Combustion instability — low-frequency (chugging), high-frequency (screaming): Worked examples

3.3.32 · D3 · Physics › Rocket Propulsion › Combustion instability — low-frequency (chugging), high-freq

Shuru karne se pehle, teen clocks aur symbols ka ek reminder jo hum baar baar use karenge, taaki kuch bhi unexplained na lage:

Recall

$τ$ physically kya mean karta hai? ::: Combustion/transit lag — propellant ke enter karne aur poora burn hone ke beech ka time (milliseconds). Ye slow clock hai jo chugging set karta hai. $a = γ R_{g a s} T_{c}$ kya mean karta hai? ::: Hot chamber gas mein speed of sound — fast clock. $γ$ = ratio of specific heats, $R_{g a s}$ = specific gas constant, $T_{c}$ = chamber temperature. Rayleigh integral $\oint p^{'} q^{'} d t$ kya measure karta hai? ::: Kya heat pulse $q^{'}$ ek cycle mein pressure pulse $p^{'}$ ke saath in phase aata hai. Positive → badhta hai, negative → khatam hota hai.

Scenario matrix

Is topic ke har question ka ek cell hota hai. Har example neeche apne cell ke saath tagged hai.

Cell	Case class	Kya test ho raha hai	Example
A	Rayleigh positive (phase $ϕ < 90°$ )	$\oint p^{'} q^{'} d t$ ka sign → growth	Ex 1
B	Rayleigh negative (phase $90° < ϕ < 270°$ )	sign → decay	Ex 2
C	Rayleigh zero (degenerate, $ϕ = 90°$ )	neutral / marginal boundary	Ex 3
D	Low-frequency band	$τ$ se $f_{c h ug}$ predict karo	Ex 4
E	Degenerate feed: $Δ p_{inj}$ limit	injector stiffness fix, ratio test	Ex 5
F	High-frequency band, 1T mode	acoustic mode frequency	Ex 6
G	Limiting behaviour: $ℓ \to$ large / pure longitudinal	mode-index limits	Ex 7
H	Real-world word problem	data se engine diagnose karo	Ex 8
I	Exam twist: kaunsa cure?	Rayleigh + reservoir reasoning	Ex 9

Cell A — Rayleigh integral positive (growth)

Worked example

Ex 1 (Cell A). Pressure fluctuation $p^{'} = cos ω t$ aur heat release $q^{'} = cos (ω t - ϕ)$ jahan $ϕ = 3 0^{\circ}$ . Kya oscillation grow karegi ya die?

Forecast: $3 0^{\circ}$ ek chhota lag hai — heat abhi bhi zyada tar high pressure ke saath overlap karti hai. Guess: grows.

Step 1. Do cosines ke liye ek full cycle par Rayleigh integral hai $\oint p^{'} q^{'} d t \propto \frac{1}{T} \int_{0}^{T} cos ω t cos (ω t - ϕ) d t = \frac{1}{2} cos ϕ .$ Ye step kyun? Product-to-sum identity $cos A cos B = \frac{1}{2} [cos (A - B) + cos (A + B)]$ ek constant $\frac{1}{2} cos ϕ$ aur ek oscillating $cos (2 ω t - ϕ)$ mein split ho jaati hai jo cycle ke upar average hokar zero ho jaati hai. Sirf constant bachta hai — isliye poora answer sirf phase ka ek clean function hai.

Step 2. $ϕ = 3 0^{\circ}$ plug karo: $\frac{1}{2} cos 3 0^{\circ} = \frac{1}{2} (0.8660) = 0.4330 > 0$ . Ye step kyun? Positive number matlab heat oscillation ko feed kar rahi hai — instability grows (neeche figure dekho, red curve zyada tar zero se upar rehti hai).

Verify: $ϕ = 0$ deta $\frac{1}{2}$ (max drive); $ϕ = 9 0^{\circ}$ deta $0$ . Hamara $3 0^{\circ}$ beech mein hai, max ke kareeb — consistent. Sign $+$ hai → grows. ✔

Cell B — Rayleigh integral negative (decay)

Cell C — Rayleigh integral zero (marginal / degenerate)

Worked example

Ex 3 (Cell C). Heat exactly quadrature mein pressure se hai, $ϕ = 9 0^{\circ}$ . Kya hota hai?

