Over - under expanded nozzle flows: Visual walkthrough

3.1.18 · D2 · Physics › Compressible Flow & Aerodynamics › Over - under expanded nozzle flows

Neeche har symbol pehle earn kiya jaata hai, phir use kiya jaata hai. Agar tumne kabhi "Mach" word ya letter $γ$ nahi suna, toh line one se padhna shuru karo aur sab theek rahega.

Step 1 — Nozzle kya hota hai, aur hum track kya kar rahe hain?

KYA. Ek nozzle ek pipe hoti hai jiska cross-section area $A$ uski length ke saath-saath badalta rehta hai. Gas iske andar se flow karti hai. Hum gas ke ek patle "slice" ko follow karte hain aur poochte hain: jab area $A$ badalta hai, toh gas ki speed $V$ aur uski pressure $p$ ka kya hota hai?

KYUN. Kisi bhi formula se pehle, hume un chaar chezon par agree karna hai jo hum measure karenge. Pressure $p$ yeh hai ki gas walls par kitna zyada sideways push karti hai (socho: balloon mein hawa). Speed $V$ yeh hai ki slice pipe mein kitni tezi se move karti hai. Density $ρ$ (Greek letter "rho") yeh hai ki molecules kitne tightly packed hain — mass per unit volume. Area $A$ wahaan pipe ki width hai.

PICTURE. Neeche di gayi pipe dekho: yeh ek kamar tak narrow hoti hai aur phir wide ho jaati hai — yeh converging–diverging (de Laval) shape hai. Green slice gas ka humara tracked chunk hai.

Step 2 — Speed of sound, aur number $M$

KYA. Sound ek choti si pressure ripple hoti hai. Yeh gas mein ek fixed speed se travel karti hai jise hum $a$ kehte hain (local speed of sound). Hum ek dimensionless number define karte hain: $M = \frac{V}{a}$ Yahan $M$ ("Mach number") gas ki speed hai apni hi sound speed se compare karke. $M < 1$ matlab sound se slower (subsonic); $M = 1$ matlab exactly sound speed (sonic); $M > 1$ matlab sound se faster (supersonic).

$V$ — gas ki speed jis par hum dhyan de rahe hain.
$a$ — news (ek pressure ripple) is gas ke andar kitni tezi se travel kar sakti hai.
$M$ — ratio; is poore chapter mein sabse important number.

KYUN sirf $V$ ki jagah yeh ratio? Kyunki pressure ki ability ki woh upstream message bhej sake, completely is baat par depend karti hai ki $V$ beats $a$ karta hai ya nahi. "300 m/s" jaisi koi number apne aap kuch nahi batati; " $M = 2$ " batata hai ki gas apni khud ki sound ko outrun kar rahi hai — yahi woh fact hai jo $p_{e}$ ko lock karega. Hume ratio chahiye, isliye hum $M$ banate hain.

PICTURE. Do runners: ek pressure-ripple runner (speed $a$ ) jo left jaane ki koshish kar raha hai, ek walkway par khada hai jo right mein speed $V$ se move kar rahi hai. Jab $V > a$ ( $M > 1$ ) toh ripple backward sweep ho jaati hai — yeh throat tak kabhi nahi pahunch sakti. Yeh "exit is deaf" ka seed hai.

Recall "Downstream se deaf" kyun already follow karta hai

Agar $M > 1$ ::: toh gas itni tezi se move karti hai ki koi bhi pressure signal upstream crawl nahi kar sakta, isliye back pressure $p_{b}$ literally exit tak message nahi bhej sakta. Exit bahar se "kuch nahi sunti."

