9.3 Combustion Analysis and Stoichiometry

Combustion analysis and stoichiometry are the tools you use to figure out how fuels burn, what products they form, and how much energy gets released. These concepts let you balance chemical equations, calculate air-fuel ratios, and determine the composition of exhaust gases.

Mastering this material is essential for optimizing combustion efficiency and reducing harmful emissions in engines, power plants, and industrial burners.

Balancing combustion reactions

Reactants and products in combustion

Combustion is the rapid oxidation of a fuel (typically a hydrocarbon) with oxygen, releasing heat and light. The reactants are the fuel and $O_2$, and the primary products are $CO_2$, $H_2O$, and energy (heat).

Nitrogen makes up about 79% of air by volume, and at high temperatures it can react with oxygen to form nitrogen oxides ($NO_x$), which are pollutants that contribute to smog and acid rain. The amount of $NO_x$ formed depends on combustion temperature, pressure, and how long the gases stay at elevated temperatures (residence time).

Balancing combustion reaction equations

The general form of a hydrocarbon combustion reaction is:

$C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O + \text{heat}$

Each carbon atom in the fuel produces one $CO_2$, and each pair of hydrogen atoms produces one $H_2O$. Balancing these reactions means adjusting coefficients so the number of atoms of each element is equal on both sides, following conservation of mass.

Complete combustion occurs when there's enough $O_2$ to convert all carbon to $CO_2$ and all hydrogen to $H_2O$:

$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{heat}$

Incomplete combustion occurs when oxygen is insufficient, producing carbon monoxide ($CO$) and/or unburned hydrocarbons alongside the usual products:

$2C_2H_6 + 5O_2 \rightarrow 3CO_2 + CO + 6H_2O + \text{heat}$

To balance a combustion equation step by step:

1. Write the unbalanced equation with fuel and $O_2$ on the left, $CO_2$ and $H_2O$ on the right.
2. Balance carbon first by setting the $CO_2$ coefficient equal to the number of carbon atoms in the fuel.
3. Balance hydrogen next by setting the $H_2O$ coefficient to half the number of hydrogen atoms in the fuel.
4. Balance oxygen last by summing the oxygen atoms needed on the product side and adjusting the $O_2$ coefficient accordingly.
5. If you end up with fractional coefficients, multiply the entire equation by the smallest integer that clears the fractions.

Air-fuel ratios for combustion

Theoretical and actual air-fuel ratios

The air-fuel ratio (AFR) is the mass of air divided by the mass of fuel in a combustion process.

The theoretical (stoichiometric) AFR is the exact amount of air needed for complete combustion, with no leftover $O_2$ and no unburned fuel. To calculate it:

1. Write and balance the combustion reaction for the fuel.
2. Determine the moles of $O_2$ required per mole of fuel from the balanced equation.
3. Convert moles of $O_2$ to moles of air. Since air is approximately 21% $O_2$ by volume (mole basis), divide the moles of $O_2$ by 0.21 to get moles of air.
4. Convert both the air and fuel moles to mass using their respective molecular weights (use 28.97 kg/kmol for air).
5. The stoichiometric AFR is the mass of air divided by the mass of fuel.

The actual AFR in a real system is typically higher than the stoichiometric value. Extra air is supplied to ensure complete combustion and to help control flame temperature.

Equivalence ratio and excess air

The equivalence ratio ($\phi$) compares the actual fuel-to-air ratio to the stoichiometric fuel-to-air ratio:

$\phi = \frac{(F/A)_{\text{actual}}}{(F/A)_{\text{stoichiometric}}} = \frac{AFR_{\text{stoichiometric}}}{AFR_{\text{actual}}}$

• $\phi > 1$: fuel-rich mixture (not enough air for complete combustion)
• $\phi = 1$: stoichiometric mixture
• $\phi < 1$: fuel-lean mixture (excess air present)

Note the inversion: because AFR is air/fuel while $\phi$ is defined in terms of fuel/air, a higher actual AFR gives a lower equivalence ratio.

Percent excess air expresses how much air is supplied beyond the stoichiometric requirement:

$\% \text{Excess Air} = \frac{AFR_{\text{actual}} - AFR_{\text{stoichiometric}}}{AFR_{\text{stoichiometric}}} \times 100\%$

For example, if the stoichiometric AFR is 15 and the actual AFR is 18:

$\% \text{Excess Air} = \frac{18 - 15}{15} \times 100\% = 20\%$

This means 20% more air than theoretically needed is being supplied.

Combustion product composition

Determining product composition using stoichiometry

Once you have a balanced combustion equation, the molar quantities of each product are directly proportional to their coefficients in that equation.

For complete combustion of a hydrocarbon with air, the products are $CO_2$, $H_2O$, and $N_2$ (which passes through largely unreacted). You get the moles of $CO_2$ and $H_2O$ from the balanced equation. The moles of $N_2$ come from the air: since air is about 79% $N_2$ by volume, for every mole of $O_2$ entering the reaction, approximately 3.76 moles of $N_2$ tag along.

For incomplete combustion, additional species like $CO$ and unburned hydrocarbons appear. Their quantities can be determined either from a balanced incomplete combustion equation or from measured exhaust gas concentrations.

