Exercises — Nozzle area ratio ε = A_e - A - — choosing for optimal performance

• Under-expanded: $p_e>p_a$. Gas exit par abhi bhi pressure se bhari hui hai; nozzle bahut chhotha hai (ε bahut chhota). Yeh figure s01 mein baaya plume hai, jo abhi bhi bahar phool rahi hai.
• Perfectly expanded (optimum): $p_e=p_a$. Figure s01 mein beech wala plume — ek saaf seedhi column.
• Over-expanded: $p_e<p_a$. Gas ambient se zyada expand ho gayi; bahari hawa plume ko andar ki taraf squeeze karti hai (figure s01 mein pinched daaya plume), pressure thrust negative ho jaata hai, separation shuru ho sakti hai.

$\epsilon =A_e/A^{\ast }=0.30/0.02=15$. Kyunki $\epsilon >1$ hai isliye diverging section hai — flow area throat ke baad badhti hai, jo already-sonic gas ko supersonic speeds tak accelerate karti hai.

Supersonic ($M_e>1$) root. Jaise figure s02 dikhati hai, $1/M_e$ front factor area–Mach curve ko U-shaped banata hai: $\epsilon =15$ par dashed line blue curve ko green (subsonic) dot aur red (supersonic) dot dono par cross karti hai. Throat (orange $M=1$ line) ke upstream flow subsonic hai; jab woh sonic throat se guzarti hai toh supersonic ho jaati hai, isliye downstream hum right-hand red ($M>1$) dot padhte hain.

Step 1 (KYA): Perfect expansion matlab $p_e=p_a=0.20$ bar. Toh $p_0/p_e=100/0.20=500$. Step 2 (YEH TOOL KYUN): Hume pressures pata hain, Mach chahiye — yeh exactly Mach–pressure relation ka kaam hai. $500={\left(1+0.1M_e^2\right)}^6,\frac{\gamma }{\gamma -1}=\frac{1.2}{0.2}=6.$ Step 3: 6th root lo: $1+0.1M_e^2=500^{1/6}=2.817$. $0.1M_e^2=1.817\Rightarrow M_e^2=18.17\Rightarrow M_e=\mathrm{4.26.}$

YEH TOOL KYUN: Ab hume $M_e$ aur $\gamma$ pata hai lekin ek geometry chahiye — exit-to-throat area ratio. Woh relation jo Mach number ko area se connect karta hai woh area–Mach relation $\epsilon =A_e/A^{\ast }$ hai; yeh hamare char tools mein se ek hi hai jisme area hai, isliye yeh sahi agla step hai. (Figure s02 padhte hue, hum red supersonic $M_e$ ko blue curve par uski $\epsilon$ height mein convert kar rahe hain.) Exponent: $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.2}{0.4}=5.5$. Front bracket factor $\frac{2}{\gamma +1}=\frac{2}{2.2}=0.909$. $\epsilon =\frac{1}{4.26}[0.909(2.817){]}^{5.5}=\frac{1}{4.26}(2.561{)}^{5.5}.$ $(2.561{)}^{5.5}=175.9$, toh $\epsilon =175.9/4.26=41.3$. Size padho: near-vacuum design ($p_a$ chhota) ko bahut expansion chahiye → bada ε (tens mein), parent note se consistent.

YEH TOOLS KYUN: "Ground par perfect expansion" $p_e=p_a=1.0$ bar fix karta hai, pressure ratio deta hai, toh hum pehle Mach–pressure relation use karte hain (pressures → $M_e$), phir $M_e$ ko area–Mach relation mein daalo taaki us Mach ko geometry ε mein badla ja sake — wahi do-step chain jaise L2.1–L2.2. $p_0/p_e=70/1.0=70$. Exponent $\gamma /(\gamma -1)=1.25/0.25=5$. $70=(1+0.125M_e^2{)}^5\Rightarrow 1+0.125M_e^2=70^{1/5}=\mathrm{2.336.}$ $0.125M_e^2=1.336\Rightarrow M_e^2=10.69\Rightarrow M_e=\mathrm{3.27.}$ Area exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.25}{0.5}=4.5$; front factor $\frac{2}{2.25}=0.889$. $\epsilon =\frac{1}{3.27}[0.889\times 2.336{]}^{4.5}=\frac{1}{3.27}(2.077{)}^{4.5}.$ $(2.077{)}^{4.5}=27.1$, toh $\epsilon =27.1/3.27=8.3$. Chhota ε — exactly wahi jo ek sea-level stage chahta hai.

