dT̂ − Yi Ri ln
To
N
T X
N
X
dT̂ −
dT̂ −
Yi Ri ln
i=1
,
(2.193)
Pi
Po
N
X
,
(2.194)
Yi Ri ln
i=1
N
X
Yi Ri ln
i=1
Pi
Po
Pi
Po
Pi
Po
,
(2.195)
,
(2.196)
.
(2.197)
All except the last term are natural extensions of the property for a single material. Consider
last term involving pressure ratios.
!
N
N
X
X
P
Pi
P
Pi
Yi Ri ln
= −
+ R ln
,
− R ln
Yi Ri ln
−
Po
Po
Po
Po
i=1
i=1
!
N
X
Ri
P
P
Pi
Yi
= −R
+ ln
,
− ln
ln
R
Po
Po
Po
i=1
!
N
X
R/Mi
P
P
Pi
Yi PN
= −R
ln
+ ln
,
− ln
Po
Po
Po
j=1 Yj R/Mj
i=1


N
X
= −R 

 i=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Yi /Mi
|
PN
!
Yj /Mj
{z
}
j=1
=yi
ln
Pi
Po
+ ln

P
P 
,
− ln 
Po
Po 

now the
(2.198)
(2.199)
(2.200)
(2.201)
91
2.3. IDEAL MIXTURES OF IDEAL GASES
= −R
= −R
= −R
= −R
= −R
= −R
= −R
= −R
!
P
P
Pi
+ ln
,
− ln
yi ln
Po
Po
Po
i=1
!
yi
N
X
Pi
P
P
ln
,
+ ln
− ln
Po
Po
Po
i=1
!
y !
N Y
Pi i
Po
P
ln
+ ln
,
+ ln
Po
P
Po
i=1
!
y !
N Po Y Pi i
P
ln
+ ln
,
P i=1 Po
Po
!
!
N
Y
P
1
Po
yi
PN
PN
,
+ ln
(Pi )
ln
Po
P i=1 yi Po i=1 yi i=1
!
y !
N Y
Pi i
P
ln
+ ln
,
P
Po
i=1
!
y !
N Y
yi P i
P
ln
+ ln
,
P
Po
i=1
!
!
N
Y
P
yi
+ ln
yi
ln
.
Po
i=1
N
X
(2.202)
(2.203)
(2.204)
(2.205)
(2.206)
(2.207)
(2.208)
(2.209)
So, the mixture entropy becomes
s
=
=
soTo
+
soTo +
|
Z
T
To
Z
T
To
!
P
,
+ ln
dT̂ − R ln
Po
T̂
i=1
!
N
Y
cP (T̂ )
P
yi
yi .
− R ln
dT̂ − R ln
Po
T̂
i=1
{z
} |
{z
}
N
Y
cP (T̂ )
classical entropy of a single body
yiyi
!
(2.210)
(2.211)
non−Truesdellian
The extra entropy is not found in the theory for a single material, and in fact is not in the form suggested
by Truesdell’s postulates. While it is in fact possible to redefine the constituent entropy definition in
such a fashion that the mixture entropy in fact takes on the classical form of a single material via the
RT
definition si = soTo ,i + To cP i (T̂ )/T̂ dT̂ − Ri ln (Pi /Po ) + Ri ln yi , this has the disadvantage of predicting
no entropy change after mixing two pure substances. Such a theory would suggest that this obviously
irreversible process is in fact reversible.
2.3.2.2
Molar basis
On a molar basis, one has equivalent relations to those found on a mass basis.
ρ =
N
X
i=1
ρi =
n
ρ
=
,
V
M
(2.212)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
92
CHAPTER 2. GAS MIXTURES
1
v = PN
1
i=1 v i
e =
N
X
=
V
1
= = vM,
n
ρ
(2.213)
yi ei = eM,
(2.214)
yi hi = hM,
(2.215)
yi cvi = cv M,
(2.216)
cv = cP − R,
N
X
cP =
y i cP i = cP M
(2.217)
h =
cv =
i=1
N
X
i=1
N
X
i=1
(2.218)
i=1
PN
cP
y i cP i
,
= Pi=1
γ =
N
cv
i=1 yi cvi
s =
N
X
(2.219)
yi si = sM,
(2.220)
i=1
ρi
,
Mi
V
v
1
=
=
=
= vi Mi ,
ni
yi
ρi
∂V V
=
=
= v = vM ,
∂ni P,T,nj |n
{z
}
ρi = yi ρ =
(2.221)
vi
(2.222)
vi
(2.223)
if ideal gas
Pi = yi P,
(2.224)
RT
,
P = ρRT =
v
{z
}
|
(2.225)
if ideal gas
RT
,
Pi = ρi RT =
vi
{z
}
|
(2.226)
if ideal gas
h = e+
P
= e + P v = e| +{zRT} = hM,
ρ
(2.227)
if ideal gas
hi =
o
hi ,
hi = ei +
if ideal gas,
Pi
= ei + Pi v i = ei + P vi = ei + RT = hi Mi ,
| {z }
ρi
if ideal gas
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(2.228)
(2.229)
93
2.3. IDEAL MIXTURES OF IDEAL GASES
hi =
si =
o
hTo ,i
|
si
|
T
cP i (T̂ ) dT̂ = hi Mi ,
{z
}
(2.230)
To
if ideal gas
soTo ,i
|
+
Z
+
Z
T
cP i (T̂ )
T̂
To
{z
=soT,i
dT̂ −R ln
}
yi P
,
Po
{z
yi P
o
= sT,i − R ln
= si Mi ,
Po
{z
}
|
(2.231)
}
if ideal gas
(2.232)
if ideal gas
s =
N
X
yi soTo ,i +
|i=1
2.3.3
Z
T
To
cP (T̂ )
T̂
dT̂ − R ln
{z
P
Po
if ideal gas
− R ln
N
Y
i=1
yiyi
!
}
= sM.
(2.233)
Amagat model
The Amagat10 model is an entirely different paradigm than the Dalton model. It is not used
as often. In the Amagat model,
• all components share a common temperature T ,
• all components share a common pressure P ,
• each component has a different volume.
Consider, for example, a binary mixture of calorically perfect ideal gases, A and B. For
the mixture, one has
P V = nRT,
(2.234)
with
n = nA + nB .
(2.235)
P VA = nA RT,
P VB = nB RT.
(2.236)
(2.237)
P VA P VB
PV
=
+
.
RT
RT
RT
(2.238)
For the components one has
Then, n = nA + nB reduces to
10
Emile Hilaire Amagat, 1841-1925, French physicist.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
94
CHAPTER 2. GAS MIXTURES
Thus
V
= VA + VB ,
VA VB
1 =
+
.
V
V
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(2.239)
(2.240)
Chapter 3
Mathematical foundations of
thermodynamics
Every mathematician knows it is impossible to understand an elementary course
in thermodynamics.
Vladimir Igorevich Arnol’d (1937-2010), “Contact Geometry: The Geometrical
Method of Gibbs’ Thermodynamics,” in Proceedings of the Gibbs Symposium,
D. Caldi and G. Mostow, eds., American Mathematical Society, 1990, p. 163.
This chapter focuses on mathematical formalism which can be applied to thermodynamics. Understanding of calculus of many variables at an undergraduate level is sufficient
mathematical background for this chapter. Some details can be found in standard sources.123
3.1
Exact differentials and state functions
In thermodynamics, one is faced with many systems of the form of the well-known Gibbs
equation:
de = T ds − P dv.
(3.1)
This is known to be an exact differential with the consequence that internal energy e is a
function of the state of the system and not the details of any process which led to the state.
As a counter-example, the work,
δw = P dv,
(3.2)
can be shown to be an inexact differential. Thus, the work depends upon the path of the
process.
1
Abbott, M. M., and Van Ness, H. C., 1972, Thermodynamics, Schaum’s Outline Series in Engineering,
McGraw-Hill, New York. See Chapter 3.
2
Borgnakke, C., and Sonntag, R. E, 2009, Fundamentals of Thermodynamics, Seventh Edition, John
Wiley, New York. See Chapters 13 and 15.
3
Vincenti, W. G., and Kruger, C. H., 1965, Introduction to Physical Gas Dynamics, John Wiley, New
York. See Chapter 3.
95
96
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Example 3.1
Show the work is not a state function.
If work were a state function, one might expect it to have the form
w = w(P, v),
provisional assumption, to be tested.
In such a case, one would have the corresponding differential form
∂w ∂w dw =
dv
+
dP.
∂v P
∂P v
Now, since dw = P dv + 0dP , one deduces that
∂w ∂v P
∂w ∂P v
(3.3)
(3.4)
=
P,
(3.5)
=
0.
(3.6)
Integrating Eq. (3.5), one finds
w = P v + f (P ),
(3.7)
where f (P ) is some function of P to be determined. Differentiating Eq. (3.7) with respect to P , one
gets
df (P )
∂w =v+
.
(3.8)
∂P v
dP
Now, use Eq. (3.6) to eliminate ∂w/∂P |v in Eq. (3.8) so as to obtain
0
=
v+
df (P )
dP
=
−v.
df (P )
,
dP
(3.9)
(3.10)
Equation (3.10) cannot be: a function of P only cannot be a function of v. So, w cannot be a state
property:
w 6= w(P, v).
(3.11)
Consider now the more general form
ψ1 dx1 + ψ2 dx2 + . . . + ψN dxN =
N
X
ψi dxi .
(3.12)
i=1
Here, ψi and xi , i = 1, . . . , N, may be thermodynamic variables. This form is known in
mathematics as a Pfaff4 differential form. As formulated, one takes at this stage
4
Johann Friedrich Pfaff, 1765-1825, German mathematician.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
97
3.1. EXACT DIFFERENTIALS AND STATE FUNCTIONS
• xi : independent thermodynamic variables,
• ψi : thermodynamic variables which are functions of xi
Now, if the differential in Eq. (3.12), when set to a differential dy, can be integrated to form
the function
y = y(x1 , x2 , . . . , xN ),
(3.13)
the differential is said to be exact. In such a case, one has
dy = ψ1 dx1 + ψ2 dx2 + . . . + ψN dxN =
N
X
ψi dxi .
(3.14)
i=1
Now, if the algebraic definition of Eq. (3.13) holds, what amounts to the definition of the
partial derivative gives the parallel result that
∂y ∂y ∂y dx1 +
dx2 + . . . +
dxN .
(3.15)
dy =
∂x1 xj ,j6=1
∂x2 xj ,j6=2
∂xN xj ,j6=N
Now, combining Eqs. (3.14) and (3.15) to eliminate dy, one gets
∂y ∂y ∂y dx1 +
dx2 + . . . +
dxN .
ψ1 dx1 + ψ2 dx2 + . . . + ψN dxN =
∂x1 xj ,j6=1
∂x2 xj ,j6=2
∂xN xj ,j6=N
(3.16)
Rearranging, one gets
!
∂y 0=
− ψ1 dx1 +
∂x1 xj ,j6=1
!
∂y − ψ2 dx2 + . . . +
∂x2 xj ,j6=2
!
∂y − ψN dxN .
∂xN xj ,j6=N
(3.17)
The variables xi , i = 1, . . . , N, are independent. Thus, their differentials, dxi , i =
1, . . . , N, are all independent in Eq. (3.17), and in general non-zero. For equality, one must
require that each of the coefficients be zero, so
∂y ∂y ∂y ψ1 =
, ψ2 =
, . . . , ψN =
.
(3.18)
∂x1 xj ,j6=1
∂x2 xj ,j6=2
∂xN xj ,j6=N
When dy is exact, one says that each of the ψi and xi are conjugate to each other.
From here on out, for notational ease, the j 6= 1, j 6= 2, . . . , j 6= N will be ignored in
the notation for the partial derivatives. It becomes especially confusing for higher order
derivatives, and is fairly obvious for all derivatives.
If y and all its derivatives are continuous and differentiable, then one has for all i =
1, . . . , N and k = 1, . . . , N that
∂2y
∂2y
=
.
(3.19)
∂xk ∂xi
∂xi ∂xk
CC BY-NC-ND. 30 March 2014, J. M. Powers.
98
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Now, from Eq. (3.18), one has
∂y ψk =
,
∂xk xj
∂y ψl =
.
∂xl xj
(3.20)
Taking the partial of the first of Eq. (3.20) with respect to xl and the second with respect
to xk , one gets
∂ψk ∂2y
∂2y
∂ψl =
=
,
.
(3.21)
∂xl xj
∂xl ∂xk
∂xk xj
∂xk ∂xl
Since by Eq. (3.19) the order of the mixed second partials does not matter, one deduces from
Eq. (3.21) that
∂ψl ∂ψk =
(3.22)
∂xl ∂xk xj
xj
This is a necessary and sufficient condition for the exact-ness of Eq. (3.12). It is a generalization of what can be found in most introductory calculus texts for functions of two
variables.
For the Gibbs equation, (3.1), de = −P dv + T ds, one has
y → e,
x1 → v,
x2 → s,
ψ1 → −P
ψ2 → T.
(3.23)
and one expects the natural, or canonical form of
e = e(v, s).
(3.24)
Here, −P is conjugate to v, and T is conjugate to s. Application of the general form of
Eq. (3.22) to the Gibbs equation (3.1) gives then
∂T ∂P =−
.
(3.25)
∂v s
∂s v
Equation (3.25) is known as a Maxwell5 relation. Moreover, specialization of Eq. (3.20) to
the Gibbs equation (3.1) gives
∂e ∂e ,
T =
.
(3.26)
−P =
∂v s
∂s v
P
If the general differential dy = N
i=1 ψi dxi is exact, one also can show
RBP
• The path integral yB − yA = A N
i=1 ψi dxi is independent of the path of the integral.
• The integral around a closed contour is zero:
I
I X
N
dy =
ψi dxi = 0.
i=1
5
James Clerk Maxwell, 1831-1879, Scottish physicist and mathematician.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.27)
3.1. EXACT DIFFERENTIALS AND STATE FUNCTIONS
99
• The function y can only be determined to within an additive constant. That is, there
is no absolute value of y; physical significance is only ascribed to differences in y.
In fact, other means, extraneous to this analysis, can be used to provide absolute
specification of key thermodynamic variables. This will be important especially for
flows with reaction.
Example 3.2
Show the heat transfer q is not a state function. Assume all processes are fully reversible. The first
law gives
de
δq
= δq − δw,
= de + δw,
(3.28)
(3.29)
= de + P dv.
(3.30)
Take now the non-canonical, although acceptable, form e = e(T, v). Then, one gets
∂e ∂e de =
dv
+
dT.
∂v T
∂T v
(3.31)
So
δq
=
=
∂e ∂e dv +
dT + P dv,
∂v ∂T v
T
∂e ∂e + P dv +
dT.
∂v T
∂T | {z v}
|
{z
}
≡M
=
(3.33)
≡N
M dv + N dT.
Now, by Eq. (3.22), for δq to be exact, one must have
∂M ∂N =
,
∂T v
∂v T
This reduces to
(3.32)
∂2e
∂P ∂ 2e
=
+
.
∂T ∂v
∂T v
∂v∂T
(3.34)
(3.35)
(3.36)
(3.37)
This can only be true if ∂P/∂T |v = 0. But this is not the case; consider an ideal gas for which
∂P/∂T |v = 0 = R/v. So, δq is not exact.
Example 3.3
Show conditions for ds to be exact in the Gibbs equation.
de
= T ds − P dv,
(3.38)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
100
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
ds
de P
+ dv,
T T
1 ∂e ∂e P
=
dv +
dT + dv,
T ∂v T
∂T
T
v
P
1 ∂e 1 ∂e +
dv +
dT.
=
T ∂v T
T
T ∂T |
| {z v}
{z
}
=
≡M
Again, invoking Eq. (3.22), one gets then
∂ 1
∂T v T
|
1 ∂e 1
1 ∂2e
− 2
+
T ∂T ∂v T ∂v T
T
1 ∂e 1
− 2
+
T ∂v T
T
(3.39)
(3.40)
(3.41)
≡N
∂e P
+
=
∂v T
T
{z
}
M
∂P P
− 2 =
∂T v T
∂P P
− 2 =
∂T
T
1 ∂e ∂ ,
∂v T T ∂T v
|
{z
}
(3.42)
N
1 ∂2e
,
T ∂v∂T
(3.43)
0.
(3.44)
v
This is the condition for an exact ds. Experiment can show if it is true. For example, for an ideal gas,
one finds from experiment that e = e(T ) and P v = RT , so one gets
0+
1R
1 RT
− 2
T v
T v
0
=
0,
(3.45)
=
0.
(3.46)
So, ds is exact for an ideal gas. In fact, the relation is verified for so many gases, ideal and non-ideal,
that one simply asserts that ds is exact, rendering s to be path-independent and a state variable.
3.2
Two independent variables
Consider a general implicit function linking three variables, x, y, z:
f (x, y, z) = 0.
(3.47)
In (x, y, z) space, this will represent a surface. If the function can be inverted, it will be
possible to write the explicit forms
x = x(y, z),
y = y(x, z),
z = z(x, y).
Differentiating the first two of the Eqs. (3.48) gives
∂x ∂x dx =
dy +
dz,
∂y z
∂z y
∂y ∂y dx +
dz
dy =
∂x z
∂z x
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.48)
(3.49)
(3.50)
101
3.2. TWO INDEPENDENT VARIABLES
Now, use Eq. (3.50) to eliminate dy in Eq. (3.49):
∂x ∂x ∂y ∂y dx =
dx +
dz +
dz,
(3.51)
∂y z ∂x z
∂z x
∂z y
{z
}
|
=dy
!
∂x ∂x ∂y ∂x ∂y dx =
+
dz,
(3.52)
1−
∂y z ∂x z
∂y z ∂z x ∂z y
!
∂x ∂y ∂x ∂y ∂x 0dx + 0dz =
dz. (3.53)
− 1 dx +
+
∂y z ∂x z
∂y z ∂z x ∂z y
{z
}
|
{z
}
|
=0
=0
Since x and y are independent, so are dx and dy, and the coefficients on each in Eq. (3.53)
must be zero. Therefore, from the coefficient on dx in Eq. (3.53)
∂x ∂y − 1 = 0,
(3.54)
∂y z ∂x z
∂x ∂y = 1,
(3.55)
∂y z ∂x z
∂x 1
(3.56)
= ∂y ,
∂y z
∂x
z
and also from the coefficient on dz in Eq. (3.53)
∂x ∂y ∂x +
= 0,
∂y z ∂z x ∂z y
∂x ∂x ∂y = −
,
∂z y
∂y z ∂z x
∂x ∂y ∂z = −1.
∂z y ∂x z ∂y x
If one now divides Eq. (3.49) by a fourth differential, dw, one gets
dx
∂x dy
∂x dz
=
+
.
dw
∂y z dw
∂z y dw
Demanding that z be held constant in Eq.
∂x ∂w z
∂x ∂w z
∂y ∂w z
∂x ∂w ∂w z ∂y z
(3.60) gives
∂x ∂y =
,
∂y z ∂w z
∂x =
,
∂y z
∂x =
.
∂y (3.57)
(3.58)
(3.59)
(3.60)
(3.61)
(3.62)
(3.63)
z
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102
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
If x = x(y, w), one then gets
∂x ∂x dx =
dy +
dw.
∂y w
∂w y
Divide now by dy while holding z constant so
∂x ∂x ∂w ∂x =
+
.
∂y z
∂y w ∂w y ∂y z
(3.64)
(3.65)
These general operations can be applied to a wide variety of thermodynamic operations.
Example 3.4
Apply Eq. (3.65) to a standard (P, v, T ) system and let
∂T ∂x =
.
∂y z
∂v s
So, T = x, v = y, and s = z. Let now e = w. So, Eq. (3.65) becomes
∂T ∂T ∂T ∂e =
+
.
∂v s
∂v e
∂e v ∂v s
Now, by definition
∂e cv =
,
∂T v
1
∂T = .
∂e v
cv
so
Now, by Eq. (3.26), one has ∂e/∂v|s = −P , so one gets
∂T ∂T P
=
− .
∂v s
∂v e cv
For an ideal gas, e = e(T ). Inverting, one gets T = T (e), and so ∂T /∂v|e = 0, thus
∂T P
=− .
∂v s
cv
(3.66)
(3.67)
(3.68)
(3.69)
(3.70)
(3.71)
For an isentropic process in a calorically perfect ideal gas, one gets
dT
dv
dT
T
=
=
=
ln
T
To
T
To
=
=
CC BY-NC-ND. 30 March 2014, J. M. Powers.
RT
P
=−
,
cv
cv v
R dv
−
,
cv v
dv
−(γ − 1) ,
v
vo
(γ − 1) ln ,
v
v γ−1
o
.
v
−
(3.72)
(3.73)
(3.74)
(3.75)
(3.76)
103
3.3. LEGENDRE TRANSFORMATIONS
3.3
Legendre transformations
The Gibbs equation (3.1), de = −P dv + T ds, is the fundamental equation of classical
thermodynamics. It is a canonical form which suggests the most natural set of variables in
which to express internal energy e are s and v:
e = e(v, s).
(3.77)
However, v and s may not be convenient for a particular problem. There may be other combinations of variables whose canonical form gives a more useful set of independent variables
for a particular problem. An example is the enthalpy:
h ≡ e + P v.
(3.78)
dh = de + P dv + vdP.
(3.79)
Differentiating the enthalpy gives
Use now Eq. (3.79) to eliminate de in the Gibbs equation, Eq. (3.1), to give
dh
| − P dv
{z − vdP} = −P dv + T ds,
(3.80)
=de
dh = T ds + vdP.
(3.81)
So, the canonical variables for h are s and P . One then expects
h = h(s, P ).
(3.82)
This exercise can be systematized with the Legendre6 transformation,7 which defines a set
of second order polynomial combinations of variables. Consider again the exact differential
Eq. (3.14):
dy = ψ1 dx1 + ψ2 dx2 + . . . + ψN dxN .
(3.83)
For N independent variables xi and N conjugate variables ψi , by definition there are 2N − 1
Legendre transformed variables:
τ1 = τ1 (ψ1 , x2 , x3 , . . . , xN ) = y − ψ1 x1 ,
τ2 = τ2 (x1 , ψ2 , x3 , . . . , xN ) = y − ψ2 x2 ,
..
.
τN = τN (x1 , x2 , x3 , . . . , ψN ) = y − ψN xN ,
(3.84)
(3.85)
(3.86)
6
Two differentiable functions f and g are said to be Legendre transformations of each other if their first
derivatives are inverse functions of each other: Df = (Dg)−1 . With some effort, not shown here, one can
prove that the Legendre transformations of this section satisfy this general condition.
7
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104
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
τ1,2 = τ1,2 (ψ1 , ψ2 , x3 , . . . , xN ) = y − ψ1 x1 − ψ2 x2 ,
τ1,3 = τ1,3 (ψ1 , x2 , ψ3 , . . . , xN ) = y − ψ1 x1 − ψ3 x3 ,
..
.
N
X
τ1,...,N = τ1,...,N (ψ1 , ψ2 , ψ3 , . . . , ψN ) = y −
ψi xi .
(3.87)
(3.88)
(3.89)
(3.90)
i=1
Each τ is a new dependent variable. Each τ has the property that when it is known as a
function of its N canonical variables, the remaining N variables from the original expression
(the xi and the conjugate ψi ) can be recovered by differentiation of τ . In general this is not
true for arbitrary transformations.
Example 3.5
Let y = y(x1 , x2 , x3 ). This has the associated differential form
dy = ψ1 dx1 + ψ2 dx2 + ψ3 dx3 .
(3.91)
Choose now a Legendre transformed variable τ1 ≡ z(ψ1 , x2 , x3 ):
z = y − ψ1 x1 .
Then
(3.92)
∂z ∂z ∂z dψ1 +
dx2 +
dx3 .
dz =
∂ψ1 x1 ,x2
∂x2 ψ1 ,x3
∂x3 ψ1 ,x2
(3.93)
Now, differentiating Eq. (3.92), one also gets
dz = dy − ψ1 dx1 − x1 dψ1 .
(3.94)
Elimination of dy in Eq. (3.94) by using Eq. (3.91) gives
dz
=
ψ dx + ψ2 dx2 + ψ3 dx3 −ψ1 dx1 − x1 dψ1 ,
|1 1
{z
}
(3.95)
=dy
=
−x1 dψ1 + ψ2 dx2 + ψ3 dx3 .
Thus, from Eq. (3.93), one gets
∂z ,
x1 = −
∂ψ1 x2 ,x3
ψ2 =
∂z ,
∂x2 ψ1 ,x3
ψ3 =
∂z .
∂x3 ψ1 ,x2
(3.96)
(3.97)
So, the original expression had three independent variables, x1 , x2 , x3 , and three conjugate variables
ψ1 , ψ2 , ψ3 . Definition of the Legendre function z with canonical variables ψ1 , x2 , and x3 allowed
determination of the remaining variables x1 , ψ2 , and ψ3 in terms of the canonical variables.
For the Gibbs equation, (3.1), de = −P dv + T ds, one has y = e, two canonical variables,
x1 = v and x2 = s, and two conjugates, ψ1 = −P and ψ2 = T . Thus, N = 2, and one can
CC BY-NC-ND. 30 March 2014, J. M. Powers.
105
3.3. LEGENDRE TRANSFORMATIONS
expect 22 − 1 = 3 Legendre transformations. They are
τ1,2
τ1 = y − ψ1 x1 = h = h(P, s) = e + P v, enthalpy,
(3.98)
τ2 = y − ψ2 x2 = a = a(v, T ) = e − T s, Helmholtz free energy, (3.99)
= y − ψ1 x1 − ψ2 x2 = g = g(P, T ) = e + P v − T s, Gibbs free energy.
(3.100)
It has already been shown for the enthalpy that dh = T ds + vdP , so that the canonical
variables are s and P . One then also has
∂h ∂h dh =
ds +
dP,
(3.101)
∂s P
∂P s
from which one deduces that
∂h T =
,
∂s P
∂h v=
.
∂P s
(3.102)
From Eq. (3.102), a second Maxwell relation can be deduced by differentiation of the first
with respect to P and the second with respect to s:
∂T ∂v =
.
(3.103)
∂P s
∂s P
The relations for Helmholtz8 and Gibbs free energies each supply additional useful relations
including two new Maxwell relations. First, consider the Helmholtz free energy
a = e − T s,
da = de − T ds − sdT,
= (−P dv + T ds) − T ds − sdT,
= −P dv − sdT.
(3.104)
(3.105)
(3.106)
(3.107)
So, the canonical variables for a are v and T . The conjugate variables are −P and −s. Thus
∂a ∂a dv +
dT.
(3.108)
da =
∂v T
∂T v
So, one gets
∂a ,
−P =
∂v T
and the consequent Maxwell relation
8
∂a −s =
.
∂T v
∂s ∂P =
.
∂T v
∂v T
(3.109)
(3.110)
Hermann von Helmholtz, 1821-1894, German physician and physicist.
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106
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
For the Gibbs free energy
g = |e +{zP v} −T s,
(3.111)
=h
= h − T s,
dg = dh − T ds − sdT,
= (T ds + vdP ) −T ds − sdT,
{z
}
|
(3.112)
(3.113)
(3.114)
=dh
= vdP − sdT.
(3.115)
So, for Gibbs free energy, the canonical variables are P and T while the conjugate variables
are v and −s. One then has g = g(P, T ), which gives
∂g ∂g dP +
dT.
(3.116)
dg =
∂P T
∂T P
So, one finds
∂g v=
,
∂P T
∂g −s =
.
∂T P
The resulting Maxwell function is then
(3.117)
∂v ∂s =−
.
∂T P
∂P T
Example 3.6
(3.118)
Canonical Form
If
h(s, P ) = cP To
P
Po
R/cP
exp
s
cP
+ (ho − cP To ) ,
(3.119)
and cP , To , R, Po , and ho are all constants, derive both thermal and caloric state equations P (v, T )
and e(v, T ).
Now, for this material
Now, since
∂h ∂s P
∂h ∂P s
=
To
=
RTo
Po
P
Po
P
Po
∂h ∂s P
∂h ∂P s
CC BY-NC-ND. 30 March 2014, J. M. Powers.
R/cP
exp
R/cP −1
s
cP
exp
,
s
cP
(3.120)
.
(3.121)
=
T,
(3.122)
=
v,
(3.123)
107
3.4. HEAT CAPACITY
one has
T
v
=
To
=
RTo
Po
P
Po
R/cP
P
Po
exp
R/cP −1
s
cP
exp
,
s
cP
(3.124)
.
(3.125)
Dividing one by the other gives
T
v
Pv
=
=
P
,
R
RT ,
(3.126)
(3.127)
which is the thermal equation of state for an ideal gas. Substituting from Eq. (3.124) into the canonical
equation for h, Eq. (3.119), one also finds for the caloric equation of state
h
h
cP T + (ho − cP To ) ,
cP (T − To ) + ho .
=
=
(3.128)
(3.129)
which is useful in itself, and shows we actually have a calorically perfect gas. Substituting in for T and
To ,
P v Po vo
+ ho .
(3.130)
−
h = cP
R
R
Using h ≡ e + P v we get
e + P v = cP
so
e
=
e
=
e
=
e
=
e
e
=
=
P v Po vo
−
R
R
+ eo + Po vo
c
P
− 1 Pv −
− 1 Po vo + uo ,
R
cR
P
− 1 (P v − Po vo ) + eo ,
cR
P
− 1 (RT − RTo ) + eo ,
R
(cP − R) (T − To ) + eo ,
c
P
(cP − (cP − cv )) (T − To ) + eo ,
cv (T − To ) + eo .
(3.131)
(3.132)
(3.133)
(3.134)
(3.135)
(3.136)
(3.137)
So, one canonical equation gives us all the information one needs. Oftentimes, it is difficult to do
a single experiment to get the canonical form.
3.4
Heat capacity
Recall that
cv =
cP =
∂e ,
∂T v
∂h .
∂T P
(3.138)
(3.139)
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108
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Then, perform operations on the Gibbs equation, Eq. (3.1):
Likewise,
de = T ds − P dv,
∂e ∂s = T
,
∂T v
∂T v
∂s cv = T
.
∂T v
dh = T ds + vdP,
∂s ∂h = T
,
∂T P
∂T P
∂s cP = T
.
∂T P
(3.140)
(3.141)
(3.142)
(3.143)
(3.144)
(3.145)
One finds further useful relations by operating on the Gibbs equation, Eq. (3.1):
So, one can then say
For an ideal gas, one has
de = T ds − P dv,
∂s ∂e = T
− P,
∂v T
∂v T
∂P − P.
= T
∂T v
e = e(T, v),
∂e ∂e de =
dT +
dv,
∂T v
∂v T
∂P − P dv.
= cv dT + T
∂T v
∂e RT
∂P R
−
,
= T
−P = T
∂v T
∂T v
v
v
= 0.
(3.146)
(3.147)
(3.148)
(3.149)
(3.150)
(3.151)
(3.152)
(3.153)
Consequently, e is not a function of v for an ideal gas, so e = e(T ) alone. Since h = e + P v
for an ideal gas reduces to h = e + RT
h = e(T ) + RT = h(T ).
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.154)
109
3.4. HEAT CAPACITY
Now, return to general equations of state. With s = s(T, v) or s = s(T, P ), one gets
∂s ∂s dT +
dv,
(3.155)
ds =
∂T v
∂v T
∂s ∂s ds =
dT +
dP.
(3.156)
∂T P
∂P T
Now, using Eqs. (3.103,3.118,3.142,3.145) one gets
cv
∂P dv,
ds =
dT +
T
∂T v
cP
∂v ds =
dP.
dT −
T
∂T P
(3.157)
(3.158)
Subtracting one from the other, one finds
∂v ∂P cv − cP
dv +
dP,
dT +
0 =
T
∂T v
∂T P
∂v ∂P dv + T
dP.
(cP − cv )dT = T
∂T v
∂T P
(3.159)
(3.160)
Now, divide both sides by dT and hold either P or v constant. In either case, one gets
∂P ∂v cP − cv = T
.
(3.161)
∂T v ∂T P
Example 3.7
So
For an ideal gas find cP − cv . For the ideal gas, P v = RT , one has
∂P ∂v R
R
= ,
= .
∂T v
v
∂T P
P
cP − cv
=
=
=
RR
,
vP
R2
,
T
RT
R.
T
(3.162)
(3.163)
(3.164)
(3.165)
This holds even if the ideal gas is calorically imperfect. That is
cP (T ) − cv (T ) = R.
(3.166)
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110
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
For the ratio of specific heats for a general material, one can use Eqs. (3.142) and (3.145)
to get
∂s T ∂T
cP
P .
γ=
(3.167)
=
∂s cv
T
∂T v
Then apply Eq. (3.56) to get
∂s ∂T .
∂T P ∂s v
γ =
Then apply Eq. (3.58) to get
∂s ∂P ∂T ∂v γ =
−
−
,
∂P T ∂T s
∂v s ∂s T
∂P ∂T ∂v ∂s ,
=
∂s T ∂P T
∂T s ∂v s
∂v ∂P =
∂P ∂v T
(3.168)
(3.169)
(3.170)
(3.171)
s
The first term can be obtained from P − v − T data. The second term is related to the
isentropic sound speed of the material, which is also a measurable quantity.
Example 3.8
For a calorically perfect ideal gas with gas constant R and specific heat at constant volume cv , find
expressions for the thermodynamic variable s and thermodynamic potentials e, h, a, and g, as functions
of T and P .
First, get the entropy:
de =
T ds =
T ds − P dv,
de + P dv,
(3.172)
(3.173)
T ds =
cv dT + P dv,
dT
P
cv
+
dv,
T
T
dv
dT
+R ,
cv
T
vZ
Z
dv
dT
+ R ,
cv
T
v
T
v
cv ln
+ R ln ,
T0
v0
T
R
RT /P
ln
+
ln
,
T0
cv RT0 /P0
R/cv
T
T P0
ln
,
+ ln
T0
T0 P
(3.174)
ds
=
=
Z
ds
=
s − s0
=
s − s0
cv
=
=
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.175)
(3.176)
(3.177)
(3.178)
(3.179)
(3.180)
111
3.5. VAN DER WAALS GAS
=
ln
=
ln
=
s
=
ln
T
T0
T
T0
T
T0
1+R/cv
1+(cp −cv )/cv
γ
s0 + cv ln
+ ln
T
T0
P0
P
γ
P0
P
R/cv
+ ln
P0
P
+ ln
γ−1
+ cv ln
,
(3.181)
(cp −cv )/cv
,
(3.182)
,
(3.183)
P0
P
γ−1
.
(3.184)
Now, for the calorically perfect ideal gas, one has
e = eo + cv (T − To ).
(3.185)
For the enthalpy, one gets
h
=
e + P v,
(3.186)
=
=
e + RT,
eo + cv (T − To ) + RT,
(3.187)
(3.188)
=
=
eo + cv (T − To ) + RT + RTo − RTo ,
eo + RTo +cv (T − To ) + R(T − To ),
| {z }
(3.189)
(3.190)
=ho
=
ho + (cv + R)(T − To ),
| {z }
(3.191)
=cp
=
ho + cp (T − To ).
(3.192)
For the Helmholtz free energy, one gets
a
= e − T s,
= eo + cv (T − To ) − T
s0 + cv ln
s0 + cv ln
T
T0
γ
T
T0
γ
+ cv ln
+ cv ln
P0
P
γ−1 !
P0
P
γ−1 !
(3.193)
.
(3.194)
For the Gibbs free energy, one gets
g
3.5
=
h − T s,
=
ho + cp (T − To ) − T
(3.195)
.
(3.196)
Van der Waals gas
A van der Waals9 gas is a common model for a non-ideal gas. It can capture some of the
behavior of a gas as it approaches the vapor dome. Its form is
P (T, v) =
9
a
RT
− 2,
v−b v
(3.197)
Johannes Diderik van der Waals, 1837-1923, Dutch thermodynamicist.
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112
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
where b accounts for the finite volume of the molecules, and a accounts for intermolecular
forces.
Example 3.9
Find a general expression for e(T, v) if
RT
a
− .
v − b v2
P (T, v) =
(3.198)
Proceed as before. First, we have
∂e ∂e dT +
dv,
de =
∂T v
∂v T
recalling from Eqs. (3.138) and (3.148) that
∂e ∂e ∂P =
c
,
=
T
− P.
v
∂T v
∂v T
∂T v
Now, for the van der Waals gas, we have
∂P =
∂T v
∂P T
−P =
∂T v
=
R
,
v−b
RT
− P,
v−b
RT
a
a
RT
−
− 2 = 2.
v−b
v−b v
v
(3.199)
(3.200)
(3.201)
(3.202)
(3.203)
So, we have
∂e ∂v T
a
,
v2
a
e(T, v) = − + f (T ).
v
=
(3.204)
(3.205)
Here, f (T ) is some as-of-yet arbitrary function of T . To evaluate f (T ), take the derivative with respect
to T holding v constant:
∂e df
= cv .
(3.206)
=
∂T v
dT
Since f is a function of T at most, here cv can be a function of T at most, so we allow cv = cv (T ).
Integrating, we find f (T ) as
Z
f (T ) = C +
T
cv (T̂ )dT̂ ,
(3.207)
To
where C is an integration constant. Thus, e is
e(T, v) = C +
Z
T
To
CC BY-NC-ND. 30 March 2014, J. M. Powers.
a
cv (T̂ )dT̂ − .
v
(3.208)
113
3.6. REDLICH-KWONG GAS
Taking C = eo + a/vo , we get
e(T, v) = eo +
Z
T
cv (T̂ )dT̂ + a
To
1
1
−
vo
v
.
(3.209)
We also find
1
1
cv (T̂ )dT̂ + a
h = e + P v = eo +
+ P v,
−
vo
v
To
Z T
RT v
1
1
a
cv (T̂ )dT̂ + a
h(T, v) = eo +
+
−
− .
vo
v
v−b v
To
Z
3.6
T
(3.210)
(3.211)
Redlich-Kwong gas
The Redlich10 -Kwong equation of state11 is
P =
RT
a
−
.
v − b v(v + b)T 1/2
(3.212)
It is modestly more accurate than the van der Waals equation in predicting material behavior.
Example 3.10
For the case in which b = 0, find an expression for e(T, v) consistent with the Redlich-Kwong state
equation.
Here, the equation of state is now
P =
a
RT
− 2 1/2 .
v
v T
Proceeding as before, we have
∂e ∂P = T
− P,
∂v T
∂T v
a
a
RT
R
+ 2 3/2 −
− 2 1/2 ,
= T
v
v
2v T
v T
3a
=
.
2v 2 T 1/2
(3.213)
(3.214)
(3.215)
(3.216)
Integrating, we find
e(T, v) = −
3a
+ f (T ).
2vT 1/2
(3.217)
10
Otto Redlich, 1896-1978, Austrian chemical engineer.
Redlich, O., and Kwong, J. N. S., 1949, “On the Thermodynamics of Solutions. V. An Equation of
State. Fugacities of Gaseous Solutions,” Chemical Reviews, 44(1): 233-244.
11
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114
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Here, f (T ) is a yet-to-be-specified function of temperature only. Now, the specific heat is found by the
temperature derivative of e:
∂e 3a
df
.
(3.218)
cv (T, v) =
=
+
∂T v
4vT 3/2 dT
Obviously, for this material, cv is a function of both T and v.
Let us define cvo (T ) via
df
≡ cvo (T ).
dT
(3.219)
Z
(3.220)
Integrating, then one gets
f (T ) = C +
T
cvo (T̂ ) dT̂ .
To
1/2
Let us take C = eo +3a/2/vo/To . Thus, we arrive at the following expressions for cv (T, v) and e(T, v):
3a
cv (T, v) = cvo (T ) +
,
4vT 3/2
Z T
3a
cvo (T̂ ) dT̂ +
e(T, v) = eo +
2
To
3.7
(3.221)
1
1/2
vo To
−
1
vT 1/2
!
.
(3.222)
Mixtures with variable composition
Consider now mixtures of N species. The focus here will be on extensive properties and
molarPproperties. Assume that each species has ni moles, and the total number of moles is
n= N
i=1 ni . Now, one might expect the extensive energy to be a function of the entropy,
the volume, and the number of moles of each species:
E = E(V, S, ni ).
(3.223)
The extensive version of the Gibbs law in which all of the ni are held constant is
dE = −P dV + T dS.
Thus
∂E = −P,
∂V S,ni
∂E = T.
∂S V,ni
(3.224)
(3.225)
In general, since E = E(S, V, ni ), one should expect, for systems in which the ni are allowed
to change that
N
X
∂E ∂E ∂E dV +
dS +
dni .
(3.226)
dE =
∂V S,ni
∂S V,ni
∂ni S,V,nj
i=1
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115
3.7. MIXTURES WITH VARIABLE COMPOSITION
Defining the new thermodynamic property, the chemical potential µi , as
∂E ,
µi ≡
∂ni S,V,nj
(3.227)
one has the important Gibbs equation for multicomponent systems:
N
X
dE = −P dV + T dS +
µi dni .
(3.228)
i=1
Obviously, by its definition, µi is on a per mole basis, so it is given the appropriate overline
notation. In Eq. (3.228), the independent variables and their conjugates are
x1
x2
x3
x4
xN +2
=
=
=
=
V,
S,
n1 ,
n2 ,
..
.
= nN ,
ψ1 = −P,
ψ2 = T,
ψ3 = µ1 ,
ψ4 = µ2 ,
..
.
ψN +2 = µN .
(3.229)
(3.230)
(3.231)
(3.232)
(3.233)
Equation (3.228) has 2N +1 −1 Legendre functions. Three are in wide usage: the extensive
analog to those earlier found. They are
H = E + P V,
A = E − T S,
G = E + P V − T S.
(3.234)
(3.235)
(3.236)
A set of non-traditional, but perfectly acceptable additional Legendre functions would be
formed from E − µ1 n1 . Another set would be formed from E + P V − µ2 n2 . There are
many more, but one in particular is sometimes noted in the literature: the so-called grand
potential, Ω. The grand potential is defined as
Ω ≡ E − TS −
N
X
µi ni .
(3.237)
i=1
Differentiating each defined Legendre function, Eqs. (3.234-3.237), and combining with
Eq. (3.228), one finds
dH = T dS + V dP +
N
X
µi dni ,
i=1
N
X
dA = −SdT − P dV +
µi dni ,
(3.238)
(3.239)
i=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
116
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
dG = −SdT + V dP +
dΩ = −P dV − SdT −
N
X
i=1
N
X
µi dni ,
(3.240)
ni dµi .
(3.241)
i=1
Thus, canonical variables for H are H = H(S, P, ni ). One finds a similar set of relations as
before from each of the differential forms:
∂H ∂E =
,
(3.242)
T =
∂S V,ni
∂S P,ni
∂A ∂Ω ∂E =−
=−
,
(3.243)
P = −
∂V S,ni
∂V T,ni
∂V T,µi
∂H ∂G V =
=
,
(3.244)
∂P S,ni
∂P T,ni
∂G ∂Ω ∂A =−
=−
,
(3.245)
S = −
∂T V,ni
∂T P,ni
∂T V,µi
∂Ω ,
(3.246)
ni = −
∂µ i V,T,µj
∂H ∂A ∂G ∂E =
=
=
.
µi =
∂ni S,V,nj
∂ni S,P,nj
∂ni T,V,nj
∂ni T,P,nj
(3.247)
Each of these induces a corresponding Maxwell relation, obtained by cross differentiation.
These are
∂T ∂P = −
,
(3.248)
∂V S,ni
∂S V,ni
∂V ∂T =
,
(3.249)
∂P S,ni
∂S P,ni
∂P ∂S =
,
(3.250)
∂T V,ni
∂V T,ni
∂V ∂S = −
,
(3.251)
∂T P,ni
∂P T,ni
∂µi ∂S = −
,
(3.252)
∂T P,nj
∂ni V,nj
∂V ∂µi =
,
(3.253)
∂P T,nj
∂ni V,nj
∂µl ∂µk =
,
(3.254)
∂nk T,P,nj
∂nl T,P,nj
CC BY-NC-ND. 30 March 2014, J. M. Powers.
117
3.8. PARTIAL MOLAR PROPERTIES
∂S =
∂V T,µi
∂ni =
∂µk V,T,µj ,j6=k
3.8
∂P ,
∂T V,µi
∂nk ∂µ (3.255)
.
(3.256)
i V,T,µj ,j6=i
Partial molar properties
3.8.1
Homogeneous functions
In mathematics, a homogeneous function f (x1 , . . . , xN ) of order m is one such that
f (λx1 , . . . , λxN ) = λm f (x1 , . . . , xN ).
(3.257)
f (λx1 , . . . , λxN ) = λf (x1 , . . . , xN ).
(3.258)
If m = 1, one has
Thermodynamic variables are examples of homogeneous functions.
3.8.2
Gibbs free energy
Consider an extensive property, such as the Gibbs free energy G. One has the canonical
form
G = G(T, P, n1 , n2 , . . . , nN ).
(3.259)
One would like to show that if each of the mole numbers ni is increased by a common factor,
say λ, with T and P constant, that G increases by the same factor λ:
λG(T, P, n1 , n2 , . . . , nN ) = G(T, P, λn1 , λn2 , . . . , λnN ).
(3.260)
Differentiate both sides of Eq. (3.260) with respect to λ, while holding P , T , and nj constant,
to get
G(T, P, n1 , n2 , . . . , nN ) =
d(λn1 )
d(λn2 )
d(λnN )
∂G ∂G ∂G +
+ ...+
,
∂(λn1 ) nj ,P,T dλ
∂(λn2 ) nj ,P,T dλ
∂(λnN ) nj ,P,T dλ
∂G ∂G ∂G =
n1 +
n2 + . . . +
nN ,
∂(λn1 ) ∂(λn2 ) ∂(λnN ) nj ,P,T
nj ,P,T
(3.261)
(3.262)
nj ,P,T
This must hold for all λ, including λ = 1, so one requires
∂G ∂G ∂G n1 +
n2 + . . . +
nN ,
G(T, P, n1, n2 , . . . , nN ) =
∂n1 nj ,P,T
∂n2 nj ,P,T
∂nN nj ,P,T
=
N
X
i=1
∂G ni .
∂ni nj ,P,T
(3.263)
(3.264)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
118
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Recall now from p. 77 the definition partial molar property, the derivative of an extensive
variable with respect to species ni holding nj , i 6= j, T , and P constant. Because the result
has units which are per unit mole, an overline superscript is utilized. The partial molar
Gibbs free energy of species i, g i is then
∂G ,
(3.265)
gi ≡
∂ni nj ,P,T
so that Eq. (3.264) becomes
G=
N
X
g i ni .
(3.266)
i=1
Using the definition of chemical potential, Eq. (3.247), one also notes then that
G(T, P, n2 , n2 , . . . , nN ) =
N
X
µi ni .
(3.267)
i=1
The temperature and pressure dependence of G must lie entirely within µi (T, P, nj ), which
one notes is also allowed to be a function of nj as well. Consequently, one also sees that the
Gibbs free energy per unit mole of species i is the chemical potential of that species:
g i = µi .
(3.268)
Using Eq. (3.266) to eliminate G in Eq. (3.236), one recovers an equation for the energy:
E = −P V + T S +
3.8.3
N
X
µi ni .
(3.269)
i=1
Other properties
A similar result also holds for any other extensive property such as V , E, H, A, or S. One
can also show that
N
X
∂V ,
(3.270)
V =
ni
∂n
i
n
,A,T
j
i=1
N
X
∂E E =
ni
(3.271)
∂n
i
n
,V,S
j
i=1
N
X ∂H ,
(3.272)
H =
ni
∂n
i
n
,P,S
j
i=1
N
X ∂A ,
(3.273)
A =
ni
∂n
i
nj ,T,V
i=1
N
X
∂S S =
ni
.
(3.274)
∂ni nj ,E,T
i=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
119
3.8. PARTIAL MOLAR PROPERTIES
Note that these expressions do not formally involve partial molar properties since P and T
are not constant.
Take now the appropriate partial molar derivatives of G for an ideal mixture of ideal
gases to get some useful relations:
G = H − T S,
∂H ∂S ∂G =
−T
.
∂ni T,P,nj
∂ni T,P,nj
∂ni T,P,nj
(3.275)
(3.276)
Now, from the definition of an ideal mixture hi = hi (T, P ), so one has
H =
N
X
nk hk (T, P ),
k=1
∂
∂H =
∂ni T,P,nj
∂ni
=
N
X
k=1
∂n
|{z}i
!
nk hk (T, P ) ,
k=1
N
X
∂nk
(3.277)
hk (T, P ),
(3.278)
(3.279)
=δik
=
N
X
δik hk (T, P ),
(3.280)
k=1
= hi (T, P ).
(3.281)
Here, the Kronecker delta function δki has been again used. Now, for an ideal gas one further
has hi = hi (T ). The analysis is more complicated for the entropy, in which
N
X
Pk
o
S =
nk sT,k − R ln
,
(3.282)
Po
k=1
N
X
yk P
o
,
(3.283)
nk sT,k − R ln
=
Po
k=1
!!
N
X
P
n
k
=
nk soT,k − R ln
− R ln PN
,
(3.284)
Po
n
q
q=1
k=1
!
N
N
X
X
P
n
k
=
nk soT,k − R ln
,
(3.285)
−R
nk ln PN
P
n
o
q
q=1
k=1
k=1
N
∂S ∂ X
P
=
nk soT,k − R ln
∂ni T,P,nj
∂ni
Po
k=1
!!
N
X
nk
∂ −R
nk ln PN
,
(3.286)
∂ni T,P,nj k=1
q=1 nq
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120
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
N
X
∂nk
P
o
=
sT,k − R ln
∂ni
Po
k=1 |{z}
=δik
!!
N
X
∂ nk
−R
nk ln PN
,
∂ni T,P,nj k=1
q=1 nq
N
X
∂ P
o
nk ln
=
−R
sT,i − R ln
Po
∂ni T,P,nj k=1
(3.287)
nk
PN
q=1
nq
!!
. (3.288)
Evaluation of the final term on the right side requires closer examination, and in fact, after
tedious but straightforward analysis, yields a simple result which can easily be verified by
direct calculation:
!!
!
N
X
∂ nk
ni
nk ln PN
= ln PN
.
(3.289)
∂ni T,P,nj k=1
n
n
q
q
q=1
q=1
So, the partial molar entropy is in fact
!
∂S P
n
i
= soT,i − R ln
,
− R ln PN
∂ni T,P,nj
Po
q=1 nq
P
o
= sT,i − R ln
− R ln yi ,
Po
Pi
o
= sT,i − R ln
,
Po
= si .
(3.290)
(3.291)
(3.292)
(3.293)
Thus, one can in fact claim for the ideal mixture of ideal gases that
g i = hi − T si .
3.8.4
(3.294)
Relation between mixture and partial molar properties
A simple analysis shows how the partial molar property for an individual species is related
to the partial molar property for the mixture. Consider, for example, the Gibbs free energy.
The mixture-averaged Gibbs free energy per unit mole is
g=
G
.
n
Now, take a partial molar derivative and analyze to get
1 ∂G G ∂n ∂g =
−
,
∂ni T,P,nj
n ∂ni T,P,nj n2 ∂ni T,P,nj
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.295)
(3.296)
121
3.9. FROZEN SOUND SPEED
1 ∂G =
n ∂ni T,P,nj
1 ∂G =
n ∂ni T,P,nj
1 ∂G =
n ∂ni T,P,nj
1 ∂G =
n ∂ni T,P,nj
1
g
gi − .
=
n
n
Multiplying by n and rearranging, one gets
!
N
X
G ∂ nk ,
− 2
n ∂ni T,P,nj k=1
N
G X ∂nk − 2
,
n k=1 ∂ni T,P,nj
N
GX
δik ,
− 2
n k=1
−
G
,
n2
(3.297)
(3.298)
(3.299)
(3.300)
(3.301)
∂g .
gi = g + n
∂ni T,P,nj
(3.302)
A similar result holds for other properties.
3.9
Frozen sound speed
Let us develop a relation for what is known as the “frozen sound speed,” denoted as c. This
quantity is a thermodynamic property defined by the relationship
∂P 2
c ≡
.
(3.303)
∂ρ s,ni
Later in Sec. 10.1.7 we shall see that this quantity describes the speed of acoustic waves, but
at this point let us simply treat it as a problem in thermodynamics. The quantity is called
“frozen” because the derivative is performed with all of the species mole numbers frozen at
a constant value. Note that since v = 1/ρ,
dv ∂P 1 ∂P ∂P 2 ∂P 2
=
=
−
=
−v
.
(3.304)
c =
∂ρ s,ni dρ ∂v s,ni
ρ2 ∂v s,ni
∂v s,ni
So, we can say
∂P (ρc) = −
.
∂v s,ni
2
(3.305)
Now, we know that the caloric equation of state for a mixture of ideal gases takes the
form, from Eq. (2.162):
e=
N
X
Yi ei (T ).
(3.306)
i=1
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122
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Employing the ideal gas law, we could say instead
e=
N
X
.
P
ρR
P
ρR
ei
Yi ei
i=1
(3.307)
In terms of number of moles, ni , we could say
e=
N
X
Mi ni
i=1
ρV
| {z }
.
(3.308)
=Yi
We can in fact generalize, and simply consider caloric equations of state of the form
e = e(P, ρ, ni ),
(3.309)
or in terms of the specific volume v = 1/ρ,
e = e(P, v, ni ).
(3.310)
These last two forms can be useful when temperature is unavailable, as can be the case for
some materials.
Now, reconsider the Gibbs equation for a multicomponent system, Eq. (3.228):
dE = −P dV + T dS +
N
X
µi dni .
(3.311)
i=1
Let us scale Eq. (3.311) by a fixed mass m, recalling that e = E/m, v = V /m, and s = S/m,
so as to obtain
N
X
µi
de = −P dv + T ds +
dni .
(3.312)
m
i=1
Considering Eq. (3.312) with dv = 0 and dni = 0, and scaling by dP , we can say
∂s ∂e =T
.
∂P v,ni
∂P v,ni
Considering Eq. (3.312) with dni = 0, and dP = 0 and scaling by dv, we can say
∂e ∂s = −P + T
.
∂v P,ni
∂v P,ni
Applying now Eq. (3.59), we find
∂P ∂s ∂v = −1.
∂v s,ni ∂P v,ni ∂s P,ni
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.313)
(3.314)
(3.315)
123
3.9. FROZEN SOUND SPEED
So, we get
∂P =−
∂v s,ni
∂s ∂v P,ni
.
∂s ∂P v,ni
Substituting Eq. (3.316) into Eq. (3.305) we find
∂s ∂v
P,n
(ρc)2 = ∂s i .
(3.316)
(3.317)
∂P v,ni
Substituting Eqs. (3.313,3.314) into Eq. (3.317) we get
1
∂e P
+
T
∂v P,ni
,
(ρc)2 =
1 ∂e =
P
So, the sound speed c is
T ∂P v,ni
∂e + ∂v
P,ni .
∂e ∂P v,ni
v
u
∂e u P + ∂v
P,ni
c = v t ∂e .
(3.318)
(3.319)
(3.320)
∂P v,ni
Eq. (3.320) is useful because we can identify the frozen sound speed from data for e, P , and
v alone. If
∂e ∂e > −P,
and
> 0,
(3.321)
∂v P,ni
∂P v,ni
we will have a real sound speed. Other combinations are possible as well, but not observed
in nature.
Example 3.11
For an ideal mixture of calorically perfect ideal gases, find the frozen sound speed.
For a calorically perfect ideal gas, we have
ei
= ei (T ) = eo + cvi (T − To ),
Pv
= eo + cvi
− To .
R
(3.322)
(3.323)
and
e =
N
X
Yi ei ,
(3.324)
i=1
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124
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
N
X
mi
=
i=1
m
ei ,
N
X
ni M i
=
i=1
m
(3.325)
ei ,
(3.326)
N
X
ni M i
Pv
eo + cvi
.
− To
m
R
i=1
=
(3.327)
Thus, we get the two relevant partial derivatives
∂e ∂v P,ni
∂e ∂P v,ni
=
N
X
ni M i
i=1
=
m
N
X
ni M i
i=1
m
N
cvi
cvi
P
P X
P cv
=
,
Yi cvi =
R
R i=1
R
N
vcv
v
v X
Yi cvi =
=
.
R
R i=1
R
Now, substitute Eqs. (3.328, 3.329) into Eq. (3.320) to get
s
P + PRcv
c = v
,
vcv
=
=
=
=
=
=
=
3.10
s
s
(3.328)
(3.329)
(3.330)
R
Pv +
cv
R
P vcv
R
RT +
s
RT
r
RT
s
RT
cv
R
,
RT cv
R
1+
cv
R
(3.331)
,
(3.332)
,
(3.333)
R + cv
,
cv
(3.334)
(cP − cv ) + cv
,
cv
(3.335)
cv
R
r
cP
RT ,
cv
p
γRT .
(3.336)
(3.337)
Irreversibility in a closed multicomponent system
Consider a thermodynamic system closed to mass exchanges with its surroundings coming
into equilibrium. Allow the system to be exchanging work and heat with its surroundings.
Assume the temperature difference between the system and its surroundings is so small that
CC BY-NC-ND. 30 March 2014, J. M. Powers.
3.10. IRREVERSIBILITY IN A CLOSED MULTICOMPONENT SYSTEM
125
both can be considered to be at temperature T . If δQ is introduced into the system, then
the surroundings suffer a loss of entropy:
dSsurr = −
δQ
.
T
(3.338)
The system’s entropy S can change via this heat transfer, as well as via other internal
irreversible processes, such as internal chemical reaction. The second law of thermodynamics
requires that the entropy change of the universe be positive semi-definite:
dS + dSsurr ≥ 0.
(3.339)
Eliminating dSsurr , one requires for the system that
dS ≥
δQ
.
T
(3.340)
Consider temporarily the assumption that the work and heat transfer are both reversible.
Thus, the irreversibilities must be associated with internal chemical reaction. Now, the first
law for the entire system gives
dE = δQ − δW,
= δQ − P dV,
δQ = dE + P dV.
(3.341)
(3.342)
(3.343)
Note because the system is closed, there can be no species entering or exiting, and so there
is no change dE attributable to dni . While within the system the dni may not be 0, the net
contribution to the change in total internal energy is zero. A non-zero dni within the system
simply re-partitions a fixed amount of total energy from one species to another. Substituting
Eq. (3.343) into Eq. (3.340) to eliminate δQ, one gets
dS ≥
1
(dE + P dV ),
{z
}
T |
(3.344)
=δQ
T dS − dE − P dV
dE − T dS + P dV
≥ 0,
≤ 0.
(3.345)
(3.346)
Eq. (3.346) involves properties only and need not require assumptions of reversibility for
processes in its derivation. In special cases, it reduces to simpler forms.
For processes which are isentropic and isochoric, the second law expression, Eq. (3.346),
reduces to
dE|S,V ≤ 0.
(3.347)
For processes which are isoenergetic and isochoric, the second law expression, Eq. (3.346),
reduces to
dS|E,V ≥ 0.
(3.348)
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126
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Now, using Eq. (3.228) to eliminate dS in Eq. (3.348), one can express the second law as
!
N
P
1X
1
≥ 0,
(3.349)
dE + dV −
µi dni T
T
T i=1
{z
}
|
=dS
E,V
−
|
1
T
N
X
i=1
{z
µi dni ≥ 0.
irreversibility
(3.350)
}
The irreversibility associated with the internal chemical reaction must be the left side of
Eq. (3.350).
Now, while most standard texts focusing on equilibrium thermodynamics go to great
lengths to avoid the introduction of time, it really belongs in a discussion describing the
approach to equilibrium. One can divide Eq. (3.350) by a positive time increment dt to get
−
N
1 X dni
≥ 0.
µ
T i=1 i dt
(3.351)
Since T ≥ 0, one can multiply Eq. (3.351) by −T to get
N
X
i=1
µi
dni
≤ 0.
dt
(3.352)
This will hold if a model for dni /dt is employed which guarantees that the left side of
Eq. (3.352) is negative semi-definite. One will expect then for dni /dt to be related to the
chemical potentials µi .
Elimination of dE in Eq. (3.346) in favor of dH from dH = dE + P dV + V dP gives
dH
| − P dV
{z − V dP} −T dS + P dV
≤ 0,
(3.353)
dH − V dP − T dS ≤ 0.
(3.354)
=dE
Thus, one finds for isobaric, isentropic equilibration that
dH|P,S ≤ 0.
(3.355)
For the Helmholtz and Gibbs free energies, one analogously finds
dA|T,V ≤ 0,
dG|T,P ≤ 0.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.356)
(3.357)
3.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM
127
The expression of the second law in terms of dG is especially useful as it may be easy in an
experiment to control so that P and T are constant. This is especially true in an isobaric
phase change, in which the temperature is guaranteed to be constant as well.
Now, one has Eq. (3.266):
G =
=
N
X
i=1
N
X
ni g i ,
(3.358)
ni µi .
(3.359)
i=1
P
One also has from Eq. (3.240): dG = −SdT + V dP + N
i=1 µi dni , holding T and P constant
that
N
X
µi dni .
(3.360)
dG|T,P =
i=1
Here, the dni are associated entirely with internal chemical reactions. Substituting Eq. (3.360)
into Eq. (3.357), one gets the important version of the second law which holds that
dG|T,P =
N
X
i=1
µi dni ≤ 0.
(3.361)
In terms of time rates of change, one can divide Eq. (3.361) by a positive time increment
dt > 0 to get
N
X
∂G dni
=
≤ 0.
(3.362)
µi
∂t T,P
dt
i=1
3.11
Equilibrium in a two-component system
A major task of non-equilibrium thermodynamics is to find a functional form for dni /dt
which guarantees satisfaction of the second law, Eq. (3.362) and gives predictions which
agree with experiment. This will be discussed in more detail in the following chapter on
thermochemistry. At this point, some simple examples will be given in which a naı̈ve but
useful functional form for dni /dt is posed which leads at least to predictions of the correct
equilibrium values. A much better model which gives the correct dynamics in the time
domain of the system as it approaches equilibrium will be presented in the following Chapter 4
on thermochemistry.
3.11.1
Phase equilibrium
Here, consider two examples describing systems in phase equilibrium.
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128
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Example 3.12
Consider an equilibrium two-phase mixture of liquid and vapor H2 O at T = 100 C, x = 0.5. Use
the steam tables to check if equilibrium properties are satisfied.
In a two-phase gas liquid mixture one can expect the following reaction:
H2 O(l) ⇌ H2 O(g) .
(3.363)
That is, one mole of liquid, in the forward phase change, evaporates to form one mole of gas. In the
reverse phase change, one mole of gas condenses to form one mole of liquid.
Because T is fixed at 100◦C and the material is a two phase mixture, the pressure is also fixed at a
constant. Here, there are two phases at saturation; g for gas and l for liquid. Equation (3.361) reduces
to
µl dnl + µg dng ≤ 0.
(3.364)
Now, for the pure H2 O if a loss of moles from one phase must be compensated by the addition to
another. So, one must have
dnl + dng = 0.
(3.365)
Hence
dng = −dnl .
(3.366)
So, Eq. (3.364), using Eq. (3.366) becomes
µl dnl − µg dnl
≤ 0,
dnl (µl − µg ) ≤ 0.
(3.367)
(3.368)
At this stage of the analysis, most texts, grounded in equilibrium thermodynamics, assert that µl = µg ,
ignoring the fact that they could be different but dnl could be zero. That approach will not be taken
here. Instead divide Eq. (3.368) by a positive time increment, dt ≥ 0 to write the second law as
dnl
(µ − µg ) ≤ 0.
dt l
One convenient, albeit naı̈ve, way to guarantee second law satisfaction is to let
(3.369)
dnl
κ ≥ 0,
convenient, but naı̈ve model
(3.370)
= −κ(µl − µg ),
dt
Here, κ is some positive semi-definite scalar rate constant which dictates the time scale of approach to
equilibrium. Note that Eq. (3.370) is just a hypothesized model. It has no experimental verification;
one can imagine that other more complex models exist which both agree with experiment and satisfy
the second law. For the purposes of the present argument, however, Eq. (3.370) will suffice. With this
assumption, the second law reduces to
−κ(µl − µg )2 ≤ 0,
κ ≥ 0,
(3.371)
which is always true.
Eq. (3.370) has two important consequences:
• Differences in chemical potential drive changes in the number of moles.
• The number of moles of liquid, nl , increases when the chemical potential of the liquid is less than that
of the gas, µl < µg . That is to say, when the liquid has a lower chemical potential than the gas, the
gas is driven towards the phase with the lower potential. Because such a phase change is isobaric and
isothermal, the Gibbs free energy is the appropriate variable to consider, and one takes µ = g. When
this is so, the Gibbs free energy of the mixture, G = nl µl + ng µg is being driven to a lower value. So,
when dG = 0, the system has a minimum G.
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129
3.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM
• The system is in equilibrium when the chemical potentials of liquid and gas are equal: µl = µg .
The chemical potentials, and hence the molar specific Gibbs free energies must be the same for
each constituent of the binary mixture at the phase equilibrium. That is
gl = gg .
(3.372)
Now, since both the liquid and gas have the same molecular mass, one also has the mass specific Gibbs
free energies equal at phase equilibrium:
gl = gg .
(3.373)
This can be verified from the steam tables, using the definition g = h − T s. From standard thermodynamic tables
kJ
kJ
kJ
gl = hl − T sl = 419.02
− ((100 + 273.15) K) 1.3068
,
(3.374)
= −68.6
kg
kg K
kg
kJ
kJ
kJ
= −68.4
− ((100 + 273.15) K) 7.3548
.
(3.375)
gg = hg − T sg = 2676.05
kg
kg K
kg
The two values are essentially the same; the difference is likely due to table inaccuracies.
3.11.2
Chemical equilibrium: introduction
Here, consider two examples which identify the equilibrium state of a chemically reactive
system.
3.11.2.1
Isothermal, isochoric system
The simplest system to consider is isothermal and isochoric. The isochoric assumption
implies there is no work in coming to equilibrium.
Example 3.13
At high temperatures, collisions between diatomic nitrogen molecules induce the production of
monatomic nitrogen molecules. The chemical reaction can be described by the model
N2 + N2 ⇌ 2N + N2 .
(3.376)
Here, one of the N2 molecules behaves as an inert third body. An N2 molecule has to collide with
something, to induce the reaction. Some authors leave out the third body and write instead N2 ⇌ 2N ,
but this does not reflect the true physics as well. The inert third body is especially important when
the time scales of reaction are considered. It plays no role in equilibrium chemistry.
Consider 1 kmole of N2 and 0 kmole of N at a pressure of 100 kP a and a temperature of 6000 K.
Find the equilibrium concentrations of N and N2 if the equilibration process is isothermal and isochoric.
Also perform a first and second law analysis with the aid of the ideal gas tables.
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130
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
The ideal gas law can give the volume.
P1
V
nN2 RT
,
V
nN2 RT
,
=
P1
kJ
(1 kmole) 8.314 kmole
=
100 kP a
= 498.84 m3 .
=
(3.377)
(3.378)
K
(6000 K)
,
(3.379)
(3.380)
Initially, the mixture is all N2 , so its partial pressure is the total pressure, and the initial partial pressure
of N is 0.
Now, every time an N2 molecule reacts and thus undergoes a negative change, 2 N molecules are
created and thus undergo a positive change, so
−dnN2 =
1
dnN .
2
(3.381)
This can be parameterized by a reaction progress variable ζ, also called the degree of reaction, defined
such that
= −dnN2 ,
1
dnN .
=
2
dζ
dζ
(3.382)
(3.383)
As an aside, one can integrate this, taking ζ = 0 at the initial state to get
ζ
=
ζ
=
nN2 |t=0 − nN2 ,
1
nN .
2
(3.384)
nN2 |t=0 − ζ,
2ζ.
(3.386)
(3.387)
(3.385)
Thus,
nN 2
nN
=
=
One can also eliminate ζ to get nN in terms of nN2 :
nN = 2 ( nN2 |t=0 − nN2 ) .
(3.388)
Now, for the reaction, one must have, for second law satisfaction, that
µN2 dnN2 + µN dnN
µN2 (−dζ) + µN (2dζ)
−µN2 + 2µN dζ
dζ
−µN2 + 2µN
dt
≤ 0,
≤ 0,
(3.389)
(3.390)
≤ 0
(3.391)
≤ 0.
(3.392)
In order to satisfy the second law, one can usefully, but naı̈vely, hypothesize that the non-equilibrium
reaction kinetics are given by
dζ
= −k(−µN2 + 2µN ),
dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
k ≥ 0,
convenient, but naı̈ve model
(3.393)
3.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM
131
Note there are other ways to guarantee second law satisfaction. In fact, a more complicated model is
well known to fit data well, and will be studied later in Chapter 4. For the present purposes, this naı̈ve
model will suffice. With this assumption, the second law reduces to
−k(−µN2 + 2µN )2 ≤ 0,
k ≥ 0,
(3.394)
which is always true. Obviously, the reaction ceases when dζ/dt = 0, which holds only when
2µN = µN2 .
(3.395)
Away from equilibrium, for the reaction to go forward, one must expect dζ/dt > 0, and then one
must have
−µN2 + 2µN ≤ 0,
2µN ≤ µN2 .
(3.396)
(3.397)
The chemical potentials are the molar specific Gibbs free energies; thus, for the reaction to go forward,
one must have
2gN ≤ g N2 .
(3.398)
Substituting using the definitions of Gibbs free
2 hN − T sN
yN P
o
2 hN − T sT,N − R ln
Po
2 hN − T soT,N − hN2 − T soT,N2
energy, one gets
≤
≤
≤
−2 hN − T soT,N + hN2 − T soT,N2
≥
−2 hN − T soT,N + hN2 − T soT,N2
≥
−2 hN − T soT,N + hN2 − T soT,N2
≥
hN 2 − T sN 2 ,
yN 2 P
o
,
hN2 − T sT,N2 − R ln
Po
yN P
yN 2 P
−2RT ln
+ RT ln
,
Po
Po
yN P
yN 2 P
− RT ln
,
2RT ln
Po
Po
2 2
yN P Po
,
RT ln
P 2 P yN
2o 2
yN P
.
RT ln
yN2 Po
(3.399)
(3.400)
(3.401)
(3.402)
(3.403)
(3.404)
At the initial state, one has yN = 0, so the right hand side approaches −∞, and the inequality holds.
At equilibrium, one has equality.
2
yN P
o
o
−2 hN − T sT,N + hN2 − T sT,N2
= RT ln
.
(3.405)
yN2 Po
Taking numerical values from Borgnakke and Sonntag’s Table A.9:
kJ
kJ
− (6000 K) 216.926
+
−2 5.9727 × 105
kmole
kg K
kJ
kJ
2.05848 × 105
− (6000 K) 292.984
kmole
kg K
2
yN P
kJ
(6000 K) ln
,
=
8.314
kmole K
yN2 Po
2
yN P
−2.87635
,
(3.406)
=
ln
| {z }
yN2 Po
≡ln KP
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132
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
0.0563399
| {z } =
≡KP
=
2
yN
P
,
yN2 Po
nN
nN +nN2
(3.407)
2
nN2
nN +nN2
(nN + nN2 )
RT
,
Po V
(3.408)
=
n2N RT
,
nN2 Po V
=
(2 ( nN2 |t=0 − nN2 )) RT
,
nN 2
Po V
=
(2 (1 kmole − nN2 )) (8.314)(6000)
. (3.411)
nN 2
(100)(498.84)
(3.409)
2
(3.410)
2
This is a quadratic equation for nN2 . It has two roots
nN2 = 0.888154 kmole
physical;
nN2 = 1.12593 kmole,
non-physical
(3.412)
The second root generates more N2 than at the start, and also yields non-physically negative nN =
−0.25186 kmole. So, at equilibrium the physical root is
nN = 2(1 − nN2 ) = 2(1 − 0.888154) = 0.223693 kmole.
(3.413)
The diatomic species is preferred.
Note in the preceding analysis, the term KP was introduced. This is the so-called equilibrium
“constant” which is really a function of temperature. It will be described in more detail later in
Sec. 4.4, but one notes that it is commonly tabulated for some reactions. Its tabular value can be
derived from the more fundamental quantities shown in this example. Borgnakke and Sonntag’s Table
A.11 gives for this reaction at 6000 K the value of ln KP = −2.876. Note that KP is fundamentally
defined in terms of thermodynamic properties for a system which may or may not be at chemical
equilibrium. Only at chemical equilibrium, can it can further be related to mole fraction and pressure
ratios.
The pressure at equilibrium is
P2
=
=
=
(nN2 + nN )RT
,
V
(0.888154 kmole + 0.223693 kmole) 8.314
498.84
111.185 kP a.
(3.414)
kJ
kmole K
(6000 K)
,
(3.415)
(3.416)
The pressure has increased because there are more molecules with the volume and temperature being
equal.
The molar concentrations ρi at equilibrium, are
ρN
=
ρN 2
=
kmole
mole
0.223693 kmole
= 4.484 × 10−4
= 4.484 × 10−7
,
498.84 m3
m3
cm3
kmole
mole
0.888154 kmole
= 1.78044 × 10−3
= 1.78044 × 10−6
.
498.84 m3
m3
cm3
(3.417)
(3.418)
Now, consider the heat transfer. One knows for the isochoric process that Q = E2 − E1 . The initial
energy is given by
E1
=
nN 2 e N 2 ,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.419)
133
3.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM
=
=
=
nN2 (hN2 − RT ),
(1 kmole) 2.05848 × 105
kJ
kmole
1.555964 × 105 kJ.
− 8.314
kJ
kmole K
(3.420)
(6000 K) ,
(3.421)
(3.422)
The energy at the final state is
E2
= nN 2 e N 2 + nN e N ,
= nN2 (hN2 − RT ) + nN (hN − RT ),
kJ
kJ
= (0.888154 kmole) 2.05848 × 105
− 8.314
(6000 K)
kmole
kmole K
kJ
kJ
(6000 K) ,
− 8.314
+(0.223693 kmole) 5.9727 × 105
kmole
kmole K
= 2.60966 × 105 kJ.
(3.423)
(3.424)
(3.425)
(3.426)
(3.427)
So
Q
= E2 − E1 ,
(3.428)
5
5
= 2.60966 × 10 kJ − 1.555964 × 10 kJ,
= 1.05002 × 105 kJ.
(3.429)
(3.430)
Heat needed to be added to keep the system at the constant temperature. This is because the nitrogen
dissociation process is endothermic.
One can check for second law satisfaction in two ways. Fundamentally, one can demand that
Eq. (3.340), dS ≥ δQ/T , be satisfied for the process, giving
Z 2
δQ
S2 − S1 ≥
.
(3.431)
T
1
For this isothermal process, this reduces to
S2 − S1
≥
Q
,
T
(3.432)
≥
Q
,
T
(3.433)
≥
Q
,
T
(3.434)
≥
Q
,
T
(3.435)
≥
Q
.
T
(3.436)
(nN2 sN2 + nN sN )|2
− (nN2 sN2 + nN sN )|1
yN 2 P
yN P
+ nN soT,N − R ln
nN2 soT,N2 − R ln
Po
Po
2
yN 2 P
yN P
− nN2 soT,N2 − R ln
+ nN soT,N − R ln
Po
Po
1
PN2
PN
nN2 soT,N2 − R ln
+ nN soT,N − R ln
Po
Po
2
PN2
PN
+ nN soT,N − R ln
− nN2 soT,N2 − R ln
Po
Po
1
RT
RT
n
n
N2
N
nN2 soT,N2 − R ln
+ nN soT,N − R ln
Po V
Po V
2

