Application of Linear Algebra to Internal Combustion Engines –
Combustion Theory
Troy Ganson
Abstract:
In the analysis of internal combustion engines, several linearly independent
equations must be simultaneously solved to obtain the products of combustion. These
products of combustion allow for mass flowrates of air and fuel to be obtained, but the
major use is in emissions testing to reduce Nitric Oxide, Nitrous Oxide, unburned HydroCarbons, and Carbon Monoxide.
Initially for a combustion reaction to take place, in an internal combustion engine,
a hydro-carbon and mixture of air must be present.
C x H y  a  (O2  3.76 N 2 )  Products of Combustion; where x and y are inherent to the
fuel used and given, and a must be solved for once the products of combustion are
realized and the oxygen and nitrogen mix make-up air with the 3.76 times nitrogen being
the molar fraction of nitrogen found commonly in air. In a simple combustion reaction,
one that does not account for: the fuel-air mix being anything besides stoichiometric
(sufficient oxygen to react fully with the fuel), different temperatures leading to nonequilibrium conditions leading to dissociations in molecular bonds, or nitrogen being
present in air leading to formation of nitric oxide and nitrous oxide.
Beginning with the case of a simple combustion reaction, the algebraic model,
 a1
a
 4
a7
a2
a5
a8
a3   x1   b1 
a6    x2   b2  will be used, but instead of x’s, C, H, O, and N will be used.
a9   x3  b3 
Using the example of complete stoichiometric combustion of methane, the following
reaction would be written: CH 4  a  (O2  3.76N 2 )  b  CO2  c  H 2O  d  N 2 . This
reaction could be written as the following:
C : 1  b 1
H :4  c2
O : a  2  b  2  c 1
N 2 : a  3.76  d  1
Now there are four equations and four unknowns allowing for the unknowns to be solved
1
0 0   a  1 
 0
 0
0
2 0   b  4



for simultaneously in:.
. Row reducing to reduced row
 2
 2  1 0   c  0 

    
0  1 d  0
3.76 0
echelon form yields: a=2, b=1, c=2, and d=7.52. This gives the products of simple
combustion per one mole of fuel.
References:
Heywood, John B. “Internal Combustion Engine Fundamentals” McGraw-Hill, Inc.,
1988.
Drallmeier, Jim PhD. “Class Notes from ME 333, Internal Combustion Engines” 2003