Combustion instability — low-frequency (chugging), high-frequency (screaming): Worked examples

Before anything, a reminder of the three clocks and symbols we will re-use, so nothing appears un-earned:

Recall

What does $τ$ mean physically? ::: The combustion/transit lag — the time between propellant entering and finishing burning (milliseconds). It is the slow clock that sets chugging. What does $a = γ R_{g a s} T_{c}$ mean? ::: The speed of sound in the hot chamber gas — the fast clock. $γ$ = ratio of specific heats, $R_{g a s}$ = specific gas constant, $T_{c}$ = chamber temperature. What does the Rayleigh integral $\oint p^{'} q^{'} d t$ measure? ::: Whether the heat pulse $q^{'}$ arrives in phase with the pressure pulse $p^{'}$ over one cycle. Positive → grows, negative → dies.

The scenario matrix

Every question this topic can throw is one of these cells. Each example below is tagged with its cell.

Cell	Case class	What is being tested	Example
A	Rayleigh positive (phase $ϕ < 90°$ )	sign of $\oint p^{'} q^{'} d t$ → growth	Ex 1
B	Rayleigh negative (phase $90° < ϕ < 270°$ )	sign → decay	Ex 2
C	Rayleigh zero (degenerate, $ϕ = 90°$ )	neutral / marginal boundary	Ex 3
D	Low-frequency band	predict $f_{c h ug}$ from $τ$	Ex 4
E	Degenerate feed: $Δ p_{inj}$ limit	injector stiffness fix, ratio test	Ex 5
F	High-frequency band, 1T mode	acoustic mode frequency	Ex 6
G	Limiting behaviour: $ℓ \to$ large / pure longitudinal	mode-index limits	Ex 7
H	Real-world word problem	diagnose an engine from data	Ex 8
I	Exam twist: which cure?	Rayleigh + reservoir reasoning	Ex 9

Cell A — Rayleigh integral positive (growth)

Worked example

Ex 1 (Cell A). Pressure fluctuation $p^{'} = cos ω t$ and heat release $q^{'} = cos (ω t - ϕ)$ with $ϕ = 3 0^{\circ}$ . Does the oscillation grow or die?

Forecast: $3 0^{\circ}$ is a small lag — heat still mostly overlaps high pressure. Guess: grows.

Step 1. The Rayleigh integral over a full cycle for two cosines is $\oint p^{'} q^{'} d t \propto \frac{1}{T} \int_{0}^{T} cos ω t cos (ω t - ϕ) d t = \frac{1}{2} cos ϕ .$ Why this step? The product-to-sum identity $cos A cos B = \frac{1}{2} [cos (A - B) + cos (A + B)]$ splits into a constant $\frac{1}{2} cos ϕ$ and an oscillating $cos (2 ω t - ϕ)$ that averages to zero over a cycle. Only the constant survives — so the whole answer is a clean function of the phase alone.

Step 2. Plug $ϕ = 3 0^{\circ}$ : $\frac{1}{2} cos 3 0^{\circ} = \frac{1}{2} (0.8660) = 0.4330 > 0$ . Why this step? A positive number means heat feeds the swing — instability grows (see figure below, red curve stays mostly above zero).

Verify: $ϕ = 0$ would give $\frac{1}{2}$ (max drive); $ϕ = 9 0^{\circ}$ gives $0$ . Our $3 0^{\circ}$ sits between, closer to max — consistent. Sign is $+$ → grows. ✔

Cell B — Rayleigh integral negative (decay)

Cell C — Rayleigh integral zero (marginal / degenerate)

Worked example

Ex 3 (Cell C). Heat is exactly quadrature to pressure, $ϕ = 9 0^{\circ}$ . What happens?

Forecast: Exactly halfway between grow and die — guess: neither, it just sits there.

Step 1. $\frac{1}{2} cos 9 0^{\circ} = \frac{1}{2} (0) = 0$ . Why this step? At $9 0^{\circ}$ the heat pulse peaks exactly where pressure crosses zero. Over half the cycle it adds energy, over the other half it removes exactly as much — net zero.

Step 2. Interpretation: this is the neutral stability boundary. In the parent's chug analysis this is the "marginal" case that gave $ω τ \approx π /2$ . Why this step? Real engines are pushed away from this line by design; it is the razor's edge that separates the two windows of Cells A and B.

Verify: $cos (9 0^{\circ}) = 0$ exactly, and it is the boundary between the $+$ region (Ex 1) and the $-$ region (Ex 2). Degenerate case handled. ✔

Cell D — Low-frequency band (chugging)

Worked example

Ex 4 (Cell D). A gas-generator engine shows a combustion lag $τ = 4.0$ ms. Predict its instability frequency and name the type.

Forecast: Milliseconds → the slow clock → chugging, so tens–hundreds of Hz. Guess: ~60 Hz.