nN2 RT
nN RT

+ nN soT,N − R ln
− nN2 soT,N2 − R ln
|{z}
Po V
Po V
=0
1
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CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
Now, at the initial state nN = 0 kmole, and RT /Po /V has a constant value of
kJ
8.314 kmole
1
RT
K (6000 K)
=
=1
,
Po V
(100 kP a)(498.84 m3 )
kmole
so Eq. (3.436) reduces to
nN2 soT,N2 − R ln
n
nN2 N
+ nN soT,N − R ln
1 kmole
1 kmole
2
n
N2
o
− nN2 sT,N2 − R ln
1 kmole
1
(3.437)
≥
(3.438)
((0.888143) (292.984 − 8.314 ln (0.88143)) + (0.223714) (216.926 − 8.314 ln (0.223714)))|2
− ((1) (292.984 − 8.314 ln (1)))|1
19.4181
kJ
kJ
≥ 17.5004
.
K
K
Q
,
T
≥
105002
,
6000
(3.439)
Indeed, the second law is satisfied. Moreover, the irreversibility of the chemical reaction is 19.4181 −
17.5004 = +1.91772 kJ/K.
For the isochoric, isothermal process, it is also appropriate to use Eq. (3.356), dA|T,V ≤ 0, to check
for second law satisfaction. This turns out to give an identical result. Since by Eq. (3.235), A = U −T S,
A2 − A1 = (E2 − T2 S2 ) − (E1 − T1 S1 ). Since the process is isothermal, A2 − A1 = E2 − E1 − T (S2 − S1 ).
For A2 − A1 ≤ 0, one must demand E2 − E1 − T (S2 − S1 ) ≤ 0, or E2 − E1 ≤ T (S2 − S1 ), or
S2 − S1 ≥ (E2 − E1 )/T . Since Q = E2 − E1 for this isochoric process, one then recovers S2 − S1 ≥ Q/T.
3.11.2.2
Isothermal, isobaric system
Allowing for isobaric rather than isochoric equilibration introduces small variation in the
analysis.
Example 3.14
Consider the same reaction
N2 + N2 ⇌ 2N + N2 .
(3.440)
for an isobaric and isothermal process. That is, consider 1 kmole of N2 and 0 kmole of N at a pressure
of 100 kP a and a temperature of 6000 K. Using the ideal gas tables, find the equilibrium concentrations
of N and N2 if the equilibration process is isothermal and isobaric.
The initial volume is the same as from the previous example:
V1
= 498.84 m3 .
(3.441)
Note the volume will change in this isobaric process. Initially, the mixture is all N2 , so its partial
pressure is the total pressure, and the initial partial pressure of N is 0.
A few other key results are identical to the previous example:
nN = 2 ( nN2 |t=0 − nN2 ) ,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(3.442)
135
3.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM
and
2gN ≤ g N2 .
Substituting using the definitions of Gibbs free
2 hN − T sN
yN P
o
2 hN − T sT,N − R ln
Po
2 hN − T soT,N − hN2 − T soT,N2
(3.443)
energy, one gets
≤
≤
≤
−2 hN − T soT,N + hN2 − T soT,N2
≥
≥
−2 hN − T soT,N + hN2 − T soT,N2
hN 2 − T sN 2 ,
yN 2 P
o
,
hN2 − T sT,N2 − R ln
Po
yN P
yN 2 P
+ RT ln
,
−2RT ln
Po
Po
yN P
yN 2 P
2RT ln
− RT ln
,
Po
Po
2 2
yN P Po
RT ln
.
Po2 P yN2
(3.444)
(3.445)
(3.446)
(3.447)
(3.448)
In this case Po = P , so one gets
−2 hN − T
soT,N
+ hN2 − T
soT,N2
≥
RT ln
2
yN
yN 2
.
(3.449)
At the initial state, one has yN = 0, so the right hand side approaches −∞, and the inequality holds.
At equilibrium, one has equality.
2 yN
o
o
= RT ln
−2 hN − T sT,N + hN2 − T sT,N2
.
(3.450)
yN 2
Taking numerical values from Borgnakke and Sonntag’s Table A.9:
kJ
kJ
5
+
− (6000 K) 216.926
−2 5.9727 × 10
kmole
kg K
kJ
kJ
2.05848 × 105
− (6000 K) 292.984
kmole
kg K
2 kJ
yN
,
=
8.314
(6000 K) ln
kmole K
yN 2
2 yN
−2.87635 = ln
,
(3.451)
| {z }
yN 2
≡ln KP
0.0563399
| {z } =
≡KP
=
2
yN
,
yN 2
nN
nN +nN2
(3.452)
2
nN2
nN +nN2
,
(3.453)
=
n2N
,
nN2 (nN + nN2 )
=
(2 ( nN2 |t=0 − nN2 ))
,
nN2 (2 ( nN2 |t=0 − nN2 ) + nN2 )
(3.454)
2
=
(2 (1 kmole − nN2 ))2
.
nN2 (2 (1 kmole − nN2 ) + nN2 )
(3.455)
(3.456)
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136
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
This is a quadratic equation for nN2 . It has two roots
nN2 = 0.882147 kmole
physical;
nN2 = 1.11785 kmole,
non-physical
(3.457)
The second root generates more N2 than at the start, and also yields non-physically negative nN =
−0.235706 kmole. So, at equilibrium the physical root is
nN = 2(1 − nN2 ) = 2(1 − 0.882147) = 0.235706 kmole.
(3.458)
Again, the diatomic species is preferred. As the temperature is raised, one could show that the
monatomic species would come to dominate.
The volume at equilibrium is
V2
(nN2 + nN )RT
,
P
(0.882147 kmole + 0.235706 kmole) 8.314
=
100 kP a
= 557.630 m3 .
(3.459)
=
kJ
kmole K
(6000 K)
,
(3.460)
(3.461)
The volume has increased because there are more molecules with the pressure and temperature being
equal.
The molar concentrations ρi at equilibrium, are
ρN
=
ρN 2
=
0.235706 kmole
kmole
mole
= 4.227 × 10−4
= 4.227 × 10−7
,
3
3
557.636 m
m
cm3
0.882147 kmole
kmole
mole
= 1.58196 × 10−3
= 1.58196 × 10−6
.
557.636 m3
m3
cm3
(3.462)
(3.463)
The molar concentrations are a little smaller than for the isochoric case, mainly because the volume is
larger at equilibrium.
Now, consider the heat transfer. One knows for the isobaric process that Q = H2 − H1 . The initial
enthalpy is given by
kJ
5
= 2.05848 × 105 kJ.
(3.464)
H1 = nN2 hN2 = (1 kmole) 2.05848 × 10
kmole
The enthalpy at the final state is
H2
=
=
=
nN2 hN2 + nN hN ,
+ (0.235706 kmole) 5.9727 × 105
(3.465)
kJ
(3.466)
,
kmole
Q = H2 − H1 = 3.22389 × 105 kJ − 2.05848 × 105 kJ = 1.16520 × 105 kJ.
(3.468)
(0.882147 kmole) 2.05848 × 105
3.22368 × 105 kJ.
kJ
kmole
(3.467)
So,
Heat needed to be added to keep the system at the constant temperature. This is because the nitrogen
dissociation process is endothermic. Relative to the isochoric process, more heat had to be added to
maintain the temperature. This to counter the cooling effect of the expansion.
Lastly, it is a straightforward exercise to show that the second law is satisfied for this process.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
3.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM
3.11.3
137
Equilibrium condition
The results of both of the previous examples, in which a functional form of a progress
variable’s time variation, dζ/dt, was postulated in order to satisfy the second law, gave a
condition for equilibrium. This can be generalized so as to require at equilibrium that
N
X
µi νi = 0.
(3.469)
i=1
| {z }
≡−α
Here, νi represents the net number of moles of species i generated in the forward reaction.
This will be discussed in detail in the following Sec. 4.4. So, in the phase equilibrium example,
Eq. (3.469) becomes
µl (−1) + µg (1) = 0.
(3.470)
In the nitrogen chemistry example, Eq. (3.469) becomes
µN2 (−1) + µN (2) = 0.
(3.471)
The negation of the term on the left side of Eq (3.469) is sometimes defined as the
chemical affinity, α:
N
X
α≡−
µi νi .
(3.472)
i=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
138
CHAPTER 3. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 4
Thermochemistry of a single reaction
These four bodies are fire, air, water, earth. Fire occupies the highest place among
them all...
Aristotle (384 BC-322 BC), Meteorology, Book 1, § 2.
————————This chapter will further develop notions associated with the thermodynamics of chemical
reactions. The focus will be on chemical reaction of N species composed of L elements
reacting in J = 1 reaction. Several sources can be consulted as references.1 2345
4.1
Molecular mass
The molecular mass of a molecule is a straightforward notion from chemistry. One simply
sums the product of the number of atoms and each atom’s atomic mass to form the molecular
mass. If one defines L as the number of elements present in species i, φli as the number of
moles of atomic element l in species i, and Ml as the atomic mass of element l, the molecular
mass Mi of species i
L
X
Mi =
Ml φli .
(4.1)
l=1
In vector form, one would say
MT = MT · φ,
or
1
M = φT · M.
(4.2)
Borgnakke, C., and Sonntag, R. E., 2009, Fundamentals of Thermodynamics, Seventh Edition, John
Wiley, New York. See Chapters 14 and 15.
2
Abbott, M. M., and Van Ness, H. C., 1972, Thermodynamics, Schaum’s Outline Series in Engineering,
McGraw-Hill, New York. See Chapter 7.
3
Kondepudi, D., and Prigogine, I., 1998, Modern Thermodynamics: From Heat Engines to Dissipative
Structures, John Wiley, New York. See Chapters 4, 5, 7, and 9.
4
Turns, S. R., 2011, An Introduction to Combustion, Third Edition, McGraw-Hill, Boston. See Chapter
2.
5
Kuo, K. K., 2005, Principles of Combustion, Second Edition, John Wiley, New York. See Chapters 1, 2.
139
140
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
Here, M is the vector of length N containing the molecular masses, M is the vector of length
L containing the elemental masses, and φ is the matrix of dimension L × N containing the
number of moles of each element in each species. Generally, φ is full rank. If N > L, φ has
rank L. If N < L, φ has rank N. In any problem involving an actual chemical reaction, one
will find N ≥ L, and in most cases N > L. In isolated problems not involving a reaction,
one may have N < L. In any case, M lies in the column space of φT , which is the row space
of φ.
Example 4.1
Find the molecular mass of H2 O.
Here, one has two elements H and O, so L = 2, and one species, so N = 1; thus, in this isolated
problem, N < L. Take i = 1 for species H2 O. Take l = 1 for element H. Take l = 2 for element O.
From the periodic table, one gets M1 = 1 g/mole for H, M2 = 16 g/mole for O. For element 1, there
are 2 atoms, so φ11 = 2. For element 2, there is 1 atom so φ21 = 1. So, the molecular mass of species
1, H2 O is
φ11
M1 = ( M1 M2 ) ·
,
(4.3)
φ21
= M1 φ11 + M2 φ21 ,
(4.4)
g g (2) + 16
(1),
(4.5)
=
1
mole
mole
g
.
(4.6)
= 18
mole
Example 4.2
Find the molecular masses of the two species C8 H18 and CO2 .
Here, the vector matrix notation is exercised. In a certain sense this is overkill, but it is useful to
be able to understand a general notation.
One has N = 2 species, and takes i = 1 for C8 H18 and i = 2 for CO2 . One also has L = 3 elements
and takes l = 1 for C, l = 2 for H, and l = 3 for O. Now, for each element, one has M1 = 12 g/mole,
M2 = 1 g/mole, M3 = 16 g/mole. The molecular masses are then given by


φ11 φ12
( M1 M2 ) = ( M1 M2 M3 ) ·  φ21 φ22  ,
(4.7)
φ31 φ32


8 1
= ( 12 1 16 ) ·  18 0  ,
(4.8)
0 2
= ( 114 44 ) .
(4.9)
That is, for C8 H18 , one has molecular mass M1 = 114 g/mole. For CO2 , one has molecular mass
M2 = 44 g/mole.
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141
4.2. STOICHIOMETRY
4.2
4.2.1
Stoichiometry
General development
Stoichiometry represents a mass balance on each element in a chemical reaction. For example,
in the simple global reaction
2H2 + O2 ⇌ 2H2 O,
(4.10)
one has 4 H atoms in both the reactant and product sides and 2 O atoms in both the reactant
and product sides. In this section stoichiometry will be systematized.
Consider now a general reaction with N species. This reaction can be represented by
N
X
νi′ χi
⇌
N
X
νi′′ χi .
(4.11)
i=1
i=1
Here, χi is the ith chemical species, νi′ is the forward stoichiometric coefficient of the ith
reaction, and νi′′ is the reverse stoichiometric coefficient of the ith reaction. Both νi′ and νi′′
are to be interpreted as pure dimensionless numbers.
In Equation (4.10), one has N = 3 species. One might take χ1 = H2 , χ2 = O2 , and
χ3 = H2 O. The reaction is written in more general form as
(2)χ1 + (1)χ2 + (0)χ3 ⇌ (0)χ1 + (0)χ2 + (2)χ3 ,
(2)H2 + (1)O2 + (0)H2 O ⇌ (0)H2 + (0)O2 + (2)H2 O.
(4.12)
(4.13)
Here, one has
ν1′ = 2,
ν2′ = 1,
ν3′ = 0,
ν1′′ = 0,
ν2′′ = 0,
ν3′′ = 2.
(4.14)
(4.15)
(4.16)
It is common and useful to define another pure dimensionless number, the net stoichiometric coefficients for species i, νi . Here, νi represents the net production of number if the
reaction goes forward. It is given by
νi = νi′′ − νi′ .
(4.17)
For the reaction 2H2 + O2 ⇌ 2H2 O, one has
ν1 = ν1′′ − ν1′ = 0 − 2 = −2,
ν2 = ν2′′ − ν2′ = 0 − 1 = −1,
ν3 = ν3′′ − ν3′ = 2 − 0 = 2.
(4.18)
(4.19)
(4.20)
With these definitions, it is possible to summarize a chemical reaction as
N
X
νi χi = 0.
(4.21)
i=1
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142
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
In vector notation, one would say
ν T · χ = 0.
(4.22)
For the reaction of this section, one might write the non-traditional form
−2H2 − O2 + 2H2 O = 0.
(4.23)
It remains to enforce a stoichiometric balance. This is achieved if, for each element, l =
1, . . . , L, one has the following equality:
N
X
φli νi = 0,
l = 1, . . . , L.
(4.24)
i=1
That is to say, for each element, the sum of the product of the net species production and
the number of elements in the species must be zero. In vector notation, this becomes
φ · ν = 0.
(4.25)
One may recall from linear algebra that this demands that ν lie in the right null space of φ.
Example 4.3
Show stoichiometric balance is achieved for −2H2 − O2 + 2H2 O = 0.
Here again, the number of elements L = 2, and one can take l = 1 for H and l = 2 for O. Also
the number of species N = 3, and one takes i = 1 for H2 , i = 2 for O2 , and i = 3 for H2 O. Then, for
element 1, H, in species 1, H2 , one has
φ11 = 2, H in H2 .
(4.26)
φ12
= 0,
H in O2 ,
(4.27)
φ13
= 2,
H in H2 O,
(4.28)
φ21
φ22
= 0,
= 2,
O in H2 ,
O in O2 ,
(4.29)
(4.30)
φ23
= 1,
O in H2 O.
(4.31)
Similarly, one gets
In matrix form then,
PN
i=1
φli νi = 0 gives
2 0
0 2
2
1


ν1
 ν2  = 0 .
0
ν3
(4.32)
This is two equations in three unknowns. Thus, it is formally underconstrained. Certainly the trivial
solution ν1 = ν2 = ν3 = 0 will satisfy, but one seeks non-trivial solutions. Assume ν3 has a known value
ν3 = ξ. Then, the system reduces to
2 0
−2ξ
ν1
=
.
(4.33)
0 2
ν2
−ξ
CC BY-NC-ND. 30 March 2014, J. M. Powers.
143
4.2. STOICHIOMETRY
The inversion here is easy, and one finds ν1 = −ξ, ν2 = − 21 ξ. Or in vector form,

ν1
 ν2  =
ν3

=

−ξ
 −1ξ  ,
2
ξ


−1
ξ  − 12  ,
1

(4.34)
ξ ∈ R1
(4.35)
Again, this amounts to saying the solution vector (ν1 , ν2 , ν3 )T lies in the right null space of the coefficient
matrix φli .
Here, ξ is any real scalar. If one takes ξ = 2, one gets
  

ν1
−2
 ν2  =  −1  ,
(4.36)
ν3
2
This simply corresponds to
−2H2 − O2 + 2H2 O = 0.
If one takes ξ = −4, one still achieves stoichiometric balance, with
  

ν1
4
 ν2  =  2  ,
ν3
−4
(4.37)
(4.38)
which corresponds to the equally valid
4H2 + 2O2 − 4H2 O = 0.
(4.39)
In summary, the stoichiometric coefficients are non-unique but partially constrained by mass conservation. Which set is chosen is to some extent arbitrary, and often based on traditional conventions from
chemistry. But others are equally valid.
There is a small issue with units here, which will be seen to be difficult to reconcile.
In practice, it will have little to no importance. In the previous example, one might be
tempted to ascribe units of moles to νi . Later, it will be seen that in classical reaction
kinetics, νi is best interpreted as a pure dimensionless number, consistent with the definition
of this section. So, in the context of the previous example, one would then take ξ to be
dimensionless as well, which is perfectly acceptable for the example. In later problems, it
will be more useful to give ξ the units of moles. Note that multiplication of ξ by any scalar,
e.g. mole/(6.02 × 1023 ), still yields an acceptable result.
Example 4.4
Balance an equation for hypothesized ethane combustion
ν1′ C2 H6 + ν2′ O2 ⇌ ν3′′ CO2 + ν4′′ H2 O.
(4.40)
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144
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
One could also say in terms of the net stoichiometric coefficients
ν1 C2 H6 + ν2 O2 + ν3 CO2 + ν4 H2 O = 0.
(4.41)
Here, one takes χ1 = C2 H6 , χ2 = O2 , χ3 CO2 , χ4 = H2 O. So, there are N = 4 species. There are also
L = 3 elements: l = 1 : C, l = 2 : H, l = 3 : O. One then has
φ11
=
2,
C in C2 H6 ,
(4.42)
φ12
φ13
=
=
0,
1,
C in O2 ,
C in CO2 ,
(4.43)
(4.44)
φ14
φ21
=
=
0,
6,
C in H2 O,
H in C2 H6 ,
(4.45)
(4.46)
φ22
φ23
=
=
0,
0,
H in O2 ,
H in CO2 ,
(4.47)
(4.48)
φ24
φ31
=
=
2,
0,
H in H2 O,
O in C2 H6 ,
(4.49)
(4.50)
φ32
φ33
=
=
2,
2,
O in O2 ,
O in CO2 ,
(4.51)
(4.52)
φ34
=
1,
O in H2 O,
(4.53)
(4.54)
So, the stoichiometry equation,
PN
i=1

φli νi = 0, is given by
2 0
6 0
0 2
  ν1   
1 0
0
ν 
0 2 2  = 0.
ν3
2 1
0
ν4
(4.55)
Here, there are three equations in four unknowns, so the system is underconstrained. There are many
ways to address this problem. Here, choose the robust way of casting the system into row echelon form.
This is easily achieved by Gaussian elimination. Row echelon form seeks to have zeros in the lower left
part of the matrix. The lower left corner has a zero already, so that is useful. Now, multiply the top
equation by 3 and subtract the result from the second to get

  ν1   
2 0 1 0
0
ν2 
 0 0 −3 2  
(4.56)
  = 0.
ν3
0 2 2 1
0
ν4
Next switch the last two equations to get

2 0
0 2
0 0
  ν1   
1 0
0
ν 
2 1 2  = 0.
ν3
−3 2
0
ν4
Now, divide the first by 2, the second by 2 and the third by −3 to get unity in the diagonal:
  ν1   

0
0
1 0 12
1   ν2 

0 1 1
=
0.
 
2
ν3
0
0 0 1 − 32
ν4
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.57)
(4.58)
145
4.2. STOICHIOMETRY
So-called bound variables have non-zero coefficients on the diagonal, so one can take the bound variables
to be ν1 , ν2 , and ν3 . The remaining variables are free variables. Here, one takes the free variable to be
ν4 . So, set ν4 = ξ, and rewrite the system as

  

1 0 12
ν1
0
 0 1 1   ν2  =  − 1 ξ  .
(4.59)
2
2
0 0 1
ξ
ν3
3
Solving, one finds
  1 
 1
ν1
−3ξ
−3
7
 ν2   − 6 ξ 
 − 67 
  =  2  = ξ  2 ,
ν3
3ξ
3
1
ξ
ν4

ξ ∈ R1 .
(4.60)
Again one finds a non-unique solution in the right null space of φ. If one chooses ξ = 6, then one gets
  

ν1
−2
 ν2   −7 
(4.61)
 =
,
ν3
4
ν4
6
which corresponds to the stoichiometrically balanced reaction
2C2 H6 + 7O2 ⇌ 4CO2 + 6H2 O.
(4.62)
In this example, ξ is dimensionless.
Example 4.5
Consider stoichiometric balance for a propane oxidation reaction which may produce carbon monoxide and hydroxyl in addition to carbon dioxide and water.
The hypothesized reaction takes the form
ν1′ C3 H8 + ν2′ O2 ⇌ ν3′′ CO2 + ν4′′ CO + ν5′′ H2 O + ν6′′ OH.
(4.63)
In terms of net stoichiometric coefficients, this becomes
ν1 C3 H8 + ν2 O2 + ν3 CO2 + ν4 CO + ν5 H2 O + ν6 OH = 0.
(4.64)
There are N = 6 species and L = 3 elements. One then has
φ11
=
3,
C in C3 H8 ,
(4.65)
φ12
φ13
=
=
0,
1,
C in O2 ,
C in CO2 ,
(4.66)
(4.67)
φ14
φ15
=
=
1,
0,
C in CO,
C in H2 O,
(4.68)
(4.69)
φ16
φ21
=
=
0,
8,
C in OH,
H in C3 H8 ,
(4.70)
(4.71)
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CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
φ22
=
0,
H in O2 ,
(4.72)
φ23
φ24
=
=
0,
0,
H in CO2 ,
H in CO,
(4.73)
(4.74)
φ25
φ26
=
=
2,
1,
H in H2 O,
H in OH,
(4.75)
(4.76)
φ31
φ32
=
=
0,
2,
O in C3 H8 ,
O in O2 ,
(4.77)
(4.78)
φ33
φ34
=
=
2,
1,
O in CO2 ,
O in CO,
(4.79)
(4.80)
φ35
φ36
=
=
1,
1,
O in H2 O,
O in OH.
(4.81)
(4.82)
The equation φ · ν = 0, then becomes

3
8
0
0 1
0 0
2 2
1 0
0 2
1 1


ν1
  ν2   
0
0
 
ν 
1 ·  3  = 0.
 ν4 
1
0
 
ν5
ν6
Multiplying the first equation by −8/3 and adding it to the second gives
 
ν1
  ν2   

0
3 0 1
1 0 0
 
ν3   
0 0 −8 −8 2 1 · 
=
0 .


3
3
 ν4 
0 2 2
1 1 1
0
 
ν5
ν6
(4.83)
(4.84)
Trading the second and third rows gives

3
0
0
0 1
2 2
0 − 38
1
1
− 38
 
ν1
  ν2   
0 0
0
 
ν 
1 1 ·  3  = 0.
 ν4 
2 1
0
 
ν5
ν6
Dividing the first row by 3, the second by 2 and the third by −8/3 gives
 
ν1


 ν2   
1
1
0
0
0
1 0 3 3
 
1
1   ν3 
0.
0 1 1 1
·
=


2
2
2
 ν4 
0 0 1 1 − 43 − 38
0
 
ν5
ν6
(4.85)
(4.86)
Take bound variables to be ν1 , ν2 , and ν3 and free variables to be ν4 , ν5 , and ν6 . So, set ν4 = ξ1 ,
ν5 = ξ2 , and ν6 = ξ3 , and get

   

− ξ31
ν1
1 0 13
 0 1 1  ·  ν2  =  − ξ1 − ξ2 − ξ3  .
(4.87)
2
2
2
0 0 1
ν3
−ξ1 + 34 ξ2 + 83 ξ3
CC BY-NC-ND. 30 March 2014, J. M. Powers.
147
4.2. STOICHIOMETRY
Solving, one finds


ν1
 ν2  =
ν3



1
8 (−2ξ2 − ξ3 )
1
.
(4ξ
1 − 10ξ2 − 7ξ3 )
8
1
8 (−8ξ1 + 6ξ2 + 3ξ3 )
For all the coefficients, one then has
 








1
ν1
0
−2
−1
8 (−2ξ2 − ξ3 )
1
 ν2 
 8 (4ξ1 − 10ξ2 − 7ξ3 ) 
 4 
 −10 
 −7 
 
1







 ν3 
 8 (−8ξ1 + 6ξ2 + 3ξ3 )  ξ1  −8  ξ2  6  ξ3  3 
  = 



+
.
=
+
ξ1
 ν4 


8  8 
8  0 
8  0 
 








ν5
8
0
ξ2
0
ν6
0
8
ξ3
0
(4.88)
(4.89)
Here, one finds three independent vectors in the right null space. To simplify the notation, take
ξˆ1 = ξ1 /8, ξ̂2 = ξ2 /8, and ξ̂3 = ξ3 /8. Then,
 






ν1
0
−2
−1
 ν2 
 4 
 −10 
 −7 
 






 −8  ˆ  6 
 ν3 
 3 
(4.90)
  = ξ̂1 
 + ξ2 
 + ξ̂3 
.
 8 
 ν4 
 0 
 0 
 






0
8
ν5
0
ν6
0
0
8
The most general reaction that can achieve a stoichiometric balance is given by
(−2ξˆ2 − ξ̂3 )C3 H8 + (4ξˆ1 − 10ξˆ2 − 7ξˆ3 )O2 + (−8ξˆ1 + 6ξˆ2 + 3ξˆ3 )CO2 + 8ξˆ1 CO + 8ξˆ2 H2 O + 8ξˆ3 OH = 0. (4.91)
Rearranging, one gets
(2ξˆ2 + ξˆ3 )C3 H8 + (−4ξˆ1 + 10ξˆ2 + 7ξˆ3 )O2 ⇌ (−8ξˆ1 + 6ξˆ2 + 3ξˆ3 )CO2 + 8ξˆ1 CO + 8ξˆ2 H2 O + 8ξˆ3 OH. (4.92)
This will be balanced for all ξ̂1 , ξ̂2 , and ξˆ3 . The values that are actually achieved in practice depend
on the thermodynamics of the problem. Stoichiometry only provides some limitations.
A slightly more familiar form is found by taking ξˆ2 = 1/2 and rearranging, giving
(1 + ξˆ3 ) C3 H8 + (5 − 4ξˆ1 + 7ξˆ3 ) O2 ⇌ (3 − 8ξˆ1 + 3ξˆ3 ) CO2 + 4 H2 O + 8ξˆ1 CO + 8ξˆ3 OH.
(4.93)
One notes that often the production of CO and OH will be small. If there is no production of CO or
OH, ξˆ1 = ξˆ3 = 0 and one recovers the familiar balance of
C3 H8 + 5 O2 ⇌ 3 CO2 + 4 H2 O.
(4.94)
One also notes that stoichiometry alone admits unusual solutions. For instance, if ξ̂1 = 100 ξˆ2 = 1/2,
and ξˆ3 = 1, one has
2 C3 H8 + 794 CO2 ⇌ 388 O2 + 4 H2 O + 800 CO + 8 OH.
(4.95)
This reaction is certainly admitted by stoichiometry but is not observed in nature. To determine
precisely which of the P
infinitely many possible final states are realized requires a consideration of the
N
equilibrium condition i=1 νi µi = 0.
Looked at in another way, we can think of three independent classes of reactions admitted by the
stoichiometry, one for each of the linearly independent null space vectors. Taking first ξ̂1 = 1/4, ξˆ2 = 0,
ξˆ3 = 0, one gets, after rearrangement
2CO + O2 ⇌ 2CO2 ,
(4.96)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
148
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
as one class of reaction admitted by stoichiometry. Taking next ξˆ1 = 0, ξ̂2 = 1/2, ξˆ3 = 0, one gets
C3 H8 + 5O2 ⇌ 3CO2 + 4H2 O,
(4.97)
as the second class admitted by stoichiometry. The third class is given by taking ξˆ1 = 0, ξ̂2 = 0, ξˆ3 = 1,
and is
C3 H8 + 7O2 ⇌ 3CO2 + 8OH.
(4.98)
In this example, both ξ and ξˆ are dimensionless.
In general, one can expect to find the stoichiometric coefficients for N species composed
of L elements to be of the following form:
νi =
N
−L
X
k=1
Dik ξk ,
i = 1, . . . , N.
(4.99)
Here, Dik is a dimensionless component of a full rank matrix of dimension N × (N − L)
and rank N − L, and ξk is a dimensionless component of a vector of parameters of length
N − L. The matrix whose components are Dik are constructed by populating its columns
with vectors which lie in the right null space of φli . Note that multiplication of ξk by any
constant gives another set of νi , and mass conservation for each element is still satisfied.
4.2.2
Fuel-air mixtures
In combustion with air, one often models air as a simple mixture of diatomic oxygen and
inert diatomic nitrogen in the
air:
′
νair
(O2 + 3.76N2 ).
(4.100)
The air-fuel ratio, A and its reciprocal, the fuel-air ratio, F , can be defined on a mass
and mole basis.
nair
mair
,
Amole =
.
(4.101)
Amass =
mf uel
nf uel
Via the molecular masses, one has
Amass =
mair
nair Mair
Mair
=
= Amole
.
mf uel
nf uel Mf uel
Mf uel
(4.102)
If there is not enough air to burn all the fuel, the mixture is said to be rich. If there is
excess air, the mixture is said to be lean. If there is just enough, the mixture is said to be
stoichiometric. The equivalence ratio, Φ, is defined as the actual fuel-air ratio scaled by the
stoichiometric fuel-air ratio:
Φ≡
Factual
Fstoichiometric
CC BY-NC-ND. 30 March 2014, J. M. Powers.
=
Astoichiometric
.
Aactual
(4.103)
149
4.2. STOICHIOMETRY
The ratio Φ is the same whether F ’s are taken on a mass or mole basis, because the ratio of
molecular masses cancel.
Example 4.6
Calculate the stoichiometry of the combustion of methane with air with an equivalence ratio of
Φ = 0.5. If the pressure is 0.1 M P a, find the dew point of the products.
First calculate the coefficients for stoichiometric combustion:
ν1′ CH4 + ν2′ (O2 + 3.76N2 ) ⇌ ν3′′ CO2 + ν4′′ H2 O + ν5′′ N2 .
(4.104)
ν1 CH4 + ν2 O2 + ν3 CO2 + ν4 H2 O + (ν5 + 3.76ν2 )N2 = 0.
(4.105)
Or
Here, one has N = 5 species and L = 4 elements. Adopting a slightly more intuitive procedure for
variety, one writes a conservation equation for each element to get
ν1 + ν3
= 0,
C,
(4.106)
4ν1 + 2ν4
2ν2 + 2ν3 + ν4
= 0,
= 0,
H,
O,
(4.107)
(4.108)
3.76ν2 + ν5
= 0,
N.
(4.109)
In matrix form this becomes

1
4

0
0
0
0
2
3.76
1
0
2
0
0
2
1
0
 
ν1

 
0
0
 ν2 
0   0
 ·  ν3  =   .
0
0
 
ν4
1
0
ν5
(4.110)
Now, one might expect to have one free variable, since one has five unknowns in four equations.
While casting the equation in row echelon form is guaranteed to yield a proper solution, one can often
use intuition to get a solution more rapidly. One certainly expects that CH4 will need to be present
for the reaction to take place. One might also expect to find an answer if there is one mole of CH4 . So
take ν1 = −1. Realize that one could also get a physically valid answer by assuming ν1 to be equal to
any scalar. With ν1 = −1, one gets

0
 0

2
3.76
1
0
2
0
0
2
1
0
    
0
1
ν2
0   ν3   4 
·  =  .
0
0
ν4
ν5
1
0
(4.111)
One easily finds the unique inverse does exist, and that the solution is

 

ν2
−2
 ν3   1 
 =
.
2
ν4
ν5
7.52
(4.112)
If there had been more than one free variable, the 4 × 4 matrix would have been singular, and no unique
inverse would have existed.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
150
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
In any case, the reaction under stoichiometric conditions is
−CH4 − 2O2 + CO2 + 2H2 O + (7.52 + (3.76)(−2))N2 = 0,
CH4 + 2(O2 + 3.76N2 ) ⇌ CO2 + 2H2 O + 7.52N2 .
(4.113)
(4.114)
For the stoichiometric reaction, the fuel-air ratio on a mole basis is
Fstoichiometric =
1
= 0.1050.
2 + 2(3.76)
(4.115)
Now, Φ = 0.5, so
Factual
= ΦFstoichiometric ,
(4.116)
= (0.5)(0.1050),
= 0.0525.
(4.117)
(4.118)
By inspection, one can write the actual reaction equation as
CH4 + 4(O2 + 3.76N2 ) ⇌ CO2 + 2H2 O + 2O2 + 15.04N2.
(4.119)
Check:
Factual =
1
= 0.0525.
4 + 4(3.76)
(4.120)
For the dew point of the products, one needs the partial pressure of H2 O. The mole fraction of
H2 O is
2
yH2 O =
= 0.0499
(4.121)
1 + 2 + 2 + 15.04
So, the partial pressure of H2 O is
PH2 O = yH2 O P = 0.0499(100 kP a) = 4.99 kP a.
(4.122)
From the steam tables, the saturation temperature at this pressure is Tsat = Tdew point = 32.88◦C. If
the products cool to this temperature in an exhaust device, the water could condense in the apparatus.
4.3
First law analysis of reacting systems
One can easily use the first law to learn much about chemically reacting systems.
4.3.1
Enthalpy of formation
The enthalpy of formation is the enthalpy that is required to form a molecule from combining
its constituents at P = 0.1 MP a and T = 298 K. Consider the reaction (taken here to be
irreversible)
C + O2 → CO2 .
(4.123)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
4.3. FIRST LAW ANALYSIS OF REACTING SYSTEMS
151
In order to maintain the process at constant temperature, it is found that heat transfer to
the volume is necessary. For the steady constant pressure process, one has
E2 − E1 =
− 1W2 ,
Z 2
= 1Q2 −
P dV,
1Q2
(4.124)
(4.125)
1
1Q2
1Q2
= 1Q2 − P (V2 − V1 ),
= E2 − E1 + P (V2 − V1 ),
= H2 − H1 ,
= Hproducts − Hreactants ,
X
X
ni hi −
=
ni hi .
(4.126)
(4.127)
(4.128)
(4.129)
(4.130)
reactants
products
In this reaction, one measures that 1Q2 = −393522 kJ for the reaction of 1 kmole of C
and O2 . That is, the reaction liberates such energy to the environment. So, measuring the
heat transfer can give a measure of the enthalpy difference between reactants and products.
Assign a value of enthalpy zero to elements in their standard state at the reference state.
Thus, C and O2 both have enthalpies of 0 kJ/kmole at T = 298 K, P = 0.1 MP a. This
enthalpy is designated, for species i,
o
o
hf,i = hTo ,i ,
(4.131)
and is called the enthalpy of formation. So, the energy balance for the products and reactants,
here both at the standard state, becomes
o
o
o
= nCO2 hf,CO2 − nC hf,C − nO2 hf,O2 ,
(4.132)
kJ
kJ
o
− (1 kmole) 0
.
−393522 kJ = (1 kmole)hf,CO2 − (1 kmole) 0
kmole
kmole
(4.133)
1Q2
o
Thus, the enthalpy of formation of CO2 is hf,CO2 = −393522 kJ/kmole, since the reaction
involved creation of 1 kmole of CO2 .
Often values of enthalpy are tabulated in the forms of enthalpy differences ∆hi . These
are defined such that
o
o
hi = hf,i + (hi − hf,i ),
| {z }
(4.134)
=∆hi
=
o
hf,i
+ ∆hi .
(4.135)
Lastly, one notes for an ideal gas that the enthalpy is a function of temperature only,
and so does not depend on the reference pressure; hence
o
hi = hi ,
o
∆hi = ∆hi ,
if ideal gas.
(4.136)
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152
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
Example 4.7
(Borgnakke and Sonntag, p. 575). Consider the following irreversible reaction in a steady state,
steady flow process confined to the standard state of P = 0.1 M P a, T = 298 K:
CH4 + 2O2 → CO2 + 2H2 O(ℓ).
(4.137)
The first law holds that
Qcv
=
X
products
ni hi −
X
ni hi .
(4.138)
reactants
All components are at their reference states. Borgnakke and Sonntag’s Table A.10 gives properties,
and one finds
Qcv
=
=
=
nCO2 hCO2 + nH2 O hH2 O − nCH4 hCH4 − nO2 hO2 ,
kJ
kJ
(1 kmole) −393522
+ (2 kmole) −285830
kmole
kmole
kJ
kJ
− (2 kmole) 0
,
−(1 kmole) −74873
kmole
kmole
−890309 kJ.
(4.139)
(4.140)
(4.141)
A more detailed analysis is required in the likely case in which the system is not at the
reference state.
Example 4.8
(adopted from Moran and Shapiro6 ) A mixture of 1 kmole of gaseous methane and 2 kmole of
oxygen initially at 298 K and 101.325 kP a burns completely in a closed, rigid, container. Heat transfer
occurs until the final temperature is 900 K. Find the heat transfer and the final pressure.
The combustion is stoichiometric. Assume that no small concentration species are generated. The
global reaction is given by
CH4 + 2O2 → CO2 + 2H2 O.
(4.142)
The first law analysis for the closed system is slightly different:
E2 − E1 = 1Q2 − 1W2 .
(4.143)
Since the process is isochoric, 1W2 = 0. So
1Q2
=
=
E2 − E1 ,
nCO2 eCO2 + nH2 O eH2 O − nCH4 eCH4 − nO2 eO2 ,
6
(4.144)
(4.145)
Moran, M. J., and Shapiro, H. N., 2003, Fundamentals of Engineering Thermodynamics, Fifth Edition,
John Wiley, New York. p. 619.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
153
4.3. FIRST LAW ANALYSIS OF REACTING SYSTEMS
=
=
=
=
=
nCO2 (hCO2 − RT2 ) + nH2 O (hH2 O − RT2 ) − nCH4 (hCH4 − RT1 ) − nO2 (hO2 − RT1 ), (4.146)
hCO2 + 2hH2 O − hCH4 − 2hO2 − 3R(T2 − T1 ),
(4.147)
o
o
o
o
(hCO2 ,f + ∆hCO2 ) + 2(hH2 O,f + ∆hH2 O ) − (hCH4 ,f + ∆hCH4 ) − 2(hO2 ,f + ∆hO2 )
−3R(T2 − T1 ),
(4.148)
−3(8.314)(900 − 298),
−745412 kJ.
(4.149)
(4.150)
(−393522 + 28030) + 2(−241826 + 21937) − (−74873 + 0) − 2(0 + 0)
For the pressures, one has
P1 V1
=
(nCH4 + nO2 )RT1 ,
V1
=
(nCH4 + nO2 )RT1
,
P1
=
=
(4.151)
(4.152)
(1 kmole + 2 kmole) 8.314
kJ
kg K
101.325 kP a
(298 K)
,
73.36 m3 .
(4.153)
(4.154)
Now, V2 = V1 , so
P2
=
=
(nCO2 + nH2 O )RT2
,
V2
(4.155)
(1 kmole + 2 kmole) 8.314
73.36 m3
kJ
kg K
(900 K)
,
= 306.0 kP a.
(4.156)
(4.157)
The pressure increased in the reaction. This is entirely attributable to the temperature rise, as the
number of moles remained constant here.
4.3.2
Enthalpy and internal energy of combustion
The enthalpy of combustion is the difference between the enthalpy of products and reactants
when complete combustion occurs at a given pressure and temperature. It is also known
as the heating value or the heat of reaction. The internal energy of combustion is related
and is the difference between the internal energy of products and reactants when complete
combustion occurs at a given volume and temperature.
The term higher heating value refers to the energy of combustion when liquid water is in
the products. Lower heating value refers to the energy of combustion when water vapor is
in the product.
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154
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
4.3.3
Adiabatic flame temperature in isochoric stoichiometric systems
The adiabatic flame temperature refers to the temperature which is achieved when a fuel and
oxidizer are combined with no loss of work or heat energy. Thus, it must occur in a closed,
insulated, fixed volume. It is generally the highest temperature that one can expect to
achieve in a combustion process. It generally requires an iterative solution. Of all mixtures,
stoichiometric mixtures will yield the highest adiabatic flame temperatures because there is
no need to heat the excess fuel or oxidizer.
Here, four examples will be presented to illustrate the following points.
• The adiabatic flame temperature can be well over 5000 K for seemingly ordinary
mixtures.
• Dilution of the mixture with an inert diluent lowers the adiabatic flame temperature.
The same effect would happen in rich and lean mixtures.
• Preheating the mixture, such as one might find in the compression stroke of an engine,
• Consideration of the presence of minor species lowers the adiabatic flame temperature.
4.3.3.1
Undiluted, cold mixture
Example 4.9
A closed, fixed, adiabatic volume contains a stoichiometric mixture of 2 kmole of H2 and 1 kmole
of O2 at 100 kP a and 298 K. Find the adiabatic flame temperature assuming the irreversible reaction
2H2 + O2 → 2H2 O.
(4.158)
The volume is given by
V
=
=
=
(nH2 + nO2 )RT1
,
P1
(2 kmole + 1 kmole) 8.314
100 kP a
74.33 m3 .
(4.159)
kJ
kmole K
(298 K)
,
(4.160)
(4.161)
The first law gives
E2 − E1
E2 − E1
nH2 O eH2 O − nH2 eH2 − nO2 eO2
nH2 O (hH2 O − RT2 ) − nH2 (hH2 − RT1 ) − nO2 (hO2 − RT1 )
CC BY-NC-ND. 30 March 2014, J. M. Powers.
= 1Q2 − 1W2 ,
= 0,
(4.162)
(4.163)
= 0,
= 0,
(4.164)
(4.165)
155
4.3. FIRST LAW ANALYSIS OF REACTING SYSTEMS
2hH2 O − 2 hH2 − hO2 +R(−2T2 + 3T1 )
|{z} |{z}
= 0,
(4.166)
2hH2 O + (8.314) ((−2) T2 + (3) (298)) = 0,
hH2 O − 8.314T2 + 3716.4 = 0,
(4.167)
(4.168)
−241826 + ∆hH2 O − 8.314T2 + 3716.4 = 0,
−238110 + ∆hH2 O − 8.314T2 = 0.
(4.170)
(4.171)
=0
=0
o
hf,H2 O + ∆hH2 O − 8.314T2 + 3716.4 = 0,
(4.169)
At this point, one begins an iteration process, guessing a value of T2 and an associated ∆hH2 O . When
T2 is guessed at 5600 K, the left side becomes −6507.04. When T2 is guessed at 6000 K, the left side
becomes 14301.4. Interpolate then to arrive at
T2 = 5725 K.
(4.172)
This is an extremely high temperature. At such temperatures, in fact, one can expect other species to
co-exist in the equilibrium state in large quantities. These may include H, OH, O, HO2 , and H2 O2 ,
among others.
The final pressure is given by
P2
nH2 O RT2
,
V
kJ
(2 kmole) 8.314 kmole
=
74.33 m3
= 1280.71 kP a.
=
(4.173)
K
(5725 K)
,
(4.174)
(4.175)
The final concentration of H2 O is
ρH2 O =
4.3.3.2
kmole
2 kmole
= 2.69 × 10−2
.
74.33 m3
m3
(4.176)
Dilute, cold mixture
Example 4.10
Consider a variant on the previous example in which the mixture is diluted with an inert, taken
here to be N2 . A closed, fixed, adiabatic volume contains a stoichiometric mixture of 2 kmole of H2 ,
1 kmole of O2 , and 8 kmole of N2 at 100 kP a and 298 K. Find the adiabatic flame temperature and
the final pressure, assuming the irreversible reaction
2H2 + O2 + 8N2 → 2H2 O + 8N2 .
(4.177)
The volume is given by
V
=
=
=
(nH2 + nO2 + nN2 )RT1
,
P1
(2 kmole + 1 kmole + 8 kmole) 8.314
100 kP a
272.533 m3 .
(4.178)
kJ
kmole K
(298 K)
,
(4.179)
(4.180)
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CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
The first law gives
E2 − E1
=
E2 − E1 =
nH2 O eH2 O − nH2 eH2 − nO2 eO2 + nN2 (eN2 2 − eN2 1 ) =
nH2 O (hH2 O − RT2 ) − nH2 (hH2 − RT1 ) − nO2 (hO2 − RT1 ) + nN2 ((hN2 2 − RT2 ) − (hN2 1 − RT1 )) =
2hH2 O − 2 hH2 − hO2 +R(−10T2 + 11T1 ) + 8( hN2 2 − hN2 1 ) =
|{z} |{z}
| {z } | {z }
=0
=0
=∆hN2
1Q2
0,
0,
− 1W2 ,
0,
0,
=0
2hH2 O + (8.314) (−10T2 + (11)(298)) + 8∆hN2 2
2hH2 O − 83.14T2 + 27253.3 + 8∆hN2 2
=
=
0,
0,
2hf,H2 O + 2∆hH2 O − 83.14T2 + 27253.3 + 8∆hN2 2
=
0,
2(−241826) + 2∆hH2 O − 83.14T2 + 27253.3 + 8∆hN2 2
=
0,
−456399 + 2∆hH2 O − 83.14T2 + 8∆hN2 2
=
0,
o
At this point, one begins an iteration process, guessing a value of T2 and an associated ∆hH2 O . When
T2 is guessed at 2000 K, the left side becomes −28006.7. When T2 is guessed at 2200 K, the left side
becomes 33895.3. Interpolate then to arrive at
T2 = 2090.5 K.
(4.181)
The inert diluent significantly lowers the adiabatic flame temperature. This is because the N2 serves as
a heat sink for the energy of reaction. If the mixture were at non-stoichiometric conditions, the excess
species would also serve as a heat sink, and the adiabatic flame temperature would be lower than that
of the stoichiometric mixture.
The final pressure is given by
P2
=
=
=
(nH2 O + nN2 )RT2
,
V
kJ
(2 kmole + 8 kmole) 8.314 kmole
3
272.533 m
637.74 kP a.
(4.182)
K
(2090.5 K)
,
(4.183)
(4.184)
The final concentrations of H2 O and N2 are
4.3.3.3
ρH2 O
=
ρN2
=
2 kmole
kmole
= 7.34 × 10−3
,
272.533 m3
m3
8 kmole
kmole
= 2.94 × 10−2
.
272.533 m3
m3
(4.185)
(4.186)
Dilute, preheated mixture
Example 4.11
Consider a variant on the previous example in which the diluted mixture is preheated to 1000 K.
One can achieve this via an isentropic compression of the cold mixture, such as might occur in an engine.
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157
4.3. FIRST LAW ANALYSIS OF REACTING SYSTEMS
To simplify the analysis here, the temperature of the mixture will be increased, while the pressure will
be maintained. A closed, fixed, adiabatic volume contains a stoichiometric mixture of 2 kmole of H2 ,
1 kmole of O2 , and 8 kmole of N2 at 100 kP a and 1000 K. Find the adiabatic flame temperature and
the final pressure, assuming the irreversible reaction
2H2 + O2 + 8N2 → 2H2 O + 8N2 .
(4.187)
The volume is given by
V
=
=
=
(nH2 + nO2 + nN2 )RT1
,
P1
(2 kmole + 1 kmole + 8 kmole) 8.314
100 kP a
914.54 m3 .
(4.188)
kJ
kmole K
(1000 K)
,
(4.189)
(4.190)
The first law gives
E2 − E1
E2 − E1
=
=
nH2 O eH2 O − nH2 eH2 − nO2 eO2 + nN2 (eN2 2 − eN2 1 ) =
nH2 O (hH2 O − RT2 ) − nH2 (hH2 − RT1 ) − nO2 (hO2 − RT1 ) + nN2 ((hN2 2 − RT2 ) − (hN2 1 − RT1 )) =
2hH2 O − 2hH2 − hO2 + R(−10T2 + 11T1 ) + 8(hN2 2 − hN2 1 ) =
2(−241826 + ∆hH2 O ) − 2(20663) − 22703 + (8.314) (−10T2 + (11)(1000)) + 8∆hN2 2 − 8(21463) =
2∆hH2 O − 83.14T2 − 627931 + 8∆hN2 2
=
1Q2
0,
− 1W2 ,
0,
0,
0,
0,
0,
At this point, one begins an iteration process, guessing a value of T2 and an associated ∆hH2 O . When
T2 is guessed at 2600 K, the left side becomes −11351. When T2 is guessed at 2800 K, the left side
becomes 52787. Interpolate then to arrive at
T2 = 2635.4 K.
(4.191)
The preheating raised the adiabatic flame temperature. Note that the preheating was by 1000 K −
298 K = 702 K. The new adiabatic flame temperature is only 2635.4 K − 2090.5 K = 544.9 K greater.
The final pressure is given by
P2
=
=
=
(nH2 O + nN2 )RT2
,
V
kJ
(2 kmole + 8 kmole) 8.314 kmole
3
914.54 m
239.58 kP a.
(4.192)
K
(2635.4 K)
,
(4.193)
(4.194)
The final concentrations of H2 O and N2 are
ρH2 O
=
ρN 2
=
2 kmole
kmole
= 2.19 × 10−3
,
3
914.54 m
m3
8 kmole
kmole
= 8.75 × 10−3
.
3
914.54 m
m3
(4.195)
(4.196)
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CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
4.3.3.4
Dilute, preheated mixture with minor species
Example 4.12
Consider a variant on the previous example. Here, allow for minor species to be present at equilibrium. A closed, fixed, adiabatic volume contains a stoichiometric mixture of 2 kmole of H2 , 1 kmole
of O2 , and 8 kmole of N2 at 100 kP a and 1000 K. Find the adiabatic flame temperature and the final
pressure, assuming reversible reactions.
Here, the details of the analysis are omitted, but the result is given which is the consequence of a
calculation involving detailed reactions rates involving equations of the form to be studied in Sec. 5.1.
One can also solve an optimization problem to minimize the Gibbs free energy of a wide variety of
products to get the same answer; for a general discussion, see Sec. 5.2.1. In this case, the equilibrium
temperature and pressure are found to be
T = 2484.8 K,
P = 227.89 kP a.
(4.197)
Equilibrium species concentrations are found to be
ρH2
minor product
(4.198)
(4.199)
minor product
ρH
=
minor product
ρO
=
ρO2
=
ρOH
=
major product
ρH2 O
=
trace product
ρHO2
=
ρH2 O2
=
ρN
=
ρN H
=
trace product
ρN H2
=
trace product
ρN H3
=
ρN N H
=
minor product
ρN O
=
trace product
ρN O2
=
trace product
ρN 2 O
=
minor product
minor product
trace product
trace product
trace product
trace product
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kmole
,
m3
kmole
,
1.9 × 10−5
m3
kmole
,
5.7 × 10−6
m3
kmole
3.6 × 10−5
,
m3
kmole
5.9 × 10−5
,
m3
kmole
2.0 × 10−3
,
m3
kmole
,
1.1 × 10−8
m3
kmole
1.2 × 10−9
,
m3
kmole
1.7 × 10−9
,
m3
kmole
3.7 × 10−10
,
m3
kmole
,
1.5 × 10−10
m3
kmole
,
3.1 × 10−10
m3
kmole
1.0 × 10−10
,
m3
kmole
3.1 × 10−6
,
m3
kmole
,
5.3 × 10−9
m3
kmole
,
2.6 × 10−9
m3
= 1.3 × 10−4
(4.200)
(4.201)
(4.202)
(4.203)
(4.204)
(4.205)
(4.206)
(4.207)
(4.208)
(4.209)
(4.210)
(4.211)
(4.212)
(4.213)
159
4.4. CHEMICAL EQUILIBRIUM
trace product
ρHN O
= 1.7 × 10−9
ρN 2
= 8.7 × 10−3
major product
kmole
,
m3
kmole
.
m3
(4.214)
(4.215)
Note that the concentrations of the major products went down when the minor species were considered.
The adiabatic flame temperature also went down by a significant amount: 2635 K −2484.8 K = 150.2 K.
Some thermal energy was necessary to break the bonds which induce the presence of minor species.
4.4
Chemical equilibrium
Often reactions are not simply unidirectional, as alluded to in the previous example. The
reverse reaction, especially at high temperature, can be important.
Consider the four species reaction
ν1′ χ1 + ν2′ χ2 ⇌ ν3′′ χ3 + ν4′′ χ4
(4.216)
In terms of the net stoichiometric coefficients, this becomes
ν1 χ1 + ν2 χ2 + ν3 χ3 + ν4 χ4 = 0.
(4.217)
One can define a variable ζ, the reaction progress. Take the dimension of ζ to be mole.
When t = 0, one takes ζ = 0. Now, as the reaction goes forward, one takes dζ > 0. And a
forward reaction will decrease the number of moles of χ1 and χ2 while increasing the number
of moles of χ3 and χ4 . This will occur in ratios dictated by the stoichiometric coefficients of
the problem:
dn1
dn2
dn3
dn4
=
=
=
=
−ν1′ dζ,
−ν2′ dζ,
+ν3′′ dζ,
+ν4′′ dζ.
(4.218)
(4.219)
(4.220)
(4.221)
Note that if ni is taken to have units of mole, νi′ , and νi′′ are taken as dimensionless, then ζ
must have units of mole. In terms of the net stoichiometric coefficients, one has
dn1
dn2
dn3
dn4
=
=
=
=
ν1 dζ,
ν2 dζ,
ν3 dζ,
ν4 dζ.
(4.222)
(4.223)
(4.224)
(4.225)
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CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
Again, for argument’s sake, assume that at t = 0, one has
n1 |t=0
n2 |t=0
n3 |t=0
n4 |t=0
=
=
=
=
n1o ,
n2o ,
n3o ,
n4o .
(4.226)
(4.227)
(4.228)
(4.229)
Then, after integrating, one finds
n1
n2
n3
n4
=
=
=
=
ν1 ζ + n1o ,
ν2 ζ + n2o ,
ν3 ζ + n3o ,
ν4 ζ + n4o .
(4.230)
(4.231)
(4.232)
(4.233)
One can also eliminate the parameter ζ in a variety of fashions and parameterize the
reaction by one of the species mole numbers. Choosing, for example, n1 as a parameter, one
gets
n1 − n1o
.
(4.234)
ζ=
ν1
Eliminating ζ, one finds all other mole numbers in terms of n1 :
n1 − n1o
+ n2o ,
ν1
n1 − n1o
= ν3
+ n3o ,
ν1
n1 − n1o
= ν4
+ n4o .
ν1
n2 = ν2
(4.235)
n3
(4.236)
n4
(4.237)
Written another way, one has
n2 − n2o
n3 − n3o
n4 − n4o
n1 − n1o
=
=
=
= ζ.
ν1
ν2
ν3
ν4
For an N-species reaction,
PN
i=1
(4.238)
νi χi = 0, one can generalize to say
dni = = νi dζ,
ni = νi ζ + nio ,
ni − nio
= ζ.
νi
Note that
CC BY-NC-ND. 30 March 2014, J. M. Powers.
dni
= νi .
dζ
(4.239)
(4.240)
(4.241)
(4.242)
161
4.4. CHEMICAL EQUILIBRIUM
Now, from the previous Chapter 3, one manifestation of the second law is Eq. (3.361):
dG|T,P =
N
X
i=1
µi dni ≤ 0.
(4.243)
Now, one can eliminate dni in Eq. (4.243) by use of Eq. (4.239) to get
dG|T,P =
N
X
i=1
µi νi dζ ≤ 0,
N
X
∂G µi νi ≤ 0,
=
∂ζ T,P
i=1
= −α ≤ 0.
(4.244)
(4.245)
(4.246)
Then, for the reaction to go forward, one must require that the affinity, defined earlier in
Eq. (3.472), be positive:
α ≥ 0.
(4.247)
One also knows from Chapter 3 that the irreversibility takes the form of Eq. (3.350):
−
N
1X
µ dni ≥ 0,
T i=1 i
(4.248)
N
1 X
µi νi ≥ 0,
− dζ
T
i=1
N
1 dζ X
−
µ νi ≥ 0.
T dt i=1 i
In terms of the chemical affinity, α = −
PN
i=1
(4.249)
(4.250)
µi νi , Eq. (4.250) can be written as
1 dζ
α ≥ 0.
T dt
(4.251)
Now, one straightforward, albeit naı̈ve, way to guarantee positive semi-definiteness of the
irreversibility and thus satisfaction of the second law is to construct the chemical kinetic rate
equation so that
N
X
dζ
= −k
µi νi = kα,
dt
i=1
k ≥ 0,
provisional, naı̈ve assumption
(4.252)
This provisional assumption of convenience will be supplanted later in Sec. 4.5 by a form
which agrees well with experiment. Here, k is a positive semi-definite scalar. In general, it is
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162
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
a function of temperature, k = k(T ), so that reactions proceed rapidly at high temperature
and slowly at low temperature. Then, certainly the reaction progress variable ζ will cease
to change when the equilibrium condition
N
X
µi νi = 0,
(4.253)
i=1
is met. This is equivalent to requiring
α = 0.
(4.254)
Now, while Eq. (4.253) is the most compact form of the equilibrium condition, it is not
the most commonly used form. One can perform the following analysis to obtain the form in
most common usage. Start by employing Eq. (3.268) equating the chemical potential with
the Gibbs free energy per unit mole for each species i: µi = g i . Then, employ the definition
of Gibbs free energy for an ideal gas, and carry out a set of operations:
N
X
g i νi = 0,
at equilibrium,
(4.255)
(hi − T si )νi = 0,
at equilibrium.
(4.256)
i=1
N
X
i=1
For the ideal gas, one can substitute for hi (T ) and si (T, P ) and write the equilibrium condition as













Z T
Z T
N


X
 o
c
(
T̂
)
y
P


P
i
i
o
h +
νi = 0,
d
T̂
c
(
T̂
)
d
T̂
s
+
ln
−T
−R


P
i
T
,i
T
,i
o
 o


P
T̂
o
T
T

o
o
i=1 


|
{z
}
{z
}


|
o

=soT,i
=∆hT,i


|
|
{z
} 
{z
}
=sT,i
o
=hT,i =hT,i
(4.257)
Now, writing the equilibrium condition in terms of the enthalpies and entropies referred to
the standard pressure, one gets
N X
yi P
o
o
νi = 0,
hT,i − T sT,i − R ln
Po
i=1
N X
i=1
|
o
hT,i
T soT,i
−
{z
=g oT,i =µoT,i
CC BY-NC-ND. 30 March 2014, J. M. Powers.
}
νi = −
(4.258)
N
X
i=1
RT νi ln
yi P
Po
,
(4.259)
163
4.4. CHEMICAL EQUILIBRIUM
−
N
X
g oT,i νi
|i=1 {z
≡∆Go
= RT
N
X
ln
i=1
}
yi P
Po
νi
,
ν
N
X
∆Go
yi P i
−
=
ln
,
Po
RT
i=1
ν !
N Y
yi P i
,
= ln
Po
i=1
ν
N Y
∆Go
yi P i
exp −
=
,
Po
RT
i=1
|
{z
}
≡KP
KP =
ν
N Y
yi P i
i=1
KP =
KP =
,
Po
PNi=1 νi Y
n
P
Po
N Y
i=1
Pi
Po
νi
yiνi ,
(4.260)
(4.261)
(4.262)
(4.263)
(4.264)
(4.265)
i=1
, at equilibrium. (4.266)
Here, KP is what is known as the pressure-based equilibrium constant. It is dimensionless.
Despite its name, it is not a constant. It is defined in terms of thermodynamic properties,
and for the ideal gas is a function of T only:
∆Go
KP ≡ exp −
,
generally valid.
(4.267)
RT
Only at equilibrium does the property KP also equal the product of the partial pressures
as in Eq. (4.266). The subscript P for pressure comes about because it is also related to
the product of the ratio of the partial pressure to the reference pressure raised to the net
stoichiometric coefficients. Also, the net change in Gibbs free energy of the reaction at the
reference pressure, ∆Go , which is a function of T only, has been defined as
o
∆G ≡
o
N
X
g oT,i νi .
(4.268)
i=1
The term ∆G has units of kJ/kmole; it traditionally does not get an overbar. If ∆Go > 0,
one has 0 < KP < 1, and reactants are favored over products. If ∆Go < 0, one gets KP > 1,
and products are favored over reactants. One can also deduce that higher pressures P push
the equilibrium in such a fashion that fewer moles are present, all else being equal. One can
also define ∆Go in terms of the chemical affinity, referred to the reference pressure, as
∆Go = −αo .
(4.269)
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CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
One can also define another convenient thermodynamic property, which for an ideal gas
is a function of T alone, the equilibrium constant Kc :
Kc ≡
Po
RT
PNi=1 νi
∆Go
,
exp −
RT
generally valid.
(4.270)
This property is dimensional, and the
units depend on the stoichiometry of the reaction.
PN
The units of Kc will be (mole/cm3 ) i=1 νi .
The equilibrium condition, Eq. (4.266), is often written in terms of molar concentrations
and Kc . This can be achieved by the operations, valid only at an equilibrium state:
KP =
νi
N Y
ρ RT
i
i=1
exp
|
Po
RT
PNi=1 νi
exp
{z
−∆Go
RT
−∆Go
RT
≡Kc
}
=
=
Kc =
RT
Po
N
Y
,
Po
PNi=1 νi Y
N
(4.271)
ρi νi ,
(4.272)
i=1
ρi νi ,
(4.273)
i=1
N
Y
ρi νi ,
at equilibrium.
(4.274)
i=1
One must be careful to distinguish between the general definition of Kc as given in Eq. (4.270),
and the fact that at equilibrium it is driven to also have the value of product of molar species
concentrations, raised to the appropriate stoichiometric power, as given in Eq. (4.274).
4.5
Chemical kinetics of a single isothermal reaction
In the same fashion in ordinary mechanics that an understanding of statics enables an understanding of dynamics, an understanding of chemical equilibrium is necessary to understand
to more challenging topic of chemical kinetics. Chemical kinetics describes the time-evolution
of systems which may have an initial state far from equilibrium; it typically describes the
path of such systems to an equilibrium state. Here, gas phase kinetics of ideal gas mixtures
that obey Dalton’s law will be studied. Important topics such as catalysis and solid or liquid
reactions will not be considered.
Further, this section will be restricted to strictly isothermal systems. This simplifies the
analysis greatly. It is straightforward to extend the analysis of this system to non-isothermal
systems. One must then make further appeal to the energy equation to get an equation for
temperature evolution.
The general form for evolution of species is taken to be
CC BY-NC-ND. 30 March 2014, J. M. Powers.
4.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION
d
dt
ρi
ω̇i
= .
ρ
ρ
165
(4.275)
Multiplying both sides of Eq. (4.275) by molecular mass Mi and using Eq. (2.43) to exchange
ρi for mass fraction Yi then gives the alternate form
dYi
ω̇i Mi
=
.
dt
ρ
4.5.1
(4.276)
Isochoric systems
Consider the evolution of species concentration in a system which is isothermal, isochoric
and spatially homogeneous. The system is undergoing a single chemical reaction involving
N species of the familiar form of Eq. (4.21):
N
X
νi χi = 0.
(4.277)
i=1
Because the density is constant for the isochoric system, Eq. (4.275) reduces to
dρi
= ω̇i .
dt
(4.278)
Then, experiment, as well as a more fundamental molecular collision theory, shows that the
evolution of species concentration i is given by
z
≡ω̇i
}|

{
!
Y
N
N

′
dρi
−E
1 Y νk 
ν


β
k
ρk
ρk ,
= νi aT exp
1 −
dt
Kc


RT
|
{z
} | k=1{z }
| k=1
{z }
≡k(T )
reverse reaction
forward reaction
|
{z
}
isochoric system
(4.279)
≡r
This relation actually holds for isochoric, non-isothermal systems as well, which will not be
considered in any detail here. Here, some new variables are defined as follows:
• a: a kinetic rate constant called the collision frequency factor. Its units will depend
on the actual reaction and could involve various combinations of length, time, and
temperature. It is constructed so that
dρ /dt has units of mole/cm3 /s; this requires
PN i ′
′
′
is a coefficient which is
it to have units of (mole/cm3 )(1−νM − k=1 νk ) /s/K β . Here, νM
present if an inert third body participates in the reaction.
• β: a dimensionless parameter whose value is set by experiments, sometimes combined
with guiding theory, to account for weak temperature dependency of reaction rates.
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CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
• E: the activation energy. It has units of cal/mole, though others are often used, and is
fit by both experiment and fundamental theory to account for the strong temperature
dependency of reaction.
Note that in Eq. (4.279) that molar concentrations are raised to the νk′ and νk powers. As
it does not make sense to raise a physical quantity to a power with units, one traditionally
interprets the values of νk , νk′ , as well as νk′′ to be dimensionless pure numbers. They are also
interpreted in a standard fashion: the smallest integer values that actually correspond to the
underlying molecular collision which has been modelled. While stoichiometric balance can
be achieved by a variety of νk values, the kinetic rates are linked to one particular set which
is defined by the community.
Equation (4.279) is written in such a way that the species concentration production rate
increases when
• The net number of moles generated in the reaction, measured by νi increases,
• The temperature increases; here, the sensitivity may be very high, as one observes in
nature,
• The species concentrations of species involved in the forward reaction increase; this
embodies the principle that the collision-based reaction rates are enhanced when there
are more molecules to collide,
• The species concentrations of species involved in the reverse reaction decrease.
Here, three intermediate variables which are in common usage have been defined. First one
takes the reaction rate to be


!
Y
N
N

1 Y νk 
−E
νk′


β
(4.280)
ρk
ρk  ,
r ≡ aT exp
1 −
Kc k=1


RT
k=1
|
{z
} | {z }
| {z }
≡k(T )
reverse reaction
forward reaction





}

N
1 Y νk′′ 

ρk  .
K

c
k=1
{z }
| {z }
| k=1
≡k(T ), Arrhenius rate
| forward reaction{z reverse reaction }
β
= aT exp
|
{z
−E
RT
N
Y
νk′
ρk
−
(4.281)
law of mass action
The reaction rate r has units of mole/cm3 /s.
The temperature-dependency of the reaction rate is embodied in k(T ) is defined by what
is known as an Arrhenius7 rate law:
−E
β
k(T ) ≡ aT exp
.
(4.282)
RT
7
Svante Arrhenius, 1859-1927, Swedish physicist.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
4.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION
167
This equation was advocated by van’t Hoff8 in 1884; in 1889 Arrhenius gave a physical justification. The units of k(T ) actually depend on the reaction. This is a weakness of the
theory,
P
′ − N
3 (1−νM
νk′ )
k=1
/s.
and precludes a clean non-dimensionalization. The units must be (mole/cm )
In terms of reaction progress, one can also take
r=
1 dζ
.
V dt
(4.283)
The factor of 1/V is necessary because r has units of molar concentration per time and ζ
has units of mole. The over-riding importance of the temperature sensitivity is illustrated
as part of the next example. The remainder of the expression involving the products of the
species concentrations is the defining characteristic of systems which obey the law of mass
action. Though the history is complex, most attribute the law of mass action to Guldberg9
and Waage10 in 1864.11 Last, the overall molar production rate of species i, often written as
ω̇i , is defined as
ω̇i ≡ νi r.
(4.284)
As νi is considered to be dimensionless, the units of ω̇i must be mole/cm3 /s.
Example 4.13
Study the nitrogen dissociation problem considered in an earlier example which was confined to
equilibrium analysis, see Sec. 3.11.2.1, in which at t = 0 s, 1 kmole of N2 exists at P = 100 kP a and
T = 6000 K. Take as before the reaction to be isothermal and isochoric. Consider again the elementary
nitrogen dissociation reaction
N2 + N2 ⇌ 2N + N2 ,
(4.285)
which has kinetic rate parameters of
a
= 7.0 × 1021
β
= −1.6,
E
cm3 K 1.6
,
mole s
(4.286)
(4.287)
cal
= 224928.4
.
mole
(4.288)
In SI units, this becomes
a
=
E
=
3 3
1.6
m3 K 1.6
1m
1000 mole
21 cm K
= 7.0 × 1018
,
7.0 × 10
mole s
100 cm
kmole
kmole s
cal
J
kJ
kJ
1000 mole
224928.4
4.186
= 941550
.
mole
cal
1000 J
kmole
kmole
(4.289)
(4.290)
8
Jacobus Henricus van’t Hoff, 1852-1922, Dutch chemist.
Cato Maximilian Guldberg, 1836-1902, Norwegian mathematician and chemist.
10
Peter Waage, 1833-1900, Norwegian chemist.
11
P. Waage and C. M. Guldberg, 1864, “Studies Concerning Affinity, Forhandlinger: Videnskabs-Selskabet
i Christiania, 35. English translation: Journal of Chemical Education, 63(12): 1044-1047.
9
CC BY-NC-ND. 30 March 2014, J. M. Powers.
168
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
At the initial state, the material is all N2 , so PN2 = P = 100 kP a. The ideal gas law then gives at
t=0
P |t=0 = PN2 |t=0 = ρN2 t=0 RT,
(4.291)
P |t=0
,
(4.292)
ρN2 t=0 =
RT
100 kP a
,
(4.293)
=
kJ
8.314 kmole
K (6000 K)
kmole
.
(4.294)
= 2.00465 × 10−3
m3
Thus, the volume, constant for all time in the isochoric process, is
V =
nN2 |t=0
1 kmole
=
ρN2 t=0
2.00465 × 10−3
kmole
m3
= 4.9884 × 102 m3 .
(4.295)
Now, the stoichiometry of the reaction is such that
−dnN2
1
dnN ,
2
1
(nN − nN |t=0 ),
2
| {z }
=
−(nN2 − nN2 |t=0 ) =
| {z }
ρN
(4.297)
=0
=1 kmole
nN
nN
V
(4.296)
2(1 kmole − nN2 ),
1 kmole nN2
,
2
−
V
V
1 kmole
2
− ρN 2 ,
4.9884 × 102 m3
−3 kmole
− ρN 2 .
2 2.00465 × 10
m3
=
=
=
=
(4.298)
(4.299)
(4.300)
(4.301)
Now, the general equation for kinetics of a single reaction, Eq. (4.279), reduces for N2 molar
concentration to
′
dρN2
′
1
−E
νN
ν
ν
ν
β
N
N
N
2
1−
(4.302)
(ρN2 ) 2 (ρN )
(ρ ) (ρN )
= νN2 aT exp
dt
Kc N2
RT
′
′
Realizing that νN
= 2, νN
= 0, νN2 = −1, and νN = 2, one gets
2
dρN2
dt
=
β
− aT exp
{z
|
−E
RT
=k(T )
}
ρ2N2
1 ρ2N
1−
K c ρN 2
Examine the primary temperature dependency of the reaction
−E
β
k(T ) = aT exp
,
RT
!
kJ
3 1.6
−941550 kmole
18 m K
−1.6
,
=
7.0 × 10
T
exp
kJ
kmole s
8.314 kmole
KT
7.0 × 1018
−1.1325 × 105
=
exp
T 1.6
T
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.303)
(4.304)
(4.305)
(4.306)
169
4.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION
k(T) (m3/kmole/s)
10 4
10 -4
10 -12
10 -20
10 -28
10 -36
1000
1500 2000
3000
5000
7000
10000
T (K)
Figure 4.1: k(T ) for Nitrogen dissociation example.
Figure 4.1 gives a plot of k(T ) which shows its very strong dependency on temperature. For this
problem, T = 6000 K, so
7.0 × 1018
−1.1325 × 105
k(6000) =
,
(4.307)
exp
60001.6
6000
m3
.
(4.308)
= 40071.6
kmole s
Now, the equilibrium constant Kc is needed. Recall
Kc =
Po
RT
PN
i=1 νi
exp
−∆Go
RT
(4.309)
PN
For this system, since i=1 νi = 1, this reduces to
−(2goN − g oN2 )
Po
Kc =
exp
,
(4.310)
RT
RT
!
o
o
−(2(hN − T soT,N ) − (hN2 − T soT,N2 ))
Po
=
exp
,
(4.311)
RT
RT
100
−(2(597270 − (6000)216.926) − (205848 − (6000)292.984))
=
exp
,(4.312)
(8.314)(6000)
(8.314)(6000)
kmole
= 0.000112112
.
(4.313)
m3
The differential equation for N2 evolution is then given by
dρN2
dt
= − 40071.6
|
m3
kmole
ρ2N2
2 2.00465 × 10−3
1
1−
ρN 2
0.000112112 kmole
m3
{z
≡f (ρN2 )
kmole
m3
− ρN2
2 !
,
}
CC BY-NC-ND. 30 March 2014, J. M. Powers.
170
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
f ( ρN) (kmole/m 3/s)
2
unstable
equilibrium
1
0.0005
0.001
0.0015
0.002
0.0025
ρ
-1
unstable
equilibrium
-2
N2
(kmole/m 3)
stable,
physical
equilibrium
Figure 4.2: Forcing function, f (ρN2 ), which drives changes of ρN2 as a function of ρN2 in
isothermal, isochoric problem.
(4.314)
= f (ρN2 ).
(4.315)
The system is at equilibrium when f (ρN2 ) = 0. This is an algebraic function of ρN2 only, and can be
plotted. Figure 4.2 gives a plot of f (ρN2 ) and shows that it has three potential equilibrium points. It
is seen there are three roots. Solving for the equilibria requires solving
2 !
−3 kmole
−
ρ
2
2.00465
×
10
m3
1
3
N
m
2
.
0 = − 40071.6
ρ2N2 1 −
kmole
ρN 2
0.000112112 kmole
m3
(4.316)
The three roots are
kmole
,
ρN 2 = 0
m3 }
| {z
unstable
0.00178121
|
{z
stable
kmole
,
m3 }
0.00225611
|
{z
kmole
.
m3 }
(4.317)
unstable
By inspection of the topology of Fig. 4.2, the only stable root is 0.00178121 kmole/m3. This root
agrees with the equilibrium value found in an earlier example for the same problem conditions. Small
perturbations from this equilibrium induce the forcing function to supply dynamics which restore the
system to its original equilibrium state. Small perturbations from the unstable equilibria induce nonrestoring dynamics. For this root, one can then determine that the stable equilibrium value of ρN =
0.000446882 kmole/m3 .
One can examine this stability more formally. Define an equilibrium concentration ρeq
N2 such that
f (ρeq
N2 ) = 0.
(4.318)
ρeq
N2 :
Now, perform a Taylor series of f (ρN2 ) about ρN2 =
df 1 d2 f
eq
eq
2
(ρ − ρeq
−
)
+
(ρ
ρ
f (ρN2 ) ∼ f (ρN2 ) +
N2
N2
N2 ) + . . .
2 dρ2N2 N2
| {z } dρN2 ρ =ρeq
=0
N2
N2
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.319)
4.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION
171
Now, the first term of the Taylor series is zero by construction. Next, neglect all higher order terms as
small so that the approximation becomes
df f (ρN2 ) ∼
(4.320)
(ρ − ρeq
N2 ).
dρN2 ρ =ρeq N2
N2
N2
Thus, near equilibrium, one can write
dρN2
df ∼
dt
dρN2 ρ
eq
N2 =ρN2
(ρN2 − ρeq
N2 ).
(4.321)
Since the derivative of a constant is zero, one can also write the equation as
df d
eq
(ρ − ρeq
(ρ − ρN2 ) ∼
N2 ).
dt N2
dρN2 ρ =ρeq N2
N2
(4.322)
N2
This has a solution, valid near the equilibrium point, of


df

(ρN2 − ρeq
t ,
N2 ) = C exp
dρN2 ρ =ρeq
N2
N2


df

ρN2 = ρeq
t .
N2 + C exp
dρN2 ρ =ρeq
N2
(4.323)
(4.324)
N2
(4.325)
Here, C is some constant whose value is not important for this discussion. If the slope of f is positive,
that is,
df > 0,
unstable,
(4.326)
dρN2 ρ =ρeq
N2
N2
the equilibrium will be unstable. That is, a perturbation will grow without bound as t → ∞. If the
slope is zero,
df = 0,
neutrally stable,
(4.327)
dρN2 ρ =ρeq
N2
N2
the solution is stable in that there is no unbounded growth, and moreover is known as neutrally stable.
If the slope is negative,
df < 0,
asymptotically stable,
(4.328)
dρN2 ρ =ρeq
N2
N2
the solution is stable in that there is no unbounded growth, and moreover is known as asymptotically
stable.
A solution via numerical integration is found for Eq. (4.314). The solution for ρN2 , along with ρN
is plotted in Fig. 4.3. Linearization of Eq. (4.314) about the equilibrium state gives rise to the locally
linearly valid
d
(4.329)
ρ − 0.00178121 = −1209.39(ρN2 − 0.00178121) + . . .
dt N2
This has local asymptotically stable solution
ρN2 = 0.00178121 + C exp (−1209.39t) .
(4.330)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
172
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
concentration (kmole/m3)
0.0020
ρ
N2
0.0015
0.0010
0.0005
ρ
N
0.001
0.002
0.003
0.004
0.005
t (s)
Figure 4.3: ρN2 (t) and ρN (t) in isothermal, isochoric nitrogen dissociation problem.
Here, C is some integration constant whose value is irrelevant for this analysis. The time scale of
relaxation τ is the time when the argument of the exponential is −1, which is
τ=
1
= 8.27 × 10−4 s.
1209.39 s−1
(4.331)
One usually finds this time scale to have high sensitivity to temperature, with high temperatures giving
fast time constants and thus fast reactions.
The equilibrium values agree exactly with those found in the earlier example. Here, the kinetics
provide the details of how much time it takes to achieve equilibrium. This is one of the key questions
of non-equilibrium thermodynamics.
4.5.2
Isobaric systems
The form of the previous section is the most important as it is easily extended to a computational grid with fixed volume elements in fluid flow problems. However, there is another
important spatially homogeneous problem in which the formulation needs slight modification: isobaric reaction, with P equal to a constant. Again, in this section only isothermal
conditions will be considered.
In an isobaric problem, there can be volume change. Consider first the problem of isobaric
expansion of an inert mixture. In such a mixture, the total number of moles of each species
must be constant, so one gets
dni
= 0,
inert, isobaric mixture.
(4.332)
dt
Now, carry out the sequence of operations, realizing the total mass m is also constant:
1 d
(ni ) = 0,
m dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.333)
4.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION
d ni dt m
d ni V
dt V m
d ρi
dt ρ
1 dρi
ρ dρ
− 2i
ρ dt
ρ dt
dρi
dt
173
= 0,
(4.334)
= 0,
(4.335)
= 0,
(4.336)
= 0,
(4.337)
=
ρi dρ
.
ρ dt
(4.338)
So, a global density decrease of the inert material due to volume increase of a fixed mass
system induces a concentration decrease of each species. Extended to a material with a
single reaction rate r, one could say either
ρ dρ
dρi
= νi r + i ,
or
dt
ρ dt
d ρi
1
=
νi r,
generally valid,
dt ρ
ρ
ω̇i
.
=
ρ
(4.339)
(4.340)
(4.341)
Equation (4.340) is consistent with Eq. (4.275) and is actually valid for general systems with
variable density, temperature, and pressure.
However, in this section, it is required that pressure and temperature be constant. Now,
differentiate the isobaric, isothermal, ideal gas law to get the density derivative.
P =
0 =
N
X
i=1
N
X
ρi RT,
RT
i=1
N
X
(4.342)
dρi
,
dt
(4.343)
dρi
,
dt
i=1
N X
ρi dρ
,
0 =
νi r +
ρ dt
i=1
0 =
N
X
(4.344)
(4.345)
N
1 dρ X
0 = r
νi +
ρi ,
ρ
dt
i=1
i=1
PN
−r i=1 νi
dρ
=
PN ρi .
dt
i=1 ρ
(4.346)
(4.347)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
174
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
P
νi
ρr N
= − PNi=1 ,
i=1 ρi
PN
ρr i=1 νi
,
= −
P
RT
= −
= −
ρRT r
ρRT r
PN
i=1
(4.348)
(4.349)
νi
P
PN
,
k=1 νk
P
(4.350)
.
(4.351)
PN
Note that if there is no net number change in the reaction,
k=1 νk = 0, the isobaric,
isothermal reaction also guarantees there would be no density or volume change. It is
convenient to define the net number change in the elementary reaction as ∆n:
∆n ≡
N
X
νk .
(4.352)
k=1
Here, ∆n is taken to be a dimensionless pure number. It is associated with the number
change in the elementary reaction and not the actual mole change in a physical system; it
is, however, proportional to the actual mole change.
Now, use Eq. (4.351) to eliminate the density derivative in Eq. (4.339) to get
P
ρi ρRT r N
dρi
k=1 νk
= νi r −
,
(4.353)
dt
ρ
P




