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

3.3.15 · D4 · Physics › Rocket Propulsion › Under-expanded nozzle — Prandtl-Meyer expansion, efficiency

Shuru karne se pehle, teen tools jo baar-baar kaam aayenge. Har symbol parent note mein build kiya gaya tha — yeh woh toolbox hai jo bench par rakhii hui hai.

Formula Teen working equations

(A) Regime test — kya exhaust squeeze ho raha hai ya over-stretch? $p_{e} > p_{\infty} \Rightarrow under-expanded, p_{e} = p_{\infty} \Rightarrow perfect, p_{e} < p_{\infty} \Rightarrow over-expanded$ Yahan $p_{e}$ = nozzle exit par gas ka static pressure, $p_{\infty}$ = bahar ka (ambient) pressure jisme plume push kar raha hai.

(B) Isentropic pressure–Mach link — total pressure $p_{0}$ (woh pressure jo gas ka hota agar use rest par laaya jaaye) smooth expansion mein constant rehta hai, isliye $\frac{p _{e}}{p _{\infty}} = \frac{( 1 + \frac{γ - 1}{2} M _{plume}^{2} ) ^{γ / (γ - 1)}}{( 1 + \frac{γ - 1}{2} M _{e}^{2} ) ^{γ / (γ - 1)}}$ $γ$ = specific heats ka ratio (gas kitna "springy" hai), $M$ = Mach number (speed ÷ local speed of sound). Dekho 3.2.7-Isentropic-flow-area-Mach-relation.

(C) Prandtl-Meyer function — Mach 1 se shuru karke, Mach $M$ tak pahunchne mein flow ne jo total angle turn kiya: $ν (M) = k arctan (\frac{1}{k} M^{2} - 1) - arctan (M^{2} - 1), k = \frac{γ + 1}{γ - 1}$ aur jo turn plume actually karta hai woh hai $θ_{turn} = ν (M_{plume}) - ν (M_{e})$ .

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

Level 1 — Recognition

Recall Solution 1.1

Kya karein: test (A) apply karo. Kyun: sirf $p_{e} - p_{\infty}$ ka sign regime fix kar deta hai. $\frac{p _{e}}{p _{\infty}} = \frac{1.8}{1.0} = 1.8 > 1 \Rightarrow p_{e} > p_{\infty}$ Toh nozzle under-expanded hai. Gas lip tak pahunchte-pahunchte abhi bhi over-pressured hai, isliye woh nozzle ke bahar Prandtl-Meyer fan se expand karta rehta hai, ek khulte hue flower ki tarah bahar ki taraf turn karta hai.

Recall Solution 1.2

(a) $0.6 < 1$ : over-expanded — shocks flow ko compress aur detach karte hain (3.3.14-Over-expanded-nozzle-shock-diamonds).
(b) $1.0$ : perfectly expanded — koi waves nahi, saara momentum axial, maximum thrust.
(c) $1.4 > 1$ : under-expanded — mild external expansion, thoda sa loss.

Perfect case (b) peak hai. Dono mismatches thrust lose karte hain, lekin kaun zyada lose karta hai woh numbers par depend karta hai — yahan (a) $0.4$ se 1 se door hai aur shocks involve karta hai, jo typically ek mild fan se zyada costly hoti hain. Efficiency ke hisab se, worst→best: (a) < (c) < (b).

Level 2 — Application

Recall Solution 2.1

Kya/kyun: formula (C) mein plug in karo; yeh Prandtl-Meyer function evaluate karne ki raw skill hai. Pehle $k = \frac{2.4}{0.4} = 6 = 2.4495$ , aur $M^{2} - 1 = 3 = 1.7321$ . $ν = 2.4495 arctan (\frac{1.7321}{2.4495}) - arctan (1.7321)$ $= 2.4495 arctan (0.7071) - 6 0^{\circ} = 2.4495 (35.26 4^{\circ}) - 6 0^{\circ} = 86.3 9^{\circ} - 6 0^{\circ} = 26.3 8^{\circ}$ Toh $ν (2.0) \approx 26. 4^{\circ}$ .

