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

This page is the drill hall for the parent topic on cryogenic propellants. The parent gave you the three heat-leak formulas and one worked example each. Here we hunt down every kind of scenario the topic can throw at you — every dominant heat path, the degenerate cases (zero conduction, radiation into a warm-side or a deep-space background), the limiting behaviour, a messy real-world word problem, and an exam-style trap.

Throughout, LOX = liquid oxygen (stored near $90\text{K}$) and LH₂ = liquid hydrogen (stored near $20\text{K}$) — the two workhorse cryogens.

Before any numbers, we lock down all four tools you'll reuse, so no symbol is ever unexplained.

Recall The four heat-leak tools (from the parent, plus convection)

Conduction (heat crawling through a solid strut) — Fourier's Law of Heat Conduction: ${\dot{Q}}_{\text{cond}}=kA\frac{\Delta T}{L}$ $k$ = how easily the material passes heat, $A$ = strut cross-section, $\Delta T$ = hot-minus-cold temperature gap, $L$ = length of the path.

Radiation (heat thrown across empty space as infrared light) — 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 carried away by a moving fluid, e.g. air on the pad) — Newton's law of cooling: ${\dot{Q}}_{\text{conv}}=hA(T_{\infty }-T_{\text{surface}})$ $h$ = convective heat-transfer coefficient (W/(m²·K)) — a single number bundling how vigorously the fluid strips heat off the surface (fast wind ⇒ big $h$, still air ⇒ small $h$); $A$ = area touched by the fluid; $T_{\infty }$ = temperature of the fluid far away (the $\infty$ subscript literally means "at infinity", i.e. the undisturbed bulk air, e.g. $295\text{K}$); $T_{\text{surface}}$ = temperature of the tank's outer skin the fluid touches. The leak is again driven by a temperature gap, so a warm-air/cold-skin gap pumps heat in.

Boil-off (heat spent boiling liquid) — Latent Heat and Phase Changes: ${\dot{m}}_{\text{boil-off}}=\frac{\dot{Q}}{L_v}$ $L_v$ = latent heat: joules needed to evaporate one kilogram.

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The scenario matrix

Every boil-off problem lives in one of these cells. The examples below are each tagged with the cell they cover, and together they hit all of them.

Cell	What makes it distinct	Covered by
A. Conduction-dominated	struts are the main leak; radiation shielded away	Ex 1
B. Radiation-dominated	good vacuum, MLI is the story; big $T^4$ gap	Ex 2
C. All three paths added	in-atmosphere pad case, convection alive	Ex 3
D. Zero / degenerate input	$\Delta T=0$, or perfect vacuum kills convection	Ex 4
E. Limiting behaviour	cold side $\ll$ hot side, so $T_{\text{cold}}^4$ vanishes; deep-space background	Ex 5
F. Compare two cryogens	same heat leak, different $L_v$ (LH₂ vs LOX)	Ex 6
G. Real-world word problem	back out allowable strut count from a boil-off budget	Ex 7
H. Exam twist	percent-per-day given, solve backwards for heat leak	Ex 8

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

Statement. A liquid-oxygen (LOX) tank hangs on 6 fibreglass struts. Each 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 is fully shielded (ignore it). Find the heat leak and the LOX boil-off in kg/day. LOX $L_v=213\times 10^3\text{J/kg}$.

Forecast: guess — will fibreglass beat the titanium of the parent's Example 1 (which leaked $5.2\text{W}$)? By how much?

The figure below sets the scene: the orange bar is the hot outer shell, the violet bar the cold cryogen wall, and the brown strut is the only bridge — follow the magenta arrow, which is the heat crawling down that bridge from hot to cold.

1. Temperature gap. $\Delta T=300-90=210\text{K}$ — the length of the double-headed arrow at the bottom of the figure. Why this step? Fourier's law is driven by that gap; kelvin and Celsius give the same difference so no conversion needed.
2. One strut. ${\dot{Q}}_1=kA\frac{\Delta T}{L}=0.04\times 1.5\times 10^{-4}\times \frac{210}{0.40}=3.15\times 10^{-3}\text{W}$. Why this step? Direct Fourier's law for the single brown path in the figure.
3. All six struts. $\dot{Q}=6\times 3.15\times 10^{-3}=1.89\times 10^{-2}\text{W}$. Why this step? Struts sit in parallel — heat can take any of them, so rates add.
4. Boil-off rate. $\dot{m}=\frac{\dot{Q}}{L_v}=\frac{1.89\times 10^{-2}}{213\times 10^3}=8.87\times 10^{-8}\text{kg/s}$. Why this step? Energy balance — every joule leaking in evaporates $1/L_v$ kg.
5. Per day. $8.87\times 10^{-8}\times 86400=7.66\times 10^{-3}\text{kg/day}$ ≈ 7.7 g/day. Why this step? $86400$ seconds in a day converts rate to a daily figure people quote.

