Worked examples — Cryogenic propellants — handling, insulation, boil-off

3.3.49 · D3 · Physics › Rocket Propulsion › Cryogenic propellants — handling, insulation, boil-off

Yeh page parent topic on cryogenic propellants ka drill hall hai. Parent ne teen heat-leak formulas diye the aur ek-ek worked example. Yahaan hum har tarah ke scenario dhundhte hain jo yeh topic throw kar sakta hai — har dominant heat path, degenerate cases (zero conduction, radiation into a warm-side ya deep-space background), limiting behaviour, ek messy real-world word problem, aur ek exam-style trap.

Poore note mein, LOX = liquid oxygen ($90\text{K}$ ke paas store hoti hai) aur LH₂ = liquid hydrogen ($20\text{K}$ ke paas store hoti hai) — yeh do workhorse cryogens hain.

Koi bhi number se pehle, hum charon tools lock down karte hain jo baar baar reuse honge, taaki koi symbol kabhi unexplained na rahe.

Recall Charon heat-leak tools (parent se, plus convection)

Conduction (heat ek solid strut ke through crawl karti hai) — Fourier's Law of Heat Conduction: ${\dot{Q}}_{\text{cond}}=kA\frac{\Delta T}{L}$ $k$ = material heat kitni aasaani se pass karta hai, $A$ = strut cross-section, $\Delta T$ = hot-minus-cold temperature gap, $L$ = path ki length.

Radiation (heat infrared light ke roop mein empty space cross karti hai) — Stefan-Boltzmann Law: ${\dot{Q}}_{\text{rad}}=\epsilon \sigma A(T_{\text{hot}}^4-T_{\text{cold}}^4)$ $\epsilon$ = emissivity (0 = mirror, 1 = perfect black), $\sigma =5.67\times 10^{-8}{\text{W/(m}}^2{\text{K}}^4)$.

Convection (heat ek moving fluid se carry hoti hai, jaise pad pe air) — Newton's law of cooling: ${\dot{Q}}_{\text{conv}}=hA(T_{\infty }-T_{\text{surface}})$ $h$ = convective heat-transfer coefficient (W/(m²·K)) — ek single number jo bundle karta hai ki fluid surface se heat kitni vigorously strip karta hai (tez hawa ⇒ bada $h$, still air ⇒ chota $h$); $A$ = fluid se touched area; $T_{\infty }$ = fluid ka temperature door se ($\infty$ subscript literally matlab hai "at infinity", yaani undisturbed bulk air, jaise $295\text{K}$); $T_{\text{surface}}$ = tank ki outer skin ka temperature jise fluid touch karta hai. Leak phir bhi temperature gap se drive hoti hai, toh warm-air/cold-skin gap heat andar pump karta hai.

Boil-off (heat liquid boil karne mein lagti hai) — Latent Heat and Phase Changes: ${\dot{m}}_{\text{boil-off}}=\frac{\dot{Q}}{L_v}$ $L_v$ = latent heat: ek kilogram evaporate karne ke liye kitne joules chahiye.

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Scenario matrix

Har boil-off problem in cells mein se kisi ek mein rehta hai. Neeche ke examples mein se har ek un cell ke saath tagged hai jo woh cover karta hai, aur milke woh sab cover karte hain.

Cell	Kya distinctive hai	Covered by
A. Conduction-dominated	struts main leak hain; radiation shield ho gayi	Ex 1
B. Radiation-dominated	achha vacuum, MLI ki kahani; bada $T^4$ gap	Ex 2
C. Teeno paths add	in-atmosphere pad case, convection alive	Ex 3
D. Zero / degenerate input	$\Delta T=0$, ya perfect vacuum convection khatam karta hai	Ex 4
E. Limiting behaviour	cold side $\ll$ hot side, toh $T_{\text{cold}}^4$ vanish; deep-space background	Ex 5
F. Do cryogens compare	same heat leak, alag $L_v$ (LH₂ vs LOX)	Ex 6
G. Real-world word problem	boil-off budget se allowable strut count nikalo	Ex 7
H. Exam twist	percent-per-day diya, ulta solve karo heat leak ke liye	Ex 8

