Worked examples — Over-expanded nozzle — oblique shocks in plume, efficiency loss

3.3.14 · D3 · Physics › Rocket Propulsion › Over-expanded nozzle — oblique shocks in plume, efficiency l

Yeh page parent topic ka drill hall hai. Yahan hum koi nayi theory nahi seekhenge — hum isko exercise karenge, physics ke har possible case ko hit karte hue. Agar koi symbol aaye jo tumne pehle nahi dekha, toh hum use wahan pe hi rebuild kar lenge.

Shuru karne se pehle, ek promise: neeche jo bhi number dikh raha hai woh sab note ke bottom mein machine se check kiya gaya hai. Kuch bhi faith pe assert nahi kiya gaya.

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Scenario matrix

Over-expansion do pressures ki comparison se govern hoti hai — nozzle ke exit plane par (woh flat circular hole jahan se gas bahar nikalti hai):

• $p_e$ = exit pressure — woh pressure jo gas ke paas hoti hai baad mein jab nozzle usse expand kar chuka hota hai. Socho isko "gas kitni compressed reh gayi jab woh bahar nikli."
• $p_a$ = ambient pressure — bahar ki hawa ka pressure jo push-back kar rahi hai. Sea level par $p_a\approx 101\text{kPa}$; vacuum mein $p_a=0$.

Ek hi quantity hai jo sab kuch decide karti hai — $(p_e-p_a)$ ka sign. Wahi hamara "quadrant" hai. Yeh raha poora matrix — har cell ko neeche ek example se cover kiya jaana chahiye.

Cell	Condition	Physical regime	Covered by
A	$p_e=p_a$	Perfectly matched (baseline)	Ex 1
B	$p_e<p_a$ (mild)	Mildly over-expanded, shocks weak	Ex 2
C	$p_e<p_a$ (severe)	Severely over-expanded, strong shocks	Ex 3
D	$p_e>p_a$	Under-expanded ("doosra sign")	Ex 4
E	$p_a=0$ (degenerate)	Vacuum limit — pressure thrust maximal	Ex 5
F	$p_e\to p_a$ (limiting)	Mismatch $\to 0$ hone par shock vanish ho jaata hai	Ex 6
G	Real-world word problem	Ascending rocket matched altitude cross karta hai	Ex 7
H	Exam twist	$\beta$ diya hai, $\theta$ nikalo aur separation check karo	Ex 8

Hum yeh tools baar baar use karte hain — pehli baar zaroorat padne par inhe re-earn kiya jaata hai:

Recall Symbols jo tumhe chahiye (tap karke reveal karo)

$\dot{m}$ ::: mass flow rate — kilograms of gas leaving per second $v_e$ ::: exhaust velocity — gas kitni tez se exit karti hai, m/s mein $v_{e,\text{ideal}}$ ::: woh exhaust velocity jo ek loss-free (isentropic, no-shock) nozzle produce karta — theoretical best $v_{e,\text{actual}}$ ::: woh exhaust velocity jo tumhe actually milti hai shocks ke baad — hamesha $\le v_{e,\text{ideal}}$ $A_e$ ::: exit area — nozzle ke munh ka area, m² mein $F$ ::: thrust — rocket par forward push, newtons mein $M_e$ ::: exit Mach number — exit speed divided by local speed of sound $\gamma$ ::: ratio of specific heats — ek gas property, hot rocket exhaust ke liye $\approx 1.2$ $\beta$ ::: shock angle — oblique shock ka tilt incoming flow se relative $\theta$ ::: deflection angle — flow kitna bend karta hai shock cross karte waqt $p_1,p_2$ ::: static pressure ek shock ke bilkul PEHLE (1) aur bilkul BAAD (2) $p_0$ ::: stagnation pressure — woh pressure jo gas tab reach karti agar use smoothly (bina shock ke) rest tak laaya jaata; flow ki stored "ordered" quality ko measure karta hai $p_{01},p_{02}$ ::: wahi $p_0$ shock se pehle (subscript 01) aur baad (subscript 02) measure ki — toh $p_{02}/p_{01}<1$ matlab quality lost ho gayi $M_{1n}$ ::: normal Mach component, $M_1\sin \beta$ — yahi akela shock strength set karta hai $C_v$ ::: velocity efficiency — actual exit speed divided by ideal exit speed (Ex 3 mein define ki gayi)

Woh master formula jo hum baar baar use karte hain:

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Ex 1 — Cell A: matched baseline

Forecast: Aage padhne se pehle guess karo — kya pressure term help karega, hurt karega, ya vanish ho jaayega?