N
X


ρ
RT
i
,
ν
−
ν
= r
k
i
 |{z}

P
reaction effect

k=1
|
{z
}
expansion effect


= r
νi
|{z}
reaction effect
−
yi ∆n
| {z }
expansion effect
.
(4.354)
(4.355)
There are two terms dictating the rate change of species molar concentration. The first, a
reaction effect, is precisely the same term that drove the isochoric reaction. The second is
due to the fact that the volume can change if the number of moles change, and this induces
an intrinsic change in concentration. Note that the term ρi RT /P = yi , the mole fraction.
Example 4.14
Study a variant of the nitrogen dissociation problem considered in an earlier example in which at
t = 0 s, 1 kmole of N2 exists at P = 100 kP a and T = 6000 K. In this case, take the reaction to be
isothermal and isobaric. Consider again the elementary nitrogen dissociation reaction
N2 + N2 ⇌ 2N + N2 ,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.356)
4.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION
175
which has kinetic rate parameters of
a
= 7.0 × 1021
β
= −1.6,
E
cm3 K 1.6
,
mole s
(4.357)
(4.358)
cal
= 224928.4
.
mole
(4.359)
In SI units, this becomes
a
=
E
=
3 cm3 K 1.6
m3 K 1.6
1m
1000 mole
7.0 × 1021
= 7.0 × 1018
,
mole s
100 cm
kmole
kmole s
cal
kJ
1000 mole
J
kJ
224928.4
.
4.186
= 941550
mole
cal
1000 J
kmole
kmole
(4.360)
(4.361)
At the initial state, the material is all N2 , so PN2 = P = 100 kP a. The ideal gas law then gives at
t=0
P
ρN 2
t=0
= PN2 = ρN2 RT,
P
,
=
RT
100 kP a
=
,
kJ
8.314 kmole
K (6000 K)
kmole
.
= 2.00465 × 10−3
m3
(4.362)
(4.363)
(4.364)
(4.365)
Thus, the initial volume is
V |t=0 =
nN2 |t=0
1 kmole
=
ρi |t=0
2.00465 × 10−3
kmole
m3
= 4.9884 × 102 m3 .
(4.366)
In this isobaric process, one always has P = 100 kP a. Now, in general
P = RT (ρN2 + ρN ),
(4.367)
therefore one can write ρN in terms of ρN2 :
ρN
=
=
=
P
− ρN 2 ,
RT
100 kP a
− ρN2 ,
kJ
8.314 kmole
K (6000 K)
kmole
2.00465 × 10−3
− ρN 2 .
m3
(4.368)
(4.369)
(4.370)
Then, the equations for kinetics of a single isobaric isothermal reaction, Eq. (4.354) in conjunction
with Eq. (4.280), reduce for N2 molar concentration to
!
′
dρN2
ρ
RT
′
−E
1
N
νN
β
ν
ν
ν
2
νN 2 −
(ρ ) N2 (ρN ) N
=
aT exp
(νN2 + νN ) .
(ρN2 ) 2 (ρN ) N 1 −
dt
Kc N2
P
RT
{z
}
|
=r
(4.371)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
176
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
′
′
Realizing that νN
= 2, νN
= 0, νN2 = −1, and νN = 2, one gets
2
dρN2
dt
=
β
aT exp
{z
|
−E
RT
=k(T )
}
ρ2N2
!
ρN2 RT
1 ρ2N
1−
−1 −
.
K c ρN 2
P
The temperature dependency of the reaction is unchanged from the previous reaction:
−E
k(T ) = aT β exp
,
RT
!
kJ
3 1.6
−941550 kmole
18 m K
−1.6
=
7.0 × 10
T
exp
,
kJ
kmole s
8.314 kmole
KT
7.0 × 1018
−1.1325 × 105
.
=
exp
T 1.6
T
(4.372)
(4.373)
(4.374)
(4.375)
For this problem, T = 6000 K, so
−1.1325 × 105
7.0 × 1018
,
exp
60001.6
6000
m3
40130.2
.
kmole s
k(6000) =
=
(4.376)
(4.377)
The equilibrium constant Kc is also unchanged from the previous example. Recall
Kc =
Po
RT
PN
i=1 νi
exp
−∆Go
RT
.
(4.378)
P
For this system, since N
i=1 νi = ∆n = 1, this reduces to
−(2goN − g oN2 )
Po
Kc =
exp
,
(4.379)
RT
RT
−(2goN − g oN2 )
Po
=
exp
,
(4.380)
RT
RT
!
o
o
−(2(hN − T soT,N ) − (hN2 − T soT,N2 ))
Po
=
exp
,
(4.381)
RT
RT
100
−(2(597270 − (6000)216.926) − (205848 − (6000)292.984))
=
exp
,(4.382)
(8.314)(6000)
(8.314)(6000)
kmole
= 0.000112112
.
(4.383)
m3
The differential equation for N2 evolution is then given by
dρN2
dt
=
40130.2
× −1 −
≡
m3
kmole
ρN 2
1
1−
0.000112112 kmole
m3
!
kJ
8.314 kmole
K (6000 K)
,
100 kP a
ρ2N2
f (ρN2 ).
CC BY-NC-ND. 30 March 2014, J. M. Powers.
2 !
2.00465 × 10−3 kmole
− ρN 2
m3
ρN 2
(4.384)
(4.385)
4.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION
f ( ρN
)
2
177
(kmole/m 3/s)
4
unstable
equilibrium
unstable
equilibrium
3
2
1
-0.002
-0.001
0.001
0.002
0.003
ρ
N2
(kmole/m 3)
-1
stable,
physical
equilibrium
stable,
non-physical
equilibrium
Figure 4.4: Forcing function, f (ρN2 ), which drives changes of ρN2 as a function of ρN2 in
isothermal, isobaric problem.
The system is at equilibrium when f (ρN2 ) = 0. This is an algebraic function of ρN2 only, and can be
plotted. Figure 4.4 gives a plot of f (ρN2 ) and shows that it has four potential equilibrium points. It is
seen there are four roots. Solving for the equilibria requires solving
2 !
− ρN 2
2.00465 × 10−3 kmole
m3
1
2
m3
0 =
40130.2
ρN 2 1 −
kmole
ρN 2
0.000112112 kmole
m3
!
kJ
8.314 kmole
ρ
K (6000 K)
× −1 − N2
.
(4.386)
100 kP a
The four roots are
ρN2 = −0.002005
|
{z
kmole
,
m3 }
stable,non−physical
kmole
0
,
m3 }
| {z
unstable
kmole
0.001583
,
3
|
{z m }
stable,physical
kmole
0.00254
.
3
|
{z m }
(4.387)
unstable
By inspection of the topology of Fig. 4.2, the only stable, physical root is 0.001583 kmole/m3 . Small
perturbations from this equilibrium induce the forcing function to supply dynamics which restore the
system to its original equilibrium state. Small perturbations from the unstable equilibria induce nonrestoring dynamics. For this root, one can then determine that the stable equilibrium value of ρN =
0.000421 kmole/m3 .
A numerical solution via an explicit technique such as a Runge-Kutta integration is found for
Eq. (4.386). The solution for ρN2 , along with ρN is plotted in Fig. 4.5. Linearization of Eq. (4.386)
about the equilibrium state gives rise to the locally linearly valid
d
ρN2 − 0.001583 = −967.073(ρN2 − 0.001583) + . . .
dt
(4.388)
This has local solution
ρN2 = 0.001583 + C exp (−967.073t) .
(4.389)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
178
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
concentration (kmole/m3)
0.0020
ρ
N2
0.0015
0.0010
0.0005
ρ
N
0.001
0.002
0.003
0.004
0.005
t (s)
Figure 4.5: ρN2 (t) and ρN (t) in isobaric, isothermal nitrogen dissociation problem.
Again, C is an irrelevant integration constant. The time scale of relaxation τ is the time when the
argument of the exponential is −1, which is
τ=
1
= 1.03 × 10−3 s.
967.073 s−1
(4.390)
Note that the time constant for the isobaric combustion is about a factor 1.25 greater than for isochoric
combustion under the otherwise identical conditions.
The equilibrium values agree exactly with those found in the earlier example. Again, the kinetics
provide the details of how much time it takes to achieve equilibrium.
4.6
Some conservation and evolution equations
Here, a few useful global conservation and evolution equations are presented for some key
properties. Only some cases are considered, and one could develop more relations for other
scenarios.
4.6.1
Total mass conservation: isochoric reaction
One can easily show that the isochoric reaction rate model, Eq. (4.279), satisfies the principle
of mixture mass conservation. Begin with Eq. (4.279) in a compact form, using the definition
of the reaction rate r, Eq. (4.281), and perform the following operations:
dρi
= νi r,
dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.391)
4.6. SOME CONSERVATION AND EVOLUTION EQUATIONS
d
dt
ρYi
Mi
= νi r,
179
(4.392)
d
(ρYi ) = νi Mi r,
dt
L
X
d
(ρYi ) = νi
Ml φli r,
dt
l=1
| {z }
(4.393)
(4.394)
=Mi
N
X
L
X
d
Ml φli νi r,
(ρYi ) =
dt
d
(ρYi ) =
dt
i=1


l=1
N X
L
X
i=1 l=1
Ml φli νi r,
N
L X
N

X
d 
 X 
Yi  =
Ml φli νi r,
ρ
dt  i=1 
l=1 i=1
| {z }
(4.395)
(4.396)
(4.397)
=1
L
N
X
X
dρ
Ml
φli νi ,
= r
dt
l=1
|i=1{z }
(4.398)
=0
dρ
= 0.
dt
Note the term
4.6.2
PN
i=1
(4.399)
φli νi = 0 because of stoichiometry, Eq. (4.24).
Element mass conservation: isochoric reaction
Through a similar series of operations, one can show that the mass of each element, l =
1, . . . , L, in conserved in this reaction, which is chemical, not nuclear. Once again, begin
with Eq. (4.281) and perform a set of operations,
dρi
= νi r,
dt
dρ
φli i = φli νi r,
dt
(4.400)
l = 1, . . . , L,
d
(φli ρi ) = rφli νi ,
l = 1, . . . , L,
dt
N
N
X
X
d
rφli νi ,
l = 1, . . . , L,
(φli ρi ) =
dt
i=1
i=1
(4.401)
(4.402)
(4.403)
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180
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
d
dt
N
X
d
dt
N
X
φli ρi
i=1
i=1
φli ρi
!
!
= r
N
X
φli νi ,
l = 1, . . . , L,
|i=1{z }
(4.404)
=0
= 0,
l = 1, . . . , L.
(4.405)
P
The term N
i=1 φli ρi represents the number of moles of element l per unit volume, by the
following analysis
N
X
i=1
N
X
moles element l moles species i
moles element l
φli ρi =
=
≡ ρl e .
moles
species
i
volume
volume
i=1
(4.406)
Here, the elemental mole density, ρl e , for element l has been defined. So, the element
concentration for each element remains constant in a constant volume reaction process:
dρl e
= 0,
l = 1, . . . , L.
(4.407)
dt
One can also multiply by the elemental mass, Ml to get the elemental mass density, ρel :
ρel ≡ Ml ρl e ,
l = 1, . . . , L.
(4.408)
Since Ml is a constant, one can incorporate this definition into Eq. (4.407) to get
dρel
= 0,
l = 1, . . . , L.
(4.409)
dt
The element mass density remains constant in the constant volume reaction. One could also
simply say since the elements’ density is constant, and the mixture is simply a sum of the
elements, that the mixture density is conserved as well.
4.6.3
Consider a simple application of the first law of thermodynamics to reaction kinetics: that
of a closed, adiabatic, isochoric combustion process in a mixture of ideal gases. One may
be interested in the rate of temperature change. First, because the system is closed, there
can be no mass change, and because the system is isochoric, the total volume is a non-zero
constant; hence,
dm
= 0,
dt
d
(ρV ) = 0,
dt
dρ
= 0,
V
dt
dρ
= 0.
dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.410)
(4.411)
(4.412)
(4.413)
181
4.6. SOME CONSERVATION AND EVOLUTION EQUATIONS
For such a process, the first law of thermodynamics is
dE
= Q̇ − Ẇ .
dt
(4.414)
But there is no heat transfer or work in the adiabatic isochoric process, so one gets
dE
= 0,
dt
(4.415)
d
(me) = 0,
dt
de
dm
m +e
= 0,
dt
dt
|{z}
(4.416)
(4.417)
=0
de
= 0.
dt
(4.418)
Thus, for the mixture of ideal gases, e(T, ρ1 , . . . , ρN ) = eo . One can see how reaction rates
affect temperature changes by expanding the derivative in Eq. (4.418)
!
N
d X
Yi ei
= 0,
(4.419)
dt i=1
N
X
d
(Yi ei ) = 0,
dt
i=1
N X
dYi
dei
= 0,
+ ei
Yi
dt
dt
i=1


(4.420)
(4.421)
N
X
 dei dT
dYi 
 Yi
 = 0,
+
e
i
 dT dt

dt
|{z}
i=1
N X
i=1
cvi
dT
dYi
Yi cvi
+ ei
dt
dt
(4.422)
= 0,
(4.423)
N
N
X
dYi
dT X
ei
Yi cvi = −
,
dt i=1
dt
i=1
| {z }
=cv
dT
cv
dt
= −
dT
dt
= −
ρcv
N
X
i=1
N
X
i=1
(4.424)
d
ei
dt
ei Mi
dρi
,
dt
Mi ρi
ρ
,
(4.425)
(4.426)
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182
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
ρcv
dT
dt
dT
dt
= −
= −
N
X
ei Mi νi r,
(4.427)
i=1
r
PN
i=1
νi ei
ρcv
.
(4.428)
If one defines the net energy change of the reaction as
∆E ≡
N
X
νi ei ,
(4.429)
dT
r∆E
=−
.
dt
ρcv
(4.430)
i=1
one then gets
The rate of temperature change is dependent on the absolute energies, not the energy differences. If the reaction is going forward, so r > 0, and that is a direction in which the net
molar energy change is negative, then the temperature will rise.
4.6.4
Solving for the reaction dynamics in an adiabatic isobaric system requires some non-obvious
manipulations. First, the first law of thermodynamics says dE = δQ−δW . Since the process
is adiabatic, one has δQ = 0, so dE + P dV = 0. Since it is isobaric, one gets d(E + P V ) = 0,
or dH = 0. So, the total enthalpy is constant. Then
d
H = 0,
dt
d
dt
N
X
Yi
i=1
Yi
dYi
dhi
+ hi
= 0,
dt
dt
dhi dT
dYi
+ hi
= 0,
dT dt
dt
|{z}
cP i
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.432)
(4.433)
(4.434)
i=1
N
X
d
(Yi hi ) = 0,
dt
i=1
i=1
N
X
d
(mh) = 0,
dt
dh
= 0,
dt
!
N
X
Yi hi
= 0,
(4.431)
(4.435)
(4.436)
(4.437)
4.6. SOME CONSERVATION AND EVOLUTION EQUATIONS
N
X
183
N
Y i cP i
i=1
dT X dYi
+
hi
= 0,
dt
dt
i=1
(4.438)
N
N
X
dT X
dYi
Y i cP i +
hi
= 0,
dt i=1
dt
i=1
| {z }
=cP
(4.439)
ρi Mi
= 0,
ρ
N
d ρi
dT X
= 0.
+
hi Mi
cP
dt
dt
ρ
i=1
N
dT X d
cP
+
hi
dt
dt
i=1
(4.440)
(4.441)
Now, use Eq (4.340) to eliminate the term in Eq. (4.441) involving molar concentration
derivatives to get
N
cP
dT X νi r
+
= 0,
hi
dt
ρ
i=1
dT
dt
= −
(4.442)
r
PN
i=1 hi νi
ρcP
.
(4.443)
So, the temperature derivative is known as an algebraic function. If one defines the net
enthalpy change as
N
X
∆H ≡
hi νi ,
(4.444)
i=1
one gets that Eq. (4.443) transforms to
dT
r∆H
.
=−
dt
ρcP
(4.445)
dT
= −r∆H.
dt
(4.446)
or
ρcP
Equation (4.446) is in a form which can easily be compared to a form to be derived later
when we add variable pressure and diffusion effects.
Now, differentiate the isobaric ideal gas law to get the density derivative.
P =
0 =
N
X
i=1
N
X
i=1
ρi RT,
(4.447)
N
dT X
dρ
ρi R
RT i ,
+
dt
dt
i=1
(4.448)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
184
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
N
N
X
ρi dρ
dT X
,
0 =
ρ +
T νi r +
dt i=1 i i=1
ρ dt
N
N
N
X
1 dT X
1 dρ X
0 =
ρ +r
νi +
ρ,
T dt i=1 i
ρ dt i=1 i
i=1
PN
PN
− T1 dT
dρ
i=1 ρi − r
i=1 νi
dt
.
=
PN ρi
dt
i=1 ρ
(4.449)
(4.450)
(4.451)
One takes dT /dt from Eq. (4.443) to get
dρ
=
dt
1 r
T
PN
i=1
hi νi
ρcP
PN
i=1 ρi −
PN ρi
i=1 ρ
r
PN
i=1
νi
.
(4.452)
Now, recall from Eqs. (2.212) and (2.218) that ρ = ρ/M and cP = cP M, so ρ cP = ρcP .
Then Equation (4.452) can be reduced slightly:
=1
PN
hi νi
cP T
i=1
dρ
= rρ
dt
= rρ
= rρ
z }| {
N
X
ρ
i
ρ
i=1
PN
PN
i=1 νi
PN
i=1
i=1 ρi
P
− N
i=1 νi
hi νi
i=1 cP T
PN
i=1
PN
−
ρi
hi
cP T
P
RT
−1
νi
,
,
,
N
ρRT X
hi
= r
νi
−1 ,
P i=1
cP T
N
X
hi
−1 ,
= rM
νi
c
T
P
i=1
(4.453)
(4.454)
(4.455)
(4.456)
(4.457)
PN
where M is the mean molecular mass. Note for exothermic reaction
i=1 νi hi < 0, so
exothermic reaction induces a density decrease as the increased temperature at constant
pressure causes the volume to increase.
Then, using Eq. (4.457) to eliminate the density derivative in Eq. (4.339), and changing
the dummy index from i to k, one gets an explicit expression for concentration evolution:
N
X
ρ
dρi
= νi r + i rM
νk
dt
ρ
k=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
hk
−1 ,
cP T
(4.458)
4.6. SOME CONSERVATION AND EVOLUTION EQUATIONS

N


ρi X
hk

νk
−1 
= r νi + M
,
ρ
c
PT
| {z } k=1

=yi
= r νi + yi
N
X
k=1
νk
hk
−1
cP T
!
.
185
(4.459)
(4.460)
PN
Defining the change of enthalpy of the reaction as ∆H ≡
k=1 νk hk , and the change of
PN
number of the reaction as ∆n ≡ k=1 νk , one can also say
dρi
∆H
= r νi + yi
− ∆n
.
(4.461)
dt
cP T
Exothermic reaction, ∆H < 0, and net number increases, ∆n > 0, both tend to decrease the
molar concentrations of the species in the isobaric reaction.
Lastly, the evolution of the adiabatic, isobaric system, can be described by the simultaneous, coupled ordinary differential equations: Eqs. (4.443, 4.452, 4.460). These require
numerical solution in general. Note also that one could also employ a more fundamental
P
treatment as a differential algebraic system involving H = H1 , P = P1 = RT N
i=1 ρi and
Eq. (4.339).
4.6.5
Consider briefly combustion in a fixed finite volume in which there is simple convective heat
transfer with the surroundings. In general, the first law of thermodynamics is
dE
= Q̇ − Ẇ .
dt
(4.462)
Because the system is isochoric Ẇ = 0. And using standard relations from simple convective
heat transfer, one can say that
dE
= −hA(T − T∞ ).
dt
(4.463)
Here, h is the convective heat transfer coefficient and A is the surface area associated with
the volume V , and T∞ is the temperature of the surrounding medium. One can, much as
before, write E in detail and get an equation for the evolution of T .
4.6.6
Entropy evolution: Clausius-Duhem relation
Now, consider whether the kinetics law that has been posed actually satisfies the second law
of thermodynamics. Consider again Eq. (3.350). There is an algebraic relation on the right
CC BY-NC-ND. 30 March 2014, J. M. Powers.
186
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
side. If it can be shown that this algebraic relation is positive semi-definite, then the second
law is satisfied, and the algebraic relation is known as a Clausius12 -Duhem13 relation.
Now, take Eq. (3.350) and perform some straightforward operations on it:
dS|E,V
dS dt E,V
N
1X
µ dni ≥ 0,
= −
T i=1 i
|
{z
}
(4.464)
irreversibility
N
V X dni 1
= −
≥ 0,
µ
T i=1 i dt V
(4.465)
N
V X dρi
µ
= −
≥ 0,
T i=1 i dt
(4.466)
!


Y
N
N
N

1 Y νk 
−E
V X
ν′


β
µi νi aT exp
ρk k 1 −
ρk  ≥ 0,(4.467)
= −
T i=1
K


RT
c
|
{z
} | k=1{z }
{z }
| k=1
≡k(T )
reverse reaction
forward reaction
|
{z
}
≡r
!
!
N
N
N
X
Y
Y
′
V
1
ν
= −
1−
(4.468)
µi νi k(T )
ρk k
ρkνk ≥ 0,
T i=1
Kc k=1
k=1
!
! N
!
N
N
X
Y
1 Y νk
V
νk′
1−
ρk
ρk
µi νi ≥ 0.
(4.469)
= − k(T )
T
Kc
i=1
k=1
k=1
|
{z
}
=−α
Change the dummy index from k back to i.
!
!
N
N
Y
dS V
1 Y νi
νi′
=
k(T )
ρi
ρi
α ≥ 0,
1−
dt E,V
T
K
c
i=1
i=1
V
rα,
T
α dζ
.
=
T dt
=
(4.470)
(4.471)
(4.472)
Consider now the affinity α term in Eq. (4.469) and expand it so that it has a more useful
form:
α=−
12
13
N
X
i=1
µi νi = −
N
X
g i νi ,
i=1
Rudolf Julius Emanuel Clausius, 1822-1888, German physicist.
Pierre Maurice Marie Duhem, 1861-1916, French physicist.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(4.473)
187
4.6. SOME CONSERVATION AND EVOLUTION EQUATIONS
N X
Pi
o
νi ,
= −
g T,i + RT ln
P
o
i=1
νi
N
N
X
X
Pi
o
= −
g T,i νi −RT
ln
,
Po
i=1
i=1
| {z }
=∆Go


N
ν
 −∆Go X
Pi i 
,
−
ln
= RT 
 RT

P
o
| {z } i=1
=ln KP
ln KP − ln
= RT
= −RT
= −RT
= −RT
= −RT
ν
N Y
Pi i
(4.474)
(4.475)
(4.476)
!
,
P
o
i=1
νi !
N Y
1
Pi
ln
,
+ ln
KP
Po
i=1
ν !
N 1 Y Pi i
,
ln
KP i=1 Po


PNi=1 νi N Po
Y ρ RT νi
i
,
ln  RT
Kc
P
o
i=1
!
N
1 Y νi
ln
ρ
.
Kc i=1 i
(4.477)
(4.478)
(4.479)
(4.480)
(4.481)
Equation (4.481) is the common definition of affinity. Sometimes, it is written equivalently
as
!
Kc
.
(4.482)
α = RT ln QN
νi
ρ
i=1 i
Another form can be found by employing the definition of Kc from Eq. (4.270) to get
α = −RT ln
Po
RT
− PNi=1 νi
exp
∆Go
RT
Y
N
i=1
ρi νi
!
!!
∆G
Po
νi
ρi
,
= −RT
+ ln
RT
RT
i=1
!
− PNi=1 νi Y
N
P
o
ν
= −∆Go − RT ln
ρi i .
RT
i=1
o
− PNi=1 νi Y
N
,
(4.483)
(4.484)
(4.485)
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188
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
To see clearly that the entropy production rate is positive semi-definite, substitute
Eq. (4.481) into Eq. (4.469) to get
!
!
!!
N
N
N
Y
Y
Y
′
V
1
1
dS ν
=
k(T )
ρi i
1−
ρi νi
−RT ln
ρi νi
≥ 0,
dt E,V
T
K
K
c
c
i=1
i=1
i=1
= −RV k(T )
N
Y
ρi
νi′
i=1
!
1−
N
1 Y νi
ρ
Kc i=1 i
!
ln
N
1 Y νi
ρ
Kc i=1 i
!
(4.486)
≥ 0.
(4.487)
Define forward and reverse reaction coefficients, R′ , and R′′ , respectively, as
′
R ≡ k(T )
R′′ ≡
N
Y
(4.488)
i=1
N
k(T ) Y
Kc
′
ρi νi ,
′′
ρi νi .
(4.489)
i=1
Both R′ and R′′ have units of mole/cm3 /s. It is easy to see that
r = R′ − R′′ .
(4.490)
Note that since k(T ) > 0, Kc > 0, and ρi ≥ 0, that both R′ ≥ 0 and R′′ ≥ 0. Since
νi = νi′′ − νi′ , one finds that
N
N
1 Y νi
1 k(T ) Y νi′′ −νi′ R′′
ρi =
ρ
= ′.
Kc i=1
Kc k(T ) i=1 i
R
Then, Eq. (4.487) reduces to
dS dt E,V
′′ R
R′′
= −RV R 1 − ′ ln
≥ 0,
R
R′
′
R
′
′′
≥ 0.
= RV (R − R ) ln
R′′
′
(4.491)
(4.492)
(4.493)
Obviously, if the forward rate is greater than the reverse rate R′ − R′′ > 0, ln(R′ /R′′ ) > 0,
and the entropy production is positive. If the forward rate is less than the reverse rate,
R′ − R′′ < 0, ln(R′ /R′′ ) < 0, and the entropy production is still positive. The production
rate is zero when R′ = R′′ .
Note that the affinity α can be written as
′
R
.
(4.494)
α = RT ln
R′′
And so when the forward reaction rate exceeds the reverse, the affinity is positive. It is zero
at equilibrium, when the forward reaction rate equals the reverse.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
189
4.7. SIMPLE ONE-STEP KINETICS
4.7
Simple one-step kinetics
A common model in theoretical combustion is that of so-called simple one-step kinetics. Such
a model, in which the molecular mass does not change, is quantitatively appropriate only
for isomerization reactions. However, as a pedagogical tool as well as a qualitative model for
real chemistry, it can be valuable.
Consider the reversible reaction
A ⇌ B.
(4.495)
where chemical species A and B have identical molecular masses MA = MB = M. Consider
further the case in which at the initial state, no moles of A only are present. Also take the
reaction to be isochoric and isothermal. These assumptions can easily be relaxed for more
general cases. Specializing then Eq. (4.240) for this case, one has
nA =
νA ζ + nAo ,
|{z}
|{z}
=−1
nB =
=no
νB ζ + nBo .
|{z}
|{z}
=1
Thus
(4.496)
(4.497)
=0
nA = −ζ + no ,
nB = ζ.
(4.498)
(4.499)
Now, no is constant throughout the reaction. Scale by this and define the dimensionless
reaction progress as λ ≡ ζ/no to get
nA
= −λ + 1,
no
|{z}
(4.500)
=yA
nB
= λ.
no
|{z}
(4.501)
=yB
In terms of the mole fractions then, one has
yA = 1 − λ,
yB = λ.
(4.502)
(4.503)
The reaction kinetics for each species reduce to
dρA
= −r,
dt
dρB
= r,
dt
ρA (0) =
ρB (0) = 0.
no
≡ ρo ,
V
(4.504)
(4.505)
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190
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
Addition of Eqs. (4.504) and (4.505) gives
d
(ρ + ρB ) = 0,
dt A
ρA + ρB = ρo ,
ρA
ρ
+ B = 1.
ρo
ρo
|{z}
|{z}
=yA
(4.506)
(4.507)
(4.508)
=yB
In terms of the mole fractions yi , one then has
yA + yB = 1.
(4.509)
The reaction rate r is then
r =
=
=
=
1 ρB
kρA 1 −
,
Kc ρA
1 ρB /ρo
ρA
1−
,
kρo
ρo
Kc ρA /ρo
1 yB
kρo yA 1 −
,
Kc y A
1 λ
.
kρo (1 − λ) 1 −
Kc 1 − λ
(4.510)
(4.511)
(4.512)
(4.513)
Now, r = (1/V )dζ/dt = (1/V )d(no λ)/dt = (no /V )d(λ)/dt = ρo dλ/dt. So, the reaction
dynamics can be described by a single ordinary differential equation in a single unknown:
1 λ
dλ
,
(4.514)
= kρo (1 − λ) 1 −
ρo
dt
Kc 1 − λ
1 λ
dλ
.
(4.515)
= k(1 − λ) 1 −
dt
Kc 1 − λ
Equation (4.515) is in equilibrium when
λ=
1
1
+ ...
1 ∼ 1−
Kc
1 + Kc
(4.516)
As Kc → ∞, the equilibrium value of λ → 1. In this limit, the reaction is irreversible. That
is, the species B is preferred over A. Equation (4.515) has exact solution
1 − exp −k 1 + K1c t
λ=
.
(4.517)
1 + K1c
CC BY-NC-ND. 30 March 2014, J. M. Powers.
191
4.7. SIMPLE ONE-STEP KINETICS
For k > 0, Kc > 0, the equilibrium is stable. The time constant of relaxation τ is
τ=
1
k 1+
1
Kc
.
(4.518)
For the isothermal, isochoric system, one should consider the second law in terms of the
Helmholtz free energy. Combine then Eq. (3.356), dA|T,V ≤ 0, with Eq. (3.239), dA =
P
−SdT − P dV + N
i=1 µi dni and taking time derivatives, one finds
!
N
X
dA|T,V = −SdT − P dV +
µi dni ≤ 0,
(4.519)
i=1
T,V
N
X dni
dA =
≤ 0,
(4.520)
µi
dt T,V
dt
i=1
N
1 dA
V X dρi
µ
−
=−
≥ 0.
T dt
T i=1 i dt
(4.521)
This is exactly the same form as Eq. (4.487), which can be directly substituted into Eq. (4.521)
to give
!
!
!
N
N
N
Y
1 Y νi
1 dA 1 Y νi
νi′
= −RV k(T )
ρi
ρ
ρ
1−
ln
≥ 0,
−
T dt T,V
Kc i=1 i
Kc i=1 i
i=1
dA dt T,V
= RV T k(T )
N
Y
ρi
νi′
i=1
!
1−
1
Kc
N
Y
i=1
ρi νi
!
ln
1
Kc
N
Y
i=1
ρi νi
!
For the assumptions of this section, Eq. (4.523) reduces to
1 λ
1 λ
dA ln
≤ 0,
= RT kρo V (1 − λ) 1 −
dt T,V
Kc 1 − λ
Kc 1 − λ
1 λ
1 λ
= kno RT (1 − λ) 1 −
ln
≤ 0.
Kc 1 − λ
Kc 1 − λ
(4.522)
≤ 0.
(4.523)
(4.524)
(4.525)
Since the present analysis is nothing more than a special case of the previous section,
Eq. (4.525) certainly holds. One questions however the behavior in the irreversible limit,
1/Kc → 0. Evaluating this limit, one finds




dA (1 − λ) ln 1 +(1 − λ) ln λ − (1 − λ) ln(1 − λ) + . . . ≤ 0.
RT
lim
=
kn
o
| {z }

1/Kc →0 dt T,V
K
| {z c }
>0
→−∞
(4.526)
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192
CHAPTER 4. THERMOCHEMISTRY OF A SINGLE REACTION
Now, performing the distinguished limit as λ → 1; that is the reaction goes to completion,
one notes that all terms are driven to zero for small 1/Kc . Recall that 1−λ goes to zero faster
than ln(1 − λ) goes to −∞. Note that the entropy inequality is ill-defined for a formally
irreversible reaction with 1/Kc = 0.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 5
Thermochemistry of multiple
reactions
This chapter will extend notions associated with the thermodynamics of a single chemical
reactions to systems in which many reactions occur simultaneously. Some background is in
some standard sources.1 23
5.1
Summary of multiple reaction extensions
Consider now the reaction of N species, composed of L elements, in J reactions. This section
will focus on the most common case in which J ≥ (N − L), which is usually the case in large
chemical kinetic systems in use in engineering models. While much of the analysis will only
require J > 0, certain results will depend on J ≥ (N − L). It is not difficult to study the
complementary case where 0 < J < (N − L).
The molecular mass of species i is still given by Eq. (4.1):
Mi =
L
X
l=1
Ml φli ,
i = 1, . . . , N.
(5.1)
However, each reaction has a stoichiometric coefficient. The j th reaction can be summarized
in the following ways:
N
X
i=1
N
X
χi νij′
⇌
N
X
χi νij′′ ,
j = 1, . . . , J,
(5.2)
i=1
χi νij
= 0,
j = 1, . . . , J.
(5.3)
i=1
1
Turns, S. R., 2011, An Introduction to Combustion, Third Edition, McGraw-Hill, Boston. Chapters 4-6.
Kuo, K. K., 2005, Principles of Combustion, Second Edition, John Wiley, New York. Chapters 1 and 2.
3
Kondepudi, D., and Prigogine, I., 1998, Modern Thermodynamics: From Heat Engines to Dissipative
Structures, John Wiley, New York. Chapters 16 and 19.
2
193
194
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Stoichiometry for the j th reaction and lth element is given by the extension of Eq. (4.24):
N
X
φli νij = 0,
l = 1, . . . , L, j = 1, . . . , J.
(5.4)
i=1
The net change in Gibbs free energy and equilibrium constants of the j th reaction are defined
by extensions of Eqs. (4.268), (4.267), (4.270):
∆Goj
Kc,j ≡
≡
KP,j ≡ exp
Po
RT
PNi=1 νij
exp
N
X
g oT,i νij ,
i=1
−∆Goj
RT
−∆Goj
RT
j = 1, . . . , J,
(5.5)
,
j = 1, . . . , J,
(5.6)
,
j = 1, . . . , J.
(5.7)
The equilibrium of the j th reaction is given by the extension of Eq. (4.253):
N
X
µi νij = 0,
j = 1, . . . , J,
(5.8)
g i νij = 0,
j = 1, . . . , J.
(5.9)
i=1
or the extension of Eq. (4.255):
N
X
i=1
The multi-reaction extension of Eq. (3.472) for affinity is
αj = −
N
X
µi νij ,
j = 1, . . . , J.
(5.10)
i=1
The multi-reaction extension of the affinity definition of Eq. (4.494) is
′
Rj
αj = RT ln
,
j = 1, . . . , J.
R′′j
(5.11)
In terms of the chemical affinity of each reaction, the equilibrium condition is simply the
extension of Eq. (4.254):
αj = 0,
j = 1, . . . , J.
(5.12)
At equilibrium, then the equilibrium constraints can be shown to reduce to the extension
of Eq. (4.266):
ν
N Y
Pi ij
KP,j =
,
j = 1, . . . , J,
(5.13)
Po
i=1
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195
5.1. SUMMARY OF MULTIPLE REACTION EXTENSIONS
or the extension of Eq. (4.274):
Kc,j =
N
Y
ρi
νij
,
j = 1, . . . , J.
(5.14)
i=1
For isochoric reaction, the evolution of species concentration i due to the combined effect
of J reactions is given by the extension of Eq. (4.279):
dρi
=
dt
z
J
X
j=1
≡ω̇i
νij aj T
|
βj
|
exp
{z
−E j
RT
}|
≡kj (T )
!
{

N

1 Y νkj 


ρk
ρk ,
1 −
Kc,j k=1


} | k=1{z
}
{z
}
|
reverse reaction
forward reaction
{z
}
N
Y
′
νkj
i = 1, . . . , N. (5.15)
≡rj =(1/V )dζj /dt
The extension to isobaric reactions, not given here, is straightforward, and follows the same
analysis as for a single reaction. Again, three intermediate variables which are in common
usage have been defined. First, one takes the reaction rate of the j th reaction to be the
extension of Eq. (4.280):
rj ≡ aj T
|
βj
exp
{z
−E j
RT
≡kj (T )
or the extension of Eq. (4.281):
rj
}|
N
Y
k=1
ν′
ρk kj
{z
!


N

1 Y νkj 


ρk  ,
1 −
Kc,j k=1


}
{z
}
|
forward reaction
−E j
RT
(5.16)
j = 1, . . . , J,
(5.17)
reverse reaction


 Y
N
N
′
′′ 
νkj
1 Y νkj


= aj T exp
ρk
−
ρk ,

Kc,j k=1

 k=1
|
{z
}
| {z }
|
{z
}
≡kj (T ), Arrhenius rate
| forward reaction {z reverse reaction }
βj
j = 1, . . . , J,
law of mass action
1 dζj
=
.
V dt
(5.18)
Here, ζj is the reaction progress variable for the j th reaction.
Each reaction has a temperature-dependent rate function kj (T ), which is and extension
of Eq. (4.282):
−E j
βj
,
j = 1, . . . , J.
(5.19)
kj (T ) ≡ aj T exp
RT
CC BY-NC-ND. 30 March 2014, J. M. Powers.
196
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
The evolution rate of each species is given by ω̇i , defined now as an extension of Eq. (4.284):
now as
ω̇i ≡
J
X
νij rj ,
i = 1, . . . , N.
(5.20)
j=1
So, we can summarize the isochoric Eq. (5.15) as
dρi
= ω̇i .
dt
(5.21)
It will be useful to cast this in terms of mass fraction. Using the definition Eq. (2.43),
Eq. (5.21) can be rewritten as
d ρYi
= ω̇i .
(5.22)
dt Mi
Because Mi is a constant, Eq. (5.22) can be recast as
d
(ρYi ) = Mi ω̇i .
dt
(5.23)
The multi-reaction extension of Eq. (4.239) for mole change in terms of progress variables
is
dni =
J
X
νij dζj ,
i = 1, . . . , N.
(5.24)
j=1
One also has Eq. (4.243):
dG|T,P =
=
N
X
i=1
N
X
µi dni ,
µi
i=1
J
X
(5.25)
νik dζk ,
k=1
N
J
X
X
∂ζk
∂G =
νik
,
µ
i
∂ζj ζp
∂ζ
j
i=1
k=1
=
N
X
µi
i=1
=
N
X
(5.26)
J
X
νik δkj ,
(5.27)
(5.28)
j=1
µi νij ,
(5.29)
i=1
= −αj ,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
j = 1, . . . , J.
(5.30)
197
5.1. SUMMARY OF MULTIPLE REACTION EXTENSIONS
For a set of adiabatic, isochoric reactions, one can show the extension of Eq. (4.430) is
dT
=−
dt
PJ
j=1 rj ∆Ej
ρcv
,
(5.31)
where the energy change for a reaction ∆Ej is defined as the extension of Eq. (4.429):
∆Ej =
N
X
ei νij ,
j = 1, . . . , J.
(5.32)
i=1
Similarly for a set of adiabatic, isobaric reactions, one can show the extension of Eq. (4.445)
to be
PJ
dT
j=1 rj ∆Hj
,
(5.33)
=−
dt
ρcP
where the enthalpy change for a reaction ∆Hj is defined as the extension of Eq. (4.444):
∆Hj =
N
X
hi νij ,
j = 1, . . . , J.
(5.34)
i=1
Moreover, the density and species concentration derivatives for an adiabatic, isobaric set can
be shown to be the extensions of Eqs. (4.457) and (4.461):
J
N
X
X
dρ
hi
= M
rj
νij
−1 ,
dt
c
T
P
j=1
i=1
J
X
dρi
∆Hj
=
rj νij + yi
− ∆nj
,
dt
cP T
j=1
(5.35)
(5.36)
where
∆nj =
N
X
νkj .
(5.37)
k=1
In a similar fashion to that shown for a single reaction, one can further sum over all
reactions and prove that mixture mass is conserved, element mass and number are conserved.
Example 5.1
Show that element mass and number are conserved for the multi-reaction formulation.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
198
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
φli
J
X
dρi
dt
=
dρi
dt
= φli
νij rj ,
(5.38)
j=1
J
X
νij rj ,
(5.39)
j=1
J
X
d
(φli ρi ) =
φli νij rj ,
dt
j=1
(5.40)
N X
J
N
X
X
d
φli νij rj ,
(φli ρi ) =
dt
i=1 j=1
i=1
!
N
J X
N
X
d X
φli ρi
φli νij rj ,
=
dt i=1
j=1 i=1
{z
}
|
(5.41)
(5.42)
=ρl e
dρl e
dt
dρl e
dt
=
J
X
j=1
= 0,
d
(Ml ρl e ) = 0,
dt
dρl e
= 0,
dt
rj
N
X
φli νij ,
(5.43)
{z
l = 1, . . . , L,
(5.44)
l = 1, . . . , L,
(5.45)
l = 1, . . . , L.
(5.46)
i=1
|
=0
}
It is also straightforward to show that the mixture density is conserved for the multi-reaction, multicomponent mixture:
dρ
= 0.
(5.47)
dt
The proof of the Clausius-Duhem relationship for the second law is an extension of the
single reaction result. Start with Eq. (4.464) and operate much as for a single reaction model.
dS|E,V
dS dt E,V
N
1X
µ dni ≥ 0,
= −
T i=1 i
|
{z
}
(5.48)
irreversibility
N
V X dni 1
= −
≥ 0,
µ
T i=1 i dt V
N
V X dρi
µ
≥ 0,
= −
T i=1 i dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.49)
(5.50)
199
5.1. SUMMARY OF MULTIPLE REACTION EXTENSIONS
= −
N
J
V X X
νij rj ≥ 0,
µi
T i=1 j=1
(5.51)
N
J
V XX
= −
µ νij rj ≥ 0,
T i=1 j=1 i
(5.52)
J
N
V X X
µ νij ≥ 0,
rj
= −
T j=1 i=1 i
(5.53)
N
J
V X Y νij′
= −
ρ
kj
T j=1 i=1 i
N
1 Y νij
ρ
1−
Kc,j i=1 i
V
= −
T
1
1−
Kc,j
J
X
kj
j=1
= −RV
J
X
j=1
N
Y
ν′
ρi ij
i=1
kj
N
Y
ν′
ρi ij
i=1
1−
N
Y
ν
ρi ij
i=1
1
Kc,j
N
Y
!
!
ν
ρi ij
i=1
N
X
i=1
!
µi νij ≥ 0,
RT ln
ln
1
Kc,j
(5.54)
N
1 Y νij
ρ
Kc,j i=1 i
N
Y
i=1
ν
ρi ij
!
!!
≥ 0,
(5.55)
≥ 0,
(5.56)
Note that Eq. (5.53) can also be written in terms of the affinities (see Eq. (5.10)) and reaction
progress variables (see Eq. (5.18)) as
J
1 X dζj
dS =
≥ 0.
(5.57)
αj
dt E,V
T j=1
dt
Similar to the argument for a single reaction, if one defines
then it is easy to show that
N
Y
′
νij
R′j
= kj
R′′j
N
kj Y νij′′
ρ ,
=
Kc,j i=1 i
ρi
,
(5.58)
i=1
rj = R′j − R′′j ,
and we get the equivalent of Eq. (4.493):
′
J
X
Rj
dS ′
′′
= RV
≥ 0.
Rj − Rj ln
′′
dt E,V
R
j
j=1
(5.59)
(5.60)
(5.61)
Since kj (T ) > 0, R > 0, and V ≥ 0, and each term in the summation combines to be positive
semi-definite, one sees that the Clausius-Duhem inequality is guaranteed to be satisfied for
multi-component reactions.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
200
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Now, let us cast this analysis with some of the language and notation similar to that of
Prigogine’s4 monograph.5 We will use a “P” to denote equations found in Prigogine’s text.
First, note that Eq. (5.53) can also be written in terms of the affinities (see Eq. (5.10)) as
dS α
T
.
(5.62)
= Vr ·
dt E,V
T
We can take Prigogine’s so-called “generalized force” to be x with
α
.
T
We can then take the so-called “generalized flux” j to be
x=
(5.63)
j = V r.
(5.64)
Therefore, Eq. (5.62) can be recast as
dS = jT · x.
dt E,V
(5.65)
Equations (5.63-5.65) can be compared with Eqs. (P4.2). Note, we know that at equilibrium
both j = 0 and x = 0 from earlier analysis: When all αj are zero, x = 0, and by Eq. (5.11),
we must have R′j = R′′j , and thus since by Eq. (5.60), rj = R′j − R′′j , rj = 0. When all rj are
zero, j = 0. Moreover, Eq. (5.20) shows all ω̇i are zero; thus, Eq. (5.21) is in equilibrium.
This proof did not rely on any linearization near equilibrium.
5.2
Equilibrium conditions
For multicomponent mixtures undergoing multiple reactions, determining the equilibrium
condition is more difficult. There are two primary approaches, both of which are essentially
equivalent. The most straightforward method requires formal minimization of the Gibbs free
energy of the mixture. It can be shown that this actually finds the equilibrium associated
with all possible reactions.
5.2.1
Minimization of G via Lagrange multipliers
P
Recall Eq. (3.357), dG|T,P ≤ 0. Recall also Eq. (3.359), G = N
Since µi = g i =
i=1 g i ni .P
PN
∂G/∂ni |P,T,nj , one also has G = i=1 µi ni . From Eq. (3.360), dG|T,P = N
i=1 µi dni . Now,
one must also demand for a system coming to equilibrium that the element numbers are
conserved. This can be achieved by requiring
N
X
i=1
4
5
φli (nio − ni ) = 0,
l = 1, . . . , L.
(5.66)
Ilya Prigogine, 1917-2003, Russian-Belgian chemist.
Prigogine, I., 1967, Introduction to Thermodynamics of Irreversible Processes, Interscience, New York.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
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5.2. EQUILIBRIUM CONDITIONS
Here, recall nio is the initial number of moles of species i in the mixture, and φli is the
number of moles of element l in species i.
One can now use the method of constrained optimization given by the method of Lagrange
multipliers to extremize G subject to the constraints of element conservation. The extremum
will be a minimum; this will not be proved, but it will be demonstrated. Define a set of L
Lagrange6 multipliers λl . Next define an augmented Gibbs free energy function G∗ , which is
simply G plus the product of the Lagrange multipliers and the constraints:
∗
G =G+
L
X
λl
l=1
N
X
i=1
φli (nio − ni ).
(5.67)
Now, when the constraints are satisfied, one has G∗ = G, so assuming the constraints can be
satisfied, extremizing G is equivalent to extremizing G∗ . To extremize G∗ , take its differential
with respect to ni , with P , T and nj constant and set it to zero for each species:
L
X
∂G ∂G∗ =
λl φli = 0.
i = 1, . . . , N.
(5.68)
−
∂ni T,P,nj
∂ni T,P,nj l=1
| {z }
=µi
With the definition of the partial molar property µi , one then gets
µi −
L
X
λl φli = 0,
i = 1, . . . , N.
(5.69)
l=1
Next, for an ideal gas, one can expand the chemical potential so as to get
X
L
Pi
o
µT,i + RT ln
−
λl φli = 0,
i = 1, . . . , N,
Po
{z
} l=1
|

=µi

! 

L
X

ni P
1
o


−
λl φli = 0,
µT,i + RT ln  PN
Po 
 l=1

k=1 nk
|
{z
}
i = 1, . . . , N.
(5.70)
(5.71)
=Pi
Recalling that
PN
k=1 nk
= n, in summary then, one has N + L equations
X
L
ni P
o
µT,i + RT ln
−
λl φli = 0,
i = 1, . . . , N,
n Po
l=1
N
X
i=1
6
φli (nio − ni ) = 0,
l = 1, . . . , L.
(5.72)
(5.73)
Joseph-Louis Lagrange, 1736-1813, Italian-French mathematician.
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CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
in N + L unknowns: ni , i = 1, . . . , N, λl , l = 1, . . . , L.
Example 5.2
Consider a previous example problem, p. 134, in which
N2 + N2 ⇌ 2N + N2 .
(5.74)
Take the reaction to be isothermal and isobaric with T = 6000 K and P = 100 kP a. Initially one has
1 kmole of N2 and 0 kmole of N . Use the extremization of Gibbs free energy to find the equilibrium
composition.
First, find the chemical potentials at the reference pressure of each of the possible constituents.
o
o
o
µoT,i = goi = hi − T soi = h298,i + ∆hi − T soi .
(5.75)
For each species, one then finds
µoN2
µoN
kJ
,
kmole
kJ
.
= 472680 + 124590 − (6000)(216.926) = −704286
kmole
= 0 + 205848 − (6000)(292.984) = −1552056
To each of these one must add
RT ln
ni P
nPo
(5.76)
(5.77)
to get the full chemical potential. Now, P = Po = 100 kP a for this problem, so one only must consider
RT = 8.314(6000) = 49884 kJ/kmole. So, the chemical potentials are
nN 2
,
(5.78)
µN2 = −1552056 + 49884 ln
nN + nN 2
nN
µN = −704286 + 49884 ln
.
(5.79)
nN + nN 2
Then, one adds on the Lagrange multiplier and then considers element conservation to get the
following coupled set of nonlinear algebraic equations:
nN 2
− 2λN = 0,
(5.80)
−1552056 + 49884 ln
nN + nN 2
nN
− λN = 0,
(5.81)
−704286 + 49884 ln
nN + nN 2
(5.82)
nN + 2nN2 = 2.
These non-linear equations are solved numerically to get
nN 2
nN
=
=
λN
=
0.88214 kmole,
0.2357 kmole,
kJ
−781934
.
kmole
These agree with results found in the earlier example problem of p. 134.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.83)
(5.84)
(5.85)
203
5.2. EQUILIBRIUM CONDITIONS
Example 5.3
Consider a mixture of 2 kmole of H2 and 1 kmole of O2 at T = 3000 K and P = 100 kP a.
Assuming an isobaric and isothermal equilibration process with the products consisting of H2 , O2 ,
H2 O, OH, H, and O, find the equilibrium concentrations. Consider the same mixture at T = 298 K
and T = 1000 K.
The first task is to find the chemical potentials of each species at the reference pressure and
T = 3000 K. Here, one can use the standard tables along with the general equation
o
o
o
µoT,i = goi = hi − T soi = h298,i + ∆hi − T soi .
(5.86)
For each species, one then finds
µoH2
µoO2
µoH2 O
µoOH
µoH
µoO
kJ
,
kmole
kJ
,
= 0 + 98013 − 3000(284.466) = −755385
kmole
= 0 + 88724 − 3000(202.989) = −520242
(5.87)
(5.88)
kJ
,
kmole
kJ
,
= 38987 + 89585 − 3000(256.825) = −641903
kmole
kJ
= 217999 + 56161 − 3000(162.707) = −213961
,
kmole
kJ
= 249170 + 56574 − 3000(209.705) = −323371
.
kmole
= −241826 + 126548 − 3000(286.504) = −974790
To each of these one must add
RT ln
ni P
nPo
(5.89)
(5.90)
(5.91)
(5.92)
to get the full chemical potential. Now, P = Po = 100 kP a for this problem, so one must only consider
RT = 8.314(3000) = 24942 kJ/kmole. So, the chemical potentials are
µH2
=
−520243 + 24942 ln
µO2
=
−755385 + 24942 ln
µH2 O
=
−974790 + 24942 ln
µOH
=
−641903 + 24942 ln
µH
=
−213961 + 24942 ln
µO
=
−323371 + 24942 ln
nH2 + nO2
nH2 + nO2
nH2 + nO2
nH2 + nO2
nH2 + nO2
nH2 + nO2
nH2
+ nH2 O + nOH
nO2
+ nH2 O + nOH
nH2 O
+ nH2 O + nOH
nOH
+ nH2 O + nOH
nH
+ nH2 O + nOH
nO
+ nH2 O + nOH
+ nH + nO
+ nH + nO
+ nH + nO
+ nH + nO
+ nH + nO
+ nH + nO
,
(5.93)
,
(5.94)
,
(5.95)
,
(5.96)
,
(5.97)
.
(5.98)
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204
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Then, one adds on the Lagrange multipliers and then considers element conservation to get the following
coupled set of nonlinear equations:
nH2
− 2λH = 0,
(5.99)
−520243 + 24942 ln
nH + nO2 + nH2 O + nOH + nH + nO
2
nO2
−755385 + 24942 ln
− 2λO = 0,
(5.100)
nH2 + nO2 + nH2 O + nOH + nH + nO
nH2 O
−974790 + 24942 ln
− 2λH − λO = 0,
(5.101)
nH2 + nO2 + nH2 O + nOH + nH + nO
nOH
− λH − λO = 0,
(5.102)
−641903 + 24942 ln
nH2 + nO2 + nH2 O + nOH + nH + nO
nH
−213961 + 24942 ln
− λH = 0,
(5.103)
nH2 + nO2 + nH2 O + nOH + nH + nO
nO
− λO = 0,
(5.104)
−323371 + 24942 ln
nH2 + nO2 + nH2 O + nOH + nH + nO
(5.105)
2nH2 + 2nH2 O + nOH + nH = 4,
(5.106)
2nO2 + nH2 O + nOH + nO = 2.
These non-linear algebraic equations can be solved numerically via a Newton-Raphson technique.
The equations are sensitive to the initial guess, and one can use ones intuition to help guide the
selection. For example, one might expect to have nH2 O somewhere near 2 kmole. Application of the
Newton-Raphson iteration yields
nH2
=
nO2
nH2 O
=
=
nOH
nH
=
=
nO
=
λH
=
λO
=
3.19 × 10−1 kmole,
(5.107)
1.10 × 10−1 kmole,
1.50 × 100 kmole,
(5.108)
(5.109)
5.74 × 10−2 kmole,
kJ
−2.85 × 105
,
kmole
kJ
.
−4.16 × 105
kmole
(5.112)
2.20 × 10−1 kmole,
1.36 × 10−1 kmole,
(5.110)
(5.111)
(5.113)
(5.114)
At this relatively high value of temperature, all species considered have a relatively major presence.
That is, there are no truly minor species.
Unless a good guess is provided, it may be difficult to find a solution for this set of non-linear equations. Straightforward algebra allows the equations to be recast in a form which sometimes converges
more rapidly:
2
nH2
λH
520243
exp
,
(5.115)
= exp
nH2 + nO2 + nH2 O + nOH + nH + nO
24942
24942
2
λO
755385
nO2
exp
= exp
,
(5.116)
nH2 + nO2 + nH2 O + nOH + nH + nO
24942
24942
2
λO
λH
974790
nH2 O
exp
exp
, (5.117)
= exp
nH2 + nO2 + nH2 O + nOH + nH + nO
24942
24942
24942
nOH
λO
λH
641903
exp
exp
,
(5.118)
= exp
nH2 + nO2 + nH2 O + nOH + nH + nO
24942
24942
24942
CC BY-NC-ND. 30 March 2014, J. M. Powers.
205
5.2. EQUILIBRIUM CONDITIONS
G (ξ) - G (0) (kJ)
40
30
20
10
-0.01
-0.005
0.005
0.01
ξ (kmole)
Figure 5.1: Gibbs free energy variation as mixture composition is varied maintaining element
conservation for mixture of H2 , O2 , H2 O, OH, H, and O at T = 3000 K, P = 100 kP a.
nH2
nH2
λH
213961
exp
,
= exp
24942
24942
λO
323371
exp
,
= exp
24942
24942
= 4,
(5.121)
= 2.
(5.122)
nH
+ nO2 + nH2 O + nOH + nH + nO
nO
+ nO2 + nH2 O + nOH + nH + nO
2nH2 + 2nH2 O + nOH + nH
2nO2 + nH2 O + nOH + nO
(5.119)
(5.120)
Then, solve these considering ni , exp (λO /24942), and exp (λH /24942) as unknowns. The same result
is recovered, but a broader range of initial guesses converge to the correct solution.
One can verify that this choice extremizes G by direct computation; moreover, this will show
that the extremum is actually a minimum. In so doing, one must exercise care to see that element
conservation is retained. As an example, perturb the equilibrium solution for nH2 and nH such that
nH2
nH
=
=
3.19 × 10−1 + ξ,
1.36 × 10−1 − 2ξ.
(5.123)
(5.124)
Leave all other species mole numbers the same. In this way, when ξ = 0, one has the original equilibrium
solution. For ξ 6= 0, the solution
P moves off the equilibrium value in such a way that elements are
conserved. Then, one has G = N
i=1 µi ni = G(ξ).
The difference G(ξ) − G(0) is plotted in Fig. 5.1. When ξ = 0, there is no deviation from the value
predicted by the Newton-Raphson iteration. Clearly when ξ = 0, G(ξ) − G(0), takes on a minimum
value, and so then does G(ξ). So, the procedure works.
At the lower temperature, T = 298 K, application of the same procedure yields very different
results:
nH2
= 4.88 × 10−27 kmole,
(5.125)
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CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
nO2
nH2 O
nOH
nH
nO
λH
λO
= 2.44 × 10−27 kmole,
(5.126)
0
= 2.00 × 10 kmole,
= 2.22 × 10−29 kmole,
(5.127)
(5.128)
= 2.29 × 10−49 kmole,
= 1.67 × 10−54 kmole,
kJ
,
= −9.54 × 104
kmole
kJ
= −1.07 × 105
.
kmole
(5.129)
(5.130)
(5.131)
(5.132)
At the intermediate temperature, T = 1000 K, application of the same procedure shows the minor
species become slightly more prominent:
nH2
nO2
nH2 O
nOH
nH
nO
λH
λO
5.2.2
= 4.99 × 10−7 kmole,
= 2.44 × 10−7 kmole,
(5.133)
(5.134)
= 2.26 × 10−12 kmole,
= 1.10 × 10−13 kmole,
kJ
,
= −1.36 × 105
kmole
kJ
.
= −1.77 × 105
kmole
(5.137)
(5.138)
= 2.00 × 100 kmole,
= 2.09 × 10−8 kmole,
(5.135)
(5.136)
(5.139)
(5.140)
Equilibration of all reactions
In another equivalent method, if one commences with a multi-reaction model, one can require
each reaction to be in equilibrium. This leads to a set of algebraic equations for rj = 0, which
Kc,j =
Po
RT
PNi=1 νij
exp
−∆Goj
RT
=
N
Y
νkj
ρk
,
j = 1, . . . , J.
(5.141)
k=1
With some effort it can be shown that not all of the J equations are linearly independent.
Moreover, they do not possess a unique solution. However, for closed systems, only one of
the solutions is physical, as will be shown in the following section. The others typically
involve non-physical, negative concentrations.
Nevertheless, Eqs. (5.141) are entirely consistent with the predictions of the N + L equations which arise from extremization of Gibbs free energy while enforcing element number
CC BY-NC-ND. 30 March 2014, J. M. Powers.
207
5.2. EQUILIBRIUM CONDITIONS
constraints. This can be shown by beginning with Eq. (5.71), rewritten in terms of molar
concentrations, and performing the following sequence of operations:
!
L
X
n
/V
P
i
o
λl φli = 0,
i = 1, . . . , N, (5.142)
−
µT,i + RT ln PN
P
n
/V
o
k
k=1
l=1
!
L
X
ρ
P
µoT,i + RT ln PN i
−
λl φli = 0,
i = 1, . . . , N, (5.143)
P
o
ρ
k=1 k
l=1
X
L
ρi P
o
λl φli = 0,
i = 1, . . . , N, (5.144)
−
µT,i + RT ln
ρ Po
l=1
X
L
RT
o
λl φli = 0,
i = 1, . . . , N, (5.145)
−
µT,i + RT ln ρi
Po
l=1
L
X
RT
o
νij µT,i + νij RT ln ρi
− νij
λl φli = 0,
i = 1, . . . , N,
Po
l=1
j = 1, . . . , J,
(5.146)
L
N
N
N
X
X
X
X
RT
λl φli = 0,
j = 1, . . . , J, (5.147)
νij
−
νij µoT,i +
νij RT ln ρi
P
o
i=1
l=1
|i=1 {z } i=1
=∆Goj
∆Goj
+ RT
N
X
i=1
X
L
N
X
RT
νij ln ρi
−
λl
φli νij = 0,
Po
l=1
|i=1 {z }
j = 1, . . . , J, (5.148)
=0
∆Goj + RT
N
X
i=1
Here, the stoichiometry for each
pliers. Continue to find
νij
N
X
RT
ln ρi
Po
i=1
νij !
N
X
RT
exp
ln ρi
Po
i=1
νij
N Y
RT
ρi
Po
i=1
PN
i=1 νij Y
N
RT
ρi νij
Po
i=1
RT
νij ln ρi
= 0,
Po
j = 1, . . . , J. (5.149)
reaction has been employed to remove the Lagrange multi-
= −
∆Goj
,
j = 1, . . . , J,
RT
∆Goj
= exp −
,
j = 1, . . . , J,
RT
∆Goj
,
j = 1, . . . , J,
= exp −
RT
∆Goj
,
j = 1, . . . , J,
= exp −
RT
(5.150)
(5.151)
(5.152)
(5.153)
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CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
N
Y
ρi
νij
=
|
i=1
N
Y
Po
RT
PNi=1 νij
∆Goj
,
exp −
RT
{z
}
j = 1, . . . , J, (5.154)
=Kc,j
ρi νij = Kc,j ,
j = 1, . . . , J.
(5.155)
i=1
Thus, extremization of Gibbs free energy is consistent with equilibrating each of the J
reactions.
5.2.3
Zel’dovich’s uniqueness proof
Here, a proof is given for the global uniqueness of the equilibrium point in the physically
accessible region of composition space following a procedure given in a little known paper
by the great Russian physicist Zel’dovich.7 The proof follows the basic outline of Zel’dovich,
but the notation will be consistent with the present development. Some simplifications were
available to Zel’dovich but not employed by him. For further background see Powers and
Paolucci.8
5.2.3.1
Isothermal, isochoric case
Consider a mixture of ideal gases in a closed fixed volume V at fixed temperature T . For
such a system, the canonical equilibration relation is given by Eq. (3.356), dA|T,V ≤ 0. So,
A must be always decreasing until it reaches a minimum. Consider then A. First, combining
Eqs. (3.235) and (3.236), one finds
A = −P V + G.
(5.156)
Now, from Eq. (3.359) one can eliminate G to get
A = −P V +
N
X
ni µi .
From the ideal gas law, P V = nRT , and again with n =
A = −nRT +
= −
N
X
N
X
ni µi ,
i=1
ni RT +
i=1
7
N
X
i=1
ni
(5.157)
i=1
µoT,i
PN
i=1
+ RT ln
ni , one gets
(5.158)
Pi
Po
,
(5.159)
Zel’dovich, Ya. B., 1938, “A Proof of the Uniqueness of the Solution of the Equations for the Law of
Mass Action,” Zhurnal Fizicheskoi Khimii, 11: 685-687.
8
Powers, J. M., and Paolucci, S., 2008, “Uniqueness of Chemical Equilibria in Ideal Mixtures of Ideal
Gases,” American Journal of Physics, 76(9): 848-855.
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209
5.2. EQUILIBRIUM CONDITIONS
= RT
= RT
N
X
i=1
N
X
i=1
ni
ni
µoT,i
RT
− 1 + ln
ni P
nPo
µoT,i
RT
− 1 + ln ni
Po V
RT
,
(5.160)
.
(5.161)
For convenience, define, for this isothermal isochoric problem no , the total number of moles
at the reference pressure, which for this isothermal isochoric problem, is a constant:
no ≡
So
A = RT
N
X
i=1
ni
Po V
.
RT
µoT,i
RT
(5.162)
− 1 + ln
n i
o
n
.
(5.163)
Now, recall that the atomic element conservation, Eq. (5.66), demands that
N
X
i=1
φli (nio − ni ) = 0,
l = 1, . . . , L.
(5.164)
As defined earlier, φli is the number of atoms of element l in species i; note that φli ≥ 0.
It is described by a L × N non-square
P matrix, typically of full rank, L. Defining the initial
number of moles of element l, εl = N
i=1 φli nio , one can rewrite Eq. (5.164) as
N
X
φli ni = εl ,
l = 1, . . . , L.
(5.165)
i=1
Equation (5.165) is generally under-constrained, and one can find solutions of the form
 

 





D1 N −L
n1
n1o
D11
D12
 n2   n2o   D21 

 D 
D
 .  =  .  +  .  ξ1 +  .22  ξ2 + . . . +  2 .N −L  ξN −L .
(5.166)
 ..   ..   .. 

 .. 

