Converging nozzle — subsonic flow, Mach 1 at exit: Worked examples

3.1.9 · D3 · Physics › Compressible Flow & Aerodynamics › Converging nozzle — subsonic flow, Mach 1 at exit

Scenario matrix

Har converging-nozzle question in cells mein se ek hai. Poora tree split karne wala ek hi decision hai: kya $p_{b} / p_{0}$ critical value $0.528$ se upar hai ya neeche?

Cell	Case class	Input signature	Kya fixed hai	Kya solve karte ho
A	No flow (degenerate)	$p_{b} = p_{0}$	$V = 0$ everywhere	Kuch flow nahi; $\overset{m}{˙} = 0$
B	Subsonic exit (unchoked)	$0.528 < p_{b} / p_{0} < 1$	$p_{e} = p_{b}$	Exit $M$ , $T_{e}$ , $V_{e}$ , $\overset{m}{˙}$
C	Exactly critical	$p_{b} / p_{0} = 0.528$	$M_{e} = 1$ boundary	Confirm karo ki ye just choke karta hai
D	Choked, under-expanded	$p_{b} / p_{0} < 0.528$	$M_{e} = 1$ , $p_{e} = 0.528 p_{0}$	$p_{e}$ , $\overset{m}{˙}$ (max), kyun $\overset{m}{˙}$ flat hai
E	Limiting: $p_{b} \to 0$ (vacuum)	$p_{b} / p_{0} \to 0$	Still $M_{e} = 1$ , $p_{e} = 0.528 p_{0}$	Dikhao ki exit change NAHI hota
F	Real-world word problem	Engine/tank numbers diye hain	mix of above	Physical answer tak poora pipeline
G	Exam twist — sizing	Target $\overset{m}{˙}$ chahiye	Area $A$ solve karo	Choked mass-flow law invert karo
H	Exam twist — non-air gas	$γ \neq = 1.4$	Naya critical ratio	$0.528 \to$ naya value recompute karo

Neeche, har example un cell(s) ke saath tagged hai jo wo cover karta hai. Milake ye poora table fill karte hain.

Sign conventions set karna (pehle build karo, phir use karo)

Definition Symbols, simple shabdon mein

Pressures aur gate

$p_{0}$ — stagnation pressure: tank mein still gas ka pressure, before it moves.
$p_{b}$ — back pressure: us room ka pressure jisme nozzle dump karta hai.
$p_{e}$ — exit pressure: gas ka actual pressure bilkul nozzle mouth par.
$r \equiv p_{b} / p_{0}$ — wo ek ratio jo sab kuch decide karta hai. Hamesha $0$ aur $1$ ke beech (sirf high se low pressure ki taraf push kar sakte ho).

Temperatures

$T_{0}$ — stagnation temperature: tank mein still gas ka temperature.
$T_{e}$ — exit temperature, nozzle mouth par gas ka temperature.

"Star" (critical) state — wo value jo ek quantity tab leti hai jab exit exactly $M = 1$ par ho. Chhota star superscript ( $*$ ) hamesha "choked value" matlab hai:

$p^{*}$ — critical (choked) exit pressure $= 0.528 p_{0}$ air ke liye.
$T^{*}$ — critical exit temperature $= \frac{2}{γ + 1} T_{0}$ .
$ρ^{*}$ — critical exit density.
$a^{*}$ — choked exit par speed of sound; jab $M = 1$ tab gas exactly is speed par move karti hai.

Motion aur flow

$M_{e}$ — exit Mach number = (exit speed) ÷ (local speed of sound). $M_{e} = 1$ matlab gas exactly speed of sound par move kar rahi hai.
$a$ — speed of sound: ek chhota pressure signal gas mein kitni fast travel karta hai, $a = γ R T$ (dekho Speed of Sound). Exit par iska value $a_{e}$ likhte hain.
$V$ — gas ki bulk flow speed ( $V_{e}$ = exit speed). Relation: $V = M a$ .
$ρ$ — density: gas ka mass per cubic metre ( $ρ_{e}$ = exit density, $ρ^{*}$ = choked density).
$A$ — nozzle exit (throat) area: narrow mouth ka cross-sectional area jisse gas bahar jaati hai, $m^{2}$ mein.
$\overset{m}{˙}$ — mass flow rate: kilograms of gas per second, $\overset{m}{˙} = ρ A V$ .
$R$ — specific gas constant of the gas (air: $287 J/kg\cdotK$ ), ideal-gas law $p = ρR T$ mein constant. Speed of sound bhi set karta hai.
$γ$ — ratio of specific heats (air ke liye $1.4$ , helium jaisi monatomic gas ke liye $5/3$ ).