Step 1 — Momentum thrust compute karo. $\dot{m}{v}_e=250\times 2400=600,000\text{N}=600\text{kN}$ Yeh step kyun? Momentum thrust hamesha present rehta hai; yeh "engine ka honest push" hai. Hum ise pehle isolate karte hain taaki pressure term alag se dikh sake.

Step 2 — Pressure thrust compute karo. $(p_e-p_a)A_e=(60-60)\times 10^3\times 0.8=0$ Yeh step kyun? Matched case ka poora point yahi hai ki yeh term zero ho jaati hai. Jab $p_e=p_a$ hota hai toh exit disc ko dono taraf se equal push milti hai, toh kuch contribute nahi hota.

Step 3 — Total. $F=600+0=600\text{kN}$

Verify: Units: $\text{kg/s}\times \text{m/s}={\text{kg⋅m/s}}^2=\text{N}$ ✓. Pressure term $\text{Pa}\times {\text{m}}^2=\text{N}$ ✓. Yeh woh ceiling hai jiske against har off-design case measure kiya jaata hai.

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Ex 2 — Cell B: mildly over-expanded

Forecast: Ab $p_e<p_a$ hai. Pressure term positive hoga ya negative?

Step 1 — Pressure term. $(60-75)\times 10^3\times 0.8=-12,000\text{N}=-12\text{kN}$ Yeh step kyun? Kyunki $p_e<p_a$ hai, bahar ki hawa exit disc ko exhaust se zyada press karti hai — yeh ek retarding force hai. Minus sign woh physics hai jo keh rahi hai "tujhe peeche push kiya ja raha hai."

Step 2 — Total thrust. $F=600-12=588\text{kN}$ Yeh step kyun? (Negative) pressure thrust ko unchanged momentum thrust mein add karo.

Step 3 — Loss. $\Delta F=588-600=-12\text{kN}(2\%\text{loss})$

Verify: Ek mild mismatch (75 vs 60 kPa) se ek mild $2\%$ hit milta hai — sensible hai. Yahan koi shock computation nahi chahiye kyunki mismatch gentle hai; plume weak oblique shocks se correct karta hai jo efficiency ko barely dent karte hain.

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Ex 3 — Cell C: severely over-expanded (with shock loss)

Forecast: Yahan do losses stack hoti hain — pressure mismatch aur shocks. Guess karo: kya $C_v$ $0.9$ se upar rahega?

Step 1 — Pressure thrust loss. $(8-101)\times 10^3\times 0.8=-74,400\text{N}=-74.4\text{kN}$ Yeh step kyun? Bahut bada mismatch (8 vs 101 kPa) matlab atmosphere plume ko crush kar raha hai — retarding force enormous hai.

Step 2 — Shock ka normal Mach component. Shock $\beta =30^{\circ }$ par tilted hai. Sirf flow ka woh component jo shock ke perpendicular hai compress hota hai; parallel part bina kisi change ke through nikal jaata hai. Woh perpendicular Mach hai $M_{1n}=M_e\sin \beta =4.0\times \sin 30^{\circ }=4.0\times 0.5=2.0$ Yeh step kyun? Ek oblique shock bas ek normal shock hai jo velocity ke perpendicular part ko dekhta hai — isliye hum velocity ko ek normal leg (length $\propto \sin \beta$) aur ek tangential leg (length $\propto \cos \beta$) mein split karte hain. Tangential leg unchanged pass through hoti hai, toh shock strength sirf $M_{1n}$ par depend karti hai. Yeh figure mein woh pink arrow hai.

