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Octane + air combustion equation

  1. Jan 28, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate the flame temperature of normal octane (liquid) burning in air at an equivalence ration of 0.5. All reactants are at 298 K and the system operates at the pressure of 1 atm.

    2. Relevant equations

    Hp = Hr

    3. The attempt at a solution
    I know that the combustion of liquid octane in the theoretical amount of air is
    C8H18 + 12.502 + 47N2 --> 8C02 + 9H20 + 47N2
    Does the equivalency ratio affect this equation? I know equivalency ratio is the actual fuel/oxidizer ratio divided by the stoichiometric fuel/oxidizer ratio, but what does this mean for the above equation? If we assume the stoichiometric fuel/oxidizer ratio is 1, this means that the actual fuel/oxidizer ratio is 1/2....

    If I can get the equation right, I know how to solve for flame temp. using Hp = Hr, JANAF tables, and iteration.

  2. jcsd
  3. May 16, 2009 #2
    well, by your definition of equivalence ratio, that would mean that you're inputting twice as much air as is theoretically needed, so your equation would go:

    C8H18 + 2*12.5 O2 + 2*47N2 --> 8CO2 + 9H2O + 2*47N2 + 12.5 O2

    and all the extra moles of air that form part of the combustion gases have to be heated too, so the adiabatic flame temperature will be less than when you're using the theoretical stoichiometric amount of air. To find the adiabatic flame temp, just equate the enthalpy of the reactants at the input temp. to the enthalpy of the produts at the adiabatic flame temp and solve for the adiabatic flame temp (of course finding the flame temp is a recursive process because you don't know the enthalpy of the products without first assuming the temp they're at).
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