Versteeg H., Malalasekra W. An Introduction to Computational Fluid Dynamics: The Finite Volume Method
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12.4 RELATIONSHIPS AND PROPERTIES OF GASEOUS MIXTURES 347
Alternatively
= (12.12)
The density of a species k is defined as
ρ
k
= (12.13)
Here R
u
is the universal gas constant. The density of the mixture
ρ
is then
ρ
=
ρ
k
==X
k
(MW )
k
= (12.14)
The concentration of a species k is defined as the number of kilomoles of the
species per unit volume, and usually denoted by symbol C
k
with units kmol/m
3
:
C
k
== (12.15)
Using the equation of state pV = n
total
R
u
T the concentration can also be
written as
C
k
==X
k
= (12.16)
Concentration may also be related to mass fraction using the relationships
between mole fraction and mass fraction:
C
k
== (12.17)
where
ρ
is the density of the mixture. Alternatively
Y
k
= (12.18)
The enthalpy of a species with respect to the reference enthalpy at standard
pressure and temperature is defined as
h
k
= c
pk
dT +∆h
k
0
(12.19)
sensible chemical
where ∆h
k
0
is the enthalpy of formation ( J/kg) of the species k at standard
pressure and temperature. It is related to the enthalpy of formation ∆h
k
f
(J/kmol) introduced earlier by means of ∆h
k
0
=∆h
k
f
/(MW )
k
.
T
T
0
C
k
(MW )
k
ρ
Y
k
ρ
(MW )
k
p
R
u
T
Y
k
(MW )
mix
(MW )
k
p
k
R
u
T
p
R
u
T
X
k
n
total
n
total
R
u
T/p
X
k
n
total
V
n
k
V
p
R
u
/(MW )
mix
T
all species
∑
k
p
R
u
T
p
k
R
u
/(MW)
k
T
all species
∑
k
all species
∑
k
p
k
R
u
/(MW )
k
T
Y
k
(MW )
k
all species
∑
k
1
(MW )
mix
!
@
1
2
3
ANIN_C12.qxd 29/12/2006 04:44PM Page 347
Properties such as the specific enthalpy h
mix
of a mixture may be calcu-
lated from one of the following two expressions:
h
mix
= Y
k
h
k
or (12.20)
P
mix
= X
k
P
k
(12.21)
where h
k
is the specific enthalpy (units J/kg) of species k and the overbar
indicates molar specific enthalpy of species k, which has units J/kmol.
In combustion calculations we often perform a simple analysis which
assumes that the fuel burns completely to form products (complete combus-
tion). Most common combustion reactions involve oxidation of hydrocarbon
fuels that have a high content of carbon (C) and hydrogen (H). When com-
bustion is complete all C atoms in the fuel are consumed to form CO
2
and all
H atoms in the fuel become H
2
O. Any other combustible elements such as
sulphur (S) etc. also combine with O
2
. A chemical equation for complete
combustion could be written by considering an atom balance. This would
yield the exact amount of oxidiser required for complete combustion: the
so-called stoichiometric oxidiser requirement. Usually the oxidiser is air,
therefore the exact air requirement for complete combustion is called the
stoichiometric air requirement for complete combustion. We can also
calculate the air to fuel ratio necessary for complete combustion, which is
called the stoichiometric air/fuel ratio. The composition of air is con-
sidered to be 79% of N
2
and 21% of O
2
by volume. For each mole of O
2
in
air there are 79/21 (= 3.76) moles of N
2
. By mass the ratio is 23.3% of O
2
and
76.7% of N
2
. The molecular weight of air is taken as 29 kg/kmol; 1 kmol of
O
2
is contained in 100/21 kmol of air (i.e. 4.76 kmol of air).
