Exercises — Nozzle thermodynamics — isentropic expansion from chamber to exit

$P_0,T_0$ ::: chamber (stagnation) pressure aur temperature — gas poori tarah se rest mein laaya gaya, saari energy thermal hai. $M$ ::: Mach number $=V/a$, flow speed ko local sound speed $a=\sqrt{\gamma RT}$ ki units mein measure kiya gaya. $\gamma$ ::: heat capacity ratio $c_p/c_v$; rocket exhaust ke liye hum $\gamma =1.2$ use karte hain. $c_p$ ::: specific heat at constant pressure — 1 kg gas ko 1 K badhane ke liye chahiye energy jab woh expand karne ke liye free ho. Ek ideal gas ke liye yeh $\gamma$ aur $R$ se fix hota hai: $c_p=\frac{\gamma R}{\gamma -1}$. $R$ ::: specific gas constant, hamare exhaust ke liye $320\text{J/(kg⋅K)}$; equals $R_u/\mathcal{M}$ jahan $R_u=8314\text{J/(kmol⋅K)}$ universal hai aur $\mathcal{M}$ molar mass hai. $g_0$ ::: standard gravity, $9.81{\text{m/s}}^2$ — woh fixed constant jo ek exhaust velocity ko specific impulse ke seconds mein convert karta hai. $A^{\ast }$ ::: throat area, sabse narrow cross-section, jahan exactly $M=1$ hota hai. starred quantities ($T^{\ast },P^{\ast }$) ::: throat par values, yaani $M=1$ par evaluate ki gayi.

Throat par, definition se $M=1$ hota hai, isliye critical ratio directly apply hoti hai: $T^{\ast }=T_0\cdot \frac{2}{\gamma +1}=3500\cdot \frac{2}{2.2}=3500\times 0.9091=\boxed{3182\text{K}}$ Humne kya kiya: temperature relation mein $M=1$ substitute kiya. Kyun: throat define hota hai sonic point ke roop mein, isliye hume $M$ guess nahi karna — woh fixed hai.

$P^{\ast }=P_0{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma }{\gamma -1}}=10\times (0.9091{)}^6$ Exponent: $\frac{\gamma }{\gamma -1}=\frac{1.2}{0.2}=6$. $P^{\ast }=10\times 0.5645=\boxed{5.645\text{MPa}}$ Exponent 6 kyun hai: pressure, temperature se zyada tezi se girta hai kyunki yeh expanding gas ke dono — density drop aur temperature drop — par depend karta hai.

Exponent $=\frac{1.4}{0.4}=3.5$; base $=\frac{2}{2.4}=0.8333$. $\frac{P^{\ast }}{P_0}=(0.8333{)}^{3.5}=\boxed{0.5283}$ Yeh kya dikhata hai: air ke liye famous "0.528" rule — ek converging nozzle tab choke hoti hai jab back pressure chamber pressure ke ~53% se neeche gir jaata hai.

Pehle bracket $1+\frac{\gamma -1}{2}{M}_e^2=1+0.1\times 9=1.9$. $T_e=\frac{3500}{1.9}=\boxed{1842\text{K}}$ $P_e=\frac{10}{1.9^6}=\frac{10}{47.05}=0.2125\text{MPa}\Rightarrow \boxed{212.5\text{kPa}}$ Ab exit par local sound speed: $a_e=\sqrt{\gamma R{T}_e}=\sqrt{1.2\times 320\times 1842}=\sqrt{707,328}=841.6\text{m/s}$. $V_e=M_e{a}_e=3\times 841.6=\boxed{2525\text{m/s}}$ $T_e$ pehle kyun: velocity ko local sound speed chahiye, jo local temperature chahti hai — isliye temperature hamesha pehla domino hota hai.

Yeh formula aakhir hai kyun? Area-Mach relation koi nayi law nahi hai — yeh woh result hai jo aapko mass conservation $\rho VA={\rho }^{\ast }{V}^{\ast }{A}^{\ast }$ likhne par milta hai (same gas har station se guzarta hai) aur phir har $\rho$, $V$ aur $T$ ko isentropic value se $M$ ke terms mein replace karne par. Momentum conservation wahan se aata hai isentropic $P/{\rho }^{\gamma }=\text{const}$ ke through jis par woh substitutions rely karti hain. Isliye yeh formula bas "constant mass flow, Mach mein expressed" hai. $\frac{A_e}{A^{\ast }}=\frac{1}{3}{\left[\frac{2}{2.2}\times 1.9\right]}^{\frac{\gamma +1}{2(\gamma -1)}}$ Exponent $=\frac{2.2}{0.4}=5.5$; bracket $=0.9091\times 1.9=1.7273$. $\frac{A_e}{A^{\ast }}=\frac{1}{3}(1.7273{)}^{5.5}=\frac{1}{3}\times 20.21=\boxed{6.74}$ Iska matlab: diverging cone ko throat area ka lagbhag 7× tak khulna hoga taaki gas Mach 3 tak pahunch sake.

