Converging nozzle — subsonic flow, Mach 1 at exit: Visual walkthrough

3.1.9 · D2 · Physics › Compressible Flow & Aerodynamics › Converging nozzle — subsonic flow, Mach 1 at exit

Step 1 se pehle humein kuch plain-language ideas chahiye. Baaki sab unhi par build hoga.

Definition Woh words jahan se hum shuru karte hain

==Speed of sound $a$ ==: kitni tezi se ek tiny pressure "tap" (ek whisper) gas mein travel karta hai. Ise samjho top speed ke roop mein jis par ek gas molecule apne agle ko bata sakta hai "kuch upstream badal gaya." Hum maante hain (from Speed of Sound) ki $a = γ R T$ (units: metres per second) — jab gas zyada hot hogi tab zyada.
==Mach number $M$ ==: flow ki apni speed $V$ ko $a$ ki units mein measure kiya jaata hai, yaani $M = V / a$ . $M = 0.5$ matlab "whisper-speed ki aadhi speed par move karna." $M = 1$ matlab "bilkul whisper-speed par move karna." Yahi last wala poori kahaani hai.
==Stagnation state (subscript $0$ )==: gas pipe se pehle bade reservoir mein still baithi hai, pressure $p_{0}$ aur temperature $T_{0}$ ke saath. "Stagnation" = "rest mein laaya gaya." (Dekho Stagnation Properties.)
==Specific heat at constant pressure $c_{p}$ ==: energy ki matra (joules mein) jo ek kilogram gas ko ek kelvin warm karne ke liye chahiye jab uska pressure fixed ho. Units: $J / (kg \cdot K)$ . To $c_{p} T$ simply hai "ek kilogram temperature $T$ par kitni thermal energy carry kar raha hai." Air ke liye, $c_{p} \approx 1005 J / (kg \cdot K)$ .

Symbols $γ$ (ek fixed number, air ke liye $1.4$ , describe karta hai ki gas heat kaise store karti hai) aur $R$ (gas constant, units $J / (kg \cdot K)$ ) sirf known numbers ke labels hain — humein inhe kabhi kholne ki zaroorat nahi.

Definition Game ke rules (assumptions jo hum ek baar banate hain)

Neeche sab kuch ye assume karta hai:

Steady flow — picture time ke saath nahi badlti; jo ek second mein enter hota hai woh barabar ek second mein nikalta hai.
One-dimensional flow — pipe ke har slice par gas ki ek single speed, pressure aur temperature hoti hai (hum values ko pipe ke along track karte hain, across nahi).
No gravity term — gas halki hai aur pipe short hai, isliye height changes (potential energy) thermal aur motion energy ke comparison mein negligible hain.
Adiabatic (walls ke across koi heat nahi jaati) aur reversible (koi friction nahi, koi shocks nahi) — dono milke isentropic.

Ye chaar humein energy conservation (Step 2), mass conservation (Step 1) aur isentropic pressure–temperature link (Step 3) ko clean equalities ke roop mein use karne dete hain.

Step 1 — Corridor draw karo, aur kyun narrow matlab faster

KYA. Ek gas reservoir ek aisi pipe ko feed karta hai jo sirf narrow hoti jaati hai. Steady, one-dimensional flow ke liye mass conservation kehta hai ki same amount of gas har slice mein har second pass karni chahiye: $ρ A V = constant$ jahan $ρ$ density hai (kg per cubic metre), $A$ slice ka cross-sectional area hai (square metres), aur $V$ flow speed hai. Agar $A$ shrink kare aur $ρ$ muskil se badle (sach hai jab flow slow ho), to $V$ zaroor rise karega product fixed rakhne ke liye.

KYUN. Yeh poore chapter ka engine hai: yahi kyun hai ki pipe squeeze karna gas ko speed up karta hai. Lekin yeh humein warn bhi karta hai — jab gas tezi se move karti hai to $ρ$ stable rehna band kar deta hai, aur yahi exactly woh jagah hai jahan se $M = 1$ ka ceiling aata hai. To hum resulting speed plot karte hain aur ceiling dhundenge.

PICTURE.

Figure — Converging nozzle — subsonic flow, Mach 1 at exit

Blue curve left-to-right climb karti hai (narrow ⇒ faster, seedha $ρ A V =$ const se). Red dashed line $M = 1$ hai, whisper-speed. Notice karo ki curve red line ko exit par touch kar sakti hai lekin inside rehte hue kabhi usse upar nahi ja sakti. Woh "touch but never cross" hi "choked" ka matlab hai, aur Steps 2–6 prove karte hain ki aisa hona kyun zaroori hai.

