Chamber-to-exit relation - all quantities as f(M_e, γ): Worked examples

3.3.12 · D3 · Physics › Rocket Propulsion › Chamber-to-exit relation - all quantities as f(M_e, γ)

Shuru karne se pehle, vocabulary ka ek reminder, taaki koi bhi symbol use hone se pehle samajh aa jaye:

Definition Is page par zaroori symbols

$M_{e}$ ::: exit Mach number = exit speed divided by local speed of sound. Ek pure number, koi units nahi.
$γ$ ::: exhaust gas ke specific heats ka ratio (rocket products ke liye lagbhag $1.2$ , air ke liye $1.4$ ). Ek pure number.
Subscript $0$ ::: stagnation (chamber) value — jo aap measure karte agar flow ko gently rest par laaya jaata.
Subscript $e$ ::: exit plane par actually present static value.
$A^{*}$ ::: throat area, wo cross-section jahan flow exactly sonic hai ( $M = 1$ ).
$R$ ::: specific gas constant, units $J/(kg\cdotK)$ .
$V_{e}$ ::: exit velocity — exhaust gas ki actual speed jab wo exit plane chodta hai, units $m/s$ .
$a_{e}$ ::: exit par local speed of sound, $a_{e} = γ R T_{e}$ , units $m/s$ . (To $M_{e} = V_{e} / a_{e}$ .)
$c_{p}$ ::: specific heat at constant pressure, $c_{p} = γ R / (γ - 1)$ , units $J/(kg\cdotK)$ . Yeh measure karta hai ki gas ka har kilogram per degree kitni thermal energy store karta hai.
$\overset{m}{˙}$ ::: mass flow rate — kilograms of exhaust jo kisi bhi cross-section se per second guzarti hai, units $kg/s$ .
$g_{0}$ ::: standard gravity, $9.81 m/s^{2}$ , exhaust speed ko Specific Impulse mein seconds mein convert karne ke liye use hota hai.

Poore page par hum ek grouping par rely karte hain jo har jagah aata hai. Ise ek naam dete hain:

Scenario matrix

Is topic mein jo bhi aa sakta hai wo inhi cells mein se ek mein aata hai. Neeche har worked example us cell ke saath tagged hai jo wo cover karta hai.

Cell	Kya cheez ise alag banati hai	Covered by
A. Subsonic exit $M_{e} < 1$	diverging section flow ko slow karega; ratios mild	Ex 1
B. Sonic / throat $M_{e} = 1$	reference state $A^{}$ khud; $A_{e} / A^{} = 1$	Ex 2
C. Supersonic exit $M_{e} > 1$	real rocket case; steep drops	Ex 3
D. Degenerate $M_{e} = 0$	chamber = exit; saare ratios $1$ pe collapse	Ex 4
E. Limiting $M_{e} \to \infty$	vacuum limit, $V_{e} \to V_{m a x}$	Ex 5
F. Two-branch trap given area ratio, find $M_{e}$	ek $A_{e} / A^{*}$ ⇒ do Mach numbers	Ex 6 (figure)
G. Real-world word problem	full chamber→thrust chain	Ex 7
H. Exam twist $M_{e}$ hidden behind $P_{e} / P_{0}$	pressure relation ko invert karo	Ex 8
I. Sensitivity to $γ$	same $M_{e}$ , different gas	Ex 9

Agar koi bhi cell shaky lage toh prerequisites: Isentropic Flow Relations, Area Ratio and Mach Number, Converging-Diverging Nozzle.

Worked example Ex 1 — Cell A: Subsonic exit (

$M_{e} = 0.5$ , $γ = 1.4$ ) Given air, $M_{e} = 0.5$ , $T_{0} = 300 K$ , $P_{0} = 2.00 bar$ . $T_{e}$ , $P_{e}$ , $ρ_{e} / ρ_{0}$ nikalo.

Forecast: Kya temperature zyada girega ya thoda? (Guess: sirf thoda — flow slow hai, isliye wo thodi si thermal energy hi leta hai.)

