Over - under expanded nozzle flows: Exercises — Hinglish deep dive

3.1.18 · D4 · Physics › Compressible Flow & Aerodynamics › Over - under expanded nozzle flows

Yeh self-testing page hai Over/Under-Expanded Nozzle Flows ke liye. Pehle har problem khud solve karo, phir solution kholo. Difficulty L1 Recognition (sirf regime ka naam batao) se L5 Mastery (kai ideas ko chain karo) tak badhti hai.

Shuru karne se pehle, ek baar pin down kar lete hain woh sirf teen tools jo is page par chahiye. Har ek parent note mein build kiya gaya tha; yahan hum bas unhe restate kar rahe hain taaki koi symbol use hone se pehle unnamed na rahe.

Formula Teen tools (kuch aur chahiye hi nahi)

(A) Stagnation-to-static pressure — chamber pressure $p_{0}$ ko exit pressure $p_{e}$ se link karta hai exit Mach number $M_{e}$ ke through (yeh "speed number" = flow speed ÷ local sound speed): $\frac{p _{0}}{p _{e}} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{\frac{γ}{γ - 1}}$ (B) Area–Mach relation — exit-to-throat area ratio ko $M_{e}$ se link karta hai; from Area-Mach Number Relation: $\frac{A _{e}}{A ^{*}} = \frac{1}{M _{e}} [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M_{e}^{2})]^{\frac{γ + 1}{2 ( γ - 1 )}}$ (C) Thrust — momentum conservation se: $F = \overset{m}{˙} V_{e} + (p_{e} - p_{b}) A_{e}$

Symbols, har ek pehle se named:

$γ$ (gamma) = ratio of specific heats, ek gas property; air/zyaadatar combustion gases ke liye $γ = 1.4$ .
$A^{*}$ = ==throat area jahan $M = 1$ == (dekho Choked Flow & the Throat (M=1)).
$p_{b}$ = ambient / back pressure nozzle ke bahar.
$\overset{m}{˙}$ ("m-dot") = mass-flow rate, exit se nikalne wale gas ke kilograms har second (kg/s).
$V_{e}$ = exit velocity, exit plane par gas ki speed (m/s).
$A_{e}$ = exit-plane area (m²).

Recall Trio figure ko kaise padhen (is page par baar baar use hota hai)

Neeche wali figure mein teen panels hain, sab ek hi de Laval nozzle (white bell) dikha rahe hain jisme gas left-to-right flow kar rahi hai (blue arrow), sirf yeh alag hai ki lip ke bilkul baad kya hota hai:

Left panel, red mein drawn — over-expanded ( $p_{e} < p_{b}$ ). Jet boundary andar ki taraf murt hai (converging red lines) kyunki ambient pressure use squeeze kar raha hai. Red dashed lines jo "X" banaate hain woh oblique shocks hain jo flow ko compress karti hain.
Middle panel, green mein drawn — perfectly expanded ( $p_{e} = p_{b}$ ). Jet boundary parallel hai (do seedhi green lines): koi waves nahi, clean exit.
Right panel, yellow mein drawn — under-expanded ( $p_{e} > p_{b}$ ). Jet boundary bahar ki taraf bulge karti hai (yellow curves flaring open) kyunki jet abhi bhi over-pressured hai; yellow dashed lines expansion fans hain.

Toh jab koi solution "red panel" kahe, matlab over-expanded/shocks; "green panel" = perfect; "yellow panel" = under-expanded/fans.

Figure — Over - under expanded nozzle flows

Recall Tools A–C use karne se pehle: kya flow supersonic bhi hai?

Tools (A), (B), (C) jaise likhe hain yeh assume karte hain ki nozzle choked hai aur diverging section supersonic chal raha hai — tabhi $p_{e}$ geometry se lock hota hai. Yeh automatic nahi hai. Yeh depend karta hai ki nozzle ko kitna hard drive kiya ja raha hai, yaani ratio $p_{0} / p_{b}$ par.

