Combustion Calculations
E-11 • The flue flue gas gas from from an indu indus strial rial furn furnace ace has the the follow followin ing g composition by volume • CO2: 11 11.7 .73 3%, CO CO: 0.2 0.2 %, SO SO2: 0.0 0.09 9%, O2: O2: 6.8 6.81 1% an and N2: 81.17 % • Calculat alculate e th the perce percent ntag age e exces excess s air em employ ployed in the the combustion if the loss of carbon in the clinker ash is 1% of the fuel used and the fuel has the following composition by weight: • C: 74 74%, H2: H2: 5 %, O2: O2: 5%, 5%, N2: N2: 1%, 1%, S: S: 1%, 1%, H2O H2O: 9%, 9%, ash : 5%
E-11 • • • • • •
Basis: 100 kg of the fuel C +O2 =>CO2 H2 +1/2 O2 =>H2O S +O2 =>SO2 Oxygen Balance: Oxygen required for complete combustion =74/12 + 5/2 X 1/2 + 1/32=7.447 kg mole Oxygen present in fuel=5/32=0.157 kg mole Net oxygen from air= 7.447-0.157=7.29 kg mole
E-11 • Carbon balance Carbon lost in clinker and ash = 1 kg Carbon burnt= 74-1= 73 kg = 73/12=6.08 kg atom Assume z kg of flue gas are formed By carbon balance (0.1173+0.002)*Z=6.08 z=50.96 kg moles So N2 in flue gas= 50.96 *0.8117=41.36 N2 from fuel= 1 kg=0.036 kg moles N2 from air = 41.36-0.036=41.324 Oxygen from air =41.324 * 21/79=10.98 kg mole Excess oxygen = 10.98-7.29= 3.69
E-11 Excess oxygen = 10.98-7.29= 3.69 kg mole Percentage excess air used =percentage excess oxygen used = excess/theoretical* 100= 3.69/7.29* 100= 50.62
E-12 • • • • • •
Octane is burnt with 10 % excess air. Calculate Air to fuel ratio by weight Air to fuel ratio by volume Weight of dry exhaust gas formed per unit weight of fuel Moles of oxygen in the exhaust gas per unit weight of fuel Moles of water vapour in exhaust gas per unit weight of fuel • Volume of exhaust gas at 1 atmosphere and 260 C per unit weight of fuel The specific gravity of octane may be taken as 0.7
E-13 • A producer gas with the composition by volume CO: 27.3 %, CO2:5.4%, O2: 0.6 %, N2 : 66.7% is burnt with 20 % excess air. If the combustion is 98% complete, calculate the composition by volume of the flue gases • Solution: • Basis: 100 kg mole of producer gas burnt Oxygen balance: O2 required for CO combustion= 27.3*0.5=13.65 kg mole O2 present in fuel = 0.6 kg mole Net O2 required =13.65 – 0.6 = 13.05 kg mole O2 supplied by 20 % excess air = 13.05*1.20= 15.66 kg mole O2 used for 98% combustion of CO=27.3*0.5*0.98=13.38 kg mole O2 excess = 15.66 -13.38 =2.28 kg mole
E-13 Nitrogen balance: N2 from air =15.66 * 79/21= 58.91 kg mole N2 from producer gas = 66.7 kg mole Total N2 in flue gas =66.70 +58.91=125.61 kg mole CO2 balance: CO2 from producer gas 5.4 kg mole CO2 from combustion of CO = 27.3 *0.98= 26.75 kg mole Total CO2 in flue gas =5.4 +26.75 =32.15 kg mole CO balance: CO burnt = 26.75 kg mole CO left =27.3 – 26.75 =0.55 kg mole
E-13 Flue gas analysis
Constituent
Kg mole
Mole %
CO2
32.15
20.02
CO
0.55
0.34
N2
125.61
78.22
O2
2.28
1.42
160.59
100
Total
E-14 • A furnace is fired with a natural gas that consists entirely of hydrocarbons( no inert or sulphur). The analysis of the flue gas is 9.5 % CO2, 2.0% O2 and 1.8% CO • What is the molar ratio of net hydrogen to carbon in the fuel? • What percentage of excess air is being used?
