Acoustic modes in combustion chamber — cause of instability: Worked examples

3.3.33 · D3 · Physics › Rocket Propulsion › Acoustic modes in combustion chamber — cause of instability

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Scenario matrix

Is topic ke har problem inhi cells mein se kisi ek mein aata hai. Neeche ke examples mein us cell ka label hai jise woh cover karta hai.

#	Cell class	Kya mushkil banata hai	Covered by
A	Baseline longitudinal $f_{n}$	plug-and-go fundamental	Ex 1
B	Overtone ( $n > 1$ )	konsa "note" hai $n$ ?	Ex 2
C	Hot vs cold gas ( $c$ change hota hai)	$T$ se $c$ recompute karo	Ex 3
D	Transverse (tangential) mode	Bessel root, $L$ nahi $R$ use hota hai	Ex 4
E	Rayleigh sign — in phase	$cos ϕ > 0$ , badhta hai	Ex 5
F	Rayleigh sign — out of phase	$cos ϕ < 0$ , damp hota hai	Ex 5
G	Degenerate phase $ϕ = 9 0^{\circ}$	exactly neutral, $cos ϕ = 0$	Ex 6
H	Limiting input: $L \to \infty$ , $L \to 0$	frequency $\to 0$ ya blow up ho jaati hai	Ex 7
I	Real-world word problem	ek structural resonance avoid karo	Ex 8
J	Exam twist: gain vs loss balance	drive ko damping se zyada hona chahiye	Ex 9

Worked examples

Worked example Example 1 — Baseline fundamental (Cell A)

Ek chamber $L = 0.40 m$ lamba hai, jisme gas bhari hai jahan $c = 960 m/s$ hai. Fundamental frequency $f_{1}$ nikalo.

Forecast: roughly $c / (2 L) \approx 960/0.8$ — thoda ek kilohertz se zyada. Aage padhne se pehle guess karo.

$f_{n} = \frac{n c}{2 L}$ use karo, $n = 1$ ke saath. Yeh step kyun? $n = 1$ woh sabse chhota pattern hai jo "fit" hota hai: exactly aadha wave chamber ko span karta hai (har rigid end par ek antinode), isliye uski wavelength $2 L$ hai aur $f_{1} = c / (2 L)$ .
Substitute karo: $f_{1} = \frac{1 \cdot 960}{2 ( 0.40 )} = \frac{960}{0.80} = 1200 Hz$ . Yeh step kyun? Jab pattern fix ho jaata hai toh sirf arithmetic bacha rehta hai.

Verify: units hain $\frac{m/s}{m} = 1/s = Hz$ ✓. Wavelength $2 L = 0.80 m$ wala aadha-wave $960 m/s$ par travel karta hua ek point se $960/0.80 = 1200$ baar per second guzarta hai ✓.

Worked example Example 3 — Hot gas khatra badhata hai (Cell C)

Ek chamber $L = 0.50 m$ hai jo $T = 2800 K$ par run karta hai. Gas mein $γ = 1.20$ , $R_{univ} = 8.314 J mol^{- 1} K^{- 1}$ , molar mass $M = 0.022 kg/mol$ hai. $c$ nikalo, phir $f_{1}$ . Thande air ke saath $c = 340 m/s$ use karne se compare karo.

Forecast: hot combustion gas → $c \approx 1000 - 1300 m/s$ , thande value se kai guna zyada.

Speed of sound: $c = \frac{γ R _{univ} T}{M}$ . Yeh step kyun? Yeh Speed of sound in gases result hai — ek squeeze stiffer, hotter, aur halke gas mein tezi se travel karti hai. Hume actual hot-gas $c$ chahiye, room-air $c$ nahi.
$c = \frac{1.20 \cdot 8.314 \cdot 2800}{0.022} = \frac{27 935}{0.022} = 1.2698 \times 1 0^{6} \approx 1127 m/s$ . Yeh step kyun? Hot numbers plug karo taaki frequency reality reflect kare.
Hot mode: $f_{1} = \frac{1127}{2 ( 0.50 )} = 1127 Hz$ .
Cold (galat) estimate: $f_{1}^{cold} = \frac{340}{1.0} = 340 Hz$ . Yeh step kyun? Classic error dikhata hai: thanda air $1127/340 \approx 3.3 \times$ se underestimate karta hai.

Verify: ratio $c / c_{cold} = 1127/340 \approx 3.3$ , parent note ki " $\sim 3 \times$ " warning se match karta hai ✓. $c$ ke units: $\frac{( J/mol/K ) ( K )}{kg/mol} = \frac{J}{kg} = m^{2} / s^{2} = m/s$ ✓.

Worked example Example 4 — Ek tangential (spinning) mode (Cell D)

Ek cylindrical chamber ka radius $R = 0.15 m$ hai, gas $c = 1000 m/s$ hai. Pehla tangential mode Bessel-derivative root $α_{11} = 1.841$ use karta hai. Uski frequency nikalo.