Step 3 — Conservation of energy → pressure $M$ se tied hai

KYA. Hum gas ko isentropic track karte hain (smooth, koi shocks nahi, koi heat lost nahi). Energy conservation kehti hai ki total energy per unit mass constant rehti hai. Ise ek "stagnation" (subscript $0$ ) quantity ki tarah likho: socho gas ko gently rest par slow karo — jो pressure tab hogi woh stagnation pressure $p_{0}$ hai (yeh chamber pressure hai). Energy + perfect-gas isentropic link $p / ρ^{γ} = const$ dete hain: $\frac{p _{0}}{p} = (1 + \frac{γ - 1}{2} M^{2})^{\frac{γ}{γ - 1}}$

Term by term:

$p_{0}$ — chamber (stagnation) pressure, "full tank" reference, ek fixed constant.
$p$ — local static pressure jo hum chahte hain.
$γ$ ("gamma") — gas ka stiffness ratio (air ke liye, $1.4$ ); yeh batata hai ki squeeze hone par pressure aur density kaise trade off karte hain.
$M^{2}$ — isliye aata hai kyunki kinetic energy speed ke square ke saath scale karti hai; faster flow ⇒ zyada energy motion mein hai ⇒ kam pressure mein hai ⇒ $p_{0}$ ke relative $p$ drop karti hai.
Poora bracket $γ / (γ - 1)$ tak raised — woh exact exponent jo isentropic (constant-entropy) rule demand karta hai.

KYUN yeh tool? Hume ek knob chahiye jo pressure ko $M$ se connect kare. Energy conservation woh knob hai: yeh Isentropic Flow Relations ki master relation hai. Ise aise padho: "mujhe $M$ batao, main bataunga ki $p_{0}$ se $p$ kitna neeche gira hai."

PICTURE. Jaise $M$ left→right climb karta hai, curve $p / p_{0}$ zero ki taraf slide karta jaata hai. High Mach = low pressure. Woh slope yaad karo.

Step 4 — Conservation of mass → throat ek reference set karta hai

KYA. Mass pile up nahi ho sakta: har station par per second se flow hone wali mass same rehti hai, $ρ A V = const$ . Ise isentropic relations ke saath combine karo aur ek special area milta hai: woh area jahan $M = 1$ hota hai, jise throat area $A^{*}$ kehte hain (star = sonic). Baaki har area ko hum $A^{*}$ ke against measure karte hain: $\frac{A}{A ^{*}} = \frac{1}{M} [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M^{2})]^{\frac{γ + 1}{2 ( γ - 1 )}}$

$A$ — interest ke station par area.
$A^{*}$ — throat area, jahan $M = 1$ ; flow wahaan choked hoti hai (dekho Choked Flow & the Throat (M=1)).
Aage $\frac{1}{M}$ + bracket — milkar $A / A^{*}$ ka minimum exactly $M = 1$ par 1 hota hai, dono sides par utha hua.

KYUN yeh tool? Kyunki hum jo geometry actually build kar sakte hain woh area hai, Mach nahi. Mass conservation "shape" ( $A / A^{*}$ ) se "speed" ( $M$ ) tak ka bridge hai. Yeh Area-Mach Number Relation hai.

PICTURE. U-shaped curve: $A / A^{*}$ $M = 1$ par $1$ tak dip karta hai aur dono sides par climb karta hai. Toh ek area ratio curve ko do Mach numbers par touch karta hai — ek subsonic, ek supersonic.

Step 5 — Correct root choose karna (edge case!)

KYA. Ek given $A_{e} / A^{*} > 1$ U-curve ko do points par milta hai: ek subsonic $M < 1$ aur ek supersonic $M > 1$ . Ek supersonic nozzle ke diverging part mein hum supersonic root $M_{e} > 1$ lete hain.

KYUN. Converging flow mein, widening gas ko slow karta hai (subsonic branch). Throat ke past jahan $M = 1$ hai, widening ab gas ko speed up karta hai — supersonic branch. Physics ne throat par humari branch choose ki; hum bas follow karte hain.

Degenerate cases — sabko cover karo:

$A_{e} / A^{*} = 1$ : sirf ek root hai $M = 1$ . Exit exactly sonic hai; bilkul bhi supersonic flow nahi.
$A_{e} / A^{*} < 1$ : impossible — definition se koi area throat se chota nahi hota. Curve kabhi 1 se neeche nahi dip karta.
$A_{e} / A^{*} \to \infty$ (huge bell): $M_{e} \to \infty$ , aur Step 3 se, $p_{e} \to 0$ . Infinite expansion, vacuum-like exit.
Agar throat choked nahi hai (chamber pressure bahut low): $M$ kabhi 1 tak nahi pahunchta, flow fully subsonic rehti hai, aur yeh poori supersonic story apply nahi hoti.

