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

3.6.33 · D3 · Physics › Spacecraft Structures & Systems Engineering › Environmental testing — thermal vacuum (TVAC), vibration, ac

Yeh parent topic ka "sleeves chadhao aur kaam karo" wala companion hai. Parent ne tumhe laws diye. Yahan hum unhe use karte hain — baar baar — jab tak har possible case — hot, cold, spinning, resonating, degenerate — routine na lag jaye.

Shuru karne se pehle, do symbols jo parent ne use kiye the lekin poori tarah explain nahi kiye. Ab hum unhe earn karte hain.

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Scenario matrix

Is topic ke har problem ko in cells mein se kisi ek mein daala ja sakta hai. Neeche ke examples mein cell label diya gaya hai.

Cell	Kya cheez ise alag banati hai	Example
A. Single-surface radiation	Heat sirf ek face se nikalti hai	Ex 1
B. Two-surface radiation	Front aur back dono se radiate karta hai	Ex 2
C. Degenerate: $\epsilon \to 0$	Perfect reflector — $T$ ka kya hota hai?	Ex 3
D. Cold-case limit	Koi sun nahi (eclipse): $P_{in}\to 0$	Ex 4
E. Resonance ($r=1$)	Input frequency natural frequency se milti hai	Ex 5
F. Below/above resonance	$r<1$ aur $r>\sqrt{2}$ dono regimes, "detuning" ke dono signs	Ex 6
G. Zero damping limit	$\zeta \to 0$ — runaway case	Ex 7
H. Random-vibe RMS	Ek flat PSD band integrate karo	Ex 8
I. Word problem	Real cubesat: panel coating $\epsilon$ solve karo	Ex 9
J. Exam twist	Thermal + fourth-power sensitivity combine karo	Ex 10

Constants jo har jagah use hoti hain: $\sigma =5.67\times 10^{-8}{\text{W/(m}}^2{\text{K}}^4)$, solar constant $S=1368{\text{W/m}}^2$.

Figure s01 — poore page ka hamara map. Left panel (thermal): ek cyan rectangle spacecraft panel hai jo temperature $T$ par hai; amber arrow jis par "aS in" likha hai wo absorbed sunlight hai jo upar-left se aa rahi hai; do white arrows jo left aur right faces se nikal rahe hain wo radiated heat $\sigma \epsilon T^4$ hain jo bahar ja rahi hai. Equilibrium ka matlab hai amber arrow ki power white arrows ki power ke sum ke barabar hai. Right panel (vibration): upar wala white bar shaker table hai (ceiling); cyan zig-zag spring hai stiffness $k$ ka; uske saath wala white box damper $c$ hai jo energy bleed karta hai; amber box mass $m$ hai; amber arrow $x(t)$ batata hai ki mass kitna move karta hai. Har thermal example left picture par play out hota hai, har vibration example right par.

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Thermal cases (A–D, I–J)

Ex 1 — Cell A: single-surface radiation

Forecast: Andaza lagao — kya yeh plate us plate se zyada hot ya cool hogi jo dono sides se radiate karti hai? (Heat ke liye kam exits… yeh soch ke rako.)

1. Balance in = out with $n=1$. $\alpha S=\sigma \epsilon T^4$. Yeh step kyun? Sirf ek face radiate karta hai, isliye $n=1$. $A$ dono sides se cancel ho jaata hai.
2. $T^4$ ke liye solve karo. $T^4=\frac{\alpha S}{\sigma \epsilon }=\frac{0.92\times 1368}{5.67\times 10^{-8}\times 0.85}$. Yeh step kyun? Root lene se pehle unknown ko isolate karo.
3. Compute karo. $T^4=2.611\times 10^{10}$, toh $T=(2.611\times 10^{10}{)}^{1/4}=401.6\text{K}$. Yeh step kyun? Fourth root fourth power ko undo karta hai — yeh poochta hai "kaun sa $T$ 4th power tak raised kar ke yeh deta hai?"

