Visual walkthrough — Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss

3.3.15 · D2 · Physics › Rocket Propulsion › Under-expanded nozzle — Prandtl-Meyer expansion, efficiency

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Woh symbols jo hum kamaenge (ek baar padh lo)

Kisi bhi picture se pehle, yeh raha chhota sa dictionary. Inme se har ek ko neeche ek figure milega — yeh sirf isliye hai taaki tum cast ko pehle se jaano.

• $a$ = gas mein speed of sound — ek tiny pressure ripple kitni tez us mein travel karta hai.
• $V$ = flow speed — gas kul milake kitni tez stream kar raha hai.
• $M$ = Mach number = $V/a$ = (gas kitni tez move kar raha hai) ÷ (us gas mein sound ki speed). $M=1$ matlab "exactly sound speed"; $M>1$ hai supersonic.
• $\gamma$ = ratio of specific heats — gas ko describe karne wala ek single number (air ≈ 1.4, hot rocket exhaust ≈ 1.2–1.3). Abhi ke liye ise ek fixed dial maano.
• $p$ = static pressure (gas jo pressure isko touch karne wali kisi bhi cheez par sideways lagata hai).
• $p_0$ = total (stagnation) pressure — woh pressure jo gas ke paas hoti agar tum use smoothly dead stop par le aao. Frictionless (isentropic) expansion mein yeh constant rehti hai chahe ordinary pressure $p$ drop ho.
• $\theta$ = woh angle jis se flow direction mudti hai.
• $\nu (M)$ = Prandtl-Meyer function — show ka star. Hum ise ek accumulated turning angle ke roop mein define karenge.
• expansion fan = Mach lines ka spread-out sheaf jis se supersonic flow guzarti hai jab woh ek corner ke around mudti hai zyada jagah mein — jaise ek hand-fan ki ribs ek single pivot se khulti hain. Hum ise Step 5 mein draw karte hain.

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Step 1 — Ek supersonic message sirf ek cone mein travel kar sakti hai

KYA. Move ho rahe gas mein baitha ek tiny disturbance (pressure ka ek pebble) ripples bahar ki taraf sound ki speed $a$ par bhejtaa hai. Agar gas khud flow speed $V$ par stream kar rahi hai jo sound se tez hai, toh woh ripples downstream sweep ho jaate hain aur ek single straight line mein pile up ho jaate hain jise Mach line kehte hain.

KYUN. Yahi wajah hai ki expansion waves ke zariye hoti hai na ki ek saath. Supersonic flow mein koi signal upstream nahi ja sakta, isliye gas sirf inhi Mach lines par "pata chalta hai ki aage pressure drop ho gayi." Neeche ka har step isi fact par chalta hai.

PICTURE.

Blue disturbance point ek time-tick mein sound ka ek circle emit karta hai (radius = sound speed $a$). Us hi tick mein flow us point ko aage distance = flow speed $V=Ma$ (kyunki $M=V/a$) par le jaata hai. Kyunki $M>1$ hai, flow distance sound circle se lambi hai, toh saare circles ek cone ke andar tuck ho jaate hain. Us cone ka half-angle Mach angle $\mu$ hai:

$\mu =\arcsin \left(\frac{1}{M}\right)$

Term by term: $\frac{1}{M}$ = (ek sound-tick, length $a$) ÷ (ek flow-tick, length $V=Ma$) = picture mein dono radii ka ratio; $\arcsin$ = "kis angle ka yeh sine hai?" — yeh angle seedha us right triangle se padhta hai jiska opposite side sound circle hai aur hypotenuse flow distance hai.

Recall

Disturbance upstream wave kyun nahi bhej sakta? ::: Flow har sound ripple ko downstream itni tez sweep kar deta hai ki ripple travel hi nahi kar sakta, isliye signals sirf Mach cone ke andar rehte hain. $\mu$ ke saath bada hota hai ya chhota? ::: Jab $M\to 1$, $1/M\to 1$, $\mu \to 90°$ (wave almost perpendicular hai). Jab $M\to \infty$, $\mu \to 0°$ (wave almost flow ke saath lie karta hai).

