Converging-diverging (de Laval) nozzle — subsonic, supersonic flow: Worked examples

3.1.10 · D3 · Physics › Compressible Flow & Aerodynamics › Converging-diverging (de Laval) nozzle — subsonic, supersoni

Kisi bhi example se pehle, hum char symbols pin karte hain jinhe ye poori page use karti hai — plain words aur pictures mein.

Definition Pehle ye char symbols khud ke karo

Stagnation (reservoir) pressure $p_{0}$ : gas ka wo pressure jab wo aram mein ho, tank/combustion chamber ke andar kaafi gehraai mein jo nozzle ko feed karta hai, jahan flow speed $V = 0$ ho. Socho ek bada shant tank; $p_{0}$ uski wall ka gauge hai. Kyunki wahan gas move nahi kar rahi, uski saari energy "stored" hai; jaise-jaise wo nozzle mein accelerate hoti hai, local pressure $p$ gir jaata hai $p_{0}$ se neeche. To $p_{0}$ fixed ceiling hai; $p$ wo hai jo kuch energy ke speed ban jaane ke baad bachi hai.
Stagnation temperature $T_{0}$ : temperature ke liye wahi idea — us shant tank mein temperature jahan $V = 0$ ho. Jaise-jaise gas speed up karti hai wo cool hoti hai, isliye local $T < T_{0}$ .
Throat / sonic area $A^{*}$ : cross-sectional area jahan flow exactly sonic ho, $M = 1$ . Ek chalte de Laval nozzle mein ye physical throat hota hai (sabse sankra point). Hum har doosri area ko iske against measure karte hain ratio $A / A^{*}$ ke roop mein — isliye master relation (AR) neeche $A / A^{*}$ ke terms mein likhi hai, raw area mein nahi.
Back pressure $p_{b}$ : wo environment ka pressure jisme nozzle exhaust karta hai — rocket ke liye atmosphere, ya steam turbine ke liye condenser. Ye nozzle ke bahar se, exit plane par, impose hota hai. $p_{b}$ ko kam karna exit par zyada sucking jaisa hai; ye wo akela dial hai jo decide karta hai ki Nature kaunsa flow regime choose kare.

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

Picture padho: baayi taraf ka shant tank gas ko $p_{0}, T_{0}$ par rakhta hai ( $V = 0$ ); hourglass duct use accelerate karta hai; daayein taraf ka environment $p_{b}$ impose karta hai. Is page par baaki sab kuch in chaar labels ke beech bookkeeping hai.

Scenario matrix

Har nozzle question asliyat mein inhi case classes mein se ek hai. Daayaan column us example ka naam batata hai jo use cover karta hai.

#	Case class	Distinguishing feature	Covered by
A	Area ratio ka subsonic root	$A / A^{*} > 1$ , $M < 1$ answer lo	Ex 1
B	Area ratio ka supersonic root	$A / A^{*} > 1$ , $M > 1$ answer lo	Ex 2
C	Degenerate area: $A / A^{*} = 1$	throat, $M = 1$ exactly	Ex 3
D	Limiting input: $A / A^{*} \to \infty$	vacuum / infinite expansion	Ex 3
E	Choking test (back pressure ka sign)	kya $p_{b}$ critical se neeche hai?	Ex 4
F	Not choked (venturi regime)	high $p_{b}$ , throat kabhi $M = 1$ nahi pahuncha	Ex 5
G	Diverging part ke andar normal shock	$p_{b}$ shock-at-exit aur design ke beech	Ex 6
H	Real-world word problem	exit state se rocket thrust	Ex 7
I	Exam twist / trap	converging-only nozzle "supersonic ho jaayega?"	Ex 8

Teen master relations jinhe har example plug karta hai. Ye isentropic ratios hain; har ek kehti hai "Mach $M$ par pahunchne par ye quantity apne calm-tank value se kitna gir gayi?"