Complete combustion of CH
4
for example is given by the equation
CH
4
+ 2(O
2
+ 3.76N
2
) → CO
2
+ 2H
2
O + 2 × 3.76N
2
(12.22)
The stoichiometric air fuel/ratio by mass is
(A/F)
st
=
=
==17.255 kg air/kg fuel (12.23)
Combustion with less air than the stoichiometric air requirement is fuel rich,
and combustion with air in excess of the stoichiometric air requirement is
called lean combustion. The equivalence ratio is used in combustion calcula-
tions to define the strength of a mixture with respect to the stoichiometric
29 × 2 × 4.76
16 × 1
(MW )
air
× number of kmol of O
2
× number of kmol of air per kmol of O
2
(MW )
CH
4
× number of kmol of CH
4
m
air
m
fuel
all species
∑
k
all species
∑
k
348 CHAPTER 12 CFD MODELLING OF COMBUSTION
Stoichiometry12.5
Equivalence ratio12.6
ANIN_C12.qxd 29/12/2006 04:44PM Page 348
12.7 ADIABATIC FLAME TEMPERATURE 349
mixture strength. Given a fuel we may calculate the stoichiometric air/fuel
ratio as explained above; say the value is (A/F)
st
. In the actual mixture
the air fuel ratio may be (A/F)
actual
. The equivalence ratio indicates the
strength of a mixture with respect to the stoichiometric air/fuel ratio. It is
defined as follows:
φ
== (12.24)
If the equivalence ratio of an air/fuel mixture is equal to unity, the mixture
is stoichiometric. A value less than 1, i.e.
φ
< 1, indicates a lean mixture, and
if
φ
> 1, the mixture is rich.
If a fuel/air mixture is considered to be burnt completely under constant
pressure and if no external heat or work transfer takes place then all the
energy liberated by the chemical reaction will be used to heat up the prod-
ucts. This process will achieve the maximum possible temperature, which is
called the adiabatic flame temperature. It can be calculated using the
first law of thermodynamics and enthalpies of reactants and products. When
there are no heat losses the total enthalpy of products is equal to the enthalpy
of reactants.
Calculate the adiabatic flame temperature for combustion of a stoichiometric
mixture of CH
4
and air at standard temperature and pressure.
The combustion equation is
CH
4
+ 2(O
2
+ 3.76N
2
) → CO
2
+ 2H
2
O + 2 × 3.76N
2
(12.25)
Application of the first law gives
H
P 2
− H
R1
= (H
P 2
− H
P 0
) + (H
P 0
− H
R0
) + (H
R0
− H
R 1
) (12.26)
For a constant pressure adiabatic process (neglecting kinetic energy changes)
H
P 2
− H
R1
= 0 (12.27)
It is given that the reactants start off at standard pressure and temperature
so no heat is taken up in heating the reactants to bring them to standard
temperature. Thus H
R0
− H
R1
= 0, and we can simplify equation (12.26) as
(H
P2
− H
P0
) =−(H
P0
− H
R0
) (12.28)
m
k
c
pk
(T
2
− T
0
) =−(H
P 0
− H
R 0
) (12.29)
Here temperature T
2
is the final temperature of the products and
H
P0
− H
R0
=∆H
0
is the heat of combustion of CH
4
.
Before we calculate the value of T
2
we need to consider that the change in
temperature during combustion will cause the values of specific heats c
v
and
c
p
to change. In simple adiabatic flame temperature calculations mean specific
heat values are used. To obtain mean c
pk
-values for products (CO
2
and H
2
O)
we need a temperature range. In a calculation like this we start by assuming
a final temperature to estimate a mean temperature for products and then
all products
∑
k
(F/A)
actual
(F/A)
st
(A/F)
st
(A/F)
actual
Adiabatic flame
temperature
12.7
Example 12.1
Solution
ANIN_C12.qxd 29/12/2006 04:44PM Page 349
repeat the calculation using the estimated product temperature after the
first calculation. We guess a final temperature T
2
, say 2000 K. Then a mean
temperature for products would be (2000 + 298)/2 = 1149 K. For gases at
different temperatures the value of c
pk
may be calculated using an equation
of the form c
p
= a + bT + cT
2
+ dT
3
(see Cengel and Boles (2002) for values
of the constants a, b, c and d):
c
p
value for CO
2
at 1149 K = 1.274 kJ/kg K (12.30a)
c
p
value for H
2
O at 1149 K = 2.384 kJ/kg K (12.30b)
c
p
value for N
2
at 1149 K = 1.1931 kJ/kg K (12.30c)
To evaluate T
2
from equation (12.29) we need the mass m
k
of each of the
product species. The combustion equation in mole form (12.25) can be con-
verted into mass form by multiplying the molecular weight of each substance
by the coefficient of the molar balance equation.
In terms of mass,
16 kg CH
4
+ 64 kg O
2
+ 210.56 kg N
2
⎯→ 44 kg CO
2
+ 36 kg H
2
O + 210.56 kg N
2
(12.31)
For CH
4
combustion the heat of combustion H
P0
− H
R0
=−50 050 kJ/kg.
