ABSTRACT
There is a need to be able to perform a range of relatively simple calculations employing mainly mass and energy conservation equations applied to the fuel combustion processes, so as to establish some of their important characteristics� These may include the energy released by chemical reactions and the associated changes to products temperature and composition following combustion� Stoichiometry is a term generally used for the consideration of such treatment of fuel combustion reactions�
As an example, let the combustion of propane, C3H8, in air be considered� Ideally, the reaction can be represented by the following overall reaction equation:
2 2C H O N CO H O+ + → + +5 79 21 3 4 1[ / ] 8 8� N2
(Note: Air is usually considered to contain approximately 21% oxygen by volume and 79% nitrogen by volume, that is, in the volumetric ratio of 1 to 3�76� The corresponding concentrations by mass can be shown to be 23�3% and 76�7%, respectively, that is, in a mass ratio of 1 to 3�29)�
Such an equation can provide information on mass basis through balancing the conserved mass of elements on either side of the equation:
[ [ � ] [ ] [ ] �3 12 8 1 5 32 3 76 28 3 12 32 4 2 16 18 8 2× + × + + × → + + + + × 8
(i�e�, 730�4 kg → 730�4 kg) Information can be obtained also on molar or volume basis� However, the
volumes may not be necessarily balanced, for example, for the above reaction:
1 5 1 3 76 3 4 18 8vol� vol� vol� vol� vol�+ + ≠ + +[ � ] �
(i�e�, 24�8 for the reactants but 25�8 for the products)� It is to be noted that the moles represent volumes only for gases and when
Avogadro’s hypothesis applies, that is, when both the reactants and products are at the same temperature and pressure�
The overall reaction equation should indicate the phase of each component species, for example, whether gaseous (g), liquid (liq�), or solid (s)� For example:
C s O g CO g( ) ( ) ( )+ →2 2
or
H g O g H O liq�2 1 2 2 2( ) ( ) ( )+ →
If the reactants and products are at the same phase, then often the phase may not be displayed in the equations�
The reaction equation may give information about the property changes during the reaction and the associated ideal energy released:
CH g O g CO g H O g4 2 2 22 2( ) ( ) ( ) ( )+ → + + Q
where ΔQ is the energy released by the fuel combustion reactions, often in the form of heat� Its magnitude varies depending on the type of chemical reaction, its reactants and associated products, the temperatures of the reactants and products, the state of the reacting species, and whether the reaction is proceeding at a constant volume or constant pressure�
In general, ΔQ is described as the heat of reaction and for combustion reactions it is also described as the heat of combustion� The value of ΔQ can be specified in a number of possible ways, such as the following:
• kJ/mol of fuel • kJ/mol of air • kJ/mol of products • kJ/mol of reactants • kJ/kg of fuel • kJ/kg of reactants • kJ/kg of products, and so on
The value of ΔQ changes as the temperature of the reactants and/or the products are changed� It is of a negative value for combustion reactions, since for the exothermic processes it represents heat leaving the reaction system and thus by convention has a negative value� When
T T T Treactants products reference K= = = 298
then, ΔQ/kgfuel is commonly taken as the heating value on mass basis of the fuel� Often, it is reported in the literature for a temperature of 298 K�
It is to be noted that the term heating value is the same as the calorific value and often it is also given at 298 K� For solid and liquid fuels the heating
values are usually quoted per unit mass, while for gaseous and vapor fuels they are usually quoted on volume basis�
When H2O in the products is still in the vapor phase, the heating value is described as the lower or net heating value� But when H2O is allowed to condense to liquid and the enthalpy of condensation is fully recovered, then the value is given as the higher or gross heating value�
