Altitude compensation methods — nozzle extension, aerospike: Worked examples

3.3.16 · D3 · Physics › Rocket Propulsion › Altitude compensation methods — nozzle extension, aerospike

Ye page Altitude compensation methods — nozzle extension, aerospike ka "drill ground" hai. Parent note ne ideas build kiye; yahan hum har tarah ke case grind karte hain jo ek problem throw kar sakti hai, taaki exam ya design meeting mein koi scenario surprise na kare.

Shuru karne se pehle, is page ke har symbol ko earn karte hain.

Woh ek equation jis par hum sabse zyada rely karte hain:

The scenario matrix

Is topic ki har problem inhi cells mein se ek mein land hoti hai. Neeche ke worked examples un cells ke saath labelled hain jo wo cover karte hain, aur milke ye sab ko hit karte hain. (Ab jo $ϵ$ aur uska $p_{e}$ se link upar define ho gaya, Cell E saaf padh'ta hai: $ϵ \to \infty$ push karna matlab mouth ko bina limit ke widen karna, jo $p_{e} \to 0$ drive karta hai.)

#	Cell (case class)	Isko distinct kya banata hai	Covered by
A	$p_{e} > p_{a}$ — under-expanded	pressure term positive; energy "left on the table"	Ex 1
B	$p_{e} = p_{a}$ — perfectly expanded	pressure term zero; design sweet-spot	Ex 2
C	$p_{e} < p_{a}$ — over-expanded	pressure term negative; atmosphere wapas push karta hai	Ex 3
D	Zero / degenerate: $p_{a} = 0$ (vacuum)	ambient term vanish ho jaata hai; upper-stage regime	Ex 4
E	Limiting value: $p_{e} \to 0$ as $ϵ \to \infty$	diminishing returns, $v_{e}$ saturate hoti hai	Ex 5
F	Geometry from length ( $ϵ$ from $L$ , $α$ )	conical-nozzle trig, extendable case	Ex 6
G	Real-world word problem: same engine, do altitudes	ek nozzle, thrust ka pressure term ka sign flip hota hai	Ex 7
H	Exam twist: aerospike vs bell, "kaun kahan jeetta hai?"	conceptual + break-even reasoning	Ex 8

Throughout use hone wale constants: $γ = 1.2$ = heat capacity ratio (ek gas property jo set karti hai ki wo kaise expand hoti hai); chamber pressure aur areas har problem mein given hain.

Example 1 — Cell A: Under-expanded ( $p_{e} > p_{a}$ )

Worked example Sea level par ek chhota nozzle

Ek nozzle $p_{e} = 150$ kPa par exit karti hai. Ye sea level par fire hoti hai, $p_{a} = 101$ kPa. Given $\overset{m}{˙} = 200$ kg/s, $v_{e} = 2600$ m/s, $A_{e} = 0.5$ m². Thrust find karo aur expansion classify karo.

Forecast: Pehle guess karo — kya pressure term help kar raha hai ya hurt? Kyunki gas zyada zyada bahar push kar rahi hai ( $150$ ) banasbat hawa ke ( $101$ ), isko thrust add karna chahiye. Kya ye badi correction hogi ya chhoti?

Step 1 — Classify. $p_{e}$ aur $p_{a}$ compare karo: $150 > 101$ , toh $p_{e} > p_{a}$ → under-expanded. Ye step kyun? $(p_{e} - p_{a})$ ka sign sab decide karta hai; pehle classify karne se compute karne se pehle answer ka flavour pata chalta hai.

Step 2 — Momentum term. $\overset{m}{˙} v_{e} = 200 \times 2600 = 520, 000$ N. Ye step kyun? Ye mass ko backward throw karne ka forward kick hai — altitude se koi fark nahi padta, isliye ye thrust ka "base" hai jis par hum build karte hain.

Step 3 — Pressure term. $(p_{e} - p_{a}) A_{e} = (150, 000 - 101, 000) \times 0.5 = 49, 000 \times 0.5 = 24, 500$ N. Ye step kyun? kPa → Pa convert karo (×1000) taaki units match karen ( $Pa \cdot m^{2} = N$ ). Term positive hai, jo hamara forecast confirm karta hai.