Forecast: Exactly grow aur die ke beech — guess: kuch nahi, waise hi rehta hai.

Step 1. $\frac{1}{2} cos 9 0^{\circ} = \frac{1}{2} (0) = 0$ . Ye step kyun? $9 0^{\circ}$ par heat pulse exactly wahan peak karta hai jahan pressure zero cross karta hai. Cycle ke aadhe hisse mein energy add hoti hai, doosre aadhe mein exactly utni hi remove hoti hai — net zero.

Step 2. Interpretation: ye neutral stability boundary hai. Parent ke chug analysis mein ye wahi "marginal" case hai jisne $ω τ \approx π /2$ diya tha. Ye step kyun? Real engines design se is line se door push kiye jaate hain; ye woh razor's edge hai jo Cells A aur B ke do windows ko alag karta hai.

Verify: $cos (9 0^{\circ}) = 0$ exactly, aur ye $+$ region (Ex 1) aur $-$ region (Ex 2) ke beech ki boundary hai. Degenerate case handled. ✔

Cell D — Low-frequency band (chugging)

Worked example

Ex 4 (Cell D). Ek gas-generator engine mein combustion lag $τ = 4.0$ ms hai. Uski instability frequency predict karo aur type batao.

Forecast: Milliseconds → slow clock → chugging, isliye tens–hundreds of Hz. Guess: ~60 Hz.

Step 1. Parent se marginal chug condition hai $ω τ \approx π /2$ , yaani $f_{c h ug} \approx \frac{1}{4 τ}$ . Ye step kyun? Pressure cycle ka quarter ( $π /2$ of phase) exactly woh delay hai jo extra burning ko pressure dip se pressure rise tak le jaata hai — Cell A ki in-phase condition feed loop par apply hoti hai.

Step 2. $f_{c h ug} = \frac{1}{4 ( 4.0 \times 1 0 ^{- 3} )} = \frac{1}{0.016} = 62.5$ Hz. Ye step kyun? Sirf slow clock plug karne se ek number milta hai, aur $62.5$ Hz, $10$ – $400$ Hz chugging band ke andar aata hai.

Verify: Units: $1/ s =$ Hz ✔. Band check: $62.5$ Hz $\in [10, 400]$ → chugging confirmed. Cross-link: ye Injector Design & Pressure Drop use karta hai feed lag origin ke liye. ✔

Cell E — Degenerate feed: injector pressure-drop limit

Worked example

Ex 5 (Cell E). Chamber pressure $p_{c} = 7.0$ MPa. Injector drop $Δ p_{inj} = 0.7$ MPa hai. Kya feed rule of thumb ke hisaab se "stiff enough" hai, aur drop ratio kya hai?

Forecast: Rule of thumb chahta hai $Δ p_{inj} ≳ 0.2 p_{c}$ . $0.7$ vs $0.2 \times 7 = 1.4$ ? Guess: too soft.

Step 1. Ratio compute karo $\frac{Δ p _{inj}}{p _{c}} = \frac{0.7}{7.0} = 0.10$ . Ye step kyun? Rayleigh Criterion feedback gain $k$ chug equation mein injector stiffness ke inversely scale karta hai. Dimensionless ratio, akela $p_{c}$ nahi, $k$ set karta hai — exactly parent ka Mistake #1.

Step 2. Threshold $0.20$ se compare karo. Hamare paas $0.10 < 0.20$ hai → injector too soft hai; chugging likely hai. Degenerate limit $Δ p_{inj} \to 0$ $\overset{m}{˙}_{in}$ ko perfectly $p_{c}$ follow karne deta ( $k \to \infty$ ) — maximum coupling. Ye step kyun? Limit check karne se pata chalta hai kaunsi direction fix karti hai: $Δ p_{inj}$ badhao jab tak ratio $0.20$ tak na pahunche, yaani $Δ p_{inj} \geq 1.4$ MPa.

