burnng1

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 Given FG Flow 712 Nm3/hr  sreekanthtm's CAL CH4 97.49 mol% Calculation of com C2H6 1.5 mol% N2 1 mol% component quantity CO2 !.!1 mol% "mol #ince $&i& w& 1!!"mol' nee( t o convert t o 31.76"mol' t here)ore N2 !.317659 CH4 3!.96*54 mol% CO2 ! C2H6 !.4764** mol% CH4 3!.96*54 N2 !.317659 mol% C2H6 !.4764** CO2 -0.00!"" mol% C3H* ! C4H1! ! CO (assumed) ! +otl !."## ,-ttin the&e in' one et& 3!3"mol #toichiometric ir )or complete $-rnin Air for 100% burnin "mol + "in 0 o) ir to $e 2*.9 ir 0 0&& o) ir i& *753.23 " ir ce&& ir % From pict-re $elow ' !.26 " wter/" (r ir O2 in with ir "mol 2275.*4 " wter N2 in with ir "mol Convert to "mol 1*.!153 ter 0 H-mi(it in "mol !$#.$% "mol wter + otl ir in "mol + otl ir in Nm3 &esulting flue gases O2 "mol CO2 "mol H2O "mol N2 "mol + otl "mol + otl Nm3/h 8&i& !00 kmol of gi ve Notice CO2 i& netive' co-l( $e (irect import o) re&-lt& )rom n nl:er nee(in $it o) cli$rtion' +here)ore' &ome mnip-ltion i& nee(e(. ie p-t to !

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burnNGGivenFG Flow712Nm3/hrsreekanthtm's CALCthorium90's CALCCH497.49mol%Calculation of combustion air required for the given NG and given excess air.Calculation of combustion air required for the given NG and given excess air.C2H61.5mol%Basis: 100 kmol of given natural gas (NG), normal conditions (MCR=100%)Basis: 100 kmol of given natural gas (NG), normal conditions (MCR=100%)N21mol%componentquantitystoichiometric O2 (*)stoichiometric O2Resulting gases after stoichiom burningcomponentquantitystoichiometric O2 (*)stoichiometric O2Resulting gases after stoichiom burningCO2-0.01mol%kmolkmol O2 / kmol compkmolCO2, kmolH2O (g), kmolN2, kmolkmolkmol O2 / kmol compkmolCO2, kmolH2O (g), kmolN2, kmolSince basis was 100kmol, need to convert to 31.76kmol, thereforeN20.317658606200000.3176586062N20.317658606200000.3176586062CH430.9685375212mol%CO2000000CO2000000C2H60.4764879093mol%CH430.9685375212261.937075042430.968537521261.93707504240CH430.9685375212261.937075042430.968537521261.93707504240N20.3176586062mol%C2H60.47648790933.51.66770768270.95297581871.4294637280C2H60.47648790933.51.66770768270.95297581871.4294637280CO2-0.0031765861mol%C3H8050000C3H8050000Notice CO2 is negative, could be direct import of results from an analyzer needing abit of calibration, Therefore, some manipulation is needed. ie: put to 0C4H1006.50000C4H1006.50000CO (assumed)
Kostas Kalaitzoglou: Just to close balance and have total = 100 kmol.You have to do something similar, not necessarily with CO.00.50000CO (assumed)
Kostas Kalaitzoglou: Just to close balance and have total = 100 kmol.You have to do something similar, not necessarily with CO.00.50000Total31.76663.631.963.40.32Total31.76663.631.963.40.32Putting these in, one gets 303kmol "Stoichiometric air for complete burning"Air for 100% burningkmol303Stoichiometric air for complete burningAir for 100% burningkmol303Stoichiometric air for complete burningTaking MW of air to be28.9Air MWMass of air is8753.2296226423kg airExcess air%20specific value to be advised by VendorExcess air%20specific value to be advised by VendorFrom picture below,0.26kg water/kg dry airO2 in with airkmol76O2 in with airkmol762275.839701887kg waterN2 in with airkmol287N2 in with airkmol287Convert to kmol18.0153Water MWHumidity inkmol0
Kostas Kalaitzoglou: You can estimate it for more precise resultsHumidity inkmol126.5
Kostas Kalaitzoglou: You can estimate it for more precise results

126.328kmol waterTotal air inkmol363Total air inkmol490Total air inNm38147Total air inNm310982Resulting flue gasesResulting flue gasesO2 kmol12.72O2 kmol12.72CO2kmol31.9CO2kmol31.9H2O (g)kmol63.4H2O (g)kmol189.9N2 kmol287N2 kmol287Totalkmol395Totalkmol522TotalNm3/h8864TotalNm3/h11699