Free preview · Thermal course v2.1

Spacecraft Thermal Control

How spacecraft stay at the right temperature, from the physics of radiation to a complete CubeSat design. Short sections, worked examples and graded practice, with every number checked by a script.

Free preview: chapter 1 of 16, complete. The full course: 16 chapters · 33 worked examples · 48 practice problems · 2894 script checks.

How to use this pack

1Read 'The idea in one minute' and the 'You will be able to' list.
2Work through the short sections and the worked examples with a pencil.
3Try each practice problem before opening the hint; open the full solution only to check.
4Do the three-question check and the spaced recap before moving on.

What is in the full course

1Heat transfer in spaceFree preview30 min
2Radiation basics: α, ε and the two bands30 min
3The orbital thermal environment35 min
4Beta angle and eclipse30 min
5View factors35 min
6Single-node equilibrium30 min
7Hot and cold cases with margins40 min
8Multi-node thermal networks45 min
9Transients and eclipse cool-down40 min
10Passive control I: coatings, MLI and radiators40 min
11Passive control II: heat pipes, louvres, doublers and straps40 min
12Active control: heaters, thermostats and pumped loops35 min
13Component limits and design margins30 min
14The thermal analysis workflow and tools30 min
15Testing: thermal vacuum and thermal balance30 min
16Worked mission case: a 6U CubeSat end to end60 min

This preview is chapter 1, complete. Your ticks are saved in this browser only. Original work, and a study aid, not assignment answers.

Chapter 1

Heat transfer in space

30 minutes

The idea in one minute

In orbit there is no air, so heat cannot be carried away by convection. Heat enters a spacecraft as radiation (sunlight, sunlight reflected off Earth, and Earth's own infrared glow), moves around inside mostly by conduction through structure, and finally leaves as infrared radiation to the cold of deep space. Thermal control is the art of shaping those paths so that every part stays inside its allowed temperature range.

Heat arrives as radiation and can only leave as radiation: there is no air to carry it away.

You will be able to

Say which heat transfer mode dominates on a given path in orbit
Compute a conductive conductance G = kA/L and combine conductances in series and parallel
Treat radiation as a conductance for small temperature differences

Conduction

Fourier's law says heat flows down a temperature gradient: Q = kA ΔT / L. It is convenient to lump the geometry and material into one number, the conductance G = kA/L in W/K. Its inverse 1/G is a thermal resistance, and resistances in series add exactly like electrical ones; conductances in parallel add directly.^{[R11, R12, R13]}

Radiation

Every surface emits εσT⁴ watts per square metre, with T in kelvin. Net exchange between surfaces depends on the fourth powers of their temperatures, so it is non-linear. For small differences you can linearise: q ≈ h_r ΔT with h_r = 4εσT³. Near room temperature h_r is about 5 W/(m² K).^[R5]

Convection, where it still exists

Convection needs a fluid. It survives inside pressurised modules (with fans, since there is no buoyancy in free fall), inside pumped fluid loops, and in the vapour of heat pipes. On the outside of a satellite it is absent, and on the ground during testing it must be removed, which is why thermal tests are done in vacuum. Before launch it works the other way: conditioned air blown through the fairing raises the film coefficient and the temperature difference, and cools powered units on the pad.^{[R1, R5]}

Key equations

Fourier's law of conduction

Q = k A ΔT / L

where

Qheat flow, W

kthermal conductivity, W/(m K)

Aarea, m²

ΔTtemperature difference, K

Llength of the conduction path, m

Conductance and resistance

G = kA/L, R = 1/G

where

Gconductance, W/K

Rthermal resistance, 1/G, K/W

Paths in series and in parallel

1/G_s = Σ 1/G_i; G_p = Σ G_i

where

G_stotal conductance of elements in series, W/K

G_ptotal conductance of elements in parallel, W/K

G_iconductance of element i, W/K

A bolted or bonded joint

G_c = h_c A

where

G_cconductance of a joint, W/K

h_ccontact conductance per unit area, W/(m² K)

Radiation treated as a conductance

h_r = 4εσT_m³, q ≈ h_r ΔT

where

h_rradiative heat-transfer coefficient, W/(m² K)

εinfrared emissivity (0 to 1), dimensionless

σStefan-Boltzmann constant, 5.670 × 10⁻⁸, W/(m² K⁴)

T_mmean of the two temperatures, K

qheat flux (heat flow per unit area), W/m²

Good when ΔT is small compared with T.

