Worked examples — Rocket staging — series staging, parallel staging

3.3.45 · D3 · Physics › Rocket Propulsion › Rocket staging — series staging, parallel staging

Yeh page Rocket staging ki practice range hai. Isme jo bhi tools use hote hain — Tsiolkovsky rocket equation, structural aur propellant fractions, thrust-weighted exhaust velocity — sab waise hi use hote hain jaise parent note ne build kiya tha. Agar koi symbol example mein aata hai, to hum pehle wahi explain karte hain, phir use karte hain.

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Scenario matrix

Hum yeh symbols poore document mein use karte hain (sab parent note se hain):

• $\Delta v$ — wo velocity jo ek stage add karta hai, metres per second (m/s) mein.
• $v_e$ — exhaust velocity: burnt gas nozzle se kitni tez nikalta hai, m/s mein. Zyada $v_e$ = fuel ke har kilogram se zyada push.
• $m_0$ — burn ke shuru mein mass (structure + uska apna fuel + jo bhi wo carry karta hai).
• $m_f$ — burn ke end mein mass (start mass minus jo fuel jala diya).
• $m_{\text{pay}}$ — payload mass: useful cargo (satellite, capsule) jo ek stage deliver karta hai; kisi bhi stage ke nazar se, isme uske upar ke saare stages bhi shamil hain. kg mein maapa jata hai.
• $\ln$ — natural logarithm, wo function jo batata hai "$e$ ko kitni power mein uthao to yeh number milti hai?" Yeh isliye aata hai kyunki Tsiolkovsky hai $\Delta v=v_e\ln (m_0/m_f)$.
• $\lambda$ (propellant fraction) aur $ϵ$ (structural fraction): ek stage ke hardware mass $S$ mein se, fuel ka hissa ($\lambda =m_p/S$) vs tanks/engines ($ϵ=m_s/S$), jahan $\lambda +ϵ=1$. Dekho Structural fraction and propellant fraction.
• $F$ — thrust: engine ya engine group jo pushing force produce karta hai, meganewtons (MN) mein. $F_c$ = core thrust, $F_b$ = ek booster ka thrust.
• $I_{sp}$ — specific impulse, exhaust velocity seconds mein: $v_e=I_{sp}g$ jahan $g=9.81{\text{m/s}}^2$. Dekho Specific impulse.

Is page ke do figures do aise ideas build karte hain jo dekhne mein mushkil hain: mass kadam-kadam kaise girta hai (figure s01) aur mass ratio — raw mass nahi — $\Delta v$ ko kaise control karta hai (figure s02). Hum inhe tab dekhte hain jab hum untak pahunchte hain.

Cell	Scenario class	Worked in
A	Clean multi-stage series $\Delta v$ sum	Example 1
B	Parallel boosters + thrust-weighted ${\bar{v}}_e$, two phases	Example 2
C	Degenerate: single stage ($n=1$) — "no staging" baseline	Example 3
D	Zero / limiting input: $ϵ\to 0$ (perfect tanks) aur $ϵ\to 1$ (dead-weight limit)	Example 4
E	Series vs parallel head-to-head same hardware pe	Example 5
F	Word problem: "target $\Delta v$ hit karne ke liye kitne boosters chahiye?"	Example 6
G	Exam twist: do students, ek galat mass ratio (error dhundho)	Example 7
H	Unit / sanity trap: $I_{sp}$ seconds mein vs $v_e$ m/s mein	Example 8

Neeche har cell A–H cover kiya gaya hai.

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Example 1 — Cell A: clean two-stage series

Forecast: Compute karne se pehle guess karo — kya yeh low orbit ke liye zaroori ~9400 m/s clear karega? Abhi haan/nahi likho.

Yaad karo wo symbols jo chahiye: $\lambda$ ek stage ke apne hardware $S$ ka fuel share hai, isliye fuel $=\lambda S$; $ϵ=1-\lambda$ structure share hai, isliye structure $=ϵS$. $m_0$/$m_f$ start/end masses hain, $m_{\text{pay}}$ payload hai (yahan stage 1 ke nazar se stage 2 bhi shamil hai), aur $\Delta v=v_e\ln (m_0/m_f)$.