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Step 2 — Flow ko thoda sa moddo aur ek Mach wave padho

KYA. Supersonic gas ko ek aise wall ke against rakho jo flow se door thode angle $d\theta$ par bend hoti hai. Gas ke paas failne ki jagah hai, toh yeh thoda sa tez ho jaata hai (Mach number $dM$ se badhta hai) aur pressure thoda drop hota hai — aur yeh exactly ek Mach line ke across karta hai.

KYUN. Ek bada smooth turn sirf aise hi lakho tiny turns ka stack hai. Agar hum ek infinitesimal turn ke liye exact bookkeeping rule dhundh sakein — kitna $M$ change hota hai per degree turned — toh hum sab ko add kar sakte hain. Yeh puri formula ka seed hai.

PICTURE.

Formula kahaan se aata hai — velocity triangle. Picture dekho. Incoming velocity ka magnitude $V$ hai jo wall ke along point kar raha hai. Single green Mach line us flow ke saath Mach angle $\mu$ par baitha hai. Ab velocity ko do pieces mein split karo jo Mach line ke relative measure ki gayi hain: line ke along ek piece, $V\cos \mu$, aur line ke across (uske perpendicular) ek piece, $V\sin \mu$.

Key physical fact: sirf across-the-line piece change hoti hai; along-the-line piece wave se untouched rahta hai. Toh flow isliye mudti hai kyunki chhota perpendicular piece $V\sin \mu$ thoda badhta hai jabki $V\cos \mu$ wahi rehta hai — total vector ko tilt karta hai. Is tilt ko chhote triangle mein chase karne par saaf milta hai,

$d\theta =\sqrt{M^2-1}\frac{dV}{V},$

kyunki $\tan \mu =1/\sqrt{M^2-1}$ toh $\cot \mu =\sqrt{M^2-1}$ exactly woh lever hai jo fractional speed change $dV/V$ ko ek turn mein convert karta hai. Yeh pure geometry hai — abhi tak koi gas physics nahi.

Ab compressibility term. Hum relation $dM/M$ ke terms mein chahte hain, $dV/V$ ke terms mein nahi, kyunki $M$ woh hai jo hamare lookup table mein hoga. Lekin $M$ aur $V$ compressible gas mein proportional nahi hain: jaise gas tez hoti hai waise thandi bhi hoti hai, isliye $a$ bhi change hota hai. $M=V/a$ se start karke aur isentropic energy balance (total temperature conserved) use karke, exact link milta hai

$\frac{dV}{V}=\frac{1}{1+\frac{\gamma -1}{2}{M}^2}\frac{dM}{M}.$

Factor $1+\frac{\gamma -1}{2}{M}^2$ exactly total temperature aur static temperature ka ratio hai; yeh isliye aata hai kyunki Mach-number increase ka kuch hissa gas ko thanda karne mein "use up" ho jaata hai na ki pure speeding-up mein. Ise geometry result mein substitute karo aur master differential relation milti hai:

$d\theta =\frac{\sqrt{M^2-1}}{1+\frac{\gamma -1}{2}{M}^2}\frac{dM}{M}$

Padho ise:

• $d\theta$ = woh tiny angle jo hum mude (left side — woh cheez jo hum control karte hain).
• $\frac{dM}{M}$ = Mach number mein fractional change (thoda mudo, thodi speed pao).
• $\sqrt{M^2-1}$ = geometry lever $\cot \mu$ velocity triangle se; yeh zero hai $M=1$ par (sonic flow bilkul bhi wave se nahi mud sakta) aur $M$ ke saath badhta hai.
• $1+\frac{\gamma -1}{2}{M}^2$ = compressibility tax = (total temperature)/(static temperature); yeh turn ko shrink karta hai kyunki $M$ badhane se gas thandi bhi hoti hai, aur $\gamma$ set karta hai kitni strongly.

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Step 3 — Saare whiskers add karo: $\nu (M)$ define karo

KYA. Integrate (sum up) karo woh differential turn jab se flow exactly sonic thi ($M=1$) kisi bhi Mach $M$ tak. Angle ka running total woh hai jo hum naam dete hain $\nu (M)$.

KYUN. Hum ek lookup table chahte hain: "agar flow Mach $M$ par hai, toh woh sonic hone ke baad se kitna total turn kar chuki hai?" Woh single running total hume kisi bhi expansion ko subtraction se handle karne deta hai, jo Step 5 hai. Hum count $M=1$ par anchor karte hain kyunki yeh woh ek Mach number hai jahan turning pehli baar possible hoti hai, aur hum $\nu (1)=0$ set karte hain taaki counter zero se shuru ho.

