Acoustic modes in combustion chamber — cause of instability: Exercises

3.3.33 · D4 · Physics › Rocket Propulsion › Acoustic modes in combustion chamber — cause of instability

Definition Is page par use hone wala notation (pehle yeh padho)

Prime $^{'}$ ka matlab hai "choti fluctuating part". Gas mein ek steady average value hoti hai aur uske upar ek tiny wobble hoti hai. Hum sirf wobble ko prime ke saath likhte hain: $p^{'}$ fluctuating pressure hai (pressure kitna momentarily average se upar/neeche jaata hai), $u^{'}$ fluctuating gas velocity hai, aur $q^{'}$ fluctuating heat-release rate hai (burn rate momentarily apne average se kitna upar/neeche jaata hai). Yahi quantities hain jo ek acoustic mode mein oscillate karti hain.
Overline $\overline{\cdot}$ ka matlab hai "ek poore oscillation cycle ka average". Toh $\overline{p^{'} q^{'}}$ product $p^{'} q^{'}$ ka ek complete period ke across average hai — yeh humein cycle ke net effect ke baare mein bataata hai, jab fast up-and-down wiggles cancel ho jaate hain.
Pressure-mode shape $P (x) = A cos (nπ x / L)$ ko parent note mein step-by-step derive kiya gaya tha the parent topic (Steps 2–3): ek resonance har jagah ek hi frequency par oscillate karti hai, uski shape $P (x)$ equation $P^{''} + k^{2} P = 0$ satisfy karti hai, aur rigid-wall boundary conditions (dono ends par zero slope) cosine ko $k_{n} = nπ / L$ ke saath force karte hain. Yahan $A$ amplitude hai aur $x$ chamber axis ke saath $0$ se $L$ tak jaata hai.
Letter $R$ ke baare mein warning. Sound-speed formula mein $R$ ka matlab hai universal gas constant ( $8.314 J mol^{- 1} K^{- 1}$ ). Baad mein, tangential-mode problem (L5.1) mein, humein chamber ka radius chahiye hoga, jo ek length hai — bilkul alag quantity. Gas constant se clash avoid karne ke liye, hum chamber radius ko $a$ (metres mein) likhenge is poore page par. Toh: $R =$ gas constant, $a =$ chamber radius.

Neeche wali figure har longitudinal-mode question ke liye hamari reference picture hai. Jab bhi koi problem "fundamental", "antinode", ya "half-wave" mention kare, ise dekho.

Figure — Acoustic modes in combustion chamber — cause of instability

Intuition Figure s01 kya dikhati hai

Do moti white vertical lines rigid chamber walls hain: injector face $x = 0$ par (left) aur throat $x = L$ par (right). Horizontal axis chamber axis ke saath position $x$ hai, $0$ se $L$ tak. Teen colored curves top se bottom tak pressure-mode shapes $P (x) = A cos (nπ x / L)$ dikhate hain $n = 1$ (blue), $n = 2$ (yellow), $n = 3$ (green) ke liye — har ek apne dashed baseline ke baare mein envelope (solid aur faint mirror curve) ki tarah draw kiya gaya hai. Red dots $n = 1$ curve ke upar dono walls par baithe hain: yeh pressure antinodes mark karte hain, jahan pressure sabse zyada swing karta hai kyunki gas velocity wall par zero pin hoti hai. Notice karo ki har curve dono walls par maximum height (ek antinode) se shuru aur khatam hoti hai — yahi boundary condition visibly dikh rahi hai.

Level 1 — Recognition

L1.1 — Kaun si end condition hai?

Ek combustion chamber ko dono ends par closed tube ki tarah model kiya gaya hai (injector face aur throat dono strongly reflect karte hain). Injector wall par gas metal ke through flow nahi kar sakti. Kya pressure oscillation wahan ek node (hamesha zero) hai ya ek antinode (maximum swing)?