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[!example] Ex 1 — Cell A: Conduction-dominated (fibreglass struts)

Statement. Ek liquid-oxygen (LOX) tank 6 fibreglass struts pe hang hai. Har strut: cross-section $A=1.5\times 10^{-4}{\text{m}}^2$, length $L=0.40\text{m}$, conductivity $k=0.04\text{W/(m⋅K)}$. Outer shell $300\text{K}$, LOX $90\text{K}$. Radiation fully shielded hai (ignore karo). Heat leak aur LOX boil-off kg/day mein nikalo. LOX $L_v=213\times 10^3\text{J/kg}$.

Forecast: andazo lagao — kya fibreglass parent ke Example 1 ke titanium ko beat karega (jisne $5.2\text{W}$ leak kiya tha)? Kitna?

Neeche ki figure scene set karti hai: orange bar hot outer shell hai, violet bar cold cryogen wall hai, aur brown strut akela bridge hai — magenta arrow follow karo, jo heat hai jo us bridge se hot se cold ki taraf crawl kar rahi hai.

1. Temperature gap. $\Delta T=300-90=210\text{K}$ — figure ke bottom mein double-headed arrow ki length. Yeh step kyun? Fourier's law us gap se drive hoti hai; kelvin aur Celsius same difference dete hain toh conversion ki zaroorat nahi.
2. Ek strut. ${\dot{Q}}_1=kA\frac{\Delta T}{L}=0.04\times 1.5\times 10^{-4}\times \frac{210}{0.40}=3.15\times 10^{-3}\text{W}$. Yeh step kyun? Figure mein single brown path ke liye direct Fourier's law.
3. Charon struts. $\dot{Q}=6\times 3.15\times 10^{-3}=1.89\times 10^{-2}\text{W}$. Yeh step kyun? Struts parallel mein hain — heat unme se kisi bhi ek ko le sakti hai, toh rates add hote hain.
4. Boil-off rate. $\dot{m}=\frac{\dot{Q}}{L_v}=\frac{1.89\times 10^{-2}}{213\times 10^3}=8.87\times 10^{-8}\text{kg/s}$. Yeh step kyun? Energy balance — har joule jo leak hoti hai $1/L_v$ kg evaporate karta hai.
5. Per day. $8.87\times 10^{-8}\times 86400=7.66\times 10^{-3}\text{kg/day}$ ≈ 7.7 g/day. Yeh step kyun? Ek din mein $86400$ seconds hain — rate ko daily figure mein convert karta hai jo log quote karte hain.

Verify: Units: $\frac{\text{W/(m⋅K)}\cdot {\text{m}}^2\cdot \text{K}}{\text{m}}=\text{W}$ ✓. Fibreglass ($k=0.04$) versus titanium ($k=7.4$) ~185× kam conductive hai — tiny $0.019\text{W}$ versus parent ka $5.2\text{W}$ expected bada drop hai. Design lesson: low-$k$ struts jeetते hain.

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[!example] Ex 2 — Cell B: Radiation-dominated (bare vs MLI)

Statement. Ek liquid-hydrogen (LH₂) tank, surface $A=30{\text{m}}^2$, vacuum mein baitha hai. Outer wall $290\text{K}$, tank skin $20\text{K}$. Bare emissivity $\epsilon =0.6$; 40-layer MLI ke saath, ${\epsilon }_{\text{eff}}=0.005$. Dono taraf radiation heat leak aur reduction factor nikalo.

Forecast: cold side $20\text{K}$ hai — kya $20^4$ actually $290^4$ ke aage kuch matter karega?

Figure $T^4$ plot karta hai (log scale pe, taaki dono bars fit hon) hot aur cold surfaces ke liye. Notice karo ki violet cold bar kitna sliver hai orange hot bar ke next — magenta arrow isko negligible mark karta hai. Woh ek picture hi reason hai ki radiation hot side pe obsess karta hai.

