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Mass and molar fractions
yi 
mi

mtot
mi
xi 
N
m
ni

ntot
j
j 1
N
ni
yi – mass fraction of species i
xi – molar fraction of species i
mi – mass of species i [kg]
ni – number of moles of species i
Mi – molar mass of species i [kg/kmol]
N – total number of species
N
n
j
j 1
N
 y  x
i
i 1
i
1
i 1
mi  ni M i
mtot  ntot M mist
N
M mist
m
 tot 
ntot
n
j
j 1
ntot
Mj
N

n
j 1
nj
N
Mj 
tot
x
j
Mj
M mist 
j 1
mtot
m
 N tot 
ntot
nj

j 1
yi 
mi
ni M i
Mi

 xi

mtot ntot M mist
M mist
xi M i
N

j 1
Thermochemistry
xj M j
xi 
ni
ntot
mtot
N
1

N
M M
mj
j
j 1
mi
yi
Mi
Mi



mtot
1
M mist
M mist
j 1
yj
j
yi
Mi
N
M
j 1
yj
j
Combustion
Mass and molar concentrations
X i   ni
[Xi] – molar concentration of species i [kmol/m3]
V
In combustion problems, chemical species may generally be treated as
perfect gases satisfying the ideal gas law
p V = n Ro T
pi V = ni Ro T
Ro=8 314.3 J/kmol K : universal gas
constant
pv=RT
pi vi = Ri T
Ri : gas constant of species i
R = Ro / Mmist
Ri = Ro / Mi
p – total pressure [atm]
V – total volume [m3]
pi – partial pressure of species i [N/m2]
v – specific volume [m3/kg]
p i ni

 xi
p
n
Thermochemistry
Combustion
Mass and molar concentrations
Relations between molar concentrations and molar fractions
 X i   xi
xi   X i 
p
Ro T
Ro T

p
X i 
 X 
N
j
j 1
Relations between molar concentrations and mass fractions
X i   p
Ro T
yi
Mi
N
yj

Mj
j 1
y
 i
Mi
yi 
X i  M i
 X  M
N
j
j
j 1
Mass concentration of species i, i [kg/m3]
i 
mi mi mtot

 yi 
V
V mtot
Thermochemistry
Combustion
Stoichiometry
• A reactive mixture of a fuel and an oxidant is referred to as stoichiometric when
the mass of oxidant is the theoretically necessary and sufficient to fully burn the
fuel in an ideal combustion process.
• Stoichiometric chemical reaction of an hydrocarbon burning in air:
CxHy + (x + y/4) (O2 + 3.76 N2)  x CO2 + (y/2) H2O + 3.76 (x + y/4) N2
(The mole fractions of O2 and N2 in air are 21% and 79%, respectively, yielding
79/21=3.76 moles of N2 per mole of O2)
• The stoichiometric air/fuel ratio for an hydrocarbon is given by


 mar 
x  y 4 M O2  3.76 M N 2
M



 4.76 x  y 4 ar
m 
M fu
M fu
 fu esteq
Thermochemistry
Combustion
Stoichiometry
• Equivalence ratio:
m fu
mox

 m fu 


m
 ox esteq
ox stands for oxidant (typically, the air), rather than oxygen
 < 1 – fuel lean mixture
 = 1 – stoichiometric mixture
 > 1 – fuel rich mixture
• When the oxidant is the air, the excess air coefficient is defined as

