Combustion Thermochemistry
On a mole or a volume basis, dry air is composed of 20.9 percent oxygen, 78.1 percent nitrogen, 0.9 percent
argon, and small amounts of carbon dioxide, helium, neon, and hydrogen.
In the analysis of combustion processes, the argon in the air is treated as nitrogen, and the gases that exist in
trace amounts are disregarded.
Then dry air can be approximated as 21 percent oxygen and 79 percent nitrogen by mole numbers. Therefore,
each mole of oxygen entering a combustion chamber is accompanied by 0.79/0.21 = 3.76 mol of nitrogen.
That is,
1 kmol O2 + 3.76 kmol N2 = 4.76 kmol air
During combustion, nitrogen behaves as an inert gas and does not react with other elements, other than forming
a very small amount of nitric oxides.
A frequently used quantity in the analysis of combustion processes to quantify the amounts of fuel and air is
the air–fuel ratio AF. It is usually expressed on a mass basis and is defined as the ratio of the mass of air to
the mass of fuel for a combustion process. That is
𝐴 𝑚𝑎𝑖𝑟
=
𝐹 𝑚𝑓𝑢𝑒𝑙
The mass m of a substance is related to the number of moles N through the relation m = NM, where M is the
molar mass.
It is often instructive to study the combustion of a fuel by assuming that the combustion is complete. A
combustion process is complete if all the carbon in the fuel burns to CO2, all the hydrogen burns to H2O, and
all the sulfur (if any) burns to SO2. That is, all the combustible components of a fuel are burned to completion
during a complete combustion process.
Conversely, the combustion process is incomplete if the combustion products contain any unburned fuel or
components such as C, H2, CO, or OH.
Insufficient oxygen is an obvious reason for incomplete combustion, but it is not the only one. Incomplete
combustion occurs even when more oxygen is present in the combustion chamber than is needed for complete
combustion.
This may be attributed to insufficient mixing in the combustion chamber during the limited time that the fuel
and the oxygen are in contact. Another cause of incomplete combustion is dissociation, which becomes
important at high temperatures.
Oxygen has a much greater tendency to combine with hydrogen than it does with carbon. Therefore, the
hydrogen in the fuel normally burns to completion, forming H2O, even when there is less oxygen than needed
for complete combustion. Some of the carbon, however, ends up as CO or just as plain C particles (soot) in the
products.
The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or
theoretical air. Thus, when a fuel is completely burned with theoretical air, no uncombined oxygen is present
in the product gases. The theoretical air is also referred to as the chemically correct amount of air, or 100
percent theoretical air.
A combustion process with less than the theoretical air is bound to be incomplete. The ideal combustion
process during which a fuel is burned completely with theoretical air is called the stoichiometric or theoretical
combustion of that fuel.
For example, the theoretical combustion of methane is
CH4 + 2(O2 + 3.76 N2) → CO2 + 2 H2O + 7.52N2
Notice that the products of the theoretical combustion contain no unburned methane and no C, H2, CO, OH, or
free O2.
In actual combustion processes, it is common practice to use more air than the stoichiometric amount to
increase the chances of complete combustion or to control the temperature of the combustion chamber. The
amount of air in excess of the stoichiometric amount is called excess air.
The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or
percent theoretical air. For example, 50 percent excess air is equivalent to 150 percent theoretical air, and
200 percent excess air is equivalent to 300 percent theoretical air.
Of course, the stoichiometric air can be expressed as 0 percent excess air or 100 percent theoretical air.
Amounts of air less than the stoichiometric amount are called deficiency of air and are often expressed as
percent deficiency of air. For example, 90 percent theoretical air is equivalent to 10 percent deficiency of air.
The amount of air used in combustion processes is also expressed in terms of the equivalence ratio ∅, which
is the ratio of the actual fuel–air ratio to the stoichiometric fuel–air ratio.
