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

What we do: convert watts to a mass rate, then multiply by the seconds in a day. $\dot{m}=\frac{\dot{Q}}{L_v}=\frac{8.9}{213\times 10^3}=4.18\times 10^{-5}\text{kg/s}$ Why: each joule that arrives can only do one job here — boil liquid — and $L_v$ is the joules-per-kilogram price tag. Over one day: $m_{\text{day}}=4.18\times 10^{-5}\times 86400=3.61\text{kg/day}$

Reasoning first: LH₂ has a higher $L_v$ ($445$ vs $213$ kJ/kg). A bigger price-per-kilogram means the same watts buy fewer kilograms boiled. So LH₂ boils off less mass for the same heat. $\dot{m}=\frac{8.9}{445\times 10^3}=2.00\times 10^{-5}\text{kg/s}$ $m_{\text{day}}=2.00\times 10^{-5}\times 86400=1.73\text{kg/day}$ Indeed smaller than the $3.61$ kg/day for LOX. ✔

Step 1 — area of the rod's cross-section. Heat flows along the rod, so the relevant area is the circle it presents, 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 (the push driving heat down the rod): $\Delta T=300-20=280\text{K}$ Step 3 — Fourier's law. Look at the figure: the steeper the temperature drop over the length, the faster heat pours through. $\dot{Q}=kA\frac{\Delta T}{L}=15\times 1.767\times 10^{-4}\times \frac{280}{0.40}$ $=15\times 1.767\times 10^{-4}\times 700=1.855\text{W}$ So one steel strut leaks about $1.86\text{W}$.

Step 1 — Stefan-Boltzmann, net form. Both surfaces glow; the net flow is hot⁴ minus cold⁴. $\dot{Q}=\epsilon \sigma A(T_{\text{hot}}^4-T_{\text{cold}}^4)$ Step 2 — the fourth powers. This is why $T^4$ matters so much: it makes the cold side almost irrelevant. $300^4=8.10\times 10^9{\text{K}}^4,80^4=4.10\times 10^7{\text{K}}^4$ Notice $80^4$ is about $200\times$ smaller — nearly negligible, but keep it. $T_{\text{hot}}^4-T_{\text{cold}}^4=8.10\times 10^9-4.10\times 10^7=8.059\times 10^9{\text{K}}^4$ Step 3 — assemble: $\dot{Q}=0.05\times 5.67\times 10^{-8}\times 12\times 8.059\times 10^9=274\text{W}$

Step 1 — add the parallel paths. Heat leaks are additive: independent doors into the tank. ${\dot{Q}}_{\text{cond}}=4\times 1.86=7.44\text{W}$ ${\dot{Q}}_{\text{total}}=7.44+274+0=281.4\text{W}$ Step 2 — dominance. Radiation ($274$ W) is $\sim 37\times$ the conduction ($7.44$ W). Radiation dominates — which is exactly why MLI (multilayer insulation) matters more than strut design here. 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. Over a 30-day loiter in orbit that is $\sim 11\%$ of the propellant gone — a serious mission constraint, and why long-duration missions obsess over insulation. This lost mass hurts the Propellant Mass Fraction directly.

Step 1 — turn the percentage spec into an allowed $\dot{Q}$. Work backwards through the boil-off chain. 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 — subtract the fixed conduction to find the radiation budget. ${\dot{Q}}_{\text{rad,allowed}}=77.3-7.44=69.8\text{W}$ Step 3 — find the required emissivity. Bare $\epsilon =1$ gives $5480$ W, and radiation scales linearly with ${\epsilon }_{\text{eff}}$: ${\epsilon }_{\text{eff}}=\frac{69.8}{5480}=0.01274$ Step 4 — solve for the number of layers. ${\epsilon }_{\text{eff}}=\frac{{\epsilon }_0}{N}\Rightarrow N=\frac{{\epsilon }_0}{{\epsilon }_{\text{eff}}}=\frac{0.03}{0.01274}=2.35$ Since layers are whole and we must beat the spec, round up to $N=3$ layers. Sanity check with $N=3$: ${\epsilon }_{\text{eff}}=0.01$, so ${\dot{Q}}_{\text{rad}}=0.01\times 5480=54.8$ W, total $=62.2$ W $<77.3$ W. ✔ Spec met.