Step 3 — Ek shock ke across stagnation pressure ratio. Symbol box se yaad karo ki stagnation pressure $p_0$ woh pressure hai jo gas tab reach karti agar use smoothly (bina shock ke) rest tak laaya jaata; yeh flow ki ordered energy ki "quality" measure karta hai, aur $p_{01}$ / $p_{02}$ bas shock se pehle aur baad measure kiya gaya $p_0$ hai. Ek shock iska kuch hissa destroy kar deta hai. Ratio $M_{1n}$ par apply kiye gaye normal-shock equations se aata hai (energy conserved, lekin entropy rises), jo deta hai: $\frac{p_{02}}{p_{01}}={\left(\frac{(\gamma +1)M_{1n}^2}{2+(\gamma -1)M_{1n}^2}\right)}^{\frac{\gamma }{\gamma -1}}{\left(\frac{\gamma +1}{2\gamma M_{1n}^2-(\gamma -1)}\right)}^{\frac{1}{\gamma -1}}$ Yeh form kyun? Pehla bracket shock ke across density/temperature build-up hai; doosra static-pressure jump hai. Unhe un $\gamma$-powers tak raise karna woh hai jaise ideal-gas entropy rise lost stagnation pressure mein turn hoti hai. $M_{1n}=2.0$, $\gamma =1.2$ plug in karo: $\frac{p_{02}}{p_{01}}\approx 0.680$ Yeh step kyun? Yeh ratio exactly woh hai ki ek shock ke baad kitna "usable push" survive karta hai. 1 se kam matlab loss hai.

Step 4 — Teen shocks compound karo (diamond pattern). $\frac{p_{0,\text{final}}}{p_{0,\text{initial}}}=(0.680{)}^3\approx 0.314$ Yeh step kyun? Shock diamonds plume mein kai baar compression repeat karte hain (do lip shocks plus Mach disk); har stage survival ratio ko multiply karta hai, toh losses compound hoti hain.

Step 5 — Velocity efficiency ($C_v$ define karo).

$C_v=\sqrt{0.314}\approx 0.560$ Square root kyun? Exhaust kinetic energy $\frac{1}{2}{v}_e^2$ available stagnation pressure track karta hai; kyunki speed${}^2$ $p_0$ follow karta hai, speed khud $\sqrt{p_0}$ follow karti hai.

Step 6 — Actual momentum thrust aur total. $v_{e,\text{actual}}=0.560\times 2400\approx 1345\text{m/s}$ $F=\dot{m}{v}_{e,\text{actual}}+(p_e-p_a)A_e=250(1345)-74,400\approx 336,300-74,400=261.9\text{kN}$

Verify: $600\text{kN}$ ke matched ceiling (Ex 1) se hum $\approx 262\text{kN}$ par crash ho gaye — yeh ek $56\%$ collapse hai. Yahi exact reason hai ki parent note warn karta hai: kabhi bhi vacuum engine ko sea level par fire mat karo. Agar yeh aur bhi worse ho jaaye toh dekho Nozzle Flow Separation.

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Ex 4 — Cell D: DOOSRA sign (under-expanded)

Forecast: Ab $p_e>p_a$ hai. Pressure term ka kya sign hai, aur kya gas shock karta hai ya expand karta hai?

Step 1 — Pressure term. $(60-40)\times 10^3\times 0.8=+16,000\text{N}=+16\text{kN}$ Yeh step kyun? $p_e>p_a$ matlab exhaust atmosphere se zyada push karta hai, toh exit disc ko ek extra forward shove milta hai — yeh ek positive contribution hai.

Step 2 — Total thrust. $F=600+16=616\text{kN}$

Step 3 — Shock ya fan? Kyunki flow ko lower ambient se match karne ke liye apna pressure drop karna hai, woh expansion fans se karta hai, jo smooth aur reversible (isentropic) hote hain — koi entropy penalty nahi. Over-expansion ke lossy shocks se contrast karo (dekho Shock Wave Fundamentals aur Isentropic Flow).

Verify: Ex 2 ke comparison mein pressure term ka sign flip ho gaya — yeh mirror matrix ke do "quadrants" confirm karta hai. Note karo ki under-expansion ne koi shock loss ke bina $16\text{kN}$ add kiya, jo parent ki key mistake-warning show karta hai: do off-design cases symmetric nahi hain.

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Ex 5 — Cell E: degenerate vacuum limit ($p_a=0$)

Forecast: $p_a=0$ ke saath, kya pressure term sabse bada hai ya sabse chota jo kabhi ho sakta hai?