Verify: $401.6=128.5°\text{C}$. Neeche wale two-sided plate se zyada hot — makes sense, heat ke liye sirf ek exit hai. 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: Heat ke liye do exits — kya $T$ exactly aadha drop karega, ya kam?

1. Balance with $n=2$. $\alpha S=2\sigma \epsilon T^4$. Yeh step kyun? Sunlight abhi bhi ek face pe padti hai, lekin heat do faces se escape karti hai.
2. Solve karo. $T^4=\frac{0.92\times 1368}{2\times 5.67\times 10^{-8}\times 0.85}=1.305\times 10^{10}$. Yeh step kyun? Same algebra, factor 2 $T^4$ ko half karta hai, $T$ ko nahi.
3. Root. $T=(1.305\times 10^{10}{)}^{1/4}=337.7\text{K}=64.5°\text{C}$.

Verify: Ex 1 se ratio: $\frac{337.7}{401.6}=0.841=(1/2{)}^{1/4}$. Extra face ise fourth root of 2 se cool karta hai, aadha nahi — yeh $T^4$ law action mein hai. ✓ (Yeh parent note ke cubesat example se match karta hai.)

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

Forecast: Ek body jo sunlight absorb karta hai lekin radiate nahi kar sakti — maths kaun sa temperature deta hai?

1. Balance likho. $\alpha S=2\sigma \epsilon T^4\Rightarrow T^4=\frac{\alpha S}{2\sigma \epsilon }$. Yeh step kyun? $\epsilon$ ko symbolic rakhte hain taaki hum ise shrink hote dekh sakein.
2. $\epsilon \to 0$ hone do. Denominator $\to 0$, toh $T^4\to \infty$, hence $T\to \infty$. Yeh step kyun? Yeh ek limit hai — hum poochte hain ki formula kis value ki taraf jaata hai, zero plug in nahi karte (zero se division undefined hai).

Verify: Physically: ek body jo absorb karta hai lekin kabhi radiate nahi karta, bina rukke heat up hota hai. Real coatings mein $\epsilon \gtrsim 0.02$ hota hai, yahi wajah hai ki thermal engineers internal radiators par pure mirrors reject karte hain. Sanity: $\epsilon =0.02$ ke saath, $T=337.7\times (0.85/0.02{)}^{1/4}=337.7\times 2.556=863\text{K}$ — dangerously hot, jo trend confirm karta hai. ✓

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

Forecast: Koi sun nahi, toh kya panel 3 K (space temperature) tak drop karega ya kahin warmer settle karega?

1. Naya balance: internal power out. $P_d=2A\sigma \epsilon T^4$. Yeh step kyun? Eclipse mein heat ka source sirf electronics hai; equilibrium hai dissipation = radiation.
2. $T^4$ ke liye solve karo. $T^4=\frac{5}{2\times 1\times 5.67\times 10^{-8}\times 0.85}=5.187\times 10^7$. Yeh step kyun? Unknown group $T^4$ ko root lene se pehle isolate karo, bilkul Ex 1 ki tarah — algebra identical hai, sirf source term $\alpha S$ se $P_d$ mein badla.
3. Root. $T=(5.187\times 10^7{)}^{1/4}=84.9\text{K}=-188.2°\text{C}$. Yeh step kyun? Fourth root fourth power ko undo karta hai taaki Kelvin mein actual temperature recover ho sake.

Verify: Yeh 3 K tak nahi girta — internal power ki trickle ise $-188°\text{C}$ ke paas rakhti hai. Phir bhi brutally cold: yahi woh cold soak hai jo TVAC recreate karta hai. Units Ex 1 ki tarah check hote hain. ✓

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Vibration cases (E–H)

Figure s02 — transmissibility curve $T(r)$. Horizontal axis frequency ratio $r=f/f_n$ hai; vertical axis (log scale) $T$ hai, part ki shaking table ki shaking se kitni baar zyada hai. Do curves: cyan wala light damping ka hai ($\zeta =0.03$), amber wala heavier damping ka ($\zeta =0.1$). Teen regions notice karo jo white guide-lines mark karti hain: far left par ($r<1$) dono curves $T=1$ ke paas hain (part table ko copy karta hai); white dashed line $r=1$ par woh spike karte hain — cyan curve kaafi zyada kyunki kam damping matlab zyada tall peak; amber dotted line $r=\sqrt{2}$ ke baad dono $T=1$ se neeche drop ho jaate hain (horizontal white dotted line), matlab part ab table se kam shake karta hai — yahi isolation hai. Neeche ke har vibration example mein is curve se ek single point read kiya jaata hai.

