Mass budgets — dry mass, wet mass, margin: Worked examples

3.6.24 · D3 · Physics › Spacecraft Structures & Systems Engineering › Mass budgets — dry mass, wet mass, margin

Yeh page mass budgets ka drill hall hai. Parent note ne aapko definitions aur key formula diye. Yahaan hum har tarah ke problem ko cover karte hain jo mass budgeting mein aate hain — small satellites, bade cargo ships, burns jo mass reduce karte hain, margin jo erode hota hai, aur rocket equation ke do directions jo sabko confuse karte hain.

Shuru karne se pehle, do tools ka ek reminder:

Yahaan $v_{e} = I_{s p} \cdot g_{0}$ with $g_{0} = 9.81 m/s^{2}$ — yeh Specific Impulse se link karta hai. Mass ratio ka idea Tsiolkovsky Rocket Equation ka dil hai.

Neeche wali figure is page ke har example ke peeche ki mental picture hai.

Figure — Mass budgets — dry mass, wet mass, margin

Intuition Woh figure kaise padhein

Horizontal axis ratio $Δ V / v_{e}$ hai — woh exponent jo $R = e^{Δ V / v_{e}}$ ko feed karta hai. Left par small (easy missions), right par large (demanding missions). Vertical axis propellant ka share hai poore vehicle ka, yaani $\frac{m _{prop}}{m _{wet}} = \frac{R - 1}{R}$ , ek number 0 aur 1 ke beech.

Coral (red) curve woh fraction hai. Notice karo yeh left par 0 ke paas start hoti hai aur tezzi se upar jhukti hai — yeh exponential ka kaatna hai.
Dashed mint line at 1.0 woh impossible ceiling hai "100 % fuel, koi ship nahi bachi" — curve approach karta hai lekin kabhi reach nahi karta.
Do lavender dots hamare worked cases mark karte hain: Example 1 ( $Δ V / v_{e} \approx 0.1$ par, sirf ~10 % fuel) aur Example 3 ( $Δ V / v_{e} = 3$ par, ek crushing ~95 % fuel). Left-to-right padhne par, picture hi woh story hai jo batati hai mass discipline kyun zaroori hai.

Scenario matrix

Har mass-budget problem in cells mein se ek (ya blend) hota hai. Neeche ke examples cell order A → B → C → D → E → F → G → H mein hain taaki tags seedhe page ke neeche padhein.

Cell	Isko distinct kya banata hai	Covered by
A. Sizing forward	Given dry mass + $Δ V$ , propellant/wet mass find karo	Ex 1
B. Flight backward	Given wet mass, burn ke baad mass find karo	Ex 2
C. Zero / tiny $Δ V$	Degenerate limit: $R \to 1$ , propellant $\to 0$	Ex 3
D. Huge $Δ V$ limit	$R ≫ 1$ , propellant dry mass par dominate karta hai	Ex 4
E. Margin erosion	Growth reserve khaata hai; $%$ margin recompute karo	Ex 5
F. Dry-mass-growth penalty	1 kg dry $\to$ kai kg propellant	Ex 6
G. Multi-burn / staged	Do burns chain karo, har ek mein mass track karo	Ex 7
H. Exam twist	Backwards solve karo ek unknown ke liye ( $I_{s p}$ ya $Δ V$ )	Ex 8

Mission $Δ V$ numbers Mission ΔV Budget se aate hain; empty-mass fraction ke ideas Structural Mass Fraction se connect hote hain.

Cell A — Sizing forward

Worked example Example 1 — Chota comms satellite, tank size karo

Given: margin ke saath dry mass $m_{dry} = 420$ kg. Life mein station-keeping ke liye $Δ V = 220$ m/s chahiye. Bipropellant thruster, $I_{s p} = 230$ s. Find: propellant mass aur launch wet mass.

Forecast: $Δ V$ $v_{e}$ ke comparison mein chhota hai, toh andaza lagao propellant dry mass se bahut chhota hai — shayad tens of kg.

Step 1 — Exhaust speed. $v_{e} = I_{s p} g_{0} = 230 \times 9.81 = 2256.3$ m/s. Yeh step kyun? Rocket equation sirf $v_{e}$ jaanta hai, $I_{s p}$ nahi. Hume seconds ko real speed mein convert karna hoga pehle (dekho Specific Impulse).