Yeh nozzle geometry jo describe ki gayi — $M=1$ par throat, $A_e/A^{\ast }=6.74$ tak exit khuli — neeche draw ki gayi hai. Note karo ki throat (teal dashed) pinch point hai aur exit (plum dashed) wahan hai jahan gas Mach 3 tak pahunchi hai.

Mass flow ko throat par evaluate karna sabse aasaan hai, jahan saari conditions fixed hoti hain. Throat: $T^{\ast }=3182\text{K}$, $P^{\ast }=5.645\text{MPa}$. Ideal gas se density: ${\rho }^{\ast }=\frac{P^{\ast }}{R{T}^{\ast }}=\frac{5.645\times 10^6}{320\times 3182}=5.545{\text{kg/m}}^3$. Velocity $V^{\ast }=a^{\ast }=\sqrt{\gamma R{T}^{\ast }}=\sqrt{1.2\times 320\times 3182}=1105.7\text{m/s}$. $\dot{m}={\rho }^{\ast }{V}^{\ast }{A}^{\ast }=5.545\times 1105.7\times 0.01=\boxed{61.3\text{kg/s}}$ Throat kyun: $\dot{m}=\rho VA$ har jagah constant hai, lekin throat par har factor $M$ solve kiye bina known hai.

Exit pressure ko ambient se compare karo: $P_e=212.5\text{kPa}>P_a=101\text{kPa}$ Kyunki jet zyada pressure par nikalta hai apne surroundings se, gas under-expanded hai — uske paas abhi bhi pressure energy convert karne ke liye bacha hai, isliye woh nozzle chhod ne ke baad bhi expand karta rehta hai. Lip par yeh expansion fans (Prandtl–Meyer waves) produce karta hai jo flow ko outward turn karte hain. Contrast: agar $P_e<P_a$ hota toh jet over-expanded hota aur oblique shocks form karta jo use inward squeeze karte. Perfect expansion $P_e=P_a$ hai. Neeche bar chart comparison ko concrete banata hai: teal bar ($P_e$) plum ambient line ($P_a$) se uuncha khada hai, jo exactly under-expansion ki signature hai. Cases ki poori family ke liye Nozzle Flow Regimes dekho.

Har ke liye, $T=T_0/(1+0.1M^2)$ phir $V=M\sqrt{\gamma RT}$. A ($M=2$): bracket $=1.4$, $T=2500\text{K}$, $V=2\sqrt{1.2\times 320\times 2500}=2\times 979.8=1960\text{m/s}$. B ($M=4$): bracket $=2.6$, $T=1346\text{K}$, $V=4\sqrt{1.2\times 320\times 1346}=4\times 718.9=2876\text{m/s}$. B jeetta hai: $\frac{2876}{1960}=\boxed{1.47\times }$ faster. Subtlety: Mach double karne se velocity double nahi hui, kyunki sound speed $a$, $T$ girne par drop ho jaati hai. Velocity $M\times a$ hai, aur $a$ high $M$ par aapke against kaam karta hai — yahi wajah hai ki hamesha bade nozzles diminishing returns dete hain.

Energy equation total energy cap karti hai: $c_{p}T+V^2/2=c_p{T}_0$. Velocity tab sabse zyada hoti hai jab saari thermal energy kinetic ban jaaye, yaani $T\to 0$: $V_{\max }=\sqrt{2c_p{T}_0}=\sqrt{2\cdot \frac{\gamma R}{\gamma -1}{T}_0}=\sqrt{2\cdot \frac{1.2\times 320}{0.2}\times 3500}$ $V_{\max }=\sqrt{2\times 1920\times 3500}=\sqrt{1.344\times 10^7}=\boxed{3666\text{m/s}}$ Jaise $T\to 0$, sound speed $a\to 0$, isliye $M=V/a\to \infty$. Maximum velocity finite hai even though Mach diverge karta hai — yeh ek khoobsoorat reminder hai ki Mach number speed nahi hai, yeh ek shrinking yardstick ke relative speed hai.

Do unknowns ($M_e,T_e$) lekin do equations hain. Inhe combine karo: $V_e=M_e\sqrt{\gamma R{T}_e},T_e=\frac{T_0}{1+\frac{\gamma -1}{2}{M}_e^2}.$ Doosre ko pehle mein substitute karo aur square karo: $V_e^2=M_e^2\gamma R\frac{T_0}{1+0.1M_e^2}.$ Numbers plug karo ($V_e^2=8.649\times 10^6$, $\gamma R{T}_0=1.2\times 320\times 3500=1.344\times 10^6$). $x=M_e^2$ let karo: $8.649\times 10^6(1+0.1x)=1.344\times 10^{6}x\Rightarrow 8.649+0.8649x=1.344x\Rightarrow 8.649=0.4791x.$ $x=18.05\Rightarrow M_e=\sqrt{18.05}=\boxed{4.25}.$ Area ratio: bracket $=1+0.1(18.05)=2.805$; $\frac{A_e}{A^{\ast }}=\frac{1}{4.25}[0.9091\times 2.805{]}^{5.5}=\frac{1}{4.25}(2.550{)}^{5.5}$. $(2.550{)}^{5.5}=173.3$, isliye $\frac{A_e}{A^{\ast }}=\frac{173.3}{4.25}=\boxed{40.8}$. Lesson: ek modest velocity increase ek bahut bada extra expansion ratio demand karta hai — high Mach par nozzle bells explosively grow karti hain.

