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

WHAT: Name the dead route and the survivor. WHY: Convection needs a fluid (air, water) to physically carry heat away by moving. In vacuum there is no fluid to move, so convection is gone. Conduction still moves heat through solid structure to other parts of the craft, but that heat still has to leave the craft eventually — and the only way to shed heat into empty space is to glow it away as radiation. Answer: Convection stops; radiation is the only route to deep space.

WHAT: Spot the fourth-power term. WHY: $\sigma$ is fixed. $\epsilon$ appears to the first power, so doubling it doubles $q$. But $T$ appears as $T^4$, so doubling $T$ multiplies $q$ by $2^4=16$. Answer: ==Temperature $T$==; doubling it multiplies radiated power by $16$.

WHAT: Plug into $Q=A\sigma \epsilon T^4$. WHY the background is dropped: deep space is $\approx 3$ K, and $3^4=81$ versus $300^4=8.1\times 10^9$ — the cold side contributes one part in a hundred million, so we ignore it. $Q=0.5\times 5.67\times 10^{-8}\times 0.85\times 300^4$ $Q=0.5\times 5.67\times 10^{-8}\times 0.85\times 8.1\times 10^9\approx 195.2\text{W}$ Answer: $\approx 195\text{W}$.

WHAT: Use $f_n=\frac{1}{2\pi }\sqrt{k/m}$. WHY this formula and not just $\sqrt{k}$: the mass resists acceleration, so a heavier part bounces slower; the spring pulls back, so a stiffer mount bounces faster. The ratio $k/m$ captures both, and the square root turns "energy-like" quantities into a frequency. $f_n=\frac{1}{2\pi }\sqrt{\frac{2\times 10^6}{0.5}}=\frac{1}{2\pi }\sqrt{4\times 10^6}=\frac{2000}{2\pi }\approx 318.3\text{Hz}$ Answer: $\approx 318\text{Hz}$.

WHAT: Compute $Q$, multiply the input. WHY $Q=1/(2\zeta )$: damping $\zeta$ is the fraction of "bounce energy" bled off each cycle. Tiny damping ($\zeta =0.02$) means almost nothing is bled, so energy piles up cycle after cycle until the response is huge — the pile-up factor is exactly $1/(2\zeta )$. $Q=\frac{1}{2\times 0.02}=25$ $a_{\text{felt}}=Q\times a_{\text{in}}=25\times 1.0=25\text{g}$ Answer: ==$25\text{g}$== — a benign $1$ g test input becomes a violent $25$ g at the part. This is why a sine sweep to locate $f_n$ is done at low level.

WHAT: Substitute the two ratios. (a) $r=0.5$ (below resonance): $T=\frac{1}{\sqrt{(1-0.25{)}^2+(2\cdot 0.02\cdot 0.5{)}^2}}=\frac{1}{\sqrt{0.5625+0.0004}}\approx 1.333$ Slightly above 1 — mild amplification, as expected below resonance. (b) $r=\sqrt{2}$ (crossover): $T=\frac{1}{\sqrt{(1-2{)}^2+(2\cdot 0.02\cdot \sqrt{2}{)}^2}}=\frac{1}{\sqrt{1+0.0032}}\approx 0.998\approx 1$ WHY $r=\sqrt{2}$ is special: ignoring damping, $(1-r^2{)}^2=1$ there, so $T=1$ exactly — the mount neither amplifies nor protects. Below it you get shaken more; above it the mass "can't keep up" with fast shaking and rides quieter (isolation). Answer: (a) $T\approx 1.33$ (amplification region); (b) $T\approx 1.0$ (crossover — isolation begins beyond).

WHAT: Balance power in against power out and solve for $T$. WHY balance: at equilibrium the panel is neither warming nor cooling, so watts absorbed = watts radiated. Power in (only the sunlit face collects): $P_{\text{in}}=\alpha SA=0.90\times 1368\times 1=1231.2\text{W}$. Power out (both faces glow, hence the factor $2$): $P_{\text{out}}=2A\sigma \epsilon T^4$. Set equal: $1231.2=2\times 1\times 5.67\times 10^{-8}\times 0.80\times T^4$ $T^4=\frac{1231.2}{2\times 5.67\times 10^{-8}\times 0.80}=\frac{1231.2}{9.072\times 10^{-8}}=1.357\times 10^{10}$ $T=(1.357\times 10^{10}{)}^{1/4}\approx 341.4\text{K}\approx 68.3^{\circ }\text{C}$ Answer: $\approx 341\text{K}(\approx 68^{\circ }\text{C})$ — hot enough that TVAC must confirm coatings/heat-paths hold it in the safe band.

WHAT: For a flat PSD, the integral is just level $\times$ bandwidth. WHY the square root: PSD $G$ has units of ${\text{g}}^2/\text{Hz}$ — it is power (amplitude-squared) per hertz. Adding up power over the band gives total mean-square (${\text{g}}^2$); the RMS is the square root to get back to $\text{g}$. Bandwidth $=800-50=750\text{Hz}$. $\int Gdf=0.04\times 750=30{\text{g}}^2$ $g_{\text{rms}}=\sqrt{30}\approx 5.48\text{g}$ Answer: $\approx 5.5\text{g}$ (the full GEVS spectrum with its sloped ends totals $\approx 7.7$ g; the flat plateau alone already contributes most of the energy).

(a) WHAT: Invert $f_n=\frac{1}{2\pi }\sqrt{k/m}$ for $k$. WHY: we are told the frequency and want the stiffness that guarantees it. $k=m(2\pi f_n{)}^2=0.8\times (2\pi \times 140{)}^2=0.8\times (879.6{)}^2\approx 6.19\times 10^5\text{N/m}$ Answer (a): $k\gtrsim 6.19\times 10^5\text{N/m}$.

(b) WHAT: $Q=1/(2\zeta )$. $Q=\frac{1}{2\times 0.05}=10$ Answer (b): $Q=10$.

(c) WHAT: judge whether the resonance sitting in the flat band ($140$ Hz is inside $50$–$800$ Hz) is acceptable. WHY / reasoning: With $Q=10$, at resonance the box feels $10\times$ the local input amplitude, and because that frequency lies inside the flat $0.04{\text{g}}^2/\text{Hz}$ plateau, the shaker will pour full broadband energy right at $140$ Hz. This is actually the intended philosophy: you want the test to hit the resonance so any weak solder joint fails on the ground. The mission's $140$ Hz floor is not there to dodge the test — it is there to keep $f_n$ above the launch-vehicle's own low-frequency transients while $Q=10$ (thanks to reasonable damping $\zeta =0.05$, not $0.02$) keeps the amplification survivable. So: placing the resonance in the band is fine and deliberate, provided damping keeps $Q$ modest. Answer (c): Acceptable — the band placement guarantees the test excites the mode; the design safety comes from adequate damping ($Q=10$) and from $f_n$ being high enough to clear launch transients, not from hiding the resonance.