Neeche decision figure wo flowchart hai jise har example follow karta hai.

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

Intuition Figure s01 kaise padhein — apni ungli se trace karo

Apni ungli top black box par rakkho: yahan tum hamesha shuru karte ho, $r = p_{b} / p_{0}$ compute karke. Ek arrow seedha neeche red diamond mein jaata hai — poori page par ek hi decision, yes/no question " $r > 0.528$ ?".

Left slide karo "yes" arrow ke saath → tum black unchoked box mein land karte ho: exit subsonic hai ( $M_{e} < 1$ ) aur exit pressure room se match karta hai, $p_{e} = p_{b}$ . Ye Cells A aur B hain.
Right slide karo "no" arrow ke saath → tum red-outlined choked box mein land karte ho: exit $M_{e} = 1$ par lock hai aur exit pressure $p_{e} = 0.528 p_{0}$ par freeze hai, room se deaf. Ye Cells C, D, E, F hain.

Note karo yahan exactly ek diamond hai: kisi bhi example ko kabhi second decision ki zaroorat nahi. Neeche har worked example literally tum wahi ho jo ek number top box mein daalo aur dekho wo kis branch mein jata hai.

$0.528$ kahan se aaya (taaki hum ise badal sakein)

Cell H mein ek alag gas hai, isliye main $0.528$ ko magic number ki tarah quote nahi kar sakta — mujhe wo machine dikhani hogi jo ise produce karti hai. Ye seedha master relation se aata hai.

Cell A — No flow (degenerate input)

Cell B — Subsonic exit (everyday case)

Worked example Cell B ·

$0.528 < r < 1$ Air reservoir $p_{0} = 500 kPa$ , $T_{0} = 300 K$ , exit area $A = 10 cm^{2}$ . Back pressure $p_{b} = 350 kPa$ . Exit Mach, exit temperature, exit speed, aur mass flow find karo.

Forecast: Exit subsonic hoga ya sonic? $M_{e}$ ek decimal tak guess karo.

Gate: $r = 350/500 = 0.70 > 0.528$ → unchoked, so $p_{e} = p_{b} = 350 kPa$ . Ye step kyun? Critical se upar, jet abhi bhi room ko "feel" karta hai aur match karta hai.
$M_{e}$ pane ke liye master relation invert karo: $\frac{p _{0}}{p _{e}} = \frac{500}{350} = 1.4286 = (1 + 0.2 M_{e}^{2})^{3.5}$ $3.5$ -th root lo: $1.428 6^{1/3.5} = 1.1073$ , so $0.2 M_{e}^{2} = 0.1073 \Rightarrow M_{e} = 0.536 \approx 0.732$ . Ye step kyun? Master relation ek measurable pressure ratio ko $M$ se tie karta hai. Invert karne se speed nikalta hai.
$T_{0} / T_{e} = 1 + 0.2 M_{e}^{2} = 1.1073$ se exit temperature: $T_{e} = 300/1.1073 = 270.9 K$ Ye step kyun? Hume exit par actual gas temperature chahiye kyunki speed of sound $a = γ R T$ usi par depend karta hai — $T_{e}$ ke bina hum $M_{e}$ ko real m/s speed mein convert nahi kar sakte.
Local speed of sound aur exit speed (dekho Speed of Sound): $a_{e} = γ R T_{e} = 1.4 \cdot 287 \cdot 270.9 = 330.1 m/s, V_{e} = M_{e} a_{e} = 0.732 \cdot 330.1 = 241.6 m/s$ Ye step kyun? Mach number ek ratio hai; physical speed m/s mein paane ke liye ise local sound speed $a_{e}$ se multiply karna padta hai, jo abhi $T_{e}$ jaanne ke baad hi compute ho sakti hai.
Ideal gas $p_{e} = ρ_{e} R T_{e}$ se exit density: $ρ_{e} = \frac{350000}{287 \cdot 270.9} = 4.502 kg/m^{3}$ Ye step kyun? Mass flow kilograms count karta hai, isliye gas kitna densely packed hai ye jaanna zaroori hai; ideal-gas law known $p_{e}, T_{e}$ ko density $ρ_{e}$ mein convert karta hai.
Mass flow $\overset{m}{˙} = ρ_{e} A V_{e} = 4.502 \cdot (10 \times 1 0^{- 4}) \cdot 241.6 = 1.088 kg/s$ . Ye step kyun? $\overset{m}{˙} = ρ A V$ literally (mass per volume)×(area)×(speed) = mass per second bahar jaana — final physical answer.

