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

WHAT compute karein: ratio $p_b/p_0$ nikalo aur $0.528$ se compare karo. $\frac{p_b}{p_0}=\frac{400}{600}=0.667$ WHY 0.528 se compare karein: choking tabhi hoti hai jab back pressure itna low ho ki exit ko $M=1$ tak le jaaye. Wo threshold hai critical ratio $0.528$. Kyunki $0.667>0.528$ hai, room itna low-pressure nahi hai. Not choked — exit subsonic hai aur $p_e=p_b=400\text{kPa}$.

$\frac{p_b}{p_0}=\frac{200}{600}=0.333<0.528\Rightarrow \text{choked.}$ WHAT exit pressure hai: jab choked ho, exit critical pressure par lock ho jaata hai, $p_b$ par nahi: $p_e=p^{\ast }=0.528p_0=0.528\times 600=316.9\text{kPa}.$ Bacha hua drop $316.9\to 200$ kPa nozzle ke bahar expansion ke roop mein hota hai.

WHAT flow ko drive karta hai: gas high pressure se low pressure ki taraf move karti hai. Push hai difference $p_0-p_b$.

• $p_b=p_0$: difference zero hai, isliye koi forward flow nahi — mass flow rate $\dot{m}=0$ aur $M=0$ har jagah. Gas bas balance hokar baithti hai.
• $p_b>p_0$: ab room tank se zyada pressure pe hai, isliye gas reservoir mein wapas flow karti (ek converging nozzle iske liye design nahi hua — practically ye sirf "koi useful forward flow nahi" hai). Hamare forward-flow formulas sirf $p_b<p_0$ ke liye apply hote hain. WHY ye matter karta hai: poora ratio logic ($0.528$, choked/unchoked) window $0\le p_b/p_0<1$ mein rehta hai. Top edge $p_b/p_0=1$ pe flow band ho jaata hai; ye picture ki boundary hai.

Pehle, kya hum flow window mein hain aur unchoked hain? $p_b/p_0=0.70$, jo $1$ se neeche hai (flow exist karta hai) aur $0.528$ se upar hai (subsonic, unchoked). Isliye $p_e=p_b=350$ kPa aur hum master relation invert karte hain. $\frac{p_0}{p_e}=\frac{500}{350}=1.4286={\left(1+0.2M^2\right)}^{3.5}.$ Exponent hatane ke liye $3.5$-th root lo (yaani $1/3.5$ ki power tak raise karo): $1.4286^{1/3.5}=1.1073\Rightarrow 0.2M^2=0.1073\Rightarrow M^2=0.5366\Rightarrow M=\pm \mathrm{0.732.}$ WHY sirf positive, subsonic root: $M^2$ solve karne se do square roots milti hain, $M=+0.732$ aur $M=-0.732$. Negative $M$ matlab reverse direction mein flow, jo hamar forward-blowing nozzle nahi hai, isliye discard karte hain. (Algebraically full nozzle problem mein ek supersonic branch bhi hoti hai, lekin converging-only duct sirf subsonic flow ko $M=1$ tak accelerate kar sakta hai — supersonic root kabhi reach nahi kar sakta. Isliye yahan sirf subsonic answer physical hai.) $M=0.732$.

$T^{\ast }$ kahaan se aata hai (unpacked): $M=1$ pe stagnation relation deta hai $\frac{T_0}{T^{\ast }}=1+\frac{\gamma -1}{2}(1{)}^2=\frac{2+\gamma -1}{2}=\frac{\gamma +1}{2}\Rightarrow T^{\ast }=\frac{2}{\gamma +1}{T}_0.$ Numbers plug karo: $T^{\ast }=\frac{2}{2.4}\times 320=266.7\text{K}.$ WHY exit speed = speed of sound: $M=1$ ka matlab hai flow speed local sound speed $a=\sqrt{\gamma RT}$ ke barabar hai. $V_e=a^{\ast }=\sqrt{1.4\times 287\times 266.7}=327.3\text{m/s}.$

WHAT build karein: exit pe $M=1$ hai, isliye starred values use karte hain. Temperature factor wahi unpacked wala hai: $T^{\ast }=\frac{2}{\gamma +1}{T}_0=\frac{2}{2.4}(300)=250\text{K},p^{\ast }=0.528(500)=264.1\text{kPa}.$ Exit pe density ideal gas law $\rho =p/(RT)$ se: ${\rho }^{\ast }=\frac{264100}{287\times 250}=3.681{\text{kg/m}}^3.$ Exit speed $V_e=a^{\ast }=\sqrt{1.4\times 287\times 250}=316.9\text{m/s}$. WHY $\dot{m}=\rho AV$: exit se har second mass = (density)(area)(speed). $\dot{m}={\rho }^{\ast }{A}_e{V}_e=3.681\times 1.0\times 10^{-3}\times 316.9=1.167\text{kg/s}.$

$p_b=264\text{kPa}$ pe ratio bilkul $0.528$ hai — exit just $M=1$ reach karta hai. Yahaan se flow choked ho jaata hai.

