Stoichiometric Combustion Calculator for Different Fuel Types in MATLAB

PROBLEM SPECIFICATION

Stoichiometric combustion offers an easy way to understand combustion concepts. When fuel and air mix they produce products as co2, co, oh, h2o, no, n2o, and many more. In stoichiometric combustion, we make a simple assumption that the by-products are only co2, h2o, and n2. 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

1. Write a program in MATLAB to calculate the stoichiometric co-efficient (ar) for each fuel

2. 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)

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NOTE: The purpose of including the theory is to understand the relation between the stoichiometric coefficient (`ar`) for different types of fuels. Henceforth, we’ll be able to understand the MATLAB coding which shows the plot that compares the effect of the number of atoms of carbons on the stoichiometric coefficient (`ar`) for different types of fuels (alkanes, alkenes, and alkynes).

THEORY

A chemical reaction during which a fuel is oxidized and a large amount of energy are released is known as combustion. In industrial combustion processes and engine combustion, the air is the oxidizer. Pure oxygen is used as an oxidizer only in some applications, such as cutting and welding.

Dry air is composed of 20.9 % oxygen, 78.1 % nitrogen, 0.9 & 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 neglected. The dry air can be approximated as 21% oxygen and 79% nitrogen by mole numbers. Therefore each mole of oxygen entering in a combustion chamber is accompanied by 0.79/0.21=3.76 mole of nitrogen, i.e. `1 kmol (O_2) + 3.76 kmol (N_2) = 4.76 kmol (air).`

During combustion, nitrogen acts as an inert and doesn’t react with other elements, but it forms a very small amount of nitric oxides. However, the presence of nitrogen greatly affects the outcome of a combustion process because it enters a combustion chamber in large quantities at low temperatures and exits at considerably higher temperatures, absorbing a large proportion of the chemical energy released during combustion.

In the combustion chamber, the air that enters has some water vapor or moisture content, which also deserves consideration because for most combustion processes; the moisture in the air and the `H_2 O` that forms during combustion can also be treated as an inert gas. But at very high temperatures, some water vapor dissociates into `H_2 & O_2` as well as into `H, O, and OH.`

During a combustion process, the components that exist before the reaction are called reactants, and the components that exist after the reaction are called products.

We already knew from our elementary chemistry textbooks, 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 a chemical reaction. That is, the total mass of each element on the right-hand side of the reaction equation (the products) must be equal to the total mass of that element on the left-hand side (the reactants) even though the elements exist in different chemical compounds in the reactants and products. Also, the total number of atoms of each element is conserved during a chemical reaction since the total number of atoms is equal to the total mass of the element divided by its atomic mass.

Stoichiometric Combustion

The ideal combustion process during which a fuel is burned completely with theoretical air is called the stoichiometric combustion of that fuel. Here, the minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air. Thus, when 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. Also as a product, we obtain only `CO_2, H_2 O, and N_2.`

To show a plot that compares the effect of the number of atoms of carbon on the stoichiometric coefficient (`ar`) for different fuel types namely, alkanes, alkenes, and alkynes; we have to find the relation for stoichiometric coefficient (`ar`) for different types of fuels; we can do that by applying mass balancing to each of the elements in the relation for each type of fuel.

1. Stoichiometric combustion reaction for alkanes:

`C_n H_(2n+2)+ar(O_2+3.76N_2 )=aCO_2+bH_2 O+cN_2`

Where, `n =`number of moles of carbon(C)

`ar =` stoichiometric coefficient

`a =` number of moles of `CO_2`

`b =` number of moles of `H_2 O`

`c =` number of moles of `N_2`

In the above reaction, at the left-hand side, the term in the parenthesis represents the composition of dry air containing 1 mole of `O_2` and 3.76 moles of `N_2`. On the other hand, the coefficients represent the number of moles of gases obtained as a product. The unknowns are calculated by applying mass balance to each of the elements of the above reaction.

For the stoichiometric combustion reaction for alkanes, we’re going to apply the mass balance to each of the elements:-

For carbon `(C)`:- `n=a` ………… (I)

For hydrogen `(H)`:- `2n+2=2b or n+1=b` ………… (II)

For oxygen `(O)`:- `2ar=2a+b` ………… (III)

For nitrogen `(N)`:- `3.76*2*ar=c` ………… (IV)

From the above balancing equation, we’re going to find `'ar'` in terms of `'n'`.

