Acoustic modes in combustion chamber — cause of instability: Worked examples

Before we start, a two-line reminder of the symbols so nothing is used unearned:

The scenario matrix

Every problem this topic throws lands in one of these cells. The examples below are labelled with the cell they cover.

#	Cell class	What makes it tricky	Covered by
A	Baseline longitudinal $f_{n}$	plug-and-go fundamental	Ex 1
B	Overtone ( $n > 1$ )	which "note" is $n$ ?	Ex 2
C	Hot vs cold gas ( $c$ changes)	recompute $c$ from $T$	Ex 3
D	Transverse (tangential) mode	Bessel root, uses $R$ not $L$	Ex 4
E	Rayleigh sign — in phase	$cos ϕ > 0$ , grows	Ex 5
F	Rayleigh sign — out of phase	$cos ϕ < 0$ , damps	Ex 5
G	Degenerate phase $ϕ = 9 0^{\circ}$	exactly neutral, $cos ϕ = 0$	Ex 6
H	Limiting input: $L \to \infty$ , $L \to 0$	frequency $\to 0$ or blows up	Ex 7
I	Real-world word problem	avoid a structural resonance	Ex 8
J	Exam twist: gain vs loss balance	drive must beat damping	Ex 9

Worked examples

Worked example Example 3 — Hot gas raises the danger (Cell C)

A chamber $L = 0.50 m$ runs at $T = 2800 K$ . The gas has $γ = 1.20$ , $R_{univ} = 8.314 J mol^{- 1} K^{- 1}$ , molar mass $M = 0.022 kg/mol$ . Find $c$ , then $f_{1}$ . Compare to using cold air at $c = 340 m/s$ .

Forecast: hot combustion gas → $c \approx 1000 - 1300 m/s$ , several times the cold value.

Speed of sound: $c = \frac{γ R _{univ} T}{M}$ . Why this step? This is the Speed of sound in gases result — a squeeze travels faster in a stiffer, hotter, lighter gas. We need the actual hot-gas $c$ , not room-air $c$ .
$c = \frac{1.20 \cdot 8.314 \cdot 2800}{0.022} = \frac{27 935}{0.022} = 1.2698 \times 1 0^{6} \approx 1127 m/s$ . Why this step? Plug the hot numbers so the frequency reflects reality.
Hot mode: $f_{1} = \frac{1127}{2 ( 0.50 )} = 1127 Hz$ .
Cold (wrong) estimate: $f_{1}^{cold} = \frac{340}{1.0} = 340 Hz$ . Why this step? Shows the classic error: cold air underestimates by $1127/340 \approx 3.3 \times$ .

Verify: ratio $c / c_{cold} = 1127/340 \approx 3.3$ , matching the parent note's " $\sim 3 \times$ " warning ✓. Units of $c$ : $\frac{( J/mol/K ) ( K )}{kg/mol} = \frac{J}{kg} = m^{2} / s^{2} = m/s$ ✓.

Worked example Example 4 — A tangential (spinning) mode (Cell D)

A cylindrical chamber has radius $R = 0.15 m$ , gas $c = 1000 m/s$ . The first tangential mode uses Bessel-derivative root $α_{11} = 1.841$ . Find its frequency.

Forecast: it's set by $R$ not $L$ ; expect a few kilohertz.

Use the transverse formula $f_{mn}^{trans} = \frac{α _{mn} c}{2 π R}$ . Why this step? A spinning wave loops around the axis, so the "length it must fit into" is set by the circumference-scale $2 π R$ , not the axial length $L$ . The Bessel root $α_{11}$ plays the role $nπ$ played for the straight tube.
$f_{11} = \frac{1.841 \cdot 1000}{2 π ( 0.15 )} = \frac{1841}{0.9425} \approx 1953 Hz$ . Why this step? Arithmetic once the geometry is chosen.

Verify: units $\frac{m/s}{m} = Hz$ ✓. Tangential modes are the Rayleigh Criterion-dangerous "most destructive" ones — worth locating precisely. See Standing waves and resonance for why only special patterns survive.

Worked example Example 5 — Rayleigh sign, both directions (Cells E & F)

A pressure sensor reads $p^{'} = cos ω t$ . Two designs are tested: (E) heat lags by $ϕ = 3 0^{\circ}$ ; (F) heat lags by $ϕ = 17 0^{\circ}$ . Which grows, which damps?

Forecast: small lag = "almost in phase" = grows; near- $18 0^{\circ}$ = "almost anti-phase" = damps.

The energy gained per cycle $\propto \overline{p^{'} q^{'}} = \frac{1}{2} cos ϕ$ . Why this step? This is the parent's Rayleigh Criterion result: the time-average of two cosines separated by $ϕ$ is $\frac{1}{2} cos ϕ$ . Only the sign of $cos ϕ$ decides growth vs decay.
Case E: $cos 3 0^{\circ} = 0.866 > 0$ → positive → mode gains energy → unstable.
Case F: $cos 17 0^{\circ} = - 0.985 < 0$ → negative → mode loses energy → stable / damped.

