Mass budgets — dry mass, wet mass, margin: Exercises

3.6.24 · D4 · Physics › Spacecraft Structures & Systems Engineering › Mass budgets — dry mass, wet mass, margin

Yeh page ek self-test hai. Har problem ko pehle khud padhkar try karo, phir collapsible Solution kholkar har step check karo. Problems L1 (sirf words pehchano) se L5 (conflicting constraints ke under poora budget design karo) tak jaate hain. Agar koi symbol unfamiliar lage, toh pehle parent note revisit karo.

Yahan sab kuch teen ideas par based hai jo parent mein build kiye gaye hain:

Recall Teen quantities jo tumhe chahiye (agar rusty ho toh kholao)

Dry mass $m_{dry}$ ::: sab kuch consumables ke siwa — structure, payload, empty tanks, wiring.
Wet mass $m_{wet}$ ::: dry mass plus propellant (aur pressurant/doosre consumables).
Mass margin ::: ek reserved fraction $ϵ$ estimate ka jo mass growth ke khilaf insurance ke taur par rakha jaata hai: $m_{margin} = ϵ m_{dry,est}$ . Yahan $m_{dry,est}$ exactly CBE (Current Best Estimate) hai — component masses ka raw sum bina kisi cushion ke. Toh is page par "estimate" aur "CBE" ek hi number ko kehte hain.

Aur ek equation jo mass ko motion se jodti hai — Tsiolkovsky Rocket Equation: $Δ V = v_{e} ln (\frac{m _{wet}}{m _{dry}})$ jahan $v_{e}$ exhaust velocity hai (gas nozzle se kitni tez nikalti hai, $m/s$ mein), jo Specific Impulse se $v_{e} = I_{s p} g_{0}$ ke zariye judi hai, jahan $g_{0} = 9.81 m/s^{2}$ .

Intuition Is page ke har formula ke peeche ka fine print

Tsiolkovsky teen cheezein assume karta hai, aur humare saare answers unhe inherit karte hain:

Constant $v_{e}$ — engine ki exhaust speed mid-burn mein nahi badalti. Real engines thodi drift karte hain; hum ise fixed treat karte hain.
Koi gravity ya drag losses nahi — equation ideal $Δ V$ deta hai (velocity change agar ship empty space mein coast kare). Planet ke paas real maneuvers mein kuch $Δ V$ gravity aur air mein kho jaata hai, isliye jo mission $Δ V$ tum plug in karte ho woh usually un losses ko cover karne ke liye padded hota hai (dekho Mission ΔV Budget).
Saara propellant exhaust ke roop mein nikalta hai — koi leaks nahi, koi unusable residuals nahi (actually kuch percent trapped ho jaata hai).

Edge cases yaad rakhne ke liye:

$Δ V \to 0$ : toh $e^{Δ V / v_{e}} \to e^{0} = 1$ , isliye $m_{wet} \to m_{dry}$ — zero maneuver ko zero propellant chahiye. Achha sanity check hai.
$ϵ = 0$ (koi margin nahi): $m_{margin} = 0$ , toh design mein koi cushion nahi — jaise hi koi component bade, budget break ho jaata hai. Isliye real programs kabhi $ϵ = 0$ pe nahi udte jab tak hardware ban aur weigh na ho jaaye.
$v_{e} \to \infty$ (ek perfect, infinitely efficient engine): $Δ V / v_{e} \to 0$ , toh phir $m_{wet} \to m_{dry}$ — tum logon ko almost koi propellant nahi chahiye hoga. Yahi wajah hai ki high Specific Impulse itni valued hai.

Level 1 — Recognition

Goal: identify karo kaunsa mass kaunse bucket mein jaata hai aur definitions sahi se padho.

Neeche ka figure is page ke har problem ke peeche ka mental picture hai — har kilogram kahan rehta hai:

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

Recall Solution 1.1

Hum kya karte hain: har item ko dry ya wet mein sort karo. Dry = sab kuch jo consumable nahi hai. Empty tank dry count hota hai (yeh hardware hai); hydrazine hi ek maatra consumable hai. $m_{dry} = 120 + 40 + 25 + 15 = 200 kg$ $m_{wet} = m_{dry} + m_{prop} = 200 + 300 = 500 kg$ Answer: dry $= 200 kg$ , wet $= 500 kg$ .

