Premixed vs diffusion flames: Worked examples — Chemistry deep dive

Before anything, let me re-earn the two tools so no symbol appears unexplained.

The scenario matrix

Think of every "case class" this topic can throw. The columns below are the cells; each worked example names the cell it covers.

#	Cell (scenario class)	Degenerate / limit checked	Example
A	Classify a flame (premixed vs diffusion)	mixed device (both regimes at once)	Ex 1
B	$δ$ from $S_{L}$ and $α$ — standard	order-of-magnitude sanity	Ex 2
C	$Z_{s t}$ for a hydrocarbon in air	air-side flame ( $Z_{s t} ≪ 1$ )	Ex 3
D	$Z_{s t}$ limit — pure oxygen oxidizer	$Y_{O, 2} \to 1$ (max)	Ex 4
E	$Z_{s t}$ limit — diluted fuel stream	$Y_{F, 1} < 1$ (degenerate input)	Ex 4b
F	Scaling response: double $α$ , halve $t_{c h e m}$	how $S_{L}$ scales (ratios)	Ex 5
G	Stabilization sign check: flashback vs blow-off	gas speed $=$ , $<$ , $>$ $S_{L}$	Ex 6
H	Real-world word problem: burner rim heat	units, area, power	Ex 7
I	Exam twist: extinction / $χ$ limit (diffusion)	mixing "too fast" degenerate	Ex 8

Every numeric result below is machine-checked in the Verify block.

Worked examples

Worked example Ex 1 — Cell A: Classify a candle

and a lit lighter left burning A candle burns yellow and tall. A butane lighter, held with the flame away from any surface, burns partly blue at the base and yellow at the tip. Classify each region.

Forecast: Guess first — is the yellow tip premixed or diffusion? Is the blue base the same kind as a stove flame?

Ask the one question: were fuel and oxidizer mixed before reaching the flame? Why this step? That single question — from the parent note — is the entire classifier.
Candle: wax melts, vaporizes, rises; it meets air only at the flame surface. No premixing. → Diffusion everywhere. Why this step? Fuel and air start separate, so the flame sits where they diffuse together.
Lighter blue base: butane jets out fast and entrains some air inside the jet before burning → premixed (partially). The high jet speed is what drags air in. Why this step? Entrainment before the reaction = premixing, exactly like a Bunsen with the air hole open.
Lighter yellow tip: leftover fuel that did not get premixed burns in room air → diffusion (sooty, incandescent).

Verify: The rule "yellow = soot from a fuel-rich diffusion zone, blue = premixed" holds for both. One device, both regimes — consistent with the parent's Bunsen discussion ✓. See Soot formation and the rich premixed zone.

Worked example Ex 3 — Cell C:

$Z_{s t}$ for propane in air $C_{3} H_{8} + 5 O_{2} \to 3 CO_{2} + 4 H_{2} O$ . Air $Y_{O, 2} = 0.233$ , pure fuel $Y_{F, 1} = 1$ . Find $Z_{s t}$ .

Forecast: Bigger or smaller than methane's $0.055$ ? Propane needs more oxygen per gram — guess the direction.

Compute $s$ (mass O₂ per mass fuel): molar mass C₃H₈ = 44, five O₂ = $5 \times 32 = 160$ . $s = \frac{160}{44} = 3.636.$ Why this step? $s$ says how many grams of O₂ one gram of fuel devours — it sets where the stoichiometric surface sits.
Plug into $Z_{s t}$ : $Z_{s t} = \frac{0.233/3.636}{1 + 0.233/3.636} = \frac{0.0641}{1.0641} \approx 0.0602.$ Why this step? The formula finds the mass fraction of fuel-origin material at the surface where fuel and O₂ arrive in the ratio $s$ .
Interpret: $Z_{s t} \approx 0.060$ , slightly higher than methane's $0.055$ .

Verify: Propane's $s$ (3.64) is smaller than methane's (4.0) per gram, so it needs slightly less O₂ mass per fuel mass → surface shifts a hair toward the fuel side → larger $Z_{s t}$ . Direction consistent ✓.

Worked example Ex 4 — Cell D:

$Z_{s t}$ limit, pure-oxygen oxidizer (oxy-fuel) Same methane, but now the oxidizer stream is pure O₂: $Y_{O, 2} = 1$ . Keep $s = 4$ , $Y_{F, 1} = 1$ . Find $Z_{s t}$ and compare with air.