Verify: $M_{e} = 0.732 < 1$ ✓ (subsonic, jaisa converging nozzle mein hona chahiye). Units: $kg/m^{3} \cdot m^{2} \cdot m/s = kg/s$ ✓. Sab neeche numerically check hai.

Cell C — Exactly critical (knife-edge)

Cell D — Choked aur under-expanded

Worked example Cell D ·

$r < 0.528$ Air reservoir $p_{0} = 500 kPa$ , $T_{0} = 300 K$ , exit area $A = 10 cm^{2}$ . Ab $p_{b} = 101 kPa$ (atmosphere). $M_{e}$ , $p_{e}$ , $T_{e}$ , $V_{e}$ , aur $\overset{m}{˙}$ find karo.

Forecast: $p_{e}$ guess karo. Trap: kya ye $101$ kPa hai?

Gate: $r = 101/500 = 0.202 < 0.528$ → choked. Ye step kyun? Critical se neeche, exit $M_{e} = 1$ par lock ho jaata hai chahe $p_{b}$ kuch bhi ho.
Exit pressure nozzle se fix hota hai, $p_{b}$ se NAHI: $p_{e} = p^{*} = 0.528 \cdot 500 = 264 kPa (not 101!)$ Ye step kyun? $M = 1$ par pressure signals upstream travel nahi kar sakte, isliye room ka $101$ kPa andar kabhi nahi pohanchta. Remaining drop $264 \to 101$ bahar expansion waves ki tarah hota hai.
Exit temperature: $T_{e} = T^{*} = \frac{2}{2.4} \cdot 300 = 250 K$ . Ye step kyun? Choked exit par $M = 1$ hai, isliye stagnation-to-static relation fixed factor $T^{*} / T_{0} = 2/ (γ + 1)$ par collapse ho jaata hai — next step mein sound speed ke liye $T^{*}$ chahiye.
Exit speed = local sound speed (kyunki $M = 1$ ): $V_{e} = a^{*} = 1.4 \cdot 287 \cdot 250 = 316.9 m/s$
Density aur mass flow (ceiling value): $ρ^{*} = \frac{264000}{287 \cdot 250} = 3.679 kg/m^{3}, \overset{m}{˙} = ρ^{*} A V_{e} = 3.679 \cdot 1 0^{- 3} \cdot 316.9 = 1.166 kg/s$ Ye step kyun? Same $\overset{m}{˙} = ρ A V$ recipe Cell B ki tarah, ab choked star-state values se; ye nozzle ka maximum mass flow hai.

Verify: $\overset{m}{˙} = 1.166 kg/s$ yahan vs $1.088 kg/s$ Cell B mein — choked flow larger hai, aur ye maximum hai (next example prove karta hai ye grow nahi karega). ✓

Cell E — Limiting case: vacuum mein exhaust

Worked example Cell E ·

$r \to 0$ Same tank, lekin ab $p_{b} = 0$ (vacuum chamber). Cell D se compare karte hue exit par kya change hota hai?

Forecast: Kya exit ab supersonic ho jaata hai jab room empty hai?