Upar ki picture (AR) padhne ki key hai: $1$ se bade kisi bhi area ratio ke liye, ek horizontal line curve ko do baar cross karti hai — ek baar falling (subsonic) branch par aur ek baar rising (supersonic) branch par. Yahi ek fact cases A aur B generate karta hai.

Example 1 — Subsonic root (cell A)

Forecast: solve karne se pehle andaza lagao — kya $M$ $1$ ke paas hoga ya $0$ ke paas? Aur kya $p / p_{0}$ $1$ ke paas hoga (barely expanded) ya tiny?

Steps.

(AR) likho $A / A^{*} = 2$ ke saath: solve karo $\frac{1}{M} [\frac{2}{2.4} (1 + 0.2 M^{2})]^{3} = 2$ . Ye step kyun? (AR) wahi akela equation hai jo geometry ko Mach se jodta hai; $2 > 1$ to do roots exist karte hain aur humein subsonic wala choose karna hai kyunki poora flow subsonic bataya gaya hai.
Numerically subsonic root hai $M \approx 0.306$ . Ye step kyun? Figure mein curve ki falling branch par, $2.0$ par horizontal line ek chhote Mach number se milti hai — ye hamara forecast match karta hai ki ek wide, non-choked venturi sirf mildly speed up hoti hai.
Pressure from (P): $\frac{p _{0}}{p} = (1 + 0.2 \cdot 0.30 6^{2})^{3.5} \approx 1.067$ , to $p / p_{0} \approx 0.937$ . Ye step kyun? Low Mach matlab thodi kinetic energy li gayi, isliye pressure barely gira — gas abhi bhi reservoir pressure $p_{0}$ ke kaafi paas hai.

Verify: plug $M = 0.306$ back into (AR): $\frac{1}{0.306} [0.833 (1 + 0.2 \cdot 0.0936)]^{3} = 3.27 \cdot 0.611 \approx 2.0$ ✅. Pressure ratio $0.937$ $1$ ke kaafi paas hai, consistent hai "subsonic aur barely expanded" ke saath. Units: sab ratios, dimensionless ✅.

Example 2 — Supersonic root (cell B)

Forecast: wahi area ratio — kya tumhe Example 1 jaisa Mach milega, ya kuch $1$ se bada?

Steps.

(AR) use karo $A / A^{*} = 2$ ke saath phir se, lekin supersonic root lo ( $M > 1$ ). Ye step kyun? Choked throat ke baad diverging section $d A / A = (M^{2} - 1) d V / V$ follow karta hai $M > 1$ ke saath, isliye area ka widening flow ko accelerate karta hai — Nature curve ki rising branch par hai.
Numerically $M \approx 2.197$ . Ye step kyun? Figure mein $2.0$ par horizontal line rising branch ko bhi strike karti hai — same area, bilkul alag speed. Ye "two-solution fact" concrete ho gaya.
Pressure: $\frac{p _{0}}{p} = (1 + 0.2 \cdot 2.19 7^{2})^{3.5} = (1.966)^{3.5} \approx 10.7$ , to $p / p_{0} \approx 0.0935$ . Ye step kyun? Supersonic flow ne gas se bahut zyada kinetic energy drain kar li hai, isliye static pressure roughly ek-tenth reservoir tak gir gayi — utterly unlike Example 1 ke $0.937$ se.

Verify: re-plug $M = 2.197$ : $\frac{1}{2.197} [0.833 (1 + 0.2 \cdot 4.827)]^{3} = 0.455 \cdot 4.40 \approx 2.0$ ✅. Note karo dono examples same equation solve karte hain $2.0$ tak opposite branches se — yahi exactly cases A aur B ka twin hona hai.

Example 3 — Degenerate + limiting inputs (cells C & D)

Forecast: (a) ke liye, sirf ek answer bachta hai — kaunsa? (b) ke liye, kya $M$ hamesha badhta rehega, ya koi ceiling hai?

Steps.