The difference between these two sets of values is due to the latent enthalpy of condensation, Hfg, of the water produced� However, for constant volume combustion the energy release will be due to changes in the internal energy of the reacting and products species, while the corresponding difference due to the condensation of water is the corresponding latent energy of condensation, Ufg� Table 5�1 shows a listing of the combustion properties of a number of common fuels including their higher and lower heating/caloric values at 298 K�
In general, when assuming ideal gases are involved the following simple relationship applies:
H U RT n nfuel fuel p R= + ∑ − ∑( )
where ∑nP and ∑nR are the moles of products and reactants, respectively, and R is the universal gas constant� For example, for the ideal combustion of propane considered earlier,
H U RT
per mol of fuel – ( � – � )
( ) =
=
25 8 24 8
When the temperature of the products is allowed to increase, the less will be the heat available that can be extracted and transferred to the surroundings of the reaction system� If no heat transfer is allowed out of the reacting medium, then the temperature of the products becomes sufficiently high and is known as the adiabatic flame temperature� Also, if some external energy is supplied to the reaction system, such as in the form of a spark or through preheating the reactants, then the temperature of the products will exceed the corresponding value of the adiabatic temperature�
In all stoichiometric calculations, it is always advisable to write the overall reaction equation in the molar form, since the reaction is assumed to be proceeding on that basis� Of course, in general, fuel combustion reactions in practical systems do not have to be carried out exclusively
in chemically correct proportions, which are usually described as stoichiometric or chemically correct proportions, where neither excess air nor excess fuel are present� Accordingly, the composition of fuel-air reactant mixtures is usually quoted in terms of the equivalence ratio, which relates the fuel to air mass ratio relative to the corresponding stoichiometric value� That is
Equivalence ratio
Mass of Fuel Mass of Air
, Φ =
Mass of Fuel Mass of Air etric
or
=
[ ] [ ]
Mass of Air Mass of Air
Fuel-air mixtures with excess air are described as being lean or weak, while those with excess fuel are called rich mixtures� Table 5�2 shows a listing of the adiabatic flame temperature values for the combustion in air of some common fuels at an initial reactant temperature of 298 K and atmospheric pressure� The values are listed for three sets of equivalence ratios of 0�8, 1�0, and 1�2� Accordingly, for lean fuel-air mixtures (with Φ < 1�0), some oxygen remains in the products unutilized, for example,
Φ + +
→ Φ + Φ + − Φ +C H 5 O
79 21
N 3 CO 4 H O [5 5 ]O 18�8N3 8 2 2 2 2 2 2
The concentration of a product such as CO2 in the wet products where the water vapor remains in the vapor phase will be
=
Φ Φ + Φ + − Φ +
3 [3 4 5 5 18�8]
The corresponding concentration in the dry products when the water vapor produced is condensed and removed out of the gaseous products will be
=
+ − +
3 3 5 5 18 8
Φ Φ Φ[ � ]
However, for fuel-rich mixtures the composition of the products cannot be determined without providing further information since incomplete and partial oxidation products such as carbon monoxide will be formed in addition to some possible unconverted fuel� When the nature and concentrations
of the products species are known or assumed then the reaction equation can be written in a balanced form�
For fuel mixtures the reaction equation can be similarly written� For example, the reaction equation of a fuel mixture of methane and propane in molar proportions of i1 to i2, respectively, with excess air is
i i A a b d f1 4 2 3 8 2 2 2 2 2 23 76CH C H O N CO H O O N+ + + → + + +( � )
where the conservation of the elements involved gives the following equations:
Carbon element mass balance:
= + 31 2a i i
Hydrogen mass balance:
= +2 41 2b i i
Oxygen mass balance:
= + +2 2 2A a b d
Nitrogen mass balance:
= 3�76f A
Equivalence ratio:
[ ] [ ] [ ]
Φ =
∑ ∑
∑ ∑
=
∑ ∑
=
+
mass mass
mass mass
mass
mass
2 5
1 2i i A
For example, for a mixture made up by volume of 15% methane, 5% propane, and 80% air the equivalence ratio becomes
