Worked examples — Environmental testing — thermal vacuum (TVAC), vibration, acoustic, EMC - EMI

This is the "roll up your sleeves" companion to the parent topic. The parent gave you the laws. Here we use them until every possible case — hot, cold, spinning, resonating, degenerate — feels routine.

Before we start, two symbols the parent used but never fully unpacked. We earn them now.

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The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	What makes it different	Example
A. Single-surface radiation	Heat leaves one face only	Ex 1
B. Two-surface radiation	Radiates from front and back	Ex 2
C. Degenerate: $\epsilon \to 0$	Perfect reflector — what happens to $T$?	Ex 3
D. Cold-case limit	No sun (eclipse): $P_{in}\to 0$	Ex 4
E. Resonance ($r=1$)	Input frequency hits natural frequency	Ex 5
F. Below/above resonance	$r<1$ and $r>\sqrt{2}$ regimes, both signs of "detuning"	Ex 6
G. Zero damping limit	$\zeta \to 0$ — the runaway case	Ex 7
H. Random-vibe RMS	Integrate a flat PSD band	Ex 8
I. Word problem	Real cubesat: solve for panel coating $\epsilon$	Ex 9
J. Exam twist	Combine thermal + fourth-power sensitivity	Ex 10

Constants used everywhere: $\sigma =5.67\times 10^{-8}{\text{W/(m}}^2{\text{K}}^4)$, solar constant $S=1368{\text{W/m}}^2$.

Figure s01 — our map for the whole page. Left panel (thermal): a cyan rectangle is the spacecraft panel at temperature $T$; the amber arrow labelled "aS in" is absorbed sunlight streaming in from the top-left; the two white arrows leaving the left and right faces are radiated heat $\sigma \epsilon T^4$ going out. Equilibrium means the amber arrow's power equals the sum of the white arrows'. Right panel (vibration): the white bar on top is the shaker table (the ceiling); the cyan zig-zag is the spring of stiffness $k$; the white box beside it is the damper $c$ that bleeds off energy; the amber box is the mass $m$; the amber arrow $x(t)$ is how far the mass moves. Every thermal example plays out on the left picture, every vibration example on the right.

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Thermal cases (A–D, I–J)

Ex 1 — Cell A: single-surface radiation

Forecast: Guess — will this be hotter or cooler than a plate that radiates from both sides? (Fewer exits for heat… hold that thought.)

1. Balance in = out with $n=1$. $\alpha S=\sigma \epsilon T^4$. Why this step? Only one face radiates, so $n=1$. $A$ cancels on both sides.
2. Solve for $T^4$. $T^4=\frac{\alpha S}{\sigma \epsilon }=\frac{0.92\times 1368}{5.67\times 10^{-8}\times 0.85}$. Why this step? Isolate the unknown before taking a root.
3. Compute. $T^4=2.611\times 10^{10}$, so $T=(2.611\times 10^{10}{)}^{1/4}=401.6\text{K}$. Why this step? The fourth root undoes the fourth power — it asks "which $T$ raised to the 4th gives this?"

Verify: $401.6=128.5°\text{C}$. Hotter than the two-sided plate below — makes sense, heat has only one exit. Units: $\left[\frac{{\text{W/m}}^2}{{\text{W/(m}}^2{\text{K}}^4)}\right]={\text{K}}^4$. ✓

Ex 2 — Cell B: two-surface radiation

Forecast: Two exits for heat — will $T$ drop by exactly half, or by less?

1. Balance with $n=2$. $\alpha S=2\sigma \epsilon T^4$. Why this step? Sunlight still hits one face, but heat escapes from two.
2. Solve. $T^4=\frac{0.92\times 1368}{2\times 5.67\times 10^{-8}\times 0.85}=1.305\times 10^{10}$. Why this step? Same algebra, the factor 2 halves $T^4$, not $T$.
3. Root. $T=(1.305\times 10^{10}{)}^{1/4}=337.7\text{K}=64.5°\text{C}$.

Verify: Ratio to Ex 1: $\frac{337.7}{401.6}=0.841=(1/2{)}^{1/4}$. The extra face cools it by the fourth root of 2, not by half — that is the $T^4$ law in action. ✓ (This matches the parent note's cubesat example.)

Ex 3 — Cell C: degenerate $\epsilon \to 0$ (perfect mirror)

Forecast: A body that absorbs sunlight but can't radiate — what temperature does the maths give?

1. Write the balance. $\alpha S=2\sigma \epsilon T^4\Rightarrow T^4=\frac{\alpha S}{2\sigma \epsilon }$. Why this step? Keep $\epsilon$ symbolic so we can watch it shrink.
2. Let $\epsilon \to 0$. The denominator $\to 0$, so $T^4\to \infty$, hence $T\to \infty$. Why this step? This is a limit — we ask what value the formula heads toward, not plug in zero (division by zero is undefined).