1. Top-down kaam karo: pehle stage 2. Sab fractions mein rakho: structure $={ϵ}_2{S}_2=(1-0.87)\times 8000=0.13\times 8000=1040\text{kg}$; fuel $={\lambda }_2{S}_2=0.87\times 8000=6960\text{kg}$. Yeh step kyun? Stage 2 sirf payload carry karta hai, stage 1 ka dead mass nahi, isliye uske numbers self-contained hain. $\lambda ,ϵ$ use karne se (raw masses nahi) hum yahi kisi bhi $S_2$ ke liye reuse kar sakte hain.

2. Stage 2 masses. Start $m_0=S_2+m_{\text{pay}}=8000+1000=9000\text{kg}$. End (fuel khatam) $m_f={ϵ}_2{S}_2+m_{\text{pay}}=1040+1000=2040\text{kg}$. Yeh step kyun? Tsiolkovsky ko before/after mass chahiye; $m_f$ mein sirf structure ${ϵ}_2{S}_2$ plus payload rehta hai jab ${\lambda }_2{S}_2=6960$ kg fuel khatam ho jaata hai.

3. Stage 2 $\Delta v$. $\Delta v_2=v_{e,2}\ln \left(\frac{m_0}{m_f}\right)=3500\ln \left(\frac{9000}{2040}\right)=3500\times 1.484\approx 5190\text{m/s}.$ Yeh step kyun? $\ln$ mass ratio ko yeh batata hai ki kitne "$e$-foldings" shed hue; $v_e$ use speed mein scale karta hai.

4. Stage 1 ko stage-2 hardware + payload uthana hai $=8000+1000=9000\text{kg}$. Structure $={ϵ}_1{S}_1=0.15\times 90000=13500\text{kg}$, fuel $={\lambda }_1{S}_1=0.85\times 90000=76500\text{kg}$. Yeh step kyun? Stage 1 ke nazar se, "payload" uske upar ki har cheez hai, isliye yahan $m_{\text{pay}}=9000$ kg hai.

5. Stage 1 masses aur $\Delta v$. $m_0=90000+9000=99000$, $m_f=13500+9000=22500$. $\Delta v_1=3000\ln \left(\frac{99000}{22500}\right)=3000\times 1.482\approx 4450\text{m/s}.$ Yeh step kyun? Hum Tsiolkovsky stage 1 pe akele apply karte hain, bilkul stage 2 ki tarah: uska start mass poora stack hai, end mass stage-1 structure plus jo kuch bhi uska apna fuel khatam hone ke baad bhi carry kar raha hai. Equation stage-agnostic hai — same law, naye masses.

6. Inhe add karo (series $\Delta v$'s add hote hain kyunki har stage wahan se shuru hota hai jahan pichle ne chodha): $\Delta v_{\text{total}}\approx 5190+4450=9640\text{m/s}.$

Recall Verify

Answer $\approx 9640$ m/s > 9400 m/s target, isliye orbit reach kar leta hai. Units: $(\text{m/s})\times (\text{dimensionless }\ln )=\text{m/s}$. ✓

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Example 2 — Cell B: parallel boosters, two phases

Koi bhi algebra shuru karne se pehle, mass ka kadam-kadam girna dekho. Figure s01 vehicle mass (vertical axis, tonnes) ko time (horizontal axis) ke against plot karta hai. Magenta line phase 1 hai (core + boosters ek saath jal rahe hain, mass $92$ t se $38$ t tak slide karta hai). Separation pe orange dashed cliff khaali booster shells drop karta hai ($38\to 32$ t). Violet line phase 2 hai (sirf core, $32\to 5$ t). Dhyan do ki timeline ek smooth slide nahi hai — jettison cliff hi staging ka poora point hai.

Forecast: Kya ${\bar{v}}_e$ boosters ke $2650$ ya core ke $4400$ ke zyada kareeb hoga? Guess karo kaun sa engine group "jeetega."

Yahan symbols: $F_c$ core thrust hai aur $F_b$ ek booster ka thrust hai (force, meganewtons MN mein), isliye do boosters $2F_b$ dete hain; $v_{e,c},v_{e,b}$ core aur booster exhaust velocities hain (m/s); ${\bar{v}}_e$ ek single "effective" exhaust velocity hai jo dono engine groups ke ek saath jalne ko represent karti hai.

