Exercises — Hybrid engines — advantages, disadvantages

Answer: (c). Definition ke hisaab se ek classical hybrid fuel ko solid grain ki tarah aur oxidiser ko fluid ki tarah rakhta hai jo central port mein inject hoti hai.

• (a) ek liquid engine hai.
• (b) ek solid motor hai (fuel aur oxidiser bake hokar ek saath hain).
• (d) reverse hybrid hai — physically possible hai lekin jo "classical" pucha gaya hai wo nahi.

Oxidiser (${\dot{m}}_{ox}$). Fuel ek solid wall hai — wo sirf tab vaporise hoti hai jab hot oxidiser uske upar se flow kare. Oxidiser band karo aur wall fuel dena band kar deti hai, toh total flow $\dot{m}\to 0$ aur thrust khatam. Flowing reactant wahi hai jiske paas control knob hai.

Tool: thrust equation $F=\dot{m}{v}_e+(p_e-p_a)A_e$, jahan (toolbox se) $p_e$ exit pressure hai, $p_a$ ambient pressure hai, aur $A_e$ nozzle exit area hai. Perfectly expanded ka matlab $p_e=p_a$ hai, toh $(p_e-p_a)=0$ aur poora doosra term zero ho jaata hai. Yeh tool kyun: thrust wo momentum hai jo har second peechhe feka jaata hai — mass per second times speed — plus ek correction jab exit pressure baahri hawa se match nahi karti. $F=\dot{m}{v}_e=(3.5)(2300)=\boxed{8050\text{N}}$

Tool: ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}$ (dekho Regression Rate and Boundary Layer Combustion). Kyun: fuel area $A_b$ ki wall se speed $\dot{r}$ par peel hoti hai; swept volume per second $A_b\dot{r}$ (m³/s) hai; density se multiply karo kg/s milta hai. ${\dot{m}}_{fuel}=920\times 0.40\times 1.2\times 10^{-3}=\boxed{0.4416\text{kg/s}}$

Tool: $G_{ox}={\dot{m}}_{ox}/A_{port}$. Kyun: flux = hole ke cross-section se squeeze hoti flow. $G_{ox}=\frac{2.0}{0.010}=\boxed{200{\text{kg⋅s}}^{-1}{\text{m}}^{-2}}$

Step 1 — old thrust. $F_0=({\dot{m}}_{ox}+{\dot{m}}_{fuel})v_e=(2.0+1.0)(2400)=7200$ N. Kyun: thrust equation ke hisaab se, har second total mass $\dot{m}={\dot{m}}_{ox}+{\dot{m}}_{fuel}$ speed $v_e$ par peechhe phenki jaati hai, peechhe $\dot{m}{v}_e$ momentum lete hue; Newton ke third law se rocket ko wahi momentum aage thrust ke roop mein milta hai. Perfect expansion ke saath pressure term zero hai, toh $F_0=\dot{m}{v}_e$ — isliye hum pehle dono flows jodte hain, phir $v_e$ se multiply karte hain. Step 2 — fuel kaise respond karta hai? (chain banao, links skip mat karo.)

• Fuel flow kabhi independent nahi hoti — wo ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}$ follow karti hai. Yahan ${\rho }_f$ aur $A_b$ fixed hain, toh ${\dot{m}}_{fuel}\propto \dot{r}$.
• Regression rate $\dot{r}=a{G}_{ox}^n$ follow karti hai, toh $\dot{r}\propto G_{ox}^{0.6}$.
• Flux $G_{ox}={\dot{m}}_{ox}/A_{port}$ hai; port fixed hone se $G_{ox}\propto {\dot{m}}_{ox}$.
• Inhe chain karo: ${\dot{m}}_{fuel}\propto \dot{r}\propto G_{ox}^{0.6}\propto {\dot{m}}_{ox}^{0.6}$. Toh oxidiser ko aadha karne se fuel $0.5^{0.6}=0.660$ se multiply hoti hai, nayi fuel $=1.0\times 0.660=0.660$ kg/s milti hai. Kaisa dikhta hai: figure mein, amber fuel bar amber oxidiser bar se kam steeply girta hai — kyunki exponent $0.6<1$ fall ko soften karta hai. Step 3 — new thrust. $F=(1.0+0.660)(2400)=3984$ N. $\frac{F_{\text{new}}}{F_{\text{old}}}=\frac{3984}{7200}=0.553\approx \boxed{55\%}$ Oxidiser aadha karne se thrust fraction-wise aadhe se zyada girta hai? Nahi — thrust $55\%$ tak girta hai, yaani lagbhag aadha hota hai. Dono terms gire, lekin fuel ne resist kiya (exponent $<1$), toh total $50\%$ tak nahi gira. Yeh coupling hi hybrid throttle karne ka poora point hai.

Arrows follow karo, ek physical link at a time:

1. $A_{port}\uparrow$ (port chauda hota hai jaise wall peeche hatti hai).
2. $G_{ox}={\dot{m}}_{ox}/A_{port}\Rightarrow G_{ox}\downarrow$ (wahi oxidiser, bada hole → patla flux).
3. $\dot{r}=a{G}_{ox}^n\Rightarrow \dot{r}\downarrow$ (kamzor flux → dheema wall recession).
4. ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}$: haalaanki $A_b$ thoda badh sakta hai, flux fall usually dominate karta hai, toh ${\dot{m}}_{fuel}\downarrow$.
5. $\text{O/F}={\dot{m}}_{ox}/{\dot{m}}_{fuel}$: numerator constant, denominator gir raha hai $\Rightarrow$ O/F badhta hai — engine oxidiser-rich drift karta hai. Optimal O/F se yeh drift $I_{sp}$ ko thoda girata hai — yeh hybrid ka signature disadvantage hai.