Step 1. The marginal chug condition from the parent is $ω τ \approx π /2$ , i.e. $f_{c h ug} \approx \frac{1}{4 τ}$ . Why this step? A quarter of the pressure cycle ( $π /2$ of phase) is exactly the delay that carries the extra burning from the pressure dip to the pressure rise — the in-phase condition of Cell A applied to the feed loop.

Step 2. $f_{c h ug} = \frac{1}{4 ( 4.0 \times 1 0 ^{- 3} )} = \frac{1}{0.016} = 62.5$ Hz. Why this step? Plugging the only slow clock we have gives a number, and $62.5$ Hz sits inside the $10$ – $400$ Hz chugging band.

Verify: Units: $1/ s =$ Hz ✔. Band check: $62.5$ Hz $\in [10, 400]$ → chugging confirmed. Cross-link: this uses Injector Design & Pressure Drop for the feed lag origin. ✔

Cell E — Degenerate feed: injector pressure-drop limit

Worked example

Ex 5 (Cell E). Chamber pressure $p_{c} = 7.0$ MPa. Injector drop is $Δ p_{inj} = 0.7$ MPa. Is the feed "stiff enough" by the rule of thumb, and what is the drop ratio?

Forecast: Rule of thumb wants $Δ p_{inj} ≳ 0.2 p_{c}$ . $0.7$ vs $0.2 \times 7 = 1.4$ ? Guess: too soft.

Step 1. Compute the ratio $\frac{Δ p _{inj}}{p _{c}} = \frac{0.7}{7.0} = 0.10$ . Why this step? The Rayleigh Criterion feedback gain $k$ in the chug equation scales inversely with injector stiffness. The dimensionless ratio, not $p_{c}$ alone, sets $k$ — exactly the parent's Mistake #1.

Step 2. Compare to the threshold $0.20$ . We have $0.10 < 0.20$ → the injector is too soft; chugging is likely. The degenerate limit $Δ p_{inj} \to 0$ would make $\overset{m}{˙}_{in}$ perfectly follow $p_{c}$ ( $k \to \infty$ ) — maximum coupling. Why this step? Checking the limit tells us which direction fixes it: raise $Δ p_{inj}$ until the ratio reaches $0.20$ , i.e. $Δ p_{inj} \geq 1.4$ MPa.

Verify: $0.20 \times 7.0 = 1.4$ MPa required; we have $0.7$ MPa, exactly half of the requirement → fails by a factor of 2. Consistent with "too soft". ✔

Cell F — High-frequency band, first tangential mode

Worked example

Ex 6 (Cell F). Chamber: $T_{c} = 3200$ K, $γ = 1.20$ , $R_{g a s} = 320$ J/(kg·K), radius $R = 0.12$ m. Find the first tangential (1T) frequency.

Forecast: Hot gas, small radius → kHz → screaming. Guess: ~3 kHz.

Step 1. Sound speed $a = γ R_{g a s} T_{c} = 1.20 \times 320 \times 3200 = 1.2288 \times 1 0^{6} \approx 1108.5$ m/s. Why this step? The acoustic clock needs the local speed of sound; it is the transit speed of the pressure wave around the Acoustic Modes of a Cylindrical Cavity.

Step 2. 1T mode uses Bessel root $α_{10} = 1.841$ , longitudinal index $ℓ = 0$ : $f_{1 T} = \frac{a}{2} \cdot \frac{α _{10}}{π R} = \frac{1108.5}{2} \cdot \frac{1.841}{π ( 0.12 )} .$ Why this step? With $ℓ = 0$ the general cavity formula collapses to the pure transverse term — a wave that sloshes side-to-side, the most destructive mode.

Step 3. $= 554.25 \times \frac{1.841}{0.37699} = 554.25 \times 4.8834 \approx 2707$ Hz. Why this step? kHz confirms screaming, driven by chamber acoustics — not plumbing.

Verify: Band check: $2707$ Hz $\in [1000, 15000]$ → screaming ✔. Cross-link: cured by Injector Baffles and Helmholtz Resonator cavities. ✔

Cell G — Limiting behaviour: pure longitudinal & large index

Worked example

Ex 7 (Cell G). Same chamber ( $a \approx 1108.5$ m/s), length $L = 0.30$ m. (a) Find the first pure longitudinal mode (transverse root = 0, $ℓ = 1$ ). (b) As $ℓ \to \infty$ , how does $f$ behave?

Forecast: Longitudinal uses length, not radius; $L > R$ so guess lower than the 1T of Ex 6, maybe ~2 kHz. And large $ℓ$ → frequency grows without bound.

Step 1. With the transverse term zero, the cavity formula reduces to $f_{00 ℓ} = \frac{a}{2} \cdot \frac{ℓ}{L} .$ Why this step? Setting $α_{mn} = 0$ kills the radial/tangential contribution, leaving the classic organ-pipe result $f = ℓ a / (2 L)$ .