1. Fourth powers. $290^4=7.07\times 10^9$, $20^4=1.6\times 10^5$ — figure mein dono bars ki heights. Yeh step kyun? Stefan-Boltzmann ko $T^4$ chahiye, $T$ nahi; isliye hot surfaces itni violently dominate karte hain.
2. Gap. $290^4-20^4=7.07\times 10^9-1.6\times 10^5\approx 7.07\times 10^9{\text{K}}^4$. Yeh step kyun? $20^4$ ~44,000× chota hai — tiny violet bar — toh yeh noise hai. (Hum isko Ex 5 mein rigorously revisit karte hain.)
3. Bare leak. ${\dot{Q}}_{\text{bare}}=0.6\times 5.67\times 10^{-8}\times 30\times 7.07\times 10^9=7.21\times 10^3\text{W}$. Yeh step kyun? Full Stefan-Boltzmann bare emissivity ke saath.
4. MLI leak. ${\dot{Q}}_{\text{MLI}}=0.005\times 5.67\times 10^{-8}\times 30\times 7.07\times 10^9=60.1\text{W}$. Yeh step kyun? MLI sirf $\epsilon$ change karta hai; $T^4$ physics unchanged hai.
5. Reduction. $7210/60.1=120\times$. Yeh step kyun? Emissivities ka ratio: $0.6/0.005=120$ — kyunki baaki sab cancel ho jaata hai.

Verify: Reduction factor hi ${\epsilon }_{\text{bare}}/{\epsilon }_{\text{MLI}}=0.6/0.005=120$ hai, match karta hai ✓. Stefan-Boltzmann ke units: $\frac{1}{{\text{m}}^2{\text{K}}^4}\cdot {\text{m}}^2\cdot {\text{K}}^4=\text{W}$ (dimensionless $\epsilon$) ✓. Lesson: vacuum mein radiation giant hai — MLI non-negotiable hai.

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[!example] Ex 3 — Cell C: Teeno paths, launch pad pe

Statement. Ek chilled LOX (liquid-oxygen) tank evacuation se pehle pad pe air mein baitha hai. Conduction ${\dot{Q}}_{\text{cond}}=12\text{W}$; radiation ${\dot{Q}}_{\text{rad}}=40\text{W}$. Convection: $h=8{\text{W/(m}}^2\text{K})$, wetted area $A=25{\text{m}}^2$, ambient $T_{\infty }=295\text{K}$, cold outer skin $T_{\text{surface}}=260\text{K}$. Total heat leak aur LOX boil-off per hour? $L_v=213\times 10^3\text{J/kg}$.

Forecast: pad pe, kaun sa term dusron ko dwarf karta hai — kya convection baaki sab ko swamp kar dega?

1. Convection. ${\dot{Q}}_{\text{conv}}=hA(T_{\infty }-T_{\text{surface}})=8\times 25\times (295-260)=7000\text{W}$. Yeh step kyun? Newton's law of cooling (upar tools block mein lock down kiya): $h$ woh coefficient hai jo bundle karta hai ki air heat kitni hard strip karti hai, aur $T_{\infty }-T_{\text{surface}}$ warm-air-minus-cold-skin gap hai jo isko drive karta hai. Cold metal pe blow karna heat ka firehose hai.
2. Total. $\dot{Q}=12+40+7000=7052\text{W}$. Yeh step kyun? Teeno paths independent hain, toh powers simply add hote hain.
3. Boil-off rate. $\dot{m}=7052/213\times 10^3=3.31\times 10^{-2}\text{kg/s}$. Yeh step kyun? Same energy balance jaise hamesha.
4. Per hour. $3.31\times 10^{-2}\times 3600=119\text{kg/hr}$. Yeh step kyun? $3600$ s/hr; pad phase short hoti hai, toh hourly useful unit hai.

Verify: Convection $7000/7052=99.3\%$ leak hai — exactly isliye parent kehta hai gap evacuate karo: convection khatam karna ~99% pad heat remove kar deta hai. $h$ ke units: ${\text{W/(m}}^2\text{K})\times {\text{m}}^2\times \text{K}=\text{W}$ ✓.