Forecast: yeh $L$ se nahi $R$ se set hota hai; kuch kilohertz expect karo.

Transverse formula use karo $f_{mn}^{trans} = \frac{α _{mn} c}{2 π R}$ . Yeh step kyun? Ek spinning wave axis ke around loop karti hai, isliye "woh length jisme use fit hona hai" circumference-scale $2 π R$ se set hoti hai, na ki axial length $L$ se. Bessel root $α_{11}$ wohi role play karta hai jo straight tube ke liye $nπ$ karta tha.
$f_{11} = \frac{1.841 \cdot 1000}{2 π ( 0.15 )} = \frac{1841}{0.9425} \approx 1953 Hz$ . Yeh step kyun? Jab geometry choose ho jaati hai toh sirf arithmetic.

Verify: units $\frac{m/s}{m} = Hz$ ✓. Tangential modes Rayleigh Criterion-dangerous "sabse zyada destructive" wale hain — inhe precisely locate karna zaroori hai. Standing waves and resonance dekho yeh samajhne ke liye ki sirf special patterns kyun survive karte hain.

Worked example Example 5 — Rayleigh sign, dono directions (Cells E & F)

Ek pressure sensor $p^{'} = cos ω t$ read karta hai. Do designs test ki jaati hain: (E) heat $ϕ = 3 0^{\circ}$ se lag karta hai; (F) heat $ϕ = 17 0^{\circ}$ se lag karti hai. Kaun badh raha hai, kaun damp ho raha hai?

Forecast: chhota lag = "almost in phase" = badhta hai; near- $18 0^{\circ}$ = "almost anti-phase" = damp hota hai.

Energy gained per cycle $\propto \overline{p^{'} q^{'}} = \frac{1}{2} cos ϕ$ . Yeh step kyun? Yeh parent ka Rayleigh Criterion result hai: $ϕ$ se alag do cosines ka time-average $\frac{1}{2} cos ϕ$ hota hai. Sirf $cos ϕ$ ka sign decide karta hai growth vs decay.
Case E: $cos 3 0^{\circ} = 0.866 > 0$ → positive → mode energy gain karta hai → unstable.
Case F: $cos 17 0^{\circ} = - 0.985 < 0$ → negative → mode energy lose karta hai → stable / damped.

Verify: signs $ϕ = 9 0^{\circ}$ par flip hote hain jaise expected ( $cos$ $9 0^{\circ}$ ke neeche positive, upar negative) ✓. Phase-versus-drive picture:

Worked example Example 6 — Exactly-neutral degenerate case (Cell G)

Heat release exactly quarter-cycle out hai: $ϕ = 9 0^{\circ}$ . Growing, damping, ya kuch nahi?

Forecast: yeh crossover point hai — "kuch nahi" guess karo.

Compute karo $cos 9 0^{\circ} = 0$ . Yeh step kyun? $9 0^{\circ}$ par heat wobble tab peak karta hai jab pressure apne average par hota hai (zero wobble), isliye ek poore cycle mein wave par koi net work nahi hota.
Isliye $\overline{p^{'} q^{'}} = \frac{1}{2} \cdot 0 = 0$ : acoustic energy har cycle mein unchanged rehti hai — neutrally stable (baaki losses ko ignore karte hue). Yeh step kyun? Yeh woh knife-edge hai jo Cells E aur F ko alag karti hai; real losses (nozzle, viscosity) phir outcome decide karti hain.

Verify: $cos 9 0^{\circ} = 0$ exactly ✓. Ex 5 ke saath consistent: $3 0^{\circ}$ (neeche) ne drive kiya, $17 0^{\circ}$ (upar) ne damp kiya, toh $9 0^{\circ}$ zero hona chahiye.

Worked example Example 7 — Limiting inputs (Cell H)

(a) Jaise chamber length $L \to \infty$ jaata hai $c$ fixed rakh ke, $f_{1}$ ka kya hoga? (b) Jaise $L \to 0$ jaata hai, kya hoga? Numbers se sanity-check karo $c = 1000 m/s$ use karke: $L = 2 m$ aur $L = 0.01 m$ .

Forecast: lamba tube → low bass note → $f \to 0$ ; tiny tube → shreell → $f \to \infty$ .

$f_{1} = \frac{c}{2 L}$ se, $L \to \infty$ lo: $f_{1} \to 0$ . Yeh step kyun? Ek wave jiska half-wavelength ek bade chamber par stretched hai bahut dheere oscillate karta hai — note silence ki taraf girta hai.
$L \to 0$ lo: $f_{1} \to \infty$ . Yeh step kyun? Ek half-wave ko vanishing length mein squeeze karna ultra-high pitch force karta hai; physically model pehle break ho jaata hai, lekin trend warning hai.
Numbers: $L = 2 m \Rightarrow f_{1} = 1000/4 = 250 Hz$ ; $L = 0.01 m \Rightarrow f_{1} = 1000/0.02 = 50000 Hz$ . Yeh step kyun? Confirm karta hai ki dono limits sahi direction mein point karte hain — bada drop, bada rise.

