Worked examples — Thermal control — multi-layer insulation (MLI), heaters, heat pipes, radiators

This page is the drill hall. The parent note gave you the four tools (MLI, heaters, heat pipes, radiators). Here we throw every kind of number at those tools — hot sides, cold sides, zero temperatures, infinite layers, gravity-hostile pipes, and one exam trap — so that no problem on a test surprises you.

Everything below leans on one master law, so let us pin it once in plain words before any symbol appears.

The master radiation formula, read aloud: heat leaving = throwing-strength × how-black × area × (hot⁴ minus cold⁴).

$Q=\epsilon \sigma A(T_{\text{hot}}^4-T_{\text{cold}}^4)$

---

The scenario matrix

Every problem this topic can ask is one of these cells. The examples below are tagged with the cell they hit.

#	Cell class	What makes it distinct	Hit by
C1	Hot side dominates	$T_{\text{hot}}\gg T_{\text{cold}}$, cold term negligible	Ex 1, Ex 6
C2	Both terms matter	$T_{\text{cold}}$ is a real fraction of $T_{\text{hot}}$, keep both	Ex 2
C3	Degenerate: cold side = 0 K	deep space $\approx 3$ K, cold term vanishes	Ex 1, Ex 6
C4	Limiting: layers $N\to \infty$	MLI diminishing returns, where the idealisation breaks	Ex 3
C5	Zero / no-source case	eclipse, no Sun, component cools by itself	Ex 4
C6	Time-dependent (rate)	how long until a limit is hit, uses $dT/dt$	Ex 4
C7	Phase-change transport	heat pipe, latent heat, no $T^4$	Ex 5
C8	Sign / direction flip	gravity helps vs. fights the wick	Ex 5b
C9	Real-world word problem	full mission sizing, pick the tool yourself	Ex 6
C10	Exam twist	a hidden unit or a "which is bigger" comparison	Ex 7

---

Ex 1 — Hot radiator into cold void · cells C1, C3

Forecast: Guess the answer to the nearest 100 W before reading on. Does the 3 K matter at all?

1. Write the net law. $Q=\epsilon \sigma A(T_r^4-T_s^4)$. Why this step? Radiation is the only heat path in vacuum; both surfaces glow, so we subtract what space glows back at us.
2. Test whether the cold term survives. $T_r^4=300^4=8.1\times 10^9$, while $T_s^4=3^4=81$. Why this step? Before crunching, check if a term is negligible — it saves arithmetic and teaches the physics. Here $81$ vs $8.1\times 10^9$ is a ratio of $10^{-8}$: the void contributes nothing. This is why we always approximate $T_s\to 0$ (cell C3).
3. Compute. $Q=0.85\times 5.67\times 10^{-8}\times 1.2\times 8.1\times 10^9\approx 468\text{W}$. Why this step? Multiply the surviving terms.

Verify: Units: $[{\text{W/m}}^2{\text{K}}^4]\cdot [{\text{m}}^2]\cdot [{\text{K}}^4]=\text{W}$. ✓. Look at the figure — the red outgoing arrow is huge, the cyan return arrow from space is a whisker. That matches the $10^{-8}$ ratio.

---

Ex 2 — Both temperatures matter (MLI leak) · cell C2

Forecast: Is $180$ K "small enough to ignore" like the 3 K was? Guess yes/no.

1. Effective emissivity of the stack. ${\epsilon }_{\text{MLI}}\approx \frac{\epsilon }{N-1}=\frac{0.03}{29}\approx 1.03\times 10^{-3}$. Why this step? Each of the 30 shiny Mylar layers re-radiates, forming a chain of resistances; the whole stack acts like one very-low-emissivity surface.
2. Check the cold term. $T_1^4=293^4=7.37\times 10^9$; $T_2^4=180^4=1.05\times 10^9$. Why this step? Now $1.05\times 10^9$ is $14\%$ of $7.37\times 10^9$ — not negligible. This is a genuine cell-C2 problem; dropping $T_2$ would overstate the leak by $14\%$.
3. Compute. $Q=1.03\times 10^{-3}\times 5.67\times 10^{-8}\times 2\times (7.37\times 10^9-1.05\times 10^9)$. $=1.03\times 10^{-3}\times 5.67\times 10^{-8}\times 2\times 6.32\times 10^9\approx 0.74\text{W}$.

Verify: If we had wrongly set $T_2=0$: $Q^{\prime }\approx 0.86\text{W}$, which is $16\%$ higher — confirming the cold term was worth keeping. Contrast with Ex 1 where keeping it changed nothing. The lesson: always compare $T_{\text{cold}}^4$ to $T_{\text{hot}}^4$ first.

