Visual walkthrough — Acoustic modes in combustion chamber — cause of instability

We are answering ONE question in pictures:

Why does hot gas in a tube pick out special frequencies, and why can burning fuel make one of them explode in loudness?

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Step 0 — The three words we must earn first

Before any symbol, three plain-word ideas, each with a picture reference coming up.

Everything below is built from just these three.

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Step 1 — A wobble that shoves its neighbour: the wave equation

WHAT. We watch a thin slice of gas. When the pressure on its left is higher than on its right, the slice gets pushed rightward. That push changes how squished the next slice is, which changes its pressure, and so on.

WHY this tool. We need a rule that says "acceleration of gas is caused by a difference in pressure across it." That difference-of-a-difference is exactly what the symbol ${\nabla }^2$ (and in 1-D, the second derivative ${\partial }^2/\partial x^2$) measures — the curvature of the pressure shape. Curvature is the right tool because a straight ramp of pressure gives a uniform push (no net squeezing change), only a bend in the pressure profile makes gas pile up or spread out.

Putting "gas accelerates because of pressure curvature" into symbols gives the wave equation:

$\underset{\text{how fast the squeeze at a point speeds up in time}}{\underbrace{\frac{{\partial }^2{p}^{\prime }}{\partial t^2}}}=\underset{\begin{array}{c}\text{speed of hand-off,}\\ \text{squared}\end{array}}{\underbrace{c^2}}\underset{\text{curvature of the squeeze along the tube}}{\underbrace{\frac{{\partial }^2{p}^{\prime }}{\partial x^2}}}$

• $\frac{{\partial }^2{p}^{\prime }}{\partial t^2}$ — the symbol $\partial /\partial t$ means "rate of change in time, holding the spot fixed"; doing it twice gives acceleration of the pressure at that spot.
• $\frac{{\partial }^2{p}^{\prime }}{\partial x^2}$ — $\partial /\partial x$ means "rate of change as you step along the tube"; twice gives curvature — how much the pressure graph bends.
• $c^2$ — the conversion factor. Bigger $c$ (hotter gas) means the same bend produces a faster response.

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Step 2 — A resonance hums at ONE pitch: separating space and time

WHAT. We guess that a resonance keeps a fixed shape and just breathes in and out at one steady rhythm. So we split $p^{\prime }$ into "a shape that depends only on position" times "a wobble that depends only on time."

WHY. A guitar string vibrating on its lowest note doesn't change shape — the middle just goes up and down. Same here: the pattern is frozen, only its amplitude pulses. Writing that as a product lets us solve the shape and the rhythm separately.

$p^{\prime }(x,t)=\underset{\begin{array}{c}\text{the frozen shape}\\ \text{along the tube}\end{array}}{\underbrace{P(x)}}\times \underset{\begin{array}{c}\text{the in-and-out}\\ \text{breathing in time}\end{array}}{\underbrace{\cos (\omega t)}}$

• $P(x)$ — the standing pattern: tells you how big the wobble is at each position $x$. It never moves.
• $\cos (\omega t)$ — a number that swings smoothly between $+1$ and $-1$ forever. $\cos$ is the tool for anything that repeats evenly, because it is the shadow of a point going round a circle at steady speed.
• $\omega$ (Greek "omega") — the angular frequency: how fast that circle spins, in radians per second. Big $\omega$ = fast breathing = high pitch.

Plug this guess into the wave equation. The time-derivative of $\cos (\omega t)$ twice pulls down a factor $-{\omega }^2$; the space part only touches $P(x)$:

$-{\omega }^{2}P=c^2{P}^{\prime \prime }⟹\underset{\text{shape equation}}{\underbrace{P^{\prime \prime }+k^{2}P=0}},\underset{\text{wavenumber}}{\underbrace{k=\frac{\omega }{c}}}$

• $P^{\prime \prime }$ — curvature of the frozen shape.
• $k$ — the wavenumber: how many radians of wave you fit per metre. Since $k=\omega /c$, a fast wobble ($\omega$) in slow gas (small $c$) crams in more waves.

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Step 3 — The walls decide which cosine survives

WHAT. The chamber has ends. At each end we must satisfy a physical rule, and only special cosines obey it.

WHY the rule "velocity = 0 at a rigid wall." Gas cannot flow through a solid wall. No flow means gas velocity is zero right at the wall. Now here is the key mechanical fact: gas velocity is driven by the slope of pressure (pressure difference pushes gas). Zero velocity therefore means zero slope of pressure — a flat top or flat bottom of the wave. A flat-topped point of a wave is its peak, i.e. a pressure antinode, NOT a node.

We need a shape with zero slope at both ends $x=0$ and $x=L$:

$P(x)=A\cos \left(\underset{\text{how tight}}{\underbrace{k}}x\right),\text{slope }{P}^{\prime }(x)=-Ak\sin (kx)$

• At $x=0$: $\sin (0)=0$ ✓ — cosine automatically starts flat.
• At $x=L$: we need $\sin (kL)=0$. That happens only when $kL$ is a whole number of half-turns:

$kL=n\pi ⟹k_n=\frac{n\pi }{L},n=1,2,3,\dots$

• $n$ — counts how many half-waves fit in the tube. This is the "fits neatly" condition.
• $A$ — just the loudness; it drops out because the equation is happy at any amplitude.

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Step 4 — Turn the fitted shapes into frequencies

WHAT. We now know the allowed $k_n$. Convert each into an actual pitch $f_n$.

WHY. $k$ counts waves per metre; humans and hardware care about wobbles per second. The bridge is $\omega =ck$ (from Step 2) and $\omega =2\pi f$ (a full circle is $2\pi$ radians, and $f$ full circles per second).

