Training Combustion Engineering — Stoichiometry, Thermochemistry & Emissions Stoichiometry & Flue Gas Analysis
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Stoichiometry & Flue Gas Analysis

60 min Combustion Engineering — Stoichiometry, Thermochemistry & Emissions

Stoichiometry & Flue Gas Analysis

Combustion stoichiometry is the quantitative accounting of mass and atomic species across a chemical reaction. Perfect stoichiometric combustion (excess air = 0%, $\phi = 1$) maximises flame temperature; real systems operate lean ($\phi < 1$, excess air $> 0\%$) to ensure complete combustion and to reduce NOx via dilution. Flue gas analysers (Orsat, NDIR, paramagnetic O₂) measure the dry-basis mole fractions of CO₂, O₂, and CO to infer $\phi$ via the Ostwald formula.

1. Fuel-Air Ratio and Equivalence Ratio

For a general hydrocarbon $C_x H_y$ burning in air (21% O₂, 79% N₂ by volume):

$$C_x H_y + a_{st}(O_2 + 3.76 N_2) \to x CO_2 + \frac{y}{2} H_2O + 3.76 a_{st} N_2$$

where $a_{st} = x + y/4$ is the stoichiometric O₂ coefficient. The stoichiometric air-fuel ratio by mass:

$$AFR_{st} = \frac{a_{st} \times M_{air}}{M_{fuel}} = \frac{a_{st} \times 28.85}{12x + y}$$

Equivalence ratio: $\phi = AFR_{st}/AFR_{actual}$ (or $\phi = (F/A)_{actual}/(F/A)_{st}$). Excess air: $EA = (\lambda - 1) \times 100\%$ where $\lambda = 1/\phi$. Rich flame: $\phi > 1$, $\lambda < 1$; lean flame: $\phi < 1$, $\lambda > 1$.

2. Wet and Dry Flue Gas Composition

For methane ($CH_4$) burning at excess air $EA$:

$$CH_4 + (1+EA/100)\cdot 2(O_2 + 3.76N_2) \to CO_2 + 2H_2O + \frac{EA}{100}\cdot 2 O_2 + 2(1+EA/100)\cdot 3.76 N_2$$

Total dry moles $= 1 + EA/50 + 7.52(1+EA/100)$. Dry-basis CO₂% $= \frac{1}{\text{total dry}} \times 100$. As $EA$ increases: CO₂% decreases, O₂% increases. The Orsat analyser measures dry-basis CO₂, O₂, CO by absorption; $\phi$ is back-calculated from these measurements.

3. Chemical Equilibrium at High Temperature

Above ~1500 K, CO₂ and H₂O partially dissociate: $CO_2 \leftrightarrow CO + \frac{1}{2}O_2$ with $K_p(T) = \exp(-\Delta G°/RT)$. This limits actual flame temperatures below the ideal adiabatic value and produces CO even at lean conditions. Equilibrium is computed by minimising Gibbs free energy of the product mixture subject to atomic balance constraints — typically solved numerically (NASA CEA code, Cantera).

Worked Example — Methane Stoichiometry

$CH_4 + 2O_2 + 7.52N_2 \to CO_2 + 2H_2O + 7.52N_2$. $M_{CH_4} = 16$. $AFR_{st} = 2 \times 32 + 7.52 \times 28)/(16) = (64 + 210.6)/16 = 17.2$ kg air / kg fuel. At $EA = 20\%$: actual AFR $= 1.2 \times 17.2 = 20.6$. $\phi = 17.2/20.6 = 0.835$. $\lambda = 1.20$. ✓

Combustion Stoichiometry Calculator
AFR_stoich =17.2kg/kg
AFR_actual =20.6kg/kg
λ (air excess ratio) =1.20
φ (equivalence ratio) =0.833
Adiabatic flame T ≈~2050K

Practice Problems

1. Propane ($C_3H_8$) burns at $EA = 15\%$. Compute $a_{st}$, $AFR_{st}$, $AFR_{actual}$, $\phi$, and $\lambda$. Write the complete balanced equation for the actual mixture.
2. A dry-basis flue gas analyser reads 10.2% CO₂, 4.5% O₂, 0.0% CO for a natural gas burner (assume pure CH₄). Back-calculate the excess air percentage and equivalence ratio using an atomic balance.
3. At 2000 K, the equilibrium constant for $CO_2 \leftrightarrow CO + \frac{1}{2}O_2$ is $K_p = 0.037$ atm^{1/2}$. For a total pressure of 1 atm with an initial mixture of 1 mol CO₂, find the equilibrium CO and O₂ mole fractions.
Adiabatic Flame Temperature & Zeldovich NOx Kinetics