Pehle, ek naya symbol: $p^{*}$ = ==woh exit static pressure jis exact moment throat pehli baar $M = 1$ reach karta hai== (the "just-choked" pressure). Jab throat choking ke bilkul edge par ho, wahan jo pressure appear hota hai woh chamber ke relative sirf $γ$ se fix hota hai. Tool (A) mein $M = 1$ rakhne se critical pressure ratio milta hai: $\frac{p _{0}}{p ^{*}} = (1 + \frac{γ - 1}{2} \cdot 1^{2})^{\frac{γ}{γ - 1}} = (\frac{γ + 1}{2})^{\frac{γ}{γ - 1}} \approx 1.893 (γ = 1.4) .$ Yeh $p_{0} / p_{b}$ par test kaise banta hai: throat ko choke karने के liye, chamber ko gas exit hone se pehle pressure kam se kam $p^{*}$ tak drop karna hoga. Agar bahar ka pressure $p_{b}$ , $p^{*}$ se zyada hai, toh flow utna neeche nahi ja sakti aur throat kabhi $M = 1$ reach nahi karta. Toh practical choking test hai:

$p_{0} / p_{b} \geq 1.893$ → throat chokes → supersonic diverging flow → $p_{e}$ geometry se lock → over/under-expansion apply hoti hai.
$p_{0} / p_{b} < 1.893$ → throat subsonic rehta hai → unchoked, poora nozzle subsonic hai, aur $p_{e} = p_{b}$ force hota hai (exit back pressure sun leta hai). Koi external shocks/fans nahi; classification apply nahi hoti.

Problems 5.3 aur 5.4 is boundary ko directly test karte hain. Baaki har problem mein $p_{0} / p_{b}$ 1.893 se kaafi upar hai, toh flow safely choked aur supersonic hai.

Level 1 — Recognition

Problem 1.1 (L1)

Ek nozzle $p_{e} = 0.6$ bar produce karta hai. Rocket us altitude par fly kar raha hai jahan $p_{b} = 0.9$ bar. Regime ka naam batao aur lip par jo wave banti hai woh state karo.

Recall Solution 1.1

KYA compare kar rahe hain: $p_{e} = 0.6$ vs $p_{b} = 0.9$ . Kyunki $0.6 < 0.9$ , hame $p_{e} < p_{b}$ mila → over-expanded. KYUN: exit pressure ambient se neeche gir gayi hai, toh surroundings ko jet ko andar push karna padega — yeh compression oblique shocks se hoti hai jo nozzle lip par anchored hoti hain. Yeh figure ka left (red) panel hai: jet andar squeeze hoti hai aur crossing shock lines banti hain. (Mnemonic check: OLE = Over → Low → Exit. Over-expanded ⇒ Low exit pressure. ✓)

Problem 1.2 (L1)

Wahi nozzle ( $p_{e} = 0.6$ bar) ab launchpad par sea level par, $p_{b} = 1.0$ bar. Regime?

Recall Solution 1.2

KYA compare kar rahe hain: $p_{e} = 0.6$ vs $p_{b} = 1.0$ . $0.6 < 1.0$ → $p_{e} < p_{b}$ → over-expanded, 1.1 se zyada strongly. "Zyada strongly" KYUN: bahar ko jo mismatch correct karna hai woh gap $p_{b} - p_{e}$ hai. 1.1 mein woh $0.9 - 0.6 = 0.3$ bar tha; yahan $1.0 - 0.6 = 0.4$ bar hai — bada gap matlab zyada strong required compression, isliye zyada strong shocks. Physically KYUN matter karta hai: agar gap itna bada ho jaye ki gentle oblique shocks akele kaam na kar sakein toh ek normal shock / Mach disk form hoti hai (dekho Normal Shock & Mach Disk). Wahi figure ka red panel.

Problem 1.3 (L1)

Ek nozzle ka $p_{e} = 0.4$ bar hai aur use vacuum chamber mein fire kiya gaya hai, $p_{b} \approx 0$ . Regime?

