Thermal control — multi-layer insulation (MLI), heaters, heat pipes, radiators: Exercises

This page is your self-testing gym for the thermal-control topic. Every problem states its level (L1 Recognition → L5 Mastery), then hides a complete worked solution inside a collapsible callout. Peek only after you have tried.

Before we start, let us pin down the tools we will keep reaching for, so no symbol appears un-earned.

Level 1 — Recognition

Problem 1.1

A radiator surface is at $T = 300 K$ with emissivity $ε = 0.85$ . How much power does one square metre radiate into cold space? (Treat space as $0 K$ .)

Recall Solution

What we do: plug straight into $Q = ε σ A T^{4}$ with $A = 1 m^{2}$ . Why: L1 is about recognising which single equation fits — no traps yet. $T^{4} = 30 0^{4} = 8.1 \times 1 0^{9} K^{4}$ $Q = 0.85 \times 5.67 \times 1 0^{- 8} \times 1 \times 8.1 \times 1 0^{9} \approx 390 W$ So each square metre at 300 K rejects about $390 W$ .

Problem 1.2

Name the thermal-control device that matches each job: (a) keeps a battery from freezing in eclipse, (b) moves 1 kW of heat 1 m with no power, (c) blocks radiation with 30 shiny layers, (d) dumps waste heat to space.

Recall Solution

(a) Heater (resistive) — adds electrical power when the part gets too cold. (b) Heat pipe — evaporation/condensation cycle, no moving parts. Relies on Phase Change Heat Transfer and Capillary Action. (c) MLI (multi-layer insulation) — stacked reflective shields, see Materials Science — Kapton & Mylar. (d) Radiator — a high-emissivity surface obeying Stefan-Boltzmann Law.

Level 2 — Application

Problem 2.1

An MLI blanket has $N = 25$ layers, each layer emissivity $ε = 0.03$ . Estimate the effective emissivity $ε_{MLI}$ of the stack.

Recall Solution

What we do: use the shield-network approximation $ε_{MLI} \approx \frac{ε}{N - 1}$ . Why $N - 1$ and not $N$ : with $N$ layers there are $N - 1$ radiation gaps between them; each gap is one $T^{4}$ barrier that re-radiates heat back. More gaps → less leak. $ε_{MLI} \approx \frac{0.03}{25 - 1} = \frac{0.03}{24} = 0.00125$ So $ε_{MLI} \approx 1.25 \times 1 0^{- 3}$ — about 700× better than the bare $ε = 0.9$ black surface.

Problem 2.2

Using that blanket, find the heat leak $Q$ through $A = 3 m^{2}$ when inside is $T_{1} = 293 K$ and outside is $T_{2} = 120 K$ .

Recall Solution

What we do: radiation between two temperatures uses the difference of fourth powers.

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

Figure (above): the warm inside wall (orange) glows outward at a rate set by

T_{1}^{4}

— that is the thick magenta arrow crossing the violet MLI stack. The cold outside wall (navy) glows back inward at the much weaker rate set by

T_{2}^{4}

— the thin navy arrow. What actually leaks is the difference of those two glows, printed at the bottom of the figure as

Q \propto (T_{1}^{4} - T_{2}^{4})

. Why the difference: the hot side sends power

\propto T_{1}^{4}

outward, the cold side sends

\propto T_{2}^{4}

back; the net leak is the gap between them.

Q = ε_{MLI} σ A (T_{1}^{4} - T_{2}^{4})

T_{1}^{4} = 29 3^{4} = 7.37 \times 1 0^{9}, T_{2}^{4} = 12 0^{4} = 2.07 \times 1 0^{8}

T_{1}^{4} - T_{2}^{4} = 7.16 \times 1 0^{9}

Q = 1.25 \times 1 0^{- 3} \times 5.67 \times 1 0^{- 8} \times 3 \times 7.16 \times 1 0^{9} \approx 1.52 W

So only about

1.5 W

leaks through — the whole point of MLI. This is pure radiation in Heat Transfer in Vacuum.

Level 3 — Analysis

Problem 3.1

A radiator must reject $Q = 500 W$ with $ε = 0.85$ . Compare the required area at (a) $T = 300 K$ and (b) $T = 350 K$ . By what factor does the area shrink?

Recall Solution

What we do: rearrange $Q = ε σ A T^{4}$ to solve for area: $A = \frac{Q}{ε σ T ^{4}}$ . Why rearrange: here $Q$ is fixed (the electronics dump 500 W no matter what), so area is the free variable.