(a) (AR) mein $A / A^{*} = 1$ set karo. Curve ki minimum value $1$ hai, sirf $M = 1$ par pahunchi. Ye step kyun? Minimum par do roots ek mein merge ho jaate hain — subsonic aur supersonic branches touch karte hain. Ye throat hai, wahi jagah jise humne sonic area $A^{*}$ define kiya; master law force karta hai $d A = 0$ jab $M = 1$ .
To sirf solution hai $M = 1$ . Koi doosra Mach area ratio $1$ ke saath nahi hai. Ye step kyun? Ye "two-solution" ambiguity ko exactly throat par khatam karta hai — ek degenerate, single-root case.
(b) (AR) lo jab $M \to \infty$ : bracket $M^{2}$ jaisa grow karta hai, power $3$ tak raise hoga, $M^{6}$ dega, $1/ M$ se multiply → $A / A^{*} \sim M^{5}$ . To $A / A^{*} \to \infty$ chahiye $M \to \infty$ . Ye step kyun? Infinitely large exit infinite Mach degi — lekin (T) se temperature, $T / T_{0} = 1/ (1 + 0.2 M^{2}) \to 0$ , to gas freeze ho jaati hai; real gases liquefy ho jaati hain aur ye sirf ek limit hai, kabhi reach nahi hoti.
Lekin maximum exit velocity finite hai: $V_{ma x} = 2 c_{p} T_{0}$ — wo speed jo tab milti hai jab tank ki saari thermal energy $c_{p} T_{0}$ kinetic energy $\frac{1}{2} V^{2}$ ban jaaye (energy line $c_{p} T_{0} = c_{p} T + \frac{1}{2} V^{2}$ mein $T \to 0$ set karo). $M \to \infty$ par bhi, $V$ capped hai kyunki local sound speed $a = γ R T \to 0$ . Ye step kyun? "Mach $\to \infty$ " (kyunki sound speed vanish hoti hai) aur "speed $\to \infty$ " (jo nahi hota) mein fark karta hai. Yahan $c_{p}$ wahi specific heat hai jo upar introduce ki, $R$ se $c_{p} = γ R / (γ - 1)$ se bandi.

Verify: (a) plug $M = 1$ into (AR): $\frac{1}{1} [0.833 \cdot 1.2]^{3} = [1.0]^{3} = 1.0$ ✅ minimum confirmed. (b) at $M = 5$ , (AR) $= \frac{1}{5} [0.833 (1 + 0.2 \cdot 25)]^{3} = 0.2 \cdot 125 = 25$ — already huge, confirming $A / A^{*}$ $M$ ke saath blow up karta hai ✅.

Example 4 — Choking test: kya back pressure kaafi low hai? (cell E)

Forecast: choking tab hoti hai jab throat pressure $p_{0}$ ke critical fraction tak gire. Andaza lagao: kya $3/8$ us threshold se upar hai ya neeche?

Steps.

$M = 1$ par critical pressure ratio: $\frac{p ^{*}}{p _{0}} = (\frac{2}{γ + 1})^{3.5} = (0.8333)^{3.5} \approx 0.528$ . Ye step kyun? Ye (P) hai $M = 1$ par evaluate hua; ye wo pressure hai jis par throat must fall kare sonic hone se pehle. Dekho Choked Flow & Mass Flow Limit.
To $p^{*} = 0.528 \times 8 = 4.23 bar$ . Ye step kyun? Dimensionless threshold ko real pressure mein convert karta hai $p_{b}$ se compare karne ke liye.
Compare karo: $p_{b} = 3 bar < p^{*} = 4.23 bar$ . Ye step kyun? Agar environment critical throat pressure se neeche ka exit pressure demand kare, to throat already $M = 1$ hit kar chuki hai — flow choked hai, aur $\overset{m}{˙}$ $\overset{m}{˙}_{ma x}$ par lock hai.
Conclusion: choked. $p_{b}$ ko $3$ se $2$ ya $1$ bar tak lower karna mass flow nahi badhayega. Ye step kyun? Ye cell E ka heart hai sign-based decision — sab kuch is par depend karta hai ki back pressure $0.528 p_{0}$ ke kis side par hai.