Worked examples

Worked example 1.1

How much heat does a bracket carry?^{[R12, R13]}

Given

An aluminium 6061 bracket (k = 167 W/(m K)) is 80 mm long with a 30 × 4 mm cross-section. One end is 15 K warmer than the other. Find its conductance and heat flow, then repeat for the same bracket in titanium (k = 6.7 W/(m K)).

Working

Cross-section A = 30 mm × 4 mm = 120 mm²1.20e-04 m²
Conductance G = kA/L = 167 × 1.20e-04 / 0.0800.2505 W/K
Heat flow Q = G ΔT = 0.2505 × 153.76 W
Titanium: G = 6.7 × 1.20e-04 / 0.080 = 0.01005 W/K, so Q0.151 W
Ratio Al/Ti = 24.9, exactly the ratio of the conductivities

Answer. Aluminium: G = 0.251 W/K, Q = 3.76 W. Titanium: G = 0.0100 W/K, Q = 0.151 W.

Sanity check. Units: (W/(m K)) × m² / m = W/K. Titanium is the classic isolator: the same part carries 25 times less heat.

Pythonedit a value, then Run

# Worked example 1.1: How much heat does a bracket carry?
k_al = 167.0       # W/(m K), aluminium 6061
k_ti = 6.7         # W/(m K), Ti-6Al-4V
L = 0.080          # m, bracket length
w = 0.030          # m, cross-section width
t = 0.004          # m, cross-section thickness
dT = 15.0          # K, end-to-end temperature difference

A = w * t                      # m^2
G_al = k_al * A / L            # W/K
Q_al = G_al * dT               # W
G_ti = k_ti * A / L
Q_ti = G_ti * dT
print(f"A = {A:.2e} m^2")
print(f"Aluminium: G = {G_al:.4f} W/K, Q = {Q_al:.3f} W")
print(f"Titanium:  G = {G_ti:.5f} W/K, Q = {Q_ti:.4f} W")
print(f"Ratio Al/Ti = {G_al / G_ti:.1f}")

A = 1.20e-04 m^2
Aluminium: G = 0.2505 W/K, Q = 3.757 W
Titanium:  G = 0.01005 W/K, Q = 0.1507 W
Ratio Al/Ti = 24.9

Worked example 1.2

Radiation as a conductance^[R5]

Given

A surface with ε = 0.85 sits at 300 K and radiates to black surroundings at 290 K. Compare the exact net flux with the linearised form using a radiative heat-transfer coefficient h_r = 4εσT_m³ at the mean temperature.

new symbols

Tabsolute temperature, K

Working

Exact: q = εσ(T₁⁴ − T₂⁴) = 0.85 × 5.6704e-08 × (300⁴ − 290⁴)49.51 W/m²
Mean temperature T_m = 295 K; h_r = 4 × 0.85 × 5.6704e-08 × 295³4.949 W/(m² K)
Linear: q = h_r ΔT = 4.949 × 1049.49 W/m²
Difference0.029 %

Answer. q = 49.5 W/m² exactly, 49.5 W/m² linearised; h_r ≈ 4.9 W/(m² K).

Sanity check. A few W/(m² K) is about the size of gentle natural convection on Earth, which is why radiation alone can cool a spacecraft when areas are large. The linear form is excellent for small temperature differences.

Pythonedit a value, then Run

# Worked example 1.2: Radiation as a conductance
sigma = 5.670374419e-8   # W/(m^2 K^4), Stefan-Boltzmann (CODATA 2018)
eps = 0.85               # emissivity
T1 = 300.0               # K, surface
T2 = 290.0               # K, black surroundings

q_exact = eps * sigma * (T1**4 - T2**4)     # W/m^2
Tm = (T1 + T2) / 2                          # K, mean temperature
h_r = 4 * eps * sigma * Tm**3               # W/(m^2 K)
q_lin = h_r * (T1 - T2)                     # W/m^2
diff = abs(q_lin - q_exact) / q_exact * 100
print(f"q exact      = {q_exact:.2f} W/m^2")
print(f"h_r          = {h_r:.3f} W/(m^2 K)")
print(f"q linearised = {q_lin:.2f} W/m^2")
print(f"difference   = {diff:.3f} %")

q exact      = 49.51 W/m^2
h_r          = 4.949 W/(m^2 K)
q linearised = 49.49 W/m^2
difference   = 0.029 %

Practice

Practice T1.1 · warm-up

A thin copper strap^[R11]

Given

A copper strap (k = 390 W/(m K)) has a cross-section of 20 mm² and a length of 150 mm. Its ends differ by 8 K. What heat does it carry?

Mark as done

Hint

Convert mm² to m² first (1 mm² = 10⁻⁶ m²), then G = kA/L.