..
nN
nN o
DN 1
DN 2
DN N −L
Here, Dik represents a dimensionless component of a matrix of dimension N ×(N −L). Here,
in contrast to earlier analysis, ξi is interpreted as having the dimensions of mole. Each of
the N − L column vectors of the matrix whose components are Dik has length N and lies in
the right null space of the matrix whose components are φli . That is,
N
X
i=1
φli Dik = 0,
k = 1, . . . , N − L, l = 1, . . . , L.
(5.167)
There is an exception to this rule when the left null space of νij is of higher dimension than
the right null space of φli . In such a case, there are quantities conserved in addition to the
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CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
elements as a consequence of the form of the reaction law. The most common exception
occurs when each of the reactions also conserves the number of molecules. In such a case,
there will be N − L − 1 free variables, rather than N − L. One can robustly form Dij from
the set of independent column space vectors of νij . These vectors are included in the right
P
null space of φli since N
i=1 φli νij = 0.
One can have Dik ∈ (−∞, ∞). Each of these is a function of φli . In matrix form, one
can say
 


 

D11 D12 . . . D1 N −L
ξ1
n1o
n1
 n2   n2o   D21 D22 . . . D2 N −L   ξ2 
 . .
 . = . + .
(5.168)
..
..
..
  .. 
 ..   ..   ..
.
.
.
nN
nN o
DN 1
DN 2
. . . DN
N −L
ξN −L
In index form, this becomes
ni = nio +
N
−L
X
k=1
Dik ξk ,
i = 1, . . . , N.
(5.169)
It is also easy to show that the N − L column space vectors in Dik are linear combinations
of N − L column space vectors that span the column space of the rank-deficient N × J
components of νij . The N values of ni are uniquely determined once N − L values of ξk
are specified. That is, a set of independent ξk , k = 1, . . . , N − L, is sufficient to describe
the system. This insures the initial element concentrations are always maintained. One
can also develop a J-reaction generalization of the single reaction Eq. (4.99). Let Ξkj ,
k = 1, . . . , N − L, j = 1, . . . , J, be the extension of ξk . Then, the appropriate generalization
of Eq. (4.99) is
N
−L
X
νij =
Dik Ξkj .
(5.170)
k=1
In Gibbs notation, one would say
ν = D · Ξ.
(5.171)
D · Ξ = ν,
D · D · Ξ = DT · ν,
(5.172)
(5.173)
One can find the matrix Ξ by the following operations
T
Ξ =
DT · D
−1
· DT · ν.
(5.174)
This calculation has only marginal utility.
Returning to the primary exercise, note that one can form the partial derivative of
Eq. (5.169):
N
−L
X
∂ni ∂ξk =
Dik
,
∂ξp ξj
∂ξp ξj
k=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
i = 1, . . . , N; p = 1, . . . , N − L,
(5.175)
211
5.2. EQUILIBRIUM CONDITIONS
N
−L
X
=
k=1
Dik δkp ,
= Dip ,
i = 1, . . . , N; p = 1, . . . , N − L,
i = 1, . . . , N; p = 1, . . . , N − L.
(5.176)
(5.177)
Here, δkp is the Kronecker delta function.
Next, return to consideration of Eq. (5.161). It is sought to minimize A while holding
T and V constant. The only available variables are ni , i = 1, . . . , N. These are not fully
independent, but they are known in terms of the independent ξp , p = 1, . . . , N − L. So, one
can find an extremum of A by differentiating it with respect to each of the ξp and setting
each derivative to zero:
!
o
N
n n X
µT,i
∂A ∂ni ∂
∂ni i
i
−1 +
ln o
= 0,
= RT
ln o + ni
∂ξp ξj ,T,V
∂ξp ξj RT
∂ξp ξj
n
∂ξp
n
i=1
= RT
p = 1, . . . , N − L,
(5.178)
!
o
N
n n X ∂nq ∂
µT,i
∂ni ∂ni i
i
−
1
+
ln
ln
+
n
= 0,
i
o
o
∂ξp ξj RT
∂ξp ξj
n
∂ξp ∂nq
n
q=1
N
X
i=1
= RT
Dip
Dip
N
X
i=1
= RT
N X
i=1
= RT
N
X
i=1
Dip
p = 1, . . . , N − L,
!
N
n X
µoT,i
1
i
− 1 + Dip ln o + ni
Dqp δiq = 0,
n
ni
RT
q=1
µoT,i
RT
− 1 + Dip ln
n i
o
n
n µoT,i
i
= 0,
+ ln o
n
RT
+ Dip
p = 1, . . . , N − L,
(5.179)
(5.180)
= 0,
p = 1, . . . , N − L,
p = 1, . . . , N − L.
(5.181)
(5.182)
Following Zel’dovich, one then rearranges Eq. (5.182) to define the equations for equilibrium:
N
X
Dip ln
,
p = 1, . . . , N − L, (5.184)
,
no
RT
i=1
! !Dip
N
N
−L
X
X
Dip µoT,i
1
= −
,
nio +
Dik ξk
no
RT
i=1
k=1
p = 1, . . . , N − L, (5.185)
ln
n Dip
= −
i
o
n
= −
N Y
ni Dip
= −
ln
i=1
i=1
N
X
Dip µoT,i
p = 1, . . . , N − L, (5.183)
i=1
ln
i
o
n
,
i=1
N
X
N
Y
n i=1
N
X
i=1
N
X
RT
Dip µoT,i
RT
Dip µoT,i
p = 1, . . . , N − L. (5.186)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
212
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Equation (5.186) forms N −L equations in the N −L unknown values of ξp and can be solved
by Newton iteration. Note that as of yet, no proof exists that this is a unique solution. Nor
is it certain whether or not A is maximized or minimized at such a solution point.
One also notices that the method of Zel’dovich is consistent with a more rudimentary
form. Rearrange Eq. (5.182) to get
∂A ∂ξp ξj ,T,V
n i
o
=
Dip µT,i + RT ln o
= 0,
n
i=1
N
X
ni P
o
=
Dip µT,i + RT ln
,
nPo
i=1
N
X
Pi
o
,
Dip µT,i + RT ln
=
Po
i=1
N
X
=
N
X
i=1
p = 1, . . . , N − L,
(5.187)
(5.188)
(5.189)
µi Dip .
(5.190)
Now, Eq. (5.190) is easily found via another method. Recall Eq. (3.239), and then operate
on it in an isochoric, isothermal limit, taking derivative with respect to ξp , p = 1, . . . , N − L:
dA = −SdT − P dV +
dA|T,V
∂A ∂ξp ξj ,T,V
=
N
X
N
X
µi dni ,
(5.191)
i=1
µi dni ,
(5.192)
∂ni
,
∂ξp
(5.193)
µi Dip .
(5.194)
i=1
=
=
N
X
i=1
N
X
i=1
µi
One can determine whether such a solution, if it exists, is a maxima or minima by
examining the second derivative, given by differentiating Eq. (5.182). We find the Hessian9 ,
H, to be
H=
9
N
n X
∂2A
∂
i
Dip
= RT
ln o ,
∂ξj ∂ξp
∂ξ
n
j
i=1
N
X
∂
∂
o
ln ni −
ln n ,
Dip
= RT
∂ξj
∂ξj
i=1
after Ludwig Otto Hesse, 1811-1874, German mathematician.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.195)
(5.196)
213
5.2. EQUILIBRIUM CONDITIONS
= RT
= RT
N
X
i=1
N
X
i=1
Dip
1 ∂ni
,
ni ∂ξj
(5.197)
Dip Dij
.
ni
(5.198)
Here, j = 1, . . . , N − L. Scaling each of the rows of Dip by a constant does not affect the
√
rank. So, we can say that the N × (N − L) matrix whose entries are Dip / ni has rank
N − L, and consequently the Hessian H, of dimension (N − L) × (N − L), has full rank
N − L, and is symmetric. It is easy to show by means of singular value decomposition, or
other methods, that the eigenvalues of a full rank symmetric square matrix are all real and
non-zero.
Now, consider whether ∂ 2 A/∂ξj ∂ξp is positive definite. By definition, it is positive definite
if for an arbitrary vector zi of length N − L with non-zero norm that the term ΥT V , defined
below, be always positive:
ΥT V =
−L
N
−L N
X
X
j=1 p=1
∂2A
zj zp > 0.
∂ξj ∂ξp
(5.199)
Substitute Eq. (5.198) into Eq. (5.199) to find
ΥT V
=
N
−L N
−L
X
X
RT
j=1 p=1
= RT
N
X
i=1
1
ni
N
X
Dip Dij
ni
i=1
N
−L N
−L
X
X
j=1 p=1
zj zp ,
Dip Dij zj zp ,
N
−L
N −L
N
X
X
1 X
= RT
Dip zp ,
Dij zj
n
i
p=1
j=1
i=1
! N −L
!
N
N
−L
X 1 X
X
= RT
Dij zj
Dip zp .
ni j=1
p=1
i=1
(5.200)
(5.201)
(5.202)
(5.203)
Define now
yi ≡
N
−L
X
j=1
Dij zj ,
i = 1, . . . , N.
(5.204)
This yields
ΥT V
= RT
N
X
y2
i
i=1
ni
.
(5.205)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
214
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Now, to restrict the domain to physically accessible space, one is only concerned with ni ≥ 0,
T > 0, so for arbitrary yi , one finds ΥT V > 0, so one concludes that the second mixed partial
of A is positive definite globally.
We next consider the behavior of A near a generic point in the physical space ξ̂ where
A = Â. If we let dξ represent ξ − ξ̂, we can represent the Helmholtz free energy by the
Taylor series
1
A(ξ) = Â + dξT · ∇A + dξT · H · dξ + . . . ,
(5.206)
2
where ∇A and H are evaluated at ξ̂. Now, if ξ̂ = ξ eq , where eq denotes an equilibrium
point, ∇A = 0, Â = Aeq , and
1
A(ξ) − Aeq = dξ T · H · dξ + . . . .
2
(5.207)
Because H is positive definite in the entire physical domain, any isolated critical point will
be a minimum. Note that if more than one isolated minimum point of A were to exist in
the domain interior, a maximum would also have to exist in the interior, but maxima are
not allowed by the global positive definite nature of H. Subsequently, any extremum which
exists away from the boundary of the physical region must be a minimum, and the minimum
is global.
Global positive definiteness of H alone does not rule out the possibility of non-isolated
multiple equilibria, as seen by the following analysis. Because it is symmetric, H can be
orthogonally decomposed into H = QT · Λ · Q, where Q is an orthogonal matrix whose
columns are the normalized eigenvectors of H. Note that QT = Q−1 . Also Λ is a diagonal
matrix with real eigenvalues on its diagonal. We can effect a volume-preserving rotation of
axes by taking the transformation dw = Q · dξ; thus, dξ = QT · dw. Hence,
1
1
1
A−Aeq = (QT ·dw)T ·H·QT ·dw = dwT ·Q·QT ·Λ·Q·QT ·dw = dwT ·Λ·dw. (5.208)
2
2
2
The application of these transformations gives in the neighborhood of equilibrium the quadratic
form
N −L
1X
λp (dwp )2 .
(5.209)
A − Aeq =
2 p=1
For A to be a unique minimum, λp > 0. If one or more of the λp = 0, then the minimum
could be realized on a line or higher dimensional plane, depending on how many zeros are
present. The full rank of H guarantees that λp > 0. For our problem the unique global
minimum which exists in the interior will exist at a unique point.
Lastly, one must check the boundary of the physical region to see if it can form an
extremum. Near a physical boundary given by nq = 0, one finds that A behaves as
lim A ∼ RT
nq →0
N
X
i=1
ni
CC BY-NC-ND. 30 March 2014, J. M. Powers.
µT,i
RT
− 1 + ln
n i
o
n
→ finite.
(5.210)
215
5.2. EQUILIBRIUM CONDITIONS
The behavior of A itself is finite because limnq →0 nq ln nq = 0, and the remaining terms in
the summation are non-zero and finite.
The analysis of the behavior of the derivative of A on a boundary of nq = 0 is more
complex. We require that nq ≥ 0 for all N species. The hyperplanes given by nq = 0 define
a closed physical boundary in reduced composition space. We require that changes in nq
which originate from the surface nq = 0 be positive. For such curves we thus require that
near nq = 0, perturbations dξk be such that
dnq =
N
−L
X
k=1
Dqk dξk > 0.
(5.211)
We next examine changes in A in the vicinity of boundaries given by nq = 0. We will restrict
our attention to changes which give rise to dnq > 0. We employ Eq. (5.187) and find that
dA =
N
−L
X
p=1
N −L
n NX
X
∂A i
o
dξp
=
Dip dξp .
µi + RT ln o
∂ξp T,V,ξj6=p
n
p=1
i=1
(5.212)
On the boundary given by nq = 0, the dominant term in the sum is for i = q, and so on this
boundary
−L
n NX
q
Dqp dξp → −∞.
(5.213)
lim dA = RT ln o
nq →0
n p=1
| {z }
>0
The term identified by the brace is positive because of Eq. (5.211). Because R and T > 0,
we see that as nq moves away from zero into the physical region, changes in A are large and
negative. So, the physical boundary can be a local maximum, but never a local minimum
in A. Hence, the only admissible equilibrium is the unique minimum of A found from
Eq. (5.186); this equilibrium is found at a unique point in reduced composition space.
5.2.3.2
Isothermal, isobaric case
Next, consider the related case in which T and P are constant. For such a system, the
canonical equilibration relation is given by Eq. (3.357) holds that dG|T,P ≤ 0. So, G must
be decreasing until it reaches a minimum. So, consider G, which from Eq. (3.359) is
G =
=
=
N
X
i=1
N
X
i=1
N
X
i=1
ni µi ,
ni
ni
µoT,i
µoT,i
(5.214)
+ RT ln
+ RT ln
Pi
Po
ni P
nPo
,
(5.215)
,
(5.216)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
216
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
= RT
N
X
ni
i=1
µoT,i
RT
+ ln
P
Po
ni
+ ni ln
PN
k=1 nk
!!
.
(5.217)
Next, differentiate with respect to each of the independent variables ξp for p = 1, . . . , N − L:
N
X
∂G
= RT
∂ξp
i=1
∂ni
∂ξp
µoT,i
RT
+ ln
P
Po
∂
+
∂ξp
N X
∂ni µoT,i
P
= RT
+ ln
∂ξp RT
Po
i=1
!!!
N
X
ni
∂nq ∂
,
ni ln PN
+
∂ξ
∂n
n
p
q
k
k=1
q=1
o
N X
µT,i
P
Dip
= RT
+ ln
Po
RT
i=1
+
N
X
q=1
ni
Dqp − PN
k=1 nk
+ δiq
1 + ln
ni
ni ln
PN
k=1
nk
!!!
−RT
= RT
N
N
X
ni X
i=1
N X
i=1
N
X
−RT
= RT
q=1
N X
i=1
= RT
= RT
N X
i=1
N X
i=1
n
q=1
Dip
Dqp
Dip
Dip
Dip
ni
PN
k=1 nk
!!!!
,
(5.220)
µoT,i
RT
i=1
P
Po
n Dqp ,
N
X
ni
µoT,i
n
µoT,i
RT
µoT,i
RT
µoT,i
RT
+ ln
+ ln
+ ln
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.221)
(5.222)
P
Po
+ 1 + ln
n i
n
,
+ ln
(5.218)
(5.219)
!
N
X
n
i
i
Dip
= RT
+ ln
+ 1 + ln
−
Dqp ,
n
n
RT
q=1
i=1
N
o
X
µT,i
P
ni
Dip
+ ln
= RT
+ 1 + ln
Po
n
RT
i=1
N
X
,
(5.223)
P
Po
P
Po
P
Po
+ 1 + ln
+ 1 + ln
+ ln
n
i
n
n i
n
n i
n
,
− RT
− RT
N
X
q=1
N
X
i=1
Dqp ,
(5.224)
Dip ,
(5.225)
(5.226)
217
5.2. EQUILIBRIUM CONDITIONS
ni P
o
,
=
Dip µT,i + RT ln
nP
o
i=1
N
X
ni P
o
=
Dip µT,i + RT ln
,
nPo
i=1
N
X
=
N
X
i=1
µi Dip ,
(5.227)
(5.228)
p = 1, . . . , N − L.
(5.229)
Note this simple result is entirely consistent with a result
Pthat could have been deduced by
commencing with the alternative Eq. (3.361), dG|T,P = N
i=1 µi dni . Had this simplification
been taken, one could readily deduce that
N
X
∂ni
∂G µi
=
,
∂ξp ξj
∂ξ
p
i=1
=
N
X
i=1
µi Dip ,
p = 1, . . . N − L,
(5.230)
p = 1, . . . N − L.
(5.231)
Now, to equilibrate, one sets the derivatives to zero to get
ni P
+ RT ln
Dip
nPo
i=1
Dip
N
X
ni P
ln
nPo
i=1
Dip
N Y
ni P
ln
nPo
i=1
Dip

PN −L
N
Dik ξk P
nio + k=1
Y


ln
PN −L
PN Po
i=1
k=1 Dqk ξk
q=1 nqo +
N
X
µoT,i
= 0, p = 1, . . . , N − L,
= −
= −
= −
N
X
i=1
N
X
i=1
N
X
i=1
Dip
Dip
Dip
µoT,i
RT
µoT,i
RT
µoT,i
RT
(5.232)
,
(5.233)
,
(5.234)
,
p = 1, . . . , N − L. (5.235)
These N −L non-linear algebraic equations can be solved for the N −L unknown values of ξp
via an iterative technique. One can extend the earlier analysis to show that the equilibrium
is unique in the physically accessible region of composition space.
We can repeat our previous analysis to show that this equilibrium is unique in the physically accessible region of composition space. By differentiating Eq. (5.231) it is seen that
N
n X
∂2G
∂
i
Dip
,
= RT
ln
∂ξj ∂ξp
∂ξj
n
i=1
(5.236)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
218
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
= RT
= RT
= RT
!
N
X
∂
∂
ln ni −
Dip
ln n ,
Dip
∂ξ
∂ξ
j
j
i=1
i=1
!
N
N
X
X
1 ∂ni
1 ∂n
Dip
,
−
Dip
ni ∂ξj
n ∂ξj
i=1
i=1
N
X
N
X
i=1
= RT
= RT
N
1 ∂
1 ∂ni X
−
Dip
Dip
ni ∂ξj
n ∂ξj
i=1
N
X
q=1
N
N
1 ∂nq
1 ∂ni X X
−
Dip
Dip
ni ∂ξj
n ∂ξj
q=1
i=1
i=1
!
N
N
N
X
1 XX
Dip Dij
−
Dip Dqj .
ni
n i=1 q=1
i=1
N
X
nq
(5.237)
(5.238)
!!
!!
,
(5.239)
,
(5.240)
(5.241)
Next consider the sum
ΥT P =
N
−L N
−L
X
X
j=1 p=1
2
∂ G
zj zp = RT
∂ξj ∂ξp
N
−L N
−L
X
X
j=1 p=1
N
X
Dip Dij
i=1
ni
N
N
1 XX
−
Dip Dqj
n i=1 q=1
!
zj zp . (5.242)
We use Eq. (5.204) and following a long series of calculations, reduce Eq. (5.242), to the
positive definite form
ΥT P
2
r
N
N r
RT X X
nj
ni
yi −
yj > 0.
=
n i=1 j=i+1
ni
nj
(5.243)
It is easily verified by direct expansion that Eqs. (5.242) and (5.243) are equivalent.
On the boundary ni = 0, and as for the isothermal-isochoric case, it can be shown that
dG → −∞ for changes with dni > 0. Thus, the boundary has no local minimum, and we
can conclude that G is minimized in the interior and the minimum is unique.
5.2.3.3
One can extend Zel’dovich’s proof to other sets of conditions. For example, consider a case
which is isochoric and isoenergetic. This corresponds to a chemical reaction in an fixed
volume which is thermally insulated. In this case, one operates on Eq. (3.357):
dE = −P |{z}
dV +T dS +
|{z}
=0
=0
0 = T dS +
N
X
µi dni ,
i=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
N
X
µi dni ,
(5.244)
i=1
(5.245)
219
5.2. EQUILIBRIUM CONDITIONS
dS = −
N
1X
µ dni ,
T i=1 i
(5.246)
N −L
N
1X X
Dik dξk ,
µ
= −
T i=1 i k=1
∂S
∂ξj
(5.247)
N N −L
1 XX
= −
µ Dik dξk ,
T i=1 k=1 i
(5.248)
N N −L
∂ξk
1 XX
µi Dik
= −
,
T i=1 k=1
∂ξj
= −
(5.249)
N N −L
1 XX
µ Dik δkj ,
T i=1 k=1 i
(5.250)
N
1X
µ Dij ,
T i=1 i
N Pi
1X o
µT,i + RT ln
−
Dij ,
T i=1
Po
N ni P
1X o
µT,i + RT ln
Dij ,
−
T i=1
nPo
N 1X o
ni RT To
Dij ,
−
µT,i + RT ln
T i=1
Po V To
N 1X o
T
ni RTo
−
µT,i + RT ln
+ RT ln
Dij ,
T i=1
To
Po V


N  o
X
T
ni RTo 

 µT,i
+ R ln
+R ln
−
 Dij ,

To
Po V 
 T
i=1
|
{z
}
= −
(5.251)
=
(5.252)
=
=
=
=
(5.253)
(5.254)
(5.255)
(5.256)
≡ψi (T )
∂2S
∂ξk ∂ξj
N X
ni RTo
= −
ψi (T ) + R ln
Dij ,
Po V
i=1
N X
R ∂ni
∂ψi (T ) ∂T
Dij ,
= −
+
∂T ∂ξk ni ∂ξk
i=1
N X
R
∂ψi (T ) ∂T
+ Dik Dij .
= −
∂T ∂ξk ni
i=1
(5.257)
(5.258)
(5.259)
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220
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Now, consider dT for the adiabatic system.
E = Eo =
N
X
q=1
nq eq (T ),
(5.260)


N
X
 ∂eq

nq
,
dT
+
dE = 0 =
e
dn
q
q
 ∂T

|{z}
q=1
(5.261)
=cvq
0 =
N
X
(nq cvq dT + eq dnq ) ,
(5.262)
q=1
PN
q=1 eq dnq
dT = − PN
q=1 nq cvq
= −
,
(5.263)
N −L
N
1 X X
eq
Dqp dξp ,
ncv q=1 p=1
(5.264)
N N −L
1 XX
= −
eq Dqp dξp ,
ncv q=1 p=1
(5.265)
N N −L
1 XX
∂T
∂ξp
= −
,
eq Dqp
∂ξk
ncv q=1 p=1
∂ξk
(5.266)
N N −L
1 XX
= −
eq Dqp δpk ,
ncv q=1 p=1
= −
N
1 X
eq Dqk .
ncv q=1
2
∂ S
∂ξk ∂ξj
=
N
X
i=1
(5.267)
N
∂ψi (T ) 1 X
R
eq Dqk − Dik
∂T ncv q=1
ni
(5.268)
!
Dij ,
N
N
N
X
X
∂ψi (T ) 1 X
R
=
Dik Dij ,
eq Dqk Dij −
∂T
nc
n
v
i
q=1
i=1
i=1
N
N
N
X
1 X X ∂ψi (T )
Dik Dij
=
.
eq Dqk Dij − R
ncv i=1 q=1 ∂T
ni
i=1
(5.269)
(5.270)
(5.271)
Now, consider the temperature derivative of ψi (T ), where ψi (T ) is defined in Eq. (5.256):
µoT,i
T
,
(5.272)
ψi ≡
+ R ln
T
To
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221
5.2. EQUILIBRIUM CONDITIONS
µoT,i
dψi
1 dµoT,i R
= − 2 +
+ .
dT
T
T dT
T
(5.273)
Now
o
µoT,i = hT,i − T soT,i ,
!
Z T
Z T
c
(
T̂
)
o
Pi
dT̂ ,
= hTo ,i +
cP i (T̂ )dT̂ −T soTo ,i +
T̂
To
To
|
{z
}
|
{z
}
o
=hT,i
Z T
dµoT,i
cP i (T̂ )
o
= cP i − sTo ,i −
dT̂ − cP i ,
dT
T̂
To
Z T
cP i (T̂ )
o
= −sTo ,i −
dT̂ ,
T̂
To
= −soT,i .
(5.274)
(5.275)
=soT,i
(5.276)
(5.277)
(5.278)
Note that Eq. 5.278 is a special case of the Gibbs equation given by Eq. 3.252. With this,
one finds that Eq. 5.273 reduces to
µoT,i soT,i R
dψi
= − 2 −
+ ,
dT
T
T
T
1
= − 2 µoT,i + T soT,i − RT ,
T
1
= − 2 g oT,i + T soT,i − RT ,
T 
= −
(5.279)
(5.280)
(5.281)

1  o

o
o
h − T s +T sT,i − RT  ,
T 2 | T,i {z T,i}
(5.282)
=goT,i
=
=
=
=
1 o
− 2 hT,i − RT ,
T
1
− 2 ei + Pi v i − RT ,
T
1
− 2 ei + RT − RT ,
T
1
− 2 ei .
T
(5.283)
(5.284)
(5.285)
(5.286)
So, substituting Eq. (5.286) into Eq. (5.271), one gets
∂2S
∂ξk ∂ξj
N
= −
N
N
X Dik Dij
1 XX
.
e
e
D
D
−
R
i
q
qk
ij
ncv T 2 i=1 q=1
n
i
i=1
(5.287)
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222
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Next, as before, consider the sum
N
−L N
−L
X
X
k=1 j=1
N −L N −L N
N
1 X X XX
∂2S
zk zj = −
ei eq Dqk Dij zk zj
∂ξk ∂ξj
ncv T 2 k=1 j=1 i=1 q=1
−R
= −
N
−L N
−L X
N
X
X
k=1 j=1 i=1
Dik Dij
zk zj ,
ni
N
N N −L N −L
1 XX X X
eq Dqk zk ei Dij zj
ncv T 2 i=1 q=1
j=1
k=1
−R
N N
−L N
−L
X
X
X
i=1 k=1 j=1
N
Dik zk Dij zj
,
ni
N
N
−L
X
1 XX
= −
ncv T 2 i=1 q=1
−R
N
X
i=1
N
X
1
−R
ni
i=1
1
= −
ncv T 2
1
ncv T 2
N
−L
X
1
ni
1
= −
ncv T 2
= −
(5.288)
k=1
k=1
Dik zk
N N
−L
X
X
i=1 j=1
N
−L
X
k=1
N N
−L
X
X
i=1 j=1
N
X
i=1
ei yi
eq Dqk zk
!
ei Dij zj
Dik zk
!
ei Dij zj
!2
(5.289)
−R
!
N
−L
X
j=1
!
Dij zj
!
q=1 k=1
j=1
Dij zj
ei Dij zj
yi2
.
ni
(5.290)
eq Dqk zk
!
!
,
!
,
N
X
1
−R
n
i=1 i
N
X
i=1
j=1
N N
−L
X
X
N
−L
X
!2
N
−L
X
(5.291)
N
−L
X
k=1
Dik zk
!2
(5.292)
,
(5.293)
Since cv > 0, T > 0, R > 0, ni > 0, and the other terms are perfect squares, it is obvious
that the second partial derivative of S < 0; consequently, critical points of S represent a
maximum. Again near the boundary of the physical region, S ∼ −ni ln ni , so limni →0 S → 0.
From Eq. (5.268), there is no formal restriction on the slope at the boundary. However, if a
critical point is to exist in the physical domain in which the second derivative is guaranteed
negative, the slope at the boundary must be positive everywhere. This combines to guarantee
that if a critical point exists in the physically accessible region of composition space, it is
unique.
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5.3. CONCISE REACTION RATE LAW FORMULATIONS
5.2.3.4
223
A similar proof holds for the adiabatic-isobaric case. Here, the appropriate Legendre transformation is H = E + P V , where H is the enthalpy. We omit the details, which are similar
to those of previous sections, and find a term which must be negative semi-definite, ΥHP :
N
−L N
−L
X
X
∂2S
zk zj
∂ξ
∂ξ
k
j
k=1 j=1
!2
2
r
N
N r
N
X
RX X
nj
ni
1
yi −
yj .
hi yi −
= −
ncP T 2 i=1
n i=1 j=i+1
ni
nj
ΥHP =
(5.294)
(5.295)
Because cP > 0 and ni ≥ 0, the term involving hi yi is a perfect square, and the term
multiplying R is positive definite for the same reasons as discussed before. Hence, ΥHP < 0,
and the Hessian matrix is negative definite.
5.3
Concise reaction rate law formulations
One can employ notions developed in the Zel’dovich uniqueness proof to obtain a more efficient representation of the reaction rate law for multiple reactions. There are two important
cases: 1) J ≥ (N − L); this is most common for large chemical kinetic systems, and 2)
J < (N − L); this is common for simple chemistry models.
The species production rate is given by Eq. (5.15), which reduces to
J
dρi
1 X dζj
νij
=
,
dt
V j=1
dt
i = 1, . . . , N.
(5.296)
Now, differentiating Eq. (5.169), one obtains
dni =
N
−L
X
k=1
Dik dξk ,
i = 1, . . . , N.
(5.297)
Comparing then Eq. (5.297) to Eq. (5.24), one sees that
J
X
νij dζj =
j=1
1
V
J
X
j=1
νij dζj =
N
−L
X
k=1
1
V
N
−L
X
k=1
Dik dξk ,
i = 1, . . . , N,
(5.298)
Dik dξk ,
i = 1, . . . , N,
(5.299)
(5.300)
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224
5.3.1
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Reaction dominant: J ≥ (N − L)
Consider first the most common case in which J ≥ (N − L). One can say the species
production rate is given
N −L
J
dρi
dξk X
1 X
Dik
=
=
νij rj ,
dt
V k=1
dt
j=1
i = 1, . . . , N.
(5.301)
One would like to invert
for dξk /dt. However, Dik is non-square and has
P and solve directly P
N
φ
D
=
0,
and
no inverse. But since N
i=1 φli νij = 0, L of the equations N equations
i=1 li ip
in Eq. (5.301) are redundant.
At this point, it is more convenient to go to a Gibbs vector notation, where there is an
obvious correspondence between the bold vectors and the indicial counterparts:
dρ
1
dξ
= D·
= ν · r,
dt
V
dt
dξ
= V DT · ν · r,
DT · D ·
dt
dξ
= V (DT · D)−1 · DT · ν · r.
dt
(5.302)
(5.303)
(5.304)
Because of the L linear dependencies, there is no loss of information in this matrix projection.
This system of N − L equations is the smallest number of differential equations that can be
solved for a general system in which J > (N − L).
Lastly, one recovers the original system when forming
D·
dξ
= V D · (DT · D)−1 · DT ·ν · r.
{z
}
|
dt
(5.305)
=P
Here, the N × N projection matrix P is symmetric, has norm of unity, has rank of N − L,
has N − L eigenvalues of value unity, and L eigenvalues of value zero. And, while application
of a general projection matrix to ν · r filters some of the information in ν · r, because the
N × (N − L) matrix D spans the same column space as the N × J matrix ν, no information
is lost in Eq. (5.305) relative to the original Eq. (5.302). Mathematically, one can say
P · ν = ν.
5.3.2
(5.306)
Species dominant: J < (N − L)
Next consider the case in which J < (N −L). This often arises in models of simple chemistry,
for example one- or two-step kinetics.
The fundamental reaction dynamics are most concisely governed by the J equations
which form
1 dζ
= r.
V dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.307)
225
5.4. IRREVERSIBILITY PRODUCTION
However, r is a function of the concentrations; one must therefore recover ρ as a function
of reaction progress ζ. In vector form, Eq. (5.296) is written as
1
dζ
dρ
= ν· .
dt
V
dt
(5.308)
Take as an initial condition that the reaction progress is zero at t = 0 and that there are an
appropriate set of initial conditions on the species concentrations ρ:
ζ = 0,
ρ = ρo ,
t = 0,
t = 0.
(5.309)
(5.310)
Then, since ν is a constant, Eq. (5.308) is easily integrated. After applying the initial
conditions, Eq. (5.310), one gets
ρ = ρo +
1
ν · ζ.
V
(5.311)
Last, if J = (N − L), either approach yields the same number of equations, and is equally
concise.
5.4
Irreversibility production
This section will consider the production of entropy due to irreversible chemical reactions. It
is closely related to the Clausius-Duhem inequality discussed on p. 198. We will first consider
the production of irreversibility in the context of what is known as Onsager10 reciprocity. This
notion relies on a linearization of the reaction rate laws in the neighborhood of equilibrium.
Note that because the Clausius-Duhem relation, Eq. (5.61) is already known to hold for
the full non-linear regime, that we find only that Onsager’s analysis simply reinforces this
notion in a simpler context. We then go on to give a more general analysis, and finish with
an illustration based on 1) the Zel’dovich mechanism for the example problem of Sec. 1.1.2,
and 2) an extended Zel’dovich mechanism including N2 and O2 dissociation. The original
Zel’dovich mechanism induces a strictly diagonal Onsager matrix, and thus does not fully
capture the essence of the analysis; the extended Zel’dovich mechanism induces a full Onsager
matrix.
5.4.1
Onsager reciprocity
There is a powerful result from physical chemistry which speaks to how systems behave near
equilibrium. The principle was developed by Onsager in the early twentieth century, and is
known as the reciprocity principle. Where it holds, one can guarantee that on their approach
to equilibrium, systems will approach the equilibrium in a non-oscillatory manner. It will
be illustrated here.
10
Lars Onsager, 1903-1976, Norwegian-American physicist.
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226
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Note that Eq. (5.18) can be written as
1 dζj
= rj = R′j − R′′j ,
j = 1, . . . , J,
V dt
R′′j
= R′j 1 − ′ ,
j = 1, . . . , J.
Rj
Note further that the definition of affinity, Eq. (5.11), gives
R′′j
−αj
exp
=
,
j = 1, . . . , J.
R′j
RT
(5.312)
(5.313)
(5.314)
Therefore, one can say
1 dζj
−αj
′
,
= rj = Rj 1 − exp
V dt
RT
j = 1, . . . , J.
(5.315)
Now, as each reaction comes to equilibrium, one finds that αj → 0, so a Taylor series
expansion of rj yields
αj
1 dζj
′
+ ...
,
j = 1, . . . , J,
(5.316)
= rj ∼ Rj 1 − 1 −
V dt
RT
αj
∼ R′j
,
j = 1, . . . , J.
(5.317)
RT
Note that R′j > 0, while αj can be positive or negative. Note also there is no summation on
j. Take now the matrix R′ to be the diagonal matrix with Rj populating its diagonal:
 ′

R1 0 . . . 0
 0 R′2 . . . 0 
R′ ≡ 
.
(5.318)
..
.. 
..
 ...
.
.
. 
0
0
0
R′J
1 ′
(5.319)
R · α.
RT
Now, the entropy production is given for multi-component systems by Eq. (5.57):
r=
dS dt E,V
J
V X
=
αj rj ≥ 0.
T j=1
Cast this entropy inequality into Gibbs notation:
V T
dS =
α · r ≥ 0.
dt E,V
T
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.320)
(5.321)
227
5.4. IRREVERSIBILITY PRODUCTION
Now, consider the definition of affinity, Eq. (5.10), in Gibbs notation:
α = −ν T · µ.
(5.322)
Now, ν T is of dimension J × N with rank typically N − L. Because ν T is typically not of
full rank, one finds only N − L of the components of α to be linearly independent. When
one recalls that ν T maps vectors µ into the column space of ν T , one recognizes that α can
be represented as
α = C · α̂.
(5.323)
Here, C is a J × (N − L)-dimensional matrix of full rank, N − L, whose N − L columns are
populated by the linearly independent vectors which form the column space of ν T , and α̂
is a column vector of dimension (N − L) × 1. If J ≥ N − L, one can explicitly solve for α̂,
starting by operating on both sides of Eq. (5.323) by CT :
CT · α = CT · C · α̂,
CT · C · α̂ = CT · α,
−1
α̂ = CT · C
· CT · α,
−1
= − CT · C
· CT · ν T · µ,
−1
α = C · α̂ = − C · CT · C
· CT ·ν T · µ.
{z
}
|
(5.324)
(5.325)
(5.326)
(5.327)
(5.328)
≡B
Here, in the recomposition of α, one can employ the J × J symmetric projection matrix B,
which has N − L eigenvalues of unity and J − (N − L) eigenvalues of zero. The matrix B
has rank N − L, and is thus not full rank.
Substitute Eqs. (5.319, 5.323) into Eq. (5.321) to get
dS dt E,V
T
=α
=α
z }| {
z
1 ′ }| {
V
T
C · α̂ ·
R · C · α̂ ≥ 0,
=
T
|RT {z
}
=r
!
!
T
α̂
V α̂
T
·C
R′ · C} ·
≥ 0.
=
| · {z
T
R T
(5.329)
(5.330)
≡L
Since each of the entries of the diagonal R′ are guaranteed positive semi-definite in the
physical region of composition space, the entropy production rate near equilibrium is also
positive semi-definite. The constant square matrix L, of dimension (N − L) × (N − L), is
given by
L ≡ CT · R′ · C.
(5.331)
The matrix L has rank N −L and is thus full rank. Because R′ is diagonal with positive semidefinite elements, L is symmetric positive semi-definite. Thus, its eigenvalues are guaranteed
to be positive and real. Note that off-diagonal elements of L can be negative, but that
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228
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
the matrix itself remains positive semi-definite. Onsager reciprocity simply demands that
near equilibrium, the linearized version of the combination of the thermodynamic “forces”
(here the affinity α) and “fluxes” (here the reaction rate r) be positive semi-definite. Upon
linearization, one should always be able to find a positive semi-definite matrix associated
with the dynamics of the approach to equilibrium. Here, that matrix is L, and by choices
made in its construction, it has the desired properties.
One can also formulate an alternative version of Onsager reciprocity using the projection
matrix B, which from Eq. (5.328), is
−1
B ≡ C · CT · C
· CT .
(5.332)
With a series of straightforward substitutions, it can be shown that the entropy production
rate given by Eq. (5.330) reduces to
V αT
α
dS T
′
=
·
B
·
R
·
B
(5.333)
·
| {z } T ≥ 0.
dt E,V
R T
≡L
Here, an alternative symmetric positive semi-definite matrix L, of dimension J × J and rank
N − L, has been defined as
L ≡ BT · R′ · B.
(5.334)
One can also express the entropy generation directly in terms of the chemical potential
rather than the affinity by defining the J × N matrix S as
S ≡ B · νT ,
= C · (CT · C)−1 · C · ν T .
(5.335)
(5.336)
The matrix S has rank N − L and thus is not full rank. With a series of straightforward
substitutions, it can be shown that the entropy production rate given by Eq. (5.333) reduces
to
dS µ
V µT
T
′
=
· |S · {z
R · S} ·
≥ 0.
(5.337)
dt E,V
T
R T
≡L
Here, an alternative symmetric positive semi-definite matrix L, of dimension N × N and
rank N − L, has been defined as
L ≡ ST · R′ · S.
(5.338)
Example 5.4
Find the matrices associated with Onsager reciprocity for the reaction mechanism given by
H2 + O2
H2 + OH
H + O2
H2 + O
CC BY-NC-ND. 30 March 2014, J. M. Powers.
⇌ 2OH,
⇌ H + H2 O,
(5.339)
(5.340)
⇌ O + OH,
⇌ H + OH,
(5.341)
(5.342)
229
5.4. IRREVERSIBILITY PRODUCTION
H +H
2OH
H2
O2
H + OH
⇌ H2 ,
(5.343)
⇌ O + H2 O,
⇌ H + H,
(5.344)
(5.345)
⇌ O + O,
⇌ H2 O.
(5.346)
(5.347)
Here, there are N = 6 species (H, H2 , O, O2 , OH, H2 O), composed of L = 2 elements (H, O), reacting
in J = 9 reactions. Here, J ≥ N − L, so the analysis of this section can be performed.
Take species i = 1 as H, i = 2 as H2 ,. . ., i = N = 6 as H2 O. Take element l = 1 as H and element
l = L = 2 as O. The full rank stoichiometric matrix φ, of dimension 2 × 6 = L × N and rank 2 = L, is
1 2 0 0 1 2
φ=
.
(5.348)
0 0 1 2 1 1
The rank-deficient matrix of stoichiometric coefficients ν, of dimension 6×9 = N ×J and rank 4 = N −L
is


0
1 −1 1 −2 0
2
0 −1
0 −1 0
0 
 −1 −1 0 −1 1


0
1 −1 0
1
0
2
0 
 0
ν=
(5.349)
.
0
0
0 −1 0 
 −1 0 −1 0


2 −1 1
1
0 −2 0
0 −1
0
1
0
0
0
1
0
0
1
Check for stoichiometric balance:


0
1 −1 1 −2 0
2
0 −1
−1 −1 0 −1 1
0 −1 0
0 



1 2 0 0 1 2
0
1 −1 0
1
0
2
0 
 0
φ·ν =
·
(5.350)
,
0 0 1 2 1 1
0
0
0 −1 0 
 −1 0 −1 0


2 −1 1
1
0 −2 0
0 −1
0
1
0
0
0
1
0
0
1
0 0 0 0 0 0 0 0 0
=
.
(5.351)
0 0 0 0 0 0 0 0 0
So, the number and mass of every element is conserved in every reaction; every vector in the column
space of ν is in the right null space of φ.
The detailed version of the reaction kinetics law is given by
dρ
dt
= ν · r,
(5.352)


r1
  r2 
0
1 −1 1 −2 0
2
0 −1
 
r 
0 −1 0
0   3
 −1 −1 0 −1 1

  r4 
0
1 −1 0
1
0
2
0   
 0
= 
 · r ,
0
0
0 −1 0   5 
 −1 0 −1 0

  r6 
2 −1 1
1
0 −2 0
0 −1
 
 r7 
0
1
0
0
0
1
0
0
1
 
r8
r9


r2 − r3 + r4 − 2r5 + 2r7 − r9
 −r1 − r2 − r4 + r5 − r7 


r3 − r4 + r6 + 2r8


= 
.
−r1 − r3 − r8




2r1 − r2 + r3 + r4 − 2r6 − r9
r2 + r6 + r9

(5.353)
(5.354)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
230
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
The full rank matrix D, of dimension 6 × 4 = N × (N − L) and rank 4 = N − L, is composed of
vectors in the right null space of φ. It is non-unique, as linear combinations of right null space vectors
suffice. It is equivalently composed by casting the N − L linearly independent vectors of the column
space of ν in its columns. Recall that some of the columns of ν are linearly dependent. In the present
example, the first N − L = 4 column vectors of ν happen to be linearly dependent, and thus will not
suffice. Other sets are not; the last N − L = 4 column vectors of ν happen to be linearly independent
and thus suffice for the present purposes. Take then


0
2
0 −1
0 
 0 −1 0


0
2
0 
 1
D=
(5.355)
.
0 −1 0 
 0


−2 0
0 −1
1
0
0
1
It is easily verified by direct substitution that D is in the right null space of φ:


0
2
0 −1
0 −1 0
0  


1 2 0 0 1 2
1
0
2
0 
0 0 0 0

φ·D=
·
.
=
0 0 1 2 1 1
0 −1 0 
0 0 0 0
 0


−2 0
0 −1
1
0
0
1
(5.356)
Since φ · ν = 0 and φ · D = 0, one concludes that the column spaces of both ν and D are one and the
same.
The non-unique concise version of the reaction kinetics law is given by
dξ
dt
= V (DT · D)−1 · DT · ν · r,


0 1
 0

 2 −1 0
= V 
0 2
 0

−1 0 0
(5.357)

−1
0 

0 

0 

−1
1
 
r1
  r2 
−1
 
r 
0   3
  r4 
0   
 ·r ,
0   5
  r6 
−1
 
 r7 
1
 
r8
r9
0
2
0

0 −2 1  0 −1 0

0
0 0 1
0
2

−1 0 0  0
0 −1

0 −1 1
−2 0
0
1
0
0

0
1 −1 1 −2 0
2
0
0 −1 0
 −1 −1 0 −1 1

0
1 −1 0
1
0
2
 0

0
0
0 −1
 −1 0 −1 0

2 −1 1
1
0 −2 0
0
0
1
0
0
0
1
0
0


−2r1 − r3 − r4 + r6
 r + r2 + r4 − r5 + r7 
= V 1
.
r1 + r3 + r8
2r1 + r2 + r3 + r4 + r9
−1

0
0
 2 −1
·
0
0
−1 0
1 0 −2
0 0
0
2 −1 0
0 0 −1

1
0
·
0
1
(5.358)
(5.359)
The rank-deficient projection matrix P, of dimension 6 × 6 = N × N and rank 4 = N − L, is
−1
P = D · DT · D
· DT ,
(5.360)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
231
5.4. IRREVERSIBILITY PRODUCTION




= 


54
61
−14
61
3
61
6
61
−4
61
−11
61
−14
61
33
61
6
61
12
61
−8
61
−22
61
3
61
6
61
51
61
−20
61
−7
61
−4
61
6
61
12
61
−20
61
21
61
−14
61
−8
61
−4
61
−8
61
−7
61
−14
61
50
61
−15
61
−11
61
−22
61
−4
61
−8
61
−15
61
35
61




.


(5.361)
The projection matrix P has 4 = N − L eigenvalues of unity and 2 = L eigenvalues of zero.
The affinity vector α, of dimension 9 × 1 = J × 1, is given by
α
= −ν T · µ,

0
1 −1 1 −2
 −1 −1 0 −1 1

0
1 −1 0
 0
= −
0
 −1 0 −1 0

2 −1 1
1
0
0
1
0
0
0

0 −1 0 −1 2
0 −1
 1 −1 0

1 −1 1
 −1 0

1
 1 −1 −1 0

= −  −2 1
0
0
0

0
1
0 −2
 0

0
0
 2 −1 0

0
0
2 −1 0
−1 0
0
0 −1


µ2 + µ4 − 2µ5
 −µ1 + µ2 + µ5 − µ6 


 µ1 − µ3 + µ4 − µ5 


 −µ1 + µ2 + µ3 − µ5 


= 
2µ1 − µ2
.


−µ
+
2µ
−
µ


3
5
6


−2µ1 + µ2




−2µ3 + µ4
µ1 + µ5 − µ6
(5.362)
T 

µ1
0
2
0 −1
0 −1 0
0   µ2 
  
1
0
2
0   µ3 
 ·  ,
0
0 −1 0   µ4 
  
µ5
−2 0
0 −1
1
0
0
1
µ6

0
1  
µ1

0
  µ2 
0  
 µ 
0 ·  3 ,
 µ 
1  4 
µ5

0
µ6

0
1
(5.363)
(5.364)
(5.365)
The full rank matrix C, of dimension 9 × 4 = J × (N − L) and rank 4 = N − L, is composed of
the set of N − L = 4 linearly independent column space vectors of ν T ; thus, they also comprise the
N − L = 4 linearly independent row space vectors of ν. It does not matter which four are chosen, so
long as they are linearly independent. In this case, the first four column vectors of ν T suffice:

C
0
 1

 −1

 1

=  −2

 0

 2

0
−1

−1 0 −1
−1 0
0 

0
1 −1 

−1 −1 0 

1
0
0 .

0
1
0 

−1 0
0 

0
2 −1
0
0
0
(5.366)
When J > N − L, not all of the components of α are linearly independent. In this case, one can
CC BY-NC-ND. 30 March 2014, J. M. Powers.
232
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
form the reduced affinity vector, α̂, of dimension 4 × 1 = (N − L) × 1 via
α̂ =
=
(CT · C)−1 · CT · α,




1 −1 1 −2
 0

−1
−1
0 −1 1


0
0
1 −1 0


−1
0
−1
0
0




0
1
 −1 −1

0
0
−1 0
=
(5.367)
−1 1 −2 0
0 −1 1 0
1 −1 0 1
−1 0
0 0

−µ1 − µ5 + µ6
 −µ2 − 2µ5 + 2µ6 
.

−µ3 + 2µ5 − µ6
−µ4 + 4µ5 − 2µ6

−1

0 −1 0 −1
0 
 1 −1 0


1 −1 
  −1 0
0 2
0 −1


 1 −1 −1 0 
0 −1 0
0  

0
0 
 ·  −2 1
1 0
2
0


0
1
0 
 0
0 0 −1 0


0 
 2 −1 0


0
0
2 −1
−1 0
0
0


µ2 + µ4 − 2µ5
 −µ1 + µ2 + µ5 − µ6 


  µ1 − µ3 + µ4 − µ5 
2
0 −1


 −µ1 + µ2 + µ3 − µ5 
−1 0
0  

2µ1 − µ2
,
·
0
2
0


 −µ3 + 2µ5 − µ6 
0 −1 0


−2µ1 + µ2




−2µ3 + µ4
µ1 + µ5 − µ6
·
(5.368)
(5.369)
The rank-deficient projection matrix B, of dimension 9 × 9 = J × J and rank 4 = N − L, is found to be
B
= C · (CT · C)−1 · CT ,
 57
9
31
94
9
 94
 31
 94
 26
 94

=  −1
94
 −17
 94
 1
 94
 6
94
8
94
94
41
94
−5
94
14
94
−15
94
27
94
15
94
−4
94
26
94
94
−5
94
35
94
−4
94
11
94
−1
94
−11
94
28
94
6
94
26
94
14
94
−4
94
30
94
−12
94
−16
94
12
94
−22
94
2
94
−1
94
−15
94
11
94
−12
94
33
94
−3
94
−33
94
−10
94
18
94
−17
94
27
94
−1
94
16
94
−3
94
43
94
3
94
18
94
24
94
1
94
15
94
−11
94
12
94
−33
94
3
94
33
94
10
94
−18
94
6
94
−4
94
28
94
−22
94
−10
94
18
94
10
94
60
94
−14
94
8
94
26
94
6
94
2
94
18
94
24
94
−18
94
−14
94
44
94







.





(5.370)
(5.371)
The projection matrix B has a set of 9 = J eigenvalues, 4 = N −L of which are unity, and 5 = J −(N −L)
of which are zero. One can recover α by the operation α = C · α̂ = −B · ν T · µ.
The square full rank Onsager matrix L, of dimension 4×4 = (N −L)×(N −L) and rank 4 = N −L,
is given by
L =
CT · R · C,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.372)
233
5.4. IRREVERSIBILITY PRODUCTION
=

0
1 −1 1 −2 0
−1
−1
0 −1 1 0


0
0
1 −1 0 1
−1 0 −1 0
0 0

0
 1

 −1

 1

·  −2

 0

 2

0
−1
=

−1 0 −1
−1 0
0 

0
1 −1 

−1 −1 0 

1
0
0 ,

0
1
0 

−1 0
0 

0
2 −1
0
0
0


R′1
 0

  0
2
0 −1

 0
−1 0
0  
· 0
0
2
0

 0
0 −1 0

 0

0
0
R′2 + R′3 + R′4 + 4R′5 + 4R′7 + R′9
−R′2 − R′4 − 2R′5 − 2R′7


−R′3 − R′4
R′3
0
R′2
0
0
0
0
0
0
0
0
0
R′3
0
0
0
0
0
0
−R′2 − R′4 − 2R′5 − 2R′7
R′1 + R′2 + R′4 + R′5 + R′7
R′4
R′1
0
0
0
R′4
0
0
0
0
0
0
0
0
0
R′5
0
0
0
0
0
0
0
0
0
R′6
0
0
0
0
0
0
0
0
0
R′7
0
0
−R′3 − R′4
R′4
′
′
R3 + R4 + R′6 + 4R′8
−R′3 − 2R′8
0
0
0
0
0
0
0
R′8
0

0
0 

0 

0 

0 

0 

0 

0
′
R9
(5.373)

R′3
′
R1

.
−R′3 − 2R′8
′
′
′
R1 + R3 + R8
(5.374)
Obviously L is symmetric, and thus has all real eigenvalues. It is also positive semi-definite. With a
similar effort, one can obtain the alternate rank-deficient square Onsager matrices L and L. Recall that
L, L, and L each have rank N − L, while L has the smallest dimension, (N − L) × (N − L), and so
forms the most efficient Onsager matrix.
5.4.2
General analysis
This section will be restricted to isothermal, isochoric reaction. It is known that Gibbs free
energy and irreversibility production rates reach respective minima at equilibrium.
Consider the gradient of the irreversibility production rate in the space of species progress
variables ξk . First, recall Eq. (5.231) for the gradient ofP
Gibbs free energy with respect to
N
the independent species progress variables, ∂G/∂ξk =
i=1 µi Dik . Now, beginning from
Eq. (4.464), define the differential irreversibility dI as
N
1X
dI = −
µ dni .
T i=1 i
(5.375)
In terms of an irreversibility production rate, one has
N
dI
1 X dni
µ
= İ = −
.
dt
T i=1 i dt
(5.376)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
234
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Take now the time derivative of Eq. (5.169) to get
N
−L
X
dξk
dni
Dik
=
.
dt
dt
k=1
(5.377)
Substitute from Eq. (5.377) into Eq. (5.376) to get
İ = −
N
N −L
dξk
1X X
Dik
,
µi
T i=1
dt
k=1
(5.378)
N −L
N
1 X dξk X
= −
µ Dik ,
T
dt i=1 i
k=1
| {z }
(5.379)
∂G
= ∂ξ
k
= −
1
T
N
−L
X
k=1
dξk ∂G
.
dt ∂ξk
(5.380)
Now, Eq. (5.304) gives an explicit algebraic formula for dξk /dt. Define then the constitutive
ˆk:
function ω̇
ˆ k (ξ1 , . . . , ξN −L ) ≡ dξk = V (DT · D)−1 · DT · ν · r.
(5.381)
ω̇
dt
So, the irreversibility production rate is
İ = −
The gradient of this field is given by
N −L
∂ İ
1 X
=−
∂ξp
T k=1
N −L
1 X ˆ ∂G
ω̇ k
.
T k=1
∂ξk
2
ˆ k ∂G
∂ ω̇
ˆk ∂ G
+ ω̇
∂ξp ∂ξk
∂ξp ∂ξk
(5.382)
!
.
(5.383)
The Hessian of this field is given by
N −L
1 X
∂ 2 İ
=−
∂ξl ∂ξp
T k=1
3
ˆ k ∂G ∂ ω̇
ˆ k ∂2G
ˆ k ∂2G
∂ 2 ω̇
∂ ω̇
ˆk ∂ G
+
+
+ ω̇
∂ξl ∂ξp ∂ξk
∂ξp ∂ξl ∂ξk
∂ξl ∂ξp ∂ξk
∂ξl ∂ξp ∂ξk
ˆ k = 0 as well as ∂G/∂ξk = 0. Thus
Now, at equilibrium, ξk = ξke , we have ω̇
= 0,
İ ξk =ξke
∂ İ = 0,
∂ξk e
ξk =ξk
!
N −L
ˆ k ∂2G
ˆ k ∂2G
1 X ∂ ω̇
∂ ω̇
∂ 2 İ = −
+
.
∂ξl ∂ξp T k=1 ∂ξp ∂ξl ∂ξk
∂ξl ∂ξp ∂ξk
e
e
ξk =ξk
CC BY-NC-ND. 30 March 2014, J. M. Powers.
ξk =ξk
!
(5.384)
(5.385)
(5.386)
(5.387)
235
5.4. IRREVERSIBILITY PRODUCTION
It can be shown for arbitrary matrices that the first term in the right hand side of Eq. (5.387)
is the transpose of the second; moreover, for arbitrary matrices, the two terms are in general
asymmetric. But, their sum is a symmetric matrix, as must be since the Hessian of İ must
be symmetric. But for arbitrary matrices, we find no guarantee that the Hessian of İ is
positive semi-definite, as the eigenvalues can take any sign. However, it can be verified by
direct expansion for actual physical systems that the two terms in the Hessian evaluated
at equilibrium are identical. This must be attributable to extra hidden symmetries in the
system. Moreover, for actual physical systems, the eigenvalues of the Hessian of İ are positive
semi-definite. So, one can say
N −L
ˆ k ∂ 2 G 2 X ∂ ω̇
∂ 2 İ = −
.
(5.388)
∂ξl ∂ξp T
∂ξ
∂ξ
∂ξ
e
p
l
k
ξ
=ξ
e
e
k
k=1
ξk =ξk
ξk =ξk
k
ˆ k and G
Where does the hidden symmetry arise? The answer lies in the fact that the ω̇
are not independent. Using notions from Onsager reciprocity, let us see why the Hessian of
İ is indeed positive semi-definite. Consider, for instance, Eq. (5.330), which is valid in the
neighborhood of equilibrium:
!
!
T
V α̂
α̂
İ =
·L·
valid only near equilibrium.
(5.389)
T
R T
Recall that by construction L is positive semi-definite. Now, near equilibrium, α̂ has a
Taylor series expansion
∂
α̂
(5.390)
α̂ = α̂eq +
· (ξ − ξ|eq ) + ....
|{z} ∂ξ eq
=0
Now, recall that at equilibrium that α̂ = 0. Let us also define the Jacobian of α̂ as Jα .
Thus, Eq. (5.389) can be rewritten as
T
V İ =
Jα · (ξ − ξ|eq ) · L · Jα · (ξ − ξ|eq ) ,
(5.391)
RT 2
V
=
(ξ − ξ|eq )T · JTα · L · Jα · (ξ − ξ|eq ).
(5.392)
RT 2
By inspection, then we see that the Hessian of İ is
Hİ =
2V T
Jα · L · Jα .
RT 2
(5.393)
Moreover, L has the same eigenvalues as the similar matrix JTα · L · Jα , so Hİ is positive
semi-definite.
Returning to the less refined formulation, in Gibbs notation, we can write Eq. (5.388) as
2
Hİ = − HG · J,
T
(5.394)
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CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
ˆ k . Consider the eigenvectors and eigenvalues of the various
where J is the Jacobian matrix of ω̇
matrices here. The eigenvalues of J give the time scales of reaction in the neighborhood of
equilibrium. They are guaranteed real and negative. The eigenvectors of J give the directions
of fast and slow modes. Near equilibrium, the dynamics will relax to the slow mode, and the
motion towards equilibrium will be along the eigenvector associated with the slowest time
scale. Now, in the unlikely circumstance that HG were the identity matrix, one would have
the eigenvalues of Hİ equal to the product of −2/T and the eigenvalues of HG . So, they
would be positive, as expected. Moreover, the eigenvectors of Hİ would be identical to those
of J, so in this unusual case, the slow dynamics could be inferred from examining the slowest
˙ Essentially the same conclusion would be reached if HG had a
descent down contours of I.
diagonalization with equal eigenvalues on its diagonal. This would correspond to reactions
proceeding at the same rate near equilibrium. However, in the usual case, the eigenvalues
of HG are non-uniform. Thus, the action of HG on J is to stretch it non-uniformly in such
a fashion that Hİ does not share the same eigenvalues or eigenvectors. Thus, it cannot be
used to directly infer the dynamics.
Now, consider the behavior of İ in the neighborhood of an equilibrium point. In the
neighborhood of a general point ξk = ξˆk , İ has a Taylor series expansion
N
−L
N
−L N
−L 2
X
X
X
∂ İ 1
∂ İ ˆ
ˆ
ˆ
+
İ = İ ξk − ξk +
ξl − ξl
ξp − ξp + . . .
ξk =ξ̂k
∂ξk 2
∂ξl ∂ξp k=1
l=1 p=1
ξk =ξ̂k
ξk =ξ̂k
(5.395)
Near equilibrium, the first two terms of this Taylor series are zero, and İ has the behavior
N
−L N
−L
2
X
X
1
∂ İ e
İ =
ξp − ξpe + . . .
(5.396)
(ξl − ξl )
2
∂ξl ∂ξp e
l=1 p=1
ξk =ξk
Substituting from Eq. (5.388), we find near equilibrium that
N −L N −L N −L
ˆ k ∂
ω̇
1 XXX
(ξl − ξle )
İ = −
T l=1 p=1 k=1
∂ξp ξk =ξke
∂ 2 G e
ξ
−
ξ
p
p + ...
∂ξl ∂ξk ξk =ξe
(5.397)
k
Lastly, let us study whether İ is a Lyapunov function. We can show that İ > 0, ξp 6= ξpe ,
and İ = 0, ξp = ξpe . Now, to determine whether or not the Lyapunov function exists, we
must study dİ/dt:
dİ
dt
=
N
−L
X
p=1
=
N
−L
X
p=1
∂ İ dξp
,
∂ξp dt
(5.398)
∂ İ ˆ
ω̇ p ,
∂ξp
(5.399)
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237
5.4. IRREVERSIBILITY PRODUCTION
=
N
−L
X
p=1
N −L
1 X
−
T k=1
N −L N −L
1 XX
= −
T p=1 k=1
2
ˆ k ∂G
∂ ω̇
ˆk ∂ G
+ ω̇
∂ξp ∂ξk
∂ξp ∂ξk
!!
2
ˆ k ∂G
∂ ω̇
ˆ p + ω̇
ˆ p ∂ G ω̇
ˆk
ω̇
∂ξp ∂ξk
∂ξp ∂ξk
ˆ p,
ω̇
!
.
(5.400)
(5.401)
ˆ k = 0, so obviously dİ/dt = 0, at equilibrium. Away from
At the equilibrium state, ω̇
equilibrium, we know from our uniqueness analysis that the term ∂ 2 G/∂ξp ∂ξk is positive
semi-definite; so this term contributes to rendering dİ/dt < 0. But the other term does not
transparently contribute.
Let us examine the gradient of dİ/dt. It is
∂ dİ
∂ξm dt
N −L N −L
1 XX
= −
T p=1 k=1
2
ˆ k ∂G
ˆ
ˆ
ˆ
∂ 2 ω̇
ˆ p + ∂ ω̇ k ∂ G ω̇
ˆ p + ∂ ω̇ k ∂G ∂ ω̇ p
ω̇
∂ξm ∂ξp ∂ξk
∂ξp ∂ξm ∂ξk
∂ξp ∂ξk ∂ξm
!
3
2
ˆ p ∂2G
ˆk
∂ ω̇
∂
ω̇
∂
G
∂
G
ˆ k + ω̇
ˆp
ˆ k + ω̇
ˆp
.
(5.402)
+
ω̇
ω̇
∂ξm ∂ξp ∂ξk
∂ξm ∂ξp ∂ξk
∂ξp ∂ξk ∂ξm
At equilibrium, all terms in the gradient of dİ/dt are zero, so it is a critical point.
Let us next study the Hessian of dİ/dt to ascertain the nature of this critical point.
Because there are so many terms, let us only write those terms which will be non-zero at
equilibrium. In this limit, the Hessian is
N −L N −L
ˆ p ∂ ω̇
ˆ k ∂ 2 G ∂ ω̇
ˆp
ˆ k ∂ 2 G ∂ ω̇
1 X X ∂ ω̇
∂ 2 dİ +
= −
∂ξn ∂ξm dt T p=1
∂ξp ∂ξm ∂ξk ∂ξn
∂ξp ∂ξn ∂ξk ∂ξm
k=1
eq
!
2
2
ˆ
ˆ
ˆ
ˆ
∂ ω̇ p ∂ G ∂ ω̇ k ∂ ω̇ p ∂ G ∂ ω̇ k
+
+
.
(5.403)
∂ξm ∂ξp ∂ξk ∂ξn
∂ξn ∂ξp ∂ξk ∂ξm
With some effort, it may be possible to show the total sum is negative semi-definite. This
would guarantee that dİ/dt < 0 away from equilibrium. If so, then it is true that İ is a
Lyapunov function in the neighborhood of equilibrium. Far from equilibrium, it is not clear
whether or not a Lyapunov function exists.
5.4.3
Zel’dovich mechanism example
Let us now consider a numerical example, building on the analysis of Sec. 1.1.2; many numerical details have already been presented and will not be repeated here. For the conditions
simulated in Sec. 1.1.2, we plot dS/dt|E,V /V = İ/V as a function of time as predicted by
the full theory of Eq. (5.62), repeated here in slightly modified form
İ
α
1 dS T
=
,
(5.404)
=r ·
V dt E,V
V
T
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238
(1/V)dS/dt (erg/K/s/cm3)
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
107
10
10-5
10-11
10-17
10-9
10-8
10-7
10-6
10-5
10-4
t (s)
Figure 5.2: Irreversibility production rate versus time for the Zel’dovich mechanism problem
of Sec. 1.1.2.
in Fig. 5.2. Clearly, İ/V is always positive and approaches zero as the system approaches
equilibrium. Note that we have made no appeal to notions of Onsager nor have we confined
ourselves to the neighborhood of equilibrium. The function İ/V is non-linear, positive, and
decreases to zero by its construction, which is consistent with the Clausius-Duhem results
shown earlier on p. 198.
Now, let us consider the system of Sec. 1.1.2 in the language of Onsager reciprocity. We
see from Eq. (1.160) that
−1 −1 1 1 0
T
ν =
.
(5.405)
1 −1 0 1 −1
Here, it is easily shown that ν T is of full rank, R = 2, and thus has two linearly independent
column vectors. Following the discussion of Eq. (5.323), let us simply choose the first two
column vectors of ν T to form the matrix C:
−1 −1
C=
.
(5.406)
1 −1
Then, by Eq. (5.332), we find the projection matrix B to be
−1
B = C · CT · C
· CT ,
−1 −1 −1
−1 1
−1 −1
−1 1
=
·
·
,
1 −1
−1 −1
1 −1
−1 −1
1 0
=
.
0 1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.407)
(5.408)
(5.409)
239
5.4. IRREVERSIBILITY PRODUCTION
Remarkably, we recover the identity matrix: B = I. For more general reaction mechanisms
which have ν T of less than full rank, this will not be true.
So, from Eq. (5.334), we find one of the Onsager matrices to reduce to
′
R1 0
T
′
T
′
′
L= B ·R ·B =I ·R ·I= R =
.
(5.410)
0 R′2
Since each Rj , j = 1, . . . , J, is positive semi-definite, the matrix R′ is positive semi-definite.
We can then cast the entropy production of Eq. (5.333) as
α
dS V αT
.
(5.411)
=
· L·
dt E,V
T
R T
{z
} | {z }
|
∼jT
=x
Comparing Eqs. (5.411) and (5.65), we see in the Onsager limit that
V αT
V
T
j ∼
· L = xT · L.
R T
R
(5.412)
Transposing both sides, we can also say
V
j ∼ LT ·
R
V
α
= LT · x.
T
R
We also can recast in the standard form
V T
dS x · L · x.
= İ =
dt E,V
R
(5.413)
(5.414)
For the conditions simulated in Sec. 1.1.2, we plot İ/V as a function of time for both the
full non-linear model as well as the prediction in the Onsager limit of Eq. (5.411) in Fig. 5.3.
Clearly at early time the two theories predict different values, but the Onsager limit become
a better approximation as equilibrium is approached.
Numerically, direct substitution of values at equilibrium shows that
0.573127
0
L=
.
(5.415)
0
0.549134
The units of the entries of L are mole/cm3 /s.
Now at equilibrium, Eq. (5.12) tells us the affinity of each reaction must be zero; thus,
at equilibrium, we must have x = 0. This then induces the fluxes j = 0 at equilibrium.
Near equilibrium, it is possible to perform Prigogine’s extremization exercise surrounding
Eq. (P6.8). Recognizing that for our problem L12 = L21 = 0, which may not hold for more
general problems, we slightly extend our analysis to consider the more general case. For such
a case, we have near equilibrium that
V LT11 LT12
x1
j1
·
.
(5.416)
=
T
T
x2
j2
R L21 L22
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240
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
(1/V)dS/dt (erg/K/s/cm3)
exact
1010
Onsager limit
9
10
108
107
106
2 ´ 10-8
5 ´ 10-8
1 ´ 10-7
2 ´ 10-7
5 ´ 10-7
1 ´ 10-6
t (s)
Figure 5.3: Irreversibility production rate versus time for the Zel’dovich mechanism problem
of Sec. 1.1.2 for the full non-linear model and the Onsager limit.
Thus,
V T
L11 x1 + LT12 x2 ,
R
V T
=
L21 x1 + LT22 x2 ,
R
j1 =
(5.417)
j2
(5.418)
(5.419)
The irreversibility production becomes
İ =
V T 2
L11 x1 + LT12 + LT21 x1 x2 + LT22 x22
R
(5.420)
Let us extremize İ. We then require at a critical point
∂ İ
V
2LT11 x1 + LT12 + LT21 x2 = 0,
=
∂x1
R
∂ İ
V
LT12 + LT21 x1 + 2LT22 x2 = 0.
=
∂x2
R
(5.421)
(5.422)
Now by its method of construction, L must be symmetric; thus, LT12 = LT21 , and we can thus
state at an extreme point that
2V T
∂ İ
L11 x1 + LT12 x2 = 2j1 = 0,
=
∂x1
R
∂ İ
2V T
L21 x1 + LT22 x2 = 2j2 = 0.
=
∂x2
R
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.423)
(5.424)
241
5.4. IRREVERSIBILITY PRODUCTION
Thus, j1 = j2 = 0 at an extremum of İ, consistent with Eq. (P6.6).
Away from equilibrium, there is no simple way to formulate İ explicitly in terms of
affinity. One can however easily formulate İ globally in terms of reduced species variables ξk
as in Eq. (5.382). While the formulation is straightforward, its mathematical form is lengthy,
and will not be repeated here. Following the analysis after Eq. (5.382), one can form the
gradient and Hessian of İ, and note that at chemical equilibrium that İ = 0 and ∇İ = 0.
Moreover, from Eq. (5.393), we see that the Hessian of İ is positive definite, so that in fact
İ is a minimum at equilibrium.
For our Zel’dovich example problem, the Hessian matrix evaluated at chemical equilibrium is from Eq. (5.388):
∂ 2 İ ∂ξl ∂ξp ξk =ξke
=
6.81073 × 1020
−1.06440 × 1021
−1.06440 × 1021
1.46438 × 1023
.
(5.425)
The units of each of the entries of the Hessian are erg/K/s/mole2 . The eigenvalues of the
Hessian are 1.464 × 1023 and 6.733 × 1020 , as expected positive definite.
5.4.4
Extended Zel’dovich mechanism example
The example of the previous section induced a diagonal Onsager matrix because the matrix
ν was full rank. Let us extend the kinetics mechanism so as to induce a ν which is less
than full rank, and thus induce a full Onsager matrix. To do so, we can add N2 and O2
dissociation mechanisms to our kinetics laws. We take our extended Zel’dovich reaction
mechanism to be
N + NO
N + O2
N2 + N2
O2 + O2
⇌
⇌
⇌
⇌
N2 + O,
NO + O,
2N + N2 ,
2O + O2 .
(5.426)
(5.427)
(5.428)
(5.429)
For this system we have N = 5 species composed of L = 2 elements reacting in J = 4
reactions. We take the reaction constants for the first two reactions from Sec. 1.1.2; the N2
and O2 dissociation kinetic rates are those given earlier on p. 167, and in Sec. 1.1.1.2.1.1,
respectively.
Applying the extended Zel’dovich model for the same conditions simulated in Sec. 1.1.2,
we plot ρN O , ρN and ρO2 as a function of time in Fig. 5.4. We note that the results for NO
and N are similar to those for the ordinary Zel’dovich mechanism, see Fig. 1.8, at early time.
But at late time there is an additional equilibration with more complicated dynamics, with
the relative concentrations of NO and N switching places in importance.
We summarize some of the key vectors and matrices involved in the analysis. We adopt
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242
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
ρNO, ρN, ρO2 (mole/cm3)
2 ´ 10-6
1 ´ 10-6
5 ´ 10-7
N
2 ´ 10-7
NO
1 ´ 10-7
5 ´ 10-8
2 ´ 10-8
O2
10-8
10-6
10-4
10 -2
Figure 5.4: NO, N, and O2 concentrations versus time for T = 6000 K, extended Zel’dovich
mechanism.
Eq. (1.154) for our vector of N = 5 concentrations:

ρN O
 ρN 



ρ=
 ρN2  .
 ρO 
ρO2

(5.430)
We have L = 2 elements, with N, (l = 1) and O, (l = 2). We adopt Eq. (1.155) for the
stoichiometric matrix φ, of dimension L × N = 2 × 5,
φ=
1 1 2 0 0
1 0 0 1 2
.
(5.431)
The first row of φ is for the N atom; the second row is for the O atom. For the J = 4
reactions, we take
 
r1
 r2 

(5.432)
r=
 r3  ,
r4
where the rates are given by the standard law of mass action kinetics, not shown in detail
CC BY-NC-ND. 30 March 2014, J. M. Powers.
243
5.4. IRREVERSIBILITY PRODUCTION
here. The matrix ν, of dimension N × J

−1
 −1

ν=
 1
 1
0
= 5 × 4 is new for this problem and is

1
0
0
−1 2
0 

0 −1 0 
.
1
0
2 
−1 0 −1
(5.433)
The matrix ν has rank R = N − L = 3. It is not full rank in this case, in contrast to the
previous example. Our stoichiometric constraint on element conservation for each reaction,
Eq. (1.22), φ · ν = 0, holds here:


−1 1
0
0
 −1 −1 2
0 


0 0 0 0
1 1 2 0 0


.
(5.434)
0 −1 0  =
φ·ν =
· 1
0 0 0 0
1 0 0 1 2
 1
1
0
2 
0 −1 0 −1
With a similar analysis to that given in Sec.

 
 
b
ρN O
ρN O
1
 ρN   b


ρ

  N   01
 ρN  =  b
 
 2   ρN2  +  − 2
 ρO   b
ρO   −1
b
0
ρO2
ρO2
| {z
} | {z
} |
b
ρ
ρ
This takes the form of Eq. (1.37):
1.1.2, we find that

0
0


ξNO
1
0 

.

