Converging-diverging (de Laval) nozzle — subsonic, supersonic flow: Visual walkthrough

3.1.10 · D2 · Physics › Compressible Flow & Aerodynamics › Converging-diverging (de Laval) nozzle — subsonic, supersoni

Hum sirf arithmetic aur yeh idea assume karte hain ki ek picture ko measure kiya ja sakta hai. Har naya symbol ko plain-word meaning milta hai aur drawing mein ek jagah milti hai, use use karne se pehle.

Step 0 — Woh picture jo hum describe kar rahe hain

Kisi bhi algebra se pehle, dekho hum kis cheez ki baat kar rahe hain.

Figure — Converging-diverging (de Laval) nozzle — subsonic, supersonic flow

Hum gas ko follow karte hain jab woh ek fat slice se ek thin slice ki taraf jaati hai. Teen cheezein badal sakti hain: width $A$ , speed $V$ , packing $ρ$ . Poori derivation bas yeh bookkeeping hai ki yeh teeno kaise trade off karti hain.

Step 1 — Conservation of mass: woh quantity jo kabhi nahi badlti

KYA. Steady flow mein, jo bhi mass fat end mein har second enter karta hai wahi thin end se har second nikalna chahiye — na gas pile up hoti hai, na gayab hoti hai. Kisi bhi slice se har second guzarne wale mass ko mass flow rate kehte hain, likha jaata hai $\overset{m}{˙}$ (dot ka matlab hai "per second").

$\overset{m}{˙} = kg per m^{3} ρ \times m^{2} A \times m / s V = constant$

YEH MULTIPLY KYUN KARTA HAI. Ek second mein gas $V$ length (metres travelled) aur cross-section $A$ ka ek plug sweep out karti hai. Us plug ka volume $A \cdot V$ cubic metres hai, aur har cubic metre mein $ρ$ kilograms hain — toh mass $ρ A V$ hai. Yeh continuity hai.

PICTURE. Neeche ke do plugs dekho: fat plug short-and-wide hai, thin plug long-and-thin hai, lekin dono mein same number of pink dots hain (same mass). Yahi " $\overset{m}{˙}$ constant" ka poora matlab hai.

Step 2 — Gas slug ke liye Newton's law (Euler equation)

KYA. Ab poochho: gas speed up kyun hoti hai? Ek force. Frictionless gas ke liye sirf ek hi push hai — ek slug ke across pressure difference. Pehle pressure ko hi name karna padega.

PRESSURE KYUN. Agar peeche ka pressure aage ke pressure se zyada hai, toh ek net forward shove hai aur slug accelerate karti hai. Use ek equation mein badalne ke liye hum ek thin gas slug ke liye carefully Newton's second law, $force = mass \times acceleration$ , likhte hain.

Formula

$F = ma$ ka har piece banana Face area $A$ aur tiny length $d x$ ki ek thin slug lo.

Slug ka mass. Uska volume $A d x$ hai aur har cubic metre mein $ρ$ kilograms hain, toh uska mass $ρ A d x$ hai.
Slug par net force. Back face ko $p A$ se forward push kiya jaata hai; front face ko $(p + d p) A$ se backward push kiya jaata hai. Unhe add karne par, net force $p A - (p + d p) A = - d p A$ hai.
Slug ka acceleration. Yahan ek key kinematic move hai. Steady flow mein speed position par depend karti hai, $V (x)$ , directly time par nahi. Jab slug $d x$ aage move karti hai toh uski speed $d V$ se change hoti hai, aur woh $d x$ ko time $d t = d x / V$ mein cover karti hai. Toh uska acceleration hai $a_{slug} = \frac{d V}{d t} = \frac{d V}{d x / V} = V \frac{d V}{d x} .$ Yahi reason hai ki acceleration $V d V / d x$ kyun nikalta hai aur simply $d V / d t$ nahi: steady flow mein ek slug speed up hoti hai ek tez region mein move karke, toh uska acceleration speed times speed-gradient hai.

Ab $F = ma$ assemble karo: $net force - d p A = mass ρ A d x \times acceleration V \frac{d V}{d x}$ Dono sides par $A$ cancel ho jaata hai aur acceleration ke andar $d x$ cancel ho jaata hai, bacha: $- d p = ρ V d V$

PICTURE. Neeche ki slug ke back face par pressure $p$ hai aur front par $p + d p$ . Agar $d p$ negative hai (pressure aage girti hai), toh back push jeet jaata hai aur blue arrow lamba hota hai — gas accelerate karti hai.

