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

3.3.15 · Physics › Rocket Propulsion

Under-Expanded Nozzle Kya Hota Hai?

Doosre cases se comparison:

Perfectly expanded: $p_{e} = p_{\infty}$ → koi waves nahi, maximum axial thrust
Over-expanded: $p_{e} < p_{\infty}$ → oblique shock waves flow ko compress karti hain, isse bhi efficiency loss hoti hai (3.3.14-Over-expanded-nozzle-shock-diamonds mein cover kiya gaya hai)
Under-expanded: $p_{e} > p_{\infty}$ → Prandtl-Meyer expansion fans flow ko bahar ki taraf turn karti hain

Prandtl-Meyer Expansion Waves Kyun Banti Hain

First Principles: Exit Plane Par Pressure Imbalance

Nozzle exit lip par, $p_{e}$ pressure ka exhaust $p_{\infty}$ par ambient air se milta hai. Agar $p_{e} > p_{\infty}$ hai, toh high-pressure exhaust external resistance mein achanak girावट "dekhta" hai. Equilibrate karne ke liye, usse expand karna hoga.

Lekin expansion waves kyun hoti hain, uniform expansion kyun nahi?

Supersonic flow mein, information upstream travel nahi kar sakti (disturbances Mach waves ke saath angle $μ = arcsin (1/ M)$ par propagate karti hain). Toh expansion ek centered expansion fan ke through honi chahiye—Mach waves ki ek continuous family jo nozzle lip se emanate karti hai.

Formula Prandtl-Meyer Function

Prandtl-Meyer function $ν (M)$ ek isentropic expansion ke through Mach number ko cumulative flow turning angle se relate karta hai: $ν (M) = \frac{γ + 1}{γ - 1} arctan (\frac{γ - 1}{γ + 1} (M^{2} - 1)) - arctan (M^{2} - 1)$ Iska matlab: Agar Mach $M_{1}$ par flow angle $θ$ se turn karti hai, toh naya Mach number $M_{2}$ satisfy karta hai: $ν (M_{2}) = ν (M_{1}) + θ$

First principles se derivation:

Ek isentropic turn ke liye, total pressure aur total temperature constant rehte hain.
Mach wave angle $μ = arcsin (1/ M)$ hai. Har infinitesimal wave flow ko $d θ$ se turn karti hai.
Mach number change $d M$ aur turn angle $d θ$ geometry aur isentropic relations se related hain.
Isentropic flow mein area-velocity relation se shuru karte hue: $\frac{d A}{A} = (\frac{1}{M ^{2}} - 1) \frac{d V}{V}$ Ek turn ke liye (fan mein khud koi area change nahi), hum velocity hodograph aur momentum balance use karte hain.
Differential relation ko integrate karte hain (oblique shock relations se derive kiya gaya jab $p_{2} / p_{1} \to 1$ ): $d θ = \frac{M ^{2} - 1}{1 + \frac{γ - 1}{2} M ^{2}} \frac{d M}{M}$
$M = 1$ se integrate karo (jahan $θ = 0$ , define $ν (1) = 0$ ) arbitrary $M$ tak: $ν (M) = \int_{1}^{M} \frac{M ^{2} - 1}{M ( 1 + \frac{γ - 1}{2} M ^{2} )} d M$
Trig substitution $M = sec ϕ$ use karke aur simplify karne par upar wala closed form milta hai.

Yeh step kyun? Prandtl-Meyer function supersonic flow mein possible maximum turning angle ko encapsulate karta hai, jab tak flow infinite Mach par "straight" nahi ho jaati. $γ = 1.4$ ke liye, $ν_{m a x} \approx 130.5°$ .