We use the so-called lower heating value, which assumes that H
2
O in the
products is in the vapour phase. Since heat of combustion values are tabu-
lated per kg of fuel we rewrite (12.31) as follows to obtain values for the mass
of products species m
k
:
1 kg CH
4
+ 64/16 kg O
2
+ 210.56/16 kg N
2
⎯→ 44/16 kg CO
2
+ 36/16 kg H
2
O + 210.56/16 kg N
2
or
1 kg CH
4
+ 4 kg O
2
+ 13.16 kg N
2
⎯→ 2.75 kg CO
2
+ 2.25 kg H
2
O + 13.16 kg N
2
(12.32)
We use equation (12.29) with the c
pk
-values of (12.30a–c) and the values of
mass of products species m
k
from (12.32) and obtain
∑m
k
c
pk
(T
2
− T
0
) = [(2.75 × 1.274) + (2.25 × 2.384)
+ (13.16 × 1.193)] × 10
3
× (T
2
− 298)
= 50 050 × 10
3
which gives T
2
= 2335 K.
As we have used an estimated T
2
at the start to obtain c
p
, the calculation
has to be repeated with the new T
2
to obtain a new mean value of tempera-
ture to determine a new set of mean c
p
-values. Another iteration using this
new mean temperature of 1316 K for properties yields a flame temperature
of 2276 K. After a few more iterations we obtain a final value of the adiabatic
flame temperature as 2288 K. This sort of calculation can be easily per-
formed using an iterative spreadsheet with polynomial fits for c
p
-values of the
species. More precise calculations require integrals for mean c
p
which could
also be incorporated into a calculation using polynomial fits for c
p
-values.
Knowing the adiabatic flame temperature for a given air/fuel ratio is import-
ant in CFD, because it constitutes an upper limit for predicted temperatures
in a combustion calculation. If a combustion calculation produces higher
temperatures than the adiabatic flame temperature, the procedure has to be
checked carefully and corrected. Sometimes such overpredictions can occur
350 CHAPTER 12 CFD MODELLING OF COMBUSTION
ANIN_C12.qxd 29/12/2006 04:44PM Page 350
12.8 EQUILIBRIUM AND DISSOCIATION 351
due to underflow and overflow errors in the computations and the computa-
tional procedure should be structured to avoid such errors.
The above analysis for adiabatic flame temperature is based on complete
combustion. In practice, however, at high temperatures some reactions occur
in the reverse direction. This phenomenon is called dissociation. Dissociation
reactions are endothermic: therefore, the actual temperature will be lower
than the adiabatic flame temperature based on complete combustion. At
typical combustion temperatures important dissociation equations are
CO
2
bcg CO + O
2
H
2
O bcg H
2
+ O
2
H
2
O bcg H + OH
H
2
bcg H + H
O
2
bcg O + O
NO bcg N
2
+ O
2
As a consequence of dissociation, species like CO, H
2
, OH, O
2
will be pre-
sent in the products. These are called minor species since they usually occur
in much smaller concentrations than the main reaction products CO
2
and
H
2
O. Later we will see that combustion mechanisms for even simple fuels
are much more complicated and result in many minor species in the prod-
ucts. The assumption of complete combustion is certainly a simplification.
When concentrations of minor species in the products are known the adia-
batic flame temperature of such a mixture can be calculated by taking into
consideration all species of reactants and products.
Given the conditions for combustion and the air/fuel ratio we would like
to know what the resulting composition of the mixture is after dissociation.
When a mixture has reached chemical equilibrium, the composition of the
gas mixture can be obtained using a parameter known as the equilibrium
constant K
p
. The equilibrium criterion is based on the second law of ther-
modynamics. See standard thermodynamics textbooks for the theory on the
applications of the second law to combusting systems. In the analysis the
thermodynamic property known as the Gibbs function is used. The Gibbs
function is defined as
g = h − Ts (12.33)
where h is specific enthalpy (units J/kg), T is temperature in K and s is
specific entropy (units J/kg.K). The units of specific Gibbs function g are
the same as the specific enthalpy, i.e. J/kg, and its values are available in
thermodynamic tables (Rogers and Mayhew, 1994).
