Exercises — Premixed vs diffusion flames

(a) candle = diffusion — wax vapour and air are separate and meet only at the flame surface. (b) SI petrol engine = premixed — fuel and air are blended in the intake before the spark. (c) Diesel = diffusion — liquid fuel is sprayed into already-hot air and burns as it mixes. (d) yellow closed-hole Bunsen = diffusion — no air is pre-entrained, so fuel mixes with room air at the flame (soot makes it yellow).

Only the premixed flame has a meaningful $S_L$. It is a self-propagating wave: the reactants are already blended, so the flame can march into fresh mixture at speed $S_L$. A diffusion flame does not travel into anything — the fuel and oxidizer are separate, so there is no pre-mixed gas to burn into. Its "rate" is set by how fast mixing delivers reactants, not by a propagation speed.

We use the derived scaling from the parent note, $\delta \sim \frac{\alpha }{S_L}$ (a wave sweeps thickness $\delta$ in the conduction time ${\delta }^2/\alpha$; matching speeds gives this). $\delta \sim \frac{2\times 10^{-5}}{0.40}=5\times 10^{-5}\text{m}=50\mu \text{m}.$ The reaction zone is tens of microns wide — that is why we model flames as surfaces.

$\delta \sim \frac{2\times 10^{-5}}{0.43}=4.65\times 10^{-5}\text{m}\approx 46.5\mu \text{m}.$ It is thinner than the $50\mu \text{m}$ case. Look at $\delta \sim \alpha /S_L$: with $\alpha$ fixed, a larger $S_L$ means a smaller $\delta$. Faster chemistry burns the fuel before heat can spread far ahead, so the whole zone is squeezed narrower.

Count masses on each side. One mole of ${\text{CH}}_4$ has mass $12+4(1)=16\text{g}$. It needs $2$ moles of ${\text{O}}_2$, mass $2\times 32=64\text{g}$. $s=\frac{{\text{mass O}}_2}{{\text{mass CH}}_4}=\frac{64}{16}=4{\text{g O}}_2/{\text{g CH}}_4.$

First $Y_{O,2}/s=0.233/4=0.05825$. Then $Z_{st}=\frac{0.05825}{1+0.05825}=\frac{0.05825}{1.05825}\approx \mathrm{0.055.}$ Only about 5.5 % of the local mass at the flame came from the fuel stream. So the flame sits deep on the air side of the mixing layer (small $Z_{st}$). Everything at $Z>Z_{st}$ (between flame and fuel core) is fuel-rich and oxygen-starved — exactly the zone where fuel pyrolyses into soot. This is why diffusion flames are tall, yellow, and sooty. See Soot formation and the rich premixed zone.

Two timescales. (1) Heat must conduct across the zone of thickness $\delta$: diffusion time $t_{diff}\sim {\delta }^2/\alpha$. (2) The gas sitting in the zone must finish reacting: residence time $t_{res}=\delta /S_L$ must match the chemical time, $t_{res}\sim t_{chem}$. From (1), a wave crossing $\delta$ in $t_{diff}$ moves at $S_L\sim \delta /t_{diff}=\alpha /\delta$, so $\delta \sim \alpha /S_L$. From (2), $\delta \sim S_L{t}_{chem}$. Set the two $\delta$ equal: $\frac{\alpha }{S_L}\sim S_L{t}_{chem}\Rightarrow S_L^2\sim \frac{\alpha }{t_{chem}}\Rightarrow \boxed{S_L\sim \sqrt{\alpha /t_{chem}}},\delta \sim \sqrt{\alpha t_{chem}}.$ Halving $t_{chem}$: $S_L\propto t_{chem}^{-1/2}$, so $S_L$ multiplies by $(1/2{)}^{-1/2}=\sqrt{2}\approx 1.41$ — a 41 % increase. $\delta \propto t_{chem}^{+1/2}$, so $\delta$ multiplies by $(1/2{)}^{1/2}=1/\sqrt{2}\approx 0.707$ — a 29 % decrease. The flame gets faster and thinner, consistent with $\delta \sim \alpha /S_L$.

At a stabilised flame the fresh gas flows out at speed $v$ while the flame tries to travel in at $S_L$; when $v=S_L$ they cancel and the flame appears frozen in place. (a) $v=0.25<S_L=0.40$: the flame wins the tug-of-war and travels back into the burner — flashback. Dangerous: the flame enters the supply tube. (b) $v=2.0>S_L$ everywhere: the gas outruns the flame at every point, so the flame is pushed off the rim and cannot anchor — blow-off (lift-off then extinction). Stability needs a point where $v=S_L$; real burners rely on the low-speed boundary layer near the rim to provide it.

(a) $t_{chem}\sim \frac{\delta }{S_L}=\frac{5\times 10^{-5}\text{m}}{0.40\text{m/s}}=1.25\times 10^{-4}\text{s}$. (b) Combustor A: $\mathrm{Da}=\frac{10^{-2}}{1.25\times 10^{-4}}=80$. Combustor B: $\mathrm{Da}=\frac{5\times 10^{-5}}{1.25\times 10^{-4}}=0.4$. (c) A has $\mathrm{Da}=80\gg 1$: chemistry is far faster than the flow/mixing — combustion completes easily, robust. B has $\mathrm{Da}=0.4<1$: the flow whisks reactants through faster than they can react, so heat release can't sustain the flame — B is at risk of extinction/blow-out.

Speed and chemistry rate are different things. What limits a flame is its slowest step (its rate-limiting timescale).

• A premixed flame is already mixed, so its only bottleneck is chemistry+conduction: it is reaction-limited, and $S_L\sim \sqrt{\alpha /t_{chem}}$ is a real propagation speed.
• A diffusion flame's chemistry may be fast, but fuel and oxidizer must first find each other. The bottleneck is mixing (transport), so it is transport-limited. It looks "slow/lazy" because it waits on diffusion, not because reactions are slow. In $\mathrm{Da}$ language: a diffusion flame typically runs at large $\mathrm{Da}$ (chemistry fast vs mixing). Push the mixing rate up — increase the scalar dissipation rate $\chi$ — and you shrink $t_{flow}$, dragging $\mathrm{Da}$ down toward 1. Reactants are then diluted and swept apart faster than they release heat, and the flame extinguishes (blows out). Same physics as L5.1's Combustor B, seen from the mixing side.