Over - under expanded nozzle flows: Worked examples

3.1.18 · D3 · Physics › Compressible Flow & Aerodynamics › Over - under expanded nozzle flows

Is page mein hum har woh case dhundhte hain jo ek nozzle-pressure problem mein aa sakta hai: zyada kam exit pressure, zyada zyada exit pressure, exact match, degenerate "no expansion" nozzle, limiting "vacuum" case, aur real-world climbing rocket. Pehle hum poore territory ka map banate hain, phir har cell ko ek full worked example se conquer karte hain.

Machinery (do master equations) ke liye parent Over - under expanded nozzle flows dekho, aur building blocks ke liye Isentropic Flow Relations, Area-Mach Number Relation, aur Choked Flow & the Throat (M=1).

Woh do tools banana jo hum har jagah reuse karte hain

Kisi bhi example se pehle, main — plain words mein — sirf woh do formulas re-state karta hoon jo humein chahiye, taaki koi symbol bina reason ke na aaye.

Verdict rule, ek baar mein:

Scenario matrix

Har nozzle-pressure problem inhi cells mein se ek hoti hai. Neeche ke worked examples [Cell n] se tagged hain.

#	Case class	Kya khaas hai	Example
1	$p_{e} < p_{b}$ (over-expanded, mild)	lip par oblique shocks	Ex 1
2	$p_{e} < p_{b}$ strongly	Mach disk / separation risk	Ex 2
3	$p_{e} = p_{b}$ (perfect)	pressure term vanish hota hai, max thrust	Ex 3
4	$p_{e} > p_{b}$ (under-expanded)	expansion fans, jet bulges	Ex 4
5	Degenerate $A_{e} / A^{*} = 1$	throat = exit, $M_{e} = 1$ (sonic)	Ex 5
6	Limiting $p_{b} \to 0$ (vacuum)	hamesha under-expanded	Ex 6
7	Same nozzle, changing $p_{b}$ (climb)	regime mid-flight mein flip hota hai	Ex 7
8	Thrust sign across regimes	$(p_{e} - p_{b}) A_{e}$ $+$ ya $-$ ho sakta hai	Ex 8
9	Exam twist: design altitude ke liye solve karo	logic ko ulta karo	Ex 9

Hum poore time $γ = 1.4$ fix karte hain jab tak kuch aur na kaha jaye, toh $\frac{γ - 1}{2} = 0.2$ aur $\frac{γ}{γ - 1} = 3.5$ .

Figure — Over - under expanded nozzle flows

Worked example Ex 1 — Mild over-expansion

[Cell 1] $p_{0} = 10$ bar, $A_{e} / A^{*} = 4.0$ , $p_{b} = 1$ bar.

Forecast: Bada area ratio (4×) matlab gas bahut zyada expand ho gayi, toh $p_{e}$ shayad chhota hoga — guess over-expanded.

Tool 1 se $M_{e}$ nikalo. Area ratio $4.0$ ⇒ supersonic root $M_{e} \approx 2.94$ . Yeh step kyun? Geometry akele exit Mach fix karti hai; throat choked hai toh sirf shape chahiye.
Tool 2 se $p_{e}$ nikalo. $\frac{p _{0}}{p _{e}} = (1 + 0.2 \cdot 2.9 4^{2})^{3.5} = (2.729)^{3.5} \approx 34.0$ . Kyun? Yeh Mach number ko us pressure mein convert karta hai jiske saath woh correspond karta hai.
$p_{e}$ compute karo. $p_{e} = 10/34.0 \approx 0.294$ bar. Kyun? Bas rearrange karo.
Verdict. $p_{e} = 0.294 < p_{b} = 1.0$ ⇒ $PR = 0.294 < 1$ ⇒ over-expanded. Lip par oblique shocks. Dekho Oblique Shocks.

Verify: Pressure ratio $p_{0} / p_{e} = 10/0.294 = 34.0$ , step 2 se match ✓. Units: bar/bar dimensionless ✓. Forecast sahi ✓.

Worked example Ex 2 — Strong over-expansion, separation risk

[Cell 2] $p_{0} = 10$ bar, $A_{e} / A^{*} = 8.0$ , $p_{b} = 1$ bar.

Forecast: Aur bada area ratio ⇒ aur bhi kam $p_{e}$ ⇒ strongly over-expanded. Pressure gap bahut bada hai, toh Mach disk ya andar creep karte shock ke liye watch karo.