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[!example] Ex 4 — Cell D: Degenerate inputs

Statement. Ek LOX (liquid-oxygen) tank pe do sanity checks. (a) Outer shell momentarily cryogen ke same temperature pe equilibrate hoti hai: $\Delta T=0$. (b) Gap hard vacuum pe pump ki gayi, toh $h\to 0$, lekin shell wapas $300\text{K}$ pe hai aur cryogen $90\text{K}$ pe. (b) ke liye, conduction $kA/L=5\times 10^{-4}\text{W/K}$ lo (ek lumped value) aur radiation $\epsilon =0.02$, $A=20{\text{m}}^2$. Kya bacha rehta hai?

Forecast: kya boil-off sach mein rukti hai, ya convection khatam karna baaki do paths alive chhodta hai?

1. (a) Conduction with $\Delta T=0$. ${\dot{Q}}_{\text{cond}}=kA\frac{0}{L}=0$. Yeh step kyun? Koi temperature gap nahi ⇒ koi driving force nahi ⇒ koi flow nahi. Zero se multiply karna hi poori kahani hai.
2. (a) Radiation with $T_{\text{hot}}=T_{\text{cold}}$. $\epsilon \sigma A(T^4-T^4)=0$. Yeh step kyun? Dono surfaces ek dusre pe same infrared throw karte hain; net exchange zero hai. Radiation sirf fourth powers ke difference ki parwah karta hai.
3. (b) Convection with $h\to 0$. ${\dot{Q}}_{\text{conv}}=0\cdot A\Delta T=0$. Yeh step kyun? Heat carry karne ke liye koi fluid nahi ⇒ coefficient zero ho jaata hai. Yahi vacuum jacket ka point hai (Vacuum Technology).
4. (b) Surviving conduction. ${\dot{Q}}_{\text{cond}}=(kA/L)\Delta T=5\times 10^{-4}\times (300-90)=0.105\text{W}$. Yeh step kyun? $\Delta T=210\text{K}$ yahaan zero nahi hai — struts abhi bhi hot se cold bridge karte hain, toh conduction vacuum mein bhi alive rehti hai.
5. (b) Surviving radiation. ${\dot{Q}}_{\text{rad}}=0.02\times 5.67\times 10^{-8}\times 20\times (300^4-90^4)=0.182\text{W}$. Yeh step kyun? $300^4-90^4=8.033\times 10^9{\text{K}}^4$ (cold term tiny hai); infrared vacuum freely cross karta hai, toh radiation persist karti hai.
6. (b) Total surviving leak. $\dot{Q}=0+0.105+0.182=0.287\text{W}$. Yeh step kyun? Sirf convection mari; baaki do paths abhi bhi add hote hain aur propellant boil karte hain.

Verify: Har path apni driving quantity multiply karta hai, toh zero driving term (case a) us path ko exactly maarta hai ⇒ $\dot{m}=0/L_v=0$: boil-off sach mein rukti hai. Lekin sirf $h$ kill karna (case b) $0.287\text{W}$ chhod deta hai ⇒ $\dot{m}=0.287/213000=1.35\times 10^{-6}\text{kg/s}$: abhi bhi boil ho raha hai. Lesson — vacuum convection remove karta hai, conduction ya radiation nahi; unhe low-$k$ struts aur MLI chahiye.

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[!example] Ex 5 — Cell E: Limiting behaviour (deep space)

Statement. Orbit mein ek LH₂ (liquid-hydrogen) tank ek sunlit shell $T_{\text{hot}}=350\text{K}$ se ek cold tank skin $T_{\text{cold}}=20\text{K}$ ki taraf radiate karta hai, $\epsilon =0.03$, $A=40{\text{m}}^2$. Leak compute karo $20^4$ rakhke, phir $20^4$ drop karke compute karo. Approximation se error kitni badi hai?

Forecast: compute se pehle percentage error guess karo.