Recall Solution

Wall kya karti hai: metal gas motion ko block karta hai, toh oscillating gas velocity $u^{'} = 0$ wall par hoti hai (yaad karo $u^{'}$ upar define ki gayi choti fluctuating velocity hai). Yeh antinode kyun hota hai: ek sound wave mein velocity aur pressure quarter-wave out of step hote hain — jahan velocity zero hoti hai, pressure sabse zyada swing karta hai. Formally velocity $\propto \partial p^{'} / \partial x$ , toh $u^{'} = 0$ force karta hai $\partial p^{'} / \partial x = 0$ , jo pressure shape ka slope-zero point hai — ek peak, yaani ek pressure antinode. Answer: ek pressure antinode. (s01 figure mein dono ends par red dots dekho — woh antinodes hain.)

L1.2 — Mode identify karo

Ek chamber ka axis ke saath pressure pattern mein ek full pressure peak har end par aur middle mein ek trough hai (ek single half-wave shape reflected). Yeh kaun sa mode number $n$ hai, aur kya yeh fundamental hai?

Recall Solution

Hum kya use karte hain: parent note mein derive kiya gaya mode shape, $P (x) = A cos (nπ x / L)$ , saath mein rigid-wall boundary conditions ( $x = 0$ aur $x = L$ par ek pressure antinode). $n = 1$ kyun ek half-wave ke liye: boundary conditions demand karte hain dono walls par $∣ P ∣$ ka peak. Cosine $cos (nπ x / L)$ automatically $x = 0$ par peak hoti hai (uski value $A$ hai) aur $x = L$ par tab hi peak hoti hai jab $nπ$ ek whole multiple of $π$ ho, yaani $n = 1, 2, 3, \dots$ . $n$ mein har increase walls ke beech ek aur half-wave add karta hai. Smallest allowed pattern — har end par ek antinode aur middle mein ek single sign-flip (trough) — exactly chamber ke across half a wavelength hai, aur yahi $n = 1$ case hai. Toh "ek half-wave fit hota hai" aur " $n = 1$ " ek hi statement hai do tarafon se padhi. Answer: $n = 1$ , fundamental (sabse lowest possible frequency). Zyada $n$ usi length mein zyada half-waves squeeze karte hain.

Level 2 — Application

L2.1 — Fundamental frequency

Ek chamber ki length $L = 0.40 m$ hai; combustion gas ki sound speed $c = 980 m/s$ hai. Fundamental frequency $f_{1}$ find karo.

Recall Solution

Kya karna hai: $f_{n} = n c / (2 L)$ mein $n = 1$ plug karo. $f_{1} = \frac{1 \times 980}{2 \times 0.40} = \frac{980}{0.80} = 1225 Hz$ $2 L$ kyun: fundamental length $L$ mein half wave fit karta hai, toh wavelength $2 L$ hai; frequency $c$ divided by wavelength hai. Answer: $f_{1} \approx 1225 Hz$ .

L2.2 — Third harmonic

Usi chamber ke liye ( $L = 0.40 m$ , $c = 980 m/s$ ), $n = 3$ longitudinal mode ki frequency find karo.

Recall Solution

$f_{3} = \frac{3 \times 980}{2 \times 0.40} = 3 \times 1225 = 3675 Hz$ Kyun: longitudinal modes fundamental ke exact integer multiples hote hain, toh $f_{3} = 3 f_{1}$ . Answer: $f_{3} \approx 3675 Hz$ .

L2.3 — Gas properties se speed of sound

$T = 3000 K$ par combustion products mein $γ = 1.20$ aur molar mass $M = 0.022 kg/mol$ hai. $c$ find karo. ( $R = 8.314 J mol^{- 1} K^{- 1}$ use karo.)

Recall Solution

Kya karna hai: $c = γ R T / M$ use karo (dekho Speed of sound in gases). $c = \frac{1.20 \times 8.314 \times 3000}{0.022} = \frac{29 930}{0.022} = 1.360 \times 1 0^{6} \approx 1166 m/s$ Yeh formula kyun: ek gas mein, ek pressure pulse springiness ( $γ$ , pressure) ke through inertia ( $M$ , mass) se ladti hui propagate karti hai. Hotter aur lighter gas ⇒ faster sound. Answer: $c \approx 1166 m/s$ .