Step 2 — Mass ratio. $R = e^{Δ V / v_{e}} = e^{220/2256.3} = e^{0.09751} = 1.1024$ . Yeh step kyun? $R$ woh single number hai jo kehta hai "loaded 1.10× hai empty ship ka."

Step 3 — Wet mass (sizing, toh multiply). $m_{wet} = m_{dry} \cdot R = 420 \times 1.1024 = 463.0$ kg. Yeh step kyun? Hume light end (dry) pata hai aur heavy end (loaded) chahiye, toh $R$ se multiply karte hain.

Step 4 — Propellant. $m_{prop} = m_{wet} - m_{dry} = 463.0 - 420 = 43.0$ kg. Yeh step kyun? Propellant simply woh mass hai jo humne dry ship ke upar add ki, toh wet se dry subtract karte hain use isolate karne ke liye — yeh woh number hai jo tank mein hona chahiye.

Verify: Wapas plug karo: $v_{e} ln (463.0/420) = 2256.3 \times ln (1.1024) = 2256.3 \times 0.09751 = 220$ m/s ✓. Propellant (43 kg) dry mass ka ~10 % hai — small- $Δ V$ forecast se match karta hai. Units: kg throughout, $v_{e}$ m/s mein log ke andar $Δ V$ ke saath cancel hota hai. ✓

Cell B — Flight backward (ek burn mass reduce karta hai)

Worked example Example 2 — Burn ke baad mass kitni bachi

Given: ek probe burn mein daakhil hota hai wet mass $m_{i} = 2500$ kg ke saath. Woh $Δ V = 1200$ m/s execute karta hai $I_{s p} = 320$ s ke saath. Find: burn ke baad baaki mass, aur propellant used.

Forecast: Burn mass bahar fenkta hai, toh answer 2500 kg se kam hai. Yahaan hum $R$ se divide karte hain, multiply nahi.

Step 1 — Exhaust speed. $v_{e} = 320 \times 9.81 = 3139.2$ m/s. Yeh step kyun? Equation real speed mein likhi hai, toh pehle datasheet ke $I_{s p}$ (seconds mein) ko $v_{e}$ (m/s mein) mein turn karte hain kisi bhi formula mein enter karne se pehle.

Step 2 — Mass ratio. $R = e^{1200/3139.2} = e^{0.38226} = 1.4656$ . Yeh step kyun? $R$ wahi climb-or-slide factor hai jaise hamesha; ise ek baar compute karne se hum ya toh multiply kar sakte hain (sizing) ya divide (flight) — yahaan hum divide karenge.

Step 3 — Final mass (flight, toh divide). $m_{f} = m_{i} / R = 2500/1.4656 = 1705.8$ kg. Yeh step kyun? Hume heavy start pata hai aur lighter end chahiye, toh lighter mass division bar ke neeche jaati hai — equation ka flight direction.

Step 4 — Propellant consumed. $2500 - 1705.8 = 794.2$ kg. Yeh step kyun? Burned propellant woh mass hai jo ship ne khoya, yaani start minus finish — yeh woh number hai jo hum tank gauge se tick karte hain.

Verify: Answer se forward equation run karo: $v_{e} ln (2500/1705.8) = 3139.2 \times ln (1.4656) = 3139.2 \times 0.38226 = 1200$ m/s ✓. Agar humne galti se multiply kiya hota, toh 3664 kg milta — burn ke baad heavier, physically absurd. ✓

Cell C — Zero / tiny ΔV (degenerate limit)

Worked example Example 3 — Jab ΔV → 0 kya hota hai?

Given: ek 420 kg satellite sirf $Δ V = 2$ m/s ki pointing tweak kar raha hai, $v_{e} = 2256$ m/s. Find: propellant, aur limiting behaviour confirm karo.

Forecast: Almost koi velocity change nahi matlab almost koi propellant nahi. Jaise $Δ V \to 0$ , propellant $\to 0$ hona chahiye.

Step 1 — Mass ratio. $R = e^{2/2256} = e^{0.000886} = 1.000887$ . Yeh step kyun? Hum dekhna chahte hain mass ratio 1 ke kitna close aata hai — yeh upar wali figure mein flat, far-left part hai.

Step 2 — Propellant. $m_{prop} = m_{dry} (R - 1) = 420 \times 0.000887 = 0.372$ kg. Yeh step kyun? $R - 1$ fractional extra mass hai; kilograms mein convert karne ke liye dry mass se multiply karo.

