Altitude compensation methods — nozzle extension, aerospike: Exercises

3.3.16 · D4 · Physics › Rocket Propulsion › Altitude compensation methods — nozzle extension, aerospike

Ye problems "kya aap picture padh sakte ho?" se shuru hokar "kya aap design rule khud bana sakte ho?" tak jaate hain. Yahan use hone wale har symbol ki definition parent topic mein di gayi hai. Agar koi symbol unfamiliar lage, toh woh ek signal hai ki aage badhne se pehle parent ko dobara padho.

Symbol refresher (sab parent se hain). Is page par jo bhi symbol aaye, pehle yahan define hai — jo na pehchano, shuru karne se pehle padh lo:

Symbol	Saral shabdon mein
$A_{t}, A_{e}$	throat area, exit area (m²)
$ϵ = A_{e} / A_{t}$	expansion ratio (gas kitna failta hai)
$p_{0}, p_{e}, p_{a}$	chamber, exit, ambient pressure
$γ$	heat capacity ratio ( $\approx 1.2$ H₂/O₂ exhaust ke liye)
$M_{e}$	exit Mach number (exit speed ÷ local sound speed)
$\overset{m}{˙}$	mass flow rate (har second mein nikalne wali gas ke kg)
$v_{e}$	exhaust velocity (m/s)
$F$	thrust (aage ki taraf push, newtons N mein)
$R_{t}, R_{e}$	throat radius, exit radius — sabse sankre point aur munh par circular cross-section ka radius (m ya cm)
$L$	nozzle (ya extension) length — bell throat se exit tak kitna stretch karti hai (m ya cm)
$α$	cone half-angle — nozzle wall ka centre-line se jhukav (degrees)
$R$	exhaust ka specific gas constant — ek kg gas har kelvin par kitni pressure–volume energy carry karta hai, H₂/O₂ products ke liye $\approx 400 J/(kg\cdotK)$
$T_{0}$	stagnation (chamber) temperature — combustion chamber mein gas ka temperature accelerate hone se pehle (K)
$g_{0}$	standard gravity, fixed reference $g_{0} = 9.81 m/s^{2}$ jo $I_{s p}$ define karne ke liye use hota hai
$I_{s p} = v_{e} / g_{0}$	specific impulse (seconds)

Neeche di gayi picture har Level 1–5 problem ke peeche ka mental model hai: $p_{e}$ ko $p_{a}$ ke against padho aur plume ki shape tumhe state batayegi. Jab bhi koi problem pooche "ye kaun sa case hai?" toh isse refer karo.

Figure — Altitude compensation methods — nozzle extension, aerospike

Teen plumes notice karo: left mein (pink) plume andar ki taraf pinch ho rahi hai kyunki ambient air exhaust ko peeche dha rahi hai (over-expanded); beech mein (yellow) plume ke kinnare parallel hain — bilkul match; right mein (blue) plume bahar ki taraf flare karti hai kyunki exhaust abhi bhi over-pressured hai aur patli hawa mein expand hoti rehti hai (under-expanded). L1.1–L1.2 literally "plume ka naam batao" hain.

Core equations jo hum baar baar use karte hain (unme har symbol upar table mein define hai):

Level 1 — Recognition

L1.1

Ek nozzle ka $p_{e} = 40$ kPa hai. Rocket sea level par hai jahan $p_{a} = 101$ kPa. Kya nozzle under-expanded, over-expanded, ya perfectly expanded hai?

Recall Solution

Hum kya compare karte hain: $p_{e}$ ko $p_{a}$ se. Yahan $p_{e} = 40 < 101 = p_{a}$ . Kyunki exit pressure ambient se neeche hai, bahar ki hawa exhaust ko peeche dhak rahi hai — gas zyada expand ho gayi. Answer: over-expanded (pinched-plume, upar figure mein left picture). (Mnemonic: over-expanded matlab tumne spreading zyada kar di, toh $p_{e}$ , $p_{a}$ se neeche gir gayi.)

L1.2

Wahi nozzle ( $p_{e} = 40$ kPa) ab vacuum mein hai ( $p_{a} = 0$ ). Ye kaun sa case hai?

Recall Solution

$p_{e} = 40 > 0 = p_{a}$ . Exit pressure ambient se upar hai ⇒ gas aur expand ho sakti thi. Under-expanded (flaring-plume, upar right picture). Yahi reason hai ek fixed nozzle har jagah perfect nahi ho sakti: wahi $p_{e}$ sirf altitude badlane se over- se under-expanded mein flip ho gayi.