Tool 1 se $M_{e}$ . $A_{e} / A^{*} = 8.0$ ⇒ $M_{e} \approx 3.65$ . Kyun? Area ratio double karne se supersonic root aur upar chali jaati hai.
Tool 2 se $p_{e}$ . $p_{0} / p_{e} = (1 + 0.2 \cdot 3.6 5^{2})^{3.5} = (3.665)^{3.5} \approx 92.5$ , toh $p_{e} = 10/92.5 \approx 0.108$ bar. Kyun? Same isentropic conversion.
Verdict + severity. $PR = 0.108/1 = 0.108$ . Bahut zyada 1 se neeche: strongly over-expanded. Kyun care karo? Ek bada required pressure rise ( $p_{b} / p_{e} \approx 9.3$ ) woh exceed kar sakta hai jo ek attached oblique shock provide karta hai, toh ek normal shock / Mach disk form hota hai aur flow diverging wall ke andar separate ho sakti hai. Dekho Normal Shock & Mach Disk aur Flow Separation in Nozzles.

Verify: $p_{b} / p_{e} = 1/0.108 \approx 9.26$ — almost ek order-of-magnitude ka jump, "strong" ke saath consistent. Ex 1 se compare karo jahan $p_{b} / p_{e} \approx 3.4$ (milder) ✓.

Worked example Ex 5 — Degenerate nozzle, throat = exit

[Cell 5] $p_{0} = 10$ bar, $A_{e} / A^{*} = 1.0$ , $p_{b} = 1$ bar.

Forecast: Area ratio 1 matlab exit hi throat hai — koi diverging section nahi. Throat choked hone par $M = 1$ par hota hai, toh guess karo $M_{e} = 1$ , sonic (supersonic nahi).

Tool 1 se $M_{e}$ . $A_{e} / A^{*} = 1$ par, ek hi root hai $M_{e} = 1$ . Kyun? Tool 1 exactly 1 ke barabar hota hai $M = 1$ par (throat area function ka minimum point hai). Dekho Choked Flow & the Throat (M=1).
Tool 2 se $p_{e}$ . $p_{0} / p_{e} = (1 + 0.2 \cdot 1^{2})^{3.5} = (1.2)^{3.5} \approx 1.893$ , toh $p_{e} = 10/1.893 \approx 5.28$ bar. Kyun? Sonic exit ka ek fixed pressure ratio hota hai lagbhag 1.89.
Verdict. $p_{e} = 5.28 > p_{b} = 1$ ⇒ under-expanded — lekin note karo flow sirf sonic hai, toh external adjustment subsonic-to-transonic character ka hai, classic supersonic fan nahi. Over/under machinery phir bhi pressure se isko label karti hai.

Verify: $(1.2)^{3.5} \approx 1.893$ ; $10/1.893 \approx 5.28$ bar ✓. $γ = 1.4$ ke liye critical pressure ratio textbook ka $0.528$ hai (yaani $p_{e} / p_{0} = 0.528$ ) ✓.

Worked example Ex 6 — Limiting case: vacuum mein fire karna

[Cell 6] Same $A_{e} / A^{*} = 4.0$ nozzle, lekin $p_{b} \to 0$ (deep space).

Forecast: $p_{e} = 0.294$ bar ek fixed positive number hai; $p_{b} = 0$ . Kyunki koi bhi positive number > 0 hoti hai, yeh under-expanded honi chahiye — aur maximally so.

$PR$ . $PR = p_{e} / p_{b} = 0.294/0 \to + \infty$ . Yeh step kyun? Jab $p_{b} \to 0^{+}$ , ratio diverge karta hai. Vacuum mein har real nozzle under-expanded hoti hai.
Physical reading. Jet physically possible hadd tak expand karta hai; fans apni limit tak khulte hain (Prandtl–Meyer maximum turning). Koi shocks kabhi form nahi hote. Kyun? Tum zero pressure tak pahunchne ke liye compression ki zaroorat nahi rakh sakte.

Verify: Kisi bhi finite $p_{e} > 0$ aur $p_{b} \to 0$ ke liye, $PR \to \infty > 1$ ⇒ under-expanded har aise nozzle ke liye ✓ (ek limit inequality ke roop mein check kiya gaya).

Worked example Ex 7 — Same rocket, poora ascent

[Cell 7] Ek booster $p_{0} = 10$ bar, $A_{e} / A^{*} = 4.0$ ( $p_{e} = 0.294$ bar) ke saath climb karta hai. Teen altitudes par classify karo: sea level $p_{b} = 1.0$ , mid $p_{b} = 0.294$ , high $p_{b} = 0.10$ bar.

Forecast: Jab hum climb karte hain, $p_{b}$ girta hai jabki $p_{e}$ same rehta hai, toh regime over-expanded → perfect → under-expanded hona chahiye.

