Visual walkthrough — De Laval nozzle geometry — conical, bell (Rao contour), 80% bell

3.3.17 · D2 · Physics › Rocket Propulsion › De Laval nozzle geometry — conical, bell (Rao contour), 80%

Yeh page parent note ka sabse important number — divergence correction factor — bilkul scratch se rebuild karta hai:

$\lambda =\frac{1+\cos \alpha }{2}$

— koi integral yaad nahi, koi symbol use nahi jab tak usse draw na kar lein. Agar tumne kabhi cosine ya integral sign nahi dekha, pehli line se shuru karo aur phir bhi final answer tak pahunch jaoge.

Hum sirf wahi assume karte hain jo Converging-Diverging Nozzle Basics ne already diya hai: hot gas ek tange throat se squeeze hoti hai aur phir ek chauda hote cone se nikalne diya jaata hai, tezi se bahar aati hai. Is poore page ka hamaara ek hi sawaal hai:

Jab gas seedha peeche jaane ki jagah baahir ki taraf fan karti hai, toh kitna push lose hota hai?

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Step 1 — "Thrust" actually hai kya, ek picture ke roop mein

KYA HAI. Thrust woh backward-shove hai jo engine ko milti hai jab woh gas ko peeche phenkhta hai. Agar gas seedha peeche jaati hai, toh saari shove rocket ko seedha aage push karti hai. Yahi ideal hai.

YAH YAHAN SE KYU SHURU KAREIN. Kisi loss ko measure karne se pehle, hume compare karne ke liye poori amount chahiye. Toh hum pehle perfect case draw karte hain: har gas particle centre line ke parallel chalti hai.

PICTURE. Figure s01 dekho. Dashed horizontal line nozzle axis hai (woh centre line jis par rocket uda karta hai). Pale-yellow arrows gas particles hain jo nikal rahi hain. Ideal case mein yeh sab seedha baayi taraf point karte hain — bilkul axis ke saath.

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Step 2 — Ek real cone gas ko fan out karta hai

KYA HAI. Ek straight-walled conical nozzle mein, walls ek fixed angle par baahir ki taraf flare karti hain. Gas walls ko follow karti hai, isliye woh seedha peeche nahi jaati — woh tilted hoke baahir jaati hai.

YEH KYUN MATTER KARTA HAI. Ek tilted arrow ke do parts hote hain: ek part seedha peeche point karta hai (useful) aur ek part sideways point karta hai (waste). Hume har arrow ko in do parts mein split karna hoga.

PICTURE. Figure s02 mein do chalk-blue walls ek funnel ki tarah khulti hain. Ek single gas arrow edge par wall ke saath jaata hai — axis se $\alpha$ angle par tilted hota hai.

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Step 3 — Ek tilted arrow ko split karna: cosine se milte hain

KYA HAI. Ek gas arrow lo jo axis se $\theta$ angle par tilted hai (main $\theta$ use karta hoon ek general arrow ke liye; wall ka arrow special case hai $\theta =\alpha$). Hum sirf uska forward-pointing part rakhte hain.

COSINE KYU, KUCH AUR KYU NAHI. Hume ek aisi machine chahiye jo angle leta hai aur wapas deta hai "is arrow ka kitna fraction aage point karta hai". Arrow ko ek right triangle ki slanted side ki tarah draw karo jiska bottom side axis ke saath lie karta ho. Forward part woh bottom side hai. Right triangle ke liye rule "bottom side ÷ slanted side" bilkul wahi hai jiska matlab $\cos \theta$ hai — cosine define hota hai adjacent over hypotenuse. Yahi woh sawaal hai jo hum pooch rahe hain, isliye cosine sahi tool hai, sine nahi (jo sideways part deta) aur tangent bhi nahi.

PICTURE. Figure s03: length $1$ (ek unit of speed) ka pink arrow tilted gas hai. Ise axis par drop karo. Neeche yellow segment, length $\cos \theta$ ka, woh forward part hai jo hum rakhte hain.

