Worked examples — Nozzle thermodynamics — isentropic expansion from chamber to exit

3.3.11 · D3 · Physics › Rocket Propulsion › Nozzle thermodynamics — isentropic expansion from chamber to

Yeh page parent nozzle note ke liye drill ground hai. Parent ne tumhe formulas diye. Yahan hum un formulas par har tarah ke cases throw karte hain — subsonic vs supersonic, throat, limits, aur tricky exam twists — taaki baad mein koi bhi scenario tumhe surprise na kare.

Kuch bhi shuru karne se pehle, parent se charon workhorse equations re-pin karte hain taaki hum koi bhi aisa symbol use na karein jise hum ne re-anchor nahi kiya hai.

Recall Woh symbols jo hum baar baar use karte rahenge

• $T_0,P_0,{\rho }_0$ ::: chamber (stagnation) values — gas ko poori tarah rest par laya gaya, saari energy thermal, koi motion nahi.
• $T,P,\rho$ ::: local static values us point par jahan gas speed $V$ se move kar rahi hai.
• $M=V/a$ ::: Mach number, gas speed divided by sound speed $a=\sqrt{\gamma RT}$. $M<1$ subsonic, $M=1$ sonic, $M>1$ supersonic.
• $\gamma =c_p/c_v$ ::: heat-capacity ratio, hot rocket exhaust ke liye $\approx 1.2$.
• $A^{\ast }$ (aur koi bhi "$\ast$" quantity) ::: throat par ki value, jahan $M=1$. $A$ kisi bhi doosre station par area hai.

Poore note mein, jab tak kuch aur na bataya jaye, hum parent ka gas use karte hain: $\gamma =1.2$, $R=320\text{J/(kg⋅K)}$, $P_0=10\text{MPa}$, $T_0=3500\text{K}$.

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Scenario matrix

Is topic ke har possible question ka answer in cells mein se ek mein hai. Neeche har worked example us cell ke saath tagged hai jise woh fill karta hai.

Cell	Ise kya distinct banata hai	Covered by
A. Chamber → throat (M=1)	sonic point, critical ratios	Ex 1
B. Subsonic station (M<1)	converging section, area $A^{\ast }$ se upar	Ex 2
C. Supersonic station (M>1)	diverging section, area $A^{\ast }$ se upar phir se	Ex 3
D. Zero / degenerate input (M→0)	chamber khud, ratios → 1, $A/A^{\ast }\to \infty$	Ex 4
E. Limiting behaviour (M→∞)	maximum possible exhaust speed	Ex 5
F. Inverse problem	area ratio di gayi, find M (do roots!)	Ex 6
G. Real-world word problem	target exit velocity ke liye nozzle design karo	Ex 7
H. Exam twist — sign/regime trap	same $A/A^{\ast }$, which branch? over/under-expanded	Ex 8

Do subtle traps hain — cell D (ek degenerate input jo naive area formula ko todta hai) aur cell F/H (area ratio do Mach numbers deta hai — tumhe physically correct wala choose karna hai). Hum inhe seedha tackle karte hain.

Figure s01 — "ek ratio, do Machs". Horizontal axis Mach number $M$ hai; vertical axis area ratio $A/A^{\ast }$ hai. Blue curve subsonic branch hai ($M<1$), red curve supersonic branch ($M>1$), aur dono throat minimum $A/A^{\ast }=1$ par $M=1$ pe milte hain (white dotted line). Dashed green horizontal line $A/A^{\ast }=4$ par curve ko do baar cross karti hai — ek baar blue par (subsonic root), ek baar red par (supersonic root), dono yellow dot se mark hain. Yahi double crossing visual reason hai ki cells B, C, F aur H sab exist karte hain: geometry akele kabhi nahi batati ki tum kaunse regime mein ho.

Regime language ke liye neeche use hone wali cheezein dekhne ke liye Nozzle Flow Regimes aur Mach Number and Sound Speed bhi dekho.

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Example 1 — Cell A: chamber se throat tak

Forecast: $M=1$ par ratios fixed numbers hote hain (woh sirf $\gamma$ par depend karte hain). Guess: temperature thodi giregii, pressure roughly half ho jaayega.

