De Laval nozzle geometry — conical, bell (Rao contour), 80% bell: Exercises

3.3.17 · D4 · Physics › Rocket Propulsion › De Laval nozzle geometry — conical, bell (Rao contour), 80%

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

Figure dekho: throat waist (magenta), chaudte cone/bell wall (violet), half-angle $α$ jo wall aur centre axis ke beech measure hota hai, aur axial length $L$ throat plane se exit plane tak. Neeche ke saare problems isi ek picture par based hain.

Level 1 — Recognition

L1.1 — Expansion ratio padhna

Ek nozzle ka throat radius $R_{t} = 5$ cm aur exit radius $R_{e} = 20$ cm hai. Iska expansion ratio $ϵ$ kya hai?

Recall Solution

Kya karte hain: Expansion ratio exit area divided by throat area hai. Circle ki area $= π R^{2}$ , aur $π$ cancel ho jaata hai: $ϵ = \frac{A _{e}}{A _{t}} = \frac{π R _{e}^{2}}{π R _{t}^{2}} = \frac{R _{e}^{2}}{R _{t}^{2}} = \frac{2 0 ^{2}}{5 ^{2}} = \frac{400}{25} = 16.$ Kyun: Humein $π$ ki kabhi zaroorat nahi kyunki woh divide out ho jaata hai — areas ka ratio radii squared ka ratio hota hai. Answer: $ϵ = 16$ .

L1.2 — Cone angle se efficiency padhna

Ek conical nozzle ka half-angle $α = 1 5^{\circ}$ hai. Iska divergence correction factor $λ$ compute karo aur thrust loss percent mein batao.

Recall Solution

$λ = \frac{1 + c o s 1 5 ^{\circ}}{2} = \frac{1 + 0.9659}{2} = 0.9830.$ Number ka matlab: $λ = 0.983$ kehta hai ki ideal momentum ka 98.3% aage point karta hai, isliye loss hai $1 - 0.983 = 0.017 = 1.7%$ . Answer: $λ \approx 0.983$ , loss $\approx 1.7%$ .

Level 2 — Application

L2.1 — Scratch se cone length

Throat radius $R_{t} = 10$ cm, expansion ratio $ϵ = 16$ , cone half-angle $α = 1 5^{\circ}$ . Exit radius aur conical diverging section ki axial length nikalo.

Recall Solution

Step 1 — exit radius: $R_{e} = R_{t} ϵ = 1016 = 10 \times 4 = 40$ cm. Step 2 — length: wall $R_{e} - R_{t} = 40 - 10 = 30$ cm vertically uthti hai jabki slope $tan 1 5^{\circ} = 0.2679$ par bahar jaati hai. Length = rise / slope: $L_{cone} = \frac{R _{t} ( ϵ - 1 )}{t a n α} = \frac{10 ( 4 - 1 )}{0.2679} = \frac{30}{0.2679} \approx 112 cm .$ Answer: $R_{e} = 40$ cm, $L_{cone} \approx 112$ cm.

L2.2 — 80% bell length

Upar wale same nozzle ke liye, ek 80% bell cone ki jagah le leta hai. Iska length kya hoga?

Recall Solution

Definition used: ek "80% bell" woh hai jiska axial length $α = 1 5^{\circ}$ par equivalent conical length ka $0.8$ guna ho. $L_{bell} = 0.8 L_{cone} = 0.8 \times 112 = 89.6 cm .$ Answer: $L_{bell} \approx 89.6$ cm — cone se 20% chhota.

L2.3 — $λ$ se thrust

Ideal thrust (perfectly axial exhaust) $\overset{m}{˙} v_{e} = 100$ kN hai. $1 5^{\circ}$ cone ( $λ = 0.983$ ) aur 80% bell ( $λ = 0.985$ ) ke actual thrust compare karo.

Recall Solution

Actual thrust $= λ \times (\overset{m}{˙} v_{e})$ , kyunki $λ$ exactly woh fraction hai jo momentum aage point karta hai.

Cone: $F = 0.983 \times 100 = 98.3$ kN.
Bell: $F = 0.985 \times 100 = 98.5$ kN. Answer: Cone 98.3 kN, bell 98.5 kN — bell 0.2 kN zyada deta hai aur 20% chhota hai.