Verify: $0.20 \times 7.0 = 1.4$ MPa required; hamare paas $0.7$ MPa hai, requirement ka exactly aadha → factor of 2 se fail. "Too soft" ke saath consistent. ✔

Cell F — High-frequency band, first tangential mode

Worked example

Ex 6 (Cell F). Chamber: $T_{c} = 3200$ K, $γ = 1.20$ , $R_{g a s} = 320$ J/(kg·K), radius $R = 0.12$ m. First tangential (1T) frequency nikalo.

Forecast: Hot gas, chhota radius → kHz → screaming. Guess: ~3 kHz.

Step 1. Sound speed $a = γ R_{g a s} T_{c} = 1.20 \times 320 \times 3200 = 1.2288 \times 1 0^{6} \approx 1108.5$ m/s. Ye step kyun? Acoustic clock ko local speed of sound chahiye; ye pressure wave ka transit speed hai Acoustic Modes of a Cylindrical Cavity ke around.

Step 2. 1T mode Bessel root $α_{10} = 1.841$ use karta hai, longitudinal index $ℓ = 0$ : $f_{1 T} = \frac{a}{2} \cdot \frac{α _{10}}{π R} = \frac{1108.5}{2} \cdot \frac{1.841}{π ( 0.12 )} .$ Ye step kyun? $ℓ = 0$ ke saath general cavity formula sirf transverse term tak collapse ho jaata hai — ek wave jo side-to-side sloshes, sabse zyada destructive mode.

Step 3. $= 554.25 \times \frac{1.841}{0.37699} = 554.25 \times 4.8834 \approx 2707$ Hz. Ye step kyun? kHz confirm karta hai screaming, chamber acoustics se driven — plumbing se nahi.

Verify: Band check: $2707$ Hz $\in [1000, 15000]$ → screaming ✔. Cross-link: Injector Baffles aur Helmholtz Resonator cavities se cure hota hai. ✔

Cell G — Limiting behaviour: pure longitudinal & large index

Worked example

Ex 7 (Cell G). Same chamber ( $a \approx 1108.5$ m/s), length $L = 0.30$ m. (a) Pehla pure longitudinal mode nikalo (transverse root = 0, $ℓ = 1$ ). (b) Jab $ℓ \to \infty$ , $f$ kaise behave karta hai?

Forecast: Longitudinal length use karta hai, radius nahi; $L > R$ isliye Ex 6 ke 1T se lower guess karo, shayad ~2 kHz. Aur large $ℓ$ → frequency bina kisi limit ke badhti hai.

Step 1. Transverse term zero ke saath, cavity formula reduce ho jaata hai $f_{00 ℓ} = \frac{a}{2} \cdot \frac{ℓ}{L} .$ Ye step kyun? $α_{mn} = 0$ set karne se radial/tangential contribution khatam ho jaata hai, classic organ-pipe result $f = ℓ a / (2 L)$ bach jaata hai.

Step 2. $ℓ = 1$ : $f_{001} = \frac{1108.5}{2} \cdot \frac{1}{0.30} = 554.25 \times 3.333 = 1847.5$ Hz. Ye step kyun? Ye chamber ki length mein fundamental standing wave hai.

Step 3. Limit: $f_{00 ℓ} = \frac{a}{2 L} ℓ$ $ℓ$ mein linear hai, isliye $f \to \infty$ jab $ℓ \to \infty$ — higher harmonics $1847.5$ Hz ke integer multiples hain. Physically, ever-shorter wavelengths thermoacoustic losses se damp ho jaate hain isliye sirf sabse neeche waale kuch matter karte hain. Ye step kyun? Limit check karne se pata chalta hai spectrum unbounded hai lekin sirf low modes dangerous hain — real losses useful set ko cap karte hain.

Verify: $ℓ = 2$ exactly $2 \times 1847.5 = 3695$ Hz hona chahiye (integer harmonic) ✔. Aur $1847.5 < 2707$ (Ex 6 1T) jaise forecast kiya, kyunki $L > R$ -scaled length. ✔

Cell H — Real-world word problem

Worked example

Ex 8 (Cell H). Ek engine test rig mein ek steady 90 Hz "put-put" sunai deta hai jo tab chala jaata hai jab team injector orifice pressure drop badhati hai. Ek doosri, faint 4.2 kHz whistle sirf full thrust par aati hai. Dono diagnose karo aur back out karo (a) combustion lag aur (b) confirm karo ki whistle ek acoustic mode hai agar $a = 1150$ m/s, $R = 0.11$ m.

