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

3.1.10 · D4 · Physics › Compressible Flow & Aerodynamics › Converging-diverging (de Laval) nozzle — subsonic, supersoni

Poori exercise mein, jab tak problem alag na kahe, air ke liye $γ = 1.4$ lo.

Definition Is page par use hone wali Notation (ek baar, yahan define ki gayi)

$V$ = kisi point par gas ki speed (m/s).
$a$ = local speed of sound, woh speed jis par ek choti si pressure ripple travel karti hai, $a = γ R T$ . Air ke liye $R = 287 \frac{J}{kg K}$ (specific gas constant) aur $γ = c_{p} / c_{v} = 1.4$ .
$M = \frac{V}{a}$ = Mach number: gas kitni guna faster than sound chal rahi hai. $M < 1$ subsonic, $M = 1$ sonic, $M > 1$ supersonic. Dekho Mach Number.
$A$ = kisi point par tube ki cross-sectional area.
$A^{*}$ = throat / sonic area — woh area jahan $M = 1$ hota hai. Star superscript ka matlab hamesha "sonic point par ki value" hai.
Subscript $0$ = stagnation value: jo still reservoir mein measure hoga jahan $V = 0$ (yani $T_{0}$ , $p_{0}$ reservoir temperature aur pressure hain).
$p^{*}, T^{*}, ρ^{*}, a^{*}$ = pressure, temperature, density, sound speed throat par ( $M = 1$ ).
$p_{b}$ = back pressure, us surrounding ka pressure jismein gas exhaust hoti hai.

Level 1 — Recognition

(Kya tum rule ko naam de sakte ho aur uska sign padh sakte ho?)

L1.1 Master law $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ mein, ek gas subsonic hai ( $M = 0.5$ ) aur hum chahte hain ki woh speed up kare ( $d V > 0$ ). Kya area badhega ya ghatega? Ye everyday mein kaunsi device se match karta hai?

Recall Solution L1.1

Bracket ka sign calculate karo: $M^{2} - 1 = 0.25 - 1 = - 0.75 < 0$ . $d V > 0$ (speed up) ke liye, right side negative hai, isliye $d A < 0$ → area ko ghattna chahiye (converging). Ye garden hose / venturi se match karta hai: ise dabao aur paani speed up ho jaata hai.

L1.2 Ek de Laval nozzle design condition par chal raha hai. $M = 1$ kahan located hai, aur woh sirf wahan hi kyun reh sakta hai?

Recall Solution L1.2

$M = 1$ sirf throat par (minimum area par) hota hai. Karan: master law mein $M = 1$ set karo → $(M^{2} - 1) = 0$ , isliye $\frac{d A}{A} = 0$ . Zero fractional area change ka matlab hai ki area momentarily flat hai — ek minimum — jo exactly throat hai.

L1.3 Sach ya jhooth: "Kisi bhi area ratio $A / A^{*} > 1$ ke liye exactly ek possible Mach number hota hai." Explain karo.

Recall Solution L1.3

Jhooth. $A / A^{*}$ curve $M = 1$ par minimum $1$ tak girti hai aur dono sides par badhti hai. Isliye har $A / A^{*} > 1$ par do Mach numbers hote hain: ek subsonic ( $M < 1$ ) aur ek supersonic ( $M > 1$ ). Nature kaun sa pick karti hai yeh back pressure par depend karta hai. Dekho Isentropic Flow Relations.

Level 2 — Application

(Isentropic formulas mein sahi se plug karo.)

L2.1 Air $M = 2.0$ par flow kar rahi hai. Temperature ratio $T_{0} / T$ aur pressure ratio $p_{0} / p$ nikalo.

Recall Solution L2.1

KYA: stagnation ratios use karo. KYUN: ye reservoir (still) state ko moving state se energy + isentropic conservation ke zariye connect karte hain. $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2} = 1 + 0.2 \times 4 = 1.8$ $\frac{p _{0}}{p} = (1.8)^{\frac{1.4}{0.4}} = 1. 8^{3.5} \approx 7.82$ Toh static pressure reservoir pressure ka lagbhag $1/7.82 \approx 0.128$ hai.

L2.2 Ek reservoir mein air $p_{0} = 8$ bar par hai. Nozzle choked hai. Throat (critical) pressure $p^{*}$ kya hai?