Answer. G = 0.0520 W/K and Q = 0.416 W.

Working

A = 20 mm² = 2.0e-05 m², L0.150 m
G = kA/L = 390 × 2.0e-05 / 0.1500.0520 W/K
Q = G ΔT = 0.0520 × 80.416 W

Answer. G = 0.0520 W/K and Q = 0.416 W.

Sanity check. Check units: W/(m K) · m² / m = W/K. Under half a watt for 8 K: thin straps are weak, so real straps use many foils or braids.

Pythonedit a value, then Run

# Practice T1.1: A thin copper strap
k = 390.0          # W/(m K), copper
A = 20e-6          # m^2 (20 mm^2)
L = 0.150          # m
dT = 8.0           # K

G = k * A / L      # W/K
Q = G * dT         # W
print(f"G = {G:.4f} W/K")
print(f"Q = {Q:.3f} W")

G = 0.0520 W/K
Q = 0.416 W

Practice T1.2 · standard

A heat path in series^{[R5, R9, R12, R13]}

Given

Heat leaves an electronics box through three elements in series: the aluminium bracket of the first worked example (G = 0.2505 W/K), a bolted joint of contact area 1 cm² with contact conductance h_c = 2000 W/(m² K), and a titanium standoff (k = 6.7 W/(m K)) of diameter 8 mm and length 20 mm. The ends of the path differ by 20 K. Find the total conductance, the heat flow, and which element controls it.

Mark as done

Hint

Conductances in series add like resistors in series: add the resistances 1/G, not the G values.

Answer. G_total = 0.0146 W/K, Q = 0.292 W; the titanium standoff controls the path (87 % of the resistance).

Working

Joint: G_j = h_c A = 2000 × 1.0e-040.200 W/K
Standoff: A = π d²/4 = 5.027e-05 m², G_s = kA/L = 6.7 × 5.027e-05 / 0.020.01684 W/K
Resistances: 1/G = 3.99 + 5.00 + 59.3968.38 K/W
G_total = 0.01462 W/K; Q = 0.01462 × 200.292 W
The standoff holds 87 % of the total resistance

Answer. G_total = 0.0146 W/K, Q = 0.292 W; the titanium standoff controls the path (87 % of the resistance).

Sanity check. The total (0.0146 W/K) is below the weakest element (0.0168 W/K), as it must be in series. To change this path, change the standoff, not the bracket.

Pythonedit a value, then Run

# Practice T1.2: A heat path in series
import math
G_bracket = 167.0 * 30e-3 * 4e-3 / 0.080   # W/K, aluminium bracket of worked example 1.1
h_c = 2000.0        # W/(m^2 K), bolted joint contact conductance
A_joint = 1.0e-4    # m^2 (1 cm^2)
k_ti = 6.7          # W/(m K), titanium standoff
d = 0.008           # m, standoff diameter
L = 0.020           # m, standoff length
dT = 20.0           # K across the whole path

G_joint = h_c * A_joint
A_s = math.pi * d**2 / 4
G_standoff = k_ti * A_s / L
R_total = 1 / G_bracket + 1 / G_joint + 1 / G_standoff   # resistances add in series
G_total = 1 / R_total
Q = G_total * dT
share = (1 / G_standoff) / R_total
print(f"G_joint = {G_joint:.3f} W/K, G_standoff = {G_standoff:.5f} W/K")
print(f"R = {1/G_bracket:.2f} + {1/G_joint:.2f} + {1/G_standoff:.2f} = {R_total:.2f} K/W")
print(f"G_total = {G_total:.5f} W/K, Q = {Q:.3f} W")
print(f"Standoff share of resistance = {share*100:.1f} %")

G_joint = 0.200 W/K, G_standoff = 0.01684 W/K
R = 3.99 + 5.00 + 59.39 = 68.38 K/W
G_total = 0.01462 W/K, Q = 0.292 W
Standoff share of resistance = 86.8 %

Practice T1.3 · challenge

When conductivity changes with temperature

Given

A support rod for a cooled detector has A = 50 mm² and L = 100 mm, with ends at 120 K and 300 K. Its conductivity varies as k(T) = k₀[1 + b(T − T₀)] with k₀ = 15 W/(m K), T₀ = 293 K and b = 1.5e-03 1/K. Find the exact heat flow, show that using k at the mean temperature gives the same answer, and find the error of using the room-temperature datasheet value k₀.

Mark as done

Hint

Fourier's law with variable k: Q = (A/L) ∫ k(T) dT between the end temperatures.

Answer. Q = 1.182 W (mean-temperature k gives the same); the datasheet value overestimates by 14.2 %.

