Alkane

Objective:

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

Here we will show the following:

1. A plot will be plotted that compares the effect of the number of atoms of carbon on the stoichiometric co-efficient "ar". Then comparison will be made on how this trend changes with the fuel type (alkane, alkene and alkyne).

2. Discussing the Outputs.

Introduction:

Here we need to develop a relationship between number of Carbon atoms, n with stoichiometric coefficient, 'ar'.

For Alkane:

`C_nH_(2n+2) + ar(O_2 + 3.76N_2) -> nCO_2 + bH_2O + cN_2`

From the equation,

Balancing H-atom on both side:`b = (2n+2)/2`

Balancing O-atom on both side:`ar = (2n + b)/2`

For Alkene:

`C_nH_(2n) + ar(O_2 + 3.76N_2) -> nCO_2 + bH_2O + cN_2`

From the equation,

Balancing H-atom on both side:`b = 2n/2`

Balancing O-atom on both side:`ar = (2n + b)/2`

For Alkyne:

`C_nH_(2n-2) + ar(O_2 + 3.76N_2) -> nCO_2 + bH_2O + cN_2`

From the equation,

Balancing H-atom on both side:`b = (2n-2)/2`

Balancing O-atom on both side:`ar = (2n + b)/2`

Code:

close all
clear all
clc

%Step:1: Defining the relation between 'n' and'ar'

carbon_moles=1:10;
n=carbon_moles;

%Case:1: Alkanes
% C_n.H_2n+2 + ar*(O_2 + 3.76*N_2) --> n*C.O_2 + b*H_2.O + c*N_2

alkane_ar = zeros(1,length(n));
for i=1:length(n)
    n=i;
    b = (2*n+2)/2;
    ar = ((2*n) + b )/2; 
    alkane_ar(i) = ar;
end

%Case:2: Alkenes
% C_n.H_2n + ar*(O_2 + 3.76*N_2) --> n*C.O_2 + b*H_2.O + c*N_2

alkene_ar = zeros(1,length(n));
for i=1:length(n)
    n=i;
    b = (2*n)/2;
    ar = ((2*n) + b )/2;
    alkene_ar(i) = ar;
end

%Case:3: Alkynes
% C_n.H_2n-2 + ar*(O_2 + 3.76*N_2) --> n*C.O_2 + b*H_2.O + c*N_2

alkyne_ar = zeros(1,length(n));
for i=1:length(n)
    n=i;
    b = (2*n-2)/2;
    ar = ((2*n) + b )/2;  
    alkyne_ar(i) = ar;
end

%Step: 2: Graph Plotting

plot(carbon_moles,alkane_ar, ':','Linewidth', 2)
hold on

plot(carbon_moles,alkene_ar, '-.','Linewidth', 2)
hold on

plot(carbon_moles,alkyne_ar, '--','Linewidth', 2)

xlim([0 10])
%ylim([0 18])

xlabel('Moles of Carbon "n"','FontSize',14)
ylabel('Stoichiometric co-efficient "ar"','FontSize',14)

hold off
legend({'Alkane','Alkene','Alkyne'},'Location','northwest','NumColumns',1)
title('Effect of number of moles of carbon on the stoichiometric co-efficient','FontSize',16,'FontWeight','bold')

Algorithm of Code:

1. First of all we have defnied number of carbon atoms in a array 'n' then we have coded the relation between 'ar' and 'n' and then stored the corresponding values of ar for various values of n in the variables 'alkane_ar', 'alkene_ar' and 'alkyne_ar'.

2. Then we have plotted the 'alkane_ar', 'alkene_ar' and 'alkyne_ar' with number of carbon atoms, n and obtained th following graph.

Results & Outputs:

Discussion of Results:

1. From the graph, we can observe that with the change of carbon atoms number the stoichiometric coefficient, 'ar' changes proportionately. So 'ar' is a linear function of 'n'.

2. For a particular value of 'n' Alkanes have the highest value of Stoichiometric Ratio 'ar' and the Alkynes have the lowest value.