Payload fraction as function of Δv and Isp: Exercises

3.3.47 · D4 · Physics › Rocket Propulsion › Payload fraction as function of Δv and Isp

Yeh page ek graded workout hai. Har problem parent topic, Tsiolkovsky Rocket Equation, Specific Impulse, Structural Coefficient, aur Mass Ratio pe build karti hai. Levels ko order mein complete karo — har ek agla unlock karta hai.

Definition Is Page Par Kaam Aane Wale Symbols Aur Do Tools

Neeche jo bhi hai, sirf inhi cheezon se solve hoga. Inhe dhyaan mein rakho.

Mission ki demand. $Δ v$ (padho "delta-vee") wo total speed change hai, m/s mein, jo mission require karta hai — rocket ko kitna speed up ya slow down karna hai apni target orbit reach karne ke liye. Yeh destination se set hota hai, rocket builder se nahi.

Masses. Liftoff par rocket ki total mass $m_{0}$ hoti hai (payload + structure + propellant). Burn ke baad, jab fuel khatam ho jata hai, bacha hua mass $m_{f}$ hota hai (sirf payload + structure). Inka ratio $m_{0} / m_{f}$ Mass Ratio hai — hamesha $\geq 1$ kyunki rocket sirf halka hi hota hai.

Exhaust speed. $v_{e} = I_{s p} g_{0}$ woh speed hai (m/s mein) jis par engine apna propellant peeche fenkta hai. Yeh Specific Impulse $I_{s p}$ (seconds mein) ko $g_{0} = 9.81 m/s^{2}$ se multiply karke real velocity mein convert kiya jata hai. Kyunki $v_{e} = I_{s p} g_{0}$ hai, neeche wala exponent dono tarah likha ja sakta hai: $\frac{Δ v}{I _{s p} g _{0}} = \frac{Δ v}{v _{e}}$ — literally "hum kitne exhaust-speeds ka velocity change demand kar rahe hain."

Characteristic mass ratio — "loaded rocket kitni baar bada hai empty rocket se" wala number: $R = e^{Δ v / (I_{s p} g_{0})} = e^{Δ v / v_{e}} = \frac{m _{0}}{m _{f}}$ Payload fraction — launch mass ka woh hissa jo actual cargo hai: $λ = \frac{1 + ϵ ( 1 - R )}{R}$ jahan $ϵ$ (Structural Coefficient) hai tank/engine ka kg per kg of fuel.

Figure s01 (neeche): solve pipeline. Upar teen boxes — Inputs ( $Δ v, I_{s p}, ϵ$ ) → Exponent ( $Δ v / (I_{s p} g_{0})$ ) → Mass ratio ( $R = e^{(\cdot)}$ ) — ek arrow curve karke ek chauthe box mein jaata hai, Payload fraction $λ = (1 + ϵ (1 - R)) / R$ . Yeh ek visual reminder hai ki $R$ woh gateway hai jisse har problem $λ$ compute karne se pehle guzarti hai.

Figure — Payload fraction as function of Δv and Isp

Level 1 — Recognition

Exercise 1.1 (L1)

Tsiolkovsky equation likhi hai $Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})$ . Charon quantities ko words mein identify karo, aur batao kaun si ek ko badhane se engine zyada efficient banti hai (same mission par zyada payload).

Recall Solution

$Δ v$ = total velocity change jo mission demand karta hai (orbital mechanics se set, humse nahi).
$I_{s p}$ = specific impulse = engine ki "fuel economy", seconds mein.
$g_{0} = 9.81 m/s^{2}$ = standard gravity, sirf ek unit-conversion constant jo $I_{s p}$ mein baki hai.
$m_{0} / m_{f}$ = Mass Ratio: loaded mass $m_{0}$ divided by dry mass $m_{f}$ (dono upar box mein define hain).

Efficiency improve karne ke liye == $I_{s p}$ == badhao. Bada $I_{s p}$ exponent $Δ v / (I_{s p} g_{0})$ ko chota karta hai, jo $R$ ko chota karta hai, jo $λ$ badhata hai.

Exercise 1.2 (L1)

Ek rocket ka $R = 1$ hai. Physically kya iska matlab hai us propellant ke baare mein jo woh carry karta hai, aur jab $ϵ = 0.10$ ho toh $λ$ kya hoga?