Recall Solution 1.3

KYA compare kar rahe hain: $p_{e} = 0.4$ vs $p_{b} \approx 0$ . $0.4 > 0$ → $p_{e} > p_{b}$ → under-expanded — figure ka right (yellow) panel, jet bahar ki taraf expansion fans se bulge karti hai. Plume itna wide KYUN spread hota hai: jet near-vacuum ke relative enormously over-pressured hai, toh woh nozzle ke bahar har outward direction mein expand karta rehta hai. Isliye rocket plumes space mein dramatically fan out karti hain — $p_{b}$ jitna chota hoga, fans ko utna zyada outward turning produce karna hoga.

Level 2 — Application

Problem 2.1 (L2)

Ek nozzle chamber pressure $p_{0} = 20$ bar aur exit Mach $M_{e} = 2.5$ ( $γ = 1.4$ ) ke saath run karta hai. $p_{e}$ find karo.

Recall Solution 2.1

KYA: tool (A) use karo — yahi ek formula hai jo $p_{0}$ , $M_{e}$ , $p_{e}$ ko link karta hai. (A) KYUN na ki (B): $M_{e}$ directly diya hua hai, toh area step skip karte hain (tool B tabhi chahiye jab $M_{e}$ geometry se find karna ho). $\frac{p _{0}}{p _{e}} = (1 + 0.2 \cdot 2. 5^{2})^{3.5} = (1 + 1.25)^{3.5} = 2.2 5^{3.5}$ Ab $2.2 5^{3.5} \approx 17.09$ . Toh $p_{e} = \frac{20}{17.09} \approx 1.17 bar .$

Problem 2.2 (L2)

2.1 wala nozzle sea level par fire kiya gaya, $p_{b} = 1.0$ bar. Over- ya under-expanded?

Recall Solution 2.2

KYA compare kar rahe hain: $p_{e} \approx 1.17$ vs $p_{b} = 1.0$ . $1.17 > 1.0$ → $p_{e} > p_{b}$ → under-expanded (mildly). Sirf mildly KYUN: gap $p_{e} - p_{b} \approx 0.17$ bar chota hai, toh sirf weak expansion fans chahiye — figure ka yellow panel, lekin dramatically bulging boundary ke bajaye gently-flaring boundary ke saath.

Problem 2.3 (L2)

$A_{e} / A^{*} = 2.0$ aur $γ = 1.4$ diya hai, supersonic exit Mach number $M_{e}$ find karo. (Numerically solve karo; equation tool (B) hai.)

Recall Solution 2.3

KYA: tool (B) hi ek equation hai jo area ratio ko $M$ se relate karta hai. Supersonic root KYUN: area–Mach curve U-shaped hoti hai — 1 se upar har area ratio ke liye DO roots hote hain, ek subsonic ( $M < 1$ ) aur ek supersonic ( $M > 1$ ). Choked throat ke baad diverging section mein flow supersonic hoti hai, toh $M > 1$ root lete hain. $γ = 1.4$ plug karke: $2.0 = \frac{1}{M _{e}} [\frac{1}{1.2} (1 + 0.2 M_{e}^{2})]^{3} .$ Numerically solve karne par $M_{e} \approx 2.197$ milta hai.

Level 3 — Analysis

Problem 3.1 (L3)

Ek nozzle ka $A_{e} / A^{*} = 6.0$ , $p_{0} = 50$ bar, $γ = 1.4$ hai. $M_{e}$ find karo, phir $p_{e}$ , phir sea level par classify karo ( $p_{b} = 1.0$ bar).