• $p_b=150\text{kPa}$: ratio $0.30<0.528$ → choked, $\dot{m}$ $1.167\text{kg/s}$ par hi rehta hai.
• $p_b=50\text{kPa}$: ratio $0.10<0.528$ → still choked, $\dot{m}$ unchanged $1.167\text{kg/s}$ par. WHAT plot dikhata hai (coral curve, dashed line ke right mein): $\dot{m}$ flat hai — ek horizontal ceiling. Reservoir "girte room pressure ko sun" nahi sakta kyunki signals sonic flow ke through upstream swim nahi kar sakti. Dekho Mass Flow Rate & Choking.

WHY shortcut exist karta hai: parent Step 4 ke choked mass-flow formula mein, $\dot{m}=A{p}_0\sqrt{\frac{\gamma }{R{T}_0}}\frac{M}{{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{\frac{\gamma +1}{2(\gamma -1)}}}$ $M=1$ pe $p_0$ aur $T_0$ ke alawa sab fixed hai. Isliye constant $T_0$ par $\dot{m}\propto p_0$. $p_0$ double karne se $\dot{m}$ double ho jaata hai: ${\dot{m}}_{\text{new}}=2\times 1.167=2.334\text{kg/s}.$ Scratch se verify: ${\rho }^{\ast }$ double ho jaata hai (kyunki $p^{\ast }=0.528p_0$ double ho jaata hai aur $T^{\ast }$ unchanged hai), $V_e=a^{\ast }$ unchanged rehta hai (sirf $T^{\ast }$ pe depend karta hai), area unchanged $\Rightarrow \dot{m}$ double ho jaata hai. ✓

Same formula se $M=1$ pe: $\dot{m}\propto p_0/\sqrt{T_0}$. $T_0$ double karne se $\dot{m}$ $\sqrt{2}$ se divide ho jaata hai: ${\dot{m}}_{\text{new}}=\frac{1.167}{\sqrt{2}}=0.825\text{kg/s}.$ WHY girata hai jabki gas faster hai: hotter gas matlab higher $a^{\ast }$ (faster exit), lekin lower density ${\rho }^{\ast }=p^{\ast }/(R{T}^{\ast })$ bhi, aur density haar jaati hai. Net result: $\dot{m}\propto 1/\sqrt{T_0}$.

(a) $p_b/p_0=300/800=0.375$, jo $1$ se neeche hai (flow exist karta hai) aur $0.528$ se bhi neeche → choked. (b) Choked hai, isliye exit $M=1$ pe hai: $p_e=p^{\ast }=0.528(800)=422.5\text{kPa}.$ Temperature factor pehle ki tarah unpack karo: $M=1$ pe, $\frac{T_0}{T^{\ast }}=1+\frac{\gamma -1}{2}=\frac{\gamma +1}{2}$, isliye $T_e=T^{\ast }=\frac{2}{\gamma +1}{T}_0=\frac{2}{2.4}(500)=416.7\text{K}.$ $V_e=a^{\ast }=\sqrt{1.4\times 287\times 416.7}=409.2\text{m/s}.$ (c) ${\rho }^{\ast }=p^{\ast }/(R{T}^{\ast })=422500/(287\times 416.7)=3.533{\text{kg/m}}^3$. $\dot{m}={\rho }^{\ast }{A}_e{V}_e=3.533\times 8\times 10^{-4}\times 409.2=1.157\text{kg/s}.$ (d) Converging-only duct mein, mass-flow factor $M/(1+\frac{\gamma -1}{2}{M}^2{)}^{(\gamma +1)/[2(\gamma -1)]}$ $M=1$ pe peak karta hai — shrinking area sirf subsonic flow ko Mach 1 tak push kar sakta hai, kabhi past nahi. $M=1.5$ paane ke liye area ko throat ke baad phir se increase hona chahiye: ek Converging-Diverging (de Laval) Nozzle. Sirf wahaan diverging section already-sonic flow ko supersonic regime mein accelerate karta rehta hai. (Agar $p_b$ itna low nahi hai, to wo supersonic flow ek normal shock ke zariye wapas jump kar sakti hai — ye ek alag kahani hai.)

WHY $p$ aur $T$ se start karein: hamare paas already $p^{\ast }/p_0$ aur $T^{\ast }/T_0$ hain; inhe combine karne ke liye ideal gas law $\rho =p/(RT)$ use karo: $\frac{{\rho }^{\ast }}{{\rho }_0}=\frac{p^{\ast }/p_0}{T^{\ast }/T_0}=\frac{{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma }{\gamma -1}}}{\frac{2}{\gamma +1}}={\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma }{\gamma -1}-1}={\left(\frac{2}{\gamma +1}\right)}^{\frac{1}{\gamma -1}}.$ Air ke liye, $\gamma =1.4$: ${\left(\frac{2}{2.4}\right)}^{1/0.4}=0.8333^{2.5}=0.634$. Parent ke "sanity anchor" ${\rho }^{\ast }=0.634{\rho }_0$ se match karta hai. ✓

Reservoir $\to$ choke pe exit: $p^{\ast }=0.528p_0$, $T^{\ast }=0.833T_0$, ${\rho }^{\ast }=0.634{\rho }_0$. Is page mein sab kuch inhi se plus $\dot{m}=\rho AV$ se bana hai.