− 21 0 
 ξN
0 −2  ξ O2
| {z }
0
1
{z
} =ξ
(5.435)
D
b + D · ξ.
ρ=ρ
(5.436)
Here, the matrix D is of dimension N × R = 5 × 3. It spans the same column space as does
the N × J matrix ν which is of rank R. Note that φ · D = 0:


1
0
0
 0
1
0 
 1

0 0 0
1 1 2 0 0
1


.
(5.437)
φ·D=
·  −2 −2 0  =
0 0 0
1 0 0 1 2
 −1 0 −2 
0
0
1
Now let us move to the Onsager analysis. The matrix C is of dimension J ×(N −L) = 4×3
and has in its columns R = N − L = 3 linearly independent columns of ν T . We easily see
that ν T is


−1 −1 1 1 0
 1 −1 0 1 −1 
.
(5.438)
νT = 
 0
2 −1 0 0 
0
0
0 2 −1
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244
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
Now, among many choices, we can take the first, second, and fourth columns of ν T to form
C:


−1 −1 1
 1 −1 1 
.
(5.439)
C=
 0
2 0
0
0 2
Then, by Eq. (5.332), we find the J × J = 4 × 4 projection matrix B to be
−1
B = C · CT · C
· CT ,
 3

1
1
1
−
−
4
4
4
4
3
 −1
− 41 14 
4
4


=  1
3
1 .
− 4 − 41
4
4
1
4
1
4
1
4
(5.440)
(5.441)
3
4
The matrix B has rank N − L = 3 and is thus not full rank. It has eigenvalues of 1, 1, 1, 0.
The Onsager matrix L is given by
L
=
=
BT · R′ · B,

9R′1 + R′2 + R′3 + R′4
1  −3R′1 − 3R′2 + R′3 + R′4

16 −3R′1 + R′2 − 3R′3 + R′4
3R′1 − R′2 − R′3 + 3R′4
(5.442)
−3R′1 − 3R′2 + R′3 + R′4
R′1 + 9R′2 + R′3 + R′4
R′1 − 3R′2 − 3R′3 + R′4
−R′1 + 3R′2 − R′3 + 3R′4
−3R′1 + R′2 − 3R′3 + R′4
R′1 − 3R′2 − 3R′3 + R′4
R′1 + R′2 + 9R′3 + R′4
−R′1 − R′2 + 3R′3 + 3R′4

3R′1 − R′2 − R′3 + 3R′4
−R′1 + 3R′2 − R′3 + 3R′4 
.
−R′1 − R′2 + 3R′3 + 3R′4
R′1 − R′2 + R′3 + 9R′4
(5.443)
One can estimate the behavior of İ with both the full non-linear model and its Onsager
approximation. As expected, the Onsager approximation works well near equilibrium. Meaningful graphical display of the comparison of the two approximations is difficult due to numerical precision issues. As expected, the Hessian matrix of İ is symmetric and positive
definite at equilibrium with eigenvalues of 9.245 × 1022 , 1.50 × 1022 and 4.025 × 1017 .
For variety, let us work with a different Onsager matrix. Let us first get the reduced
affinity by Eq. (5.326):
−1
α̂ = CT · C
· CT · α,
(5.444)
 

 1
 
α1
1
α̂1
0 0
−2
2


 α̂2  =  − 1 − 1 3 1  ·  α2  ,
(5.445)
8
8
8
8
 α3 
1
1
1
3
α̂3
8
8
8
8
α4


− α21 + α22
α
=  − 81 − α82 + 3α8 3 + α84  .
(5.446)
3α4
α1
α2
α3
+
+
+
8
8
8
8
The (N − L) × (N − L) = 3 × 3 Onsager matrix L is given by Eq. (5.331):
L = CT · R′ · C,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(5.447)
245
5.4. IRREVERSIBILITY PRODUCTION

 
−1 −1 1
R′1 0
0
0
−1 1 0 0
 1 −1 1 
 0 R′2 0
0 
,



·
=
−1 −1 2 0  · 
 0
2 0
0 R′3 0   0
1
1 0 2
0
0 2
0
0
0 R′4

 ′
R1 + R′2
R′1 − R′2
−R′1 + R′2
′
′
′
′
′
−R′1 − R′2  .
=  R1 − R2 R1 + R2 + 4R3
−R′1 + R′2
−R′1 − R′2
R′1 + R′2 + 4R′4



Then, we can write Eq. (5.330) as
 ′
R1 + R′2
dS V
=
( α̂1 α̂2 α̂3 ) ·  R′1 − R′2
dt E,V
T
−R′1 + R′2
R′1 − R′2
R′1 + R′2 + 4R′3
−R′1 − R′2
(5.448)
(5.449)
  
α̂1
−R′1 + R′2
′
′


−R1 − R2
· α̂2  .
R′1 + R′2 + 4R′4
α̂3
(5.450)
In summary, we have spent considerable time exposing notions of production of irreversibility because it is associated with an ongoing debate on whether additional principles
beyond the traditional laws of thermodynamics reflect nature. One of the more common
contentions is given by Prigogine,11 who notes
“We shall see that stationary states may be characterized by an extremum
principle which states that in the stationary state, the entropy production has
its minimum value compatible with some auxiliary conditions to be specified in
each case.”
Notions such as these have spurred much activity in a variety of disciplines, including combustion, with the goal of using new principles to better codify observed behaviors. One must
note, however, that these notions have also attracted widespread skepticism. A thorough
review and discussion is given by Grandy,12 who notes on p. 170,
“...these observations do not in themselves seem to support the notion of a
general extremum principle for dissipation rates in steady-state processes, either
linear or nonlinear, near or far from equilibrium.”
Grandy concludes on p. 172 that
“After all is said and done, the issue with which we began this discussion—
the role of additional principles in non equilibrium thermodynamics—does not
appear to be much closer to a definite resolution. In those cases where we are
able to apply a principle of minimum (or maximum) rate of energy dissipation,
it seems that we already have sufficient tools in the conservation laws and constitutive equations. Although the conservation laws are exact, the constitutive
11
Prigogine, I., 1967, Introduction to Thermodynamics of Irreversible Processes, Interscience, New York,
p. 75.
12
Grandy, W. T., 2008, Entropy and the Time Evolution of Macroscopic Systems, Oxford, Oxford.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
246
CHAPTER 5. THERMOCHEMISTRY OF MULTIPLE REACTIONS
relations are of necessity approximated to some degree, so it may be that the
choice of which route to follow very much depends on the details of any specific
problem. Rather than focus on some general and improbable principle, it may be
more reasonable to search for general conditions or classes of problems for which
extremum rate principles might be both valid and productive.”
Consistent with Grandy, the analysis presented in these notes does not explicitly rely on
invocation of additional extremum principles beyond the ordinary laws of thermodynamics,
though one cannot ignore the possibility that such principles may be in play at a deeper level
in the construction of constitutive models. But at this stage, we are not prepared to accept
as proven that additional laws of nature based on extremizing entropy production rates have
been firmly established.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 6
Reactive Navier-Stokes equations
Double, double toil and trouble;
Fire burn, and caldron bubble.
William Shakespeare (1564-1616), Macbeth, Act IV, § i, ll. 10-11.
————————Up to this point we have exclusively focused on systems which are spatially homogeneous.
We now turn to important systems for which the state variables are functions of both time
and space. Here, we will focus on a general model for a mixture of gases which are allowed to
react, advect, and diffuse. As such, we will present the compressible reactive Navier1 -Stokes2
equations for an ideal mixture of N ideal gases. We will not give detailed derivations. We
are guided in part by the excellent derivations given by Aris3 and Merk.4
6.1
Evolution axioms
6.1.1
Conservative form
The conservation of mass, linear momenta, and energy, and the entropy inequality for the
mixture are expressed in conservative form as
∂ρ
+ ∇ · (ρu) = 0,
∂t
∂
(ρu) + ∇ · (ρuu + P I − τ ) = 0,
∂t
∂
1
1
q
ρ e+ u·u
+ ∇ · ρu e + u · u + j + (P I − τ ) · u
= 0,
∂t
2
2
1
(6.1)
(6.2)
(6.3)
Claude-Louis Navier, 1785-1836, French engineer.
George Gabriel Stokes, 1819-1903, Anglo-Irish mathematician.
3
R. Aris, 1962, Vectors, Tensors, and the Basic Equations of Fluid Mechanics, Dover, New York.
4
H. J. Merk, 1959, “The Macroscopic Equations for Simultaneous Heat and Mass Transfer in Isotropic
Continuous and Closed Systems,” Applied Scientific Research, 8(1): 73-99.
2
247
248
CHAPTER 6. REACTIVE NAVIER-STOKES EQUATIONS
jq
∂
≥ 0.
(ρs) + ∇ · ρsu +
∂t
T
(6.4)
New variables here are the velocity vector u, the viscous shear tensor τ , and the diffusive
heat flux vector jq . Note that these equations are precisely the same one would use for a
single fluid. We have neglected body forces.
The evolution of molecular species is dictated by an evolution axiom, an extension of
Eq. (5.23) to account for reaction, and now additionally, advection and diffusion:
∂
(ρYi ) + ∇ · (ρYi u + jm
i ) = Mi ω̇i ,
∂t
i = 1, . . . , N − 1.
(6.5)
Here, the diffusive mass flux vector of species i is jm
i . Note that together, Eqs. (6.1) and
N − 1 of Eq. (6.5) form N equations for the evolution of the N species. We insist that the
species diffusive mass flux be constrained by
N
X
jm
i = 0.
(6.6)
i=1
Recalling our earlier definitions of Mi and ω̇i , Eqs. (5.1) and (5.20), respectively, Eq. (6.5)
can be rewritten as
! J
L
X
X
∂
Ml φli
νij rj .
(6.7)
(ρYi ) + ∇ · (ρYi u + jm
i ) =
∂t
j=1
l=1
Let us sum Eq. (6.7) over all species to get

! J
N
N
N
L
X
X
X
X
X
∂
(ρYi ) +
∇ · (ρYi u + jm
Ml φli
νij rj ,
i ) =
∂t
i=1
i=1
i=1
j=1
l=1



L X
N
N
N
N X
J

 X

X
X
∂ 

 X 
m
Y
Y
j
ρ
+
∇
·
ρu
+
=
Ml φli νij rj ,


i
i
i 
∂t  i=1 
 i=1

i=1 l=1 j=1
| {z }
| {z } |i=1{z }
=1
=1
(6.8)
(6.9)
=0
=
J
X
j=1
∂ρ
+ ∇ · (ρu) = 0.
∂t
rj
L
X
l=1
Ml
N
X
i=1
φli νij ,
| {z }
(6.10)
=0
(6.11)
This is the extension of the spatially homogeneous Eq. (4.399). So, the summation over all
species gives a redundancy with Eq. (6.1).
CC BY-NC-ND. 30 March 2014, J. M. Powers.
249
6.1. EVOLUTION AXIOMS
We can get a similar relation for the elements. Let us multiply Eq. (6.5) by our stoichiometric matrix φli to get
∂
(ρYi ) + φli ∇ · (ρYi u + jm
i ) = φli Mi ω̇i ,
∂t
J
X
∂ ρφli Yi
ρφli Yi
jm
i
+∇·
= φli
νij rj .
u + φli
∂t
Mi
Mi
Mi
j=1
(6.12)
φli
(6.13)
Now sum over all species to get
X
N
N
N
J
X
X
X
∂ ρφli Yi
ρφli Yi
jm
i
+
∇·
=
φli
νij rj ,
u + φli
∂t
M
M
M
i
i
i
i=1
i=1
i=1
j=1
∂
∂t
∂
∂t
∂
Ml
∂t
∂
∂t
ρMl
ρ
N
X
φli Yi
Mi
i=1
!
+∇·
!
N
X
φli Yi
+∇·
Mi
i=1
!
N
X
φli Yi
ρ
+ Ml ∇ ·
Mi
i=1
N
X
i=1
ρ
φli Yi
Mi
!
+∇·
ρMl
ρ
N
X
φli Yi
Mi
i=1
ρ
ρ
N
X
i=1
N
X
i=1
N
X
i=1
u+
N
X
i=1
φli Yi
u+
Mi
φli Yi
u+
Mi
φli Yi
u + Ml
Mi
N
X
i=1
N
X
i=1
N
X
i=1
jm
φli i
Mi
!
jm
i
Mi
!
φli
jm
φli i
Mi
!
jm
i
Mi
!
φli
(6.14)
=
J
X
j=1
rj
N
X
i=1
= 0,
φli νij ,
| {z }
=0
(6.15)
(6.16)
= 0, l = 1, . . . , L,
(6.17)
= 0, l = 1, . . . , L,
(6.18)
Let us now define the element mass fraction Yle , l = 1, . . . , L, as
Yle
≡ Ml
N
X
φli Yi
i=1
Mi
.
(6.19)
Note that this can be expressed as
N
Yle
mass element l X moles element l mass species i mole species i
, (6.20)
=
species i | total{zmass } mass species i
{z
}l i=1 |moles {z
|moles element
}
{z
}
|
=Ml
mass element l
=
.
total mass
=φli
=Yi
=1/Mi
(6.21)
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250
CHAPTER 6. REACTIVE NAVIER-STOKES EQUATIONS
Similarly, we take the diffusive element mass flux to be
jel ≡ Ml
N
X
φli
i=1
jm
i
,
Mi
l = 1, . . . , L.
(6.22)
Substitute Eqs. (6.19,6.22) into Eq. (6.18) to get
∂
(ρYle ) + ∇ · (ρYle u + jel ) = 0.
∂t
(6.23)
We also insist that
L
X
jel = 0,
(6.24)
l=1
so as to keep the total mass of each element constant. This is easily seen to be guaranteed.
Sum Eq. (6.22) over all elements to get
L
X
jel
=
l=1
=
=
L
X
Ml
N
X
jm
i
,
Mi
(6.25)
jm
i
,
Mi
(6.26)
Ml φli ,
(6.27)
φli
i=1
l=1
L
N
XX
i=1 l=1
N
X
jm
i
i=1
Mi
Ml φli
L
X
|l=1 {z
=Mi
N
X
jm
i
Mi ,
=
Mi
i=1
=
N
X
jm
i ,
}
(6.28)
(6.29)
i=1
= 0.
(6.30)
In summary, we have L − 1 conservation equations for the elements, one global mass conservation equation, and N − L species evolution equations, in general. These add to form N
equations for the overall species evolution.
6.1.2
Non-conservative form
It is often convenient to have an alternative non-conservative form of the governing equations.
Let us define the material derivative as
∂
d
=
+ u · ∇.
(6.31)
dt
∂t
CC BY-NC-ND. 30 March 2014, J. M. Powers.
251
6.1. EVOLUTION AXIOMS
6.1.2.1
Mass
Using the product rule to expand the mass equation, Eq. (6.1), we get
∂ρ
+ ∇ · (ρu) = 0,
∂t
(6.32)
∂ρ
+ u · ∇ρ +ρ∇ · u = 0,
∂t
|
{z
}
(6.33)
=dρ/dt
dρ
+ ρ∇ · u = 0.
dt
6.1.2.2
(6.34)
Linear momentum
We again use the product rule to expand the linear momentum equation, Eq. (6.2):
∂ρ
∂u
+ u + u∇ · (ρu) + ρu · ∇u + ∇P − ∇ · τ = 0,
∂t ∂t
∂ρ
∂u
ρ
+ u · ∇u +u
+ ∇ · (ρu) +∇P − ∇ · τ = 0,
∂t
∂t
{z
}
{z
}
|
|
ρ
=du/dt
(6.35)
(6.36)
=0
ρ
du
+ ∇P − ∇ · τ = 0.
dt
(6.37)
The key simplification was effected by using the mass equation Eq. (6.1).
6.1.2.3
Energy
Now, use the product rule to expand the energy equation, Eq. (6.3).
1
1
1
∂ρ
∂
e + u · u + ρu · ∇ e + u · u + e + u · u
+ ∇ · (ρu)
ρ
∂t
2
2
2
∂t
{z
}
|
=0
+∇ · jq + ∇ · (P u) − ∇ · (τ · u) = 0,
∂
1
1
ρ
e + u · u + ρu · ∇ e + u · u + ∇ · jq + ∇ · (P u) − ∇ · (τ · u) = 0.
∂t
2
2
(6.38)
(6.39)
We have once again used the mass equation, Eq. (6.1), to simplify. Let us expand more using
the product rule:
∂u
∂e
+ u · ∇e +ρu ·
+ u · ∇u + ∇ · jq + ∇ · (P u) − ∇ · (τ · u) = 0, (6.40)
ρ
∂t
∂t
{z
}
|
=de/dt
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252
CHAPTER 6. REACTIVE NAVIER-STOKES EQUATIONS
∂u
de
+ u · ∇u + ∇P − ∇ · τ +∇ · jq + P ∇ · u − τ : ∇u = 0,
ρ +u· ρ
dt
∂t
|
{z
}
(6.41)
=0
ρ
de
+ ∇ · jq + P ∇ · u − τ : ∇u = 0.
dt
(6.42)
We have used the linear momentum equation, Eq. (6.37), to simplify.
From the mass equation, Eq. (6.34), we have ∇·u = −(1/ρ)dρ/dt, so the energy equation,
Eq. (6.42), can also be written as
ρ
de
P dρ
+ ∇ · jq −
− τ : ∇u = 0.
dt
ρ dt
(6.43)
This energy equation can be formulated in terms of enthalpy. Use the definition, Eq. (3.78),
h = e + P/ρ to get an expression for dh/dt:
P
1
dρ + dP,
2
ρ
ρ
de P dρ 1 dP
dh
=
−
+
,
dt
dt ρ2 dt ρ dt
dh 1 dP
de P dρ
− 2
=
−
,
dt ρ dt
dt
ρ dt
de P dρ
dh dP
ρ −
= ρ −
.
dt
ρ dt
dt
dt
dh = de −
(6.44)
(6.45)
(6.46)
(6.47)
So, the energy equation, Eq. (6.43), in terms of enthalpy is
ρ
6.1.2.4
dP
dh
+ ∇ · jq −
− τ : ∇u = 0.
dt
dt
(6.48)
Second law
Let us expand Eq. (6.4) to write the second law in non-conservative form.
∂
jq
≥ 0,
(ρs) + ∇ · ρsu +
∂t
T
q
∂s
∂ρ
j
ρ
≥ 0,
+ u · ∇s +s
+ ∇ · (ρu) +∇ ·
∂t
∂t
T
{z
}
{z
}
|
|
=ds/dt
=0
ds
ρ +∇·
dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
q
j
≥ 0.
T
(6.49)
(6.50)
(6.51)
253
6.2. MIXTURE RULES
6.1.2.5
Species
Let us expand Eq. (6.5) to write species evolution in non-conservative form.
∂
(ρYi ) + ∇ · (ρYi u + jm
i ) = Mi ω̇i ,
∂t
∂ρ
∂Yi
= Mi ω̇i ,
ρ
+ u · ∇Yi +Yi
+ ∇ · (ρu) +∇ · jm
i
∂t
∂t
{z
}
{z
}
|
|
=dYi /dt
i = 1, . . . , N − 1, (6.53)
=0
ρ
6.1.2.6
i = 1, . . . , N − 1, (6.52)
dYi
+ ∇ · jm
= Mi ω̇i ,
i
dt
i = 1, . . . , N − 1. (6.54)
Elements
Let us expand Eq. (6.23) to write element conservation in non-conservative form.
∂
(ρYle ) + ∇ · (ρYle u + jel ) = 0,
∂t
e
∂ρ
∂Yl
e
e
ρ
+ u · ∇Yl +Yl
+ ∇ · (ρu) +∇ · jel = 0,
∂t
∂t
{z
}
{z
}
|
|
=dYle /dt
(6.55)
l = 1, . . . , L,
(6.56)
l = 1, . . . , L.
(6.57)
=0
ρ
6.2
l = 1, . . . , L,
dYle
+ ∇ · jel = 0,
dt
Mixture rules
We adopt the following rules for the ideal mixture:
P =
1 =
ρ =
e =
h =
N
X
i=1
N
X
i=1
N
X
i=1
N
X
i=1
N
X
Pi ,
(6.58)
Yi ,
(6.59)
ρi =
N
X
ρi
,
M
i
i=1
(6.60)
Yi ei ,
(6.61)
Yi hi ,
(6.62)
i=1
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254
CHAPTER 6. REACTIVE NAVIER-STOKES EQUATIONS
N
X
s =
i=1
N
X
cv =
i=1
N
X
cP =
Y i si ,
(6.63)
Yi cvi ,
(6.64)
Y i cP i ,
(6.65)
i=1
V = Vi ,
T = Ti .
6.3
(6.66)
(6.67)
Constitutive models
The evolution axioms do not form a complete set of equations. Let us supplement these by
a set of constitutive model equations appropriate for a mixture of calorically perfect ideal
gases that react according the to the law of mass action with an Arrhenius kinetic reaction
rate. We have seen many of these models before, and repeat them here for completeness.
For the thermal equation of state, we take the ideal gas law for the partial pressures:
Pi = RT ρi = RT
ρYi
ρi
= RT
= Ri T ρi .
Mi
Mi
(6.68)
So, the mixture pressure is
P = RT
N
X
ρi = RT
i=1
N
X
ρYi
N
X
ρi
= RT
.
Mi
Mi
i=1
i=1
(6.69)
For the ideal gas, the enthalpy and internal energy of each component is a function of T at
most. We have for the enthalpy of a component
hi =
hoTo ,i
+
Z
T
cP i (T̂ )dT̂ .
(6.70)
To
So, the mixture enthalpy is
h =
N
X
i=1
Z
o
Yi hTo ,i +
T
cP i (T̂ )dT̂
To
.
(6.71)
We then use the definition of enthalpy to recover the internal energy of component i:
ei = hi −
Pi
= hi − Ri T.
ρi
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(6.72)
255
6.3. CONSTITUTIVE MODELS
So, the mixture internal energy is
Z
N
X
o
e =
Yi hTo ,i − Ri T +
=
=
=
=
i=1
N
X
i=1
N
X
i=1
N
X
i=1
N
X
Yi
Yi
Yi
Yi
i=1
hoTo ,i
cP i (T̂ )dT̂
To
− Ri (T − To ) − Ri To +
hoTo ,i
− Ri To +
hoTo ,i
eoTo ,i
T
− Ri To +
+
Z
Z
Z
(6.73)
T
cP i (T̂ )dT̂
To
T
To
Z
,
(cP i (T̂ ) − Ri )dT̂
T
cvi (T̂ )dT̂
To
T
cvi (T̂ )dT̂
To
,
,
(6.74)
(6.75)
,
(6.76)
.
(6.77)
The mixture entropy is
s =
N
X
i=1
Yi soTo ,i
+
Z
T
To
cP (T̂ )
T̂
dT̂ −
N
X
i=1
Yi Ri ln
Pi
Po
.
(6.78)
The viscous shear stress for an isotropic Newtonian fluid which satisfies Stokes’ assumption is
∇u + (∇u)T
1
τ = 2µ̂
− (∇ · u)I .
(6.79)
2
3
Here, µ̂ is the mixture viscosity coefficient which is determined from a suitable mixture rule
averaging over each component.
The energy flux vector jq is written as
N
N
X
X
Mi ∇P
DiT ∇yi
q
m
.
(6.80)
j = −k∇T +
ji hi − RT
+ 1−
M
y
M
P
i
i
i=1
i=1
Here, k is a suitably mixture averaged thermal conductivity. The parameter DiT is the
so-called thermal diffusion coefficient. Recall yi is the mole fraction.
We consider a mass diffusion vector with multicomponent diffusion coefficient Dik :
N
X
Mi Dik Yk ∇yk
Mk ∇P
∇T
m
− DiT
− 1−
.
(6.81)
ji = ρ
M
y
M
P
T
k
k=1, k6=i
We adopt, as before, the reaction rate of creation of species i, Eq. (5.20):
ω̇i =
J
X
νij rj .
(6.82)
j=1
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256
CHAPTER 6. REACTIVE NAVIER-STOKES EQUATIONS
Here, rj is given by the law of mass action, Eq. (5.16):
rj = k j
N
Y
ν′
ρkkj
!
N
1 Y νkj
1−
ρk . ,
Kc,j
(6.83)
k=1
k=1
The term kj is given by the Arrhenius kinetics rule, Eq. (5.19):
−E j
βj
kj = aj T exp
,
RT
(6.84)
and the equilibrium constant Kc,j for the ideal gas mixture is given by Eq. (5.7):
Kc,j =
6.4
Po
RT
PNi=1 νij
exp −
∆Goj
RT
.
(6.85)
Temperature evolution
Because temperature T has an important role in many discussions of combustion, let us formulate our energy conservation principle as a temperature evolution equation by employing
a variety of constitutive laws.
Let us begin with Eq. (6.48) coupled with our constitutive rule for h, Eq. (6.71):



Z T
N
d 
dP
X

o
ρ 
cP i (T̂ )dT̂  +∇ · jq −
Yi hTo ,i +
− τ : ∇u = 0,
dt  i=1
dt

To
|
{z
}
=hi
|
{z
}
(6.86)
=h
N
d X
ρ
Yi hi + ∇ · jq −
dt i=1
N X
dYi
dhi
+ ∇ · jq −
+ ρhi
ρYi
dt
dt
i=1
N X
dYi
dT
+ ∇ · jq −
+ ρhi
ρYi cP i
dt
dt
i=1
ρ
dP
− τ : ∇u = 0,
dt
(6.87)
dP
− τ : ∇u = 0,
dt
(6.88)
dP
− τ : ∇u = 0,
dt
(6.89)
N
N
X
dP
dT X
dYi
+∇ · jq −
hi ρ
Y i cP i +
− τ : ∇u = 0,
dt i=1
dt
dt
| {z }
| {z } i=1 =Mi ω̇i −∇·jm
(6.90)
i
=cP
N
dP
dT X
q
+
hi (Mi ω̇i − ∇ · jm
− τ : ∇u = 0.
ρcP
i )+∇·j −
dt
dt
i=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(6.91)
257
6.4. TEMPERATURE EVOLUTION
Now, let us define the “thermal diffusion” flux jT as
N
X
DiT ∇yi
Mi ∇P
T
j ≡ −RT
.
+ 1−
Mi
yi
M
P
i=1
(6.92)
With this, our total energy diffusion flux vector jq , Eq. (6.80), becomes
q
j = −k∇T +
N
X
T
jm
i hi + j .
(6.93)
i=1
Now, substitute Eq. (6.93) into Eq. (6.91) and rearrange to get
N
dT X
ρcP
+
hi (Mi ω̇i − ∇ · jm
i )+∇·
dt
i=1
−k∇T +
N
X
T
jm
i hi + j
i=1
!
dP
+ τ : ∇u,(6.94)
dt
N
dT X
dP
m
T
ρcP
+
(hi Mi ω̇i − hi ∇ · jm
+ τ : ∇u,(6.95)
i + ∇ · (ji hi )) = ∇ · (k∇T ) − ∇ · j +
dt
dt
i=1
=
N
dP
dT X
T
+
(hi Mi ω̇i + jm
+ τ : ∇u,(6.96)
ρcP
i · ∇hi )) = ∇ · (k∇T ) − ∇ · j +
dt
dt
i=1
N
ρcP
dP
dT X
T
+
(hi Mi ω̇i + cP i jm
+ τ : ∇u.(6.97)
i · ∇T ) = ∇ · (k∇T ) − ∇ · j +
dt
dt
i=1
So, the equation for the evolution of a material fluid particle is
N
X
dT
dP
T
ρcP
=−
(hi Mi ω̇i + cP i jm
+ τ : ∇u.
i · ∇T ) + ∇ · (k∇T ) − ∇ · j +
dt
dt
i=1
(6.98)
Let us impose some more details from the reaction law. First, we recall that hi = hi Mi ,
so
N
ρcP
N
X
X
dT
dP
T
=−
+ τ : ∇u.
hi ω̇i −
cP i jm
i · ∇T + ∇ · (k∇T ) − ∇ · j +
dt
dt
i=1
i=1
(6.99)
Impose now Eq. (6.82) to expand ω̇i
N
J
N
X X
X
dT
dP
T
=−
+ τ : ∇u,(6.100)
ρcP
hi
νij rj −
cP i jm
i · ∇T + ∇ · (k∇T ) − ∇ · j +
dt
dt
i=1
j=1
i=1
ρcP
N
J
N
X
X
X
dT
dP
T
hi νij −
cP i jm
=−
rj
+ τ : ∇u.(6.101)
i · ∇T + ∇ · (k∇T ) − ∇ · j +
dt
dt
i=1
j=1
i=1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
258
CHAPTER 6. REACTIVE NAVIER-STOKES EQUATIONS
Now, recall Eq. (5.34):
∆Hj ≡
N
X
hi νij .
(6.102)
i=1
With this, Eq. (6.101) becomes, after small rearrangement,
J
N
X
dP X
dT
T
=−
rj ∆Hj +
−
cP i jm
ρcP
i · ∇T + ∇ · (k∇T ) − ∇ · j + τ : ∇u . (6.103)
dt
dt
j=1
|i=1
{z
}
diffusion effects
By comparing Eq. (6.103) with Eq. (5.33), it is easy to see the effects of variable pressure and
diffusion on how temperature evolves. Interestingly, mass, momentum, and energy diffusion
all influence temperature evolution. The non-diffusive terms are combinations of advection,
reaction, and spatially homogeneous effects.
6.5
Shvab-Zel’dovich formulation
Under some restrictive assumptions, we can simplify the energy equation considerably. Let
us assume
• the low Mach5 number limit is applicable, which can be shown to imply that pressure changes and work work due to viscous dissipation are negligible at leading order,
dP/dt ∼ 0, τ : ∇u ∼ 0,
• the incompressible limit applies, dρ/dt = 0,
• thermal diffusion is negligible, DiT ∼ 0,
• all species have identical molecular masses, so that Mi = M,
• the multicomponent diffusion coefficients of all species are equal, Dij = D,
• all species possess identical specific heats, cP i = cP , which is itself a constant, and
• energy diffuses at the same rate as mass so that k/cP = ρD.
With the low Mach number limit, the energy equation, Eq. (6.48), reduces to
ρ
5
dh
+ ∇ · jq = 0.
dt
Ernst Mach, 1838-1916, Austrian physicist.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(6.104)
259
6.5. SHVAB-ZEL’DOVICH FORMULATION
With DiT = 0, the diffusive energy flux vector, Eq. (6.80), reduces to
q
j = −k∇T +
N
X
jm
i hi .
(6.105)
i=1
Substituting Eq. (6.105) into Eq. (6.104), we get
dh
ρ
+∇·
dt
−k∇T +
N
X
jm
i hi
i=1
!
= 0.
(6.106)
With all component specific heats equal and constant, cP i = cP , Eq. (6.71) for the mixture
enthalpy reduces to
h =
N
X
i=1
Yi hoTo ,i + cP (T − To ) ,
= cP (T − To ) +
N
X
(6.107)
Yi hoTo ,i .
(6.108)
i=1
Similarly for a component, Eq. (6.70) reduces to
hi = hoTo ,i + cP (T − To ).
(6.109)
With Eqs. (6.108,6.109), Eq. (6.106) transforms to
!
!
N
N
X
dT
d X
ρcP
jm
= 0,
Yi hoTo ,i − ∇ · k∇T −
hoTo ,i + cP (T − To )
+ρ
i
dt
dt i=1
i=1
dT
d
ρcP
+ρ
dt
dt
N
X
Yi hoTo ,i
i=1
!


(6.110)
N
N


X
X

m o
m
ji hTo ,i − cP (T − To )
ji  = 0,
− ∇ · k∇T −


i=1
i=1
| {z }
=0
ρcP
ρcP
d
dT
+ρ
dt
dt
d
dt
T+
N
X
Yi hoTo ,i
i=1
PN
Yi hoTo ,i
cP
i=1
!
!
−∇·
−∇·
k∇T −
N
X
k∇T −
N
X
o
jm
i hTo ,i
i=1
i=1
o
jm
i hTo ,i
!
!
(6.111)
= 0,
(6.112)
= 0,
(6.113)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
260
CHAPTER 6. REACTIVE NAVIER-STOKES EQUATIONS
With M = Mi , we recover mass fractions to be identical to mole fractions, Yk = yk .
Using this along with our assumptions of negligible thermal diffusion, DiT = 0 and equal
multicomponent diffusion coefficients, Dij = D, the mass diffusion flux vector, Eq. (6.81),
reduces to
jm
i
= ρ
N
X
k=1,k6=i
= ρD
D∇Yk ,
N
X
k=1,k6=i
= ρD∇
(6.114)
∇Yk ,
N
X
(6.115)
Yk ,
(6.116)
k=1,k6=i


N


X


Yk  ,
= ρD∇ −Yi +


|k=1
{z }
(6.117)
=1
= ρD∇ (−Yi + 1) ,
= −ρD∇Yi .
Now, substitute Eq. (6.119) into Eq. (6.113)
ρD = k/cP , to get
!
PN
o
d
i=1 Yi hTo ,i
T+
−∇·
ρcP
dt
cP
!
PN
o
Y
h
d
i
T
,i
o
ρcP
T + i=1
−∇·
dt
cP
!
PN
o
k
d
i=1 Yi hTo ,i
T+
−
∇·
dt
cP
ρcP
(6.118)
(6.119)
and use our equidiffusion rate assumption,
k∇T + ρD
N
X
i=1
N
X
∇Yi hoTo ,i
!
= 0,
!
k
k∇T +
∇Yi hoTo ,i
= 0,
cP i=1
!!
PN
o
i=1 Yi hTo ,i
= 0.
∇ T+
cP
(6.120)
(6.121)
(6.122)
Now, iff
P
o
• there is a spatially uniform distribution of the quantity T +(1/cP ) N
i=1 Yi hTo ,i at t = 0,
and
P
o
• there is no flux of T + (1/cP ) N
i=1 Yi hTo ,i at the boundary of the spatial domain for
all time,
then, Eq. (6.122) can be satisfied for all time by the algebraic relation
PN
PN
o
Yi (0)hoTo ,i
i=1 Yi hTo ,i
T+
= T (0) + i=1
,
cP
cP
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(6.123)
261
6.5. SHVAB-ZEL’DOVICH FORMULATION
where T (0) and Yi (0) are constants at the spatially uniform initial state. We can slightly
rearrange to write
cP (T − To ) +
|
{z
N
X
i=1
=h(T,Yi )
Yi hoTo ,i
}
= cP (T (0) − To ) +
|
{z
N
X
i=1
=h(T (0),Yi (0))
h(T, Yi ) = h(T (0), Yi (0)).
Yi (0)hoTo ,i ,
(6.124)
}
(6.125)
This simply says the enthalpy function is a constant, which can be evaluated at the initial
state.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
262
CHAPTER 6. REACTIVE NAVIER-STOKES EQUATIONS
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 7
Linear combustion:
In order to acquire a better understanding of the fundaments of combustion, a model problem is employed here to illustrate the basic ideas. Consider the following linear advectiondiffusion-reaction problem:
∂
∂
ψ(x, t) + u ψ(x, t)
∂t
∂x
ψ(x, 0)
ψ(0, t)
∂ψ
(∞, t)
∂x
∂2
ψ(x, t) − aψ(x, t),
∂x2
= ψu ,
= ψu ,
= D
→ 0,
(7.1a)
(7.1b)
(7.1c)
(7.1d)
where the independent variables are t > 0 and x ∈ (0, ∞). Here, ψ(x, t) is a general scalar,
u > 0 is a constant advective wave speed, D > 0 is a constant diffusion coefficient, and a > 0
is the chemical consumption rate constant.
The spatially homogeneous version of Eqs. (7.1) is
dψh (t)
= −aψh (t),
dt
ψh |t=0 = ψu ,
(7.2a)
which has the solution
ψh (t) = exp (−at) ψu .
(7.2b)
The time scale τ over which ψh evolves is
1
τ= .
a
(7.3)
This time scale serves as an upper bound for the required time step to capture the dynamics
in a numerical simulation. Since there is only one dependent variable in this problem, the
temporal spectrum contains only one time scale. Consequently, this formulation of the
system is not temporally stiff.
263
264
A simple means to determine the relevant length scales, and consequently, an upper
bound for the required spatial grid resolution, is to obtain the steady structure ψs (x), which
is governed by the time-independent version of Eqs. (7.1):
d2 ψs (x)
dψs dψs (x)
→ 0.
(7.4a)
=D
− aψs (x),
ψs |x=0 = ψu ,
u
dx
dx2
dx x→∞
Assuming solutions of the form ψs = Ceλx , we are led to a characteristic polynomial of
uλ = Dλ2 − a,
(7.4b)
which has roots
u
λ=
2D
1±
r
1+ 2
u
!
.
(7.4c)
Taking λ+ to denote the + root, for which λ+ > 0, and λ− to denote the minus root, for
which λ− < 0, the solution takes the form
ψs (x) = C1 eλ+ x + C2 eλ− x ,
(7.4d)
where C1 and C2 are constants. Thus, taking the spatial derivative, we get
dψs
= C1 λ+ eλ+ x + C2 λ− eλ− x ,
dx
(7.4e)
In the limit of large positive x, the boundary condition requires the derivative to vanish
giving
dψs
= 0 = lim C1 λ+ eλ+ x + C2 λ− eλ− x .
x→∞ dx
x→∞
lim
(7.4f)
Because λ+ > 0, we must insist that C1 = 0 in order to satisfy the far field boundary
condition. Then, enforcing that ψs (0) = ψu , we find the solution of Eq. (7.4a) is
ψs (x) = exp (λ− x) ψu ,
where
u
λ− =
2D
1−
r
1+ 2
u
(7.4g)
!
.
(7.4h)
Here, there is one length scale in the system, ℓ ≡ 1/|λ− |; this formulation of the system is
not spatially stiff. By examining Eq. (7.4h) in the limit aD/u2 ≫ 1, one finds that
r
D √
ℓ≈
= Dτ ,
(7.5)
a
CC BY-NC-ND. 30 March 2014, J. M. Powers.
265
where τ = 1/a is the time scale from spatially homogeneous reaction. So, this length scale
ℓ reflects the inherent physics of coupled reaction-advection-diffusion. In the opposite limit
of aD/u2 ≪ 1, one finds λ → 0, ℓ → ∞, and ψs (x) → ψu , a constant.
Now, for Eqs. (7.1), it is possible to find a simple analytic expression for the continuous spectrum of time scales τ associated with a particular linearly independent mode of
wavenumber k̂. A linearly independent mode with wavenumber k̂ has wavelength Λ = 2π/k̂.
Assume a solution of the form
ψ(x, t) = exp ik̂x Ψ(t),
(7.6)
where Ψ(t) is the time-dependent amplitude of the chosen linearly independent mode. Substituting this into Eq. (7.1a) gives
dΨ(t)
= −λΨ(t),
dt
Ψ(0) = Ψo ,
(7.7)
!
(7.8)
where
D k̂ 2 ik̂u
λ=a 1+
+
a
a
.
This has a solution of the form
Ψ(t) = exp (−λt) Ψo .
(7.9)
We see that the continuous time scale spectrum for amplitude growth or decay is given by
τ=
1
1
,
= D k̂ 2
|Re(λ)|
a 1+ a
0 < k̂ ∈ R.
(7.10)
From Eq. (7.10), it is clear that for D k̂ 2 /a ≪ 1, i.e. for sufficiently small wavenumbers
or long wavelengths, the time scales of amplitude growth or decay will be dominated by
reaction:
1
lim τ = .
(7.11a)
a
k̂→0
However, for D k̂ 2 /a ≫ 1, i.e. for sufficiently large wavenumbers or small wavelengths, the
amplitude growth/decay time scales are dominated by diffusion:
2
1 Λ
1
.
(7.11b)
=
lim τ =
D 2π
k̂→∞
D k̂ 2
p
From Eq. (7.10), we see that a balance between reaction and diffusion exists for k̂ = a/D.
In terms of wavelength, and recalling Eqs. (7.5), we see the balance at
r
1
D √
Λ
(7.12)
= =
= Dτ = ℓ,
2π
a
k̂
CC BY-NC-ND. 30 March 2014, J. M. Powers.
266
where ℓ = 1/k̂ is proportional to the wavelength.
The oscillatory behavior is of lesser importance. The continuous time scale spectrum for
oscillatory mode, τO is given by
τO =
1
1
.
=
|Im(λ)|
k̂u
(7.13)
As k̂ → 0, τO → ∞. While τO → 0 as k̂ → ∞, it approaches at a rate ∼ 1/k̂, in contrast to
the more demanding time scale of diffusion which approaches zero at a faster rate ∼ 1/k̂ 2 .
Thus, it is clear that advection does not play a role in determining the limiting values of the
time scale spectrum; reaction and diffusion are the major players. Lastly, it is easy to show
in the absence of diffusion, that the length scale where reaction effects balance advection
effects is found at
u
ℓ = = uτ,
(7.14)
a
where τ = 1/a is the time scale from spatially homogeneous chemistry.
As an illustration, we examine the behavior of the system quantitatively by choosing
arbitrary numerical values, loosely motivated by parameters of gas phase systems, of a =
108 1/s, D = 101 cm2 /s, u = 102 cm/s. For these values, we find the estimate from Eq. (7.5)
for the length scale where reaction balances diffusion as ℓ = Λ/(2π) = 3.16228 × 10−4 cm.
A plot of τ versus ℓ = Λ/(2π) from Eq. (7.10) is given in Fig. 7.1. For long wavelengths,
the time scales are determined by reaction; for fine wavelengths, the√time scale’s falloff is
dictated by diffusion, and our simple formula for the critical ℓ = Dτ , illustrated as a
dashed line, predicts the transition well. For large k̂, it is seen that a one decade decrease
in ℓ induces a two decade decrease in τ , consistent with the prediction of Eq. (7.11b):
limk̂→∞ (ln τ ) ∼ 2 ln (ℓ) − ln (D) . Lastly, over the same range of ℓ, the oscillatory time scales
induced by advection are orders of magnitude less demanding, and are thus not included in
the plot.
The results of this simple analysis can be summarized as follows:
• Long wavelength spatial disturbances have early time dynamics which are dominated
by chemistry; each spatial point behaves as an isolated spatially homogeneous reactor.
Late time dynamics are dictated by diffusion.
• Short wavelength spatial disturbances have early time dynamics which are dominated
by diffusion; diffusion first smears the disturbance and then reaction acts over late
time scales, such that at late time the reaction is spatially homogeneous over the entire
spatial domain.
• Intermediate wavelength spatial disturbances have time dynamics determined by fully
coupled combination diffusion and chemistry.
The critical intermediate length scale
√
where this balance exists is given by ℓ = Dτ .
CC BY-NC-ND. 30 March 2014, J. M. Powers.
267
−8
10
1/a = τ = 10−8 s
−9
τ (s)
10
2
−10
10
1
−11
ℓ=
10
√
Dτ = 3.2 × 10−4 cm
−12
10
−5
10
−3
10
−1
10
1
10
10
3
ℓ = Λ/(2π) (cm)
Figure 7.1: Time scale spectrum versus wavelength for the simple reaction-advectiondiffusion model.
• A fully resolved computation of a combustion process with advection-reaction-diffusion
requires
∆t < τ,
√
Dτ .
∆x <
(7.15)
(7.16)
Less restrictive choices will not capture time dynamics and spatial structures inherent
in the continuum model.
• Advection usually plays a secondary role in determining time dynamics.
We show analogous results for a detailed kinetics reaction-advection-diffusion combustion
system in Fig. 7.2. The analysis necessary to generate these results is detailed and given by
Al-Khateeb, and Powers, et al.1 Fig. 7.2 has an analog in Al-Khateeb and Powers, but was
analyzed in a slightly different fashion. This is inconsequential for our present discussion.
The results of Fig. 7.2 were obtained for an N = 9-species, L = 3-element, J = 19reaction model of hydrogen-air kinetics augmented by both advection and multi-component
mass and energy diffusion. To obtain the figure, an equilibrium state was subjected to linear
perturbation, and the corresponding eigenvalues and eigenfunctions were determined. The
time scales are the reciprocals of the real parts of the eigenvalues.
1
Al-Khateeb, A. N. and Powers, J. M., 2010, “Analysis of the Reaction-Advection-Diffusion Spectrum of
Laminar Premixed Flames,” 48th AIAA Aerospace Sciences Meeting, Orlando, Florida, AIAA-2010-0954.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
268
−4
10
finest
8.4 × 10−4 cm
−6
τf undamental (s)
τslowest = 1.85 × 10−4 s
=
10
−8
τf astest = 1.03 × 10−8 s
10
−10
1
10
2
−12
10
=
1.1 × 10−1 cm
coarsest
−14
10
10
−6
10
−4
10
−2
10
0
10
2
2L/π (cm)
Figure 7.2: Time scales associated with the fundamental modes for a nine-species, nineteen
step stoichiometric, p = 1 atm, To = 800 K hydrogen–air reaction-advection-diffusion system
versus the length 2L/π.
For long length scales, the time scales are effectively identical to those obtained by analysis
of a spatially homogeneous system. Thus we see clearly N − L = 6 independent time scales.
The fastest is τf astest = 1.03 × 10−8 s and the slowest is τslowest = 1.85 × 10−4 s. As the
length scale is decreased, one finds that diffusive effects first become important in describing
dynamics associated with the slowest reaction model. As the length scale continues to
decrease, diffusion becomes more and more important for each progressively faster reaction
mode, until the fastest reaction mode is reached. With the diffusion coefficient estimated
independently as Dmix = 68.91 cm2 /s, the coarsest and finest transition length scale, ℓcoarsest
and ℓf inest are found to be
p
ℓcoarsest =
D τ
= 1.1 × 10−1 cm,
(7.17)
p mix slowest
−4
ℓf inest =
Dmix τf astest = 8.4 × 10 cm.
(7.18)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 8
Simple solid combustion:
reaction-diffusion
Here, we will modify the simple thermal explosion theory of Sec. 1.2.1 which balanced unsteady evolution against reaction to include the effects of diffusion. Starting from a fully
unsteady formulation, we will focus on cases which are steady, resulting in a balance between
reaction and diffusion; however, we will briefly consider a balance between all three effects.
Such a theory is known after its founder as Frank-Kamenetskii1 theory. In particular, we
will look for transitions from a low temperature reaction to a high temperature reaction,
mainly in the context of steady state solutions, i.e. solutions with no time dependence. We
draw upon the work of Buckmaster and Ludford for guidance.2
8.1
Simple planar model
Let us consider a slab of solid fuel/oxidizer mixture. The material is modelled to be of infinite
extent in the y and z directions and has length 2L in the x direction. The temperature at
each end, x = ±L, is held fixed at T = To .
The slab is initially unreacted. Exothermic conversion of solid material from reactants to
products will generate an elevated temperature within the slab T > To , for x ∈ (−L, L). If
the thermal energy generated diffuses rapidly enough, the temperature within the slab will
be T ∼ To , and the reaction rate will be low. If the energy generated by local reaction diffuses
slowly, it will accumulate in the interior, accelerate the local reaction rate, and induce rapid
energy release and high temperature.
Let us assume
• The material is an immobile, incompressible solid with constant specific heat,
– Thus, ρ is constant,
1
David Albertovich Frank-Kamenetskii, 1910-1970, Soviet physicist.
Buckmaster, J. D., and Ludford, G. S. S., 1983, Lectures on Mathematical Combustion, SIAM, Philadelphia. See Chapter 1.
2
269
270
CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
– Thus, cP = cv is a constant,
– Thus, there is no advective transport of mass, momentum, or energy: u = 0.
• The material has variation only with x and t,
• The reaction can be modelled as
A → B,
(8.1)
where A and B have identical molecular masses.
• The reaction is irreversible,
• The reaction is exothermic,
• Initially only A is present,
• The thermal conductivity, k is constant.
Let us take, as we did in thermal explosion theory, Sec. 1.2.1
YA = 1 − λ,
YB = λ.
(8.2)
(8.3)
We interpret λ as a reaction progress variable which has λ ∈ [0, 1]. For λ = 0, the material
is all A; when λ = 1, the material is all B.
8.1.1
Model equations
Our simple model for reaction is
∂λ
= ae−E/R/T (1 − λ),
∂t
∂e
∂q
ρ
= − ,
∂t
∂x
∂T
q = −k ,
∂x
e = cv T − λq.
(8.4)
(8.5)
(8.6)
(8.7)
Equation (8.4) is our reaction kinetics law. It is the equivalent of the earlier derived
Eq. (1.281) in the irreversible limit, Kc → ∞. Equation (8.5) is our energy conservation expression. It amounts to Eq. (6.42), with q playing the role of the heat flux jq . Equation (8.6)
is the constitutive law for heat flux; it is the equivalent of Eq. (6.80) when mass diffusion
is neglected and amounts to Fourier’s3 law. Equation (8.7) is our caloric equation of state;
it is Eq. (1.314) with q = eoTo ,A − eoTo ,B and YB = λ. Equations (8.4-8.7) are completed by
initial and boundary conditions, which are
T (−L, t) = T (L, t) = To ,
3
T (x, 0) = To ,
Jean Baptiste Joseph Fourier, 1768-1830, French mathematician.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
λ(x, 0) = 0.
(8.8)
271
8.1. SIMPLE PLANAR MODEL
8.1.2
Simple planar derivation
Let us perform a simple “control volume” derivation of the planar energy equation, Eq. (8.7).
Consider a small volume of dimension A by ∆x. At the left boundary x, we have heat flux
q, in which we notate as qx . At the right boundary, x + ∆x, we have heat flux out of qx+∆x .
Recall the units of heat flux are J/m2 /s.
The first law of thermodynamics is
change in total energy = heat in − work
| {z out} .
(8.9)
total energy @ t + ∆t − total energy @ t = energy flux in − energy flux out .
{z
} |
{z
}
|
and
diffusion
| {z }
(8.10)
=0
There is no work for our system. But there is heat flux over system boundaries. In a
combination of symbols and words, we can say
=0
Mathematically, we can say
E|t+∆t − E|t = − (Ef lux out − Ef lux in ) ,
ρA∆x e|t+∆t − e|t = − (qx+∆x − qx ) |{z}
A∆t ,
| {z } |
{z
} 2
|
{z
}
kg
J/kg
ρ
J/m2 /s
e|t+∆t − e|t
qx+∆x − qx
= −
∆t
∆x
(m
(8.11)
(8.12)
s)
(8.13)
Now, let ∆x → 0 and ∆t → 0, and we get
ρ
∂e
∂qx
=−
.
∂t
∂x
(8.14)
Now, standard constitutive theory gives Fourier’s law to specify the heat flux:
q = −k
∂T
.
∂x
(8.15)
So, Eq. (8.15) along with the thermal state equation, (8.7) when substituted into Eq. (8.14)
yield
∂
∂2T
(cv T − λq) = k 2 ,
∂t
∂x
∂λ
∂2T
∂T
− ρq
= k 2,
ρcv
∂t
∂t
∂x
2
∂T
∂
T
− ρqae−E/R/T (1 − λ) = k 2 ,
ρcv
∂t
∂x
∂T
∂2T
ρcv
= k 2 + ρqae−E/R/T (1 − λ).
∂t
∂x
ρ
(8.16)
(8.17)
(8.18)
(8.19)
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
So, our complete system is two equations in two unknowns with appropriate initial and
boundary conditions:
∂T
∂2T
ρcv
= k 2 + ρqae−E/R/T (1 − λ),
∂t
∂x
∂λ
= ae−E/R/T (1 − λ),
∂t
T (x, 0) = To ,
λ(x, 0) = 0.
T (−L, t) = T (L, t) = To ,
8.1.3
(8.20)
(8.21)
(8.22)
Let us consider an ad hoc approximation to a system much like Eqs. (8.20-8.22), but which
has the advantage of being one equation and one unknown.
8.1.3.1
Planar formulation
If there were no diffusion, Eq. (8.14) would yield ∂e/∂t = 0 and would lead us to conclude
that e(x, t) = e(x). And because we have nothing now to introduce a spatial inhomogeneity,
there is no reason to take e(x) to be anything other than a constant eo . That would lead us
to
e(x, t) = eo ,
(8.23)
eo = cv T − λq.
(8.24)
so
Now, at t = 0, λ = 0, and T = To , so eo = cv To ; thus,
cv To = cv T − λq,
cv (T − To )
λ =
.
q
(8.25)
(8.26)
We also get the final temperature at λ = 1 to be
T (λ = 1) = To +
q
.
cv
(8.27)
We shall adopt Eq. (8.26) as our model for λ in place of Eq. (8.21). Had we admitted species
diffusion, we could more rigorously have arrived at a similar result, but it would be more
difficult to justify treating the material as a solid.
Let us use Eq. (8.26) to eliminate λ in Eq. (8.20) so as to get a single equation for T :
cv (T − To )
∂2T
∂T
−E/R/T
1−
,
(8.28)
= k 2 + ρqae
ρcv
∂t
∂x
q
|
{z
}
reaction source term
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273
8.2. NON-DIMENSIONALIZATION
The first two terms in Eq. (8.28) are nothing more than the classical heat equation. The
non-classical term is an algebraic source term due to chemical reaction. Note when T = To ,
the reaction source term is ρqa exp(−E/R/T ) > 0, so at the initial state there is a tendency
to increase the temperature. When λ = 1, so T = To + q/cv , the reaction source term is
zero, so there is no local heat release. This effectively accounts for reactant depletion.
Scaling the equation by ρcv and employing the well known formula for thermal diffusivity
α = k/ρ/cP ∼ k/ρ/cv for the solid, we get
∂T
∂2T
q
(8.29)
= α 2 +a
− (T − To ) e−E/R/T ,
∂t
∂x
cv
T (−L, t) = T (L, t) = To ,
T (x, 0) = To .
(8.30)
From here on, we shall consider Eqs. (8.30) and cylindrical and spherical variants to be the
full problem. We shall consider solutions to it in various limits. We will not return here to
the original problem without the ad hoc assumption, but that would be a straightforward
exercise.
8.1.3.2
More general coordinate systems
We note that Eq. (8.30) can be extended to general coordinate systems via
∂T
q
2
= α∇ T + a
− (T − To ) e−E/R/T .
∂t
cv
(8.31)
Appropriate initial and boundary conditions for the particular coordinate system would be
necessary.
For one-dimensional solutions in planar, cylindrical, and spherical coordinates, one can
summarize the formulations as
q
1 ∂
∂T
m ∂T
x
+a
= α m
− (T − To ) e−E/R/T .
(8.32)
∂t
x ∂x
∂x
cv
Here, we have m = 0 for a planar coordinate system, m = 1 for cylindrical, and m = 2 for
spherical. For cylindrical and spherical systems, the Dirichlet boundary condition at x = −L
would be replaced with a boundedness condition on T at x = 0.
8.2
Non-dimensionalization
Let us non-dimensionalize Eqs. (8.31). The scaling we will choose is not unique.
Let us define non-dimensional variables
T∗ =
cv (T − To )
,
q
x∗ =
x
,
L
t∗ = at.
(8.33)
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
With these choices, Eqs. (8.31) transform to
q 1 −m ∂
qa ∂T∗
m ∂T∗
x∗
=
α x
cv ∂t∗
cv L2 ∗ ∂x∗
∂x∗
q
1
q
E
+a
,
(8.34)
− T∗ exp −
cv cv
R To + (q/cv )T∗


α −m ∂
∂T∗
E
∂T∗
1
 . (8.35)
xm
+ (1 − T∗ ) exp −
=
x∗
∗
2
∂t∗
aL
∂x∗
∂x∗
RTo 1 + q T∗
cv To
Now, both sides are dimensionless. Let us define the dimensionless parameters
D=
aL2
,
α
Θ=
E
,
RTo
Q=
q
.
cv To
(8.36)
Here, D is a so-called Damköhler4 number. If we recall a diffusion time scale τd is
L2
,
τd =
α
(8.37)
and a first estimate (which will be shown to be crude) of the reaction time scale, τr is
1
τr = .
a
(8.38)
We see that the Damköhler number is the ratio of the thermal diffusion time to the reaction
time (ignoring activation energy effects!):
D=
thermal diffusion time
τd
L2 /α
=
=
.
1/a
τr
reaction time
(8.39)
We can think of Θ and Q as ratios as well with
Θ=
activation energy
,
ambient energy
Q=
exothermic heat release
.
ambient energy
(8.40)
The initial and boundary conditions scale to
T∗ (−1, t∗ ) = T∗ (1, t∗ ) = 0,
T∗ (x∗ , 0) = 0,
if
T∗ (1, t∗ ) = 0,
T∗ (0, t∗ ) < ∞,
T∗ (x∗ , 0) = 0,
4
Gerhard Damköhler, 1908-1944, German chemist.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
m = 0,
if
m = 1, 2.
(8.41)
(8.42)
275
8.2. NON-DIMENSIONALIZATION
8.2.1
Diffusion time discussion
As an aside, let us consider in more depth the thermal diffusion time. Consider the dimensional heat equation
∂2T
∂T
=α 2,
∂t
∂x
(8.43)
with a complex representation of a sinusoidal initial condition:
T (x, 0) = Aeik̂x .
(8.44)
Here, A is a temperature amplitude, k̂ is the wave number, and i2 = −1. If one wants, one
can insist we only worry about the real part of the solution, but that will not be important
for this analysis. Note that since
Aeik̂x = A cos(k̂x) + Ai sin(k̂x),
we can think of the initial condition as a signal with wavelength λ = 2π/k̂. Note when the
wave number k̂ is large, the wavelength λ is small, and vice versa.
Let us assume a separation of variables solution of the form
T (x, t) = B(t)eik̂x .
(8.45)
∂T
dB ik̂x
=
e ,
∂t
dt
(8.46)
Thus,
and
∂T
∂x
∂2T
∂x2
= ik̂Beik̂x ,
(8.47)
= −k̂ 2 Beik̂x .
(8.48)
So, our partial differential equation reduces to
dB
= −αk̂ 2 Beik̂x ,
dt
dB
= −αk̂ 2 B,
B(0) = A,
dt
2
B(t) = Ae−αk̂ t .
eik̂x
(8.49)
(8.50)
(8.51)
Note that as t → ∞, B → 0 with diffusion time constant
τd =
1
αk̂ 2
.
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
So, a large thermal diffusivity α causes modes to relax quickly. But high wave number k̂
also induces modes to relax quickly. Note also that in terms of wavelength of the initial
disturbance,
1 2
λ .
τd =
α4π 2
So, small wavelength disturbances induce fast relaxation.
Importantly, recall that an arbitrary signal will have a Fourier decomposition with an
infinite number of modes, each with a different wave number. The preceding analysis shows
that each mode will decay with its own diffusion time constant. Indeed some of the modes,
even at the initial state, may have negligibly small amplitude. But many will not, and we
cannot know a priori which ones will be large or small.
Lastly, we note that in a finite domain, we will find a discrete spectrum of wave numbers;
in an infinite domain, we will find a continuous spectrum.
8.2.2
Final form
Let us now drop the ∗ notation and understand that all variables are dimensionless. So, our
equations become
∂T
−Θ
1 −m ∂
m ∂T
x
+ (1 − T ) exp
,
=
x
∂t
D
∂x
∂x
1 + QT
T (−1, t) = T (1, t) = 0,
T (x, 0) = 0,
if
m = 0,
T (1, t) = 0,
T (0, t) < ∞,
T (x, 0) = 0,
if
m = 1, 2.
(8.52)
Note that the initial and boundary conditions are homogeneous. The only inhomogeneity
lives in the exothermic reaction source term, which is non-linear due to the exp(−1/T ) term.
Also, it will prove to be the case that a symmetry boundary condition at x = 0 suffices,
though our original formulation is more rigorous. Such equivalent boundary conditions are
T (1, t) = 0,
8.2.3
∂T
(0, t) = 0,
∂x
T (x, 0) = 0,
m = 1, 2, 3.
(8.53)
Integral form
As an aside, let us consider the evolution of total energy within the domain. To do so we
integrate a differential volume element through the entire volume. We recall dV ∼ xm dx,
for m = 0, 1, 2, (planar, cylindrical, spherical).
−Θ
1 m −m ∂
m ∂T
m ∂T
m
x
x
dx + x (1 − T ) exp
dx,
dx =
x x
∂t
D
∂x
∂x
1 + QT
(8.54)
Z 1
Z 1
Z 1
∂T
∂T
−Θ
1 ∂
xm
dx +
dx,
xm
xm (1 − T ) exp
dx =
∂t
∂x
1 + QT
0
0
0 D ∂x
CC BY-NC-ND. 30 March 2014, J. M. Powers.
277
(8.55)
Z 1
∂
1 ∂T −Θ
dx .
xm T dx
+
=
xm (1 − T ) exp
∂t 0
D ∂x x=1
1 + QT
0
|
{z
}
|
{z
}
| {z }
thermal energy change
internal conversion
boundary heat flux
(8.56)
Z
1
Note that the total thermal energy in our domain changes due to two reasons: 1) diffusive
energy flux at the isothermal boundary, 2) internal conversion of chemical energy to thermal
energy.
8.2.4
Infinite Damköhler limit
Note for D → ∞ diffusion becomes unimportant in Eq. (8.52), and we recover a balance
−Θ
dT
,
T (0) = 0.
(8.57)
= (1 − T ) exp
dt
1 + QT
This is the problem we have already considered in thermal explosion theory, Sec. 1.2.1. We
recall that thermal explosion theory predicts significant acceleration of reaction when
t→
8.3
eΘ
.
QΘ
(8.58)
Let us seek solutions to the planar (m = 0) version of Eqs. (8.52) that are formally steady,
so that ∂/∂t = 0, and a balance between reaction and energy diffusion is attained. In that
limit, Eqs.(8.52) reduce to the following two point boundary value problem:
1 d2 T
+ (1 − T ) exp
0 =
D dx2
0 = T (−1) = T (1).
−Θ
1 + QT
,
(8.59)
(8.60)
This problem is difficult to solve analytically because of the strong non-linearity in the
reaction source term.
8.3.1
High activation energy asymptotics
Motivated by our earlier success in getting approximate solutions to a similar problem in
spatially homogeneous thermal explosion theory, let us take a similar approach here.
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
Let us seek low temperature solutions where the non-linearity may be weak. So, let us
define a small parameter ǫ with 0 ≤ ǫ ≪ 1. And now let us assume a power series expansion
of T of the form
T = ǫT1 + ǫ2 T2 + . . .
(8.61)
Here, we assume T1 (x) ∼ O(1), T2 (x) ∼ O(1), . . .. We will focus attention on getting an
approximate solution for T1 (x).
With the assumption of Eq. (8.61), Eq. (8.59) expands to
ǫ
d2 T1
+ . . . = −D(1 − ǫT1 − . . .) exp(−Θ(1 − ǫQT1 + . . .)).
dx2
(8.62)
Ignoring higher order terms and simplifying we get
ǫ
d2 T1
= −D(1 − ǫT1 ) exp(−Θ) exp(ǫΘQT1 ).
dx2
(8.63)
Now, let us once again take the high activation energy limit and insist that ǫ be defined such
that
1
.
QΘ
(8.64)
D
d2 T1
=
−
(1 − ǫT1 ) exp(−Θ) exp(T1 ).
dx2
ǫ
(8.65)
ǫ≡
This gives us
We have gained an analytic advantage once again by moving the temperature into the numerator of the argument of the exponential.
Now, let us neglect ǫT1 as small relative to 1 and define a new parameter δ such that
δ≡
D
exp(−Θ) = DQΘ exp(−Θ).
ǫ
(8.66)
Our governing equation system then reduces to
d2 T1
= −δeT1 ,
dx2
T1 (−1) = T1 (1) = 0.
(8.67)
(8.68)
It is still not clear how to solve this. It seems reasonable to assume that T1 should have
symmetry about x = 0. If so, we might also presume that the gradient of T1 is zero at x = 0:
dT1 = 0,
(8.69)
dx x=0
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279
which induces T1 (0) to take on an extreme value, say T1 (0) = T1m , where m could denote
maximum or minimum. Certainly T1m is unknown at this point.
Let us explore the appropriate phase space solution. To aid in this, we can define q as
q≡−
dT1
.
dx
(8.70)
We included the minus sign so that q has a physical interpretation of heat flux. With this
assumption, Eq. (8.67) can be written as two autonomous ordinary differential equations in
two unknowns:
dq
= δeT1 ,
dx
dT1
= −q,
dx
q(0) = 0,
(8.71)
T1 (0) = T1m .
(8.72)
This is slightly unsatisfying because we do not know T1m . But we presume it exists, and is
a constant. Let us see if the analysis can reveal it by pressing forward.
Let us scale Eq. (8.72) by Eq. (8.71) to get
dT1
q
= − T1 ,
dq
δe
T1 |q=0 = T1m .
(8.73)
Now, Eq. (8.73) can be solved by separating variables. Doing so and solving we get
−δeT1 dT1 = qdq,
q2
+ C.
−δeT1 =
2
(8.74)
(8.75)
Applying the initial condition, we get
m
−δeT1 = C,
q2
m
= δ(eT1 − eT1 ),
2
q
m
2δ(eT1 − eT1 ).
q =
(8.76)
(8.77)
(8.78)
The plus square root is taken here. This will correspond to x ∈ [0, 1]. The negative square
root will correspond to x ∈ [−1, 0].
We can now substitute Eq. (8.78) into Eq. (8.72) to get
q
dT1
m
= − 2δ(eT1 − eT1 ).
(8.79)
dx
Once again separate variables to get
dT1
p
= −dx.
m
2δ(eT1 − eT1 )
(8.80)
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
Computer algebra reveals this can be integrated to form
q
√
T
2 tanh−1
1 − eT1m1
e
√
= −x + C.
−
m
eT1 /2 δ
(8.81)
Now, when x = 0, we have T1 = T1m . The inverse hyperbolic tangent has an argument of
zero there, and it evaluates to zero. Thus, we must have C = 0, so
q
√
T
2 tanh−1
1 − eT1m1
e
√
= x,
(8.82)
m /2
T
1
e
δ
!
r
r
T1
e
δ
m
1 − Tm
tanh−1
= eT1 /2
x.
(8.83)
e 1
2
Now, recall that
tanh−1
so Eq. (8.83) becomes
p
1 − β 2 = sech−1 β,
r
T1 −T1m
δ
m
x,
= eT1 /2
sech−1 e 2
2
e
T1 −T1m
2
e
T1
2
(8.84)
!
δ
x ,
= sech e
2
r !
δ
m
m
= eT1 /2 sech eT1 /2
x ,
2
r
T1m /2
T1m /2
T1 = 2 ln e
r
T1m /2
sech e
(8.85)
(8.86)
!!
δ
x
.
2
(8.87)
So, we have an exact solution for T1 (x). This is a nice achievement, but we are not sure it
satisfies the boundary condition at x = 1, nor do we know the value of T1m . We must choose
T1m such that T1 (1) = 0, which we have not yet enforced. So
r !!
δ
m
m
.
(8.88)
0 = 2 ln eT1 /2 sech eT1 /2
2
Only when the argument of a logarithm is unity does it map to zero. So, we must demand
that
r !
δ
m
m
eT1 /2 sech eT1 /2
= 1,
(8.89)
2
r
δ
m
T1m /2
(8.90)
= sech−1 e−T1 /2 ,
e
2
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281
m
T1
8
6
4
2
0.2
0.4
0.6
0.8
δ
Figure 8.1: Plot of maximum temperature perturbation T1m versus reaction rate constant δ
for steady reaction-diffusion in the high activation energy limit.
T1m /2
e
r
δ
m
= cosh−1 eT1 /2 ,
r2
δ
m
m
= e−T1 /2 cosh−1 eT1 /2 .
2
(8.91)
(8.92)
Equation (8.92) gives a direct relationship between δ and T1m . Given T1m , we get δ explicitly.
Thus, we can easily generate a plot; see Fig 8.1. The inverse cannot be achieved analytically,
but can be done numerically via iterative techniques. We notice a critical value of δ, δc =
0.878458. When δ = δc = 0.878458, we find T1m = 1.118684. For δ < δc , there are two
admissible values of T1m . That is to say the solution is non-unique for δ < δc . Moreover,
both solutions are physical. For δ = δc , there is a single solution. For δ > δc , there are no
Note also that since we have from Eq. (8.61) that T ∼ ǫT1 and from Eq. (8.64) that
ǫ = 1/Θ/Q, we can say
Tm ∼
T1m
.
ΘQ
(8.93)
Presumably re-introduction of neglected processes would aid in determining which solution is realized in nature. We shall see that transient stability analysis aids in selecting the
correct solution when there are choices. We shall also see that re-introduction of reaction
depletion into the model induces additional physical solutions, including those for δ > δc .
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
T1
3.0
2.5
2.0
1.5
1.0
0.5
-1.0
0.5
-0.5
1.0
x
Figure 8.2: Plot of T1 (x) for δ = 0.4 showing the two admissible steady solutions.
For δ = 0.4 < δc we get two solutions. Both are plotted in Fig 8.2. The high temperature
solution has T1m = 3.3079, and the low temperature solution has T1m = 0.24543. At this
point we are not sure if either solution is temporally stable to small perturbations. We shall
later prove that the low temperature solution is stable, and the higher temperature solution
is unstable. Certainly for δ > δc , there are no low temperature solutions. We will see that
upon re-introduction of missing physics, there are stable high temperature solutions. To
prevent high temperature solutions, we need δ < δc .
Now, recall from Eq. (8.66) that
δ = DQΘe−Θ .
(8.94)
Now, to prevent the high temperature solution, we demand
δ = DQΘe−Θ < 0.878458.
(8.95)
Let us bring back our dimensional parameters to examine this criteria:
DQΘe−Θ < 0.878458,
L2 a q
E
−E
< 0.878458.
exp
α cv To RTo
RTo
|{z}
|{z } |{z}
D
Q
Θ
Thus, the factors that tend to prevent thermal explosion are
• High thermal diffusivity; this removes the thermal energy rapidly,
• Small length scales; the energy thus has less distance to diffuse,
• Slow reaction rate kinetics,
• High activation energy.
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(8.96)
(8.97)
283
8.3.2
Method of weighted residuals
The method of the previous section was challenging. Let us approach the problem of calculating the temperature distribution with a powerful alternate method: the method of weighted
residuals, using so-called Dirac5 delta functions as weighting functions. We will first briefly
review the Dirac delta function, then move on to solving the Frank-Kamenetskii problem
with the method of weighted residuals.
The Dirac δD -distribution (or generalized function, or simply function), is defined by
Z β
0
if a 6∈ [α, β]
f (x)δD (x − a)dx =
(8.98)
f
(a)
if a ∈ [α, β]
α
From this it follows by considering the special case in which f (x) = 1 that
Z
∞
−∞
δD (x − a) = 0 if x 6= a
δD (x − a)dx = 1.
(8.99)
(8.100)
d2 T1
+ δeT1 = 0,
dx2
T1 (−1) = T1 (1) = 0.
(8.101)
(8.102)
We shall approximate T1 (x) by
T1 (x) = Ta (x) =
N
X
ci fi (x).
(8.103)
i=1
At this point, we do not know the so-called trial functions fi (x) nor the constants ci .
8.3.2.1
One-term collocation solution
Let us choose fi (x) to be linearly independent functions which satisfy the boundary conditions. Moreover, let us consider the simplest of approximations for which N = 1. A simple
function which satisfies the boundary conditions is a polynomial. The polynomial needs to
be at least quadratic to be non-trivial. So, let us take
f1 (x) = 1 − x2 .
(8.104)
Note this gives f1 (−1) = f1 (1) = 0. So, our one-term approximate solution takes the form
Ta (x) = c1 (1 − x2 ).
5
(8.105)
Paul Adrien Maurice Dirac, 1902-1984, English physicist.
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
We still do not yet have a value for c1 . Let us choose c1 so as to minimize a residual error
of our approximation. The residual error of our approximation r(x) will be
d2 Ta
r(x) =
+ δeTa (x) ,
2
dx
d2
c1 (1 − x2 ) + δ exp(c1 (1 − x2 )),
=
2
dx
= −2c1 + +δ exp(c1 (1 − x2 )).
(8.106)
(8.107)
(8.108)
If we could choose c1 in such a way that r(x) was exactly 0 for x ∈ [−1, 1], we would be done.
That will not happen. So, let us choose c1 to drive a weighted domain-averaged residual to
zero. That is, let us demand that
Z 1
ψ1 (x)r(x) dx = 0.
(8.109)
−1
We have introduced here a weighting function ψ1 (x). Many choices exist for the weighting
function. If we choose ψ1 (x) = f1 (x) our method is known as a Galerkin6 method. Let
us choose instead another common weighting, ψ1 (x) = δD (x). This method is known as a
collocation method. We have chosen a single collocation point at x = 0. With this choice,
Eq. (8.109) becomes
Z 1
δD (x) −2c1 + δ exp(c1 (1 − x2 )) dx = 0.
(8.110)
−1
The evaluation of this integral is particularly simple due to the choice of the Dirac weighting.
We simply evaluate the integrand at the collocation point x = 0 and get
−2c1 + +δ exp(c1 ) = 0.
(8.111)
δ = 2c1 e−c1 .
(8.112)
Thus,
Note here δ is a physical parameter with no relation the Dirac delta function δD . Note that
with our approximation Ta = c1 (1 − x2 ), that the maximum value of Ta is Tam = c1 . So, we
can say
m
δ = 2Tam e−Ta .
(8.113)
We can plot Tam as a function of δ; see Fig 8.3. Note the predictions of Fig. 8.3 are
remarkably similar to those of Fig. 8.1. For example, when δ = 0.4, numerical solution of
Eq. (8.113) yields two roots:
Tam = 0.259171,
6
Tam = 2.54264.
Boris Gigoryevich Galerkin, 1871-1945, Russian-Soviet engineer.
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(8.114)
285
m
Ta
8
6
4
2
0.1
0.2
0.3
0.4
0.5
0.6
0.7
δ
Figure 8.3: Plot of maximum temperature perturbation Tam versus reaction rate constant δ
for steady reaction-diffusion using a one-term collocation method.
Thus, we get explicit approximations for the high and low temperature solutions for a oneterm collocation approximation:
Ta (x) = 0.259171(1 − x2 ),
Ta (x) = 2.54264(1 − x2 ).
(8.115)
(8.116)
Plots of the one-term collocation approximation Ta (x) for high and low temperature solutions
are given in Fig 8.4.
We can easily find δc for this approximation. Note that differentiating Eq. (8.113) gives
dδ
m
= 2e−Ta (1 − Tam ).
m
dTa
(8.117)
A critical point exists when dδ/dTam = 0. We find this exists when Tam = 1. So, δc =
2(1)e−1 = 0.735769.
8.3.2.2
Two-term collocation solution
We can improve our accuracy by including more basis functions. Let us take N = 2. We
have a wide variety of acceptable choices for basis function that 1) satisfy the boundary
conditions, and 2) are linearly independent. Let us focus on polynomials. We can select the
first as before with f1 (x) = 1 − x2 as the lowest order non-trivial polynomial that satisfies
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
Ta
3.5
3.0
2.5
2.0
1.5
1.0
0.5
-1.0
-0.5
0.5
1.0
x
Figure 8.4: Plots of high and low temperature distributions Ta (x) using a one-term collocation method.
both boundary conditions. We could multiply this by an arbitrary constant, but that would
not be of any particular use. Leaving out details, if we selected the second basis function as
a cubic polynomial, we would find the coefficient on the cubic basis function to be c2 = 0.
That is a consequence of the oddness of the cubic basis function and the evenness of the
solution we are simulating. So, it turns out the lowest order non-trivial basis function is a
quartic polynomial, taken to be of the form
f2 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 .
(8.118)
We can insist that f2 (−1) = 0 and f2 (1) = 0. This gives
a0 + a1 + a2 + a3 + a4 = 0,
a0 − a1 + a2 − a3 + a4 = 0.
(8.119)
(8.120)
We solve these for a0 and a1 to get a0 = −a2 − a4 , a1 = −a3 . So, our approximation is
f2 (x) = (−a2 − a4 ) − a3 x + a2 x2 + a3 x3 + a4 x4 .
(8.121)
Motivated by the fact that a cubic approximation did not contribute to the solution, let us
select a3 = 0 so to get
f2 (x) = (−a2 − a4 ) + a2 x2 + a4 x4 .
(8.122)
R1
Let us now make the choice that −1 f1 (x)f2 (x) dx = 0. This guarantees an orthogonal
basis, although these basis functions are not eigenfunctions of any relevant self-adjoint linear
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287
operator for this problem. Often orthogonality of basis functions can lead to a more efficient
capturing of the solution. This results in a4 = −7a2 /8. Thus,
7 4
1
2
(8.123)
f2 (x) = a2 − + x − x .
8
8
Let us select a2 = −8 so that
f2 (x) = 1 − 8x2 + 7x4 = (1 − x2 )(1 − 7x2 ).
(8.124)
and
f1 (x) = 1 − x2 ,
f2 (x) = (1 − x2 )(1 − 7x2 ).
(8.125)
(8.126)
So, now we seek approximate solutions TA (x) of the form
TA (x) = c1 (1 − x2 ) + c2 (1 − x2 )(1 − 7x2 ).
(8.127)
This leads to a residual r(x) of
d2 TA
+ δeTA (x) ,
2
dx
d2
2
2
2
c
(1
−
x
)
+
c
(1
−
x
)(1
−
7x
)
=
1
2
dx2
+δ exp(c1 (1 − x2 ) + c2 (1 − x2 )(1 − 7x2 )),
= −2c1 + 56c2 x2 − 2c2 (1 − 7x2 ) − 14c2 (1 − x2 )
+δ exp(c1 (1 − x2 ) + c2 (1 − x2 )(1 − 7x2 )).
r(x) =
Now, we drive two weighted residuals to zero:
Z 1
ψ1 (x)r(x) dx = 0,
−1
Z 1
ψ2 (x)r(x) dx = 0.
(8.128)
(8.129)
(8.130)
(8.131)
(8.132)
−1
Let us once again choose the weighting functions ψi (x) to be Dirac delta functions so that we
have a two-term collocation method. Let us choose unevenly distributed collocation points
so as to generate independent equations taking x = 0 and x = 1/2. Symmetric choices would
lead to a linearly dependent set of equations. Other unevenly distributed choices would work
as well. So, we get
Z 1
δD (x)r(x) dx = 0,
(8.133)
−1
Z 1
δD (x − 1/2)r(x) dx = 0.
(8.134)
−1
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
or
r(0) = 0,
r(1/2) = 0.
(8.135)
(8.136)
−2c1 − 16c2 + δ exp (c1 + c2 ) = 0,
3
9
−2c1 + 5c2 + δ exp
c1 − c2
= 0.
4
16
(8.137)
Expanding, these equations are
(8.138)
For δ = 0.4, we find a high temperature solution via numerical methods:
c1 = 3.07054,
c2 = 0.683344,
(8.139)
c2 = 0.00149141.
(8.140)
and a low temperature solution as well
c1 = 0.243622,
So, the high temperature distribution is
TA (x) = 3.07054(1 − x2 ) + 0.683344(1 − x2 )(1 − 7x2 ).
(8.141)
The peak temperature of the high temperature distribution is TAm = 3.75388. This is an
improvement over the one term approximation of Tam = 2.54264 and closer to the high
temperature solution found via exact methods of T1m = 3.3079.
The low temperature distribution is
TA (x) = 0.243622(1 − x2 ) + 0.00149141(1 − x2 )(1 − 7x2 ).
(8.142)
The peak temperature of the low temperature distribution is TAm = 0.245113. This in
an improvement over the one term approximation of Tam = 0.259171 and compares very
favorably to the low temperature solution found via exact methods of T1m = 0.24543.
Plots of the two-term collocation approximation TA (x) for high and low temperature
solutions are given in Fig 8.5. While the low temperature solution is a very accurate representation, the high temperature solution exhibits a small negative portion near the boundary.
8.3.3
high activation energy limit, Eq.(8.60):
−Θ
1 d2 T
,
(8.143)
+ (1 − T ) exp
0 =
D dx2
1 + QT
0 = T (−1) = T (1).
(8.144)
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289
TA
4
3
2
1
-1.0
0.5
-0.5
1.0
x
Figure 8.5: Plots of high and low temperature distributions TA (x) using a two-term collocation method.
Rearrange to get
d2 T
dx2
= −D (1 − T ) exp
−Θ
1 + QT
,
QΘ exp(−Θ)
= −D
(1 − T ) exp
QΘ exp(−Θ)
(8.145)
−Θ
1 + QT
.
(8.146)
Now, simply adapting our earlier definition of δ = DQΘ exp(−Θ), which we note does not
imply we have taken any high activation energy limits, we get
exp(Θ)
−Θ
d2 T
,
(8.147)
= −δ
(1 − T ) exp
dx2
QΘ
1 + QT
T (−1) = T (1) = 0.
(8.148)
Equations (8.147-8.148) can be solved by a numerical trial and error method where we
demand that dT /dx(x = 0) = 0 and guess T (0). We keep guessing T (0) until we have
satisfied the boundary conditions.
When we do this with δ = 0.4, Θ = 15, Q = 1 (so D = δeΘ /Q/Θ = 87173.8 and
ǫ = 1/Θ/Q = 1/15), we find three steady solutions. One is at low temperature with T m =
0.016. This is essentially equal that predicted by our high activation energy limit estimate
of T m ∼ ǫT1m = 0.24542/15 = 0.016. We find a second intermediate temperature solution
with T m = 0.417. This is a little higher than that predicted in the high activation limit of
T m ∼ ǫT1m = 3.3079/15 = 0.221. And we find a high temperature solution with T m = 0.987.
There is no counterpart to this solution from the high activation energy limit analysis. Plots
of T (x) for high, low, and intermediate temperature solutions are given in Fig 8.6.
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
T
1.0
0.8
0.6
0.4
0.2
-1.0
0.5
-0.5
1.0
x
Figure 8.6: Plots of high, low, and intermediate temperature distributions T (x) for δ = 0.4,
Q = 1, Θ = 15.
We can use a one-term collocation approximation to estimate the relationship between δ
and T m . Let us estimate that
Ta (x) = c1 (1 − x2 ).
(8.149)
With that choice, we get a residual of
δ
Θ
r(x) = −2c1 +
(1 − c1 (1 − x2 )).
exp Θ −
QΘ
1 + c1 Q(1 − x2 )
We choose a one term collocation method with ψ1 (x) = δD (x). Then, setting
0 gives
δ
Θ
r(0) = −2c1 +
(1 − c1 ) = 0.
exp Θ −
QΘ
1 + c1 Q
(8.150)
R1
−1
ψ1 (x)r(x)dx =
(8.151)
We solve for δ and get
2c1 QΘ
δ=
exp
1 − c1 eΘ
Θ
1 + c1 Q
.
(8.152)
The maximum temperature of the approximation is given by Tam = c1 and occurs at x = 0.
A plot of Tam versus δ is given in Fig 8.7. For δ < δc1 ∼ 0.2, one low temperature solution
exists. For δc1 < δ < δc2 ∼ 0.84, three solutions exist. For δ > δc2 , one high temperature
solution exists.
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291
m
Ta
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
1.2
1.4
δ
Figure 8.7: Plots of Tam versus δ, with Q = 1, Θ = 15 from a one-term collocation approximate solution.
8.4
Let us know study the effects of time-dependency on our combustion problem. Let us
consider the planar, m = 0, version of Eqs. (8.52):
∂T
1 ∂2T
−Θ
,
=
+ (1 − T ) exp
∂t
D ∂x2
1 + QT
T (−1, t) = T (1, t) = 0,
T (x, 0) = 0.
(8.153)
8.4.1
Linear stability
We will first consider small deviations from the steady solutions found earlier and see if those
deviations grow or decay with time. This will allow us to make a definitive statement about
the linear stability of those steady solutions.
8.4.1.1
Formulation
First, recall that we have independently determined three exact numerical steady solutions
to the time-independent version of Eq. (8.153). Let us call any of these Te (x). Note that by
construction Te (x) satisfies the boundary conditions on T .
Let us subject a steady solution to a small perturbation and consider that to be our
initial condition for an unsteady calculation. Take then
T (x, 0) = Te (x) + ǫA(x),
A(−1) = A(1) = 0,
A(x) = O(1),
0 < ǫ ≪ 1.
(8.154)
(8.155)
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
Here, A(x) is some function which satisfies the same boundary conditions as T (x, t).
Now, let us assume that
T (x, t) = Te (x) + ǫT ′ (x, t).
(8.156)
T ′ (x, 0) = A(x).
(8.157)
with
Here, T ′ is an O(1) quantity. We then substitute our Eq. (8.156) into Eq. (8.153) to get
1 ∂2
∂
(Te (x) + ǫT ′ (x, t)) =
(Te (x) + ǫT ′ (x, t))
2
∂t
D ∂x
′
+ (1 − Te (x) − ǫT (x, t)) exp
−Θ
.
1 + QTe (x) + QǫT ′ (x, t)
(8.158)
From here on we will understand that Te is Te (x) and T ′ is T ′ (x, t). Now, consider the
exponential term:
!
−Θ
1
−Θ
exp
,
(8.159)
= exp
Q
1 + QTe + QǫT ′
1 + QTe 1 + 1+QT
ǫT ′
e
Q
−Θ
′
,
(8.160)
1−
ǫT
∼ exp
1 + QTe
1 + QTe
−Θ
ǫΘQ
′
∼ exp
exp
(8.161)
T ,
1 + QTe
(1 + QTe )2
ǫΘQ
−Θ
′
1+
(8.162)
T .
∼ exp
1 + QTe
(1 + QTe )2
So, our Eq. (8.158) can be rewritten as
∂
1 ∂2
(Te + ǫT ′ ) =
(Te + ǫT ′ )
∂t
D ∂x2
′
+ (1 − Te − ǫT ) exp
−Θ
1 + QTe
ǫΘQ
′
1+
T ,
(1 + QTe )2
1 ∂2
(Te + ǫT ′ )
2
D ∂x ǫΘQ
−Θ
′
′
(1 − Te − ǫT ) 1 +
T ,
+ exp
1 + QTe
(1 + QTe )2
1 ∂2
=
(Te + ǫT ′ )
D ∂x2
(1 − Te )ΘQ
−Θ
′
2
(1 − Te ) + ǫT −1 +
+ O(ǫ )
+ exp
1 + QTe
(1 + QTe )2
=
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293
1 ∂ 2 Te
−Θ
ǫ ∂2T ′
(1 − Te )
+
+ exp
=
D ∂x2
D ∂x2
1 + QTe
{z
}
|
=0
(1 − Te )ΘQ
−Θ
′
2
ǫT −1 +
+ exp
+ O(ǫ ) .
1 + QTe
(1 + QTe )2
(8.163)
Now, we recognize the bracketed term as zero because Te (x) is constructed to satisfy the
steady state equation. We also recognize that ∂Te (x)/∂t = 0. So, our equation reduces to,
neglecting O(ǫ2 ) terms, and canceling ǫ
∂T ′
(1 − Te )ΘQ
1 ∂2T ′
−Θ
−1 +
=
+ exp
T ′.
(8.164)
2
2
∂t
D ∂x
1 + QTe
(1 + QTe )
Equation (8.164) is a linear partial differential equation for T ′ (x, t). It is of the form
1 ∂2T ′
∂T ′
=
+ B(x)T ′ ,
∂t
D ∂x2
with
B(x) ≡ exp
8.4.1.2
−Θ
1 + QTe (x)
(1 − Te (x))ΘQ
−1 +
.
(1 + QTe (x))2
(8.165)
(8.166)
Separation of variables
Let us use the standard technique of separation of variables to solve Eq. (8.165). We first
assume that
T ′ (x, t) = H(x)K(t).
(8.167)
So, Eq. (8.165) becomes
1
d2 H(x)
dK(t)
= =
K(t)
+ B(x)H(x)K(t),
dt
D
dx2
1 dK(t)
1 1 d2 H(x)
=
+ B(x) = −λ.
K(t) dt
D H(x) dx2
H(x)
(8.168)
(8.169)
Since the left side is a function of t and the right side is a function of x, the only way the
two can be equal is if they are both the same constant. We will call that constant −λ.
Now, Eq. (8.169) really contains two equations, the first of which is
dK(t)
+ λK(t) = 0.
dt
(8.170)
K(t) = C exp(−λt),
(8.171)
This has solution
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
where C is some arbitrary constant. Clearly if λ > 0, this solution is stable, with time
constant of relaxation τ = 1/λ.
The second differential equation contained within Eq. (8.169) is
1 d2 H(x)
+ B(x)H(x) = −λH(x),
D dx2
1 d2
+ B(x) H(x) = −λH(x).
D dx2
(8.172)
(8.173)
This is of the classical eigenvalue form for a linear operator L; that is L(H(x)) = −λH(x).
We also must have
H(−1) = H(1) = 0,
(8.174)
to satisfy the spatially homogeneous boundary conditions on T ′ (x, t).
This eigenvalue problem is difficult to solve because of the complicated nature of B(x).
Let us see how the solution would proceed in the limiting case of B as a constant. We will
generalize later.
If B is a constant, we have
d2 H
+ (B + λ)DH = 0,
dx2
H(−1) = H(1) = 0.
(8.175)
The following mapping simplifies the problem somewhat:
y=
x+1
.
2
(8.176)
This takes our domain of x ∈ [−1, 1] to y ∈ [0, 1]. By the chain rule
dH
dH dy
1 dH
=
=
.
dx
dy dx
2 dy
So
1 d2 H
d2 H
=
.
dx2
4 dy 2
So, our eigenvalue problem transforms to
d2 H
+ 4D(B + λ)H = 0,
dy 2
H(0) = H(1) = 0.
(8.177)
This has solution
p
p
H(y) = C1 cos ( 4D(B + λ) )y + C2 sin ( 4D(B + λ)) y .
(8.178)
H(0) = 0 = C1 (1) + C2 (0),
(8.179)
At y = 0 we have then
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295
so C1 = 0. Thus,
p
H(y) = C2 sin ( 4D(B + λ)) y .
At y = 1, we have the other boundary condition:
p
H(1) = 0 = C2 sin ( 4D(B + λ)) .
Since C2 6= 0 to avoid a trivial solution, we must require that
p
sin ( 4D(B + λ)) = 0.
(8.180)
(8.181)
(8.182)
For this to occur, the argument of the sin function must be an integer multiple of π:
p
4D(B + λ) = nπ,
n = 1, 2, 3, . . .
(8.183)
Thus,
n2 π 2
λ=
− B.
4D
(8.184)
We need λ > 0 for stability. For large n and D > 0, we have stability. Depending on the
value of B, low n, which corresponds to low frequency modes, could be unstable.
8.4.1.3
Numerical eigenvalue solution
Let us return to the full problem where B = B(x). Let us solve the eigenvalue problem via
the method of finite differences. Let us take our domain x ∈ [−1, 1] and discretize into N
points with
∆x =
2
,
N −1
xi = (i − 1)∆x − 1.
(8.185)
Note that when i = 1, xi = −1, and when i = N, xi = 1. Let us define B(xi ) = Bi and
H(xi ) = Hi .
We can rewrite Eq. (8.172) as
d2 H(x)
+ D(B(x) + λ)H(x) = 0,
dx2
H(−1) = H(1) = 0.
(8.186)
Now, let us apply an appropriate equation at each node. At i = 1, we must satisfy the
boundary condition so
H1 = 0.
(8.187)
At i = 2, we discretize Eq. (8.186) with a second order central difference to obtain
H1 − 2H2 + H3
+ D(B2 + λ)H2 = 0.
∆x2
(8.188)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
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CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
We get a similar equation at a general interior node i:
Hi−1 − 2Hi + Hi+1
+ D(Bi + λ)Hi = 0.
∆x2
(8.189)
At the i = N − 1 node, we have
HN −2 − 2HN −1 + HN
+ D(BN −1 + λ)HN −1 = 0.
∆x2
(8.190)
At the i = N node, we have the boundary condition
HN = 0.
These represent a linear tri-diagonal system

2
1
(− D∆x
0
2 + B2 )
D∆x2
1
2
1

(− D∆x2 + B3 ) D∆x
2
D∆x2

1

0
.
.
.
D∆x2


...
...
...
0
0
...
{z
|
=L
(8.191)
of equations of the form




H2
H2
0 ... 0
H 
 H3 
0 ... 0 
 . 
  .3 
.
. 