1. Har $v_e$ ko uske thrust se weight karo. Boosters kaafi zyada push karte hain ($2F_b=25$ vs $F_c=5.5$ MN), isliye average mein unhi ka domination hai. Yeh step kyun? Momentum bookkeeping: total thrust $={\dot{m}}_{\text{tot}}{\bar{v}}_e$, aur har engine ${\dot{m}}_i{v}_{e,i}=F_i$ contribute karta hai. Total $F$ ko total $\dot{m}$ se divide karne se thrust-weighted mean milta hai.

2. Plug in: ${\bar{v}}_e=\frac{F_c{v}_{e,c}+2F_b{v}_{e,b}}{F_c+2F_b}=\frac{5.5\times 4400+25\times 2650}{5.5+25}=\frac{24200+66250}{30.5}=\frac{90450}{30.5}\approx 2966\text{m/s}.$

3. Geometry padho (figure s01): annotation "uses vbar_e" magenta phase-1 line pe hai — yeh ${\bar{v}}_e$ poore combined burn ko govern karta hai. Orange cliff ke baad, annotation "uses v_e core" violet line pe hai — phase 2 core ke apne $v_{e,c}$ pe wapas aa jaata hai.

Recall Verify

${\bar{v}}_e\approx 2966$ m/s, $2650$ aur $4400$ ke beech hai, forecast ke mutabik $2650$ ke zyada kareeb — boosters zyada thrust carry karte hain. ✓

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Example 3 — Cell C: single stage (degenerate baseline)

Forecast: Kam mass carry karne ke saath, kya single-stage Example 1 ke $4450$ m/s stage 1 se zyada dega? Higher ya lower guess karo.

Yaad dilate hain: $S$ stage ka apna hardware mass hai; $ϵ=0.15$ matlab $S$ ka $15\%$ structure hai, isliye fuel $=\lambda S=(1-ϵ)S$; $m_{\text{pay}}$ cargo hai.

1. Masses. Fuel $=(1-ϵ)S=0.85\times 90000=76500$, structure $=ϵS=0.15\times 90000=13500$. $m_0=S+m_{\text{pay}}=90000+1000=91000$, $m_f=ϵS+m_{\text{pay}}=13500+1000=14500$. Yeh step kyun? $n=1$ Tsiolkovsky hai bina kisi cheez ko jettison kiye — sabse simple possible case, wo baseline jise baaki sab beat karna chahiye.

2. $\Delta v$: $\Delta v=3000\ln \left(\frac{91000}{14500}\right)=3000\times 1.837\approx 5510\text{m/s}.$

3. Interpret karo. Ek single stage yahan ~$5500$ m/s tak pahuncha — orbital $9400$ se kaafi door. Isliye staging exist karta hai: structural dead weight se bachne ke liye jo single stage ko cap kar deta hai.

Recall Verify

$5510$ m/s two-stage rocket ke $9640$ m/s se kaafi kam hai, staging advantage confirm ho gaya. Units m/s ✓.

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Example 4 — Cell D: zero aur limiting inputs

Figure s02 is example ka visual core hai. Yeh $\Delta v$ (vertical, m/s) ko structural fraction $ϵ$ (horizontal, $0$ se $1$ tak) ke against stage 2 ke liye plot karta hai. Violet curve ko left-to-right trace karo: $ϵ=0$ pe yeh magenta ceiling ($\approx 7690$ m/s) ko touch karta hai, real design orange dot pe hai ($ϵ=0.13$, $\approx 5190$ m/s), aur $ϵ=1$ pe curve navy dot zero pe land karta hai. Curve steep hai — dead structure mein chhoti si bhi baadhna bahut saara $\Delta v$ kha jaati hai.

Forecast: Case (a) mein kya $\Delta v$ infinity tak ja sakta hai ya ceiling pe hit karta hai? Case (b) mein $\Delta v$ kya hona chahiye?

Yaad dilate hain: $m_0=S+m_{\text{pay}}$ yahan kabhi nahi badlega (hardware size $8000$ pe fixed hai); sirf $m_f=ϵS+m_{\text{pay}}$ move karta hai jab $ϵ$ change hota hai.