Gate: $r = 0/500 = 0 < 0.528$ → abhi bhi choked. Ye step kyun? Gate sirf care karta hai ki $r$ $0.528$ se neeche hai ya nahi. Zero usse neeche hai, isliye exit ke baare mein kuch nahi badlta.
Exit locked rehta hai: $M_{e} = 1$ , $p_{e} = 264 kPa$ , $T_{e} = 250 K$ , $\overset{m}{˙} = 1.166 kg/s$ — Cell D ke identical. Ye step kyun? Yahi choking ka pura point hai: sonic hone ke baad exit $p_{b}$ se deaf ho jaata hai. $264 \to 0$ tak extra expansion bilkul bahar hoti hai.

Verify: Mass-flow formula ka $M$ -factor $\frac{M}{( 1 + 0.2 M ^{2} ) ^{3}}$ $M = 1$ par maximise hota hai (wahan derivative zero). Kyunki exit $M = 1$ par pinned hai, $\overset{m}{˙}$ rise nahi kar sakta. Perfect vacuum bhi same $1.166 kg/s$ deta hai. ✓ (neeche numerically proven)

Mass-flow "ceiling" wo shape hai jo memory mein burn karni chahiye:

Intuition Figure s02 kaise padhein — curve par right-to-left chalo

Horizontal axis gate ratio $r = p_{b} / p_{0}$ hai, aur maine ise reversed draw kiya hai (left par 1, right par 0) taaki back pressure kam karna ka physical act tumhe steadily right ki taraf move kare — bottom ke saath black arrow follow karo.

Bilkul left se shuru karo ( $r = 1$ ): floor par flat, $\overset{m}{˙} = 0$ (ye Cell A hai).
Right chalo ( $p_{b}$ drop karo): black curve climb karta hai — unchoked regime, Cells B. Zyada pressure drop, zyada flow, bilkul garden hose ki tarah.
Tum red dashed vertical line at $r = 0.528$ par pohanchte ho: exit $M = 1$ par just hit ho gaya.
Iske past right chalte raho: curve flat aur red ho jaata hai — ek dead-level plateau. Wo red plateau hi choking hai. $p_{b}$ ko right par vacuum tak drop karo aur height kabhi nahi badlti.

Picture ka ek hi message: red flat top ek ceiling hai. $p_{b}$ ko $0.528$ se neeche karna tumhe zero extra mass flow deta hai — exactly isliye Cell D aur Cell E identical $1.166 kg/s$ dete hain.

Cell F — Real-world word problem

Worked example Cell F · applied

Ek pressurised nitrogen tank ( $γ = 1.4$ , $R = 297 J/kg\cdotK$ ) ek small thruster ko converging nozzle ke through feed karta hai jiska exit area $A = 2 cm^{2}$ hai. Tank: $p_{0} = 800 kPa$ , $T_{0} = 290 K$ . Ye space mein fire karta hai ( $p_{b} \approx 0$ ). Propellant per second kitna nikalta hai?

Forecast: Space mein choked hai ya nahi? Aur $\overset{m}{˙}$ chhota hai ya bada?

Gate: $r = 0/800 = 0 < 0.528$ → choked (space critical se kaafi neeche hai). Ye step kyun? Same universal gate; nitrogen ka $γ = 1.4$ hai isliye $0.528$ number abhi bhi apply hota hai.
Exit conditions: $p_{e} = 0.528 \cdot 800 = 422.5 kPa$ , $T_{e} = \frac{2}{2.4} \cdot 290 = 241.7 K$ . Ye step kyun? Choked matlab $M = 1$ , isliye exit pressure aur temperature $p_{0}, T_{0}$ ke fixed star-state fractions par locked hain — speed aur density ke liye dono chahiye.
Exit speed: $V_{e} = a^{*} = 1.4 \cdot 297 \cdot 241.7 = 316.9 m/s$ . Ye step kyun? $M_{e} = 1$ isliye exit speed hi local sound speed hai — Stagnation Properties ke liye ye "throat" state hai.
Density aur mass flow: $ρ^{*} = \frac{422500}{297 \cdot 241.7} = 5.885 kg/m^{3}, \overset{m}{˙} = 5.885 \cdot (2 \times 1 0^{- 4}) \cdot 316.9 = 0.373 kg/s$ Ye step kyun? $\overset{m}{˙} = ρ A V$ exit state ko actual kilograms-per-second propellant mein turn karta hai — physical answer jo thruster designer chahta hai.