𝐴
(𝐹 )
𝑠
∅=
𝐴
(𝐹 )
𝑎
This parameter can have either of three values :
𝐴 𝐴
∅ < 1 this means the (𝐹 ) > (𝐹 ) i.e. the amount of air actually supplied is greater than that which is needed
𝑎 𝑠
for complete combustion, or, for the same amount of air, the amount of fuel supplied is less than the
stoichiometric one. This means that the mixture has excess oxygen. This is called Lean Mixture.
𝐴 𝐴
∅ = 1 this means the (𝐹 ) = (𝐹 ) i.e. the amount of air actually supplied is equal to that which is needed for
𝑎 𝑠
complete combustion, or, for the same amount of air, the amount of fuel supplied equal to the stoichiometric
one.
𝐴 𝐴
∅ > 1 this means the (𝐹 ) < (𝐹 ) i.e. the amount of air actually supplied is less than that which is needed
𝑎 𝑠
for complete combustion, or, for the same amount of air, the amount of fuel supplied is greater that the
stoichiometric one. This means that the mixture has deficient oxygen. This is called Rich Mixture
Example :
𝐴
When we say that the stoichiometric (𝐹 ) for gasoline (IsoOctane) = 15.1:1. This means that 15.1 kg of air is
𝑠
needed to completely burn 1 kg of this fuel such that all C is converted to CO2, H is converted to H2O and no
unused oxygen appears in the products.
𝐴
For lean mixture (𝐹 ) > 15.1, say 17. Which means we supplied 17 kg of air to burn 1 kg of fuel, or we
𝑎
supplied 0.888 kg of fuel to be burned with 15.1 kg of air. In both cases, we supplied air more than required.
The extra O2 will appear in the products.
𝐴
For rich mixture (𝐹 ) < 15.1, say 14. Which means we supplied 14 kg of air to burn 1 kg of fuel, or we
𝑎
supplied 1.08 kg of fuel to be burned with 15.1 kg of air. In both cases, we supplied air less than required.
Hence, carbon will not burn completely, rather, is will appear as CO2 + CO.
The amount of excess/deficient air can be found from the following equation :
𝐴 𝐴
(𝐹 ) − (𝐹 ) 1
𝑎 𝑠
% 𝑒𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟 = ( ) ∗ 100% = ( − 1) ∗ 100%
𝐴 ∅
(𝐹 )
𝑠
The amount of theoretical air supplied can be found by adding 100% to the excess air or :
𝐴
(𝐹 ) 1
𝑎
% 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑎𝑖𝑟 = ( ) ∗ 100% = ( ) ∗ 100%
𝐴 ∅
( )
𝐹 𝑠
The general combustion equation for any hydrocarbon fuel can be written as follows :
𝐵𝑡ℎ
𝐴(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + (𝑂2 + 3.76 𝑁2 ) → 𝑥1 (𝐶𝑂2 ) + 𝑥2 (𝐶𝑂) + 𝑥3 (𝐻2 𝑂) + 𝑥4 (𝑁2 ) + 𝑥5 (𝑂2 ) + 𝑥6 (𝑆𝑂2 )
∅
Where “x”, “y” , “z” and “l” are the number of atoms as in the fuel chemical formula.
(I) Stoichiometric combustion:
In this combustion all the Carbon in the fuel gets burned to CO2, Hydrogen converted to H2O, Sulfur to SO2
and no unused Oxygen remains or is present in the exhaust. This is also called chemically correct, Theoretical,
100% Theoretical air.