Step 3 — The limit check. Tiny $x$ ke liye, $e^{x} \approx 1 + x$ . Toh $m_{prop} \approx m_{dry} \cdot \frac{Δ V}{v _{e}} = 420 \times \frac{2}{2256} = 0.372$ kg — exponential ek straight line mein collapse ho jaata hai. Yeh step kyun? Yeh prove karta hai formula zero par smooth hai: koi division-by-zero nahi, koi blow-up nahi. Exactly $Δ V = 0$ par, $R = 1$ , $m_{prop} = 0$ — aap koi fuel nahi lete kyunki aap kuch nahi karte.

Verify: Linear estimate (0.372 kg) aur exact exponential (0.372 kg) 3 decimals tak agree karte hain, bilkul expected jab $Δ V ≪ v_{e}$ . ✓

Cell D — Huge ΔV limit

Worked example Example 4 — High ΔV par propellant runaway hota hai

Given: ek upper stage jo $Δ V = 9000$ m/s supply karna chahiye (near single-stage-to-orbit), dry mass $m_{dry} = 1000$ kg, $v_{e} = 3000$ m/s (chemical). Find: wet mass, aur dekhein propellant kitna badly dominate karta hai.

Forecast: $Δ V$ $v_{e}$ ka teen guna hai. Upar wali figure mein, yeh horizontal axis par $Δ V / v_{e} = 3$ par baithta hai, jahaan coral curve mint 100 % ceiling ke paas already upar jhuk chuki hai. Guess karo wet mass kai tonnes hai.

Step 1 — Mass ratio. $R = e^{9000/3000} = e^{3} = 20.09$ . Yeh step kyun? Exponent $Δ V / v_{e} = 3$ hai, aur $e^{3} \approx 20$ — yeh single number batata hai loaded ship empty ka bees guna hai, toh hum isse pehle compute karte hain.

Step 2 — Wet mass. $m_{wet} = 1000 \times 20.09 = 20, 090$ kg. Yeh step kyun? Hume light (dry) end pata hai aur heavy (loaded) end chahiye, toh — "heavy multiplies" follow karte hue — dry mass ko $R$ se multiply karte hain.

Step 3 — Propellant. $m_{prop} = 20090 - 1000 = 19, 090$ kg. Yeh step kyun? Dry ko wet se subtract karne par propellant isolate hota hai; yahaan yeh poore vehicle ka 95 % hai. Isliye hum rockets stage karte hain instead of ek giant tank banane ke (Launch Vehicle Performance).

Verify: Propellant ka fraction $= 19090/20090 = 0.9502$ , yaani 95 % — woh right-hand lavender dot jo figure $Δ V / v_{e} = 3$ par mark karta hai. Ex 1 (10 %) se compare karo: $Δ V / v_{e}$ ko 0.1 se 3 par le jaane par fuel fraction tiny se overwhelming ho gayi — exponential ki signature. ✓

Cell E — Margin erosion

Worked example Example 5 — Reserve shrink hote dekhna

Given: lander CBE $= 800$ kg, policy margin 25 %, toh MEV $= 800 \times 1.25 = 1000$ kg. Design review ke baad teen growths aati hain: structure +40 kg, avionics +18 kg, harness +12 kg. Find: baaki margin kg aur percent mein.

Forecast: Total growth 200 kg reserve mein se 70 kg hai — margin 25 % se mid-teens ki taraf girna chahiye.

Step 1 — New CBE. $800 + 70 = 870$ kg. Yeh step kyun? Growth ship par real hardware hai; yeh CBE (actual mass) raise karta hai, fixed MEV ceiling ko nahi.

Step 2 — Baaki margin (kg). MEV ceiling 1000 kg par fixed rehti hai. Reserve $= 1000 - 870 = 130$ kg. Yeh step kyun? Margin ceiling (MEV) aur current mass (CBE) ke beech gap ke roop mein defined hai; bade hue CBE ko fixed MEV se subtract karne par baaki reserve milta hai.

Step 3 — Margin percentage. $\frac{130}{870} = 0.1494 = 14.9%$ . Yeh step kyun? Margin percent hamesha reserve ÷ current-best-estimate hota hai, toh dono numerator aur denominator mass grow hone par move karte hain.