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

Forecast: 1 g in — kitne g out? Ek order of magnitude guess karo.

1. $r=1$. $r=f/f_n=300/300=1$. Yeh step kyun? Frequencies ka match karna resonance ki definition hai — danger zone. ($f/f_n$ use karna legal hai kyunki $2\pi$ factors cancel ho jaate hain, jaise upar bataya.)
2. $T$ mein plug in karo. $r=1$ par: $(1-r^2)=0$, toh $T=\frac{1}{2\zeta r}=\frac{1}{2\zeta }=Q$. Yeh step kyun? $r=1$ par pehla term vanish ho jaata hai; formula collapse karke $Q$, magnification factor, ban jaata hai.
3. Compute karo. $Q=\frac{1}{2\times 0.03}=16.67$. Output $=16.67\times 1\text{g}=16.67\text{g}$.

Verify: Ek gentle 1 g, part par 16.7 g ban jaata hai — exactly yahi wajah hai ki sine sweeps random vibe se pehle resonances dhundhte hain. ✓

Ex 6 — Cell F: resonance ke dono sides ($r<1$ aur $r>\sqrt{2}$)

Forecast: Inme se ek barely amplify karega, doosra isolate karega (output input se chhota). Kaun sa kaun sa?

1. (a) $r=0.5$. $(1-0.25{)}^2=0.5625$; $(2\times 0.03\times 0.5{)}^2=0.0009$. Sum $=0.5634$; $T=1/\sqrt{0.5634}=1.332$. Yeh step kyun? Resonance se neeche structure almost rigidly input follow karta hai — mild amplification.
2. (b) $r=2$. $(1-4{)}^2=9$; $(2\times 0.03\times 2{)}^2=0.0144$. Sum $=9.014$; $T=1/\sqrt{9.014}=0.333$. Yeh step kyun? $r=\sqrt{2}\approx 1.414$ se upar part fast shaking ke saath keep up nahi kar sakta — woh isolate karta hai, output < input.

Verify: $T(0.5)=1.33>1$ (amplify), $T(2)=0.33<1$ (isolate). Crossover $T=1$ $r=\sqrt{2}$ par hai — soft-isolators mount karo taaki equipment $r>\sqrt{2}$ zone mein rahe. ✓

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

Forecast: Resonance par energy bleed karne ke liye koi damping nahi — amplitude kidhar jaata hai?

1. $r=1$ par, $T=1/(2\zeta )$. Jaise $\zeta \to 0$, denominator $\to 0$. Yeh step kyun? Yeh phir ek limit hai — hum dekh rahe hain $T$ kya hota hai jab damping vanish hoti hai.
2. Conclude $T\to \infty$. Yeh step kyun? Undamped resonance bina bound ke badhta hai — theoretical infinite amplitude.

Verify: Real structures mein hamesha $\zeta \ge 0.01$ hota hai, $Q$ ko ~50 par cap karta hai. Lekin yeh limit warn karta hai: kabhi bhi ek lightly-damped part ko launch spectrum ke andar $f_n$ ke saath design mat karo. Sanity: $\zeta =0.01\Rightarrow Q=50$; $\zeta =0.001\Rightarrow Q=500$ — clearly diverging. ✓

Ex 8 — Cell H: random-vibration RMS

Forecast: PSD ke neeche ka area, phir square root. Roughly kitne g?