Putting the value of (I) and (II) in equation (III); we get:

`2ar=2n+n+1`

Stoichiometric coefficient, `ar=(3n+1)/2`

This above equation shows the relationship between the stoichiometric coefficient (`ar`) and the number of moles of carbon (`n`) for alkanes.

2. Stoichiometric combustion reaction for alkenes:

`C_n H_2n+ar(O_2+3.76N_2 )=aCO_2+bH_2 O+cN_2`

Where, `n =`number of moles of carbon(C)

`ar =` stoichiometric coefficient

`a =` number of moles of `CO_2`

`b =` number of moles of `H_2 O`

`c =` number of moles of `N_2`

Applying mass balance to each of the elements of the above reaction:-

For carbon `(C)`:- `n=a` ………… (V)

For hydrogen `(H)`:- `2n=2b or n=b` ………… (VI)

For oxygen `(O)`:- `2ar=2a+b` ………… (VII)

For nitrogen `(N)`:- `3.76*2*ar=c` ………… (VIII)

From the above balancing equation, we’re going to find `'ar'` in terms of `'n'`.

Putting the value of (V) and (VI) in equation (VII); we get:

`2ar=2n+n`

Stoichiometric coefficient, `ar=(3n)/2`

This above equation shows the relationship between the stoichiometric coefficient (`ar`) and the number of moles of carbon (`n`) for alkenes.

3. Stoichiometric combustion reaction for alkynes:

`C_n H_(2n-2)+ar(O_2+3.76N_2 )=aCO_2+bH_2 O+cN_2`

Where, `n =`number of moles of carbon(C)

`ar =` stoichiometric coefficient

`a =` number of moles of `CO_2`

`b =` number of moles of `H_2 O`

`c =` number of moles of `N_2`

Applying mass balance to each of the elements of the above reaction:-

For carbon `(C)`:- `n=a` ………… (IX)

For hydrogen `(H)`:- `2n-2=2b or n-1=b` ………… (X)

For oxygen `(O)`:- `2ar=2a+b` ………… (XI)

For nitrogen `(N)`:- `3.76*2*ar=c` ………… (XII)

From the above balancing equation, we’re going to find `'ar'` in terms of `'n'`.

Putting the value of (IX) and (X) in equation (XII); we get:

`2ar=2n+n-1`

Stoichiometric coefficient, `ar=(3n-1)/2`

This above equation shows the relationship between the stoichiometric coefficient (`ar`) and the number of moles of carbon (`n`) for alkynes.

MATLAB program to calculate the stoichiometric coefficient (`ar`) for each fuel and to show a plot that compares the effect of the number of atoms of carbon on the stoichiometric co-efficient `ar`

clear all
close all
clc
 
% creating an array for 20 carbon atoms
% number of moles of carbon is denoted by ‘n’
% stoichiometric coefficient is denoted by ‘ar’
n=linspace(1,20,20);
i=1:length(n);

for i=1:length(n)
    ar_alkanes(i)=((3*n(i))+1)/2
    ar_alkenes(i)=(3*n(i))/2
    ar_alkynes(i)=((3*n(i))-1)/2
end
 
% plotting the graph b/w stoichiometric coefficient (ar) vs. number of moles of carbons (n) for different fuels
 
figure(1)
plot(n,ar_alkanes,'linewidth',2,'marker','o')
hold on
grid on
plot(n,ar_alkenes,'linewidth',2,'marker','x')
hold on
plot(n,ar_alkynes,'linewidth',2,'marker','*')
title('Stoichiometric coefficien(ar) vs. Number of moles(n)')
xlabel('Number of moles(n)')
ylabel('Stoichiometric coefficient(ar)')
legend('alkanes','alkenes','alkynes')

RESULTS

CONCLUSION

From the above plot, we can conclude that the variation of the stoichiometric coefficient (`ar`) vs. the number of moles of carbons (`n`) is linear. Here all three different fuels, named alkanes, alkenes, and alkynes depicting different y-intercepts. It is understood from the above plot and as well as from the ‘ar_alkynes’ values, that y-intercept of alkynes are showing the least y-intercept as compared to the other two fuels; alkynes are showing smaller values of stoichiometric coefficient (`ar`). Since, stoichiometric coefficient simply denotes, the number of moles of oxygen required to completely burn the fuel. Henceforth, alkynes require a small amount of oxygen (air) to get completely burned in comparison to alkenes which requires little more oxygen and alkanes the most amount of oxygen to get burned.