Worked examples — De Laval nozzle geometry — conical, bell (Rao contour), 80% bell

3.3.17 · D3 · Physics › Rocket Propulsion › De Laval nozzle geometry — conical, bell (Rao contour), 80%

Yeh page nozzle-geometry topic ke liye ek drill sheet hai. Hum teen formulas lete hain jo parent note ne banaye the — divergence factor, cone length, aur bell length — aur inhe har tarah ke inputs ke through push karte hain: normal cases, extreme angles, "flat" degenerate cone, limiting bell, ek real launch-vehicle word problem, aur ek exam twist. Kuch bhi naya assume nahi kiya gaya hai: neeche har symbol ko pehli baar explain kiya gaya hai.

Pehle sum se pehle, woh symbols mile jinhein hum har jagah reuse karte hain:

Figure s01 dekho: $R_t$ narrow throat radius hai, $R_e$ wide exit radius hai, $\alpha$ straight wall ka tilt hai, aur amber arrow exhaust hai — $\alpha$ se tilt hai, isliye sirf iska horizontal shadow (cyan dashed arrow) actually rocket ko push karta hai.

Agle do figures dikhate hain ki do knobs badalne se shape aur exhaust vector par kya asar padta hai — examples se pehle inhe study karo taaki neeche ke numbers ke saath ek picture attached ho.

Figure s02 teen cones ko same radii par lekin alag $\alpha$ ($15^{\circ }$, $30^{\circ }$, aur near-$0^{\circ }$) par overlay karta hai: dekho wall steep hoti hai, length shrink hoti hai, aur exhaust arrow aur zyada off-axis tilt hota hai — amber arrow ka forward shadow shrink hota hai, yehi $\lambda$ ka girna hai.

Figure s03 $\alpha$ fix karta hai aur $ϵ$ grow karta hai: bada expansion ratio ⇒ wider exit ⇒ longer nozzle. Yeh picture Example 6 aur Example 8 mein over-expansion trap ke peeche hai.

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Scenario matrix

Is topic ka har problem inhi cells mein se ek hoti hai. Neeche ke worked examples ko cell(s) ke saath tag kiya gaya hai jo woh cover karte hain.

Cell	Case class	Kya tricky hai	Example
A	Normal cone (12–18°)	plug-and-chug baseline	Ex 1
B	Large half-angle (30°+)	$\lambda$ tezi se girta hai — loss feel karo	Ex 2
C	Degenerate $\alpha \to 0$	wall axis ke parallel: $L\to \infty$, $\lambda \to 1$	Ex 3
D	Degenerate $ϵ=1$	bilkul bhi diverging section nahi ($R_e=R_t$)	Ex 3
E	80% bell vs uska parent cone	length-scaling rule, weight saving	Ex 4
F	Limiting bell (${\theta }_e\to 0$)	perfectly axial exit, $\lambda \to 1$	Ex 5
G	Real-world word problem	vacuum upper-stage sizing	Ex 6
H	Exam twist (inverse)	given $L$ aur $\lambda$, $\alpha ,ϵ$ ke liye solve karo	Ex 7
I	Over-expansion trap	zyada length ≠ zyada thrust	Ex 8

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Worked examples

Example 1 — Cell A: standard cone

Forecast: compute karne se pehle guess karo — kya $L$ 50 cm ke kareeb hoga ya 150 cm? Kya $\lambda$ $0.98$ se upar ya neeche hoga?

1. Exit radius. $R_e=R_t\sqrt{ϵ}=10\sqrt{16}=40$ cm. Yeh step kyun? Area radius squared ke saath scale hoti hai, isliye radius sirf $\sqrt{ϵ}$ se grow karta hai, $ϵ$ se nahi.
2. Length. Straight wall $R_t$ se $R_e$ tak $\tan \alpha$ slope par chadhti hai (rise over run). "Rise $=\text{run}\times \tan \alpha$" rearrange karne par: $L=\frac{R_e-R_t}{\tan \alpha }=\frac{R_t(\sqrt{ϵ}-1)}{\tan \alpha }=\frac{10(4-1)}{\tan 15^{\circ }}=\frac{30}{0.2679}\approx 112\text{ cm}.$ Yeh step kyun? $\tan \alpha$ (tangent = wall ke right triangle par opposite/adjacent) ek height ko jo hume chhadni hai ek horizontal length mein convert karta hai jo hume travel karni hai. Steep wall (bada $\alpha$) ⇒ bada $\tan \alpha$ ⇒ chhota nozzle.
3. Efficiency. $\lambda =\frac{1+\cos 15^{\circ }}{2}=\frac{1+0.9659}{2}=0.983$.