PICTURE.

Curve hai $\nu$ jo $M$ badhne ke saath climb karta hai. Har shaded sliver Step 2 ka ek $d\theta$ hai; ek given $M$ par curve ki height uske left ke saare slivers ka sum hai — total accumulated turn.

$\nu (M)={\int }_1^M\frac{\sqrt{{\bar{M}}^2-1}}{\bar{M}\left(1+\frac{\gamma -1}{2}{\bar{M}}^2\right)}d\bar{M}$

Yahan $\bar{M}$ sirf woh dummy "sweeping" Mach number hai jo 1 se target $M$ tak chalta hai. Yeh integral substitution $\bar{M}=\sec \varphi$ se karne par (jo ugly square root ko clean $\tan \varphi$ mein badle) page ke top ki closed form mein collapse ho jaata hai. Tumhe dobara integrate karna nahi padega — do $\arctan$ terms is integral ka answer hain.

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Step 4 — Ceiling: infinite Mach par maximum turn

KYA. $\nu (M)$ mein $M\to \infty$ push karo. Dono $\arctan$ terms $90°=\pi /2$ hit karte hain, aur poori cheez ek finite number par settle ho jaati hai — woh sabse bada angle jis se supersonic gas kabhi bhi turn kar sakti hai.

KYUN. Yeh degenerate/limiting case hai, aur physically matter karta hai: ek plume bahar ki taraf forever nahi mud sakta. Agar pressure mismatch ne ${\nu }_{\max }$ se zyada turn demand ki, toh flow vacuum mein peel away ho jaati aur model fail ho jaata. Ceiling jaanna humein batata hai kab formula par trust karna band karein.

PICTURE.

$\nu$ curve ek horizontal dashed ceiling ke against flatten ho jaata hai. $\gamma =1.4$ ke liye:

${\nu }_{\max }=\frac{\pi }{2}\left(\sqrt{\frac{\gamma +1}{\gamma -1}}-1\right)\approx 130.5°$

Term by term: $\frac{\pi }{2}$ woh hai jahan har $\arctan$ max out hota hai; bracket hai (gas-constant factor) minus 1 kyunki doosra $\arctan$ apna $90°$ subtract karta hai. Hotter exhaust ke liye ($\gamma$ chhota) ceiling zyada oonchi hai — gas zyada mud sakti hai.

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Step 5 — Plume par use karo: exit expansion

KYA. Nozzle lip par exhaust Mach $M_e$ par pressure $p_e$ ke saath hai. Bahar ambient $p_{\infty }$ baitha hai. Agar $p_e>p_{\infty }$ toh gas ek expansion fan (hamare dictionary ki Mach lines ki sheaf) ke zariye expand karti hai, $M_{\text{plume}}$ tak speed up hoti hai aur bahar ${\theta }_{\text{turn}}$ se mudti hai. Turn sirf $\nu$ counter par ek subtraction hai:

${\theta }_{\text{turn}}=\nu (M_{\text{plume}})-\nu (M_e)$

KYUN. Yahi reason hai kyun humne Step 3 mein ek cumulative counter banaya. Hum re-integrate nahi karte — counter do Mach numbers par padhte hain aur subtract karte hain. Jitna bada $M_{\text{plume}}$ hoga, utna ooncha counter, toh flow ko utna zyada turn karna pada.

$M_{\text{plume}}$ KAHAAN SE aata hai? Gas tab tak expand hoti hai jab tak uski static pressure $p_{\infty }$ ke barabar na ho jaaye. Kyunki expansion isentropic hai, total pressure $p_0$ (dictionary mein defined — woh pressure jo gas ke paas hoti agar use smoothly roka jaaye) constant rehti hai, jo ordinary pressure $p$ ko Mach number se link karti hai:

$\frac{p_e}{p_{\infty }}=\frac{{\left(1+\frac{\gamma -1}{2}{M}_{\text{plume}}^2\right)}^{\gamma /(\gamma -1)}}{{\left(1+\frac{\gamma -1}{2}{M}_e^2\right)}^{\gamma /(\gamma -1)}}$

(Yeh 3.2.7-Isentropic-flow-area-Mach-relation se standard isentropic pressure-Mach link hai; dono sides sirf $p_0/p$ ratios hain, toh shared $p_0$ cancel ho jaata hai.) Ise $M_{\text{plume}}$ ke liye solve karo, dono Machs ko $\nu$ mein feed karo, subtract karo — ho gaya.