Verify: $(2/2.4)^{3.5}$ : $2/2.4 = 0.83333$ ; $0.8333 3^{3.5} = e^{3.5 l n 0.83333} = e^{3.5 (- 0.18232)} = e^{- 0.6381} = 0.5283$ ✅. Aur $0.528 \cdot 8 = 4.23$ bar, aur $3 < 4.23$ ✅ → choked.

Example 5 — Not choked: venturi regime (cell F)

Forecast: $p_{b}$ itna $p_{0}$ ke paas hai, kya throat sonic hai? Nozzle mein fastest flow kahan hai?

Steps.

Choke check: $p_{b} / p_{0} = 7/8 = 0.875 > 0.528$ . Ye step kyun? Critical ratio se upar throat $M = 1$ reach nahi kar sakta — nozzle ek subsonic venturi ki tarah act karta hai, supersonic accelerator ki tarah nahi.
Kyunki sab subsonic hai, geometric throat kabhi $M = 1$ reach nahi karta, to koi station sonic nahi aur label $A^{*}$ (sonic area) ka is flow mein koi physical location nahi — ye sirf (AR) ke liye bookkeeping reference ke roop mein exist karta hai. Flow converging part mein speed up hoti hai, narrowest point par peak (subsonic) karta hai, phir diverging part mein slow down hota hai. Ye step kyun? Subsonic flow obey karta hai $d A < 0 \Rightarrow d V > 0$ ; diverging section mein $d A > 0$ hai isliye ye decelerate karta hai — venturi ki tarah pressure recovery.
Exit pressure back pressure ke barabar: $p_{e} = p_{b} = 7$ bar. (P) se: $\frac{p _{0}}{p _{e}} = \frac{8}{7} = 1.143 = (1 + 0.2 M_{e}^{2})^{3.5}$ . Ye step kyun? Fully subsonic flow mein exit pressure set hota hai environment ke back pressure se; $M_{e}$ padhne ke liye (P) invert karo.
Solve: $1 + 0.2 M_{e}^{2} = 1.14 3^{1/3.5} = 1.0388$ , to $M_{e}^{2} = 0.194$ , $M_{e} \approx 0.441$ . Ye step kyun? Ek modest subsonic exit confirm karta hai — highest speed narrowest point par baith gayi, exit par nahi, supersonic case se bilkul ulta.

Verify: $1.14 3^{1/3.5} = e^{(1/3.5) l n 1.143} = e^{0.2857 \cdot 0.1335} = e^{0.03816} = 1.0389$ ; $(1.0389 - 1) /0.2 = 0.1944$ ; $0.1944 = 0.441$ ✅. Sanity: $M_{e} < 1$ , aur $0.875 > 0.528$ ⇒ genuinely un-choked ✅.

Example 6 — Diverging section mein normal shock (cell G)

Forecast: shock ke across, kya flow subsonic ho jaati hai ya supersonic rehti hai? Kya pressure badhta hai ya ghadta hai?

Steps.

Post-shock Mach normal-shock relation se: $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{1}^{2}}{γ M _{1}^{2} - \frac{γ - 1}{2}} .$ Ye step kyun? Shock ek discontinuous, non-isentropic jump hai — smooth (T)/(P)/(AR) relations iske across apply nahi hote; humein dedicated conservation-across-shock formula chahiye.
Plug $M_{1} = 2$ : numerator $= 1 + 0.2 \cdot 4 = 1.8$ ; denominator $= 1.4 \cdot 4 - 0.2 = 5.4$ . To $M_{2}^{2} = 1.8/5.4 = 0.3333$ , $M_{2} \approx 0.577$ . Ye step kyun? Confirm karta hai ki flow subsonic ho gayi — ek shock hamesha supersonic flow ko $M = 1$ se neeche le jaata hai.
Static pressure ratio: $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) = 1 + \frac{2.8}{2.4} (4 - 1) = 1 + 1.1667 \cdot 3 = 4.5$ . Ye step kyun? Shock ke downstream pressure sharply jump up karta hai — isliye shock ke baad ki flow ek decelerating, higher-pressure subsonic stream hai jo back pressure $p_{b}$ se match karta hai.
Shock ke baad bachi diverging section ek subsonic diffuser ki tarah act karti hai (aur decelerate karti hai), $p_{b}$ se exit par milti hai. Ye step kyun? Shock case ko Ex 5 ke venturi behaviour se jodta hai — wahi subsonic rule, bas mushkil raaste se pahunche.

