Acoustic modes in combustion chamber — cause of instability: Question bank

3.3.33 · D5 · Physics › Rocket Propulsion › Acoustic modes in combustion chamber — cause of instability

In traps ke peeche ki machinery ke liye, dekho Standing waves and resonance, Speed of sound in gases, aur Rayleigh Criterion.

Is bank mein use hone wale Symbols (pehle ye padho)

Traps se pehle, neeche har symbol kahin na kahin answers mein aata hai. Inhe assume nahi kiya gaya hai — yahan bilkul clearly likha hai ki har ek ka kya matlab hai.

Figure — Acoustic modes in combustion chamber — cause of instability

True ya False — justify karo

Ek rigid chamber wall pressure node force karta hai.

False — rigid wall gas velocity ko zero force karta hai, jo pressure antinode (maximum) hai. Motion block karne se pressure wall par pile up hoti hai, cancel nahi hoti.

Zyada bada heat release hamesha engine ko zyada unstable banata hai.

False — phase

ϕ

decide karta hai ki energy transfer ka sign kya hoga

cos ϕ

ke zariye; amplitude sirf ye set karta hai ki kitni tezi se ek already-growing mode grow karta hai.

18 0^{\circ}

out of phase release ki heat actively damp karti hai, chahe kitni bhi badi ho.

Agar Rayleigh integral

R

positive hai, toh mode ka badhna guaranteed hai.

False — positive

R

drive hai, lekin real chambers mein losses bhi hoti hain (nozzle radiation, viscosity). Growth ke liye drive ko losses se zyada hona chahiye, sirf positive hona kaafi nahi.

Cold-air value

c \approx 340 m/s

use karna chamber mode frequency ka ek reasonable estimate deta hai.

False — combustion products

3000 K

ke paas hote hain, toh

c \approx 900

–

1300 m/s

. Kyunki

f_{n} \propto c

hai, cold-air

c

frequency ko roughly teen ke factor se kam estimate karta hai.

Tangential modes usually sabse zyada destructive hote hain.

True — tangential mode ka pressure antinode chamber wall par baithta hai aur injector face ke around ghoomta hai, injectors ke sabse dense ring ke directly upar se guzarta hai, isliye ye axial mode se kahin zyada strongly combustion se couple karta hai jiska antinode gas ke andar ek flat sheet hai.

Pressure ke saath exactly

9 0^{\circ}

out of phase add ki gayi heat mode ko na drive karti hai na damp.

True — cycle-averaged drive

cos ϕ

ke anusaar hota hai, aur

cos 9 0^{\circ} = 0

, toh ek cycle mein net work exactly zero hai (neutral).

Fundamental longitudinal mode chamber mein ek poori wavelength fit karta hai.

False — fundamental (

n = 1

) do pressure antinodes ke beech half wavelength fit karta hai;

f_{n} = n c /2 L

L = nλ /2

se aata hai.

Do closed ends aur do open ends ek jaisi longitudinal frequency formula dete hain.

True (form mein) — dono ek jaisi ends dete hain

f_{n} = n c /2 L

jahan

n = 1, 2, 3, \dots

; fark ye hai ki antinodes vs nodes kahan baithte hain, spacing mein nahi. Mixed (ek-closed-ek-open) tube woh hai jo formula badalta hai.

Ek-closed, ek-open tube wahi

f_{n} = n c /2 L

use karta hai jo symmetric tube use karta hai.

False — mixed tube sirf odd quarter-wavelengths fit karta hai, deta hai

f_{n} = \frac{( 2 n - 1 ) c}{4 L}

(

n = 1, 2, 3, \dots

): closed end par pressure antinode aur open end par node. Iska fundamental closed-closed tube se aadha hota hai.

Error dhundo

"Injector face par gas rush karta hai andar, toh woh pressure node hona chahiye."

Reflecting face zyada rigid wall ki tarah behave karta hai (velocity ≈ 0), deta hai pressure antinode, node nahi. Steady inflow mean flow hai, acoustic velocity fluctuation

u^{'}

nahi.

"Kyunki

c = γ R T / M

, temperature double karne par mode frequency double ho jaati hai."

Nahi —

c \propto T

hai, toh

T

double karne par

c

(aur

f_{n}

) sirf

2 \approx 1.41

se multiply hota hai,

2

se nahi.

"Chota chamber mode frequency kam karta hai kyunki hilane ke liye gas kam hai."

Ulta hai —

f_{n} = n c /2 L

, toh chota

L

f_{n}

badhata hai. Chota organ pipe zyada uunchi awaaz deta hai.

ϕ = 18 0^{\circ}

matlab heat aur pressure ek dusre se fight karte hain, jo engine ko zyada shake karta hai."

Ye oppose zaroor karte hain, lekin oppose karne ka matlab hai ki heat wave se remove hoti hai jab pressure high hota hai —

cos 18 0^{\circ} = - 1

deta hai damping, sabse tez case nahi, sabse shant case.

"Wave equation

\partial_{t}^{2} p^{'} = c^{2} \nabla^{2} p^{'}

sirf bade, violent pressure swings ke liye hold karta hai."

Iska ulta — ye mean state ke around linearize karne se aata hai, toh ye chote perturbations

p^{'}

ke liye hold karta hai. Genuinely bade swings ke liye poori nonlinear equations chahiye.

"Transverse (tangential/radial) modes chamber length

L

par depend karte hain."

Ye radius

R

par depend karte hain Bessel roots ke zariye (

f_{mn} \propto α_{mn} c /2 π R

); sirf longitudinal modes

L

se set hote hain.