Verify: $200/800 = 25%$ se shuru kiya, ab $14.9%$ — forecast ke anusaar gira. Sanity: $870 + 130 = 1000$ kg = unchanged MEV ceiling. ✓

Cell F — Dry-mass-growth penalty

Worked example Example 6 — Ek extra kilogram asal mein kitna costly hai

Given: Mars transfer stage, $Δ V = 6000$ m/s, $v_{e} = 3000$ m/s. Dry mass $Δ m_{dry} = 12$ kg badhti hai. Find: us growth ki wajah se extra propellant.

Forecast: Bada $Δ V$ matlab badi penalty — 12 kg growth dozens of kg extra fuel mein blow up honi chahiye.

Step 1 — Is page par penalty factor derive karo. $Δ V$ fixed rakho (yeh mission requirement hai). Wet mass hamesha $m_{wet} = m_{dry} R$ hai $R = e^{Δ V / v_{e}}$ ke saath, toh propellant hai $m_{prop} = m_{dry} R - m_{dry} = m_{dry} (R - 1)$ . Dry mass ko $Δ m_{dry}$ badhane par propellant badalta hai $Δ m_{prop} = (m_{dry} + Δ m_{dry}) (R - 1) - m_{dry} (R - 1) = Δ m_{dry} (R - 1) .$ Yeh step kyun? Sab kuch cancel ho jaata hai sivaaye $Δ m_{dry} (R - 1)$ ke — extra fuel directly extra dry mass ke proportional hai, multiplier $R - 1 = e^{Δ V / v_{e}} - 1$ ke saath. Yeh penalty factor hai, yahaan derive kiya, borrowed nahi.

Step 2 — Factor evaluate karo. $R - 1 = e^{6000/3000} - 1 = e^{2} - 1 = 7.389 - 1 = 6.389$ . Yeh step kyun? Mission numbers ko us factor mein plug karo jo humne abhi banaya.

Step 3 — Extra propellant. $Δ m_{prop} = 12 \times 6.389 = 76.7$ kg. Yeh step kyun? Per-kg penalty ko actual 12 kg growth se scale karo.

Verify: 12 kg problem "grow" karke $12 + 76.7 = 88.7$ kg extra wet mass ban gaya — ek $88.7/12 = 7.39 \times$ amplification, jo $e^{2}$ , full mass ratio ke barabar hai. Consistent. ✓

Cell G — Multi-burn / staged bookkeeping

Worked example Example 7 — Ek ke baad ek do burns

Given: cargo ship $m_{0} = 45000$ kg par launch hota hai. Burn 1 (trans-Mars injection): $Δ V_{1} = 3800$ m/s. Burn 2 (Mars orbit insertion): $Δ V_{2} = 2100$ m/s. Dono $I_{s p} = 450$ s use karte hain. Dry mass $= 5000$ kg. Find: har burn ke baad mass; check karo kaafi propellant bacha hai ya nahi.

Forecast: Do burns successively divide karte hain — har division ship ko shrink karta hai. Dono ke baad, hume still 5000 kg dry se upar rehna chahiye, warna mission fail.

Step 1 — Exhaust speed. $v_{e} = 450 \times 9.81 = 4414.5$ m/s. Yeh step kyun? Dono burns same engine use karte hain, toh iska $I_{s p}$ ek baar $v_{e}$ mein convert karo aur reuse karo — har burn ke liye redo karne ki zaroorat nahi.

Step 2 — Burn 1 (divide). $R_{1} = e^{3800/4414.5} = e^{0.86081} = 2.3650$ . $m_{1} = 45000/2.3650 = 19, 027$ kg. Yeh step kyun? Flight direction: burn ke baad ship lighter hai, toh divide karte hain.

Step 3 — Burn 2 ( $m_{1}$ se phir divide). $R_{2} = e^{2100/4414.5} = e^{0.47569} = 1.6091$ . $m_{2} = 19027/1.6091 = 11, 824$ kg. Yeh step kyun? Doosra burn jo mass burn 1 se bacha usi se shuru hota hai — aap hamesha pichle result se chain karte hain.

