Stoichiometric Combustion Calculator

Objective: -

In this challenge, we will be writing a code for a Stoichiometric Combustion Calculator for Alkane, Alkene and Alkyne fuels using Matlab.

Solution: -

Stoichiometric or Theoretical Combustion is the ideal combustion process where fuel is burned completely.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

`C_nH_(2n+2) + ar(O2 + 3.76N2) = aCO2 + bH2O + cN2`

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

Alkanes, alkenes and alkynes are simple hydrocarbon chains with no functional groups. Alkanes are identified because the carbon chain has only single bonds. Common alkanes include methane (natural gas), propane (heating and cooking fuel), butane (lighter fluid) and octane (automobile fuel). Alkenes have at least one double bond and alkynes have at least one triple bond. The most common alkyne is ethyne, better known as acetylene. The generic formula for alkanes is `C_nH_(2n+2)`, where n is the number identified by the prefix. Alkenes have the formula `C_nH_(2n)` and alkynes use the formula `C_nH(2n-2)`.

Writing the formulas for simple alkanes, alkenes and alkynes is as simple as determining how many carbons are in the formula and then putting that number into the generic formula for that hydrocarbon. Since alkanes do not have any real parts to identify, unlike all other organic molecules, there is no need to number the carbons.

Code:

clear all 
close all 
clc 

% first 20 carbon atoms
x = linspace(1,20,20);

for i=1:length(x)          
 Alkane(i)=3/2*i+1/2;     
 Alkene(i)=3/2*i;     
 Alkyne(i)=3/2*i-1/2; 
end 

% plotting the effect
hold on
plot(x,Alkane) 
plot(x,Alkene) 
plot(x,Alkyne) 

xlabel('Stochiometric Co-efficent') 
ylabel('Number of Carbon Atoms')

axis([1 20 1 20]); 
title('Effect of number of moles of carbon on the stoichiometric co-efficient ') 
legend('Alkane','Alkene','Alkyne')

Results:

From the above figure, we can say that there is a linear trend in the stoichiometric ratio of carbon atoms. Having the same number of carbon atoms, alkanes require more amount of time to combust when compared with Alkenes and Alkynes.