1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Feed Ratio and Mole fraction outut

  1. Nov 27, 2014 #1
    1. The problem statement, all variables and given/known data
    The chemical analysis of a fuel is determined to be: 84.60 wt% C, 11.12 wt% H, 1.95 wt % S and the balance, inerts. The fuel is consumed in a steam generating furnace with 33% excess theoretical air. The air is preheated to 190degreesC and enters the furnace at a gauge pressure of 225 mm Hg. The sulfur and hydrogen are oxidized to SO2 and H2O ; 95% of the C forms CO2 and the balance goes to CO. (Assume ideal gas)
    Calculate the feed ratio of (m3 air)/ (kg oil) and the output on a mole fraction basis, dry and wet.


    2. The attempt at a solution
    Just going to give this a try as to how I should attempt this.
    Have the fuel - 84.60% c
    - 11.12% H
    - 1.95% S
    - 2.33% Inerts
    33% excess air, thus we have 77% air consumed which is preheated to 190 degrees C at 255 mm HG
    So we know the S + H -> SO2 and H2O while 95% C forms CO2 and 5% C forms CO

    Could we say that we have a theoretical 100 moles which can then be divided down to 84.60 moles C and so forth.
    We could then assume that we have 100 g of material and knowing that we have 77% of the air consumed work towards finding the ratio of this 77% to the 100g? I am slightly confused...
     
  2. jcsd
  3. Nov 27, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Is 84.60 wt% C the same as saying that for 100 moles of fuel 84.60 moles are C?

    The feed ratio should be per kg fuel. Why not assume you have 1 kg of fuel? How much oxygen would that correspond to? And how much air in theory?

    And you should probably start by writing down all the chemical reactions down to figure out the stoichiometry.
     
  4. Nov 27, 2014 #3
    Wait I just realized now that you pointed that out.... we are dealing with wt% so it is the mass percent and would not be on the mole ratio right?
    So if we assumed 1kg of fuel we know that 77% of the air is consumed in theory.
    We don't know what the starting mixture is of the fuel we just know that it contains C, S, H, and inerts.
    So to work out chemical equations how would that work?
    Would we say S + H2 + 2O2 -> SO2 + 2H2O and 3C + 2O2 -> CO2 + 2CO
    Could we look at it in terms of input and output from here knowing the %C formed.
    So if we assume and input of 1kg of the fuel
    Our input would then be
    846g of C
    111.2g of H
    19.5g of S
    23.3g of inerts
    Our output would then be
    SO2
    H2O
    CO2
    CO
    Of these we know that 95% of the C forms CO2 and 5% forms CO. Could we then say that 803.7g of CO2 is formed and 42.3g of CO is produced.
    But then where would I go from here. Calculate moles and determine the amount of O2 originally reacted with the carbon?
     
  5. Nov 28, 2014 #4

    DrClaude

    User Avatar

    Staff: Mentor

    Right. There are many ways of attacking this kind of problem, and I think that I would start by converting these mass ratios into mole ratios. [Edit: or rather moles per kg, because of the presence of inerts)]

    I was a bit sloppy in my wording. I meant something like what you did there. You will need to know for each mole of each atom how many moles of O2 you need.

    That's another way to proceed, instead of starting with mole ratios. But be careful: does having 846g of C and forming 95% CO2 mean that you form 803.7g of CO2?
     
  6. Nov 28, 2014 #5
    If we start with mole ratios I would then have:
    846 g of C / 12.01 g/mol = 70.44 moles C
    111.2g of H / 1.00 g/mol = 111.2 moles H
    19.5g of S / 32.07 g/mol = 0.61 moles S
    23.3g of inerts -> don't know moles because we don't know what the inerts contain.

    the 846 g of C or 70.44 moles would mean that within the CO2 we have 803.7 g of C, but we don't know how much O2.

    So if we look at the equations then
    S + H2 + 2O2 -> SO2 + 2H2O
    3C + 2O2 -> CO2 + 2CO

    Could we say that we need 4O2 molecules to react in order to get the products described. I am not sure where to proceed from here though. We know the S and H in the first equation so we could determine using a mole ratio the amount of moles of O2 on a 1:2 ratio. Same with the 2nd equation on a 3:2 ratio from C. Then we could determine the total moles of O2 reacted and use that to find the volume of air? Could we then use PV=nRT since we know the temp is 190degreesC and enters the furnace at a gauge pressure of 225 mm Hg. Now that we have moles we solve for volume? And since we initally started with a basis of 1kg of fuel, that would represent the amount of O2 in 1 kg oil?
    Then the outputs we could again use ratios to determine the amount produced from the inputs?
     
  7. Nov 28, 2014 #6

    DrClaude

    User Avatar

    Staff: Mentor

    Good.

    I don't understand what you are trying to do here. I suggest you keep everything in moles for now.

    Separate all the "reactions":
    S + O2 → SO2
    2H + 1/2 O2 → H2O
    C + O2 →CO2
    C + O →CO

    Can you figure out how many moles of O2 are needed per kg fuel?

    Did you mean "that would represent the amount of O2 needed for 1 kg oil"? Otherwise, that sounds good! And don't forget that you are given the gauge pressure and that there is excess air.
     
  8. Nov 28, 2014 #7
    So if we have:
    S + O2 → SO2 S=0.61 moles thus 1:1 ratio so 0.61 moles of O2
    2H + 1/2 O2 → H2O H=111.2 moles thus 2:1/2 ratio so 27.8 moles O2
    C + O2 →CO2 here we know 95% of the C went to CO2 and 1:1 ratio so 66.92 moles O2
    C + O →CO here 5% of C went to CO so 1:1 ratio so 3.52 moles O2
    So total we know there is 0.61 + 27.8 + 66.92 + 3.52 =98.85 total moles of O2 per kg of fuel

    PV=nRT
    V=nRT/P
    V=(98.85moles)*(0.0821 liter·atm/mol·K)*(463K)/ (0.296 atm)
    V=12694.3 L
    V=12.69m^3 O2 per kg of oil

    Does that seem right?
     
  9. Dec 2, 2014 #8

    DrClaude

    User Avatar

    Staff: Mentor

    You made an error here.

    What does "gauge pressure" mean?
     
  10. Dec 2, 2014 #9
    You forgot to account for the nitrogen in the air, and you forgot to multiply by 4/3 to account for the excess air.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Feed Ratio and Mole fraction outut
  1. Vapor Fraction (Replies: 1)

  2. Gear Ratio (Replies: 1)

Loading...