$ϕ$ compute karo. $ϕ = 1 + \frac{0.4}{2} (0.5)^{2} = 1 + 0.2 \cdot 0.25 = 1.05$ . Yeh step kyun? Har ratio $ϕ$ ki ek power hai; ise ek baar nikaal lo.
Temperature. $T_{e} = T_{0} ϕ^{- 1} = 300/1.05 = 285.7 K$ . Kyun? Energy conservation se $T_{e} / T_{0} = ϕ^{- 1}$ .
Pressure. Exponent $- γ / (γ - 1) = - 1.4/0.4 = - 3.5$ . $P_{e} = 2.00 \cdot (1.05)^{- 3.5} = 2.00 \cdot 0.84261 = 1.685 bar$ . Yeh step kyun? Pressure isentrope ke along temperature follow karta hai, lekin extra exponent $- γ / (γ - 1)$ ke saath $P V^{γ} = const$ se — isliye pressure temperature se faster girta hai.
Density. Exponent $- 1/ (γ - 1) = - 2.5$ . $ρ_{e} / ρ_{0} = (1.05)^{- 2.5} = 0.88474$ . Yeh step kyun? Density fix ho jaati hai jab $P$ aur $T$ pata ho, ideal-gas law $ρ \propto P / T$ se; iska exponent exactly $γ / (γ - 1) - 1 = 1/ (γ - 1)$ hai.

Verify: Ideal-gas identity se cross-check: $ρ_{e} / ρ_{0} = (P_{e} / P_{0}) / (T_{e} / T_{0}) = 0.84261/0.95238 = 0.88474$ — step 4 se exactly match ✓. Units: saare ratios dimensionless ✓; $T_{e} < T_{0}$ aur $P_{e} < P_{0}$ jaise accelerating flow mein expect karte hain ✓. Forecast ke mutabiq, changes chote hain ( $T$ mein < 6%).

Worked example Ex 2 — Cell B: Sonic throat (

$M_{e} = 1$ , $γ = 1.2$ ) Given $M_{e} = 1$ . Har ratio aur area ratio $A_{e} / A^{*}$ nikalo.

Forecast: Jab exit khud sonic point ho toh $A_{e} / A^{*}$ kya hoga? (Guess: exactly 1 — exit hi throat hai.)

$M = 1$ par $ϕ$ . $ϕ^{*} = 1 + \frac{0.2}{2} (1)^{2} = 1.1$ . Kyun? Yeh throat "starred" reference state define karta hai.
Temperature ratio. $T^{*} / T_{0} = 1/1.1 = 0.9091$ . Yeh step kyun? $T / T_{0} = ϕ^{- 1}$ hamaara master temperature relation hai; yahan sonic point par evaluate karke throat temperature milti hai.
Pressure ratio. exponent $- 1.2/0.2 = - 6$ : $P^{*} / P_{0} = (1.1)^{- 6} = 0.5645$ . Yeh step kyun? Yeh particular value ( $M = 1$ par pressure) woh "critical pressure ratio" hai jo batata hai ki nozzle kab choke hoti hai — ise khud compute karna zaroori hai.
Area ratio. $M_{e} = 1$ ko area–Mach formula mein daalo: $\frac{A _{e}}{A ^{*}} = \frac{1}{1} [\frac{2}{2.2} \cdot 1.1]^{\frac{2.2}{0.4}} = [1]^{5.5} = 1.$ Yeh step kyun? Bracket ban jaata hai $\frac{2}{γ + 1} ϕ^{*} = \frac{2}{2.2} \cdot 1.1 = 1$ , aur $1$ kisi bhi power par $1$ hota hai.

Verify: $A_{e} / A^{*} = 1$ exactly — exit throat ke saath coincide karta hai, jaise forecast kiya tha ✓. Sanity: $P^{*} / P_{0} = 0.5645$ $γ = 1.2$ ke liye classic "critical pressure ratio" hai (choked-flow condition) ✓.

Worked example Ex 3 — Cell C: Supersonic exit (

$M_{e} = 2.5$ , $γ = 1.4$ ) Given air, $M_{e} = 2.5$ , $T_{0} = 600 K$ , $R = 287 J/(kg\cdotK)$ . $T_{e}$ , $P_{e} / P_{0}$ , $V_{e}$ , $A_{e} / A^{*}$ nikalo.

Forecast: Ex 1 se bada $M_{e}$ — bada temperature drop aur mota area ratio expect karo. Kitna mota? Compute karne se pehle guess karo.