L1.3

Do conical nozzles ek hi throat radius $R_{t}$ share karte hain. Nozzle A, nozzle B se longer hai (same half-angle $α$ ). Kiski expansion ratio $ϵ$ zyada hai?

Recall Solution

$R_{e} = R_{t} + L tan α$ : zyada $L$ se zyada bada exit radius $R_{e}$ milta hai. Kyunki $ϵ = (R_{e} / R_{t})^{2}$ , $R_{e}$ ke saath badhta hai, longer nozzle A ka $ϵ$ zyada hai. Yahi extendable nozzle ka poora principle hai — length badhaao, $ϵ$ badhaao.

Level 2 — Application

L2.1

Ek conical nozzle ka throat radius $R_{t} = 5$ cm, half-angle $α = 1 5^{\circ}$ , length $L_{1} = 40$ cm hai. Ek extension ise $L_{2} = 90$ cm tak le jaati hai. $ϵ_{1}$ aur $ϵ_{2}$ nikaalo.

Neeche di gayi geometry figure is problem mein har symbol — $R_{t}$ , $R_{e}$ , $L$ aur half-angle $α$ (sab symbol table mein hain) — fix karti hai, toh compute karte waqt har length uس par trace karo.

Recall Solution

Step 1 — exit radii ( $R_{e} = R_{t} + L tan α$ figure se seedha padho). $tan 1 5^{\circ} = 0.2679$ . $R_{e, 1} = 5 + 40 (0.2679) = 5 + 10.72 = 15.72 cm$ $R_{e, 2} = 5 + 90 (0.2679) = 5 + 24.11 = 29.11 cm$ Step 2 — radius ratio ko square karo. $ϵ_{1} = (15.72/5)^{2} = (3.144)^{2} \approx 9.88$ $ϵ_{2} = (29.11/5)^{2} = (5.821)^{2} \approx 33.9$ Answer: $ϵ_{1} \approx 9.9$ , $ϵ_{2} \approx 33.9$ . 50 cm length add karne se expansion ratio teen guna se zyada ho gayi — kyunki $ϵ$ radius ke square par depend karta hai.

L2.2

$γ = 1.2$ , $p_{0} = 3$ MPa ke liye, ek nozzle $M_{e} = 3.8$ par exit karti hai. Exit pressure $p_{e}$ nikalo.

Recall Solution

Kaun sa tool aur kyun: isentropic pressure–Mach relation, kyunki ye "kitna fast" ( $M_{e}$ , geometry se set hota hai) ko "kitna pressure bachi hai" mein convert karta hai. $\frac{p _{e}}{p _{0}} = (1 + \frac{0.2}{2} (3.8)^{2})^{- 1.2/0.2} = (1 + 0.1 \cdot 14.44)^{- 6} = (2.444)^{- 6}$ $(2.444)^{- 6} \approx 0.00470 \Rightarrow p_{e} = 3000 kPa \times 0.00470 \approx 14.1 kPa$ Answer: $p_{e} \approx 14.1$ kPa.

L2.3

Ek rocket mein $\overset{m}{˙} = 10$ kg/s, $v_{e} = 3900$ m/s, $p_{e} = 14.1$ kPa, $A_{e} = 0.21$ m² hai. Vacuum thrust ( $p_{a} = 0$ ) nikalo.

Recall Solution

Pehle convert karo (upar unit reminder): $p_{e} = 14.1 kPa = 14100 Pa$ . $F = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e} = 10 (3900) + (14100 - 0) (0.21)$ $F = 39000 + 2961 = 41961 N$ Answer: $F \approx 42.0$ kN. Dhyan do pressure term (2961 N) momentum term (39 000 N) ka sirf $\sim 7.6%$ hai — momentum dominate karta hai.

Level 3 — Analysis

L3.1

Boxed $v_{e}$ formula use karo $γ = 1.2$ , specific gas constant $R = 400 J/(kg\cdotK)$ , stagnation temperature $T_{0} = 3300$ K, $p_{0} = 3$ MPa ke saath; $p_{e} = 14.1$ kPa aur $p_{e} = 0.43$ kPa ke liye $v_{e}$ compare karo. Deeper expansion se $v_{e}$ mein kitna percent gain milta hai?

