Haariskhan Posted January 10, 2016 Share Posted January 10, 2016 I'm really stuck on this question: A fuel gas consists of methane and ethane in unknown proportions is burned in a furnace with air. Combustion is known to be complete and an analysis of the flue gas on a dry basis gives 85.99% N2 and 10.88% CO2. Determine a) composition of the fuel gas, b) the percentage excess air used. My basis I've chosen(Don't know if this is an appropriate basis) is 1 kmol of the fuel gas. I've let x=kmol CH4 and 1-x=kmol C2H6. From there I wrote equations for CO2 formed and theoretical O2 needed, but now I am extremely stuck. Can someone give me a method of approaching these sorts of questions? I don't even know if what I have done above is right. Link to comment Share on other sites More sharing options...
pavelcherepan Posted January 11, 2016 Share Posted January 11, 2016 (edited) I would try to solve it differently. When you burn 1 mol of CH4 completely you get 1 mol of co2 and 2 mol of co2 when you burn butane. Normal co2 concentration is .4% then you get the first equation you need to solve: x + 2y = (10.88-0.4) Then you're also down on oxygen. You use 2 mol of o2 to completely burn methane and 3.5 mol to completely burn 1 mol of butane. Normal o2 concentration is 22% then: 2x + 3.5y = (22-(100-85.99-10.88)) Solving thaw two equations you get molar concentrations of methane and buthane in original gas mix. I get 1.22 mol of butane and 8.04 mol of methane. Actually, this is all wrong. Nitrogen concentration is 86% meaning that all ratios have changed. Edited January 11, 2016 by pavelcherepan Link to comment Share on other sites More sharing options...
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