Step 4 — Feasibility. $m_{2} = 11, 824$ kg $> 5000$ kg dry, toh $11824 - 5000 = 6824$ kg propellant baad ke maneuvers ke liye bacha hai. Mission viable. Yeh step kyun? Mass budget ka poora point go/no-go check hai: surviving mass ko dry mass se compare karna batata hai koi propellant bacha hai ya nahi, aur isliye mission fly ho sakta hai ya nahi.

Verify: Ek single combined ratio se total $Δ V$ : $R_{1} R_{2} = 2.3650 \times 1.6091 = 3.806$ , aur $v_{e} ln (3.806) = 4414.5 \times 1.3365 = 5900$ m/s $= 3800 + 2100$ ✓. Ratios chain karne par $Δ V$ s ka sum reproduce hota hai — log products ko sums mein turn karta hai. ✓

Cell H — Exam twist (backwards solve karo)

Worked example Example 8 — Given fuel, required

$I_{s p}$ find karo Given: aapko zaroor $Δ V = 4000$ m/s achieve karna hai. Aapka dry mass 600 kg fixed hai aur aap sirf 900 kg propellant load kar sakte ho, toh start (wet) mass $m_{i} = 600 + 900 = 1500$ kg hai aur finish (dry) mass $m_{f} = 600$ kg hai. Exam puchta hai: minimum $I_{s p}$ kya hai jo yeh possible banata hai? Find: $I_{s p}$ .

Forecast: Yeh har pichle example ko invert karta hai — hume masses aur $Δ V$ pata hai, hum engine chahte hain. Ek demanding $Δ V$ sirf modest fuel-to-dry ratio ke saath matlab hume high-performance engine chahiye, toh expect karo $I_{s p}$ cold-gas value se kaafi upar.

Step 1 — Masses se mass ratio. $R = m_{i} / m_{f} = 1500/600 = 2.5$ . Yeh step kyun? Masses fixed hain, toh achievable ratio fixed hai — koi bhi engine choice ise nahi badal sakti. Yeh woh number hai jo engine ko velocity ke roop mein "cash in" karna hai.

Step 2 — Required exhaust speed. $Δ V = v_{e} ln R$ se shuru karo aur $v_{e}$ ko isolate karo dono sides ko $ln R$ se divide karke: $v_{e} = \frac{Δ V}{l n R} = \frac{4000}{l n 2.5} = \frac{4000}{0.91629} = 4365.5 m/s .$ Yeh step kyun? $ln$ exactly woh operation hai jo $R = e^{Δ V / v_{e}}$ mein exponential ko undo karta hai; $Δ V$ ko $ln R$ se divide karne par woh exhaust speed milti hai jo engine ko deliver karni hai.

Step 3 — $I_{s p}$ mein convert karo. $I_{s p} = v_{e} / g_{0} = 4365.5/9.81 = 445.0$ s. Yeh step kyun? Engine datasheets $I_{s p}$ seconds mein quote karte hain, $v_{e}$ nahi; $g_{0}$ se divide karne par hamaari exhaust speed shopping-list number mein turn hoti hai.

Verify: $I_{s p} = 445$ s ke saath forward check: $v_{e} = 445 \times 9.81 = 4365.5$ m/s, $Δ V = 4365.5 \times ln 2.5 = 4365.5 \times 0.91629 = 4000$ m/s ✓. ~445 s ka $I_{s p}$ matlab high-efficiency bipropellant engine — ek cold-gas thruster ( $I_{s p} \approx 65$ s) hopelesssly short padta. ✓

Recall Quick self-test

In reveal lines mein, triple colon se pehle sab question hai; uske baad sab hidden answer hai — right-hand side cover karo aur pehle khud try karo.

Sizing vs flight: dono ke liye kaun si operation? ::: Sizing dry mass ko $R$ se multiply karta hai; flight wet mass ko $R$ se divide karta hai. Jaise $Δ V \to 0$ , propellant kya approach karta hai? ::: Zero — $R \to 1$ , toh $m_{prop} = m_{dry} (R - 1) \to 0$ . $Δ V_{1}$ phir $Δ V_{2}$ ke do burns: unke mass ratios kaise combine hote hain? ::: Woh multiply hote hain, $R = R_{1} R_{2}$ , kyunki $Δ V$ ke logs add hote hain. Dry mass growth ke har kg par penalty factor? ::: $e^{Δ V / v_{e}} - 1$ .

Related deep pages: Propellant Management, Center of Mass Control, Mass Properties Measurement, Systems Engineering V-Model.