1


• Excess air (in percentage):
Thermochemistry
mar
m fu
 mar 


m 
 fu esteq
mar  mar 

m fu  m fu 
1
esteq
e % 
 100% 
 100%

 mar 


m 
 fu esteq
Combustion
Enthalpy of formation
• Enthalpy of formation of a compound, hf [J/kg], is the net change of enthalpy that
occurs in the reaction of formation of a unit of mass of that compound from its
constituent elements in their natural states (at given pressure and temperature).
• Molar enthalpy of formation of a compound, h f [J/kmol], is the net change of
enthalpy that occurs in the reaction of formation of one mole of that compound
from its constituent elements in their natural states (at given pressure and
temperature).
• The natural state of an element is the most stable state (gas, liquid or solid) of
that element, i.e., the state at which that element is found in nature.
• Standard reference state conditions: Tref = 25 ºc, pref=po = 1 atm (superscript
denotes atmospheric pressure)
o
• The standard molar enthalpy of formation h fo Tref  for several substances is given
in Table A1.1 of Appendix 1 of the book.
Thermochemistry
Combustion
Enthalpy of formation
• The enthalpy of formation of chemical elements in their natural state is
arbitrarily set to zero, e.g., O2(g) , H2(g) , N2(g) , Hg(l) , C(s)
• The enthalpy of formation of a compound is negative when the chemical
reaction of that compound from its constituent elements releases energy, e.g.,
C (s) + O2 (g)  CO2 (g)
h fo (Tref=25 ºC) = -393 522 kJ/kmol (see Table A1.1
for CO2)
• In contrast, the enthalpy of formation is positive when the reaction of formation
absorbs energy, e.g.,
O2 (g)  2 O
(g)
h fo (Tref=25 ºC) = 249 173 kJ/kmol (see Table
A1.1 for O)
• A substance with high enthalpy of formation is more chemically reactive than
another one with a lower enthalpy of formation
Thermochemistry
Combustion
Absolute enthalpy
• The absolute enthalpy (specific or molar) of a substance is defined as:
 
hi T   h fo,i Tref   hs ,i T 
hi T   h of ,i Tref  hs ,i T 
•
In the case of a perfect gas, the sensible enthalpy (specific or molar) is
given by
hs ,i T   
T
hs ,i T   
T
Tref
Tref
•
hs – Sensible enthalpy, defined as the
difference between the enthalpy at T and at Tref
c p,i T * dT *
cp,i [J/kg.K]– specific heat capacity at constant
pressure
c p,i T * dT *
The specific and the molar enthalpies of an ideal mixture are given by
N
hmist   y i hi
i 1
Thermochemistry
N
hmist   xi hi
i 1
Combustion
Enthalpy of reaction and enthalpy of
combustion
• In systems with constant pressure or constant volume, the change of energy in
a chemical reaction depends only on the initial and final states of the system.
• In an open system at constant pressure, the change of energy in a chemical
reaction is equal to the change of enthalpy of the mixture (neglecting variation
of potential and kinetic energy, as well as the work done on the system or by
the system, except work due to pressure forces).
When the reactants and the products are at the same temperature and
pressure, that change of energy is referred to as enthalpy of reaction and, in
the case of combustion reactions, by enthalpy of combustion.
hR T   h prod (T )  hreag T 
Specific enthalpy of reaction/combustion [J/kg]
hR T   h prod (T )  hreag T 
Molar enthalpy of reaction/combustion [J/kmol]
H R T   H prod (T )  H reag T 
Total enthalpy of reaction/combustion [J]
Thermochemistry
Combustion
Enthalpy of reaction and enthalpy of
combustion
• The enthalpy of combustion is always negative, since the combustion reactions
are exothermic.
• At standard reference state conditions (T = 25 ºC and p = 1 atm), we have the
standard enthalpy of reaction and the standard enthalpy of combustion.
• Hess’s law: the heat released or absorbed in a chemical process is the same
whether the process takes place in one or in several steps. Example:
Given
C (s) + O2 (g)  CO2 (g)
hRo = -393 522 kJ/kmol
and
CO (g) + ½ O2 (g)  CO2 (g)
hRo = -282 995 kJ/kmol
Then
C (s) + ½ O2 (g)  CO (g)
hRo = -393 522 + 282 995
= -110 527 kJ/kmol
• The enthalpy of reaction/combustion depends on the temperature
Thermochemistry
Combustion
Enthalpy of reaction and enthalpy of
combustion
• The enthalpies of combustion at two different temperatures may be related as
follows:
 