1. Case (a): $ϵ=0$. Tab structure $=ϵS=0$, isliye $m_f=0+1000=1000$ kg aur $m_0=8000+1000=9000$. $\Delta v=3500\ln \left(\frac{9000}{1000}\right)=3500\times 2.197\approx 7690\text{m/s}.$ Yeh step kyun? $ϵ=0$ ek given tank size ke liye theoretical best hai — ceiling. Yeh finite hai kyunki $1000$ kg payload $m_f$ ko $0$ tak pahunchne nahi deta (log tabhi diverge hota jab $m_f\to 0$).

2. Case (b): $ϵ=1$. Koi fuel nahi, isliye $m_f=S+m_{\text{pay}}=m_0=9000$. $\Delta v=3500\ln (1)=0\text{m/s}.$ Yeh step kyun? $\ln (1)=0$ — kuch nahi jalaane se kuch nahi badalta. Yeh degenerate zero-input check hai, figure s02 mein navy dot.

3. Trend padho (figure s02): jaise-jaise $ϵ$ $0$ se $1$ ki taraf barhta hai, $\Delta v$ smoothly $7690$ se $0$ tak slide karta hai; real design ($ϵ=0.13$, orange dot) us curve pe $5190$ m/s pe upar baitha hai. Yeh Payload fraction optimization mein low structural fraction chase karne ka visual argument hai.

Recall Verify

(a) $\ln 9=2.197$, deta hai $7690$ m/s (finite, real $5190$ m/s se upar). (b) bilkul $0$ kyunki $\ln 1=0$. Dono limits behave kar rahe hain. ✓

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Example 5 — Cell E: series vs parallel, same hardware

Forecast: Series dead weight co-lift karne se bachta hai — compute karne se pehle bet lagao kaun jeetega.

Yaad dilate hain: yahan har unit ke liye $ϵ=0.1$, isliye structure $=ϵS=3000$ kg aur fuel $=(1-ϵ)S=27000$ kg per $30000$ kg unit; $m_{\text{pay}}=2000$ kg final cargo hai.

Series layout (unit ek bottom stage hai jo core se pehle drop hoti hai):

1. Bottom stage (core + payload) $=30000+2000=32000$ ko apna payload maanta hai. $m_0=S_b+32000=30000+32000=62000$; $m_f=ϵS_b+32000=3000+32000=35000$. $\Delta v_1=3400\ln \left(\frac{62000}{35000}\right)=3400\times 0.5716\approx 1943\text{m/s}.$
2. Core stage: $m_0=S_c+2000=32000$; $m_f=ϵS_c+2000=3000+2000=5000$. $\Delta v_2=3400\ln \left(\frac{32000}{5000}\right)=3400\times 1.856\approx 6311\text{m/s}.$
3. Series total $\approx 1943+6311=8254\text{m/s}$. Yeh step kyun? Har dropped stage agli mass ratio ko improve karta hai — series ka compounding advantage.

Parallel layout (dono boosters + core phase 1 mein jalte hain; boosters drop hote hain; core finish karta hai): 4. Phase 1: sab engines fire karte hain. Har jagah same $v_e$ hai isliye ${\bar{v}}_e=3400$. Hum ek cross-feed design stipulate karte hain jisme core ke tanks separation tak full rahte hain — boosters core ko feed karte hain isliye phase 1 mein sirf booster propellant draw hota hai. Start mass $=$ core $30000$ + do boosters $60000$ + payload $2000=92000$ kg. Boosters apna saara fuel $=2\times 27000=54000$ kg jalate hain; core abhi kuch nahi jalata: $m_1=92000-54000=38000$. $\Delta v_1=3400\ln \left(\frac{92000}{38000}\right)=3400\times 0.8842\approx 3006\text{m/s}.$ Yeh step kyun? "Core phase 1 mein kuch nahi jalata" assumption arbitrary nahi hai — yeh explicit cross-feed rule hai jo humne choose kiya taaki dono layouts core burn same full tanks se shuru karein, comparison fair bane. (Cross-feed ke bina core bhi phase 1 mein drain hota, ek alag lekin similarly computable split deta.) 5. Separation. Boosters ab khaali shells hain; inhe jettison karo. Unka dropped mass $2\times ϵS_b=2\times 3000=6000$ kg hai, isliye $m_2=m_1-6000=38000-6000=32000$ kg. Yeh step kyun? Separation sirf booster structure remove karta hai — fuel phase 1 mein already exhaust ke roop mein nikal gaya. Yahi cliff hai jo parallel staging buy karta hai. 6. Phase 2: core apna pura $27000$ kg jalata hai. Start $m_2=32000$; final $m_{2,f}=ϵS_c+m_{\text{pay}}=3000+2000=5000$. $\Delta v_2=3400\ln \left(\frac{32000}{5000}\right)=3400\times 1.856\approx 6311\text{m/s}.$ Yeh step kyun? Jettison ke baad vehicle sirf core hai payload carry karta hua, isliye Tsiolkovsky core pe akele apply hota hai apne untouched fuel ke saath. 7. Parallel total $\approx 3006+6311=9317\text{m/s}$.