Recall Solution 3.1

Step 1 (tool B) — geometry to Mach. $6.0 = \frac{1}{M _{e}} [\frac{1}{1.2} (1 + 0.2 M_{e}^{2})]^{3}$ ko supersonic root ke liye solve karo: $M_{e} \approx 3.37.$ Tool B pehle KYUN: hume geometry ( $A_{e} / A^{*}$ ) diya gaya hai, Mach nahi, toh pressure touch karne se pehle area ratio → Mach convert karna hoga. Answer bada KYUN hai: bada area ratio matlab gas zyada area mein expand karti hai, zyada accelerate hoti hai, toh $M_{e}$ bada hota hai (6 ek bada ratio hai ⇒ $M_{e} > 3$ ). Step 2 (tool A) — Mach to pressure. $\frac{p _{0}}{p _{e}} = (1 + 0.2 \cdot 3.3 7^{2})^{3.5} = (1 + 2.272)^{3.5} = 3.27 2^{3.5} \approx 63.1.$ $p_{e} = \frac{50}{63.1} \approx 0.792 bar .$ Tool A doosra KYUN: ab jab $M_{e}$ pata hai, tool (A) hi ek relation hai jo use pressure mein convert karta hai. $p_{e}$ itna $p_{0}$ se neeche KYUN: high Mach ka matlab hai gas ne apna almost poora pressure speed gain karne mein diya, toh $p_{e}$ 50-bar chamber ka ek tiny fraction hai. Step 3 — classify. $p_{e} = 0.792 < p_{b} = 1.0$ → over-expanded (figure ka red panel). Shocks KYUN: exit pressure ambient se neeche baith gayi, toh bahar ko use compress karke upar laana hoga — yeh oblique shocks lip par karti hain.

Problem 3.2 (L3)

3.1 wale nozzle ke liye woh back pressure $p_{b}^{⋆}$ kya hai jis par nozzle perfectly expanded ho? Phir: us ambient pressure se neeche ya upar, kaun sa regime?

Recall Solution 3.2

KYA: perfect expansion ka matlab $p_{e} = p_{b}$ , aur $p_{e}$ geometry se fix hai. Toh $p_{b}^{⋆} = p_{e} \approx 0.792$ bar. KYUN: wahi ek ambient pressure hai jo geometry-locked exit pressure se exactly match karta hai, toh jet cleanly nikalta hai (figure ka green centre panel). Altitude ke baare mein reasoning: rocket climb karta hai toh $p_{b}$ girti hai.

Jab $p_{b} > 0.792$ bar (low altitude, e.g. sea level) → $p_{e} < p_{b}$ → over-expanded.
Jab $p_{b} < 0.792$ bar (high altitude) → $p_{e} > p_{b}$ → under-expanded. Kyun flip hota hai: $p_{e}$ apni jagah rehti hai (geometry kabhi nahi badalti), toh $p_{b}$ jaise $0.792$ bar se neeche slide karta hai waise $p_{e} - p_{b}$ ka sign reverse ho jaata hai. Nozzle us altitude se guzarte hue regime flip karta hai — yeh Rocket Nozzle Design & Thrust Optimization mein core design tension hai.

Level 4 — Synthesis

Problem 4.1 (L4)

Ek rocket engine: $p_{0} = 70$ bar, $A_{e} / A^{*} = 6.0$ , $γ = 1.4$ , mass-flow rate $\overset{m}{˙} = 250$ kg/s, exit velocity $V_{e} = 2900$ m/s, exit area $A_{e} = 0.90$ m². Sea level ( $p_{b} = 101, 325$ Pa) aur vacuum ( $p_{b} = 0$ ) par thrust compute karo. Vacuum mein jaane par thrust kitna badhta hai?