Verify: Units kg/s tak chain karte hain ✓. Ek small thruster ke liye sensible (hundreds of grams per second). Neeche check kiya.

Cell G — Exam twist: target flow ke liye nozzle size karo

Worked example Cell G · choked mass-flow law ko area ke liye invert karo

Tumhe ek choked air nozzle chahiye jo $\overset{m}{˙} = 0.5 kg/s$ pass kare $p_{0} = 600 kPa$ , $T_{0} = 320 K$ se. Exit area $A$ kya hoga?

Forecast: $cm^{2}$ se bada hoga ya chhota? Pehle guess karo.

Stagnation terms mein likhey general mass-flow law se shuru karo (Mass Flow Rate & Choking se): $\overset{m}{˙} = A p_{0} \frac{γ}{R T _{0}} \frac{M}{( 1 + \frac{γ - 1}{2} M ^{2} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}}$ Kyunki bola gaya hai ye choked hai, $M = 1$ set karo. $M$ -factor fixed choked constant par collapse ho jaata hai: $\overset{m}{˙} = A p_{0} \frac{γ}{R T _{0}} (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}}$ Ye step kyun? Choked flow $M = 1$ fix karta hai, isliye $\overset{m}{˙}$ ab sirf $A, p_{0}, T_{0}, γ$ par depend karta hai — ek unknown $A$ ke siwa har quantity. Isse ye ek-line solve ban jaata hai.
Unknown area isolate karne ke liye rearrange karo: $A = \frac{m ˙}{p _{0} \frac{γ}{R T _{0}} ( \frac{2}{γ + 1} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}}$ Ye step kyun? Hume $A$ chahiye, isliye target $\overset{m}{˙}$ ko baaki sab se divide karte hain. Ye literally matrix ka "choked law invert karna" hai.
Air ke liye do constant factors evaluate karo. Pehle exponent: $\frac{γ + 1}{2 ( γ - 1 )} = \frac{2.4}{0.8} = 3$ . Phir $\frac{1.4}{287 \cdot 320} = 1.524 \times 1 0^{- 5} = 3.904 \times 1 0^{- 3}, (\frac{2}{2.4})^{3} = 0.5787$ Ye step kyun? $A$ touch karne se pehle har gas/reservoir constant pin down karo, taaki final division pure arithmetic ho.
Per unit area nozzle pass karta hai: $\frac{m ˙}{A} = p_{0} \frac{γ}{R T _{0}} (\frac{2}{γ + 1})^{3} = 600000 \cdot 3.904 \times 1 0^{- 3} \cdot 0.5787 = 1355 kg/(s\cdotm^{2})$ Ye step kyun? Ye "flow per square metre" exactly wahi hai jisme hum target divide karte hain.
Solve karo: $A = 0.5/1355 = 3.69 \times 1 0^{- 4} m^{2} = 3.69 cm^{2}$ . Ye step kyun? Demanded $\overset{m}{˙}$ ko per-square-metre flow se divide karne par required throat area milta hai — design answer.

Verify: $A$ wapas choked law mein daalo: $1355 \cdot 3.69 \times 1 0^{- 4} = 0.500 kg/s$ ✓. Units: $Pa \cdot \frac{1}{J/kg} \cdot m^{2} = kg/s$ (kyunki $J/kg = m^{2} / s^{2}$ ) ✓. Neeche numerically check kiya.

Cell H — Exam twist: alag gas ( $γ \neq = 1.4$ )

Worked example Cell H · monatomic gas, gate recompute karo

Helium ( $γ = 5/3$ ) $p_{0} = 500 kPa$ se $p_{b} = 250 kPa$ wale room mein flow karta hai. Kya ye choked hai? Helium ke liye critical ratio kya hai?

Forecast: Helium ka magic number air ke $0.528$ se bada hai ya chhota? Computing se pehle guess karo.