1. Temperature. $T^{\ast }=T_0\cdot \frac{2}{\gamma +1}=3500\cdot \frac{2}{2.2}=3181.8\text{K}$. Yeh step kyun? $T_0/T=1+\frac{\gamma -1}{2}{M}^2=1+0.1=1.1$ mein $M=1$ set karo, toh $T=T_0/1.1$ — yeh general relation ko special throat value mein convert karta hai.
2. Pressure. $P^{\ast }=P_0{\left(\frac{2}{2.2}\right)}^6=10\times 0.9091^6=5.645\text{MPa}$. Yeh step kyun? Pressure formula ka exponent $\gamma /(\gamma -1)=1.2/0.2=6$ hai; $M=1$ par bracket $2/(\gamma +1)$ mein collapse ho jaata hai.
3. Sound speed = exit speed. $a^{\ast }=\sqrt{\gamma R{T}^{\ast }}=\sqrt{1.2\times 320\times 3181.8}=1105.6\text{m/s}$. Yeh step kyun? Throat par $M=1$ ka matlab hai $V^{\ast }=a^{\ast }$ exactly, toh sound speed compute karna hi wahan gas speed compute karna hai.

Verify: temperature sirf $9\%$ gira lekin pressure $44\%$ gira — pressure kaafi zyada sensitive hai kyunki uska exponent (6) temperature ke exponent (1) se bahut bada hai. $a^{\ast }$ ki units: $\sqrt{(\text{J/kg⋅K})(\text{K})}=\sqrt{\text{J/kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✓.

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Example 2 — Cell B: ek subsonic station

Forecast: subsonic hai, toh abhi tak koi khaas acceleration nahi — expect karo $T,P$ chamber values ke kareeb, aur area throat se badi.

1. Bracket factor. $1+\frac{\gamma -1}{2}{M}^2=1+0.1(0.25)=1.025$. Yeh step kyun? Yeh ek number charon formulas mein jaata hai — ise ek baar compute karo aur reuse karo, taaki ek arithmetic slip teen jagah chhup na sake.
2. Temperature. $T=T_0/1.025=3500/1.025=3414.6\text{K}$. Yeh step kyun? Temperature sirf $T_0$ ko bracket se divide karna hai — gas ne sirf thodi si thermal energy di hai kyunki woh barely move kar rahi hai.
3. Pressure. $P=P_0/1.025^6=10/1.1596=8.624\text{MPa}$. Yeh step kyun? Same bracket, ab pressure exponent $6$ tak raise kiya; pressure temperature se zyada tezi se girta hai, jaise hamesha hota hai.
4. Area ratio. $\frac{A}{A^{\ast }}=\frac{1}{0.5}[\frac{2}{2.2}(1.025){]}^{5.5}=2\times (0.9318{)}^{5.5}=2\times 0.6807=1.361$. Yeh step kyun? Inside term $\frac{2}{2.2}(1.025)=0.9318$ hai jo 1 se neeche hai, toh $5.5$ tak raise karne se yeh chhota ho jaata hai — $1/M=2$ out front phir ratio ko 1 se upar lift karta hai, duct throat se wider milta hai.

Verify: $A/A^{\ast }=1.36>1$ ✓ — converging section throat se wider hai, subsonic ke liye sahi. Ex 3 se compare karo: same value of area ratio supersonically bhi appear ho sakta hai, yahi matrix ka poora point hai.

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Example 3 — Cell C: ek supersonic station

Forecast: badi acceleration — expect karo temperature roughly half, pressure do orders of magnitude neeche, velocity near $2500\text{m/s}$, aur wide exit bell.

1. Bracket factor. $1+0.1(9)=1.9$. Yeh step kyun? Pehle ki tarah same master number; $M^2=9$ ise subsonic case se kaafi bada banata hai, badi energy conversion signal karta hai.
2. Temperature. $T_e=3500/1.9=1842.1\text{K}$. Yeh step kyun? $T_0$ ko bracket se divide karo — aadhi thermal energy already motion ban chuki hai.
3. Pressure. $P_e=10/1.9^6=10/47.05=0.2125\text{MPa}=212.5\text{kPa}$. Yeh step kyun? Bracket ko $6$th power tak; bada drop hi woh cheez hai jo acceleration drive karti hai.
4. Velocity. $V_e=3\sqrt{1.2\times 320\times 1842.1}=3\times 841.4=2524\text{m/s}$. Yeh step kyun? $V=Ma$ with $a=\sqrt{\gamma R{T}_e}$ — exhaust speed jo ultimately thrust banati hai.
5. Area ratio. Pehle inside term: $\frac{2}{2.2}(1.9)=0.9091\times 1.9=1.7273$. $5.5$ tak raise karo: $1.7273^{5.5}=20.22$. Phir $\frac{A_e}{A^{\ast }}=\frac{1}{3}\times 20.22=6.74$. Yeh step kyun? Yahan inside term $>1$ hai, toh $5.5$ power ise strongly amplify karti hai (to $20.22$); $M=3$ se divide karne par wide expansion ratio $6.74$ milti hai. Yahi woh exponent hai jise sablog galat evaluate karte hain — note karo $1.7273^{5.5}\ne 14.14$.