H R T2  
mi
T1
i
reag 
 
H R T2  
ni
i
reag 
T2
T2
T1
c p,i T  dT  H R T1  


mj
T1
j
 prod 
c p,i T  dT  H R T1  
T2
 
nj
T2
T1
j
 prod 
c p, j T  dT
c p, j T  dT
• If T1= 25 ºC and p = 1 atm:
H R T2  
j
 prod 

m j  h of , j Tref 

  T 2
ref
  
 H Ro Tref 
mj
j
 prod 
Thermochemistry
T
T2
Tref

c p , j T  dT  

c p , j T  dT 
i
reag 

mi  h of ,i Tref 

  T 2
ref
i Tref
mi
T2
T

c p ,i T  dT 

c p ,i T  dT
reag 
Combustion
Internal energy of formation,
reaction and combustion
• The enthalpies of formation, reaction and combustion are used to determine
changes of energy in systems at constant pressure (e.g., combustion
chambers, boilers, furnaces, gas turbine combustors)
• The internal energies of formation, reaction and combustion are used to
determine changes of energy in systems at constant volume (e.g., spark
ignition engines)
• The enthalpy and the internal energy of reaction may be related as follows:

H R T   U R T    pV   U R T   RoT n prod  nreag

U R T   U prod T   U reag T  

j
 prod 

m j  u of , j Tref 

  T
ref
T
   m j T
ref
T
 U Ro Tref 
j
 prod 
Thermochemistry

cv , j T  dT  

cv, j T  dT 
i
reag 

mi  u of ,i Tref 

  T
ref
i Tref
mi
T
T

cv ,i T  dT  

cv ,i T  dT
reag 
Combustion
Heating value of a fuel
• Lower heating value at constant pressure = symmetric of the standard enthalpy
of combustion, per unit mass of fuel, when there is water vapour in the
combustion products (always positive)
• Lower heating value at constant volume = symmetric of the standard internal
energy of combustion, per unit mass of fuel, when there is water vapour in the
combustion products (always positive)
• Higher heating value at constant pressure = symmetric of the standard
enthalpy of combustion, per unit mass of fuel, when there is liquid water in the
combustion products (always positive)
• Higher heating value at constant volume = symmetric of the standard internal
energy of combustion, per unit mass of fuel, when there is liquid water in the
combustion products (always positive)
• The difference between the higher and the lower heating value is equal to
product of the mass of water (liquid or gas) in the combustion products by the
latent heat of vaporization (hfg for constant pressure or ufg for constant volume)
Thermochemistry
Combustion
Adiabatic flame temperature
• The adiabatic flame temperature is the temperature of the combustion products
in an adiabatic combustion process with negligible variation of the kinetic and
potential energy and with no work exchange with the surroundings.
• The adiabatic flame temperature is an upper limit of the temperature in a
combustion process, the actual temperature of the combustion products being
generally lower due to heat losses by conduction, convection or radiation.
• The adiabatic flame temperature in a combustion process at constant pressure
with the reactants at temperature T1 is obtained from:
H prod Tad   H reag T1 



o
o
 m j h f , j Tref   Trefad c p , j T  dT   mi h f ,i Tref   Tref1 c p ,i T dT
j
 prod 
T
i
 reag 
T