Let us have an unknown amount of fuel “A” that is burned with stoichiometric amount of air “𝐵𝑡ℎ ”. We need
to develop equation for both the amount of air and products, as follows:
𝐴(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + 𝐵𝑡ℎ (𝑂2 + 3.76 𝑁2 ) → 𝑥1 (𝐶𝑂2 ) + 𝑥3 (𝐻2 𝑂) + 𝑥4 (𝑁2 ) + 𝑥6 (𝑆𝑂2 )
Carbon Balance: (A * 12) * 𝑥 = (𝑥1 *12) * 1 ➔ 𝑥1 = 𝐴 𝑥
𝑦
Hydrogen Balance: (A * 1) * 𝑦 = (𝑥3 * 1) * 2 ➔ 𝑥3 = 𝐴 2
Sulfur Balance : (A * 32) * 𝑙 = (𝑥6 * 32) * 1 ➔ 𝑥6 = 𝐴 𝑙
Nitrogen Balance: (𝐵𝑡ℎ * 3.76 * 14) * 2 = (𝑥4 * 14) * 2 ➔ 𝑥4 = 3.76 ∗ 𝐵𝑡ℎ
Oxygen Balance: A * 16 * 𝑧 + 𝐵𝑡ℎ * 16 * 2 = 𝑥1 *16 * 2+ 𝑥3 * 16 * 1 + 𝑥6 * 16 * 2
𝑦
= 𝐴 𝑥 *16 * 2+ 𝐴 2 * 16 * 1 + 𝐴 𝑙 * 16 * 2
∴ we can write
𝑦 𝑧
𝐵𝑡ℎ = 𝐴 (𝑥 + 4 − 2 + 𝑙)
Hence, we can write the stoichiometric equation as follows:
𝑦 𝑧 𝑦 𝑦 𝑧
𝐴(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + 𝐴 (𝑥 + − + 𝑙) (𝑂2 + 3.76 𝑁2 ) → 𝐴𝑥(𝐶𝑂2 ) + 𝐴 (𝐻2 𝑂) + 3.76𝐴 (𝑥 + − + 𝑙) (𝑁2 ) + 𝐴𝑙(𝑆𝑂2 )
4 2 2 4 2
Divide all terms by 𝐴
𝑦 𝑧 𝑦 𝑦 𝑧
(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + (𝑥 + − + 𝑙) (𝑂2 + 3.76 𝑁2 ) → 𝑥(𝐶𝑂2 ) + (𝐻2 𝑂) + 3.76 (𝑥 + − + 𝑙) (𝑁2 ) + 𝑙(𝑆𝑂2 )
4 2 2 4 2
𝑦 𝑧
𝐴 (𝑥+ − +𝑙)∗(32+3.76∗28)
4 2
The Stoichiometric Air-Fuel ratio is given by: (𝐹 ) = on mass basis
𝑠 (12∗𝑥+1∗𝑦+16∗𝑧)
𝑦 𝑧
𝐴 (𝑥+ − )∗(1+3.76)
4 2
The Stoichiometric Air-Fuel ratio is given by: (𝐹 ) = on mole basis
𝑠 1
In actual practice the amount of air supplied to the system can be GREATER THAN 𝐵𝑡ℎ (means lean mixture)
, EQUAL TO 𝐵𝑡ℎ (means stoichiometric mixture) or LESS THAN 𝐵𝑡ℎ (means rich mixture).
Rich mixture is when the amount of air is less than stoichiometric for the same amount of fuel or for same
amount of air, the fuel is more than the air can burn.
Lean mixture is when the amount of air is more than stoichiometric for the same amount of fuel or for same
amount of air, the fuel is less than the air can burn.
Example :
Let us take the combustion of 1 mole of C8H18 with stoichiometric air. Applying equation (1) we get : x = 8, y
18
= 18 and z = 0. Hence; 𝐵𝑡ℎ = (8 + 4 ) = 12.5 the stoichiometric equation can be written as :
C8H18 + 12.5 (O2 + 3.76 N2) → 8CO2 + 9H2O + 47 N2
𝐴 (12.5)∗(32+3.76∗28)
The Stoichiometric Air-Fuel ratio is given by: (𝐹 ) = = 15.05 on mass basis
𝑠 (12∗8+1∗18)
𝐴 (12.5)∗(1+3.76)
The Stoichiometric Air-Fuel ratio is given by: (𝐹 ) = = 59.5 on mole basis
𝑠 1
Now for lean mixture we say that:
For 1 mole of fuel the amount of air is GREATER THAN 15.05
OR, for an amount of air = 15.05, the amount of fuel is LESS THAN 1 mole.