Verify: $P_e=212\text{kPa}>P_a=101\text{kPa}$, toh jet ambient se upar nikalta hai → nozzle under-expanded hai (aur expand ho sakta tha). Note karo $A_e/A^{\ast }=6.74>1$ phir se, bilkul subsonic Ex 2 ki tarah — prove karta hai ek area ratio, do regimes. Links: Thrust Equation Derivation, Real Nozzle Losses.

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Example 4 — Cell D: degenerate input, M → 0 (chamber khud)

Forecast: koi motion nahi, toh static = stagnation, isliye teeno thermodynamic ratios → 1. Lekin area ratio mein $1/M$ out front hai — kya hoga?

1. Thermodynamic ratios. $1+\frac{\gamma -1}{2}(0{)}^2=1$, toh $T/T_0=1$, $P/P_0=1$, aur $\rho /{\rho }_0=1^{-1/(\gamma -1)}=1$. Yeh step kyun? Koi kinetic energy nahi yaani koi conversion nahi — gas hi stagnation par hai. Density ratio matter karta hai kyunki woh $\rho$ hai jo mass flow $\rho VA=\text{const}$ mein appear hota hai; $M=0$ par yeh bilkul chamber density ${\rho }_0$ ke barabar hai, jise hum agale step mein use karte hain.
2. Area ratio limit. $\frac{A}{A^{\ast }}=\frac{1}{M}[\dots {]}^{5.5}\to \infty$ as $M\to 0$. Yeh step kyun? $1/M$ blow up karta hai jabki bracket term finite rahta hai (yeh $(2/2.2{)}^{5.5}$ tend karta hai). Physically: mass flow $\rho VA=\text{const}$ mein $\rho \to {\rho }_0$ (step 1 se) aur $V\to 0$ area $\to \infty$ force karta hai. Real chamber approximately aisa hi hai — throat ke compare mein huge area.

Verify: $M=0.05$ par, inside term hai $\frac{2}{2.2}(1+0.1\times 0.0025)=0.9091\times 1.00025=0.9093$, $5.5$ tak raise karne par $0.6318$ milta hai, $1/0.05=20$ se multiply karne par $A/A^{\ast }=12.64$ milta hai — pehle se large hai aur $M$ shrink hone ke saath climb kar raha hai, blow-up confirm karta hai. Yahi degenerate-input case hai jo naive user bhool jaata hai: chamber par area formula invert nahi kar sakte, sirf throat ke paas aur uske baad.

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Example 5 — Cell E: limiting behaviour, M → ∞

Forecast: infinite speed nahi mil sakti — saari thermal energy finite hai. Guess: ek fixed number jo $c_p{T}_0$ se set hota hai.

1. Energy equation. Parent se steady-flow energy equation, $h+\frac{V^2}{2}=h_0$, ideal gas ke liye ($h=c_{p}T$, $h_0=c_p{T}_0$) ban jaata hai: $c_{p}T+\frac{V^2}{2}=c_p{T}_0$. Yeh stagnation-enthalpy conservation hi hai — yeh hold karta hai kyunki flow adiabatic hai (wall se koi heat cross nahi karti) aur steady with no shaft work, exactly parent ke isentropic-flow assumptions. Phir jab $T\to 0$ (saari thermal energy kharach), $V\to V_{\max }=\sqrt{2c_p{T}_0}$. Yeh step kyun? Yahi physical ceiling hai — tum $c_p{T}_0$ total thermal energy se zyada kinetic energy extract nahi kar sakte jo tumhare paas shuru mein thi, kyunki streamline ke saath total enthalpy conserved hai.
2. $c_p$ compute karo. $c_p=\frac{\gamma R}{\gamma -1}=\frac{1.2\times 320}{0.2}=1920\text{J/(kg⋅K)}$. Yeh step kyun? Temperature ko velocity mein convert karne ke liye $c_p$ absolute units mein chahiye.
3. Ceiling. $V_{\max }=\sqrt{2\times 1920\times 3500}=\sqrt{1.344\times 10^7}=3666\text{m/s}$. Yeh step kyun? Numbers ko seedha $\sqrt{2c_p{T}_0}$ mein plug karo.

