Worked examples — Payload fraction as function of Δv and Isp

3.3.47 · D3 · Physics › Rocket Propulsion › Payload fraction as function of Δv and Isp

Yeh page parent note payload-fraction master equation ka "har case cover karo" workbook hai. Agar neeche koi bhi symbol unfamiliar lage, toh pehle parent note dobara padho — lekin woh ek formula hum yahan phir se likhte hain taaki tumhe is page se kahin aur jaana na pade.

Aage ka sab kuch bas usi box mein numbers daalna hai, isliye interesting part yeh jaanna hai ki kaunse numbers ise tod dete hain, saturate kar dete hain, ya koi trap chhupa dete hain. Yahi hai scenario matrix.

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The scenario matrix

Har rocket problem ko ek grid mein ek point samjho. Do knobs sabse zyada matter karte hain: $R$ kitna bada hai ($\Delta v$ aur $I_{sp}$ se drive hota hai), aur $ϵ$ kitna bada hai. Us grid ke corners aur edges alag-alag behave karte hain.

Cell	Kya special hai	Example jo ise hit karta hai
A. Easy regime	chhota $R$ ($R$ close to 1): tiny burn, kaafi bacha hua	Ex 1
B. Hard regime	bada $R$: $\lambda$ negative ho jaata hai → single-stage mission impossible	Ex 2
C. Zero-$\Delta v$ degenerate	$\Delta v=0\Rightarrow R=1$	Ex 3
D. Zero-structure ideal	$ϵ=0$: theoretical best case	Ex 4
E. The break-even boundary	exact $R$ dhundho jahan $\lambda =0$ ho	Ex 5
F. $I_{sp}\to \infty$ limit	infinite exhaust speed, $\lambda$ kahan approach karta hai	Ex 6
G. Real-world word problem	woh engine chunno jo required payload mass deliver kare	Ex 7
H. Exam twist	diya gaya $\lambda$ aur $ϵ$, ulta $\Delta v$ nikalo	Ex 8

Neeche ka map $\lambda$ ko $R$ ke against plot karta hai kuch $ϵ$ values ke liye, taaki hum kuch bhi compute karne se pehle dekh sakein ki har cell kahan rehta hai.

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The worked examples

Forecast: Chhota $\Delta v$ hai, toh $R$ close to 1 hona chahiye aur $\lambda$ bada hona chahiye (50% se zyada). Aage padhne se pehle guess karo.

1. Exponent compute karo. $\frac{\Delta v}{I_{sp}{g}_0}=\frac{1500}{320\times 9.81}=\frac{1500}{3139.2}=0.4778$. Yeh step kyun? Poora formula is dimensionless number par chalta hai; yeh pehle aana chahiye.
2. $R$ nikalo. $R=e^{0.4778}=1.6125$. Yeh step kyun? Ek baar $ϵ$ fix ho jaaye toh $\lambda$ sirf $R$ par depend karta hai.
3. Box mein daalo. $\lambda =\frac{1+0.10(1-1.6125)}{1.6125}=\frac{1-0.06125}{1.6125}=\frac{0.93875}{1.6125}$. Yeh step kyun? Direct substitution — yahi master equation hai.
4. Result. $\lambda =0.582=58.2\%$.

Verify: Units — exponent $\text{(m/s)}/(\text{s}\cdot {\text{m/s}}^2)=$ dimensionless ✓. $R>1$ ✓. $\lambda <1$ aur forecast ke anusaar comfortably positive ✓.

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Forecast: Yeh parent note ka SSTO case hai. Bahut bada $\Delta v$ → bahut bada $R$ → hum expect karte hain $\lambda$ negative hoga, yaani impossible.

1. Exponent. $\frac{9400}{350\times 9.81}=\frac{9400}{3433.5}=2.7378$. Yeh step kyun? Wahi pehla move; bada $\Delta v$ ise bada banata hai.
2. $R$. $R=e^{2.7378}=15.454$. Yeh step kyun? 15+ ka mass ratio matlab launch mass ka 94%+ propellant hai.
3. Daalo. $\lambda =\frac{1+0.10(1-15.454)}{15.454}=\frac{1-1.4454}{15.454}=\frac{-0.4454}{15.454}$.
4. Result. $\lambda =-0.0288$.

