Exercises — Acoustic modes in combustion chamber — cause of instability

WHAT the wall does: metal blocks gas motion, so the oscillating gas velocity $u^{\prime }=0$ at the wall (recall $u^{\prime }$ is the small fluctuating velocity defined above). WHY that means an antinode: in a sound wave the velocity and pressure are a quarter-wave out of step — where velocity is zero, pressure swings hardest. Formally velocity $\propto \partial p^{\prime }/\partial x$, so $u^{\prime }=0$ forces $\partial p^{\prime }/\partial x=0$, which is the slope-zero point of the pressure shape — a peak, i.e. a pressure antinode. Answer: a pressure antinode. (Look at the red dots at both ends of the s01 figure — those are the antinodes.)

WHAT we use: the mode shape derived in the parent note, $P(x)=A\cos (n\pi x/L)$, together with the rigid-wall boundary conditions (a pressure antinode at $x=0$ and at $x=L$). WHY $n=1$ for one half-wave: the boundary conditions demand a peak of $|P|$ at both walls. The cosine $\cos (n\pi x/L)$ automatically has a peak at $x=0$ (its value is $A$) and a peak at $x=L$ only when $n\pi$ is a whole multiple of $\pi$, i.e. $n=1,2,3,\dots$. Each increase in $n$ adds one more half-wave between the walls. The smallest allowed pattern — one antinode at each end with a single sign-flip (trough) in the middle — is exactly half a wavelength across the chamber, and that is the $n=1$ case. So "one half-wave fits" and "$n=1$" are the same statement read two ways. Answer: $n=1$, the fundamental (lowest possible frequency). Higher $n$ squeeze more half-waves into the same length.

WHAT: plug into $f_n=nc/(2L)$ with $n=1$. $f_1=\frac{1\times 980}{2\times 0.40}=\frac{980}{0.80}=1225\text{Hz}$ WHY the $2L$: the fundamental fits half a wave in length $L$, so the wavelength is $2L$; frequency is $c$ divided by wavelength. Answer: $f_1\approx 1225\text{Hz}$.

$f_3=\frac{3\times 980}{2\times 0.40}=3\times 1225=3675\text{Hz}$ WHY: longitudinal modes are exact integer multiples of the fundamental, so $f_3=3f_1$. Answer: $f_3\approx 3675\text{Hz}$.

WHAT: use $c=\sqrt{\gamma RT/M}$ (see Speed of sound in gases). $c=\sqrt{\frac{1.20\times 8.314\times 3000}{0.022}}=\sqrt{\frac{29930}{0.022}}=\sqrt{1.360\times 10^6}\approx 1166\text{m/s}$ WHY this formula: in a gas, a pressure pulse propagates by the springiness ($\gamma$, pressure) fighting the inertia ($M$, mass). Hotter and lighter gas ⇒ faster sound. Answer: $c\approx 1166\text{m/s}$.

WHAT: the cycle-average (the overline, defined above) of two cosines with a phase difference $\varphi$ is $\overline{p^{\prime }{q}^{\prime }}=\frac{1}{2}\cos \varphi .$ WHY: expand $\cos \omega t\cos (\omega t-\varphi )=\frac{1}{2}[\cos \varphi +\cos (2\omega t-\varphi )]$; the fast $\cos (2\omega t-\varphi )$ term averages to zero over a full cycle, leaving $\frac{1}{2}\cos \varphi$. $\overline{p^{\prime }{q}^{\prime }}=\frac{1}{2}\cos 150^{\circ }=\frac{1}{2}(-0.8660)=-0.4330$ Negative ⇒ heat is being removed from the wave. Answer: $\overline{p^{\prime }{q}^{\prime }}\approx -0.433<0$, the mode is damped / stable.

Neutral means $\overline{p^{\prime }{q}^{\prime }}=\frac{1}{2}\cos \varphi =0$, i.e. $\cos \varphi =0$. Between $0^{\circ }$ and $180^{\circ }$ the only solution is $\varphi =90^{\circ }$. Answer: $\varphi =90^{\circ }$. Below $90^{\circ }$ (more in-phase) the mode grows; above $90^{\circ }$ it damps. This is why designers push the phase past quarter-cycle.