Figure (above): the magenta curve is the required area

A

plotted against radiator temperature

T

, all for the same fixed 500 W. It plunges steeply because

A

falls as

1/ T^{4}

. The two dashed drop-lines mark our two cases: the orange point at

T = 300 K

needs

A \approx 1.28 m^{2}

, while the violet point at

T = 350 K

needs only

A \approx 0.69 m^{2}

— a visibly smaller radiator for the very same heat load. (a)

T = 300

T^{4} = 8.10 \times 1 0^{9}

A_{a} = \frac{500}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 8.10 \times 1 0 ^{9}} = \frac{500}{390} \approx 1.28 m^{2}

(b)

T = 350

T^{4} = 1.50 \times 1 0^{10}

A_{b} = \frac{500}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 1.50 \times 1 0 ^{10}} = \frac{500}{723} \approx 0.69 m^{2}

Factor:

A_{a} / A_{b} = 1.28/0.69 \approx 1.85

. Why: raising

T

from 300 to 350 is a ratio

350/300 = 1.167

; to the fourth power that is

1.16 7^{4} \approx 1.85

. The area shrinks by exactly that factor. Running hotter lets you use a smaller radiator.

Problem 3.2

A single ammonia heat pipe carries vapour at $\overset{m}{˙} = 8 \times 1 0^{- 4} kg/s$ , latent heat $L_{v} = 1.37 \times 1 0^{6} J/kg$ . How much power does it transport? Compare to a solid copper rod of the same $A = 7.85 \times 1 0^{- 5} m^{2}$ , length $1 m$ , conductivity $k = 400 W/m\cdotK$ , across a $Δ T = 10 K$ .

Recall Solution

Heat pipe carries heat as latent heat of the moving vapour: $Q = \overset{m}{˙} L_{v}$ (uses Phase Change Heat Transfer). $Q_{pipe} = 8 \times 1 0^{- 4} \times 1.37 \times 1 0^{6} = 1096 W \approx 1.1 kW$ Copper rod carries heat by conduction (Fourier): $Q = \frac{k A Δ T}{L}$ . $Q_{Cu} = \frac{400 \times 7.85 \times 1 0 ^{- 5} \times 10}{1} = 0.314 W$ Ratio: $Q_{pipe} / Q_{Cu} = 1096/0.314 \approx 3500$ . The heat pipe wins by three-and-a-half thousand times per unit cross-section — because latent heat carries far more energy than a slow conductive trickle.

Level 4 — Synthesis

Problem 4.1

A $5 kg$ battery ( $c = 1000 J/kg\cdotK$ , so heat capacity $C = 5000 J/K$ ) starts at $20° C$ and loses $Q_{loss} = 6 W$ during a $95$ -minute eclipse. (a) With no heater, what is its final temperature? (b) Does it stay above its $0° C$ limit? (c) What steady heater power holds it exactly at $10° C$ ?

Recall Solution

(a) Cooling with no heater. The battery's stored heat drains at a steady rate: $\frac{d T}{d t} = - \frac{Q _{loss}}{C} = - \frac{6}{5000} = - 1.2 \times 1 0^{- 3} K/s$ Why divide by $C$ : heat capacity says "how many joules to change temperature by 1 K". Draining $Q_{loss}$ joules per second, the temperature slides down at $Q_{loss} / C$ per second. Eclipse time $t = 95 min = 5700 s$ : $Δ T = - 1.2 \times 1 0^{- 3} \times 5700 = - 6.84 K$ Final temperature $= 20 - 6.84 = 13.16° C$ . (b) $13.16° C > 0° C$ → yes, it survives this eclipse with no heater. The battery's own thermal mass is the buffer. (c) Heater to hold $10° C$ . To keep temperature constant, power in must equal power out. So the heater exactly replaces the loss: $P_{heater} = Q_{loss} = 6 W$ This heater power is drawn from Spacecraft Power Systems, so engineers usually spec a $10$ – $12 W$ heater on a duty cycle for margin.

Problem 4.2

The same eclipse, but now the battery starts at $2° C$ and the loss is $10 W$ . Show it would freeze without a heater, and find the heater power needed to keep it from dropping below $0° C$ over the $5700 s$ eclipse.

Recall Solution

Without heater: $Δ T = - \frac{10}{5000} \times 5700 = - 11.4 K$ , landing at $2 - 11.4 = - 9.4° C$ — well below freezing, fails. With heater: to hold the starting temperature (worst case: don't drop below 0°C, and it starts at 2°C, so simplest is to prevent any net cooling), the heater must supply at least the full loss: $P_{heater} = 10 W$ Refinement: we are allowed to fall to 0°C, a drop of $2 K$ available as buffer. The buffer absorbs $C Δ T = 5000 \times 2 = 1 0^{4} J$ over $5700 s$ , i.e. it can offset $1 0^{4} /5700 = 1.75 W$ . So the minimum heater is: $P_{heater,min} = 10 - 1.75 = 8.25 W$ Why the subtraction: the 2 K of "temperature room" is free stored energy; the heater only needs to cover the remaining loss.