Recall Solution 4.1

Step 1 — exit pressure. 3.1 ki geometry se ( $A_{e} / A^{*} = 6$ ), $M_{e} \approx 3.37$ aur $p_{0} / p_{e} \approx 63.1$ . 3.1 reuse KYUN: area ratio identical hai, toh Mach aur pressure ratio identical hain — sirf $p_{0}$ badla. $p_{0} = 70$ bar $= 7.0 \times 1 0^{6}$ Pa ke saath, $p_{e} = \frac{7.0 \times 1 0 ^{6}}{63.1} \approx 1.109 \times 1 0^{5} Pa (\approx 1.11 bar) .$ Step 2 — momentum thrust (dono cases mein same). $\overset{m}{˙} V_{e} = 250 \times 2900 = 7.25 \times 1 0^{5} N .$ Yeh piece altitude-independent KYUN: $\overset{m}{˙}$ aur $V_{e}$ internal supersonic flow se set hote hain, jo $p_{b}$ sun nahi sakti; sirf pressure term atmosphere feel karta hai. Step 3 — pressure thrust, sea level. $(p_{e} - p_{b}) A_{e} = (1.109 \times 1 0^{5} - 1.01325 \times 1 0^{5}) (0.90) = (9.575 \times 1 0^{3}) (0.90) \approx 8.62 \times 1 0^{3} N .$ Yahan positive KYUN: $p_{e} > p_{b}$ (under-expanded), toh exit atmosphere se zyada hard push out karta hai — yeh term thrust mein add hoti hai. $F_{sea} = 7.25 \times 1 0^{5} + 8.62 \times 1 0^{3} \approx 7.336 \times 1 0^{5} N .$ Step 4 — pressure thrust, vacuum ( $p_{b} = 0$ ): $(p_{e} - 0) A_{e} = 1.109 \times 1 0^{5} \times 0.90 \approx 9.981 \times 1 0^{4} N .$ Yeh badhta KYUN: atmosphere hatane se inward $- p_{b} A_{e}$ push hatt jaata hai, toh full exit pressure ab contribute karta hai. $F_{vac} = 7.25 \times 1 0^{5} + 9.981 \times 1 0^{4} \approx 8.248 \times 1 0^{5} N .$ Step 5 — gain. $Δ F = F_{vac} - F_{sea} \approx 8.248 \times 1 0^{5} - 7.336 \times 1 0^{5} \approx 9.12 \times 1 0^{4} N .$ Exactly yahi KYUN: difference exactly $p_{b} A_{e} = 1.01325 \times 1 0^{5} \times 0.90 \approx 9.12 \times 1 0^{4}$ N hai — woh atmospheric back-push jo gayab ho jaata hai. Vacuum mein jaane par thrust lagbhag $91$ kN badhta hai.

Problem 4.2 (L4)

Usi engine ke liye, sea level par over- ya under-expanded hai? Aur kya sea level par bada area ratio (6.0) chalana achha idea hai?

Recall Solution 4.2

KYA compare kar rahe hain: $p_{e} \approx 1.11$ bar vs $p_{b} = 1.013$ bar. $1.11 > 1.013$ → $p_{e} > p_{b}$ → sea level par under-expanded (barely). Sea level par nearly perfect KYUN: gap tiny hai, toh sea-level penalty choti hai aur altitude ke saath sirf improve hoti hai ( $p_{b}$ girti hai toh positive pressure-thrust term badhta hai). Bada area ratio phir bhi sea level par risky KYUN: ratio aur badha do aur $p_{e}$ $p_{b}$ se neeche gir jaata hai, over-expansion hoti hai aur, agar strong ho toh, flow separation aur side-loads. Badi bells vacuum mein efficient hain lekin launch par hazardous.

Level 5 — Mastery

Problem 5.1 (L5)

Design task. Tum chahte ho ek nozzle jo us altitude par perfectly expanded ho jahan $p_{b} = 0.30$ bar, $p_{0} = 40$ bar, $γ = 1.4$ use karke. Required $M_{e}$ aur phir required area ratio $A_{e} / A^{*}$ find karo.

Recall Solution 5.1

Step 1 — required $p_{e}$ . Perfect expansion ⇒ $p_{e} = p_{b} = 0.30$ bar. Toh $p_{0} / p_{e} = 40/0.30 = 133.3$ . KYUN: perfect expansion defined hai $p_{e} = p_{b}$ se, jo woh pressure ratio pin karta hai jo hume engineer karna hai. Step 2 — tool (A) ko $M_{e}$ ke liye invert karo. Exponent ko $1/3.5$ tak raise karke undo karte hain: $1 + 0.2 M_{e}^{2} = (133.3)^{1/3.5} = 133. 3^{0.2857} \approx 4.05.$ $M_{e}^{2} = \frac{4.05 - 1}{0.2} = 15.25 \Rightarrow M_{e} \approx 3.91.$ Invert KYUN: hume pressure ratio pata hai aur woh Mach chahiye jo use produce kare — Problem 2.1 ka ulta. Step 3 — area ratio ke liye tool (B). $M_{e} = 3.91$ plug karo: $\frac{A _{e}}{A ^{*}} = \frac{1}{3.91} [\frac{1}{1.2} (1 + 0.2 \cdot 3.9 1^{2})]^{3} = \frac{1}{3.91} [\frac{4.057}{1.2}]^{3} = \frac{1}{3.91} (3.381)^{3} .$ $(3.381)^{3} \approx 38.64$ , toh $A_{e} / A^{*} \approx 38.64/3.91 \approx 9.88$ . Answer: $M_{e} \approx 3.91$ , $A_{e} / A^{*} \approx 9.9$ . Ek bada, high-altitude nozzle — exactly isliye upper-stage engines ke huge bells hote hain.