It can be shown that, as a consequence of the second law, a system seeks
to maximise its entropy, so a reacting system is in equilibrium when the total
Gibbs function (units J) attains a minimum value. Mathematically, the cri-
terion is expressed in terms of
(dG)
T, p
= 0 (12.34)
1
2
1
2
1
2
1
2
Equilibrium and
dissociation
12.8
ANIN_C12.qxd 29/12/2006 04:44PM Page 351
When this is applied to a combustion reaction in the form
aA + bB bcg cC + dD (12.35)
where a, b, c and d are stoichiometric coefficients of the species participating
in the reaction, and these species are perfect gases and the reaction is isother-
mal, it is possible to show that the condition for equilibrium is (see e.g.
Kuo, 2005)
∆G
o
T
=−R
u
T ln K
p
(12.36)
where ∆G
o
T
is called the standard state Gibbs function change, K
p
is the
equilibrium constant and R
u
is the universal gas constant.
Furthermore, it can be shown that
K
p
=
where p
A
, p
B
, p
C
etc. are partial pressures of species A, B, C etc.
We note that each partial pressure is raised to a power corresponding to
the stoichiometric coefficient. In equation (12.36) the standard state Gibbs
function change ∆G
o
T
may be calculated from
∆G
o
T
=∆H
o
− T∆S
o
(12.37)
These quantities may be obtained from thermodynamic property tables.
From equation (12.36) K
p
= exp(−∆G
o
T
/R
u
T), so K
p
-values may be obtained
from thermodynamic property tables or calculated from the standard state
Gibbs function change using equations (12.37) and (12.36).
If there are many reactants and products involved in the mixture in the
form
aA + bB + cC + ...bcg eE + fF + . . . (12.38)
the equilibrium constant K
p
is given by
K
p
== (12.39)
In the equation for K
p
products appear in the numerator and reactants
appear in the denominator. The equation also shows that partial pressures
can be written in terms of mole fractions.
In practical equilibrium combustion calculations many independent
reactions are involved, so we solve balance equations for overall mass or mole
conservation along with expressions written for K
p
in terms of mole fractions
for each independent reaction to give a set of relationships between product
and reactant mole fractions. Additional relationships are obtained from atom
balances for each of the atomic species participating in the reaction and from
the air/fuel ratio. This set of algebraic equations is solved to obtain mole
fractions of all the molecular species. An illustrative example of this type of
equilibrium chemistry calculation is given below.
Dedicated computer codes based on the principle of Gibbs function
minimisation are used for equilibrium calculations in systems with complex
reaction mechanisms involving many species (see e.g. Morley, 2005; Kuo,
2005). Further literature on equilibrium calculations can be found in Turns
(2000) and Warnatz et al. (2001). The interested reader is also referred to
the NASA equilibrium calculation program documented by Gordon and
X
e
E
X
f
F
...
X
a
A
X
b
B
...
p
e
E
p
f
F
...
p
a
A
p
b
B
...
p
c
C
p
d
D
p
a
A
p
b
B
352 CHAPTER 12 CFD MODELLING OF COMBUSTION
ANIN_C12.qxd 29/12/2006 04:44PM Page 352
12.8 EQUILIBRIUM AND DISSOCIATION 353
McBride (1994) and available via the Internet at the CEA website (see URL
under McBride, 2004).
Methane and air at an equivalence ratio of
φ
= 1.25 burn at 1600 K and 1.0 atm
pressure. Determine the composition of the products, which include H
2
and CO.
The stoichiometric combustion equation is
CH
4
+ 2(O
2
+ 3.76N
2
) → CO
2
+ 2H
2
O + 2 × 3.76N
2
(12.40)
Equation (12.23) gives the stoichiometric air/fuel ratio (A/F )
st
= 17.255.
Equation (12.24) allows us to calculate the actual air/fuel ratio for the given
equivalence ratio
φ
= 1.25:
(A/F) ===13.804
The actual combustion process includes dissociation. The following dissoci-
ation reactions are responsible for the creation of the species H
2
and CO:
CO + O
2
bcg CO
2
H
2
+ O
2
bcg H
2
O
When the fuel mixture is richer than stoichiometric (
φ
> 1) it can be argued
that O
2
is very low as it will be consumed by H
2
. The two chemical equations
above can be combined into
CO + H
2
O bcg CO
2
+ H
2
This is called the water–gas equilibrium equation.