Picture ko dhyan mein rakh kar extremes check karo:

• $\theta =0$ (seedha peeche): $\cos 0=1$ → sab kuch rako. ✓
• $\theta =90°$ (seedha sideways): $\cos 90°=0$ → kuch nahi rako. ✓

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Step 4 — Lekin saare arrows ek angle share nahi karte

KYA HAI. Ek cone ka exit ek poori ring hai. Centre ke paas ki gas almost seedhi jaati hai ($\theta$ near $0$); bilkul edge ki gas poore wall angle par jaati hai ($\theta =\alpha$). $0$ se $\alpha$ tak har angle present hai.

KYU SIRF $\cos \alpha$ USE NAHI KAR SAKTE. Sirf $\cos \alpha$ use karna maanenge ki har arrow edge angle par nikalta hai. Yeh loss ko zyada count karta hai. Hume $\cos \theta$ ko saare arrows ke spray par average karna hoga — aur average fair honi chahiye: har angle ko us hisaab se weight karna chahiye ki kitni gas actually us angle par jaati hai.

PICTURE. Figure s04 exit ko ek shallow spherical cap ki tarah dikhata hai (socho gas ek curved dome se nikal rahi hai). Us dome par ek thin ring angle $\theta$ par hai. Rim ke paas bade rings se zyada gas flow hoti hai centre ke paas choti rings se — yahi weighting hai jise hume respect karna hai.

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Step 5 — Infinitely many rings add karna: integral

KYA HAI. "$\cos \theta$ ka average, ring size $\sin \theta$ se weighted" = ($\cos \theta \times \text{ring area}$ ka sum) ÷ (ring areas ka sum). Integral sign $\int$ bas "infinitely many thin rings add karo" hai.

INTEGRAL KYU. Angle $0$ se $\alpha$ tak smoothly change hota hai — 5 ya 50 rings nahi hain balki unka ek continuous fan hai. Ek ordinary sum "infinitely many, each infinitely thin" handle nahi kar sakta; integral woh tool hai jo exactly ek smooth continuous sum ke liye bana hai. Angle $\theta$ par har ring ka area $2\pi R\sin \theta d\theta$ ke proportional hai ($d\theta$ = angle ka ek infinitely thin slice).

PICTURE. Figure s05 thin rings ko $\theta =0$ (centre) se $\theta =\alpha$ (rim) tak stack karta hai, har ek apne weight $\sin \theta$ se shaded hai — centre ke paas faint, rim ke paas bold.

$2\pi R$ cancel karne par:

$\lambda =\frac{{\int }_0^{\alpha }\cos \theta \sin \theta d\theta }{{\int }_0^{\alpha }\sin \theta d\theta }$

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Step 6 — Do sums karna (dono simple curves ke neeche areas hain)

KYA HAI. Top aur bottom sums evaluate karo.

YEH PARTICULAR ANTIDERIVATIVES KYU. Integrating matlab hai "un-differentiating": hum poochte hain kaun sa function hai, jab uska slope liya jaaye, toh $\sin \theta \cos \theta$ (ya $\sin \theta$) wapas aaye?

• $\frac{1}{2}{\sin }^2\theta$ ka slope $\sin \theta \cos \theta$ hai → isliye top sum $[\frac{1}{2}{\sin }^2\theta {]}_0^{\alpha }$ hai.
• $-\cos \theta$ ka slope $\sin \theta$ hai → isliye bottom sum $[-\cos \theta {]}_0^{\alpha }$ hai.

PICTURE. Figure s06 dono curves ko $0$ se $\alpha$ tak dikhata hai; unke neeche shaded areas exactly yeh do sums hain.

Limits plug in karne par ($\theta =0$ par: $\sin 0=0$, $\cos 0=1$):

$\text{top}=\frac{1}{2}{\sin }^2\alpha -\frac{1}{2}{\sin }^{2}0=\frac{1}{2}{\sin }^2\alpha$ $\text{bottom}=(-\cos \alpha )-(-\cos 0)=1-\cos \alpha$

Isliye

$\lambda =\frac{\frac{1}{2}{\sin }^2\alpha }{1-\cos \alpha }$

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Step 7 — Clean-up jo famous formula reveal karta hai

KYA HAI. Ek identity use karke simplify karo taaki koi messy squares na rahe.