Step 4 — Total. $F = 520, 000 + 24, 500 = 544, 500$ N. Ye step kyun? Thrust gas par act karne wale do haathon ka sum hai: momentum push plus pressure imbalance, exactly jaise master equation kehti hai.

Verify: Units: $kg/s \cdot m/s = kg\cdotm/s^{2} = N$ ✓, $Pa \cdot m^{2} = N$ ✓. Pressure term sirf thrust ka $\sim 4.5%$ hai — under-expansion kuch potential velocity "waste" karta hai, lekin kam se kam ye thrust kabhi subtract nahi karta. Sanity: total > momentum term, as expected for $p_{e} > p_{a}$ .

Example 2 — Cell B: Perfectly expanded ( $p_{e} = p_{a}$ )

Worked example Design point

Wahi engine, ab us altitude par jahan redesigned nozzle ke liye $p_{a}$ equals $p_{e} = 66$ kPa hai. Given $\overset{m}{˙} = 200$ kg/s, $v_{e} = 2700$ m/s, $A_{e} = 0.6$ m², $p_{e} = 66$ kPa, $p_{a} = 66$ kPa. Thrust find karo.

Forecast: $p_{e} = p_{a}$ ke saath, dono haath equally push karte hain — pressure term vanish ho jaana chahiye. Predict karo $F = \overset{m}{˙} v_{e}$ exactly.

Step 1 — Classify. $p_{e} = p_{a} = 66$ kPa → perfectly expanded. Ye is altitude ke liye design sweet-spot hai. Ye step kyun? Perfect expansion ka matlab hai saari pressure energy velocity ban gayi; kuch bhi waste nahi hua, kuch bhi fight nahi hua.

Step 2 — Pressure term. $(66, 000 - 66, 000) \times 0.6 = 0$ N. Ye step kyun? "Perfect" ka poora point yahi hai ki ye term zero ho jaaye — thrust pure momentum hai.

Step 3 — Total. $F = 200 \times 2700 + 0 = 540, 000$ N. Ye step kyun? Pressure hand exactly balanced hone ke saath, master equation sirf momentum term tak collapse ho jaati hai — ye sabse cleanest possible thrust hai.

Verify: Units ✓. Kyunki pressure term zero hai, $F$ exactly momentum term hai — sabse cleanest possible case. Ek fixed nozzle ye sirf EK altitude par hit karti hai; wahi ek point exactly hai kyun altitude compensation exist karti hai.

Example 3 — Cell C: Over-expanded ( $p_{e} < p_{a}$ )

Worked example Ek vacuum nozzle sea level par kheechi gayi

Ek badi nozzle $p_{e} = 40$ kPa par exit karti hai lekin sea level par fire hoti hai $p_{a} = 101$ kPa. Given $\overset{m}{˙} = 200$ kg/s, $v_{e} = 2900$ m/s, $A_{e} = 1.2$ m². Thrust find karo, aur danger par comment karo.

Forecast: Ab hawa ( $101$ ) gas ( $40$ ) ke mukable zyada push karti hai. Predict karo pressure term negative hai — thrust momentum term se neeche gir jaata hai.

Step 1 — Classify. $40 < 101$ , toh $p_{e} < p_{a}$ → over-expanded. Ye step kyun? Ye dangerous regime hai jiske baare mein parent note ne warn kiya tha — atmosphere nozzle ke andar push karta hai.

Step 2 — Momentum term. $200 \times 2900 = 580, 000$ N. Ye step kyun? Wahi as always — mass eject karne ka forward kick; ise bahar ki hawa se koi fark nahi padta, isliye hum ise independently compute karte hain pressure fight fold karne se pehle.

Step 3 — Pressure term. $(40, 000 - 101, 000) \times 1.2 = (- 61, 000) \times 1.2 = - 73, 200$ N. Ye step kyun? Minus sign poora lesson hai: atmosphere yahan thrust subtract karta hai.

Step 4 — Total. $F = 580, 000 - 73, 200 = 506, 800$ N. Ye step kyun? Hum dono haathon ko master equation ke mutabiq add karte hain; yahan ek haath aage push karta hai aur ek peeche kheechta hai, toh total momentum term se neeche gir jaata hai.

