Numerical problems concerning combustion

1. Solved numerical problem on
Combustion thermodynamics

2. 2
Combustion Thermodynamics
Combustion is a chemical reaction between substances usually including oxygen
and accompanied by the generation of heat and light in the form of flame. The
practical goal in combustion study is the prediction of its performance for a safe,
efficient operation of fire-making devices. It is known that in various heat engines,
gas turbines, and steam power plants, the heat is obtained from combustion
processes, using either solid fuel (e.g. coal or wood). liquid fuel (e.g. gasolene,
kerosine, or diesel fuel), or gaseous fuel (e.g. natural gas or propane). Here,
chemistry and thermodynamics of combustion of generic hydrocarbon fuels -
(CxHy) is introduced in which the oxydizer is the oxygen contained in atmospheric
air.

3. 3
The basic combustion process can be described by the fuel (the hydrocarbon)
plus oxydizer (air or oxygen) called the Reactants, which undergo a chemical
process while releasing heat to form the Products of combustion such that mass
is conserved. The simplest combustion process, is known as Stoichiometric
Combustion. In combustion, All the carbon in the fuel forms carbon dioxide
(CO2) and all the hydrogen forms water (H2O) in the products, thus we can write
the chemical reaction as follows:
Note that this reaction yields five unknowns: z, a, b, c, d, thus we need five
equations to solve. Stoichiometric combustion assumes that no excess oxygen
exists in the products, then d = 0.

4. 4
Question. One kmol of octane (C8H18) is burned with air that contains 20 kmol of
O2. Assuming the products contain only CO2, H2O, O2, and N2, determine the mole
number of each gas in the products and the air–fuel ratio for this combustion
process.
Solution: The amount of fuel (C8H18) and the amount of oxygen in the air are
given. The combustion products contain CO2, H2O, O2, and N2 as per the question.
The molar mass of air = 29 kg/kmol
The molar mass of carbon = 12 kg/kmol
The molar mass of Hydrogen = 1 kg/kmol
(It is known that 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 79/21 = 3.76 moles of nitrogen. Therefore, 1 kmol O2
+ 3.76 kmol N2 = 4.76 kmol air) Note that, it is assumed that the nitrogen will not
normally undergo any chemical reaction.

5. 5
Now,
The chemical equation for this combustion process can be written as
C8H18 + 20(O2 + 3.76N2) → xCO2 + yH2O + zO2 + wN2
where the terms in the parentheses represent the composition of dry air that
contains 1 kmol of O2 and x, y, z, and w represent the unknown mole numbers of
the gases in the products. These unknowns are determined by applying the mass
balance to each of the elements, i.e., by equating mole the number of atoms of
each element in the reactants (carbon, hydrogen, oxygen and nitrogen) with the
number of atoms of those elements in the products :
C: 8 = x;
H: 18 = 2y => y = 9;
O: 20×2 = 2x + y + 2z => z = 7.5;
N: 20×3.76 = w => w = 75.2

6. 6
The chemical equation after substitution becomes
C8H18 + 20(O2 + 3.76N2) → 8CO2 + 9H2O + 7.5O2 + 75.2N2
So, number of moles of CO2, H2O, O2, and N2 in the products are 8 kmol, 9
kmol, 7.5 kmol and 75.2 kmol respectively.
The air–fuel ratio (AF) is determined by taking the ratio of the mass of the air
and the mass of the fuel, i.e.,
AF =
𝑚𝑎𝑖𝑟
𝑚𝑓𝑢𝑒𝑙
=
(𝑁𝑀)𝑎𝑖𝑟
(𝑁𝑀)𝐶+(𝑁𝑀)𝐻
=
(20× 4.76 𝑘𝑚𝑜𝑙)×(29
𝑘𝑔
𝑘𝑚𝑜𝑙
)
8 𝑘𝑚𝑜𝑙 × 12
𝑘𝑔
𝑘𝑚𝑜𝑙
+ (18 𝑘𝑚𝑜𝑙)×(1
𝑘𝑔
𝑘𝑚𝑜𝑙
)
= 24.2 kg of
air/kg of fuel
Where, N is the number of moles and M is the molar mass.
This means 24.2 kg of air is used to burn each kg of fuel during this combustion
process.

7. 7
Questions
Suggestions