Where Turbulence Stops Speeding Up a Flame — Damköhler's Wrinkling Hypothesis and Bending

A wildfire spreads faster as the wind picks up. At 5 m/s the flame front outruns a sprinting person. But keep raising the wind and past a certain speed the fire blows itself out. Firefighters call it blow-off. Measure a turbulent premixed flame on a laboratory burner and the curve has exactly the same shape.

Raise the turbulence intensity $u'$ and the turbulent flame speed $S_T$ climbs almost linearly at first. Then it bends over. That flattening is the bending effect. Push further and the flame quenches. The straight part is area. The bend is a cap on wrinkles. Quenching is the time of the smallest eddies.

The slider $u'/S_L$ deepens the wrinkles, and the measured $\Sigma$ falls away from the straight-line prediction.

Push $u'/S_L$ to 6 and measured $\Sigma$ stalls near 4 while the straight line reaches 7. Drop $l_t/\delta_L$ to 2 and push $u'/S_L$ to 12: the badge flips to quenching and $\Sigma$ falls from 3.5 to 2.0.

Why turbulence speeds a flame up#

For a methane-air premixed flame the laminar flame speed (the speed at which the front propagates on its own into a quiescent mixture) $S_L$ is about 0.40 m/s at room conditions. The flame thickness $\delta_L$ is around 0.5 mm. No engine runs on those numbers. Crossing a 40 mm bore in 2 ms takes 20 m/s. That is 50 times $S_L$.

Turbulence fills the gap. That is why we define a turbulent flame speed $S_T$. The definition comes out of an integral of the reaction rate. Integrate the mean fuel mass fraction transport equation across the flame brush along its normal, cancel the diffusion terms, and this is what is left.

$\rho_u$ is the fresh gas density, $Y_F^u$ and $Y_F^b$ the fuel mass fractions on the fresh and burnt sides, $\overline{\dot{\omega}_F}$ the mean fuel reaction rate, and $n$ the coordinate normal to the mean flame brush.

The point worth holding onto is that this is a direct extension of the relation used for laminar flames — the laminar version carries the instantaneous reaction rate in place of $\overline{\dot{\omega}_F}$. So $S_T$ is not a quantity defined as a velocity. It is an integrated measure of how much burning happens per unit area. That single fact explains everything that follows.

The wrong intuition — turbulence mixes the fuel better?#

The most common explanation goes like this. Turbulence mixes fuel and air better, so the reaction goes faster. For a premixed flame that is wrong.

Premixed means what it says: fuel and air were already mixed before ignition. The gas ahead of the front is a perfectly uniform mixture. There is nothing left for turbulence to mix. Mixing governs the speed of diffusion flames, where fuel and oxidizer arrive separately, not of premixed flames.

There is a second version. Turbulence raises the effective diffusivity, so the flame goes faster. The laminar flame speed scales as $S_L \sim \sqrt{D/\tau_c}$. Swap $D$ for a turbulent diffusivity $u' l_t$ and you get $S_T/S_L \sim \sqrt{u' l_t / D}$. A square root. But the initial slope the experiments show is linear. The exponent does not match.

There is a deeper problem with both. Each assumes the flame leaves the turbulence alone. The truth runs the other way. Right after ignition the temperature jumps from 300 K to above 2000 K. The kinematic viscosity $\nu$ rises roughly as $T^{1.7}$, so it grows by more than a factor of 20. The local Reynolds number $u' l_t / \nu$ drops by the same factor. A well-developed turbulent flow can relaminarize after ignition. While the turbulence is shaking the flame, the flame is killing the turbulence.

Damköhler's answer: it is area, not speed#

Damköhler proposed a different picture. Turbulence does not make the local combustion faster. It folds the flame sheet.

There is one assumption. Every point on the wrinkled front still propagates locally at $S_L$. Chemistry is untouched; only geometry changes. Mass conservation then hands over the answer. The fresh gas passing through the true wrinkled area $A_T$ has to equal the fresh gas passing through the projected area $\bar{A}$ of the mean flame brush.

$\Sigma$ is the wrinkling factor, the flame surface area per unit projected area. The left side is an area ratio, the right side a speed ratio. They are the same number.

Turbulence sets the size of the wrinkles. While an eddy of size $l_t$ pushes the front at speed $u'$, the front pushes back at $S_L$. The ratio of those two displacements sets the slope of the wrinkle, and the slope sets the area gain. Out comes Damköhler's linear law.

$u'$ is the RMS velocity fluctuation of the fresh gas and $S_L$ the laminar flame speed. At $u' = 0$ it collapses back to $S_T = S_L$.

For this picture to hold, the front has to be thinner than the eddies. Two dimensionless numbers measure that condition.

$Da$ is the Damköhler number (turbulent time over chemical time) and $Ka$ the Karlovitz number (chemical time over Kolmogorov time). $\tau_t$ is the eddy turnover time, $\tau_c$ the chemical time, $\tau_\eta$ the timescale of the smallest eddies. When $Da \gg 1$ the flame reacts faster than the eddies turn. When $Ka \lt 1$ not even the smallest eddy fits inside the flame thickness. Where both hold is the flamelet regime, and Damköhler's linear law is only valid there.

