Over-expanded nozzle — oblique shocks in plume, efficiency loss: Visual walkthrough

3.3.14 · D2 · Physics › Rocket Propulsion › Over-expanded nozzle — oblique shocks in plume, efficiency l

Yeh page parent result ko — the over-expanded nozzle story — bilkul scratch se rebuild karta hai. Koi bhi pehla symbol assume nahi kiya gaya. End tak aap thrust equation ka har term draw kar chuke honge aur shocks ko form hote dekh chuke honge.

Hum sirf yeh ideas building blocks ke roop mein use karte hain (har ek define kiya jaata hai jaise hi woh aata hai): pressure, ek supersonic jet, aur ek simple triangle. Baaki sab yahan build karte hain.

Step 1 — "Pressure pushing on a surface" ka matlab kya hai?

WHAT. Rockets ki baat karne se pehle, hum woh ek fact build karte hain jis par sab kuch tika hai: ek gas kisi bhi wall ko dabata hai jo use touch karti hai, aur woh push ek force hai.

WHY. Poori efficiency loss ek push ka doosre par jeerna hai. Agar aap pressure ko ek force ke roop mein dekh nahi sakte, to thrust equation ka koi bhi hissa real nahi lagega.

PICTURE. Figure mein, area $A$ (square metres mein measure kiya gaya, $m^{2}$ ) ki ek flat plate gas mein pressure $p$ par baithi hai (pascals mein measure kiya gaya, $Pa$ — yani newtons per square metre). Amber arrows on the left notice karo: har ek gas ka plate par hammering hai, sab ek hi taraf point kar rahe hain. Ab single white arrow on the right follow karo — woh tab milta hai jab aap har ek chhote hammer-blow ko add karte hain: ek clean total force.

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

Step 2 — Thrust kahan se aata hai: do alag alag pushes

WHAT. Hum ek rocket ke total thrust ko do clean pieces mein split karte hain aur har ek draw karte hain.

WHY. Parent ke master formula mein do terms hain. Beginners inhe ek saath blur kar dete hain. Hum inhe alag rakhte hain taaki baad mein loss ka ek unambiguous ghar ho.

PICTURE. Figure nozzle ka circular exit plane (white ellipse) dikhata hai, area $A_{e}$ (chhota $e$ matlab "exit par"). Do arrows trace karo jo isse nikl rahe hain: fat cyan arrow left point karta hai woh mass hai jo peeche flung kiya ja raha hai — momentum push; thin amber arrow iske thoda neeche exit disk par gas pressure hai — pressure push. Yeh dono yahan ek hi taraf point karte hain, lekin Step 3 mein amber wala flip ho sakta hai.

Yeh full Thrust Equation se link karta hai.

Step 3 — Pressure term ke teen signs (har case cover karo)

WHAT. $p_{e} - p_{a}$ ke sign mein exactly teen possibilities hain. Hum teeno draw karte hain.

WHY. Contract ki demand hai ki har case ho. "Over-expanded" ek specific sign hai, aur aap ise tab hi samajhte hain jab apne do siblings ko dekho.

PICTURE. Teen mini-panels. Har ek mein, cyan bar ( $p_{e}$ , exhaust) aur amber bar ( $p_{a}$ , atmosphere) compare karo. Left panel: cyan taller hai → push helps. Middle: equal hain → koi push nahi. Right panel (hamaara villain): amber cyan ke upar tower karta hai → atmosphere jeetta hai aur push reverse ho jaata hai.

Case	$p_{e} - p_{a}$ ka Sign	Pressure push	Naam
$p_{e} > p_{a}$	$+$	thrust help karta hai	under-expanded
$p_{e} = p_{a}$	$0$	zero — pure momentum	matched (ideal)
$p_{e} < p_{a}$	$-$	thrust ko hurt karta hai	over-expanded

Step 4 — Atmosphere sirf gently squeeze kyun nahi karta: supersonic flow

WHAT. Exhaust sound se faster nikalta hai. Hum dikhate hain ki yeh kya forbid karta hai.