Recall Solution

Kyun ye tool: exhaust speed energy conservation se aati hai, $ϵ$ shortcuts se nahi — $ϵ$ sirf $p_{e}$ set karke enter karta hai. Prefactor: $\frac{2 γ}{γ - 1} R T_{0} = \frac{2 ( 1.2 )}{0.2} (400) (3300) = 12 \times 400 \times 3300 = 1.584 \times 1 0^{7}$ . Exponent $(γ - 1) / γ = 0.2/1.2 = 0.1667$ .

Case A, $p_{e} / p_{0} = 14.1/3000 = 4.700 \times 1 0^{- 3}$ : $(4.700 \times 1 0^{- 3})^{0.1667} \approx 0.4004, v_{e} = 1.584 \times 1 0^{7} (1 - 0.4004) = 9.498 \times 1 0^{6} \approx 3082 m/s$ Case B, $p_{e} / p_{0} = 0.43/3000 = 1.433 \times 1 0^{- 4}$ : $(1.433 \times 1 0^{- 4})^{0.1667} \approx 0.2300, v_{e} = 1.584 \times 1 0^{7} (1 - 0.2300) = 1.220 \times 1 0^{7} \approx 3492 m/s$ Gain: $(3492 - 3082) /3082 = 13.3%$ . Lesson: $p_{e}$ mein $\sim 33 \times$ giraaawat se sirf $\sim 13%$ zyada speed mili — bracket $[1 - (p_{e} / p_{0})^{0.1667}]$ , $p_{e} \to 0$ ke saath 1 ki taraf saturate ho jaata hai. Diminishing returns, bilkul waisa hi jaisa parent ne warn kiya tha.

L3.2

Sea level par ( $p_{a} = 101$ kPa) ek nozzle jiska $A_{e} = 0.21$ m² hai, $p_{e} = 14.1$ kPa tak over-expanded hai. Thrust mein pressure-term contribution calculate karo. Uska sign tumhe kya batata hai?

Recall Solution

Pehle Pa mein convert karo: $p_{e} = 14100$ Pa, $p_{a} = 101000$ Pa. $(p_{e} - p_{a}) A_{e} = (14100 - 101000) (0.21) = (- 86900) (0.21) = - 18249 N$ Answer: $- 18.2$ kN. Negative sign matlab atmospheric pressure exhaust column ko peeche dhak rahi hai — ek ~18 kN penalty. Haqeeqat mein itna bada mismatch flow separation bhi cause karta (edge-case callout), toh asli loss is saaf number se bhi zyada messy hoti. Yahi reason hai vacuum-tuned nozzle sea level par buri hoti hai, aur kyun extendable extensions near-vacuum tak stowed rehti hain.

L3.3

Ek aerospike ka kaam hai exhaust boundary ko $p_{a}$ ke saath "self-adjust" karne dena. Agar ek ideal aerospike hamesha $p_{e} = p_{a}$ achieve karta hai, toh uski thrust equation likho aur ek sentence mein explain karo ki ye ek fixed nozzle ko ascent bhar kyun beat karta hai.

Recall Solution

$p_{e} = p_{a}$ hone par pressure term gayab ho jaata hai: $F = \overset{m}{˙} v_{e} + = 0 (p_{e} - p_{a}) A_{e} = \overset{m}{˙} v_{e}$ Kyun jeetat hai: fixed nozzle ek penalty term $(p_{e} - p_{a}) A_{e}$ carry karti hai jo neeche negative hota hai (over-expanded, separation ka risk bhi), aur upar energy waaste karta hai; aerospike us term ko har altitude par zero rakhta hai, toh koi thrust atmosphere se ladne ya ignore karne mein waaste nahi hoti.

Level 4 — Synthesis

L4.1

Ek conical extension design karo. Tumhe deployed exit radius chahiye jo $R_{t} = 5$ cm throat radius, $α = 1 5^{\circ}$ half-angle se $ϵ_{2} = 84$ de. Kaun sa deployed length $L_{2}$ chahiye?

