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Homework Help: Combustion analysis

  1. Mar 7, 2012 #1
    1. The problem statement, all variables and given/known data

    A fuel oil burned in a steam generator has a composition that may be represented as C15H32.

    A dry basis analysis of the flue gas yields the following molar composition: CO2 10.111%,
    CO 1.000%, O2
    5.611%, and N2 83.278%.

    Determine the following:
    (a). The percent excess air used to burn the fuel oil;
    (b). The flue gas molar composition on a wet basis, reporting all constituents, including
    water vapor, to the nearest 0.001%; and
    (c). The dew point temperature (°C) of the flue gas, if the pressure exiting the combustion
    stack is 180 kPa

    2. Relevant equations
    Excess air = (actual air- stoichiometric air)/actual air

    3. The attempt at a solution
    Since there is CO in the product, it means that part of the fuel undergoes incomplete combustion. I am trying to find the percentage of the fuel that undergoes the full combustion and the percentage that undergoes partial combustion.
    C5H32 + 13 O2 ->5CO2 + 16 H2O ------>complete combustion
    C5H32 + 10.5 O2 ->5CO2 + 16 H2O ---------> Partial combustion
    I am not sure where to go next.
  2. jcsd
  3. Mar 8, 2012 #2

    You should add nitrogen to the left hand side of your equation. It would be entered in the correct ratio as it exits in air. It must then appear on the right hand product side.
  4. Mar 8, 2012 #3
    I added the N2 to both equations:
    C5H32 + 13 O2 + 13.(3.76) N2 -->5CO2 + 16 H2O+48.88 N2 ------>complete combustion
    C5H32 + 10.5 O2+10.5.(3.76) N2 -->5CO2 +16 H2O+39.48 N2-------> Partial combustion

    Now how can I find out how much excess air was used and how much of the fuel was in complete and how much of the fuel was used as incomplete combustion?
  5. Mar 9, 2012 #4
    Your second equation is not balanced and you are not indicating any CO or O2 in the products of combustion.


    Write the second equation as

    aC5H32 + bO2 + cN2 = 10.111CO2 + 1.000CO + 5.611O2 + 83.278N2 + dH2O

    and determine correct values for a,b,c,d so that equation balances. Then proceed from there.
    Last edited: Mar 9, 2012
  6. Mar 9, 2012 #5
    Do you think we should exclude H2O since the molar composition is given in dry basis?
  7. Mar 9, 2012 #6
    H2O is a product ofcombustion. In order for the equation to balance, it should be present.
  8. Mar 9, 2012 #7
    I did this:
    Since the amount of N2 in the reactant has to be equal the amount of N2 in products, therefore and using the ratio of N2/O2 = 3.76 I set up two equations:
    x.y=5.6 (amount of O2); x = theoretical air and y = 1+excess air
    x(1+y) (3.76) = 83.278 (amount of N2) and I got x=16.568 and y=1.338 which means there is 33.8% of excess air. The amount of O2 therefore is: 16.568*1.338=22.168 and the amount of N2 = 16.58*1.338*3.76 = 83.278. The balanced equation becomes:
    0.674 C15H32+ 22.168 O2 + 83.278 N2 = 10.111 CO2 + 1 CO + 83.278 N2 + 10.784 H2O
    which is the molar composition on wet base. Does this look ok? By the way, there is about 0.5 mole of O2 more in right side of the equation than the left side which I assumed is because I only used three decimal points. Is it reasonable?

    Thanks alot
  9. Mar 9, 2012 #8
    My balanced equation is:

    aC15H32 +bO2 + cN2 = 10.111CO2 + 1.000CO + 5.611O2 + 83.278N2 + dH2O



    It balances well.
    Last edited: Mar 9, 2012
  10. Mar 9, 2012 #9
    So this answers the part (a) and (b) of the equation, correct?
  11. Mar 9, 2012 #10
    To get excess air you should write the balanced actual equation for 1 mole of fuel. Then write another equation for theoretical air and perfect combustion for same 1 mole of fuel. Using the atomic weights compute the air to fuel ratio for each. Then form a ratio of the ratios. This provides the percent theoretical air so anything over 100% is excess.. The above also gives you what you seek for part b.
  12. Mar 9, 2012 #11
    Here is my calculation:
    C15H32+29.899 O2+112.42 N2=13.649 Co2 + 1.350 CO +7.574 O2+112.425 N2+ 15.998 H2O
    And for the theoretical air:
    C15H32+ 23 O2+ 86.48 N2=15 CO2+ 16 H2O+86.48 N2
    AFR(theoretical) = (28.8*(23+86.48))/212 = 14.872
    AFR(stoich) = (28.8*(29.899+112.42))/212 = 19.333
    Excess air = 19.333/14.872 = 1.3 or 30%

    Looks good?
  13. Mar 9, 2012 #12
    That's it, 30% excess air for the combustion.
  14. Mar 9, 2012 #13
    Thank you
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