Verify: signs flip across $ϕ = 9 0^{\circ}$ as expected ( $cos$ positive below $9 0^{\circ}$ , negative above) ✓. The phase-versus-drive picture:

Worked example Example 7 — Limiting inputs (Cell H)

(a) As chamber length $L \to \infty$ with $c$ fixed, what happens to $f_{1}$ ? (b) As $L \to 0$ , what happens? Sanity-check with numbers using $c = 1000 m/s$ : $L = 2 m$ and $L = 0.01 m$ .

Forecast: long tube → low bass note → $f \to 0$ ; tiny tube → shrill → $f \to \infty$ .

From $f_{1} = \frac{c}{2 L}$ , take $L \to \infty$ : $f_{1} \to 0$ . Why this step? A wave with half-wavelength stretched over a huge chamber oscillates very slowly — the note drops toward silence.
Take $L \to 0$ : $f_{1} \to \infty$ . Why this step? Squeezing a half-wave into a vanishing length forces an ultra-high pitch; physically the model breaks before this, but the trend is the warning.
Numbers: $L = 2 m \Rightarrow f_{1} = 1000/4 = 250 Hz$ ; $L = 0.01 m \Rightarrow f_{1} = 1000/0.02 = 50000 Hz$ . Why this step? Confirms the two limits point the right way — big drop, big rise.

Verify: $1000/4 = 250$ ✓ and $1000/0.02 = 50000$ ✓; longer $L$ gave lower $f$ , shorter gave higher $f$ ✓.

Worked example Example 8 — Real-world word problem (Cell I)

An engine's turbopump feed line has a mechanical resonance at $2000 Hz$ . The chamber runs at $c = 1100 m/s$ . The designer must choose a chamber length $L$ so that no longitudinal mode $f_{1}, f_{2}, f_{3}$ sits within $\pm 50 Hz$ of $2000 Hz$ . If $L = 0.55 m$ is proposed, is it safe?

Forecast: compute the first few modes and see if any lands near $2000 Hz$ .

$f_{1} = \frac{1100}{2 ( 0.55 )} = \frac{1100}{1.10} = 1000 Hz$ . Why this step? The fundamental sets the whole ladder; every overtone is a multiple of it.
$f_{2} = 2000 Hz$ , $f_{3} = 3000 Hz$ . Why this step? Overtones are $n \times f_{1}$ for a both-ends-closed chamber.
Check the danger band $2000 \pm 50 Hz = [1950, 2050]$ : $f_{2} = 2000 Hz$ sits dead centre → NOT safe. The second mode would couple straight into the feed-line resonance. Why this step? This is exactly the overlap the parent note warns designers to avoid.

Verify: $2 \times 1000 = 2000$ , and $2000 \in [1950, 2050]$ ✓ — the proposal fails. A fix: nudge $L$ so $f_{2}$ clears the band, or use Injector design and baffles / Nozzle flow and acoustic damping to add loss at that frequency.

Worked example Example 9 — Exam twist: gain must beat loss (Cell J)

A mode has a Rayleigh gain proportional to $G = cos ϕ$ with $ϕ = 4 0^{\circ}$ , but the nozzle and viscous damping remove energy at a rate proportional to $D = 0.90$ (same units). Net growth requires $G > D$ . In phase-in ( $ϕ = 4 0^{\circ}$ ), is the mode unstable? What is the largest $ϕ$ that still barely goes unstable?

Forecast: $cos 4 0^{\circ}$ is about $0.77$ ; if damping is $0.90$ the drive may not win.

Gain: $G = cos 4 0^{\circ} = 0.766$ . Why this step? Rayleigh drive depends only on the phase, per Ex 5.
Compare to loss $D = 0.90$ : since $0.766 < 0.90$ , net energy change is negative → stable, even though heat is in-phase-ish. Why this step? Instability is a race: positive Rayleigh integral is necessary but must also exceed all losses. This is the subtlety the parent flags — amplitude/phase alone isn't the last word.
Break-even phase: solve $cos ϕ = 0.90 \Rightarrow ϕ = arccos (0.90) = 25. 8^{\circ}$ . Why this step? Any lag smaller than $25. 8^{\circ}$ makes $G > D$ and the mode grows; larger lag is safe.

Verify: $cos 4 0^{\circ} = 0.766 < 0.90$ ✓ (stable); $arccos (0.90) \approx 25.8 4^{\circ}$ , and $cos (25.8 4^{\circ}) \approx 0.90$ ✓.

Recall Quick self-test across the matrix

Fundamental of $L = 0.40 m$ , $c = 960 m/s$ ? ::: $1200 Hz$ . Third mode of that same chamber? ::: $3600 Hz$ (= $3 f_{1}$ ). Hot-gas $c$ for $γ = 1.20$ , $T = 2800 K$ , $M = 0.022$ ? ::: $\approx 1127 m/s$ . Is $ϕ = 3 0^{\circ}$ driving or damping? ::: Driving ( $cos 3 0^{\circ} > 0$ ). Is $ϕ = 17 0^{\circ}$ driving or damping? ::: Damping ( $cos 17 0^{\circ} < 0$ ). What happens to $f_{1}$ as $L \to \infty$ ? ::: It goes to $0$ (very low note). With gain $cos 4 0^{\circ}$ vs loss $0.90$ , stable or not? ::: Stable, because $0.766 < 0.90$ .