Level 3 — Analysis

L3.1 — Rayleigh sign check

Ek pressure signal $p^{'} = cos ω t$ padhta hai. Heat release $q^{'} = cos (ω t - ϕ)$ ke saath $ϕ = 15 0^{\circ}$ lag ke saath fluctuate karta hai. Time-averaged Rayleigh product $\overline{p^{'} q^{'}}$ compute karo aur batao ki mode grow karta hai ya decay.

Recall Solution

Kya karna hai: phase difference $ϕ$ ke saath do cosines ka cycle-average (overline, upar define kiya) hai $\overline{p^{'} q^{'}} = \frac{1}{2} cos ϕ .$ Kyun: $cos ω t cos (ω t - ϕ) = \frac{1}{2} [cos ϕ + cos (2 ω t - ϕ)]$ expand karo; fast $cos (2 ω t - ϕ)$ term ek full cycle par average hokar zero ho jaata hai, sirf $\frac{1}{2} cos ϕ$ bachta hai. $\overline{p^{'} q^{'}} = \frac{1}{2} cos 15 0^{\circ} = \frac{1}{2} (- 0.8660) = - 0.4330$ Negative ⇒ heat wave se remove ho rahi hai. Answer: $\overline{p^{'} q^{'}} \approx - 0.433 < 0$ , mode damped / stable hai.

L3.2 — Neutral phase find karo

$0^{\circ}$ aur $18 0^{\circ}$ ke beech kaun sa lag $ϕ$ ho toh heat release mode ko na drive kare na damp — exact stability boundary?

Recall Solution

Neutral ka matlab $\overline{p^{'} q^{'}} = \frac{1}{2} cos ϕ = 0$ , yaani $cos ϕ = 0$ . $0^{\circ}$ aur $18 0^{\circ}$ ke beech single solution $ϕ = 9 0^{\circ}$ hai. Answer: $ϕ = 9 0^{\circ}$ . $9 0^{\circ}$ se neeche (zyada in-phase) mode grow karta hai; upar damp hota hai. Isliye designers phase ko quarter-cycle se aage push karte hain.

L3.3 — Kitna shift engine bachata hai?

Ek engine currently $ϕ = 6 0^{\circ}$ ke saath run karta hai (unstable, kyunki $cos 6 0^{\circ} = + 0.5 > 0$ ). Ek baffle redesign flame response mein extra lag $Δ ϕ$ add karta hai. Minimum $Δ ϕ$ kya hai jo just stability boundary tak pahunch jaye?

Recall Solution

Humein new phase $ϕ + Δ ϕ$ ko neutral point $9 0^{\circ}$ tak pahunchana hai. $6 0^{\circ} + Δ ϕ = 9 0^{\circ} \Rightarrow Δ ϕ = 3 0^{\circ} .$ Sirf $9 0^{\circ}$ tak kyun pahunchna hai: exactly $9 0^{\circ}$ par Rayleigh integral zero hai (neutral). Koi bhi push $9 0^{\circ}$ se aage ise negative (damped) banata hai, jo safer hai, lekin $3 0^{\circ}$ growth rokne ke liye minimum hai. Answer: $Δ ϕ_{m i n} = 3 0^{\circ}$ .

Level 4 — Synthesis

L4.1 — Chamber ko resonance se door design karo

Ek injector plate mein $f_{struct} = 1500 Hz$ par ek structural resonance hai jise fundamental acoustic mode ke saath avoid karna hai. Gas sound speed $c = 1100 m/s$ par fixed hai. Tum chamber length $L$ choose kar sakte ho. Agar tum chahte ho ki acoustic fundamental $f_{1}$ structural resonance se kam se kam $20%$ neeche ho, toh minimum length $L$ kya honi chahiye?

Recall Solution

Hum kya chahte hain: $f_{1} \leq 0.80 \times 1500 = 1200 Hz$ . Setup: $f_{1} = c / (2 L)$ , aur ek longer chamber lower $f_{1}$ deta hai, toh constraint $f_{1} \leq 1200$ ek minimum length ban jaati hai. $\frac{c}{2 L} \leq 1200 \Rightarrow L \geq \frac{c}{2 \times 1200} = \frac{1100}{2400} = 0.4583 m$ Answer: $L_{m i n} \approx 0.458 m$ . Koi bhi lambi length aur safe hai ( $f_{1}$ ko structural note se aur neeche push karti hai). Standing waves and resonance dekho kyun matching frequencies damaging resonance cause karte hain.