. . . . . . . . .   .  = −λ 
 ..  (8.192)
.
 .. 
. . . . . . . . .   .. 
.
..
..
... ... ...
.
} | {z }
=h
This is of the classical linear algebraic eigenvalue form L · h = −λh. All one need do is
discretize and find the eigenvalues of the matrix L. These will be good approximations to the
eigenvalues of the differential operator L. The eigenvectors of L will be good approximations
of the eigenfunctions of L. To get a better approximation, one need only reduce ∆x.
Note because the matrix L is symmetric, the eigenvalues are guaranteed real, and the
eigenvectors are guaranteed orthogonal. This is actually a consequence of the original problem being in Sturm-Liouville form, which is guaranteed to be self-adjoint with real eigenvalues
and orthogonal eigenfunctions.
8.4.1.3.1 Low temperature transients For our case of δ = 0.4, Q = 1, Θ = 15 (so
D = 87173.8), we can calculate the stability of the low temperature solution. Choosing
N = 101 points to discretize the domain, we find a set of eigenvalues. They are all positive,
so the solution is stable. The first few are
λ = 0.0000232705, 0.000108289, 0.000249682, 0.000447414, . . . .
(8.193)
The first few eigenvalues can be approximated by inert theory with B(x) = 0, see Eq. (8.184):
λ∼
n2 π 2
= 0.0000283044, 0.000113218, 0.00025474, 0.00045287, . . . .
4D
(8.194)
The first eigenvalue is associated with the longest time scale τ = 1/0.0000232705 =
42972.9 and a low frequency mode, whose shape is given by the associated eigenvector,
plotted in Fig. 8.8. This represents the fundamental mode. Shown also in Fig. 8.8 are the
first and second harmonic modes.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
297
0.14
0.12
0.10
0.10
0.05
0.05
0.10
0.08
0.06
-1.0
0.5
-0.5
0.04
1.0
x
-1.0
0.5
-0.5
-0.05
-0.05
-0.10
-0.10
1.0
x
0.02
-1.0
0.5
-0.5
1.0
x
Figure 8.8: Plots of fundamental, first, and second harmonic modes of eigenfunctions versus
x, with δ = 0.4, Q = 1, Θ = 15, low temperature steady solution Te (x).
8.4.1.3.2 Intermediate temperature transients For the intermediate temperature
solution with T m = 0.417, we find the first few eigenvalues to be
λ = −0.0000383311, 0.0000668221, 0.000209943, . . .
(8.195)
Except for the first, all the eigenvalues are positive. The first eigenvalue of λ = −0.0000383311
is associated with an unstable fundamental mode. All the harmonic modes are stable. We
plot the first three modes in Fig. 8.9.
0.14
0.12
0.10
0.10
0.10
0.05
0.05
0.08
-1.0
0.06
0.04
0.02
-1.0
-0.5
0.5
1.0
0.5
-0.5
1.0
x
-1.0
0.5
-0.5
-0.05
-0.05
-0.10
-0.10
1.0
x
Figure 8.9: Plot of fundamental, first, and second harmonic modes of eigenfunctions versus
x, with δ = 0.4, Q = 1, Θ = 15, intermediate temperature steady solution Te (x).
8.4.1.3.3 High temperature transients For the high temperature solution with T m =
0.987, we find the first few eigenvalues to be
λ = 0.000146419, 0.00014954, 0.000517724, . . .
(8.196)
All the eigenvalues are positive, so all modes are stable. We plot the first three modes in
Fig. 8.10..
8.4.2
Full transient solution
We can get a full transient solution to Eqs. (8.153) with numerical methods. We omit details
of such numerical methods, which can be found in standard texts.
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x
298
CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
-1.0
0.5
-0.5
1.0
0.15
x
0.15
0.10
0.10
-0.05
0.05
-1.0
0.05
0.5
-0.5
1.0
x
-0.10
-1.0
-0.05
0.5
-0.5
1.0
-0.05
-0.10
-0.15
-0.15
-0.10
Figure 8.10: Plot of fundamental, first, and second harmonic modes of eigenfunctions versus
x, with δ = 0.4, Q = 1, Θ = 15, high temperature steady solution Te (x).
300 000
t
T
200 000
100 000
0.015
0
0.015
T
0.010
0.010
0.005
0.000
-1.0
0.005
-0.5
0.0
x
0.5
1.0
50 000
100 000
150 000
200 000
250 000
300 000
t
Figure 8.11: Plot of T (x, t) and plot of T (0, t) along with the long time exact low temperature
centerline solution, Te (0), with δ = 0.4, Q = 1, Θ = 15.
8.4.2.1
Low temperature solution
For our case of δ = 0.4, Q = 1, Θ = 15 (so D = 87173.8), we show a plot of the full transient
solution in Fig. 8.11. Also seen in Fig. 8.11 is that the centerline temperature T (0, t) relaxes
to the long time value predicted by the low temperature steady solution:
lim T (0, t) = 0.016.
t→∞
8.4.2.2
(8.197)
High temperature solution
We next select a value of δ = 1.2 > δc . This should induce transition to a high temperature
solution. We maintain Θ = 15, Q = 1. We get D = δeΘ /Θ/Q = 261521. The full transient
solution is shown in Fig. 8.12 Also shown in Fig. 8.12 is the centerline temperature T (0, t).
We see it relaxes to the long time value predicted by the high temperature steady solution:
lim T (0, t) = 0.9999185.
t→∞
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(8.198)
x
299
T
1.5 106
t
6
1. 10
1.0
5. 105
1.0
0.8
0
0.6
T
0.5
0.4
0.0
0.2
-1. 0
-0. 5
0.0
x
0.5
1.0
4 ´ 105
8 ´ 105
12´ 105
t
Figure 8.12: Plot of T (x, t) and plot of T (0, t) along with the long time exact low temperature
centerline solution, Te (0), with δ = 1.2, Q = 1, Θ = 15.
It is clearly seen that there is a rapid acceleration of the reaction for t ∼ 106 . This compares
with the prediction of the induction time from the infinite Damköhler number, D → ∞,
thermal explosion theory of explosion to occur when
t→
e15
eΘ
=
= 2.17934 × 105 .
QΘ
(1)(15)
(8.199)
The estimate under-predicts the value by a factor of five. This is likely due to 1) cooling
of the domain due to the low temperature boundaries at x = ±1, and 2) effects of finite
activation energy.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
300
CHAPTER 8. SIMPLE SOLID COMBUSTION: REACTION-DIFFUSION
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 9
Laminar flames:
There is no more open door by which you can enter into the study of natural
philosophy than by considering the physical phenomena of a candle.
Michael Faraday (1791-1867), The Chemical History of a Candle, Griffin and
Bohn, London, 1861.
————————Here, we will consider premixed one-dimensional steady laminar flames. Background
is available in many sources.1 23 This topic is broad, but we will restrict attention to the
simplest cases. We will consider a simple reversible kinetics model
A ⇌ B,
(9.1)
where A and B have identical molecular masses, MA = MB = M, and are both calorically
perfect ideal gases with the same specific heats, cP A = cP B = cP . Because we are modelling
the system as premixed, we consider species A to be composed of molecules which have their
own fuel and oxidizer. This is not common in hydrocarbon kinetics, but is more so in the
realm of explosives. More common in gas phase kinetics are situations in which cold fuel and
cold oxidizer react together to form hot products. In such a situation, one can also consider
non-premixed flames in which streams of fuel first must mix with streams of oxidizer before
significant reaction can commence. Such flames will not be considered in this chapter.
We shall see that the introduction of advection introduces some unusual mathematical
difficulties in properly modelling cold unreacting flow. This will be overcome in a way
which is æsthetically unappealing, but nevertheless useful: we shall introduce an ignition
temperature Tig in our reaction kinetics law to suppress all reaction for T < Tig . This
will serve to render the cold boundary to be a true mathematical equilibrium point of the
1
Buckmaster, J. D., and Ludford, G. S. S., 2008, Theory of Laminar Flames, Cambridge, Cambridge.
Williams, F. A., 1985, Combustion Theory, Benjamin-Cummings, Menlo Park, California.
3
Law, C. K., 2006, Combustion Physics, Cambridge, Cambridge.
2
301
302
model; this is not a traditional chemical equilibrium, but it is an equilibrium in the formal
mathematical sense nonetheless.
9.1
9.1.1
Governing Equations
Evolution equations
We will first consider conservation and evolution equations which are independent of any
specific material.
9.1.1.1
Conservative form
Let us commence with the one-dimensional versions of Eqs. (6.1-6.3) and (6.5):
∂ρ
∂
+
(ρu)
∂t ∂x
∂
∂
ρu2 + P − τ
(ρu) +
∂t
∂x
∂
1 2
1 2
∂
q
ρ e+ u
ρu e + u + j + (P − τ )u
+
∂t
2
∂x
2
∂
∂
(ρYB ) +
(ρuYB + jBm )
∂t
∂x
9.1.1.2
= 0,
(9.2)
= 0,
(9.3)
= 0,
(9.4)
= M ω̇B .
(9.5)
Non-conservative form
Using standard reductions similar to those made to achieve Eqs. (6.34),(6.37),(6.42), (6.54),
the non-conservative form of the governing equations, Eqs. (9.2-9.5) is
∂ρ
∂ρ
∂u
+u
+ρ
∂t
∂x
∂x
∂u ∂P
∂τ
∂u
+ ρu
+
−
ρ
∂t
∂x
∂x
∂x
∂e
∂e ∂j q
∂u
∂u
ρ + ρu
+
+P
−τ
∂t
∂x
∂x
∂x
∂x
∂YB ∂jBm
∂YB
+ ρu
+
ρ
∂t
∂x
∂x
= 0,
(9.6)
= 0,
(9.7)
= 0,
(9.8)
= M ω̇B .
(9.9)
Equations (9.6-9.9) can be written more compactly using the material derivative, d/dt =
∂/∂t + u∂/∂x:
∂u
dρ
+ρ
= 0,
dt
∂x
du ∂P
∂τ
ρ +
−
= 0,
dt
∂x
∂x
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(9.10)
(9.11)
303
9.1. GOVERNING EQUATIONS
ρ
9.1.1.3
∂u
∂u
de ∂j q
+
+P
−τ
= 0,
dt
∂x
∂x
∂x
dYB ∂jBm
+
= M ω̇B .
ρ
dt
∂x
(9.12)
(9.13)
Formulation using enthalpy
It will be more useful to formulate the equations using enthalpy. Use the definition of
enthalpy Eq. (3.78), h = e + P v = e + P/ρ, to get an expression for dh:
P
1
dρ + dP,
2
ρ
ρ
de P dρ 1 dP
dh
=
−
+
,
dt
dt ρ2 dt ρ dt
de P dρ
dh 1 dP
− 2
=
−
,
dt ρ dt
dt
ρ dt
dh dP
de P dρ
= ρ −
.
ρ −
dt
ρ dt
dt
dt
dh = de −
(9.14)
(9.15)
(9.16)
(9.17)
Now, using the mass equation, Eq. (9.10) to eliminate P ∂u/∂x in the energy equation,
Eq. (9.12), the energy equation can be rewritten as
ρ
de ∂j q
P dρ
∂u
+
−
−τ
= 0,
dt
∂x
ρ dt
∂x
(9.18)
Next use Eq. (9.17) to simplify Eq. (9.18):
ρ
9.1.1.4
dP
∂u
dh ∂j q
+
−
−τ
= 0.
dt
∂x
dt
∂x
(9.19)
Low Mach number limit
In the limit of low Mach number, one can do a formal asymptotic expansion with the reciprocal of the Mach number squared as a perturbation parameter. All variables take the form
ψ = ψo + M2 ψ1 + . . ., where ψ is a general variable, and M is the Mach number, not to be
confused with the molecular mass M. In this limit, the linear momentum equation can be
shown to reduce at leading order to ∂Po /∂x = 0, giving rise to Po = Po (t).
We shall ultimately be concerned only with time-independent flows where P = Po .
We adopt the constant pressure assumption now. Also in the low Mach number limit, it
can be shown that viscous work is negligible, so τ ∂u/∂x ∼ 0. Our evolution equations,
Eqs. (9.10),(9.19),(9.13), then in the low Mach number, constant pressure limit are
∂ρ
∂ρ
∂u
+u
+ρ
= 0,
∂t
∂x
∂x
(9.20)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
304
∂h ∂j q
∂h
+ ρu
+
= 0,
∂t
∂x
∂x
∂YB ∂jBm
∂YB
+ ρu
+
= M ω̇B .
ρ
∂t
∂x
∂x
(9.21)
ρ
(9.22)
Note that we have effectively removed the momentum equation. It would re-appear in a
non-trivial way were we to formulate the equations at the next order.
Equations (9.20-9.22) take on the conservative form
∂ρ
∂
+
(ρu) = 0,
∂t ∂x
∂
∂
(ρh) +
(ρuh + j q ) = 0,
∂t
∂x
∂
∂
(ρYB ) +
(ρuYB + jBm ) = M ω̇B .
∂t
∂x
9.1.2
(9.23)
(9.24)
(9.25)
Constitutive models
Equations (9.20-9.22) are supplemented by simple constitutive models:
Po = ρRT,
h = cP (T − To ) − YB q,
∂YB
,
jBm = −ρD
∂x
∂YB
∂T
+ ρDq
,
j q = −k
∂x
∂x
β −E/(RT )
ω̇B = aT e
Kc = e
.
(9.28)
(9.29)


ρ
1

(1 − YB ) 1 −
Kc
|M {z
}
=ρA
q/(RT )
(9.26)
(9.27)
YB
1−Y
| {z B
=ρB /ρA



 H(T − Tig ),

}
(9.30)
(9.31)
Thus, we have nine equations for the nine unknowns, ρ, u, h, T , j q , YB , jBm , ω̇B , Kc .
Many of these are obvious. Some are not. First, we note the new factor H(T − Tig ) in
our kinetics law, Eq. (9.30). Here, H is a Heaviside unit step function. For T < Tig it takes
a value of zero. For T ≥ Tig , it takes a value of unity:
0, T < Tig ,
H(T − Tig ) =
(9.32)
1, T ≥ Tig .
Next, let us see how to get Eq. (9.28) from the more general Eq. (6.81). We first take
the thermal diffusion coefficient to be zero, DiT = 0. We also note since MA = MB = M
CC BY-NC-ND. 30 March 2014, J. M. Powers.
305
9.1. GOVERNING EQUATIONS
that yk = Yk ; that is mole fractions are the same as mass fractions. So, Eq. (6.81) simplifies
considerably to
jim
=ρ
N
X
Dik
k=1,k6=i
∂Yk
.
∂x
(9.33)
Next we take Dik = D and write for each of the two diffusive mass fluxes that
∂YB
,
∂x
∂YA
.
= ρD
∂x
jAm = ρD
(9.34)
jBm
(9.35)
Since YA + YB = 1, we have also
∂YA
,
∂x
∂YB
.
jBm = −ρD
∂x
Note that jAm + jBm = 0, as required by Eq. (6.6).
Under the same assumptions, Eq. (6.80) reduces to
jAm = −ρD
∂T
+ jAm hA + jBm hB ,
∂x
∂T
+ jAm (hA,To + cP (T − To )) + jBm (hB,To + cP (T − To )),
= −k
∂x
∂T
+ jAm hA,To + jBm hB,To + (jAm + jBm ) cP (T − To ),
= −k
| {z }
∂x
j q = −k
(9.36)
(9.37)
(9.38)
(9.39)
(9.40)
=0
∂T
− jBm hA,To + jBm hB,To ,
= −k
∂x
∂T
= −k
− jBm (hA,To − hB,To ),
|
{z
}
∂x
(9.41)
(9.42)
=q
∂T
− jBm q,
j q = −k
∂x
∂T
∂YB
j q = −k
+ ρDq
.
∂x
∂x
Here, we have defined a heat release per unit mass, q, as
q = hA,To − hB,To .
(9.43)
(9.44)
(9.45)
For the equilibrium constant Kc , we specialize Eq. (4.270), recalling νA + νB = 1 − 1 = 0 for
our simple reaction kinetics, and get
−∆Go
,
(9.46)
Kc = exp
RT
CC BY-NC-ND. 30 March 2014, J. M. Powers.
306
=
=
=
=
=
−(g oB − g oA )
exp
,
RT
o
gA − gBo
,
exp
RT
o
hA − T soA − (hoB − T soB )
,
exp
RT
o
o
hA − hoB
sB − soA
exp
exp
,
RT
R
o
o
sB,To − soA,To
hA,To − hoB,To
exp
.
exp
RT
R
(9.47)
(9.48)
(9.49)
(9.50)
(9.51)
Here, because of constant specific heats for A and B, many terms have canceled. Let us take
now soB,To = soA,To and our definition of the heat release, Eq. (9.45), to get
Kc = exp
9.1.3
q .
RT
(9.52)
Alternate forms
Further analysis of the evolution equations combined with the constitutive equations can
yield forms which give physical insight.
9.1.3.1
Species equation
If we combine our species evolution equation, Eq. (9.22) with the constitutive law, Eq. (9.28),
we get
∂YB
∂YB
∂YB
∂
ρ
ρD
= M ω̇B .
(9.53)
+ ρu
−
∂t
∂x
∂x
∂x
Note that for flows with no advection (u = 0), constant density (ρ = constant), and no
reaction (ω̇B = 0), this reduces to the classical “heat” equation from mathematical physics
∂YB /∂t = D(∂ 2 YB /∂x2 ).
9.1.3.2
Energy equation
Consider the energy equation, Eq. (9.21), using Eqs. (9.27,9.29) to eliminate h and j q :
∂
∂
∂
ρ (cP (T − To ) − YB q) +ρu (cP (T − To ) − YB q) +
{z
}
{z
}
|
|
∂t
∂x
∂x
=h
CC BY-NC-ND. 30 March 2014, J. M. Powers.
=h
∂T
∂YB
= 0,
−k
+ ρDq
∂x
∂x
|
{z
}
=j q
(9.54)
307
9.1. GOVERNING EQUATIONS
q
q
∂
∂
k ∂T
∂YB q
T − YB
+ ρu
T − YB
−
= 0,
− ρD
cP
∂x
cP
∂x cP ∂x
∂x cP
(9.55)
2
∂T
∂YB
∂YB
∂T
k ∂ T
q
∂YB
∂
ρ
= 0,
ρ
ρD
+ ρu
−
−
+ ρu
−
2
∂t
∂x
cP ∂x
cP
∂t
∂x
∂x
∂x
{z
}
|
∂
ρ
∂t
=M ω̇B
(9.56)
We notice that the terms involving YB simplify via Eq. (9.53) to yield an equation for
temperature evolution
ρ
∂T
∂T
k ∂2T
q
+ ρu
−
−
M ω̇B = 0,
∂t
∂x
cP ∂x2
cP
∂T
k ∂2T
q
∂T
+u
=
+
M ω̇B .
2
∂t
∂x
ρcP ∂x
ρcP
|{z}
(9.58)
k
.
ρcP
(9.59)
(9.57)
=α
We note, see p. 273, that thermal diffusivity α is
α=
We note here that this definition, convenient and in the combustion literature, is slightly
non-traditional as it involves a variable property ρ. So, the reduced energy equation is
∂T
∂T
+u
∂t
∂x
∂2T
q
= α 2 +
M ω̇B .
∂x
ρcP
(9.60)
Equation (9.60) is closely related to the more general Eq. (6.103). Note that for exothermic
reaction, q > 0 accompanied with production of product B, ω̇B > 0, induces a temperature
rise of a material particle. The temperature change is modulated by energy diffusion. In the
inert zero advection limit, we recover the ordinary heat equation ∂T /∂t = α(∂ 2 T /∂x2 ).
9.1.3.3
Shvab-Zel’dovich form
Let us now adopt the assumption that mass and energy diffuse at the same rate. This is
not difficult to believe as both are molecular collision phenomena in gas flames. Such an
assumption will allow us to write the energy equation in the Shvab-Zel’dovich form such
as studied in a more general sense in Sec. 6.5. We note the dimensionless ratio of energy
diffusivity α to mass diffusivity D is known as the Lewis4 number, Le:
Le =
4
α
.
D
(9.61)
Warren Kendall Lewis, 1882-1975, American chemical engineer.
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308
If we insist that mass and energy diffuse at the same rate, we have Le = 1, which gives
D=α=
k
.
ρcP
With this assumption, the energy equation, Eq. (9.55), takes the form
q
∂
q
∂
k ∂T
k ∂YB q
∂
T − YB
+ ρu
T − YB
−
= 0,
−
ρ
∂t
cP
∂x
cP
∂x cP ∂x
cP ∂x cP
∂
q
∂
q
k ∂
q
∂
ρ
T − YB
+ ρu
T − YB
−
T − YB
= 0,
∂t
cP
∂x
cP
cP ∂x ∂x
cP
∂
q
∂
q
k ∂
q
∂
T − YB
+u
T − YB
−
T − YB
= 0.
∂t
cP
∂x
cP
ρcP ∂x ∂x
cP
We rewrite in terms of the material derivative
q
k ∂2
q
d
T − YB
=
.
T − YB
dt
cP
ρcP ∂x2
cP
(9.62)
(9.63)
(9.64)
(9.65)
(9.66)
Equation (9.66) holds that for a material fluid particle, the quantity T − YB q/cP changes
only in response to local spatial gradients. Now, if we consider an initial value problem in
which T and YB are initially spatially uniform and there are no gradients of either T or
YB at x → ±∞, then there will no tendency for any material particle to have its value of
T − YB q/cP change. Let us assume that at t = 0, we have T = To and no product, so YB = 0.
Then, we can conclude that the following relation holds for all space and time:
q
= To .
(9.67)
T − YB
cP
Solving for YB , we get
YB =
cP (T − To )
.
q
(9.68)
For this chapter, we are mainly interested in steady waves. We can imagine that our steady
waves are the long time limit of a situation just described which was initially spatially
uniform. Compare Eq. (9.68) to our related ad hoc assumption for reactive solids, Eq. (8.26).
They are essentially equivalent, especially when one recalls that λ plays the same role as YB
and for the reactive solid cP ∼ cv .
We can use Eq. (9.68) to eliminate YB in the species equation, Eq. (9.53) to get a single
equation for temperature evolution. First adopt the equal diffusion assumption, Eq. (9.62),
in Eq. (9.53):
k ∂YB
∂YB
∂
∂YB
= M ω̇B .
(9.69)
+ ρu
−
ρ
∂t
∂x
∂x cP ∂x
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309
9.1. GOVERNING EQUATIONS
Next use Eq. (9.68) to eliminate YB :
∂ cP (T − To )
∂ cP (T − To )
∂
k ∂ cP (T − To )
ρ
+ ρu
−
= M ω̇B .
∂t
q
∂x
q
∂x cP ∂x
q
(9.70)
Simplifying, we get
ρcP
∂T
∂T
+u
∂t
∂x
∂2T
−k 2
∂x
= qM ω̇B .
(9.71)
Now, let us use Eq. (9.30) to eliminate ω̇B :
∂T
∂2T
∂T
ρcP
−k 2
+u
∂t
∂x
∂x
β −E/(RT )
= ρqaT e
1 YB
H(T − Tig ).
(1 − YB ) 1 −
Kc 1 − Y B
(9.72)
Now, use Eq. (9.52) to eliminate Kc and Eq. (9.68) to eliminate YB so to get
∂2T
∂T
∂T
−k 2
+u
ρcP
∂t
∂x
∂x
!
cP (T −To )
c
(T
−
T
)
P
o
q
= ρqaT β e−E/(RT ) 1 −
1 − e−q/(RT )
H(T − Tig ). (9.73)
cP (T −To )
q
1−
q
Equation (9.73) is remarkably similar to our earlier Eq. (8.28) for temperature evolution in
a heat conducting reactive solid. The differences are as follows:
• We use cP instead of cv ,
• We have accounted for finite β,
• We have accounted for reversible reaction.
• We have imposed and ignition temperature.
9.1.4
Equilibrium conditions
We can gain further insights by examining when Eq. (9.73) is in equilibrium. There is one
potential equilibria, the state of chemical equilibrium where Kc = YB /(1 − YB ) = YB /YA .
This occurs when
−q/(RT )
1−e
1
cP (T −To )
q
cP (T −To )
−
q
= 0.
(9.74)
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310
Parameter
Value
Fundamental Dimensional
To
300
cP
1000
R
287
q
1.5 × 106
E
1.722 × 105
αo
10−5
a
105
uo
1.4142
Po
105
Tig
600
Secondary Dimensional
ρo
1.1614
k
0.0116114
Fundamental Dimensionless
γ
1.4025
Q
5
Θ
2
D
2
TIG
0.2
Units
K
J/kg/K
J/kg/K
J/kg
J/kg
m2 /s
1/s
m/s
Pa
K
kg/m3
J/m/s
Table 9.1: Numerical values of parameters for simple laminar flame calculation.
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311
This is a transcendental equation for T . Let us examine how the equilibrium T varies for
some sample parameters. For this and later calculations, we give a set of parameters in
Table 9.1. Let us consider how the equilibrium temperature varies as q is varied from near
zero to its maximum value of Table 9.1. The chemical equilibrium flame temperature and
product mass fraction are plotted as functions of q in Figure 9.1. For q = 0, the equilibrium
1.0
2000
0.8
T
eq
YB
eq
(K)
1500
1000
0.6
0.4
500
0.2
0
750 000
0
1 500 000
750 000
q (J/kg)
1 500 000
q (J/kg)
Figure 9.1: Variation of chemical equilibrium temperature and product mass fraction with
heat release.
temperature is the ambient temperature. As q increases, the equilibrium temperature increases monotonically. When q = 0, the equilibrium mass fraction is YB = 0.5; that is, there
is no tendency for either products or reactants. As q increases, the tendency for increased
product mass fraction increases, and YB → 1.
There is also the state of complete reaction when YB = 1, which corresponds to
1−
cP (T − To )
= 0,
q
T = To +
(9.75)
q
.
cP
(9.76)
Numerically, at our maximum value of q = 1.5 × 106 J/kg, we get complete reaction when
T = 1800 K. Note this state is not an equilibrium state unless Kc → ∞. As x → ∞,
we expect a state of chemical equilibrium and a corresponding high temperature. At some
intermediate point, the temperature will have its value at ignition, Tig . We will ultimately
transform our domain so the ignition temperature is realized at x = 0, which we can associate
with a burner surface. We say that the flame is thus anchored to the burner. Without such
a prescribed anchor, our equations are such that any translation in x will yield an invariant
solution, so the origin of x is arbitrary.
9.2
Let us consider an important problem in laminar flame theory: that of a burner-stabilized
flame. We shall consider a doubly-infinite domain, x ∈ (−∞, ∞). We will assume that as
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312
x → −∞, we have a fresh unburned stream of reactant A, YA = 1, YB = 0, with known
velocity uo , density ρo and temperature To . The values of ρo and To will be consistent with
a state equation so that the pressure has a value of Po .
9.2.1
Formulation
Let us consider our governing equations in the steady wave frame where there is no variation
with t: ∂/∂t = 0. This, coupled with our other reductions, yields the system
d
(ρu) = 0,
dx
dT
cP (T − To )
d2 T
β −E/(RT )
ρucP
1−
− k 2 = ρqaT e
dx
dx
q
!
cP (T −To )
× 1 − e−q/(RT )
Po = ρRT.
1−
q
cP (T −To )
q
H(T − Tig ),
(9.77)
(9.78)
(9.79)
Now, we have three equations for the three remaining unknowns, ρ, u, T .
For a burner-stabilized flame, we assume at x → −∞ that we know the velocity, uo , the
temperature, To , and the density, ρo . We can integrate the mass equation to get
ρu = ρo uo .
(9.80)
We then rewrite the remaining differential equation as
d2 T
dT
−k 2
ρo uo cP
dx
dx
cP (T − To )
Po
β−1 −E/(RT )
1−
=
qaT
e
R
q
!
cP (T −To )
× 1 − e−q/(RT )
1−
q
cP (T −To )
q
H(T − Tig ).
(9.81)
Our boundary conditions for this second order differential equation are
dT
→ 0,
dx
x → ±∞.
(9.82)
We will ultimately need to integrate this equation numerically, and this poses some difficulty
because the boundary conditions are applied at ±∞. We can overcome this in the following
way. Note that our equations are invariant under a translation in x. We will choose some
large value of x to commence numerical integration. We shall initially assume this is near
the chemical equilibrium point. We shall integrate backwards in x. We shall note the value
of x where T = Tig . Beyond that point, the reaction rate is zero, and we can obtain an
exact solution for T . We will then translate all our results so that the ignition temperature
is realized at x = 0.
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313
Let us scale temperature so that
T = To 1 +
q
T∗ .
cP To
(9.83)
Inverting, we see that
T∗ =
cP (T − To )
.
q
(9.84)
We get a dimensionless ignition temperature TIG of
TIG =
cP (Tig − To )
.
q
(9.85)
Let us also define a dimensionless heat release Q as
Q=
q
.
cP To
(9.86)
Thus
T
= 1 + QT∗ .
To
(9.87)
Note that T∗ = 1 corresponds to our complete reaction point where YB = 1. Let us also
restrict ourselves, for convenience, to the case where β = 0. With these scalings, and with
Θ=
E
,
RTo
(9.88)
our differential equations become
dT∗
d2 T∗
ρo uocP To Q
− kTo Q 2
dx
dx
1 − T∗
=
To (1 + QT∗ )
!
!
γ
Q
T∗
γ−1
H(T∗ − TIG ),(9.89)
× 1 − exp −
1 + QT∗ 1 − T∗
Po
qa exp
R
dT∗
→ 0,
dx
−Θ
1 + QT∗
x → ±∞.
(9.90)
Let us now scale both sides by ρo uo cP To to get
dT∗
k 1 d2 T∗
Po
q a
Q
Q 2 =
exp
−
dx
ρo cP uo dx
ρ RT c T u
| o{z o} | P{z o} o
=1
−Θ
1 + QT∗
=Q
× 1 − exp −
γ
Q
γ−1
1 + QT∗
!
T∗
1 − T∗
1 − T∗
1 + QT∗
!
H(T∗ − TIG ). (9.91)
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314
Realizing that Po = ρo RTo , the ambient thermal diffusivity, αo = k/(ρo cP ) and canceling Q,
we get
dT∗
1 − T∗
1 d2 T∗
a
−Θ
− αo
=
exp
dx
uo dx2
uo
1 + QT∗
1 + QT∗
!
!
γ
Q
T∗
γ−1
H(T∗ − TIG ).
(9.92)
× 1 − exp −
1 + QT∗ 1 − T∗
Now, let us scale x by a characteristic length, L = uo /a. This length scale is dictated by
a balance between reaction and advection. When the reaction is fast, a is large, and the
length scale is reduced. When the incoming velocity is fast, uo is large, and the length scale
is increased.
x
ax
x∗ = =
.
(9.93)
L
uo
With this choice of length scale, Eq. (9.92) becomes
dT∗
−Θ
a d2 T∗
1 − T∗
= exp
− αo 2
dx∗
uo dx2∗
1 + QT∗
1 + QT∗
!
!
γ
Q
T
∗
γ−1
H(T∗ − TIG ).
× 1 − exp −
1 + QT∗ 1 − T∗
(9.94)
Now, similar to Eq. (8.36), we take the Damköhler number D to be
D=
u2
aL2
= o .
αo
aαo
(9.95)
In contrast to Eq. (8.36) the Damköhler number here includes the effects of advection. With
this choice, Eq. (9.94) becomes
dT∗
−Θ
1 d2 T∗
1 − T∗
= exp
−
dx∗ D dx2∗
1 + QT∗
1 + QT∗
!
!
γ
Q
T
∗
γ−1
H(T∗ − TIG ).
(9.96)
× 1 − exp −
1 + QT∗ 1 − T∗
Lastly, let us dispose with the ∗ notation and understand that all variables are dimensionless. Our differential equation and boundary conditions become
dT
1−T
1 d2 T
−Θ
−
= exp
dx D dx2
1 + QT
1 + QT
!
!
γ
Q
T
γ−1
H(T − TIG ),
(9.97)
× 1 − exp −
1 + QT 1 − T
dT → 0.
(9.98)
dx x→±∞
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315
Equation (9.97) is remarkably similar to our Frank-Kamenetskii problem embodied in
Eq. (8.59). The major differences are reflected in that Eq. (9.97) accounts for
• variable density,
• reversible reaction,
• doubly-infinite spatial domain.
9.2.2
Solution procedure
There are some unusual challenges in the numerical solution of the equations for laminar
flames. Formally we are solving a two-point boundary value problem on a doubly-infinite
domain. The literature is not always coherent on solutions methods or the interpretations
of results. At the heart of the issue is the cold boundary difficulty. We have patched
this problem via the use of an ignition temperature built into a Heaviside5 function; see
Eq. (9.30). We shall see that this patch has advantages and disadvantages. Here, we will
not dwell on nuances, but will present a result which is mathematically sound and offer
interpretations. Our result will use standard techniques of non-linear analysis from dynamic
systems theory. We will pose the problem as a coupled system of first order non-linear
differential equations, find their equilibria, use local linear analysis to ascertain the stability
of the chemical equilibrium fixed point, and use numerical integration to calculate the nonlinear laminar flame structure.
9.2.2.1
Model linear system
Before commencing with the difficult Eq. (9.97), let us consider a related linear model system,
given here:
dT
d2 T
dT 3
→ 0.
(9.99)
− 2 = 4(1 − T )H(T − TIG ),
dx
dx
dx x→±∞
The forcing is removed when either T = 1 or T < TIG .
Defining q ≡ −dT /dx, we rewrite our model equation as a linear system of first order
equations:
dq
= 3q + 4(1 − T )H(T − TIG ),
dx
dT
= −q,
q(−∞) = 0.
dx
5
q(∞) = 0,
(9.100)
(9.101)
Oliver Heaviside, 1850-1925, English engineer.
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316
The system has an equilibrium point at q = 0 and T = 1. The system is also in equilibrium
when q = 0 and T < TIG , but we will not focus on this. Taking T ′ = T − 1, the system near
the equilibrium point is
d
3 −4
q
q
.
(9.102)
=
′
′
−1
0
T
T
dx
The local Jacobian matrix has two eigenvalues λ = 4 and λ = −1. Thus, the equilibrium is
Considering the solution away from the equilibrium point, but still for T > TIG , and
returning from T ′ to T , the solution takes the form
1 4x 4C1 − 4C2
1 −x C1 + 4C2
0
q
+ e
.
(9.103)
=
+ e
C1 + 4C2
−C1 + C2
T
1
5
5
We imagine as x → ∞ that T approaches unity from below. We wish this equilibrium to be
achieved. Thus, we must suppress the unstable λ = 4 mode; this is achieved by requiring
C1 = C2 . This yields
1 −x 5C1
q
0
1
0
−x
=
+ e
.
(9.104)
=
+ C1 e
T
1
1
5C1
1
5
We can force T (xIG ) = TIG by taking C1 = −(1 − TIG )exIG . The solution for x > xIG is
T = 1 − (1 − TIG )e−(x−xIG ) ,
q = −(1 − TIG )e−(x−xIG ) .
(9.105)
(9.106)
At the interface x = xIG , we have T (xIG ) = TIG and q(xIG ) = −(1 − TIG ).
For x < xIG , our system reduces to
dT
d2 T
dT 3
− 2 = 0,
T (xIG ) = TIG ,
= 0.
dx
dx
dx x→−∞
(9.107)
Solutions in this region take the form
T = TIG e3(x−xIG ) + C 1 − e3(x−xIG ) ,
3(x−xIG )
q = −3TIG e
3(x−xIG )
+ 3Ce
.
(9.108)
(9.109)
All of them have the property that q → 0 as x → −∞. One is faced with the question of how
to choose the constant C. Let us choose it to match the energy flux predicted at x = xIG .
At the interface x = xIG we get T = TIG and q = −3TIG + 3C. Matching values of q at the
interface, we get
−(1 − TIG ) = −3TIG + 3C.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(9.110)
317
Solving for C, we get
C=
4TIG − 1
.
3
So, we find for x < xIG that
3(x−xIG )
T = TIG e
+
4TIG − 1
3
(9.111)
1 − e3(x−xIG ) ,
q = −3TIG e3(x−xIG ) + (4TIG − 1) e3(x−xIG ) .
9.2.2.2
(9.112)
(9.113)
System of first order equations
Let us apply this technique to the full non-linear Eq. (9.97). First, again define the nondimensional Fourier heat flux q as
dT
.
(9.114)
dx
Note this is a mathematical convenience. It is not the full diffusive energy flux as it makes no
account for mass diffusion affects. However, it is physically intuitive. With this definition,
we can rewrite Eq. (9.114) along with Eq. (9.97) as
1−T
−Θ
dq
= Dq + D exp
dx
1 + QT
1 + QT
!
!
γ
Q
T
γ−1
H(T − TIG ),
(9.115)
× 1 − exp −
1 + QT 1 − T
q=−
dT
= −q.
dx
We have two boundary conditions for this problem:
(9.116)
q(±∞) = 0.
(9.117)
We are also going to fix our coordinate system so that T = TIG at x = xIG ≡ 0. We will
require continuity of T and dT /dx at x = xIG = 0.
Our non-linear system has the form
d
q
f (q, T )
=
.
(9.118)
g(q, T )
dx T
9.2.2.3
Equilibrium
Our equilibrium condition is f (q, T ) = 0, g(q, T ) = 0. By inspection, equilibrium is found
when
1 − exp −
γ
Q
γ−1
1 + QT
!
q = 0,
T
1−T
= 0
(9.119)
or
T < TIG .
(9.120)
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318
We will focus most attention on the isolated equilibrium point which corresponds to traditional chemical equilibrium. We have considered the dimensional version of the same equilibrium condition for T in an earlier section. The solution is found via solving a transcendental
equation for T .
9.2.2.4
Linear analysis near equilibrium
Linear analysis near the equilibrium point can be achieved by first examining the eigenvalues
of the Jacobian matrix J:
∂f ∂f ∂f D
∂T
∂T eq
=
J = ∂q
.
(9.121)
∂g
∂g
−1
0
∂q
∂T
eq
This Jacobian has eigenvalues of
D
λ=
2
For D ≫
1±
s
!
4 ∂f 1− 2
.
D ∂T eq
(9.122)
p
∂f /∂T , the eigenvalues are approximated well by
1 ∂f .
λ1 ∼ D,
λ2 ∼
D ∂T eq
This gives rise to a stiffness ratio, valid in the limit of D ≫
2
λ1 ∼ D .
λ2 ∂f ∂T
(9.123)
p
∂f /∂T , of
(9.124)
We adopt the parameters of Table 9.1. With these, we find the dimensionless chemical
equilibrium temperature at
T eq = 0.953.
(9.125)
This corresponds to a dimensional value of 1730.24 K. Our parameters induce a dimensional
length scale of uo /a = 1.4142 × 10−5 m. This is smaller than actual flames, which actually
have much larger values of D. So as to de-stiffen the system, we have selected a smaller than
normal value for D. For these values, we get a Jacobian matrix of
2 −0.287
J=
.
(9.126)
−1
0
We get eigenvalues near the equilibrium point of
λ = 2.134,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
λ = −0.134.
(9.127)
319
The equilibrium is a saddle point. Note the negative value of ∂f /∂T at equilibrium gives rise
to the saddle character of the equilibrium. The actual stiffness ratio is |2.134/(−0.134)| =
15.9. This is well predicted by our simple formula which gives |22 /(−0.287)| = 13.9.
In principle, we could also analyze the set of equilibrium points along the cold boundary,
where q = 0 and T < TIG . Near such points, linearization techniques fail because of the
nature of the Heaviside function. We would have to perform a more robust analysis. As an
alternative, we shall simply visually examine the results of calculations and infer the stability
of this set of fixed points.
9.2.2.5
Laminar flame structure
9.2.2.5.1 TIG = 0.2. Let us get a numerical solution by integrating backwards in space
from the equilibrium point. Here, we will use the parameters of Table 9.1, in particular
to contrast with a later calculation, the somewhat elevated ignition temperature of TIG =
0.2. We shall commence the solution near the isolated equilibrium point corresponding to
chemical equilibrium. We could be very careful and choose the initial condition to lie just off
the saddle along the eigenvector associated with the negative eigenvalue. It will work just
as well to choose a point on the correct side of the equilibrium. Let us approximate x → ∞
by xB , require q(xB ) = 0, T (xB ) = T eq − ǫ, where 0 < ǫ ≪ 1. That is, we perturb the
temperature to be just less than its equilibrium value. We integrate from the large positive
x = xB back towards x = −∞. When we find T = TIG , we record the value of x. We
translate the plots so that the ignition point is reached at x = 0 and give results.
Predictions of T (x) and q(x) are shown in in Figure 9.2. We see the temperature has
T
q
1.0
T = T eq =0.953
0.8
-5
-0.02
0.6
-0.04
5
10
15
20
25
x
-0.06
0.4
-0.08
0.2
-5
-0.10
0
5
10
15
20
25
x
-0.12
Figure 9.2: Dimensionless temperature and heat flux as a function of position for a burnerstabilized premixed laminar flame, TIG = 0.2.
T (0) = 0.2 = TIG . As x → ∞, the temperature approaches its equilibrium value. For
x < 0, the temperature continues to fall until it comes to a final value of T ∼ 0.132. Note
that this value cannot be imposed by the boundary conditions, since we enforce no other
condition on T except to anchor it at the ignition temperature at x = 0. The heat flux q(x)
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320
is always negative. And it clearly relaxes to zero as x → ±∞. This indicates that energy
released in combustion makes its way back into the fresh mixture, triggering combustion
of the fresh material. This is the essence of the laminar diffusion flame. Predictions of
ρ(x) ∼ 1/(1 + QT (x)) and u(x) ∼ 1 + QT (x) are shown in in Figure 9.3. We are somewhat
u
ρ
0.8
6
5
0.6
4
0.4
3
2
0.2
1
5
0
5
10
15
20
25
30
x
-5
0
5
10
15
20
25
30
x
Figure 9.3: Dimensionless density ρ and fluid particle velocity u as functions of distance for
a burner-stabilized premixed laminar flame, TIG = 0.2.
troubled because ρ(x) nowhere takes on dimensionless value of unity, despite it being scaled
by ρo . Had we anchored the flame at an ignition temperature very close to zero, we in fact
would have seen ρ approach unity at the anchor point of the flame. We chose an elevated
ignition temperature so as to display the actual idiosyncrasies of the model. Even for a flame
anchored, in dimensional variables, at Tig ∼ To , we would find a significant decay of density
in the cold region of the flow for x < xig . Similar to ρ, we are somewhat troubled that the
dimensionless velocity does not take a value near unity. Once again, had we set Tig ∼ To ,
we would have found the u at x = xig would have taken a value very near unity. But it
also would modulate significantly in the cold region x < xig . We also would have discovered
that had Tig ∼ To that the temperature in the cold region would drop significantly below
To , which is a curious result. Alternatively, we could iterate on the parameter Tig until we
found a value for which limx→−∞ T → 0. For such a value, we would also find u and ρ to
approach unity as x → −∞.
We can better understand the flame structure by considering the (T, q) phase plane as
shown in Fig. 9.4. Here, green denotes equilibria. We see the isolated equilibrium point at
(T, q) = (0.953, 0). And we see a continuous one-dimensional set of equilibria for {(T, q)|q =
0, T ∈ (−∞, TIG ]}. Blue lines are trajectories in the phase space. The arrows have been
associated with movement in the −x direction. This is because this is the direction in which
it is easiest to construct the flame structure.
We know from dynamic systems theory that special trajectories, known as heteroclinic,
are those that connect one equilibrium point to another. These usually have special meaning.
The heteroclinic orbit shown here in the thick blue line is that of the actual flame structure. It
CC BY-NC-ND. 30 March 2014, J. M. Powers.
321
q
0.2
0.4
0.6
0.8
1.0
1.2
T
-0.05
-0.10
-0.15
Figure 9.4: (T ,q) phase plane for a burner-stabilized premixed laminar flame, TIG = 0.2.
connects the saddle at (T, q) = (0.953, 0) to a stable point on the one-dimensional continuum
of equilibria at (T, q) = (0.132, 0). It originates on the eigenvector associated with the
negative eigenvalue at the equilibrium point. Moreover, it appears that the heteroclinic
trajectory is an attracting trajectory, as points that begin on nearby trajectories seem to be
drawn into the heteroclinic trajectory.
The orange dashed line represents our chosen ignition temperature, TIG = 0.2. Had
we selected a different ignition temperature, the heteroclinic trajectory originating from the
saddle would be the same for T > TIG . However it would have turned at a different location
and relaxed to a different cold equilibrium on the one-dimensional continuum of equilibria.
Had we attempted to construct our solution by integrating forward in x, our task would
have been more difficult. We would likely have chosen the initial temperature to be just
greater than TIG . But we would have to had guessed the initial value of q. And because of
the saddle nature of the equilibrium point, a guess on either side of the correct value would
cause the solution to diverge as x became large. It would be possible to construct a trial and
error procedure to hone the initial guess so that on one side of a critical value, the solution
diverged to ∞, while on the other it diverged to −∞. Our procedure, however, has the clear
advantage, as no guessing is required.
As an aside, we note there is one additional heteroclinic orbit admitted mathematically;
however, its physical relevance is far from clear. It seems there is another attracting trajecCC BY-NC-ND. 30 March 2014, J. M. Powers.
322
tory for T > 0.953. This trajectory is associated with the same eigenvector as the physical
trajectory. Along this trajectory, T increases beyond unity, at which point the mass fraction
becomes greater than unity, and is thus non-physical. Mathematically this trajectory continues until it reaches an equilibrium at (T, q) → (∞, 0). These dynamics can be revealed using
the mapping of the Poincaré sphere; see Fig. 9.5. Details can be found in some dynamic
q’
1.0
0.5
1.0
0.5
0.5
1.0
T’
0.5
1.0
Figure 9.5: Projection of trajectories on Poincaré sphere onto the (T ′ , q′ ) phase plane for a
burner-stabilized premixed laminar flame, TIG = 0.2.
systems texts.6 In short the mapping from (T, q) → (T ′ , q′ ) via
T′ = p
q′ = p
T
1 + T 2 + q2
q
1 + T 2 + q2
,
(9.128)
,
(9.129)
p
is introduced. Often a third variable in the mapping is introduced Z = 1/ 1 + T 2 + q2 .
This induces T ′2 + q′2 + Z 2 = 1, the equation of a sphere, known as the Poincaré sphere.
As it is not clear that physical relevance can be found for this, we leave out most of the
mathematical details and briefly describe the results. This mapping takes points at infinity
in physical space onto the unit circle.
6
Perko, L., 2001, Differential Equations and Dynamical Systems, Third Edition, Springer, New York.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
323
Equilibria are marked in green. Trajectories are in blue. While the plot is somewhat
incomplete, we notice in Fig. 9.5 that the saddle is evident around (T ′ , q′ ) ∼ (0.7, 0). We
also see new equilibria on the unit circle, which corresponds to points at infinity in the
original space. One of these new equilibria is at (T ′ , q′ ) = (1, 0). It is a sink. The other
two, one in the second quadrant, the other in the fourth, are sources. The heteroclinic
trajectories which connect to the saddle from the sources in the second and fourth quadrants
form the boundaries for the basins of attraction. To the left of this boundary, trajectories
are attracted to the continuous set of equilibria. To the right, they are attracted to the point
(1, 0).
9.2.2.5.2 TIG = 0.076. We can modulate the flame structure by altering the ignition
temperature. In particular, it is possible to iterate on the ignition temperature in such a
way that as x → −∞, T → 0, ρ → 1, and u → 1. Thus, the cold flow region takes on
its ambient value. This has some æsthetic appeal. It is however unsatisfying in that one
would like to think that an ignition temperature, if it truly existed, would be a physical
property of the system, and not just a parameter to adjust to meet some other criterion.
That duly noted, we present flame structures using all the parameters of Table 9.1 except
we take Tig = 414 K, so that TIG = 0.076.
Predictions of T (x) and q(x) are shown in in Figure 9.6. We see the temperature has
T
q
1.0
T = T eq =0.953
5
0.8
10
15
20
25
x
-0.02
-0.04
0.6
-0.06
0.4
-0.08
-0.10
0.2
-0.12
0
5
10
15
20
25
x
-0.14
Figure 9.6: Dimensionless temperature and heat flux as a function of position for a burnerstabilized premixed laminar flame, TIG = 0.076.
T (0) = 0.076 = TIG . As x → ∞, the temperature approaches its chemical equilibrium value.
For x < 0, the temperature continues to fall until it comes to a final value of T ∼ 0. Thus,
in dimensional terms, the temperature has arrived at To in the cold region.
Predictions of ρ(x) ∼ 1/(1 + QT (x)) and u(x) are shown in in Figure 9.7. We see that
this special value of TIG has allowed the dimensionless density to take on a value of unity as
x → −∞. In contrast to our earlier result, for this special value of TIG , we have been able
to allow u = 1 as x → −∞. We can better understand the flame structure by considering
CC BY-NC-ND. 30 March 2014, J. M. Powers.
324
u
ρ
5
1.2
6
1.0
5
0.8
4
0.6
3
0.4
2
0.2
1
0
5
10
15
20
25
30
x
-5
0
5
10
15
20
25
30
x
Figure 9.7: Dimensionless density ρ and fluid particle velocity u as functions of distance for
a burner-stabilized premixed laminar flame, TIG = 0.076.
the (T, q) phase plane as shown in Fig. 9.8. Figure 9.8 is essentially the same as Figure 9.4
except the ignition point has been moved.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
325
q
0.2
0.4
0.6
0.8
1.0
1.2
T
-0.05
-0.10
-0.15
Figure 9.8: (T ,q) phase plane for a burner-stabilized premixed laminar flame, TIG = 0.076.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
326
9.2.3
Detailed H2-O2 -N2 kinetics
We give brief results for a premixed laminar flame in a mixture of calorically imperfect
ideal gases which obey mass action kinetics with Arrhenius reaction rates. Multi-component
diffusion is modelled as is thermal diffusion. We consider the kinetics model of Table 1.2. The
solution method is described in detail in a series of reports from Reaction Design, Inc.78910
The methods employed in the solution here are slightly different. In short, a large, but
finite domain defined. Then, the ordinary differential equations describing the flame structure are discretized. This leads to a large system of non-linear algebraic equations. These
are solved by iterative methods, seeded with an appropriate initial guess. The temperature
is pinned at To at one end of the domain, which is a slightly different boundary condition
than we employed previously. At an intermediate value of x, x = xf , the temperature is
pinned at T = Tf . Full details are in the report.
For To = 298 K, Po = 1.01325 × 105 P a, and a stoichiometric unreacted mixture of
2H2 + O2 + 3.76 N2 , along with a very small amount of minor species, we give plots of
Yi (x) in Fig. 9.9. There is, on a log-log scale, an incubation period spanning a few orders of
0
10
−5
Yi
10
H2
−10
10
O2
H 2O
OH
H
O
HO 2
−15
10
H 2O 2
N2
−20
10
−2
10
−1
10
10
0
1
10
x (cm)
Figure 9.9: Mass fractions as a function of distance in a premixed laminar flame with detailed
H2 -O2 -N2 kinetics.
7
Kee, R. J., et al., 2000, “Premix: A Fortran Program for Modeling Steady, Laminar, One Dimensional
Premixed Flames,” from the Chemkin Collection, Release 3.6, Reaction Design, San Diego, CA.
8
Kee, R. J., et al., 2000, “Chemkin: A Software Package for the Analysis of Gas-Phase Chemical and
Plasma Kinetics,” from the Chemkin Collection, Release 3.6, Reaction Design, San Diego, CA.
9
R. J. Kee, et al., 2000, “The Chemkin Thermodynamic Data Base,” part of the Chemkin Collection
Release 3.6, Reaction Design, San Diego, CA.
10
Kee, R. J., Dixon-Lewis, G., Warnatz, J., Coltrin, M. E., and Miller, J. A., 1998, “A Fortran Computer
Code Package for the Evaluation of Gas-Phase Multicomponent Transport Properties,” Sandia National
Laboratory, Report SAND86-8246, Livermore, CA.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
327
magnitude of length. Just past x = 2 cm, a vigorous reaction commences, and all species
relax to a final equilibrium. Temperature, density, and fluid particle velocity as functions of
distance are shown in Fig. 9.10.
−4
x 10
1600
2500
1500
1000
500
0
−4
u (cm/s)
ρ (g/cm3)
T (K)
2000
8
6
4
4
8
x - xf (cm)
12
−4
800
400
2
0
1200
0
4
8
x - xf (cm)
12
−4
0
4
8
12
x - xf (cm)
Figure 9.10: Temperature, density, and fluid particle velocity as functions of distance in a
premixed laminar flame with detailed H2 -O2 -N2 kinetics.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
328
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 10
Simple detonations:
Plato says, my friend, that society cannot be saved until either the Professors
of Greek take to making gunpowder, or else the makers of gunpowder become
Professors of Greek.
Undershaft to Cusins in George Bernard Shaw’s, 1856-1950, Major Barbara, Act
III, § iv.
————————Let us consider aspects of the foundations of detonation theory. Detonation is defined as a
shock-induced combustion process. Most major features of detonations are well modelled by
considering only the mechanisms of advection and reaction. Diffusion, which we will neglect
here, plays a more subtle role. Its inclusion is not critical to the main physics; however,
among other things it is potentially important in determining stability boundaries, as well
as in correctly capturing multi-dimensional instabilities. The definitive text is that of Fickett
and Davis.1 The present chapter is strongly influenced by this monograph.
10.1
Reactive Euler equations
10.1.1
One-step irreversible kinetics
Let us focus here on one-dimensional planar solutions in which all diffusion processes are
neglected. This is known as a reactive Euler2 model. We shall also here only be concerned
with simple one-step irreversible kinetics in which
A → B.
1
2
Fickett, W., and Davis, W. C., 1979, Detonation, U. California, Berkeley.
Leonhard Euler, 1707-1783, Swiss mathematician.
329
(10.1)
330
We will adopt the assumption that A and B are materials with identical properties; thus,
MA = MB = M,
cP A = cP B = cP ,
(10.2)
but A is endowed with chemical energy which is released as it forms B. So here, we have
N = 2 species and J = 1 reaction. The number of elements will not be important for this
analysis. Extension to multi-step kinetics is straightforward, but involves a sufficient number
of details to obscure many of the key elements of the analysis. At this point, we will leave the
state equations relatively general, but we will soon extend them to simple ideal, calorically
perfect relations. Since we have only a single reaction, our reaction rate vector rj , is a scalar,
which we will call r. Our stoichiometric matrix νij is of dimension 2 × 1 and is
−1
νij =
.
(10.3)
1
Here, the first entry is associated
PJ with A, and the second with B. Thus, our species production rate vector, from ω̇i = j=1 νij rj , reduces to
ω̇A
ω̇B
=
−1
1
(r) =
−r
r
.
(10.4)
Now, the right side of Eq. (6.5) takes the form
Mi ω̇i .
(10.5)
Let us focus on the products and see how this form expands.
MB ω̇B = MB r,
= MB kρA ,
= MA kρA ,
ρYA
= MA k
,
MA
= ρkYA ,
= ρk(1 − YB ).
(10.6)
(10.7)
(10.8)
(10.9)
(10.10)
(10.11)
For a reaction which commences with all A, let us define the reaction progress variable λ as
λ = YB = 1 − YA . And let us define r for this problem, such that
r = k(1 − YB ) = k(1 − λ).
Expanding further, we could say
E
E
β
β
(1 − λ) = aT exp −
(1 − λ),
r = aT exp −
RT
RT
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(10.12)
(10.13)
331
10.1. REACTIVE EULER EQUATIONS
but we will delay introduction of temperature T . Here, we have defined E = E/M. Note
that the units of r for this problem will be 1/s, whereas the units for r must be mole/cm3 /s,
and that
ρ
r.
(10.14)
r=
MB
The use of two different forms of the reaction rate, r and r is unfortunate. One is nearly
universal in the physical chemistry literature, r; the other is nearly universal in the one-step
detonation chemistry community, r. It is important that the two be distinguished.
10.1.2
Sound speed and thermicity
Let us specialize the definition of frozen sound speed, Eq. (3.320), for our one-step chemistry:
v
u
∂e u P + ∂v
P,λ
c = v t ∂e .
(10.15)
∂P v,λ
This gives, using v = 1/ρ,
ρ2 c2 =
P+
∂e ∂v P,λ
∂e ∂P v,λ
(10.16)
Let us, as is common in the detonation literature, define the thermicity σ via the relation
∂e ∂λ P,v
(10.17)
ρc2 σ = − ∂e .
∂P v,λ
We now specialize our general mathematical relation, Eq. (3.59), to get
∂P ∂e ∂λ = −1,
∂e v,λ ∂λ p,v ∂P v,e
so that one gets
−
∂e ∂λ P,v
∂e ∂P v,λ
∂P =
.
∂λ v,e
Thus, using Eq. (10.19), we can rewrite Eq. (10.17) as
∂P 2
,
ρc σ =
∂λ v,e
1 ∂P σ =
.
ρc2 ∂λ v,e
(10.18)
(10.19)
(10.20)
(10.21)
For our one step reaction, we see that as the reaction moves forward, i.e. as λ increases, that
σ is a measure of how much the pressure increases, since ρ > 0, c2 > 0.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
332
Parameter
γ
M
R
Po
To
ρo
vo
q
E
a
β
Value
1.4
20.91
397.58
1.01325 × 105
298
0.85521
1.1693
1.89566 × 106
8.29352 × 106
5 × 109
0
Units
kg/kmole
J/kg/K
Pa
K
kg/m3
m3 /kg
J/kg
J/kg
1/s
Table 10.1: Numerical values of parameters which roughly model H2 -air detonation.
10.1.3
Parameters for H2 -air
Let us postulate some parameters which loosely match results of the detailed kinetics calculation of Powers and Paolucci 3 for for H2 -air detonations. Rough estimates which allow
one-step kinetics models with calorically perfect ideal gas assumptions to approximate the
results of detailed kinetics models with calorically imperfect ideal gas mixtures are given in
Table 10.1.
10.1.4
Conservative form
Let us first consider a three-dimensional conservative form of the governing equations in the
inviscid limit:
∂ρ
+ ∇ · (ρu) = 0,
∂t
∂
(ρu) + ∇ · (ρuu + P I)
∂t ∂
1
1
P
ρ e+ u·u
+ ∇ · ρu e + u · u +
∂t
2
2
ρ
∂
(ρλ) + ∇ · (ρuλ)
∂t
e
r
(10.22)
= 0,
(10.23)
= 0,
(10.24)
= ρr,
(10.25)
= e(P, ρ, λ),
= r(P, ρ, λ).
(10.26)
(10.27)
For mass conservation, Eq. (10.22) is identical to the earlier Eq. (6.1). For linear momentum conservation, Eq. (10.23) is Eq. (6.2) with the viscous stress, τ = 0. For energy
3
Powers, J. M., and Paolucci, S., 2005, “Accurate Spatial Resolution Estimates for Reactive Supersonic
Flow with Detailed Chemistry,” AIAA Journal, 43(5): 1088-1099.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
333
10.1. REACTIVE EULER EQUATIONS
conservation, Eq. (10.24) is Eq. (6.3) with viscous stress τ = 0, and diffusive energy transport jq = 0. For species evolution, Eq. (10.27) is Eq. (6.5) with diffusive mass flux jm
i = 0.
Equations (10.22-10.27) form eight equations in the eight unknowns, ρ, u, P , e, λ, r. Note
that u has three unknowns, and Eq. (10.23) gives three equations. Note also that we assume
the functional forms of e and r are given.
The conservative form of the equations is the most useful and one of the more fundamental
forms. It is the form which arises from the even more fundamental integral form, which
admits discontinuities. We shall take advantage of this in later forming shock jump equations.
The disadvantage of the conservative form is that it is unwieldy and masks simpler causal
relations.
10.1.5
Non-conservative form
We can reveal some of the physics described by Eqs. (10.22-10.24) by writing them in what
is known as a non-conservative form.
10.1.5.1
Mass
We can use the product rule to expand Eq. (10.22) as
∂ρ
+ u · ∇ρ +ρ∇ · u = 0.
}
|∂t {z
(10.28)
=dρ/dt
We recall the well known material derivative, also known as the derivative following a material
particle, the total derivative, or the substantial derivative:
d
∂
≡
+ u · ∇.
dt
∂t
Using the material derivative, we can rewrite the mass equation as
dρ
+ ρ∇ · u = 0.
dt
(10.29)
Often, Newton’s notation for derivatives, the dot, is used to denote the material derivative.
We can also recall the definition of the divergence operator, div, to be used in place of ∇·,
so that the mass equation can be written compactly as
ρ̇ + ρ div u = 0.
(10.30)
The density of a material particle changes in response to the divergence of the velocity field,
which can be correlated with the rate of volume expansion of the material region.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
334
10.1.5.2
Linear momenta
We can use the product rule to expand the linear momenta equations, Eq. (10.23), as
∂u
∂ρ
+u
+ u(∇ · (ρu)) + ρu · ∇u + ∇P = 0,
∂t
∂t
∂u
∂ρ
ρ
+u
+ ∇ · (ρu) +ρu · ∇u + ∇P = 0,
∂t
∂t
{z
}
|
=0
∂u
ρ
+ u · ∇u +∇P = 0,
∂t
|
{z
}
ρ
(10.31)
(10.32)
(10.33)
=du/dt
ρ
du
+ ∇P = 0.
dt
(10.34)
Or in terms of our simplified notation with v = 1/ρ, we have
u̇ + v grad P = 0.
(10.35)
The fluid particle accelerates in response to a pressure gradient.
10.1.5.3
Energy
We can apply the product rule to the energy equation, Eq. (10.24) to get
∂
1
1
1
∂ρ
ρ
e+ u·u + e+ u·u
+ e + u · u ∇ · (ρu)
∂t
2
2
∂t
2
1
+ρu · ∇ e + u · u + ∇ · (P u) = 0,
2
1
1
∂ρ
∂
e+ u·u + e+ u·u
+ ∇ · (ρu)
ρ
∂t
2
2
∂t
{z
}
|
=0
1
+ρu · ∇ e + u · u + ∇ · (P u) = 0,
2
∂
1
1
ρ
e + u · u + ρu · ∇ e + u · u + ∇ · (P u) = 0.
∂t
2
2
(10.36)
(10.37)
(10.38)
This is simpler, but there is more that can be done by taking advantage of the linear momenta
equations. Let us continue working with the product rule once more along with some simple
rearrangements
∂ 1
1
∂e
u · u + ρu · ∇
u · u + ∇ · (P u) = 0,
(10.39)
ρ + ρu · ∇e + ρ
∂t
∂t 2
2
CC BY-NC-ND. 30 March 2014, J. M. Powers.
335
10.1. REACTIVE EULER EQUATIONS
ρ
∂u
∂e
+ ρu · ∇e + ρu ·
+ ρu · (u · ∇)u + u · ∇P + P ∇ · u = 0,
∂t
∂t
∂u
∂e
ρ + ρu · ∇e +u · ρ
+ ρ(u · ∇)u + ∇P +P ∇ · u = 0,
∂t
}
| ∂t {z
{z
}
|
=ρ de/dt
(10.40)
(10.41)
=0
ρ
de
+ P ∇ · u = 0.
dt
(10.42)
Now, from Eq. (10.29), we have ∇ · u = −(1/ρ)dρ/dt. Eliminating the divergence of the
velocity field from Eq. (10.42), we get
de P dρ
−
= 0,
dt
ρ dt
de P dρ
−
= 0.
dt ρ2 dt
(10.43)
ρ
(10.44)
Recalling that ρ = 1/v, so dρ/dt = −(1/v 2 )dv/dt = −ρ2 dv/dt, our energy equation becomes
dv
de
+P
= 0.
dt
dt
(10.45)
Or in terms of our dot notation, we get
ė + P v̇ = 0.
(10.46)
The internal energy of a fluid particle changes in response to the work done by the
pressure force.
10.1.5.4
Reaction
Using the product rule on Eq. (10.27) we get
∂λ
∂ρ
+ λ + ρu∇λ + λ∇ · (ρu) = ρr,
∂t
∂t
∂ρ
∂λ
= ρr,
ρ
+ u · ∇λ +λ
+ ∇ · (ρu)
∂t
∂t
{z
}
{z
}
|
|
ρ
=dλ/dt
(10.47)
(10.48)
=0
dλ
= r,
dt
λ̇ = r.
(10.49)
(10.50)
The mass fraction of a fluid particles product species changes according to the forward
reaction rate.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
336
10.1.5.5
Summary
In summary, our non-conservative equations are
ρ̇ + ρ div u
ė + P v̇
λ̇
e
r
0,
0,
0,
r,
e(P, ρ, λ),
r(P, ρ, λ),
1
.
ρ =
v
10.1.6
=
=
=
=
=
=
(10.51)
(10.52)
(10.53)
(10.54)
(10.55)
(10.56)
(10.57)
One-dimensional form
Here, we consider the equations to be restricted to one-dimensional planar geometries.
10.1.6.1
Conservative form
In the one-dimensional planar limit, Eqs. (10.22-10.27) reduce to
∂ρ
∂
+
(ρu)
∂t ∂x
∂
∂
ρu2 + P
(ρu) +
∂t
∂x
∂
1 2
1 2 P
∂
ρ e+ u
ρu e + u +
+
∂t
2
∂x
2
ρ
∂
∂
(ρλ) +
(ρuλ)
∂t
∂x
e
r
10.1.6.2
= 0,
(10.58)
= 0,
(10.59)
= 0,
(10.60)
= ρr,
(10.61)
= e(P, ρ, λ),
= r(P, ρ, λ).
(10.62)
(10.63)
Non-conservative form
The non-conservative Eqs. (10.51-10.57) reduce to the following in the one-dimensional planar
limit:
∂u
∂x
∂P
u̇ + v
∂x
ė + P v̇
λ̇
ρ̇ + ρ
CC BY-NC-ND. 30 March 2014, J. M. Powers.
= 0,
(10.64)
= 0,
(10.65)
= 0,
= r,
(10.66)
(10.67)
337
10.1. REACTIVE EULER EQUATIONS
e = e(P, ρ, λ),
r = r(P, ρ, λ),
1
.
ρ =
v
(10.68)
(10.69)
(10.70)
Equations (10.64-10.70) can be expanded using the definition of the material derivative:
∂ρ
∂u
∂ρ
+ρ
+u
= 0,
(10.71)
∂t
∂x
∂x
∂u
∂P
∂u
+v
+u
= 0,
(10.72)
∂t
∂x
∂x
∂e
∂v
∂e
∂v
+P
= 0,
(10.73)
+u
+u
∂t
∂x
∂t
∂x
∂λ
∂λ
= r,
(10.74)
+u
∂t
∂x
e = e(P, ρ, λ),
(10.75)
r = r(P, ρ, λ),
(10.76)
1
.
(10.77)
ρ =
v
10.1.6.3
Reduction of energy equation
Let us use standard results from calculus of many variables to expand the caloric equation
of state, Eq. (10.75):
∂e ∂e ∂e de =
dP +
dv +
dλ.
(10.78)
∂P v,λ
∂v P,λ
∂λ P,v
Taking the time derivative gives
∂e ∂e ∂e ė =
v̇ +
Ṗ +
λ̇.
∂P v,λ
∂v P,λ
∂λ P,v
(10.79)
Note this is simply a time derivative of the caloric state equation; it says nothing about energy
conservation. Next let us use Eq. (10.79) to eliminate ė in the first law of thermodynamics,
Eq. (10.66) to get
∂e ∂e ∂e v̇ +
Ṗ +
λ̇ +P v̇ = 0,
(10.80)
∂P v,λ
∂v P,λ
∂λ P,v
|
{z
}
=ė
!
∂e ∂e P + ∂v
P,λ v̇ + ∂λ
P,v λ̇ = 0,
(10.81)
Ṗ +
∂e ∂e |
∂P v,λ
{z
=ρ2 c2
}
∂P v,λ
| {z }
=−ρc2 σ
CC BY-NC-ND. 30 March 2014, J. M. Powers.
338
Ṗ + ρ2 c2 v̇ − ρc2 σ λ̇ = 0,
Ṗ = −ρ2 c2 v̇ + ρc2 σ λ̇,
Ṗ = ρc2 (σr − ρv̇) .
(10.82)
(10.83)
(10.84)
Now, since v̇ = −(1/ρ2 )ρ̇, we get
Ṗ = c2 ρ̇ + ρc2 σr.
(10.85)
Note that if either r = 0 or σ = 0, the pressure changes will be restricted to those from
classical isentropic thermo-acoustics: Ṗ = c2 ρ̇. If σr > 0, reaction induces positive pressure changes. If σr < 0, reaction induces negative pressure changes. Moreover, note that
Eq. (10.85) is not restricted to calorically perfect ideal gases. It is valid for general state
equations.
10.1.7
Characteristic form
Let us obtain a standard form known as characteristic form for the one-dimensional unsteady
equations. We follow the procedure described by Whitham (1927-).4 For this form let us use
Eq. (10.85) and generalized forms for sound speed c and thermicity σ to recast our governing
equations, Eq. (10.71-10.77) as
∂ρ
∂ρ
∂u
+u
+ρ
∂t
∂x
∂x
∂u 1 ∂P
∂u
+u
+
∂t
∂x ρ ∂x
∂P
∂ρ
∂ρ
∂P
2
+u
−c
+u
∂t
∂x
∂t
∂x
∂λ
∂λ
+u
∂t
∂x
c2
r
σ
= 0,
(10.86)
= 0,
(10.87)
= ρc2 σr,
(10.88)
= r,
(10.89)
= c2 (P, ρ),
= r(P, ρ, λ),
= σ(P, ρ, λ).
(10.90)
(10.91)
(10.92)
Let us write the differential equations in matrix form:
  ∂ρ
  ∂ρ  

u
ρ 0 0
1
0 0 0
∂t
∂x
1
  ∂u
  ∂u   0
 0
0
u
1
0
0
ρ
  ∂x
  ∂t  
 2
 −c 0 1 0   ∂P  +  −c2 u 0 u 0   ∂P
∂t
∂x
∂λ
∂λ
0
0 0 1
0
0 0 u
∂t
∂x
These take the general form
Aij
4

0
  0 
 =  2 .
  ρc σr 
r

∂wj
∂wj
+ Bij
= Ci .
∂t
∂x
Whitham, G. B., 1974, Linear and Nonlinear Waves, Wiley, New York.
CC BY-NC-ND. 30 March 2014, J. M. Powers.