---

Ex 3 — The limit $N\to \infty$: where MLI lies to you · cell C4

Forecast: The formula says ${\epsilon }_{\text{MLI}}\to 0$. Guess whether a 100-layer blanket really beats a 30-layer one in practice.

1. Plug in. $N=2:0.03/1=0.030$; $N=11:0.03/10=0.0030$; $N=31:0.03/30=0.0010$; $N=101:0.03/100=0.0003$. Why this step? To see the shape of the returns.
2. Take the limit. As $N\to \infty$, ${\epsilon }_{\text{MLI}}=\frac{0.03}{N-1}\to 0$. Why this step? Mathematically insulation looks perfect. The idealisation says infinite layers = zero leak.
3. Find where the model breaks. From $N=31$ to $N=101$ we spent 70 extra layers to drop $\epsilon$ from $0.0010$ to $0.0003$ — a factor of only $3.3$, at large mass cost. Worse, real layers compress under launch load and touch, opening solid-conduction shortcuts the formula ignores. Why this step? Cell C4 is about knowing where your equation stops being true. The curve in the figure flattens: this is diminishing returns.

Verify: Ratio check: $\epsilon (31)/\epsilon (101)=0.0010/0.0003=3.33$ while layer count rose $\times 3.26$ — nearly one-for-one, i.e. no leverage left. That is why flight blankets sit at 10–30 layers.

---

Ex 4 — Eclipse cooldown: the zero-source, time-dependent case · cells C5, C6

Forecast: Guess the cooldown rate in °C per minute, then the time to reach 0°C.

1. Heat capacity. $C=mc=5\times 1000=5000\text{J/K}$. Why this step? $C$ tells us how many joules to change the battery by 1 K — the thermal "inertia".
2. Cooling rate. Energy balance with no source: $C\frac{dT}{dt}=-Q_{\text{loss}}$, so $\frac{dT}{dt}=-\frac{Q_{\text{loss}}}{C}=-\frac{5}{5000}=-1.0\times 10^{-3}\text{K/s}=-0.06\text{K/min}.$ Why this step? We use a rate (a derivative) because the question is about time, not a steady value. This is what makes it cell C6. In vacuum with no Sun (Orbital Thermal Environment) the source term is exactly zero — cell C5.
3. Time to the limit. Allowed drop $\Delta T=20\text{K}$ (from $20°$ to $0°$). Time $t=\frac{\Delta T}{0.06\text{K/min}}=\frac{20}{0.06}\approx 333\text{min}$. Why this step? Constant rate ⇒ time = drop ÷ rate.
4. Compare to eclipse length. $333\text{min}\gg 90\text{min}$, so a 90-min eclipse only cools it by $90\times 0.06=5.4\text{K}$, landing at $14.6°\text{C}$ — safe, no heater needed for this eclipse.

Verify: $90\text{min}\times (-0.06\text{K/min})=-5.4\text{K}$; $20-5.4=14.6°\text{C}>0$. ✓. If the eclipse were $>333$ min, a heater (Spacecraft Power Systems) drawing $\ge 5$ W would be mandatory.

---

Ex 5 — Heat pipe capacity: phase change, no $T^4$ · cell C7

Forecast: Order-of-magnitude guess: how many times better than copper?

1. Heat pipe transport. $Q=\dot{m}{L}_v=0.001\times 1.37\times 10^6=1370\text{W}$. Why this step? A heat pipe carries energy as latent heat (phase-change): the fluid soaks up $L_v$ joules per kg boiling at the hot end and dumps them condensing at the cold end. No $T^4$ appears — this is transport, not rejection.
2. Copper conduction. $Q_{\text{cu}}=\frac{kA\Delta T}{L}=\frac{400\times 7.85\times 10^{-5}\times 10}{1}=0.314\text{W}$. Why this step? Solids move heat by conduction, Fourier's law — the honest competitor.
3. Ratio. $1370/0.314\approx 4360\times$. Why this step? Shows why every large spacecraft uses heat pipes: same pencil-thin cross-section, thousands of times the throughput.

Verify: Units: $[\text{kg/s}]\cdot [\text{J/kg}]=\text{J/s}=\text{W}$ ✓; $[\text{W/m⋅K}]\cdot [{\text{m}}^2]\cdot [\text{K}]/[\text{m}]=\text{W}$ ✓.

