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Flame Temperature

  1. Nov 20, 2014 #1
    C4H10 + 4.5O2 -> 4Co2 + 5H2O1. The problem statement, all variables and given/known data

    A fuel gas consists of 75% butane (C4 H10 ), 10% propane (C3 H8 ) and butene (C4 H8 ) by volume.
    It is to be fed to the combustion chamber in 10% excess air at 25ºwhere 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 90ºC.

    Determine the maximum flame temperature.

    2. Relevant equations
    Actual oxygen supplied = theoretical x excess air
    Heat content = enthalpy x no of moles

    3. The attempt at a solution

    lower valvue for propane 2046 KJ/mol, butane 2660 KJ/mol

    (.75*2660)+(.1*2046)+(.15*2660)
    1995+204.6+399=2598.6KJ = 2598.6KJ/mol fuel

    C4H10 + 4.5O2 -> 4Co2 + 5H2O
    1mol 4.5mol 4 mol 5 mol
    .75 mol 3.375mol 3mol 3.75mol

    C3H8 + 5O2 -> 3Co2 + 4H2O
    1mol 5mol 3 mol 4 mol
    .1 mol .5mol .3mol .4mol

    C4H8 + 6O2 -> 4Co2 + 4H2O
    1mol 6mol 4 mol 4 mol
    .15 mol .9mol .6mol .6mol

    4.775 O2 Req
    actual O2 = theo x excess air
    4.775 x 1.1 = 5.2525 mol
    N2 = 5.2525 x 3.76=19.749 moles in flue gas (3.76 is 79%/21% conversion factor from O2 to air to N2 )

    the problem is the next bit once i work out the heat content the enthalpy table goes up to 2200 Deg C however when i work out the heat content it it below what i worked out in the first steps (2598.6KJ/mol fuel) could someone check my calcs above to ensure there are correct as i believe it should be somewhat less to tie into the enthalpy table to give me a max flame temp.

    thanks
     
  2. jcsd
  3. Nov 20, 2014 #2

    Bystander

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    Plus combustion products?
     
  4. Nov 21, 2014 #3
    Hi bystander
    I'm not sure what you mean, do you need to add more values to the nitrogen?
    Thanks
     
  5. Nov 21, 2014 #4

    Bystander

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    What is the composition of the flue gas?
     
  6. Nov 21, 2014 #5
    O2=0.4775
    N2=19.749
    H2O=4.75
    CO2=3.9moles
     
  7. Nov 21, 2014 #6

    Bystander

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    And the heat capacity?
     
  8. Nov 21, 2014 #7
    I believe it to be 4.18 KJ/kg/k
    Obtained from steam tables
     
  9. Nov 21, 2014 #8

    Bystander

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    That's for liquid water --- we're still in the firebox rather than the boiler. What's the heat capacity of the flue gas?
     
  10. Nov 21, 2014 #9
    I'm not sure how to obtain this value
    With the flue gas being a gas I'm assuming this is going to be a really low value?
     
  11. Nov 21, 2014 #10

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    Just checked through your original post, and the heat of combustion for butene is NOT the same as that of butane, it's about 95% that of butane, which reduces your total enthalpy of combustion by 1-2%.

    That enough to fit your table? And we'll forget I started trying to get you to do things the hard way.
     
  12. Nov 21, 2014 #11
    No problem, the lecturer has stated we can use the same value for both butane and butene although these are slightly different. 1-2% would not change my final value so it would fit that of the table.
    Is it possible it is above 2200 deg C or is this too high ?
    The question stated max temp so I'm assusung it is below 2200?
     
  13. Nov 21, 2014 #12

    Bystander

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    What I've been trying to do is enthalpy divided by heat capacity for del T. This ignores the fact that CO2 and H2O are significantly dissociated at high temperatures. If you hit "flame temperature" on Wiki, it comes up around 2000 C, give or take. Whereas, total enthalpy of combustion which is measured by cooling products to 25 C ------- and all of a sudden the golden age brain death becomes transparent!!!!!

    Total enthalpy is as you've calculated MINUS enthalpy of vaporization of water for the 4.75 moles of water since they are not condensed in the flue gas --- that'll be around 200 kJ you can take from the total heat of combustion since it has not been recovered by condensation.

    Sorry, Man --- I'm up too late.
     
  14. Nov 21, 2014 #13

    Ah I get this now I will re do this and hopefully it will work out- thanks for your help
    Much appreciated
     
  15. Nov 23, 2014 #14
    Hi I was wondering if you would know what equation to use for this ...

    1. (i) 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.

      Can I just deduct 5% from the heat available then input that into an equation to give the amount of steam raised per hr?
      Thanks mate
     
  16. Nov 23, 2014 #15

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    Should do it.
     
  17. Mar 21, 2016 #16
    Mitch, could you help me out by letting me know what equation you are referring to here?
     
  18. Mar 21, 2016 #17
    Hi
    I will dig out my notes and let you know. I am not 100% off the top of my head
    Regards
     
  19. Mar 22, 2016 #18
    appreciate that!


    cheers
     
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