Question

Question 4 746 Ethane (C2H6) is burned with 50% excess air. The percentage conversion of the ethane is 90% Of the ethane burned, 25% reacts to form CO and the balance to form CO2. Calculate the composition of the fuel gas and the ratio of water to dry fuel gas. (40) [40 marks]

Please explain

Answer 1

C2H6 + 3.5 02 2 CO2+ 3 H2O Rxn-1 C2H6 +2.5 02 2 C0+ 3 H2O Rxn-2 Let basis 100 mol of C2H6 Conversion=90 %=90/100=0.9 Total mole of C2H6 reacted=Mole of C2H6*Conversion=100 mol*0.9=90 mol % converted to CO=25%=25/100=0.25 Mole of C2H6 reacted in rxn-2=% converted to CO* Total mole of C2H6 reacted=90 mo1*0.25=22.5 mol Mole C2H6 reacted in rxn-1= Total mole of C2H6 reacted- Mole of C2H6 reacted in rxn-2=90 mol-22.5 mol=67.5 mol Mole of Oxygen required=3.5* Mole C2H6 reacted in rxn-1+2.5* Mole C2H6 reacted in rxn-2=3.5*67.5 mol+2.5*22.5 mol=292.5 mol Assume air contain 21% oxygen and 79% nitrogen Mole of air required=Mole of oxygen required/0.21=292.5 mol/0.21=1392.86 mol % excess of air=50 %=50/100=0.5 Mole of air supplied=1.5* Mole of air required=15*1392.86=2089.286 mol Mole of oxygen supplies=0.21* Mole of air supplied=0.21*2089.286 mol=438.75 mol Mole of Nitrogen supplies=0.79* Mole of air supplied=0.79*2089.286 mol=1650.54 mol

Fuel gas Mole of C2H6=Mole of C2H6 supplied-Total mole C2H6 reacted=100 mol-90 mol=10 mol -a Mole of Oxygen=Mole of Oxygen supplied-Mole of oxygen reacted=438.75 mol-292.5 mol=146.25 mol -b Mole of nitrogen=Mole of nitrogen supplied=1650.54 mol Mole of CO2 produced=2*Mole of C2H6 reacted in rxn-1=2*67.5 mol=135 mol-d Mole of Co produced=2*Mole of C2 H6 reacted in rxn-2=2*22.5 mol=45 mol -e Mole of water produced=3* Mole of C2H6 reacted in rxn-1+3* Mole of C2H6 reacted in rxn-2=3*67.5 mol+3*22.5 mol=270 mol -f Total mole of wet fuel gas=a +b+c+d+e+f=10 mol+146.25 mol+ 1650.54 mol+ 135 mol+45 mol+270 mol=2256.79 Total mole of dry fuel gas=a+b+c+d+e=10 mol+146.25 mol+1650.54 mol+ 135 mol+45 mol=1986.79 mol Composition on wet basis Mole fraction of C2H6= Mole of C2H6/ Total mole of wet fuel gas=10 mol/2256.79 mol=0.00443 Mole fraction of CO2= Mole of C2H6/ Total mole of wet fuel gas=135 mol/2256.79 mol=0.0598

Mole fraction of CO=Mole of CO/ Total mole of wet fuel gas=45 mol/2256.79 mol=0.02 Mole fraction of N2= Mole of N2/ Total mole of wet fuel gas=1650.54 mol/2256.79 mol=0.73 Mole fraction of 02= Mole of 02/ Total mole of wet fuel gas=146.25 mol/2256.79 mol=0.0648 Mole fraction of H2O= Mole of H20/ Total mole of wet fuel gas=270 mol/2256.79 mol=0.11964 Composition on dry basis Mole fraction of C2H6= Mole of C2H6/ Total mole of dry fuel gas=10 mol/1986.79 mol=0.005 Mole fraction of CO2= Mole of C2H6/ Total mole of dry fuel gas=135 mol/1986.79 mol=0.068 Mole fraction of CO=Mole of CO/ Total mole of dry fuel gas=45 mol/1986.79 mol=0.0226 Mole fraction of N2= Mole of N2/ Total mole of dry fuel gas=1650.54 mol/1986.79 mol=0.831 Mole fraction of 02= Mole of 02/ Total mole of dry fuel gas=146.25 mol/1986.79 mol=0.074 Molar ratio of water to dry fuel gas Mole of water/Mole of dry fuel gas=270 mol/1986.79 mol=0.136