YEH IDENTITY KYU. Hum chahte hain ki $(1-\cos \alpha )$ top par aaye taaki bottom cancel ho. Identity ${\sin }^2\alpha =1-{\cos }^2\alpha =(1-\cos \alpha )(1+\cos \alpha )$ exactly yahi karta hai — yeh top ko ek piece mein factor karta hai jo bottom se match karta hai aur ek leftover piece.

$\lambda =\frac{\frac{1}{2}(1-\cos \alpha )(1+\cos \alpha )}{1-\cos \alpha }=\frac{1+\cos \alpha }{2}$

$(1-\cos \alpha )$ top aur bottom cancel ho jaata hai, aur woh result bachta hai jo parent note ne quote kiya tha.

Teen cases mein term-by-term sanity (sab cover karo):

• $\alpha \to 0$ (bilkul seedha pipe): $\cos 0=1$, isliye $\lambda =\frac{1+1}{2}=1$. Koi fanning nahi, koi loss nahi. ✓
• $\alpha =15°$ (ek typical cone): $\cos 15°=0.9659$, isliye $\lambda =\frac{1.9659}{2}=0.983$ → ek 1.7% loss. ✓
• $\alpha \to 90°$ (walls flat khuli, gas sideways spray karti hai): $\cos 90°=0$, isliye $\lambda =\frac{1+0}{2}=0.5$ → aadhi thrust gayi. ✓

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Ek-picture summary

Figure s07 poori chain ek board par rakhta hai: ek tilted arrow → $\cos \theta$ se split → $\sin \theta$ se weighted rings par average → do areas → boxed answer, $15°$ number pinned ke saath.

Recall Feynman retelling — plain words mein wapas kaho

Engine gas ko seedha peeche phenkhna chahta hai. Ek cone yeh nahi kar sakta: woh gas ko ek spray mein fan kar deta hai. Koi bhi sprayed arrow sirf thoda help karta hai — sirf woh part jo peeche point karta hai, aur woh part exactly $\cos \theta$ hai, kyunki "backward-part over whole-arrow" yahi hai jo cosine ka matlab hai right triangle par. Gas ke alag-alag bits alag-alag tilts par nikaalte hain, centre par zero se leke rim par poore wall angle $\alpha$ tak, aur rim par zyada gas hoti hai (bade rings, $\sin \theta$ se weighted). Toh main $\cos \theta$ ko un saari rings par fairly average karta hoon — woh averaging hi integral hai. Do sums aate hain $\frac{1}{2}{\sin }^2\alpha$ top par aur $1-\cos \alpha$ bottom par, ek identity mess cancel kar deti hai, aur bahar aata hai $\lambda =\frac{1+\cos \alpha }{2}$: woh fraction of thrust jo fanning ke baad bachta hai. Seedha pipe → $1$, ek $15°$ cone → $0.983$, walls wide open → $0.5$. Aur punchline: rim ko wapas seedha mod do aur woh khoye hue $1.7\%$ free mein wapas milte hain — yahi bell hai.

Recall Quick self-check

Hum $\cos \alpha$ use karne ki jagah $\cos \theta$ average kyun karte hain? ::: Kyunki gas har angle par $0$ se $\alpha$ tak jaati hai, sirf rim par nahi; sirf $\cos \alpha$ use karna loss ko overstate karta. Har ring ko $\sin \theta$ se kyun weight kiya jaata hai? ::: Angle $\theta$ par ek ring ki radius $\propto \sin \theta$ hai, isliye bade (outer) rings zyada gas carry karte hain aur average mein zyada count hone chahiye. $\lambda =1$ physically kya matlab hai? ::: Saari gas axis ke parallel nikaalti hai ($\alpha \to 0$) — zero divergence loss.