L4.2 — Combined thermal + geometric shift

Ek test engine do tarafon se redesign kiya gaya hai: chamber length $L_{1} = 0.30 m$ se $L_{2} = 0.45 m$ tak badhi, aur flame temperature itni badh gayi ki sound speed $c_{1} = 1000 m/s$ se $c_{2} = 1150 m/s$ ho gayi. Kya fundamental frequency upar ya neeche jaati hai, aur kis factor se?

Recall Solution

Kya karna hai: $f_{1} = c / (2 L)$ , toh ratio hai $\frac{f _{1}^{new}}{f _{1}^{old}} = \frac{c _{2} / ( 2 L _{2} )}{c _{1} / ( 2 L _{1} )} = \frac{c _{2}}{c _{1}} \cdot \frac{L _{1}}{L _{2}} = \frac{1150}{1000} \times \frac{0.30}{0.45} .$ $= 1.15 \times 0.6667 = 0.7667$ Kyun: hotter gas frequency upar push karta hai ( $\times 1.15$ ) lekin longer chamber use neeche push karta hai ( $\times 0.667$ ); length change jeet jaata hai. Answer: fundamental apni purani value ka roughly $0.767$ tak drop karta hai (lagbhag 23% decrease).

Level 5 — Mastery

L5.1 — Tangential vs longitudinal mode competition

Ek cylindrical chamber ki length $L = 0.50 m$ aur radius $a = 0.10 m$ hai (hum radius ke liye $a$ use karte hain taaki yeh gas constant $R$ se kabhi clash na kare), $c = 1100 m/s$ ke saath. Pehla tangential mode frequency hai $f_{T} = \frac{α _{11} c}{2 π a}, α_{11} = 1.8412$ (Bessel function $J_{1}$ ki derivative ka pehla root). Pehla longitudinal mode $f_{L} = c / (2 L)$ hai. Kaun sa mode higher frequency rakhta hai, aur kis ratio mein? (Tangential modes usually sabse zyada destructive hote hain — parent note dekho.)

Intuition Denominator

$2 π a$ kyun hai — circumference wavelength set karta hai Ek tangential mode ek pressure wave hai jo chamber ke around run karti hai, circular wall ko hug karti hai, axis ke upar-neeche jaane ke bajaye. Ek baar around jaane ka path circle ka circumference hai, $2 π a$ (yeh sirf "radius $a$ wale circle ke around distance" hai). Wave ke liye ek self-consistent standing pattern hona zaroori hai — iska matlab hai ki use us loop mein "neatly fit" karna hoga — wahi fits-neatly idea jaise parent note mein longitudinal cosine, lekin ab closing condition ring ke around hai, do walls ke beech ke bajaye. Toh characteristic length jo wavelength set karti hai woh circumference $2 π a$ hai, aur frequency $c$ divided by us length ki tarah scale karta hai: $f \sim c / (2 π a)$ . Extra pure number $α_{11} = 1.8412$ woh hai jo flat-tube problem ke tidy integer $nπ$ ki jagah leta hai. Kyunki chamber cross-section ek disc hai (straight line nahi), wave shape ek simple cosine nahi balki Bessel function $J_{1}$ hai — disc par natural "cosine". Rigid-wall condition (wall par zero radial velocity, yaani wahan pressure ka zero slope) sirf un special radii par poori hoti hai jahan $J_{1}$ ki derivative zero hoti hai; $α_{11}$ esa pehla root hai. Yeh exactly wahi role play karta hai jo $nπ$ straight tube ke liye play karta tha: yeh circular boundary ke liye "fits-neatly" number hai. Standing waves and resonance dekho.

