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Combusting Natural gas calculate Q

  1. Feb 14, 2014 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    A natural-gas fuel contains 85 mol-% methane, 10 mol-% ethane, and 5 mol-% nitrogen.
    a) What is the standard heat of combustion (kJ mol-1) of the fuel at 25°C with H2O(g) as a
    product?
    b) The fuel is supplied to a furnace with 50% excess air, both entering at 25°C. The products
    leave at 600°C. If combustion is complete and if no side reactions occur, how much heat
    (kJ mol-1 of fuel) is transferred in the furnace?


    2. Relevant equations



    3. The attempt at a solution
    For this problem, I am unsure if in general, the exiting temperature is the reaction temperature. I mean, initially the gases are at 25 C, but when they exit they are 600C. That doesn't mean it reacted at 600 C, the result of the reaction just happened to be that temperature increase, right?
     

    Attached Files:

  2. jcsd
  3. Feb 14, 2014 #2
    Yes. You have the right idea. I didn't check every last detail of what you did, but you certainly knew to use Hess's Law. The heat given off by the reaction at 25C minus the heat loss from the reactor is enough to raise the temperature of the products from 25C to 600C.

    chet
     
  4. Feb 18, 2014 #3

    Maylis

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    Does nitrogen combust with oxygen?

    Also, is my part (a) correct? A friend of mine is trying it my doing .85ΔHf,methane + .10ΔHf,ethane + .05ΔHf,N2 - ΔHf, CO2 - ΔHf,H2O(g)
     
    Last edited: Feb 18, 2014
  5. Feb 18, 2014 #4
    The N2 does not react.

    I don't follow what your friend did, but in part (a) you should have taken 0.85 times the heat of reaction 1 plus 0.10 times the heat of reaction 2. There is no change in the enthalpy of the N2 at 25C.

    In part (b), I really like the way you did it. That's what I would have done.

    Chet
     
  6. Feb 18, 2014 #5

    Maylis

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    Gold Member

    Yes, I did that. Once you get the answer to the heat of combustion to part (a), you just add up the sensible heat changes. However, I actually changed my basis from 1 mol in part (a) to 100 moles in part (b), so I went back, changed the basis to 1 mol, and multiplied all the values I got for the sensible heats of the species by the number of moles coming out of the reactor.

    It gives the right answer, but I was a little shaky why I had to do that. It seems like it has to do with the wording based on the fact that its per mole of fuel.
     
  7. Feb 19, 2014 #6
    Yes. That's correct.

    Chet
     
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