$(p_e-p_a)A_e=(0.20-1.0)\times 10^5\times 1.2=(-0.80\times 10^5)(1.2)=-9.6\times 10^4\text{N}.$ Yeh term −96 kN hai: yeh thrust se subtract karta hai. Nozzle ground par badly over-expanded hai (uski $p_e$ moti sea-level hawa se bahut neeche hai) — figure s03 ka red region, matched point se kaafi daaye. Ambient hawa exit plane par push back karti hai; practically yeh severe mismatch flow separation bhi trigger karta (Flow Separation in Over-expanded Nozzles).

$dF/d{A}_e=p_e-p_a=(0.6-1.0)\times 10^5=-0.4\times 10^5<0$. Area add karna (bada ε) thrust decrease karega — tum over-expanded ho. Toh tumhe ε shrink karna chahiye jab tak $p_e$ upar aakar $p_a=1.0$ bar se na mile. Figure s03 exactly yahi dikhati hai: over-expanded (red) side par thrust curve neeche slope karti hai, toh tum green matched peak ki taraf baaye slide karo.

Engine A: $p_0/p_e=80/1.0=80\Rightarrow (1+0.1M_e^2{)}^6=80\Rightarrow 1+0.1M_e^2=80^{1/6}=2.076\Rightarrow M_e^2=10.76\Rightarrow M_e=\mathrm{3.28.}$ Engine B: $p_0/p_e=80/0.10=800\Rightarrow (1+0.1M_e^2{)}^6=800\Rightarrow 1+0.1M_e^2=800^{1/6}=3.06\Rightarrow M_e^2=20.6\Rightarrow M_e=\mathrm{4.54.}$ Kyunki area–Mach relation $M_e$ ke saath badhti hai (supersonic branch ke liye — figure s02 mein blue curve ka badhta hua right arm), aur B ka $M_e$ zyada hai, isliye ${\epsilon }_B>{\epsilon }_A$. Physically: patla bahari hawa ⇒ chhota target $p_e$ ⇒ zyada expansion ⇒ bada ε. High-altitude engines "bell-flared" giants hote hain.

Step 1 — $p_e$ choose karo: representative ambient se match karo, $p_e=0.55$ bar. $p_0/p_e=90/0.55=163.6$. Step 2 — $M_e$: $\gamma /(\gamma -1)=1.22/0.22=5.545$. $1+\frac{0.22}{2}{M}_e^2=163.6^{1/5.545}=163.6^{0.1803}=\mathrm{2.508.}$ $0.11M_e^2=1.508\Rightarrow M_e^2=13.71\Rightarrow M_e=\mathrm{3.70.}$ Step 3 — ε: area exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.22}{0.44}=5.045$; front factor $\frac{2}{2.22}=0.9009$. $\epsilon =\frac{1}{3.70}[0.9009\times 2.508{]}^{5.045}=\frac{1}{3.70}(2.259{)}^{5.045}.$ $(2.259{)}^{5.045}=60.1$, toh $\epsilon =60.1/3.70=16.2$. Design comment: liftoff par ($p_a=1.0$) yeh nozzle over-expanded hai ($p_e=0.55<1.0$); burnout par ($p_a=0.30$) yeh under-expanded hai ($p_e=0.55>0.30$). Middle se match karke, hum puri burn mein total penalty minimize karte hain — classic single-nozzle compromise, aur isliye Altitude Compensation — Aerospike Nozzles invent kiye gaye.