Step 3 — Speed of sound enter karta hai (aur kyun karna hi padega)

KYA. Hamare paas ab do equations hain, lekin teen unknowns change ho rahe hain ( $ρ$ , $A$ , $V$ , $p$ — actually chaar). Humein ek aur link chahiye. Yeh is sawaal se aata hai: is gas mein pressure changes aur density changes ka kya relation hai?

YEH EXACT TOOL KYUN. Ek small pressure ripple gas mein travel karti hai woh ek sound wave hai. Us ripple ki speed, speed of sound $a$ , defined hai exactly is ratio se ki "pressure change per unit density change kitna hai":

$a^2 \;=\; \left.\frac{dp}{d\rho}\right|_{\text{isentropic}}\qquad\Longrightarrow\qquad d\rho = \frac{dp}{a^2}\tag{3}$

PICTURE. Figure mein ek squeeze dikhaya gaya hai: gas push karo (p badhao) aur woh tighter pack hoti hai ( $ρ$ badhao). Ek stiff gas (bada $a$ ) barely compress hoti hai; ek soft gas (chhota $a$ ) bahut compress hoti hai. Number $a^{2}$ us squeeze curve ki slope hai, no-heat-loss (isentropic) path ke along measure ki gayi.

Step 4 — Teenon equations ko ek saath fold karo

KYA. Hum (2) aur (3) ko combine karte hain taaki density change ko speed change ke terms mein likhein, phir mass balance (1) mein substitute karein.

Step 4a — (2) ko (3) mein daalo. (2) se, $d p = - ρ V d V$ . Use (3) mein daalo: $d ρ = \frac{- ρ V d V}{a ^{2}} ⟹ \frac{d ρ}{ρ} = - \frac{V ^{2}}{a ^{2}} \frac{d V}{V}$

Step 4b — ratio ko naam do. Clump $\frac{V ^{2}}{a ^{2}}$ hai (speed / sound-speed) squared. Us ratio ka ek naam hai:

Step 4c — mass balance (1) mein substitute karo. $from d ρ / ρ - M^{2} \frac{d V}{V} + \frac{d A}{A} + \frac{d V}{V} = 0$

Do $\frac{d V}{V}$ terms group karo: $\frac{d A}{A} + (1 - M^{2}) \frac{d V}{V} = 0$

Cross karke move karo aur sign flip karo:

PICTURE. Neeche ki figure $(M^{2} - 1)$ multiplier ko $M$ ke against plot karti hai. Dhyan do yeh zero exactly $M = 1$ par cross karta hai. Left mein yeh negative hai (area aur speed opposite ways move karti hain); right mein yeh positive hai (area aur speed same way move karti hain). Woh single sign flip nozzle ka poora raaz hai.

Step 5 — Case A: subsonic gas ( $M < 1$ )

KYA. Jab $M < 1$ , factor $(M^{2} - 1)$ negative hota hai. Gas accelerate karne ke liye hum $d V > 0$ chahte hain, toh $\frac{d V}{V} > 0$ . Tab: $\frac{d A}{A} = negative (M^{2} - 1) \cdot positive \frac{d V}{V} < 0$ toh $d A < 0$ : area shrink honi chahiye. Speed up karne ke liye squeeze karo.

YEH EVERYDAY LIFE SE KYUN MATCH KARTA HAI. Slow flow ke liye, density barely change hoti hai (gas almost incompressible act karti hai, paani jaisi). Toh $ρ A V$ constant rakhne ke liye: $A$ shrink karo, aur $V$ gap fill karne ke liye rise karna chahiye. Yeh garden-hose intuition hai, aur yeh sahi hai — lekin sirf Mach 1 se neeche.

PICTURE. Converging pipe, blue arrow walls ke close aane ke saath lamba hota ja raha hai.

Step 6 — Case B: supersonic gas ( $M > 1$ )

KYA. Jab $M > 1$ , factor $(M^{2} - 1)$ positive hota hai. Accelerate karne ke liye ( $\frac{d V}{V} > 0$ ) ab humein chahiye: $\frac{d A}{A} = positive (M^{2} - 1) \cdot positive \frac{d V}{V} > 0$ toh $d A > 0$ : area badhni chahiye. Speed up karne ke liye widen karo. Everyday intuition ka ulta.

KYUN. Step 4b dekho: $\frac{d ρ}{ρ} = - M^{2} \frac{d V}{V}$ . $M > 1$ ke saath density speed se tez plunge karti hai. Mass balance $ρ A V = const$ mein, agar $ρ$ crash kare toh $A V$ climb karna chahiye — aur yeh nikalta hai ki dono $A$ aur $V$ saath badhte hain. Thinning gas, widening pipe se zyada thin hoti hai, aur speed baki ka hissa pura karti hai.