Verify: Units ✓. Thrust below momentum term hai — exactly over-expansion penalty. Danger note: agar $p_{e}$ kaafi neeche gir jaaye, toh exhaust wall se separate ho jaayegi (flow separation) jo side loads cause karti hai. Rule of thumb: separation risks tab appear hote hain jab $p_{e} ≲ 0.4 p_{a}$ . Yahan $40/101 = 0.40$ , bilkul edge par — ek real warning flag.

Example 4 — Cell D: Vacuum / degenerate input ( $p_{a} = 0$ )

Worked example Wahi over-expanded nozzle, ab space mein

Example-3 nozzle lo ( $p_{e} = 40$ kPa, $A_{e} = 1.2$ m², $\overset{m}{˙} = 200$ kg/s) lekin ab vacuum mein: $p_{a} = 0$ . Vacuum mein exit velocity thodi zyada hai, $v_{e} = 2950$ m/s (kam back-pressure gas ko thoda aur accelerate karne deta hai). Thrust find karo.

Forecast: $p_{a} = 0$ ke saath, atmosphere ka haath bilkul gayab ho jaata hai. Pressure term purely positive ho jaata hai ( $p_{e} \cdot A_{e}$ ). Example 3 ke muqable ek bada jump predict karo.

Step 1 — $p_{a} = 0$ set karo. Ye degenerate limit hai; master equation $F = \overset{m}{˙} v_{e} + p_{e} A_{e}$ tak simplify ho jaati hai. Ye step kyun? Vacuum mein kuch bhi wapas push nahi kar raha — Ex 3 ka "over-expanded" nozzle ab ek acha nozzle hai.

Step 2 — Momentum term. $200 \times 2950 = 590, 000$ N. Ye step kyun? Hum ise recompute karte hain kyunki $v_{e}$ thoda bada ho gaya (koi back-pressure fight karne ko nahi), isliye forward kick sea level se thoda bada hai.

Step 3 — Pressure term. $(40, 000 - 0) \times 1.2 = 48, 000$ N. Ye step kyun? Wahi nozzle, lekin $p_{a} = 0$ ke saath sign $- 73, 200$ N (Ex 3) se $+ 48, 000$ N ho gaya — altitude ki kahaani ek number mein.

Step 4 — Total. $F = 590, 000 + 48, 000 = 638, 000$ N. Ye step kyun? Dono haath ab aage push kar rahe hain, toh master equation ka sum momentum kick plus full exit-pressure push hai — ye sabse bada thrust hai jo ye nozzle de sakti hai.

Verify: Units ✓. Compare karo: Ex 3 ne sea level par $506, 800$ N diya, Ex 4 ne vacuum mein $638, 000$ N diya — identical nozzle space mein $\sim 26%$ zyada thrust produce karti hai. Yahi exactly hai kyun bade- $ϵ$ nozzles upper stages par rehte hain.

Example 5 — Cell E: Limiting value ( $p_{e} \to 0$ , diminishing returns)

Numbers se pehle, teen naye symbols earn karte hain jo is example ko chahiye:

Ab formula itself earn karte hain use karne se pehle — ye kahan se aata hai aur strange exponent kyun appear hota hai.

Intuition Exhaust-velocity formula kahan se aata hai (aur uske do assumptions)

Parent note ka boxed formula magic nahi hai — ye gas ke liye energy accounting hai, exactly do assumptions par resting hai:

Ideal gas — exhaust $p = ρR T$ obey karta hai, toh uski energy per kilogram clean $c_{p} T$ hai. Ye hume "thermal energy" sirf temperature ki tarah treat karne deta hai.
Isentropic (loss-free, adiabatic) expansion — walls se koi heat leak nahi hoti aur koi friction energy waste nahi karta; gas jo thermal energy deti hai uska har joule jet ki ordered kinetic energy ban jaata hai.

In dono ke under, chamber (subscript $0$ , almost still) se exit (subscript $e$ , fast) tak energy conservation: $chamber mein thermal energy c_{p} T_{0} = exit par thermal bacha c_{p} T_{e} + kinetic gained \frac{1}{2} v_{e}^{2} \Rightarrow \frac{1}{2} v_{e}^{2} = c_{p} (T_{0} - T_{e}) .$ Force balance kyun nahi, energy conservation kyun? Hume ek speed chahiye, aur "kitni heat motion mein convert hui" ki sabse clean bookkeeping energy in = energy out hai. Force balance ko nozzle ke neeche pura pressure profile chahiye hoga; energy ko sirf do endpoints chahiye.

