3.5 Combustion reactions

Combustion reactions describe the burning of fuels with oxygen to produce heat. They show up everywhere in chemical engineering, from power plants to industrial furnaces. Understanding how to set up material balances around combustion systems lets you calculate air requirements, predict flue gas compositions, and optimize fuel use while minimizing emissions.

Combustion Reactions for Fuels

Balanced Combustion Reactions

Combustion is a chemical reaction between a fuel and an oxidant that produces heat and light. The most common oxidant is oxygen, supplied through air.

Common fuels you'll encounter:

• Hydrocarbons: natural gas (methane), propane, butane
• Alcohols: methanol, ethanol
• Biomass: wood, agricultural waste

A balanced combustion reaction has the correct stoichiometric coefficients so that atoms are conserved on both sides. For methane:

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

One mole of methane reacts with exactly two moles of $O_2$, producing one mole of $CO_2$ and two moles of $H_2O$. Every combustion problem starts with getting this equation balanced correctly.

Complete and Incomplete Combustion

Complete combustion happens when enough oxygen is available to fully convert all the carbon to $CO_2$ and all the hydrogen to $H_2O$. For propane:

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Incomplete combustion happens when oxygen is insufficient. Carbon monoxide ($CO$), unburned hydrocarbons, and other byproducts form instead. For example, incomplete combustion of ethanol:

$2C_2H_5OH + 3O_2 \rightarrow 2CO + 4H_2O + 2CH_4$

The key distinction: complete combustion maximizes energy release and minimizes harmful emissions, while incomplete combustion wastes fuel and produces toxic $CO$.

Air Requirements for Combustion

Theoretical and Excess Air

Theoretical (stoichiometric) air is the exact minimum amount of air needed for complete combustion, based directly on the balanced equation. For methane combustion, the equation tells you that 2 moles of $O_2$ are needed per mole of $CH_4$. That's the theoretical oxygen requirement.

In practice, furnaces and boilers always supply more than the theoretical amount. This extra supply is called excess air, and it ensures the fuel actually burns completely (since real mixing is never perfect).

If you use 20% excess air for methane combustion, you supply 1.2 times the theoretical oxygen:

$2 \times 1.2 = 2.4 \text{ moles } O_2 \text{ per mole } CH_4$

Air-Fuel Ratio and Percent Excess Air

The air-fuel ratio (AFR) is the mass of air divided by the mass of fuel in a combustion process. The stoichiometric AFR corresponds to theoretical air; the actual AFR is higher when excess air is used.

Percent excess air tells you how much extra air you're supplying beyond the theoretical requirement:

$\% \text{ excess air} = \frac{\text{actual air} - \text{theoretical air}}{\text{theoretical air}} \times 100$

So if actual air is 1.5 times theoretical air, the percent excess air is 50%.

To calculate air requirements for any combustion problem:

1. Write and balance the combustion reaction
2. Determine the theoretical $O_2$ from stoichiometry
3. Convert to air (air is approximately 21% $O_2$ and 79% $N_2$ by volume)
4. Multiply by $(1 + \% \text{ excess}/100)$ to get actual air

Composition of Combustion Products

Complete and Incomplete Combustion Products

The gas mixture leaving a combustion chamber is called flue gas. Its composition depends on the fuel, the air-fuel ratio, and whether combustion is complete or incomplete.

• Complete combustion of hydrocarbons produces $CO_2$, $H_2O$ vapor, leftover $O_2$ (from excess air), and $N_2$ (from air)
• Incomplete combustion also yields $CO$, $H_2$, and possibly unburned fuel

The nitrogen from air doesn't react (at typical combustion temperatures for these problems), so it passes straight through to the flue gas. Since air is roughly 79% $N_2$, nitrogen usually makes up the largest fraction of flue gas by volume.

Calculating Composition of Combustion Products

Here's how to find the mole fraction of each species in the flue gas. Take the complete combustion of methane with 20% excess air as an example:

1. Write the balanced reaction: $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$

2. Calculate actual $O_2$ supplied: $2 \times 1.2 = 2.4$ moles $O_2$ per mole $CH_4$

3. Calculate $N_2$ entering with the air: Air is 79% $N_2$ and 21% $O_2$ by mole, so for every mole of $O_2$ there are $79/21 = 3.76$ moles of $N_2$. Total $N_2$: $2.4 \times 3.76 = 9.024$ moles

4. List all species in the flue gas (per mole of $CH_4$):

   • $CO_2$: 1 mole
   • $H_2O$: 2 moles
   • Excess $O_2$: $2.4 - 2 = 0.4$ moles
   • $N_2$: 9.024 moles

5. Total moles of flue gas: $1 + 2 + 0.4 + 9.024 = 12.424$ moles

6. Mole fraction of $CO_2$: $1 / 12.424 = 0.0805$

Material Balances in Combustion

Conservation of Mass in Combustion Reactions

Every combustion material balance rests on conservation of mass: the total mass of reactants entering equals the total mass of products leaving. This applies to the overall system and to each individual element (carbon, hydrogen, oxygen, nitrogen).

For combustion problems, the "system" is typically the combustion chamber. Inputs are fuel and air. Outputs are flue gas components.

Solving Combustion Material Balance Problems

Follow these steps for any combustion material balance:

1. Write and balance the combustion reaction for the given fuel
2. Identify what's given: fuel composition, flow rate, percent excess air, etc.
3. Calculate the theoretical air requirement from stoichiometry
4. Determine actual air supplied: multiply theoretical air by $(1 + \% \text{ excess}/100)$
5. Calculate moles (or mass) of each product using stoichiometry and accounting for excess $O_2$ and $N_2$
6. Apply conservation of mass to find unknown flow rates or compositions

Worked example: Find the mass flow rate of flue gas from burning 100 kg/h of propane with 30% excess air.

1. Balanced reaction: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

2. Molar flow rate of propane: Molecular weight of $C_3H_8$ is 44 kg/kmol, so $100 / 44 = 2.273$ kmol/h

3. Theoretical $O_2$: $5 \times 2.273 = 11.36$ kmol/h

4. Actual $O_2$ supplied: $11.36 \times 1.3 = 14.77$ kmol/h

5. $N_2$ from air: $14.77 \times 3.76 = 55.55$ kmol/h

6. Flue gas components (kmol/h):

   • $CO_2$: $3 \times 2.273 = 6.82$
   • $H_2O$: $4 \times 2.273 = 9.09$
   • Excess $O_2$: $14.77 - 11.36 = 3.41$
   • $N_2$: 55.55

7. Convert each to mass (kg/h):

   • $CO_2$: $6.82 \times 44 = 300.1$
   • $H_2O$: $9.09 \times 18 = 163.6$
   • Excess $O_2$: $3.41 \times 32 = 109.1$
   • $N_2$: $55.55 \times 28 = 1555.4$

8. Total flue gas mass flow rate: $300.1 + 163.6 + 109.1 + 1555.4 = 2128.2$ kg/h

You can verify this with an overall mass balance: mass in (fuel + air) should equal mass out (flue gas). Air mass = $(14.77 \times 32) + (55.55 \times 28) = 472.6 + 1555.4 = 2028.0$ kg/h. Fuel + air = $100 + 2028.0 = 2128.0$ kg/h, which matches the flue gas total. That check is a good habit for catching errors.