Verify: $R_e/R_t=40/10=4=\sqrt{16}$ ✓. Units: cm/(dimensionless tan) = cm ✓. $\lambda <1$ jaisa kisi bhi real cone ke liye hona chahiye ✓. Toh hum $1-0.983=1.7\%$ thrust sideways leakage se kho dete hain.

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Example 2 — Cell B: wide-angle warning

Forecast: nozzle chhota hoga — lekin kitna thrust kharcha karna hoga?

1. Length. $L=\frac{30}{\tan 30^{\circ }}=\frac{30}{0.5774}\approx 51.96$ cm. Yeh step kyun? $\alpha$ double karne se $\tan \alpha$ zyada se zyada double ho gaya, isliye length almost half ho gayi — yehi temptation hai.
2. Efficiency. $\lambda =\frac{1+\cos 30^{\circ }}{2}=\frac{1+0.8660}{2}=0.9330$. Yeh step kyun? $\cos$ angle badhne par tezi se girta hai, isliye axial fraction steeply drop karta hai.
3. Penalty. Loss $1.7\%$ se $1-0.933=6.7\%$ ho gayi — almost chaar guna bura.

Verify: $\lambda (30^{\circ })=0.933<\lambda (15^{\circ })=0.983$ ✓ (bada angle, bura aim). $L(30^{\circ })=52\text{cm}<L(15^{\circ })=112\text{cm}$ ✓ (steep wall, chhota). Trade confirm: aadhi length, chaar guna loss. Yehi wajah hai real cones 12–18° par hote hain. (Figure s02 dekho: $30^{\circ }$ cone middle, stubbier funnel hai jisme sabse zyada tilted exhaust arrow hai.)

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Example 3 — Cells C & D: do degenerate limits

Forecast: inme se ek infinitely long nozzle deta hai; ek koi nozzle nahi deta. Kaun sa kaun sa hai?

1. Cell C, $\alpha \to 0$ (wall axis ke parallel). Pehle symbolic rakhte hain, general throat radius $R_t$ aur general expansion ratio $ϵ>1$ ke liye: $\lambda =\frac{1+\cos 0^{\circ }}{2}=\frac{1+1}{2}=1,L=\frac{R_t(\sqrt{ϵ}-1)}{\tan 0^{\circ }}.$ Kyunki $\tan 0^{\circ }=0$ jabki numerator $R_t(\sqrt{ϵ}-1)$ ek fixed positive number hai, hum ek positive constant ko kuch aise divide kar rahe hain jo zero ki taraf shrink ho raha hai, isliye $L\to +\infty$. (Example-1 numbers $R_t=10$, $ϵ=16$ ke liye numerator exactly $30$ hai, yaani $30/\tan 0^{\circ }\to \infty$.) Yeh step kyun? Zero tilt wali wall gas ko perfectly forward aim karti hai ($\lambda =1$), lekin $R_t$ se $R_e$ tak zero-slope line hamesha ke liye run karni padti hai. Perfect aim ki cost infinite length hai — yeh nozzle design ka fundamental tension hai. Yeh figure s02 mein rightmost, near-flat cone hai.
2. Cell D, $ϵ=1$ ($A_e=A_t$). Phir general rakhte hain: kisi bhi $R_t$ aur kisi bhi half-angle $\alpha$ ke liye, $R_e=R_t\sqrt{1}=R_t,L=\frac{R_t(\sqrt{1}-1)}{\tan \alpha }=\frac{R_t\cdot 0}{\tan \alpha }=0.$ Yeh step kyun? Agar exit area throat area ke barabar hai, toh expand karne ko kuch nahi — diverging cone ki zero length hai. Formula $L=0$ return karta hai, jo correctly bata raha hai "yahan koi diverging section nahi hai."

Verify: Jab $\alpha \to 0^{\circ }$, $\cos \alpha \to 1$ toh $\lambda \to 1$ ✓ aur $\tan \alpha \to 0$ toh $L\to +\infty$ ✓. $ϵ=1$ par, $\sqrt{ϵ}-1=0\Rightarrow L=0$ regardless of $\alpha$ ✓. Dono limits physically sensible hain, errors nahi.

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Example 4 — Cell E: 80% bell apne parent cone ke against

Forecast: bell chhoti hai — kya iska thrust bhi kam hai, ya somehow zyada?