H Ro Tref    m j Trefad c p , j T  dT   mi Tref1 c p ,i T  dT
T
j
 prod 
Thermochemistry
T
i
 reag 
Combustion
Adiabatic flame temperature
• Knowledge of the chemical composition of the combustion products is required
to determine the adiabatic flame temperature.
• At typical adiabatic flame temperatures of hydrocarbons (close to 2300 K),
dissociation of the combustion products occurs, and the chemical composition
of the products must be determined together with the adiabatic flame
temperature. For the time being, dissociation of the combustion products will be
ignored.
• The specific heat capacities as a function of temperature may be accurately
approximated by polynomials, which are given in table A2.1 for several
common species. In the case of a manual calculation, it is sufficient to take
constant values evaluated at the average temperature (Tref+Tad)/2. Instead, the
sensible enthalpies given in tables A2.2 to A2.13 may be used.
• The adiabatic flame temperature is a function of the temperature of the
reactants and of the equivalence ratio. If there is no dissociation, the adiabatic
flame temperature is maximum for a stoichiometric mixture.
Thermochemistry
Combustion
Adiabatic flame temperature
• The adiabatic flame temperature in a combustion process at constant volume
with the reactants at temperature T1 is obtained from:
U prod Tad   U reag T1 
H prod Tad   H reag T1   U prod Tad   U reag T1   V  p 2  p1 
H prod Tad   H reag T1   Ro n prod T prod  nreag Treag 
• The ideal gas equation of state must be solved together with the above
equation, since the pressure of the combustion products is unknown, and it
influences the chemical composition of the combustion products
Thermochemistry
Combustion
Dissociation
• Chemical species such as CO2, H2O, N2 and O2 are generally stable up to
about 1250 K. At higher temperatures, dissociation occurs, yielding other
species such as CO, H2, OH, H, O, etc., which are formed in reactions like
CO2  CO + ½ O2
(symbol  denotes a reversible reaction)
• The proportion of CO2, CO and O2 at a particular temperature and pressure
adjusts itself such that, after some time, the reaction rates of the direct and
inverse reactions are equal, i.e., the number of moles of CO2 that dissociate,
according to the direct reaction, is equal to the number of moles that are
formed, according to the inverse reaction, per unit of time.
A chemical equilibrium state is then achieved,
with the direct and inverse reactions proceeding
at the same rate.
• Mass balances are sufficient to determine the
chemical composition of the combustion products
only if no dissociation occurs.
Thermochemistry
Combustion
Chemical equilibrium criteria
• The chemical equilibrium condition is obtained by combining the 1st and 2nd
laws of thermodynamics:
dU = dQ – p dV
 dQ 
dS  

T

 rev
• Therefore,
(1st law for a closed system, assuming that there is only
work done by pressure forces)
 dQ 
dS  

T

irrev
dS U ,V
0
(2nd law)
dU S ,V
0
• It can be shown that for a reactive system in chemical
equilibrium the reactions are reversible. It follows that:
The equilibrium condition for a process with U and V constants is the
maximization of the entropy of the system;
The equilibrium condition for a process with S and V constants is the
minimization of the internal energy of the system.
Thermochemistry
Combustion
Chemical equilibrium criteria
• Processes with U and V constant or with S and V constants do not generally
occur in combustion systems. In these systems, the Gibbs free energy, G,
and the Helmholtz function, F, are more common:
G = H –T S
F = U –T S
dG p,T
dF V ,T
0
0
• The equilibrium condition for a process with p and T constants is the
minimization of the Gibbs free energy of the system.
• The equilibrium condition for a process with V and T constants is the
minimization of the Helhmoltz function of the system.
• All chemical equilibrium criteria are equivalent, i.e., all criteria yield the same
solution, i.e., the same chemical equilibrium composition of a reactive system.
Thermochemistry
Combustion
Chemical equilibrium
dG  dH  TdS  SdT  dU  pdV  Vdp  TdS  SdT  dQ  Vdp  TdS  SdT
• Free Gibbs energy for an isothermal system in chemical equilibrium
dG  V dp 
n Ro T
dp  n Ro T d ln p 
p

GT , p   G o T   n Ro T ln p p o
GT , p  
N
n
i
g i T , p  
i 1
 n g
N
i
o
i
Go [J] -free Gibbs energy at po=1 atm

g io T  [J/kmol] - molar Gibbs energy at T and po
T   Ro T ln pi
po

i 1
 
g io T   g io, f Tref  g io,s T 
gio, s T    d gio
T
Tref
Thermochemistry
g io, f
(Tref) [J/kmol] –standard free Gibbs energy of formation
g io, s [J/kmol] – sensible molar Gibbs free energy
Combustion
Chemical equilibrium
• The Gibbs free energy of formation is equal to zero for chemical elements in
their natural state
• The Gibbs free energy of formation of several species is given in Tables A2.2
to A2.13
Differentiation of the free Gibbs energy yields
dGT , p  
 