Chain karo, ek earned symbol at a time.

1. $G_{ox}=\frac{{\dot{m}}_{ox}}{A_{port}}=\frac{1.6}{8.0\times 10^{-3}}=200$ kg·s⁻¹·m⁻².
2. $\dot{r}=a{G}_{ox}^n=3.0\times 10^{-5}\times 200^{0.5}=3.0\times 10^{-5}\times 14.142=4.243\times 10^{-4}$ m/s.
3. ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}=920\times 0.50\times 4.243\times 10^{-4}=0.1952$ kg/s.
4. $\text{O/F}=\frac{{\dot{m}}_{ox}}{{\dot{m}}_{fuel}}=\frac{1.6}{0.1952}=8.20$.
5. $\dot{m}={\dot{m}}_{ox}+{\dot{m}}_{fuel}=1.6+0.1952=1.7952$ kg/s.
6. $F=\dot{m}{v}_e=1.7952\times 2350=\boxed{4219\text{N}}$ (lagbhag $4.22$ kN). O/F $8.2$ notice karo — hybrids typically oxidiser-rich run karte hain; yeh normal hai.

Yeh formula kyun: $I_{sp}$ thrust measure karta hai propellant ke unit weight-flow per — seconds mein fuel-economy score. $I_{sp}=\frac{F}{\dot{m}{g}_0}=\frac{4219}{1.7952\times 9.81}=\frac{4219}{17.61}=\boxed{239.6\text{s}}$ Kyunki humne perfect expansion assume kiya, $F=\dot{m}{v}_e$, toh equivalently $I_{sp}=v_e/g_0=2350/9.81=239.6$ s — ek neat cross-check. Real hybrids $250$–$350$ s range mein aate hain; hamara simplified number low end par hai, jaise model ke liye expected hai.

"Bas $5\times$ oxidiser" kyun fail hota hai: Thrust $\approx \dot{m}{v}_e$ aur $\dot{m}$ mainly ${\dot{m}}_{ox}$ se dominate hoti hai, toh naively $5\times$ oxidiser thrust raise karta hai — lekin fixed geometry par yeh O/F ko optimal se bahut door le jaata hai. Yahan ek sentence mein derivation hai kyun O/F badhta hai: shuru karo $\text{O/F}={\dot{m}}_{ox}/{\dot{m}}_{fuel}$ se; fixed port area par ${\dot{m}}_{fuel}\propto \dot{r}\propto G_{ox}^n\propto {\dot{m}}_{ox}^n$, toh substitute karne se $\text{O/F}\propto {\dot{m}}_{ox}/{\dot{m}}_{ox}^n={\dot{m}}_{ox}^{1-n}$ milta hai, aur kyunki $n<1$ exponent $1-n>0$ hai iska matlab O/F oxidiser badhne ke saath badhta hai. Bahut oxidiser-rich run karna fuel waste karta hai, combustion efficiency girata hai, aur nozzle ko jalata hai. Toh sirf thrust hi nahi badalta — mixture degrade hota hai. Recommended fix (parent note se): ek multi-port grain — kaafi parallel ports khaodo taaki total burn area $A_b$ badhe, ${\dot{m}}_{fuel}$ ko ${\dot{m}}_{ox}$ ke saath pace rakhne do aur O/F optimal ke paas rakho. Ek port high thrust par itna $A_b$ nahi de sakta. Fixed O/F par burn-area estimate: fixed O/F par, ${\dot{m}}_{fuel}$ bhi $\times 5$ scale hona chahiye. Kyunki ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}$ hai, agar hum $\dot{r}$ hold karein (enlarged geometry mein roughly fixed flux $G_{ox}$ rakh ke) toh $A_b$ ko $\times 5$ jaana chahiye: $A_b^{\text{new}}\approx 5\times 0.50=\boxed{2.5{\text{m}}^2}$ Utna extra wall area exactly wahi hai jo ek multi-port (wagon-wheel) grain provide karta hai.

Limit link by link lo (with $n>0$):

• $G_{ox}={\dot{m}}_{ox}/A_{port}\to 0$.
• $\dot{r}=a{G}_{ox}^n\to a\cdot 0^n=0$ ($n>0$ ke liye).
• ${\dot{m}}_{fuel}={\rho }_f{A}_b\dot{r}\to 0$ — solid vaporise hona band ho jaati hai kyunki koi hot oxidiser use nahi nahlaa raha.
• $\dot{m}={\dot{m}}_{ox}+{\dot{m}}_{fuel}\to 0+0=0$.
• $F=\dot{m}{v}_e+(p_e-p_a)A_e\to 0$ (chamber depressurise hota hai, toh pressure term bhi collapse ho jaata hai). Words mein: valve band karna har term ko smoothly zero tak le jaata hai. Koi residual burn nahi hoti, kyunki fuel ka koi independent oxidiser nahi hai (unlike ek solid motor). Valve dobara khoolo aur oxidiser flux wapas aata hai → $\dot{r}$ zero se upar chadhta hai → combustion resume hoti hai. Yeh ek clean stop aur restart hai — mathematically $\dot{r}=a{G}_{ox}^n$ ke $G_{ox}=0$ par vanish hone se forced.

Kaunsa ek physical fact tha jisne is ladder par har advantage aur disadvantage generate kiya? ::: Fuel aur oxidiser alag phases mein store hote hain aur physically separated hote hain, jisme oxidiser valved hota hai — har cheez (safety, throttling, O/F shift, low regression, multi-port scaling) isi se flow karti hai.