Level 3 — Analysis

L3.1 — Divergence formula kahaan se aata hai

Dikhaao ki $λ = \frac{1 + cos α}{2}$ exit cone par $cos θ$ average karne se nikalta hai, aur limit $α \to 0$ evaluate karo.

Recall Solution

Kya average karte hain: angle $θ$ par nikalta exhaust ka har ring forward momentum $\propto cos θ$ contribute karta hai, ring ki area $\propto sin θ d θ$ se weighted (cone ki surface ring at angle $θ$ ki circumference $\propto sin θ$ hoti hai). Isliye $λ = \frac{\int _{0}^{α} c o s θ s i n θ d θ}{\int _{0}^{α} s i n θ d θ} .$ Top integral: $\int_{0}^{α} cos θ sin θ d θ = \frac{1}{2} sin^{2} α$ . Bottom integral: $\int_{0}^{α} sin θ d θ = 1 - cos α$ . Combine, phir use karo $sin^{2} α = (1 - cos α) (1 + cos α)$ : $λ = \frac{\frac{1}{2} ( 1 - c o s α ) ( 1 + c o s α )}{1 - c o s α} = \frac{1 + c o s α}{2} .$ Limit $α \to 0$ : $cos 0 = 1$ , isliye $λ \to \frac{1 + 1}{2} = 1$ . Perfectly axial flow kuch nahi khoata — bilkul wahi jo hum expect karte hain. Answer: derivation confirmed; $λ (0^{\circ}) = 1$ .

L3.2 — Cone angle double karne ki kitni cost hai?

$α = 1 5^{\circ}$ versus $α = 3 0^{\circ}$ par thrust loss compare karo. Kya loss $α$ mein linear hai?

Recall Solution

$λ (1 5^{\circ}) = \frac{1 + c o s 1 5 ^{\circ}}{2} = 0.9830 \Rightarrow loss 1.70%.$ $λ (3 0^{\circ}) = \frac{1 + c o s 3 0 ^{\circ}}{2} = \frac{1 + 0.8660}{2} = 0.9330 \Rightarrow loss 6.70%.$ Kya yeh linear hai? $α$ double karne par (15→30) loss $6.70/1.70 \approx 3.9$ guna ho gaya, 2 guna nahi. Linear nahi — loss chhhote $α$ ke liye roughly $α^{2}$ ki tarah badhta hai (kyunki $1 - cos α \approx α^{2} /2$ ). Isliye designers cone angles chhote rakhte hain aur bells (jo flow ko axial karte hain) jeet jaate hain. Answer: losses 1.70% vs 6.70%; strongly non-linear (near-quadratic).

Curve dikhata hai ki $λ$ jaise jaise $α$ badhta hai tezi se girta hai — $α = 0$ ke paas flat top dikhata hai ki chhote angles almost free hain, aur steep drop wide cone ki saza hai.

Level 4 — Synthesis

L4.1 — Poora 80% bell design

$ϵ = 25$ , $R_{t} = 8$ cm ke liye ek 80% bell design karo. (a) $α = 1 5^{\circ}$ par equivalent cone length, (b) bell length, (c) exit radius nikaalo.

Recall Solution

(a) $L_{cone} = \frac{8 ( 25 - 1 )}{tan 1 5 ^{\circ}} = \frac{8 ( 5 - 1 )}{0.2679} = \frac{32}{0.2679} \approx 119.5$ cm. (b) $L_{bell} = 0.8 \times 119.5 \approx 95.6$ cm. (c) $R_{e} = 825 = 8 \times 5 = 40$ cm. Answer: $L_{cone} \approx 119.5$ cm, $L_{bell} \approx 95.6$ cm, $R_{e} = 40$ cm.

L4.2 — Parabolic contour fit karo

Ek simplified parabolic wall hai $R (x) = R_{t} + a x + b x^{2}$ over $x \in [0, L_{bell}]$ . $R_{t} = 8$ cm, $L_{bell} = 95.6$ cm, $R_{e} = 40$ cm, aur chosen quadratic coefficient $b = 0.0015$ use karke $a$ nikalo aur exit wall angle $θ_{e}$ check karo.

