Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss: Worked examples

3.3.15 · D3 · Physics › Rocket Propulsion › Under-expanded nozzle — Prandtl-Meyer expansion, efficiency

Yeh page 3.3.15 Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss (Hinglish) ke English parent ka "koi bhi case na chhoota jaaye" wala companion hai. Hum har tarah ki situation se guzarte hain jo ek under-expanded nozzle problem mein aa sakti hai — pressure gap ka har sign, degenerate "bas abhi abhi" cases, extreme limits, ek real flight problem, aur ek exam trap — aur har ek ko poori tarah solve karte hain.

Pehli line se pehle, ek vaada: koi bhi cheez use nahi hogi jab tak woh build na ho. Aao teen tools build karte hain jinpar hum rely karte hain.

Teen tools (zero se build kiye gaye)

Scenario matrix

Har under-expanded problem inhi cells mein se ek (ya blend) hoti hai. Neeche har example uss cell ke saath tagged hai jise woh fill karta hai.

Cell	Kya special hai	Example
A. Baseline	$r > 1$ , ordinary supersonic exit	Ex 1
B. Degenerate $r = 1$	pressure tie — "zero turn" boundary	Ex 2
C. Wrong-sign trap	student ko $r < 1$ diya gaya hai — P-M method reject karna hoga	Ex 3
D. Small mismatch limit	$r \to 1^{+}$ , tiny angle, thrust loss $\to 0$	Ex 4
E. Large mismatch limit	high altitude, bada $r$ , bada turn	Ex 5
F. Thrust-force accounting	$F = \overset{m}{˙} v_{e} + A_{e} (p_{e} - p_{\infty})$ use karta hai	Ex 6
G. Real-world word problem	rocket chadhta hai, $p_{\infty}$ altitude ke saath girta hai	Ex 7
H. Exam twist	alag $γ$ , loss factor $1 - cos θ$ maangta hai	Ex 8

Worked examples

Worked example Example 1 — Cell A: baseline under-expanded jet

Diya gaya: $M_{e} = 3.5$ , $p_{e} = 1.5$ atm, $p_{\infty} = 1.0$ atm, $γ = 1.3$ . Dhundhna hai: $M_{plume}$ aur turning angle $θ_{turn}$ .

Forecast: Andaza lagao — kya $M_{plume}$ , $3.5$ se bada hoga ya chhota? Kya $θ$ kuch degrees hoga ya tens of degrees?

Regime confirm karo. $r = 1.5/1 = 1.5 > 1$ → under-expanded. Yeh step kyun? Poora method tabhi apply hota hai jab gas abhi bhi over-pressured ho; pehle check karo.
Plume Mach. Exponent $γ / (γ - 1) = 1.3/0.3 = 4.333$ ; aur $\frac{γ - 1}{2} = 0.15$ . Denominator bracket $= 1 + 0.15 (3. 5^{2}) = 1 + 0.15 (12.25) = 2.8375$ . $1.5 = (\frac{1 + 0.15 M _{plume}^{2}}{2.8375})^{4.333}$ Toh $1 + 0.15 M_{plume}^{2} = 2.8375 \cdot 1. 5^{1/4.333} = 2.8375 (1.0964) = 3.111$ . Is tarah $M_{plume}^{2} = (3.111 - 1) /0.15 = 14.07$ , deta hai $M_{plume} \approx 3.75$ . Yeh step kyun? Pressure ka $p_{\infty}$ tak girna flow ko force karta hai tez hone ke liye — isentropic relation hi pressure ratio se Mach number tak pahunchne ka ek maatra bridge hai.
Turning angle $(γ + 1) / (γ - 1) = 2.3/0.3 = 2.769$ ke saath: $ν (3.5) = 2.769 arctan (\frac{1}{2.769} 11.25) - arctan 11.25 = 66.3°$ $ν (3.75) = 2.769 arctan (\frac{1}{2.769} 13.06) - arctan 13.06 = 71.0°$ $θ_{turn} = ν (3.75) - ν (3.5) = 71.0° - 66.3° = 4.7°$ Yeh step kyun? Turning bank account: $M_{e}$ se $M_{plume}$ tak jaane ka cost dono totals ka difference hai.

Verify karo: $M_{plume} > M_{e}$ ✓ (tez, kyunki pressure gira). $θ \approx 4.7°$ "kuch degrees" hai — parent ke $r = 1.5$ ke liye $3$ – $5°$ estimate se match karta hai. Neeche fan geometry dekho.