N
dni g io
i 1
T   Ro T ln pi
  n d g
N
p
o
i
o
i
T   Ro T ln pi
po

i 1
The last term is zero for T and p constant (why?)
Thermochemistry
Combustion
Chemical equilibrium
Hence, the chemical equilibrium condition may be written as
dGT , p  
 dn g
N
i
o
i
T   Ro T ln pi

po  0
i 1
Let’s consider the following general reaction in chemical equilibrium
a A + b B + ...  e E + f F +...

A, B, ..., E, F, ... – chemical species
a, b, …, e, f, … – stoichiometric coefficients
dn A
dn
dn
dn
  B  E  F 
a
b
e
f

 
   
 e g Eo T   Ro T ln pE p o  f g Fo T   Ro T ln pF p o    0

o
T   Ro T ln p A p o  b g Bo T   Ro T ln pB p o
a gA
Thermochemistry
Combustion
Chemical equilibrium
Chemical equilibrium condition:


 p po
o
o
o
o
a g A T   b g B T     e g E T   f g F T     Ro T ln  E
 p A p o
 p
 p
e
F
po
B
po
a
 
  
f
b
G o T   e g Eo T   f g Fo T     a g Ao T   b g Bo T   
Equilibrium constant:
pEe pFf 
Kp  a b
p A pB 
(with pA, pB, …, pE, pF, … in atm)
The equilibrium constants for the reactions of formation of several species from
their constituent elements in their natural state are given in table A4.1
The equilibrium constants of other reactions may be determined from the
tabulated equilibrium constants by means of algebraic manipulations.
Thermochemistry
Combustion
Equilibrium constants
Example: calculate the equilibrium constant of reaction
H2O  H + OH
The following equilibrium constants are tabulated:
K p1 
H2 + ½ O2  H2O
½ H2  H
K p2 
½ H2 + ½ O2  OH
K p3 
p H 2O
pH2
pO2
pH
pH2
pOH
pH2
pO2
The equilibrium constant of the reaction given above is then
K p4 

K p2
p H pOH

p H 2O
Thermochemistry
pH2
K
p3
pH2
K p1 p H 2 pO2
pO2
 K
p2
K p3
K p1
Combustion
Equilibrium constants
• The equilibrium constant may be explicitly obtained from
 G T   Ro T ln K p
o
as
 G o T  

K p  exp  
Ro T 

• The equilibrium constant, as well as the change of the Gibbs free energy,
indicates the trend of conversion of reactants into products or, equivalently, the
conversion of products into reactants.
• If Kp<1, then Go(T)>0  there are more reactants than products in equilibrium.
If Kp>1, then Go(T)<0  there are more products than reactants in equilibrium.
Accordingly, the larger Kp, the smaller Go(T), and the greater the amount of
products relatively to reactants under equilibrium conditions.
• The equilibrium constant Kp is independent of the total pressure, being only a
function of the temperature
Thermochemistry
Combustion
Equilibrium constants
• The change of the Gibbs free energy may be expressed as
G o T   H o T   T S o T 
yielding
 H o T  
 S o T  
 exp 

K p  exp  
Ro T 

 Ro 
• Assuming that Ho(T) and So(T) change little with the temperature (this is a
good approximation, particularly for small or moderate temperature
changes), then ln(Kp) changes linearly with the inverse of temperature.
The relation between the equilibrium constants at two different temperatures
may then be expressed as
o
 K p T2  
T   1  1 