Step 1 — Degenerate value par pressure term. $(p_e-0)A_e=8000\times 0.8=+6,400\text{N}=+6.4\text{kN}$ Yeh step kyun? Ambient variable ko uski floor, zero, par set karo. Ab har leftover exit pressure pure forward thrust ban jaati hai kyunki bahar kuch nahi hai jo push back kare.

Step 2 — Total. $F=600+6.4=606.4\text{kN}$

Verify: Ek given $p_e$ ke liye pressure term yahan maximum tak pahunch sakti hai: kuch push back nahi karta, toh saari exit pressure help karti hai. Yeh bhi dikhata hai ki vacuum nozzles bade kyun banaye jaate hain — vacuum mein low $p_e$ ka koi penalty nahi, sirf reward hai. Altitude Compensation aur Nozzle Area Ratio se ties hain.

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Ex 6 — Cell F: limiting case $p_e\to p_a$

Forecast: Almost-zero mismatch ke saath, shock kitna strong hai?

Step 1 — Pressure jump needed. $\frac{p_2}{p_1}=\frac{p_a}{p_e}\approx 1.0000017\approx 1$ Yeh step kyun? Shock ka ek hi kaam hai — $p_e$ ko bridge karke $p_a$ tak pahunchana. Agar woh gap essentially zero hai, toh shock ke paas essentially kuch karne ko nahi hai.

Step 2 — Static-pressure relation se normal Mach. Ek shock ke across static-pressure jump uske normal Mach component se set hota hai: $\frac{p_2}{p_1}=1+\frac{2\gamma }{\gamma +1}\left(M_{1n}^2-1\right)$ Yeh form kyun? Yeh shock ke across mass aur momentum conservation se aata hai: jab $M_{1n}=1$ tab bracket zero hai aur koi jump nahi; jab $M_{1n}$ badhta hai tab jump badhta hai. Left side $\to 1$ hone par $M_{1n}$ solve karo: $M_{1n}^2-1\to 0\Rightarrow M_{1n}\to 1$ Yeh step kyun? $M_{1n}=1$ ek Mach wave hai — sabse weak possible disturbance, koi real compression nahi.

Step 3 — Poora deflection ($\theta$–$\beta$–$M$) relation, aur $\theta \to 0$ kyun.

Numerator hai $M_1^2{\sin }^2\beta -1=M_{1n}^2-1$. Step 2 se yeh $\to 0$ hai, aur denominator finite aur positive rehta hai, toh $\tan \theta \to 2\cot \beta \times \frac{0}{\text{finite}}=0\Rightarrow \theta \to 0^{\circ }$ Yeh step kyun? Jab perpendicular Mach 1 hit karta hai, numerator vanish ho jaata hai aur flow kuch bhi bend nahi hota — "shock" sirf ek whisper hai. Yeh smoothly matched case (Ex 1) ko hand-off karta hai.

Step 4 — Limit par shock angle. $\sin \beta =\frac{M_{1n}}{M_e}\to \frac{1}{4.0}=0.25\Rightarrow \beta \to 14.48^{\circ }$ Yeh step kyun? $M_{1n}\to 1$ ke saath tilt Mach angle par settle ho jaata hai, is Mach number ke liye sabse shallow possible wave.

Verify: Mismatch $\to 0$ hone par: $\beta \to$ Mach angle ($\approx 14.5^{\circ }$), $\theta \to 0$, loss $\to 0$. Physics continuous hai — perfect matching cross karte waqt koi sudden jump nahi. Yeh Cell B aur Cell A ke beech exact boundary hai.

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Ex 7 — Cell G: real-world ascent word problem

Forecast: Kya yeh stage pad par zyada push karta hai, ya upar jaake?

Step 1 — Liftoff pressure term (over-expanded). $(70-101)\times 10^3\times 3.0=-93,000\text{N}=-93\text{kN}$ Yeh step kyun? Pad par thick atmosphere push back karti hai; term negative hai (Cell C-type over-expansion).

Step 2 — Liftoff thrust. $F_{\text{SL}}=2000-93=1907\text{kN}$ Yeh step kyun? (Negative) pressure thrust ko constant momentum thrust mein add karo.

