Exercises — Choked flow — condition M = 1 at throat, maximum mass flow

Step 1 (WHAT): Back-pressure ratio compute karo: $p_b/p_0=600/800=0.750$. Step 2 (WHY): Critical ratio $0.5283$ se compare karo. Throat tabhi $M=1$ reach kar sakta hai jab exit ko pressure tak ya usse neeche drop karne ko kaha jaaye $p^{\ast }/p_0$. Step 3: $0.750>0.5283$, toh required exit ratio threshold se upar hai. Flow poori jagah subsonic rehta hai; exit pressure simply $p_b$ se match karta hai. Answer: Choked nahi hai. $p_b$ ko aur neeche laane par bhi $\dot{m}$ badhega (jab tak hum $0.528$ hit nahi kar lete).

Step 1: $p_b/p_0=300/800=0.375$. Step 2: $0.375<0.5283$: back pressure critical se neeche hai. Ek converging nozzle nahi de sakta exit pressure $p^{\ast }$ se neeche, valid subsonic flow rehte hue — throat $M=1$ par lock ho jaata hai. Step 3: Toh exit pressure $p_b$ nahi hai; ye lock ho jaata hai: $p^{\ast }=0.5283\times 800\text{kPa}=422.6\text{kPa}.$ Answer: Choked hai. Exit pressure $=422.6\text{kPa}$ ($p_b$ se zyada); $300\text{kPa}$ tak baaki expansion nozzle ke bahar ek under-expanded jet ke roop mein hoti hai.

Step 1 — bracket (WHAT/WHY): exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.4}{0.8}=3$, aur $\frac{2}{\gamma +1}=0.8333$. Toh bracket hai $0.8333^3=0.5787$. Step 2 — square-root factor: $\sqrt{\frac{\gamma }{R{T}_0}}=\sqrt{\frac{1.4}{287\times 290}}=\sqrt{1.6816\times 10^{-5}}=4.101\times 10^{-3}.$ Step 3 — assemble karo: ${\dot{m}}_{\max }=A^{\ast }{p}_0\times (4.101\times 10^{-3})\times \mathrm{0.5787.}$ $A^{\ast }{p}_0=1.5\times 10^{-4}\times 8\times 10^5=120$ ke saath: ${\dot{m}}_{\max }=120\times 4.101\times 10^{-3}\times 0.5787=0.285\text{kg/s}.$ Answer: ${\dot{m}}_{\max }\approx 0.285\text{kg/s}$.

Step 1: Ye already choked hai ($0.375<0.528$), toh throat $M=1$ par hai aur wahan hi rehta hai. Step 2 (WHY): ${\dot{m}}_{\max }$ sirf $p_0,T_0,A^{\ast },\gamma ,R$ par depend karta hai — formula mein koi $p_b$ nahi hai. Downstream signals sonic throat se upstream nahi ja sakte, toh throat naye $p_b$ ko kabhi "sunता" nahi. Answer: $\dot{m}$ unchanged hai: abhi bhi $0.285\text{kg/s}$. $\Delta \dot{m}=0$.

Step 1 — Predict karo (WHY): Formula mein ek hi $T_0$ hai: ${\dot{m}}_{\max }\propto 1/\sqrt{T_0}$. Zyada hot reservoir ⇒ throat par gas kam dense hogi; faster sound speed ($\propto \sqrt{T}$) compensate nahi karta. Toh mass flow girta hai. Step 2 — Compute karo: $\frac{{\dot{m}}_{\max }(580)}{{\dot{m}}_{\max }(290)}=\sqrt{\frac{290}{580}}=\sqrt{0.5}=\mathrm{0.7071.}$ Answer: Mass flow apni value ka $70.7\%$ ho jaata hai, yaani $0.285\times 0.7071=0.202\text{kg/s}$. $T_0$ double karne par ${\dot{m}}_{\max }$ lagbhag $29\%$ girta hai.

Step 1 — choked? $p_b/p_0=200/500=0.400<0.5283$ ⇒ choked ⇒ exit $M=1$ par hai? Dhyan se: purely converging nozzle ke liye, haan — apne exit (throat/exit same section hai) par jo maximum Mach reach kar sakta hai wo $M=1$ hai. Step 2 (WHY $M=1$ aur zyada nahi): Converging duct subsonic flow ko sirf $M=1$ tak accelerate kar sakta hai; isse exceed karne ke liye diverging section chahiye (dekho Converging–Diverging (de Laval) Nozzle). Exit pressure $p^{\ast }=0.5283\times 500=264\text{kPa}$ par lock ho jaata hai. Answer: $M_{\text{exit}}=1$. (Gas $264\text{kPa}$ se $200\text{kPa}$ tak nozzle ke bahar expand hoti rehti hai.)