PICTURE. Diverging pipe: walls flare out karti hain, pink dot-cloud dramatically thin hoti hai, aur blue arrow sab se lamba hota hai.

Step 7 — Degenerate case: exactly sonic ( $M = 1$ )

KYA. $M = 1$ par factor $(M^{2} - 1) = 0$ exactly, toh law ban jaata hai: $\frac{d A}{A} = 0 \cdot \frac{d V}{V} = 0 ⟹ d A = 0$

YEH THROAT KYUN PIN KARTA HAI. $d A = 0$ ka matlab hai area na badh rahi hai na ghut rahi — yeh ek turning point par hai, ek local minimum. Woh location throat hai. Isliye gas exactly sound ki speed se sirf throat par hi hit kar sakti hai, kabhi bhi converging ya diverging run ke beech mein nahi. Yahi reason hai ki ek de Laval nozzle ka ek throat hona zaroori hai: yeh subsonic aur supersonic worlds ke beech ka ek maatra darwaaza hai.

PICTURE. Hourglass throat: left par converging, pinch jahan $M = 1$ aur $d A = 0$ , right par diverging.

Ek-picture summary

Upar sab kuch ek single hourglass mein compress ho jaata hai jisme $(M^{2} - 1)$ ka sign har region ko colour karta hai. Summary figure ko left se right padho: gas left par blue converging cone mein slow enter karti hai (wahan $M < 1$ hai, aur factor $(M^{2} - 1)$ negative hai, toh tube ko shrink karne se woh speed up hoti hai); woh yellow pinch par beech mein exactly sound ki speed tak pahuncha hai (wahan $M = 1$ hai aur factor zero hai, toh area minimum par honi chahiye — yahi throat hai); aur woh right par pink diverging cone se supersonic bahar nikarti hai (wahan $M > 1$ hai, factor positive hai, toh tube ko widen karne se woh speed up hoti hai). Ek equation teeno regions ko govern karti hai, aur sirf bracket ka sign change hota hai.

Recall Feynman retelling — plain words mein poora walkthrough

Humne ek gas packet ko pipe se slide karte follow kiya. Pehle humne notice kiya ki har second har slice se same amount of gas pass hoti hai — yahi hamara unbreakable rule hai ( $ρ A V$ kabhi nahi badlta). Humne use teen fractional changes ki tarah likha jo zero mein add hone chahiye: width, speed, aur packing. Phir humne poochha ki gas ko faster kya push karta hai: ek pressure drop. Pressure mein "downhill" girna speed khareedta hai. Agle humein jaanna tha ki pressure change hone par gas kitna zyada pack hoti hai — aur woh "stiffness" number literally speed of sound squared hai. Unhe fold karo aur packing change $- M^{2}$ times speed change ke barabar nikalta hai, jahan $M$ bas "speed of sound ka kitna times" hai. Use wapas never-changing rule mein plug karo aur ek clean equation bahar aati hai: $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ . Sound ki speed se neeche bracket negative hai, toh pipe squeeze karne se gas speed up hoti hai (hose trick). Sound ki speed se upar bracket flip ho jaata hai positive, toh speed up karne ke liye tumhe pipe widen karni padegi — kyunki gas pipe ke badhne se tez thin hoti hai. Aur exactly sound ki speed par bracket zero hai, toh pipe wahan width change nahi kar sakti: yahi throat hai, slow se fast ka ek maatra darwaaza. Ek hourglass tube: squeeze in, Mach 1 pinch par hit karo, phir supersonic blast karne ke liye flare out karo. Yahi ek rocket nozzle hai.

Recall Quick self-check

$M = 1$ throat par hi kyun hona chahiye aur kahi nahi? ::: Kyunki $M = 1$ par factor $(M^{2} - 1) = 0$ force karta hai $\frac{d A}{A} = 0$ , yaani area minimum par hai (throat). Subsonic flow mein, gas accelerate karne ke liye area kis direction mein change honi chahiye? ::: Shrink honi chahiye ( $d A < 0$ ), kyunki $(M^{2} - 1) < 0$ hai. Supersonic flow mein, pipe widen karne se gas speed up kyun hoti hai? ::: Kyunki density area se tez girti hai, toh $ρ A V$ constant rakhne ke liye speed badhni padti hai. Kaunsi ek equation momentum, mass, aur sound ki physics ko link karti hai? ::: Area–Velocity law $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ .

Yeh kahan le jaata hai: yahan sign flip choking, rocket aur steam-turbine nozzles ki shapes, aur diverging section mein normal shock kahan stand kar sakti hai, ka reason hai.