Exercises — Cryogenic propellants — handling, insulation, boil-off

Hum kya karenge: watts ko mass rate mein convert karo, phir din ke seconds se multiply karo. $\dot{m}=\frac{\dot{Q}}{L_v}=\frac{8.9}{213\times 10^3}=4.18\times 10^{-5}\text{kg/s}$ Kyun: har jo joule aati hai woh yahan sirf ek kaam kar sakti hai — liquid boil karna — aur $L_v$ joules-per-kilogram ki price tag hai. Ek din mein: $m_{\text{day}}=4.18\times 10^{-5}\times 86400=3.61\text{kg/day}$

Pehle reasoning: LH₂ ka higher $L_v$ hai ($445$ vs $213$ kJ/kg). Zyada price-per-kilogram matlab hai ki wahi watts kam kilograms boil kar paate hain. Toh LH₂ same heat ke liye kam mass boil off karta hai. $\dot{m}=\frac{8.9}{445\times 10^3}=2.00\times 10^{-5}\text{kg/s}$ $m_{\text{day}}=2.00\times 10^{-5}\times 86400=1.73\text{kg/day}$ LOX ke $3.61$ kg/day se wakai kam hai. ✔

Step 1 — rod ke cross-section ka area. Heat rod ke along flow hoti hai, toh relevant area woh circle hai jo woh present karta hai, radius $r=0.75\text{cm}=0.0075\text{m}$. $A=\pi r^2=\pi (0.0075{)}^2=1.767\times 10^{-4}{\text{m}}^2$ Step 2 — temperature difference (woh push jo heat ko rod ke neeche drive karta hai): $\Delta T=300-20=280\text{K}$ Step 3 — Fourier's law. Figure dekho: length par temperature drop jitna steep hoga, heat utni hi tezi se andar aayegi. $\dot{Q}=kA\frac{\Delta T}{L}=15\times 1.767\times 10^{-4}\times \frac{280}{0.40}$ $=15\times 1.767\times 10^{-4}\times 700=1.855\text{W}$ Toh ek steel strut lagbhag $1.86\text{W}$ leak karta hai.

Step 1 — Stefan-Boltzmann, net form. Dono surfaces glow karti hain; net flow hot⁴ minus cold⁴ hai. $\dot{Q}=\epsilon \sigma A(T_{\text{hot}}^4-T_{\text{cold}}^4)$ Step 2 — fourth powers. Isi liye $T^4$ itna matter karta hai: yeh cold side ko almost irrelevant bana deta hai. $300^4=8.10\times 10^9{\text{K}}^4,80^4=4.10\times 10^7{\text{K}}^4$ Notice karo $80^4$ lagbhag $200\times$ chhota hai — almost negligible, lekin rakhna. $T_{\text{hot}}^4-T_{\text{cold}}^4=8.10\times 10^9-4.10\times 10^7=8.059\times 10^9{\text{K}}^4$ Step 3 — assemble karo: $\dot{Q}=0.05\times 5.67\times 10^{-8}\times 12\times 8.059\times 10^9=274\text{W}$

Step 1 — parallel paths add karo. Heat leaks additive hoti hain: tank mein independent doors. ${\dot{Q}}_{\text{cond}}=4\times 1.86=7.44\text{W}$ ${\dot{Q}}_{\text{total}}=7.44+274+0=281.4\text{W}$ Step 2 — dominance. Radiation ($274$ W) conduction ($7.44$ W) se $\sim 37\times$ zyada hai. Radiation dominate karta hai — exactly isi liye MLI (multilayer insulation) yahan strut design se zyada matter karta hai. Step 3 — boil-off. LH₂: $L_v=445\times 10^3\text{J/kg}$. $\dot{m}=\frac{281.4}{445\times 10^3}=6.32\times 10^{-4}\text{kg/s}$ $m_{\text{day}}=6.32\times 10^{-4}\times 86400=54.6\text{kg/day}$

$\text{Boil-off \%}=\frac{m_{\text{day}}}{m_{\text{total}}}\times 100\%=\frac{54.6}{15000}\times 100\%=0.364\%\text{per day}$ Interpretation: roughly $0.36\%$ per day. 30-day orbit loiter par yeh propellant ka $\sim 11\%$ uda deta hai — ek serious mission constraint, aur isi liye long-duration missions insulation par itna obsess karte hain. Yeh lost mass Propellant Mass Fraction ko directly hurt karta hai.

Step 1 — percentage spec ko allowed $\dot{Q}$ mein convert karo. Boil-off chain ke through backwards kaam karo. Allowed mass/day: $m_{\text{day,max}}=0.0010\times 15000=15.0\text{kg/day}$ Allowed mass/second: ${\dot{m}}_{\max }=\frac{15.0}{86400}=1.736\times 10^{-4}\text{kg/s}$ Allowed total watts: ${\dot{Q}}_{\max }={\dot{m}}_{\max }{L}_v=1.736\times 10^{-4}\times 445\times 10^3=77.3\text{W}$ Step 2 — fixed conduction subtract karo radiation budget nikalne ke liye. ${\dot{Q}}_{\text{rad,allowed}}=77.3-7.44=69.8\text{W}$ Step 3 — required emissivity nikalo. Bare $\epsilon =1$ pe $5480$ W milti hai, aur radiation ${\epsilon }_{\text{eff}}$ ke saath linearly scale karti hai: ${\epsilon }_{\text{eff}}=\frac{69.8}{5480}=0.01274$ Step 4 — layers ki number solve karo. ${\epsilon }_{\text{eff}}=\frac{{\epsilon }_0}{N}\Rightarrow N=\frac{{\epsilon }_0}{{\epsilon }_{\text{eff}}}=\frac{0.03}{0.01274}=2.35$ Kyunki layers whole honi chahiye aur hume spec beat karna hai, up round karo $N=3$ layers tak. Sanity check $N=3$ ke saath: ${\epsilon }_{\text{eff}}=0.01$, toh ${\dot{Q}}_{\text{rad}}=0.01\times 5480=54.8$ W, total $=62.2$ W $<77.3$ W. ✔ Spec meet ho gayi.