Step 3 — Matched-altitude thrust. $(70-70)\times 10^3\times 3.0=0\Rightarrow F_{\text{match}}=2000+0=2000\text{kN}$ Yeh step kyun? Matched altitude par pressure term vanish ho jaata hai (Cell A), toh thrust full momentum value tak rise ho jaati hai.

Step 4 — Climb ke dauran gain. $\Delta F=2000-1907=93\text{kN}\text{climbing se gain kiya.}$

Verify: Rockets altitude ke saath zyada efficient hote hain matched hone tak, phir uske upar under-expanded ho jaate hain (Cell D). Yahi real reason hai Rocket Staging aur Altitude Compensation ka — koi bhi ek fixed nozzle har height par optimal nahi hota. Numbers internally consistent hain: $-93+93=0$ baseline par wapas ✓.

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Ex 8 — Cell H: exam twist ($\theta$ nikalo, phir separation test karo)

Forecast: Mach 3 par ek shallow-ish $40^{\circ }$ shock — bada turn ya chota turn? Separation hogi ya nahi?

Step 1 — $\theta$–$\beta$–$M$ relation se deflection. Hum Ex 6 mein stated formula reuse karte hain: $\tan \theta =2\cot \beta \frac{M_1^2{\sin }^2\beta -1}{M_1^2(\gamma +\cos 2\beta )+2}$ $M_1=3$, $\beta =40^{\circ }$, $\gamma =1.2$ plug in karo:

• $M_1^2{\sin }^2\beta =9\times {\sin }^{2}40^{\circ }=9\times 0.4132=3.719$
• numerator $=3.719-1=2.719$
• $\cos 2\beta =\cos 80^{\circ }=0.1736$, toh denominator $=9(1.2+0.1736)+2=9(1.3736)+2=14.36$
• $\cot 40^{\circ }=1.1918$

$\tan \theta =2(1.1918)\frac{2.719}{14.36}=2.3836\times 0.1894=0.4514$ $\theta =\arctan (0.4514)\approx 24.3^{\circ }$ $\arctan$ kyun? Hum formula se turn angle ka tangent jaante hain, aur hume angle chahiye — $\arctan$ woh inverse hai jo jawaab deta hai "kis angle ka yeh tangent hai?"

Step 2 — Pressure ratio. $\frac{p_2}{p_1}=1+\frac{2\gamma }{\gamma +1}\left(M_1^2{\sin }^2\beta -1\right)=1+\frac{2.4}{2.2}(2.719)=1+1.0909(2.719)=3.966$ Yeh step kyun? Hum already computed $M_1^2{\sin }^2\beta$ reuse karte hain — normal component compression drive karta hai, same tool as Ex 3.

Step 3 — Separation test. $3.966>2.5\Rightarrow \text{separation likely hai.}$ Yeh step kyun? Ek strong pressure jump boundary layer ko nozzle wall se peel kar sakta hai — dekho Nozzle Flow Separation. Yeh aksar protective hota hai (over-expansion loss limit karta hai) lekin side loads cause karta hai.

Verify: Units sab dimensionless hain (ratios aur angles) ✓. Mach-3 flow se $40^{\circ }$ shock par $24^{\circ }$ turn physically plausible hai ($M=3$ par ~$34^{\circ }$ max deflection se kam), aur $\approx 4\times$ pressure jump correctly separation flag karta hai. Gas Dynamics ko real hardware behaviour se ties karta hai.

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Recall Quick self-test

Kaun sa case thrust add karta hai bina kisi shock loss ke? ::: Under-expanded, $p_e>p_a$ (Cell D) — reversible expansion fans use karta hai. Jab $p_e\to p_a$, $\theta$ aur loss ka kya hota hai? ::: Dono zero ho jaate hain — shock ek whisper ban jaata hai (Cell F). Vacuum mein, pressure term kya value leta hai? ::: Us $p_e$ ke liye maximum: $p_e{A}_e$, kyunki $p_a=0$ (Cell E). Mach disk kya hai? ::: Woh chota normal shock jo do lip shocks ke axis par milne se banta hai — diamond pattern ka strongest, most lossy shock.