Recall Solution

$R = e^{Δ v / (I_{s p} g_{0})} = 1$ exponent ko $0$ force karta hai, yani $Δ v = 0$ . Rocket ko velocity change nahi karni, isliye woh koi propellant nahi jalata. $R = 1$ ke saath: $λ = \frac{1 + ϵ ( 1 - 1 )}{1} = \frac{1 + 0}{1} = 1.$ Payload fraction 100% hai — poori launch mass cargo hai, kyunki na fuel hai na (fuel-proportional) structure. Yeh degenerate limiting case hai.

Level 2 — Application

Exercise 2.1 (L2)

Ek lunar-transfer stage ko $Δ v = 3, 100 m/s$ chahiye, hydrolox engine use karta hai jiska $I_{s p} = 450 s$ hai, aur $ϵ = 0.08$ hai. $R$ aur $λ$ compute karo.

Recall Solution

Exponent: $\frac{3100}{450 \times 9.81} = \frac{3100}{4414.5} = 0.7022$ . $R = e^{0.7022} = 2.018.$ $λ = \frac{1 + 0.08 ( 1 - 2.018 )}{2.018} = \frac{1 - 0.0814}{2.018} = \frac{0.9186}{2.018} = 0.4552.$ Payload fraction $\approx 45.5%$ .

Exercise 2.2 (L2)

LEO mein ek spacecraft jiska total mass $12, 000 kg$ hai, use $Δ v = 1, 900 m/s$ burn karna hai $I_{s p} = 320 s$ , $ϵ = 0.12$ ke saath. Actual payload ke kitne kilograms deliver honge?

Recall Solution

Exponent: $\frac{1900}{320 \times 9.81} = \frac{1900}{3139.2} = 0.6053$ . $R = e^{0.6053} = 1.8318.$ $λ = \frac{1 + 0.12 ( 1 - 1.8318 )}{1.8318} = \frac{1 - 0.09982}{1.8318} = \frac{0.90018}{1.8318} = 0.4914.$ Payload mass $= λ m_{0} = 0.4914 \times 12000 = 5897 kg$ (lagbhag 5.9 tonnes).

Level 3 — Analysis

Exercise 3.1 (L3)

Fixed engine ( $I_{s p} = 350 s$ ) aur fixed $ϵ = 0.10$ ke liye, woh $Δ v$ nikalo jis par payload fraction exactly zero ho jaye. Physically interpret karo.

Recall Solution

Set $λ = 0 \Rightarrow 1 + ϵ (1 - R) = 0 \Rightarrow R = 1 + \frac{1}{ϵ}$ . $ϵ = 0.10$ ke saath: $R = 1 + 10 = 11$ . Ab $R = e^{Δ v / (I_{s p} g_{0})}$ ko invert karo: $Δ v = I_{s p} g_{0} ln R = 350 \times 9.81 \times ln (11) = 3433.5 \times 2.3979 = 8234 m/s .$ Matlab: $Δ v = 8234 m/s$ par fuel-plus-its-structure pura rocket exactly consume kar leta hai — zero cargo. Isse bada koi bhi $Δ v$ (jaise LEO ke liye ~9,400 m/s) negative $λ$ deta hai: physically impossible ek stage mein. Yahi mathematical wajah hai ki chemical Staging kyun mandatory hai.

Exercise 3.2 (L3)

Same $I_{s p} = 350 s$ , $ϵ = 0.10$ use karte hue, $Δ v = 4, 000$ aur $Δ v = 8, 000 m/s$ par $λ$ compute karo, aur dikhao ki yeh drop linear se worse hai (ek explicit straight-line prediction se compare karo).

Recall Solution

$I_{s p} g_{0} = 3433.5$ .

$Δ v = 4000$ par: exponent $= 1.1650$ , $R = 3.2059$ . $λ = \frac{1 + 0.10 ( 1 - 3.2059 )}{3.2059} = \frac{0.77941}{3.2059} = 0.2431 = 24.3%.$ $Δ v = 8000$ par: exponent $= 2.3300$ , $R = 10.278$ . $λ = \frac{1 + 0.10 ( 1 - 10.278 )}{10.278} = \frac{0.07220}{10.278} = 0.007025 = 0.70%.$