Figure — Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss

Red waves centered expansion fan hain; violet arrow exit flow hai, orange arrow deflected plume — unke beech ka chhota angle tumhara $θ_{turn}$ hai.

Worked example Example 2 — Cell B: degenerate tie

$p_{e} = p_{\infty}$ Diya gaya: $M_{e} = 3.5$ , $p_{e} = 1.0$ atm, $p_{\infty} = 1.0$ atm, $γ = 1.3$ . Dhundhna hai: $M_{plume}$ aur $θ_{turn}$ .

Forecast: Jab koi pressure gap nahi, kya gas bilkul bhi turn karti hai?

Regime. $r = 1.0$ exactly → perfectly expanded, hamare topic ki boundary.
Plume Mach. Right side $1$ ke barabar hona chahiye, toh numerator bracket = denominator bracket, yaani $M_{plume} = M_{e} = 3.5$ . Yeh step kyun? Koi pressure shed nahi karna → aur expansion nahi.
Turn. $θ_{turn} = ν (3.5) - ν (3.5) = 0°$ . Yeh step kyun? Bank account mein dono sides same Mach → zero withdrawal.

Verify karo: $θ = 0$ aur $M_{plume} = M_{e}$ — flow bilkul seedhi nikalti hai, maximum axial thrust, zero loss. Yahi limit hai jo har real design chase karta hai (3.3.13-Optimal-expansion-ratio).

Worked example Example 3 — Cell C: wrong-sign trap

Diya gaya: $M_{e} = 3.5$ , $p_{e} = 0.7$ atm, $p_{\infty} = 1.0$ atm, $γ = 1.3$ . Ek student blindly Prandtl-Meyer apply karta hai. Dhundhna hai: Actually kya ho raha hai, aur P-M method yahan illegal kyun hai.

Forecast: Kya yeh under-expanded nozzle bhi hai?

Regime. $r = 0.7 < 1$ → $p_{e} < p_{\infty}$ → over-expanded, under-expanded nahi. Yeh step kyun? $r - 1$ ka sign decide karta hai kaunsi physics chalegi. Yahan ambient jeetta hai.
P-M fail kyun hota hai. Expansion fans pressure girate hain. Yahan gas ko $p_{\infty}$ tak pahunchne ke liye apna pressure badhana hai. Supersonic flow mein pressure badhana oblique shocks ke through hota hai, expansion fans ke through nahi — dekho 4.1.3-Oblique-shock-theory aur 3.3.14-Over-expanded-nozzle-shock-diamonds.
Plume Mach direction. Agar tumne force karke isentropic formula lagaya: $1 + 0.15 M_{plume}^{2} = 2.8375 \cdot 0. 7^{1/4.333} = 2.8375 (0.9192) = 2.608$ , deta hai $M_{plume}^{2} = (2.608 - 1) /0.15 = 10.72$ , $M_{plume} \approx 3.27 < M_{e}$ . Yeh step kyun? Woh Mach jo gira hai, compression ka mathematical fingerprint hai — formula chupke se bata raha hai ki fan ki jagah shock chahiye.

Verify karo: $M_{plume} = 3.27 < 3.5$ ✓ — deceleration = compression = shock regime. Yahan turning angle ke liye $ν (M)$ use mat karo. Sahi tool: oblique shock relations.

Worked example Example 4 — Cell D: vanishing-mismatch limit

Diya gaya: $M_{e} = 3.5$ , $p_{e} = 1.02$ atm, $p_{\infty} = 1.0$ atm, $γ = 1.3$ . Dhundhna hai: $θ_{turn}$ aur confirm karo ki thrust loss $\to 0$ .

Forecast: Jab $r \to 1^{+}$ , kya $θ$ smoothly zero tak shrink hota hai?

Plume Mach. $1 + 0.15 M_{plume}^{2} = 2.8375 \cdot 1.0 2^{1/4.333} = 2.8375 (1.00458) = 2.8505$ . $M_{plume}^{2} = (2.8505 - 1) /0.15 = 12.337$ , $M_{plume} \approx 3.513$ . Yeh step kyun? Tiny pressure drop → tiny Mach rise.
Turn. $ν (3.513) - ν (3.5)$ . Kyunki change chhota hai, $θ \approx 0.24°$ . Yeh step kyun? Almost equal bank balances → almost zero withdrawal.
Loss factor. $1 - cos (0.24°) = 1 - 0.9999912 \approx 8.8 \times 1 0^{- 6}$ → lagbhag $0.0009%$ . Yeh step kyun? Thrust loss $(1 - cos θ)$ ke saath scale karta hai, jo chhote $θ$ ke liye $\sim θ^{2} /2$ hai — yeh quadratically khatam hota hai.