Recall Solution

Geometry invert karo. $ϵ_{2} = (R_{e, 2} / R_{t})^{2}$ se: $R_{e, 2} = R_{t} ϵ_{2} = 584 = 5 (9.165) = 45.83 cm$ Ab $R_{e, 2} = R_{t} + L_{2} tan α$ se $L_{2}$ solve karo: $L_{2} = \frac{R _{e, 2} - R _{t}}{t a n α} = \frac{45.83 - 5}{0.2679} = \frac{40.83}{0.2679} \approx 152.4 cm$ Answer: $L_{2} \approx 1.52$ m. Fairing ke neeche wo length yahi reason hai ki ise launch par stow karna padta hai — "high $ϵ$ " aur "rocket mein fit ho" ka synthesis hi extendable nozzle hai.

L4.2

Ek stage $t = 700$ s tak constant thrust $F = 41361$ N dete hue $\overset{m}{˙} = 10$ kg/s se jalti hai. Nikalo (a) total propellant mass used, (b) specific impulse $I_{s p}$ diya $v_{e} = 4100$ m/s.

Recall Solution

(a) Propellant $= \overset{m}{˙} t = 10 \times 700 = 7000$ kg. (b) $I_{s p} = v_{e} / g_{0} = 4100/9.81 = 417.9$ s (standard gravity $g_{0} = 9.81 m/s^{2}$ use karke). Answer: 7000 kg jala; $I_{s p} \approx 418$ s. (Bada deployed $ϵ$ , $v_{e}$ ko thoda upar dhakelta hai — yahi wajah hai extendable upper stages $I_{s p}$ ka har extra second dhoondti hain.)

Level 5 — Mastery

L5.1

Ek vehicle utha aur $p_{a}$ linearly-ish 101, 50, 10, 0 kPa se girta hai. Ek fixed nozzle $p_{e} = 30$ kPa ke liye tuned hai. Har ambient value ke liye, expansion state classify karo aur pressure-thrust term ka sign do. Phir argue karo ki ye nozzle kis single altitude ke liye "optimal" hai.

Recall Solution

Fixed $p_{e} = 30$ kPa ko har $p_{a}$ se compare karo:

$p_{a}$ (kPa)	$p_{e} - p_{a}$	state	pressure-term sign
101	$- 71$	over-expanded	negative (penalty)
50	$- 20$	over-expanded	negative
30	$0$	perfectly expanded	zero
10	$+ 20$	under-expanded	positive lekin sub-optimal
0	$+ 30$	under-expanded	positive, zyada waasted potential

Optimal altitude: jahan $p_{a} = p_{e} = 30$ kPa — roughly ~9 km upar. Neeche penalty (aur agar mismatch badhta hai, flow separation); upar waasted expansion potential. Ek single fixed nozzle sirf is table ki ek line par perfect ho sakti hai — aerospikes (har row par perfect) aur extendable nozzles (mid-flight mein $ϵ$ badlao) ki exact motivation yahi hai.

L5.2

Thrust equation se derive karo wo condition on $ϵ$ jo ek diye gaye altitude par thrust maximize kare, aur explain karo kyun ye $p_{e} = p_{a}$ tak reduce hoti hai. ( $\overset{m}{˙}$ , $T_{0}$ , $γ$ fixed maano; $ϵ$ dono $p_{e}$ aur $A_{e}$ set karta hai.)

Recall Solution

Setup. $F (ϵ) = \overset{m}{˙} v_{e} (p_{e}) + (p_{e} - p_{a}) A_{e}$ . Jab $ϵ$ badhta hai, exit area $d A_{e}$ ka ek patla ring add hota hai; isse $p_{e}$ girta hai (zyada expansion) aur $A_{e}$ badhta hai. Maximum dhundhne ke liye $d F / d ϵ = 0$ set karo.

Step 1 — term by term differentiate karo. Chain rule use karke (kyunki $v_{e}$ , $ϵ$ par sirf $p_{e}$ ke zariye depend karta hai): $\frac{d F}{d ϵ} = \overset{m}{˙} \frac{d v _{e}}{d p _{e}} \frac{d p _{e}}{d ϵ} + \frac{d p _{e}}{d ϵ} A_{e} + (p_{e} - p_{a}) \frac{d A _{e}}{d ϵ} .$