1. Bell length. Definition ke hisaab se ek "80% bell" ka $L_{\text{bell}}=0.8L_{\text{cone}}=0.8\times 112=89.6$ cm hai. Yeh step kyun? "80%" label hi length rule hai — iska matlab hai "equivalent 15° cone ka 80% utna lamba."
2. Thrust har ek ka. Kyunki nozzle matched hai, pressure term zero hai, isliye puri thrust equation sirf momentum piece mein collapse ho jaati hai: $F=\dot{m}{v}_e\lambda$. $F_{\text{cone}}=100\times 0.983=98.3\text{ kN},F_{\text{bell}}=100\times 0.985=98.5\text{ kN}.$ Yeh step kyun? $\lambda$ literally woh fraction hai $100$ kN raw push ka jo axial thrust ke roop mein survive karta hai jab sideways leakage remove ho jaati hai.
3. Result. Bell $98.5-98.3=0.2$ kN zyada thrust deta hai jabki $112-89.6=22.4$ cm chhota hai.

Verify: $0.8\times 112=89.6$ ✓. $\Delta F=0.2$ kN $>0$ ✓ — chhota aur stronger, kyunki curved wall exhaust ko axial re-aim karta hai jabki cone usse sideways leak karta hai. Length saving $=22.4/112=20\%$ ✓, "80%" naam se match karta hai.

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Example 5 — Cell F: perfectly-axial limit

Forecast: kya $7^{\circ }$ exit hamare paas woh $6.7\%$ loss aayegi jo humne 30° cone ke liye dekhi thi?

1. Ideal exit (${\theta }_e=0^{\circ }$). ${\lambda }_{\text{exit}}=\frac{1+\cos 0^{\circ }}{2}=1$. Yeh step kyun? Seedha axis ke saath nikalta flow sideways kuch waste nahi karta — yeh wahi hai jo bell ki curve-back achieve karti hai jo cone nahi kar sakta.
2. Real 80% bell exit (${\theta }_e=7^{\circ }$). ${\lambda }_{\text{exit}}=\frac{1+\cos 7^{\circ }}{2}=\frac{1+0.99255}{2}=0.99627$. Yeh step kyun? Exit par ek gentle $7^{\circ }$ tilt almost axial hai, isliye exit-plane loss sirf $\approx 0.37\%$ hai — cone se kaafi kam, kyunki cone apni poori wall ko $\alpha$ se tilt karta hai, sirf exit lip ko nahi.

Verify: $\cos 0^{\circ }=1\Rightarrow \lambda =1$ ✓. $\lambda (7^{\circ })=0.9963>\lambda (15^{\circ })=0.983$ ✓: bell ka near-axial exit cone ke 15° ko har jagah beat karta hai. Dekho Method of Characteristics for Nozzle Design ki curve actually kaise solve hoti hai.

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Example 6 — Cell G: real-world vacuum upper stage

Forecast: exit radius $\sqrt{25}=5\times$ throat tak grow hota hai — kya length ek metre se kam ya zyada hogi?

1. Equivalent cone length reference $\alpha =15^{\circ }$ par: $L_{\text{cone}}=\frac{R_t(\sqrt{ϵ}-1)}{\tan 15^{\circ }}=\frac{8(5-1)}{0.2679}=\frac{32}{0.2679}\approx 119.4\text{ cm}.$ Yeh step kyun? 80% rule relative to 15° cone define kiya gaya hai, isliye hume pehle woh reference cone banana hoga.
2. Bell length. $L_{\text{bell}}=0.8\times 119.4\approx 95.6$ cm. Yeh step kyun? 80% bell ki length-scaling definition apply karo.
3. Exit radius. $R_e=R_t\sqrt{ϵ}=8\times 5=40$ cm. Yeh step kyun? Exit area $25\times$ throat hai, isliye radius $\sqrt{25}=5\times$ bada hai. (Yeh bada $ϵ$ figure s03 ke right mein tall funnel hai.)

Verify: $\sqrt{25}=5$ ✓; $8\times 4/0.2679=119.4$ ✓; $0.8\times 119.4=95.6$ ✓; $R_e=40$ cm deta hai $A_e/A_t=(40/8{)}^2=25=ϵ$ ✓. Bada $ϵ$ vacuum ke liye sahi hai: low ambient pressure ⇒ zyada expand karo.

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Example 7 — Cell H: inverse (exam twist)

Forecast: same radii ke liye yeh cone Example 1 se chhota hai — toh kya $\alpha$ 15° se bada hai ya chhota?

1. Angle recover karo. $L=\frac{R_e-R_t}{\tan \alpha }$ se, $\tan \alpha$ ke liye solve karo: $\tan \alpha =\frac{R_e-R_t}{L}=\frac{40-10}{80}=0.375\Rightarrow \alpha =\arctan (0.375)=20.56^{\circ }.$ Yeh step kyun? $\arctan$ (arctangent) ka jawab hai "kis angle ka yeh tangent hai?" — yeh $\tan$ ko undo karta hai, hamare measured slope $0.375$ ko wapas ek angle mein turn karta hai. Humein iska zaroorat hai kyunki geometry ne hume ratio diya, angle directly nahi.
2. Efficiency. $\lambda =\frac{1+\cos 20.56^{\circ }}{2}=\frac{1+0.9361}{2}=0.9680$. Yeh step kyun? Same divergence formula, ab recovered $\alpha$ ke saath.