$ϕ$ . $ϕ = 1 + 0.2 \cdot (2.5)^{2} = 1 + 0.2 \cdot 6.25 = 2.25$ . Yeh step kyun? Hamesha pehle master group nikalo taaki baad ke har ratio ek one-line power ho.
Temperature. $T_{e} = 600/2.25 = 266.7 K$ . Kyun? Lagbhag $2.25 \times$ thanda — ab kinetic energy dominate kar rahi hai.
Pressure. exponent $- 3.5$ : $P_{e} / P_{0} = (2.25)^{- 3.5} = 0.0585$ . Yeh step kyun? Humein pressure ratio chahiye "50-to-1 drop jo thrust banata hai" feel karne ke liye aur yeh check karne ke liye ki flow fully expanded hai; yeh $ϕ$ ko $- γ / (γ - 1)$ power par raise karna hai.
Velocity. $V_{e} = M_{e} \frac{γ R T _{0}}{ϕ} = 2.5 \frac{1.4 \cdot 287 \cdot 600}{2.25}$ . Andar: $1.4 \cdot 287 \cdot 600 = 241, 080$ ; $/2.25 = 107, 147$ ; $= 327.3$ . To $V_{e} = 2.5 \cdot 327.3 = 818.4 m/s$ . Kyun? $V_{e} = M_{e} a_{e}$ with $a_{e} = γ R T_{e}$ ; denominator mein $ϕ$ chamber temperature ko exit temperature mein convert karta hai.
Area ratio. bracket $= \frac{2}{2.4} \cdot 2.25 = 1.875$ ; exponent $\frac{2.4}{0.8} = 3$ : $A_{e} / A^{*} = \frac{1}{2.5} (1.875)^{3} = \frac{6.592}{2.5} = 2.637$ . Yeh step kyun? Flow relations thermodynamics set karte hain; area ratio unhe geometry mein translate karta hai jo tumhe actually machine karni hai — $M_{e} > 1$ ke liye diverging cone ko is factor se chauda karna padega.

Verify: Standard isentropic tables $M = 2.5$ , $γ = 1.4$ par $A / A^{*} = 2.637$ dete hain ✓. Speed check $a_{e}$ se: $a_{e} = 1.4 \cdot 287 \cdot 266.7 = 327.2$ , aur $V_{e} = 2.5 \cdot 327.2 = 818$ ✓ (step 4 jaisa). $V_{e}$ ke units: $J/kg = m^{2} / s^{2} = m/s$ ✓.

Worked example Ex 4 — Cell D: Degenerate zero-Mach (

$M_{e} = 0$ ) Given $M_{e} = 0$ , koi bhi $γ$ . Saare ratios kya hain?

Forecast: Agar exit flow move hi nahi kar rahi, toh exit chamber se alag kaise hoga? (Guess: bilkul nahi.)

$ϕ$ . $ϕ = 1 + \frac{γ - 1}{2} (0)^{2} = 1$ . Kyun? Koi motion nahi ⇒ koi kinetic-energy term nahi.
Saare ratios. $ϕ^{- 1} = 1$ , $ϕ^{- γ / (γ - 1)} = 1$ , $ϕ^{- 1/ (γ - 1)} = 1$ . To $T_{e} = T_{0}, P_{e} = P_{0}, ρ_{e} = ρ_{0}$ . Yeh step kyun? Har ratio $ϕ$ ki ek power hai, aur $1$ ki koi bhi power $1$ hoti hai — isliye koi flow na hone par exit state simply chamber state hi hai.
Velocity. $V_{e} = 0 \cdot (\dots) = 0$ . Yeh step kyun? $V_{e} = M_{e} a_{e}$ ; $M_{e} = 0$ ke saath exhaust move hi nahi kar raha, jo "chamber = exit" ko physically confirm karta hai.
Area ratio. $A_{e} / A^{*} = \frac{1}{0} [\dots]$ → infinity ki taraf diverge karta hai. Kyun? Zero speed par same mass flow carry karne ke liye infinitely wide passage chahiye. Physically, $M_{e} = 0$ real supersonic nozzle mein nahi pahuncha ja sakta — yeh mathematical edge hai.

Verify: $ϕ = 1$ par $1$ ki har power $1$ hai, to exit = chamber, "koi expansion nahi hua" se consistent ✓. $1/ M_{e}$ ka blow-up sahi se flag karta hai ki $M_{e} = 0$ flowing nozzle ke liye unphysical hai ✓.

Worked example Ex 5 — Cell E: Vacuum limit (

$M_{e} \to \infty$ ) Given $γ = 1.2$ , $T_{0} = 3500 K$ , $R = 520 J/(kg\cdotK)$ . Maximum possible exit velocity kya hai?