Sea level. $PR = 0.294/1.0 = 0.294 < 1$ ⇒ over-expanded (shocks). Kyun? Moti hawa jet ko andar push karti hai.
Mid. $PR = 0.294/0.294 = 1$ ⇒ perfect. Kyun? Hum design altitude par aa gaye.
High. $PR = 0.294/0.10 = 2.94 > 1$ ⇒ under-expanded (fans). Kyun? Patli hawa jet ko expand karte rehne deti hai.

Verify: Ratios $0.294, 1.000, 2.94$ order mein 1 ke aas-paas hain ✓ — ek nozzle, teen regimes, sirf $p_{b}$ se drive hote hain.

Worked example Ex 8 — Regimes ke across thrust sign

[Cell 8] $F = \overset{m}{˙} V_{e} + (p_{e} - p_{b}) A_{e}$ use karte hue. Lo $A_{e} = 0.10 m^{2}$ , $p_{e} = 0.294$ bar $= 29.4$ kPa. Sirf pressure-thrust term compute karo $p_{b} = 1$ bar (100 kPa) aur $p_{b} = 0.10$ bar (10 kPa) par.

Forecast: Over-expanded mein pressure term negative (drag-jaisa) hona chahiye; under-expanded mein positive.

Sea level (over-expanded). $(p_{e} - p_{b}) A_{e} = (29.4 - 100) \times 1 0^{3} \times 0.10 = - 7.06 \times 1 0^{3} N = - 7.06$ kN. Negative kyun? Ambient jet ko outpush karta hai; pressure thrust subtract karta hai. Units: Pa·m² = N ✓.
High altitude (under-expanded). $(29.4 - 10) \times 1 0^{3} \times 0.10 = + 1.94 \times 1 0^{3} N = + 1.94$ kN. Positive kyun? Jet ambient ko outpush karta hai; pressure thrust add karta hai.

Verify: $(29.4 - 100) \cdot 1 0^{3} \cdot 0.10 = - 7060$ N aur $(29.4 - 10) \cdot 1 0^{3} \cdot 0.10 = + 1940$ N ✓; sign regime ke saath flip karta hai ✓.

Worked example Ex 9 — Exam twist: ek target altitude ke liye design karo

[Cell 9] "Tumhe $p_{e} = 0.30$ bar chahiye (ek chosen altitude par perfect hone ke liye). $p_{0} = 10$ bar, $γ = 1.4$ . Kya $M_{e}$ aur roughly kya $A_{e} / A^{*}$ ?"

Forecast: Hum tools ko ulta run karte hain: pressure se shuru karo, $M_{e}$ lo, phir area ratio lo.

$M_{e}$ ke liye Tool 2 invert karo. $p_{0} / p_{e} = 10/0.30 = 33.33 = (1 + 0.2 M_{e}^{2})^{3.5}$ . Toh $1 + 0.2 M_{e}^{2} = 33.3 3^{1/3.5} = 33.3 3^{0.2857} \approx 2.717$ , giving $M_{e}^{2} = (2.717 - 1) /0.2 = 8.585$ , $M_{e} \approx 2.930$ . Kyun? Exponent undo karne ke liye $1/3.5$ power lo, phir $M^{2}$ mein linear relation solve karo.
Area ratio ke liye Tool 1 forward. $M_{e} = 2.93$ plug karo: $A_{e} / A^{*} \approx 3.97 \approx 4.0$ . Kyun? Yeh designer ko batata hai ki throat ke relative exit kitna bada banana hai.
Design par verdict. Construction se $p_{e} = p_{b} = 0.30$ bar wahan ⇒ perfect.

Verify: $33.3 3^{1/3.5} \approx 2.717$ ; $M_{e} \approx 2.93$ ; forward Tool 1 gives $A_{e} / A^{*} \approx 3.97$ — poore used $A_{e} / A^{*} = 4$ nozzle ke saath consistent ✓.

Recall Quick self-test

Ek nozzle $p_{e} = 0.30$ bar deti hai. Yeh $p_{b} = 0.30$ bar par fire ki gayi. Regime? ::: Perfectly expanded ( $PR = 1$ ), koi waves nahi, us $V_{e}$ ke liye max thrust. Same nozzle, $p_{b} = 0.05$ bar. Regime aur wave type? ::: Under-expanded ( $PR = 6 > 1$ ); Prandtl–Meyer expansion fans, jet bahar nikalta hai. Same nozzle, $p_{b} = 2$ bar. Regime aur wave type? ::: Over-expanded ( $PR = 0.15 < 1$ ); oblique shocks, possibly Mach disk / separation. Vacuum mein fire karna — over ya under? ::: Hamesha under-expanded, $PR \to \infty$ .