What the experiments showed — bending and quenching#

The experimental data draws three regimes. Nearly linear at first. Then flat. Then quenching.

Three effects overlap to make it flatten. Wrinkles are created and destroyed. The front propagates along its own normal and keeps shaving off convex cusps. When creation and destruction balance, $\Sigma$ saturates. Once $Ka$ exceeds 1, small eddies enter the preheat zone, the flame thickens, and the local $S_L$ drops. Stronger strain punches holes in the sheet.

One expression carries all three.

$\Sigma_\infty$ is the wrinkling saturation ceiling (6.0 here), $Ka_q$ the quench Karlovitz number (25.0 here), and $q$ a quench attenuation factor between 0 and 1. The saturating denominator produces the bending; $q$ produces the quenching.

Honesty is required here. This expression is not a derived law. It is a phenomenological form fitted to reproduce the observed shape of the curve. $\Sigma_\infty = 6.0$ and $Ka_q = 25.0$ are tuning values, not physical constants, and they shift with fuel, pressure and burner geometry. The source says the same thing: predicting the bend of the curve is difficult, and predicting the quenching limits is "almost impossible". Only the linear regime came out of physics. The other two are still correlations.

Walk one row by hand. At the large scale, $l_t/\delta_L = 10$ and $u'/S_L = 6$, the Damköhler number is turbulent time over chemical time, so $10/6 \approx 1.67$. The Karlovitz number ties the same two ratios with different exponents. Divide $6^{1.5}$ by $\sqrt{10}$ and the table's 4.65 comes back. Divide that by the quench value 25, square it, and the exponential gives $q \approx 0.966$. Near 1. On this row the quench factor barely bends the curve. The denominator does. Dividing $u'/S_L$ by the cap of 6 adds 1, the denominator becomes 2, and the numerator 6 shrinks to 3. Then $1 + 0.966 \times 3 = 3.90$. The linear law writes 7.00. The gap of 3.1 is not slower chemistry. The wrinkle area hit the cap.

Multiply by the laminar speed 0.40 m/s and the turbulent speed is the printed 1.559 m/s. Crossing a 40 mm bore at that speed is $0.040/1.559 \approx 0.026$ s, about 26 ms. The 2 ms an engine wants is still more than ten times shorter. Raise the ratio to 12 and the large-scale wrinkling factor is only 4.03, so the speed stays near $0.40 \times 4.03 \approx 1.61$ m/s. The 13-fold, 5.2 m/s promised by the linear law never shows up in the bending column.

Drop only the scale, from 10 to 2, and the same ratio of 6 jumps $Ka$ from 4.65 to 10.39. The square root of $10/2$ is about 2.24, and the ratio of those two Karlovitz numbers matches it. The small-eddy time got shorter. In a cascade that hands energy down the scales, that time falls below the chemical time first. At ratio 12 and scale 2, $Ka$ is 29.39. Divided by 25 it exceeds 1, $q$ drops to about 0.25, and the wrinkling factor becomes 2.00. The sheet did not gain area. The attenuation erased area that had already been folded.

The flame changes the turbulence too — baroclinic torque#

Gas expands as it crosses the front. Mass flux has to be conserved, so

$\rho_b$ is the burnt gas density and $u_b$ the gas speed leaving the front. The flame front is a device that accelerates the flow sevenfold. That acceleration acts directly on the turbulence.

The four terms of the vorticity transport equation show the route.

$\boldsymbol{\omega}$ is the vorticity vector, $\mathbf{u}$ the velocity, $\rho$ the density, $p$ the pressure. I is vortex stretching, II is dilatation, III is viscous dissipation, and IV is the baroclinic torque (the torque that appears when the density gradient and the pressure gradient are misaligned).

In cold turbulence II and IV are zero. Add a flame and both come alive. At the front $\nabla \cdot \mathbf{u} \gt 0$, so II removes vorticity. IV, on the other hand, creates new vorticity. The $\nabla \rho$ across the front is large and normal to the surface, while the $\nabla p$ from the acceleration points elsewhere. Their cross product is nonzero. That is the source of what is called flame-generated turbulence.

A density ratio near 7 is not only a speed story. With fresh-gas $S_L = 0.40$ m/s the gas leaves the front at about 2.8 m/s. Once temperature jumps from 300 K past 2000 K, kinematic viscosity follows $T^{1.7}$ and grows by more than 20, so the Reynolds number $u' l_t / \nu$ drops by the same factor. Term II removes vorticity where the gas expands. Term IV puts vorticity back in proportion to how far $\nabla\rho$ and $\nabla p$ miss each other.

This feedback carries an annoying side effect. It makes the "$u'$" in an $S_T(u')$ curve ambiguous. Measured in the fresh gas, in the burnt gas, or across the whole flame brush, it is three different numbers. A brush average taken without density weighting counts the intermittent switching between fresh and burnt states as fluctuation and comes out inflated. Models rarely make this distinction. It is one reason RANS codes show non-physical behavior near a flame.