Verify karo: $θ \to 0$ aur loss $\to 0$ jab $r \to 1$ ✓ — continuously Example 2 ke exact zero se match karta hai. Physics ka boundary par koi jump nahi hai.

Worked example Example 5 — Cell E: big-mismatch (high-altitude) limit

Diya gaya: $M_{e} = 3.5$ , $p_{e} = 1.0$ atm, $p_{\infty} = 0.1$ atm (bahut upar), $γ = 1.3$ . Toh $r = 10$ . Dhundhna hai: $M_{plume}$ aur $θ_{turn}$ .

Forecast: Bada pressure ratio — kya turn bada hoga (tens of degrees)?

Plume Mach. $1 + 0.15 M_{plume}^{2} = 2.8375 \cdot 1 0^{1/4.333} = 2.8375 (1.6885) = 4.7912$ . $M_{plume}^{2} = (4.7912 - 1) /0.15 = 25.27$ , $M_{plume} \approx 5.03$ . Yeh step kyun? Dashguna pressure drop ek bahut tez plume maangta hai.
Turn. $ν (5.03) \approx 88.9°$ , $ν (3.5) = 66.3°$ , toh $θ = 88.9 - 66.3 \approx 22.6°$ . Yeh step kyun? Bada $M_{plume}$ → bank account se bahut bada withdrawal.
Loss factor. $1 - cos (22.6°) = 1 - 0.9232 = 0.0768$ → extra expansion momentum ka $\sim 7.7%$ sideways phenkha gaya. Yeh step kyun? Ab $θ$ bada hai; small-angle shortcut invalid hai, toh full cosine use karo.

Verify karo: $θ$ $4.7°$ (Ex 1) se $22.6°$ tak jump kiya jab $r$ , $1.5 \to 10$ gaya ✓ — bada mismatch, bada loss, exactly wahi trend jo parent warn karta hai. Yahi reason hai ki altitude-compensation nozzles exist karti hain (3.3.16-Altitude-compensation-nozzles).

Worked example Example 6 — Cell F: full thrust-force accounting

Diya gaya: $\overset{m}{˙} = 200$ kg/s, $v_{e} = 2800$ m/s, $A_{e} = 0.5 m^{2}$ , $p_{e} = 1.5 \times 1 0^{5}$ Pa, $p_{\infty} = 1.0 \times 1 0^{5}$ Pa. Dhundhna hai: Thrust $F$ , aur momentum aur pressure terms ka split.

Forecast: Kya over-pressure yahan thrust mein add karta hai ya subtract?

Momentum term. $\overset{m}{˙} v_{e} = 200 \times 2800 = 560, 000$ N. Yeh step kyun? Thrust ka bulk nozzle se nikalti momentum flux hai.
Pressure term. $A_{e} (p_{e} - p_{\infty}) = 0.5 \times (1.5 \times 1 0^{5} - 1.0 \times 1 0^{5}) = 0.5 \times 5 \times 1 0^{4} = 25, 000$ N. Yeh step kyun? Under-expanded ka matlab $p_{e} > p_{\infty}$ hai, toh yeh term positive hai — over-pressure exit plane par forward push karta hai.
Total. $F = 560, 000 + 25, 000 = 585, 000$ N $= 585$ kN. Yeh step kyun? Momentum theorem: dono contributions ko sum karo.

Verify karo: Units: $[\overset{m}{˙} v_{e}] = \frac{kg}{s} \cdot \frac{m}{s} = \frac{kg\cdotm}{s ^{2}} =$ N ✓; $[A_{e} p] = m^{2} \cdot Pa = m^{2} \cdot \frac{N}{m ^{2}} =$ N ✓. Pressure term positive hai ✓ (under-expansion ke liye sahi sign). Note: gain $25$ kN real hai lekin free lunch nahi — yahan $v_{e}$ perfectly-expanded nozzle se chhota hai. Dekho 2.5.12-Thrust-coefficient-definition.

Worked example Example 7 — Cell G: ek climbing rocket (word problem)

Diya gaya: Ek rocket $p_{e} = 0.8$ atm fixed rakhta hai (fixed nozzle, fixed chamber). Launch par $p_{\infty} = 1.0$ atm; $11$ km altitude par $p_{\infty} = 0.22$ atm. Exit $M_{e} = 4$ , $γ = 1.2$ . Dhundhna hai: Kis altitude par nozzle under-expanded hai, aur turn roughly kaise badalta hai.

Forecast: Ek fixed nozzle har jagah perfect nahi ho sakta — yeh over- se under-expanded mein kab flip karta hai?

