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# Combustion Senatorlibya

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### Combustion Senatorlibya

1. 1. The Theory of Combustion Benzene as a Fuel
2. 2. Definition <ul><li>Benzene , also known as benzol , is an organic chemical compound with the formula C 6 H 6 . </li></ul>
3. 3. 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
4. 4. 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%
5. 5. Fuel Lean Mixture <ul><li>Fuel-air mixtures with more than stoichiometric air (excess air) can burn </li></ul><ul><li>With excess air you have fuel lean combustion </li></ul><ul><li>At low combustion temperatures, the extra air appears in the products in </li></ul><ul><li>unchanged form: </li></ul><ul><li>for a fuel lean mixture have excess air, so  > 1 </li></ul><ul><li>Above reaction equation has two unknowns ( d, e ) and we have two atom balance equations ( O, N ) so can solve for the unknowns </li></ul>
6. 6. <ul><li>Fuel-air mixtures with less than stoichiometric air (excess fuel) can burn. </li></ul><ul><li>With less than stoichiometric air you have fuel rich combustion, there is </li></ul><ul><li>insufficient oxygen to oxidize all the C and H in the fuel to CO 2 and H 2 O. </li></ul><ul><li>Get incomplete combustion where carbon monoxide (CO) and molecular </li></ul><ul><li>hydrogen (H 2 ) also appear in the products. </li></ul>where for fuel rich mixture have insufficient air   < 1 Fuel Rich Mixture <ul><li>Above reaction equation has three unknowns ( d, e, f ) and we only have </li></ul><ul><li>two atom balance equations ( O, N ) so cannot solve for the unknowns </li></ul><ul><li>unless additional information about the products is given. </li></ul>
7. 7. Matlab <ul><li>>> gama=0.6:0.1:1.6; </li></ul><ul><li>>> afratio=(gama*240+gama*789.6)/78; </li></ul><ul><li>>> plot(gama,afratio) </li></ul>
8. 8. % in Fuel Lean Mixture <ul><li>gama=1.5 </li></ul><ul><li>Products: 6+3+1.5x7.5x3.76+3.75=55.05 mole </li></ul><ul><li>A/F ratio=19.8 </li></ul><ul><li>N 2 (%) :76.84% </li></ul><ul><li>CO 2 (%): 10.9% </li></ul><ul><li>H 2 O(%): 5.45% </li></ul>
9. 9. N 2 % in Fuel Rich Mixture <ul><li>Assume H 2 gives H 2 O </li></ul><ul><li>Gama =0.8 </li></ul><ul><li>Prodcts : 3 + 3 +6x3.76+3=31.56 mol </li></ul><ul><li>A/F =(6x32+6x3.76x28)/78=10.56 </li></ul><ul><li>N 2 (%)= 6x3.76/31.56=71.033% </li></ul><ul><li>CO 2 =3/31.56=9.45% </li></ul><ul><li>H 2 O=3/31.56=9.45% </li></ul>
10. 10. Plotting <ul><li>>> afr=[10.56 13.2 19.8]; </li></ul><ul><li>>> n2per=[71.033 75.81 76.84]; </li></ul><ul><li>>> plot(afr,n2per) </li></ul><ul><li>>> hold on </li></ul><ul><li>>> co2per=[9.45 16.3 10.9]; </li></ul><ul><li>>> plot(afr,co2per) </li></ul><ul><li>>> grid </li></ul>
11. 11. Dew Point <ul><li>Partial pressure of H 2 O = % (H 2 O)X total pressure </li></ul><ul><li>At each case of 2,4,6 bar </li></ul><ul><li>And each case of A/F ratio </li></ul><ul><li>We can find the dew point temperature from steam table which is corresponding to the partial pressure in saturated case. </li></ul><ul><li>Also we can use interpolation scheme to find an approximation temperature. </li></ul>
12. 12. 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