Combustion Chemistry and Flame Temperature

Stoichiometry of kerosene, fuel–air and equivalence ratios, heating values, the burner energy balance, and adiabatic flame temperature with dissociation

More fuel, no hotter

A kilogram of Jet A releases about 43 MJ when it burns — about ten times the energy a kilogram of TNT releases when it explodes. Feed a flame more and more of it and you would expect it to get hotter and hotter. It does not. Burn kerosene in air arriving at 850 K and 40 bar — a modern combustor's conditions — and the flame peaks at about 2700 K, just rich of the chemically correct mixture; add more fuel and it cools. Burn it in ambient air and the peak is about 2300 K. And even at the chemically correct mixture the flame falls about 180 K short of what complete burning would give. Three things set those numbers: how much air each kilogram of fuel can use, where the energy goes, and the products themselves falling apart in the heat. Lab 2 computes all three with Cantera.

Jet A is a blend of hydrocarbons; for engineering it is represented by an average molecule, about C₁₂H₂₃. Burned completely it gives carbon dioxide and water, and the oxygen comes with nitrogen, 3.76 molecules of N₂ for every O₂ in air. The stoichiometric mixture contains exactly the oxygen needed:

\mathrm{C_{12}H_{23}} + 17.75\,(\mathrm{O_2} + 3.76\,\mathrm{N_2}) \;\longrightarrow\; 12\,\mathrm{CO_2} + 11.5\,\mathrm{H_2O} + 66.74\,\mathrm{N_2}.

By mass that is

f_{st} = 0.0686

kg of fuel per kg of air — an air–fuel ratio of 14.6.

f_{st} = \frac{M_{fuel}}{17.75\,(M_{O_2} + 3.76\,M_{N_2})} = 0.0686, \qquad \phi = \frac{f}{f_{st}}

The stoichiometric fuel–air ratio of C12H23 and the equivalence ratio φ: below 1 lean (excess air), above 1 rich (excess fuel).

Derivation

Balancing the reaction

Carbon and hydrogen fix the products: 12 CO₂ from 12 carbon atoms, 11.5 H₂O from 23 hydrogen atoms. Oxygen must then balance:

2a = 2\cdot 12 + 11.5 \quad\Longrightarrow\quad a = 17.75,

and the nitrogen,

3.76a = 66.74

molecules, passes through unchanged (at least in this approximation; Lesson 7-4 is about the part that does not). With molar masses of 167.3 kg/kmol for the fuel and 137.3 kg/kmol for each "mole of air oxygen" with its nitrogen,

f_{st} = 167.3/(17.75 \times 137.3) = 0.0686

💡Two heating values

Each kilogram of Jet A makes 1.24 kg of water. If the products are cooled until that water condenses, its latent heat is recovered too: the higher heating value, about 46 MJ/kg. Engines exhaust their water as steam, so cycle analysis uses the lower heating value,

h_{PR} \approx 43

MJ/kg. A combustor at a turbine entry of about 1700 K runs at

f \approx 0.025

— an equivalence ratio of only 0.36. Most of the air is there to cool, not to burn.

Adiabatic flame temperature. If no heat escapes, the products leave with the same total enthalpy the reactants brought in, chemical energy included. The temperature at which that happens is the adiabatic flame temperature. Two effects keep it lower than a simple estimate:

The products' $c_p$ rises with temperature (Lesson 4-3), so each extra joule buys less temperature.
Above about 2000 K the products dissociate: CO₂ and H₂O partly break back into CO, H₂, OH, O and H, storing chemical energy instead of releasing it as heat.

Just rich of

\phi = 1

the products hold some CO and H₂ in place of CO₂ and H₂O. Their lower heat capacity more than offsets the energy left unreleased, so the flame peaks slightly rich, near

\phi = 1.06

. Richer still, the unreleased energy wins and the flame cools.

📐Flame temperatures from Cantera

Equilibrium flame temperatures of Jet A (Lab 2):

Air at 850 K and 40 bar: 1775 K at $\phi = 0.4$ , 2472 K at 0.8, 2679 K at 1.0; the peak is 2695 K at $\phi \approx 1.06$ , falling to 2578 K at 1.3.
Without dissociation the stoichiometric flame would reach 2856 K: dissociation costs 178 K.
With ambient air at 1 atm the peak is about 2290 K.

📐How much fuel does Tt4 really need?

The cycle analysis of Unit 4 used an energy balance with two constant specific heats,

f = (c_{pt}T_{t4} - c_{pc}T_{t3})/(h_{PR} - c_{pt}T_{t4})

. Against Cantera's equilibrium combustion from 850 K (Lab 3):

$T_{t4} = 1500$ K: formula 0.0213, Cantera 0.0184 — the formula asks for 16 % too much fuel.
1700 K: 0.0271 against 0.0248, 9 % too much.
1900 K: 0.0329 against 0.0316, 4 % too much.

Good enough to rank designs, not to size a fuel system — another reason production cycle codes use real-gas tables.

There is no more open door by which you can enter into the study of natural philosophy than by considering the physical phenomena of a candle.

– Michael Faraday, The Chemical History of a Candle (1860) (Wikiquote)

💡Why burn so lean?

A stoichiometric flame at 2700 K would destroy any turbine. Real combustors burn a part of the air near stoichiometric to keep the flame stable, then mix in the rest to bring the gas down to 1600–1900 K — the zones of Lesson 7-3. The overall equivalence ratio of 0.3–0.45 is the result.

💡Where you will use this

Lesson 7-2 asks how fast these reactions go; 7-3 designs the zones; 7-4 follows the minor species — NO, CO, soot — that matter for emissions. The same equilibrium tools design rocket engines and industrial furnaces.

📚Further reading

Stephen R. Turns, An Introduction to Combustion: Concepts and Applications (3rd ed., McGraw-Hill, 2012), chapters 2–3.
Arthur H. Lefebvre and Dilip R. Ballal, Gas Turbine Combustion (3rd ed., CRC Press, 2010).
The Cantera documentation, for equilibrium and reactor calculations.

↓But there is a problem

Equilibrium tells you where a reaction ends up. It says nothing about how fast it gets there — and in a combustor the gas has only a few thousandths of a second. On 24 June 1982 a British Airways 747 flew into a cloud of volcanic ash over Java at 37,000 feet, and all four of its flames went out.