We need the new phase $\varphi +\Delta \varphi$ to hit the neutral point $90^{\circ }$. $60^{\circ }+\Delta \varphi =90^{\circ }\Rightarrow \Delta \varphi =30^{\circ }.$ WHY only reaching $90^{\circ }$: at exactly $90^{\circ }$ the Rayleigh integral is zero (neutral). Any push beyond $90^{\circ }$ makes it negative (damped), which is safer, but $30^{\circ }$ is the minimum to stop growth. Answer: $\Delta {\varphi }_{\min }=30^{\circ }$.

WHAT we want: $f_1\le 0.80\times 1500=1200\text{Hz}$. Set up: $f_1=c/(2L)$, and a longer chamber gives a lower $f_1$, so the constraint $f_1\le 1200$ becomes a minimum length. $\frac{c}{2L}\le 1200\Rightarrow L\ge \frac{c}{2\times 1200}=\frac{1100}{2400}=0.4583\text{m}$ Answer: $L_{\min }\approx 0.458\text{m}$. Any longer is even safer (pushes $f_1$ further below the structural note). See Standing waves and resonance for why matching frequencies is what causes damaging resonance.

WHAT: $f_1=c/(2L)$, so the ratio is $\frac{f_1^{\text{new}}}{f_1^{\text{old}}}=\frac{c_2/(2L_2)}{c_1/(2L_1)}=\frac{c_2}{c_1}\cdot \frac{L_1}{L_2}=\frac{1150}{1000}\times \frac{0.30}{0.45}.$ $=1.15\times 0.6667=0.7667$ WHY: the hotter gas pushes frequency up ($\times 1.15$) but the longer chamber pushes it down ($\times 0.667$); the length change wins. Answer: the fundamental drops to about $0.767$ of its old value (a ~23% decrease).

WHAT — longitudinal: $f_{\text{L}}=\frac{1100}{2\times 0.50}=1100\text{Hz}.$ WHAT — tangential: $f_{\text{T}}=\frac{1.8412\times 1100}{2\pi \times 0.10}=\frac{2025.3}{0.6283}\approx 3223\text{Hz}.$ Ratio: $\frac{f_{\text{T}}}{f_{\text{L}}}=\frac{3223}{1100}\approx \mathrm{2.93.}$ WHY it matters: the tangential mode is nearly 3× higher in frequency here, spinning around the axis. Because it sweeps the whole injector face, it couples strongly to combustion and is the mode engineers most fear. Answer: $f_{\text{T}}\approx 3223\text{Hz}$, about $2.93\times$ the longitudinal fundamental.

(a) — WHAT: apply the sign rule $\Delta E\propto \cos \varphi$ (positive drives the mode, negative damps it — this is the Rayleigh result from L3).

• Injector A: $\cos {\varphi }_A=\cos 40^{\circ }=+0.766>0$ ⇒ energy is added each cycle ⇒ the mode grows ⇒ unstable.
• Injector B: $\cos {\varphi }_B=\cos 100^{\circ }=-0.174<0$ ⇒ energy is removed each cycle ⇒ the mode decays ⇒ stable. WHY: $40^{\circ }$ is below the neutral $90^{\circ }$ boundary (from L3.2), so it lies in the driving region; $100^{\circ }$ is just past $90^{\circ }$, in the damping region. Answer to (a): injector B is stable, A is not.

(b) — WHAT: we want A's shifted phase ${\varphi }_A^{\prime }={\varphi }_A+\Delta \varphi =40^{\circ }+\Delta \varphi$ to give the same energy gain as B, i.e. $\cos {\varphi }_A^{\prime }=\cos {\varphi }_B=\cos 100^{\circ }$. WHY the angles must match directly: on the interval $0^{\circ }$ to $180^{\circ }$ the cosine function is one-to-one (it decreases steadily from $+1$ to $-1$ with no repeats), so equal cosines force equal angles there. Hence ${\varphi }_A^{\prime }=100^{\circ }$. $\Delta \varphi ={\varphi }_A^{\prime }-{\varphi }_A=100^{\circ }-40^{\circ }=60^{\circ }.$ WHAT IT MEANS: adding $60^{\circ }$ of lag carries injector A from $40^{\circ }$ (driving) straight across the neutral $90^{\circ }$ line to $100^{\circ }$ (damping), so it now behaves exactly like the stable injector B. Answer to (b): $\Delta \varphi =60^{\circ }$.

Frequencies come from geometry and $c$ ($f_n=nc/2L$, tangential from Bessel roots and radius $a$ via the circumference $2\pi a$); stability comes from phase (Rayleigh's $\cos \varphi$: in-phase drives, out-of-phase damps, $90^{\circ }$ is the boundary).