E-14 • Solution: Basis: 100 kg of dry flue gas N2 =100 –(9.5+1.8+2.0) =86.7 Oxygen balance: O2 supplied by air = 86.7 *21/79 =23.05 kg mole O2 in dry flue gas= 9.5+1.8/2 +2 =12.4 kg mole O2 unaccounted ( reacted with H2)= 23.05-12.4 =10.65 kg mole Moles of H2 reacted = 10.65 * 2 =21.3 Amount of carbon = 9.5 + 1.8 = 11.3 kg atoms Moles of H2/Moles of C= 21.3/11.3 =1.18 Moles of O2 required for complete combustion=moles for H2 +moles for C =10.65 + 11.3=21.95 Amount of excess O2 = 23.05 -21.95=1.1 kg mole % excess air = 1.1/21.95*100 = 5
E-15 • The exhaust gas from a hydrocarbon fuel oil fired furnace show 10.2% CO2, 7.9% O2 and 81.9% N2. Calculate • %excess air used • Kg of dry air supplied per kg of oil burnt
E-16 • Determine the flue gas analysis and air/fuel ratio by weight when a fuel oil with 84.9% carbon, 11.4% hydrogen, 3.2% sulphur, 0.4% oxygen and 0.1% ash by weight is burnt with 20 % excess air. Assume complete combustion.
E-17 • A boiler is fired using 200 kg/hr of a pure saturated hydrocarbon gas CnHm at atmospheric pressure and 20 C. The dry analysis of the flue gas which leaves the boiler at atmospheric pressure and 300 C is CO2: 12%, O2: 3% and N2 :85%. Estimate the formula of the fuel and total volumetric flow rate of the gas • Solution: • Basis: 100 kg mole of dry flue gas • Oxygen Balance N2 in flue gas = 85 kg mole O2 supplied by air = 85 *21/79=22.59 kg mole
E-17 • O2 reported in • flue gas =O2 in CO2 +O2 as O2 =12 + 3.0 =15.0 kg mole O2 unaccounted (reacted with H2) =22.59 – 15.0 = 7.59 kg mole H2 reacted = 7.59 * 2=15.18 kg mole = 30.36 kg atoms Amount of carbon = 12 kg atoms Amount of hydrogen=30.36 kg atoms Ratio=Atoms of H/Atoms of C=30.36/12=2.53 Paraffin formula=CnH2n+2 (2n+2)/n=2.53 So n =3.77 ~4 Hence the fuel is C4H10 Volumetric flow rate =? Amount of fuel = 200/58 3.45 kg mole Assuming ideal gas law, volume at 20 C and 1 atmosphere is : V=nRT/P=3.45*0.08206*293/1 = 82.95 m3/hr
E-18 • A furnace is fired with a fuel having the volumetric composition H2: 52%, CH4: 30 %, CO: 8%, C3H6: 3.6%, CO2: 2%, O2: 0.4% and rest N2. • Using a certain quantity of air in excess over stoichiometric , complete combustion of the gas is archived giving a dry waste gas of 5 m3 per m3 of fuel burned. Estimate • Composition by volume of dry waste gas formed • Per cent excess air used • Weight of water formed per m3 of gas burned
E-18 Solution: Basis: 100 kg mole of fuel gas CH4 + 2 O2 => CO2 + 2H2O CO + 1/2O2 => CO2 C3H6 + 4.5 O2 => 3 CO2 + 3H2O H2 + ½ O2 =>H2O Oxygen balance O2 required for CO=8*1/2=4 kg mole O2 required for CH4=30*2=60 kg mole O2 required for C3H6=3.6*4.5=16.2 kg mole O2 required for H2 = 52*1/2=26 kg mole Total O2 required =106.2 kg mole O2 in fuel =0.4 kg mole
E-18 Net O2 required from air =106.2 -0.4 =105.8 kg mole DRY Flue Gas formed with theoretical air: CO2 formed=8.0(from CO) + 30.0 (from CH4) + 10.8 (from C3H6) +2.0(present in fuel) =50.8 kg mole N2 from air = 105.8 * 79/21= 398 kg mole Total N2 =398 +4 (present in fuel) =402 kg mole Total amount of dry flue gas (with theoretical air) =402+50.8 =452.8 kg mole Flue gas actually produced=5 m3/m3 of fuel Total dry flue gas produced=100*5 =500 kg mole Excess air=500 – 452.8 =47.2 kg mole Theoretical air =105.8 *100/21=503.81 kg mole