Step 2 — Energy gayab nahi ho sakti: stagnation-temperature relation

KYA. Gas ke ek packet ko reservoir se fast stream mein follow karo. Koi cheez bahar se use heat nahi karti (adiabatic), koi paddle-wheel us par kaam nahi karta, aur height changes ignore kiye jaate hain. To sum "stored thermal energy + motion ki energy" har jagah same hai:

$stored (thermal) c_{p} T + of motion \frac{1}{2} V^{2} = all stored, at rest c_{p} T_{0}$

Term by term: $c_{p} T$ temperature $T$ par per kilogram thermal energy hai (upar define ki gayi); $\frac{1}{2} V^{2}$ speed $V$ par per kilogram kinetic energy hai; $c_{p} T_{0}$ reservoir value hai jahan $V = 0$ to saari energy thermal hai.

KYUN. Yeh sirf "energy conserved hai" hai, lekin ek moving gas ke liye likha gaya. Yeh humein batata hai ki gas speed ke liye cool down hoke pay karti hai — motion energy thermal energy se kharidi jaati hai. Woh trade hi speed limit ka seed hai.

PICTURE.

Green thermal bar shrink karti hai aur yellow motion bar grow karta hai jaise hum pipe ke neeche jaate hain, lekin dono bars hamesha same total height tak stack hote hain (dashed line). Woh fixed ceiling $c_{p} T_{0}$ hai.

Ab key algebraic move — aur kyun hum ise choose karte hain. Equation mein teen quantities hain jo pipe ke neeche change hoti hain ( $T$ , $V$ , aur implicitly $a$ ). Hum ek single clean variable chahte hain jo pehle se sound-speed ceiling ke baare mein "jaanta" ho, aur woh variable hai $M = V / a$ . To hum $M$ ko equation mein engineer karte hain:

Har term ko $c_{p} T$ se divide karo. Isse left side dimensionless ho jaati hai aur reservoir term ratio $T_{0} / T$ ban jaata hai — bilkul wahi cheez jo hum dhundh rahe hain.
Motion term rewrite karo. Divide karne ke baad, yeh $\frac{V ^{2}}{2 c _{p} T}$ read karta hai. Substitute karo $V^{2} = M^{2} a^{2}$ aur $a^{2} = γ R T$ , to yeh $\frac{M ^{2} γ R T}{2 c _{p} T}$ ban jaata hai — aur $T$ cancel ho jaata hai.
$c_{p} = \frac{γ R}{γ - 1}$ substitute karo. Ab $γ R$ bhi cancel ho jaata hai, sirf pure number $\frac{γ - 1}{2} M^{2}$ bachta hai.

Steps 1–3 ka point: $T$ , $R$ aur $c_{p}$ ki har appearance deliberately cancel ho jaati hai, $M$ ko sole survivor ke roop mein isolate karti hai. Isliye yeh particular substitution aur koi nahi.

Step 3 — Isentropy temperature ko pressure ke liye trade karti hai

KYA. Flow bhi reversible hai (koi friction nahi, koi shocks nahi). Ek reversible adiabatic (isentropic) perfect gas ke liye, temperature aur pressure ek fixed rule se saath move karte hain:

$\frac{T _{0}}{T} = (\frac{p _{0}}{p})^{\frac{γ - 1}{γ}}$

Yahan $p_{0} / p$ yeh hai ki reservoir pressure local pressure se kitna zyada hai, aur exponent $\frac{γ - 1}{γ}$ (air ke liye, $0.286$ ) sirf gas ka fixed "conversion rate" hai temperature ratio aur pressure ratio ke beech.

KYUN. Gauges pressure padhte hain, temperature nahi. Step 2 ne humein temperature $M$ ke terms mein diya; yeh step humein sab kuch us quantity mein restate karne deta hai jo hum actually exit par measure aur control kar sakte hain — pressure.

PICTURE.

Curve dikhata hai: jaise gas cool hoti hai (right move karti hai), uska pressure aur bhi tezi se drop karta hai, kyunki exponent relationship ko bend karta hai. Axes se padhna ek temperature drop ko pressure drop mein convert karta hai.