Recall Solution 1.2

CBE (Current Best Estimate) bina margin ka raw sum hai. MEV margin add karta hai: $MEV = CBE \times (1 + ϵ) = 800 \times 1.25 = 1000 kg$ Answer: $1000 kg$ .

Level 2 — Application

Goal: rocket equation mein numbers plug karo, dono directions mein.

Recall Solution 2.1

Exponential form kyun? Hum dry (final) mass jaante hain aur wet (initial) mass jaanna chahte hain, isliye hum $m_{wet}$ solve karne ke liye Tsiolkovsky rearrange karte hain. $ln$ ko undo karo exponentiation se: $m_{wet} = m_{dry} e^{Δ V / v_{e}} = 200 e^{1.2/3.0} = 200 \times e^{0.4}$ Yahan $Δ V$ aur $v_{e}$ dono already same unit (km/s) mein hain, isliye ratio $1.2/3.0 = 0.4$ ek clean pure number hai — koi conversion nahi chahiye. $= 200 \times 1.4918 = 298.4 kg$ $m_{prop} = m_{wet} - m_{dry} = 298.4 - 200 = 98.4 kg$ Answer: propellant $\approx 98.4 kg$ , wet mass $\approx 298.4 kg$ .

Recall Solution 2.2

Step 1 — $v_{e}$ nikalo. Specific Impulse $I_{s p}$ (seconds mein) exhaust velocity mein convert hota hai $g_{0} = 9.81 m/s^{2}$ se multiply karke: $v_{e} = I_{s p} g_{0} = 220 \times 9.81 = 2158.2 m/s$ Dhyan do yeh m/s mein aata hai. Step 2 — divide karne se pehle units match karo. Exponent $Δ V / v_{e}$ ek pure number hona chahiye, isliye dono quantities same unit mein honi chahiye. Hamara $v_{e}$ m/s mein hai, isliye $Δ V$ bhi m/s mein convert karte hain: $Δ V = 0.5 km/s = 0.5 \times 1000 = 500 m/s$ Yeh step kyun matter karta hai: agar hum $Δ V$ ko $0.5 km/s$ hi rakhte $2158.2 m/s$ ke upar, toh hum $0.5/2158.2 = 0.00023$ compute karte $500/2158.2 = 0.2317$ ke bajaye — 1000 ka factor off aur nonsense answer. Step 3 — propellant. $m_{wet} = 50 e^{500/2158.2} = 50 e^{0.2317} = 50 \times 1.2607 = 63.03 kg$ $m_{prop} = 63.03 - 50 = 13.03 kg$ Answer: $v_{e} \approx 2158 m/s$ , propellant $\approx 13.0 kg$ .

Level 3 — Analysis

Goal: track karo ki ek change (mass growth, ek burn) kaise propagate hota hai, aur number interpret karo.

Recall Solution 3.1

Yeh formula kyun? Parent derivation se, $Δ V$ fixed rakhte hue dry mass badhne par wet mass proportionally badhta hai, aur extra propellant dry growth hai amplified penalty factor se: $Δ m_{prop} = Δ m_{dry} (e^{Δ V / v_{e}} - 1)$ Pehle penalty factor compute karo (dono speeds km/s mein, isliye $6/3 = 2$ ek clean pure number hai): $e^{6/3} - 1 = e^{2} - 1 = 7.389 - 1 = 6.389$ $Δ m_{prop} = 8 \times 6.389 = 51.1 kg$ Interpretation: structure growth ka har kilogram $\approx 6.4 kg$ propellant cost karta hai — yahi wajah hai ki margins itni carefully guard ki jaati hain. Neeche ke figure mein red curve dikhata hai ki yeh penalty factor $Δ V / v_{e}$ badhne ke saath kitni violently climb karta hai.

Figure (upar): horizontal axis dimensionless ratio $Δ V / v_{e}$ hai; vertical axis dikhata hai ki dry mass ke extra kilogram par kitne kilogram propellant onboard force hote hain. Black dots ek GEO-class maneuver ( $Δ V / v_{e} = 0.5$ , penalty $\approx 0.65$ ) aur is Mars case ( $Δ V / v_{e} = 2$ , penalty $\approx 6.4$ ) ko mark karte hain. Dhyan do curve left par almost flat hai aur $Δ V / v_{e} \approx 1$ ke baad upar rocket karta hai — yeh steepening exponential ka high- $Δ V$ missions ko punish karna hai.

Answer: $\approx 51.1 kg$ extra propellant.