"Kyunki drive

cos ϕ

ke anusaar scale hoti hai,

ϕ = 27 0^{\circ}

ka phase unstable hai."

cos 27 0^{\circ} = 0

, toh ye neutral hai, unstable nahi. Sirf woh phases jo

cos ϕ > 0

dete hain (roughly

- 9 0^{\circ} < ϕ < 9 0^{\circ}

) growth drive karte hain.

"Wavenumber

k

aur frequency ek hi cheez hai."

Nahi —

k = 2 π / λ

wave-cycles per metre count karta hai (space), jabki frequency cycles per second count karti hai (time). Ye

ω = c k

se linked hain, lekin

k

ek spatial quantity hai.

Why questions

Combustion chamber mein discrete "notes" kyun hote hain — har frequency kyun nahi?

Sirf woh wavelengths jo boundaries se reflect karke ek saath wapas aati hain standing waves ke roop mein survive karti hain; baaki sab destructively interfere karte hain. Boundary condition

k L = nπ

(

L

mein fit hone wale half-wavelengths ki poori sankhya) ek discrete set select karta hai.

Heat release aur pressure ke beech phase

ϕ

, heat ki matra nahi, stability ka deciding factor kyun hai?

Wave ko di gayi cycle-averaged energy

\propto cos ϕ

hai; agar heat out of step aati hai, toh positive aur negative contributions cancel ya reverse ho jaate hain, chahe har burst kitna bhi bada kyun na ho.

Pressure peak par heat add karna oscillation ko amplify kyun karta hai?

Ye swing ko arc ke top par push karne jaisa hai — energy exactly tab enter karti hai jab wave sabse zyada "stretched" hoti hai (

p^{'}

maximal,

ϕ = 0

), toh har cycle zyada amplitude ke saath wapas aata hai (constructive drive).

Engineers baffles aur acoustic cavities se phase

ϕ

18 0^{\circ}

ki taraf kyun push karte hain?

cos ϕ

ko negative banane ke liye, taaki combustion har cycle mein acoustic energy remove kare (dekho Injector design and baffles aur Nozzle flow and acoustic damping) — ek potential driver ko damper mein badal dete hain.

Nozzle throat stability ke liye kyun important hai jab instability flame zone mein shuru hoti hai?

Throat waves ko partly reflect karta hai (boundary condition aur mode shape set karta hai) aur acoustic energy ko radiate out hone deta hai — ye ek key loss channel hai jise Rayleigh drive se aage rehna chahiye.

Ek jaisi size ke do engines ki mode frequencies alag kyun ho sakti hain?

Kyunki

f_{n} \propto c

aur

c = γ R T / M

temperature aur composition (

γ

M

) par depend karta hai; alag propellants ya mixture ratios

c

aur isliye har mode ko shift karte hain.

Tangential modes radial ya longitudinal ones se zyada combustion se strongly couple kyun karte hain?

Unka pressure antinode outer wall ke saath chalte hue injector elements ki ring ke paas se rotate karta hai jahan fresh propellant enter karta hai, toh pressure fluctuation directly wahan burning ko modulate karti hai jahan combustion sabse zyada sensitive hoti hai — Rayleigh integral mein

p^{'} q^{'}

overlap maximize karta hai.

Edge cases

Agar heat-release fluctuation

q^{'}

har jagah exactly zero hai, toh kya mode unstable ho sakta hai?

Nahi — koi fluctuating heat nahi toh wave ko feed karne ke liye kuch nahi hai; Rayleigh integral

R

zero hai aur koi bhi acoustic energy sirf losses ke zariye decay hogi.

Jab chamber length

L \to 0

ki limiting case mein mode frequencies ka kya hoga?

f_{n} = n c /2 L \to \infty

— modes arbitrarily high frequency ki taraf push karte hain aur effectively dangerous structural band se bahar ho jaate hain; ek vanishing chamber mein koi low acoustic resonance nahi hoti.

Agar drive aur losses exactly equal hain, toh mode ka behaviour kya hai?

Ye neutral (marginal) stability boundary par baitha hai — amplitude na badhti hai na ghatti, ek steady limit oscillation hold karta hai. Koi bhi choti extra drive ise unstable kar deti hai.

Exactly

ϕ = 9 0^{\circ}

ke phase ke liye, kya mode hamesha safe hai?

Sirf marginally — drive zero hai toh badhta nahi, lekin combustion se damp bhi nahi ho raha; real losses tab decide karte hain, aur

ϕ

mein koi bhi drift

0^{\circ}

ki taraf ise unstable bana deti hai.

Agar mean gas sach mein rest par hota aur perfectly lossless hota in-phase heat ke saath, toh model kya predict karta?

Unbounded exponential growth — positive Rayleigh drive aur zero losses ke saath kuch bhi mode ko check nahi karta, toh linear model blow up ho jaata hai (real nonlinearities eventually ise limit cycle par cap karte hain).

Pressure node par, acoustic velocity kya kar rahi hai?

Ye apne antinode (maximum) par hai — pressure aur velocity quarter-wave apart hain, toh jahan pressure vanish hota hai wahan gas sabse tez slosh kar rahi hoti hai.

Recall Har trap ka ek-line summary

Primes = ripples ( $p^{'} = p - \overset{p}{ˉ}$ ); node = kabhi nahi hilta, antinode = sabse bada swing; wall velocity zero force karta hai → pressure antinode; $k = 2 π / λ$ , $ω = c k$ ; stability phase $ϕ$ se set hoti hai $R = \oint \int p^{'} q^{'} d V d t \propto cos ϕ$ ke zariye, amplitude se nahi; hot $c = γ R T / M \sim 1000$ m/s use karo; $f_{n} = n c /2 L$ (closed-closed), $f_{n} = (2 n - 1) c /4 L$ (mixed); tangential modes sabse zyada bite karte hain.