Compare karo: Series $8254$ vs parallel $9317$ m/s — parallel yahan jeet jaata hai.

Recall Verify

Series $8254$ vs parallel $9317$ m/s. Yahan parallel jeet jaata hai kyunki series case mein core ko doosri unit ka poora fuel pehle burn mein payload ki tarah uthana pada, uske ratio ko hurt karte hue; cross-feed parallel burn us fuel ko jaldi aur neeche drain karta hai. "Series always more efficient hai" wala slogan tabhi kaam karta hai jab stages alag-alag optimise ki jaayein — yeh dikhata hai kyun tumhe actually compute karna chahiye. ✓

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Example 6 — Cell F: word problem, "kitne boosters chahiye?"

Forecast: $n$ guess karo: kya $1$ enough hai, ya kai chahiye — ya possible hi nahi?

Yaad dilate hain: $m_0$ = liftoff mass (fixed core $60000$ + $25000$ per booster); $m_1$ = booster burnout pe mass ($m_0$ minus $20000n$ kg booster fuel).

1. Hume $\Delta v_1\ge 9400-7000=2400\text{m/s}$ chahiye. Yeh step kyun? Core already $7000$ supply karta hai; boosters ko gap cover karna hai.

2. $n=3$ try karo: $m_0=60000+75000=135000$; fuel burnt $=3\times 20000=60000$; $m_1=75000$. $\Delta v_1=1000\ln \left(\frac{135000}{75000}\right)=1000\times 0.5878=588\text{m/s}.$ Bahut kam.

3. $n=8$ try karo: $m_0=60000+200000=260000$; fuel $=160000$; $m_1=100000$. $\Delta v_1=1000\ln \left(\frac{260000}{100000}\right)=1000\times 0.9555=956\text{m/s}.$ Phir bhi kam — aur note karo ki yeh slowly badha. Diminishing returns dikh raha hai.

4. Limit $n\to \infty$ lo. Ratio explicitly likho aur upar-neeche $n$ se divide karo: $\frac{m_0}{m_1}=\frac{60000+25000n}{60000+5000n}=\frac{60000/n+25000}{60000/n+5000}\underset{n\to \infty }{\to }\frac{25000}{5000}=5.$ Yeh step kyun? Jaise $n$ barta hai, fixed core mass $60000$ booster terms ke saamne negligible ho jaata hai jo $n$ ke saath scale karte hain, isliye ratio purely booster fuel-to-total se govern hota hai. $60000/n$ terms vanish ho jaate hain, sirf pure booster ratio $25000:5000=5$ bchta hai. Yahi ceiling hai. $\Delta v_1^{\max }=1000\ln 5=1000\times 1.609=1609\text{m/s}.$

5. Conclusion: kyunki $n\to \infty$ bhi sirf $1609<2400$ m/s deta hai, in boosters ki koi bhi sankhya target reach nahi kar sakti. Tumhe booster $v_e$ badhana hoga ya unka structural mass kam karna hoga, aur nahi jodna.

Recall Verify

$n=3\to 588$, $n=8\to 956$, limit $\to 1609$ m/s, sab zaroori $2400$ se neeche. Sahi conclusion: in boosters se achieve nahi hoga. ✓

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Example 7 — Cell G: exam twist, galat mass ratio pakdo

Forecast: Kaun sa student phase 1 mein already jala hua fuel double-count kar raha hai?

Yaad dilate hain: $m_{p,c}^{(1)}$ = phase 1 mein jala propellant; $m_{p,c}^{(2)}$ = phase 2 ke liye bacha propellant; $m_2$ = phase 2 ke start pe mass (boosters ja chuke hain).