1. PSD integrate karo. $g_{rms}=\sqrt{{\int }_{f_1}^{f_2}Gdf}$. Flat band ke liye yeh hai $\sqrt{G(f_2-f_1)}$. Yeh step kyun? PSD (${\text{g}}^2/\text{Hz}$) times bandwidth (Hz) deta hai ${\text{g}}^2$ — mean-square. Square root mean-square ko RMS mein badalta hai.
2. Area compute karo. $G(f_2-f_1)=0.04\times (800-50)=0.04\times 750=30{\text{g}}^2$.
3. Root. $g_{rms}=\sqrt{30}=5.48\text{g}$.

Verify: Units: $\sqrt{\frac{{\text{g}}^2}{\text{Hz}}\cdot \text{Hz}}=\text{g}$. ✓ (Parent ke sloped 20–50 Hz aur 800–2000 Hz skirts add karne par full-spectrum figure ~7.7 g tak jaata hai, parent note se consistent.)

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Word problem & exam twist (I, J)

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

Forecast: Zyada cool karne ke liye, kya humein higher ya lower $\epsilon$ chahiye? Solve karne se pehle guess karo.

1. Balance, $\epsilon$ ke liye solve karo. $\alpha S=2\sigma \epsilon T^4$ se: $\epsilon =\frac{\alpha S}{2\sigma T^4}$. Yeh step kyun? Hum $\epsilon$ ko unknown design knob treat karte hain aur $T=T_{max}$ set karte hain.
2. Plug in karo. $\epsilon =\frac{0.92\times 1368}{2\times 5.67\times 10^{-8}\times 300^4}=\frac{1258.6}{2\times 5.67\times 10^{-8}\times 8.1\times 10^9}$. Yeh step kyun? Worst-case (sabse hot) allowed temperature substitute karo; woh value sabse zyada radiating capability demand karti hai, isliye minimum acceptable $\epsilon$ deti hai.
3. Compute karo. Denominator $=918.5$; $\epsilon =1258.6/918.5=1.370$. Yeh step kyun? Arithmetic carry out karne se numerical emissivity milti hai jo design ko require karega — woh number jise hum phir physical ceiling $\epsilon \le 1$ ke against sanity-check karte hain.

Verify: $\epsilon =1.37$ impossible hai — emissivity kabhi 1 se zyada nahi ho sakta! Interpretation: is sun-facing panel par koi bhi passive coating 300 K hold nahi kar sakti; extra radiator area, heat pipe, ya sun-shielding chahiye. Negative-space answer bhi ek answer hai. ✓

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

Forecast: 10% temperature rise — power 10%, 20%, ya 40% se zyada badhlegi?

1. Ratio lo. $\frac{Q_2}{Q_1}=\frac{\sigma \epsilon A{T}_2^4}{\sigma \epsilon A{T}_1^4}={\left(\frac{T_2}{T_1}\right)}^4$. Yeh step kyun? $T$ ke alawa sab cancel ho jaata hai — poora answer fourth power mein hai.
2. Compute karo. ${\left(\frac{330}{300}\right)}^4=(1.1{)}^4=1.4641$. Yeh step kyun? Do temperatures substitute karne se symbolic ratio ek concrete number ban jaata hai — wo factor jis se heat rejection climb karta hai.

Verify: 10% temperature bump → heat rejection mein 46.4% jump. Woh non-linearity poore topic ki headline hai: chhote $T$ changes, bade power swings. ✓

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Recall Self-test

Two-sided vs one-sided panel: kaun sa cooler hai? ::: Two-sided — extra radiating face $T$ ko factor $(1/2{)}^{1/4}\approx 0.84$ se lower karta hai. Resonance $r=1$ par, transmissibility kya hoti hai? ::: $Q=1/(2\zeta )$. Kis $r$ se upar ek mount isolate karna shuru karta hai? ::: $r>\sqrt{2}\approx 1.414$. Flat PSD $0.04{\text{g}}^2/\text{Hz}$ over 750 Hz kya RMS deta hai? ::: $\sqrt{30}\approx 5.48$ g. 10% temperature rise radiated power ko kis factor se change karta hai? ::: $(1.1{)}^4=1.464$, matlab $+46\%$ jump.

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