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Flue gas flow rate

  1. Nov 8, 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.

    If 5% of the heat available for steam production is lost to the atmosphere, determine the amount of steam raised per hour when the total flow of flue gases is 1400 kmol h–1.

    2. Relevant equations


    3. The attempt at a solution

    Cp(mean) = (Cp x T2) - (Cp1 x T1) / T2 - T1

    = (2.934 x 2000) - (1.89 x 90) / 2000 - 90

    = 2.98kJ/kG K then .......

    delta (T) x Cp(mean) x Q x 0.95
    = (198790) x 2.98 x 1400 x 0.95 = 7518579kJ/hr
    https://www.physicsforums.com/file:///page9image15296 [Broken] https://www.physicsforums.com/file:///page9image15456 [Broken]
    latent heat of evaporation @ 90°C = 2282 kJ / kg k

    so....

    7518579 / 2282
    = 3295kmol/hr

    I have been told that my answer is incorrect.

    any help to see where i have gone wrong is appreciated.

    thanks.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Nov 8, 2016 #2
    Where did the 2000 come from?
     
  4. Nov 8, 2016 #3
    Hi chestermiller.

    Its the flame temperature.
    I calculated the flame temperature to be 1987°C. (which might also be incorrect).

    what do you make it?

    Regards.
     
  5. Nov 9, 2016 #4
    Please show your work.

    How can the average heat capacity be greater than either of the values at the upper temperature and lower temperature?
     
  6. Nov 9, 2016 #5
    Butane = -2660 kJ / mol
    Propane = -2046 kJ / mol
    Butane = -2540 kJ / mol

    Heat released by combustion of 1 mole of the fuel :

    = (0.75x2660) + (0.1x2046) + (0.15x2540) = 1995 + 204.6 + 381
    = 2581 kJ

    To burn 1 mole of fuel mix we require 6.3 mole of O2
    and produce in our flue gas
    3.9 mole of CO2
    4.75 mole of H2O
    with 10% excess air 1.1x6.3 = 6.93 mole O2

    6.3 mole is used for the combustion 0.63 is present in the flue gas

    N2 associated with the O2 supply = 6.93x3.76
    =26.06 moles
    N2 is unchanged through the process
    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
    Value is slightly high so assume a flame temp of 1900°C

    For N2, 62.46 x 26.06 = 1627.7kJ CO2, 102.13 x 3.9 = 398.3 kJ O2, 65.81 x 0.63 = 41.5kJ H2O, 81.17 x 4.75 = 385.6 kJ

    Total = 2453.1 kJ
    This shows temperature must sit between 1900°C — 2000°C

    I used a graph and made it to be 1987°C
     
  7. Nov 9, 2016 #6
    I'm having trouble figuring out what you are doing here. What is the 66.1 for N2?
     
  8. Nov 9, 2016 #7
    N2 = 66.1 kj / mol is the enthalpy at 2000°C
    This is how I have derived the maximum flame temp. to be 1987°C. Then from this obtained the heat capacity.
    what do you make it?
     
  9. Nov 9, 2016 #8
    How did you determine this enthalpy value (per mole)? Please provide the equation and data values used.
     
  10. Nov 10, 2016 #9
    my course work supplied me with a table for gas enthalpy at various temperatures. I didn't do ant equations here. I just simply used the data from the table.
    then used the theory of heat in = heat out. My heat in I calculated to be 2581kj. Then i tabulated the output estimates which were within my heat content onto a graph and for a balanced combustion the flame temperature would need to be 1987°C.
    I think this must be incorrect?????
     
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