Figure dhyan se padho: white walls throat tak pinch karte hain (dotted line, $M = 1$ ), flow supersonically accelerate hoti hai (amber region, $M_{1} = 2.0$ ), phir vertical amber bar normal shock hai. Dekho kaise labels us bar ke across flip karte hain — baayen supersonic $M_{1} = 2.0$ se daayein subsonic $M_{2} = 0.577$ tak, static pressure $4.5$ se multiply hoke. Amber bar ke downstream sab kuch Example 5 ke subsonic venturi ki tarah behave karta hai.

Verify: $M_{2}^{2} = 1.8/5.4 = 0.3333$ ✅, $0.3333 = 0.5774$ ✅. $p_{2} / p_{1} = 1 + (2.8/2.4) \cdot 3 = 4.5$ ✅. Sanity: $M_{2} < 1 < M_{1}$ aur $p_{2} > p_{1}$ — dono compression shock ke signatures ✅.

Example 7 — Real-world word problem: rocket exit state (cell H)

Forecast: $A_{e} / A^{*} = 4$ ke saath (Example 1 ke $2$ se bada), kya $M_{e}$ $2.94$ se upar hoga ya neeche? Kya exit hot hogi ya cold?

Steps.

(AR) solve karo $A / A^{*} = 4$ ke saath, supersonic root: $M_{e} \approx 2.94$ . Ye step kyun? Rocket supersonic chalne ke liye design hai (thrust high exit speed demand karta hai), isliye hum rising branch lete hain — Ex 1 ke $2.0$ se bada area ratio ⇒ bada Mach, forecast match karta hai.
Relation (T) se exit temperature: $T_{0} / T_{e} = 1 + 0.2 M_{e}^{2} = 1 + 0.2 \cdot 8.644 = 2.729$ . To $T_{e} = 3000/2.729 \approx 1099 K$ . Ye step kyun? Kinetic energy thermal energy se kharidi jaati hai; gas dramatically cool hoti hai accelerate hote waqt — $3000$ K calm chamber ( $T_{0}$ ) mein se karib $1100$ K tak.
(P) se exit pressure: $p_{0} / p_{e} = (2.729)^{3.5} \approx 34.1$ , to $p_{e} = 50/34.1 \approx 1.47 bar$ . Ye step kyun? Design condition ise maximum thrust ke liye ambient se match karta hai; ye temperature ratio ka (P) counterpart hai.
Exit velocity: $a_{e} = γ R T_{e} = 1.4 \cdot 320 \cdot 1099 = 492, 300 \approx 701.6 m/s$ , phir $V_{e} = M_{e} a_{e} = 2.94 \cdot 701.6 \approx 2063 m/s$ . Ye step kyun? $V = M a$ abstract Mach ko physical exhaust speed mein convert karta hai jo thrust produce karta hai — nozzle ka poora point.

Verify: $M_{e}^{2} = 8.644$ , $1 + 0.2 \cdot 8.644 = 2.729$ ; $T_{e} = 3000/2.729 = 1099.3$ K ✅. $2.72 9^{3.5} = e^{3.5 l n 2.729} = e^{3.5 \cdot 1.0040} = e^{3.514} = 33.6$ (≈34) ✅. $a_{e} = 1.4 \cdot 320 \cdot 1099.3 = 701.6$ m/s; $V_{e} = 2.94 \cdot 701.6 = 2062.7$ m/s ✅. Units: $(J/kg) = m^{2} / s^{2} = m/s$ ✅.