Intuition Exponent

$(γ - 1) / γ$ kyun hai — isentropic relation Hume abhi bhi pressures ke terms mein $T_{e}$ chahiye. Ek isentropic (constant-entropy) ideal-gas process ke liye, temperature aur pressure ek saath locked hain: $\frac{T _{e}}{T _{0}} = (\frac{p _{e}}{p _{0}})^{(γ - 1) / γ} .$ Ye exact power kyun? Entropy constant rehna force karta hai $p V^{γ} = const$ (jo gas expand hoti hai wo $γ$ ke set rule se cool hoti hai). Ise ideal-gas law $p V = R T$ se combine karo volume $V$ eliminate karne ke liye, toh temperature pressure se leftover exponent $(γ - 1) / γ$ ke saath tied ho jaata hai. Toh odd-looking power simply "what's left of $γ$ after you trade volume for temperature" hai. Bada $γ$ (stiffer gas) matlab chhota exponent, toh gas same pressure drop ke liye kam cool hoti hai.

Is $T_{e}$ ko $c_{p} = γ R / (γ - 1)$ ke saath $\frac{1}{2} v_{e}^{2} = c_{p} (T_{0} - T_{e})$ mein substitute karo to neeche ka boxed result milta hai — har symbol ab earned hai.

Worked example Expansion do baar double karna

Parent note ka boxed formula use karte hue $v_{e} = \frac{2 γ}{γ - 1} R T_{0} [1 - (\frac{p _{e}}{p _{0}})^{(γ - 1) / γ}],$ $T_{0} = 3500$ K, $R = 400$ J/(kg·K), $p_{0} = 3$ MPa, $γ = 1.2$ ke saath, teen exit pressures ke liye $v_{e}$ compute karo: $p_{e} = 30$ kPa, phir $15$ kPa, phir $7.5$ kPa (har baar halving). Dikhao ki $v_{e}$ gains har baar shrink hote hain.

Forecast: Har baar $p_{e}$ half karo — kya equal velocity jumps milte hain? Guess: nahi, gains chhote hote jaate hain kyunki $v_{e}$ saturate hoti hai.

Step 1 — Formula recall karo (abhi upar derive hua). $v_{e} = \frac{2 γ}{γ - 1} R T_{0} [1 - (\frac{p _{e}}{p _{0}})^{(γ - 1) / γ}]$ Ye step kyun? Ye correct scaling hai — naive $v_{e} \propto ϵ$ galat hai. Jo kuch vary karta hai wo sab us bracket ke andar hai; baaki ek fixed constant hai. Bracket literally chamber thermal energy ka wo fraction hai jo motion mein convert hua (upar ke energy balance se).

Step 2 — Prefactor. $\frac{2 γ}{γ - 1} R T_{0} = \frac{2.4}{0.2} \times 400 \times 3500 = 12 \times 1, 400, 000 = 16, 800, 000$ . Ye step kyun? Constant ek baar compute karo; sirf bracket $p_{e}$ ke saath change hota hai, toh ye kaam teen baar repeat karne se bachata hai.

Step 3 — Har $p_{e}$ par bracket (exponent $(γ - 1) / γ = 0.2/1.2 = 1/6$ , upar se isentropic power):

$p_{e} = 30$ kPa: $p_{e} / p_{0} = 0.01$ , bracket $= 1 - 0.0 1^{1/6} = 1 - 0.4642 = 0.5358$
$p_{e} = 15$ kPa: $p_{e} / p_{0} = 0.005$ , bracket $= 1 - 0.00 5^{1/6} = 1 - 0.4136 = 0.5864$
$p_{e} = 7.5$ kPa: $p_{e} / p_{0} = 0.0025$ , bracket $= 1 - 0.002 5^{1/6} = 1 - 0.3686 = 0.6314$ Ye step kyun? Bracket wo fraction hai jo thermal energy motion mein convert hua; ye $p_{e}$ girne par $1$ ki taraf badhta hai, isliye iska growth track karte hain.