Forecast: 90 Hz + injector se cure = chugging. 4.2 kHz = screaming. Guess $τ \approx 2.8$ ms.

Step 1. 90 Hz low band mein hai aur injector stiffness se respond karta hai → chugging. $f = \frac{1}{4 τ}$ invert karo: $τ = \frac{1}{4 f} = \frac{1}{4 ( 90 )} = 2.78 \times 1 0^{- 3}$ s $= 2.78$ ms. Ye step kyun? Cure (injector $Δ p$ ) feed reservoir ka fingerprint hai, isliye hum $τ$ recover karne ke liye chug clock use karte hain.

Step 2. 4.2 kHz high band mein hai → screaming, likely 1T. 1T predict karo: $f_{1 T} = \frac{a}{2} \frac{1.841}{π R} = \frac{1150}{2} \cdot \frac{1.841}{π ( 0.11 )} = 575 \times \frac{1.841}{0.34558} = 575 \times 5.328 = 3063$ Hz. Ye step kyun? Predicted 1T ( $\approx 3.06$ kHz) observed 4.2 kHz ke same order ka hai — whistle ek chamber acoustic mode hai, plumbing nahi (observed value zyada hona ek coupled 1T+1L mode suggest karta hai).

Verify: $τ = 2.78$ ms physical few-ms range mein hai ✔; recomputed $f = 1/ (4 \times 2.78 ms) = 89.9 \approx 90$ Hz ✔. Screaming order-of-magnitude (kHz) confirmed ✔. ✔

Cell I — Exam twist: sahi cure chunno

Worked example

Ex 9 (Cell I). Exam: "Ek chamber apne 1T mode par screams karta hai. Ek engineer propellant feed lines ko deepen karne ka proposal deta hai lag add karne ke liye. Kya isse fayda hoga? Actually kya cure karta hai, aur kyun — Rayleigh se argue karo."

Forecast: Feed changes slow reservoir ko touch karte hain; screaming fast wale mein rehta hai. Guess: nahi, wrong reservoir.

Step 1. Reservoir identify karo. Screaming = chamber-gas acoustics, clock = acoustic transit (μs). Feed lag $τ$ (ms) ise influence karne ke liye ek hazaar guna slow hai. Ye step kyun? Parent ka core lesson: same Rayleigh rule, different time-delay reservoir. Wrong reservoir fix karne se kuch nahi hota.

Step 2. Real cure $\oint p^{'} q^{'} d t$ ko net-negative banata hai mode frequency par acoustic loss add karke: Injector Baffles transverse (1T) pattern ko chop karte hain, aur liner mein tuned Helmholtz Resonator cavities $f_{1 T}$ par energy absorb karti hain. Ye step kyun? Distributed flame ka phase aasani se change nahi kiya ja sakta, isliye aap magnitude par attack karo — itna damping pile karo ki negative-loss term Cell A ki positive drive ko overwhelm kar de.

Verify: Consistency check — ye exactly parent Mistake "Baffles fix chugging" ulta hai: baffles/resonators screaming fix karte hain, injector $Δ p$ chugging fix karta hai. Reservoir logic Ex 4/Ex 5/Ex 6 se match karta hai. ✔

Recall

250 Hz instability jo injector stiffen karne par gayab ho jaati hai, kaunsi type hai? ::: Chugging (low-frequency, feed-system reservoir). 3 kHz instability jisme baffles aur Helmholtz cavities chahiye, kaunsi type hai? ::: Screaming (high-frequency, chamber-acoustic reservoir). Heat pressure se $6 0^{\circ}$ lag karti hai — grow ya die? ::: Grow, kyunki $cos 6 0^{\circ} = + 0.5 > 0$ ( $\pm 9 0^{\circ}$ window ke andar). Heat pressure se $20 0^{\circ}$ lag karti hai — grow ya die? ::: Die, kyunki $cos 20 0^{\circ} < 0$ (damping window ke andar).