Problem 5.2 (L5)

Woh altitude-optimized nozzle lo ( $A_{e} / A^{*} \approx 9.9$ , $p_{e} = 0.30$ bar) aur sea level par fire karo ( $p_{b} = 1.0$ bar). Classify karo, ratio $p_{b} / p_{e}$ compute karo, aur physical danger explain karo.

Recall Solution 5.2

KYA compare kar rahe hain: $p_{e} = 0.30$ vs $p_{b} = 1.0$ . $0.30 < 1.0$ → $p_{e} < p_{b}$ → strongly over-expanded (figure ka red panel, apni extreme par pushed). Required pressure jump jo bahar impose karna hai woh hai $p_{b} / p_{e} = 1.0/0.30 \approx 3.33$ . Yeh dangerous KYUN hai: itna bada compression gentle oblique shock akele handle nahi kar sakta; ek Mach disk (jet ke across ek normal shock, dekho Normal Shock & Mach Disk) banti hai, aur adverse pressure ek shock ko diverging section mein andar push kar sakta hai, boundary layer ko walls se tear kar deta hai — flow separation (dekho Flow Separation in Nozzles). Isse side-loads aate hain jo ignition par nozzle ko physically damage kar sakte hain — yeh launchpad-vs-altitude dilemma hai jo dual-bell aur altitude-compensating nozzles se solve kiya jaata hai.

Problem 5.3 (L5)

Ek converging–diverging nozzle ko sirf gently drive kiya gaya hai: chamber pressure $p_{0} = 1.5$ bar against ambient $p_{b} = 1.0$ bar ( $γ = 1.4$ ). Kya throat choked hai? Exit pressure kya hai, aur kya over/under-expanded classification apply bhi hoti hai?

Recall Solution 5.3

Step 1 — choking check karo. Driving ratio ko critical ratio $p_{0} / p^{*} \approx 1.893$ se compare karo: $\frac{p _{0}}{p _{b}} = \frac{1.5}{1.0} = 1.5 < 1.893.$ Yeh test KYUN: throat ko sonic-choke karne ke liye chamber ko pressure $p^{*}$ (just-choked pressure) tak drop karna hoga. Kyunki $p_{b}$ , $p^{*}$ se upar baith gayi hai jiske liye $p_{0} / p_{b} \geq 1.893$ chahiye, aur hum us se neeche hain, throat kabhi $M = 1$ reach nahi karta — nozzle unchoked hai. Step 2 — consequence. Sonic throat ke bina, poora nozzle subsonic rehta hai. Isse exit pressure fix KYUN hoti hai: subsonic flow pressure upstream transmit kar sakti hai, toh exit ambient se match karne ke liye forced hai: $p_{e} = p_{b} = 1.0$ bar. Diverging section diffuser ki tarah kaam karta hai (subsonic flow decelerate karta hai), supersonic accelerator ki tarah nahi. Step 3 — classification. Over-/under-expanded sirf supersonic exit ke liye defined hai jab $p_{e}$ geometry se locked ho. Yahan apply KYUN nahi hoti: $p_{e}$ locked nahi hai — woh $p_{b}$ maanta hai. Toh koi external shocks ya fans nahi, sirf smooth subsonic flow. Yahi woh regime hai jiske baare mein parent note ka [!mistake] warn karta hai — naive intuition "back pressure controls exit pressure" yahan sahi hai exactly kyunki flow subsonic hai.

