Combustion Performance

Lefebvre frames performance as three questions a designer must answer at every point in the flight envelope:

1. Is the fuel's energy actually released? (efficiency)
2. Will the flame stay lit? (stability)
3. If it goes out, can we light it again? (ignition and relight)

All three turn out to be the same contest, between how long the gas stays and how long the chemistry takes.

One ratio to rule them: the Damköhler number

Gas spends a finite time in the primary zone, the residence time $\cair{\tau_\text{flow}} \sim L/\cair{U}$. The chemistry needs its own time to complete, $\cflame{\tau_\text{chem}}$, and from Lesson 2 we know that time depends exponentially on temperature. Their ratio is the Damköhler numberDamköhler numberDa = τ_flow / τ_chem, residence time over chemical time. Above 1 the flame survives; below 1 it blows out.:

$\text{Da} = \frac{\cair{\tau_\text{flow}}}{\cflame{\tau_\text{chem}}}$

If $\text{Da} > 1$, the chemistry finishes before the gas leaves: the flame lives. If $\text{Da} < 1$, reacting gas is swept out before it can sustain the reactions behind it: the flame dies.

Now watch what happens when we combine this with the two ingredients from Lesson 2. The chemical time scales like $e^{+E_a/RT_{ad}(\phi)}$, and $T_{ad}(\phi)$ is our bell curve. So the velocity at which the flame blows out is

$U_\text{blow-off}(\phi) \;\propto\; L\, P^{\,n}\, \exp\!\left(-\frac{E_a}{R\,T_{ad}(\phi)}\right)$

A bell-shaped temperature, pushed through an exponential, gives a sharply peaked curve. Plot the boundary $\text{Da} = 1$ in the plane of velocity and mixture and you get this:

This is Lefebvre's stability loop, the "peaked hat", and you just derived it. Everything inside is stable. The lower edge is lean blowoutLean blowoutFlame-out caused by too little fuel: the chemistry slows until it can no longer keep up with the flow., the upper edge rich blowout, and the tip is the maximum velocity at which the combustor can burn at all, found slightly rich of stoichiometric where the flame is hottest.

Now use the sliders to see why designers care about each variable:

• Pressure enlarges the loop (faster chemistry). That makes high altitude, where pressure is low, the dangerous corner of the envelope.
• Inlet temperature lifts $T_{ad}$ and enlarges the loop, so a cold ground start is harder than a hot one.
• Combustor volume raises the residence time and the whole boundary moves out. Stability is bought with size and weight.

Modern lean-burn combustors deliberately live with smaller lean-blowout margins than older designs, and manage the risk with pilot flames (a small, richer, very stable zone that keeps the main zone lit), fuel staging schedules, and fast control logic.

Combustion efficiency

Combustion efficiency is the fraction of the fuel's chemical energy actually released:

$\eta_c = \frac{\Delta T_\text{actual}}{\Delta T_\text{ideal}} \approx 1 - \frac{\dot m_\text{CO}\, H_\text{CO} + \dot m_\text{UHC}\, H_\text{UHC}}{\dot m_f\, H_f}$

At take-off, $\eta_c > 99.9\%$. At idle it can fall to 97–99%. That sounds small, but it is almost entirely CO and unburned hydrocarbons, precisely the pollutants regulated around airports.

Why does efficiency fall at low power? Pressure and inlet temperature drop (chemistry slows), the fuel spray coarsens (Lesson 6) and the combustor is suddenly too small for the job. Lefebvre condensed all of that into one loading parameterLoading parameter θLefebvre’s P^1.75·A·D^0.75·e^(T₃/300)/ṁ: bundles pressure, temperature, size and airflow into one predictor of efficiency.:

$\theta = \frac{P_3^{1.75}\, A_\text{ref}\, D_\text{ref}^{0.75}\, \exp(T_3/300)}{\dot m_A}$

Pressure (to a high power), temperature (exponentially) and size in the numerator; air flow in the denominator. Large $\theta$ is a lightly loaded, efficient combustor. Remarkably, data from very different combustors collapse onto one S-shaped curve of efficiency against $\theta$:

Try the presets. Take-off sits comfortably on the plateau. Cruise does too. Idle begins to slip down the shoulder, and a windmilling relight at altitude is far down the slope. This is the sizing equation: if the engine's air flow doubles and the combustor doesn't grow, $\theta$ halves.

Ignition: a race against heat loss

Starting a combustor happens in three acts:

1. Kernel formation: a spark creates a small ball of very hot gas in a combustible pocket of mixture.
2. Kernel growth: the ball must grow. It releases heat in proportion to its volume and loses heat to the cold surroundings through its surface. Small kernels lose; big kernels win. So there is a critical size, roughly the quenching distance, below which the kernel dies.
3. Propagation: the flame spreads through the primary zone and around the annulus (remember the light-around in Lesson 1).

Now go to altitude. The air is thinner and colder, the fuel is colder and more viscous, and the spray is coarser. Every one of these effects works against you twice: the spark's energy makes a smaller kernel, and the critical size gets bigger. The minimum ignition energy climbs roughly as $P^{-2}$.

The result is a finite relight envelope: a region of altitude and flight speed inside which the engine must demonstrate reliable relight for certification. It is one of the most demanding requirements a combustor faces, and it often fights with emissions. A primary zone tuned lean for low NOₓ at cruise is exactly the wrong primary zone for lighting at 30,000 ft.

All of this assumed the fuel was already vapour, mixed with air. In reality it arrives as a liquid, and turning it into a fine mist is the next problem.