The actual combustion equation should therefore include these additional
product species H
2
and CO:
CH
4
+ a(O
2
+ 3.76N
2
) bcg bCO
2
+ cCO + dH
2
O + eH
2
+3.76aN
2
The number of moles of air a can now be calculated from the air/fuel ratio:
(A/F ) ==13.804
a = 1.6
Thus,
CH
4
+ 1.6(O
2
+ 3.76N
2
) bcg bCO
2
+ cCO + dH
2
O + e H
2
+ 3.76 × 1.6N
2
An atom balance for all three atomic species that participate in the combus-
tion reaction yields three equations:
Carbon (C) balance: 1 = b + c (12.41)
Oxygen (O) balance: 3.2 = 2b + c + d (12.42)
Hydrogen (H) balance: 4 = 2d + 2e (12.43)
Since there are four unknown b, c, d and e we need a further equation. As we have
seen, the equilibrium constant (12.39) connects the mole fractions of reactants
and products, so the equilibrium constant K
p
supplies the fourth equation.
a × 4.76 × 29
1 × 16
1
2
1
2
17.255
1.25
(A/F)
st
φ
Example 12.2
Solution
ANIN_C12.qxd 29/12/2006 04:44PM Page 353
K
p
for the water–gas equilibrium equation is
K
p
== (12.44)
K
p
= (12.45)
where n is the total number of moles in the mixture.
Values for K
p
for the water–gas equilibrium equation are tabulated as a
function of temperature in thermodynamic data tables (Rogers and Mayhew,
1994; or the JANAF tables in Chase et al., 1985). For a combustion temper-
ature of 1600 K, the tables show ln(K
p
) =−1.091, which gives K
p
= 0.3358
and, hence,
0.3358 = (12.46)
Now equations (12.41)–(12.43) and (12.46) are solved to obtain values of b,
c, d and e. In this case by simple elimination a quadratic equation for b can be
obtained
0.6642b
2
+ 0.8746b − 0.7387 = 0
The solution gives
b = 0.5848
c = 0.4152
d = 1.6152
e = 0.3848
As mentioned earlier, dissociation reactions are endothermic and result
in incomplete combustion. At equilibrium, reactants such as CO and H
2
are
present in the products. This means that the fuel has not been utilised in full
and the flame temperature will be lower than the adiabatic flame temperature
estimated on the basis of complete combustion. However, it takes time for
combustion products to achieve equilibrium. In practice, therefore, equilib-
rium calculations will give good predictions of product composition if the
residence time is long. For example, in long furnaces which operate at high
temperature the products might achieve equilibrium.
In many other combustion systems, where the flow or operating con-
ditions do not provide adequate time for equilibrium to be reached, the
product composition cannot be predicted from equilibrium alone. Other
factors, such as finite rate reaction kinetics, turbulence, transient effects etc.,
have to be taken into account. The internal combustion engine is an import-
ant example of combustion taking place in a highly transient environment.
Peak temperatures and pressures at which dissociation takes place occur
only for a very short period. The rapid drop in pressures and temperatures
immediately after the combustion phase causes the product composition to
be ‘frozen’ before equilibrium is reached. Another important non-equilibrium
phenomenon in engine combustion is pollutant formation, which is strongly
influenced by chemical kinetics. These processes will be discussed in the
next sections. Nevertheless, for many combustion systems equilibrium cal-
culations give a fair representation of final products. Such details are very
useful for the design of combustion equipment. Where applicable, CFD
be
cd
be
cd
(b/n)(e/n)
(c/n)(d/n)
p
CO2
p
H
2
p
H
2
O
p
CO
354 CHAPTER 12 CFD MODELLING OF COMBUSTION
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12.10 OVERALL REACTIONS AND INTERMEDIATE REACTIONS 355
calculations of combustion can be carried out with equilibrium chemistry
models (Jones and Priddin, 1978; Nazha et al., 2001). For situations where
equilibrium models are not applicable more elaborate combustion models are
required. General-purpose CFD codes obviously need to include capabilities
to deal with any situation, so composition calculations need to take into
account effects such as departure from equilibrium, turbulence, finite rate
chemistry and radiative heat transfer. The relevant modelling concepts are
discussed in sections to follow.
As mentioned earlier, combustion of a fuel does not occur in a single reac-
tion, but may involve a number of different steps. In the above discussion it
was mentioned that it takes time for products to reach equilibrium. Chemical
kinetics ultimately determines how long it takes for a system to reach its
final equilibrium state. Essentially, chemical kinetics is the study of reaction
mechanisms and reaction rates.