Verify: Units: $\frac{\text{W/(m⋅K)}\cdot {\text{m}}^2\cdot \text{K}}{\text{m}}=\text{W}$ ✓. Fibreglass ($k=0.04$) versus titanium ($k=7.4$) is ~185× less conductive — the tiny $0.019\text{W}$ versus the parent's $5.2\text{W}$ is the expected huge drop. Design lesson: low-$k$ struts win.

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

Statement. A liquid-hydrogen (LH₂) tank, surface $A=30{\text{m}}^2$, sits in vacuum. Outer wall $290\text{K}$, tank skin $20\text{K}$. Bare emissivity $\epsilon =0.6$; with 40-layer MLI, ${\epsilon }_{\text{eff}}=0.005$. Find radiation heat leak both ways and the reduction factor.

Forecast: the cold side is $20\text{K}$ — will $20^4$ matter at all next to $290^4$?

The figure plots $T^4$ (on a log scale, so both bars fit) for the hot and cold surfaces. Notice how the violet cold bar is a sliver next to the orange hot bar — the magenta arrow marks it as negligible. That single picture is the reason radiation obsesses over the hot side.

1. Fourth powers. $290^4=7.07\times 10^9$, $20^4=1.6\times 10^5$ — the heights of the two bars in the figure. Why this step? Stefan-Boltzmann needs $T^4$, not $T$; that's why hot surfaces dominate so violently.
2. The gap. $290^4-20^4=7.07\times 10^9-1.6\times 10^5\approx 7.07\times 10^9{\text{K}}^4$. Why this step? $20^4$ is ~44,000× smaller — the tiny violet bar — so it's noise. (We revisit this rigorously in Ex 5.)
3. Bare leak. ${\dot{Q}}_{\text{bare}}=0.6\times 5.67\times 10^{-8}\times 30\times 7.07\times 10^9=7.21\times 10^3\text{W}$. Why this step? Full Stefan-Boltzmann with the bare emissivity.
4. MLI leak. ${\dot{Q}}_{\text{MLI}}=0.005\times 5.67\times 10^{-8}\times 30\times 7.07\times 10^9=60.1\text{W}$. Why this step? MLI only changes $\epsilon$; the $T^4$ physics is unchanged.
5. Reduction. $7210/60.1=120\times$. Why this step? Ratio of emissivities: $0.6/0.005=120$ — because everything else cancels.

Verify: The reduction factor is ${\epsilon }_{\text{bare}}/{\epsilon }_{\text{MLI}}=0.6/0.005=120$, matching ✓. Units of Stefan-Boltzmann: $\frac{1}{{\text{m}}^2{\text{K}}^4}\cdot {\text{m}}^2\cdot {\text{K}}^4=\text{W}$ (dimensionless $\epsilon$) ✓. Lesson: radiation is the giant in vacuum — MLI is non-negotiable.

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[!example] Ex 3 — Cell C: All three paths, on the launch pad

Statement. A chilled LOX (liquid-oxygen) tank sits on the pad in air before evacuation. 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 and LOX boil-off per hour? $L_v=213\times 10^3\text{J/kg}$.

Forecast: on the pad, which term dwarfs the others — will convection swamp the rest?

1. Convection. ${\dot{Q}}_{\text{conv}}=hA(T_{\infty }-T_{\text{surface}})=8\times 25\times (295-260)=7000\text{W}$. Why this step? Newton's law of cooling (locked down in the tools block above): $h$ is the coefficient bundling how hard the air strips heat, and $T_{\infty }-T_{\text{surface}}$ is the warm-air-minus-cold-skin gap that drives it. Air blowing on cold metal is a firehose of heat.
2. Total. $\dot{Q}=12+40+7000=7052\text{W}$. Why this step? The three paths are independent, so the powers simply add.
3. Boil-off rate. $\dot{m}=7052/213\times 10^3=3.31\times 10^{-2}\text{kg/s}$. Why this step? Same energy balance as always.
4. Per hour. $3.31\times 10^{-2}\times 3600=119\text{kg/hr}$. Why this step? $3600$ s/hr; the pad phase is short, so hourly is the useful unit.

Verify: Convection is $7000/7052=99.3\%$ of the leak — exactly why the parent says evacuate the gap: killing convection removes ~99% of pad heat. Units of $h$: ${\text{W/(m}}^2\text{K})\times {\text{m}}^2\times \text{K}=\text{W}$ ✓.