$\frac{A}{A^{\ast }}(M)=25$ solve karo area-Mach formula se, $\gamma =1.2$, exponent $5.5$: $25=\frac{1}{M}{\left[0.9091(1+0.1M^2)\right]}^{5.5}.$ $M=3.60$ try karo: bracket $=0.9091(2.296)=2.087$; $2.087^{5.5}=62.8$; $/3.6=17.4$ — bahut kam. $M=3.90$ try karo: bracket $=0.9091(2.521)=2.292$; $2.292^{5.5}=90.1$; $/3.9=23.1$ — abhi bhi kam. $M=3.98$ try karo: bracket $=0.9091(2.584)=2.349$; $2.349^{5.5}=99.5$; $/3.98=25.0$. ✔ $\boxed{M_e\approx 3.98}$ Iterate kyun: area-Mach relation elementary functions mein invertible nahi hai — tum $M$ sweep karte ho jab tak geometry match na ho. Real design tables (aur Specific Impulse Optimization tools) exactly yahi karte hain.

Seconds mein specific impulse $I_{sp}=V_e/g_0$ hai jab pressure term vanish ho (perfect vacuum expansion, Thrust Equation Derivation dekho). Yahaan $g_0=9.81{\text{m/s}}^2$ woh fixed standard gravity hai jo velocity ko seconds mein turn karta hai. Maximum $V_e$ Exercise 3.3 ka $T\to 0$ limit hai, $c_p=\frac{\gamma R}{\gamma -1}$ use karte hue: $V_{\max }=\sqrt{2c_p{T}_0}=\sqrt{2\cdot \frac{\gamma R}{\gamma -1}{T}_0}=3666\text{m/s}.$ $I_{sp,\max }=\frac{3666}{9.81}=\boxed{374\text{s}}.$ Yeh ceiling kyun hai: koi bhi nozzle, chahe kitna bhi lamba ho, total thermal enthalpy $c_p{T}_0$ se zyada extract nahi kar sakta. Yeh single number tumhe batata hai ki koi propellant pursue karne layak hai ya nahi, koi bhi hardware design karne se pehle. Real losses (Real Nozzle Losses) achieved value ko iske ~95% tak kheench laate hain.

$V_{\max }=\sqrt{2c_p{T}_0}$ se $c_p$ fixed rakho: $V\propto \sqrt{T_0}$. $T_0$ double karne par $V$ mein $\sqrt{2}=\boxed{1.41}$ ka multiplication hota hai. Ab $R=R_u/\mathcal{M}$ likho jahan $R_u=8314\text{J/(kmol⋅K)}$ universal hai aur $\mathcal{M}$ molar mass hai. Phir $V_{\max }=\sqrt{\frac{2\gamma }{\gamma -1}\cdot \frac{R_u}{\mathcal{M}}{T}_0}\propto \sqrt{\frac{T_0}{\mathcal{M}}}.$ Velocity $1/\sqrt{\mathcal{M}}$ ke scale par hai. Molar mass half karne par same $\sqrt{2}$ boost milta hai jaise temperature double karne par — lekin $\mathcal{M}$ half karna (e.g. hydrogen jalana, $\mathcal{M}\approx 2$) ek already 3500 K flame double karne se kaafi aasaan aur safe hai jo chamber melt kar de. Design punchline: issi liye hydrogen–oxygen engines high-$I_{sp}$ upper stages mein dominate karte hain — light exhaust, hot exhaust ko beat karta hai. Verify karo: $\mathcal{M}=10$ vs $20$ aur same $T_0$ ke saath, $V$ ratio $=\sqrt{20/10}=1.41$.

Thrust $F=\dot{m}{V}_e+(P_e-P_a)A_e$. Jaise cone lambi hoti hai, $A_e/A^{\ast }\uparrow$, $M_e\uparrow$, isliye $V_e\uparrow$ — lekin diminishing returns ke saath kyunki $V_e\to V_{\max }=3666\text{m/s}$ (ek hard ceiling) hai. Meanwhile $P_e\to 0$, aur jaise hi $P_e<P_a$ hota hai pressure term $(P_e-P_a)A_e$ negative ho jaata hai — atmosphere hamesha bade hote exit par push back karta hai. Numerically, $M=3.8$ ($A/A^{\ast }=25$) se $M=4.25$ ($A/A^{\ast }=41$) tak jaane par ~$65\%$ zyada area milta hai lekin $V_e$ sirf ~$2900$ se ~$3010\text{m/s}$ tak uthta hai, sirf $4\%$. Isliye thrust ek asymptote ki taraf badhta hai aur pressure term eventually subtract karta hai: ek optimum length hoti hai, infinite growth nahi. $\boxed{\text{Refuted.}}$ Giant bell ki extra structural mass bhi vehicle acceleration cut karti hai — sachchi optimisation Specific Impulse Optimization mein rahti hai.