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Material Balance
GATE 2011
Hard
Methane combustion: CH4+2O2→CO2+2H2O and 2CH4+3O2→2CO+4H2O. 20% excess air is supplied. The conversion of methane is 80% and the molar ratio of CO to CO2 in the flue gas is 1:3. Assume air is 80mol% N2, rest O2. The O2 consumed as a PERCENTAGE of O2 entering the combustor is
- A20
- 62.5
- C80
- D83.3
Explanation
Basis 100 mol CH4 fed: 80 mol reacted. CO:CO2=1:3 with CO+CO2=80 (carbon balance) gives CO=20, CO2=60. O2 consumed=60×2(for CO2 rxn)+20×1.5(for CO rxn)=120+30=150 mol. Theoretical O2 (for complete combustion of all 100 mol CH4)=200 mol; with 20% excess, O2 supplied=240 mol. O2 consumed %=150/240×100=62.5%.
Source: GATE CH 2011 (2011)
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