Ex 5b — Sign flip: gravity helps or fights the wick · cell C8

1. Gravity head. $\Delta P_g=680\times 9.81\times 1=6670\text{Pa}\approx 6.67\text{kPa}$. Why this step? This is the pressure gravity imposes across 1 m of liquid.
2. Case (a): condenser above, liquid must fall down to evaporator. Gravity assists the return: capillary pressure $1$ kPa plus gravity — the wick easily wins. Sign of gravity term: $+$.
3. Case (b): condenser below, liquid must climb 1 m against gravity. Now the wick's $1\text{kPa}$ must overcome $6.67\text{kPa}$ — it cannot (Capillary Action pump too weak). Sign: $-$, and $|-6.67|>1$ ⇒ dry-out, pipe fails.
4. Microgravity resolution. In orbit $g\approx 0$, so $\Delta P_g\to 0$ and either orientation works — the whole sign problem disappears.

Verify: $1\text{kPa}<6.67\text{kPa}$ ⇒ ground failure in the uphill case; $1\text{kPa}>0$ ⇒ orbit success. This is why ground tests must orient the pipe "condenser-up".

---

Ex 6 — Real-world sizing: reject 500 W · cells C1, C3, C9

Forecast: Guess the area in m². Then guess: does raising the allowed temperature by 20 K shrink it a little or a lot?

1. Rearrange the law for area. With $T_s\approx 0$ (cell C3): $A=\frac{Q}{\epsilon \sigma T^4}$. Why this step? We know the heat and the temperature limit; area is the unknown to solve for.
2. Plug in at 313 K. $313^4=9.60\times 10^9$; $A=\frac{500}{0.85\times 5.67\times 10^{-8}\times 9.60\times 10^9}=\frac{500}{462.6}\approx 1.08{\text{m}}^2$. Why this step? Denominator $462.6{\text{W/m}}^2$ is how much each square metre radiates at 313 K.
3. Raise the limit to 333 K. $333^4=1.23\times 10^{10}$; per-m² emission $=0.85\times 5.67\times 10^{-8}\times 1.23\times 10^{10}=593{\text{W/m}}^2$; $A=500/593\approx 0.84{\text{m}}^2$. Why this step? Shows the $T^4$ leverage: only +20 K but area shrinks by $\sim 22\%$.

Verify: Ratio of areas $=(313/333{)}^4=0.94^4=0.78$, and indeed $0.84/1.08=0.78$ ✓. The $T^4$ dependence means running hotter is a strong lever on radiator size — the classic thermal trade.

---

Ex 7 — Exam twist: the hidden unit / "which is bigger" · cell C10

Forecast: Guess the size of the student's error — factor of 2? 10? 1000?

1. Spot the trap. $T^4$ needs kelvin, not celsius. $40°\text{C}=313\text{K}$, not $40$. Why this step? The Stefan–Boltzmann law is defined from absolute zero; celsius has a false origin.
2. Student's wrong number. $\sigma \times 40^4=5.67\times 10^{-8}\times 2.56\times 10^6=0.145{\text{W/m}}^2$. Why this step? To measure how catastrophic the slip is.
3. True number. $\sigma \times 313^4=5.67\times 10^{-8}\times 9.60\times 10^9=544{\text{W/m}}^2$. Why this step? The correct emission.
4. The comparison twist. $-40°\text{C}=233\text{K}$; its emission $=\sigma \times 233^4=5.67\times 10^{-8}\times 2.95\times 10^9=167{\text{W/m}}^2$. Twice that is $334{\text{W/m}}^2$. Since $544>334$, the $40°\text{C}$ radiator beats double the cold one. Why this step? Exam loves "is it more than double?" — the $T^4$ curve means an 80 K rise more than doubles emission.

Verify: Error factor $=544/0.145\approx 3750\times$ — a mission-killing mistake. And $544/167=3.26>2$ ✓, confirming the hot radiator exceeds twice the cold one. Cross-check with Cryogenics intuition: cold surfaces barely radiate, exactly the $167$ vs $544$ gap.

---

Recall Self-test before you close this page

Which matrix cell needs a derivative? ::: C6 (time-dependent), Ex 4 — cooling rate $dT/dt=-Q/C$. When is the 3 K space term safe to drop? ::: When $T_{\text{cold}}^4\ll T_{\text{hot}}^4$; here $3^4=81$ vs $\sim 10^9$, a $10^{-8}$ ratio (Ex 1). Why do MLI returns flatten past ~30 layers? ::: ${\epsilon }_{\text{MLI}}=\epsilon /(N-1)$ falls as $1/N$ (weak leverage) plus compression conduction shorts (Ex 3). Ground heat-pipe rule of thumb? ::: Keep the condenser above the evaporator so gravity assists return; wick alone (~1 kPa) loses to a 1 m ammonia column (~6.7 kPa) (Ex 5b). A radiator "at 40°" — what must you convert first? ::: Celsius → kelvin (313 K); forgetting it errs by ~3750× (Ex 7).