Exercises — Combustion instability — low-frequency (chugging), high-frequency (screaming)

150 Hz 10–400 Hz chugging band mein aata hai. Toh yeh chugging hai. Delay store karne wala reservoir feed system hai (propellant lines + injector pressure drop): disturbance ko plumbing se guzarna padta hai aur combustion lag $\tau$ ka intezaar karna padta hai, jo ek slow clock hai (milliseconds). Slow clock ⇒ low frequency. ✔

4200 Hz 1000–15000 Hz band mein hai ⇒ screaming (screech). Reservoir chamber gas khud hai, ek organ pipe ki tarah bajta hua. Iska clock chamber ke aakross acoustic transit time hai — microseconds — feed lag se hazaar guna tez, isliye kHz. ✔

$f_{chug}\approx \frac{1}{4\tau }$ use karo. $\frac{1}{4}$ factor isliye aata hai kyunki delayed heat ko jab pressure phir se badh rahi ho tab pahunchne ke liye quarter-cycle of phase chahiye. $f_{chug}=\frac{1}{4(3.2\times 10^{-3})}=\frac{1}{0.0128}\approx 78.1\text{Hz}.$ Yeh chugging band mein baithta hai. ✔

$a=\sqrt{\gamma R_{gas}{T}_c}$ — yeh woh speed hai jis par ek pressure blip hot gas se guzarta hai, aur yahi woh clock hai jo acoustic frequencies set karta hai. $a=\sqrt{1.20\times 320\times 2800}=\sqrt{1,075,200}\approx 1036.9\text{m/s}.$ ✔

$\ell =0$ ke saath longitudinal term vanish ho jaata hai, toh $f_{1T}=\frac{a}{2}\cdot \frac{{\alpha }_{10}}{\pi R}=\frac{1036.9}{2}\cdot \frac{1.841}{\pi (0.12)}.$ Numerator: $518.45\times 1.841=954.5$. Andar denominator: $\pi (0.12)=0.3770$. $f_{1T}=\frac{954.5}{0.3770}\approx 2532\text{Hz}.$ kHz ⇒ screaming. ✔ Neeche mode picture dekho.

Ek full cycle mein cross-term integral $\cos \varphi$ ke proportional hota hai (sine-cross terms average to zero ho jaate hain). Toh $\oint p^{\prime }{q}^{\prime }dt\propto \cos 60^{\circ }=+0.5>0.$ Positive ⇒ heat mostly jab pressure high ho tab aa rahi hai ⇒ badhti hai. Growth window hai $|\varphi |<90^{\circ }$; $60^{\circ }$ uske andar hai. ✔ Phase figure dekho.

Integral $\propto \cos \varphi$ sign tab badalta hai jab $\cos \varphi =0$, yaani $\varphi =90^{\circ }$ par (aur phir $270^{\circ }$ par). Toh growth window hai $0\le \varphi <90^{\circ }$ aur $270^{\circ }<\varphi \le 360^{\circ }$; damping window hai $90^{\circ }<\varphi <270^{\circ }$. $\varphi =135^{\circ }$ par: $\cos 135^{\circ }=-0.7071<0$ ⇒ dampen hoti hai. Design goal: $\varphi$ ko damping window mein dhakel do. ✔

Required minimum: $0.2\times 6.0=1.2$ MPa. Hamare paas sirf $0.9$ MPa hai, toh yeh rule fail karta hai — injector bahut "soft" hai, ${\dot{m}}_{in}$ ko $p_c$ ke peeche bhaagne deta hai aur chug feedback loop band karta hai. Shortfall $=1.2-0.9=0.3$ MPa. Humein $\Delta p_{inj}$ kam se kam $0.3$ MPa badhana hoga. ✔