WHY. Agar flow slow hota, atmosphere ise smoothly compress kar sakta aur koi shocks, koi loss nahi hota. Poora drama exist karta hai kyunki flow supersonic hai.

PICTURE. Cyan arrow follow karo — speed $v_{e}$ par gas flow, lamba draw kiya gaya kyunki fast hai. Ab neeche wala chhota amber arrow ek sound signal hai jo speed $a$ par travel kar raha hai, yeh message upstream (left side) carry karne ki koshish kar raha hai ki "yahan bahar pressure bahut zyada hai!" Yeh nahi kar sakta: cyan arrow ise overtake kar leta hai. Message nozzle tak smoothly nahi pahunch sakta, isliye flow ko hard way se pata karna padta hai.

Ek over-expanded rocket ke exit par, $M_{e} > 1$ hota hai (aksar $4$ ke aas paas) — supersonic. Iske baare mein aur Gas Dynamics aur Isentropic Flow mein.

Step 5 — Shock oblique kyun hai, head-on kyun nahi

WHAT. Compression sides se aata hai (atmosphere plume ke edges ko squeeze kar raha hai), isliye shock ek slant par baithta hai.

WHY. Ek head-on normal shock (Step 4 mein defined) sabse violent, sabse wasteful compression hogi. Nature sabse sasti legal option use karta hai — ek tilted (oblique) shock.

PICTURE. Figure mein slant karta hua amber shock line dhundho — woh shock hai, angle $β$ par tilted incoming flow se. Cyan arrow incoming flow $M_{1}$ hai (Step 4 ka upstream Mach number). Dekho kaise do white arrows us cyan arrow ko ek right triangle mein split karte hain: ek white arrow shock line ke seedha across point karta hai ( $M_{1 n}$ ), doosra iske along slide karta hai ( $M_{1 t}$ ). Sirf across-piece squeeze hota hai.

Step 6 — Turn: flow $θ$ se inward deflect hota hai, aur jab koi shock attach nahi ho sakta

WHAT. Oblique shock se guzarne ke baad, flow centreline ki taraf angle $θ$ se bend karta hai. Hum yeh bhi dikhate hain ki har $θ$ ke liye usually do possible shock angles $β$ hote hain, aur ek maximum $θ$ jiske baad koi attached oblique shock exist nahi karta.

WHY. Plume boundary ko squeeze fit karne ke liye andar fold karna padta hai — yahi aapke thrust ko axis se tilt karta hai. Aur "two-branch / maximum-angle" behaviour exactly decide karta hai ki aapko ek clean oblique shock milegi ya ek detached, aur bhi zyada wasteful, normal-like shock.

PICTURE (left). Streamline shock par kink karta hai: horizontal enter karta hai, $θ$ se neeche tilted nikalti hai. PICTURE (right). Fixed $M_{1}$ ke liye $θ$ – $β$ – $M$ curve: $θ$ bottom se read karo, $β$ side se upar. Peak se neeche kisi bhi $θ$ ke liye curve do $β$ values par hit hoti hai — chhota wala weak shock hai (small $β$ , flow supersonic rehta hai, jo plume actually pick karta hai), bada wala strong shock hai (large $β$ , flow subsonic ho jaata hai). Peak par $θ_{m a x}$ baitha hai; usse steeper turn maango aur curve ka koi solution nahi — shock detach ho jaata hai aur ek curved, normal-like front ke roop mein bow out ho jaata hai.

Step 7 — Asli cost: stagnation pressure girta hai, aur wahi lost exit velocity hai

WHAT. Har shock kuch ordered kinetic energy ko random heat mein convert karta hai. Hum ise stagnation pressure $p_{0}$ mein drop se measure karte hain, exact $p_{02} / p_{01}$ formula likhte hain, aur precisely dikhate hain ki yeh exhaust speed ko kaise cap karta hai.