Verify: hamare $M=3$ exit ne $2524\text{m/s}$ diya — comfortably $3666\text{m/s}$ ceiling ke neeche ✓. Koi bhi finite nozzle $V_{\max }$ tak nahi pahunchta (infinite area chahiye hogi), toh real exhaust speeds hamesha is bar ke neeche rehti hain. Dekhein Specific Impulse Optimization aur Entropy and Reversibility.

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Example 6 — Cell F: inverse problem, area ratio → Mach (do roots)

Forecast: figure s01 se, $4.0$ par ek horizontal line curve ko do baar cut karti hai — ek subsonic root, ek supersonic root. Dono mathematically valid hain.

1. Equation set up karo. $4.0=\frac{1}{M}[\frac{2}{2.2}(1+0.1M^2){]}^{5.5}$ numerically solve karo. Yeh step kyun? Koi closed-form inverse exist nahi karta — yeh inherently ek root-finding problem hai, exactly yahi reason hai ki cell F apna alag case hai.
2. Supersonic root (bisection). Try $M=2.6$: inside $=\frac{2}{2.2}(1+0.1\times 6.76)=0.9091\times 1.676=1.5236$, $1.5236^{5.5}=10.30$, $/2.6=3.96$ (thoda low). Try $M=2.62$: inside $=0.9091\times 1.6864=1.5331$, ${}^{5.5}=10.60$, $/2.62=4.046$ (thoda high). Value $4.0$ beech mein hai, $M\approx 2.611$ par. Yeh step kyun? Bisection root ko bracket karta hai: ek trial $4.0$ se thoda neeche deta hai, agla thoda upar, toh answer $2.60$ aur $2.62$ ke beech trapped hai.
3. Subsonic root (bisection). Try $M=0.15$: inside $=0.9091\times 1.00225=0.9111$, ${}^{5.5}=0.6024$, $/0.15=4.016$ (thoda high). Try $M=0.151$: inside $=0.9091\times 1.00228=0.9112$, ${}^{5.5}=0.6026$, $/0.151=3.991$ (thoda low). Toh subsonic root $M\approx 0.1506$ hai. Yeh step kyun? Blue branch par same bracketing, jahan $1/M$ term dominate karta hai toh chhota $M$ badi ratio produce karta hai.

Verify: $M=2.611$ back plug karo: inside $=0.9091\times (1+0.1\times 6.817)=0.9091\times 1.6817=1.5288$; $1.5288^{5.5}=10.44$; $/2.611=3.999\approx 4.0$ ✓. $M=0.1506$ plug karo: inside $=0.9091\times 1.00227=0.9112$; ${}^{5.5}=0.6025$; $/0.1506=4.001\approx 4.0$ ✓. Do Mach numbers ek area ratio share karte hain, aur sirf physics (agla example) real wala choose karta hai.

Figure s02 — hardware mein har cell kahan rehta hai. Blue outline nozzle wall hai: yeh wide chamber (left) se converge karke throat tak jaata hai (yellow dashed line, $M=1$, area $A^{\ast }$) aur phir wide exit tak diverge karta hai (right). Axis ke saath red arrow flow direction hai. Green text subsonic converging region mark karta hai (cell B, Ex 2), red text supersonic diverging region (cells C, G, H), aur far-left label near-stagnant chamber mark karta hai ($M\to 0$, cell D, Ex 4). Picture ko left-to-right padhna gas ko rest se accelerate hote, throat par Mach 1 se, exit par supersonic tak padhna hai.

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Example 7 — Cell G: real-world design

Forecast: $2900$ Ex 5 ki $3666$ ceiling ke kareeb hai, toh fairly high Mach chahiye hoga (3 se kaafi upar) aur large area ratio.