Step 1 (a) — mass flow throat se set hota hai: $A^{\ast }=4{\text{cm}}^2=4\times 10^{-4}{\text{m}}^2$ convert karo, $p_0=10^6\text{Pa}$. Bracket $=0.8333^3=0.5787$; $\sqrt{\gamma /(R{T}_0)}=\sqrt{1.4/(287\times 350)}=\sqrt{1.3937\times 10^{-5}}=3.733\times 10^{-3}$. ${\dot{m}}_{\max }=4\times 10^{-4}\times 10^6\times 3.733\times 10^{-3}\times 0.5787=0.864\text{kg/s}.$ Step 2 (b) — continuity (WHY): Steady flow matlab $\dot{m}=\rho AV$ har cross-section par identical hai — wahi $0.864\text{kg/s}$ throat aur exit dono se guzarta hai. Bada exit area wahan faster, less dense supersonic gas se balance hota hai (Area–Mach Relation (A/A*) fix karta hai ki exit Mach kya hoga). Mass flow number sirf throat par decide hota hai, kyunki wahan $M=1$ hota hai. Answer: ${\dot{m}}_{\max }\approx 0.864\text{kg/s}$ dono sections se.

Step 1 (WHY dono scalings combine karein): ${\dot{m}}_{\max }\propto p_0/\sqrt{T_0}$ (baaki sab factors identical). Ratio banao: $\frac{{\dot{m}}_B}{{\dot{m}}_A}=\frac{p_{0,B}}{p_{0,A}}\sqrt{\frac{T_{0,A}}{T_{0,B}}}=\frac{900}{600}\sqrt{\frac{300}{675}}.$ Step 2: $\frac{900}{600}=1.5$; $\sqrt{300/675}=\sqrt{0.4444}=0.6667$. Product $=1.5\times 0.6667=1.000$. Answer: Dono same mass flow pass karte hain (${\dot{m}}_B/{\dot{m}}_A=1.000$). B ka zyada pressure exactly uske zyada temperature se cancel ho jaata hai.

Step 1 — choking ki onset (WHAT): choking tab shuru hoti hai jab $p_b/p_0=0.5283$, yaani $p_{b,\text{choke}}=0.5283\times 400=211.3\text{kPa}.$ Step 2 — do regimes (WHY):

• $p_b$ $400$ se $211.3\text{kPa}$ tak: subsonic, unchoked. $\dot{m}$ badhta hai jaise $p_b$ girta hai (bada pressure drop ⇒ zyada flow), $0$ se ${\dot{m}}_{\max }$ ki taraf charhta hai.
• $p_b$ $211.3$ se $50\text{kPa}$ tak: choked. $\dot{m}$ ${\dot{m}}_{\max }$ par flat rehta hai — throat $p_b$ ke liye behra hai. Step 3 — ${\dot{m}}_{\max }$: bracket $=0.5787$; $\sqrt{1.4/(287\times 320)}=\sqrt{1.5244\times 10^{-5}}=3.905\times 10^{-3}$. ${\dot{m}}_{\max }=3\times 10^{-4}\times 4\times 10^5\times 3.905\times 10^{-3}\times 0.5787=0.2712\text{kg/s}.$ Answer: Choking $p_b=211.3\text{kPa}$ par shuru hoti hai; ${\dot{m}}_{\max }\approx 0.271\text{kg/s}$. $\dot{m}$–$p_b$ curve pehle rise karti hai phir knee par flat ho jaati hai — neeche figure dekho.

Step 1 — formula invert karo (WHY): ${\dot{m}}_{\max }=A^{\ast }{p}_0\sqrt{\gamma /(R{T}_0)}(0.5787)$ ko $A^{\ast }$ ke liye solve karo: $A^{\ast }=\frac{{\dot{m}}_{\max }}{p_0\sqrt{\gamma /(R{T}_0)}(0.5787)}.$ Step 2 — numbers: $\sqrt{1.4/(287\times 400)}=\sqrt{1.2195\times 10^{-5}}=3.492\times 10^{-3}$. Denominator $=6\times 10^5\times 3.492\times 10^{-3}\times 0.5787=1212.5$. $A^{\ast }=\frac{0.500}{1212.5}=4.124\times 10^{-4}{\text{m}}^2.$ Answer: $A^{\ast }\approx 4.12\times 10^{-4}{\text{m}}^2$ ($\approx 4.12{\text{cm}}^2$, yaani throat diameter $\approx 2.29\text{cm}$).

Step 1 (WHY claim galat hai): Throat shock se upstream hai aur $M=1$ par rehta hai; downstream shock, sonic throat se information peeche nahi bhej sakta. Mass flow throat par set hota hai. Step 2: Ek normal shock adiabatic hota hai — ye $\dot{m}$ aur $T_0$ ko apne across conserve karta hai; ye sirf entropy badhata hai (shock ke downstream $p_0$ girta hai). Ye throat ko touch nahi karta. Answer: Jhooth. $\dot{m}$ unchanged rehta hai. Shock downstream pressure/velocity change karta hai aur uske baad stagnation pressure reduce karta hai, lekin choked throat $\dot{m}$ fix karta hai chahe kuch bhi ho.

$p_0=400$ kPa air ke liye onset-of-choking back pressure? ::: $0.5283\times 400=211$ kPa. Kya choked nozzle mein downstream normal shock $\dot{m}$ change karta hai? ::: Nahi — throat sonic rehta hai. C-D nozzle mein ${\dot{m}}_{\max }$ kaun sa area set karta hai? ::: Throat area $A^{\ast }$. $T_0$ double karo: ${\dot{m}}_{\max }$ ka kya hota hai? ::: $1/\sqrt{2}=0.707$ factor se girta hai. Air ke liye choked-flow bracket exponent? ::: $\frac{\gamma +1}{2(\gamma -1)}=3$.