Visual walkthrough — Premixed vs diffusion flames

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Step 0 — What is a flame, as a picture?

Before any symbol, look at the thing itself.

A premixed flame is a thin bright sheet with cold fresh gas on the left and hot burned gas on the right. The sheet does not stay still in the gas — it creeps left, eating into the cold gas, one layer at a time. The single number we want is the speed of that creep.

The two temperatures we will keep referring to:

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Step 1 — Freeze the flame by riding along with it

WHAT. Instead of watching the flame move through still gas, we hop onto the flame and let the gas flow past us. Now the flame is stationary and the fresh gas rushes in from the left at speed $S_L$.

WHY. A moving thing is hard to draw and hard to do bookkeeping on. A stationary thing with stuff flowing through it is the easiest possible picture — everything that enters must leave, nothing piles up. This "sit on the wave" trick is the whole reason the algebra will be short.

PICTURE.

In this frame, cold gas enters the left at speed $S_L$ and hot gas exits the right. The flame is now a fixed structure we can measure with a ruler.

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Step 2 — Draw the two zones and name the thickness $\delta$

WHAT. Zoom into the sheet. It is not infinitely thin. It has an inner structure:

1. a preheat zone — gas getting warmer by conduction, not yet reacting;
2. a reaction zone — the actual burning, where fuel turns into heat.

We call the total width $\delta$.

WHY. The flame advances only because heat leaks leftward into cold gas and warms it up to the point where it will ignite. That leaking-forward of heat happens across a finite width. If we do not give that width a name, we cannot write down "how long does heat take to cross it." So we name it $\delta$.

PICTURE.

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Step 3 — The messenger of heat: thermal diffusivity $\alpha$

WHAT. Heat spreads through gas by conduction — hot molecules jostle their cold neighbours. The number that says how fast heat spreads is the thermal diffusivity: $\alpha =\frac{k}{\rho c_p}.$

WHY this tool and not another? We need a single quantity whose units are ${\text{m}}^2/\text{s}$ — "area per time" — because that is exactly the shape of a spreading question: "how much area of gas gets warm per second?" A raw conductivity $k$ alone won't do; it has the wrong units. Dividing by $\rho c_p$ (how much heat it takes to warm a lump) converts "energy flow" into "how fast the warm region grows." That is precisely $\alpha$.

Term by term:

=\frac{\overbrace{k}^{\text{conductivity: how well heat flows}}} {\underbrace{\rho}_{\text{density}}\ \cdot\ \underbrace{c_p}_{\text{heat needed per degree}}}$$ **PICTURE.** ![[deepdives/dd-chemistry-5.3.05-d2-s04.png]] A warm blob planted in cold gas grows outward; after a time $t$ its radius has grown by roughly $\sqrt{\alpha t}$. Big $\alpha$ = fast-growing warm patch = heat reaches the next cold layer sooner. > [!intuition] Why $\sqrt{\alpha t}$ and not $\alpha t$? > Diffusion is a *random-walk* spread, not a straight march. In a random walk the distance covered > grows like the **square root** of time, not linearly. That single fact ($\text{distance}\sim\sqrt{\alpha t}$) > is the seed of the square root in the final flame-speed formula. --- ## Step 4 — Timescale #1: how long heat takes to cross the zone **WHAT.** Turn "distance $\sim \sqrt{\alpha t}$" around. If heat must cross the whole width $\delta$, the time it needs is $$t_{diff}\sim\frac{\delta^2}{\alpha}.$$ **WHY.** Set the spread distance equal to the width: $\delta\sim\sqrt{\alpha\,t_{diff}}$, then square and solve for the time. This $t_{diff}$ is the **heat-delivery clock**: how long before the cold gas just ahead has been warmed up enough to catch fire. Term by term: $$\underbrace{t_{diff}}_{\text{heat-delivery time}} \sim\frac{\overbrace{\delta^2}^{\text{(width)}^2}} {\underbrace{\alpha}_{\text{spread rate}}}$$ **PICTURE.** ![[deepdives/dd-chemistry-5.3.05-d2-s05.png]] Bigger $\delta$ ⇒ longer crossing (and it grows as the *square*, because of the random walk). Bigger $\alpha$ ⇒ shorter crossing. --- ## Step 5 — Timescale #2: how long the fuel takes to burn **WHAT.** The chemistry has its own clock. Once a lump of gas is hot enough, it takes a time $t_{chem}$ to actually complete the reaction: $$t_{chem}\sim\frac{\rho}{\dot\omega}=\frac{1}{\dot\omega/\rho}.$$ **WHY a separate clock?** Heat delivery (Step 4) and burning are two *different* physical events. Delivery is transport; burning is chemistry. They can be fast or slow independently. Here $\dot\omega$ is the reaction rate — mass of fuel consumed per volume per second — so $\dot\omega/\rho$ is "fraction of the local gas consumed per second," and its inverse is "seconds to consume it." Term by term: $$\underbrace{t_{chem}}_{\text{burning time}} \sim\frac{\overbrace{\rho}^{\text{mass available per volume}}} {\underbrace{\dot\omega}_{\text{mass burned per volume per second}}}$$ **PICTURE.** ![[deepdives/dd-chemistry-5.3.05-d2-s06.png]] Two clocks side by side: a **heat clock** ($t_{diff}$) and a **chemistry clock** ($t_{chem}$). The next step is the whole trick — they must read the same time. --- ## Step 6 — The steady-flame condition: the two clocks must match **WHAT.** For a flame to sit still and neither thicken nor collapse, the gas must spend *just enough* time inside the zone to finish burning. The time a gas parcel spends inside the width $\delta$ while flowing through at $S_L$ is the **residence time** $$t_{res}=\frac{\delta}{S_L},$$ and steadiness demands $$t_{res}\sim t_{chem}\quad\text{and}\quad t_{res}\sim t_{diff}.$$ **WHY.** Think of it as a race between the heat clock and the chemistry clock: - If chemistry were *slower* than heat delivery, heat would race ahead, spread out, and never get any one spot hot enough — the flame **quenches**. - If chemistry were *faster*, burning finishes before heat has spread, the zone shrinks — $\delta$ gets thinner until the clocks re-balance. A stable flame is the self-adjusted state where both clocks read the same. This matching is the physical heart of the derivation. **PICTURE.** ![[deepdives/dd-chemistry-5.3.05-d2-s07.png]] > [!intuition] This matching is a [[Damköhler number|Damköhler number]] of one > The ratio $\text{Da}=t_{diff}/t_{chem}$ hovering near 1 *is* the balance we just drew. Premixed > flames are the tidy case where these two clocks lock together. --- ## Step 7 — Solve the two equations for $S_L$ **WHAT.** We now have two independent statements. From $t_{res}\sim t_{diff}$: $$\frac{\delta}{S_L}\sim\frac{\delta^2}{\alpha} \;\;\Rightarrow\;\; S_L\sim\frac{\alpha}{\delta}. \tag{A}$$ From $t_{res}\sim t_{chem}$: $$\frac{\delta}{S_L}\sim t_{chem} \;\;\Rightarrow\;\; \delta\sim S_L\,t_{chem}. \tag{B}$$ **WHY.** We have two unknowns we care about — $S_L$ and $\delta$ — and exactly two equations. That is solvable. We eliminate the one we care about less, $\delta$. **Substitute (A) into (B):** replace $\delta$ in (B) with $\alpha/S_L$ from (A): $$\frac{\alpha}{S_L}\sim S_L\,t_{chem} \;\;\Rightarrow\;\; S_L^2\sim\frac{\alpha}{t_{chem}}.$$ **PICTURE.