Level 5 — Mastery

Problem 5.1

A spacecraft bay generates $Q = 450 W$ continuously. It is wrapped in $30$ -layer MLI ( $ε = 0.03$ per layer, wall area $A_{wall} = 4 m^{2}$ , inside $293 K$ , outside $150 K$ ) and vents its waste heat through a radiator ( $ε_{r} = 0.88$ , seeing $0 K$ space) held at $310 K$ . (a) How much heat leaks out through the MLI? (b) What radiator area is required to reject the remaining heat? (c) A heat pipe delivers the 450 W to the radiator with $\overset{m}{˙} = 3.5 \times 1 0^{- 4} kg/s$ of ammonia — is this flow enough?

Recall Solution

(a) MLI leak (a bonus heat path, small). What we do: treat the blanket as one effective emissivity and use the two-temperature radiation law. Why: the MLI sits between the warm inside (293 K) and cold outside (150 K), so its leak follows the same difference-of-fourth-powers form as Problem 2.2. $ε_{MLI} = \frac{0.03}{30 - 1} = \frac{0.03}{29} = 1.034 \times 1 0^{- 3}$ $T_{1}^{4} - T_{2}^{4} = 29 3^{4} - 15 0^{4} = 7.37 \times 1 0^{9} - 5.06 \times 1 0^{8} = 6.87 \times 1 0^{9}$ $Q_{MLI} = 1.034 \times 1 0^{- 3} \times 5.67 \times 1 0^{- 8} \times 4 \times 6.87 \times 1 0^{9} \approx 1.61 W$ So the MLI leaks only about $1.6 W$ — negligible next to 450 W. (b) Radiator area. What we do: rearrange the single-surface radiation law $Q = ε_{r} σ A T_{r}^{4}$ to $A = Q / (ε_{r} σ T_{r}^{4})$ . Why apply $Q = ε σ A T^{4}$ here: the radiator is one hot surface glowing into 0 K space, so the plain single-surface form (no back-radiation term) is exactly the right tool. It must reject the net heat the pipe delivers. The MLI leak (1.6 W) already sheds a tiny amount, so the radiator handles $450 - 1.6 \approx 448.4 W$ . (We could conservatively use the full 450 W.) $T_{r}^{4} = 31 0^{4} = 9.24 \times 1 0^{9}$ $A_{r} = \frac{448.4}{0.88 \times 5.67 \times 1 0 ^{- 8} \times 9.24 \times 1 0 ^{9}} = \frac{448.4}{461} \approx 0.97 m^{2}$ So $A_{r} \approx 0.97 m^{2}$ . (c) Heat-pipe check. What we do: compute the pipe's carrying power from $Q = \overset{m}{˙} L_{v}$ and compare it to the 450 W load. Why: the pipe transports heat as latent heat of the flowing ammonia vapour, so its capacity is just mass flow times latent heat. With $L_{v} = 1.37 \times 1 0^{6} J/kg$ : $Q_{pipe} = \overset{m}{˙} L_{v} = 3.5 \times 1 0^{- 4} \times 1.37 \times 1 0^{6} = 479.5 W$ Since $479.5 W > 450 W$ , yes — the flow carries the load with about $30 W$ of margin.

Problem 5.2 — Reasoning

On the ground, an engineer tests this same ammonia heat pipe vertically, with the condenser $0.5 m$ below the evaporator. The wick can pump a maximum capillary pressure of $Δ P_{cap} = 1500 Pa$ . Ammonia liquid density $ρ \approx 600 kg/m^{3}$ , $g = 9.81 m/s^{2}$ . Will the pipe work in this orientation?

Recall Solution

What we do: compare the wick's maximum pumping pressure against the gravity head the returning liquid must climb. Why: the liquid must be pushed from the condenser back up to the evaporator. The only pump is capillary suction; the only opposing force here is gravity. If gravity's pressure demand beats the wick's supply, the wick dries out. Gravity head over height $h = 0.5 m$ : $Δ P_{grav} = ρ g h = 600 \times 9.81 \times 0.5 = 2943 Pa$ Compare: $Δ P_{cap} = 1500 Pa < Δ P_{grav} = 2943 Pa$ Verdict: since the capillary pumping ( $1500 Pa$ ) is less than the gravity head ( $2943 Pa$ ), the wick cannot lift the liquid back up. The evaporator starves, the pipe dries out and fails in this orientation. $The pipe does NOT work.$ Why it still works in orbit: in microgravity $g \approx 0$ , so $Δ P_{grav} \to 0$ and the modest capillary pressure (from Capillary Action) easily returns the liquid in any orientation. This is exactly why ground testing must respect gravity.

Recall Quick self-check reveals

Radiator power per m² at 300 K, ε=0.85 ::: about 390 W Effective emissivity of 25-layer MLI (ε=0.03) ::: 0.03/24 ≈ 0.00125 Why we use $T_{1}^{4} - T_{2}^{4}$ not $(T_{1} - T_{2})^{4}$ ::: each surface radiates power ∝ its own $T^{4}$ ; net leak is the difference of those powers Heat a heat pipe carries ::: $Q = \overset{m}{˙} L_{v}$ (mass flow × latent heat) Why heat pipes can fail on the ground ::: gravity head $ρ g h$ can exceed the wick's capillary pressure