$0.528$ reuse mat karo — wo $γ = 1.4$ se derive hua tha. $γ = 5/3$ ke saath boxed critical-ratio formula se gate recompute karo: $\frac{p ^{*}}{p _{0}} = (\frac{2}{γ + 1})^{\frac{γ}{γ - 1}} = (\frac{2}{8/3})^{(5/3) / (2/3)} = (0.75)^{2.5} = 0.4871$ Ye step kyun? Gate value sirf $γ$ ka pure function hai. Gas badlo, number badlo — Cell H ka pura point yahi hai.
Helium ke liye gate apply karo: $r = 250/500 = 0.50$ . $0.487$ se compare karo. Kyunki $0.50 > 0.487$ → unchoked. Ye step kyun? Monatomic gas kam ratio par choke karta hai, isliye wahi $r$ jo air choke karta ( $0.50 < 0.528$ ) helium ko subsonic chhod deta hai.
Unchoked hone se, $p_{e} = p_{b} = 250 kPa$ ; exit Mach paane ke liye master relation helium ke $γ$ ke saath invert karo. Yahan $\frac{γ - 1}{2} = \frac{1}{3}$ aur $\frac{γ}{γ - 1} = 2.5$ : $\frac{p _{0}}{p _{e}} = \frac{500}{250} = 2 = (1 + \frac{1}{3} M_{e}^{2})^{2.5}$ $2.5$ -th root lo: $2^{1/2.5} = 1.3195$ , so $\frac{1}{3} M_{e}^{2} = 0.3195 \Rightarrow M_{e} = 0.9585 = 0.979$ . Ye step kyun? Cell B jaisi hi inversion machinery, lekin har $γ$ -dependent exponent change ho gaya. Note karo $M_{e} = 0.979 < 1$ — barely subsonic, consistent hai kyunki $r$ helium gate ke just upar hai.

Verify: $0.487 < 0.50$ ⇒ unchoked, aur inversion $M_{e} < 1$ deta hai ✓. Same setup air mein ( $r = 0.50 < 0.528$ ) choked hota — proof ki gas matter karta hai aur gate recompute karna zaroori hai. Neeche check kiya.

Recall Main kis cell mein hoon? (self-test)

$r = p_{b} / p_{0}$ compute karo, phir jawab do. Agar $r = 1$ toh kya flow karta hai? ::: Kuch nahi; $\overset{m}{˙} = 0$ (Cell A). Agar $r$ air ke liye $0.528$ se thoda neeche hai, exit Mach? ::: $M_{e} = 1$ , choked (Cell D). Air ke liye choked exit pressure? ::: $p_{e} = 0.528 p_{0}$ , $p_{b}$ se independent. $p_{b}$ ko $0.2$ se $0$ karne par $\overset{m}{˙}$ change hoga? ::: Nahi — exit $M = 1$ par pinned hai (Cell E). General critical-ratio formula? ::: $p^{*} / p_{0} = (2/ (γ + 1))^{γ / (γ - 1)}$ . Monatomic gas $γ = 5/3$ ke liye critical ratio? ::: $(0.75)^{2.5} \approx 0.487$ (Cell H). Pehle se choked hone par zyada $\overset{m}{˙}$ pass karne ke liye kya change karna hoga? ::: Bada area $A$ ya zyada $p_{0}$ — lower $p_{b}$ nahi (Cell G).

Connections

Converging-Diverging (de Laval) Nozzle — parent; $M = 1$ ceiling todne ke liye diverging section add karo.
Isentropic Flow Relations — master relation jo Cells B, C, H mein invert hoti hai aur critical ratio derive karne mein use hoti hai.
Mass Flow Rate & Choking — choked $\overset{m}{˙}$ formula jo Cells D–G mein invert hota hai.
Speed of Sound — kyun exit speed $M = 1$ par $a^{*}$ ke equal hoti hai.
Stagnation Properties — $p_{0}, T_{0}$ reference state jo har example feed karta hai.
Normal Shock Waves — kya hota hai agar tum diverging section attach karo aur over-compress karo.