STOICHIOMETRIC COMBUSTION COEFFICIENT CALCULATOR USING MATLAB

AIM:

1. Write functions to calculate the stoichiometric co-efficient (ar) for each fuel

2. Show a plot that compares the effect of number of atoms of carbon on the stoichiometric co-efficient "ar". Compare how this trend changes with the fuel type (alkane, alkene and alkyne).

THEORY:

Combustion or burning is a high temperature exothermic reaction. It happens between a fuel and oxygen (oxidant), giving out gaseous products, also termed as smoke.

Alkane is referred to as a saturated open chain hydrocarbon consisting of carbon-carbon single bonds. For a generic alkane fuel the combustion process will following the below equation:

C_nH_2n+2 + ar(O₂ + 3.76N₂) `->` aCO₂ + bH₂O + cN₂

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

The stoichiometric coefficient is given by:

ar = `(2a+b)/2` `->` Eq 1

n=a `->` Eq 2

2n+2 = 2b `->` n=b-1 `->` b=n+1 `->` Eq 3

Eq 2 and 3 in 1 gives:

ar = `(3n+1)/2`

Alkene is referred to as a saturated open chain hydrocarbon consisting of carbon-carbon double bonds. For a generic alkene fuel the combustion process will following the below equation:

C_nH_2n + ar(O₂ + 3.76N₂) `->` aCO₂ + bH₂O + cN₂

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

The stoichiometric coefficient is given by:

ar = `(2a+b)/2` `->` Eq 1

n=a `->` Eq 2

n=b `->` Eq 3

Eq 2 and 3 in 1 gives:

`ar = (3n)/2`

Alkyne is referred to as a saturated open chain hydrocarbon consisting of carbon-carbon triple bonds. For a generic alkyne fuel the combustion process will following the below equation:

C_nH_2n-2 + ar(O₂ + 3.76N₂) `->` aCO₂ + bH₂O + cN₂

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

The stoichiometric coefficient is given by:

ar = `(2a+b)/2` `->` Eq 1

n=a `->` Eq 2

n=b+1 `->` Eq 3

Eq 2 and 3 in 1 gives:

`ar = (3n-1)/2`

MATLAB Code to compares the effect of number of atoms of carbon on the stoichiometric co-efficient "ar"

function stoichiometric_calculator(max_number_c_atoms)
n=linspace(1,max_number_c_atoms,20);

for i=1:length(n)
    ar1(i) = ((3*(n(i)))+1)/2;   %stoichiometric coeficient for alkanes ar = 3n+1/2
    ar2(i) = (3*n(i))/2;         %stoichiometric coeficient for alkenes ar = 3n/2
    ar3(i) = ((3*n(i))-1)/2;     %stoichiometric coeficient for alkanes ar = 3n-1/2

end
plot(n,ar1,n,ar2,n,ar3)
grid on
xlabel('No: of Carbon Atoms')
ylabel('Stoichiometric Coefficient')
legend('Alkanes','Alkenes','Alkynes')

end

Here, the user of the calculator can determine what should be the maximum number of carbon atoms.

We can enter the following line of code to call on the fuction

stoichiometric_calculator(20) 

or

stoichiometric_calculator(30) 

or

stoichiometric_calculator(50) 

depending on user descretion.

OBSERVATIONS:

OUTPUT FOR stoichiometric_calculator(20) 

OUTPUT FOR stoichiometric_calculator(30) 

OUTPUT FOR stoichiometric_calculator(50) 

CONCLUSION:

• From the above plots of Stoichiometric coefficients(ar) vs Number of Carbon Atoms we can observe that the variation is linear.
• The order of decrease of slopes is:   Alkanes>Alkenes>Alkynes
• The above remark means that, for a lesser stoichiometric coefficient less amount of oxygen or air is required for complete combustion. Hence, alkynes requires less oxygen than alkenes which requires less oxygen than alkanes for complete combustion.

 LINK TO CODE: https://drive.google.com/file/d/1YQC7wmVjQhqQngXkzD26-1dUWjQIS4SJ/view?usp=sharing