Forecast: Jaise $M_{e}$ badhta hai, $V_{e}$ badhta rehta hai — lekin kya yeh forever badhta hai ya level off ho jaata hai? (Guess: level off ho jaata hai, kyunki temperature absolute zero se neeche nahi ja sakti.)

$V_{e}$ ka limit lo. $V_{e} = M_{e} \frac{γ R T _{0}}{1 + \frac{γ - 1}{2} M _{e}^{2}}$ . Bade $M_{e}$ ke liye " $1 +$ " negligible hai: $V_{e} \approx M_{e} \frac{γ R T _{0}}{\frac{γ - 1}{2} M _{e}^{2}} = \frac{2 γ R T _{0}}{γ - 1}$ . Yeh step kyun? Andar ka $M_{e}^{2}$ bahar ke $M_{e}$ ko cancel karta hai, ek finite ceiling $V_{m a x}$ chodta hai.
Number. $V_{m a x} = \frac{2 \cdot 1.2 \cdot 520 \cdot 3500}{0.2} = \frac{4 , 368 , 000}{0.2} = 21, 840, 000 = 4673 m/s$ . Yeh step kyun? Actual chamber numbers plug karne se abstract ceiling ek concrete "is gas se koi exhaust 4673 m/s nahi beat kar sakta" ban jaati hai — performance ki hard physical limit.
Interpretation. Saari thermal energy directed kinetic energy ban gayi; $T_{e} \to 0$ . Yeh step kyun? Yeh explain karta hai ki ceiling kyun exist karti hai: aap stored thermal energy $c_{p} T_{0}$ se zyada speed extract nahi kar sakte, aur woh pool finite hai.

Verify: Energy check: max KE per unit mass $= \frac{1}{2} V_{m a x}^{2} = c_{p} T_{0}$ with $c_{p} = γ R / (γ - 1) = 1.2 \cdot 520/0.2 = 3120$ . To $\frac{1}{2} (4673)^{2} = 1.092 \times 1 0^{7}$ aur $c_{p} T_{0} = 3120 \cdot 3500 = 1.092 \times 1 0^{7}$ ✓ — perfect match. Real nozzles shayad iska $70%$ hi reach karti hain; forecast ke mutabiq ek finite ceiling exist karti hai ✓.

Worked example Ex 6 — Cell F: Two-branch trap (given

$A_{e} / A^{*}$ , $M_{e}$ nikalo) Given $γ = 1.4$ aur area ratio $A_{e} / A^{*} = 2.0$ . $M_{e}$ nikalo.

Forecast: Ek area ratio — kitne Mach numbers ise satisfy karte hain? (Neeche figure dekhne se pehle guess karo.)

Neeche figure area ratio $A_{e} / A^{*}$ (red curve) ko exit Mach number $M_{e}$ (horizontal axis) ke against plot karta hai. Dhyan do ki curve $M_{e} = 1$ par exactly $1$ ki ek lowest point par dip karta hai (black square) aur dono sides par chahta hai. Dashed black line hamaara target height $A_{e} / A^{*} = 2.0$ hai; jahan yeh red curve ko cross karta hai (do black dots) wahi answers hain.

Shape dekho. Figure mein, red curve $A_{e} / A^{*}$ vs $M_{e}$ ka minimum 1 hai $M = 1$ par aur dono sides par rise karta hai. To height $2.0$ par ek horizontal line ise do baar kaategi. Kyun? Ek converging-diverging nozzle same mass ko same area se ya subsonically ( $M < 1$ ) ya supersonically ( $M > 1$ ) pass karta hai.
Subsonic root. Numerically solve karne par, $M_{e} \approx 0.306$ . Yeh step kyun? Left crossing subsonic solution hai — ek nozzle jo poori tarah Mach 1 se neeche rehti hai woh bhi yeh area ratio produce kar sakti hai, isliye hume ise bhi report karna chahiye.
Supersonic root. Numerically solve karne par, $M_{e} \approx 2.197$ . Kyun? Right crossing supersonic solution hai; formula genuinely two-valued hai, isliye geometry akele branch nahi chun sakti.
Physically pick karo. Thrust ke liye design ki gayi rocket nozzle supersonic branch par chalti hai: $M_{e} = 2.197$ . Yeh step kyun? Extra physics (nozzle "running full" hai / chamber pressure high hai) supersonic root select karta hai; us context ke bina dono mathematically valid hain.