Forecast: Air gave $0.055$ . Will pure O₂ push the flame toward the fuel side or the air side?

Substitute the extreme input $Y_{O, 2} = 1$ : $Z_{s t} = \frac{1/4}{1 + 1/4} = \frac{0.25}{1.25} = 0.20.$ Why this step? We are testing the maximum oxidizer richness — the limiting corner of the matrix.
Interpret: $Z_{s t}$ jumps from $0.055$ to $0.20$ — the flame surface moves much closer to the middle, deep away from the air side. Why this step? With no inert nitrogen diluting the O₂, less oxidizer mass is needed at the surface, so more fuel-origin mass is tolerated.

Verify (degenerate endpoint): Let $Y_{O, 2} \to 0$ (no oxygen at all): numerator $\to 0$ , so $Z_{s t} \to 0$ — the flame would need to sit at pure oxidizer, i.e. no flame. Sensible limit ✓. And $Y_{O, 2} = 1$ giving $0.20 > 0.055$ confirms richer oxidizer ⇒ larger $Z_{s t}$ .

Worked example Ex 4b — Cell E:

$Z_{s t}$ with a diluted fuel stream (degenerate input) Now the "fuel" stream is only 50% methane by mass (rest inert): $Y_{F, 1} = 0.5$ . Oxidizer is air $Y_{O, 2} = 0.233$ , $s = 4$ . Find $Z_{s t}$ .

Forecast: Diluting the fuel — does the flame move toward more or less fuel-origin mass?

Substitute $Y_{F, 1} = 0.5$ : $Z_{s t} = \frac{0.233/4}{0.5 + 0.233/4} = \frac{0.05825}{0.55825} \approx 0.1043.$ Why this step? $Z$ counts all mass from the fuel stream, including its inert; a weaker fuel stream must supply more of its mass to deliver the same fuel.
Interpret: $Z_{s t}$ rises from $0.055$ to $0.104$ — nearly doubles.

Verify: As $Y_{F, 1} \to 0$ (no real fuel in the "fuel" stream), denominator $\to Y_{O, 2} / s$ and $Z_{s t} \to 1$ — the surface retreats to the pure fuel-stream side, meaning no reachable flame. Correct limiting behaviour ✓.

Worked example Ex 5 — Cell F: Scaling — how

$S_{L}$ responds to $α$ and $t_{c h e m}$ A design change doubles thermal diffusivity ( $α \to 2 α$ ) and, separately, a hotter mixture halves the chemical time ( $t_{c h e m} \to t_{c h e m} /2$ ). By what factor does $S_{L}$ change in each case, and if both happen together?

Forecast: Guess: does doubling $α$ double $S_{L}$ ? (Careful — there's a square root.)

Recall the form $S_{L} \sim α / t_{c h e m}$ . Why this step? $S_{L}$ is a geometric mean, so factors enter under a square root — ratios, not additions.
Double $α$ : $S_{L} \propto α$ , so factor $= 2 \approx 1.414$ . Why this step? Only $α$ changed; pull it out of the root.
Halve $t_{c h e m}$ : $S_{L} \propto 1/ t_{c h e m}$ , so $t_{c h e m} \to t_{c h e m} /2$ gives factor $2 \approx 1.414$ .
Both together: $S_{L} \to 2 α / (t_{c h e m} /2) = 4 S_{L} = 2 S_{L}$ . Factor $= 2$ . Why this step? Multiply the two independent ratios inside the root: $2 \times 2 = 4$ , root $= 2$ .

Verify: $2 \cdot 2 = 2$ ✓. A designer learns: to double flame speed you can't just tweak one lever a little — you need the product $α / t_{c h e m}$ to quadruple. Links to Damköhler number (the $t_{f l o w} / t_{c h e m}$ ratio).

Worked example Ex 6 — Cell G: Stabilization signs — flashback, anchored, blow-off

A burner has flame speed $S_{L} = 0.40$ m/s. Test three exit gas velocities: $v = 0.25, 0.40, 0.80$ m/s. What happens in each?

Forecast: In which case does the flame crawl back down the tube (dangerous)?