Step 2. $ℓ = 1$ : $f_{001} = \frac{1108.5}{2} \cdot \frac{1}{0.30} = 554.25 \times 3.333 = 1847.5$ Hz. Why this step? This is the fundamental standing wave down the length of the chamber.

Step 3. Limit: $f_{00 ℓ} = \frac{a}{2 L} ℓ$ is linear in $ℓ$ , so $f \to \infty$ as $ℓ \to \infty$ — higher harmonics are integer multiples of $1847.5$ Hz. Physically, ever-shorter wavelengths are damped by thermoacoustic losses so only the lowest few matter. Why this step? Checking the limit tells us the spectrum is unbounded but only low modes are dangerous — real losses cap the useful set.

Verify: $ℓ = 2$ should be exactly $2 \times 1847.5 = 3695$ Hz (integer harmonic) ✔. And $1847.5 < 2707$ (Ex 6 1T) as forecast, since $L > R$ -scaled length. ✔

Cell H — Real-world word problem

Worked example

Ex 8 (Cell H). An engine test rig hears a steady 90 Hz "put-put" that goes away when the team increases the injector orifice pressure drop. A second, faint 4.2 kHz whistle appears only at full thrust. Diagnose both and back out (a) the combustion lag and (b) confirm the whistle is an acoustic mode if $a = 1150$ m/s, $R = 0.11$ m.

Forecast: 90 Hz + cured by injector = chugging. 4.2 kHz = screaming. Guess $τ \approx 2.8$ ms.

Step 1. 90 Hz is in the low band and responds to injector stiffness → chugging. Invert $f = \frac{1}{4 τ}$ : $τ = \frac{1}{4 f} = \frac{1}{4 ( 90 )} = 2.78 \times 1 0^{- 3}$ s $= 2.78$ ms. Why this step? The cure (injector $Δ p$ ) is the fingerprint of the feed reservoir, so we use the chug clock to recover $τ$ .

Step 2. 4.2 kHz is in the high band → screaming, likely 1T. Predict 1T: $f_{1 T} = \frac{a}{2} \frac{1.841}{π R} = \frac{1150}{2} \cdot \frac{1.841}{π ( 0.11 )} = 575 \times \frac{1.841}{0.34558} = 575 \times 5.328 = 3063$ Hz. Why this step? Predicted 1T ( $\approx 3.06$ kHz) is the same order as the observed 4.2 kHz — the whistle is a chamber acoustic mode, not plumbing (the observed value being higher suggests a coupled 1T+1L mode).

Verify: $τ = 2.78$ ms lies in the physical few-ms range ✔; recomputed $f = 1/ (4 \times 2.78 ms) = 89.9 \approx 90$ Hz ✔. Screaming order-of-magnitude (kHz) confirmed ✔. ✔

Cell I — Exam twist: choose the correct cure

Worked example

Ex 9 (Cell I). Exam: "A chamber screams at its 1T mode. An engineer proposes deepening the propellant feed lines to add lag. Will it help? What actually cures it, and why — argue from Rayleigh."

Forecast: Feed changes touch the slow reservoir; screaming lives in the fast one. Guess: no, wrong reservoir.

Step 1. Identify the reservoir. Screaming = chamber-gas acoustics, clock = acoustic transit (μs). The feed lag $τ$ (ms) is a thousand times too slow to influence it. Why this step? The parent's core lesson: same Rayleigh rule, different time-delay reservoir. Fixing the wrong reservoir does nothing.

Step 2. The real cure makes $\oint p^{'} q^{'} d t$ net-negative by adding acoustic loss at the mode frequency: Injector Baffles chop up the transverse (1T) pattern, and tuned Helmholtz Resonator cavities in the liner absorb energy at $f_{1 T}$ . Why this step? You cannot change the phase of a distributed flame easily, so you attack the magnitude — pile on damping so the negative-loss term overwhelms the positive drive of Cell A.

Verify: Consistency check — this is exactly the parent Mistake "Baffles fix chugging" turned around: baffles/resonators fix screaming, injector $Δ p$ fixes chugging. Reservoir logic matches Ex 4/Ex 5/Ex 6. ✔

Recall

A 250 Hz instability that vanishes when you stiffen the injector is which type? ::: Chugging (low-frequency, feed-system reservoir). A 3 kHz instability requiring baffles and Helmholtz cavities is which type? ::: Screaming (high-frequency, chamber-acoustic reservoir). Heat lags pressure by $6 0^{\circ}$ — grow or die? ::: Grow, since $cos 6 0^{\circ} = + 0.5 > 0$ (inside the $\pm 9 0^{\circ}$ window). Heat lags pressure by $20 0^{\circ}$ — grow or die? ::: Die, since $cos 20 0^{\circ} < 0$ (inside the damping window).