PICTURE. Same U-curve, $A_{e} / A^{*} = 4$ par ek horizontal line, do intersection dots highlighted — hum supersonic (right) wale ko circle karte hain.

Step 6 — Combine karo: geometry se akele exit pressure

KYA. Supersonic $M_{e}$ (Step 5 mein $A_{e} / A^{*}$ se mila) ko Step 3 ke pressure law mein daalo: $p_{e} = \frac{p _{0}}{( 1 + \frac{γ - 1}{2} M _{e}^{2} ) ^{γ / (γ - 1)}}$

$M_{e}$ — sirf $A_{e} / A^{*}$ se aata hai (Step 4–5).
$p_{0}$ — chamber pressure (ek chosen constant).
Result $p_{e}$ — isliye sirf geometry aur $p_{0}$ par depend karta hai. Back pressure $p_{b}$ kahin nahi dikhta.

KYUN yeh punchline hai. Notice karo kya missing hai: $p_{b}$ . Is poori chain mein koi ek line nahi hai jahan bahar ki hawa enter karti ho. Yeh mathematical proof hai ki ek supersonic exit khud ko atmosphere se match nahi kar sakta — mismatch bahar shocks ya fans ki tarah spillout hona hi chahiye.

PICTURE. Logic ka ek flow-chart: shape $\to$ $M_{e}$ $\to$ $p_{e}$ , jismein $p_{b}$ ko ek locked door par knocking karte dikhaya gaya hai.

Step 7 — Consequence: teen regimes

KYA. Ab geometry-locked $p_{e}$ ko atmosphere $p_{b}$ se compare karo:

$p_{e} = p_{b}$ → perfectly expanded, clean parallel jet.
$p_{e} < p_{b}$ → over-expanded, atmosphere Oblique Shocks se squeeze karke andar aati hai (ya agar severe ho toh Mach disk / separation).
$p_{e} > p_{b}$ → under-expanded, jet Prandtl-Meyer Expansion Fans se baahaar ki taraf expand hoti rehti hai.

KYUN. Kyunki $p_{e}$ fixed hai, sirf variable $p_{b}$ hai. Atmosphere ko lower karo (altitude par chadho) aur ek nozzle over- se perfectly- se under-expanded mein slide kar sakti hai — Rocket Nozzle Design & Thrust Optimization ki physics.

PICTURE. Teen jets side by side: shocks andar pinch kar rahe hain (over), clean column (perfect), fans baahaar bulge kar rahe hain (under).

Ek-picture summary

Upar sab kuch ek logic ki arrow mein compress ho jaata hai: geometry → Mach → pressure → regime, bahar ki duniya (atmosphere) pressure calculation se permanently locked out hai.

Recall Feynman: poora walkthrough retell karo

Socho ek water slide ek magic tube mein sealed hai. Humne gas ki ek bucket ko iske andar ride karte dekha. Pehle humne ise measure karna seekha: tube kitni wide hai ( $A$ ), gas kitni tezi se jaati hai ( $V$ ), yeh kitna push karti hai ( $p$ ). Phir humne ek smart number invent kiya, $M$ — gas ki speed divided by iske andar ek shout ki speed se. Badi surprise: jab gas ek shout se tezi ho jaati hai ( $M > 1$ ), toh bahar se koi bhi shout slide ke upar travel nahi kar sakti. Aage, energy bookkeeping ne bataya "faster gas = lower pressure," aur mass bookkeeping ne bataya "tube ka widest-vs-throat ratio speed pick karta hai." Dono ko saath rako aur exit pressure nikal aati hai — poori tube ki shape aur tank pressure se bani. Bahar ki hawa ko kabhi vote nahi mila. Toh jab gas finally bahar nikarti hai aur hawa ko different pressure par paati hai, use mismatch wahan khule mein fix karna padta hai: andar squish ho kar shocks mein agar hawa bahut pushy hai, ya bahar fan hokar agar hawa bahut weak hai. Yeh poori kahaani hai — slide mid-ride mein shape nahi badal sakti, isliye bahar ki duniya adjust karti hai.