Now for Rich mixture we say that:
For 1 mole of fuel the amount of air is LESS THAN 15.05
OR, for an amount of air = 15.05, the amount of fuel is GREATER THAN 1 mole.
For this purpose we define a term called EQUIVALENCE RATIO (∅). This is defined as the ratio between the
Stoichiometric to actual air-fuel ratio.
𝐴
(𝐹 )
𝑆
∅ =
𝐴
(𝐹 )
𝐴
This can be of three cases: LESS THAN 1 (for lean mixture), EQUAL TO 1 (for Stoichiometric) and
GREATER THAN 1 (for rich mixture).
(II) Lean mixture combustion [𝐵𝑎 > 𝐵𝑡ℎ ]
In this combustion all the Carbon in the fuel gets burned to CO2, Hydrogen converted to H2O, Sulfur to SO2
and some Oxygen remains unused or is present in the exhaust. This is also called excess air, greater than 100%
𝐵
air or lean mixture combustion. Keeping in mind that the actual amount of air 𝐵𝑎 = ∅𝑡ℎ
𝐵
Let us have an unknown amount of fuel 𝐴 that is burned with amount of air ∅𝑡ℎ at an equivalence ratio ∅ < 1.
We need to develop equation for both the amount of air and products, as follows:
𝐵𝑡ℎ
𝐴(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + (𝑂2 + 3.76 𝑁2 ) → 𝑥1 (𝐶𝑂2 ) + 𝑥3 (𝐻2 𝑂) + 𝑥4 (𝑁2 ) + 𝑥5 (𝑂2 ) + 𝑥6 (𝑆𝑂2 )
∅
Where “x”, “y” and “z” are the number of atoms as in the fuel chemical formula.
Carbon Balance: (A * 12) * 𝑥 = (𝑥1 *12) * 1 ➔ 𝑥1 = 𝐴 𝑥
𝑦
Hydrogen Balance: (A * 1) * 𝑦 = (𝑥3 * 1) * 2 ➔ 𝑥3 = 𝐴 2
Sulfur Balance : (A * 32) * 𝑙 = (𝑥6 * 32) * 1 ➔ 𝑥6 = 𝐴 𝑙
𝐵 𝐵
Nitrogen Balance: ( ∅𝑡ℎ * 3.76 * 14) * 2 = (𝑥4 * 14) * 2 ➔ 𝑥4 = 3.76 ∗ ∅𝑡ℎ
𝐵𝑡ℎ
Oxygen Balance: A * 16 * 𝑧 + * 16 * 2 = 𝑥1 *16 * 2+ 𝑥3 * 16 * 1 + 𝑥6 * 16 * 2
∅
𝑦
= 𝐴 𝑥 *16 * 2+ 𝐴 * 16 * 1 + 𝐴 𝑙 * 16 * 2
2
∴ we can write
1
𝑥5 = 𝐵𝑡ℎ (∅ − 1) = (𝐵𝑎 − 𝐵𝑡ℎ )
𝑦 𝑧
𝐵𝑡ℎ = 𝐴 (𝑥 + − + 𝑙)
4 2
Hence, we can write the stoichiometric equation as follows:
𝑦 𝑧
𝐴 (𝑥 + − + 𝑙) 𝑦 𝐵𝑡ℎ 1
𝐴(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + 4 2 (𝑂2 + 3.76 𝑁2 ) → 𝐴𝑥(𝐶𝑂2 ) + 𝐴 (𝐻2 𝑂) + 3.76 ∗ (𝑁2 ) + 𝐵𝑡ℎ ( − 1) (𝑂2 ) + 𝐴𝑙(𝑆𝑂2 )