Recall Solution L2.2

YEH FORMULA KYUN: "choked" ka matlab hai throat ne exactly $M = 1$ reach kar liya hai. Toh hum general pressure ratio $\frac{p _{0}}{p} = (1 + \frac{γ - 1}{2} M^{2})^{γ / (γ - 1)}$ lete hain aur $M = 1$ set karte hain — woh special case critical (starred) pressure hai. Yeh hume woh particular throat pressure batata hai jis par choking shuru hoti hai. $M = 1$ par: $\frac{p ^{*}}{p _{0}} = (1 + \frac{γ - 1}{2})^{- \frac{γ}{γ - 1}} = (\frac{2}{γ + 1})^{\frac{γ}{γ - 1}} = (\frac{2}{2.4})^{3.5} \approx 0.5283.$ Is tarah $p^{*} = 0.5283 \times 8 \approx 4.23$ bar. Dekho Choked Flow & Mass Flow Limit.

L2.3 Reservoir temperature $T_{0} = 600$ K, air. Throat temperature $T^{*}$ aur throat speed of sound $a^{*}$ nikalo.

Recall Solution L2.3

RATIO KO INVERT KYUN KARTE HAIN: formula $T_{0} / T$ deta hai (still-reservoir over local), lekin hum $T_{0}$ jaante hain aur local throat $T^{*}$ chahte hain. Isliye hum $T^{*} = T_{0} / (T_{0} / T^{*})$ rearrange karte hain — jaani hui reservoir temperature ko $M = 1$ par evaluate ki gayi ratio se divide karte hain. $M = 1$ par: $\frac{T _{0}}{T ^{*}} = 1 + 0.2 \times 1 = 1.2$ , isliye $T^{*} = 600/1.2 = 500$ K. Wahan speed of sound: $a^{*} = γ R T^{*} = 1.4 \times 287 \times 500 = 200900 \approx 448 m/s$ . Kyunki throat par $M = 1$ hai, flow speed equals $a^{*} \approx 448$ m/s. Dekho Speed of Sound in a Gas.

Level 3 — Analysis

(Competing effects compare karo; sahi branch chuno.)

L3.1 Ek nozzle ka exit area $A_{e} = 4 A^{*}$ hai, air, supersonic flow. Exit Mach number $M_{e}$ aur ratio $p_{0} / p_{e}$ nikalo.

Recall Solution L3.1

KYA: $A / A^{*} = 4$ ko $M$ ke liye solve karo. SUPERSONIC ROOT KYUN: choked throat ke baad diverging section mein $d A > 0$ ke saath $M > 1$ hota hai, isliye hum supersonic root lete hain. $\frac{1}{M} [\frac{2}{2.4} (1 + 0.2 M^{2})]^{3} = 4$ ko numerically solve karne par $M_{e} \approx 2.94$ milta hai. Phir $\frac{p _{0}}{p _{e}} = (1 + 0.2 \times 2.9 4^{2})^{3.5} = (1 + 1.729)^{3.5} \approx 33.0$ , isliye $p_{e} \approx p_{0} /33$ .

L3.2 Figure 1 dekho. Same area ratio $A / A^{*} = 2$ ke liye, do Mach numbers list karo aur batao ki kaun si physical situation mein se kaun sa select hota hai.

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

Recall Solution L3.2

$A / A^{*} = 2$ solve karne par ek subsonic root $M \approx 0.31$ aur ek supersonic root $M \approx 2.20$ milta hai.

Subsonic root ( $0.31$ ) tab hota hai jab back pressure high ho: diverging part diffuser ki tarah kaam karta hai aur flow ko slow karta hai (venturi mode).
Supersonic root ( $2.20$ ) tab hota hai jab back pressure design value par ya usse neeche ho: diverging part already-sonic throat flow ko accelerate karta hai. Nature kaun sa choose karti hai yeh exit back pressure $p_{b}$ se tay hota hai.

L3.3 Ek converging-only nozzle ko reservoir se feed kiya ja raha hai aur back pressure ko baar baar neeche giraya ja raha hai. Achievable maximum exit Mach number kya hai, aur yeh use kyun exceed nahi kar sakta?

Recall Solution L3.3

Maximum $M_{e} = 1$ hai. Converging-only duct mein sabse choti area exit khud hai. Master law $M = 1$ ko sirf wahan cross karne deta hai jahan $d A = 0$ ho. Kyunki koi diverging ( $d A > 0$ ) section nahi hai, flow zyada se zyada exit par sonic reach kar sakta hai; $p_{b}$ aur neeche girane se sirf choking hoti hai — exit $M = 1$ par rehta hai aur $\overset{m}{˙}$ maximum par lock ho jaata hai. Dekho Choked Flow & Mass Flow Limit.