${\omega }_n=c{k}_n=\frac{n\pi c}{L},f_n=\frac{{\omega }_n}{2\pi }=\frac{n\pi c}{2\pi L}$

The $\pi$'s cancel, leaving the parent note's headline result:

Link: this is the Standing waves and resonance result specialised to a hot-gas tube.

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Step 5 — Same idea, two extra directions (the mode zoo)

WHAT. A real chamber is a cylinder, not a 1-D line. Waves can also bounce across it and spin around it.

WHY it matters. The axial (length-wise) waves we just built are the Longitudinal modes. But the most dangerous ones slosh sideways.

• Longitudinal (L): along the axis — set by $L$, the mode we derived.
• Tangential (T): spinning around the axis — usually the most destructive, because the peak sweeps the injector face.
• Radial (R): breathing in and out from the centre line.

For the round cross-section the "fits neatly" condition uses Bessel-function roots instead of plain cosines:

$f_{mn}^{\text{trans}}=\frac{{\alpha }_{mn}c}{2\pi R}$

• $R$ — chamber radius.
• ${\alpha }_{mn}$ — the sideways version of $n\pi$: the special numbers where a round drumhead's wave has zero slope at the rim. Same logic (zero velocity at the wall), curved geometry.

You don't need to compute Bessel roots here — just know the pattern is identical: walls impose zero velocity, only certain shapes fit, each fitted shape is one mode.

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Step 6 — Why one mode can EXPLODE: the swing and the flame

WHAT. Now the fuel burns. Burning dumps heat $q^{\prime }$ (fluctuating heat-release). We ask: does that heat feed the wobble or fight it?

WHY the swing analogy is exact. A swing gains height only if you push while it moves in your push's direction. Push at the wrong moment and you slow it. For the acoustic wave, "pushing" means adding heat while the gas is already compressed (pressure high). Heat added to squished gas expands it further — a shove in the direction it's already going.

We measure the energy handed to the wave over one full cycle. Take pressure $p^{\prime }\propto \cos (\omega t)$ and heat lagging by a phase $\varphi$: $q^{\prime }\propto \cos (\omega t-\varphi )$.

$\Delta E\propto {\int }_0^{2\pi /\omega }\underset{\text{pressure}}{\underbrace{\cos (\omega t)}}\underset{\text{heat}}{\underbrace{\cos (\omega t-\varphi )}}dt\propto \cos \varphi$

• The integral ${\int }_0^{2\pi /\omega }\dots dt$ — adds up the push over one whole cycle (the $\int$ symbol just means "grand total of").
• $\varphi$ — the phase lag: how many degrees behind the pressure the heat arrives.
• The whole thing collapses to $\frac{1}{2}\cos \varphi$, because the average of two cosines separated by $\varphi$ is $\frac{1}{2}\cos \varphi$.

Reading the $\cos \varphi$ dial:

$\varphi$	$\cos \varphi$	Meaning
$0^{\circ }$	$+1$	heat exactly on the peak → max drive, unstable
$90^{\circ }$	$0$	heat on the up-swing → neutral
$180^{\circ }$	$-1$	heat on the trough → damps, stable

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Step 7 — The edge cases (never leave the reader stranded)

WHY a whole step. The formula $f_n=nc/2L$ and the dial $\cos \varphi$ hide three degenerate cases you WILL meet.

• $n=0$? Not a mode — $\cos (0)=1$ everywhere means uniform pressure with no wobble, zero frequency. The lowest real mode is $n=1$.
• Cold-gas mistake. Using $c\approx 340$ m/s (room air) instead of $\sim 1000$ m/s (3000 K products) undershoots every $f_n$ by roughly $3\times$. Always use the hot $c=\sqrt{\gamma RT/M}$.
• $\varphi =90^{\circ }$ knife-edge. $\cos 90^{\circ }=0$: the mode neither grows nor decays from heat alone. Tiny extra losses then tip it stable; tiny extra drive tips it unstable — a designer never wants to sit here.
• Amplitude ≠ danger. A giant $q^{\prime }$ at $\varphi =180^{\circ }$ actively stabilises. Big heat with the wrong sign of $\cos \varphi$ is your friend.

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Worked examples (re-derived on the pictures)

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The one-picture summary

The whole chain, left to right: wobble shoves neighbour ($p^{\prime \prime }$) → standing shape fits the walls ($k_n=n\pi /L$) → pitch ($f_n=nc/2L$) → flame's timing ($\cos \varphi$) decides grow or die.

Recall Feynman: the whole walkthrough in plain words

Squeeze the gas in one spot and it shoves its neighbour — that shove travels at the speed of sound, and if the shape bends the gas piles up faster, which is the wave equation. A resonance keeps its shape and just breathes in and out at one steady beat. The walls won't let gas pass through, so the gas can't move right at them — and "no motion at the wall" means the pressure is bunched up highest there, an antinode. Only a few cosine shapes have flat tops exactly at both ends; those are the modes, and turning each one into a pitch gives $f_n=nc/2L$. Longer tube, lower hum; hotter gas, higher hum. Then the fire joins in. If the flame flares brightest exactly when the gas is most squeezed, it's like pushing a swing at the top of its arc — the hum grows and grows until the engine shakes apart. If the flame flares at the wrong moment instead, it kills the swing. The whole story is one dial: $\cos \varphi$, the timing of the flame against the pressure.

Recall Quick self-test

Lowest real mode number? ::: $n=1$ (n=0 is just uniform pressure, no wobble). What is zero at a rigid wall — pressure or velocity? ::: Velocity; that makes pressure an antinode. What single quantity decides grow vs. damp? ::: $\cos \varphi$, the phase of heat vs. pressure. Longer chamber → higher or lower fundamental? ::: Lower (L in the denominator of $f_n$).