∅ 2 ∅ ∅
Divide all terms by 𝐴
𝐵𝑡ℎ 𝑦 𝐵𝑡ℎ
(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + (𝑂2 + 3.76 𝑁2 ) → 𝑥(𝐶𝑂2 ) + (𝐻2 𝑂) + 3.76 ∗ (𝑁2 ) + (𝐵𝑎 − 𝐵𝑡ℎ )(𝑂2 ) + 𝑙(𝑆𝑂2 )
∅ 2 ∅
Example :
Let us take the combustion of 1 mole of C8H18 with equivalence ratio 0.8. Applying equation (1) we get : x =
18
8, y = 18 and z = 0. Hence; 𝐵𝑡ℎ = (8 + 4 ) = 12.5 the Ba = (12.5/0.8) = 15.625 equation can be written as :
C8H18 + 15.625 (O2 + 3.76 N2) → 8CO2 + 9H2O + 58.75 N2+3.125 O2
𝐴 (15.625)∗(32+3.76∗28)
The Actual Air-Fuel ratio is given by: (𝐹 ) = = 18.815 on mass basis
𝑎 (12∗8+1∗18)
𝐴 (15.625)∗(1+3.76)
The Actual Air-Fuel ratio is given by: (𝐹 ) = = 74.375 on mole basis
𝑎 1
1 1
The amount of excess air supplied in this reaction = (∅ − 1) ∗ 100% = (0.8 − 1) ∗ 100% = 25%
𝐴 𝐴
( ) −( ) 18.815−15.05
𝐹 𝑎 𝐹 𝑠
Another method of calculating the excess air is = ( 𝐴 ) ∗ 100% = ∗ 100% = 25%
( ) 15.05
𝐹 𝑠
This means that we gave the fuel the 12.5 amount of air and 25% extra (0.25*12.5) with it.
𝐴
1 ( )
𝐹 𝑎
The percentage theoretical air supplied in this reaction is found by = (∅) ∗ 100% = ( 𝐴 ) ∗ 100% = 125%
( )
𝐹 𝑠
(III) Rich mixture combustion [𝐵𝑎 < 𝐵𝑡ℎ ]
In this combustion part of the Carbon in the fuel gets burned to CO2 and the other part gets converted to CO
because the amount of Oxygen supplied is less than needed for complete burning of fuel This is also called
deficient air, less than 100% air or rich mixture combustion. Keeping in mind that the actual amount of air
𝐵
𝐵𝑎 = ∅𝑡ℎ
𝐵
Let us have an unknown amount of fuel 𝐴 that is burned with amount of air ∅𝑡ℎ at an equivalence ratio ∅ > 1.
We need to develop equation for both the amount of air and products, as follows:
𝐵𝑡ℎ
𝐴(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + (𝑂2 + 3.76 𝑁2 ) → 𝑥1 (𝐶𝑂2 ) + 𝑥2 (𝐶𝑂) + 𝑥3 (𝐻2 𝑂) + 𝑥4 (𝑁2 ) + 𝑥6 (𝑆𝑂2 )
∅
Where “x”, “y” and “z” are the number of atoms as in the fuel chemical formula.