Explicit "linear guess" banana. Linear model ke liye ek slope chahiye. Pehle interval se lo: $Δ v = 0$ par bhi $λ$ compute karo, jo $λ = 1$ hai (100%, Exercise 1.2 se). Do points $(0, 100%)$ aur $(4000, 24.3%)$ se seedhi line ka slope: $slope = \frac{24.3% - 100%}{4000 - 0} = - 0.0189% per m/s .$ Usi line ko $Δ v = 8000$ tak extrapolate karo: $λ_{linear} (8000) = 100% + (- 0.0189%/ m/s) (8000) = 100% - 151.4% = - 51.4%.$ Linear guess wildly impossible $- 51.4%$ predict karta hai, jabki honest exponential curve $0.70%$ par barely positive hai. Linear thinking fall ko catastrophically overshoot karta hai kyunki true fall $e^{- Δ v}$ se dominate hoti hai: 4000 se 8000 jaane par $λ$ ek factor of ~35 se cut hua, fixed amount se nahi.

Figure s02 (neeche): exponential tyranny. Ek magenta curve $λ$ (% mein) versus $Δ v$ (km/s mein) $I_{s p} = 350 s, ϵ = 0.10$ ke liye. Yeh $Δ v = 0$ par 100% se steeply girta hai, "4 km/s → 24.3%" aur "8 km/s → 0.70%" marked violet dots se guzarta hai, dashed zero line ko orange square " $λ = 0$ , 8.23 km/s" par cross karta hai, aur baad mein negative hota chala jata hai — clearly seedhi line nahi hai.

Level 4 — Synthesis

Exercise 4.1 (L4)

Ek engineer apna budget do mein se ek tarike se kharch kar sakta hai $Δ v = 5, 000 m/s$ mission ke liye, baseline $I_{s p} = 340 s$ , $ϵ = 0.12$ se start karte hue:

Option A: engine improve karke $I_{s p} = 380 s$ karo ( $ϵ = 0.12$ rakho).
Option B: structure halka karke $ϵ = 0.06$ karo ( $I_{s p} = 340 s$ rakho).

Kaun sa choice zyada payload fraction deta hai?

Recall Solution

Baseline ( $I_{s p} = 340, ϵ = 0.12$ ): $I_{s p} g_{0} = 3335.4$ , exponent $= 1.4991$ , $R = 4.4776$ . $λ_{0} = \frac{1 + 0.12 ( 1 - 4.4776 )}{4.4776} = \frac{0.58269}{4.4776} = 0.1301 = 13.0%.$

Option A ( $I_{s p} = 380, ϵ = 0.12$ ): $I_{s p} g_{0} = 3727.8$ , exponent $= 1.3413$ , $R = 3.8241$ . $λ_{A} = \frac{1 + 0.12 ( 1 - 3.8241 )}{3.8241} = \frac{0.66111}{3.8241} = 0.1729 = 17.3%.$

Option B ( $I_{s p} = 340, ϵ = 0.06$ ): $R = 4.4776$ (same, engine nahi badla). $λ_{B} = \frac{1 + 0.06 ( 1 - 4.4776 )}{4.4776} = \frac{0.79134}{4.4776} = 0.1767 = 17.7%.$ Verdict: Option B (17.7%) Option A (17.3%) se thoda aage hai yahan. Structure lightening isliye help karta hai kyunki $R \approx 4.5$ par structural penalty $ϵ (1 - R)$ badi hai; $ϵ$ ko aadha karne se woh penalty directly aadhi ho jaati hai. Lekin margin thin hai — zyada $Δ v$ par $I_{s p}$ ka exponential dominance answer flip kar deta.

Exercise 4.2 (L4)

4.1 ke baseline ke liye, kis $Δ v$ par dono options equal $λ$ denge? Equation set up karo aur use hand-runnable numerical method (bisection) se solve karo, intermediate bracketing steps dikhao. Interpret karo.

Recall Solution

Maano $x = Δ v$ . Define karo $R_{A} (x) = e^{x /3727.8}, R_{B} (x) = e^{x /3335.4},$ aur difference function $f (x) = λ_{A} \frac{1 + 0.12 ( 1 - R _{A} )}{R _{A}} - λ_{B} \frac{1 + 0.06 ( 1 - R _{B} )}{R _{B}} .$ Hum root $f (x) = 0$ chahte hain. Bisection ke liye ek bracket chahiye jahan $f$ sign change kare. Trial points par dono payload fractions evaluate karo:

$x$ (m/s)	$λ_{A}$	$λ_{B}$	$f (x)$	sign
$5000$	$0.1729$	$0.1767$	$- 0.0038$	$-$
$10000$	$0.0248$	$0.0116$	$+ 0.0133$	$+$

