Exercises — Combustion instability — low-frequency (chugging), high-frequency (screaming)

150 Hz lands inside the 10–400 Hz chugging band. So it is chugging. The delay-storing reservoir is the feed system (propellant lines + injector pressure drop): the disturbance has to travel through the plumbing and wait out the combustion lag $\tau$, which is a slow clock (milliseconds). A slow clock ⇒ a low frequency. ✔

4200 Hz is in the 1000–15000 Hz band ⇒ screaming (screech). The reservoir is the chamber gas itself, ringing like an organ pipe. Its clock is the acoustic transit time across the chamber — microseconds — a thousand times faster than the feed lag, hence kHz. ✔

Use $f_{chug}\approx \frac{1}{4\tau }$. The factor $\frac{1}{4}$ comes from the quarter-cycle of phase the delayed heat needs to arrive as pressure is rising again. $f_{chug}=\frac{1}{4(3.2\times 10^{-3})}=\frac{1}{0.0128}\approx 78.1\text{Hz}.$ That sits in the chugging band. ✔

$a=\sqrt{\gamma R_{gas}{T}_c}$ — this is the speed at which a pressure blip travels through the hot gas, which is exactly the clock that sets acoustic frequencies. $a=\sqrt{1.20\times 320\times 2800}=\sqrt{1,075,200}\approx 1036.9\text{m/s}.$ ✔

With $\ell =0$ the longitudinal term vanishes, so $f_{1T}=\frac{a}{2}\cdot \frac{{\alpha }_{10}}{\pi R}=\frac{1036.9}{2}\cdot \frac{1.841}{\pi (0.12)}.$ Numerator: $518.45\times 1.841=954.5$. Denominator inside: $\pi (0.12)=0.3770$. $f_{1T}=\frac{954.5}{0.3770}\approx 2532\text{Hz}.$ kHz ⇒ screaming. ✔ See the mode picture below.

Over a full cycle the cross-term integral is proportional to $\cos \varphi$ (the sine-cross terms average to zero). So $\oint p^{\prime }{q}^{\prime }dt\propto \cos 60^{\circ }=+0.5>0.$ Positive ⇒ heat is arriving mostly while pressure is high ⇒ grows. The growth window is $|\varphi |<90^{\circ }$; $60^{\circ }$ is inside it. ✔ See the phase figure.

The integral $\propto \cos \varphi$ changes sign where $\cos \varphi =0$, i.e. at $\varphi =90^{\circ }$ (and again at $270^{\circ }$). So the growth window is $0\le \varphi <90^{\circ }$ and $270^{\circ }<\varphi \le 360^{\circ }$; the damping window is $90^{\circ }<\varphi <270^{\circ }$. At $\varphi =135^{\circ }$: $\cos 135^{\circ }=-0.7071<0$ ⇒ damps. Design goal: shove $\varphi$ into the damping window. ✔

Required minimum: $0.2\times 6.0=1.2$ MPa. We have only $0.9$ MPa, so it fails the rule — the injector is too "soft," letting ${\dot{m}}_{in}$ chase $p_c$ and closing the chug feedback loop. Shortfall $=1.2-0.9=0.3$ MPa. We must raise $\Delta p_{inj}$ by at least $0.3$ MPa. ✔

(a) Predicted chug: $f=\frac{1}{4\tau }=\frac{1}{4(2.6\times 10^{-3})}=96.2$ Hz. Observed 95 Hz matches within ~1%. ⇒ Yes, this is feed-system chugging keyed to $\tau$. ✔ (b) Required $0.2p_c=0.2(5.0)=1.0$ MPa; we have $0.8$ MPa ⇒ too soft, short by $0.2$ MPa. (c) Stiffen the injector so $\Delta p_{inj}\ge 1.0$ MPa. Why it works via Rayleigh: a stiff injector makes ${\dot{m}}_{in}$ nearly independent of $p_c$, so a pressure dip no longer summons extra propellant. The delayed heat release $q^{\prime }$ stops tracking $p^{\prime }$, the phase overlap collapses, and $\oint p^{\prime }{q}^{\prime }dt$ falls below zero ⇒ the loop can no longer feed itself. ✔

Screaming lives in the chamber acoustics, not the feed loop. Lowering $\tau$ (i) only shifts the chugging clock — it changes $f_{chug}$, not the acoustic mode $f_{1T}$ (which depends on $a$, $R$, $L$). So (i) does essentially nothing to screech. Option (ii), Helmholtz cavities tuned to 3000 Hz, sit at the exact frequency and absorb acoustic energy there — they add loss so that $\oint p^{\prime }{q}^{\prime }dt$ goes net-negative for the 1T mode. That is the correct cure. ✔ (For deeper mechanism see Helmholtz Resonator and Injector Baffles.)

From $f_{chug}=\frac{1}{4\tau }$, requiring $f_{chug}\le 60$ Hz gives $\tau \ge \frac{1}{4\times 60}=\frac{1}{240}\approx 4.17\times 10^{-3}\text{s}=4.17\text{ms}.$ Trade-off: a longer lag pushes chugging lower, but a long combustion lag means propellant lingers unburned — poorer mixing, lower c* efficiency, and a larger required chamber volume. So you cannot chase stability purely through $\tau$: the real lever is usually injector stiffness (breaking gain $k$), not slowing the burn. ✔

With $\ell =0$: $f=\frac{a}{2}\cdot \frac{\alpha }{\pi R}$, and $\frac{a}{2\pi R}=\frac{1100}{2\pi (0.10)}=1750.7$ Hz per unit $\alpha$.

• 1T: $1750.7\times 1.841=3223$ Hz.
• 2T: $1750.7\times 3.054=5347$ Hz.
• 1R: $1750.7\times 3.832=6708$ Hz.
• 1L ($\alpha =0,\ell =1$): $f=\frac{a}{2}\cdot \frac{1}{L}=\frac{1100}{2}\cdot \frac{1}{0.30}=1833$ Hz.

Ranked low→high: 1L (1833) < 1T (3223) < 2T (5347) < 1R (6708) Hz. Most dangerous = 1T: its pressure sloshes side-to-side, scrubbing hot gas against the wall and driving the fastest heat loss into the metal. ✔ (See Acoustic Modes of a Cylindrical Cavity.)

$f=\frac{a}{2}\sqrt{{\left(\frac{1.841}{\pi (0.10)}\right)}^2+{\left(\frac{1}{0.30}\right)}^2}.$ Inside: $\frac{1.841}{0.31416}=5.859$, squared $=34.33$. And $\frac{1}{0.30}=3.333$, squared $=11.11$. Sum $=45.44$, root $=6.741$. $f=\frac{1100}{2}\times 6.741=550\times 6.741\approx 3708\text{Hz}.$ Higher than pure 1T (3223 Hz), as adding a longitudinal half-wave must raise it. ✔