Verify: Physically: a body that absorbs but never radiates heats without bound. Real coatings have $\epsilon \gtrsim 0.02$, which is why thermal engineers reject pure mirrors on internal radiators. Sanity: with $\epsilon =0.02$, $T=337.7\times (0.85/0.02{)}^{1/4}=337.7\times 2.556=863\text{K}$ — dangerously hot, confirming the trend. ✓

Ex 4 — Cell D: cold-case limit (eclipse)

Forecast: With no sun, does the panel drop to 3 K (space temperature) or settle somewhere warmer?

1. New balance: internal power out. $P_d=2A\sigma \epsilon T^4$. Why this step? In eclipse the only heat source is the electronics; equilibrium is dissipation = radiation.
2. Solve for $T^4$. $T^4=\frac{5}{2\times 1\times 5.67\times 10^{-8}\times 0.85}=5.187\times 10^7$. Why this step? Isolate the unknown group $T^4$ before rooting, exactly as in Ex 1 — the algebra is identical, only the source term changed from $\alpha S$ to $P_d$.
3. Root. $T=(5.187\times 10^7{)}^{1/4}=84.9\text{K}=-188.2°\text{C}$. Why this step? The fourth root undoes the fourth power to recover the actual temperature in Kelvin.

Verify: It does not fall to 3 K — the trickle of internal power holds it near $-188°\text{C}$. Still brutally cold: this is exactly the cold soak TVAC recreates. Units check as in Ex 1. ✓

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Vibration cases (E–H)

Figure s02 — the transmissibility curve $T(r)$. The horizontal axis is the frequency ratio $r=f/f_n$; the vertical axis (log scale) is $T$, how many times the part's shaking exceeds the table's. Two curves: the cyan one is light damping ($\zeta =0.03$), the amber one is heavier damping ($\zeta =0.1$). Notice three regions the white guide-lines mark: at the far left ($r<1$) both curves sit near $T=1$ (the part copies the table); at the white dashed line $r=1$ they spike — the cyan curve much higher because less damping means a taller peak; past the amber dotted line $r=\sqrt{2}$ both drop below $T=1$ (the horizontal white dotted line), meaning the part now shakes less than the table — this is isolation. Each vibration example below is a single point read off this curve.

Ex 5 — Cell E: exactly at resonance ($r=1$)

Forecast: 1 g in — how many g out? Guess an order of magnitude.

1. $r=1$. $r=f/f_n=300/300=1$. Why this step? Matching frequencies is the definition of resonance — the danger zone. (Using $f/f_n$ is legal because the $2\pi$ factors cancel, as shown above.)
2. Plug into $T$. With $r=1$: $(1-r^2)=0$, so $T=\frac{1}{2\zeta r}=\frac{1}{2\zeta }=Q$. Why this step? At $r=1$ the first term vanishes; the formula collapses to $Q$, the magnification factor.
3. Compute. $Q=\frac{1}{2\times 0.03}=16.67$. Output $=16.67\times 1\text{g}=16.67\text{g}$.

Verify: A gentle 1 g becomes 16.7 g at the part — precisely why sine sweeps hunt resonances before random vibe. ✓

Ex 6 — Cell F: both sides of resonance ($r<1$ and $r>\sqrt{2}$)

Forecast: One of these should barely amplify, the other should isolate (output smaller than input). Which is which?

1. (a) $r=0.5$. $(1-0.25{)}^2=0.5625$; $(2\times 0.03\times 0.5{)}^2=0.0009$. Sum $=0.5634$; $T=1/\sqrt{0.5634}=1.332$. Why this step? Below resonance the structure follows the input almost rigidly — mild amplification.
2. (b) $r=2$. $(1-4{)}^2=9$; $(2\times 0.03\times 2{)}^2=0.0144$. Sum $=9.014$; $T=1/\sqrt{9.014}=0.333$. Why this step? Above $r=\sqrt{2}\approx 1.414$ the part can't keep up with fast shaking — it isolates, output < input.

Verify: $T(0.5)=1.33>1$ (amplify), $T(2)=0.33<1$ (isolate). The crossover $T=1$ sits at $r=\sqrt{2}$ — mount soft-isolators so equipment lives in the $r>\sqrt{2}$ zone. ✓

Ex 7 — Cell G: zero-damping limit ($\zeta \to 0$)

Forecast: With no damping to bleed off energy at resonance, where does the amplitude head?