Sign 5000 aur 10000 ke beech flip hota hai, toh us interval mein root hai. Bisect karo (midpoint test karo, woh half rakho jo abhi bhi zero ko straddle kare):

step	midpoint $x$	$f (x)$	keep
1	$7500$	$+ 0.0053$	left half $[5000, 7500]$
2	$6250$	$+ 0.0011$	left half $[5000, 6250]$
3	$5625$	$- 0.0011$	right half $[5625, 6250]$
4	$5937$	$+ 0.0000$	root yahan

$Δ v \approx 5930 m/s$ par converge ho raha hai. Interpretation: ~5930 m/s se neeche lighter structure (Option B) jeetata hai; usse upar better engine (Option A) jeetata hai, kyunki exponential exponent $\propto 1/ I_{s p}$ importance gain karta hai jaise $Δ v$ badhta hai. Yeh "high- $Δ v$ stages love high $I_{s p}$ " ka quantitative chehra hai. Dekho Optimal Staging.

Level 5 — Mastery

Exercise 5.1 (L5)

Ek two-stage vehicle ko $Δ v_{total} = 9, 200 m/s$ produce karna hai. Har stage ka $I_{s p} = 360 s$ aur $ϵ = 0.09$ hai, aur split equal hai: $Δ v_{1} = Δ v_{2} = 4, 600 m/s$ . Stage 1 ka "payload" poora stage 2 hai (iska apna payload fraction phir multiply hota hai). Overall payload fraction $λ_{total} = λ_{1} λ_{2}$ compute karo, aur single-stage result se compare karo.

Recall Solution

Har stage $Δ v = 4600$ , $I_{s p} g_{0} = 360 \times 9.81 = 3531.6$ dekhta hai. Exponent $= 4600/3531.6 = 1.30253$ , toh $R = e^{1.30253} = 3.6787$ . $λ_{stage} = \frac{1 + 0.09 ( 1 - 3.6787 )}{3.6787} = \frac{1 - 0.24108}{3.6787} = \frac{0.75892}{3.6787} = 0.20630.$ Kyunki dono stages identical hain, $λ_{1} = λ_{2} = 0.20630$ , isliye $λ_{total} = 0.2063 0^{2} = 0.04256 = 4.26% .$ Compare: ek single stage jo saare 9,200 m/s kare in numbers ke saath, exponent $9200/3531.6 = 2.6051$ , $R = 13.531$ , deta hai $λ_{single} = \frac{1 + 0.09 ( 1 - 13.531 )}{13.531} = \frac{- 0.12779}{13.531} = - 0.00944 < 0,$ jo impossible hai. Same total $Δ v$ ko do stages mein split karna ek impossible mission ko viable ~4.3% payload fraction mein convert kar deta hai. Negative se positive ka yahi jump poori wajah hai jiske liye Staging exist karta hai.

Exercise 5.2 (L5)

Payload-fraction formula $λ (Δ v) = \frac{1 + ϵ ( 1 - R )}{R}$ lo jahan $R = e^{Δ v / (I_{s p} g_{0})}$ hai. Analytically dikhao ki $\frac{d λ}{d ( Δ v )} < 0$ sabhi $Δ v \geq 0$ ke liye (payload fraction strictly decreasing hai), given $0 < ϵ < 1$ .

Recall Solution

$k = 1/ (I_{s p} g_{0}) > 0$ use karo toh $R = e^{k Δ v}$ aur $λ = \frac{1 + ϵ}{R} - ϵ = (1 + ϵ) e^{- k Δ v} - ϵ .$ $Δ v$ ke respect mein differentiate karo: $\frac{d λ}{d ( Δ v )} = - (1 + ϵ) k e^{- k Δ v} .$ Har factor positive hai: $(1 + ϵ) > 0$ , $k > 0$ , aur $e^{- k Δ v} > 0$ sabhi real $Δ v$ ke liye. Isliye derivative strictly negative hai har jagah. $λ$ monotonically girta hai — koi bump nahi, koi rise nahi — aur iska shape ek pure decaying exponential hai jo $ϵ$ se neeche shift hai. Woh constant $- ϵ$ floor hi wajah hai kyun $λ$ eventually negative ho jata hai: bade $Δ v$ par $e^{- k Δ v}$ term vanish hoti hai aur $λ \to - ϵ$ . Dekho Propellant Mass Fraction complementary view ke liye.