Step 4 — Velocities $v_{e} = 16, 800, 000 \times bracket$ :

$v_{e} (30) = 16.8 M \times 0.5358 \approx 3000$ m/s
$v_{e} (15) = 16.8 M \times 0.5864 \approx 3139$ m/s → gain +139 m/s
$v_{e} (7.5) = 16.8 M \times 0.6314 \approx 3257$ m/s → gain +118 m/s Ye step kyun? Square root lena energy bracket ko actual speed mein convert karta hai, aur successive speeds compare karna shrinking gains expose karta hai.

Verify: Do gains ( $139$ phir $118$ m/s) equal-ratio halvings ke liye hain, phir bhi doosra gain chhota hai — diminishing returns, confirm karta hai ki $v_{e}$ $p_{e} \to 0$ as saturate hoti hai. Units: $J/kg = m^{2} / s^{2} = m/s$ ✓. Neeche flattening curve dekho.

Figure walkthrough (neeche): Horizontal axis exit pressure $p_{e}$ kPa mein hai; vertical axis exhaust velocity $v_{e}$ m/s mein hai. Blue curve right par (high $p_{e}$ ) steeply rise karti hai lekin left ki taraf $p_{e} \to 0$ as flatten ho jaati hai. Teen coloured dots (green $30$ , orange $15$ , red $7.5$ kPa) height mein ever closer hote jaate hain chahe har baar pressure half ho — woh visual bunching hi diminishing return hai. Blue curve ko leftward follow karo: ye ceiling approach karti hai aur kabhi shoot up nahi karti, yahi reason hai ki $ϵ$ forever double karna less aur less speed buy karta hai.

Figure — Altitude compensation methods — nozzle extension, aerospike

Example 6 — Cell F: Geometry — cone length se expansion ratio

Worked example Extendable nozzle: stowed vs deployed

Ek conical nozzle ka throat radius $R_{t} = 0.10$ m aur half-angle $α = 1 5^{\circ}$ hai. Stowed length $L_{1} = 1.0$ m; deployed length $L_{2} = 2.5$ m. $ϵ_{1}$ aur $ϵ_{2}$ find karo.

Forecast: Lamba cone → wide mouth → bada $ϵ$ . Guess karo ki deploying (2.5× length) roughly $2.5 \times$ , zyada, ya $2.5 \times$ se kam expansion ratio deta hai.

Step 1 — Exit radius linearly grow karta hai length ke saath. $R_{e} = R_{t} + L tan α$ . Ye step kyun? Ek straight-walled cone mein, axis ke saath $L$ distance walk karo aur wall $L tan α$ outward climb kar chuki hogi. Yahan $tan α$ = opposite/adjacent = (wall kitni dur rise karti hai) / (tum kitna chale) — woh ratio hi wall ki steepness hai. Figure mein red wall dekho.

Step 2 — $tan 1 5^{\circ} = 0.2679$ compute karo.

$R_{e, 1} = 0.10 + 1.0 \times 0.2679 = 0.3679$ m
$R_{e, 2} = 0.10 + 2.5 \times 0.2679 = 0.7698$ m Ye step kyun? Do lengths ko linear rule mein plug karne se har configuration mein mouth radius milta hai — wo raw geometry jo square karne se pehle chahiye.

Step 3 — Expansion ratio area ratio hai, aur area $\propto R^{2}$ : $ϵ = (\frac{R _{e}}{R _{t}})^{2}$ Ye step kyun? $ϵ = A_{e} / A_{t} = π R_{e}^{2} / π R_{t}^{2}$ ; $π$ cancel ho jaata hai, squared radius ratio bachta hai.

$ϵ_{1} = (0.3679/0.10)^{2} = 3.67 9^{2} = 13.5$
$ϵ_{2} = (0.7698/0.10)^{2} = 7.69 8^{2} = 59.3$

Verify: Deploying ne length $2.5$ se multiply ki lekin $ϵ$ $59.3/13.5 = 4.4 \times$ se — linear se zyada, kyunki $ϵ$ radius squared ke saath scale karta hai. Units: ratio dimensionless ✓. Geometry neeche.