Nozzle Exit Par Physical Process

Exit lip par ( $x = x_{e}$ , $r = r_{e}$ ): Flow Mach $M_{e}$ aur pressure $p_{e} > p_{\infty}$ par exit karti hai।
Expansion fan nucleate hoti hai: Mach waves lip se spread hoti hain, har ek flow ko thoda bahar ki taraf turn karti hai।
Flow accelerate aur turn hoti hai: Jab exhaust fan se guzarta hai, Mach number $M_{plume}$ tak increase hota hai, pressure $p_{\infty}$ tak drop hoti hai, aur velocity vector angle $θ_{turn}$ se rotate hota hai।
Fan ke baad: Flow dobara uniform ho jaati hai zyada Mach par, lower pressure par, lekin ab ek radial velocity component hai।

Total turning angle $θ_{turn}$ satisfy karta hai: $ν (M_{plume}) = ν (M_{e}) + θ_{turn}$ jahan $M_{plume}$ isentropic relation se milta hai (total pressure $p_{0}$ conserved hai): $\frac{p _{e}}{p _{\infty}} = \frac{( 1 + \frac{γ - 1}{2} M _{plume}^{2} ) ^{γ / (γ - 1)}}{( 1 + \frac{γ - 1}{2} M _{e}^{2} ) ^{γ / (γ - 1)}}$ yaani plume tab tak expand hota hai jab tak uski static pressure $p_{\infty}$ se match nahi kar leti, zyada Mach $M_{plume} > M_{e}$ tak reach karke।

Efficiency Loss: Thrust Reduction

Axial vs. Radial Momentum

Thrust coefficient measure karta hai ki ideal momentum flux ka kitna axial thrust mein convert hota hai। Under-expanded nozzle ke liye:

$C_{F} = \frac{F}{m ˙ v _{e}^{ideal}} < C_{F, optimum}$

Thrust kyun drop hota hai?

Momentum redirection: Expansion fan velocity ko $θ_{turn}$ se bahar ki taraf turn karta hai। Sirf axial component $v cos θ_{turn}$ thrust mein contribute karta hai।
React karne ke liye koi nozzle wall nahi: Perfectly expanded nozzle mein, poori expansion andar hoti hai jahan walls flow ko redirect karti hain। Yahan, bahar ki expansion "free" hai — koi wall push back nahi karti।

Formula Under-Expanded Nozzle ka Thrust

Standard rocket thrust equation, physical exit plane par apply kiya gaya: $F = \overset{m}{˙} v_{e} + A_{e} (p_{e} - p_{\infty})$ Derivation:

Momentum theorem apply karo ek control volume par jo exhaust ko exit plane tak bound karta hai। Momentum flux bahar hai $\overset{m}{˙} v_{e}$ ; exit plane par net pressure force hai $A_{e} (p_{e} - p_{\infty})$ ।
Jab $p_{e} > p_{\infty}$ (under-expanded), pressure term $A_{e} (p_{e} - p_{\infty}) > 0$ thrust mein add karta hai, lekin $v_{e}$ fully-expanded exit velocity se kam hai kyunki gas nozzle ke andar $p_{\infty}$ tak abhi expand nahi hui।
Optimum tab hota hai jab $p_{e} = p_{\infty}$ : pressure term zero ho jaata hai lekin $v_{e}$ maximum hoti hai, aur yeh dikhaया ja sakta hai ki $\partial F / \partial A_{e} = 0$ wahan — perfectly-expanded case sabse bada $F$ deta hai।

Efficiency loss perfectly-expanded ideal ke relative: $Δ F = F_{opt} - F_{actual} > 0$ Physically, external Prandtl-Meyer expansion momentum ko off-axis turn karti hai: plume mein axial momentum $\overset{m}{˙} v_{plume} cos θ_{turn}$ hai, aur factor $(1 - cos θ_{turn})$ additional expansion momentum ka woh fraction quantify karta hai jo radial direction mein waste hota hai। Yeh term $p_{e} / p_{\infty}$ ke saath badhta hai (bada $θ_{turn}$ ), toh bade pressure mismatches zyada costly hote hain।

Practical numbers: Agar $p_{e} / p_{\infty} = 1.5$ aur $M_{e} = 3$ , typical $θ_{turn} \approx 3 - 5°$ , jo $≲ 1%$ thrust loss deta hai। Bade pressure ratios → bade angles → zyada loss।

Worked Examples

Worked example Example 1: Sea-Level Under-Expansion

Given: Nozzle exit Mach $M_{e} = 3.5$ , exit pressure $p_{e} = 1.5 atm$ , ambient $p_{\infty} = 1 atm$ , $γ = 1.3$ ke saath। Find: Plume Mach $M_{plume}$ , turning angle $θ_{turn}$ , aur efficiency loss।

Solution:

