Matlab Code to Estimate the Values Stoichiometric Co-Efficient during Combustion Process
Aim
The major objective of this project was to calculate the values of all the unknown Constant that is the Stoichiometric coefficient that are there in Combustion Equation Using Matlab.
Introduction
Stoichiometric combustion offers an easy way to understand combustion concepts. When fuel and air mix they produce products as `COunderset{2}{} ,Co,OH,Hunderset{2}{}O, NO,Nunderset{2}{}O` and many more. In stoichiometric combustion, we make a simple assumption that the by-products are only `COunderset{2}{},Hunderset{2}{}O,Nunderset{2}{}`. For a generic alkane fuel, the combustion process will following the below equation.
`Cunderset{n}{}Hunderset{2n+2}{}+ar(ounderset{2}{}+3.76Nunderset{2}{})=aCounderset{2}{}+bHunderset{2}{}"o"+cNunderset{2}{}`
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
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Balancing the Equation
Alkane
`Cunderset{n}{}Hunderset{2n+2}{}+ar(ounderset{2}{}+3.76Nunderset{2}{})=aCounderset{2}{}+bHunderset{2}{}"o"+cNunderset{2}{}`
Balancing the Carbon
`"Lhs"=Cunderset{n}`
`"Rhs"=aC`
`"So finally " n=a" ""Equation 1 "`
Balancing the Hyrdogen
`"Lhs "=Hunderset{2n+2}`
`"Rhs"=bHunderset{2}`
`2n+2=2b`
`n+1=b" Equation 2"`
Balancing the Oxygen
`"Lhs"=arOunderset{2}`
`"Rhs"=aOunderset{2}+bO`
`2ar=2a+b`
On substituting the value of a and b
`ar=frac{3n+1}{2}`
Alkene
`Cunderset{n}{}Hunderset{2n}{}+ar(ounderset{2}{}+3.76Nunderset{2}{})=aCounderset{2}{}+bHunderset{2}{}"o"+cNunderset{2}{}`
Balancing the Carbon
`"Lhs"=Cunderset{n}`
`"Rhs"=aC`
`"So finally " n=a" ""Equation 1 "`
Balancing the Hyrdogen
`"Lhs "=Hunderset{2n}`
`"Rhs"=bHunderset{2}`
``2n=2b``
``n=b`" Equation 2"`
Balancing the Oxygen
`"Lhs"=arOunderset{2}`
`"Rhs"=aOunderset{2}+bO`
`2ar=2a+b`
So finally we get on Substituting the value os a and b
`ar=frac{3n}{2}`
Alkyne
`Cunderset{n}{}Hunderset{2n-2}{}+ar(ounderset{2}{}+3.76Nunderset{2}{})=aCounderset{2}{}+bHunderset{2}{}"o"+cNunderset{2}{}`
Balancing the Carbon
`"Lhs"=Cunderset{n}`
`"Rhs"=aC`
`"So finally " n=a" ""Equation 1 "`
Balancing the Hyrdogen
`"Lhs "=Hunderset{2n-2}`
`"Rhs"=bHunderset{2}`
`n-1=b" Equation 2"`
Balancing the Oxygen
`"Lhs"=arOunderset{2}`
`"Rhs"=aOunderset{2}+bO`
`2ar=2a+b`
On substituting the value of a and b
`ar=frac{3n-1}{2}`
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MatLab Code:
%MATLAB CODE TO ESTIMATE THE VALUES STOICHIOMETRIC CO-EFFICIENT COMBUSTION PROCESS
clc
clear all
close all
n=0:1:10;
%STOICHIOMETRIC CO-EFFICIENT For Alkane
alkane=(3*n+1)/2;
%STOICHIOMETRIC CO-EFFICIENT For Alkene
alkene=(3*n/2);
%STOICHIOMETRIC CO-EFFICIENT For Alkyne
alkyne=(3*n-1)/2;
figure(1)
plot(n,alkane);
hold on
plot(n,alkene);
plot(n,alkyne);
hold off
% add title of the plot
title("Effect of no of atoms of carbon on the Stoichiometric Coeff.")
% add legend in the plot
legend("Alkane","Alkene","Alkyne")
ylabel("STOICHIOMETRIC CO-EFFICIENT(ar)");
xlabel("Number of Moles of Carbon")
Conclusion
At last it can be concluded that the alkane has higher stoichiometric coefficient as compared to Alkyne and Alkene where as Alkyne had least as the number of moles of carbon increases