Combustion Air & Flue Gas
Finds the air a fuel needs to burn completely, the mass of flue gas it makes, and the CO2, O2 and N2 in the dry flue gas, at a chosen excess air. The fuel is given by its ultimate analysis, the mass fractions of its elements. Boiler and burner engineers use it to size fans and ducts and to check analyzer readings.
Equations
O2,st = 2.664·C + 7.937·H + 0.998·S − O
Ast = O2,st/0.232 A = Ast·(1 + EA/100)
C, H, S, O and N are the mass fractions of carbon, hydrogen, sulfur, oxygen and nitrogen in the fuel. O2,st is the oxygen each kilogram of fuel needs, in kg/kg. The factors come from the reactions C + O2 → CO2, 2H2 + O2 → 2H2O and S + O2 → SO2 [1], [2]. Air is 23.2 % oxygen by mass, so Ast is the stoichiometric air–fuel ratio. EA is the excess air in percent, and A the actual air–fuel ratio.
The flue gas holds 3.664·C of CO2, 8.937·H of water, 1.998·S of SO2, 0.768·A + N of nitrogen and 0.232·(A − Ast) of unused oxygen, per kilogram of fuel. The dry analysis leaves out the water, in mole (volume) percent.
Inputs
- Choosing a pure fuel fills in its analysis from its formula [5]. The page then also gives its heating values at 25 °C, from enthalpies of formation [4].
- For coal, oil or biomass, enter the ultimate analysis from the fuel’s lab report. Moisture and ash take no part. So the fractions may add up to less than 1, but never more.
Limits and common mistakes
- Combustion is taken as complete: no CO, soot or unburned fuel. Real burners need some excess air to get close to that [3].
- Analyzers read the dry gas, so compare them with the dry results.
- Atmospheric argon is counted as nitrogen.
- Don’t enter a proximate analysis (fixed carbon, volatiles, ash). The page needs the elements.
Worked example
A boiler burns No. 2 fuel oil with 87 % carbon, 12.5 % hydrogen and 0.5 % sulfur by mass, at 15 % excess air. How much air does each kilogram of oil need, and what should the flue-gas analyzer read?
- Keep Fuel at Custom ultimate analysis.
- Enter Carbon mass fraction 0.87, Hydrogen mass fraction 0.125, Sulphur mass fraction 0.005, and 0 for oxygen and nitrogen.
- Enter Excess Air 15 % and press Solve. The Solution shows:
Stoichiometric O₂ demand: 3.3148 kg/kg fuel
Stoichiometric air–fuel ratio: 14.2879 kg/kg
Actual air–fuel ratio: 16.4311 kg/kg
Flue gas (wet) mass: 17.4311 kg/kg fuel
CO₂ in dry flue gas: 13.4468 vol %
O₂ in dry flue gas: 2.88466 vol %
N₂ in dry flue gas: 83.6395 vol %
Check by hand: the oil needs 0.87/12.011 + 0.125/4.032 + 0.005/32.06 = 0.10358 kmol of O2 per kg, which is 3.3147 kg. The flue gas is the air plus the fuel, 16.4311 + 1 = 17.4311 kg. With air as 21 % O2 by volume, CO2 comes out at 13.5 % and O2 at 2.9 %, typical of an oil-fired boiler.