Recall Solution 3.2

Divide kyun, multiply nahi? L2 mein humne ek tank size kiya: hum final (dry) mass jaante the aur poochha ki burn se pehle ship kitna bada hona chahiye, isliye humne multiply karke heavier initial mass tak gaye. Yahan ulta sawaal hai — hum already initial (heavier) mass jaante hain aur poochhte hain ki fuel kharcha hone ke baad kya bachega. Tsiolkovsky $Δ V = v_{e} ln (m_{i} / m_{f})$ se shuru karke $m_{f}$ ke liye solve karo: exponentiate karo $m_{i} / m_{f} = e^{Δ V / v_{e}}$ paane ke liye, phir flip karo, $m_{f} = m_{i} / e^{Δ V / v_{e}}$ . Final (lighter) mass denominator mein hai, isliye hum jaani hui initial mass ko ratio $R$ se divide karte hain. Divide karna ship ko lighter banata hai — exactly wahi jo fuel burn karna karna chahiye. $mass ratio R = e^{3800/4414} = e^{0.8609} = 2.3654$ (dono speeds already same km/s unit mein hain, isliye $3800/4414$ kaam karta hai chahe hum unhe m/s ya km/s mein likhein — ratio identical hai.) $m_{after} = \frac{45000}{2.3654} = 19, 024 kg$ $m_{prop,burned} = 45000 - 19024 = 25, 976 kg$ $m_{prop,remaining} = m_{after} - m_{dry} = 19024 - 5000 = 14, 024 kg$ Answer: burn ke baad $\approx 19, 024 kg$ ; burn hua $\approx 25, 976 kg$ ; bacha hua propellant $\approx 14, 024 kg$ .

Level 4 — Synthesis

Goal: margin bookkeeping ko rocket equation ke saath ek budget mein combine karo.

Yeh figure trace karta hai ki raw component mass ka ek kilogram launch pad tak jaate jaate kaise swells — pehle margin, phir rocket-equation mass ratio:

Recall Solution 4.1

Step 1 — dry CBE. Kyun: CBE hamara raw best estimate hai, koi cushion add karne se pehle starting point; hum component masses add karte hain. $CBE = 1.2 + 0.8 + 1.0 = 3.0 kg$ Step 2 — dry margin apply karo. Kyun: real hardware badhta hai (adhesive, harness, coating), isliye hum estimate ko $ϵ = 0.20$ se inflate karte hain design dry mass paane ke liye jiske around hum actually tank build karte hain. $m_{dry} = 3.0 \times 1.20 = 3.6 kg$ Step 3 — $I_{s p}$ ko $v_{e}$ mein convert karo. Kyun: rocket equation ko seconds nahi velocity chahiye; $g_{0}$ se multiply karo. $v_{e} = 65 \times 9.81 = 637.65 m/s$ Yahan $Δ V = 50 m/s$ already m/s mein hai aur $v_{e}$ ki unit se match karta hai, isliye koi conversion nahi chahiye. Step 4 — propellant (size up). Kyun: hum design dry (final) mass jaante hain aur wet (initial) mass chahte hain, isliye hum mass ratio se multiply up karte hain, bilkul L2 jaisi. $m_{wet} = 3.6 e^{50/637.65} = 3.6 e^{0.07841} = 3.6 \times 1.08157 = 3.8937 kg$ $m_{prop} = 3.8937 - 3.6 = 0.2937 kg$ Step 5 — propellant margin. Kyun: residuals, valve leakage aur off-nominal burns bhi propellant khaate hain, isliye hum propellant par hi $10%$ reserve add karte hain. $m_{prop,tot} = 0.2937 \times 1.10 = 0.3231 kg$ Step 6 — final wet mass. Kyun: deliverable pad par total mass hai = design dry + total propellant. $m_{wet,final} = 3.6 + 0.3231 = 3.923 kg$ Check: $3.923 kg ≪ 12 kg$ ✓ — kaafi headroom hai (dekho Structural Mass Fraction ki uska kitna frame hai). Answer: wet mass $\approx 3.92 kg$ ; comfortably fit karta hai.