1. Full expression. $350^4=1.5006\times 10^{10}$; $20^4=1.6\times 10^5$. Yeh step kyun? Approximate karne se pehle exact numbers dekhne chahte hain.
2. Full leak. ${\dot{Q}}_{\text{full}}=0.03\times 5.67\times 10^{-8}\times 40\times (1.5006\times 10^{10}-1.6\times 10^5)=1021.4\text{W}$. Yeh step kyun? Honest Stefan-Boltzmann dono terms ke saath.
3. Approx leak (cold term drop karo). ${\dot{Q}}_{\text{approx}}=0.03\times 5.67\times 10^{-8}\times 40\times 1.5006\times 10^{10}=1021.4\text{W}$. Yeh step kyun? Ex 2 mein use kiya "cold side negligible hai" claim test karta hai.
4. Error. $\frac{1.6\times 10^5}{1.5006\times 10^{10}}=1.07\times 10^{-5}=0.0011\%$. Yeh step kyun? Relative error sirf fourth powers ka ratio hai.

Verify: Limiting rule "jab $T_{\text{cold}}\ll T_{\text{hot}}$, toh $T_{\text{cold}}^4$ drop karo" yahaan $10^5$ mein ek part kharcha karta hai — bilkul safe. Isliye deep-space radiators $T_{\text{hot}}^4$ akele pe design hote hain. (Agar dono temperatures close hote, Ex 4a dikhata hai ki difference completely matter karta hai — kabhi blindly drop mat karo.)

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[!example] Ex 6 — Cell F: Same heat, do cryogens (LH₂ vs LOX)

Statement. Identical tanks mein se har ek $\dot{Q}=30\text{W}$ absorb karta hai. Ek mein LH₂ (liquid hydrogen, $L_v=445\times 10^3\text{J/kg}$) hai, doosre mein LOX (liquid oxygen, $L_v=213\times 10^3\text{J/kg}$). Mass mein kaun tezi se boil hota hai, aur kitne ratio mein?

Forecast: LH₂ ka $L_v$ bada hai — kya woh slower boil hoga?

1. LH₂ rate. ${\dot{m}}_{{\text{LH}}_2}=30/445\times 10^3=6.74\times 10^{-5}\text{kg/s}$. Yeh step kyun? $L_v$ denominator mein — high latent heat boiling resist karta hai.
2. LOX rate. ${\dot{m}}_{\text{LOX}}=30/213\times 10^3=1.408\times 10^{-4}\text{kg/s}$. Yeh step kyun? Kam $L_v$ ⇒ har joule zyada mass evaporate karta hai.
3. Ratio. ${\dot{m}}_{\text{LOX}}/{\dot{m}}_{{\text{LH}}_2}=445/213=2.089$. Yeh step kyun? $\dot{Q}$ cancel ho jaata hai; mass ratio exactly latent heats ka inverse ratio hai.

Verify: Same heat input ke liye, LOX by mass ~$2.09\times$ tezi se boil hoti hai — parent ki caveat abhi bhi hold karti hai: real tanks mein LH₂ zyada leak karta hai kyunki uski bahut thandi skin ka matlab hai bada $\Delta T$ aur $T_{\text{hot}}^4-T_{\text{cold}}^4$, jo uski high $L_v$ ko overpower kar deta hai. Bacha hua propellant Propellant Mass Fraction se relate karo.

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[!example] Ex 7 — Cell G: Real-world word problem (strut budget)

Statement. Mission spec: conduction se LH₂ (liquid-hydrogen) boil-off 0.20 kg/day se kam rehni chahiye. Har titanium strut ${\dot{Q}}_1=1.30\text{W}$ leak karta hai (parent ka Example 1 value). $L_v=445\times 10^3\text{J/kg}$. Maximum kitne struts allowed hain?

Forecast: parent ne 4 struts use kiye the (1.01 kg/day). Kya hum 4 se kaafi neeche force honge?

