a) The flue gas volume formed by combustion of 2,5 tons peat with a moisture con
ID: 1294452 • Letter: A
Question
a) The flue gas volume formed by combustion of 2,5 tons peat with a moisture content of 40% and excess air 30%. The flue gas temperature is 1600C and pressure 100kPa, abs. Assume ideal gas, dry air and complete combustion.
b) The sulphur dioxide concentration in the flue gas (ppm)
PS - In total theres 3000 points for this question. If you do it, please also submit your answer on another link (its just that theres max. of 1500 points to one question).
http://www.chegg.com/homework-help/questions-and-answers/calculate-flue-gas-volume-formed-combustion-2-5-tons-peat-moisture-content-40-excess-air-3-q6255965
LFO (light fuel oil) Peat Wood %-m %-m . C 85,7 56.0 51.2 H2 13.7 5.5 6.0 O2 0.5 33.3 41.5 S 0.1 0.2 0 LHV 42.7 (MJ/kg Calculate: a) The flue gas volume formed by combustion of 2,5 tons peat with a moisture content of 40% and excess air 30%. The flue gas temperature is 1600C and pressure 100kPa, abs. Assume ideal gas, dry air and complete combustion. b) The sulphur dioxide concentration in the flue gas (ppm) PS - In total theres 3000 points for this question. If you do it, please also submit your answer on another link (its just that theres max. of 1500 points to one question).Explanation / Answer
a)Mass of peat in 2.5 ton sample=0.60*2.5=1.5 ton
Mass of C=0.56*1.5=0.84 ton=840*1000/12 moles=70000 moles
Mass of H2=0.055*1.5=0.0825 ton=82.5*1000/2=41250 moles
Mass of O2=0.333*1.5=0.4995 ton=499.5*1000/32=15609.4 moles
Mass of S=0.002*1.5=0.003 ton=3*1000/32=93.75 moles
Moles of O2 required for complete combustion of this sample=Moles for C+Moles for H2+Moles for S- Moles of O2 in sample=70000*1+41250*0.5+93.75-15609.4=75109.4 moles
Moles of O2 actually used (assuming 30% excess)=1.3*75109.4=97642.2 moles
Moles of gases formed=Moles of CO2+Moles of H2O+moles of SO2 +moles of O2 left=70000+41250+93.75+0.3*97642.2=140636.4 moles
Also, the moisture content of peat converts to water vapour.
Moles of water vapour=1000*1000/18=55555.6
total moles=140636.4+55555.6=196192 moles
V=nRT/P=196192*8.314*(273+160)/10^5=7062.8m^3
b)Conc. of SO2=93.75*10^6/196192=477.8 ppm
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