Combustion
Every power plant, engine and furnace in this course has so far taken its heat input as a given number, delivered from a reservoir with no questions asked. Combustion is where that number gets made, and it turns out to be an ordinary chemical reaction, balanced the way any equation is balanced, with the first law applied across it exactly as it was applied to everything else.
Combustion is a rapid oxidation reaction that releases energy, and treating it thermodynamically means treating it as two separate jobs done in sequence. First, work out exactly what goes in and what comes out: a matter of balancing a chemical equation, no different in kind from balancing any other. Second, apply the first law across the reaction to find out how much energy the rearrangement released. Neither step needs anything beyond what this course has already built.
Fuels and stoichiometric combustion
Most fuels burned for power are hydrocarbons, compounds of carbon and hydrogen written generically as \(C_xH_y\), from methane at the simple end through propane, octane and diesel to coal, which is better described by an ultimate analysis than a single formula but obeys the same bookkeeping. The oxidiser is almost always air rather than pure oxygen, because pure oxygen is expensive and air is free, and air's composition matters enough to the arithmetic that it is fixed once for the whole topic: approximately \(21\)% oxygen and \(79\)% nitrogen by mole, the same approximation used in gas mixtures and psychrometrics. Every mole of oxygen supplied therefore drags along \(0.79/0.21 = 3.76\) moles of nitrogen, whether or not the nitrogen does anything once it arrives.
And for an ordinary combustion temperature, it does not. Nitrogen is treated as chemically inert through this entire topic. It enters with the air and leaves in the products completely unchanged, a bystander that nonetheless has to be carried through every balance because it absorbs some of the reaction's energy on its way through, warming up like everything else in the exhaust even though it took no part in the reaction that warmed it. That combination, chemically silent, thermally present, is the single fact that makes air a worse oxidiser than pure oxygen and shapes most of what follows.
Stoichiometric, or theoretical, combustion is the case where the air supplied is exactly enough to burn the fuel completely, with no leftover oxygen in the products and no unburned fuel either. For a general hydrocarbon,
and the coefficient \(a\) is not a free choice: it is fixed by the oxygen balance alone, since carbon and hydrogen have already told the products how many oxygen atoms they need: \(2x\) for the carbon dioxide and \(y/2\) for the water, for a total of \(2x + y/2\) oxygen atoms, supplied \(2a\) at a time:
For propane, \(C_3H_8\), that gives \(a = 3 + 8/4 = 5\), and the whole reaction is fixed before a single mass has been computed. Every quantity used in the rest of this topic (the air required, the products formed, the energy released) traces back to getting this one coefficient right.
A hydrocarbon fuel, CxHy, is the reactant that actually carries chemical energy.
Air-fuel ratio and excess air
The air-fuel ratio is the mass of air supplied per unit mass of fuel burned,
computed from the balanced equation by converting the mole coefficients to mass with each species' molar mass. For propane's stoichiometric equation, with \(a = 5\) moles of oxygen bringing \(5(4.76) = 23.8\) moles of air along with it, and air's apparent molar mass taken as \(28.97\) kg/kmol,
which says that burning one kilogram of propane completely, with nothing wasted and nothing left over, needs nearly sixteen kilograms of air alongside it, a ratio worth remembering roughly, because most hydrocarbon fuels land in the same neighbourhood, from about \(14\) for petrol to a little over \(17\) for methane.
No real burner is run at exactly the theoretical ratio, and the reason is a mixing problem rather than a chemistry one. Getting every last molecule of fuel to encounter enough oxygen at the instant it is ready to react is not achievable with a real burner, real turbulence and a real residence time, so combustion at the theoretical air-fuel ratio leaves some fuel unburned or only partially burned: carbon monoxide instead of carbon dioxide, or soot, both a waste of fuel and, for CO, a serious hazard. Practical combustion is therefore always run with excess air, more than theory demands, so that oxygen is available everywhere the fuel actually needs it.
