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Help with college question

  • 15-04-2003 1:38pm
    #1
    stereo_steve
    Registered Users, Registered Users 2 Posts: 1,227 ✭✭✭


    I've got my exams coming up soon and i'm going over past questions. The questions are pretty much the same every year. So I'm trying to get the answers to last years paper. Everything was going well until I came to a question. Can anyone help me with it? Any help would be much appreciated!

    b) Determine the ideal or stoichrometric air/fuel ratio for octane (C8H18).

    Would that just be 4.5 O2 to make water with the hydrogen and 4 O2 to make CO? Making a total of 9.5 O2's?

    But where does the ratio come in ??


Comments

  • #2
    seamus
    Registered Users, Registered Users 2 Posts: 68,317 ✭✭✭✭seamus


    I don't know much about this kind of thing, but perhaps they're looking for the compression ratio? 9.5:1 / 11.5:1 etc

    :)


  • #3
    Ciaran
    Registered Users, Registered Users 2 Posts: 939 ✭✭✭Ciaran


    Originally posted by stereo_steve
    But where does the ratio come in ??
    I think (and Christ I should remember after doing this only last year) that the ratio is the mass of air (inc. N2) over the mass of fuel.


  • #4
    stereo_steve
    Registered Users, Registered Users 2 Posts: 1,227 ✭✭✭stereo_steve


    Thanks for all the help!

    So going by what you said ciaran,

    9.5 oxygen molecules @ ~ 20% of air means ~ 80% N2

    Therefore ~ 38 molecules of N2

    Relative molecular mass of the amount of air above is then
    (O2(16 * 9.5) + N2(14 * 38) ) / Relative molecular mass of Fuel


    => 684 / 106 = ratio = 6.45 : 1



    Can someone confirm if this is right or not, cos its worth 10% which is HUGH cos I may very well fail this exam!


  • #5
    mirv
    Registered Users, Registered Users 2 Posts: 848 ✭✭✭mirv


    How's your 3650 steve? Mine's great, except for the video camera app being slightly jerkier than when it's installed on a 7650, you noticed that?

    (on topic btw, no really!)


  • #6
    SOL
    Registered Users, Registered Users 2 Posts: 1,155 ✭✭✭SOL


    C8H18 + 12.5O2 ==> 8 CO2 + 9 H2O

    so 12.5 moles of O2 for every mole of C8H18,
    12.5*32(molecular weight of O2) grams of O2 for every
    (8*12+18*1)(Mol weight C8H18)grams of Octane

    400grams O2
    114grams Oct
    so thats your Ratio then make the allowance that air is 20% O2
    (but you have to know that it is 20% by weight not no. of molecules.)

    (Stereo i think you have failed to realise the O2 has a molevular mas of 32 as it is diatomic and nitrogen 28)

    I also may be wrong though but I have neither the time nor the inclination to check so lambaste away :)

    Also just out of interest, what course / year /college is this question from?


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