(a) Predicted chug: $f=\frac{1}{4\tau }=\frac{1}{4(2.6\times 10^{-3})}=96.2$ Hz. Observed 95 Hz ~1% se match karta hai. ⇒ Haan, yeh feed-system chugging hai jo $\tau$ se key hai. ✔ (b) Required $0.2p_c=0.2(5.0)=1.0$ MPa; hamare paas $0.8$ MPa hai ⇒ bahut soft, $0.2$ MPa kam. (c) Injector stiffen karo taki $\Delta p_{inj}\ge 1.0$ MPa ho. Rayleigh se kyun kaam karta hai: ek stiff injector ${\dot{m}}_{in}$ ko $p_c$ se almost independent bana deta hai, toh pressure dip extra propellant nahi bulaata. Delayed heat release $q^{\prime }$ $p^{\prime }$ ko track karna band kar deta hai, phase overlap collapse ho jaata hai, aur $\oint p^{\prime }{q}^{\prime }dt$ zero se neeche gir jaata hai ⇒ loop khud ko feed nahi kar sakta. ✔

Screaming chamber acoustics mein rehti hai, feed loop mein nahi. $\tau$ ghatana (i) sirf chugging clock shift karta hai — $f_{chug}$ badalta hai, acoustic mode $f_{1T}$ nahi (jo $a$, $R$, $L$ par depend karta hai). Toh (i) screech par essentially kuch nahi karta. Option (ii), Helmholtz cavities 3000 Hz par tune kiye hue, exact frequency par baithte hain aur wahaan acoustic energy absorb karte hain — woh loss add karte hain taki 1T mode ke liye $\oint p^{\prime }{q}^{\prime }dt$ net-negative ho jaaye. Yahi sahi ilaaj hai. ✔ (Deeper mechanism ke liye dekho Helmholtz Resonator aur Injector Baffles.)

$f_{chug}=\frac{1}{4\tau }$ se, $f_{chug}\le 60$ Hz require karne par $\tau \ge \frac{1}{4\times 60}=\frac{1}{240}\approx 4.17\times 10^{-3}\text{s}=4.17\text{ms}.$ Trade-off: lamba lag chugging ko neeche dhakelti hai, lekin lamba combustion lag matlab propellant unburned reh kar late karta hai — poor mixing, lower c* efficiency, aur zyaada bada chamber volume chahiye. Toh tum stability sirf $\tau$ ke zariye nahi chase kar sakte: asli lever usually injector stiffness hai (gain $k$ todna), jalanaa slow karna nahi. ✔

$\ell =0$ ke saath: $f=\frac{a}{2}\cdot \frac{\alpha }{\pi R}$, aur $\frac{a}{2\pi R}=\frac{1100}{2\pi (0.10)}=1750.7$ Hz per unit $\alpha$.

• 1T: $1750.7\times 1.841=3223$ Hz.
• 2T: $1750.7\times 3.054=5347$ Hz.
• 1R: $1750.7\times 3.832=6708$ Hz.
• 1L ($\alpha =0,\ell =1$): $f=\frac{a}{2}\cdot \frac{1}{L}=\frac{1100}{2}\cdot \frac{1}{0.30}=1833$ Hz.

Low→high rank: 1L (1833) < 1T (3223) < 2T (5347) < 1R (6708) Hz. Sabse dangerous = 1T: iska pressure side-to-side sloshes karta hai, hot gas ko wall ke against scrub karta hai aur metal mein fastest heat loss drive karta hai. ✔ (Dekho Acoustic Modes of a Cylindrical Cavity.)

$f=\frac{a}{2}\sqrt{{\left(\frac{1.841}{\pi (0.10)}\right)}^2+{\left(\frac{1}{0.30}\right)}^2}.$ Andar: $\frac{1.841}{0.31416}=5.859$, squared $=34.33$. Aur $\frac{1}{0.30}=3.333$, squared $=11.11$. Sum $=45.44$, root $=6.741$. $f=\frac{1100}{2}\times 6.741=550\times 6.741\approx 3708\text{Hz}.$ Pure 1T (3223 Hz) se zyaada, kyunki ek longitudinal half-wave add karna use badhata hi hai. ✔