Step 2 — added ring ke liye momentum theorem use karo. Nozzle exit par one-dimensional momentum balance standard differential result deta hai: $\overset{m}{˙} \frac{d v _{e}}{d p _{e}} = - A_{e},$ matlab har $d p_{e}$ drop se exactly $\overset{m}{˙} d v_{e} = - A_{e} d p_{e}$ momentum milta hai (ye sirf $\overset{m}{˙} d v_{e} + A_{e} d p_{e} = 0$ hai). Pehle term mein substitute karo: $\frac{d F}{d ϵ} = momentum gain (- A_{e}) \frac{d p _{e}}{d ϵ} + pressure-area change \frac{d p _{e}}{d ϵ} A_{e} + (p_{e} - p_{a}) \frac{d A _{e}}{d ϵ} .$

Step 3 — cancellation dekho. Pehle do terms hain $- A_{e} \frac{d p _{e}}{d ϵ} + A_{e} \frac{d p _{e}}{d ϵ} = 0$ exactly. Ye cancel ho jaate hain kyunki $p_{e}$ girane se jo velocity gain hoti hai woh exactly wahi momentum hai jo pressure term lose karta hai. Sirf last term bachta hai: $\frac{d F}{d ϵ} = (p_{e} - p_{a}) \frac{d A _{e}}{d ϵ} .$

Step 4 — derivative zero set karo. Kyunki $d A_{e} / d ϵ > 0$ (zyada $ϵ$ ka matlab hamesha zyada exit area, toh ye factor kabhi zero nahi hota), $d F / d ϵ = 0$ hone ka ek hi tarika hai: $p_{e} = p_{a}$

Step 5 — confirm karo ki ye maximum hai, aur physically WHY batao. $\frac{d F}{d ϵ} = (p_{e} - p_{a}) \frac{d A _{e}}{d ϵ}$ ka sign dekho:

Jab $p_{e} > p_{a}$ (nozzle abhi under-expanded hai): $d F / d ϵ > 0$ , toh nozzle add karne se thrust badhti hai.
Jab $p_{e} < p_{a}$ (nozzle ab over-expanded): $d F / d ϵ < 0$ , toh nozzle add karne se thrust ghatti hai.

Derivative exactly $p_{e} = p_{a}$ par $+$ se $-$ flip hoti hai — sign change ek maximum ki pehchaan hai. Physically: exit par nozzle wall ka har extra sliver andar se $p_{e}$ aur bahar se $p_{a}$ feel karta hai. Jab $p_{e} > p_{a}$ , net outward push thrust add karta hai; jis moment $p_{e}$ , $p_{a}$ se neeche girta hai, wo sliver ek dead weight ban jaata hai jise atmosphere peeche dha rahi hai. Break-even — peak — exactly $p_{e} = p_{a}$ hai. Yahi ek condition sab altitude-compensation hardware ka north star hai.

L5.3

L5.2 ki spirit ka numerical check: L2/L3 data ke saath, $p_{a} = 14.1$ kPa par ek nozzle perfectly expanded hai ( $p_{e} = 14.1$ kPa). Confirm karo ki uska pressure term zero hai aur $F$ compute karo $\overset{m}{˙} = 10$ , $v_{e} = 3082$ m/s ke saath.

Recall Solution

$(p_{e} - p_{a}) A_{e} = (14100 - 14100) (0.21) = 0$ $F = \overset{m}{˙} v_{e} = 10 (3082) = 30820 N$ Answer: pressure term $= 0$ , $F = 30.82$ kN — pure momentum thrust, ek fixed nozzle jitna clean ho sakta hai, aur wo state jo ek aerospike har jagah mimic karta hai.

Recall Self-test — one-line reveals

Over-expanded matlab $p_{e}$ hai ::: $p_{a}$ se neeche (zyada expand ho gaya) Kisi bhi altitude par maximum thrust ki single condition ::: $p_{e} = p_{a}$ $ϵ$ , exit radius ke saath scale karta hai ::: $ϵ = (R_{e} / R_{t})^{2}$ (radius squared) Kyun extendable nozzles launch par stowed rehti hain ::: fairing ke neeche length bachane ke liye; ye near-vacuum mein fire hoti hain, sea-level thrust ke liye nahi Kyun $ϵ$ double karne se $v_{e}$ mein diminishing returns milte hain ::: bracket $1 - (p_{e} / p_{0})^{(γ - 1) / γ}$ , $p_{e} \to 0$ ke saath 1 ki taraf saturate ho jaata hai Kya galat hota hai badly over-expanded nozzle mein low altitude par ::: flow wall se separate ho jaata hai, ek internal shock banta hai, thrust loss, side-loads aur vibration hoti hai