Level 4 — Synthesis

(Mass flow, shocks aur geometry ko combine karo.)

L4.1 Reservoir: $p_{0} = 10$ bar, $T_{0} = 500$ K, air. Throat area $A^{*} = 5 cm^{2} = 5 \times 1 0^{- 4} m^{2}$ . Nozzle choked hai. Maximum mass flow rate $\overset{m}{˙}_{m a x} = ρ^{*} A^{*} a^{*}$ nikalo.

Recall Solution L4.1

Step 1 — throat temperature: $T^{*} = T_{0} /1.2 = 500/1.2 = 416.7$ K. Step 2 — throat pressure: $p^{*} = 0.5283 \times 10 \times 1 0^{5} = 5.283 \times 1 0^{5}$ Pa. Step 3 — throat density (ideal gas $ρ = p / R T$ ): $ρ^{*} = \frac{5.283 \times 1 0 ^{5}}{287 \times 416.7} = 4.418 kg/m^{3}$ . Step 4 — throat sound speed: $a^{*} = 1.4 \times 287 \times 416.7 = 409.2$ m/s (yeh $V$ ke equal hai kyunki $M = 1$ ). Step 5 — combine: $\overset{m}{˙}_{m a x} = ρ^{*} A^{*} a^{*} = 4.418 \times 5 \times 1 0^{- 4} \times 409.2 \approx 0.904 kg/s$ . $p_{b}$ ko choke se neeche girane par yeh nahi badhta. Dekho Conservation of Mass (Continuity).

L4.2 Air ek normal shock se $M_{1} = 2.0$ par enter karti hai (jo diverging section mein hai). Normal-shock relations use karo: $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{1}^{2}}{γ M _{1}^{2} - \frac{γ - 1}{2}}, \frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1),$ downstream Mach number $M_{2}$ aur static pressure jump $p_{2} / p_{1}$ nikalo.

Recall Solution L4.2

SHOCK KYUN AATA HAI: jab back pressure full supersonic exit ke liye too high ho lekin pure subsonic flow ke liye too low ho, tab gas ko abruptly subsonic mein wapas jump karna padta hai — ek normal shock. $M_{2}^{2} = \frac{1 + 0.2 \times 4}{1.4 \times 4 - 0.2} = \frac{1.8}{5.4} = 0.3333 \Rightarrow M_{2} \approx 0.577.$ $\frac{p _{2}}{p _{1}} = 1 + \frac{2 \times 1.4}{2.4} (4 - 1) = 1 + 1.1667 \times 3 = 4.50.$ Toh flow $M_{2} \approx 0.58$ par aa jaati hai aur static pressure shock ke across $4.5 \times$ badh jaata hai.

Edge-case warning — stagnation pressure shock ke through conserved NAHI hota. Shock irreversible hota hai (entropy badhti hai), isliye jab static pressure upar jaata hai, stagnation pressure girata hai: $p_{0, 2} < p_{0, 1}$ . $M_{1} = 2$ ke liye ratio $p_{0, 2} / p_{0, 1} \approx 0.721$ hai — lagbhag $28%$ loss. Yeh matter karta hai jab tum ek flow path ke along relations chain karo: shock tak sirf upstream $p_{0}$ use karo; uske baad reduced $p_{0, 2}$ par switch karo. (Shock ke across static-only ratios $T, p, ρ$ theek hain; reservoir/stagnation reference reset hota hai.)

Level 5 — Mastery

(Poore regime mein design-level reasoning.)

L5.1 Ek rocket nozzle design kiya gaya hai taaki exit exactly ambient pressure $p_{b} = p_{e} = 0.5$ bar se match kare, reservoir $p_{0} = 50$ bar ke saath, air. (a) Design exit Mach number nikalo. (b) Required area ratio $A_{e} / A^{*}$ nikalo.

