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Homework Help: Net calorific value (CV) per kmol of an fuel/air mix

  1. Nov 7, 2017 #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

    Calculate:

    The net calorific value (CV) per kmol of the fuel/air mix at 25°C?

    Data:
    Net calorific value (MJ m–3) at 25°C of:
    Butane (C4H10) = 111.7 MJ m-3
    Butene (C4H8) = 105.2 MJ m-3
    Propane (C3H8) = 85.8 MJ m-3

    Air is 21% oxygen, 79% nitrogen by volume



    2. Relevant equations

    PV = nRT


    3. The attempt at a solution

    C4H10 + 6½O2 ⇒ 4CO2 + 5H2O
    C3H8 + 5O2 ⇒ 3CO2 + 4H2O
    C4H8 + 6O2 ⇒ 4CO2 + 4H2O

    n = PV / RT

    C4H10:

    n = (100 x 0.75) / (8.314 x 298)

    n = 0.0303 kmol

    Amount of 10% excess air reacted with:

    0.0303 x 6.5 = 0.197 kmol of O2

    10% excess of O2 = 0.197 x 1.1 = 0.217 kmol

    Therefore N2 = 0.217 x (79/21) = 0.816 kmol

    Total kmol of air: 0.217 + 0.816 = 1.033 kmol

    Similar calculations for C3H8 & C4H8 give:

    C3H8: 0.00404 Kmol & Air: 0.1057 kmol

    C4H8: 0.00605 Kmol & Air: 0.19 kmol

    Total amount of kmol of fuel = 0.0404 kmol
    Total amount of kmol of air = 1.3287 kmol

    Total kmol fuel/air mix = 1.3691 kmol

    Therefore the net calorific value (CV) per kmol of the fuel/air mix at 25°C:

    This is where I now get stuck. (But I will attempt to finish the question)

    C4H10: [0.0303 x (1.3287 / 1.3691)] x 111.7 = 3.28 MJ m-3

    C3H8: [0.00404 x (1.3287 / 1.3691)] x 105.2 = 0.41 MJ m-3

    C4H8: [0.00605 x (1.3287 / 1.3691)] x 85.8 = 0.54 MJ m-3

    Therefore the net calorific value (CV) per kmol of the fuel/air mix at 25°C:

    3.28 + 0.41 + 0.54 = 4.23 MJ m-3 per kmol.

    Close?
     
  2. jcsd
  3. Nov 8, 2017 #2
    This exact same problem has been addressed on Physics Forums previously. Please use the Search to find it.
     
  4. Nov 8, 2017 #3
    Evening,

    I have just read though this thread:

    https://www.physicsforums.com/threads/net-calorific-value.888755/#post-5590572

    But all that is answered there is "the net calorific value (CV) per m^3 of the fuel/air mix at 25°C"

    I have already worked this out for myself. I can't see a solution to work out the net CV per kmol.

    Best regards,
     
  5. Nov 8, 2017 #4
    Actually there is a further post on page 2 I didn't see.

    "so per kmol is 40.89 mols / 3.1804 = 0.013 Mj / Kmol"

    Is it as simple as that?
     
  6. Jan 3, 2018 #5
    Is there any further feedback on this?
     
  7. Jan 3, 2018 #6
    DM me and I'll give you my findings.
     
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