Regime check karo: $p_{e} / p_{\infty} = 1.5 > 1$ → under-expanded। Sahi hai।
Plume Mach isentropic relation se। Total pressure $p_{0}$ conserved hai, toh: $\frac{p _{e}}{p _{\infty}} = \frac{( 1 + \frac{γ - 1}{2} M _{plume}^{2} ) ^{γ / (γ - 1)}}{( 1 + \frac{γ - 1}{2} M _{e}^{2} ) ^{γ / (γ - 1)}}$ $γ = 1.3$ ke saath: exponent $= γ / (γ - 1) = 1.3/0.3 = 4.333$ , aur $\frac{γ - 1}{2} = 0.15$ । $1 + 0.15 M_{e}^{2} = 1 + 0.15 (12.25) = 2.8375$ । $1.5 = (\frac{1 + 0.15 M _{plume}^{2}}{2.8375})^{4.333}$ Yeh step kyun? Plume ko aur expand karna hoga (Mach badhta hai, static pressure $p_{\infty}$ tak girती hai), toh numerator $M_{plume}$ carry karta hai। Solve karo: $1. 5^{1/4.333} = 1.0964$ , toh $1 + 0.15 M_{plume}^{2} = 1.0964 \times 2.8375 = 3.111$ । $0.15 M_{plume}^{2} = 2.111 \Rightarrow M_{plume}^{2} = 14.07 \Rightarrow M_{plume} \approx 3.75$
Turning angle Prandtl-Meyer se ( $γ = 1.3$ , $(γ + 1) / (γ - 1) = 2.3/0.3 = 2.769$ ): $ν (3.5) = 2.769 arctan (\frac{1}{2.769} 12.25 - 1) - arctan 11.25$ $= 2.769 arctan (1.2111) - arctan (3.354) = 2.769 (50.46°) - 73.42° = 66.3°$ $ν (3.75) = 2.769 arctan (\frac{1}{2.769} 14.06 - 1) - arctan 13.06$ $= 2.769 arctan (1.3055) - arctan (3.614) = 2.769 (52.55°) - 74.53° = 71.0°$ $θ_{turn} = 71.0° - 66.3° = 4.7°$
Thrust loss: $cos (4.7°) \approx 0.9966$ , toh extra expansion par off-axis momentum loss $\sim 0.3%$ hai। Mild under-expansion → chhota penalty।

Yeh step kyun? Modest pressure ratios (1.5:1) chhote turning angles dete hain—under-expansion yahan catastrophic nahi hai।

Worked example Example 2: Extreme Under-Expansion

Given: $p_{e} / p_{\infty} = 5$ , $M_{e} = 2.5$ , $γ = 1.4$ । Find: Plume Mach aur turning angle।

Solution:

Exponent $γ / (γ - 1) = 1.4/0.4 = 3.5$ , aur $\frac{γ - 1}{2} = 0.2$ । Phir $1 + 0.2 M_{e}^{2} = 1 + 0.2 (6.25) = 2.25$ । $5 = (\frac{1 + 0.2 M _{plume}^{2}}{2.25})^{3.5}$ $5^{1/3.5} = 1.5849$ , toh $1 + 0.2 M_{plume}^{2} = 1.5849 \times 2.25 = 3.566$ । $0.2 M_{plume}^{2} = 2.566 \Rightarrow M_{plume}^{2} = 12.83 \Rightarrow M_{plume} \approx 3.58$
Prandtl-Meyer ( $γ = 1.4$ , $2.4/0.4 = 6 = 2.449$ ): $ν (2.5) = 2.449 arctan (\frac{1}{2.449} 5.25) - arctan 5.25$ $= 2.449 arctan (0.9354) - arctan (2.291) = 2.449 (43.09°) - 66.42° = 39.1°$ $ν (3.58) = 2.449 arctan (\frac{1}{2.449} 11.82) - arctan 11.82$ $= 2.449 arctan (1.4038) - arctan (3.438) = 2.449 (54.53°) - 73.78° = 59.8°$ $θ_{turn} = 59.8° - 39.1° = 20.7°$
Momentum loss: $cos (20.7°) \approx 0.9354$ → additional expansion par $\sim 6.5%$ off-axis loss; net thrust penalty (pressure-thrust partial compensation ke baad) kai percent hai। Yeh kyun matter karta hai: $p_{e} / p_{\infty} = 5$ par, turning angle bada hai aur losses significant hain। Altitude-optimized nozzles sea level par buri tarah suffer karte hain।

Common Mistakes

Active Recall Flashcards

#flashcards/physics

"Under-expanded nozzle" ka matlab kya hai?