WHY. Lost stagnation pressure exhaust speed banane ki lost ability hai. Yahi woh number hai jo finally jawab deta hai "humne kitna thrust pheka?"

PICTURE. Shock se pehle, $p_{01}$ ka ek tall cyan bar (usable "push potential"); baad mein, ek chhota amber bar $p_{02}$ . Dotted white line unke tops ko connect karta hai taaki aap dekh sako missing slice — arrow uski taraf point karta hai: woh slice forever gone hai, disorder mein turn ho gayi.

Definition Velocity coefficient — har symbol

$C_{v} = \frac{v _{e, actual}}{v _{e, ideal}} = \frac{p _{0, actual}}{p _{0, chamber}}$

$v_{e, actual}$ — woh exhaust speed jo aap actually lete hain shocks ke kuch $p_{0}$ khaane ke baad ( $m/s$ ).
$v_{e, ideal}$ — woh exhaust speed jo aap lete ek loss-free (isentropic) expansion se, koi shocks nahi ( $m/s$ ).
$p_{0, chamber}$ — chamber stagnation pressure: full push-potential jo gas ne combustion chamber ke andar deep se start ki (yeh pehle shock ka $p_{01}$ hai).
$p_{0, actual}$ — woh push-potential jo exhaust tak survive karti hai har shock ke apna cut lene ke baad.
$C_{v}$ — velocity coefficient: ideal exhaust speed ka woh fraction jo aap rakhte hain.

Square root kyun: speed $\propto energy \propto p_{0}$ , upar ke velocity formula se. Yeh model assume karta hai ki shock $p_{0}$ -loss akela spoiler hai (friction aur heat leakage neglect kiya — standard first-order).

Step 8 — Shocks repeat karte hain: diamond pattern (edge case: over-correction)

WHAT. Ek shock pressure match ko overshoot ya undershoot karta hai, isliye plume phir se expand hota hai, phir re-compress hota hai — baar baar. Isse visible shock diamonds bante hain.

WHY. Steps 5–7 ka single shock end nahi hai; losses compound hote hain. Contract chahta hai ki yeh limiting/degenerate behaviour dikhayi de.

PICTURE. Plume ko left se right trace karo. Cyan envelope plume boundary hai; amber diamonds crossing oblique shocks hain. Pehle crossing par, axis par chhota white vertical line dhundho — woh Mach disk hai (ek normal shock, Step 4 mein defined). Uske baad flow over-compress hota hai, phir $p_{a}$ se neeche fan back out karta hai, aur cycle repeat hota hai. Diamonds ko right ki taraf shrink hote dekho — har ek weaker hai kyunki $p_{0}$ (aur $M_{1}$ ) gir gaya hai.

Worked example — RL-10 sea level par fire kiya gaya (har number checked)

Pressure thrust loss. $Δ F = (p_{e} - p_{a}) A_{e} = (5.5 - 101) \times 1 0^{3} \times 0.5 = - 47.75 kN$ Ek retarding $47.75 kN$ — atmosphere Step 1 ka tug-of-war jeet raha hai.

Shock par normal Mach (Step 5). Pehle shock ke liye $M_{1} = M_{e} = 4.2$ , isliye $M_{1 n} = 4.2 sin 2 5^{\circ} \approx 1.775$

Pehle shock ka stagnation loss (Step 7). $\frac{p _{02}}{p _{01}} \approx 0.84$

Teen diamonds ke baad (Step 8). $(0.84)^{3} \approx 0.60 \Rightarrow C_{v} = 0.60 \approx 0.77$

Aap ideal exhaust speed ka sirf lagbhag $77%$ rakhte hain. Terrible — exactly parent ki warning.