Step 1 — wall se kitni heat nikalni hai. Yeh sensible heat hai (ek temperature change, koi phase change nahi), toh hum $mc\Delta T$ use karte hain: $Q_{\text{wall}}=mc\Delta T=200\times 900\times (300-20)=5.04\times 10^7\text{J}$ Yeh formula kyun: aluminum phase change nahi kar raha — woh sirf thanda ho raha hai — toh uski energy change hai mass × specific heat × temperature drop. Step 2 — woh heat LH₂ boil karti hai. Wall se nikli heat hydrogen mein jaati hai aur use boil karti hai. Har kilogram LH₂ $L_v$ absorb karta hai jab woh gas mein turn hota hai (constant temperature par ek phase change), toh hum $L_v$ se divide karte hain: $m_{{\text{LH}}_2}=\frac{Q_{\text{wall}}}{L_v}=\frac{5.04\times 10^7}{445\times 10^3}=113.3\text{kg}$ Result: lagbhag $113\text{kg}$ liquid hydrogen sirf empty wall chill karne ke liye boil ho jaata hai. Yeh kyun matter karta hai: $100$ kg se zyada hydrogen ek bhi usable propellant ki drop store hone se pehle uda jaata hai — har cryogenic fill ki ek real, budgeted cost. (Practice mein thanda nikalta vapour bahar jaate hue pre-cooling se kuch recover karta hai; yeh figure worst case hai bina kisi vapour recovery ke.)

Step 1 — har tank mein heat leak. Cold-side fourth power dono mein tiny hai (yahi $T^4$ ki khoobsurti hai: $20^4$ aur $90^4$ dono $300^4=8.10\times 10^9$ se dab jaate hain). $20^4=1.6\times 10^5,90^4=6.56\times 10^7$ LH₂: ${\dot{Q}}_H=0.02\times 5.67\times 10^{-8}\times 30\times (8.10\times 10^9-1.6\times 10^5)=275.5\text{W}$ LOX: ${\dot{Q}}_O=0.02\times 5.67\times 10^{-8}\times 30\times (8.10\times 10^9-6.56\times 10^7)=273.3\text{W}$ Yeh almost identical hain — kyunki warm shell dono par dominate karta hai. Step 2 — boil-off mass rate, har fluid ke apne $L_v$ use karke: ${\dot{m}}_H=\frac{275.5}{445\times 10^3}=6.19\times 10^{-4}\text{kg/s}\Rightarrow 53.5\text{kg/day}$ ${\dot{m}}_O=\frac{273.3}{213\times 10^3}=1.283\times 10^{-3}\text{kg/s}\Rightarrow 110.9\text{kg/day}$ Step 3 — verdict. Chahe heat leaks almost equal hain, LOX per day double se zyada mass lose karta hai ($111$ vs $53$ kg/day). Iski wajah sirf latent heat hai: LOX ka low $L_v$ matlab har watt zyada kilograms boil kar deta hai. Lesson: "colder therefore worse" ek myth hai jab tum dono effects alag kar lo. Yahan warmer fluid (LOX) zyada mass-loser hai, purely apni sasti latent heat ki wajah se.

Step 1 — better insulation se daily propellant saving. Dono boil-off rates compare karo: $\Delta \dot{m}=54.6-15.0=39.6\text{kg/day saved}$ Kyun: har din, improved tank $39.6$ kg hydrogen rakhta hai jo otherwise vent ho jaata. Step 2 — break-even condition. MLI "free" ho jaata hai jab cumulative saved propellant $120\text{kg}$ dry-mass penalty ke barabar ho jaaye: $\Delta \dot{m}\times t_{\text{break-even}}=120\text{kg}$ $t_{\text{break-even}}=\frac{120}{39.6}=3.03\text{days}$ Step 3 — interpretation. Agar mission lagbhag 3 din se zyada loiter kare, toh heavy MLI jeet jaata hai: woh jitni propellant mass bachaata hai woh us dry mass se zyada hai jo usne add ki. 3 din se kam ke quick sub-orbital hop ke liye, woh MLI dead weight hai aur Propellant Mass Fraction ko hurt karta hai. Yahi woh thermal-vs-structural trade hai jo Structural Design - Pressure Vessels se couple karta hai.

Master chain, warm shell se lost kilograms tak ::: $\dot{Q}={\dot{Q}}_{cond}+{\dot{Q}}_{rad}+{\dot{Q}}_{conv}$, phir $\dot{m}=\dot{Q}/L_v$, phir $m_{day}=\dot{m}\times 86400$, phir $\%=m_{day}/m_{total}\times 100$. Higher $L_v$ kyun fixed heat ke liye lower boil-off mass deta hai? ::: Kyunki $L_v$ joules-per-kilogram hai; zyada price per kilogram matlab wahi joules kam boiled kilograms khareed paate hain. Radiation usually conduction ko kyun beat karta hai dominant leak ke roop mein? ::: Warm shell ka $T^4$ term enormous hai, aur small emissivity bhi large area aur $\sigma T_{hot}^4$ se multiply hokar kuch thin struts se zyada ho jaati hai.