stoichiometric co-efficient (ar) calculator

Aim:

To create a stoichiometric coefficient (ar) calculator code using MATLAB.

Objective:

• To calculate the stoichiometric co-efficient (ar) for alkane, alkene and alkyne type of fuels.
• To create a plot that compares the effect of number of atoms of carbon on the stoichiometric co-efficient "ar" in different types of fuels.

Stoichiometric combustion:

Stoichiometric combustion is an ideal combustion process in which the fuel is burned completely. The complete combustion is happened in which the carbon(C) is converted into carbon-di-oxide (C02) and all hydrogen atom is converted into H2O similarly all the sulphur (S) is converted into SO2. In this combustion process, when the air and fuel mixed, they produce co2, co, oh, h2o, no, n2o and many more as a product. Here we made an assumption that the products are co2, h2o and n2 only. This stoichiometric combustion can be represented by the following chemical equation,

where,

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

The above mentioned equation represent the combustion of the alkane fuel. The alkane fuel is represented by CnH2n+2. Similarly, the alkene fuel can be represented by CnH2n and the alkyne fuel by CnH2n-2.

The "ar" term is different for different types of fuels. This term is used to find the nature of the air-fuel mixture.

Calculation:

The equation of different types of fuels are solved to get the relation between stoichiometric co-efficient value (ar) and the number of moles of C (n).

Equating C term:

Number of C atom in L.H.S = 1*n = n

Number of C atom in R.H.S = a*1 = a

By equating L.H.S and R.H.S, n=a (eqn., 1)

Equating H term:

Number of H atom in L.H.S = 1*(2n+2) = 2n+2

Number of H atom in R.H.S = b*2 = 2b

By equating L.H.S and R.H.S, 2n+2=2b => n+1 = b (eqn., 2)

Equating Oxygen term:

Number of O atom in L.H.S = 2*ar = 2ar

Number of O atom in R.H.S = (a*2) + (b*1)= 2a+b

By equating L.H.S and R.H.S, 2ar = 2a+b

by substituting eqn.,1 and eqn.,2 2ar = 2n+(n+1)

2ar = 3n+1

ar = (3n+1)/2

By solving this equation for several values of n, we can found the relation between stoichiometric co-efficient value (ar) and the number of moles of C (n).

Similarly for alkene (CnH2n) => ar = 3n/2

alkyne (CnH2n-2) => ar = (3n-1)/2

Code:

clear all
close all
clc

%Initializing the different values for the number of atoms of carbon
n=1:10;

%By equating the alkane equation, ar and n is related with the 
% equation ar = (3n+1)/2
ar_alkane = (1.5*n)+0.5;

% For alkene ar = 3n/2
ar_alkene = 1.5*n;

% For alkyne ar = 3n-1/2
ar_alkyne = (1.5*n)-0.5;

%plot that compares the effect of number of atoms of carbon on 
% the stoichiometric co-efficient "ar"
plot(n,ar_alkane,'linewidth',2,'color','r')
hold on
plot(n,ar_alkene,'linewidth',2,'color','b')
hold on
plot(n,ar_alkyne,'linewidth',2,'color','g')
xlabel('number of atoms of carbon')
ylabel('stoichiometric co-efficient "ar"')
legend('Alkane','Alkene','Alkyne')

Explanation:

• Initially, the n value is assigned from 1 to 10.
• Solving the relation between ar and n for alkane fuel.
• Similarly, the relation for other two types of fuels are also solved.
• Finally, the ar for different n values are plotted for different types of fuels and the plot is analysed.

Output:

Conclusion:

From the plot obtained, it can observe that the stoichiometric value increases with increase in number of moles of C atom. This slope has a higher value for alkane, which shows that the complete combustion of fuel take place in alkane is better when compared with other two types of fuels. Thus, alkane will be the best fuel to achieve better stoichiometric combustion. The fuel preference is of the order of alkane>alkene>alkyne.