Recall Solution 2.2

Kya/kyun: turning $ν$ mein add hota hai (equation C). Hum Mach numbers ki ladder par upar chadh rahe hain. $ν (M_{plume}) = ν (M_{e}) + θ_{turn} = 26.3 8^{\circ} + 6^{\circ} = 32.3 8^{\circ}$ Ab invert karo: woh $M$ dhundho jiska $ν$ $32.3 8^{\circ}$ ke barabar ho. $M = 2.2$ try karte hain: $M^{2} - 1 = 1.9596$ , $ν = 2.4495 arctan (1.9596/2.4495) - arctan (1.9596) = 2.4495 (38.6 6^{\circ}) - 62.9 6^{\circ} = 94.7 1^{\circ} - 62.9 6^{\circ} = 31.7 5^{\circ} .$ Thoda kam; $M = 2.23$ try karo: $M^{2} - 1 = 1.9917$ , $ν = 2.4495 (39.1 1^{\circ}) - 63.3 4^{\circ} = 95.8 0^{\circ} - 63.3 4^{\circ} = 32.4 6^{\circ}$ . Bahut close, toh $M_{plume} \approx 2.23$ .

Level 3 — Analysis

Recall Solution 3.1

Kya/kyun: plume tab tak expand karta hai jab tak uska static pressure $p_{\infty}$ ke barabar na ho jaaye; equation (B) us pressure ratio ko $M_{plume}$ se link karta hai. $γ = 1.3$ ke saath: exponent $\frac{γ}{γ - 1} = \frac{1.3}{0.3} = 4.3333$ , aur $\frac{γ - 1}{2} = 0.15$ . Denominator: $1 + 0.15 (3.5)^{2} = 1 + 0.15 (12.25) = 2.8375$ . $1.5 = (\frac{1 + 0.15 M _{plume}^{2}}{2.8375})^{4.3333}$ $4.3333$ -th root lo: $1. 5^{1/4.3333} = 1.0964$ , toh $1 + 0.15 M_{plume}^{2} = 1.0964 \times 2.8375 = 3.1110 \Rightarrow M_{plume}^{2} = \frac{2.1110}{0.15} = 14.07 \Rightarrow M_{plume} \approx 3.75 .$

Recall Solution 3.2

$M = 3.5$ par: $M^{2} - 1 = 11.25 = 3.3541$ . $ν (3.5) = 2.7689 arctan (3.3541/2.7689) - arctan (3.3541) = 2.7689 (50.4 6^{\circ}) - 73.4 2^{\circ} = 139.7 2^{\circ} - 73.4 2^{\circ} = 66.3 0^{\circ} .$ $M = 3.75$ par: $M^{2} - 1 = 13.0625 = 3.6142$ . $ν (3.75) = 2.7689 arctan (3.6142/2.7689) - arctan (3.6142) = 2.7689 (52.5 5^{\circ}) - 74.5 3^{\circ} = 145.5 1^{\circ} - 74.5 3^{\circ} = 70.9 8^{\circ} .$ Toh $θ_{turn} = 70.9 8^{\circ} - 66.3 0^{\circ} \approx 4. 7^{\circ}$ .

Recall Solution 3.3

Kya/kyun: sirf axial component $v cos θ_{turn}$ hi rocket ko push karta hai; sideways wala hissa waste ho jaata hai. $1 - cos (4. 7^{\circ}) = 1 - 0.99664 = 0.00336 \approx 0.34% .$ Nozzle ke bahar expand hone se jo extra momentum milta hai, woh sideways motion mein sirf ek-tihaayi percent se bhi kam lose karta hai — yeh ek mild mismatch ke liye ek chhota sa penalty hai. Thrust coefficient se $C_{F}$ ke zariye connect karo.

Level 4 — Synthesis

Recall Solution 4.1

Point A: $p_{e} / p_{\infty} = 1.0$ → perfectly expanded, koi waves nahi, koi loss nahi. Point B: $p_{e} / p_{\infty} = 2.0 > 1$ → under-expanded.

