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Calculate the Flame temperature

  1. Oct 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume.
    It is to be fed to the combustion chamber in 10% excess air at 25°C, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90°C.

    Determine the maximum flame temperature ?.

    2. Relevant equations

    heat input to process = heat out of process
    Heat out of process = ∑ (mass × mean specific heat capacity × TF ) for each product of combustion


    3. The attempt at a solution

    2581 Kj

    1987°C

    This answer is incorrect.
    Any help is much appreciated.
     
  2. jcsd
  3. Oct 20, 2016 #2

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    C'mon --- you know the "drill" --- sum the heat capacities, do the division, ....
     
  4. Oct 20, 2016 #3
    butane = -2660 kj/mol
    propane = -2046 kj/mol
    butene = -2540 kj/mol
    heat released by 1 mole of fuel
    = (0.75x2660)+(0.1x2046)+(0.15x2540)
    =2581 kj (heat in)

    is this correct so far?
     
    Last edited: Oct 20, 2016
  5. Oct 20, 2016 #4

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    What are the products, and what are their heat capacities?
     
  6. Oct 20, 2016 #5
    Flue gas =

    H2O = 4.75 moles
    CO2 = 3.9 moles
    O2 = 0.63 moles
    N2 = 26.06 moles
    Total = 35.34 moles

    assuming a flame temperature of 2000°C

    Heat content = enthalpy x no. of moles


    For N2, 66.10 x 26.06 = 1722.6 kJ

    CO2, 108.32 x 3.9 = 422.5 kJ

    O2, 69.65 x 0.63 = 43.9 kJ

    H2O, 86.24 x 4.75 = 409.6 kJ

    Total = 2598.6 kJ

     
  7. Oct 21, 2016 #6
    how does this look?
     
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