Many combustion processes are physically controlled, i.e. the rate of com-
bustion depends on flow, turbulence and diffusion processes. Some examples
are: (i) wick flames such as candle flames and oil lamps, (ii) combustion of
pools of liquids, pool fires, (iii) droplet combustion, burning of liquid fuels
in furnaces, (iv) diesel engine combustion, (v) gas turbine combustion at low
altitude, (vi) rocket motors, (vii) laminar and turbulent jet diffusion flames,
(viii) combustion in boilers and furnaces, (ix) turbulent premixed combus-
tion in petrol engines, aeroengines etc. In situations such as these diffusion
and turbulence dominate mixing and subsequent combustion.
However, chemical kinetics can play an important role under certain con-
ditions, e.g. when the pressure is low or the supply of oxygen is restricted.
Combustion is never independent of physical processes, so ‘kinetically
influenced’ is the correct way to describe these processes. Some examples
of kinetically influenced situations are: (i) propagation of laminar flames
through premixed fuel and oxidant, (ii) aerated laminar flames such as gas
cooker and domestic boiler flames, (iii) ignition processes such as spark
ignition in petrol engines, auto-ignition in diesel engines, ignition in domes-
tic appliances, (iv) extinction processes such as extinction of gas turbines
at high altitude, petrol engines running too weak or too rich, extinction
of stabilised flames, (v) complex situations where kinetics competes with
mixing, e.g. burning of very small droplets or particles, insufficient avail-
ability of oxygen in pulverised coal flames, situations with large radiative heat
losses from coal particles and highly sheared flow situations.
In the combustion calculations which we discussed earlier we wrote the com-
bustion equation as a single overall reaction or global reaction:
Fuel + Oxidant → Products (12.47)
For example,
CH
4
+ 2O
2
→ CO
2
+ 2H
2
O (12.48)
In practice this reaction does not occur in this single-step fashion, since
it would require the simultaneous meeting of three different reactant
Mechanisms of
combustion and
chemical kinetics
12.9
Overall reactions
and intermediate
reactions
12.10
ANIN_C12.qxd 29/12/2006 04:44PM Page 355
molecules. Chemical reactions more commonly occur due to collisions of
pairs of molecules, where chemical bonds are broken during impact and new
bonds are formed. In the process many intermediate bonds and intermediate
species are formed. Such reactions are called elementary reactions and
involve stable intermediate species and radicals. For example, some common
radicals involved in reactions are H, O, OH, CH, HO
2
, C
2
. These radicals are
unstable and highly reactive because they contain unpaired electrons. The
simplest and most well-documented reaction is H
2
combustion:
H
2
+ O
2
→ H
2
O (12.49)
Even this simple reaction involves many intermediate steps such as
H
2
+ O
2
bcg 2OH (12.50)
H
2
+ OH bcg H
2
O + H (12.51)
H + O
2
bcg OH + O (12.52)
H
2
O + O bcg OH + OH (12.53)
H + OH + M bcg H
2
O + M (12.54)
The symbol ‘M’ represents a third body that can be any of the species
present in the system. The subject of chemical kinetics deals with details
of how these reactions occur. There are many mechanisms and paths of
reactions, and chemical kinetics has been a research area which has attracted
a great deal of attention. We do not intend to go on into the details here, and
the reader is referred to Turns (2000), Kuo (2005), Warnatz et al. (2001),
Bartok and Sarofim (1991), Gardiner (1984) and further references therein
for more details. Below we outline some of the basic concepts useful for CFD
combustion procedures.
In combustion modelling it is required to determine rates of reactant con-
sumption and product formation. These are used as source terms in transport
equations for each of the species. It has already been mentioned that many
intermediate equations are involved in combustion, and one particular species
may be formed and consumed in a number of different reactions where some
of these reactions may be reversible. For illustration purposes let us consider
a scheme of (forward-only) reactions represented by the stoichiometric equation
ν
′
kj
M
k
→
ν
″
kj
M
k
for j = 1, 2,...,m (12.55)
ν
′
kj
are stoichiometric coefficients of reactant species M
k
in the reaction j.
ν
″
kj
are stoichiometric coefficients of product species M
k
in the reaction j. N is
the total number of species involved and m is the total number of reactions
in the scheme.
Mass conservation enforces that
ν
′
kj
(MW )
k
=
ν
″
kj
(MW )
k
for j = 1, 2,...,m (12.56)
N
∑
i=1
N
∑
k=1
N
∑
k=1
N
∑
k=1
1
2
356 CHAPTER 12 CFD MODELLING OF COMBUSTION
Reaction rate12.11
ANIN_C12.qxd 29/12/2006 04:44PM Page 356