Plume Mach (equation B, $γ = 1.4$ ): exponent $= 1.4/0.4 = 3.5$ , $\frac{γ - 1}{2} = 0.2$ . Denominator $1 + 0.2 (9) = 2.8$ . $2.0 = (\frac{1 + 0.2 M _{plume}^{2}}{2.8})^{3.5} \Rightarrow 2. 0^{1/3.5} = 1.2190,$ $1 + 0.2 M_{plume}^{2} = 1.2190 \times 2.8 = 3.4132 \Rightarrow M_{plume}^{2} = \frac{2.4132}{0.2} = 12.07 \Rightarrow M_{plume} \approx 3.47 .$

Turning angle ( $k = 6 = 2.4495$ ): $ν (3.0)$ : $8 = 2.8284$ , $ν = 2.4495 arctan (2.8284/2.4495) - arctan (2.8284) = 2.4495 (49.1 1^{\circ}) - 70.5 3^{\circ} = 120.2 9^{\circ} - 70.5 3^{\circ} = 49.7 6^{\circ} .$ $ν (3.47)$ : $3.4 7^{2} - 1 = 11.04 = 3.3227$ , $ν = 2.4495 arctan (3.3227/2.4495) - arctan (3.3227) = 2.4495 (53.6 0^{\circ}) - 73.2 4^{\circ} = 131.3 0^{\circ} - 73.2 4^{\circ} = 58.0 6^{\circ} .$ $θ_{turn} = 58.0 6^{\circ} - 49.7 6^{\circ} \approx 8. 3^{\circ} .$ Toh design altitude se twice-the-ratio wale Point B tak climb karne par pressure mismatch double ho jaata hai aur ~ $8. 3^{\circ}$ outward turn produce hota hai.

Level 5 — Mastery

Recall Solution 5.1

Constants: $γ = 1.2$ , toh $\frac{γ - 1}{2} = 0.1$ , exponent $= 1.2/0.2 = 6.0$ , $k = 2.2/0.2 = 11 = 3.3166$ .

(a) Sea level, $p_{e} / p_{\infty} = 1.6$ . Denominator $1 + 0.1 (16) = 2.6$ . $1.6 = (\frac{1 + 0.1 M _{plume}^{2}}{2.6})^{6} \Rightarrow 1. 6^{1/6} = 1.0812,$ $1 + 0.1 M_{plume}^{2} = 1.0812 \times 2.6 = 2.8111 \Rightarrow M_{plume}^{2} = 18.11 \Rightarrow M_{plume} = 4.256.$ Ab angles: $ν (4.0)$ : $15 = 3.8730$ , $ν = 3.3166 arctan (3.8730/3.3166) - arctan (3.8730) = 3.3166 (49.4 2^{\circ}) - 75.5 0^{\circ} = 163.9 0^{\circ} - 75.5 0^{\circ} = 88.4 0^{\circ} .$ $ν (4.256)$ : $4.25 6^{2} - 1 = 17.11 = 4.1364$ , $ν = 3.3166 arctan (4.1364/3.3166) - arctan (4.1364) = 3.3166 (51.2 8^{\circ}) - 76.4 1^{\circ} = 170.0 7^{\circ} - 76.4 1^{\circ} = 93.6 6^{\circ} .$ $θ_{turn} = 93.6 6^{\circ} - 88.4 0^{\circ} \approx 5. 3^{\circ} .$

(b) Vacuum limit. Jaise $p_{\infty} \to 0$ hota hai, ratio $p_{e} / p_{\infty} \to \infty$ , isliye plume ko infinite Mach tak expand karna padega. Lekin $ν (M)$ bina bound ke nahi badhta — yeh ek ceiling $ν_{max}$ tak approach karta hai. $γ = 1.2$ ke liye: $ν_{max} = \frac{π}{2} (k - 1) \cdot \frac{180}{π} = 9 0^{\circ} (k - 1) = 9 0^{\circ} (3.3166 - 1) = 208. 5^{\circ} .$ Toh $M_{e} = 4.0$ se maximum possible turn hai $ν_{max} - ν (4.0) = 208. 5^{\circ} - 88.4 0^{\circ} \approx 120. 1^{\circ}$ . Plume isse zyada nahi turn kar sakta — physically iska matlab hai ki gas eventually utni fast ho jaati hai jitni ho sakti hai aur fan ka expansion "khatam" ho jaata hai. Plume boundary ek hard maximum-turn cone ban jaati hai.