Verify: $M_{e} = 2.197$ plug karo: $ϕ = 1 + 0.2 (2.197)^{2} = 1.9654$ ; bracket $= \frac{2}{2.4} (1.9654) = 1.638$ ; $^{3} = 4.394$ ; $/2.197 = 2.000$ ✓. Aur $M_{e} = 0.306$ : $ϕ = 1.01874$ ; bracket $= 0.8489$ ; $^{3} = 0.6117$ ; $/0.306 = 1.999$ ✓. Dono branches confirm.

Worked example Ex 7 — Cell G: Real-world word problem (full chain to thrust)

Given ek upper-stage engine: $γ = 1.2$ , $T_{0} = 3200 K$ , $P_{0} = 50 bar$ , $R = 390 J/(kg\cdotK)$ , design exit $M_{e} = 3.5$ , mass flow $\overset{m}{˙} = 8.0 kg/s$ , matched exit pressure tak expand (ambient $= P_{e}$ ). $V_{e}$ aur ideal thrust $F = \overset{m}{˙} V_{e}$ nikalo.

Forecast: Exhaust roughly kitni tezi se nikalta hai — hundreds ya thousands of m/s? (Yeh Specific Impulse drive karta hai.)

$ϕ$ . $ϕ = 1 + 0.1 (3.5)^{2} = 1 + 1.225 = 2.225$ . Yeh step kyun? Har downstream quantity ( $T_{e}$ , phir $V_{e}$ , phir thrust) $ϕ$ par built hai, isliye ise pehle ek baar compute karo.
Exit temperature. $T_{e} = 3200/2.225 = 1438 K$ . Yeh step kyun? Humein $T_{e}$ chahiye kyunki exit speed local speed of sound $a_{e} = γ R T_{e}$ se set hoti hai — exit temperature ke bina velocity nahi nikal sakte.
Exit velocity. $V_{e} = 3.5 \frac{1.2 \cdot 390 \cdot 3200}{2.225}$ . Andar: $1.2 \cdot 390 \cdot 3200 = 1, 497, 600$ ; $/2.225 = 673, 079$ ; $= 820.4$ . $V_{e} = 3.5 \cdot 820.4 = 2871 m/s$ . Yeh step kyun? Thrust momentum flow $\overset{m}{˙} V_{e}$ hai, to $V_{e}$ pivotal number hai; yahan hum $V_{e} = M_{e} a_{e}$ ko step 2 ke temperature ke saath combine karte hain.
Thrust (pressure-matched). $F = \overset{m}{˙} V_{e} = 8.0 \cdot 2871 = 22, 970 N \approx 23.0 kN$ . Yeh step kyun? Jab exit pressure ambient ke barabar ho, Thrust Equation sirf momentum term $\overset{m}{˙} V_{e}$ tak reduce ho jaata hai.

Verify: Units: $kg/s \cdot m/s = kg\cdotm/s^{2} = N$ ✓. $V_{e}$ cross-check $a_{e} = 1.2 \cdot 390 \cdot 1438 = 820.3$ , $V_{e} = 3.5 \cdot 820.3 = 2871$ ✓. Specific impulse sanity: $I_{s p} = V_{e} / g_{0} = 2871/9.81 = 293 s$ , ek believable upper-stage value ✓.

Worked example Ex 8 — Cell H: Exam twist (

$M_{e}$ hidden behind $P_{e} / P_{0}$ ) Given $γ = 1.3$ , aur sirf $P_{e} / P_{0} = 0.02$ bataya gaya hai. $M_{e}$ aur $T_{e} / T_{0}$ nikalo.

Forecast: Mach number seedha nahi diya — tumhe pressure relation invert karni hogi. Kya $M_{e}$ 1 se upar ya neeche hoga? (Guess: bahut upar, kyunki pressure 50 guna gir gaya.)