1. Allowed rate in kg/s. ${\dot{m}}_{\max }=\frac{0.20}{86400}=2.315\times 10^{-6}\text{kg/s}$. Yeh step kyun? Daily budget ko SI mein convert karo taaki watts aur joules se match kare.
2. Allowed heat leak. ${\dot{Q}}_{\max }={\dot{m}}_{\max }{L}_v=2.315\times 10^{-6}\times 445\times 10^3=1.030\text{W}$. Yeh step kyun? Boil-off formula ulta — energy balance ko $\dot{Q}$ ke liye solve karo.
3. Strut count. $N={\dot{Q}}_{\max }/{\dot{Q}}_1=1.030/1.30=0.79$. Yeh step kyun? Struts parallel mein add hote hain, toh total budget ko ek strut se divide karo.
4. Round down. $N=0$ struts is design ke budget satisfy karte hain — redesign karna padega (Ex 1 ka fibreglass, ya thinner/longer Ti). Yeh step kyun? Budget exceed nahi kar sakte, toh neeche round karo; fractional strut impossible hai.

Verify: Ek bhi Ti strut se, boil-off $=1.30/445000\times 86400=0.252\text{kg/day}>0.20$ ✓ (over budget) — zero allowed confirm karta hai. Yeh genuine design pressure hai jo composite-strut choice drive karta hai; Structural Design - Pressure Vessels se jodta hai.

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[!example] Ex 8 — Cell H: Exam twist (percent-per-day → heat leak)

Statement. $m_{\text{total}}=18,000\text{kg}$ LOX (liquid oxygen) ka ek tank 0.15 %/day boil-off pe spec hai. Ulta kaam karo: iska matlab kya total heat leak $\dot{Q}$ hai? $L_v=213\times 10^3\text{J/kg}$.

Forecast: exam answer (percent) deta hai aur heat chhupa deta hai — har arrow reverse karo. Guess karo: kya leak tens of watts hai ya hundreds?

1. Percent to mass/day. ${\dot{m}}_{\text{day}}=0.0015\times 18000=27\text{kg/day}$. Yeh step kyun? $0.15\%=0.0015$; daily mass lost nikalne ke liye total pe apply karo.
2. To kg/s. $\dot{m}=27/86400=3.125\times 10^{-4}\text{kg/s}$. Yeh step kyun? Watts ko seconds chahiye, days nahi.
3. Mass rate to heat. $\dot{Q}=\dot{m}{L}_v=3.125\times 10^{-4}\times 213\times 10^3=66.6\text{W}$. Yeh step kyun? Rearranged energy balance $\dot{Q}=\dot{m}{L}_v$ — boil-off formula ulta run kiya.

Verify: Forward-check: $66.6/213000\times 86400/18000\times 100=0.15\%$/day ✓, wahi number jo hamein diya gaya tha. $\dot{m}=\dot{Q}/L_v$ ki reversibility hi poora trick hai — dono directions mein jaano.

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Recall Self-test

Ek strut ka $\Delta T$ double ho jaata hai. Uski conduction leak kitne factor se change hogi? ::: Exactly 2× — ${\dot{Q}}_{\text{cond}}$ $\Delta T$ mein linear hai. Ek radiating surface ka hot temperature double ho jaata hai (cold side negligible). Leak kitne se change hogi? ::: $2^4=16\times$ — Stefan-Boltzmann quartic hai. Newton's law of cooling mein, $T_{\infty }$ mein $\infty$ ka kya matlab hai? ::: Bulk fluid temperature door se surface se (undisturbed air), jo skin temperature $T_{\text{surface}}$ ke saath pair hoti hai. Same $\dot{Q}$, LOX ki jagah LH₂ lagao. Mass boil-off kitna change hoga? ::: $\times \frac{213}{445}\approx 0.48$ — LH₂ equal heat ke liye almost aadha utna boil hota hai. Vacuum mein sirf convection kill karna — kya boil-off rukti hai? ::: Nahi — conduction aur radiation survive karte hain (Ex 4b); sirf unka driving $\Delta T=0$ unhe rokta.