H

ln 
 K T  
Ro  T2 T1 
 p 1 
Thermochemistry
Exothermic reaction: Ho(T) < 0, T   Kp 
Endothermic reaction: Ho(T) > 0, T   Kp 
Combustion
Equilibrium constants
• The equilibrium constant may also be expressed in terms of mass fractions,
molar fractions, molar concentrations and total number of moles of every
species, e.g.
e
f

E  F  
Kc 
Aa Bb 
nEe nFf 
Kn  a b
nA nB 
 Ro T 
K p  K c  o 
 p 
e  f  a  b 
 p 

 K n 
o 
 ntot p 
e  f  a  b 
• In contrast with Kp, the equilibrium constants Kc and Kn depend on both
temperature and total number of moles, i.e., total pressure.
Thermochemistry
Combustion
Le Châtelier´s principle
• If a system in chemical equilibrium is disturbed (e.g., by changes in
temperature, pressure, chemical composition), the system will tend to shift its
equilibrium state so as to minimize or counteract the effect of the disturbance.
• If there is an increase of pressure, keeping the temperature unchanged, the
system will reduce the pressure by producing fewer moles.
Example:
CO2  CO + ½ O2
An increase of pressure will favour the
inverse reaction.
The system will shift to an equilibrium
state with lower CO and O2 and higher
CO2 mole fractions than originally.
Thermochemistry
Combustion
Le Châtelier´s principle
• If there is an increase of temperature, keeping the pressure unchanged, the
system will reduce the temperature by favouring the endothermic reaction.
Similarly, a decrease of temperature favours the exothermic reaction.
p = 1 atm
p=10 atm
xCO2
0.9994
0.9997
xCO
3.601×10-4
1.672×10-4
xO2
1.801×10-4
8.357×10-5
xCO2
0.9777
0.9895
T = 2000 K xCO
0.0149
6.96×10-3
xO2
0.0074
3.48×10-3
T =1500 K
Thermochemistry
Combustion
Calculation of chemical equilibrium
composition
• Whenever dissociation is present, the calculation of the adiabatic flame
temperature and the determination of the chemical equilibrium composition
require the specification of all the chemical species
• At least the following species are present in the combustion of hydrocarbons:
CO2, H2O, O2, CO, H2, OH, O, H, and for combustion in air, N2, N and NO.
Other species are also generally present, such as HO2, H2O2, N2O, NO2,
C(s), HCN, CH4 or C2H2.
• Formulation of the chemical equilibrium problem for a system at constant
pressure:
N species  N+1 equations (to determine molar fractions and adiabatic
flame temperature)
1 equation for energy conservation:
H prod Tad   H reag T1 
K equations for atomic mass balances (typically O, H, C, N)
N-K linearly independent equations to be found from chemical equilibrium
Thermochemistry
Combustion
Calculation of chemical equilibrium
composition
• K independent species out of the N species are selected in such a way that
none of those independent species may be transformed in another one of
them by means of chemical reaction. Moreover, the remaining N-K species
should be obtained by chemical reaction from the selected K independent
species.
The N-K equations are determined from the condition of chemical equilibrium
of the reactions of formation of the N-K species from the selected K
independent species.
• Example: combustion of methane in air with N = 11 species present under
equilibrium - CO2, H2O, O2, CO, H2, OH, O, H, N2, N and NO.
nCH 4 CH4  nar O2  3.76 N 2   nCO 2 CO2  nH 2 O H 2O  nO 2 O2  nCO CO  nH 2 H 2 
 nOH OH  nO O  nH H  nN 2 N 2  nN N  nNO NO
n CH 4 and nar are given, so that there are N = 11 unknown mole fractions
Thermochemistry
Combustion
Calculation of chemical equilibrium
composition
nCH 4 CH4  nar O2  3.76 N 2   nCO 2 CO2  nH 2 O H 2O  nO 2 O2  nCO CO  nH 2 H 2 
 nOH OH  nO O  nH H  nN 2 N 2  nN N  nNO NO
4 equations may be written for the atomic mass balances:
nCH 4  nCO 2  nCO
4 nCH 4  2 nH 2 O  2 nH 2  nOH  nH
2 nar  2 nCO 2  nH 2O  2 nO 2  nCO  nOH  nO  n NO
3.76  2 nar  2 nN 2  nN  nNO
The following 4 species are selected as independent: O2, H2, N2 and CO
The remaining 7 equations are obtained from the condition of chemical
equilibrium of the reactions of formation of the other 7 species from the 4
independent species
Thermochemistry
Combustion
Calculation of chemical equilibrium
composition
nCH 4 CH4  nar O2  3.76 N 2   nCO 2 CO2  nH 2 O H 2O  nO 2 O2  nCO CO  nH 2 H 2 
 nOH OH  nO O  nH H  nN 2 N 2  nN N  nNO NO
CO + ½ O2  CO2
H2 + ½ O2  H2O
½ H2 + ½ O2  OH
½ O2  O
½ H2  H
½ N2  N
½ N2 + ½ O2  NO
Thermochemistry
Kp 
 p 
 o