Ab combine karo. Step 2 deta hai $T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}$ ; Step 3 deta hai $T_{0} / T = (p_{0} / p)^{(γ - 1) / γ}$ . Dono right-hand sides equal set karne par: $(\frac{p _{0}}{p})^{\frac{γ - 1}{γ}} = 1 + \frac{γ - 1}{2} M^{2}$ $p_{0} / p$ ko uske fractional exponent $\frac{γ - 1}{γ}$ se free karne ke liye, inverse power $\frac{γ}{γ - 1}$ dono sides par apply karo — dono exponents left par multiply hokar $1$ dete hain, $p_{0} / p$ bare reh jaata hai, jabki right side us exponent ko inherit kar leti hai:

Step 4 — Dial $M = 1$ par ghुमao: number 0.528 appear hota hai

KYA. Hum exact pressure ratio chahte hain jis par exit just whisper-speed tak pahunchti hai. To master relation mein $M = 1$ set karo:

$\frac{p _{0}}{p ^{*}} = (1 + \frac{γ - 1}{2})^{\frac{γ}{γ - 1}} = (\frac{γ + 1}{2})^{\frac{γ}{γ - 1}} \Rightarrow \frac{p ^{*}}{p _{0}} = (\frac{2}{γ + 1})^{\frac{γ}{γ - 1}}$

Starred $p^{*}$ "critical" exit pressure hai — woh pressure jo gas ke paas us waqt hoti hai jab woh $M = 1$ hit karta hai.

KYUN. $M = 1$ Step 1 ka ceiling hai. Ise plug in karne se humein woh ek pressure ratio milta hai jis par nozzle choke hone par lock ho jaata hai. Air ke liye ( $γ = 1.4$ ):

$\frac{p ^{*}}{p _{0}} = (\frac{2}{2.4})^{3.5} \approx 0.528$

PICTURE.

Master curve $M \mapsto p / p_{0}$ ek baar draw ki gayi hai. Hum $M = 1$ par ek vertical line daalte hain; jahan woh curve se milti hai woh height $0.528$ hai. Us line ke left ka sab kuch subsonic hai (exit back pressure feel karta hai); line par wala point locked, choked state hai.

Step 5 — Back pressure kam karna kyun help karna band kar deta hai: flow rate $M = 1$ par peak karti hai

KYA. Mass flow $\overset{m}{˙} = ρ A V$ hai — density × area × speed. Ise reservoir terms mein rewrite karna (density aur temperature ke liye Steps 2–3 use karke) deta hai:

$\overset{m}{˙} = A p_{0} \frac{γ}{R T _{0}} \cdot ise f (M) kahte hain \frac{M}{( 1 + \frac{γ - 1}{2} M ^{2} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}}$

$f (M)$ ke aage sab kuch reservoir aur fixed exit area se fixed hai. To $\overset{m}{˙}$ versus $M$ ki shape $f (M)$ ki shape hai.

KYUN. Parent claim tha " $\overset{m}{˙}$ ek ceiling hit karta hai." Yahi exactly claim hai " $f (M)$ ka ek maximum hai." Kahan? $f^{'} (M) = 0$ lo — ek hump ke top par zero slope hota hai — aur woh exactly $M = 1$ par land karta hai.

PICTURE.

$f (M)$ $0$ se rise karta hai, arch over karta hai, aur uska peak exactly $M = 1$ line par baithta hai. $M = 1$ ke baad curve fall karega — lekin ek converging pipe anyway us region tak nahi pahunch sakti (Step 6). To reachable maximum flow peak hai, $M = 1$ par. Yeh choking ka mathematical chehra hai: jab aap peak par ho, back pressure neeche kheenchna aapko aur upar nahi lift kar sakta. (Zyada detail mein Mass Flow Rate & Choking.)

Step 6 — Har case, degenerate ones samet

KYA AUR KYUN. Hume sure hona hai ki koi scenario skip nahi hua. Back pressure $p_{b}$ ko "reservoir ke barabar" se "vacuum" tak walk karo.

Figure exit Mach aur $\overset{m}{˙}$ ko $p_{b} / p_{0}$ ke against $1$ se $0$ tak plot karta hai.

$p_{b} / p_{0} = 1$ (degenerate: koi push nahi). Koi pressure difference nahi, to $V = 0$ , $M = 0$ , $\overset{m}{˙} = 0$ . Dono curves ka flat left end.
$0.528 < p_{b} / p_{0} < 1$ (subsonic branch). Exit unchoked: $p_{e} = p_{b}$ , aur $M$ master relation invert karke aata hai. Jaise $p_{b}$ girta hai, $M$ badhta hai aur $\overset{m}{˙}$ badhta hai — "intuitive" straw-sucking region.
$p_{b} / p_{0} = 0.528$ (critical point). Exit just $M = 1$ reach karta hai. Dono curves apne corner tak pahunchti hain.
$0 \leq p_{b} / p_{0} < 0.528$ (choked branch). Exit frozen $M = 1$ par, $p_{e}$ frozen $0.528 p_{0}$ par, $\overset{m}{˙}$ frozen apne peak par. $0.528 p_{0}$ se $p_{b}$ tak ka extra pressure drop nozzle ke bahar expansion waves ke roop mein hota hai. Flat right ends.
$p_{b} / p_{0} = 0$ (degenerate: perfect vacuum). Exit par exactly $M = 1$ abhi bhi — nozzle "critical se thoda neeche" aur "total vacuum" mein fark nahi bata sakta, kyunki signals sonic flow ke against upstream swim nahi kar sakte.