Figure walkthrough (neeche): Horizontal axis nozzle axis ke saath distance $x$ (m) hai; vertical axis radius (m) hai, centre-line ke upar aur neeche draw kiya gaya hai taaki poora funnel dikh sake. $x = 0$ par blue double-arrow throat width $2 R_{t}$ mark karta hai. Red wall stowed cone hai (throat se $L_{1}$ tak); dashed orange wall deployed extension hai ( $L_{1}$ se $L_{2}$ tak). Green arrow stowed mouth radius $R_{e, 1}$ measure karta hai aur black arrow deployed mouth radius $R_{e, 2}$ — notice karo mouth roughly double ho jaata hai radius mein lekin, kyunki area radius squared ke saath jaata hai, $ϵ$ quadruple se zyada ho jaata hai.

Example 7 — Cell G: Real-world word problem (ek nozzle, do altitudes)

Worked example Sea level vs 30 km — same fixed nozzle

Ek first-stage engine ka fixed nozzle hai: $p_{e} = 70$ kPa, $A_{e} = 0.9$ m², $\overset{m}{˙} = 300$ kg/s. Exhaust velocity $\approx 2800$ m/s (yahan roughly constant treat karo). Sea level ( $p_{a} = 101$ kPa) aur 30 km altitude ( $p_{a} \approx 1.2$ kPa) par thrust compare karo.

Forecast: Sea level par $p_{e} < p_{a}$ (over-expanded, penalty). Upar, $p_{e} > p_{a}$ (under-expanded, bonus). Predict karo thrust is fixed nozzle ke liye altitude ke saath badhega.

Step 1 — Momentum term (dono jagah same). $300 \times 2800 = 840, 000$ N. Ye step kyun? $\overset{m}{˙}$ aur $v_{e}$ altitude ke saath barely change karte hain, isliye ye forward kick ek fixed baseline hai; sirf pressure term move karta hai jab hum climb karte hain.

Step 2 — Sea level pressure term. $(70, 000 - 101, 000) \times 0.9 = - 31, 000 \times 0.9 = - 27, 900$ N. → over-expanded, negative. Ye step kyun? Sea level par dense air exit gas ko out-push karta hai, toh ye haath thrust neeche kheechta hai.

Step 3 — 30 km pressure term. $(70, 000 - 1, 200) \times 0.9 = 68, 800 \times 0.9 = 61, 920$ N. → under-expanded, positive. Ye step kyun? Upar hawa thin hai, isliye wahi exit gas ab ise out-push karta hai aur ye haath thrust add karta hai — sign flip ho gaya.

Step 4 — Totals.

Sea level: $840, 000 - 27, 900 = 812, 100$ N.
30 km: $840, 000 + 61, 920 = 901, 920$ N. Ye step kyun? Fixed momentum kick ko har altitude ke pressure term ke saath add karna (master equation se) sign-flip ko real thrust mein convert karta hai jo har jagah deliver hoti hai.

Verify: Thrust $812, 100$ N se $901, 920$ N tak climb karta hai jaise rocket ascend karta hai — ek real effect jo har launch mein dikhta hai. Gain $= 89, 820$ N ( $\sim 11%$ ). Ye exactly woh inefficiency hai jo altitude compensation attack karta hai: fixed nozzle sirf beech ke ek crossover altitude par "right" hai (jahan $p_{a} = 70$ kPa). Units ✓.

Example 8 — Cell H: Exam twist — aerospike vs bell break-even

Worked example "Kaun sa nozzle jeeetta hai, aur kahan?"

Do engines same momentum term $\overset{m}{˙} v_{e} = 800, 000$ N aur same $A_{e} = 1.0$ m² share karte hain. Bell fixed hai $p_{e} = 20$ kPa ke saath (vacuum-optimised). Aerospike self-adjust karta hai taaki $p_{e} \approx p_{a}$ har altitude par ho (har jagah perfectly expanded, ideally). Kis ambient pressure par bell aerospike se catch up karta hai, aur sea level par kaun jeetta hai ( $p_{a} = 101$ kPa)?

Forecast: Aerospike ka pressure term hamesha $\approx 0$ hai (perfectly expanded), toh uski thrust $\approx 800, 000$ N hamesha. Bell ka pressure term $(20, 000 - p_{a}) \times 1.0$ hai. Guess karo: bell sirf tab tie/beat karta hai aerospike ko jab $p_{a}$ kaafi low ho.