1. Propellant split karo. $m_{p,c}^{(1)}=\frac{1}{3}\times 36000=12000$ kg. Remaining $m_{p,c}^{(2)}=36000-12000=24000$ kg. Yeh step kyun? Phase 2 sirf wahi jala sakta hai jo bacha hai — Student A ka formula pretend karta hai ki $12000$ kg abhi bhi onboard hai.

2. Sahi phase-2 masses (Student B): $m_2=m_{s,c}+m_{p,c}^{(2)}+m_{\text{upper}}=4000+24000+6000=34000$; final $m_{2,f}=m_{s,c}+m_{\text{upper}}=4000+6000=10000$. $\Delta v_2=v_{e,c}\ln \left(\frac{m_2}{m_{2,f}}\right)=3200\ln \left(\frac{34000}{10000}\right)=3200\times 1.2238\approx 3916\text{m/s}.$

3. Student A ki galat value. Saare $36000$ kg use karte hue: $m_0=4000+36000+6000=46000$, giving $3200\ln \left(\frac{46000}{10000}\right)=3200\times 1.526=4884\text{m/s}$ — $\approx 968$ m/s phantom fuel se inflated. Yeh step kyun? Dono numbers compare karne se error pin hoti hai: surplus bilkul wahi $\Delta v$ hai jo pehle se khatam ho chuke $12000$ kg falsely provide karte.

Recall Verify

Sahi $\Delta v_2\approx 3916$ m/s (Student B). Student A ka $4884$ m/s already spent $12000$ kg overcount karta hai — bilkul wahi "whole-propellant" error jo parent note flag karta hai. ✓

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Example 8 — Cell H: unit trap ($I_{sp}$ seconds vs $v_e$ m/s)

Forecast: Student kya multiply karna bhool gaya?

Yaad dilate hain: $I_{sp}$ exhaust velocity seconds mein expressed hai; Tsiolkovsky ko feed karne ke liye $v_e$ m/s mein chahiye, jo relate hota hai $v_e=I_{sp}g$ se jahan $g=9.81{\text{m/s}}^2$. Dekho Specific impulse.

1. $I_{sp}$ ko exhaust velocity mein convert karo. $v_e=I_{sp}g=452\times 9.81=4434\text{m/s}$. Yeh step kyun? Tsiolkovsky ko $v_e$ m/s mein chahiye, $I_{sp}$ seconds mein nahi; $452$ seedha use karna seconds$\times$(dimensionless) mein answer deta hai — galat units, $g$ factor miss hai.

2. Ab Tsiolkovsky apply karo. Mass ratio $m_0/m_f=200000/60000=10/3\approx 3.333$. $\Delta v=v_e\ln \left(\frac{m_0}{m_f}\right)=4434\ln \left(\frac{200000}{60000}\right)=4434\times 1.2040\approx 5339\text{m/s}.$

3. Student ki raw error. $452\times 1.204\approx 544$ — yeh bilkul $5339/9.81$ hai, yaani sahi answer missing $g$ ki wajah se $9.81$ se shrunk hua. Yeh step kyun? Yeh recognize karna ki galat number sahi answer divided by $9.81$ ke barabar hai, ek single cause confirm karta hai — dropped $g$ factor.

Recall Verify

$v_e=452\times 9.81=4434.12$ m/s; $\Delta v\approx 5339$ m/s. Galat answer $544\approx 5339/9.81$ missing $g$ confirm karta hai. Units check: $(\text{s})\times ({\text{m/s}}^2)=\text{m/s}$ $v_e$ ke liye. ✓

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Recall Self-test

Example 1 se series total $\Delta v$ ::: lagbhag 9640 m/s Example 2 mein thrust-weighted ${\bar{v}}_e$ ::: lagbhag 2966 m/s $ϵ\to 1$ se $\Delta v=0$ kyun hota hai? ::: koi fuel nahi, isliye $m_0=m_f$ aur $\ln 1=0$ Example 8 mein unit slip ::: bhool gaye $v_e=I_{sp}\times g$

Related: Tsiolkovsky rocket equation · Gravity losses · Falcon Heavy · Saturn V · Payload fraction optimization.