Φ = × + ×
× =
(15 2 5 5) 80 0�21
3�27
which is a very rich mixture� Often in certain applications such as sometimes with fuel-lean mixtures,
the inverse of [] is used and it is given the symbol λ, that is,
λ =
∑ ∑
∑ ∑
mass of air mass of fuel mass of air mass of fuel
Quoted values of the fuel to air ratio need to be specified as to whether they are based on mass, volume, or a molar basis, since their values will be numerically different� Similarly, the presence of a component in the fuel-air mixture that does not take part in the reaction may be treated as a diluent such as N2, He, and CO2, and can be considered in the calculations, similarly, for example, when considering a mole of fuel mixture made up of 8% CH4, 16% H2 and 76% N2, the simultaneous solution of the elemental mass
balance equations will show that (0�08 × 2 + 0�16 × 0�5) moles of oxygen will be needed, which corresponds to requiring 1�142 moles of air for complete combustion of one mole of the fuel:
0 08 0 16 0 76 1 142 0 21 0 79 0 08
4 2 2 2 2� � � � ( � � ) �
CH H N O N+ + + + → CO H O N2 2 20 32 1 66+ +� �
The variation in the calculated temperature rise following the adiabatic combustion of methane-air mixtures at atmospheric pressure with changes in the values of the fuel to air mass ratio for different values of initial mixture temperature is shown in Figure 5�1� It can be seen that the peak value of the temperature rise is obtained with mixtures of around the stoichiometric value� Also, the higher the initial temperature, the lesser is the relative temperature rise�
Figure 5�2 shows typically the calculated adiabatic flame temperature for methane-air mixtures as a function of initial temperature for lean (Φ = 0�50) stoichiometric and rich (Φ = 1�50) mixtures displaying an almost linear relationship� It is seen that the stoichiometric mixture region produces the highest product temperature with a lower slope than that for lean or rich mixtures that produce relatively low products temperatures� An increase in the inlet mixture temperature will raise the products temperature but to a lesser extent� This is due to a number of reasons that include the non-linear changes
in the thermodynamic properties and the increased effects of dissociation with temperature�
As an example, consider the combustion of a fuel gas-air mixture of a known composition made up of methane, carbon dioxide, water vapor, and nitrogen initially at a temperature T, burning under constant atmospheric pressure in normal air to produce products at higher temperature of T2, while transferring heat to the surroundings to an extent of Δq per mole of fuel� Then, through thermodynamic analysis employing the first law of energy equation, while assuming the products to be in thermodynamic equilibrium, the temperature and composition of the products can be established according to the following procedure�
Let the overall reaction equation be represented by the following equation that assumes more realistically the presence of some unreacted methane and oxygen while producing some partial oxidation products of carbon monoxide and hydrogen� The products temperature expectedly is higher than that for the initial feed mixture:
+ + + + + + → +
+ + + + + +
CH CO H O N O (O 3�76N ) CH CO
A B D E G F a b
d e f g h Y q
In this equation there are seven unknown individual concentrations of the products (a, b, d, e, f, g and h) that may be assumed to be in thermodynamic equilibrium plus the unknown products temperature, Tprod, and the associated heat release, YΔq� The requisite independent equations come about from considering the four elemental mass balance equations, which are in
this case carbon, hydrogen, oxygen, and nitrogen, plus three suitably independent equilibrium constant equations and the energy equation for the changes due to the reaction process� The solution procedure without the aid of a suitable computer software tends to be very lengthy and would require a trial-and-error procedure with a knowledge of the appropriate thermodynamic properties as functions of temperature� For example, the four elemental mass balance equations in this case are