Problem 5.4 (L5)

Ek actual subsonic exit calculation karo. Ek purely converging nozzle (koi diverging section nahi, exit = throat) ka $A_{e} = 1.5 A_{some upstream section where M = 0.20}$ … simpler: ek subsonic nozzle ke exit par flow $M_{e} = 0.50$ reach karti hai chamber pressure $p_{0} = 1.6$ bar ke saath, $γ = 1.4$ . (a) Tool (A) use karke exit static pressure $p_{e}$ find karo. (b) Alag se, exact critical case $p_{0} / p_{b} = 1.893$ analyse karo: tab exit Mach number aur exit pressure kya hai, aur isme kya special hai?

Recall Solution 5.4

Part (a) — subsonic exit pressure. Tool (A) phir bhi KYUN kaam karta hai: tool (A) sirf isentropic energy relation hai (from Isentropic Flow Relations); yeh kisi bhi Mach number par valid hai, subsonic ya supersonic. Sirf woh tools off-limits hain jo sonic throat assume karte hain ( $A^{*}$ , choking) jab unchoked ho. Yahan hume simply $M_{e}$ diya gaya hai, toh plug in karo: $\frac{p _{0}}{p _{e}} = (1 + 0.2 \cdot 0. 5^{2})^{3.5} = (1 + 0.05)^{3.5} = 1.0 5^{3.5} \approx 1.186.$ $p_{e} = \frac{1.6}{1.186} \approx 1.349 bar .$ $p_{e}$ , $p_{0}$ ke itna close KYUN: low Mach par gas barely accelerate hui, toh usne apna almost poora pressure rakha — supersonic cases ke exact opposite jahan $p_{e} ≪ p_{0}$ . Part (b) — exact critical case $p_{0} / p_{b} = 1.893$ . Yeh unchoked aur choked ke beech ki razor's edge hai. KYUN: exactly critical ratio par throat just $M = 1$ reach karta hai, toh exit (agar throat hi exit hai) ka $M_{e} = 1$ hoga aur $p_{e} = p^{*} = p_{0} /1.893$ . Maano $p_{0} = 1.893$ bar toh $p_{e} = 1.0$ bar $= p_{b}$ — flow choked hai aur perfectly matched, exactly sonic ( $M_{e} = 1$ ) par baitha hai. Yeh unique boundary point hai: koi bhi $p_{0} / p_{b}$ zyada hone par throat sonic ho jaata hai aur (diverging section ke saath) supersonic; koi bhi kum hone par subsonic rehta hai. Toh critical case woh ek value hai jahan "just choked," " $M = 1$ at the exit," aur " $p_{e} = p_{b}$ " teeno ek saath coincide karte hain.

Recall Jaane se pehle ek-line self-check

Kisi bhi classification se pehle pehla sawaal ::: Kya $p_{0} / p_{b}$ critical ratio ( $γ = 1.4$ ke liye $\approx 1.893$ ) se upar hai? Agar nahi, toh flow unchoked/subsonic hai aur $p_{e} = p_{b}$ — koi over/under-expansion nahi. $p^{*}$ kya hai? ::: Just-choked exit pressure, yaani $p_{0}$ divided by critical ratio; tool (A) mein $M = 1$ set karne par $p_{0} / p^{*} = ((γ + 1) /2)^{γ / (γ - 1)} \approx 1.893$ milta hai. Regime rule ::: Jab choked & supersonic ho, geometry se $p_{e}$ compute karo (tools B phir A), $p_{b}$ se compare karo: $p_{e} < p_{b}$ over-expanded (shocks), $p_{e} > p_{b}$ under-expanded (fans), $p_{e} = p_{b}$ perfect. Kya $p_{b}$ supersonic flow mein $p_{e}$ badalta hai? ::: Nahi — $p_{e}$ $A_{e} / A^{*}$ aur $p_{0}$ se locked hai; regime sirf isliye flip hota hai kyunki $p_{b}$ fixed $p_{e}$ ko cross karta hai. Over-expanded hone par pressure-thrust term ka sign? ::: Negative — yeh thrust se subtract hoti hai.