Step 1 — Aerospike thrust. $F_{s p ik e} = 800, 000 + (p_{a} - p_{a}) \times 1.0 = 800, 000$ N sabhi altitudes par. Ye step kyun? Iski defining property — $p_{e} = p_{a}$ — har altitude par pressure term kill kar deti hai. Aerospike ka poora selling point yahi hai.

Step 2 — Bell thrust. $F_{b e l l} = 800, 000 + (20, 000 - p_{a}) \times 1.0$ . Ye step kyun? Bell ka fixed $p_{e} = 20$ kPa hai, isliye uska pressure term $p_{a}$ ke saath swing karta hai instead of track karne ke — isiliye wo sirf ek altitude par right ho sakta hai.

Step 3 — Break-even. $F_{b e l l} = F_{s p ik e}$ set karo: $20, 000 - p_{a} = 0 \Rightarrow p_{a} = 20$ kPa (roughly $12$ km altitude). Ye step kyun? Dono thrusts equate karna crossover ambient pressure isolate karta hai — $20$ kPa se upar bell haarta hai, neeche tie/win karta hai, kyunki wahan uska fixed $p_{e}$ finally ambient se match karta hai.

Step 4 — Sea level. $p_{a} = 101$ kPa par: $F_{b e l l} = 800, 000 + (20, 000 - 101, 000) \times 1.0 = 800, 000 - 81, 000 = 719, 000$ N, aerospike $800, 000$ N ke muqable. Ye step kyun? Sea-level ambient ko bell ke swinging pressure term mein plug karne se concrete penalty dikhti hai jo fixed nozzle neeche pay karta hai, jahan aerospike shine karne ke liye design hua hai.

Verify: Sea level par aerospike bell ko $800, 000 - 719, 000 = 81, 000$ N ( $\sim 11%$ ) se beat karta hai — iska headline advantage. Break-even $p_{a} = 20$ kPa par $p_{e} = 20$ kPa se match karta hai (bell exactly wahan perfectly expanded hai, Cell B). $12$ km se upar fixed bell under-expanded hai (Cell A) aur effectively ideal aerospike se match karta hai. Units ✓.

Conclusion / interpretation: Ye standard exam takeaway hai. Aerospike poori low-altitude climb jeetta hai kyunki wo self-compensate karta hai aur kabhi pressure penalty nahi pay karta; fixed bell sirf upar rival karta hai, jab thin air finally uske single design pressure se match karta hai. Yahi reason hai ki aerospikes single-stage-to-orbit vehicles ke liye attractive hain jo pad se space tak efficient hone chahiye, jabki extendable sections wale bells (Examples 4–6) upper stages ke liye practical choice hain jo sirf near vacuum mein fire karte hain.

Recall Jaane se pehle self-test

Pressure term ka sign jab $p_{e} < p_{a}$ ho ::: Negative (over-expanded — atmosphere thrust subtract karta hai) Vacuum mein $(p_{e} - p_{a}) A_{e}$ ka kya hota hai ::: $p_{a} = 0$ , toh ye $+ p_{e} A_{e}$ ban jaata hai (hamesha positive) $ϵ$ double karne se $v_{e}$ gains kyun shrink hote hain ::: Bracket $[1 - (p_{e} / p_{0})^{(γ - 1) / γ}] \to 1$ , toh $v_{e}$ saturate ho jaati hai Cone geometry se $ϵ$ ::: $ϵ = (R_{t} + L tan α)^{2} / R_{t}^{2}$ Fixed-nozzle thrust altitude ke saath kyun badhta hai ::: $p_{a}$ girta hai, pressure term negative se positive flip ho jaata hai Har altitude par aerospike ka pressure term ::: Approximately zero, kyunki ye $p_{e} \approx p_{a}$ maintain karta hai $v_{e}$ formula ke peeche do assumptions ::: Ideal gas aur isentropic (loss-free, adiabatic) expansion Exponent $(γ - 1) / γ$ ki origin ::: Isentropic relation $T_{e} / T_{0} = (p_{e} / p_{0})^{(γ - 1) / γ}$