E-18 Theoretical air =105.8 *100/21=503.81 kg mole a) % excess air =47.2/503.8*100=9.39 b) Composition of flue gas: Air used = 503.81 + 47.2 = 551.01 kg mole N2 from air = 551.01 *79/100 = 435.29 kg mole Total N2 =435.29 +4.0 =439.29 kg mole Excess O2 = 47.2 *21/100 =9.91 kg mole Flue gas analysis: constituent
Amount kg mole
Vol%
CO2 O2 N2
50.8 9.91 439.29
10.16 1.98 87.86
total
500
100
E-18 • (C) Amount of water formed =52.0 (from H2) +60 (from CH4) + 10.8 ( from C3H6) = 122.8 kg mole = 122.8 *18 =2210.4 kg • Amount of gas burned=100 kg mole=22.4*100 m3 at NTP • Weight of water formed/m3 gas burned=2210.4 kg /2240 m3 =0.987 kg of water/m3 of gas
E-19 • The dry flue gas from an oil fired furnace has the composition of 11.2 % CO2, 5.8% O2, 83% N2. Calculate • % excess air • Weight of combustion air per kg of oil fired • Assume fuel has 82% C, 12 % H, 3 % S and balance is impurities. • Solution: • Basis: 100 kg of oil fired • Oxygen balance: • O2 required for 82 kg of carbon=82/12=6.833 kg mole • O2 required for 12 kg of H =12/2*0.5=3 kg mole • O2 required for 3 kg of S =0.094 kg mole • total O2 required = 6.833 + 3.0 +0.094 =9.927 kg mole
E-19 • • • • • • • • • • •
Let z kg mole of dry flue gas is formed Carbon balance: 0.112* z (out) =6.833 (in) So z =61.0 kg mole Amount of N2 in flue gas =61 * 0.83 =50.83 kg mole O2 from air = 50.83 *21/79 =13.46 kg mole O2 excess =13.46 – 9.927 =3.533 kg mole % excess air =% excess O2 =3.533/9.927*100=35.6% Amount of combustion air =50.63*100/79=64.09 kg mole Molecular weight of air =28.84 Amount of air in kg =64.09*28.84=1848.36 kg Mass of air in kg/mass of fuel in kg=1848.36/100=18.48
E-30 • A fuel gas containing 97% methane and 3% N2 by volume is burned in boiler furnace with 200% excess air, 85% of methane goes to CO2, 10% to CO and 5% remains unburnt. • Calculate the composition of stack gas Solution: Basis: 100 kg mole of fuel gas CH4 +2 O2 =>CO2 +2H2O ….. (1) 2 CH4 + 3 O2 => 2 CO + 4 H2O …..(2) O2 required for complete combustion= 97*2=194 kg mole O2 supplied by 200 % excess air = 3*194 = 582 kg mole Reaction (1) is 85% complete and (2) is 10% complete
E-30 Methane converted by reaction (1) = 97 *0.85=82.45 kg mole Methane converted by reaction (2) = 97*0.1=9.7 kg mole Oxygen used in reaction (1) & 2= 82.5*2+9.7*3/2=179.45 kg mole stack gas:
CO2 formed = 82.45 kg mole CO formed =9.7 kg mole CH4 unconverted = 5 kg mole N2 from air = 582 * 79/21=2189.43 kg mole N2 from fuel = 3.0 kg mole Total N2 in flue gas =2189.43 +3.0=2192.43 kg mole H2O formed = 82.25*2(by reaction 1) + 9.7*4/2(by reaction 2) =184.3 kg mole Excess oxygen =582 – 179.45 = 402.55 kg mole
E-30
stack gas analysis
constituent
kg mole
Mole %
CO2
82.45
2.87
CO
9.70
0.34
CH4
5
0.17
O2
402.55
13.99
N2
2192.43
76.22
H2O
184.30
6.41
Total
2676.28
100
