๐ฅ Combustion Products Analysis
Calculate wet/dry molar compositions of flue gas products and theoretical air for various hydrocarbon fuel mixtures and excess air values.
Tools
๐ Configuration
Oโ_stoich = C/12 + H/4 โ O/32 + S/32
T_ad = T_air + Hc / ฮฃ(nแตขยทCpแตข)
๐ Results
Configure inputs and click Compute.
๐ Methodology
Stoichiometry
Complete combustion stoichiometry computes the theoretical Oโ demand from the fuel's ultimate analysis (C, H, O, S mass fractions). Excess air ensures complete combustion and determines the Oโ content in flue gas.
Adiabatic Flame Temperature
T_ad = T_air + Hc / ฮฃ(nแตขยทCpแตข), where Hc is the lower heating value and nแตข are the product moles. This is a simplified first-law estimate assuming no dissociation or heat loss.
Air-Fuel Ratio
AFR = actual_air ร MW_air / fuel_mass. Higher excess air lowers flame temperature but ensures complete combustion. Typical values: 10-20% excess for gas, 20-50% for solid fuels.
๐ Calculation Methodology: Flue Gas Combustion Products & Stoichiometric Air
Mathematical Model & Theory
Combustion stoichiometry calculates the theoretical oxygen and atmospheric air required to oxidize hydrocarbon fuels $C_x H_y O_z$, predicting wet and dry flue gas composition:
Assumptions
- Complete combustion without unburned hydrocarbons or carbon monoxide.
- Standard dry air composition: $21\%\ O_2, 79\%\ N_2$.
Academic References
- Turns, S. R.: An Introduction to Combustion, McGraw-Hill.
- ASME PTC 4: Fired Steam Generators Performance Test Code.
Worked Engineering Example
Propane ($C_3H_8$) burns with $20\%$ excess air ($\lambda = 1.20$). Calculate stoichiometric air-to-fuel ratio $(AF)_{stoich}$ and total moles of flue gas per mole fuel.
Step-by-step Solution:
1. $n_{th} = 3 + 8/4 = 5.0\text{ moles } O_2$.
2. $(AF)_{stoich} = \frac{5.0 \times 32 + 5.0 \times 3.76 \times 28.02}{44.09} = \frac{160 + 526.8}{44.09} = 15.58\text{ kg air/kg fuel}$.
3. Flue gas per mole $C_3H_8$: $3 CO_2 + 4 H_2O + (1.2-1)(5) O_2 + 1.2(5)(3.76) N_2 = 3 + 4 + 1.0 + 22.56 = 30.56\text{ moles}$.
Final Result:
Stoichiometric air/fuel ratio is $\mathbf{15.58\text{ kg/kg}}$ producing $\mathbf{30.56\text{ mol flue gas/mol fuel}}$.