Example 8 — Exam trap: "converging nozzle ko supersonic banao" (cell I)

Forecast: verdict guess karo — kya zyada suction eventually sound barrier todega?

Steps.

Purely converging nozzle mein sabse chhota area exit khud hai; $d A < 0$ hamesha, kabhi $d A = 0$ nahi sivaay bilkul exit par. Ye step kyun? Master law $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ $M = 1$ cross karna forbid karta hai kahin bhi sivaay jahan $d A = 0$ , isliye sirf jagah jahan $M = 1$ ho sakta hai wo exit hai.
Exit best jo reach kar sakta hai wo $M = 1$ hai (choked). (P) mein $M_{e} = 1$ set karo: $p_{e} / p_{0} = (1 + 0.2 \cdot 1)^{3.5}^{- 1} = (1.2)^{- 3.5} = 0.528$ , to $p_{e} = 0.528 \times 10 = 5.28 bar$ . Ye step kyun? Jab exit $M = 1$ par ho nozzle choked hai; wo critical exit pressure $5.28$ bar ceiling hai — dekho Choked Flow & Mass Flow Limit.
Ab $p_{b}$ $5.28 bar$ se neeche karo. Nozzle internally respond nahi kar sakta — exit $M = 1$ par aur $p_{e} = 5.28 bar$ par rehta hai; $p_{b}$ tak extra expansion nozzle ke bahar hoti hai expansion fans ki tarah (ek under-expanded jet). Ye step kyun? Koi diverging section exist nahi karta jo $M > 1$ ke liye $d A > 0$ satisfy kare, isliye andar supersonic flow geometrically impossible hai; gas sirf khule mein expand kar sakti hai.
Verdict: disproved. Ek converging-only nozzle exactly sonic exit par max out karti hai ( $M_{e} = 1$ , $p_{e} = 5.28$ bar); koi bhi $p_{b}$ neeche expansion ko sirf bahar shift karta hai. Supersonic flow ke liye de Laval nozzle ka diverging section chahiye. (Contrast karo Steam Turbine Nozzles se, jo exactly isi reason ke liye divergence add karte hain.) Ye step kyun? Yahi poora reason hai de Laval shape ka exist karna — cell I topic ka signature trap hai.

Verify: ceiling pressure $(1/1.2)^{3.5} = (0.8333)^{3.5} = 0.528$ , times $10 = 5.28$ bar ✅ (Ex 4 ke critical value jaisa ratio — hona hi tha — dono " $M = 1$ " hain). Koi formula all-shrinking area se $M_{e} > 1$ produce nahi karta, impossibility confirm karta hai.

Recall Matrix par quick self-test

Same area ratio do Mach numbers deta hai — kaunsa choose karte ho aur kyun? ::: Subsonic root agar poora flow subsonic hai (high $p_{b}$ , venturi); supersonic root agar throat choked hai aur diverging section gas accelerate karta hai (low $p_{b}$ , design). Back pressure $p_{b} = 0.6 p_{0}$ ke saath $γ = 1.4$ — choked hai ya nahi? ::: Critical ratio $0.528$ hai; $0.6 > 0.528$ , isliye not choked (throat par abhi bhi subsonic). $M_{1} = 2$ par normal shock ke baad flow sound se faster hai ya slower? ::: Slower — $M_{2} \approx 0.577 < 1$ ; har shock supersonic flow ko subsonic tak jump karta hai aur pressure badhata hai. Suction akele converging nozzle ko supersonic kyun nahi bana sakta? ::: Area sirf shrink karti hai, isliye $M = 1$ sirf exit par ho sakta hai (choked); koi $d A > 0$ region nahi, to $M > 1$ geometrically impossible hai.