Working

Q = (A/L) k₀ [ (T_b − T_a) + (b/2)((T_b − T₀)² − (T_a − T₀)²) ]
= (5.0e-05/0.1) × 15 × [180 + 7.50e-04 × (49 − 29929)]1.1819 W
Mean temperature 210 K: k = 13.1325 W/(m K); Q = (A/L) k ΔT = 1.1819 W, identical because k is linear in T
Room-temperature k₀ = 15: Q = 1.3500 W, an error of +14.2 %

Answer. Q = 1.182 W (mean-temperature k gives the same); the datasheet value overestimates by 14.2 %.

Sanity check. The rod is mostly colder than room temperature, so its real k is lower and the room value must overestimate. A sign check like this catches most k(T) mistakes.

Pythonedit a value, then Run

# Practice T1.3: When conductivity changes with temperature
A = 50e-6          # m^2
L = 0.10           # m
Ta = 120.0         # K, cold end
Tb = 300.0         # K, warm end
k0 = 15.0          # W/(m K) at T0
T0 = 293.0         # K
b = 1.5e-3         # 1/K

def k(T):
    return k0 * (1 + b * (T - T0))

# Q = (A/L) * integral of k(T) dT from Ta to Tb
Q_exact = A / L * k0 * ((Tb - Ta) + b / 2 * ((Tb - T0)**2 - (Ta - T0)**2))
Tm = (Ta + Tb) / 2
Q_mean = A / L * k(Tm) * (Tb - Ta)
Q_room = A / L * k0 * (Tb - Ta)
err = (Q_room - Q_exact) / Q_exact * 100
print(f"Q exact        = {Q_exact:.4f} W")
print(f"Q with k(Tm)   = {Q_mean:.4f} W  (k = {k(Tm):.4f} W/(m K))")
print(f"Q with k0      = {Q_room:.4f} W, error {err:+.2f} %")

Q exact        = 1.1819 W
Q with k(Tm)   = 1.1819 W  (k = 13.1325 W/(m K))
Q with k0      = 1.3500 W, error +14.22 %

Common mistakes

Leaving lengths in mm: G comes out 1000 times wrong. Convert everything to metres first.
Adding conductances that are in series. Add resistances (1/G) in series, conductances in parallel.
Using °C inside σT⁴. Radiation always uses kelvin.
Forgetting joints: a bolted interface can be the weakest link in an otherwise good path.

Sources for this chapter

R1Day, G. R. and Lewis, K. D., "Thermal Effects", Ch. 10 in Sarafin, T. P. (ed.), Spacecraft Structures and Mechanisms: From Concept to Launch, Microcosm / Kluwer, 1995, pp. 295-307. Book; not open access.
R5Liles, K. and Amundsen, R. (compilers), NASA Passive Thermal Control Engineering Guidebook, Rev. 5.1, NASA Engineering and Safety Center Passive Thermal TDT, 19 Dec 2024 (distribution: public).
R9NASA Small Spacecraft Systems Virtual Institute, State-of-the-Art of Small Spacecraft Technology, 2024 edition, NASA/TP-20250000142, NASA Ames, Feb 2025, Ch. 7 Thermal Control.
R11NIST Cryogenics Technologies Group, "Material Properties" pages: 6061-T6 Aluminum; Ti-6Al-4V; OFHC Copper (curve fits 4-300 K). Accessed 2026-11-04.
R12ASM Aerospace Specification Metals Inc., "Aluminum 6061-T6; 6061-T651" data sheet (MatWeb data, Aluminum Association typical values), 2023.
R13ASM Aerospace Specification Metals Inc., "Titanium Ti-6Al-4V (Grade 5), Annealed" data sheet, 2023.

Where in each source, and links: References.

I have finished chapter 1

Full course

15 more chapters in the full course

The full Spacecraft Thermal Control course continues with 15 more chapters (550 minutes of study and 45 more graded problems), the formula sheet, the glossary, every assumption with its source, and a printable PDF. It comes with the founding membership.

Founding membership

Founding membership opens soon. Join the free weekly email to hear first.

Reference

Answers

Problem	Answer
T1.1	G = 0.0520 W/K and Q = 0.416 W.
T1.2	G_total = 0.0146 W/K, Q = 0.292 W; the titanium standoff controls the path (87 % of the resistance).
T1.3	Q = 1.182 W (mean-temperature k gives the same); the datasheet value overestimates by 14.2 %.

Reference

References

Every reference the pack cites. "Where" names the section, table or page that holds the cited value; an entry says so when the source was not read in full or is not openly available.