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

Statement. Two sanity checks on a LOX (liquid-oxygen) tank. (a) The outer shell momentarily equilibrates to the same temperature as the cryogen: $\Delta T=0$. (b) The gap is pumped to hard vacuum, so $h\to 0$, but the shell is back at $300\text{K}$ and the cryogen at $90\text{K}$. For (b), take conduction $kA/L=5\times 10^{-4}\text{W/K}$ (a lumped value) and radiation $\epsilon =0.02$, $A=20{\text{m}}^2$. What survives?

Forecast: does boil-off truly stop, or does killing convection leave the other two paths alive?

1. (a) Conduction with $\Delta T=0$. ${\dot{Q}}_{\text{cond}}=kA\frac{0}{L}=0$. Why this step? No temperature gap ⇒ no driving force ⇒ no flow. Multiplying by zero is the whole story.
2. (a) Radiation with $T_{\text{hot}}=T_{\text{cold}}$. $\epsilon \sigma A(T^4-T^4)=0$. Why this step? Both surfaces throw the same infrared at each other; net exchange is zero. Radiation only cares about the difference of fourth powers.
3. (b) Convection with $h\to 0$. ${\dot{Q}}_{\text{conv}}=0\cdot A\Delta T=0$. Why this step? No fluid to carry heat ⇒ the coefficient collapses to zero. This is the point of the vacuum jacket (Vacuum Technology).
4. (b) Surviving conduction. ${\dot{Q}}_{\text{cond}}=(kA/L)\Delta T=5\times 10^{-4}\times (300-90)=0.105\text{W}$. Why this step? $\Delta T=210\text{K}$ is not zero here — the struts still bridge hot to cold, so conduction stays alive even in vacuum.
5. (b) Surviving radiation. ${\dot{Q}}_{\text{rad}}=0.02\times 5.67\times 10^{-8}\times 20\times (300^4-90^4)=0.182\text{W}$. Why this step? $300^4-90^4=8.033\times 10^9{\text{K}}^4$ (the cold term is tiny); infrared crosses vacuum freely, so radiation persists.
6. (b) Total surviving leak. $\dot{Q}=0+0.105+0.182=0.287\text{W}$. Why this step? Only convection died; the two remaining paths still add up and still boil propellant.

Verify: Each path multiplies its driving quantity, so a zero driving term (case a) kills that path exactly ⇒ $\dot{m}=0/L_v=0$: boil-off truly stops. But killing only $h$ (case b) leaves $0.287\text{W}$ ⇒ $\dot{m}=0.287/213000=1.35\times 10^{-6}\text{kg/s}$: still boiling. Lesson — vacuum removes convection, not conduction or radiation; those need low-$k$ struts and MLI.

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

Statement. In orbit a LH₂ (liquid-hydrogen) tank radiates from a sunlit shell $T_{\text{hot}}=350\text{K}$ toward a cold tank skin $T_{\text{cold}}=20\text{K}$, $\epsilon =0.03$, $A=40{\text{m}}^2$. Compute the leak keeping $20^4$, then compute it dropping $20^4$. How large is the error from the approximation?

Forecast: guess the percentage error before computing.

1. Full expression. $350^4=1.5006\times 10^{10}$; $20^4=1.6\times 10^5$. Why this step? We want to see the exact numbers before approximating.
2. Full leak. ${\dot{Q}}_{\text{full}}=0.03\times 5.67\times 10^{-8}\times 40\times (1.5006\times 10^{10}-1.6\times 10^5)=1021.4\text{W}$. Why this step? Honest Stefan-Boltzmann with both terms.
3. Approx leak (drop cold term). ${\dot{Q}}_{\text{approx}}=0.03\times 5.67\times 10^{-8}\times 40\times 1.5006\times 10^{10}=1021.4\text{W}$. Why this step? Tests the "cold side is negligible" claim used in Ex 2.
4. Error. $\frac{1.6\times 10^5}{1.5006\times 10^{10}}=1.07\times 10^{-5}=0.0011\%$. Why this step? The relative error is just the ratio of the fourth powers.

Verify: The limiting rule "when $T_{\text{cold}}\ll T_{\text{hot}}$, drop $T_{\text{cold}}^4$" costs about one part in $10^5$ here — utterly safe. This is why deep-space radiators are designed on $T_{\text{hot}}^4$ alone. (If both temperatures were close, Ex 4a shows the difference matters completely — never blindly drop it.)

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

Statement. Identical tanks each absorb $\dot{Q}=30\text{W}$. One holds LH₂ (liquid hydrogen, $L_v=445\times 10^3\text{J/kg}$), the other LOX (liquid oxygen, $L_v=213\times 10^3\text{J/kg}$). Which boils faster in mass, and by what ratio?

Forecast: LH₂ has the bigger $L_v$ — does that make it boil slower?

