Combustion

Counting atoms

A chemical reaction rearranges atoms; it never creates or destroys them. So a combustion equation is balanced by conserving each element. For methane burning completely in air — which we model as 21% oxygen and 79% nitrogen by moles, or 3.76 moles of N₂ for every mole of O₂:

$\mathrm{CH_4} + 2\,(\mathrm{O_2} + 3.76\,\mathrm{N_2}) \rightarrow \mathrm{CO_2} + 2\,\mathrm{H_2O} + 7.52\,\mathrm{N_2}.$

That is the theoretical or stoichiometricStoichiometric (theoretical) airExactly enough air to burn all the fuel completely, with no oxygen left over. amount of air: exactly enough oxygen for complete combustion. Engineers describe mixtures by the air–fuel ratio (17.2 kg of air per kg of methane at stoichiometric), the percent of theoretical air, or the equivalence ratio $\varphi$ — actual fuel–air ratio divided by stoichiometric — below 1 for lean mixtures with excess air, above 1 for rich ones.

Notice how much of the product gas is nitrogen, carried through the flame unchanged. It absorbs a large share of the heat — which is why the flame temperature depends so strongly on how much excess air you use. Notice, too, the dew point of the products: around 55–60 °C for natural gas. Cool the exhaust below it, and the water vapour condenses, releasing its latent heat. That is the trick of a condensing boiler, and the difference between the two heating values below.

Energy in chemical bonds

Up to now, every enthalpy had an arbitrary zero for each substance — fine, as long as the substance didn't change. With chemical reactions, substances do change, so all enthalpies need a common reference. The convention: elements in their stable form at 25 °C and 1 atm have zero enthalpy. The enthalpy of a compound at that state is its enthalpy of formation, $\bar h_f^\circ$, the heat released or absorbed when it is formed from its elements. CO₂'s is −393 520 kJ/kmol: forming it releases a lot of energy. At any other temperature,

$\bar h(T) = \bar h_f^\circ + \left[\bar h(T) - \bar h(T_\text{ref})\right].$

The energy balance for a reactor in steady state, with no work, then reads

$\frac{\dot Q_\text{cv}}{\dot n_F} = \sum_P n_e\,\bar h_e - \sum_R n_i\,\bar h_i.$

Heating values

The heat released when a fuel burns completely, with products cooled back to 25 °C, is its heating valueHeating valueThe heat released by complete combustion of a unit of fuel. The higher heating value counts the water in the products as liquid, the lower as vapour.. If the water in the products ends up as liquid, it is the higher heating value (HHV); as vapour, the lower heating value (LHV). The difference is the latent heat of that water: about 10% for natural gas, 18% for hydrogen. Power-plant efficiencies in Europe are usually quoted on the LHV, in the United States on the HHV — always check which.

The adiabatic flame temperature

If a combustor lost no heat and burned completely, all the chemical energy would go into heating the products. The temperature they would reach is the adiabatic flame temperatureAdiabatic flame temperatureThe temperature the products reach if all the chemical energy goes into heating them., found by setting the products' total enthalpy equal to the reactants':

$\sum_P n_e\,\bar h_e(T_\text{ad}) = \sum_R n_i\,\bar h_i(T_\text{in}).$

With temperature-dependent heat capacities this needs a numerical solution — which the widget below does for five fuels:

The flame is hottest near stoichiometric and cools on either side: lean mixtures waste heat on excess air, rich ones leave fuel unburned. That gives engineers a powerful lever. A gas turbine can't run its combustor at 2300 K, so it burns lean — or burns near stoichiometric and dilutes with extra air — to bring the gas down to the 1600–1900 K the turbine blades can stand. Preheating the air raises the flame temperature degree for degree, which is how glass furnaces and steel reheating furnaces reach their temperatures.

Next: real flames never quite reach the adiabatic temperature, and their exhaust always contains traces of CO and NO. Why doesn't burning ever quite finish?