1. $V_e$ ko $M$ se relate karo. $V_e=M\sqrt{\gamma RT}$ with $T=T_0/(1+0.1M^2)$. Toh $V_e=M\sqrt{\frac{\gamma R{T}_0}{1+0.1M^2}}$. Yeh step kyun? Ek equation, ek unknown $M$ — velocity Mach fix karti hai.
2. Solve karo. $\gamma R{T}_0=1.2\times 320\times 3500=1.344\times 10^6$. Numerically $M\approx 4.20$. Yeh step kyun? Velocity equation iterate karne par yahan converge hota hai; high value forecast confirm karta hai.
3. Area ratio. inside term $\frac{2}{2.2}(1+0.1\times 17.64)=0.9091\times 2.764=2.5127$; $5.5$ tak raise karo: $2.5127^{5.5}=158.8$; $M=4.20$ se divide karo: $\frac{A_e}{A^{\ast }}=\frac{158.8}{4.20}=37.8$. Yeh step kyun? Ex 3 ki tarah same careful exponent handling — inside term ab 1 se kaafi upar hai, toh $5.5$ power ise enormously inflate karti hai, ek vacuum engine ki badi bell deti hai.

Verify: $M=4.20$ back-substitute karo: $T=3500/2.764=1266.3\text{K}$, $V_e=4.20\sqrt{1.2\times 320\times 1266.3}=4.20\times 697.5=2930\text{m/s}\approx 2900$ ✓ (rounding). $\sim 38$ ka ratio exactly woh type ka huge bell hai jo upper-stage vacuum engines par dikhta hai. Dekhein Nozzle Flow Regimes ki bade ratios vacuum ke liye kyun suit karte hain.

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Example 8 — Cell H: exam twist (regime trap)

Forecast: geometry force karti hai supersonic root (jab flow throat par choke ho jaata hai aur attached rehta hai), toh $M_e$ aur $P_e$ sirf geometry se fixed hain — back-pressure sirf shock/plume behaviour decide karta hai, $M_e$ nahi.

1. Branch choose karo. Attached supersonic flow → $M_e=3.0$ use karo, same area ratio ka subsonic root nahi. Yeh step kyun? Yahi trap hai: subsonic root sirf purely converging duct ya unchoked nozzle mein hota hai. Full chala hua supersonic diverging nozzle upper branch par rehta hai.
2. Exit pressure. bracket $=1+0.1(9)=1.9$; $P_e=10/1.9^6=0.2125\text{MPa}=212.5\text{kPa}$. Yeh step kyun? $P_e$ purely $M_e$ se set hota hai, jise geometry ne fix kiya — Ex 3 ka number reuse karo.
3. Case (a) $P_a=250$ kPa. $P_e=212.5<P_a=250$ → over-expanded, ambient jet ko squeeze karta hai, lip par oblique shocks form hote hain. Yeh step kyun? Over-expanded ka matlab nozzle ne pressure ambient se neeche drop kar diya.
4. Case (b) $P_a=20$ kPa. $P_e=212.5>P_a=20$ → under-expanded, plume bahar expand karta rehta hai. Yeh step kyun? Same fixed $P_e$, lekin ab yeh ambient se zyada hai — label sirf isliye flip hota hai kyunki $P_a$ badla.

Verify: $P_e$ sirf fixed $M_e$ par depend karta hai, toh dono cases mein same $212.5$ kPa hai — back-pressure ne kabhi $P_e$ nahi badla ✓. Sirf comparison $P_e$ vs $P_a$ label flip karta hai (over-expanded at $250$ kPa, under-expanded at $20$ kPa). Dekhein Real Nozzle Losses ki woh shocks aage kya karte hain.

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Recall Quick self-test

Ek area ratio kitne Mach numbers deta hai, aur tum choose kaise karte ho? ::: Do (ek subsonic, ek supersonic). Flow regime se choose karo: converging/unchoked → subsonic root; diverging full-running → supersonic root. Exhaust velocity par absolute ceiling kya set karta hai? ::: $V_{\max }=\sqrt{2c_p{T}_0}$ — total thermal energy fully converted, sirf $M\to \infty$ par reach hota hai. $M=0$ par, $A/A^{\ast }$ kya equals hota hai? ::: $\infty$ — is idealisation mein chamber throat ke compare mein infinitely wide hai.