Mole and mass fractions of combustion products

Mole fraction of a product is its moles divided by the total moles of all products.

Consider methane burning with stoichiometric air. The balanced equation including nitrogen is:

$CH_4 + 2O_2 + 7.52N_2 \rightarrow CO_2 + 2H_2O + 7.52N_2$

(The 7.52 comes from $2 \times 3.76$.)

Total product moles: $1 + 2 + 7.52 = 10.52$

• Mole fraction of $CO_2$: $\frac{1}{10.52} = 0.095$
• Mole fraction of $H_2O$: $\frac{2}{10.52} = 0.190$
• Mole fraction of $N_2$: $\frac{7.52}{10.52} = 0.715$

Mass fraction requires one more step. Multiply each component's moles by its molecular weight, then divide by the total product mass:

• Mass of $CO_2$: $1 \times 44 = 44$ kg
• Mass of $H_2O$: $2 \times 18 = 36$ kg
• Mass of $N_2$: $7.52 \times 28 = 210.56$ kg
• Total mass: $44 + 36 + 210.56 = 290.56$ kg

Mass fraction of $CO_2$: $\frac{44}{290.56} = 0.151$

(Note: this differs slightly from simply multiplying the mole fraction by the molecular weight of one species and dividing by a weighted average. Always compute total mass from all components to avoid errors.)

Adiabatic flame temperature

The adiabatic flame temperature is the maximum temperature combustion products would reach if no heat were lost to the surroundings. It represents an upper bound on the actual flame temperature.

To calculate it, you apply an energy balance: the total enthalpy of the reactants (including enthalpies of formation and sensible enthalpy at the inlet temperature) equals the total enthalpy of the products at the flame temperature. Since product enthalpies depend on temperature through their specific heats, this typically requires iteration:

1. Write the enthalpy balance: $H_{\text{reactants}} = H_{\text{products}}(T_{ad})$
2. Compute the reactant enthalpy using standard enthalpies of formation and sensible enthalpies at the inlet conditions.
3. Guess an adiabatic flame temperature.
4. Compute the product enthalpy at that temperature using enthalpy tables or $c_p$ data.
5. Adjust your guess and repeat until the two sides balance.

For methane burning with stoichiometric air at standard conditions (298 K, 1 atm), the adiabatic flame temperature is approximately 2230 K. With excess air, this temperature drops because the extra $N_2$ and $O_2$ absorb energy without contributing fuel energy.

Combustion efficiency analysis

Factors affecting combustion efficiency

Combustion efficiency measures how effectively the chemical energy in the fuel is converted to thermal energy in the products.

• Excess air is needed to ensure complete combustion, but too much dilutes the hot products with cool air, lowering flame temperature and increasing stack (exhaust) heat losses.
• Incomplete combustion from insufficient oxygen or poor fuel-air mixing produces $CO$ and unburned hydrocarbons. These species still contain chemical energy that wasn't released as heat, representing a direct efficiency loss. They're also harmful pollutants.
• There's a trade-off: too little air causes incomplete combustion, while too much air causes excessive heat loss. The optimal operating point balances these effects.

Calculating combustion efficiency

One common way to express combustion efficiency based on the air-fuel ratio is:

$\eta_{\text{comb}} = \frac{AFR_{\text{stoichiometric}}}{AFR_{\text{actual}}} \times 100\%$

This formulation captures the idea that supplying more air than necessary dilutes the products and wastes energy heating inert nitrogen. For example, if the stoichiometric AFR is 14.7 and the actual AFR is 16:

$\eta_{\text{comb}} = \frac{14.7}{16} \times 100\% = 91.9\%$

Keep in mind that this is a simplified metric. A more rigorous efficiency calculation would account for the actual enthalpy of the exhaust gases relative to the fuel's heating value, including losses from both excess air and incomplete combustion products.

Exhaust gas analysis for incomplete combustion

Exhaust gas analysis reveals how completely the fuel burned. Two key indicators:

• Carbon monoxide ($CO$) concentration, measured in ppm or as a volume percentage. Higher $CO$ means more incomplete combustion. An exhaust reading of 1000 ppm $CO$ indicates significantly worse combustion than 100 ppm.
• Unburned hydrocarbon (UHC) concentration, measured in ppm of carbon (ppmC). This represents fuel that passed through without fully reacting. An exhaust with 500 ppmC is burning much less completely than one with 50 ppmC.

Both metrics are used in practice to tune combustion systems and to verify compliance with emissions regulations.

Optimizing combustion efficiency

Three main strategies improve combustion efficiency:

• Optimize the air-fuel ratio. Run close to stoichiometric with just enough excess air to ensure complete combustion. Continuous monitoring of exhaust $O_2$ and $CO$ levels helps maintain this balance.
• Improve fuel-air mixing. Better atomization (e.g., fuel injectors that produce fine droplets) increases the surface area of fuel exposed to oxygen, promoting faster and more complete reactions.
• Maintain proper combustion temperatures. Turbulators or swirl vanes in the combustion chamber create turbulence that enhances mixing and sustains high local temperatures, both of which drive the reaction toward completion.