Carbon Balance: (A * 12) * 𝑥 = (𝑥1 *12) * 1 + (𝑥2 *12) * 1 ➔ 𝑥1 + 𝑥2 = 𝐴 𝑥
𝑦
Hydrogen Balance: (A * 1) * 𝑦 = (𝑥3 * 1) * 2 ➔ 𝑥3 = 𝐴 2
Sulfur Balance : (A * 32) * 𝑙 = (𝑥6 * 32) * 1 ➔ 𝑥6 = 𝐴 𝑙
𝐵 𝐵
Nitrogen Balance: ( ∅𝑡ℎ * 3.76 * 14) * 2 = (𝑥4 * 14) * 2 ➔ 𝑥4 = 3.76 ∗ ∅𝑡ℎ
𝐵𝑡ℎ
Oxygen Balance: A * 16 * 𝑧 + * 16 * 2 = 𝑥1 *16 * 2+ 𝑥2 *16 *1+ 𝑥3 * 16 * 1 + 𝑥6 * 16 * 2
∅
Also we have another equation 𝑥1 + 𝑥2 = 𝐴 𝑥
∴ Solving for 𝑥1 or 𝑥2 we get
1
𝑥2 = 2𝐵𝑡ℎ (1 − ∅) = 2(𝐵𝑡ℎ − 𝐵𝑎 )
𝑥1 = 𝐴 𝑥 − 2(𝐵𝑡ℎ − 𝐵𝑎 )
𝑦 𝑧
𝐵𝑡ℎ = 𝐴 (𝑥 + 4 − 2 + 𝑙)
Hence, we can write the stoichiometric equation as follows:
𝑦 𝑧
𝐴 (𝑥 + − + 𝑙)
𝐴(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + 4 2 (𝑂2 + 3.76 𝑁2 )
∅
1 1 𝑦 𝐵𝑡ℎ
→ (𝐴 𝑥 − 2𝐵𝑡ℎ (1 − )) (𝐶𝑂2 ) + 2𝐵𝑡ℎ (1 − ) (𝐶𝑂) + 𝐴 (𝐻2 𝑂) + 3.76 ∗ (𝑁2 ) + 𝐴𝑙(𝑆𝑂2 )
∅ ∅ 2 ∅
Divide all terms by 𝐴
𝑦 𝑧
(𝑥 + − + 𝑙) 𝑦 𝐵𝑡ℎ
(𝐶𝑥 𝐻𝑦 𝑂𝑧 𝑆𝑙 ) + 4 2 (𝑂2 + 3.76 𝑁2 ) → (𝑥 − 2(𝐵𝑡ℎ − 𝐵𝑎 )) (𝐶𝑂2 ) + 2(𝐵𝑡ℎ − 𝐵𝑎 )(𝐶𝑂) + (𝐻2 𝑂) + 3.76 ∗ (𝑁2 ) + 𝑙(𝑆𝑂2 )
∅ 2 ∅
Example :
Let us take the combustion of 1 mole of C8H18 with equivalence ratio 1.1. Applying equation (1) we get : x =
18
8, y = 18 and z = 0. Hence; 𝐵𝑡ℎ = (8 + 4 ) = 12.5 the Ba = (12.5/1.1) = 11.36 equation can be written as :
C8H18 + 11.36 (O2 + 3.76 N2) → 5.72CO2 + 2.27CO + 9H2O + 42.71 N2
𝐴 (11.36)∗(32+3.76∗28)
The Actual Air-Fuel ratio is given by: (𝐹 ) = = 13.68 on mass basis
𝑎 (12∗8+1∗18)
𝐴 (11.36)∗(1+3.76)
The Actual Air-Fuel ratio is given by: (𝐹 ) = = 54.073 on mole basis
𝑎 1
1 1
The amount of excess air supplied in this reaction = (∅ − 1) ∗ 100% = (1.1 − 1) ∗ 100% = −9.09%
𝐴 𝐴
( ) −( ) 11.36−15.05
𝐹 𝑎 𝐹 𝑠
Another method of calculating the excess air is = ( 𝐴 ) ∗ 100% = ∗ 100% = −9.09%
( ) 15.05
𝐹 𝑠
The minus sign means that we gave deficient air (less than stoichiometric) by 9.09%.
𝐴
1 ( )
𝐹 𝑎
The percentage theoretical air supplied in this reaction is found by = (∅) ∗ 100% = ( 𝐴 ) ∗ 100% = 90.9%
( )
𝐹 𝑠
Test your knowledge :
Try the above three cases for C12H26 , CH4, C2H6O and compare your results with the table supplied.