Pressure relation likho. $P_{e} / P_{0} = ϕ^{- γ / (γ - 1)}$ , exponent $- 1.3/0.3 = - 4.333$ . Yeh step kyun? Yeh ek hi master relation hai jo diye gaye ( $P_{e} / P_{0}$ ) ko $ϕ$ se connect karta hai, jo $M_{e}$ ko chhupaata hai; rearrange karne se pehle ise likho.
$ϕ$ isolate karo. $ϕ = (P_{e} / P_{0})^{- (γ - 1) / γ} = (0.02)^{- 0.3/1.3} = (0.02)^{- 0.2308}$ . Compute: $ln 0.02 = - 3.912$ ; $\times (- 0.2308) = 0.9029$ ; $e^{0.9029} = 2.467$ . To $ϕ = 2.467$ . Yeh step kyun? Dono sides ko reciprocal power par raise karne se exponent undo ho jaata hai, $ϕ$ expose ho jaata hai.
$M_{e}$ nikalo. $ϕ = 1 + \frac{γ - 1}{2} M_{e}^{2} \Rightarrow M_{e} = \frac{2 ( ϕ - 1 )}{γ - 1} = \frac{2 ( 1.467 )}{0.3} = 9.78 = 3.127$ . Yeh step kyun? $ϕ$ sirf $M_{e}$ mein ek rearranged quadratic hai; ise solve karne se woh Mach number milta hai jo problem ne chhupaane ki koshish ki thi.
Temperature ratio. $T_{e} / T_{0} = ϕ^{- 1} = 1/2.467 = 0.4054$ . Yeh step kyun? Jab $ϕ$ pata ho, temperature ratio free hai — yeh $ϕ$ ki sabse simple power hai ( $ϕ^{- 1}$ ).

Verify: Forward-check: $ϕ = 2.467$ , $ϕ^{- 4.333}$ : $ln 2.467 = 0.9029$ ; $\times (- 4.333) = - 3.912$ ; $e^{- 3.912} = 0.0200$ ✓ diya gaya $P_{e} / P_{0}$ recover karta hai. $M_{e} = 3.13 > 1$ forecast ke mutabiq ✓.

Worked example Ex 9 — Cell I:

$γ$ ke baare mein sensitivity (same $M_{e}$ , do gases) Given $M_{e} = 3.0$ fixed. $γ = 1.2$ vs $γ = 1.4$ ke liye $T_{e} / T_{0}$ aur $P_{e} / P_{0}$ compare karo.

Forecast: Same Mach number par kaun sa gas exhaust ko zyada garam rakhta hai? (Guess: halka wala, low $γ$ — woh energy zyada molecular modes mein store karta hai.)

Har ek ke liye $ϕ$ . $γ = 1.2$ : $ϕ = 1 + 0.1 \cdot 9 = 1.9$ . $γ = 1.4$ : $ϕ = 1 + 0.2 \cdot 9 = 2.8$ . Yeh step kyun? Poora comparison really "kaise $ϕ$ $γ$ ke saath badalta hai?" hai — baaki sab iska power hai, isliye pehle yahan difference isolate karo.
Temperature ratio. $γ = 1.2$ : $1/1.9 = 0.5263$ . $γ = 1.4$ : $1/2.8 = 0.3571$ . Kyun? Zyada $γ$ ⇒ bada coefficient $\frac{γ - 1}{2}$ ⇒ bada $ϕ$ ⇒ thanda exit.
Pressure ratio. $γ = 1.2$ : exponent $- 6$ , $(1.9)^{- 6} = 0.02128$ . $γ = 1.4$ : exponent $- 3.5$ , $(2.8)^{- 3.5} = 0.02722$ . Yeh step kyun? $γ$ ke saath base $ϕ$ aur exponent $- γ / (γ - 1)$ dono change hote hain, isliye pressure sensitivity temperature se genuinely alag calculation hai — apni line ki haqdar hai.
Interpretation. Lower $γ$ same $M_{e}$ par zyada thermal energy rakhta hai — ek wajah ki H₂-rich exhaust (low $γ$ ) high Specific Impulse aur achha Characteristic Velocity c-star kamaata hai. Yeh step kyun? Yeh do abstract ratios ko ek design rule mein turn karta hai: low- $γ$ propellant chuno jab aap garam, energetic exhaust aur high specific impulse chahte ho.

Verify: Parent note ke Worked Example 2 table se exactly match karta hai: $0.526, 0.357, 0.0213, 0.0272$ ✓. Forecast ke mutabiq, low- $γ$ gas zyada garam rehta hai ( $0.526 > 0.357$ ) ✓.

Recall Har example ne kaun sa cell hit kiya?

A subsonic ::: Ex 1 B sonic throat ( $A_{e} / A^{*} = 1$ ) ::: Ex 2 C supersonic ::: Ex 3 D degenerate $M_{e} = 0$ ::: Ex 4 E limit $M_{e} \to \infty$ (vacuum $V_{m a x}$ ) ::: Ex 5 F two-branch area-ratio inversion ::: Ex 6 G word problem → thrust ::: Ex 7 H invert $P_{e} / P_{0}$ to find hidden $M_{e}$ ::: Ex 8 I sensitivity to $γ$ ::: Ex 9