3.3.33 · D5Rocket Propulsion
Question bank — Acoustic modes in combustion chamber — cause of instability
For the machinery behind these traps, see Standing waves and resonance, Speed of sound in gases, and Rayleigh Criterion.
Symbols this bank uses (read this first)
Before the traps, every symbol below appears somewhere in the answers. None of them are assumed — here is exactly what each one means.
True or false — justify
A rigid chamber wall forces a pressure node.
False — a rigid wall forces gas velocity to zero, which is a pressure antinode (maximum). Blocking motion piles pressure up at the wall, it doesn't cancel it.
A bigger heat release always makes the engine more unstable.
False — the phase decides the sign of energy transfer via ; amplitude only sets how fast an already-growing mode grows. Heat released out of phase actively damps, no matter how large.
If the Rayleigh integral is positive, the mode is guaranteed to grow.
False — positive is the drive, but real chambers also have losses (nozzle radiation, viscosity). Growth needs drive to exceed losses, not just be positive.
Using the cold-air value gives a reasonable chamber mode frequency.
False — combustion products sit near , so –. Since , cold-air underestimates the frequency by roughly a factor of three.
Tangential modes are usually the most destructive.
True — the pressure antinode of a tangential mode sits on the chamber wall and sweeps around the injector face, riding directly over the densest ring of injector elements, so it couples to combustion far more strongly than an axial mode whose antinode is a flat sheet inside the gas.
Heat added exactly out of phase with pressure neither drives nor damps the mode.
True — the cycle-averaged drive goes as , and , so the net work over a cycle is exactly zero (neutral).
The fundamental longitudinal mode fits one full wavelength in the chamber.
False — the fundamental () fits half a wavelength between the two pressure antinodes; comes from .
Two closed ends and two open ends give the same longitudinal frequency formula.
True (in form) — both ends alike gives with ; the difference is where the antinodes vs nodes sit, not the spacing. A mixed (one-closed-one-open) tube is what changes the formula.
A one-closed, one-open tube uses the same as a symmetric tube.
False — a mixed tube fits only odd quarter-wavelengths, giving (): a pressure antinode at the closed end and a node at the open end. Its fundamental is half that of the closed-closed tube.
Spot the error
"At the injector face the gas rushes in, so that must be a pressure node."
The reflecting face behaves closer to a rigid wall (velocity ≈ 0), giving a pressure antinode, not a node. The steady inflow is the mean flow, not the acoustic velocity fluctuation .
"Since , doubling the temperature doubles the mode frequency."
No — , so doubling multiplies (and ) by only , not .
"A shorter chamber lowers the mode frequency because there's less gas to shake."
Backwards — , so smaller raises . A shorter organ pipe plays a higher note.
" means the heat and pressure fight each other, which shakes the engine harder."
They do oppose, but opposing means heat is removed from the wave when pressure is high — gives damping, the calmest case, not the harshest.
"The wave equation only holds for large, violent pressure swings."
Opposite — it comes from linearizing about a mean state, so it holds for small perturbations . Genuinely large swings need the full nonlinear equations.
"Transverse (tangential/radial) modes depend on chamber length ."
They depend on the radius via Bessel roots (); only the longitudinal modes are set by .
"Because the drive scales as , a phase of is unstable."
, so it's neutral, not unstable. Only phases with (roughly ) drive growth.
"The wavenumber and frequency are the same thing."
No — counts wave-cycles per metre (space), while frequency counts cycles per second (time). They are linked by , but is a spatial quantity.
Why questions
Why does a combustion chamber have discrete "notes" at all rather than every frequency?
Only wavelengths that reflect off the boundaries and return in step survive as standing waves; all others interfere destructively. The boundary condition (a whole number of half-wavelengths fitting ) selects a discrete set.
Why is the phase between heat release and pressure, not the amount of heat, the deciding factor for stability?
The cycle-averaged energy fed to the wave is ; if heat arrives out of step, positive and negative contributions cancel or reverse regardless of how big each burst is.
Why does adding heat at a pressure peak amplify the oscillation?
It's like pushing a swing at the top of its arc — energy enters exactly when the wave is most "stretched" ( maximal, ), so each cycle returns with more amplitude (constructive drive).
Why do engineers push the phase toward with baffles and acoustic cavities?
To make negative, so combustion removes acoustic energy each cycle (see Injector design and baffles and Nozzle flow and acoustic damping) — turning a potential driver into a damper.
Why does the nozzle throat matter for stability even though instability starts in the flame zone?
The throat partly reflects waves (setting the boundary condition and mode shape) and lets acoustic energy radiate out — it's a key loss channel that must outpace the Rayleigh drive.
Why can two engines of identical size have different mode frequencies?
Because and depends on temperature and composition (, ); different propellants or mixture ratios shift and thus every mode.
Why do tangential modes couple to combustion more strongly than radial or longitudinal ones?
Their pressure antinode rides along the outer wall and rotates past the ring of injector elements where fresh propellant enters, so the pressure fluctuation directly modulates the burning where combustion is most sensitive — maximizing the overlap in the Rayleigh integral.
Edge cases
If the heat-release fluctuation is exactly zero everywhere, can the mode be unstable?
No — with no fluctuating heat there is nothing to feed the wave; the Rayleigh integral is zero and any acoustic energy only decays through losses.
What happens to the mode frequencies in the limiting case where the chamber length ?
— the modes push to arbitrarily high frequency and effectively out of the dangerous structural band; a vanishing chamber has no low acoustic resonance.
If drive and losses are exactly equal, what is the mode's behaviour?
It sits at the neutral (marginal) stability boundary — the amplitude neither grows nor decays, holding a steady limit oscillation. Any small extra drive tips it unstable.
For a phase of exactly , is the mode safe forever?
Only marginally — the drive is zero so it doesn't grow, but it's not being damped by combustion either; real losses then decide, and any drift in toward makes it unstable.
If the mean gas were truly at rest and perfectly lossless with in-phase heat, what does the model predict?
Unbounded exponential growth — with positive Rayleigh drive and zero losses nothing checks the mode, so the linear model blows up (real nonlinearities eventually cap it at a limit cycle).
At a pressure node, what is the acoustic velocity doing?
It is at its antinode (maximum) — pressure and velocity are a quarter-wave apart, so where pressure vanishes the gas is sloshing fastest.
Recall One-line summary of every trap
Primes = ripples (); node = never moves, antinode = biggest swing; wall forces velocity zero → pressure antinode; , ; stability set by phase via , not amplitude; use hot m/s; (closed-closed), (mixed); tangential modes bite hardest.