For carbon:
A B a b d+ = + + (5�1)
For hydrogen:
4 2 4 2 2A D a e f+ = + + (5�2)
For oxygen:
2 2 2 2 2B D G F b d e g+ + + = + + + (5�3)
For nitrogen:
3 76� F h= (5�4)
The assumption of having a state of equilibrium in the products implies that the proportion of the concentrations of the different species relative to one another will be fixed in accordance with the value of the corresponding equilibrium constants at that temperature� By reference to the relevant thermodynamic texts, the three independent equilibrium reactions can be chosen to be the following, which incorporate only species present in the products:
CH H O CO 3H4 2 2+ ↔ + (I)
CH CO 2CO 2H4 2 2+ ↔ + (II)
CO CO O2 2↔ + 12 (III)
Their corresponding equilibrium equations in terms of the partial pressures of species in atmospheres are as follows:
Kp P P
P P f d ae
P ni
= =
∑
2
(5�5)
Kp P P
P P f d ab
P ni
= =
∑
2
2
(5�6)
Kp P P
P dg
b P ni
= =
∑
2
(5�7)
These can be shown to be also equal to the following:
Kp
Kp Kp Kp1
Kp
Kp Kp KpII
= 2
and
Kp
Kp KpIII
where ∑ni is the summation of all the moles of the products appearing in the overall reaction equation, while P is the pressure in atmospheres� Kp Kp Kp KpCO CO CH H Oand2 4 2, , , are the corresponding equilibrium constants for the formation reactions of the species shown� For example, for carbon dioxide:
C O CO2 2+ ↔
and for water vapor:
H O H O2 12 2 2+ ↔
The equilibrium constant may be listed in terms of molar concentrations instead of in terms of partial pressures� For ideal gases these two types of values are mutually convertible easily� For example, the equilibrium constant on molar basis Kn in terms of Kp for a typical reaction
a b c dA B C D+ ↔ +
will be given as
K K
P n
p c= ∑
Table 5�3 shows a listing of the formation reactions for a number of common species together with their corresponding enthalpy of formation at 298 K� Table 5�4 shows a listing of values of the equilibrium constants of the formation reactions of some common species at a temperature of 298 K�
Figure 5�3 shows the approximately linear logarithmic variation of the equilibrium constant with the inverse of absolute temperature for a range of reactions commonly encountered in fuels and combustion applications�
The values of the relevant equilibrium constants are to be obtained from the literature chosen for the assumed product temperature T2, yet to be determined by invoking the energy equation:
n h n h Y qi i
R ∑ ∑− =
where hi, the specific molar enthalpy of each species, (i), at temperature T, is given as
h h p TT i
= + ∫fio c d hfio is the corresponding enthalpy of formation at reference temperature and cpi is the molar specific heat at constant pressure, a function of temperature� ni is the number of moles�
For the simple case of having the products containing neither unutilized oxygen nor fuel, the solution will be simplified to include the elemental mass balance equations and the common water gas shift reaction, arising from a combination of reactions (i), (ii), and (iii):
CO H O CO H+ ↔ +2 2 2
The equilibrium constant equation for this reaction then becomes
Kp bf de
Kp
Kp Kp f
= = CO
where Kpf is the equilibrium constant of the formation reaction of species “i�” In these types of calculations the air is often considered mainly for con-
venience to be dry� The presence of humidity that displaces some air can be easily accounted once the prevailing values of temperature, pressure, and relative humidity of the moist air are known�
For example, for humid air at 25°C and 100 kPa with a relative humidity of 60%, the value of the partial pressure of the water vapor will be 0�60 of the corresponding saturation value, which is under these conditions 3�169 kPa, as taken from the steam tables� Accordingly,
PH O kPa2 0 60 3 169 1 9014= × =� � �
The mass ratio of the water vapor to that of dry air, while assuming the low concentration vapor to behave as an ideal gas and the air to have a molar mass of 28�9 kg/kmol will be
mass mass
×
P
P
M
M
=
×
× =
1 9014 18 98 10 28 9 0 0128
� � �
�
Again, it can be seen from steam tables that the corresponding dew point in this case is around 16�5°C, which indicates that some of the water vapor will begin to condense as the temperature is lowered below this value�
It is to be noted that most of the combustion reactions considered so far assumed that the fuel was initially in a vapor phase� If the fuel is initially in the liquid phase, then the energy equation must account for the consequences of changing into vapor� This would be done through knowledge of the corresponding value of hfg for the fuel under the prevailing conditions�
A summary of the procedural steps to be followed for calculating the products temperature, pressure, and composition following combustion under products equilibrium conditions is the following:
• Specify what are the initial components of the reacting mixture and the relative concentration of all the items present, even those that may not undergo a change, such as nitrogen when considered as a diluent�
• Decide what items are going to be of significant concentrations in the products� This listing is dependent on the accuracy desired in the results, the composition of the reactive mixture, and the initial and final expected temperatures and pressures�
• Fuel rich mixtures require often the consideration of many partial oxidation and dissociated products� Also, the higher the final temperature, the more extensively diverse is the composition of the products that may need to be specified and considered�
• Write the overall reaction equation in moles, listing the minimum number of different likely important products for the degree of accuracy required in the final answer�
• Set the necessary mass conservation equations of the elements involved to solve for the unknown products concentrations� If there were X unknown items in the products involving Y elements, then Y independent equations will be those for balancing the mass of the elements� The remaining (X − Y) equations will be those of the relevant independent equilibrium constants needed to be identified and set�
• If the products temperature is not known beforehand, then a value needs to be guessed and values of the corresponding thermodynamic properties need to be predicted, for example, those of the equilibrium constants and the associated specific heat values, enthalpies, and so on from the relevant literature� Using the values of thermodynamic properties at the guessed products temperature, the X unknowns are then solved�
• Once the concentrations of the individual X products are determined at the guessed temperature, then check that they are the correct values by ensuring that they satisfy the corresponding energy equation for the guessed products temperature� If they do not, then repeat the procedure with a better guessed value for the temperature and try again� Repeat until a satisfactory agreement is obtained� Often a linear interpolation between the resulting deviation and the assumed temperature can lead quickly to a sufficiently adequate answer�
• For non-isobaric combustion reactions the equation of state needs to be employed to verify that the products pressure used in the calculations is the correct one� Otherwise, a new pressure value is assumed and the whole procedure is repeated�
This procedure can be seen to be very laborious and time-consuming when no appropriate solution software is employed� Of course, the answer will relate be relating to the assumption that the products are indeed in chemical
equilibrium, which may not be necessarily the case� Then, resort needs to be made to consider the even more complex approaches involving chemical kinetics where reaction rates are considered� This more realistic approach will lead perhaps to orders of magnitude increasing in complexity and the need to know the corresponding additional chemical kinetics data� Fortunately, specialized software is increasingly made widely available to obtain convenient solutions for the combustion of relatively complex fuel-air mixtures� Moreover, for most combustion process calculations, the products species are assumed to behave adequately as ideal gases� Otherwise, suitable equations of state for real gases need to be employed� This would add to the complexity of the solution�
Figure 5�4 shows the calculated equilibrium molar concentrations of the products of a stoichiometric mixture with air of a hydrocarbon fuel having a molar H/C ratio of 2�25 (e�g�, iso-octane, C8H18) as a function of equilibrium temperature and at 50 atm� It can be seen that the concentrations of stable molecules such as those of carbon dioxide and water vapor decline substantially with the increase in temperature� Those of oxides of nitrogen increase correspondingly almost logarithmically� Moreover, the concentrations of species produced by dissociated products such as H, O, and OH increase significantly and almost logarithmically with the increase in temperature� Such a rise in the concentrations of dissociated products produced via endothermic reactions will reduce substantially the net energy release by the exothermic combustion process�
In such thermodynamic calculations of combustion reaction processes it is the relative masses of the reacting elements that enter into the calculations for a specific set of product species� Accordingly, the initial reactants may be conveniently grouped on molar basis rather than having to represent them individually as components of the reactants� For example, a certain reacting mixture having the following composition:
0 1 0 5 2 0 1 0 8 5 32 04 6 6 2 2 2 2� � � � � �CH C H H O CO O N+ + + + +
may be considered as equivalent to an elemental representation of the type
C H O Na b d f
For this example a, b, d, and f become 4�1, 7�4, 2�1, and 6�4, respectively� Such an approach often represents a useful and a more general formulation�
Figure 5�5 shows the variation in the calculated adiabatic flame temperature with changes in the equivalence ratio for methane containing various concentrations of the diluents carbon dioxide, water vapor, and nitrogen for a range of diluent concentrations in the fuel mixture at an initial mixture temperature of 300 K and atmospheric pressure� The maximum values of the flame temperature always appear associated with mixtures in the vicinity of the stoichiometric value� The relative increased presence of the diluent in the fuel mixture for any equivalence ratio reduces markedly the resulting flame temperature� The decrease is particularly significant when carbon dioxide is the diluent as compared to the corresponding values with water vapor or nitrogen� On this basis the mixing of some exhaust products with the intake mixture reduces the level of temperatures achieved during combustion� Exhaust
gas recirculation (EGR) is thus an approach increasingly employed to reduce the level of the oxides of nitrogen pollutant production during combustion�
An example of standard calorimeters normally employed to determine the heating value of liquid and solid fuels is the standard bomb calorimeter, shown schematically in Figure 5�6� It is used to determine the heating values under constant volume condition, that is, ΔEfuel� For gaseous fuels a steady flow combustion calorimeter operating at ambient pressure and temperature may be used to yield the enthalpy of combustion, ΔHfuel (Figure 5�7)�
1� Find the net and gross heating value at constant pressure per kilogram of a mixture at 298 K for a stoichiometric mixture of benzene vapor and air� The enthalpy of combustion at 298 K is 3�1695 GJ/kmol of fuel when water in the products is in the vapor phase�
Answer: The overall reaction equation with ideal products in moles is
C H O N CO H O N6 6 2 2 2 2 23 76+ + → + +λ( � ) a b c
Through elemental mass balance equations for C, O2, H2, and N2, λ is found to be 7�5 and the stoichiometric reaction equation becomes
C H O N CO H O N6 6 2 2 2 2 27 5 3 76 6 3 28 2+ + → + +� ( � ) �
Find the total mass of the mixture 1�0 kmol of C6H6 equals 78 kg of the fuel (6 × 12 + 6 × 1)
Total mass of mixture = + + × =
78 7 5 32 37 6 28 11
� ( � ) 08kg of mixture per mol of fuel
The lower heating value (net) given is the enthalpy of combustion per kg of fuel when reactants and products are at 298 K with H2O in the vapor phase:
Lower heating value = −3�1695/78 = −40,630 kJ/kgf
Also, the lower heating value per kg of mixture = −3�1695(GJ)/1108 = −2860 kJ/kgmix� To obtain the corresponding higher heating value, consider H2O in the products of combustion:
1 kmol of fuel produces 3 × 18 kg of H2O or 1 kg of mixture contains 3 × 18/1108 = 0�045 kg kgH O mix2 /
From the steam tables, the enthalpy of condensation hfg at 298 K equals −2442
2 kJ kgH O/ � The negative value indicates heat leaving the system upon
condensation�
Higher heating value Lower heating value= + −( 2442 × = − − × = −
0 045 2860 2442 0 045 2980
� ) � /kg kgmix
For constant volume the corresponding values of ΔU need to be determined:
H U PV U nRT= + = +
For this reaction
n = + + − + + = +( � ) ( � � ) � /6 3 28 2 1 7 5 28 2 0 5 kmol kmolf
= − = − × × = −U H nRT H H298 8 31 0 5 1238� � ( ) kJ/kmolf
U
H mix
mix mixkJ kg=
−
= − − = −
1238 1108
2860 1 12 2861� /
For the gross higher heating value, the H2O is in the liquid state and will have zero gaseous volume producing a Δn = −2�5:
= − × × − = − + = −U H 298 8 31 2 5 2980 5 6 2974 4� ( � ) � � kJ/kmolf
2� An engine operates on diesel fuel supplemented by propane vapor� Calculate the rate of excess air employed when 0�255 kg/h diesel fuel, 2�12 kg/h propane, and 49�53 kg/h air were used� Assume that the diesel fuel has to be represented by cetane, C16H34�
Answer:
Mwt of C H kg kmol16 34 16 12 34 226= × + = /
Mwt of C H kg kmol3 8 3 12 8 44= × + = /
Mwt of air (effective) kg km= × + × =0 21 32 0 79 28 28 9� � � / ol
Overall reaction equation for a stoichiometric mixture in air (in moles) is
0 255 226
2 12 44
0 210 0 7916 34 3 8 2 2 � �
( � � )C H C H N+ + + →λ a b dCO H O N2 2 2+ +
Employ elemental mass balance equations to find the unknowns, λ, a, b, and d�
Carbon balance: a = 0�255 × 16/226 + 2 × 12 × 3/44 = 1�654 Hydrogen balance: b = 0�255 × 17/226 + 72/44 = 1�769
Oxygen balance: 0�21λ = a + b/2 = 0�1626 + 0�10595 = 0�2686 where λ = 1�279� Mass of stoichiometric air: 1�279 (0�21 × 32 + 0�79 × 28) = 36�87 kg/h Air supplied = 49�53 kg/h Excess air = 49�53 – 36�87 = 12�66 kg/h Air supplied = 49�53 kg/h Equivalence ratio:
φ =
m m
m m
= 36�87/49�58 = 0�744
3� An unknown hydrocarbon fuel is burned in air such that the dry products have the following volumetric composition: CO2: 9�0%, CO: 1�1%, O2: 8�8%, and N2: 81�1%� Determine
i� Hydrogen to carbon molar ratio ii� Air to fuel mass ratio iii� Amount of excess air employed iv� Temperature at which the water vapor begins to condense at an
ambient pressure of 86�2 kPa (use steam tables)
Answer:
i� Consider 100 kmol of dry products; the overall molar reaction equation for the hydrocarbon fuel (CxHy) combustion in air becomes
C H O N CO CO Ox y a+ + → + + +λ( � ) [( � � � �2 2 2 23 76 9 0 1 1 8 8 81 1 2 2N H O) ]+ b
Note : When one mole of fuel is considered then the multiplier “a” is needed to balance the equation with the addition of the condensed water, since the products composition given is for 100 moles of dry gas�
There are five unknowns but only four elemental balance equations� Hence we can only obtain the ratio of y/x and not the individual values of x or y�
Oxygen balance equation: λ = [9�0 + 0�55 + 8�8 + b/2]/a Nitrogen balance equation: 3�76λ = 81�1a
81 1 3 76
9 0 0 55 8 8 2
21 57 �
� � � � �= + + + =
b
b = × =( � – � – � – � ) �21 57 9 0 8 8 0 55 2 6 44
ii� x/y = (9�0 + 1�1)/(2 × 6�44) = 10�1/12�88 = 0�785 From the exhaust composition:
m m
m m
m m
m
=
+
+
= + +
( ) ( )
( � � �
2 9 0 8 8 0 55 3 44 32 587 2
81 1 28 2271
+ × =
= × =
=
� ) �
�
m
m N
air 71 587 2 2858 2 9 0 1 1 12 6 44 2 1
+ =
= + × + × =
� � ( � � ) �mfuel 21 2 12 88 134 08� � �+ =
iii� (mair/mfuel)used = 2858�2/134�08 = 21�32 From consideration of the composition of the products:
Stoichiometric O needed2 9 0 1 1 2 16 6 44= + × × + ×( � � ) � 16 426 2
426 2 12
[ ] = = +
�
�Stoichiometric N needed 3 76 28 32
1828 4� � × = =% excess air air a
– ( � – � ) �
m m
( ) =
=
2858 2 1828 4 1828 4
56 3� % of stoich air
iv� The water vapor in the products begins to condense� At what temperature will its partial pressure be the same as the saturation pressure?