PICTURE.

Straight nozzle wall khatam hoti hai; green expansion fan Mach lines lip se hand-fan ki ribs ki tarah khulti hai; flow neeche (bahar) ${\theta }_{\text{turn}}$ se bend hoti hai; Mach $M_e$ se $M_{\text{plume}}$ tak badhta hai jabki pressure $p_e$ se $p_{\infty }$ tak girti hai.

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Step 6 — Turn thrust kyun khota hai

KYA. Sirf plume ke momentum ka axial slice rocket ko aage push karta hai. Flow ko bahar ${\theta }_{\text{turn}}$ se modna useful part ko $\cos {\theta }_{\text{turn}}$ se multiply kar deta hai, toh lost fraction $(1-\cos {\theta }_{\text{turn}})$ hai.

KYUN. Thrust backward momentum ka reaction hai. Sideways pheka gaya momentum axisymmetric plume mein khud cancel ho jaata hai aur kuch nahi karta. Yeh woh efficiency loss hai jiske baare mein parent note ne warn kiya tha, aur yeh ${\theta }_{\text{turn}}$ ke saath badhta hai — isliye pressure mismatch $p_e/p_{\infty }$ ke saath.

PICTURE.

Plume velocity vector split hoti hai ek axial (forward, useful) leg $v\cos {\theta }_{\text{turn}}$ aur ek radial (sideways, wasted) leg $v\sin {\theta }_{\text{turn}}$ mein.

Hamare example ke liye, ${\theta }_{\text{turn}}\approx 4.7°$, toh extra-expansion momentum lost hai $1-\cos (4.7°)=1-0.99664\approx 0.0034\Rightarrow \text{lagbhag }0.3\%\text{ extra push ka.}$ Yahan chhota hai, lekin yeh tez climb karta hai jaise $p_e/p_{\infty }$ badhta hai — exactly yahi reason hai kyun 3.3.16-Altitude-compensation-nozzles exist karte hain aur kyun thrust coefficient $C_F$ apne optimum se neeche girta hai. (Mirror-image loss $p_e<p_{\infty }$ ke liye, fans ki jagah shocks ke zariye, 3.3.14-Over-expanded-nozzle-shock-diamonds hai; sweet spot $p_e=p_{\infty }$ 3.3.13-Optimal-expansion-ratio hai.)

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Ek-picture summary

Sab kuch ek canvas par: Mach cone (Step 1) → ek tiny turn (Step 2) → accumulated $\nu$ curve (Step 3) apni ceiling ke saath (Step 4) → plume fan ek $\nu$-difference padhta hua (Step 5) → velocity split jo thrust khota hai (Step 6).

Recall Feynman retelling — plain words mein kaho

Supersonic gas sirf slanted Mach lines ke along khud se whisper kar sakti hai — yeh hai cone. Wall ko ek hair door moddo aur gas ek line ke across speed ka ek hair gain karti hai; exact exchange rate master differential relation hai — pure velocity-triangle geometry ($\cot \mu =\sqrt{M^2-1}$) times ek compressibility tax kyunki tez hone se gas thandi bhi hoti hai. Sound speed se shuru hoke turning ka har hair add karo aur tumhe $\nu$ kaha jaane wala running counter milta hai — "yeh flow total mein kitna mudi hai?" ka lookup table. Woh counter lagbhag $130°$ par top out hota hai: gas kitni bend ho sakti hai iska ek limit hai. Ab ise ek rocket par aim karo: exhaust still over-pressured hoke nozzle chodti hai, isliye woh bahar expansion fan ke zariye expand karti hai, uska Mach counter $\nu (M_e)$ se $\nu (M_{\text{plume}})$ tak climb karta hai, aur difference outward turn angle hai. Flow ko sideways modne ka matlab hai sirf $\cos {\theta }_{\text{turn}}$ naya momentum abhi bhi backward point karta hai — baaki waste hai, aur yeh hai tumhara efficiency loss.