Specific impulse $I_{sp}=F/(\dot{m}{g}_0)$ hai, jahan $g_0=9.81{\text{m/s}}^2$ standard gravity hai (ek fixed constant jo weight-flow ko force mein convert karne ke liye use hoti hai, local gravity nahi). $\dot{m}$ aur $g_0$ fixed hone se, zyada $F$ ⇒ zyada $I_{sp}$. Perfectly expanded nozzle mein $p_e=p_a$ hai, toh uska pressure-thrust term zero hai aur momentum thrust se kuch bhi subtract nahi hota; over-expanded mein negative $(p_e-p_a)A_e$ hota hai. Isliye perfectly expanded nozzle same $\dot{m}$ par zyada thrust produce karta hai aur isliye zyada $I_{sp}$ deta hai. Reasoning route matter karta hai: $I_{sp}$ "sirf exhaust speed" nahi hai — yeh unit weight flow par total thrust hai, aur total thrust mein pressure term shamil hai.

Fixed chamber aur $\dot{m}$ ke parent ke incremental result se shuru karo: exit area ka ek patal ring $d{A}_e$ add karne se thrust badlata hai $dF=p_{e}d{A}_e-p_{a}d{A}_e=(p_e-p_a)d{A}_e.$ (Internal momentum aur pressure terms combine hokar local wall pressure $p_e$ mein aa jaate hain jo added area par act karta hai; ambient usi added area par push back karta hai.) Extremum wahan hai jahan $dF/d{A}_e=0$: $p_e-p_a=0\Rightarrow \boxed{p_e=p_a}.$ Sign check karo: $dF/d{A}_e>0$ jab $p_e>p_a$ (area add karo → abhi bhi gain ho raha → under-expanded), $<0$ jab $p_e<p_a$ (over-expanded). Toh $p_e=p_a$ ek maximum hai, minimum nahi. Altitude link: geometry $p_e$ fix karti hai, lekin $p_a$ altitude ke saath girta hai, toh single ε jo $p_e=p_a$ satisfy karta hai woh exactly ek height par perfect hota hai.

(a) Perfect expansion ⇒ $p_e=p_a=0.12$ bar $=1.2\times 10^4$ Pa. $p_0/p_e=120/0.12=1000$. $\gamma /(\gamma -1)=1.18/0.18=6.556$. Toh $1+0.09M_e^2=1000^{1/6.556}=1000^{0.1526}=2.858$. $0.09M_e^2=1.858\Rightarrow M_e^2=20.65\Rightarrow M_e=\mathrm{4.544.}$ Area exponent $\frac{2.18}{0.36}=6.056$; front factor $\frac{2}{2.18}=0.9174$. $\epsilon =\frac{1}{4.544}[0.9174\times 2.858{]}^{6.056}=\frac{1}{4.544}(2.622{)}^{6.056}=\frac{1}{4.544}(346.6)=\mathrm{76.3.}$ (b) Design thrust ($p_e=p_a$, pressure term zero): $F=\dot{m}{v}_e+(p_e-p_a)A_e=480\times 3100+0=1.488\times 10^6\text{N}=1488\text{kN}.$ (c) $p_a=0.40$ bar $=4.0\times 10^4$ Pa par, $p_e=1.2\times 10^4$ Pa abhi bhi: $(p_e-p_a)A_e=(1.2\times 10^4-4.0\times 10^4)(2.0)=(-2.8\times 10^4)(2.0)=-5.6\times 10^4\text{N}.$ $F=1.488\times 10^6-5.6\times 10^4=1.432\times 10^6\text{N}=1432\text{kN}.$ Comment: moti hawa ($p_a$ upar) is vacuum-optimised engine ko over-expand karti hai aur thrust se 56 kN chheen leti hai — lagbhag 3.8% — high-ε bell ko bahut neeche fire karne ki concrete price. Isliye upper-stage engines (huge ε jaise 76) kabhi first stage ke roop mein use nahi hote.

Optimum thrust condition ::: $p_e=p_a$ (perfect expansion). Over-expanded hone par $dF/d{A}_e$ ki sign ::: Negative — ε shrink karo. High altitude ko bada ε kyun chahiye ::: Chhota $p_a$ ⇒ chhota target $p_e$ ⇒ zyada expansion ⇒ bada $A_e/A^{\ast }$. Throat ke baad kaunsa Mach root ::: Supersonic $M_e>1$ root.