 p ntot 
nCO 2
nCO nO 2
1 2
Kp 
nOH
Kp 
nH 2 nO 2
Kp 
nH
nH 2
 p

 po n
tot

n H 2O
nH 2
Kp 
nO
Kp 
nN
nO 2
nO 2
12




Kp 
nN2
 p 


 po n 
tot 

1 2
12
 p

 po n
tot





 p

 po n
tot





12
nNO
nN 2 nO 2
Combustion
Chemical equilibrium software
• PER (Olikara and Borman, 1975) – Calculates the chemical equilibrium
composition and the adiabatic flame temperature for the combustion of
CxHyOzNw in air considering the 11 species of the previous example (a
modified version of this code is available in the book by Turns)
• The determination of the equilibrium composition using the reaction constants
has a few disadvantages (e.g., the need to specify the chemical reactions in
equilibrium, numerical difficulties when there are very small concentrations,
difficulty in dealing with problems with solid or liquid phases)
• CEA (Gordon and McBride, 1994) – code developed by NASA and based on
the minimization of the Gibbs free energy using Lagrange multipliers. See:
http://www.grc.nasa.gov/WWW/CEAWeb/ceaHome.htm
• STANJAN (Reynolds, 1986) – code based on the method of chemical
potentials of the elements. Online version available at
http://navier.engr.colostate.edu/~dandy/code/code-4/index.html
Thermochemistry
Combustion
Examples of application of chemical
equilibrium software
Example of
application of
the PER code
DADOS DE ENTRADA
Número de átomos de carbono no combustível
Número de átomos de hidrogénio no combustível
Número de átomos de oxigénio no combustível
Número de átomos de azoto no combustível
Razão de equivalência
Temperatura (K) – estimativa inicial
Pressão (Pa)
Entalpia dos reagentes (kJ/kmol combustível)
3
8
0
0
1.000
2000.0
101325.0
-103847.0
DADOS DE SAÍDA
Temperatura da chama e propriedades dos produtos da combustão
Temperatura adiabática de chama [K]
= 2266.78
Entalpia da mistura [J/kg]
= -0.1420E+06
Calor específico da mistura cp [J/kg-K]
= 0.227230E+04
Razão entre os calores específicos, cp/cv
= 1.1655
Massa molar da mistura [kg/kmol]
= 28.0638
Número de moles de combustível por mole de produtos
= 0.03838622
Fracções molares das espécies em condições de equilíbrio
H:
0.00046610 O:
H2: 0.00331509 OH:
NO: 0.00247177 O2:
CO2: 0.10273075 N2:
Thermochemistry
0.00031115
0.00320690
0.00589086
0.72078613
N: 0.00000002
CO: 0.01242792
H2O: 0.14839331
Combustion
Examples of application of chemical
equilibrium software
Thermochemistry
Combustion
Examples of application of chemical
equilibrium software
Thermochemistry
Combustion
Examples of application of chemical
equilibrium software
Thermochemistry
Combustion