Recall Solution L5.1

(a) KYUN: "perfectly matched" ka matlab hai isentropic exit static pressure ambient ke equal ho — koi shocks nahi, koi expansion fans nahi. $\frac{p _{0}}{p _{e}} = \frac{50}{0.5} = 100 = (1 + 0.2 M_{e}^{2})^{3.5} .$ Dono sides ko $1/3.5$ power par le jaao: $10 0^{1/3.5} = 1 + 0.2 M_{e}^{2}$ . Ab $10 0^{0.2857} = 3.727$ , isliye $0.2 M_{e}^{2} = 2.727$ , jisse $M_{e}^{2} = 13.63$ aur $M_{e} \approx 3.69$ milta hai. (b) YEH BRANCH AUR YEH FORM KYUN: area–Mach relation $\frac{A}{A ^{*}} = \frac{1}{M} [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M^{2})]^{\frac{γ + 1}{2 ( γ - 1 )}}$ kisi bhi station ki area ko sonic throat area se compare karta hai; rocket cone mein supersonic run karta hai, isliye hum supersonic $M_{e} = 3.69$ (bada root) isme feed karte hain. $\frac{A _{e}}{A ^{*}} = \frac{1}{3.69} [\frac{2}{2.4} (1 + 0.2 \times 13.63)]^{3} = \frac{1}{3.69} [0.8333 \times 3.727]^{3} = \frac{( 3.106 ) ^{3}}{3.69} \approx 8.12.$ Toh exit throat area ka lagbhag $8.1 \times$ hai.

L5.2 Figure 2 use karke, de Laval nozzle ke paanch back-pressure regimes ko decreasing $p_{b}$ ke order mein explain karo, aur batao ki har transition kya trigger karta hai.

Recall Solution L5.2

Pressure-vs-position curves ko upar (high $p_{b}$ ) se neeche (low $p_{b}$ ) padhte hue:

Venturi (high $p_{b}$ ): fully subsonic; converging speed up karta hai, diverging slow karta hai. Throat choked nahi hai.
Choke onset: $p_{b}$ tab tak girta hai jab tak throat par $M = 1$ na aa jaaye. $\overset{m}{˙}$ ab maxed out hai — transition tab hota hai jab throat sonic reach karta hai.
Shock in diverging section: thoda aur low $p_{b}$ ; flow supersonic ho jaati hai phir ek normal shock use subsonic mein jump karta hai taaki still-high $p_{b}$ se match ho sake. $p_{b}$ ghataane par shock exit ki taraf move karta hai.
Design (perfectly expanded): $p_{e} = p_{b}$ ; exit tak smooth supersonic flow, koi shocks nahi.
Under-expanded ( $p_{b} <$ design): flow bahar expansion waves ke zariye expand karna jaari rakhti hai. Controlling knob hamesha $p_{b}$ hota hai; throat sonic condition (choke) woh pivot hai jo purely-subsonic operation ko saare supersonic regimes se alag karti hai. Turbomachinery mein isi picture ke liye dekho Steam Turbine Nozzles.

L5.3 Ek nozzle choked hai jisme $\overset{m}{˙}_{m a x}$ fixed hai. Reservoir temperature $T_{0}$ double kar di jaati hai (pressure $p_{0}$ fixed, same throat area). Kya $\overset{m}{˙}_{m a x}$ badhega, ghatega, ya same rahega? Scaling se justify karo.

Recall Solution L5.3

$\overset{m}{˙}_{m a x} = ρ^{*} A^{*} a^{*}$ . $p_{0}$ fixed hone par, $p^{*}$ fixed hai (yeh sirf $0.528 p_{0}$ hai), isliye $ρ^{*} = p^{*} / (R T^{*}) \propto 1/ T^{*} \propto 1/ T_{0}$ . Jabki $a^{*} = γ R T^{*} \propto T_{0}$ . Product: $\overset{m}{˙}_{m a x} \propto \frac{1}{T _{0}} \times T_{0} = \frac{1}{T _{0}}$ . $T_{0}$ double karne par $\overset{m}{˙}_{m a x}$ , $1/ 2 \approx 0.707$ se multiply hota hai — mass flow ghat kar lagbhag $70.7%$ ho jaata hai. Hotter gas utni hi zyada thin ho jaati hai jitni fast hoti hai, isliye kam mass flow hoti hai.

Recall Ek-line self-test

Kaun si single quantity, jab throat $M = 1$ reach karke ek baar fix ho jaaye, mass flow ko "freeze" kar deti hai chahe back pressure kitni bhi neeche jaaye? ::: Mass flow rate $\overset{m}{˙}_{m a x} = ρ^{*} A^{*} a^{*}$ — flow choked hai.