Verify: Negative $\lambda$ physically meaningless hai (negative payload nahi ho sakta). Math hume bata raha hai ki structure akela hi empty-mass budget se zyada hai. Sanity check: yeh parent note ke $-0.029$ se match karta hai. Real mein fix = staging. ✓

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Forecast: Koi burn nahi matlab koi propellant nahi, toh poora launch mass payload hi ho sakta hai → $\lambda =1$? Dekhte hain $ϵ$ ise kharaab karta hai ya nahi.

1. Exponent = 0. $\frac{0}{I_{sp}{g}_0}=0$. Yeh step kyun? Formula ki degenerate boundary test kar raha hai.
2. $R=e^0=1$. Yeh step kyun? Koi burn nahi ⇒ initial mass equals final mass ⇒ ratio exactly 1.
3. Daalo. $\lambda =\frac{1+ϵ(1-1)}{1}=\frac{1+0}{1}=1$.
4. Result. $\lambda =1=100\%$.

Verify: Zero propellant ke saath, $ϵ\cdot m_{\text{prop}}=0$ structure bhi, toh sab kuch payload hai. $\lambda =1$ bilkul sahi hai aur $ϵ$ se independent hai — $ϵ$ term vanish ho jaata hai kyunki $(1-R)=0$. ✓

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Forecast: Koi structural tax nahi, toh $\lambda$ ek clean $1/R$ hona chahiye — sirf reciprocal mass ratio, jo kisi bhi real design se kabhi approach ho sakta hai.

1. Exponent. $\frac{3800}{450\times 9.81}=\frac{3800}{4414.5}=0.8608$.
2. $R$. $R=e^{0.8608}=2.3651$.
3. $ϵ=0$ ke saath daalo. $\lambda =\frac{1+0}{R}=\frac{1}{2.3651}$. Yeh step kyun? $ϵ=0$ set karne par box collapse ho jaata hai $\lambda =1/R=e^{-\Delta v/(I_{sp}{g}_0)}$ mein.
4. Result. $\lambda =0.4228=42.3\%$.

Verify: Yeh mass ratio ka reciprocal hai aur realistic $ϵ=0.08$ answer (parent ke Ex mein 37.7%) se zyada hona chahiye. $42.3\%>37.7\%$ ✓ — tax hamesha $\lambda$ ko girata hai.

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Forecast: Ex 2 mein humne dekha $R=15.45$ already negative ho gaya; break-even us se thoda neeche hona chahiye. Guess ~11.

1. Numerator ko zero set karo. $\lambda =0\Rightarrow 1+ϵ(1-R)=0$. Yeh step kyun? $\lambda$ ek fraction hai jisme $R>0$ denominator mein hai, toh woh zero hoga exactly jab top zero hoga.
2. $R$ ke liye solve karo. $1+0.10-0.10R=0\Rightarrow R=\frac{1+ϵ}{ϵ}=\frac{1.10}{0.10}=11$. Yeh step kyun? Clean algebra se break-even mass ratio $R^{\star }=(1+ϵ)/ϵ$ milta hai.
3. $\Delta v$ back out karo. $\Delta v=I_{sp}{g}_0\ln R=350\times 9.81\times \ln 11=3433.5\times 2.3979=8233\text{m/s}$. Yeh step kyun? Critical $R$ ko mission speed mein translate karne ke liye Tsiolkovsky invert karo.

Verify: $R=11$ wapas plug karo: $\lambda =\frac{1+0.10(1-11)}{11}=\frac{1-1}{11}=0$ ✓. Aur $8233<9400$ (Ex 2 ise cross karke negatives mein chala gaya) ✓. "~11" ka forecast sahi nikla.

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Forecast: Infinite $I_{sp}$ matlab almost koi propellant nahi chahiye, toh intuitively $\lambda \to 1$. Lekin kya $ϵ$ term koi nishaan chhod jaata hai? Guess karo.

1. Exponent 0 ho jaata hai. Jab $I_{sp}\to \infty$, $\frac{\Delta v}{I_{sp}{g}_0}\to 0$. Yeh step kyun? Denominator explode karta hai, exponent ko kill karta hai.
2. $R\to e^0=1$. Yeh step kyun? Infinitely efficient engine ⇒ negligible propellant ⇒ mass ratio ≈ 1.
3. Limit lo. $\lambda \to \frac{1+ϵ(1-1)}{1}=1$. Yeh step kyun? Wahi collapse jaise zero-$\Delta v$ case mein — $(1-R)\to 0$ se $ϵ$ erase ho jaata hai.
4. Result. $\lambda \to 1=100\%$.

