1

The Theory of CombustionBenzene as a Fuel

2

Definition

• Benzene, also known as benzol, is an organic chemical compound with the formula C6H6 .

3

Combustion Stoichiometry

Air contains molecular nitrogen N2, 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 CH with air is:

22222 )76.3( dNOcHbCONOaHC

The above equation defines the stoichiometric proportions of fuel and air.

Example: For benzene (C3H8) = 6 and = 6

22222 476.3

2)76.3(

4NOHCONOHC

C balance: = bH balance: = 2c c = /2O balance: 2a = 2b + c a = b + c/2 a = + /4N balance: 2(3.76)a = 2d d = 3.76a/2 d = 3.76( + /4)

C6H6 + (7.5)(O2+3.76N2) 6CO2+3H2O+3.76(7.5)N2

4

Combustion Stoichiometry

Note above equation only applies to stoichiometric mixture

For benzene (C6H6), (A/F)s = 13.2

The stoichiometric mass based air/fuel ratio for CH fuel is:

HC

NO

fuelii

airii

fuel

airs MM

MM

Mn

Mn

m

mFA

22 476.3

4/

The products of benzene:

6CO2+3H2O+3.76(7.5)N2

=6 + 3 + 3.76x7.5 = 37.2 mol

N2(%)=75.81%

H2O(%)=8.1%

CO2(%)=16.13%

5

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 unchanged form:

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

222222 2)76.3)(

4( eOdNOHCONOHC

6

• 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 CO2 and H2O.

• Get incomplete combustion where carbon monoxide (CO) and molecular hydrogen (H2) also appear in the products.

222222 2)76.3)(

4( fHeCOdNOHCONOHC

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.

7

Matlab• >> gama=0.6:0.1:1.6;• >> afratio=(gama*240+gama*789.6)/78;• >> plot(gama,afratio)

0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.66

8

10

12

14

16

18

20

22

gamma

air/

fuel

rat

io

8

% in Fuel Lean Mixture

• gama=1.5

• Products: 6+3+1.5x7.5x3.76+3.75=55.05 mole

• A/F ratio=19.8

• N2 (%) :76.84%

• CO2(%): 10.9%

• H2O(%): 5.45%

9

N2 % in Fuel Rich Mixture

Assume H2 gives H2O

• Gama =0.8

• Prodcts : 3 + 3 +6x3.76+3=31.56 mol

• A/F =(6x32+6x3.76x28)/78=10.56

• N2 (%)= 6x3.76/31.56=71.033%

• CO2=3/31.56=9.45%

• H2O=3/31.56=9.45%

10

Plotting• >> afr=[10.56 13.2 19.8];• >> n2per=[71.033 75.81 76.84];• >> plot(afr,n2per)• >> hold on• >> co2per=[9.45 16.3 10.9];• >> plot(afr,co2per)• >> grid

10 11 12 13 14 15 16 17 18 19 200

10

20

30

40

50

60

70

80

A/F ratio

N2

%C

O2

%

N2 %

CO2 %

11

Dew Point

• Partial pressure of H2O = % (H2O)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.

12

Plotting

10 11 12 13 14 15 16 17 18 19 2045

50

55

60

65

70

75

80

85

A/F ratio

dew

poi

nt te

mp.

(c)

p=2 bar

p=4 bar

p=6 bar

>> 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