Ek-picture summary

Yeh top par promised composite hai: ek blueprint mein poora over-expanded plume. Left par nozzle se start karo — amber inward arrows atmosphere ( $p_{a} > p_{e}$ ) hai jo plume edges ko squeeze kar raha hai. Woh edges slanted amber edge shocks angle $β$ par throw off karti hain. Shocks axis par chhote white vertical Mach disk par cross karte hain, aur cyan envelope phir fading diamonds ke roop mein repeat hota hai. Left-to-right read karo aur aapne har step replay kar liya: squeeze → oblique shock → sirf $M_{1 n} = M_{1} sin β$ compress karta hai → $θ$ se turn → lost $p_{0}$ → chhota $C_{v}$ aur chhota thrust.

Recall Feynman retelling — plain words mein wapas bolo

Ek rocket do tarike se push karta hai: gas ko peeche phenk kar, aur gas ke exit disk par press karne se. Agar nozzle over-expand karta hai, uska exhaust surrounding air se lower pressure par nikalta hai, isliye hawa inward zyada harder press karti hai exhaust ke out press karne se — woh mismatch thrust subtract karta hai. Kyunki gas sound se faster fly kar rahi hai, woh smoothly adjust nahi kar sakti; balki woh slanted compression walls mein slam karti hai jise oblique shocks kehte hain. Motion ka sirf woh part jo straight into the shock point karta hai squeeze hota hai, isliye ek tilted shock head-on se cheaper hai — lekin "cheaper" matlab "free" nahi. Kisi bhi required turn ke liye do shock angles hote hain (ek gentle weak aur ek violent strong), aur agar maanga gaya turn flow se handle se zyada steep ho, shock give up kar deta hai aur ek bowed front mein detach ho jaata hai — worst case. Har shock same teen bookkeeping laws follow karta hai — conserve mass, momentum, aur energy — aur woh laws dono ek pressure jump aur disorder mein ek unavoidable rise force karte hain, jo lost push-potential (stagnation pressure) ke roop mein appear hota hai. Kyunki exhaust speed us surviving push-potential ke square root ki tarah badhti hai, lost $p_{0}$ directly lost speed hai. Pattern plume ke neeche glowing diamonds ke roop mein repeat hota hai, aur losses multiply hote hain, isliye sea level par test kiya gaya ek vacuum engine pressure penalty ke upar apni exhaust speed ka fifth part throw away kar sakta hai.

Recall Quick self-test

Normal shock aur oblique shock mein kya difference hai? ::: Ek normal shock flow par perpendicular baithta hai (head-on, maximum loss); ek oblique shock ek slant par baithta hai isliye motion ka sirf ek part ise head-on cross karta hai (gentler). $M$ , $M_{e}$ aur $M_{1}$ mein se har ek ka kya matlab hai? ::: $M = v / a$ generic Mach number hai; $M_{e}$ nozzle exit par uski value hai; $M_{1}$ ek given shock ke just upstream uski value hai (pehle shock ke liye $M_{e}$ ke equal). Fixed $M_{1}$ aur turn $θ$ ke liye, relation ko kitne shock angles $β$ solve karte hain? ::: Usually do — ek weak branch (small $β$ , supersonic rehta hai) aur ek strong branch (large $β$ , subsonic ho jaata hai); agar $θ > θ_{m a x}$ , koi nahi — shock detach ho jaata hai. Shock relations produce karne wale teen conservation laws kaun se hain? ::: Thin shock sheet ke across mass, momentum aur energy ka conservation. Sirf $M_{1 n} = M_{1} sin β$ shock strength kyun set karta hai? ::: Sirf motion jo shock par perpendicular hai woh compress hota hai; parallel motion untouched slide karta hai. Over-expanded nozzle ke liye $(p_{e} - p_{a})$ ka sign kya hai, aur thrust par kya karta hai? ::: Negative — yeh thrust se subtract karta hai. Lost stagnation pressure lost thrust kyun hai, aur $C_{v}$ mein square root kyun? ::: Entropy mein rise $p_{0}$ mein drop ke roop mein appear hota hai; kyunki kinetic energy $\propto v^{2}$ hai, speed $\propto p_{0}$ hai, isliye $C_{v} = p_{0, actual} / p_{0, chamber}$ .