14.6 Combustion and Combustion Products

14.6 Combustion and Combustion Products

NCEES lists "Combustion and combustion products (e.g., CO, CO₂, NOₓ, ash, particulates)" under Thermodynamics and Heat Transfer. Expect one balancing or air-fuel-ratio calculation and one conceptual item on which pollutant comes from where.

The Composition of Air

By volume (equivalently, by moles), dry air is 21% O₂ and 79% N₂. So per mole of oxygen:

$\frac{79}{21} = 3.76\ \text{moles of N}_2$

$\text{Air} = \text{O}_2 + 3.76\,\text{N}_2 \qquad\text{(the working unit for every combustion equation)}$

Molar masses: air 28.97, O₂ 32.00, N₂ 28.01, CO₂ 44.01, H₂O 18.02, CO 28.01, SO₂ 64.06 kg/kmol.

Nitrogen is inert but not irrelevant. It does not react (except in trace NOₓ formation), but it absorbs a large fraction of the released heat and leaves up the stack, which is why flue-gas heat loss is dominated by nitrogen and why oxygen-enriched combustion runs so much hotter.

Stoichiometric (Theoretical) Combustion

For a hydrocarbon $\text{C}_x\text{H}_y$ burning completely in theoretical air:

$\text{C}_x\text{H}_y + a(\text{O}_2 + 3.76\,\text{N}_2) \rightarrow x\,\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} + 3.76a\,\text{N}_2$

$a = x + \frac{y}{4}$

Methane ($x=1$, $y=4$, so $a = 2$):

$\text{CH}_4 + 2(\text{O}_2 + 3.76\,\text{N}_2) \rightarrow \text{CO}_2 + 2\,\text{H}_2\text{O} + 7.52\,\text{N}_2$

Propane ($x=3$, $y=8$, so $a = 5$):

$\text{C}_3\text{H}_8 + 5(\text{O}_2 + 3.76\,\text{N}_2) \rightarrow 3\,\text{CO}_2 + 4\,\text{H}_2\text{O} + 18.8\,\text{N}_2$

Air-Fuel Ratio

$AFR_{\text{mass}} = \frac{m_{\text{air}}}{m_{\text{fuel}}} = \frac{a(4.76)(28.97)}{M_{\text{fuel}}}$

The equivalence ratio $\phi$ compares actual fuel to stoichiometric fuel:

$\phi = \frac{AFR_{\text{stoich}}}{AFR_{\text{actual}}}$

$\phi > 1$ is fuel-rich (produces CO and soot); $\phi < 1$ is fuel-lean (excess oxygen in the products); $\phi = 1$ is stoichiometric.

Excess Air

$\%\text{ excess air} = \frac{a_{\text{actual}} - a_{\text{theoretical}}}{a_{\text{theoretical}}}\times100\%$

With $e$ as the fractional excess air, the reaction becomes:

$\text{C}_x\text{H}_y + (1+e)a(\text{O}_2+3.76\,\text{N}_2) \rightarrow x\,\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} + e\,a\,\text{O}_2 + 3.76(1+e)a\,\text{N}_2$

Excess air appears as O₂ in the products — which is precisely how it is measured in practice. A flue-gas oxygen analyzer is the standard combustion-tuning instrument, and the relationship between flue O₂ and excess air is the reason:

The optimization trade-off. Too little excess air produces CO and unburned fuel — a safety hazard and an efficiency loss. Too much excess air carries heat up the stack as hot nitrogen and oxygen, wasting fuel. Typical practice is 10–20% excess air for gas and 20–30% for solid fuels. Efficiency is maximized just above the CO breakthrough point, which is why the tuning target is stated as a flue O₂ percentage.

Worked Example: Excess Air Combustion of Propane

Propane burns with 25% excess air. Write the balanced equation and find the air-fuel ratio and the dry volumetric CO₂ percentage.

Theoretical $a = 5$; actual $= 1.25(5) = 6.25$.

$\text{C}_3\text{H}_8 + 6.25(\text{O}_2+3.76\,\text{N}_2) \rightarrow 3\,\text{CO}_2 + 4\,\text{H}_2\text{O} + 1.25\,\text{O}_2 + 23.5\,\text{N}_2$

Check oxygen: supplied $= 6.25(2) = 12.5$ atoms; used $= 3(2) + 4(1) = 10$; excess $= 2.5$ atoms $= 1.25$ mol O₂ ✓

Air-fuel ratio by mass:

$m_{\text{air}} = 6.25(4.76)(28.97) = 861.6\ \text{kg}, \qquad M_{\text{C}_3\text{H}_8} = 3(12.01)+8(1.008) = 44.09$

$AFR = \frac{861.6}{44.09} = \boxed{19.5\ \text{kg air/kg fuel}}$

Confirm against the stoichiometric value: $15.7 \times 1.25 = 19.6$ ✓

Dry product analysis (excluding H₂O, as flue-gas analyzers report):

$n_{\text{dry}} = 3 + 1.25 + 23.5 = 27.75\ \text{mol}$

$\%\text{CO}_2 = \frac{3}{27.75} = 10.8\%, \qquad \%\text{O}_2 = \frac{1.25}{27.75} = 4.5\%$

The 4.5% flue oxygen is exactly what an analyzer would read to confirm 25% excess air.

Dew point of the products. The wet total is $27.75 + 4 = 31.75$ mol, so the water mole fraction is $4/31.75 = 0.126$. At 101.3 kPa the water partial pressure is $0.126(101.3) = 12.8$ kPa, whose saturation temperature is about 51 °C.

Why the dew point matters: cooling flue gas below ~51 °C condenses water in the stack. For sulfur-bearing fuels the acid dew point is far higher (120–150 °C) because SO₃ forms sulfuric acid, which is why conventional boilers keep stack temperatures well above it and why condensing boilers require corrosion-resistant materials.

Combustion Products and Their Origins

The central asymmetry, and a favorite exam item. CO and NOₓ respond in opposite directions to the same lever. Raising flame temperature and excess air burns out CO but increases thermal NOₓ; lowering peak temperature suppresses NOₓ but risks CO breakthrough. This is why combustion tuning is a genuine optimization rather than a maximization, and why low-NOₓ burners must be carefully commissioned against CO limits.

Note also that thermal NOₓ nitrogen comes from the combustion air, not the fuel — so burning a nitrogen-free fuel such as pure methane or hydrogen in air still produces NOₓ. Only the flame temperature controls it.