1. At $r=1$, $T=1/(2\zeta )$. As $\zeta \to 0$, the denominator $\to 0$. Why this step? This is a limit again — we watch $T$ as damping vanishes.
2. Conclude $T\to \infty$. Why this step? Undamped resonance grows without bound — theoretical infinite amplitude.

Verify: Real structures always have $\zeta \ge 0.01$, capping $Q$ at ~50. But the limit warns us: never design a lightly-damped part with $f_n$ inside the launch spectrum. Sanity: $\zeta =0.01\Rightarrow Q=50$; $\zeta =0.001\Rightarrow Q=500$ — clearly diverging. ✓

Ex 8 — Cell H: random-vibration RMS

Forecast: The area under the PSD, then a square root. Roughly how many g?

1. Integrate the PSD. $g_{rms}=\sqrt{{\int }_{f_1}^{f_2}Gdf}$. For a flat band this is $\sqrt{G(f_2-f_1)}$. Why this step? PSD (${\text{g}}^2/\text{Hz}$) times bandwidth (Hz) gives ${\text{g}}^2$ — the mean-square. The square root turns mean-square into RMS.
2. Compute area. $G(f_2-f_1)=0.04\times (800-50)=0.04\times 750=30{\text{g}}^2$.
3. Root. $g_{rms}=\sqrt{30}=5.48\text{g}$.

Verify: Units: $\sqrt{\frac{{\text{g}}^2}{\text{Hz}}\cdot \text{Hz}}=\text{g}$. ✓ (Adding the parent's sloped 20–50 Hz and 800–2000 Hz skirts pushes the full-spectrum figure up to ~7.7 g, consistent with the parent note.)

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Word problem & exam twist (I, J)

Ex 9 — Cell I: design a coating (solve for $\epsilon$)

Forecast: To cool more, do we need higher or lower $\epsilon$? Guess before solving.

1. Balance, solve for $\epsilon$. From $\alpha S=2\sigma \epsilon T^4$: $\epsilon =\frac{\alpha S}{2\sigma T^4}$. Why this step? We treat $\epsilon$ as the unknown design knob and set $T=T_{max}$.
2. Plug in. $\epsilon =\frac{0.92\times 1368}{2\times 5.67\times 10^{-8}\times 300^4}=\frac{1258.6}{2\times 5.67\times 10^{-8}\times 8.1\times 10^9}$. Why this step? Substitute the worst-case (hottest) allowed temperature; that is the value that demands the most radiating capability, so it yields the minimum acceptable $\epsilon$.
3. Compute. Denominator $=918.5$; $\epsilon =1258.6/918.5=1.370$. Why this step? Carrying out the arithmetic gives the numerical emissivity the design would require — the number we then sanity-check against the physical ceiling $\epsilon \le 1$.

Verify: $\epsilon =1.37$ is impossible — emissivity can never exceed 1! Interpretation: no passive coating on this sun-facing panel can hold 300 K; you need extra radiator area, a heat pipe, or sun-shielding. A negative-space answer is still an answer. ✓

Ex 10 — Cell J: exam twist (fourth-power sensitivity)

Forecast: A 10% temperature rise — will power rise by 10%, 20%, or more than 40%?

1. Take the ratio. $\frac{Q_2}{Q_1}=\frac{\sigma \epsilon A{T}_2^4}{\sigma \epsilon A{T}_1^4}={\left(\frac{T_2}{T_1}\right)}^4$. Why this step? Everything except $T$ cancels — the whole answer lives in the fourth power.
2. Compute. ${\left(\frac{330}{300}\right)}^4=(1.1{)}^4=1.4641$. Why this step? Substituting the two temperatures turns the symbolic ratio into a concrete number — the factor by which heat rejection climbs.

Verify: A 10% temperature bump → a 46.4% jump in heat rejection. That non-linearity is the headline of the whole topic: small $T$ changes, big power swings. ✓

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Recall Self-test

Two-sided vs one-sided panel: which is cooler? ::: Two-sided — the extra radiating face lowers $T$ by a factor $(1/2{)}^{1/4}\approx 0.84$. At resonance $r=1$, transmissibility equals what? ::: $Q=1/(2\zeta )$. Above what $r$ does a mount start to isolate? ::: $r>\sqrt{2}\approx 1.414$. Flat PSD $0.04{\text{g}}^2/\text{Hz}$ over 750 Hz gives what RMS? ::: $\sqrt{30}\approx 5.48$ g. A 10% temperature rise changes radiated power by what factor? ::: $(1.1{)}^4=1.464$, i.e. a $+46\%$ jump.

See also: Spacecraft Thermal Control Systems · Launch Vehicle Dynamics · Structural Mechanics · Reliability Engineering · Quality Assurance in Aerospace · Electromagnetic Wave Propagation