Set up the tug-of-war: flame tries to travel into the gas at $S_{L}$ ; gas flows out at $v$ . Net advance $= S_{L} - v$ . Why this step? The parent's key insight — a "sitting" flame is a cancelled race, not a still object.
Case $v = 0.25 < S_{L}$ : net $= 0.40 - 0.25 = + 0.15$ m/s (flame wins) → flashback (burns back into the tube).
Case $v = 0.40 = S_{L}$ : net $= 0$ → anchored / stationary flame.
Case $v = 0.80 > S_{L}$ : net $= 0.40 - 0.80 = - 0.40$ m/s (gas wins) → flame pushed away; if this holds everywhere → blow-off. Why this step? Covering all three signs of $(S_{L} - v)$ is exactly the "all cases" requirement.

Verify: Signs: $+$ , $0$ , $-$ → flashback, anchored, blow-off. Matches Flashback and blow-off limits ✓.

Worked example Ex 7 — Cell H: Real-world word problem — heat from a burner strip

A stove ring burns methane with laminar flame speed $S_{L} = 0.40$ m/s over a total flame area $A = 8 \times 1 0^{- 4} m^{2}$ . Unburned gas density $ρ_{u} = 1.1 kg/m^{3}$ ; each kg of methane/air mixture releases $q = 2.9 \times 1 0^{6} J/kg$ . Estimate the heat-release power.

Forecast: Kilowatts or megawatts for a kitchen ring? Guess the order.

Mass flow into the flame: $\overset{m}{˙} = ρ_{u} S_{L} A$ . Why this step? The flame consumes a sheet of gas: how much mass crosses per second is (density)(speed)(area) — exactly $S_{L}$ 's physical meaning. $\overset{m}{˙} = 1.1 \times 0.40 \times 8 \times 1 0^{- 4} = 3.52 \times 1 0^{- 4} kg/s .$
Power: $P = \overset{m}{˙} q = 3.52 \times 1 0^{- 4} \times 2.9 \times 1 0^{6} \approx 1021 W \approx 1.0 kW .$ Why this step? Power = mass rate × energy per mass; units $(kg/s) (J/kg) = W$ ✓.

Verify: ~1 kW for a modest stove ring is exactly right (home rings are 1–3 kW). Units and magnitude both pass ✓.

Worked example Ex 8 — Cell I: Exam twist — diffusion flame extinction (mixing "too fast")

A diffusion flame extinguishes when the scalar dissipation rate $χ$ (a measure of mixing intensity, units s⁻¹) exceeds a critical $χ_{q}$ . Given $χ_{q} = 12 s^{- 1}$ , a burner runs at $χ = 9 s^{- 1}$ ; a designer then triples the strain, $χ \to 27 s^{- 1}$ . Does the flame survive each time? Why does more mixing kill a diffusion flame?

Forecast: Premixed flames like fast transport (bigger $α$ ⇒ faster). Does a diffusion flame also like more mixing?

State the rule: survive if $χ < χ_{q}$ . Why this step? A diffusion flame is transport-limited: reactants must dwell long enough to react. $χ$ is inversely a residence time.
Case $χ = 9$ : $9 < 12$ → burns (survives).
Case $χ = 27$ : $27 > 12$ → extinguishes (blows out). Why this step? Too-fast mixing sweeps heat and radicals out of the reaction zone faster than chemistry can replace them — the flame is diluted/quenched, not fed.
Contrast with premixed: there mixing is already done, so faster thermal diffusion helps ( $S_{L} ↑$ ). Different regimes, opposite intuition — that's the trap the exam sets.

Verify: $9 < 12$ (survive) and $27 > 12$ (extinct) ✓. This is the diffusion analogue of blow-off, tied to the Damköhler number: extinction when $D a = t_{mi x} / t_{c h e m}$ drops below a critical value.

Active recall

Recall Test yourself (hide the answers)

Which cell shows the flame surface moving toward the middle, and what caused it? ::: Cell D — pure-oxygen oxidizer ( $Y_{O, 2} = 1$ ) raised $Z_{s t}$ from 0.055 to 0.20. To double $S_{L}$ , by what factor must $α / t_{c h e m}$ change? ::: By a factor of 4 (because of the square root). Gas exits a burner slower than $S_{L}$ — what happens? ::: Flashback: the flame burns back down the tube. Why does more mixing extinguish a diffusion flame but speed up a premixed one? ::: Diffusion is transport-limited (fast mixing sweeps out heat/radicals); premixed is reaction-limited (faster thermal diffusion preheats fresh gas quicker). A diluted fuel stream ( $Y_{F, 1} < 1$ ) does what to $Z_{s t}$ ? ::: Raises it — more fuel-stream mass (including inert) must reach the surface.