Ek-picture summary

Poori derivation ko ek clean frame mein kaise padhen: ek curve, teen labelled milestones.

Blue curve master relation $p / p_{0}$ versus exit Mach number $M$ hai (Steps 2–4).
Green stretch ( $M < 1$ ) subsonic branch hai, jahan exit back pressure obey karta hai ( $p_{e} = p_{b}$ ).
Yellow dot $M = 1$ par critical point hai: height $0.528$ — air ke liye choking pressure ratio.
Red dashed line sonic ceiling mark karta hai jise flow converging pipe ke andar touch to kar sakta hai lekin cross nahi.

Curve ko left-to-right trace karo: energy conservation aur isentropy ise build karte hain (Steps 2–3), aur woh single dot jahan yeh $M = 1$ se milti hai poori punchline hai — air $0.528$ par choke karti hai, Mach 1 par locked.

Recall Feynman: poora walkthrough plain words mein

Ek bade tank se still gas ko ek aisi pipe mein push karo jo narrow hoti jaati hai. Kyunki same gas ko har second har slice mein pass karna hai, pipe tight karna gas ko hurry karne par majboor karta hai (Step 1). Jaise woh hurry karti hai, woh apni warmth speed ke liye trade karti hai — accelerate karte waqt cool hoti hai (Step 2). Kyunki flow clean aur reversible hai, woh cooling ek matching pressure drop ke saath aati hai, aur hum ek clean rule likh sakte hain jo connect karta hai ki gas kitni tezi se move kar rahi hai isse ki uska pressure kitna gir gaya (Step 3). Ab pucho: exactly us waqt pressure kitna gira hai jab woh bilkul utni tezi se move kar rahi hai jitni tezi se ek whisper travel karta hai — Mach 1? $M = 1$ rule mein plug karo aur air ke liye $0.528$ nikal aata hai (Step 4). Flow us se aage speed up kyun refuse karta hai? Kyunki gas ka amount jo per second squeeze ho raha hai ek hump hai jo exactly Mach 1 par peak karta hai (Step 5) — tum summit par khade ho, aur bahar pressure neeche karna tumhe aur upar nahi lift kar sakta. Aur chahe bahar thoda neeche ho ya total vacuum, exit Mach 1 par pinned rehta hai, apna leftover pressure sirf pipe chodne ke baad dump karta hai (Step 6). Aur tezi se jaane ke liye corridor ko pinch ke baad phir se widen karna hoga — lekin woh alag nozzle hai.

Recall

Woh single dial jo sab kuch decide karta hai ::: back-pressure ratio $p_{b} / p_{0}$ compare kiya $0.528$ se. Mass-flow hump ka peak kahan hai ::: exactly $M = 1$ par (uska slope wahan zero hai). Gas accelerate hone par speed ke liye kya "trade" hota hai ::: uski thermal energy (woh cool hoti hai), total fixed rakhte hue. Narrow matlab faster kyun hota hai ::: mass conservation $ρ A V =$ const ke saath $ρ$ low speed par almost fixed. Woh do laws jis par poori derivation resti hai ::: energy conservation (Step 2) aur isentropy (Step 3).

Connections

Converging nozzle — subsonic flow, Mach 1 at exit — parent result jo yeh page picture by picture rebuild karta hai.
Isentropic Flow Relations — $c_{p} = \frac{γ R}{γ - 1}$ aur Steps 2–3 mein use hone wala $T$ – $p$ link supply karta hai.
Speed of Sound — $a = γ R T$ define karta hai aur kyun $M = 1$ ceiling hai.
Stagnation Properties — reservoir subscript- $0$ state.
Mass Flow Rate & Choking — Step 5 ka hump-peaks-at-one argument.
Converging-Diverging (de Laval) Nozzle — $M = 1$ ke aage jaane ke liye zaruri widening section.
Normal Shock Waves — kya appear ho sakta hai downstream jab supersonic ho.