Recall Solution

Kya karna hai — longitudinal: $f_{L} = \frac{1100}{2 \times 0.50} = 1100 Hz .$ Kya karna hai — tangential: $f_{T} = \frac{1.8412 \times 1100}{2 π \times 0.10} = \frac{2025.3}{0.6283} \approx 3223 Hz .$ Ratio: $\frac{f _{T}}{f _{L}} = \frac{3223}{1100} \approx 2.93.$ Kyun yeh matter karta hai: tangential mode yahan nearly 3× higher frequency par hai, axis ke around spin karta hua. Kyunki yeh poore injector face ko sweep karta hai, yeh combustion ke saath strongly couple karta hai aur yahi mode hai jo engineers sabse zyada fear karte hain. Answer: $f_{T} \approx 3223 Hz$ , longitudinal fundamental ka lagbhag $2.93 \times$ .

L5.2 — Growth-rate reasoning (full synthesis)

Do candidate injectors same acoustic mode dete hain lekin alag flame phases ke saath: injector A mein $ϕ_{A} = 4 0^{\circ}$ hai, injector B mein $ϕ_{B} = 10 0^{\circ}$ hai. Energy gained per cycle scale karti hai $Δ E \propto cos ϕ$ ke saath. (a) Kaun sa injector stable hai? (b) Acoustic cavities add karke tum injector A mein aur lag add kar sakte ho; kitna extra lag $Δ ϕ$ add karna chahiye taaki injector A ka per-cycle energy gain injector B ke barabar ho jaye (yaani $cos ϕ_{A} = cos ϕ_{B}$ ho)?

Recall Solution

(a) — Kya karna hai: sign rule $Δ E \propto cos ϕ$ apply karo (positive mode ko drive karta hai, negative damp karta hai — yeh L3 se Rayleigh result hai).

Injector A: $cos ϕ_{A} = cos 4 0^{\circ} = + 0.766 > 0$ ⇒ har cycle mein energy add hoti hai ⇒ mode grows ⇒ unstable.
Injector B: $cos ϕ_{B} = cos 10 0^{\circ} = - 0.174 < 0$ ⇒ har cycle mein energy remove hoti hai ⇒ mode decays ⇒ stable. Kyun: $4 0^{\circ}$ neutral $9 0^{\circ}$ boundary (L3.2 se) se neeche hai, toh yeh driving region mein hai; $10 0^{\circ}$ just $9 0^{\circ}$ se aage hai, damping region mein. (a) ka Answer: injector B stable hai, A nahi.

(b) — Kya karna hai: hum chahte hain ki A ka shifted phase $ϕ_{A}^{'} = ϕ_{A} + Δ ϕ = 4 0^{\circ} + Δ ϕ$ wahi energy gain de jaise B, yaani $cos ϕ_{A}^{'} = cos ϕ_{B} = cos 10 0^{\circ}$ . Angles seedhe kyun match karne chahiye: $0^{\circ}$ se $18 0^{\circ}$ ke interval par cosine function one-to-one hai (yeh steadily $+ 1$ se $- 1$ tak decrease karta hai bina kisi repeat ke), toh equal cosines wahan equal angles force karte hain. Hence $ϕ_{A}^{'} = 10 0^{\circ}$ . $Δ ϕ = ϕ_{A}^{'} - ϕ_{A} = 10 0^{\circ} - 4 0^{\circ} = 6 0^{\circ} .$ Iska kya matlab hai: $6 0^{\circ}$ lag add karna injector A ko $4 0^{\circ}$ (driving) se neutral $9 0^{\circ}$ line ke seedhe across $10 0^{\circ}$ (damping) par le jaata hai, toh ab yeh exactly stable injector B ki tarah behave karta hai. (b) ka Answer: $Δ ϕ = 6 0^{\circ}$ .

Recall Master check — yaad rakhne wala ek sentence

Frequencies geometry aur $c$ se aati hain ( $f_{n} = n c /2 L$ , tangential Bessel roots aur radius $a$ se circumference $2 π a$ ke through); stability phase se aati hai (Rayleigh ka $cos ϕ$ : in-phase drive karta hai, out-of-phase damp karta hai, $9 0^{\circ}$ boundary hai).