Verify: Wapas plug karo: $L=(40-10)/\tan (20.56^{\circ })=30/0.375=80$ cm ✓. Example 1 ke 112 cm se chhota aur indeed $\alpha =20.6^{\circ }>15^{\circ }$ ✓ (steep wall = chhota cone), aur $\lambda =0.968<0.983$ ✓ (steepness ki keemat). Yeh Manufacturing Tolerances in Nozzles territory hai: measure karo, phir verify karo ki tumne actually kaun sa angle paya.

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Example 8 — Cell I: over-expansion trap

Forecast: parent note ka Mistake 1 ek hint hai — "longer = better" intuition nahi, full thrust equation trust karo.

1. Exit area. $A_e=\pi R_e^2=\pi (0.40{)}^2=0.5027{\text{ m}}^2$. Yeh step kyun? Pressure term $(p_e-p_a)A_e$ ko woh area chahiye jis par pressure act karta hai, aur area $=\pi R^2$.
2. Nozzle X (matched). Pressure term $=(p_e-p_a)A_e=(101,325-101,325)\times 0.5027=0$. Isliye $F_X=\dot{m}{v}_e\lambda +0=100,000\times 0.983=98,300\text{ N}=98.3\text{ kN}.$ Yeh step kyun? Jab $p_e=p_a$ toh pressure piece exactly zero hai — "matched" target hone ki poori wajah yehi hai: koi pressure penalty nahi, sirf momentum thrust.
3. Nozzle Y (over-expanded). Same momentum piece, lekin ab pressure piece negative hai kyunki $p_e<p_a$: $(p_e-p_a)A_e=(60,000-101,325)\times 0.5027=(-41,325)\times 0.5027=-20,774\text{ N}.$ $F_Y=100,000\times 0.983-20,774=98,300-20,774=77,526\text{ N}\approx 77.5\text{ kN}.$ Yeh step kyun? Over-expansion ka matlab hai bahar ki hawa exhaust se zyada andar push kar rahi hai, isliye pressure term subtract ho jaata hai. Lamba nozzle lagbhag $20.8$ kN kho deta hai.
4. Decision. $F_X=98.3$ kN, $F_Y=77.5$ kN se jeet jaata hai. Lamba nozzle Y sea level par bura hai, lagbhag $21$ kN se. Yeh step kyun? Yeh directly Mistake 1 quantify karta hai: length tab tak hi help karta hai jab tak $p_e$, $p_a$ tak pahunche; uske baad extra length over-expand karta hai aur shocks thrust khaate hain. Y sirf altitude par jeetta hai, jahan $p_a$ $60,000$ Pa ki taraf girta hai aur iska pressure term wapas zero ki taraf chadh ta hai.

Verify: Geometry akele dono nozzles ko $\lambda =0.983$ deti hai, yeh prove karta hai ki $\lambda$ loss detect nahi kar sakta — tumhe zaroor pressure term use karna hoga ✓. Sign check: $p_e-p_a=-41,325<0$ ⇒ negative pressure thrust ✓. Units: Pa $\times$ m${}^2$ = N ✓. $F_Y<F_X$ despite Y ke lamba hone ke bawajood ✓ — exactly wahi trap hai. Poora treatment Nozzle Exit Pressure and Altitude Compensation mein.

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Recall Self-test (try karne ke baad reveal karo)

Cone with $R_t=10$, $ϵ=16$, $\alpha =15^{\circ }$: length? ::: $112$ cm Same radii lekin $L=80$ cm — half-angle? ::: $\alpha =\arctan (0.375)=20.56^{\circ }$ $\lambda$ ek $30^{\circ }$ cone ke liye? ::: $0.933$ (ek $6.7\%$ loss) Jab $\alpha \to 0^{\circ }$, $L$ aur $\lambda$ ka kya hota hai? ::: $L\to \infty$, $\lambda \to 1$ (perfect aim, infinite length) 80% bell ki length jo ek 112 cm cone replace kare? ::: $89.6$ cm Sea level par matched vs over-expanded ($p_e=60$ kPa, $A_e=0.5027$ m²) nozzle ka net thrust? ::: $98.3$ kN vs $77.5$ kN Ek lamba nozzle kam sea-level thrust kyun de sakta hai? ::: over-expansion $p_e<p_a$ banata hai, isliye $(p_e-p_a)A_e$ negative hai