Exit pressure

p_{e}

, ambient

p_{\infty}

se zyada hoti hai; exhaust Prandtl-Meyer waves ke zariye nozzle ke bahar expand hota hai, flow outward turn hoti hai aur axial thrust loss hoti hai।

Prandtl-Meyer function

ν (M)

kya hai?

Ek function jo Mach number ko expansion fan mein cumulative isentropic turning angle se relate karta hai; turn

θ

ke liye

ν (M_{2}) = ν (M_{1}) + θ

।

Under-expansion mein nozzle exit par expansion waves kyun banti hain?

Pressure imbalance (

p_{e} > p_{\infty}

) expansion force karti hai, lekin supersonic flow uniformly adjust nahi ho sakti—disturbances Mach waves ki tarah propagate hoti hain, ek centered expansion fan banti hai।

Under-expansion thrust kyun reduce karta hai?

Expansion fan velocity ko

θ

angle se bahar ki taraf turn karta hai; sirf

v cos θ

axial thrust mein contribute karta hai, aur

v_{e}

fully-expanded optimum se below hai।

Under-expanded vs. over-expanded nozzle?

Under-expanded:

p_{e} > p_{\infty}

, expansion waves, flow bahar ki taraf turn hoti hai। Over-expanded:

p_{e} < p_{\infty}

, oblique shocks, flow andar ki taraf turn hoti hai। Dono efficiency lose karte hain; optimum

p_{e} = p_{\infty}

hai।

Exit plane par use ki jaane wali thrust equation kya hai?

F = \overset{m}{˙} v_{e} + A_{e} (p_{e} - p_{\infty})

; maximize hoti hai jab

p_{e} = p_{\infty}

।

Prandtl-Meyer expansion ke liye key assumption kya hai?

Isentropic flow (entropy constant, total pressure constant)। Expansion smooth aur reversible hoti hai, shocks se unlike।

Mnemonic

Feynman Recap

Recall 12-saal ke bachche ko explain karo

Socho tum ek balloon phula rahe ho aur tum use chhor dete ho—hawa bahar nikalti hai, hai na? Ab socho balloon ki neck ek rocket nozzle hai। Agar andar ki hawa tight squeezed hai (high pressure) aur bahar ki hawa loose hai (low pressure), toh hawa nozzle chhodne ke baad aur spread out hona chahti hai।

Lekin yeh baat hai: rocket mein, hum chahte hain ki poori hawa seedha peeche shoot kare rocket ko aage push karne ke liye। Agar hawa exit hote waqt abhi bhi "bahut tight" hai, toh woh nozzle ke bahar expand karti rehti hai, aur ek fan ki tarah sideways spread hoti hai। Woh sideways motion rocket ko push karne mein help nahi karta—woh waste ho jaata hai।

Scientists is cheez ko "under-expanded" kehte hain kyunki nozzle ne gas ko enough expand nahi karne diya andar jahan hum isko control kar sakte hain। Gas yeh smooth wave patterns (Prandtl-Meyer waves) banati hai jab woh spread out hoti hai। Jitna zyada sideways spread hogi, utna kam push milega। Isliye rocket engineers exit pressure ko bahar ki hawa se match karne ki koshish karte hain—toh poori expansion nozzle ke andar ho, seedha peeche push karte hue!

Connections

3.3.13-Optimal-expansion-ratio — Kyun $p_{e} = p_{\infty}$ thrust maximize karta hai
3.3.14-Over-expanded-nozzle-shock-diamonds — Opposite case: $p_{e} < p_{\infty}$
3.2.7-Isentropic-flow-area-Mach-relation — Expansion math ki foundation
4.1.3-Oblique-shock-theory — Contrast: expansion waves vs. shocks
3.3.16-Altitude-compensation-nozzles — Designs (aerospike, dual-bell) jo under/over-expansion mitigate karte hain
2.5.12-Thrust-coefficient-definition — Kaise $C_{F}$ efficiency capture karta hai