Percentage excess air and percentage theoretical air report the same fact two ways:
Run the propane example with 20% excess air and \(AF_{\text{actual}} = 1.20 \times 15.64 = 18.77\) kg air/kg fuel. The trade is a real one and it runs both ways: too little excess air risks incomplete combustion and its efficiency losses and safety hazards; too much excess air wastes fuel heating nitrogen and unreacted oxygen that carry no chemical energy of their own and simply leave the stack hot, a loss that shows up directly in the adiabatic flame temperature calculation at the end of this topic. Typical practice runs somewhere between \(10\)% and \(100\)% excess air depending on the fuel and the equipment, gas burners needing the least and solid fuel the most.
Propane's stoichiometric air comes straight from the oxygen balance: a = 3 + 8/4 = 5.
Balancing the combustion equation
With excess air, or with an assumed product split for incomplete combustion, the balanced equation is no longer read straight off the stoichiometric formula and has to be built one atom at a time. The method does not change from ordinary chemistry: conserve carbon, conserve hydrogen, conserve oxygen, conserve nitrogen, in that order, because each element's balance is easiest to close once the ones before it are fixed.
Take propane again, now with 20% excess air, so \(a = 1.20 \times 5 = 6\) moles of oxygen are supplied where only \(5\) are needed:
Carbon fixes the CO₂ coefficient immediately: three carbon atoms in the fuel can only become three molecules of \(CO_2\). Hydrogen does the same for water: eight hydrogen atoms, two per water molecule, becomes four. Nitrogen never reacts, so its coefficient on the right is whatever came in on the left, \(6(3.76) = 22.56\), unchanged by anything else happening in the equation.
Oxygen is the balance that closes the books, and it is the one that reveals the excess air directly in the products. Six moles of \(O_2\) supply twelve oxygen atoms; the carbon dioxide and water together need \(3(2) + 4(1) = 10\) of them, leaving two oxygen atoms, or one full mole of \(O_2\), unreacted in the exhaust. That leftover oxygen is not a separate calculation bolted onto the stoichiometric answer. It is what excess air looks like once the equation is actually balanced: a term that would not exist at all if the air-fuel ratio had been exactly theoretical.
The same four-step method handles incomplete combustion, with one adjustment: a problem that says some fraction of the carbon forms \(CO\) rather than \(CO_2\) gives you that split as data, and the oxygen balance is worked exactly as before, just with an extra product term and one fewer oxygen atom claimed by every mole of \(CO\) compared with a mole of \(CO_2\). Nothing about the method changes; there is simply one more unknown to carry through it.
Carbon in the fuel has nowhere to go but CO2 — count the carbons first.
The first law for combustion
A balanced equation says what happens to the atoms. It says nothing yet about energy, and applying the first law across a reaction needs one extra idea that has not been needed anywhere else in this course: a common zero. Internal energy and enthalpy have always been used as differences, with the reference state cancelling out of every calculation. But a reaction changes what the substance is, so the reactants' and products' enthalpies can no longer be measured from an arbitrary reference each, the way steam tables set \(h_f = 0\) for liquid water at the triple point in properties of pure substances. Both sides of the equation need to be measured from the same zero, or the difference between them means nothing.
That common zero is the enthalpy of formation, \(h^\circ_f\): the enthalpy change when one mole of a compound forms from its elements in their most stable form, all at the standard reference state of \(25\) °C and \(1\) atm. Elements in their stable form ( \(O_2\), \(N_2\), graphite) are assigned \(h^\circ_f = 0\) at that state, by the same kind of convention as the steam tables' triple point, and every compound's formation enthalpy is measured relative to that zero. Carbon dioxide's is \(-393{,}520\) kJ/kmol and water vapour's is \(-241{,}820\) kJ/kmol, both strongly negative, which says both compounds are considerably more stable than the separated elements they were built from, and that difference in stability is precisely where combustion's energy comes from.