Recall Solution 4.2

(a) Original MEV. Kyun: MEV woh CBE hai jo margin fraction se inflate hua ho — woh ceiling jise program commit karta hai. $MEV = 1200 \times 1.25 = 1500 kg$ (b) Remaining margin. Kyun: real growth ka har kilogram fixed MEV ceiling ko eat karta hai. Pehle growth sum karo, ise old CBE mein add karo new CBE paane ke liye, phir dekho ceiling uske kitna upar hai. $growth = 80 + 30 + 15 = 125 kg, new CBE = 1200 + 125 = 1325 kg$ $remaining margin = 1500 - 1325 = 175 kg$ (c) Margin percentage. Kyun: margin tabhi meaningful hai jab cheez abhi kitna weigh karti hai ke relative ho — hamara best current estimate, new CBE — frozen ceiling ke nahi. $\frac{175}{1325} = 0.1321 = 13.2%$ Interpretation: margin $25%$ se gir ke $13.2%$ ho gaya. Typical $\sim 15%$ PDR floor se neeche — mass-reduction "tiger team" ka time aa gaya. Answer: MEV $= 1500 kg$ ; remaining $175 kg$ ; $13.2%$ .

Level 5 — Mastery

Goal: ek hard constraint ke under design karo — margin, $Δ V$ , aur payload ko ek doosre ke against trade karo.

Recall Solution 5.1

Step 1 — unknown naam do. Kyun: payload woh hai jiske liye hum solve kar rahe hain, isliye ise ek symbol do aur dry mass ko uske terms mein express karo. Maano payload $= P$ , toh total dry CBE $= 150 + P$ . Step 2 — margin apply karo. Kyun: tank ko design dry mass ke around size karna hai, jo CBE ko $20%$ se inflate karke milti hai: $m_{dry} = 1.20 (150 + P) .$ Step 3 — rocket equation se launch cap se link karo. Kyun: mission jo wet mass demand karta hai woh design dry mass hai mass ratio se multiply up karke (hum final jaante hain, initial chahiye — wahi "multiply up" jaisi L2), aur woh wet mass launcher jo lift kare usse exceed nahi kar sakta: $m_{wet} = 1.20 (150 + P) e^{Δ V / v_{e}} \leq 600, e^{2/3} = 1.9477$ (dono speeds km/s mein, isliye $2/3$ ek clean pure number hai.) Step 4 — inequality solve karo. Kyun: $P$ isolate karo do amplifying factors divide out karke. $1.20 \times 1.9477 (150 + P) \leq 600 \Rightarrow 2.3372 (150 + P) \leq 600$ $150 + P \leq \frac{600}{2.3372} = 256.72 \Rightarrow P \leq 106.72 kg$ Answer: maximum payload $\approx 106.7 kg$ . Sanity: $P = 106.7$ par, dry CBE $= 256.7$ , design dry $= 308.1$ , wet $= 308.1 \times 1.9477 = 600$ ✓ — exactly cap par.

Recall Solution 5.2

Step 1 — baseline wet mass ( $P = 90$ ). Kyun: humein "before" number chahiye change measure karne ke liye. Wahi margin-then-mass-ratio chain apply karo jaisi 5.1 mein hai. $dry CBE = 240, m_{dry} = 1.20 \times 240 = 288, m_{wet} = 288 \times 1.9477 = 560.94 kg$ Step 2 — post-swap wet mass ( $P = 110$ ). Kyun: heavier payload ke saath identical chain redo karo "after" number paane ke liye. $dry CBE = 260, m_{dry} = 1.20 \times 260 = 312, m_{wet} = 312 \times 1.9477 = 607.68 kg$ (a) Wet mass mein rise. Kyun: before ko after se subtract karo. $607.68 - 560.94 = 46.74 kg$ Dhyan do $20 kg$ payload swap $46.7 kg$ wet mass cost karta hai — $\times 1.20$ margin aur $\times 1.9477$ mass ratio dono ise amplify karte hain ( $20 \times 1.20 \times 1.9477 = 46.74$ ). (b) Cap ke under slack. Kyun: cap minus demanded wet mass batata hai ki tum still fit karte ho ya nahi. $600 - 607.68 = - 7.68 kg (negative \to launcher se 7.7 kg over)$ (c) Slack negative hone ka consequence. Tumhe kahin aur se mass recover karna hoga — $Δ V$ reduce karo (ek chota ΔV budget, matlab kam orbit capability), structure halka karo, ya bade (mehnge) launch vehicle par jaao. Answer: (a) $+ 46.7 kg$ ; (b) $- 7.7 kg$ ( $7.7 kg$ over); (c) e.g. $Δ V$ cut karo / descope / launcher upsize karo.