Launch. $r = 0.8/1.0 = 0.8 < 1$ → over-expanded (shocks, Ex 3 regime). Yahan yeh hamare page ka kaam nahi. Yeh step kyun? Sea level par ambient strong hai; $r$ se classify karo.
Altitude. $r = 0.8/0.22 = 3.64 > 1$ → under-expanded (fan banta hai). Yeh step kyun? Jab rocket chadhta hai, $p_{\infty}$ collapse karta hai; eventually exhaust use dominate kar leta hai.
Crossover. Perfect expansion tab hoti hai jab $p_{\infty} = p_{e} = 0.8$ atm. Barometric-style drop use karke, yeh $11$ km se neeche kahin hoga. Iske upar, plume khulta hai: $γ = 1.2$ ke saath, $\frac{γ - 1}{2} = 0.1$ , exponent $= 6$ , $(γ + 1) / (γ - 1) = 11 = 3.317$ . Denominator bracket $= 1 + 0.1 (16) = 2.6$ . $1 + 0.1 M_{plume}^{2} = 2.6 \cdot 3.6 4^{1/6} = 2.6 (1.2405) = 3.225$ , $M_{plume}^{2} = 22.25$ , $M_{plume} \approx 4.72$ . Yeh step kyun? Nayi $γ$ aur $r$ ke saath same isentropic bridge track karo.

Verify karo: Regime over- ( $r < 1$ ) se under-expanded ( $r > 1$ ) mein flip karta hai jab yeh chadhta hai ✓ — exactly yahi reason hai ki altitude-compensating designs pursue ki jaati hain (3.3.16-Altitude-compensation-nozzles). $M_{plume} = 4.72 > 4$ ✓ (upar tez).

Worked example Example 8 — Cell H: exam twist, air-like

$γ$ ke saath loss factor Diya gaya: $M_{e} = 2.5$ , $p_{e} / p_{\infty} = 2.0$ , $γ = 1.4$ . Dhundhna hai $θ_{turn}$ aur loss factor $1 - cos θ_{turn}$ .

Forecast: Springier gas ( $γ = 1.4$ ) ke saath kya same $r$ se $γ = 1.3$ cases se bada ya chhota turn milta hai?

Constants. $\frac{γ - 1}{2} = 0.2$ , exponent $= 1.4/0.4 = 3.5$ , $(γ + 1) / (γ - 1) = 2.4/0.4 = 6 = 2.449$ .
Plume Mach. Denominator bracket $= 1 + 0.2 (6.25) = 2.25$ . $1 + 0.2 M_{plume}^{2} = 2.25 \cdot 2^{1/3.5} = 2.25 (1.2190) = 2.7428$ , $M_{plume}^{2} = 8.714$ , $M_{plume} \approx 2.952$ . Yeh step kyun? Same isentropic bridge, air ka $γ$ .
Turn. $ν (2.5) = 2.449 arctan (\frac{1}{2.449} 5.25) - arctan 5.25 = 39.12°$ . $ν (2.952) = 2.449 arctan (\frac{1}{2.449} 7.714) - arctan 7.714 = 48.78°$ . $θ_{turn} = 48.78 - 39.12 = 9.66°$ .
Loss factor. $1 - cos (9.66°) = 1 - 0.98583 = 0.01417$ → lagbhag $1.4%$ . Yeh step kyun? Exam fraction of extra momentum jo sideways waste hua chahta hai, sirf angle nahi.

Verify karo: $M_{plume} = 2.95 > 2.5$ ✓; loss $\approx 1.4%$ moderate $r = 2$ ke liye modest hai ✓, moderate mismatch par "few-percent" losses se consistent. Springier air, $γ = 1.3$ exhaust se thoda zyada turn karta hai per unit pressure drop — ek subtle lekin examinable point.

Recall Quick self-test

Kaun sa regime hai $p_{e} / p_{\infty} = 1.0$ ? ::: Perfectly expanded — zero turn, zero loss (boundary case). $p_{e} < p_{\infty}$ ke liye kaun sa tool? ::: Oblique shocks, Prandtl-Meyer NAHI. Turning angle formula ek line mein? ::: $θ_{turn} = ν (M_{plume}) - ν (M_{e})$ . Thrust loss kaise scale karta hai? ::: $(1 - cos θ_{turn}) \approx θ^{2} /2$ chhote $θ$ ke liye. Under-expanded hone par pressure thrust term ka sign? ::: Positive, $A_{e} (p_{e} - p_{\infty}) > 0$ .