1. LH₂ rate. ${\dot{m}}_{{\text{LH}}_2}=30/445\times 10^3=6.74\times 10^{-5}\text{kg/s}$. Why this step? $L_v$ in the denominator — high latent heat resists boiling.
2. LOX rate. ${\dot{m}}_{\text{LOX}}=30/213\times 10^3=1.408\times 10^{-4}\text{kg/s}$. Why this step? Lower $L_v$ ⇒ each joule evaporates more mass.
3. Ratio. ${\dot{m}}_{\text{LOX}}/{\dot{m}}_{{\text{LH}}_2}=445/213=2.089$. Why this step? $\dot{Q}$ cancels; the mass ratio is exactly the inverse ratio of latent heats.

Verify: For the same heat input, LOX boils ~$2.09\times$ faster by mass — the parent's caveat still holds: in real tanks LH₂ leaks more because its far colder skin means a bigger $\Delta T$ and $T_{\text{hot}}^4-T_{\text{cold}}^4$, overpowering its higher $L_v$. Relate leftover propellant to Propellant Mass Fraction.

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

Statement. Mission spec: LH₂ (liquid-hydrogen) boil-off from conduction must stay under 0.20 kg/day. Each titanium strut leaks ${\dot{Q}}_1=1.30\text{W}$ (the parent's Example 1 value). $L_v=445\times 10^3\text{J/kg}$. What is the maximum number of struts allowed?

Forecast: the parent used 4 struts (1.01 kg/day). Will we be forced well below 4?

1. Allowed rate in kg/s. ${\dot{m}}_{\max }=\frac{0.20}{86400}=2.315\times 10^{-6}\text{kg/s}$. Why this step? Convert the daily budget to SI so it matches watts and joules.
2. Allowed heat leak. ${\dot{Q}}_{\max }={\dot{m}}_{\max }{L}_v=2.315\times 10^{-6}\times 445\times 10^3=1.030\text{W}$. Why this step? Invert the boil-off formula — solve the energy balance for $\dot{Q}$.
3. Strut count. $N={\dot{Q}}_{\max }/{\dot{Q}}_1=1.030/1.30=0.79$. Why this step? Struts add in parallel, so divide the total budget by one strut.
4. Round down. $N=0$ struts of this design satisfy the budget — you must redesign (fibreglass from Ex 1, or thinner/longer Ti). Why this step? You cannot exceed the budget, so round down; a fractional strut is impossible.

Verify: With even one Ti strut, boil-off $=1.30/445000\times 86400=0.252\text{kg/day}>0.20$ ✓ (over budget) — confirming zero allowed. This is a genuine design pressure driving the composite-strut choice; ties to Structural Design - Pressure Vessels.

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

Statement. A tank of $m_{\text{total}}=18,000\text{kg}$ LOX (liquid oxygen) is spec'd at 0.15 %/day boil-off. Work backwards: what total heat leak $\dot{Q}$ does this imply? $L_v=213\times 10^3\text{J/kg}$.

Forecast: the exam gives the answer (percent) and hides the heat — reverse every arrow. Guess: is the leak tens of watts or hundreds?

1. Percent to mass/day. ${\dot{m}}_{\text{day}}=0.0015\times 18000=27\text{kg/day}$. Why this step? $0.15\%=0.0015$; apply it to the total to get daily mass lost.
2. To kg/s. $\dot{m}=27/86400=3.125\times 10^{-4}\text{kg/s}$. Why this step? Watts need seconds, not days.
3. Mass rate to heat. $\dot{Q}=\dot{m}{L}_v=3.125\times 10^{-4}\times 213\times 10^3=66.6\text{W}$. Why this step? Rearranged energy balance $\dot{Q}=\dot{m}{L}_v$ — the boil-off formula run in reverse.

Verify: Forward-check: $66.6/213000\times 86400/18000\times 100=0.15\%$/day ✓, the number we were handed. Reversibility of $\dot{m}=\dot{Q}/L_v$ is the whole trick — know it both directions.

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Recall Self-test

A strut's $\Delta T$ doubles. By what factor does its conduction leak change? ::: Exactly 2× — ${\dot{Q}}_{\text{cond}}$ is linear in $\Delta T$. A radiating surface's hot temperature doubles (cold side negligible). Leak changes by? ::: $2^4=16\times$ — Stefan-Boltzmann is quartic. In Newton's law of cooling, what does the $\infty$ in $T_{\infty }$ mean? ::: The bulk fluid temperature far from the surface (the undisturbed air), paired with the skin temperature $T_{\text{surface}}$. Same $\dot{Q}$, swap LOX for LH₂. Mass boil-off changes by? ::: $\times \frac{213}{445}\approx 0.48$ — LH₂ boils about half as fast for equal heat. Killing only convection in vacuum — does boil-off stop? ::: No — conduction and radiation survive (Ex 4b); only their driving $\Delta T=0$ would stop them.