A furnace is charged with 100 kmols of fuel gas, the composition of which is: 50% CO, 30% H2, 5% CH4, 5% C2H6 and 10%

Question

A furnace is charged with 100 kmols of fuel gas, the composition of which is: 50% CO, 30% H2, 5% CH4, 5% C2H6 and 10% O2. Dry air is supplied 50% in excess for combustion. Assuming complete combustion. What's the orsat analysis?

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Answer: Orsat Analysis of Flue Gas %CO2 = 10.56% %N2 = 81.08% %O2 = 8.36% Step-by-step explanation On the basis of 100kmol fuel, we compute the moles of the following: nCO = 100kmol(0.50) = 50kmol nH2 = 100kmol (0.30) = 30kmol nC2H6 = 100kmol(0.05) = 5kmol nO2 = 100kmol(0.10) = 10kmol Reactions based on complete combustion 2CO + O2 →2CO2 2C2H6 + 14O2 → 4CO2 + 3H20 The moles of CO and C2H6 converted to CO2 nCO2 = 50kmol CO (2 kmol CO2/ 2kmol CO) + (5kmol C2H6) (4 kmol CO2/ 2kmol CO) nCO2 = 60 kmol Theoretical oxygen in order to attain complete combustion nTheo O2 = 50 kmol ( 1 kmol O2 / 2 kmol CO) + (30 kmol H2) (1kmol O2 / 2kmol H2) + (5 kmol C2H6 (14kmol O2 / 2 kmol C2H6) nTheo O2 =75 kmol Moles of oxygen fed nO2total= 10 kmol O2 + 75 kmol O2 (1.5) nO2total= 122.5 kmol Moles of unreacted O2 nO2 = nO2total - nTheo O2 nO2 =122.5 kmol - 75 kmol nO2 = 47.5 kmol Moles of nitrogen nN2 = 122.5 kmol O2 (79 kmol N2 / 21 kmol O2) nN2 = 460.83 kmol Total moles of flue gas nT = nN2 + nO2 +nCO2 nT = 460.83 kmol + 47.5 kmol + 60 kmol nT = 568.33 kmol %CO2 =(nCO2 /nT )(100%) %CO2 =(60 kmols / 568.33 kmol )(100%) %CO2 = 10.56% %N2 =(nN2/nT )(100%) %N2 =(460.83 kmol/ 568.33 kmol)(100%) %N2 = 81.08% %O2 = (nO2/nT )(100%) %O2 = (47.5kmol/568.33kmol)(100%) %O2 = 8.36%

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A furnace is charged with 100 kmols of fuel gas, the composition of which is: 50% CO, 30% H2, 5% CH4, 5% C2H6 and 10% O2

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