Step 1 — heat that must leave the wall. This is sensible heat (a temperature change, not a phase change), so we use $mc\Delta T$: $Q_{\text{wall}}=mc\Delta T=200\times 900\times (300-20)=5.04\times 10^7\text{J}$ Why this formula: the aluminum is not changing phase — it is simply getting colder — so its energy change is mass × specific heat × temperature drop. Step 2 — that heat boils LH₂. The heat leaving the wall enters the hydrogen and boils it. Each kilogram of LH₂ absorbs $L_v$ as it turns to gas (a phase change at constant temperature), so we divide by $L_v$: $m_{{\text{LH}}_2}=\frac{Q_{\text{wall}}}{L_v}=\frac{5.04\times 10^7}{445\times 10^3}=113.3\text{kg}$ Result: about $113\text{kg}$ of liquid hydrogen boils away just to chill the empty wall. Why this matters: over 100 kg of hydrogen vanishes before a single drop of usable propellant is stored — a real, budgeted cost of every cryogenic fill. (In practice the cold escaping vapour recovers some of this by pre-cooling on its way out; this figure is the worst case with no vapour recovery.)

Step 1 — heat leak into each tank. The cold-side fourth power is tiny either way (that's the beauty of $T^4$: $20^4$ and $90^4$ are both dwarfed by $300^4=8.10\times 10^9$). $20^4=1.6\times 10^5,90^4=6.56\times 10^7$ 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: ${\dot{Q}}_O=0.02\times 5.67\times 10^{-8}\times 30\times (8.10\times 10^9-6.56\times 10^7)=273.3\text{W}$ They are almost identical — because the warm shell dominates both. Step 2 — boil-off mass rate, using each fluid's own $L_v$: ${\dot{m}}_H=\frac{275.5}{445\times 10^3}=6.19\times 10^{-4}\text{kg/s}\Rightarrow 53.5\text{kg/day}$ ${\dot{m}}_O=\frac{273.3}{213\times 10^3}=1.283\times 10^{-3}\text{kg/s}\Rightarrow 110.9\text{kg/day}$ Step 3 — verdict. Even though the heat leaks are nearly equal, LOX loses more than twice the mass per day ($111$ vs $53$ kg/day). The reason is entirely the latent heat: LOX's low $L_v$ means each watt boils away more kilograms. The lesson: "colder therefore worse" is a myth once you split the two effects. Here the warmer fluid (LOX) is the bigger mass-loser, purely because of its cheaper latent heat.

Step 1 — daily propellant saving from the better insulation. Compare the two boil-off rates: $\Delta \dot{m}=54.6-15.0=39.6\text{kg/day saved}$ Why: each day, the improved tank keeps $39.6$ kg of hydrogen it would otherwise have vented. Step 2 — break-even condition. The MLI is "free" once the cumulative saved propellant equals the $120\text{kg}$ dry-mass penalty it cost: $\Delta \dot{m}\times t_{\text{break-even}}=120\text{kg}$ $t_{\text{break-even}}=\frac{120}{39.6}=3.03\text{days}$ Step 3 — interpretation. If the mission loiters longer than about 3 days, the heavy MLI wins: it saves more propellant mass than the dry mass it added. For a quick sub-orbital hop under 3 days, that MLI is dead weight and hurts the Propellant Mass Fraction. This is exactly the thermal-vs-structural trade that couples to Structural Design - Pressure Vessels.

The master chain, warm shell to lost kilograms ::: $\dot{Q}={\dot{Q}}_{cond}+{\dot{Q}}_{rad}+{\dot{Q}}_{conv}$, then $\dot{m}=\dot{Q}/L_v$, then $m_{day}=\dot{m}\times 86400$, then $\%=m_{day}/m_{total}\times 100$. Why does higher $L_v$ mean lower boil-off mass for fixed heat? ::: Because $L_v$ is joules-per-kilogram; a higher price per kilogram means the same joules buy fewer boiled kilograms. Why does radiation usually beat conduction as the dominant leak? ::: The warm shell's $T^4$ term is enormous, and even small emissivity times a large area times $\sigma T_{hot}^4$ outweighs a few thin struts.