Benzene , also known as benzol , is an organic chemical compound with the formula C 6 H 6 .
Combustion Stoichiometry Air contains molecular nitrogen N 2 , when the products are low temperature the nitrogen is not significantly affected by the reaction, it is considered inert . The complete reaction of a general hydrocarbon C H with air is: The above equation defines the stoichiometric proportions of fuel and air. Example: For benzene (C 3 H 8 ) = 6 and = 6 C balance: = b H balance: = 2c c = / 2 O balance: 2a = 2b + c a = b + c / 2 a = + /4 N balance: 2 ( 3.76)a = 2d d = 3.76a/2 d = 3.76( + / 4 ) C 6 H 6 + (7.5)(O 2 +3.76N 2 ) 6CO 2 +3H 2 O+3.76(7.5)N 2
Combustion Stoichiometry Note above equation only applies to stoichiometric mixture For benzene (C 6 H 6 ), (A/F) s = 13.2 The stoichiometric mass based air/fuel ratio for C H fuel is: The products of benzene: 6CO 2 +3H 2 O+3.76(7.5)N 2 =6 + 3 + 3.76x7.5 = 37.2 mol N 2 (%)=75.81% H 2 O(%)=8.1% CO 2 (%)=16.13%
Fuel Lean Mixture
Fuel-air mixtures with more than stoichiometric air (excess air) can burn
With excess air you have fuel lean combustion
At low combustion temperatures, the extra air appears in the products in
for a fuel lean mixture have excess air, so > 1
Above reaction equation has two unknowns ( d, e ) and we have two atom balance equations ( O, N ) so can solve for the unknowns
Fuel-air mixtures with less than stoichiometric air (excess fuel) can burn.
With less than stoichiometric air you have fuel rich combustion, there is
insufficient oxygen to oxidize all the C and H in the fuel to CO 2 and H 2 O.
Get incomplete combustion where carbon monoxide (CO) and molecular
hydrogen (H 2 ) also appear in the products.
where for fuel rich mixture have insufficient air < 1 Fuel Rich Mixture
Above reaction equation has three unknowns ( d, e, f ) and we only have
two atom balance equations ( O, N ) so cannot solve for the unknowns
unless additional information about the products is given.
% in Fuel Lean Mixture
Products: 6+3+1.5x7.5x3.76+3.75=55.05 mole
N 2 (%) :76.84%
CO 2 (%): 10.9%
H 2 O(%): 5.45%
N 2 % in Fuel Rich Mixture
Assume H 2 gives H 2 O
Prodcts : 3 + 3 +6x3.76+3=31.56 mol
N 2 (%)= 6x3.76/31.56=71.033%
CO 2 =3/31.56=9.45%
H 2 O=3/31.56=9.45%
>> afr=[10.56 13.2 19.8];
>> n2per=[71.033 75.81 76.84];
>> hold on
>> co2per=[9.45 16.3 10.9];
Partial pressure of H 2 O = % (H 2 O)X total pressure
At each case of 2,4,6 bar
And each case of A/F ratio
We can find the dew point temperature from steam table which is corresponding to the partial pressure in saturated case.
Also we can use interpolation scheme to find an approximation temperature.
Plotting >> dewpoint1=[58.53 54.666 47.087]; >> dewpoint2=[74.5 70.77 61.765]; >> dewpoint3=[84.431 80.5 71]; >> plot(afr,dewpoint1) >> hold on >> plot(afr,dewpoint2) >> hold on >> plot(afr,dewpoint3) >> grid