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CV calculation

  1. Feb 11, 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

    i) Calculate the net CV per M^3 of the fuel/air mix @ 25°C

    ii) Calculate the net CV per kmol of the fuel/air mix @ 25°C



    2. Relevant equations


    3. The attempt at a solution

    i) 1m^3 @ 25°C

    Butane = 111.7 MJ m^-3
    Propane = 85.8 MJ m^-3
    Butene = 105.2 MJ m^-3

    CV in M^3

    (0.75*111.7)+(0.1*85.8)+(0.15*105.2)

    ANS= 108.135 MJ M^-3


    ----------------------------------------------------------------------------------

    ii)

    Butane @ 75%

    C4H10 + 4.5O2 → 4CO2 + 5H2O
    Stoichiometric value 1 MOL 4.5 MOL
    Actual Value 0.75 MOL 3.375 MOL

    Propane @ 10%
    C3H8 + 5O2 → 3CO2 + 4H2O
    Stoichiometric value 1 MOL 5 MOL
    Actual Value 0.10 MOL 0.5 MOL

    Butene @ 15%
    C4H8 + 6O2 → 4CO2 + 4H2O
    Stoichiometric value 1 MOL 6 MOL
    Actual Value 0.15 MOL 0.9 MOL

    TOTAL MOLES in 1M^3

    Butane = 0.75 MOL
    Propane = 0.1 MOL
    Butene = 0.15 MOL
    O2 = (3.375+0.5+0.9) * 1.1 (excess air) = 5.253 MOL
    N2 = 5.253*3.76 = 19.751 MOL
    ------------------------------------------------------
    TOT = 26.004 MOL
    ------------------------------------------------------

    n = PV / RT
    =(101300*1) / (8.314*298.15)
    =40.86
    =40.9 MOL

    26/40.9
    =0.64M^3

    Butane = 0.75 / 0.64 = 1.17MOL
    Propane= 0.10 / 0.64 = 0.15MOL
    Butene = 0.15 / 0.64 = 0.23MOL
    Oxygen = 5.253 / 0.64 = 8.20MOL
    Nitrogen = 19.751 / 0.64 = 30.86MOL

    V= (n*RT) / P

    Butane = 1.17 * (8.314*298.15) / 101300 = 0.028 MJ / M^3
    Propane = 0.15 * (8.314*298.15) / 101300 = 0.003 MJ / M^3
    Butene = 0.23 * (8.314*298.15) / 101300 = 0.005 MJ / M^3

    Net CV

    Butane = 0.028 * 111.7 = 3.127 MJ M^3
    Propane = 0.003 * 85.8 = 0.257 MJ M^3
    Butene = 0.005 * 105.2 = 0.526 MJ M^3
    -----------------------------------------------------
    TOT = 3.910 MJ M^3
    -----------------------------------------------------

    ANSWER = 40.9/3.910

    ANSWER = 0.0104 MJ / KMOL
     
    Last edited: Feb 11, 2017
  2. jcsd
  3. Feb 11, 2017 #2
    What does CV stand for? Is this happening at a specific pressure?
     
  4. Feb 11, 2017 #3
    Hi Chester, thanks for the reply, iv had a breakthrough- ill just type it up now! 2 secs
     
  5. Feb 11, 2017 #4
    Amended my original effort, I now have an answer! How does it look?
     
  6. Feb 11, 2017 #5

    BvU

    User Avatar
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    Homework Helper
    Gold Member

    It asks for the CV of the fuel/air mixture in part i). Where's the air ?
     
  7. Feb 11, 2017 #6
    I havent got one unfortunately!
     
  8. Feb 11, 2017 #7
    Anyone????
     
  9. Feb 12, 2017 #8
    CV - Calorific Value
     
  10. Feb 12, 2017 #9

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You haven't got one - what ?

    You calculate CV for the fuel mixture, not for the fuel/air mixture as the exercise asks.
     
  11. Feb 12, 2017 #10
    Like I said im struggling with the question, a bit of guidance would be appreciated.

    Thanks
     
  12. Feb 12, 2017 #11
    I would do this differently. I would do part ii first, and then part i is easy.

    I would start out with:

    0.75 moles of Butane
    0.10 moles of Propane
    0.15 moles of Butene

    Then determine the heat for each. Then do the stoichiometry to find the number of moles of air. Then I would get the heat per mole of mixture.
     
  13. Feb 13, 2017 #12
    Butane = 111.7 MJ m^-3
    Propane = 85.8 MJ m^-3
    Butene = 105.2 MJ m^-3

    Gross CV in M^3 (not accounting for water vapour produced)

    (0.75*111.7)+(0.1*85.8)+(0.15*105.2)

    ANS= 108.135 MJ M^-3

    Net CV

    1. Calculate number of H2O moles produced by each gas

    Butane @ 75%
    C4H10 + 4.5O2 → 4CO2 + 5H2O
    0.75*5 = 3.75mol of H2O

    Propane @ 10%
    C3H8 + 5O2 → 3CO2 + 4H2O
    0.10*4 = 0.4mol of H2O

    Butene @ 15%
    C4H8 + 6O2 → 4CO2 + 4H2O
    0.15*4 = 0.6mol of H20

    n = 100,000 * (3.75+0.4+0.6) / (8.314*288) = 57.4mol = 0.0574 Kmol

    KG = Kmol * Mol Mass
    = 0.0574 * 18
    = 1.033kg of water are produced

    Heat released = (mass * latent heat) + ( mass*specific heat capacity*fall in temp)
    = (1.033 * 2258 ) + (1.033*4.20*85)
    = 2701KJ
    = 2.701MJ

    Heat in the water vapour = 1.033 * 1.88 * 85 = 165KJ

    TOTAL DIFFERENCE = 2701KJ - 165KJ
    =2536 KJ
    =2.536 MJ

    NET CV = GROSS CV - HEAT LOST DUE TO VAPORISATION
    = 108.135MJ - 2.536 MJ
    = 105.599 MJ / m^-3
    = 0.105 MJ / M^3

    Is this any better?
     
  14. Feb 13, 2017 #13
    What are the heats of combustion in kJ/mole of the three species at 25 C and 1 atm? (lower heating value)
     
  15. Feb 13, 2017 #14

    Butane = 0.75 * - 2874 = -2155.5 kj/mol
    Propane = 0.10 * - 2220 = -222 kj/mol
    Butene = 0.15 * - 2718 = -407.7 kj/mol

    TOTAL = - 2785.2 KJ/MOL
     
  16. Feb 13, 2017 #15
    The table of lower heating values in Wiki does not agree with your input data: https://en.wikipedia.org/wiki/Heat_of_combustion

    Are you sure you are not using higher heating values?
     
  17. Feb 13, 2017 #16
    Butane = 0.75 * - 2660 = -1995 kj/mol
    Propane = 0.10 * - 2046 = -204 kj/mol
    Butene = 0.15 * - CANT FIND VALUE = -??? kj/mol
     
  18. Feb 13, 2017 #17
    It's right in the table under monoolefins.
     
  19. Feb 13, 2017 #18

    Thats MJ/KG, im looking for KG/MOL
     
  20. Feb 13, 2017 #19
    Are you saying that you don't know how to convert between the two? What is the molecular weight of Butene?
     
  21. Feb 13, 2017 #20
    Butene LLV = MJ/KJ * Mol (w) = 45.334 * 56.106 = 2543

    Butane = 0.75 * - 2660 = -1995 kj/mol
    Propane = 0.10 * - 2046 = -204 kj/mol
    Butene = 0.15 * - 2543 = -381 kj/mol

    ????????????????????????????????????????????????????
     
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