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Combustion Senatorlibya
 

Combustion Senatorlibya

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    Combustion Senatorlibya Combustion Senatorlibya Presentation Transcript

    • The Theory of Combustion Benzene as a Fuel
    • Definition
      • 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
      • 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
      • 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.
    • Matlab
      • >> gama=0.6:0.1:1.6;
      • >> afratio=(gama*240+gama*789.6)/78;
      • >> plot(gama,afratio)
    • % in Fuel Lean Mixture
      • gama=1.5
      • Products: 6+3+1.5x7.5x3.76+3.75=55.05 mole
      • A/F ratio=19.8
      • 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
      • Gama =0.8
      • Prodcts : 3 + 3 +6x3.76+3=31.56 mol
      • A/F =(6x32+6x3.76x28)/78=10.56
      • N 2 (%)= 6x3.76/31.56=71.033%
      • CO 2 =3/31.56=9.45%
      • H 2 O=3/31.56=9.45%
    • 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
    • Dew Point
      • 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