** ![[deepdives/dd-chemistry-5.3.05-d2-s08.png]] > [!formula] The result, born from two clocks > $$\boxed{\,S_L\sim\sqrt{\dfrac{\alpha}{t_{chem}}}\;=\;\sqrt{\alpha\,\dot\omega/\rho}\,} > \qquad > \boxed{\,\delta\sim\dfrac{\alpha}{S_L}=\sqrt{\alpha\,t_{chem}}\,}$$ > Reading the boxes: > - $S_L$ is the **geometric mean** (square-root of a product) of a *transport* number $\alpha$ and a > *chemistry* rate $1/t_{chem}$. Speed up either clock ⇒ faster flame. > - $\delta$ is the geometric mean of $\alpha$ and $t_{chem}$: faster chemistry ⇒ *thinner* flame. > > The square root is not a coincidence — it is the random-walk $\sqrt{\alpha t}$ of Step 3 surfacing > at the very end. This scaling drives the whole family: it fixes [[Laminar flame speed (S_L)|$S_L$]] itself, sets the [[Flashback and blow-off limits|flashback and blow-off limits]] (compare gas speed to $S_L$), and feeds the [[Diesel vs SI engine combustion|SI-engine flame-propagation]] story. --- ## Step 8 — Edge and degenerate cases (never leave a gap) **WHAT & WHY — walk every limit of the two boxes:** > [!example] Case A — Chemistry gets very slow ($t_{chem}\to\infty$) > $S_L\sim\sqrt{\alpha/t_{chem}}\to 0$ and $\delta\sim\sqrt{\alpha t_{chem}}\to\infty$. > **Picture:** the flame slows to a crawl and smears out into an infinitely fat, faint band — this is > the mathematical face of **quenching**. A flame cooled below its ignition chemistry simply cannot > propagate. > [!example] Case B — Chemistry gets very fast ($t_{chem}\to 0$) > $S_L\to\infty$ (formally) and $\delta\to 0$. **Picture:** the reaction zone collapses toward a true > mathematical surface — the "flame = thin sheet" idealization. In reality $S_L$ saturates because > transport ($\alpha$) can't keep up, but the *trend* is a razor-thin, fast flame. > [!example] Case C — No diffusivity ($\alpha\to 0$) > $S_L\to 0$: with no way to conduct heat forward, no cold gas ever reaches ignition, so the flame > **cannot advance at all**. This proves conduction is the *messenger* — kill the messenger and the > wave dies even with infinitely fast chemistry. > [!example] Case D — A diffusion flame has no $S_L$ at all > Set fuel and oxidizer *apart* (not premixed). There is no fresh premixed gas to "eat into," so the > speed $S_L$ is undefined. Its structure is set by the [[Mixture fraction Z and conserved scalars|mixture fraction $Z$]], > and its life-or-death knob is mixing, not a propagation speed. Push mixing too hard and it > [[Flashback and blow-off limits|blows out]] — the diffusion analogue of quenching. > [!mistake] "Doubling $t_{chem}$ halves $S_L$." > **Why it feels right:** $t_{chem}$ is in the denominator, so more of it should mean less speed — > linearly. **The fix:** it's under a *square root*. Doubling $t_{chem}$ multiplies $S_L$ by > $1/\sqrt2\approx 0.71$, not $0.5$. The random walk of Step 3 softens every dependence to a half-power. --- ## Worked check — put numbers on the boxes > [!example] Methane/air flame thickness > Given $S_L\approx 0.40\ \text{m/s}$ and $\alpha\approx 2\times10^{-5}\ \text{m}^2/\text{s}$: > $$\delta\sim\frac{\alpha}{S_L}=\frac{2\times10^{-5}}{0.40}=5\times10^{-5}\ \text{m}=50\ \mu\text{m}.$$ > Fifty microns — thinner than a hair. That is *why* we are allowed to draw the flame as a line. > [!example] Back out the chemical clock > Using $\delta\sim\sqrt{\alpha\,t_{chem}}$ with the same numbers: > $$t_{chem}\sim\frac{\delta^2}{\alpha}=\frac{(5\times10^{-5})^2}{2\times10^{-5}} > =1.25\times10^{-4}\ \text{s}\approx 0.13\ \text{ms}.$$ > The whole burn happens in about a tenth of a millisecond — the chemistry clock ticks fast, exactly > why premixed flames are "reaction-limited but still quick." --- ## The one-picture summary ![[deepdives/dd-chemistry-5.3.05-d2-s09.png]] Everything on this page is one diagram: cold gas flows in at $S_L$, heat conducts forward over a width $\delta$ (clock $t_{diff}=\delta^2/\alpha$), the gas burns in its own time $t_{chem}$, and the flame is steady exactly when the *residence time* $\delta/S_L$ equals both clocks — which pins $S_L\sim\sqrt{\alpha/t_{chem}}$. > [!recall]- Feynman retelling — the whole walkthrough in plain words > Picture a line of cold gas and a burning wall creeping into it. To move forward, the wall has to > warm up the next slice of cold gas by leaking heat forward — that leaking has a *speed* (that's > $\alpha$, "how fast warmth spreads"). It takes the heat a certain time to cross the warm band — > call it the heat clock. But the gas also needs time to actually *finish burning* once it's hot — > that's the chemistry clock. A flame that sits still and neither fattens nor dies is the one where > both clocks read the same. Write "same time" for both clocks, do a tiny bit of shuffling, and out > pops the answer: the flame speed is the square-root of (spread-speed divided by burn-time). The > square root is there because heat spreads like a wandering drunk — distance grows as the square > root of time. Make the chemistry faster or the gas conduct better, and the flame both speeds up > and gets thinner. Take diffusion away and it dies; take chemistry away and it dies. Beautiful and > symmetric. > [!recall]- Test yourself > - Why do we ride along with the flame instead of watching it move? ::: To make it stationary so "in = out" bookkeeping is trivial. > - What are the two clocks, in words? ::: The heat-delivery clock $t_{diff}=\delta^2/\alpha$ and the burning clock $t_{chem}$. > - What condition makes the flame steady? ::: Residence time $\delta/S_L$ equals both clocks. > - Where does the square root come from physically? ::: From random-walk diffusion, distance $\sim\sqrt{\alpha t}$. > - What happens to $S_L$ and $\delta$ as $\alpha\to 0$? ::: $S_L\to 0$; no heat is carried forward so the flame cannot advance.