For a steady-flow combustion chamber with no work extracted, the energy balance is a straightforward extension of the steady-flow energy equation, with every enthalpy now built from a formation value plus a sensible correction for sitting away from the reference temperature:
If both reactants and products happen to sit at the reference temperature, every \(\Delta h\) term vanishes and the heat released collapses to a difference of formation enthalpies alone, which is the standard definition of a fuel's enthalpy of combustion. For propane burned completely with stoichiometric air, both sides at \(25\) °C,
With \(h^\circ_{f,\mathrm{C_3H_8}} = -103{,}850\) kJ/kmol, this comes to \(-103{,}850 - \left[3(-393{,}520) + 4(-241{,}820)\right] \approx 2{,}044{,}000\) kJ per kilomole of fuel burned, roughly \(46.4\) MJ per kilogram, which is within the ordinary range quoted for liquid petroleum gas.
One choice inside that number is worth naming, because it changes the answer by a fixed and predictable amount. Water leaves a real flue as vapour, and the calculation above assumed exactly that, which is the lower heating value, LHV. If the water is instead imagined condensed back to liquid (recovering its latent heat as part of the released energy), the result is the higher heating value, HHV, larger by exactly \(N_{\mathrm{H_2O}}\,h_{fg}\) at the reference temperature. Furnace and boiler efficiencies quoted against HHV and against LHV for the same equipment differ by several percentage points for no reason connected to how well the equipment was built, which is why the figure is always worth checking before two efficiencies are compared.
Every enthalpy of formation is measured from the same zero: the elements, at 298 K.
Adiabatic flame temperature
Run the same energy balance with the opposite assumption: no heat allowed to leave at all, an insulated chamber rather than one feeding a boiler. With \(Q = 0\) and no work done, every joule the reaction releases has nowhere to go but into raising the temperature of the products themselves, and the energy balance becomes
with the reactants entering at some known inlet temperature \(T_i\), often the reference \(298\) K, and \(T_{ad}\), the adiabatic flame temperature, the single unknown. Solving for it is an iteration rather than an algebra problem, because each product's \(\Delta h\) term is a nonlinear function of \(T_{ad}\) read from an ideal-gas table. Guess a temperature, sum the products' enthalpies, compare against the reactants' fixed total, and adjust the guess until the two sides agree, but the physical content is entirely contained in the one equation above.
\(T_{ad}\) is a ceiling in the same sense the reversible efficiencies elsewhere in this course are ceilings, and for the same reason: any heat lost to the surroundings, any incomplete combustion, and any dissociation of the products at high temperature (CO₂ splitting partly back into \(CO\) and \(O\), which absorbs energy rather than releasing it), all pull the real flame temperature below the adiabatic, ideal-combustion value. For a stoichiometric hydrocarbon-air flame at atmospheric pressure, that ceiling typically lands somewhere in the low thousands of kelvin, commonly quoted in the range of \(2000\) to \(2300\) K depending on the fuel, hot enough to matter enormously to what a combustor can be built from and how quickly nitrogen oxides form, both of which are the reasons a gas turbine or a furnace designer needs this number rather than merely finding it interesting.
Excess air lowers \(T_{ad}\), and the mechanism is worth stating plainly because it is easy to guess backwards. Extra air brings no additional fuel and releases no additional chemical energy. It only adds inert mass, nitrogen and unreacted oxygen, that has to be heated by the same fixed quantity of released energy as before. The same numerator, spread over a larger denominator's worth of matter to warm, gives a smaller temperature rise. This is precisely why the choice of excess air in the second section was described as a trade rather than a free safety margin: every percentage point added to protect against incomplete combustion is also a percentage point taken off the flame temperature, and gas turbine combustors in particular are designed around that trade with some care, using exactly enough excess air to control the temperature the first turbine blades will see rather than to guarantee complete combustion alone.