Verify: $1000/4 = 250$ ✓ aur $1000/0.02 = 50000$ ✓; bade $L$ ne chhota $f$ diya, chhote ne bada $f$ diya ✓.

Worked example Example 8 — Real-world word problem (Cell I)

Ek engine ki turbopump feed line mein $2000 Hz$ par mechanical resonance hai. Chamber $c = 1100 m/s$ par run karta hai. Designer ko chamber length $L$ choose karni hai taaki koi bhi longitudinal mode $f_{1}, f_{2}, f_{3}$ $2000 Hz$ ke $\pm 50 Hz$ ke andar na aaye. Agar $L = 0.55 m$ propose kiya gaya hai, toh kya yeh safe hai?

Forecast: pehle kuch modes compute karo aur dekho koi $2000 Hz$ ke paas aata hai ya nahi.

$f_{1} = \frac{1100}{2 ( 0.55 )} = \frac{1100}{1.10} = 1000 Hz$ . Yeh step kyun? Fundamental poori ladder set karta hai; har overtone uska multiple hota hai.
$f_{2} = 2000 Hz$ , $f_{3} = 3000 Hz$ . Yeh step kyun? Overtones $n \times f_{1}$ hote hain dono-taraf-band chamber ke liye.
Danger band check karo $2000 \pm 50 Hz = [1950, 2050]$ : $f_{2} = 2000 Hz$ dead centre mein hai → SAFE NAHI. Doosra mode seedha feed-line resonance se couple ho jaayega. Yeh step kyun? Yahi woh overlap hai jise parent note designers ko avoid karne ki warning deta hai.

Verify: $2 \times 1000 = 2000$ , aur $2000 \in [1950, 2050]$ ✓ — proposal fail ho jaata hai. Ek fix: $L$ ko nudge karo taaki $f_{2}$ band clear kare, ya Injector design and baffles / Nozzle flow and acoustic damping use karo us frequency par loss add karne ke liye.

Worked example Example 9 — Exam twist: gain ko loss se beat karna hoga (Cell J)

Ek mode mein Rayleigh gain $G = cos ϕ$ ke proportional hai, $ϕ = 4 0^{\circ}$ ke saath, lekin nozzle aur viscous damping energy ko $D = 0.90$ ke proportional rate se remove karte hain (same units). Net growth ke liye zaroori hai $G > D$ . Phase-in ( $ϕ = 4 0^{\circ}$ ) mein, kya mode unstable hai? Woh sabse bada $ϕ$ kya hai jo barely unstable hoga?

Forecast: $cos 4 0^{\circ}$ lagbhag $0.77$ hai; agar damping $0.90$ hai toh drive nahi jeet sakti.

Gain: $G = cos 4 0^{\circ} = 0.766$ . Yeh step kyun? Rayleigh drive sirf phase par depend karti hai, Ex 5 ke according.
Loss $D = 0.90$ se compare karo: kyunki $0.766 < 0.90$ , net energy change negative hai → stable, chahe heat in-phase-ish ho. Yeh step kyun? Instability ek race hai: positive Rayleigh integral zaroori hai lekin saari losses se bhi zyada hona chahiye. Yahi woh subtlety hai jo parent flag karta hai — amplitude/phase akela last word nahi hai.
Break-even phase: solve karo $cos ϕ = 0.90 \Rightarrow ϕ = arccos (0.90) = 25. 8^{\circ}$ . Yeh step kyun? $25. 8^{\circ}$ se chhota koi bhi lag $G > D$ banata hai aur mode grow karta hai; bada lag safe hai.

Verify: $cos 4 0^{\circ} = 0.766 < 0.90$ ✓ (stable); $arccos (0.90) \approx 25.8 4^{\circ}$ , aur $cos (25.8 4^{\circ}) \approx 0.90$ ✓.

Recall Matrix ke across quick self-test

$L = 0.40 m$ , $c = 960 m/s$ ka fundamental? ::: $1200 Hz$ . Usi chamber ka teesra mode? ::: $3600 Hz$ (= $3 f_{1}$ ). $γ = 1.20$ , $T = 2800 K$ , $M = 0.022$ ke liye hot-gas $c$ ? ::: $\approx 1127 m/s$ . Kya $ϕ = 3 0^{\circ}$ driving hai ya damping? ::: Driving ( $cos 3 0^{\circ} > 0$ ). Kya $ϕ = 17 0^{\circ}$ driving hai ya damping? ::: Damping ( $cos 17 0^{\circ} < 0$ ). $L \to \infty$ hone par $f_{1}$ ka kya hoga? ::: Yeh $0$ ho jaata hai (bahut low note). Gain $cos 4 0^{\circ}$ vs loss $0.90$ ke saath, stable hai ya nahi? ::: Stable, kyunki $0.766 < 0.90$ .