Recall Solution

Exit radius match karo: $R (L) = R_{t} + a L + b L^{2} = 40$ . $8 + a (95.6) + 0.0015 (95.6)^{2} = 40.$ Compute karo $0.0015 \times 95. 6^{2} = 0.0015 \times 9139.4 = 13.71$ . Isliye $8 + 95.6 a + 13.71 = 40 \Rightarrow 95.6 a = 18.29 \Rightarrow a = 0.1913$ . Exit wall angle: wall slope hai $R^{'} (x) = a + 2 b x$ ; exit par $R^{'} (L) = 0.1913 + 2 (0.0015) (95.6) = 0.1913 + 0.2868 = 0.4781$ . Angle hai $θ_{e} = arctan (R^{'} (L))$ — arctan jawaab deta hai "kis angle ka yeh slope hai?" $θ_{e} = arctan (0.4781) \approx 25. 6^{\circ} .$ Result padhna: itna bada $b$ wall ko exit par abhi bhi tezi se chadhaata rehta hai — ek real 80% bell bada $b$ use karegi (zyada downward curvature) taaki $θ_{e}$ ko $\approx 7^{\circ}$ ki taraf mod sake. Exercise mechanism dikhata hai: $b$ badhaao taaki exit flatten ho. Answer: $a \approx 0.191$ , exit slope $0.478$ , $θ_{e} \approx 25. 6^{\circ}$ (bahut steep — curvature $b$ badhaao).

Figure straight cone wall (violet dashed) aur parabolic bell wall (magenta) overlay karta hai: bell throat ke paas zyada steep shuru hoti hai (tezi se expansion) aur ideally exit ke paas flatten ho jaati hai (axial flow). Note karo ki dono alag lengths par same exit radius tak pahunchti hain.

Level 5 — Mastery

L5.1 — Length–efficiency trade study

Tum koi bhi bell length fraction $f \in {0.6, 0.8, 1.0}$ bana sakte ho jinka efficiencies $λ = {0.96, 0.985, 0.99}$ hain. Equivalent cone $L_{cone} = 112$ cm hai aur ideal thrust $\overset{m}{˙} v_{e} = 100$ kN. Nozzle mass length ke proportional hai $2.0$ kg per cm wall. Kaun sa fraction nozzle ke har kilogram par thrust maximize karta hai?

Recall Solution

Har option ke liye length, thrust, mass:

$f$	$L = 112 f$ (cm)	$F = 100 λ$ (kN)	mass $= 2.0 L$ (kg)	$F / mass$ (kN/kg)
0.6	67.2	96.0	134.4	0.714
0.8	89.6	98.5	179.2	0.550
1.0	112.0	99.0	224.0	0.442
Padho: thrust per kilogram sabse zyada $f = 0.6$ par hai (0.714), kyunki mass length ke saath straight-line badhti hai lekin thrust 96 kN ke baad barely hilta hai.
Subtlety: raw thrust abhi bhi $f = 1.0$ par sabse zyada hai (99 kN). Isliye "best" objective par depend karta hai. Agar total thrust sabse zyada mayne rakhta hai (vacuum mein upper stage), lamba jao; agar mass budget dominate karta hai (compact stage), chhota bell thrust-per-kg par jeet jaata hai. 80% bell woh compromise hai jo tab use hota hai jab koi extreme forced nahi hota.
Answer: thrust-per-kg $f = 0.6$ par maximize hota hai ( $0.714$ kN/kg); absolute thrust $f = 1.0$ par; 80% bell balanced middle hai.

L5.2 — Over-expansion sanity check

Vacuum-optimized bell ka itna bada $ϵ$ hai ki sea level par iska exit pressure ambient se neeche hai. Ek line mein explain karo ki nozzle ko aur lambaana sea-level thrust kyun reduce karega, aur woh vault topic batao jo yeh resolve karta hai.

Recall Solution

Kyun: Lamba nozzle gas ko aur bhi kam exit pressure tak expand karta hai; jab exit pressure ambient se neeche gir jaata hai (over-expansion), bahar ki hawa wapas push karti hai aur bell ke andar flow separation / shocks trigger kar sakti hai jo exit momentum kharaab kar dete hain — isliye thrust badhne ki jagah girta hai. Yeh exactly altitude-matching problem hai jo Nozzle Exit Pressure and Altitude Compensation mein cover hai; design tools Method of Characteristics for Nozzle Design mein hain aur real-wall effects Manufacturing Tolerances in Nozzles mein hain. Answer: over-expansion → separation/shocks → thrust loss; altitude compensation se resolve hota hai.

Yeh bhi dekho Converging-Diverging Nozzle Basics aur Thrust Vectoring with Bell Nozzles.