STOICHIOMETRIC COMBUSTION CALCULATOR FOR HYDROCARBONS

Stoichiometric Combustion Calculator for Alkane, Alkene and Alkyne fuels (Hydrocarbons) to compare the effect of carbon atoms on the stoichiometric coefficient.

Problem Specification:-

For a generic alkane fuel, the combustion process will following the below equation

Here

n = Number of moles of C

ar = Stoichiometric co-efficient

a = Number of moles of Carbon-di-oxide

b = number of moles of water

c = number of moles of nitrogen

Show a plot that compares the effect of the number of atoms of carbon on the stoichiometric co-efficient "ar". Compare how this trend changes with the fuel type (alkane, alkene, and alkyne)

THEORY:-

Combustion:-

A chemical reaction during which a fuel is oxidized and a large amount of energy is released is called combustion. the oxidizer most often used in the combustion process is air, for obvious reason-it is free and readily available. The dry air can be approximated as 21 percent of oxygen and 79 percent of nitrogen by mole numbers. Therefore each mole of oxygen entering the 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

1 mol O₂ + 3.76 mol N₂ = 4.76 mol 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. However, even then the presence of nitrogen greatly affects the outcome of a combustion process since nitrogen usually enters a combustion chamber in large quantities at low temperatures and exits at considerably higher temperatures, absorbing the large proportion of chemical energy released during combustion.

Air that enters a combustion chamber normally contains some water vapor or moisture, which also deserves consideration. For most combustion processes, the moisture in the air and the H2o that forms during combustion can be treated as an inert gas, like nitrogen. At very high temperatures, however, some water vapor dissociates into H2 and O2 as well as into H, O, and OH.

Chemical Reaction Balancing Principle:-

As we recall from our chemistry courses, chemical equations are balanced based on the conservation of mass principle or the mass balance, which can be stated as follows: The total mass of each element is conserved during the chemical reaction. That is, the total mass of each element on the right-hand side of the chemical reaction is equal to the total mass of that element on the left-hand side even though the elements exist in the different chemical compounds in the reactants and product. Also, the total number of atoms is conserved during a chemical reaction since the total number of atoms is equals to the total mass of the element divided by its atomic mass.

Stoichiometric combustion:-

Stoichiometric combustion offers an easy way to understand combustion concepts. When fuel and air mix they produce products as CO2, CO, OH, HO, NO, N2O, and many more. In stoichiometric combustion, we make a simple assumption that the by-products are only CO2, H2O, and N2.

To Show a plot that compares the effect of the number of atoms of carbon on the stoichiometric co-efficient "ar" for different fuel types (alkane, Alkenes, and Alkynes), we have to find the relation for stoichiometric co-efficient "ar" for different types of fuels. we can do that by applying mass balancing to each of the elements in the reactions for each type of fuel.

Combustion reactions for Alkane, Alkenes, and Alkynes assuming complete combustion and for 1 mole of reactant hydrogen are discussed below.

1. Analysis of Combustion Reaction for Alkanes:-

Chemical reaction: C_nH_2n+2 + a_r ( O₂ + 3.76 N₂ ) = a.CO₂ + b.H₂O + c.N₂

In the above reaction, the term in the parenthesis represents the composition of dry air that contains 1 mol of O2 and a, b, c represents Unkown mole numbers of gases in the product. The unknowns are determined by applying the mass balance to each of the elements- that is by requiring that the total mass or mole number of each element in the reactants be equal to that of the product.

So, by applying the mass balance to each of the elements of the above reaction.

For carbon, C: n=a,

For hydrogen, H: 2n+2=2b,

For nitrogen, N: 2*3.76*ar = 2c

For oxygen, O: 2*ar = 2a+b

Therefore stoichiometric coefficient ar = (3n+1)/2

2. Analysis of Combustion Reaction for Alkenes:-

Chemical reaction: C_nH_2n + a_r ( O₂ + 3.76 N₂ ) = a.CO₂ + b.H₂O + c.N₂

As done before, by applying the mass balance to each of the elements of the above reaction.

For carbon, C: n=a,

For hydrogen, H: 2n=2b,

For nitrogen, N: 2*3.76*ar = 2c

For oxygen, O: 2*ar = 2a+b

Therefore stoichiometric coefficient ar = (3n)/2

3. Analysis of Combustion Reaction for Alkynes:-

Chemical reaction: C_nH_2n-2 + a_r ( O₂ + 3.76 N₂ ) = a.CO₂ + b.H₂O + c.N₂

As done before, by applying the mass balance to each of the elements of the above reaction.

For carbon, C: n=a,

For hydrogen, H: 2n-2=2b,

For nitrogen, N: 2*3.76*ar = 2c

For oxygen, O: 2*ar = 2a+b

Therefore stoichiometric coefficient ar = (3n-1)/2

The program that Shows a plot that compares the effect of the number of atoms of carbon on the stoichiometric co-efficient "ar" for different the fuel type (alkane, alkene, and alkyne)

clear all
close all
clc

% 1 mole of alkanes, alkenes, and alkynes burned with air that contains "ar" mol of O2;
% considering the first 20 carbon atoms;

ar=zeros(1,20);
x=linspace(1,20,20)

for i=1:length(x)

Stoichiometric_coefficients_for_alkanes(i)=((3*x(i))+1)/2;
Stoichiometric_coefficients_for_alkenes(i)=(3*x(i))/2;
Stoichiometric_coefficients_for_alkynes(i)=((3*x(i))-1)/2;

end

% plotting the graph that shows the effect of the number of atoms of carbon on the stoichiometric co-efficient "ar" for different fuel types.

figure(1)
plot(x,Stoichiometric_coefficients_for_alkanes)
hold on
plot(x,Stoichiometric_coefficients_for_alkenes,'r')
hold on
plot(x,Stoichiometric_coefficients_for_alkynes,'g')
title('Effect of number of atoms of carbon on the stoichiometric co-efficient "ar"')
xlabel('No. of carbon atoms')
ylabel('Stoichiometric Coefficients ar')
legend('Stoichiometric coefficients for alkanes','Stoichiometric coefficients for alkenes','Stoichiometric coefficients for alkynes')

output:-

1. From the plot, we can see that all the curves are linear and make different y-intercept depicting the change in the type of hydrocarbon. i.e. Alkanes, Alkenes, and Alkynes.

2. Alkynes have the smallest Y-intercept and have smaller values of stoichiometric co-efficient "ar" for corresponding values of carbon atoms as compared to alkanes and alkenes.

3. Stoichiometric co-efficient simply denotes how many mols of oxygen will require to completely burn the hydrocarbon i.e. Alkanes, Alkenes, and Alkynes.

4. So, Alkynes required a smaller amount of oxidizer ( air or oxygen) to completely get burned as compared to alkanes and alkenes.