Id	Reference
R1	Day, G. R. and Lewis, K. D., "Thermal Effects", Ch. 10 in Sarafin, T. P. (ed.), Spacecraft Structures and Mechanisms: From Concept to Launch, Microcosm / Kluwer, 1995, pp. 295-307. Book; not open access.Where: §10.1 eqs 10.1-10.5, Fig. 10.1 (k ranges), Fig. 10.3 (α/ε map), Fig. 10.4; §10.2 eqs 10.6-10.9, Table 10.1 (p. 304), §10.2.2-10.2.4 (pp. 305-307)not openly available (book, no URL)Cited in: Chapter 1
R5	Liles, K. and Amundsen, R. (compilers), NASA Passive Thermal Control Engineering Guidebook, Rev. 5.1, NASA Engineering and Safety Center Passive Thermal TDT, 19 Dec 2024 (distribution: public).Where: p. 40 (margin example), p. 43 (natural convection ~5 W/m²K), p. 45 (contact conductance 10-5000, nominal ~250 W/m²K), p. 48 (MLI ε* 0.005 to 0.05-0.1; model 0.01 best / 0.05 worst), p. 68 (correlated contact 1000-5000 W/m²K), p. 119 (heater duty cycle 70 % GOLD / 80 % MIL-STD-1540 = 25 % margin), §5.12 p. 142 (louvre ε 0.75 / 0.15, turn-down ~5; VCHP as thermal switch), §5.13 (cryocooler 100 % lift margin), pp. 166-167 (thermal cycling, 11 °C uncertainty example)https://ntrs.nasa.gov/api/citations/20220006584/downloads/NASAPassiveThermalGuidebookv5%201Public.pdfCited in: Chapter 1
R9	NASA Small Spacecraft Systems Virtual Institute, State-of-the-Art of Small Spacecraft Technology, 2024 edition, NASA/TP-20250000142, NASA Ames, Feb 2025, Ch. 7 Thermal Control.Where: §7.2.1 (MLI performs worse on small spacecraft; edge and attachment losses, pp. 209-210), Table 7-6 (bolted-joint conductance per screw, p. 213, after Gilmore ch. 8), §7.2.6 louvres (p. 214), §7.2.8 heat pipes (p. 215)https://www.nasa.gov/wp-content/uploads/2025/02/soa-2024.pdfCited in: Chapter 1
R11	NIST Cryogenics Technologies Group, "Material Properties" pages: 6061-T6 Aluminum; Ti-6Al-4V; OFHC Copper (curve fits 4-300 K). Accessed 2026-11-04.Where: fitted k(T), c_p(T), E(T), expansion; evaluated at 293 K by script: Al 6061-T6 k 154 W/mK, c_p 943 J/kgK, E 70.4 GPa; Ti-6Al-4V k 7.4 W/mK; OFHC Cu (RRR 100) k 397 W/mKhttps://trc.nist.gov/cryogenics/materials/materialproperties.htmCited in: Chapter 1
R12	ASM Aerospace Specification Metals Inc., "Aluminum 6061-T6; 6061-T651" data sheet (MatWeb data, Aluminum Association typical values), 2023.Where: k 167 W/mK at 77 °F; c_p 0.896 J/g°C; CTE 23.6 µm/m°C (20-100 °C); E 68.9 GPa; yield 276 MPa; density 2.70 g/cchttps://www.aerospacemetals.com/wp-content/uploads/2023/06/Aluminum-6061-T6-6061-T651.pdfCited in: Chapter 1
R13	ASM Aerospace Specification Metals Inc., "Titanium Ti-6Al-4V (Grade 5), Annealed" data sheet, 2023.Where: k 6.7 W/mK; CTE 8.6 µm/m°C (20-100 °C)https://www.aerospacemetals.com/wp-content/uploads/2023/07/Titanium-Ti-6Al-4V-Grade-5-Annealed.pdfCited in: Chapter 1

The references cited by this preview. The full course lists every source it uses.

Spacecraft Thermal Control

How to use this pack

What is in the full course

You will be able to

Conduction

Radiation

Convection, where it still exists

Key equations

Worked examples

How much heat does a bracket carry?[R12, R13]

Radiation as a conductance[R5]

Practice

A thin copper strap[R11]

A heat path in series[R5, R9, R12, R13]

When conductivity changes with temperature

Common mistakes

Check yourself

Sources for this chapter

15 more chapters in the full course

Answers

References

How much heat does a bracket carry?^{[R12, R13]}

Radiation as a conductance^[R5]

A thin copper strap^[R11]

A heat path in series^{[R5, R9, R12, R13]}