(10.93)
(10.94)
339
10.1. REACTIVE EULER EQUATIONS
Let us attempt to cast this the left hand side of this system in the form ∂wj /∂t + µ∂wj /∂x.
Here, µ is a scalar which has the units of velocity. To do so, we shall seek vectors ℓi such
that
∂wj
∂wj
∂wj
∂wj
ℓi Aij
.
(10.95)
+ ℓi Bij
= ℓi Ci = mj
+µ
∂t
∂x
∂t
∂x
For ℓi to have the desired properties, we will insist that
ℓi Aij = mj ,
ℓi Bij = µmj .
(10.96)
(10.97)
Using Eq. (10.96) to eliminate mj in Eq. (10.97), we get
ℓi Bij = µℓi Aij ,
0 = ℓi (µAij − Bij ).
(10.98)
(10.99)
Equation (10.99) has the trivial solution ℓi = 0. For a non-trivial solution, standard linear
algebra tells us that we must enforce the condition that the determinant of the coefficient
matrix be zero:
|µAij − Bij | = 0.
Specializing Eq. (10.100) for Eq. (10.93), we find
µ−u
−ρ
0
0
1
0
0
µ − u −ρ
−c2 (µ − u)
0
µ−u
0
0
0
0
µ−u
(10.100)
= 0.
(10.101)
Let us employ standard co-factor expansion operations to reduce the determinant:
µ−u
−ρ
0 0
µ − u − ρ1 = 0,
(µ − u) (10.102)
−c2 (µ − u)
0
µ−u
2
−c (µ − u)
2
= 0,
(10.103)
(µ − u) (µ − u) (µ − u) + ρ
ρ
(µ − u)2 (µ − u)2 − c2 = 0.
(10.104)
There are four roots to this equation; two are repeated:
µ
µ
µ
µ
=
=
=
=
u,
u,
u + c,
u − c.
(10.105)
(10.106)
(10.107)
(10.108)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
340
Let us find the eigenvector ℓi associated with the eigenvalues µ = u ± c. So, Eq. (10.99)
reduces to


(u ± c) − u
−ρ
0
0


0
0
(u ± c) − u
− ρ1
 = (0 0 0 0).
( ℓ1 ℓ2 ℓ3 ℓ4 ) 

 −c2 ((u ± c) − u)
0
(u ± c) − u
0
0
0
0
(u ± c) − u
(10.109)
Simplifying,
( ℓ1
ℓ2
ℓ3

±c
 0
ℓ4 ) 
 ∓c3
0
−ρ 0
±c − 1ρ
0 ±c
0
0

0
0 
 = (0 0 0 0).
0 
±c
(10.110)
We thus find four equations, with linear dependencies:
±cℓ1 ∓ c3 ℓ3
−ρℓ1 ± cℓ2
1
− ℓ2 ± cℓ3
ρ
±cℓ4
= 0,
= 0,
(10.111)
(10.112)
= 0,
(10.113)
= 0.
(10.114)
Now, c 6= 0, so Eq. (10.114) insists that
ℓ4 = 0.
(10.115)
We expect a linear dependency, which implies that we are free to set at least one of the
remaining ℓi to an arbitrary value. Let us see if we can get a solution with ℓ3 = 1. With
that Eqs. (10.111-10.113) reduce to
±cℓ1 ∓ c3 = 0,
−ρℓ1 ± cℓ2 = 0,
1
− ℓ2 ± c = 0.
ρ
(10.116)
(10.117)
(10.118)
Solving Eq. (10.116) gives
ℓ 1 = c2 .
(10.119)
Then, Eq. (10.117) becomes −ρc2 ± cℓ2 = 0. Solving gives
ℓ2 = ±ρc.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(10.120)
341
10.1. REACTIVE EULER EQUATIONS
This is redundant with solving Eq. (10.118), which also gives ℓ2 = ±ρc. So, we have for
µ = u ± c that
ℓ i = ( c2
±ρc 1 0 ) .
(10.121)
So, Eq. (10.93) becomes after multiplication by ℓi from Eq. (10.121):
( c2

1
 0
±ρc 1 0 ) 
 −c2
0
0
1
0
0
0
0
1
0
  ∂ρ
0
∂t
 ∂u
0
∂t
  ∂P
0   ∂t
∂λ
1
∂t


 + ( c2


  ∂ρ 
u
ρ 0 0
∂x
 0
 ∂u 
u ρ1 0 
∂x 



±ρc 1 0 )  2

−c u 0 u 0   ∂P
∂x
∂λ
0
0 0 u
 ∂x 
0

0 

= ( c2 ±ρc 1 0 ) 
 ρc2 σr 
r
(10.122)
Carrying out the vector-matrix multiplication operations, we get
 ∂ρ 

( 0 ±ρc 1 0 ) 

∂t
∂u
∂t
∂P
∂t
∂λ
∂t



 + ( 0 ±ρc(u ± c) u ± c 0 ) 


= ( c2
∂ρ
∂x
∂u
∂x
∂P
∂x
∂λ
∂x





0
 0 

±ρc 1 0 ) 
 ρc2 σr  .
r

(10.123)
Simplifying,
±ρc
∂u
∂u
+ (u ± c)
∂t
∂x
+
∂P
∂P
+ (u ± c)
∂t
∂x
= ρc2 σr.
(10.124)
Now, let us confine our attention to lines in x − t space on which
dx
= u ± c.
dt
(10.125)
On such lines, Eq. (10.124) can be written as
±ρc
∂u dx ∂u
+
∂t
dt ∂x
+
∂P
dx ∂P
+
∂t
dt ∂x
= ρc2 σr.
(10.126)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
342
Consider now a variable, say u, which is really u(x, t). From calculus of many variables, we
have
∂u
∂u
dt +
dx,
∂t
∂x
∂u dx ∂u
du
=
+
.
dt
∂t
dt ∂x
du =
(10.127)
(10.128)
If we insist dx/dt = u ± c, let us call the derivative du/dt± , so that Eq. (10.126) becomes
±ρc
dP
du
+
= ρc2 σr.
dt± dt±
(10.129)
In the inert limit, after additional analysis, Eq. (10.129) reduces to the form dψ/dt± = 0,
which shows that ψ is maintained as a constant on lines where dx/dt = u ± c. Thus,
one can say that a signal is propagated in x − t space at speed u ± c. So, we see from
the characteristic analysis how the thermodynamic property c has the added significance of
influencing the speed at which signals propagate.
Let us now find ℓi for µ = u. For this root, we find


0 −ρ 0 0
 0 0 −1 0 
ρ
 = (0 0 0 0).
( ℓ1 ℓ2 ℓ3 ℓ4 ) 
(10.130)
0 0
0 0
0 0
0 0
It can be seen by inspection that two independent solutions ℓi satisfy Eq. (10.130):
ℓi = ( 0 0 1 0 ) ,
ℓi = ( 0 0 0 1 ) .
(10.131)
(10.132)
These two eigenvectors induce the characteristic form which was already obvious from the
initial form of the energy and species equations:
∂P
∂ρ
∂P
∂ρ
2
= ρc2 σr,
(10.133)
+u
−c
+u
∂t
∂x
∂t
∂x
∂λ
∂λ
+u
= r.
(10.134)
∂t
∂x
On lines where dx/dt = u, that is to say on material particle pathlines, these reduce to
Ṗ − c2 ρ̇ = ρc2 σr and λ̇ = r.
Because we all of the eigenvalues µ are real, and because we were able to find a set of
four linearly independent right eigenvectors ℓi so as to transform our four partial differential
equations into characteristic form, we can say that our system is strictly hyperbolic. Thus, it
• is well posed for initial value problems given that initial data is provided on a noncharacteristic curve, and
CC BY-NC-ND. 30 March 2014, J. M. Powers.
343
10.1. REACTIVE EULER EQUATIONS
• admits discontinuous solutions described by a set of Rankine-Hugoniot jump conditions
which arise from a more primitive form of the governing equations.
In summary we can write our equations in characteristic form as
dP
du
+ ρc
dt+
dt+
dP
du
− ρc
dt−
dt−
dP
dρ
− c2
dt
dt
dλ
dt
10.1.8
= ρc2 σr,
on
= ρc2 σr,
on
= ρc2 σr,
on
= r
on
dx
= u + c,
dt
dx
= u − c,
dt
dx
= u,
dt
dx
= u.
dt
(10.135)
(10.136)
(10.137)
(10.138)
Rankine-Hugoniot jump conditions
As described by LeVeque5 the proper way to arrive at what are known as Rankine6 -Hugoniot7
jump equations describing discontinuities is to use a more primitive form of the conservation
laws, expressed in terms of integrals of conservative form quantities balanced by fluxes and
source terms of those quantities. If q is a set of conservative form variables, and f(q) is
the flux of q (e.g. for mass conservation, ρ is a conserved variable and ρu is the flux), and
s(q) is the internal source term, then the primitive form of the conservation form law can
be written as
Z x2
Z
d x2
s(q(x, t))dx.
(10.139)
q(x, t)dx = f(q(x1 , t)) − f(q(x2 , t)) +
dt x1
x1
Here, we have considered flow into and out of a one-dimensional box for x ∈ [x1 , x2 ]. For
our reactive Euler equations we have



 

ρu
0
ρ
2





ρu
+
P
0
ρu 




,
f(q)
=
,
s(q)
=
q=
 ρu e + 1 u2 + P 
 0  . (10.140)
 ρ e + 1 u2 
2
2
ρ
ρr
ρλ
ρuλ
If we assume there is a discontinuity in the region x ∈ [x1 , x2 ] propagating at speed U, we
can find the Cauchy8 principal value of the integral by splitting it into the form
Z
Z
−
d x1 +U t
d x2
q(x, t)dx +
q(x, t)dx
dt x1
dt x1 +U t+
Z x2
= f(q(x1 , t)) − f(q(x2 , t)) +
s(q(x, t))dx.
(10.141)
x1
5
LeVeque, R. J., 1992, Numerical Methods for Conservation Laws, Birkhäuser, Basel.
William John Macquorn Rankine, 1820-1872, Scottish engineer.
7
Pierre Henri Hugoniot, 1851-1887, French mathematician.
8
Augustin-Louis Cauchy, 1789-1857, French mechanician.
6
CC BY-NC-ND. 30 March 2014, J. M. Powers.
344
Here, x1 + Ut− lies just before the discontinuity and x1 + Ut+ lies just past the discontinuity.
Using Leibniz’s rule, we get
Z x2
Z x1 +U t−
∂q
∂q
+
−
dx + 0 − q(x1 + Ut , t)U +
dx (10.142)
q(x1 + Ut , t)U − 0 +
∂t
x1 +U t+ ∂t
x1
Z x2
s(q(x, t))dx.
= f(q(x1 , t)) − f(q(x2 , t)) +
x1
Now, if we assume that x2 −x1 → 0 and that on either side of the discontinuity the volume of
integration is sufficiently small so that the time and space variation of q is negligibly small,
we get
q(x1 )U − q(x2 )U = f(q(x1 )) − f(q(x2 )),
U (q(x1 ) − q(x2 )) = f(q(x1 )) − f(q(x2 )).
(10.143)
(10.144)
Note that the contribution of the source term s is negligible as x2 − x1 → 0. Defining next
the notation for a jump as
Jq(x)K ≡ q(x2 ) − q(x1 ),
(10.145)
the jump conditions are rewritten as
U Jq(x)K = Jf(q(x))K .
(10.146)
If U = 0, as is the case when we transform to the frame where the wave is at rest, we
simply recover
0 = f(q(x1 )) − f(q(x2 )),
f(q(x1 )) = f(q(x2 )),
Jf(q(x))K = 0.
(10.147)
(10.148)
(10.149)
That is, the fluxes on either side of the discontinuity are equal. We also get a more general
result for U 6= 0, which is the well-known
U=
f(q(x2 )) − f(q(x1 ))
Jf(q(x))K
=
.
q(x2 ) − q(x1 )
Jq(x)K
(10.150)
The general Rankine-Hugoniot equation then for the one-dimensional reactive Euler equations across a non-stationary jump is given by

 

ρ2 u2 − ρ1 u1
ρ2 − ρ1
ρ2 u22 + P2 − ρ1 u21 − P1
 

ρ2 u2 − ρ1 u1

.



=
U
ρ2 e2 + 21 u22 − ρ1 e1 + 21 u21   ρ2 u2 e2 + 21 u22 + Pρ22 − ρ1 u1 e1 + 12 u21 + Pρ11 
ρ2 λ2 − ρ1 λ1
ρ2 u2 λ2 − ρ2 u1 λ1
(10.151)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
345
10.1. REACTIVE EULER EQUATIONS
Note that if there is no discontinuity, Eq. (10.139) reduces to our partial differential
equations which are the reactive Euler equations. We can rewrite Eq. (10.139) as
Z x2
Z x2
d
s(q(x, t))dx, (10.152)
q(x, t)dx + (f(q(x2 , t)) − f(q(x1 , t))) =
dt x1
x1
Now, if we assume continuity of all fluxes and variables, we can use Taylor series expansion
and Leibniz’s rule to say
Z x2
Z x2
∂
∂f
s(q(x, t))dx,
q(x, t)dx +
f(q(x1 , t)) +
(x2 − x1 ) + . . . − f(q(x1 , t))
=
∂x
x1 ∂t
x1
let x2 → x1
Z x2
Z x2
∂f
∂
s(q(x, t))dx,
q(x, t)dx +
(x2 − x1 )
=
∂x
x1
x1 ∂t
Z x2
Z x2
Z x2
∂
∂f
s(q(x, t))dx.
q(x, t)dx +
dx =
x1 ∂t
x1 ∂x
x1
(10.153)
Combining all terms under a single integral, we get
Z x2 ∂q ∂f
+
− s dx = 0.
∂t
∂x
x1
(10.154)
Now, this integral must be zero for an arbitrary x1 and x2 , so the integrand itself must be
zero, and we get our partial differential equation:
∂q ∂f
+
− s = 0,
∂t
∂x
∂
∂
q(x, t) +
f(q(x, t)) = s(q(x, t)),
∂t
∂x
(10.155)
(10.156)
which applies away from jumps.
10.1.9
Galilean transformation
We know that Newtonian9 mechanics have been constructed so as to be invariant under
a so-called Galilean10 transformation which takes one from a fixed laboratory frame to a
constant velocity frame with respect to the fixed frame. The Galilean transformation is such
that our original coordinate system in the laboratory frame, (x, t), transforms to a steady
travelling wave frame, (x̂, t̂), via
x̂ = x − Dt,
t̂ = t.
9
10
(10.157)
(10.158)
after Isaac Newton, 1643-1727, English polymath.
after Galileo Galilei, 1564-1642, Italian polymath.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
346
We now seek to see how our derivatives transform under this coordinate change. We shall
arrive at the same result via two methods: 1) a direct approach, and 2) a more formal
approach based on notions from differential geometry. The first is more straightforward; the
second gives greater cohesion with a more general theory of coordinate transformations.
10.1.9.1
Direct approach
For the direct approach, we first take differentials of Eqs. (10.157,10.158) and get
∂ x̂
dx +
∂x
∂ t̂
dt̂ =
dx +
∂x
dx̂ =
∂ x̂
dt = dx − Ddt,
∂t
∂ t̂
dt = dt.
∂t
(10.159)
(10.160)
Scaling dx̂ by dt̂ gives us then
dx̂
dx
− D.
=
dt
dt̂
(10.161)
Taking as usual the particle velocity in the fixed frame to be u = dx/dt and defining the
particle velocity in the laboratory frame to be û = dx̂/dt̂, we see that
û = u − D.
(10.162)
Now, a dependent variable ψ has a representation in the original space of ψ(x, t), and in the
transformed space as ψ(x̂, t̂). And they must both map to the same value of ψ and the same
point. And there is a differential of ψ in both spaces which must be equal:
∂ψ ∂ψ ∂ψ ∂ψ dt̂,
(10.163)
dψ =
dx̂ +
dx +
dt =
∂x t
∂t x
∂ x̂ t̂
∂ t̂ x̂
∂ψ ∂ψ =
dt,
(10.164)
(dx − Ddt) +
∂ x̂ t̂
∂ t̂ x̂
∂ψ ∂ψ ∂ψ dx +
dt.
(10.165)
=
−D
∂ x̂ t̂
∂ x̂ t̂
∂ t̂ x̂
Now, consider Eq. (10.165) for constant x, thus dx = 0, and divide by dt:
∂ψ ∂ψ ∂ψ −D
.
=
∂t x
∂ x̂ t̂
∂ t̂ x̂
More generally the partial with respect to t becomes
∂ ∂ ∂ .
= −D
∂t x
∂ x̂ t̂
∂ t̂ x̂
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(10.166)
(10.167)
347
10.1. REACTIVE EULER EQUATIONS
Now, consider Eq. (10.165) for constant t, thus dt = 0, and divide by dx:
∂ψ ∂ψ .
=
∂x t
∂ x̂ t̂
(10.168)
More generally,
10.1.9.2
∂ ∂ .
=
∂x t
∂ x̂ t̂
(10.169)
Differential geometry approach
For the differential geometry approach, we view the transformation in light of notions from
linear theory where such concepts as the Jacobian, metric tensor, co- and contravariant representations are often used. These ideas are traditionally confined to spatial systems. Here, we
will adapt them in a non-traditional way to account for our spatio-temporal transformation.
First, we invert Eqs. (10.157,10.158) to form
x = x̂ + D t̂,
t = t̂.
(10.170)
(10.171)
Now, physically, x and t have different units. For dimensional consistency between the two
independent variables, let us choose to consider the independent variables in the untransformed and transformed coordinates to be (x, co t), and (x̂, co t̂), respectively. It is thus trivial
to see that Eqs. (10.170),10.171) can be rewritten as
x = x̂ +
co t = co t̂.
D
co t̂ ,
co
(10.172)
(10.173)
In matrix form, we could say
x
1 cDo
x̂
=
.
·
co t
co t̂
0 1
| {z } | {z } | {z }
J
ξ
ξ̂
(10.174)
Here, we have defined a vector containing our untransformed independent variables as
1 x
ξ
=
.
(10.175)
ξ=
co t
ξ2
The superscripts are not exponents, but indices. The transformed independent variables are
1 ξˆ
x̂
ξ̂ = ˆ2 =
.
(10.176)
co t̂
ξ
CC BY-NC-ND. 30 March 2014, J. M. Powers.
348
These are linked by the linear transformation
ξ = J · ξ̂,
where the constant Jacobian matrix, J is
 ∂ξ1 ∂ξ  ∂ ξ̂1 ξ̂2
J=
=
∂ξ 2 ∂ ξˆ
(10.177)

∂ξ 1 ∂ ξ̂ 2 ξ̂ 1
=
∂ξ 2 ∂ ξ̂ 2 ξ̂ 1
∂ ξ̂ 1 ξ̂ 2
1
0
D
co
1
.
(10.178)
It is easily shown that the ratio of generalized areas between the two coordinate systems is
given by J = det J. Here, we have
J = det J = 1,
(10.179)
so that generalized areas are preserved in the mapping. Moreover, it can be shown that
because J > 0, the transformation is orientation-preserving.
We also have, by differentiating Eqs. (10.175-10.177), the differentials
dx
dx̂
dξ =
,
dξ̂ =
,
(10.180)
co dt
co dt̂
as well as the transformation rule
dξ = J · dξ̂.
(10.181)
The metric tensor G of the transformation is
1
1 cDo
1 0
T
·
= D
G=J ·J= D
1
0 1
co
co
D
co
1+
D2
co
.
(10.182)
A generalized distance in the untransformed space is deduced from the Pythagorean formula
dx
T
2
(ds) = dξ · dξ = ( dx co dt )
= dx2 + c2o dt2 .
(10.183)
co dt
In the transformed space, Eq. (10.183) becomes, after using Eqs. (10.181,10.182),
T (ds)2 = J · dξ̂ · J · dξ̂ ,
(10.184)
T
T
= dξ̂ · J
| {z· J} ·dξ̂,
(10.185)
G
T
= dξ̂ · G · dξ̂ .
| {z } |{z}
(10.186)
G = gij ,
(10.187)
dξ̂j
dξ̂ j
Using upstairs-downstairs index notation, we can define our covariant metric tensor as
CC BY-NC-ND. 30 March 2014, J. M. Powers.
349
10.1. REACTIVE EULER EQUATIONS
the contravariant dξˆj as
dξ̂ =
and the covariant dξˆj as
T
dξ̂ · G = ( dx̂ co dt̂ )
D
co
1
D
co
1+
D2
co
dx̂
co dt̂
=
= dξˆj ,
(10.188)
D
dx̂
co
dx̂ + Ddt̂
+ 1+
For a general function ψ, the chain rule tells us

 
 ∂ψ   ∂ξ1 ∂ψ ∂ξ 2 1
1
1
∂ξ 1 2
 ∂ ξ̂ ξ̂2  =  ∂ ξ̂1 ξ̂2 ∂ ξ̂2 ξ̂2  · 
ξ  .
∂ξ ∂ψ ∂ψ ∂ξ ∂ξ 2 1
∂ ξ̂ 2 ξ̂ 1
∂ ξ̂ 2 ξ̂ 1
∂ ξ̂ 2 ξ̂ 1
{z
} | {z ξ }
| {z } |
JT
∇ ψ
ξ̂
D2
c2o
co dt̂
= dξˆj .
(10.189)
(10.190)
∇ξ ψ
Equation (10.190) can be expressed in terms of the space-time gradients ∇ and ∇ξ
T
= J · ∇ξ ψ.
∇ ψ
ξ̂
ξ̂
(10.191)
Inverting Eq. 10.191), we get
∇ξ ψ
=
JT
−1
(10.192)
· ∇ ψ.
ξ̂
Transposing then inverting J from Eq. (10.178), we get
1
0
T −1
.
J
=
− cDo 1
(10.193)
For our particular independent variables, Eq. (10.192) becomes
! !
∂ψ ∂ψ ∂x cot
∂ x̂ cot̂
1
0
.
=
·
D
∂ψ ∂ψ − co 1
∂(c t) ∂(c t̂)
o
x
In scalar form, Eq. (10.194) becomes
∂ψ =
∂x co t
∂ψ =
∂(co t) x
o
(10.194)
x̂
∂ψ ,
∂ x̂ co t̂
∂ψ D ∂ψ .
−
∂(co t̂) x̂ co ∂ x̂ co t̂
And because co is a constant, Eqs. (10.195,10.196) reduce to
∂ψ ∂ψ ,
=
∂x t
∂ x̂ t̂
∂ψ ∂ψ ∂ψ .
=
−D
∂t x
∂ x̂ t̂
∂ t̂ x̂
(10.195)
(10.196)
(10.197)
(10.198)
Equations (10.197,10.198) are identical to those found using the direct method, Eqs. (10.168,10.166).
CC BY-NC-ND. 30 March 2014, J. M. Powers.
350
10.1.9.3
Transformed equations
Applying either approach, we can quickly write our non-conservative form, Eqs. (10.64-10.67)
in the transformed frame. First, let us consider how a material derivative transforms then
∂ ∂ ∂ ∂ ∂ −D
+ (û + D)
,
+u
=
∂t x
∂x t
∂ x̂ t̂
∂ x̂ t̂
∂ t̂ x̂
∂ ∂ =
+ û
.
∂ x̂ ∂ t̂ x̂
(10.199)
(10.200)
t̂
With this, we can then say
∂ û
∂ρ
∂ρ
+ρ
+ û
∂ x̂
∂ x̂
∂ t̂
∂ û 1 ∂P
∂ û
+ û
+
∂ t̂ ∂ x̂ ρ ∂ x̂
∂e
∂v
∂v
∂e
+ û
+ û
−P
∂ x̂
∂ x̂
∂ t̂
∂ t̂
∂λ
∂λ
+ û
∂ x̂
∂ t̂
= 0,
(10.201)
= 0,
(10.202)
= 0,
(10.203)
= r.
(10.204)
Moreover, we can write these equations in conservative form. Leaving out the details, which
amounts to reversing our earlier steps which led to the non-conservative form, we get
∂ρ
∂
(ρû)
+
∂ t̂ ∂ x̂
∂
∂
(ρû) +
ρû2 + P
∂t̂ ∂ x̂
1 2 P
∂
∂
1 2
ρû e + û +
+
ρ e + û
2
∂ x̂
2
ρ
∂ t̂
∂
∂
(ρûλ)
(ρλ) +
∂ x̂
∂ t̂
10.2
= 0,
(10.205)
= 0,
(10.206)
= 0,
(10.207)
= ρr.
(10.208)
Let us consider a steadily propagating disturbance in a one dimensional flow field. In the
laboratory frame, the disturbance is observed to propagate with constant velocity D. Let us
analyze such a disturbance.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
10.2.1
351
In the steady frame, our jump conditions, Eq. (10.151) have U = 0 and reduce to

  
ρ2 û2 − ρ1 û1
0
ρ2 û22 + P2 − ρ1 û21 − P1
0 


 =
(10.209)
1
P
P
1
2
2
 0   ρ2 û2 e2 + û2 + 2 − ρ1 û1 e1 + û1 + 1  .
2
ρ2
2
ρ1
0
ρ2 û2 λ2 − ρ1 û1λ1
The mass jump equation can be used to quickly simplify the energy and species jump equations to get a revised set
ρ2 û2
+ P2
1 2 P2
e2 + û2 +
2
ρ2
λ2
ρ2 û22
= ρ1 û1 ,
= ρ1 û21 + P1 ,
1
P1
= e1 + û21 + ,
2
ρ1
= λ1 .
10.2.2
Ordinary differential equations of motion
10.2.2.1
Conservative form
(10.210)
Let us now assert that in the steady laboratory frame that no variable has dependence on
t̂, so ∂/∂ t̂ = 0, and ∂/∂ x̂ = d/dx̂. With this assumption our partial differential equations of
motion, Eqs. (10.205-10.208), become ordinary differential equations:
d
(ρû)
dx̂
d
ρû2 + P
dx̂
d
1 2 P
ρû e + û +
dx̂
2
ρ
d
(ρûλ)
dx̂
e
r
= 0,
(10.211)
= 0,
(10.212)
= 0,
(10.213)
= ρr,
(10.214)
= e(P, ρ, λ),
= r(P, ρ, λ).
(10.215)
(10.216)
We shall take the following conditions for the undisturbed fluid just before the shock at
x̂ = 0− :
ρ(x̂ = 0− )
û(x̂ = 0− )
P (x̂ = 0− )
e(x̂ = 0− )
λ(x̂ = 0− )
=
=
=
=
=
ρo ,
−D,
Po ,
eo ,
0.
(10.217)
(10.218)
(10.219)
(10.220)
(10.221)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
352
Note that in the laboratory frame, this corresponds to a material at rest since u = û + D =
(−D) + D = 0.
10.2.2.2
Unreduced non-conservative form
We can gain insights into how the differential equations, Eqs. (10.211-10.214) behave by
writing them in a non-conservative form. Let us in fact write them, taking advantage of
the reductions we used to acquire the characteristic form, Eqs. (10.86-10.89) to write those
equations after transformation to the steady Galilean transformed frame as
dû
dρ
+ û
= 0,
(10.222)
dx̂
dx̂
dû 1 dP
= 0,
(10.223)
û +
dx̂ ρ dx̂
dP
dρ
û
− c2 û
= ρc2 σr,
(10.224)
dx̂
dx̂
dλ
= r.
(10.225)
û
dx̂
Let us attempt next to get explicit representations for the first derivatives of each equation.
In matrix form, we can say our system is


  dρ  
û
ρ 0 0
0
dx̂
 0
 dû   0 
û ρ1 0 
dx̂ 



  dP
(10.226)
 −c2 û 0 û 0    =  ρc2 σr  .
dx̂
dλ
r
0
0 0 û
dx̂
ρ
We can use Cramer’s11 rule to invert the coefficient matrix to solve for the evolution of each
state variable. This first requires the determinant of the coefficient matrix ∆.
û
ρ
0
0
1
0
û ρ 0 ∆ = (10.227)
2
,
−c û 0 û 0 0
0 0 û û
ρ 0 û ρ1 ,
= û 0
(10.228)
−c2 û 0 û c2 û
2
,
(10.229)
= û û(û ) − ρ
ρ
= û2 (û2 − c2 ).
(10.230)
As ∆ appears in the denominator after application of Cramer’s rule, we see immediately that
when the waveframe velocity û becomes locally sonic (û = c) that our system of differential
equations is potentially singular.
11
Gabriel Cramer, 1704-1752, Swiss mathematician.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Now, by Cramer’s rule, dρ/dx̂ is found via
0
ρ
0
0
1
0
û ρ 0 ρc2 σr 0 û 0 r
0 0 û dρ
=
,
dx̂
∆
0
ρ 0 û ρ1 û 0
ρc2 σr 0 û ,
=
û2 (û2 − c2 )
û(−ρ)(−c2 σr)
=
,
û2 (û2 − c2 )
ρc2 σr
.
=
û(û2 − c2 )
353
(10.231)
(10.232)
(10.233)
(10.234)
Similarly, solving for dû/dx̂ we get
dû
=
dx̂
=
=
=
For dP/dx̂, we get
dP
dx̂
=
=
û
0
0
0
1
0
0 0
ρ
−c2 û ρc2 σr û 0 0
r
0 û ,
∆
û
0
0 1 0
û 0
ρ −c2 û ρc2 σr û ,
û2 (û2 − c2 )
û2 (−c2 σr)
,
û2 (û2 − c2 )
c2 σr
− 2
.
û − c2
ρ
0
0 û
0
0 2
0 ρc σr 0 0
r
û ,
∆
û
ρ
0 0
û
0 2
−c û 0 ρc2 σr ,
û2 (û2 − c2 )
û
0
−c2 û
0
û (10.235)
(10.236)
(10.237)
(10.238)
(10.239)
(10.240)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
354
ûρc2 σrû2
,
û2 (û2 − c2 )
ûρc2 σr
.
= 2
û − c2
(10.241)
=
(10.242)
Lastly, we have by inspection
dλ
r
= .
dx̂
û
(10.243)
We can employ the local Mach number in the steady wave frame,
M̂2 ≡
û2
,
c2
(10.244)
to write our system of ordinary differential equations as
dρ
dx̂
dû
dx̂
dP
dx̂
dλ
dx̂
= −
ρσr
û(1 − M̂2 )
σr
=
,
1 − M̂2
ρûσr
= −
,
1 − M̂2
r
=
.
û
,
(10.245)
(10.246)
(10.247)
(10.248)
We note the when the flow is locally sonic, M̂ = 1, that our equations are singular. Recalling an analogy from compressible flow with area change, which is really an application of
l’Hôpital’s rule, in order for the flow to be locally sonic, we must insist that simultaneously
the numerator must be zero; thus, we might demand that at a sonic point, σr = 0. So,
an end state with r = 0 may in fact be a sonic point. For multi-step reactions, each with
their own thermicity and reaction progress, we require the generalization σ · r = 0 at a local
sonic point. Here, σ is the vector of thermicities and r is the vector of reaction rates. This
condition will be important later when we consider so-called eigenvalue detonations.
10.2.2.3
Reduced non-conservative form
Let us use the mass equation, Eq. (10.211) to simplify the reaction equation, Eq. (10.214),
then integrate our differential equations for mass momentum and energy conservation, apply
the initial conditions, and thus reduce our system of four differential and two algebraic
equations to one differential and five algebraic equations:
ρû = −ρo D,
P + ρû2 = Po + ρo D 2 ,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(10.249)
(10.250)
355
1 2 P
ρû e + û +
2
ρ
dλ
dx̂
e
r
1 2 Po
= −ρo D eo + D +
,
2
ρo
1
=
r,
û
= e(P, ρ, λ),
= r(P, ρ, λ).
(10.251)
(10.252)
(10.253)
(10.254)
With some effort, we can unravel these equations to form one ordinary differential equation in
one unknown. But let us delay that analysis until after we have examined the consequences
of the algebraic constraints
10.2.3
Rankine-Hugoniot analysis
Let us first analyze our steady mass, momentum, and energy equations, Eqs. (10.249-10.251).
Our analysis here will be valid both within, (0 < λ < 1) and at the end of the reaction zone
(λ = 1).
10.2.3.1
Rayleigh line
Let us get what is known as the Rayleigh12 line by considering only the mass and linear
momentum equations, Eqs. (10.249-10.250). Let us first rewrite Eq. (10.250) as
P+
ρ2 û2
ρ2 D 2
= Po + o .
ρ
ρo
(10.255)
Then, the mass equation, Eq. (10.22) allows us to rewrite the momentum equation, Eq. (10.255)
as
P+
ρ2o D 2
ρ2 D 2
= Po + o .
ρ
ρo
(10.256)
Rearranging to solve for P , we find
P = Po −
ρ2o D 2
1
1
−
ρ ρo
.
(10.257)
In terms of v = 1/ρ, and a slight rearrangement, Eq. (10.257) can be re-stated as
D2
(v − vo ) ,
vo2
D2
v
= 1−
−1 .
Po vo vo
P = Po −
P
Po
(10.258)
(10.259)
Note:
12
John William Strutt, 3rd Baron Rayleigh, 1842-1919, English physicist.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
356
P HPaL
8. ´ 106
6. ´ 106
4. ´ 106
2. ´ 106
O
0.5
1.0
1ρ Hm3 kgL
Figure 10.1: Plot of Rayleigh line for parameters of Table 10.1 and D = 2800 m/s. Slope
is −ρ2o D 2 < 0.
• This is a line in (P, 1/ρ) space, the Rayleigh line.
• The slope of the Rayleigh line is strictly negative.
• The magnitude of the slope of the Rayleigh line is proportional the square of the wave
speed; high wave speeds induce steep slopes.
• The Rayleigh line passes through the ambient state (Po , 1/ρo ).
• Small volume leads to high pressure.
• These conclusions are a consequence of mass and momentum conservation alone. No
consideration of energy has been made.
• The Rayleigh line equation is valid at all stages of the reaction: the inert state, a
shocked stated, an intermediate reacted state, and a completely reacted state. It is
always the same line.
A plot of a Rayleigh line for D = 2800 m/s and the parameters of Table 10.1 is shown
in Fig. 10.1. Here, the point labeled “O” is the ambient state (1/ρo , Po ).
10.2.3.2
Hugoniot curve
Let us next focus on the energy equation, Eq. (10.251). We shall use the mass and momentum
equations (10.249,10.250) to cast the energy equation in a form which is independent of both
CC BY-NC-ND. 30 March 2014, J. M. Powers.
357
velocities and wave speeds. From Eq. (10.249), we can easily see that Eq. (10.251) first
reduces to
P
1
Po
1
= eo + D 2 + .
(10.260)
e + û2 +
2
ρ
2
ρo
Now, the mass equation (10.249) also tells us that
ρo
û = − D,
ρ
(10.261)
so Eq. (10.260) can be recast as
2
1 ρo
P
1
Po
e+
D2 +
= eo + D 2 + ,
2 ρ
ρ
2
ρo
!
2
1
Po
ρo
P
e − eo + D 2
= 0,
−1 + −
2
ρ
ρ
ρo
P
Po
1 2 (ρo − ρ)(ρo + ρ)
+ −
= 0.
e − eo + D
2
2
ρ
ρ
ρo
Now, the Rayleigh line, Eq. (10.257) can be used to solve for D 2 ,
−1
Po − P 1
ρρo
1
Po − P
2
D =
.
−
=
ρ2o
ρ ρo
ρ2o
ρo − ρ
Now, use Eq. (10.265) to eliminate D 2 in Eq. (10.264):
ρρo
P
Po
(ρo − ρ)(ρo + ρ)
1 Po − P
+ −
e − eo +
2
2
2
ρo
ρ −ρ
ρ
ρ
ρo
|
{z o
}
=D 2
1 Po − P
P
ρo + ρ
Po
e − eo +
+ −
2
ρo
ρ
ρ
ρo
1
P
1
1
Po
e − eo + (Po − P )
+ −
+
2
ρ ρo
ρ
ρo
1P
P
Po
1 Po 1 Po 1 P
+
−
−
+ −
e − eo +
2 ρ
2 ρo
2ρ
2 ρo
ρ
ρo
1 Po Po P
P
e − eo +
−
+ −
2 ρ
ρo
ρ
ρo
1
1
1
e − eo + (P + Po )
−
2
ρ ρo
Rearranging, we get
P + Po
=
−
×
e − eo
| {z }
2
| {z }
change in energy -average pressure
1
1
−
.
ρ ρo
| {z }
change in volume
(10.262)
(10.263)
(10.264)
(10.265)
= 0,
(10.266)
= 0,
(10.267)
= 0,
(10.268)
= 0,
(10.269)
= 0,
(10.270)
= 0.
(10.271)
(10.272)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
358
This is the Hugoniot equation for a general material. It applies for solid, liquid, or gas.
Note that
• This form of the Hugoniot does not depend on the state equation.
• The Hugoniot has no dependency on particle velocity or wave speed.
• The Hugoniot is valid for all e(λ); that is it is valid for inert, partially reacted, and
totally reacted material. Different degrees of reaction λ will induce shifts in the curve.
Now, let us specify an equation of state. For convenience and to more easily illustrate
the features of the Rankine-Hugoniot analysis, let us focus on the simplest physically-based
state equation, the calorically perfect ideal gas for our simple irreversible kinetics model.
With that we adopt the perfect caloric state equation studied earlier, Eq. (8.7):
e = cv T − λq.
(10.273)
We also adopt an ideal gas assumption
P = ρRT.
(10.274)
Solving for T , we get T = P/ρ/R. Thus, cv T = (cv /R)(P/ρ). Recalling that R = cP − cv ,
we then get cv T = (cv /(cP − cv ))(P/ρ). And recalling that for the calorically perfect ideal
gas
cP
γ= ,
(10.275)
cv
we get
cv T =
1 P
.
γ −1 ρ
(10.276)
Thus, our caloric equation of state, Eq. (10.273), for our simple model becomes
e(P, ρ, λ) =
1 P
− λq.
γ −1 ρ
(10.277)
e(P, v, λ) =
1
P v − λq.
γ−1
(10.278)
In terms of v, Eq. (10.277) is
As an aside, we note that for this equation of state, the sound speed can be deduced from
Eq. (10.16) as
∂e 1
P
+
P + γ−1
P
γP
∂v
P,λ
2 2
=
.
(10.279)
=
ρc =
1
∂e v
v
∂P v,λ
c2 = γ
CC BY-NC-ND. 30 March 2014, J. M. Powers.
γ−1
P
P 1
= γP v = γ .
2
vρ
ρ
(10.280)
359
For the thermicity, we note that
R
R e + λq
P = ρRT = T =
,
v
v
c
v
Rq
∂P =
= (γ − 1)ρq,
∂λ v,e cv v
1 ∂P 1
q
γ − 1 ρq
σ= 2
= 2 (γ − 1)ρq = (γ − 1) 2 =
.
ρc ∂λ v,e ρc
c
γ P
(10.281)
(10.282)
(10.283)
Now, at the initial state, we have λ = 0, and so Eq. (10.277) reduces to
eo =
1 Po
.
γ − 1 ρo
(10.284)
We now use our caloric state relations, Eqs. (10.277,10.284) to specialize our Hugoniot
relation, Eq. (10.272), to
P + Po 1
P
Po
1
1
− λq = −
.
(10.285)
−
−
γ−1 ρ
ρo
2
ρ ρo
Employing v = 1/ρ, we get a more compact form:
1
1
(P v − Po vo ) − λq = − (P + Po ) (v − vo ) .
γ −1
2
(10.286)
Let us next operate on Eq. (10.285) so as to see more clearly how the Hugoniot is represented
in the (P, v) = (P, 1/ρ) plane.
Note that as
1
1
(P v − Po vo ) + (P v + Po v − P vo − Po vo ) = λq,
γ −1
2
1
1
1
1
1
1
P
v + v − vo − Po
vo − v + vo
= λq,
γ−1
2
2
γ−1
2
2
P γ+1
Po γ + 1
v − vo −
vo − v
= λq,
2 γ−1
2 γ−1
γ+1
vo − v
2λq + Po γ−1
P =
.
γ+1
v
−
v
o
γ−1
v→
γ −1
vo ,
γ+1
P → ∞.
(10.287)
(10.288)
(10.289)
(10.290)
(10.291)
For γ = 7/5, this gives v → vo /6 induces and infinite pressure. In fact an ideal gas cannot
be compressed beyond this limit, known as the strong shock limit. Note also that as
v → ∞,
−
γ −1
Po < 0.
γ+1
(10.292)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
360
So, very large volumes induce non-physical pressures.
Let us continue to operate to get a more compact form for the calorically perfect ideal
gas Hugoniot curve. Let us add a common term to both sides of Eq. (10.290):
γ+1
v
−
v
2λq
+
P
o γ−1 o
γ−1
γ −1
P+
Po =
Po , (10.293)
+
γ+1
γ+1
γ+1
v
−
v
o
γ−1
γ−1
γ+1
γ+1
P+
Po
v − vo
= 2λq + Po
vo − v
γ+1
γ−1
γ −1
γ+1
γ−1
Po
v − vo ,
(10.294)
+
γ+1
γ−1
γ+1
Po vo − Po v
= 2λq +
γ−1
γ−1
+Po v −
Po vo ,
(10.295)
γ+1
γ−1
γ+1
Po vo −
Po vo ,
(10.296)
= 2λq +
γ−1
γ+1
4γ
Po vo ,
(10.297)
= 2λq + 2
γ −1
γ+1 v
γ−1
2λq
4γ
P
+
−1
=
+ 2
,
(10.298)
Po γ + 1
γ − 1 vo
Po vo γ − 1
P
v
γ − 1 2λq
γ−1
γ−1
4γ
=
+
−
+
.
(10.299)
Po γ + 1
vo γ + 1
γ + 1 Po vo (γ + 1)2
Equation (10.299) represents a hyperbola in the (P, v) = (P, 1/ρ) plane. As λ proceeds from
0 to 1, the Hugoniot moves. A plot of a series of Hugoniot curves for values of λ = 0, 1/2, 1
along with two Rayleigh lines for D = 2800 m/s and D = 1991.1 m/s is shown in Fig. 10.2.
The D = 1991.1 m/s Rayleigh line happens to be exactly tangent to the λ = 1 Hugoniot
curve. We will see this has special significance.
Let us define, for convenience, the parameter µ̂2 as
µ̂2 ≡
γ−1
.
γ+1
With this definition, we note that
2
4γ
γ 2 + 2γ + 1 − γ 2 + 2γ − 1
γ−1
4
=
.
=
1 − µ̂ = 1 −
2
γ+1
(γ + 1)
(γ + 1)2
With this definition, Eq. (10.299), the Hugoniot, becomes
λq
v
P
2
2
= 2µ̂2
+ µ̂
− µ̂
+ 1 − µ̂4 .
Po
vo
Po vo
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(10.300)
(10.301)
(10.302)
361
P HPaL
8. ´ 106
6. ´ 106
N
S
4. ´ 106
N
2. ´ 106
C
W
0.5
1.0
O
1ρ Hm3 kgL
Figure 10.2: Plot of λ = 0, 1/2, 1 Hugoniot curves and two Rayleigh lines for D = 2800 m/s,
D = 1991.1 m/s and parameters of Table 10.1.
Now, let us seek to intersect the Rayleigh line with the Hugoniot curve and find the points
of intersection. To do so, let us use the Rayleigh line, Eq. (10.259), to eliminate pressure in
the Hugoniot, Eq. (10.302):
v
λq
v
D2
2
2
= 2µ̂2
− 1 + µ̂
− µ̂
+ 1 − µ̂4 .
1−
Po vo vo
vo
Po vo
(10.303)
We now structure Eq. (10.303) so that it can be solved for v:
D2 v
λq
D2
v
2
2
1 + µ̂ −
= 2µ̂2
+
− µ̂
+ 1 − µ̂4 .
Po vo vo Po vo
vo
Po vo
(10.304)
Now, let us regroup to form
2 D2
D2
v
D2
2
2
2
−
+
(1 + µ̂ ) 1 +
− µ̂ 1 + µ̂ +
Po vo
vo
Po vo
Po vo
λq
−2µ̂2
− 1 + µ̂4 = 0,
Po vo
(10.305)
2 D2
D2
D2
λq
v
v
−
+
(1 + µ̂2 ) 1 +
− µ̂2 1 +
− 2µ̂2
− 1 = 0.
vo
Po vo
vo
Po vo
Po vo
Po vo
(10.306)
v
vo
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362
This quadratic equation has two roots:
D2
1 + Po vo (1 + µ̂2 )
v
=
2
vo
2 PDo vo
r
2
2
2
1 + PDo vo (1 + µ̂2 )2 − 4 PDo vo 1 + 1 +
±
2
2 PDo vo
D2
Po vo
µ̂2 − 2µ̂2 Pλq
o vo
.
(10.307)
For a given wave speed D, initial undisturbed conditions, Po , vo , and material properties,
µ̂, q, Eq. (10.307) gives the specific volume as a function of reaction progress λ. Depending
on these parameters, we can mathematically expect two distinct real solutions, two repeated
solutions, or two complex solutions.
10.2.4
Shock solutions
We know from Eq. (10.210) that λ does not change through a shock. So, if λ = 0 before
the shock, it has the same value after. So, we can get the shock state by enforcing the
Rankine-Hugoniot jump conditions with λ = 0:
v
vo
=
±
1+
D2
Po vo
(1 + µ̂2 )
2
2 PDo vo
r
2
2
2
1 + PDo vo (1 + µ̂2 )2 − 4 PDo vo 1 + 1 +
D2
Po vo
2
2 PDo vo
µ̂2
.
(10.308)
With considerable effort, or alternatively, by direct calculation via computational algebra,
the two roots of Eq. (10.308) can be shown to reduce to:
v
= 1,
vo
Po vo
v
= µ̂2 + 2 (1 + µ̂2 ).
vo
D
(10.309)
(10.310)
The first is the ambient solution; the second is the shock solution. The shock solution can
also be expressed as
v
γ−1
2γ Po vo
=
+
.
vo
γ + 1 γ + 1 D2
(10.311)
In the limit as Po vo /D 2 → 0, the so-called strong shock limit, we find
γ−1
v
→
.
vo
γ+1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(10.312)
363
The reciprocal gives the density ratio in the strong shock limit:
γ+1
ρ
→
.
(10.313)
ρo
γ−1
With the solutions for v/vo , we can employ the Rayleigh line, Eq. (10.259) to get the
pressure. Again, we find two solutions:
P
= 1,
(10.314)
Po
P
D2
=
(1 − µ̂2 ) − µ̂2 .
(10.315)
Po
Po vo
The first is the inert solution, and the second is the shock solution. The shock solution is
rewritten as
2 D2
γ−1
P
=
−
.
(10.316)
Po
γ + 1 Po vo γ + 1
In the strong shock limit, Po vo /D 2 → 0, the shock state reduces to
2 D2
P
→
.
Po
γ + 1 Po vo
(10.317)
Note, this can also be rewritten as
2γ D 2
P
→
,
Po
γ + 1 γPo vo
2γ D 2
→
.
γ + 1 c2o
(10.318)
(10.319)
Lastly, the particle velocity can be obtained via the mass equation, Eq. (10.261):
ρo
û
= − ,
(10.320)
D
ρ
v
= − ,
(10.321)
vo
2γ Po vo
γ−1
−
.
(10.322)
= −
γ + 1 γ + 1 D2
In the strong shock limit, Po vo /D 2 → 0, we get
û
γ−1
→−
.
(10.323)
D
γ+1
Note that in the laboratory frame, one gets
γ −1
u−D
→ −
,
(10.324)
D
γ+1
u
γ −1
→ 1−
,
(10.325)
D
γ+1
2
u
→
.
(10.326)
D
γ+1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
364
10.2.5
Equilibrium solutions
Our remaining differential equation, Eq. (10.252) is in equilibrium when r = 0, which for
one-step irreversible kinetics, Eq. (10.12), occurs when λ = 1. So, for equilibrium end states,
we enforce λ = 1 in Eq. (10.307) and get
D2
(1 + µ̂2 )
1
+
Po vo
v
=
2
vo
2 PDo vo
r
2
2
q
D2
D2
D2
2
2
2
1 + Po vo (1 + µ̂ ) − 4 Po vo 1 + 1 + Po vo µ̂ − 2µ̂ Po vo
.
(10.327)
±
2
2 PDo vo
This has only one free parameter, D. There are two solutions for v/vo at complete reaction.
They can be distinct and real, repeated and real, or complex, depending on the value of
D. We are most interested in D for which the solutions are real; these will be physically
realizable.
10.2.5.1
Chapman-Jouguet solutions
Let us first consider solutions for which the two roots of Eq. (10.327) are repeated. This
is known as a Chapman13 -Jouguet14 (CJ) solution. For a CJ solution, the Rayleigh line is
tangent to the Hugoniot at λ = 1 if the reaction is driven by one-step irreversible kinetics.
We can find values of D for which the solutions are CJ by requiring the discriminant
under the square root operator in Eq. (10.327) to be zero. We label such solutions with a
CJ subscript and say
2
2
2
2
DCJ
DCJ
DCJ
2 2
2
2 q
1+
1 + µ̂ − 4
1+ 1+
µ̂ − 2µ̂
= 0.
(10.328)
Po vo
Povo
Po vo
Po vo
2
Equation (10.328) is quartic in DCJ and quadratic in DCJ
. It has solutions
q
2q
4
2 q
2
−
µ̂
±
2
µ̂2 (1 + 2µ̂2 − µ̂4 )
1
+
4µ̂
DCJ
Po vo
Po vo
=
.
Po vo
(µ̂2 − 1)2
(10.329)
The “+” root corresponds to a large value of DCJ . This is known as the detonation branch.
For the parameter values of Table 10.1, we find by substitution that DCJ = 1991.1 m/s
for our H2 -air mixture. It corresponds to a pressure increase and a volume decrease. The
“-” root corresponds to a small value of DCJ . It corresponds to a pressure decrease and
a volume increase. It is known as the deflagration branch. For our H2 -air mixture, we
find DCJ = 83.306 m/s. Here, we are most concerned with the detonation branch. The
13
14
David Leonard Chapman, 1869-1958, English chemist.
Jacques Charles Émile Jouguet, 1871-1943, French engineer.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
365
deflagration branch may be of interest, but neglected mechanisms, such as diffusion, may
be of more importance for this branch. In fact laminar flames in hydrogen move much
slower than that predicted by the CJ deflagration speed. We also note that for q → 0, that
2
2
DCJ
/(Po vo ) = (1 + µ̂2 )/(1 − µ̂2 ) = γ. Thus, for q → 0, we have DCJ
→ γPo vo , and the wave
speed is the ambient sound speed.
2
Taylor series expansion of the detonation branch in the strong shock limit, Po vo /DCJ
→0
shows that
2
DCJ
→ 2q(γ 2 − 1),
γ
vCJ →
vo ,
γ+1
q
PCJ → 2(γ − 1) ,
vo
γ p
ûCJ → −
2q(γ 2 − 1),
γ+1
p
2q(γ 2 − 1)
.
uCJ →
γ+1
(10.330)
(10.331)
(10.332)
(10.333)
(10.334)
Importantly the Mach number in the wave frame at the CJ state is
û2CJ
,
c2CJ
û2CJ
,
=
γPCJ vCJ
M̂2CJ =
=
=
=
(10.335)
(10.336)
γ2
2q(γ 2 − 1)
(γ+1)2
γ ,
γ v2qo (γ − 1)vo γ+1
γ2
(γ 2
(γ+1)2
− 1)
γ
γ(γ − 1) γ+1
γ2
(γ
(γ+1)2
= 1.
(10.337)
,
+ 1)(γ − 1)
γ
γ(γ − 1) γ+1
(10.338)
,
(10.339)
(10.340)
In the strong shock limit, the local Mach number in the wave frame is sonic at the end of
the reaction zone. This can be shown to hold away from the strong shock limit as well.
10.2.5.2
Weak and strong solutions
For D < DCJ there are no real solutions on the detonation branch. For D > DCJ , there are
two real solutions. These are known as the weak and strong solution. These solutions are
represent the intersection of the Rayleigh line with the complete reaction Hugoniot at two
points. The higher pressure solution is known as the strong solution. The lower pressure
CC BY-NC-ND. 30 March 2014, J. M. Powers.
366
D HmsL
6000
5000
4000
3000
2000
D CJ
1000
0
1000
2000
3000
4000
u f HmsL
Figure 10.3: Plot of detonation wave speed D versus piston velocity uf ; parameters from
Table 10.1 loosely model H2 -air detonation.
solution is known as the weak solution. Equilibrium end state analysis cannot determine
which of the solutions, strong or weak, is preferred if D > DCJ .
We can understand much about detonations, weak, strong, and CJ, by considering how
they behave as the final velocity in the laboratory frame is changed. We can think of the final
velocity in the laboratory frame as that of a piston which is pushing the detonation. While
we could analyze this on the basis of the theory we have already developed, the algebra is
jump equations, Eqs. (10.249-10.251) with caloric state equation, Eq. (10.277) with the final
state, denoted by the subscript f , being λ = 1 and the initial state being λ = 0, Eq. (10.249)
being used to simplify Eq. (10.251), and the laboratory frame velocity u used in place of û:
ρf (uf − D) = −ρo D,
Pf + ρf (uf − D)2 = Po + ρo D 2 ,
1 Pf
1
1 Po 1 2 Po
Pf
− q + (uf − D)2 +
=
+ D + .
γ − 1 ρf
2
ρf
γ − 1 ρo 2
ρo
| {z }
{z
}
|
ef
(10.341)
(10.342)
(10.343)
=eo
Let us consider the unknowns to be Pf , ρf , and D. Computer algebra solution of these three
equations yields two sets of solutions. The relevant physical branch has a solution for D of


v
!2
u
u P
γ + 1 γ − 1 q

tγ o + γ + 1 + γ − 1 q
D = uf 
+
(10.344)
+
.
2
2
2
4
2 uf
ρo uf
4
2 uf
We give a plot of D as a function of the supporting piston velocity uf in Figure 10.3.
We notice on Fig. 10.3 that there is a clear minimum D. This value of D is the CJ value of
DCJ = 1991.1 m/s. It corresponds to a piston velocity of uf = 794.9 m/s.
A piston driving at uf = 794.9 m/s will just drive the wave at the CJ speed. At this
piston speed, all of the energy to drive the wave is supplied by the combustion process itself.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
367
As uf increases beyond 794.9 m/s, the wave speed D increases. For such piston velocities,
the piston itself is supplying energy to drive the wave. For uf < 794.9 m/s, our formula
predicts an increase in D, but that is not what is observed in experiment. Instead, a wave
propagating at DCJ is observed.
We last note p
that in the inert, q = 0, small piston velocity, uf → 0, limit that Eq. (10.344)
reduces to D = γPo /ρo . That is, the wave speed is the ambient sound speed.
10.2.5.3
Summary of solution properties
Here is a summary of the properties of solutions for various values of D that can be obtained
by equilibrium end state analysis:
• D < DCJ : No Rayleigh line intersects a complete reaction Hugoniot on the detonation
branch. There is no real equilibrium detonation solution.
• D = DCJ : There is a two repeated solutions at a single point, which we will call C, the
Chapman-Jouguet point. At C, the Rayleigh line is tangent to the complete reaction
Hugoniot. Some properties of this solution are
– ûCJ /cCJ = 1; the flow is sonic in the wave frame at complete reaction.
– This is the unique speed of propagation of a wave without piston support if the
reaction is one-step irreversible.
– At this wave speed, D = DCJ , all the energy from the reaction is just sufficient
to drive the wave forward.
– Because the end of the reaction zone is sonic, downstream acoustic disturbances
cannot overtake the reaction zone.
• D > DCJ ; Two solutions are admitted at the equilibrium end state, the strong solution
at point S, and the weak solution at point W .
– The strong solution S has
∗ û/c < 1, subsonic.
∗ piston support is required to drive the wave forward
∗ some energy to drive the wave comes from the reaction, some comes from the
piston.
∗ if the piston support is withdrawn, acoustic disturbances will overtake and
weaken the wave.
– The weak solution W has
∗ û/c > 1, supersonic
∗ often thought to be non-physical, at least for one-step irreversible kinetics
because of no initiation mechanism.
∗ exceptions exist.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
368
10.2.6
ZND solutions: One-step irreversible kinetics
We next consider the structure of the reaction zone. Structure was first considered contemporaneously and independently by Zel’dovich, von Neumann15 and Döring16 Equation (10.307)
gives v(λ) for either the strong or weak branches of the solution. Knowing v(λ) and thus
ρ(λ), since v = 1/ρ, we can use the integrated mass equation, Eq. (10.249) to write an
explicit equation for û(λ). Our Rankine-Hugoniot analysis also give T (λ). These can be
employed in the reaction kinetics equation, Eq. (10.252, 10.13) to form a single ordinary
differential equation for the evolution of λ of the form
E
β
a(T
(λ))
exp
−
(1 − λ)
RT (λ)
dλ
=
,
λ(0) = 0.
(10.345)
dx̂
û(λ)
We consider the initial unshocked state to be labeled O. We label the point after the shock
N for the Neumann point, named after von Neumann, one of the pioneers of detonation
theory. Recall λ = 0 both at O and at N. But the state variables, e.g. P , ρ, û, change from
O to N.
Before we actually solve the differential equations, we can learn much by considering
how P varies with λ in the reaction zone by using Rankine-Hugoniot analysis. Consider the
Rankine-Hugoniot equations, Eqs. (10.249-10.251) with caloric state equation, Eq. (10.277),
Eq. (10.249) being used to simplify Eq. (10.251):
ρû = −ρo D,
P + ρû2 = Po + ρo D 2 ,
P
1 Po 1 2 Po
1 P
1
− λq + û2 +
=
+ D + .
γ −1 ρ
2
ρ
γ−1ρ
2
ρo
|
| {z o}
{z
}
e
(10.346)
(10.347)
(10.348)
=eo
Computer algebra reveals the solution for P (λ) to be



s
2
2
1 
2(γ − 1)λq 
γPo
P (λ) =
Po + ρo D 2 1 ±
.
−
1−
2
γ+1
ρo D
D2
(10.349)
In Fig. (10.4), we plot P versus λ for three different values of D: D = 2800 m/s > DCJ ,
D = 1991.1 m/s = DCJ , D = 1800 m/s < DCJ .
We can also, via detailed algebraic analysis get an algebraic expression for M̂ as a function
of λ. We omit that here, but do get a plot for our system. In Fig. (10.5), we plot M̂
versus λ for three different values of D: D = 2800 m/s > DCJ , D = 1991.1 m/s = DCJ ,
D = 1800 m/s < DCJ . The plot of M̂ versus λ is log-log so that the sonic condition may be
clearly exhibited.
15
16
John von Neumann, 1903-1957, Hungarian-American genius.
Werner Döring, 1911-2006, German physicist.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
369
P HPaL
N
5. ´ 106
D > D CJ
S
4. ´ 106
3. ´ 106
N
D = D CJ
N
2. ´ 106
D < D CJ
C
1. ´ 106
W
D > D CJ
O
0.2
0.4
0.6
0.8
1.0
λ
Figure 10.4: P versus λ from Rankine-Hugoniot analysis for one-step irreversible reaction
for D = 2800 m/s > DCJ , D = 1991.1 m/s = DCJ , D = 1800 m/s < DCJ for H2 /air-based
parameters of Table 10.1.
`
M
5.0
O
O
D > D CJ
O
3.0
D = D CJ
D < D CJ
W
2.0
1.5
C
1.0
D < D CJ
S
N
0.01
D > D CJ
0.02
0.05
0.10
0.20
0.50
1.00
λ
Figure 10.5: M̂ versus λ from Rankine-Hugoniot analysis for one-step irreversible reaction
for D = 2800 m/s > DCJ , D = 1991.1 m/s = DCJ , D = 1800 m/s < DCJ for H2 /air-based
parameters of Table 10.1.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
370
For D = 2800 m/s > DCJ , there are two branches, the strong and the weak. The
strong branch commences at N where λ = 0 and proceeds to decrease to λ = 1 where the
equilibrium point S is encountered at a subsonic state. There other branch commences at
O and pressure increases until the supersonic W is reached at λ = 1.
For D = 1991.1 m/s = DCJ , the behavior is similar, except that the branches commencing at N and O both reach complete reaction at the same point C. The point C can be shown
to be sonic with M̂ = 1. We recall from our earlier discussion regarding Eqs. (10.245-10.248)
that sonic points are admitted only if σr = 0. For one-step irreversible reaction, r = 0 when
λ = 1, so a sonic condition is admissible.
For D = 1800 m/s < DCJ , the strong and weak branches merge at a point of incomplete
reaction. At the point of merger, near λ = 0.8, the flow is locally sonic; however, this is not
a point of complete reaction, so there can be no real-valued detonation structure for this
value of D.
Finally, we write an alternate differential-algebraic equations which can be integrated for
the detonation structure:
ρû = −ρo D,
P + ρû2 = Po + ρo D 2 ,
1
P
1
Po
e + û2 +
= eo + D 2 + ,
2
ρ
2
ρo
1 P
− λq,
e =
γ −1 ρ
P = ρRT,
dλ
1−λ
E
.
=
a exp −
dx̂
û
RT
(10.350)
(10.351)
(10.352)
(10.353)
(10.354)
(10.355)
We need the condition λ(0) = 0. These form six equations for the six unknowns ρ, û, P , e,
T , and λ.
10.2.6.1
CJ ZND structures
We now fix D = DCJ = 1991.1 m/s and integrate Eq. (10.345) from the shocked state N
to a complete reaction point, C, the Chapman-Jouguet detonation state. We could also
integrate from O to C, but this is not observed in nature. After obtaining λ(x̂), we can use
our Rankine-Hugoniot analysis results to plot all state variables as functions of x̂.
In Fig. (10.6), we plot the density and pressure versus x̂. The density first jumps discontinuously from O to its shocked value at N. From there it slowly drops through the reaction
zone until it relaxes near x̂ ∼ −0.01 m to its equilibrium value at complete reaction. Thus,
the reaction zone has a thickness of roughly 1 cm. Similar behavior is seen for the pressure.
The wave frame-based fluid particle velocity is shown in Fig. 10.7. Since the unshocked
fluid is at rest in the laboratory frame with u = 0 m/s, the fluid in the wave frame has
velocity û = 0 − 1991.1 m/s = −1991.1 m/s. It is shocked to a lower velocity and then
relaxes to its equilibrium value.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
371
ρ Hkgm3 L
5
P HPaL
4
2.5 106
3
2. 106
1.5 106
2
1. 106
1
0.015
5. 105
0.010
0.005
0.000
x (m)
0.010
0.005
0.000
x (m)
Figure 10.6: ZND structure of ρ(x̂) and P (x̂) for D = DCJ = 1991.1 m/s for one-step
irreversible reaction for H2 /air-based parameters of Table 10.1.
u HmsL
u (m/s)
0.010
0.005
0.000
x (m)
1500
500
1000
1000
500
1500
2000
0
0.010
0.005
0.000
x (m)
Figure 10.7: ZND structure of wave frame-based fluid particle velocity û(x̂) and laboratory
frame-based particle velocity u(x̂) for D = DCJ = 1991.1 m/s for one-step irreversible
reaction for H2 /air-based parameters of Table 10.1.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
372
M
5
T HK L
2500
4
2000
3
1500
2
1000
1
500
0.010
0.005
0.000
x (m)
0.015
0.010
0.005
0.000
x (m)
Figure 10.8: ZND structure of T (x̂) and wave frame-based Mach number M̂ for D = DCJ =
1991.1 m/s for one-step irreversible reaction for H2 /air-based parameters of Table 10.1.
The structure of the velocity profiles is easier to understand in the laboratory frame, as
also shown in Fig. 10.7. Here, we see the unshocked fluid velocity of u = 0 m/s. The fluid is
shocked to a high velocity, which then decreases to a value at the end of the reaction zone.
The final velocity can be associated with that of a supporting piston, uf = 794.9 m/s.
The temperature and wave frame-based Mach number are plotted in Fig. 10.8. It is
shocked from 298 K to a high value, then continues to mainly increase through the reaction
zone. Near the end of the reaction zone, there is a final decrease as it reaches its equilibrium
value. The Mach number calculated in the wave frame, M̂, goes from an initial value of
M̂ = 4.88887, which we call the CJ detonation Mach number, MCJ = 4.88887, to a postshock value of M = 0.41687. Note this result confirms a standard result from compressible
flow that a standing normal shock must bring a flow from a supersonic state to a subsonic
state. At equilibrium it relaxes to M = 1. This relaxation to a sonic state when λ = 1 is
what defines the CJ state. We recall that this result is similar to that obtained in so-called
“Rayleigh flow” of one-dimensional gas dynamics. Rayleigh flow admits heat transfer to a
one-dimensional channel, and it is well known that the addition of heat always induces the
flow to move to a sonic (or “choked”) state. So, we can think of the CJ detonation wave as
a thermally choked flow.
The reaction progress variable λ is plotted in Fig. 10.9. Note that it undergoes no shock
jump and simply relaxes to its equilibrium value of λ = 1 near x̂ = 0.01 m. Lastly, we
plot T (−x̂) on a log-log scale in Fig. (10.10). The sign of x̂ is reversed so as to avoid the
plotting of the logarithms of negative numbers. We notice on this scale that the temperature
is roughly that of the shock until −x̂ = 0.001 m, at which point a steep rise begins. We call
this length the induction length, ℓind . When we compare this figure to Fig. 2 of Powers and
Paolucci, 2005, we see a somewhat similar behavior. However the detailed kinetics model
shows ℓind ∼ 0.0001 m. The overall reaction zone length ℓrxn is predicted well by the simple
model. Its value of ℓrxn ∼ 0.01 m is also predicted by the detailed model. Some of the
final values at the end state are different as well. This could be due to a variety of factors,
CC BY-NC-ND. 30 March 2014, J. M. Powers.
373
λ
1.0
0.8
0.6
0.4
0.2
0.0
0.015
0.010
0.005
x (m)
0.000
Figure 10.9: ZND structure of λ(x̂) for D = DCJ = 1991.1 m/s for one-step irreversible
reaction for H2 /air-based parameters of Table 10.1.
T HK L
2600
2400
2200
2000
1800
`
- x (m)
10-5
10-4
10-3
10-2
Figure 10.10: ZND structure on a log-log scale of T (x̂) for D = DCJ = 1991.1 m/s for
one-step irreversible reaction for H2 /air-based parameters of Table 10.1.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
374
parameter
ℓrxn
ℓind
DCJ
Ps
PCJ
Ts
TCJ
ρs
ρCJ
M̂o
M̂s
M̂CJ
simple
detailed
−2
10 m
10−2 m
−3
10 m
10−4 m
1991.1 m/s
1979.7 m/s
6
2.80849 × 10 P a 2.8323 × 106 P a
1.4553 × 106 P a 1.6483 × 106 P a
1664.4 K
1542.7 K
2570.86 K
2982.1 K
4.244 kg/m3
4.618 kg/m3
1.424 kg/m3
1.5882 kg/m3
4.88887
4.8594
0.41687
0.40779
1
0.93823
Table 10.2: Numerical values of parameters which roughly model CJ H2 -air detonation.
especially differences in the state equations. Comparisons between values predicted by the
detailed model of Powers and Paolucci, 2005, against those of the simple model here are
given in Table 10.2.
10.2.6.2
Strong ZND structures
We increase the detonation velocity D to D > DCJ and obtain strong detonation structures.
These have a path from O to N to the equilibrium point S. These detonations require piston
support to propagate, as the energy supplied by heat release alone is insufficient to maintain
Similar to our plots of the CJ structures, we give plots of the strong, D = 2800 m/s
structures of ρ(x̂), P (x̂), û(x̂), u(x̂), T (x̂), M̂(x̂), λ(x̂) in Figs. 10.11-10.14, respectively.
The behavior of the plots is qualitatively similar to that for CJ detonations. We see however
that the reaction zone has become significantly thinner, ℓrxn ∼ 0.0001 m. This is because
the higher temperatures associated with the stronger shock induce faster reactions, thus
thinning the reaction zone. Comparison with Fig. 10.2 reveals that the shocked and final
values of pressure agree with those of the Rankine-Hugoniot jump analysis. We also note
the final value of M̂ is subsonic. This allows information to propagate from the supporting
piston all the way to the shock front.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
375
ρ Hkgm3 L
5
P HPaL
4
5. 106
3
4. 106
3. 106
2
2. 106
1
0.0002
1. 106
0.0001
0
x (m)
0.0001
0
x (m)
Figure 10.11: ZND structure of ρ(x̂) and P (x̂) for strong D = 2800 m/s > DCJ for one-step
irreversible reaction for H2 /air-based parameters of Table 10.1.
u HmsL
u (m/s)
0.0001
500
0
x (m)
2000
1500
1000
1000
1500
500
2000
2500
0
0.0001
0
x (m)
Figure 10.12: ZND structure of wave frame-based fluid particle velocity û(x̂) and laboratory frame-based fluid particle velocity u(x̂) for strong D = 2800 m/s > DCJ for one-step
irreversible reaction for H2 /air-based parameters of Table 10.1.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
376
M
7
T HK L
4000
6
5
3000
4
2000
3
2
1000
1
0.0001
x (m)
0
0.0002
0.0001
0
x (m)
Figure 10.13: ZND structure of T (x̂) and M̂(x̂) for strong D = 2800 m/s > DCJ for one-step
irreversible reaction for H2 /air-based parameters of Table 10.1.
λ
1.0
0.8
0.6
0.4
0.2
0.0
0.0002
0.0001
0
x (m)
Figure 10.14: ZND structure of λ(x̂) for strong D = 2800 m/s > DCJ for one-step irreversible reaction for H2 /air-based parameters of Table 10.1.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
377
λ
0.08
0.06
0.04
0.02
3.2249 1021
1.61245 1021
x (m)
Figure 10.15: ZND structure of λ(x̂) for unshocked, weak D = 2800 m/s > DCJ for onestep irreversible reaction for H2 /air-based parameters of Table 10.1. The galactic distance
scales are far too large to be realistic representations of reality!
10.2.6.3
Weak ZND structures
For the simple one-step irreversible kinetics model, there is no path from O through the
shocked state N to the weak solution W . There is a direct path from O to W ; however, it
is physically unrealistic. For D = 2800 m/s, we plot λ versus x̂ in Fig. 10.15. Numerical
solution was available only until λ ∼ 0.02. Numerical precision issues arose at this point.
Note, importantly, that ℓrxn ∼ 1021 m is unrealistically large! Note the distance from Earth
to the Large Magellanic Cloud, a dwarf galaxy orbiting the Milky Way, is 1021 m. Our
combustion model is not well calibrated to those distances, so it is entirely unreliable to
predict this class of weak detonation!
10.2.6.4
Piston problem
We can understand the physics of the one-step kinetics problem better in the context of
a piston problem, where the supporting piston connects to the final laboratory frame fluid
velocity up = u(x̂ → −∞). Let us consider pistons with high velocity and then lower them
and examine the changes of structure.
• up > up,CJ . This high velocity piston will drive a strong shock into the fluid at a speed
D > DCJ . The solution will proceed from O to N to S and be subsonic throughout.
Therefore changes at the piston face will be able to be communicated all the way to
the shock front. The energy to drive the wave comes from a combination of energy
released during combustion and energy supplied by the piston support.
• up = up,CJ At a critical value of piston velocity, up,CJ , the solution will go from O to
CC BY-NC-ND. 30 March 2014, J. M. Powers.
378
N to C, and where it is locally sonic.
• up < up,CJ For such flows, the detonation wave is self-supporting. There is no means to
communicate with the supporting piston. The detonation wave proceeds at D = DCJ .
We note that DCJ is not a function of the specific kinetic mechanism. So, for a one-step
irreversible kinetics model, the conclusion the DCJ is the unique speed of propagation of
an unsupported wave is verified. It should be noted that nearly any complication added
to the model, e.g. reversibility, multi-step kinetics, multi-dimensionality, diffusion, etc.,
will alter this conclusion.
10.2.7
Detonation structure: Two-step irreversible kinetics
Let us consider a small change to the one-step model of the previous sections. We will now
consider a two-step irreversible kinetics model. The first reaction will be exothermic and
the second endothermic. Both reactions will be driven to completion, and when they are
complete, the global heat release will be identical to that of the one-step reaction. All other
parameters will remain the same from the one-step model. This model is discussed in detail
by Fickett and Davis, and in a two-dimensional extension by Powers and Gonthier.17
We will see that this simple modification has profound effects on what is a preferred
detonation structure. In particular, we will see that for such a two-step model
• the CJ structure is no longer the preferred state of an unsupported detonation wave,
• the steady speed of the unsupported detonation wave is unique and greater than the
CJ speed,
• there is a path from the unshocked state O to the shocked state N through a sonic
incomplete reaction pathological point P to the weak equilibrium end state W ,
• there is a strong analog to steady compressible one-dimensional inert flow with area
change, i.e. rocket nozzle flow.
Let us pose the two step irreversible kinetics model of
1 : A → B,
2 : B → C.
(10.356)
(10.357)
Let us insist that A, B, and C each have the same molecular mass, MA = MB = MC = M,
and the same constant specific heats, cP A = cP B = cP C = cP . Let us also insist that both
reactions have the same kinetic parameters, E1 = E2 = E, a1 = a2 = a, β1 = β2 = 0.
Therefore the reaction rates are such that
k1 = k2 = k.
17
(10.358)
Powers, J. M., and Gonthier, K. A., 1992, “Reaction Zone Structure for Strong, Weak Overdriven, and
Weak Underdriven Oblique Detonations,” Physics of Fluids A, 4(9): 2082-2089.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
379
Let us assume at the initial state that we have all A and no B or C: ρA (0) = ρ̂A , ρB (0) = 0,
ρC (0) = 0. Since J = 2, we have a reaction vector rj of length 2:
r1
rj =
.
(10.359)
r2
Our stoichiometric matrix νij has dimension 3 × 2 since N = 3 and J = 2:


−1 0
νij =  1 −1  .
0
1
P
For species production rates, from ω̇i = Jj=1 νij rj we have








ρA /ρ
ω̇A
−1 0
−r1
1
d 
1
1
r1
ρB /ρ  =  ω̇B  =  1 −1 
=  r1 − r2  .
r2
dt
ρ
ρ
ρ
ρC /ρ
ω̇C
0
1
r2
(10.360)
(10.361)
We recall here that d/dt denotes the material derivative following a fluid particle, d/dt =
∂/∂t + û∂/∂ x̂. For our steady waves, we will have d/dt = ûd/dx̂. We next recall that
ρi /ρ = Yi /Mi , which for us is Yi /M, since the molecular masses are constant. So, we have








YA
ω̇A
−1 0
−r1
M
M
M
d 
r1
YB  =
ω̇B  =
1 −1 
=
r1 − r2  .
(10.362)
r
dt
ρ
ρ
ρ
2
YC
ω̇C
0
1
r2
Elementary row operations gives us the row echelon form




YA
−1 0
d 
M
 0 −1  r1 .
YA + YB  =
r2
dt
ρ
YA + YB + YC
0
0
(10.363)
We can integrate the homogeneous third equation and apply the initial condition to get
YA + YB + YC = 1.
This can be thought of as an unusual matrix equation:


YA
( 1 1 1 )  YB  = ( 1 ) .
YC
We can perform an analogous exercise to finding the form ρ = ρ̂ + D · ξ and get

   

YA
1
−1 0
 YB  =  0  +  1 −1  λ1 .
λ2
YC
0
0
1
{z
}
|
(10.364)
(10.365)
(10.366)
=F
CC BY-NC-ND. 30 March 2014, J. M. Powers.
380
The column vectors of F are linearly independent and lie in the right null space of the
coefficient matrix (1, 1, 1). The choices for F are not unique, but are convenient. We can
think of the independent variables λ1 , λ2 as reaction progress variables. Thus, for reaction
1, we have λ1 , and for reaction 2, we have λ2 . Both λ1 (0) = 0 and λ2 (0) = 0. The mass
fraction of each species can be related to the reaction progress via
Y A = 1 − λ1 ,
Y B = λ1 − λ2 ,
Y C = λ2 .
(10.367)
(10.368)
(10.369)
When the reaction is complete, we have λ1 → 1, λ2 → 1, and YA → 0, YB → 0, YC → 1.
Now, our reaction law is








YA
−r1
−kρA
−YA
d 
M
M
YB  =
r1 − r2  =
kρA − kρB  = k  YA − YB  .
(10.370)
dt
ρ
ρ
YC
r2
kρB
YB
Eliminating YA , YB and YC in favor of λ1 and λ2 , we get




1 − λ1
−(1 − λ1 )
d 
λ1 − λ2  = k  (1 − λ1 ) − (λ1 − λ2 )  .
dt
λ2
λ1 − λ2
This reduces to




λ1
1 − λ1
d 
λ1 − λ2  = k  1 − 2λ1 + λ2  .
dt
λ2
λ1 − λ2
(10.371)
(10.372)
The second of these equations is the difference of the first and the third, so it is redundant
and we need only consider
d λ1
1 − λ1
=k
.
(10.373)
λ1 − λ2
dt λ2
In the steady wave frame, this is written as
dλ1
= (1 − λ1 )k,
dx̂
dλ2
= (λ1 − λ2 )k.
û
dx̂
û
(10.374)
(10.375)
Because the rates k1 = k2 = k have been taken to be identical, we can actually get λ2 (λ1 ).
Dividing our two kinetic equations gives
λ1 − λ2
dλ2
=
.
dλ1
1 − λ1
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(10.376)
381
Because λ1 (0) = 0 and λ2 (0) = 0, we can say that λ2 (λ1 = 0) = 0. We rearrange this
differential equation to get
1
λ1
dλ2
+
λ2 =
.
dλ1 1 − λ1
1 − λ1
(10.377)
This equation is first order and linear. It has an integrating factor of
Z
dλ1
1
exp
= exp (− ln(1 − λ1 )) =
.
1 − λ1
1 − λ1
Multiplying both sides by the integrating factor, we get
2
λ1
1
1 dλ2
+
.
λ2 =
1 − λ1 dλ1
1 − λ1
(1 − λ1 )2
Using the product rule, we then get
λ2
λ1
d
=
,
dλ1 1 − λ1
(1 − λ1 )2
Z
λ1
λ2
=
dλ1 .
1 − λ1
(1 − λ1 )2
(10.378)
(10.379)
(10.380)
Taking u = λ1 and dv = dλ1 /(1 − λ1 )2 and integrating the right side by parts, we get
Z
λ1
dλ1
λ2
=
−
,
(10.381)
1 − λ1
1 − λ1
1 − λ1
λ1
=
+ ln(1 − λ1 ) + C,
(10.382)
1 − λ1
λ2 = λ1 + (1 − λ1 ) ln(1 − λ1 ) + C(1 − λ1 ).
(10.383)
Now, since λ2 (λ1 = 0) = 0, we get C = 0, so
λ2 (x̂) = λ1 (x̂) + (1 − λ1 (x̂)) ln(1 − λ1 (x̂)).
(10.384)
Leaving out details of the derivation, our state equation becomes
e(T, λ1 , λ2 ) = cv (T − To ) − λ1 q1 − λ2 q2 .
(10.385)
We find it convenient to define Q(λ1 , λ2 ) as
Q(λ1 , λ2 ) ≡ λ1 q1 + λ2 q2 .
(10.386)
So, the equation of state can be written as
e(T, λ1 , λ2 ) = cv (T − To ) − Q(λ1 , λ2 ).
(10.387)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
382
Parameter
γ
M
R
Po
To
ρo
vo
q1
q2
E
a
β
Value
1.4
20.91
397.58
1.01325 × 105
298
0.85521
1.1693
7.58265 × 106
−5.68698 × 106
8.29352 × 106
5 × 109
0
Units
kg/kmole
J/kg/K
Pa
K
kg/m3
m3 /kg
J/kg
J/kg
J/kg
1/s
Table 10.3: Numerical values of parameters for two-step irreversible kinetics.
The frozen sound speed remains
c2 = γP v = γ
P
.
ρ
(10.388)
There are now two thermicities:
σ1
σ2
1 ∂P γ ρq1
=
,
=
ρc2 ∂λ1 v,e,λ2
γ−1 P
γ ρq2
1 ∂P .
=
=
2
ρc ∂λ2 v,e,λ1
γ−1 P
(10.389)
(10.390)
Parameters for our two step model are identical to those of our one step model, except for
the heat releases. The parameters are listed in Table 10.3. Note that at complete reaction
Q(λ1 , λ2 ) = Q(1, 1) = q1 + q2 = 1.89566 × 106 J/kg. Thus, the overall heat release at
complete reaction λ1 = λ2 = 1 is identical to our earlier one-step kinetic model.
Let us do some new Rankine-Hugoniot analysis. We can write a set of mass, momentum,
energy, and state equations as
ρû = −ρo D,
P + ρû2 = Po + ρo D 2 ,
1
P
1
Po
e + û2 +
= eo + D 2 + ,
2
ρ
2
ρo
1 P
− λ1 q1 − λ2 q2 ,
e =
γ−1 ρ
λ2 = λ1 + (1 − λ1 ) ln(1 − λ1 ).
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(10.391)
(10.392)
(10.393)
(10.394)
(10.395)
383
P HPaL
N
5. ´ 106 N
S

D>D
4. ´ 106
S
N

D =D
3. ´ 106

D <D

D =D
P

D <D

D =D
2. ´ 106

D>D
1. ´ 106
W
O
0.2
0.4
0.6
0.8
1.0
λ1
Figure 10.16: Pressure versus λ1 for two-step kinetics problem for three different values of
D: D = 2800 m/s > D̃, D = 2616.5 m/s = D̃, D = 2500 m/s < D̃; parameters are from
Table 10.3.
Let us consider D and λ1 to be unspecified but known parameters for this analysis. These
equations are five equations for the five unknowns, ρ, û, P , e, and λ2 . They can be solved
for ρ(D, λ1 ), û(D, λ1 ), P (D, λ1), e(D, λ1 ), and λ2 (λ1 ).
The solution is lengthy, but the plot is revealing. For three different values of D, pressure
as a function of λ1 is shown in in Fig. 10.16. There are three important classes of D, each
shown in Fig. 10.16, depending on how D compares to a critical value we call D̃.
• D > D̃. There are two potentially paths here. The important physical branch starts
at point O, and is immediately shocked to state N, the Neumann point. From N the
pressure first decreases as λ1 increases. Near λ1 = 0.75, the pressure reaches a local
minimum, and then increases to the complete reaction point at S. This is a strong
solution. There is a second branch which commences at O and is unshocked. On this
branch the pressure increases to a maximum, then decreases to the end state at W .
While this branch is admissible mathematically, its length scales are unphysically long,
• D = D̃. Let us only consider branches which are shocked from O to N. The unshocked
branches are again non-physical. On this branch, the pressure decreases from N to
the pathological point P . At P , the flow is locally sonic, with M̂ = 1. Here, the
pressure can take two distinct paths. The one chosen will depend on the velocity of
the supporting piston at the end state. On one path the pressure increases to its final
value at the strong point S. On the other the pressure decreases to its final value at
the weak point W .
CC BY-NC-ND. 30 March 2014, J. M. Powers.
384
`
M
O
5.0 O
3.0

D >D
W
2.0

D <D
1.5

D =D
P
1.0

D <D

D =D

D >D
S
0.50
1.00
N
0.01
0.02
0.05
0.10
0.20
λ1
Figure 10.17: Mach number M̂ versus λ1 for two-step kinetics problem for three different
values of D: D = 2800 m/s > D̃, D = 2616.5 m/s = D̃, D = 2500 m/s < D̃; parameters
are from Table 10.3.
• D < D̃. For such values of D, there is no physical structure for the entire reaction
zone 0 < λ1 < 1. This branch is discarded.
Mach number in the wave frame, M̂ as a function of λ1 is shown in in Fig. 10.17. The
results here are similar to those in Fig. 10.16. Note the ambient point O is always supersonic,
and the Neumann point N is always subsonic. For flows originating at N, if D > D̃, the flow
remains subsonic throughout until its termination at S. For D = D̃, the flow can undergo
a subsonic to supersonic transition at the pathological point P . The weak point W is a
supersonic end state.
The important Fig. 10.16 bears remarkable similarity to curves of P (x) in compressible
inert flow in a converging-diverging nozzle. We recall that for such flows, a subsonic to
supersonic transition is only realized at an area minimum. This can be explained because the
equation for evolution of pressure for such flows takes the form dP/dx ∼ (dA/dx)/(1 − M 2 ).
So, if the flow is locally sonic, it must encounter a critical point in area, dA/dx = 0 in order
to avoid infinite pressure gradients. This is what is realized in actual nozzles.
For us the analogous equation is the two-step version of Eq. (10.247), which can be shown
to be
ρû(σ1 r1 + σ2 r2 )
dP
=−
.
dx̂
1 − M̂ 2
(10.396)
Because the first reaction is exothermic, we have σ1 > 0, and because the second reaction
CC BY-NC-ND. 30 March 2014, J. M. Powers.
385
is endothermic, we have σ2 < 0. With r1 > 0, r2 > 0, this gives rise to the possibility that
σ1 r1 + σ2 r2 = 0 at a point where the reaction is incomplete. The point P is just such a point;
it is realized when D = D̃.
Finally, we write our differential-algebraic equations which are integrated for the detonation structure:
ρû = −ρo D,
P + ρû2 = Po + ρo D 2 ,
1
P
1
Po
e + û2 +
= eo + D 2 + ,
2
ρ
2
ρo
1 P
e =
− λ1 q1 − λ2 q2 ,
γ−1 ρ
P = ρRT,
λ2 = λ1 + (1 − λ1 ) ln(1 − λ1 ),
1 − λ1
E
dλ1
.
=
a exp −
dx̂
û
RT
(10.397)
(10.398)
(10.399)
(10.400)
(10.401)
(10.402)
(10.403)
We need the condition λ1 (0) = 0. These form seven equations for the seven unknowns ρ,
û, P , e, T , λ1 , and λ2 . We also realize that the algebraic solutions are multi-valued and
must take special care to be on the proper branch. This becomes particularly important for
solutions which pass through P .
10.2.7.1
Strong structures
Here, we consider strong structures for two cases: D > D̃ and D = D̃. All of these will
proceed from O to N through a pressure minimum, and finish at the strong point S.
10.2.7.1.1 D > D̃ Structures for a strong detonation with D = 2800 m/s > D̃ are given
in Figs. 10.18-10.21. The structure of all of these can be compared directly to those of the
one-step kinetics model at the same D = 2800 m/s, Figs. 10.11-10.14. Note the shock values
are identical. The reaction zone thicknesses are similar as well at ℓrxn ∼ 0.0001 m. The
structures themselves have some differences; most notably, the two-step model structures
display interior critical points before complete reaction.
We take special note of the pressure plot of Fig. 10.18, which can be compared with
Fig. 10.16. We see in both figures the shock from O to N, followed by a drop of pressure
to a minimum, followed by a final relaxation to an equilibrium value at S. Note that the
two curves have the opposite sense of direction as λ1 commences at 0 and goes to 1, while x̂
commences at 0 and goes to −0.0002 m.
We can also compare the M̂ (x̂) plot of Fig. 10.20 with that of M̂ (λ1 ) of Fig. 10.17. In
both the supersonic O is shocked to a subsonic N. The Mach number rises slightly then falls
in the reaction zone to its equilibrium value at S. It never returns to a supersonic state.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
386
ρ Hkgm3 L
5
P HPaL
4
5. 106
3
4. 106
3. 106
2
2. 106
1
0.0002
1. 106
0.0001
0
x (m)
0.0001
0
x (m)
Figure 10.18: ZND structure of ρ(x̂) and P (x̂) for strong D = 2800 m/s > D̃ for two-step
irreversible reaction with parameters of Table 10.3.
u HmsL
u (m/s)
0.0001
0
x (m)
500
2000
1500
1000
1000
1500
500
2000
0
2500
0.0001
0
x (m)
Figure 10.19: ZND structure of û(x̂) and u(x̂) for strong D = 2800 m/s > D̃ for two-step
irreversible reaction with parameters of Table 10.3.
M
7
T (K)
5000
6
4000
5
3000
4
3
2000
2
1000
1
0.0001
0
x (m)
0.0002
0.0001
0
x (m)
Figure 10.20: ZND structure of T (x̂) and M̂ (x̂) for strong D = 2800 m/s > D̃ for two-step
irreversible reaction with parameters of Table 10.3.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
387
λ1
1.0
λ2
1.0
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.0
0.0
0.0002
0.0001
0
x (m)
0.0002
0.0001
0
x (m)
Figure 10.21: ZND structure of λ1 (x̂) and λ2 (x̂) for strong D = 2800 m/s > D̃ for two-step
irreversible reaction with parameters of Table 10.3.
P HPaL
5. 106
4. 106
3. 106
2. 106
1. 106
0.0001
0
x (m)
Figure 10.22: ZND structure of P (x̂) for strong D = 2616.5 m/s = D̃ for two-step irreversible reaction with parameters of Table 10.3.
10.2.7.1.2 D = D̃ For D = D̃ = 2616.5 m/s, we can find a strong structure with a path
from O to N to S. Pressure P and Mach number M̂ are plotted in Figures 10.22-10.23.
Note that at an interior point in the structure, a cusp in the P and M̂ profile is seen. At
this point, the flow is locally sonic with M̂ = 1.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
388
M
7
6
5
4
3
2
1
0.0002
0.0001
0
x (m)
Figure 10.23: ZND structure of M̂(x̂) for strong D = 2616.5 m/s = D̃ for two-step
irreversible reaction with parameters of Table 10.3.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
389
ρ Hkgm3 L
5
P HPaL
5. 106
4
4. 106
3
3. 106
2
2. 106
1
0.000292911
1. 106
0
x (m)
0.000585822
0.000292911
0
x (m)
Figure 10.24: ZND structure of ρ(x̂) and P (x̂) for weak D = 2616.5 m/s = D̃ for two-step
irreversible reaction with parameters of Table 10.3.
10.2.7.2
Weak, eigenvalue structures
Let us now consider weak structures with D = D̃ = 2616.5 m/s. Special care must be
taken in integrating the governing equations. In general one must integrate to very near the
pathological point P , then halt. While there are more sophisticated techniques involving
further coordinate transformations, one can record the values near P on its approach from
N. Then, one can perturb slightly all state variables so that the M̂ is just greater than unity
and recommence the integration.
Figures of the structures which commence at O, are shocked to N, pass through sonic
point P , and finish at the supersonic W are shown in Figs. 10.24-10.27. One may again compare the pressure and Mach number plots of Figs. 10.24,10.26 with those of Figs. 10.16,10.17.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
390
u HmsL
u (m/s)
0.000585822
0.000292911
0
x (m)
2000
500
1500
1000
1000
1500
500
2000
0
2500
0.000585822
0.000292911
0
x (m)
Figure 10.25: ZND structure of û(x̂) and u(x̂) for weak D = 2616.5 m/s = D̃ for two-step
irreversible reaction with parameters of Table 10.3.
M
7
T HK L
6
4000
5
3000
4
3
2000
2
1000
1
0.000585822
0.000292911
0
x (m)
0.000585822
0.000292911
0
x (m)
Figure 10.26: ZND structure of T (x̂) and M̂ (x̂) for weak D = 2616.5 m/s = D̃ for two-step
irreversible reaction with parameters of Table 10.3.
λ2
1.0
λ1
1.0
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.0
0.0
0.000585822
0.000292911
0
x (m)
0
`
x (m)
Figure 10.27: ZND structure of λ1 (x̂) and λ2 (x̂) for weak D = 2616.5 m/s = D̃ for two-step
irreversible reaction with parameters of Table 10.3.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
10.2.7.3
391
Piston problem
We can understand the physics of the two-step kinetics problem better in the context of
a piston problem, where the supporting piston connects to the final laboratory frame fluid
velocity up = u(x̂ → −∞). Let us consider pistons with high velocity and then lower them
and examine the changes of structure.
• up > ũps . This high velocity piston will drive a strong shock into the fluid at a speed
D > D̃. The solution will proceed from O to N to S and be subsonic throughout.
Therefore changes at the piston face will be able to be communicated all the way to
the shock front. The energy to drive the wave comes from a combination of energy
released during combustion and energy supplied by the piston support.
• up = ũps . At a critical value of piston velocity, ũps , the solution will go from O to N
to P to S, and be locally sonic. This is analogous to the “subsonic design” condition
for a converging-diverging nozzle.
• up ∈ [ũps , ũpw ]. Here, the flow can be complicated. Analogous to flow in a nozzle, there
can be standing shock waves in the supersonic portion of the flow which decelerate
the flow so as to match the piston velocity at the end of the reaction. Such flows will
proceed from O to N through P , and then are shocked back onto the subsonic branch
to terminate at S.
• up = ũpw . This state is analogous to the “supersonic design” condition of flow in a
converging-diverging nozzle. The fluid proceeds from O to N through P and terminates
at W . All of the energy to propagate the wave comes from the reaction.
• up < ũpw . For such flows, the detonation wave is self-supporting. There is no means
to communicate with the supporting piston. The detonation wave proceeds at D = D̃.
We note that D̃ is a function of the specific kinetic mechanism. This non-classical result
contradicts the conclusion from CJ theory with simpler kinetics in which the wave speed
of an unsupported detonation is independent of the kinetics. Note for our problem that
D̃ = 2616.5 m/s. This stands in contrast to the CJ velocity, independently computed
of DCJ = 1991.1 m/s for the same mixture.
10.2.8
Detonation structure: Detailed H2 − O2 − N2 kinetics
These same notions for detonation with simple kinetics and state equations can easily be
extended to more complex models. Let us consider a one-dimensional steady detonation in
a stoichiometric hydrogen-air mixture with the detailed kinetics model of Table 1.2. We
shall consider a case almost identical to that studied by Powers and Paolucci.18 The mixture
is thus taken to be 2H2 + O2 + 3.76N2 . As we did in our modelling of the same mixture
18
Powers, J. M., and Paolucci, S., 2005, “Accurate Spatial Resolution Estimates for Reactive Supersonic
Flow with Detailed Chemistry,” AIAA Journal, 43(5): 1088-1099.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
392
under spatially homogeneous isochoric, adiabatic conditions in Sec. 1.2.2, we will take the
number of moles of each of the minor species to be a small number near machine precision.
This has the effect of removing some numerical roundoff errors in the very early stages of
reaction. Our model will be the steady one-dimensional reactive Euler equations, obtained
q
by considering the reactive Navier-Stokes equations in the limit as τ , jm
i , j all go to zero.
Our model then is 1) the integrated mass, momentum, and energy equations, Eqs. (10.24910.251) with an opposite sign on D to account for the left-running wave, 2) the onedimensional, steady, diffusion-free version of species evolution, Eq. (6.54), 3) the calorically
imperfect ideal gas state equations of Eqs. (6.69), (6.77), and 4) the law of mass action with
Arrhenius kinetics of Eqs. (6.82-6.85). In order to cleanly plot the results on a logarithmic
scale, we will, in contrast to the previous right-running detonations, consider left-running
waves with detonation velocity D. The governing equations are as follows:
ρû
P + ρû2
1 2 P
ρû e + û +
2
ρ
dYi
ρû
dx̂
= ρo D,
= Po + ρo D 2 ,
1 2 Po
= ρo D eo + D +
,
2
ρo
(10.404)
(10.405)
(10.406)
= Mi ω̇i ,
(10.407)
N
X
Yi
,
P = ρRT
Mi
i=1
Z
N
X
o
e =
Yi eTo ,i +
i=1
ω̇i =
J
X
(10.408)
T
cvi (T̂ )dT̂
To
,
νij rj ,
(10.409)
(10.410)
j=1
rj
kj
Kc,j
!
N
1 Y νkj
= kj
1−
ρ . ,
Kc,j k=1 k
k=1
−E j
βj
,
= aj T exp
RT
PNi=1 νij
∆Goj
Po
.
exp −
=
RT
RT
N
Y
ν′
ρkkj
(10.411)
(10.412)
(10.413)
This differential-algebraic system must be solved numerically. One can use standard methods
to achieve this. Alternatively, and with significant effort, one can remove all of the algebraic
constraints. Part of this requires a numerical iteration to find certain roots. After this effort,
one can in principle write a set of N − L differential equations for evolution of the N − L
independent species.
Here, we chose D ∼ DCJ = 1979.70 m/s. Because this model is not a simple one-step
model, we cannot expect to find the equilibrium state to be exactly sonic. However, it is
CC BY-NC-ND. 30 March 2014, J. M. Powers.
393
possible to slightly overdrive the wave and achieve a nearly sonic state at the equilibrium
state. Here, our final Mach number was M̂ = 0.9382. Had we weakened the overdrive
further, we would have encountered an interior sonic point at a non-equilibrium point, thus
inducing a non-physical sonic singularity.
Numerical solution for species mass fraction is given in Fig. (10.28). The figure is plot0
10
−5
Yi
10
OH
H 2O
H2
O2
H
O
HO 2
H 2O2
N2
−10
10
−10
10
−5
10
x (m)
0
10
Figure 10.28: Detailed kinetics ZND structure of species mass fractions for near CJ detonation, DCJ ∼ D = 1979.70 m/s, in 2H2 + O2 + 3.76N2 , Po = 1.01325 × 105 P a, To = 298 K,
Ts = 1542.7 K,Ps = 2.83280 × 106 P a.
ted on a log-log scale because of the wide range of length scales and mass fraction scales
encountered. Note that the minor species begin to change at a very small length scale. At
a value of x̂ ∼ 2.6 × 10−4 m, a significant event occurs, known as a thermal explosion. This
length is known as the induction length, ℓind = 2.6 × 10−4 m. We get a rough estimate
of the induction time by the formula tind ∼ ℓind /ûs = 7.9 × 10−7 s. Here, ûs is the post
shock velocity in the wave frame. Its value is ûs = 330.54 m/s. All species contribute to the
reaction dynamics here. This is followed by a relaxation to chemical equilibrium, achieved
around 0.1 m.
The pressure profile is given in Fig. (10.2.8). We artificially located the shock just away
from x̂ = 0, so as to ease the log-log plot. The pressure is shocked from its atmospheric value
to 2.83280 × 106 P a (see Table 10.2). After the shock, the pressure holds nearly constant for
several decades of distance. Once the thermal explosion commences, the pressure relaxes to
its equilibrium value. This figure can be compared with its one-step equivalent of Fig. 10.11.
Similar behavior is seen in the temperature plot of Fig. 10.30. The temperature is shocked
CC BY-NC-ND. 30 March 2014, J. M. Powers.
394
6
P (Pa)
10
10
5
10
−10
10
−5
10
0
x (m)
Figure 10.29: Detailed kinetics ZND structure of pressure for near CJ detonation, DCJ ∼
D = 1979.70 m/s, in 2H2 + O2 + 3.76N2 , Po = 1.01325 × 105 P a, To = 298 K, Ts =
1542.7 KPs = 2.83280 × 106 P a.
to Ts = 1542.7 K, stays constant in the induction zone, and then increases to its equilibrium
value after the thermal explosion. This figure can be compared with its one-step equivalent
of Fig. 10.13.
It is interesting to compare these results to those obtained earlier in Sec. 1.2.2. There, an
isochoric, adiabatic combustion of precisely the same stoichiometric hydrogen-air mixture
with precisely the same kinetics was conducted. The adiabatic/isochoric mixture had an
initial temperature and pressure identical to the post-shock pressure and temperature here.
We compare the induction time of the spatially homogeneous problem, tind = 6.6 × 10−7 s,
Eq. (1.373) to our estimate from the detonation found earlier, tind ∼ ℓind /ûs = 7.9 × 10−7 s.
The two are remarkably similar!
We compare some other relevant values in Table 10.4. Note that these mixtures are
identical at the onset of the calculation. The detonating mixture has reached the same initial
state after the shock. And a fluid particle advecting through the detonation reaction zone
with undergo a thermal explosion at nearly the same time a particle that was stationary
in the closed vessel will. After that the two fates are different. This is because there is
no kinetic energy in the spatially homogeneous problem. Thus, all the chemical energy
is transformed into thermal energy. This is reflected in the higher final temperature and
pressure of the spatially homogeneous problem relative to the detonating flow. Because the
final temperature is different, the two systems relax to a different chemical equilibrium, as
reflected in the different mass fractions. For example, note that the cooler detonating flow
CC BY-NC-ND. 30 March 2014, J. M. Powers.
395
Parameter
To
Ts
Po
Ps
tind
T eq
P eq
ρeq
ûeq
YOeq2
YHeq
eq
YOH
YOeq
YHeq2
YHeq2 O
eq
YHO
2
YHeq2 O2
YNeq2
Spatially Homogeneous
1542.7 K
−
2.83280 × 106 P a
−
6.6 × 10−7 s
3382.3 K
5.53 × 106 P a
4.62 kg/m3
0 m/s
1.85 × 10−2
5.41 × 10−4
2.45 × 10−2
3.88 × 10−3
3.75 × 10−3
2.04 × 10−1
6.84 × 10−5
1.04 × 10−5
7.45 × 10−1
Detonation
298 K
1542.7 K
1.01325 × 105 P a
2.83280 × 106 P a
7.9 × 10−7 s
2982.1 K
1.65 × 106 P a
1.59 kg/m3
1066 m/s
1.38 × 10−2
2.71 × 10−4
1.48 × 10−2
1.78 × 10−3
2.57 × 10−3
2.22 × 10−1
2.23 × 10−5
3.08 × 10−6
7.45 × 10−1
Table 10.4: Comparison of relevant predictions of a spatially homogeneous model with those
of a near CJ detonation in the same mixture.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
396
3500
3000
2500
T (K)
2000
1500
1000
500
0
10
−10
10
−5
10
0
x (m)
Figure 10.30: Detailed kinetics ZND structure of temperature for near CJ, DCJ ∼ D =
1979.70 m/s, detonation in 2H2 + O2 + 3.76N2 , Po = 1.01325 × 105 P a, To = 298 K,
Ts = 1542.7 K, Ps = 2.83280×106 P a. Shock at x̂ = 0 artificially translated to x̂ = 10−11 m.
has a higher final mass fraction of H2 O.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Chapter 11
Blast waves
Here, we will study the Taylor1 -Sedov2 blast wave solution. We will follow most closely two
papers of Taylor34 from 1950. Taylor notes that the first of these was actually written in
1941, but was classified. Sedov’s complementary study5 is also of interest. One may also
consult other articles by Taylor for background.67
Consider a point source of energy which at t = 0 is released into a calorically perfect
ideal gas. The point source could be the combustion products of an intense reaction event.
We shall follow Taylor’s analysis and obtain what is known as self-similar solutions. Though
there are more general approaches which may in fact expose more details of how self-similar
solutions are obtained, we will confine ourselves to Taylor’s approach and use his notation.
The self-similar solution will be enabled by studying the Euler equations in what is known
as the strong shock limit for a spherical shock wave. Now, a shock wave will raise both the
internal and kinetic energy of the ambient fluid into which it is propagating. We would
like to consider a scenario in which the total energy, kinetic and internal, enclosed by the
strong spherical shock wave is a constant. The ambient fluid is initially at rest, and a point
source of energy exists at r = 0. For t > 0, this point source of energy is distributed to the
mechanical and thermal energy of the surrounding fluid.
1
Geoffrey Ingram Taylor, 1886-1975, English physicist.
Leonid Ivanovitch Sedov, 1907-1999, Soviet physicist.
3
Taylor, G. I., 1950, “The Formation of a Blast Wave by a Very Intense Explosion. I. Theoretical Discussion,” Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences 201(1065):
159-174.
4
Taylor, G. I., 1950, “The Formation of a Blast Wave by a Very Intense Explosion. II. The Atomic
Explosion of 1945,” Proceedings of the Royal Society of London. Series A, Mathematical and Physical
Sciences, 201(1065): 175-186.
5
Sedov, L. I., 1946, “Rasprostraneniya Sil’nykh Vzryvnykh Voln,” Prikladnaya Matematika i Mekhanika
10: 241-250.
6
Taylor, G. I., 1950, “The Dynamics of the Combustion Products Behind Plane and Spherical Detonation
Fronts in Explosives,” Proceedings of the Royal Society of London. Series A, Mathematical and Physical
Sciences, 200(1061): 235-247.
7
Taylor, G. I., 1946, “The Air Wave Surrounding an Expanding Sphere,” Proceedings of the Royal Society
of London. Series A, Mathematical and Physical Sciences, 186(1006): 273-292.
2
397
398
CHAPTER 11. BLAST WAVES
Let us follow now Taylor’s analysis from his 1950 Part I “Theoretical Discussion” paper.
We shall
• write the governing inert one-dimensional unsteady Euler equations in spherical coordinates,
• reduce the partial differential equations in r and t to ordinary differential equations in
an appropriate similarity variable,
• solve the ordinary differential equations numerically, and
• show our transformation guarantees constant total energy in the region r ∈ [0, R(t)],
where R(t) is the locus of the moving shock wave.
We shall also refer to specific equations in Taylor’s first 1950 paper.
11.1
Governing equations
The non-conservative formulation of the governing equations is as follows:
∂ρ
∂u
2ρu
∂ρ
+u +ρ
= −
,
∂t
∂r
∂r
r
∂u
∂u 1 ∂P
+u
+
= 0,
∂t
∂r ρ ∂r
P ∂ρ
∂e
∂ρ
∂e
− 2
= 0,
+u
+u
∂t
∂r
ρ
∂t
∂r
1 P
,
e =
γ−1 ρ
P = ρRT.
(11.1)
(11.2)
(11.3)
(11.4)
(11.5)
For review, let us look at the energy equation in a little more detail. Recall the material
derivative is d/dt = ∂/∂t + u∂/∂r, so the energy equation, Eq. (11.3) can be rewritten as
de P dρ
−
= 0.
dt ρ2 dt
Let us now substitute the caloric energy equation, Eq. (11.4)
Eq. (11.6):
P dρ
1 d P
− 2
=
γ − 1 dt ρ
ρ dt
1 P dρ
1 1 dP
P dρ
−
+
− 2
=
2
γ − 1 ρ dt γ − 1 ρ dt
ρ dt
P dρ
P dρ 1 dP
+
− (γ − 1) 2
=
− 2
ρ dt ρ dt
ρ dt
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(11.6)
into the energy equation,
0,
(11.7)
0,
(11.8)
0,
(11.9)
399
11.2. SIMILARITY TRANSFORMATION
P dρ
1 dP
−γ 2
ρ dt
ρ dt
P dρ
dP
−γ
dt
ρ dt
1 dP
P dρ
− γ γ+1
γ
ρ dt
ρ
dt
d P
dt ργ
∂ P
∂ P
+u
∂t ργ
∂r ργ
11.2
= 0,
(11.10)
= 0,
(11.11)
= 0,
(11.12)
= 0,
(11.13)
= 0.
(11.14)
Similarity transformation
We shall next make some non-intuitive and non-obvious choices for a transformed coordinate
system and transformed dependent variables. These choices can be systematically studied
with the techniques of group theory, not discussed here.
11.2.1
Independent variables
Let us transform the independent variables (r, t) → (η, τ ) with
r
,
R(t)
τ = t.
η =
(11.15)
(11.16)
We will seek solutions such that the dependent variables are functions of η, the distance
relative to the time-dependent shock, only. We will have little need for the transformed time
τ since it is equivalent to the original time t.
11.2.2
Dependent variables
Let us also define new dependent variables as
P
= y = R−3 f1 (η),
Po
ρ
= ψ(η),
ρo
u = R−3/2 φ1 (η).
(11.17)
(11.18)
(11.19)
These amount to definitions of a scaled pressure f1 , a scaled density ψ, and a scaled velocity
φ1 , with the assumption that each is a function of η only. Here, Po , and ρo are constant
ambient values of pressure and density, respectively.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
400
CHAPTER 11. BLAST WAVES
We also assume the shock velocity to be of the form
U=
dR
= AR−3/2 .
dt
(11.20)
The constant A is to be determined.
11.2.3
Derivative transformations
By the chain rule we have
∂η ∂
∂τ ∂
∂
=
+
.
∂t
∂t ∂η
∂t ∂τ
(11.21)
Now, by Eq. (11.15) we get
∂η
r dR
= − 2
,
∂t
R dt
η dR
= −
,
R(t) dt
η
= − AR−3/2 ,
R
Aη
= − 5/2 .
R
(11.22)
(11.23)
(11.24)
(11.25)
From Eq. (11.16) we simply get
∂τ
= 1.
∂t
(11.26)
Thus, the chain rule, Eq. (11.21), can be written as
∂
Aη ∂
∂
= − 5/2
+
.
∂t
R ∂η ∂τ
(11.27)
As we are insisting the ∂/∂τ = 0, we get
∂
Aη d
= − 5/2 .
∂t
R dη
(11.28)
In the same way, we get
∂η ∂
∂τ ∂
∂
=
+
,
∂r
∂r ∂η ∂r |{z}
∂τ
(11.29)
=0
1 d
.
=
R dη
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(11.30)
401
11.3. TRANSFORMED EQUATIONS
11.3
Transformed equations
Let us now apply our rules for derivative transformation, Eqs. (11.25,11.30), and our transformed dependent variables, Eqs. (11.17-11.19), to the governing equations.
11.3.1
Mass
First, we shall consider the mass equation, Eq. (11.1). We get
1 d
1 d
2
Aη d
− 5/2
(ρo ψ) + R−3/2 φ1
(ρo ψ) + ρo ψ
R−3/2 φ1 = − ρo ψ R−3/2 φ1 . (11.31)
{z
}
|
|{z}
{z
}
{z
}
|
|
R dη
R dη
R dη | {z }
r |{z} | {z }
| {z } =ρ
| {z } =ρ
=u
=u
=ρ | {z }
=ρ
=u
=∂/∂t
=∂/∂r
=∂/∂r
Realizing the R(t) = R(τ ) is not a function of η, canceling the common factor of ρo , and
eliminating r with Eq. (11.15), we can write
−
Aη dψ
φ1 dψ
ψ dφ1
2 ψφ1
+ 5/2
+ 5/2
=−
,
5/2
R dη
R dη
R
dη
η R5/2
dψ
dφ1
2
dψ
+ φ1
+ψ
= − ψφ1 ,
−Aη
dη
dη
dη
η
dψ
dψ
dφ1 2
−Aη
+ φ1
+ψ
+ φ1
= 0,
dη
dη
dη
η
(11.32)
(11.33)
mass.
(11.34)
Equation (11.34) is number 9 in Taylor’s paper, which we will call here Eq. T(9).
11.3.2
Linear momentum
Now, consider the linear momentum equation, Eq. (11.2), and apply the same transformations:
∂
1 ∂
∂
R−3/2 φ1 + R−3/2 φ1
R−3/2 φ1 +
Po R−3 f1 = 0, (11.35)
∂t | {z } | {z } ∂r | {z } ρo ψ ∂r | {z }
|{z}
=u
=u
=u
=P
=1/ρ
∂φ1 3 −5/2 dR
∂
1 ∂
R−3/2
R−3/2 φ1 +
Po R−3 f1 = 0, (11.36)
− R
φ1 +R−3/2 φ1
∂t {z2
dt }
∂r
ρo ψ ∂r
|
=∂u/∂t
R
∂
dφ1 3 −5/2
R−3/2 φ1
− R
AR−3/2 φ1 + R−3/2 φ1
dη
2
∂r
1 ∂
+
Po R−3 f1 = 0, (11.37)
ρo ψ ∂r
1 ∂
∂
Aη dφ1 3 A
R−3/2 φ1 +
Po R−3 f1 = 0, (11.38)
−
φ1 + R−3/2 φ1
− 4
4
R dη
2R
∂r
ρo ψ ∂r
−3/2
Aη
− 5/2
R
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402
CHAPTER 11. BLAST WAVES
−
1 d
1 1 d
Aη dφ1 3 A
R−3/2 φ1 +
Po R−3 f1 = 0, (11.39)
−
φ1 + R−3/2 φ1
4
4
R dη
2R
R dη
ρo ψ R dη
φ1 dφ1
Po 1 df1
Aη dφ1 3 A
−
φ1 + 4
+
= 0, (11.40)
− 4
4
R dη
2R
R dη
ρo ψ R4 dη
dφ1 3
dφ1
Po df1
−Aη
− Aφ1 + φ1
+
= 0. (11.41)
dη
2
dη
ρo ψ dη
Our final form is
3
dφ1 Po 1 df1
dφ1
−A
+ φ1
φ1 + η
+
= 0,
2
dη
dη
ρo ψ dη
linear momentum.
(11.42)
Equation (11.42) is T(7).
11.3.3
Energy
Let us now consider the energy equation. It is best to begin with a form in which the
equation of state has already been imposed. So, we will start by expanding Eq. (11.11) in
terms of partial derivatives:
∂P
∂ρ
∂P
P ∂ρ
+u
+u
−γ
= 0,
(11.43)
ρ ∂t
∂r
|∂t {z ∂r}
|
{z
}
=dP/dt
=dρ/dt
∂
∂
Po R−3 f1 + R−3/2 φ1
Po R−3 f1
∂t
∂r
Po R−3 f1 ∂
∂
−3/2
−γ
(ρo ψ) + R
φ1 (ρo ψ)
ρo ψ
∂t
∂r
∂
∂
R−3 f1 + R−3/2 φ1
R−3 f1
∂t
∂r
R−3 f1 ∂ψ
∂ψ
−3/2
−γ
+R
φ1
ψ
∂t
∂r
∂f
dR
∂
1
R−3
R−3 f1
− 3R−4 f1 + R−3/2 φ1
∂t
dt
∂r
−3
∂ψ
R f1 ∂ψ
−3/2
+R
φ1
−γ
ψ
∂t
∂r
Aη df1
∂
R−3 − 5/2
R−3 f1
− 3R−4 (AR−3/2 )f1 + R−3/2 φ1
R
dη
∂r
−3
∂ψ
R f1 ∂ψ
−3/2
+R
φ1
−γ
ψ
∂t
∂r
= 0,
(11.44)
= 0,
(11.45)
= 0,
(11.46)
= 0.
(11.47)
Carrying on, we have
−
A
1 df1
Aη df1
− 3 11/2 f1 + R−3/2 φ1 R−3
11/2
R
dη
R
R dη
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403
11.4. DIMENSIONLESS EQUATIONS
Aη dψ
1 dψ
R−3 f1
−3/2
− 5/2
= 0,
+R
φ1
−γ
ψ
R
dη
R dη
Aη df1
A
φ1 df1
f1
dψ
dψ
− 11/2
= 0,
− 3 11/2 f1 + 11/2
−γ
−Aη
+ φ1
R
dη
R
R
dη
ψR11/2
dη
dη
dψ
df1
f1
dψ
df1
−Aη
= 0.
− 3Af1 + φ1
−γ
+ φ1
−Aη
dη
dη
ψ
dη
dη
Our final form is
df1
f1
dψ
df1
A 3f1 + η
+ γ (−Aη + φ1 )
− φ1
= 0,
dη
ψ
dη
dη
energy.
(11.48)
(11.49)
(11.50)
(11.51)
Equation (11.51) is T(11), correcting for a typographical error replacing a r with γ.
11.4
Dimensionless equations
Let us now write our conservation principles in dimensionless form. We take the constant
ambient sound speed co to be defined for the calorically perfect ideal gas as
c2o ≡ γ
Po
.
ρo
(11.52)
Note, we have used our notation for sound speed here; Taylor uses a instead.
Let us also define
c 2
o
f1 ,
f ≡
A
φ1
φ ≡
.
A
11.4.1
(11.53)
(11.54)
Mass
With these definitions, the mass equation, Eq. (11.34) becomes
dψ
dψ
dφ 2
−Aη
+ Aφ
+ψ A
+ Aφ
= 0,
dη
dη
dη η
dψ
dφ 2
dψ
+φ
+ψ
+ φ
= 0,
−η
dη
dη
dη η
dψ
dφ 2
(φ − η) = −ψ
+ φ ,
dη
dη η
dφ
+ 2φ
1 dψ
dη
η
=
,
mass.
ψ dη
η−φ
(11.55)
(11.56)
(11.57)
(11.58)
Equation (11.58) is T(9a).
CC BY-NC-ND. 30 March 2014, J. M. Powers.
404
11.4.2
CHAPTER 11. BLAST WAVES
Linear momentum
With the same definitions, the momentum equation,
dφ Po 1 A2 df
dφ
3
+ A2 φ
Aφ + Aη
+
−A
2
dη
dη
ρo ψ c2o dη
3
dφ 1 1 df
dφ
−
+φ
φ+η
+
2
dη
dη γ ψ dη
3
1 df
dφ
(φ − η) − φ +
dη
2
γψ dη
dφ
(η − φ)
dη
Eq. (11.42) becomes
= 0,
(11.59)
= 0,
(11.60)
= 0,
(11.61)
=
1 df
3
− φ, momentum. (11.62)
γψ dη 2
Equation (11.62) is T(7a).
11.4.3
Energy
The energy equation, Eq. (11.51) becomes
2
A2 df
f A2
dψ
A2 df
A
+γ
(−Aη
+
Aφ)
φ
−
A
= 0,
(11.63)
A 3 2 f +η 2
co
co dη
ψ c2o
dη
c2o dη
df
f
dψ
df
3f + η + γ (−η + φ)
−φ
= 0,
(11.64)
dη
ψ
dη
dη
1 dψ
df
df
f (−η + φ) − φ
= 0, energy. (11.65)
3f + η + γ
dη
ψ dη
dη
Equation (11.65) is T(11a).
11.5
Reduction to non-autonomous form
Let us eliminate dψ/dη and dφ/dη from Eq. (11.65) with use of Eqs. (11.58,11.62).
!
2φ
dφ
+
df
df
dη
η
3f + η + γf
(−η + φ) − φ
= 0,
dη
η−φ
dη

 1 df − 3 φ
2φ
γψ dη
2
+
df
df
η−φ
η 
3f + η + γf 
(−η + φ) − φ
= 0,
dη
η−φ
dη
!
1 df
− 32 φ 2φ
df
γψ dη
= 0,
+
3f + (η − φ) − γf
dη
η−φ
η
3
2φ
1 df
2 df
− γf
− φ + (η − φ)
= 0,
3f (η − φ) + (η − φ)
dη
γψ dη 2
η
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(11.66)
(11.67)
(11.68)
(11.69)
405
11.5. REDUCTION TO NON-AUTONOMOUS FORM
3
2γφ
f df
2
− f −3(η − φ) − γφ +
(η − φ)
= 0,
(η − φ) −
ψ dη
2
η
f df
3
2γφ2
2
(η − φ) −
= 0,
+ f 3η − 3φ + γφ − 2γφ +
ψ dη
2
η
f df
1
2γφ2
2
(η − φ) −
= 0.
+ f 3η − φ 3 + γ +
ψ dη
2
η
Rearranging, we get
1
2γφ2
f df
2
.
= f −3η + φ 3 + γ −
(η − φ) −
ψ dη
2
η
(11.70)
(11.71)
(11.72)
(11.73)
(11.74)
Equation (11.73) is T(14).
We can thus write an explicit non-autonomous ordinary differential equation for the
evolution of f in terms of the state variables f , ψ, and φ, as well as the independent variable
η.
2γφ2 1
f
−3η
+
φ
3
+
γ
− η
2
df
.
(11.75)
=
dη
(η − φ)2 − ψf
Eq. (11.75) can be directly substituted into the momentum equation, Eq. (11.62) to get
dφ
=
dη
1 df
γψ dη
− 32 φ
η−φ
.
(11.76)
Then, Eq. (11.76) can be substituted into Eq. (11.58) to get
dφ
+ 2φ
dψ
dη
η
=ψ
.
dη
η−φ
(11.77)
Equations (11.75-11.77) form a non-autonomous system of first order differential equations of the form
df
= g1 (f, φ, ψ, η),
dη
dφ
= g2 (f, φ, ψ, η),
dη
dψ
= g3 (f, φ, ψ, η).
dη
(11.78)
(11.79)
(11.80)
They can be integrated with standard numerical software. One must of course provide
conditions of all state variables at a particular point. We apply conditions not at η = 0, but
CC BY-NC-ND. 30 March 2014, J. M. Powers.
406
CHAPTER 11. BLAST WAVES
at η = 1, the locus of the shock front. Following Taylor, the conditions are taken from the
Rankine-Hugoniot equations applied in the limit of a strong shock. We take the subscript s
to denote the shock state at η = 1. For the density, one applies Eq. (10.313) and finds
ρs
γ+1
=
,
ρo
γ−1
ρo ψs
γ+1
=
,
ρo
γ−1
γ+1
.
ψs = ψ(η = 1) =
γ−1
(11.81)
(11.82)
(11.83)
For the pressure, one finds, by slight modification of Eq. (10.319) (taking D 2 = (dR/dt)2 ),
that
dR 2
dt
c2o
2 −3
=
AR
=
c2o
A2 R−3
=
c2o
1 =
fs = f (η = 1) =
γ + 1 Ps
,
2γ Po
γ + 1 −3
R f1s ,
2γ
γ + 1 −3 A2
fs ,
R
2γ
c2o
γ+1
fs ,
2γ
2γ
.
γ+1
(11.86)
2
,
γ+1
(11.89)
(11.84)
(11.85)
(11.87)
(11.88)
For the velocity, using Eq. (10.326), one finds
us
dR
dt
=
2
R−3/2 φ1s
=
,
−3/2
AR
γ+1
R−3/2 Aφs
2
=
,
AR−3/2
γ+1
2
.
φs = φ(η = 1) =
γ+1
(11.90)
(11.91)
(11.92)
Equations (11.83, 11.88,11.92) form the appropriate set of initial conditions for the integration of Eqs. (11.75-11.77).
11.6
Numerical solution
Solutions for f (η), φ(η) and ψ(η) are shown for γ = 7/5 in Figs. 11.1-11.3, respectively.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
407
11.6. NUMERICAL SOLUTION
f
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.2
0.4
0.6
0.8
1.0
η
Figure 11.1: Scaled pressure f versus similarity variable η for γ = 7/5 in Taylor-Sedov blast
wave.
φ
1.0
0.8
0.6
0.4
0.2
0.0
0.2
0.4
0.6
0.8
1.0
η
Figure 11.2: Scaled velocity φ versus similarity variable η for γ = 7/5 in Taylor-Sedov blast
wave.
ψ
7
6
5
4
3
2
1
0.0
0.2
0.4
0.6
0.8
1.0
η
Figure 11.3: Scaled density ψ versus similarity variable η for γ = 7/5 in Taylor-Sedov blast
wave.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
408
CHAPTER 11. BLAST WAVES
So, we now have a similarity solution for the scaled variables. We need to relate this to
physical dimensional quantities. Let us assign some initial conditions for t = 0, r > 0; that
is, away from the point source. Take
u(r, 0) = 0,
ρ(r, 0) = ρo ,
P (r, 0) = Po .
(11.93)
We also have from the ideal gas law
T (r, 0) =
Po
= To .
ρo R
(11.94)
For the calorically perfect gas we further have
e(r, 0) =
11.6.1
1 Po
= eo .
γ − 1 ρo
(11.95)
Calculation of total energy
Now, as the point source expands, it will generate a strong shock wave. Material which
has not been shocked is oblivious to the presence of the shock. Material which the shock
wave has reached has been influenced by it. It stands to reason from energy conservation
principles that we want the total energy, internal plus kinetic, to be constant in the shocked
domain, r ∈ (0, R(t)], where R(t) is the shock front location.
Let us recall some spherical geometry so this energy conservation principle can be properly
formulated. Consider a thin differential spherical shell of thickness dr located somewhere in
the shocked region: r ∈ (0, R(t)]. The volume of the thin shell is
dV =
dr
4πr 2
|{z}
|{z}
(surface area) (thickness)
(11.96)
The differential mass dm of this shell is
dm = ρdV,
= 4πr 2 ρdr.
(11.97)
(11.98)
Now, recall the mass-specific internal energy is e and the mass-specific kinetic energy is u2 /2.
So, the total differential energy, internal plus kinetic, in the differential shell is
1 2
dE =
e + u dm,
2
1 2 2
= 4πρ e + u r dr.
2
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(11.99)
(11.100)
409
11.6. NUMERICAL SOLUTION
Now, the total energy E within the shock is the integral through the entire sphere,
Z R(t)
Z R(t)
1 2 2
(11.101)
E=
dE =
4πρ e + u r dr,
2
0
0
Z R(t)
1 P
1 2 2
=
4πρ
(11.102)
+ u r dr,
γ−1 ρ
2
0
Z R(t)
Z R(t)
4π
2
=
(11.103)
ρu2 r 2 dr .
P r dr + 2π
γ −1 0
0
{z
} |
{z
}
|
thermal energy
kinetic energy
We introduce variables from our similarity transformations next:
Z 1
Z 1
4π
2 2
−3
ρo ψ R−3 φ21 R2 η 2 Rdη ,
E =
Po R f1 R η Rdη +2π
|{z}
| {z } | {z } |{z}
γ − 1 0 | {z } | {z } |{z}
0
Z
P
1
r2
ρ
dr
Z
u2
dr
1
4π
Po f1 η 2 dη + 2π
ρo ψφ21 η 2 dη,
γ−1 0
0
Z 1
Z 1
2
4π
Po A
2
=
f η dη + 2π
ρo ψA2 φ2 η 2 dη,
2
γ − 1 0 co
0
Z 1
Z
ρo 1 2 2
Po
2
2
f η dη +
ψφ η dη ,
= 4πA
c2o (γ − 1) 0
2 0
Z 1
Z
1
1 1 2 2
2
2
= 4πA ρo
f η dη +
ψφ η dη .
γ(γ − 1) 0
2 0
{z
}
|
dependent on γ only
=
r2
(11.104)
(11.105)
(11.106)
(11.107)
(11.108)
The term inside the parentheses is dependent on γ only. So, if we consider air with γ = 7/5,
we can, using our knowledge of f (η), ψ(η), and φ(η), which only depend on γ, to calculate
once and for all the value of the integrals. For γ = 7/5, we obtain via numerical quadrature
1
1
2
E = 4πA ρo
(0.185194) + (0.185168) ,
(11.109)
(7/5)(2/5)
2
= 5.3192ρoA2 .
(11.110)
Now, from Eq. (11.17, 11.52,11.53, 11.110) with γ = 7/5, we get
A2
,
c2o
ρo 2
= Po R−3 f
A ,
γPo
1
= R−3 f ρo A2 ,
γ
P = Po R−3 f
(11.111)
(11.112)
(11.113)
CC BY-NC-ND. 30 March 2014, J. M. Powers.
410
CHAPTER 11. BLAST WAVES
= R−3 f
1
7
5
E
,
5.3192
= 0.1343R−3Ef,
E
r
.
P (r, t) = 0.1343 3 f
R (t)
R(t)
(11.114)
(11.115)
(11.116)
The peak pressure occurs at η = 1, where r = R, and where
f (η = 1) =
2(1.4)
2γ
=
= 1.167.
γ+1
1.4 + 1
(11.117)
So, at η = 1, where r = R, we have
E
.
(11.118)
R3
The peak pressure decays at a rate proportional to 1/R3 in the strong shock limit.
Now, from Eqs. (11.19,11.54,11.110) we get for u:
P = (0.1343)(1.167)R−3E = 0.1567
u = R−3/2 Aφ,
s
= R−3/2
u(r, t) =
s
E
φ,
5.319ρo
E
r
1
.
φ
5.319ρo R3/2 (t)
R(t)
(11.119)
(11.120)
(11.121)
Let us now explicitly solve for the shock position R(t) and the shock velocity dR/dt. We
have from Eqs. (11.20, 11.110) that
dR
= AR−3/2 ,
dt
s
1
E
,
=
3/2
5.319ρo R (t)
s
E
R3/2 dR =
dt,
5.319ρo
s
E
2 5/2
R
=
t + C.
5
5.319ρo
Now, since R(0) = 0, we get C = 0, so
s
E
t,
5.319ρo
2 5/2 p
t =
R
5.319ρo E −1/2 ,
5
−1/2
= 0.9225R5/2 ρ1/2
.
o E
2 5/2
R
=
5
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(11.122)
(11.123)
(11.124)
(11.125)
(11.126)
(11.127)
(11.128)
411
11.6. NUMERICAL SOLUTION
Equation (11.128) is T(38). Solving for R, we get
1
tρ−1/2 E 1/2 ,
0.9225 o
R(t) = 1.03279ρ−1/5
E 1/5 t2/5 .
o
R5/2 =
(11.129)
(11.130)
Thus, we have a prediction for the shock location as a function of time t, as well as point
source energy E. If we know the position as a function of time, we can easily get the shock
velocity by direct differentiation:
dR
= 0.4131ρ−1/5
E 1/5 t−3/5 .
o
dt
(11.131)
If we can make a measurement of the blast wave location R at a given known time t, and
we know the ambient density ρo , we can estimate the point source energy E. Let us invert
Eq. (11.130) to solve for E and get
ρo R5
,
(1.03279)5t2
ρo R5
= 0.85102 2 .
t
E =
11.6.2
(11.132)
(11.133)
Comparison with experimental data
Now, Taylor’s Part II paper from 1950 gives data for the 19 July 1945 atomic explosion at
the Trinity site in New Mexico. We choose one point from the photographic record which
finds the shock from the blast to be located at R = 185 m when t = 62 ms. Let us assume
the ambient air has a density of ρo = 1.161 kg/m3 . Then, we can estimate the energy of the
device by Eq. (11.133) as
kg
(185 m)5
1.161 m
3
E = 0.85102
,
(0.062 s)2
= 55.7 × 1012 J.
(11.134)
(11.135)
Now, 1 ton of the high explosive TNT is known to contain 4.25 × 109 J of chemical energy.
So, the estimated energy of the Trinity site device in terms of a TNT equivalent is
T NTequivalent =
55.7 × 1012 J
= 13.1 × 103 ton.
J
4.25 × 109 ton
(11.136)
In common parlance, the Trinity site device was a 13 kiloton bomb by this simple estimate.
Taylor provides some nuanced corrections to this estimate. Modern estimates are now around
20 kiloton.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
412
11.7
CHAPTER 11. BLAST WAVES
Contrast with acoustic limit
We saw in Eq. (11.118) that in the expansion associated with a strong shock, the pressure
decays as 1/R3 . Let us see how that compares with the decay of pressure in the limit of a
weak shock.
Let us first rewrite the governing equations. Here, we 1) rewrite Eq. (11.1) in a conservative form, using the chain rule to absorb the source term inside the derivative, 2) repeat the
linear momentum equation, Eq. (11.2), and 3) re-cast the energy equation for a calorically
perfect ideal gas, Eq. (11.11) in terms of the full partial derivatives:
∂ρ
1 ∂ 2 r ρu = 0,
+ 2
∂t r ∂r
∂u 1 ∂P
∂u
+u
+
= 0,
∂t
∂r ρ ∂r
∂P
∂P
P ∂ρ
∂ρ
= 0.
+u
−γ
+u
∂t
∂r
ρ ∂t
∂r
(11.137)
(11.138)
(11.139)
Now, let us consider the acoustic limit, which corresponds to perturbations of a fluid at
rest. Taking 0 < ǫ ≪ 1, we recast the dependent variables ρ, P , and u as
ρ = ρo + ǫρ1 + . . . ,
P = Po + ǫP1 + . . . ,
u = uo +ǫu1 + . . . .
|{z}
(11.140)
(11.141)
(11.142)
=0
Here, ρo and Po are taken to be constants. The ambient velocity uo = 0. Strictly speaking,
we should non-dimensionalize the equations before we introduce an asymptotic expansion.
However, so doing would not change the essence of the argument to be made.
We next introduce our expansions into the governing equations:
1 ∂ 2
∂
r (ρo + ǫρ1 ) (ǫu1 ) = 0,
(ρo + ǫρ1 ) + 2
∂t
r ∂r
∂
∂
1
∂
(ǫu1 ) + (ǫu1 ) (ǫu1 ) +
(Po + ǫP1 ) = 0,
∂t
∂r
ρo + ǫρ1 ∂r
∂
∂
(Po + ǫP1 ) + (ǫu1 ) (Po + ǫP1 )
∂t
∂r
Po + ǫP1 ∂
∂
−γ
(ρo + ǫρ1 ) + (ǫu1 ) (ρo + ǫρ1 )
= 0.
ρo + ǫρ1 ∂t
∂r
(11.143)
(11.144)
(11.145)
Now, derivatives of constants are all zero, and so at leading order the constant state satisfies
the governing equations. At O(ǫ), the equations reduce to
1 ∂
∂ρ1
+ 2 (r 2 ρo u1 ) = 0,
∂t
r ∂r
CC BY-NC-ND. 30 March 2014, J. M. Powers.
(11.146)
413
11.7. CONTRAST WITH ACOUSTIC LIMIT
1 ∂P1
∂u1
+
= 0,
∂t
ρo ∂r
Po ∂ρ1
∂P1
−γ
= 0.
∂t
ρo ∂t
(11.147)
(11.148)
Now, adopt as before c2o = γPo /ρo , so the energy equation, Eq. (11.148), becomes
∂P1
∂ρ1
= c2o
.
∂t
∂t
(11.149)
Now, substitute Eq. (11.149) into the mass equation, Eq. (11.146), to get
1 ∂P1
1 ∂ 2
+
(r ρo u1 ) = 0.
c2o ∂t
r 2 ∂r
We take the time derivative of Eq. (11.150) to get
1 ∂ 2
∂
1 ∂ 2 P1
r ρo u1
= 0,
+
c2o ∂t2
∂t r 2 ∂r
∂u1
1 ∂
1 ∂ 2 P1
2
r ρo
= 0.
+ 2
c2o ∂t2
r ∂r
∂t
(11.150)
(11.151)
(11.152)
We next use the momentum equation, Eq. (11.147), to eliminate ∂u1 /∂t in Eq. (11.152):
1 ∂ 2 P1
1 ∂P1
1 ∂
2
r ρo −
= 0,
(11.153)
+ 2
c2o ∂t2
r ∂r
ρo ∂r
1 ∂ 2 P1
1 ∂
2 ∂P1
r
= 0,
(11.154)
− 2
c2o ∂t2
r ∂r
∂r
1 ∂ 2 P1
1 ∂
2 ∂P1
r
.
(11.155)
= 2
c2o ∂t2
r ∂r
∂r
This second-order linear partial differential equation has a well-known solution of the
D’Alembert form:
1
1
r
r
P1 = g t −
+ h t+
.
(11.156)
r
co
r
co
Here, g and h are arbitrary functions which are chosen to match the initial conditions. Let
us check this solution for g; the procedure can easily be repeated for h.
If P1 = (1/r)g(t − r/co), then
r
1 ′
∂P1
,
(11.157)
=
g t−
∂t
r
co
∂ 2 P1
r
1 ′′
,
(11.158)
=
g t−
∂t2
r
co
CC BY-NC-ND. 30 March 2014, J. M. Powers.
414
CHAPTER 11. BLAST WAVES
and
∂P1
r
1
11 ′
r
− 2g t −
.
= −
g t−
∂r
co r
co
r
co
(11.159)
With these results, let us substitute into Eq. (11.155) to see if it is satisfied:
11 ′
1 ∂
1
r
r
r
1 1 ′′
2
= 2
r −
− 2g t −
,
(11.160)
g t−
g t−
c2o r
co
r ∂r
co r
co
r
co
1 ∂
r
r ′
r
= − 2
g t−
+g t−
,
(11.161)
r ∂r co
co
co
r
1 ′
r
1 ′
r
r ′′
1
+ g t−
− g t−
(11.162)
,
= − 2 − 2g t −
r
co
co
co
co
co
co
1 r ′′
r
= 2
g t−
,
(11.163)
r
c2o
co
r
1 1 ′′
.
(11.164)
= 2 g t−
co r
co
Indeed, our form of P1 (r, t) satisfies the governing partial differential equation. Moreover,
we can see by inspection of Eq. (11.156) that the pressure decays as does 1/r in the limit of
acoustic disturbances. This is a much slower rate of decay than for the blast wave, which
goes as the inverse cube of radius.
CC BY-NC-ND. 30 March 2014, J. M. Powers.
Bibliography
Here, we give a summary of some of the major books relevant to these lecture notes. Specific
journal articles are not cited, though the main text of the lecture notes occasionally draws
on such articles where relevant. The emphasis is on combustion, physical chemistry, and
thermodynamics, with some mathematics. Some general works of historic importance are
also included. The list is by no means comprehensive.
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This is written in the style of all the Schaum’s series. It has extensive solved problems and a crisp
rigorous style that is readable by undergraduate engineers. It has a chemical engineering emphasis,
but is also useful for all engineers.
R. Aris, 1962, Vectors, Tensors, and the Basic Equations of Fluid Mechanics, Dover, New
York.
This book is a jewel of continuum mechanics theory. It is brief and incisive. While the focus is on
the inert Navier-Stokes equations, there is a short, excellent chapter on the reactive Navier-Stokes
equations. Highly recommended.
J. Bebernes and D. Eberly, 1989, Mathematical Problems from Combustion Theory, Springer,
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This monograph focuses on mathematical analysis of combustion systems, focusing on blowup phenomena.
A. Bejan, 2006, Advanced Engineering Thermodynamics, Third Edition, John Wiley, Hoboken, New Jersey.
This is an advanced undergraduate text. It gives a modern treatment of the science of classical
thermodynamics. It does not confine its attention to traditional engineering problems, and considers
applications across biology and earth sciences as well. The thermodynamics of irreversible processes
are discussed in detail.
R. S. Berry, S. A. Rice, and J. Ross,, 2000, Physical Chemistry, Second Edition, Oxford
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This is a rigorous general text in physical chemistry at a senior or first year graduate level. It has a
full treatment of classical and statistical thermodynamics as well as quantum mechanics.
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R. B. Bird, W. E. Stewart, and E. N. Lightfoot, 2006, Transport Phenomena, Second
Edition, Wiley, New York.
This acknowledged classic of the chemical engineering literature does not focus on reaction, but has
an exemplary treatment of mass, momentum, and energy advection-diffusion phenomena.
L. Boltzmann, 1995, Lectures on Gas Theory, Dover, New York.
This is a detailed monograph by the founding father of statistical thermodynamics.
C. Borgnakke, and R. E. Sonntag, 2009, Fundamentals of Thermodynamics, Seventh Edition, John Wiley, New York.
This classic and popular undergraduate mechanical engineering text has stood the test of time and
has a full treatment of most classical problems.
R. Boyle, 2003, The Sceptical Chymist, Dover, New York.
This is the original work of the famous early figure of the scientific revolution.
J. D. Buckmaster and G. S. S. Ludford, 1983, Lectures on Mathematical Combustion, SIAM,
This is set of lecture notes which serves as an excellent graduate-level introduction to some of the
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This seminal monograph, first published in 1982, highlights rigorous mathematical methods applied
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with a few chapters devoted to quantum and statistical foundations.
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This is the original source of Carnot’s foundational work on the heat engines. Also included is a paper
of Clausius.
N. Chigier, 1981, Energy, Combustion, and Environment, McGraw-Hill, New York.
This advanced undergraduate text gives an integrated science-based discussion of the topic matter
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416
This seminal mathematical treatment of inviscid compressible flow gives an outstanding and rigorous
discussion of the application of partial differential equations to fluid mechanics. There is only a small
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S. R. de Groot and P. Mazur, 1984, Non-Equilibrium Thermodynamics, Dover, New York.
This is an influential monograph which summarizes much of the work of the famous Belgian school
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chemical reactions.
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This short classic clearly and efficiently summarizes the fundamentals of thermodynamics. It is based
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W. Fickett and W. C. Davis, 1979, Detonation, U. California, Berkeley.
This distinctive monograph has a rich and focused discussion on detonation science. It opens with
the relevant physical chemistry, then gives detailed exposition on the theory of detonation. Careful
discussion of experiment is also included.
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This well known monograph gives a good description of important combustion phenomena.
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This monograph gives an enlightening description, fully informed by both historical and modern
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This is a graduate level text in combustion.
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This beginning graduate text has a rigorous development of classical thermodynamics.
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This comprehensive tome is a valuable addition to any library of thermal science. Its wide ranging
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remains original.
417
W. Hundsdorfer and J. G. Verwer, 2003, Numerical Solution of Time-Dependent AdvectionDiffusion-Reaction Equations, Springer, Berlin.
This nice graduate level monograph focuses on modern numerical techniques for paradigm reactive
flow systems. The systems themselves are easy to pose which allows effective exposition of the key
challenges in numerical modelling.
A. M. Kanury, 1975, Introduction to Combustion Phenomena, Gordon and Breach, Amsterdam.
This introductory graduate level text gives practical problems and good discussion of combustion
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This short introductory monograph gives lecture notes on asymptotic analysis as applied to combustion.
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This comprehensive text gives an introductory graduate level discussion of fluid mechanics, thermochemistry, and finite rate chemical kinetics. The focus is on low Mach number flows, and there is
significant discussion of how to achieve computational solutions.
C. Kittel and H. Kroemer, 1980, Thermal Physics, Second Edition, Freeman, San Francisco.
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D. Kondepudi and I. Prigogine, 1998, Modern Thermodynamics: From Heat Engines to
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Detailed modern exposition which exploits the authors’ unique vision of thermodynamics with both
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This is a graduate level monograph.
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K. K. Kuo, 2005, Principles of Combustion, Second Edition, John Wiley, New York.
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This classic comprehensive text of fluid physics contains a small discussion of reactive flow.
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For many years this was the graduate text in combustion. It has a fine description of scientific
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This is a standard undergraduate thermodynamics text, and one of the more popular. Note: some
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I. Müller and T. Ruggeri, 1998, Rational Extended Thermodynamics, Springer-Verlag, New
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This modern monograph gives a rigorous treatment of some of the key issues in modern continuum
mechanics and classical thermodynamics.
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This is a unique treatise on fundamental concepts in thermodynamics. The author provide mathematical rigor, historical perspective, and examples from a diverse set of scientific fields.
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The author gives a readable text at an undergraduate level which highlights some of the many controversies of thermodynamics, both ancient and modern.
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This monograph by two of the leading figures of computational combustion gives a good discussion
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This is a monograph on thermodynamics by a well known physicist.
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New York.
This monograph gives no explicit discussion of combustion but does give an outstanding treatment of
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This brief book gives many unique insights from a great physicist.
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This is a useful graduate level introduction to combustion.
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This is a beginning graduate text in combustion engineering.
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This graduate level monograph gives a chemical engineering perspective on reaction engineering in
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Advanced undergraduate text in thermodynamics from a chemical engineering perspective with a
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This classic two volume set has a comprehensive treatment of the subject of its title. It has numerous
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This is probably the most common undergraduate text in thermodynamics in chemical engineering.
It is rigorous and has went through many revisions.
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This unique monograph links irreversible thermodynamics with phase transition, ignition and detonation.
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This entry level graduate text in thermodynamics is written from a chemical engineer’s perspective.
It has a strong mathematical development for both classical and statistical thermodynamics.
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This is an entry level graduate text on combustion science.
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This idiosyncratic monograph has a lucid description of the history of nineteenth century thermal
science. It is written in an erudite fashion and the reader who is willing to dive into a difficult subject
will be rewarded for diligence by many new insights.
C. A. Truesdell, 1984, Rational Thermodynamics, Springer-Verlag, New York.
This is a modern update on the evolution of classical thermodynamics in the twentieth century. It is
at a graduate level and contains several articles by some of the present leaders of the field.
S. R. Turns, 2011, An Introduction to Combustion, Third Edition, McGraw-Hill, Boston.
This is a popular senior-level undergraduate text on combustion which uses many notions from thermodynamics of mixtures.
H. C. van Ness, 1983, Understanding Thermodynamics, Dover, New York.
This is a short readable monograph from a chemical engineering perspective.
W. G. Vincenti and C. H. Kruger, 1965, Introduction to Physical Gas Dynamics, John
Wiley, New York.
This graduate text on high speed non-equilibrium flows contains some nice descriptions of the interplay
of classical and statistical mechanics. There is an emphasis on aerospace applications.
J. Warnatz, U. Maas, and R. W. Dibble, 2006, Combustion, Fourth Edition, Springer, New
York.
This short graduate level text is nonetheless comprehensive and gives a good introduction to combustion science.
422
F. A. Williams, 1985, Combustion Theory, Benjamin-Cummings, Menlo Park, California.
This influential monograph was first published in 1965. It was one of the first texts to give a comprehensive theoretical treatment of reactive fluid mechanics for model and practical systems.
L. C. Woods, 1975, The Thermodynamics of Fluid Systems, Clarendon, Oxford.
This graduate text gives a good, detailed survey of the thermodynamics of irreversible processes,
especially related to fluid systems in which convection and diffusion play important roles.
I. B. Zel’dovich and A. S. Kompaneets, 1960, Theory of Detonation, Academic Press, New
York.
This is a readable monograph whose first author is one of the founding fathers of modern detonation
theory.
Ya. B. Zel’dovich and Yu. P. Raizer, 2002, Physics of Shock Waves and High-Temperature
Hydrodynamic Phenomena, Dover, New York.
This classic tome of the Russian school is a tour de force of reactive flow physics.
423
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