2 replies to this topic

[zee]

I have been stuck with this question for a day now.. and can't figure out how to solve it .. I tried to start with 1 mol/day of waste stream and 1.20 per day of air.... tried doing a mass balance on elements but have gotten no where.. any help would be appreciated at this point

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[S.AHMAD]

1. Please show your calculations - if possible in EXCEL for review by experienced engineers.
2. Otherwise we do not know what went wrong.

[kkala]

I agree with S.AHMAD, besides only through hard individual labor (even with errors) can knowledge be gained. Following remark could be helpful to this direction, in case that confusion is due to CO2/CO mass ratio = 4.
1. Consider 100 kgmol of the mixture and count kgrat of C: 2x3+14x4+36x1+20x2+8x1= 146 kgat C.
2. These are converted into 146 kgmol of CO2 + CO, CO2/CO mass ratio=4. So 104.82 kgmol CO2 and 41.18 kgmol CO will be contained in flue gases (resulting from 100 kgmol of mixture).
3. It is not a matter what compound is burnt to CO2 and what to CO, partially or totally, this does not affect stoichiometry. Consider all compounds (of the 100 kgmol mixture) burnt into 146 kgmol CO at first stage. Then 104.82 kgmol CO is further burnt into CO2 at second stage.
4. As problem points out, excess air will be based on complete burning to CO2.
This may be a way to proceed, present calculations in case of difficulty. Other members may offer useful hints.