Verify: Numerically $I_{sp}=10^6\text{s}$, $\Delta v=3800$ par: exponent $=3800/(10^6\cdot 9.81)=3.87\times 10^{-4}$, $R=1.000387$, $\lambda =\frac{1+0.10(1-1.000387)}{1.000387}=0.99958\approx 1$ ✓. Isliye chemistry-beating drives (ion, nuclear) itne prized hain — dekho Isp.

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Forecast: Parent ke Ex 2 se, is configuration mein $\lambda =37.7\%$ aata hai. Toh $m_0$ roughly $2000/0.377\approx 5300$ kg hona chahiye. Compute karne se pehle guess karo.

1. $\lambda$ nikalo. Exponent $=\frac{3800}{450\times 9.81}=0.8608$, $R=e^{0.8608}=2.3651$. Yeh step kyun? Mass mein invert karne se pehle fraction chahiye.
2. Daalo. $\lambda =\frac{1+0.08(1-2.3651)}{2.3651}=\frac{1-0.10921}{2.3651}=\frac{0.89079}{2.3651}=0.3767$. Yeh step kyun? Yahi 37.7% hai jo parent ne nikala — acha cross-check.
3. Definition invert karo. $\lambda =m_{\text{pay}}/m_0\Rightarrow m_0=m_{\text{pay}}/\lambda =2000/0.3767$. Yeh step kyun? $\lambda$ ek ratio hai; $m_0$ ke liye solve karna bas divide karna hai.
4. Result. $m_0=5309\text{kg}$.

Verify: Payload check $0.3767\times 5309=2000$ kg ✓. Forecast (~5300) se match karta hai. Units: kg ÷ (dimensionless) = kg ✓.

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Forecast: $\lambda =25\%$ middling hai, toh $R$ moderate hoga (easy cases se bada, break-even $R^{\star }=(1.08)/0.08=13.5$ se kaafi chhota). Kuch km/s $\Delta v$ expect karo.

1. Box ko $R$ ke liye rearrange karo. $\lambda R=1+ϵ-ϵR\Rightarrow R(\lambda +ϵ)=1+ϵ\Rightarrow R=\frac{1+ϵ}{\lambda +ϵ}$. Yeh step kyun? Log lene ke liye $R$ akela chahiye; $R$ terms group karo.
2. Numbers daalo. $R=\frac{1.08}{0.25+0.08}=\frac{1.08}{0.33}=3.2727$. Yeh step kyun? Diye gaye $\lambda$ ko mass ratio mein convert karta hai.
3. Tsiolkovsky invert karo. $\Delta v=I_{sp}{g}_0\ln R=340\times 9.81\times \ln (3.2727)=3335.4\times 1.1857=3955\text{m/s}$. Yeh step kyun? $\Delta v$ exponent ke andar hai; natural log use karke undo karo.
4. Result. $\Delta v\approx 3955\text{m/s}$.

Verify: Forward-check: $R=e^{3955/3335.4}=e^{1.1857}=3.273$ ✓, phir $\lambda =\frac{1+0.08(1-3.273)}{3.273}=\frac{0.8182}{3.273}=0.250$ ✓. Aur $R=3.27<R^{\star }=13.5$, toh $\lambda >0$ jaisa hona chahiye ✓.

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Recall Quick self-test

Break-even mass ratio formula ::: $R^{\star }=(1+ϵ)/ϵ$ Negative $\lambda$ physically kya matlab hai? ::: Un parameters par koi valid single-stage rocket exist nahi karta — staging chahiye. Jab $I_{sp}\to \infty$, $\lambda \to$ ::: $1$ (100%), kyunki $R\to 1$. $ϵ=0$ ke saath, formula collapse ho jaata hai ::: $\lambda =1/R=e^{-\Delta v/(I_{sp}{g}_0)}$ mein.

Related: Tsiolkovsky Rocket Equation · Mass Ratio · Optimal Staging · Oberth Effect.