Drawing the bending curve in Python#

Here is the model above transcribed straight into code. It prints the linear and bending predictions side by side at two turbulence scales.

import math
 
SIGMA_INF = 6.0     # wrinkling saturation ceiling
KA_QUENCH = 25.0    # quench Karlovitz number
 
 
def damkohler_number(u_ratio, l_ratio):
    """Da = turbulent time / chemical time."""
    return l_ratio / u_ratio
 
 
def karlovitz_number(u_ratio, l_ratio):
    """Ka = chemical time / Kolmogorov time."""
    return u_ratio**1.5 * l_ratio**-0.5
 
 
def quench_factor(ka):
    """Quench attenuation factor, falling to 0 as Ka grows."""
    return math.exp(-(ka / KA_QUENCH) ** 2)
 
 
def wrinkling_factor(u_ratio, l_ratio):
    """Sigma = S_T / S_L, with saturation and quenching."""
    ka = karlovitz_number(u_ratio, l_ratio)
    return 1.0 + quench_factor(ka) * u_ratio / (1.0 + u_ratio / SIGMA_INF)
 
 
def turbulent_flame_speed(s_laminar, u_ratio, l_ratio):
    """Turbulent flame speed [m/s]."""
    return s_laminar * wrinkling_factor(u_ratio, l_ratio)
 
 
U_RATIOS = [0.5, 1, 2, 3, 4, 6, 8, 10, 12]
 
for l_ratio in (10.0, 2.0):
    print(f"l_t/delta_L = {l_ratio:.0f}")
    print(f"{'u/SL':>6}{'Da':>9}{'Ka':>9}{'linear':>9}{'bending':>9}")
    for u_ratio in U_RATIOS:
        da = damkohler_number(u_ratio, l_ratio)
        ka = karlovitz_number(u_ratio, l_ratio)
        lin = 1.0 + u_ratio                      # Damkohler linear law
        ben = wrinkling_factor(u_ratio, l_ratio)  # saturation + quenching
        print(f"{u_ratio:6.1f}{da:9.2f}{ka:9.2f}{lin:9.2f}{ben:9.2f}")
    print()
 
print("S_T at S_L = 0.40 m/s, u'/S_L = 6, l_t/delta_L = 10 :",
      round(turbulent_flame_speed(0.40, 6.0, 10.0), 3), "m/s")

The output looks like this.

l_t/delta_L = 10
  u/SL       Da       Ka   linear  bending
   0.5    20.00     0.11     1.50     1.46
   1.0    10.00     0.32     2.00     1.86
   2.0     5.00     0.89     3.00     2.50
   3.0     3.33     1.64     4.00     2.99
   4.0     2.50     2.53     5.00     3.38
   6.0     1.67     4.65     7.00     3.90
   8.0     1.25     7.16     9.00     4.16
  10.0     1.00    10.00    11.00     4.20
  12.0     0.83    13.15    13.00     4.03
 
l_t/delta_L = 2
  u/SL       Da       Ka   linear  bending
   0.5     4.00     0.25     1.50     1.46
   1.0     2.00     0.71     2.00     1.86
   2.0     1.00     2.00     3.00     2.49
   3.0     0.67     3.67     4.00     2.96
   4.0     0.50     5.66     5.00     3.28
   6.0     0.33    10.39     7.00     3.52
   8.0     0.25    16.00     9.00     3.28
  10.0     0.20    22.36    11.00     2.68
  12.0     0.17    29.39    13.00     2.00
 
S_T at S_L = 0.40 m/s, u'/S_L = 6, l_t/delta_L = 10 : 1.559 m/s

Set $u_\theta/S_L$ near 2 and press Release: the vortex breaks on the front and opposite vorticity appears. Toward 10 the vortex crosses and only a cusp remains. Drop $\rho_u/\rho_b$ to 1 and $\nabla\rho$ vanishes, so the baroclinic term switches off.

Compare the two columns and the parting point shows up. Out to $u'/S_L = 1$ the linear and bending values agree within 15%. By $u'/S_L = 6$ they are already 7.00 against 3.90. And at the large scale ($l_t/\delta_L = 10$) pushing to $u'/S_L = 12$ still leaves $\Sigma$ near 4. At the small scale ($l_t/\delta_L = 2$), the same $u'$ drives $Ka$ past 29 and $\Sigma$ down to 2.0. Same turbulence intensity, but the smaller scale quenches first.

Drop Ka and the same u' hides the quench#

One cell at $u'/S_L = 12$ holds both $\Sigma = 4.03$ and $2.00$. What splits them is $l_t/\delta_L$, and the Karlovitz numbers are 13.15 and 29.39. The linear law writes 13.00 for both. If $S_T$ stalls on a burner and the reaction constants get turned first, a capped area gets read as slow chemistry.

When a RANS field looks wrong near a flame, where $u'$ is sampled matters more than the model constants. Fresh gas, burnt gas, and a brush average are three different numbers. A Germano average that drives the dynamic coefficient negative is the same kind of averaging trap. $\Sigma_\infty = 6$ and $Ka_q = 25$ were tuned for this burner. They are not constants to carry to the next fuel.